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Full text of "Resistance of materials, for beginners in engineering"

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 



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0000 






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5 * 

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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). 



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xiii 



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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 






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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 



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