I ! '^''iw ,^ '^^. v..,Wv ■S' ,\ ^cK - y s' V V ^.^ :i^'" .N' * % A^ "bo* v*^ ^/,?^ ■^ -^^ ^ ..^-■ c^. s'^'^ - ^ ^ .^'^^ ^.c.<i^ v^' V </> XN^ "^.. .>"^-" -:*-, ^t.- v^^ '%, <^^'^ WORKS OF PROF. I. P. CHURCH PUBLISHED BY JOHN WILEY & SONS. Mechanics of Engineering. Comprising Statics and Dynamics of Solids, the Mechanics of the Materials of Construction or Strength and Elasticity of Beams, Columns, Shafts, Arches, etc., and the Principles of Hydraulics and Pneumatics with Applications. For the Use of Technical Schools. 8vo, xxvi + 854 pages,4 half tones, 656 figures, cloth, $6.00. Notes and Examples in Mechanics. With an Appendix on the Graphical Statics of Mechanism. 8vo, 167 pages, 178 figures, cloth, $2.00. Diagrams of Mean Velocity of Water in Open Channels. Based on the Formula of Ganguillet and Kutter. gX 12 inches, paper, $1.50. Hydraulic flotors ; with related subjects ; includ- ing Centrifugal Pumps, Pipes, and Open Channels. 8vo, 279 pages, 125 figures, cloth, $2.00. One of the two " clinometers " in use in the Testing Laboratory of the College of Civil En^- neering at Cornell University (see p. 241). The main barrel or sleeve of the instrument encircles the horizontal shaft or rod (in testing machine) whose angle of torsion is to be obtained, near one extremity of the same. At each end of the barrel are four brass screws having smooth rounded ends where they bear on the shaft. These are used for centering the barrel on the shaft, but do not grip it. The four steel " gripping screws," at the middle of the barrel, are thumb-screws with flat heads and hardened sharp points. They serve to grip the shaft after the centering is completed. After the shaft has thus been gripped at a certain transverse section, the collar carrying the graduated arc is clamped upon the barrel, the plane of the arc and its vernier arm being that of the points of the gripping screws. By taking a reading of the vernier on the arc at any stage of the test (the vernier-arm being adjusted each time so that the bubble of the spirit level carried by this arm is brought to the center ot its scale) and subtract- ing its initial reading, the angle through which the transverse section has turned from its initial position becomes known. The second clinometer is placed at another transverse section, near the other end of the shaft, and serves to measure its turning movement. The difference of these movements is the angle of torsion. The verniers read to single minutes. (The shaft in above figure is H in. in diameter). Frontispiece. Mechanics of Engineering COMPRISING Statics and Kinetics of Solids ; the Mechanics ov the Materials of Construction, or Strength and Elasticity OF Beams, Columns, Shafts, Arches, etc. ; and thk Principles of Hydraulics and Pneumatics, WITH Applications. FOE USB IN TECHNICAL SCHOOLS. IRVING P. OHUROH, O.E, Professor of Applied Mechanics and Hydraulics, College of Civil Enginkkring, Cornell University. REVISED EDITION, PARTLY REWRITTEN total issue, fifteen thousand. NEW YORK: JOHN WILEY & SON& T^KDON CHAPMAN & HALL. Limitid. 1908 r ^^'o^ .<t \ O)- USHARV of CONGRESS IwoCooies Kecetvb« JUL 1 2908 ouvj^iiKiii collar COPY d; / Up. C!opyrisht, 1890. 1908 BT iHviNG P. Church. SUibert BrumnwnJi ani> ©ompaMg N«n Inrk PREFACE. In presenting a revised edition of this work for the use of technical schools the writer would call attention to the principal changes that have been made; omissions as well as additions. The chapter on "Continuous Girders by Graphics" has been omitted in its entirety, while the graphic treatment of the horizontal straight girder, formerly a part of the chapter on "Arch Ribs," has been removed to the appendix, in which will also be found varioiis paragraphs involving special problems in flexure, once located in the body of the book. Former chap- ters V and VI in Part III, on beams under oblique forces and on columns, respectively, have been merged in one (Chapter VI), -^e matter having been largely rewritten and more fully illustrated, with introduction of the more modern formiiisB for columns and some treatment of the problem of eccen- tric loading. Chapter V in Part III of the revised book, on "Flexure of Reinforced Concrete Beams," is entirely new and presents both theory and numerical ' illustration; as also diagrams aiding in practical design. New matter will also be found in an analytical treatment of "Circular Ribs and Hoops," placed at the end of the chapter on "Arch Ribs." Two other new chapters in Part III, are XII, on the flexure of beams treated by a geometrical method (which, however, does not call for the use of drafting instruments) leading to a very simple and available form of the Theorem of Three Moments ; and XIII, which gives the analysis of stresses in thick hollow cylinders and spheres. A few pages on the strength of plates have also been added in Chapter III. In Part IV additional matter is presented relating to the differential manometer, gas- and oil-engines, the Cippolletti weir, losses of heads in pipes and bends, the hydraulic grade-line, the Venturi meter, current-meters, Pitot's tube, use of Kutter's formula, etc. In Parts I and II numerous additional examples and illustrations are introduced while many pages have been rewritten throughout the book, aside from the new chapters already referred to. Tables of logarithms, trigonometric functions, and hyperbolic sines and cosines, will be found in the appendix. Grateful acknowledgment is again due to Dr. H. T. Eddy for the use of his methods* in treating arch ribs; to Prof. C. L. Crandall for the chapter on retaining- walls ; and to Col. J. T. Fanning for the table of coefficients of friction of water in pipes. The writer would also extend his thanks to Messrs. Buff and Buff of Boston, for the half-tone cut of their current-meter; and to Builders Iron Foundry of Providence, R. I., for the engravings illustrating the Venturi meter. Cornell University, Ithaca, N. Y, June, 1908. Note .^Additional matter involving many examples and forming an appendix to the present work, but too bulky to be incorporated with it, was issued in a separate volume in 1892 and entitled "Notes and Examples in Mechanics." A second edition, revised and enlarged, was published in 1897. * See pp. 14 and 25 of " Researches in Ghaphical Statics." by Prof. H. T. Eddy, C.E., Ph.D., publi.shed by D. Van Nostrand, New York, 1878, reprinted from Van Nostrand's Magazine for 1877; or the German translation of the same, " Neue Constructionen aus der Graphischen Statik," published by Teubner u. Cie., I.eipsic, 1880. INTRODUCTORY NOTES. Preparation. — Prior to tlie use of this book the student is supposed to have had the usual training given in technical schools in analjrtical geometry and in the differential and integral calculus ; and also a year of college physics. Gravitation Measure of a Force. Mass and Weight. — Since the gravitation measure of a force is the one almost exclusively used by engineers, a brief resume of its nature is here given, aside from the paragraph of p. 835, Appendix. The amount of matter in a certain piece of platinum, kept by the British government, is called by the physicist a pound of mass, but the engineer understands by the word "pound" the force of gravitation, or weight, exerted by the earth on this piece of metal at London; and if this piece of metal be supported, at London, by a spring balance, the scale of which is so grad- uated that the pointer now stands at unity, such a balance constitutes a standard instrument with which to measure forces for the purposes of the engineer. According to the indications of such an instrument the same piece of metal, if suspended on the same balance at the equator, at sea-level, would be found to weigh only 0.997 lbs. (force) on account of the diminished intensity of gravitation; the difference, however, being only about three parts in a thousand, or one-third of one per cent. For ordinary engineering problems involving the strength of structures, this difference is of no prac- tical importance. A unit of force based on this gravitation method is called a gravitation measure of force. The mass of the piece of platinum, has, of course, suffered no change in the transit from London to the equator, and since the fraction obtained by dividing the weight (obtained from the spring balance) by the acceleration of gravity, g, is constant, regardless of the place where the two quantities are measured, it is convenient (though not essential) for the engineer to give the name "mass" to this fraction when it occurs in the equations of kinetics. For instance, since g (for foot and second) =32.18 at 1.000 0.997 London, and 32.09 at the equator (at sea-level), we note that „^ „ = „^ ^„ =0.03108. Arithmetic. — In arithmetical operations the student should remember that the degree of refinement attained or employed does not depend on the number of decimal places used, but upon the number of significant figures. Thus, each of the quantities 0.0003674 and 510.4 contains four significant figures. For instance, let us suppose that the value of x is to be obtained from the relation x = a-b, where a = 0.0000568 and b = 0.0000421. Should the student conclude that five decimal places would be accurate enough and^ thus write 0.00005 for a, and 0.00004 for b, he would obtain a: = 0.00001, con- taining only one significant figure; whereas the true result is a; = 0.0000147. Hence the former result is seen to be in error to the extent of 47 parts in 147, or 32 parts in 100, i.e., 32 per cent. ; which is a very gross and totally unnecessary error. Values obtained from the ordinary 10-inch slide rule usually contain only three significant figures (four if near left of scale). Logarithms. — The following facts and operations are not usually fresh in the student's mind. The logarithm of a number less than unity is a negative quantity but is usually expressed as the algebraic sum of a positive mantissa (or decimal part) and a negative characteristic which is a whole number; thus, the common logarithm of 0.20 is f.301030 . . . , that is, log. 0.20 = -1-0.301030-1.000000 (or, -1-9.301030-10). This should be borne in mind in raising such a number to any power. For example: required the value of Solution.—^^OMQl and log. 0.8461 = 1.9274, i.e., =0.9274-1.0000. Hence 0.71 X log. 0.846_1 = 0.71(0.9274 -1.0000), =0.6584 -0.7100, = -0.0516 = 1.9484 = log. 0.8880; therefore a;=0.8880. Note that, according to the definition of a logarithm, the statement en=m is equivalent to the statement n = loge m. iv MATHEMATICAL DATA. Trigonometry. cos^A +sin^A = 1. cos^ A — sin^ A ^ cos 2 A sin 2A = 2 sin A cos A cos 2A =cos^A — sin^A sin A 1 — cos A Solution of Oblique Triangles, etc. B 2 sinM = l — cos 2A. 2cos2A = l + cos2A. tan A 1 cot iA = . sin A = cos A = Vl+tan^A cosecA 1 1 Given a, b, B; to find A : a, b, C; a,b,C; a, b, c: A: c: C: sin A = tan A = Vl+tan^A sec A sin A _ sin B _ sin C a b c ' d = a sin C d = c sin A m = c cos A n = o cos C d = mtanA d=n tan C a sin B a sin C cos C b — a cos C c' = a^ + &^ — 2a& cos C a^ + b'^ — c^ 2ab ^—■{a- r (the Mensuration. Area of a circle =;rr'; ^ circumference = 2n,r. Area of sector ^ / a° A5C0A=(3gQ, latter a ia radians). Vol. of sphere = 4 ■^Ttr^. Area of the segment, ABC DA, of a circle, W = (area of sector ABCOA)- (area of triangle ACO). Area of rightsegment of a para6oto= two- thirds that of circumscribing rectangle, = f(2/ia). Equation to curve OA is v^ X — p = -. Distance OC, of center of gravity, 3 is X = ?a, from vertex 0. ti o VI MATHEMATICAL DATA. Integral Forms. — (Each integral to be taken between limits, or to have a constant added and determined). (See also p. 480.) ixndx xn+1 J n + 1 cos X da; = sin x; j* dx J X dx \/.T^ ± a' dx = sin— ix; = loge(a; + v'x^ 1 isin x dx= — cos x ; ' dx 1 + x^ '-'y' \^a tan— ix; dx + hx — cx^ , y/ab + bx loge -7=- l'=-2^'^oge{a-bx^). \a-bx^ SVab'"* Vab-bx' ja-bx^ Numerical Constants. — The acceleration of gravity, g, (for the English foot and second) is 32.16 for the latitude of Philadelphia at sea-level, and for any latitude ji, and elevation h above sea-level, is 32.1723-0.0833 cos 2/?-0.000003/i. For ordinary problems in mechanics, however, in the northern United States g may be taken as 32.2, for which value we have \/23 = 8.025; - = 0.03105; 9 and j^ =0.01553. 22 The ratio 7r = 3. 141592, or approx. 3J-, i.e., -=-; -=0.31831; ;r2 = 9.86960; ^=0.10132; V'7r= 1 .77245. 1° = 0.01745 radians. One radian = 57° 17' 44.8". If n denote, any number, then Login (n) = 0.43429 X logs (r>) ; and loge (n) =2.30258 Xlogio (n). Base of nat. logs. = e, =2.71828; base of Briggs system = 10. GREEK ALPHABET. A a B p ry E € z c H V Odd I I K K A X Mix Names, Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mu Letters. Names. N V Nu E a Xi o Omicroa Htt Pi P P Rho 2 (T s Sigma T t Tau Tv UpsiloQ Phi Xx Chi W^ ip Psi n CO Omega TABLE OF CONTENTS. [Mjechanics of solids.} PRELIMINARY CHAPTER PACS §§ l-'lS'i. Definitions. Kinds of Quanlily. Homogeneous Equa- tions. Parallelogram of Forces ,... 1 PAET I. STATICS. Chapter I. Statics of a Material Point. IP 16-19. Composition and Equilibrium of Concurrent Forces. ... 8 Chapter II. Parallel Forces and the Centre op Gravity. §§ 20-22. Parallel Forces 1^ §§23-276. Centre of Gravity. Problems. Centrobaric Method. . . 18= Chapter III. Statics of a Rigid Body. g§ 28-34. Couples 27 ^§35-39. Composition and Equilibrium of Non-concurrent Forces. 31 Chapter IV. Statics of Flexible Cords. §§ 40-48. Postulates. Suspended Weights. Parabolic Cord. Cat- enary 4r PART II. KINETICS. Chapter I. Rectiijnear AIotion of a Material Poini ^§ 49-55. Uniform Motion. Falling Bodies. Newton's Laws, Mass .... 49 g§ 56=60= Uniformly Accelerated Motion. Graphic Representa- tions. Variably Accelerated Motions. Impact. . . 5^- V Chapter II. Virtual Velocities. SS Q1-6&, Definitions and Propositions. Connecting-rod. Prob lems 67 yiil CONTENTS., •^ChAPTBR III. CUKVILINEAR MOTION OP A MATERIAL POINT. PAOS §§ 70-74. Composition of Motions, of Velocities, etc. General Equations 72 §§ 75-84 Normal Acceleration. Centripetal and Centrifugal Forces. Simple Pendulums. Projectiles. Rela- tive Motion 77 Chapter IV. Moment of Inertia. %% 85-94. Plane Figures. Rigid Bodies. Reduction Formulae. The Rectangle, Triangle, etc. Compound Plane Figures. Polar Moment of Inertia 9t §S 95-104. Rod. Tliin Plates. Geometric Solids 98 ■§§105-107. Numerical Substitution. Ellipsoid of Inertia ICo Chapter V. Kinetics of a Rigid Body. §§108-115. Translation. Rotation about a Fixed Axis. Centre of Percussion 105 |§ 117-121. Compound Pendulum. The Fly-wheel 118 §§ 122-123. Uniform Rotation. " Centrifugal Action," Free Axes. 125 |§ 124-136. Rolling Motions. Parallel Rod of Locomotive 130 ^ Chapter VI. Work, Energy, and Power. §§ 127-134. Work. Power, Horse-power. Kit etic Energy...... 133 §§ 135-138. Steam-hammer. Pile-driving. Inelastic Impact -. 188 §§ 139-141. Rotary Motion. Equivalent Systems of Forces. Any Motion of a Rigid Body .„ 143 §§ 142-146. Work and Kinetic Energy in a Moving Machine of Rigid Parts 147 §§ 147-155. ^ Power of Motors. Potential Energy. Heat, etc. Dy- namometers. Boat-rowing. Examples 153 Chapter VII. Friction. §§ 156-164. Sliding Friction. Its Laws. Bent Lever 164 §§ 165-171. Axle-friction. Friction Wheels. Pivots. Belting. Transmission of Power by Belting 175 |§ 172-177. Rolling Friction. Brakes. Friction of Car Journals; and of Well-lubricated Journals. Rigidity of Cordage. Examples 186 CONTENTS. IX PART III. STRENGTH OF MATERIALS. (or mechanics of matekials.) OHAPTER I. ELEMENTARY STRESSES AND STRAINa ^§ 178-183. Stress and Strain i of Two Bands. Oblique Section of Rod in Tension . 195 §§ 183a-190o Hooke's Law. Elasticity. Safe Limit. Elastic Limit. Rupture. Modulus of Elasticity. Isotropes. Resilience. Ellipse of Stress. Classification of Cases 301 TENSION. Hooke's Law by Experiment. Strain Diagrams. Lateral Contraction. Modulus of Tenacity. . . . 307 Besilience of Stretched Prism. Load Applied Sud- denly. Prism Under Its Own Weight. Solid of Uniform Strength. Temperature Stresses . . 313 §§ 191-195. i§ 196-199b §§ 300-30Sp §§ 303-306. Tables 3§ so^-aia COMPRESSION OF SHORT BLOCKS. Short and Long Columns. Remarks on Grusliing, . . EXAMPLES IN TENSION AND COMPRESSION. Factor of Safety. Practical 318 Examples. Notes. .... SHEARING. Eivets. Shearing Distortion. Examples. ............... 230 Punching. cooo..^ .. 335 ^ CHAPTER II. TORSION, §§ 314-330. Angle and Moment of Torsion. Torsional Strengtila, Stiffness, and Resilience. Non-Circular Shafts 383 §§ 331-333. Transmission of Power. Autographic Testing Ma- chine. Torsion Clinometers. Examples 238 OHAPTER in. FLEXURE OF HOMOGENEOUS PRISMS UNDER PERPENDICULAR FORCES IN ONE PLANE. §§ 334r-333ao The Common Theory. Elastic Forces. Neutral Axis. The "Shear" and "Moment." Flex- lu-al Strength and Stiffness. Radius of Curva- Uire. Resilience c ... c . ^ .-..., ^ - .. . 344 ELASTIC CURVES. ^§ 333-338. Single Central Load ; at Rest, and Applied Suddenly. Eccentric Load. Uniform Load. Cantilever. . 353 X co:n'tents. SAFE LOADS. §§ 239-246. Maximum Mom^ent. Shear = x-Derivative of the Moment. Simple Beams With Various Loads. Comparative Strength and Stiffness of Rectan- gular Beams 262 §§ 247-252. Moments of Inertia. Rolled Steel I-Beams, etc. Cantilevers. Tables. Numerical Examples . . . 273^ SHEARING STRESSES IN FLEXURE. §g 253-257. Shearing Stress Parallel to Neutral Surface ; and in Cross Section. Web of I-Beam. Riveting of Built Beams . , 284 SPECIAL PROBLEMS IN FLEXURE. §§ 258-265. Designing Sections of Built Beams. Moving Loads. Special Cases of Quiescent Loads. Hydrostatic Load 395 §§ 266-270. Strength of Flat Plates. Weight Falling on Beam. Crank Shaft. Other Shafts. Web of I-Beam . 310^ §§ 271-276. CHAPTER IV. FLEXURE; CONTINUED. CONTINUOUS GIRDERS. Analytical Treatment of Symmetrical Cases of Beams on Three Supports ; also, Built in ; (see p. 499.) . 320' THE DANGEROUS SECTION IN NON-PRISMATIC BEAMS. §§ 277-279. Double Truncated Wedge, Pyramid, and Cone 332 NON-PRISMATIC BEAMS OF UNIFORM STRENGTH. §§ 280-289. Parabolic and Wedge-Shaped Beams. Elliptical Beams. (See Appendix, p. 841, for Cantilevers). 335 CHAPTER V. FLEXURE OF REINFORCED CONCRETE BEAMS. §§ 284-287. Concrete, and "Concrete-Steel" Beams. Beams of Rectanaiular Section. Horizontal Shearing Stresses in Latter. Examples 33& §§ 288-294. Concrete-Steel Beams of T-Form Section. Deflection of Concrete-Steel Beams. Practical Fotmulse and Diagrams 346 CONTENTS. XI CHAPTER VI. FLEXURE. COLUMNS AND HOOKS. OBLIQUE LOADS. §§ 294-300. Oblique Cantilever. Moment, Thrust and Shear. Experimental Proof. Common Theory of Crane-Hooks. Winkler-Bach Theory 352 |§ 301-314. Long Columns. End-Conditions. Euler's and Ran- kine's Formulae. Examples. Radii of Gyra- tion. Built Columns. Other Formul:e: Merri- man-Ritter ; ' ' Straight-Line ; ' ' Parabolic . Wooden Posts. Eccentric Loading of Columns. Eccentric Loading Combined with Uniform Transverse Pressures. . Buckling of Web of Plate Girder 360 CHAPTER VII. LINEAR ARCHES (OF BLOCKWORK.) §§ 315-823. Inverted Catenary. Parabolic Arch. Circular Arch. Transformed Catenary as Arch 886 CHAPTER VIII. ELEMENTS OF GRAPHICAL STATICS §§ 831-326. Force Polygons. Concurrent and Non-Concurrent Forces in a Plane. Force Diagrams. Equili- brium Polygons 897 §§ 327-832. Constructions for Resultant, Pier-Reactions, and Stresses in Roof Truss. Bow's Notation. The Special Equilibrium Polygon 403 CHAPTER IX. GRAPHICAL STATICS OF VERTICAL FORCES. §g 333-336. Jointed Rods. Centre of Gravity 413 §§ 337-848. Useful Relations Between Force Diagrams and Their- Equilibrium Polygons 415 CHAPTER X. RIGHT ARCHES OF MASONRY. §§ 344-353. Definitions. Mortar and Friction. Pressure in Joints. Conditions of Safe Equilibrium. True Linear Arch 421 §§ 353-357. Arrangement of Data for Graphic Treatment 428 §§ 358-363. Graphical Treatment of Arch. Symmetrical and Unsymmetrical Cases 431 CHAPTER XI. ARCH-RIBS. §§ 334-374, Mode of Support. Special Equilibrium Polygon and its Force Diagram. Change in Angle Between Rib Tangents. Displacement of Any Point on Rib 438 §§ 374a-378a. Graphical Arithmetic. Summation of Products. . Moment of Inertia by Graphics. Classification of Arch-Ribs 450 XU CONTENTS. §§ 379-388. Prof. Eddy's Graphical Method for Arch-Ribs of Hinged Ends ; and of Fixed Ends. Stress Diagrams. Temperature Stresses. Braced Arches 461 §§ 389-391. Cu-cular Ribs and Hoops 479 CHAPTER XII. FLEXURE OF BEAMS, SIMPLE AND CONTIN- UOUS, GEOMETRICAL TREATMENT. §§392-398. Geometrical Treatment Defined. Angle between End- tangents. Relative Displacement of Any Point of Elastic Curve. Deflections and Slopes by Calculus. Examples. Properties of Moment Diagrams. Deflections and Slopes by Geomet- ric Method. Examples 485 §§ 399-405. The "Normal Moment Diagram." The Theorem of Three Moments. Values of Products of Moment- Areas by "Gravity x's." Continuous Girders Treated by the Theorem of Three Moments. Continuous Beam with "Built-in" Ends. Deflections Found by Theorem of Three Mo- ments 494 CHAPTER XIII. THICK HOLLOW CYLINDERS AND SPHERES. §§ 405a-405sr. General Relations between Stress and Strain. The " Elongation Theory" of Safety. Thick Hollow Cylinder under Internal Fluid Pressure; also under External Fluid Pressure. Equalization of Hoop Stress in Compound Cylinder. Equa- tion of Continuity for Cylinder and Sphere. Thick Hollow Sphere under Internal Fluid Pressure 507 [CONTENTS OF ''MECHANICS OF FLUIVS,"] PAET lY.— HYDKAULICS. CHAPTER I.— DEFINITIONS. FLUID PRESSURE. HYDRO= STATICS BEGUN. PAGE §§ 40&-417. Perfect fluids. Liquids and Gases. Principle of ' ' Equal Transmission of Pressure." Non-planar Pistons .. . 515 §§ 418-427. Hydraulic Press. Free Surface of Liquid. Barometers and Manometers. The Differential Manometer. Safety-valves. Strength of Thin Hollow Cylinders against Bursting and Collapse 526 CHAPTER II.— HYDROSTATICS CONTINUED. PRESSURE OF LIQUIDS IN TANKS AND RESERVOIRS. §§428-434. Liquid in Motion, but in "Relative Equilibrium." Pressure on Bottom and Sides of Vessels. Centre of Pressure of Rectangles, Triangles, etc 540' §§ 435-444. Stability of Rectangular and Trapezoidal Walls against Water Pressure. High Masonry Darns. Proposed Quaker Bridge Dam. Earthwork Dam. Water Pressure on both Sides of a Gate 554 CHAPTER IIL -EARTH PRESSURE AND RETAINING WALLS. §§ 445-455. Angle of Repose. Wedge of Maximum Thrust. Geo- metrical Constructions. Resistance of Retaining Walls. Results of Experience 573 CHAPTER IV.— HYDROSTATICS CONTINUED. IMMERSION AND FLOTATION. §§456-460. Buoyant Effort. Examples of Immersion. Specific Gravity. Equilibrium of Flotation. Hydrometer.. 586- §§ 461-465. Depth of Flotation. Draught and Angular Stability of Ships. The Metacentre 593. xiii XIV CONTENTS. CHAPTER v.— HYDROSTATICS CONTINUED. GASEOUS FLUIDS. §§466-478. Thermometers. Absolute Temperature. Gases and Va- pors. Critical Temperature. Law of Charles. Closed Air-manometer. Mariotte's Law. Mixture of Gases. Barometric Levelling. Adiabatic Change.. G04 §§479-489. Work Done in Steam-engine Cylinders. Expanding Steam. Graphic Representation of Change of Stat? of Gas. Compressed-air Engine. Air-compressor. Hot-air Engines. Gas-engines. Heat- efficiency. Duty of Pumping-engines. Buoyant Effort of the Atmosphere 624 CHAPTER VL— HYDROKINETICS BEGUN. STEADY FLOW OF LIQUIDS THROUGH PIPES AND ORIFICES. §§ 489a-495. Phenomena of a " Steady Flow." Bernoulli's Theorem for Steady Flow without Friction, and Applications, Orifice in " Thin Plate" 646 §§ 496-500. Rounded Orifice. Various Problems involving Flow through Orifices. Jet from Force-pump. Velocity and Density; Relation. Efflux under Water. Efflux from Vessel in Motion. Barker's Mill 663 §§ 501-508. Efflux from Rectangular and Triangular Orifices. Pon- celet's Experiments. Perfect and Complete Con- traction, etc. Overfall Weirs. Experiments of Francis, Fteley and Stearns, and Bazin. The Cip- poUetti Weir. Short Pipes or Tubes 672 §§ 509-513. Conical Tubes. Venturi's Tube Fluid Friction. Fronde's Experiments. Bernoulli's Theorem with Friction. Hydraulic Radius. Loss of Head. Prob- lems involving Friction Heads in Pipes. Accumu- lator 693 §§ 513«-518. Loss of Head in Orifices and Short Pipes. Coefficient of Friction of Water in Pipes. Fanning's Tabic. Petroleum Pumping. Flow through Long Pipes . . 703 §§519-526. Chezy's Formula. The "Hydraulic Grade-Line." Pressure-energy. Losses of Head due to Sudden Enlargement of Section; Borda's Formula. Dia- phragm in Pipe. Venturi Water-meter 71'! .?§ 537-536. Sudden Diminution of Section. Losses of Head due to Elbows, Bends, Valve-gates, and Throttle-valves. Examples, Capt. Bellinger's Experiments on Elbows. Siphons. Branching Pipes. Time of Emptying Vessels of Various Forms; Prisms, Wedges Pyra- mids, Cones, Paraboloids, Spheres, Obelisks, and Volumes of Irregular Form using Simpson's Rule. . 727 CONTEXTS. XV CHAPTER VII.— HYDROKINETICS, CONTINUED; STEADY FLOW OF WATER IN OPEN CHANNELS. PAGE §§ 538-542a. Nomenclature. Velocity Measurements and Instru- ments for the same. Ritchie-Haskell Direction Current-meter. Change of Velocity with Depth. Pitot's Tube and W. M. White's Experiments. Current-meters. Gauging Streams. Chezy's For- mula for Uniform Motion in Open Channel. Experiments 749 §§ 54Ji6-647. Gutter's Formula. Sections of Least Resistance. Trape- zoidal Section of Given Side Slope and Minimum Friction. Variable Motion in Open C'hauuel. Bends. Formula introducing Depths at End Sec- tions. Backwater 758 CHAPTER VIII— KINETICS OF GASEOUS FLUIDS. §§ 548-556. Theorem for Steady Flow of Gases without Friction. Flow through Orifices by Water-formula; with Isothermal Expansion; with Adiabatic Expansion. Maximum Flow of Weight. Experimental Co- efficients for Orifices and Short Pipes. Flow con- sidering Velocity of Approach - W3 §^ 567-561a. Transmission of Compressed Air through Long Pipes. Experiments in St. Gothard Tunnel. Pipes of Vari- able Diameter. Tiie Piping of Natural Gas. ...... 786 J CHAPTER IX.— IMPULSE AND RESISTANCE OF FLUIDS §§ 562-569. Reaction of a Jet of Liquid. Impulse of Jet on Curved Vanes, Fixed and in Motion. Pitot's Tube. The California "Hurdy-gurdy." Impulse on Plates. Plates Moving in a Fluid. Plates in Currents of Fluid .... 798 B§ 57C'-575. Wind-pressure. Smitlisoniau Scale. Mechanic-;- of the Sail-boat. Resistance of Still Water to ImmersL'-'l Solids in Motion. Spinning Ball, Deviation fiom Vertical Plane. Robinson's Cup anemometer. Re- sistance of Ships. Transporting Power of a Cur- rent. Fire-streams, Hose- friction, etc - 818 Appendix. Miscellaneous Addenda and Tables 835-854 MEOHAmCS OF ENGINEERING. PEELIMINARY CHAPTER. 1. Mechanics treats of the nnitual actions and relative mo- tions of material bodies, solid, liquid, and gaseous ; and by Mechanics of Engineering is meant a presentment of those principles of pnre raeclianics, and their applications, which are of special service in engineering problems. 2. Kinds of Quantity. — Mechanics involves the following fundamental kinds of quantit}' : Space, of one, t\vo, or three dimensions, i.e., length, surface, or volume, respectively ; time, which needs no definition here; force and mass, as defined be- low; and abstract numbers, whose values are independent of arbitrary units, as, for example, a ratio. 3. Force. — A force is one of a pair of equal, opposite, and simultaneous actions between two bodies, by which the state* of their motions is altered or a change of form in the bodies themselves is effected. Pressui-e, attraction, repulsion, and traction are instances in point. Muscular sensation conveys the idea of force, while a spring-balance gives an absolute measure of it, a beam-balance only a relative measure. In accordance with Newton's third law of motion, that action and reaction are equal, opposite, and simultaneous, foi'ces always occur in pairs; thus, if a pressure of 4:0 11)S. exists between bodies A and B, if A is considered by itself (i.e., " free"), apart from all other bodies whose actions upon it are called forces, among these forces will be one of 40 lbs. directed from B toward A. Similarly, if B is under consideration, a force * The state of motion of a small body under the action of no force, or of balanced forces, is cither absolute rest, or uniform motion in a right line. If the motion is different from this, the fact is due to the action of an un- balanced force (§ 54), 2 MECHANICS OF ENGINEBRINO. of 40 lbs. di]-ected from A toward £ takes its place ;imoiig the forces acting on £. This is the interpretation of Newton's third law, [Note.— In some common phrases, such as " The tremendous force '' o^ a heavy Dody in rapid motion, the word force is not used in a technical sense, but signifies energy (as ex- plained in Chap. VI.). The mere fact that a body is in motion, whatever its mass and velocity, does not imply that it is under the action of any force, necessarily. For instance, at any point in the path of a cannon ball through the air, the only forces acting on it ara the resistance of the air and the attraction of the earth, the latter having a vertica in(J downward direction.] 4. Mass is the quantity of matter in a body. The masses of several bodies being proportional to their weights at the same locality on the earth's surface, in physics the weight is taken as the mass, but in practical engineering another mode is used for measuring it (as explained in a subsequent chapter), viz.'. the mass of a body is equal to its weight divided by the ac- celeration of gravity in the locality where the weight is taken, or, symbolically, M= G -r- g. This quotient is a constant quantity, as it should be, since the mass of a body is invariable wherever the body be carried. 6. Derived Quantities. — All kinds of quantity besides the fundamental ones just mentioned are compounds of the latter, formed by multiplication or division, such as velocity, accele- ration, momentum, work, energy, moment, power, and force- disti'ibution. Some of these are mej-ely names given for convenience to certain combinations of factors which come together not in dealing with first principles, but as a result of common algebraic transformations. 6. Homogeneous Equations are those of such a form that they are true for any arbitrary system of units, and in which all terms combined by algebraic addition are of the same kind. of Thus, the equation s = ~ (in which g = the acceleration of gravity and t the time of vertical fall of a body in vacuo, from rest) will give the distance fallen through, «, whatevei units be adopted for measuring time and distance. But if foi PRELIMIlSrARY CHAPTER. S g we write the niimerlcal value 32.2, which it assumes when time is measured in seconds and distance in feet, the equation s = IQ.lf is true for those units alone, and the equation is not of liomogeneous form. Algebraic combination of homogeneous equations should always produce homogeneous equations ; if not, some error has been made in the algebraic woi'k. If any equation derived or proposed for practical use is not homogene ous, an explicit statement should be made in the context as to the proper units to be employed. 7. Heaviness. — By heaviness of a substance is meant tlie weight of a cubic unit of the substance. E.g. the heaviness of fresh water is 62.5, in case the unit of force is the pound, and the foot the unit of space; i.e., a cubic foot of fresh water weighs 62. 5 lbs.* In case the substance is not uniform in composition, the heaviness varies from point to point. If the weight of a homogeneous body be denoted by G, its volume by F", and the heaviness of its substance by y, then G = Yy, Weight in Pounds of a Cubic Foot (i.e., the heaviness) of vakious MATEIIIAL& Anthracite, solid 100 " broken 57 Brick, common hard 125 " soft 100 Brick-work, common 112 Concrete 125 Earth, loose 72 " as mud 102 Granite 164 to 172 Ice 58 Iron, cast 450 " wrought 480 Masonr}^ dry rubble 138 " dressed granite or limestone 165 Mortar 100 Petroleum _ 55 Snow 7 " wet 15 to 50 Steel 490 Timber 25 to 60 Water, fresh 62. 5 sea 64.0 8. Specific Gravity is the ratio of the heaviness of a material to that of water, and is therefore an abstract number. 9. A Material Point is a solid body, or small particle, whose dimensions are practically nothing, compared with its range of motion. - Or, we may write 62.5 lbs. /cub. ft.; or 62.5 Ibs./ft.^ 4 MECHANICS OF ENGINEERING. 10. A Eigid Body is a solid, M-liose distortion or change of form under anj system of forces to be brought upon it in practice is, for certain purposes, insensible. 11. Equilibrium. — When a system of forces applied to a body produces the same effect as if no force acted, so far as the state of motion of the body is concerned, they are said to be balanced, or to be in equilibrium. [If no force acts on a material point it remains at rest if already at rest ; but if already in motion it continues in motion, and uniformly (equal spaces in equal times), in a right line in direction of its original motion. See § 54.] 12. Division of the Subject. — ^to^^'c* will treat of bodies at rest, i.e., of balanced forces or equilibrium; kinetics, of bodies in motion ; strength of materials will treat of the effect of forces in distorting bodies ; hydraulics, of the mechanics of liquids and gases (thus mcXxx&mg j)7ieumatics). 13. Parallelogram of Forces. — Ducliayla's Proof. To fully determine a force we must have given its amount, its direc- tion, and its point of application in the body. It is generally denoted in diagrams by an arrow. It is a matter of experience that besides the point of application already spoken of any other may be chosen in the line of action of the force. This is called the transmissibility of force; i.e., so far as the state of motion of the body is concerned, a force may be applied any- where in its line of action. The Resultant of two forces (called its components) applied at a point of a body is a single force applied at the same point, which will replace them. To prove that this resultant is given in amount and position by the diagonal of the parallelogram formed on the two given forces (conceived as laid off to some scale, so many pounds to the inch, say), Duchayla's method requires four postulates, viz. : (1) the resultant of two forces must lie in the same plane with them ; (2) the resultant of two equal forces must bisect the angle between them ; (3) if one of the two forces be increased, the angle between the other force and the resultant will be greater than before; and (4) the trans- missibility of force, already mentioned. Granting these, we proceed ns follows (Fig. 1) : Given the two foi'ces P and Q - PRELIMINARY CHAPTER. 5 P' + P" {P' and P" being each equal to P, so that Q = 2P), applied at 0. Transmit P'[ to A. Draw the parallelograms OP and AP ; OP will also be a parallelogram. By postulate (2), since OP is a rhombus, P and P' at may be replaced by a single force P acting through P. Transmit P' to P and replace it by P and P\ Transmit P from P to A, P' from P to i?. Similarly P and i-*", at A, may be replaced by a single force P" passing through P ; transmit it there and re- solve it into P and P" . P' is already at P, Hence P and P' -\- P'\ acting at J?, are equivalent to P and P' -f- P" act- ing at {?, in their I'espective directions. Therefore the result- ant of P and P' -\- P" must lie in the line OP^ the diagonal of the parallelogram formed on P and Q = 2P at O. Similarly SLg-a C/ FV /B --:^...N^.;::JD H\E Fig. 2. this may be proved (that the diagonal gives the direction of the resultant) for any two forces P and mP ; and for any two forces nP and mP, m and i^ being any two whole numbei-s, i.e., for any two commensurable forces. When the forces are incommensurable (Fig. 2), P and Q being the given forces, we may use a reductio ad ahsurdum^ thus : Form the parallelo- gram OP on P and Q applied at 0. Snppose for an instant that P the resultant of P and Q does not follow the diagonal OP, but some other direction, as OP'. Note the intersection H, and draw HG parallel to PP. Divide P Into equal parts, each less than HP ; then in laying off parts equal to these from O along OP, a point of division will come at some point F between C and P. Complete the parallelogram OFEG. The force Q" = OF is commensurable with P, and hence their 6 MECHANICS OF ENGINEERING. resultant acts along OE. Now Q is greater than Q'\ while R makes a less angle with P than OE^ which is contrary to pos- tulate (3); therefore R cannot lie outside of the line OD. Q. E. D. It still remains to prove that the resultant is represented in amount, as well as position, by the diagonal. OD (Fig. 3) is ••. /p' the direction of M the resultant of P and /F ''\^ Q ; required its amount. If P' be a force ^"~— "^^r:;^ y equal and opposite to P it will balance P ■••• i^ "''nD/. ^^^ Q 5 ^'^j tl^^ resultant of P' and P P l^'-< must lie in the line QO prolonged (besides ^^**' ^' being equal to Q). We can therefore de- termine P' by drawing PA parallel to DO to intersect QO prolonged in A ; and then complete the parallelogram BF on BO and BA as sides. Since OFAB and AODB are paraUelograms, OF must=5A and BA must = OL'. Hence OF and OD are equal and lie on the same right line. Evidently if R^ were any shorter or any longer than OF the resultant of it and OB(=P) would not take the direction QOA. Hence R^ must = 0F, i.e., =0D', and hence R=zOD in amount. Q. E. D. Corollary. — The resultant of three forces applied at the same point is the diagonal of the parallelopiped formed on the three forces. 14. Concurrent forces are those whose lines of action intersect in a common point, while non-concurrent forces are those which do not so intersect ; results obtained for a system of concurrent forces are really derivable, as particular cases, from those per- taining to a system of non-concurrent forces. 15. Resultant. — A single force, the action of which, as re- gards the state of motion of the body acted on, is equivalent to that of a number of forces forming a system, is said to be the Resultant of that system, and may replace the system ; and con- versely a force which is equal and opposite to the resultant of a system will balance that system, oi', in other words, when it is combined with that system there will result a new system in equilibrium ; this (ideal) force is called the Anti-resultant. In general, as will be seen, a given system of forces can al- PRELIMINARY CHAPTER. 7 ways De I'eplaced by two single forces, but tliese two can be combined into a single resultant only in particular cases. 15a. Equivalent Systems are those which may be replaced by the same set of two single forces — or, in other words, those which have the same effect, as to state of motion, upon the given body. 15b. Formulae. — If in Fig. 3 the forces P and $ and the angle or = PO Q are given, we have, for the resultant. JS = OD = V-f" + §' + 2 Pg cos tx. (If a is > 90° its cosine is negative.) In general, given any three parts of either plane triangle D Q, or D B, the other three may be obtained by ordinary trigonometry. Evidently if a = 0, R = P + Q; ifa = 180°, i? = P - ^ ; and if a =t 90°, R = V -?" + Q"- 15c. Varieties of Forces. — Great care should be used in deciding what may properly be called forces. The latter may be divided into ac- tions by contact, and actions at a distance. If pressure exists between two bodies and they are perfectly smooth at the surface of contact, the pressure (or thrust, or compressive action), of one against the other constitutes a force, whose direction is normal to the tangent plane at any point of contact (a matter of experience) ; while if those surfaces are not smooth there may also exist mutual tangential actions or friction. (If the bodies really form a continuous substance at the surface considered, these tangential actions are called shearing forces.) Again, when a rod or wire is subjected to tension, any portion of it is said to exert a pull or tensile force upon the remainder ; the ability to do this depends on the property of cohesion. The foregoing are examples of actions by contact. Actions at a distance are exemplified in the mysterious attractions, or re- pulsions, observable in the phenomena of gravitation electricity, and mag- netism, where the bodies concerned are not necessarily in contact. By the term weight we shaU always mean the force ot the earth's attraction on the body in question, and not the amount of matter in it. lad. Example 1. — If OD, = R, is given, =40 lbs., while the angle BOD is 110° and QOD = 40° (also = ODB), find the components P and Q. Solution. — From the triangle BOD, OB.OD: :sin 40°: sin 30°; whence P, or OB, = (40X0.6428) -^ 0.5000 = 51.42 lbs. Similarly, from triangle BOD, we have BD:OD: :sin 110°: sin 30°, .-. Q, or 5Z), = (40X0.9397) ^0.05000 = 75.17 lbs. Example 2.— Given P = 20 lbs., Q = 30 lbs., and angle a{ = POQ), =115°, find the resultant R in amount and direction. As to amount R^s/{20)'+{30) +2X 20X30 X( -0.4226) = V792:88 = 28.16 lbs. As to direction, let /? denote the angle ODB,==QOD; we then have, from triangle OBD, 20:28.16: :sin /?:sin 65°; whence, solving, sin/? =(20X0.9063)^-28.16 = 0.6437; i.e., angle ^ = 40° 4'. PART I.-STATICS. CHAPTER I. STATICS OF A MATERIAL POINT. 16. Composition of Concurrent Forces. — A system of forces acting on a material point is necessarily composed of concurrem: forces. Case I. — All the forces in One Plane. Let be the material point, the common point of apph'catiou of all the forces ; Pj, P^, etc., the given forces, making ""j?z angles tVj, a^^ etc., with the axis X. By the -■/A -p^p, parallelogram of forces P, may be resolved ^/J2^4^i i i^to and replaced by its components, P^ cos or, -^-* — '- — *^— ^ acting along JT, and P^ sin a^ along Y. Fig. 4. Similarly all the remaining forces may be re< placed by their X and Y components. We have now a new system, the equivalent of that first given, consisting of a set of ^forces, having the same line of application (axis X^^ and a set of I^ forces, all acting in the line Y. The resultant of the X forces being their algebraic snm (denoted by "^X^ (since they have the same line of application) we have ^X=^ P^ cos a, -\- P^ COS fl'j + etc. = '2{^P cos «), and similarly ^Y =^ P^ sin ar, + P^ sin a^ -|- etc. = 2{P sin a). These two forces, 2X and ^Y^ may be combined by the parallelogram of forces, giving P = VCSXY -\- i^Y^ ^^ ^^^^ single resultant of the whole system, and its direction is deter- :sY mined by the angle or; thus, tan a = ^^r^-; see Fig. 5. For eQuilibrium to exist, R must = 0, which requires, sejparately^ STATICS OF A MATERIAL POINT. 9 '2X^=0, and ^1^= (for tlie two squares {2X^y and {2 Yy can neither of them be negative quantities). Case II. — The forces having any directions in space, but all applied at 0, the material point. Let ^j, P^, etc., be the given forces, jP^ making the angles a^, ^j, and y^, respectively, with tliree arbitrary axes, X^ T^, and Z (Fig. 6), at right angles to each other and intersecting at 0, the origin. Siniilai-ly let a^, /3^, y^, be the angles made by jP^ with these axes, and so on for all the forces. By the parallelepiped of forces, 7^1 may be replaced by its components. Xi = Pi cos ofj, Yi = Pi cos /3i, and Z^ = JP^ cos ;/, ; and Y 2Y' R ^^ : X sx Fig. 5. Fig. 6. Fig. r. similai-ly for all the forces, so that the entire system is now- replaced by the tliree forces, 2X = F, cos a^ + J\ cos a^ -\- etc ; 2 T = P, cos ^, + P, cos 13, + etc ; ^Z = P, cos y, + P^ cos /^ + etc ; and finally by the single resultant R = V{2Xf + [2 ry + {2zy. Therefore, for eqnilibrinm we mnst have separately, :SX= 0, :SY = 0, and 2Z= 0. ^s position may be determined l)y its direction cosines, viz., cos 2x , ^r ^z a — —^ ; cos // == -jy- ; cos ;k = -^. 17. Conditions of Equilibrium. — Evidently, in dealing with a system of concurrent forces, it would be a simple matter to 10 . MECHANICS OF ENGINEERING. replace any two of the forces by their resultant (diagonal formed on them), then to combine this resultant with a third force, and so on until all the forces had been combined, the last resultant being the resultant of the whole system. The foregoing treatment, however, is useful in showing that for equilibrium of concurrent forces in a plane onlj' two conditions are necessary, viz., ^ JT = and 2 JT = 0; while in space there are three, 2^= 0, 2 Y = 0, and 2Z = 0. In Case I., then, we have conditions enough for determining two unknown quantities ; in Case II., three. 18. Problems involving equilibrium of concurrent forces. (A rigid body in equilibrium under no more than three forces may be treated as a material point, since the (two or) three forces are necessarily concurrent.)* ■ Problem 1. — A body weighing G lbs. rests on a horizontal table: required the pressure between it and the table. Fig. 8. Consider the body free, i.e., conceive all other bodies removed , (the table in this instance), being replaced by the forces which they exert on the first body. Taking the axis J" vertical and positive upM^ard, and not +X assuming in advance either the amount or drrec- |IM tion of JV, the pressure of the table against the I body, but knowing that G, the action of the earth Fig. 8. ^^ ^j^^ body, is vertical and downward, we have here a system of concurrent forces in equilibrium, in which the ^ and Y components of G are known (being and — G respectively), while those, iVx ^"^ -^^ of JV are unknown. Putting 2^ = 0, we have JV^ -|- = ; i.e., iVhas no hori- zontal component, .'. iV is vertical. Putting 2 Y = 0, we have iVy — G = 0, .". JV^ =^ -\- G; or the vertical component of JV, i.e., JV itself, is positive (upward in this case), and is numerically equal to G. Peoblem 2. — Fig. 9. A body of weight G (lbs.) is moving in a straight line over a rough horizontal table with a uniform velocity v (feet per second) to the right. The tension in an oblique cord by which it is pulled is given, and = P (lbs.), * Three parallel forces form an exception ; see §§ 20, 21, etc. STATICS OF A MATERIAL POINT. 11 •which remains constant, the cord making a given angle of elevation, a^ with the patli of the body. Required the vertical pressure iV (lbs.) of the table, and also its ^y' horizontal action F (friction) (lbs.) against the body Referring by anticipation to Newton's fii'st law of motion, viz., a material point acted •on by no force or by balanced forces is either fig. 9. .at rest or moving uniformly in a straight line, we see that this problem is a case of balanced forces, i.e., of equilibrium. Since there are only two unknown quantities, iV and F, we may •detei-mine tliem by the two equations of Case I., taking tlie axes Xand Y as before. Here let us leave the direction of iVas well as its amount to be determined by the analysis. As ^must evidently point toward the left, treat it as negative in summing the X components ; the analysis, therefore, can be •expected to give only its numerical value. 2X = gives P 0,0^ a — F = 0. .-. F = P cos a. ^^I^= gives iV+P sin «- G = 0. .-. I^= G - Psin a. .". iV is upward or downward according as 6^ is > or < P sin a. For i\^ to be a downward pressure upon tlie body would require the surface of the table to be above it. The ratio of the friction F to the pressure iV" which produces it can now be •obtained, and is called the " coefficient of friction." It may Tary somewhat with the velocity. (See p. 168.) This problem may be looked npon ns arising fi'om an experi- ment made to determine tlie coefiicieiit of friction between the given surfaces at the given uniform velocity. 19. The Eree-Body Method. — The foregoing rather labored so- lutions of very simple problems have been made such to illus- trate what may be called the "free-body method" of treating any problem involving a body acted on by a system of forces. It consists ill conceiving the body isolated from all others which act * on it in any way, those actions being introduced as so many forces known or unknown, in amount and position. The sys- tem of forces thus formed may be made to yield certain equa- tions, whose character and number depend on circumstances, such as the behavior of the body, whether the forces are confined to * That is, in any "force-ahle" way. 12 MECHANICS OF ENGINEERING. a plane or not, etc. , and which are therefore theoretically avail- able for determining an equal number of unknown quantities. li'a. Examples. — 1. A cast-iron cylinder, with axis horizontal, rests against two smooth inclined surfaces, as shown in Fig. 9a. Its length, I, is 4 ft., diameter, d, is 10 in., and "heaviness" (p. 3) 480 lbs. /cub. ft. Required the pressures (or "reactions," or " supportivg forces"), P and Q at the two points of contact A and B. (Points, in the end view.) These pressures on the cylinder are shown pointing normal to the sur- faces {smooth surfaces) and hence pass through the center of the body, p; i \ 40° , / ,'^ "■^^ -^20° J^ ;i§p|| G \'?0'' ^] X 4X480= 1047.6 lbs. Fig. 9a. Fig. 9h. C, where we may consider the resultant weight, G, of the body to act. These' three forces, then, form a concurrent system, and the body is in equilibrium under their action. 4~''^4 . i:X = gives: +P cos 40°-Q cos 20° + = 0; (1) IY = Q " +Psin40° + Q sin 20°-G = 0; . (2) that is, numerically, 0.7660P-0.9396Q = 0; (3) and 0.6428P + 0.3420Q = 0.1047.6 lbs (4) From (3) we have P=1.227Q, which in (4) gives (0.7887 + 0.3420)Q = 1047.6 lbs. ; and hence Q = 926.4 lbs. \ . Therefore P, =1.227Q, =1127.6 lbs. / Example 2. — Fig. 96. The 4-ton weight is suspended on the bolt C, which passes through the ends of boom OC^ and tie-rod DC. Bolt C is also subjected to a horizontal pull tov/ard the left, due to the 2-ton weight, suspended as shown. . Find the pull P in the tie and the thrust Q in the boom. Note that the boom is pivoted at both ends and hence (if we neglect its weight) is under only two pressures; both of which, therefore (for the equilibrium of the boom), m,ust point along its length.. Hence the thrust Q on bolt C makes an angle of 41° with the horizontal. Similarly, P, the action of tie-rod on C, is at 15°. Solution. — At (?>) we see the bolt as a "free body"; in equilibrium under the four concurrent forces. 2X = Qcos41°-Pcosl5°-(?2-0 = 0; (5) iF = Q sin41°-Psin 15°-Gi-0 = 0; (6) or, numerically, 0.7547Q-0.9659P-2 = 0, (7) and 0.6560Q-0.2588P-4 = (8) From (7), Q = 2.6514- 1.279P, which in (8) gives 0.6560(2.651 + 1.279P) - 0.2588P = 4 ; that is, 1.740-H0.8390P-0.2588P = 4; and hence, finally, P= 2.260 ^0.5802 = 3.896 tons, and .-. Q = 7.633 tons. Ans. PARALLEL FORCES AND THE CENTRE OF GRAVITY. 13 CHAPTER 11. PARALLEL FORCES AND THE CENTRE OF GRAVITY. 20. Preliminary Remarks. — Althongli by its title tliis section sliould be restricted to a treatment of tiie equilibrium of forces, certain propositions involving the composition and resolution of forces, without reference to the behavior of the body under their action, will be found necessary as preliminary to the prin- cipal object in view. As a rigid body possesses extension in three dimensions, to deal with a system of forces acting on it we require three co- ordinate axes : in other words, tlie system consists of " forces in space," and in general the forces are non-concurrent. In most problems in statics, however, the forces acting are in one plane: we accordingly begin by considering non-concurrent forces in a plane, of which the simplest case is that of two parallel forces. For the present the body on which the forces act will not be shown in the figure, but must be understood to be there (since we have no conception of forces independently of material bodies). The device will frequently be adopted of introducing into the given system two opposite and equal forces acting in the same line : evidently this will not alter the e£fect of the given system, as regards the rest or motion of the body. 21. Resultant of two Parallel Forces. Case I. — The two forces have the same d'u'ection. Fig. 10. Let P and Q be the given forces, and AJB a line perpendicular to them {P and Q are supposed to have sL---3/---Js been transferred to the intersections ^^<*- ^^^ A and B). Put in at A and B two equal and opposite forces 8 and S^ combining them with P and Q to form P' 6 "S A TP t ^ i Q Q < VOr -"^ "f; ■X--H D /B S 14 MECHANICS OF ENGINEERING. shaded by dots, .•, iave* -7^ == Q X and Q'. Transfer P' and Q' to tlieir intersection at C, and thera resolve them again into S and P, /S'and Q. 8 and /iS^ annul each other at C', therefore P and Q^ acting along a common line CD, replace the P and Q first given ; i.e., the resultant of the origi- nal two forces is a force R =^P -\- Q, acting parallel to them through the point P, whose position must now be determined. The triangle CAP is similar to the triangle shaded by lines, .'. P : S :: GP : a?; and CPB being similar to the triangle 8 '. Q :: a — a? : CP. Combining these, we ■'• "^ ^ '^TO "^ %• ^^^ write this Px = Qa, and add Pc, i.e., i^c-j- Qg, to each member, c being the distance of (Fig. 10), any point in AP produced, from A. This wull give P{x -\- c) = Pc -\-Q{a -{- c), in which c, ■a-\~ c, and x -\- c are respectively the lengths of perpendiculars let fall from upon P, Q, and their resultant P. Any one of these products, such asPc, is for convenience (since products of this form occur so frequently in Mechanics as a result of alge- braic transformation) called the Moment of the force about the arbitrary point 0. Hence the resultant of two parallel forces of the same direction is equal to their sum, acts in their plane, in a line parallel to them, and at such a distance from any arbi- trary point in their plane as may be determined by writing its moment about equal to the sum of the moments of the two forces about 0. O is called a centre of moments, 'dud each of the perpendiculars a lever-arm. Case II. — Two parallel forces i^ and Q of opposite direc- 11. By a process similar to the foregoing, we obtain P =P- Q and {P — Q)x = Qa, i.e., Px = Qa. Subtract each member of the last equation from Pc (i.e., Pc—Qc), in wliicli c is the distance, from A, of any arbi- trary point in A£ produced. This gives P{c — x) = Pc — Q{a -j- c). But ((? — «), G, and {a-{-G) are re- FiG. 11. spectively the perpendiculars, from * That is, the resultant of two parallel forces pointing in the same direc- tion divides the distance between them, in the inverse ratio of those foi'ces. tions. PARALLEL FORCES AND THE CENTRE OF GRAVITY. 15 O., upon i?, P^ and Q. That is, i?(c — x) is the moment of R about 0\ Pc, that of P aboiit 0; and ^(«+c), that of Q about 0. But the moment of Q is subtracted from that of P, which corresponds with the fact that Q in tliis figure would produce a rotation about opposite in direction to that of P. Havi.jg in view, tlien, this imaginary rotation, we may define the moment of a force && positive when tlie indicated direction about the given point is against the hands of a watch; as nega- tive when with the hands of a watch.* Hence, in general, the resultant of any two parallel forces is, in amount, equal to their algebraic sum, acts in a parallel direc- tion in the same plane, while its moment, about any arbitrary point in the plane, is equal to the algebraic sum of the mo- ments of the two forces about the same point. Corollary. — If each term in the preceding moment equations be multiplied by the secant of an angle {a, Fig, 12) thus; p.. %-'^"" ^^^ ..^'i' Of^- — -a-i- jt 1 0, y j k — - fta 'A Fig. 13. Fig. 13. (using tlie notation of Fig. 12), we have Pa sec a = Pxai sec a+P^jii sec a, i.e., P6 = Pi6i -I-P2&25 in which, h, h\ and 62 are tHe oblique distances of the three lines of action from any point in tlieir plane, and lie on the same straight line ; P is the resultant of the parallel forces P^ and P2' 22. Resultant of any System of Parallel Forces in Space. — LetP*i, ^2? Pii 6tC'5 t>e the forces of the system, and a?,, y„ s„ a?,, ^j, Sj, etc., the co-ordinates of their points of application as referred to an arbitrary set of three co-ordinate axes X, Y^ and Z, perpendicular to each other. Each force is here re * These two directions of rotation are often called counter clockwise, and clockwise, rescectively. 16 MECHANICS OF ENGINEERING. stricted to a definite point of application in its line of action (with reference to establishing more directly the fundamental equations for the co-ordinates of the centre of gravity of a body). The resultant P' of any two of the forces, as Pj and /*„ is = P, + P^ ^^^ ™^y be applied at C, the in- tersection of its own line of action with a line BD joining the points of application of P^ and P^^ its components. Produce the latter line to^, where it pierces the plane ^Y^ and let 5„ &', and 5^, respectively, be the distances of B^ (7, D^ from A. \ then from the corollary of the last article we have p'y^Ph^p^K', but from similar triangles V \\\\\\z' : z, : 0„ .-. P'z' = P,z, + P,3,. Now combine P., applied at C^ with P^^ applied at E^ calling their resultant P" and its vertical co-ordinate z'\ and we obtain P"z" = P'z' + P3S3, i-e., P"z" = Pa + P.\ + ^3^3, also P-=P' + P3 = P,+P, + P,. Proceeding thus until all the forces have been considered, we shall have finally, for the resultant of the whole system, P-P. + i'.+ i^s + etc.; and for the vertical co-ordinate of its point of application, which we may "write 3, Rz — P,z, + P,s, + P3S3 + etc ; - P,z, + P,z, + P,z,... _^{Pz)^ ..e.,2 _ p^_^p^_^p^:^ - ^p , and similarly for the other co-ordinates. In these equations, in the general case, such products as P,j!!i» etc., cannot strictly be called moments. The point whose co» PARALLEL FORCES ANB THE CENTRE OF GRAVITY. 11 ordinates are the x, y, and b, just obtained, is called the Centre of Parallel Forces, and its position is independent of the {com- mon) direction of the forces concerned. ExaTYijple. — If the parallel forces are contained in one plane, and the axis I^be assumed parallel to the direction of the forces, then each product like P^x^ will be a moment, as de- fined in § 21 ; and it will be noticed in the accompanying nu- merical example, Fig. 14, that a detailed substitution in the equation R □ iY R ra _ t f^ L-i-l 1 i?a?=:P,a?, + P,a?,+ etc., . . . (1) U -i..i^ |_ having regard to the proper sign of each ^, 0| +X force and of each abscissa, gives the same fig. i4. result as if each product Px were first obtained numerically, and a sign affixed to the product considered as a moment about the point 0. Let P^ = — 1 lb.; P, = + 2 lbs.; P^ = + 3 lbs.; P^^~-& lbs.; a?^ = + 1 ft.; a;^ = -f 3 ft.; a?, = — 2 ft.; and a?^ = — 1 ft. Required the amount and position of the resultant R. In amount R = -SP =— 1 + 2 + 3 — 6 = — 2 lbs.; i.e., it is a downward force of 2 lbs. As to its position, Rx= 2{Px) gives ( — 2)« = ( - 1) X (+ 1) + 2 X 3 + 3 X (- 2) + (-6) X(-l) = -l + 6-6 + 6. Now from the figure, by inspection, it is evident that the moment of P, about is negative {with the hands of a watch), and is numer- ically = 1, i.e., its moment = — 1 ; similarly, by inspection, that of Pj is seen to be positive, that of P^ negative, that of P, positive; which agree with the results just found, that (- 2)^ = - 1 + 6 - 6 + 6 = + 5 ft. lbs. (Since a moment is a product of a force (lbs.) by a length (ft.), it may be called so many foot-pounds.) Next, solving for a?, we obtain X = (+ 5) -f- ( — 2) = — 2.5 ft.; i.e., the resultant of the given forces is a downward force of 2 lbs, acting in a vertical line 2.5 ft. to the left of the origin. Hence, if the body in question be a horizontal rod whose weight has been already included in the statement of forces, a support placed 2.5 ft. to the left of and capable of resisting at least 2 lbs. downward pressure will preserve equilibrium ; and the pressure which it exerts 18 MECHANICS OF ENGINEERING. against the rod must be an upward force, P^, of 2 lbs., i e. tiie equal and opposite of the resultant of P^, P^? P^^ Pa- Fig. 15 shows the rod as a fi-eo body in equilibrium under the live forces. P^ = -|- 2 lbs. — the reaction of the support. Of course P^ is one of a pair of equal and opposite forces ; the other one J is the pi'essure of the rod against the I' "—I \ ' ^ ^ ?^\ — -gi-s- iO support, and would take its place among Fig. 15. ■ the forces acting on the support. 23. Centre of Gravity. — Among the forces acting on any rigid body at the surface of the earth is the so-called attraction of the latter (i.e., gravitation), as shown by a spring-balance, which indicates the weight of the body hung upon it. The weights of the different particles of any rigid body constitute a system of parallel forces (practically so, though actually slightly convergent). The point of application of the resultant of these forces is called the centre of gravity of the body, and may also be considered the centre of onass, the body being of very small dimensions compared with the earth's radius. If a?, y, and z denote the co-ordinates of the centre of gravity of a body referred to three co-ordinate axes, the equations derived for them in § 22 are directlj' applicable, with slight changes in notation. Denote tlie weight of any particle * of the body by dG, its volume by d F, by ;^its heaviness (rate of weight, see § 7) and its co-ordinates by a?, y, and z ; then, using the integral sign as indicating a summation of like terms for all the particles of the body, vs^e have, \v: heterogeneous bodies (see also p. 119, Notes). r~_fy^dy^ -_frydV_, - _fr^. ,-,x ^- fydV' y ~ fydV' ^ - fydV^ * ' ^^^ while, if the body is homogeneous, y is the same for all its ele- ments, and being therefore placed outside the sign of sumnu\- tion, is cancelled out, leaving for homogeneous bodies {Y de- noting the total volume) -„ _-«I. ^ -Ml., and I --^I f2) a? — p:— , y — y , ana z — y . . . \^z) * Any subdivision of the body may be adopted for use of equations (1) and (2), etc.; but it must be remembered that the ai (or y, or s) in each term of the summations, or integralSj is the co-ordinate of the center of gravity of the subdivision employed. PARALLEL FOECES AND THE CEN"TRE OF GRAVITY. 19 Corollary. — It is also evident that if a homogeneous body is for convenience considered as made up of several finite parts, whose volumes are y^, F^? etc., and whose gravity co-ordinates are a?„ y„ z^ ; «„ y,, z^ ; etc., we may write . = -^-j-^-^-— .... (3) If the body is heterogeneous, put G^ (weights), etc., instead Df T^i, etc., in equation (3). If the body is an infinitely thin homogeneous shell of uni- form thickness = h, then dV =^ hdF {dF dawoimg an element, and J^the whole area of one surface) and equations (2) become, after cancellation, ^-f^il. z-Ml. -,-fi^ u) For a thin homogeneous plate, or shell, of uniform thick- ness, and composed of several finite parts, of area Fi, F2, etc., wdth gravity co-ordinates Xi, X2, etc., we may write _ FiXi+F2X2+ . . . , . - ^= F,+F2+... • • • • (^«) Similarly, for a homogeneous wi?'e of constant small cross- section (i.e.. a geometrical line, having weight), its length being s, and an element of length ds, we obtain 3=^;^=>^»;i=-^. ... ^5) 24. Symmetry. — Considerations of symmetry of form often determine the centre of gravity of homogeneous solids without analysis, or limit it to a certain line or plane. Tlius the centre of gravity of a sphere, or any regular polyedron, is at its centre of figure;, of a right cylinder, in the middle of its axis; of a thin plate of the form of a circle or regular polygon, in the centre of figure ; of a straight wire of uniform cross-section, in the middle of its length. Again, if a homogeneous body is symmetrical about a plane, the centre of gravity must lie in that plane, called a plane of 20 MECHANICS OF ENGINEEUING. gravity; if about a line, in that line called a line of gravity; if about a point, in that point. 25. By considering certain modes of subdivision of a homo- geneous body, lines or planes of gravity are often made appar- ent. E.g., a line joining the middle of the bases of a trape- zoidal plate is a line of gravity, since it bisects all the strips of uniform width determined by drawing parallels to the bases; similarly, a line joining the apex of a triangular plate to the middle of the opposite side is a line of gravity. Other cases can easily be suggested by the student. 26. Problems.— (1) Required the position of the centre of A, gravity of %fine homogeneous wire of the ,,. r-g^ form of a circular arc, A£, Fig. 16. Take the origin at the centre of the circle, and the axis ^ bisecting the wire. Let the length of the wire, s, = 2Si ; ds = ele- ment of arc. We need determine only the X, since evidently y ^ 0. Equations (5), fxds Fig. 16. 23, are applicable here, i.e., x From similar triangles we liave 7 ^^V ds : dy :: r : x; .-. ds = — -; ,?/ = + a ^ra :^ I dy — -^r—, i.e., = chord X radius -r- length of 2s wire. For a semicircular M'ire, this reduces to x == 2r -~ 7t. Problem 2. Centre of gravity of trapezoidal {and trian- giilar) thin plates, homogeneous, etc. — Prolong the non-parallel sides of the trapezoid to intersect at 0, which take as an origin, making the axis X perpendicular to the bases h and &,. We may here use equations (4), § 23. and may take a vertical strip for our element of area, dF, in determining x; for each point of such a strip has the same x. Now dF ^ {y -f- y')dx. and * The two triangles meant {m being any point of the wire) are the finite triangle Omc, and the infinitely small one at m formed by the infinitesimal lengths dy, dx, and ds. PARALLEL POKCES AND THE CENTKE OF GRAVITY. 21 from similar triangles y-{-y' = jx. 'NowF, = -{hh — bji,),"^ can be written ^ , {^^ ~ K'), and x = --^ — becomes = Ji UK -2a(^-^0 = 3;,. -K for the trapezoid. For a ti •iangle h^ = — 2 0, and we liave x = h ; that centre of gravity of a triangle is one tliird the altitude from the base. The centre of gravity is finally determined by knowing Fig. 17. Fig. 18. that a line joining the middles of h and h^ is a line of gravity; or joining O and the middle of h in the case of a triangle. Problem 3. Sector of a circle. Thin plate, etc. — Let the notation, axes, etc., be as in Fig. 18. Angle of sector = 2<ar; a? = ? Using polar co-ordinates, the element of area dF (a small rectangle) = pdqi . dp, and its a? = p cos qj ; hence the total area = 9 F= J'^'^X fpdiP\dcp = y^+"^ r'd^ = i.e., F:= T^a. From equations (4), § 23, we have - \ n X = -jp i xdF ♦Note that h\•.'h^'.•.'h'.h, so that ?)ifei=(&-^^)i'ii'. 22 MECHANICS OF ENGINEERING. {Note on double integration. — The quantity cos (p J p' dp \dq), is that portion of the summation / / cos cpp'dpdq) which belongs to a single elementary sector (triangle), since all its elements (rectangles), from centre to circumference, have the same q) and dcp.) That is, — 1 r^ n + a. 'p^ r+a 2 y sin or a? = ^^-o / cos ^c?ffi> = 5-^ sinffl = -5-. : o c^ T 1 ^ — 4 7* sin -I /? or, putting p =z 'za z= total angle 01 sector, a? = -^ -z • — 4:7' For a semicircular plate this reduces to a? = 7;—. \_Mote. — In numerical substitution the arcs a and /? used above (unless sin or cos is prefixed) are understood to be ex- pressed in circular measure (;r-measure) ; e.g., for a quad- rant, yS = I = 1.5707*+ ; for 30°, /? = ^ ; or, in general, if fi m degrees := , then p in ;r-measure = — . ° n n J Problem 4. Sector of a flat ring ; thin _ -^^ plate, etc. — Treatment similar to that of ,^y \.---^...\\- Problem 3, the difference being that the #^^P^ If - _ ... . P' limits of the interior integrations are instead of | . Result, FiQ, 19. 1-0 - _ 4 T^ — r^ sin ^/? ^""l- r/ - r: ' ~~W~° * "Radians." PARALLEL FORCES AND THE CENTRE OF GRAVITY. 23 Pkoblem 5. — Segment of a circle ; thin plate, etc. — Fig. 20. Since each rectangular element of any ver- tical strip has the same x, we may take the strip as dF \w finding x, and use y as the half-height of the strip. dF = 2ydx, and from similar triangles x : y :: {— dy):dx,^ i.e., xdx = — ydy. Hence from eq. (4), - ^/(vdF ^ /x^ydx - 2XVVZy _2__ F F F SF but a = the half-chord, hence, finally, x = 12F. Fio. 21. Problem 6. — Trajpezoid ', thin plate, etc., by the method in the corollary of § 23 ; equa- tion (4a). Kequired the distance x from the base AB. Join BB^ thus dividing the trape- zoid ABCB into two triangles ABB = F^ and BBC = F^, whose gravity a?'s are, re- spectively, x^ = ^h and x^ = |A. Also, F^ = iMj, F^ = ^hh^, and F (area of trape- zoid) = ih{h, + h,). Eq. (4a) of § 23 gives Fx = F^x^ -\- F^x^ ; hence, substituting, (6i +62)^ = k^ih^lhji. _^h (61 + 2&2) •'• ^-3 * 61+62 • The line joining the middles of 5, and h^ is a line of gravity, and is divided in such a ratio by the centre of gravity that the fol- lowing construction for finding the latter holds good : Prolong each base, in opposite directions, an amount equal to the other base; join the two points thus found: the intersection with the other, line of gravity is the centre of gravity of the trape- zoid. Thus, Fig. '21, with BF= h &ndBF= \, join FF, etc. * The minus sign is used for dy since, as we progress from left to right in bringing into account all the various strips, x increases while y diminishes; i.e., dx is an increment and dy a decrement. At the point of beginning of the summation, on left, y= +a; while at the extreme right, y = 0. 24 MECHANICS OF ElSTGnsrEERING. Peoblem 1. Homogeneous ohlique cone or pyramid. — ■ Take the origin at the vertex, and the axis X perpendicular to the base (or bases, if a frustum). In finding x we may put dY^ =^ vohirae of any lamina parallel to YZ, ^ being the base •of such a lamina, each point of the lamina having the same x. Hence, (equations (2), § 23), (see also Fig. 22). x= ^fxdV, V=/dV=/Fdxi but, from the geometry of similar plane figures, F:F, :: x' : h,% .-.F. F and ^=i-^»'*'^^=§'' -',fxdY=^Jx'dx=-!^ F K LI* Q Z 4 Z, 4 -For a frustum, x = 7 • , '3 ~ y\ I while for a pyramid, Aj, be- — 3 ing = 0, a? = jA. Hence the centre of gravity of a pyramid is one fourth the altitude from the base. It also lies in the line joining the vertex to the centre of gravity of the base. Pkoblem 8. — If the heaviness of the ma- terial of the above cone or pyramid varied directly as x, y^ being its heaviness at the base F^, we should use equations (1)5 § 23, putting y = j^ x\ and finally, for the frustum, - 4 h:-h: Fig. 23. r« h:-hr and for a complete cone m = — A,. 27. The Centrobaric Method. — ^If an elementary area dF he revolved about an axis in its plane, through an angle a < 'Itt. PAEALLEL FORCES AND THE CENTRE OF GKAVITY. 25 the distance from the axis being = x, tlie volume generated is ^Y =z axdF^ and the total volume generated by all the dF''% of a finite plane figure whose plane con- ^^,g tains the axis and which lies entirely on one ;side of the axis, will be T^ = fd V = afxdF. But from §23, afxdF^aFx\ ax being the length of path described by the centre of gravity of the ])lane iigure, Ym. 23. we may write : The vohime of a solid of revolution generated hi/ a plane figure, lying on one side of the axis, equals the area of the figure multiplied hy the length of curve descrihed hy the centre of gravity of the figure. A corresponding statement may be made for the surface generated by the revolution of a line. The arc a must be ex- pressed in It measure in numerical work. 27a. Centre of Gravity of any Cluadrilateral. — Fig. 23a. Construction', ABOD being any quad- rilateral. Draw the diagonals. On the long segment DK of DB lay off BE = BK, the shorter, to determine E\ simi- larly, determine iV^on the other diagonal, by making GN = AK. Bisect FK in H and KN in M. The intersection of FM and NH\& the centre of gravity, C. p.poof—R being the middle of BB, and AH and HG Slaving been joined, I the centre of gravity of the triangle ABB is found on AH, by making ^/= i-^iZ; similarlj^, by makino- HB = ^HG, B is the centre of gravity of triangle BBG. . ' . IB is parallel to AG and is a gravity-line of the whole figure; and the centre of gravity Cmay be found on it if we can make CB : CI :: area ABB : area BBG (§ 21). But since these triangles have a common base BB, their areas are proportional to the slant heights (equally inclined to BB) AK and KG, i.e., to GN and NA. Hence HN, which di- vides IB in the required ratio, contains C, and is .'. a gravity- line. By similar reasoning, using tlie other diagonal, AG, and Fig. 23a. 26 MECHANICS OF ENGINEERING. the two triangles into wliicb it divides the whole figure, we may prove E2i to be a gravity-line also. Hence the construc- tion is proved. 27b. Examples. — 1. Required the volume of a sphere bj the centrobaric method. A sphere may be generated by a semicircle revolving about its diameter through an arc a = 27r. The length of the path descj'ibed by its centre of gravity is = Stt^— ("see Prob. 3, § 26), while the area of the semicircle is |-7^r^ Hence by § 27, 4r 4 Yolume venerated = 27r . 7-— . ^nr^ = — nr'. 2. Tiequired the position of the centre of gravity of the sectoi* of a flat ring in which '}\ = 21 feet, r^ = 20 feet, and /3 = 80° (see Fig. 19', and § 26, Prob. 4). /3 . sin — = sin 40° = 0.64279, and y5 in circular measure =• 80 4 T^7^=-q7r = 1.3962 radians. By using ri and r2 in feet, X will be obtained in feet. • I _ 4 ri^-r2^ ^^^- 2 4 1261 0.64279 .'.T=— — — = - -, • - = 18 87 feet "^ S'n^-r^^- /? 3*41 '1.3962 ^^-^'leei. 3. Find tlie height (z, = OC) of the center of gravity of 05" uo.e'tj the plane figure in Fig. 23& "I above its base OX. L This figure is bounded i by straight lines and is an j_ approximation to the shape: '^ o -15- — -»«5-i of the cross-section of a steel ^'^- 2^^- , "channel" (see p. 275). Dividing it into three rectangles and two triangles (see dotted lines in figure) and applying eq. (4a) of p. 19, we have i.r J. 15X.6X.3-f2[3.4x.6x2.3]+2 z = — 71 1.7X.5X^ — = 0.882 in. 15X0.6 + 2[3.4X0.6] + 2[1.7X0.5] (The student should carefully verify these numerical details.) STATICS OF A RIGID BODY. 27 CHAPTER III. STATICS OF A RIGID BODY. 28. Couples. — On account of the peculiar properties and utility of a system of two equal forces acting in parallel lines and in opposite directions, it is specially ^^ considered, and called a Couple. The z^::::^^^^^^^^ t ar7n of a couple is the perpendicular « fr^^V^^ ""'^'^^^^ distance between the forces ; its TwomeTi^, P S'* lQJ<i J^ the product of this arm, by one of the |<^ ^^'"'^^^ .^^^^^^^^^ forces. The axis of a couple is an ^"^''^^:>^^,^^<^^^^ imaginary line drawn perpendicular to y\q. 24. its plane on tliat side from which the rotation appears positive (against the hands of a watch). (An ideal rotation is meant, suggested by the position of the arrows ; any actual rotation of the rigid body is a subject for future consideration.) In dealing with two or more couples the lengths of their axes are made proportional to their moments; in fact, by selecting a proper scale, numerically equal to these moments. E.g., in Fig. 24, the moments of the two couples there shown are Pa and Qh\ their axes p and q so laid off that Pa : Qh '.: p : q, and that the ideal rotation may appear positive, viewed from the outer end of the axis. For example, if each force P of a couple is 60 lbs., and the arm is a=6 ft., its moment is 360 jbot-pounds; or 0.180 foot- tons; or 4320 inch-pounds; or 2.16 inch-tons. 29. No single force can halance a couple. — For suppose the couple P^ P, could be balanced by a force P', then this, acting ?f at some point C, ought to hold the couple ni /..:■-. -P- -Q in equilibrium. Draw CO throuo-h 0, the tT /p p^f centre of symmetry of the couple, and Fig. 25. make OD = OC. At D put in two op- posite and equal forces, S and T, equal and parallel to P', The supposed equilibrium is undisturbed. But if P\ P, and "^S MECHANICS OF ENGINEERING. P are in equilibrium, so ought (by symmetry about 0) S, jP, and P to be iu equilibrium, and they may be removed without disturbing equilibrium. But we have left Tand P', which are evidently not in equilibrium ; .•. the proposition is proved by this reductio ad absurdum. Conversely a couple has no singlo resultant. 30, A couple may he transferred anywhere in its own plane. — First, it may be turned through any angle «', about any p* point of its arm, or of its arm produced. Gt- -; j^- -|- Let {P^ /*')be a couple, G any point of its \-.-'-j^-4 yp' arm (produced), and a. any angle. Make ^^^■.,_ i 0G= GA, CD — AB, and put in at G, '^ \ ^ \; I P^ and P^ equal to P {or P'), opposite to --* "^ each other and perpendicular to GC; and '® "" \ R*' P^ and P^ similarly at P. IS^ow apply and Fig. 26. combine P and P, at 0, P' and P, at 0'\ then evidently P and R' neutralize each other, leaving P^ and P^ equivalent to the original couple {P^ P'). The arm CD = AB. Secondly, if G be at infinity, and or = 0, the same proof applies, i.e., a couple may be moved parallel to itself in its own plane. Therefore, by a combination of the two traiisferrals, the proposition is established for any trans- ferral in the plane. 31. A coujple. may he replaced hy another of equal moment in- a parallel plane. — Let {P, P') be a couple. - Let CD, in a parallel plane, be parallel to AB. At D put in a pair of equal and opposite forces, ^3 and S^., parallel to P and each = ^=:iP. ED Similarly at (7, 8^ and 8^ parallel to P and each = ==-P. sLkj But, from similar triangles, ^ — ^. . o _ c _ o _ e pjjy — PC'' ' ' "^ — 5 — ' — *' * See Fig. 27, which is a perspective view. The arm of the couple (P, P') is AB, in the background. The length of CD, which is in the foreground, may be anything whatever. STATICS OF A KIGID BODY. 29- [Note. — The above values are so chosen that the intersection point E may be the point of application of (P' -|- JS2), the resultant of F and /6a;. and also of {P-\- Sa), the resultant of Pand S3, as follows from § 21; thus (Fig. 28), Ji, the resultant of the two parallel forces Pand iSs, is = P-f-xSg, and its moment about any centre of moments, as E, its own point of ap- plication, should equal the (algebraic) sum of the moments of its com- AhJ ponents about E; i.e., B X zero = P . AE — Sz . DE; .-.83 = == . P.] UE lA S,| I hk-ff" I R! Fig. 27. D E Fig. 28. I Replacing P' and S, by {P' + S,\ and P and S, hj {P -f- ^,), the latter resultants cancel each other at E^ leaving the couple {S^, 8^ with an arm CD^ equivalent to the original couple P, P' vi^ith an arm AB. But, since 8^ = ===. P = MjL/ -=r. . P, we have S.xOP = PxAB ; that is, their moments ai'e equal. 32. Transferral and Transformation of Couples. — In view of the foregoing, we may state, in general, that a couple acting on a rigid body may be transferred to any position in any parallel plane, and may have the values of its forces and arm changed in any way so long as its moment is kept unchanged, and still have the same eifect on the rigid body (as to rest or motion, not in distorting it). Corollaries. — A couple may be replaced by another in any position so long as their axes are equal and parallel and simi- larly situated with respect to their planes. A couple can be balanced only by another couple whose axis is equal and parallel to that of the jfirst, and dissimilarly situ- ated. For example. Fig. 29, Pa being = Qb^ the rigid body AB (here supposed without weight) is in equilibrium in each :30 MECHANICS OF ENGINEERING. case shown. By " reduction of a couple to a certain arm «" is meant that for the original couple whose arm is a' ^ with forces each = P\ a new couple is substituted whose arm shall be = «, and the value of whose forces P and P must be com- puted from the condition Pa = P'a\ i.e., P = P'a' -^ a. Fig. 29. Fig. 30. 33. Composition of Couples. — Let (P, P') and {Q, Q') be two -couples in different planes reduced to the same arm AB = a, which is a portion of the line of intersection of theii' planes. That is, whatever the original values of the individual forces and arms of the two couples were, they have been transferred and replaced in accordance with § 32, so that P . AP, the moment of the first couple, and the direction of its axis, p, have remained unchanged ; similarly for the other couple. Combining P with Q and P' with Q', we have a resultant couple {P, i?')M^hose arm is also AP. The axes p ^.nd q of the component couples are proportional to P . AP and Q . AB, i.e., to P and Q, and contain the same angle as P and Q. "Therefore the parallelogram p . . . q\& similar to the parallelo- gram P . . . Q\ whence p '. q : r'.'.P '. Q : P, or p : q : ri: Pa : Qa : Pa. Also r is evidently perpendicular to the plane of the resultant couple (P, i?'), whose moment is Pa. Hence r, the diagonal of the parallelogram on p and q, is the axis of the resultant couple. To combine two couples, therefore, we have only to combine their axes, as if they were forces, by v> parallelogram, the diagonal being the axis of the resultant couple ; the plane of this couple will be perpendicular to tlie STATICS OF A RIGID BODY. 31 axis just found, and its moment bears the same relation to the moments of the component couples as the diagonal axis to the two component axes. Thus, if two couples, of moments Pa and Q}}^ lie in planes perpendicular to each other, their result- ant couple has a moment Re = ^{Paf + {Qbf' It three couples in different planes are to be combined, the axis of their resultant couple is the diagonal of the parallelo- piped formed on the axes, laid off to tliesame scale 2a\d point- ing in the proper directions, the proper direction of an axis being away from the plane of its couple, on the side from which the couple appears of positive rotation. 34. If several couples lie in the same plane their axes are parallel and the axis of tlie resultant couple is their algebraic sum ; and a similar relation holds for the moments : thus, in Fig. 24, the resultant of the two couples has a moment = Qh — Pa, which shows us that a convenient way of combining couples, when all in one plane, is to call the moments positive or negative, according as the ideal rotations are against, or with, the hands of a watch, as seen from the same side of the plane ; the sign of the algebraic sum will then show the ideal rotation of the resultant couple. 35. Composition of Non-concurrent Forces in a Plane. — Let Pj, Pa, etc., be the forces of the system ; x^, y^, x^, y^, etc., the y: ■-x^, -/-^■y^^ / J' y;'' Fig. 31. co-ordinates of their points of application ; and a^, a^, . , . etc., their angles with the axis X. Replace P^ by its components Xj and ]rj, parallel to the arbitrary axes of reference. At the origin put in two forces, opposite to each other and equal and parallel to X^ ; si'milarly iovY ^. (Of course X^ = P^ cos oc and Y^ = P^ sin a.) We now have P^ replaced by two forces X. 32 MECHANICS OF ENGINEERING. and y, at the origin^ and two couples, in the same plane, whose moments are respectively — X{y^ and + ^ x^x-: ^"^ ^^^ there- fore (§34) equivalent to a single couple, in the same plane with a moment = {Y ^x^—X^y^. Treating all the remaining forces in the same way, the whole system of forces is replaced by the force :2{X) =X, +X, + . . . attlie origin, along the axisX; the force ^{Y) = Y,^ Z, + . . . at the origin, along the axis Y- and the couple whose mom. G= ^ { Yx — Xy), which may be called the couple C (see Fig. 32), and may be placed anywhere in the plane. Now -5'(X) and 2( Y) may be combined into a force jR i i.e., , ... ^X R = V[^Xf -\- 2 Yy and its direction-cosine is cos a = —p-. Since, then, the whole system reduces to C and i?, we must have for equilibrium B = 0, and G = ; i.e., for equilibrium 2X= 0, ^r= 0, and ^{Yx-Xy) = 0. . eq. (1) If i? alone = 0, the system reduces to a couple whose mo- ment is 6^ = ^( Yx—Xy) ; and if G alone = the system re- duces to a single force i?, applied at the origin. If, in general, neither I^ nor G = 0, the system is still equivalent to a single force, but not applied at the origin (as could hardly be ex- pected, since the origin is arbitrary) ; as follows (see Fig. 33) : Replace the couple by one of equal moment, G, with each G force = jR. Its arm will therefore be -^. Move this couple in the plane so that one of its forces i? may cancel the i? al- ready at the origin, thus leaving a single resultant i? for the whole system, applied in a line at a perpendicular distance, G c = -p , from the origin, and making an angle or whose cosine = 2X ■^ , wdtb the axis X. It is easily preyed that the " moment" Re, of tbe single resultant, about tbe origin 0, is equal to the algebraic sum of those of its " components " (i.e., the forces of the system. 36. More convenient form for the equations of equilibrium of non-concurrent forces in a plane. — In (I.), Fig. 34, being STATICS OF A EIGID BODY. 33 any point and a its perpendicular distance from a force P\ put in at two equal and opposite forces P and P' = and || to P, and we have P replaced by an equal single force P' at 0, and a couple whose moment is -|- Pa. (II.) shows a simi- lar cunstrtcrion, dealing- with the JTand 1^ components of P, so that in (II.) P is replaced by single forces ^' and Y' at ^.....^......Liix ^_ 1! „ X X' Y^ (TI.) (I.) Fig. 34. (and they are equivalent to a resultant P', at 0, as in (I.), and two couples whose moments are -|- Yx and — ^y. Hence, being the same point in both cases, the couple Pa is equivalent to the two last mentioned, and, their axes being parallel, we must have Pa = Yx — Xy. Equations (1), § 35, for equilibrium, may now be written* :SX- 0, 2Y = 0, and :S{Pa) = 0. . . (2) In problems involving the equilibrium of non-concurrent forces in a plane, we have three independent conditions^ or equations.^ and can determine at most tliree unknown quantities. For practical solution, then, the rigid body having been made free (by conceiving the actions of all other bodies as repre- sented by forces), and being in equilibrium (which it must be if at rest), we apply equations (2) literally ; i.e., assuming an origin and two axes, equate the sum of the JT components of all the forces to zero; similarly for the 1^ components ; and then for the "moment-equation," having dropped a perpen- dicular from the origin upon each force, write the algebraic sum of the products {moments) obtained by multiplying each force by its perpendicular, or " lever-armj^'' equal to zero, call- ing each product + or — according as the ideal rotation ap- pears against, or with, the hands of a watch, as seen from the same side of the plane. (The converse convention would do as ■well.) * Another proof is given on p. 15 of th« " Notes and Examples in Mechanics," 34 MECHANICS OF ENGHNEERING. Sometimes it is convenient to use three moment equations, takkig a new origin each time, and then the 2X =i and 2Y = are superfluous, as they would not be independent equa- tions. 37. Problems involving Non-ooncurrent Forces in a Plane. — Remarks. The weight of a rigid body is a vertical force through its centre of gravity, downwards. If the surface of contact of two bodies is smooth the action (pressure, or force) of one on the other is perpendicular to the surface at the point of contact. If a cord must be imagined cut, to make a body free, its tension must be inserted in the line of the cord, and in such a direction as to keep taut the small portion still fastened to the body. In case tiie pin of a hinge must be removed, to make the body free, its pressure against the ring being unknown in direction and amount, it is most convenient to represent it by its unknown components X and J^, in known directions. In the following problems there is supposed to be no friction. If the line of action of an un- known force is known, but not its direction (forward or back ward), assume a direction for it and adhere to it in all the three equations, and if the assumption is correct the value of the force, after elimination, will be positive ; if incorrect, negative. * ProhleTTh 1. — Fig. 35. Given an oblique rigid rod, with two loads (xj (its own weight) and G^ ; required the reaction of the smooth vertical wall at A, and the direction and amount of the A^^^e-pressure at 0. The reaction at A must be horizontal ; call it X'. The pres- ^1 sure at 0, being unknown in direction, will have both its X and ]P" components un- known. The three unknowns, then, are ^^^^^^'^^ ^M ^' 1 and J^o, while G^, G^, «„ a„ and ■^''^ h are known. The figure shows the rod Fis. 35. ^g ^ y^^^ hody, all the forces acting on it have been put in, and, since the rod is at rest, constitute a sys- tem of non-concurrent forces in a plane, ready for the condi' tions of equilibrium. Taking origin and axes as in the figure, * That is, the force must point in a direction opposite to that first assumed for it. 6 STATICS OF A RIGID BODY. 35 2X= gives +X„ - X' = ; :S Y = gives -^ T, - G, — G, = 0; while 2{Fa) = 0, about 0, gives + XA — (?,«j — G^a^ = 0. (Tiie moments of JC^ and Y^ about are, each, = zero.) By elimination we obtain Y^ ^ ^i 4" 6^2 ; Xj = X' = [6^iaj -f- ^2*^2] -i- '^^j while the pressure at = VX^ -[" ^o^ ^"d makes with the horizontal an angler, whose tan = I^o -r- X,,. [N.B. A special solution for this problem consists in this, that the result- ant of the two known forces Oi and O2 intersects llie line of X' in a point which is easily found by § 21. The hinge-pressure must puss through this point, since three forces in equilibrium must be concurrent.] IN'ote that the line of action of the pressure at 0, i.e. , of the resultant of Xq and Yq, does not coincide with the axis of the rod; the rod being subjected to more than the two forces at its extremities. The case therefore differs from that presented by the boom in Ex. 2 of p. 12. Problem, 2. — Given two rods with loads, three hinges (or " pin -joints"), and all dimensions: required the three hinge- Fig. 36. FiGf. sr. pressures; i.e., there are six unknowns, viz., three Xand three Y components. We obtain three equations from each of the two free bodies in Fig. 37. The student may fill out the de- tails. Notice the application of the principle of action and reaction at B (see § 3). ProMem 3. — A Warren bridge-truss rests on the horizontal smooth abutment-surfaces in Fig. 38. It is composed of equal isosceles triangles ; no piece is continuous beyond a joint, each of which is a, pin connection. All loads are considered as acting at the joints, so that each piece will lj j j i^i be subjected to a simple tension pjo ^g. ' or compression. " Two-force ipieces ; Bee -p. 18, Notes.) MECHANICS OF ENGINEERING. First, required the reactions of the supports "Fj and T^' these and the loads are called the external forces. '^[Pd) about ■=■ ^ gives (the whole truss is the free body) F,3« - P, \a I\.%a- P3.f« = 0; while ^{Pa) about K =^ gives and - F, . 3« + P3 . i« + P^^a + P,|a = 0; V. = iCA + 3P. + 5^3]. Secondly, required the stress (thrust or pull, compression or tension) in each of the pieces A, P, and Cent by the imaginary line PP. The stresses in the pieces are called internal forces. These appear in a system of forces acting on a free body only when a portion of the truss or frame is conceived separated from the remainder in such a way as to expose an internal plane of one or more pieces. Consider as a free body the por- tion on the left of PE (that on the right would serve as well, I p p but the pulls or thrusts in A, P, and 6^ would be found to act in directions opposite to those they have on the other portion ; see § 3). Fig. 39. The ^iIq arrows (forces) A, B, and C, are as- sumed to 23oint, respectively, in the directions shown in the figure. They, with Vi, P\, and P2, form a system holding the body in. equilibrim. For this system, I (Pa) about (9=0 gives 0+Ah-Vi2a+Pi ■ fa+P2 • ia = 0; and hence A = {ia^h)[Wi-3Pi^P2l which is positive; since, (see above), 4Fi is >3Pi+P2. Therefore the assumption that A points to the left is con- firmed and A is a thrust, or compression ; (its value as above.) Again, taking moments about Oi (intersection of A and B), we have an equation in which the only unknown is (7, viz. , C/i-7i|a+Pia=0; /. C=(ia-J-/i)[37i-2PJ, Fig. 39. STATICS OF A RIGID BODY. 37 a positive value since 37i is >2Pi; :.C must point to the right as assumed; i.e., is a tension, and=— [STi— 2Pi]. Finally, to obtain 5, j)ut 2'(vert. comp8.)=0; i.e. 5 cos <j6 + 7i-Pi-P2 = 0. .*. B cos 9S =Pi +P2— Vi ; but, (see foregoing value of V\) we may write K = (P. + PJ - (iP, + IP^ + ^P3, /. P cos cp will be + (upward) or — (downward), and P will be compression or tension^ as ^P^ is < or > [^P, -j- iPj. P = [P,+ P,-FJ--cos9> ^^-V.+P.-rj. Problem 4. — Given the weight G^ of rod, the weight G^, and all the geometrical elements (the student will assume a w\|Pi m Gi AG, Fig. 40. ■ Fig. 41. convenient notation); required the tension in the cord, and the amount and direction of pressure on hinge-pin. Fig. 41. Problem 5. — Roof-truss; pin-connection; all loads at joints ; wind-pressures W and TF, normal to OA ; required the three reactions or supporting forces (of the two horizontal surfaces and one vertical surface), and the stress in each piece. All geomet- rical elements are given ; also P, P,P„Tr(Fig.40). 38. Composition of Non-concur- rent Forces in Space. — Let P„ P„ etc., be the given forces, and a?j, y^^ 2!„ 35,, y,, s,, etc., their points of ap- plication referred to an arbitrary origin and axes; a^^ /?j, y^^ etc., Fig. 42. the angles made by their lines of application with Xs. Y^ and Z, 38 MECHANICS OF ENGINEERING. 1 Considering tlie first force* i^j, replace it by its three com- ponents parallel to the axes, J^^ = P^ cos a^^ Y^ = P^ cos /?,; and Z, = P^ cos y^ {P^ itself is not shown in the figure). At (?, and also at A^ put a pair of equal and opposite forces, each equal and parallel to Z^ ; Z, is now replaced by a single force Z^ acting upward at the origin, and two couples, one in a plane parallel to YZ and having a moment = — Z^y^ (as we see it looking toward from a remote point on the axis -\- JT), the other in a plane parallel to XZ and having a mo- ment := -f~ ^\^\ (seen from a remote point on the axis -^ Y^. Similarly at and G put in pairs of forces equal and parallel to -Z^, and we have X^^ at B^ replaced by the single force X^ at the origin, and the couples, one in a plane parallel to XY^ and having a moment -|- X{y^^ seen from a remote point on the axis -|- Z, the other in a plane parallel to XZ. and of a moment ■= — ^i^i, seen from a remote point on the axis -\-Y\ and finally, by a similar device, Y^ at B is replaced by a force Y^ at the origin and two couples, parallel to the planes XY and YZ^ and having moments — Y^x^ and -f^ i^2j, respective* ly. (In Fig. 42 the single forces at the origin are broken lines, while the two forces constituting any one of the six couples may be recognized as being equal and parallel, of opposite di- rections, and both continuous, or both dotted.) We have, therefore, replaced the force P^ by three forces Xj, y"j, Z^, at 0, and six couples (shown more clearly in Fig. 43; the couples have been transferred to symmetrical posi- tions). Combining each two couples whose axes are parallel to X^ Y^ or Z, they can be reduced to three, viz., one with an X axis and a moment = Y^z^ — Z^y^ ; one with a T^axis and a moment = Z-^o^ — X^z^\ one with a Z axis and a moment =: X^y^ — Y^x^. * This "first force," Pj, is applied at the point B, whose co-ordinates are Xi, y^, and 2i, and is typical of all the other forces of the system. Fig. 43. STATICS OF A RIGID BODY. 39 Dealing with each of the other forces P^, P^, etc., in the same manner, the whole system maj finally be replaced by three forces 2X, ^Y, and 2Z, at the origin and three couples whose moments are, respectively, (ft-lbs., for example) Z, = 2( Yz — Zy) with its axis parallel to X\ M = 2{Zx — Xs) with its axis parallel to JT; JV = -2'(.Zy — Yx) with its axis parallel to Z. The " axes" of tliese couples, being parallel to the respective co-ordinate axes JT, Y, and Z, and proportional to tl]^ mo- ments Z, 2f, and JV, respectively, tlie axis of their resultant C, whose moment is G, must be the diagonal of a parallelo- pipedon constructed on the three component axes (propor- tional to) Z, M, and iT. Therefore, G = VZ' ~{- M' -^ ]V% while the resultant of ^X, -2 Y, and 2Z is p = Vi^xy + (^ Yy + {:szy acting at the origin. If a, y5, and y are the direction-angles ^X , 2Y 2Z of P, we have cos oc = — ^, cos p = ~ti-, and cos ;^ i= -^ ; while if A, ju, and r are those of the axis of the couple C, we- Z J^ . ^ have cos A, = -p, cos >u = --^, and cos r = --^. For equilibrium we have both G = and ^ = ; i.e., separately, six conditions, viz., :^X= 0, 2 r = 0, 2Z=:0 ; and Z=0, Jf=0, JV=0 . (1) Now, noting that :SX = 0,:2Y= 0, and ^(Xy - Yx) = are the conditions for equilibrium of the system of non-concur- rent forces which would be formed by projecting each force of our actual system upon the plane XY, and similar relations for the planes YZ and XZ, we may restate equations (1) in another form, more serviceable in practical problems, viz. : Note. — I]f a system of non-concurrent forces in space is in equilibrium, the plane systems formed hy projecting the given systein upon each of three arbitrary co-ordinate planes will eaah be in equilibrium. But we car obtain only six independent 40 MECHANICS OF ENGINEERING. equations in any case, available for six unkno'wns. If H alone ^ 0, we have the system equivalent to a couple C^ whose moment = ^ ; if 6^^ alone = 0, the system has a single re- sultant R applied at the origin. In general^ neither i? nor G being = 0, we cannot further combine i? and G (as was done with non-concurrent forces in a plane) to produce a single re- sultant unless B, and C happen to be in parallel planes ; in which case the system may be reduced to a single resultant by use •of the device explained near foot of p. 32. Remark. — In general, R and C not being in parallel planes, the system may be reduced to two single forces not in the same plane, b^ assigning any value we please to P, one of the forces of the couple C, computing the corresponding arm a = G-i-P, transferring C until one of the P's has the same point of application as R, and then combining these two forces into a single resultant. This last force and the second P are, then, the equivalent of the original system, but are not in the same plane. (See §§ 15 and 15a.) Again, if a reference plane be chosen at right angles to R, and the couple C be decomposed into two couples, one in the reference plane and the other in a plane at right angles to it, this second couple and R may be replaced by a single force (as on p. 32) and we then have the whole system replaced by a single force and a couple situated in a plane perpen- dicular to that force; (and this may be called a "screwdriver action.") Example. — A shaft, with crank and drum attached and supported horizontally on two smooth cylindrical bearings, constitutes a hoisting device. See Fig. 43a. | „ ,,^ A force P is to be applied to the crank handle at 30° with the horizontal (and T to the crank), and acting in a plane at right angles to the shaft; and is to be of such value as to preserve equilibrium when the weight of 800 lbs. is sustained, as shown. The weight of the shaft, etc., is 200 lbs., and its center of grav- ity is at C in the axis of the shaft. (Counterpoise for crank not shown.) The reactions at the two bearings will lie in planes T to the axis of the shaft {smooth cylindrical surfaces), making unknown angles with the vertical; and will be represented by their X and Z componentsi 8001 Fig. 43o. STATICS OF A EIGID BODY. 41 as shown. It is required to find the proper value for P and the amount and position of the two reactions. Solution. — The seven forces shown in the figure (of which five are un- known) constitute a non-concurrent system of forces in space; in equi- librium. Since there are no Y-components the condition -- F = is already satisfied. Let us now apply the statement of the "note" on p. 39, first projecting the forces on the plane ZX (vertical plane T to the shaft). (That is, we take an "end-view" of the system.) Each of the seven forces projects in full length, or value, since they are all parallel to that plane. Treating the plane system so formed as in equilibrium and taking mo- ments about the point 0, we find (feet and lbs.) + PX1.5-800XH0 = 0; .-. P= 177.77 lbs. . . . (1) Next projecting on the vertical plane ZY, containing axis of shaft (i.e., taking a "side-view" of the system) we note that#the projection of P is P sin 30° and those of Xq and X^ each zero, while Z^, Z^, and the 200 and 800 lbs., project in full length; hence taking moments about we have -|-200Xlf+800X2-ZiX3-PX0.50X4 + 0=0 ... (2) while moms, about Oigives+ZoX 3 -Px0.5Xl-200Xli- 800- 1 = (3) Finally, projecting on the horizontal plane XZ ("top-view"), the forces in this projection are P cos 30°, X^, and X^; so taking moms. about point Oj, +ZoX3-Px0.8660Xl = 0; .-. X„= +51.34 lbs.; . (4) whne from ^X = 0, X^-X^ = 0, or X^=Xq; i.e., Xi= +51.34 lbs. . (5) From (2) and (3) we have Zi= +525.93 lbs., and Zo= +385.18 lbs. All these + signs show that the arrows for X^, X^, Zg, and Z^ have been correctly assumed (Fig. 43a) as to direction. Combining results, we find that the pressure or reaction at O is Rq, =VXo^ + Zo^ = 388.6 lbs- and makes an angle whose tang, is Xq-hZq, (i.e., 0.1333), Viz., 7° 36', on the left of the vertical; also that the pressure or reaction at 0^ is Pi, =V'Xi^ + Zi^ = 528.4 lbs., at an angle on the right of the vertical whose tang., =X^^Z^, ( = 0.09763); i.e., 5° 34'. 39. Problem (Somewhat similar to the foregoing.) — Given all geo- metrical elements (including a, /?, ;-, angles of P) , also the weight of Q, and weight of apparatus G; A being a hinge whose pin is in the axis F, a ball-and- socket joint -.vequiredthe amount of P (lbs.) to preserve equi- librium, also the pressures (amount and direction) at A and O; no friction. Replace P by its X, Y, and Z com- ponents. The pressure at A will have Z and X components; that at 0, X, Y, and Z com- ponents. [Evidently there are six unknowns; Yq will come out negative. Fig. 44. 42 MECHANICS OF ENGINEERING. CHAPTER lY. STATICS OF FLEXIBLE CORDS. 40. Postulate and Principles. — The cords are perfectly flexi- ble and inexteiisible. All problems will be restricted to one plane. Solutions of problems are based on two principles, viz.: Pein. I. — The strain or tension, in a cord at any point can act only along the cord, or along the tangent if it be curved. Pein. II. — We may apply to flexible cords in equilibrium all the conditions for the equilibrium of rigid bodies ; since, if the system of cords became rigid, it would stiU, with greater reason, be in equilibrium. 41. The Simple Pulley. — A "simple pulley" is one that is acted on by only one cord (or belt) and the reaction of the bearing supporting its axle (or "journal"). A cord in equilibrium over a simple pulley whose axle is smooth is under equal tensions on both sides; for, Fig. 46, Fig. 46. Fig. 47. considering the pulley and its portion of cord free 2(Pa) — about the centre of axle gives I^'r =^ Pr, i.e., J*' = I^ = ten- sion in the cord. Hence the pressure i? at the axle bisects the angle ex, and therefore if a weighted pulley rides upon a cord ABC, Fig. 47, its position of equilibrium, B, may be found by cutting the vertical through A by an arc of radius CD = length of cord, and centre at C, and drawing a horizon- tal through the middle of AD to cut CD in B. A smooth ring would serve as well as the pulley ; this would be a slip- knot. From Fig. 46, R = 2P cos Ja. STATICS OF FLEXIBLE COKDS. 43 42. If tliree cords meet at 2i fixed Tcnot, and are in equilib- rium, the tension in any one is the equal and ymy/»y^///.. ^opposite of the resultant of those in the other y _ two. 43. Tackle. — If a cord is continuous over a number of sheaves in blocks forming a tackle, neglecting the weight of the cord and blocks and friction of any sort, we may easily find the ratio between the cord-tension P and the weight to be sustained. E.g., Fig. 48, regarding all the straight cords as vertical and considering the block B free, we have, Fig. 49 (from •2Y=%^P- G ri = 0, .*. P = -T-. The stress on the support G will = 5P. 4 Fig. 49^ Fig. 49c. G Fig. 49c?. Other designs of tackle are presented in Figs. 49a, 496, 49c, and 49<^, and should be worked out as exercises by the student. In each case the weight G is supposed to be given and [a value of the smaller weight (or pull) P must be determined for the equilibrium of the tackle. Friction, and the weights of the pulleys and cords, are neglected and all straight parts of cords (or chains) are considered vertical. . All of the pulleys shown are "simple pulleys," except the one at A in Fig. 49d, which represents a "differential pulley" tackle. Pulley A consists of two ordinary pulleys fastened together, the groove in each being so rough, or furnished with "sprocket-teeth" in case a chain is used, that slipping of the cord or chain is prevented. The chain or cord is endless, the loop C being slack. B is a simple pulley. In this case, for equilibrium the pull P must =W(r^~r)-r-ri. The other results .are p = iG for Fig. 49a; iG for 496; and IG for 49c. 44 MECHANICS OF ENGINEERING. 44. Weights Suspended at Fixed Knots.— Given all the geo. metrical elements in Fig. 50, iind one weight, G^\ required the re- maining weights and the forces iZoi 1^05 Hn ^i'^ y^ni at the points of support, that equilibrium may obtain. H^^ and Fq are the hori- zontal and vertical components of the tension in the cord at 0\ simihvrlj 11^ and Y^ those at n. There are ^ -f 2 unknowns.. (The solution of this problem is deferred. See p. 420.) Fig. 50a. 45. Example. — The boom OD, tie-rod RT), with four simple pulleys and a cable, form a crane as shown in Fig. 50a. Find the necessary vertical force P to be exerted on the piston at H, that the load of 800 lbs. may be sustained. Also find the pressure of pulley B on its bearing, the pull T in the tie-rod and the pressure Pq (amount and position) at pin O; neglecting all friction and rigidity (p. 192) and the weights of the members. Dimensions as in figure. Since all puUeye are "simple" the tension in cable is the same at all points; and is. = 400 lbs. since the straight parts of cable adjoining pulley A are parallel. For a similar reason P = 800 lbs. Pressure at B bisects the angle (50°) between adjoining straight parts of cable; i.e., is 25° with vertical, and =2X400Xcos 25° = 725 lbs., (§ 41). Next take the free body in Fig. 506 (boom and pulley '5 together with a part of cable.) Three unknowns and three equations. ^(moms.)o = 0; .-. -F TXQXtan 40° + 400X 1-400X9-400X7 = (1) i.e., TX 9 X 0.8391 = 6000 ft.-lbs.; .-. 2" = 794.3 lbs., (tension in tie-rod.) IX = Xi, .•.Xo-400cos40°-r = 0; .-. Xo = 794.3-H400X .766= 1100.7 lbs. ^F=0, .-. Yo-^OO sin40°-400-400 = 0; .-. y„ = 400 X. 6428 + 800 =1057. 12 lbs. Hence P„, = VZo^ -F Fo^ = 1 526 lbs. at tan-* Fo/Zq, or 43° 50', with horiz.. STATICS OF FLEXIBLE CORDS. 45 Note. — If the weight 800 lbs. were attached directly to cable on right. of pulley B, the value of P would need to be 1600 lbs. 46. Loaded Cord as Parabola. — If the weights are equal and inlinitelj small, and are intended to be uniformly spaced along the horizontal, when equilib- rium obtains, the cord having no weight, it will form a parabola. Let q = weight of loads per horizontal linear unit, O be the vertex of the curve in which the cord hangs, and Tn any point. We may consider the portion Om as a free body, if the reactions of the contiguous portions of the cord are put in, J3q and T, and these (from Prin. I.) must act along the tangents to the curve at O and m, respectively ; i.e., Sq is horizontal, and T makes some angle cp (whose tangent = —, etc.) with the axis X! Applying Prin. II., 2X = gives Tcos, cp — Hq = ; i.e., T^~^ = jSq ; ds 2 Y= gives T sin (p — qx = ', i.e., T-j- = . (1) qx. . . (2) Dividing (2) by (1), member by member, we have — = -^ ; q dy = -^xdx, the differential equation of the curve ;, B, (/ -_-._ / tJuCLvO — -r~r or X y, the equation oi a parabola whose vertex is at (9 and axis vertical. Note. — The same result, ~ = %- , mav be obtained by considering that we have here (Prin. II.) ?ifree rigid hody acted on by three forces, T, Hq, and R = qx, acting verti- cally through the middle of the abscissa x; the resultant of Hq and R must be equal and oppo- site to T, Fig. 53. R dy qx tan o) = — -, or -f =-- Ho dx H^ Evidently also the tangent-line bisects the ab- scissa X. (Try moments about m.) Example. — Let g = 800 lbs. "per foot run" and .'c = 100 ft., with 2/ = 20 ft. Then we have, for the value of the tension at the vertex of the parabola, F = ?x= ^ 2j/ = 800 X (100) => -^ 40 = 200,000 lbs. 46 MECHANICS OF ENGINEERING. 47. Problem under § 46. [Case of a suspension-bridge m which the suspension-rods are vertical, the weight of roadway is uniform per horizontal foot, and large compared with that of the cable and rods. Here the roadway is the only load : it is generally furnished with a stiffening truss to avoid deforma- tion under passing loads.] — Given the span = 2J, Fig. 53, Y; vf 71 ^^1® deflection = a, and the rate of loading j. ^_ —j^y^^ = q lbs. per horizontal foot ; required the tension in the cable at 0, also at m ; and ^ the length of cable needed. From the equation of the parabola qx^ = 'iH^y, put- ting a? = 5 and y = a, we have Mq = qjf -r- 2« — the tension at 0. From -S'T' = we have Y^ = qb, while ^^= gives qr--x- Fis. 53. M, = U,\ .-. the tension at m = \^ H^ + Y^= ^-{_qb V4:a'-\- b']. Act The semi-length, Om , of cable (from p. 88, Todhunter's In- tegral Calculus) is (letting n denote Hq -t- 2^', = 5^ -f- 4a) Otn = Vna -\- a" -{- n . log^ [( Va -j- Vn -j- a) -^- Vti]. 48. The Catenary.* — A flexible, inextensible cord or chain, of uniform weight per unit of length, hung at two points, and supporting its own weight alone^ forms a curve called the catenary. Let the tension Hq at the lowest point or vertex be represented (for algebraic convenience) by the weight of an imaginary' length, c, of similar cord weighing q lbs. per unit of length, i.e., 11^=. qG\ an actual portion of the cord, of length 5, weighs qs lbs. Fig. 54 shows -as, free and in equilib- -f-^^ ^ rium a portion of the curve of any length s, reckoning from the vertex. Requii-ed the equation of the curve. The load is uniformly spaced along the curve, and not horizontally, as in §§ 46 and 47. Fig. 54. 2T gives 2-J^ = ^s; while rdx SX - gives T-^- = qG. squaring, c^«?2/'' = ^do^. . Hence, by division, cdy = sdx, and • ■ (1) * For the " transformed catenary," see p. 395. STATICS OF FLEXIBLE COEDS. 47 Put dy^ = ds^ — dx^, and we have, after solving for dx ods /*s (is rs and X =0 . log, [(s + V? + c') -^ c], . . . (2) a relation between the horizontal abscissa and length of curve Again, in eq. (1) put dx^ = ds^ — dy"^, and solve for dy. This gives dy = , = ^r . \ , o cr. Therefore . ^ ^ i/c^ -I- 5« 2 (c^+5')* • 2/ = iX'{o' + 5')~*c?(c' + «') = i '2(c^ + sy, and finally y = |/*^ 4- c" _ c (3) Clearing of radicals and solving for c, we have c = («'-y')-% (4) Kow T, the tension at any point, = V(qs)~ + (56)2, and from (3) we obtain T=q(y + c) (4a) Example. — A 40-foot chain weighs 240 lbs., and is so hung from two points at the same level that the deflection is 10 feet. Here, for s=20ft., ?/ = 10; hence eq. (4) gives the parameter, c = (400 — 100) -^- 20 = 15 feet. 5 = 240-^-40 = 6 lbs. per foot. .•. the tension at the middle is if (; = gc =6X15 = 90 lbs.; while the greatest tension is at either support and = \/90M^120'=150 lbs. Knowing c=15 feet, and putting s = 20 feet = half length of chain, we may compute the corresponding value of x from eq. (2) ; this will be the half-span. That is, .x = 15 . loge 3 = 15 X2.303 X 0.4771 = 16.48 ft. To derive s in terms of x, transform eq. (2) in the way that ?i = logg m may be transformed into e^ = m, clear of radicals and solve for s, obtaining "^ or, s = c.sinh(— j. . . (5) Again, eliminate s from (2) by substitution from (3), trans- form as above, clear of radials, and solve for y+c, whence y +c = lc[ee -{-e « J; or, ?/-l-c = c. cosh|— j. (6) which is the equation of a catenary with axes as in Fig. 54. If the horizontal axis be taken a distance = c below the vertex, * sinh and cosh denote "hyperbolic" sine and cosine; see table, appendix. 48 MECHANICS OF ENGINEERING. the new ordinate z = y-\-c, while x remains the same; the last equation is simplified. See figure below. If the span and length of chain are given, or if the span and deflection are given, c can be determined from (5) or (6) only by successive assumptions and approximations. 48a. Catenary (Chain or Cable) with Supports at Different Levels. — Given the span k, the difference of elevation d of the two supports, and the whole length of chain, Z; it is required to find x' and y' (see Fig. 54a) and thus deter- mine the position of the vertex, or lowest point, 0, of the cate- nary. By applying the equa- tions of p. 47 to parts A'O and B'O, in turn, and combining, we may finally deduce (see p. 179 of Rankine's Applied Mechanics) Fig. 54a. -d^ = 2c.sinh( k" (7) (8) -l/P-d^ = ch2c-e 2cJ; i.e., ^/P- and also the relation x' — x" = c-\oge\ y—i • From (7) we find the "parameter," c, by trial; then the value of x' — x" from (8) ; whence, finally, we obtain x' and x" separately (since x'-f a;" = fc). With x' known y' is found from (6) ; i.e., [x' x'-i ^ . - ^- ec +e c\ or, y' + c-- =c-coshl — (9) Thus the position of the vertex is located. The greatest tension will be at the highest point A', viz., TA' = q{y' + c) (10) [The expression ^-[e^— g-w] is called the "hyperbolic sine" of the number u, or sinh (u); and ^[e^+e-u] the "hyperbolic cosine" of w, or cosh (u); e being the Naperian base 2.71828 . . . Tables of sinh u and cosh u will be found in the appendix.] Example. — A chain 100 ft. long is supported at two points 80 ft. apart horizontally and 30 ft. vertically; find the position of its lowest point. That is, (Fig. 5 4a) giv en Z=100 ft., k = 80 ft., d = 30 ft. Solution.— ■\/P-d^ = \/9100 = 95A ft., the left-hand member of (7). Assuming c = 20 ft. as a first trial, we find fc-r-2c = 2.00 and sinh (2.00) = 3.6269, so that 2c sinh (2.00) is 145.176, which is much larger than 95.4. Next, with c assumed as 40, 39, and 38.3 ft., we find 2c sinh (A;-r-2c) to be 2X40X1.1745 = 93.96; 2X39X1.2153 = 94.8; and 2X38.3 X 1.2444 = 95.3, respectively; and hence conclude that c=38.2 ft. will satisfy eq. (7) with sufficient accuracy. Eq. (8) now becomes a;'-x" = 38.2X2.303X0.2689 = 23.66 ft.; and finally we obtain x' = 51.83 and x" = 28.17 ft. From eq. (9) we now have 2/' + 38.2 = 38.2X2.07 13 = 79. 10 ft. and .-. j/' = 40.9 ft. With ?=1.5 lbs. per foot, the tension at A'=l. 5X79. 1 = 118.6 lbs. PART II.-KINETICS. CHAPTEE I. EECTILINEAR MOTION OF A MATERIAL POINT, 49. Uniform Motion implies that the moving point passes over equal distances in equal times ; variable motion, that un- equal distances are passed over in equal times. In uniform motion the distance passed over in a unit of time, as one sec- ond, is called tlie velocity (= v), which may also be obtained by dividing the length of any portion (= s) of the path by the time (= t) taken to describe that portion, however small or great ; in variable motion, however, the velocity varies from point to point, its value at any point being expressed as the quotient of ds (an infinitely small distance containing the given point) by dt (the infinitely small portion of time in which ds is described). 49«. By acceleration is meant the rate at which the velocity of a variable motion is changing at any point, and may be a uniform acceleration, in which case it equals the total change of velocity between any two points, however far apart, divided by the time of passage ; or a variable acceleration, having a different value at every point, this value then being obtained by dividing the velocity -increment, dv, or gain of velocity in passing from the given point to one infinitely near to it, by dt, the time occupied in acquiring the gain.* (Acceleration must be understood in an algebraic sense, a negative accelera- tion implying a decreasing velocity, or else that the velocity in a negative direction is increasing.) The foregoing applies to motion in a path or line of any form whatever, the distances mentioned being portions of the path, and therefore measured along tlie path. (Sue p. 43 in the ' ' Notes, ' ' etc. * See addendum on p. 836. 49 50 MECHANICS OF EJSGINEERING. 50. Eectilinear Motion, or motion in a straight line. — The general relations of the quantities involved may be thus stated (see Fig. 55) : Let v = velocity of the body at any instant ; -s g .as as, t^ ^^^^^ ^^ ~ ^^^^ ^^ velocity • ? \ \ I in an instant of time dt. Let idtidt'k i i^ = time elapsed since the body left a given fixed point, which will be taken as an origin, 0. Let s = distance (-f- or — ) of the body, at any instant, from the origin ; then ds = distance traversed in a time dt. Let^ = acceleration = rate at which v is increasing at any instant. All these may be variable ; and t is taken as the independent variable, i.e., time is conceived to elapse by eqical small increments, each = dt ; lience two consecutive dsh will not in general be equal, their difference being called d^s. Evidently d^t is = zero, i.e., dt is constant. Since -,- = number of instants in one second, the velotity at any instant (i.e., the distance which would be described at that •IN . 7 1 ^^ /TV rate m one second) \q v =■ ds . -n-', .' . v ■= -^- (L) Similarly, J> =^ dv . -t> and I since dv = di^^J =~^ J' dv d^s Eliminating dt, we have also vdv = pd^. (HI.) Tliese are the fundamental differential formulae of rectilinear motion (for curvilinear motion we have these and some in ad- dition) as far as kinematics, i.e., as far as space and time, is concerned. The consideration of the mass of the material point and the forces acting upon it will give still another rela- tion (see § 55). Example. — If we have given s = [6<^ + 3<^ + 2<] ft., for a certain motion, then the velocity, v, at any time, =ds^^dt, =[18i'*+6< + 2] ft. per sec; and the acceleration, p, =dv-^dt, =[36t + 6]ft. per sec. per sec. 61. Rectilinear Motion due to Gravity. — If a material point fall freely in vacuo, no initial direction other than vertical having been given to its motion, many experiments have RECTILINEAR MOTION OF A MATERIAL POINT. 51 sho^*•n that this is a uniformly accelerated rectilinear motion in a vertical line having an acceleration (called the accelera- tion of gravity) equal to 32.2 feet per square second,* or 9.81 metres per square second ; i.e., the velocity increases at this constant rate in a downward direction, or decreases in an up- ward direction. [Note. — By " square second " it is meant to fay stress on tlie fact that an acceleration (being = d^s -h df^) is in quality equal to one dimension of length divided by two dimensions of time. E.g., if instead of using the foot and second as units of space and time we use the foot and the minute, g will = 33.3 X 3600; whereas a velocity of say six feet per second would = 6 X 60 feet per minute. The value of g = 33.3 implies the units foot and second, and is sufficiently exact for practical purposes.] 52. Free Fall in Vacuo.— Fig. 56. Let the body start at v,dth an initial downward velocitj^ = c. The accelera- _s tion is constant and = -\- g. Reckoning both time and I distance (-|- downwards) from O, required the values of „i^\ the variables s and v after any time t. From eq. (II.), \ \ c § 50, we have -{- g = dv -^ dt; .'. dv = gdt, in which the " k two variables are separated. I 1 dv = gj dtt\ i.e., v ^=^ g\ t\ ox v — c =^ ""j gt — ^\ and finally, -y = c + (/^ (1) fig. 56. (ISTotice the correspondence of the limits in the foregoing operation ; when ^ = 0, -y = -f- <^') From eq. (I.), § ^^^v ^= ds -^ dt\ .'. substituting from (1), ds ^ {c -{- gt)dt, in which the two variables s and t are sepa- rated. ds = cj^ dt + gj^ tdt ; i.e., [_^5 = e\j + ^[^ ^ ' or 5 = c? + ^gf (2) Again, eq. (Ill-), § 50, vdv = gds, in which the variables v and ,9 are already separated. /v ps r"u r~s vdv = gj^ ds; or \ iv' = g s; i.e., ^{v' — c') = gs, ■_.c L-0 * Or, 82.3 "feet per second per second." 52 MECHANICS OF ENGINEERING. If the initial velocity = zero, i.e., if the body falls from rest, t eq. (3) gives s=^-audv= V^gs. [From the frequent re- currence of these forms, especially in hydraulics, ^is called the "height due to the velocity t;," i.e., the vertical height through which the body must fall from rest to acquire the velocity v ; while, conversely, V'2gs is called the velocity due to the height or "head" s.] By eliminating g between (1) and (3), we may derive another formula between three variables, s, v, and t, viz., S = i{G-i-V)t . (4:) Example. — A leaden ball occupies 4.6 seconds in falling from the eaves of a tall building to the sidewalk; initial velocity zero. Find the height fallen through, =s'. We have from eq. (2) s' = + i(32.2)(4.6) = = 341 ft. 53. Upward Throw. — If the initial velocity were in an up- ward direction in Fig. 56 we might call it — c, and introduce it with a negative sign in equations (1) to (4), just derived ; but for variety let us call the upward direction -|-, in which case an upward initial velocity would = -|- c, while the acceleration = — g, constant, as before. (The motion is supposed confined within such a small range that g does not sensibly vary.) Fig. i 67. From p = dv -^ dt we have dii ■= — gdt and J^ dv = — gj^ dt; r.v — G = — gt\orv = G — gt. {l)a From V ^=ds -^ dt, ds = cdt — gtdt, ns r*t r>t i.e., J^ ds = cj^ dt — gj^ tdt ; or s=Gt — ^gt\ (2)a O i ' p' p^ — S vdv = pds gives / vdv = — a I ds, whence Fig. 57. .^ & «/c if Jo ^(^' — c') = — gs, or finally, s = . . (3)a And by eliminating g from {l)a and (3)a, s = i{G-{-v)f (4)a The following is now easily verified from these equations ; the body passes the origin again (6' = 0) with a velocity = — c, after a lapse of time =2g -r- g. The body comes to rest (for if EECTILINEAE MOTION OF A MATERIAL POINT. 53 an instant) (put v = 0) after a time = c -^ ^, and at a distance s = c^ -^ 2g (" height due to velocity c") from 0. For t > G -^ g, V is negative, sliowing a downward motion ; for t > 2g -^ g, s is negative, i.e., the body is below the starting-point while the rate of change of v is constant and = — ^ at all points. Example.— Let c be 40 ft./sec. Then in a time = 40 -r- 32.2, =1.24 sec, the body will reach its maximum height, (40)2^-2X32.2 = 24.84 ft. above the start. After 3 sec. the body will be found a distance S3=40x3-i(32.2)(3)2=-24.9 ft. from the origin, i.e., below it. 54. Newton's Laws. — As showing the relations existing in general between the motion of a material point and the actions (forces) of other bodies upon it, experience furnishes the fol- lowing three laws or statements as a basis for kinetics : (1) A material point under no forces, or under balanced forces, remains in a state of rest or of uniform motion in a right line. (This property is often called Inertia^ (2) If the forces acting on a material point are unbalanced, ' an acceleration of motion is produced, proportional to the re- sultant force and in its direction. (3) Every action (force) of one body on another is always accompanied by an equal, opposite, and simultaneous reaction. (This was interpreted in § 3.) As all bodies are made up of material points, the results ob- tained in Kinetics of a Material Point serve as a basis for the Kinetics of a Rigid Body, of Liquids, and of Gases. 55. Mass.- — If a body is to continue moving in a right line, the resultant force P at all instants must be directed along that line (otherwise it would have a component deflecting the body from its straight course). (See addendum on p.' 835.) In accordance with Newton's second law, denoting by j? the acceleration produced by the resultant force {O- being the body's weight), we must have the proportion P \ G : : jp \ g \ i.e., P = — .p , orP=i^. . . (lY.) Eq. lY. and (I.), (II.), (III.) of § 50 are the fundamental equations of Dynamics. Since the quotient G ■— g \s, invaria- 54 MECHANICS OF ENGINEERING. ble, wherever the body be moved on the earth's surface {O and g changing in the same ratio), it will be used as the measure of the mass M ov quantity of matter in the body. In this way it will frequently happen that the quantities G and g will ap- pear in problems where the weight of the body, i.e., the force of the earth's attraction upon it, and the acceleration of gravity have no direct connection with the circumstances. No name will be given to the unit of mass, it being always understood that the fraction G -^ g will be put for M before any numeri- cal substitution is made. From (lY.) w'e have, in words, accelerating force = mass X acceleration^ also, acceleration — accelerating force -=- Quass. 56. Uniformly Accelerated Motion. — If the resultant force is constant as time elapses, the acceleration must be constant (from eq. (IV.), since of course J/" is constant) and = P -^ M. The motion therefore will be uniformly accelerated, and we have only to substitute + pi, (constant) , ior g in eqs. (1) to (4) of § 52 for the equations of this motion, the initial velocity being = c (in the line of the force). v = G-]-pit . . . (1); s = ct-i-ipit^; , . . (2) ('y*-_c'') If the force is in a negative direction, the acceleration will be negative, and may be called a retardation; the initial veloc- ity should be made negative if its direction requires it. 57. Examples of Unif. Ace. Motion. — Exomh;ple 1. Fig. 58. A small block whose weight is \ lb. has already described a — S g P M ^ g distance Ao = 4-8 inches over a A SMOOTH — 7-> ^" -0^~ 1 smooth portion of a horizontal Fig. 58. table in two seconds ; at 6^ it en- counters a rough portion, and a consequent constant friction of 2 oz. Required the distance described beyond 0, and the time occupied in coming to rest. Since we shall use 32.2 for g, times must be in seconds, and distances in feet ; as to the unit 8 = ^-^;, . . (3), and5 = K^ + ^)^5 ... (4) KECTILINEAK MOTION OF A MATERIAL POINT. 55 -t force, as that is still arbitrary, say ounces. Since AO was smooth, it must have been described with a uniform motion (the resistance of the air being neglected); hence with a veloc- ity = 4 ft. -^ 2 sec. = 2 ft. per sec. The initial velocity for the retarded motion, then, is c = -|- 2 at (9, At any point be- yond the acceleration = force -— mass = ( — 2 oz.) -r- (8 oz. -f- 32.2) = — 8.05 ft. per square second, i.e., p = — 8.05 = constant ; hence the motion is uniformly accelerated (retarded here), and we may use the formulae of § 56 with g = + 2, pi = — 8.05. At the end of the motion v must be zero, and the corresponding values of s and t may be found by putting v = in equations (3) and (1), and solving for s and t respectively : thus from (3), 5 =^(-4)-4- (— 8.05), i.e.,s = 0.248 +, which must be feet ; while from (1), t={—2)-^{— 8.05) = 0.248 +, which must be seconds. Examjple 2. (Algebraic.) — Fig. 59. The two masses J/", = G^ -~ g and M ^ G -^ g are connected by a flexible, inexten- sible cord. Table smooth. Required the acceleration common to the two rectilinear motions, and the tension in the string S,. i^s. Fig. 59. Fig. 60. there being no friction under G^, none at the pulley, and no mass in the latter or in the cord. At any instant of the mo- tion consider G^ free (Fig. 60), iV being the pressure of the table against G^. Since the motion is in a horizontal right line ^(vert. compons.)= 0, i.e., iV— G^ = 0, which determines iV„ S, the only horizontal force (and resultant of all the forces) = M,p, i.e., S= G,p-^g. (i; At the same instant of the motion consider G free (Fig. 61);. the tension in the cord is the same value as above = S. The accelerating force is ^ — 8, and .*. = mass X ace, or G — 8 ^= {G -^ g)p. . (2) y 56 MECHANICS OF ENGINEERING. j |g From equations (1) and (2) we obtain p — {Gg) ~ \ y, {G -^ G^ = a constant ; hence each motion is uniformly j r S accelerated, and we may employ equations (1) to (4) of i ^ § 56 to find the velocity and distance from the starting- ^ points, at the end of any assigned time t, or vice versa. ■ The initial velocity must be known, and may be zero. Also, from (1) and (2) of this article, S = {GG,) -^ ((9 + G,) = constant. Example 3. — A body of 2J (short) tons weight is acted on during ^ minute by a constant force P. It had previously de- scribed 316f yards in 180 seconds under no force ; and subse- quently, under no force, describes 9900 inches in -^ of an hour, Eequired the value of P. Ans. P = 22.1 lbs. . / Example 4. — A body of 1 ton weight, having an initial ''velocity of 48 inches per second, is acted on for \ minute by a force of 400 avoirdupois ounces. Required the final velocity, Ans. 10.03T ft. per sec. Exa/mjple 5. — Initial velocity, 60 feet pei second ; body weighs 0.30 pf a ton. A resistance of 112|- lbs. retards it for -^-^ of a ininute. Required the distance passed over during this time. Ans. 286.8 feet. Example 6. — ^Required the time in which a force of 600 avoir- dupois ounces will increase the velocity of a body weighing \\ tons from 480 feet per minute to 240 inches per second. Ans. 30 seconds. Example 7. — What distance is passed over by a body of (0.6) tons weight during the overcoming of a constant resistance (friction), if its velocity, initially 144 inches per sec, is reduced to zero in 8 seconds. Required, also, the friction. Ans. 48 ft. and 55 lbs. Example 8. — Before the action of a force (value required) a body of 11 tons had described uniformly 950 ft. in 12 minutes. Afterwards it describes 1650 feet uniformly in 180 seconds. The force acts 30 seconds. P — % Ans. P = 178 lbs. KliCTILINEAR MOTION OF A MATERIAL POINT. 67 58. Graphic Eepresentations. Unif. Ace. Motion. — With the initial velocity = 0, tlie equations of § 56 become V =pit,. = V 2pi, (1) (3) s = ipit",. and (2) (4) Eqs. (]), (2), and (3) contain each two variables, which may graphically be laid off to scale as co-ordinates and thus give a curve corresponding to the equation. Thus, Fig. 62, in (I.), we A V J s . 0- t- !t MQ- 0^ J^-y( (II.) (HI.) Fig. 63. have a right line representing eq. (I.) ; in (II.), a parabola with axis parallel to s, and vertex at the origin for eq. (2) ; also a parabola similarly situated for eq. (3). Eq. (4) contains tlii'ee variables, s, -y, and t. Tliis relation can be shown in (I.), s be- ing represented by the a?'ea of the shaded triangle = ^vt. (11.) and (III.) have this advantage, that the axis OS may be made the actual path of the body. [Let the student determine how the origin shall be moved in each case to meet the supposi^- tion of an initial velocity = -|- c or — c] (SeelSTotes, p. 120.) 59. Variably Accelerated Motions. — We here restate the equa- tion s ( differential) 9) = ds dt dv d^s •(i-);p = ^ = dt df . (XL); v<?'y=^^5..(III.) and resultant force = P=Mp,. , . . o . (lY.); which are the only ones for general use in rectilinear motion and involve the five variables, s, t, v, p, and P. Problem 1. — In pulling a mass M along a smooth, horizon- tal table, by a liorizontal cord, the tension is so varied that i? = 4:f {not a homogeneous equation / the units are, say, the foot and second). Required by what law the tension varies. 58 MECHANICS OF ENGINEERING. From (L) , = ^^ = A_l ^ i^f ; from (IL), p = ^-^ = 24:t', and (lY.) the tension = P = 2fp = ^^Mt, i.e., varies, directly as the time. Pkoblem 2. " Harmonic Motion," Fig. 63. — A small block Tig. 63 on a smooth horizontal table is attached to two horizontal elastic cords (and they to pegs) in snch a way that when the- block is at (9, each cord is straight but not tense ; in any other position, as ^y?., one cord is tense, the other slack. The coi'ds are alike in every respect, and, as with springs, the tension varies directly with the elongation (= 5 in figure). If for an elongation s^ the tension is Tj, then for any elongation s it is- r = riSH-Si.v If the block be given an initial velocity =c at 0, it begins to execute an oscillatory motion on both sides of 0; m is any point of its motion. The tension T is the accelerating force ; variable and always pointing toward 0. The acceleration at any point m, then, is p — — {T -^ M) = — {T^s -f- J/Sj), which for brevity put ^ = — as, a being a constant. Required the equations of motion, the initial velocity being = -[- c, at 0. From eq. (til.)- vdv = — asds ; .'. / vdv := — a I sds, i.e., ^(y' — c') = — ^as^ ; or, y' = 6'' — as\ . (1)- From (I.), dt = ds^v',\ CK^ r\, .- -, \ ,-,^ > I dt= I [ds^yc^—as^]: or, hence from (1), }Jo Jo 1 . JsV^\ Va ^ = ~-^sm-^[-—j (2), RECTILINEAR MOTION OF A MATERIAL POINT. 59 Inverting (2), we have 5 = (c -r- Va) sin (t Va), ... (3) iLgain, by differentiating (3), see (I.), -y = c cos (tVa) (4) Differentiating (4), see (II.), 2^ = — cVa sin {t Va). . . (5) These are the relations required, each between two of the four variables, s, t, v, and p; but the peculiar property of the motion is made apparent by inquiring the time of pass- ing from (? to a state of rest ; i.e., put i) = in equation (4), we obtain i =z ^tt -i- Va, or ^tc — Va, or ^n -^ V~a, and so on, while the cori-esponding vakies of s (from equation (3)), are '\-{g -^ Va), — (c -^ Va), -f- (c -^ Va), and so on. This shows that tlie body vibrates equalW on both sides of in a cycle or period whose duration = 2;r -^ Va, and is indejpendent of the initial velocity given it at 0. Each time it passes the velocity is either -|- c, or — c, the acceleration = 0, and the time since the start is = nn -f- \/a^ in which n is any whole number. At the extreme point ^9 = :f c j/a, from eq. (5). If then a different amplitude be given to the oscillation by changing c, the duration of the period is still the same, i.e., the vibration is isochronal.'^ The motion of an ordinary pen- dulum is nearly, that of a cycloidal pendulum exactly, harmonic. If the crank-pin of a reciprocating engine moved uniformly in its circular path, the piston would have a harmonic motion if the connecting-rod were infinitely long, or if the design in •< 2r > Fig. 64. Fig. 64 were used. (Let the student prove this from eq. (3).) Let 2r = length of stroke, and c = the uniform velocity of the crank-pin, and M = mass of the piston and rod A£. Then the velocity of M at mid-stroke must = c, at the dead-points, zero; its acceleration at mid-stroke zero; at the dead-points the ace. = c Va, and s = r = c -7- Va (from eq. (3)) ; .*. V~a =: c -^ r, and the ace. at a dead point (the maximum ace.) * See illustrations and example on pp. 47, 48, of the "Notes." 60 MECHANICS OF ENGINEEEING- := c* -4- r. Hence on account of the acceleration (or retarda- tion) of J!/^in the neighborhood of a dead-point a pressure will be exerted on the crank-pin, equal to mass X ace. = M& -^ r at those points, independently of the force transmitted due to steam-pressure on the piston-liead, and makes the resultant pressure on the pin at G smaller, and at D larger than it would be if the '"''inertia)'' of the piston and rod were not thus taken into account. We may prove this also by the free-body method, considering AB free immediately after passing the dead-point P, A Fig. 65. \ ?' C, neglecting all friction. See Fig. 65. The forces acting are : G, the weight ; W, the pressures of the guides ; P, the known effective steam- pressure on piston-head ; and P', the unknown pressure of crank-pin on side of slot. There is no change of motion ver- tically ; .-. iT' + i\^— 6^ = 0, and the resultant force is P — P' = mass X accel. = Mc" -^ r^ hence P' ^= P — M& — r. Similarly at the other dead-point we would obtain P' =^ P -{- Mc^ -^ r. In high-speed engines with heavy pistons, etc., Mg^ -^ r is no small item. [The upper half-revol., alone, is here considered.] (See example on p. 68, "Notes.") Problem 3. — Supposing the earth at rest and the resistance of the air to he null, a body is given an initial upward vertical velocity = c. Required the velocity at any distance s from the centre of the earth, whose attraction varies in- versely as the square of the distance s. See Fig. Q^. — The attraction on the body at the surface of the earth where s = r, the radius, is its weight G; at any point m it will be P = G(r2-=- s^),* while its mass = G -7- g. Hence the acceleration at m = j? = ( — P) -j- J/ = — ^(r° -=- s°). Take equation III., vdv = j)ds, and we have vdv = — gr^s ~^ds°, .'. Fig. 66. I vdv =z — gr^ I s 'ds or. 4^,« fv = — gr I.e. ii^' ^')=-^r^{l-~ (1) * That is, the force of attraction, {P, lbs.) at any distance, s, from O is to the force at the surface (viz., G lbs.) as r'' is to s^. BECTILINEAR MOTION OF A MATERIAL POINT. 61 Evidently v decreases, as it should. Now inquire how small A value c may have that the body shall never return/ i.e., that V shall not = until 5 = oo. Put v = and s =s 00 in (1) and solve for g ; and we have c = V2ffr = V2 X 32.2 X 21000000, = about 36800 ft. per sec. or nearly 7 miles per sec. Con- versely, if a body be allowed to fall, from rest, toward the earth, the velocity with which it would strike the surface would be less than seven miles per second through whatever distance it may have fallen. If a body were allowed to fall through a straight opening in the earth passing through the centre, the motion would be har- monic, since the attraction and consequent acceleration now vary directly with the distance from the centre. See Prob. 2. . This supposes the earth homogeneous. Problem 4. — Steam working expansively and raising a weight. Fig. 67. — A piston works without friction in a vertical cylinder. Let S = total steam-pressure on the underside of the piston ; the weight G, of the mass G — g (which in- cludes the piston itself) and an atmospheric pressure = A^ con- stitute a constant back-pressure. Through the portion OB = s^, of the stroke. Sis constant = S^, while beyond B, boiler com- munication being " cut off," S diminishes with Boyle's law, i.e., in this case, for any point above JB, we have, neglecting the " clearance", 2^ being the cross-section of the cylinder, * S:S,::Fs,: Fs; or S=S,s,-^s. (Which gives *S as a function of s at any point above B.) Full length of stroke = ON" = s^. Given, then, the forces S^ and A, the distances s, and 5„, and the velocities at and at ^both = (i.e., the mass M = 6^^ -j- ^ is to start from rest at O. and to come to rest at iV), required the proper weight G to * See p. 627 for meaning of "clearance." Fig. 67. €2 MECHANICS OF ENGINEERING. fulfil these conditions, ^S* varying as already stated. The accel- eration at any point will be j9= [_S-A-G^~M. . . . . (1) Hence (eq. III.) Mvdv = [S — A — G'jds, and .*. for the whole stroke J/j^ vdv=J^ [S-A- G]ds; i.e., = Sj^ & + <Sa/ -^-A^de-oX ds, or 8A[l + iog/fJ = As,+ es,.. . . (2) Since S = /S^ ^ constant, from O to £, and variable, = ^i^i -^ 5, from ^ to JV, we have had to write the summation X Sds in two parts. From (2), G becomes known, and .". J!f also {= G — g). Required, further, the time occupied in this upward stroke. From to B (the point of cut-off) the motion is uniformly accelerated, since p is constant {8 being = 8^ in eq. (1) ), with the initial velocity zero; hence, from eq. (3), § 56, the velocity 2it B = v^ z= V'i, \_8^ — A — G'\s^ -^ M'k known ; .'. the time ^j = 2Sj -i- v^ becomes known (eq. (4), § 56) of de- scribing OB. At any point beyond B the velocity v may be ob- tained thus : From (III.) vdv — _pds, and eq. (1) we have, summing between B and any point above, M^vd. = S,s.£ ~-iA + <?)/&; i.e., G (v" — V') S f A y n^ f N -- — 2—^ = ^1*1 log. --{A^G){s- 5,). This gives the relation between the two variables v and 8 anywhere between B and iV"; if we solve for -y and insert its value in dt ^= ds -~- v, we shall have dt = a. function of s and ds, which is not integrable. Hence we may resort to approxi- KECTILINEAU aiOTION OF A MATERIAL POINT. 63 mate methods for the time from B to iT. Divide the space BN'mio an imeveii nnuiber of equal parts, say five; the dis- tances of the points of division from will be s„ s,, s^, s^, s^, and Sn- For these values of s compute (from above equation) Wi (already known), v„ -y,, v,, v,, and v^ (knowTi to be zero). To the first four spaces apply Simpson's Eule,* and we have the time from JS to the end of 5,, t 1 , 4 , 3 . 4: . 1 : / - ; approx. = ^. - - -\ L- while regarding the motion from 5 toiTas uniformly retarded (approximately) with initial velocity = t\ and the final = zero, we have (eq. (4), § 56), -TV t = 2{S^ - S,) -r- V,. —6 By adding the three times now found we have the whole time of ascent. In Fig. 67 the dotted curve on the left shows by horizontal ordinates the variation in the velocity as the distance s increases ; similarly on the right are ordinates showing the variation of jS. The point ^, where the velocity is a maximum = v,n, may be found by putting p = 0, i.e., for S ■= A-\- G, the acelerating force being = 0, see eq. (1). Below ^the ac- celerating force, and consequently the acceleration, is positive; above, negative (i.e., the back-pressure exceeds the steam- pressure). The horizontal ordinates between the line IIEKL and the right line HT hve proportional to the accelerating force. If by condensation of the steam a vacuum is produced be- low the piston on its arrival at iV^, the accelerating force is downward and ^ A-\- G. [Let the student determine how the detail of this problem would be changed, if the cylinder were horizontal instead of vertical.] 60. Direct Central Impact. — Suppose two masses J/j and Jf, to be moving in the same right line so that their distance apart continually diminishes, and that when the collision or impact takes place the line of action of the mutual pressure coincides with the line joining their centres of gravity, or centres of * See p. 13 of the "Notes and Examples." 64 MECHANICS OF ENGINEERING. mass, as they may be called in this connection. This is called a direct central impact, and the motion of each mass is varia- bly accelerated and rectilinear during their contact, the only force being the pressure of the other body. The whole mass of each body will be considered concentrated in the centre of mass, on the supposition that all its particles undergo simul- taneously the same change of motion in parallel directions. (This is not strictly true ; the effect of the pressure being gradually felt, and transmitted in vibrations. These vibrations endure to some extent after the impact.) When the centres of mass cease to approach each other the pressure between the bodies is a maxinmm and the bodies have a common velocity; after this, if any capacity for restitution of form (elasticity) exists in either body, the pressure still continues, but dimin- ishes in value gradually to zero, when contact ceases and the bodies separate with different velocities. Reckoning the time from the first instant of contact, let t' = duration of the first period, just mentioned ; t" that of the first -(- the second (resti- tution). Fig. 68. Let Jf^ and Jf^ be the masses, and at any <..?i. I — ^..3?2.^ instant during the contact let v^ and v^ ~1)T be simultaneous values of the velocities -^ p Ml M2 of the mass-centres respectively (reckon- ^^®- ^^- ing velocities positive toward the right), and f* the pressure (variable). At any instant the acceleration of J[fj is j?j = — (P -^ J!/i), while at the same instant that of M, is j)^ = -\- {P -^ M^ ; Jfj being retarded, M^ accelerated, in velocity. Hence (eq. 11.,^ = dv -f- dt) we have M,dv,= — Pdt\ and M^dv^— -\- Pdt. . . (1) Summing all similar terms for the first period of the impact, we have (calling the velocities before impact c^ and c^, and the common velocity at instant of maximum pressure G) ^Jcy^^= - H'P^i^ ^-e-' ^^(^' -^a) = - S! Pdt ; (2) EECTILINEAK MOTION OF A MATERIAL POINT. 66 The two integrals * are identical, numerically, term by term, since the pressure which at any instant accelerates J/, is nu- merically equal to that which retards J/, ; lience, though we do not know liow P varies with the time, we can eliminate the definite integral between (2) and (3) and solve for C. If the impact is inelastic (i.e., no power of restitution in either body, eitlier on account of their total inelasticity or damaging effect of the pressure at the surfaces of contact), they continue to move with tliis common velocity, which is therefore their final velocity. Solving, \Ne have ^- M,+M, ^^^ Next, supposing that the impact \q partially elastic, ihid, the bodies are of the same material, and that the summation I Pdt for the second period of the impact bears a ratio, e, to that / Pdt, already used, a ratio peculiar to the material, if the impact is not too severe, we have, summing equations (1) for the second period (letting Y^ and Y^ = the velocities after impact), ^^ X""^^^ = - S"^^*^ ^•^•' ^'( ^- ^) = - ^/*^^^; (s) M, / V = + X'Pdt, i.e., M,{ Y- 0) = + ej^dt. (6) 6 is called the coefficient of restitution. Having determined the value of / Pdt from (2) and (3) in terms of the masses and initial velocities, substitute it and that of (7, from (4), in (5), and we have (for the final velocities) y. = W^: + M^c- eMlc-c:j\ - [if, + JfJ; . (7) and similarly F, = [JfA + ^.c.+ ^^,(^-^.)]-W + ^,]- • (8) For d = 0, i.e., for inelastic impact^ Y^= Y,= C m eq. (4) ; for * That is, the right-hand members of eqs. (2) and (3). 66 MECHAISriCS OF ENGINEERING. <g= 1, or elastic impact^ (7) :ind (8) become somewhat simpli- fied. To deterinino e experimeiitallj, let a ball (-3/,) of the sub- stance fall upon a very large slab [M^ of the same substance, noting both the height of fall h^. and the height of rebomid H^. Considering M^ as = cc, with Ci= ^ ^yh^, V = —V ^H,, and c, = o, eq. (7) gives — ^/ "^gH^, =— eV "Igh, ; .-. e = V^, -^ h^. Let the student prove the following from equations (2), (3), (5), and (6) : {a) For any direct central impact whatever, [The product of a mass by its velocity being sometimes called its momentum^ this result may be stated thus : In any direct central impact the snm of the momenta before impact is eqnal to that after impact (or at any instant during impact). This principle is called the Conservation of Momen- tum. The present is only a particular case of a more genei'al proposition. It can. be proved that C, eq. (4), is the velocity of the centre of gravity of the two masses before impact ; the conservation of momentum, then, asserts that this velocity is unchanged by the impact, i.e., by the mutual actions of the two bodies.] (h) The loss of velocity of Jf,, and the gain of velocity of J/j, are twice as great when the impact is elastic as when in- elastic. (c) If e = 1, and M, = M„ then V, = + c^, and V^ = c,. Example. — Let Mi and M^ be perfectly elastic, having weights = 4 and 6 lbs. respectively, and let Ci = 10 ft. per sec. and C2 = — 6 ft. per see. (i. e., before impact M^ is moving in a direction contrary to that of Mi). By substituting in eqs. (7) and (8), with 6 = 1, Mi — A-^ g, and Jfa = 5 -i- g, we have Vi = ir4 X 10 + 5 X (- 6) - 5 (lO - (- 6))]= - 7.7 ft. per sec. Vi = lr4 X 10 + 5 X (- 6) + 4 Ao - (- 6))]= + 8.2 ft. per sec. as the velocities after impact. Notice their directions, as indicated by thdr "virtual velocities." 67 CHAPTER II. "VIRTUAL velocities; 61. Definitions. — If a material point is moving in a direction not coincident with that of the resultant force acting (as in cnrvihnear motion in the next chapter), and any element of its path, ds, projected npon this force;* the length of this projec- tion, du, Fig. 69, is called the "Virtual Yelocity" of the force, since du -^ dt uvAy be considered the veloc- ity of the force at this instant, just as ds -=- dt is ,7,, that of the point. The product of the force by q^ its du will be called its mrtucd moment, reckoned -f- or — according as the direction from (9 to Z^ is the same as that of the force or opposite. 62. Prop. I. — The virtual moment of a force equals the algebraic sum of those of its components. Tig. YO. Take the p direction of ds as an axis JT; let P^ and P, '^ be components of P\ a^, a^, and a their angles with X. Then (§ 16) P cos a. =^ cos a^-\-P^ cos a^. Hence P{ds cos ar)= P^{ds cos a^) -\- PJids cos a^. But ds cos a = the projection of ds upon P, i.e., ^ du ; Fig. 70. ^^ gQg ^^ _ ^/^^^^ g^g^ . _._ J^^^^ _ p^g.^^ _j_ P^du^. If in Fig. 70 a^ M'ere > 90°, evidently we would have Pdu = — Pfi.u^ -|- P^du^^ i.e., Pflu^ would then be negative, and OD^ would fall behind 0; lience the definition of -\- and — in § 61. For any number of components tlie proof would be similar, and is equally applicable whether they are in one plane or not. 63. Prop. II. — The sum of the mrtual moments equals zero, for concurrent forces in equilihrium. , * We should rather say " projected ou the line of action of the force ;* but the phrase used may be allowed, for brevity. 68 MiCCHANICS OF ENGINEERINGo (If the forces are balanced, the material point is moving in a straight line if moving at all.) The resultant foi'ce is zero. Hence, from § 62, P^du^ -f- P^du^ -\- etc. = 0, having proper regard to sign, i.e., ^{Pdv) ■=■ 0. 64. Prop. III. — The sum of the mrtual moinents equals zero for any small displacement or motion of a rigid hody in equi- librium under non-concurrent forces in a plane ^ all points oj the hody moving parallel to this plane. (Although the kinds of motion of a given rigid body which are consistent with balanced non-concurrent forces have not yet been investigated, we mav ima^ne any slio-ht motion for the sake of the alo;'e- braic relations between the different du^^ and forces.) • First, let the motion be a translation^ all points of the body describing equal parallel ..^ lengths = ds. Take ^parallel to ds ; let aTj, J^.3.^^ \jr etc., be the angles of the forces with X. ^^v Then (§ 35) '2{P cos «') = ; .-. ds^{P cos a) * = ; but ds cos a^ = du^ ; ds cos a^ = du^ ; ^^ ^ etc. ; .-. :2{Pdu) ^ 0. Q. E. D. Secondly, let the motion be a rotation Fig. 71. through a small angle dd in the plane of the forces about any point in that plane, Fig. 72. With (9 as a pole let /Oj be the radius-vector of the point of application of P^. and «j its lever-arm from 0\ similarly for the p other forces. In the rotation each point of application describes a small arc, ds.^^ ds^, / / ' ' etc., propoi'tional to Pj, Pg, etc., since ds^ //'^^ ..--''^\ ^ p,dd, ds, = p,dd, etc. From § 36, ^ ..^'-r^'.'-i^^ ^/^ P^a^ -\- etc. = ; but from similar triangles '^""-•-d ds^ : du^ :: Pi : a, ; .*. «, = p^d^i-^ -^ ds^ '"-■' = du^ -f- dd ; similarly a^ = du, -=- dO, etc. ^^'^' ^^" Hence we must have [P^du^ -\- P^du, -j- . . .] -f- dd = 0, i.e., ^{Pdu) = 0. Q. E. D. Now since any slight displacement or motion of a body may be conceived to be accomplished by a small translation fol- lowed by a rotation through a small angle, and since the fore- "VIRTUAL VELOCITIES. 69 going deals only with projections of paths, the proposition is established and is called the Principle of Virtual Velocities. [A simihxr proof may be used for any slight motion what- ever in space when a system of non-concurrent forces is bal- anced.] Evidently if the path {ds) of a point of application is perpendicnlar to the force, the virtual velocity {du), and con- sequently the virtual moment {Pdu) of the force are zero. Hence we may frequently make the displacement of such a character in a problem that one or more of the forces may be excluded from the summation of virtual moments. 65. Connecting-Rod by Virtual Velocities. — Let the effective steam-pressure P be the means, througli the connecting-rod and crank (i.e., two links), of raising the weight G very slowly ^ neglect friction and the weight of the links themselves. Con- sider AB as free (see (5) in Fig. 73), BC also, at (c) ; let the Bi B/- \ N "^^X nN^^>\ Fig. 73. "small displacements" of both be simidtaneous small portions of their ordinary motion in the apparatus. A has moved to A^ througli dx ; B to ^i, through ds, a small arc ; C has not moved. The forces acting on AB are P (steam-pressure), N (vertical reaction of guide), and N' and :7^(the tangential and normal components of the crank-pin pressure). Those on BC are N' and T (reversed), the weight (7, and the oblique pressure of bearing P'. The motion being slow (or rather the accelera- tion being small), each of these two systems will be considered ais balanced. Now put 2{Pdu) = for AB, and we have Pdx -\-]^xO-\-]V' XO-Tds = 0. . . (1) For the simultaneous and corresponding motion of BC, ^(Pdu) = gives 70 MECHANICS OF ENGINEERING. iV^' X + Tds - Gdh + P' X = 0, . . (2) (Zh being the vertical projection of {j's motion. From (1) and (2) we Lave, easily, Pdx — Gdh = 0, . (3) ./Bi which is the same as we mio'ht have I _^^>'--^k(^^ -?^ 'i obtained by putting 2{Pdu) = Ofor i^.-.-.-::."^-''' .... Jq\ p' the two links together^ regarded col- ^ ' \ lectively as a free hody^ and describ- ^i**- 74. iiig a small portion of the motion they really have in the mechanism, viz., (Fig. 74,) Pdx+WxO- Gdh-^P' X0 = 0. , o (4) We may therefore announce the — 66. Generality of the Principle ofVirtual Velocities. — If ci^y mechanism of jtexible inextensible cords, or of rigid bodies jointed together, or hoth, at rest, or in motion with very son all accelerations, he considered free collectively {or any portion of it), and all the external forces put in ; then {disregarding mutual f'ictions) for a small portion of its prescribed motion, 2{Pdu) must = 0, in which the du, or virtual velocity, of each force, P, is the projection of the path of the point of application upon the force (the product, Pdu, being -j- or — according to § 61). 67. Example. — In the problem of § 65, having given the weight G, required the proper steam-pressure (effective) P to hold G in equilibrium, or to raise it uniformly, if already in motion, for a given position of the links. That is. Fig. 75, given a, r, c, a, and /?, re- \^^/\p--a \^ quired the ratio dh \ dx; for, ^^■■-'^^^■:'''^ ' from equation (3), § 65, P ^^.i^:::^^^^^''''''''''^ = G{dh : dx). The projec- p^_^^.^-::::^^^y^ /! "^^§^1 tions of dx and ds upon AP dx A, ^ r -•>- will be equal, since A£ = ^^^- '^^■ A^B^, and makes an (infinitely) small angle with A^B^, i.e., 6^ cos a = ds cos (/3 — a). Also, dh = {c '. r)ds sin J3. "virtual velocities." 71 Eliminating ds, we have, dfi c sin /? cos a dx r cos (/i — oc) ' P= 6^ <? sin /? cos a 7" cos (/? — a)' 68. When the acceleration of the parts of the mechanism is riot practically zero, 2{Pdu) will not = 0, but a function of the masses and velocities to be explained in the chapter on Work, Energy, and Power. If friction occurs at moving joints, enough '• free bodies" should be considered that no free body extend beyond such a joint ; it will be found that this friction cannot be eliminated in the way T and N' were, in § 65. 69. Additional Problems; to be solved by "virtual velocities." Problem 1. — Find relations between the forces acting on a straight lever in equi- librium; also, on a bent lever. Problem 2. — When an ordinary copying-press is in equilibrium, find ■the relation between the force applied horizontally and tangentially at Lhe circumference of the wheel, and the vertical resistance under the screw-shaft. See Fig. 75a. Solution. — Considering free the rigid body consisting of the wheel and screw-shaft, let B be the resistance at the point of the shaft (poiuting along the axis of the shaft), and Pthe required horizontal tangential force at edge of wheel. Let radius of wheel be r. Besides R and P there are also acting on this body certain pressures, or "supporting forces," consist- ing of the reactions of the collars, and reactions of the threads of nut against the threads of screw. Denote by s the " pitcJi" of the screw, i.e., the dis- tance the shaft would advance for a full turn of the wheel. Then if we imagine the wheel to turn through a small angle dB, the corresponding advance, ds, of the shaft would be x<-, from the proportion s : ds :: %Tt: dB . The path of the point of application of P is AS, a small portion of a helix, the projection of which on the line of P is rdQ, while d& projects, in its full length on the line of the force R. In the case of each of the other forces, however, the path of the point of application is per- pendicular to the line of the force (which is normal to the rubbing surfaces, , friction being disre- garded). Hence, substituting in I{Pdu) = 0, we have + P . rdd-R . ds + + 0=0; whence ds P= rdO R- 'Inr R. Fig. 75a. 72 MECH ATTICS OF EWGINJfiEKINO. CHAPTER HI. CURYILINEAR MOTION OF A MATERIAL POINT. A° f ° D° Ln R'/7 ^ AC^-^ /'/ ° o A// bA 1 -ii/ 1 \ kJ \ Fig. 76. [Motion in a plane, only, will be considered in this chapter.] 70. Parallelogram of Motions. — It is convenient to regard the curvilinear motion of a point in a plane as compounded, or made up, of two independent rectilinear motions parallel respectively to two co-ordinate axes X. and 7^ as may be ex- plained thus : Fig. 76. Consider . the drawing-board CD as fixed, and let the head of a T-square move from A toward B along the edge according to any law whatever, while a pencil moves from M toward Q along the blade. The result is a curved line on the board, whose form depends on the character of the two ^ and IT component motions, ^^ they may be called. If m a time z5, the 2^-square head has moved an ^distance = J/7V, and the pencil simultaneously a Y distance = MP, hy com- pleting the parallelogram on these lines, we obtain li, the position of the point on the board at the end of the time t^. Similarly, at the end of the time t^ we find the point at R'. 71. Parallelogram of Velocities. — Let the X and T" motions be uniform, required the resulting motion. Fig. 77. Let g„ and Cy be the constant uniform X and Y velocities. Then in any time, t, we have a? = c^,^ and y = v Y/ €yt ; whence we have, eliminating t, as ~r y = c„ -^ Cy =. constant, i.e., x is proportional to y, i.e., the path is a O-^, straight line. Laying off OA = c^, I — ^. /<*.- and AB = Cy, ^ is a point of the path, Fig. n. and OB is the distance described by the point in the first CUKVILINEAR MOTION OF A MATERIAL POINT. 73 second. Since by similar triangles OR : x i: OB : c„ we have also OH = OB . t ; hence the resultant motion is uniform, and its velocity, OB = g, is the diagonal of the parallelograTn, formed on the two component velocities. Corollary. — If the resultant motion is curved, the direction and velocity of the motion at any point will be given by the diagonal formed on the component velocities at that instant. The direction of motion is, of course, a tangent to the curve. 72. TJniformly Accelerated X and Y Motions. — The initial velocities of hath heing zero. Required the resultant motion. Fig. Y8. From § 56, eq. (2) (both c^, andCj, / ^ being = 0), we have x = ^pj^ and y = //'" ^<^ ^yf, whence x -i- y = Px-^Py=^ constant, wVy'^^^i / and the resultant motion is in a straight j/^^ ^h -r- L — y^ line. Conceive lines laid off from 6^ on ^ ^* ^ and Y to represent j?a. and^j, to scale, and ^®' '^ ' form a parallelogram on them. From similar triangles {OB being the distance described in the resultant n^^otion in any time t), OB : x :: 'OB : p^ ; .-. Oir= ^OBf\ Hence, from the form of this last equation, the resultant motion is uniformly accelerated, and its acceleration is OB =pi, (on the same scale as^^ and^j,). This might be called the parallelogram of accelerations, but is really a parallelogram of forces, if we consider that a free material point, initially at rest at 0, and acted on simulta- neously by constant forces P^ and Py (so that p^. = P^ -^ M 2a\^ Py — Py -=- J[/), must begin a uniformly accelerated recti- linear motion in the direction of the resultant force, {having no initial velocity in any direction.) 73. In general, considering the point hitherto spoken of as a free material point, under the action of one or more forces, in view of the foregoing, and of Newton's second law, given the initial velocity in amount and direction, the starting-point, the initial amounts and directions of the acting forces and the 74 MECHANICS OF ENGINEERING-. laws of their variation if they are not constant, we can resolve them into a single ^ and a single T^ force at any instant, determine the ^ and T' motions independently, and afterwards the resultant motion. Note. — The resultant force is never in the direction of the tangent to the path (except at a point of infiection). The relations which its amount and position at any instant bear to the velocity, rates of change of that velocity (i.e., accelerations), both as to amount and position, and the radius of curvature of the path, will now be treated (§ 74). In Fig. 79, A, B, and C are three ' 'consecutive" positions of the moving point, AB and BC being two short chords of the curve. When dt is taken smaller and smaller (position B remaining unchanged) and finally becomes zero, the points A and C merge into B and the chords AB and BC becomes tangents at B; and hence the results to be obtained only apply to a single point, B. But note that before dt becomes zero each equation [except (7)] is divided through by dt (or dt^) and therefore the individual terms do not necessarily become zero also. 74. General Equations for the curvilinear motion of a ma- terial point in a plane. — The motion will be considered result- ,.K I ing from the composition of ,H'''' "'•- independent JT and 1^ motions, ..-0' r^ ^' \ C. - , ' ' 5^ tv ^ X and Y being perpendicular to ^2\L..--" eacli other. Fig. 79. In two consecutive equal times (each = dt), let dx and dx' = small spaces due to the X motion ; and dy and CK ^ dy\ due to the Y motion. Then ds and ds' are two consecutive elements of the curvilinear motion. Pro- long ds, making BE =^ ds; then EE = d'x, dF= d^y, and 00 = d^s {EO being perpendicular to BE). Also draw CL perpendicular to BG and call CL d^n. Call the velocity of the JT motion v^.-, its acceleration 'p^; those of the J" motion, Vy and fy. Then, dx dy dv. For the velocity along the curve (i.e., tangent) V =■ ds -T dt, we shnll have, since ds^ = dx^ -\- dy'' Fig. 79. _ d'^x , _ d^y _ ^V ~ df' ^^^^'>~~dt~ d¥° dsV _ fdxV [dy\ + Vdl) dt dtl 'Vx + V (1) CURVILINEAR MOTION OF A MATERIAL POINT. 75 Hence v is the diagonal formed on v^^ and Vy (as in § 71). Let pf=thG acceleration of v, i.e., tlie tangential acceleration. then Pt = ^'"5 -J- ^^? ai^d, since d's = the sum of the projec- tions of ^^and CI^ on £C, i.e., d''s = d^x cos a -\- d^y sin a, we have d^s d^x , «^'2/ . . , . /^v ^ = ^ cos «? +^ sin or; i.e.,_^j =^^ cos or +^^s]n o'. (2) By Normal Acceleration we mean the rate of cliaisge of the velocity in the direction of the normal. In describing the ele- ment AB = ds^ no progress lias been made in the direction of the normal JBHi.e.^ there is no velocity \\\ the direction of the normal; bnt in describing ^C' (on account of the new direc- tion of path) the point has progressed a distance GL (call it d^n) in the direction of the old iiormal BH (though none in that of the new normial (7/). Hence, just as the tang. ace. ds' — ds d^s ^ - , CL — zero d^7i = 775 = -m, so the normal accel. = ■ ^, = ^-,-. df df dt df It now remains to express this normal acceleration (^j^^) in terms of tlie X and Y accelerations. From the figure, CL = CM- ML, i.e., d^n = d'y cos a — d^x sin a | since EF = d?x\ ; dj'n d^y d^x . -df^df "^' ""-df '^" ''• Hence ^^^^j^cosor — ^a.sino' (3) The norm. ace. may also be expressed in terms of the tang. Telocity -y, and the radius of curvature r, as follows : ds' =. rda, or da = ds' ~ r ; also d^n = ds'da, = ds'''' ~ r, . d'n [ds'yi v' '■''■^df-\-dtl P ^^ ^- = r (^) If now, Fig. 80, we resolve the forces jf = Mp^ and Y Y = Mpy, which at this instant account for the JT and Y accelerations (M = mass of tlie \ /\.'^^ material point), into components along the ■.\^-'''\a \ tangent and normal to the curved path, w^e ■-Jt. ,£-^^-^M\ ,.---'" shall have, as their eqioivalent, a tangential -^ T = Mj>x cos ^ + ^Pv sin or, 76 MECHANICS OF ENGINEERING. and a normal force J7 = Mjpy COS a — Mjpx sin or. But [see equations (2), (3), and (4)] we maj also write r = Jf?>, = J/-^; and N^M^^^m"^. . (5) Hence, if a free material point is moving in a curved path^ the sum of the tangential components of the acting forces must equal (the mass) X tang, accel.; that of the normal components, =^ (the mass) X normal accel. = (mass) X (square of veloc. ia path) H- (rad. curv.). It is evident, therefore, that the resultant force (= diagonal on T and N or on JTand Y, Fig. 80) does 7iot act along the tan- gent at any point, but toward the concave side of the path ; un- less r = oo. Hadius of curvature. — From the line above eq. (4) we have d'^n = ds'^ -~ r ; hence (line above eq. (3) ), ds''' -r- r = d^y cos a — d^x sin a ; but cos a-=:dx-^ ds, and sin a = dy ^ ds, T dx -,„ dy I-; dX-f as ds ds'^ds - = dVij7 - ^''^^o ' I.e. = dx^d ds ds , , or ' = dx r dy- _dx_ ds'''ds'^ w 'dxd'y — dyd"x dx^ =■ dx^d (tan a-), ~ I dx\' di2iCQ. a" 'dij dT' or. y = -y -^ r 2 ^ *^" ^' (6) whicli is equally true if, for v^ and tan or, we put Vy and tan (90° — a;). I'espectively. Change in the velocity square. — Since the tangential accelera^ dv ^ ^ dv , . tion -.- ^Pf, we have ds-^- =pfd8\ i.e., -^-dv-=ptds, or vdv=.ptds and /. — r — = / pfds. (7) having integrated between any initial point of the curve where V = c, and any other point where v = v. This is nothing more than equation (HI.), of § 50. CURVILINEAR MOTION OF A MATERIAL POINT. 77 75. Normal Acceleration. Another Method. — Fig. 81. Consider a material point TO describing a circular path ABC, with constant velocity = v; the center of the curve being at and the radius = r. The velocity v is always tangent to the curve. Let the linear arc BC be described in the small time dt, the angle at the center being da. At B the velocity is directed along the tangent BT, while at C it is 1 to OC and makes an angle da with a line parallel to BT. As m moves along the curve from B to C, the point n, which is the foot of the T dropped from any position Fig. 81. of TO upon the normal BO, moves from B toward D; whUe the foot, a, of the T let fall from to upon the tangent BT, moves along BT with an average velocity = v', a little less than v. Now the motion of to may be regarded as compounded of these two motions, viz., that of n and that of a. The motion of n is called the "motion of to along the normal." The velocity of n is zero at B, where i" is T to the normal, and is v sin da at the point D; hence in the time dt the gain of n's velocity is v sin da — 0, and the rate of gain, or acceleration, is pn=v sin da-i-dt. But sin da = CD-^r and CD = BC' = v'dt. Substituting, we have pn=vv'^r. Now make dt equal to zero and we have v' = v; and finally pn=v^-i-r, as the value of the normal acceleration, just at the point B. 76. TTniform Circular Motion. Centripetal Force. — The ve- locity being constant, j!?^ must be = 0, and .'. T{ov 2Tii there are severalforees) must = 0. The resultant of all the forces, therefore, must be a normal force = {Mc^ -i- r) = a con- stant (eq. 5, § 74). This is called the " deviating force," or " centripetal force ;" without it the body would continue in a straight line. Since forces always occur in pairs (§ 3), a " centrifugal force," equal and opposite to the " centri- petal" (one being the reaction of the other), will be found among the forces acting on the body to whose constraint the deviation of the first body from its natural straight course is due. For example, the attraction of the earth on the moon acts as a centripetal or deviating force on the latter, while the equal and opposite force acting on the earth may be called the centrifugal. If a small block moving on a smooth horizontal table is gradually turned from its straight course ^^ by a fixed circular guide, tangent to AB at ^, the pressure of the guide against the block is the centripetal force M&-^ r directed toward the centre of curvature, wliile 78 MECHANICS OF ENGINEERING. y/////////^,/// Fig. 83. the cent "■'^■ugal force Mc^ -^ r is tlie pressure of the bloct against th«. uide^ directed away from that centre. Note. — One is not justified, therefore, in saying that a body descrifeing a circular path is under the action of a "centrifugal force.' The Conioal Pendulum^ or governor- ball. — Fig. 83. If a material point of mass z= M =^ G ^ g^ suspended on a cord of leLgth ^ Z, is to maintain a uniform cir- cular motion in a horizontal plane, with a given radius r, under the action of gravity and the cord, required the velocity c to be given it. At B we have the body free. The only forces acting are G and the cord- tension P. The sum of their normal com- ponents, i.e., -5'i\^, must = Mc^ -r- r, i.e., P sin a = Md^ -f- r ; but, since -2" (vert, comps.) = 0, /^ cos a =. G. Hence G tan a = Gc^-^ gr; .•. c = Vgr tan a. Let u = number of revolutions per unit of time, then u = c -v- 27rr = Vg -i- 27r Vh ; i.e., is inversely proportional to the square root of the vertical projection of the length of cord. The time of executing one revolution is =1 -hw. JElevation of the outer rail on raih'oad curves (considera- tions of traction disregarded). — Consider a single car as a material point, and free^ having a given velocity = c. J^ is the rail-pressure r^ against the wheels. So long as the car $^_ -t — R-H;; follows the track the resultant P of P ' "■^■■ and 6r must point toward the centre of curvature and have a value = Md^ -^ r. But ^=:= 6^ tan a, whence tan or = c^-f- gr. If therefore the ties are placed at this angle a with the horizontal, the pressure will come upon the tread and not on the flanges of the wheels ; in other words, the car will not leave the track. (This is really the same problem as the preceding) Apparent weight of a body at the equator. — This is less than the true weight or attraction of the earth, on account of the uniform circular motion of the body with the earth in its diurnal rotation. If the body hangs from a spring-balancej Fig. 84. CURVILINEAR MOTION OF A MATERIAL POINT. 79 whose indication is G lbs. (apparent weight), while the true attraction is G' lbs., we have G' — G == M& -^ r. For M we may use G ^ g (apparent values); for r about 20,000,000 ft.; for c, 25,000 miles in 24 hrs., reduced to feet per second. It results from this that 6^ is < ^' by -^G' nearly, and (since 17^ = 289) hence if the earth revolved on its axis seven- teen times as fast as at present, G would = 0, i.e., bodies would apparently have no weight, the earth's attraction on them being jnst equal to the necessary centripetal or deviating force necessary to keep the body in its orbit. Centripetal force at any latitude. — If the earth were a ho- mogeneous liquid, and at rest, its form would be spherical ; but when revolving uniformly about the polar diameter, its form of relative equilibrium (i.e., no motion of the particles relatively to each other) is nearly ellipsoidal, the pohir diameter being an axis of symmetry. Lines of attraction on bodies at its surface do not intersect in a common point, and the centripetal force requisite to keep a suspended body in its orbit (a small circle of the ellipsoia), at any latitude /? is the resultant, iT, of the attraction or true weight G' directed (nearly) toward the centre, and of G tiie tension of the string. Fig. 85. ^ = the apparent weight, in- dicated by a spring-balaTice and MA is its ..^G line of action (plumb-line) normal to the y^r.:.....L.:'i:\.:^^U ocean surface. Evidently the apparent weiglit, and consequently g, are less than the true values, since N must be perpen- x eq^^j^— dicular to the polar axis, while the true values themselves, varying inversely as ^^igTssT the square of MC, decrease toward the equator, hence the ap- parent values decrease still more rapidly as the latitude dimin- ishes. The apparent g for any latitude /5, at h ft. above sea- level, is (Chwolson, 1901), for foot-second units,* ^ = 32.lY23-0.083315cos2,5-0.000003/i. (The value 32.2 is accurate enough for practical purposes.) Since the earth's axis is really not at rest, but moving about * At the equator, g^ = 32.09 at sea-level but decreases to 32.06 at an ele- vation of 10,000 ft. above the sea. 80 MECHANICS OF ENGINEERING. the sun, and also about the centre of gravity of the moon and earth, the form of the ocean surface is periodically varied, i.e., the phenomena of the tides are produced. 77. Cycloidal Pendulum. — This consists of a material point at the extremity of an imponderable, flexible, and inextensible cord of length = Z, confined to the arc of a cycloid in a ver^ tica] plane by the cycloidal evolutes shown in Fig. 86. Let the oscillation begin (from rest) at A, a height = h above *;he vertex. On reaching any lower point, as JS (height = 3 above 0), the point has acquired some velocity v, which is at this instant increasing at some rate = Pf l^ow consider the point free, Fig. 87; the forces acting are I^ the cord- tension, normal to path, and G the M^eight, at an angle (p with the path. From § 74, eq. (5), ^T = Mpt gives 6^ cos ^ + P cos 90° = {G -^ g)2Jt\ :. Jpt = ^ cos ^ Hence (eq. (7). § 74), 'vdv = p4s gives qidv = g cos cpds ; bnt ds 0,0% cp =. — dz\ .'. vdv = — gdz. Summing between A and _5, we have ¥^' = - ^A^^; ^'' ^' = 2^(^^ ~ ^)5 ♦-.he same as if it had fallen freely from rest through the height h — z. {This result evidently applies to any form of path when, he sides the weight G, there is hut one other force, and that always norwal to the path, y^ From ^iV" = J/v" -^ r,, we have P — G sin q) = Mv* -j-ri, * Compare with lower part of p. 83. CURVILINEAR MOTION OF A MATElilAL POINT. 81 whence P^ the cord-tension at any point, may be found (here r{=- the radius of curvature at any point = length of straight portion of the cord). To find the time of passing from ^ to (9, a half-oscillation, substitute the above value of -y^ in v ^ ds -^ dt^ putting ds^ = dx" -j- ds"", and we have df = {dx" + dz'') -=- [2^(A — s)]. To find dx in terms of dz, differentiate the equation of the curve, which in this position is a? = r ver. sin.~^ (0 -^- r) -j- V'irz — z^ ; whence dx = .'. dx^ = rdz (r^— z)dz {2r — z)dz V2rz ~2r V2rz — z^ V2: rz - 1 dz' {r = radius of the generating circle). Substituting, we have ^r (— dz) g ' \/hz - z"" ' dt = {> = V ^-/ dz rh ver. sin. ^^0 y/is — s^ y g Hence the whole oscillation occupies a time = rr Vl ^ g (since I = 4r). This is independent of A, i.e., the oscillations are isochronal. This might have been proved by showing that 'pt is proportional to OB measured^ along the curve j i.e., that the motion is harmonic. (§ 59, Prob. 2.) 78. Simple Circular Pendulum. — If the material point oscil- lates in the arc of a circle, Fig. 88. proceeding as in the preceding problem, we have finally, after integration by series, as the time of a full ■ \ ■■< oscillation, in one direction,* L.t^ _ \§-''' ^ "\ gjL 1 + 9 ^ 256' P 225 h^ 18432 *F + .. Hence for a small h the time is nearly tt Vl -=- g. and the os- * See p. 651 of Coxe's translation of Weisbach's Mechanics. 82 MECHANICS OF ENGINEERING.. dilations nearly isochronal. (For the Compound Pendulum, see § 117.) [Note. — While the simple pendulum is purely ideal, the conception is a very useful one. A sphere of lead an inch in diameter and suspended by a silk thread, or very fine wire, more than 2 ft. in length, makes a close approximation to a simple pendulum; the length I being measured from the point of suspension to the middle of the sphere (strictly it should be a little greater). The length of the simple pendulum beating seconds (small amplitude) is about 3.26 ft.; (see p. 120)]. 79. Change in the Velocity Square. — From eq. (7), § 74, we have ^{v'^ — g'^) z=Jpfds. But, from similar triangles, du be- ing the projection of any ds of the path upon the resultant fo]-ce S at that instant, Hdu = Tds (or, Prin. of Yirt. Yels. § 62, lidu = Tds + iV^ X 0). T and iV^are the tangential and normal components of ^. Fig. 89. Hence, finally, IMv' -\MG^^fRdu, («) for all elements of the curve between any two points. In ^gen- -^^ -v eneral R is different in amount and direc- -/ '""" -4. . '" tion for each ds of the path, but du is the ~'--.jp distance through which R acts, in its own N Fig. 89. direction, while the body describes any ds ; Rdu is called tlie work done by R wlien ds is described by the body. The above equation is read : The difference hetween the initial and final hinetio energy of a hody = the work done hy the residtant force in that portion of the path. (These phrases will be fui-ther spoken of in Chap. YI.) Application of equation (a) to a planet in its orhit about the sun. — Fig. 90. Here the only force at any iTistant is the at traction of the sun R =^ O -^ u"^ (see Prob. 3, § 59), where (7 is a constant and u the variable I'adius Ns. vector. As u diminishes, v increases, therefore \ " dv and du have contrary signs ; hence equation i c??^X^® {a) gives {p being the velocity at some initial 1 / \ point 0) LJr X^ Ji 2} Juo U ■1 1 u, u„ \Q>) ^Cri In ^su- •. -y, =A / c^'-f ijF — — — .which is independ- fig. 90. ent of the direction of the initial velocity c. CUKVlLlNEAli MOTION OF A MATKKIAL POINT. 83 Ajpplication of eq. (a) to a projectile in vacuo body's weight, is the only force acting, and therefore = i?, while M= G -^ g. Tliere- fore equation {a) gives ■-■ . -^- = Gj dy = Gy,', ■G, the dy-dii G=R """^\}yl dy] .V ^r<i G^-' Fig 93. ,". -y. = \/c' ~\- 2^2/,' which is independent of fiq. 91 the angle, a^ of projection. Application of equation (a) to a body sliding, wit/iovt fric* tion, on a iixed curved guide in a vertical plane, initial velo- city = c at 0. — Since there is some pressure at each point be- tween the body and the guide, to consider the bod)' free in space, we must consider the guide removed and that the body describes the given curve as a re- sult of the action of tlie two forces, its weight G, and the pressure /*, of the guide against the body. G is constant, wdiile jP varies from point to point, though always (since ^ there is no friction) normal to curve. At any point, H being the resultant of G and jP, project ds upon i?, thus obtaining du ; on G, thus obtaining dy ; on I^, thus obtaining zero. But by the principle of virtual velocities (see § 62) we have Rdu = Gdy +P X zero"^ = Gdy, which substituted in eq. (a) gives ~l{v,^ - 0^) =f''Gdy=Gfyy=Gy:., .-.v,^ VT^W, and therefore depends only on the vertical distance fallen througli and the initial velocity, i.e., is independent of the fo7")n of the guide. As to the value of P, the mutual pressui-e betw'een the guide and body at any point, since ^iVinust equal 3fv'^ ~ r, r being the variable radius of curvature, we have, as in §77, P — G&m q) = Mv" -^ r ; .. P = G\fix\ cp-^^v" ~ gr)\ As, in general, q) and r are different from point to point of * It is quite essential that the guide be fixed, as well as smooth, in order that this projection be zero; sinci if the guide were in motion, the *orce P, although 1 to the guide, would not be T to the ds or element of the path oi the body, for that path would then be different from the curve of the guide. 84 MECHANICS OF ENGINEERING. the path, P is not constant. Should the curve at the point in question be convex upward (instead of concave upward as in Fig. 92) we must write G &m.(f)—P=Mv^-i-r; etc. 80. Projectiles in Vacuo. — A ball is projected into the air (whose resistance is neglected, hence the '"""f!^ ,'--"'Tf plirase in vacuo) at an angle = a^ with the /•i" G'''^ horizontal ;* required its path ; assuming it \ i i „ coniined to a vertical plane. Resolve the 'oil Cx ±\ ... . , -^ ZZTl^'ZIZl^""' motion into independent horizontal {X) Fig. 93. and vertical {Y) motions, G, the weight, the only force acting, being correspondingly replaced b}^ its horizontal component = zero, and its vertical component = — G. Similarly the initial velocity along X = 0^^-=^ c cos <x^, along y, = Cy = csin a^. The JT acceleration =j?a; = -^ J/ = 0, i.e., the X motion is uniform, the velocity v^. remains = c^ = c cos a^ at all points, hence, reckoning the time from 0^ at the end of any time t we have X = c(cos a^t (1) In the Y motion, j?y = ( — G) -^ M=. — g^ i.e., it is uniformly retarded, the initial velocity being Cy =^ c sin a^ ; hence, after any time t, the Y velocity will be (see § 56) v^ = c sin a^ — gt, while the distance y = c(sin a^)t - ^gf (2) Between (1) and (2) we may eliminate t, and obtain as the equation of the trajectory or path y =: X tan a. — —-z — . ^ " 2c^ cos' o'o For brevity put c' = 2gh, h being the ideal height due to the velocity c, i.e., c^ -f- 2g (see § 53 ; if the ball were directed ver- tically upward, a height h = o^ -^ 2g would be actually at- tained, oTfl being = 90°), and we have y = xt^na,--^j^^^- (3) This is easily shown to be the equation of a parabola, with its axis vertical. * And with a velocity of c ft. per sec. CURVILINEAR MOTION OF A MATERIAL POINT. 85 The ho7'izontal range.- tion (3), we obtain a? X tan ar„ 4A cos^ «„ -Fig. 94. Putting y = in equa- 0, Fig. 94. which is satisfied both by a? = (i.e., at the origin), and by a? = 4:A cos a^ sin a^. Hence the horizontal range for a given g and a^ is x^ = 4A cos <arg sin a^ = 2A sin 2a^. For afg =: 45° this is a maximum (c remaining the same), being then = 2A. Also, since sin 2a^ = sin (180° — 2a^) = sin 2(90° — a^), therefore any two complementary angles of projection give the same horizontal range. Greatest height of ascent / that is, the value of y maximum, = y^. — Fig. 94. Differentiate (3), obtaining dy X dx ~ " 2A cos'' or ' which, put = 0, gives a? = 2A sin a^ cos ar„, and this value of X in (3) gives y^=i h sin" oc^. (Let the student obtain this more simply by considering the Y motion separately.) 81. Actual Path of Projectiles. — Small jets of water, so long as they remain unbroken, give close approximations to parabolic paths, as also any small dense object, e.g., a ball of metal, hav- ing a moderate initial velocity. The course of a cannon-ball, however, with a velocity of 1200 to 1400 feet per second is much affected by the resistance of the air, the descending branch of the curve being much steeper than the ascending; see Fig. 96(2. The equation of this curve has not yet been determined, but only the expression for the slope (i.e., dy : dx) at any point. See Professor Bart- lett's Mechanics, § 151 (in which the body is a sphere having no motion of rotation). Swift rotation about an axis, as well as an unsymmetrical form with reference to the direction of motion, alters the trajectory still further, and may deviate it from a vertical plane, presence of wind would occasion increased irregularity Fig. 96a. The See Johnson's Encyclopaedia, article " Gunnery." (See p. 823.) 86 MECHANICS OF ENGINEERING. 82. Special Problem (imaginary ; from Weisbach's Mechan- ics. The equations are not homogeneous). — Suppose a ma- terial point, mass = M^ to start from the point 0, Fig. 97, with a velocity = 9 feet per second along the — Y axis, beiiig snb- ilJ_J^,^ jected thereafter to a constant attractive JT force, of a valne X = 12M, and to a variable Y force increasino; with the time Fig. 97. (in scconds, reckoned from 0), viz., Y = 8Mi. Required the path, etc. For the JC motioa we Imvepx = X -^ If = 12, and hence dvy. = I ])Jit = 13 / dt\ i.e., 'o^ = 12^; dx ■=■ j v^dt ; i-.e., a? = 12 / tdt = 6f. . (1) For the Y motion ^j, = Y~3f=St, ..f 'd/Vy=%f tdt ; /y pt dy =^ I Vydtj .'. y = ^f fdt — 9^ dt, or y = |f — 9f. . . (2) Eliminate t between (1) and (2), and we have, as the equa- tion of the path, 4:fx\^ (x\i which indicates a curve of the third order. The velocity at any jpoint is (see § 74, eq. (1) ) (3) -y=|/^,^ + 'U/=4if + 9 (4) The length of cu7've measured from will be (since v = ds -i- dt) s —Tds =f vdt = 4c f fdt -^9jdt = ^f + 9^. (5) The slojpe, tan a, at any point = -y^^ -^ -Va, = {Aff — 9) -^ 12^, d tan a U^ -^9 and .*. dt l^f CUKVILINEAK MOTION OF A MATERIAL POINT. 87 TTie radius of curvature at any point (§ 74, eq. (6) ), sub- stituting Vg. = 12f, also from (4) and (6), is r = v^ -i- \vj d tan a 1 ^^j=i#^'+9r, . . (7) and the normal acceleration = v^ — r (eq. (4), § 74), becomes from (4) and (7)^^= 12 (ft, per square second), a constant. Hence the centripetal or deviating force at any point, i.e., the SW of tiie forces X and Y, is the same at all points, and = Mv' -^r = 12M. From equation (3) it is evident that the curve is sjunmetrical about the axis X. Negative vahies of ?! and s would apply to points on the dotted portion in Fig. 97, since the body may be considered as having started at any point whatever, so long as ill the variables have their proper values for that point. (Let the student determine how the conditions of this motioa could be approximated to experimentally.) 83. Relative and Absolute Velocities. — Fig. 98. Let JUT be a material point having a uniform motion of velocity v^ along a straight groove cut in the deck of a steamer, which itself has a uniform motion of translation, of velocity v^, over the bed of a river. In one second 3f ad- \ / vances a distance v^ along the \ / groove, which simultaneously has z"^^-' iVi72^II^f^^^^----..._ moved a distance v, = AJB with I I ///\/ ll tlie vessel. The absolute path of — ~ d^r'[^~J^ r:::. M during the second is evidently fig. 98. w (the diagonal formed on -y^ and -y^), which may therefore be called the ahsolute velocity of the body (considering the bed of the river as fixed) ; while v^ is its relative velocity, i.e., rela- tive to the vessel. If the motion of the vessel is not one of translation, the construction still holds good for an instant of time, but V^ is then the velocity of that point of the deck over which JSTis passing at this instant, and v^ is Jff's velocity rela- tively to that point alone. Conversely, if M be moving over the deck with a given absolute velocity = lu, v^ being that oi the vessel, the relative velocity v,^ may be found by resolving w into two components^ one of which shall be v^ ; the other will be v^. 'A„..-"B 'S8 MECHAJMICS OF EJS'GlNEElllNG. If w is the absolute velocity and direction of the wind., the vane on rfie mast-head mmII be parallel to 3£T, i.e., to v^ the relative velocity; while if the vessel be rollins^ and the mast- head therefore describing a sinuous path, the direction of tha vane varies periodically. Evidently the effect of the wind on the sails, if any, will 'depend on v^ the relative, and not directly on w the absolute, velocity. Similarly, if w is the velocity of a jet of water, and Vj that of a water-wheel channel, which the water is to enter without sudden deviation, or impact, the channel-partition ■should be made tangent to v^ and not to w. Again, the aberration of light of the stars depends on the ;Same construction ; v^ is the absolute velocity of a locality of the ■earth's surface (being practically equal to that of the centre) ; w is the absolute direction and velocity of the light from a certain star. To see the star, a telescope must be- directed .2,\o\)g MT, i.e., parallel to v^ the relative velocity; just as in the case of the moving vessel, the groove must have the direc- tion MT. if the moving material point, having an absolute velocity w, is to pass down the groove without touching its sides. Since the velocity of light = 192,000 miles per second =: w, and that of the earth in its orbit = 19 miles per second = -Wj, the angle of aberration SMT, Fig. 98, Avill not exceed 20 seconds of arc ; while it is zero when w and v^ are parallel. Returning to the wind and sail-boat,"^ it will be seen from Fig. 98 that when 'y, = or even > w, it is still possible for -y, to be of such an amount and direction as to give, on a sail properly placed, a small wind-pressure, having a small fore-and aft component, which in the case of an ice-boat may exceed the small fore-and-aft resistance of such a craft, and thus v^ will be still further increased ; i.e., an ice-boat may sometimes travel faster than the wind which drives it. This has often been proved experimentally on the Hudson Hiver. (See p. 819.) 84. Examples. — 1. A platform-car on a straight I-evel track carries a vertical smooth pole loosely encircled by an iron ring weighing 30 lbs., and is part of a train having a uniform northward motion with velocity of 20 miles/ hour. The ring, at first fastened at the top of pole, 10 ft. above floor, is set free. Find its absolute velocity just before striking the floor and the distance the car has progressed during the fall of ring. Solution. — The X-motion (horizontal) of the ring is that of the car and has a constant velocity Cx = 20X5280^3600 = 29.34ft./sec. Its F-motion CUKVILINEAR MOTION OF A MATERIAL POINT. 89 ■(i.e., along pole) has initial velocity =0 and a constant downward acceler- ation py=g (since the only force acting on ring is vertical and is its own weight) . Hence from § 56 the time of the 10-ft. fall = i/2X10h-32.2 = 0.788 sec. and the F-velocity generated at end of that time is Vy~gt = 25A ft./sec. This is now combined with the simultaneous X- velocity of riag, i.e., 29.34, to give V, =l/ca;^ + %^ =38.8 ft./sec, for the required final absolute velocity of ring, which is therefore at this instant moving obliquely north- ward and downward at an angle of 40° 52' with the horizontal (since ^j,-=-Ca;= 25.4-^29.34 = 0.8655 =tan 40° 52'). Example 2. — Pole and car, etc., as in example 1, but the train now has a uniformly accelerated motion, gaining 25 velocity-units (ft. per sec.) in each 5 sees, of time. The ring begins to drop when the train already has a velocity of 6 ft./sec. Find the final absolute velocity of ring; also the final pressure of pole on ring. Solution. — The ^''-motion of ring is the same as before, since the pressure on the ring from the pole (smooth vertical sides) must be horizontal and hence does not affect the F-motion. Hence the time of descent is, as before, €.788 sec. During this time the velocity of the train has increased to a value of i;x = 6-h (25^5)0.788 = 9.94 ft./sec, which is the velocity of the JC-motion of the ring at the final instant, whence its final absolute velocity, -w, =1/(9.94)2+ (25.4)2, =27.3 ft./sec, directed obliquely downward and northward at an angle of 68° 38' with horizontal (9.94-^27.3) =0.3642 = cos 68° 38'. The pressure of the pole on ring is constant and =Px = Mpx = (30^32.2) X 5 = 4.65 lbs. Example 3. — Conical pendulum. Fig. 83, p. 78. Given G = 8 lbs. and 1=2 ft., at what angle a will the cord finally place itself with the vertical if a steady rotation is kept up at the rate of 50 revs./min. ; and what will then be the tension in the cord? Solution. — ^With the ft., lb., and sec. as units we have w = 0.8333 revs. /sec, = 8, 1 = 2, a=? Hence from v? = g-^ {4:n%) , we find /i = 1.174 ft. and cos a, =h-^l, =0.5873; /. a = 54°0'. As for the tension in cord, P=G-^ cos a = 13. 62 lbs. \Note. — In this example, if the assigned value of u, or of the cord-length Z, had been so small as to make lu'''^g^-{i7i^), we should have obtained for cos a a value ^1.00; which is, of course, impossible for a cosine. That is, the value assigned for u must be ^i/g-^(27r]/Z), in order that the cord may depart at all from its original vertical position.] Example 3. — Compute the length Z of a simple pendulum which is to oscillate 4500 times in an hour. Amplitude small; 5°. Solution. — For small oscillations we have, from p. 81, t = Tt\/l-^g as the time of one oscillation ; that is, for the foot and second as units, 3600-^500 = 7r-j/Z-^32.2; and therefore Z = 2.089 ft. Example 4. — A leaden ball weighing ^ ounce, and of diameter 0.53 in., is allowed to slide down the inside of a fixed and rigid hemispherical bowl, of perfectly smooth internal surface and with its upper edge in a horizontal plane. Its radius is 18 in. The ball is to start from rest at upper edge. Find the time occupied by the ball in reaching the lowest point, and the pressure under it as it passes that point; also the pressure in passing the 45° point. 90 MECHANICS OF ENGINEERING. Solution. — Regarding the ball as a material point we note that its motion is practically that of a simple pendulum with ^ = [18 — ^(0. 53)] in., =1.478 ft., for which (see Fig. 88, p. 81) the ratio h-r-l=1.00. Hence (§ 78, p. 81) the time of a half oscillation (applicable here) will be (ft. and sec), (For a small amplitude this would be only -|7Z-(.02143)JL00] = 0.337 sec.) At the bottom the velocity will be (p. 83), v = '\/2gl, whence v'^-hgl = 2; and the pressure (see foot p. 83, with sin = 1.0, is P=-K1 +2) = 1.5 ou nces. As the ball passes the 45° point its velocity is v' = '\/2gX0.707l; i.e., ^'^-|-3i = 1.414, while sin 45° = 0.707; whence, for the pressure, P', P' = i[0.707 + 1.414]= 1.06 ounces. Example 5. — A body at latitude 41° weighs apparently (i.e., by spring balance) 10 lbs.; what is the amount and direction of its real weight? (Fig. 85.) That is, we have given G = 10 lbs. and angle /? = 41°; and desire the value of force G' and of the angle which it makes with MA (plumb line) . (This angle, 0, =that at vertex G of the parallelogram in Fig. 85). Solution. — At the equator the earth's radius is r = 20,920,000 ft. and the velocity of objects at the surface is c=1521 ft. /sec. The radius of the small circle at M is r' = r cos 41° = 15,780,000 ft., and hence the velocity of the 10-lb. body at M is c', =(r'H-r)c, =1148 ft./ sec. Therefore the result- ant iV, = Mc^~r', = [(10 --32.2)(1148)2]H- 15,780,000 = 0.0259 lbs. Call the projection of N on GM prolonged, T, and its projection on a 1 to GM, S; then T, =N cos /?, = 0.01954 lbs., and S, =Nsm /?, =0.01699 lbs. We have also tan = 5-4- [G + T] = 0.0016957; hence = 0° 5' 48". Then G', ={G + T) sec 0, =10.01955 lbs. [By a somewhat more refined process we obtain 10.01964 lbs. (Du Bois).] Example 6. — A small compact jet of water (see Fig. 94, p. 85) issues obliquely from a nozzle. It strikes the horizontal plane of nozzle at 6 ft. from the latter, and its highest point is 26.4 in. above that plane. Find c, the velocity at nozzle, and the angle of projection a^. Solution. — From p. 85 (foot and second units) we have Ah cos ao sin «„ = 6 ft., and h sin^ q;o = 2.2 ft.; whence, by division (tan «(,-=- 4) = (2.2 -=-6), or tan ctg = 1 .4666 ; and therefore ao = 55°43'. a^ being now known we find from /i sin^ ao = 2.2 that A = 3.22 ft. But h simply stands for the ex- pression c^^2g, and hence, finally, we obtain for the velocity of the jet where it leaves the nozzle c = 14.4 ft. per sec. Example 7. — If in Fig. 98, the absolute velocity of the air-particles (wind) is w = 10 miles/ hour and directly from the northwest, the boat's velocity being =12 miles /hour toward the east, in what direction and with what velocity does the wind appear to come, to a man on the boat? Ans. From a direction 34° 52' east of north, and at 8.62 ft. /sec. Example 8. — If to a passenger on board a boat going eastward at 15 miles /hour, the wind appears to come from the northeast and to have a velocity 10 miles/ hour, what is the true or "absolute" velocity of the wind, and what is its true direction (angle with north and south line)? Ans. 10.63 ft./ sec, and from a point 41° 44' east of north. MOMENT OF INEKTIA. 91 CHAPTER lY. MOMENT OF INERTIA. [Note, — For the propriety of this term and its use in Mechanics, see ■§§ 114, 216, and 229 ; for the present we deal only with the geometrical Jiature of these two kinds of quantity.] 85. Plane Figures. — Just as in dealing with the centre of gravit}' of a plane figui-e (§ 23), we had occasion to suni tlie ■&eY\ei fzdF, 3 being the distance of any element of area, dF, from an axis ; so in subsequent chapters it Mall be necessarj' to know the value of the seriesysW-^for plane figures of various shapes referred to various axes. This summation J'z'^dF of the products arising from multiplying each elementary area of the figure by the square of its distance from an axis is called the moment of inertia of the plane figure rcith respect to the ■axis ill question / its symbol will be I. If the axis is perpen- dicular to the plane of the figure, it may be named the polar mom. of inertia (§94); if the axis lies in the plane, the rec- tangular mom. of inertia (§§ 90-93). Since the / of a plane figure evidently consists oi four dimensions of length., it inay always be resolved into two factors, thus /= Fk^^ in which i^= total area of the figure, while h = Vl-r- F, is called the Tadius of gyration, because if all the elements of area were situated at the sa^ne radial distance, Jc, from the axis, the moment of inertia would still be the same, viz., I = fk'dF = kfdF = Fh\ For example, if the moment of inertia of a certain plane figure about a specified axis is 248 biquadratic inches (i.e., four-dimension inches; or in.'*), while its area is 12 sq. in. (or in. 2), the corresponding radius of gyration is A; = 1/248-^12 = 4.55 in. 86. Rigid Bodies. — Similarly, in dealing with the rotary motion of a rigid body, we shall need the sura of the series fp'^dM, meaning the summation of the products arising from multiplying the mass dM oi each elementary volume dY oi &. 92 MECHANICS 0¥ ENGINEEKING. rigid body bj the square of its distance from a specified axis.. This will be called the moment of inertia of the hody with respect to the particular axis mentioned (often indicated by a. subscript), and will be denoted by /. As before, it can of tea be conveniently written Mh^^ in which 3£ is the whole muss, and h its "radius of gyration" for the axis used, h being = Vl -T- M. If the body is homogeneous, the heaviness, y, of all its particles will be the same, and we may write . I =fp''dM ={r~ g)fp^d V={y~g) Vl\ 87. If the body is a homogeneous plate of an infinitely smaW. thickness = r, and of area = F, we have Z = (/ -f- g^fp'd K '"= {y -^ 9YfP^'^^'-, i-e-, = {y -^ g) X thickness X mom. iner- tia of the plane figure. 88. Two Parallel Axes. Reduction Formula.* — Fig. 99. Let Z and Z' be two parallel axes. Tiien Ig =fp'dM, and I^.^fp'^dM. Bu t d being the distance between the axes, so that a^' -f lf= d\ we have p'^= {x - af-\-{y-hf = (£»' + y"") -\-d^ — 2aa? — %y, and .-. I^, =fp'dM-\-dYdM- ^afxdM -%fydM. . (1) Fig. m. V>\\ifp''dM = /^, fdM= M, and from the- theory of tlie centre of gravity (see§23, eq. (1), knowing that dJI =yd V~ g, and .-. that S^fyd F] -^ g=M) we \\2,YefxdM = Mx dindifydM = My\ hence (1) becomes /^, = I,-\- Mid' - 2«^ - %y\ .... (2)^ in which a and h are the x and y of the axis Z'\ x and y refer to the centre of gravity of the body. If Z is a. gravity-axis- (call it g), both x and y = 0, and (2) becomes I,.=I^ + Md\... or h,'^k;+d\ . . (3) It is therefore evident that the mom. of inertia about a grav- ity-axis is smaller than about any other parallel axis. Eq. (3) includes the particular case of a plane figure, by * The particle of mass = dM, shown in Fig. 99, is typical of the vast number of particles which form the rigid body. That is, o, 6, and d are constants, but x, y, z, p, and p' are variables. MOMENT OF INEKTIA. 93 writing area instead of mass, i.e., wlien Z (now g) is a gravitj- axis, I,,=I^-\-Fd\ (4) 89. Other Reduction Formulse ; for Plane Figures. — (The axes here mentioned lie in the plane of the figure.) For two sets of rectangtdar axes, having the smne origin, the following holds good. Fig. 100. Since I^=fifdF, and ly^fx'dF, we have Ix + Iy =/(«;' + y')dF. Similarly, I^ + Iy =f{v' + %iyF. But since the x and y of any <^^have the same hypothennse as the u and v, we have v^ -{- v^ = ic^-J- y"; . ". -^x + -^r = A;- + -^r- Fig. 100. Fig. 100a. Let Xl)e an axis of symmetry ', then, given Ix and Iy {0 is anywhere on X). required Ijj, JJheing an axis through and maJcing any angle a with X. See Fig. 100a. I^ -^fv^dF^fiy co&oc —X sin dfdF\ i.e., Ijj = cos^ af/dF — 2 sin a. cos afxydF-\- sin* afx^dF. But since the area is symmetrical about X, in summing up the products xydF, for every term x{ -\- y)dF, there is also a term K — y)dF to cancel it ; which gives fxydF =: 0. Hence Ijj^ — cos* al^ -f- sin* aly. The student may easily prove that if two distances a and h be set ofE from on X and Y^ respectively, made inversely proportional to Vix and VTy, and an ellipse described on a and h as semi-axes ; then the moments of inertia of the figure about 94 MECHANICS OF ENGINEERING. im dz ^n any axes through are inversely proportional to the squares of the corresponding semi-diameters of this ellipse ; called therefore the Ellijpse of Inertia. It follows therefore that the moments of inertia about all gravity-axes of a circle, or a regular polygon, are equal ; since their ellipse of inertia must be a circle. Even if the plane figure is not symmetrical, an " ellipse of inertia" can be located at any point, and has the properties already mentioned ; its axes are called t\\e principal axes for that point. 90. The Rectangle. — First, ahout its hase. Fig. 101. Since all points of a strip parallel to the base -j,....^ ^ ?) -^ have the same co-ordinate, 0, we may take dz the area of such a strip for dF ■=^ hds\ .-. Ib= z'dF= I / z'dz L-o Secondly, about a gravity-axis parallel to hase. z'dz = -^W. -ih Hence the radius of gyration =k = h-^\' 12. Thirdly, about any other axis in its plane. Use the results already obtained in connection with the reduction-formulae of §§ 88, 89. 90a. The Triangle. — First, about an axis through the vertex and parallel to the base ; i.e., 1-^ .^ .1, » ^ .j,, in Fig. 103. Here the length of the strip is variable ; call it y. ^ From similar triangles l _i. .\j^. _v AZ i V 2/ = (& -i- h)z ; Fig. 103. Fig. 104. Fig 101. Fig. 102. Uh' .'. ly ^fz'dF^ fz'ydz = {h-^ h)f z'dz = I Secondly, about g, a gravity -axis parallel to the hase. 104. From § 88, eq. (4), we have, since F=^ ^hh and d = |A, Ig = Iy- Fd' = IW - Ihh . ^h' = ^W. Fig MOMENT OF INEKTIA. 95 Thirdly^ Fig. 104, about the hase • Ijb = 1 From § 88, eq. (4), Ib=^ Ig-\- Fd^, with 6? = -JA ; hence I J, = ^^hh' + ^hh . \h' = -^hkK 91. The Circle. — About any diameter, as g, Fig. 105. Polar co-ordinates, Ig =^ fz^dF. Here we take dF=^ area of an ele- mentary rectangle = pdq) . dp, while z=^ p sin cp. ■» h — -• 1 f-i^r hi ^1 -- *•- c ibi- Fig. 105. Fig. 106. Ig= I I {p sin (pypdcpdp = I I sin' cpdcpj p^dp J = — / sin'' 9?<^^ = T / "^^"^ ~" ^^^ '^(p)dqi ^* /.S'^fl 1 ~| = :f y^ 1^2^^ - J . cos 2(?)^(2^)J _o)-(0-0)_. 1 .r^Vl = r = r 2;r 2" »^. = 4'^^*- Hence the radius of gyration =\r. 92. Compound Plane Figures. — Since I =^ fz^dF is an in- finite series, it may be considered as made up of separate groups or subordinate series, combined by algebraic addition, corresponding to the subdivision of the compound figure into component figures, each subordinate series being the moment of inertia of one of these component figures ; but these separate moments Tnust all he referred to the same axis. It is con- venient to remember that the {rectangular) / of a plane figure remains unchanged if we conceive some or all of its elements shifted any distance parallel to the axis of refer- ence. E.g., in Fig. 106, the sum of the Is of the rectangle CE, and that of FD is = to the Ib of the imaginary rectangle 96 MECHANICS OF ENGINEERING. formed by shifting one of tliem parallel to B, until it touches the other ; i.e., I^ of CE-^ Ib of FD = ^hX (§ 90). Hence the Ib of the T shape in Fig. 106 will be = I^ of rectangle AD - Ib of rect. CE- Ib of rect. FIf. That is, /^ of T = i[5/''' - \Kl • • • (§ 90). . . (1) Ahoiit the gravity-axis, g, Fig. 106. To find the distance d from the base to the ceiirre of gravity, we may make use of eq. (3) of § 23, wu-iting areas instead of volumes, or, experi- mentally, having cut the given shape out of sheet-metal or card-board, we may balance ition a knife-edge. Supposing d to be known by some such method, we have, from eq. (4) of § 88, since the area E= bh — hjt„ Ig=: Ib— Fd^ ; i.e., Ig = l\hk' - hji,'-] - {hh - lji,)d' (2) The doiihle-'Y (on), and the hox forms of Fig. 106a, if syminetrical about the gravity- axis g, have moments of inertia alike in form. Here the grav- ity-axis (parallel to base) of the compound figure is also a grav- FiG. 106a. ity axis (parallel to base) of each of the tw-o component rectangles, of dimensions h and A, h^ and Aj, respectively. Hence by algebraic addition we have (§ 90), for either com- pound figure, I,= i,\}h^-\h.n (3) (If there is no axis of symmetry parallel to the base we must proceed as in dealing with the T-form.) Similarly for the ring, Fig. 107. Fig. 108. Fig. 107, or space between two concentric circumferences, we have, about any diam-eter or ^ (§ 91), Io = \«-r:) (4) MOMENT OF INERTIA. 97 The rhorribus about a gravity-axis, g, perpendicular to a diagonal, Fig. 108. — This- axis divides the figure into two equal triangles, symmetrically jplaced, hence the Ig of the rhombus equals double the moment of inertia of one triangle about its base ; hence (§ 90a) /, = 2 . ^li^Kf = -i^W (5) (The result is the same, if either vertex, or both, be shifted •any distance parallel to AB.) For practice, the student may derive results for the trapezoid ^ for the forms in Fig. 106, when the inner corners are rounded into equal quadrants of circles; for the double- "f, when the lower flanges are shorter than the upper; for the regular polygons, etc. (See table in the Cambria Steel Co.'s hand-book. ) 93. If the plane figure be bounded, wholly or partial]}', by curves, it may be subdivided into an infinite number of strips,, and the moments of inertia of these (referred to the desired axis) added by integration, if the equations of the curves are hnown I if not, Simpson's Rule,* for a finite even number of strips, of equal width, may be employed for an approximate result. If these strips are parallel to the axis, the / of any one strip = its length X its width X square of distance from axis; while if perpendicular to, and terminating in, the axis, its /= -J its width X cube of its length (see § 90). A graphic method of determining the moment of inertia of any irregular figure will be given in a subsequent chapter.* 94. Polar Moment of Inertia of Plane Figures (§ 85). — Since the axis is now perpendicular to the plane of the figure, inter- secting it in a point, (9, the distances of the ele- ments of area will all radiate from this point, and would better be denoted by p instead of s; hence, Fig. 109, fp^dF'^s the polar moment, of | inertia of any plane figure about a specified point ; this may be denoted by Ip. But p^ Fiq. 109. = a?" -|- 2/^ for each dJ^', hence 4 =f{x' + f)dF=fx^dF+fy^dJ^= /^+ /^. 7 * See pp. 13, 79, 80, and 81 of the author's "Notes and Examples in Mechanics," and p. 454 of this book. 98 MECHANICS OF ENGINEERING. i.e., the polar Tnoment of inertia ahout any gwen point i/n the plane equals the sum of the rectangular moine^its of iner- tia about any two axes of the plane figure^ which intersect ai right angles in the given point. We liave therefore for the circle about its centre 7p = \7ir' -}- ^Ttr" = ^Ttr" ; For a ring of radii r^ and r,, 4 = k7t{r: - r:) ; For the rectangle about its centre^ For the square, this reduces to -^p — 6" • (See §§90 and 91.) 95. Slender, Prismatic, Homogeneous Rod. — Returning tcv the moment of inertia of rigid bodies, or solids, we begin with tliat of a material line, as it uiiglit be called, about ^^^yy"^' an axis througli its extremity making some an- r y^^i ./"'' gle oi with the rod. Let I = length of the rod, y^--^" X i^its cross-section (very small, the result being y'' strictly true only when F = 0). Subdivide Fig. 110. the rod into an inlinite number of small prisms, each having _^as a base, and an altitude = ds. Let y = the heaviness of the material ; then the mass of an elementary prism, or dJif, = (r "^ 9)F'ds, while its distance from the axis Z \% p ^= s sin a. Hence the moment of inertia of the rod with respect to Z as an axis is Jz = fp'dM= {y -^ ^)i^sin^ afs^^ds = ^{y H- g)FT sin' a. But yFl -~ g =z mass of rod and I sin a z= a, the distance ot the further extremity from the axis ; lience Iz = ^Ma^ and tiie radius of gyration, or Jc, is found by writing-|-J!/a'''= Mh^ ; .-. ]c' = ^a\ or h = V^a (see § 86). If or = 90°, a = l. 96. Thin Plates. Axis in the Plate. — Let the plates be homo- geneous and of small constant thickness = t. If the surface of MOMENT OF INEETIA. 99 the plate be = F, and its lieaviness y, then its mass = yFr — g. From § 87 we have for the plate, about anj axis, I ^^ {y -^ g)r X m,oin. of inertia of the j)lane fgure formed hy the shape of the plate (1) Rectangular 'plate. Gravity-axis parallel toJ)ase. — Dimen- sions h and h. From eq. (1) and § 90 we have Similarly, if the base is the axis, I^ = \MU, .'. Tc^ = -|A^ Triangular plate. Axis through vertex parallel to hase. — From eq. (1) and § 90a, dimensions being h and A, ly = {y -^ g)rlW = {yihhr ^ g)^h' = ^Mh'; .'. ¥ = \h\ Circular plate, with any diameter as axis. — From eq. (1) and § 91 we have Ig = {y -^ gy^Ttr" — {yytr'^t -^ g)^r^ = ^Mr^; ¥ = ^\ Fig. 111. 97. Plates or Right Prisms of any Thickness (or Altitude). Axis Perpendicular to Surface (or Base). — As before, the solid is homogeneous, i.e., of constant heaviness y, let the altitude = h. Consider an elementary pi'ism, Fig. Ill, whose length is parallel to the axis of reference Z. Its altitude = h = that of the whole solid ; its base = dF = an element \ of i^the area of the base of solid ; and each point of it has the same p. Hence we may take its mass, = yhdF -^ g, as the dMin summing the series l^=fp^dM', .'.Iz={yh^g)fp-'dF = {yh -^ g) X polar mom. of inertia of base. . . (2) By the use of eq. (2) and the results in § 94 we obtain the following: Circular plate, or right circular cylinder, about the geo- metrical axis, r ■= radius, h = altitude. Ig = {yh -^ g)^7tr' = (yhrrr' -i- g)ir' = ^Mr'; .'. ¥ — \r\ Right parallelopiped or rectangular plate. — Fig. 112, I. = (r^ - g)^M^' + ^1 = -^iV^'; ••• ^^ = ^^- loo MECHANICS OF EKGINEEKING. For a hollow cylinder^ about its geometric axis, — "A — ~] (n b:::::^- V Fig 112. Fig. 113. 98. Circular Wire. — Fig. 113 (perspective). Let Z be a gravity-axis pei-pendicular to tlie plane of the wire ; X and Y lie in this plane, intersecting at right angles in the centre 0. The wire is hoitiogeneons and of constant (small) cross-section. Since, referred to Z, each dM has the same p — r, we have /^ ^fr''dM= Mr\ ]N"ow I^ must equal 7^, and (§ 94) their sum = Iz-, .-. 7x5 or Iy, = iMr\ and ^x^ ov Ity = ^^■ 99. Homogeneous Solid Cylinder, dboxd a diameter of its base. — Fig. 114. 7x ^ ? Divide the cylinder into an infinite num- ber of larainse, or thin plates, parallel to the base. Each is some distance s from X, of thickness ds, and of radius r (constant). In each draw a gravity-axis (of its own) parallel to Fig. 114. X. We may now obtain the I^ of the whole cylinder by adding the IxS of all the laminae. The Ig of any one lamina (§96, circular plate) = its mass X i^''; hence its Ix (eq- (3), § 88) = its ^ -(- (its mass) X ^^ Hence for the whole cylinder Ix= f\{ydznT'^^g){\r'^^z-^)-\ I/O i.e., Jx = {jtr-'hy - g\lr^ + W) == M^kr' + W)- 100. Let the student prove (1) that if Fig. 114 represent any right prism, and hp denote the radius of gyration of any one lamina, referred to its gravity-axis parallel to X^ then the Ix of whole prism = M{]i^ -|- \li^) ; and (2) that the moment MOMENT OF INERTIA. 101 of inertia of the cylinder about a gravitj-axis parallel to the base is = M{ir' + J^^')- 101. Homogeneous Right Cone. — Fig. 115. First, about an axis F, through the vertex and jparallel to the base. As before, divide into laminae parallel to the base. Each is a circular thin plate, but its radius, x, is not = r, but, 1^ from proportion, is a? = (r -^ h)z. \ \i 'rX:^ The /of any lamina referred to its own gravity- ^ axis parallel to "Fis (§96) = (its mass) X ia?^, and |_ its Iv (eq. (3), §88) is .-. = its mass X i^i^ + fig. lis. its mass X s\ Hence for the whole cone, ly— I {nx^dzy -^ g)[iaf -j- s'] ^ /Secondly, about a gravity-axis parallel to the hase. — From eq. (3), § 88, with d = ^h (see Prob. 7, § 26), and the result just obtained, we have /= J^-i-o[_r" -\- ^h^^. Thirdly, about its geometric axis, Z. — Fig. 116. Since the axis is perpendicular to each circular lamina through the centre, its Iz (§ 97) is = its mass X i-(rad.)° = {ynx^dz -^ g')^. Now a? = (r H- Ti)z, and hence for the w^hole cone Iz = \{yitr' - gU) t z'dz = {litr^hy - g)i-^r' = M^r\ Fig. 116. Fig. 118. 102. Homogeneous Eight Pyramid of Rectangular Base. — About its geometrical axis. Proceeding as in the last para- 102 MECHANICS OF ENGHSTEERINft. graph, we deri^^e Iz = M^^d\ in which d is the diagonal of the base. 103. Homogeneous Sphere. — About any diameter. Fig. 118. Iz = ? Divide into lamiuge perpendicular to Z. By § 97, and noting that a?' = r'— z% we have finally, for the whole sphere, Iz = {yTt ^ 2g) r+r {r'^ - ¥'^' + ¥1 = T^r^r^ - ^ For a segment, of one or two bases, put proper limits for s in the foregoing, instead oi -\- r and — r. 104. Other Cases. Y Fig. 119. Fig. 120. -Parabolic plate, Fig. 119, homogeneous and of (any) constant thickness, about an axis through 0, the middle of the )-X chord, and perpendicular to the plate. This is The area of the segment is = f As. For an elliptic plate. Fig. 120, homogeneous and of any constant thickness, semi-axes a and h, we have about an axis through 0, normal to surface Iq = M^[a^ -f- h^'] ; while for a very small constant thickness I^=Mih% and Iy=Mia\ The area of the ellipse = rrah. Considering Figs. 119 and 120 as plane figures, let the student determine tlieir polar and rectangular moments of inertia about various axes. For numerous other cases Kent's Mechanical Engineers' Pocket-Book may be consulted ; also Trautwine's Civil Engi- neers' Pocket-Book. 105. liTumerical Substitution. — The momsnts of inertia of 'plane figures involve dimensions of length alone, and will be utilized in the problems involving flexure and torsion of beams, where the inch is the most convenient linear unit. E.g., the MOMENT OF INEKTIA. 103 polar moment of inertia of a circle of two inches radius about its centre is ^Ttr* = 25.1 o -[-Mquadralie, or four-dimension^ inches, as it may be called. Since this quantity contains iowv dimensions of length, the use of the foot instead of the inch would diminish its numerical value in the ratio of the fonrtli power of twelve to unity. The moment of inertia of a rigid hody, or solid, liowever, = MTc* = (G- -^ ffWi ill which G, the weight, is expressed in units oi force, g involves both time and space (length), while W involves length (two dimensions). Hence in any homogeneous formula in whicli the / of a solid occurs, we must be careful to employ units consistently ; e.g., if in substituting G -^ g for M (as will always be done numerically) we put g = 32.2, we should use the second as unit of time, and the foot as linear unit. 106. Example. — Hequired the moment of inertia, about the axiS of rotation, of a pulley consisting of a rim, four parallelo- pipedical arms, and a cylindrical hub which may be considered solid, being filled by a portion of the shaft. Fig. 121. Call the weight of the hub G, its radius t\ similarly, for the rim, {r^, r^ and 7*2 ; the weight of one arm being = G^. The total / will be the sum of the /'s of the component parts, referred to the same axis, viz. : Those of the hub and rim will be {G ~ g)hy and {G, ~ ^)K^.= + r;), respectively (§ 97), while if the arms are ^^cj- 1-^- not very thich compared with their length, we have for them (§§ 95 and 88) 4 (^1- g) [i(^. - ry ~ i(r, -. ry + [r + l{r, - r)]'], i.e., 4((7i-^g)[Kr2-r)2+rr2] .... (4) as an approximation (obtained by reduction from the axis at the extremity of an arm to a parallel gravity-axis, then to the required axis, then multiplying by four). In most fly-wheels, the rim is proportionally so heavy, besides being the farthest removed from the axis of rotation, that the moment of inertia of the other parts is only a small part of the whole. Numerically let us have given r = 4, r2 = 36, and r^ = ^7 inches; the 104 MECHANICS OF ENGINEERING. respective weights being <?2 = 500 lbs. for the rim, (ri = 48 lbs. for each arm, and ^ = 120 lbs. for the hub. The quantity ^ will be retained as a mere symbol. Using the foot-pound-second system of units we then have for the moment of inertia of the huh (120-^gf)^[^]^= 6.66 -i-g'; for that of the four arms [by substitution in eq. (4) above] 4 36' -S) 2-/36 _£Y 3\12 12/ '^ 12 12 1 while for the rim we obtain (500-f-gr)— Sry /36 12J ^ll2 1647.2 ^q; = 4627.0 ^g. ^^=(^-^j^(^) = 7.91 sq.ft., These results are seen to be approximately in the ratio of the numbers 1, 100, and 700; showing that the neglect of the hub and arms in com- puting the moment of inertia would give a result about ^ too small. Adding, we find for the total moment of inertia of the body about the axis of rotation the quantity / = 5280.8 -^gr, for the units foot and pound. The unit of time is still involved in the quantity g. We are now ready to compute the square of the corresponding radius of gyration, viz., k"^, by dividing / by the whole mass M, =668 -i-g (see § 86) ; whence and therefore k itself = 2.82 ft. This is seen to be a little less than the 3.04 ft. value for k which would be implied in the approximate assumption that the moment of inertia is the same as if the whole mass were concentrated at the mid-point of the thickness of the rim, which assumption would be very nearly true if the masses of the hub and arms could be neglected. 107. Ellipsoid of Inertia. — The moments of inertia about all axes ptissing through any given point of any rigid body whatever may be proved to be inversely proportional to the squares of the diameters which they intercept in an imaginary ellipsoid, whose centre is the given point, and whose position in the body depends on the distribution of its mass and the location of the given point. The three axes which contain the three principal diameters of the ellipsoid are called the Princi- pal Axes of the body for the given point. This is called the ellipsoid of inertia. (Compare §89.) Hence the moments of inertia of any homogeneous reguhir polyedron about all gravity- axes are equal, since then the ellipsoid becomes a sphere. It can also be proved that for any rigid body, if the co-ordinate axes .X^ T", and .^, are taken coincident with the three principal axes at any point, we shall have fxydM = ; fyzdM = ; and fsxdM = 0. Note. — These three siimmations are called the "-products of inertia" and will occur in § 114 of this book. KINETICS OF A RIGID BODY. 105 CHAPTER Y. KINETICS OF A RIGID BODY. 108. General Method. — Among the possible* motions of a figid body the most important for practical purposes (and for- tunately the most simple to treat) are : a motion of translation, in which the particles move in parallel right lines with equal accelerations and velocities at any given instant; and rotation about a fixed axis, in which the particles describe circles in parallel planes with velocities and accelerations proportional (at any given instant) to their distances from the axis. Other motions will be mentioned later. To determine relations, or -equations, between the elements of the motion, tlie mass and form of the body, and the forces acting (which do not neces- sarily form an unbalanced system), the most direct method to be employed is that of two equivalent systems of forces (§ 15), one consisting of the actual forces acting on the body, con- sidered free, the otlier imaginary, consisting of the infinite number of forces which, applied to the separate material points composing the body, would account for their individual mo- tions, as if they were an assemblage of particles without mutual actions or coherence. If the body were at rest, then considered J'ree, and the forces referred to three co-ordinate axes, they would constitute a balanced system, for which the six summa- tions ^X, 2Y, ^Z, ^(mom.)x. ^''(mom.)y, and -^'(mom.)^. would each = ; but in most cases of motion some or all of these sums are equal (at any given instant), not to zero, but to the corresponding summation of the imaginary equivalent system, i.e., to expressions involving the masses of the particles (or material points), their distribution in the body, and the elements* of the motion. That is, we obtain six equations by putting the IX of the actual system equal to the IX of the imaginary, and so on ; for a definite instant of time (since some of the quantities may be variable), * Motions of such character that the particles of the body do not .change their relative positions. In other words, the body remains rigid. 106 MECHANICS OF ENGINEERING. 108a. The "Imaginary System." — In conceiving the imagi- nary equivalent system in § 108, applied to the material points or particles (supposed destitute of mutual action, and not exposed to gravitation), which make up the rigid body, we employ the simplest system of forces that is capable, by the Mechanics of a Material Point, of producing the motion, which the particles actually have. If now the mutual actions, co- herence, etc., were suddenly re-established, there would evi- dently be no change in the motion of the assemblage of parti- cles ; that is, in what is now a rigid body again, hence the imagi- nary system is equivalent to the actual system. In applying this logic to the motion of translation of a rigid body (see § 109 and Fig. 122,) we reason as follows : If the particles or elementary masses did not cohere together, being altogether without mutual action and not subjected to gravitation, their actual rectilinear motion in parallel lines, each having at a given instant the same velocity and also the sams acceleration, p, as any other, could be maintained only by the application, to each particle, of a force having a value = its mass X p, directed in the line of motion. In this way system (II.) is conceived to be formed and is evidently composed of parallel forces all pointing one way, whose resultant must be equal to "^eir sum, viz. I dMXp. But since at this instant p is common to the motion of all the particles, this sum can be written p i dM, =the whole mass Mxp. If now the mutual coherence of contiguous particles were sud- denly to be restored, system (II.) still acting, the motion of the assemblage of particles would not he affected (precisely as the fall- ing motion in vacuo of two wooden blocks in contact is just the same whether they are glued together or not) and consequently we argue that the imaginary system (II.) is the equivalent of whatever system of forces the body is actually subjected to, viz. system (I.), (in which the body's own weight belongs) producing the actual motion. Since the resultant of system (II.) is a single force, = ikfp, parallel to the direction of the acceleration, and in a line passing through the center of gravity of the body, it follows that th& resultant of the actual system is the same. KINETICS OF A RIGID BODY, 107 • 109. Translation. — Fig. 122. At a given instant all the par- ticles liave the same velocity = v, in parallel right lines (par- allel to the axis >^, say), and the same acceleration p. Required the 2^ of the acting forces, shown at (I.). (II.) shows the imaginary equivalent system, con- sisting of a force = mass X ace. = dMp applied parallel to 21 to each particle, since such a force would be necessary (from eq. {YY.) § 55) to account for the accelerated rectilinear motion of the particle, independently of the others. Putting {'2X)i-={'2X)ii, we have Fig. 122. (^X)j =fpdM =j)fdM = Mp. (^•) It is evident that the resultant of system (II.) must be paral- lel to X; hence* that of (I.), which = (2X)j and may be de- noted by -S, must also be parallel to X; let a = perpendicular distance from H to the plane YX; a will be parallel to Z. Now put [-2'(mom.)y]j = \_2 (mom. y)]ii, (T'is an axis perpen- dicular to paper through 0) and we have — lia = —fdMjpz = —pfdMz = — pMz (§88), i.e., a := 2. A similar result may be proved as regards y. Hence, if a rigid hody has a motion of translation., the resultant force m,ust act in a line through the centre of gravity (here more ])roperly called the centre of mass), and parallel to the direction of motion. Or, practically, in dealing with a rigid body having a motion of translation, we may consider it concentrated at its centre of mass. If the velocity of translation is uniform, R =■ M X = 0, i.e., the forces are bnlanced. 109a. Example. — The symmetrical rigid body in Fig. 122a weighs (G = ) 4 tons, and touches a smooth horizontal floor at the two points and B, symmetrically situated. Its center of gravity, C, is 6 ft. above the floor; and it is required to find the effect of applying a horizontal orce of P=l ton, pointing to the right and 4 ft. below the level of the center of gravity C. Evidently a motion of translation will ensue from left to right, with some acceleration p, unless the body should begin to overturn about or 5 as a pivot. The latter would be proved to * The forces of system (I.) cannot form a couple; since those of system (II.) do not reduce to a couple, all pointing one way. 108 MECHANICS OF ENGINEERING. J^G. 122a. be the case if either reaction, Vg or V, of the floor against the body at O and B, is found to be negative as the result of an analysis which assumes translation to occur. The actual forces acting on the body are only four, viz.: G and P, and the unknown vertical reactions V and Fq. A special device (very convenient for the present case) will now be used as a means of solution. The resultant of the "equivalent system," II, in this case of translation (see Fig. 122), is R, = Mp, lbs., acting through the center of gravity in a line parallel to that of the motion and in the direction of the acceleration, and hence is also the resultant of the actual system (just described). If, there- fore, we annex to the actual system its anti-resultant (which is a force, R', of the same value, Mp, as R, and in same line, but pointing in the opposite direction) we thereby form a system under which the body would be in equilibrium; which would justify our writing iX = 0, IY=0, i'(moms.) = 0, etc. (R' is called the "reversed inertia force" and is, of coiKse, fictitious). With this system, then, in view, putting IX = we obtain P—R' = 0; i.e., R', =Mp, =lton; whence the acceleration p=lH- (G-h^) = 1-^(4-^32.2) = 8.05 ft./sec.2 By T(moms. about point A) = we find E'X4'-F(?X2'-FX4' = or 7 = 2.5 tons; and, by -Z = 0, ¥ + ¥^-0 = 0, or Fo = 4-2.5=1.5 tons. Since neither V nor Fq is found to be negative the body does not tend to overturn but moves parallel to itself (i.e., translation) with a uniformly accelerated motion, the value of the acceleration being p = 8.05 ft. /sec. ^; so that at the end of the first second the body would be 4.025 ft. from the start (no initial velocity); at the end of the second second, 16.1 ft. If P were zero, or if P Were applied horizontally through the center of gravity, F and Fq would each be one haK of G, i.e. 2 tons. It appears, therefore, that the effect of the eccentric application of P (viz. 4 ft. below the center of gravity C) is to increase F by 0.5 ton and diminish Fq by an equal amount. If P acted 4 ft. above C, F and Fq would change places in this respect. For F to be just zero, P (in its present position) would need to have a value of 2 tons, and the body would be on the point of overturning toward the left. Or, again, with P = 1 ton, its line of application would have to be 8 ft. below or above C, for one of the reactions to be just zero. In fact, in the fictitious equilibrated system which includes R', since P=R' (in this simple case) they form a couple; and hence the three forces G, F, and Fq are equiva- lent to a couple of equal and opposite moment (viz. 4 ft.-tons in Fig. 122a). From the above it is seen that in the case of the last car of a railroad train, when it has an accelerated motion (just leaving a station), the pressures under the front and rear trucks will be slightly different from their values when the motion is uniform or zero, if the pull in the coupling does not pass through the center of gravity of the car. KINETICS OF A RIGID BODY. 109 110. Rotation about a Fixed Axis. — First, as to the elements of space and time involved. Fig. 123. Let be the axis of rotation (perpendicular to paper), OY d. fixed e ^"— --- x,w line of refei-ence, and OA a convenient line of (^ y^V\ the rotating body, passing through the axis and / X.^ \ perpendicular to it, accompanying the body in / — — 1~ its angular motion, Mdiich is the same as that of V_^_,,^-__^ — ^ OA. Just as in linear motion we dealt with ^^*^- ■'^■ linear space (.§), linear velocity (-y), and linear acceleration {^jp)y so here M'e distinguish at any instant ; a, the angular space between OY and OA, (radians; or de- grees, or revolutions) ; a) = -TT^ the angular velocity, or rate at which a is changing, (such as radians per sec. , or revolutions per minute, etc.); and ^ = -^ = -77^, the angular acceleration, or rate at which 0/ is changing (radians per sec. per sec, e.g.) These are all in angular measure and may be + or — , ac- cording to their direction against or with the hands of a watch. da. is a small increment of a, while d^a is the difference be- tween two da^s, described in two consecutive small and equal time-intervals, each= dt. (Let the student interpret the following cases : (1) at a cer- tain instant gd is -|-, and — ; (2) go is — , and 6 -{-; (3) a is — , GO and 9 l)otli -f ; (4) a -{-, go and 6 both — .) For rotary motion we have therefore, in general, and .•. (by elimination) codco — 6da; (VIIL) corresponding to eqs. (I.), (II.), and (III.) in § 50, for rectilinear motion. . Hence, for uniform rotary motion, go being constant and ^ = 0, we have a = Got, t being reckoned from the instant when a = 0. * See pp. 132, 133, of the "Notes," etc, for further illustration. 110 MECHANICS OF ENGINEERING. For uniformly accelerated rotary motion Q is constant, and if (jjQ denote tlie initial angular velocity (when a and t = 0) we may derive as in § 5G, denoting the constant d by 6'i, ,Go= Go^-\-dit; . . (1) a= Go^t + ^dxt'-', . . (2) a = — ^ — -"- ; . . (3) and a = i{oa^ + ^y- - • (4) If in an_y problem in rotary motion 0, go, and a have been determined for any instant, the corresponding linear vakies for any point of the body whose radial distance from tlie axis is p, will be 5= o'p (= distance described by the point measured along its circular path from its initial position), v = cop = its velocity, and j?^ = dp its tangential acceleration, at the instant in question, ii a, w and d, are expressed in radians. Example. — (1) What value of co, the angular velocity, is implied in the statemient that a pulley is revolving at the rate of 100 revolutions per minute if the radian is unit angle? 100 revolutions per minute is at the rate of 2;rXl00 = 628.32 radian units of angular space per minute = 10.472 per second. .". (y = 628.32 radians per minute or 10.472 radians per second. (2) A grindstone whose initial speed of rotation is 90 revo- lutions per minute is brought to rest in 30 seconds, the an- gular retardation (or negative angular acceleration) being con- stant; required the angular acceleration, di, and the angular .space a described. Use the second and radian as units. a»o = 27r|| = 9.4248 radians per second; .'. from eq. (1) /?!= — - — = — 9. 424 -=-30= —0.3141 radians per sec. per sec. t The angular space, from eq. (2) is a =a;o^ + J^ii2 = 30X9. 42-1(0.314)900 = 141. 3 radians; that is, the stone has made 22.4 revolutions in <3oming to rest and a point 2 ft. from the axis has described a distance s = ap = 141. 3 X 2 = 282. 6 ft. in its circular path. 111. Rotation. Preliminary Problems. Axis Fixed. — For clearness in subsequent matter we now consider the following KINETICS OF A RIGID BODY, 111 80 lbs. problem. Fig. 124 shows a rigid homogeneous right cylin- der A of weight G = 200 lbs. and radius r = 2 ft., mounted on a horizontal axle and concentric with the same. The center of gravity of the cyUnder is in the axis of rotation (Z). The axle carries a Ught and concentric drum, of 10 in. radius, from which a light inextensihle cord may unwind as the attached weight B descends, thus imparting an accelerated rotary motion to the cylinder. The weights and masses of the drum, cord, and axle, and all friction, will be neglected; and the two journals will be considered as one. The cylinder being originally at rest we wish to deter- mine its motion as produced by a constant down- ward pull or ten- sion of 80 lbs. in the vertical cord. (I) ^V.^L^--^B"(n> (T^' necessary y Fig. 124. \t weight, G', of the body to be used at B*, to secure this 80 lbs. tension in the cord, will be found later.) During this motion the real system of forces (system (I)) acting on a body A consists of the weight 200 lbs., always acting through Z, the fixed axis of rotation; the downward pull of 80 lbs. at 10 in. from the axis; the ver- tical component V of the reaction of the bearing; and the horizontal component (if any), H. At (II), Fig. 124, is shown an imaginary equivalent system capable of producing the same motion in the particles, each of mass = dM, if they were inde- pendent. Since each particle is moving in a circle of some radius p with some linear (tangential) acceleration pt at any instant, the cylinder having at that same instant some an- gular velocity co and some angular acceleration 6, we have v = a)p atid pt = ^P- (^ ^nd 6 in radians.) This circular motion of each particle could be produced (see eq. (5), p. 76) by a tangential force dT lbs., =dMpt, = 6dMp, accompanied by a normal force dN lbs., =dMv^-^p, = co^dMp. Our equivalent system, then, in (II), consists of a dT and a dN of proper value applied to each particle of body A at a given instant. Axes X and Y are shown in Fig. 124, * The body B is not shown in the figure. 112 MECHANICS OF ENGINEERING. axis Z, the axis of rotation, being i to the paper through origin 0. Let us now, for any instant of the motion, equate I (moms.)^ of the actual system, (I), to J (moms.)^ in sys- tem (II) ; using the integral sign to denote a summation which extends over all the particles of body A (for this instant; the integral might therefore be called an instantaneous integral). This gives, if we note that each of the normal forces dN of system (II) has no moment about axis Z, and that d is common to all the particles at this instant, (with ft.-lb.-sec. units), + 80x10/12= +/dT.p=+d/dMp^=^ +61,. . (1) The summation (instantaneous) /dMp^ is seen to be the quantity called "moment of inertia," about axis Z, of the body A and remains constant, since the p's do not change in value as the motion proceeds. For a solid homogeneous cylinder Ig = ^Mr^ (p. 99), and hence 800 = 6^[200-^32.2](2)2; i.e., ^ = 7.376 rads./sec.2 That is, 6 is constant and the rotary motion of the cylinder is uniformly accelerated. (N. B. — From eq. (1) we note that, in general, in order to obtain the angular acceleration, 6, of the rotary motion [of a rigid body about a fixed axis Z we have only to treat the body as a "free body" and write J! (moms.) about axis of rotation = angul. accel.Xmom. of inertia about Z.) 112. Further Results in Preceding Problem. — As to the necessary weight, G', of body B (suspended on the cord and causing the motion of both bodies), in order to produce the 80 lbs. tension in the cord, we note that body B has the same motion (only in a right line), as a point in the circumference of the drum, where the acceleration is p' = dX ^ = 4.48 ft. /sec. That is, the 'motion of B wUl be uniformly accelerated, with an acceleration of 4.48 ft. /sec. ^ Hence the weight of B must not only produce the 80 lbs. tension in the cord but also accelerate the mass of B, {M' = G'-^g) with an acceleration of 4.48 ft./ sec. ^ I.e., we have G' = 80+ {G'^g)p'; which is nothing more than saying that the net accelerating force, G' — 80, =massXaccel. ; whence we find, on solving, G' = 92.9 lbs. for the weight of the body to be used at B. For example, in the first 3 sec. of time, starting from rest, B will descend a distance (see p. 54), s^ = ip'(3y = 20.16 ft. and will have ac- quired a (linear) velocity of V3 = p'X3 = 13.44 ft. /sec. ; while body A will have turned through an angle of a3 = Ji9(3)^ = 33.19 radians, (or 5.283 revolutions) and will possess an angular velocity of a>3 = 5X3 = 72.13 rads./sec. or, (33.19 ^2;r = ), 3.525 revs./sec. Reaction of the bearing; (two journals considered as one). To find the two components H and V of this reaction, we again have recourse to the two equivalent systems of Fig. 124, acting on body A. (N.B. — The upward 80 lbs. and the force G' do not belong to system (I), since they act on body B.) During the motion, the coordinates x and y of each particle (of mass = dM) are continually changing, as also the angle ^ KINETICS OF A RIGID BODY. 113 between the p of the particle and axis Y (but not p itself). At any given instant we note that x = p sin 4> ^^d y — p cos (f>, for each particle. Let us now put i'F of system (I) equal io lY oi system (II). This gives us V -2m-m= fdT sin ^-/dN cos <^ . .... (2) As before, these integrals are "instantaneous integrals," being extended over all the particles of the body at a given instant of time, [so that in general the value of each may change with the progress of the motion. Substituting for dT and dN, etc., this may be written F - 200 - 80 = efdMp sin ^ - u?fdMp cos ^S ... (3) or, F-200-80=5/dM:c-w2/dMy, (4) Note that the value of (9, and also of w, at this single instant are common to all the particles and have been factored out, as shown. But the summation of fdMx is nothing more than Mx\ where M is the mass of the whole cylinder A ( = 200-^?) [see p. 18, eq. (1)], and X is the X coordinate of its center of gravity; and, similarly, J~dMy = My. We may therefore write V-2m-SQ = eMx-w'^My (5) But in the present case, since the center of gravity of body A is in the axis of rotation at all times, we have both x and 2/ = zero at all times; and hence finally 7-200-80=0; or F = 280 lbs. As to the horizontal component, H, of the bearing reaction, we place 2X of system (I) equal to IX of system (II) and obtain H = -fdT cos ^ -fdN sm4>=- OfdMp cos ^ - oJ^fdMp sin 4>, (6) i.e., H=-efdMy-w^fdMx,= -eMy-w^Mx (7) But since x and y are zero at all times, H must be zero, from (7) ; and we therefore conclude that in this case the reaction of the bearing is purely vertical at all times and is V, = 280 lbs. 113. Centre of Percussion of a Rod suspended from one End. — > Fig. 126. The rod is initially at rest (see (I.) in figure), is sti-aight, homogeneous, and of constant (small) cross-section. Neglect its weight. A horizontal force or pressure, P, due to a blow (and varying in amount during the blow), now acts upon it from the left, perpendicularly to the axis, Z, of suspension. An accelerated rotary motion begins about the fixed axis Z. °y° n ^^dt (in.) (TI.) Fig. 126. (II.) shows the rod free, at a certain instant, with the reactions X^ and Y,, put in at 0„. (III.) shows an imaginary system which would produce the same effect at this instant, and consisting of a dT = dMOp, and a <^iV = oo^dMp applied to each dM, the rod being composed of an infinite number of dM^s, eacli at some distance p from tlie axis. Considering that the rotation has just begun, go, the 114 MECHANICS OF ENGINEERING. angular velocity is as yet small, and will be neglected. Re- quired Yo tlie horizontal reaction of the support at in terms of P. By putting lYji= lYm, we have P-Yo =/dT = e/pdM = eu'p. /. J^o = -P — OM p ; p is the distance of the centre of gravity from the axis (IST.B. J'pdM = M p is only true when all the p's are parallel to each other). But the value of the angular acceleration 6 at this instant depends on P and a, for 2 (mom.)> in (IL) = :2 (luom.)^ in (III.), whence Fa = dfp'dM^ diz, where Iz is the moment of inertia of the rod about Z, and from § 95 = \Ml\ Now p = i^ ; hence, finally, "F — pfl _ ?- -' U. ± J. i) ' 1 If now J^u is to = 0, i.e., if there is to be no shook between the rod and axis, we need only apply P at a point whose dis- tance a = f / from the axis ; for then Y^ = 0. This point is called the centre of percussion for the given rod and axis. It and the point of suspension are interchangeable (see § 118). (Lay a pencil on a table; tap it at a point distant one third of the length from one end ; it will begin to rotate about a vertical axis through the farther end. Tap it at one end ; it will begin to rotate about a vertical axis through the point first mentioned. Such an axis of rotation is called an axis of instantaneous rota- tion, and is different for each point of impact — ^just as the point of contact of a wheel and rail is the one point of the wheel which is momentarily at rest, and about which, therefore, all the others are turning for the instant. Tap the pencil at its centre of gravity, and. a motion of translation begins; see § 109.) 114. Rotation. Axis Fixed. General Formulae. — Consider Fig. 127. Fig. 128. |.ng now a rigid body of any shape whatever, let Fig. 12Y indi- cate the system of forces acting at any given instant.^ Z being KINETICS OF A RIGID BODY. 115 the fixed axis of rotation, go and 6 tlie angular velocity and angular acceleration, at the given instant. X^ and IT are two axes, at right angles to each other and to Z^ fixed in space. At this instant eacii clM oi the body has a definite x, y, and q) (see Fig. 128), which will change, and also a p, and 0, which \v ill not change, as the motion progresses, and is pursuing a circu- lar path with a velocity = cop and a tangential acceleration = dp. Hence, if to each dM of the body (see Fig. 128) we imagine a tangential force dT =^ dMO p -Audi a normal force — dJf{oopy -^ p = QJ^dMp to be applied (eq. (5), §74), and these alone, we have a system comprising an infinite number of forces, all parallel to XJ^, and equivalent to the actual system in Fig. 127. Let ^JT, etc., represent the sums (six) for Fig. 127, whatever they may be in any particular case, while for 128 we shall write the corresponding sums in detail, looting that fdli cos cp = GoYdMp cos cp = coydMy = g9^J/^(§88); that/6^iV^sin (p = coydMp sin cp = coydMx = go'Mx; and similarly, that /dT cos cp = dfdMp cos q) = 6 My, and fdT sin q) = OMx; while in the moment sums (the moment of dT cos 9? about J^, for example, being — dT cos (p . z ■=■ — OdMp (cos (p)s= — 6dMyz, the sum of the moms, y of all the (^rcos 9>)'s = - QfdMyz) fdTeo% (pz = dfdMyz,fdN^m cpz = aoydMxz, etc., W6 have, since the systems are equivalent, :sX=-{-6My-Go'Mx; . . . . (IX.) :SY :=-6Mx-co'Ify, . . . . (X.^ 2Z= 0; (XL) 2 moms.x = - 0/dMxz — coydMyz ; . (XIL) :S moms.y = - e/dllyz + JfdMxz ; . (XIII.) :S moms.^ = OfdMp' = 61^. . • . (XIY.) These hold good for any instant. As the motion proceeds x and y change, as also the sums fdMxz and fdMyz. If the Ijody, however, is homogeneous, and symmetrical about the plane XY, fdMxz and fdMyz would always = zero ; since 116 MECHANICS OF ENGINEERING. the z of any <^JI/'does not change, and for every term dMy{-\-z\ there would be a term dMy{ — z) to cancel it ; similarly for fdMxz. The eq. (XIY,), ^ (moms, about axis of rotat.) = fdTp = QJdMff = {angular accel.) X {mom. of inertia oj hody about axis of rotat.), shows how the snvafdMp^ arises in problems of this chapter. That a iovce dT :=^ dMdp should be necessary to account for the acceleration (tang-ential) dfj of the mass dM, is due to the so-called inertia of the mass (§ 54), and its moment dTp, or OdMp^, might, with some reason, l>e called t\\e moment of inertia oi the dll, imdf6dMp^= OfdMp' that of the whole body. But custom has restricted the nanse to the snmfdMp^, Mdiich, being without the 0, has no term to suggest the idea of inertia. For want of a better the name is still retained, and is generally denoted by /. (See §§86, etc. ) 115. Example of the Preceding. — A liomoofeneous rig lit par- FiG. 129. allelopiped is mounted on a vertical axle (no friction), as in figure. is at its centre of gravity, hence hoth X and y are zero. Let its henviness be y, its dimensions A, 5„ and b (see § 97). XY is a plane of symmetry, hence both fdMxz and fdMyz are zero at all times (see above). The tension P in the (inextensible) cord is caused by the hanging weight P^ (but is not = /^j, unless the rotation is uniform). The figure shows both rigid bodies ^r^e. P^ will have a motion of trans- lation ; the parallelopiped, one of rotation about a fixed axis. No masses are considered except P^ -^ g. and bhb^y -^ g. The Iz = MTc^ of the latter = its mass X tV(^i' + ^')' § ^T. At any instant, the cord being taut, if ^ = linear acceleration of ^., we have jp = da. eq. (<?) From (XIY.), Pa = 61^ ; .: P = Olz ~ a. . . . (1) For the free mass P^ -i- g we have (§ 109) P^ — P = mass X ace, = {P.'^9)p = {P.-^g)ea; .:P = P,{l-ea^g). (2) Equate these two values of P and solve for 6^, whence Mkl-^{P,-^g)a' ^^^ 6 = KINETICS OF A RIGID BODY. 117 All the terms here are constant, hence d is constant ; there- fore the rotary motion is uniformly accelerated, as also the translation of P,. The formulae of § 56, and (1), (2), (3), and (4) of §110, are applicable. The tension P is also constant; see eq. (1). As ior the five unknown reactions (components) at (?i and 0^, the bearings, we shall find that they too are con- stant ; for from (IX.) we have' Zi + Z2 = 0; (4) from (X.) we have P + ri+F2 = 0; (5) from (XI.) we have Z^-G = 0; (fi) from (XII.) we have P . AO + Y, .0,0-Y^ .0^0 = 0; (7) from (XIII.) we have -X^ . Ofi + X^ ."0^ = 0. (8) Numerical substitution in the above problem. — Let the parallelepiped be of wrought-iron ; let Pi = 48 lbs.; a = 6 in. = J ft.; 6 = 3 in. = J ft. (see Fig. 112); 6i = 2 ft. 3 in. = |^ft.; ^.nd /i = 4 in.=i ft. Also let 0^0 = 020 = 18 in.= | ft., and AO = S in. = i ft. Selecting the foot- pound second system of units, in which g' = 32.2, the linear dimensions must be used in feet, the heaviness, ^, »of the iron must be used in lbs. per cubic foot, i.e., ^' = 480 (see § 7), and all forces in lbs., times in seconds. The weight of the iron will be G = Fr = ^f'i^r = i • I ■ iX480 = 90 lbs.; its mass = 90 -T- 32.2 = 2.79; and its moment of inertia about Z=/z = MA;2^ =Mxi^(V_-|-62) = 2.79X0.426 = 1.191. (That is, the radius of gyration, kz, ="i/0.426 = 0.653 ft.; or the moment of inertia, or any result depend- ing solely upon it, is just the same as if the mass were concentrated in a thin shell, or a line, or a point, at a distance of 0.653 feet from the axis.) We can now compute the angular acceleration, d, from eq. (3) ; 48 X + 24 1.191 + (48H-32.2)Xi'" 1.191+0.372" radians per sec. per sec. The linear acceleration of Pi is p = 0a = 7.68 feet per sec. per sec. for the uniformly accelerated translation. Nothing has yet been said of the velocities and initial conditions of the motions; for what we have derived so far applies to any point of time. Suppose, then, that the angular velocity <y = zero when the time, t=0; and correspondingly the velocity, v = o^a, of translation of Pj, be also = when t = Q. At the end of any time t, a> = 9t (,§§ 56 and 110) and v = pt = 6at; also the angular space, a = ^dt^, described by the parallelopiped during the time t, and the linear space s = \pt'^ = ^Qat'^, through which the weight P^ has sunk vertically. For example, during the first second the parallelopiped has rotated through an angle a = ^9t' = iX 15.36 XI = 7.68 radians, i.e., (7.68 -^2;r) = 1.22 revolutions, while P^ has sunk through s = ^9at^ = 3.84: ft., vertically. 118 MECHANICS OF ENGINEERING. The tension in the cord, from (2), is P = 48(l-15.36Xi-^?) =48(1-0.24) = 36.48 lbs. The pressures at the bearings will be as follows, at any instant: from (4) and (8), X^ and X2 must individually be zero; from (6) Z2 = G=Vj- = 90 lbs.; while from (5) and (7), Yi= -21.28 lbs., and 1^2= -15.20 lbs., and should point in a direction opposite to that in which they were assumed in Fig. 129 (see last lines of Jj- 39). 117. The Compound Pendulum is any rigid body allowed to oscillate without friction under the action of gravity when mounted on a horizontal axis. Fig. 131 shows the body /"ree, in any position during the progress of the oscillation. C is the centre of gravitj^; let OG = s. From (XI Y.), § 114, we have 2 (mom. about fixed axis) = angul. ace. X mom. of inertia. .-. — Gs sin a = 61^, and = — Gs sin a -^ I^ = — Mgs sin a -f- MTcl, i.e., = — ^s sin «r -^ ^/ (1) Hence d is variable, proportional to sin a. Let us see what the length I = OJT, of a simple circular pendulum, must be, to have at this instant (i.e., for this value of a) the same angular acceleration as the rigid body. The linear (tangential) accelera- tions of ^ the extremity of the required simple pendulum would be (§ 77) Pt = — 9 sin a, and hence its angular accelera- tion* would = — gsina-^l. "Writing this equal to d in eq. (i), we obtain ^ = ^0^^^ (2) Bat this is independent of a ; therefore the length of the sim- ple penduhim having an angular acceleration equal to that of the oscillating body is the same in all positions of the latter, and if the two begin to oscillate simultaneously from a position of rest at any given angle oc^ with the vertical, they will keep abreast of each other during the whole motion, and hence have * Most easily obtained by considering that if the body shrinks into a mere point at K, and thus becomes a simple pendulum, we have both ka and s equal to I ; which in (1) gives B — — g sin a -i- 1. KINETICS OF A RIGID BODY. 119 the same duration of oscillation ; which is .*. , for small ampli- tudes (§ 78), t' = 7t VT^ = 7t Vk; -^ gs, .... (3) j^is called the centre of oscillation corresponding to the given centre of suspension 0, and is identical with the cenl/re of per- cussion (§ 113). Example. — Required the time of oscillation of a cast-iron cylinder, whose diameter is 2 in. and length 10 in., if the axis of suspension is taken 4 in. above its centre. If we use 32.2 for g, all linear dimensions should be in feet and times in seconds. From § 100, we have Jf(f From eq. (3), § 88, .-. i; = 0.170 sq. ft.; 1 7,2\ — llffi 1 II 10 0\ — ]\ f 1 103 M[^.\^^ + i-]=Mx0.m; t'= 7t VO.ITO -^ ^32.2 Xi) = 0.395 sec. 118. The Centres of Oscillation and Suspension are Inter- changeable. — (Strictly speaking, these centres are points in the line through the centre of gravity perpendicular to the axis of suspension.) Refer the centre of oscillation K to the centre of gravity, thus (Fig. 132, at (I.) ) : = l-s = Ms s = MJ ic' + Ms' Ms s = —- (1) s ^ ^ !N'ow invert the body and suspend it at K', required CK^, or s^i to find the centre of oscillation corresponding to K as centre of suspension. By analogy from (1) we have s s^ = he -^ -Si ; but from (1). k^ -^ s^ ^ s .'. s^ = s\ in other words, ^j is identical with 0. Hence the proposition is proved. Advantage may be taken of this to determine the length X of the theoretical simple pendulum vibrating seconds, and thus finally the acceleration of gravity from formula (3), § 117, viz.. (I.) (11.) Fig. 132. 120 MECHANICS OF ENGINEEEING. when i! = 1.0 and I (now = Z) has been determined experi- mentally, we have g (in ft, per sq. second) = X (in ft.) X tt*. . . (2) This most accurate method of determining g at any locality requires the use of a bar of metal, furnished with a sliding weight for shifting the centre of gravity, and with two project- ing blocks provided with knife-edges. These blocks can also be shifted and clamped. By suspending the bar by one knife- edge on a proper support, the duration of an oscillation is com- puted by counting the total number in as long a period of time as possible; it is then reversed and suspended on the other with like observations. By shifting the blocks between successive experiments, the duration of the oscillation in one position is made the same as in the other, i.e., the distance be- tween the knife-edges is the length, I, of the simple pendulum vibrating in the computed time (if the knife-edges are not equi- distant from the centre of gravity), and is carefully measured. The I and t' of eq. (3), § 117, being thus known, g may be com- puted. The length, in feet, of the simple pendulum vibrating seconds, at any latitude /?, and at a height of h ft. above sea- level, is (Chwolson, 1902). L = 3.25974-0.008441 cos 2/3-0.0000003/i. 119. Isochronal Axes of Suspension. — In any compound 'pendulum., for any axis of suspension, there are always three others, parallel to it in the same gramty-plane, for which the oscillations are Tnade in the same time as for the first. For any assigned time of oscillation t', eq. (3), § 117, compute the corresponding distance CO = s oi O from O; . Mk: 7r\MkJ + Ms^ i.e.,from t = ^ ^^- = ^- , we have s= {gt"-^27r')± V{g'r-r-4:7r') — kJ. . . (1) Hence for a given f, there are two positions for the axis O parallel to any axis through C, in any gravity-plane, on both sides; i.e., four parallel axes of suspe?ision, in any gravity- plane, giving equal times of vibration ; for two of these axes KINETICS or A RIGID BODY. 121 we must reverse the body. E.g., if a slender, homogeneous, prismatic rod be marked off into thirds, tlie (small) vibrations will be of the same duration, if the centre of suspension is taken at either extremity, or at either point of division. Examjple. — Required the positions of the axes of suspension, parallel to the base, of a right cone of brass, whose altitude is six inches, radius of base, 1.20 inches, and weight per cubic inch is 0.304 lbs., so that the time of oscillation may be a half- second. (N.B. For variety, use the inch-pound-second system of units, first consulting § 51.) 120. The Fly-Wheel in Fig. 133 at any instant experiences a pressure P' against its crank-pin from the connecting-rod and a resisting pressure P" from the teeth of a spur-wheel with Fi& 133. which it gears. * Its weight G acts through C (nearly), and there are pressures at the bearings, but these latter and G have no moments about the axis C (perpendicular to paper). The figure shows it free^ P" being assumed constant (in practice this depends on the resistances met by the machines which D drives, and the fluctuation of velocity of their moving parts). P\ aiivl therefore T its tangential component, are '.variable, depending on the effective steam-pressure on the piston at any instant, on the obliquity of the connecting-rod, and in high- speed engines on the masses and motions of the piston and con- necting-rod. Let r. = radius of crank-pin circle, and a the perpendicular from G on P" . From eq. (XIY.), § 114, we have Tr - P"a = Qlg, .: 6 = {Tr - P"a) -^ Ig, • (1 *Bearings at C not shown. P is the thrust in the piston-rod due to steam pressure on piston. 122 MECHANICS OF ENGHSTEEKING. as the angular accelei-atioii at any instant ; substituting wliicliin the general equation (VIII.), § 110, we obtain IqWcLoo = Trda — P"ada (2) From (1) it is evident that if at any position of the crank-pin the variable Tr is equal to the constant P"a^ 6 is zero, and consequently the angular velocity a) is either a maximum or a minimum. Suppose this is known to be the case both at m and n\ i.e., suppose T, which was zero at the dead-point A, has been gradually increasing, till at n, Tr = P"a\ and there- after increases still further, then begins to diminish, until at m Tr again = P"a^ and continues to diminish toward the dead- point P. The angular velocity go, whatever it may have been on passing the dead-point A, diminishes, since 6 is negative^, from A to n, where it is c»^, a minimum ; increases from n to- ??^. where it reaches a maximum value, c»,^. n and m being known points, and supposing co^ known, let us inquire what Go^ will be. From eq. (2) we have IcJ^ Goda>=J^^ Trda-P"J^^ ada. . . (3> But rda = 6/s = an element of the path of the crank-pin, and also the " virtual velocity" of the force T, and ada = ds", an. element of the path of a point in the pitch-circle of the fly- wheel, the small space through which P" is overcome in dt. Hence (3) becomes /ci(c»J - GD^') =J^ Tds - P" X linear arc EF. (4> To determine / Tds we might, by a knowledge of the vary- ing steam-pressnre, the varying obliqnit}^ of the connecting-rod, etc., determine T for a number of points equally spaced along- the cnrve nm^ and obtain an approximate value of this sum by Simpson's Rule; but a simpler method is possible by noting (see eq. (1), § 65) that each term Tds of this sum = the corre- sponding term Pdx in the series / Pdx, in which P = the^ KINETICS OF A RIGID BODY. 123 effective steam-pressure on the piston in the cylinder at any in- stant, dx the small distance described by the piston while the crank-pin describes any ds^ and n' and m' the positions of the piston (or of cross-head, as in Fig. 133) when the crank-pin is at n and m respectively. (4) may now be written Ic\{po^ - O =J^^, Pdx - P" X linear arc EF, (5) from which c»^ may be found as proposed. More generally, it is available, alone (or with other equations), to determine any one (or more, according to the number of equations) unknown quantity. This problem, in rotary motion, is analogous to that in §59 (Prob, 4) for rectilinear motion. Friction and the in- ertia of piston and connecting-rod liave been neglected. As to the time of describing tlie arc wm, from equations similar to (5), we may determine values of co for points along nm, divid- ing it into an even number of equal parts, calling them cw^, &?„ etc., and then employ Simpson's Rule* for an approximate value pm n,m g^ of the sum \ t= I — (from eq. (YL), § 110) ; e.g., with four parts, we would have f^l ri4241~ ^ = T-=: (angle wC'w, in rads.) — I 1 1 1- — Ln 12 ^ ° Lo^n ' Ci5, C^j 0^3 <^m-^ .(6) 121. Numerical Example. Fly-Wheel.— (See Fig. 133 and the equations of § 120.) Suppose the engine is non-condensing and non-expansive (i.e., that P is constant), and that P = 5500 lbs., r = 6in. = ift., a = 2ft., and also that the wheel is to make 120 revolutions per minute, i.e., that its inean angular velocity is to be oo' = ^^- X 27r, i.e., oa' = 47r " radians" per sec. First, required the amount of the resistance P" (constant) that there shall be no permanent change of speed, i.e., that the angular velocity shall have the same value at the end of a com- plete revolution as at the beginning. Since an equation of the form of eq, (5) holds good for any range of the motion, let * See p. 13 of the "Notes and Examples in Mechanics." 134 MECHANICS OF ENGINEERING. that range be a complete revolution, and we shall have zero as the left-hand member ; fPdx = P X 2 f t. = 5500 lbs. X 2 ft., or 11,000 foot-pounds (as it may be called); while P" is un- known, and instead of lin. arc EF we have a whole circumfer- ence of 2 ft. radius, i.e., 4;r ft.; .-. = 11,000 - P" X 1 X 3.1416; whence P" = 875 lbs. Secondly, required the pi'oper mass to be given to the fly- wheel of 2 ft. radius that in the forward stroke (i.e., while the crank-pin is describing its upper semicircle) the max. angular velocity g?^ shall exceed the minimum go^ by only ^L-g?', assum- ing (which is nearly true) that ^{oj^ -j- go^) = go'. There be- ing now three unknowns, we require three equations, which are, including eq. (5) of § 120, viz.: J^^C i-(^m + COn){(^m " ^n) =J^^ Pdx - P" X linear arc EF\ (5) \{<^m-\- (^n)= co'=4:7r; (7) and go^- go^ = -^gj' = ^n. (8) The points n and m are found most easily and with sufficient accuracy by a graphic process. * Laying off the dimensions to scale, by trial such positions of the crank-pin are found that T, the tangential component of the thrust P' produced in the connecting-rod by the steam-pressure P (which may be resolved into two components, along the connecting-rod and a normal to itself) is =(a -^r)P'\ i.e., is = 3500 lbs. These points will be n and m (and two others on the lower semicircle). The positions of the piston n' and W, corresponding to n and m of the crank-pin, are also found graphically in an obvious manner. "We thus determine the angle nCm to be 100°, so that linear arc EF= \^7t X 2 ft. = ^tt ft., while nm' pin' / Pdx = 5500 lbs. X / dx= 5500x^/iW=5500x 0.77 ft., n'm' being scaled from the draft. Xow substitute from (7) and (8) in (5), and we have, with Jcq = 2 ft. (which assumes that the mass of the fly-wheel is con- centrated in the rim), * See p. 85, "Xotes and Examples," etc. KINETICS OF A RIGID BODY. 125 {G-^g)X4:X4:7tx^7c = 5500 X C.77 - 875 X ^^, which being solved for G (with ff = 32.2 ; since we have used the foot and second), gives G = 600.7 lbs. The points of max. and min. angular velocity on the back- stroke may be found similarly, and their values for the fly- wheel as now determined ; they will differ but slightly from the Go^ and co^ of the f orwai'd stroke. Professor Cotterill says- that the rim of a fly-wheel should never have a max. velocity > 80 ft. per sec; and that if made in segments, not more than 4rO to 50 feet per second. In the present example M^e have for the forward stroke, from eqs. (7) and (8), gl;^= 13.2 (7r-measure units) per second; i.e., the corresponding velocity of the wheel- rim is VJ,^ = co^a = 26.4 feet per second. 122. Angular Velocity Constant. Fixed Axis. — If co is con- stant, the angular acceleration, 6, must be = zero at all times^ which requires 2 (mom.) about the axis of rotation to be = (eq. (XIY.), § 114). An instance of this occurs when the only forces acting are the reactions at the bearings on the axis, and the body's weight, parallel to or intersecting the axis ; the values of these ^"\2i2'EA^y' reactions are now to be determined for dif- /' ^ — ^ ferent forms of bodies, in various positions fig. 134. relatively to the axis. (The opposites and equals of these reac- tions, i.e., the forces with which the axis acts upon the bearings, are sometimes stated to be due to the " centrifugal forces^'' or " centrifugal action," of the revolving body.) Take the axis of rotation for Z, then, with = 0, the equa- tions of § 114 reduce to — 00" Mx ; — oo'My ; 0: . . ^ moms.x = — GofdMyz ; '2 moms.y = -{- a^fdMxz ; '2 moms.^ = 0. . . . (IX«.) (X«..). (XIa.) (X1I«.) (Xllla.) (XIYa.) 126 MECHANICS OF ENGIITEEEIWa. For greater convenience, let ns suppose the axes ^ and Y {since tlieir position is arbitrary so long as tliey are perpen- dicular to each other and to Z) to revolve with the hody in its uniform rotation. 122a. If a homogeneous hody have a plane of symmetry ■and rotate uniformly about any axis Z perpendicular to that plane {intersecting it at 0)^ then the acting forces are equiva- . lent to a single force^ = co'^Mp, applied at and acting in a groA^ity-line, hut directed away from the centre of gravity, it is evident that such a /^- ^ . - . force P :3= ofMp, applied as stated ^"'- '^''- (see Fig. 135), will satisfy all six con- ditions expressed in the foregoing equations, taking ^through the centre of gravity, so that x = p. For, from (IX«.), i^must ■ = afMp, while in each of tlie other summations the left- hand member will be zero, since P lies in the axis of ^; and as their right-hand meinbers will also be zero for the present body (y = 0; and each of the sums fdMyz and fdlfxB is zero, since for each term dMy{ -\- z) there is another dMy{^ — z) to cancel it ; and similarly, for fdMxz), they also are satisfied ; Q.E.D. Hence a single point of support at will suffice to maintain the uniform motion of the body, and the pressui-e against it will be equal and opposite to P.* First Example. — Fig. 136. Supposing (for greater safety) that the uniform rotation of 210 revolutions per minute of each segment of a fly-wheel is .^--""' maintained solely by the tension in the cor- <f - f' responding arm, P ; required the value of P * P if the segment and arm together weigh -J^- of a ton, and the distance of their centre of ^^<^- ^3^- gravity from the axis is p = 20 in., i.e., = |- ft. "With the foot- ton-second system of units, with g = 32.2, we have P = co'Mp = [^ X 27cY X [^ ^ 32.2] X f = 0.83 tons^ or 1660 lbs. * That is, neglecting gravity. The body's weight, if considered, will take its place among the actual forces acting on the body. KINETICS OF A RIGID BODY. 127 Second Exmnple. — Fig. 137. Suppose the Tiniform rotation of the same ily-wheel depends solely on the tension in the rim, required its amount. The figure shows the half- rim fi-ee, with the two equal tensions, ]r*\ put in at the surfaces exposed. Here it is assumed tliat the arms exert no tension on the rim. Erom §122a we have 2P' = oo'^Mp^ where J/^is themass of the half- p' rim, and p its gravity co-ordinate, which may be ob- fig. 137. tained approximately by § 26, Problem 1, considering the rim as a circular wire, viz., p = 2r -^ rt. Let M = (180 lbs.) -^- g, with r = 2 ft. We have then P' = i(22)Xl80 ~ 32.2)(4 -^ n) = 1718.0 lbs. (In realit}^ neither the arms nor the rim sustain the tensions just computed ; in treating the arms we have supposed no duty done by the rim, and vice versa. The actual stresses are less, and depend on the yielding of the parts. Then, too, we have supposed the wheel to take no part in the transmission of mo- tion by belting or gearing, which would cause a bending of the arms, and have neglected its weight.) 122b.' If a homogeneous hody have a line of symmetry and rotate uniformly dboxit an axis parallel to it [0 being the foot of the perpendicular from the centre of gravity on the axis)^ then the acting forces are equivalent to a single force P = Go'Mp, applied at O and acting in a gravity-line away from the centre of gravity. Taking the axis X^ tiirough the (JM centre of gravity, Z being the "H axis of rotation, Fig. 138, while j Z' is the line of symmetry, pass an auxiliary plane Z' IT' parallel to ZY. Then the sum fdMxz may be written fdM{p -\- x')z which = JfdMz + fdMx'z. Fi»- 138. ButfdMz = Mz = 0, since 1 = 0, and every term dJif{-\- x')z is cancelled by a numerically 128 MECHANICS OF ENGINEERIISrG. equal term dM{— x')3 of opposite sign. 'H.eme fdMxz = 0. Also ydMyz = 0, since each positive product is annulled by an equal negative one (from symmetry about Z'). Since, also, 3/ r= 0, all six conditions in § 122 ai-e satisfied. Q. E. D. If the lioiiiogeneous body is any solid of revolution whose geometrical axis is jparallel to the axis of rotation, the forego- ing is directly applicable. 122c. If a homogeneous hody revolve uniformly about any axis lying in a plane of symmetry, the acting forces are equiv- alent to a single force P = oa'Mp, acting parallel to the grav- ity-line which is perpendicular to the axis (Z), and away from the centre of gravity, its distance from any origin in the axis Z being = [fdMxB] -^ Up {the plane ZX being a gravity plane). — Fig. 139. From the position of the body we have p z=z X, and y = \ hence if a value cD^Mp be given to P and it be made to act through Z and parallel to X, and away from the centre of gravity, all the conditions of § 122 are satisfied except {Xlla.) and (XIII«.). But symmetry about the plane XZ makes fdMyz = 0, and satisfies (XJI«.), and by placing P at a distance a =fdMxz -^ Mp from along Z we satisfy (XIII«.). Q. E. D. Example. — A slender, homogeneous, prismatic rod, of length = I, is to have a uniform motion, about a ver- q tical axis passing through one extremity, maintained by a cord-connection with a fixed p point in this axis. Fig. 140. Given oo, (p, I, (p =: \l cos cp), and F the cross-section of the rod, let s = the distance from to any dJif of the rod, dM being = Fyds -^ g. The x of any dM =^ s cos q); its s = s . sin qj ; .\fdMx3 = {Fy -T- g) sin (p cos q) / s^ds — -^{Fyl -r- g)r sin cp cos (p = iJifl' sin q> cos <p. Fig. 139. Fig. 140. KINETICS OF A RIGID BODY. 129 HeiKje a, =^fdMxz -^ Mp, is = |Z sin 9?, and the line of ac- tion of P ( = oo'Mp = gl)' (i'VZ -^- ^) ^Z cos q)) is therefore higher up than the middle of the rod. Find the intersection D of G and the horizontal drawn tlirongh ^ at distance <zfroni 0. Determine P' by completing the parallelogram GP', at- taching the cord so as to nudvc it coincide with P'^ for this will satisfy the condition of maintaining the motion, Mdien once be- gun, viz., that the acting forces G, and the cord-tension P', shall be equivalent to a force P = oo'^Mp, applied horizontally through Z at a distance a from 0. 123. Free Axes. Uniform Rotation. — Referring again to § 122 and Fig. 134, let us inquire under what circumstances the lateral forces, J^^. 1^„ ^^. Y^, with which the bearings pi'ess the axis, to maintain the motion, are individually zero, i.e., that the hearings are not needed, and may therefore he removed (except a sniooth horizontal plane to sustain the body's weight), leaving the motion undisturbed like that of a top "asleep." For this, not only must 2X and 2 Y both be zero, but also (since otherwise X^ and X^ might form a couple, or Y^ and Y^ similarly) ^ (moms.)^and 2 (moms.)y must each = zero. The necessary peculiar distribution of the body's mass about the axis of rotation, then, must be as follows (see the equations of §122): First, X and y each = 0, i.e., the axis must he a gravity-axis. Secondly, fdMyz — 0, KndfdMxz = 0, the origin being any- where on Z, the axis of rotation. An axis {Z) (of a body) fulfilling these conditions is called a Free Axis, and since, if either one of the three Principal Axes for the centre of gravity (see § 107) be made an axis of rotation (the other two being taken for X and Y), the conditions ^ = 0, y = 0, fdMxz = 0, and fdMyz = 0, are all satisfied, it follows that every rigid hody has at least three free axes, which are the Principal Axes * of Inertia of the centre of gramit/y at right angles to each other. In the case of homogeneous hodies free axes can often be . determined by inspection : e.g.. any diameter of a sphere ; any * See § 107. p. 104. IBO MECHANICS OF ENGINEP:RI Nti. transverse diameter of a right circular cylinder througli its centre of gravity, as well as its geometrical axis; the geomet- rical axis of aTi}^ solid of revohition ; etc. 124. Rotation about an Axis which has a Motion of Translation, — Take only the particular case where the moving axis is a fjj^ gravity-axis. At any instant, let the dM d? velocity and acceleration of axis Z be?; and p ; the angular velocity and accelera- tion about that axis, oo and 6. Then, since '1^ the actual motion of a dM in any dt is compounded of its motion of rotation about the gravity-axis and the motion of translation in common with that axis, Fig. 141. we may, in forming the imaginary equiva- lent system in Fig. 141, consider each dM as subjected to the simultaneous action of dP = dMjp parallel to ^, of the tan- gential dT = dMdp^ and of the normal dN ■= dMioopf -^ p = Go^dMp. Take ^in the direction of translation, Z (perpen- dicular to paper through 0) is the moving gravity-axis ; Y perpendicular to both. At any instant we shall have, then, the following conditions for the acting forces (remembering that /> sin 9? = y.fdMy = 3fy ~ ; etc.) : :2X = fdP - fdT sin cp - fdN Qo&<p = Mp; . (1) 2Y=/dTcos(p -/dJ^siu (p =0;. . (2) ^ moms.^ =/dTp -fdPy ^ dfdMp" = dl^ ^ OMlcz', (3) and three other equations not needed in the following example. Example. — A homogeneous solid of revolution rolls {with- out slipping) down a rough inclined plane. Investigate the motion. Con- sidering the \>o^j free., the acting forces are G (known) and N and P., the un- ^ ...- known normal and tangential compo- ''V nents of the action of the plane on the roller. If slipping occurs, then P is the ^'*^- ^^^ gliding friction due to the pressure Ni^ 156); here, however, it is less by hypothesis (perfect rolling). At any instant the four unknowns are found by the equations KINETICS OF A RIGID BODY. 131 JSX, i.e., G sin fi - P, = {G-^ g)p ; . (1) :SY, i.e., (? cos ^ - if, = ; . . . . (2) ^ moms.^, i.e., Pa, = OMkz ; • • (^) while on account of the perfect rolKng, da = p . . • (4) Solving, we have, for the acceleration of translation, _2? = ^ sin /i - [1 + (/.■/ - a')l (If the body slid wirhont friction, j9 would = p'sin /?.) Hence for a cylinder (§ 97), kz" beiiio- = -^a", we have^ = f^ sin /3 ; and for a sphere (§ 103) j> = ^g sin /3. (If the plane is so steep or so smooth that both rolling and slipping occur, then da no longer = _p, but the ratio of i-* to iV is known from experiments on sliding friction ; hence there are still four equations.) The motion of translation being thus found to be uniformly accelerated, we may use the equations of § 56. Numerically, if a homogeneous solid sphere took 1.20 sec. to descend (from rest() 10 ft. along a rough inclined plane, with /? = 30°, did any- slipping occur, or was the motion perfect rolling? From p. 54 we have s = -|-p/^ that is, 10 = -|- . ^ .^ sin 30° . t^, for perfect rolling; from which we obtain i=1.32 seconds, which is >1.20 sec. Hence some slipping must have occurred. (The time of descent would have been only i = "|/2s-=-^g = 1.114 sec, if the surfaces had been perfectly smooth; and the sphere would have had simply a motion of translation, the force P being zero). N.B. — A hollow sphere would occupy a longer time than a solid one in descending the plane (if rough) ; since the ratio kz^a is greater for the former. 125. Parallel-Rod of a Locomotive. — "When the locomotive moves uniformly, each dJf of the rod between the two (or three) driving-wheels rotates with j \ ; uniform velocity about a centre of its own on the line j5i>, Fig. 143, and with a velocity -y* and radius r common ^._.^ to all, and likewise has a horizontal ( * js : , ; m t ) ■M/l'^/or■7;^ motion ot transhition. Hence (ii.) if we inquire what are the reactions P ^^^- ^^3. * This velocity is that which the dM ?ias relatively to the frame of the locomotive, in a circular path. E.g., if the locomotive (frame; has a velocity of GO miles per hour and the radius r is one-third of the radius of the driver, then V is 20 miles per hour. 132 MECHANICS OF ENGI]S'EEEING. of its supports, as induced solely T)y its weight and motion^ w'.ien in its lowest position (independently of any thrust along the rod), we put JSJT of (I.) = 2Y of (II.) (II. shows the imaginary equivalent system), and obtain 2P - G =fdN =fdMo' -^r^iv'-^ r)fdM = Mv' :- r. Example. — Let the velocity of translation = 50 miles per hour, the radius* of the pins be 18 in. = f ft., and = half that of the driving -wheels, while the weight of the rod is 200 lbs. With g = 32.2, we must use the foot and second, and obtain V = i[60 X 5280 ^ 3600] ft. per second = 36.6; while Jf = 200 ^ 32.2 = 200 X .0310 = 6.20 ; and finally P = i[200 + 6.2(36.6)^-=- f] = 2868.3 lbs., or nearly If tons, about thirty times that due to the weight alone. 126. So far in this chapter the motion has been prescribed, and the necessary conditions determined, to be fulfilled by the acting forces at any instant. Problems of a converse nature, i.e., where the initial state of the body and the acting forces are given while the resulting motion is required, are of much greater complexity, but of rare occurrence in practice. For further study in this direction the reader is referred to Routh's "Rigid Dynamics," Rankine's "Applied Mechanics," Sehell's " Theorie der Bewegung und der Kraefie," and Worthington's "Dynamics of Rota- tion" (this last being a small but clearly written and practical book). In Wood's "Analytical Mechanics" will be found the proof of "Euler's Equations," which are the basis of the treatment of the gyroscope in the book of that name by Gen. J. G. Barnard (Van Nostrand's Science Series, No. 90). The article on the gyroscope in Johnson's Cyclopaedia is by Gen. Barnard. Perry's "Spinning Tops" is an interesting popular book. The Brennan "Monorail Car" (model) is described in the Engineering Record for Aug. 31, 1907, p. 226, and depends for its stability (there being but one rail under the car) upon two gyroscope wheels revolving at 7000 revs, per min. in a vertical plane parallel to the rail. See also McClure's Magazine for Dec, 1907, p. 163. * Or, rather, the radius of the circular path of the pin-centre, whose velocity in this path is 25 miles per hour. WORK, ENEKGY, AND POWER. 18b CHAPTER YI. WORK, ENERGY, AND POWER. 127. Remark. — These quantities as defined and developed in this cliapter, though compounded of the fundamental ideas of matter, force, space, and time, enter into theorems of such wide application and practical use as to more than justify their consideration as separate kinds of quantity, 128, Work in a XJniform Translation. Definition of Work. — Let Fig. 144 represent a rigid body having a motion of trans- lation parallel to X, acted on by a yrp^ system of forces P^, P^, H^, and Ii^, ^f — ''~'&</<VY^ which remain constant.* ("'^y\,...s..-~^ ''•'2_s-..:.-\J_ Let s be any distance described by f ^^'"'''sa.., g \ ^ the body during its motion ; then ^ JT^+VP^" '"*"" '^'\\P/j must be zero (§ 109), i.e., noting that Rf- ^-v,^_^ /\y^^ R^ and R^ have negative X com- ^"^^ ^fi ponents (the supplements of tlieir ^^'^- '^^'^• angles with ^are used), P^ cos a-j -J- P^ cos a^ — R^ cos a^ — R^ cos o'^ = ; or, multiplying by s and transposing, we have (noting that s cos cfj = Sj the projection of s on P^^ that s cos a^ = s^, the •projection of s on P^^ and so on), P,s, + P,5, - R^s^ + R,s^ {a) The projections s^^ s^, etc., may be called the distances de- scribed in their respective directions by the forces Pj, P^, etc; Pj and P^ having \\\Q)\%^ forward^ since 5, and s^ fall in front of the initial position of their points of application ; R^ and R^ backward, since s^ and a\ fall behind the initial positions in their case, (By forward and backward we refer to the direc- * Constant in direction as well as amount. 134 MECHANICS OF ENGINEERING. tion of each force in turn.) The name Work is given to tlie product of a force hy the distance described in the direction of the force l)y the point of application. If the force moves forward (see above), it is called a worhingforce, and is said ta do the work (e.g., P^s^ expressed by this prodnct ; while if hachioard, it is called a resistance, and is then said to have the worh (e.g., R^s^, done upon it, in overcoming it through the distance mentioned (it might also be said to have done nega- tive work). Eq. {a) above, then, proves the theorem that : In a uniform translation, the ivorhing forces do an amount of work which is entii'ely applied to overcoming the resistances. 129. Unit of Work. — Since the work of a force is a product of force by distance, it may logically be expressed as so many foot-pounds, inch-pounds, kilogram-meters, according to the system of units employed. The ordinary English unit is the foot-pound, or ft.-lb. It is of the same quality as a force- moment. 130. Power. — Work as already defined does not depend on the time occupied, i.e., the work P^s^ is the same whether per- formed in a long or short time ; but the element of time is of so great importance in all the applications of dynamics, as well as in such practical commercial matters as water-supply, con- sumption of fuel, fatigue of animals, etc., that the rate of worh is a consideration both of interest and necessit}'. Power is the rate at which work is done, and one of its units is one foot-pound per second in English practice ; a larger one will be mentioned presently. The power exerted by a working force, or expended upon a resistance, may be expressed symbolically as L = P,s, -■ t, or ^,^3 -^ t, in which t is the time occupied in doing the work P^s, or P^s, (see Fig. 144) ; or if v^ is the component in the direction of the force P^ of the velocity v of the body, we may also write L=P\V\, ft. -lbs. per sec {h} WOEK, ENERGY, AND POWER. 135 131. Example. — Fig. 145, shows as a free hody a sledge which is being drawn uniformly up a rough inclined plane by a cord parallel to the plane. Required the total power exerted (and expended), if the tension in the cord is P^ = 100 lbs., the weiglit of sledge ^3 = 160 Fig. 145. lbs., P = 30°, and the sledge moves 240 ft. each minute. iV^ and J^^ are the normal and parallel (i.e., P^ = friction) com- ponents of the reaction of the plane on the sledge. From eq. (1), § 128, the work done while the sledge advances through s = 240 ft. may be obtained either from the working forces, which in this case are represented by _Pj alone, or from the resistances JR, and P^. Take the former method first. Pro- jecting s upon ^j we have s^ = s. Hence P,s, or 100 lbs. X 240 ft. = 24,000 ft.-lbs. of work done in 60 seconds. That is, the power exerted hy the working forces is L = P,5, -=- ^ = 400 ft.-lbs. per second. As to the other method, we notice that ^g and R^^ are resist- ances, since the projections s^= s sin ^, and s^ — s, would fall back of their points of application in the initial position, while JV is neutral, i.e., is neither a working force nor a resistance, since the projection of s upon it is zero. From :SX = we have — £,— R3 sin /? + Pi = 0, ■And from 2 T = (§ 109) JV — Ji, cos /3 =0; whence /^, the friction = 20 lbs., and JV = 138.5 lbs. Also,, since s, = 5 sin /6f = 240 X i = 120 ft., and s^ = s, = 240 ft., we have for the work done upon the resistances (i.e., in over- coming them) in 60 seconds B^s, + ^,6', = 160 X 120 + 20 X 240 = 24,000 ft.-lbs., and the power expended in overcoming resistances, L = 24,000 -^ 60 = 400 ft.-lbs. per second, as already derived. Or, in words the power exerted by the tension in the cord is expended entirely in raising the weight a vertical height of 2 feet, and overcoming the friction through 136 MECHANICS OF EJ^iGlJ^EEKIlJJG. a distance of 4 feet along tlie plane, every second ; the motion heing a uniform translation. 132. Horse-Power.- — As an average, a horse can exerts a trac- tive effort or pull of 100 lbs., at a uniform pace of 4 ft. per sec- ond, for ten hours a day without too great fatigue. This gives a power of 400 ft.-lbs. per second ; but Boulton & Watt in rating their eiigines, and experimenting with the strong drav- liorses of London, fixed upon 550 ft.-lbs. per second, or 33,000 ft.-lbs. per minute, as a convenient large unit of power. (The Prench horse-power, or cheval-vapeur^ is slightly less than the English, being 75 kilogrammeters per second, or 32,550 ft.-lbs. per minute.) This value for the horse-power is in common use. In the example in § 131, then, the power of 400 ft.-lbs, per second exerted in raising the weight and overcoming fric- tion may be expressed as (400-^550 =) yjof a horse-power. A man can work at a rate equal to about J^ of a liorse-powe» , with proper intervals for eating and sleeping, 133. Kinetic Energy. Retarded Translation. — In a retarded translation of a rigid body whose mass = Jf, suppose thei-e are no working-forces, and that the resistances are constant and their resultant is H. (E.g., Fig. 146 shows such a case ; a sledge, having an initial velocity c and slid- —7 ^^ ing on a rough horizontal plane, is gnidu- ^^T ally retarded by the friction H.) i?is par- allel to the direction of translation (§ 109) Fig. 146. and the acceleration is j? = — M -^ M ] hence from vdv =-pds we have '; \ ) fvdv = - (1 -^ M)fRds. .... (1) But the projection of each ds of the motion upon R \q ■= ds itself ; i.e. (§ 128), Rds is the work done upon jR, in overcoin ing it through the small distance ds, and /Rds is the snm of ,all such amounts of work throughout any definite portion ol the motion. Let the range of motion be between the points where the velocity = c, and where it = zero (i.e., the mass lias come to rest). "With these limits in eq. (1), (0 and s' be- ina; the corresponding 1 M(? C^'t^ . limits for s), we have J 2 *Jo ^ ' WORK, ENERGY, AND POWEK. 137 Til at is, in giving up all its velocity c the hody has heen ahle to do the worh fRds (this, if R remains constant, reduces to JR.s') or its equal —^7—. If, then, bv energy we designate tlie ability to ^perform work, we give the name kinetic energy of a niuving body to the product of its mass hy half the square . fJ'fv^\ cf its velocity \~h~"); ,i-e., energy due to motion. Example. — If the sledge in Fig. 146 has initially a velocity oi c=^j ft./sec. and its weight is G = 322 lbs. (so that its mass in the ft.-lb.-sec. system is M = 10) its initial kinetic energy is ilfc^J_2=500 ft.-lbs. If the friction or resistance, R, is constant and has a value of 20 lbs., we compute s' = 25 ft. (from 500 = Rs') as the distance the sledge will go in overcoming this resistance; i.e., in giving up all its kinetic energy. If the sledge goes 40 ft. we conclude the average resistance to have been only 12.5 lbs.; since 500-^-40 = 12.5. Now suppose R variable, say = (20 + 4s) lbs., (s in ft.), and we have 500= / [20 + 4s]ds = 20s'+2s'2; .-. s' = 11.6ft. 134, Work and Kinetic Energy in any Translation. — Let P be the resultant of the working forces at any instant, R that of the resistances ; they (§ 109) will both M u act in agravity-line* parallel to the di- <- rection of translation. The acceleration C— '_ -§■'. ----*o' at any instnnt is ^ = {^A. -^ M) fig. 147. =: {^P — R) -T- M\ hence from vdv ^ pds we have Mvdv = Pds — Rds (1) Integrating between any two points of the motion as and 0' where the velocities are 'o^ and v\ we have after transposition / Pds= Rds 4- 'Mv" Mv, . 2 ~ 2 J {d) But P being the resultant of P^, P^, etc., and R that oi ^,, ^5,. etc., we may prove, as in § 62. that if dii^. du,, etc.. be the respective projections of any ds upon P^. P^, etc., while dw^, dv\, etc., are those upon R^, R^, etc., then Pds=.P,du^-\-P^du^-\- and Rds=R,dw^-\-R,dw, ; and (d) may be rewritten * That is, a line passing through the centre of gravity. Ig8 MECHANICS OF ENGINEERING. £ P^du, +y ' P^du, + . . . . P,dw, +y P,dw, + + I -^ ~2^ J 5 (^) ° or, in words : In any translation, a portion of the worh done hy the working forces is applied in overcoming the resistances lohile the remainder equals the change in the kinetic energy of the l)ody. It will be noted that the bracket in {e) depends only on the initial and final velocities, and not upon any intermediate values ; hence, if the initial state is one of rest, and also the final, the total change in kinetic enei'gy is zero, and the work of the working forces has been entirely expended in the work of overcoming the resistances ; but at intermediate stages the former exceeds the work so far needed to overcome resistances, and this excess is said to be stored in the moving mass ; and as the velocity gradually becomes zero, this stored energy becomes available for aiding the working forces (which of themselves are then insufficient) in overcoming the resistances, and is then said to be restored. (The function of a fly-wheel might be stated in similar terms, but as that involves rotary motion it will be deferred.) Work applied in increasing the kinetic energ}' of a body is sometimes called " work of inertia," as also the work done by a moving body in overcoming resistances, and thereby losing speed. 135. Example of Steam-Hammer. — Let us apply eq. {e) to determine the velocity v' attained by a steam-hammer at the lower end of its stroke (the initial velocity being = 0), just before delivering its blow upon a forging, supposing that the steam-pressure i-*^ ^^ "^ stages of the downward stroke is given by an indicator. Fig. 148. Weight of moving mass is 322 lbs.; .-. J!/' =10 (foot-pound-second system), Z = 1 foot. The working forces at any instant are P^^ O ^= 322 lbs.; P^, which is variable, but whose values at the s,&wQn equally spaced WORK, ENERGY, AND POWER. 139 fZs joints a, h, c, d, e, f, g, are 800, 900, 900, 800, 600, 500, 450 lbs., respectively. R^ the exhaust-pressure (16 lbs. per sq. inch X 20 sq. inches piston-area) = 320 lbs., is the only resistance, and is constant. Hence fi*om eq. {e), since here the projections du^^ etc., of any ds upon the respective forces i are equal to each other and = ds, Pj ds +y P,ds = Rj ds + ^-. (1) 'The term fP^ds can be obtained approximately * ^Qj Simpson's Rule, using tlie above values for six equal divisions, vi^hich gives J^[800 + 4(900 + 800 + 500) -f 2(900 -(- 600) + 450] .= 725 ft.-lbs. of work. Hence, making all the substitutions. we have, since I ds =^1 ft., 322 X 1 + 725 = 320 X 1 + IMv"; .-. ^Mv" = 727 ft.-lbs. of energy to be expended on the forging. (Energy is evi- dently expressed in the same kind of unit as work.) We may then say that the forging receives a blow of 727 ft.-lbs. •energy. The pressure actually felt at the surface of the ham- mer varies from instant to instant during the compression of the forging and the gradual stopping of the hammer, and -depends on the readiness with which the hot metal yields. If the mean resistance encountered is R^, and the depth of ■compression s", we would have (neglecting the force of gravity, and noting that now the initial velocity is v', and the final zero), from eq. (c), ^Mv" = Rrr^s"; i.e., R^ = [727 -^ s" (ft.)] lbs. E.g., if s" = I of an inch = -gL of a foot, R^ = 43620 lbs., and the maximum value of R would probably be about double this near the end of the impact. If the anvil also sinks during the impact a distance s'", we must substitute s'" -\- s" instead •of s" ; this will give a smaller value for ^^. * See p. 13 of "Notes and Examples in Mechanics." 140 MECHANICS OF ENGINEEKINa. By mean value for B, is meant [eq. (c)] that value, B,^^ which satisfies the relation BJ =f Bds. This may be called more explicitly a space-average, to dis- tinguish it from a time-average, which might appear in some problems, viz,, a value Bt^, to satisfy the relation {t' being the duration of the impact) I^tnt' = / Bdt, and is different from B^. From \Mv'^ = T2T ft.-lbs., we have v' = 12.06 ft. per sec, whereas for a free fall it would have been 4/2x32.2x1 = 8.03. (This example is virtually of the same kind as Prob. 4, § 59, differing chiefly in phraseology.) 136. Pile-Driving.* — The safe load to be placed upon a pile after the driving is finished is generally taken us a fraction (from ^ to ^) of the resistance of the earth to the passage of the pile as indicated by the effect of the last few blows of the ram, in ac- cordance with the following approximate theory : Toward the * end of the driving the resistance B encountered by ! the pile is nearly constant, and is assumed to be that ^ met by the I'am at the head of the pile; the distance i s' through which the head of the pile sinks as an M^^ effect of the last blow is observed. If G, then, is the weight of the ram, = 3Ig, and h the height of free fall, the velocity due to h, on sti-iking the pile, is c = V2gh (§ 52), and we have, from eq. (c), iMc\ i.e., GL = f Bds = Bs' . . (1) {B being considered constant) ; hence B = Gh -f- s' . and the safe load (for ordinary wooden piles), P = from \ to ^ oi Gh^s' (2) Maj. Sanders recommends |- from experiments made at Fort Fig. 149. * See also p. 87 of the author's Notes and Examples in Mechanics. WORK, ENERGY, AND POWER. 141 Delaware in 1851; Molesworth, |-; General Barnard, ^, from extensive experiments made in Holland. Of course from eq. (2), given J*, we can compute s'. (Owing to the uncertainty as to how much of the resistance H is due to friction of the soil on the sides of the pile, and how mucli to the inertia of the soil around the shoe, the more elaborate theories of Weisbach and Rankine seem of little practical account.) 137. Example. — In preparing the foundation of a bridge-pier it is desired that each pile (placing them 4 ft. apart) shall bear safely a load of 72 tons. If the ram weighs one ton, and falls 12 ft., what should be the effect of the last blow on each pile? Using the foot-ton-second systein of units, and Molesworth factor \, eq. (2) gives s' = 1(1 X 12 -j- 72) = ^- of a foot = J of an inch. That is, the pile should be driven until it sinks only J incih ^nder each of the last few blows. 138. Kinetic Energy Lost in Inelastic Direct Central Impact. — Referring to § 60, and using the same notation as there given, we find that if the united kinetic energy possessed by two in- elastic bodies after their impact, viz., ^Jf^C -j" i^^^O^ C' hav- ing the value {M,c, + M^o^) -^ {M^ + i/,), be deducted from the amount before impact, viz., ^M^c^ -{- ^M^e^. the loss of Tcinetic energy dxiring iwijpact of two inelastic hodies is * An equal amount of energy is also lost by partially elastic bodies during the first period of the impact, but is partly re- gained in the second. If the bodies were perfectly elastic, we would find it wholly regained and the resultant loss zero, from the equations of § 60 ; but this is not quite the reality, on account of internal vibrations. The kinetic energy still remaining in two inelastic hodies after impact (they move together as one mass) is * See Eng, News, July, 1888, pp. 33 and 34. 142 MECHANICS OF ENGINEERING. •|^(J/"j + ^^C-, or, after inserting the value of C = {M,G, + M^c^) -- {M, + M^, we have 2 (2) M,| Exainjple 1. — The weight ^^ = M^g falls freely through a height A, impinging upon a weight 6^, = JI/2^, which was initially at rest. After their {in- elastic) impact they move on together with the com- bined kinetic energy just given in (2), which, since Cj and (?25 the velocities before impact, are respectively \^'2gh and 0, may be reduced to a simpler form. This energy is soon absorbed in overcoming the flange-pressure R^ which is proportional (so long as the elasticity of the ]-od is not impaired) to the elongation 6', as with an ordinary spring. If from Fig. 150. previous experiment it is known tliat a force R^ produces an elongation «„, then the variable R = (^„ -^ s^)s. Keglecting the weight of the two bodies as a working force, we now have, from eq. (d), + 4^ s ids R. I.e. R, 0=^ f sds + (3) When s = s\ i.e., when the masses are (momentarily) at rest in the lowest position, the flange-pressure or tensile stress in the rod is a maximum, R' = {R^ -^ s^)s', whence s' = R's^ ~- R^; and (3) may be written M:gh or R' 2 2R. s = M^gh (4) (5) Eq. (3) gives the final elongation of the rod. and (5) the greatest tensile force upon it, provided the elasticity of the rod is not W(3KK, ENERGY, AND POWKK. 143 impaired. The forin ^R's' in (4) may be looked upon as a direct integration of / jRds, viz., the mean resistance {^H') multi- plied by the whole distance {s') gives the work done in over- coming the variable R through the successive ds's,. If the elongation is considerable, the working-forces G, and G^ cannot be neglected, and would appear in the term-|-(^i -f- G^s' in the right-hand members of (3"). (4), and (5). The upper end of the rod is firmly fixed, and the rod itself is of small mass compared with M^ and M^. Exmnple 2. — Two ears, Fig. 151, are connected by an elastic chain on a horizontal track. Yelocities before impact (i.e., before the stretching of the chain be- ^g^ o ^ci gins, by means of which they are l_-[~.,_____^ H brought to a common velocity at the M^ Mi instant of greatest tension R', and Fig. 151. elongation s' of tlie chain) are <?j = g^, and c^ = 0. During the stretching, i.e., the first period of the impact, the kinetic enei'gy lost by the masses has been expended in sti-etch- ing the chain, i.e., in doing the work ^i?V ; hence we may write (the elasticity of the chain not being impaired) (see eq. (1) ) M,M,e,^ _ 1 _ R, ^^_^.„ in which the different symbols have the same meaning as in Example 1, in which the rod corresponds to the chain of this example. In this case the mutual accommodation of velocities is due to the presence of the chain, whose stretching corresponds to the compression (of the parts in contact) in an ordinary impact. In numerical substitution, 32.2 for g requires the use of the units foot and second for space and time, while the unit of force may be anything convenient. 139. Work and Energy in Rotary Motion. Axis Fixed. — The rigid body being considered free, let an axis through O perpendicular to the paper be the axis of rotation, and resolve all forces not intersecting the axis into components parallel and perpendicular to the axis, and the latter again into com- ponents tangent and normal to the circular path of the point 144 MECHANICS OF ENGINEEKING. of application. These tangential com- ponents are evidently the only ones of the three sets mentioned which have moments about the axis, those having moments of the same sign as 00 (the angular velocity at any instant) being called working forces^ T^, T„ etc. ; those of opposite sign, resist- ances, T^', T^', etc.; for when in time dt the point of application ^j, of T^, describes the small arc ds^ =: a^da, whose projection on T^ is = ds^, this projection falls ahead (i.e., in direction of force) of the position of the point at the beginning of dt, while the reverse is true for T/. From eq. (XIY.), § 114, we have for 6 (angul. accel.) 6 = ' J , (1) which substituted in codco = Qda (from § 110) gives (remem- bering that a^doi ■=. ds^, etc.), after integration and transposition, T,ds,+J^ TA + etc. T^dsi-^j^ T:ds: ^^i^.-\-\_\oo^^i -koo:i\ (2) where and n refer to any two (initial and final) positions of the rotating body. Eq. (4), § 120, is an example of this. Now \oo^I— \Qo^fdMp' =f^dM{GOnpf, which, since go^P is the actual velocity of any dM ^i this (final) instant, is nothing more than the sum of the amounts of kinetic energy possessed at this instant by all the particles of the body ; a similar state- ment may be made for \oa^I. (a»o a^nd ojn in radians.) Eq. (2) therefore may be put into words as follows : Between any two positions of a rigid hody rotating about a fixed axis, the worh done hy the working forces is partly used in overcoming the resistances, and the remainder in changing the kinetic energy of the individual particles. If in any case this remainder is negative, the final kinetic energy is less than the initial, i.e., the work done by the working forces is less than that necessary to overcome the resistances through their respec- tive spaces, and the deficiency is made up by the restoring of WORK, ENERGY, AND POWER. 145 some of the initial kinetic energy of the rotating body. A moving fly-wheel, then, is a reservoir of kinetic energy. Example. — The 668-lb. pulley of p. 104 was found to have a radius of gyration of |/7.91 ft., and a moment of inertia about its axis, Z, of ikfA;^ = (668-7-3) 7.91. Let us suppose it mounted on a short shaft of (ro = ) 2 in. radius (whose 7z may be neglected) supported in proper bearings. The pulley and shaft are in contact with nothing except the bearings, which offer a friction T/, tangent to outer surface of shaft, of 120 lbs. If the pulley has an initial rotary speed of 300 revs./min., in how many turns, n, will it be brought to rest? Evidently ^^ = 0, while o^Q, = 27r300 -;- 60, =31.41 rads./sec. That is, the initial kinetic energy is i^'Mk^ =a(31.41)2(668-h32.2) 7.91, =80,810 ft.-lbs.: and the final, zero. T'i' = 120 lbs., constant, and the work done on T/ is, T^' I "'ds^' = 120 .n{2n) .lr== 125. Qn ft.-lbs. Hence from eq. (2) we. JO have = 125.6n + [0-80,810]; i.e., n-643 turns, Ans. 140. Work of Equivalent Systems the Same. — If two plane systems of forces acting on a rigid hody are equivalent (§ 1 oa), the aggregate worh done hy either of them during a given slight displacement or motion of the hody parallel to their plane is the same. By aggregate work is meant what Las ah'eady been defined as the sum of the " virtual moments" (§§ 61 to 64), iu any sniall displacement of tlie body, viz., the algebraic sum of the products, 2 [Pdu), obtained by multiplying eacli foi'ce by the pi'ojection {du) of the displacement of (or small space described by) its point of application upon the force. (We here class resistances as negative working forces.) Call the systems A and B] then, if all the forces of B were reversed in direction and applied to the body along Avith those of A. the compound system would be a balanced system, and lience we should have (§ 64), for a small motion parallel to the plane of the forces, :2{Pdii) = 0, i.e.. 2{Fdu) for A - :S{Fdu) for ^ = 0, or . + 2{Pdtc) for A = -{- 2{Fd'u) for B. But -f- 2 (Fdu) for A is the aggregate work done by the forces of A dui-ing the given motion, and -f 2(Fdu) for B is a similar quantity for the forces of B (not reversed) during the same small motion if B acted alone. Hence the theorem is p)-oved, and could easily be extended to space of three dimen= sions. 10 - 146 MECHANICS OP ENGINEERING. Fig. 153. £>, of the body; a final, n 141. Relation of Work and Kinetic Energy for any Extended Motion of a Rigid Body Parallel to a Plane. — (If at any instant any of the forces acting are not parallel to the plane mentioned, their components lying in or parallel to that plane, will be used instead, since the other compo- nents obviously would be neither working forces nor resistances.) Fig. 153 shows an initial position, and anj' intermediate, as q. The forces of the system acting may vary in any manner during the motion. In this motion each dM describes a curve of its own with varying velocity v, tangential acceleration j^t-, ^^^d radius of curvature r ; hence in any position ^, an imaginary system JB (see Fig. 154), equivalent to the actual system A (at q in Fig. 153), would be formed by applying to each dM a tangential force dT =^ dMpt, and a normal force dN' = dMv'^ -V- r. By an infinite number of con- secutive small displacements, the body passes from o to n. In the small displacement of which q is the initial position, each 6? J/^ describes a space ds^ and dT does the work dTds = dMvdv, while dJV does the work- dJV X = 0. Hence the total work done by £ in the small displacement at q would be dN dT dM'v'dv' + dM"n}"dv" -f etc., (1) including all the dM^& of the body and their respective veloci- ties at this instant. But the work at q in Fig. 153 by the actual forces (i.e., of system A) during the same small displacement must (by § 140) be equal to that done by B. hence P,du, -f P,du^ + etc. = dM'v'dv' + dM"v"dv" -f etc. (^) Now conceive an equation like {g) written out for each of WORK, ENERGY, AND POWER. 147 the small consecutive displacements between positions o and yi and corresponding terms to be added ; this will give P/hc^ -\- 1 P^du^ -\- etc. = dW / v'dii' + dM" / v"dv" + etc. = \dM'{v^^ - ^;=) + i^^Jf'X V - <'') + etc. The second member may be rewritten so as to give, finally, / P,dit,+ P,du,-^etG.=:S{idMv,')-:S{^dMv,'),{XY.) or, in words, the worTi, done hy the acting forces {treating a re- sistance as a negative worhing force) between any two posi- tions is equal to the gain {or loss) in the aggregate Icinetic energy of the particles of the hody hetwee7i the tioo positions. To avoid confusion, 2 has been used instead of the sign y in one member of (XY.), in which v^ is the final velocity of any dM {not the same for all necessarily) and v^ the initial. (The same method of proof can be extended to three dimen- sions.) Since kinetic energy is always essentially positive, if an ex- pression for it comes out negative as the solution of a problem, some impossible conditions have been imposed. 142. Work and Kinetic Energy in a Moving Machine. — Defining a mechanism or machine as a series of rigid bodies jointed or connected together, so that working-forces applied to one or more may be the means of overcoming resistances occurring anywhere in the system, and also of changing the amount of kinetic energy of the moving masses, let us for simplicity consider a machine the motions of whose parts are all parallel to a plane, and let all the forces acting on any one piece, considered free, at auy instant be parallel to the same plane. Now consider each piece of the machine, or of any series of its pieces, as a free body, and write out eq. (XY.) for it be- tween any two positions (whatever initial and final positions are selected for the first piece, those of the others must be corre- sponding initial and corresponding final positions), and it will 348 MEOIIA.MCS OF EXGINEEIilNG. be found, on adding np corresponding members of these equa- tions, that the terms involving tliose components of the mutual pressures (between the pieces considered) which are normal to the rubbing surfaces at any instant will cancel out, while their components tangential to the rubbing surfaces {i.e., fric- tion, since if the surfaces are perfectly smooth there can be no tangential action) will appear in the algebraic addition as resistances multiplied by the distances rubbed through, meas- ured on the rubbing surfaces. For example. Fig. 155, where one I'otating piece both presses and rubs on another. Let the normal pressure between them at A be R^ = P^ ; it is a work- ing force for the body of mass M" , but a resistance for M' , hence the separate symbols for the numerically equal forces (action and reaction). Similarly, the f liction at ^ is i?3 = ^Pg ; a resistance for M' , a working-force for M" . (In some cases, of course, friction may be a resistance for both bodies.) For a small motion, A describes tlie small arc AA' abont 0' in dealing with M\ but for M" it describes the arc AA" about 0" . A' A" being parallel to the surface of contact AD, while AB is perpen- FiG. 156. Fig. 157. Fig. 155. dicular to A' A" . In Figs. 156 and 15Y we see M' and M" free, and their corresponding small rotations indicated. During these motions the kinetic energy (K. E.) of each mass has clianged by amounts <f(K. E.)j,f/ and (i(K. E.)j/// respectively, and hence eq. (XY.) gives, for each free body in turn, P\a^' - R,AB - R,A^ = di^. E.)^. . (1) - RW+ P.AB + P^JJ^ = d(K. 'E.)m". . (2) Now add (1) and (2), member to member, remembering that P^ = P^ and P^ = P^ = P^ = friction, and we have P,aa' - F,A'A" - R^jb" = d{K. E.)^' + d{lL E.)m", (3) WUllK, ENERGY, AND POWER. 149 in which the mutual actions of M' and M" do not appear, except the friction, the work done in overcoTning which, when the t'loo hodies are thus considered collectively, is the product of the friction hy the distance A' A" of actual nibbing meas- tired on the rubbing sttrface. For any number of pieces, then, considered free collectively, the assertion made at the beginning of this article is true, since any finite motion consists of an infinite number of small motions to each one of wliich an equa- tion like (3) is applicable. Summing the corresponding terms of all such equations, we have f" P,du, -{-fF,du,+ etc. = :^(K.E.),,- :^(K. E.)o.(XYI.) This is of the same form as (XY.), but instead of applying to a single rigid body, deals with any assemblage of rigid pai-ts forming a machine, or any part of a machine (a similar proof will apply to thi-ee dimensions of space); but it must be remem- bered that it excludes all the mutual actions* of the pieces con- sidered except friction, which is to l)e introduced in the manner just illustrated. A flexible inextensible cord may be considered as made up of a great number of short rigid bodies jointed M'ithout friction, and hence may form part of a machine with- out vitiating the truth of (XVI.). ^(K. E.)„ signifies the sum obtained by adding the amounts of kinetic energy {^dMv^ for each elementary mass) possessed by all the particles of all the rigid bodies at their final posi- tions ; ^(K. E.)„, a similar sum at their initial positions. For example, the K. E. of a rigid body having amotion of transla- tion of velocity -y, =^ ^vfdM =^ ^Mv^ ', that of a rigid body having an angular velocity go about a fixed axis Z, = ^oo'^Iz (§ 139) ; while, if it has an angular velocity w about a gravity- axis Z, which has a velocity Vz of translation at right angles to itself, the (K. E.) at this instant may be j)roved to be (§ 143) the sum of the amounts due to the two motions separately. * These mutual actions consist only of actions by contact (pressure, ruo, etc.) . No magnetic or electrical attractions or repulsions are here considered. 150 MECHANICS OF ENGINEERING. 143. K. E. of Combined Rotation and Translation. — The last statement may be thus proved. Fig. 158. At a given instant the velocity of any dJf is V, the diagonal formed on the velocity Vz of translation, and the rotaiy velocity oop rela- tively to the moving gravity-axis Z (per- pendicnlar to paper) (see § Yl), Fig. 158. i-©., v' = Vz + {oopY — ^{Gop)vz COS 9? ; hence vv^e have K. E., at tliis instant, = f^dMv' = \v^fdM + WfdMp" - GovzfdMp cos ^, but p cos q) ^=:y, and fdMy = My = 0, since Z is a gravity- axis, .-. K. E. = iMvz' + WIz- Q. E. D. It is interesting to notice that the K. E. due to rotation, viz., \go^Tz = \M{w]tY^ is the same as if the whole mass were con- centrated in a point, line, or thin shell, at a distance ^,-the radius of gyration, from the axis. Example. — A solid homogeneous sphere of radius r = 6 in. and weight = 322 lbs. is rolling down an incline. At a certain instant the velocity of its centre is 10 ft. per sec. and hence, i'/ no slipping occurs, its angular velocity about its centre is co, ==Vz-^r, =10-7- J, =20 radians/ sec. Con- sequently, at this instant (see § 103, p. 102) its total kinetic energy is i(322 4_32.2)[(10)2 + (20)2 . f(i)2] = 700 ft.-lbs. 144. Example of a Machine in Operation. — Fig. 159. Con- sider the four consecutive moving masses, M\ M'\ M"\ and M'^^ (being tlie piston ; connecting-rod ; fly-wheel, crank, drum, and chain ; and weiglit on inclined plane) as free, collectively. Let us apply eq. (XYI.), the initial and final positions being taken when the crank-pin is at its dead-points o and n\ i.e., we deal with the progress of the pieces made while the crank-pin describes its upper semicircle. Remembering that the mutual actions between any two of these four masses can be left out of account (except friction), the only forces to be put in are the actions of other bodies on each one of these four, and are WORK, ENERGY, AND POWER. 151 shown in the figure. The only mutual friction considered will be at the crank- pin, and if tliis as an average — F'\ the work done on it between o and n = F"7tr" ^ where r" = radius of cranlv-pin. The work done by jP^ the effective steam- pressure (let it be constant) daring this period is = I^^l' ; that done in overcoming J^j, the friction between piston and cylinder, = I^^l' ; that done upo?i the weight G'oi connecting-rod is cancelled by the work done by it in the descent following ; the work done Fig. 159, upon G''', = G'^Tta sin /?, where a = radius of drum ; that upon the friction i^^, = J^^rra. The pressures JV, W, N'^, and N'", and weights G' and G'", are neutral, i.e., do no work either positive or negative. Hence the left-hand member of (XVI.) becomes, between o and n, P,V - F,V - F"7tr" - G'^Tta sin /? - Fjta, . . (1) provided the respective distances are actually described by these forces, i.e., if the masses have sufficient initial kinetic energy to carry the crank-pin beyond the point of minimum velocity, with the aid of the working force P^^ whose effect is small up to that instant. As for the total initial kinetic energy, i.e., ^(K. £.)„, lei; us express it in terms of the velocity of crank-pin at o, viz., Y^. The (K.E.)„ of M' is nothing ; that of M" , which at this in- stant is rotating about its right extremity {fixed ioix \\\& instant) with angular velocity oo" = F„ ^ l'\ is \(^"^1^' \ that oiM'" = \o!}"'^I^'\ in which oo'" = V^^r; that of M''' (translation) ^ iJf ^X''"? in which v;^ = {a-^r) V,. 2(K. E.)„ is expressed 152 MECHANICS OF ENGINEERING. in a corresponding manner with F^ (final velocity of crank-pin) instead of Y^. Hence the right-hand member of (XYI.) will give (potting the radius of gyration of Jf about 0" = Jc", and that of Jf about G = Jc) i( K' - F;)[j/-|^ + M^-~ +M-y~']. . . (2) By writing (1) = (2), we have an equation of condition, capa- ble of solution for any one unknown quantity, to be satisfied for the extent of motion considered. It is understood that the chain is always taut, and. that its weight and mass are neg- lected. 145. Numerical Case of the Foregoing. — (Foot-pound-second system of units for space, force, and time ; this 'requires g = 32.2.) Suppose the following data : Feet. Lbs. Lbs. Mass Units. V = 2.0 I" = 4.0 a = 1.5 r = 1.0 k = 1.8 k" = 2.3 r" = 0.1 Pi = Fi = F" (av'ge) = F,= 6000 200 400 300 0' = 60 G" = 50 0'" = 400 0'^ = 3220 (and .-.) M' = 1.86 M" = 1.55 M'" = 12.4' M^^ = 100.0 Also let Fo = 4 ft. per sec; /:/=30'' Denote (1) by TTand the large bracket in (2) by M (this l)y some is called* the total mass ^'- reduced''^ to the crank-pin). Putting (1) = (2) we have, solving for the unknown Vn, K = 2Tf i-v:. (3) For above values, W = 12,000 - 400 - 125.T - Y590.0 - 141 Y.3 = 2467 foot-pounds ; while ^ = 0.5 + 40.3 -f 225.0 = 265.8 mass-units; whence F„ = 4/18.56 + 16 = VsU^ = 5.88 ft. per second. As to whether the crank-pin actually reaches the dead-point n, requires separate investigations to see whether F becomes zero or negative between o and n (a negative value is inad- WORK, ENERGY, AND POWER. 153 inissible, since a reversal of direction Implies a different value for W), i.e., whether the proposed extent of motion is realized ; and these are made by assigning some othei' inter- mediate position 771, as a final one, and computing F^, remem- bering that when m is not a dead-point the (K. E.),^ of M' is not zero, and must be expressed in terms of F^, ;uid that the (K. E.)to of the connecting-rod J/'''^raust be obtained from § 143. 146. Eegulation of Machines. — As already illustrated in several exauiples (§ 121), a fly-wheel of sufficient weight and radius may prevent too great fluctuation of speed in a single •stroke of an engine ; but to prevent a permanent change, which must occur if the work of the working force or forces (such as the steam-pressure on a piston, or water-impulse in a turbine) exceeds for several successive strokes or revolutions the work required to overcome resistances (such as friction, gravity, re- sistance at the teeth of saws, etc., etc.) through their respective spaces, automatic governors are employed to diminish the working force, or the distance tlu-ough wliich it acts per stroke, until the normal speed is restored ; or vice versa, if the speed slackens, as when new resistances are temporarily brought into play. Hence when several successive periods, strokes (or other -cycle), are considered, the kinetic energy of the moving parts will disappear from eq. (XYI.), leaving it in this form : work of' worhing-forces = work done upon resistances. 147. Power of Motors. — In a mill Avhere the same number of machines are run continuously at a constant speed proper for theii- work, turning out per hour tlie same number of barrels of flour, feet of himber, or other coujmodity, the motor (e.g., a steam-engine, or turbine) woi'ks at a constant rate, i.e., de- velops a definite horse-power (H.P.), which is thus found in the case of reciprocating steam-engines (doubie-actingj, II.P. = total mean effective \ l distance in feet ] steam-pressure on I X •< travelled by pis- I ~- 550, piston in lbs. ) ( ton per second. ) i.e., the work (in ft.-lbs) done per second by the working force 164 MECHANICS OF ENGINEERING. divided by 550 (see § 132). The total effective pressure at auj" instant is the excess of the forward over the back-pressure^ and by its mean vakie (since steam is nsnally used expansively) is meant such a vahie^' as, multiplied by the length of stroke- I, shall give P'l=.J Pdx, where P is the variable effective pressure and dx an elemenfc of its path. If u is tlie number of strokes per second, we may also write {foot-jpound- second system) H.P. = P'lu -^ 550 = f Pdx u -^ 550. (XYII.)' Yery often the number of revolutions _^er minute, m, of th&: crank is given, and then H.P. = P' (lbs.) X 2Z (feet) X m ^ 33,000. II P= area of piston we may also write P' ^Pp', where j?' is the mean effective steam-pressure per unit of area. Evi- dently, to obtain P' in lbs., we multiply i^in sq. in. byj?' ia lbs. per sq. in., or P m sq. ft. hj p' in lbs. per sq. foot ; the former is customary, p^ in practice is obtained by measurements and computations from " indicator- cards " (see § 152, p. 159, where the value of P' is found by Simpson's Rule) ; or P'7, i.e., / Pdx, may be computed theoretically as in § 59, Problem 4.. The power as thus found is expended in overcoming the- friction of all moving parts (which is sometimes a large item), and the resistances peculiar to the kind of work done by the ma- chines. The work periodically stored in the increased kinetic energy of the moving masses is restored as they periodically resume their minimum velocities. Example. — In the steady running of a (reciprocating) steam-engine operating a mill, the value of the mean total effective pressure on the piston is P' = 16,800 lbs. and the radius of the circle described by the crank-pin is 10 in. (so that the length of stroke is 1 = 20 inches). The^ fly-wheel turns at rate of 330 revs./min. Find the horse-power developed.. Substituting, we find H.P. = [16,800X2 XffXSSO]-^ 33,000 = 560 H.P. WORK, ENEIJGY, AND POWEU. 155 148. Potential Energy. — There are other ways in which work or energy is stored and then restored, as follows : First. In raising a weight G through a height h, an amount of work = Gh is done ujyon G, as a resistance, and if at any subsequent time the weight is allowed to descend through the same vertical distance Ti (the form of path is of no account), G, now a worMng force, does the work Gh, and thus in aiding the motor repays, or restores, the Gh expended by tlie motor in raising it. If h is the vertical height tlirough which the centre of gravity rises and sinks periodically in the motion of the machine, the force G may be left out of account in i-eckoning the expenditure of the motor's work, and the body when at its liighest point is said to possess an amount Gh of potential energy, i.e., energy of jposition, since it is capable of doing the work Gh in sinking through its vertical range of motion. Second. So far, all bodies considered have been by express stipulation rigid, i.e., incapable of changing shape. To see the effect of a lack of rigidity as affecting the principle of work and energy in machines, -^---.JC^^^^^^^rr^ A take the simple case in Fig. 160. ^ Y^^'^''^^^ ? A helical spring at a given in- ^f---^^;^;^;|^^^^j~^ Btant is acted on at each end by f'^'^'i r ti>?' ^ a force jP in an axial direction ' ' ' / j (they are equal, supposing the Fig. leo. mass of the spring small). As the machine operates of which it is a member, it moves to a new consecutive position J^, suffering a further elongation dX in its length (if F is increas- ing). P on the right, a woi'king force, does the work Pdx'', how is this expended ? ^ on the left has the work Pdx done upon it, and the mass is too small to absorb kinetic energy or to bring its weight into consideration. The remainder, Pd'x' — Pdx = Pdx, is expended in stretching the spring an addi- tional amount dX, and is capable of restoration if the spring retains its elasticity. Hence the work done in changing the form of bodies if they are elastic is said to be stored in the form of potential energy. Tliat is, in the operation of ma- chines, the name potential energy is also given to the energy stored and restored periodically in the changing and regaining of form of elastic bodies. 156 MECHANICS OF ENGINEERING. EKatiif)le. — A given helical spring, when held stretched s'=J ft. beyond its "natural" (or unstrained) length, exerts a pull of i2' = 1200 lbs. at its two ends; and the "potential energy" residing in it is — mean forceX distance, =^R's\ = (i) 1200 (^), =300 ft.-lbs. If such a stretched spring be placed on a car of 644 lbs. weight on a level track and properly connected with a driving-wheel, which does not slip on the track, its recovery of natural length may be made the means of starting the car into motion and causing it to attain a final velocity of v = 5A7 ft. /sec. (if no friction is met with); from i(644-f-32.2)i;2 = 300. 149. Other Forms of Energy. — Numerous experiments witli various kinds of apparatus have proved that for every 7Y8 (about) ft.-lbs. of work spent in overcoming friction, one British unit of heat is produced (viz., tlie quantity of lieat necessary to raise tlie temperature of one pound of water from 32° to 33° Fahrenheit); while from converse experiments, in which the amount of heat used in operating a steam-engine was all carefully estimated, the disappearance of a certain portion of it could only be accounted for by assuming that it liad been converted into work at the same rate of (about) 778 ft.-lbs. of vs^ork to each unit of heat (or 427 kilogrammetres to each French unit of lieat). This number 778 or 427, according to the system of units employed, is called the Mechanical Equivalent of Htot Heat then is energy, and is supposed to be of the kinetic form due to tJie rapid motion or vibration of the molecules of a substance. A similar agitation among the molecules of the (hypothetical) ether diffused through space is supposed to pro- duce the phenomena of light, electricity, and magnetism. Chemical action being also considered a method of transform- ing energy (its possible future occurrence as in the case of coal and oxygen being called potential energy), the well-known doctrine of the Conser nation of Energy^ in accordance with which energy is indestructible, and the doing of work is simply the conversion of one or more kinds of energy into equivalent amounts of others, is now an accepted hypothesis of physics. Work consumed in friction, though practically lost, still re- mains in the universe as heat, electricity, or some other subtile form of energy. 150. Power Required for Individual Machines. Dynamome- ters of Transmission. — If a machine is driven by an endless belt from the main-shaft, A^ Fig. 161, being the driving-pulley WORK, ENERGY, AND POWER. 167 Fig. lUl. on the machine, the working force which drives the machine,, in other words tlie " grip" with which the belt takes hold of the pulley tangentially^ = /^ — P' ^ P and P' being the tensions in the "driving" and ''following" sides of the belt respectively. The belt is supposed not to slip on the pulley. If v is the ve- locity of the pulley -circumference, the work expended on the machine per second, i.e., \\\q lyower, is L = (P-POv, ft.-lbs. per sec. .... (1) To measure the force {P — P')^ an apparatus called a Dy- nainometer of Transmission may be placed between the main shaft and the machine, and the belt made to pass through it in such a way as to measure the tensions P and P' ^ or princi- pally their difference, without meeting any resistance in so do- ing ; that is, the power is transmitted^ not absorbed, by the apparatus. One invention for this purpose (mentioned in the Journal of the FranMin Institute some years ago) is shown {in principle) in Fig. 162. A ver- tical plate carrying four pulleys and a scale-pan is first balanced on the pivot C. The belt being then ad- justed,- as shown, and the power turned on, a sufficient weight G is placed in the scale-pan to balance Fig. 163. tlie plate again, for whose equilib- rium we must have Gh = Pa — P'a, since the P and P' on the right have action-lines passing through C. The velocity of belt, V, is obtained by a simple counting device. Hence (P —P') and V become known, and .'. L from (1). Many other forms of transmission-dynamometers are in use, some applicable whether the machine is driven by belting or gearing from the main shaft. Emerson's Hydrodynamics de- scribes his own invention* on p. 283, and gives results of meas- m-ements with it ; e.g., at Lowell, Mass., the power required to drive 112 looms, weaving 3 6 -inch sheetings, No. 20 yarn, 60 threads to the inch, speed 130 picks to the minute, was found to be 16 H.P., i.e., \- H.P., to each loom (p. 335). * Prof . Flather's '^Dynamometers" is a standard book (1907). 158 MECHANICS OF ENGINEERING. Example. — The endless belt connecting the pulley (running at n=180 revs./min., with a radius of r = 2 ft.) of an engine shaft with that of a planing machine is led over the idle pulleys of the apparatus in Fig. 162, as there shown (engine pulley on left, and that of machine on right; but neither shown in figure). To balance the plate in position shown (with a = 2 ft. and 6 = 4 ft.) is found to require a weight G = 210 lbs. We have, therefore, from {P-P')a = Gb, P-P' = 210X4 --2 = 420 lbs. as the net working force operating the machine; while the velocity of the belt is v, =n2;rr, =(180-^60) 2;r2= 18.85 ft. /sec. Hence the power transmitted through the dynamometer of transmission is L, = {P-P')v, =420X18.85 = 79,170 ft.-lbs. per sec, or 14.4 H.P 151. Dynamometers of Absorption. — These are so named .since they furnish in themselves the resistance (friction or a weight) in tlie overcoming (or raising) of which the power is expended or absorbed. Of these the Prony Friction Brake is the most common, and is nsed for measuring the power developed by a given motor (e.g., a steam-engine or turbine) mot absorbed in the friction of the motor itself. Fig:. 163 «hows one fitted to a vertical pulley driven by the motor. By tightening the bolt B^ the velocity i) of pulley-rim may be made constant at any desired value (within certain limits) by the consequent friction, -y is measni-ed by a counting appara- tus, while the friction (or tangential components of action be- tween pulley and brake), = I\ becomes known by noting the weight G which must be placed in the scale pan to balance the arm between the checks; then with G^''= weight of brake and h' =tlie horizontal distance of its center of gravity from the center of pulley, we have, for the equilibrium of the brake (moments about pulley center), Fa=Gb + GV', (1) and the work done on F per unit of time, or power, is L=i^y, ft.-lbs. per sec (2) (In case the pulley is horizontal, a bell-crank must be inter- posed between the arm and the scale-pan.) WORK, ENERGY, AND POWER. 159 Example. — A vertical pulley of a = 2 ft. radius and run by a turbine water-wheel, is gripped by a Prony brake, as in Fig. 163, with arm fe = 4 ft. 9 in. A load of G = lQO lbs. is placed in the scale pan, the water turned on, and the bolt B screwed up until the friction F of pulley-rim on brake is just sufficient to lift the weight and hold the brake in equi- librium Weight of brake is (r' = 40 lbs., with centre of gravity 6' = 1.5 ft. on right of pulley centre. The speed to which the pulley now adjusts itself is at rate of 210 revs./min. The friction is F ={Gb + Gb')^a = (160X4.75+40X1.5)-h2 = 410 lbs.; the velocity of pulley-rim is v = (210^60) 27rX2 = 44 ft. /sec; hence the power developed is F'?; = 410X44 = 18,040 ft.-lbs. per sec. ; or 32.8 H.P. Note. — For an account of various modern designs of absorption and transmission dynamometers, the reader is referred to Prof. Flather's book, already mentioned in the foot-note on p. 157. This is a recent and a standard work. In the "Notes and Examples in Mechanics" by the present writer, brief descriptions are given (pp. 96 and 97) of the Appold and the Carpentier dynamometers of absorption, with the theory of the same; both of these are "automatic" or "self-adjusting." It must be carefully noted by the student that in the absorption dyna- mometer, which for purposes of test temporarily takes the place of useful machines, the power is absorbed and converted into heat, necessitating cooling devices if the parts are to run smoothly and lubricants are to remain unaffected; whereas in the dynamometer of transmission the power simply passes through without heating effect. 152, The Indicator, used with steam and other fluid engines, is a special kind of dynamometer in which the automatic mo- tion of a pencil describes a curve on paper whose ordinates are proportional to the fluid pres- sures exerted in the cylinder at successive points of the stroke. Thus, Fig. 164, the back-pres- sure being constant and = P^, fig. lu. the ordinates P^, P^, etc.. represent the effective pressures at equally spaced points of division. The mean effective pressure P' (see § 147) is, for this figure, by Simpson's Rule (six equal spaces), P' = tV[^o + 4(P, + P3 + P.) + 2(P. + P.) + Pe]. This gives a near approximation. The power is now found by §147^ 153. Mechanical Efficiency of a Steam or Vapor Engine (gas, petroleum, gasoline, or alcohol vapor, etc.). By the term *' mechanical efficiency " is meant the ratio of the power obtained < -- l ! ^\/ fk< P1 P2 P3 P4 Pa T 1 Pe l^b 2ERC 3 LINE X 160 MECHANICS OF ENGINEERING. at tlie rim of tlie pnlley or fly-wheel on the main shaft of the engine (where it would be connected with machinery to be operated or where in a test the resistance of brake -friction wonld be overcome) to the power exerted directly on the piston of the engine by the pressure of the fluid concerned. This latter item becomes known through the use of the ' ' indicator ' ' (see preceding paragraph) and is hence often called the 'indi- cated 'power ; " the power spent on friction provided by a Prony brake, for testing purposes, being called the " brake-power.^^ Example. — If from indicator-cards the value of P', or total mean effective pressure on the piston of an engine, is found to be 12,000 lbs., the piston speed being at the (mean) rate of 6 ft. per sec, the "indi- cated power" of the engine is = 72,000 ft.-lbs./sec. Now, when the engine is running under these same conditions of pressure and speed, if it is found by the use of a Prony friction brake that the power spent on brake friction consists of overcoming a friction of 6000 lbs. through 10 ft. each second, and that therefore the power obtained at the brake, or "brake power," is equal only to 60,000 ft.-lbs./sec, the mechanical efficiency of the engine (in this test) is 60,000-^72,000, =0.833 or 83 J per cent. In other words, 16t per cent of the power exerted by the fluid pressure on the piston, or "indicated power," is lost in the overcoming of the friction of the engine itself, i.e., among the moving parts situated between the piston and the rim of the test pulley. 153a. Efficiency of Power Transmission. — In transmitting power through a long line of shaftmg, or by ropes or belts, or water in pipes, or by electric current, the efficiency is the ratio of the power put in at the sending station to that obtained at at the receiving station. For example, Example. — An engine exerts power at the rateof (say) 600,000 ft.-lbs./sec, in running a "dynamo" at the sending or power station. The electric current so generated is conducted 60 miles through wires to a receiving station, where by operating an electric motor it enables a pulley to be run within a Prony brake from whose indications it is found that a power of 360,000 ft.-lbs./sec. is there obtainable. Hence the efficiency of transmission is 360,000-^600,000, =60 per cent. 154. Boat-Rowing. —x^'ig. 166. During the stroke proper, let /* = mean pressure on one oar-handle ; hence the pressures on the foot-rest are 2P, resistances. Let J!f"= mass of boat and load, v^ and Vn its velocities at beginning and end of stroke. Pj = pressures between oar-blade and water. Ji = mean re- sistance of water to the boat's passage at this (mean) speed. These are the only (horizontal) forces to be considered as act- ing on the boat and two oars, considered free collectively. During the stroke the boat describes the space s^ = CD, the oar-liandle tlie space s„ = AB, wliile tlie oar-blade slips back- WORK, ENERGY, AND POWER. 161 ward through the small space (the " slip") = s^ (average). Hence by eq. (XYI.), § 142, i.e., 2P{s,-s,)=2FxAJi=2Fs =Bs,+2P,s,-\- iMiv^'-v,'); or, in words, the product of the oar-handle pressures into the distance described by them measured on the hoat, i.e., the work- done by these pressures relatively to the hoat, is entirely ac- counted for in the work of slip and of liquid-resistance, and in^ ■■..si>^ Fig. 166. creasing the kinetic energy of the mass. (The useless work due to slip is inevitable in all paddle or screw propulsion, as well as a certain amount lost in machine-friction, not considered in the present problem.) During the '' recover" the velocity decreases again to v^. (See example on p. 9T, Notes, etc.) 155. Examples. — 1. What work is done* on a level traci^, in bringing up the velocity of a train weighing 200 tons, from zero to 30 miles per hour, if the total frictional resistance (at any velocity, say) is 10 lbs. per ton, and if the change of speed is accomplished in a length of 3000 feet ? {Foot-ton-second system.) 30 miles per hour = 44 ft. per sec. The mass - 200 -^ 32.2 = 6.2 ; .'. the change in kinetic energy, (= Wv-' -iM X 0% = i(6.2) X 44* = 6001.6 ft.-tons. * That is, what work is done by the pull, or tension, P, in the draw-bar between the locomotive and the "tender." 162 MECHANICS OF ENGlNEEKliMG. The work done in overcoming friction = Fs, i.e., = 200 X 10 X 3000 = 6,000,000 ft.-lbs. = 3000 ft.-tons ; .-. total work = 6001.6 + 3000 = 9001.6 ft.-tons. (If the track were an up-grade, 1 in 100 say, the item of 200 X 30 = 6000 ft.-tons would be added.) Exmnjple 2. — Required the rate of work, or power, in Ex- ample 1. The power is variable, depending on the velocity of the train at any instant. Assume the motion to be uniformly accelerated, then the working force is constant ; call it P. The acceleration (§ 56) will be ^='y'-=- 2^=1936-^6000 = 0.322 ft. per sq. sec; and since P — J^= Mjp^ we have P = 1 ton + (200 -=- 32.2) X 0.322 = 3 tons, which is 6000 -^ 200 = 30 lbs. per ton of train, of which 20 is due to its inertia, since when the speed becomes uniform the work of the engine is expended on friction alone. Hence when the velocity is 44 ft. per sec, the engine is working at the rate of P'o = 264,000 ft.-lbs. per sec, i.e., at the rate of 480 H. P.; At i of 3000 ft. from the start, at the rate of 240 H. P., half as much ; At a uniform speed of 30 miles an hour the power would be simply 1 X 44 = 44 ft. -tons per sec. = 160 H. P. Example 3. — The resistance offered by still water to the passage of a certain steamer at 10 knots an hour is 15,000 lbs. What power must be developed by its engines, at this uniform speed, considering no loss in " slip" nor in friction of ma- chinery ? A71S. 461 H. P. Example 4, — Same as 3, except that the speed is to be 15 knots (i.e., nautical miles ; each = 6086 feet) an hour, assum- ing that the resistances are as the square of the speed (approxi- mately true). Ans. 1556 H. P. Example 5. — Same as 3, except that 12^ of the power is ab- soi-bed in the " slip" (i.e., in pushing aside and backwards the water acted on by the screw or paddle), and 8^ in friction of machinery. Ans. 576 H. P. Example 6. — In Example 3, if the crank-shaft makes 60 "WORK, ENERGY, AND POWEE. 163 revolutions per minute, the crank-pin describing a circle of 15 incbes radius, required the average* value of the tangential component of the thrust (or pull) of the connecting-rod against the crank-pin. Ans. 26890 lbs. Example 7. — A solid sphere of cast-iron is rolling up an in- cline of 30°, and at a certain instant its centre has a velocity of 36 inches per second. Neglecting friction of all kinds, how much further will the ball mount the incline (see § 143) % Ans. 0.390 ft. \ Example 8.— In Fig. 163, with J = 4 f t. and a = 16 inches, it is found in. one experiment that the friction which keeps the speed of the pulley at 120 revolutions per minute is balanced by a weight G — 160 lbs. Eequired the power thus measured.. Ans. 14.6 H. R Although in Examples 1 to 6 the steam cylinder is itself in motion, the work per stroke is still ■=■ mean effective steam- pressure on piston X length of stroke, for this is the final form to which the separate amounts of work done by, or upon, the two cylinder heads and the two sides of the piston will re- duce, when added algebraically. See § 154. * By " average value" is meant such a value, Tm, as multiplied into the length of path described by the crank-pin per unit of time shall give the power exerted. 164 MECHANICS OF EWGINEEKING. CHAPTEK YIL FRICTION. 156. Sliding Friction. — When the surfaces of contact of two bodies are perfectly smooth, the direction of the pressure or pair of forces between them is normal to these surfaces, i.e., to their ; tangent-plane ; but when thej are rough, and 'Y"'f \ moving one on the other, the forces or ac- pV 4N tions between them incline awaj from the i \ I ij -, P- normal, each on the side opposite to the di- WmJ^ S^ W//Jw///M ^^ction of the (relative) motion of the body /-. ^q/ ///M ^^ which it acts. Thus, Fig. 167, a block Fig. 167. whose weight is G, is drawn on a rough horizontal table by a horizontal cord, the tension in which is P. On account of the roughness of one or both bodies the ac- tion of the table upon the block is a force P^^ inclined to the normal (which is vertical in this case) at an angle = (p away from the direction of the relative velocity -y. This angle q) is called the angle of friction^ while the tangential component of P^ is called the friction = F. The normal component N^ which in this case is equal and opposite to G the weight of tlie body, is called the normal pressure. Obviously i^= iV^tan <p, and denoting tan cp hjf we have F=fJ^. (1) /"is called the coefficient of friction, and may also be defined as the ratio of the friction ^to the normal pressure JSf which produces it. In Fig. 167, if the motion is accelerated (ace. =J?), we have (eq. (lY.), § 55) P - i^ = J!/^ ; if uniform, P - F= ; from vv'hich equations (see also (I))/" may be computed. In the latter case/" may be found to be different with different veloci- ties (the surfaces retaining the same character of course), and then a uniformly accelerated motion is impossible unless P — P were constant. As for the lower block or table, forces the equals and op- posites of iV^andP(or a single force equal and opposite to P^) are comprised in the system of forces actirg upon it. As to whether i^ is a worMng force or a resistance, when FRICTION. 166 either of the two bodies is considered free, depends on the cir- cumstances of its motion. For example, in friction-gearing the tangential action between the two pulleys is a resistance for one, a working force for the other. If the force P^ Fig. 167, is just sufficient to start the body, or is just on the point of starting it (this will be called impending fnotion), F\& called ihe friction of rest. If the body is at rest and P is not sufficient to start it, the tangential component will then be < the friction of rest, viz., just =^ P. As P increases, this component continually equals it in value, and P^^ acquires a. direction more and more inclined from the normal, until the instant of impending motion, when the tangential component =/'-ZV"= the friction of rest. When motion is once in prog- ress, the friction, called then the friction of motion., = fJV, in which/" is not necessarily the same as in the friction of rest. 157. Variation of Friction and of tlie Coefficient of Friction, f. — Careful distinction must be made between the friction of dry surfaces and of those that are lubricated; though in the latter case as the supply of lubri- cant (oil, soap, graphite, etc.) is reduced from the extreme state of "flooding," the friction approximates in variation and magnitude to that of dry surfaces. Also, if the pressure is very great, the lubri- cant may be pressed out and the phenomena reduced to those of dry surfaces, which imder great pressures "seize," i.e., abrade, one another. With dry surfaces the amount of friction, F (lbs.), depends on the nature of the materials and their initial roughness, being somewhat reduced as they become more polished, when a sliding motion has been long continued. With the surfaces in a given condition it is found (unless the pressure is very low) that increase of velocity diminishes the friction, as is unfortunately the case with railroad brakes, the friction between a brake-shoe and the rim of the car-wheel being least ^t the first application of the brakes, when the velocity of rubbing is greatest (see p. 168). The friction increases with the normal pressure N (the coefficient /, itself, increasing with N when N is large) and is some- what smaller after motion begins than when motion is "impending" (friction of rest). With well lubricated surfaces, however, the following may be said: The nature of the materials of the two bodies has but slight influence on the amount of friction, F, and when motion has begun, the friction is very much less than that of "impending motion." The friction is practically independent of the pressure when the lubrication is very ■copious (bearings "flooded") (resembling, therefore, "fluid friction;" see p. 695), the coefficient / being as small as 0.001 or under (Tower); but with more scanty lubrication conditions may approach those of dry surfaces. As to the effect of velocity (Goodman), the frictional resistance 166 MECHANICS OF ENGINEERING. varies directly as the speed for low pressures. For high pressures, how- ever, it is relatively great at low velocities, a minimum at about 100 ft./min., and afterwards increases approximately as the square root of the speed. A rise of temperature has a very important influence in diminishing the viscosity of the oil and enlarging the diameter of the bearing of a shaft more than that of the shaft itself. In the problems of this chapter the coefficient / will be considered as constant; so that where it really varies (as when the velocity changes) an average value will be understood. 158. Experiments on Sliding Friction. — These may be made with simple aj)pai'atus. If a block of weight = G, Fig. 168, be placed on an inclined plane of uniformly rough surface, and the latter be gradually more and more inclined from the horizontal until the block begins to move,, the value of fS at Fig. 168. Fig. 169. this instant =; cp, and tan cp ■= f ■=^ coefficient of friction of rest. For from -2'X = we have F, \.%.^ fN^ = G sin ySj from ^1^= 0, iV^=: G cos fi ; whence tan /? =y, .-, /? must = cp. Suppose /? so great that the motion is accelerated, the body starting from rest at o, Fig. 169. If it is found that the distance x varies as the square of the time, then (§ 56) the motion is uniformly accelerated (along the axis X). (Notice in the figure that G is no longer equal and opposite to Pi, the^ resultant of N and P, as in Fig. 168.) We have, then lY = 0, which gives N- (? cos ^= ; JZ = Mpi, whicli gives G sin /?- fN = {G ■^g)pi ; while (from § 56) pi = '2x^t^. Hence, by elimination, x and the corresponding time t having been observed, we have for the coefficient of friction of motion, 2x ' '^ gt^ cos /? as an average (since the acceleration may not be uniform). FRICTION. 167 In view of (3), § 157, it is evident that if a value /3^ has been found experimentally for /? such that the block, once started hy hand, preserves a uniform motion down the plane, then, since tan /3^ =.f for friction of motion, /?^ may be less than the /? in Fig. 168, for friction of rest. 159. Another apparatus consists of a horizontal plane, apul- lej^, cord, and two weights, as shown in Fig. 59. The masses of the cord and pulley being small and hence neglected, the analysis of the problem when G is so large as to cause an ac- celerated motion is the same as in that example [(2) in § 57], except in Fig. 60, where the frictional resistance yW^ should be put in pointing toward the left. iT still = G^^ and .*. 8-fG^^{G,^g):p', (1) while for the otlier free body in Fig. 61 we have, as before, G-8={G-^g):p (2). From (1) and (2), 8 the cord-tension can be eliminated, and solving for p, writing it equal to 2^ -^ f, s and t being the ob-^ served distance described (from rest) and corresponding time,, we have finally for friction of motion (as an average) '^- G, - G, ' gf ^^> If G, Fig. 59, is made just sufficient to start the block, or sledge, G^, we have for the friction of rest . /=|.. ....... ii) 160. Results of Experiments on Sliding Friction. — For accounts of recent experiments (and others) and deductions therefrom, the reader may consult the Engineer's "Pocket-books" of Kent and Trautwine; also Thurston's "Friction and Lost Work," Barr and Kimball's "Machine Design," and " Lubrication and Lubricants, ' ' by Archbutt and Deely. The following table gives a few values for the coefficient of friction, /, for slow motion, taken from the results obtained by Morin and others. Small changes in the condition of the surfaces may produce considerable variation in the value of /. Our knowledge is still quite incomplete in this respect. 168 MECHANICS OF ENGINEEKING. TABLE FOR FRICTION OF SLOW MOTION. No. Surfaces. Unguent. Angle <t). / = tan ^. 1 Wood oil wood. None. 14° to 36i° 0.25 to 0.5U 2 Wood on wood. iSoap. 2° to nr 0.04 to 0.20 3 Mcilal on wood. None. 26r to 3ir 0.50 lo 0.60 4 Metal on wood. Water. 15° to 20° 0.25 10 0.35 5 Metal on wood. Soap. 1U° 0.20 6 Leather on metal. None. 29;5r° 0.56 7 Leather on metal. Greased. 13° 0.23 8 Leather on metal. Water. 20° 0.36 9 Leather on metal. Oil. sr 0.15 10 Metal on metal Mone 8*° to 18° 0.15 to 0.30 11 Metal on metal Water 18° (average) 0.30 12 Metal on metal Oil 0.001 to 0.080 For the coefficient of friction of rest, the above values might be in- creased from 20 to 40 per cent., roughly spealdng. As showing the effect of velocity in diminishing the friction of dry surfaces, we may note that in the Galton-Westinghouse experiments with railroad brakes (cast-iron brake-shoes on steel-tired car- wheels), values for / were found as follows: When the velocity of rubbing was 10 miles per hour, / = 0.24; for 20 miles per hour, / = 0.19; for 30, 40, 50, and 60 miles per hour / was found to be 0.164, 0.14, 0.116, and 0.074, respectively. The foregoing values of / were obtained imme- diately on the application of the brake, but when the brake-shoe and wheel had been in contact some five seconds at a constant velocity, / was reduced some 20 or 30 per cent. ; while for a contact of 15 seconds still further reduction ensued. The value of / for a "skidding" car- wheel (i.e., held fast by the brake pressure) sliding or "skidding" on the rail, was reduced from 0.25 for impending skidding, to 0.09 at a velocity of 7 miles per hour; and to 0.03 for 60 miles per hour. (See p. 190.) That increasing the velocity of lubricated surfaces diminishes the co- efficient of friction is well shown in the results obtained by Mr. Welling- ton, in 1884, with journals revolving at different speeds, viz., For velocity = 0.00 2.16 4.86 21.42 53.01 106ft /min. Coeff. / =0.118 0.094 0.069 0.047 0.035 0.026 For a sledge on dry ground Morin found / = 0.66. For stone on stone, see p. 555 of this book. 161. Cone of Friction. — Fig. 170. Let A and B be two a-ougli blocks, of which jB is immovable, and P the resultant ■■.-.. 0/ of all the forces acting on A. except the pres- sure from £. jB can furnish any required normal pressure JV to balance P cos /?, but the limit of its tangential resistatice is /"iV^. So long then as /? is < cp the angle of fric- tion, or in other words, so long as the line of Fig. 170. action of P is within the " cone offriGtion" FRICTIOIT. 169 generated by revolving 6^6* about ON^ the block A Mall not slip on B^ and the tangential resistance of B is simplj P sin /? ; but if ^ is > cp, this tangential resistance being oiiij fN and < P sin /?, A will begin to slip, with an acceleration. 162. Problems in Sliding Friction. ^ — In the following prob- lemsy is supposed known at points where rubbing occurs, or is impending. As to the- pressure iV^ to which the friction is due, it is generally to be considered unknown until determined by the conditions of the problem. Sometimes it may be an advantage to deal with the single unknown force P\ (resultant of N 'Aw^fN^ acting in a line making the known angle 9? with the normal (on the side away from the motion). Problem 1. — Keqnired the value of the weight P^ Fig. lYl, the slightest addition to which will cause motion of the hori- zontal rod OB, resting on rough planes at 45°. The weight G of the rod may be applied at the ys^^y middle. Consider the rod free ; at €ach point of contact there is an un- known JSf and a friction due to it fN\ the tension in the cord will be = P, since there is no acceleration and no friction at pulley. Notice fig. 171. the direction of the frictions, both opposing the impending motion. Take axes OX and OF as shown, and note the inter- sections, A and C, of the line of G with axes OX. and OY . The student should not rush to the conclusion that, if G were transferred to A and there resolved into components along AO and AB, the value of 'N (and A^i) would be equal to that of one of these components, viz., mG (where m denotes sin 45°). Few problems in mechanics are so simple as to admit of an immediate mental solution, and guess-work should be care- fully avoided. It will be found of advantage to take C as a center of moments. The line of P at is considered as passing through point C , as also the lines of /AT and /iVi- For equilibrium (impending slipping) we have, therefore, i-Z^O; i.e.,/iVi+mG-iV-P = 0; . . . (1) 27 = 0; i.e., i\ri + /Ar-mG = 0; .... (2) i'(moms.)c=0; i.e., iVa-A^'ia = 0. ... (3) 170 MECHANICS OF ENGINEERING. The three unknowns P, N, and Ni, can now be founds From (3) we have N = Ni, which in (2) gives A^ = t-71- Also from (1) we now find P=^^jN; and hence finally P 1 +/"!+/■ G. Peoblem 2. — Fig. 1Y2. A rod, centre of gravity at middle,., leans against a rough wall, and rests on an equally rough floor; how small may the angle a become before it slips ? Let a = the half-length. The figure- p shows the rod free, and following the sugges- P tion of § 162, a single unknown force ^, p making a known angle (p (whose tan =/") P with the normal D^, is put in at D, leaning^ away from the direction of the impending motion, instead of an JV and /"iV; similarly 7^2 acts at C. The present system consisting^ of but three forces, the most direct method of finding or, with- out introducing the other two unknowns jP^ and P*^ at all, is to use the principle that if three forces balance, their lines of action must intersect in a point. That is, P*^ must inter- sect the vertical containing G, the weight, in the same point as Pi, viz., A. ISTow, since CF is 1 to FD and the two angles <f> are equal, CA is T to DA and therefore DAC is a right triangle. "We also note that if CA be prolonged to meet DF in some point K, A must be the mid-point of CK, since B is the mid-point of CF ; and therefore it follows that in triangle DCK not only does DA bisect the side KC but is 1 to it. In other words, KDC is an isosceles triangle, with DK and DC as the two equal sides, and therefore the line DA bisects the angle KDC. Hence the angle KDC = '2(p. That is to say, the angle a, which was to be determined, is the complement of double the friction-angle, or a: = 90°- 20. PKICTION. 171 Pkoblem 3. — Fig. 173. Given the resistance Q, acting parallel to tlie fixed guide C, the angle a, and tlie (equal) co- efficients of friction at the rubbing surfaces, required the Fig. 173. amount of the horizontal force P, at the head of the block A (or wedge), to overcome Q and the frictions. D is fixed, and ah is perpendicular to cd. Here we have four unknowns, viz., JP, and the three pressures iV^, JV^, and JV^, between the blocks. Consider ^ and 5 as free bodies, separately (see Fig. 174), re- membering l^ewton's law of action and reaction. The full values {e.g.,f]V) of the frictions are put in, since we suppose .a slow uniform motion taking place. For A, 2X= and :S r = give i^i — iVcos a -|-y^sin a — J^ sin a = ; . f¥^ + N sin a -[-/iTcos a — P cos a = 0. . 'FoyB, :SXand ^Zgive 'Q-JSr,+fJV, = 0;....iS) and W,-fJ^, = 0. Solve (4) for JV^ and substitute in (3), whence j^.o^-r) = Q (1) (2) m (5) Solve (2) for JV, substitute the result m (1), as also the value of i\r^ from (5), and the resulting equation contains but one un- known, P. Solving for P, putting for brevity ycos a-\- sin a ■we have P = m and cos a — /"sin a = n, {w,-\-fn)Q or {n . cos a -\- m . sin a)(l — yy {T} 172 MECHANICS OF ENGINEERING. Numerical Examjple of Problem 3. — If Q = 120 lbs., /" = 0.20 {an abstract number, and .*. the same in any system of units), while a = 14°, whose sine = 0.240 and cosine = .970,, then m = 0.2 X. 97 + 0.24 = 0.43 and t^ = .97 — .2X.24 = 0.92,. whence P = OMQ = 76.8 lbs. While the wedge moves 2 inches P does the work (or exerts an energy) of 2 X 76.80 = 153.6 in.-lbs. = 12.8 ft.-lbs. For a distance of 2 inches described by the wedge horizon- tally, the block P (and .•. the resistance Q) has been moved through a distance = 2 X sin 14° = 0.48 in. along the guide- G, and hence the work of 120 X 0.48 = 57.6 in.-lbs. has been done upon Q. Therefore for the supposed portion of the motion 153.6 — 57.6 = 96.0 in.-lbs. of work has been lost in friction (converted into heat). For the "mechanical efficiency" of this machine (see § 153) we have 57.6-153.6=0.375. Also note that for / = 1.0(> P = '^ ; andfor a: = 90°, P = Q. Problem 4. Numerical. — With what minimum pressure P should the pulley A be held against P, which it drives by ■^ n x riA " frictioiial gearing," to transmit 2 H. p.;. ^ - p -(--- ^ > if a = 45°, f for impending (relative) «y 7>! ^ motion, i.e., for impending slipping = Fig. 175. 0.40, and the velocity of the pulley-rim; is 9 ft. per second ? The limit-value of the tangential " grip" T = 2/i\^= 2 X 0.40 X P sin 45°, 2 H. P. = 2 X 550 = 1100 ft.-lbs. per second. Putting T X 9 ft. = 1100, we have* 2 X 0.40 X 4/5 X P X 9 = 1100 ; .: P = 215 lbs.. Problem 6. — A block of weight G lies on a rough plane, inclined an angle ^ from the horizontal ; find the pull P, mak- ing an angle a with the first plane, which will maintain a uni- form motion up the plane. * In this problem the student should note that, in general, when a is not 45°, we have N = iP -v- cos a (since in such a case the parallelogram of. forces i& not a square). FRICTION. 173 Pkoblem 7. — Same as 6, except that the pull P is to permit a uniform motion down the plane. Pkoblem 8. — The thrust of a screw-propeller is 15 tons. Tlie ring against which it is exerted has a mean radius of 8 inches, the shaft makes ©ne revolution per second, andy = 0.06. Required the H. P. lost in friction from this cause. Ans. 13.7 H. P. 163. The Bent-Lever with Friction. Worn Bearing. — Fig. 176. Neglect the weight of the lever, and suppose the plumb- er-block so worn that there is contact along one element only of the shaft. Given the amount and line of action of the resistance R^ and the line of action of jP, re- quired the amount of the latter for impending slipping in the direction of the dotted arrow. As P grad- ually increases, the shaft of the lever (or gear-wheel) rolls on its fig. 176. bearing until the line of contact has reached some position Ay when rolling ceases and slipping begins. To find A^ and the value of P^ note that the total action of the bearing upon the lever is some force ^,, applied at A and making a known angle q) (^f =: tan q)) with the normal A C. P^ must be eqnal and opposite to the resultant of the known P and the unknown P, and hence graphically (a graphic is much simpler here than an analytical solution) if we describe about C 2i circle of radius = r sin 9?, r being the radius of shaft (or gudgeon), and draw a tangent to it from P, we determine PA as the line of action of P^. If PG is made = P, to scale, and ^^ drawn parallel to P . . . P, P is determined, being = PP, while P^ = PK If the known force P is capable of acting as a working force, by drawing the other tangent PP from P to the " friction- circle," we have P = PP, and P^ = PK, for impending rotation in an opposite direction. If P and P are the tooth-pressures upon two spur-wheels, keyed upon the same shaft and nearly in the same plane, the 174 MECHANICS OF ENGINEERING. y = [Pj sin (p]27rr. same constructions hold good, and for a continuous uniform motion, since the friction = P, sin cp, the work lost in friction per revolution, It is to be remarked, that without friction Pj would pass through 0, and that the moments of i? and I^ would balance about C (for rest or uniform rotation) ; whereas with friction thej balance about the proper tangent-point of the friction- circle. Another way of stating this is as follows : So long as the resultant of I^ and P falls within the " dead-angle" BDA, motion is impossible in either direction. If the weight of the lever is considered, the resultant of it and the force M can be substituted for the latter in the fore- going. 164. Bent-Lever with Friction. Triangular Bearing. — Like the preceding, the gudgeon is much exaggerated in the figure (1Y7). For impending rotation in direction of the force P, the total actions at A^ and A^ must lie in known directions, making angles = cp with the respective normals, and in- clined away from tlie shpping. Join the intersections D and L. Since the resultant of P and R i^i D must act along PL to balance that of P^ and P^^ having given one force, say Fig. 177. B, wc easily iind P = PE, wliile P^ and P^ = ZJf^and ZiV respectively, LO having been made = PP, and the parallelogram completed. (If the direction of impending rotation is reversed, the change in the construction is obvious.) If P^ = 0, the case reduces to that in Fig. 176 ; if the construction gives P^ negative, the supposed contact at A^ is not realized, and the angle A^OA^ should be increased, or shifted, until P^ is positive. As before, P and P may be the tooth-pressures on two FRICTION. 175 spur-wheels nearly in tiie same plane and on the same shaft ; if so, then, for a uniform rotation. "Work lost in fric. per revol. = [P^ sin cp -\- P^ sin cp\^7tr. 165. Axle-Friction. — Tiie two foregoing articles are intro- ductory to the subject of axle -friction. When the bearing is new, or nearly so, the elements of the axle which are in contact with the bearing are iniinite in number, thus giving an infinite number of unknown forces similar to P^ and P^ of the last paragraph, each making an angle cp with its normal. Refined theories as to the law ox distribution of these pressures are of little use, considering tne uncertainties as to the value of y ( ^ tan <p) ; hence tor practical purposes axle-friction may be written in which f is a coejficieni of axle-friction derivable from experiments with axles, and JR the resultant pressure on the bearing. In some cases Jti may be partly due to the tightness of the bolts with which the cap of the bearing is fastened. As before, the work lost in overcoming axle-friction j)6r revolution is ■=.fR'^7tr^ in wliich r is the radius of the axle. /'', like y, is an abstract number. As in Fig. 176, a " friction' circle," of radius =fr, majr be considered as subtending the " dead-angle." 166. Experiments with Axle-Friction. — Prominent among recent experiments liave been those of Professor Thurston (1872-73), who invented a special instrument for that purpose, shown (in princi- ple only) in Fig. 178. By means of an internal spring, tne amount of whose compression is reaa on a scale, a weighted bar or pendulum i5 oaused to exert pressure on a projecting axle from which it is suspended. Tlie axle is made to rotate at any desired velocity by some source of power, the axle-friction causing 176 MECHANICS OF ENGINEERING. the pendulum to remain at rest at some angle of deviation from the vertical. The figure shows the pendulum free, the action of gravity upon it being (r, that of the axle consisting of the two pressures,* each = i?, and of the two frictions (each being F =^f R\ due to them. Taking moments about (7, we have for equilibrium ^f'Rr = Gh, in which all the quantities except jT are known or observed. The temperature of the bearing is also noted, with reference to its effect on the lubricant employed. Thus the instrument covers a wide range of relations. General Morin's experiments as interpreted by Weisbach give the following practical results: (See also p. 192). 0.054 for well-sustained lubrication ; 0.07 to .08 for ordinary lubrication. For iron axles, in iron or brass bearings /' = By "pressure per square inch on the bearing" is commonly meant the quotient of the total pressure in lbs. by the area in square inches obtained by multiplying the width of the axle by the length of bearing (this length is quite commonly four times, the diameter) ; call it j?, and the velocity of rubbing m feet per minute, v. Then, according to Rankine, to prevent overheat- ing, we should have p{v + 20) < 44800 . . . (not homog.). Still, in marine-engine bearings pv alone often reaches 60,000^ as also in some locomotives (Cotterill). Good practice keeps P within the limit of 800 (lbs. per sq. in.) for other metals than steel (Thurston), for which 1200 is sometimes allowed. With ^ = 200 (feet per min.) Professor Thurston found that for ordinary lubricants p should 2iot exceed values ranging, from 30 to 75 (lbs. per sq. in.). The product pv is obviously proportional to the power ex- pended in wearing the rubbing surfaces, per unit of area. * The weight O being small compared with the compressive force Hia. the spring, each pressur*^ is practically equal to B. FRICTION. 177 167. Friction-Wheels — (Or, rather, anti-friction wheels). In Fig. 179, M and M (and two more behind) are the "fric- tion-wheels," with axles Ci and C\ in fixed bearings. G is the weight of a heavy wheel, Pi is a known vertical resistance (tootli-pressnre), and P an unknown vertical working force, whose value is to be determined to maintain a uniform rotation. The utility of the friction-wheels is also to be shown. The resultant of P^, G, and J-* is a vertical force P, pass- ing nearly through the centre C of the main axle which rolls on the four friction-wheels. J?, resolved along €A and CB, produces (nearly) equal pressures, each being J^ =: P -r- 2 cos (x, at the two axles of the friction- wheels, which rub against their fixed plumber- blocks. P ^ P -\- P^-\- 6^„ and .*. contains the unknown P, but approximately ■= G-{- 2P^, i.e., is nearly the same (in this case) whether friction-wheels are employed or not. When G makes one revolution, the friction /'''iV^ at each axle C^ is overcome through a distance = (r, : a^) 27rr, and Work lost per revol. \ Fig. 179. T T \ " n. n. cc «, COS oc fP^Ttr. with friction-wheels, Whereas, if C revolved in a fixed bearing, Work lost per revol. ) without V =f'P'i7tr. friction-wheels, ) Apparentl)^, then, there is a saving of work in the ratio r^ : a, cos o', but strictly the P is not quite the same in the two cases ; for with friction-wheels the force P is less than without, and P depends on P as well as on the known G and P^. By dimin- ishing the ratio r^ : a^^ and the angle or, the saving is increased. If a were so large that cos or < r, : a^, there would be no saving, but the reverse. As to the value of P to maintain uniform rotation, we have 12 178 MECHANICS OF ENGINEERING. foi' equilibrium of moments about (7, with fri-ction-wlieels (con^ sidering the large wheel and axleyV'ee), P5 = PA + 21>, ....... (1) in which T is the tangential action, or "grip," between one pair of friction-wheels and the axle C which rolls upon them. ^ would noL equal yiV unless slipping took place or were im- ])ending at E^ but is known bj considering a pair of friction- wheels free, when ^ (-P«) about C^ gives 2 ' cos «' which in (1) gives finally b T T P = iP,^ ^-f'R. (2) ' ' »! cos a*^ ^ ' Without friction-wheels, we would have P^\p,^fR\ : . (3) The last term in (2) is seen to be less than that in (3) (unless a is too large), in the same ratio as already found for the saving of work, supposing the jS's equal. If P^ were on the same side of C as P^ it would be of an opposite direction, and the pressure i? would be diminished. Again, if P were horizontal, R would not be vertical, and the friction-wheel axles would not bear equal pressures. Since P depends on Pj, G^ and the frictions^ while the friction depends on R^ and R on P^^ G, and P, an exact analysis is quite complex, and is not warranted by its practical utility. Example. — If an empty vertical water-wheel weighs 25,000 lbs., required the force P to be applied at its circumference to maintain a uniform motion, with « = 15 ft., and r = 5 inches. Here P^ = 0, and R = G (nearly ; neglecting the influence of P on R), i.e., R = 25,000 lbs. Eirst, iDiihout friction-wheels (adopting the foot-pound-sec- ond system of units), withy =: .07 (abstract number). Frona eq. (3) we have P =: + 0.07 X 25,000 X (tV "^ 1^) = 48.6 lbs. FRICTION, 179 The work lost in friction per revolution is f'B^Tir = O.or X 25,000 X 2 X 3.14 X A = ^580 ft.-lbs. Secondly, with friction-wheels, in which r^ '. a^ =: ^ and cos a = 0.80 (i.e., a = 36°). From eq. (2) J> = 0-^^.\^X 48.6 = only 12.15 lbs., while the work lost per revolution = 1. . jMi X 4580 = 1145 ft.-lbs. Of course with friction- wheels the wheel is not so steady as without. In this example the force J* has been simply enough to overcome friction. In case the wheel is in actual use, JP is the weight of water actually in the buckets at any instant, and does the work of overcoming I^^, the resistance of the mill machinery, and also the friction. By phicing J*^ pointing upward on the same side of C as P, and making h^ nearly ■=!), H will = G nearly, just as when the Avheel is running empty; and the foregoing numerical results will still hold good for practical purposes. 168. Friction of Pivots. — In the case of a vertical shaft or axle, and sometimes in other cases, the extremity requires sup- port against a thrust along tlie axis of the axle or pivot. If the end of the pivot \%flat and also the surface against which it rubs, we may consider the pressure, and therefore the friction, as uniform over the surface. With a flat circular pivot, then. Fig, 180, the frictions on a small sector of the circle form a system of parallel foices whose resultant is equal to their sum, and is ^^'^' "^^°" applied a distance of -|r from the centre. Hence the sum of the moments of all the frictions about the centre =y!^|r, in which .^ is the axial pressure. Therefore a force P necessary to overcome the friction with uniform rotation must have a moment Pa =fR\r, 180 MECHAl^ICS OF ENGINEERING, and the work lost in friction per revolution is ^fR^Tt .\T = ^,7tfRr. . . . . (1) As the pivot and step become worn, the resultant frictioii* in the small sectors probably approach the centre; for the greatest wear occurs first near tlie outer edge, since there the product ^J)^> is greatest (see § 166). Hence for \r we may more reasonably put ^\ Exam]jle. — A vertical flat-ended pivot presses its step with a force of 12 tons, is 6 inches in diameter, and makes 40 revolu- tions per minute. Required the H. P. absorbed by the friction. Supposing the pivot and step new, and /"for good lubrication = 0.07, we have, from eq. (1) {foot-lb. -second), "Work lost per revolution = .07 X 24,000 X 6.28 X I • i = 1758.4 ft.-lbs., and .*. work per second = 1758.4 X |-t = 1172.2 ft.-lbs., which -i- 550 gives 2.13 H. P. absorbed in friction. If ordi- nary axle-friction also occurs its effect must be added. If the flat-ended pivot is hollow, with radii r^ and r^, we may put ^{i\-\-')\) instead of the fr of the preceding. It is obvious that the smaller the lever-arm given to the resultant friction in each sector of the rubbing surface the smaller the power lost in friction. Hence pivots should be made as small as possible, consistently with strength. For a conical pivot and step, Fig. 181, the resultant friction in each sector of the conical bearing surface has a lever-arm = f r, about tlie axis A, and a value >- than for a flat-ended pivot ; for, on account of the wedge-like action of the bodies, the pressure causing friction is greater. The sum of the moments of these resultant frictions about A is the same as if only two elements of the cone received pressure (each = iV= ^R -f- sin or). Hence the FRICTION. 181 moment of friction of the pivot, i.e., the moment of the force necessary to maintain uniform rotation, is '^ 3 ' "^ sm or 3 " 4 B and work lost per revolution = o'^f~ '^v o sin oi By making r^ small enough, these values may be made less than those for a flat-ended pivot of the same diameter = 2r. In Schiele's " anti-friction" pivots the outline is designed according to the following theory for securing uniform vertical wear. Let j!? = the pressure per r— — ^^ — _ — horizontal unit of area (i.e., Ip jA ip = -^ -r- horizontal projection of .^ ! ■ L<c^^ ^^^^ the actual rubbing surface) ; "^^^^ I "^^t^^^^ this is assumed constant. Let i!?C..'^^<n z^^^. tlie unit 01 area be small, for M ~ic ^ algebraic simplicity. The fric- fig. 182. tion on the rubbing surface, whose horizontal projection = unity, is = yiV =y (^ -f- sin a) (see Fig. 182; the horizontal com- ponent of J9 is annulled by a corresponding one opposite). The work per revolution in producing wear on this area = fN^ny. But the vertical depth of wear per revolution is to be the same at all parts of the surface ; and this implies that the same volume of material is worn away under each horizontal unit of area. HenceyiTSTr^/, i.e.,y-r^^ — B^ry, is to be constant for all values of y ; and since ^7^ and 27r are constant, we must have, as the law of the curve, y , i.e., the tangent BC = the same at all potnts. sm a This curve is called the " tractrixP Schiele's pivots give a very uniform wear at high speeds. The smoothness of wear prevents leakage in the case of cocks and faucets. 169. Normal Pressure of Belting. — "When a perfectly flexible cord, or belt, is stretched over a smooth cylinder, both at rest, 182 MECHANICS OF ENGINEERING. the action between them is normal at every point. As to its j^ \_\ t t s ^^0^1^ ^3 i^j P^r linear unit of arc, the fol- C\\^-d:::^ > lowing will determine. Consider a semi- circle of the 2'2vd tree, neglecting its weight. Fig. 183. The forces holding it in equilib- rium are tlie tensions ar the two ends (these are equal, manifestly, the cylinder being „ smooth ; for tnev are the only two forces * / 7/1 having moments about c/, and each has the Fig. 183. smus lever-ariTi^. and the normal pressures, which are infinite in number, but nave an intensity, p^ per linear unit, which must be constant along the curve since 8 is the same at all points. The normal pressure on a single ele- ment, ds, of the cord is = //Jsr. aiid its JT component = pds cos 6 — prdd cos ^. Hence S.X = gives cos BdQ — 2/S' = 0. i.e., rjp\ sin d = 28; .-in .'. rp[l — (— 1)] = 26' or z> = S (1) 170. Belt on Eough Cylinder. ImDending Slipping. — If fric- tion is possible between the two bodies, the tension may vary »k)ug the arc of contact, so that ^also varies, and consequently Fig. 184. Fig. 185. the friction on lui element (^s being =^ds =f{8-^ r)ds, also varies. If slipping is impendina. the law of variation of the tension S may be found, as follows ° Fig. 184, in which the ERICTIWN. 183 impending slipping is toward the left, shows the cord free. For any element, ds, of the cord, we have, putting 2 (moms, about 0) = Q (Fig. 185), {8+ dSy = Sr + dFr ; i.e., dF= dS, or (see above) dS =f{S -^ r)ds. But ds = rdO ; hence, after ti-ansfonning, fde = §. (1) In (1) the two variables and S are separated ; (1) is there- fore ready for integration. fa = loge 8n — l0g« 8, = l0ge[_^J. (2) Or, by inversion, 8^ef"- — 8n, (3) <?, denoting the Naperian base, = 2.71828 -{-; a of course is in TT-measure. Since 8n evidently increases very rapidly as oc becomes larger, 8^ remaining the same, we have the explanation of the well-known fact that a comparatively small tension, 8^, exerted by a man, is able to prevent the slipping of a rope around a pile-head, when the further end is under the great tension 8^ due to the stopping of a moving steamer. For example, with ^ = ^, we have (Weisbach) for or = J turn, or ^r = ^tt, 8^ = l-BQ-SI, ; = "I turn, or a = 7t^ 8n = '2.S68„ ; = 1 turn, or a- = 2;r, 8^ = S.lS^Su ; = 2 turns, or a = 4c7t, 8^ = 65.94^„ ; = 4 turns, or a = Stt, 8^ = 4348.56/S'„. If slipping actually occurs, we must use a value of f for fric- tion of motion. Examjple. — A leather belt drives an iron pulley, covering one half the circumference. What is the limiting value of the 184 MECHANICS OF ENGINEERING. ratio of Sn (tension on driving-side) to S^ (tension on follow- ing side) if tlie belt is not to slip, taking the low value of y = 0.25 for leather on iron ? We have given ya: = 0.25 X ^r = .T854, which by eq. (2) is the Naperian log. of {S^ '. /So) when slipping occurs. Hence the common log. of {S^ : /S,) = 0.7854 X 0.43429 = 0.34109 ; i.e., if (5;:/S;) = 2.193,say2.2, the belt will (barely) slip (for/= 0.25). (0.43429 is the modulus of the common system of loga- rithms, and = 1 : 2.30258. See example in § 48.) At very high speeds the relation^ = /S' -i- r (in § 169) is not strictly true, since the tensions at the two ends of an element ds are partly employed in furnishing the necessary deviating force to keep the element of the cord in its circular path, the remainder producing normal pressure. 171. Transmission of Power by Belting or Wire Eope. — In the simple design in Fig. 186, it is required to find the motive weight Gy necessary to overcome the given resistance ^ at a DRIVING SIDE Fig. 186. uniform velocity = v^; also the proper stationary tension weight G„ to prevent slipping of the belt on its pulleys, and the amount of power, Z, transmitted. In other words. Given : j B, a, y, a^, r^; a z= n for both pulleys, ) ( -y,; andy for both pulleys ; ) andy for both pulleys; -p • ^ . j Z ; ^, to furnish Z ; G^ for no slip ; 'y the velocity '\<A G\ v' that of belt ; and the tensions in belt. FRICTION. 185 Neglecting axle-friction and the rigidity of the belting, tlie power transmitted is that required to overcome i? through a distance = v^ every second, i.e., Z = Bv, ,. (1) Since (if the belts do not slip) we have a : r::v' : V, and a^ : r^iiv' : v^, V = —v., and v = v.. (2) I^egleeting the mass of the belt, and assuming that each pul- ley revolves on a gravity-axis, we obtain the following, by con- siderino^ the free bodies in Fig'. 187 : CA free) (B free) Fig. 187. (B and tr_u.ck fr.e.e) 2 (moms.) = in A free gives Er^ = {8^ — S,)a, ; . (3) 2 (moms.) = in ^ free gives Gr = (Sn — Sa)a ; . (4) whence we readily find r a. Evidently JR and G are inversely proportional to their velo- cities v^ and v ; see (2). This ought to be true, since in Fig. 186 G is the only working-force, ^ the only resistance, and the motions are uniform ; hence (from eq. (XYI.), § 142) Gv - Ev, = 0. 2J^ = 0, for _5 and truck free, gives G, = S^+S„ (5) while, for impending slip, ^n = ^0^^' (6) 186 MECHANICS OF ENGINEERING-. By elimination between (4), (5), and (6), we obtain and ^n = -/ • ^aZII: (8) Hence G^ and 8^, vary directly as the power transmitted and inversely as the velocity of the belt. For safety G^ should be made > the above value in (T) ; corresponding values of the two tensions may then be found from (5), and from the rela- tion (see § 150) {8^-8y = L (64 These new values of the tensions will be found to satisfy the- condition of no slip, viz., (^,:xS'„)<^-(§170). For leather on iron, ef"" = 2.2 (see example in § lYO), as a. low value. The belt should be made strong enough to with- stand 8n safely. As the belt is more tightly stretched, and hence elongated,, on the driving than on the following side, it ^' creeps'^ back- ward on the driving and forward on the driven pulley, so that the former moves slightly faster than the latter. The loss of work due to this cause does not exceed 2 per cent with ordinary belting (Cotterill). In the foregoing it is evident that the sum of the tensions in the two sides = G„, i.e., is the same, whether the power is^ being transmitted or not ; and this is found to be true, both in theory and by experiment, when a tension-weight is not used, viz., when an initial tension S is produced in the whole belt before transmitting the power, then after turning on the latter the sum of the two tensions (driving and following) always = ^S, since one side elongates as much as the other contracts ; it being understood that the pulley-axles preserve a constant distance apart. 172. Rolling Friction. — The few experiments which have- been made to determine the resistance offered by a level road- FRICTION. 187 "Way to the uniform motion of a roller or wheel rolling upon it corroborate approximately the following theory. The word friction is hardly appropriate in this connection (except when the roadway is perfectly elastic, as will be seen), but is sanctioned by usage. Firsts let the roadway or track be compressible, but inelastic, O the weight of the roller and its load, and P the horizontal force necessary to preserve a uniform motion ^ — ~\ — > (both of translation and rotation). The track / |g \ (or roller itself) being compressed just in h"^ j'"*'^ front, and not reacting symmetrically from \ q[\ ,_^^ behind, its resultant pressure against the //////////mmW^y^-'^'^^^- roller is not at vertically under the centre, ^^**" ^^^' but some small distance, OD = h, in front. (The successive crushing of small projecting particles has the same effect.) Since for this case of motion the forces have the same relations as if balanced (see § 124), we may put 2 moms, about D = 0, .'.Fr=Gh; or, P = ~G (1) According to Professor Goodman we have the following values of b, approximately : Inches. Iron or steel wheels on iron or steel rails. . 6 = 0.007 to 0.015 " " " " " wood 0.06 " 0.10 " " " " " macadam 0.05 " 0.20 " " " " " soft ground 3.0 "5.0 Pneumatic tires on good road, or asphalt.. 0.02 " 0.022 '' " heavy mud 0.04 " 0.06 Solid rubber tires on good road, or asphalt 0.04 " " heavy mud 0.09 " 0.11 According to the foregoing theory, P, the " rolling fi-iction" (see eq. (1)), is directly proportional to G, and inversely to the radius, if h is constant. The experiments of General Morin and others confirm this, while those of Dupuit, Poiree. and Sauvage indicate it to be proportional directly to G, and inversely to the square root of the radius. 188 MECHAlSriCS OF ENGINEEEING. Although J is a distance to be expressed in linear units, and not an abstract number like the /"and f for sliding and axle- friction, it is sometimes called a " coefficient of rolling fric- tion." In eq. (1), h and r should be expressed in the same unit. Of course if P is applied at the top of the roller its lever- arm about D is 2r instead of r^ with a corresponding change in eq. (1). With ordinary railroad cars the resistance due to axle and rolling frictions combined is about 8 lbs. per ton of weight on a level track. For wagons on macadamized roads & = |- inch, but on soft ground from 2 to 3 inches. Secondly^ when the roadway is jperfectly elastic. This is chiefly of theoretic interest, since at first sight no force would be considered necessary to maintain a uniform rolling motion. But, as the material of the roadway is compressed under the roller its surface is first elongated and then recovers its former state ; hence some rubbing and consequent sliding friction must occur. Fig. 189 gives an exaggerated view of the circum- stances, P being the horizontal force applied at the centre necessary to maintain a uniform motion. The roadway (rub- ber for instance) is heaped up both in front and behind the roller, being the })oint of greatest pressure and elongation of the surface. The forces acting are ^, P^ the normal pressures, and the frictions due to them, and must form a balanced system. Hence, since G and P^ and also the normal pressures, pass through C^ the resultant of the frictions must also pass through G\ therefore the frictions, or tangential actions, on the roller must be some forward and some backward (and not all in one direction, as seems to be asserted on p. 260 of Cotterill's Applied Mechanics, where Professor Reynolds' FRICTION. 189 explanation is cited). The resultant action of the roadwaj upon the roller acts, then, through some point J9, a distance OD = h ahead of (9, and in the direction DC, and we have as before, with 2? as a centre of moments, Pr=Gh, or P=-G. If rolling friction is encountered above as well as helow the rollers, Fig. 190, the student may easily prove, by considering three separate free bodies, that for uniform motion p = '-^<^' Fia. 190. where h and h^ are the respective " coefficients of rolling fric- tion ' ' for the upper and lower contacts. (See Kent's " Pocket- Book for Mechanical Engineers'' for "friction-rollers,'^ "ball-bearings," and "roller-bearings." Exa/mjple 1. — If it is found that a train of cars will move uniformly down an incline of 1 in 200, gravity being the only working force, and friction (both rolling and axle) the only resistance, required the coefficient, f\ of axle-friction, the diameter of all the wheels being 2f = 30 inclies, that of the journals 2a = S inches, taking h = 0.02 inch for the rolling friction. Let us use equation (XYI.) (§ 142), noting that while the train moves a distance s measured on the incline, its weight 1 A G does the work G ^z s, the rolling friction — G (at* the axles) has been overcome through the distance s, and the axle-friction (total) through the (relative) distance — sin the journal boxes j whence, the change in kinetic energy being zero, 1 ^ b ^ a Gs -^Ga-^Gs--fas = 0. Gs cancels out, the ratios h : r and a : r are = tAtt ^"<5 iV* respectively (being ratios or abstract numbers they have the * That is, the ideal resistance, at centre of axles and || to the incline, equiV' alent to actual rolling resistance. 190 MECHANICS OF ENGINEERING. same numerical values, whether the inch or foot is used), an (J solving, we have / = 0.05 - 0.0133 = 0.036. Examjple 2. — How many pounds of tractive effort per ton of load would the train in Example 1 require foi- uniform mo- tion on a level track ? Ans. 10 lbs. 173. Eailroad Brakes.* — During the uniform motion of a railroad car the tangential action between the track and each wheel is small. Thus, in Example 1, just cited, if ten cars of eight wheels each make up the train, each car weighing 20 tons, the backward tangential action of the rails upon each wheel is only 25 lbs. When the brakes are applied to stop the train this action is much increased, and is the only agency by which the rails can retard the train, directly or indirectly : directly^ when the pressure of the brakes is so great as to prevent the wheels from turning, thereby causing them to "skid" (i.e., slide) on the rails ; indirectly^ when the brake-pressure is of such a value as still to permit perfect rolling of the wheel, in which case the rubbing (and heating) occurs between the brake and wheel, and the tangential action of the rail has a value equal to or less than the friction of rest. In the first case, then (skidding), the retarding influence of the rails is the/r^c- iion of motion between rail and wheel; in the second, a force which may be made as great as the friction of rest between rail and wheel. Hence, aside from the fact that skidding produces objectionable flat places on the wheel-tread, the brakes are more effective if so applied that skidding is impending, but not actually produced ; for the friction of rest is usually greater than that of actual slipping (§160). This has been proved experimentally in England. The retarding effect of axle and rolling friction has been neglected in the above theory. Example 1. — A twenty-ton car with an initial velocity of 80 feet per second (nearly a mile a minute) is to be stopped on a level within 1000 feet ; required the necessary friction on each of the eight wheels. Supposing the wheels not to skid, the friction will occur * See statement on p. 168, as to diminution of the coefficient / with ; speed. iJ'RICTION. 191 between the brakes ana wheels, and is overcome through the (relative) distance 1000 feet. Eq. (XYI.), § 14:2, gives (foot- Ib.-second system) 1 40000 - 8i^'X 1000 = - 1 ^^(80)^ from which F { = friccion at circumference of each wheel) = 496 lbs. Note. — This result of 496 lbs. must be looked upon as only an average value. For a given pressure, A'^, of brake-shoe on wheel-rim on account of the variation of the coefficient /' with changing speed (see p. 168) the friction will be small at first and gradually increase. This same remark applies to Examples 3 and 4, also. 1/ Examjple 2.— Supoose sisiddins^ to be impending in the fore- going, and tlie coefhcient of friction of rest (i.e., impending slipping) between ran ana wneel to be/'=0,20o In what -distance will the. car oe stopped? Ans. 496 ft Example 3. — Supoose tne car in Example 1 to be on an up» grade of 60 feet to tne mile. Qn applying eq. (XVI.) here, the weight 20 tons win enter as a resistance.) Ans. 439 lbs. Example 4. — In Ji,xample 3. consider all four resistances, viz., gravitj^, rolling triction. and brake and axle frictions, the distance being 1000 ft., and \F the unknown quantity. (Take the wheel dimensions of p. 189.) Ans. 414 lbs. 174. Friction of Car-journals in Brass Bearings. — :(Prof. J. E. Denton, in Vol. xii Transac. Am. See, Mecli. Engs., p. 405; also Kent's Pocket-Book.) A new brass dressed with an emery wheel, loaded with 5000 lbs., may have an actual bearing surface on the journal, as shown by the polish of a portion of the surface, of only one sq. inch. "With this pressure of 5000 Ibs./sq.in. the coefficient of friction may be 0.060 and the brass may be overheated, scarred, and cut ; or, on the contrary, it may wear down evenly to a smooth bearing, giving a highly polished area of contact of 3 sq. in., or more, inside of two hours of running, gradually decreasing the pressure per sq. in. of contact, and showing a coefficient of friction of less than 0.005. A reciprocating motion in the direction of the axis is of importance in reducing the friction. With such polished surfaces any oil will lubricate and the 192 MECHANICS OF ENGINEERING. coefficient of friction then depends on the viscosity of the oil. With a pressure of 1000 lbs, per sq. in. , revolutions from 170 to 320 per min., and temperature of 75° to 113° Fahr., with both sperm and parraffine oils, a coefficient as low as 0.0011 has been obtained, the oil being fed continuously by a pad. 175. Well Lubricated Journals. Laws of Friction. — In the Proc. Inst. Civ. Engs. for 1886 (see also Engineering News for Mar. 31, April 7 and 14, 1888) Prof. Goodman presents the conclusions arrived at by him as to the laws of friction of well lubricated journals as based on the experiments made by Thurston, Beauchamp Tower, and Stroudley. They are as follows : 1 . The coefficient friction with the surfaces efficiently lubri- cated is from ^ to ^ that for dry or scantily lubricated surfaces. 2. The coefficient of friction for moderate pressures and speeds varies approximately inversely as the normal pressure ; the frictional resistance varies as the area- in contact, the normal pressure remaining constant. 3. At very low journal speeds the coefficient of friction is abnormally high, but as the speed of sliding increases from about 10 to 100 ft. per min. the friction diminishes; and again rises when that speed is exceeded, varying approximately as the square root of the speed. 4. The coefficient of friction varies approximately inversely as the temperature, within certain limits, viz., just before abrasion takes place. In one of Mr. Tower's experiments it was found that when the lubrication was made by a pad under the journal (which received pressure on its upper surface) the coefficient was some seven times as large as when an ' ' oil bath, ' ' or copious supply of oil, was provided; (0.0090 as against 0.0014). 176. Rigidity of Ropes. — If a rope or wire cable passes over a pulley or sheave, a force J-* is required on one side greater than the resistance Q on the other for uniform motion, aside from axle-friction. Since in a given time botli I^ and Q describe the same space s, if ^ is > Q, then I^sis > Qs, i.e., the work done by i^ is > than that expended upon Q. This is because some of the work J*s has been expended in bending the stiff rope or cable, and in overcoming friction between the strands, both where the rope passes upon and where it leaves FRICTION. 198 the pulley. With hemp ropes, Fig. 191, the material being nearly inelastic, the energy spent in bending it on at D is nearly all lost, and energy must also be spent in straightening Fig. 191. it at E\ but with a wire rope or cable some of this energy is restored by the elasticity of the material. The energy spent in friction or rubbing of strands, however, is lost in both cases. The iigure shows geometrically why P must be > ^ for a uniform motion, for the lever-arm, a, of P is evidently < h that of Q. If axle-friction is also considered, we must have Pa=qb^f{P-^Q)r, r being the radius of the journal. Experiments with cordage have been made by Prony, Cou- lomb, Eytelwein, and Weisbach, with considerable variation in the results and forraulse proposed. (See Coxe's translation of vol. i., "Weisbach's Mechanics.) With pulleys of large diameter the effect of rigidity is very slight. For instance, Weisbach gives an example of a pulley five feet in diameter, with which, Q being = 1200 lbs., P = 1219. A wire rope f in. in diameter was used. Of this difference, 19 lbs., only 5 lbs. was due to rigidity, the remainder, 14 lbs., being caused by axle-friction. When a hemp-rope 1.6 inches in diameter was substituted for the wire one, P — ^=27 lbs., of which 12 lbs. was due to the rigidity. Hence in one case the loss of work was less than \ of \%. in the other about 1^, caused by the rigidity. For very small sheaves and thick ropes the loss is probably much greater. 13 194 MECHANICS OF ENGINEERING. /Vl'^, Miscellaneous Examples. — Example 1. Tiie end of a \ /shaft 12 inches ill diameter and making 50 revolutions per min- 1/ ute exerts against its bearing an axial pressure of 10 tons and / a lateral pressure of 40 tons. With /" ^y = 0.05, required the H. P. lost in friction. Ans. 22.2 H. P. Example 2. — A leather belt passes over a vertical pulley, covering half its circumference. One end is held by a spring balance, which reads 10 lbs. vi'hile the other end sustains a vreight of 20 lbs., the pulley making 100 revolutions per min- ute. Required the coefficient of friction, and the H. P. spent in overcoming the friction. Also suppose the pulley turned in the other direction, the weight remaining the same. The diameter of the pulley is 18 inches. . {f = 0.22 ; ^^' I 0.142 and .284 H. P. Example 3. — A grindstone with a radius of gyration = 12 inches has been revolving at 120 revolutions per minute, and at a given instant is left to the influence of gravity and axle friction. The axles are 1|- inches in diameter, and the wheel makes 160 revolutions in coming to rest. Required the coeffi- (jient of axle-friction. (Average.) Ans. / = 0.039. Example 4. — A board A, weight 2 lbs., rests horizontally on another B:, coefficient of friction of rest between them being f = 0.30. B is now moved horizontally with a uniformly accelerated motion, the acceleration being = 15 f set per " square seco'id ;" will A keep company with it, or not ? Ans, " if o." y V STRENGTH OF MATERTALSo [Or Mechanics of Materials] = CHAPTEE I. ELEMENTARY STRESSES AND STRAINS. 178. Beformation of Solid Bodies. — In tlie preceding por- tions of this work, what was called technically a " rigid body," was supposed incapable of changing its form, i.e., the positions of its particles relatively to each other, under the action of any forces to be brought upon it. This sup- position was made because the change of form which must actually occur does not appreciably alter the distances, angles, etc., measured in any one body, among most of the pieces of a properly designed structure or machine. To show how the individual pieces of such constructions should be designed to avoid undesirable deformation or injury is the object of this division of Mechanics of En- gineering, viz., the Strength of Materials, ,^6 '€5. D Fis. 193. § 178. AS perhaps tne simplest instance of the deformation or distortion of a solid, let us consider the case of a prismatic rod in a state of tension, Eig. 192 (eye-bar of a bridge 195 196 MECHANICS OF EI>J GINBERmG. truss, e.g.). The pull at each end is P, and the body is said to be under a tension of P (lbs., tons, or other unit), not 2P. Let ABGD be the end view of an elementary parallelopiped, originally of square section and with faces at 45° with the axis of the prism. It is now deformed, the four faces perpendicular to the paper being longer"^ than before, while the angles BAD and BCD, originally right angles, are now smaller by a certain amount d, ABC and ADG larger by an equal amount d. The element is said to be in a state of strain, viz.: the elongation of its edges (parallel to paper) is called a tensile strain, while the alter- ation in the angles between its faces is called a shearing strain, or angular distortion (sometimes also called a slid- ing, or tangential, strain, since BG has been made to slide, relatively to AD, and thereby caused the change of angle). [This use of the word strain, to signify change of form and not the force producing it, is of recent adoption among many, though not all, technical writers.] 179. Strains. Two Kinds Only. — Just as a curved line may be considered to be made up of small straight-line ele- ments, so the substance of any solid body may be consid- ered to be made up of small contiguous parallelopipeds, whose angles are each 90° before the body is subjected to the action of forces, but which are not necessarily cubes. A line of such elements forming an elementary prism is sometimes called a> fibre, but this does not necessarily imply a fibrous nature in the material in question. The system of imaginary cutting surfaces by which the body is thus subdivided need not consist entirely of planes ; in the sub- ject of Torsion, for instance, the parallelopipedical ele- ments considered lie in concentric cylindrical shells, cut both by transverse and radial planes. Since these elements are taken so small that the only possible changes of form in any one of them, as induced by a system of external forces acting on the body, are * When a is nearly 0° (or 90°) BG and AD (or AB and DG) are shorter than before, on account of lateral contraction; see § 193. ELEMENTARY STRESSES, ETC, 197 elongations or contractions of its edges, and alteration of its angles, there are but two kinds of strain, elongation (contraction, if negative) and shearing. 180. Distributed Forces or Stresses. — In the matter preced- ing this chapter it has sufficed for practical purposes to consider a force as applied at a point of a body, but in reality it must be distributed over a definite area ; for otherwise the material would be subjected to an infinite force per unit of area. (Forces like gravity, magnetic at- traction, etc., we have already treated as distributed over the mass of a body, but reference is now had particularly to the pressure of one body against another, or the action •of one portion of the body on the remainder.) For in- stance, sufficient surface must be provided between the end of a loaded beam and the pier on which it rests to avoid injury to either. Again, too small a wire must not be used to sustain a given load, or the tension per unit of area of its cross section becomes sufficient to rupture it. Stress is distributed force, and its intensity at any point of the area is • o e (1) "where dF is a small area containing the point and dP the force coming upon that area. If equal dP^s (all parallel) act on equal dF'soi a plane surface, the stress is said to be of uniform intensity, which is then i>=p . . . . (2) where P=-= total force and ^the total area over which it acts. The steam pressure on a piston is an example of stress of uniform intensity. X98 MECHANICS OF ENGINEEKING. For example, if a force P= 28800 lbs, is uniformly dis- tributed over a plane area of ^=72 sq. inches, or ^ of a sq. foot, the intensity of the stress is 28800 ,^^,, . , p= =400 lbs. Der sq. inch, (or jp = 28800^ >^ =57600 lbs. per sq. foot, or p=14400-j' ^=28.8 tons per sq. ft,, etc... 181. Stresses on an Element : of Two Kinds Only. — When a solid body of any material is in eauiiibrium under a sys- tem of forces which do not rupture it. not only is its shape altered (i.e. its elements are strained), and stresses pro- duced on those planes on which the forces act, but other stresses also are induced on some or all internal surfaces which separate element from element, f over and above the forces with which the elements mav have acted on each other before the application of the external stresses or " applied forces "). So long as the whole solid is the "free body " under consideration, these internal stresses, being the forces with which the portion on one side of an imag- inary cutting plane acts on the portion on the other side, do not appear in any equation of eauiiibrium (for if intro- duced they would cancel out); but if we consider free a portion only, some or all of whose bounding surfaces are cutting planes of the original bodv. the stresses existing on these planes are brought into the eauations of equilib- rium. Similarly, if a single element of the body is treated by itself, the stresses on all six of its faces, together with its weight, form a balanced system of forces, the body being supposed at rest. FiS. 138. ELEMENTAE.T STRESSES, ETC. 199 As an example of internal stress, consider again the case of a bar in tension ; Fig. 193 shows tlie whole bar (or eye- bar) free, the forces P being the pressures of the pins in. the eyes, and causing external stress (compression here) on the surfaces of contact. Conceive a right section made through BS, far enough from the eye, (7, that we may con- sider the internal stress to be uniform * in this section, and consider the portion BSG as a free body, in Fig. 194. The stresses on R8, now one of the bounding surfaces of the free body, must be parallel to P, i.e., normal to B8; (otherwise they would have components perpendicular to P, which is precluded by the necessity oi lY being = 0, and the supposition of uniformity.) Let .^ = the sec- FlG. 194, Fig. 195. tional area RS, and p = the stress per unit of area ; then. P IX-= gives P= Fp, i.e., p= F .(2). The state of internal stress, then, is such that on planes perpendicular to the axis of the bar the stress is tensile and normal (to those planes). Since if a section were made oblique to the axis of the bar, the stress would still be parallel to the axis for reasons as above, it is evident that on an oblique section, the stress has components both nor- mal and tangential to the section, the normal component being a tension. * As will be shown later (§ 295) the line of the two P's in Fig. 193 must pass through the centre of gravity of the cross-section RS (plane figure) of the bar, for the stress to be uniform over the section. 200 MECHANICS OF EISTGINEERLNG. The presence of the tangential or shearing stress in ob- lique Sections is rendered evident by considering that if an oblique dove-tail joint were cut in the rod, Fig. 195, the shearing stress on its surfaces may be sufficient to over- come friction and cause sliding along the oblique plane. If a short prismatic block is under the compressive ac- tion of two forces, each =P and applied centrally in one base, we may show that the state of internal stress is the same as that of the rod under tension, except that the nor- mal stresses are of contrary sign, i.e., compressive instead of tensile, and that the shearing stresses (or tendency to slide) on oblique planes are opposite in direction to those in the rod. Since the resultant stress on a given internal plane of a body is fully represented by its normal and tangential components, we are therefore justified in considering but iwo kinds of internal stress, normal or direct, and tangen- tial or shearing. 182. Stress on Obliq[ue Section of Rod in Tension, — Consider free a small cubic element whose edge =a in lengthy it has two faces parallel to the paper, being taken near the middle of the rod in Fig. 192. Let the angle which the face AB, Fig. 196, makes with the axis of the rod be = a. This angle, for our present purpose, is considered to remain the same while the two forces P are acting, as before their action. The re- sultant stress on the face AB hav- ing an intensity p=P-h-F, (see eq. 2) per unit of transverse section of rod, is = jp (a sin a) a. Hence its component normal to AB is ■pa^ sin^ a ; and the tangential or shearing component along AB '=*pa^ sin a cos a. Dividing by the area, a^, we have the following : For a rod in simple tension we have, on a plane making an angle, a, with the axis : ELEMENTARY STRESSES, ETC. 201 a Normal Stress =p Bin? a per unit of area . . (1) and a Shearing Stress =p sin a cos a per unit of area . (2) " Unit of area " here refers to the oblique plane in ques- tion, while p denotes the normal stress per unit of area of a transverse section, i.e., when a=90°. Fig. 194. The stresses on CD are the same in value as on AB, while for BG and AD wq substitute 90° — a for a. Fig. 197 shows these normal and shearing stresses, and also, much exaggerated, the strains or change of form of the element (see Fig. 192). 182a, Eelation between Stress and Strain. — Experiment shows that so long as the stresses are of such moderate value that the piece recovers its original form completely when the external forces which induce the stresses are re- moved, the following is true and is known as Hoohe's Law (stress proportional to strain). As the forces P in Fig. 193 (rod in tension) are gradually increased, the elonga- tion, or additional length, of BK increases in the same ratio as the normal stress, p, on the sections BS and KI^^ per unit of area [§ 191]. As for the distorting effect of shearing stresses, considei in Fig. 197 that since p sin a cos a = p cos (90° — a) sm (90° — a) the shearing stress per unit of area is of equal value on all four of the faces (perpendicular to paper) in the elementary block, and is evidently accountable for the shearing strain, i.e., for the angular distortion, or difference, d, between 90° and the present value of each of the four angles. Ac- cording to Hooke's Law then, as P increases within tlvg limit mentioned above, d varies proportionally to p sin a cos a, i.e. to the stress. 182b. Example. — Supposing the rod in question were of a kind of wood in which a shearing stress of 200 lbs. per sq. inch along the grain, or a normal stress of 400 lbs. per 8q. inch, perpendicular to a fibre-plane will produce rup- ture, required the value of a the angle which the grain must make with the axis that, as P increases, the danger of rupture from each source may be the same. This re- 202 MECHANICS OE ENGIl>rEBEI2fG. quires that 200:400::p sin a cos a :p sin^a, i.e. tan. a must = 2.000.-.a=63i^°. If the cross section of the rod is 2 sq. inches, the force P at each end necessary to produce rup- ture of either kind, when a=63^°, is found by putting p sin a cos ^=^00. '.^=500.0 lbs. per sq. inch. "Whence, since p=P-^F, P=1000 lbs. (Units, inch and pound.) 183. Elasticity is the name given to the property which most materials have, to a certain extent, of regaining their original form when the external forces are removed. If the state of stress exceeds a certain stage, called the Elastic Limit, the recovery of original form on the part of the ele- ments is only partial, the permanent deformation being called the Set. Although theoretically the elastic limit is a perfectly defi- nite stage of stress, experimentally it is somewhat indefi- nite, and is generally considered to be reached when the permanent set becomes well marked as the stresses are in- creased and the test piece is given ample time for recovery in the intervals of rest. The Safe Limit of stress, taken well within the elastic limit, determines the working strength or safe load of the piece under consideration. E.g., the tables of safe loads of the structural steel beams for floors, made by the Cambria Steel Co. , at Johnstown, Pa, , are computed on the basis that the greatest normal stress (tension -or compression) occurring on any internal plane shall not exceed 16,000 lbs. per sq. inch; and, again, by the building laws of Philadelphia, the greatest shearing stress to be permitted in ' ' web plates " of " mild steel" is 8750 lbs. /in. 2 The tJltimate Limit is reached when rupture occurs. 184. The Modulus of Elasticity (sometimes called co-efficient of elasticity) is the number obtained by dividing the stress per unit of area by the corresponding relative strain. Thus, a rod of wrought iron ^ sq. inch sectional area being subjected to a tension of 2^ tons =5,000 lbs., it is ELEMENTARY STRESSES, ETC. 203 iound that a length whicli was six feet before tension is »= 6.002 ft. during tension. The relative longitudinal strain or elongation is then= (0.002)-^6= 1 : 3,000 and the corres- ponding stress (being the normal stress on a transverse plane) has an intensity of i?t=P^i^= 5,000-^ 1^=10,000 lbs., per sq. inch. Hence by definition the modulus of elasticity is (for ten- sion), if we denote the relative elongation by s, Bt-=Pt'^ £=10,000^ g-^ =30,000,000 lbs. per sq. inch, (the sub-script " t " refers to tension). It will be noticed that since £ is an abstract number, Et is of the same quality as p^, i.e., lbs. per sq. inch, or one di- mension of force divided by two dimensions of length. (In the subject of strength of materials the inch is the most convenient English linear unit, when the pound is the unit of force ; sometimes the foot and ton are used to- gether.) The foregoing would be called the modulus of elasticity of lorought iron in tension in the direction of the fibre, as given by the experiment quoted. • But by Hooke's Law p and £ vary together, for a given direction in a given ma- terial, hence ivithin the elastic limit E is constant for a given direction in a given material. Experiment confirms this approximately. Similarly, the modulus of elasticity for compression E^ in a given direction in a given material may be determined by experiments on short blocks, or on rods confined lat- erally to prevent flexure. As to the modulus of elasticity for shearing, E^, we divide the shearing stress per unit of area in the given direction by (? (in radians) the corresponding angular strain or distortion; e.g., for an angular distortion of 0.10° or ,^ = .001T4, and a shearing stress of 15,660 lbs. per sq. inch, we have £;=^^ = 9,000,000 lbs. per sq. inch. 204 MECHANICS OF ENGINEERING. 184a. Young's Modulus is a name frequently given toEf and Ec, it being understood that in the experiments to determine these moduli the elastic limit is not passed, and also that the rod or prism tested is not subjected to any stress on the sides. See p. 507. 185. Isotropes. — This name is given to materials which are homogenous as regards their elastic properties. In such a material the moduli of elasticity are individually the same for all directions. E.g., a rod of rubber cut out of a large mass will exhibit the same elastic behavior when subjected to tension, whatever its original position in the mass. Fibrous materials like wood and wrought iron are not isotropic ; the direction of grain in the former must always be considered. The " piling " and welding of nu- merous small pieces of iron prevent the resultant forging from being isotropic. 186. Resilience refers to the potential energy stored in a body held under external forces in a state of stress which does not pass the elastic limit. On its release from con- straint, by virtue of its elasticity it can perform a certain amount of work called the resilience, depending in amount upon the circumstances of each case and the nature of the material. See § 148. 187. General Properties of Materials. — In viev/ of some defi- nitions already made we may say that a material is ductile when the ultimate limit is far removed from the elastic limit ; that it is brittle like glass and cast iron, when those limits are near together. A small modulus of elasticity means that a comparatively small force is necessary to produce a given change of form, and vice versa, but implies little or nothing concerning the stress or strain at the elastic limit ; thus Weisbach gives E^, lbs. per sq. inch for wrought iron = 28,000,000= double the E^ for cast iron while the compressive stresses at the elastic limit are the same for both materials (nearly). ELEMENTARY STRESSES, ETC. 205 188. Element with Normal Stress on Sides as well as on End-Faces. Ellipse of Stiess.— In Fig. 193, p. 198, the parallelopiped RKNS is sub- jected to stress on the two end-faces only. Let us now consider a small square-cornered element of material subjected to a normal stress p^ (tension) on the two vertical end-faces, while on the horizontal side faces there acts a normal (also tensile) stress of pj Ibs./in.^; (but no stress on the vertical side ^- ^n'---^ / faces). In Fig. 197a '''' ' ^""- '^ is shown, as a free body in equilibrium, a triangular prism ABC, which is the upper right-hand half of such an ele- ment; obtained by passing the cutting plane AC along a diagonal of the side plane (plane of paper) on which there is no stress, and 1 to it. The angle 6 may have any value and it is desired to deter- mine the unit stress "iG- 197a. 5o induced on the oblique plane AC by the normal stresses Pi and Pz acting respectively on the end face BC and on the side face AB. The unit stress q^ on the face AC is not 1 to that face but makes with it some angle ^. Let AB = h inches, BC = n in., and AC = c in.; each of the rectangular areas having a common dimension, =d in., T to the paper. Then the total (oblique) stress on face AC is q^cd lbs., that on AB is P'pd, and that on BC is p-jid lbs. Since the total stress on AC is the anti-resultant of the other two, and these are T to each other, we have {,q^cdy={,PTndy+{p^hdy; i.e., ?o^ = fpiyj + (pj- But, since ?i^c = sin d, and fe-^c = cos 6, this may be written q-'={p,smey+{p,coBdy (i) Eq. (1) gives the magnitude of q^ for any value of angle d; but both position and magnitude are best shown by a geometrical construction. being any point on AC, draw a circle with center at and radius, OH^, equal by scale to the unit stress p^. Similarly, with radius OH 2, equal (on same sea'"") to the unit stress P2, draw the circle H2E2. Through draw EfiN normal to the face AC on which the stress 50 is to be deter- mined, and note the intersections E^ and £'2 (both on left of 0) with the two circles respectively. A vertical line through E^ and a hori- zontal through E2 intersect at some point m. Cm is the magnitude and position of the stress q^; since mD2 = EiDi = pi sin 0, and ODj = P2COSO; hence from eq. (1) Om=5o. 206 MECHANICS OF ENGINEERING. T he po int m is a point in an ellipse whose semi-principal axes are OH^ and OH 2, i.e., p^ and pj. This ellipse is called the Ellipse of Stress; Om being a semi-diameter, determined in the way indicated. (Similarly, if the elementary right parallelepiped is subjected to the action of three normal stresses, Pi, p2, and p^, on all three pairs of faces, respectively, the unit stress on any oblique plane is a semi-diameter of an Ellipsoid of Stress). The unit shearing stress on the oblique face AC is qs=^qo cos ,u; and the unit normal stress is q=qo sin /;. In case the normal stress P2 on the face AB were compressive, p^ being tensile, a horizontal would be drawn through E'2 on the circle of radius OH2, instead of through E2, to meet the vertical through E^, and would thus determine Om', instead of Om, as the stress on AC. If, in such a case, P2 were numerically equal to p^, and d were 45°, go would = Pi = pj, and would lie in the surface AC (pure shear; compare with Exam. 5, p. 242). With Pi = P2, and both tensUe, or both compressive, 50 would be equal to Pi, =P2, for all values of d. 189. Classification of Cases. — Althougli in almost any case whatever of the deformation of a solid body by a balanced system of forces acting on it, normal and shearing stresses are both, developed in every element which is affected at all (according to the plane section considered,) still, cases where the body is prismatic, and the external system con- sists of two equal and opposite forces, one at each end of the piece a,nd directed away from each other, are commonly called cases of Tension; (Fig, 192); if the piece is a short prism with the same two terminal forces directed toward each other, the case is said to be one of Compression ; a case similar to the last, but where the prism is quite long (" long column "), is a case of Flexure or bending, as are also most cases where the " applied forces " (i.e., the external forces), are not directed along the axis of the piece. Rivet- ed joints and " pin-connections " present cases of Shearing; a twisted shaft one of Torsion. When the gravity forces due to the weights of the elements are also considered, a combination of two or more of the foregoing general cases may occur. In each case, as treated, the principal objects aimed at are, so to design the piece or its loading that the greatest stress,* in whatever element it may occur, shall not exceed a safe value ; and sometimes, furthermore, to prevent too great deformation on the part of the piece. The first ob- ject is to provide sufficient strength; the second sufficient stiffness. * See § 405b for mention of the "elongation theory" of safety. This is based on considerations of strain, or deformation, instead of stress. TMNSION. 207 te:nsion. 191. Hooke's Law by Experiment. — As a typical experiment in the tension of a long rod of ductile metal sncli as wrought iron and the mild steels, the following table is quot- ed from Prof. Cotterill's " Applied Mechanics." The experi- ment is old, made by Hodgkinson for' an English Railway Commission, but well adapted to the purpose. From the great length of the rod, which was of wrought iron and 0.517 in. in diameter, the portion whose elongation was observed being 49 ft. 2 in. long, the small increase in length below the elastic limit was readily measured. The succes- sive loads were of such a value that the tensile stress p=P^F, or normal stress per sq. in. in the transverse section, was made to increase by equal increments of 2657.5 lbs. per sq. in., its initial value. After each application of load the elongation was measured, and after the removal of the load, the permanent set, if any. Table of Elongations of a Wrought Iron Rod, of a Length = 49 Ft. 2 In. p X JA e^X^l X' Load (lbs square ii . per Elongation, ich.) (inches.) Increment of Elongation. s, the relative elongation, (ab- stract number.) Permanent Set. (inches.) 1X266 7.5 .0485 .0485 0.000082 2X ' . 1095 .061 .000186 3X ' . 1675 .058 .000283 0.0015 4X ' .224 .0565 .000379 .002 5X ' .2805 .0565 .000475 .0027 6X ' .337 .0565 .000570 .003 7X ' .393 .056 .004 8X ' .452 .059 .000766 .0075 9X ' .5155 .0635 .0195 lOX ' .598 .0825 .049 IIX ' .760 .162 .1545 12X ' 1.310 .550 .667 etc. 208 MECHANICS OF E]SrGIl!fEEEIl!fG. Referring now to Fig. 198, the notation is evident. P is the total load in any experiment, F the cross section of the rod ; hence the normal stress on the transverse section is p=P-r-F. When the loads are increased by equal in- crements, the corresponding increments of the elongation a should also be equal if Hooke's law is true. It will be noticed in the table that this is very nearly true up to the 8th loading, i.e., that JX, the difference between two con- secutive values of }., is nearly constant. In other words the proposition holds good ; if P and Pi are any two loads below the 8th, and X and ki the corresponding elongations. The permanent set is just perceptible at the 3d load, and increases rapidly after the 8th, as also the increment of elongation. Hence at the 8th load, which produces a ten- sile stress on the cross section of j9= 8x2667.5= 21340.0 lbs. per sq. inch, the elastic limit is reached. As to the state of stress of the individual elements, if we conceive such sub-division of the rod that four edges of each element are parallel to the axis of the rod, we find that it is in equilibrium between two normal stresses on its end faces ''^^ (Fig. 199) of a value ^pdF== {P^F)dF where dF is the hor- izontal section of the element. If dx was the original length, and dX the elongation produced by pdF, we shall have, since all the dx's of the length are equally elongated at the dX X same time, w" ^ T where Z = total (original) length. But dX^dx is the relative elongation e, and by definition (§ 184) the Modulus of Elas- ticity for Tension, Ei, = jp^e, {Young's Modulus, § 184a). TENSION. 209 .'.E.=-4rr or E,=^^ .... (1) Jblq. (1) enables us to solve problems involving the elonga- tion of a prism under tension, so long as the elastic limit is not surpassed. The values of E^ computed from experiments like those just cited should be the same for any load under the elas- tic limit, if Hooke's law were accurately obeyed, but in reality they differ somewhat, especially if the material lacks homogeneity. In the present instance (see Table) we have from the 2d Exper. ^=^-^£=28,680,000 lbs. per sq. in. 5th " Ec= " =28,009,000 8th " ^t= " =27,848,000 192. Stress-Strain Diagrams. — If the relative elongations or "strains " (s) corresponding to a series of values of the tensile unit-stresses (p) (Ibs./in.^) to which a rod of metal has been subjected in a testing machine, are plotted as abscissae, and the unit-stresses themselves (p) as ordinates, we have in the curve joining these points a useful graphic, representation of the results of experiment. Fig. 200 shows some of these curves, giving average re- sults for the principal "ferrous " metals. On the left, in (I), the scale adopted (horizontal) for the "strain " (e) or "unit-elongation " is one hundred times as great as that used in the right-hand diagram, (II) ; while the vertical scale for stress (p) in (I) is only twice as great as that in ' (II) . The change of form within the elastic limit is so small compared with that beyond, that this difference in scale is quite necessary in order that diagram (I) may show what occurs within the elastic limit and a Httle beyond. Diagram (II) shows the remainder of the curves of wrought iron and soft steel, up to the point of rupture. We have here the means of comparing the properties of the four typical metals represented, as to elasticity and tenacity. Up to the respective elastic hmits, B, B' , B" , and B'" , stress is fairly proportional to strain, and a straight line is the result; 210 MECHANICS OF ENGINEERING. the "true elastic limit " being regarded as the point where such proportionahty ceases. In the case of wrought iron and soft steel there is a point Y, called the "yield point," a little above the true elastic limit, and sometimes called the "apparent elastic limit/' or "commercial elastic limit/' immediately be- yond which further slight increments of stress produce rela- tively great increments of strain, permanent set becoming very marked; i.e., the part YD of the curve is almost hori- zontal. Beyond D the curve rises again, more steeply, but just before rupture [see (II)] may descend somewhat; since, 50,000 r r--y^ — ^ 40,000 10,000 Ibs./iv} Fig. 200. on account of the lateral contraction mentioned in the next paragraph, here plotted, the stress being computed by dividing the total pull by the original sectional area, is less toward rupture than at stages closely preceding. If at any point beyond the elastic limit, as at C (see curve for wrought iron) in (I), the stress be gradually removed, the relations of stress and strain during this gradual diminution of stress, are shown by the straight line CC. The position of the point C indicates that there is in the rod (now under no stress) a permanent set, or relative elongation, of £ = 0.0015, or 15 parts in 10,000, an elastic recovery having occurred from 0.0028 to 0.0015 (see horizontal scale). Since by definition the modulus of elasticity E = p^s, the values of the respective moduli for the metals in diagram (I) are propor- TENSION. 211 tional to the tangent of the angle which the corresponding straight portions OB, OB' , etc., make with the horizontal axis. From the various ordinates and abscissae for the points B, B', etc., we find E for cast iron to be 14,000,000 Ibs./in.^ and for the other three metals 28,000,000, 30,000,000, and 40,000,000, respectively. The curve for the "harder steel" is not shown in (II), being beyond the limits of the diagram, as to stress; and the complete curve for cast iron is contained within the limits of diagram (I), since the elongation at rupture is very small in the case of this metal, only about 3/10 of one per cent, or 3 parts in 1000; whereas that for wrought iron or soft steel is 300 parts in 1000 (or 30 per cent). In the case of cast iron the elastic limit is very ill- defined and the proportion of carbon and the mode of manu- facture have much influence on its behavior under test. "Soft steel" is another name for "structural steel," used in construction on a large scale, as in buildings and bridge trusses; "medium steel " being a somewhat harder grade of the same. Many grades of steel are made which are much stronger and harder than these, such as tool steel, nickel steel, and piano wire (whose rupturing stress may be as high as 300,000 Ibs./in.^). Wrought iron in the form of wire is much stronger than in bars. Note. — Such a line as CC, showing the relation of stress and strain as the stress is gradually removed, will be called an "elasticity line" on p. 241. In § 206 some mention will be made of the phenomena of "overstraining" a test-piece of iron or steel, showing that on re-applying stress after a certain period of rest the plotted results of stress and strain relations show that the line C'C is retraced to C and continues in the same straight line prolonged, to a new elastic limit higher than C, before curving off to the right. 193. Lateral Contraction. — In the stretching of prisms of nearly all kinds of material, accompanying the elongation of length is found also a diminution of width whose rela- tive amount in the case of the three metals just treated is about ^ or i^ of the relative elongation (within elastic limit). Thus, in the third experiment in the table of § 191, this relative lateral contraction or decrease of diameter ~ H ^^ /i ^^ ^> ^■^•> about 0.00008. In the case of cast iron and hard steels contraction is not noticeable ex- 212 MECHANICS OF ENGINEEBING csept by very delicate measurements, both within and with- out the elastic limit ; but the more ductile metals, as wrought iron and the soft steels, when stretched beyond the elastic limit show this feature of their deformation in a very marked degree. Fig. 201 shows by dotted lines the original contour of a wrought iron rod, while the con- tinuous lines indicate that at rupture. At the cross section of rupture, whose position is determined by some local weakness, the drawing out is peculiarly pronounced. The contraction of area thus produced is some- times as great as 50 or 60% at the fracture. 194. "Flow of Solids." — When the change in re- lative position of the elements of a solid is ex- treme, as occurs in the making of lead pipe, I drawing of wire, the stretching of a rod of duc- j tile metal as in the preceding article, we have Fig. 201. instances of what is called the Flow of Solids, in- teresting experiments on which have been made by Tresca. 195. Moduli of Tenacity. — The tensile stress per square inch (of original sectional area) required to rupture a prism of a given material will be denoted by T and called the modulus of ultimate tenacity ; similarly, the modulus oj safe tenacity, or greatest safe tensile stress on an element, by T' ', while the tensile stress at elastic limit may be called T". The ratio of T' to T" is not fixed in practice but depends upon circumstances (from j4, to ^). Hence, if a prism of any material sustains a total pull or load P, and has a sectional area=jP, we have P= FT for the ultimate or breaking load. \ P'=FT' " " safe load. f ' ' (^^ P"=FT" " " load at elastic limit. ) Of course T' should always be less than T". (The hand-, book of the Cambria Steel Co. , in quoting from the building laws of various cities of the U. S., gives allowable unit- stresses for ordinary materials, both in tension and com- pression.) TENSION. 213 196. Resilience of a Stretched Prism. — ^Fig. 202. In the gradual stretcliing of a prism, fixed at one extremity ^ the value of the tensile force P at the other necessarily de- pends on the elongation A at each stage of the lengthening, according to the relation [eq. (1) of § 191.] ' ^^ (8) FE, within the elastic limit. (If we place a weight G on the ^^ flanges of the unstretched prism and then leave ^ it to the action of gravity and the elastic action of the prism, the weight begins to sink, meeting an increasing pressure P, proportional to l, from the flanges). Suppose the stretching to continue until P reaches some value P" (at elastic limit \\ say), and I a value X'. Then the work done so N^ far is (see p. 155) Fig 802 ^7= mean force X space = ^ P" /I" . . (4) But from (2) F'=FT", and (see §§ 184 and 191) K"=e"l .-. (4) becomes XJ=y2 T e". Fl=}^ T e" V . . (6) where Fis the volume of the prism. The quantity }4T"£", or work done in stretching to the elastic limit a cubic inch (or other unit of volume) of the given material, may be called the Modvlus of Resilience for tension. From (5) it appears that the amounts of work done in stretching to the elastic limit prisms of the same material but of differ- ent dimensions are proportional to their volumes simply. The quantity }4T"e" is graphically represented by the area of one of the triangles such as OA'B, OA'B" in Fig. 200 ; for (in the curve for wrought iron for instance) the modulus of tenacity at elastic limit is represented by ^'P, and e" (i.e., e for elastic limit) by OA'. The remainder of 214 MEOHAE^ICS OF ENGINBERI^a. the area OBG included between the curve and the hori- zontal axis, i.e., from B to G, represents the work done in stretching a cubic unit from the elastic limit to the point of rupture, for each vertical strip having an altitude =p and a width =de, has an area ^pde, i.e., the work done by the stress p on one face of a cubic unit through the dis- tance de, or increment of elongation. If a weight or load = (r be " suddenly "applied to stretch the prism, i.e., placed on the flanges, barely touching them, and then allowed to fall, when it comes to rest again it has fallen through a height X^, and experiences at this instant some pressure P\ from the flanges; Pi=:?. Apply- ing to this body the "Work and Energy" method (p. 138), noting that its initial and final kinetic energy are each zero and that the force G is constant, while the upward force P (from the flanges) is variable, with an average value of JPi, we have GAi = iPiAi + 0-0; whence Pi = 2(?. Since Pi = 2G, i.e., is >(t, the body does not remain in this position but is pulled upward by the elasticity of the prism. In fact, the motion is harmonic (see §§ 59 and 138). Theoretically, the elastic limit not being passed, the oscillations should continue indefinitely. Hence a load O " suddenly applied " occasions double the tension it would if compelled to sink gradually by a sup- port underneath, which is not removed until the tension is just = Q, oscillation being thus prevented. If the weight G sinks through a height —h before strik- ing the flanges. Fig. 202, we shall have similarly, within elastic limit, if ^i= greatest elongation, (the mass of rod being small compared with that of G). G{h^K)=%P,K .... (6) If the elastic limit is to be just reached we have from eqs. (5) and (6), neglecting ^ compared with h, Gh=%T"B"V . . . (7> TBNSIOX. 215 nn equation of condition that tlie prism shall not be in- jured. Example. — If a steel prism have a sectional area of i/ eq. inch and a length ^=10 ft. =120 inches, what is the greatest allowable height of fall of a weight of 200 lbs., that the final tensile stress induced may not exceed T" = 30,000 lbs. per sq. inch, if e" z=.002 ? From (7), using tha inch and pound, we have h= T"e"V 30,000 X. 002x1^x120. 2^ 2x200 :4.5 inches. 197. Stretching of a Prism by Its Own Weight. — In the case of a very long prism such as a mining- pump rod, its weight must be taken into account as well as that of the terminal load P , see Fig. 203. At (a.) the prism is shown in its unstrained condition ; at (&) strained by the load P^ and its own weight. Let the cross section be =jP, the heaviness of the prism =y. Then the rela- tive extension of any element at a distance Fig. 203. jy from o is "^ ^_dX {P,+rFx) dx' FE, (1) {See eq. (1) § 191) ; since P^-^-Fjx is the load hanging upon the cross section at that locality. Equal c?a?'s, therefore, are unequally elongated, x varying from to I. The total elongation is =/*=jk/f^''"+''^"'"^=/i; '/2GI FE, Le., k= the amount due to Pi, plus an extension which half the weight of the prism would produce, hung at the lower extremity. PI * In A. = put dX for A, dx for I, and (P, + j^^.r) for P, FEt 216 MECHANICS OF ENGINEERING. The foregoing relates to tlie deformation of the piece, and is therefore a problem of stiffness. As to the strength of the prism, the relative elongation e=dA-h-dx [see eq. (1)], which is variable, must nowhere exceed a safe value e'= T'^E, (from eq. (1) § 191, putting P=FT', and X=X), Now the greatest value of the ratio dX : dx, by inspecting eq. (1), is seen to be at the upper end where x=l. The proper cross section F, for a given load Pj, is thus found. Putting ^]^-~^ ^e have F =^^ . (2) 198. Solid of Uniform Strength in Tension, or hanging body of minimum material supporting its own weight and a terminal load Pj. Let it be a solid of revolution. If every cross-section P at a distance =x from the lower extrem- ity, bears its safe load FT', every element of the body is doing full duty, and its form is the most economical of material. The lowest section must have an area Fi(j.204. Fo=P^-^T', since Pi is its safe load. Fig. 204. Consider any horizontal lamina ; its weight is yFdx, (j= heaviness of the material, supposed homogenous), and its lower base Pmust have Pi-\-G for its safe load, i.e. G+F,=FT' ... a) in which G denotes the weight of the portion of the solid below F. Similarly for the upper base F-\-dF, we have G+P,+r^dx={F+dF)T' . , (2) By subtraction we obtain rFdsc=T'dFi ie. -l.dx^ ^ T F TENSION. 217 in whicli the two variables x and F are separated. By in- tegration we now have ;or^,=log.e^ . . (3) . \x p yx 1.6., F=Foer =-1, e~ (4) from which i^may be computed for any value of x. The weight of the portion below any F is found from (1) and (4); i.e. while the total extension ^ will be ^=^"^1 (6) "the relative elongation dX-i-dx being the same for every dx and bearing the same ratio to e" (at elastic limit), as T' does to T". 199. Tensile Stresses Induced by Temperature. — If the two ends of a prism are immovably fixed, when under no strain and at a temperature t, and the temperature is then low- ered to a value t', the body suffers a tension proportional to the fall in temperature (within elastic limit). If for a rise or fall of 1° Fahr. (or Cent.) a unit of length of the material would change in length by an amount t^ (called the co-efficient of expansion) a length =1 would be con- tracted an amount X=-fjl(t-t') during the given fall of tem- perature if one end were free. Hence, if this contraction is prevented by fixing both ends, the rod must be under a :tension P, equal in value to the force which would be 218 MECHANICS OF exgi]s:eeeing. necessary to produce the elongation X, just stated, under ordinary circumstances at the lower temperature. !From eq. (1) §191, therefore, we have for this tension dtite to fall of temperature For 1° Cent, we may write For Cast iron -f] = .0000111 ; « Wrought iron = .0000120 ; « Steel = .0000108 to .0000114 J « Copper yj = .0000172 ; « Zinc 7^ = .0000300, COMPRESSION OF SHORT BLOCKS. 200, Short and Long Columns. — In a prism in tension, its- own weight being neglected, all the elements between thl jocaiities of application of the pair of external forces pro- ducing; the stretching are in the same state of stress, if the external forces act axially (excepting the few elements in the immediate neighborhood of the forces ; these suffering local stresses dependent on the manner of application of the external forces), and the prism may be of any length without vitiating this statement. But if the two external forces are directed toivard each other the intervening ele- ments will not all be in the same state of compressive stress unless the prism is comparatively short (or unless numerous points of lateral support are provided). A long prism will buckle out sideways, thus even inducing tensile stress, in some cases, in the elements on the convex side. Hence the distinction between sTiort UocTcs and long columns. Under compression the former yield by crush- ing or splitting, while the latter give way by flexure (i.e. bending). Long columns, then will be treated separately COMPRESSION OF SHORT BLOCKS. 219 In a subsequent chapter. In the present section tlie blocks treated being about tliree or four times as long as wide, ^11 tlie elements will be considered as being under -equal compressive stresses at tbe same time. 201. Notation for Compression. — By using a subscript c, we may write E^= Modulus of Elasticity;* i.e. tlie quotient of the compressive stress per unit of area divided by the relative shortening. (Young's Modulus; no stress on sides); C= Modulus of crushing ; i.e. the force per unit of sec- tional area necessary to rupture the block by crushing ; G'= Modulus of safe compression, a safe compressive stress per unit of area ; and G"= Modulus of compression at elastic limit. For the absolute and relative shortening in length we may still use X and e, respectively, and within the elastic limit may write equations similar to those for tension, F being the sectional area of the block and F one of the ter- minal forces, while p = compressive stress per unit of area of Ff viz.: . (1) v-F _ -dx X- rF_ ■A ~ .PI ~FX ithin the elastic limit. Also for a short block Crushing force =FG Compressive force at elastic limit =iFG" }• . (2]f Safe compressive force =FG' limit r=FG" \ . 7' ) 202. Remarks on Crushing. — As in § 182 for a tensile stress, so for a compressive stress we may prove th<at a *[NoTE. — It must be remembered that the modulus of elasticity, whether for normal or shearing stresses, is a number indicative of stiff- ness, not of strength, and has no relation to the elastic limit (except that experiments to determine it must not pass that limit).] 220 MECHANICS OF ENGINEERING. shearing stress = p.sin. a cos a is produced on planes at an angle a with the axis of the short block, p being the com- pressive stress per unit of area of transverse section. Experi- ment shows, however, that, although the above value for the shea,riug stress is a maximum for a = 4: 5°, in the crushing of shoi^t blocks or rather brittle materials like cast iron and stone, the surface along which separation takes place makes an angle smaller than 45° with the axis (35° for cast iron, according to Hodgkinson's experiments) ; but the block must be two or three times as long as wide to enable this phenomenon to take place. This seems to show that the presence of the com- pressive stress on the 45° plane is sufficient to strengthen the material against rupture by shearing on that plane, causing the separation to occur along a plane on which the compressive stress is considerably less. Crushing by splitting into pieces parallel to the axis sometimes occurs. Blocks of ductile material, however, yield by swelling out, or bulging, laterally, resembling plastic bodies some- what in this respect. The elastic limit is more difficult to locate than in ten- sion^ but seems to have a position corresponding to that in tension, in the case of wrought iron and steel. With cast iron, however, the relative compression at elastic limit is about double the relative extension (at elastic limit in tension), but the force producing it is also double. For all three metals it is found that E^^=E^ quite nearly, so that the single symbol U m.aj be used for both. EXAMPLES IIS" TENSION AND COMPRESSION. 203. Tables for Tension and Compression. — The round numbers in the following tables are to be taken as rude aver- ages only ; the scope and design of the present work admitting of nothing more. For abundant details of the more import- ant experiments and researches of recent years, the reader is referred to Professor J. B. Johnson's "Materials of Con- struction ' ' and the works of Professors Thurston, Burr, and Lanza ; also to ' ' Testing of Materials ' ' by Unwin, and Martens' work of similar title. Another column might have been added giving the Modulus of Resilience, viz. : ie"T'\ {~^T"^^2E; see §196). e is an abstract num- EXAMPLES Ilf TEirSIOlS' A^D COMPRESSION 221 ber, and =X-^l, while E^, T", and T are given in pounds per square incli: TABLE OF THE MODULI, ETC., OP MATEEIALS IN TENSION. e" £ K rpn T Material. (Elastic limit.) At Rupture. Mod. of Elast. Elastic limit. Eupture. abst. number. abst. number. lbs. per sq. in. lbs. per sq. in. lbs. per sq. in. Soft Steel, .00120 .3000 30,000,000 35,000 60,000 Hard Steel, .00200 .0500 40,000,000 60,000 120,000 Cast Iron, Wro't Iron, Brass, .00066 .00080 .00100 .0020 .3000 14,000,000 28,000,000 10,000,000 9,000 22,000 f 7,000 to L 19,000 18,000 • ■ 45 000 to 60,000 16.000 to 50,000 Glass, 9,000,000 3,500 Wood, with the fibres. ( .00200 K to ( .01100 .0070 to .0150 200,000 to 2,000,000 3,000 to 19,000 6,000 to 28,000 Hemp rope, 7,000 [N.B.— Expressed in kilograms per square centim., E^, T and T" would be nu merically about V]4 as large as above, while € and e" would be unchanged.] TABLE OF MODULI, ETC.; COMPEESSION OP SHORT BLOCKS. e" £ E, G" C Material. Elastic limit. At lupture- Mod. of Elast. Elastic limit. Rupture. abst. number abst. number. lbs. per sq. in. lbs. per sq. in. lbs. per sq. in. Soft Steel, 0.00100 30,000,000 30,000 Hard Steel. 0.00120 0.3000 40,000,000 50,000 200,000 Cast Iron, 0.00150 14,000,000 20,000 90,000 Wro't Iron, 0.00080 0.3000 28,000,000 24,000 40,000 Glass, 20,000 Granite, Sandstone, Brick, See §213a 10,000 5,000 3,000 Wood, with the fibres. I 0.0100 < to 1 0.0400 350,000 to 2,000,000 2,000 to 10,000 Portland 1 Cement, f (§ 213a) 4,000 "222 MECHANICS OF ENGINBEEING. 204. Examples. No. 1. — A bar of tool steel, of sectional area =0.097 sq. inches, is ruptured by a tensile force of 14,000 lbs. A portion of its length, originally ^ a foot, is now found to have a length of 0.532 ft. Required T, and e at rupture. Using the inch and pound as units (as in the foregoing tables) we have T= 1^=144326 lbs. per Bq. in.; (eq. (2) § 195) ; while e=(0.532— 0.5)x12h-(0.50x12)=0.064. Example 2. — Tensile test of a bar of " Hay Steel " for the Glasgow Bridge, Missouri. The portion measured was originally 3.21 ft. long and 2.09 in. X 1.10 in. in section. At the elastic limit P was 124,200 lbs., and the elongation was 0.064 ins. Required E^, T''^ and e" (for elastic limit). e"=^ =,-M54^=.00166 at elastic limit. I 3.21x12 r"=124,200--(2.09xl.l0)=54,000 lbs. per sq. in. Nearly the same result for E^ would probably have been ^obtained for values of p and e below the elastic limit. The Modulus of Resilience of the above steel (see § 196) would be ^2 e" :!r"= 44.82 inch -pounds of work per cubic inch of metal, so that the whole work expended in stretch- ing to the elastic limit the portion above cited is Cr= y^ e" T" r=3968. inch -lbs. An equal amount of work will be done by the rod in re- -covering its original length. Example 3. — ^A hard steel rod of ^ sq. in. section and :20 ft. long is under no stress at a temperature of 130' EXAMPLES IN TENSION AND COMPEESSION. 223 Cent., and is provided witli flanges so that tlie slightest contraction of length will tend to bring two walls nearer together. If the resistance to this motion is 10 tons how low must the temperature fall to cause any motion ? tj be- ing =.0000110 (Cent, scale). From § 199 we have, ex- pressing P in lbs. and F in sq. inches, since E^= 40,000,000 ^hs. per sq. inch, 10x2,000 =40,000,000 x }4 X(1304') x 0,000011 ; whence ^'=39.0° Centigrade. Example 4. — If the ends of an iron beam bearing 5 tons at its middle rest upon stone piers, required the necessary bearing surface at each pier, putting C for stone =200 lbs. per sq. inch. 25 sq. in., Ans. Example 5. — How long must a wrought iron wire* be, supported vertically at its upper end, to break with its own weight ? 216,000 inches, Ans. Example 6, — One voussoir (or block) of an arch-ring presses its neighbor with a force of 50 tons, the joint hav- ing a surface of 5 sq. feet ; required the compression per sq. inch. 138.8 lbs. per sq. in., Ans. 205. Factor of Safety. — When, as in the case of stone, the value of the stress at the elastic limit is of very uncertain determination by experiment, it is customary to refer the value of the safe stress to that of the ultimate by making it the w'th portion of the latter, n is called a factor o/ safety, and should be taken large enough to make the safe stress come within the elastic limit. For stone, n should not be less than 10, i.e, C'^G-^n; (see Ex. 6, just given). 206. Practical Notes. — It was discovered independently by Commander Beardslee and Prof. Thurston, in 1873, that if wrought iron rods were strained considerably beyond the elastic limit and allowed to remain free from stress * Take T = 60,UU0 lbs. per square incli. 224 MECHANICS OF ENGINEERING. for at least one day thereafter, a second test would sliow higher limits both elastic and ultunate. In 1899 Mr. James Muir discovered that this recovery of elasticity and raising of both the yield-point and ultimate strength, in the case of iron and steel after "overstraining," may be brought about by sim23ly heating the metal for a few minutes in a bath of boiling water. In one experiment a bar of a kind of mild steel which under ordinary tests broke at 39 tons/in.2 with 20% elongation on 8 in., was stretched just to its yield-point, then relieved and heated for a few minutes to 100° Cent., then stretched just to its new yield-point, then relieved and heated as before; and so on, for three times more. The first yield -point was at 27, the others at 33, 38, 43i, and 47 tons/in.2 The bar was then broken at 49 tons/in.2 with total extension of 12%. The diminished ultimate extension shows the hardening effect of the treatment. (See Prof. Ewing's ''Strength of Materials,'' pp. 38 and 40.) 'Bj fatigue of metals we understand the fact, recently dis- covered by Wohler in experiments made for the Prussian Ci-overnment, that rupture may be produced by causing the stress on the elements to vary repeatedly between two limiting values, the highest of which may be considerably below T (or G), the number of repetitions necessary to produce rupture being dependent both on the range of variation and the higher value. For example, in the case of Phoenix iron in tension, Tupture was produced by causing the stress to vary from to 52,800 lbs. per sq. inch, 800 times ; also, from tc 44,000 lbs. per sq. inch 240,858 times ; while 4,000,000 va- liations between 26,400 and 48,400 per sq. inch did not cause rupture. Many other experiments were made and the following conclusions drawn (among others): Unlimited repetitions of variations of stress (lbs. per ^q. in.) between the limits given below will not injure the paetal (Prof. Burr's Materials of Engineering). ^ , , . j From 17,600 Comp. to 17,600 Tension, roug iron. | ^^ ^ ^^ ^^^^^^ ( From 30,800 Comp. to 30,800 Tension. Axle Cast Steel. ^ " to 52,800 ( " 38500 Tens, to 88,000 « (See p, 5i3'2 for an addendum to this paragraph.) SHEARING. 225 SHEARING. 207. Rivets. — The angular distortion called shearing strain in the elements of a body, is specially to be provided for in the case of rivets joining two or more plates. This distortion is shown, in Figs. 205 and 206, in the elements near jhe plane of contact of the plates, much exaggerated. jT^ i > >r* T Fig. 205. Fig, 206. In Fig. 205 (a lap-joint) the rivet is said to be in single shear ; in Fig. 206 in double shear. If P is just great enough to shear off the rivet, the modulus of ultimate shear- ing, which may be called S, (being the shearing force per unit of section when rupture occurs) is F iTid^ (1) in which i^==the cross section of the rivet, its diameter being =d. For safety a value S'= }{ to ^ of >S' should be taken for metal, in order to be within the elastic limit. As the width of the plate is diminished by the rivet hole the remaining sectional area of the plate should be ample to sustain the tension P, or 2P, (according to the plate considered, see Fig. 206), P being the safe shearing force for the rivet. Also the thickness t of the plate should be such that the side of the hole shall be secure against crushing ; P must not be > C'td, Fig. 205. Again, the distance a, Fig. 205, should be such as to prevent the tearing or shearing out of the part of the plate between the rivet and edge of the plate. 226 MECHANICS OF E:NGINEEKING. For economy of material tlie seam or joint sliould be no more liable to rupture by one tban by another, of the o o ^ o ■.t Fig. 307. four modes just mentioned. The relations which must then subsist will be illustrated in the case of the " butt= joint " with two cover-plates, Fig. 207. Let the dimen- sions be denoted as in the figure and the total tensile force on the joint be = Q. Each rivet (see also Fig. 206) is ex- posed in each of two of its sections to a shear of I2 Q} hence for safety against shearing of rivets we put 12 Q- -% Tides' (1) Along one row of rivets in the main plate the sectional area for resisting tension is reduced to {b — ^d)t,, hence for safety against rupture of that plate by the tension Q, we put Q=(h—3d)t,T' , (2) Equations (1) and (2) suffice to determine d for the rivets and ^1 for the main plates, Q and b being given ; but the values thus obtained should also be examined with refer- ence to the compression in the side of the rivet hole, i.e., J^ Q must not be > C't.d. [The distance a, Fig. 205, to the edge of the plate is recommended by different authorities to be from d to 3d.] Similarly, for the cover -plate we must have and 12^ not > G'td, ^4QoT(b—dd)tT' < (8) SHEARi:srG. 227 If the rivets do not fit their holes closely, a large margin should be allowed in practice. Again, in boiler work, the pitch, or distance between centers of two consecutive rivets may need to be smaller, to make the joint steam-tight, than would be required for strength alone. 208, Shearing Distortion. — The change of form in an ele- ment due to shearing is an angular deformation and will be measured in tt -measure. This angular change or dif- ference between the value of the corner angle during strain and ^4i'^, its value before strain, will be called d, and is proportional (within elastic limit) to the shearing stress per unit of area, p^, existing on all the four faces whose angles with each other have been changed. Fig. 208. (See § 181). By § 184 the Modulus of Shearing Elasticity is the quotient obtained by dividing p^hj d \ i.e. {elastic limit not passed)^ ^s=^ . . . . (1) or inversely, d=p^-^E^ (1)' The value of E^ for different substances is most easily I determined by experiments on torsion in which shearing is the most promi- / nent stress.* (This prominence depends y\ on the position of the bounding planes j^ of the element considered ; e.g., in Fig. _ __.L 208, if another element were considered / '---cix->, within the one there shown and with Fig. 208. its plaues at 45° with those of the first, we should find tension alone on one pair of opposite faces, compression alone on the other pair.) It will be noticed that shearing stress cannot be present on two opposite faces only, but exists also on another pair of faces (those perpendicular to the stress on the first), forming a couple of equal and opposite moment to the first, this being necessary for the equilibrium of the element, even when * For instance, see numerical example on p. 237, giving a value of Es as resulting from a torsion test made by students in the Civil Engi- neering Laboratory at Cornell University, April, 1904. 228 MECHAIflCS OF ENGINBERIKG. tensile or compressive stresses are also present on the faces considered. 209. Shearing Stress is Always of the Same Intensity on the Four Faces of an Element. — (By intensity is meant per unit of area ; and the four faces referred to are those perpen- dicular to the paper in Fig. 208, the shearing stress being parallel to the paper.) Let dx and dz be the width and height of the element in Fig. 208, while dy is its thickness perpendicular to the paper. Let the intensity of the shear on the right hand face be =q^, that on the top face =Ps. Then for the ele- ment aw a free body, taking moments about the axis per- pendicular to paper, we have q^ dz dy X dx — ^g dx dy x dz =0 .•. qs =p^ {dx and dz being the respective lever arms of the forces q^ dz dy and p^ dx dy.) Even if there were also tensions (or compressions) on one or both pairs of faces their moments about would balance (or fail to do so by a differential of a higher order) independently of the shears, and the above result would still hold. 210. Table of Moduli for Shearing. d" ^s 8" s Material. i.e. 6 at elastic limit. Mod. of Elasticity for Shearing. (Elastic limit.) (Rupture.) arc in radians. lbs. per sq. in. lbs. per sq. in lbs. per sq. in. Soft Steel, 9,000,000 30,000 70,000 Hard Steel, 0.0033 14,000,000 45,000 90,00C Cast Iron, 0.0021 7,000,000 15,000 30,00C Wrought Iron, 0.0022 9,000,000 20,000 50,000 Brass, 5,000,000 Glass, Wood, across ( fibre, 1 1,500 to 8,000 Wood, along ( fibre, ( 500 to 3,200 SHEAKIISG. 229 As in tlie tables for tension and compression, tlie above values are averages. The true values may differ from these as mucli as 30 per cent, in particular cases, accord-* ing to the quality of the specimen. 211. Punching rivet holes in plates of metal requires the overcoming of the shearing resistance along the convex surface of the cylinder punched out. Hence ii d = diam- eter of hole, and t= the thickness of the plate, the neces- sary force for the punching, the surface sheared being F= tjvd^ is P=8t^d (2)- Another example of shearing action is the " stripping " of the threads of a screw, when the nut is forced off lon- gitudinally without turning, and resembles punching in its nature. 212. EandEgj Theoretical Relation. — In case a rod is in iension within the elastic limit, the relative (linear) lateral contraction (let this =m) is so connected with E^ and E^ ihat if two of the three are known the third can be de- duced theoretically. This relation is proved as follows, by Prof. Burr. Taking an elemental cube with four of its faces at 45° with the axis of the piece, Fig. 209, the axial half-diagonal AD becomes of a length AD'=AD-\-s.AD under stress, while the transverse half diagonal contracts to a length B'D'=AD — m.AD, The angular distortion d .-\U.S . • X<'>^' /^<\^^^ A< ^ D D/^ ^^' ,. \>6-^ Fig. 209. § 212. Fig. 210. 230 MECHAIflCS OF EI^GINEEKING. is supposed very small compared with 90° and is due to the shear j9g per unit of area on the face BG (or BA\ From the figure we have tan(45°— -) = __^,=__=1— m— s, approx. [But, Fig. 210, tan(45° — x)=l — 2x nearly, where a; is a small angle, for, taking CA=unitj=AE, ian AD=AF= AE—EF. Now approximately EF= EG, y2 and EG= BDa^^=x^^ .'. AF= 1 — 2a7 nearly.] Hence 1 — d= 1 — m — £ ; or d= m+e . . (2) Eq. (2) holds good whatever the stresses producing the deformation, but in the present case of a rod in tension, if it is an isotrope, and if ^ = tension per unit of area on its transverse section, (see § 182, putting «=45°), we have ^t==p-=-£ and E^={psOJx BG')-^d=}^p-^d. Putting also (m : £)= h, whence m=k£, eq. (2) may finally be written* >4-==(^ + l)4-; i.e., ^s=-^^ . . (3) Prof, Bauschinger, experimenting with cast iron rods,, found that in tension the ratio m: £was =m} as an average, which in eq. (3) gives ^=12^^,= !^, nearly. , , . (4) 246 5 ^ ^ ^ His experiments on the torsion of cast iron rods gave ^,= 6,000,000 to 7,000,000 lbs. per sq. inch. By (4), then, E, should be 15,000,000 to 17,500,000 which is approxi- mately true (§ 203). Corresponding results may be obtained for short blocks in compression, the lateral change being a dilatation in- stead of a contraction. * This ratio, m-^e, denoted by k, is called Poisson's Ratio. For metals its value lies approximately between 0.20 and 0.35. See also p. 507. 313. Examples in Shearing. — Example 1. — Kequired the^ proper length, a, Fig. 211, to guard against the shearing off, along the grain, of the portion ah, of a wooden tie-rod, the force P being = 2 tons, and the width of the tie = 4 inches. Using a value of S' = 100 lbs. per sq. irL.s we put 6a/S''= 4,000 cos 45° ; i.e. ■Fi^m. a= (4,000x0. 707) --(4x100)= 7.07 inches. Example 2.— A ^ in. rivet of wrought iron, in single shear (see Eig. 205) has an ultimate shearing strength P= FS=}(7T(PS= %7t{ Yq y X 50,000= 30,050 lbs. For safety, putting aS"= 8,000 instead of aS',P'=4,800 lbs. is its safe shearing strength in single shear. The wrought iron plate, to be secure against the side- crushing in the hole, should have a thickness t, computed thus I P'=tdC' ; or 4,800=^.^ x 12,000 ,-. ^=0.46 in. If the plate were only 0.23 in. thick the safe value of P would be only ^ of 4,800. Example 3. — Conversely, given a lap-joint, Fig. 205, in which the plates are ^ in. thick and the tensile force on the joint = 600 lbs. per linear inch of seam, how closely must ^ inch rivets be spaced in one row, putting jS"=8,000 and 6" =12,000 lbs. per sq. in. ? Let the distance between centres of rivets be =x (in inches), then the force upon each rivet =600a7, while its section P=0.44 sq. in. Having regard to the shearing strength of the rivet we put 600cc= 0.44x8,000 and obtain a?=5.86 in.; but considering that the safe crushing resistance of the hole is =1^-^.12,000= 2,250 lbs., 600aj=2,250 gives a;=3.75 inches, which is the pitch to be adopted. What is the tensile strength of the. reduced sectional area of the plate, with this pitch '? 232 MECHAXIC3 OF EKGIISTEBIilNG. Example 4 — Double butt-joint ; (see Fig. 207) ; ^s iiich plate; ^ in. rivets; F =C' =12,000 ; S' =8,333; width of plates=14 inches. Will one row of rivets be sufficient at each, side of joint, if ^=30,000 lbs.? The number of rivets ^^ ? Here each rivet is in double shear and has therefore a double strength as regards shear. In double shear the safe strength of each rivet =2i^>S"= 7,333 lbs. Now 30,000^ 7,333=40 (saj). With the four rivets in one row the re- duced sectional area of the main plate is =[14 — 4x ^] X^s =4,12 sq. in., whose safe tensile strength is =i^J"=4„12x 12,000=49,440 lbs.; which is > 30,000 lbs. .% main plate is safe in this respect. But as to side-crushing in holes in main plate we find that G't^d (i,e, 12,000 X Vs >^ M^^'^'^^ lbs.) is <.%Q i,e. <7,500 lbs., the actual force on side of hole. Hence four rivets in one row are too few unless thickness of maiiL plate be doubled. Will eight in one row be safe ? 213a, (Addendum to § 206.) Elasticity of Stone and Cements. — Experiments by Gen. Gill more with the large Watertowi» testing-machine in 1883 resulted as follows (see p. 221 for notation) : With cubes of Haverstraw Freestone (a homogeneous brown- stone) from 1 in. to 12 in. on the edge, E^ was found to be from 900,000 to 1,000,000 lbs. per sq. in. approximately ; and C about 4,000 or 5,000 lbs. per sq. in. Cubes of the same range of sizes of Djckerman's Portland cement gave E^ from 1,350,000 to 1,630,000, and G from 4,000 to 7,000, lbs. per sq. in. Cubes of concrete of the above sizes, made with the Newark Cc.'s Rosendale cement, gave E^ about 538,000^ while cubes of cement-mortar, and some of concrete, both made with National Portland cement, showed E^ from 800,000 to 2,000,- OOO lbs. per sq. in. The compressibility of hrick jpiers 12 in. square in section and 16 in. high was also tested. They were made of common North River brick with mortar joints f in. thick, and showed a value for E„ of about 300,000 or 400,000, while at elastic limit C" was on the average 1,000, lbs. per sq. in. TORSIOM. 233 CHAPTER IL TOKSION. S14. Angle of Torsion and of Helix. "When a cylindrical beam or shaft is subjected to a twisting or torsional action, I. e. when it is the means of holding in equilibrium two couples in parallel planes and of equal and opposite mo- ments, the longitudinal axis of symmetry remains straight and the elements along it exper- lience no stress (whence it may be I called the "line of no twist"), while the lines originally parallel to Fig. 212. i^ assume the form of helices, each element of which is distorted in its angles (originally right angles), the amount of distortion being assumed pro- portional to the radius of the helix. The directions of the faces of any element were originally as follows : two radial, two in consecutive transverse sections, and the other two tangent to two consecutive circular cylinders whose com- mon axis is that of the shaft. E.g. in Fig. 212 we have an unstrained shaft, while in Fig. 213 it holds the two 234 MECHANICS OF ENGINEEHmG. couples (of equal moment Pa = Qh) in equilibrium. These couples act in parallel planes perpendicular to the axis of the prism and a distance, ?, apart. Assuming that the transverse sections remain plane and parallel during tor- sion, any surface element, m, which in Fig. 212 was entire- ly right-angled, is now distorted. Two of its angles have been increased, two diminished, by an amount d, the angle between the helix and a line parallel to the axis. Suppos- ing m to be the most distant of any element from ihe axis, this distance being e, any other element at a distance s from the axis experiences an angular distortion =- <§„ If now we draw B' parallel to 0' A the angle B B', =a, is called the Angle of Torsion, while d may be called the helix angle', the former lies in a transverse plane, the latter in a plane tangent to the cylinder. Now tan d = (linear arc B B')-t-1; but lin. arc B B' =' ea; hence, putting d for tan d, (3 being small) (1) (d and « both in radians). 215. Shearing Stress on the Elements. The angular distor- tion, or shearing strain, d, of any element (bounded as al- ready described) is due to the shearing stresses exerted on it by its neighbors on the four faces perpendicular to the tangent plane of the cylindri- cal shell in which the element is situated. Consider these neighboring elements of an outside element removed, and the stresses put in ; the latter are accountable for the dis- ^®- ^*^ tortion of the element and -pdF TOKSiou". 235 hold it in equilibrium. Fig. 214 shows this element "free." Within the elastic limit ^ is known to be propor- tional to jOg, the shearing stress per unit of area on the faces whose relative angular positions have been changed. That is, from eq. (1), § 208, S -^^p^-r-Us; whence, see (1) of § 214, In (2) Ps, and e both refer to a surface element, e being the radius of the cylinder, and p^ the greatest intensity of shearing stress existing in the shaft. Elements lying nearer the axis suffer shearing stresses of less intensity in pro- portion to their radial distances, i.e., to their helix-angles. That is, the shearing stress on that face of the element which forms a part of a transverse section and whose dis- tance from the axis is z, is p, =— p^, per unit of area, and the total shear on the face is pdF, c?^ being the area of the face. 216. Torsional Strength. — ^We are now ready to expose tlia full transverse section of a shaft under torsion, to deduce formulae of practical utility. Making a right section of the shaft of Fig. 213 anywhere between the two couples and considering the left hand portion as a free body, the forces holding it in equilibrium are the two forces P of the bft-hand couple and an infinite number of shearing forces, each tangent to its circle of radius s, on the cross section exposed by the removal of the right-hand portion. The cross section is assumed to remain plane during tor- sion, and is composed of an infinite number of dF's, each being the area of an exposed face of an element | see !Fig. 236 elementary shearing force = S p^dF, and s is its lever arm about the axis Oo . For equilibrium, S (mom.), about the axis Oo must =0 ; i.e. in detail _p^o^p^a+ f ( -£ p,dF)%^ii ©r, redncing. h rz^dF=Pa\ or, A.^Pa eJ e (3) Eq, (3) relates to torsional strength, since it contains ^s, tha greatest shearing stress induced by the torsional couple, whose moment Pa is called the Moment of Torsion, the stresses in the cross section forming a couple of equal and opposite moment. Pa is also called the "torque." Ip is recognized as the Polar Moment of Inertia of the cross section, discussed in § 94 ; e is the radial distance of the outermost element, and = the radius for a circular shafto 217. Torsional Stiffness. — In problems involving the angle of torsion, or deformation of the shaft, we need an equa- tion connecting Pa and a, which is obtained by substitut- ing in eq. (3) the value of p^ in eq. (2), whence I =Pa. (4) From this is appears that the angle of torsion, a, is pro- portional to the moment of torsion, or " torque," Pa inch -lbs., within the elastic limit; a must be expressed in radians. TORSION. 237 Example. — A portion 3.4 ft. long, of a solid cylindrical shaft of soft steel, of diam. = 1.5 in., is found by the use of "Torsion Clinometers" (see frontispiece) to be held at an angle of torsion of a = 5.41°, =0.0944 radians, just before the elastic limit is reached, by a "torque," =Pa, of 10,200 in. -lbs. Compute the Modulus of Elasticity for Shearing. Substituting in eq. (4), with I-p=Tzr^l2, (§94), =7r(0.75)^^2, =0.497 in.*, and Z = 3.4X 12 = 40.8 in., we have 10,200X40 .8 Qg^nnnniv. ' ^ 0.0944 X 0.4 97 ^ 8,870,000 lbs. per sq. m. 218. Torsional Resilience is the work done in twisting a shaft from an unstrained state until the elastic limit is reached in the outermost elements. If in Fig. 213 we imagine the right-hand extremity to be fixed, while the other end is gradually twisted through an angle each force P of the couple must be made to increase grafdually from a zero value up to the value Pj, corresponding to ai. In this motion each end of the arm a describes a space = ^aai, and the mean value of the force = }4Pi (compare § 196). Hence the work done in twisting is Ui=}4FiX}4aaiX2=}4Piaaj^ . . (5) By the aid of preceding equations, (5) can be written If for ps "^e write 8' (Modulus of safe shearing) we have for the safe resilience of the shaft U'=4r^ -. . . . (7) If the torsional elasticity of an originally unstrained shaft is to be the means of arresting the motion of a moving mass whose weight is O, (large compared with the parts intervening) and velocity =v, we write (§ 133) g 2' as tlie condition that the shali shall not be injnrecL 238 mecha:^ics of engineering. 21U. roiar Moment of Inertia. — ^For a shaft of circular cross section (see § 94) /p=i^7rr*; for a hollow cylinder /p=i^7r(ri* — r^) ', while for a square shaft If=yih^, h being "the side of the square ; for a rectangular cross-section sides h and li, I^=lJbh{lf-\-¥). For a cylinder e=r; if hoi- low, e=r , the greater radius. For a square, e=i^6y'2. 220. Ifon-Circular Shafts. — If the cross-section is not cir- cular it becomes warped, in torsion, instead of remaining plane. Hence the foregoing theory does not strictly ap- ply. The celebrated investigations of St. Tenant, how- ever, cover many of these cases. (See § 708 of Thompson and Tait's Natural Philosophy ; also, Prof. Burr's Elas- ticity and Strength of the Materials of Engineering). His results give for a square shaft (instead of the ab'E.^ Pa of eq. (4) of § 217), Ql Pa=OMl^t . . . . (1) and Pa=Jffi^p^f instead of eq. (3) of § 216, 2?s being the greatest shearing stress. The elements under greatest shearing strain are found at the middles of the sides, instead of at the corners, when the prism is of square or rectangular cross-section. The warping of the cross-section in such a case is easily veri ' fied by the student by twisting a bar of india-rubber in his fingers. 221. Transmission of Power. — Fig. 216. Suppose the cog- wheel B to cause A, on the same shaft, to revolve uni- formly and overcome a resis- tance Q, the pressure of the teeth of another cog-wheel, P 5 being driven by still another Fig. 216. wheel. The shaft AB is un- der torsion, the moment of torsion being =Pa= Qh. (Pi and ^1 the bearing reactions have no moment about the axis of the shaft). If the shaft makes u revolutions per unit-time, the work transmitted {transmitted ; not expend^ TOESION. 239 ed in twisting the shaft whose angle of torsion remains constant, corresponding to Fa) per unit-time, i.e. the Power, is X/=P.27ra.u=27ruPa . , , (8) To reduce L to Horse Power (§ 132), we divide by N, the number of units of work per unit -time constituting vOne H. P. in the system of units employed, i.e., Horse Power =H. p=?!E!^ JSl For example JSf =33,000 ft. -lbs. per minute, or =396,000 inch -lbs. per minute ; or = 550 ft. -lbs. per second. Usually the rate of rotation of a shaft is given in revolutions per minute. But eq^. (8) happens to contain Fa the moment of torsion acting to maintain the constant value of the angle of tor- sion, and since for safety (see eq. (3) § 216) Fa= S' I.^-^ e, with -ZJ,= y^TiT^ and e=r for a solid circular shaft, we have for such a shaft (Safe),H.P.=?f^ . . . (9) N which is the safe H. P., which the given shaft can trans- mit at the given speed. S' may be made 7,000 lbs. per sq. inch for wrought iron ; 10,000 for steel, and 5,000 for cast- iron. If the value of Pa fluctuates periodically, as when a shaft is driven by a connecting rod and crank, for (H. P.) we put toX(H. p.), m being the ratio of the maximum to the mean torsional moment; m= about 172 under ordi- nary circumstances (Cotterill). With a hollow cylindrical shaft, of outer radius = rj, and inner = r 2 the r^ of eq. (9) must be replaced by (?*i*— /•2*)-^'"i- If> furthermore, the thickness of metal is small, we may proceed thus, taking numerical data: Let the radius to the middle of the thickness be ro = 10 in., the thickness t=\ in., and the (steel) shaft make m=120 revs./min. ; with *S' = 5000 Ibs./in.^; then the total safe shearing stress in the cross-section is- 12' = 27rroi5;' = 27rlOXiX 5000 = 78,540 lbs., whHe the velocity of the mid-thickness is v' = 27rroU = 27: 10X2 = 125.6 in./sec. = 10.47 ft. /sec. Hence the (safe) power that may be transmitted at given speed is L = R'v' = 78,540X10.47 = 822,100 ft.-lbs. per sec; or, (-^550), =1495 H.P. 240 MECHAITICS OF ElifGINEEEING. 222. Autographic Testing Machine. — Tlie principle of Prof Thurston's invention bearing this name is shown in Fig • Fie. 217. 217. The test-piece is of a standard shape and size, its central cylinder being subjected to torsion. A jaw, carry- ing a handle (or gear-wheel turned by a worm) and a drum on which paper is wrapped, takes a firm hold of one end of the test-piece, whose further end lies in another jaw rigidly connected with a heavy pendulum carrying a pen- cil free to move axially. By a continuous slow motion of the handle the pendulum is gradually deviated more and more from the vertical, through the intervention of the test-piece, which is thus subjected to an increasing tor- sional moment. The axis of the test-piece lies in the axis of motion. This motion of the pendulum by means of a proparly curved guide, WH, causes an axial (i.e., parallel to axis of test-piece) motion of the pencil A, as well as an angular deviation /9 equal to that of the pendulum, and this axial distance CF,=^sT, of the peiicil from its initial position measures the momenr of torsion =i^«=:^(? sin )5.. As the piece twists, the drum and paper move relatively to the pencil through an angle sUo equal to the angle of torsion a so far attained. The abscissa so and ordinate sT oi the curve thus marked on the paper, measure,, when the paper is unrolled, the values of a and Pa through. TOESrOK 241 all the stages of the torsion. Fig. 218 shows typical Fig. 218. Jurves thus obtained. Many valuable indications are given by these strain diagrams as to homogeneousness of composition, ductility, etc., etc. On relaxing the strain at any stage within the elastic limit, the pencil retraces its path ; but if beyond that limit, a new path is taken called an " elasticity-line," in general parallel to the first part of the line, and showing the amount of angular re- CO very, BC, and the permanent angular set, OB. 2222i.. Torsion Clinometers. — ^When the test-piece used in the Thurston testing machine is short, the indicated angles of torsion below the elastic limit are far in excess of the actual values, on account of the initial yielding of the wedges in the jaws. By the use of " torsion clinometers," however (see frontispiece) the angle of torsion can be measured accurately within one minute of arc. 223. Examples in Torsion. — The modulus of safe shearing strengtn. S', as given in § 221, is expressed in pounds per square inch ; hence these two units should be adopted throughout in any numerical examples where one of the above values for S' is used. The, same statement applies to the modulus of shearing elasticity, E^, in the table of § 210. - Example 1.— Fig. 216. With P = 1 ton, a = 3 ft., I ^ 10 ft. , and the radius of the cylindrical shaft r=2.5 inches, required the max. shearing stress per sq. inch, ps, the shaft being of wrought iron. From eq. (3) § 216 Pae 2,000x36x2.5 o oomi, • -u ^----T^' V..X(2.5)^ =2,930 lbs. per sq. inch, which is a safe value for any ferrous raetaL 242 MECHANICS OF EIN GINEEEIXG. Example 2. — What H. P. is the shaft in Ex. 1 transmit- ting, if it makes 50 revolutions per minute ? Let u = number of revolutions per unit of time, and N = the num- ber of units of work per unit of time constituting one horse-power. Then H. V.^Pu^na-^N, which for the foot» pound-minute system of units gives H. P.=2,000x50x27rx3--33,000=57i4: H. P. Example 3. — What different radius should be given t(- the shaft in Ex. 1, if two radii at its extremities, originally parallel, are to make an angle of 2° when the given moment of torsion is acting, the strains in the shaft remaining con- stant. From eq. (4) § 217, and the table 210, with a=i|^c;r=* 0.035 radians (i.e. ;7-measure), and I^=^j^T^^ we have y^ 2,000x36x120 >^7r0.035x 9,000,000 -—=17.45 .-. r=2.04 inches. (This would bring about a different p,, but still safe.) The foregoing is an example in stiffness. Example 4. — A working shaft of steel (solid) is to tran:^- mit 4,000 H. P. and make 60 rev. per minute, the maximum twisting moment being 1^ times the average; requireil its diameter. • c^=14.74 inches. Ans. Example 5. — In example 1, p = 2,930 lbs. per square inch ; what tensile stress does this imply on a plane at 45° with the pair of planes on which Ps acts ? Fig. 219 shows p,dx dx^'Ps 'dx^ps Via. 220. TOESiOE^. 243 a small cube, of edge =dx, (taken from the outer helix of Fig. 215,) free and in equilibrium, tbe plane of the paper being tangent to the cylinder ; while 220 shows the portion BD 0, also free, with the unknown total tensile stress jorfa;^,^/^ acting on the newly exposed rectangle of area =dxxdx^% p being the unknown stress per unit of area. From sym- metry the stress on this diagonal plane has no shearing component. Putting 2' [components normal to^-D]=0, we have pdx^^2=2dx'^p^Gos4:5°=dx^p^^/2.'.p=ps . (1) That is, a normal tensile stress exists in the diagonal plane BD of the cubical element equal in intensity to the shearing stress on one of the faces, i.e., =2,930 lbs. per sq. in. in this case. Similarly in the plane AG will be found a compressive stress of 2,930 lbs. per sq. in. If a plane surface had been exposed making any other angle than 45° with the face of the cube in Fig. 219, we should have found shearing and normal stresses each less than p^ per sq. inch. Hence the interior dotted cube in 219, if shown " free " is in tension in one direction, in compression in the other, and with no shear, these normal stresses having equal intensities. Since S' is usually less than T' or C, ii Ps is made = S' the tensile and compressive actions are not injurious. It follows therefore that when a cylinder is in torsion any helix at an angle of 45° with the axis is a line of tensile, or of compressive stress, according as it is a right or left handed helix, or vice versa. Example 6. — A solid and a hollow cylindrical shaft, of equal length, contain the same amount of the same kind of metal, the solid one fitting the hollow of the other. Compare their torsional strengths, used separately. The solid shaft has only ^ the strength of the hollow one, Ans. Example 7. — Compare the shafts of Example 6 as to tor- sional stiffness (i. e. , the angles of torsion due to equal moments) . The solid shaft is only one-third as stiff as the other ; an equal moment produces three times the angle. Ans. 244 MECHANICS OF El!fGrNEi:KrN:G. CHAPTER in. FliEXURE OF HOMOGENEOUS PRISMS UHDEK PERPENDICULAK FORCES IN ONE PLANE. 224. Assumptions of tlie Common Theory of Plexure. — Wlien a prism is bent, under tlie action of external forces per- pendicular to it and in tlie same plane witli each otlier, it may be assumed tliat the longitudinal fibres are in tension on the convex side, in compression on the concave side, and that the relative stretching or contraction of the ele- ments is proportional to their distances from a plane in- termediate between, with the understanding that the flex- ure is slight and that the elastic limit is not passed in any element. i This " common theory " is sufficiently exact for ordinary engineering purposes if the constants employed are prop- erly determined by a wide range of experiments, and in- volves certain assumptions of as simple a nature as possi- ble, consistently with practical facts. These assumptions are as follows, (for prisms, and for solids with variable cross sections, when the cross sections are similarly situated as regards a central straight axis) and are approximately borne out by experiment : (1.) The external or " applied " forces are all perpendicu- lar to the axis of the piece and lie in one plane, which may be called the force-plane ; the force-plane contains the axis of the piece and cuts each cross-section symmetri- cally ; (2.) The cross-sections remain plane surfaces during flexure ; (3.) There is a surface (or, rather, sheet of elements) which is parallel to the axis and perpendicular to the force-plane, and along which the elements of the solid ex- FLEXURE. 245 perience no tension nor compression in an axial direction, this being called tlie Neutral Surface; (4.) The projection of the neutral surface upon the force plane (or a || plane) being called the Neutral Line or Elastic Curve, the bending or flexure of the piece is so slight that an elementary division, ds, of the neutral line may be put ^dx, its projection on a line parallel to the direction of the axis before flexure ; (5.) The elements of the body contained between any two consecutive cross-sections, whose intersections with the neutral surface are the respective Neutral Axes of the sections, experience elongations (or contractions, accord- ing as they are situated on one side or the other of the neutral surface), in an axial direction, whose amounts are proportional to their distances from the neutral axis, and indicate corresponding tensile or compressive stresses ; , (6.) E,=E,; (7.) The dimensions of the cross-section are small com- pared with the length of the piece ; (8.) There is no shear perpendicular to the force plane on internal surfaces perpendicular to that plane. In the locality where any one of the external forces is Applied, local stresses are of course induced which demand separate treatment. These are not considered at present. 225. Illustration. — Consider the case of flexure shown in Fig. 221. The external forces are three (neglecting the Fig. 22i. 246 MECHANICS OF EXGIXEBRING. weight of the beam), viz.: P^, Pg, and P3. P^ and P3 are loads, P2 the reaction of the support. The force plane is vertical. N^L is the neutral line or elastic curve. NA is the neutral axis of the cross-section at m / this cross-section, originally perpendicular to the sides of the prism, is during flexure ~| to their tangent planes drawn at the intersection lines ; in other words, the side view QNB, of any cross-section is perpendicular to the neutral line. In considering the whole prism free we have the system Pj, P2, and P3 in equilibrium, whence from 2^=0 we have P2=Pi+P3j and from 2" (mom. about P) =0, P3?3=Pi?i. Hence given Pi we may determine the other two external forces. A reaction such as Pg is some- times called a supporting force. The elements above the neutral surface NiOLS Sive in tension ; those below in com- pression (in an axial direction). 226. The Elastic Forces. — Conceive the beam in Fig. 221 separated into two parts by any transverse section such as QA, and the portion NiOJSf, considered as a free body in Fig. 222. Of this free body the surface QAB is one of ^^dx T«e. 222. FLEXURE. 247 i tlie bounding surfaces, but was originally an internal sur- face of tlie beam m Fig. 221. Hence in Fig. 222 we must put in the stresses acting on all the dF^^ or elements of area of QAB. These stresses represent the actions of the bodj taken away upon the body which is left, and according to assumptions (5), (6) and (8) consist of normal stresses (ten- sion or compression) proportional per unit of area, to th© distance, z, of the cZi^'s from the neutral axis, and of shear- ing stresses parallel to the force-plane (which in most cases will be vertical). The intensity of this shearing stress on any dF varies with the position of the dF with respect to the neutral axis, but the law of its variation will be investigated later (§§ 253 and 254). These stresses, called the Elastic Forces of the cross-section exposed, and the external forces Pj and P2, form a system in equilibrium. We may therefore ap- ply any of the 3onditions of equilibrium proved in § 38. 227. The Neutral Axis Contains the Centre of Gravity of the Cross-Section. — Fig. 222. Let e— the distance of the outer- most elem.ent of the cross-section from the neutral axis, and the normal stress per unit of area upon it be =p, whether tension or compression. Then by assumptions (5) and (6), § 224, the intensity of nprmal stress on any dF is = -1 p and the actual normal stress on any dFis= — pdF , {1} This equation is true for dF's having negative «'s, i.e. on the other side of the neutral axis, the negative value of the force indicating normal stress of the opposite char- acter ; 'for if the relative elongation (or contraction) of two axial fibres is the same for equal g's, one above, the other below, the neutral surface, the stresses producing the changes in length are also the same, provided ^t=:^^; see§§ 184 and 201. For this free body in equilibrium put 2'X=0 (Xis a horizontal axis). Put the normal stresses equal to their X components, the flexure being so slight, and the X com- 248 MECHANICS OF ENGINEEEIKG. ponent of the shears = for the same reason. This gives (see eq. (1) ) r± pdF= ; i.e. Z PdFz^ ; or, ^ i^i=0 (2) Ih which z— distance of the centre of gravity of the cross- section from the neutral axis, from which, though un- known in position, the g;'s have been measured (see eq. (4) § 23). In eq. (2) neither p-^e nor F can be zero .•. z must = ; i.e. the neutral axis contains the centre of gravity. Q. E. D. [If the external forces were not all perpendicular to the beam this result would not be obtained, necessarily.] 228. The Shear. — The " total shear," or simply the '* shear," in the cross-section is the sum of th.e vertical shearing stresses on the respective dF's. Call this sum J, and we shall have from the free body in Fig. 222, by putting ^y=0 (F being vertical) P,—F,—J=0.:J=F,—Pi . . (3) That is, the shear equals the algebraic sum of the ex- ternal forces acting on one side (only) of the section con- sidered. This result implies nothing concerning its mode of distribution over the section. 229. The Moment. — By the "Moment of Flexure" or simply the Moment, at any cross- section is meant the sum of the moments of the elastic forces of the section, taking ihe neutral axis as an axis of moments. In this summa- tion the normal stresses appear alone, the shear taking no part, having no lever arm about the axisiVA. Hence, Fig. 222, the moment of flexure (or "moment of resistance") =J(ipdF).=f/dF.^=£^ (*) This function, CdFz^, of the cross-section or plane figure FLEXURE. 249 is the quantity called Moment of Inertia of a plane figure, § 85. For the free body in Fig. 222, by putting 2'(mom.3 about the neutral axis NA)=0, we have then ^ — PiX^-]rP^X2=Q, or in general^ lL=zM . (5) e e in which M signifies the sum of moments,* about the neutral axis of the section, of all the forces acting on the free body considered, exclusive of the elastic forces of the exposed section itself. M is also called the "Bending Moment." Example. — In Fig. 222 let Pi = 3 and P2 = 4 tons, Xi = l ft. 8 in. and ^2 = 5 in.; the section of the beam being a rectangle, with NA=b = 3 in. and QB=^h = Q in. Then I about axis NA is, (p. 94), fo/i^n- 12 = 54 in.*; and e=3 in. Hence the "bending moment," M, =3X20-4X5 = 40 in. -tons. Equating M to the "moment of resistance" [or moment of the "stress couple" (see § 230)] we obtain, from eq. (5), p = Me-^I = 40 X 3 H- 54 = 2.22 tons/in.^ for the unit normal stress in the outer fibre at Q, or B. We find also, for the shear at section QB, / = 4— 3 = 1 ton. 230. Strength in Flexure. — Eq. (5) is available for solving problems involving the Strength of beams and girders, since it contains p, the greatest normal stress per unit of area to be found in the section. In the cases of the present chapter, where all the exter- nal forces are perpendicular to the prism or beam, and have therefore no components parallel to the beam, i.e. to the axis X, it is evident that the normal stresses in any section, as QB Fig. 222, are equivalent to a couple ; for the condition I!X=0 falls entirely upon them and cannot be true unless the resultant of the tensions is equal, parallel, and opposite to that of the compressions. These two equal and parallel resultants, not being in the same line, form a couple (§ 28), which we may call the stress -couple. The moment of this couple is the " moment of flexure " '~ , and it is further evident that the remaining forces in Fig. 222, viz.: the shear J and the external forces Pj and Pg* are equivalent to a couple of equal and opposite moment to the one formed by the normal stresses, * It is evident, therefore, that J!f (ft.-lbs., or in. -lbs.) is numerically equal to the "moment of flexure," or moment of the " stress couple " ; so that occasionally it maybe convenient to use "Jf" to denote the value of the latter momeut also. 250 MECHAKICS OF EXGHirBEIiriirG. 231. Flexural Stiffness. — Tiie neutral line, or elastic curvo^ containing the centres of gravity of all tlie sections, was originally straight ; its radius of curvature at any point, as N, Fig. 222, c'uring flexure may be introduced as fol- lows. QB and U'V are two consecutive cross -sections, originally parallel, but now inclined so that the intersec- tion G, found by prolonging them sufficiently, is the centre of curvature of the ds (put =dx) which separates them, at JSf, and CG=p= the radius of curvature of the elastic curve at N. From the similar triangles U' TIG and GNG we have dk'.dx'.:e;Pf in which dX is the elongation, U' U^ of a portion, originally =c?cc, of the outer fibre. But the rela- tive elongation £=-t— of the latter is, by §184, within the elastic limit, =^.\ -:^ =— and eq. (5) becomeL E E p ^ ^ EI =M (6) AXIS X From (6) the radius of curvature can be computed. E~ the value of E^—E^, as ascertained from experiments in bending. ~ To obtain a differential equation of the elastic curve, (6) may be transformed thus, Fig. 223. The curve being very flat, consider two consecutive (is's with equal dx's ; they may be put = their c^x's. Produce the first to intersect the dy of the second, thus cutting off the d^y^ i'e. the difference between two ^■^Jfy consecutive dy'^. Drawing a per- pendicular to each ds at its left extremity, the centre of curva- ture G is determined by their in- tersection, and thus the radius of curvature p. The two shaded Fig. 223. triangles have their small angles equal, and d^y is nearly perpen- dicular to the prolonged ds ; hence, considering them sim- ilar, we have \p,dx:'.dx'.d^y :.-^J^^, FLEXURE. 251 and hence from eq. (6) we ) , , ^A^V may write | (approx.) ±EI^=M . (7) as a differential equation of the elastic curve. From this the equation of the elastic curve may be found, the de- flections at different points computedj and an idea thus formed of the stiffness. All beams in the present chap- ter being prismatic and Jiomogeneous both jE' and / are the same (i.e. constant) at all points of the elastic curve^ In using (7) the axis Xmust be taken parallel to the length of the beam before flexure, which must be slight ; the minus sign in (7) provides for the case when d^y-r-dx^ ises"= sentialiy negative. 232. Resilience of Flexure. — If the external forces are made to increase gradually from zero up to certain maximum Yalues,. some of them may do work, by reason of their points of application moving through certain distances due to the yielding, or flexure, of the body. If at the be- ginning and also at the end of this operation the body is at rest, this work has been expended on the elastic resis- tance of the body, and an equal amount, called the work of resilience (or springing-back), will be restored by the elasticity of the body, if released from the external forces, provided the elastic limit has not been passed. The energy thus temporarily stored is of the potential kind; see §§ 148, 180, 196 and 218, 232a. Distinction. Between Simple, and Continuous, Beams (or ■** Girders "). — The external forces acting on a beam consist generally of the loads and the " reactions " of the sup* ports. If the beam is horizontal and rests on two supports only, the reactions of those supports are easily found by elementary statics [§ 36] alone, without calling into ac- count the theory of flexure, and the beam is said to be a Simple Beam, or girder ; whereas if it is in contact with more than two supports, being " continuous,'* therefore, over some of them, it is a Continuous Girder (§ 271). The Temainder of this chapter will deal only with simple 252 MECHANICS or E:NGi:tfEEIl]:NG. ELASTIC CURVES. 233. Case I. Horizontal Prismatic Beam, [Supported at Both Ends, With a Central Load, Weight of Beam Neglected. — Fig. 224. First considering the whole beam free, we find eack k- -Vd- % .—x- -I- :^ Fig. 324. § 233. reaction to be =%P. AOB is the neutral line ; required the equation of the portion OB referred to as an origin, and to the tangent line through as the axis of X To do this consider as free the portion mB between any sec- tion, m on the right of and the near support, in Fig. 225 The forces holding this free body in equilibrium Fig. S25. Fis. S26. nre the one external force ^P, and the elastic forces act- ing on the exposed surface. The latter consist of J, the shear, and the tensions and compressions represented in the figure by their equivalent " stress-couple." Selecting N, the neutral axis of m, as an axis of moments (that J may not appear in the moment equation) and putting 2 (mom) =0 we have P (I 2V"2 — X j rd^y-i 'y -P (I dx" dx' 2 \2 / (1) Fig. 226 shows the elastic curve OB in its purely geomet- rical aspect, much exaggerated. For axes and origin as in. figure d^y-^doc^ is positive. ELASTIC CURVES. 253 Eq. (1) gives the second a;-deriYative of y equal to a function of x. Hence tlie first fl?-derivative of y will be equal to tlie a?-anti-derivative of tliat function, plus a con- stant, (7. (By anti -derivative is meant tlie converse of de- rivative, sometimes called integral though not in the sense of summation). Hence from (1) we have (^/ being a con- stant factor remaining undisturbed) M^=~(Lx — -\+G . . (2)* dx 2 V2 2r (2)' is an equation between two variables c?2/-i-c?a; and a?, and holds good for any point between and B; dy-^dx de- noting the tang, of a, the slope, or angle between the tan- gent line and X At the slope is zero, and x also zero ; nencs at (2)' becomes ^7x0=0— 0+C which enables us to determine the constant C, whose value must be the same at as for all points of the curve. Hence C=0 and (2)' becomes EI ay _ r ( I xf\ ..^ ^~2"i-2'^2j • • ' ^^' from which the slope, tan. «, (or simply a, ir jt -measure: since the angle is small) may be found at auy point. Thus at B we have x=}4l and dy-^dx=ai, and . _ 1 PI' ••"^""16" M Again, taking the cp-anti-derivative of both, members of eq, (2) we have ^i2/=-f-(^-^)+C" . . . (3)' and since at both x and y are zero, G' is zero. Hence the equation of the elastic curve OB is 254 MECHANICS OF ENGIXEEEIFG. ^^^=f (^f) • • • <«' To compute the deflection of from the right line joiii'^ ing A and ^ in Fig. 224, i.e. BK, =c?, we put x^}^lm{^), a being then =d, and obtain ^^=■^=©•5 • • • <** Eq. (3) does not admit of negative values for x ; for if the free body of Fig. 225 extended to the left of 0, the ex- ternal forces acting would be P, aownward, at ; and y^P, upward, at B, instead of the latter alone ; thus altering the form of eq. (1). From symmetry, however, we know that the curve AO, Fig. 224, is symmetrical with OB about the vertical through Q. Numerical Illustration. — Let [the beam shown in Fig. 224, resting on two unyielding supports at the same level, be of white oak timber and bear a load of P = 200 lbs. at the middle, its length being Z=12 ft. and cross-section rectangular with a width (horizontal) of 6 = 2 in. and height /i = 6 in. The modiilus of elasticity E will be taken as 1,600,000 lbs./ in. ^ Required the radius of curvature, p, or the elastic curve at a point 4 ft. from the right-hand pier (or left). From the free body in Fig. 225 we have, using the form El-i-p for the moment of the stress-couple in the section, and putting i'(moms.)j\r = 0, with x = 2 ft., £7 -^J0= 100X48, the inch and pound being selected as units. Now I=bh^-irl2 (p. 94) which = 36 iu.^j whence, solving, (0 = 1,600, 000 X 36 -H 4800 = 4000 in. The cin-ve is evidently very flat. The smallest radius of curvature is found at the middle of the beam and is 2666 in.; at either extremity, A or B, it is infinite, since at each of these points the moment of the stress-couple is zero. At the same point (4 ft. from B) the "slope" of the elastic curve, viz., dy^dx, is found by putting x = 2 ft. = 24 in!, in eq. (2) from which is derived tan a = dy I dx = Q.Q025, corresponding to an angle of 0° 8' 36". At the extremity B we find, from ai = PP-7-16£'/, the slope of the tangent line to be ai = 0.0045; which is the tangent of 0° 15' 29". The deflection of the middle point is known from eq. (4), viz., d = PP^48EI; i.e., d=(200X 144 X 144 X 144) -v- (48X1,600,000X36) = 0.216 in. It now remains to ascertain if the elastic limit is passed in any fibre of the beam. If we put the form p/-^e (for moment of stress-couple) in place of the present left-hand member of eq. (1), and solve for the unit (normal) stress in outer fibre, we findp=JPe(J Z— 2;)-f-/, which shows that p is greatest in the outer fibre of the section for which ^l—x is greatest, within the limits of the half-length; and this occurs at the middle of the beam, where x = 0. With this substitution we obtain p(max.) = pm = Pie h- (47) ; or pm = (200 X 12 X 12 X 3) -f- (4 X 36) = 600 lbs./ in. ^, which is well within the elastic limit, for tension or compression in white oak. ELASTIC CURVES. 255 233a. Load Suddenly Applied. — Eq. (4) gives the deflection d corresponding to the force or pressure P applied at the middle of the beam, and is seen to be proportional to it If a load G hangs at rest from the middle of the beam, P=G', but if the load G, being initially placed at rest .upon the unbent beam, is suddenly released from the ex- ternal constraint necessary to hold it there, it sinks and dehects the beam, the pressure P actually felt by the beam varying with the deflection as the load sinks. What is the maximum deflection d^ ? and what the pressure P^^ between the load and the beam at the instant of maximum deflection? In this motion of the body, or "load," it is acted on by two forces, the constant downward force G (its weight) and the variable upward force P, whose average vahie is |Pni ; while its initial and final kinetic energy are each zero. G does the work Gd^, while the work done upon P is ^Pmdm \ hence, by the theorem of '' Work and Energy " (p. 138), v/e have Gd^^hPrJra + 0-Q (5) That is, Pm = 2(r. Since at this instant the load is sub- jected to an upward force of 2 (r and to a downward force of only G (gravity) it immediately begins an upward mo- tion, reaching the point whence the motion began, and thus the oscillation continues. We here suppose the elas- ticity of the beam unimpaired. This is called the " sud- den " application of a load, and produces, as shown above, double the pressure on the beam which it does when grad- ually applied, and a double deflection. The work done by the beam in raising the weight again is called its re- silience. Similarly, if the weight G is allowed to fall on the mid- dle of the beam from a height Ji, we shall have Gx(h+d^, or approx., Gh,= ^P^d^i and hence, since (4) gives d,^ in terms of P^, e;i=i .^P oreA=2i^^ . (6) 256 MECHANICS OIT EXGINEEEIXG. This theory suppcs5es the mass of the beam small com- pared with the falling weight. 234. Case II. Horizontal Prismatic Beam, Supported at Both End? Bearing a Single Eccentric Load. Weight of Beam Neg- p p lected. — Fig. 227. The reactions 4 t . of the -points of support, Pn and O I AXIS X I B "■ . X i ' >^ y'^^^^i^^J?^ Jvm ^i^^^^^^^^fe -^1' ^^^ easily found by consider- j, — -]Ei^._-_J^^^^^^'^ I ing the whole beam free, and put- j Ip j ting first 2'(mom.)o=0, whence i'l ^\ '^' 1 =PZh-Zi, and then J(mom.)B=0, ri«227. whence Po= An— O^^i- i'o and Pi will now be treated as known quantities. The elastic curves 0(7 and OP, though having a comm on tangent line at (and hence the same slope a^, and a com- mon ordinate at 0, have separate equations and are both referred to the same origin and axes, as shown in the figure. The slope at 0, «o> and that at P,«i, are unknown constants, to be determined in the progress of the work. Eq[uation of OC. — Considering as free a portion of the beam extending from P to a section made anywhere on OC, X and y being the co-ordinates of the neutral axis of that section, we conceive the elastic forces put in on the exposed surface, as in the preceding problem, and put 2'(mom. about neutral axis of the section) =0 which gives (remembering that here d?y-~dx^ is negative.) Ei^^=p{y-x)—p,{k—x)', , . (1) (X OC whencO;. by taking the x anti-derivatives of both members ■ M ^ =P(lx—^)-F,{lx— -^)+ C ax 2 2 To find 0, write out this equation for the point 0, where dy-^dx=aQ and a;=0, and we have 0=P/«o> hence the equation for slope is FLEXURE ELASTIC CUKVES. 257 EI^=P{lx—^)-P,{l,x-^)+EIa^ .. (2) Again taking the x anti-derivatives, we have from (2) Ely =P (^^|1_|^._P,^^|^_^ YEIa,x+{C'=0) (3) '^at Oboth X and 2/ are —0 .°. C'=0). In equations (1), (2), and (3) no value of x is to be used <0 or >Z, since for points in CB different relations apply, thus Equation of CB. — Fig. 227. Let the free body extend from ^ to a section made anywhere on (7^.2'(moms.), as before, =0, gives (see foot-note on p, 322) ^^^=-^^^1-^) . . . (4) (N.B. In (4), as in (1), Eld^y—dx^ is written equal to a neg- ative quantity because itself essentially negative ; for the curve is concave to the axis X in the first quadrant of the co-ordinate axes.) From (4) we have in the ordinary way (aj-anti-deriv.) EI"^ =-Pil,x -J^)+C" . . (5X ax 2 To determine C", consider that the curves CB and OG have the same slope (dy-r-dx) at G where x=l; hence put x~l in the right-hand members of (2) and of (5)' and equate tha results. This gives C" = %PV-\-EIaQ and .-. ^it-^ + ^I'^PS.^t^ . (5) A A o 258 MECHANICS OE E2fGlNEEIlIN&. At C, where J5 = ?, botli curves have the same ordinate; hence, by putting x — l in the right members of (3) and (6)' and equating results, we obtain C'"— — }iPl^. .'. (6)' bo comes Mly = y2Pl'x+EIa^7>—P^ ~2 6" ~6" (6) as the Equation of CB, Fig. 227. But ag is still an unknown constant, to find which write out (6) for the point B where X = li, and y = 0, whence we obtain - 1 ^Fl'—3Fl\-{-2P,l,^] . . , (7) " 6m\ «!= a similar form, putting Pq ^^t P,, and (l^ — I) for I. 235. Maximum Deflection in Case II — Fig. 227. The or- dinate ?/„ of the lowest point is thus found. Assuming ^> /4 k> it will occur in the curve G. Hence put the dy-h-dx of that curve, as expressed in equation (2), =0. Also for O.Q write its value from (7), having put Pi=P?-r-Zij and we have whence [a? for max. y] = ^yi(2k—l) ■ Now substitute this value of x in (3), also ao from (7), and putPi =P?-T-Zi, whence Max. Deflec.=2/max=^% . -^ ll'—3l%+W,'] ^M^^O- 236. Case III. Horizontal Prismatic Beam Supported at Both Ends and Bearing a Uniformly Distributed Load along its Whole Length. — (The weight of the beam itself, if considered, FLEXUEE. ELASTIC CUltVES. 259 constitutes a load of this nature.) Let 1= the length of the beam and w= the weight, per unit of length, of the loading ; then the load coming upon any length x will be =ivx, and the whole load ^=ui. By hypothesis w is constant. Fig. 228. From symmetry we know that the W=«)? Ulli I 1 1 1 lU Fig. 228. reactions at A and B are each =}4iol, that the middle of the neutral line is its lowest point, and the tangent line at is horizontal. Conceiving a section made at any point m of the neutral line at a distance x from 0, consider as free the portion of beam on the right of m. The forces holding this portion in equilibrium are yz'^h ^^^ reaction at B ; the elastic forces of the exposed surface at m, viz.: the tensions and compressions, forming a couple, and J the total she?r ; and a portion of the load, iv(^/2l — x). The sum of the mc ments of these latter forces about the neu- tral axis of m, is the same as that of their resultant; (i.e., their sum, since they are parallel), and this resultant acts in the middle of the length ^Z — x. Hence the sum of these moments =w(}4l — x)^[}4l — x). Now putting 2' (mom. about neutral axis of w)=0 for this free body, we have BI dx^ }4wl{}4l—x)—}^w(}41^xy i.e.,^/g- = > ^t^(l^?2_^) (1) 260 MECHANICS OF ENGINEEEIN&. Taking tlie cc-anti-derivative of both sides of (1), ^^Tx =yM}{l^'^—}i^')+{G=0) (2) as the equation of slope. (The constant is =0 since at both dy-i-dx and x are =0.) From (2), my=-^i}il'x'-%x')+[C'=0] . . (3) which is the equation of the elastic curve ; throughout, i.e., it admits any value of x from x=-\-y2^ to x= — yil. This is an equation of the fourth degree, one degree high- er than those for the Curves of Cases I and II, where there were no distributed loads. If w were not constant, but proportional to the ordinates of an inclined right line, eq. (3) would be of the fifth degree ; if lo were propor- tional to the vertical ordinates of a parabola with axis vertical, (3j would be of the sixth degree ; and so on. By putting x=y^l in (3) we have the deflection of be- low the horizontal thro ugh A and B, viz.: (with W=^ total load ^wl) 384 ' m S84: ' FI ' ' ^ ^ 237. Case IV. Cantilevers. — A horizontal beam whose only support consists in one end being built in a wall, as in Fig. 229(a), or supported as in Fig. 229(&) is sometimes called a canti- lever. Let the student prove that in Fig. 229(a) with a single end load P, the deflection of^ below the tangent at Ois d=j/Pl^-i-£^I;the same state- ment applies to Fig. 229(&), but the tangent at is not horizontal if the beam was originally so. It can also be proved that the slope at B, Fig. 229(a) (from the tangent at 0) is FLEXUEE ELASTIC CURVES. 261 «i= 2^7" The greatest deflection of the elastic curve from the right line Joining AB, in Fig. 229(6), is evidently given by the equation for y max. in § 235, by writing, instead of P of that equation, the reaction at in Fig. 229(&). This assumes that the max. deflection occurs between A and 0. If it occurs between and B put (li—l) for I. If in Fig. 229(a) the loading is uniformly distributed along the beam at the rate of w pounds per linear unit, the student may also prove that the deflection of B below the tangent at is 238. Case V. Horizontal Prismatic Beam Bearing Equal Ter- minal Loads and Supported Symmetrically at Two Points.— Fig. 231. Weight of beam neglected. In the preceding cases we have made use of the approximate form Eld'^y-r-dx^ in determining the forms of elastic curves. In the present ris~T% p\ 1-^ '- < 1- Pig. 231. Fig. )i^Z. case the elastic curve from to (7 is more directly dealt with by employing the more exact expression EI-^f> (see § 231) for the moment of the stress-couple in any sectioUo The reactions at and Care each =P, from symmetry. Considering free a portion of the beam extending from A to any section m between and C (Fig. 232) we have, by putting 2 (mom. about neutral axis of m)=0, P{i+x)- ^~Px==o .-. p^ 4r 262 MECHANICS OF EXGINEBKIJfG. That is, the radius of curvature is the same at all points of OG ] in other words 0(7 is the arc of a circle with the above radius. The upward deflection of F from the right line joining and G can easily be computed from a knowl- edge of this fact. This is left to the student as also the value of the slope of the tangent line at (and G). The deflection of D from the tangent at G=^l^Pf-^EL as ip Fig, 229(a), SAFE LOADS IIS^ FLEXUKE. 239. Maximum Moment. — As we examine the different sec- tions of a given beam undar a given loading we find differ- ent values oi p, the normal stress per unit of area in the outer element, as obtained from eq. (5) § 229, viz.: ^=il/. . . . , (1) e in which I is the " Moment of Inertia " (§ 85) of the plane figure formed by the section, about its neutral axis, e the distance of the most distant (or outer) fibre from the neu< tral axis, and ilf the sum of the moments, about this neu- tral axis, of all the forces acting on the free body of which the section in question is one end, exclusive of the stresses on the exposed surface of that section. In other words Jf is the sum of the moments of the forces which balance the stresses of the section, these moments being taken about the neutral axis of the section under examination. For the prismatic beams of this chapter e and /are the same at all sections, hence p varies with M and becomes a maximum when J/ is a maximum. In any given case the location of the " dangerous section" or section of maximum M, and the amount of that maximum value may be deter- mined by inspection and trial, this being the only method (except by graphics) if the external forces are detached. FLEXUEE SAFE LOADS. 263 If, however, the loading is continuous according to a de- finite algebraic law the calculus may often be applied, taking care to treat separately each portion of the beam between two consecutive reactions of supports^ or detached loads. As a graphical representation of the values of 31 along the beam in any given case, these values may be conceived laid off as vertical ordinates (according to some definite scale, e.g. so many inch-lbs. of moment to the linear inch of paper) from a horizontal axis just below the beam. If the upper fibres are in compression in any portion of the beam, so that that portion is convex downwards, these or- dinates will be laid off below the axis, and vice versa ; for it is evident that at a section where ilf=0, p also =0, i.e., the character of the normal stress in the outermost fibre changes (from tension to compression, or vice versa) when if changes sign. It is also evident from eq. (6) § 231 that the radius of curvature changes sign, and consequently the curvature is reversed, when J/ changes sign. These mo- ment ordinates form a Moment Diagram, and the extremities a Moment Curve. The maximum riioment, ilf^, being found, in terms of the loads and reactions, we must make the p of the " dan- gerous section," where M= M^^ equal to a safe value R', and thus may write ^=M^ . . . o (2) e Eq. (2) is available for finding any one unknown quanti- ty, whether it be a load, span, or some one dimension of the beam, and is concerned only with the Strength, and not with the stiffness of the beam. If it is satisfied in any given case, the normal stress on all elements in all sections is known to be = or <i?', and the design is therefore safe in that one respect. As to danger arising from the shearing stresses in any 264: MECHAXrCS OF ENGINEEKING. section, the consideration of the latter will be taken up in 11 subsequent chapter and will be found to be necessary only in beams composed of a thin web uniting two flanges. The total shear, however, denoted by J, bears to the mo- ment ilf, an important relation of great service in deter- mining M^. This relation, therefore, is presented in the next article. pd? t-pdF ^ 240, The Shear is the First x-Derivative of the Moment. — ^ Fig. 233. {x is the distance of any section, measured parallel wdx ij' to the beam from an arbitrary p'dp origin). Consider as free a ver- tical slice of the beam included between any two consecutive vertical sections whose distance apart is dx. The forces acting are the elastic forces of the two internal surfaces now laid bare, and, possibly, a portion, tvdx, of the loading, which at this part of the beam has some intensity =w lbs. per running linear unit. Putting 2'(mom. about axis .iV)=0 we have (noting that since the tensions and compressions of section JSf form a couple, the sum of their moments about N' is just the same as about N,) i- ^^ — + Jdx+ivdx :0 But P^=M, the Moment of the left hand section,^ =31% 6 e that of the right ; whence we may write, after dividing through by dx and transposing. M'—M dx -r , (jjdu . dM r dx (3) for w -2 vanishes when added to the finite J, and M*^ — M= d3l= increment of the moment corresponding to the incre- ment, dx, of X. This proves the theorem. FLEXURE. SAFE LOADS, 265 Now the value of a? wliich renders M a maxininm or minimum would be obtained by putting the derivative dM ~ dx = zero; hence we may state as a Corollary. — At sections whe7'e the 'tnoment is a maximum ■or Tninimiwrn the shear passes through the value zero. The shear J at any section is easily determined by con° sidering free the portioiL of beam from the section to either end of the beam and putting 2'( vertical components) = 0. In this article the words maximum and minimum are used in the same sense as in calculus ; i.e., graphically, they are the ordinates of the moment curve at points where tie tangent line is horizontal. If the moment curve be reduced to a straight line, or a series of straight lines, it ias'no maximum or minimum in the strict sense just stated ; nevertheless the relation is still practically borne out by the fact that at the sections of greatest and least ordinates in the moment diagram the shear changes sign suddenly. This is best shown by drawing a shear diagram, whose ordinates are laid off vertically from a horizontal axis and under the respective sections of the beam. They will be laid off upward or downward according as J" is found to be upward or downward, when the free body con- sidered extends from the section toward the right. In these diagrams the moment ordinates are set off on an arbitrary scale of somany inch-pounds, or foot-pounds, to the linear inch of paper ; the shears being simply pounds, or some other unit oi force, on a scale of so many pounds to the inch of paper. The scale on which the beam is drawn is so many feet, or inches, to the inch of (paper. 241. Safe Load at the Middle of a Prismatic Beam Support- ed, at the Ends. — Fig. 234. The reaction at each support is ^P. Make a section n at any dis.tance cc<-L from B. Consider the portion nB free, putting in the proper elas- tic and external forces. The weight of beam is neglected. From i'(mom. about %)=0 we have 2G6 MECHAIN'ICS OF EXGrSTEEKING. pL=^x; i.e., M=%Px e 2 Evidently Tlfis proportional to x, and tlie ordinates repre^ senting it will 4;lierefore be limited by the straight line Fig. 234. B'Bt forming a triangle B'BA'. From symmetry, another triangle OR A' forms the other half of the moment dia- gram. Frqm inspection, the maximum iHf is seen to be in the middle where cc= }4l, and hence (il/max.)=7!/;„=i^P? . (1) Again by putting 2'(vert. compons.)=0, for the free body nB we have and must point downward since ~ points upward. Hence the shear is constant and = i^P at any section in the right hand half. If n be taken in the left half we would have, nB being free, from J(vert. com.)=0, FLEXUKE. SAFE LOADS, 267 tlie same numerical value as before ; but J" must point up- ward, since | at 5 and J at n must balance tbe downward P at A. At A, then, the shear changes sign suddenly, that is, passes through the value zero; also at A, Mis a maximum, thus illustrating the statement in § 240. Notice the shear diagram in Fig. 234. To find the safe load in this case we write the maximum value of the normal stress, p,^=R% a safe value, (see table in a subsequent article) and solve the equation for P. But the maximum value of p is in the outer fibre at A, since Jf for that section is a maximum. Hence S^^%Pl (2) is lh.e equation for safe loading in this case, so far as the normal stresses in any section are concerned. Example. — If the beam is of wood and has a rectangu- lar section with width &= 2 in., height h-= 4 in., while its length 1= 10 ft., required the safe load, if the greatest nor- mal stress is limited to 1,000 lbs. per sq. in. Use the pound and inch. From § 90 1=^1^ M^=Vi2X 2x64= 10.66 biquad. inches, while e=l=2 in. .-. P- ifiZ-ixiM^O^^lTT.T lbs. le 120x2 ■^ 242. Safe Load Uniformly Distributed along a Prismatic Beam Supported at the Ends.— Let the load per lineal unit of the length of beam he =w (this can be made to include the weight of the beam itself). Fig. 235. From symmetry, each reaction = yiwl. For the free body wO we have, put'' ting 2'(mom. about n)=Q, pi wl / X a? ii/r w ,1 9\ ^ = -^x— (tax) - .-. Jf= j-ilx-^3ty) 268 MECHANICS OF ENGINEEKIN^ wliicli gives Jf for any section by making x vary from (F to I. Notice tliat in this case tlie law of loading is con- tinuous along tlie wliole length, and that hence the mo- ment curve is continuous for the whole length. W=«;? Fig. 235. To find the shear J, at n, we may either put 2'(vert. com pons.)=0 for the free body, whence e7= YiWl — wx^ and mus therefore_be downward for a small value of x ; or, employ ing § 240, we may write out dM-~dx, which gives J= dM dx (l—2x) (1/ the same as before. To find the max. 31, or Jfn,, put J- O.- which gives cc^^L This indicates ajnaximum, forwliaB substituted in d^3I-^dx\ i.e., in — iv, a negative result TB obtained. Hence ilf^ occurs at the middle of the beam and its value is = iiwl'; .'. ^=yiwV=%Wl m the equation of safe loading. W= total load=tyl- It can easily be shewn that the moment curve is 2 por« FLEXURE. SAFE LOADSo 26& lion of a parabola, whose vertex is at A" under the mid- Jls of the beam, and axis vertical. The shear diagram consists of ordinates to a single straight line inclined to its axis and crossing it, i.e., giving a zero shear, under the middle of the beam, where we find the max. 31. If a frictionless dove-tail joint with vertical faces were introduced at any locality in the beam and thus divided the beam into two parts, the presence of J" would be made manifest "by the downward slipping of the left hand part oji the right hand part if the joint were on the right of the middle, and vice versa if it were on the left of the middle. This shows why the ordinates in the two halves of the shear diagram have opposite signs. The greatest shear is close to either support and is Jj^=^wl. 243, Prismatic Beam Supported at its Extremities and Loaded in any Manner. Equation for Safe Loading. — Fig. 236. Given p .p^ p the loads Pj, P^, and P3, whose g I ' I I Q distances from the right sup- ^^.. lLi port are l^, l^, and ^ ; ,required the equation for safe loading ; i.e., find ilf^ and write it = If the moment curve were continuous, i.e., if M were a continuous function of x from end to end of the beam, we could easily find Jf^ by making Fig. 236. dM-^dx=0, i.e., J=0, and sub- stitute the resulting value of x in the expression for M. But in the present case of detached loads, J is not zero, necessarily, at any section of the beam. Still there is sore J one section where it changes sign, i.e., passes sud- denly through the value zero, and this will be the section of greatest moment (though not a maximum in the stric^j sense used in calculus). By considering any portion n '^ as free, «/is found equal to the Reaction at Diminished by the Loads Occurring Between n and 0. The reaction at B is 270 MECHAls'ICS or ENGi:JfEEEING. obtained by treating the whole beam as free (in which case no elastic forces come into play) and putting 2'(mom. about O)=0; while that at 0,=Pf,=Py-^P^-\-P^—Ps If n is taken anywhere between and E, J=Pq E " F,J=Po-Pi F " H,J^P^-P^-P^ H " B, J=Po-P\-P2-Ps This last value of j/also = the reaction at the other support,^. Accordingly, the shear diagram is seen to consist of a number of horizontal steps. The relation J=dM-^dx is such that the dope of the moment curve is proportional to the ordinate of the shear diagram, and that for a sudden change in the slope of the moment curve there is a sudden change in the shear ordinate. Hence in the present instance, J being constant between any two consecutive loads, the moment curve reduces to a straight line between the same loads, this line having a different inclination under each of the portions into which the beam is divided by the loads. Under each load the slope of the moment curve and the ordinate of the shear diagram change suddenly. In Fig. 236 the shear passes through the value zero, i.e., changes sign, at E; or algebraically we are sup- posed to find that Pq—P^ is + while PQ—P1—P2 is — , in the present case. Considering EO, then, as free, we find Jf;„ to be Mai=Poh~Pi{h~^i) and the equation for safe loading is ?^-Pol-P.{k-k) (1) (i.e., if the max. il/is at F). It is also evident that the greatest shear is equal to the reaction at one or the other support, whichever is the greater, and that the moment at either support is zero. The student should not confuse the moment curve, which FLEXUKE. SAFE LOADS. 271 is entirely imaginary, with the neutral line (or elastic curve) of the beam itself. The greatest moment is not necessarily at the section of maximum deflection of the neutral line (or elastic curve). For the case in Fig. 236 we may therefore state that the max. moment, and consequently the greatest tension or compression in the outer fibre, will be found in the sec- tion under that load for which the sum of the loads (in- cluding this load itself) between it and either support first equals or exceeds the reaction of that support. The amount of this moment is then obtained by treating as free either of the two portions of the beam into which this section divides the beam. 244. Numerical Example of the Preceding Article. — Fig. 237. Given Pi, Pg* -Ps* equal to i/^ ton, 1 ton, and 4 tons, re- spectively ; <i =5 feet, ^2= 7 feet, and ^3= 10 feet ; while ike total length is 15 feet. The beam is of timber, of rectan- gular cross-section, the horizontal width being b=10 inches, and the value of B' (greatest safe normal stress), = ^ ton per sq. inch, or 1,000 lbs. per sq inch. 272 ' MECHANICS Of^ ENGINEERING. Requirea the proper deptli k lor the beam, for safe load- ing. Solution. — Adopting a definite system of units, viz., the inch-ton-second system, we must reduce all distances such as I, etc., to inches, express all forces in tons, write K'= ^^ (tons per sq. inch), and interpret all results by the same sys- tem. Moments will be in inch-tons, and shears in tons. [N. B. In problems involving the strength of materials the inch is more convenient as a linear unit than the foot, since any stress expressed in lbs., or tons, per sq. inch, is . numerically 144 times as small as if referred to the square foot.] Making the whole beam free, we have from moms, about O, Pb~ [>^X 60+1x84+4x120] ^3.3 tons .-. Po=5.5— 3.3=2.2 tons. The shear anywhere between O and ^is J= Po=2.2 tons. ^ and i^ is e/ =2.2— 1^=1.7 tons. The shear anywhere between i^ and His J =2.2 — ^ — 1 = 0.7 tons. The shear anywhere between H and B is J = 2.2 — }4 — 1 —4 =—3.3 tons. Since the shear changes sign on passing H, .-. the max. moment is at ^; whence making HO free, we have M at H=M,,, =2.2 x 120— ^^ x 60—1 x 36 =198 inch -tons. For safety M,„ must = , in which B'='^ ton per sq. inch, e = }4^ — }4 of unknown depth of beam, and /, §90, = I bM, with & = 10 inches ,vi. >^ .|-Xl0.¥^198; or 71^-237.6 .-. h=15A inches. 245. Comparative Strength of Rectangular Beams. — For such a beam, under a given loading, the equation for safe load- ing is ^=3i;„ i. e. ye E bh'=M^ .... (1) « FLEXUEE. SAFE LOADS. 273 whence the following is evident, (since for the same length, mode of support, and distribution of load, M^ is propor- tional to the safe loading.) For rectangular prismatic beams of the same length, same material, same mode of support and same arrange- ment of load : (1) The safe load is proportional to the width of beams having the same depth (A). (2) The safe load is proportional to the square of the depth of beams having the same width (h). (3) The safe load is proportional to the depth of beams having the same volume (i. e. the same hh] (It is understood that the sides of the section are hori- zontal and vertical respectively and thai the materia] \^ homogeneous.) 246. Comparative Stiflfness of Rectangular Beams.— Taking tli*. deflection under the same loading as an inverse me^-sure of the stiffness, and noting that in §§ 233, 235, and 236, this deflection is inversely proportional to I—k hh^ = the " moment of inertia "of the section about its neutral axis, we may state that : For rectangular prismatic beams of the same length, same material, same mode of support, and same loading .• (1) The stiffness is proportional to the width for beams of the same depth. (2) The stiffness is proportional to the cube of the height for beams of the same width (&). (3) The stiffness is proportional to the square of the ciepth for beams of equal volume (hhl), (4) It the length alone vary, the stiffness is inversely proportional to the cube of the length. 247. Table of Moments of Inertia. — These are here recapitu- lated for the simpler cases, and also the values of *?. the distance of the outermost fibre from the axis. Since the stiffness varies as /(other things being equal). 274 MECHANICS OF ENGINEERING. while tlie strength, varies* as I~-e, it is evident that a square beam has the same stiffness in any position (§89), while its strength is greatest with one side horizontal, for then e is smallest, being —^6. Since for any cross-section 1= j dF z^^ in which «=the distance of any element, dF, of area from the neutral axis, a beam is made both stiffer and stronger by throwing most of its material into two flanges united by a vertical web, thus forming a so-called " I-beam " of an I shape. But not without limit, for i;he web must be thick enough to cause the flanges to act together as a solid of continuous substance, and, if too high, is liable to buckle sideways^ thus requiring lateral stiffening. These points will be treated later. SECTION. / e Rectangle, width = b, depth = h (vertical) Vm bh^ %h Bollow Rectaiigle, symmet. about neutral axia. See 1 Fig. 238 (a) f Vi» [6i h,»-b^ h\^ %h, •Triangle, width =6, height = h, neutral axis parallel ^_ to base (horizontal). ) Vse M3 %h Circle of radius r %^r^ r Eing of concentric circles. Fig. 238 (b) }in(r\~r*^) Ti Ehombus; Fig. 238 (c) h = diagonal which is vertical. V4e 5AS %h Square with side b vertical. Via b* %b " " " 6 at 45° with horiz. Vn ** HbVS 248. Moment of Inertia of I-beams, Box-girders, Etc. — In common with other large companies, the Cambria Steel * This function, /-r-e, of the plane figure formed by the cross-section of a beam is evidently of three dimensions of length (cubic inches, for example), and is tabulated in the handbooks of the steel companies for different shapes of section; it is called the "section-modulus." See next page. FLEXURE. SAFE LOADS. 275 Co. of Johnstown, Pa., manufactures prismatic rolled beams and other "shapes," of structural steel, which are variously called I-beams, deck-beams (or " bulb -beams "), rails, angles, T-bars, channels, Z-bars, etc., according to the form of their sections. See Fig. 239 for some of these forms. The company d- ^ T CHANNEL. DECK-BEAM. RAIL. Fie. 239. publishes a pocket-book giving tables of quantities rela- ting to the strength and stiffness of beams, such as the safe loads for various spans, moments of inertia of their sections in various positions, etc., etc„ The moments of inertia of /-beams and deck -beams are computed accord- ing to §§ 92 and 93, with the inch as linear unit. The /-beams range from 4 in. to 24 inches deep, the deck- beams being about 7 and 8 in. deep. (See foot-note, p. 274.) For beams of still greater stiffness and strength com- binations of plates, channels, angles, etc., are riveted to- gether, forming " built-beams,"' or " plate girders." The proper design for the riveting of such beams will be ex- amined later. For the present the parts are assumed to act together as a continuous mass. For example, Fig. 240 shows a " box -girder," formed of two " channels " and two plates riveted together. If the axis of symmetry, JV, h h \ is to be horizontal it becomes the neu- 3 ff"* tral axis. Let (7= the moment of iner- tia of one channel (as given in the pocket-book mentioned) about the axis iV perpendicular to the web of the chan- nel. Then the total moment of inertia oj the combination is (nearly) ^ m ^ Fig. 340. 4 =W^'iUd?—ld't'{d^y2if (1> 276 MECHANICS OF EKGIXEBRING. In (1), &, t, and d are the distances given in Fig. 240 {d ex- tends to the middle of plate) while d' and t' are the length and width of a rivet, the former from head to head (i.e., d' and t' are the dimensions of a rivet-hole). For example, a box-girder of structural steel is formed of two 15 -in. channels (35 lbs, per foot) and two plates 10 in. wide and 1 in. thick ; the rivet-holes | in, wide and If in. long. That is, 5-10; i = l; d=8; f = f; and d' = l| in. Also from the hand-book we find that for the channel in question C==320 in,-* (i.e., biquad. in.). Hence, eq. (1), 7^=640 + 2X10X1X64-4x1. 1(8-^)^ = 1625 in.4^ In this instance e=8^in, ; and if 15,000 lbs, /in.^ ( = 7,5 tons/in.2) be taken as the value of R' (greatest safe normal stress in the extreme fibre of any section) as used by the Cambria Steel Co. for box-girders in buildings, we have " R'l 15000X1625 . . ., — = ^^ = 2, 867, 500 inch-lbs. e 8.5 ' ' That is, the max, safe ^'moment of resistances^ of the box- girder is M^ = 2,867,00O inch-lbs. = 1433.7 inch-tons; this quantity having to do with normal stresses in the section. The greatest " hending-momenf^ due to the amount, and mode, of loading on the beam, must not exceed this. Proper provision for the shearing stresses in the section, and in the rivets, wiU be considered later,) 249. Strength of Cantilevers. — In Fig. 241 with a single , I ^ w^w? concentrated load P at the i::!::^:::::::;;^ Q_i_J_J_J_J_Jl projecting extremity, we r° easily find the moment at 71 to be Jf =Px, and the / max. moment to occur at the section next the wall, j ^f its value being M^=Pl. The shear, J", is constant, Fi«-24i. Fig. 242. ^mj = P at all sections. The moment and shear diagrams are drawn in accordance with these results. FLEXUKE. SAFE LOADS. 277 If the load W= wli^ uniformly distributed on tlie can- tilever, as in fig. 242, by making nO free we have, putting -2'(mom. about n) = 0, pi ^-^=ivx . I .'.M=}iwaP.'. J4=;^w;Z2= ^ Wl. Hence the moment curve is a parabola, whose vertex is at 0' and axis vertical. Putting I (vert, compons.) = we obtain J = ivx. Hence the shear diagram is a triangle, and the max. J = wl = TV. 250. Resume'ofthe Four Simple Cases. — The following table shows the values of the deflections under an arbitrary load P, or W, (within elastic limit), and of the safe load ; Deflection J Safe load (from ?!I I = Mm) Relative strength j Relative stiffness 1 under same load j Relative stiffness I under safe load J Max. shear = Jm,(.wa.d t location, Cantilevers. With one end loadP Fii?. 241 EI B'l P, (at wall) With unif . load Fig. 242 Beams with two end supports. Load P in middle Fig. 234 EI B'l J. 3 W, (at wall) iS'EI ,B'l ViP, (at supp). Unif. loaa W=wl Fig. 235 5 y\/fi 384' EI 8 128 6 16 5 W, (at suppi) also the relative strength, the relative stiffness (under the same load), and the relative stiffness under the safe load, for the same beam. The max. shear will be used to determine the proper web-thickness for /-beams and " built-girders." The stu- dent should carefully study the foregoing table, noting especially the relative strength, stiffness, and stiffness under safe load, of the same beam. Thus, a beam with two end supports will bear a double 278 MECHANICS OF ENGINEERING. load, if uniformly distributed instead of concentrated in the middle, but will deflect ^;( more ; whereas with a given load uniformly distributed the deflection would be only 5/^ of that caused by the same load in the middle, provided <-he elastic limit is not surpassed ii? either case. 261. E', etc. For Various Materials.— The formula& = Jf^, e from which in any given case of flexure we can compute the value of p^, the greatest normal stress in any outer element, provided all the other quantities are known, holds good theoretically within the elastic limit only. Still, some experimenters have used this formula for the rupture of beams by flexure, calling the value of p^ thus obtained the Modulus of Rupture, B. R may be found to differ considerably from both the ^ or C of § 203 with some materials and forms, being frequently much larger. This might be expected, since even supposing the relative extension or compression (i.e., strain) of the fibres to be proportional to their distances from the neutral axis as the load increases toward rupture, the corresponding bLi'esses, not being proportional to these strains beyond the elastic limit, no lono^'er vary directly as tlie distances from the neutral axis ; and the neutral axis does not pass through the centre of gravity of the section, necessarily. The following table gives average values for R, R', R", and E for the ordinary materials of construction.'^ E, the modulus of elasticity for use in the formulsB for deflection, is given as computed from experiments in flexure, and is nearly the pame as E^^ and E^. In any example involving R', e is usually written equal to the distance of the outer fibre from the neutral axis, whether that fibre is to be in tension or compression ; since in most materials not only is the tensile equal to the compressive stress for a given strain (relative extension or contraction) but the elastic limit is reached at about the same strain both in tension and compression. * Wet, or unseasoued, timber is very cousiderably weal^er than that (such as ordinary " dry" timber) containing only 12 per cent, of moisture. Large pieces of timber talie a much longer time to season than small ones. (Johnson.) FLEXUKE. SAFE LOADS. 279 Table foe Use in Examples in Flexure. Timber. Cast Iron. Wro't Iron, Structural Steel. Max. safe stress in outer fi- ) bre— if'dbs. per sq. inch). ) 600 to 1,200 6,000 in tens. 12,000 in comp. 12,000 15,000 Stress in outer fibre at Elas. ) limit = j?''(lbs. per sq. in.) ) 17,000* to 35,000 30,000 and upward. " Modul. of Rupture " 1 =i?=lbs. per sq. inch. ) 4,000 to 10,000 40,000 50,000 60,000 E^Mod. of Elasticity, j =lbs. per sq. inch. ) 1,000,000 to 2,000,000 17,000,000 25,000,000 29,000,000 In the case of cast iron, however, (see § 203) the elastic limit is reached in tension with a stress =9,000 lbs. per sq. inch and a relative extension of ^^ of one per cent., while in compression the stress must be about double to reach the elastic limit, the relative change of form (strain) being also double. Hence with cast iron beams, once extensively used but now largely replaced by rolled beams of structural steel, an economy of material was effected by making the outer fibre on the compressed side twice as far from the neutral axis as that on the stretched side. Thus, Fig. 243, cross-sections with unequal flanges were used, so proportioned that the centre of gravity was twice as near to the outer fibre in tension as to that in compression, i.e., e2=2ej; in other words more material J is placed in tension than in compression. The fibre A being in tension (within elas- tic limit), that at B, since it is twice as far from the neu- tral axis and on the other side, is contracted twice as much as A is extended ; i.e., is f.nder a compressive strain double the tensile strain at A, but in accordance with the above figures its state of stress is proportionally as much within the elastic limit as that of A. * In the tests by U. S. Gov. in 1879 with I-beams, B" ranged from 25,000 to 38,000, and tlie elastic limit was reached with less stress in the large than in the smaller beams. Also, for the same beam, U" decreased with larger spans. 1 N i 1 ^ A f. Fig. 243. 280 MECHANICS OP ENGIJfEERING. The great range of values of R for timber is due not only to the faet that the various kinds of wood differ widely in strength, while the behavior of specimens of any ona kind depends somewhat on age, seasoning, etc., but also to the circumstance that the size of the beam un- der experiment has much to do with the result. The ex- periments of Prof. Lanza at the Mass. Institute of Tech- nology in 1881 were made on full size lumber (spruce), of dimensions such as are usually taken for floor beams in buildings, and gave much smaller values of R (from 3,200 to 8,700 lbs. per sq. inch) than had previously been ob- tained. The loading employed was in most cases a con- centrated load midway between the two supports. These low values are probably due to the fact that in large specimens of ordinary lumber the continuity of it& substance is more or less broken by cracks, knots, etc., the higher values of most other experimenters having been obtained with small, straight-grained, selected pieces, from one foot to six feet in length. See footnote p. 278* Yaluable information and tables relating to timber beams may be found in the hand-book of the Cambria Steel Co. The value R' = 1^,000 lbs. per sq. inch is employed by the Cambria Steel Co. in computing the safe loads for their rolled I-beams of structural steel ; but with the stipulation that the beams (which are high and of narrow width) must be secure against yielding sideways. If such is not the case the ratio of the actual safe load to that computed with i2' = 16,000 is taken less and less as the span increases. The lateral security referred to may be furnished by the brick arch-filling of a fire-proof floor, or by light lateral bracing with the other beams. 252. Numerical Examples. — Example 1. — A square bar of wrought iron, 1^ in. in thickness is bent into a circular :|: arc whose radius is 200 ft,, the plane of bending being par- allel to the side of the square. Bequired the greatest nor^ mal stress p^ in any outer fibre. Solution. From §§ 230 and 231 we may write —t =£— .'. p=eU-T-p, i.e., is constant. FLEXURE. SAFE LOADS. 281 For the units incli and pound (viz. tliose of the table in § 251) we have e=% in., /> =2,400 in., and ^=25,000,000 lbs. per sq. inch, ani .•. i>=i>m=^x 25,000,000-^2,400 =7,812 lbs. per sq. in.^ which is quite safe. At a distance of ^ inch from tne neutral axis, the normal stress is =\_%-^}i.^Pm — %Pm— 5,208 lbs. per sq. in. (If the force-plane (i.e., plane of bending) were parallel to the diagonal of the square, e would =}4x 1.5^^2 inches, giving p^ = \l ,S12x ^/2 ] H^s. per sq. in.) § 238 shows an instance where a portion, 0C7, Fig. 231, is bent in a circular arc. Example 2. — A hollow cylindrical cast-iron pipe of radii 3 y2 and 4 inches* is supported at its ends and loaded in middle (see Fig. 234). Eequired the safe load, neglecting the weight oi the pipe. From the table in § 250 we have for safety P=4^ le From § 251 we put i?'= 6,000 lbs. per sq. in.; and from § 247/=^(ri* — rf)\ and with these values, r2 being =4> '"'i — 4l, e=ri=4, 7i=-B- and Z=144 inches (the inch must be the unit of length since i?' = 6,000 lbs. per sq. inch) we have 7>=4x6,000x;^- ^(256-150)-r-[144x4] .-. P=3,470 lbs. The weight of the beam itself is (r= Vy, (§ 7), i.e., Q^^^r,'^ri)lr= f-(16-12i^)144Xjg=448 lbs. (Notice that y, here, must be lbs., per cubic inch). This weight being a uniformly distributed load is equivalent to half as much, 221 lbs., applied in the middle, as far as the strength of the beam is concerned (see § 250), .*. P must be taken =3,249 lbs. when the weight of the beam is consid- ered. * And length of 13 feet, should be added. 282 MECHANICS OF ENGINEERING. Example 3. — A Cambria I-beam, of structural steel, is to be placed horizontally on two supports at its extremities and is to be loaded imiformly (Fig. 235), the span being ^ = 20 ft. 5g„_^ Its cross-section. Fig. 244, has a depth T ^ V^ parallel to the web, of 15 in. In the lA W f\ ^ Fig. 244. handbook of the Cambria Steel Co. it -\ n" is designated as B 53, 15 in. in depth, and weighing 42 lbs. per foot of length; its section having a moment of inertia /i = 442 in.4 about a gravity axis per- pendicular to the web (for use when the web is vertical; the strongest position) and 72 = 14.6 in.^ about a gravity axis parallel to the web (i.e., when the web is placed horizontally). First, placing the web vertically, we have from § 250, 7->/7 Wi = safe load, distributed, = 8-^. With R' = 16,000, 1 1 = 442, I = 240 inches, and ei --^ 7 J inches, this gives * TFi = (8 X 16,000 X 442) - (240 X 7.5) - 31,430 lbs. But this includes the weight of the beam, =20x42 = 840 lbs.; hence a distributed load of 30,600 lbs., or 15.3 tons, may be placed on the beam (secured against lateral yielding). The handbook of the Cambria Steel Co. referred to gives 15.7 tons as the safe load.) With the web placed horizontally, we find as safe load Tf 2 = 8^-^^ = (8 X 16,000 X 14.6) ^ (^240 X ^) = 2830 lbs. ; or less than 1/10 of Wi. Hence in this position the beam could carry safely only 1990 lbs. above its own weight. Example 4. — Required the deflection at the middle in the first case of Ex. 3. From § 250 this deflection is , _A ^'"384 * The handbook of the Cambria Steel Co. also gives in a separate column the quantity /j^ej, called the "section-modulus," S, (cub. in. or in.^); so that the formula for the safe load would be TFj = 8JS'*S -e- i, S having the value 58.9 in.' in the present instance. Wil^ 5 8R'h l^ 5 R' Z2 Ell "384" lei 'Ell "48 E ' ei FLEXUliE. SAFP] LOADS. 283 . , 5 16,000 (240)2 '-'•' ^^^48 • 29,000,000 ' V ^ ^'^^ .-. d,=0. 4:4:1 in. Example 5. — A rectangular beam of yellow pine, of widtli 6=4 inches, is 20 ft. locg, rests on two end supports, and is to carry a load of 1,200 lbs. at the middle ; required the proper depth h. From § 250 le T 12 ' ^h .: h?=6Pl-^4:B'b. For variety, use the inch and ton. For this system of units P=0.60 tons, i?'=0.50 tons per sq. in., 1=24:0 inches and 6= 4 inches. .-. /i2=(6x0.6x240)4-(4x0.5x4)=108sq. in. .-. ^=10.4 in. Example 6. — Suppose the depth in Ex. 5 to be deter- mined by the conditien that the deflection shall be = Ygoo •f the span or length. "We should then have from § 250 d= k 1=1 ^' 600 48 EI Using the inch and ton, with ^=1,200,000 lbs. per sq. in., which = 600 tons per sq. inch, and /=Yl2^^^ we have ;^3^ 500x0.60x240x240x12 ^^^ _. ^^^^^ ^ 48x600x4 As this is > 10.4 the load would be safe, as well. Example 7. — Required the length of a wro't iron pipe supported at its extremities, its internal radius being 2}^ in., the external 2.50 in., that the deflection under its oivn weight may equal Yioo of the length. 579.6 in. Ans. Example 8. — Fig. 245. The wall is 6 feet high and one foot thick, of common brick work (see § 7) and is to be borne by an 7-beam in whose outer fibres no n greater normal stress than 8,000 *£ lbs. per sq. inch is allowable. If Pio. 245. a number of I-beams is available, 284 MECHANICS OF ENGINEERLNG. ranging in height from 6 in. to 15 in. (by whole inches), which one shall be chosen in the present instance, if their cross-sections are Similar Figures, the moment of inertia of the 15-inch beam being 800 biquad. inches ? The 12-inch beam. Ans. SHEARIXa STRESSES IN FLEXURE. 253. Shearing Stresses in Surfaces Parallel to the Neutral Surface. — If a pile of boards (see Fig. 246) is used to sup- port a load, the boards being free to slip on each other, it is noticeable that the ends overlap, although the boards Fig. 246. are of equal length (now see Fig. 247) ; i.e., slipping has occurred along the surfaces of contact, the combina- tion being no stronger than the same boards side by side. If, however, they are glued together, piled as in the former figure, the slipping is prevented and the deflection is much less under the same load P. That is, the com- pound beam is both stronger and stiifer than the pile of loose boards, but the lendency to slip still exists and is known as the " shearing stress in surfaces parallel to the neutral surface." Its intensity per unit of area will now be determined by the usual " free-body " method. In Fig. 248 let AN' be a portion, considered free, on the left of any N N FiQ. S48. SHEAB, US FLEXURE, 285 section N', of a prismatic beam slightly bent under forces in one plane and perpendicular to the beam. The moment equation, about the neutral axis at JSf', gives ■^=M' ; whence «'= — =^ e 1 (1) Similarly, with AN as a free body, NN' being =^dx, t—=M; whence ^=- . e I . (2) p and p' are the respective normal stresses in the outer fibre in the transverse sections N and N' respectively. Now separate the block NN', lying between these two consecutive sections, as a free body (in Fig. 249). And ^W %f^^ I BART OF J furthermore remove a portion of the top of the latter block, the portion lying above a plane passed parallel to the neu- tral surface and at any distance z" from that surface. This latter free body is shown in Fig. 250, with the system of forces representing the actions uj)on it of the portions taken away. The under surface, just laid bare, is a portion of a sur- face (parallel to the neutral surface) in which the above men- tioned slipping, or shearing, tendency exists. The lowfer por- tion (of the block NN') which is now removed exerted this 286 MECHANICS OF ENGINEEEIKG. rubbing, or sliding, force on the remainder along the under surface of the latter. Let the unknown intensity of this shearing force be X(per unit of area) ; then the shearing force on this under surface is =Xy"dx, (y",= oa in figure, being the horizontal width of the beam a.t this distance z" from the neutral axis of N') and takes its place with the other forces of the system, which are the normal stresses between , and portions of J and J', the respective _z=z" total vertical shears. (The manner of distribution of J over the vertical section is as yet unknown ; see next arti- cle.) Putting 2 (horiz. compons.) = in Fig. 250, we have P -p'dF— r -pdF—Xy"dx=0 ,'.Xy"dx=P'—P fzdF ^ z" But from eqs. (1) and (2), p'— p = (iHf — Jf)J-=^ dM, while from § 240 dM = Jdx ; .■.Xrd.=^-^jlaF.:X=^fliF (3) z " z as the required intensity per unit of area of the shearing force in a surface parallel to the neutral surface and at a distance z" from it. It is seen to depend on the " shear " J and the moment of inertia I of the whole vertical section; upon the horizontal thickness* y'' of the beam at the sur- face in question ; and upon the integral / zdF, ^«" which (from § 23) is the product of the area of that part of the vertical section extending from the surface in question to the outer fibre, by the distance of the centre of gravity of that part from the neutral surface. * Thickness of actual substance. SHEAR IN FLEXURE. 287 It now follows, from § 209, that the intensity (per unit area) of the shear on an elementary area of the vertical cross section of a bent beam, and this intensity we may call Z, is equal to that X, just found, in the horizontal section which is at the same distance (z") from the neutral axis. 254. Mode of Distribution of J, the Total Shear, over the Verti- cal Cross Section. — The intensity of this shear, Z (lbs. per sq. inch, for instance) has just been proved to be Z=X=^, CzdF (4) ly To illustrate this, required the value of Z two inches above the neu- tral axis, in a cross section close to the abutment, in Ex. 5, § 252. Fig. 251 shows this section. From it we have for the shaded portion, lying above the locality in question, y" = 4 inches, and C ~ ' sdF = (area ^ z"= 2 of shaded portion) X (distance of its centre of gravity from J^A) = Fia.aoi. (12.8 sq. in.) x (3.6 in.) = 46.08 cubic inches. The total shear J = the abutment reaction = 600 lbs., while 1= L bM = ^ X 4 X (10.4)^ = 375 biquad. inches. Boih J Siud I refer to the whole section. 600x46.08 ift.oiK = 18.42 lbs. per sq. m.. ..Z- 375x4 •qui+e insignificant. In the neighborhood of the neutral axis, where z" = 0, we have y'" = 4 and r^ zdF= r^ zdF=^ 20.8 X 2.6=54.8, J z"=0 J z''=0 J wh__e J and I of course are the same as before, for z" =0 Hence 288 MECHANICS OF ENGINEERING. Z=^o^ 21.62 lbs. per sq. in. At the outer fibre since f^ zdF^O, %" being = e, ^ is = for a beam of any shape. For a solid rectangular section like the above, Z and 2" bear the same relation to each other as the co-ordinates of the para- bola in Fig. 252 (axis horizontal). Since in equation (4) the horizontal thickness, y" , from side to side ef the sec- tion of the locality where Z is desired, ^iq, 25? occurs in the denominator, and since / %dF increases as g" grows numerically smaller, the following may be stated, as to the distribution of J, the shear, in any vertical section, viz.: The intensity (lbs. per sq. in.) of the shear is zero at the outer elements of the section, and for beams of ordi- nary shapes is greatest where the section crosses the neu- tral surface. For forms of cross section having thin webs its value may be so great as to require special investiga- tion for safe design. Denoting by Z^ the value of ^at the neutral axis, (which =Xo in the neutral surface where it crosses the vertica section in question) and putting the thickness of the sub- stance of the beam = &o at the neutral axis, we have, Zq—Xq — J Iho X area above neutral axis (or below) X the dist. of its cent, grav.from that axis (5) 255. Values of Zo for Special Forms of Cross Section. — From the last equation it is plain that for a prismatic beam the value of Zo is proportional to J, the total shear, and hence to the ordinate of the shear diagram for any particular case of loading. The utility of such a diagram, as obtain- SHEAR ll>r FLEXURE. 289 ed in Figs. 234-237 inclusive, is therefore evident, for by- locating tlie greatest shearing stress in the beam it enables us to provide proper relations between the load- ing and the form and material of the beam to secure safety against rupture by shearing. The table in § 210 gives safe values which the ^^s-^ maximum Zq in any case should not exceed. It is \ only in the case of beams with thin webs (see Figs. 238 and 240) however, that Zq is likely to need at- tention. For a Rectangle we have, Fig. 253, (see eq. 5, § ! l4 Fig. 253. 254) 6o=&, I=>/xib'h\ and C\dF=%'h'h , yih^yiW .'.Zn — Xn- 2 ~ (total shear) -7- (whole area) Hence the greatest intensity of shear in the cross-section is A as great per unit of area as if the total shear were uniformly distributed over the section. Fig. 254. Fig. 233. Fig. ^5 For a Solid Circular section Fig. 254 Fig. 257. Z,= -^f^dF = IhffJo l^nr^ . 2r Stt 3 Tir^ [See § 26 Prob. 3]. For a Hollow Circular section (concentric circles) • Fig. 255, we have similarly, 290 Zo=- MECHATTICS OF BKGIKEERIIJ^G. J Xri*-r2*)2(ri-n) J{r{'—r.i') 3 ;r(ri*— r2*)(ri— rg) A-pplying this formula to Example 2 § 252, we first have as the max. shear J„, = ^P =1,735 lbs., this being the abut- ment reaction, and hence (putting tc = (22 -=- 7)) ^0 max. = _ 4x7xl735[64-42.8] 3x22[256-150](4-3.5) = 294 lbs. per sq. in. which cast iron is abundantly able to withstand in shear- ing. For a Hollow Rectangular Beam, symmetrical about its neutral surface, Fig. 256 (box girder) ;7_ 3}i{hJh'-hJi i) _3 J[5,V-5,V ] ' %ihji,'-hjh%h-h,) 2* [6A-^-M/][6;-&,] The same equation holds good for Fig. 257 (I-beam with square corners) but then &2 denotes the sum of the widths of the hollow spaces. 256. Shearing Stress in the Web of an I-Beam. — It is usual to consider that, with I-beams (and box- beams) with the web vertical the shear J, in any vertical section, is borne exclusively by the web and is uniformly distributed ■ over its section. That this is nearly true may be proved as follows, the flange area being comparatively large. Fig. 258.- Let Fi be the area of one flange, and F^ that of the half web. Then since _i = N b< Fig. 258. /=^ W+2i^, (f ) SHEAE IK PLiEXUEE. 291 (tlie last term approximate, ^ /iq being taken for the radi- us of gyration of jPi,) while r zdF=Fi ±.-{-Fq-^, (the first term approx.) we have J r\dF Jo _ J%K{^F,+F,) J if we write (2i^i+i^o) ^ (6i^i+2i^o)=K • ^ut &o^o is the area of the whole web, .'. the shear per unit area at the neutral axis is nearly the same as if J were uniformly dis- tributed over the web. E. g., with jPi = 2 sq. in., and Fq = 1 sq. in. we obtain Zq = 1.07 (J-r-hoh^. Similarly, the shearing stress per unit area at n, the upper edge of the web, is also nearly equal to e7-f- JqAo (see eq.,4(§254) for then \ T {zdF)'\ ^ F^.y^K nearly, while / remains as before. The shear per unit area, then, in an ordinary I-beam 1& obtained by dividing the total shear J by the area of the web section.* Example. — It is required to determine the proper thick- ness to be given to the web of the 15- inch structural steel rolled I-beam of Example 3 of p. 282, the height of web being 13 in., and the maximum safe shearing stress being taken as 8750 lbs. /in. ^ (as prescribed by the Philadelphia building laws for mild steel). The web is vertical. The greatest total shear, J^^ which occurs at either support, is equal to half the load, i.e., to 15,715 lbs.; and hence, with 6o denoting the thickness of web, we have J 15 715 Zomax.=^; i.e., 8750= ,-^—j^ ; .-. 6o = 0.138 in. * That, is, for the vertical, or horizontal, section of web. The shear on bome oblique plane may be somewhat larger than this. . See §§ 270a and 314. 292 MECHANICS OF ENGINEERING. (Units, inch and pound). The 15-incli I-beam in question of the Cambria Steel Co.), weighing 42 lbs. to the linear foot, has a web 0.41 in. thick, which provides a very ample resist- ance to shearing stress. In the middle of the span, Zo = 0, since J = 0. ' 257. Designing of Riveting for Built Beams. — The latter are generally of the I-beam and box forms, made by riveting together a number of continuous shapes, most of the ma- terial being thrown into the flange members. E. g., in fig. 259, an I-beam* is formed by riveting together, in the manner shown in the figure, a " vertical stem plate " or web, four "angle-bars," and two "flange-plates," each of Fig. 259. Fig. 260. these seven pieces being continuous through the whole length of the beam. Fig 260 shows a box-girder. If the riveting is well done, the combination forms a single rigid beam whose safe load for a given span may be found by foregoing rules ; in computing the moment of inertia, how- ever, the portion of cross section cut out by the rivet holes must not be included. (This will be illustrated in a subsequent paragraph.) The safe load having been com- puted from a consideration of normal stresses only, and the web being made thick enough to take up the max. total shear, J",,, with safety, it still remains to design the riveting, through whose agency the web and flanges are caused to act together as a single continuous rigid mass. It will be on the side of safety to consider that at a given * Such a built I-beam is usually designated a " plate-girder. '\ See handbook of the Cambria Steel Co. SHEAK m FLEXURE. 293 locality in the beam the shear carried by the rivets con- necting the angles and flanges, per unit of length of beam, is the same as that carried by those connecting the angles and the web ("vertical stem -plate"). The amount of this shear may be computed from the fact that it is equal to that occurring in the surface (parallel to the neutral sur- face) in which the web joins the flange, in case the web and flange were of continuous substance, as in a solid I- beam. But this shear must be of the same amount per horizontal unit of length as it is per vertical linear unit in the web itself, where it joins the flange ; (for from § 254 Z =X) But the shear in the vertical section of the web, being uniformly distributed, is the same per vertical linear unit at the junction with the flange as at any other part of the web section (§ 256,) and the whole shear on the ver- tical section of web = J, the " total shear " of that section of the beam. Hence we may state the following : The riveting connecting the angles with the flanges, (or the web with the angles) in any locality of a built beam, must safely sustain a shear equal to J on a horizontal length eqtial to the height of web. The strength of the riveting may be limited by the re- sistance of the -rivet to being sheared (and this brings into account its cross section) or upon the crushing resist- ance of the side of the rivet hole in the plate (and this in- volves both the diameter of the rivet and the thickness of the metal in the web, flange, or angle. In its hand-book, the Cambria Steel Co. gives tables for the safe strength of rivets, and compressive resistance of plates ; based on unit shearing stresses from 6,000 to 10,000 Ibs./sq. in. for shearing stress in the rivets, and 12,000 to 20,000 Ibs./sq. in. compressive re- sistance, in the side of the rivet hole, the axial plane section of the hole being the area of reference. In fig. 259 the rivets connecting the web with the angles are in double shear, which should be taken into account in considering their shearing strength, which is then double ; those connecting the angles and the flange plates are in 294 MECHANICS OF ENGINEERING. single shear. In fig. 260 (box-beam) where the beam is built of two webs, four angles, and two flange plates, all the rivets are in single shear. If the web plate is very high compared with its thickness, vertical stiffeners in the form of "angles " may need to be riveted upon them laterally [see § 314]. Example. — ^A built I-beam of structural steel (fig. 259) is to support a uniformly distributed load of 40 tons, its ex- tremities resting on supports 20 feet apart, and the height and thickness of web being 20 ins. and J in. respectively. How shall the rivets, which are | in. in diameter, be spaced between the web and the angles which are also ^ in. in thick- ness? Let the unit stresses taken be 7500 for shearing, and 12,500 for side compression (Ibs./in.^). Referring to fig. 235 we find that J = | W = 20 tons at each support and diminishes regularly to zero at the middle, where no riveting will there- fore be required. Each rivet, having a sectional area of J7r(|)2 = 0.60 sq. inches, can bear a safe shear of 0.60x7500 = 4500 lbs. in single shear, and hence of 9000 lbs. in double shear, which is the present case. But the safe compressive resistance of the side of the rivet hole in either the web or the angle is only I in. Xj in. X 12500 = 5470 lbs., and thus determines the spacing of the rivets as follows : Near a support the riveting must sustain a shear equal to 40,000 lbs. on a horizontal length equal to the height of web, i.e., to 20 ins., and the safe compression for each rivet is 5470 lbs. Hence 4000 h- 5470, or 7.2, rivets will be needed for the 20-inch length. In other words, they must be spaced 20-^7.2 = 2.7 in. apart, center to center, near the supports; 5.4 in. apart at ^ the span from a support; none at all in the middle. By the Cambria handbook, this distance apart should never be less than 3 diameters of the rivet; and, in connecting plates in compression, should not exceed 16 times the thickness of the plate. As for the rivets connecting the angles and flange plates, being in two rows and opposite (in" pairs) the safe shear- FLEXURE, BUILT BEAM. 295 ing resistance of a pair (eacli in single shear) is 9,000 lbs., while the safe compressive resistance of the sides of the two rivet holes in the angle bars (the flange plate being much thicker) is =10,940 lbs. Hence the former figure (9,000) divided into 40,000, gives 4.44 as the number of pairs of rivets for 20 in. of length of the beam; i.e., the rivets in one row should be 20^-4.44 = 4.5 in. apart, centre to centre, near a support ; the interval to be increased in inverse ratio to the distance from the middle of span, (^bearing in mind the practical limitation just given). If the load is concentrated in the middle of the span, instead of uniformly distributed, e/is constant along each half-span, (see fig. 234) and the rivet spacing must accord- ingly be made the same at all localities of the beam. SPECIAL, PROBLEMS IN FLEXURE. 258. Designing Cross Sections of Built Beams. — The last par- agraph dealt with the riveting of the various plates ; we now consider the design of the plates themselves. Take for instance a plate-girder, fig. 261 ; one vertical stem= Fig. 261. 296 MECHANICS OF ENGINEERING, plate, four angle bars, (each of sectional area = A, re- maining after the holes are punched, with a gravity axis parallel to, and at a distance = a from its base), and two flange plates of width = h, and thickness = t. Let the whole depth of girder = 7^, and the diameter of a rivet hole =f. To safely resist the tensile and compressive forces induced in this section by M,^ inch-lbs. (itf^ being the greatest moment in the beam which is prismatic) we have from § 239, ifn. = — - (1) e E' for mild steel = 15,000 lbs. per sq. inch, e is = ^ ^ while i, the moment of inertia of the compound section, is obtained as follows, taking into account the fact that the rivet holes cut out part of the material. In dealing with the sections of the angles and flanges, we consider them concentrated at their centres of gravity (an approx-. imation, of course,) and treat their moments of inertia about N as single terms in the series fdF z^ (see § 85). The subtractive moments of inertia for the rivet holes in the web are similarly expressed ; let 6o = thickness of web. j It, for web = ^h, (h—2tf—2b,t' [^—t—a'Y „•, ■} I^ for four angles = 4 A ['l — t — aY ( In for two flanges = 2(6— 2f) t ('^f the sum of which makes the ^ of the girder. Eq. (1) may now be written which is available for computing any one unknown quan- tity. The quantities concerned in /^ are so numerous and they are combined in so complex a manner that in any numerical example it is best to adjust the dimensions of the section to each other by successive assumptions and FLEXUEE BUILT BEAM. 297 trials. (The hand-book of the Cambria Steel Co. gives tables of safe loads of beam box-girders and plate-girders for a large variety of sizes and distances between supports ; but attention- is called to the fact that the loads given in the tables are based on the assumption that the girder is supported laterally, and that otherwise a proper reduction in the allowable safe load must be made, as explained elsewhere in the hand-book. The value of 15,000 Ibs./sq. in. for R' has been used in computing these tables.) Example. — (Units, inch and pound) . A plate-girder with end supports, of span = 20 ft. = 240 inches, is to support a uniformly distributed load of 45 tons = 90,000 lbs. If f inch rivets are used, angle bars 3" X 3'' X 2'% vertical web I" in thickness, and plates 1 inch thick for flanges, how wide (6 = ?) must these flange-plates be ? taking h = 22 inches = total height of girder. Solution. — From the table in § 250 we find that the max. 31 for this case is ^ Wl, where W = the total distributed load (including the weight of the girder) and I = span- Hence the left hand member of eq. (2) reduces to Wl h 90000 X 240 X 22 16 • R' - 16 X 15000 "■^^^^• That is, the total moment of inertia of the section must be = 1,980 biquad. inches, of which the web and angles supply a known amount, since &o = >^", t = l''> t'= )'i" , a' = 1^", A= 2.0 sq. in., a = 0.9', and h = 22", are known, while the remainder must be furnished by the flanges, thus determining their width b, the unknown quantity. The elective area. A, of an angle bar is found thus : The full sectional area for the size given, = 3 X ^ + 2>^ X % = 2.75 sq. inches, from which deducting for two rivet holes we have A= 2.75—2 X ^ X >^_ 2.0 sq. in. The value a = 0.90" is found by cutting out the shape 298 MECHANICS OF ENGINEEKING. X3 of two angles from sheet iron, tlius : I and balancing it on a knife edge* (The gaps left by the rivet holes may be ignored, without great error, in finding or). Hence, fig!^ a substituting we have Ih for web =A. . 1^x20^— 2x>^ . ^ l8}(Y=282.d In for four angles =4x2x [9.10]2=662.5 In for two flanges=2(6— |)xlx(10>^)2=220.4(&— 1.5) .-. 1980=282.3+662.5+(Z^1.5)220.4 whence b = 4.6 + 1.5 = 6.1 inches the required total width of each of the 1 in. flange plates. This might be increased to 6.5 in. so as to equal the United width of the two angles and web. The rivet spacing can now be designed by § 257, and the assumed thickness of web, )4 in., tested for the max, total shear by § 256. The latter test results as follows ; The max. shear J^„ occurs near either support and = )4 ^=45,000 lbs. .-., calling 6'o the least allowable thickness of web in order to keep the shearing stress as low as 8, 000 lbs. per sq. inch, 6'o X 20" X 8000 = 45000 .-. 6'o=0.28 in. showing that- the assumed width of )4 in. is safe. This girder will need vertical stiffeners near the ends, as explained subsequently, and is understood to be sup- ported laterally, f Built beams of double web, or box- form, (see Fig. 260) do not need this lateral support, 259. Set of Moving Loads. — When a locomotive passes over a number of parallel prismatic girders, each one of which experiences certain detached pressures corresponding to the dijfferent wheels, by selecting any definite position of the wheels on the span, we may easily compute the reac- tions of the supports, then form the shear diagram, and finally as in § 243 obtain the max. moment, Jf^s and the * The Cambria hand-book gives values of / and a for sections of angle- bars. t See § 314. FliEXUEE. MOVING LOADS. 299 max. shear J^, for this particular position of the wheels. But the values of Jf^ and J^ for some other position may be greater than those just found. We therefore inquire "which will be the greatest moment among the infinite number of {M^Js, (one for each possible position of the wheels on the span). It is evident from Fig. 236 from the nature of the moment diagram, that when the pressures or loads are detached, the 31^ for any position of the loads, which of course are in this case at fixed distances apart, must occur under one of the loads (i.e. under a wheel). We begin .*. by asking : What is the position of the set of moving loads when the moment under a given wheel is greater than will occur under that wheel in any other po- sition? For example, in Fig. 262, in what position of the Fig 263. loads Pi, P2> stc. on the span will the moment ilfa* i-6., under Pn, be a maximum as compared with its value under Pg in any other position on the span. Let P be the resultant of the loads which are now on the span, its variable distance from be = cc, and Unfixed distance from Pg = a'', while a, h, c, etc., are the fixed distances between the loads (wheels). For any values oi x , as the loading moves through the range of motion within which no wheel of the set under consideration goes off the span, and no new wheel comes on it, we have Pi =--^ P, and the mament under Pg =M^=RS-(x-a'y]—P^h-Pih-\-c) ' i.e. M2=j(l^—'^^a'x)—P,b—P,(b-\-c) (1) 300 MECHANICS OF EXGINEBRING. In (1) we liave M2 as a function of x, all tlie other quan- tities in tlie right hand member remaining constant as the loading moves ; x may vary from x=^a-\-d to x=l—{c-\-h—a). For a max. M2, we put dMi-h-dx=0, i. e. m j{l-2x+a)=0 .'. x{ioT Max M.,)=}4l+}^(^' (For this, or any other value of x, (FM^-^daf is negative, hence a maximum is indicated). For a max. M2, then, B must be as as far {%Oj') on one side of the middle of the span as P2 is on the other ; i.e., as the loading moves, the moment under a given wheel becomes a max. when that wheel and the centre of gravity of all the loads {then on the span) are equi-distant from the middle of the span. In this way in any particular case we may find the- respective max. moments occurring under each of the wheels during the passage, and the. greatest of these is the 3I„^ to be used in the equation ilt/,„ =^R'I-^e for safe loading:.* As to the shear J, for a given position of the wheels this will be the greatest at one or the other support, and equals the reaction at that support. When the load moves toward either support the shear at that end of the beam evidently increases so long as no wheel rolls completely over and beyond it. To find J" max., then, dealing with each support in turn, we compute the successive reactions at the support when the loading is successively so placed that consecutive Avheels, in turn, are on the point of roll- ing ofi^ the girder at that end ; the greatest of these is the max. shear, J^^. As the max. moment is apt to come under the heaviest load it may not be necessary to deal with more than one or two wheels in finding M,„. Example. — Given the following wheel pressures, ^< . . 8' . . >B< . . 5' . . >C< . . 4 . . <D 4 tons. 6 tons. 6 tons. 5 tons, on one rail which is continuous over a girder of 20 ft. span, under a locomotive. * Since this may be regarded as a case of " sudden application" of a load, it is customary to make R' much gmaller than for a dead load; from one-third to one-half smaller. FLEXURE. MOVIl!lG LOADS. 301 1. Required the position of the resultant of A, B, and (7' 2. " " " " A, B, C, and I) ; 3. " " " " B, G, and I). 4.. In what position of the wheels on the span will the moment under ^ be a max. ? Ditto for wheel C? Required the value of these moments and which is M^^ ? 5. Required the value of J^, (max. shear), its location and the position of loads. Results.— (1.) 7.8' to right of A. (2.) 10' to right of A. (3.) 4.4' to right of B. (4.) Max. M^ = 1,273,000 inch lbs. with all the wheels on ; Max. Jfc = 1,440,000 inch-lbs. with wheels B, C, and JD on. (5.) J^ = 13.6 tons at right sup- port with wheel I) close to this support. 260. Single Eccentric Load. — In the following special cases of prismatic beams, peculiar in the distribution of the loads, or mode of support, or both, the main objects sought are the values of the max. moment M^] for use in the equation il4^:?y(see§239); e and of the max. shear J,^, from which to design the web riveting in the case of an I or box-girder. The modes of support will be such that the reactions are independent of the form and material of the beam (the weight of beam being neglected). As before, the flexure is to be slight, and the forces are all perpendicular to the beam. The present problem is that in fig. 263, the beam being prismatic, supported at the ends, with a single eccentric load, P. We shall first disregard the weight of the beam itseli Let the span = ?i4-?2- First considering the whole beam free we have the reactions Bi = PI, ^ I and B2 = PI, -f- Z. Making a section at m and having Om free, x being < I2, S (vert, compons,) == gives Fig. 263. 302 MECHANICS OF ENGINEEELNG. i?2 — J=0, i.e., e7=i?2 ; while from H (moni.),„=0 we have P-^-R.,x= .-. Jf = B,x=?}}x e I These values of J and M hold good between and 0, J being constant, while 31 is proportional to x. Hence for C the shear diagram is a rectangle and the moment dia^ gram a triangle. By inspection the greatest M for C is for X =■ I2, and = FI1I2 -4- I. This is the max. M for the beam, since between G and B, M is proportional to the disr tance of the section from B. .'.M^=?}^a.Tid.^=I}i' ... (1) lei is the equation for safe loading. J = A\ ill any section along OB, and is opposite in sign to what it is on 0(7; i.e., practically, if a dove-tail joint existed anywhere on 0(7 the portion of the beam on the right of such section would slide downward relatively to the left hand portion ; but vice versa on GB. Evidently the max. shear «/„, = ^x 01* ^2> as I2 or Z^ is the greater segment. It is also evident that for a given span and given beam the safe load P', as computed from eq. (1) above, becomes very large as its point of application approaches a sup- port ; this would naturally be expected but not without limit, as the shear for sections between the load and the support is equal to the reaction at the near support and may thus soon reach a limiting value, when the safety of the web or the spacing of the rivets, if any, is considered. Secondly, considering the weight of the heam, or any uniformly distributed loading, weigliing w lbs. per unit of length of beam, in addition to P, Fig. 264, we have the reactions ' iJ,=^+|'; and B,=i^+^ Let 1-2 he >?i ; then for a portion Om of length a?<^ moments about m give FLEXURE. SPECIAL PROBLEMS. 303 ^ — BoX + ivx.~ =0 e " 2 Le., on 00, M—B-iX — y^ lox^ . . . , (2) Evidently for x = (i.e. at 0) M = 0, while for x = Iz (i.e. at (7) we have, putting w = W -r- I Mr=B>,l,— y: wll- Z 2 ^^ J (3) (4) it remains to be seen whether a value of M may not exist in some section between- and G, (i.e., for a value of x <l2 in eq. (2)), still greater than Mq. Since (2) gives Ji" as a continuous function of x between and C, we put dM-r- dx = 0, and obtain, substituting the value of the con- stants B2 and w, ( max, B^—ivx^O .'. Xa -< for M or ( min. This must be for 31 max., since d^M -^ dx^ is negative when this value of x is sub- stituted. If the particular -—J value of X given by (4) is 'P <l2, the corresponding vahie of 31 (call it iSiJ from eq. (2) will occur on 00 and will be greater than 3Iq (Dia- grams II. in fig. 264 show this case) ; but if x„ is> h, we are not concerned with the corresj)onding value of 31, and the greatest 31 on 00 would be 3Ic. For the short portion BG, which has moment and shear diagrams of its own not con- tinuous with those for 00, it "' may easily be shown that 3£c is the greatest moment of P,gj_ge4. any section. Hence the 31 304 MECHANICS OF EXGIXEERING. max., or Ji|„, of tlie whole beam is either Mq or J^, according as x^ is > or < I2. This latter critPT.ion may be expressed thus, [with h — yi I denoted by l^, the distance of P from the middle of the span] : From (eq. 4) and since from (4) and (2) £ii_L.i/n>7 IS equiv ?F^^'7<- alent to L tf)<OjJ if. -Pk W ■PI, w (5) The equation for safe loading is and e JV li — =Jf„,when^is < ^ e W k . . . . (6) Seeeqs. (3) and (5) for M, and Jf„ If either P, W, \, or \ is the unknown quantity sought, the criterion of (6) cannot be applied, and we .•. use both equa- tions in (6) and then discriminate between the two results. The greatest shear is J^=Bi, in Fig. 264, where l^ is 281. Two Equal Terminal Loads, Two Symmetrical Supports Fig. 265. [Same case as in Fig. 231, § 238]. Neglect weight of beam. The reaction at each support being =P, (from symmetry), we have for a free body Om with a; < Z, .Pl .0 Px—^. e M=Px while where a? > Zi and < ?i+?2 Px-P {x--\)—^=0 .: M=P\ (1> (2) That is, see (1), ilf varies directly with x between and C, while between G and D it is constant. Hence for safe loading i.e., — ^Pl , . (3) FLEXURE. SPECIAL PROBLEMS. 305 a( i 11 illlllllk.i MOMS. illlllllllllllllllll 1 SHEARS Tlie construction of the B moment diagram is evident ^l^ ^1 ^^ p from equations (1) and (2). \ ! As for J", tlie shear, the same free bodies give, from I, (vert. forces)=0. On OG . J=^P ... (4) On CD . J=P—P==zerol5) (4) and (5) might also be ob- Pig 265 tained from (1) and (2) by- writing J=d M-T-dx, but the former method is to be preferred in most cases, since the latter requires M to be expressed as a function of x while the former is applicable for examining separate sections without making use of a variable. If the beam is an I-beam, the fact that J is zero any- where on G D would indicate that we may dispense with a web along G D io unite the two flanges ; but the lower flange being in compression and forming a " long column " would tend to buckle out of a straight line if not stayed by a web connection with the other, or some equivalent brac- ing. 282. XTniform Load over Part of the Span. Two End Supports. Fig. 266. Let the load= W, extending from one support over a portion =c, of the span, (on the left, say,) so that W= IOC, w being the load per unit of length. Neglect wei ght o f beam. For a 'free body Dm of any length X <, B (i.e. < c), 2" moms^=0 gives pi wx- 2 -^icc=0 .'.M= (1) which holds good for any section on B. As for sections on B (7 it is more simple to deal with the free body m'G, of leiigth ' x' < G B from which we have M^R^ x' . (2) MECHANICS OF ENGINE EEH^TG. Fig. 2G6. wMch shows the moment curve for B G tohe a. straight line DC, tangent at D to the parabola 0' D representing eq. (1.) (If there were a con- centrated load at B, CD would meet the tangent at D at an angle instead of co- inciding with it ; let the stu- dent show why, from the shear diagram). The shear for any value of ic on -S is : On 5 while on B C . e/= Bo= constant (3) The shear diagram is constructed accordingly. To find the position of the max. ordinate of the para- bola, (and this from previous statements concerning the tangent at the point D must occur on B, as will be seen and will .'. be the M^ for the whole beam) we put e/=--0 in eq (3) whence X (for JC)= i?i_JF[?— |] ^_(? w tu (5) W and is less than c, as expected. [The value oi Bi^--j- (l — '^\ —[wc ~-T) (I — 2), (the whole beam free) has been substi- tuted]. This value of x substituted in eq. (1) gives is the equation for safe loading. The max. shear J^ is found at and is evidently >i?2j at C. Bx, which is FLEXUEE. SPECIAL PROBLEMS. 30^ 263. XTniforin Load Over Whole length With Two Symmetricj Supports. Fig. 267. — With the notation expressed in the fig- ure, the following results may be obtained, after having divided the length of the beam into three parts for sepa- rate treatment as necessitated by the external forces, which are the distributed load W, and and the two reactions, each = }^ W. The moment curve is made up of parts of three dis- tinct parabolas, each with its axis vertical. The central par- abola maj sink below the hori- zontal axis of reference if the supports are far enough apart, in which case (see Fig.) the elas- tic curve of the beam itself becomes concave upward be- tween the points E and F of " contrary flexure." At each of these points the moment must be zero, since the radius of curvature is co and M = EI ^ p (see § 231) at any sec- tion ; that is, at these points the moment curve crosses its horizontal axis. As to the location and amount of the max. moment M^, inspecting the diagram we see that it will be either at H, the middle, or at both of the supports B and C (which from symmetry have equal moments), i.e., (with I = total length,) Fig. 267. w Mr .[and.-.— J= ( either ~ \ %li-l,^-] at ^ or Ell' at ^ and a 2Z according to which is the greater in any given case ; i.e. according as I2 is > or < l^ y'g. The shear close on the left oi B = ivl^, while close to the right oiBit=}4 W — id^. (It will be noticed that in this case since the beam overhangs, beyond the support, the shear near the support is not equal to the reaction there, as it was in some preceding cases.) 308 MECHANICS OF ENGINEEEHsTG. Hence (/„= wli /2 ^-t^Zi P^^^^^^^g ^^ ^1 <^ 264, Hydrostatic Pressure Against a. Vertical Plank. — From elementary hydrostatics we know that the pressure, per unit area, of quiescent water against the vertical side of a tank, varies directly with the depth, x, below the surface, and equals the weight of a prism of water whose altitude = X, and whose sectional area is unity. See Fig. 268. Fig. 26S. *Tt& plank is of rectangular cross section, its constant breadth, — b, being r~ to the paper, and receives no sup- port excepi at its two extremities, and B, being level with the water surface. The loading,' or pressure, per unit of length of the beam, is here variable and, by above defini- nition, is = w= yxb, where ;' = weight of a cubic unit (i.e. the heaviness, see § 7) of water, and x = Om =■ depth of any section m below the surface. The hydrostatic pres- sure on dx = ivdx. These pressures . for equal dx's, vary as the ordinates of a triangle ORiB. Consider Onti free. Besides the elastic forces of the ex- posed section m, the forces acting are the reaction Bq, and the triangle of pressure OEm. The total of the latter is W. 0(? = I wdx = yb I xdx = 'fb-^ (1) and the sum of the moments of these pressures about m is equal to that of their resultant ( = their sum, since they are parallel) about m, and .% ==: jb -^ , ^, A o rLEXHRB. SPECIAL PEOBLEMS. 309 [From (1) wh«n x==1,wq have for tlie total water pres- sure on the beam Wi = jb ^ and since one-third of this will be borne at we have i?o =^}i T^^^-~\ Now putting i'( moms, about the neutral axis of wi)=0, for Om free, we have Box—JK . ^—^=0 .-. 31= /eyb {Vx—:j(?) O 6 (2) (which holds good from x = Oto x — I). From I (horiz. forces) = we have also the shear J=R,— W^=% yh {P—Sx') .... (3) as might also have been obtained by differentiating (2), since J = dM -^ dx. By putting e7 = (§ 240, corollary) we have for a max. M, x = I -i- V3, which is less than I and hence is applicable to the problem. Substitute this in eq. 2, and reduce, and we have Efl ,, . R'l 1 1 -^=Ji^, i.e. — =g "^^•rbV' . (4) as the equation for safe loading. 265. Example. — If the thickness of the plank is h, re- quired 7i = ?, if B' is taken = 1,000 lbs. per sq. in. for timber (§ 251), and I = 6 feet. For the inch-pound-second system of units, we must substitute B' = 1,000 ; ? = 72 inches ; y = 0.036 lbs. per cubic inch (heaviness of water in this system of units); while I =h¥ -4- 12, (§ 247), and e — }i h. Hence from (4) we have 1000 &A3 0.0366x723 ,„ ..^ . n 07 • ^rs 7T— — n /- 1 ••• ^^=5.16 .'. h = 2.27 m. It will be noticed that since x for ilfm = I -^ Vs, and not ^ I, ifm does not occur in the section opposite the resul- tant of the water pressure ; see Fig. 268. The shear curve is a parabola here ; eq. (3). 310 MECHANICS OF ENGINEERING. ^ ^ ^ ^'^ ^T £r ^ 1 wTtr- Fig. 289. Fig. 269a. 266. Flat Circular Plate, Homogeneous and of Uniform Thick- ness, Supported all Eound its Edge and Subjected to Uniform Fluid Pressure of w lbs. per sq. in. A strict treatment of this case being very complicated, an approximate method, due to Prof. C. Bach, will be presented."* Fig. 269 shows a top view of the circular plate, in a horizontal position and covering a circular I opening, its edge being supported C^) continuously on the edge of the opening (but not clamped to it). Let the radius of the plate be r and its thickness h. Under the plate is the atmosphere, while on its upper surface, acting uniformly over the whole of that surface, is a fluid pressure whose excess over that of the at- mosphere is w Ibs./sq. in. The particles near the upper surface are under compressive stress, which is obviously greater near the center of the circle ; those near the lower surface are in tension. Let now the half-plate, CODE, (cutting along the diameter CD) be considered as a "free body" in Fig. 269a; the tensile and compressive stresses in the section COD being assumed to form a stress-couple, as in previous case of flexure, the unit- stress varying as the distance from the middle of the thick- ness, with the stress in the outermost fiber denoted by p. Then the moment of this couple will be written pi — e, as before, where e = ^h and I =2rh^ -i- 12. On the upper surface of the free body we find a total pressure of | W7rr^ lbs., covering a semicircular surface ; so that (p. 22) the distance of the re- sultant from is 4 r -^ Stt. The upward reaction from the supporting edge is also | wirr'^ lbs., but its resultant acts 2r/7r in. from (center of gravity of a semicircular "wire," p. 20). Taking moments, then, about we have wrrr^ r2 r 4 rl _ prh^ ~2~ [V~3^J^ ""3" * Elasticitaet und Festigkeit, by C. Bach ; Berlin, 1898. or, tv = -,« (0) FLEXUEE. PLATES. 311 Notwithstanding the imperfections of this analysis, the experimental work of Prof. Bach shows that a modification of eq. (0), viz. : — - (1) 5 K" „, r may be used with safety for the design of a plate under these circumstances ; R% a safe unit working stress for the material, having been substituted for p. For example, let the plate (e.g., cylinder-head of a loco- motive) be of mild steel with h = 1 in. and r = 8 in. Putting R' = 16,000 lbs. /sq. in., we have from eq. (1) a safe w = 1 (16000) X (1 H- 8)' =. 208 lbs. per sq. in. [N.B. If the plate is clamped all round the edge, we may write f instead of the |. (Bach.) ] 266a. Homogeneous Circular Plate of Uniform Thickness, h, Supported all Round the Edge, with Concentrated Load (P lbs.) in Center. By the same method as before we may here derive P = 1 Trh^p ; but from his experiments in this case Prof. Bach concludes that the formula for safe design should be written P lirh'R'. o (2) It is seen from eq. (2) that the value of P is independent of the radius of the plate; depending only on the material and the thickness, h. 266b. Homogeneous Elliptical and Rectangular Plates of Con- stant Thickness, h , Supported all Round the Periphery. According to Prof. Bach's approximate analysis, as supplemented by his experimental researches, we may use the following formulae for Fig. 269&. safe design of elliptical and rectangular plates, supported (not clamped) around the whole periphery. See Fig. 2696 for notation of dimensions ; h being the uniform thickness in each 312 MECHANICS OF ENGINEERING. case, and a being > 6. R' = max. safe unit stress for the material. For the elliptical plate under unifornily distributed pressure (over whole area) of w lbs. / sq. in., denoting the ratio 6 -r- a by m, we have w = ^ {1 +m')J^,.R'; .... (3) and under a central concentrated load of P lbs., 3^ 3 + 2m^ + 3w^ , ... (N.B. If the edge is clamped down all round we may use values of w and P about 50 per cent, greater than the above.) With rectangular plates under a uniformly distributed pressure, denoting the ratio 6 -;- a by m, we have w = l{l +m')f,.R' ..... (5) and for a concentrated central load P, with n denoting the ratio P = i- (1 + n') . h'R' .'.... (6) on In the particular case of the square plate, the side of the square being a, eqs. (5) and (6) reduce to 7 2 (uniform pressure) w = S.6 — .R'; (7) (central load) P =o h'R' (8) o 266c. Homogeneous Flat Circular Plate, of TTniforin Thickness, used as Piston of an Engine. In such a case we have fluid pressures ou both sides of the plate or disc, neither of which is necessarily one atmosphere ; while at the center we have acting the concentrated pull or thrust, P lbs., of the piston rod. (Fric- tional forces around the edge may be disregarded.) If w denote the greatest difference between the (uniform) fluid pressures (per sq. in.) on the two sides, we may write (according to Grashof's analysis, as quoted by Unwin), for safe design in this case : — ^-If.-^' . (9) (E', h, and r, have the same meaning as before.) FLEXUEE SPECIAL PROBLEMS. 313 267. ResilienceofBeamWithEndSupports.— Fig. 270. If a 9g mass whose weight is G {G large com- !^ I pared with that of beam) be allowed to ^^ J __^^'- l_p fall freely through a height = h upon g J I - 1^ ^j^^ centre of a beam supported at its a-.y TPm extremities, the pressure P felt by the Fig. 270. beam increases from zero at the first instant of contact up to a maximum P^, as already stated in §233a, in which the equation was derived, d^ being small compared with h, The elastic limit is supposed not passed. In order that the maximum normal stress in any outer fibre shall at most be=^', a safe value, (see table §251) we must put =-7^ [according to eq. (2) §241,] i.e. in equation (a) above, substitute F^= [4 Ii'l]-^Ie, which gives having put I==FJi? {h being the radius of gyration §85) and Fl= V the volume of the (prismatic) beam. From equation (&) we have the energy, Gh, (in ft. -lbs., or inch- lbs.) of the vertical blow at the middle which the beam of Pig. 270 will safely bear, and any one unknown quantity can be computed from it, (but the mass of G shaiili not be small compared with that of the beam.) The energy of this safe impact, for two beams of the same material and similar cross-sections (similarly placed), is seen to be proportional to fheii volumes; while if further- more their cross-sections are the same and similarly placed, the safe Gh is proportional to their lengths. (These same relations hold good, approximately, beyond the elas' tic limit.) It will be noticed that the last statement is just the re- 314 MECHANICS OF ENGINEEEING. verse of wliat was found in §245 for static loads, (the pressure at tlae centre of the beam being then equal to the weight of the safe load) ; for there the longer the beam (and .°. the span) the less the safe load, in inverse ratio. As appropriate in this connection, a quotation will be given from p. 186 of " The Strength ' of Materials an^ Structures," by Sir John Anderson, London, 1884, viz.: "It appears from the published experiments and state- ments of the Railway Commissioners, that a beam 12 feet long will only support )4 of the load that a beam 6 feet long of the same breadth and depth will support, but that it will bear double the weight suddenly applied, as in the case of a weight falling upon it," (from the same height, should be added) ; " or if the same weights are used, the longer beam will not break by the weight falling upon it unless it falls through twice the distance required to frac- ture the shorter beam." 268. Combined Flexure and Torsion. Crank Shafts. Fig. 271. Let OiB be the crank, and NOi the portion projecting beyond the nearest bearing N. P is the pressure of the connecting-rod against the crank-pin at a definite in- stant, the rotary motion be- ing uniform. Let a= the perpendicular dropped from the axis OOi of the shaft upon P, and 1= the distance of P, along the axis Oj from the cross-section iV^TwiV^' of the Let NW be a diameter of this In considering the portion NOiB free, and thus exposing the circular section iVmZV^, we may assume that the stresses to be put in on the ele- ments of this surface -are the tensions (above NN') and the compressions (below NN') and shears "| to NN', due to the bending action of P ; and the shearing stress tan= shaft, close to the bearing, section, and parallel to a. FLEXURE. SPECIAL PROBLEMS. 315 gent to tlie circles which have as a common centre, and pass through the respective dF's or elementary areas, these latter stresses being due to the twisting action of P. In the former set of elastic forces let p = the tensile stress per unit of area in the small parallelopipedical ele- ment m of the helix which is furthest from NN' (the neu- tral axis) and /= the m.oment of inertia of the circle about NN'-, then taking moments about NN' for the free body, (disregarding the motion) we have as in cases of flexure (see §239) pT T.7 . .. . .. Plr .= PI ; i.e., p- («) [None of the shears has a moment about iVW.] Next taking moments about OOi, (the flexure elastic forces, both normal and shearing, having no moments about OOi) we have as in torsion (§216) ■^^^-i^= Pa ; i.e., p^= Par ~I7 Q>) in which p^ is the shearing stress per unit of area, in the torsional elastic forces, on any outermost dF, as at m ; and 7p the polar moment of inertia of the circle about its centre 0. Next consider free, in Fig. 272, a small parallelopiped taken from the helix at m (of Fig. 271.) The stresses [see §209] acting on the four faces p" to the paper in Fig. 272 are there represented, the dimensions (infinitesimal) being n " to NN, & II to 00,, and d -] io the paper in Fig. 272. /pnd pM' ^Pgticl — "pM p^na pnd - H -J) M P,M^, -p/id Fig. 272. qcd- ./"" Fiff. 273. pnd Sl(i MECHANICS OF EXGINEEKllJTG. By altering the ratio of 6 to % we may make the angle 6 what we please. It is now proposed to consider free the triangular prism, GUT, to find the intensity of normal stress q, per unit of area, on the diagonal plane GH, (oi length — c,) which is a bounding face of that triangulai' prism. See Fig, 273. By writing 2" (compons. in direc-^ tion of normal to GII)=0, we shall have, transposing, qcd=pnd sin d+pjbd sin d+pjid cos d ; and solving for q q=jp -- sin d+p, -sin<?+-. cos 6j ; . (1; but n : c= sin d and b : c= cos 6 .*. q=p sin^^+Ps2 sin d cos d . . (2) This may be written (see eqs. 63 and 60, O. W. J. Trigo- nometry) q^}4pO—Gos2d)+p,sm2d . . (3) As the diagonal plane GH is taken in different positions (i.e., as d varies), this tensile stress q (lbs. per sq. in. for instance) also varies, being a function of d, and its max, value may be >^. To find 6 for q max. we put tJ, =j9sin2^4-2^sCos2(?, . . (4) = 0, and obtain: tan[2(^ for q max)]=> — ~ . . • (5) Call this value of 6, 6'. Since tan 2d' is negative, 2d' lies either in the second or fourth quadrant, and hence sin2^^=± , ^" and cos 2^'= rp— 7=^= (6) [See equations 28 and 29 Trigonometry, O. W. J.] The FliEXUBE. CEANK SHAFT. 317 apper signs refer to tlie second quadrant, the lower to tlie fourth. If we now differentiate (4), obtaining ^=2i)cos2^-4p,sin2^ . . . (7) W8 laote that if the sine and cosine of the [2^'] of the 2nd quadrant [upper signs in (6)] are substituted in (7) the re- sult is negative, indicating a maximum ; that is, g is a max- imum for 6= the d' of eq. (6) when the U2>per signs are taken (2nd quadrant). To iind q max., then, put 6' for 6 in (3) substituting from (6) (upper signs). We thus find * g-max =;^[p+Vy+4^] . . (8) A similar process, taking components parallel to GH, Fig. 273, will yield q^ max., i.e., the max. shear per unit of area, ^hich for a given p and p^ exists on the diagonal plane GH in any of its possible positions, as d varies. This max. shearing stress is g^max =^yp2_|_4^^2 ^ ^ (9j In the element diametrically opposite to m in Fig. 211, p is compression instead of tension ; q maximum will also be compression but is numerically the same as the q max. of eq. 8. 269. Example.— In Fig. 271 suppose P=2 tons = 4,000 lbs., a=Q in., 1=5 in., and that the shaft is of wrought iron. Required its radius that the max. tension or com- pression may not exceed i^' = 12,000 lbs. per sq. in.; nor the max. shear exceed /S" = 7,000 lbs. per sq. in. That is, we put 5'=12,000 in eq. (8) and solve for r : also ^,,=7,000 in (9) and solve for r. The greater value of r should be taken. From equations (a) and (5) we have (see §§ 219 and 247 for /p and i) * According to the conceptions of § 405&, safe design would require that we put the max. '^ strain" in this case equal to a safe value, as determined by simple tensile or compressive tests. Here the max. strain (tensile) is £=[|p + |-\/p^ + 4ps^]-^-E'- (Grashof's method.) 318 MECHANICS OF EKGINEEEIITG. \P= r ^^cl p^= -. •which in (8) and (9) give mas. g=>^~ [4?+|/(4^)=+4(2«)"] ^ . . (8^ p and max. g's=^— 3A/(4!)H4(2a)2 . . • (9«) With max. g= 12,000, and the values of P, a, and Z, already given, (units, inch and pound) we have from (8a), r^=2.72 cubic inches .*. r=1.39 inches. Next, with max. 5's=7,000; P, a, and I as before; from (9a), r^=2.84 cubic inches .*. r=1.41 inches. The latter value of r, 1.41 inches, should be adopted. It is here supposed that the crank-pin is in such a position (when P= 4,000 lbs., and a=Q in.) that q max. (and q^ max.) are greater than for any other position ; a number of trials may be necessary to decide this, since P and a are different with each new position of the connecting rod. If the shaft and its connections are exposed to shocks, H and S' should be taken much smaller. 270. Another Example of combined torsion and flexure is shown in Fig. 274. The '" ' /^^ "^^ "^i< ^B 'wo^k of the working force Pi( vertical cog-pressure) is B expended in overcoming the resistance (another vertical cog-pressure) Q^. ^la- 27'4. That is, the rigid body consisting of the two wheels and shaft is employed to transmit power, at a uniform angular velocity, and since it is symmetrical about its axis of rotation the forces act- ing on it, considered free, form a balanced system. (See § 114). Hence given Pi and the various geometrical quan- TLEXUEE. CEAXK SHAFT. 319 titles «!, 5i, etc., we may obtain Q^, and the reactions Pq and Pr, in terms of Pj. The greatest moment of flexure in the shaft will be either FJi, at G; or PJ3, at B. The portion CD is under torsion, of a moment of torsion =Piai= Qih^. Hence we proceed as in the example of § 269, simply put- ting Poll (or Pb4, whichever is the greater) in place of Fl, and PiCTi in place of Pa. We have here neglected the weight of the shaft and wheels. If Qi were an upuard ver- tical force and hence on the same side of the sh:it as Pj, the reactions Pq and Pg would be less than before, and on© or both of them might be reversed in directioji. 270a. Web of I-Beam. Maximum Stresses on an Oblique Plane. — The analysis of pp. 315, 316, etc., also covers the case of an element of the web of a horizontal I-beam under stress, when this element is taken near the point of junction with the flange. Supposing that the thickness of web has already been designed such that the shearing stress on the vertical (and therefore also on the horizontal) edges of such an element is at rate of 8000 lbs. per sq. inch ; and that the horizontal tension at each end of this element (since it is not far from the outer fibre of the whole section) is at rate of 10,000 lbs. per sq. in.; we note that Fig, 272 gives us a. side view of this element, with p^ = 8000, and p = 14,000, lbs. per sq. inch. GTis the lower edge of the upper flange, corresponding (in an end view) to the point n in Fig. 258 on p. 290. (We here suppose the upper flange to be in tension ; of course, an illustration taken from the compression side would do as well.) Substitution in equations (8) and (9) of p. 317 results in giving as maximum stresses on internal oblique planes : q max. = 17,630 lbs. per sq. in. tension; and g^ max. =10,630 " " " " shearing. These two values are seen to be considerably in excess of the respective safe values for shearing and tensile stresses in the case of structural steel, and the necessity is therefore em- phasized of adopting values for shearing stress in webs some- what lower than the 8000 lbs./in.2 used above ; to avoid the occurrence of excessive stress on internal oblique planes. See p. 291. 320 MECHANICS OF ENGINEEHIiN^a. CHAPTER IV. FLEXURE, CONTINUED. CONTINUOUS GIRDERS. 271. Definition. — A continuous girder, for present pur« poses, may be defined to be a loaded straight beam sup- ported in more than two points, in which case we can no longer, as heretofore, determine the reactions at the sup- ports from simple Statics alone, but must have recourse to the equations of the several elastic curves formed by its neutral line, which equations involve directly or indirect- ly the reactions sought ; the latter may then be found as if they were constants of integration. Practically this amounts to saying that the reactions depend on the man- ner in which the beam bends ; whereas in previous cases, with only two supports, the reactions were independent of the forms of the elastic curves (the flexure being slight, however). As an Illustration, if the straight beam of Fig. 275 is placed on three supports 0, B, and (7, at the same level, the reactions of these supports seem at first sight indeterm- inate ; for on considering the p -i ^ whole beam free, we have three \'^_~^~ 1. '^ j;* $ unknown quantities and only bZT""^ Z^° ^ — -^ two equations, viz : S (vert. fig. 275. compons.) = and S (moms, about some point) = 0. If now be gradually lowered, it receives less and less pres- FLEXUBE. CONTIJJUOUS GIKDEBS. 321 sure, until it finally readies a position where the beam barely touches it ; and then O's reaction is zero, and B and C support the beam as if were not there. As to how low must sink to obtain this position, depends on the stiffness and load of the beam. Again, if be raised above the level of B and C it receives greater and greater pressuTt., until the beam fails to touch one of the other supports. Still another consideration is that if the beam were tapering in form, being stiffest at 0, and pointed at B and (7, the three reactions would be different from their values for a prismatic beam. It is therefore evident that for more than two supports the values of the reactions de- pend on the relative heights of the supports and upon the form and elasticity of the beam, as well as upon the load. The circumstance that the beam is made continuous over the support 0, instead of being cut apart at into two independent beams, each covering its own span and hav- ing its own two supports, shows the significance of the term " continuous girder." All the cases here considered will be comparatively simple, from the symmetry of their conditions. The beams will all be prismatic, and all external forces (i.e. loads and reactions) perpendicular to the beam and in the same plane. All supports at the same level, 272. Two Equal Spans; Two Concentrated Loads, One in the Mid- ^e of Each Span. Prismatic Beam. — Fig. 275. Let each half- Bpan = i^ /i. Neglect the weight of the beam. Required the reactions of the three supports. Call them P^, Pq and p \.. From symmetry P^ = Pc, and the tangent to the elastic curve at is horizontal ; and since the supports are on a level the deflection of C (and B) below O's tangent is zero. The separate elastic curves OD and DC have a common slope and a common ordinate at D. For the equation of OD, make a section n anywhere be- tween and Z>, considering n(7 a free body. Fig. 276 (a) 322 MECHANICS OF ENGINEERIXG. Y —X ^>| (6) Fig. 276. •with origin and axes as there indicated. * From H (moms about neutral axis oi n) = we have (see § 281) Ei'^^=p{y2i—x)—Pc{i—x) eA =F{y2ix—% dx (1 (2) The constant = 0, for at both x, and dy -^ dx, = 0. Taking the x-anti-derivative of (2) we have ^/2/=P(^_^')-Pe[^-|'] . . (3) Here again the constant is zero since at 0,x and y both =0. (3) is the equation of OD, and allows no value of cc <0 or>^. It contains the unknown force P^. For the equation of BC, let the variable section n be made anywhere between D and C, and we have (Fig. 276 ih\ j x may now range between J^Z and T) ^^^^— ^^(^-) ^jdy_ dx Ix-t^+C (4) (5)' To determine C\ put x = }4l both in (5)' and (2), and equate the results (for the two curves have a common tangent line at D) whence C" = ^ PV Elp.^yiP¥—Pjl 0[?\ 2~j (5) * These are such that XOY is our "first quadrant"; in which, for points in a part of a curve convex toward the axis of X, d^yldx^ is essentially positive; and vice versa. It will be seen that both eqs. (1) and (4) are on this basis. They must be on the same basis; otherwise, later com- parisons of equations would result in error. FLBXUEE. CONTINUOUS GIRDEES. 323 Hence Ely ^ % PT?x-PA^^—^'\j^O" . . (6)' At D tlie curves have the same y, hence put a? = i- in the right hand member both of (3) and (6)', equating results, and we derive C"= — ^ Pf EIy = y,PVa>~P^\^__p^^XPf . . (6) which is the equation of DC, but contains the unknown reaction P^. To determine P^ we employ the fact that O's tangent passes through G, (supports on same level) and hence when a? = Hn (6), y is known to be zero. Making these substitutions in (6) we have Q=y,pf-y,p^f-±pi^ ... P=^P From symmetry P^ also = —P, while Pq must = ~P, since P^ + P^ + P<7 = 2 P (whole beam free). [Note. — If the supports were not on a level, but if, (for instance) the middle support were a small distance = Ag below the level line joining the others, we should put x = I and y = — Iiq in eq. (6), and thus obtain P^ = Pc= -^^ P + SET—, which depends on the material and form of the prismatic beam and upon the length of one span, (whereas with supports all on a level, P^ — P^ = -| P is independent of the material and form of the beam so long as it is ho- mogeneous and prismatic.) If Pq, which would then = ?| P — 6 EI {Jiq-^F'), is found to be negative, it shows that requires a support from above, instead of below, to cause it to occupy a position 7^o below the other supports, i.e. the beam must be " latched down " at 0.] The moment diagram, of this case can now be easily con- structed ; Fig. 277, For any free body nC, n lying in BG, we have 324 MECHANICS OF ENGINEEEING. i.e., varies directly as cc, un- c til X passes D wlien, for any point on DO, I wliicli is =0, (point of in- flection of, elastic curve) I for .T=yii? (note that x is Fig. 277. measured from C in this figure) and at 0, where x= I, becomes — ^^Pl •'• K=—lPl; M^^O; 3I^=LPl; andif„=0 Hence, since if max. =^Ply the equation for safe loading is B'l 6 -PI (7) The shear at (7 and anywhere on CD=~Pf while on DO it =^^P in the opposite direction .•.j;„=;ip . " . . . (8) The moment and shear diagrams are easily constructed, as shown in Fig. 277, the former being svmmetrical about a vertical line through 0, the latter about the point 0" Both are bounded by right lines. 273. Two Equal Spans. IJniformly Distributed Load Over „, , Whole Length. Prismatic Beam. ^ y\ ^_:^ ^ c —Fig. 278. Supports B, 0, bQIM I I 1 \ [\\ 111 C, on a level. Total load V~- — o| 1 ^ -- ^ ^---1 = 2 W= ^icl and may include I |po j ^'^k^'^ j that of the beam P. ^ , , "w IS con- I I li I I 11 stant. Asbefore, from sym- metry P^=P^, the unknown i Pc| reactions at the extremi- VMt. 278. ties. FLBXUKE. CONTINUOUS GIRDERS. 125 Let On=x ; then with wC free, 2" moms, about n= gives rdy IV EI'^^- lFx-lx'+^]-F,[lx- ^]+[Coiist=0] (2) [Const. =0 for at both dy-i-dx the slope, and x, are =0] ... EIy= '^[}4Fc^-j4M+Vi2^]-P.[}4lx^-y6^]+{G=0) (3) [Const. =0 for at both x and y are =0]. Equations (1), (2), and (3) admit of any value of x from to I, i.e., hold good for any point of the elastic curve OC, the loading on ■which follows a continuous law (viz. : w= constant). But when x=l, i.e., at G, y is known to be equal to zero, since 0, B and G are on the axis of X, (tangent at 0). "With these values of x and y in eq. (3) we have 0= |L . t-y^pj? ... p,=y8wi=yQW .-. PB=^^and Po=27r— 2Pe=? W The Moment and Shear Diagrams can now be formed since ;j jovv all tli6 external forces are Lw^ known. In Fig. 279 meas- ure X from G. Then in any section n the moment of the "stress-couple " is M^yQWoo- wxr . (1) j which holds good for any value of x on GO, i.e., from 07=0 up to x=l. By inspec- PiG S79. tion it is seen that for 07=0, M=0 ; and also for x=yi, M=0, at the in/lection point' G, beyond which, toward 0, the upper fibres are in tension 326 MECHANICS or engixeert^tg. the lower in compression, whereas between C and G the;^ are vice versa. As to the greatest moment to be found on CG, put dM-i-dx—0 and solve for x. This gives ^ W—wx=0 .'. [X for if max.]=^Z . (2) i^rhich in eq. (1) gives Jfu(at JV, seefigure)=+^jr? . . (2) But this is numerically less than Mo{=—}i Wl) hence the stress in the outer fibre at being T/ Wle /Q\ Po=%—j-, ... (3) the equation for safe loading is B'l _., Wl . . . . (4) the same as if the beam were cut through at 0, each half, of length I, retaining the same load as before [see § 242 eq. (2)]. Hence making the girder continuous over the mid- dle support does not make it any stronger under a uni- formly distributed load ; but it does make it considerably stiffer. As for the shear, J, we obtain it for any section by tak- ing the x-derivative of M in eq. (1), or by putting ^(ver- tical forces) =0 for the free body nG, and thus have for any section on GO J=z/qW—wx ... (5) j/is zero for x='^l (where M reaches its calculus maxi- mum M^ ; see above) and for x=l it = — Yq fF" which is nu merically greater than yi W, its value at G. Hence Jm=y8w . . . ". (6) FLEXURE. CONTINUOUS GIRDERS. 327 The moment curve is a parabola (a separate one for each span), the shear curve a straight line, inclined to the hor- izontal, for each span. Problem. — How would the reactions in Fig. 278 be changed if the support were lowered a (small) distance Iiq below the level of the other two ? 274. Prismatic Beam Fixed Horizontally at Both Ends (at Same Level). Single Load at Middle. — Fig. 280. [As usual j /^~x p-| the beam is understood to 1^^ — ■ V^py ■ — — - ' I be homogeneous so that E E ^P~ I '-' ^ is the same at all sections]. IJ I * j The building in, or fixing, i lyij ji of the two ends is supposed objr^-Jt:;;™--— 1^ — --—^^^ — (c* to be of such a nature as to Yi ' Br — —- — Vi-T--^ cause no 'horizontal con- FiG. 280. straint ; i.e., the beam does not act as a cord or chain, in any manner, and hence the sum of the horizontal components of the stresses in any section is zero, as in all preceding cases of flexure. In other words the neutral axis still contains the centre of gravity of the section and the tensions and compressions are equivalent to a couple (the stress-couple) whose mo- ment is the " moment of flexure." If the beam is conceived cut through close to both wall faces, and this portion of length=Z, considered free, the forces holding it in equilibrium consist of the downward force P (the load) ; two upward shears J^ and J^ (one at each section) ; and two " stress-couples " one in each sec- tion, whose moments are 31^ andJ/g. From symmetry we know that J, — J„ and that M^=M,. From I Y=Q for the free body just mentioned, (but not shown in the figure), and from symmetry, we have «/„= % P and J^-= % P ', but to determine M^ and M„ the form of the elastic curves B and B G must be taken into account as follows : Equation of OB, Fig. 280. I [mom. about neutral axis of any section n on 5] = (for the free body nC which 528 MBCHAXics or EXGi:^rEEEi:srG. lias a section exposed at each end, n being tlie variable section) will give BI^y-= P(y2 l-x) + M,- -y2P{i~x) (1) J^Note. In forming this moment equation, notice that M^ is the sum of the moments of the tensions and com- pressions at G about the neutral axis at n, just as much as about the neutral axis of 0', for those tensions and com- pressions are equivalent to a couple, and hence the sum of their moments is the same taken about any axis whatever "I to the plane of the couple (§32).] Taking the a:-anti-derivative of each member of (1), EI^=P(% I x—% a^)-f- if, x—y PQ x—% x") ax (2) (The constant is not expressed, as it is zero). Now from symmetry we know that the tangent-line to the curve B s>i B is horizontal, *.e., for x^y^l, dy-^dx^Q, and these values in eq. (2) give us 0=yi Pf+ y^I^l—f^PV; whence M,=M,=}i PI , (3) Safe Loading. Fig. 281. Having now all the forces which act as external forces in straining the beam 00, we are ready to draw the moment diagram and find M^^. For con- venience measure x from 0. For the free body nO, we have [see eq. (3)] y2Px-M, + P~=0.'.M=}iPl-}4Px ... (4) p Eq. (4) holds good for any J section on OB. By put- f7^ ting x=0 we have M=M^= y% PI; \&j oEEO'=M, to scale (so many inch-pounds moment to the inch of pa- per). At B, for x^y I, M^= — y^ PI ; hence lay offB'I)=ys PI on theop- FiG. 281. posite side of the axis O'O' c FLEXUEE. CONTIGUOUS GIEDERS. 329 from HG', and join DH. DK, symmetrical witli i>^ about B'D, completes the moment curves, viz.: two riglit lines. The max. iHf is evidently =yi Fl and the equation of safe loading ?Li=upi (5) Hence the beam is twice as strong as if simply supported at the ends, under this load ; it may also be proved to be four times as stiff. The points of inflection of the elastic curve are in the iniddles of the half-spans, while the max. shear is J.n = y2P (8) 275. Prismatic Beam Fixed Horizontally at Both Ends [at Same Level]. Uniformly Distributed Load Over the Whole Length. Pig. 282. As in the preceding problem, we know from symmetry that e/o=^c=/^ ^=/^ *^^> ^^^ tl^s-^ Mq=M^, and determine the latter quantities by the equation of the curve OG, there being but one curve in the present in- stance, instead of two, as there is no change in the law of loading between and C. "With nO free, I (mom^)=0 .gives ax 2 o X 9 (1) (2) ^N^wl & i C \n J opr J i I 11 I I H i L Fig. 282. MBgHAITlCS OF ENGIXEEKIlira. The tangent line at being horizontal we have for x=0, dx 0, .'. C=0. But since the tangent line at C is also hori- zontal, we may for x=l put dy-^dx=0, and obtain O^—i^Wl'+lIol+yewV; whence Mo=^Wl (3) a.s the moment of the stress-couple close to the wall at and at 0. Hence, Fig. 283, the equation of the moment curve (a single continuous curve in this case) is found by putting 2' (moma)=0 for the free body nO, of length x, thus obtaining w y^=wl lUi I ] j lull Fig. 283. lL+i4Wx-Mo iva^ =0 I.e. M=lWl+ ^-}4Wx , .(4) an equation of the second degree, indicating a conic. At 0, M=Mq of course,= 4- ^^/ ati?by putting a; = i^ Z in (4), we have M^— — }^ Wl, which is less than Jig, although M^ is the calculus max. (negative) for 31, as may be shown by writ- iijg the expression for the shear {J=% W — wx) equal to zero, etc. FiiEXUKE. coxTrsruous giudees. 331 Hence 31^=^ Wl, and tlie equation for safe loading is --^Wl (5) B'l __^ Since (with this form of loading) if the beam were not built in but simply rested on two end supports, the equa- tion for safe loading would be \_R'I-^e\ = yi Wl,isee §242), it is evident that with the present mode of support it is 50 per cent, stronger as compared with the other ; i.e., as re- gards normal stresses in the outer elements. As regards shearing stresses in the web if it has one, it is no stronger, since t/m = j^ JFin both cases. As to stiffness under the uniform load, the max. deflec- tion in the present case may be shown to be only i- of that in the case of the simple end supports. Eiieiit It is noteworthy that the shear diagram in Fig. 283 is identical with that for simple end supports §242, under uniform load ; while the moment diagrams differ as fol- lows : The parabola KB'A^ Fig. 283, is identical with tha*- in Fig. 235, but the horizontal axis from which the ordi- nates of the former are measured, instead of joining the extremities of the curve, cuts it in such a way as to have equal areas between it and the curve, on opposite sides i.e., areas [^C"^'+^i6^'0']=area R'G'B' In other words, the effect of fixing the ends horizontally is to shift the moment parabola upward a distance = 3Ic (to scale), = i Wl, with regard to the axis of reference, 0'^', in Fig. 235. 276. Remarks. — The foregoing very simple cases of con- tinuous girders illustrate the means employed for deter- mining the reactions of supports and eventually the max. moment and the equations for safe loading and for deflec- tions "When there are more than three supports, with spans of unequal length, and loading of any description the analysis leading to the above results is much more complicated and tedious, but is considerably simplified 332 MECHAXTCS OF ENGINEERING. and systematized by tlie use of tlie remarkable theorem of three moments, the discovery of Clapeyron, in 1857. By this theorem, given the spans, the loading, and the vertical heights of the supports, we are enabled to write out a rela- tion between the moments of each three consecutive sup- ports, and thus obtain a sufficient number of equations to determine the moments at all the supports [p. 641 Eankine'a Applied Mechanics.] From these moments the shears close to each side of each support are found, then the reactions, and from these and the given loads the moment at any section can be determined ; and hence finally the max. moment ilf^,,, and the max. shear J^„. The treatment of the general case of continuous girdera hy algebraic methods founded on the properties of familiar geo- metrical figures, however, is comparatively simple ; and will be developed and applied in another part of this book. (See Chap. XII, pp. 485, etc.) THE DANGEROUS SECTIOIS^ OF ]S^O]?^-PRIS- MATIC BEAMS. 277. Eemarks. By " dangerous section " is meant that sec- tion (in a given beam under given loading with given mode of support) where p, the normal stress in the outer fibre, at distance e from its neutral axis, is greater than in the outer fibre of any other section. Hence the elasticity of the material will be first impaired in the outer fibre of this section, if the load is gradually increased in amount (but not altered in distribution). In all preceding problems, the beam being prismatic, I, the moment of inertia, and e were the same in all sections, hence when the equation P—=M [§289] was solved for », e Me .... giving i>=— . . . . (1) we found that p was a max., = p^, for that section whose ili" was a maximum, since p varied as M, or the moment FLEXUEE NON-PEISMATIC BEAMS. 333 of the stress-couple, as successive sections along the beam were examined. But for a non-prismatic beam Zand e change, from sec- tion to section, as well as 31, and the ordinate of the moment diagram no longer shows the variation of p, nor is ^ a max. where ilf is a max. To find the dangerous section, then, for a non-prismatic beam, we express the ilf, the I, and the e of any section in terms of x, thus obtain- ing ^=func. {x), then writing dp-~dx=0, and solving for x. 278. Dangerous Section in a Double Truncated Wedge. Two End Supports. Single Load in Middle. — The form is shown in Fig. 284. Neglect weight of beam ; measure x from one sup- port 0. The r e a c tion a t each support is i^ F. The width of the beam == 5 at all sections, while its height, v, varies, being = h at 0. To express thee = }4 v, and the /= 1 hv^ (§247) of any section on 0(7, in terms of x, conceive the sloping faces of the truncated wedge to be prolonged to their intersec- tion A, at a known distance = c from the face at 0. We then have from similar triangles [Tpl Fig. 284. V : X -{- c : : h 3, .: V = ~ (x -\- c) c and e = h (x-{-G) and I = K^ -^x-^rcf For the free body nO, H (moms.^) Px—tL± e [That is, the M = )4 Fx.] p=SF and^= 3F ^ ax hH ' {x+cf ' By putting dp -t- dx = (1) (2) (3) we find X = + c\ showing a gives Fxe '~w • • • But from (2), (3) becomes dp_ o-p & {x-\-cy — 2a;(a;-t-c) p- 334: MECHAXICS OF ENGINEERING. maximum for p (since it will be found to give a negative result on substitution in d^p 4- dotf). Hence tlie dangerous section is as far from tlie support 0, as the imaginary edge^ A, of tlie completed wedge, but of course on the opposite side. This supposes that the half -span, )4l, is, > ^; if not, the dangerous section will be at the middle of the beam, as if the beam were prismatic. tx -xi, ) the equation for safe { ■Dfh'h'2 Hence, with L ^^.^1^ .^^ ^f^>^^,^^^,\ ^-%Pl (5) A'' <^ ) at middle) ( ^ while with )tl^e equation for safe j ^,j ^A]^ , , p ,.. 1/7 ^ r h loading is : (put x=c ■{ ^-^= V2 Pc (6) /^^ > ^ ) and_p=i?'in [3]) ( ^ (see §239.) 279. Double Truncated Pyramid and Cone. Fig. 285. For Fig. 285. the truncated pyramid both width = u, and height = v^ are variable, and if h and Ji are the dimensions at 0, and c = QJ[ = distance from to the imaginary vertex A, we shall have from similar triangles u=~ (a;+c)and v= ~ (x-\-c). G c Hence, substituting 6=^^ and 7=1 uv^, in the moment equation £^_^=0.weW^=34,.^-|-,. . (7) . dp ^ op <^ (x+c)^ — 3x [x+cy ' * dx bW" ' {x-\-cf (8) FLEXURE NON-PRISMATIC BEAMS. 335 Putting the derivative =0, for a maximum p, we liave x = -h ^ c, hence the dangerous section is at a distance a? = ^ c from 0, and the equation for safe loading is either :?^= 14: PZ if >^ Z is < >^ c . . . , (9) (in which V and h' are the dimensions at mid-span) or MM)lhlf=y^P,,iy^i 6 is > >^ c ... (10) For the truncated cone (see Fig. 285 also, on right) where e = the variable radius r, and / = i^ ;r r*, we also have /=[Const.] .,—^.3 (11) and hence j9 is a max. for a? = ^ c, and the equation for safe loading either 5£i^ = % Fl,iox %l <% c , . , , . (12) (where r' = radius of mid-span section) ; ^^ ^-R' (l^o)' ^%Fc,ioxy2l> %c (13) (where r^ = radius of extremity.) IS^ON-PMSMATIC BEAMS OF "UNIFORM . STRE]?^GTH." 380. Eemarks. A beam is said to be of " uniform strength " when its form, its mode of support, and the dis- tribution of loading, are such that the normal stress ^ has the same value in all the outer fibres, and thus one ele- ment of economy is secured, viz. : that all the outer fibres may be made to do full duty, since under the safe loading, p will be = to B' in all of them. [Of course, in all cases of flexure, the elements between the neutral surface and 336 MECHANICS OF EIS^GIJ^JEEEmG. fclie outer fibres being under tensions and compressions less than R' per sq. inch, are not doing full duty, as tegards economy of material, unless perhaps with respect to shearing stresses.] In Fig. 265, §261, we have already had an instance of a body of uniform strength in flexure, viz. : the middle segment, CD, of that figure ; for the moment is the same for all sections of CD [eq. (2) of that §], and hence the normal stress p in the outer fibres (the beam being prismatic in that instance). In the following problems the weight of the beam itself is neglected. The general method pursued will be to find an expression for the outer -fibre-stress p, at a definite sec- tion of the beam, where the dimensions of the section are known or assumed, then an expression for p in the varia- ble section, and equate the two. For clearness the figures are exaggerated, vertically. ' 281. Parabolic Working Beam. UnsymmetricaL Fig. 286 i. Pig. 286, CBO is a working beam or lever, B being the fixed fulcrum or bearing. The force P^ being given we may compute P^ from the mom. equation Pq^o = PJ^u while the fulcrum reaction is P^^P^-^-P^^. All the forces are ~\ to the beam. The beam is to have the same width h at all points, and is to be rectangular in section. Ilequir6*d first, the proper height hx, at B, for safety. From the free body BO, of length = l^, we have I (momss) = ; i.e., -^ rX,oxp^- -— ... (1) FLEXUEE. IfON-PIlISMATIC BEAMS. 337 Hence, putting jo^ = B', h^ becomes known from (1). Required, lecondly, tlie relation between the variable height V (at any section n) and the distance x ot n from 0. For the free body nO, we have (2 momSu = 0) 3iL=F,x ; or ^" ^^ ^^' =P,x and .-. p, =^l^ (2) But • for " uniform strength " p^ must = p^ \ hence equate their values from (1) and (2) and we have ^ = — 1, which may be written {% vj = .>'^p' x (3) so as to make the relation between the abscissa x and the ordinate }4 v more marked; it is the equation of a para- bola, whose vertex is at 0. The parabolic outline for the portion BC is found simi- larly. The local stresses at G, B, and must be proper- ly provided for by evident means. The shear J = Pq, at 0, also requires special attention. This shape of beam is often adopted in practice for the working beams of engines, etc. The parabolic outlines just found may be replaced by trapezoidal forms, Fig. 287, without using much more ma- terial, and by making the slop- ing plane faces tangent to the parabolic outline at points Tq and Ti, half-way between and ^^"^^b^^"'^'^ ° B, and C and B, respectively. It fis. 287. can be proved that they contain minimum volumes, among all trapezoidal forms capable of circumscribing the given parabolic bodies. The dangerous sections of these trape- zoidal bodies are at the tangent points Tq and Ti. This is as it should be, (see § 278), remembering that the subtan- gent of a parabola is bisected by the vertex. 338 MECHANICS OF ENGINEEKING. 283. Rectang. Section. Height Constant. Two Supports (at Ex- tremities). Single Eccentric Load. — Fig. 289. h and Ji are tlie dimensions of the section at B. "With BO free we have Pah ■Folo=0.\pj, ■ 6Po?o (1) Fig. 289. At any other section on BO, as n, where the width is u, the variable whose relation to x is required, we have for wOfree P^^.F,x;ovPll/^=P,x QP,,x Pn = Equating pu and ^„ we have u :h :: x :Iq „ That is, BO must be wedge-shaped ; edge at 0, vertical. (2) (3) ■k- 1 ^k-^wl Fig. 289 a. 283a. Sections Rectangular and Similar. Otherwise as Before. — Fig. 289a. The dimensions at B are b and h; at any other section n, on BO, the height V, and width u, are the variables whose relation to x is desired, and by hypothesis are connected by the relation u:v::b:h (since the section at m is a rectangle similar to that at B). By the same method as before, putting pB = Pn, we obtain lf^^bh'^ = x-i-uv^; in which placing u^bv^h, we have finally v^=(h^^lf))x; and similarly, u^={b^-^lQ)x; . . . (4) i.e., the width u, and height v, of the different sections are each pro- portional to the cube root of the distance x from the support. (The same relation would hold for the radii, in case all sections were circular.) 283b. Beam of Uniform Strength under Uniform Load. Two End Supports. Sections Rectangular with Constant Width. — Fig. 289&. WeigM of beam neglected. How should the height v vary, (the height and width at middle being h and b) ? As before, we equate pB and pn ', whence finally (ivy = [h^^P](lx-x') (5) This relation between the half-height ^v (as ordinate) and the abscissa. X is seen to be the equation to an ellipse with origin at vertex. FLEXURE OF BEINFOKCED CONCRETE BEAMS. 339 CHAPTER V. Flexure of Reinforced Concrete Beams. 284. Concrete and "Concrete-Steel" Beams. Concrete is an artificial stone composed of broken stone or gravel (sometimes cinders), cement and sand, properly mixed and wet beforehand and then rammed into moulds or " forms " and left to harden or " set." This material, after thorough hardening or " setting," thouglT fairly strong in resisting compressive stress is compara- tively weak in tension. When it is used in the form of beams to bear transverse loads (i. e., under " transverse stress ") the side of the beam subjected to tensile stress is frequently " re- inforced " by the imbedding of steel rods on that side. In this way a composite beam may be formed which is cheaper than a beam of equal strength composed entirely of concrete or one composed entirely of steel. Of course the steel rods are placed in the mixture when wet, and previous to the ramming and compacting, and their aggre- gate sectional area may not need to be more than about one per cent, of that of the concrete. No reliance being placed on the tensile resistance of the concrete (on the tension side of the beam) it is extremely important that there should be a good adhesion, and consequent resistance to shearing, between the sides of the steel rods and the adjacent concrete, for without this adhesion the -rods and the concrete would not act together as a beam of continuous substance.* In some specifications, for instance, it is required that the shearing stress, or tendency to slide, between the steel rods and the concrete shall not exceed 64 lbs. per sq. in. Sometimes the steel rods are provided with projecting shoulders, or ridges, or corrugations, along their sides, to secure greater resistance to sliding. * * For an account of tests of this adhesion see Engineering News, Aug. 15, 1907, p. 169, and also p. 120 of the Engineering Record for Aug. 3, 1907. 340 MECHANICS OF ENGINEERBsTG. Fig. 290 gives a perspective view of a concrete-steel beam of rectangular section, placed in a horizontal position on two supports at its extremities, and thus fitted to sustain vertical loads or weights ; while Fig. 291 shows a concrete-steel beam flange- teeb, or stem- of T-section, in which the flange is intended to resist compres- sion, while the steel rods in the lower part of the " stem " are to take care of the tension. These two shapes of beam will be the only ones to be considered here, in a theoretical treatment. The ratio of the Modulus of Elasticity of steel (viz. — about 30,000,000 lbs. per sq. in.) to that of concrete (say, from 1,000,000 to 4,000,000 lbs. per sq. in., according to the propor- tions of ingredients used) is of great importance in the theory, since in general the stresses induced in two materials for a given percentage change of length are directly proportional to the modulus of elasticity (for same sectional area). Generally the diameter of a steel rod is so small compared with the full height of the beam that the stress in the rod is taken as uniform over the whole of its section. 285. Concrete-Steel Beam of Rectangular Section. Flexure Stresses. — As in the common theory of flexure of homogeneous beams, it will be assumed that cross-sections plane before flexure are still plane when the beam is slightly bent, so that changes of length occurring in the various fibers are propor- FLEXUEE OF EEINFORCED CONCRETB BEAMS. 541 tional to the distances of those fibers from a certain neutral axis of the cross-section, and upon the amount of any such change of length (relative elongation) can be based an expression for the accompanying stress. Now in the case of concrete it is not strictly true that stresses are proportional to changes of length (" strains " or deformations) ; in other words its modulus of elasticity, E, is not constant for different degrees of shortening under compressive stress. Nevertheless, since this modulus does not vary much, within the limits of stress to which the concrete is subjected in safe design, it' will be considered con- stant, the resulting equations being sufficiently accurate for practical purposes. Let us now take as a " free body " any portion, ON, of the beam in Fig. 290, extending from the left-hand support to any section, at any distance x from that support. In the plane section terminating this body on the right, BNS (see now Fig. 292, in which we have also, at the right-hand, an end-view of < 6 > t 1 ] h \ 1 -4-- 1 AS-;>:-iy:V;-:;?il! N. axis — • — • — •— End View.. " the body), we note that the fibers of concrete from Z> down to a neutral axis iV^are in a state of compression, while below iVthe steel rod alone is considered as under stress, viz., a total tensile stress of F'p', where F' is the aggregate sectional area of all the steel rods, these rods being at a common distance a' above the lower edge of the section, and p' is the unit (tensile) stress in the steel rods. The distance BN, or " ^," of the neutral axis N below the " outer fiber " i), is to be determined. Let p denote the unit compressive stress in the fiber at I) (outer fiber) of the concrete ; then the unit stress in any fiber of the concrete at distance z from iV will be - j9, lbs. per sq. in., and the total stress on any 342 MECHANICS OF ENGLNEERING. such fiber is — »c?i^, lbs. (where c?jPis the sectional area of the e fiber). All the (horizontal) fibers between the two consecutive cross sections DS and D' 8' were originally dx inches long, but now (during stress), we find that the fiber at D has been shortened an amount d\ and the steel rod "fibers " elongated an amount d\'^ so that we have the proportion d\ : d\' : :e: a — e; d\ e , ,». or, ■-— = (0) dV a- e ^ ^ For the free body in Fig. 292 we have, for equilibrium, the sum of horizontal components of forces = (the shear J" has no horizontal component); that is, remembering that below iVno tensile forces are considered as acting on the concrete, but simply the total tensile stress F'p' in the steel rods, t/n - pdF - F'p' = 0. e But here =^ is a constant ; and for the rectangular cross -section, dF = b . dz, and 'iI>'^-'T4-'-'T-^y • • • W But from the definition of modulus of elasticity (F for the concrete and E^ for the steel), we have (§ 191) F = p —- (relat. elongation), or F = p -i- [dX/dx) ; and similarly, F' = p' -^ {dV /dx) ; whence d\' p'' E ^ ^ But, from eq. {\.)^ p -— p' = IF' -^ he, combim'ng which with eqs. (0) and (2), and denoting the ratio F' -^ Fhj n, we mid = — ^ — ■. . (3) a — e be The ratio n may have a value from 10 to 25 for " rock- concrete," and still higher for " cinder-concrete ; " see § 284. Now solve eq. (3) for the distance e, obtaining F'nfj2ab ^ A FLEXURE OF REINFORCED CONCRETE BEAMS. 34S This locates the neutral axis, iV". [See, later, eq. (29), § 291.] Returning to the free body OJV in Fig. 292 above, we note that the resultant compression in the concrete between iV^ and D, viz., ^ p .he, lbs. [see eq. (1)], is equal in value to the total tension F'p', lbs., in the steel rods at G', and that they are parallel. Consequently they form a couple (the " stress- couple " of the section) whose moment is equal to the product of one of these forces, say F'p\ by the perpendicular | j^ \ distance = a", between Gr' and a point Cr (see now Fig. 293) whose distance from the " outer fiber " i> is one-third of e. The "arm" of this couple is a' a- -■ For Fig. 293. equilibrium of the free body ON in Fig. 292 the shear J and the two forces V (reaction) and P^ (load) must be equivalent to a couple of opposite and equal moment to that of the stress couple. Call this moment M [in this case it has a value of Yx — P^{x — a;J]; it is the " bending moment " of the section at DS. We may therefore write (see Fig. 293) : M = F'p' [« — i e] ; and .-. p' = M F\a-\e) (5) which will give the unit-stress p\ induced in the steel rods at section BS. It is seen to depend on the position of the neutral axis N (i.e., upon e); upon the bending moment, M, at that section; upon the sectional area F' of the steel rods (aggre- gate); and on the distance, a, at which they are placed from the compression edge, B, of the beam. But since the resultant compression, h p -he, is equal to the resultant tension, .F'p', we may also write 2M M= ^p .be [« — i e] and .-. p = he (a — I e) (6> which gives the unit-stress (compression) in the outer " fiber " at i), of the concrete, for this section BS. 1 e ..i G ■^— N N" V'(p + clp'l s" 344 MECHANICS OF ENGINEERING. 286. Horizontal Shear in the Foregoing Case (Rectangular Sec- tion). The shear per sq. in. along the sides of the steel rods, I ^^ and also along the horizontal ,/' " Neutral Surface,'' NN" (see Fig. 294), may be obtained as follows : — Let dx be the length of a small portion of the beam ,, , , „ (of Fig'. 290) situated between ■^ r^^^^n '- — two vertical sections Do and D"S". Fig. 294 shows this ^^*^- 294. portion as a " free body." The forces acting consist of the tension F'p' on the left-hand end of the steel; the tension on the right-hand end of these rods [being something greater (say) and expressed by F' (^p' -\- dp'^ in which dp is the difference between the unit-tensions at the two ends of the steel " re-inforcement "] ; the resultant com- pression, i he.p, in the concrete on the left; and that, 1 5g . (j? -f dp^, on the right ; and, finally, the two vertical shears, J'and J". Here p is the unit compressive stress (lbs. per sq. in.) in outer fiber of concrete at the left-hand extremity of the same, while p -\- dp expresses the unit compressive stress in the same outer fiber at the right-hand extremity. Evidently the difference between the total tensile stresses at the extremities of the steel rods will give the total horizontal shearing stress on the sides of those rods and this may be written pjl^dx (lbs.), where pj = unit shearing stress between the steel and concrete and l^ == aggregate perimeter of the steel rods (so that l^dx = total area of the outside surface of rods in Fig. 294); hence p/l^dx ^ F' {p' j^- dp') — F'p' .... (7) But if, for the free body of Fig. 294, we put 2 moms. = about the point Cr (a distance ^ e from upper fiber) we find Jdx = [F' (p' + dp') — F'p'']{a -\ e)\ . . (8) and hence ) , J" /q\ see (7), \ ^' ^ l^ (a -^ e) ^ ^ FLEXURE OP EElNFOliCED CONCRETE BEAMS. 346 p bdx Also, if we let p^ denote the unit shearing stress (or tendency to slide) along the horizontal sur- face NN" or neutral surface, the total amount is pjbdx (lbs.). In Fig. 295, which shows as a free body the portion NN"S"S of Fig. 294, we see this horizontal force (of concrete on concrete) act- ing toward the left. The other ^^°- ^^s. forces acting on the free body are as shown in Fig. 295 and, by putting 2 horiz. compons. = 0, we find and finally, see eq. (8), F' {f + df) — F'p' = pJbdx; J Ps (9a) (10) 5(a-ie) • This (unit) shearing stress in the concrete along NN", the "neutral surface," should nowhere exceed a certain value \Q.g., 64 lbs. per sq. in.). For horizontal planes above NN" it is smaller than along NN". Similarly, the unit stress pj should not exceed a proper limit. 287. Numerical Example of a Concrete-Steel Beam of Bectangular Section. (See foregoing equations.) Fig. 296 sliows the section [8 by 11 inches] of the beam. Four round steel rods are imbedded near the under (tension) side, their centers being 10 in. from W=600 Ihs. .- d=0.45 4L p-^? Fig. 296. the top of section (a = 10 in.). This beam is to be placed on two supports at the same level and 8 feet apart, and is to support a concentrated load P, lbs., at the middle of the span as well as its own weight, which is K = 600 lbs. P is to be determined of such a safe value that the greatest stress in the steel rods shall not exceed 16,000 lbs. per sq. in. The compressive stress in concrete is not to exceed 700 lbs. per sq. in., nor the greatest shear either in the concrete or between the steel and the concrete, 64 lbs. per sq. in. Each steel rod is continuous throughout the whole span and has a diameter of 0.45 in., from which we easily compute the aggregate perimeter of the rods 346 MECHANICS OF ENGHSTEERING. to be 5.65 inches (=io)> ^^^ ^^^ aggregate sectional area to be 0.64 sq. in (= Fy The ratio of the modulus of elasticity for steel to that of the concrete will be taken as 15 to 1 ; i.e., n = 15. The first step is to locate the neutral axis by finding the value of e from eq. (4), thus: — 64 15 / / 2 X 10 X 8 , , , \ „ J, . . ^-iOQ-T(V .64X15 +l-l)=3.84m. Next, if for p' we write 16,000 (using inch and pound) and substitute in eq. (5), solving for M, we obtain the greatest bending moment to which any section of the beam should be exposed, so far as the steel is concerned, viz: — M = p'F'la- I) = 16,000 x 0.64 (10 - 1.28) = j e\ ..nnn.. n«..in 1 oe^ _ i 89,000 in.-lbs. i.e., max. moment is to be 89,300 in.-lbs. For the mode of loading of the present beam the max. moment occurs at the section at the middle of the span and has a value (with I denoting the span, PI Wl or 96 in.) of — + -tt- • We therefore write PX96 ^ 6_00x9_6 ^ gg 3^^_ ^^^^^ ^ ^ g^^^O lbs. 4 o To find the accompanying maximum compressive stress in the concrete, eq. (6) gives (for outer "fiber") 2Jf 2 x 89,300 __ -, P = T-, ; — -s = 5 — FTTH s-^ = 666 lbs. per sq. m., •^ le{a-\e) 8 x 3.84 x 8.72 ±' ^ > which is within the limit set (700 lbs. per sq. in.). As for the max. shearing unit stresses Ps and ps, they are greatest where the vertical shear, J, is a max., which is close to one of the supjjprts. Here we note that J is equal to J of 3,420 + i of 600 = 2,010 lbs. Hence, from eq. (9), 2,010 2,010 ^' - 5.65 X (10 - 1.28) = 5.65 x 8.72 " = ^^"^ ^^'- ^^^ "I- '''■' while ) 2,010 from (10) i^^ = 8"3r8^ = ^^'^ ^^^- P^^ ^*1- ^"•' These shearing stresses are seen to be well within the limit set, of 64 lbs. per sq, in. As to compressive stress, the building laws of most cities put 500 lbs. per sq. in. as max, safe limit forp, the compressive stress in concrete. 288. Concrete-Steel Beam of T-Form Section. See Fig. 297. In this form of beam, to secure simplicity in treatment, it will be considered that the flange {TK') alone is subjected to com- pressive stress [although strictly a small portion of "stem" between the flange and the neutral axis of a section is under that kind of stress] . The part of stem below the neutral axis (as before) is not considered to offer any tensile resistance, all FLEXUEE OF HEINFOECED CONCEETE BEAMS. 347 tension being borne by the steel rods or " re-inforcement." Fig. 297 shows a side view and also an end-view of a portion of the beam in Fig. 291 extending from the left-hand support np to any section DjS (or up to W in Fig. 291). - As before, sections plane before flexure are considered to be still plane during flexure, so that the elongations or shortenings of any horizontal *' fiber," w^hether steel or concrete, are proportional to the dis- Fig. 297. tances from a neutral axis iV, at some distance e from the top fiber of the flange, where the unit compressive stress has some value p. Also, since the U for concrete in compression is to be taken as constant the stresses in the concrete will also be proportional to the distances of the " fibers " from H the neutral axis. Let p'^ denote the unit-stress in the concrete at H, the bottom fiber of the flange ; then, by proportion, p :p^' : : e: e — d, where d is the thickness of the flange. Since the compressive stresses in the concrete between H and D are distributed over a rectangle their average unit-stress is (p-\-p'')/2, and their resultant, which acts horizontally through some point Gr, has a value of bd.(p -\- p")l2\ or, as it may be written (see above for y ), '{^p {1 e - d) .Id) -^ {1 e\ . The total tensile stress in the steel rods will be ¥'p', as before, where ¥' is the aggregate sectional area of the rods and p' the unit stress in them at section DS. Besides the stresses just mentioned the other forces acting on the free body in Fig. 297 are all vertical ; viz., the shear J'and the pier reaction and certain loads between and D ; hence by summing the horizon- tal components we note that the compressive stress in the con- crete is equal to the tensile stress F'p' in the steel, so that this 348 MECHANICS OF ENGINEERING. tensile force F'p' and the resultant compressive stress form a couple (^^ stress-couple '^ of the section; with an arm = 6r6r', = a"), and we have pi2e-d)bd ^ • 2e -^ ^ ^ Consequently the shear and the other vertical forces acting on the free body form a couple also, and the moment of this couple (equal to that of the " stress-couple ") will be called M. In the figure these vertical forces are not shown, but simply an equiva- lent couple (on the left). If at this part of the beam a length dx of the steel has stretched an amount dX' and an equal length, dx, of the outer fiber at D has shortened an amount dX, we have from eq. (2) of previous work dX'-p'"E' ^^^^' where E' and E are the moduli of elasticity of the steel and concrete, respectively. But, from (11), p' (e - ^d)bd' ^ ^ and from similar triangles dX : dX' : : e : (a — e) . . . .(14) Eqs. (33), (14), and (12), with E' -i- E = n, give F-n.a+—- '= bd + En ' ^^^^' and thus the neutral axis, iV, is located. It will now be necessary to locate the point of application,, between E and E, of the resultant compressive stress on EE ; that is, the point (r in Fig. 298 which gives a side view of these stresses alone, forming, as they do, a trapezoidal figure whose center of gravity, U, projected horizontally on EE gives- the desired point, G: The lower base EC^^ of this trapezoid FLEXURE OF JREINFOECED CONCRETE BEAMS. 349 represents the unit stress jw"; the upper, DC", represents the unit stress p. The distance, call it c, of G- from N, is to be determined. Let the trapezoid be divided into a rectangle BD'" C" H and a triangle D'"G"'Q". The center of gravity of the latter is at a vertical distance of \ d from a line WW" drawn horizontally at distance \ d from D . H" H'" passes through the center of gravity of the rectangle. Let us now find the distance GrR" by writing the moment of the resultant stress about point W equal to the sum of those of its two parts, or components, represented by the rectangle and the triangle ; whence we have Fig. 298. \{p^p").uy. aw = o + ^P-^P^'^ .'^ (16) Noting that ji?'' = d-" - d p, we have, solving, aw=\ 6 2e- d NCr, i.e., c, = e , and therefore, measuring from N, d 1 2 "^ 6 ■ 2e - cZ (17) Now that both e and c have been determined in any given case it remains to find expressions for the unit stresses p' and j? (in steel and fiber I) of concrete}. Since (r is the point of application of the resultant compres- sion in flange, the arm of the stress-couple, a" (Fig. 297), is the distance from G- to Gr' (see Fig. 297); that is, a" = c -{- [a— e); and hence we may write Pf(c + a-e)~M;.:f = ^,^/^^_^^ .(18) Also, by eliminating the ratio d\:d\' from eqs. (12) and (14) we have, solving for^, p = p e n (a — e) (19) 350 MECHANICS OF ENGINEERING. 289. Shearing Stresses in T-Form Concrete-Steel Beams. As regards the unit shearing stress, p'^ induced on the sides of the steel rods, in this case of the concrete steel beam of T-form section, an analysis similar to the corresponding one in the case of the beam of rectangular section leads to the result J ( where «7is the total vertical shear at ) ,oa\ Iq{c -\- a— e) I the section BS, and l^ the aggregate ) perimeter of the steel rods. And, similarly, for the unit shearing stress on the horizontal surface separating the flange from the "web" or "stem" (see Fig. 297 ) at H, where the width of the web is h", we find for this unit horizontal shear, p^, P'= V-(c + a-e) (^^' 290. Deflections of Concrete-Steel Beams. The deflection of a loaded prismatic concrete-steel beam resting on two supports at its extremities, may be obtained for the cases dealt with in §§ 233-236 inclusive, in connection with homogeneous beams ; provided the product EI occurring in the expressions for these deflections be replaced by ■ [ a— -^), for concrete-steel beams of rectangular section; and by E' F' {a — e) (c + a — e), for those of T-form section. 291. Practical Formulae and Diagrams for use with Concrete-Steel Beams of Rectangular Section. The equations of the foregoing theory will now be put into convenient form for practical use in designing these beams. Let us denote the ratio of p' (stress in steel at section of max. moment) to p (stress in outer fiber of concrete) by r ; i.e., r = p'/p ; while n = E'/E, as before. Also let ?n, = M -^h, denote the max. bending moment per inch of width (6) of beam ; and let F' (area of steel) -h & be called/, i.e., steel areajjer mc/i of width {b). In other words, we have the notation r =-?:_; n= -=-;™ = ^; and /'= -=-; . . . . (22) Fig. 299. p E If we now substitute e = 2 r/, from eq. (1), in eq. (3), we have 2Tf {r + n)=an (23) Now e = 2 r/, which from (23) =an-^{r + n) ; Hence eq. (5) -will give 3?n(r + Ti) = a/'p'(3r + 2n) (24) REINFORCED CONCRETE BEAMS DIAGRAM 1 ^ rectangular'). 20 25 30 40 L/ 1 r^ M 1 ir^ III .!,• ; - -r -> 1 H - ■i:: :-_!_: '.!.'. .-H ::f. i-t.. 60 s = /«- » :;-r-:-?: - T-- ^^ ~ «->«./'„ . „ \ ' - _l- _ :.-^y-. in which ^c ::z5:::t: ,?-i- -i- I'.Tl'i - -\-^ ■ J-. y y ..2f_ -■^•- 50 — — 1 / — 1 — ^. 1 ^1 - - -L V- ^ y J --J&- --1-- .._,.(« y /'^ 1 .,2^ y ■'-y r- «:^- \ d ah 1 xc / : _Ly 1 "[ \ J ^5 ^ ^ n Q '"i"" '/ / - -y-- 1 I'' ' ---1--- ' y - -k-. An y \l%L"'-X\ '"^ *HJ CO p°^ • ^^ '^^ in <• o / O "^ /I y I y m ^' ~"i~ ... — y K 1 -- / 1 --y- 1 '^ .:i!:: 1 / 1 y ' y * ' ^ ' , i! ' .. oc ' / ^*^ ■-r-- / \y ! > ' Z- y T ■ 35 1 / y y y "Xy ' y ?^ --y/ -■ y ~ c - ^ / 1 y/j.. -A- ■>- y / 1 / V y ^-' 1 ± oo ^ / ' y y / yi yi ^ T ■ +■■ 30 54^ y ^ 1 1 / ^; / 1 y ' \y 1 1 i' ' / - 'A / /I -\y' y > ,y^.. 1 \f' . ._... y y /\ J ^ 1 / / y * ' ..±t: OR / ' > > y V ?*■ y y -1- -^'- -=r 25 ., / j/^ _/ / -^ J y / 1 .''^; y\ "i::' ^ /• ^_ y y_ y- - / ^^- y y ' 1 -^ .<:: / /■ >■ / y / ' y y 1 1 .^ -y'T >'' ; y y y / ,x ^ ^ y^ ' ^.>' ' 1 ■■ 20 1 1 / /y /^ y 1 / y '\ y" y- .-^ ;- 4:j s / y j / / y y- 1 y y y y. y ' *;'x. 1 { / y/^ y y /K ^ y ; y' 1 . _L Its y ^ \y X y y >i^ y /> -y -T ^. y^-1.. ..4... ]g ■ r ' ■ /.. y\ y X y y ^y y ^ y <^- - -I- - ---T f -H -- -1- 1 1 •I 2 / .^r\. _ . ^ > y y y - 4- - .. J . l_ 1 y -y(-y > /^ y_ 1 1 - -- ■ 1 - - r - 1 ' A ■ - 1 1 1 - - 1 ■ 1 ■ 1 ' . .'. . . L. 1 1 1 i > 6 7 8 9 10 12 15 20 25 30 VALUES OF n, =E'^E ijj 1 ly 40 \ To face page 350. REINFORCED CONCRETE BEAMS 10 12 15 20 (rectangular ^ « nil 1 » .75 .80 ^ i .85 O .90 CO u D ' _l < > .95 1 1 1 , , UIMbKMm li. \ sj 1 1 3r -t-2^ ^ V t \ s V 1 1 ■-+-- 1 1 -i 3 (r -^n) \ sj \ s. \ ^, 1 25 30 in the foftiiula j — , - Q />'« 50 60 s k > \ .p. 1 1 ..J... m ^ V ' N s ^^^ H 1 or \ \ \ \ S ■4^ , 1 s ' m • \ \ \ 1 » 1 . 1 -X-.-i-.i.- T y "p uj + ..,x.._l 1 \ \ s. ^s Pv . ' ;,i.:4.. I.. ',.J m N s ^ \ s 1 \ ^^ i •• !..i.|.. \ t \ s ^^' ^4. r'ii IS \ I) .^ \> N i S, : 'K,\ ^ \ Nj \ \ 1 ^ 1 i. ^\ 'K i i. :i..J..i. K N, \ 1 t ^ T ^ -- -!- \ \ 1 1 1 s 1 ^^ 'v 1 I **;;.. 1 1 ^^<, kH 's 1 1 1^ L_. 1 1 1 _ J_ - ' ..jitJ... D -'-- - - - 1 1 -1 — i 1 1 > 1 T ^N -• 1 12 15 20 25 30 4( VALUES OF r^ D 50 60 [To face page 351, FLEXURE OF EEINFOECED CONCRETE BEAMS. 351 If now S denote the quotient /'-=-a (i.e., /S=area of steel section j3er unit area of concrete section above the steel), (23) may be written in the form S = n-h[2r{r+n)] (25) 3 J'4-2 n Asain, let ^— r- be denoted by Q - . . . (26) ° 3(r + n) ^ ' Prom (24), (25) and (26) we may then derive M=qSp'a^b .... (27); andi m = QSp'a^ ...... (28) for practical use in design; in connection with Diagrams I and II (opp. pp. 350 and 351) from which we may obtain values of Q and 8, respectively, for any given values of r and n. As to unit shearing stresses, which should not exceed 64 lbs. per sq. in. (say), use is made of eqs. (9) and (10), § 286, in which the maximxim total vertical shear Jm should be substituted. In computing "e " for use in (9) and (10), it is simplest to employ the relation [derived from eqs. (1), (23) and (25)] e = 2 Sra (29) 292. Numerical Examples. Rectangular Section. Let us suppose that in the cross-section of maximum moment the stresses in both materials are to attain their greatest safe values; viz., 16,000 lbs. per sq. in. tension for the steel, and 600 for the compressive stress in outer fiber of the (rock) concrete. Also suppose that ^' = 30,000,000 and £' = 2,000,000 lbs. per sq. in. That is, we have n=15 and r = 16,000 H- 500, = 32 ; and from Diagram I find S = 0.005. In other words the necessary area of steel section is J of one per cent, of the area of the concrete (above steel rods). "We also find, from Diagram II, that Q = 0.894, If now " a " be taken as 10 in., eq. (28) gives : — m = 0.894 X 0.005 X 16,000 X 100, = 7,152 inch-lbs. bending moment that could safely be withstood by each inch in the width &; so that if "&" were 8 inches we should have M = mb = 7,152 x 8 = 57,216 inch-lbs., safe bending moment for the section of the beam. Again, with r and n still equal to 32 and 15, respectively, and hence with Q and /S as before, viz., 0.894 and 0.005, if b is assumed as 10 in., and the max. bending moment to be sustained is If = 80,000 inch-lbs. (so that m = 8,000), we find from eq. (28) = \/^ « = ^/ -.894x0^07x16,000 = ^^''^ ^^^^^^' as necessary value of a ; while the total area of steel section needed is F', = S . a &= 0.005 X 10. 58 X 10 = 0. 5290 sq. in. which is seen to be one half of one per cent, of the area [10x10.58] of the concrete above steel (i.e., S = 0.005, as found from Diagram I originally). 293. Cost of Beams of Rectangular Section. While in a general sense economy in cost is favored by having the width " &" of the rectangular section small compared with the height " a," a limit to narrowness of width is set by the unit shearing stress in the neutral surface which would be found to exceed a safe limit if the beam were too narrow. The thickness "a'" of concrete below the steel rods (see Fig. 292) might be made -^5 of " a." 362 MECHANICS OF ENGINEERING. CHAPTER VI. Flexure. Columns and Hooks. Oblique Loads. 294. Oblique Prismatic Cantilever. In Fig. 301, at (a), (on p. 354) we have a prismatic beam built in at K, projecting- out obliquely, and carrying -a vertical load P at upper end ; the line of action of P passing through the center of gravity of the upper base of the prism. In such a case the fibers of the beam where they cross any transverse plane mg will evidently be subjected to compressive stress (called a ^'■thrust'''') due to the component of P parallel to the axis OKoi the prism ; to a shear J" due to the component of P at right angles to that axis ; and also to additional stresses, both tensile and compressive, formings a " stress-couple^^ due to the moment of P (i.e., Pu) about ^, the center of gravity of the cross-section m'm. More in detail, consider in Fig. 300 a portion AB of the prism, being the part lying above a cross-section mm' near the top, so that the portion gO of axis is practically perpen- dicular to the section mm' which is a plane both before and after flexure, g being the center of gravity of the plane figure formed by the cross-section. Let the unit stress on the end of the extreme fiber at m be represented by the length sm and that [also com- pression (say)] on the other extreme fiber, at m', by s'm'. Draw the straight line ss' ; then by the common theory of flexure the stress on any inter- mediate fiber, at c, would be the intercept, or ordinate, ac to this line. Now the unit stress p^, on the fiber g at the center of gravity of cross-section, being gr, draw through r a line t'rt Fia. 300. FLEXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 353 parallel to m'm, and we now have the stress on any fiber as o divided into two parts be, or p^, the same for all the fibers ; and ab, different for the different fibers but proportional to the distance z of the fiber from g. Hence we have : the unit stress on any fiber c is P = Pi -\ P2 (ibs. per sq. in.) . . . . (1) where p^ is st and e the distance of the extreme fiber ??? from g ; and hence the total stress on fiber c is pdF =p^dF H — Pnli^F, lbs. ; where dF is the area (sq. in.) of section of fiber, or element of area of the cross-section, F being the total area of the cross- section, mm' . Geometrically, we note that while the system of normal stresses on all the fibers forms a trapezoid, m's'sm in this side-view, and that they are all compressive, they are equivalent to a rectangle, m't'tm, of stress of uniform compressive unit-stress p^ ; and two triangles, one, rst, of compressive stress, and the other, rs't', of tensile stress.* It will now be shown that the sum of the moments of the stresses of the rectangle about center g is zero, and that the two triangles of stress form a couple. ^(moms.) of stresses in triangle = I (p^dF)z = pi dFz =p^Fz = zero ; since z =zero, the 2's being measured from the center of gravity, g, of section mm' [§ 23, eq. (4)]. Again, if we sum (algebraically) the stresses of the two triangles, we have / - p dF = — I zdF = ^Fz, = zeYo Jz = -e' e e J e that is, the resultant of the compressive stresses in rts equals that of the tensile stresses in rs't' ; hence they form a couple. If, therefore, we have occasion to sum the moments about g, of all the stresses acting on the fibers in section wm' we are to note that this moment-sum involves the stresses of the triangles alo7ie (that is, of the couple), and is in.-lbs. ; where I^ is the " moment of inertia " of the cross-section * These plane figures are the side views of geometric solids. 354 MECHANICS OF ENGINEERING referred to an axis through g (its center of gravity) and perpen= dicular to the "force plane " (plane of paper here). If, again, we sum the components of all the stresses (on plane mm') parallel to the axis gO we note that this sum is zero for the couple and also for the shear J and hence reduces simply to fp^dF =_Pj CdF=p^F, lbs. (the Thrust) . . (3) (corresponding to the rectangle, fm). The sum of components perpendicular to axis ^ is of course simply the shear, J, lbs. Evidently the unit stress (normal) in fiber at m is expressed e' as jp,„ = j9j 4-^2' ^^^^ ^^^^ ^^ *^' '^^^ Pm'=Pi P2' ^^ ^^ ^^^J case the latter is negative it indicates that the actual stress in this fiber is tension. 295. Oblique Cantilever. Fig. 301, (a) and (5). At (h) is shown as a " free body," a portion [of the cantilever at (a)] of any length X from top. The forces acting are the vertical load P at 0, and the stresses on the ends of the fibers in the section m'm.; and these stresses are now indicated as consisting of a thrust, T, of uniform intensity p^, the total thrust being ^9^^', lbs., (where Y is the total area of section) ; of a stress-couple, ' '" C, whose moment is -^ in.- FiG. 301. e lbs., in which pj = Pm — P^ ^^^ I is the " moment of inertia " of the cross-section (about an axis through its center of gravity g at right angles to the plane (" force plane ") containing Og and force P ; the same I that has been used in previous cases of flexure) ; and the total shear, J, lbs., parallel to force plane and perpendicular to gO. The lever arm of P about g i^ u which practically = x sin a (unless the beam is considerably bent or is nearly vertical). FLEXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 355 For this free bjdy (in order to find p^, p^ and J ) Xxr A • Ti -n f\ Pcosa Jl = Ogives: /^ cos a— p^i^ =0 ; .•. p^ = F X^ X A Vol r> A -^^^ (moms.)^ = . .-^ Fu = 0; .-. p^ = — — • (4) . (5) and Xi'=o P sin a — J" = (9 ; .-.J — P sin a, (6) As X varies, from to Z, we note that p^ and J remain unchanged but tliat p., increases ivith u ; so that the maximum value of the unit stress jo,,,, Avhicli = p^ + p,' will be found in the section at K, where x = I ; and if this stress is not to exceed a safe value, R', for the material, we put p^i,^^ K) +p^ = R', (as the equation of safe loading) ; ^^nan ^, .... (7) or, P "cos a ~'f~ Pn (N. B. For a cross-section of unusual shape the stress e' , = p^ P2, at K, might happen to be numerically greater than Pj^„ and thus govern the design). 296. Experimental Proof of Foreg^oing. A stick or test piece of straight-grained pine wood, 12 inches in length and of square cross-section (one inch square), originally straight and planed smooth and with bases perpendicular to ^the length, was placed in a testing machine ; steel shoes, with (outside) spherical bearing surfaces, being centered on the ends. See Fig. 302, where AB is the stick and S, S\ the two steel shoes. The stick was gradually compressed between the two horizontal plates B,B'^ of the machine and bent progressively in a smooth curve under increasing force. From the nature of the "end conditions," as the stick changed form, the line of action of the two end pressures P,P, always passed through the centers of gravity, a and h, of the respective bases. When the force P had reached the value 4500 lbs. a fine wrinkle was observed to be forming on the right-hand surface Fie. 302. 356 MECHANICS or ENGINEERING. of the stick at the outside fiber m of the middle section gm. The other tibers of this section were evidently uninjured. At m then, the unit-stress must have been about 8000 lbs. per sq. in., the crushing stress (as known from previous experiments with sticks of similar material and equal section but only three or four inches long; these were too short to bend, and wrinkles formed around the whole 'perimeter^ showing incipient crushing in all the fibers). The distance gc at this time was found to be \ in. ; i.e., the lever arm, w, of the force P about g, the center of gravity of the section. In this case, then, it is to be noted that the value of %i was entirely due to the bending of tit e piece. Substituting, in eqs. (4) and (5) of § 295, the values w = |- in., a = 0, cos a = l, e = e', =i inch, i^=l sq. in., hh^ 1x1^ 1 and i; = — , = ^^ = — in/, we find p^ = 4500 lbs. per sq. in. and p^= 3375 lbs. per sq. in. Hence stress at m, = Pi-\- P21 == 7875 lbs. per sq. in., which is about 8000, as should be expected. On the fiber at 0, how- ever, we find a stress of p^ — p., or of only 1125 lbs. per sq. in. compression. We find, then, that in the section om, when P reached the value of 4500 lbs., there was a total-thrust (p,F^ of 4500 lbs.; a unit-thrust (w^) of 4500 lbs. per sq. in. ; and a stress-couple pi having a moment of Pw, = ^— , = 562.5 in .-lbs., (implying a separate stress oi p^^^'^l^ lbs. per sq. in. in the outer fibers, to be combined with that due to the thrust). Also that /, the shear, was zero. 297. Crane-Hooks. First (Imperfect) Theory. Fig. 303 shows a common crane-hook of iron or steel. Early writers (Brix and others) treated this problem as follows : — The load being P, if we make a horizontal section at AB^ about whose gravity axis, gr, P has its greatest moment, and con- sider the lower portion C as a free body, in Fig. 304), we find, using the notation and subdivision of stresses already set forth in § 294 for an oblique prism, that the uniformly distributed pull (or " negative thrust ") on the fibers is p^F = P, lbs. ; FI^EXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 357 P while the moment of the stress-couple is ^- = Pa ft.-lbs.; and e that the shear, /, is zero. Hence on the ex- treme fiber at B we have a total unit tensile stress of Pae T' which for safe de- sign must not ex- ceed the safe unit- stress for the ma- terial, R' lbs. per sq. in. ; whence we should have p[l + f] = iS' . . as the equation of safe loading.* Example: Safe P = ?, if section AB is a circle of radius 2 in., while a = 4 in. ; the material being mild steel for which (in view of the imperfection of the theory) a low value, say 6000 lbs. per sq. in., should be taken for R'. With these data we obtain: — 1 4x2" Fig. 303. Fig. 304. Fig. 305. (8) P = 6000 -[ 12.56 ^ 50.24 h 25130 lbs. The simple crane in Fig. 305, being practically an inverted hook, may be treated in the same manner. 298. Crane-Hooks. Later, More Exact, Theories. The most exact and refined theory of hooks yet produced is that of Andrews and Pearson,! but it is very complicated in practical application and far too elaborate and extended to be given here. The next best (and fairly satisfactory) treatment is that of Winkler and Bach, of which the principal practical features and results will now be presented. * See experiments by Prof. Goodman, in Engineering, vol. 72, p. 537. Re- sults are irregular, due probably to the use of this imperfect theory. t Drapers' Company Research Memoirs. Technical Series I. London, 1904. 358 MECHANICS OF ENGINEERING. In AB, Fig. 306, we have again the free body of Fig. 304, but the vertical stresses acting on the cross-section m'm are proportional to the ordinates of a curve instead of a straight line. The imperfection of the early theories lies in the fact that the sides of a hook are curved, and not straight and par- allel as in the prismatic body of Fig. 301 ; and the variation of stress from fiber to fiber on the cross-section must follow a dif- ferent law, as may thus be illustrated : As preliminary, the student should note, from the expres- P EX sion— = — -of p. 209, that in the case of two fibers under ten- F I sion, with the same sectional area F, the unit-stress P -^ F (or p) is not proportional to the elongation }. of the fiber unless the two lengths I are equal. In Fig. 306 the center of gravity of the cross-section is g, and is the center of curvature of the curved axis gk of this part of the hook (or other curved body). The two consecutive radial sections m'm and ft are assumed to remain plane during stress, and hence the changes of length, due to stress, of the (verti- cal) fiber lengths between them are proportional to the ordinates of a straight line ; and if these fiber lengths were equal in length (as would be the case for a prismatic beam) the unit-stresses acting would also be proportional to the ordinates of a straight line (this is the case in Fig. 301). But in the present case these fiber-lengths are un- equal, so that the unit- stresses in action are (in general) proportional to the ordinates of a curved line. Such a curved line we note in vCi Fig. 306, the ordinates between which and the horizontal line hi represent the unit-stresses, p, acting on the upper ends Fig. 306. FLEXUEE. COLUMNS AND HOOKS. OBLIQUE FORCES. 359 of the vertical fibers from m' to m. Tlius, the stress on the fiber mt is p^ = ei (tension); and that on the other extreme fiber, (at m') is p^, = hv (compression). If now we compute the average unit-stress p^ =^ P -i- F and lay it off, == is, upward from hi, and draw the horizontal 6s, we thereby re-arrange the stresses into a uniformly distributed pull (or " negative thrust ") p^F ^hs., represented by the rectangle hsih, and a stress-couple formed by the ordinates lying between the curve and the axis hs. It will be noted in Fig. 306 that there is a fiber at some point n (on right of g) where the stress is zero ; i.e., the " neu- tral axis " of the section is at n, ~1 to paper. Also, at some point n', the actual stress is equal to the average, p^, and an axis ~] to paper through this point would be the neutral axis if the forces acting on this free body, other than the fiber stresses, consisted, not of a single force P, but of a couple, with a mo- ment = Pa. This axis through n' might be called the neutral axis for " pure bending ", since then the whole system of fiber stresses would reduce to a couple and the stresses would be measured by the ordinates between hs and the curve. 299. Crane-Hooks. Winkler-Bach Theory. Formula for Stress. In Fig. 306, let F be the area of the plane figure formed by the section m'm, dF an element of this area, and z its distance (reckoned positive toward the right) from the gravity axis, g^ of the section. The radius of curvature of gk is r, and a is the lever arm of P, the load, about g. Let gm = e and gm' = e' (dis- tances of extreme fibers) and let /Vr r'=+^ I dF S denote the quantity ( t, / ( an abstract number depending on the area, shape, and position, of the cross- section m'ni ; and upon the radius of curvature r. Its value may be obtained by the calculus (or Simpson's Eule) for ordinary cases. For instance, if the section is a rectangle of width b, and altitude = A, = m'm, we find -1 ; ■ (1) '-jh-'^l)-' (^: From the Winkler-Bach theory it results that the unit-stress on any fiber between m and m', at a distance z from the gravity axis g (on the right, toward the center of curvature, 0; if on the left, z is negative) is a /. Z 1"' J" ^^F I r\ r — z S/. (3) lbs. per sq. inch. A positive result from (3) indicates tension; a negative, com- pressive stress. Of course, for P -=- Pwe might write the symbol p^, or " aver- age stress." If p were set = zero, a solution of (3) for z would locate the neu- 360 MECHANICS OF ENGINEERING. tral axis, n, of Fig. 306; while by placing p — Pj = 0, a solution for z would locate the point n', or neutral axis for "pure bending." 300. Numerical Example. Let the cross-section be a trapezoid, of base 6 = 3 in. at m, and upper base 6' = 1 in. at ?/i', both | to paper ; the altitude /i, =• m/m, being 4 in. This brings g f in. (= e ) from m and | in. (= e') from m'. Let N be in the same vertical as and Om = 2 in. Hence r=a = 2 + |.= y in. The material is mild steel and the load P is 8 tons ; find p^ and -pm' . From above dimensions we find area ^=8 sq. in,, while from eq. (1), (using the calculus), S= 0.0974. For p^ w^ put z = + f in. in eq. (3) ; and for Pm', 3 = - I in. ; obtaining, finally, p,« = 17,120 lbs. per sq. in. (tension) ; and Pm' = — 7,980 (compression). Evidently the elastic limit is not passed. Using the imperfect theory of § 297, we should have obtained pm = 12,000 lbs. per sq. in., only ; which is seen to be about 30 per cent, in error, compared with the above value of 17,120. The reason for taking a low value for the safe unit-stress, B', in the example of § 297 is now apparent, an additional reason being the fact that loads are sometimes "suddenly applied " on hooks. 301. By "column" or "long column" is meant a straight beam, usaally prismatic, which is acted on by two com- pressive forces, one at. each extremity, and whose length is so great compared with its diameter that it gives way (or " fails ") by buckling sideways, i.e. by flexure, instead of by crushing or splitting like a short block (see § 200). The pillars or columns used in buildings, the compression members of bridge-trusses and roofs, the " bents " of a trestle work, and the piston-rods and connecting-rods of steam-engines, are the principal practical examples of long columns. That they should be weaker than short blocks of the same material and cross-section is quite evident, but their theoretical treatment is much less satisfactory than in other cases of flexure, experiment being very largely relied on not only to determine the physical constants which theory introduces in the formulae referring to them, but even to modify the algebraic form of those formulae, thus rendering them to a certain extent empirical. 302. End Conditions. — The strength of a column is largely dependent on whether the ends are free to turn, or are fixed and thus incapable of turning. The former condi- tion is attained by rounding the ends,' or providing them with hinges or ball-and-socket-joints ; the latter by facing off' each end to an accurate plane surface, the bearing on which it rests being plane also, and incapable of turning. In the former condition the column is spoken of as having FLEXURE. LONG COLUMNS. 361 round ends ; * Fig. 311, (a) ; in the latter as having fixed ends, (ov flat bases ; or square ends), Fig. 311, (&). Fig. 312. Sometimes a coliimn is fixed at one end while the othei end is not only round but incapable of lateral deviation from the tangent line of the other extremity ; this state of end conditions is often spoken of as "Pin and Square," Fig. 311, (c). If the rounding * of the ends is produced by a hinge or ** pin joint," Fig. 312, both pins lying in the same plane and having immovable bearings at their extremities, the column is to be considered as round-ended as regards flex- ure in the plane 1 to the pins, but as square-ended as re- gards flexure in the plane containing the axes of the pins. The " moment of inertia " of the section of a column will be understood to be referred to a gravity axis of the sec- tion which is "I to the plane of flexure (and this corres- ponds to the " force-plane " spoken of in previous chap- ters), or plane of the axis of column when bent. 303. Euler's Formula. — Taking the case of a round-ended column, Fig. 313 (a), assume the middle of the length as an origin, with the axis X tangent to the elastic curve at that point. The flexure being slight, we may use the form EI (Py-^dx^ for the moment of the stress-couple in any * With round ends, or pin ends, it should be understood tliiit the force at each end must be so applied as to act through the centre of gravity of the base (plane figure) of the prismatic column at that end ; and continue to do so as the column b^ nds. 562 MECHANICS OF ENGINEERING. dp dy dy dx- —'r- y J/c dx- " / / dx 7 f -a-y. 1 if 4+- -Or— 1 "1 \x 1 1 1 1 1 J 1 I Y Fig. 313. Fig. 314. ^section w, remembering tliat with this notation the axis X must be || to the beam, as in the figure (313). Considering the free body nC, Fig, 313 (h), we note that the shear is zero, that the uniform thrust =P, and that 2'(moms.n)=0 gi'ves (a being the deflection at 0) EI d'y dx^ --F{ar-y) Multiplying each side by dy we have El dx" dy (Fy=Pa dy — Fy dy (1) (2) ' Since this equation is true for the y, dx, dy, and d^y of any element of arc of the elastic curve, we may suppose it written out for each element from where ?/=0, andc''y=0, up to any element, (where dy=dy and y=y) (see Fig. 314) and then write the sum of the left hand members equal to itliat of the right hand members, remembering that, since dx is assumed constant, l-^dsc^ is a common factor on the left. In other words, integrate between and any point of the curve, n. That is. f[dy]d[dy] =Fa f dy—P T ydy (3) The product dy d^y has been written {dy)d(dy\ (for d^y m EI da? FLEXURE. LONG COLUMNS. 363 the differential or increment of dy) and is of a form like xdx, or ydy. Performing the integration we have EI d_l y^ .... (-1) dx' 2^2 ■ ^ which is in a form applicable to any point of the curvej and contains the variables x and y and their increments dx and dy. In order to separate the variables, solve for dxy and we have di dx=l^-JL==^OTdx=^ I EI, \aJ ... d(y) '^ (X \C(/ / i.e.,a!=±y-p- (vers, sin ^^j , , . (6) (6) is the equation of the elastic curve DOG^ Fig. 313 (a), and contains the deflection a. If P and a are both given, y can be computed for a given cc, and vice versa, and thus the curve traced out, but we would naturally suppose a to depend on P, for ineq. (6)whena7=^Z, y should —a. Mak- ing these substitutions we obtain - ^ 'A^= V^ (^^^"- ^^^ "' ^-^^^ ' ^•^- >^^= 7^ I ^^^ Since a has vanished from eq. (7) the value for P ob- tained from this equation, viz.: i\=EI ^ .... (8) is independent of a, and is ,\ to be regarded as that force (at each end of the round' ended column in Fig. 313) which will hold the column at any small deflection at which it may previously have been set. 364 MECHANICS OF ENGINEERING. In other words, if the force is less than Pq no flexure at all will be produced, and hence P,, is sometimes called the force producing " incipient flexure." [This is roughly ver- ified by exerting a downward pressure with the hand on the upper end of the flexible rod (a T-squai e-blade for in- stance) placed vertically on the floor of a room ; the pres- sure must reach a definite value before a decided buckling takes place, and then a very slight increase of pressure oc- casions a large increase of deflection.] It is also evident that a force slightly greater than P^ would very largely increase the deflection, thus gaining for itself so great a lever arm about the middle section as to cause rupture. For this reason eq. (8) may be looked upon as giving the Breaking Load of a column with round ends, and is called Euler^ s fornfiula. Referring now to Fig. 311, it will be seen that if the three parts into which the flat-ended column is di- vided by its two points of inflection A and B are considered free, individually, in Fig. 315, the forces acting will be as there shown, viz.: At the points of inflection there is no stress- couple, and no shear, but only a thrust, =P, and hence the portion AB is in the condition of a round-ended column. Also, the tangents to the elastic curves at and G being pre- served vertical by the f rictionless guide-blocks and guides (which are introduced here simply as a theoretical method of preventing the ends from turning, but do not interfere with verti- cal freedom) OA is in the same state of flex- ure as half of AB and under the same forces. Hence the length AB must = one half the total length I of the flat-ended column. In other words, the breaking load of a round- ended column of length =^Z, is the same as that of a flat-ended column of length —I. Hence for the I oi eq. (8) write %l and we have as the breaking load of a column with flat-ends and of length =1. }il f/MWM Fig. 315. TLEXURE. LONG COLUMNS. 365 r.^4.m^ .... (9) Similar reasoning, applied to tlie " pin-and-square " mode of support (in Fig. 311) where the points of inflec- tion are at B, approximately y^ I from G, and at the extremity itself, calls for the substitution of ^ I for I in eq. (8), and hence the breaking load of a ^'pin-and-square " column, of length = I, is P^=l ^/^ . . . (10) Comparing eqs. (8), (9), and (10), and calling the value of Pi (flat-ends) unity, we derive the following statement : The breaking loads of a given column are as the numbers 1 flat-ends 9/16 pin-and-square y^ j according to the round-ends \ mode of support. These ratios are approximately verified in practice. Euler's Formula [i.e., eq. (8) and those derived from it, (9) and (10)] when considered as giving the breaking load is peculiar in this respect, that it contains no reference to the stress per unit of area necessary to rupture the material of the column, but merely assumes that the load producing " incipient flexure ", i.e., which produces any bending at all, will eventually break the beam because of the greater and greater lever arm thus gained for itself. In the canti- lever of Fig. 241 the bending of the beam does not sensibly affect the lever-arm of the load about the wall-section, but with a column, the lever- arm of the load about the mid- section is almost entirely due to the deflection produced. It is readily seen, from the form of eqs. (8), (9) and (10), that when I is taken quite small the values obtained for Po, Pi, and P2 become enormous, and far exceed what would be found from the formula for crushing load of a short block, viz., P = FC (see p. 219), with F denoting the area of section of the prism and C the crushing unit-stress of the material. The degree of slenderness a column must have to justify the use of Euler's relations will appear in the next paragraph. 36G MECHANICS OF ENGINEERING. 304. Euler's Formula Tested by Experiment. — Since the "moment of inertia," /, (referred to a certain axis) of the cross- section of the column may be written I = Fk^, where k is the "radius of gyration " (see p. 91), and F the area of the plane figure, eq. (8), for ''round 1 Pq tz^E ends," may be written \ F~~(l^ky ' * Here Po^F is the average unit-stress (compressive) on the cross-section and l^k is a ratio measuring the slenderness of the column. (Of course, when the column actually gives way by buckhng, the unit-stress on the concave side at the middle of the length is much greater than the average). In the ex- periments by Christie, described on p. 112 of the Notes and Examples, the value of the ratio l-i-k ranges from 20 to 480. As an example consider a 3"x3"Xi" angle-bar (or "angle") of wrought iron, with Z = 15 ft., to be used as a column. Fig. 315a shows the cross-section of this shape, with di- mensions. Q is the center of gravity of this plane figure. Let the force be applied at each end of the column according to Christie's mode of "round ends," i.e., by a ball-bearing device. Fig. 3l5a. the force always passing through the point C of the section at each extremity of the column. Since the ends are free to turn in any plane, the axis of the column will deflect in the plane CN 1 to the axis 2 ... 2 (of the plane figure) about which the values of / and of k are least. For this shape, we find from the hand- book of the Cam- bria Steel Co., that k about axis 2 ... 2 is the least radius of gyration and =0.58 in. ; also that FLEXURE. LONG COLUMNS. 367 the area of the figure is F = 2.75 sq. in. Hence the "slender- ness-ratio-;' l^k, is 180" -^ 0.58" = 310; and from eq. (11) we have, with E for wrought iron taken as 25,000,000, lbs./in.2 (p. 279), I (Po -rF)=Ti^X 25,000,000 -^ (310)2 = 2570 lbs. / in.2 ; while from the Christie experiments we find (Po^P)=2650 lbs. /in.2 as the average unit-stress at rupture; a fairly close agreement with the Euler result. The total rupturing load would then be Po = 2570X2.75 = 7070 lbs., and the safe load, with the "factor of safety " of 8 recommended in the Christie report, would be 884 lbs. In this way it may be ascertained that for values oi l^k from 200 to 400 for "round ends " and from 300 to 400 for fixed ends there is an approximate agreement between Euler's equations and the Christie experiments. But most of the columns used in engineering practice involve values oil^k less than 200, so that Euler's formulae are not adapted to actual columns (though used to some extent in Germany). A formula of such nature as to be available for all degrees of slenderness has therefore been established (Rankine's, see next paragraph), based partly on theory and partly on experiment, which has obtained a very wide acceptance among engineers. In Fig. 3156 is shown a curve, Er, resulting from plotting as abscissa and ordinate the values of Po-Hi^ and Z-h A:, as related in Euler's formula (8) for columns with round ends, for "medium" structural steel; with £' = 30,000,000 lbs./ in- ^ Ej is a similar curve plotted from Euler's formula (9) for fixed ends for the same material. Each of these Euler curves is tangent to both axes at infinity. The other curves wUl be referred to later. 305. Rankine's Formula for Columns. — The formula of this name (some times called Gordon's, iu some of its forms) has a somewhat more rational basis than Euler's, in that it in- troduces the maximum normal stress in the outer fibre and is applicable to a column or block of any length, but stili contains assumptions not strictly borne out in theory, thus introducing some co-efficients requiring experimental de- termination. It may be developed as follows : Since in the flat-ended column in Fig. 315 the middle portion AB, between the inflection points A and B, is acted on at each end by a thrust = P, not accorapanied by any shear or stress-couple, it will be simpler to treat thai 368 MECHANICS OF ENGINEERING. p., portion alone Fig. 316, (a), since the thrust and stresa- couple induced in tlie section at R, the middle of AB, will be equal to those at the flat ends, and G, in Fig. 315. Let a denote the de- flection of R from the straight line AB. Now consider the portion AR as a free body in Fig. 316, (b), putting in the elastic forces of the section at R, which may be clas- sified into a uniform thrust = PiF, and a stress couple of moment Fiq. sie. 294). (The shear is evidently zero, from = L:_, (see e I (hor comps.) = 0). Here p^ denotes the uniform pres- sure (per unit of area), due to the uniform thrust, and jpg the pressure or tension (per unit of area), in the elastic forces constituting the stress-couple, on the outermost element of area, at a distance e from the gravity axis (~| to plane of flexure) of the section. F is the total area of the section. / is the moment of inertia about the said gravity axis, g 1 (vert, comps.) = gives P == p^F , . (Tj 2' (moms.j,) = gives Pa = ■^— .... (2) For any section, n, between A and R, we should evidently have the same^j as at R, but a smaller pi, since Py < Pa while e, /, and F, do not change, the column being pris- matic. Hence the max. (pi+JJa) is oil the concave edge at R and for safety should be no more than G -^ n, where G is the Modulus of Crushing (§ 201) and w is a " factor of safety." Solving (1) and (2) for j9i and^gj and putting their sum = C -V- %, we have P.Pae G (3) We might now solve for P and call it the safe load, biat a§ FLEXURE. LONG COLUMNS. 369 is customary to present the formula in a form for giving the breaking load, the factor of safety being appHed after- ward. Hence, we shall make n=l, and solve for P, calling it then the breaking load. Now the deflection a. is unknown, but may be expressed approximately, as follows, in terms of e and l. If we now consider ARB to be a circular arc, of radius = |0, we have from geometry (similar triangles) a=(Z-^- 4)^^2/9; and if we equate the two expressions for the moment of the EI V2I stress-couple at R there results — = — (see pp. 249 and 250) . A combination of these two relations gives ae=(p2^S2E)P, Now under a safe load the total stress, pi + p2, in the outer fibre (concave side) at R will have reached a safe value, R', for the material, and is therefore constant for this material, and if the rude assumption is made that the portion p2 of this stress is also constant, it follows that the fraction (p2 h- S2E) = a constant; which may be denoted by /?, (an abstract number). Let us also write, for convenience, I = Fk^, (k being the radius of gyration of the cross-section about a (gravity) axis through ^ 1 to paper). Hence finally, we have, from eq. (3), Breaking load 1 FC forflatends J ^^r+^5(m)2 • • • (4) By the same reasoning as in § 303,' for a round-ended column we substitute 21 for I; for a column with one end round and the other '^fiat " or ''fixed " (i.e., for a " pin-and-square " column), ^l for I; and obtain Breaking load for a round- 1 FC ^ ended column \^^^TTW(hW' • • • (^) Breaking load for a ''pin- 1 FC and-square " column J ^^^l + 1.78/9(Z-^/c)^' ' ' • ^^^ Each of these equations (4), (5), and (6), is known as Ran- kine's Formula, for the respective end-conditions mentioned. They find a very extended use among engineers in English- speaking countries; with some variation, however, in the 370 MECHANICS OF ENGINEERING. numerical values used for quantities C and /?, which are con- stants for a given material; and also in the fraction of the breaking load which should be taken as the safe, or working, load (the reciprocal of this fraction being called the "factor of safety,") =n. A set of fair average values for these con- stants, as recommended by Rankine and others, is here pre- sented : Hard steel. Medium Steel. Soft Steel. Wrought Iron. Cast Iron. Timber, C (lbs./in.2) 70,000 50,000 45,000 36,000 70,000 7,200 /? (abstract number) .... 1 1 1 1 36,000 1 1 25,000 36,000 36,000 6,400 3,000 The factor of safety, n, usually employed with the fore- going formulae and constants, is n = 4 for wrought iron and steel in quiescent structures; and 5 under moving loads, as in bridges; while n = 10 should be used for timber and 8 for cast iron. In Fig. 315?) are two dotted curves, plotted for round ends (Rr) and fixed ends (Rj) in the case of medium steel; the above equations (Rankine), with the above values of C and /?, having been used. The ''slenderness ratio," l^k, i& the abscissa; and Po^F, or Pi^F, (the average breaking unit-stress), is the ordinate, of any point. These curves may now be com- pared with the Eule'r curves, E^. and £/, (in the same figure), .already mentioned as having been plotted for structural steel (of modulus of elasticity £? = 30,000,000 lbs./in.2) 306. Examples; under the Rankine Formulae. — Example 1. Let it be required to compute the breaking load of awrought- iron solid cylinder, used as a column, of length 1 = 8 ft. and diameter, =d, =2.4 inches; with round ends, i.e., the pressure acting at each end at the center of the circular base, the ends being free to turn in any direction. The "end conditions " call for the employment of the ''least k," but here k is the same for any gravity axis of the circular section. That is we have A;2 = / ^^ = 17,^4 ^ ;,^2_i ^2 _ 1(1, 2)2 _ 0.36 in.2; .-. /(; = 0.6 in. FLEXURE. LONG COLUMNS. 371 and (Z-^A;) = "slenderness-ratio " = 96-j-0.6 = 160. Hence from eq. (5) ■KT^C 1 Po = j^pX^Tieop' ^^^^ '^^ 36 000 ^^^ <^ = 36,000 lbs./in.2; Le., ;r( 1.2)236,000 162,800 ,<, .nn ik It is seen that, on account of the degree of slenderness of the column, the breaking load is about one quarter of what it would be for a short prism of same section. With a factor of safety of 5 we should take 5 of 42,300, i.e., 8460 lbs., as safe load. Example 2. — It is required to compute the diameter, d, . of a solid cast-iron cylinder, 16 ft. in length, to serve as a column with fiat ends, whose safe load is to be 6 tons, the factor of safety being 6. This calls for the use of eq. (4) in which we put Pi = 6x12,000 = 72,000 lbs., the required break- ing load ; with C = 70,000 lbs. / in.2 and /3 = 1 -^ 6400. The least radius of gryation should be used, but in this case the k"^ is constant for all axes of the section, viz., k'^ ^\7tii^ ^nr^ ^d^ ^IQ. Hence from eq. (4) we have (for inch and pound) „ \Kd^C 54,980^2 ^onnmu ^' = l + 7ra-/cl2 =- r^ = 72,000 lbs. This on reduction leads to the bi-quadratic equation #-1.309^2 = 120.7; which being solved for d^ gives d2 = o.645± 11.01. The upper sign being taken we have, finally, c? = 3.41 in. as the required diameter. The "slenderness ratio," therefore, proves to be 192-^0.85 = 225, which though seemingly high is not extreme for a flat- ended column; corresponding, as it does, to 112 for a round-'^ ended column. Example 3. — A prism of medium steel, of uniform rec- tangular section (solid) with dimensions 6 = 3 in. and /i = l in., is to be subjected to a thrust (connecting-rod of a steam- engine). Its ends are provided with pins (see Fig. 312) capable^ of turning in firm bearings, the axis of each pin being T to 372 MECHANICS OF ENGINEERING. ' the "b" dimension of the rectangular section. The length between axes of pins, is Z = 6 ft. It is required to find the breaking load by the Rankine formula). Since the end conditions would be ''round-ends" if the axis of the column were to bend in a plane T to the axes of the pins (as in Fig. 312), but ''flat-ends" [Fig. 311(6)] in case it bent in the plane containing the axes of the pins; and since the k of the section is different for the two cases, it will be necessary to make each supposition in turn and take the smaller of the two results for breaking load (i.e., as the one to which the factor of safety should be applied). For round-ended buckling the value of k^ is I^F = [hh^^l2]^hh = 0.75 in.^; and, with the values of C and ^ for medium steel, we have from eq. (5), p 50,000X3.0 150,000 ^,^^^,. ^36,000' 0.75 while for flat-ended buckling, in the other plane, the P to be used would be A;^ = [6/^3^ 12] -6/i = 0.0833 in.2, and hence from eq. (4) p 50,000X3.0 150,000 -, ^.q lu ^^36,000' 0.0833 It is seen that Pi is smaller than Pq, so that with a factor of safety of 6 we have for the safe, or working, load, I of 54,933, =9,155 lbs. 307. Radii of Gyration. — The following table, taken from p. 523 of Eankine's Civil Engineering, gives values of ^'^, the square of the least radius of gyration of the given cross- eection about a gravity-axis. By giving the least value oi h^ it is implied that the plane of flexure is not determined by the end-conditions of the column (i. e., it is implied that the column has either flat ends or round ends). If either end (or both) is a pin-Joint the column may need to be treated as having a flat-end as regards flexure in a plane containing the axis of the column and the axis of the pin, if the bearings of the pin are firm ; while as regards flexure in a plane perpendicular to the pin it is to be considered round-ended at that extremity. FLEXURE. LONG COLUMNS. 373 In the case of a " thin cell " the value of h"^ is strictly true for metal infinitely thin and of uniform thieJcness ; still, if that thickness does not exceed ^ of the exterior diame- ter, the form given is sufficiently near for practical pur- poses; similar statements apply to the branching forms. f f^mm h i h <—-h > (a) (5) wmmm, Fig. 317. *— -.. \W (6) (cj Fia. 818. Solid Eectangle. %— least side. Thin Squfire Cell. Side— In. Thin Kectangular Cell. Yia 317 fc> -" /i^ /i+36 h^=- least side. Solid Circular Section. Diameter —d. Thin Circular Cell. Exterior diam. = d. Fig. 317(a). ]z''=l^}i^ Fig. 317(6). A;2 = |/i2 p = 12'/i+6 Fig. 317 (c?). p^i^2 Id Fig. 317(e). A;2 = Jd2 Angle-Iron of Equal j^ig. 317^^) A;2=^-62 ribs F: 62^* A^ngle-Iio-n of unequal , -^-g g^g^^^^ ^= 12(F+3?) Cross of equal arms. Fig. 318 (6). ^=4^' I-Beam as a pillar. Let area of web =5, j.- 313 (c). F= -. . -^-j-^ « « &o^A flanges & ^ ^ 12 A-^-H =A. Channel ^ig. 318(^). ^=^^^ [l2T^)+iT^^] Let area of web =B; of flanges =A (both). ^ extends from edge of flange to middle of web. 374 MECHANICS OF ENGINEERING. 308. Built Columns. — The "compression members" of bridge trusses, and columns in steel framework buildings are generally composed of several pieces of structural steel riveted together, each column being thus formed of a combination of plates, channels, angles, Z-bars, etc. In Figs. 319 and 320 PHCENIX COLUMN. Fig. 319. ^^**- are shown examples of these compound shapes. The Phoenix column is seen to consist of four quadrantal segments riveted together. In Fig. 319 is a combination of two channels and one plate, these three pieces being continuous along the whole length of the column. On the side opposite to the plate are seen lattice bars, arranged in zig-zag, which serve to stiffen the column on that side. The center of gravity of the cross- section of this column is nearer to the edge carrying the plate than to the lattice edge; and if the ends of the column are provided with pins 1 to the webs of the channels the axis of each of these pins should be so placed as to contain the center of gravity of the cross-section of the column at that point. The handbooks of the various steel companies present formulae and tables enabling the breaking loads to be found for their various designs of built columns, and for single I-beams used as columns. For example, the tables given in the hand- book of the Cambria Steel Co. for built columns of "medium steel " are stated to be computed from the following formulae (which are evidently of the Rankine type). The breaking load for a column of length I and with cross- section of area F and least radius of gyration k is (in pounds) : Square Bearing, Pin and Square Bearing. Pin Bearings. 50,000i^ „ 50,000i^ _ 50,000i^ Pi = 1 + 36,000 W P2 = 1 + 24,000 ar Po = 1+ 18,000VA: FLEXURE. LONG COLUMNS. 375 In these formulae I and k should be in the same unit (both feet, or both inches; since (l^k) is a ratio) and the proper k to be used for the case of "pin and square bearing " (i.e., one end provided with a pin and the other with a square bearing) should be ascertained as in example 3, p. 371. To obtain the total safe load for the column: "For quiescent loads, as in buildings, divide by 4. For moving loads, as in bridges, divide by 5." Considerable variety will be found among the formulae of the Rankine type proposed by different engineers as best satisfying the results of experiment. For accounts of ex- periments beyond those already quoted in the author's "Notes and Examples in Mechanics," the reader is referred to special works. Kent's Pocket Book for Mechanical Engineers contains much valuable matter on the subject of columns. The handbooks of the Carnegie Steel Co., the Pencoyd Iron Works, and the Phoenix Iron Co., give ex- tensive data relating to steel columns. Osborne's Tables of moments of inertia and radii of gyration of compound sec- tions is a valuable book in this connection. 309. Moment of Inertia of Built ColumiL Example.— It is pro- posed to form a column by joining two I-beams by lattice- work, Fig. 321, (a). (While the lattice-work is relied upon to cause the beams to act together as one piece, it is not regarded in estimating the area F, or the moment of iner- tia, of the cross section). It is also required to find the proper distance apart = x, Fig. 321, at which these beams must be placed, from centre to centre of webs, that the liability to flexure shall be equal in all axial planes, i.e. that the 1 of the compound section shall be the same about all gravity axes. This condition will be ful- filled if Iy can be made ~i^* (§89), being the centre of gravity of the compound section, and X perpendicular to the parallel webs of the two equal I-beams. Let F' = the sectional area of one of the I-beams, Fx Tsee Fig. 321(a) its moment of inertia about its web-axis, that about an axis ~[ to web. (These quantities can 1)8 * That is, with flat ends or ball ends ; but with pin ends, Fig. 313, if the pin is II to X. put 4/y = Ixi if II to Y, put 47x = Ir . 376 MECHANICS OF ENGINEERING. found in tlie hand-book of the iron company, for each size of rolled beam). Then the total 7x = 2rx ; and total I^ = 2ri'v -f W-Yl (see §88 eq. 4.) If these are to be equal, we write them so and solve for a?, obtaining X V jr, (1) 310. Numerically; suppose each girder to be a 10}4 inch light I-beam, 105 lbs. per yard, of the N. J. Steel and Iron Co., in whose hand-book we find that for this beam I'x = 185.6 biquad. inches, and I'r = 9.43 biquad. inches, while F' = 10.44 sq. inches. "With these values in eq. (1) we have „^J± (185.6-9.43) Vera = 8.21 inchea. V 1 0.4-4. n^ V 7^ ■^ r'l^ -a;— H ^ iai ^ ^ ^^ -P- ^11:=. (6) Fig. 331. The square of the radius of gyration will be F=2Px-^2i^'= 371.2 -^20.88=17.7 sq. in. . (2) and is the same for any gravity axis (see § 89). As an additional example, suppose the two I-beams united by plates instead of lattice. Let the thickness of the plate ■= t, Fig. 321, (&). Neglect the rivet-holes. The distance a is known from the hand-book. The student may derive a formula for x, imposing the condition that (total /x)= /y- FLEXURE. LONG COLUMNS. 377 310a. Design of Columns. — General considerations governing economy and efficiency in the design of built columns are that the various pieces, besides being continuous for the whole length, should be placed as far from the axis of the column as possible, in order to increase the value of k the (least) radius of gyration, thus leading to a larger value of the safe load for a given amount of material, or to a minimum amount of material for a given required safe load; and that the parts should be well fastened together by rivets, preventing all relative motion. The economy secured by placing the material as far from the center as possible also holds, of course, for single pieces used as columns. For example, if the safe load of a hollow cylindrical cast-iron flat-ended column, 20 ft. long, is to be 40 tons, i.e., 80,000 lbs., and the thickness of metal is not to be less than \ in., we find, after a few trials with Rankine's formula eq. (4), p. 369, taking a factor of safety of 8 (so that the breaking load would be 640,000 lbs.) that an outside diameter of d = 8 in. is the largest permissible. Thus, taking the least k"^, {^(F^8), from p. 373, for a thin cylindrical cell, with Z = 240 in., with the sectional area, F, as the quantity to be solved for, we have \^'^^^.94QN2 =640,000 lbs.; .-. i^ = 19.43 sq. in. 1 + : 6400 [82-8] Let ^2 denote the internal diameter of the section; then j(82 — d22) = 19.43; whence ^2 = 6.26 in.; i.e., the thickness of metal==^(d — ^2) =0.87 in., or practically | in. 310b. The Merriman-Ritter-Formula for Columns was de- rived independently by Professors Merriman and Ritter (see Engineering News, July 19, 1894) and has a mathematical basis as follows. In Fig. 315& curves have been plotted for the Euler and Rankine formulae for medium steel, both for flat and round ends; and it is seen that each of the Rankine curves is tangent to the horizontal line through V and is roughly parallel to, and not very distant from, the corresponding Euler curve on the extreme right. Professor Merriman de- rives the equation (of the same form as Rankine's) for a curve which has a horizontal tangent at V, and is exactly tangent 378 MECHANICS OF ENGINEERING. to the Euler curve at some point on the extreme right (at infinity, in fact) and thus secures a more rational value for the constant called ^ in Rankine's formula. With P' denoting the safe load for the column and C the safe compressive unit-stress for the material, this formula may be written . . . P' = — Tvrpp^i • • '^ • (M) where C" denotes the unit compressive stress at elastic limits E the modulus of elasticity, F the sectional area, and n an abstract number whose value (as before, in the Rankine for- mulae) is 1, 16/9 (or 1.78), and 4, for flat ends, pin-and-square, and round ends, respectively. If for Q we write P, the breaking load, and correspondingly C for C, and plot values oi P-^F and l-^k, the curve would not differ greatly from the Rankine curve in Fig. 120 for medium steel; and similarly for wrought iron; but for timber and cast iron the variation is considerable, and hence Prof. Merriman does not recommend the use of his formula for the latter two materials. (Crehore's formula differs from the above only in replacing C" by C.) 310c. The "Straight-Line Formula."— It will be noticed that in Fig. 315& the straight line connecting points A and C (medium steel, round ends) or A' and C (medium steel, flat ends) would not vary widely from the Rankine curve, so that on account of its simpKcity, when restricted to proper Hmiting values of the ratio l-^k,& straight line, or hnear relation, between the quantity P-^F and ratio l-i-k was proposed by Mr. T. H. Johnson (see Transac. Am. Soc. C. E., 1886, p. 530) for the breaking loads of columns of various materials. Among them are the following : Wrought iron: Hinged ends, Po = [42,000- 157/'^]li^; " Flat ends. Pi = [42,000- 128(|-)lP; Mild steel : Hinged ends, Po = [52,000 - 220(77)]^; " " Flat ends, Pi = [52,000 -179(^)lp. FLEXURE. LONG COLUMNS. 379 In these formulae Pq, or Pi, is breaking load in lbs., F= sectional area (in sq. in.), Z = tlie length, and k is the least radius of gyration of the cross-section for flat ends (as for hinged ends, see example 3, § 306) ; I and k in same unit. 310d. The J. B. Johnson Parabolic Formula for Columns. — If in. Fig. 315a a parabola be plotted with its axis vertical (and downward) and vertex at the point V of the two Rankine curves, and also made tangent to the Euler curve for the end conditions concerned, the points on such a curve for values of l^k between zero and the point of tangency to the Euler curve are found to agree fairly well with experiment; and the corre- sponding formula, or the equation to the curve, is of much simpler form than that of the Rankine types, being almost as simple as the straight line formula. Such a formula was proposed by the late Prof. J. B. Johnson, those for mild steel and wrought iron being given below (breaking load in lbs.). Mild steel: Pin ends, Pq = [42,000 -0.97(^HF; U not >150 Flat ends. Pi = [42,000- 0.62(^ MP; 1^ not >190 Wrought iron: Pin ends, Po = [ 34,000 -O.erQHp; (^ not >170 Flat ends. Pi = [34,000 -0.43(^ MP; U not >210 The notation is the same as in the preceding article. The limiting values mentioned for l^k refer to the points of tan- gency with the Euler curve. In Fig. 3156 the curve FTT^A^ is a parabola fulfilling the above mathematical condition for medium steel, with flat ends. 311. Solid Wooden Columns and Posts. Formula of U. S. Dept. of. Agriculture, Division of Forestry. — This formula was derived by Johnson from the results of experiments sonducted by the Division of Forestry and appUes to solid wooden columns provided with '' square ends," the constraint due to which, however, is not to be considered as fully equivalent to that of "fixed ends." The breaking load being denoted by Pi, 380 MECHANICS OF ENGINEERING. the sectional area by F, the ratio of length I to the ^^ least dimension,'^ d,' oi the cross section, by m (i.e., l^d = m), and the unit crushing stress for the material by C, the formula is J700 + 15m)FC ^ 700 + 15m + m2 ^^ The values of C to be used for different kinds of timber are given as follows : White oak and Georgia yellow pine 5000 Ibs./in.^ Douglas fir and short-leaf yellow pine 4500 '' Red pine, spruce, hemlock, cypress, chestnut, CaU- fornia redwood, and Cahfornia spruce 4000 ' ' White pine and cedar 3500 ' ' The fraction of Pi to be taken as the safe load depends on the wood and the degree of moisture present, four classes being designated in this respect; from Class A (18 per cent of moisture; timber exposed to weather), to Class D (10 per cent;, timber at all times protected from the weather). For yellow pine the safe load should be from 0.20Pi for Class A to O.SlPi for Class D. For all other timbers, from 0.20Pi for- Class A to 0.25Pi for Class D. 312. Column under Eccentric Loading. — In Fig. 322 let the load P be applied at i, at a distance or "eccentricity" =c from the center of gravity 1^1 A oi the upper base of the column, the reaction at j^\ j the. other end (at k) having an equal eccentricity yj i from B; the ends of the column being free to turn. ^^ / j (In an extreme case Ai and Bk might be brackets / [ "^ fastened to the ends of the column.) / . I AOB is the elastic curve, or bent condition of the 1^1 \ j^ axis of the column, originally straight. With as )"r" I I origin, any point n in the elastic curve has a vertical I I I co-ordinate x and a horizontal co-ordinate y. The \ I I unknown lateral deflection of the point from AB \ I T is a. With n cs any point in the elastic curve, and \i I nAi as free body, we have for the moment of the ■^ Bj I stress couple in section at n £'/[d2?/^dx^] = P(c -I- a—?/);. !^_c_J which is seen to differ from eq. (1) of p. 362 only in I having the constant c + a in place of the constant a. Fig. 322. y^^ ^^j therefore use eq. (6) of p. 363 for the present case, after replacing a by c + a; and hence, denoting 's/P^EI by h, remem- bering that vers. sin. = 1 — cos, we may write, as the equation to the elastic curve, y=(c + a)[l-cos (6x)] (1) For x = ^l, y should = the deflection o; on substituting which values in. (1) there results finally FLEXURE. LONG COLUMNS. 381 o=c sec (-^1—1 . . (2); and c+o=c- sec (-^j. • Hence the moment of the stress couple at is M^ = P{c+ a) = Pc- sec I -^\ (3) bl\ and the unit stress in outer fibre on concave side at is p- P_ M,e P Pc secQftZ) "y^ I ■ (4) (In this case of eccentric loading, then, the deflection a is not indeter- minate as was the case in deriving Euler's formula on p. 363. Note that ^bl is an angle in radians.) Example. — Let the value of P be 10,000 lbs., the length of the colunin be Z=20 ft. = 240 in., and the cross-section be a square cell [see Fig. 317 (6)] 4 inches being the side of the outer square ; area F = 7 m? and / = 14.58 in.^ Let the eccentricity be c = 2 in., each force P being applied in the middle of a side of the 4 in. square. Let £' = 30,000,000 Ibs./in.^; material, medium steel. "With this position of the force plane, e = 2 in. Here we have hhl =*(/^ 10,000 ) X 240 = 0.5736 radians, corre- V 30, 000, 000 X 14. 58^ spending to 32° 52', whose sec. = 1.190; and therefore a = 2X (1.190-1) = 0.380 in., and Mo=10,OOOX2X 1.190 = 23,800 in.-lbs. Finally ■p-. 10,000 23,800X2 ■ + - = 1430 + 3265 = 4695 lbs./ in. ^ 7 14.58 With P= 20,000 lbs., we should obtain iW= 0.811 radians (46° 30'), a=0.906 in., Mo = 57,120 in.-lbs., and p = 2860 + 7835 =10,695 Ibs./in.^ This latter unit stress is seen to be only moderate in value for the mate- rial, leading to the conclusion that 20,000 lbs. for P is a safe load; but on account of the possible original lack of straightness in the column, and of lack of homogeneity, both of which causes might increase a and Mq, it would be better to limit the load to 15,000 lbs. ; considering, also, the fact that Rankine's Formula for round ends (with a safety factor of 4) applied to this column for the case of no eccentricity would give about 22,000 lbs. as safe load. Fig. 322a. 318. Beam or Column with Eccentric End Pressures and also under Uniform Transverse Loading. — For example, in Fig. 322a let AB he the -bent axis of a beam, or column (originally straight), the longitudinal forces P and P being applied at an eccentricity c from A and B, while there is at the same time a vertical loading W, =wl, uniformly distributed along the 382 MECHANICS OF ENGINEERING. whole length at rate of w lbs. per running inch. The reactions of the two end supports will therefore be each ^W. The ends of the column are free to turn. It is required to find the deflection a, the moment Mq of the couple at middle section 0, and the unit stress p on the concave side at O. Take the free body iAn, n being any point of the elastic curve AOB, with co- ordinates X and y referred to the horizontal and vertical axes through as an origin, as shown. Then the moment of stress couple at n is EI{d'y^dx') = P{c + a-y) + {i)w{P-4x^) .... (5) Since (d^y-r-dx^) is a variable, let us denote i—EI-i-P)(d^y-T-dx^) by u, as an auxiliary variable; and eq. (5) wUl now read y-u=c+a+[{i)w{P-4:X^)]^P (6) Differentiating (6) twice, with respect to x, we have d^y d^u w . d^u P w = ; that is, — = u-\ — (7) dx' dx' P' ' dx' EI P ^^ Multiplying (7) by 2du, and denoting P-i-EI by b^ and 2w-^P by h, we have by integration, {dx'^ is a constant, x being the independent variable), (dM)^-7-(dx)^= — 6^M^ + /iM+C, where C is a constant of integration; and hence dx = dM-f- (VC + Zim— 6V), which integrates into a;=A.sm-M , — +C^ (8) where C is a constant. Transfo rmation of (8) gives (Vh' + ACb') sin [b(x-C')]+h = 2b^u (9) Eliminating u by aid of eqs. (6) and (9) we have 2b'y==\/h' + 4Cb'-sin [b(x-C')] + h + 2b\c+a) + (i)b^h{P-4x^) (10) from which 2b\dyldx) = bVh^ + 4C6^ • cos [b {x- C')]-b^hx . . . (11) To determine the three constants C, C", and a, we now make use of the facts that in (10) when x = 0, y also =0, and for x = ^l, y = a; and that in (11) for x=0, dy/dx, must =0. The three equations thus obtained, con- taining constants only, enable us to determine C, C, and a, and insert their values in (10) ; thus giving us as the equation to the elastic curve AOB, 2/ = (, h \ri-cos (6a: "1 ,, , ,^„^ ^+26-^jL^os(i60-J"*^'^' ^^'^ as also the value of the deflection a^[c+{h^2b^)lsec{^bl)-l]-{i^)hV .... (13) To find the moment of stress couple, Mq, at 0, we have now only to sub- stitute a; = and y = in eq. (5), and for a its value from (13); and thus obtain [fc+^Ysec(i60-^j (14) With F as the sectional area of the cross-section of the (prismatic) column (or beam, as it might also be called in this connection), and e as the dis- tance of the outer fibre from the gravity axis of the section, we now have for p, the stress in outer fibre on concave side at 0, V-l^^f (15) M„=P FLEXURE. LONG COLUMNS. 383 Since 6 and h denote VP-i-EI and 2w-^P, respectively, it is seen that when w is zero, h is zero and eq. (13) reduces to eq. (2) of the previous article. Again, if the two forces P are central, i.e., [applied at A and B, we put c = 0; in which case an approximate result may be reached by writing for the deflection a the value it would have if the 5 WP end forces P were not present, i.e., ^^ • -^j, as due to the uniform load W alone (see p. 260). On this basis the value of M^ is Pa+ (,^)Wl. (In case the vertical load on the beam or column in Fig. 322a is a single load Q concentrated in the middle at 0, a treatment similar to the foregoing may be applied, but is somewhat more complicated. For details of such a case the reader is referred to Mecanique AppUquee, by Bresse, Tome I. p. 384.) 314. Buckling of Web-Plates in Built Girders. — In §257 men- tion was made of the fact that very high web plates in built beams, such as /beams and box-girders, might need to be stiffened by riveting " angles " on the sides of the web. (The girders here spoken of are horizontal ones, such as might be used for carrying a railroad over a short sj^an of 20 to 50 feet. An approximate method of determining whether such stiffening is needed to prevent lateral buckling of the web, may be based upon Rankine's formula for a long column and will now be given. In Fig. 323 we have, free, a portion of a bent I-beam, between two vertical sections at a distance apart= Ai = the height of the web. In such a beam under forces L ^o its axis it has been proved (§256) that we may consider the web to sustain all the shear, J, at any section, and the flanges to take all the tension and compression, which form the " stress -couple" of the section. These couples and the two shears are shown in Fig. 323, for the two exposed sections. There is supposed to be no load on this portion of the beam, hence the shears at the two ends are Fig. 324. 384 MECHANICS OF ENGINEERING, equal. Now tlie shear acting between eacli flange and tlie horizontal edge of the web is equal in intensity per square inch to that in the vertical edge of the web ; hence if the web alons, of Fig. 323, is shown as a free body in Fig. 324,, we must insert two horizontal forces = J, in opposite directioii^,, on its upper and lower edges. Each of theM ^ J since we have taken a. horizontal length hi = height of web. In this figure, 324, we notice that the effect oi the acting forces is to lengthen the diagonal BD ancj shorten the diagonal AG, both of those diagonals making an angle of 45° with the horizontal. Let us now consider this buckling tendency along ^(7, by treating as free the strip ^(7, of small width = \. This is shown in Fig. 325. The only forces acting in the direc- tion of its length AG&ie the components along AG oi the four forces J' at the extremities. "VVe may therefore treat the strip as a long column of a length I = hi ^2, of a sec- tional area F = bb^, (where b is the thickness of the web plate), with a value of F = Yjg 6^ (see § 309), and with fixed (or flat) ends. Now the sum of the longitudinal components of the two J'.'s &t A is. Q = 2 J' ]/?, V2 = J' V2 ; but J' itself =■ rr. b j4 bi \'% since the small rectangle on which J' acts has an area = b )4 h^ ^2, and ihe shearing stress on it has an intensity of (J -r- bh{) per unit of area. Hence the longitudinal force at each end of this long column iis 'i-r/ w According to eq. (4) and the table in § 305, the safe load (factor of safety = 4) for a medium steel column of this form, with flat ends, would be (pound and inch) ib6i50,000 _ 12,500&&i 1 1 2/ii2 1 h^ • • (2) ^36,000* V1262 1 + 1,500' 62 If, then, in any particular locality of the girder (of medium steel) we find that Q is >Pi, i.e. FLEXUEE. LONG COLUMNS. 385 12,500& l+rl,.-'^ (pound and inch). (3) W =40 TONS 1,500 62 then vertical stiffeners will be required laterally. When these are required, they are generally placed at inter- vals equal to hi, (the depth of web), along that part of the girder where Q is >Pi. Example Fig. 326.— Will stiffening pieces be required in a plate 'girder of 20 feet span, bearing a uniform load of 40 tons, and having a web 24 in. deep and I in. thick? From § 242 we know that the -|t greatest shear, J max., is close to either pier, and hence we investigate that part of the girder first. J max. = iTF = 20 tons -40,000 lbs. .'. (inch and lb.), see (3), J _ 40,000 hi -10^ — -^ Trn Fig. 326. 24 = 1666.6 while, see (3), (inch and pound), 12,500X1 1 + . 242 1270 (4) (5) 1,500 (1)2 which is less than 1666.66. Hence stiffening pieces will be needed near the extremities of the girder. Also, since the shear for this case of loading diminishes uniformly toward zero at the middle they will be needed from each end up to a distance of ^ of 10 ft. from the middle. ^86 MECHANICS OF ENGLNEEBIlfG. CHAPTER Vn. UJTEAK ARCHES (OF BLOCKWOKI^. 815. A Blockwork Arch is a structure, spanning an openicg or gap, depending, for stability, upon the resistance to compresssion of its blocks, or voussoirs, the material ot which, such as stone or brick, is not suitable for sustain- ing a tensile strain. Above the voussoirs is usually placed a load of some character, (e.q. a roadway,) whose pressure upon the voussoirs will be considered as vertical, only. This condition is not fully realized in practice, unless the load is of cut stone, with vertical and horizontal joints resting upon voussoirs of corresponding shape (see Fig. 327), but sufficiently so to warrant its assumption in theory. Symmetry of form about a vertical axis will also be assumed in the following treatment. 316. Linear Arches. — For purposes of theoretical discussion the voussoirs of Fig. 327 may be considered to become Fig. 327. infinitely small and infinite in number, thus forming a " linear arch," while retaining the same shapes, their depth "1 to the face being assumed constant that it may not appear in the formulae. The joints between them are "1 to the curve of the arch, i.e., adjacent voussoirs can exert pressure on each other only in the direction of the tangent-line to that curve. LINEAR AKCHES. 387 317. Inverted Catenary, or Linear Arch Sustaining its Own Weight Alone. — Suppose tlie infinitely smalJ voussoirs to have weight, uniformly distributed along the curve, weigh- ing q lbs. per running linear unit. The eqiii]ibrium of such a structure, Fig. 328, is of course unstable but theo- retically possible. Required the form of the curve when equilibrium exists. The conditions of equilibrium are, obviously : 1st. The thrust or mutual pressure T between any two adjacent voussoirs at any point. A, of the curve must be tangent to the curve ; and 2ndly, considering a portion BA as a free body, the resultant of Hq the pres- FiQ. 328. Fig. 329. Fig. 330. sure at B the crown, and T &i A, must balance R the re- sultant of the il vertical forces (i.e.,weights of the elementary voussoirs) acting between B and A. But the conditions of equilibrium of a flexible, inexten- sible and uniformly loaded cord or chain are the very same (weights uniform along the curve) the forces being reversed in, direction. Fig. 329. Instead of compression we have tension, while the || vertical forces act toward in- stead of away from, the axis X. Hence the curve of equi- librium of Fig. 328 is an inverted catenary (see § 48) whose equation is y+c= e -\- e . (1) See Fig. 330. e = 2.71828 the Naperian Base. The "par- ameter " c may be determined by putting x = a, the half span, and y= Y, the rise, then solving for c by successive 388 MEOHAJSriCS OF ENGINEBKING. approximations. The " horizontal thrust" or H^^, is = yc, while if s = length, of arch OA, along the curve, the thrust T at any point A is T=^I Riffs' (2..) From the foregoing it may be inferred that a series ot vcui»' soirs of finite dimensions, arranged so as to contain the catenary curve, with joints "I to that curve and of equal weights for equal lengths of arc will be in equilibrium, and moreover in stable equilibrium on account of friction, and the finite width of the joints ; see Fig. 331. FIG. 331. 318. Linear Arches under Given Loading. — The linear arches to be considered further will be treated as without weight themselves but as bearing vertically pressing loads (each voussoir its own). Problem. — Given the form of the linear arch itself, it is required to find the law of vertical depth of loading under which the given linear arch will be in equilibrium. Fig. 332, given the curve ABC, i.e., the linear arch itself, re- quired the form of the curve MON, or upper limit of load- ing, such that the linear arch ABC shall be in equilibrium under the loads lying between the two curves. The load- ing is supposed homogeneous and of constant depth "^ to paper ; so that the ordinates z between the two curves are proportional to the load per horizontal linear unit. Assume a height of load z^ at the crown, at pleasure ; then required the z of any point m as a function of ^ and the curve ABC. LINEAB ARCHES. 389 Practical Solution. — Since a linear arcli under vertical pressures is nothing more than the inversion of the curve assumed by a cord loaded in the same way, this problenj might be solved mechanically by experimenting with a light cord, Fig. 333, to which are hung other heavy cords, or bars of uniform weight per unit length, and at equal horizontal distances apart ivhen in equilibrium,. By varying the lengths of the bars, and their points of attachment, we may finally find the curve sought, MON. (See also § 343.) Analytical Solution. — Consider the structure in Fig. 334 A number of rods of finite length, in the same plane, are in equilibrium, bearing the weights P, P^ etc., at the con- FiG. 334. Tig. 335. necting joints, each piece exerting a thrust T against the adjacent joint. The joint A, (the " pin " of the hinge), im- agined separated from the contiguous rods and hence free, is held in equilibrium by the vertical force P (a load) and the two thrusts T and T', making angles = d and d' with the vertical ; Fig. 335 shows the joint -4 fi'^e. From 2'(hor«^ izontal comps.)=0, we have. That is, the horizontal component of the thrust in any ro J is the same for all ; call it H^. ,\ T^ H. Bin (1) 390 MECHANICS OF ENGINEEKING. Now draw a line As *i to T' and write 2* ( compons. I to As)=0; whence F sin ^'=2^ sin ^, and [see (1)] . p_ jgp sin /? sm 6 sm (2) Let the rods of Fig. 334 become infinitely small a,nd infi- nite in number and the load continuous. The length of each rod becomes =ds an element of the linear arch, fi is the angle between two consecutive ds's, d is the angle be- tween the tangent line and the vertical, while P becomes the load resting on a single dx, or horizontal distance be- tween the middles of the two cZs's. That is, Fig. 336, if Y= weight of a cubic unit of the loading, P-=yzdx. (The lamina of arch and load considered is unity, 1 to paper, in thickness.) -Ho=a constant = thrust at crown ; 6=6', and sin /3=ds-^p, (since the' angle between two consecutive tan- gents is = that between two con- secutive radii of curvature). Hence eq. (2) becomes Yzdx = Kds p BID? 6 but dx—ds sin 6y Fia. 336. ,\yz- H. p siii^/? (3) Call the radius of curvature at the crown. /?o» and since there z=Zq and ^*=90°, (3) gives x^qPq~3^', hence (3) may be written sin^ d (4) This is the law of vertical depth of loading required. For a point of the linear arch where the tangent line is verti- cal, sin 6 =0 and z would == oo ; i.e., the load would be in- LINEAR ARCHES. 391 finitely high. Hence, in practice, a full semi-circle, for in- stance, could not be used as a linear arcli. 319. Circular Arc as Linear Arch. — ^As an example of the preceding problem let us ap- ply eq. (4) to a circular arc, Fig. 337, as a linear arcb. Since for a circle p is con- stant — r, eq. (4) reduces to sin^ 6 (5) Fig. 337. Hence tlie deptb of loading must vary inversely as the cube of tbe sine of the angle d made by the tangent line (of the linear arch) with the ver- tical. To find the depth z by construction.— Having z^ given, C being the centre of the arch, prolong Ga and make ob = go ; at 5 draw a 1 to Gb, intersecting the vertical through a at some point d ; draw the horizontal dc to meet Ga at some point c. Again, draw ce "| to Gc, meeting ad m e\ then ae= z required ; a being any point of the linear arch. For, from the similar right triangles involved, we have z„=ab=ad sin 0=ac sin d. sin ^=ae sin d sin d sin d ae= — ^— ; i.e., ae=2. Q.E.D. mn'd [-gee (5.)] 320. Parabola as Linear Arch. — To apply eq. 4 § 318 to a parabola (axis vertical) as linear arch, we must find values of p and po the radii of curvature at any point and the crown respectively. That is, in the general formula. -M dy\ dx) _ dx we must substitute the forms for the first and second dif- ferential co-efficients, derived from the equation of the 392 MECHANICS OF EI*f G [2f EERIN^G. Fig. 338. Fig. 339. curve (parabola) in Fig. 338, i.e. from x^ =^ 2 py; whence we obtain ~2.,or cot 0,= — ana-^=— ax p dor p Hence ^=i3^°M=^ ^ l^P cosec. I.e. p sin^^ . (6) At tbe vertex d = 90** ,*. />„ = p. Hence by substituting for p and p^ in eq. (4) of § 318 we obtain g=s^= constant [Fig. 339) (7) for a parabolic linear arcli. Therefore tbe depth of homo- geneous loading must be the same at all points as at the crown ; i.e., the load is uniformly distributed with respect to the horizontal. This result might have been antici- pated from the fact that a cord assumes the parabolic form when its load (as approximately true for suspension bridges) is uniformly distributed horizontally. Sae § 46 in Statics and Dynamics. 321. Linear Arch for a Given Tipper Contour of Loading, the arch itself being the unknown lower contour. Given the upper curve or limit of load and the depth z^ at crown, re- quired the form of linear arch which will be in equili- brium under the homogenous load between itself and that upper curve. In Fig. 340 let MON be the given upper contour of load, z^ is given or assumed,s' and z" are the respective ordinates of the two curves -S^ (7 and MON, Required the eqation of BAG. LINEAR ARCHES. 393 Fig. mo. Fig. 341. As before, tlie loading is homogenous, so that the weights of any portions of it are proportional to the corresponding areas between the curves. (Unity thick- ness "I to paper.) Now, Fig. 341, regard two consecutive ds's oi the linear arch as two links or consecutive blocks bearing at their junction w the load dP =y (^z -\- z"} dx in which Y denotes the heaviness of weight of a cubic unit of the loading. If T and T' are the thrusts exerted on these two blocks by their neighbors (here supposed removed) we have the three forces dP, T and T', forming a system in equilibrium. Hence from IX =0, T cos <p = T' cos cp' (1) and 1*7=0 gives T' sin cp'— T sin <p = dP ... (2) From (1) it appears that T cos f is constant at all points of the linear arch (just as we found in § 318) and hence .= the thrust at the crown, = Jff, whence we may write T=H-^ cos <p and r^E~ cos q)' . . . (3) Substituting from (3) in (2; we obtain H (tan <f' — tan <p)=dP (4) ■But tan <p =-^ and tan ip' = ^ "J ^ , {dx constant) while dP = y {z' -\- z") dx. Hence, putting for convenience H = yo?, (where a = side of an imaginary square of the 394 MECHANICS OP ENGINEERING. loading, whose thickness = unity and whose weight = IT) we have. ^=^'+'"'> <^> as a relation holding good for any point of the linear arch which is to be in equilibrium under the load included between itself and the given curve whose ordinates are «", Fig. 340. 322. Example of Preceding. Tipper Contour a Straight Line.— Fig. 342. Let the upper contour be a right line and hor- izontal ; then the a" of eq. 5 becomes zero at all points of ON. Hence drop the accent of z' in eq. (5) and we have dot? €^ Multiplying which by dz we obtain dz dh 1 do(? zdz (6) This being true of the z, dz, d?z and dx of each element of the curve O'B whose equation is desired, conceive it writ- ten out for each element between 0' and any point m, and put the sum of the left-hand members of these equations = to that of the right-hand members, remembering that a,^ and dx'^ are the same for each element. This gives dz=dz z=z d^ I « / ** 2 al2 2j nJ (te— %/ z=Zo adz ^ [ zj .... (7.) d .'. da; = -T^==«« LINEAR ARCHES. 395 Fig. 342. Pis 343 Integrating (7.) between 0' and any point m /, f =«[:iog..(^+,/(-i)-i) . . (8) i.e., fl?=a log. D-^]= or %= gg r i -E.-1 (8.) (9.) This curve is called the transformed catenary since we may obtain it from a common catenary by altering all the ordi- nates of the latter in a constant ratio, just as an ellipse may be obtained from a circle. If in eq. (9) a were = z^ the curve would be a common catenary. Supposing Sj and the co-ordinates x^ and gj of the point B (abutment) given, we may compute a from eq. 8 by put- ting X =Xi and z = g„ and solving for a. Then the crown- thrust H = ya^ becomes known, and a can be used in eqs. (8) or (9) to plot points in the curve or linear arch. From eq. (9) we have (10) area 00' mn Fig. 343. Call this area, A. As for the thrusts at the different joints of the linear arch, see Fig. 343, we have crown- thrust = ZT = ^a' . . . ; • - . • (11) and at any joint m the thrust T^VH'+irAf =rV^^^' .... (12} 396 MECHANICS OF ENGINEERING. 323. Remarks. — The foregoing results may be utilized with arches of finite dimensions by making the arch-ring contain the imaginary linear arch, and the joints 1 to the curve of the same. Questions of friction and the resist- ance of the material of the voussoirs are reserved for a succeeding chapter, (§ 344) in which will be advanced ^ more practical theory dealing with approximate linear arches or " equilibrium polygons " as they will then be called. Still, a study of exact linear arches is valuable on many accounts. By inverting the linear arches so far pre- sented we have the forms assumed by flexible and inexten- sible cords loaded ini the same way- GliAPHICAIi STATICS, tS9? CHAPTER VrX ELEMENTS OE GRAPHICAIi STATICS. 324. Definition. — In many respects graphical processes titve duvantages over tlie purely analytical, whicli recom- mend their use in many problems where celerity is desired without refiiied accuracy. One of these advantages is that gross errors are more easily detected, and another that the relations of the forces, distances, etc., are made so apparent to the eye, in the drawing, that the general effect of a given change in the data can readily be predicted at a glance. Graphical Statics in the system of geometrical construc- tions by which prt^blems in Statics may be solved by the use of drafting iixsiruments, forces as well as distances being represented in amount and direction by lines on the paper, of proper length and position, according to arbi- trary scales ; so many fest of distance to the linear inch of paper, for example, for distances ; and so many pounds or tons to the linear inch of paper for forces. Of course results should be interpreted by the same scale as that used for the data. The parallelogram of forces is the basis of all constructions for combining and resolving forces. 325. Force Polygons and Concurrent Forces in a Plsuae. — If a material point is in equilibrium under three forces Pi P, P3 (in the same plane of course) Fig. 344, any one of them, 398 MECHANICS OF ENGIXEERING. as Pi, must be equal and opposite to B the resultant of the other two (diagonal of their parallelogram). If now we lay off to some convenient scale a line in Fig. 345 = Pi and II to Pi in Fig. 344 ; and then from the pointed end of Pi a line equal and || to Pg and laid off pointing the same ivay, we note that the line remaining to p close the triangle in Fig. 345 must be = and || to Pg, since that tri- angle is nothing more than the left-hand half -parallelogram of Fig. 345. Fig. 344. Also, in 345, to close the triangle properly the directions of the arrows must be continuous Point to Butt, round the periphery. Fig. 345 is called a force polygor ; of three sides only in this case. By means of it, given any two of the three forces which hold the point in equilibrium, the third can be found, being equal and 1| to the side necessary to " close " the force polygon. Similarly, if a number of forces in a plane hold a mate- rial point in equilibrium, Fig. 346, their force polygon. FiG.344. Fig. 347, must close, whatever be the order in which its sides are drawn. For, if we combine Pj and P2 into a re- sultant Oa, Fig. 346, then this resultant with P3 to form a resultant Oh, and so on ; we find the resultant of Pi, P2, Ps? and P4 to be Oc, and if a fifth force is to produce equilib- rium it must be equal and opposite to Oc, and would close the polygon OdabcO, in which the sides are equal and par- GEAPHICAL STATICS. 399 allel respectively to the forces mentioned. To utilize tliis fact we can dispense witli all parts of tlie parallelograms in Pig. 346 except tlie sides mentioned, and tlien proceed as follows in Fig. 347 : If P5 is the unknown force which is to balance the other four (i.e, is their anti-resultant), we draw the sides of the force polygon from A round to B, making each line paral- lel and equal to the proper force and pointing the same way ; then the line BA represents the required F^ in amount and direction, since the arrow BA must follow the continuity of the others (point to butt). If the arrow BA were pointed at the extremity B, then it gives, obviously, the amount and direction of the result^ ant of the four forces Pj . . . P4. The foregoing shows that if a system of Concurrent Forces in a Plane is in equi- librium, ii^ force polygon must close. 326. Non-Concurrent Forces in a Plane. — Given a system of non-concurrent forces m a plane, acting on a rigid body, required graphic means of finding their resultant and anti- resultant ; also of expressing conditions of equilibrium. The resultant must be found in amount and direction ; and also in position (i.e., its line of action must be determined). E. g., Fig. 348 shows a curved rigid beam fixed in a vise at T, and also under the action of forces Pi P2 P3 and P^ {besides the action of the vise); required the resultant of By the ordinary parallelogram of forces we com- bine Pi and P2 at a, the intersection of their lines of PjQ 34g action, into a re- sultant Pa, ; then Pa with Pg at b, to form PbJ and finally P,, with P4 at c to form B^ which is .*. the resultant required, ie., of Pi . . . . P4 ; and c . , . P is its line of action. 400 MSCHAXICS OF EXGIXEEKIXG. Fig. 349. The separate force triangles (half-parallelograms) by wliich. the successive partial resultants B^^, etc., were found, are again drawn in Fig. 349. Now since B^ acting in the line C..F, Fig. 348, is the resultant of Pi . . Fi, it is plain that a force FJ equal to B,. and act- ing along c . . i^.but in the opposite di- rection, would balance the system Pi . . . P4, (is their anti- resultant). That is, the forces Pi P2 P3 P4 and BJ would form a system in equilibrium. The force B^' then, repre- sents the action of the vise T upon the beam. Hence re- place the vise by the force B/ acting in the line . . . F . . .c • to do which requires us to imagine a rigid prolongation of that end of the beam, to intersect F . . . c. This is shown in Fig. 350 where the whole beam is free, in equilibrium, under the forces shown, and in precisely the same state of stress, part for part, as in Fig. 348. Also, by combining in one force diagram, in Fig. 351, all the force triangles of Fig. 349 (by making their common sides coincide, and putting B/ instead of B^., and dotting all forces other than those of Fig. 350), we have a figure to be interpreted in connection with Fig. 350. A "poL^^iQH J' SPACE DIAGRAM Fig. 350. FORCE DIAGRAM Fig. 351. Here we note, first, that in the figure called a force-dia- gram, P1P2P3P4 and R/ form a closed polygon and that Gr^APHlCAli STATICS. 401 their arrows follow a continuous order, point to butt, around the jperimeter ; which proves that one condition of equilibrium of a system of non-concurrent forces ir^ a, plane is that its force polygon must close. Secondly, note that ah is II to Oa', and be to Oh' ; hence if the force-diagram has been drawn (including the rays, dotted) in order to deter- mine the amount and direction of HJ, or any other one force, we may then find its line of action in the space-diagram, as follows: (N. B. — By space diagram is meant the figure show- ing to a true scale the form of the rigid body and the lines of action of the forces" concerned). Through a, the intersec- tion of Fi and F-j, draw a line || to Oa' to cut P3 in some point b ; then through b a line || to Ob' to cut F^ at some point c; cF drawn || to Oc' is the required line of action of RJ, the anti- resultant of Pi, F2, P3, and P4. abc is called an equilibrium polygon; this one having but two segments, ab and bo (sometimes the lines of action of F^ and RJ may conveniently be considered as segments.) The segments of the equilibrium polygon are parallel to the respect- ive rays of the force diagram. Hence for the equilibrium of a system of no;ti-conciirrent forces in a plane not only must its force polygon close, but also the first and last segments of the corre- sponding equilibrium polygon must coincide with the resultants of the first two forces, and of the last two forces, respectively, of the system. E.g., ab coin- cides with the line of action of the resultant of F^ and F^, I he with that of F^ and E'c- Evidently the equil. polygon "will be different with each different order of forces in the force polygon or different choice of a pole, 0. But if the order of forces be taken as above, as they occur along the beam, or structure, and the pole taken at the " butt " of the first force in the force polygon, there will be only one j (and this one will be called the special equilibrium polygon in the chapter on arch-ribs, and the " true linear arch " in dealing with the stone arch.) After the rays (dotted in Fig. 351) have been added, by joining the pole to each 402 MEOn Allies OF ENGrKEEEi:srG. vertex with wliicli it is not already connected, tJbe finai figure may be called the/brce diagram. It may sometimes be convenient to give tlie name of rays to tlie two forces of tlie force polygon which, meet at the pole, in which case the first and last segments of the corresponding equil. polygon will coincide with the lines of action of those forces in the space-diagram (as we may call the representation of the body or structure on which the forces act). This " space diagram " shows the real field of action of the forces, while the force diagram, which may be placed in any convenient position on the paper, shows the magnitudes and directions of the forces acting in the former diagram, its lines being interpreted on a scale of so many lbs. or tons to the inch of paper ; in the space-diagram we deal with a scale of so many/ee^ to the inch of paper. We have found, then, that if any vertex or corner of the closed force polygon be taken as a pole, and rays drawn from it to all the other corners of the polygon, and a cor- responding equil. polygon drawn in the space diagram., the first and last segments of the latter polygon must co-incide with the first and last forces according to the order adopted (or with the resultants of the first two and last two, if more convenient to classify them thus). It remains to utilize this principle. 327. To Find the Resultant of Several Forces in a Plane. — This might be done as in § 326, but since frei^^uently a given set of forces are parallel, or nearly so, a special method will now be given, of great convenience in such cases. Fig. 352. Let Pi Pg and Pa be the given forces whose resultant is re- quirsd. Let us first find their and -' resultant, or force which Fig. 352. Pm. 353. will balance GEAPHICAL STATICS. 403 them. This anti-resultant may be conceived as decom- posed into two components P and P' one of which, say P, is arbitrary in amount and position. Assuming P, then, at convenience, in the space diagram, it is required to lind F'. The live forces must form a balanced system ; hence if beginning at Oi, Fig. 353, we lay off a line O^A = P by scale, then Al = and || to P,, and so on (point to butt), the line POi necessary to close the force polygon is = P' re- quired. Now form the corresponding equil. polygon in the space diagram in the usual way, viz.: through a the intersection of P and P^ draw ab || to the ray 0, . . . 1 (Avhich connects the pole Oi with the point of the last force mentioned). From h, where ab intersects the line of Pg* draw he, || to the ray O^ . . 2, till it intersects the line of Pg. A line mc drawn through c and || to the P' of the force diagram is the line of action of P'. Now the resultant of P and P' is the anti-resultant of Pi, P2 and P3; .'. d, the intersection of the lines of P and P', is a point in the line of action of the anti-resultant re- quired, while its direction and magnitude are given by the line BA in the force diagram ; for BA forms a closed poly- gon both with Pi P2 P3, and with PP'. Hence a line through (i || to BA, viz., de, is the line of action of the anti- resultant (and hence of the resultant) of Pj, P2, P3. Since, in this construction, P is arbitrary, we may first choose Oi, arbitrarily, in a convenient position, i.e., in such a position that by inspection the segments of the result- ing equil. polygon shall give fair intersections and not pass off the paper. If the given forces are parallel the device of introducing the oblique P and P' is quite neces- sary. 328. — The result of this construction may be stated as follows, (regarding Oa and cm as segments of the equil. polygon as well as ah and he): If any tivo segments of an equU. polygon he prolonged, their intersection is a point in the line of action of the resultant of those forces acting at 404 MECHANICS OF ENGINEERING. the vertices intervening between the given segments, the resultant of Pi P2 P3 acts through d. Here, 329. Vertical Reaction of Piers, etc. — Fig. 354. Given the vertical forces or loads Pj P2 and P3 acting on a rigid body (beam, or truss) which is supported by two piers having smooth horizontal surfaces (so that the reactions must be vertical), required the reactions Fq and V^ of the piers. For an instant suppose V^ and V^ known ; they are in iVn Fig. 354. equil. with Pi Pg and P3. The introduction of the equal and opposite forces P and P' in the same line will not dis- turb the equilibrium. Taking the seven forces in the order P Vq Pj Pg P3 V^ and P', a force polygon formed with them will close (see (h) in Fig. where the forces which really lie on the same line are slightly separated). With Oy the butt of P, as a pole, draw the rays of the force dia- gram OA, OB, etc. The corresponding equil. polygon begins at a, the intersection of P and V^ in {a) (the space diagram), and ends at n the intersection of P' and V^. Join an. Now since P and P' act in the same line, an must be that line and must be || to P and P' of the force diagram. Since the amount and direction of P and P' are arbitrary, the position of the pole is arbitrary, while Pi, P2, and P3 are the only forces known in advance in the force diagram. Hence Vq and V^ may be determined as follows : Lay off the given loads Pi, P2, etc., in the order of their occur- rence in the space diagram, to form a " load-line " AD GRAPHICAL STATICS. 405 (see (h.) Fig. 854) as a beginning for a force-diagram ; take any convenient pole 0, draw the rays OA, OB, 00 and OD. Tlien beginning at any convenient point a in the vertical line containing the unknown Vq, draw ab || to OA, be II to OB, and so on, until the last segment [dn in this case) cuts the vertical containing the unknown V„ in some point n. Join an (this is sometimes called a closing line) and draw a || to it through 0, in the force-diagram. This last line will cut ths " load-line " in some point n', and divide it in two parts n' A and i>w', which are respectively Vq and Va required. Corollary. — Evidently, for a given system of loads, in given vertical lines of action, and for two given piers, or abut- ments, having smooth horizontal surfaces, the location of the point n' on the load line is independent of the choice of a •pole. Of course, in treating the stresses and deflection of the ligid body concerned, P and P' are left out of account, as being imaginary and serving only a temporary purpose. 330. Application of Foregoing Principles to a Roof Truss- Fig. 355. Wi and W.^ are wind pressures. Pi and P. are loads, while the lemaining external forces, viz., the re- 406 MECHANICS OF ENGINEERING. actions, or supporting forces. To, F'„ and H^i niay be fonnd by preceding §§. (We here suppose that the right abut- ment furnishes all the horizontal resistance ; none at the left). Lay off the forces (known) Wi, W2, Pi, and P2 in the usual way, to form a portion of the closed force polygon. To close the polygon it is evident we need only draw a horizontal through 5 and limit it by a vertical through 1. This determines H^ but it remains to determine ?^' the point of division between F^ and V^. Select a convenient pole Oi, and draw rays from it to 1, 2, etc. Assume a con- venient point a in the line of V„ in the space diagram, and through it draw a line || to Oil to meet the line of W^ in some point b ; then a line || to Oi2 to meet the line of W2 in some point c ; then through c || to OjS to meet the line of Pi in some point d ; then through d || to Oi4 to meet the line of P2 in some point e, (e is identical with d, since Pi and P2 are in the same line) ; then efW to Oi5 to meet Hj^ in some point/; then fg \\ to OS to meet V^ in some point g. abcdefg is an equilibrium polygon corresponding to the pole Oj. Now join ag, the " closing-line," and draw a || to it through Oi to determine n', the required point of division between Vo and V„ on the vertical 1 6. Hence F^ and V^ are now determined as well as H^^. [The use of the arbitrary pole Oi implies the temporary employment of a pair of opposite and equal forces in the line ag, the amount of either being = Oiti']. Having now all the external forces acting on the truss, and assuming that it contains no " redundant parts," i.e., parts unnecessary for rigidity of the frame-work, we proceed tc find the pulls and thrusts in the individual pieces, on the following plan. The truss being pin-connected, no piece extending beyond a joint, and all loads being con- sidered to act at joints, the action, pull or thrust, of each piece on the joint at either extremity will be in the direction of the piece, i.e., in a knoivn direction, and the pin of each GKAPHICAL STATICS. 407 joint is in equilibrium under a system of concurrent forces consisting of the loads (if any) at tlie joint and the pulls or thrusts exerted upon it by the pieces meeting there.. Hence we may apply the principles of § 325 to each joint in turn. See Fig. 356. In constructing and interpreting the various force polygons, Mr. E.. H Bow'g convenient notation will be used; this is as follows: In the space diagram a capital letter [ABC, etc.] is placed in each tri- angular cell of the truss, and also in each angular space in the outside outline of the truss between the external forces and the adjacent truss-pieces. In this way we can speak of the force Wi as the force BC, of W2 as the force C-E, the stress in the piece a/3 as the force QI), and so on. That is, the stress in any one piece can be named from the letters in the spaces bordering its two sides. Corresponding to these capital letters in the spaces of the space-dia~ gram, small letters will be used at the vertices of the closed force-polygons (one polygon for each joint) in such a way that the stress in the piece CD, for example, shall be thQ forc3 cd of the force polygon belonging to any joint in which that piece terminates ; the stress in the piece FO by the force fg, in the proper force polygon, and so on. In Fig. 356 the whole truss is shown free, in equili- brium under the external forces. To find the pulls or thrusts (i.e., tensions or compressions) in the pieces, con- sider that if all but two of the forces of a closed force polygon are known in magnitude and direction,, while the directions, only, of those two are known, the wliole force polygon may he drawn, thus determining the amounts of those two forces by the lengths of the corresponding sides. We must .'. begin with a joint where no more than two pieces nieet, as at a ; [call the joints a, /9, y, d, and the cor- corresponding force polygons a', /9' etc. Fig. 356.] Hence at a' (anywhere on the paper) make oh \\ and = (by scale) to the known force AB (i.e., V^) pointing it at the upper end, and from this end draw he — and || to the known force BG (i.e., Wj) pointing this at the lower end. iU8 MECHANICS OF E:i^GINEEBraG. Fig. 356. To close the polygon draw througli c a || to the piece CD, and through a a || to AD ; their intersection deter- mines d, and the polygon is closed. Since the arrows must be point to butt round the periphery, the force with which the piece CD acts on the pin of the joint a is a force of an amount = cd and in a direction from c toward d ; hence the piece CD is in compression ; whereas the action of the piece DA upon the pin at a is from d toward a (direction of arrow) and hence DA is in tension. Notice that in constructing the force polygon «' a right-handed (or clock-wise) rotation has been observed in considering in turn the spaces ABC and D, round the joint a. A similar order will be found convenient in each of the other joints. Knowing now the stress in the piece GD, (as well as in DA) all but two of the forces acting on the pin at the joint /? are known, and accordingly we begin a force polygon, /3', for that joint by drawing dc,= and || to the dc of polygon a', hut pointed in the opposite direction, since the action of OD on the joint /? is equal and opposite to its action on the joint a (this disregards the weight of the piece). Through c draw ce = and || to the force CE (i.e., W^ and GRAPHICAL STATICS. 409 pointing tlie same way ; tlien ef, = and || to tlie load EF (i.e. Pj) and pointing downward. Througli f draw a || to tlie piece FG and through d, a || to the piece OB, and the polygon is closed, thus determining the stresses in the pieces FG and GT>. Noting the pointing of the arrows, we readily see that FG is in compression Avhile GD is in tension. Next pass to the joint (5, and construct the polygon o' , thus determining the stress gli in GB. and that ad in AD ; this last force ad should check with its equal and oppo- site ad already determined in polygon a'. Another check consists in the proper closing of the polygon y', all of whose sides are now known. [A compound stress-diagram may be formed by super- posing the polygons already found in such a way as to make equal sides co-incide ; but the character of each stress is not so readily perceived then as when they are kept separate]. In a similar manner we may find the stresses in any pin- connected frame-work (in one plane and having no redun- dant pieces) under given loads, provided all the support- ing forces or reactions can be found. In the case of a braced-arch (truss) as shown in Fig. 357, hinged to the abutments at both ends and not free to slide laterally upon them, the reactions at and B de- FiG. 357. pend, in amount and direc- tion, not only upon the equations of Statics, but on the form and elasticity of the arch-truss. Such cases will be treated later under arch -ribs, or curved beams. 332. The Special Equil. Polygon. Its Relation to the Stresses in the Rigid Body. — Eeproducing Figs. 350 and 351 in Figs. 358 and 359, (where a rigid curved beam is in equilibrium under the forces P^ Pg, P3, P4 and B'^) we call a . . b . , MECHANICS OF Eis^GINEERIKG. tiie special equil. polygon because it corresponds to a force diagram in which the same order of forces has been ob- S3rved as that in which they occur along the beam (from left to right here). From the relations between the force SPACE DIAGRAM Fig. 358. FORCE DIAGRAM Pig, 359. diagram and equil. polygon, this special equil. polygon in the space diagram has the following properties in connec- tion with the corresponding rays (dotted lines) in the force diagram. The stresses in any cross-section of the portion O'A of the beam, are due to P^ alone ; those of any cross-section on AB to Pi and P2, i.e., to their resultant R , whose mag- nitude is given by the line Oa' in the force diagram, while its liLe of action is ah the first segment of the equil. poly- gon. Similarly, the stresses in BC are due to P^, P^ and P., i.e., to their resultant R^, acting along the segment &c, its magnitude being =^0h' in the force diagram. E.g., if the section at m be exposed, considering O'ABm as a free body, we have (see Fig. 360) the elastic stresses (or inter- P3 Fig. 360. Fig. 361. nal forces) at m balancing the exterior or " applied forces " Pi, Pj and P3. Obviously, then, the stresses at m are just GEAPHIOAL STATICS. 411 the same as if B^, tlie resultant of Pj, P^ and Pg, acted upon an imaginary rigid prolongation of tlie beam intersecting he (see Fig. 361).i?i, might be called the " anti-stress-resuU- ant" ior the portion PC of the beam. We may .•. state the following : If a rigid body is in equilibrium under a sys- tem of Hon-Concurrent Forces m a plane, and the special equi- librium polygon has been draivn, then each ray of tlie force diagram is the anti-stress-resultant of that portion of the beam which corresponds to the segment of the equilibrium polygon to which the ray is parallel ; and its line of action is the seg- ment just mentioned. Evidently if the body is not one rigid piece, but com- posed of a ring of uncemented blocks (or voussoirs), it may be considered rigid only so long as no slipping takes place or disarrangement of the blocks; and this requires that the " anti-stress-resultant " for a given joint between two blocks shall not lie outside the bearing surface of the joint, nor make too small an angle with it, lest tipping or slipping occur. For an example of this see Fig. 362, show- ing a line of three blocks in equilibrium under five forces. The pressure borne at the s^2 joint MN, is = Pa ^^ the force -diagram and acts in the line ab. The con- struction supposes all the forces given except Fig. 362. oue, in amount and posi- tion, and that this one could easily be found in amount, as being the side remaining to close the force polygon, while its position would depend ox^ I;he equil. polygon. But in practice the t^m forces Pj and B\ are generally unknown, hence the point 0, or pole of the force diagram, can not be fixed, nor the special equil. polygon located, until other considerations, outside of those so far presented, are brought into play. In the progress of such a problem, as will be seen, it will be necessary to use arbitrary trial po- sitions for the pole 0, and corresponding ^rmZ equilibrium polygons. 412 MECHANICS OF BJUGmBBlWSGc, CHAPTER IX. GRAPHICAL. STATICS OF VERTICAL. FORCES, 333. Remarks. — (Witli the excoption of § 378 a) in prob- lems to be treated subsequently (either the stiff arch-rib, or the block-work of an arch-ring, of masonry) when the body is considered free all the forces holding it in equil. will be vertical (loads, due to gravity) except the reactions at the two extremities, as in Eig, 363 ; but for convenience each reaction will be replaced by its horizontal and verti- cal components (see Fig. 364). The two fi^'s are of course pqual, since they are the only horizontal forces in the system. Henceforth, aU equil. polygons under discussion will he understood to imply this kind of system of forces. Pi, Pz, r r f A t\ U v„ Fig. 363. Fig. 364a. Fig. 364. etc. , will represent the ' ' loads " ; Vq and F„ the vertical components of the abutment reactions; H the value of either horizontal component of the same. (We here sup- pose the pressures To and Tn resolved along the horizontal and vertical.) JRAPHICAIi STATICS. 413 334. Concrete Conception of an Equilibrium Polygon. — Any equilibrium polygon has this property, due to its mode of construction, viz.: If the ab and be of Fig. 358 were im- ponderable straight rods, jointed at b without frictioiij they would be in equilibrium under the system of forces there given. (See Fig. 364a). The rod ah suffers a compression equal to the H^ of the force diagram, Fig. 359, and be a -^compression = B^^. In some cases these rods might be in tension, and would then form a set of links playing the part of a suspension-bridge cable. (See § 44). 335. Example of EcLuilibrium Polygon Drawn to Vertical Loads — Fig. 365. [The structure bearing the given loads is not shown, but simply the imaginary rods, or segments of an equilibrium polygon, which would support the given loads in equilibrium if the abutment points A and B, to which the terminal rods are hinged, were firm. In the present case this equilibrium is unstable since the rods form a standing structure ; but if they were hanging, the equilibri- um would be stable. Still, in the present case, a very light bracing, or a little friction at all joints would make the equilibrium stable. 2 FT. TO aNOH Fig. S65. Given three loads Pi, F2, and P3, and two " abutment verticals " A' and B', in which we desire the equil. poly- gon to terminate, lay off as a "load-line," to scale, Pj, P2, and P« end to end in their order. Then selecting any pole. 414 MBCHAXICS OF EjS^GI:N^EEIII]^G. 0, draw the rays 01, 02, etc., of a force diagram (tlie F's and P's, though really on the same vertical, are separated slightly for distinctness ; also the H's, which both pass through and divide the load-line into V^ and F^). We determine a corresponding equilibrium polygon by draw- ing through A (any point in A') a line || to . . 1, to inter- sect Pi in some point b ; through 6 a || to . . 2, and so ou> until B'' the other abutment-vertical is struck in some point B. AB is the " abutment-line " or " closing -line." By choosing another point for 0, another equilibrium polygon would result. As to which of the infinite number (which could thus be drawn, for the given loads and the A' and B' verticals) is the special equilibrium poly' gon for the arch-rib or stone-arch, or other structure, on which the loads rest, is to be considered hereafter. In any of the above equilibrium polygons the imaginary series of jointed rods would be in equilibrium. 336. Useful Property of an Equilibrium Polygon for Vertical Loads. — (Particular case of § 328). See Fig. 366. In any equil. polygon, supporting vertical loads, consider as free any number of consecutive segments, or rods, with the loads at their joints, e. g., the 5th and 6th and portions of C/r^.^ the 4th and 7th which, we sup- /g i ,> ^ ^6"--. ^ -^^ pose cut and the compressive — ~S<^^ forces in them put in, T^ and ^^ Tj, in order to consider 4 5 6 7 "^^ ^ as a free body. For equil,, ~^~[-^\ according to Statics, the lines ' ' 'Pe "\ of action of Ti and Ty (the com- i^iG. 366. pression in those rods) must in- tersect in a point, C, in the line of action of the resultant of Fi, P5, and Pq ; i.e., of the loads occurring at the inter- vening vertices. That is, the point C must lie in the ver- tical containing the centre of gravity of those loads. Since the position of this vertical must be independent of the particular equilibrium polygon used, any other (dotted lines in Fig. 366) for the same loads will give the same re- GKAPHICAL STATICS. 415 suits. Hence tlie vertical CD, containing the centre of gravity of any number of consecutive loads, is easily found by drawing tlie equilibrium polygon corresponding to any convenient force diagram having the proper load-line. This principle can be advantageously applied to finding a gravity -line of any plane figure, by dividing the latter into parallel strips, whose areas may be treated as loads applied in their respective centres of gravity. If the strips are quite numerous, the centre of gravity of each may be considered to be at the centre of the line joining the mid- dles of the two long sides, while their areas may be taken as proportional to the lengths of the lines drawn through these centres of gravity parallel to the long sides and lim- ited by the end-curves of the strips. Hence the " load- line " of the force diagram may consist of these lines, or of their halves, or quarters, etc., if more convenient (§ 376). USEFUL, RELATIONS BETWEEN FORCE DIA- GRAMS AND EQUILIBRIUM POLYGONS,, (for vertical loads,) 237. Il6sum6 of Construction.— Fig. 367. Given the loads Pi, etc., 'their verticals, and the two abutment verticals ^4' and B', in which the abutments are to lie ; we lay off a load-line 1 ... 4, take any convenient pole, 0, for a force- diagram and complete the latter. For a corresponding equilibrium polygon, assume any point A in the vertical" A', for an abutment, and draw the successive segments Al, 2, etc., respectively parallel to the inclined lines of the force diagram (rays), thus determiDiDg finally the abut- ment B, in B', v/hich {B) will not in general lie in the hor- izontal through A. Now join AB, calling AB the abutment-line, and draw a parallel to it through 0, thus fixing the point n' on the 416 MECHANICS or ENGINEERIKG. Pi P. t l' Fig. load-line. This point %' , as above determined, is indepen' dent of the location of the pole, 0, (proved in § 329) and divides the load-line into two portions ( V'o = 1 . . . n\ and V'n = n' .. .4:) which are the vertical pressures which two supports in the verticals A' and B' would sustain if the given loads rested on a horizontal rigid bar, as in Fig. 368. See § 329. Hence to find the point n' we may use any convenient pole 0. [N. B.— The forces V^ and V^ of Fig. 367 are not identi- cal with F'o and V'„, but may be obtained by dropping a "I from to the load-line, thus dividing the load-line into two portions which are V^ (upper portion) and F^. However, if A and B be connected by a tie-rod, in Fig. 367, the abutments in that figure will bear vertical press- ures only and they will be the same as in Fig, 368, while the tension in the tie -rod will be = On'.^ 338. Theorem. — The vertical dimensions of any two equili- brium polygons, drawn to the same loads, load-verticals, and abutment -verticals, are inversely proportional to their H^s {or "pole distances "). We here regard an equil. polygon and its abutment-line as a closed figure. Thus, in Fig. 369, we have two force-diagrams (with a common load-line, for convenience) and their corresponding equil. polygons, for the same loads and verticals. From § 337 we know that On' is II to AB and OqW' is || to A^B^. Let CD be any ver- tical cutting the first segments of the two equil. polygons. GRAPHICAL STATICS. 417 SB| Denote the intercepts thus determined by z' and %\, respect- gC r ively. From the parallelisms just mentioned, and others more famil~ ,/ iar, we have the triangle \n' sim* ilar to the triangle Az' (shaded), and the triangle O^n' similar to the tri- angle Ajz,^,. Hence c 1 P. u-^ / hi 1 . p^ ^. A. ^y 1 —- 1-.. -^-N^ Fig. 369. the proportions between ( \n' bases and altitudes ( H h and .*. z' : z\ : : H^ '■ H. The same kind of proof may easily be applied to the vertical intercepts in any other segments, e. g., 2" and z'\. Q. E. D. 339. Corollaries to the foregoing. It is evident that : (1.) If the pole of the force-diagram be moved along a vertical line, the equilibrium polygon changing its form in a corresponding manner, the vertical dimensions of the equilibrium polygon remain unchanged ; and (2.) If the pole move along a straight line which con- tains the point n\ the direction of the abutment-line remains constantly parallel to the former line, while the vertical dimensions of the equilibrium polygon change in inverse proportion to the pole distance, or H, of the force- diagram, [^is the "1 distance of the pole from the load- line, and is called the pole-distance]. § 340. Linear Arch as Equilibrium Polygon. — (See § 316.) If the given loads are infinitely small with infinitely small horizontal spaces between them, any equilibrijim polygon becomes a linear arch. Graphically we can not deal with these infinitely small loads and spaces, but from § 336 it is evident that if we replace them, in successive groups. 418 MECHAXICS OF ENGINEERING. Fig. 370. bj finite forces, eacli of wliicli = tlie STim of those com^. I I I I I I I I P°''^^ """^ ^^""^P ^""^ ^' .M i V ..M M. .^^ M I ,M I + /, applied tlirougli the cen- tre of gravity of that group, we can draw an equilibrium polygon whose segments will be tangent to the curve of the corresponding linear arch, and indicate its posi- tion with siifficient exactness for practical purposes. (See Fig. 370), The successive points of tangency A, m, n, etc.. lie vertically under the points of division between the groups. This relation forms the basis of the graphical treatment of voussoir, or blockwork, arches. 341. To Peas an Equilibrium Polygon Through. Three Arbitrary Points. — (In the present case the forces are vertical. For a construction dealing with any plane system of forces see construction in § 378a.) Given a system of loads, it is re- ^ quired to draw r /^l ^^ equilibrium polygon for t h e m through -anythree points, two of which may be consid- ered as abut- ments, outside of the load-verticals, the third point being between the verticals of the first two. See Fig. 371. The loads Pi, etc., are given, with their verticals, while A, p, and B are the three points. Lay oft the load-line, and with any convenient pole, Oj, construct a force-diagram, then a corresponding preliminary equilibrium polygon beginning at A. Its right abutment P,, in the vertical through B, is thus found. Oj n' can now be drawn || to AB^, to determine n\ Draw n'O \\ to BA. The pole of the required equilibrium polygon must lie on n'O (§ 337} Fig. 371. GEAPHICAL STATICS. Draw a vertical throiigli jp. The E. of tlie required equili- brium polygon must satisfy the proportion H : H^ : : rs i pm. (See § 338). Hence construct or compute H from the proportion and draw a vertical at distance H from the load-line (on the left of the load-line here) ; its inter- section with n' gives the desired pole, for which a force diagram may now be drawn. The corresponding equilibrium polygon beginning at the first point A will also pass through p and B ; it is not drawn in the figure. 342. Symmetrical Case of the Foregoing Problem.— If two points A and B are on a level, the third, p, on the middle vertical between them ; and the loads (an even number) symmetrically disposed both in position and magnitvde, about p, we may proceed more simply, as follows : (Fig. 372). From symmetry n' must occur in the mid- dle of the load-line, of which we need lay off only the upper half. Take a convenient pole ■piG. 372. Oi, in the horizontal through n', and draw a half force diagram and a corres- ponding half equilibrium polygon (both dotted). The up- per segment he of the latter must be horizontal and being prolonged, cuts the prolongation of the first segment in a point d, which determines the vertical CD containing the centre of gravity of the loads occurring over the half -span on the left. (See § 336). In the required equilibrium poly- gon the segment containing the point p must be horizon- tal, and its intersection with the first segment must lie in CD. Hence determine this intersection, C, by drawing the vertical CD and a horizontal through p ; then join CA, which is the ^rst segment of the required equil. polygon. A parallel to GA through 1 is the Jirst ray of the corres- ponding force diagram, and determines the pole on the horizontal through n'. Completing the force diagram foi 420 MECHAXIOS OF El^J^GINEBEIN^G. Fig. 373. this pole (half of it only here), the required equil. poly- gon is easily finished afterwards. 343. To Find a System of Loads Under Which a Given Equi- librium Polygon Would be in Equilibrium, — Fig. 373. Let AB he the given equilibrium polygon. Through any point as a pole draw a parallel to each segment of the equilibrium polygon. Any vertical, as V, cutting these lines will have, intercepted upon it, a load-line 1, 2, 3, whose parts 1 . . 2, 2 . . 3, etc., are proportional to the successive loads which, placed on ih@ corresponding joints of the equilibrium polygon would be supported by it in equilibrium (unstable). One load may be assumed and the others constructed. A hanging, as well as a standing, equilibrium polygon may be dealt with in hke manner, but will be in stable equilibrium. The problem in § 44 may be solved in this way, the various steps and final re- sults being as follows (Fig. 50 is here re- peated) : — Let weight Gi be given, =66 lbs., and the positions of the cord segments be as in Fig. 50. We first lay of! (see Fig. 373a) vertically, a6 = 66 lbs., by some convenient scale, and prolong this vertical fine indefi- d nitely downward. aO is then drawn parallel to and bO parallel to 1 ... 2. Their intersection determines a pole, 0, through which Oc and Od, parallel respectively to 2 ... 3 and 3 . . . n, are drawn, to intersect ad in c and d. We also draw the horizontal On, in Fig. 373a. By scaling, we now find the results: — G2 = bc = 42 lbs.; G3 = cd = 50 lbs.; H = 58.5 lbs., ( = Ho and'^„ of Fig. 50); while 70 = ^^=100 lbs. and y„ = 58 lbs. Fig. 373a. .1 of Fig. 50, AECHES OF MASOifBT. 421 CHAPTER X. RIGHT ARCHES OF MASONRY. Note. — The treatment given in this chapter is by many engineers considered sufficiently exact for ordinary masonry arches, the mOie refined methods of the "elastic theory" being reserved for arches of fairly continuous material, such as those of metal and of concrete (re- inforced or otherwise); and is accordingly retained in this revised edition. 844. — In an ordinary "right" storce-arcli (i.e., one in which the faces are "[ to the axis of the cylindrical soffit, or under surface), the successive blocks forming the arch- ring are called voussoirs, the joints between them being planes which, prolonged, meet generally in one or more horizontal lines; e.g., those of a three-centred arch in three II horizontal lines ; those of a circular arch in one, the axis of the cylinder, etc. Elliptic arches are sometimes used. The inner concave surface is called the soffit, to which the radiat- ing joints between the voussoirs are made perpendicular. The curved line in which the soffit, is intersected by a plane Fig. 374. H to the axis of the arch is the Intrados. The curve in the same plane as the intrados, and bounding the outer ex- tremities of the joints between the voussoirs, is called the Extrados. Fig. 374 gives other terms in use in connection with, a 422 MECHAKICS OF ElSGIXEEIxiKG, stone arch, and explains those already given. " springing-line." AB is the 345o Mortar and Friction. — As 'common mortar hardens very slowly, no reliance should be placed on its tenacity as an element of stability in arches of any considerable size ; though hydraulic mortar and thin joints of ordinary mortar can sometimes be depended on. Friction, however, between the surfaces of contiguous voussoirs, plays an essential part in the stability of an arch, and will there- fore be considered. The stability of voussoir-arches must .•. be made to depend on the resistance of the voussoirs to compresssion and to sliding upon each other ; as also of the blocks composing the piers, the foundations of the latter being firm. 346. Point of Application of the Eesultant Pressure between two consecutive voussoirs ; (or pier blocks). Applying Navier's principle (as in flexure of beams) that the press- ure per unit area on a joint varies uniformly from the extremity under greatest comj)ression to the point of least compression (or of no compression); and remembering that negative pressures (i.e., tension) can not exist, as they might in a curved beam, we may represent the pressure per unit area at successive points of a joint (from the intra- dos toward the extrados, or vice versa) by the ordinates of a straight line, forming the surface of a trapezoid or tri- angle, in which figure the foot of the ordinate of the cen- tre of gravity is the point of application of the resultant pressure. Thus, where the least compression is supposed Fig. 575. Fig. 376. Fig. 377. Fig. 378, MASONEY ARCHES. 423 to occur at the intrados A, Fig. 375, tlie pressures vary as tlie ordinates of a trapezoid, increasing to a maximum value at B, in the extrados. In Fig. 376, where the pressure is zero at B, and varies as the ordinates of a triangle, the result- ant pressure acts through a point one-third the joint- length from A. Similarly in Fig. 377, it acts one-third the joint-length from B. Hence, when the pressure is not zero at either edge the resultant pressure acts within the middle third of the joint. Whereas, if the resultant press- ure falls loitliout the middle third, it shows that a portion -4m of the joint, see Fig. 378, receives no pressure, i.e., the joint tends to open along Am. Therefore that no joint tend to open, the resultant press- ure must fall within the middle third. It must be understood that the joint surfaces here dealt with are rectangles, seen edgewise in the figures. 347. Friction. — By experiment it has been found the angle of friction (see § 156) for two contiguous voussoirs of stone or brick is about 30° ; i.e., the coefficient of fric- tion is / = tan. 30°. Hence if the direction of the press- ure exerted upon a voussoir by its neighbor makes an angle a less than 30° with the normal to the joint surface, there is no danger of rupture of the arch by the sliding of one on the other. (See Fig. 379). 348. Resistance to Crushing. — When the resultant pressure falls at its extreme allowable limit, viz. : the edge of the middle third, the pressure per unit of area at n, Fig. 380, iy double the mean pressure per unit of area. Hence, in de- signing an arch of masonry, we must be assured that at every joint (taking 10 as a factor of safety) ( Double the mean press- | ^^^^ ^^ j^^^ ^^^^ y g I ure per unit oi area \ ' Fig. 379. Fig. 380. 424 MECHA]srics of engineekikg. C being tlie ultimate resistance to crushing, of tlie material emj)loyed (§ 201) (Modulus of Crushing). Since a lamina one foot thick will always be considered in what follows, careful attention must be paid to the units employed in applying the above tests. Example. — If a joint is 3 ft. by 1 foot, and the resultant pressure is 22.5 tons the mean pressure per sq. foot is p=22.5-^3=7.5 tons per sq. foot .'. its double=15 tons per sq. foot=208.3 lbs. sq. inch, which is much less than '/lo of C for most building stones ; see § 203, and below. At joints where the resultant pressure falls at the middle, the max. pressure per square inch would be equal to the mean pressure per square inch ; but for safety it is best to assume that, at times, (from moving loads, or vibrations) it may move to the edge of the middle third, causing the max. pressure to be double the mean (per square inch). Gem Gillmore's experiments in 1876 gave the following results, among many others : NAME OF BUILDING STONE. C IN LBS. PER SQ. INCH. Berea sand-stone, 2-inch cube, - - - - 8955 4 " " - - - - 11720 Limestone, Sebastopol, 2-inch cube {chalk)^ - - 1075 Limestone from Caen, France, - - . . 3650 Limestone from Kingston, ]^. Y., - - . - 13900 Marble, Vermont, 2-inch cube, - - 8000 to 13000 Granite, New Hampshire, 2-inch cube, 15700 to 24000 349. The Three Conditions of Safe E(iiiilibriiim for an arch of uncemented voussoirs. Recapitulating the results of the foregoing paragraphs, we may state, as follows, the three conditions which must be satisfied at every joint of arch -ring and pier, for each of any possible combination of loads upon the structure : (1). The resultant pressure must pass within the middle- third, (2). The resultant pressure must not make an angle > 30° with the normal to the joint. (3). The m'^.an pressure per unit of area on the surface AKCH OF MASOK^RT. 425 of tlie joint must not exceed Ygo of the Modulus of crush- ing of the material. 350. The True Linear-Arch, or Special Equilibrium Polygon; and the resultant pressure at any joint. Let the weight of each voussoir and its load be represented by a vertical force passing through the centre of gravity of the two, as in Fig. 381o Taking any two points A and JB, A being in the first joint and B in the last ; also a third point, p, in the crown joint (supposing such to be there, although gener- ally a key-stone occupies | the crown), through these fig. ssi. three points can be drawn [§ 341] an equilibrium polygon for the loads given ; suppose this equil. polygon nowhere passes outside of the arch-ring (the arch-ring is the por- tion between the intrados, mn, and tha (dotted) extrados m'n') intersecting the joints at h, c, etc. Evidently if such be the case, and small metal rods (not round) were insert- ed at A, h, c, etc., so as to separate the arch-stones slight- ly, the arch would stand, though in unstable equilibrium, the piers being firm ; and by a different choice of A, p, and B, it might be possible to draw other equilibrium poly- gons with segments cutting the joints within the arch- ring, and if the metal rods were shifted to these new inter- sections the arch would again stand (in unstable equilib- rium). In other words, if an arch stands, it may be possible to draw a great number of linear arches within the limits of the arch-ring, since three points determine an equilibrium polygon (or linear arch) for given loads. The question arises then : luMch linear arch is the locus of the actual re- sultant pressures at the successive joints ? [Considering the arch-ring as an elastic curved beam inserted in firm piers (i.e., the blocks at the springing-line 426 MEOHAKIOS OF ENGIKEEEING. are incapable of turning) and Jbaving secured a close fit at all joints before the centering is lowered, the most satisfac- tory answer to this question is given in Prof. Greene's " Arches," p. 131 ; viz., to consider the arch-ring as an arch rib of fixed ends and no hinges ; see § 380 of next chapter;, but the lengthy computations there employed (and the method demands a simple algebraic curve for the arch) may be most advantageously replaced by Prof. Eddy's graphic method (" New Constructions in Graphical Statics," published in Van Nostrand's Magazine for 1877)„ which applies to arch curves of any form. This method will be given in a subsequent chapter, on Arch Eibs, or Curved Beams ; but for arches of masonry a much simpler procedure is sufficiently exact for practical purposes and will now be presented]. If two elastic blocks of an arch-ring touch at one edge. Fig. 382, their adjacent sides making a small angle with each •"iG- ^82. Fig. 383. other, and are then grad- ually pressed more and more forcibly together at the edge m, as the arch-ring settles, the centering being gradually lowered, the surface of contact becomes larger and larger, from the compression which ensues (see Pig. 383); while the resultant pressure between the blocks, first applied at the extreme edge m, has now probably advanced nearer the middle of the joint in the mutual adjustment of the arch- stones. With this in view we may reasonably deduce the following theory of the location of the true linear areh (sometimes called the " line of pressures " and " curve of pressure ") in an arch under given loading and with ^rm piers. (Whether the piers are really unyielding, under the oblique thrusts at the springing-line, is a matter for sub- sequent investigation. 351. Location of the True Linear Arch. — Granted that the voussoirs have been closely fitted to each other over the ARCH OF MASOXKT. 427 ■centering (sheets of lead are sometimes used in tlie joints to make a better distribution of pressure); and tliat the piers are firm ; and that the arch can stand at all without the centering ; then we assume that in the mutual accom- modation between the voussoirs, as the centering is low- ered, the resultant of the pressures distributed over any joint, if at first near the extreme edge of the joint, advances nearer to the middle as the arch settles to its final posi- tion of equilibrium under its load ; and hence the follow- ing 352. Practical Conclusions. I. If for a given arch and loading, with firm piers, an •equilibrium polygon can be drawn (by proper selection of the points A, p, and B, Fig. 381) entirely within the mid^ die third of the arch ring, not only will the arch stand, but the resultant pressure at every joint will be within the middle third (Condition 1, § 349) ; and among all possible equilibrium polygons which can be drawn within the mid- dle third, that is the " true " one which most nearly coin- •cides with the middle line of the arch-ring. II. If (with firm piers, as before) no equilibrium poly- rgon can be drawn Avithin the middle third, and only one within the arch-ring at all, the arch may stand, but chip- ping and spawling are likely to occur at the edges of the joints. The design should .*. be altered. III. If no equilibrium polygon can be drawn within the arch-ring, the design of either the arch or the loading .must be changed ; since, although the arch may standi from the resistance of the spandrel walls, such a stability must be looked upon as precarious and not countenanced in any large important structure. (Very frequently, in small arches of brick and stone, as they occur in buildings, the cement is so tenacious that the whole structure is vir- tually a single continuous mass). When the " true " linear arch has once been determined, the amount of the resultant pressure on any joint is given by the length of the proper ray in the force diagram. 428 MECHANICS OF EXGIXEEHING. ARRANGEMENT OF DATA FOR GRAPHIC TREATMENT. 353. Cli-aracter of Load. — In most large stone arch bridges the load (permanent load) does not consist exclusively of masonry up to tlie road-way but partially of earth filling above the masonry, except at the faces of the arch where the spandrel walls serve as retaining walls to hold the earth. (Fig. 384). If the intrados is a half circle or half- Fig. 385. Fig. 384. ellipse, a compactly -built masonry backing is carried up beyond the springing-line to AB about 60° to 45° from the crown. Fig. 385 ; so that the portion of arch ring below AB may be considered as part of the abutment, and thus AB is the virtual springing-line, for graphic treatment. Sometimes, to save filling, small arches are built over the haunches of the main arch, with earth placed over them, as shown in Fig. 386. In any of the preceding cases Fig. 387. it is customary to consider that, on account of the bond- ing of the stones in the arch shell, the loading at a given distance from the crown is uniformly distributed over the width of the roadway. AECHES OF MASONET. 429 354, Reduced Load-Contour. — In the graphical discussioa of a proposed arch we consider a lamina one foot thick, this lamina being vertical and "| to the axis of the arch ; i.e., the lamina is || to the spandrel walls. For graphical treatment, equal areas of the elevation (see Fig, 387) of this lamina must represent equal weights. Taking the material of the arch-ring as a standard, we must find for each point "p of the extrados an imaginary height z of the arch-ring material, which would give the same pressure (per running horizontal foot) at that point as that due to the actual load above that point. A number of such or- dinates, each measured vertically upward from the extra- dos determine points in the "Reduced Load-Contour," i.e., the imaginary line, AM, the area between which and the extrados of the arch-ring represents a homogeneous load of the same density as the arch -ring, and equivalent to the actual load (above extrados), vertical hy vertical. 355. Example of Reduced Load-Contour. — Fig, 388. Given an arch-ring of granite (heaviness = 170 lbs. per cubic foot) with a dead load of rubble (heav. = 140) and earth (heav. = 100), distributed as in figure. At the point p, of the extrados, the depth 5 feet of rubble is equivalent to a depth of [^^ x5]=4.1 ft. of granite, while the 6 feet of earth is equivalent to [l°?x6]=3.5 feet of granite. Hence the Reduced Load-Contour has an ordinate, above p, of 7.6 feet. That is, for each of several points of the arch -ring extrados- reduce the rubble ordinate in the ratio of 170 : 140, and the earth ordinate in the ratio 170 : 100 and add the re- sults, setting off the sum vertically from the points in the ■extrados*. In this way Fig. 389 is obtained and the area *TUs is most conveniently done by graphics, thus : On a right-line set off 17 equal. parts (of any convenient magnitude.) Call this distance OA. Through t> draw another right line at any convenient angle (30° to 60°) with OA, and on it from O set off OB equal to 14 (for the ruhble ; or 10 for the earth) of the eame egaal parts. Join AB. From O toward A set off* all the rubble ordiaates to be reduced^ (each being set off from 0} and through the other extremity of each draw a Bne par- allel to AB. The reduced ordinates will be the respective lengths, from O, along OB, to the intersections of these parallels ynth OB. * Witli the dividers. 430 MBCHAISICS OF ENGINEERING. :/EART.HV; Av,*//%V;*i*»'v5i;'i!lV;?V/;i;;*'uVf/-^;;-';^^ there given is to be treated as representing liomogeneous granite one foot thick. This, of course, now includes the arch-ring also. AB is the " reduced load- contour." 356. Live Loads. — In discussing a railroad arch bridge the " live load " (a train of locomotives, e.g., to take an ex- treme case) can not be disregarded, and for each of its po- sitions we have a separate Reduced Load-Contour. Example. — Suppose the arch of Fig. 388 to be 12 feet wide (not including spandrel walls) and that a train of lo- comotives weighing 3,000 lbs. per running foot of the track covers one half of the span. Uniformly * distributed later- ally over the width, 12 ft., this rate of loading is equiva- lent to a masonry load of one foot high and a heaviness of 250 lbs. per cubic ft., i.e., is equivalent to a height of 1.4 ft. of granite masonry [since ^[|- X 1.0 — 1.4] over the half span considered. Hence from Fig. 390 we obtain Fig. 391 in an obvious manner. Fig. 391 is now ready for graphic treatment. Fig. 390. Fig. 391. 357. Piers and Abutments. — In a series of equal arches the pier between two consecutive arches bears simply the weight of the two adjacent semi-arches, plus the load im- * If the earth-filling is sLallcw, the Icminse directly under the track prob* aibly receive a greater pressure than the others. AKCHES OF MASONRY. 431 mediately above the pier, and .-. does not need to be as large as the abutment of the first and last arches, since these latter must be prepared to resist the oblique thrusts of their arches without help from the thrust of another on the other side. In a very long series of arches it is sometimes customary to make a few of the intermediate piers large enough to act as abutments. These are called " abutment piers," and in case one arch should fall, no others would be lost except those occurring between the same two abutment piers as the first. See Fig. 392. A is an abutment-pier. Fig. 39;^. GRAPHICAL. TREATMENT OF ARCH. 358. — Having found the " reduced load-contour," as in preceding paragraphs, for a given arch and load, we are ready to proceed with the graphic treatment, i.e., the first given, or assumed, form and thickness of arch-ring is to be investigated with regard to stability. It may be necessary to treat, separately, a lamina under the spandrel wall, and one under the interior loading. The constructions are equally well adapted to arches of all shapes, to Gothic as well as circular and elliptical. 359.— Case I. Symmetrical Arch and Symmetrical Loading.— (The " steady " (permanent) or " dead " load on an arch is usually symmetrical). Fig. 393. From symmetry we need Fig. Fig. 394. Fis. 395. i33 MECHAJTICS OF ENGINEERmG. deal witli only one half (say the left) of tlie arch and load. Divide this semi-arch and load into six or ten divisions by vertical lines ; these divisions are considered as trape- zoids and should have the same horizontal width = 6 (a convenient whole number of feet) except the last one, LKN, next the abutment, and this is a pentagon of a different v\ridth hy, (the remnant of the horizontal distance LC). The weight of masonry in each division is equal to (the area of division) X (unity thickness of lamina) x (weight of a cu- bic unit of arch-ring). For example for a division having an area of 20 sq. feet, and composed of masonry weighing 160 lbs. per cubic foot, we have 20x1x160=3,200 lbs., applied through the centre of gravity of the division. The area of a trapezoid. Fig. 394, is^&(7ii+7i2), audits cen- tre of gravity may be found. Fig. 395, by the construction of Prob. 6, in § 26 ; or by § 27a. The weight of the pen- tagon LN, Fig, 393, and its line of application (through centre of gravity) may be found by combining results for the two trapezoids into which it is divided by a vertical through K. See § 21. Since the weights of the respective trapezoids {excepts ing LN) are proportional to their middle vertical in- tercepts [such as ^(^1+7^2) Fig- 394] these intercepts (trans- ferred with the dividers) may be used directly to form the load-line, Fig. 396, or proportional parts of them if more convenient. The force scale, which this implies, is easily computed,, and a proper length calculated to represent the weight of the odd division LN ; i.e., 1 ... 2 on the load- line. Now consider A, the middle point of the abutment joint. Fig. 396, as the starting point of an equilibrium polygon (or abutment of a linear arch) for a given loading, and re- quire that this equilibrium polygon shall pass through j>, the middle of the crown joint, and through the middle of the abutment joint on the right (not shown in figure). Proceed as in § 342, thus determining the polygon Ap for the half-arch. Draw joints in the arch-ring through those points where the extrados is intersected by the ver- ABCHES OF MASONRY. 433 Jig. 396. Fig. 397. Heal separating tlie divisions (not the gravity verticals), Tlie points in which these joints are cut by the segments of the equilibrium polygon, Fig. 397, are (very nearly, if th« joint is not more than 60° from jp, the crown) the points of application in these joints, respectively, of the resultant pressures on them, (if this is the " true linear arch " for this arch and load) while the amount and direction of each such pressure is given by the proper ray in the force -dia- gram. If at any joint so drawn the linear arch (or equilibrium polygon) passes outside the middle third of the arch-ring, the point A, or p, (or both) should be judiciously moved (within the middle third) to find if possible a linear arch which keeps within limits at all joints. If this is found impossible, the thickness of the arch -ring may be increased at the abutment (giving a smaller increase toward the crown) and the desired result obtained ; or a change in the distribution or amount of the loading, if allowable, may gain this object. If but one linear arch can be drawn within the middle third, it may be considered the " true " one ; if several, the one most nearly co-inciding with the middles of the joints (see §§ 351 and 352) is so considered. 360.— Case II Unsymmetrical Loading on a Symmetrical Arch,' (e.g., arch with live load covering one half -span as in Figs. 390 and 391). Here we must evidently use a full force diagram, and the full elevation of the arch -ring and load* 434 MBCHAXICS OF EXGINEEELNG. See Fig. 398. Select three points A, p, and B, as follows, to determine a trial equilihriu'm ])6lygon : ' Select A at the Joicer limit of the middle third of tLa Fig. 398. abutment-joint at the end of the span -vihich is the more heavily -loaded ; in the other abutment-joint take B at tht upper limit of the middle third ; and take p in the middle of the crown-joint. Then by § 341 draw an equilibrium polygon (i.e., a linear arch) through these three points for the given set of loads, and if it does not remain within the middle third, try other positions for A, p, and B, within the middle third. As to the " true linear arch " alterations of the design, etc., the same remarks apply as already given in Case I. Very frequently it is not necessary to draw more than one linear arch, for a given loading, for even if one could be drawn nearer the middle of the arch- ring than the first, that fact is almost always apparent on mere inspection, and the one already drawn (if within middle third) will furnish values sufiiciently accurate for the pressures on the respective joints, and their direction angles. 360a. — The design for the arch-ring and loading is not to be considered satisfactory until it is ascertained that for the dead load and any possible combination of live-load '(in addition) the pressure at any joint is ARCHES OF MASONRY. , 435 1.) Witliin the middle third of that joint ; {2.) At an angle of < 30° with the normal to joint- SYirface. (3.) Of a mean pressure per square inch not > thanVa) of the ultimate crushing resistance. (See § 348.) § 361. Abutments. — The abutment should be compactly and solidly built, and is then treated as a single rigid mass. The pressure of the lowest voussoir upon it (considering a lamina one foot thick) is given by the proper ray of the force diagram (0 .. 1, e. g., in Fig. 396) in amount and direc- tion. The stability of the abutment will depend on the amount and direction of the resultant obtained by com- bining that pressure P^ with the weight G of the abutment and its load, see Fig. 399. Assume a probable width RS for the abutment and compute the weight G of the corresponding abutment OBRS and MNBO, and find the centre of gravity of the whole mass G. Apply G in the vertical through C, and combine it with P„ at their in- tersection D. The resultant P should not cut the base R8m. a point beyond the middle third p/^ / " (or, if this rule gives too massive a pier, take I / / ° such a width that the pressure per square I// inch at 8 shall not exceed a safe value as ^ Fis. 399. computed from § 362.) After one or two trials a satisfactory width can be obtained. We should also be assured that the angle PD G is less than 30°. The horizontal joints above RS should also be tested as if each were, in turn, the lowest base, and if nscessary may be inclined (like mn) to prevent slipping. On no joint should the maximum pressure per square inch be > than y^o the crushing strength of the cement. Abut- ments of firm natural rock are of course to be preferred where they can be had. If water penetrates under an abutment its buoyant effort lessens the weight of the lat- ter to a considerable extent. 436 MECHANICS OF ENGINEEKIifG. 362. Maximum Pressure Per Unit of Area When the Resultant Pressure Falls at Any Given Distance from the Middle ; according to Navier's theory of the distribution of the pressure ; see § 346. Case I. Let the resultant pressure P, Fig. 400, (a). Fig. 400. Fig. 401. fall within the middle third, a distance = wc? ( < ^ d) from the middle of joint [d = depth of Joint.) Then we have the following relations : p (the mean press, per.- sq. in.,),^,,, (max. press, persq. in.), and p^ (least press, per sq. in.) are proportional to the lines h (mid. width), a (max. base), and c (min. base) respectively, of a trapezoid. Fig. 400, (&), through whose centre of gravity P acts. But (§ 26) • nd=---. i.e., n = y^ — = — or a=h (6w+l) 6 a-\-c n . ''• Pm—JP (6w+l). Hence the following table: If 7id= j4> d Ya d press. Pn,= 2 y^ Vs then the max. times the mean pressure. Case II. Let P fall outside the mid. third, a distance = "nd {^ )4 d) from the middle of joint. Here, since the joint is not considered capable of withstanding tension, we have a triangle, instead of a trapezoid. Fig. 401. First compute the mean press, per sq. in. V - P (lbs.) (1— 2w) 18 d inches foot thick). or from this table : (lamina ona AKOHES OF MASOlfEY. 437 For nd = ^d ■hd ■hd T\d ^d ^d P = 1 P 10* d 1 P 8 * d 1 P 6 * d' 1 P 4 <^ 1 P 2 6^ infinity. {d in inches and P in lbs.; with arch lamina 1 ft. in thickness.) Then the maximum pressure (at A, Fig. 401) />,„, = 2p, becomes known, in lbs. per sq. in. 362a. Arch-ring under Uon- vertical Forces. — An example of this occurs when a vertical arch-ring is to support the pressure of a liquid on its extrados. Since water-pressures are always at right angles to the surface pressed on, these pressures on the extradosal surface of the arch-ring form a system of non-paral~ lei forces which are normal to the curve of the extrados at; their respective points of application and lie in parallel vertical planes, parallel to the faces of the lamina. "We here assume that the extradosal surface is a cylinder (in the most general sense) whose rectilinear elements are 1 to the faces of the lamina. If, then, we divide the length of the extrados, from crown to each abutment, into from six to ten parts, the respective pressures on the corresponding surfaces are obtained by multiplying the area of each by the depth of its centre of gravity from the upper free surface of the liquid, and this product by the weight of a unit of volume of the liquid ; and each such pressure may be considered as acting through the centre of the area. Finally, if we find the resultant of each of these pressures and the weight of the corresponding portion of the arch-ring, these resultants form a series of non-vertical forces in a plane, for which an equilibrium polygon can then be passed through three assumed points by § 378a, these three points being taken in the crown-joint and the two abutment- joints. As to the " true linear arch" see § 359. As an extreme theoretic limit it is worth noting that if the extrados and intrados of the arch-ring are concentric circles ; if the weights of the voussoirs are neglected ; and if the rise of tb« arch is very small compared with the depth of the crown ^/elow the water surface, then the circularGentre-line of the wrch-ring is the " true linear archP 4:38 MECHAIiflCS OF ENGLNiiJEJiLNG. CHAPTER XI. ARCH-RIBS. Note. — The methods used in this chapter for the treatment of the "elastic arch" are practically the equivalent of those based on the theory of "Least Work." 364. Definitions and Assumptions. — An arcli-rib (or elastic- arch, as distinguished from a block-work arcb) is a rigid curved beam, either solid, or built up of pieces like a truss (and then called a braced arch) the stresses in which, under a given loading and with prescribed mode of sup- port it is here proposed to determine. The rib is sup- posed symmetrical about a vertical plane containing its axis or middle line, and the Moment of Inertia of any cross section is understood to be referred to a gravity axis of the section, which (the axis) is perpendicular to the said vertical plane. It is assumed that in its strained condi- tion under a load, the shape of the rib differs so little from its form when unstrained that the change in the ab- scissa or ordinate of any point in the rib axis (a curve) may be neglected when added (algebraically) to the co- ordinate itself ; also that the dimensions of a cross-section are small compared with the radius of curvature at any part of the curved axis, and with the span. 365. Mode of Support. — Either extremity of the rib may be hinged to its pier (which gives freedom to the end-tangent- line to turn in the vertical plane of the rib when a load is applied); or may 'hef,xed, i.e., so built-in, or bolted rigid- ly to the pier, that the en^-tangent-line is incapable of changing its direction when a load is applied. A hinge may be inserted anywhere along the rib, and of course ARCH BIBS. 439 destroys the rigidity, or resistance to bending at that point. (A hinge having its pin horizontal "1 to the axis of the rib is meant). Evidently no more than three such hinges could be introduced along an arch- rib between two piers ; unless it is to be a hanging structure, acting as a suspension-cable. 366. Arch Rib as a Free Body. — In considering the whole rib free it is convenient, for graphical treatment, that no section be conceived made at its extremities, if fixed ; hence in dealing with that mode of support the end of the rib will be considered as having a rigid prolongation reach- ing to a point vertically above or below the pier junction, an unknown distance from it, and there acted on by a force of such unknown amount and direction as to preserve the actual 'extremity of the rib and its tangent line in the same position and direction as they really are. As an illustra- tion of this Fig. 402 shows free an arch rib. ONB, with its extremi- ties and BJixed in the piers, with no hinges, q and bearing two loads P. and P^. The other . :ces of the sys- tem holding it in equi- librium are che horizontal and vertical components, of the pier reactions {H, V, H,„ and V^), and in this case of fixed ends each .of these two reactions is a single force not in- tersecting the end of the rib, but cutting the vertical through the end in some point F (on the left ; and in G on the right) at some vertical distance c, (or d), from the end. Hence the utility of these imaginary prolongations OQF, and BRG, the pier being supposed removed. Compare Figs. 348 and 350. The imaginary points, or hinges, F and G, will be called ctbutments being such for the special equilibrium polygon Fig. 402. 440 MBCHAlSriCS OF ENGINEEKING. (dotted line), while and B are the real ends of the curved beam, or rib. In this system of forces there are five unknowns, viz.: V, V,„ H = H^, and the distances c and d. Their determina- tion by analysis, even if the rib is a circular arc, is ex-, tremely intricate and tedious ; but by graphical statics (Prof, Eddy's method ; see § 350 for reference), it is com- paratively simple and direct aiid applies to any shape of rib, and is sufficiently accurate for practical purposes. This method consists of constructions leading to the loca- tion of the " special equilibrium polygon " and its force diagram. In case the rib is hinged to the piers, the re- actions of the latter act through these hinges, Fig. 403, i e., the abutments of the special equilibrium polygon coincide with the ends of the rib and B, and for a given rib and load the unknown quantities are only three V, F'n, and H; (strictly there are four ; but IX "^^ = gives H^ = H). The solution fig. 403. by analytics is possible only for ribs of simple algebraic curves and is long and cumbrous ; 'whereas Prol Eddy's graphic method is comparatively brief and simple and ia applicable to any shape of rib whatever. 367. Utility of the Special Equilibriun Polygon and its force diagram. The use of locating these will now be illustrated [See § 832]. As proved in §§ 332 and 334 the compres- sion in each " rod " or segment of the '* special equilibrium polygon" is the anti-stress resultant of the cross sections in the corresponding portion of the beam, rib, or other struc- ture, the value of this compression (in lbs. or tons) being measured by the length of the parallel ray in the force diagram. Suppose that in some way (to be explained sub- sequently) the special equilibrium polygon and its force diagram have been drawn for the arch -rib in Fig. 404 hav- ing fixed ends, and B, and no hinges ; required the elastic stresses in any cross-section of the rib as at m. Let the ARCH RIBS. Ml FiG. 404. of the force-diagram on the right be 200 lbs. to the inch, say, and that of the space-diagram (on the left) 30 ft. to the inch. The cross section m lies in a portion TK, of the rib, cor- responding to the rod or segment he of the equilibrium polygon; hence its anti-stress-resultant is a force R2 acting in the line 6c, and of an amount given in the force-diagram. Now i?2 is the resultant of V, H, and Pj, which with the elastic forces at m form a system in equilibrium, shown in ?ig. 405 ; the portion FO Tm being considered free. Hence Pig. 405. Fio. 406. taking the tangent line and the normal at m as axes we should have I (tang, comps.) = ; -T (norm, comps.) = j and 2* (moms, about gravity axis of the section at w) = Oj and could thus find the unknowns pi, "p^, and J", which ap- pear iu the expressions 'p^F the thrust, ^ the moment* of 442 MECHANICS OF ENGINEERING. the stress-couple, and J the shear. These elastic stresses are classified as in § 295, which see. p^ and jpa are ^hs. per square inch, J is lbs., e is the distance from the horizontal gravity axis of the section to the outermost element of area, (where the compression or tension is p^ lbs. per sq. in., as due to the stress-couple alone) while I is the " mo- ment of inertia " of the section about that gravity axis. [See §§ 247 and 295 ; also § 85]. Graphics, however, gives us a m.ore direct method, as follows : Since i?2> i^ the line he, is the equivalent of V, H, and Pj, the stresses at m will be just the same as if ^2 acted directly upon a lateral pro- longation of the rib at T (to intersect ScFig. 405) as shown in Fig, 406, this prolongation Tb taking the place of TOF in Fig. 405. The force diagram is also reproduced here. Let a denote the length of the "] from m's gravity axis upon he, and 2 the vertical intercept between m and Jc. For this imaginary free body, we have, from I (tang. compons.)=0, i?2 cos a=^piF and from 2' (norm. compons.)=0,i?2 sin «=«/ while from J' (moms, about) ) ^ ^ rj P2I ,-, ., ^- 4; \ A }-we have ixott =^ -^ • the gravity axis 01 to)=0, j ^ e But from the two similar triangles (shaded ; one of them is in force diagram) a :z :; Zf:i?2 .•. R^a^ Hz, wIh&tigq we may rewrite these relations as follows (with a general state- ment), viz.: If the Special Equilibrium Polygon and Its Force Diagram Have iBeen Drawn for a given arch-rib, of given mode of support, p.nd under a given loading, then in any cross-section of the J ib, we have {F = area of section): The projection of the proper i \ \ rri.^T>.,m.f i^ ^ rr- J ^«2/ (of tlie force diagram) up- (L) The Thrust. i.e.,i>,i^-^ ^^^^^ ^^^^^^^ ^.^^ ^^ ^-^^ ^i^, drawn at the given section. ARCH KIBS. 443 (2.) The Shear, i.e., J", = C / 1-11 1 J.1 The proieetion of me proper (upon which dependstne , » ,i « -,. ^ ^, . , . ,1 ray (oi the lorce diaarram) up- shearmg stress m the-^ "^ .; ? , i^i -i web). (See §§ 253 and 256). on the normal to the rib curve at the given section. (3.) The Moment of the stress couple, i.e.,-^ , = " 6 The product {Hz) of the fl (or pole-distance) of the force- diagram by the vertical dis- tance of the gravity axis of the section from the spec, equilib- rium polygon. By the ** proper ray " is meant that ray which is parallel to the segment (of the equil. polygon) immediately under or above which the given section is situated. Thus in Fig. 404, the proper ray for any section on TK is B2 ; on KB, i?3 ; on TO, Bi. The projection of a ray upon any given tangent or normal, is easily found by drawing through each end of the ray a line ^ to the tangent (or normal) ; the length between these "I's on the tangent (or normal) is the force required (by the scale of the force diagram). We may thus construct a shear diagram, and a thrust diagram for a given case, while the successive vertical intercepts between the rib and special equilibrium polygon form a moment diagram. For example if the s of a point m is ^ inch in a space diagram drawn to a scale of 20 feet to the inch, while Zf measures 2.1 inches in a force diagram con- structed on a scale of ten tons to the inch, we have, for the moment of the stress-couple at m, J!f=^s= [2.1x10] tons X [ ^ X 20] ft. =210 ft. tons. 368. — It is thus seen how a location of the special equili Ibrium polygon, and the lines of the corresponding iovoi diagram, lead directly to a knowledge of the stresses in al the cross-sections of the curved beam under consideration, bearing a given load; or, vice versa, leads to a stateme?^^ of conditions to be satisfied by the dimensions of the rir for proper security. It is here supposed that the rib has sufficient latei 444 MECHANICS OF ENGlNEElil^rG, bracing (witli others wliicli lie parallel witli it) to prevent buckling sideways in any part like a long column. Before proceeding to the complete graphical analysis of the differ- ent cases of arch-ribs, it will be necessary to devote the next few paragraphs to developing a few analytical rela- tions in the theory of flexure of a curved beam, and to giving some processes in " graphical arithmetic." 369. Change in the Angle Between Two Consecutive Rib Tan- gents when the rib is loaded, as compared with its value before loading. Consider any small portion (of an arch rib) included between two consecutive cross- sections ; Fig. 407. KHG W is its unstrained form. Let EA, = ds, be the original length of this portion of the rib axis. The length of all the fibres (ii to rib-axis) was originally =ds (nearly) and the two consecutive tangent-lines, at E and Ay made an angle = dO originally, with each other. While under strain, however, all the fibres are shortened equally an amount dX^, by the uniformly distributed tangential thrust, but are unequally shortened (or lengthened, accord- ing as they are on one side or the other of the gravity axis E, or A, of the section) by the system of forces making what we call the " stress couple," among v/hich the stress at the distance e from the gravity axis A of the section is called p-i per square inch ; so that the tangent line at A' now takes the direction A'D ~j to H'A'G' instead of A'G (we suppose the section at E to remain fixed, for coUTezii- ^6! =^ t/"/^'^' cIp.^' .-r- "'v-^.^f** ^pd? AECH lilBS. 445 ence, since tlie change of angle between tlie two tangents depends on the stresses acting, and not on tlie new posi- tion in space, of this part of the rib), and hence the angle between the tangent-lines at E and A (originally = dd) is now increased by an amount GA'D = d(p (or O'A'R = dip); G'H' is the new position of GH. We obtain the value of d(p as follows : That part {dk^ of the shortening of the fibre at Q, at distance e from A due to the force p.^dFy is § 201 eq. (1), dX.2 = ^ft' But, geometrically, J^ also ~edf, Eedcp-^pzds ■ (1.) But, letting ilf denote the moment of the stress-couple at section A (ilf depends on the loading, mode of support, etc., in any particular case) we know from § 295 eq. (6) that Jf=-^j and hence by substitution in (1) we. have •, Mds r^x '^^^I . • . . (2) [If the arch-rib in question has less than three hinges, the equal shortening of the due to the thrust (of the block in last figure) p^F, will have an indirect effect on the angle d(p. This will be considered later.] 370. Total Change i.e. CcU in the Angle Between the End Tangents of a Rib, before and after loading. Take the ex- ample in Fig. 408 of a rib fixed at one end and hinged at Fig. 408. 446 MECHAT^ICS OF ENGINEEEIJIG. the other. When the rib is unstrained (as it is supposed to be, on the left, its own weight being neglected ; it is not supposed sprung into place, but is entirely without strain) then the angle between the end-tangents has some value 6' = j dd— the sum of the successive small angles dd for do each element ds of the rib curve (or axis). After loading., [on the right, Fig. 408], this angle has increased having now a value d'-\- r d(p, i.e., a value = d'+ C -—jr (I.) Fig. 409. There must oe no hinge between and B. § 371. Example of Eq[iiation (I.) in Anal" ysis. — ^A straight, homogeneous, pris- matic beam, Fig. 409, its own weight neglected, is fixed obliquely in a wall. After placing a load P on the free end, required the angle between the end- tangents. This was zero before load- ing .'. its value after loading is =o+f'=o+ 4r r'-^^^^ UIJo By considering free a portion between and any da of the beam, we find that M=Fx=mom.. of the stress couple. The flexure is so slight that the angle between any ds and its dx is still practically =a (§ 364), and .*. ds=dx sec a. Hence, by substitution in eq. (I.) we have ^'=A rms= l^rxdx= ^ EI Jo EI Jo P sec ar'*'^* L2 ' EI ... ^'=?^^^l [Compare with § 237]. ARCH EIBS. 447 It is now apparent that if hoth ends of an arcli rib are fixed, wlien unstrained, and the rib be then loaded (within elastic limit, and deformation slight) we must have r {Mds-^Eiy zero, since (p'=0. 372. Projections of the Displacement of any Point of a Loaded Uib Relatively to Another Point and the Tangent Line at the Lat- ter. — (There must be no hinge between and B). Let be the point whose displacement is considered and B the other point. Fig. 410. If ^'s tangent-line is fixed while the extremity is not supported in any way (Fig. 410) then a load P put on, is displaced to a new position 0^, Fig. 410. Fig. 411, Fig; 413. With as an origin and OB as the axis of X, the projec- tion of the displacement OOj, upon X, will be called Ja?, that upon Y, Ay. In the case in Fig.. 410, O's displacement with respect to B and its tangent-line BT, is also its absolute displacement in space, since neither B nor BT has moved as the rib changes form under the load. In Fig. 411, however, the extremities and B are both hinged to piers, or supports, the dotted line showing its form when deformed under a load. The hinges are supposed immovable, the rib being free to turn about them without friction. The dotted line is the changed form under a load, and the absolute dis- placement of is zero ; but not so its displacement rela- tively to B and j5's tangent BT, for BT has moved to a new position BT'. To find this relative displacement con- ceive the new curve of the rib superposed on the old in a way that B and BT m&j coincide with their original po- 448 MECHANICS OF ENGIIsTEEEING. sitions. Fig. 412. It is now seen that O's displacement relatively to B and BT is not zero but =00„, and lias a small Jx but a comparatively large zly. In fact for this case of hinged ends, piers immovable, rib continuous be- tween 'them, and deformation slight, we shall write Jx = zero as compared with Jy, the axis X passing through OB). 373. Values of the X and Y Projections of O's Displacement Rela- tively to Band B's Tangent; the origin being taken at 0. Fig. 413. Let the co- ordinates of the dif- ferent points jE, I), G, etc., of the rib, re- ferred to and an arbitrary X axis, be X and y, their radial distances from be- . ing u (i.e., u for G, u' for D, etc.; in gener- al, ^0- OEDG is the J unstrained form of the rib, (e.g., the form it would assume if it lay flat on its side on a level platform, under no straining forces), while 0,,E"B'GB is its form under some loading, i.e., under strain. (The superposi- tion above mentioned (§ 372) is supposed already made if necessary, so that BT i^ tangent at B to both forms). Now conceive the rib OB to pass into its strained condi- tion by the successive bending of each ds in turn. The straining or bending of the first ds, BC, through the small angle d(p (dependent on the moment of the stress couple at G in the strained condition) causes the whole finite piece OG io turn about (7 as a centre through the same small angle d(p ; hence the point describes a small linear arc 00'=ov, whose radius = u the hypothenuse of the x and y of G, and whose value .*. is dv=ud(p. Next let the section B, now at D\ turn through its proper angle d(p' (dependent on its stress-couple) carrying Fig. 413. AECH KIBS. ' 449 with it the portion D'O', into the position D'O", making 0' describe a linear arc O'O" = {dvy =u'd(p', in which u' = the hypothenuse on the x' and y' (of D), (the deformation is so slight that the co-ordinates of the different points referred to and X are not appreciably affected). Thus, each section having been allowed to turn through the an- gle proper to it, finally reaches its position, 0„, of dis- placement. Each successive dv, or linear arc described by 0, has a shorter radius. Let dx, {dx)', etc., represent the projections of the successive (^v)'s upon the axis X; and similarly dy, (dy)' etc., upon the axis Y. Then the total X projection of the curved line . . . . 0^ will be Jx= / (5j? and similarly J?/ = / dy , , , (1) But d V = u d (f, and from similar -right-triangles, 3 x: dv : : y : u and dy : 8v :: x : u .'. 8x = yd<p and dy=xd<p ; whence, (see (1) and (2) of §369) Ax = fS. = fyd^=£JMl... (IL) and Ay = C dy = C xd(p = C ^^ . . . , (III.^ If the rib is homogeneous E is constant, and if it is of constant cross-section, all sections being similarly cut by the vertical plane of the rib's axis (i.e., if it is a " curved prism ") /, the moment of inertia is also constant. 374. Hecapitulation of Analytical Relations, for reference* (Not applicable if there is a hinge between and E) Total Change in Angle between ) _ ^^Mds tai Lfiiaiige lu Aiigie oerween / _ p>"m.as .j ^ tangent-lines and ^ [ ~ Jo ^ ' o • • W The X-Projection of O's Displacement "] Relatively to B and B's tangent- I A^Myds /tt ^ line ; {the origin being at 0) I — / — ^fjr- • • • (JJ-J and the axes X and F 1 to [ ^o i^i each other) • 450 MECHAKICS OF EKGINEEKING. The Y-Prqjection of O's Displacement, | _ etc., as above. . (m.) Hviie X anv.. y are the co-ordinates of points in the rib- curve, ds an element of that curve, M the moment of the stress-couple in the corresponding section as induced by the loading, or constraint, of the rib. (The results already derived for deflections, slopes, eto„, for straight beams, could also be obtained from these formulae, I., 11. and III. In these formulae also it must be remembered that no account has been taken of the shortening of the rib-axis by the thrust, nor of the effect of a change of temperature.) 374a. R^sumfe of the Properties of Ec^mlibiiiim Polygons and their Force Diagrams, for Systems of Vertical Loads. — See §§ 335 to 343. Given a system of loads or vertical forces, P^, P2, 1 etc., Eig. 414, and two abutment verti- cals, F' and G' ; if "we lay off, vertically, to form a " load- line," 1 .. 2 - P^, 2. . . 8=P2> etc., select any Pole, Oi, and join 0^ ... 1, Oi . . . 2, etc. ; also, beginning at any point F^ in the vertical P', if we draw i^i . . . a I| to Oj . . 1 to intersect the line of Pi ; then ah \\ to Oi . . 2, and so on until finally a point G]y in G', is determined; then the figure Pj ,ahc G^iis an equilibrium polygon for the given loads and load verti- cals, and Oi . . . 1234 is its ". force diagram." The former is so called because the short segments PjCt oh, etc., if considered to be rigid and imponderable rods, in a vertical plane, hinged to each other and the terminal ones to abut- ments Pi and G^, would be in equilibrium under the given loads hung at the joints. An infinite number of equilil> Fig. 414. AECH-HIBS. 451 rium polygons may be drawn for tlie given loads and abntment-verticals, by choosing different poles in the force diagram. [One other is shown in the fignie ; O2 is its pole. {Fi Gi and F2 U-^ are abutment lines.)] For all of these the following statements are true : (1.) A line through the pole, i| to the abutiifint line cutii the load-line in the same point n', whichever equilibrium polygon be used ( /. any one will serve to determine n' (2.) If a vertical GI) he drawn, giving an intercept z' in each of the equilibrium polygons, the product Hz' is the same for all the equilibrium polygons. That is, (see Fig., 414) for any two of the polygons we have H,:H,:: z/ : z,' ; or H,z,' = H, z,'. (3.) The compression in each rod is given by thai " ray " (in the force diagram) to which it is parallel. (4.) The " pole distance " H, or ~| let fall from the pole upon the load-line, divides it into two parts which are the vertical components oi the compressions in the abutment- rods respectively ( the other component being horizontal) ; H is the horizontal component of each (and, in fact, of each of the compressions in all the other rods). The compressions in the extreme rods may also be called the abutment reactions (oblique) and are given byti^e extreme rays. (5.) Three Points [not all in the same segment (or rod)] determine an equilibrium polygon for given loads. Hav- ing given, then, three points, we may draw the eaailibrium polygon by §341. « 375. Summation of Prcducts. Before proceedini^ to treat graphically any case of arch-ribs, a few processes in graphical arithmetic, as it may be called, must be pre- sented, and thus established for future use. To make a summation of products of two factors in each by means of an equilibrium polygon. 452 MECHANICS OF ENGINEERING. Construction, Suppose it required to make the summa- tion I {x z) {. e., to sum the series Xi %+ X2 Z2 + x^z^ 4. bj graphics. Having first arranged the terms in the order of magnf- tude of the ic's, we proceed as follows : Supposing, for illustration, that two of the s's (% and z^ are negative (clotted in figure) see Fig. 415. These quantities x and z may be of any nature whatever, anything capable of being represented by a length, laid off to scale. First, in Fig 416, lay off the s's in their order, end to end, on a ver- tical load-line taking care to "^ lay off % and .. % upuard in ^« their turn. Take any con- FiG. 416. venient j-ola ; draw the rays ... 1, ... 2, etc.; then, having pre- viously drawn vertical lines whose horizontal distances from an extreme left-hand vertical F' are made = x^, x-, Xs, etc., respectively, we begin at any point F, in the verti- cal F', and draw a line 11 to ... 1 to intersect the Xi ver- tical in some point ; then 1' 2' II to . . . % and so on, fol- lowing carefully the proper order. Produce the last seg- ment (6' ... (x in this case) to intersect the vertical F' in some point K. Let KF =k (measured on the same scale as the i»'s), then the summation required is J/ {xz) = m. H is measured on the scale of the 2's, which need, not be the same as that of the aj's ; in fact the 2's may not be the same kind of quantity as the a;'s. [Peoof. — From similar triangles H: z^v.x^^: h^, .'. x^z^ — IHc^ \ and " " " H\Zo :: x^ : ^2> •*• x.^Zi=Hki . Fig. 415. AECH-KIBS. 453 and so on. But H {h,+h+eiG.)^HxFK=Hh']. 376. Gravity Vertical. — From the same construction in Fig. 415 we can determine tlie line of action (or gravity vertical) of tlie resultant of the parallel vertical forces 2i, Z2, etc. (or loads); by prolonging the first and last segments to their intersection at 0. The resultant of the system of forces or loads acts through C and is vertical in this case ; its value being = ^ (2), that is, it = the length 1 ... 7 in the force dia- gram, interpreted by the proper scale. It is now supposed that the 2's represent forces, the x'b, being their respective lever arms about F. If the ?'s represent the areas of small finite por- tions of a large plane figure, we may find a gravity -line (through C) of that figure by the above construction; each z being-applied through the centre of gravity of its own portion. Calling the distance X between the verticals through C and F, we have also x . I [z) = I (xz) because I (z) is the resultant of the || z's. ^' This is also evident from the proportion (similar triangles) H : (1. .7)::x:Jc. 454 MECHANICS OF ENGINEERING. 376a. Moment of Inertia (of Plane Figure) by Graphics.* — rig. 416a. /n= ? First, for the portion on right. Divide OR into equal parts each = ^x. Let «i, Z2, etc., be the middle ordinates of the strips thus obtained, and x^, etc. their abscissas (of middle points). Then we have approximately /n for 0R=Ax.ZyX^-\-Ax.Z2xi-{- =Ax[{z^Xi)x^+{z2X^X2-\- ...]..(!) But by §375 we may construct the products ZyXi,Z2X2, etc., taking a convenient H\ (see Fig. 416, (&)), and obtain \, ki, etc., such that z^x^ = H'ki, z^x^ = H'k2i etc. Hence eq. (1) becomes : li,ioT OR a.-p-prox.=II'^x[kiXi-\-k2X2-{- ...]... (2) By a second use of § 375 (see Fig. 416 c) we construct Z, such that kiX^ + kzXz +....= £["l \^H" taken at con- venience]. .'. from eq. (2) we have finally, (approx.), In for OR=H'H"lAx (3) For example if OR — 4 in., with four strips. Ax would = 1 in.; and if ^' = 2 in., H" = 2 in., and I = 5.2 in., then Jn for OR = 2x2x5.2x1.0=20.8 biquad. inches. The 7x for OL, on the left of N, is found in a similar manner and added to 7^ for OR to obtain the total I^. The position of a gravity axis is easily found by cutting the shape out of sheet metal and balancing on a knife edge ; or may be obtained graphically by § 336 ; or 376. 377. Construction for locating a line vw (Fig. 417) at (a), in the polygon FG in such a position as to satisfy the two following conditions with reference to the vertical inter- cepts at 1, 2, 3, 4, and 5, between it and the given points 1„ 2, 3, etc., of the perimeter of the polygon. * Another graphic method for this purpose will be found in § 76 (p. 80), of the author's Notes and Examples in Mechanics. ARCH-EIBS. 455 Condition I. — (Calling these intercepts u^, u^, etc., and tlieir horizontal distances from a given vertical F, x^, x.^, etc.) 2" (u) is to = ; i.e., the sum of the positive u's must be numerically — - that of the negative (which here are at 1 and 5). An infinite number of positions of vm will satisfy condition I. Condition II. — 2* (ux) is to = ; i.e., the sum of the ij 1 1 . r-— ;^?n moments of the positive u'^ • — ■' ■ — "^^^^ ' about F must = that of the '' negative -m's. i.e., the moment of the resultant of the posi- tive w's must = that of the resultant of the negative ; and .*. (Condit. I being already satisfied) these two resultants must be directly opposed and equal. But the ordinates u in (a) are indi- vidually equal to the difiFer- ence of the full and dotted ordinates in (&) with the same cc's .'. the conditions may be rewritten : I. 2 (full ords. in (6))= 2" (dotted ords. in (&)) II. 2 [each full ord. in (h) X its £c] = - [each dotted ord. in (b) x its x] i.e., the Fig. 4ir. Centres of gravity of the full and of the dotted in (6) must lie in the same vertical Again, by joining ^(x, we may divide the dotted ordi- nates of (b) into two sets which are dotted, and broken, re- spectively, in (c) Then, finally, drawing in (d), B, the resultant of full ords. of (c) T, " " " broken " " " T', " " " dotted " " " 456 MECHANICS OF ENGINEERING. we are prepared to state in still another and final form tlie conditions wliicli vm must fulfil, viz. : (I.) T+T must = i?; and (II.) The resultant of T and T' must act in the same vertical as R. In short, the quantities T, T', and R must form a bal- anced system, considered as forces. All of which amounts practically to this : that if the verticals in which T and T' act are known and R be conceived as a load supported by a horizontal beam (see foot of Fig. 417, last figure) resting on piers in those verticals, then T and T' are the respec- tive reac^'^ons o/" ^Aose jjiers. It will now be shown that the verticals of T and T' are easily found, being independent of the position of vm; and that both the vertical and the mag- nitude of R, being likewise independent of vm, are deter- mined with facility in advance. For, if v be shifted up or down, all the broken ordinates in (c) or {d) will change in the same proportion (viz. as vF changes), while the dotted ordinates, though shifted along their verticals, do not change in value ; hence the shifting of v affects neither the vertical nor the value of T', nor the vertical of T. The value of T, however, is proportional to vF. Similar- ly, if m be shifted, up or down, T' will vary proportionally to mG, but its vertical, or line of action, remains the same. T is unaffected in any way by the shifting of m. R, de- pending for its value and position on the full ordinates of (c) Fig. 417, is independent of the location of vm. We may .*. proceed as follows : 1st. Determine R graphically, in amount and position, by means of § 376. 2ndly. Determine the verticals of T and T' by any trial position of vm (call it v-im.,), and the corresponding trial values of T and T' (call them T, and T',). 3rdly. By the fiction of the horizontal beam, construct (§ 329) or compute the true values of T and T', and then determine the true distances vF and w6^ by the propor- tions vF : v.F :: T : T. and mG : m,G : : T' i T^. AECH-EIBS. 457 Example of this. Fig. 418. (See Fig. 417 for s and t.) From A tovi^ard B in (e) Fig. 418, lay off the lengths (or lines proportional to them) of the full ordinates 1, 2, etc., of (/). Take any pole Oi, and draw the equilibrium poly- gon {/y and pro- long its extreme seg- ments to find C and thus determine ^'s vertical. JR is repre- sented by AB. In (g) [same as (/) but shifted to avoid complexity of lines] draw a trial VoWi and join V2 G2. Deter- mine the sum T2 of the broken ordi- TFig. 418. nates (between V2G2 ana ^^2^2) and its vertical line of ap- plication, precisely as in dealing with B ; also T'2 that of the dotted ordinates (five) and its vertical. Now the true T=Btj-{s+t) and the true T'=Bs^(s+t). Hen ce com- pute vF={T^T2) ^2 and ?^^=(T'-^^^) m^G^., and by laying them off vertically upward from F and G respec- tively we determine v and m, i.e., the line vm to fulfil the conditions imposed at the beginning of this article, rela- ting to the vertical ordinates intercepted between vm and given points on the perimeter of a polygon or curve. Note (a\ If the verticals in which the intercepts lie are equidistant and quite numerous, then the lines of action of T2 and T'2 will divide the horizontal distance between F and G into three equal parts. This will be exactly true in the application of this construction to § 390. Note (b). Also, if the verticals are symmetrically placed about a vertical line, (as will usually be the case) VjWg is 458 MECHANICS OF ENGINEERING. best drawn parallel to FG, for then T^ and T'^ will be equal and equi-distant from said vertical line. 378, Classification of Arch-Eibs, or Elastic Arches, according to continuity and modes of support. In the accompany^ ing figures Htxefull curves show the unstrained form of the rib (before any load, even its own weight, is permitted to come upon it) ;the dotted curve shows its shape (much ex- aggerated) when bearing a load. For a given loading Three Conditions must be given to determine the special equilibrium polygon (§§ 366 and 367). Class A. — Continuous rib, free to slip laterally on the piers, which have smooth horizontal surfaces. Fig. 420. This is chiefly of theoretic interest, its consideration being therefore omitted. The pier reactions are neces- sarily vertical, just as if it were a straight horizontal beam. Class B. Rib of Three Hinges, two at the piers and one intermediate (usually at the crown) Fig. 421. Fig. 36 also is an example of this. That is, the rib is discontinuous and of two segments. Since at each hinge the moment of tlie stress couple must be zero, the special equilibrium polygon must pass through the hinges. Hence as three points fully determine an equilibrium polygon for given load, the special equilibrium is drawn by § 341. Fig. 420. Fig. 421. [§ 378a will contain a construction for arch-ribs of three hinges, when the forces are not all vertical.] Class C. Rib of Two Hinges, these being at the piers, the rib continuous between. The piers are considered im- movable, i.e., the span cannot change as a consequence of loading. It is also considered that the rib is fitted to its AKCH RIBS. 459 hinges at a definite temperature, and is tlien under no con- straint from the piers (as if it lay flat on the ground), not even its own weight being permitted to act when it is fi- nally put into position. When the " false works " or temporary supports are removed, stresses are in- duced in the rib both by its loading, including its own weight, and by a change of temperature. Stresses due to temperature may be ascertained separately and then combined with those duo to the loading. [Classes A and B are not subject to temperature stresses.] Fig. 422 shows a rib of two hinges, at ends. Conceive the dotted curve (form and position un- der strain) to be superposed on the continuous curve (form before strain) in such a way that B and its tangent line (which has been dis- placed from its original position) may occupy their pre- vious position. This gives us the broken curve O^B. 00,^ is .*. O's displacement relatively to B and -S's tangent, Now the piers being immovable OqB (right line) =05 ; i.e., the X projection (or Jx) of OOn upon OB (taken as an axis of X) is zero compared with its Jy. Hence as one condi- tion to fix the special equilibrium polygon for a given load- ing we have (from § 373) Fig. 422. r^[Myds-^EI^=0 (1) The other two are that the [ must pass through . (2) special equilibrium polygon ) " " " B . (3) Class D. Bill with Fixed Ends and no hinges, i.e., continu- ous. Piers immovable. The ends may be ^xed by being inserted, or built, in the masonry, or by being fastened to large plates which are bolted to the piers. [The St. Louis Bridge and that at Coblenz over the Rhine are of this class.] Fig. 423. In this class there being no hinges we 460 MECHANICS OF ENGINEERING. Fig. 423. have no point given in advance througli whicli the special equilibrium polygon must pass. However, since O's dis- placement relatively (and absolutely) to B and ^'s tangent is zero, both z/a:; and z/^[see § 373] = zero, AIsq the tan- gent-lines both at and B being fixed in direction, the angle be- tween them is the same under loading, or change of temperature, as when the rib was first placed in position under no strain and at a definite temperature. Hence the conditions for locating the special equilibrium polygon are p^ Mds _ Q . p Myds ^ ^ . n^ Mxds _ q Jo ^jT ' Jo '~m~ ' Jo EI In the figure the imaginary rigid prolongations at the ends are shown [see § 366]. Other designs than those mentioned are practicable (such as : one end fixed, the other hinged ; both ends fixed and one hinge between, etc.), but are of unusual occur- rence. 378a. Eib of Three Hinges, Forces not all Vertical,* If the given rib of three hinges upholds a roof, the wind-press- ure on which is to be considered as well as the weights of the materials composing the roof-covering, the forces will not all be vertical. To draw the special equil, polygon in such a case the following construction holds : Re- quired to draw an equilib- rium polygon, for any plane system of forces, through three arbitrary xs^ points. A, p and B ; Fig, B 423a. Find the line of action of B^, the resultant of all the forces occurring between A and p; also, Fig. 423a. * See p. 117 of the author's "Notes and Examples io Mechanics" for > detailed example of the following construction. ARCH-EIBS. 4C1 that of R,, tlie resultant of all forces between ^p and B ; also the line of action of B, tlie resultant of B^ and B.2, [see § 328.] Join any point iH^ in ^ witli A and also witli B, and join the intersections iVand 0. Then A iV will be the direction of the first segment, B that of the last, and NO itself is the segment corresponding to p (in the de- sired polygon) of an equilibrium polygon for the given forces. See § 328. If A N' p 0' B are the corresponding segments (as yet unknown) of the desired equil. polygon, we note that the two triangles MNO and M'N' O, having their vertices on three lines which meet in a point [i.e., B meets Bi and B^ in C], are homological [see Prop. YII. of Introduc. to Modern Geometry, in Chauvenet's Geometry,] and that . • . the three intersections of their corresponding sides must lie on the same straight line. Of those inter-r sections we already have A and B, while the third must be at G, found at the intersection of AB and NO. Hence by connecting C and p, we determine N and 0'. Joining N'A aiid O'B, the first ray of the required force diagram will be II to NA, while the last ray will be || to O'B, and thus the pole of that diagram can easily be found and the cor- responding equilibrium polygon, beginning at A, will pass through p and B. (This general case includes those of §§ 341 and 342.) 379. Arch-Rib of two Hinges; by Prof. Eddy's Method.* [It is understood that the hinges are at the ends.] Re- quired the location of the special equilibrium polygon. "VVe here suppose the rib homogeneous (i.e., the modulus of Elasticity E is the same throughout), that it is a " curved prism " (i.e., that the moment of inertia / of the cross- section is constant), that the piers are on a level, and that the rib-curve is symmetrical about a vertical line. Fig. 424. For each point m of the rib curve we have an x and y (both known, being the co-ordinates of the point), and also a z (intercept between rib and special equilib- FiG. 424. ^ rium polygon) and a z' (intercept *P. S5 of Prof. Eddy's book ; see reference in preface of this work. 462 MECHANICS OF ENGINEEEING. between tlie spec. eq. pol. and the axis X (whicli is OB). The first condition given in § 378 for Class C may be transformed as follows, remembering [§ 367 eq. (3)] that M = Hz at any point m of the rib (and that EI is con- stant). 1^ EI H r Myds = 0, i.e., — C zyds = . • . f zyds do El c/o t/o ^^y _^, \-''J^(y~ ^')yds=0', i.e., J^ yyds =J^ yz'ds . (1) In practical graphics we can not deal with infinitesimals ; hence we must substitute As a small finite portion of the rib-curve for ds', eq. (1) now reads I^ yy As = 2'^ yz' As. But if we take all the As's equal, As is a common factor and cancels out, leaving as a final form for eq. (1) I^\yy) = I^^{yz') . . . (1/ The other two conditions are that the special equilibrium polygon begins at and ends at B. (The subdivision of the rib-curve into an even number of equal As's will be ob- served in all problems henceforth.) 379a. Detail of the Construction. Given the arch-rib B, Fig. 425, with specified loading. Divide the curve into Fig. 425. ARCH RIBS. 4G; eight equal ^s's and draw a vertical through the middle of each. Let the loads borne by the respective ^s's be Pi, P2, etc., and with them form a vertical load-line A C to some convenient scale. With any convenient pole 0" draw a trial force diagram 0" AC, and a corresponding trial equilibrium polygon F G, beginning at any point in the vertical F. Its ordinates %", 22", etc., are propor- tional to those of the special equil. pol. sought (whose abutment line is OB) [§ 374a (2)]. We next use it to de- termine n^ [see § 374a]. We know that OB is the " abut- ment-line " of the required special polygon, and that . ' . its pole must lie on a horizontal through n'. It remains to determine its H, or pole distance, by equation (1)' just given, viz. : IJ^ yy = Sfyz'. First by § 375 find the value of the summation Ii{yy), which, from symmetry, we may write = 2i'/(2/2/) =2 [2/12/1+2/22/2+2/32/3+2/42/4] Hence, Fig. 426, we obtain 11 {yy)=2 [HM Next, also by § 375, see Fig. 427, using the same pole dis- tance Ho as in Fig. 426, we find I\{yz")=HA"; i.e., +2/22:2'' + 2/3%" +2/4^'.= Again, since II {yz") = ysz/' + 2/7^7" + 2/6^6" + 2/5^5" which from symmetry (of rib) =2/i%"+2/2^7"+2/326"+2/'<', we obtain, Fig. 428, l"! (2/O = ^oV', (same /^,); and .-. Jf {yz")=ff, {\"+hJ'). If now we find that /fc/'+/b/'=2yfc. 464 MECHANICS or engineeking. the condition 2^1 (yy) = II {y^'') is satisfied, and the pole distance of our trial polygon in Fig. 425, is also that of the special polygon sought; i.e., the z" 's.are identical in value with the s"s of Fig. 424. In general, of course, we do not find that ky'~{-k/' = 2A;. Hence the z" 's must all be increased in the ratio 2k: {lc^"-\~k/') to become equal to the g"s. That is, the pole distance H of the spec, equil* polygon must be 7j-_ ki'-\-'k/' jT,, (in which W = the pole distance of the 2^c trial polygon) since from §339 the ordi- nates of two equilibrium polygons (for the same loads) are inversely as their pole distances. Having thus found the if of the special polygon, knowing that the pole must lie on the horizontal through n', Fig. 425, it is easily drawn, beginning at 0. As a check, it should pass through B. For its utility see § 367, but it is to be remembered that the stresses as thus found in the different parts of the rib under a given loading, must afterwards be combined with those resulting from change of temperature and the shortening of the rib axis due to the tangential thrusts, before the actual stress can be declared in any part. Note. — Variable Moment of Inertia. If the / of the rib section is dif- ferent at different sections we may proceed as follows: For eq. (1); we PB ds i'B ds now write I yy y = I yz'—-. Taking the I of the crown section (say) Jo i Jo I as a standard of reference, denoting it by /', we may write for any other section I = nl', where n is a variable ratio, or abstract number; whence eq. (1) becomes, after putting Js for ds, y / 2/J/— ="77 / V^—- If now the length of each successive Js, from the crown down, be made directly prop>ortional to the number n at that part of the rib, the quantity ^ s^n will have the same value in all the terms of each summation and may be factored out ; and we then have a relation identical in form with eq. (1)', but with the understanding that the j/'s and 2"s concerned are those in the successive verticals drawn through the mid-points of the unequal -s's, or subdivisions along the rib, obtained by following the above plan that each As is proportional to the value of the moment of inertia at that part of the rib. For instance, if the / of a section near the hinge 0, or B, is three times that (/') at the (3rown, then the length of the Js at the former point must be made three tim,es the length of the As first assumed at the crown when the subdivision is begun. By a little preliminary investigation, a proper value for this crown , s may be decided upon such that the total .number of As's shall be sufficient for accuracy (sixteen or twenty in all) ABCn-KIBS. 465 f-$BO. Arch Rib of Fixed Ends and no Hinges,— Example of Class D. Prof. Eddy's Method.* As before, E and / are constant along tlie rib Piers immovable. Rib curve symmetrical about a vertical line. Fig. 429 shows such a rib under any loading. Its span is OB, wliicli is taken as an axis X. The co-ordinate of any point m' of the rib curve are x and y, and z is the vertical intercept between w' and the special equilibrium polygon (as yet unknown, but to be constructed). Prof. Eddy's method will now be ■given for finding tha spe- cial equil. polygon. The three conditions it must satisfy (see § 378, Class D, remembering that E and /are constant and that M — ITz from § 367) are H-« ^^ Fig. 429. / zds=^ ; / xzds— ; and / yzds =0 e/o e/o e/o (1) Now suppose the auxiliary reference line (straight) vm to have been drawn satisfying the requirements, with respect to the rib curve that / z'ds—0 ; and / xz'ds=Q e/o c/o (2) in which z' is the vertical distance of any point m' from vm and x the abscissa of mf from 0. From Fig. 429, letting z" denote the vertical intercept (corresponding to any m') between the spec, polygon and the auxiliary line vm, we have z=z'—z", hence the three conditions in (1.) become r{z'-z")ds=0; i.e., see eqs. (2) C^. z"ds=0 , .. (3) * p. 14 of Prof. Eddy's book ', see reference in preface of this work. 466 mecha:nics of engineeeing. B B fx {z'—^')ds=0 ; i.e., see eqs. (2) f xz"ds=o (4/ ^nifh'-^)ds=;0,^7^^Zl-f}.'ds=fkds . (5) provided vm has been located ^s prescribed. For graphical purposes, having subdivided the rib curve into an even number of small equal J.s's, and drawn a verti- cal through the middle of each, we first, by § 377, locate vm to satisfy the conditions ll{z')=0 and l^,{xz')=0 . . (6) (see ec[. (2) ; the di cancels out) ; and then locate the special equilibrium polygon, with vm as a reference-line, by making it satisfy the conditions. :EI{z'')=0 . (7); Il{xz")=Q . (8); I^Xyz")^l^Xyz') . (9) (obtained from eqs. (3), (4), (5) by putting ds = ^s, and can- celling). Conditions (7) and (8) may be satisfied by an infinite number of polygons drawn to the given loading. Any one of these being drawn, as a trial polygon, we determine for it the value of the sum l'f^(yz") by § 375, and compare it with the value of the sum l'^,{yz') which is independent of ihe special polygon and is obtained by § 375. [N.B. Itmist be understood that the quantities (lengths) x, y, z, z\ and z" , kere dealt with are thosa pertaining to the verticals drawn through the middles of the respective ^s's, which must be sufficiently numerous to obtain a close result, and not to the verticals ia which the loads act, necessarily, since these latter may be few or many according to circumstances, see Fig. 429]. If these sums are not equal, the pole distance of the trial equil. polygon must be altered in the proper ratio (and thus change the 2;'"s in the inverse ratio) neces- sary to make these sums equal and thus satisfy conditicn (9). The alteration of the 2'"s, all in the same ratio, will AECH-EIBS. 4G7 aot interfere with conditions (7) and (8) whicli are alreadj^ satisfied. 381. Detail of Construction of Last Problem. Symmetrical Arch- Rib of Fixed Ends. — As an example take a span of the St. Louis Bridge (assuming /constant) "with. " live load'' cov- sring the half span on the left. Fig. 430, where the verticaJ feME:^^d==Ui Fig. 430. scale is much exaggerated for the sake of distinctness*. Divide into eight equal Js's. (In an actual example sixteen or twenty should be taken.) Draw a vertical through the * Each arch -rib of the St. Louis bridge is a built up or trussed, rib of steel about 53i ft. span and 52 ft. rise, ia the form of a segment of a circle . Its moment of inertia, however, is not strictly constant, the portions near each pier, of a length equal to one twelfth of the span, having a value of / one-half greater than that of the remainder at the arc. 468 MECHANICS OF ENGINEERING. middle of each ^s. P^ , etc., are the loads coming upon the respective Js's. First, to locate vm, by eq. (6) ; from symmetry it must be horizontal. Draw a trial vm (not phown in the figure), and if the (-{- 8')'^ exceed the (— 2')'s by an amount z^, the true vm will lie a height —z' above the trial vm (or below, if vice versa) ; n = the number of z/s's. Now lay off the load-line on the right (to scale), take any convenient trial pole 0'^' and draw a correspond- ing trial equil. polygon F'"G"\ In r"G"', by §377, locate a straight line v"'m!" so as to make 2^(2'") = and ^l(xz!") = (see Note (&) of § 377). [We might now redraw F'" G'" in such a way as to bring v"'m!" into a horizontal position, thus : first determine a point n'" on the load-line by drawing 0"'n"' \ to v"'m"' , take a new pole on a horizontal through n'" , with the same II'" , and draw a corresponding equil. polygon ; in the lat- ter v"'m"' would be horizontal. We might also shift this new trial polygon upward so as to make v"'m!" and vm. coincide. It would satisfy conditions (7) and (8), having the same %'"'''& as the first trial polygon ; but to satisfy con- dition (9) it must have its 2""s altered in a certain ratio, which we must now find. But we can deal with the individ- ual 2""s just as well in their present positions in Fig. 430.] The points ^and L in vm, vertically over E'" and L'" in v"'m'", are now fixed ; they are the intersections of the special polygon 7'equired, ivith vm. The ordinates between v"'m"' and the trial equilibrium polygon have been called z'" instead of z" ; they are pro- portional to the respective g"'s of the required special polygon. The next step is to find in what ratio the (s'")'s need to be altered (or H'" altered in inverse ratio) in order to be- come the {z"^^ ; i.e., in order to fulfil condition (9), viz. : AUCH-EIBS. 469 ^.{yz")=I\{yz') . (9) This may be done pre- cisely as for tlie rib with two hinges, but the nega- tive (s'")'s must be prop- erly considered (§ 375) See Fig. 431 for the de- tail. Negative 2;"s or g""s point upward. From Fig. 431a [j ' .*. from symmetry I\{yz')=2H^h From Fig. 4315 we have riyz"')=HX Pig. 431. and from Fig. 431c Il{yn=Ho^ [The same pole distance H^ is taken in all these construc- tions] .♦. I\yz")=H,{k,-\-\). If, then, Ho {\-\-k,) = 2HJc condition (9) is satisfied by the z""s. If not, the true pole dista-nce for the special equil. polygon of Fig. 430 will be 2k ' With this pole distance and a pole in the horizontal through n'" (Fig. 430) the force diagram may be completed for the required special polygon ; and this latter may be con- structed as follows : Beginning at the point E, in vm, through it draw a segment || to the proper ray of the force diagram. In our present figure (430) this " proper ray " would be the ray joining the pole with the point of meet- ing of P2 and Pi on the load-line. Having this one seg- 470 MECHANICS OF EXGIXEEEING. ment of the special polygon the others are added in an obvious manner, and thus the whole polygon completed. It should pass through L, but not and B. For another loading a different special equil. polygon would result, and in each case we may obtain the tkrusty shear, and moment of stress couple for any cross-section of the rib, by § 367. To the stresses computed from these, should be added (algebraically) those occasioned by change of temperature and by shortening of the rib as occasioned by the thrusts along the rib. These " temperature stresses," and stresses due to rib-shortening, will be con- sidered in a subsequent paragraph. They have no exist- ence for an arch-rib of three hinges. Note. — If the moment of inertia of the rib section is variable, instead of dividing the rib axis into equal Js's, we should make them unequal, following the plan indicated in the note on p. 464, the As being made proportional to the values of the moment of inertia along the rib. After such subdivision is made, and a vertical drawn through the mid- point of each Js, the various ^'s, z^'s, etc., in these verticals are dealt with in the same manner as just shown for the case of constant moment of inertia. 381a. Exaggeration of Vertical Dimensions of Both Space and Force Diagrams. — In case, as often happens, the axis of the given rib is quite a flat curve, it is more accurate (for find- ing M) to proceed as follows : After drawing the curve in its true proportions and pass- ng a vertical through the middle of each of the equal z/s's, compute the ordinate (y) of each of these middle points from the equation of the curve, and multiply each y by four (say). These quadruple ordinates are then laid off from the span upward, each in its proper vertical. Also multiply each load, of the given loading, by four, and then with these quadruple loads and quadruple ordinates, and the upper extremities of the latter as points in an exagge- rated rib-curve, proceed to construct a special equilibrium polygon, and the corresponding force diagram by the proper method ( for Class B, C, or D, as the case may be) for this exaggerated rib -curve. The moment, Hz, thus found for any section of the ex- AECn-KIBS. 471 aggerated rib-curve, is to be divided by four to obtain the moment in tlie real rib, in tlie same vertical line. To find the thrust and shear, however, for sections of the real rib, besides employing tangents and normals of the real rib W9 must draw, and use, another force diagram, obtained from the one already drawn (for the exaggerated rib) by re- ducing its vertical dimensions (only), in the ratio of four to one. [Of course, any other convenient number besides four, may be adopted throughout.] 382. Stress Diagrams. — Take an arch -rib of Class D, § 378,. i.e., of fixed ends, and suppose that for a given loading (in- cluding its own weight) the special , ^^^^ ^thrust equil. polygon and its force diagram have been drawn [§ 381]. It is re- "^^^^ " —coopte- quired to indicate graphically the variation of the three stress-elements for any section of the rib, viz., the thrust, shear, and mom. of stress- couple. / is constant. If at any point TO of the rib a section is made, then the stresses in that section are classified into three sets (Fig. 432). (See §§ 295 and 367) and from § 367 eq. (3) we see that the ver- tical intercepts between the rib and the special equil. polygon being proportional to the products Hz or moments of the stress-couples in the corresponding sec- tions form a moment diagram, on inspection of which we can trace the change in this moment, Hz = ^ , and e hence the variation of the stress per square inch, jJjj (as. due to stress couple alone) in the outermost fibre of any section (tension or compression) at distance e from the gravity axis of the section), from section to section along the rib. By drawing through lines On' and OV parallel re- spectively to the tangent and normal at any point m of the rib axis [see Fig. 433] and projecting upon them, in turn, the proper ray (B^ in Fig. 433) (see eqs. 1 and 2 of § 367) tJ Fig. 432. 472 MECHANICS OF ENGINEElimG. we obtain the values of the thrust and shear for the sec- tion at m. When found in this way for a number of points along the rib their values may be laid off as vertical lines from a horizontal axis, in the verticals containing the re- spective points, and thus a thrust diagram and a shear dia- gram may be formed, as constructed in Fig. 433. Notice that where the moment is a maximum or minimum the shear changes sign (compare § 240), either gradually or Fig. 433. suddenly, according as the max. or min. occurs between two loads or in passing a load ; see m', e. g.' Also it is evident, from the geometrical relations involv- ed, that at those points of the rib where the tangent-line is parallel to the " proper ray " of the force diagram, the thrust is a maximum (a local maximum) the moment (of ARCH RIBS 47S stress couple) is either a maximum or a minimum and the shear is zero. From the moment, Hz = ^, p2 — — - e 1 may be computed. From the thrust = Fp^^, pi=- , (F = area of cross-section) may be computed. Hence the greatest compression per sq. inch (Pi+p^) may be found in each section. A separate stress-diagram might be con- structed for this quantity (pi+p^)- Its max. value (after adding the stress due to change of temperature, or to rib- shortening, for ribs of less than three hinges), wherever it occurs in the rib, must be made safe by proper designing of the rib. The maximum shear J,,^ can be used as in §256 to determine thickness of web, if the section i?^ I-shaped, or box-shaped. See § 295. 383. Temperature Stresses.— In an ordinary bridge truss and straight horizontal girders, free to expand or contract longitudinally, and in Classes A and B of § 378 of arch- ribs, there are no stresses induced by change of tempera- ture ; for the form of the beam or truss is under no constraint from the manner of support ; but with the arch- rib of two hinges (hinged ends, Class C) and of fixed ends (Class D) having immovable piers which constrain the dis- tance between the two ends to remain the same at all tem- peratures, stresses called " temperature stresses '* are in- duced in the rib whenever the temperature, t, is not the same as that, t^, when the rib was put in place. These may be determined, as follows, as if they were the only ones, and then combined, algebraically, with those due to the loading. 384. Temperature Stresses in the Arch-Rib of Hinged Ends,— (Class C, § 378.) Fig. 434. Let E and /be constant, with i74 MECHANICS OF ENGINEBErNG. Fig. 434. oilier postulates as in § 379. Let t^, = temperature of erection, and i — any other temperature ; also let I = length of span = OB (in- variable) and 7^ "CO -efficient of linear expansion of the material of the curved beam or rib (see § 199), At tempeia- fcure t there must be a horizontal reaction H at each hinge to prevent expansion into the form O'B (dotted cuive), which is the form natural to the rib for temperature t and without constraint. We may /. consider the actual form OB as having resulted from the unstrained form O'B by displacing 0' to 0, i.e., producing a horizontal displace- ment O'O =1 {t-Q-/j. But O'O = Jx (see §§ 373 and 374) ; (KB. B'% tangent has moved, but this does not affect Jx, if the axis X is horizontal, as here, coinciding with the span ;) and the ordinate y of any point m of the rib is identical Avith its z or intercept between it and the spec, equil. polygon, which here consists of one segment only, viz. : OB, Its force diagram consists of a single ray Oi n' • see Fig. 434. Now (§ 373) .J B Ja? = -A j3Iyds ; and M=Hz = in this case, Hy H .:l{t-Qrj=—Jfds; hence for graphics, and equal Js's, we have Ell{t-t,)y^=HJs I^y' . . . . (1) From eq. (1) we determine H, having divided the rib-curve into from twelve to twenty equal parts each called Js . For instance, for wrought iron, t and t^,, being expressed In Fahrenheit degrees, -/j = 0.0000066. If E is expressed in lbs. per square inch, all linear quantities should be la inches and H will be obtained in pounds. 2'o?/^ may be obtained by § 375, or may be computed. B being known, we find the moment of stress-couple = Hy, AKCH-KIBS. 475 at any section, while the thrust and shear at that section are the projections of //, i.e., of O^n' upon the tangent and normal. The stresses due to these may then be determined in any section, as already so frequently explained, and then combined with those due to loading. 385. Temperature Stresses in the Arch-Ribs with Fixed Ends,— See Fig, 435. (Same postulates as to symmetry, E and J constant, etc., as in § 380.) t and t^ have the same meaning as in § 384. Here, as before, we consider the rib to have reached its ac- tual form under tem- perature t by having had its span forcibly shortened from the length natural to temp, t, viz. : O'B', to the actual length OB, which the immovable piers compel it to assume. But here, since the tangents at and B are to he the same in direction under constraint as before, the two forces H, representing the action of the piers on the rib, must be considered as acting on imaginary rigid prolonga- tions at an unknown distance d above the span. To find H and d we need two equations. From § 373 we have, since M=Hz=H {y—d), Ax, i.e., WO-VBW, i.e., \t-t:)r^,=-^J{y-^yds . (2) or, graphically, with equal As's Fig. 435. EIl{t- -Qr- -HAs I-f-dSly (3) Also, since there has been no change in the angle betweeij end-tangents, we must have, from § 374, ^_rMds=0; i.e., — / 2c?s=0;i.e., ny-d)ds=0 476 MECHANICS OF ENGINEEEING. or for graphics, witli equal jU 's, I'^y = nd . , • (^\ in wMcli n denotes tli6 number of J.s's. From (4) wq determine d, and tlien from (3) can compute U. Drawing the horizontal F G, it is the special equilibrium polygon (of but one segment) and the moment of the stress-couple at any section = Hz, while the thrust and shea\' are the projections of H=^0{ii' on the tangent and normal respect- ively of any point m of rib. For example, in one span, of 550 feet, of the St. Louis Bridge, having a rise of 55 feet and fixed at the ends, the force H of Fig. 435 is = 108 tons, when the temperature is 80° Fahr. higher than the temp, of erection, and the en- forced span is 3^ inches shorter than the span natural to iliat higher temperature. Evidently, ;f the actual temp- ■erature I is lower than that ^„, of erection, ^must act in a direction opposite to that of Figs. 435 and 434, and th& "'thrust " in any section will be negative, i.e., a pull. 386. Stresses Due to Rib-Shortening — In § 369, Fig. 407, the shortening of the element AE to a length A'E, due to the uniformly distributed thrust, PiF, was neglected as pro- ducing indirectly a change of curvature and form in the rib axis ; but such will be the case if the rib has less than three hinges. This change in the length of the different, portions of the rib curve, may be treated as if it were due to a change of temperature. For example, from § 199 we- see that a thrust of 50 tons coming upon a sectional area. of i^ = 10 sq. inches in an iron rib, whose material has a modulus of elasticity — E = 30,000,000 lbs. per sq. inch, and a coefficient of expansion yj = .0000066 per degree Fahrenheit, produces a shortening equal to that due to a fall of temperature {to—t) derived as follows: (See § 199) (units, inch and pound) ^° ^ FEri 10 X 30,000,000 X. 0000066" Fahrenheit. Practically, then, since most metal arch bridges of ©lasses G and D are rather flat in curvature, and the thrusts. AECH-RIBS. 477 due to ordinary modes of loading do not vary more than 20 or 30 per cent, from each other along the rib, an imagin- ary fall of temperature corresponding to an average thrust in any case of loading may be made the basis of a con- struction similar to that in § 384 or § 385 (according as the ends are hinged, ov fixed) from which new thrusts, shears, and stress-couple moments, may be derived to be combin- ed with those previously obtained for loading and for change of temperature. 387. Resume — It is now seen how the stresses per square inch, both shearing and compression (or tension) may be obtained in all parts of any section of a solid arch-rib or curved beam of the kinds described, by combining the re- sults due to the three separate causes, viz.: the load, change of temperature, and rib-shortening caused by the thrusts due to the load (the latter agencies, however, com- ing into consideration only in classes G and D, see § 378). That is, in any cross-section, the stress in the outer fibre is, [letting J',/, T-^", T^"', denote the thrusts due to the ihree causes, respectively, above mentioned ; {H&)', {Hz)"y {Hz)'", the moments] ^T}}.^I}l'^I^±tUHz)'±{Hzy'±{Hzy"'\ . . . (1) i.e., lbs. per sq. inch compression (if those units are used). The double signs provide for the cases where the stresses in the outer fibre, due to a single agency, may be tensile. Fig. 436 shows the meaning of e (the same used heretofore) /is the moment of in- ertia of the section about the gravity axis (horizontal) (7. i^ = area of cross- section. [Ci = e ; cross section symmet- rical about (7]. For a given loading we may find the maximum stress in a given rib, or design the rib so that this maximum stress shall be safe for the ma- terial employed. Similarly, the resultant shear (total, not 478 MECHANICS OF ENGINEERrNG. per sq. inch) = «/' ± J" ± 3'" is obtained for any section to compute a proper thickness of web, spacing of rivets, etc. 388 The Arch-Truss, or braced arch. An open-work truss, if of homogeneous design from end to end, may be treated as a beam of constant section and constant moment of inertia, and if curved, like the St. Loi*is Bridge and the Coblenz Bridge (see § 378, Class D), may be treated as an arch-rib.* The moment of inertia may be taken as r=2i^, A (I) where F^ is the sectional area of one of the pieces II to the curved axis midway between them. Fig. 437, and h = dis« fcance between them. Fig. 438. Fig. 437. Treating this curved axis as an arch-rib, in the usual way (see preceding articles), we obtain the spec, equil. pol. and its force diagram for given loading. Any plane ~| to the rib -axis, where it crosses the middle m of a " web- member," cuts three pieces, A^ B and 6', the total com- *The St Louis Bridge 13 not strictly of constant moment of inertia, being somewha* strengthened near eaoli pier, ARCH-EIBS. 479 pressionB (or tensions) in which are thus found : For the point m, of rib-axis, there is a certain moment = Hz, a thrust = Th, and a shear = J, obtained as previously ex- plained. We may then write Psin/9 = J . . . • (1) and thus determine whether P is a tension or compres- sion ; then putting P'+P" ± P cos /? = T,, 2 (in which P is taken with a plus sign if a compression, and mlQus if tension), and (P'-P")^=Rz ...... (3) we compute P* and P", whi(5h are assumed to be both com-' pressions here. /9 is the angle between the web member and the tangent to rib-axis at m, the middle of the piece. See Fig. 406, as an explanation of the method just adopted. Circular Ribs akd Hoops. 389. Deflections and Changes of Slope of Curved Beams. Analyt- ical Method. For finding these quantities we may use eqs. (I.), (II.) and (III.) of § 374. For example, we have in Fig. 439, a curved beam of the form of the ox k — quadrant of a circle, fixed vertically ^ at lower extremity p, and carrying a single concentrated load, P, at the free end 0. [Its own weight neglected.] As a consequence of the load- i /^^^' \q \ ing, the extremity is displaced to kfl__ 1__^J^ some position, ^\, but the bending M^^^ is slight. Required, the projections ^^^' ^^^' Ax and Ay of this displacement and also the angle OKOn or <^, which the tangent-line at 0„ makes with its former fhorizon- tal) position OX. The beam is homogeneous and of constant cross-section ; i.e., E and / are constants. To use the equations for Ax and Ay we must take as an origin (since is the point whose displacement is under con- 480 MECHANICS OF ENGINEERING. sideration). Hence the co-ordinates x and y of any point, m, of the axis of the beam are as shown in the figure. Taking now polar co-ordinates, as shown, we note tliat x = r cos_fi-| y = r ( 1 — sin ^) ; and ds = rdO. We must also put down the following integral forms for reference ; viz. : — fsin ^ . d^ = — cos ^ ; f^ . cos 6* . di9 = e . sin ^ + cos ^ ; fcos e.de = + sin d ; fsin^ 6 .dd = ^ 6 ~ \s,m2 d ; fsin e.cos e .de = \ sin^ e ; C c,o%^ 6 ,de = ^ 6 + \sva.20. Taking the portion 0„m {m being any point on curved axis of beam) as a free body, we have, for the moment of the stress couple a.t m, M — Px, = Pr cos 6, and hence derive, for the angle ^, Also 5 and z z 1 r^ Pr^ r /'^ z*^ 1 Pr^ ^^'=^X^-^''^' = ^[Jo «<^«^-^^-X -^^•--^•^^] = 2^z- (3>- It must be understood that the elastic limit is not passed in any fiber and that the bending is very slight. A simple curved crane and a ship's davit are instances of this problem, provided the cross-section has the same moment of inertia, /, about a gravity axis perpendicular to the plane of the paper in Fig. 439, at all parts of the beam. 390. Semi-Circular Arch-Rib. Hinged at the Two Piers or Sup- ports, and Continuous Between. Fig. 440. The supports are at the same level. The arch-rib, or curved beam, is homogeneous and has a constant I at all sections. (It is a " curved prism".) It is stipulated that no constraint is necessary in fitting the rib upon the hinges at the piers before any load is placed on the rib ; that is, that the distance apart of the piers (which are unyielding') is just equal to the distance between the ends of the rib when entirely free from strain. In other words, after the rib is in position it is under no stress until a load is put upon it. Its own weight is neglected and the load is a concen- trated one of 2 P lbs. placed at the " crown ", B. As a conse- CIRCULAR RIBS AND HOOPS. 481 quence of the gradual placing of the load the crown B settles slightly, but on account of symmetry the tangent-line to the curved axis at B remains horizontal. Also the extremities and A tend to spread further apart, but this is prevented by the fact that the piers are immovable (or we may express it "the span is invariable"). Hence the reaction at each hinge support will have a horizontal component ^as well as a ver- tical component, V, lbs. Fig. 2 shows the axis of the rib. Taking the whole rib as a free body we easily find (by putting ^ vert, comps. = zero, and from symmetry) that each y=P; the whole load being called 2 P ; but for de- termining the value of H (same at each hinge ; from 2 (horiz. comps.) =zero) we must have recourse to the theory of elasticity ; i.e., must depend on the following fact, viz. : — that in the gradual settling of point B under the load, B remains in the same vertical, and the tang, line at B remains horizontal, and hence (since moves neither horizontally nor vertically in actual space) the horizontal projection of O's dis- placement relatively to B and ^B's tangent is zero (or Ax^O), while the vertical projection of O's displacement relatively to B and 5's tangent (A?/) equals the distance B has settled in actual space. Here we must take as origin for x and y (as in figure) for any point m between and B\ and note that the X = r {1 — cos 6), and y = r mi 6; while ds = r. d6. With Om as a free body (m being any point between and 5) we have for the moment of the stress couple at m, M.^Vx- Hy, = Px- Hy. 1 rB r^ Ax, = — j Myds, =0 ; .'. I [P(X- rco^ 6) - Hr sm e'jr^ sm Odd =^0', Fig. 440. .-. P C sin0.de - P C smdcosddd-H C sin'6'.d^ = 0: Jo Jo Jo and hence, (see integrals in § 389), 482 MECHANICS or ENGINEERING. (-[- COS 6 Slll^ H ■0 _ sin 2 6h\"^ 2"" 4 = 0. Inserting the limits, we have pr_0 + l-i + 0J-//|^|-0-0-(-0)" = 0; .:H = 2 P load Also we may obtain, for the settlement of the crown, at B 1 C^ Pr'^ r3 TT 1 "I Aw of relatively to 5, = =7 / Mxds = -77^ -. — 2 ^ "^ EI Jo EI I 4: TT j while the tangent-line at 0, originally vertical, now makes with the vertical (on the outside) 'eT This is a '■'statically indeterminate structure " ; that is, one in which a solution is impossible by ordinary statics but must depend on the theory of the elastic change of form of the beam or body in question. If the load were not placed at the crown, or highest point, we should be obliged to put an angle EI Jo Mds = - +1- jj£Myds^O EI for the Ax of relatively to A (instead of to jB). 391. Cylindrical Pipe Loaded on Side. A cylindrical pipe of homogeneous material and small uniform thickness of pipe-wall, i, and length I, (so that the moment of inertia of the cross-section of wall is for present purposes I = It^ -i- 12) rests in a horizontal position on a firm horizontal floor and bears a concentrated load of 2P at the highest point, or crown, jB. See Fig. 441. It is to be considered as a continuous curved beam or " hoop ", without hinges. We neglect the weight of the pipe itself. The dotted circle shows the original unstrained form of the pipe-wall, or hoop, while the full line is its (slightly deformed) shape when it bears the load. The elastic limit is of course not to be passed. The upward force 2P at iV"is the reaction of the floor. Required, the maximum moment of stress-couple ; and also the increase in the length of the horizontal diameter, and the decrease in that of the vertical diameter. Consider as a free body the upper left-hand quadrant of the hoop, viz., OB, in Fig. 442, cutting just on the loft of the load at L', a horizontal section being made at 0. At each end of this body we must indicate a stress-couple, a shear, and a thrust. But at it is evident, after a little consideration, that the shear (which would be horizontal) must be zero ; there being at 0, .•. only CIECULAE EIBS AND HOOPS. 483 a thrust Tg and a stress-couple of unknown moment M^. At the other section the shear must he equal to one half of the load 2P (from considerations of symmetry) i.e., J" at B = P ; while the thrust at B is soon shown to be zero (since S (horiz. compons.) must = zero, and this thrust if it existed would b» Fig. 441. Fig 442. Fig. 443. the only horiz. force besides those formhig the stress-couple at B). At B, therefore, we find only a stress-couple, of an unknown moment Ms, and a shear Jb of direction shown in Fig. 442. By writing S (vert, compons.) = zero for this free body we find that the thrust, Tg, at 0, must have a value P. To determine M^ we make use of the fact (evident from Fig. 441) that in the deformed condition of the ' ' hoop ' ' the tangent-lines at points and B are still vertical and horizontal, respectively ; in other words that the angle be- tween them has not changed, i.e., is still' 90°. Hence the value of 0, or change of angle between tangents at and B is zero. Apply this fact to Fig. 442. Take as origin for the x and y of any point m on OB (using 6 later). From a consideration of the free body Om shown in Fig. 443 we have for the stress- couple-moment M at any section m the value M = Px — M^. We have also a; = r (1 — cos d) ; y = r sin 6 ; and ds = rdd. Since 1 r^ r^ 0, = ^ j Jfds, = 0, .-. j \_Pr^ - Pr^ cos 6 - Mf\ dd = ; i.e., {Pr^ [0 - sin e\ - M^rd) 2 = ; or, Pr^ f^ ~ ^ 1 ~ -^^^l ^ ^ '' whence, finally, we have Jlfg = Pr 1 I (1) Now that Jlfg is known, we may find Mb by taking moments about the lower section, 0, in Fig. 442, with OB as fj-ee body whence Mb = (2 -=- tt) Pr, which is greater than M^. Hence the equation for safe loading is {R'l -=- e) = 2Pr -^ TT, where R' is the maximum safe unit-stress for the material, and e the distance of the extreme fiber from the gravity axis of a section. (If, how- ever, the radius, r, of the cylinder is not large compared with the radial thick- ness of the section, see §§ 298 and 299.) Evidently the horizontal diameter has been lengthened by an amount 2 Ax, if Ax denote the horiz. proj. of O's displacement relatively to B and 5's tan- gent ; and similarly, the shortening of the vertical diameter is 2 Ay, if Ay denote the vert. proj. of O's displacement with regard to B and JS's tangent-line. 484 MECHANICS OF ENGINEERING. Hence ^ ix =. =y j Myds = J- j [Pr^ sin 6 .dO - Pr^ cos 6 sin edd — M^r'' sin edd]; from which we have, with M^ = Pr [1 — {2 -t- tt)], TT ^y = ^C -^a^s, = -^ r ^ [Pr3 (1 - COS ey de - M,r^ {dd - COS edd)-], Pr^ ^2 _ 8 It will be noted that the results obtained in this problem apply also to the case where the hoop is a circular link of a chain under a tension 2 P, except that the moments will be of opposite character and shears and thrusts of oppo- site direction. Also, the change of length 2 Ax of the horiz. diameter will be a shortening, that of the vertical diameter, a lengthening. (See Prof. Filkins' article on p. 99 of Vol. IV of the Transac. of Assoc. C. E. of Cornell Univ. and Engineering News, Dec. 1904, p. 547.) BTumerical Example. Fig. 441. The length of a cast iron pipe is 10 ft., the thickness of wall ^ inch, and the radius of the pipe (measured* to the middle of the thickness) is 6 inches. Kequired, the value of the safe load at crown, 2 P when the pipe is supported horizontally on a firm smooth bed or floor; the max. safe unit-stress being taken at the low figure 2000 lbs. per sq. inch. Solution. "We have only to substitute these values in M^ = R' I -i- e and *9000 V 120 V C^y Obtain (since I =1. P ^ 12), frrr_iL^Lil^== (0.6366 P X 6) ; hence safeload (^) X (2) X 1^ = 2 P, = 5236. lbs. ; that is, 43.63 lbs. per running inch of pipe length. If now the thickness be doubled, i.e., t = 1", with other data unchanged, we find the safe load to be four times as great, i.e., 2 P = 20,944 lbs. ; or 174.5 lbs. per running inch of pipe-length. Although the load is called "concentrated" as regards the end-view of the pipe, it must be understood to be uniformly distributed along the length. FLEXURE OF BEAMS: GEOMETEICAL TREATMENT. 485 CHAPTER XII. Flexure of Beams ; both Simple and Continuous. Geometrical Treatment. 392. By Geometrical Treatment is meant making use of the properties of geometrical figures to deduce algebraic relations. This does not necessitate the use of drafting instruments ; but the graphic ideas involved greatly simplify the algebraic detail of finding deflections, angles, moments, shears, etc., in the case of horizontal beams originally straight and slightly bent under vertical loads and reactions. In the case of " continuous beams", or "girders", (p. 320), this mode of treatment leads to conceptions and methods which are remarkably clear and simple. 393. Angle Between End-Tangents of a Portion of a Bent Beam. If the cantilever beam of Fig. 443a (slender and originally o |C B .ax I norvuilto CD '_ ^^ — T" ^. ,, \d(p ^■'' » ' -* 1 ^1 Fig. 443a. Fig. 4436. straight) be loaded as shown, and the beam thus slightly bent, the two cross-sections, AH and CD, at the two ends of any dx of the axis of beam, are no longer parallel but become in- clined at a small angle d<^ which is also the angle between the normals to these sections, in their new (relative) position (see now Fig. 4436). AZT now occupies the position A'W (relatively to (7Z)). The outer fiber AC (originally of length = dx) is longer by some amount dX ; and evidently the value of angle d^ may be written = dX -j- g. But, by the definition of the modulus of elasticity of the material, J5', we have also E = p H- — , (p. 209) ; whence d^ = '^-^ dx hiQ (1) 486 MECHANICS OF ENGINEERING. ' Now if M denote the moment of the stress-couple to which the tensions and compressions on the ends of the fibers in the section A^ H' are equivalent (M would equal Px in this simple case) we may combine the relation, (§ 229), M = pi -r- e with eq. (1) and thus derive, as a fundamental relation : , . . d4> = — — (2) EI for the angle between two tangents to the elastic curve, one at each end of the elementary length, dx, of the curve ; since the two normals to the sections A'H' and C D in Fig. 4436 are tangents to the ends of the short length dx of the elastic curve. (This value of the angle d^ is in 7r-measure ; i.e., radians.) It follows, therefore, that when the cantilever of Fig. 443a is gradually bent from its original condition (in which the tangent lines at the two extremities and B Avere coincident, i.e., made with each other an angle of zero) into its final form, by the gradual application of the load P at 0, the angle between the tangent to elastic curve at 0^ (the final position of 0) and that at B (which tangent, in this ,,. case, has not moved) will have a value Fig. 444. \ _ -^ obtained by summing up all the small values of d^, one for each of the dx'^ between and B (these dx'^ making up the length of curve between those points). Or, in general, if 0„ and B are any two points of an elastic curve (of axis of bent beam, originally straight and now only slightly bent, x beiag measured along the beam) we have for the angle between the ^ _ ., _ T^ Mdx ,on tangents at 0„ and B\^ Jo EI (See Fig. 444 for case of cantilever.) This may. be called the angle hstween end-tangents of any portion of such elastic curve. The beam must be continuous between these two j)oints and only slightly bent. Usually the beam in question is homogeneous and then E may be taken outside of the integral sign. Also, if the beam be prismatic in form (i.e., sides par- allel to a central axis, originally straight) the moment of inertia, FLEXURE OF BEAMS ; GEOMETEICAL TREATMENT, 487 /, of the cross-section is the same for each dx, and may be placed outside of the / sign. 394. (Relative) Displacement of any Point, 0, of Elastic Curve of a Bent Beam. In the case of the simple cantilever of Fig. 443a let us consider that the axis of the beam, originally straight and in position OB, passes gradually into its final form or elastic curve 0„ A'" A'' . . . B hj the successive change of form of each small block, or elementary length dx ; beginning Successive bending of each dx of Cantilever. Fig. 444a. at the end B. When the section at A^ turns through its angle d(f)^, as due to the lengthening and shortening of the fibers forming the block (i.e., to the stress-couple in section A', of moment M') it carries with it all the portion OA' (still straight) into position O^A' so that the extremity describes a small distance (practically vertical) 00 ^ = OA' . d(f>^. Similarly when, next in order, the section at A^' turns through its small angle d(f>^, the left-hand end of the beam describes a further small distance Ofi^ ^ ^^" • ^^2 ' ^^^ ^° ^^^ ' "i^^til finally the extremity has arrived at its final position 0„, having executed a total (vertical) displacement OOn-, which will be called Ay. If, now, any one of the elementary vertical displacements (like OjOj? as typical) be called 8y, we note that Ay is the sum of all these small 8i/'s, each of which is practically a small cir- cular arc described with a radius x swinging through a small angle d(fi, (the successive x's being successively smaller for the Sy's lower in the series), so that By = xd(}>; hence Ay, =- fSy, = Cxd(j>. But, from eq. (2), d(f> = Mdx -^ AT; ( Displacement of point 0) _ _ T ^ Mxdx I relatively to 5's tangent ) Jo -^-^ (4) 488 MECHANICS OF ENGINEEEING. (N.B. In the use of this relation the x of each dx must he measured from the point whose displacement is desired.) Although the special case of the cantilever has been in mind in the figure used in this connection, this result in eq, (4) may- be generalized by stating that it gives the displacement A?/ of any point from the tangent-line drawn at any other point, B, of the elastic curve formed by the axis of a beam originally straight and slightly bent under the action of vertical forces and reactions. In order to use it, the value of the moment M of the stress-couple in each successive dx must be expressed as a function of x. If, in addition, the beam has a constant moment of inertia, /, of the cross-sections, the " I " may be taken outside of the sign of integration. An integration is then generally possible. (For example, in the above cantilever, for M we should write Px.) 395. Deflections and Slopes of Straight Homogeneous Prismatic Beams Slightly Bent under Vertical Loads and Eeactions. (Beam Horizontal.) If the beam is a prism and homogeneous, both U and I are constant along its length and may be taken outside of the in- D^ tegral sign in eqs. (3) B,.-*':^^'' and (4), and these two J \ '^T^v----^c'""°''^ equations may now be I [dxi applied to a portion of a beam situated between any two points and B Fig. 4446. of the elastic curve as- sumed by the (originally straight) axis of the beam (Fig. 4446) under some load. The tangent-lines at 0„ and B were origi- nally coincident, and hence the angle between these tangents when the beam is bent is the total change in angle between the 1 r^ tangents, and consequently may be written (^ = __ / Mdx and is 0„ 00 in the figure. Again, if a vertical be drawn through the point 0„ to B's tangent-line OB, the length 00,^ is evi- dently O's displacement relatively to 5's tangent-line, since originally the point 0„ was situated in 5's tangent itself. 1 r^ That is, 00„, or A?/, = -— I Mxdx,\n which M is the mo- hil Jo PLEXUBE OF BEAMS : GEOMETRICAL TREATMENT. 489 ment of the stress-couple in the cross-section at any distance, x, from 0. Note that in general Mis a variable; also that the x must be measured from the point whose displacement is under consideration. Example I. Simple cantilever (Fig. 444,), huilt Inliorizontally at B and bear- ing a concentrated load = P lbs. at the free extremity. Both E and / are con- sfcant (homogeneous prism). Pind the deflection 00« and the slope (p. Solution. From the free body OhVI (m being any point between and B) we have M = Px as mom. of stress-couple at ?n. PP 2ET 1 /.5 p p! the "slope" at On- For Jy we have Pl^ Fig. 444i. Fig. 4442- Example II. Prismatic beam on two end-supports. Concentrated load P, V)S., in middle, Fig. 4442- The two supports being at same level we note that from symmetry the tangent at the middle point B of the elastic curve is hori- zontal. Hence the displacement OOn of the extremity from this tangent is «qual to the deflection of B itself below the horizontal line 0„G. To find OOn or Jy, Pl^ 40/ Example III. Prismatic beam on two end-supports at same level, the load being uniformly distributed over the whole span, I. Fig. 4443. That is, W = wl, 1 r-B 1 /-S pP -| P /•■!■= 2 Jy = -=-:- I Mx dc = -r— 1 TT X \ xdx = -— , ^ I x^dx ^ ^ ElJo EI Jo L2 J 2ElJx=^ W = ivl Fig. 4443. Fig. 4444. w being the load per running inch. As before, the tangent-line at middle point B of the elastic curve must be horizontal, so that the displacement of extremity On from this tangent will also give the deflection of B from the horizontal OnC. Measuring x from (as must always be done in these cases) we note that M at any point m W wx- I pBr-WX'^ WX3 -| 1 r WX^ WXn 5 °^« = MJo I ^^"- -2-^^J= ^[-6-- Xjo = 5 584 EI 490 MECHANICS OF ENGINEEEIKG. Example IV. Prismatic beam on end-supports, hearing two equal loads, each = P, symmetrically placed on the span. Fig. 444^. Required, tlie deflection of the middle point, B, of the elastic curve, below the horizontal OnC. Length = 4a. Solution. In previous problems of this article the expression for M, the mom. of stress-couple for any point m betvreen the points O and B, has been a single function of x, applying to all such points m. But in the present problem, having found the reaction at O to be = P lbs., we note by considering a free body Onin (where m is any point between On and B) that the value of 31 is M= Pz ; whereas if the free body extends into the portion DB the expression for M (the free body being now Onm') is M=Px —P{x — a) which reduces to M=Pa, a, different function of x; (in fact a constant). Therefore, in making the summation 00« = (l-=-^J) ( Mx dx for all the dx^s between and B, this summation must be divided into two parts, viz. : one from to D, involving for X the limits x=0 and x=a; and the other from D to B, for which the limits for X are x=a and x=2a. Hence (The student should verify all details of this operation, noting that each sum- mation or integi'al contains the proper value of M, as a function of x, for the proper portion of the elastic curve. As before, it should be said that on ac- count of symmetry the tangent-line at the middle point B is horizontal, and parallel to OnC. Otherwise OnO would not he equal to the deflection of B.) 396. Non-prismatic Beam. VariableMoment of Inertia, I. If the J is vari- able, (e.g., if the beam tapers) it must be retained on the right of the integral sign in the expressions for cj) and Jy and then expressed as a function of x be- fore the integration can be proceeded with. In some cases I may be constant within the limits of definite portions of the beam and then the procedure is simple. For instance, if the beam in Fig. 444^ has a constant value, = I^, for 3 the portions OB and FC, and a larger (but constant) value, of J,) = h ^i» for all the sections from B to F, the following takes the place of eq. (1) above : . P fa Pa /'2a 4 Pa^ 397. Properties of Moment Diagrams (Moment-Areas and Cen- ters of Gravity). Prismatic Beams in Horizontal Position. Vertical Loads and Reactions. In Fig. 445 let AI) be the bent condition (i.e., elastic curve) of the axis of a straight prismatic homo- geneous beam supported on supports at, or nearly at, the same level (so that all tangent lines to the elastic curve deviate but slightly from the horizontal. That is, the bending is slight). Also, let A"D"B"' 0'" be the corresponding moment diagram (as defined and illustrated on pp. 265 to 309). For instance, for any point m of the elastic curve the moment of stress-couple (or "bending moment ") in that section of the bent beam is repre- FLEXURE OF BEAMS; GEOMETRICAL TREATMENT, 491 sented (to scale) by the ordinate m"m"' or M, in the same vertical as m. If now a small horiz. distance dx^ or m'V, be laid off from m" and a vertical r . .8 be drawn through r, the pro- duct M . dx would be proj)or- tional to, and may be repre- sented by, the area of the vertical strip m"rsm"'. Now dx being inches (say) and M being inch-lbs., this product might be called so many "sq. inch-lbs." of moment-area (as it will be called). But the angle <^ between the tangent- i^. ,^ ^ Mom. Diagraml Fig. 445. lines drawn at an}- two points On and B of the elastic curve is 1 r^ equal to — - / Mdx ; and hence we may write <!> [total " moment-area between and B ]' EI (la] or, for brevity, ((> ={A^^ ) ^ EI . . . . . . . . . . (1) This " moment-area," then, between and B is the pro- duct of the base O'^B" (inches) by the average moment be- tween and B regarded as the average altitude of the figure, 0"B"B"V", this altitude being inch-lbs. Again, if the elementary " moment-area " Mdx be multiplied by x, its horizontal distance from 0" (i.e. from and 0„), and these products summed up for all the dx'^ between and 5, there results the expression I (M , dx) . x which may be written {A^^.x, where x denotes the horiz. distance of the center of gravity of the moment-area O^'B'" from 0''0"' (since, from the theory of the center of gravity, the sum of the pro- ducts of each ^tri]} of an area by its x co-ordinate is equal to the product of the whole area by the distance of its center of gravity from the same axis). In the figure the center of grav- ity of the moment-area 0"B"' is shown at C" •, and the cor- responding X is marked. 492 MECHANICS OF ENGINEERING. But we have ^y, or OOn, - \J Mxdx 1 -5- E/ =r r Mdx ,x'\^ EI\ Sv — - and hence we may write 00,,, or AV, = [(A^) ,i] -^ E7 . . . . (2) which furnishes us with a simple means of determining the dis- placement of any point 0^ in the elastic curve of the bent beam from the tangent-line at any other point B in that elastic curve. Evidently, from equations (1) and (2) we have A?/, = 00^^ = j>x ; and can therefore state that the intersection of the two tangent-lines, one drawn at 0, the other at B, lies in the same vertical as the center of gravity of the intervening moment-area- (N.B. Instead of the product {Ao)'X, we may, of course, use the algebraic sum of similar products for any component parts into which it may be convenient to subdivide the total moment-area.) 398. Examples of TJse of Eqs. (1) and (2) of Preceding Paragraph. Example I. Simple Cantilever. Concentrated load at free end. Fig. 444i. Constant E and I. {Prism.) Here the moment-diagram for whole length is a triangle (§ 249) whose base is I inches and whose altitude is PI inch-lbs. • Hence, with and B taken as in Fig. 445,, we note that AB-. '?=('•?). ...,= [.?]. 2 and that x = ^ Z. o E.I.== at On', while OOn = T^ 1 Fig. 445i. I.e., OOn^^ EI ['•"] PP ^ 2 EI' 1_ EI pn 2 3 for the slope i^l)' ^> Z = PP 3. EI Example II. Prismatic Beam on Two End-Supports, Load Uniformly dis- tributed over the whole span or length, I', W = wl. From p. 268 we know that the moment-diagram (Fig. 4462) is a symmetrical segment of a parabola with axis vertical, and that the moment at the middle section is Wl -^ 8. Also, from p. 12 of Notes, etc., in Mechanics, we find the x of the left-hand half of this moment-figure, measured from the left-hand extremity On, is ^ Z — | of 1 1; i.e., i = f of 1 1. The area of this semi-pardbolic-segment is two-thirds that of the circum- scribing rectangle. From symmetry, the tangent-line drawn at B, the middle point of the elastic curve, is parallel to On D, so that the displacement OnO of from that tangent is equal to the deflection of B from OnD. Hence -2 IWl 5 I- 3 •2* 8 OnO, (Ao).x EI ' a^ 384 EI ELBXUEE OF BEAMS: GEOMETRICAL TREATMENT. 493 Example III. Prismatic Beam. Ends Supported. Two Concentrated Loads Equidistant from Supports. Fig. 4453. Here, as before, from symmetry the tangent at B is horizontal, parallel to V yvYvVVvVYYVVYVVVYVV On C • D. -^ j^- FiG. 4452. Fig. 4453. OnD; so that On equals the deflection of B from C (its position before load- ing of beam). Each load P is in middle of a half -span. Required OnO ;, i.e., CB=? In this case the moment-diagram is easily shown to consist of a triangle at each end with a central rectangle of altitude = Pa (inch-lbs.). To find OnO we need the product (A^).x. But this A^ consists of the triangle 0"A"N with its center of gravity distant f of a from and of the rectangle A"B"KN whose center of gravity is at a distance of | of a from 0. Utilizing, therefore the principle stated in the N.B. of § 397, we write Q-Q _ (^)O- 1 r Pa 2a -r. 3 a --l-a.Pa-^ n Pa^ ~(S-Er' Example IV. Prismatic Beam on End Supports. Single Eccentric Load, P. Fig. 445^. Here a tangent drawn to the elastic curve at the load-point B, not being horizontal, is not par- allel to OnCn, and hence OnO does not = the deflection, 5, otB. However, the displace- ments ( = (^1 andd.,), of Ofrom 5's tangent and of C from B's tangent, are easily found, the moment-diagram 0"NC" having been drawn, in which 'B"N = {Pa,a, -^ I) inch-lbs. (§260). Call" the "moment- Fig. 445,. area" of triangle 0"B"N, A' ■ and that on right of load, viz. of C"B"N, A". Then, from eq. (2) of § 397, we may write Eld^ = A'z,; and Eld., = A"x.,. If now we draw a horizontal line, HI)., through the point B of the elastic curve, we note, from the similar triangles thus formed, the proportion -J — -r = — . From these three equations d^ and d., may be eliminated and d 494 MECHANICS OF ENGlISEEillNG. obtained; (since A and Xj = I a, ■) Pa{a^ ~ I, and A" = ^ Pa^^ -i- I; while Xj = ja^, We thus obtain 5 = (J Pa^^a.^-) -^ {EI, I) (3) This is for the load-point. For the maximwn deflection see next example. Example V. Maximum Deflection of Prismatic Beam. End Supports. Single Eccentric Load. Fig. 4i5g. To locate the lowest point D of elastic curve and determine its deflection, d, below the horizontal 0„B. Draw a tangent at D, also at B whose distance n from D is, as yet, unknown. Note that the tan- gent at D is horizontal. The moment-diagram is a tri- angle of altitude ik' ; (M' = Pah^l); denote the moment at B by m'. We have m' =.(n -h 6) .M'. Now the angle (p =d' -h I, and d' = (AI) .x-^EI = Fig. 4455. ^ M'l (a+| [J (a + &) - a])--E'I. . •. 6 EI4> = M' (2 a -j- 6). But0, = (^^) -- ^I, = n . ?n' ^ 2EI, and ^ = 0^ ; .-., finally, we have n = \^\b{2a+ b), which locates the point R. Now note that the intersection C lies in the vertical through the center of gravity of the shaded triangle (§ 397). Hence CB = In and therefore from similar triangles B8 = \ns. But RB, =d, = BS- BS, and BS = <p^.CB = <i>.\n. Hence d = |0n and finally by substitution, and with M' placed = Pah -H I, we have (with h> a) . Pah ^-\ EI [2a + 6] Vifi (2a + h) same as on page 258 399. The "Normal Moment Diagram.'' If a portion, OB, of a horizontal beam carrying loads, be conceived separated from the remainder of beam and placed on two supports at its extremities and B, while carrying the loads [say P^ and PJ originally lying between and B, the corresponding moment- diagram, 0'"TB"' of Fig. 446 may be called the "normal-moment diagram" for portion* OB (of original beam) and its load. If Vq is the pier reaction at left, we have for any section t (say between P^ and P^) x ft. from 0, the moment of stress-couple [call it Mn or " normal moment "] M„ Vf^x — P,(x — a) (1) Now consider OB in its original condition (see lower part of Fig. 446) when forming part of a much longer beam sup- FLEXURE OF BEAMS: GEOMETKICAL TREATMENT. 495 ported in any manner. If we consider OB, now, as a " free body," M^e must put in, besides the loads P^ and P^, a shear Jq and a stress-couple of moment Mq in section at 0, and Jq and couple of moment Mb at B. The moment in any section t of OB is now M = Mo + /o^ — Pi (x — a). Let 7 = difference between J^ and Vq, I.e., then M = Mo + 7x + [7o:c-P,(a;-a)] . (2) i.e. [see (1)], M = M^ + Vx + M„ (3) N, ~T^° :- Hence the moment M {=kwin. Fig. 446) of any section of OB is made up of a constant part Mw a part proportional to X, and a third part equal to the " normal moment " of th&,t section. Therefore, if, in the moment-diagram 0'B'B"wO" for OB we join 0" and B" by a straight line, and also draw a hori- zontal through 0'^ the ver- tical intercepts [such as uvo] between the line 0"B" [or "chord"] and the broken line 0"wB" are the normal moments for OB and its load, and the area (mom.- area) of the figure formed by these intercepts is equal to that of the normal moment diagram. It is also evident that the center of gravity of the figure 0"B"w lies in the same vertical as that of the normal moment diagram. (In the next paragraph the trapezoid 0'B'B"0" will be divided into two triangles, instead of into a triangle and a rectangle. ) Fig. 446. 496 MECHANICS OF ENGINEERING. 400. The Theorem of Three Moments. Let 0, B, and C, Fig. 446a, be any three points in the elastic curve of a homogeneous, continuousy and prismatic beam, origi- nally straight and hori- zontal but now slightly bent under vertical forces (some of which are reac- tions of supports; no loads or forces are shown in the figure). Let Mq, Mj, and M^ be the moments of the couples in sections 0, B, and C. The moment-diagram for portion OBC is 0'C'C"T^B"TP". Join 0"B' and C"B' ; also 0"B" and C"B" . At the point B of elastic curve draw a tan- gent mjn^ and join OC. Then Omo, or d^, is the displacement of point from 5's tangent, and d^ = Cm^, is the displacement of Cfrom the same tangent; while S is the deflection of Bfrcm the straight line joining and C. ■ Now the vertical displacement d^ = [mom .-area O'B'T^ X distance of its cent. grav. from 00"'\ -^ El. But the moment- figure O'B'T^, under OB, is composed of the two triangles shown and the '■'■normal moment-diagram ''for OB, viz. : 0"B"T^, whose mom.-area may be called A^ and whose center of gravity is x^ ft. from 00", while the corresponding distances for the triangles are i a and | a . Hence, from eq. (2), § 397, we have: (1) and similarly, with corresponding notation, for the right-hand portion, or- segment, BC-, of OBC (denoting the " normal mom.- area " C"B"T^ by A^ and reckoning x^, etc., from CC"), Eld. = i M^a^ ¥ ^2 + i M,a, . 7 a, + A.x^. (2) If now a straight line be conceived to be drawn 'through B parallel to OC, we have, from the similar triangles so formed. FLEXURE OF BEAMS ; GEOMETRICAL TREATMENT. 497 (as ill Fig. 4454), {d, -h) ^ a, = {h- d,) -^ a, this with eqs. (1) and (2) we have finally Combining Mpg, M^(a, + aS) M^a^ A^ A^x + + + + EI8 (4) 6 ' 3 '6 1 which is the "Theorem of Three Moments." E is the modalus of elasticity of material of beam, I the " moment of inertia " of its cross-section ; M^, M^, M^, the moments of stress-couples (" bending-moments ") at 0, B, and C respectively. Distances a^ and a^ are shown in Fig. 446a, while A^, A^, x^, and x^ are as above ; 8 being the deflection of point B from the straight line joining and C. U.B. It should be carefully noted that eq. (4) does not apply unless the part of beam from to C is continuous and pris- matic ; also that in its derivation, the elastic curve is considered concave upward throughout ; hence if a negative number is ob- tained for Mq, M,, or M^, in any example by the use of eq. (4), it implies that at that section the beam is convex upward, instead of concave ; in other words that the upper fibers are in tension and the lower in compression (instead of the 'reverse, as in Fig. 446a). 401. Values of ^,1-, and A2X2 iii Special Cases. The Theorem of Three Moineuts involves the use of the (imaginary) normal mom.-area of each of the two portions (left and right "spans", or "panels"), OB and BC ; i.e. of the products ^jXi and A^x.^, where Xj is measured from the left end of the left panel, and x, from the right end of the right panel. "We are now to determine values of ^jX, and A^x., for several ordinary cases of loading. I. Single Cen- tral Concentrated Load, P, Fig. 446j. Here, for a left-hand panel, PI I I 4 2 2' A^x,= — ; and for a right- hand panel, AoX., = PP 16"' 1^ X Fig. 446i. Case II. Single Non-central Concentrated Load, P. case as a left-hand panel, Pbc (I + c) A,x^= [- Fig. 4462. Fig. 4463. For this while, as a right-hand panel, -42X2= 498 MECHANICS OF ENGIKEEKING. Case III. Two (or more) Concentrated Loads. Fig. 4463. A^, = I [P'b'C (I + b') + P"b"c" {I + &")] ; and for each load more than two add a proper term in the bracket. For A.^2 interchange b' and c', b" and c", etc. ' B o -hi ^%- Fig 446,. Fig. 4464. Case IV. Any Continuous Load over a Part or the Whole of the Span; of w ibs. per linear foot, w being variable or constant. Fig. 446^. The load on a length dx (of loaded part) is wdx lbs. ; comparing which with the P of Case II, (or one of the P's of Case III), we note that x corresponds to b, and l—x to c ; — 1 /^X=Ci 1 /»Cl hence AjX^ = ^ t wdx {I — x)x{l + x) =g- t wx {P — x') dx. If w is variable it must first be expressed in terms of x. (For A^^ we measure x from the right-hand end, B.) Case V. Uniformly Distributed Load over Whole Span ; (i. e. , w is constant). Let W, = wl, = whole load, lbs. Fig. 4465. o liilUIIIUiUlB oliiiUUU A^x, = A.x^ -2 I Wl I ^WP ~ 24* Case VI. Uniformly Distributed Load Ad- joining one End of Span; (left end for example). Fig. 446g. Total load =W= wb. Applying method of Case IV, withCj = &, and b^ =0, we have A^^ = J^ Wbd^ — \ b^). Also from Case IV, now measuring x from B,A.j^, = ^\W{l'-c^) (l + c). Parabola Fig. 446s. FLEXUJRE OF BEAMS ; GEOMETRICAL TREATMENT. 499 Case VII. Uniformly Distributed Load Not Adjoining either End of the Span. Fig^. 446. Whole load = W= w (e- 6). By Case IV, we tind A,x, — W{e + b) -[''- e2 + 62 ■ ■A^2 ~~ 12 , L ' 2 J ' 12 ■D'-^l It is now seen how A^x^^ and ^2*2 ™^y be obtained for any loading. 402. Continuous Girders Treated by the Theorem of Three Moments. This theorem is of special advan- tage in solving continuous beams (p. 271) ; and examples will now be given. Example I. Fig. 447^. A straight, homogeneous, prismatic beam or girder, 35 feet long, is placed upon three supports at the same level, forming two spans of 15' and 20'; two concen- trated loads in the left span, a uniformly distributed load on part of right span. Required the maximum moment, and maximum shear. (Neglect weight of beam.) Take 0, B, and C, as the three sections where the three moments Mo, M^, and M^ are situated [respectively] used in the theorem of § 400. But both Mo and M^ are zero in this case, and 8 (deflection of point B from line joining and C) is also zero (since the supports are on the same level). Hence M^ (i.e., at B) is the only unknown quantity in applying the theorem of § 400 (eq. (4) ) to this problem. Taking the A^x^ from Case III, and A^x^ from Case VII (with /= 0), of § 398, we obtain (using the foot and ion as units), Mr^^ + ^O) _^ ^ _^_ _1 1-6x4x11x19+8x10x5x251 3 6 X 15 •- -" + 16x16 12x20 202- 16^ 1=0; and .'. M, = - 39.2 ft.-tons. The negative sign shows that at section B the elastic curve is convex on its upper side (see N. B. in § 400). To follow up the solution from this point, let us draw the actual moment- diagram somewhat differently from that in Fig. 446a, (which see), where the actual moment for any section is measured from 500 MECHANICS OF ENGINEERING. a continuous horizontal line, O'C, as an axis. Let the « chords " 0"B" and B"C" of the two normal moment fig- AVo >'E -f.^>-^6'- >'F Vb 10 urns Hy i y V I i l i i I i i I i AVo k -^—i5'-j- — M-t-- —16'— >-l: Fig. 447i. ures, 0"B"T^ and C"B"T^, be made a contin- uous horizontal line by an up- ward shifting {each in its own vertical) of the intercepts of those figures. The intercepts 0"0', B"B', and C"C', and the four tri- angles involved with them, now extend upward from that hori- zontal line. But in our present problem both M^^ and M^ are zero; hence the two upper triangles disappear and the two inner triangles project above the horizontal line, with M^ as a com- mon base. The actual moments of the points of the elastic curve are now measured (in general) by the vertical intercepts between the lower boundary of the normal moment figures and the upper edges of the two triangles ; but since in the present case ilf J is negative, M^ must be laid off below the (new) hori- zontal line so that the lines 0"B" and C"B" will cross the lower boundaries of the normal figures ; the actual moments being now measured by the vertical intercepts between these two oblique lines and the curved (or broken^ boundaries of the normal figures. This re-arrangement has been observed in the moment-diagram 0"B" G" of Fig. 447j, where line-shaded areas correspond to the parts of the elastic curve which are concave upward; and the dot- shaded areas, to parts convex upward (or upper fibers in tension). Before determining the shears, J, along the beam, we must first determine the reactions at the support, viz. Vq, Fgj and V^ Consider the portion OB as a " free body", cutting just on left FLEXURE OF BEAMS: GEOMETRICAL TREATMENT. 501 of support B, and put ^ (moments) = about the neutral axis of section at B -, deriving 6x11 + 8x5- 39.2 _ 7^ x 15 = 0; and .-. Vo= 4.5 tons. Similarly, with EC as free body, taking moments about B, 16 x 12 - 39.2 - y^, X 20 = 0; and .-. Vo = 7.6 tons; and hence, since V o + Vb + Vc = ^^ tons, Vg = l'^-^ tons. The shear-diagram is now easily formed (see Fig. 417^) ; the Jiiax- imum shear being evidently 9.55 tons, occurring in the section just on the left of support B. We note that the shear changes sign three times, corre- sponding to the three (local) maximum moments (at E, B, and ■K). To locate, and determine, M^, note that the change of sign of the shear at I) is gradual and that hence the shear is exactly zero at K; which requires that "the load between K and the support be equal to the reaction at C, (from the free body concerned). Since w along HC is one ton per foot, the dis- tance ^Cmust be 7.64 tons -v- 1.00, = 7.6 ft. From this free body, KC, we now find, by moments, that ilf^ = 29.2 ft.-tons. As to the other maximum moment, i.e., at E, we note that the moment at E in the normal moment^figure would be E^'' E" = 28.2; from which by deducting t*^ of Mb (i.e., of 39.2) we obtain M^, = 17.7 ft.-tons. Hence, the greatest moment to be found in the beam is that at B, viz. 39.2 ft.-tons, and upon this depends the choice of a safe and economical beam. . Example II. Fig. 4472, Con- tinuous prismatic beam OG. Three supports at same level. Find maximum moment, etc., under given loading, the 12 tons being uniformly distributed over whole of right-hand span. Neg- lect weight of beam. i/ max. =ilfi,= 16.6 ft.-tons. Ans. rj ions -^5 ■ "■B- us 20- -40- ''10 if 10 tons Fig. 4473. ■^—16'— 4^ > ' 16 tons Fig. 4472. Example III. Continuous prismatic beam on three supports O, B, C, at same level. Three concentrated loads. Neglect weight of beam. Find Mb and maximum moment, etc. Mb= - 92.6 ft.-tons. Max. if = 116 ft.-tons, at D. Ans. 502 MECHANICS OF ENGINEERING. Example IV. Continuous prismatic beam, 40 ft. long and extending ovei four supports at the same level. The loading is symmetrical, as shown (Fig. lUlUiUiUl UllilUllUio 4474). Here we note that from symmetry Jf^ must = Mc; also Mo and Mj) each = 0. Applying the Three-Moment Theorem to O, B, and C, (with 5=0) we find + 1 Jf5X26+lilfBXl2+ l.?|xl7+ J-^.i X 12^ =0; and . •. Ms = — 23.7 ft. -tons, ( = Mc, also). Completing the mom. -diagram we find that Mb(= Mc), or 23.7, is greater than any other moment along the beam; .-. M max. =23.7 ft.-tons. The reactions of the supports are found to be : Fq (and Vd) =8.3 tons ; and Vb 25 tons (and Vc) = 16.6 tons. Evi- illliJ,!! dently the elastic curve is con- vex up, over both B and C. ' 10 tons -<r—:is!- >\ ^-6': \<—7'- S! ^ Fig. 4476- The maximum shear is 11.8 tons, close on the left of sup- port B, (or close on right of C) . Example V. Continuous prismatic beam, 38 ft. long, on three supports at the same level. Fig. 447^. Uniformly distributed loads over portions of the length. Find the maximum moment and maximum shear, (J). Max. if = - 39.6 ft.-tons (at B) ; Max. J = 10.2 tons (close at right of B) Example VI. Fig. 447„. Continuous prismatic beam, 50 ft. long, on /our supports at same level ; but the arrangement of loading and span-lengths is non- symmetrical. Fig. 4476. Find the max. M and max. J. In this case Mo and Md are each = 0, but Mb is not = Mc- We are therefore compelled to apply the Theorem of Three Moments twice, viz. : first to the three points 0, B, and C; and then to the three points B, C, and B ; whence we have - Mb (14 + 20) McX 20 16 X 6 X 8 (14 + 6) 20 x 2D» Ans. 6 X 14 24 X20 Mb X 20 Mc (20 + 16) + + 20 X 203 16 X 7 X 9 qe + 7) _ 24 X 20 6X16 FLEXURE OF BEAMS; GEOMETRICAL TREATMENT. 503 (1) (2) or, 68Jlf5 + 20 if c+ 1097.1 + 2000 = 0; .... and 20 Mb + 72 if c + 2000 + 1358.3 = ; .... two simultaneous equations, foi- determining ifs and Mc- Solving, we find if 5 = - 34.6, and Mo = - 37.0, ft.-tons. The three "normal mom. -diagrams " having been drawn to the common base 0"B"C"D" in the figure, we lay off B"B' downward from B" and aiiiuiiiiuiiiic ■''-\-s'- 20'. 9^ —7^ Iq" Ic" Id" 1^— -:J4 16'-\- Fig. 4476. = 34.6 ft.-tons; and C"C', also downward, = 37.0; and draw the straight lines 0"B\ B'C, and C'B" ; thus completing the mom. -diagram, in which, as before, the differently shaded portions show whether the elastic curve is con- cave up (line-shading), or convex up (dot-shading), in the corresponding part of beam. The four reactions of supports are then found, viz. : To, F^, Fc, and Vd, = 6.7, 19.2, 19.0, and 6.1 tons, respectively. Shears are now easily found and are shown in the shear diagram, the max. J being 9.8 tons, occurring just on the right of support B. The max. if is found to occur at E, and to have a value of 42.7 ft.-tons. 402a. Continuous Beam with One Span Unloaded. In Fig. 448 we have a continuous prismatic beam supported at three points 0, S, and C at the same level ; but the support at is above the beam, instead of below; since that end tends to rise, there being no load in the left-hand span. (Case of a drawbridge with the left-end " latched down "). Neglect weight of beam. By the Theorem of Three Moments applied to 0, B, and C, with Mq and Mc = 0, and Mb unknown, we find Mb = — 4.9 ft.-tons. In forming the moment-diagram here, we note that since there is no load from to B the lower edge of the "normal mom.-diagram" 604 MECHANICS OF ENGINEERING. J), tons ■ Fig. 448. for OB coincides with its upper edge, i.e., with the axis itself, viz. 0"B". The "normal mom.-diagram " for BG is, a. triangle, with. B"Q" as base. Laying off B"B' = 4.9 downward from B', and joining 0" B' and B'C'^-, we complete the actual (shaded) mom.- diagram, as shown. Max. moment is found under the load and = + 9.55 ft. -tons. To find the reaction at 0, take OB as " free body", cutting close on left of ^. (See (i^) in the figure. Note the position of stress-couple at right-hand end of this body.) By moments about B we have V a X 10 — 4.9 = 0, whence V q = (say) 0.5 tons. The other reactions and the shear diagram are now easily d'etermined. 403. Supports out of Level. In the foregoing examples the quantity S has been zero in each instance of the application of the Theorem of Three Moments ; but when such is not the case the quantities E and I are brought into play. In this con- nection it must be remembered that any unequal settling of the supports (originally at same level) after the beam has been put in place, may cause considerable changes in the values of the various moments and shears, and consequent stresses in the material. (See lower half of p. 323.) 404. Continuous Beam with " Built-in" Ends. Fig. 449. As a case for illustration take the prismatic beam in Fig. 449, " huilt-in " or clamped, horizontally, at B and at C ; at the same level. A load P is placed as shown. On account of the mode of support the tangents to the elastic curve at B and will be horizontal and are coincident ; so that portions of the curve near the ends are convex up a-> Fig. 449. Now conceive the beam to be sustained at i? by a simple FLEXURE OF BEAMS: GEOMETRICAL TREATMENT. 505 support underneath and to extend toward the left a length Qq at the end of which a support, 0, is placed above the beam, and at same level as B and C (allowing for thickness of beam). This makes an additional span (with M q = zero) ; and the tan- gent to elastic curve at B will no longer be horizontal. But it may he made as nearly horizontal as we please, by taking ttg small •enough (supposing no limit to strength of beam). When a^ = zero the tangent at B will be in its actual position (horizontal). We may therefore apply the Theorem of Three Moments (§400) to 0, B, and C, [noting that there is no load on 5], if we write both Mq, and a^, = 0, whence (see also Case II of § 401), a(Sa + 2a) ^^ Ms (0 + 3a) ^ Mc^ ^ Q ^ P^a 0. 3 ' 6 ' ' 6 x,3a Similarly, by conceiving the beam extended to tlie right, a dis- tance a' to a point D, for another support, etc., we may apply the theorem to the three points, B, C, and D in like fashion, with Mo and a' = 0, obtaining Ms^a , M^(3a + 0) 6 3 + + P2a. a (3a + a) 6 X 3a + = 0. Elimination gives Mb = — t: Pa-, and Mc = V -Pa, ft. -tons. y 405. Deflections found by the Theorem of Three Moments for Prismatic Beams. Since this theorem contains h (see Fig. 446a) the deflection of the point B of the elastic curve of a con- tinuous prismatic beam from the line joining the two others, and C, we may use the theorem in many cases for determining deflec- tions when the three mcments are known. Example I. Fig. 450. Case of two end-supports and a single non-central load, P ; (with weight of beam neglected). Taking 0, B, C, as the three points (i.e., OB is the left-hand, and BO the right-hand, span: icith no lead on either span) we have, with Mq and Mc = 0, and Mg = Pa^a, -i- I, Fig. 450. + ~ i^ - + -h I 3 = ElS a J d = Pa,%^ SEI ' 506 MECHANICS OF ENGINEERING. Example n. If n is any point between and 5, at x ft. from 0, and 0, n, and C are taken as the three points for the theorem, we may find 8„, the deflection of n below OC^ That is, with Mo and Mc = and Mn = P{a^ ^ I) x, +^^L£-ii + 4- + -P^2(c^t -^)i{l -^) +^ 2] 3Z ' Q{1 — x) = EI8„ X I — xj or, after reduction, ^"^eif^ [z^-<-a:^> (4) Now if ttj > a^, we may find the distance x\ from 0, of the point of maximum deflection, by putting " ■ = : whence is dx obtained x' = V ^ {l^ - a,') C^) and this substituted in (4) gives Pa )EI (Compare with pp. 258 and 494.) max. deflection, = -^^ (f - a}) V\ {I' - a/) ... (6) (The following example is the one referred to at the loot of page 514.) Example. — A hollow sphere of mild steel, of thickness 2 in. and internal radius of ?'o = 4 in., contains fluid at a pressure of 2 tons/in.^ Find max. stress and max. strain; with £'=15,000 tons/in.^ and A; = 0.30. Here »i = 1.5; and by substitution in eq. (30) we obtain max. hoop stress to be §■0= — 2.26 tons/in.^ (tension), while from eqs. (22) and (23) the tangential, or hoop strain, at inner surface is found to be £3=" 0-000145 (elongation), and the radial strain to be £1= +0.000224 (shortening). The latter strain, £■^, is seen to be the greater and the ideal ^'equivalent simple stress" (see § 4056) is ££1=4-3.35 tons/in.^, compression, i.e., much larger than the actual max. stress (2.26) in this case. On the "elongation theory" (see § 4056), the 3.36, and not the 2.26, tons/in.^, is the figure that, for safety, should not pass a prescribed unit stress ais Liferred from com- pressive tests with an ordinary testing machine. THICK HOLLOW CYLINDERS AND SPHERES. 507 A' d\ CHAPTER XIII. THICK HOLLOW CYLINDERS AND SPHERES. 405a. General Relations between Stress and Strain. — (Elas- tic limit not passed.) If a small cube of homogeneous and isotropic material, dx inches long on each edge, is subjected to a compressive stress of pi Ibs./in.^ on two opposite faces, not only is its length in direction of the stress diminished, and by an amount dX, but its lateral dimensions are increased by an amount d/' which is a certain fraction (from 0.20 to 0.35 for metals) of dL This ratio, or fraction, is called Poisson's Ratio, and will be denoted by k. Thus, in Fig. 450a we have such a cube, AD being its unstrained form. Axes 1 and 2 are in the plane of the paper while axis 3 is 1 to the paper. On the left and right faces is shown acting the compressive unit-stress pi Ibs./in.^, A'D being the form of the cube under this stress. If now E represent the modulus of elasticity (Young's) of the material, we have (see p. 203) denoting the ratio dX-^dx, or relative decrease in length, by si, ei = px -r-E; so that if dX" is the increase in length of the vertical edges we havedX" -i-dx (call this ratio £2=^—kpi^E; while the relative increase of length in the horizontal edges 1 to paper will be an equal amount, viz., e^^—kpi-^E. These ratios £i, £2, and £3 are called the strains along the three axes 1, 2, and 3, respectively, and are abstract numbers. Hence the three strains produced by the stress pi acting alone are P\ kpi kpx . £i = ^; ^2=-~-^; and £3=--^. . . (1) Now if while pi is still in action a compressive stress of P2 lbs./in.2 acts on the two horizontal faces, and also a com- pressive stress of ps on the two vertical faces which are parallel to the paper, the total strain in the direction of axis 1 (that is, the relative shortening of the cube in that direction) will be, by superposition, £r 1- dx Fig. 450a. Pi k{p2 + Pz) J • M 1 • +U -^ ^ — ; and similarly, m the directions of the other two axes, we have £0 — ^ — P2 kipi + ps) E E and £3 = Ts — Ps k(pi + p2) E E (2) 508 MECHANICS OF ENGINEERING. (This form of stress-strain relation is due to Grashof.) Note that if either p^, p2, or ps is a tensile stress, a negative number must be substituted for it; and that if a negative num- ber is obtained for si, £2, or £3, in any problem, it indicates a lengthening instead of a shortening. Similarly, if the con- dition is imposed that the strain £2 (say) shall be a relative elongation of 0.00020,-0.00020 must be substituted for it in above relation. 405b. "Elongation Theory" of Safety. — In all preceding chapters the criterion of safety has been that the unit-stress in the element of the elastic body where the stress is highest, regardless of stress on the side faces, should not exceed a cer- tain value, or working stress, =R' Ibs./in.^, as determined upon by a consideration of the stress at "elastic limit; " this "elastic limit " being itself determined by the ordinary ex- periments on "simple " tension or compression of rods of the material in question, there being no stress on the sides of the rod. In such experiments, however, an element with four faces parallel to the axis is subjected to stress, say p, on two (end) faces only; and the question naturally arises whether the elastic limit would be reached for the same value of p as before, in case there were also present tensile or compressive stress acting on the side faces of the element. Experiments which would throw much light on this point are unfortunately wanting, and some authorities,, notably on the continent of Europe, contend that the extreme limit of safety, as regards state of stress in isotropic materials, is when the greatest rela- tive strain (elongation or shortening), say £1, is as great as would be produced at the elastic limit in an experiment involving only "simple " tension, or compression (as above described), in an ordinary testing machine. This view would make the greatest "strain," or deformation (change of form), the cri- terion of safety instead of the greatest stress. Now if a stress of "simple " tension, =p', (no side stresses) acts on an element, the highest strain produced is in thQ direction of this stress and has a value e' = p' ^E, since Young's modulus, E, is deter- mined by experiments of this very nature; that is, p' = Ee'. Hence if the greatest strain in an element in some compound state of stress, as in § 405a, is £1, and it is desired to place it equal to | (say) of the £1 in simple tension at elastic limit, we may write £i = f £' = |(p'-^£'); or E£i = lp'. If now we THICK HOLLOW CYLINDEES AND SPHERES. 509 denote |p' by p" we may write Eei==p" and describe j/' , or Eei, as the ideal tensile stress which would produce a strain, or relative elongation, equal to £i in case there were no side stresses; Cotterill calls this ideal stress (Esi) the ^'equivalent simple stress." For instance, if on an element of the shell of a cylindrical steam-boiler of soft steel the "hoop stress " (p. 537) is pi on two end faces and the stress on two of the other faces is p2, = -|pi, (the stress on the remaining two opposite faces being ps = practically zero in this connection) we have for the strain in direction 1, by eq. (2), Eei = pi-k{hpi + 0). . . . (2') Let p", =—15,000 lbs./in.2, tension, be the safe working stress for the metal in simple tension; with £' = 30,000,000 lbs./in.2, and Poisson's ratio = A; = 0.30. Then according to the view of preceding chapters the greatest safe value for the stress pi would be —15,000 Ibs./in.^ But according to the new view now being illustrated the safe value of pi must be determined by limiting the strain si to a value which would be produced by 15,000 Ibs./in.^ in simple tension, i.e., —0.00050, (which =^p" -^E); which amounts to the same thing as re- quiring that the ''equivalent simple stress" shall =15,000 lbs./in.2 Hence, substituting in eq. (2'), we have - 15,000 = pi(l- J. (0.30)); i.e., pi = -17,600 lbs./ in.2 tension; which is considerably greater than the 15,000 allowed by the older theory. The relation thus brought out in this case that the tenacity of a material is increased by the presence of lateral tension "can hardly be considered as intrinsically probable, and such direct experimental evidence as exists is against the supposition " (Cotterill). But in many cases the results of this "elongation theory " are more probable than those based on the older theory ; hence the former is much favored by continental writers. 405c. Thick Hollow Cylinder. Stresses and Strains. — Fig. 4506 shows a longitudinal section of a thick hollow cyUn- der of homoge- neous and isotropic material (say steel or iron) provided with end stoppers (^no iriciion nor vj/,,,,,,,^,,,,,,. „y ,,^yy„,,^//,/,,///////,/// leakage); and Fig. Fig. 4506. 450c a transverse section, giving dimensions. Fig. 450c. ro is the inner 510 MECHANICS OF ENGINEERING. radius and nro the outer radius {n is a ratio). Fig. 450c also shows (dotted Knes) an elementary hoop, or shell, of inner radius r and outer radius r + dr. The interior of the cylinder is filled with fluid under a high pressure, po Ibs./in.^; and it is required to determine the stresses and strains in a cubic element in any elementary hoop or sign such as ABC, Fig. 450c. Let the half hoop ABC of Fig. 450c be considered as a "free body " in Fig. 450d, showing also at 3 a small cubic element, as mentioned above. The compressive stress P+dpk \ A-j-,'^ on the inner surf ace of the hoop isp (radial), \l jK^^'^^jT exerted on it by the adjacent inner hoop; while on the outer surface of the elementary \ I hoop, and exerted on it by the adjacent r___i-Ji— a'- outer hoop, is the compressive stress p+dp. The thickness of the hoop is dr. The stress on the edges, A and C, of this free body (half hoop), will be taken as compressive /\ ^h-y^l" at first, of intensity q Ibs./in.^ Let the 21 / / f s ^ hoop or thin shell have a length =Z, 1 to Fig. 450d. ^^^^^ -^ ^ ^ longitudinally (see Fig. 4506). Now for this free body put IX = and we have (see pp. 525 and 526) {p + dp)i2r + 2dr)l-2ql . dr~p{2rl)=0 . .(3) i.e., pr + r . dp + p . dr + dp . dr — q . dr—pr = . . (4) and hence, omitting the term dp . dr of the second order, r . dp + p . dr^q . dr (5) which is a differential equation of stress. Next consider the relations of stress and strain to be found in the small cube at 3, Fig. 450d. It is subjected to a compressive stress p along the radial axis 1 ; to a compressive stress q a long the tangential axis 2; while on the front and back faces the stress is p3 = zero, parallel to axis 3 (t to paper). Let now oi denote the radial strain, £2 the tangential, and £3 the axial strain, this latter being parallel to the axis of the cylinder. All of these strains are supposed to be shortenings for the present; and from the circumstances of the case the third strain £3 (axial) is con- sidered constant (i.e., the same for all values of the variable r), since the cylinder is under no constraint as to longitudinal change of form. We may therefore write the relations (see § 405a) Ee, = p-k.q, (6); Ee2 = q-k.p, (7); Ee3 = 0-k(p + q), (g) THICK HOLLOW CYLINDERS AND SPHERES. 511 From (8) we have q=—'p—{Eez-^k), which in (5) gives r .dp + 2pr .dr=-{Ee3^k)dr .... (9) Now multiply by r (integrating factor) and denote Eos-^k by A, an unknown constant, (unknown since it contains the strain £3) and we have r2. dp + 2pr . dr=—Ar . dr; . . . (10) that is to say, d[r^p]= —Ar . dr; . .' . (11) which may be integrated; giving, r^p=—^Ar^-\-C; . . (12) where C is a constant of integration. The two constants A and C may now be determined by substituting in (11) the values To and po which the two variables r and p assume at the inside surface of the cylinder. Similarly, at the outside surface r and p have the values nro and (atmospheric pressure relatively small and hence neglected); which being placed in (11) give rise to a second equation, which like the first contains constants only. From these two equations we easily find . 2po , „ n^ro^po A = -^ — r; and C = ^ — r-; n^ — r n^ — 1 and hence finally, from equations (12) and (8), P=^[^-'l (13);and,= -J^j[5>Vl]. (14) From (13) and (14) we may find the stresses p and q for any value of the variable distance, r, from the axis. The negative sign for q shows that it is in reality a tensile stress, the reverse of the character assigned to it at first; i.e., it is a "negative compressive " stress for this case of fluid pressure acting inside the cylinder. Both p and q have maximum values at the inner surface and diminish toward the outside. Example. — With inner radius rQ = 4 in.; and outer, =5 in. (hence w = 5/4 or 1.25); and j9o = 800 lbs. /in. ^; we find q max. (or q^), at inner surface, to be 3644 Ibs./in.^; while at outer surface g= — 2844 lbs. /in. ^, tension. If the metal is cast iron we may put ^ = 0.23 (see p. 230) and £' = 15,000,000 lbs. /in. ^, and thus obtain for the radial strain at the inner surface, e,= + 0.000109, indicating a shortening; and for the "hoop" strain (or tan- gential strain) at the same place, £2= ~ 0-000255, i.e., a relative elongation of about 2^ parts in 10,000. (The student should verify all the details of this example, carefully noting the proper signs to be used). 405(1. Thick Hollow Cylinder under External Fluid Pressure. — If the cylinder is siirrounded by fluid under high pressure, pn lbs. /in. ^, the internal pressure Po being practically zero (atmosphere) in comparison, we must determine the constants A and C of eq. (12) on the basis that p = for r = ro and that p = pn for r = nr^ ; whence, finally, we obtain ^=5^('-'^)^ ■ ■ ■ ('« "-^^ii^^')) a« for the stresses at any distance r from the axis. Here both p and q are com- 512 MECHANICS OF ENGINEERING. pressive stresses; the latter increasing, and the former diminishing, toward the center. Evidently if the cylinder were not hollow, but solid, r^ would = 0, and n = cc ; and both p and q would be constant, =pn, at all points. 405e. Approximate equalization of the tensile hoop stress in successive rings of a thick hollow cylinder under internal fluid pressure may be brought about by using two or more separate, cylinders of which fhe outer ones are successively "shrunk on" before the fluid is introduced. For instance, with only two cylinders, the outer one is first heated to such an extent that it barely fits over the inner one, the latter being cold. When the compound cylinder has cooled the outer one has shrunk and is in a state of hoop tension, while the inner one is in a state of hoop compression. The radial pressure between them, at their siirface of contact, is an internal or bursting pressure for the outside cylinder, and an external or collapsing pressure for the inside cylinder. The original and final temperatures being known, we are able to make use of the foregoing equations [(6) to (16)] to compute all stresses and strains thus induced before the fluid is intro- duced into the interior of the smaller cylinder. When the internal pressure is eventually produced, the hoop stresses in the smaller cylinder, initially compressive, will be con- verted into moderate tensions and the tensile stresses in the external cylinder will be increased; but the maximum ten- sion is not so great as if the complete cyhnder had been origi- nally continuous.. (See Prof. Ewing's Strength of Materials). In the case of thick hollow cylinders subjected to the severe internal pressures needed in the manufacture of lead pipe, to produce ''flow" of the metal, it is well known (Cotterill) that a permanent increase in the internal diameter takes place, showing that in the inner layers of the cylinder the elastic limit has been passed. In this way an approach is made to equalization in the hoop stresses of all the layers and the above formuliE no. longer hold; but the cylinder as a whole is not injured, having simply adapted itself better to its function. Cast-iron hydraulic press cylinders are sometimes subjected to the high internal pressure of 3 tons/in.^ If the cylinder is short, its resistance is doubtless much increased by connec- tion with the end plates, or ''domes." 405f. Equation of Continuity for Thick Hollow Cylinder under Stress. — (Cotterill.) In Fig. 450e we have mABCD the form and position of a " cubical" element (of an elementary hoop) between the two radial planes ABO and Z)CO, in its con- THICK HOLLOW CYLINDERS AND SPHERES. 513 dition of no stress. After it is subjected to stress it will still be iound*between the same radial planes and will occupy some position A'B'C'D'. The radial' thickness AB = dr, or t, f i will have changed to t', / /L ? ' BC has changed to B'C, etc. With axes 1 and 2 as shown, 1 being radial, and 2 tangential (or cir- I" cumferential), we note that the whole circumfer- ence of which BC is a part has shortened from a length 2Tzr to 27rs, so that 27rs=2ffr(l— Sj), £2 being the value of the tan- gential strain, or "hoop" strain, at distance r from center 0; and similarly 2ns' = 27zr'{l— £'2), where £3' is the hoop strain at distance r' (i.e., r+dr) from 0. But £2 varies with r, so £2'= ^2 + t ■ -j^- Hence, subtracting. FiG. 450e. t' = t-t£,-r't s=(r'- ds2 dr ■r){l~e2)- dr' -r't-r. dr or. whence £1 — e. = r d£. -r't'^- dr ' dr (17) (18> .(19) which is the "equation of continuity of substance," of the cylinder. Since in the foregoing cases of thick hollow cylinder, under bursting or' collapsing fluid pressure, both s^ and $2 may be expressed as functions of^ r, it is a simple matter for the student to show that (19) is verified in those cases, and that hence the solutions given are adequate. Evidently eq. (19) also holds in the case of the thick hollow sphere, where there is a hoop strain £3, q to paper in Fig. 450, equal to that, £2? along axis 2. lOcg. Thick Hollow Sphere under Internal Fluid Pressure. — (For thin- walled spheres see p. 536). As thick hollow spheres are sometimes used' to hold fluids under high pressure, and the halves of such spheres fre- quently form the ends of thick hollow cylinders, the following treatment will be of practical value. In Fig. 450/ we have a cross-section of the sphere through the center. The inner radius is Tq; the outer, nro (where n is a ratio) ; while the (variable) r is the inner radius of any thin spherical shell, Fig. 450/. Fig. 450?. of thickness dr, an infinite number of which make up the complete sphere. Though each of these shells is under a "hoop tension," when the internal fluid pressure is in action, we shall at first deal with this stress as if compressive, for uniformity in applying the principles of § 450a. 514 MECHANICS OF ENGINEERING. Consider as a free body any hemispherical shell as shown in Fig. 450g. The pressure (radial) on the inside, from the adjacent shell, is p lbs. /in.*; and that from the adjacent shell on the outside is p + dp, or p'. The radius of the outside will be called r', { = r + dr). The "hoop compression" on the thin edge of the shell is q Ibs./in.^ These three quantities, p, r, and q, are variables, i.e., refer to any infinitesimal shell of the sphere. The strains affecting any small "cubical" block in any shell are p, radial; q, tangential; and 53, =q, along a tangent T to the first mentioned. Taking an axis X through the center and T to sectional plane AB, and putting 2 (X-components) = 0, for equilibrium, we have 7ir'^p' — 7rr^p—27:r.dr.q = 0; i.e., r'^p' — r^p = 2qr . dr. But the difference, r'^p' — r^p, is nothing more than the increment accruing to the product r^p when r takes the increment dr, and is therefore the differential of the quantity or product r^p ; hence we may write d{r^p)=^2qr . dr; or, r^ . dp+2pr . dr = 2qr . dr . . . (20) We next consider the relations of stress and strain in the small cubical element shown in Fig. 450/i, having four radial and two tangential faces. The unit stresses on the faces are as there shown; p on the two tan- gential, and q on the four radial, faces. Radial strain = e^ and hoop strain (tangential) = £2 = same for any tangent. It has already been proved in §405/ that e,-e, = r.^; (21) and we also have, from § 405a, Eei = p—2kq; (22) and Ee2 = q-k{p + q) (23) From the four equations, (20), (21), (22) and (23), containing the five variables p, q, £1, £2) and r, we may by elimination and integration finally determine p and q, each as a function of r; as follows: (C, C", Cj, C^, etc., will denote constants of integration, or, constant products involving E and k. From (22) and (23) we obtain a value for £1— £n which in (21) gives rise to an expression for p—q. Another expression for p—q is obtained from (20) . Equating these two expressions we derive —dp = Ci . dsi', that is, by integration, — p = Ci£2 + C2 (24) By elimination of £2 between (24) and (23) we obtain q in terms of p which substituted in (20) produces r^ . dp + 2>pr . dr = C^r . dr .... (25) Eq. (25) is made integrable by multiplying by r (integrating factor); whence r^ . d'p + 3pr^ . dr = C^r'^ . dr The left-hand member is evidently d[r^p\ Therefore (^[r^^?] = C^r^ . dr ; C" or, integrating, Hp = ^Cor^ + C^ ; that is, p =C'-\ — 3-, . . (26) dp 3C" C" whence, also, -f^= ^; which in (20) gives rise to q = C' — -^. . (27) We may now determine the two constants C and C" substituting in eq. (26), first p = P(, and r = rg ; and then p = with r = nr^. The values of C" and C" so obtained are placed in (26) and (27), resulting finally in the relations P=„-^5?^-l)^ • (28) and ,_--.?i-.("^+l). . (29) The negative sign of q shows that it is a tensile stress, or "hoop tension." It is evidently a maximum for r=rf), this maximum value being Pi ^(^+0 <^« (For a numerical example see foot of p. 506). go, =gmax.,= . ^^ PART IV. HYDRAULICS. CHAPTEE I. DEFINITIONS— FLUID PRESSURE-HYDROSTATICS BEGUN. 406. A Perfect Fluid is a substance the particles of which are capable of moving upon each other with the greatest free dom, absolutely without friction, and are destitute of mutual attraction. In other words, the stress between any two con- tiguous portions of a perfect fluid is always one of comjpression and normal to the dividing surface at every point ; i.e., no shear or tangential action can exist on any imaginary cutting plane. Hence if a perfect fluid is contained in a vessel of rigid ma- terial the pressure experienced by the walls of the vessel is normal to the surface of contact at all points. For the practical purposes of Engineering, water, alcohol, mercury, air, steam, and all gases may be treated as perfect fluids within certain limits of temperature. 407. Liq[uids and Gases. — A fluid a definite mass of which occupies a definite volume at a given temperature, and is in- capable both of expanding into a larger volume and of being compressed into a smaller volume at that temperature, is called a Liquid, of which water, mercury, etc., are common examples; whereas a Gas is a fluid a mass of which is capable of almost indefinite expansion or compression, according as the space within the confining vessel is made larger or smaller, and al- ways tends to fill the vessel, which must therefore be closed in. every direction to prevent its escape. 515 616 MECHANICS OF ENGINEERING. Liquids are sometimes called inelastic fluids, and gases elastic fluids. 408. Eemarks. — Though practically we may treat all liquids as incompressible, experiment shows them to be compressible to a slight extent. Thus, a cubic inch of water under a pres- sure of 15 lbs. on each of its six faces loses only fifty millionths (0.000050) of its original volume, while remaining at the same temperature ; if the temperature be sufficiently raised, how- ever, its bulk will remain unchanged (provided the initial tem- perature is over 40° Fahr.). Conversely, by heating a liquid in a rigid vessel completely filled by it, a great bursting pressure may be produced. The slight cohesion existing between the particles of most liquids is too insignificant to be considered in the present connection.* The property of indefinite expansion, on the part of gases, by which a confined mass of gas can continue to fill a confined space which is progressively enlarging, and exert pressure against its walls, is satisfactorily explained by the " Kinetic Theory of Gases," according to which the gaseous particles are perfectly elastic and in continual motion, impinging against each other and the confining walls. Nevertheless, for prac- tical purposes, we may consider a gas as a continuous sub- stance. Although by the abstraction of heat, or the application of great pressure, or both, all known gases may be reduced to liquids (some being even solidified); and although by con- verse processes (imparting heat and diminishing the pressure) liquids may be transformed into gases, the range of tempera- ture and pressure in all problems to be considered iu this work is supposed kept within such limits that no extreme changes of state, of this character, take place. A gas approaching the point of liquefaction is called a Vapor. Between the solid and the liquid state we find all grades of intermediate conditions of matter. For example, some sub- stances are described as soft and plastic solids, as soft putty, moist earth, pitch, frosh mortar, etc.; and others as viscous and sluggish liquids, as molasses and glycerine. In sufficient bulk, * Water has recently been subjected to a pressure of 65,000 Ibs./in.^; resulting in a reduction of 10 per cent in the volume. See Engineering News, Oct. 1900, p. 236. DEFINITIOlSrS — FLUID PRESSURE — HYDROSTATICS. 517 however, the latter may still be considered as perfect fluids. Even water is slightly viscous. 409. Heaviness of Fluids. — The weight of a cubic unit of a homogeneous fluid will be called its heaviness,* or rate of weight (see § 7), and is a measure of its density. Denoting it hy y, and the volume of a definite portion of the fluid by Vy Ive have, for the weight of that portion, 6^= Fr. a) This, like the great majority of equations used or derived in this work, is of homogeneous form (§ 6), i.e., admits of any sys- tem of units. E.g., in the metre-kilogram-second system, if y is given in kilos, per cubic metre, Y must be expressed in cubic metres, and G will be obtained in kilos.; and similarly in any other system. The quality of 7, = 6^ -f- "F, is evidently one dimension of force divided by three dimensions of length. In the following table, in the case of gases, the temperature and pressure are mentioned at which they have the given heaviness, since under other conditions the heaviness would be different ; in the case of liquids, however, for ordinary pur- poses the effect of a change of temperature may be neglected (within certain limits). HEAVINESS OF VARIOUS FLUIDS.* [In ft. lb. sec. system; y = weight in lbs. of a cubic foot.] Liquids. Freshwater, y =^ 63.5 Sea water 64.0 Mercury 848.7 Alcohol 49.3 Crude Petroleum, about 55.0 (N.B. — A cubic inch of water weighs 0.0361 lbs.; and a cubic foot 1000 av. oz.) p(„„„„ J At temp, of melting ice; and 14.7 ^^tdbet, j jjjg pgj, gq jj^_ tension. Atmospheric Air 0807C Oxygen 0.0893 Nitrogen ...0.0786 Hydrogen 0.0056 Illuminating ) from 0. 0300 Gas, )to 0.040(^ Natural Gas, about 0.0500 * Sometimes called "specific weight;" while its reciprocal, or \^x may be styled the "specific volume" of the substance, i.e., the volume of a unit of weight. 518 MECHANICS OF ENGINEEEING. For use in problems where needed, values for the heaviness of pure fresh water are given in the following table (from Rossetti) for temperatures ranging from freezing to boiling ; as also the relative density, that at the temperature of maxi- mum density, 39°. 3 Fahr. being taken as unity. The temper- atures are Fahr., and y is in lbs. per cubic foot. Temp. Rel. Dens. 7- Temp. Rel. Dens. V- Temp. Rel. Dens. V- 32° .99987 62.416 60° .99907 62.366 140° .98338 61.886 35° .99996 62.421 70° .99802 62.300 150° .98043 61.203 39°. 3 1.00000 62.424 80° .99669 62.217 160° .97729 61.006 40° .99999 62.423 90° .99510 62.118 170° .97397 60.799 43° .99997 62.422 100° .99318 61.998 180° .97056 60.586 45° .99992 62.419 110° .99105 61.865 190° .96701 60.365 50° .99975 62.408 120° .98870 61.719 200° .96333 60.135 55° .99946 62.390 130° .98608 61.555 212° .95865 59.843 From D. K. Clark's] for temp. = " Manual." \ y =^ 230° 59 4 250° 270° 58.7 58.2 290° 57.6 298° 57.3 338° 56.1 366° 55.3 390° 54.5 Example 1. What is the heaviness of a gas, 432 cub. in. of which weigh 0.368 ounces? Use ft.-lb.-sec. system. 432 cub. in. = \ cub. ft. and 0.368 oz, = 0.023 lbs. .*. y =. -z=: -'— — = 0.092 lbs. per cub. foot. y i Example 2. Required the weight of a right prism of mer- cury of 1 sq. inch section and 30 inches altitude. 30 F=30 X 1 = 30 cub. in table, y for mercury = 848.Y lbs. per cub. ft. 30 ^ ^^ - cub. feet : while from the 1Y28. ' its weight = ^ = Yy 17! X 848 r ■= 14.73 lbs. 410. Definitions. — By Hydraulics we understand the me- <'iianics of fluids as utilized in Engineering. It may be divided i7Jto Hydrostatics^ treating of fluids at rest ; and Hydrokinetics^ which deals with fluids in motion. (The name Pneumatics is sometimes used to cover both the statics, and kinetics of gaseous fluids.) DEFINITIONS — FLUID PRESSURE — HYDROSTATICS. 519 411. Pressure per Unit Area, or Intensity of Pressure. — As in § 180 in dealing with solids, so here with fluids we indicate the pressure per unit area between two contiguons portions of fluid, or between a fluid and the wall of the containing vessel, by J?, so that if dP is the total pressure on a small area dF^ we have dP ... ^^dF (^) as the pressure per unit area, or intensity of pressure (often, though ambiguously, called the tension in speaking of a gas) on the small surface dF. If pressure of the same intensity exists over a finite plane surface of area = F, the total pres- sure on that surface is P = fpdF=pfdF= Fp, 1 P ^ .... (2) or i? = p. J (N.B. — For brevity the single word " pressure" will some- times be used, instead of intensity of pressure, where no am- biguity can arise.) Thus, it is found that, under ordinary con- ditions at the sea level, the atmosphere exerts a normal pressure (normal, because fluid pressure) on all surfaces, of an intensity of about p = 14:7 lbs. per sq. inch (= 2116. lbs. per sq. ft.). This intensity of pressure is called pne atmosjphere. For ex- ample, the total atmospheric pressure on a surface of 100 sq. in. is [inch, lb., sec] P=i^p=. 100X14.7 = 1470 lbs. ( = 0.735 tons.) The quahty of p is evidently one dimension of force divided by two dimensions of length. By one ^' atmosphere,^' then (or "standard atmosphere;" an arbitrary unit), is to be understood a unit-pressure of 14.70 Ibs./sq. in., or 2116.8 Ibs./sq. ft. This would be the weight of a prismatic column of w^ater one sq. in. in section and 33.9 ft. high (commonly considered 34 ft. for ordinary computations); or of a prismatic column of mercury 30 in. high and one sq. in. section. These numbers, 14.70, 34, 520 MECHANICS OF ENGINEERING. and 30, with their meanings as above, should be memorized by the student; as they are to be of very frequent service in this study. At high altitudes the actual pressure of the air is much smaller than the conventional "atmosphere." E.g. (see p. 621) at 7000 ft. above the sea the height of a mercury column measuring the air pressure- is only about 24 in., instead of the 30 in. above cited; varying somewhat, of course, with the weather. 412. Hydrostatic Pressure; per Unit Area, in the Interior of a Fluid at Rest. — In a body of fluid of uniforin heaviness, at rest, it is required to find the mutual pressure per unit area be- tween the portions of fluid on opposite sides of any imaginary cutting ])]ane. As customary, we shall consider portions of the fluid as free bodies, by supplying the forces exerted on them by all contiguous portions (of fluid or vessel wall), also those of the earth (their weights), and then apply the condi- tions of equilibrium. First, cutting plane horizontal. — Fig. 451 shows a body of homogeneous fluid confined in a rigid vessel closed at the top with a small air- tight but frictionless piston (a horizontal disk) of weight = G and exposed to at- mospheric pressure {=Pa per ^ii^it area) on its upper face. Let the area of piston- face be = ^. Then for the equilibrium of the piston the total pressure between its under surface and the fluid at must be WSM Fm. 451. F=G + Fpa, and hence the intensity of this pressure is ' G i^o F -\-Pa' (1) It is now required to find the intensity, p, of fluid pressure between the portions of fluid contiguous to the horizontal cut- ting plane BCa.t a vertical distance = A vertically below the pis- ton 0. In Fig. 452 we have as a free body the right parallel©- FLUID PEESSUEE. 521 piped OBC oi Fig. 451 with vertical sides (two || to paper and four ~j to it). The pressures acting on its six faces are normal to them respectively, and the weight of the prism is = vol. X;k = FhVi supposing y to have the same value at all parts of the column (which is practically true for any height of liquid and for a small height of gas). Since the prism \2 in equilibrium under the forces shown in the figure, and would still be so were it to become rigid, we may put (§ 36) >i 2 (vert, compons.) = and hence obtain *" j Fp-F:p,- Fhy = 0. . . (2) r ! ^^^ >JT-i-T— rx" (In the figure the pressures on the ver- ^ tical faces i| to paper have no vertical com- ^P ponents, and hence are not drawn.) From fig. 452, (2) we have JP =P. + hy. (3) {hy, being the weight of a column of homogeneous fluid of unity cross-section and height A, would be the total pressure on the base of such a column, if at rest and with no pressure on the upper base,, and hence might be called intensity due to weigJd.) Secondly, cutting plane oblique. — Fig. 453. Consider free an infinitely small right triangular prism bed, whose bases are li to the paper, while the three side faces (rectangles), having areas = dF, dF^ , and dF^ , are respectively hori- zontal, vertical, and oblique ; let angle cbd = a. The surface he is a portion ^_ V^h of the plane BC oi Fig. 452. Given H — V j? (= intensity of pressure on dF) and " ) Of, required ^2? the intensity of pressure on the oblique face hd, of area dF^. SJS,. B. — The prism is taken very small in order that the intensity of pressure may be considered con- stant over any one face ; and also that the weight of the prism may be neglected, since it involves the volume (three dimen- Fm. 453. 522 MECHANICS OF ENGINEEEIFG. sions) of the prism, while the total face pressures involve onlj two, and is hence a differential of a higher order.] From ^ (vert, compons.) = we shall have p^dF^ cos a —j>dF= ; but dF-^ dF^ = cos or ; I>.=P, (4) which is independent of the angle a. Hence, the mtensity of fluid 'pressure at a given point is the same on all imaginary cutting planes containing the point. This is the most important property of a fluid, and is true whether the liquid is at rest or has any kind of motion ; for, in case of rectilinear accelerated motion, e.g., although the sum of the force-components in the direction of the accelera- tion does not in general = 0, but = mass X ace, still, the mass of the bodj in question is = weight -i- g, and therefore the term mass X ace. is a dijfferential of a higher order than the other terms of the equation, and hence the same result follows as when there is no motion (or uniform rectilinear motion). 413. The Intensity of Pressure is Equal at all Points of any Horizontal Plane in a body of homogeneous fluid at rest. If we consider a right prism of the fluid in Fig. 451, of small vertical thickness, its axis lying in any horizontal plane £0^ its bases will be vertical and of equal area dF. The pressures on its sides, being normal to them, and hence to the axis, have no components |1 to the axis. The weight of the prism also has no horizontal component. Hence from 2 (hor. comps. II to axis) = 0, we have,^i smdp^ being the pressure-intensi- ties at the two bases, p,dF-p,dF=0; .:p=p,, .... (1) which proves the statement at the head of this article. It is now plain, from this and the preceding article, that the pressure-intensity p at any point in a homogeneous fluid at rest is eqiial to that at any higher point, plus the weight FLUID PRESSURE. 523 {hy) of a column of the jiuid of section unity and of altitude \fi) = vertical distance between the joints. ^.(^., p =p. + hy, (2) whether they are in the same vertical or not^ and whatever he the shajpe of the containing ^ vessel {or pipes), provided the fluid is continuous hetween the two points I for, Fig, 454, bj considering a series of small prisms, alternately ver- tical and horizontal, ohcde, we know that Pd=Pc — Ky ; and Pc—Pd'i hence, finally, by addition we have (in which A = Aj — h^. If, therefore, upon a small piston at <?, of area = ^o, a force jP„ be exerted, and an inelastic fluid (liquid) completely fills the vessel, then, for equilibrium, the force to be exerted upon the pis- ton at 6, viz., Pg , is thus computed : For equilibrium of fluid p^ =.p^ -\- hy ; and for equil. of piston o, j?„ = P„ -^ F^ ; also, P, = ^^P,-\-FM. (3) From (3) we learn that if the pistons are at the same level {h, = 0) the total pressures on their inner faces are directly proportional to their areas. If thie fluid is gaseous (2) and (3) are practically correct if h is not > 100 feet (for, gas being compressible, the lower i^^trata are generally more dense than the upper), but in (3) the pistons must be fixed, and P^ and P„ refer solely to the in- terior pressures. 524 MECHANICS OF EISTGIlSrEERIlS'G. Again, if A is small or jp^ very great, the term hy may be omitted altogether in eqs. (2) and (3) (especially with gases, since for them y (heaviness) is usually small), and we then have, from (2), i^=i?o; (4) being the algebraic form of the statement: A l)ody of Jkiid at 7'est transmits pressure with equal intensity in every direc- tion and to all of its parts. [Principle of "Equal Transmis. sion of Pressure."] 414. Moving Pistons. — If the fluid in Fig. 454 is inelastic and the vessel walls rigid, the motion of one piston (c) through a distance s^ causes the other to move through a distance s^ de- termined by the relation F^s^ = F^s^, (since the volumes de- scribed by them must be equal, as liquids are incompressible) ; but on account of the inertia of the liquid, and friction on the vessel walls, equations (2) and (3) no longer hold exactly, still are approximately true if the motion is very slow and the vessel short, as with the cylinder of a water-pressure engine. But if the fluid is compressible and elastic (gases and vapors ; steam, or air) and hence of small density, the effect of inertia and friction is not appreciable in short wide vessels like the cylinders of steam- and air-engines, and those of air-compres- sors ; and eqs. (2) and (3) still hold, practically, even with high piston-speeds. For example, in the space ABy Fig. 455, between the piston and cylinder-head of a steam-engine (piston moving toward the right) the intensity of pressure, ^, of the steam against the moving piston B is prac- FiG. 455. tically equal to that against the cylinder-head A at the same instant. 415. An Important Distinction between gases and liquids (i.e., between elastic and inelastic fluids) consists in this: A liquid can exert pressure against the walls of the contain- ing vessel only by its weight, or (when confined on all sides) by transmitted pressure coming from without (due to piston pressure, atmospheric pressure, etc.); whereas — FLUID PRESSUEE. 625 A gas, confined, as it must be, on all sides to prevent dif- fusion, exerts pressure on the vessel not onlj by its v^eiglit, but by its elasticity or tendency to expand. If pressure from without is also applied, the gas is compressed and exerts a still greater pressure on the vessel walls. 416. Component, of Pressure, in a Given Direction. — Let A BCD, whose area = c.Zi^ be a small element of a surface, plane or curved, and^ the intensity of A fluid pressure upon this element, then ^ap\ /i\ the total pressure upon it is pdJF, and is \/^ I \q of course normal to it. Let A'B' CD be / ---'''^^ > the projection of the element dJc upon cc X \|b'^,--' a plane CDM making an :?.ngle a with y" ^X,-^-'"'' the element, and let it be required to j ' find the value of the component oi jpdF ^^^' '*°^" in a direction normal to this last plane (the other component being understood to be 1| to the same plane). We shall have Compon. ofpdF ~\ to CDM = pdF cos a =j>{dF. Goa a). (1) But dF . cos a = area A'B' CD ^ the projection of <i^upon the plane CDM, .', Compon. 1 tojplane CDM ■=p X {project, of dF on CDM)\ i.e., the component of fluid pressure (on an element of a sur- face) in a given direction (the other component being ~1 to the first) is found hy midtiplying the intensity of the pressure hy the area of the projection of the element xpon a plane 1 to the given direction. It is seen, as an example of this, that if the fluid pressures on the elements of the inner surface of one hemisphere of a hollow sphere containing a gas are resolved into components ~| and II to the plane of the circular base of the hemisphere, the sum of the former components simply = n'r^p, where r is the radius of the sphere, and^ the intensity of the fluid pressure ; for, from the foregoing, the sum of these components is just the same as the total pressure would be, having an intensity p., 526 MECHANICS OF ENGINEERING, Oil a great circle of the sphere, the area, Trr^, of this circle being the sun) of the areas of the projections, upon this circle as a base, of all the elements of the hemispherical surface. (Weight of fluid neglected.) A similar statement may be made as to the pressures on the inner curved surface of a right cylinder. 417. Non-planar Pistons. — From the foregoing it follows that the sum of the components || to the piston-rod, of the fluid pressures upon the piston at A, Fig. 457, is just the same as at _5, if the cylinders are of equal size and the steam, or air, is at the same tension. For the sum of the projections of all the elements of the curved surface of A upon a plane ~\ to the piston-rod is always = Ttr'^ = area of section of cylinder-bore. Fig. 457. If the surface of A is symmetrical about the axis of the cylin- der the other components (i.e., those ~] to the piston-rod) will neutralize each other. If the line of intersection of that sur- face with the surface of the cylinder is not symmetrical about the axis of the cylinder, the piston may be pressed laterally against the cylinder-wall, but the thrust along the rod or " working force'' (§ 128) is the same (except for friction in- duced by the lateral pressure), in all instances, as if the surface were plane and 1 to piston-rod. 418. Bramah, or Hydraulic, Press. — This is a familiar instance of the principle of transmission of fluid pressure. Fig. 458. Let the small piston at O have a diameter <^ = 1 inch = -^ ft., while the plunger E, or large piston, has a diameter d' = AB = CD=lh in. = I ft. The lever MJ^ weighs <?, = 3 lbs., and a weight G — 4S) lbs. is hung at M. The lever-arms of these forces about the fulcrum N are given in the figure. The apparatus being full of water (oil is often used), the fluid pressure P„ against the small piston is found by putting FLUID PRESSUEE. 527 5(moms. about JV) == for the equilibrium of the leverj whence [ft., lb , sec] P„ X 1 - 40 X 3 - 3 X 2 = 0. /, P„ = 126 IbSo Fig. 458. But, denoting atmospheric pressure by ^„, and that of the water against the piston by p^ (per unit area), we may also write Solving for p^ , we have, putting p^, = 14.7 X 144 lbs, per aq. ft., p, = [126 -^ ~ {-^yl + 14.Y X 144 = 25286 lbs. per sq. ft. Hence at e the press, per unit area, from § 409, and (2), § 4185 m p^ = j>„ 4- A;j/ = 25236 + 3 X 62.5 = 25423 lbs. per sq. ft. = 175.6 lbs. per sq. inch or 11.9 atmospheres, and the total upward pressure at e on base of plunger is P = FePe =7t'±-p, = i 7r{iy X 25423 = 81194 lbs., or almost 16 tons (of 2000 lbs. each). The compressive force upon the block or bale, C, = P less the weight of the plunger and total atmos. pressure on a circle of 15 in. diameter. 528 MEGHAlSriCS OF ENGINEERING. 419, The Dividing Surface of Two Fluids (which do not mix) in Contact, and at Sest, is a Horizontal Plane, — For, Fig. 459, sup- posing any two points e and O of this, sur- face to be at different levels (the pressure at being ^o, that at ejCg, and the teavi- nesses of the two fluids 7, and y^ respec- tively), we would have, from a considera- tion of the two elementary prisms eb an to (vertical and horizontal;, the relation Fig. 459. while from the prisms eo and gO^ the relation These equations are conflicting, hence the aoove supposition is absurd. Therefore the proposition is true. For stable equilibrium, evidently, the heavier fluid must oc- cupy the lowest position in the vessel, and if there are several fluids (which do not mix), they will arrange themselves vertically, in the order of their den- sities, the heaviest at the bottom, Fig. 460. On account of the property called diffusion the par- ticles of two gases placed in contact soon inter- mingle and form a uniform mixture. This fact gives strong support to the " Kinetic Theory of Gases" (§ 408). Fig. 460, 420. Free Surface of a Liquid at Rest. — The surface (of a liquid) not in contact with the walls of the containing vessel ,...,._.,...,,.., is called a free surface^ and is necessarily ^fi^'^^^^^^^^m^:^'^ T' horizontal (from § 419) when the liquid is at :..;,(.:,.- w-i;: .,•' yq%\.. Fig. 461. (A gas, from its tendency to indefinite expansion, is incapable of hav- ing a free surface.) This is true even if the space above the liquid is vacuous, for if the surface were inclined or curved, points in the body of the liquid and in the same horizon- tal plane would have different heights (or " heads") of liquid Fig. 461. TWO LIQUIDS IIS'^ BEISTT TUBE. 529 between ttein and the surface, producing different intensities of pressure in the plane, which is contrary to § 413. When large bodies of liquid like the ocean are considered, grayitj can no longer be regarded- as acting in parallel lines ; consequently the free surface of the liquid is curved, being ~\ to the direction of (apparent) gravity at all points. For ordi- nary engineering purposes (except in Geodesy) the free surface of water at rest is a horizontal plane. 421. Two Liquids (whicli do not mix) at Rest in a Bent Tube open at Both Ends to the Air, Fig. 460 ; water and mercury, for instance. Let their heavinesses be y^ and y^ respectively. The pressure at e may be written (§ 413) either or according as we refer it to the water column or the mercury column and their respective free surfaces where the pressure j?Oj =i?Og = Pa = atmos. press. € is the surface of contact of the two liquids. _.-Xv_>« Hence we have l>a + Kr,=Pa + Kn; i.e., ^, : K-'-n- r^- • (3) le., the heights of the free surfaces of the two liquids above the surface of contact are inversely proportional to their respec- tive heamnesses. ExamJ'le. — If the pressure at ^ = 2 atmospheres (§ 896) we shall have from (2) (inch-lb.-sec. system of units) KYx = JPe — /?a = 2 X 14.7 — 14.7 = 14.7 lbs. per sq. inch. .-. \, must = 14.7 -j- [848.7 -H 1728] = 30 inches (since, foi mercury, y^ = 848.7 lbs. per cub. ft.). Hence, from (3), . , h,y, 30 X [848.7 -- 1728] , „g . , „ . ._^ ^' = 7r" 6275-^1728 = ^^^ ^^'^"' = ^^ ^- 530 MECHANICS OF ENGINEEKING. i.e., for equilibrium, and that j?e may = 2 atmospheres, k^ aud Aj (of mercarj and water) must be 30 in. and 34 feet respec- tively. 422. City Water-pipes. — If h = vertical distance of a point ^ of a water-pipe below the free surface of reservoir, and tlie water be at rest, the pressure on the inner surface of the pipe at B (per unit of area) is p =p^-\- hy ; and here j!?o =^„ = atmos. press. Example. — If h = 200 ft. (using the inch, lb., and second) P = 14.7 + [200 X 12][62.5 -=- 1T28] = 101.5 lbs. per sq. in. The term hy, alone, = 86.8 lbs. per sq. inch, is spoken of as the 'hydrostattG press-ure due to 200 feet height, or "Head," of water. (See Trautwine's Pocket Book for a table of hydro- static pressures for various depths.) If, however, the water {^flowing through the pipe, the pres- sure against the interior wall becomes less (a problem of Hy- drokinetics to be treated subsequently), while if that motion is suddenly checked, the pressure becomes momentarily much greater than the hydrostatic. This shock is called '■ water- ram" and " water-hammer," and may be as great as 200 to 300 lbs. per sq. inch.* 423. Barometers and Manometers for Fluid Pressure. — If a tube, closed at one end, is filled with water, and ihe other ex- tremity is temporarily stopped and afterwards opened under water, the closed end being then a (vertical) height = h above the surface of the water, it is required to find the intensity, jp^ , of fluid pressure at the top of the tube, sup- posing it to remain filled with water. Fig. 463. At E inside . the tube the pressure is 14.7 lbs. per sq. inch, the same as tljat outside at the same level (§ 413) ; hence, from Pe=^ P<s Fig. 463. H-Vj P.=I>E-hy. w * See pp. 203-214 of the author's "Hydraulic Motors." BAROMETERS. 531 Example. — Let A = 10 feet (with inch-lb.-sec. system) ; then f^ = 14.Y - 12