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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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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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
ANNOUNCEMENTS
TEXT-BOOK ON THE STRENGTH OF
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