RESISTANCE OF MATERIALS FOE BEGINNERS IN ENGINEERING BY S. E. SLOCUM, B.E., PH.D. PROFESSOR OF APPLIED MATHEMATICS IN THE UNIVERSITY OF CINCINNATI GINN AND COMPANY BOSTON NEW YORK CHICAGO LONDON x *r s t-> COPYRIGHT, 1914, BY S. E. SLOCUM ALL BIGHTS RESERVED 414.10 gftc gtfttnaeum jprrgg GINN AND COM^ \NY PRO- PRIETORS BOSTON U.S.A. PREFACE The chief feature which distinguishes this volume from other American textbooks on the same subject is that the Principle of Moments is used consistently throughout in place of the usual calculus processes. By basing the work on this principle it has been found practicable to give a simple and obvious treatment of many topics for which the calculus is usually thought to be indispensable, such as the calculation of moments of inertia, the deflection of beams, the buckling of columns, and the strength of thick cylinders. Experience has shown conclusively that the average engineering graduate, and even the practicing engineer, is deficient in the ability to apply the Principle of Moments readily, but when thus used as the central and coordinating principle, it must necessarily make an indelible impression on the mind of the student and go far toward remedying this deficiency. The mechanics of materials is of such fundamental importance in all branches of technology that it is important to begin its study as early in the course as possible. Heretofore it has been necessary to defer it - - awaiting the completion of the calculus until junior year, when the curriculum is already crowded with technical sub- jects requiring its application. This text makes it possible for the course to parallel or even to precede the calculus. In addition, it makes the subject available for trade or architectural schools where no calculus is taught. Although simple and obvious, the treatment is adequate, and its simplicity in no way limits its range or generality. The text is supplemented by a variety of engineering applications, giving practical information as well as a mastery of the principles involved. S. E. SLOCUM CONTENTS SECTION I STRESS AND DEFORMATION PAGES Elastic resistance, or stress. Varieties of strain. Strain diagram. Hooke's law. Elastic limit. Working stress. Resilience. Poisson's ratio. Temperature stress. Applications 1-14 SECTION II FIRST AND SECOND MOMENTS Static moment. Fundamental theorem of moments. Center of gravity, -r- Centroid. Centroid of triangular area. Centroid of circular arc. Centroid of circular sector and segment. Centroid of parabolic segment. Axis of symmetry. Centroid of composite figures. Moment of inertia. I for rectangle. I for triangle. I for circle. I for composite figures. Applications 15-34 SECTION III BENDING-MOMENT AND SHEAR DIAGRAMS Conditions of equilibrium. Vertical shear. Bending moment. Bending-moment and shear diagrams. Relation between shear and moment diagrams. Properties of shear and moment diagrams. General directions for sketching diagrams. Applications . . 35~48 SECTION IV STRENGTH OF BEAMS Nature of bending stress. Distribution of stress. Fundamental formula for beams. Calculation and design of beams. Applications 49-59 vi CONTENTS SECTION V DEFLECTION OF CANTILEVER AND SIMPLE BEAMS PAGES General deflection formula. Cantilever bearing concentrated load. Cantilever bearing uniform load. Cantilever under constant moment. Simple beam bearing concentrated load. Simple beam bearing uniform load. Applications 60-69 SECTION VI CONTINUOUS BEAMS Theorem of three moments for uniform loads. Theorem of three moments for concentrated loads. Effect of unequal settlement of supports. Applications 70-79 SECTION VII RESTRAINED, OR BUILT-IN, BEAMS Uniformly loaded beam fixed at both ends. Beam fixed at both ends and bearing concentrated load at center. Single eccentric load. Uniformly loaded beam fixed at one end. Beam fixed at one end and bearing concentrated load at center. Beam fixed at one end and bearing a concentrated eccentric load. Applications 80-90 SECTION VIII COLUMNS AND STRUTS Nature of compressive stress. Euler's theory of long columns. Effect of end support. Modification of Euler's formula. Ran- kine's formula. Values of the empirical constants in Rankine's formula. Johnson's parabolic formula. Johnson's straight-line formula. Cooper's modification of Johnson's straight-line formula. Eccentrically loaded columns. Applications 91-105 SECTION IX TORSION Maximum stress in circular shafts. Angle of twist in circular shafts. Power transmitted by circular shafts. Combined bending and torsion. Resilience of circular shafts. Non-circular shafts. Elliptical shaft. Rectangular and square shafts. Triangular shafts. Angle of twist for shafts in general. Applications . . 106-117 CONTENTS vii SECTION X SPHERES AND CYLINDERS UNDER UNIFORM PRESSURE PAGES Hoop stress. Hoop tension in hollow sphere. Hoop tension in hollow circular cylinder. Longitudinal stress in hollow circular cylinder. Thick cylinders. Lamp's formulas. Maximum stress in thick cylinder under uniform internal pressure. Bursting pres- sure for thick cylinder. Maximum stress in thick cylinder under uniform external pressure. Comparison of formulas for the strength of tubes under uniform internal pressure. Thick cylinders built up of concentric tubes. Practical formulas for the collapse of tubes under external pressure. Shrinkage and forced fits. Applications 118-135 SECTION XI FLAT PLATES Theory of flat plates. Maximum stress in homogeneous circu- lar plate under uniform load. Maximum stress in homogeneous circular plate under concentrated load. Dangerous section of ellip- tical plate. Maximum stress in homogeneous elliptical plate under uniform load. Maximum stress in homogeneous square plate under uniform load. Maximum stress in homogeneous rectangular plate under uniform load. Applications 136-145 SECTION XII RIVETED JOINTS AND CONNECTIONS Efficiency of riveted joint. Boiler shells. Structural steel. Unit stresses. Applications 146-155 SECTION XIII REENFORCED CONCRETE Physical properties. Design of reenforced-concrete beams. Calculation of stirrups, or web reinforcement. Reenforced- concrete columns. Radially reenf orced flat slabs. Diameter of top. Efficiency of the spider hoops. Maximum moment. Thick- ness of slab. Area of slab rods. Application of formulas. Dimension table. Dimensions of spider. Applications. . . . 156-175 viii CONTENTS SECTION XIV SIMPLE STRUCTURES PAGES Composition and resolution of forces Conditions of equilibrium of a system of coplanar forces. Equilibrium polygon. Applica- tion of equilibrium polygon to determining reactions. Equilibrium polygon through two given points. Equilibrium polygon through three given points. Application of equilibrium polygon to calcula- tion of stresses. Relation of equilibrium polygon to bending- moment diagram. Structures : external forces. Structures : joint reactions. Structures : method of sections. Applications . . . 176-204 ANSWERS TO PROBLEMS 205-207 INDEX . 209-210 TABLES OF PHYSICAL AND MATHEMATICAL CONSTANTS I. AVERAGE VALUES OF PHYSICAL CONSTANTS II. PROPERTIES OF VARIOUS SECTIONS III. PROPERTIES OF STANDARD I-BEAMS IV. PROPERTIES OF STANDARD CHANNELS V. PROPERTIES OF STANDARD ANGLES VI. PROPERTIES OF BETHLEHEM GIRDER BEAMS VII. PROPERTIES OF BETHLEHEM I-BEAMS VIII. MOMENTS OF INERTIA AND SECTION MODULI : RECTANGULAR CROSS SECTION IX. MOMENTS OF INERTIA AND SECTION MODULI : CIRCULAR CROSS SECTION X. FOUR-PLACE LOGARITHMS OF NUMBERS XI. CONVERSION OF LOGARITHMS XII. FUNCTIONS OF ANGLES XIII. BENDING-MOMENT AND SHEAR DIAGRAMS XIV. MENSURATION XV. FRACTIONAL AND DECIMAL EQUIVALENTS XVI. WEIGHTS OF VARIOUS SUBSTANCES XVII. STRENGTH OF ROPES AND BELTS ix TABLES XI HH ffi If? * 2.23J .^ *i 7^ . i s s s 1 <N <N O * 8 8 8 8 0000 2 8 S 9 5 8 g 8 s g a -3 * ~- o s C s o 5 * W 02 >> '3 a 111 Xll RESISTANCE OF MATERIALS TABLE I (Continued) 2. POISSON'S RATIO MATERIAL AVERAGE VALUES OF 1 m Steel hard 295 " structural 299 .277 Brass . .... 357 Copper 340 Lead .375 Zinc . .205 3. FACTORS OF SAFETY STEADY VARYING REPEATED OR MATERIAL STRESS : BUILDINGS, STRESS : BRIDGES, REVERSED STRESS : ETC. ETC. MACHINES Steel, hard 5 8 15 " structural 4 6 10 Iron wrought ... 4 6 10 " cast 6 10 20 Timber 8 10 15 Brick and stone .... 15 25 30 The only rational method of determining the factor of safety is to choose it sufficiently large to bring the working stress well within the elastic limit (see Article 6). OB to Q IT Se o ai E w 02 w H ^ W -, Se ^ ^ CO 1 <M w a ^o *^ *c> l ^ > 1 '^ a 2 o S o fc II H SJ ^ w <S 1 <N r< 1 <M .O 1 CM O X S II II II S: *^ j^ H H I |, s * 1 K 0^ ! 1 j :< M 1* < * I a <) a CO 1 1 r i i I i i ! It fo. ,1 i I -*-- 1 1 xiii v< H 8 J a (N ICO II H <N ICO II H I \\* J KM II I i h*-- h-3 OQ w S ^ H C_j ^ H 3 o A- 1 PH s? El ICO > % XV1U RESISTANCE OF MATERIALS TABLE III PROPERTIES OF STANDARD I-BEAMS DEPTH OF BEAM WEIGHT PER FOOT AREA OF SECTION THICK- NESS OF WEB WIDTH OF FLANGE MOMENT OF INERTIA Axis 1-1 SECTION MODU- LUS Axis 1-1 RADIUS OF GYRA- TION Axis 1-1 MOMENT OF INERTIA Axis 2-2 RADIUS OF GYRA- TION Axis 2-2 d A t b I S r I' r' Inches Pounds Sq. Inches Inches Inches Inches* Inches & Inches Inches * Inches 3 55 1.63 .17 2.33 2.5 1.7 1.23 .46 53 ii 6.5 1.91 .26 2.42 2.7 1.8 1.19 53 52 " 7.5 2.21 .36 252 2.9 1.9 1.15 .60 52 4 75 2.21 .19 2.66 6.0 3.0 1.64 .77 59 85 2.50 .26 2.73 6.4 3.2 1.59 .85 58 II 9.5 2.79 .34 2.81 6.7 3.4 154 .93 58 II 105 3.09 .41 2.88 7.1 3.6 152 1.01 57 5 9.75 2.87 21 3.00 12.1 4.8 2.05 1.23 .65 ii 12.25 3.60 !36 3.15 13.6 5.4 1.94 1.45 .63 " 14.75 4.34 50 3.29 15.1 6.1 1.87 1.70 .63 6 12.25 3.61 .23 3.33 21.8 7.3 2.46 1.85 .72 14.75 4.34 .35 3.45 24.0 8.0 2.35 2.09 .69 " . 17.25 5.07 .47 357 26.2 8.7 2.27 2.36 .68 7 15.0 4.42 .25 3.66 36.2 10.4 2.86 2.67 .78 175 5.15 .35 3.76 39.2 11.2 2.76 2.94 .76 ii 20.0 5.88 .46 3.87 42.2 12.1 2.68 3.24 .74 8 17.75 5.33 .27 4.00 56.9 14.2 3.27 3.78 .84 i 20.25 5.96 .35 4.08 60.2 15.0 3.18 4.04 .82 it 22.75 6.69 .44 4.17 64.1 16.0 3.10 4.36 .81 " 25.25 7.43 .53 4.26 68.0 17.0 3.03 4.71 .80 21.0 6.31 .29 4.33 84.9 18.9 3.67 5.16 .90 ii 25.0 7.35 .41 4.45 91.9 20.4 3.54 5.65 .88 ii 30.0 8.82 57 4.61 101.9 22.6 3.40 6.42 .85 " 35.0 10.29 .73 4.77 111.8 24.8 3.30 7.31 .84 10 25.0 7.37 .31 4.66 122.1 24.4 4.07 6.89 57 II 30.0 8.82 .45 4.80 134.2 26.8 3.90 7.65 .93 " 35.0 10.29 .60 4.95 146.4 29.3 3.77 852 51 " 40.0 11.76 .75 5.10 158.7 31.7 3.67 950 50 12 325 9.26 .35 5.00 215.8 36.0 4.83 950 1.01 " 35.0 10.29 .44 5.09 228.3 38.0 4.71 10.07 .99 " 40.0 11.76 56 5.21 245.9 41.0 457 10.95 56 15 42.0 12.48 .41 550 441.8 58.9 5.95 14.62 1.08 " 45.0 13.24 .46 555 455.8 60.8 5.87 15.09 1.07 < 50.0 14.71 56 5.65 483.4 645 5.73 16.04 1.04 ii 55.0 16.18 .66 5.75 511.0 68.1 5.62 17.06 1.03 " 60.0 17.65 .75 5.84 538.6 71.8 552 18.17 1.01 18 55.0 15.93 .46 6.00 795.6 88.4 7.07 21.19 1.15 60.0 17.65 56 6.10 841.8 935 6.91 22.38 1.13 ii 65.0 19.12 .64 6.18 8815 97.9 6.79 23.47 1.11 " 70.0 20.59 .72 6.26 921.2 102.4 6.69 24.62 1.09 20 65.0 19.08 50 6.25 11695 117.0 7.83 27.86 1.21 " 70.0 20.59 58 6.33 1219.8 122.0 7.70 29.04 1.19 " 75.0 22.06 .65 6.40 1268.8 126.9 758 30.25 1.17 24 80.0 23.32 .50 7.00 2087.2 173.9 9.46 42.86 1.36 i* 85.0 25.00 57 7.07 2167.8 180.7 9.31 44.35 1.33 " 90.0 26.47 .63 7.13 2238.4 1865 9.20 45.70 1.31 " 95.0 27.94 .69 7.19 2309.0 192.4 9.09 77.10 1.30 100.0 29.41 .75 7.25 2379.6 198.3 859 4855 1.28 TABLES TABLE IV PROPERTIES OF STANDARD CHANNELS XIX .( a- ff ^i ^ s i 03 , ^ <N 09 . JH DEPTH OF CHANNEL WEIGHT PER FOOT AREA OF SECTK THICKNESS OF WEB WIDTH OF FLANGE MOMENT OF INERTIA Axis 1 I SECTION MODUL Axis 1-1 RADIUS OF GYR TION AXIS 1-1 MOMENT OF INERTIA Axis 2 SECTION MODUL Axis 2-2 RADIUS OF GYR TION Axis 2-2 DISTANCE OF CENTER OF GRA ITY FROM OUTSIJ OF WEB d A t b I S r r S' r' X Inches Pounds Sq.In. Inches Inches Inches* Inches 3 Inches Inches* Inches 3 Inches Inches 3 4.00 1.19 .17 1.41 1.6 1.1 1.17 .20 .21 .41 -.44 5.00 1.47 .26 1.50 1.8 1.2 1.12 .25 .24 .41 .44 " 6.00 1.76 .36 1.60 2.1 1.4 1.08 .31 .27 .42 .46 4 5.25 1.55 .18 158 3.8 1.9 156 .32 .29 .45 .46 6.25 1.84 .25 1.65 4.2 2.1 1.51 .38 .32 .45 .46 " 7.25 2.13 .33 1.73 4.6 2.3 1.46 .44 .35 .46 .46 5 6.50 1.95 .19 1.75 7.4 3.0 1.95 .48 .38 50 .49 " 9.00 2.65 .33 1.89 8.9 3.5 1.83 .64 .45 .49 .48 it 11.50 3.38 .48 2.04 10.4 4.2 1.75 .82 .54 .49 51 6 8.00 2.38 .20 1.92 13.0 4.3 2.34 .70 50 .54 52 " 1050 3.09 .32 2.04 15.1 5.0 2.21 .88 57 .53 50 " 13.00 3.82 .44 2.16 17.3 5.8 2.13 1.07 .65 53 52 " 15.50 4.56 56 2.28 19.5 6.5 2.07 1.28 .74 .53 55 7 9.75 2.85 .21 2.09 21.1 6.0 2 72 .98 .63 59 .55 < 12.25 3.60 .32 2.20 24.2 6.9 259 1.19 .71 57 53 14.75 4.34 .42 2.30 27.2 7.8 2.50 1.40 .79 .57 53 " 17.25 5.07 .53 2.41 30.2 8.6 2.44 1.62 .87 56 55 " 19.75 5.81 .63 2.51 33.2 9.5 2.39 1.85 .96 56 58 8 11.25 3.35 .22 2.26 32.3 8.1 3.10 1.33 .79 .63 58 " 13.75 4.04 .31 2.35 30.0 9.0 2.98 1.55 .87 .62 56 < 16.25 4.78 .40 2.44 39.9 10.0 2.89 1.78 .95 .61 .56 18.75 551 .49 2.53 43.8 11.0 2.82 2.01 1.02 .60 .57 " 21.25 6.25 2.62 47.8 11.9 2.76 2.25 1.11 .60 59 9 13.25 3.89 .23 2.43 47.3 10.5 3.49 1.77 .97 .67 .61 " 15.00 4.41 .29 2.49 50.9 11.3 3.40 1.95 1.03 .66 59 " 20.00 5.88 .45 2.65 60.8 13.5 3.21 2.45 1.19 .65 58 " 25.00 7.35 .61 2.81 70.7 15.7 3.10 2.98 1.36 .64 .62 10 15.00 4.46 .24 2.60 66.9 13.4 3.87 2.30 1.17 -.72 .64 " 20.00 5.88 .38 2.74 78.7 15.7 3.66 2.85 1.34 .70 .61 25.00 7.35 53 2.89 91.0 18.2 3.52 3.40 1.50 .68 .62 " 30.00 8.82 .68 3.04 103.2 20.6 3.42 3.99 1.67 .67 .65 " 35.00 10.29 .82 3.18 115.5 23.1 3.35 4.66 1.87 .67 .69 12 20.50 6.03 .28 2.94 128.1 21.4 4.61 3.91 1.75 .81 .70 25.00 7.35 .39 3.05 144.0 24.0 4.43 4.53 1.91 .78 .68 30.00 8.82 51 3.17 161.6 26.9 4.28 5.21 2.09 .77 .68 " 35.00 10.29 .64 3.30 179.3 29.9 4.17 5.90 2.27 .76 .69 " 40.00 11.76 .76 3.42 196.9 32.8 4.09 6.63 2.46 .75 .72 15 33.00 9.90 .40 3.40 312.6 41.7 5.62 8.23 3.16 .91 .79 < 35.00 10.29 .43 3.43 319.9 42.7 5.57 8.48 3.22 .91 .79 40.00 11.76 3.52 347.5 46.3 5.44 9.39 3.43 .89 .78 " 45.00 13.24 .62 3.62 375.1 50.0 5.32 10.29 3.63 .88 .79 50.00 14.71 .72 3.72 402.7 53.7 5.23 11.22 3.85 .87 .80 " 55.00 16.18 .82 3.82 430.2 57.4 5.16 12.19 4.07 .87 .82 XX RESISTANCE OF MATERIALS TABLE V PROPERTIES OF STANDARD ANGLES, EQUAL LEGS < \ 1 1 , \ \ j 1 < \ 1 \ t 1 "V-^^v*! \/^ DIMENSIONS THICKNESS WEIGHT PER FOOT AREA OF SECTION DISTANCE OF CENTER OF GRAVITY FROM BACK OF FLANGE MOMENT OF INERTIA Axis 1-1 SECTION MODULUS Axis 1-1 : RADIUS OF GYRA- TION Axis 1-1 DISTANCE OF CENTER OF GRAVITY FROM EXTERNAL APEX ON LINE INCLINED AT 45 TO FLANGE LEAST MOMENT OF INERTIA Axis 2-2 SECTION MODULUS Axis 2-2 LEAST RADIUS OF GYRATION Axis 2-2 Inches Inches Pounds Sg.In. Inches Inches* Inches 3 Inches Inches Inches* Inches 5 Inches fx 2 k .58 .17 .23 .009 .017 .22 .33 .004 .011 .14 1 x 1 i .80 .23 .30 .022 .031 .30 .42 .009 .021 .19 " I 1.49 .44 .34 .037 .056 .29 .48 .016 .034 .19 li x 1J i 1.02 .30 .36 .044 .049 .38 .51 .018 .035 .24 i 1.91 .56 .40 .077 .091 .37 .57 .033 .057 .24 ** u 1 2.34 .69 .47 .14 .134 .45 .66 .058 .088 .29 i 3.35 .98 .51 .19 .188 .44 72 .082 .114 .29 1| X 1| j 2.77 .81 .53 .23 .19 .53 .75 .094 .13 .34 3.98 1.17 57 .31 .26 .51 .81 .133 .16 .34 2x2 i 3.19 .94 .59 .35 .25 .61 .84 .14 .17 .39 " i 4.62 1.36 .64 .48 .35 .59 .90 .20 .22 .39 2* x 2 4.0 1.19 .72 .70 .39 .77 1.01 .29 .28 .49 5.9 1.73 .76 .98 .57 .75 1.08 .41 .38 .48 " 7.7 2.25 .81 1.23 .72 .74 1.14 52 .46 .48 3 X3 4.9 1.44 .84 1.24 .58 .93 1.19 .50 .42 .59 " 7.2 2.11 .89 1.76 .83 .91 1.26 .72 57 .58 " ^ 9.4 2.75 .93 2.22 1.07 .90 1.32 .92 .70 58 " i 11.4 3.36 .98 2^62 1.30 .88 1.38 1.12 .81 58 3* x 3 i 8.4 2.48 1.01 2.87 1.15 1.07 1.43 1.16 .81 .68 ^ 11.1 3.25 1.06 3.64 1.49 1.06 1.50 1.50 1.00 .68 " 13.5 3.98 1.10 4.33 1.81 1.04 1.56 1.82 1.17 .68 " 2 15.9 4.69 1.15 4.96 2.11 1.03 1.62 2.13 1.31 .67 i x4 3 9.7 2.86 1.14 4.36 1.52 1.23 1.61 1.77 1.10 .79 < 1 12.8 3.75 1.18 5.56 1.97 1.22 1.67 2.28 1.36 .78 < | 15.7 4.61 1.23 6.66 2.40 1.20 1.74 2.76 1.59 .77 " 2 18.5 5.44 1.27 7.66 2.81 1.19 1.80 3.23 1.80 .77 6x6 1 19.6 5.75 1.68 19.91 4.61 1.86 2.38 8.04 3.37 1.18 " 24.2 7.11 1.73 24.16 5.66 1.84 2.45 9.81 4.01 1.17 a 28.7 8.44 1.78 28.15 6.66 1.83 2.51 11.52 4.59 1.17 " j 33.1 9.73 1.82 31.92 7.63 1.81 2.57 13.17 5.12 1.16 TABLES XXI TABLE V PROPERTIES OF STANDARD ANGLES, UNEQUAL LEGS W * 65 DIMENSIONS THICKNESS WEIGHT PER FOOT AREA OF SECTION DISTANCE OF CENTE OF GRAVITY FROM BACK OF LONGER FLANGE MOMENT OF INERTI. Axis 1-1 SECTION MODULUS Axis 1-1 RADIUS OF GYRA- TION Axis 1-1 DISTANCE OF CENTE OF GRAVITY FROM BACK OF SHORTER FLANGE MOMENT OF INERTI. Axis 2-2 SECTION MODULUS Axis 2-2 RADIUS OF GYRA- TION Axis 2-2 Inches Inches Pounds Sq.In. Inches Inches* Inches^ Inches Inches Inches* Inches 3 Inches 2$ x 2 3.6 1.06 .54 .37 .25 .59 .79 .65 .38 .78 " 5.3 1.55 .58 .51 .36 .58 .83 .91 .55 .77 " 6.8 2.00 .63 .64 .46 6 .88 1.14 .70 .75 3 x 2J 4.5 1.31 .66 .74 .40 75 .91 1.17 .56 .95 " 6.5 1.92 .71 1.04 .58 .74 .96 1.66 .81 .93 " 8.5 2.50 75 1.30 .74 .72 1.00 2.08 1.04 .91 3$ x 2J 4.9 1.44 .61 .78 .41 .74 1.11 1.80 .75 1.12 " 7.2 2.11 .66 1.09 .59 .72 1.16 2.56 1.09 1.10 9.4 2.75 .70 1.36 .76 .70 1.20 3.24 1.41 1.09 " 11.4 3.36 .75 1.61 .92 .69 1.25 3.85 1.71 1.07 3$ x 3 | 7.8 2.30 .83 1.85 .85 .90 1.08 2.72 1.13 1.09 " i 10.2 3.00 .88 2.33 1.10 .88 1.13 3.45 1.45 1.07 " 1 12.5 3.67 .92 2.76 1.33 .87 1.17 4.11 1.76 1.06 " 1 14.7 4.31 .96 3.15 1.54 .85 1.21 4.70 2.05 1.04 4x3 1 8.5 2.48 .78 1.92 .87 .88 1.28 3.96 1.46 1.26 11.1 3.25 .83 2.42 1.12 .86 1.33 5.05 1.89 1.25 " 13.6 3.98 .87 2.87 1.35 .85 1.37 6.03 2.30 1.23 " 15.9 4.69 .92 3.28 1.57 .84 1.42 6.93 2.68 1.22 5x3 9.7 2.86 .70 2.04 .89 .84 1.70 7.37 2.24 1.61 " 12.8 3.75 .75 2.58 1.15 .83 1.75 9.45 2.91 1.59 " 15.7 4.61 .80 3.06 1.39 .82 1.80 11.37 3.55 1.57 " 2 18.5 5.44 .84 3.51 1.62 .80 1.84 13.15 4.16 1.55 5 x 3 10.4 3.05 .86 3.18 1.21 1.02 1.61 7.78 2.29 1.60 " 13.6 4.00 .91 4.05 1.56 1.01 1.66 9.99 2 99 1.58 " t 16.7 4.92 .95 4.83 1.90 .99 1.70 12.03 3.65 1.56 a 19.8 5.81 1.00 5.55 2.22 .98 1.75 13.92 4.28 1.55 " i 22.7 6.67 1.04 6.21 2.52 .96 1.79 15.67 4.88 1.53 6x3$ 11.6 3.42 .79 3.34 1.23 .99 2.04 12.86 3.24 1.94 15.3 4.50 .83 4.25 1.59 .97 2.08 16.59 4.24 1.92 18.9 5.55 .88 5.08 1.94 .96 2.13 20.08 5.19 1.90 ' 22.3 6.56 .93 5.84 2.27 .94 2.18 23.34 6.10 1.89 " 25.7 7.55 .97 6.55 2.59 .93 2.22 26.39 6.98 1.87 6x4 12.3 3.61 .94 4.90 1.60 1.17 1.94 13.47 3.32 1.93 16.2 4.75 .99 6.27 2.08 1.15 1.99 17.40 4.33 1.91 < 19.9 5.86 1.03 7.52 2.54 1.13 2.03 21.07 5.31 1.90 23.6 6.94 1.08 8.68 2.97 1.12 2.08 24.51 6.25 1.88 " 27.2 7.98 1.12 9.75 3.39 1.11 2.12 27.73 7.15 1.86 XX11 RESISTANCE OF MATERIALS TABLE VI PROPERTIES OF BETHLEHEM GIRDER BEAMS DEPTH WEIGHT AREA THICK- WIDTH NEUTRAL Axis PERPEN- NEUTRAL Axis COINCIDENT WITH OF BEAM PER FOOT OF SECTION NESS OF WEB OF FLANGE DICULAR TO WEB AT CENTER CENTER LINE OF WEB Inches Pounds Square Inches Inches Inches Moment of Inertia Radius of Gyration Section Modulus Moment of Inertia Radius of Gyration 30 200.0 58.71 .750 15.00 9150.6 12.48 610.0 630.2 3.28 u 180.0 53.00 .690 13.00 8194.5 12.43 546.3 433.3 2.86 28 180.0 52.86 .690 14.35 7264.7 11.72 518.9 533.3 3.18 K 165.0 48.47 .660 12.50 6562.7 11.64 468.8 371.9 2.77 26 160.0 46.91 .630 13.60 5620.8 10.95 432.4 435.7 3.05 u 150.0 43.94 .630 12.00 5153.9 10.83 396.5 314.6 2.68 24 140.0 41.16 .600 13.00 4201.4 10.10 350.1 346.9 2.90 120.0 35.38 .530 12.00 3607.3 10.10 300.6 249.4 2.66 20 140.0 41.19 .640 12.50 2934.7 8.44 293.5 348.9 2.91 112.0 32.81 .550 12.00 2342.1 8.45 234.2 239.3 2.70 18 92.0 27.12 .480 11.50 1591.4 7.66 176.8 182.6 2.59 15 140.0 41.27 .800 11.75 1592.7 6.21 212.4 331.0 2.83 104.0 30.50 .600 11.25 1220.1 6.32 162.7 213.0 2.64 u 73.0 21.49 .430 10.50 883.4 6.41 117.8 123.2 2.39 12 70.0 20.58 .460 10.00 538.8 5.12 89.8 114.7 2.36 55.0 16.18 .370 9.75 432.0 5.17 72.0 81.1 2.24 10 44.0 12.95 .310 9.00 244.2 4.34 48.8 57.3 2.10 9 38.0 11.22 .300 8.50 170.9 3.90 38.0 44.1 1.98 8 32.5 9.54 .290 8.00 114.4 3.46 28.6 32.9 1.86 TABLES xxin TABLE VII PROPERTIES OF BETHLEHEM I-BEAMS DEPTH OF BEAM WEIGHT PER FOOT AREA OF SECTION THICK- NESS OF WEB WIDTH OF FLANGE NEUTRAL Axis PERPEN- DICULAR TO WEB AT CENTER NEUTRAL Axis COINCIDENT WITH CENTER LINE OF WEB Inches Pounds Square Inches Inches Inches Moment of Inertia Radius of Gyration Section Modulus Moment of Inertia Radius of Gyration 30 120.0 35.30 .540 10.500 5239.6 12.18 349.3 165.0 2.16 28 105.0 30.88 .500 10.000 4014.1 11.40 286.7 131.5 2.06 26 90.0 26.49 .460 9.500 2977.2 10.60 229.0 101.2 1.95 24 84.0 24.80 .460 9.250 2381.9 9.80 198.5 91.1 1.92 83.0 24.59 .520 9.130 2240.9 9.55 186.7 78.0 1.78 u 73.0 21.47 .390 9.000 2091.0 9.87 174.3 74.4 1.86 20 82.0 24.17 .570 8.890 1559.8 8.03 156.0 79.9 1.82 u 72.0 21.37 .430 8.750 1466.5 8.28 146.7 75.9 1.88 u 69.0 20.26 .520 8.145 1268.9 7.91 126.9 51.2 1.59 (( 64.0 18.86 .450 8.075 1222.1 8.05 122.2 49.8 1.62 (( 59.0 17.36 .375 8.000 1172.2 8.22 117.2 48.3 1.66 18 59.0 17.40 .495 7.675 883.3 7.12 98.1 39.1 1.50 (( 54.0 15.87 .410 7.590 842.0 7.28 93.6 37.7 1.54 u 52.0 15.24 .375 7.555 825.0 7.36 91.7 37.1 1.56 (( 48.5 14.25 .320 7.500 798.3 7.48 88.7 36.2 1.59 15 71.0 20.95 .520 7.500 796.2 6.16 106.2 61.3 1.71 u 64.0 18.81 .605 7.195 664.9 5.95 88.6 41.9 1.49 u 54.0 15.88 .410 7.000 610.0 6.20 81.3 38.3 1.55 u 46.0 13.52 .440 6.810 484.8 5.99 64.6 25.2 1.36 u 41.0 12.02 .340 6.710 456.7 6.16 60.9 24.0 1.41 u 38.0 11.27 .290 6.660 442.6 6.27 59.0 23.4 1.44 12 36.0 10.61 .310 6.300 269.2 5.04 44.9 21.3 1.42 32.0 9.44 .335 6.205 228.5 4.92 38.1 16.0 1.30 it 28.5 8.42 .250 6.120 216.2 5.07 36.0 15.3 1.35 10 28.5 8.34 .390 5.990 134.6 4.02 26.9 12.1 1.21 u 23.5 6.94 .250 5.850 122.9 4.21 24.6 11.2 1.27 9 24.0 7.04 .365 5.555 92.1 3.62 20.5 8.8 1.12 u 20.0 6.01 .250 5.440 85.1 3.76 18.9 8.2 1.17 8 19.5 5.78 .325 5.325 60.6 3.24 15.1 6.7 1.08 " 17.5 5.18 .250 5.250 57.4 3.33 14.3 6.4 1.11 XXIV RESISTANCE OF MATERIALS TABLE VIII MOMENTS OF INERTIA AND SECTION MODULI : RECTANGULAR CROSS SECTION 2 HEIGHT h MOMENT OF INERTIA Mi* ~ 12 SECTION MODULUS , bh* 6 1 BREADTH b HEIGHT h MOMENT OF INERTIA , bh* 12 SECTION MODULUS ,. BREADTH b HEIGHT h MOMENT OF INERTIA /= lf SECTION MODULI'S s=M! 1 2 3 4 5 6 7 8 9 10 11 12 .0833 .66 2.25 5.33 10.42 18 28.58 42.66 60.75 83.33 110.92 144 .166 .66 1.5 2.66 4.16 6 8.16 10.66 13.5 16.66 20.16 24 4 4 5 6 7 8 9 10 11 12 21.33 41.66 72 114.33 170.66 243 333.33 443.66 576 10.66 16.66 24 32.66 42.66 54 66.66 80.66 96 8 9 8 9 10 11 12 13 14 15 16 341.33 486 666.66 887.33 1152 1464.66 1829.33 2250 2730.66 85.33 108 133.33 161.33 192 225.33 261.33 300 341.33 5 5 6 7 8 9 10 11 12 52.08 90 142.92 213.33 303.75 416.66 554.58 720 20.83 30 40.83 53.33 67.5 83.33 100.83 120 9 10 11 12 13 14 15 16 17 18 546.75 750 998.25 1296 1647.75 2058 2531.25 3072 3684.75 4374 121.5 150 181.5 216 253.5 294 337.5 384 433.5 486 2 3 4 5 6 7 8 9 10 11 12 1.33 4.5 10.66 20.83 36 57.16 85.33 121.5 166.66 221.85 288 1.33 3 5.33 8.33 12 16.33 21.33 27 33.33 40.33 48 6 6 7 8 9 10 11 12 108 171.5 256 364.5 500 665.5 864 36 49 64 81 100 121 144 10 10 11 12 13 14 15 16 17 18 19 20 833.33 1109.16 1440 1830.83 2286.66 2810 3413.33 4094.17 4860 5715.83 6666.66 166.66 201.66 240 281.66 326.66 375 426.66 481.66 540 601.66 666.66 3 3 4 5 6 7 8 9 10 11 12 6.75 16 31.25 54 85.75 128 182.25 250 332.75 432 4.5 8 12.5 18 24.5 32 40.5 50 60.5 72 7 7 8 9 10 11 12 13 14 200.08 298.66 425.25 583.33 776.42 1008 1281.58 1600.66 57.16 74.66 94.5 116.66 141.16 168 197.16 228.66 TABLES XXV TABLE IX MOMENTS OF INERTIA AND SECTION MODULI : CIRCULAR CROSS SECTION g S5 3S s MOMENT OF INERTIA SECTION MODULUS DIAM- ETER MOMENT OF INERTIA SECTION MODULUS DIAM- ETER MOMENT OF INERTIA SECTION MODULUS 1 2 3 4 5 6 7 8 9 10 .0491 .7854 3.976 12.57 30.68 63.62 117.9 201.1 322.1 490.9 .0982 .7854 2.651 6.283 12.27 21.21 33.67 50.27 71.57 98.17 35 36 37 38 39 40 73,662 82,448 91,998 102,354 113,561 125,664 4,209 4,580 4,973 5,387 5,824 6,283 69 70 1,112,660 1,178,588 32,251 33,674 71 72 73 74 75 76 77 78 79 80 1,247,393 1,319^67 ,393,995 ,471,963 ,553,156 ,637,662 ,725,571 1,816,972 1,911,967 2,010,619 35,138 36,644 38,192 39,783 41,417 43,096 44,820 46,589 48,404 50,265 41 42 43 44 45 46 47 48 49 50 138,709 152,745 167,820 183,984 201,289 219,787 239,531 260,576 282,979 306,796 6,766 7,274 7,806 8,363 8,946 9,556 10,193 10,857 11,550 12,270 11 12 13 14 15 16 17 18 19 20 718.7 1,018 1,402 1,886 2,485 3,217 4,100 5,153 6,397 7,854 130.7 169.6 215.7 269.4 331.3 402.1 482.3 572.6 673.4 785.4 81 82 83 84 85 86 87 88 89 90 2,113,051 2,219,347 2,329,605 2,443,920 2,562,392 2,685,120 2,812,205 2,943,748 3,079,853 3,220,623 52,174 54,130 56,135 58,189 60,292 62,445 64,648 66,903 69,210 71,569 51 52 53 54 55 56 57 58 59 60 332,086 358,908 387,323 417,393 449,180 482,750 518,166 555,497 594,810 636,172 13,023 13,804 14,616 15,459 16,334 17,241 18,181 19,155 20,163 21,206 21 22 23 24 25 26 27 28 29 30 9,547 11,499 13,737 16,286 19,175 22,432 26,087 30,172 34,719 39,761 909.2 1,045 1,194 1,357 1,534 1,726 1,932 2,155 2,394 2,651 91 92 93 94 95 96 97 98 99 100 3,366,165 3,516,586 3,671,992 3,832,492 3,998,198 4,169,220 4,345,671 4,527,664 4,715,315 4,908,727 73,982 76,448 78,968 81,542 84,173 86,859 89,601 92,401 95,259 98,175 61 62 63 64 65 66 67 68 679,651 725,332 773,272 823,550 876,240 931,420 989,166 1,049,556 22,284 23,398 24,548 25,736 26,961 28,225 29,527 30,869 31 32 33 34 45,333 51,472 58,214 65,597 2,925 3,217 3,528 3,859 TABLE X FOUR-PLACE LOGARITHMS OF NUMBERS 1 1 2 3 4 5 6 7 8 9 0000 0000 3010 4771 6021 6990 7782 8451 9031 9542 1 0000 0414 0792 1139 1461 1761 2041 2304 2553 2788 2 3010 3222 3424 3617 3802 3979 4150 4314 4472 4624 3 4771 4914 5051 5185 5315 5441 5563 5682 5798 5911 4 6021 6128 6232 6335 6435 6532 6628 6721 6812 6902 5 6990 7076 7160 7243 7324 7404 7482 7559 7634 7709 6 7782 7853 7924 7993 8062 8129 8195 8261 8325 8388 7 8451 8513 8573 8633 8692 8751 8808 8865 8921 8976 8 9031 9085 9138 9191 9243 9294 9345 9395 9445 9494 9 9542 9590 9638 9685 9731 9777 9823 9868 9912 9956 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 50 1 2 3 4 5 6 7 8 9 xxvi TABLES XXVll 50 1 2 3 4 5 6 7 8 9 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 72 ' 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 100 1 2 3 4 5 6 7 8 9 XXV111 RESISTANCE OF MATERIALS TABLE XI CONVERSION OF LOGARITHMS KEDUCTION OF COMMON LOGARITHMS TO NATURAL LOGARITHMS Rule for using Table. Divide the given common logarithm into periods of two digits and take from the table the corresponding numbers, having regard to their value as decimals. The sum will be the required natural logarithm. Example. Find the natural logarithm corresponding to the common logarithm .497149. COMMON LOGARITHMS NATURAL LOGARITHMS .49 .0071 .000049 .497149 1.1282667 .016348354 .00011282667 1.14472788067 COM- MON LOGA- RITHM NATURAL LOGARITHM COM- MON LOGA- RITHM NATURAL LOGARITHM COM- MON LOGA- RITHM NATURAL LOGARITHM COM- MON LOGA- RITHM NATURAL LOGARITHM 1 2.30259 26 59.86721 51 117.43184 76 174.99647 2 4.60517 27 62.16980 52 119.73442 77 177.29905 3 6.90776 28 64.47238 53 122.03701 78 179.60164 4 9.21034 29 66.77497 54 124.33959 79 181.90422 5 11.51293 30 69.07755 55 126.64218 80 184.20681 6 13.81551 31 71.38014 56 128.94477 81 186.50939 7 16.11810 32 73.68272 57 131.24735 82 188.81198 8 18.42068 33 75.98531 58 133.54994 83 191.11456 9 20.73327 34 78.28789 59 135.85252 84 193.41715 10 23.02585 35 80.59048 60 138.15511 85 195.71973 11 25.32844 36 82.89306 61 140.45769 86 198.02232 12 27.63102 37 85.19565 62 142.76028 87 200.32490 13 29.93361 38 87.49823 63 145.06286 88 202.62749 14 32.23619 39 89.80082 64 147.36545 89 204.93007 15 34.53878 40 92.10340 65 149.66803 90 207.23266 16 36.84136 41 94.40599 66 151.97062 91 209.53524 17 39.14395 42 96.70857 67 154.27320 92 211.83783 18 41.44653 43 99.01116 68 156.57579 93 214.14041 19 43.74912 44 101.31374 69 158.87837 94 216.44300 20 46.05170 45 103.61633 70 161.18096 95 218.74558 21 48.35429 46 105.91891 71 163.48354 96 221.04817 22 50.65687 47 108.22150 72 165.78613 97 223.35075 23 52.95946 48 ' 110.52408 73 168.08871 98 225.65334 24 55.26204 49 112.82667 74 170.39130 99 227.95592 25 57.56463 50 115.12925 75 172.69388 100 230.25851 TABLES xxix TABLE XII FUNCTIONS OF ANGLES ANGLE SIN TAN SEC COSEC COT Cos 0. 0. .0 CO GO 1. 90 1 0.0175 0.0175 1.0001 57.299 57.290 0.9998 89 2 .0349 .0349 1.0006 28.654 28.636 .9994 88 3 .0523 .0524 1.0014 19.107 19.081 .9986 87 4 .0698 .0699 .0024 14.336 14.301 .9976 86 5 .0872 .0875 1.0038 11.474 11.430 .9962 85 6 0.1045 0.1051 1.0055 9.5668 9.5144 0.9945 84 7 .1219 .1228 .0075 8.2055 8.1443 .9925 83 8 .1392 .1405 .0098 7.1853 7.1154 .9903 82 9 .1564 .1584 .0125 6.3925 6.3138 .9877 81 10 .1736 .1763 1.0154 5.7588 5.6713 .9848 80 11 0.1908 0.1944 1.0187 5.2408 5.1446 0.9816 79 12 .2079 .2126 1.0223 4.8097 4.7046 .9781 78 13 .2250 .2309 1.0263 4.4454 4.3315 .9744 77 14 .2419 .2493 1.0306 4.1336 4.0108 .9703 76 15 .2588 .2679 "1.0353 3.8637 3.7321 .9659 75 16 0.2756 0.2867 1.0403 3.6280 3.4874 0.9613 74 17 .2924 .3057 1.0457 3.4203 3.2709 .9563 73 18 .3090 .3249 1.0515 3.2361 3.0777 .9511 72 19 .3256 .3443 1.0576 3.0716 2.9042 .9455 71 20 .3420 .3640 1.0642 2.9238 2.7475 .9397 70 21 0.3584 0.3839 1.0712 2.7904 2.6051 0.9336 69 22 .3746 .4040 1.0785 2.6695 2.4751 .9272 68 23 .3907 .4245 1.0864 2.5593 2.3559 .9205 67 24 .4067 .4452 .0946 2.4586 2.2460 .9135 66 25 .4226 .4663 .1034 2.3662 2.1445 .9063 65 26 0.4384 0.4877 .1126 2.2812 2.0503 0.8988 64 27 .4540 .5095 .1223 2.2027 1.9626 .8910 63 28 .4695 .5317 .1326 2.1301 1.8807 .8829 62 29 .4848 .5543 .1434 2.0627 1.8040 .8746 61 30 .5000 .5774 .1547 2.0000 1.7321 .8660 60 31 0.5150 0.6009 .1666 1.9416 1.6643 0.8572 59 32 .5299 .6249 .1792 1.8871 1.6003 .8480 58 33 .5446 .6494 1.1924 1.8361 1.5399 .8387 57 34 .5592 .6745 1.2062 1.7883 .4826 .8290 56 35 .5736 .7002 1.2208 1.7435 .4281 .8192 55 36 0.5878 0.7265 1.2361 1.7013 .3764 0.8090 54 37 .6018 .7536 1.2521 1.6616 .3270 .7986 53 38 .6157 .7813 1.2690 1.6243 .2799 .7880 52 39 .6293 .8098 1.2868 1.5890 .2349 .7771 51 40 .6428 .8391 1.3054 1.5557 .1918 .7660 50 41 0.6561 0.8693 1.3250 1.5243 .1504 0.7547 49 42 .6691 .9004 1.3456 1.4945 .1106 .7431 48 43 .6820 .9325 1.3673 1.4663 1.0724 .7314 47 44 .6947 .9657 1.3902 1.4396 1.0355 .7193 46 45 .7071 1. 1.4142 1.4142 1. .7071 45 Cos COT COSEC SEC TAN SIN ANGLE XXX RESISTANCE OF MATERIALS TABLE XIII BENDING MOMENT AND SHEAR DIAGRAMS Jf a = Jf c = 0. l/t MOMENT B t = B 2 = P. Jf a = Jf c = 0. M b = Jf E = Pa 4SEI 384 JZ M mx =R l d-P(d-a). TABLES XXXI i : i ^ I ' I 2 MOMENT SUEAR 3 Pd 2 3 El P Pi .R = P. M A =-PL p/ 3 u = wl + P. D = XXX11 RESISTANCE OF MATERIALS M B = & PL "An Wl MOMENT SHEAR\ \OjL wl* 384^1 MOMENTS' | i r> SHEAR M b =-Pd. M c =-Pd. Ik i4\i f i' TABLES xxxiu TABLE XIV MENSURATION CIRCULAR MEASURE Circumference of circle = diameter x 3.1416. Diameter of circle = circumference x 0.3183. Side of square of same periphery as circle = diameter x 0.7854. Diameter of circle of same periphery as square = side x 1.2732. Side of inscribed square = diameter of circle x 0.7071. Length of arc = number of degrees x diameter x 0.008727. Circumference of circle whose diameter is 1 = TT. v = r \ r 2 , or very nearly = r = radius c = chord v versine o = ordinate or very nearly = NUMBER ?r = 3.14159265 VTT = 1.772454. 7T 2 = 9.869604 - = 0.318310. TT -L = 0.101321. = 0.564190 COMMON LOGARITHM 0.4971499 0.24857494 0.99429995 9.50285013 - 10 9.00570025 - 10 9.75142506- 10 1 radian = angle subtended by circular arc equal in length to the radius of the circle ; TT radians = 180 degrees ; 1 radian = ( - 57.29577951. y TT i Segment of circle area of sector less triangle ; also for flat segments very 4 / r z nearly = -*/ 0. 388 v 2 + - . o \ 4 Side of square of same area as circle = diameter x 0.8862 ; also = circumference x 0.2821. Diameter of circle of same area as square = side x 1.1284. Area of parabola = base x height. Area of ellipse = long diameter x- short diameter x 0.7854. Area of regular polygon = sum of sides x half perpendicular distance from center to sides. XXXIV KESISTANCE OF MATERIALS TABLE XV FRACTIONAL AND DECIMAL EQUIVALENTS FRACTIONS OF A LINEAL INCH IN DECIMALS FRAC- TIONS DECIMALS OF AN INCH FRAC- TIONS DECIMALS OF AN INCH FRAC- TIONS DECIMALS OF AN INCH FRAC- TIONS DECIMALS OF AN INCH A 0.015625 u 0.265625 fl 0.515635 If 0.765625 A 0.03125 A 0.28125 II 0.53125 M 0.78125 A' 0.04687 H 0.296875 fl 0.546875 tt 0.796875 TV 0.0625 0.3125 & 0.5625 H 0.8125 A 0.078125 ti 0.328125 fl- 0.578125 fl 0.828125 A 0.09375 si 0.34375 it 0.59375 II 0.84375 & 0.109375 || 0.359375 if 0.609375 II 0.859375 4 0.125 I 0.375 i 0.625 i 0.875 A 0.140625 II 0.390625 ft 0.640625 fl 0.890625 A 0.15625 41 0.40625 ' Si 0.65625 il 0.90625 0.171875 H 0.421875 II 0.671875 a 6 4 0.921875 0.1875 A 0.4375 w 0.6875 if 0.9375 jt 0.203125 H 0.453125 fl 0.703125 If 0.953125 A 0.21875 1 5 0.46875 11 0.71875 0.96875 II 0.234375 II 0.484375 0.734375 ft 0.984375 i 0.25 i 0.5 i 0.75 1 1.000 LINEAL INCHES IN DECIMAL FRACTIONS OF A LINEAL FOOT LINEAL INCHES LINEAL FOOT LINEAL INCHES LINEAL FOOT LINEAL INCHES LINEAL FOOT A 0.001302083 0.00260416 2 0.15625 0.1666 7* 0.5625 0.5833 iV 0.0052083 2J 0.177083 7 ? 0.60416 ^ 0.010416 21 0.1875 7 i 0.625 T\ 0.015625 2| 0.197916 7| 0.64583 -1 0.02083 2 i 0.2088 8 0.66667 t 0.0260416 0.03125 0.0364583 2| 2 I 0.21875 0.22916 0.239583 4 8| 0.6875 0.7083 0.72916 0.0416 3 0.25 9 0.75 T\ 0.046875 31 0.27083 91 0.77083 f V it 0.052083 0.0572916 0.0625 0.0677083 3| 4 0.2916 0.3125 0.33333 0.35416 4 9| 10 101 0.7916 0.8125 0.83333 0.85416 0.072916 4 i 0.375 10J 0.875 ' 0.078125 0.0833 4| 5 0.39583 0.4166 11 0.89583 0.9166 h- ' 1 ' 1 ' 1 ' 1 JtCnNHolWiUMOcH 0.09375 0.10416 0.114583 0.125 0.135416 6| 6 0.4375 0.4583 0.47916 0.5 0.52083 11J 114 111 12 0.9375 0.9583 0.97916 1.000 I 0.14583 si 0.5416 4 TABLES xxxv TABLE XVI WEIGHTS OF VARIOUS SUBSTANCES MATERIAL WEIGHT IN LB./FT.3 Pressed brick 150 Brick and brickwork Common hard brick Soft inferior brick Good pressed-brick masonry 125 100 140 Ordinary brickwork 125 Gneiss, solid 168 Gneiss, loose piles 96 Granite 170 Limestone and marble 165 Limestone and marble, loose, broken 96 Sandstone, solid 151 Stone and Sandstone, quarried and piled 86 masonry Shale 162 Slate 175 Granite or limestone masonry, well dressed 165 Granite^or limestone masonry, mortar rubble 154 Granite or limestone masonry, well-scabbled dry rubble 138 Sandstone masonry, well dressed 144 Common loam, dry, loose 76 Common loam, moderately rammed 95 Earth,- sand, and gravel Soft flowing mud Dry hard mud Gravel or sand, dry, loose 110 80-110 90-106 . Gravel or sand, well shaken 99-117 Gravel or sand, wet 120-140 Aluminium 162 Brass, cast (copper and zinc) 504 Brass, rolled 524 Bronze (copper 8, tin 1) 529 Copper, cast 542 Copper, rolled 555 Metals Iron, cast 450 Iron, wrought 480 Lead 710 Platinum 1342 Steel 489.6 Tin, cast 459 Zinc, spelter 437.5 Anthracite, solid Pennsylvania 93 Anthracite, broken, loose (heaped bushel 80 Ib.) 54 Anthracite, broken, shaken 58 Coal Bituminous, solid 84 Bituminous, loose (heaped bushel 74 Ib.) 49 Bituminous, broken, shaken 51-56 Coke, loose (heaped bushel 40 Ib.) 26 XXXVI RESISTANCE OF MATERIALS WEIGHTS OF VARIOUS SUBSTANCES Continued MATERIAL WEIGHT IN LB./FT.3 Quicklime, loose (struck bushel 66 Ib.) 53 Quicklime, well shaken 75 American Louisville, loose 50 Lime American Rosendale, loose 56 and hydraulic American Cumberland, loose 65 cement American Cumberland, well shaken 85 English Portland 90 American Portland, loose 88 American Portland, well shaken 110 Ash, American white 38 Cherry 42 Chestnut 41 Cypress 64 Ebony 76 Elm 35 Hemlock 25 Hickory 53 Lignum-vitae 83 Locust 44 Mahogany, Spanish 53 Dry wood Mahogany, Honduras 35 Maple 49 Oak, live 59 Oak, red or black 32-45 Oak, white 48 Pine, white 25 Pine, yellow Northern 34 Pine, yellow Southern 45 Poplar 29 Sycamore 37 Spruce 25 Walnut, black 38 TABLES XXXVll TABLE XVII STRENGTH OF ROPES AND BELTS TENSION TESTS OF STEEL WIRE ROPE CIRCUMFER- ENCE in, NUMBER OF STRANDS WIRES PER STRAND MEAN DIAMETER OF WIRES in. CORE SECTIONAL AREA OF WIRE in. 2 TENSILE STRENGTH Total Ib. Total lb./in.2 1.5 1.75 2 2.125 2.25 2.50 3 3.50 4.50 CO CO CO CO CO CO CO CO CO 18 18 18 18 18 18 18 18 18 .0321 .0349 .0420 .0456 .0488 .0544 .0598 .0718 .0980 Hemp Hemp Hemp Hemp Hemp Hemp Hemp Hemp Hemp .0876 .1031 .1499 .17G6 .2021 .2510 .3024 .4380 .8151 12,898 15,736 20,780 24,430 30,960 33,270 46,370 65,120 138,625 147,236 153,893 138,360 138,383 148,650 132,500 153,340 148,675 170,075 STRENGTH OF IRON WIRE ROPE* (Rope composed of six strands and a hemp center, seven or twelve wires in each strand) DIAMETER in. CIRCUMFERENCE in. APPROXIMATE BREAK- ING STRENGTH CIRCUMFERENCE IN INCHES OF NEW MANILA ROPE Ib. OF EQUAL STRENGTH 1.75 5.50 88,000 11 1.625 5.00 72,000 10 1.50 4.75 64,000 9.5 1.375 4.25 52,000 8.5 1.25 4.00 46,000 8.0 1.125 3.50 36,000 6.5 1.000 3.00 26,000 5.75 .875 2.75 22,000 5.25 .750 2.25 14,600 4.75 .500 1.50 6,400 3.00 .375 1.125 3,600 2.25 ,250 .75 1,620 1.50 STRENGTH OF CAST STEEL WIRE ROPE* (Rope composed of six strands and a hemp center, seven or nineteen wires in each strand) DIAMETER in. CIRCUMFERENCE in. APPROXIMATE BREAK- ING STRENGTH Ib. CIRCUMFERENCE IN INCHES OF NEW MANILA ROPE OF EQUAL STRENGTH 1.25 2.125 1.00 .875 .750 .625 .500 .375 4.00 3.50 3.00 2.75 2.25 2.00 2.50 1.125 106,000 82,000 62,000 52,000 35,200 28,000 16,200 9,000 13 11 9 8.5 7.0 6.0 4.75 3.75 * As given by John A. Roebling. xxxviii BESISTANCE OF MATERIALS TABLE XVII (Continued) TESTS OF MANILA AND SISAL ROPE MANILA ROPE SIZE OF ROPE DIAMETER in. SECTIONAL, AREA in.z TENSILE STRENGTH TOTAL LOAD Ib. lb./in.2 Per Yarn Ib. 6-thread . . ... .27 .30 .38 .43 .49 .56 .61 .62 .74 .79 .78 .85 .96 1.00 .99 1.13 1.19 1.29 1.28 1.39 1.34 1.41 1.59 1.61 1.66 1.76 2.25 2.52 2.83 3.35 3.70 .0567 .0750 .114 .153 .192 .259 .288 .299 .41 .478 .462 .557 .715 .782 .746 .970 1.07 1.27 1.26 1.46 1.36 1.51 1.88 1.99 2.04 2.35 3.82 4.86 6.22 8.37 10.06 13,360 14,180 12,920 14,250 11,610 11,970 10,800 11,500 9,200 12,900 11,900 12,470 12,810 13,630 13,750 12,470 12,190 11,990 11,610 10,080 11,790 9,890 10,360 10,480 10,740 9,940 8,260 9,400 8,600 7,500 7,300 126 118 123 145 125 148 130 128 114 148 138 136 153 146 151 144 132 134 132 117 130 113 121 134 128 118 112 125 118 108 102 750 1,064 1,473 2,180 2,242 3,100 3,120 3,455 3,775 6,207 5,509 6,947 9,160 10,663 10,260 12,093 13,050 15,227 14,640 14,723 16,017 14,943 19,577 20,873 21,903 23,360 31,570 45,647 54,000 62,717 73,910 9-thread . . 12-thread . . 15-thread . . - 1 25-in 1 50-in 1 625-in 1.75-in 2-in 2.25-in 2 25-in. ... 2 50-in 2 75-in 3-in . 3-in 3 25-in 3 50-in 3 75-in. . 3 75-in. 4-in. 4-in 4.25-in 4 50-in 4 50-in 4 75-in 5-in. . 6-in. . . 7-in. . . . 8-in 9-in 10-in SISAL ROPE SIZE OF ROPE DIAMETER In. SECTIONAL AREA in.2 TENSILE STRENGTH TOTAL LOAD Ib. Ib./in.a Per Yarn Ib. .27 .33 .39 .45 .56 .63 .70 .81 .95 1.01 1.22 .0567 .082 .126 .129 .254 .302 .395 .416 .691 .780 1.128 7,700 7,300 7,500 10,810 8,100 7,600 7,200 9,500 8,300 7,500 7,200 72 67 79 93 99 96 97 94 101 104 102 432 605 944 1397 2067 2315 2925 3966 5733 5917 8230 12-thread . . . . 1 25-in 1 50-in 1 75-in. ... 2-in . ... 2.25-in 2.75-in 3-in 3.50-in TABLES XXXIX TABLE XVII (Continued) TESTS OF RUBBER BELTING DIMENSIONS TENSILE STRENGTH in. SECTIONAL DESCRIPTION AREA Tliick- in. 2 Pound per Length Width I16SS Ib./in.z Inch of Width 2-in., 4-ply . 60.17 2.02 .26 .525 3276 851 6-in., 4-ply . . . 60.17 6.08 .26 1.58 3227 839 6-in., 4-ply . . . 60.12 6.13 .26 1.59 3773 979 6-in., 4-ply . . . 60.17 6.05 .26 1.57 2739 711 12-in. , 4-ply . . . 60.02 12.08 .27 3.26 3037 819 12-in., 4-ply . . . 60.14 12.24 .26 3.18 2987 776 2-in., 6-ply . . . 60.17 2.14 .36 .770 3104 1116 6-in., 6-ply . . . 59.98 6.26 .37 2.32 2737 1014 6-in., 6-ply . . . 60.08 6.27 .36 2.26 3770 1358 12-in., 6-ply . . . 60.15 12.04 .36 4.33 3436 1236 12-in., 6-ply . . . 60.17 12.16 .34 4.13 3862 1311 24-in., 6-ply . . . 60.13 24.11 .41 9.89 2381 977 30-in., 6-ply . . . 60.04 30.18 .40 12.07 2808 1123 TESTS OF LEATHER BELTING DIMENSIONS TENSILE STRENGTH in. SECTIONAL DESCRIPTION * AREA Thick- in.z Pounds per Length Width lb./in.* Inch of Width 2-in., single . . . 60.00 1.98 .20 .396 5045 1091 6-in., single . . . 60.20 6.07 .22 1.34 2537 560 6-in., single (w) 60.11 6.08 .24 1.46 2119 633 12-in., single . . . 60.11 12.05 .18 2.17 3917 705 4-in., double . . 59.55 3.98 .33 1.31 4931 1623 6-in., double . . 60.18 5.91 .47 2.78 4309 2027 6-in., double (w) . 59.93 6.00 .40 2.40 6166 2066 12-in., double . . 59.90 11.90 .39 4.64 4090 1595 12-in., double (w) . 60.06 11.93 .36 4.29 4424 1591 24-in., double (w) . 60.00 23.90 .47 11.23 2760 1297 30-in., double . . 59.90 29.95 .43 12.88 2717 1169 * The letter w in the table stands for waterproofed. RESISTANCE OF MATERIALS SECTION I STRESS AND DEFORMATION 1. Elastic resistance, or stress. The effect of an external force acting upon an elastic body is to produce deformation, or change of shape. For example, if a bar is placed in a testing machine and a tensile load applied, it will be found that the length of the bar is increased and the area of its cross section diminished. Similarly, if a compressive load is applied, the length of the bar is diminished and the area of its cross section increased. All solid bodies offer more or less resistance to the deformation, or change of shape, produced by external force. This internal resist- ance, when expressed in definite units, is called stress. A body under the action of stress is said to be strained. In general the stress is not the same at all points of a body, but varies from point to point. The intensity of the stress at any par- ticular point is therefore expressed as the force in pounds which would be exerted if the stress were uniform and acted over an area one square inch in extent. That is to say, whatever the actual extent of the area considered, whether finite or infinitesimal, the stress is expressed in pounds per square inch (abbreviated into lb./in. 2 ). For example, suppose that a wire ^ in. in diameter is pulled with a force of 50 Ib. Then for equilibrium the total stress acting on any cross section of the wire must also be 50 Ib. But since the area of the cross section is only .049 in. 2 , the intensity of the stress 50 is -7777: > or 1000 lb./in. 2 In other words, if the wire were 1 sq. in. .U4y in cross section, the strain under a load of 1000 Ib. would be the same as that produced by a load of 50 Ib. on a wire ^ in. in diameter. 1 RESISTANCE OF MATERIALS Taking any plane section of a body under strain, the stress act- ing on this plane section may in general be resolved, like any force, into two components, one perpendicular to the plane and the other lying in the plane. The perpendicular, or normal, component is called direct stress and is either tension or compression. The tangential component, or that lying in the plane of the cross section, is called shear. In what follows, the letter p will always be used to denote normal, or direct, stress, and q to denote tan- gential stress, or shear. Tension Compression Buckling d The effect of a normal stress is to produce extension or com- pression, that is, a lengthening or shortening of the fibers, thereby changing the dimensions of the body ; whereas shear tends to slide any given cross section over the one adjacent to it, thus producing angular deformation, or change in shape, of the body without altering its dimensions. 2. Varieties of strain. The nature of the deformation produced by external forces acting on an elastic body depends on where and how these forces are applied. Although only two kinds of stress can occur, namely, normal stress (tension or compression) and STRESS AND DEFOBMATION 3 shear, these may arise in various ways. In general, five different cases of strain may be distinguished, each of which must be con- sidered separately. These are as follows : 1. If the forces act along the same line, toward or away from one another, the strain is called compression or tension (Fig. 1, a). 2. If the forces tend to slice off a portion of the body by sepa- rating it along a surface, the strain is called shear (Fig. 1, ft). 10 12 14 16 EXTENSION, PER CENT FIG. 2 18 20 22 24 3. If the forces act transverse to the length of the body (usually perpendicular to the long axis of the piece), so as to produce lateral deflection, the strain is called bending, or flexure (Fig. 1, e). 4. If one dimension of the body is large as compared with the other two, and the forces act in the direction of the long dimension and toward one another, the strain is called buckling, or column flexure (Fig. 1, d). 5. If the forces exert a twist on the body, the strain is called torsion (Fig. 1, e). Two or more of these simple strains may occur in combination, as illustrated in Fig. 1, /. 4 RESISTAHCE OF MATERIALS 3. Strain diagram. In the case of tension or compression it is easy to show graphically the chief features of the strain. Thus, suppose that a test bar is placed in a testing machine, and that the total load on the bar at any instant is read on the scale beam of the machine, and its corresponding length in inches is measured with an extensometer. Assuming that the stress is uniformly dis- tributed over any cross section of the bar, the unit stress is obtained by dividing the total load in pounds acting on the bar by the area of its cross section in square inches. That is, ., total load in pounds (1) p = unit stress = - : ~ area ot cross section in square inches Also, the total deformation, or change in length, is divided by the original unstrained length of the bar, giving the unit deformation in inches per inch. Let this be denoted by s ; that is, let ., -, . ,. change in length (2) s = unit deformation = . original length The unit deformation is therefore an abstract number. Moreover, both the unit stress and the unit deformation are independent of the actual dimensions of the test bar and depend only on the physical properties of the material. If, now, the unit stresses are plotted as ordinates and the corre- sponding unit deformations as abscissas, a strain diagram is obtained, as shown in Fig. 2. Such a diagram shows at a glance the physical properties of the material it represents, as explained in what follows. 4. Hooke's law. By inspection of the curves in Fig. 2 it is evident that the strain diagram for each material has certain char- acteristic features. For instance, in the case of wrought iron the strain diagram from to A is a straight line ; this means that for points between and A the stress is proportional to the correspond- ing deformation. That is to say, within certain limits the ratio of p to s is constant, or (3) f = *. where the constant E denotes the slope of the initial line. This important property is known as Hooke's law, and the constant E is called Young's modulus of elasticity. STRESS AND DEFORMATION 5 The upper end A of the initial line, or point where the diagram begins to curve, is called the elastic limit. The point B, where the deformation becomes very noticeable, is called the yield point. As these two points occur close together, no distinction is made between them in ordinary commercial testing. The maximum ordinate to the strain diagram represents the greatest unit stress preceding rupture, and is called the ultimate strength of the material. In the case of shear let q denote the unit shearing stress and </> the corresponding angular deformation expressed in circular measure. Then, by Hooke's law, (4) = G, where G is a constant for any given material, called the modulus of rigidity, or shear modulus. For steel and wrought iron G = A E, approximately. Average values of E and G for various materials are given in Table I. 5. Elastic limit. It is found by experiment that as long as the stress does not pass the elastic limit, the deformation disappears when the external forces are removed. If the unit stress (or, more properly, the unit deformation) exceeds the elastic limit, however, then the deformation does not entirely disappear upon removal of the load, but the body retains a permanent set. At the elastic limit, therefore, the body begins to lose its elastic properties, and hence, in constructions which are intended to last for any length of time, the members should be so designed that the actual stresses lie well below the elastic limit. It has also been found by experiment that, for iron and steel, if the stress lies well within the elastic limit, it can be removed and repeated indefinitely without causing rupture ; but if the metal is stressed beyond the elastic limit, and the stress is repeated or alter- nates between tension and compression, it will eventually cause rupture, the number of changes necessary to produce failure decreasing as the difference between the upper and lower limits of the strain increases. This is known as the fatigue of metals, 6 RESISTANCE OF MATERIALS and indicates that in determining the resistance of any material the elastic limit is much more important than the ultimate strength. Overstrain of any kind results in a gradual hardening of the material. Where this has already occurred, the elastic properties of the material can be partially or wholly restored by annealing ; that is, by heating the metal to a cherry redness and allowing it to cool slowly. 6. Working stress. The stress which can be carried by any material without losing its elastic properties is called the allowable stress or working stress, and must always lie below the elastic limit. The ratio of the assumed working stress to the ultimate strength of the material is called the factor of safety ; that is, ultimate strength (5) Working stress = factor of safety Average values of the ultimate strength, factors of safety, and other elastic constants for the various materials used in construction are given in Table I. Since for wrought iron and steel the elastic limit can be definitely located, the working stresses for these materials is usually assumed as a certain fraction, say i to |, of the elastic limit. Materials like cast iron, stone, and concrete have no definite elastic limits ; that is, they do not conform perfectly to Hooke's law. For such materials, therefore, the working stress is usually assumed as a small fraction, say from i to ^V' ^ the ultimate strength. Under repeated loads, where the stress varies an indefinite num- ber of times between zero and some large value, the working stress may be assumed as f of its value for a static load. If the stress alternates between large positive and negative values, that is, between tension and compression, the working stress may be assumed as J of its value for a static load. For example, if the elastic limit for mild steel is 35,000 lb./in. 2 , the working stress for a static load may be taken as 18,000 lb./in. 2 ; for repeated loads, either tensile or compressive, as 12,000 lb./in. 2 ; and for loads alternating between tension and compression, as 6000 lb./in. 2 STRESS AND DEFORMATION ALLOWABLE UNIT STRESSES IN LB./IN. 2 (Also called working stress or skin stress) DEAD LOAD ELASTIC LIMIT TENSION COMPRES- SION FLEXURE SHEAR TORSION Structural steel 32,000 16,000 16,000 16,000 12,000 9,000 Machinery steel . 36,000 18,000 18,000 18,000 13,500 13,500 Crucible O.H. steel 40,000 20,000 20,000 20,000 16,000 16,000 Cast steel . . 30,000 12,000 18,000 15,000 13,500 10,000 Wrought iron . . 25,000 12,500 12,500 12,500 9,000 6,000 Cast iron 4,500 15,000 7,500 4,500 4,500 Phosphor bronze . 20,000 10,000 10,000 10,000 8,000 8,000 Timber .... 3,000 1,500 1,000 1,200 120 LIVE LOAD Structural steel 10,600 10,600 10,600 8,000 6,000 Machinery steel . 12,000 12,000 12,000 9,000 9,000 Crucible O.H. steel 13,300 13,300 13,300 10,600 10,600 Cast steel . . . 8,000 12,000 10,000 9,000 6,600 Wrought iron . 8,400 8,400 8,400 6,000 4,000 Cast iron . . . 3,000 10,000 5,000 3,000 3,000 Phosphor bronze . 6,600 6,600 6,600 5,400 5.400 Timber .... 1,000 660 800 80 REVERSIBLE LOAD Structural steel 5300 5300 5300 4000 3000 Machinery steel . 6000 6000 6000 4500 4500 Crucible O.H. steel 6600 6600 6600 5300 5300 Cast steel . . 4000 6000 5000 4500 3300 Wrought iron . 4200 4200 4200 3000 2000 Cast iron . . . 1500 5000 2500 1500 1500 Phosphor bronze . 3300 3300 3300 2700 2700 Timber .... 500 330 400 40 8 RESISTANCE OF MATERIALS In actual practice the unit working stresses are usually specified, and the designer simply follows his specifications without reference to the factor of safety. Where a large number of men are employed, this method eliminates the personal equation and Insures uniformity of results. The student should become familiar with both methods, however, as it is no small part of an engineer's training to know what relation his working stress should bear to the elastic limit and ultimate strength of the material. 7. Resilience. The work done in straining a bar up to the elastic limit of the material is called the resilience of the bar. The area under the strain diagram from the origin up to the elastic limit evidently represents the work done on a unit volume of the material, say one cubic inch, in straining it up to the elastic limit. This area therefore represents the resilience per unit volume, and is called the modulus of elastic resilience of the material. Thus, if p denotes the unit stress and s the unit deformation at the elastic limit, then, since the strain diagram up to this point is a straight line, the area subtended by it, or modulus of elastic resili- ence, is i ps. Since = E, the expression for the modulus may o therefore be written 2 (6) Modulus of elastic resilience = 2 E Hence, if V denotes the volume of the test piece, its total resilience is P 2 V (7) Total resilience = Z E The resilience of a bar is a measure of its ability to resist a blow or shock without receiving a permanent set. If a load W is applied gradually, as in a testing machine, the maximum stress when the W load is all on is p = , where A denotes the area of the cross A section of the bar. If, however, the load W is applied suddenly, as in falling from a height ^, it produces a certain deformation of the bar, say AZ, and consequently the total external work done on the bar is External work = (h + AZ) W. STRESS AND DEFORMATION 9 If, now, the unit stress p produced by the impact lies below the elastic limit, the total internal work of deformation is Internal work = | (Ap) AZ. In the case of a suddenly applied, or impact, load, like that due to a train crossing a bridge at high speed, h = 0, and, equating the expressions for the internal and external work, the result is 2W whence p = -- ^\. Comparing this with the expression for the stress produced by a static load, namely, p , it is evident that a suddenly applied A load produces twice the stress that would be produced by the same load if applied gradually. 8. Poisson's ratio. Experiment shows that when a bar is sub- jected to tension or compression, its lateral, or transverse, dimen- sions are changed, as well as its length. Thus, if a rod is pulled, it increases in length and decreases in diameter ; if it is compressed, it decreases in length and increases in diameter. It was found by Poisson that the ratio of the unit lateral defor- mation to the unit change in length is constant for any given material. This constant is usually denoted by , and is called Pois j m son's ratio. Its average value for metals such as steel and wrought iron is .3. Thus, suppose that the load on a steel bar produces a certain unit deformation s lengthways of the bar. Then its unit lateral deformation will be approximately .3s. Hence, the total lateral deformation is found by multiplying this unit deformation by the width, or diameter, of the bar. Values of Poisson's ratio for various materials are given in Table I. 9. Temperature stress. A property especially characteristic of metals is that of expansion and contraction with rise and fall of temperature. The proportion of its length which a bar free to move expands when its temperature is raised one degree is called its 10 RESISTANCE OF MATERIALS coefficient of linear expansion, and will be denoted by C. Values of this constant for various materials are given in Table I. If a bar is prevented from expanding or contracting, then change in temperature produces stress in the bar, called temperature stress. Thus, let I denote the original length of a bar, and suppose its temperature is raised a certain amount, say T degrees. Then, if C denotes the coefficient of linear expansion for the material, and AZ the amount the bar would naturally lengthen if free to move, we have AZ = CTl, and consequently the unit deformation is - Therefore, if p denotes the unit temperature stress, (8) p = sE APPLICATIONS 1. A 5-in. copper cube supports a load of 75 tons. Find its change in volume. Solution. Area of one face 5 x 5 = 25 in. 2 Unit compressive stress p on 75 x 2000 this area is then p = - = 6000 lb./in. 2 Modulus of elasticity for copper E = 15,000,000 lb./in. 2 Therefore unit vertical contraction s = = - ; total E 2500 vertical contraction AZ = J s = 5 -- = in. Since Poisson's ratio for copper 1 340 is = .340, the unit deformation laterally is .340 s = - -- and the total lateral m 2500 .340 deformation is 5 x - - = .00068 in. The three dimensions of the deformed cube 2500 are therefore 5.00068, 5.00068, 4.998 ; its volume is 124.984 in. 3 , and the decrease in volume is .016 in. 3 2. A f -in. wrought-iron bolt has a head | in. deep. If a load of 4 tons is applied longitudinally, find the factors of safety in tension and shear. Solution. Area of body of bolt at root of thread = .442 in. 2 Unit tensile stress in bolt is p = 4 X 200 = 18,000 lb./in. 2 Factor of safety in tension = 5^2. = 2 .7. Area in shear = TT ---- = 1.47 in. 2 Unit shearing stress is - - = 5440 lb./in. 2 40 000 Factor of safety in shear = - - = 7.3. J 5440 STRESS AND DEFOBMATION 11 3. A steel ring fits loosely over a cylindrical steel pin 3 in. in diameter. How much clearance, or space between them, should there be in order that, when the pin is subjected to a compressive load of 60 tons, the ring shall fit tightly ? Solution. Unit compressive stress in pin is = 17,000 Ib./in.* 4 Unit longitudinal deformation s = = - - - -- = .000566. Unit transverse def- E 30,000,000 ormation = .295 x .000566 = .000167. Radial clearance = 1.5 x .000167= .00025 in. 4. A post 1 ft. in diameter supports a load of 1 ton. Assuming that the stress is uniformly distributed over any cross section, find the unit normal stress. 5. A shearing force of 50 Ib. is uniformly distributed over an area 4 in. square. Find the unit shear. 6. A steel rod 500 ft. long and 1 in. in diameter is pulled by a force of 25 tons. How much does it stretch, and what is its unit elongation ? 7. A copper wire 10 ft. long and .04 in. in diameter is tested and found to stretch .289 in. under a pull of 50 Ib. What is the value of Young's modulus for copper deduced from this experiment ? 8. A round cast-iron pillar 18 ft. high and 10 in. in diameter supports a load of 12 tons. How much does it shorten, and what is its unit contraction ? 9. A wrought-iron bar 20 ft. long and 1 in. square is stretched .266 in. What is the force acting on it ? 10. What is the lateral contraction of the bar in problem 9 ? 11. A soft-steel cylinder 1 ft. high and 2 in. in diameter bears a weight of 40 tons. How much is its diameter increased ? 12. A copper wire 100 ft. long and .025 in. in diameter stretches 2.16 in. when pulled by a force of 15 Ib. Find the unit elongation. 13. If the wire in problem 12 was 250 ft. long, how much would it lengthen under the same pull ? 14. A vertical wooden post 30 ft. long and 8 in. square shortens .00374 in. under a load of half a ton. What is its unit contraction ? 15. How great a pull can a copper wire .2 in. in diameter stand without breaking ? 16. How large must a square wrought-iron bar be made to stand a pull of 3000 Ib. ? 17. A mild-steel plate is ^ in. thick. How wide should it be to stand a pull of 10 tons ? 18. A round wooden post is 6 in. in diameter. How great a load will it bear ? 19. A wrought-iron bar is 20 ft. long at 32 F. How long will it be at 95 F. ? 20. A cast-iron pipe 10 ft. long is placed between two heavy walls. What will be the stress in the pipe if the temperature rises 25 ? 21. Steel railroad rails, each 30 ft. long, are laid at a temperature of 40 F. What space must be left between them in order that their ends shall just meet at 100 F. ? 22. In the preceding problem, if the rails are laid with their ends in contact, what will be the temperature stress in them at 100 F. ? 23. A f-in. wrought-iron bolt failed in the testing machine under a pull of 15,000 Ib. Find its ultimate tensile strength. 12 RESISTANCE OF MATERIALS n n 24. Four i-in. steel cables are used with a block and tackle on the hoist of a crane whose capacity is rated at 6000 Ib. What is the factor of safety ? (Use tables for ultimate strength of rope.) 25. A vertical hydraulic press weighing 100 tons is supported by four 2^-in. round structural-steel rods. Find the factor of safety. 26. A block and tackle consists of six strands of flexible l-in. steel cable. What load can be supported with a factor of safety of 5 ? 27. A vertical wooden bar 6 ft. long and 3 in. in diameter is found to lengthen .013 in. under a load of 2100 Ib. hung at the end. Find the value of E for this bar. 28. A copper wire | in. in diameter and 500 ft. long is stretched with a force of 100 Ib. when the temperature is 80 F. Find the pull in the wire when the tem- perature is 0F., and the factor of safety. 29. An extended shank is made for a -i^-in. drill by boring a ^|-in. hole in FIG. 3 the end of a piece of |-in. cold-rolled steel, fitting the shank into this, and putting a steel taper pin through both (Fig. 3). Standard pins taper 1 in. per foot. What size pin should be used in order that the strength of the pin against shear may equal the strength of the drill shank in compression around the hole ? 30. The head of a steam cylinder of 12-in. inside diameter is held on by ten wrought-iron bolts. How tight should these bolts be screwed up in order that the cylinder may be steam tight under a pres- sure of 180 lb./in. 2 ? 31. Find the depth of head of 'a wrought- iron bolt in terms of its diameter in order that the tensile strength of the bolt may equal the shearing strength of the head. 32. The pendulum rod of a regulator used in an astronomical observatory is made of nickel steel in the proportion of 35.7 per cent nickel to 64.3 per cent steel. The coefficient of expansion of this alloy is approximately 0.0000005. The rod carries two compensation tubes, A and B (Fig. 4), one of copper and the other of alloy, the length of the two together being 10 cm. If the length of the rod to the top of tube A is 1m., find the lengths of the two compensation tubes so that a change in temperature shall not affect the length of the pendulum. 33. Refer to the Watertown Arsenal Reports (United States Government Reports on Tests of Metals)., and from the experimental results there tabulated draw typical strain dia- grams for mild steel, wrought iron, cast iron, and timber, and compute E in each case. 34. A steel wire \ in. in diameter and a brass wire 1 in. in diameter jointly support a load of 1200 Ib. If the wires were of the same length when the load was applied, find the proportion of the load carried by each. FIG. 4 STEESS AND DEFORMATION 13 35. An engine cylinder is 10 in. inside diameter and carries a steam pressure of 801b./in.- Find the number and size of the bolts required for the cylinder head for a working stress in the bolts of 2000 lb./in. 2 36. Find the required diameter for a short piston rod of hard steel for a piston 20 in. in diameter and steam pressure of 125 lb./in. 2 Use factor of safety of 8. 37. A rivet i in. in diameter connects two wrought-iron plates each ' in. thick. Compare the shearing strength of the rivet with the crushing strength of the plates around the rivet hole. 38. In the United States government tests of rifle-barrel steel it was found that for a certain sample the unit tensile stress at the elastic limit was 71, 000 lb./in. 2 , and that the ultimate tensile strength was 118,000 lb./in. 2 What must the factor of safety be in order to bring the working stress within the elastic limit ? 39. In the United States government tests of concrete cubes made of Atlas cement in the proportions of 1 part of cement to 3 of sand and 6 of broken stone, the ultimate compressive strength of one specimen was 883 lb./in. 2 , and of another specimen was 3256 lb./in. 2 If the working stress is determined from the ultimate strength of the first specimen by using a factor of safety of 5, what factor of safety must be used to determine the same working stress from the other specimen ? 40. An elevator cab weighs 3 tons. With a factor of safety of 5, how large must a steel cable be to support the cab ? 41. A hard-steel punch is used to punch holes in a wrought-iron plate | in. thick. Find the diameter of the smallest hole that can be punched. 42. A mild-steel plate 10 in. square and ^in. thick is stretched 0.002 in. in one direction by a certain pull. What pull must be applied at right angles to reduce the first stretch to 0.0014 in. ? 43. A structural steel tie rod of a bridge is to be 25 ft. long when the bridge is completed. What should its original length be if the maximum stress in it when loaded is 18,000 lb./in. 2 ? 44. A hard-steel punch is used to punch a circular hole ^ in. in diameter in a wrought-iron plate ^ in. thick. Find the factor of safety for the punch when in use. 45. A cast-iron flanged shaft coupling is bolted together with 1-in. wrought- iron bolts, the distance from the axis of each bolt to the axis of the shaft being 6 in. If the shaft transmits a maximum torque of 12,000 ft.-lb., find the number of bolts required. 46. A steam cylinder of 16 in. inside diameter carries a steam pressure of 150 lb./in. 2 Find the proper size for the hard-steel piston rod, and the number of 5-in. wrought-iron bolts required to hold on the cylinder head. 47. A horizontal beam 10ft. long is suspended at one end by a wrought-iron rod 12 ft. long and i in. in diameter, and at the other end by a copper rod 12 ft. long and 1 in. in diameter. At what point on the beam should a load be placed if the beam is to remain horizontal ; that is, if each rod is to stretch the same amount ? 48. When a bolt is screwed up by means of a wrench, the tension Tin the bolt in terms of the pull P on the handle of the wrench is found to be given approximately by the empirical formula T 75 P for a wrench of maximum length of from 15 to 16 times the diameter of the bolt. What is the largest wrench that should be used on a -in. wrought-iron bolt, and what is the maximum pull that should be exerted on the handle ? 14 RESISTANCE OF MATERIALS 49. For a steam-tight joint the pitch (or distance apart) of studs or bolts in cylinder heads is determined by the empirical formulas High-pressure cylinders, pitch = 3.5d, Intermediate-pressure cylinders, pitch = 4.5 d, Low-pressure cylinders, pitch = 5.5 d ; or, in general, pitch = where d = diameter of studs or bolts, t = thickness of head or cover in sixteenths of an inch, w = steam pressure in lb./in. 2 Calculate the number, size, and pitch of steel studs for a steam cylinder 20 in. inside diameter under a pressure of 150 lb./in. 2 (high pressure). FIG. 5 50. The standard proportions for a cott-ered joint with wrought-iron rods and steel cotter of the type shown in Fig. 5 are as indicated on the figure. Show that these relative proportions make the joint practically of uniform strength in tension, compression, and shear. SECTION II FIRST AND SECOND MOMENTS 10. Static moment. If a force acts upon a body having a fixed axis of rotation, it will in general tend to produce rotation of the body about this axis. This tendency to rotate becomes greater as the magnitude of the force increases, and also as its distance from the axis of rotation increases. The numerical amount of this tend- ency to rotate is thus measured by the product of the force by its perpendicular distance from the given axis, or center of rotation. This product is called the first moment, or static moment, of the force with respect to the given axis, or point. Thus, let F denote any force, P the fixed axis of rotation, as- sumed to be at right angles to the plane of the paper, and d the perpendicular distance of F from P. Then d is called the lever arm of the force, and its moment about P is defined as Moment = force X lever arm, or, if the moment is denoted by M, (9) M = Fd. P (Center or axis of rotation) It is customary to call the moment positive if it tends to produce rotation in a clockwise direction, and negative if its direction is counter-clockwise (Fig. 6). 11. Fundamental theorem of moments. When two concur- rent forces act on a body simul- taneously, their joint effect is the same as that of a single force, given in magnitude and direction by the diagonal of the parallelogram formed on the two given 15 FIG. 6 16 RESISTANCE OF MATERIALS N W W FIG. 7 forces as adjacent sides (Fig. 7). This single force, which is equiv- alent to the two given forces, is called their resultant. Any number of concurrent forces may be thus combined by \ finding the resultant of any two, \ X P combining this with the third, \ etc. Or, what amounts to the \ same thing, the given forces may \ be placed end to end, forming a 1 polygon, and their final resultant will then be the closing side of this polygon (Fig. 8). Now, in Fig. 9, let F 1 and F 2 be any two concurrent forces, and F their resultant. Also let be any given point, and 0^ , c, the angles between OA and the forces F^, Ff F, respectively. Then, taking moments about 0, Moment of F l about = F l x OAsin Q v Moment of F z about = F x OA sin 2 . The sum of these moments is = F l xOA sin l + F 2 x OA sin # 2 = OA (^ sin l + F 2 sin 2 ). But, since F is the resultant of F l and F# F sin < =F l sin l +F 2 sin 0^ and consequently 2} M=OA xFsm(j>. The right member, however, is the moment of the resultant \^~---A F with respect to 0. Therefore, FlG<9 since is arbitrary, the sum of the moments of any two concurrent forces with respect to a given point is equal to the moment of their resultant with respect to this point. FIRST AND SECOND MOMENTS 17 If the forces F l and F 2 are parallel, introduce two equal and opposite forces H, H, as shown in Fig. 10, and combine the iTs with F l and F 2 into resultants F^, F' r Transferring these resultants FV FZ to their point of intersection 0, they may now be resolved into their original components, giving two equal and opposite forces, -f- H and H, which cancel, and a resultant F l + F 2 parallel to F l and F 2 . Moreover, applying the theorem of moments proved above to the concurrent forces F^, F% at 0, the sum of their moments about any point is equal to the moment of their resultant F l + F 2 about the same point. But the moment of F[ is equal to the sum of the moments of F l and H, and, simi- larly, the moment of F^ is equal to the sum of the moments of F z and + H. Since the forces + H and - H have the same line of action, their moments about any point cancel, and therefore the theorem of moments is also valid for parallel forces. This theorem may obviously be extended to any number of forces by combining the moments of any two of them into a resultant moment, combining this resultant moment with the moment of the third force, etc. Hence, The sum of the moments of any number of forces lying in the same plane with respect to a given point in this plane is equal to the moment of their resultant with respect to this point. 12. Center of gravity. An important application of the theorem of moments arises in considering a system of particles lying in the same plane and rigidly connected. The weights w^ w^ , w n of the particles are forces directed toward the center of the earth. Since this is relatively at an infinite distance as compared with the dis- tances between the particles, their weights may be regarded as a system of parallel forces. FIG. 10 ..V 18 RESISTANCE OF MATERIALS The total weight W of all the particles is W=w + iv z + . . . 4- w n = w ; that is, W is the resultant of the n parallel forces w^ w^ ., w n . The location of this resultant W may be determined by applying the theorem of moments. Thus, let x^ # 2 , -, x n denote the perpen- dicular distances of w^ w , , w n from any fixed point (Fig. 11). Then, if X Q denotes the perpendicular distance of the resultant W from 0, by the theorem of moments whence FIG. 11 2 'o- w > or, since W = ^ w, this may also be written (10) .=4^- 7, iv This relation determines the line of action of W for the given position of the system. If, now, the system is turned through any angle in its plane, and the process repeated, a new line of action for W will be determined. The point of intersection of two such lines is called the center of gravity of the system. From the method of determining this point it is evident that if the entire weight of the system was concentrated at its center of gravity, this single weight, or force, would be equivalent to the given system of forces, no matter what the position of the system might be. If the particles do not all lie in the same plane, a reference plane must be drawn through instead of a reference line. In this case ">-\ the equation X Q = ^ determines the position of a plane in which the resultant force W must lie. The intersection of three such planes corresponding to different positions of the system of particles will then determine a point which is the required center of gravity. FIRST AND SECOND MOMENTS 19 If m^ m^ - ., m n denote the masses of the n particles, and M their sum, then, since W=Mg, Wl = m^ w 2 = m 2 g, , w n = m n g, where g denotes the acceleration due to gravity, the above relations for determining the center of gravity become or, since g is constant, (11) M ill The point determined from these relations by taking the system of particles in two or more positions is called the center of inertia or center of mass. Since these relations are identical with those given above, it is evident that the center of mass is identical with the center of gravity. 13. Centroid. It is often necessary to determine the point called the center of gravity or center of mass without reference to either the mass or weight of the body, but simply with respect to its geometric form. For a solid body let Av denote an element of volume, Am its mass, and D the density of the body. Then, since mass is jointly proportional to volume and density, Therefore the formulas given above may be written or, since the density D is constant, these become (12) ' r Since the point previously called the center of gravity or center of mass is now determined simply from the geometric form of the body, it is designated by the special name centroid. 20 RESISTANCE OF MATERIALS Evidently it is also possible to determine the centroid of an area or line, although neither has a center of gravity or center of mass, since mass and weight are properties of solids. For a plane area the centroid is determined by the equations (13) where A# denotes an element of area and^l the total area of the figure. Similarly, for a line or arc the centroid is given by (14) where A? denotes an element of length and L the total length of the line or arc. 14. Centroid of triangular area. To find the centroid of a triangle, divide it up into narrow strips parallel to one side AC (Fig. 12). Since the centroid of each strip PQ is at its middle point, the centroid of the en- tire figure must lie somewhere on the line BD joining these middle points ; that is, on the median of the triangle. Simi- larly, by dividing the triangle up into strips parallel to another side BC, it is proved that the centroid must also lie on the median AE. The point of inter- section G of these two medians must therefore be the centroid of the triangle. Since the triangles DEG and ABG are similar, DG DE and since DE = 1 A B, this gives The centroid of a triangle therefore lies on a median to any side at a distance of one third the length of the median from the opposite vertex. From this it also follows that the perpendicular distance FIRST AND SECOND MOMENTS 21 of the centroid G from any side is one third the distance of the opposite vertex from that side. 15. Centroid of circular arc. For a circular arc CD (Fig. 13) the centroid G must lie on the diameter OF bisecting the arc. Now suppose the arc divided into small segments, and from the ends of any segment PQ draw PR parallel to the chord CZ>, and QR perpendicular to this chord. Since the moment of the entire arc with respect to a line AB drawn through perpendicular to OF must be equal to the sum of the moments of the small seg- ments PQ with respect to this line, the equation determining the centroid is X x. FIG. 13 But from the similarity of the triangles PQR and OQE we have PQ OQ r PQ - x = PR Therefore PQ - x = PR - r, and consequently or, since the radius r is constant, ^PQ. x = r^ PR = r . chord CD. The position of the centroid is therefore given by (15) chord arc . radius. If the central angle COD is denoted by 2 a, then arc=2ra and chord = 2 r sin a, and therefore the expression for the centroid may be written For a semicircle 2 a = TT, and consequently 22 RESISTANCE OF MATERIALS 16. Centroid of circular sector and segment. To determine the cen- troid of a circular sector OCB (Fig. 14), denote the radius by r and the central angle COB by 2 a. Then any small element OPQ of the sector may be regarded as a triangle the centroid of which is on its median at a distance of Jr from 0. The centroids of all these elemen- tary triangles therefore lie on a concentric arc DEF oi radius | r, and the centroid of the entire sector coincides with the centroid of this arc DEF. Therefore, from the results of the pre- ceding article, the centroid of the entire sector OCB is given by FIG. 14 (16) 2 sin a = r 3 a For a semicircular area of radius r the distance of the centroid from the diameter, or straight side, is 37T To determine the centroid of a circular segment CBD (Fig. 15), let G denote the centroid of the entire sector OCBD, G Q of the segment CBD, and G 1 of the triangle OCD. Then the position of G Q may be deter- mined by noting that the sum of the moments of the triangle OCD and the segment CBD about any point, say 0, is equal to the moment of the entire sector about this point. Thus, FlG ' 15 if A Q , A^ A denote the areas of the segment, triangle, and sector, re- spectively, and X Q , x^ x, the distances of their centroids from 0, then whence FIRST AND SECOND MOMENTS Ax - A Now let c denote the length of the chord CD and a the length of the arc CBD. Then, from the results of this and the preceding articles, 2 re - 3 a 2 * <? *i = 8Y-4' and also, from geometry, 1 1 A = -ar, A i = 2 Inserting these values in the expression for # , the result is Trr* ^ A For a semicircle, A Q = and c = 2 r. Therefore, in this case, as also shown above, 17. Centroid of para- u bolic segment. For a parabolic segment with _i^ vertex at A (Fig. 16) the position of the cen- troid G is given by (18) *. = ! where a and b denote the sides of the circumscribing rectangle. Also, FIG. 16 (19) Area^LBC = - ab. 3 For the external segment ABD (Fig. 16) the centroid is given by 3 3, (20) CC Q = a, y = T & > and the area of the external segment is (21) Area ABD = - ab. 3 24 RESISTANCE OF MATERIALS FIG. 17 18. Axis of symmetry. If a figure has an axis of symmetry, then to any element of the figure on one side of the axis there must correspond* an equidistant element on the opposite side, and since the moments of these equal elements about the axis of symmetry are equal in amount and oppo- site in sign, their sum is zero (Fig. 17). Since the moment of each pair of ele- ments with respect to the axis of symmetry is identically zero, the total moment is also zero, and hence the centroid of the figure must lie on the axis of symmetry. When a figure has two or more axes of symmetry, their inter- section completely determines the centroid. 19. Centroid of composite figures. To determine the centroid of a figure made up of several parts, the centroid of each part may first be determined separately. Then, assuming that the area of each part is concentrated at its centroid, the centroid of the entire figure may be deter- mined by equating its moment to the sum of the moments of the several parts. To illustrate this method, let it be required to find the centroid of the I -shape shown in Fig. 18. Since the figure has an axis of symmetry MN, the centroid must lie some- where on this line. To find its position, divide the / into three rectangles, as indicated by the dotted lines in the figure. The \M I CKJ) ..}__ i i \N FIG. 18 centroids of these rectangles are at their centers a, 5, c. Therefore, denoting these three areas of the rectangles by A, B, C, respectively, and taking moments with respect to the base line, the distance of the FIRST AND SECOND MOMENTS 25 centroid of the entire figure from the base is found to be _A x ad+ B x bd+ C X cd / ___ . ^ As another example, consider the circular disk with a circular hole cut in it, shown in Fig. 19. Here also the centroid must lie somewhere on the axis of symmetry (7 1? C 2 . Therefore, denoting the radii of the circles by R, r, as shown, and taking moments about the tangent perpendicular to the line of centers, the distance X Q of the centroid from this tan- gent is found to be or, since # 2 = R and x l = R e, where the hole, or distance between centers, TrR' 2 ?rr 2 denotes the eccentricity of R 3 -r\R- e) 20. Moment of inertia. In the analysis of beams, shafts, and columns it will be found necessary to compute a factor, called the moment of inertia, which depends only on the shape and size of the cross section of the member. This shape factor is usually denoted by 7, and is defined as the sum of the products obtained by multiplying each element of area of the cross section by the square of its distance from a given line or point. Thus, in Fig. 20, if A^4 denotes an element of area and y its distance from any given axis 00, then the moment of inertia of the figure with respect to this axis is defined as FIG. 20 26 RESISTANCE OF MATERIALS Since an area is not a solid and therefore does not possess inertia, the shape factor / should not be called moment of inertia, but rather the second moment of area, since the distance y occurs squared. To compute I for any plane area, divide the area up into small elements A^ (Fig. 21). -H ] Then the first (or static) moment of each element with respect to any axis 00 is yA^4, where y denotes the distance of this ele- ment from the given axis. Now erect on A^4 as base a prism of height y. If this is done for every element of the plane area, the result will be a solid, or truncated cylinder, as shown in Fig. 21, the planes of the upper and lower bases intersecting in the axis 00 at an angle of 45. Let V denote the volume of this moment solid, as it will be called, and y the distance of its centroidal axis from 00. Then, by the theorem of moments, \ FIG. 21 Since A V = #A A, the right member becomes Hence (22) I = Ft/ . 21. I for rectangle. FIG. 22 Let it be required to find I for a rectangle of breadth b and height h with respect to an axis through its centroid, or middle point, and parallel to the base (Fig. 22). The moment solid in this case consists of a double wedge, FIRST AND SECOND MOMENTS 27 as shown in Fig. 22, the base of each wedge being , its height - , and its volume V = - base x altitude = 2 8 Since the centroid of a triangular wedge, like that of a triangle, is at a distance of | its altitude from the vertex, _2 h_h ^~3 X 2~3* Therefore 1=2 Vii^ = - For any plane area the /'s with respect to two parallel axes are related as follows : Let 00 denote an axis through the centroid of the figure, AA any parallel axis, _O and d their distance apart (Fig. 23). Also let I denote the / of the figure with re- spect to the axis 00, and I A with respect to the axis AA. Then, from the definition of 7, FIG. 23 But since 00 is a centroidal axis, ^ykA = for this axis. There- fore, since V/ 2 A^4 = / , the above expression becomes (23) I A = TO + <?A. From this relation it is evident that the / for a centroidal axis is less than for any parallel axis. As an application of this formula, find the I for a rectangle with -j -i q 7 respect to its base. From what precedes, I = Also, d= - and I L- 2 A = bh. Hence the / for a rectangle with respect to its base is 3 28 RESISTANCE OF MATEEIALS 22. I for triangle. Consider a triangle of base b and altitude h and compute first its /with respect to an axis A A through its vertex and parallel to the base (Fig. 24). The moment solid in this case is a pyramid of base bh and altitude A, the volume of which is V=- base x altitude = o o Since the centroid of this pyramid is at a distance ?/ = ^ h from the vertex, we have 3 FIG. 24 To find /for the triangle with respect to an axis 00 through its centroid and parallel to AA, apply the theorem bh* bh , , 2 Since in the present case I A = , ^4 = -, and d = - h, we have therefore Similarly, for the axis ^^ we have r 1 B 23. I for circle. In computing the /for a circle, it is convenient to determine it first with respect to an axis through the center of the circle and perpendicular to its plane (the so-called polar moment of inertia of the circle) . FIRST AND SECOND MOMENTS 29 Consider the circle as made up of a large number of elementary triangles OAB with common vertex at (Fig. 25). Since the alti- tude of each of these triangles is the radius R of the circle, from the preceding article the / for each with respect to the point is -- For the entire circle, therefore, or, since T AB = circumference = 2 irR, this becomes (24) If D denotes the diameter of the circle, then R = -~ and we also have (35) *>- If XX and YY are two rectan- gular diameters of the circle, and r is the distance of any element of area AA from their point of inter- section (Fig. 25), then I = I Y Hence (26) Since a circle is symmetrical about all diameters, we have I x = I Y . Therefore the I of a circle with respect to any diameter is or (27) 7TD* 64 24. I for composite figures. When a plane figure can be divided into several simple figures, such as triangles, rectangles, and circles, the / of the entire figure with respect to any axis may be found by adding together the J's for the several parts with respect to this axis. Thus, in Fig. 26 each area may be regarded as the difference 30 RESISTANCE OF MATERIALS of two rectangles a large rectangle of base B and height 77, and a smaller rectangle of base b and height h. Consequently the / for either figure, with respect to its cen- troidal axis GG, in, is given by W 12' 12 Similarly, the figures shown in Fig. 27 may each be regarded as the sum of two rect- angles, and hence the / for either of these figures with respect to its centroidal axis GG is given by FIG. 26 FIG. 27 1 = BH " h 12 !2 For the angle, or tee, shown in Fig. 28 the 7" about the base line 00 is the sum of the J's for the two rectangles into which the figures are divided by the dotted lines ; that is, _BH S W ~~ir + ir ' I B *_ B G Q t Hh- 1 <- b >j I O T O FIG. 28 The position of the centroidal axis GG may then be determined by taking moments about the base 00. That is to say, since the FIRST AND SECOND MOMENTS 31 total area is A = BH + bh, we have, by the principle of moments, TT ~L x(BH + bJi) = BH x + bh X - , whence L 2 BH* + IV Having found a? , the /for the centroidal axis GG is determined by the relation T _ T A * G L O -^^o* APPLICATIONS 51. A uniform rod 18 in. long weighs 8 lb. and has weights of 2 lb., 3 lb., 4 lb., and 5 lb. strung on it at distances of 6 in. apart. Find the point at which the rod will balance. Solution. Since the rod is uniform, its weight may be assumed to be concen- trated at its center. If, then, x denotes the distance of the center of gravity from the end at which the 2-lb. weight is hung, by taking moments about this end _ 2 x 0+ 3x6+8x9+4x12 + 5x18 XQ 2+3+8+4+5 = I<T 52. A t = ^ in. .4,J_r SP section like that shown in Fig. 29 has the dimensions 6 = 3 in., d = 5 in., Locate its center of gravity, or centroid. Solution. To locate the gravity axis 1 1, take moments about any parallel line as a base, say A B. Then, dividing the figure into two rectangles, since the center of gravity of each rectangle is at its center, we have _ D Similarly, to determine the gravity axis 2 2, by taking moments about CD we have = 4 3 X i+4ix i 2 FIG. 29 53. The section shown in Fig. 30 is made up of two 10-in. channels 30 Ib./ft. and a top plate 9 in. x \ in. Locate its gravity axes and determine its moment of inertia with respect to the axis 11. Solution. From Table IV the area of each channel is 8.82 in. 2 To determine the gravity axis 1 1, take moments about the lower edge of the section. Then 2 x 8.82 + 9 x = 0? 32 RESISTANCE OF MATERIALS From the table, the moment of inertia of each channel with respect to an axis perpendicular to the web at center is 103.2 in. 4 , and the distance from this axis to la j t ffl ! 3 FIG. 30 the gravity axis of the entire section is 6.07 5 = 1.07 in. Also, the moment of inertia of the top plate with respect to its gravity axis is X **' = in. 4 , 12 12 32 and the distance of this axis from the gravity axis of the entire section is 4.18 in. Therefore Ii_i= 2 [103.2 + 8.82 x (1.07) 2 ] + [fa + 4.5 x (4.18) 2 ] =305 in. 4 For the net section the rivet holes must be deducted from this value. Assuming two | -in. rivets, the amount to be deducted is approximately 24 in. 4 , giving for the net section I^ = 281 in.* 54. In problem 53 determine the mo- ment of inertia of the net section with respect to the gravity axis 22. 55. The section shown in Fig. 31 is made up of four angles 4 x 3 x ^ in., with the longer leg horizontal, and a web plate 12 x | in., with f-in. rivets. Find the moment of inertia for its net section with respect to the gravity axis 1 1. l f FIG. 32 56. The section shown in Fig. 32 is built up of two 8-in. channels 18.751b./ft. and two plates 9 x f in. Find the moment of inertia of its net section about the gravity axis 11, deducting the area of four J-in. rivet holes. FIKST AND SECOND MOMENTS 33 57. In problem 56 find the moment of inertia of the net section with respect to the gravity axis 22. 58. The section shown in Fig. 33 has the dimensions 6=10 in., d = 4 in., t = 1 in. Locate the gravity axis 1 1. 59. In problem 58 find the moment of inertia of the section with respect to the gravity axis 2 2. 60. The section shown in Fig. 34 has the dimensions f L U A . ^__ t ^_4__ v * FIG. 33 tj = t 2 = t s = 1 in. Locate the gravity axis 1 1. 61. The section shown in Fig. 32 is composed of two 12-in. channels, 20.5 lb./ft., and two ^-in. plates. How wide must the plates be in order that the moments of inertia of the section shall be the same about both gravity axes ? 62. The section shown in Fig. 35 has the dimensions 6 = 8 in., h = 10 in., 6' = 5 in., h' = 6 in. Find its moments of inertia about both gravity axes. 63. Two6-in. channels 10. 5 lb./ft. are connected by latticing. How far apart should they be placed, back to back, in order that the moments of inertia may be the same about both gravity axes ? 64. A hollow cast-iron column is 6 in. external diameter and 1 in. thick. Find the moment of inertia of its cross section with respect to a diameter. 65. The section shown in Fig. 31 is made up of a web plate 9 x $ in. and four angles 3 x 3 x in. Find its moment of inertia o with respect to both gravity axes. 66. The top chord of a bridge truss has a section like that shown in Fig. 36, with top plate 20 x f ", two web plates each 18 x f", and four angles 3 x 3 x f". Find the eccentricity of the section ; that is, the dis- tance from center of figure to gravity axis 1 1. | 67. In problem 66 find the moment of iner- J tia of the net section with respect to the axis ~~^ 11, deducting for four |-in. rivets. I 68. A circular table rests on three legs placed I at the edge and forming an equilateral triangle. J, Find the least weight which will upset the table FIG. 35 when hung from its edge. 34 RESISTANCE OF MATERIALS 69. A horizontal beam 20 ft. long and weighing 120 Ib. rests on two supports 10 ft. apart. A load of 75 Ib. is hung at one end of the beam and 150 Ib. at the other end. How must the beam be placed so that the pressure on the supports may be equal ? 70. Explain how a clock hand on a smooth pivot can be made to show the time by means of clockwork concealed in the hand and carrying a weight around. 71. A brick wall is 12 in. thick and 40 ft. high. What uniform wind pressure will cause - - - it to tip over ? Weight of ordinary brick masonry is 125 Ib./f t. 3 72. A masonry dam is 30 ft. high, 6 ft. wide at top, and 30 ft. wide at bottom, with upstream face vertical. Assuming the masonry to weigh 160 lb./ft. 3 , compute the moment of the weight of the dam about the toe of the base. 73. In problem 72 the resultant water pressure for a vertical strip 1 ft. wide is 46,800 Ib. and is applied at a point 12ft. above the base of the dam. Determine its stability against overturning. 74. The casting for a gas-engine piston is a hollow cylinder of uniform thick- ness, with one end closed. The external diameter is 5 in., length over all 6 in., thickness of cylinder shell Jin., thickness of end l^in. Find the distance of its center of gravity from the closed end. 75. A cast-iron pulley weighs 50 Ib. and its center of gravity is 0.1 in. out of center. To balance the pulley, a hole is drilled in the light side, 6 in. from the center of the pulley and in line with its center of gravity, and rilled with lead. How much iron must be removed, the specific gravity of lead being 11.35 and of iron 7.5 ? FIG. 36 SECTION III BENDING-MOMENT AND SHEAR DIAGRAMS 25. Conditions of equilibrium. In order that any structure may be in equilibrium, the external forces acting on it must satisfy two conditions : 1. The sum of the forces acting in any given direction must be zero. 2. The sum of the moments of the forces about any point must be zero. If force and moment are denoted by F and Jf, respectively, these conditions are expressed more briefly in the form f2>=0; (28) For equilibrium \ ~; If the forces all lie in one plane, the condition VJP = is expressed more conveniently in the form (29) ]>} vertical forces = O ; V horizontal forces = O. To illustrate the application of these conditions, consider a simple beam AB of length Z, supported at the ends and bearing a single concentrated load P at a distance d from one end A IP (Fig. 37). Let the reac- tions of the supports at A and B be denoted by R v R , and, to find the value 2 FH;. 37 01 these reactions, apply the condition VTJf = ; that is, equate to zero the sum of the moments of all the external forces with respect to any convenient moment center, say A. Then Pd - RJ, = 0, Pd whence R^ = 35 36 RESISTANCE OF MATERIALS Now, applying the condition ^ vertical forces = 0, we have ^ + R z - P = 0, and inserting in this equation the value just found for R 2 and then solving for E^ the result is 26. Vertical shear. By applying the conditions V^ = 0,^Jf=0, as just explained, all the external forces acting on the beam may be found. The beam may then be supposed to be cut in two at any point and these conditions applied to the portion on either side of the section. In general, the sum of the external forces on one side of any arbitrary cross section will not be identically zero. If, then, the condition of equilibrium V F is satisfied for the portion of the beam on one side of the section, the stress in the material at this point must supply a force equal in amount and opposite in direc- tion to the resultant of the external forces on one side of this point. This resisting force, or resultant of the vertical stresses in the plane of the cross section, which balances the external forces on one side of the section, is called the vertical shear. Therefore The vertical shear on any cross section = the algebraic sum of the external vertical forces on either side of the section. For instance, suppose that a beam 10 ft. long bears a uniform load of 300 lb./ft., and it is required to find the vertical shear on a section 4 ft. from the left support. In this case the total load on the beam is 3000 lb., and, since the load is uniform, each reac- tion is 1500 lb. The load on the left of the given section is then 4 x 300 = 1200 lb., and therefore -the shear at the section is 1500 - 1200 = 300 lb. 27. Bending moment. In applying the condition Vlf = to the portion of a beam on either side of any cross section, the center of moments is taken at the centroid of the section. Since the position of the cross section is arbitrary, it is obvious that the sum of the mo- ments of the forces on one side of the section about its centroid will not in general be zero. Therefore, to satisfy the condition V Jf = 0, the normal stresses in the beam at the section considered must BENDING-MOMENT AND SHEAR DIAGRAMS 37 P, supply a moment which balances the sum of the moments of the external forces on either side of the point. This resisting moment in the beam is called the stress couple or bending moment, and is evidently equal to the resultant external moment at the point in question. Consequently The bending moment at any cross section of a beam is equal to the sum of the moments of the external forces on one side of this point, about the centroid of the section. For example, in Fig. 38, consider a cross section mn at an arbitrary distance x from the left support. Then for the portion of the beam on the left of mn the mo- ment of R 1 about the cen- troid of the section is Rjc, and the moment of P^ about the same point is 7J (x d^). A Therefore the total bend- R\ ing moment at> the section mn is M=Rx-P l (x-d\ A R^T- - x - As another example, con- sider a beam of length I bearing a uniform load of amount w per unit of length. Then the total load on the beam is wl, and each reaction is Therefore, 2 taking a section at a distance x from the left support and consider- ing only the forces on the left of the section, the total bending moment at this point is \B Pi n - FIG. 38 Wl X WX = - x wx = From this relation it is evident that M = when x = or x = Z, and attains its maximum value when x = - ; that is, the bending moment is zero at each end of the beam and a maximum at the center. 28. Bending-moment and shear diagrams. Since in general the bending moment and shear vary from point to point along a beam, it is desirable to show graphically the moment and shear at 38 RESISTANCE OF MATERIALS each point of the beam. This may be done by means of a bending- moment diagram and a shear diagram, obtained by plotting the general expressions for the moment and shear, such as those given in the examples in the preceding paragraph. Thus, the shear diagram is obtained by plotting the shear at any arbitrary section mn as ordinate and the distance x of this section from a fixed origin as abscissa. Similarly, the moment diagram is obtained by plotting the moment at any arbitrary section mn as ordinate and the distance x as abscissa. The following simple applica- tions illustrate the method of drawing the diagrams. 1. Simple beam bearing a single concentrated load P at its center (Fig. 39). From symmetry, the reactions R^ and R 2 are each equal to Let mn denote any section A of the beam at a distance x from the left support, and consider the portion of the beam on the left of this section. Then the moment at P 9 mn s = x] and the shear = ). For a section on the FIG. 39 IS right of the center the bending moment is R^ (I x) and the shear is R^. Consequently, the bending moment varies as the ordinates of a triangle, being zero at either support and attaining a maximum PI value of at the center, while the shear is constant from A to B, and also constant, but of opposite sign, from B to C. The diagrams in Fig. 39 represent these variations in bending moment and shear along the beam under the assumed loading. Consequently, if the ordinates vertically beneath B are laid off to scale to represent the bending moment and shear at this point, the bending moment and shear at any other point D of the beam are found at once from the diagram by drawing the ordinates EF and HK vertically beneath Z>. BEKDING-MOMENT AND SHEAK DIAGRAMS 39 2. Beam bearing a sin- gle concentrated load P at a distance c from one support. The reactions in this case are and ~ Hence, the bending mo- ment at a distance x from the left support is provided x < c, and Pc(l-x) RtQ-x) = - ~ * if x > c. If x = c, each of these moments becomes PC (l-c) ~r and consequently the bend- ing-moment and shear dia- grams are as shown in Fig. 40. 3. Beam bearing several separate loads. In this case the bending- moment diagram may be obtained by constructing the diagrams for each load separately and then adding their ordinates, as indicated in Fig. 41. SHEAR FIG. 41 40 RESISTANCE OF MATERIALS 4. Beam bearing a contin- uous uniform load. Let the load per unit of length be denoted by w. Then the total load on the beam is ?rZ, and the reac- tions are Hence, at a distance x from the left support the bending moment M x is wl x M x = x-wx.- FIG. 42 The ben ding-moment diagram is therefore a parabola. When x = -< wl 2 1.1. ., M x = , which is its 8 maximum value. The bending-moment and shear diagrams are therefore as repre- sented in Fig. 42. 5. Beam bearing uni- form load over part of the span. Let the load ex- tend over a distance c and be of amount w per unit of length. Then the total load is we. The reactions of the supports are the same as though the load were concentrated FIG. 43 BENDING-MOMENT AND SHEAR, DIAGRAMS 41 at its center of gravity G. Therefore, if d denotes the distance of G from the left support, ,... SI J\ and ft = I Also, the bending-moment diagrams for the portions AB and CD are the same as though the load were concentrated at G, and are therefore the straight lines A'H and D'K, intersecting in the point T vertically beneath G (Fig. 43). From B to C there is an additional bending moment due to the uniform load on this portion of the beam. Thus, if LMN is the parabolic moment diagram for a beam of length LN or c, the ordi- nates to the line HK must be increased by those to the parabola LMN, giving as a complete moment diagram the line A'HJKD'. Analytically, if x denotes the distance of any section from the left support, the equations of the three portions A'H, HJK, and KD' of the moment diagram are we (I d} x 7 c M AB = MI X = *-y- - ' or ^ x ^ d - - ; s. n , = wc(l-d)x "(*-'* + 1 2 I for V wed (I x) c -y-A for d + ^xsl 29. Relation between shear and moment diagrams. Consider a beam bearing any number of concentrated loads ^, ! PI P%, , jfJJ, at distances d^ c? 2 , , d n from one end A (Fig. 44). Then the mo- _ ^^ , . , ^ ment M at any section ?rm, distant x from the origin * 4 is F '- 44 42 RESISTANCE OF MATERIALS where the summation includes only the loads on the left of the section. For an adjacent section distant A# from mn, that is, at a distance x -f Az from the origin, the moment is Let &M denote the difference between these two moments. Then AJf = M' - M = Rx - T PAz, -- But, by definition, the shear S at the given section mn is S = S 1 -^ / P.- Consequently, This relation also holds for a beam uniformly loaded. Thus, if w denotes the uniform load per foot of length, and I is the span in feet, the moment in this case at any section distant x from the left support is wl and at a section distant Ax from this it is w x 2 Therefore the change in the moment is now = M' - M=kx-wx . Az If Ao: is assumed to be small, its square may be neglected in com- parison with the other terms. In this case, dropping the last term, we have AJf wl Evidently the same relation holds for any combination of uniform and concentrated loads. The general fundamental relation between the shear and moment diagrams is therefore BENDING-MOMENT AND SHEAB, DIAGRAMS 43 Since - represents the rate at which the ordinate to the moment A# diagram is changing, this relation may be expressed in words by saying that The rate of change of the moment is equal to the shear. From this result important properties of the two diagrams may be deduced, as explained in the next paragraph. 30. Properties of shear and moment diagrams. Consider the highest point of any given moment diagram for instance, of those shown in Figs. 39-43. Since the moment increases up to this point and decreases after it passes it, the change in the moment AJf, corresponding to an increase A# in the abscissa, must be positive on one side of the point and negative on the other. Since Az is positive in both cases, the ratio changes sign in passing the point. But since - = S, this means that the shear changes from positive to negative in pass- ing the given point, and therefore must pass through zero at the point in question. The same reasoning evidently holds for the lowest point of the moment diagram. Therefore, at the section where the moment is greatest or least the shear is either zero or passes through zero in passing the point. By referring to the diagrams in the preceding article or in Table XIII it will be observed that this is true in each case. If the moment is constant, then Alf= and consequently S = 0. That is to say, where the moment is constant the shear is zero. For a system of concentrated loads the equations for moment and shear, as shown in article 29, are The first of these represents an inclined straight line, and the second a horizontal straight line. Therefore, for concentrated loads the moment diagram is a broken line and the shear diagram is a series of horizontal lines or steps. 44 RESISTANCE OF MATERIALS For a uniform load the expressions for moment and shear, as shown in article 29, are wl wx 2 wl S = ~-wx. The first of these equations evidently represents a parabola, and the second an inclined straight line of slope = w. From these results it follows that for any combination of uniform and concentrated loads the moment diagram is a connected series of parabolic arcs, and the shear diagram is a succession of inclined lines or sloping steps. Since S&x represents an elementary vertical strip of the shear diagram, the area subtended by the shear diagram between any two given points is V$A:r. Making use of the relation AjM"=A#, and summing between two points ^ and P^ we have where M^ and M 2 denote the moments at the two points in question. Hence the difference between the moments at any two given points is equal to the area of the shear diagram between these points. At the ends of a simple beam the moment is always zero. There- fore, by the theorem just proved, for a simple beam the area of the shear diagram from one end to any point is equal to the moment at this point. 31. General directions for sketching diagrams. To economize time and effort it is important to follow a definite program in drawing the diagrams and determining the expressions for shear and moment. The following outline of procedure for either cantilever beams or simple beams resting on two supports is therefore suggested. 1. Find each reaction by summing the moments of all the ex- ternal forces about a point on the opposite reaction as moment center. Check this calculation by noting that the sum of the reac- tions must equal the sum of the loads. 2. Note that the expressions for moment and shear both change whenever a concentrated load is passed. Consequently, there will BENDING-MOMENT AND SHEAR DIAGRAMS 45 inear- foo :ale: Idiv 2100 II A fc> Scae 200 Shea lagram Dlajr Ibs. FIG. 45 be as many different segments of the moment and shear diagrams as there are segments of the beam between concentrations. 3. For a simple beam, take the origin at the left end of the beam. For a cantilever beam, take the origin at the free, or unsupported, end of the beam. Keep the origin at this point throughout the calculations. 46 RESISTANCE OF MATERIALS 4. Take a section between the origin and the first concentration, let x denote the distance of this section from the origin, and find the gen- eral expressions for the moment and shear at this section in terms of x. 5. Proceed in the same way for a section between each pair of consecutive concentrations. 6. Plot these equations, checking the work by means of the general relations stated in the preceding article. 7. Plot the shear diagram first. In plotting this diagram it is convenient to follow the direction in which the forces act. Thus, in Fig. 45 the shear at the left end is equal to the reaction and may be laid off in the same direction, that is, upwards. Proceeding to the right, drop the shear diagram by an amount equal to each load as it is met, until the reaction at the right end is reached, which will bring the shear diagram back to the base line. By following this method the shear diagram will always begin and end on the base line, which serves as a check on the work. 8. Note that as long as the shear diagram lies above the base line the shear is positive and therefore A M is also positive ; that is to say, the moment is increasing. Where the shear diagram crosses the axis, the moment diagram must attain its highest or lowest point. When the shear diagram lies below the base line, the moment is decreasing. 9. Compute numerical values of the moment and shear at the critical points of the diagrams, and indicate these numerical values on the diagrams. A sample set of diagrams as they should be drawn by the student is shown in Fig. 45. APPLICATIONS 76. A beam 16 ft. long is supported at the left end and at a point 4 ft. from the right end, and carries a uniform load of 200 Ib./ft. over its entire length and a concentrated load of 1 ton at a point 4 ft. from the left end. Sketch the shear and moment diagrams and note the maximum shear and maximum moment. Solution. On cross-section paper indicate the loading as shown in Fig. 45. To find either reaction, take moments about the other point of support. Thus, for the left reaction R l we have R l 12 - 3200 . 4 + 2000 -8 = 0, whence R^ = 2400 Ib. Similarly, for R 2 , E 2 . 12 - 3200 . 8 - 2000 -4 = 0, whence R 2 = 2800 Ib. As a check on the correctness of these results, sum of loads is 3200 + 2000 = 5200, and sum of reactions is 2400 + 2800 = 5200. BENDING-MOMENT AND SHEAR DIAGRAMS 47 To obtain the shear diagram, start at the left end and lay off the reaction of 2400 Ib. upward. Since the load is 200 lb./ft., at 4 ft. from the left end the shear will be 2400 4 x 200 = 1600 Ib. As we pass this point the concentrated load of 1 ton will cause the shear to drop to 1600 2000 = 400 Ib. The shear then continues to drop 200 lb./ft., until at the right support it becomes 400^ 8 x 200 = 2000 Ib. As this point is passed, the reaction, which is equivalent to a concentrated load of 2800 Ib. upward, causes the shear to change suddenly to 2000 + 2800 = 800 Ib. It then gradually drops again and becomes zero at the end of the beam. On account of the uniform load the moment diagram will be segments of para- bolas. To plot these parabolas the values of the moment at a number of points along the beam may be calculated. Thus, at points 2, 4, 10, 12, and 14 ft. from the left end the moments are M 2 = 2400 2 - 400 : I = 4400 f t.-lb. ^ M 4 = 2400 . 4 - 800 2 = 8000 f t.-lb. M 1Q = 2400 . 10 - 2000 5 - 2000- 6 = 2000 f t.-lb. M 12 = 2400 12 - 2400 , 6 - 2000 - 8 = - 1600 f t.-lb. M u = 2400 14 - 2800 . 7 - 2000 . 10 = - 400 ft.-lb. The maximum moment is evidently at the 1-ton load, and the maximum shear at the left support. 77. A simple beam 10 ft. long is supported at the ends and carries a load of 800 Ib. at a point 4 ft. from the left end. Draw the shear and moment diagrams.** 78. A simple beam 20 ft. long, supported at the ends, carries a uniform load of 50 lb./ft. and a 'concentrated load of 600 Ib. at 5 ft. from the right end. Draw the shear and moment diagrams. 79. A simple beam of 15 ft. span is supported at the ends and carries a uniform load of 100 lb./ft. and concentrated loads of 500 Ib. at 4 ft. from the left end and 1000 Ib. at 8 ft. from the left end. Plot the shear and moment diagrams. 80. A simple beam of 16ft. span carries concentrated loads of 200 lb.,.400 Ib., and 100 Ib. at distances of 4 ft., 8 ft., and 12 ft., respectively, from the left support. Neglecting the weight of the beam itself, sketch the shear and moment diagrams. 81. A simple beam of 9 ft. span carries a total uniform load of 400 Ib. over the middle third of the span. Neglecting the weight of the beam, draw the shear and moment diagrams for this loading. 82. The total load on a car axle is 8 tons, equally divided between the two .wheels. .Distance between centers of wheels is 4^ ft., and distance between centers o| journals is 5^ ft. Draw the shear and moment diagrams for the axle so loaded. A"83. Draw the shear and moment diagrams for a simple beam 10 ft. long, bear- ing ale*al uniform load of 100 lb./ft. and concentrated loads of 1 ton at 4 ft. from the left end and 2 tons at 3 ft. from the right end. 84. A [beam 12 ft. long is supported at the ends and carries loads of 4000 Ib. and 1000 lb! at 2 ft. and 4 ft., respectively, from the left end. No uniform load. Sketch the shear and moment diagrams. 85. A beam 20 ft. long, supported at the ends, bears a uniform load of 100 lb./ft. extending from the left end to the center, and a concentrated load of 1000 Ib. at 5 ft. from the right end. Plot the shear and moment diagrams. 86. A beam 16 ft. long, supported at the ends, carries a uniform load of 200 lb./ft. extending 10 ft. from the left end, and concentrated loads of 1 ton and \ ton at 8 ft. and 12 ft., respectively, from the left end, Draw the shear and moment diagrams. 48 RESISTANCE OF MATERIALS 87. A simple beam of 8 ft. span bears a distributed load which varies linearly from zero at one end to a maximum at the other. The total load on the beam is 1200 Ib. Plot the shear and moment diagrams. 88. A cantilever beam extends 9 ft. from a wall and bears a uniform load of 60 Ib./ft. and a concentrated load of 175 Ib. at the free end. Draw the shear and moment diagrams. 89. A cantilever beam projects 6 ft. and supports a uniform load of 100 Ib./f t. and concentrated loads of 90 Ib. and 120 Ib. at points 2ft. and 4ft., respectively, from the free end. Draw the shear and moment diagrams. 90. A cantilever beam projects 10 ft. and carries a concentrated load of 100 Ib. at the free end and also concentrated loads of 90 Ib. and 60 Ib. at 3ft. and 5ft., respectively, from the free end. Sketch the shear and moment diagrams. 91. A cantilever beam projects 6 ft. from its support and bears a concentrated load of 50 Ib. upward at the free end and 50 Ib. downward at 2 ft. from the free end. Draw the shear and moment diagrams, neglecting the weight of the beam. 92. A cantilever beam projects 8 ft. from its support and bears a distributed load which varies linearly from zero at the free end to a maximum at the fixed end. The total load is ^ ton. Draw the shear and moment diagrams. 93. Sketch the shear and moment diagrams for a cantilever 12 ft. long, carrying a total uniform load of 50 Ib./ft. and concentrated loads of 200 Ib., 150 Ib., and 400 Ib. at distances of 2 ft., 4 ft., and 7 ft., respectively, from the fixed end. 94. An overhanging beam of length 30 ft. carries concentrated loads of 1 ton at the left end, 1.5 tons at the center, and 2 tons at the right end, and rests on two supports, one 4 ft. from the left end and the other 6 ft. from the right end. Draw the shear and moment diagrams. 95. An overhanging beam 20 ft. in length bears a uniform load of 100 Ib./ft. and rests on two supports 10 ft. apart and 5 ft. from the ends of the beam. Sketch the shear and moment diagrams. 96. An overhanging beam 25 ft. in length carries a uniform load of 200 Ib./ft. over its entire length, and rests on two supports, one at the right end and the other at 10 ft. from the left end. Plot the shear and moment diagrams. 97. An overhanging beam 40 ft. in length is supported at points 4 ft. from the left end and 8 ft. from the right end. It carries concentrated loads of 4 tons at the left end, 3 tons at 6 ft. from the left end, 2 tons at 14 ft. from the left end, and 1 ton at the right end. Draw the shear and moment diagrams. 98. Draw the shear and moment diagrams for an overhanging beam 18 ft. in length, supported at points 4 ft. from each end, and carrying a uniform load of 50 Ib./ft. over its entire length and a concentrated load of 800 Ib. at the middle. 99. Draw the shear and moment diagrams for an overhanging beam 20 ft. in length, supported at points 3 ft. from the left end and 5 ft. from the right end, which carries a uniform load of 80 Ib./ft. between the supports and concentrated loads of 600 Ib. at each end. 100. Draw the shear and moment diagrams for an overhanging beam 16 ft. in length, supported at points 2 ft. from the left end and 4 ft. from the right end, which carries a load of 200 Ib./ft. distributed uniformly over 12 ft. from the left end, and a concentrated load of 1600 Ib. at the right end. SECTION IV STRENGTH OF BEAMS 32. Nature of bending stress. For a horizontal beam carrying a set of vertical loads the method just explained for drawing the moment and shear diagrams is to combine the forces on one side of any cross section into a single force, arid the moments of these forces about the centroid of the section into a single moment. For equilibrium the stresses in the beam at the given section must therefore also reduce to a single force and moment, called the shear and bend- ing moment, respectively, equal in amount and op- posite in direction to the external resultant force and moment. By considering a few simple cases the nature of the shearing and bend- ing stresses will be ap- parent. Thus, in Fig. 46, suppose that a small ver- tical slice is cut out of the beam, as shown ; then there will evidently be a tendency for the top of the cut to close up and for the lower side to spread apart. This might be prevented by placing a small block in the upper edge of the cut and connect- ing the lower edges with a link. Supposing this to be done, there will, in general, still be a tendency for the part on one side of the cut to slide up or down past the part on the other side. To pre- vent this vertical motion, it would be necessary to introduce a vertical support, as shown in the lower diagram of Fig. 46. 49 ^ Compression O O Tension \J Vertical Shear FIG. 46 50 RESISTANCE OF MATERIALS From this illustration it is evident that the resisting stress in a beam required to equilibrate any system of external forces is of two kinds : 1. A compressive stress on one side, normal (that is, perpen- dicular) to the plane of the cross section. A tensile stress on the opposite side, also normal to the plane of the cross section. 2. A vertical shearing stress in the plane of the cross section. 33. Distribution of stress. The effect of the external bending moment on a beam originally straight is to cause its axis to become ^ bent into a curve, called the elastic curve. fi j \ Considering the beam to be composed of / \/ \ single fibers parallel to its axis, it is found / /\ \ by experiment that when a beam is bent, the fibers on one side are lengthened and those on the other side are shortened. Between these there must evidently be a layer of fibers which are neither lengthened nor shortened, but retain their original length. The line in which this unstrained layer of fibers inter- sects any cross section is called the neutral axis (Fig. 47). It is also found by experiment that a cross section of the beam which was plane before flexure (bending) is plane after flexure. This is known as Bernoulli's assumption.* As a consequence of Bernoulli's assumption it is evident from Fig. 47 that the lengthening or shortening of any longitudinal fiber is pro- portional to its distance from the neutral axis. But by Hooke's law the stress is proportional to the deformation produced. There- fore the normal stress at any point in the cross section is likewise proportional to the distance of this point from the neutral axis. If, then, the normal stresses are plotted for every point of any vertical strip MN (Fig. 48), their ends will all lie in a straight line. This distribution of stress is therefore called the straight-line law. * St. Venant has shown that Bernoulli's assumption is rigorously true only for certain forms of cross section. If the bending is slight, however, as is the case in all structural work, no appreciable error is introduced by assuming it to be true whatever the form of cross section. STRENGTH OF BEAMS 51 Since the normal, or bending, stresses are the only horizontal forces acting on the portion of the beam considered, in order to satisfy the condition of equilibrium horizontal forces = we must have Resultant tensile stress = resultant compressive stress. Therefore, since the tensile and compressive stresses act in opposite directions (that is, are of opposite sign), the algebraic sum of all the normal stresses acting on the section must be zero. Thus, if AA denotes an element of area of the section, and p the intensity of the normal stress acting on it, the total stress on this area is jt?A^4, and consequently = 0. FIG. 48 Now, if the normal stress at a variable distance y from the neutral axis is denoted by JK>, and that at some fixed distance y' is denoted by p', then, from the straight-line law, , = ,-> or p' y' Inserting this value of p in the above condition of equilibrium, it becomes , Therefore, since p' and y 1 are definite quantities different from zero, we have VyA^4 = 0. But, from article 13, the distance of the centroid from the neutral axis is given by and if ^ytA = 0, then also / = 0. Therefore the neutral axis passes through the centroid of the cross section ; that is, the neu- tral axis coincides with the horizontal centroidal axis. 34. Fundamental formula for beams. For equilibrium the result- ant moment of the normal stresses acting on any cross section must be equal to the resultant moment of the external forces on one side of the section, taken with respect to the neutral axis of the 52 RESISTANCE. OF MATERIALS section. Now, if A^4 denotes an element of area of the cross section, and p f the intensity of the normal stress acting on it, the total stress on this area is p'&A. If, then, y is the distance of this stress, or internal force, from the neutral axis of the section, and M denotes the resultant moment of the external forces about this axis, for equilibrium Now let p denote the stress on the extreme fiber and e the distance of this fiber from the neutral axis. Then, by the straight-line law, .'=, y e ' and, inserting this value of p' in the above equation, it becomes The quantity V?/ 2 A^4, however, is the moment of inertia, /, of the cross section (article 20). Therefore (33) M = e The right member of this equation, ^ , is the resultant internal e stress couple, and is called the moment of resistance of the beam. Since e denotes the distance of the extreme fiber of the beam from the neutral axis, the ratio - is also a function of the shape and size of the cross section, and is therefore called the section modulus. Let this section modulus be denoted by Z. Then Z == - , and the fundamental formula becomes (34) M = pZ. Since this is an equality between the resultant external moment M and the product of the working stress p by the section modulus Z, it expresses the fact that the strength of a beam depends jointly on the shape and size of the cross section and the allowable stress for the material. 35. Calculation and design of beams. For a beam of given size and loading the maximum external moment M, acting at any point along the beam, is first determined by the methods explained in STRENGTH OF BEAMS 53 Section III. The section modulus Z is then calculated from the given dimensions. For ordinary rolled shapes of structural steel the section moduli are given in Tables III- VII. The stress in the extreme fiber (or skin stress, as it is called) is then found by substi- tuting these numerical values of M and Z in the equation . M P = ~Z' By comparing this calculated value of p with the allowable unit stress for the material, it is determined whether or not the beam is safe. For a beam of given size and shape the maximum external moment it can carry safely is found by calculating its moment of resistance. Thus, if p denotes the allowable, or working, stress for the material in lb./in. 2 , and the section modulus Z is calculated from the given dimensions of the cross section, the maximum external moment M which this beam can carry with safety is found by inserting these numerical values in the equation M=pZ. In designing a beam to carry a given loading, the maximum external moment M due to this loading is first calculated. Then, for any specified unit working stress p, the required section modulus is found from the relation ,, " P' This section modulus Z may then be looked up in Tables III- VI, thus determining the exact dimensions of the beam. APPLICATIONS 101. Find the safe moment of resistance for an oak beam 8 in. deep and 4 in. wide. Solution. In this case 1= 170.7 in. 4 and e = 4 in. Therefore the section modu- xus is From Table I the safe stress for timber may be assumed as p 1000 lb./in, Consequently, the moment of resistance for this beam is 54 RESISTANCE OF MATERIALS 102. In an inclined railway the angle of inclination with the horizontal is 30. The stringers are 10 ft. 6 in. apart, inside measurement, and the rails are placed 1 ft. inside the stringers. The ties are 8 in. deep and 6 in. wide, and the maximum load transmitted by each rail to one tie is 10 tons. Calculate the maximum normal stress in the tie. Solution. The bending moment is the same for all points of the tie be- tween the rails, and is 20,000 ft.-lb. The components of the moment with respect to the axes of the section FIG. 49 (Fig. 49) are M z = 240,000 and My = 240,000 (^) in.-lb., and the section moduli with respect to these axes are Z z = 64 in. 3 and Z y = 48 in. 3 Therefore the maximum normal stress is Pmax = 240,000 --) 240,000 - (54 48 = 57441b./in. 2 103. A rectangular cantilever projects a distance I from a brick wall and bears a single concentrated load P at its end. How far must the inner end of the cantilever be em- bedded in the wall in order that the pressure between this end and the wall shall not exceed the crushing strength of the brick? Solution. Let b denote the width of the beam and x the distance it extends into the wall. For equilibrium the re- action between the beam and the wall must con- sist of a vertical force and a moment. If p a denotes the intensity of the vertical stress, and it is assumed to be uniformly distributed over the area 6x, p then p a hx = P ; whence p a = (see Fig. 50, a). Similarly, let p b denote the maximum intensity of the stress forming the stress couple. Then, taking moments about the center C of the portion AB, since the stress forming the couple is also distributed over the area &x, we have 12 and STRENGTH OF BEAMS Therefore, substituting in the formula p , we have 55 bx 2 12 6P Consequently, whence and = Pb (-1) ,. ta 2 fee ' 2P/ 3z\ = (2+ , bx \ x/ 2P As a numerical example, let I = 5ft., P = 200 lb., 6 = 4 in., and p = 600 lb./in. 2 (for ordinary brickwork). Then, solving the above equation by the formula for quadratics, X = 2P bp whence, by substituting the above numerical values, x = 5.6 in. 104. Find the required dimen- sions for the arms of a cast-iron pulley of external diameter D for a tension in the belt of T^ on the tight side and T 2 on the slack side. Arms assumed to be elliptical in cross section, of dimensions h = 2b (Fig. 51). Solution. According to Bach the load may be assumed to be carried by one third of the spokes, and the working stress taken as 4500 lb./in.' 2 Let n denote the number of spokes. Then the maximum moment on one spoke is approximately FIG. 51 The moment of inertia of an ellipse about its minor axis is , and its section h 2 M modulus is Z = Therefore, substituting these values in the formula p = , , 16 ^ we have whence h = 56 RESISTANCE OF MATERIALS 105. Derive a formula for the pitch of a cast-iron gear to carry safely a driving force F. Solution. Circular pitch is defined as the distance between corresponding points on two successive teeth, measured along the pitch circle. Let P denote the circular pitch for the case in question (Fig. 52). Then, if h denotes the depth of the tooth, b its breadth, and t its thickness at the root, the relative proportions FlG ' 52 ordinarily used are h=.7P, = .5P BC = .47 P, i 6 = 2Pto3P, AB = .53 P, Height above pitch circle (called addendum) = .3 P, Depth within pitch circle = .4 P. The driving force -F is ordinarily applied tangent to the pitch circle. Assume, how- ever, that by reason of the gear being worn, or from some other cause, it reaches the tip of the tooth, as shown in the figure. Then, considering the tooth as a cantilever beam, the maximum moment is M=Fh, and its section modulus at the root is 6 Therefore, assuming a working stress for cast iron of p = 4500 lb./in. 2 , we have 4500 " = Fh, i< 8 'L J . L ! . i L. and, inserting the values I . . . . I 1__ this becomes k" 106. Find the moment of resist- > J' ^ ^ 12" ance for the section given in problem 53, assuming the working stress for i structural steel as 16,000 lb./in. 2 107. Find the section modulus and moment of resistance for the section given in problem 55. p 108. Find the section modulus | I t and moment of resistance for the FIG. 53 section given in problem 56. 109. Find the moment of resistance of a circular cast iron beam 6 in. in diameter. 110. Find the moment of resistance of a 24-in. steel I-beam weighing 80 Ib./ft. 111. Compare the moments of resistance of a rectangular beam 8 in. x 14 in. in cross section, when placed on edge and when placed on its side. STRENGTH OF BEAMS 57 112. Find the section moduli for the sections given in problems 58, 60, and 62. 113. Design a steel I-beam, 10 ft. long, to bear a total uniform load of 1500 Ib./f t. including its own weight. 114. A built beam is to be composed of two steel channels placed -on edge and connected by latticing. What must be the size of the channels if the beam is to be 18 ft. long and bear a load of 10 tons at its center, for a working stress of 16,000 lb./in. 2 ? 115. Compare the strength of a pile of 10 boards, each 14 ft. long, 1 ft. wide, and 1 in. thick, when the boards are piled horizontally and when they are placed close together on edge. 116. Design a rectangular wooden cantilever to , project 4 ft. from a wall and bear a load of 500 Ib. ( at its end, the factor of safety being 8. 117. A wooden girder supporting the bearing par- titions in a dwelling is made up of four 2-in. by 10-in. joists set on edge and spiked together. Find the size of a steel I-beam of equal strength. 118. A factory floor is assumed to carry a load of 200 Ib./f t. 2 and is supported by steel I-beams of 16-f t. span and spaced 4 ft. apart on centers. What size I-beam is required for a working stress of 16,000 lb./in. 2 ? 1 1 T ra G> S.I.X i 1 }. 55 58 RESISTANCE OF MATERIALS 119. Find the required size of a square wooden beam of 14-ft. span to carry an axial tension of 2 tons and a uniform load of 100 Ib./ft. 120. A floor designed to carry a uniform load of 200 Ib./ft. 2 is supported by 10-in. steel I-beams weighing 30 Ib./ft. How far apart may they be placed for a span of 16 ft. and a working stress of 16,000 lb./in. =3 T p D \ n ^ ' V \ ^ * \ [l -i h A "-! f \ \ 8g Shaft A -- ) 6= !* Y J\ J -W 1 \ i - 2Q--> 3-3.-. r > r // J u^ PLAN END ELEVATION FIG. 56 121. A floor is supported by wooden joists 2 in. x 12 in. in section and 16ft. span, spaced 16 in. apart on centers. Find the safe load per sq. ft. of floor area for a working stress of 800 lb./in. 2 122. A floor is required to support a uniform load of 150 Ib./ft. 2 and is supported by steel I-beams, 18 ft. span and spaced 5 ft. apart on centers. What size I-beam is required for a working stress of 16,000 lb./in. 2 ? 123. A structural-steel built beam is 20 ft. long and has the cross section shown in Fig. 53. Compute its mo- ment of resistance and find the safe uniform load it can carry per linear foot for a factor of safety of 5. 124. The cast-iron bracket shown in Fig. 54 has at the dangerous section the dimen- sions shown in the figure. Find the maximum concen- trated load it can carry with a factor of safety of 15. 125. Find the proper dimensions for a wrought-iron crank of dimensions shown in Fig. 55 for a crank thrust of 1500 Ib. and a factor of safety of 6. 126. A wrought-iron pipe 1 in. in external diameter and T T g- in. thick projects 6 ft. from a wall. Find the maximum load it can support at the outer end. FIG. 57 STRENGTH OF BEAMS 59 FIG. 58 127. The yoke of a hydraulic press used for forcing gears on shafts is of the form and dimensions shown in Fig. 56. The yoke is horizontal, with groove up, so that the shaft to be fitted lies in the groove, as shown in plan in the figure. The ram is 32 in. in diameter and under a water pressure of 250 lb./in. 2 Find the dangerous section of the yoke and the maximum stress at this section. 128. A 10-in. I-bar weighing 40 Ib./ft. is supported on two trestles 15 ft. apart. A chain block carrying a 1-ton load hangs at the center of the beam. Find the factor of safety. 129. The hydraulic punch shown in Fig. 57 is designed to punch a jj-in. hole in a f-in plate. The dimensions of the dangerous section AB are as given in the figure. Find the maximum stress at this section. 130. The load on a car axle is 8 tons, equally distributed between the two wheels (Fig. 58). The axle is of cast steel. Find its diameter for a factor of safety of 15. 131. The floor of an ordinary dwelling is assumed to carry a load of 50 Ib./ft. 2 and is supported by wooden joists 2 in. by 10 in. in section, spaced 16 in. apart on centers. Find the greatest allowable span for a factor of safety of 10. 132. An engine shaft of machinery steel rests in bearings 6 ft. apart between centers and carries a 12-ton flywheel midway between the bearings. Find the required size of shaft. 133. A cast-iron flange coupling is connected with ten wrought-iron bolts. Distance from axis of each bolt to axis of shaft is 6 in. Total torque (twisting moment) transmitted is 12,000 ft.-lb. If the flanges are accidentally separated 2 in. and the bolts are a drive fit, find the bending stress produced in each bolt. 134. In the carriage clamp shown in Fig. 59 the screw is of wrought iron, | in. diameter, square thread, 5 threads per inch, and the casting has the dimensions given in the figure. Find what load on the screw will cause fail- ure by shearing the threads, and find the maximum stress in the casting under this load, due to combined bending and tension. 135. In the joiner's clamp shown in Fig. 60 the bar is of carbon steel, 11 in. x Jin., tensile strength 70,000 lb./in. 2 , and the screw is steel, fin. diameter, square threads, 5 threads to the inch. Find the dimensions of the cast-iron handle so that it shall be light enough to act as the breaking piece, FIG. 59 FIG. 60 SECTION V DEFLECTION OF CANTILEVER AND SIMPLE BEAMS 36. General deflection formula. By a simple beam is meant one which is simply supported at the ends. The only external forces acting on it in addition to the loads are, then, the two vertical reac- tions at the supports. A cantilever is a beam which overhangs, or projects outward from the support, the loads on it being equili- brated by the moment at the support and by the vertical reaction at this point. The results of applying the general deflec- tion formula, derived below, to these two classes of beams will be made the basis of the treatment of continuous and restrained beams in the sec- tions which follow. Taking a vertical longi- tudinal section of a beam, the line in which this plane intersects the neutral-fiber surface is called the elastic curve. Any small segment, A#, of the elastic curve may be considered as a circular arc with center at some point (Fig. 61). This point is therefore called the center of curvature for the arc Az. The radius of curvature is not constant, but changes from point to point along the beam. Evidently the radius of curvature is least where the beam is curved most sharply. Any two adjacent plane sections, AB and DH (Fig. 61), origi- nally parallel, intersect after flexure in the center of curvature 0. Let KC = Az denote the original length of the fibers, and draw DEFLECTION OF CANTILEVER AND SIMPLE BEAMS 61 through C a line EF parallel to AB. Then DF denotes the short- ening of the extreme fiber on one side, and EH the lengthening of the extreme fiber on the other. Now, since the triangles KOC and ECH are similar, we have the proportion EH _ CH e ~KC~~OK~~r 77*77" Moreover, the left member, , is the change in length of the KG extreme fiber divided by its original length, which is by definition the unit deformation s of this fiber. Also, by Hooke's law, = E, where jo denotes the unit nor- mal stress on this fiber. Hence the above pro- portion becomes p e M Me or, since p = = , as shown in the pre- ceding section, Me e whence < 35 > F.o.62 Now let AB denote any segment of the elastic curve, and AA f , BB' the tangents at A and B respectively (Fig. 62). If AB is divided up into small segments A#, and A< denotes the angle which each subtends at the center of curvature 0, as shown in Fig. 62, then Az = 7*A<, or A</> = , and, inserting in this the value of r obtained above, it becomes El Hence, by summation, the total angular deflection </> is 62 RESISTANCE OF MATERIALS Now for any small arc, A#, the deflection Ac? at any point at a dis- tance #, measured from the tangent to the arc at the initial point (Fig. 62), is AcZ = Hence the total deflection for any finite portion of the arc AB, measured from one end A to the tangent at the other end B, is But M&x denotes the area of a small vertical strip of the moment diagram of altitude M and base Arc, and V(JfA#)# is the sum of the static moments of all these elements of area with respect to the point A. From the results of Section II, however, this is equal to the area of the moment diagram between A and B multiplied by the distance of its centroid from A. That is, if A ab denotes the area of the moment diagram between the points A and B, and X Q is the distance of the centroid of A ab from A, then Therefore d = -_ A ab . x, Jbl or, in general, (36) d = - - (static moment of the moment diagram). El The angular deflection $ between any two points A and B, that is, the angle between the tangents to the elastic curve at these two points, is given by Therefore, since ^ Mtx denotes the area A^ of the moment diagram between the two points in question, and since for the small deflec- tions which actually occur in practice we may assume <f> = tan cf> without introducing any appreciable error, (37) El DEFLECTION OF CANTILEVER AND SIMPLE BEAMS 63 37. Cantilever bearing concentrated load. For a cantilever bear- ing a concentrated load at the end, the moment diagram is a triangle, as shown in Fig. 63. The area of the moment dia- gram is therefore PI 2 A = , and the dis- tance of its centroid from the free end 2 is X Q = - L Therefore o the deflection at the free end is FIG. 63 (38) d = E I PI* - 3 El The deflection d may also be expressed in terms of the stress on the extreme fiber. Thus, since %-=M=Pl, by substituting this value of PI in the expression for c?, it becomes (39) 3Ee Also, the angular deflection at the load is found to be (40) tan^ = _ M= FIG. 64 A El _ PI* ~ 2 El If the load is at a distance a from the fixed end and b from the free end (Fig. 64), then the deflection at the load, as shown above, is Pa s (41) d = 64 RESISTANCE OF MATERIALS and similarly, from equation (40), the angular deflection at the load is fa 2 ~ 2 El' (42) Consequently, the additional deflection d' from the load to the free end of the cantilever is (43) d f = b tan <j> = 2 El The total deflection D at the free end is therefore It is often convenient to let b = M, where k denotes a proper fraction. Then in the present case a= I b = l(~L &), and the expression for the deflection at the end becomes p/3 (45) For instance, if the load is at the middle of the cantilever, then k = i, and the deflection at the free end becomes 38. Cantilever bear- ing uniform load. For a uniformly loaded cantilever the moment diagram is a parabola and the moment at the support is M=wl>- ^- (Fig. 65). Also, from article 17, -f 72 73 the area of the moment diagram is A = - - I = , and the dis- o 2 Q " tance of its centroid from the free end is X Q = -I. Therefore the deflection at the free end is (47) DEFLECTION OF CANTILEVER AND SIMPLE BEAMS 65 From the relation * = M = the expression for the deflection in terms of the maximum fiber stress p is found to be C") d = fWe Also, the total angular deflection at the end of the beam is (49) A wl 3 tan = = El GEI 39. Cantilever under constant moment. If a cantilever is sub- jected to a couple, that is, a pair of equal and opposite parallel forces, as shown in Fig. 66, the moment is constant for the entire length of the beam. The moment diagram is therefore a rectangle of area Ml, and the de- flection at the free end is (50) d = The angular deflection at the free end in this case s Ml (51) 40. Simple beam bearing concentrated load. To apply the deflec- tion formula to a simple beam, the deflection must be measured at one end A from a tangent at the middle C. For a concentrated load P at the middle (Fig. 67), the area of the moment diagram 7372 2 I I from A to C is A = -^-, and x. = -.- = -. Hence, in this case Ib O A O (52) Fl 66 RESISTANCE OF MATERIALS Also, the total angular deflection for half the beam is found to be (53) . A pi 2 tan 9 = = EI !<>/;/ The deflection may also be obtained by considering the moment diagram as representing the load on the beam, and then taking moments about the point at which the deflection is meas- ured, say the center C (Fig. 67). Since the total area of the moment diagram is 1 n j_PP is regarded as the load on the beam each reaction will be PI' 2 Then, taking Fi. 07 moments about the center, the result is ~ KJ i pp i 2 ~ Hi" ' ( PI* 48 El ' From the relation ^ = M= ', the deflection at the center may e 4 be expressed in terms- of the maximum liber stress p. Thus, PI . . pi replacing - by its equal - in the expression for <7, the result is (55) d = 16 16 FIG. 68 If the concentrated load P is not at the center but divides the span / into two unequal segments a and >, the reactions are , -, DEFLECTION OF CANTILEVER AND SIMPLE BEAMS 67 and the moment at the load is . Also, the area of the moment diagram between the load and one end is - - , and the distance .2 of the centroid of this segment from the end is :r = ^ a. Hence the o deflections of the ends from the tangent at the point of application C of the load are j,^ ,, . 7!5 ZEIl ZKli and the deflection of C below the level of the supports is I'uW pah (56) XEIl where p denotes the maximum fiber stress. 41. Simple beam bearing uniform load. For a simple beam uni- formly loaded the moment diagram is a parabola, the maximum wl* ordinate being - 8 From article 17, the area of this parabola is 2 wl -. wl A ='T' l =w To apply the general formula for deflec- tion, consider d as measured from one end A to the tan- gent at the center C (Fig. () ( .)). Then, since the area of one half the moment diagram is -, and the distance; of the centroid of f> / f> / this half from a vertical through A is X Q - - - - . tlie deflection is 8 ii KJ (57) El El To express the deflection in terms of the maximum fiber stress />, make use of the relation ^ = M= ^. Then, replacing - , in the e o o 68 RESISTANCE OF MATERIALS expression for d, by its equal , it becomes > The total angular deflection for half the beam in this case is A wl* (59} tan 9 = - = - . El 24 E I The deflection may also be obtained by regarding the moment diagram as representing the load on the beam. Since the total area wl s wl 8 of the moment diagram is , each reaction will then be - , and \2t 2< therefore, taking moments about the center to find the deflection at this point, the result is, as before, ,__!_ /wP L_wP 3 \ 5 wl* ~ ' ' 2 24 16 884JBT APPLICATIONS 136. A 15-in. .I-beam weighing 60 Ib./ft. carries a 25-ton load at the center of a 12-ft. span. Find the maximum deflection. 137. In building construction the maximum allowable deflection for plastered ceilings is ^^ of the span. A floor is supported on 2 in. x 10 in. wooden joists of 14-ft. span and spaced 16 in. apart on centers. Find the maximum load per square foot of floor surface, in order that the deflection may not exceed the amount specified. 138. Determine the proper spacing, center to center, for 12-in. steel I-beams weighing 351b./ft., for a span of 20ft. and a uniform floor load of 100 Ib./ft. 2 , in order that the deflection shall not exceed ^^ of the span. 139. A structural steel shaft 8 in. in diameter and 5 ft. long between centers of bearings carries a 25-ton flywheel midway between the bearings. Find the maxi- mum deflection of the shaft, considering it as a simple beam. 140. A wrought-iron bar 2 in. square is bent to a right angle 4ft. from one end. The other end is then embedded in a concrete block so that it stands upright with the 4ft. length horizontal. If the upright projects 12 ft. above the concrete, and a load of 300 Ib. is hung at the end of the horizontal arm, find the deflection at the end of this arm. 141. A wooden cantilever 2 in. x 10 in. in section, with the longer side vertical, projects 10 ft. from the face of a wall and carries a concentrated load of 600 Ib. at a point 6 ft. from the wall. Find the deflection at the free end of the beam. 142. A 10-in. steel I-beam weighing 40 Ib./ft. spans an opening 16 ft. wide and supports a total load of 40 tons. Find how much greater the maximum deflection of the beam is when this load is concentrated at its center than when it is distri- buted uniformly over the beam. DEFLECTION OF CANTILEVER AND SIMPLE BEAMS 69 143. A built beam is composed of two 10-in. steel channels, 40 lb./ft., placed on edge and connected with latticing. The span is 20 ft. Find what uniform load per linear foot the beam can carry under the condition that the maximum deflection shall not exceed ^ in. 144. A 15-in. steel I-beam, 42 lb./ft., spans an 18-ft. opening. Find the maxi- mum deflection for a maximum fiber stress of 16,000 lb./in. 2 145. The total load on a car axle is 10 tons, equally distributed between the two wheels. Distance between centers of wheels is 56 in., and between centers of bear- ings is 68 in. Find the maximum deflection of the axle measured from a horizontal line joining the centers of bearings. 146. A 10-in. steel I-beam weighing 30 lb./ft. rests on two supports 16 ft. apart and carries a uniform load of 200 lb./ft. in addition to its own weight. A third support just touches the beam at the center. How much must this central support be raised so that it shall carry all the weight, and the beam just touch the two end supports ? 147. A cast-iron pipe 20 in. internal diameter and 1 in. thick rests on supports 30 ft. apart. Find the maximum deflection when the pipe is full of water. 148. A beam of uniform section, carrying a concentrated load at the center, has a maximum deflection equal to 1 per cent of the span. Find the slope of the beam at its ends. 149. Three beams of the same material are laid side by side across an opening of 12-ft. span, and a load of 1000 Ib. rests across them at the center of the span so that they must all bend together. The beams are each 2 in. wide, but two of them are 6 in. deep, while the third is 12 in. deep. How much of the weight is carried by each beam ? 150. A steel bar 2 in. square rests on knife edges 5 ft. apart, and its maximum deflection under a central load of 1000 Ib. is found to be .1125 in. Calculate from this experiment the modulus of elasticity of the bar. SECTION VI CONTINUOUS BEAMS 42. Theorem of three moments for uniform loads. A continuous beam, or girder, is one which is supported at several points of its length. The reactions and moments in this case are statically indeterminate ; that is to say, the ordinary static conditions of equilibrium, ^f= 0, ^Tj/ = 0, are insufficient to determine them. To solve the problem it is necessary also to take into account the deflections of the beam. FIG. 70 The simplest method of finding the reactions and moments at the supports for a continuous beam is by applying what is known as the theorem of three moments. This theorem establishes a relation between the moments at three consecutive supports of a continuous beam and the loads on the two included spans, and was first published by Clapeyron in 1857. The following proof of the theorem, however, is very much simpler than any previously given, 70 CONTINUOUS BEAMS 71 For a continuous beam bearing a uniform load let A, B, C denote any three consecutive points of support, assumed to be in the same line, and let M^ M^ M 3 ; JK 1? R^ E 9 denote the moments and reac- tions at these three points respectively. Also let l^ Z 2 denote the lengths of the two spans considered, w^ w 2 the unit loads on them, and /S f f, /Sf the shears on the right and left of E^ respectively (Fig. 70), with a similar notation for the other points of support. Now consider a portion of the beam cut off by planes just inside the supports at A and (7, as shown in Fig. 71. Then, con- sidering the end B as fixed, the deflec- tion at A from the tangent at B consists of three parts : that due to the moment to the shear S? considered as a load, and to the uniform load Calling these deflections d^ c? 2 , t? 3 , respectively, we have (Eq. (50), Art. 39) (Eq. (38), Art. 37) CifijS \ l i SIS I (Eq. (47), Art. 38) Hence the total deflection D A of the point A measured from a tangent at the point B is (61) D A = - To eliminate the shear f, form a moment equation by taking moments about the point B. Then whence (62) 2 ' 72 RESISTANCE OF MATERIALS and, substituting this value of $f in the above expression for D A , it reduces to Similarly, by considering the span BC and calculating the deflec- tion D c of the point C measured from the same tangent at B, we obtain the equation D - Also, forming a moment equation with C as center of moments, we have j|f a = j|f 8 -$7 a + _|i., and, eliminating /Sf between these relations, the result is (65) D c= -^jL Now, since these deflections lie on opposite sides of the tangent at B, we have, from similar triangles, Therefore, substituting the expressions for D A and D c in this rela- tion, combining terms, and transposing, we obtain the relation (66) M& + 2 M 2 ft + Q + M^ = . In this relation, M^ M 2 , and M B are stress couples acting on the beam. The external moments at the supports are equal in amount but opposite in sign to the stress couples, or internal moments. There- fore, calling Jfj, Jf 2 , M s the external moments at the supports, the sign of the expression is changed ; that is (67) This is the required theorem of three moments for uniform loads. 43. Theorem of three moments for concentrated loads. Consider a continuous beam bearing a single concentrated load in each span. CONTINUOUS BEAMS 73 The distance of the load in any span from the adjoining support on the left will be denoted by M, where I is the length of the span and k is a proper fraction ; that is, kl is some fractional part of the span (Fig. 72). Thus, if the load is at the middle of the span, k = i ; if it is at the quarter point, k = ^, etc. Now consider a portion of the beam extending over three con- secutive supports A, J5, and (7, and let M^ M^ Jf 3 denote the moments, and R^ R^ R^ the reactions, at these supports. Then, to obtain the theorem of three moments, calculate the deflections of A and C measured from the tangent to the elastic curve at B. To calculate the deflection of A, suppose the beam to be cut by a plane just inside the sup- port at A, and call the shear on the section S*. Then, considering the end B as fixed, calculate the deflection of A by treating the part AB as a cantilever subjected to the moment M^ the shear $f regarded as a load, and the concentrated load J^. Call- ing these three partial deflections d^ d^ d^ respectively, we have (Eq. (50), Art. 39) (Eq. (38), Art. 37) , 3 El 2 El (Eq. (44), Art. 37) In the present notation the quantities a and b in the expression for d z are a = distance from fixed end = \ k^, b = distance from free end = k^. Substituting these values of a and , the equation for d 3 becomes T>73 /"OQ"\ fl 1 1 /^O O 7n i i"3"\ vy 8 ?77irv i"iiy* 74 RESISTANCE OF MATERIALS Therefore, by addition, the total deflection of the end A with respect to the tangent at B is Now, forming a moment equation for the portion AB, taking center of moments at B, we have whence S* = (Jf, - Jf 2 ) + ^(1 - &,), and eliminating /Sf between this equation and the expression for D A , the result is Similarly, to find the deflection at (7, measured from the tangent to the elastic curve at J5, treat the portion BC as a cantilever fixed at B and subjected to the moment M^ the shear >Sf considered as a load, and the concentrated load ^. Then, calling these partial deflections d l9 d 2 , d^ we have (Eq. (50), Art. 39) (Eq. (38), Art. 37) d >=- + (Eq- (44X Art. 37) l or, since in the present case a = kj, 2 , b = Z 2 (l & 2 ), the expression for d g becomes (71) Therefore the total deflection D c from the tangent at B is CONTINUOUS BEAMS 75 Now, forming a moment equation for the portion BC, taking center of moments at B, we have whence (73) Sf = 2 and, eliminating /Sj between this equation and the expression for Z> c , the result is (74) D c = - =- - p 4. _ (2 & 3 & 2 2 + & 2 3 ). 6 JT ZEISEI^ Since the deflections at ^4 and C lie on opposite sides of the tan- gent at B, we have, from similar triangles, Substituting in this relation the values of D A and D c just found, combining like terms, and transposing, we obtain the relation (75) 2- 8 In this relation Jlf 1? M z , M 3 are the stress couples acting on the beam. The external moments at the supports are equal in amount but opposite in sign to the stress couples. Therefore, calling M^ M^ M z the external moments at the supports, the sign of the expression is changed ; that is, (76) M& + 2 M&i + 1 2 ) + M Z 1 2 = - PJ*fa - kl) which is the required theorem of three moments for a single con- centrated load in each span. For a single concentrated load at the center of each span, each k = i. In this case the theorem becomes (77) . o If there are a number of concentrated loads in each span, an equation like (76) can be written for each load separately. By 76 RESISTANCE OF MATERIALS adding these equations the general theorem of three moments for any number of concentrated loads is found to be (78) MJ t + 2 M 2 (l, + 1 2 ) + M 3 1 2 = - ]?lVi a (*i - *') -]lV.'(2fc,- 3 *; + *). 44. Effect of unequal settlement of supports. In deriving the theorem of three moments the supports were assumed to be at a fixed elevation in the same line. If their relative elevation changes, owing to unequal settlement of the supports or to other causes, the effect in general is to increase the stress in the member. To take account of this effect in applying the theorem, suppose that the supports were originally in line, and denote the settlement of three consecu- tive supports A, B, C from their original level by h^ h 2 , h 3 , respec- tively. Then the difference in elevation between A and B is li l h^ and between B and C is h^ li^. Thus, if A settles more than B, \ ~ \ * s P os itive and the deflection at A is increased by this amount. If A settles less than B, Ji^ h 2 is negative and the deflection at A is decreased by this amount, etc. In general, then, equations (70) and (74) for the deflections at A and C become Substituting these values of D A and D c in the relation the result, after combining terms and changing the signs of M, M, is = - 2) p Ji (*i - *') - S r ^ CONTINUOUS BEAMS 77 This relation is therefore the most general form of the theorem of three moments for any number of concentrated loads, including the effect of unequal settlement of the supports, or other change in their relative elevation. APPLICATIONS 151. A continuous beam of four equal spans is uniformly loaded. Find the sending moments at the supports. Solution. The system of simultaneous equations to be solved in this case is M 1 = M 5 = 0, 7/ ,/2 ~ the solution of which gives 152. A continuous beam of n equal spans carries a uniform load of the same amount in each span. Check the moments at the supports given in the following table for values of n from 2 to 7. The tabular values here given are the numerical coefficients of wl 2 . MOMENTS AT SUPPORTS FOR EQUAL SPANS AND UNIFORM LOAD 8 t 1 ft t 2 l To t ! 1 JL 28 t 2 2 28 t 3 3 28 t t A t 2 i t 3 I 4 38" i 1 1 * I 2 t 3 9 ToT f 4 t 5 A t ft 1 2 ft 1 3 12 T42 1 4 I 5 ft 1 :, i 142 78 RESISTANCE OF MATERIALS SHEARS AT SUPPORTS FOR EQUAL SPANS AND UNIFORM LOAD 1 2 1 ^k 2 2 .oJjL 8 I 5J5 3)0 8 8 2 3 .o|* e|s 10 10 1 2 il _iLl 10 10 3 4 p|ll 17J15 28 28 1 2 13[l3 is|l7 ll|o 28 28 28 345 of 15 23J20 18J19 38 38 38 1 2 3 19J18 20J23 is[o 38 38 38 456 o|41 63J55 .49 1 51 104 104 104 123 53|53 51 J49 55| 63 41JO 104 104 104 104 4567 o|56 86|75 67 170 72 J71 142 142 142 142 71J72 70|67 7o|86 56J 142 142 142 142 153. Calculate the reactions of the supports in problem 151. Solution. The reaction at any support may be found by finding the shears close to the support on each side. The sum of these two shears is then equal to the re- action. Thus, in the present case, to find any given reaction, say # 2 , consider the portion of the beam between R l and R 2 , as shown in Fig. 70, and form the moment equation for this segment. Then and therefore, since M l = and M 2 = ^ 3 g wZ 2 , Sf = ^| wl. Similarly, the moment equation for the segment of the beam between R 2 and R 3 is whence, by substituting~3f 3 = ^ wP and M 2 = ^ wZ 2 , we have Consequently, E 2 = 8} + Sf = w l = wl. 28 28 This method applies when it is required to find one reaction only, independently of the others. If all the reactions are required, it is simpler to calculate them in succession, starting at one end, without reference to the shears. For instance, to find JR 1 , take a section through R 2 and consider the loads on the left of the section. Then the moment equation for this portion is whence CONTINUOUS BEAMS 79 To find JR 2 , take a section through R 3 and consider all the loads on the left of the section. Then the moment equation is and inserting the value of B l just obtained, it is found that By this method each reaction may be obtained in terms of those already found, without calculating the shears. 154. In problem 152 determine the shears and reactions at the supports and check the results with the values tabulated on page 78. The tabular values are the numerical coefficients of wL 155. An 18-in. steel I-beam, 601b./ft., is continuous over four supports, the lengths of the three spans, beginning at the left, being 25ft., 40ft., and 35ft., respectively. What uniform load per foot run would produce a maximum fiber stress in the beam of 16,000 lb./in. 2 ? SECTION VII RESTRAINED, OR BUILT-IN, BEAMS 45. Uniformly loaded beam fixed at both ends. By a restrained, or built-in, beam is meant one which is fixed in direction at certain points of its length, usually the ends as, for example, beams built into a wall or forming a part of monolithic concrete construction. The simplest form of restrained beam is a cantilever, which can be treated by ordinary methods, as explained in articles 37, 38, and 39. Consider first a uni- formly loaded beam fixed at both ends, as shown in Fig. 73. Let B, F denote the points of inflection of the elastic curve ; that is, the points at which the bending moment is zero. Then the central portion BF may be con- sidered as a simple beam of length 2 x bearing a total uniform load of FIG. 73 amount 2 wx, and each of the ends, AB and FE, as a cantilever uniformly loaded and carrying a concentrated load wx at the end, equal to one of the reactions for the portion BF. If, then, d denotes the deflection of the point F with respect to A or E, assumed to be at the same level, the value of d, computed from the segment AF, is, from (38) and (47), articles 37 and 38, (80) d = 1 80 BESTBAINED, OB BUILT-IN, BEAMS 81 and, computed from the segment FE, is w /i__ v (81) d= \?/ + - Equating these two values of the deflection d and solving for #, the result is 7 2V3 The length of the central portion BF is therefore 2 x = - , and the maximum moment, which occurs at the center (7, is iv I 2 (82) M c = Similarly, the negative moment at the support A or A is 2 / a and, since a; = 7= > this reduces to 2V3 (83) M A =-^L. The maximum deflection for the central portion BF, considered as a simple beam, is, from (57), article 41, V3/ 9 djrF " 384 AY "384 A and for one end, say AB, considered as a cantilever, is, from (38) and (47), articles 37 and 38, I * \* 4 -wl* . - -, - - 2 2V3/ _9 -"" 8 AY 3 AY 384 AY* Therefore, since the total deflection of the center C below the sup- ports at A or A is the sum of these two, we have 82 RESISTANCE OF MATERIALS 46. Beam fixed at both ends and bearing concentrated load at center. Following the method of the preceding article, let B and D denote the points of in- flection of the elastic curve, or positions of zero moment (Fig. 74). Then the equilibrium would not be disturbed if the beam was hinged or jointed at B and D, and it may therefore be con- sidered as a simple beam of length BD suspended from the ends of two can- tilevers AB and DE. Now consider the seg- ment AD and compute from (44), article 37, the the deflection of D below A. Then, deflection at D due to the load P is (87) Pa 2 /a El V3 2 where in the present case a = and b = x, and consequently (88) d p = But from (38), article 37, the load , acting upward at D, produces a deflection upward of amount P (89) Consequently the total deflection of D below A is RESTRAINED, OR BUILT-IN, BEAMS 83 Similarly, for the portion DE, the deflection of D below E is P (91) d DE = ^ Equating these two values of the deflection and solving for a;, the result is and consquently the length of the central portion BD is .-j- Therefore the moment at the center C and also at each end is of numerically the same amount, namely, ^- or (92) 3f = f. The maximum deflection for the central portion BD is the same as for a simple beam of span - , namely, (93) and for either end AB or D^ is the same as for a cantilever of I P length - carrying a load at the end, namely, 2 \4 (94) d AB = - 3 7?/ 384 El Therefore the total deflection of the center C below the level of the supports at A and D is the sum of these two, that is, (95) 84 RESISTANCE OF MATERIALS 47. Single eccentric load. For a beam fixed at both ends and bearing a single concentrated eccentric load the simplest method of computing the unknown reactions and moments at the sup- ports is as follows: Consider the beam as fixed at one end E only (Fig. 75) and carrying, in addition to the concen- trated load P, the shear R l at the left support and the restraining moment M l at this point. Then, from (50) and (51), article 39, the de- flection from the tangent at E due to the moment M I is MJ, From (38) and (40), article 37, that due to the shear R 1 is , Rf Rf . and from (42) and (44), article 37, that due to the load P is , Pll PIJ 2 2 d =- ^--_Ll, tancf>=- Since the total vertical deflection of the point A with respect to the point E is zero, and since the total angular deflection is also zero, these two conditions furnish the equations (96) Mf Rf 2 El + 3 El PIJ? 2 El R,l 2 Pll 2 El 2 El = 0. From the second equation, PI* RESTRAINED, OR BUILT-IN, BEAMS 85 and, inserting this value in the first equation, we have (97) 1 = J P^1 + 1). Also, inserting this value of R 1 in the expression for the moment, it is found that / QQ\ J^A _ J> 1 2 Similarly, by forming the expressions for the total deflection of the point E with respect to the tangent at A, we obtain the equations R 7 s P/ 3 P/ 2 7 _^_ M 2 i ^ ri^ i 2 _ n 3 El 3 El 2 El (99) _MJ, Rf_ PI? El 2 El 2 El and, solving these equations simultaneously for M 2 and R^ as above, the results are (wo) If AC is the longer of the two segments, the maximum deflection will occur somewhere between A and C. Also, since the tangent at this point must be horizontal, the total angular deflection from one end, say A, to the point of maximum deflection is zero. Let x denote the distance of this point of maximum deflection from A. Then, computing the total angular deflection up to this point and equating it to zero, we have _ _ ' EI + 2EI~ 2Jf, whence x = s i Inserting the values just obtained for M l and 7^, this becomes (102) 86 RESISTANCE OF MATERIALS The maximum deflection D is then found to be EI and, inserting in this expression the values of obtained, it reduces to and #, just (103) 48. Uniformly loaded beam fixed at one end. In this case (Fig. 76) the deflection of the end A with respect to the fixed end B consists of two parts: that due to the reaction R is Rl* and that dne to the total uniform load wl is SJSI Since the ends A and B are assumed to be at the same level, the total deflection of A from the tangent at B must be zero ; that is, FIG. 76 3 El 8 El whence the reaction at the unrestrained end is (104) O The reaction R r at the restrained end B is therefore 5 tvl (105) = ivl R = 8 The maximum moment, which in this case occurs at the fixed end B, is then (106) M=Kl- = _ RESTRAINED, OR BUILT-IN, BEAMS 87 At the point where the maximum deflection occurs the tangent is horizontal. Let the distance of this point from the fixed end B be denoted by x. Then, from (40), (49), and (51), articles 37, 38, and 39, the condition that the total angular deflection for this length x shall be zero is MX R'x* wx 8 _ n ~~EI + YEI~^EI = Inserting in this expression the values of M and R 1 obtained above, and solving for a?, the result is (107) = . (15 - >/33) = .578 I. The maximum deflection is then found by finding the total deflection for the length x with respect to the tangent at B. Hence, from (38), (47), and (50), articles 37, 38, and 39, D =- 2 El R'x* wx or, inserting the values of M and A*', (108) wx," 48JEI (3 1 2 - lOto + The numerical value of the deflection is most easily found by first calculating the numerical value of x and then substituting in this formula. 49. Beam fixed at one end and bearing concentrated load at center. The deflection of the end A (Fig. 77) with respect to the fixed end B in this case consists of two parts (38), article 37, that due to the reaction R is , Rl* Moment diagram FIG. 77 from 88 RESISTANCE OF MATERIALS and from (44), article 37, that due to the load P is 22 3 El Since A and B are assumed to be at the same level, the total deflection of the end A with respect to the tangent at B must be zero. Consequently, Rl a PI 3 PI 3 3 El 24 El 16 El 5 = 0; whence (109) 1 Therefore (110) R' = P-R = ^P. The maximum moment, which in this case occurs at B, is then (in) M = RI- P- 16 The position and amount of the maximum deflec- tion may be found as in the preceding article. 50. Beam fixed at one end and bearing a concen- trated eccentric load. The deflection of the end A with respect to the fixed end B (Fig. 78) also consists of two parts in this case. From (38), article 37, that due to the reaction R is Rf FIG. 78 and from (44), article 37, that due to the load P is 3 El 2 El RESTRAINED, OR BUILT-IN, BEAMS 89 Since the supports are assumed to be at the same level, the total deflection of A with respect to the tangent at B is zero. Con- sequently, 3 El 3 El 2 El whence (112) = (a^ Also F*7 (us) ' = f - R = |J [2 i(* + g _ i*\. The moment at the fixed end B is then (114) M B = l - .PI, = - ^5 (I + O, / and the moment at the load C is (115) M C = ^ 1 = ^1(2 ? + ?,). I The maximum deflection may be found by the method explained in article 48. APPLICATIONS 156. One end of a beam is built into a wall, and the other end is supported at the same level by a post 12 ft. from the wall. The beam carries a uniform load of 100 Ib. per linear foot. Find the position and amount of the maximum moment and also of the maximum deflection. 157. One end of a beam is built into a wall, and the other end rests on a prop 20ft. from the wall at the same level. The beam bears a concentrated load of 1 ton at a point 8 ft. from the wall. Find the position and amount of the maximum moment and also of the maximum deflection. 158. A cantilever of length I is loaded uniformly. At what point of its length should a prop be placed, supporting the beam at the same level as the fixed end, in order to reduce the bending stress as much as possible, and what proportion of the load is then carried by the prop ? 159. A 20-in. steel I-beam weighing 65 Ib./ft. is built into a wall at one end and rests on a support 20 ft. from the wall at the other end. A load of 25 tons rests on the beam at a point distant 15 ft. from the wall. Find the reaction of the support and the maximum deflection. 160. A beam of uniform section is built into a wall at one end, projecting 16 ft. from the face of the wall, and rests on a column at 12 ft. from the wall. The beam carries a uniform load of 5 tons per foot run. Find the load on the column. 90 RESISTANCE OF MATERIALS 161. A beam of uniform section is built into walls at both ends, the distance between walls being 25 ft. Two concentrated loads, each of 5 tons, rest on the beam at points 5 ft. from each wall. Find the maximum bending moment in the beam, and also the position of zero bending moment. 162. A continuous beam of two spans, each of 40ft., carries a uniform load of 1 ton per foot run. Find the reactions of the supports by the method of article 48, and also the maximum moment and maximum deflection. 163. A continuous beam of two equal spans is uniformly loaded. Find the bend- ing moment over the middle support when the three supports are at the same level, and also when the middle support is raised or lowered an amount h. 164. A uniform beam of 20 ft. span is fixed at both ends and carries a load of 4 tons at the center and two loads of 3 tons each at 5 ft. from each end. Find the maximum moment and the position of zero moment. 165. A beam of length 2 I is supported at the center, one end being anchored down to a fixed abutment and the other end carrying a concentrated load W. Neglecting the weight of the beam, find the deflection of the free end. SECTION VIII COLUMNS AND STRUTS 51. Nature of compressive stress. When a prismatic piece of length equal to several times its breadth is subjected to axial com- pression, it is called a column, or strut, the word column being used to designate a compression member placed vertically and bear- ing a static load, all other compression members being called struts. If the axis of a column or strut is not perfectly straight, or if the load is not applied exactly at the centers of gravity of its ends, a bending moment is produced which tends to make the column deflect sideways, or " buckle." The same is true if the material is not per- fectly homogeneous, causing certain parts to yield more than others. Such lateral deflection increases the bending moment and conse- quently increases the tendency to buckle. A compression member is therefore in a different condition of equilibrium from one subjected to tension, for in the latter any deviation of the axis from a straight line tends to be diminished by the stress instead of increased. The oldest theory of columns is due to Euler, and his formula is still the standard for comparison. Euler's theory, however, is based upon the assumptions that the column is perfectly straight, the material perfectly homogeneous, and the load exactly centered at the ends assumptions which are never exactly realized. For practical purposes, therefore, it has been found necessary to modify Euler's formula in such a way as to bring it into accord with the results of actual experiments, as explained in the following articles. 52. Euler's theory of long columns. Consider a long column sub- jected to axial loading, and assume that the column is perfectly straight and homogeneous and that the load is applied exactly at the centers of gravity of its ends. Assume also that the ends of the column are free to turn about their centers of gravity, as would be the case, for example, in a column with round or pivoted ends. 91 92 RESISTANCE OF MATERIALS Now suppose that the column is bent sideways by a lateral force, and let P be the axial load which is just sufficient to cause the column to retain this lateral deflection when the lateral force is removed. Let OX and OY be the axes of X and Y respectively (Fig. 79). Then it can be shown that the elastic curve OCX is a sine curve. For simplicity, however, it will be assumed to be a parabola. Since the deflection at any point C is the lever arm of the load P, the moment at C is Py. The moment at any point is therefore P times as great as the deflection at that point, and con- sequently the moment diagram will also be a parabola (Fig. 80). p Moment Diagram FIG. 79 FIG. 80 Now let d denote the maximum deflection, which in this case is at the center. Then the maximum ordiiiate to the moment diagram is Pd. Therefore, from article 17, the area of one half the diagram 2 I Pdl is A = ^ (Pd) - = , and the distance of its centroid from one 5 I 5 1 end is X Q = - = - . Hence, from the general deflection formula, the deflection at the center will be (116) J_ J_ (Pld\ 5J _ "El *~ EI\ 3 )lQ~ Canceling the common factor d and solving for P, the result is 9.6 E I (117) 48 "" COLUMNS AND STRUTS 93 If the elastic curve had been assumed to be a sine curve instead of a parabola, the result would have been the well-known equation ir^EI 9.87 El (118) P = __ = __, which is Euler's formula for long columns in its standard form. Under the load P given by this formula the column is in neutral equilibrium ; that is to say, the load P is just sufficient to cause it to retain any lateral deflection which may be given to it. For this reason P is called the critical load. If the load is less than this critical value, the column is in stable equilibrium, and any lateral deflection will disappear when its cause is removed. If the load exceeds this critical value, the column is in unstable equilibrium, and the slightest lateral deflection will rapidly increase until rupture occurs. 53. Effect of end support. The above deduction of Euler's formula is based on the assumption that the ends of the column are free to turn, and therefore formula (118) applies only to long columns with round or pivoted ends. If the ends of a column are rigidly fixed against turning, the elastic curve has two points of inflection, say B and D (Fig. 81). From symmetry, the tangent to the elastic curve at the center C must be parallel to the original position of the axis of the column AE, and therefore the portion AB of the elastic curve must be symmetrical with BC, and CD with DE. Consequently, the points of inflection, B and D, occur at one fourth the length of the column from either end. The critical load for a column with fixed ends is therefore the same as for a column with free ends of half the length ; whence, for fixed ends, Euler's formula becomes (119) Columns with flat ends, fixed against lateral movement, are usually regarded as coming under formula (119), the terms fixed ends and flat ends being used interchangeably. 94 RESISTANCE OF MATERIALS If one end of the column is fixed and the other end is free to turn, the elastic curve is approximately represented by the line BCDE in Fig. 81. Therefore the critical load in this case is ap- proximately the same as for a column with both ends free, of length BCD, that is, of length equal to | BE or 1 1 ; whence, for a column with one end fixed and the other free, Euler's formula becomes (120) P = 9 IT 2 El 4 I 2 approximately. One end I Ends fixed n direction; One end If the lower end is fixed in direction but the upper end is entirely free (that is, if there is no horizontal reaction to prevent it from bending out sideways), it may be regarded as half of a column with round or pin ends and of length 2 1. Consequently, in this case Euler's formula becomes 7T 2 EI Round ends; Position fixed but not direc- tion. end The general expres- sion for Euler's formula is then (122) P = K TT^EI 7 Fm. 82 where the constant k is determined by the way in which the ends of the column are supported. The values of k corresponding to various end conditions are given in Fig. 82. 54. Modification of Euler's formula. It has been found by ex- periment that Euler's formula applies correctly only to very long columns, and that for short columns or those of medium length it gives a value of P considerably too large. Very short columns or blocks fail solely by crushing, the tend- ency to buckle in such cases being practically zero. Therefore, if p denotes the crushing strength of the material and A the area COLUMNS AND STRUTS 95 of a cross section, the breaking load for a very short column is P=pA* For columns of ordinary length, therefore, the load P must lie somewhere between pA and the value given by Euler's formula. Consequently, to obtain a general formula which shall apply to columns of any length, it is only necessary to express a continuous relation between pA and Such a relation is furnished by the equation das) r = 1+pA 7T 2 EI For when I = 0, P=pA, and when I becomes very large, P approaches IT 2 El the value Moreover, for intermediate values of I this formula c gives values of P considerably less than those given by Euler's formula, thus agreeing more closely with experiment. 55. Rankine's formula. Although the above modification of Euler's formula is an improvement on the latter, it does not yet agree closely enough with experiment to be entirely satisfactory. The reason for the discrepancy between the results given by this formula and those obtained from actual tests is that the assumptions upon which the formula is based, namely, that the column is perfectly straight, the material perfectly homogeneous, and the load applied exactly at the centers of gravity of the ends, are never actually realized in practice. To obtain a more accurate formula, two empirical constants will be introduced into equation (123). Thus, for fixed ends, let (124) where / and g are arbitrary constants to be determined by experi- ment, and t is the least radius of gyration of a cross section of the column. This formula has been obtained in different ways by * As Euler's formula is based upon the assumption that the column is of sufficient length to buckle sideways, it is evident a priori that it cannot be applied to very short columns, in which this tendency is practically zero. Thus, in formula (1 1 8) , as I approaches zero P approaches infinity, which of course is inadmissible. 96 RESISTANCE OF MATERIALS Gordon, Rankine, Navier, and Schwarz.* Among German writers it is known as Schwarz's formula, but in English and American textbooks it is called Rankine 's formula. For I = 0, P = gA, and, since short blocks fail by crushing, g is therefore the ultimate compressive strength of the material. For different methods of end support Rankine' s formula takes the following forms : (125) = /7\2 Flat endS 1 _j_ / / _ ] (fixed in direction) (126) = 9 2 Round ends " -^ _i_ A f I _\ (direction not fixed) w (127) - = 9 r- 2 Hinged ends ' 1 _J_ 9 /" / _ \ (position fixed, but not J 1 I direction) (128) = ry^ One end flat and the 1+1.78/Y-) other round W 56. Values of the empirical constants in Rankine 's formula. The values of the empirical constants, / and g, in Rankine's formula have been experimentally determined by Hodgkinson and Christie, with the following results : For hard steel, g = 69,000 lb./in. 2 , /= !^. -i For mild steel, g = 48,000 lb./in. 2 , / = For wrought iron, g = 36,000 lb./in. 2 , f = 30000 1 36000 For cast iron, g = 80,000 lb./in. 2 , /= - For timber, g= 7,200 lb./in. 2 , /= 6400 1 3000 * Rankine's formula can be derived independently of Euler's formula either by assuming that the elastic curve assumed by the center line of the column is a sinusoid or by assuming that the maximum lateral deflection D at the center of the column is li given by the expression D = n , where I is the length of the column, b its least width, and M an empirical constant. COLUMNS AND STRUTS 97 These constants were determined by experiments upon columns for which 20 < - < 200, and therefore can only be relied upon to t/ furnish accurate results when the dimensions of the column lie within these limits. As a factor of safety to be used in applying the formula, Rankine recommended 10 for timber, 4 for iron under dead load, and 5 for iron under moving load. 57. Johnson's parabolic formula. From the manner in which equation (123) was obtained and afterwards modified by the intro- duction of the empirical constants / and g, it is clear that Rankine's formula satisfies the requirements for very long or very short col- umns, while for those of intermediate length it gives the average values of experimental results. A simple formula which fulfills these same requirements has been given by Professor J. B. Johnson, and is called Johnson's parabolic formula. If equation (124) is written = - 9 A * and then y is written for p, and x for - , Rankine's formula becomes I/ 1+/^ 2 For this cubic equation Johnson substituted the parabola (129) y = 8 - ex\ in which x and y have the same meaning as above, and 8 and e are empirical constants. The constants 8 and e are then so chosen that the vertex of this parabola is at the elastic limit of the material on the axis of loads (or F-axis), and the parabola is also tangent to Euler's curve. In this way the formula is made to satisfy the theoretical requirements for very long or very short columns, and for those of intermediate length it is found to agree closely with experiment. 98 RESISTANCE OF MATERIALS For different materials and methods of end support Johnson's parabolic formulas, obtained as above, are as follows : KIND OF COLUMN FORMULA LIMIT FOR USE Mild steel Hinged ends = 42,000 - .97 (l) 2 |< 15 p /7\2 7 Flat ends - = 42,000 - .62 -^190 Wrought iron Hinged ends = 34,000 - .67 A. tT z _ Flat ends = 34,000 - .43 ' 1^210 Cast iron p 95 /7\2 I _ Round ends - = 60,000- - A 4 (9 Flat ends = 60,000 - ^ A 4 w 1^120 6 Timber (flat ends) White pine = 2,500- .6 A tr -,^ 60 Short-leaf yellow pine = 3,300- .7 1? 60 t' Long-leaf yellow pine = 4,000- .8 A. ' p? 60 White oak = 3,500- .8 0' ^60 The limit for use in each case is the value of x I = - J at the point where Johnson's parabola becomes tangent to Euler's curve. For greater values, of - Euler's formula should therefore be used. v A graphical representation of the relation between Euler's for- mula, Rankine's formula, J. B. Johnson's parabolic formula, and T. H. Johnson's straight-line formula (considered in the next article) is given in Fig. 83 for the case of a wrought-iron column with hinged ends.t * In the formulas for timber tf is the least lateral dimension of the column. t For a more extensive comparison of these formulas see Johnson's framed Structures, 8th ed., 1905, pp. 159-171; also Trans. Amer. Soc. Civ, Eng., Vol. XV, pp. 518-536. COLUMNS AND STRUTS 99 58. Johnson's straight-line formula. By means of an exhaustive study of experimental data on columns Mr. Thomas H. Johnson has shown that for columns of moderate length a straight line can be made to fit the plotted results of column tests as exactly as a curve. He has therefore proposed the formula (130) P I = V <7 A t 50 100 150 200 FIG. 83. Wrought-Iron Column (Pin Ends) 250 300 1, Euler's formula ; 2, T. H. Johnson's straight-line formula ; 3, J. B. Johnson's parabolic formula; 4, Rankine's formula or, in the notation of the preceding article, (131) y = v-ax, in which v and cr are empirical constants, this being the equation of a straight line tangent to Euler's curve. This formula has the merit of great simplicity, the only objection to it being that for short columns it gives a value of P in excess of the actual break- ing load. The relation of this formula to those which precede is shown in Fig. 83. 100 RESISTANCE OF MATERIALS The constants v ando- in formula (130) are connected by the relation (132) _v_ I 4v ~3\3n7r 2 ^' where for fixed ends n = 1, for free ends n = 4, and for one end fixed and the other free n = 1.78. The following table gives the special forms assumed by Johnson's straight-line formula for various materials and methods of end support : * KIND OF COLUMN FORM i 'LA LIMIT FOR USE Hard steel Flat ends = 80,000 -3371 - P 158.0 A t i ^ Hinged ends - = 80,000 -4141 - P 129.0 A t t ^ Round ends - = 80,000 -534 - - ^ 99.9 A t t Mild steel Flat ends = 52,500 -179! - ^ 195.1 'A t t ^ Hinged ends = 52,500 -220 - - P 159.3 A t t Round ends = 52,500 -284- - ^ 123.3 A t t Wrought iron Flat ends = 42,000 - 128 - ip 218.1 A t i Hinged ends = 42,000 -157! 7^178.1 A t i Round ends = 42,000 -203 - - P 138.0 A t i ^ Cast iron Flat ends - 80,000 -438 - - ^ 121.6 A t i ^ Hinged ends = 80,000 -537! - ^ 99.3 A i i Round ends - = 80,000 - 693 - 7< 77.0 A t t Oak Flat ends = 5,400 A - 28 1 ^ - ^ 128.1 * Trans. Amer. Soc. Civ. Eng., 1886, p. 530. COLUMNS AND STRUTS '101 The limit for use in each case is the value of x I = - j f or the point at which Johnson's straight line becomes tangent to Euler's curve. 59. Cooper's modification of Johnson's straight-line formula. In his standard bridge specifications Theodore Cooper has adopted Johnson's straight-line formulas, modifying them by the introduc- tion of a factor of safety. Thus, for medium steel, Cooper specifies that the following formulas shall be used in calculating the safe load. For chords (133) For posts (134) - = 8,000 - 30 - for live-load stresses, A t P I = 16,000 60 - for dead-load stresses. A t P I = 7,000 40- for live-load stresses, A t P I = 14,000 80 - for dead-load stresses, jGL 6 - = 10,000 - 60 - for wind stresses. A t For lateral struts (135) ' P I = 9,000 - 50 - for initial stresses. By initial stress in the last formula is meant the stress due to the adjustment of the bridge members during construction. 60. Eccentrically loaded columns. In a column that carries an eccentric load (for example, a column carrying a load on a bracket or the post of a crane) there is a definite amount of bending stress due to the eccentricity of the load in addition to the column stress. As the nature of column stress is such that it is impossible to de- termine its amount, the simplest method of handling a problem of this kind is to determine its relative security against failure as a column and failure by bending. That is to say, first determine its factor of safety against failure as a column under the given column load. Then consider it as a beam and find the equivalent bending 102 RESISTANCE OF MATERIALS P = A 52,500 - 179 - the factor of safety \ '/ against column failure is moment which would give the same factor of safety. Finally, com- bine this equivalent bending moment with that due to the eccentric load, and calculate the unit stress from the ordi- nary beam formulas. To illustrate the method, suppose that a col- umn 18 ft. long is composed of two 12-in. I-beams each weighing 40 lb./ft., and carries a column load of 20 tons at its upper end and also an eccentric load of 10 tons with eccentricity 2 ft., as shown in Fig. 84. Assuming that the column has flat ends, and using Johnson's straight-line formula, 1 FIG. 84 ^(52,500-179- 6 = 2 (11.76) (52,500 -179 (47.3)) = . 60,000 60,000 Now consider the column as a beam and find the equivalent central load K corresponding to the factor of safety just found, namely, 17.3. The maximum moment in a simple beam bearing a concen- trated load K at the center is M= Hence, from the beam Kl pi. 4 JP/ whence Assuming formula M= we have = e 4 e le the ultimate strength of the material to be 60,000 lb./in. 2 , we have 60,000 ., .. 2 P = ~^r Win. 2 , 7=2(245.9) in. 4 , e = 6 in., 17.3 Z = 216 in., and, inserting these values, the equivalent load K is found to be 4 x 60,000 x 491.8 17.3 x 216 x 6 = 5220 Ib. Now the eccentric load P^ acting parallel to the axis of the column, produces the same bending effect as a horizontal reaction TTat either end, where HI = P^d. The bending moment at the center, due to a TTJ reaction H perpendicular to the axis of the beam, is, however, ft COLUMNS AND STRUTS 103 Hence the total equivalent moment at the center now becomes _Kl Hl_Kl P^d __ 5220 x 216 20,000 x 24 T""T = : T" ~Y : ~T~ ~2~ = 521,880 in.-lb. Consequently, the maximum unit stress in the member becomes M 521,880 which corresponds to a factor of safety of about .9. If this factor of safety is larger than desired, assume a smaller I-beam and repeat the calculations. A method substantially equivalent to the above is to assume that the stress in a column is represented by the empirical factor in the column formula used. Thus, for a short block the actual compressive stress p is given by the relation P = pA, whereas in the column formula used above, namely, P = A ( 52,500 179 - ) , the stress p is replaced by the empirical factor 52,500 179- Consequently, the fraction -, 52,500-179- v where u c denotes the ultimate compressive strength of the material, represents the reduction in strength of the member due to its slim- ness and method of loading ; or, what amounts to the same thing, the equivalent unit stress in the column is (136) ; Pl U ~ ^'52,500-179- T/ Applying this method to the numerical problem given above, we have A = 23.52, - = 47.3, u _ 60,000 =136 52,500 - 179 pW- 179x47.3- ' ' 104 RESISTANCE OF MATERIALS Hence the equivalent stress in the column is ^5^x1.86 = 3470 Win-' Also, the bending stress, produced by the eccentricity of the load, is Consequently, by this method, the total stress in the column is found to be < 347Q + 29 28 = 6398 lb./in. 2 If a formula of the Rankine-Gordon type is used, namely, P 9 the equivalent stress p e in the column, due to the given load P, is 9 where u c denotes the ultimate compressive strength of the material, as above. APPLICATIONS 166. A solid, round, cast-iron column with flat ends is 15ft. long and Gin. in diameter. What load may be expected to cause rupture ? 167. A square wooden post 12 ft. long is required to support a load of 15 tons. With a factor of safety of 10, what must be the size of the post ? 168. Two 8-in. steel I-beams, weighing 25.25 Ib. /ft., are joined by latticework to form a column 25 ft. long. How far apart must the beams be placed, center to center, in order that the column shall be of equal strength to resist buckling in either axial plane ? 169. Four medium steel angles, 5 x 3 x f in., have their 3-in. legs riveted to a |-in. plate so as to form an I-shaped built column. How wide must the plate be in order that the column shall be of equal strength to resist buckling in either axial plane ? 170. A hollow wrought-iron column with flat ends is 20 ft. long, 7 in. internal diameter, and 10 in. external diameter. Calculate its ultimate strength by Rankine's and Johnson's formulas and compare the results. 171. Compute the ultimate strength of the built column in problem 168 by Rankine's and by Johnson's formulas and compare the results. 172. Compute the ultimate strength of the column in problem 169 by Rankine's and by Johnson's straight-line formulas and compare the results. COLUMNS AND STRUTS 105 FIG. 85 173. A column 18 ft. long is formed by joining the legs of two 10-in. steel chan- nels, weighing 30 lb./ft., by two plates each 10 in. wide and ^ in. thick, as shown in Fig. 85. Find the safe load for this column by Johnson's straight-line formula, ^_^ _~ using a factor of safety of 4. 174. A wrought-iron pipe 10 ft. long, and of inter- nal and external diameter 3 in. and 4 in. respectively, bears a load of 7 tons. What is the factor of safety ? 175. What must be the size of a square steel strut 8 ft. long, to transmit a load of 5 tons with safety ? 176. Design a column 16 ft. long to be formed of two channels joined by two plates and to support a load of 20 tons with safety. 177. Using Cooper's formula for live load, design the inclined end post of a bridge which is 25 ft. long and bears a load of 30 tons, the end post to be com- posed of four angles, a top plate, and two side plates. 178. A strut 16 ft. long, fixed rigidly at both ends, is needed to support a load of 80,000 Ib. It is to be composed of two pairs of angles united with a single line of 1-in. lattice bars along the central plane. Determine the size of the angles for a factor of safety of 5. (Note that the angles must be spread I 2 'm. to admit the latticing.) 179. For short posts or struts, such as are ordinarily used in building construc- tion, it is customary to figure the safe load as 12,000 lb./in. 2 of cross-section area for lengths up to 90 times the radius of gyration ; that is, for - = 90. To what factor of safety does this correspond, by Johnson's straight-line formula ? 180. The posts used to support a girder in a building are 8 in. x 8 in. timbers 8 ft. long. Find the diameter of a solid cast-iron column of equal strength. If a wrought-iron pipe 4 in. in external diameter is used, what must be its thick- ness to be equally safe ? 181 . At what ratio of diameter to length would a round mild-steel strut have the same tendency to crush as to buckle ? 182. A load of 100 tons is carried jointly by three cast-iron columns 20 ft. long. What saving in material will be effected by using a single column instead of three, the factor of safety to be 15 in both cases ? 183. Determine the proper size of a hard- steel piston rod 48 in. long for a piston 18 in. in diameter and a steam pressure of 80 lb./in. 2 Consult table for proper factor of safety. 184. The side rod of a locomotive is 9ft. long between centers, 4 in. deep, and 2 in. wide. The estimated thrust in the rod is 12 tons, and the transverse inertia and gravity load 20 Ib. per inch of length. Determine the factor of safety. 185. The vertical post of a crane (Fig. 86) is to be made of a single I-beam. The post is pivoted at both ends so as to revolve about its axis. Find the size of I-beam required for factor of safety of 4 and for dimensions and loading, as shown. 8 Tons FIG. 86 SECTION IX TORSION 61. Maximum stress in circular shafts. When a uniform circular shaft, such as is shown in Fig. 87, is twisted by the application of moments of opposite signs to its ends, every straight line AB paral- lel to its axis is deformed into part of a helix, or screw thread, AC. The strain in this case is one of pure shear and is called torsion. The angle (f> is called the angle of shear and is proportional to the radius BD of the shaft. The angle 6 is called the angle of twist and is proportional to the length AB of the shaft. Consider a section of length A# cut from a circular shaft by planes perpendicular to its axis (Fig. 87). Let A0 denote the angle of twist for this section. Then, since the angle of twist is propor- tional to the length of the shaft, A0 : 6 = A# : I ; whence Also, if <f) and A0 are expressed in circular measure, BC=$'AB = $kx, and BC = A<9 - BD= Therefore <f> = r-- From Hooke's law, -= G. Hence A# I <> (!37) ,-* Therefore q is proportional to r; that is to say, the unit shear is proportional to its distance from the center, being zero at the center and attaining its maximum value at the circumference. 106 TOKSION 107 If q' denotes the intensity of the shear at the circumference, and a denotes the radius of the shaft, then the shear q at a distance r from the center is given by the formula Now if q denotes the intensity of the shear on any element of area A^4, the total force acting on this element is q&A, and its mo- ment with respect to the center is qAAr. Therefore the total internal moment of resistance is ^qkAr, where the summation extends over the entire cross section ; and since this must be equal to the exter- nal twisting moment M t , we have Inserting for q its value in terms of the radius, q = , this becomes or, since by definition ^r 2 AJ = I v , the polar moment of inertia of the cross section, . M t = &. a For a solid circular shaft of diameter 7), I p and a ; consequently, (138) q' -- p For a hollow circular shaft of external diameter D and internal diameter d, I p = (D 4 c? 4 ) and a = ; hence (1 3 9) *' = =4^nr 62. Angle of twist in circular shafts. From equation (137), e-JL-JL. Gr Ga Therefore, for a solid circular shaft, from equation (138), (140) 108 RESISTANCE OF MATERIALS and for a hollow circular shaft, from equation (139), . 32 M t l (141) = If M t is known and 6 can be measured, equations (140) and (141) can be used for determining G. If G is known and 6 measured, these equations can be used for finding M t ; in this way the horse power can be determined from the angle of twist. 63. Power transmitted by circular shafts. Let H denote the number of horse power being transmitted by a circular shaft, n its speed in revolutions per minute (R.P.M.), and M t the torque, or twisting moment, acting on it, expressed in inch-pounds. Then, since the angular displacement of M t in one minute is 2 ?m, the work done by the torque in one minute is 2 irnM t . Also, since one horse power = 33,000 ft.-lb./min. = 396,000 in.-lb./min., the total work done by the shaft in one minute is 396,000 H. Therefore 2 irnM t = 396,000 H\ whence (148) M t = ***** = 63,030 in.-,b. 277 n Therefore,' if it is required to find the diameter D of a solid circular shaft which shall transmit a given horse power H with safety, then, from equation (138), ,_ 16 M t _ 321,000 H m ~~ whence (143) J> = 68.5 ^ H As safe values for the maximum unit shear q', Ewing recom- mends 9000 lb./in. 2 for wrought iron, 13,500 lb./in. 2 for steel, and 4500 lb./in. 2 for cast iron. Inserting these values of q' in formula (143), it becomes (144) D = where for steel n = 2.88, for wrought iron p = 3.29, and for cast iron a = 4.15. TORSION 109 Expressed- in kilowatts instead of horse power, this formula becomes (145) Z> = where for steel p = 3.175, for wrought iron p = 3.627, and for cast iron p = 4.576. 64. Combined bending and torsion. In many cases shafts are subjected to combined bending and torsion, as, for instance, when a shaft transmits power by means of one or more cranks or pulleys. In this case the bending moment M b at any point of the shaft pro- duces a normal stress p in accordance with equation (33), article 34, that is, and the torque M t produces a shearing stress q given by (138), article 61, namely, In more advanced works on the strength of materials, however, it is shown that the maximum and minimum normal and shearing stresses, resulting from any such combination as the above, are given by the relations * (148) p m&x min (149) q m&x min Therefore, inserting in these expressions the values of p and q as given above, the maximum and minimum normal and shearing stresses in terms of the bending and twisting moments are found to be (150) p pUp =fc "/Jffr + -MT 2 ) (called Rankine's formula), 16 / (151) tf max = ^5 VM^J + M? (called Guest's formula). min * Slocum and Hancock, Strength of Materials, Revised Edition, pp. 24-25. 110 RESISTANCE OF MATERIALS Note that Rankine's formula gives the principal normal stresses, that is, tension or compression, whereas Guest's formula gives shear. Since the ultimate strength in tension or compression is usually different from that in shear,* in designing circular shafts carrying combined stress both formulas should be tried with the same working stress (or factor of safety), and the one used which gives the larger dimensions. 65. Resilience of circular shafts. In article 7 the resilience of a body was denned as the internal work of deformation. For a solid circular shaft this internal work is where M t is the external twisting moment and 6 is the angle of twist. From equation (137), 6 = -%- = ^- , (jrT (jrCL and from equation (138), Therefore the total resilience of the shaft is and consequently the mean resilience per unit of volume is 0*3) ,.-$-& 66. Non-circular shafts. The above investigation of the distribu- tion and intensity of torsional stress applies only to shafts of circular section. For other forms of cross section the results are entirely different, each form having its own peculiar distribution of stress. For any form of cross section whatever, the stress at the boundary must be tangential, for if the stress is not tangential, it can be resolved into two components, one tangential and the other normal to the boundary ; but a normal component would necessitate forces parallel to the axis of the shaft, which are excluded by hypothesis. Since the stress at the boundary must be tangential, the circular section is the only one for which the stress is perpendicular to a radius vector. Therefore the circular section is the only one to * The shearing strength of ductile materials, both at the elastic limit and at the ulti- mate stress, is about four fifths of their tensile strength at these points. TORSION 111 which the above development applies, and consequently is the only form of cross section for which Bernoulli's assumption holds true. That is to say, the circular section is the only form of cross section which remains plane under a torsional strain. The subject of the distribution of stress in non-circular shafts has been investigated by St. Venant, and the results of his investigations are summarized below (articles 67~70). 67. Elliptical shaft. For a shaft the cross section of which is an ellipse of semi-axes a and >, the maximum stress occurs at the ends of the minor axis instead of at the ends of the major axis, as might be expected. The unit stress at the ends of the minor axis is given by the formula (154) and the angle of twist per unit of length is 9 Mt(n * + &2) 81 = The total angle of twist for an elliptical shaft of length I is therefore 68. Rectangular and square shafts. For a shaft of rectangular cross section the maximum stress occurs at the centers of the longer sides, its value at these points being (157) ? max . lib VV + b 2 in which h is the longer and b the shorter side of the rectangle. The angle of twist per unit of length is, in this case, For a square shaft of side b these formulas become (159) and (160) 112 RESISTANCE OF MATERIALS The value of q for a square shaft found from this equation is Mr about 15 per cent greater than if the formula q = were used, and p the torsional rigidity is about .88 of the torsional rigidity of a circular shaft of equal sectional area. 69. Triangular shafts. For a shaft whose cross section is an equilateral triangle of side <?, (161) c and the angle of twist per unit of length is The torsional rigidity of a triangular shaft is therefore .73 of the torsional rigidity of a circular shaft of equal sectional area. 70. Angle of twist for shafts in general. The formula for the angle of twist per unit of length for circular and elliptical shafts can be written (163) I G A* in which I p is the polar moment of inertia of a cross section about its center, and A is the area of the cross section. This formula is rigorously true for circular and elliptical shafts, and St. Venant has shown that it is approximately true whatever the form of cross section. APPLICATIONS 186. A steel wire 20 in. long and .182 in. in diameter is twisted by a moment of 20 in.-lb. The angle of twist is then measured and found to be 6 18 3r. What is the value of G determined from this experiment ? Solution. From equation (140), article 62, where, in the present case, using pound and inch units, M t = 20, I = 20, D = 182, and 6 = 1831 / = .3232 radians. Substituting these numerical values, the result is G = 11,490,000 lb./in. 2 187. A steel shaft 5 in. in diameter is driven by a crank of 12-in. throw, the maximum thrust on the crank being 10 tons. If the outer edge of the shaft-bearing is 11 in. from the center of the crank pin, what is the stress in the shaft at this point ? TOESION 113 Solution. Referring to Fig. 88, the dimensions in the present case are ^ i dj = 12 in., d 2 = 11 in. S Consequently, M t =lQ- 2000 12 = 240,000 in.-lb. M b = 10 . 2000 . 11 = 220,000 in.-lb. Therefore, from equation (150), article 64, = _ (220,000 + V220,000 2 + 240,000 2 ) 7T 5 = 22,200 lb./in. 2 , and similarly, from equation (151), 16 / FIG. 88 188. If P and Q denote the unit stresses at the elastic limits of a material in p tension and shear respectively, show that when < 1 the material will fail in ten- P ^ sion, whereas when > 1 it will fail in shear, when subjected to combined bending and torsion, irrespective of the relative values of the bending and twisting moments. Solution. Combining Rankine's and Guest's formulas, we have 16Jf 6 p - q = P Consequently, if the bending moment is zero, p' = q', or = 1, whereas if it is not q' zero, p' > q'. Similarly, if the twisting moment is zero, = 2. q' Now let F t and F 8 denote the factors of safety in tension and shear respectively. Then Ft F s Since / T) Jjl , the fraction ^=1. Consequently, if <1 also, then <1; that P' Q F 8 is, F t < F s , and the material is weaker in tension than in shear. The second part of the theorem is proved in a similar manner. For a complete discussion of this question see article by A. L. Jenkins, En- gineering (London, November 12, 1909), pp. 637-639. 189. Three pulleys of radii 8, 4, and 6 in. respectively are keyed on a shaft as shown in Fig. 89. Pulley No. 1 is the driving pulley and transmits 30 H.P. to the shaft, of which amount 10 H.P. is taken off from pulley No. 2 and the remaining 20 H.P. from pulley No. 3. The speed is 50 R.P.M., the belts are all parallel, and the tension in the slack side of each belt is assumed to be one half the tension in the tight side. Find the required size of the shaft for a working stress of 12,000 lb./in. 2 in tension and 9000 lb./in. 2 in shear. 114 RESISTANCE OF MATERIALS Solution. The first step is to find the tensions in the belts. Since power is the rate of doing work, and 1 H.P. = 550 ft.-lb./sec., the formula for power may be written xi Fv Horse power = 55O k S 1 --^- -?-' FIG. 89 where F denotes the effective force, or difference in tension in the two sides of the belt, expressed in pounds, and v is the belt speed in ft. /sec. Hence, for the pulley of 8 in. radius transmitting 30 H.P., we have 60 . 550 whence F = 4730 Ib. Since by assumption the tension on the tight side of the belt is twice that on the slack side, their values are Tension on tight side = 9460 Ib. Tension on slack side = 4730 Ib. The belt tensions for the other pulleys are calculated in a simi- lar manner, the results being indicated on Fig. 89. Considering the shaft as a beam, the load at each pulley is equal to the sum of the belt tensions for that pulley, as shown in Fig. 90. The reac- tions of the bearings and the bending-moment diagram are next obtained, the results being given in Fig. 90. The maximum bending and twisting moments thus occur at pulley No. 1, their numerical values being M b = 364,230 in.-lb., M t = 37,840 in.-lb. Moment diagram FIG. 90 TOKSION 115 Therefore, substituting in equation (150), we have whence D = 6.725 in. and similarly, from equation (151), 9000 = V364,230 2 + 37,840 2 ; whence D = 5.842 in. The proper diameter for the shaft is then the larger of these two values, say 6f in. 190. If the angle of twist for the wire in problem 186 is 6 = 40, how great is the torsional moment acting on the wire ? 191. Compare the angle of twist given by St. Venant's general formula with the values given by the special formulas in articles 67, 68, and 69. 192. A steel shaft is required to transmit 300 H.P. at a speed of 200 revolutions per minute, the maximum moment being 40 per cent greater than the average. Find the diameter of the shaft. 193. Under the same conditions as in problem 192, find the inside diameter of a hollow circular shaft whose outside diameter is 6 in. Also compare the amount of metal in the solid and hollow shafts. 194. The semi-axes of the cross section of an elliptical shaft are 3 in. and 5 in. respectively. What is the diameter of a circular shaft of equal strength ? 195. An oak beam 6 in. square projects 4ft. from a wall and is acted upon at the free end by a twisting moment of 25,000 ft.-lb. How great is the angle of twist ? 196. A steel shaft 10 ft. long between centers of bearings and 4 in. in diameter carries a pulley 14 in. in diameter at its center. If the driving tension in the belt is 250 lb., and the following side runs slack, what is the maximum stress in the shaft, and how many H.P. is it transmitting when running at 80 R.P.M. ? 197. Find the required diameter of a solid wrought-iron circular shaft which is required to transmit 150 H.P. at a speed of 60 R.P.M. 198. Find the angle of twist in problem 192. 199. Find the angle of twist in problem 193 and compare it with the angle of twist for the solid shaft in problem 198. 200. How many H.P. can a hollow circular steel shaft of 15 in. external diam- eter and 11 in. internal diameter transmit at a speed of 50 R.P.M. if the maximum allowable unit stress is not to exceed 12,000 lb./in. 2 ? 201. Find the diameter of a structural-steel engine shaft to transmit 900 H.P. at 75 R.P.M. with a factor of safety of 10. 202. Find the factor of safety for a wrought-iron shaft 5 in. in diameter which is transmitting 60 H.P. at 125 R.P.M. 203. A structural-steel shaft is 60 ft. long and is required to transmit 500 H.P. at 90 R.P.M. with a factor of safety of 8, and to be of sufficient stiffness so that the angle of to'rsion shall not exceed .5 per foot of length. Find its diameter. 204. Under the same conditions as in problem 203, find the size of a hollow shaft if the external diameter is twice the internal. 116 RESISTANCE OF MATERIALS * 1 i 1 3 6" J - 10- 10 HH FIG. 91 205. A hollow wrought-iron shaft 9 in. in external diameter and 2 in. thick is required to transmit 600 H.P. with a factor of safety of 10. At what speed should it be run ? 206. A horizontal steel shaft 4 in. in diameter and 10 ft. long between centers of bearings carries a 300-lb. pulley 14 in. in diameter at its center. The belt on the pulley has a tension of 50 Ib. on the slack side and 175 Ib. on the driving side. Find the maximum stress in the shaft, assum- ing that the belt exerts a horizontal pull on the shaft. 207. An overhung steel crank, like that shown in Fig. 88, carries a maximum thrust on the crank pin of 2 tons. Length of crank, 9 in. ; distance from center of pin to center of bearing, 5 in. Determine the size of crank and shaft for a factor of safety of 5. 208. A propeller shaft 9 in. in diameter transmits 1000 H.P. at 90 R.P.M. If the thrust on the screw is 12 tons, determine the maxi- mum stress in the shaft. 209. A round steel bar 2 in. in diameter, supported at points 4 ft. apart, deflects .029 in. under a central load of 300 Ib. and twists 1.62 in a length of 2| ft. under a twisting moment of 1500 ft.-lb. Find E and G for the material. 210. A steel shaft sub- \ \ ! C I" 1 I 1 B *i ! r FIG. 92 E jected to combined bending and torsion has an elastic limit in tension of 64,600 lb./in. 2 and an elastic limit in shear of 29,170 lb./in. 2 Show that Guest's formula, I rather than Rankine's, applies to p this material. j^, 211. A shaft subjected to com- bined bending and twisting is made of steel for which the elastic limit in tension is 28,800 lb./in. 2 and the elastic limit in shear is 16,000 lb./in. 2 Show that if the bending moment is one half the twisting moment, the shaft will be weakest in shear, whereas if the bending moment is twice the twisting moment, it will be weakest in tension. 212. A cast-iron flanged shaft coupling is connected by eight 1^ in. bolts, the axis of each bolt being 6 in. from the axis of the shaft. The diameter of the * 1 if i jf B' /^A- * 8 ->+< 8 FIG. 93 TORSION 117 shaft is 5 in. Find the shear on each bolt when the maximum shearing stress in the shaft is 9000 lb./in. 2 213. A crank shaft revolves in bearings at A and B, as indicated in Fig. 91. The cranks are in the same plane, and the crank-pin pressures P and Q are assumed to act at right angles to the cranks. If P = 2500 lb., find Q and the reactions of the bearings at A and B. Find also the maximum stress in the crank pin C and draw the bending moment and shear diagrams. 214. A crank shaft revolves in bearings at B and F (Fig. 92) and carries two cranks C and E in the same plane. The shaft transmits a pure torque at the left end, and the crank-pin pressures are assumed to act perpendicular to the plane of the cranks. Find the stresses in the cranks C and E, and in the shaft at B and Z), and draw the bending moment and shear diagrams. 215. A crank shaft revolves in bearings at B and D (Fig. 93), the planes of the two cranks being 90 apart. Taking the dimensions given in the figure, assume P 3000 lb. and find Q, the reactions of the bearings, and the stress in the crank pin C7. SECTION X SPHERES AND CYLINDERS UNDER UNIFORM PRESSURE 71. Hoop stress. When a hollow sphere or cylinder is subjected to uniform pressure, as in the case of steam boilers, standpipes, gas, water, and steam pipes, fire tubes, etc., the effect of the radial pres- sure is to produce stress in a circumferential direction, called hoop stress. In the case of a cylinder closed at the ends, the pressure on the ends produces longitudinal stress in the side walls in addition to the hoop stress. If the thickness of a cylinder or sphere is small compared with its diameter, it is called a shell. In analyzing the stress in a thin shell subjected to uniform pressure, such as that due to water, steam, or gas, it may be assumed that the hoop stress is distributed uniformly over any cross section of the shell. This assumption will be made in what follows. 72. Hoop tension in hollow sphere. Consider a spherical shell subjected to uniform internal pressure, and suppose that the shell is cut into hemispheres by a- diametral plane (Fig. 94). Then, if iv denotes the pressure per unit of area within the shell, the resultant force act- ing on either hemisphere is P = - ? where FIG. 94 d is the radius of the sphere. If p denotes the unit tensile stress on the circular cross section of the shell, the total stress on this cross section is irdlip, approximately, where h is the thickness of the shell. Consequently, ^^ = Trdhp ; whence wd (165) p = , which gives the hoop tension in terms of the radial pressure, 118 SPHERES AND CYLINDERS 119 73. Hoop tension in hollow circular cylinder. In the case of a cylindrical shell, its ends hold the cylindrical part together in such a way as to relieve the hoop tension at either extremity. Suppose, then, that the portion of the cylinder considered is so far removed from either end that the influence of the end constraint can be assumed to be zero. Suppose the cylinder cut in two by a plane through its axis, and consider a sec- tion cut out of either half cylinder by two planes perpendicular to the axis, at a dis- tance apart equal to c (Fig. 95). Then the FlG 95 resultant internal pressure P on the strip under consideration is P = cdw, and the resultant hoop tension is 2 chp, where the letters have the same mean- ing as in the preceding article. Consequently, cdw = 2 clip ; whence (166) P = ^' This result is applicable to shells under both inner and outer pressure, if p is taken to be the excess of the internal over the external pressure. 74. Longitudinal stress in hollow circular cylinder. If the ends of a cylinder are fastened to the cylin- drical part, the internal pressure against the ends produces longitudi- nal stresses in the side walls. In this case the cylindrical part is subjected both to hoop tension and to longitudinal tension. To find the amount of the longitudinal tension, consider a cross section of the cylinder near its center, where the influence of the end restraints can be assumed to be zero (Fig. 96). Then the re- 72 sultant pressure on either end is P = and the resultant longi- tudinal stress on the cross section is Trdlip. Therefore = Trdhp ; whence (167) p = \A IB FIG. 96 ivd 120 RESISTANCE OF MATERIALS This is the same formula as for the sphere, which was to be expected, since the cross section is the same in both cases. 75. Thick cylinders. Lame's formulas. Consider a thick circular cylinder of external radius a and internal radius 5, subjected to either internal or external uniform pressure, or to both simultane- ously, and suppose that a section is cut out of the cylinder by two planes perpen- dicular to the axis at a unit distance FIG. 97 apart (Fig. 97). Now consider a thin ring of the material anywhere in the given section, of external radius r e and internal radius r { . Then, under the strain, r e will become where s e denotes the unit deformation of the fiber, which never exceeds T -oVo ^ or sa ^ e working stresses. Similarly, r i will become where the unit deformation t . is also very small. Since any safe strain produces no appreciable change in the sectional area of the thin ring here considered, by equating its sectional areas before and after strain we have Canceling out the common factor IT and reducing, this becomes .*(?+ 20 = (<? + *O; or, since the unit deformations s e and t . are very small, their squares may be neglected in comparison with their first powers, and con- sequently this expression further reduces to SPHERES AND CYLINDERS 121 Since by Hooke's law the unit stress is proportional to the unit deformation within the elastic limit, if p e denotes the unit stress on the outside fiber of the thin ring, and p { on the inside fiber, then Pe *e r or Pir?=p e r?. Hence, if p h denotes the hoop stress on any element of the thin ring and r the radius of this element, then (168) Pn 1 ^ constant, say C. It should be noted that this relation applies only to a thin ring and not to the thick cylinder as a whole. It may be used, however, to find the change in the hoop stress corresponding to a small change in the radius, that is to say, the difference in the hoop stress on two adjacent fibers, as explained in what follows. Now again consider a thin ring of the material and let its inter- nal radius be r and its thickness Ar. Also, let p h denote the hoop stress in this thin ring, p r the radial stress acting on its inner sur- face, and p r + A/? r the radial stress acting on its outer surface. Then the difference in pressure on the inside and outside of the ring must be equal to the total force holding the ring together ; that is, O r + Ap r ) 2 (r + Ar) - 2 rp r = 2pA r 5 but, since Aft. Ar is infinitesimal in comparison with the other terms, this reduces to (169) P h =Pr + r-j. If the ends of the cylinder are free from restraint, or if the cyl- inder is subjected to a uniform longitudinal stress, the longitudinal deformation must be constant throughout the cylinder. But the lateral action of p r and p h produce longitudinal deformation in accordance with Poisson's law (article 8). Thus, if denotes m Poisson's ratio, the longitudinal deformation due to the action of p r and p h is .&- + -&-, or - (p f +Ph)- Therefore, in order that mE 122 RESISTANCE OF MATERIALS this expression may be constant, p r + p h must be constant. Denoting this constant by &, we have (170) P r + P h = k. Now eliminating p h between equations (168) and (170), we have (t-p,X = G As the radius r increases, the stress p r increases or decreases ac- cording to whether the constant C is positive or negative ; that is, whether the internal pressure is greater or less than the external. Since the sign of C has no effect on the result, we may say that for a point at a distance r -f- Ar from the axis the radial stress is of amount p r Ap r , such that \k - (> r + A?,.)] (r + Ar) 2 = C. Simplifying this expression, it becomes (k - p r ) r 2 + 2 r Ar (k - p r ) - r^p r = C, and subtracting from it the original relation, namely, we have 2 rAr (Tc p r ) r*&p r = 0, Ap r ZrQc-pr) Zp h whence -^ = - ~-^- = -^ ; Ar r 2 r C and, since p h = 5 this becomes Substituting equation (171) in equation (169) and making use of equation (170), we have 2 C 2 C P* ss Pr + -f = *-p* + -jri whence and therefore, from (170), SPHERES AND CYLINDERS 123 If, therefore, the cylinder is subjected to a uniform internal pres- sure of amount w { per unit of area, and also to a uniform external pressure of amount w e per unit of area, then p r = w e when r = a, and p r = tv { when r = b. Substituting these simultaneous values in equation (173), k C k C w = --- w. = --- 1 e 2 a 2 l 2 W a?b 2 (w e w.) k wa 2 wtf whence c= i J. - Hence, substituting these values of C and in equations (172) and (173), they become Pr = (^ -b 2 )r 2 (174) Ph = -^ 7T + (a 2 -b 2 )r 2 which give the radial and hoop stresses in a thick cylinder subjected to internal and external pressure. Equations (174) are known as Lame's formulas. 76. Maximum stress in thick cylinder under uniform internal pressure. Consider a thick circular cylinder which is subjected only to internal pressure. Then w e = 0, and equations (174) become (175) =_!_(-- 1 2 2 2 w,b 2 p= -- ^- r 2 / a 2 b 2 Since p h is negative, the hoop stress in this case is tension. Since p r and p h both increase as r decreases, the maximum stress occurs on the inner surface of the cylinder, where r = b and p r = w { . Hence Clearing; the latter of fractions, we have = - - 5 whence the b 2 p h - w { thickness of the tube, h a b, is given by 124 RESISTANCE OF MATERIALS 77. Bursting pressure for thick cylinder. Let u t denote the ulti- mate tensile strength of the material of which the cylinder is com- posed. Then, from equation (176), the maximum allowable internal pressure w i is obtained from the equation whence (178) w i = "*\ ~ 2 Equation (178) gives the maximum internal pressure u\ which the cylinder can stand without bursting. 78. Maximum stress in thick cylinder under uniform external pressure. Consider a thick circular cylinder subjected only to external pressure. In this case w { = and equations (174) become wa' Since p h is positive, the hoop stress in this case is compression. For a point on the inner surface of the cylinder r = b,p r = Q, and (179) 79. Comparison of formulas for the strength of tubes under uni- form internal pressure. I. Thin Cylinder f-=.O23j. From article 73 the formula for the hoop (or circumferential) stress in a thin circular cylinder is (180) Pft = g, and from article 74, when the ends of the cylinder are closed, the longitudinal stress is (18!) SPHERES AND CYLINDERS 125 The actual stress in a thin cylinder, due to the combination of these two stresses and based on a value of Poisson's ratio = .3, is then found to be* (182) 1>=.425 . II. Thick Cylinder. Lame's Formula. In article 76 the maxi- mum stress in a thick cylinder under uniform internal pressure is given by equation (176) in terms of the radii a and b. If the internal and external diameters of the tube are denoted by d and D respec- tively, then d 2>, D= 2 a, and the formula becomes (183) p = III. Barloitfs Formula. This formula, which is widely used because of its simplicity, assumes that the area of cross section of the tube remains constant under the strain, and that the length of the tube also remains unaltered. As neither of these assump- tions is correct, the formula can give only approximate results. In the notation previously used Barlow's formula is (184) p It is therefore of the same form as the formula for the hoop stress in a thin cylinder, except that it is expressed in terms of the out- side diameter D inside of the inside diameter. From the results of their experience in the manufacture and testing of tubes, the National Tube Company asserts that for any ratio of < .3 Barlow's formula " is best suited for all ordinary calculations pertaining to the bursting strength of commercial tubes, pipes, and cylinders." For certain classes of seamless tubes and cylinders, however, and for critical examination of welded pipe, where the least thickness of wall, yield point of the material, etc. are known with accuracy, and close results are desired, they recommend that the following for- mulas, due to Clavarino and Birnie, be used rather than Barlow's. * Slocum and Hancock, Strength of Materials, Revised Edition, p. 156. 126 RESISTANCE OF MATERIALS IV. Clavarino's formula. In this formula each particle of the tube is assumed to be subjected to radial stress, hoop stress, and longitudinal stress, due to a uniform internal pressure acting jointly on the tube wall and its closed ends. The formula also involves Poisson's ratio of lateral contraction, and is theoretically correct, provided the maximum stress does not exceed the elastic limit of the material. Assuming a value of Poisson's ratio = .3 and using the same notation as above, Clavarino's formula is - * whence (186) V. Birnies formula. This formula is based upon the same assumptions as Clavarino's, except that the longitudinal stress is assumed to be zero. Using the same notation as before and assum- ing Poisson's ratio for steel to be .3, Birnie's formula is 7<l 2 ) whence (188) 10 1> 80. Thick cylinders built up of concentric tubes. From equations (174) it is evident that in a thick cylinder subjected to internal pressure the stress is greatest on the inside of the cylinder and decreases toward the outside. In order to equalize the stress throughout the cylinder and thus obtain a more economical use of material, the device used consists in forming the cylinder of several concentric tubes and producing an initial compressive stress on the inner ones. For instance, in constructing the barrel of a cannon or the cylinder of a hydraulic press the cylinder is built up of two or more tubes. The outer tubes in this case are made of somewhat smaller diameter than the inner tubes, and each is heated until it has expanded sufficiently to be slipped over the one next smaller. In cooling, the metal of the outer tube contracts, thus producing a compressive stress in the inner tube and a tensile SPHERES AND CYLINDERS 127 stress in the outer tube. If, then, this composite tube is subjected to internal pressure, the first effect of the hoop tension thus pro- duced is to relieve the initial compressive stress in the inner tube and increase that in the outer tube. Thus the resultant stress in the inner tube is equal to the difference between the initial stress and that due to the internal pressure, whereas the resultant stress in the outer tube is equal to the sum of these two. In this way the strain is distributed more equally throughout the cylinder. It is evident that the greater the number of tubes used in building up the cylinder, the more nearly can the strain be equalized. The preceding discussion of the stress in thick tubes can also be applied to the calculation of the stress in a rotating disk. For ex- ample, a grindstone is strained in precisely the same way as a thick tube under internal pressure, the load in this case being due to centrifugal force instead of to the pressure of a fluid or gas. 81. Practical formulas for the collapse of tubes under external pressure. A rigorous analysis of the stress in thin tubes, due to external pressure, using Poisson's ratio of transverse to longi- tudinal deformation, gives the formula * = * (-} 4fi-.i:>w or, in terms of the diameter D = 2 , w ~ iU 1 m 2 This formula, however, is based on the assumptions that the tube is perfectly symmetrical, of uniform thickness, and of homogeneous material conditions which are never fully realized in commercial tubes. From recent experiments on the collapse of tubes, f how- ever, it is now possible to determine the practical limitations of this formula and to so modify it, by a method similar to that by *Love, Mathematical Theory of Elasticity, Vol. II, pp. 308-316. t Carman, " Resist, of Tubes to Collapse," Univ., III. Bull., Vol. Ill, No. 17 ; Stewart, "Collap. Press. Lap- Welded Steel Tubes," Trans. A.S.M.E., 1906, pp. 730-820. 128 RESISTANCE OF MATERIALS which the Gordon-Rankine column formula was deduced from Euler's formula (articles 54, 55), as to obtain a rational formula which shall, nevertheless, conform closely to experimental results. By determining the ellipticity, or deviation from roundness, and the variation in thickness of the various types of tubes covered by the tests mentioned above, it is found that by introducing empirical constants the rational formulas can be made to fit experimental re- sults as closely as any empirical formulas, with the advantage of being unlimited in their range of application.* The formula so obtained is , 1QQ , 2EC /h\ 3 f for thin tubes (^ley; w = 1--V~' I -^.023 m? [ D where h = average thickness of tube in inches, D = maximum outside diameter in inches, = Poisson's ratio = .3 for steel, 771 C = .6 9 for lap-welded steel boiler flues, = .76 for cold-drawn seamless steel flues, = .78 for drawn seamless brass tubes. By a similar procedure for thick tubes ( >.023j a practical rational formula has been obtained from Lame's formula (article 75) for this case also, namely, I v [for thick tubes ( D" where u c = ultimate compressive strength of the material, 7jT = .89 for lap-welded steel boiler flues. Only one value of K is given, as the experiments cited were all made on one type of tube. The correction constants C and K include corrections both for ellipticity, or flattening of the tube, and for variation in thickness. * Slocum, " The Collapse of Tubes under External Pressure," Engineering. London, January 8, 1909. Also abstract of same article in Kent, 8th ed., 1910, pp. 320-322. SPHERES AND CYLINDERS 129 Thus, if the correction for ellipticity is denoted by C l and the correction for variation in thickness by C 2 , we have _ minimum outside diameter 1 maximum outside diameter _ minimum thickness average thickness and the correction constants C and K are therefore denned as c=c*c*, By an experimental determination of C 1 and C 2 the formulas can therefore be applied to any given type of tube. 82. Shrinkage and forced fits. In machine construction shrink- age and forced, or pressed, fits are frequently employed for connect- ing certain parts, such as crank disk and shaft, wheel and axle, etc. To make such a connection the shaft is finished slightly larger than the hole in the disk or ring in which it belongs. The shaft is then either tapered slightly at the end and pressed into the ring cold, or the ring is enlarged by heating until it will slip over the shaft, in which case the shrinkage due to cooling causes it to grip the shaft. To analyze the stresses aris- A I D 9 I D l ing from shrinkage and forced fits, let D l denote the diameter of the hole in the ring or disk, j, 98 and D 2 the diameter of the shaft (Fig. 98). When shrunk or forced together, D l must increase slightly and Z> 2 decrease slightly; that is, D l and Z> 2 must of necessity take the same value D. Consequently, the circumference of the hole changes from TT D l to TT D, and hence the unit deformation 8 1 of a fiber on the inner surface of the hole is - 7rD l _D D l i i I) 130 RESISTANCE OF MATERIALS Similarly, the unit deformation s 2 of a fiber on the surface of the shaft is From Hooke's law, E^ we have, therefore, for the unit stress s p 1 on the inside of the disk and for the unit stress p 2 on the surface of the shaft s = Adding these two equations to eliminate the unknown quantity Z>, the result is where K denotes the allowance, or difference in diameter of shaft and hole. For a thick disk or heavy ring this allowance K may be determined from the nominal diameter D of the shaft by means of the following empirical formulas : * For shrinkage fits, K = 1&- 2.. 9 For pressed fits, K = For driven fits, K = For thin rings, however, the allowance given by these formulas will be found to produce stresses in the ring entirely too large for safety. In deciding on the allowance for any given class of work the working stresses in shaft and ring may first be assigned and the allowance then determined from the formulas given below, so that the actual stresses shall not exceed these values. * S. H. Moore, Trans. Am. Soc. Mech. Eng.,Vo\. XXIV. SPHERES AND CYLINDERS 131 From Lame's formulas the stresses p l and p 2 may be obtained in terms of the unit pressure between the surfaces in contact. Thus, from formula (183), the stress on the inside of the hole is 00 where Z> g denotes the outside diameter of the ring, while, by substi- tuting r = a and b = in the equations of article 78, the stresses on 'the outer surface of the shaft are found to be Ph = W > Pr= W ' and consequently P, = w - Eliminating w between these expressions for p l and p^ we have (192) H- fi Now, to simplify the solution, let the coefficient of p 2 be denoted by H; that is, let A'+A* ^ = 5F^' in which case Eliminating p 1 between this relation and the above expression for the allowance /f, we have finally (193) K P 2 = + = Hp In applying these formulas the constant If is first computed from the given dimensions of the parts. If the allowance K is given, the unit stresses p l and p z in ring and shaft are then found from the above. If K is to be determined, a safe value for the stress in the ring p l is assigned, and p z is calculated from the second equation. This value is then substituted in the first equation, and K is calculated. RESISTANCE 'OF MATERIALS APPLICATIONS 216. The outside diameter of a pipe is 4 in., and thickness of wall ^ in. Find the safe internal fluid pressure by Clavarino's formula for a working stress in the steel of 10,000 lb./in. 2 Solution. The thickness ratio in this case is = 2 = 0.125 in. Also, Z> = 4 in., d - 3 in., p = 10,000 lb./in. 2 , and consequently 10(16-9) _ 2 _ lb>/in> 13 x 16 + 4 x 9 217. A cast-iron gear, 8 in. external diameter, 3 in. wide, and lin. internal diameter, is to be forced on a steel shaft. Find the stresses developed, the pressure required to force the gear on the shaft, and the tangential thrust required to shear the fit, that is, to produce relative motion between gear and shaft. Solution. From the formula K = * the allowance is found to be .004 in., J.OUO making the diameter of the shaft J> 2 = 1.754 in. Also, since D l 1.75 in. and J> 3 = 8 in., we have H= 1.1005. Hence, assuming E l = 15,000,000 lb./in. 2 and E 2 = 30,000,000 lb./in. 2 , we have p l = 23,550 lb./in. 2 , p 2 = 21,400 lb./in. 2 To find the pressure required to force the gear on the shaft it is first necessary to calculate the pressure between the surfaces in contact. From the relation p- =w this amounts to w = 21,400 lb./in. 2 The coefficient of friction depends on the nature of the surfaces in contact. As- suming it to be fj. = .15 as an average value, and with a nominal area of contact of TT x If x 3 = 16.497 in. 2 , the total pressure P required is P = 16.497 x 21,400 x .15 = 52,955 Ib. = 26.5 tons. To find the torsional resistance of the fit, we have, as above, Bearing area = 16.497 in. 2 , Unit pressure = 21,400 lb./in. 2 , fjL = .15, radius of shaft = .875 in. Hence the torsional resistance is M t = 16.497 x 21,400 x .15 x .875 = 46,336 in.-lb. Consequently the tangential thrust on the teeth of the gear necessary to shear the 43iiL = n,584 Ib. = 5.8 tons. 218. The outside diameter of a steel pipe is 5^ in., thickness of wall 1 in., and internal fluid pressure 1500 lb./in. 2 Find by Clavarino's formula the maximum fiber stress in the wall of the pipe. 219. The outside diameter of a steel pipe is 8 in., the internal fluid pressure is 2000 lb./in. 2 , and the allowable stress in the steel is 15,000 lb./in. 2 Find the required thickness of pipe wall. 220. Solve problem 216 by the other four formulas listed in article 79 and com- pare the results. SPHERES AND CYLINDERS 133 FIG. 99 221. Find the thickness necessary to give to a steel locomotive cylinder of 22 in. internal diameter if it is required to withstand a maximum steam pressure of 1501b./in. 2 with a factor of safety of 10, using both Lamp's and Clavarino's formulas. 222. In a four-cycle gas engine the cylinder is of steel with an internal diameter of 6 in., and the initial internal pressure is 200 lb./in. 2 absolute. With a factor of safety of 15, how thick should the walls of the cylinder be made, according to Lamp's formula ? 223. The steel cylinder of a hydraulic press has an internal diameter of 5 in. and an external diameter of 7 in. With a factor of safety of 3, how great an internal pressure can the cylinder withstand, according to Lamp's formula ? 224. In a fire-tube boiler the tubes are of drawn steel, 2 in. internal diameter and | in. thick. What is the factor of safety for a working gauge pressure of 200 lb./in. 2 ? 225. How great is the stress in a copper sphere 2 ft. in diameter and .25 in. thick under an internal pressure of 175 lb./in. 2 ? 226. A cast-iron water pipe is 24 in. in diameter and 2 in. thick. What is the greatest internal pressure which it can withstand, according to the formula for thin cylinders ? 227. A wrought-iron cylinder is 8 in. in external diame- ter and 1 in. thick. How great an external pressure can it withstand ? 228. An elevated water tank is cylindrical in form, with a hemispherical bottom (Fig. 99). The diameter of the tank is 20 ft. and its height 52 ft. (exclusive of the bottom). If the tank is to be built of wrought iron and the factor of safety is taken to be 6, what should be the thickness of the bottom plates and also of those in the body of the tank near its bottom ? NOTE. Formulas (165) and (166) give the required thick- ness of the plates, provided the tank is without joints. The bearing power of the rivets at the joints, however, is, in general, the consideration which determines the thickness of the plates (article 90). 229. A marine boiler shell is 16 ft. long, 8 ft. in diam- eter, and 1 in. thick. What is the stress in the shell for a working gauge pressure of 160 lb./in. 2 ? 230. The air chamber of a pump is made of cast iron of the form shown in Fig. 100. If the diameter of the air chamber is 10 in. and its height 24 in., how thick must the walls of the air chamber be made in order to stand a pres- sure of 500 lb./in. 2 , with a factor of safety of 4 ? 231. The end plates of a boiler shell are curved out to a radius of 5 ft. If the plates are f in. thick, find the tensile stress due to a steam pressure of 175 lb./in. 2 232. If the thickness of the end plates in problem 231 is changed to ^ in., the steam pressure being the same, to what radius should they be curved in order that the tensile stress in them shall remain the same ? FIG. 100 134 RESISTANCE OF MATERIALS 233. The cylinder of an hydraulic press is 12 in. inside cliam. How thick must it be in order to stand a pressure of 1500 lb./in. 2 if it is made of cast steel and the factor of safety is 10 ? 234. A high-pressure cast-iron water main is 4 in. inside diameter and carries a pressure of 800 lb./in. 2 Find its thickness for a factor of safety of 15. 235. The water chamber of a tire engine has a spherical top 18 in. in diameter and carries a pressure of 250 lb./in. 2 It is made of No. 7 B. and S. gauge copper, which is reduced in manufacture to a thickness of about .1 in. Determine the fac- tor of safety. 236. A cast-iron ring 3 in. thick and 8 in. wide is forced onto a steel shaft 10 in. in diameter. Find the stresses in ring and shaft, the pressure required to force the ring onto the shaft, and the torsional resistance of the fit. NOTE. Since the ring in this case is relatively thin, assume an allowance of about half the amount given by Moore's formula. Then, having given 7) 2 10 in., /> 3 = 16 in., and having computed the allowance A", we have also D l D 2 K, and, inserting these values in the formulas of article 82, the required quantities may be found, as explained in problem 217. 237. The following data are taken from Stewart's experiments on the collapse of thin tubes under external pressure, the tubes used for experiment being lap- welded steel boiler flues. Compute the collapsing pressure from the rational formula for thin tubes, given in article 81, for both the average thickness and least thickness, and note that these two results lie on opposite sides of the value obtained directly by experiment. OUTSIDE DIAMETER IN INCHES THICKNESS h IN INCHES ACTUAL At place of collapse At place of collapse ING Average Greatest = D Least = <1 Greatest Least Ib./in.a 8. (504 8.610 8.580 0.219 0.230 0.210 870 8.664 8.670 8.625 0.226 0.227 0.204 840 8.665 8.670 8.660 0.212 0.240 0.211 880 8.653 8.665 8.590 0.208 0.220 0.203 970 8.688 8.715 8.605 0.274 0.280 0.266 1430 8.664 8.695 8.635 0.258 0.261 0.248 1320 8.645 8.665 8.635 0.263 0.268 0.259 1590 8.674 8.675 8.675 0.273 0.282 0.270 2030 8.638 8.645 8.615 0.289 0.298 0.280 2200 10.055 10.180 9.950 0.157 0.182 0.150 210 238. The following data are taken from Stewart's experiments on the collapse of thick tubes under external pressure. The ultimate compressive strength of the material was not given by the experimenter, but from the other elastic properties given it is here assumed to be u c = 38,500 lb./in. 2 Compute the collapsing pres- sure from the rational formula for thick tubes, given in article 81, for both average and least thickness, and compare these results with the actual collapsing pressure obtained by experiment. SPHERES AND CYLINDERS 135 OUTSIDE DIAMETER IN INCHES THICKNESS h IN INCHES ACTUAL At place of collapse At place of collapse ING Greatest =D Least = (I Greatest Least lb./in.2 4.010 4.020 3.980 0.173 0.203 0.140 2050 4.014 4.050 3.990 0.178 0.277 0.158 2225 4.012 4.050 3.960 0.173 0.200 0.170 2425 4.018 4.050 4.010 0.184 0.192 0.165 2540 2.997 3.010 2.980 0.147 0.151 0.138 3350 2.987 3.010 2.970 0.139 0.139 0.125 2575 2.990 3.010 2.970 0.190 0.218 0.166 4200 2.996 3.020 2.980 0.191 0.216 0.176 4200 2.997 3.020 2.960 0.190 0.215 0.161 4175 3.000 3.020 2.960 0.182 0.192 0.165 3700 239. What is the maximum external pressure which a cast-iron pipe 18 in. in diameter and \ in. thick can stand without crushing ? 240. Solve problem 226 by Birnie's and Barlow's formulas. SECTION XI FLAT PLATES 83. Theory of flat plates. The analysis of stress in flat plates is at present the most unsatisfactory part of the strength of materials. Although flat plates are of frequent occurrence in engineering con- structions (as, for example, in manhole covers, cylinder ends, floor panels, etc.), no general theory of such plates has as yet been given. Each form of plate is treated by a special method, which in most cases is based upon an arbitrary assumption either as to the danger- ous section or as to the reactions of the supports, and therefore leads to questionable results. Although the present theory of flat plates is plainly inadequate, it is nevertheless of value in pointing out the conditions to which such plates are subject, and in furnishing a rational basis for the estimation of their strength. The formulas derived in the following paragraphs, if used in this way, with a clear understanding of their approximate nature, will be found to be invaluable in designing, or in determining the strength of flat plates. The following has come to be the standard method of treatment and is chiefly due to Bach.* 84. Maximum stress in homogeneous circular plate under uni- form load. Consider a flat, circular plate of homogeneous ma- terial, which bears a uniform load of amount w per unit of area, and suppose that the edge of the plate rests freely on a circular rim slightly smaller than the plate, every point of the rim being maintained at the same level. The strain in this case is greater than it would be if the plate was fixed at the edges, and conse- quently the formula deduced will give the maximum stress in all cases. * For an approximate method of solution see article by S. E. Slocum entitled " The Strength of Flat Plates, with an Application to Concrete-Steel Floor Panels," Engineer- ing News, July 7, 1904. 136 FLAT PLATES 137 Now suppose a diametral section of the plate taken, and regard either half of the plate as a cantilever (Fig. 101). Then if r is the 2 radius of the plate, the total load on this semicircle is - w, and 2 its resultant is applied at the center of gravity of the semicircle, which is at a distance of - from AB. The moment of this result- O 7T 2 A f) 3 ant about the support AB is therefore -w , or - Simi- 2 67T O larly, the resultant of the supporting forces at the edge of the plate is of amount -^- w and is applied at the center of gravity of 2r the semi-circumference, which is at a distance of from AB. The moment of this resultant about AB is therefore irr w z r , or r*w. Hence the total external 2 7T moment M at the support is 2 r*w r 8 w M^Sw- -. Now assume that the stress at any point of the plate is independent of the distance of this point from the center. Under this arbi- trary assumption the stress in the plate is given by the fundamental formula in the theory of beams, namely, Me FIG. 101 If the thickness of the plate is denoted by A, then, since the breadth of the section is b 2 r, bh* rW h Consequently, r\v h Me ~3~" 2 p = = rh whence (194) 138 RESISTANCE OF MATERIALS Foppl has shown that the arbitrary assumption made in deriving this formula can be avoided, and the same result obtained, by a more rigorous analysis than the preceding, and Bach has verified the formula experimentally. Formula (194) is therefore well established both theoretically and practically. 85. Maximum stress in homogeneous circular plate under con- centrated load. Consider a flat, circular plate of homogeneous mate- rial, and suppose that it bears a single concentrated load P which is distributed over a small circle of radius r Q concentric with the plate. Taking a section through the center of the plate and regarding either half as a cantilever, as in the preceding article, the total rim pres- ~p 2 T sure is and it is applied at a distance of from the center. The P * total load on the semicircle of radius r Q is , and it is applied at a dis- 4r tance of - from the section. Therefore the total external moment M 6 7T at the section is _ Pr 2 Pr _ Pr L 2 r 7T 3 7T 7T \ 3r Assuming that the stress is uniformly distributed throughout the plate, the stress due to the external moment M is given by the formula Me p= -. If the thickness of the plate is denoted by A, then rW h /=_ and e = -. Therefore p / % \ -L _[ 1 _ ) Me TT \ 3r/2 whence (195) 3 If r Q = 0, that is to say, if the load is assumed to be concentrated at a single point at the center of the plate, formula (195) becomes (196) FLAT PLATES 139 If the load is uniformly distributed over the entire plate, then r Q = r and P = TTT^W, where w is the load per unit of area. In this case formula (195) becomes p = 2\ IT -s r w d which agrees with the result of the preceding article. 86. Dangerous section of elliptical plate. Consider a homogeneous elliptical plate of semi-axes a and b and thickness 7i, and suppose that an axial cross is cut out of the plate, composed of two strips AB and CD, each of unit width, in- tersecting in the center of the plate, as shown in Fig. 102. Now suppose that a single concen- trated load acts at the intersection of the cross and is distributed to the support in such a way that the two beams AB and CD each deflect the same amount at the center. Since AB is of length 2 a, from article 40, equation (54), the deflection P ( 2 aV at the center of AB is D l ^ ~~- From symmetry, the reac- tions at A and B are equal. Therefore, if each of these reactions is denoted by R^ 2 R^ = P and, consequently, _ ~ Similarly, if R^ denotes the equal reactions at C and D, the deflec- tion D Z of CD at its center is If the plate remains intact, the two strips AB and CD must deflect the same amount at the center. Therefore D l = Z> 2 , and hence (197) -* 140 RESISTANCE OF MATERIALS For the beam AB of length 2 a the maximum external moment is H^t. Also, since AB is assumed to be of unit width, 1= and e = Hence the maximum stress p' in AB is Similarly, the maximum stress p" in CD is I Consequently, p" or, since from equation (197) = R n CL /_ By hypothesis a > b. Therefore p" > p' \ that is to say, the maxi- mum stress occurs in the strip CD (that is, in the direction of the shorter axis of the ellipse). In an elliptical plate, therefore, rupture may be expected to occur along a line parallel to the major axis a result which has been confirmed by experiment. 87. Maximum stress in homogeneous elliptical plate under uniform load. The method of finding the maximum stress in an elliptical plate is to consider the two limiting forms of an ellipse, namely, a circle and a strip of infinite length, and express a continuous relation between the stresses for these two limiting forms. The method is therefore similar to that used in Article 54 in obtaining the modified form of Euler's column formula. Consider first an indefinitely long strip with parallel sides, supported at the edges and bearing a uniform load of amount w per unit of area. Let the width of the strip be denoted by 25 and its thickness by h. Then, if this strip is cut into cross strips of unit width, each of these cross strips can be regarded as an independent beam, the load on one of these unit cross strips being 2bw and the maximum moment at the center being - - FLAT PLATES 141 Consequently, the maximum stress in the cross strips, and therefore in the original strip, is 4 (198) In the preceding article it was shown that the maximum stress in an elliptical plate occurs in the direction of the minor axis. There- fore equation (198) gives the limiting value which the stress in an elliptical plate approaches as the ellipse becomes more and more elongated. For a circular plate of radius b and thickness h the maximum stress was found to be Comparing equations (198) and (199), it is evident that the maxi- mum stress in an elliptical plate is given, in general, by the formula where k is a constant which lies between 1 and 3. Thus, for - = 1 , a (that is, for a circle) k= 1 ; whereas, if - = (that is, for an infinitely long ellipse), k = 3. The constant k may therefore be assumed to have the value 7 *=8-2-, a which reduces to the values 1 and 3 for the limiting cases, and in other cases has an intermediate value depending on the form of the plate. Consequently, (3a- (200) which is the required formula for the maximum stress p in a homo- geneous elliptical plate of thickness h and semi-axes a and b. 88. Maximum stress in homogeneous square plate under uniform load. In investigating the strength of square plates the method of taking a section through the center of the plate and regarding the 142 RESISTANCE OF MATERIALS portion of the plate on one side of this section as a cantilever is used, but experiment is relied upon to determine the position of the dangerous section. From numerous experiments on flat plates Bach has found that homogeneous square plates under uniform load always break along a diagonal.* Consider a homogeneous square plate of thickness h and side 2 a, which bears a uniform load w per unit of area. Suppose that a diagonal section of this plate is taken, and consider either half as a canti- lever, as shown in Fig. 103. Then the total load on the plate is 4wa 2 , and the reac- tion of the support under each edge is wa 2 . If d denotes the length of the diagonal AC, the resultant pres- sure on each edge of the plate is applied at a distance from AC, 7 and therefore the moment of these resultants about AC is 2 (wa 2 ) -, 2 ? 4 or The total load on the triangle ABC is 2 wa 2 , and its result- Li ant is applied at the center of gravity of the triangle, which is at a distance of - from AC. Therefore the moment of the load about b , 2 , AC is (2 wa 2 )-, or - Therefore the total external moment M b o FIG. 103 at the section A C is wa 2 d wa 2 d wa 2 d 236 Hence the maximum stress in the plate is wa 2 d h 2 from which (201) The maximum stress in a square plate of side 2 a is therefore the same as in a circular plate of diameter 2 a. * Bach, Elasticitdt und Festigkeitslehre, 3d. ed., p. 561. FLAT PLATES 143 89. Maximum stress in homogeneous rectangular plate under uniform load. In the case of rectangular plates experiment does not indicate so clearly the position of the dangerous section as 'it does for square plates. It will be assumed in what follows, how- ever, that the maximum stress occurs along a diagonal of the rec- tangle. This assumption is at least approximately correct if the length of the rectangle does not exceed two or three times its breadth. Let the sides of the rectangle be denoted by 2 a and 2 b, and the thickness of the plate by h (Fig. 104). Also let d denote the length of the diagonal AC, and c o a the altitude of the triangle ABC. Now suppose that a diagonal sec- tion AC of the plate is taken, and consider the half plate ABC as a cantilever, as shown in Fig. 104. If w denotes the unit load, the total load on the plate is 4 abw, and consequently the resultant of the reactions of the supports along AB and BC is of amount 2 abw and s* is applied at a distance - from A C. Therefore the moment of the sup- porting force about AC is abwc. Also, the total load on the triangle ABC is 2 abw, and it is applied at the center of gravity of the triangle, which is at a distance of o from AC. Consequently, the total moment of the load about AC is . Therefore the total external moment M at the section AC is FIG. 104 M = abwc 2 abwc abwc 3 3 and the maximum stress in the plate is abwc h _ Me ~3~ ' 2 2 wabc I dtf 12 dtt 144 RESISTANCE OF MATERIALS or, since cd = 4 ab, (202) p = w-, which gives the required maximum stress. For a square plate a = b and c = a V2, and formula (202) reduces to formula (201) for square plates, obtained in the preceding article. APPLICATIONS 241. The cylinder of a locomotive is 20 in. internal diameter. What must be the thickness of the steel end plate if it is required to withstand a pressure of 160 lb./in. 2 with a factor of safety of 6 ? 242. A circular cast-iron valve gate ^ in. thick closes an opening 6 in. in diam- eter. If the pressure against the gate is due to a water head of 150 ft., what is the maximum stress in the gate ? 243. Show that the maximum concentrated load which can be borne by a circular plate is independent of the radius of the plate. 244. A cast-iron manhole cover 1 in. thick is elliptical in form and covers an ellip- tical opening 3 ft. long and 18 in. wide. How great a uniform pressure will it stand ? 245. What must be the thickness of a wrought-iron plate covering an opening 4 ft. square in order to carry a load of 200 lb./ft. 2 with a factor of safety of 5 ? 246. A wrought-iron trap door is 5 ft. long, 3 ft. wide, and | in. thick. How great a uniform load will it bear ? 247. The steel diaphragm separating two expansion chambers of a steam turbine is subjected to a pressure of 150 lb./in. 2 on one side and 80 lb./in. 2 on the other. Find the required thickness for a factor of safety of 10. 248. The cylinder of a hydraulic press is made of cast steel, 10 in. inside diam- eter, with a flat end of the same thickness as the walls of the cylinder. Find the required thickness for a factor of safety of 20. Also find how much larger the factor of safety would be if the end was made hemispherical. Assume w = 1200 lb./in. 2 249. The cylinder of a steam engine is 16 in. inside diameter and carries a steam pressure of 125 lb./in. 2 If the cylinder head is mild steel, find its thickness for a factor of safety of 10. 250. A cast-iron valve gate 10 in. in diameter is under a pressure head of 200 ft. Find its thickness for a factor of safety of 15. 251. A cast-iron elliptical manhole cover is 18 in. x 24 in. in size and is designed to carry a concentrated load of 1000 Ib. If the cover is ribbed, how thick must it be for a factor of safety of 20, assuming that the ribs double its strength ? 252. Thurston's rule for the thickness of cylinder heads for steam engines is h = .00035 wD, where h = thickness of head in inches, D = inside diameter of cylinder in inches, w = pressure in lb./in. 2 Compare this formula with Bach's, assuming the material to be wrought iron and using the data of problem 249. FLAT PLATES 145 253. Show that Thurston's rule for thickness of cylinder head, given in problem 252, makes thickness of head = 1^ times thickness of walls. 254. Nichols's rule for the proper thickness of unbraced flat wrought-iron boiler heads is h = Aw IQp where h = thickness of head in inches, A = area of head in square inches, w = pressure per square inch, 50,000 ultimate strength in tension p = working stress = 10 factor of safety Compare this empirical rule with Bach's formula, using the data of problem 249 and assuming the material to be wrought iron. 255. Nichols's rule for the collapsing pressure of unbraced flat wrought-iron boiler heads is 1U llUt where w = collapsing pressure in lb./in. 2 , h = thickness of head in inches, u t = ultimate tensile strength in lb./in. 2 , A = area of head in square inches. Show that Nichols's two formulas are identical and that therefore they cannot be rational. 256. The following data are taken from Nichols's experiments on flat wrought- iron circular plates. DIAMETER THICKNESS ACTUAL BURSTING IN INCITES IN INCHES PRESSURE LB./IN. S 34.5 T 9 (T 280 34.5 1 200 28.5 3 5 300 26.5 3 370 Using these data, compare Bach's and Grashof's rational formulas with Nichols's and Thurston's empirical formulas, as given below : Circular plate, supported at edge and uniformly loaded. Bach, Grashof, fw h = r\- = .5D \ = .4564D X /-, \p Nichols, Thurston, where 10 p p h = .00035 w D, h = thickness of head in inches, D = diameter of head in inches = 2 r, w = pressure in lb./in. 2 , p = working stress in lb./in. 2 , A = area of head in square inches = Note that the Nichols and Thurston formulas apply only to wrought iron. SECTION XII RIVETED JOINTS AND CONNECTIONS 90. Efficiency of riveted joint. In structural work such as plate girders, trusses, etc., and also in steam boilers, standpipes, and similar constructions, the connections between the various members are made by riveting the parts together. Since the holes for the rivets weaken the members so joined, the strength of the structure is determined by the strength of the joint. Failure of a riveted joint may occur in various ways ; namely, by shearing across the rivet, by crushing the rivet, by crushing the plate in front of the rivet, by shearing the plate (that is, pulling out the rivets), or by tearing the plate along the line of rivet holes. Experience has shown, however, that failure usually occurs either by shearing across the rivet or by tearing the plate along the line of rivet holes. The strength of any given type of riveted joint is expressed by what is called its efficiency, denned as strength of joint Efficiency of riveted joint = strength of unriveted member Thus, in Fig. 105, if d denotes the diameter of a rivet and c the distance between rivet holes, or pitch of the rivets, as it is called, the efficiency, e, of the joint against tearing of the plate along the line of rivet holes is c-d e = c To determine the efficiency of the joint against shearing across the rivets, let q denote the ultimate shearing strength of the rivet and p the ultimate tensile strength of the plate. Then, for a single-riveted lap joint (Fig. 105), if h denotes the thickness of the plate, the area corresponding to one rivet is lie, and the area in shear for 146 RIVETED JOINTS AND CONNECTIONS 147 each rivet is - Consequently the efficiency of this type of joint against rivet shearing is 72 e = 4 chp For an economical design these two efficiencies should be equal. For practical reasons, however, it is not generally possible to make these 1 ,1 >c at*- SINGLE-RIVETED LAP JOINT EFFICIENCY 50-60 PER CENT SINGLE-RIVETED BI:TT JOINT EFFICIENCY 76-78 PER CENT DOUBLE-RIVETED LAP JOINT DOUBLE-RIVETED BUTT JOINT EFFICIENCY 70-72 PER CENT EFFICIENCY 82-83 PER CENT FIG. 105 exactly equal, and in this case the smaller of the two determines the strength of the joint. For a double-riveted lap joint the efficiency against tearing of the plate is G d 148 RESISTANCE OF MATERIALS as above, but since in this case there are two rivets for each strip of length <?, the efficiency against rivet shear is Similarly, for a single-riveted butt joint with two cover plates the efficiency of the joint against tearing of the plate is c d e = - , c and against rivet shear is = ird*q ~ 2chp' For a double-riveted butt joint with two cover plates the efficiency against tearing of the plate is _ c d and against rivet shear is chp The average efficiencies of various types of riveted joints as used in steam boilers are given in Fig. 105. 91. Boiler shells. In designing steam-boiler shells it is customary in this country to determine first the thickness of shell plates, by the following rule : To find the thickness of shell plates, multiply the maximum steam pressure to be carried (safe working pressure in lb./ in. 2 ) by half the diameter of the boiler in inches. This gives the hoop stress in the shell per unit of length. Divide this result by the safe working stress (working stress = ultimate strength, usually about 60,000 lb./in. 2 , divided by the factor of safety, say 4 or 5), and divide the quotient by the average efficiency of the style of joint to be used, expressed as a decimal. The result will be the thickness of the shell plates expressed in decimal fractions of an inch. Having determined the thickness of shell plates by this method, the diameter of the rivets is next found from the empirical formula KIVETED JOINTS AND CONNECTIONS 149 where k= 1.5 for lap joints and k= 1.3 for butt joints with two cover plates. The pitch of the rivets is next determined by equating the strength of the plate along a section through the rivet holes to the strength of the rivets in shear and solving the resulting equation for c. To illustrate the application of these rules, let it be required to design a boiler shell 48 in. in diameter to carry a steam pressure of 125 lb./in. 2 with a double-riveted, double-strapped butt joint. By the above rule for thickness of shell plates we have 125 x - 4 - 5 The diameter of rivets is then d = 1.3 > /T:=.73, say fin. To determine the pitch of the rivets, the strength of the plate for a section of width c on a line through the rivet holes is (c-)hp = (c- |) ^ x 60,000, and the strength of the rivets in shear for a strip of this width is 7T/7 2 Q 4x-^-2 = 7Tp X 40,000. Equating these two results and solving for <?, we have (c _ |) _5_ x 60,000 = TT T 9 g x 40,000, whence c = 4.5 in. As a check on the correctness of our assumptions the efficiency of the joint is found to be c-d 4.5 -.75 e = - = - :-= = .00. c 4.5 92. Structural steel. For bridge and structural work the following empirical rules are representative of American practice : * The pitch (or distance from center to center) of rivets should not be less than 3 diameters of the rivet. In bridge work the pitch should not exceed 6 in. or 16 times the thickness of the thinnest * Given by Cambria Steel Co. 150 RESISTANCE OF MATERIALS outside plates except in special cases hereafter noted. In the flanges of beams and girders, where plates more than 12 in. wide are used, an extra line of rivets with a pitch not greater than 9 in. should be driven along each edge to draw the plates together. At the ends of compression members the pitch should not ex- ceed 4 diameters of the rivet for a length equal to twice the width or diameter of the member. In the flanges of girders and chords carrying floors the pitch should not exceed 4 in. For plates in compression the pitch in the direction of the line of stress should not exceed 16 times the thickness of the plate, and the pitch in a direction at right angles to the line of stress should not exceed 32 times the thickness, except for cover plates of top chords and end posts, in which the pitch should not exceed 40 times their thickness. The distance between the edge of any piece and the center of the rivet hole' should not be less than 1| in. for |-in. and J-in. rivets, except in bars less than 2| in. wide ; when practicable it should be at least 2 diameters of the rivet for all sizes, and should not exceed 8 times the thickness of the plate. Typical illustrations of riveted connections in structural steel work are shown in Figs. 106 and 107. 93. Unit stresses. In structural-steel work it is customary to proportion the various members on the basis of certain specified unit stresses. The following specifications for the greatest allow- able unit stresses represent the best American practice : Axial tension, net section Axial compression, gross section, but not to exceed 17,000 lb./in. 2 I = length of member in inches, r least radius of gyration of member in inches. Compression in flanges of deck plate girders and built or rolled beams, but not to exceed 17,000 lb./in. 2 Compression in flanges of through plate girders, but not to exceed 17,000 lb./in. 2 I = unsupported length of flange in inches, r = least radius of gyration of flange section laterally in inches, 18,000 lb./in. 2 16,000-17- r 18,000 - 70 - 18,000- 70- 177) W KIVETED JOINTS AND CONNECTIONS 151 D = depth from top of girder to bottom of floor beams, d = depth of floor beams back to back of angles or flanges of beams, W '= width center to center of girders. Shear in webs of plate girders, net section Shear in pins and shop-driven rivets Shear in field-driven rivets Tension in extreme fiber of flanges of beams proportioned by moment of inertia, net section Tension or compression in the extreme fiber of pins, assum- ing the stresses to be applied in the centers of bearings . . . Bearing on pins in members not subject to reversal of stress, Bearing on pins in members subject to reversal of stress, using the greater of the two stresses Bearing on shop-driven rivets and stiffeners of girders, and other parts in contact Bearing on concrete masonry Bearing on sandstone and limestone masonry Bearing on expansion rollers in pounds per lineal iiirh, where d = diameter of roller in inches . 13,500 Ib./in. 2 13,500 Ib. /in. 2 10,800 Ib./in. 2 18,000 Ib./in. 2 27,000 Ib./in. 2 24,000 Ib./in. 2 12,000 Ib./in. 2 27,000 Ib./in. 2 500 Ib. /in. 2 400 Ib./in. 2 GOO d. APPLICATIONS 257. In a single-riveted lap joint calculate the pitch of the rivets and the dis- tance from the center of the rivets to the edge of the plate under the assumption that the diameter of the rivets is twice as great as the thickness of the plate. Solution. Consider a strip of width equal to the rivet pitch, that is, a strip con- taining one rivet. Let q denote the unit shearing strength of the rivet and p the unit tensile strength of the plate. Then if h denotes the thickness of the plate, in order that the shearing strength of the rivet may be equal to the tensile strength of the plate along the line of rivet holes, we must have rrd 2 q = (c-d)hp. Since the rivet is usually of better material than the plate, we may assume that the ultimate shearing strength of the rivet is equal to the ultimate tensile strength of the plate ; that is, assume that p q. Under this assumption the above relation becomes whence c = 2.5d, approximately. Similarly, in order that the joint may be equally secure against shearing off the rivet and pulling it out of the plate, that is, shearing the plate in front of the rivet, the condition is FIG. 106. Detail of column riveting with Bethlehem I-beams and H columns 152 ^ i i , O O F 1 oojiq 1 1 q '|oo oo]!"o~!|b ijoo ---O |i O \ c --'^ o Ijoi 7 OHO FIG. 107. Riveted joints in shop-building construction with Bethlehem wide-flange beams used for columns and crane girders 153 154 RESISTANCE OF MATERIALS where a denotes the margin or distance from center of rivets to edge of plate, and q' denotes the ultimate shearing strength of the plate. Assuming that q' = | q and h = - , and solving the resulting expression for a, we have a = 1.5d. 258. Find the required rivet pitch for a single-riveted lap joint with l-in. steel plates and J-in. steel rivets, in order that the joint shall be of equal strength in shear and tension. 2 Solution. For a strip of length c the strength in shear is - g, and in tension is (c d)hp. Hence, to satisfy the given condition, q, Also, efficiency of joint is then sectional area through rivet holes c d .75 e= - = 50 per cent. sectional area unriveted plate c 1.50 259. Determine the maximum diameter of rivet in terms of thickness of plate so that the crushing strength of the joint shall not be less than its shearing strength. Solution. It is customary to assume the ultimate crushing strength of the material as about twice its ultimate shearing strength, or say 100,000 lb./in. 2 Let p c = ultimate crushing strength, p t = ultimate tensile strength, q = ultimate shearing strength. Then, if the shearing strength of the rivets is to equal their crushing strength, we have for lap joints q - dhp c and p c = 2 q ; whence d = 2.54 h. A larger rivet than this will crush before it shears, whereas a smaller one will shear before it crushes. Therefore the crushing strength of a lap joint need not be considered when d < 2.54 A, as it usually is. For a rivet in double shear, assuming that the strength of the rivet in double shear is twice that of the same rivet in single shear,* we have for butt joints jrd 2 2 . q = dhp c and p c = 2 q ; 4 whence d = 1.273ft. If the diameter exceeds this value, the rivet will fail by crushing, whereas if it is smaller, it will fail by shear. Consequently the crushing strength of a butt joint need not be considered when d < 1.273 h. * This seems to be substantiated by experiment, although the English Board of Trade specifies that a rivet in double shear shall be assumed to be only 1.75 times as strong as if in single shear. EIVETED JOINTS AND CONNECTIONS 155 260. Design a double-riveted butt joint to have an efficiency of 75 per cent, using 1-in. steel plates and steel rivets. Solution. For any strip along the joint of length equal to the pitch c, two rivets are in double shear. Hence for equal strength in tension and shear we have Also, = e, whence d = c (I c) and c d = ce. Hence, substituting these values of c d and d in the first equation, the result is cchp - 7rc 2 (l c)' 2 q; whence c = - - -- = 4.586 in., 7T(l-e) 2 2 or say c = 4% in. Then d = c (1 e) = 1.150, or say d = l^V in. From the results of problem 259 it is apparent that the crushing strength of the joint need not be considered, since here d < 1.273/i. The butt straps should apparently each be half as thick as the plate, but when so designed they are found to be the weakest part of the joint. It is therefore customary to make the thickness of the butt straps about | A, or, in the present case, | in. 261. In a double-riveted lap joint the plates are ^ in. thick, rivets ? in. in diameter, and pitch 3 in. Calculate the efficiencies of the joint and determine how it will fail. 262. A boiler shell is to be 4 ft. in diameter, with double-riveted lap joints, and is to carry a steam pressure of 90 lb./in. 2 with a factor of safety of 5. Determine the thickness of shell plates and diameter and pitch of rivets. Also calculate the efficiency of the joint. 263. A cylindrical standpipe is 75 ft. high and 25 ft. inside diameter, with double-riveted, two-strap butt joints. Determine the required thickness of plates near the bottom for a factor of safety of 5, and also the diameter and pitch of rivets. 264. A boiler shell ^ in. thick and 5ft. in diameter has longitudinal, single- riveted lap joints, with 1-in. rivets and 2^-in. rivet pitch. Calculate the maximum steam pressure which can be used with a factor of safety of 5. 265. A cylindrical standpipe 80 ft. high and 20 ft. inside diameter is made of J-in. plates at the base, with longitudinal, single-riveted, two-strap butt joints, connected by 1-in. rivets with a pitch of 3J, in. Compute the factor of safety when the pipe is full of water. 266. Determine the diameter and pitch of rivets required to give the strongest single-riveted lap joint, using |-in. steel plates and steel rivets, and calculate the efficiency of the joint. SECTION XIII REENFORCED CONCRETE 94. Physical properties. The use of concrete dates from the time of the Romans, who obtained a good artificial stone from a mixture of slaked lime, volcanic dust, sand, and broken stone. The modern use of concrete, however, is of comparatively recent develop- ment, its universal use being a matter of only the last quarter of a century, while reenforced concrete is of still more recent origin. Concrete is made by mixing broken stone, varying in size from a walnut to a hen's egg, with clean, coarse sand and Portland cement, using enough water to make a mixture of the consistency of heavy cream. The proportion of these three materials depends on their relative size ; in general, enough sand being needed to fill the voids in the broken stone and enough cement to fill the voids in the sand. The cement and water cause the mass to begin to stiffen in about half an hour, and in from ten to twenty-four hours it becomes hard enough to resist pressure with the thumb. In a month the mixture becomes thoroughly hard, although the hardness continues gradually to increase for some time. Portland cement was invented by Joseph Aspdin of Leeds, Eng- land, who took out a patent for its manufacture in 1824, the name Portland being due to its resemblance to a popular limestone quarried in the Isle of Portland. Its manufacture was begun in 1825, but its use did not become general until 1850, when the French and the Germans became active in its scientific production and succeeded in greatly improving both the method of manu- facture and the quality of the finished product. Portland cement was first brought to the United States in 1865, but not until 1896 did its annual domestic production reach a million barrels. When ordinary limestone (calcium carbonate) is heated to about 800 F., carbon dioxide is driven off, leaving an oxide of calcium called quicklime. This has a great affinity for water, and when 156 REENFORCED CONCRETE 157 combined with it is said to be slaked. Slaked lime when dry falls into a fine powder. Lime mortar is formed by mixing slaked lime with a large pro- portion of sand. Upon exposure to the air this mortar becomes hard by reason of the lime combining with carbon dioxide and forming again calcium carbonate, the product being a sandy lime- stone. Lime mortar is used in laying brick walls and in structures where the mortar will not be exposed to water, since it will not set, that is, combine with carbon dioxide, under water. When limestone contains a considerable amount of clay, the lime produced is called hydraulic lime, for the reason that mortar made by using it will harden under water. If the limestone contains about 30 per cent of clay and is heated to 1000 F., the carbon dioxide is driven off, and the resulting product, when finely ground, is called natural cement. When about 25 per cent of water is added, this cement hardens because of the formation of crystals of calcium and aluminium compounds. If limestone and clay are mixed in the proper proportions, usually about three parts of lime carbonate to one of clay, and the mixture roasted to a clinker by raising it to a temperature approaching 3000 F., the product, when ground to a fine powder, is known as Portland cement. The proper proportion of limestone and clay is determined by finding the proportions of the particular clay and stone that will make perfect crystallization possible. In the case of natural cement the lime and clay are not present in such propor- tions as to form perfect crystals, and consequently it is not as strong as Portland cement. The artificial mixing of the limestone and clay in the manufac- ture of Portland cement is accomplished in different ways. Through- out the north central portion of the United States large beds of marl are found, and also in the same localities beds of suitable clay. This marl is nearly pure limestone and is mixed with the clay when wet. (These materials are also mixed dry.) Both the marl and clay are pumped to the mixer, where they are mixed in the proper proportions. The product is then dried, roasted, and ground. Most American Portland cements, however, are made by grinding a clay-bearing limestone with sufficient pure limestone to give the 158 RESISTANCE OF MATERIALS proper proportions. After being thoroughly mixed, the product is roasted and ground to a powder. Slag cement (puzzolan) is made by thoroughly mixing with slaked lime the granulated slag from an iron blast furnace and then grind- ing the mixture to a fine powder. Slag cements are usually lighter in color than the Portland cements and have a lower specific gravity, the latter ranging from 2.7 to 2.8. They are also somewhat slower in setting than the Portland cements and have a slightly lower tensile strength. They are not adapted to resist mechanical wear, such as would be necessary in pavements and floors, but are suitable for foundations or any work not exposed to dry air or great strain. True Portland cement may be made from a mixture of blast- furnace slag and finely powdered limestone, the mixture being burned in a kiln and the resultant clinker ground to powder. Both the Portland and the puzzolan cements will set under water; that is, they are hydraulic. Gravel or broken stone forms the largest part of the mass of a good concrete and is called the coarse aggregate. Its particles may be from - in. to |- in. in diameter for thin walls or where reenforcement is used, or up to 2i in. for heavy foundations or walls over a foot thick. The coarse aggregate should always be clean and hard. The sand, or fine aggregate, should be clean and coarse ; that is, a large proportion of the grains should measure ^ to 1 in. in diam- eter. All should pass through a screen of i-in. mesh. Too fine a sand weakens the mixture and requires a larger proportion of cement. The following standard proportions may be taken as a guide to the proper mixture for various classes of work:* 1. A rich mixture for columns and other structural parts sub- jected to high stresses or required to be exceptionally water-tight. Proportions 1 : 1| : 3 ; that is, one barrel (4 bags) of packed Portland cement to one and one half barrels (5.7 cu. ft.) of loose sand to three barrels (11.4 cu. ft.) of loose gravel or broken stone. * Taylor and Thompson, " Concrete Plain and Reinforced " ; also, Atlas Portland Ce- ment Co., " Concrete Construction," and Turneaure and Manrer, " Principles of Rein- forced Concrete Construction," p. 10. REENFORCED CONCRETE 159 2. A standard mixture for reenforced floors, beams, and columns, for arches, for reenforced engine or machine foundations subject to vibrations, and for tanks, sewers, conduits, and other water-tight work. Proportions 1:2:4; that is, one barrel (4 bags) of packed Portland cement to two barrels (7.6 cu. ft.) of loose sand to four barrels (15.2 cu. ft.) of loose gravel or broken stone. 3. A medium mixture for ordinary machine foundations, re- taining walls, abutments, piers, thin foundation walls, building walls, ordinary floors, sidewalks, and sewers with heavy walls. Pro- portions 1 : 2J : 5 ; that is, one barrel (4 bags) of packed Portland cement to two and one half barrels (95 cu. ft.) of loose sand to five barrels (19 cu. ft.) of loose gravel or broken stone. 4. A lean mixture for unimportant work in masses, for heavy walls, for large foundations supporting a stationary load, and for backing for stone masonry. Proportions 1:3:6; that is, one barrel (4 bags) of packed Portland cement to three barrels (11.4 cu. ft.) of loose sand to six barrels (22.8 cu. ft.) of loose gravel or broken stone. 95. Design of reenforced concrete beams. Since concrete is a mate- rial which does not conform to Hooke's law and moreover does not obey the same elastic law for tension as for compression, the exact analysis of stress in a plain or reenforced concrete beam would be much more complicated than that obtained under the assump- tions of the common theory of flexure. The physical properties of concrete, however, depend so largely on the quality of material and workmanship, that for practical purposes the conditions do not war- rant a rigorous analysis. The following simple formulas, although based on approximate assumptions, give results which agree closely with experiment and practice. Consider first a plain concrete beam, that is, one without reen- forcement. The elastic law for tension is in this case (see Fig. 108) and for compression ^- = E c . S c To simplify the solution, however, assume the straight-line law of distribution of stress ; that is, assume m l = m 2 = 1. Note, however, 160 RESISTANCE OF MATERIALS that this does not make the moduli equal. Assume also that cross sections which were plane before flexure remain plane after flexure (Bernoulli's assumption), which leads to the relation where e c and e t denote the distances of the extreme fibers from the neutral axis (Fig. 108). Now let the ratio of the two moduli be denoted by n ; that is, let it Then s = ^s = w ^. Pt *& e t For a section of unit width the resultant compressive stress R c on the section is R c = \p c e c , and similarly the resultant tensile stress R t is B t =p t e t . Also, since ^ R c and R t form a couple, R c = R t . Hence FIG. 108 v e p c e c = p t e t , or & = and, equating this v ^ t to the value of the ratio obtained above, we have Pt Since the total depth of the beam h is h = e c 4- e p we have, therefore, e c = h e c Vw, whence , and similarly e t = h 'j= , whence h ^^ Vw Now, by equating the external moment Mto the moment of the stress couple, we have or ^= REENFORCED CONCRETE whence, by solving for the unit stresses p c and p t 161 7 n or, solving one of these two relations for A, say the first, we have 7 ItJ -LVI. /_, / \ ^ = ^ (1 + Vra). For ordinary concrete ^ may be taken as 25. Also, using a factor of safety of 8, the working stress p c becomes p c = 300 lb./in. 2 Substi- tuting these numerical values in the above, the formula for the depth of the beam in terms of the external moment takes the simple form FIG. 109 h = 4 h being expressed in inches, and M in inch-pounds per inch of width of beam. For a reenforced concrete beam the tensile strength of the concrete may be neglected. Let E c and E s denote the moduli of elasticity for con- 771 crete and steel respectively, and let = n. Then, if x denotes the distance of the neutral axis from the top fiber (Fig. 109), the assumptions in this case are expressed by the relations h x and = whence or, solving for x, xli P. + n Pc Now if A denotes the area of steel reenforcement per unit width of beam, then E s = p 8 A and R c = \p c x ; and consequently, since R c = JB g , 162 KESISTANCE OF MATEEIALS Moreover, equating the external moment M to the moment of the stress couple, we have Substituting the value of x in either one of these expressions, say the first, we have whence, solving for For practical work assume w = 15, p c = 500 lb./in. 2 (factor of safety of 5), and p s = 15,000 lb./in. 2 (factor of safety of 4). Substituting these numerical values in the above, the results take the simple form h A = - , x = 6O A, 180 where H denotes the total depth of the beam in inches, d is the diameter of the reenforcement in inches, and M is the external moment in inch-pounds per inch of width. In designing beams by these formulas first find A, then A, and finally H. 96. Calculation of stirrups, or web reenforcement. For a beam reenforced with horizontal rods only, that is, having no vertical or web reenforcement, the ultimate shearing strength is found to be about 100 lb./in. 2 , calculated as the average shearing stress on the cross section. The working stress in shear for the concrete is there- fore assumed to be 25 or 30 lb./in. 2 , equivalent to a factor of safety of 3 or 4. If the average shear on any cross section exceeds 30 lb./in. 2 , vertical, or web, reenforcement is required, usually supplied in the REENFORCED CONCRETE 163 form of stirrups, or loops (Fig. 110). It can be shown that the maximum shear in a beam is inclined at an angle of 45 to the axis of the beam.* Therefore to be effective, vertical stirrups cannot be spaced farther apart than the depth of the beam. In actual prac- tice it is customary to make the distance apart about one half this amount, or - , where h denotes the depth of the beam. Since the maximum shear is inclined at an angle of 45 to the vertical, the effective area of the stirrups is V2 times their cross- sectional area ; but since the maximum shear is also approximately equal to V2 times the average shear, t it is usual simply to design the stirrups to carry the average shear. Longitudinal Section Through Beam and Columns Column Transverse Section Through Beams and Slab FIG. 110 For instance, suppose that the maximum shear on any cross sec- tion of a beam has an average value of 100 lb./in. 2 , which, as shown by the results of tests, is about the maximum limit for good work. Then, assuming that the concrete carries 30 lb./in. 2 of this shear, the vertical stirrups must be designed to carry the remainder, or 70 lb./in. 2 Therefore if the beam is of breadth b and depth A, the total shear to be carried by the stirrups is 70 bh, and consequently for a working stress of 15,000 lb./in. 2 in the steel, the required area is 70 bh 15,000 or .47 per cent of the total area of the cross section. Since the stirrups are usually in the form of a double loop, the required * Slocum and Hancock, Strength of Materials, revised edition, article 28, p. 25, Ginn and Company. t Ibid., article 55, pp. 59 and 60. 164 BESISTANCE OF MATERIALS cross-sectional area of each stirrup rod is .23 per cent of the total area of the cross section. Inclined reenf orcing rods, formed by bending up part of the hori- zontal bottom rods at an angle of 45, are usually too large and too far apart to form an effective web reenf orcement, but a combination of the two, as shown in Fig. 110, constitutes the most effective design. 97. Reenf orced concrete columns. It is seldom necessary to design reenforced concrete columns by the formulas for long columns. In ordinary construction the ratio of length to least width seldom ex- ceeds 12 or 15, while actual tests show that they may be practically considered as short blocks for ratios up to 20 or 25. The strength of a reenforced concrete column considered as a short block will therefore first be determined, and in exceptional cases this result may then be corrected by applying a general column formula. The method of reenforcing concrete columns is either 1. by means of longitudinal rods extending the full length of the column ; 2. by means of hoops or spiral bands ; 3. by a combination of longitudinal rods and hoops or spirals. Let A denote the total cross-sectional area of the column ; A e the area of the concrete ; A s the area of the steel ; and p c , p s the safe unit stresses in the concrete and steel respectively. Then the safe load P for the column is given by P=P r A c +p s A s . The unit deformations of the concrete and steel corresponding to these stresses are -! -I- where E c and E s denote Young's moduli for the concrete and steel respectively. Since the concrete and steel must deform the same amount, s c = *,, and, consequently, Ps ES = lf = n > PC EC or p s = np c , where n denotes the ratio of the i^ro moduli, ordinarily assumed to be 15. REENFORCED CONCRETE 165 It is desirable to express the load P in terms of the total area of the cross section A. For this purpose let Jc denote the percentage of reenforcemeiit, or the ratio of the area of the steel to the total area ; that is, let Then A c = A - A s = A- JcA = A(l- F). Therefore P = p c A ( . + p^A s = p c A (1 k) -f npJcA ; whence 1> = p c A [1 + (n - 1) k] . If the column was plain concrete without reinforcement, its safe load would be P' = p c A. The relative strength of a plain concrete column as compared with one reenforced is therefore Thus, if k = 1 per cent and n = 15, we have that is, a reenforcemeiit of 1 per cent of metal increases the strength 14 per cent. In the case of reenforcemeiit in the form of hoops or spirals, the increase in strength depends on the effect of the hoops or coils in preventing lateral deformation. The results of tests show that this effect is very slight for loads up to the ultimate strength of plain concrete, but beyond this point there is a notable increase in the ultimate strength of the column. Tests on hooped columns made under the direction of Professor A. N. Talbot at the University of Illinois showed that the ultimate strength of the column in terms of the percentage of steel reeiiforcement may be calculated by the formulas for mild steel, p = 1600 + 65,000 k, for high steel, p = 1600 -f 100,000 Jc ; where p denotes the stress per square inch, and k is the percentage of steel with reference to the concrete core inside the hoops. The compressive strength of plai concrete is here assumed to be 1600 lb./in. 2 As regards ultimate strength, the effect of the 166 RESISTANCE OF MATERIALS hoop reenforcement was found to be from 2 to 4 times as great as for the same amount of metal in the form of longitudinal rods. From extensive investigations and experiments on hooped col- umns, Considere has derived the formula where P = ultimate strength of the column, p c = ultimate strength of the concrete, p s = elastic limit of the steel. This formula indicates that hoops or spirals are 2.4 times as effec- tive as the same amount of metal in the form of longitudinal rods. When longitudinal rods are used without hoops, it is necessary to tie them together at intervals to prevent them from buckling and pulling away from the concrete. The distance apart for these horizontal ties may be determined by considering the longitudinal reenf orcing rods as long columns and applying Euler's formula ; namely, ^ EI ~lT Assuming a factor of safety of 5, and taking E= 25,000,000 lb./in. 2 for wrought iron, we have 10 x 25,000,000 x 2 _ T^EI _ _ 64 5P ~ Trd' 2 where d denotes the diameter of the reenforcing rods, from which the unsupported length I of the rods, or distance between ties, is found to be _1750f* For instance, suppose that a concrete column 12 in. square, carry- ing a load of 80,000 lb., is reenforced with four rods | in. in diam- eter placed in the four corners, 1 in. from the outer faces. Then A = 144 in. 2 , A s = 4 (.6013) = 2.4052 in. 2 , REENFORCED CONCRETE 167 Therefore the distance between ties should be 1750- -18 in. 98. Radially reenforced flat slabs. A system of floor construction without the use of beams or ribs, called the " mushroom system," has been devised by Mr. C. A. P. Turner (Fig. 111). The essen- tial features of this system are that the floor slab is of uniform thickness throughout, the reenforcement is radial, and the col- umn top is enlarged and reenforced with hoops. This type of con- struction is best adapted to large areas with few large openings. The following is a simple analysis of the chief features of the design. f* -D- (PI n) (Vertical If || Section) FIG. Ill 99. Diameter of top. In the case of a continuous floor slab sup- ported by several columns, it is obvious that the slab will be con- cave downwards over the columns and concave upward in the center of each panel. Between these two extremes there must be a bound- ary at which there is no curvature, that is, a line of inflection. In a restrained beam of length /, bearing a uniform load, the two points of inflection occur at a distance of 0.212 I from each support (article 45). For a continuous slab, therefore, the line of inflection for square panels may be assumed to be approximately a circle of radius between - I and ^ Z, where I denotes the span, or distance 168 RESISTANCE OF MATERIALS center to center of columns. For practical purposes the diameter D of the column top may therefore be assumed as which is approximately the mean of the above values. There is another condition, however, which also affects the diameter of the top, namely, the distribution of slab reeiiforce- ment. If the column top is too small, there will be portions of the slab which contain no reeiiforcement, as shown by the triangular areas a, 6, c, d (Fig. 112). The arrangement shown in Fig. 113, however, has no such gaps ; it requires that D 2 + D 2 = (I - D) 2 , o] o :o :o FIG. 112 FIG. 113 or D = = = 0.414 Z, which determines the minimum diameter of column top. If, then, D is assumed as L> = l = 0.4375^, 16 a slight overlap of the reenforcing rods is assured. 100. Efficiency of the spider hoops. The efficiency of tensile re- enforcement in the form of hoops as compared with direct reen- forcement may be obtained approximately as follows : In the case of column tops as here considered, take a diametral section of the top and consider half of one hoop and the portion of the material reenforced by this segment, as shown in Fig. 114. Let w denote the radial stress, or pressure on the inside of the hoop per unit of length, expressed in pounds per linear inch of hoop. Also let r denote the radius of the hoop, A its cross-sectional area, and p the unit stress in the metal. Then the radial force acting on any portion of the hoop of length As is wAs, and the component of this force perpendicular to the plane of the section is wAs sin a. Or, if A# denotes the projection of As on the diameter, then As sin a = A#, and this component of the force therefore becomes BEENFORCED CONCRETE 169 'w&x. Therefore, equating the tension in the hoop to the sum of the components of the radial stress perpendicular to the plane of the section, we have whence wr P Now let A' denote the cross-sectional area of an equivalent amount of radial reenforcement. Then, since the length of the arc considered is TTT and the radial stress is of amount w per unit of length, the total amount of radial reenforcement re- quired would be given by the equation pA' = TTTW ; whence A 1 = 7TTW P Comparing these expres- sions for A and J/, it is found that A i _ FIG. 114 Consequently, the theoretical efficiency of tensile reenforcement in the form of hoops is 3.14 times as great as the same cross- sectional area of direct, or radial, reenforcement. The amount of metal in a hoop of radius r, however, is 2 irrA, whereas that in the radial reenforcement is A r 2 r, and since A r irA, these volumes are equal. Consequently, there is no saving in material effected by making the reenforcement in the form of hoops. But when there is such a complex system of reenforcement as that shown in Fig. Ill, some of the metal may be used to better advantage in the form of hoops, as this lessens somewhat the congestion of metal at the columns. 101. Maximum moment. For a continuous beam of span Z, carrying a total uniform load of amount W, the moment at the Wl supports is ; whereas the moment at midspan is one half this 170 RESISTANCE OF MATERIALS Wl amount, or . Assuming that each of the four sets of slab rods 2i< carries one fourth the total load, the bending moment from which to determine the thickness of the slab and the amount of reenf orce- ment in the head becomes Wl where W denotes the total load on the panel, and I is the distance center to center of columns. The formula determined by experi- Wl 7 ment and used in practice is M = . For a top of diameter D = Z, oO 16 Wl 7 the moment per foot of width, say Jtf , is therefore M = -s- Z, 50 16 Wl 7 7 W . ,, Jf ='= ft - lb - 102. Thickness of slab. Let p s unit working tensile stress in reenforcement, p c = unit working compressive stress in concrete, 7? n = j- = ratio of elastic moduli, M l = bending moment in foot-pounds per foot of width. Then the thickness of slab h from the outer fiber in compres- sion to the center of the reenforcement is given by the formula (article 95) For working values of p g = 16,000 lb./in. 2 , p c = 600 lb./in. 2 , and n = 15 this becomes h= 0.1026 Vjf r W Since M l = , as explained above, this may also be written 22 h= 0.02187 VJF, where h is expressed in inches and W denotes the total load on the panel in pounds. KEENFORCED CONCRETE 171 The total thickness of slab is then found by adding to this value of h the amount needed for the reenf orcing rods plus a small amount for bond below the bottom rods. The amount of reenforcement for the unit stresses assumed above is then found from the relation (article 95) A = 0.081 A, where h is expressed in inches and A in square inches per foot of width. 103. Area of slab rods. As stated above, the moment at the center of the slab is half as great as at the supports. The effect of this on the required dimensions of the slab and reenf orcement would be to divide both h and A by V2. But since the slab is necessarily of the same thickness throughout and hence is thicker at the center than necessary, the cross-sectional area of the reenforcement in the slab may be lessened, so as to make the moment of the stress in the reenforcement equal to the moment required for the thinner slab. If, then, p denotes the unit stress in the metal, where p is assumed to be the same in both cases, and A' denotes the cross-sectional area of metal actually required, the condition that the moment shall be constant is , kpA'li = kp ; r V2 V2 whence A' = . Consequently, the design is made up by placing half the required cross-sectional area, obtained from the formula A = 0.081 A, in the slab rods and the other half in the hoops and spider in the column top. 104. Application of formulas. To illustrate the use of these formulas, consider a floor system with panels 20 ft. square, carry- ing a live load of 200 lb./ft. 2 By a preliminary calculation it is found that the floor slab will be about 9 in. thick, giving a dead load of 115 lb./ft. 2 The total live and dead load is therefore 315 lb./ft. 2 Consequently, W=W x 315-126,000 lb., and h = 0.02187 Vl26,000 - 7.76 in. 172 RESISTANCE OF MATERIALS The total area of reinforcement in the head for each radial sys- tem is then A = 0>ogl h = Q>628 sq> in> per foot Since the diameter of the top is assumed as D = -^ 1= 8.75 ft., the total area required for one radial system is 8.75^4 = 5.5 sq. in. The required area of slab rods in one system is then -^-= 2.75sq. in., equivalent to 18 rods T 7 g in. in diameter, spaced 5 in. apart. Since the four sets of rods overlap where they cross the column top, and since h denotes the distance from the extreme fiber in com- pression to the center of the reenforcement, the total thickness of^ slab becomes h + 2 d 4- |- in. = 9 in. Since the hoops around the column head are assumed to be TT times as effective as the same cross-sectional area of radial reen- forcement, the total area of the hoops, neglecting the spider, is 2 75 YJ= 0.88 sq. in. If two hoops are used, the diameter of each hoop rod may therefore be assumed as | in., giving a total cross- sectional area of 0.88 sq. in. It should be noted, however, that the effectiveness of hoop reen- forcement depends on the hoop being placed where it can carry the tensile stress in the column top. Since the outermost hoop of the top is placed as nearly as possible on the line of inflection, there is practically no stress in the slab at this point, except shear, and hence the cross-sectional area of the outer hoop should be neglected in dimensioning the top hoops. 105. Dimension table. By the use of the above formulas as just explained, the accompanying table has been calculated, giving the required thickness of slab and also the size and number of slab rods for various spans and loads. In this table the thickness of concrete below the bottom of the rods has been assumed as about i in., which has been proved by experiment to be sufficient for fireproofing purposes. Some fire- proofing specifications require more, however, in which case it will be necessary to increase the thickness of slab given in the table to the required amount. &EENFORCED CONCRETE 173 ' r4 - - ^' ,; s ^ s I Q _ 2 - i .a 1 q ' e - -% p t rJ ' .S ,jQ 4t o T* r*- 0^ rH ^ -h .3 ^ ?' o' ?' H<M (7<| TfH "7 ' 1 PH|(^ S"g i - j -H ^ ^ HJ * - H r; v. * . ^ P a 1 o .a' -3 j -7 te a j d s 7 5 7 X 7 j i=c 7 s 7^ "v !-,' 8 ri* CO CO rH t^ rH CO rH rH OS rH rH !1" * r4 fH ^S " i .a ^ P a ,0 *""* '7 "Tc _ -7 'cc ^-j |- ^ ,Q t,^ .g d ccK H* O <M CD CO rH J i s .s o ^r \ o *o GO r-i rH . - LJ ' . M - h4 ^ ^ c3 ^ M 08 H M C3 f-H ^ /^ r* \ "IS P ^ P a -3 p -3 'cc " -3 O . t * B *^p |co ,Q .to r 1 .3 5 Kl H ^Q c^|oo 2 ^ '^ |r- .2 ^ ^ | HM 10 10 ^ht 1 o t-- IO !> rH O 00 TH t^ OS rH CO rH rH N i H !J i H - ^ H *' ^ ^ "V ^ P '7 "oS p d ^5 P a W "7 CO H .3 ^ ^H S ,0 ' S ,Q ^^ ^ H* O O IO t- rH 4S ^ CO CO rH '7 o co t^ OS rH '7 i | COW 10 CO t* OS rH fH . ,0 ,4 -4 - r4 ^* V, "I ^' 1 3 P a CO . -rH 1 q ' .S a P -7 rH ' t, 1 ^ 5 1 1 ,Q *~'\-+ . ^O jr x _S ,Q ,v;Jx I '7 10 oo CO t- rH -<l?< lO CO CO CO rH t^ OS rH , ^ . J tl . tl N - o ^j c^ ^"1 4S . O ^"^ " X. "oS . "to 12 .3 ^ P .d 3 P "7 02 . to j5 ^0 C H .3 -d ^' x .3 rd rtl* '7 * | 00 3 S S lO t^- rH H-* O iO CO CO rH 5 i if -^ ^ 43 ^J -(J ^ 4^ Q rH CO ^ ^4H o_| rH CO 3 x f x 2* - f x 3 r* rH a rH a rH a 5 rH s a .g ( S S ' a a S 1 PH cJ 3 PH Q a r^ c3 O PH (< a "^^ S 'o ^ PH 174 RESISTANCE OF MATERIALS 106. Dimensions of spider. The size and number of rods in the spider are determined from the condition that the total area shall be sufficient to carry the shear, as will now be shown. It is customary to place a fillet at the top of each column, as shown in Fig. Ill, the depth of the fillet being not less than the thickness of the slab, and its diameter about twice the diameter of the column. The area of concrete in shear at the face of the column is then 7r^(2f), where d denotes the diameter of the column and t the thickness of the slab, and at the outside of the fillet is 7r(2 d)t, which is the same amount. Consequently, a fillet of these dimen- sions doubles the area of concrete in shear. The chief purpose of the fillet, however, is to avoid the weakening effect of an angle and thus enable the slab to develop its full strength at its junction with the column. Since there are more slab rods to carry the shear at the outside of the fillet than at the face of the column, the latter section is weakest in shear. To illustrate the method of dimensioning for shear, consider the same numerical problem as above ; namely, a panel 20 x 20 ft. with a total live and dead load of 320 lb./ft. 2 The total load on each column, or the total shear, is then 320 X 400=128,000 Ib. Assuming that the columns supporting the floor are 1 ft. in diameter, and taking a cylindrical section at the face of the column, the total area of concrete in shear, includ- ing the fillet, will be TT x 12 x 2 x 9.5 = 716 sq. in. For a working stress in shear of 30 lb./in. 2 , the shearing strength of the concrete alone is therefore 716 x 30 = 21,480 Ib. The amount of metal in the slab rods was determined previously as 0.628 in. 2 /ft. Since the column is 1 ft. in diameter, and since each set of rods is in double shear and there are four sets of rods, the total area of metal in shear at the surface of the column is 8 x 0.628 = 5 sq. in. Assuming the working stress in shear of the metal as 10,000 lb./in. 2 , the shearing strength developed by the slab rods alone is 5 x 10,000 = 50,000 Ib. Since the total load is 126,000 Ib., there still remains to be taken care of 126,000 - (21,480 + 50,000)- 54,520 Ib. of shear. To design the spider to carry this shear, assume that it is made up of 8 rods, as shown in Fig. 111. Then the amount of shear on REENFOBCED CONCRETE 175 each rod will be - = 6815 lb., and hence the required area of 8 each rod will be ^77777:7: =0.68 sq. in., giving a rod slightly less than 1 in. in diameter. If the fillet under the head is neglected in computing the shearing strength of the concrete, a spider made up of eight 1-in. rods will still give sufficient area to develop the required shearing strength. APPLICATIONS 267. A plain concrete beam 6 in. x 6 in. in cross section, and with a 68-in. span, is supported at both ends and loaded in the middle. The load at failure is 1008 lb. Find the maximum fiber stress. 268. A concrete building block 24 in. in length and having an effective cross section of 8 in. x 10 in. minus 4 in. x 6 in. is tested by being supported at points 2 in. from each end and loaded in the middle. The load at failure is found to be 5000 lb. Find the maximum fiber stress, the height of the block being 10 in. 269. A reenforced concrete beam 10 in. wide and 22 in. deep has four 1^-in. round bars with centers 2 in. above the lower face. The span is 16 ft. The beam is simply supported at the ends. Find the safe load per linear foot for a working stress in the concrete of 500 lb./in. 2 , and also find the tensile stress in the reenforcement. 270. A reenf orced-concrete beam of 16 ft. span is 12 in. wide and has to support a uniform load of 1000 lb. per linear foot. Determine the total depth and amount of steel reenforcement required, bars to have centers 2 in. above the lower face of beam. 271. A reenforced concrete beam 8 in. x 10 in. in cross section, and 15 ft. long, is reenforced on the tension side by six i-in. plain steel rounds. The steel has a modulus of elasticity of 30,000,000 lb./in. 2 and the center of the reenforcement is placed 2 in. from the bottom of the beam. Assuming that E c = 3,000,000 lb./in. 2 , and_p c = 600 lb./in. 2 , find the position of the neutral axis and the moment M. 272. For a stress p c 2700 lb./in. 2 on the outer fiber of concrete in the beam given in problem 271, find the stress p s in the steel reenforcement. 273. A concrete beam is 10 x 16 in. in cross section and 20 ft. long. It is reen- forced with four 5-in. steel rods with centers 2 in. above the lower face of the beam. The safe compressive strength of the concrete is 600 lb./in. 2 , and the steel used has an elastic limit of 40,000 lb./in. 2 What single concentrated load will the beam carry at its middle ? What tension will be developed in the steel ? What shearing stress along the reenforcement ? 274. Find what load, uniformly distributed, the beam in the preceding prob- lem will carry and find the tension in the steel and bond for this case. 275. A reenforced concrete floor is to carry a load of 200 lb./ft. 2 over panels 14 ft. square. Find the required thickness of the slab and the area of the reenforce- ment for working stresses of 500 lb./in. 2 in the concrete and 15,000 lb./in. 2 in the reenforcement. 276. Design a floor panel 14 ft. square, to be made of reenforced concrete and to sustain a total uniform load of 120 lb./ft. 2 , with a factor of safety of 4. SECTION XIV SIMPLE STRUCTURES 107. Composition and resolution of forces. It will now be neces- sary to recall some of the results previously obtained concerning the composition and resolution of forces. It was shown in article 11 that any number of concurrent forces may be combined by means of a vector triangle or vector polygon into a single resultant. Also that, conversely, any force may be resolved into components forming with the given force a closed triangle or polygon. In finding the resultant of several forces it is usually more convenient to resolve each of the given forces into components parallel to a set of rectangular axes, then take the algebraic sum of the components along each axis, and, finally, recombine these into the required resultant. Thus, in Fig. 115, if F^ F^ denote two forces and R their result- ant, resolve F 1 into rectangular components x^ y^ and F Z into com- ponents x^ y z . Then, if x, y, denote the components of the resultant J?, we have and, consequently, FIG. 115 In article 10 the moment of a force with respect to any point was defined as the product of the force by its perpendicular distance from the point in question ; that is, Moment = force X lever arm. 176 SIMPLE STRUCTUEES 177 It was also proved that the sum of the moments of any number of forces lying in the same plane with respect to a point in this plane is equal to the moment of their resultant with respect to this point. It remains to consider the case when the system of forces lie in the same plane but are not concurrent, that is, do not all meet in a point. This involves the properties of a force couple, denned as two equal and opposite parallel forces F, F, not acting in the same line (Fig. 116). For any couple F, F, let x denote the distance of any point in its plane from the nearest force of the couple, and d the lever arm of the couple (Fig. 116). Then the moment M of the couple with respect to the point is Therefore the moment of the couple is constant and equal to Fd with respect to any point in its plane. Moreover, since the moment of the couple involves FlG 116 only the magnitude of the forces and their distance apart, it is evident that the couple can be revolved through any angle without altering its value. A couple may, there- fore, be moved about anywhere in its plane without altering its numerical value or changing its effect in any way. It is also obvious that the forces of a couple may be altered in amount, provided that the lever arm is at the same time changed so as to keep their product constant. Two or more couples may therefore be combined by first reducing them to equivalent couples having the same lever arm and then taking the algebraic sum of the forces, thus giving a single resultant couple with this same lever arm. Now consider any number of forces F^ F^ F 8 , lying in the same plane but not concurrent. At any arbitrary point 0(Fig. 117), intro- duce two forces F^, F", opposite in direction, but each equal in amount to F^ Since Ff and F" are equal and opposite they will not disturb the equilibrium of the system. But F l and F" together form a couple of moment F I d^ leaving the single force F^, equal 178 RESISTANCE OF MATERIALS to F^ acting at 0. Similarly, each of the other forces is equivalent to a couple plus a single force (equal and parallel to the given force) acting at 0. The given force system is therefore equivalent to a system of equal but concur- rent forces acting at 0, and an equal number of couples, the moment of each couple being equal to the moment of the cor- responding given force with re- spect to the point 0. This concurrent force system, however, may now be combined . 117 into a single resultant force, and the couples also combined into a single resultant couple, as just explained. Consequently, we have the following general theorem: Any system of forces lying in the same plane is equivalent to a single force acting at any assigned point in this plane plus a couple whose moment is equal to the sum of the moments of the given forces with respect to this point. 108. Conditions of equilibrium of a system of coplanar forces. When a body acted upon by two or more forces is at rest or in uniform motion relative to any system of coordinate axes, it is said to be in equilibrium, and the forces acting on it are said to equilibrate. The conditions for equilibrium are, therefore, that the resultant force acting on the body must be zero, and that the result- ant moment or couple acting on it must also be zero. That is to say, the algebraic sum of all the forces acting on the body must be zero, and the algebraic sum of the moments of these forces with respect to any point must also be zero. Expressed symbolically the conditions of equilibrium are In general it is convenient in applying these conditions to resolve each force F into rectangular components X, Y, and replace the single condition ^ F = by the two independent conditions V X = 0, SIMPLE STRUCTURES 179 These three conditions, ^X= 0> 2) Y== ^ ^ M== ^ are obviously both necessary and sufficient to assure equilibrium. For if the first two are satisfied, the system will be in equilibrium as regards translation, and if ^?M= 0, it will also be in equilibrium as regards rotation ; and, furthermore, it will not be in equilibrium un less all three are satisfied. The conditions for equilibrium of a system of forces lying in the same plane may then be reduced to the following convenient form : 1. For equilibrium against translation, I V horizontal components = O, j V vertical components = O. 2. For equilibrium against rotation, ^P moments about any point = O. When a body is acted on by only three forces, lying in the same plane, the conditions for equilibrium are that these three forces shall meet in a point, and that one of them shall be equal and opposite to the resultant of the other two. 109. Equilibrium polygon. As explained above, the resultant of any system of forces lying in the same plane may be found by means of a vector force polygon, the resultant being the closing side of the polygon formed on the given system of forces as adjacent sides. Although this construction gives the magnitude and direction of the resultant, it does not determine its position or its line of action. The most convenient way to determine the line of action of the resultant is to introduce into the given system two equal and oppo- site forces of arbitrary amount and direction, such as P' and P" (Fig. 118). Since P' and P" balance one another, they will not affect the equilibrium of the given system. To find the line of action of the resultant R, combine P' and P into a resultant R l acting along B'A', parallel to the corresponding ray OB of the force polygon. Prolong A'B' until it intersects P 2 and then combine R I and P z into a resultant R Z acting along C'B', parallel to the corre- sponding ray OC of the force polygon, etc. Proceed in this way until the last partial resultant R is obtained. Then the resultant 180 RESISTANCE OF MATERIALS of P' and R^ will give the line of action, as well as the magnitude, of the resultant of the original system P^ P 2 , P z , P v The closed figure A'B' C'D'E'F 1 obtained in this way is called an equilibrium polygon. E FIG. 118 For a system of parallel forces the equilibrium polygon is con- structed in the same manner as above, the only difference being that in this case the force polygon becomes a straight line (Fig. 119). FIG. 119 Since P' and P" are entirely arbitrary in both magnitude and direction, the point 0, called the pole, may be chosen anywhere in the plane. Therefore, in constructing an equilibrium polygon SIMPLE STRUCTURES 181 corresponding to any given system of forces, the force polygon ABODE (Figs. 118 and 119) is first drawn, then any convenient point is chosen and joined to the vertices A, B, C, D, E, of the force polygon, and finally the equilibrium polygon is constructed by drawing its sides parallel to the rays OA, OB, OC, etc. of the force diagram. Since the position of the pole is entirely arbitrary, there is an infinite number of equilibrium polygons corresponding to any given set of forces. The position and magnitude of the resultant E, however, is independent of the choice of the pole, and will be the same no matter where is placed. For a system of concurrent forces (that is, forces which all pass through the same point) the closing of the force polygon is the necessary and sufficient condition for equilibrium. If, however, the forces are not concurrent, or if they are parallel, this condition is necessary but not sufficient, for in this case the given system of forces may be equivalent to a couple, the effect of which would be to produce rotation. To assure equilibrium against rotation, there- fore, it is also necessary that the equilibrium polygon shall close. The graphical and analytical conditions for equilibrium are then as follows : CONDITIONS OF EQUILIBRIUM Analytical Graphical Translation 2>=o Force polygon closes Rotation 2^=o Equilibrium polygon closes 110. Application of equilibrium polygon to determining reactions. One of the principal applications of the equilibrium polygon is in de- termining the unknown reactions of a beam or truss. To illustrate its use for this purpose, consider a simple beam placed horizontally and bearing a number of .vertical loads P, P 2 , etc. (Fig. 120). To determine the reactions E^ and R^ the force diagram is first con- structed by laying off the loads I, 7J, etc. to scale on a line AF, choosing any convenient point as pole and drawing the rays OA, OB, etc. The equilibrium polygon corresponding to this force diagram is then constructed, starting from any point, say A', in R^. 182 RESISTANCE OF MATERIALS Now the closing side A' G' of the equilibrium polygon determines the line of action of the resultants P' and P" at A' and G' respec- tively. For a simple beam, however, the reactions are vertical. Therefore, in order to find these reactions, each of the forces P' and P" must be resolved into two components, one of which shall be vertical. To accomplish this, suppose that a line OH is drawn from the pole in the force diagram parallel to the closing side G'A' of the equilibrium polygon. Then HO (or P') may be replaced by its components HA and A 0, parallel to R^ and A'B' respectively ; B FIG. 120 and similarly, OH may be replaced by its components FH and OF, parallel to R Z and F f G' respectively. HA and FH are therefore the required reactions. 111. Equilibrium polygon through two given points. Let it be required to pass an equilibrium polygon through two given points, say M and N (Fig. 121). To solve this problem a trial force diagram is first drawn with any arbitrary point as pole, and the corresponding equilibrium polygon M A'B' C'D'E' constructed, starting from one of the given points, say M. The reactions are then determined by drawing a line OH parallel to the closing side ME' of the equilibrium polygon, as explained in the preceding article. The reactions, however, are independent of the choice of the pole in the force diagram, and, consequently, they must be of amount AH and HE, no matter where is placed. Moreover, if the equilibrium SIMPLE STRUCTURES 183 polygon is to pass through both M and N, its closing side must coincide with the line MN, and therefore the pole of the force dia- gram must lie somewhere on a line through H parallel to MN. FIG. 121 Let 0' be any point on this line. Then, if a new force diagram is drawn with 0' as pole, the corresponding equilibrium polygon starting at M will pass through N. 112. Equilibrium polygon through three given points. Let it be required to pass an equilibrium polygon through three given points, say M, N, and L (Fig. 122). Fig. 122 As in the preceding article, a trial force diagram is first drawn with any point as pole, and the corresponding equilibrium polygon constructed, thus determining the reactions R l and R^ as HA and EH respectively. 184 RESISTANCE OF MATERIALS Now, if the equilibrium polygon is to pass through N, the pole of the force diagram must lie somewhere on a line HK drawn through H parallel to MN, as explained in the preceding article. The next step, therefore, is to determine the position of the pole on this line HK, so that the equilibrium polygon through M and N shall also pass through L. This is done by drawing a vertical LS through L and treating the points M and L exactly as M and N were treated. Thus OABCD is the force diagram for this portion of the original figure, and MA'B' C' S is the corresponding equilibrium polygon, the reactions for this partial figure being H'A and DH'. If, then, the equilibrium polygon is to pass through L, its closing side must be the line ML, and consequently the pole of the force diagram must lie on a line H'K' drawn through H' parallel to ML. The pole is therefore completely determined as the intersection 0' of the lines HK and H'K'. If, then, a new force diagram is drawn with 0' as pole, the corresponding equilibrium polygon starting from the point M will pass through both the points L and N. Since there is but one position of the pole 0', only one equi- librium polygon can be drawn through three given points. In other words, an equilibrium polygon is completely determined by three conditions. 113. Application of equilibrium polygon to calculation of stresses. Consider any structure, such as an arch or arched rib, supporting a system of vertical loads, and suppose that the force diagram and equilibrium polygon are drawn as shown in Fig. 123. Then each ray of the force diagram is the resultant of all the forces which precede it and acts along the segment of the equilibrium polygon parallel to this ray. For instance, OC is the resultant of all the forces on the left of P z and acts along C'D'. Consequently, the stresses acting on any section of the structure, say mn, are the same as would result from a single force OC acting along C'D'. Let 9 denote the angle between the segment C'D' of the equi- librium polygon and the tangent to the arch at the point S. Then the stresses acting on the section mn at S are due to a tangential thrust of amount OC cos 6 ; a shear at right angles to this, of amount OC sin 6 ; and a moment of amount OC d, where d is the perpen- dicular distance of C'D' from S. SIMPLE STRUCTURES 185 From Fig. 123 it is evident that the horizontal component of any ray of the force diagram is equal to the pole distance OH. There- fore, if OC is resolved into its vertical and horizontal components, the moment of the vertical component about S is zero, since it passes through this point ; and hence the moment OC d = OH z, where z is the vertical intercept from the equilibrium polygon to the center of moments S. Having determined the moment at any given point, the stresses at this point can easily be calculated. / B 1) 114. Relation of equilibrium polygon to bending moment diagram. In the preceding article it was proved that the moment acting at any point of a structure is equal to the pole distance of the force diagram multiplied by the vertical intercept on the equilibrium polygon from the center of moments. For a system of vertical loads, however, the pole distance is a constant. Consequently, the moment acting on any section is proportional to the vertical intercept on the equilibrium polygon from the center of moments. Therefore, if the equilibrium polygon is drawn to such a scale as to make this factor of proportionality equal to unity, the equilibrium polygon will be identical with the bending moment diagram for the given system of loads. 115. Structures : external forces. The external forces acting upon any stationary structure must be in equilibrium. Hence they may be found, in general, by applying the conditions of equilibrium given in article 109. The conditions of equilibrium may be applied either analytically or graphically. The former method has the ad- vantage of being available under all circumstances ; whereas the 186 RESISTANCE OF MATERIALS FIG. 124 latter method requires the accurate use of instruments, and is therefore confined chiefly to office work. Both methods are illus- trated in what follows. 1. Analytical method. Consider first the analytical determination of the external forces acting on a simple structure, such as the loaded jib crane, shown in Fig. 124. This consists of a vertical >A mast ED, supported by a collar B and footstep (7, and carrying a jib AD, supported by the guy AEF. The external forces acting on the crane are the load W, the counterweight W^ (including hoisting engine and machinery), and the reactions at B and C. The reaction of the collar B can have no vertical component, as the collar is made a loose fit so that the crane may be free to swivel. For convenience, the reaction of the footstep C may be replaced by its horizontal and vertical components H and V. Applying the conditions of equilibrium to the structure as a whole, we have, therefore, vertical forces = 0, W + ~tt\ -\- weight of crane V= 0, horizontal forces =0, H 1 + Jf 2 = 0, moments = (taken about IT), Wl z W^ -H H^c = 0. From the first condition the vertical reaction of the footstep is found to be equal to the entire weight of the structure and its loads. In applying the last condition, moments are taken about B, since one unknown H z is thus eliminated, leaving the resulting moment equation with only one unknown H^ The other unknown H 2 is then found from the second condition, H z = Jf^ The moment of the counterweight W^ should, when possible, be made equal to ^i where W is the maximum load the crane is designed to lift. The mast will then never be subjected to a bending moment of more than one half that due to the lifted load ; that is to SIMPLE STRUCTURES 187 say, the horizontal reactions H I and Jf 2 will never have more than one half the value they would have if the crane was not counterweighted. 2. Graphical method. To illustrate this method, consider the Pratt truss, shown in Fig. 125. Assume the loads in this case to be the weight of the truss IF, a uniform load of amount Tf^, assumed for present purposes to be concentrated at its cen- ter of gravity, and two concentrated loads JJ, P r Since the only other ex- ternal forces acting on the truss are the reactions R^ R z , they must hold the loads in equilibrium, and hence the force polygon must close. The force polygon, however, con- sists in the present case simply of a straight line 12345, and therefore does not suffice to deter- mine the values of E l and R f For this purpose an equilibrium polygon must be drawn. Thus, choose any pole on the force diagram and draw the rays 1, 02, 03, etc. ; then construct the corresponding equilibrium polygon by starting from any point a in R and drawing ab parallel to 1, from I drawing be parallel to 2, etc. Having found the closing side af of the equilibrium poly- gon, draw through the ray 6 parallel to <//, thereby determining 7^ as 50 and R z as 61. If, for any reason, it is desired to draw the equilibrium polygon through two fixed points, say a and/' in the figure, the reactions are first determined as above. Then a line is drawn through 6 parallel to af, and a pole 0' is chosen somewhere on this line. The closing side of the equilibrium polygon will then be parallel to 0'6 (or #/'), and hence if the polygon starts at a, it must end at/'. FIG. 125 188 RESISTANCE OF MATERIALS 116. Structures : joint reactions. Since all parts of a structure at rest are in equilibrium, the conditions of equilibrium may evi- dently be applied to the forces acting upon any portion of the structure. This portion may be a single joint, a single member or part of a member, or it may include several joints and members. The forces acting upon the part considered may be partly external forces and partly internal forces, or stresses, or they may be wholly stresses. As in finding external reactions, the conditions of equilibrium may be applied either analytically or graphically. Plan FIG. 126 1. Analytical method. To illustrate this method, as applied to the joints of a structure, let it be required to find the stresses in the members of the shear legs, shown in Fig. 126. Starting with the joint A, the forces acting at this point are the weight W, the tension P in the guy AC, and the reaction of the legs of the A frame. To simplify the solution the latter may be assumed for the present equivalent to a single force E acting along the center line A'C between the legs of the A frame. The condi- tions of equilibrium applied to this joint are then V vertical forces = 0, W+ P cos (0 + <) - R cos 9 = 0, V horizontal forces = 0, P sin (0 + $) R sin 6 0, giving two simultaneous equations for R and P. SIMPLE STKUCTUKES 189 Since R is by assumption equivalent to the combined action of the two shear legs, the thrust T in each may be found by resolving forces along R. Thus T cos a = 1 R, which determines T, since R has already been found. I Similarly, the force at the bottom of the shear legs tending to make them spread is T sin a. At the point C the forces acting are the up- ward pull V on the an- chorage, the horizontal pull H 011 it, and the tension P in the guy. Hence, applying the con- ditions of equilibrium, we have IT- P cos ft F=Psinft 2. G-raphical method. To illustrate the graph- ical calculation of stresses from joint reactions, con- sider the roof truss shown in Fig. 127. Since the loading in this case is symmetrical, the reactions of the supports will each be equal to half the weight on the truss. The most convenient notation is to letter the spaces between the various lines of the diagram. Each member of the truss and each external force will then be designated by the adjoining letters on opposite sides of it, as the member AH, the load BC, etc. Starting with the left support, we have three forces meeting at a point. The magnitude of one, namely R^ or AB, is known, and the directions of all three are known. Hence the other two can be determined by means of a triangle of forces. Thus, if ab is laid off to scale to represent R^ and aj, bj, are drawn from a and b parallel FIG. 127 190 KESISTAXCE OF MATERIALS to AJ and BJ, they will represent the stresses in these members to the same scale as that to which R^ was laid off (Fig. 127, /). Proceeding to the next joint, BJIC, we have four forces meeting at a point, one of which, BJ, has just been determined, and another, EC, is known. Hence the other two are found by drawing a force polygon, bjic, giving the stresses in CI and IJ (Fig. 127, //). Similarly, passing to the next joint, AJIH, the stresses in AJ and JI having been found, those in Iff and AHm&y be determined from the force polygon ajih (Fig. 127, ///), and finally for the joint If CD the remaining stresses are determined from the force polygon gJticd (Fig. 127, IV). Since each force polygon contains one side of each of the others, by placing these sides together they may all be combined into one figure, as shown in 9 Tons 15 Tons Fig- 127, F. Ill the pres- ent case separate diagrams T , , . . , wQTQ drawn i or each joint to illustrate the method. In practice, however, but one diagram, the combined one, is drawn, as it affords a F 128 saving in time and space and produces a neater and more compact appearance. Such a figure is called a Maxwell diagram. 117. Structures : method of sections. If a section is passed through a structure, cutting not more than two members whose stresses are unknown, the single condition that the force polygon, drawn for the forces acting upon the portion of the structure on one side of the section, must close, will enable the stresses in these members to be found. Commencing at one end of a structure and passing a section cutting but two members, the stresses in these can thus be determined. Then, passing a section cutting three members, one of which has already been treated, the stresses in the other two can be found, etc. Thus, by means of successive sections, all of the stresses can be determined by simple force polygons. 1. Analytical method. To illustrate the analytical application of this method, consider a Warren truss used as a deck bridge, as SIMPLE STRUCTURES 191 shown in Fig. 128. Let the depth of truss and panel length be each 15 ft., and the loads carried a,t the joints of the upper chord be 7, 10, 9, and 15 tons respectively. The reactions at B and J are found by taking moments about J and B to be 17| tons and 23| tons respectively. Since this form of truss has parallel chords and a single web system, it is not necessary to begin at any particular point, but a section may be taken anywhere, provided it cuts both chords and a single web member. Taking any section xy, and considering only the portion of the structure on one side of the section, the external forces acting on this portion will be in equilibrium with the stresses P, Q, R, in the mem- bers cut (Fig. 129). Since Q is the only stress having a vertical compo- nent, it must equilibrate the external forces at B and C. That is to say, from the condition of equilibrium ^ vertical forces = 0, we have Q sin 03 20' = 17| 7 ; whence Q = 11.854 tons and is compressive. To find P take moments about D. Then since Q and R botli pass through D, their moments about this point are zero; therefore P x 15 = 17-f x 15 - 7 x 7.5 ; whence P = 14i tons. By observing the signs of the moments of the external forces at B and C about I), P is found to act in the direction shown by the arrow, that is, in compression. Similarly, to find the stress R in J>F, take the section vy just to the left of E, then take moments about E. Since P and Q pass through E, their moments about this point are zero, and hence .flxl5=17f x 22.5 -7x15; whence R = 19.44 tons. Since the loads are vertical, R might also have been found from P and Q by the condition 2\ horizontal forces = ; that is, P + Q cos 63 26' = R ; whence R = 19.427 tons. 192 RESISTANCE OF MATERIALS 2. Graphical method. Before proceeding with the explanation of the graphical method it will be necessary to show how the moment of any number of forces with respect to a given point may be obtained from the equilibrium polygon. Let PV P 2 , ^, P denote any set of forces and B the given point about which their moment is required (Fig. 130). First draw the force polygon for these forces, choose any pole 0, and construct the corresponding equilib- rium polygon abcde. Now in the force diagram, drop a perpendicular oh from the pole on the resultant R. This is called the pole dis,- tance of R and will be denoted by H. Also, in the equilibrium diagram draw through the given point B a line par- allel to R, making the intercept xy on the equilibrium polygon. Then the triangle OAE in the force diagram is similar to the triangle xey in the equilibrium diagram, and hence r : xy = H : AE, D or Rr = H x xy. But Rr is the moment of the result- ant R about B and is equal to the sum of the moments of all the given forces about this point. The following moment theorem may therefore be stated : The moment of any system of forces about a given point is equal to the pole distance of their resultant multiplied by the intercept made by the equilibrium polygon on a line drawn through the given point parallel to the resultant. The moment of a part of the given set of forces about any point may also be found by this theorem. For example, let it be required to find the moment of P^ and P z about B. The resultant of P v P^, is given in amount by AC and acts through the point/, as shown. Hence draw through B a line parallel to this partial resultant, SIMPLE STRUCTURES 193 making the intercept mn on the equilibrium polygon. Then, since the triangles fmn and OA C are similar, we have r' : mn = H' : E', or E'r' = H' x mn, which is the expression required by the theorem. For a system of parallel forces the pole distance H is constant and hence the equilibrium polygon is similar to the moment diagram for the forces on either side of any given point. Therefore the moment of all the forces on one side of a given point, taken with re- spect to this point, is equal to the constant pole distance H multiplied by the intercept made by the equilibrium poly- gon on a vertical through the point in question. To apply this method to the roof truss shown in Fig. 131, for example, draw the force polygon and the correspond- ing equilibrium polygon, as shown in the figure. Now take any section of the truss, such as xy in the figure, and take moments of the stresses in the members cut about one of the joints, say B. Then the condition of equilibrium V moments about B = may be written Moment of stress in AF-\-^ moments of P^, P 2 , But by the above theorem ^moments of J\, P 2 , A\, about B = bb r x Oh. W X Oh FIG. 131 Hence Stress in AF ' = 194 RESISTANCE OF MATERIALS Similarly, by taking moments about A the stress in BF is found to be aa' x Oh Stress in BF = AT and the stress in J5(7, with center of moments at F, is ccf X Oh Stress in BC . By observing the signs of the moments the stresses in AB, BC, and BF are found to be compressive and that in AF tensile. In the present case, from symmetry, the stresses in the remaining members of the truss are the same as in those already found. For unsymmetrical loading it would be necessary to apply the above method to each individual member. APPLICATIONS 277. Two equal weights of 50 Ib. each are joined by a cord which passes over two pulleys in the same horizontal line, distant 12 ft. between centers. A weight of 5 Ib. is attached to the string midway between the pulleys. Find the sag. 278. The rule used by the makers of cableways for finding the stress in the cable is to calculate a factor = one half the span and multiply the load, assumed /- \ twice the sag to be at the middle, by this factor. Show how this formula is obtained. 279. It is usual to allow a sag in a cable equal to one twen- tieth of the span. What does the nu- merical factor in the preceding problem become in this case, and how does the tension in the cable compare with the load? 280. Find the relation between F and fP and the total pull on the upper support in the systems of pulleys shown in Fig. 132. 281. In a Weston differential pulley two sheaves, of radii a and 6, are fastened together, and by means of a continuous cord passing around both and also around a movable pulley, support a weight W. Find the relation between F and TF, neg- lecting friction (Fig. 133). SIMPLE STRUCTURES 195 282. In a Weston differential pulley the diameters of the sheaves in the upper block are 8 in. and 9 in. Find the theoretical advantage. 283. In the differential axle shown in Fig. 134 the rope is wound in opposite directions around the two axles so that it unwinds from one and winds up on the other at the same time. Find its mechanical advantage if the radius of the large drum is #, of the small drum is r, and of the crank is c. 284. A differential screw consists of two screws, one inside the other. The outer screw works through a fixed block, and is turned by means of a lever. This screw is cored out and tapped for a smaller screw of less pitch which works through another block, free to move along the axis of the screw, but prevented from rotat- ing. Find the mechanical advantage of such a differential screw if the lever arm is 3 ft. long, the outer screw has 8 threads to the inch, and the inner screw has 10 threads to the inch FIG. 135 285. The bed of a straight river makes an angle a with the horizontal. Taking a cross section perpendicular to the course of the river, the sides of the valley are inclined at an angle to the horizontal. Find the angle which the tributaries of the river make with it. 286. Find the least horizontaHorce necessary to pull a wheel 30 in. in diameter, carrying a load of 500 lb., over an obstacle 4 in. high. 287. A steelyard weighs 61b. and has its center of gravity in the short arm at a distance of 1 in. from the fulcrum, and the center of suspension is 3 in. from the fulcrum. The movable weight weighs 4 lb. Find the zero graduation, and the dis- tance between successive pound graduations. 288. A ladder 50 ft. long, weighing 75 lb., rests with its upper end against a smooth vertical wall and its lower end on rough horizontal ground. Find the reactions of the supports when the ladder is inclined 20 to the vertical. 289. A circular, three-legged table, 4 ft. in diameter, weighs 50 lb. and carries a load of 100 lb. 10 inches from the center and in a line joining the center and one leg. Find the pressure between each foot and the floor. Find also the smallest load which when hung from the edge of the table will cause it to tip over. 290. The average turning moment exerted on the handle of a screw driver is 120 in.-lb. The screw has a square slot, but the point of the screw driver is beveled to an angle of 10 (Fig. 135). If the point of the screw driver is 1 in. wide, find the vertical force tending to raise the screw driver out of the slot. 196 RESISTANCE OF MATERIALS 291. Three smooth cylindrical water mains, each weighing 500 lb., are placed in a wagon box, two of them just filling the box from side to side and the third being placed on top of these two. Find the pressure between the pipes and also against the bottom and sides of the wagon. 292. An engine is part way across a bridge, the weights and distances being as shown in Fig. 136. Find the reactions of the abutments. r 64-0 FIG. 136 FIG. 137 293. In the letter scales shown in Fig. 137 the length of the parallel links is 3 in., and the distance of the center of gravity of the moving parts below the pivot is 2 in. If the radius of the scale is 8 in. and the weight of the moving parts is 12 oz., find the distance between successive ounce graduations on the scale. 294. A scale is arranged as shown in Fig. 138. Determine the relation between the load and the weight P. (Quintenz scales, Strassburg, 1821.) FIG. 138 FIG. 139 Solution. With the given dimensions we have, by the principle of moments, and whence Pa = Pa = Wb - W X (b -- cV l\ e + d I SIMPLE STRUCTURES 197 Since this is independent of x, the position of the load on the platform does not affect the result. Let the dimensions be so proportioned that - W e + d and also a = 10 6. Then P = A scale so arranged is called a decimal scale. 295. In the toggle-joint press shown in Fig. 139, the length of the hand lever is = 3 ft., and 1 2 = 4 in. If the pull P = 100 lb., find the pressure between the jaws of the press when the toggle is inclined at 10 to the vertical. 296. In the crab hook shown in Fig. 140, assume that the load Q = 300 lb. and the coefficient of friction /* = .5, and determine the kind and amount of strain in the members ED, AD, and A 13 for a = 30, /3 = 90, a = 6 in., 6=3 in., and / = 18 in. Show also that in order to hold the weight with- out slipping, the condition which must be satisfied is 2 a sin a FIG. 140 297. A steam cylinder is 20 in. in diameter, the steam pressure is 150 lb./in. 2 , the crank is 18 in. long, and the connecting rod is 5 cranks long. Find the stress in the connecting rod, pressure on cross-head guides, and tangen- tial pressure on crank pin when the crank makes an angle of 45 with the horizontal on the ff in end " of the stroke. Find also the maximum tangential pressure on the crank pin. 298. Calculate the stresses in all the members of the dockyard crane shown in Fig. 141 when carrying a load of 40 tons. 299. Calculate analytically the stresses in the members of the jib crane shown in Fig. 142 when lift- ing a load of 28 tons., the dimen- sions being as given in the figure. FIG. 141 300. In the locomotive crane shown in Fig. 143, calculate the stresses in boom, mast, back stays, hoisting line, and boom line when the boom is in its lowest position, the dimensions for this case being as given in the figure. 301. Calculate the maximum stresses in all the members of the stiff -leg derrick shown in Fig. 144 when lifting a load of 10 tons, the dimensions being as follows : mast = 25 ft., boom = 38 ft., each leg = 40 ft. It is customary to load the sills with stone to give the necessary stability.. Find the amount of stone required on each sill when lifting the 10-ton load. FIG. 142 FIG. 143 FIG. 144 198 200 RESISTANCE OF MATERIALS 302. The testing machine shown in Fig. 145 is designed for a maximum load on the platform of 100,000 Ib. The dimensions of the various levers are as shown in Fig. 146, the lever C being V-shaped, with the lever D hung inside. Neglecting the Extreme position of Weight - Plan, levers and D FIG. 146 weights of the arms, compute the weight of the slider when in its extreme posi- tion required to balance a load of 100,000 Ib. on the platform. 303. Determine analytically the stresses in the members CD, DE, and EF of the curved-chord Pratt truss shown in Fig. 147, assuming the load at each panel point to be 50,000 Ib. / 304. The roof truss shown in Fig. 148 is anchored at one end A, and rests on rollers at the other end B. The span Z = 80 ft., rise h = 30 ft., distance between trusses b = 18 ft. The weight of the truss is given approximately by the for- mula W = g 1 ? bl' 2 ', the wind load, as- sumed to be from the left, is taken as 451b./ft. 2 of roof surface, and the snow load is 30 lb./ft. 2 of hori- zontal projection. Calculate analyt- ically the reactions of the supports due to all loads acting on the truss. 305. In the saw-tooth type of roof truss shown in Fig. 149, deter- mine analytically the stress in FH. 306. In the Pratt truss shown in Fig. 150, the dimensions and loads are as follows: span = 150ft., height = 30ft., number of panels = 6. The dead load per linear foot in pounds for single-track bridge of this type is given by the formula w = 5 1 + 350, where I denotes the span in feet ; the weight of single track FIG. 147 SIMPLE STRUCTURES 201 may be taken as 400 Ib. per linear foot, and live load as 3500 Ib. per linear foot. Calculate analytically the stresses in all the members. NOTE. Each truss carries one half the total load. In the present case, therefore, the total load per linear foot per truss is 1100 + 400 + 3500 2 = 25001b. 307. In the saw-tooth type of roof truss shown in Fig. 149, obtain graphically the stresses in all the members, the dimensions being as follows : span = 25 ft., dis- tance apart of trusses = 15 ft., and pitch of roof = , making the inclination of the longer leg to the horizontal = 2148 / . FIG. 149 As the span is short and the roof comparatively flat, it is sufficiently accurate to assume that the combined action of wind and snow is equivalent to a uniform vertical load, which in the present case may be assumed as 25 lb./ft. 2 of roof. The weight of this type of truss will be taken as 1.5 lb./ft. 2 of roof, and the weight of roof covering as 7 lb./ft. 2 of roof. As the top-chord panel length is 8 ft., each panel load will be 8 x 15 x 33.5 = 4020 Ib. A 308. Analyze graphically for both dead and snow loads the French type of roof truss shown in Fig. 151, the dimensions being as follows : span D F G E H K ^M N O P Q R f 4 I III \R m L J I = 100 ft., rise h = 30 ft., d = 5 ft., and distance apart of trusses b = 20 ft. The weight of truss in pounds for FIG. 150 this type is given by the formula W = 5*4 W 2 , where b and I are expressed in feet. The weight of the roof covering may be assumed as 15 lb./ft. 2 of roof surface, and the snow load as 20 lb./ft. 2 of horizontal projection. First calculate the dead load carried at each joint, due to weight of truss and roof covering, and draw the diagram for this system of loads. The diagram for 202 RESISTANCE OF MATERIALS snow load will be a similar figure, and the stresses in the two cases will be propor- tional to the corresponding loads. Hence the snow-load stresses may be obtained by multiplying each dead-load stress by a constant factor equal to the ratio of the loads. Graphical analysis of French truss G FIG. 161 In drawing the diagram start at one abutment, say the left, and take the joints in order, thus determining the stresses in 0-4, AM, AB, PB, BC, and CM. At the middle of the rafter, where the load PQ is applied, there will be three unknowns, and since these cannot all be determined simultaneously, one of the three must be ob- tained by some other means before the construction can proceed. For this purpose, SIMPLE STRUCTURES 203 proceed to the load QE and determine the stress in EF by the auxiliary construc- tion shown by the dotted lines in the diagram. Then determine the stress in ED by the same auxiliary construction. Having found the stress in ED, we may then go back to the load PQ and complete the diagram. Graphical analysis of Howe truss Wind load stress diagram, wind from left Dead load stress diagram FIG. 152 309. Analyze graphically for both dead and wind loads the Howe roof truss shown in Fig. 152 for a span of 50 ft. and of one-half pitch, that is, with a rise of ^ = 12.5 ft. The trusses are spaced 16 ft. apart ; the weight of each truss may be taken as 2.5 lb./ft. 2 of horizontal area, the roof covering as 10 lb./ft. 2 of roof, and the snow load as 20 lb./ft. 2 of roof. The wind load, based on a pressure of 204 RESISTANCE OF MATERIALS 30 lb./ft. 2 of vertical projection, gives for a roof of one-half pitch an equivalent load of 22.4 lb./ft. 2 of roof surface. Assume the left end of the truss to be on rollers and the right end fixed. The total horizontal thrust due to wind load is then carried by the right abutment. In drawing the wind-load diagram, first calculate the reactions R^ and R 2 by the method of moments. Having thus determined the point M, the remainder of the diagram is easily drawn. 310. Solve problem 304 graphically. ANSWERS TO PROBLEMS The following list comprises answers to about two thirds of the problems given at the close of each section as applications of the text. This is ample to enable the student to verify the correctness of his numerical applications, while enough remain unanswered to cultivate self-reliance and independence of thought. 1, 2. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 5. 3.1 lb./in. 2 6. .002122; 12. 73 in. Answer given. 3. Answer given. Answer given. 4. 17. 7 lb./in. 2 16,500,000 lb./in. 2 approximately. 31. t = .0044 in. approximately^ =.00002037. 32. Copper, 5.73cm.; alloy, 4.27cm. 34. Brass, 791 Ib. ; steel, 409 Ib. 35. Eight -in. bolts. 36. 2 in. crushing _ 3 shearing TT 38. 1.7 approximately. 39. 181. 40. f-in. steel-wire rope. 54. 95.45 in. 4 55. 441 in. 4 56. 337 in. 4 57. Ill in. 4 58. 1^ in. above base. 59. 83^ in. 4 60. 3.36 in. from top. 61. 171 in. 62. 576| ; 364. 63. 3.28 in. 64. 51 in. 4 68. Weight equal to that of table. 69. First support, If ft. from 150 Ib. weight. 77. Maximum moment = 1920 ft.-lb. ; maximum shear = 480 Ib. 78. Maximum moment = 4225 ft.-lb. ; maximum shear = 950 Ib. 79. Maximum moment = 4666| ft.-lb. ; maximum shear = 833^ Ib. 80. Maximum moment = 2200 ft.-lb. ; maximum shear = 375 Ib. 81. Maximum moment = 750 ft.-lb. ; maximum shear = 200 Ib. 82. Maximum moment = 3333^ ft.-lb. ; maximum shear = 8000 Ib. 83. Maximum moment = 11,850 ft.-lb. ; maximum shear = 4100 Ib. 93. Maximum moment = 7400 ft.-lb. ; maximum shear = 1350 Ib. 94. Maximum moment = 24,000 ft.-lb. ; maximum shear = 4000 Ib. 95. Maximum moment = 1250 ft.-lb. ; maximum shear = 500 Ib. 205 27,708 Ib. .000307 in. .0005076 in. s = .0018. 5. 4 in. .0000104. 1005 Ib. .245 in. square. 5 in. for p = 16,000 lb./in. 2 113 tons at failure. 20.0086 ft. 2363 lb./in. 2 .13 in. 10,980 lb./in. 2 75,280 lb./in. 2 for root area. 10|. 5.8. 17,440 Ib. 1,645,300 lb./in. 2 100 + 554 Ib. ^-in. pin. 2036 Ib. tension in each. 206 RESISTANCE OF MATERIALS 109. 110, ill, 35,280 ft.-lb. 869,500 ft.-lb. 7 to 4. H2. L ; 113, 128. 129. 130. 131. 136. 166. 167. 168. 169. 170. 171. 173, 174, 175. 205. 206, 207. 222. 223. 224. 225. 226. 228. 229. 230. 231. 232. 233. 234. 235. 241. 6 24 32 Sin. I, 20|lb. 18.4 including weight of beam. Tension = 2480 + 500 lb./in. 2 Compression = 4100 500 lb./in. 2 114. 15-in. channels, 33 Ib. 115. 1 to 12. 116. Section modulus =24. 117. 9 in. I, 21 Ib. 118. 9 in. I, 25 Ib. 119. 6 in. 51 in. 16.3ft. .192 in. 350 tons-. 9 1 in. square. 5.82 in. 123. 830in.4; 2767 Ib./ft. 124. 3967 Ib. 125. 4i x f in. 126. 28.1 Ib. 127. Stress = 51 70 lb./in. 2 137. 10111b./ft. 2 138. 12.68 ft. apart. 139. .0373 in. 156, Moment at wall = 1800 ft.-lb.; D maximum = .0776 in. at 5.06 ft. from free end. 176, 4-in. channels, 5 Jib.; 185. 20 in. I, 75 Ib. Rankine, 616 tons ; Johnson, 627 tons. Rankine, 268 tons ; Johnson, 267 tons. 132 tons. 15. 2 1 in. square. d = 4.25 in. for stiffness ; .d = 6.6 in. for strength. Minimum speed, 73 R. P. M. 2500 lb./in. 2 Shaft, 3 in. diameter. .13 in. 5911 lb./in. 2 97J. 2940 lb./in. 2 3333 lb./in. 2 Side, ^j in.; plates, 6 x \ in. 177. Top plate, 10 x i side plates, 8^ x ^ in angles, 3 x 3 x | in 180. 3.4 in.; .9 in. 181. 1= 17. Id. 182. Diameters, 10^ and 71 in. 183. About 3 in. 184. 5 + . bottom, ^ 9 in. 6528 lb./in. 2 5880 lb./in. 2 4ft. lin. 2 in. 4. 3677 lb./in. 2 ; 98.9 tons ; 13,550 lb./in. 2 ; 494.6in.-tons. 1.25 in. 221. ^ 242. 2344 lb./in. 2 244. 139 lb./in. 2 245. .28 in. 246. 17001b./ft. 2 247. t = .108r. 248. Flat, t = 3. 16 in.; hemispherical, = .7 in. 250. 1.28 in. 261. 70.8%; 52.4%. 262. J-in. plates, |-in. rivets, 3-in. pitch ; e = 75 to 78 %. 263. ^-in plates ; 1-in. rivets, 5.2-in. pitch. 264. 82.5 lb./in. 2 267. 476 lb./in. 2 268. 450 lb./in 2 . 190. 43.24 in.-lb. 192. 3.68 in. 193. 5.63 in. 194. 7. 114 in. 195. 32 28'. 196. 2| H. P. 197. 4.465 in. 200. 4484 H. P. 201. 9.2 in. 202. 32. 208. p = 4235 lb./in. 2 ; q = 4283 lb./in. 2 209. .E = 30,346,935 lb./in. 2 ; G = 12,158,485 lb./in. 2 in. 269. 7681b./ft. 277. .3ft. 279. 5 times load. W 280. F= " _ W / = 2\ a, - b 281. F 282. 1 to 18. 283 . F = E/r -0: 2 \ c 284. W= 90478 J?, neglect- ing friction. sin a 286. 463.2 Ib. 288. Upper end, 13.6 Ib.; lower end, 76.2 Ib. 289. 22flb.,63flb.,63|lb.; 50 Ib. 290. 82 Ib. ANSWERS 207 291. 288.7 Ib. between pipes ; 298. AK = 86 tons, AB = 76.3 tons, 144.35 Ib. against sides. EF = 121.6 tons, ED = 82.9 tons. 292. Left, 174,110 Ib.; 299. AD = 99.2 tons, BC = 95.6 tons, right, 235,992 Ib. AB = 87. 7 tons, A C = 85.5 tons, 295. 6.83 tons. DE = 49.5 tons, FC = 117.0 tons, 297. Connecting rod, 47,580 Ib. ; DF = 51.7 tons. cross head, 6730 Ib. 303. CD = 296,500 Ib., DE = 12,842 Ib., Maximum on crank pin 48,030 Ib. EF = 288,500 Ib. 304. BH = 49,400 Ib., BE = 33,450 Ib., HE = 31,090 Ib., HC = 33,500 Ib., CE = 24,657 Ib., CK = 41,094 Ib., KE = 10,000 Ib., KA = 51,094 Ib., AE= 16,574 Ib. 305. 1390 Ib. 306. AB = 203,100 Ib., AD = 208,300 Ib., AG = 234,375 Ib., BM = CN = 130,200 Ib., #0 = 208,300 Ib., BC = 62,500 Ib., DE = 31,250 Ib., CX> = 121,875 Ib., .EJF = 40,640 Ib. INDEX Allowable stresses, 7 Allowance for shrinkage and forced fits, 130 Annealing, 6 Bach's formula for flat plates, 145 Barlow's formula for thick cylinders, 125 Beams, built-in, 80 continuous, 70 deflection of, 60 design of, 52 fundamental formula for, 51 restrained, 80 strength of, 52 Bending and torsion, combined, 110 Bending moment, 36 Bending stress, 49 Bending-rnoment and shear diagrams, 37 Bernoulli's assumption, 50 Birnie's formula for thick cylinders, 126 Boiler shells, 148 Built-in beams, 80 Cantilever beams, 60 Cantilevers, deflection of, 63 Cement, natural, 157 Portland, 157 standard mixtures of, 158 Center of curvature, 60 of gravity, 17 of mass, 19 Centroid, 19 of circular arc, 21 of circular sector, 22 of circular segment, 22 of composite figures, 24 of parabolic segment, 23 of triangle, 20 Clapeyron, 70 Clavarino's formula for thick cylinders, 126 Coefficient of linear expansion, 10 Column formulas, comparison of, 99 Columns, 91 eccentrically loaded, 101 reenforced-concrete, 164 Compression, 2 Concrete, reenforced, 156 Conditions of equilibrium, 35 Continuous beams, 70 Considered formula for column hoops, 166 Cooper's column formulas, 101 Cottered joints, 14 Couple, definition of, 177 Curvature of beams, 60 Cylinders, built up, 126 rational formulas for, 127 thick, stress in, 120-126 thin, stress in, 119 Deflection of beams, 60 angular, 61, 62 Deformation, unit, 4 Eccentrically loaded columns, 101 Efficiency of riveted joints, 146 Elastic curve, 50, 60 Elastic limit, 5 Elastic resilience, 8 Equilibrium, conditions of, 35, 178, 181 Equilibrium polygon, 179, 181-185 Euler's column formula, 92 Expansion, coefficient of linear, 10 Factor of safety, 6 Fatigue of metals, 5 Flat plates, empirical formulas for, 145 stress in, 136-143 theory of, 136 Flat slabs, reenforced-concrete, 167, 173 Forced fits, 129 Forces, composition and resolution of, 176 Graphical methods, 187, 189, 192 Grashof s formula for flat plates, 145 Guest's formula for shafts, 109 Hooke's law, 4 Hoop stress, 118 Horse-power formula, 114 Impact, 9 Johnson's parabolic column formula, 97 Johnson's straight-line column formula, 99 Joint reactions, calculation of, 188 Joints, riveted, 146 209 210 RESISTANCE OF MATERIALS Lamp's formulas for thick cylinders, 120 Lime, 157 Maxwell diagrams, 190 Modulus of elasticity, 4 of resilience, 8 of rigidity, 5 of shear, 5 Moment, definition of, 15, 176 Moment of inertia, 25 for circle, 28 for composite figures, 29 for rectangle, 26 for triangle, 28 theorems on, 27 Moment of resistance, 52 Moment and shear diagrams, 37 directions for sketching, 44 properties of, 43 relations between, 41 Moment solid, 26 Moments, in continuous beams, 77 fundamental theorem of, 15 Mushroom system of reenforced- concrete construction, 167 Neutral axis, 50 Nichols's formula for flat plates, 145 Poisson's ratio, 9 Portland cement, 156 Power transmitted by shafts, 108 Quicklime, 156 Radially reenforced flat slabs, 167 Rankine's column formula, 95 Rankine's formula for shafts, 109 Reenforced concrete, 156 Reenforced-concrete beams, design of, 159 Reenforced-concrete columns, 164 Reenforcement, calculation of, in reen- forced-concrete beams, 162 Resilience, 8 of shafts, 110 Resistance, moment of, 52 Restrained beams, 80 Resultant, 16 Rivet pitch, 150 Riveted joints, 146 efficiency of, 146 Settlement of supports, effect of, 76 Shafts, angle of twist in, 107, 112 elliptical, 111 power transmitted by, 108 rectangular, 111 resilience of, 110 square, 111 stress in, 106 triangular, 112 Shear, 2 angle of, 106 in continuous beams, 78 vertical, 36 Shear modulus, 5 Shrinkage and forced fits, 129 Simple beams, deflection of, 65 Slag cement, 158 Specifications for structural steel, 150 Spheres, thin, stress in, 118 Spider, dimensions of, 174 Spider hoops, 168 Static moment, 15 Stirrups in reenforced-concrete beams, 162 Straight-line law, 50 Strain, definition of, 1 varieties of, 2 Strain diagram, 4 Stress, definition of, 1 in beams, 50 unit, 4 Stresses, in structures, 185-194 Structural-steel riveting, 149 Structures, stresses in, 176, 185-194 Struts, 191 Symmetry, axis of, 24 Temperature stress, 9 Tension, 2 Theorem of three moments, 70 Three moments, theorem of, 70 Thurston's formula for flat plates, 145 Torsion, 3, 106 Twist, angle of, 106 Ultimate strength, 5 Unit deformation, 4 Unit stress, 4 Unit stresses in structural steel, 150 Web reenforcement in reenforced- concrete beams, 162 Working stress, 6 Section modulus, 52 Sections, method of, 190 Yield point, 5 Young's modulus of elasticity, 4 ANNOUNCEMENTS TEXT-BOOK ON THE STRENGTH OF MATERIALS (Revised Edition) By S. E. SLOCUM, Assistant Professor of Mathematics in the University of Illinois, and E. L. HANCOCK, Instructor in Applied Mechanics in Purdue University 8vo, cloth, xii + 314 pages, illustrated, $3.00 A TEXT-BOOK which is thoroughly representative of the best modern theory and practice, and at the same time sufficiently ele- mentary for the use of students of a junior grade in technical and engineering schools. The first part is devoted to a theoretical discus- sion of the Strength of Materials for use in the class room, while the second part consists of a description of the physical properties of materials, with methods of testing, etc., and is intended for use as a laboratory manual in connection with the class work. ALBERT E. GREENE, Junior Professor of Civil Engineering, University of Michigan: The book contains a great deal of infor- mation which is available nowhere else in English. It will be a valu- able book of reference to all interested in the subject. 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