Department of Civil and Environmental Engineering
Spring 2023
Instructor: Dr. Hessam Yazdani, PE
Solved Select Problems
PROBLEM 1.1
Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that d1 = 30 mm and d 2 = 50 mm,
find the average normal stress at the midsection
of (a) rod AB, (b) rod BC.
SOLUTION
(a)
Rod AB:
Force:
P = 60  103 N tension
Area:
A=
Normal stress:
(b)
 AB =

4
d12 =

4
(30  10−3 )2 = 706.86  10−6 m 2
P
60  103
=
= 84.882  106 Pa
−6
A 706.86  10
 AB = 84.9 MPa 
Rod BC:
Force:
P = 60  103 − (2)(125  103 ) = −190  103 N
Area:
A=
Normal stress:
 BC =

4
d 22 =

4
(50  10−3 ) 2 = 1.96350  10−3 m 2
P
−190  103
=
= −96.766  106 Pa
A 1.96350  10−3
 BC = −96.8 MPa 
2
PROBLEM 1.2
Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that the average normal stress must
not exceed 150 MPa in either rod, determine
the smallest allowable values of the diameters
d1 and d2.
SOLUTION
(a)
Rod AB:
Force:
P = 60  103 N
Stress:
 AB = 150  106 Pa
A=
Area:
 AB

4

d12
4
P
P
=
 A=
A
 AB
d12 =
d12 =
P
 AB
4P
 AB
=
(4)(60  103 )
= 509.30  10 −6 m 2
 (150  106 )
d1 = 22.568  10−3 m
(b)
Rod BC:
Force:
Stress:
Area:
P = 60  103 − (2)(125  103 ) = −190  103 N
 BC = −150  106 Pa
A=
 BC

d 22
4
P
4P
=
=
A  d 22
d 22 =
4P
 BC
=
(4)(−190  103 )
= 1.61277  10−3 m 2
 (−150  106 )
d 2 = 40.159  10−3 m
3
d1 = 22.6 mm 
d2 = 40.2 mm 
PROBLEM 1.14
Two hydraulic cylinders are used to control the position of
the robotic arm ABC. Knowing that the control rods
attached at A and D each have a 20-mm diameter and
happen to be parallel in the position shown, determine the
average normal stress in (a) member AE, (b) member DG.
SOLUTION
Use member ABC as free body.
M B = 0: (0.150)
4
FAE − (0.600)(800) = 0
5
FAE = 4  103 N
Area of rod in member AE is
Stress in rod AE:
A=
 AE =

4
d2 =

4
(20  10−3 ) 2 = 314.16  10−6 m 2
FAE
4  103
=
= 12.7324  106 Pa
−6
A
314.16  10
(a)
 AE = 12.73 MPa 
Use combined members ABC and BFD as free body.
4

4

M F = 0: (0.150)  FAE  − (0.200)  FDG  − (1.050 − 0.350)(800) = 0
5
5




FDG = −1500 N
Area of rod DG:
Stress in rod DG:
A=

4
d2 =
 DG =

4
(20  10−3 )2 = 314.16  10−6 m 2
FDG
−1500
=
= −4.7746  106 Pa
A
3.1416  10−6
(b)
4
 DG = −4.77 MPa 
PROBLEM 1.25
Knowing that a force P of magnitude 750 N is
applied to the pedal shown, determine (a) the
diameter of the pin at C for which the average
shearing stress in the pin is 40 MPa, (b) the
corresponding bearing stress in the pedal at C,
(c) the corresponding bearing stress in each
support bracket at C.
SOLUTION
Since BCD is a 3-force member, the reaction at C is directed toward E, the intersection of the lines of action of
the other two forces.
CE = 3002 + 1252 = 325 mm
From geometry,
From the free body diagram of BCD,
Fy = 0 :
 pin
(a)
125
C−P=0
325
C = 2.6 P = 2.6 ( 750 N ) = 1950 N
1
1
C
C
2C
2
=
= 2
=
 2 d2
AP
d
4
2C
d =
 pin
=
2 (1950 N )
(
 40  106 Pa
)
= 5.57  10−3 m
d = 5.57 mm 
(b)
b =
(1950 )
C
C
=
=
= 38.9  106 Pa
−3
−3
Ab
dt
5.57  10
9  10
(
)(
)
 b = 38.9 MPa 
(c)
5
1
C
(1950 )
C
b = 2 =
=
= 35.0  106 Pa
Ab
2dt
2 5.57  10−3 5  10−3
(
)(
)
 b = 35.0 MPa 
PROBLEM 1.32
Two wooden members of uniform cross section are joined by the simple
scarf splice shown. Knowing that the maximum allowable tensile stress in the
glued splice is 75 psi, determine (a) the largest load P that can be safely
supported, (b) the corresponding shearing stress in the splice.
SOLUTION
A0 = (5.0)(3.0) = 15 in 2
 = 90 − 60 = 30
6
 =
P cos 2 
A0
(a)
P=
 A0
(75)(15)
=
= 1500 lb
2
cos 
cos 2 30
(b)
 =
P sin 2
(1500)sin 60
=
2 A0
(2)(15)
P = 1.500 kips 
 = 43.3 psi 
PROBLEM 1.43
Two wooden members are joined by plywood splice plates that are fully
glued on the contact surfaces. Knowing that the clearance between the
ends of the members is 6 mm and that the ultimate shearing stress in the
glued joint is 2.5 MPa, determine the length L for which the factor of
safety is 2.75 for the loading shown.
SOLUTION
 all =
2.5 MPa
= 0.90909 MPa
2.75
On one face of the upper contact surface,
A=
L − 0.006 m
(0.125 m)
2
Since there are 2 contact surfaces,
 all =
0.90909  106 =
P
2A
16  103
( L − 0.006)(0.125)
L = 0.14680 m
7
146.8 mm 
PROBLEM 1.44
For the joint and loading of Prob. 1.43, determine the factor of safety
when
L = 180 mm.
PROBLEM 1.43 Two wooden members are joined by plywood splice
plates that are fully glued on the contact surfaces. Knowing that the
clearance between the ends of the members is 6 mm and that the
ultimate shearing stress in the glued joint is 2.5 MPa, determine the
length L for which the factor of safety is 2.75 for the loading shown.
SOLUTION
Area of one face of upper contact surface:
A=
0.180 m − 0.006 m
(0.125 m)
2
A = 10.8750  10−3 m 2
Since there are two surfaces,
 all =
P
16  103 N
=
2 A 2(10.8750  10−3 m 2 )
 all = 0.73563 MPa
F.S. =
8
u
2.5 MPa
=
= 3.40
 all 0.73563 MPa

PROBLEM 1.61
For the assembly and loading of Prob. 1.60, determine (a) the average
shearing stress in the pin at C, (b) the average bearing stress at C in
member BC, (c) the average bearing stress at B in member BC.
PROBLEM 1.60 Two horizontal 5-kip forces are applied to
pin B of the assembly shown. Knowing that a pin of 0.8-in.
diameter is used at each connection, determine the maximum
value
of
the
average
normal
stress
(a) in link AB, (b) in link BC.
SOLUTION
Use joint B as free body.
Law of Sines:
FAB
FBC
10
=
=
sin 45 sin 60 sin 95
(a)
Shearing stress in pin at C.
 =
AP =
FBC = 8.9658 kips
FBC
2 AP

d2 =

(0.8) 2 = 0.5026 in 2
4
8.9658
 =
= 8.92
(2)(0.5026)
9
4
 = 8.92 ksi 
PROBLEM 1.61 (Continued)
(b)
Bearing stress at C in member BC.  b =
FBC
A
A = td = (0.5)(0.8) = 0.4 in 2
b =
(c)
Bearing stress at B in member BC.  b =
8.9658
= 22.4
0.4
 b = 22.4 ksi 
FBC
A
A = 2td = 2(0.5)(0.8) = 0.8 in 2
b =
10
8.9658
= 11.21
0.8
 b = 11.21 ksi 
PROBLEM 2.2
A control rod made of yellow brass must not stretch more than 18 in. when the tension in the wire
is 800 lb. Knowing that E = 15  106 psi and that the maximum allowable normal stress is 32 ksi,
determine (a) the smallest diameter rod that should be used, (b) the corresponding maximum
length of the rod.
SOLUTION
(a) Stress:
Area:
=
A=
P
, or
A
d2
4
, or
A=
P

d=
=
800 lb
= 25  10−3 in 2
3
2
32  10 lb/in
4A

=
4(25  10−3 in 2 )

= 0.178412 in.
 = 0.1784 in. 
(b) Strain:
PL
=
, or
AE
(
)(
)
25  10−3 in 2 15  106 lb/in 2 ( 0.125 in.)
AE
L=
=
P
800 lb
L = 58.6 in. 
11
PROBLEM 2.13
Rod BD is made of steel ( E = 29  106 psi) and is used to brace the axially
compressed member ABC. The maximum force that can be developed in member
BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in length
of BD must not exceed 0.001 times the length of ABC, determine the smallestdiameter rod that can be used for member BD.
SOLUTION
FBD = 0.02P = (0.02)(130) = 2.6 kips = 2.6  103 lb
Considering stress,  = 18 ksi = 18  103 psi
=
FBD
A
 A=
FBD

=
2.6
= 0.14444 in 2
18
Considering deformation,  = (0.001)(144) = 0.144 in.
=
FBD LBD
AE

A=
FBD LBD
(2.6  103 )(54)
=
= 0.03362 in 2
E
(29  106 )(0.144)
Larger area governs. A = 0.14444 in 2
A=
12

4
d2

d=
4A

=
(4)(0.14444)

d = 0.429 in. 
PROBLEM 2.14
The 4-mm-diameter cable BC is made of a steel with E = 200 GPa. Knowing that
the maximum stress in the cable must not exceed 190 MPa and that the elongation
of the cable must not exceed 6 mm, find the maximum load P that can be applied
as shown.
SOLUTION
LBC = 62 + 42 = 7.2111 m
Use bar AB as a free body.
 4

3.5P − (6) 
FBC  = 0
 7.2111

P = 0.9509 FBC
 M A = 0:
Considering allowable stress,  = 190  106 Pa
A=

d2 =
4
FBC
=
A

4
(0.004)2 = 12.566  10−6 m 2
 FBC =  A = (190  106 )(12.566  10−6 ) = 2.388  103 N
Considering allowable elongation,  = 6  10−3 m
=
FBC LBC
AE
 FBC =
AE (12.566  10−6 )(200  109 )(6  10−3 )
=
= 2.091 103 N
LBC
7.2111
Smaller value governs. FBC = 2.091  103 N
P = 0.9509 FBC = (0.9509)(2.091  103 ) = 1.988  103 N
13
P = 1.988 kN 
PROBLEM 2.29
A homogeneous cable of length L and uniform cross section is suspended from one end. (a) Denoting by  the
density (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of the
cable due to its own weight. (b) Show that the same elongation would be obtained if the cable were horizontal
and if a force equal to half of its weight were applied at each end.
SOLUTION
(a)
For element at point identified by coordinate y,
P = weight of portion below the point
=  gA(L − y )
Pdy  gA( L − y )dy  g ( L − y )
d =
=
=
dy
EA
EA
E
=

L
Total weight:
dy =
g 
L
1 
Ly − y 2 

E 
2 0
g 
2
L 
2
L
−


E 
2 
=
1  gL2

2 E
W =  gAL
F=
14
E
0
=
(b)
 g (L − y)
EA EA 1  gL2 1
=

=  gAL
L
L 2 E
2
1
F= W 
2
PROBLEM 2.30
A vertical load P is applied at the center A of the upper section of a homogeneous
frustum of a circular cone of height h, minimum radius a, and maximum radius b.
Denoting by E the modulus of elasticity of the material and neglecting the effect of
its weight, determine the deflection of point A.
SOLUTION
Extend the slant sides of the cone to meet at a point O and place the origin of the coordinate system there.
tan  =
From geometry,
a1 =
b−a
h
a
b
, b1 =
,
tan 
tan 
r = y tan 
At coordinate point y, A =  r 2
Deformation of element of height dy:
d =
Pdy
AE
d =
P dy
P
dy
=
2
2
E r
 E tan  y 2
Total deformation:
P
A =
 E tan 2 

b1
a1
b1
 1
dy
P
P
=
−  =
2
2
y
 E tan   y  a  E tan 2 
1
=
15
b1 − a1 P (b1 − a1 )
P
=
2
 Eab
 E tan  a1b1
1 1
 − 
 a1 b1 
A =
Ph
 Eab
 
PROBLEM 2.70
The block shown is made of a magnesium alloy, for
which E = 45 GPa and v = 0.35. Knowing that
 x = −180 MPa, determine (a) the magnitude of  y
for which the change in the height of the block will
be zero, (b) the corresponding change in the area of
the face ABCD, (c) the corresponding change in the
volume of the block.
SOLUTION
(a)
y = 0
y = 0
z = 0
1
( x − v y − v z )
E
 y = v x = (0.35)(−180  106 )
y =
= −63  106 Pa
 y = −63.0 MPa 
1
v
(0.35)(−243  106 )
( z − v x − v y ) = − ( x +  y ) = −
= +1.890  10−3
E
E
45  109
 x − v y
1
157.95  106
 x = ( x − v y − v Z ) =
=−
= −3.510  10−3
9
E
E
45  10
z =
(b)
A0 = Lx Lz
A = Lx (1 +  x ) Lz (1 +  z ) = Lx Lz (1 +  x +  z +  x z )
 A = A − A0 = Lx Lz ( x +  z +  x z )  Lx Lz ( x +  z )
 A = (100 mm)(25 mm)(−3.510  10−3 + 1.890  10−3 )
(c)
 A = −4.05 mm2 
V0 = Lx Ly Lz
V = Lx (1 +  x ) Ly (1 +  y ) Lz (1 +  z )
= Lx Ly Lz (1 +  x +  y +  z +  x y +  y  z +  z  x +  x y  z )
V = V − V0 = Lx Ly Lz ( x +  y +  z + small terms)
V = (100)(40)(25)(−3.510 10−3 + 0 + 1.890 10 −3 )
16
V = −162.0 mm3 
PROBLEM 2.74
In
many
situations,
physical
constraints prevent strain from
occurring in a given direction. For
example,  z = 0 in the case shown,
where longitudinal movement of the
long prism is prevented at every
point. Plane sections perpendicular
to the longitudinal axis remain
plane and the same distance apart.
Show that for this situation, which
is known as plane strain, we can
express  z ,  x , and  y as follows:
 z = v( x +  y )
1
[(1 − v 2 ) x − v(1 + v) y ]
E
1
 y = [(1 − v 2 ) y − v(1 + v) x ]
E
x =
SOLUTION
z = 0 =
1
(−v x − v y +  z ) or  z = v( x +  y )
E
1
( x − v y − v z )
E
1
= [ x − v y − v 2 ( x +  y )]
E
1
= [(1 − v 2 ) x − v(1 + v) y ]
E

x =
1
(−v x +  y − v z )
E
1
= [−v x +  y − v 2 ( x +  y )]
E
1
= [(1 − v 2 ) y − v(1 + v) x ]
E

y =
17

PROBLEM 3.3
A 1.75-kN · m torque is applied to the solid cylinder shown. Determine (a)
the maximum shearing stress, (b) the percent of the torque carried by the
inner 25-mm-diameter core.
SOLUTION
(a)
 max =
Given shaft:
 max =
Tc
J
(1.75  103 N  m)(0.025 m)

2
( 0.025 m )
4
= 71.301  106 Pa
 max = 71.3 MPa 
(b)
At surface of core:
1
m
2
1
=
71.301  106
2
 =
(
)
= 35.651  106 Pa
Corresponding torque:
  c4 
3
 
 
T=
=   2  =   c3  = 35.651 106 Pa   ( 0.0125 m )
 c 
c
2 
2


J
(
)
T = 109.376 N  m
%T =
109.376 N  m
(100 )
1750 N  m
6.25% 
18
PROBLEM 3.7
The solid spindle AB is made of a steel with an allowable shearing stress of
12 ksi, and sleeve CD is made of a brass with an allowable shearing stress of
7 ksi. Determine (a) the largest torque T that can be applied at A if the
allowable shearing stress is not to be exceeded in sleeve CD, (b) the
corresponding required value of the diameter d s of spindle AB.
SOLUTION
(a)
Analysis of sleeve CD:
1
1
d o = (3) = 1.5 in.
2
2
c1 = c2 − t = 1.5 − 0.25 = 1.25 in.
c2 =
J=
T=

(c
2
4
2
)
− c14 =

2
(1.54 − 1.254 ) = 4.1172 in 4
J
(4.1172)(7  103 )
=
= 19.21  103 lb  in.
c2
1.5
T = 19.21 kip  in. 
(b)
Analysis of solid spindle AB:
=
Tc
J
J

T 19.21  103
= c3 = =
= 1.601 in 3
3
c
2

12  10
c=3
(2)(1.601)

= 1.006 in.
d s = 2c
d = 2.01 in. 
19
PROBLEM 3.40
The solid spindle AB has a diameter d s = 1.5 in. and is made of a steel with
G = 11.2  106 psi and  all = 12 ksi, while sleeve CD is made of a brass with
G = 5.6  106 psi and  all = 7 ksi. Determine the angle through which end A
can be rotated.
SOLUTION
Stress analysis of solid spindle AB:
c=
1
ds = 0.75 in.
2
 =
Tc
J
T =
Stress analysis of sleeve CD:
T =
Deformation of spindle AB:
(c
2
4
2
)
− c14 =

2
(1.54 − 1.254 ) = 4.1172 in 4
J
(4.1172)(7  10−3 )
=
= 19.21  103 lb  in.
c2
1.5

2
c 4 = 0.49701 in 4 , L = 12 in., G = 11.2  106 psi
TL
(7.95  103 )(12)
=
= 0.017138 radians
GJ
(11.2  106 )(0.49701)
J = 4.1172 in 4 , L = 8 in., G = 5.6  106 psi
CD =
20

c = 0.75 in.
 AB =
Total angle of twist:
(12  103 )(0.75)3 = 7.95  103 lb  in.
T = 7.95  103 lb  in.
J =
Deformation of sleeve CD:
J

=  c3
c
2
2
1
1
c2 = d o = (3) = 1.5 in.
2
2
c1 = c2 − t = 1.5 − 0.25 = 1.25 in.
J =
The smaller torque governs.

T =
TL
(7.95  103 )(8)
=
= 0.002758 radians
GJ
(5.6  106 )(4.1172)
 AD =  AB + CD = 0.019896 radians
 AD = 1.140 
PROBLEM 3.55
Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and
to fixed supports at A and C. For the loading shown, determine (a) the reaction
at each support, (b) the maximum shearing stress in shaft AB, (c) the maximum
shearing stress in shaft BC.
SOLUTION
Shaft AB:
1
d = 25 mm = 0.025 m
2


T L
= c 4 = (0.025)4 = 613.59  10−9 m 4  B = AB AB
2
2
GJ AB
T = TAB , LAB = 0.200 m, c =
J AB
TAB =
Shaft BC:
GJ AB
(77.2  109 )(613.59  10−9 )
B =
 B = 236.847  103  B
LAB
0.200
1
d = 19 mm = 0.019 m
2


T L
= c 4 = (0.019)4 = 204.71  10−9 m 4
 B = BC BC
2
2
GJ BC
T = TBC , LBC = 0.250 m, c =
J BC
TBC =
GJ BC
(77.2  109 )(204.71  10−9 )
B =
= 63.214  103 B
LBC
0.250
Equilibrium of coupling disk.
T = TAB + TBC
1.4  103 = 236.847  103 B + 63.214  103 B
 B = 4.6657  10−3 rad
TAB = (236.847  103 )(4.6657  10−3 ) = 1.10506  103 N  m
TBC = (63.214  103 )(4.6657  10−3 ) = 294.94 N  m
(a)
TA = TAB = 1105 N  m 
Reactions at supports.
TC = TBC = 295 N  m 
(b)
Maximum shearing stress in AB.
 AB =
(c)
 AB = 45.0 MPa 
Maximum shearing stress in BC.
 BC =
21
TABc
(1.10506  103 )(0.025)
=
= 45.0  106 Pa
J AB
613.59  10−9
TBC c (294.94)(0.019)
=
= 27.4  106 Pa
J BC
204.71  10−9
 BC = 27.4 MPa 
PROBLEM 3.60
A torque T is applied as shown to a solid tapered shaft AB. Show by integration
that the angle of twist at A is
 =
7TL
12 Gc 4
SOLUTION
Introduce coordinate y as shown.
r =
cy
L
Twist in length dy:
d =
Tdy
Tdy
2TL4dy
=
=

GJ
 Gc 4 y 4
G r4
2
 = L
2L
2TL4 dy
2TL 2 L dy
=

4 4
 Gc y
 Gc 4 L y 4
2L
2TL4  1 
2TL4  1
1 
=
−
=
−
+ 3


4
3
4 
3
 Gc  3 y L
 Gc  24L
3L 
=
2TL4  7 
7TL
=
4 
3
 Gc  24 L  12 Gc 4

22
PROBLEM 3.74
Three shafts and four gears are used to form a gear train that will transmit
power from the motor at A to a machine tool at F. (Bearings for the shafts
are omitted in the sketch.) The diameter of each shaft is as follows:
d AB =16 mm, dCD = 20 mm, d EF = 28mm. Knowing that the frequency
of the motor is 24 Hz and that the allowable shearing stress for each shaft
is 75 MPa, determine the maximum power that can be transmitted.
SOLUTION
 all = 75 MPa = 75  106 Pa
Shaft AB:
c AB =
Tall =
1
d AB = 0.008 m
2

2
c3AB  all =

2
 =
Tc AB
2T
=
J AB
 c3AB
(0.008)3 (75  106 ) = 60.319 N  m
f AB = 24 Hz Pall = 2 f ABTall = 2 (24)(60.319) = 9.10  103 W
Shaft CD:
1
dCD = 0.010 m
2
Tc
2T
 3

 = CD = 3  Tall = cCD
 all = (0.010)3 (75  106 ) = 117.81 N  m
J CD
2
2
 cCD
cCD =
fCD =
Shaft EF:
rB
60
f AB =
(24) = 9.6 Hz
rC
150
cEF =
Tall =
Pall = 2 fCDTall = 2 (9.6)(117.81) = 7.11  103 W
1
d EF = 0.014 m
2

3
cEF
 all =

(0.014)3 (75  106 ) = 323.27 N  m
2
2
r
60
f EF = D fCD =
(9.6) = 3.84 Hz
rE
150
Pall = 2 f EF Tall = 2 (3.84)(323.27) = 7.80 103 W
Maximum allowable power is the smallest value.
23
Pall = 7.11 103 W = 7.11 kW 
PROBLEM 3.86
Knowing that the stepped shaft shown transmits a torque of magnitude
T = 2.50 kip  in., determine the maximum shearing stress in the shaft when
the radius of the fillet is (a) r = 18 in., (b) r = 163 in.
SOLUTION
D
2
=
= 1.33
d
1.5
D = 2 in. d = 1.5 in.
1
d = 0.75 in.
T = 2.5 kip  in.
2
Tc
2T
(2)(2.5)
=
=
= 3.773 ksi
3
J
c
 (0.75)3
c=
(a)
r =
1
in. r = 0.125 in.
8
r
0.125
=
= 0.0833
d
1.5
K = 1.42
From Fig. 3.28,
 max = K
(b)
r =
Tc
= (1.42)(3.773)
J
3
in.
16
 max = 5.36 ksi 
r = 0.1875 in.
r
0.1875
=
= 0.125
d
1.5
From Fig. 3.28,
 max = K
24
K = 1.33
Tc
= (1.33)(3.773)
J
 max = 5.02 ksi 
PROBLEM 4.1
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
SOLUTION
I =

(r
4
o
4
)
− ri 4 =

4
( 20
4
)
− 154 = 85.903  103 mm 4
I = 85.903  10−9 m 4
(a)
y A = 20 mm = 0.020 m
A = −
My A
(500 N  m)(0.020 m)
=−
= −116.4  106 Pa
I
85.903  10−9 m 4
 A = −116.4 MPa 
(b)
yB = 15 mm = 0.015 m
B = −
MyB
(500 N  m)(0.015 m)
=−
= 87.3  106 Pa
I
85.903  10−9 m 4
 B = −87.3 MPa 
25
PROBLEM 4.2
Knowing that the couple shown acts in a vertical plazne, determine
the stress at (a) point A, (b) point B.
SOLUTION
For rectangle:
I1 =
For one circular cutout:
1 3 1
3
bh = ( 4.8 )( 2.4 ) = 5.5296 in 4
12
12
I2 =

4
d4 =

4
( 0.75 )4 = 0.24850 in 4
For cross sectional area:
I = I1 − 2I 2 = 5.5296 − 2 ( 0.24850 ) = 5.0326 in 4
(a)
y A = 1.2 in.
(b)
yB = −0.75 in.
26
A = −
B = −
My A
(25 kip  in.)(1.2 in.)
=−
I
5.0326 in 4
MyB
(25 kip  in.)(−0.75 in.)
=−
I
5.0326 in 4
 A = −5.96 ksi 
𝜎𝐵 = −3.73 ksi 
PROBLEM 4.3
Using an allowable stress of 155 MPa, determine the largest bending
moment M that can be applied to the wide-flange beam shown. Neglect the
effect of fillets.
SOLUTION
Moment of inertia about x-axis:
I1 =
1
(200)(12)3 + (200)(12)(104) 2
12
= 25.9872  106 mm 4
I2 =
1
(8)(196)3 = 5.0197  106 mm 4
12
I 3 = I1 = 25.9872  106 mm 4
I = I1 + I 2 + I 3 = 56.944  106 mm 4 = 56.944  10−6 m 4
Mc
1
with c = (220) = 110 mm = 0.110 m
I
2
I
M =
with  = 155  106 Pa
c
 =
Mx =
(56.944  10−6 )(155  106 )
= 80.2  103 N  m
0.110
M x = 80.2 kN  m 
27
PROBLEM 4.8
Two W4 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used  U = 58 ksi and using a factor of safety of 3.0, determine the largest couple
that can be applied when the assembly is bent about the z axis.
SOLUTION
Properties of W4  13 rolled section.
(See Appendix E.)
Area = 3.83 in 2
Depth = 4.16 in.
I x = 11.3 in 4
For one rolled section, moment of inertia about axis a-a is
I a = I x + Ad 2 = 11.3 + (3.83)(2.08)2 = 27.87 in 4
For both sections,
I z = 2 I a = 55.74 in 4
c = depth = 4.16 in.
M all
28
U
58
= 19.333 ksi
F .S . 3.0
 I (19.333)(55.74)
= all =
c
4.16
 all =
=
=
Mc
I
M all = 259 kip  in. 
PROBLEM 4.7
Two W4 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used  Y = 36 ksi and  U = 58 ksi and using a factor of safety of 3.0, determine the
largest couple that can be applied when the assembly is bent about the z axis.
SOLUTION
Properties of W4  13 rolled section.
(See Appendix E.)
Area = 3.83 in 2
Width = 4.060 in.
I y = 3.86 in 4
For one rolled section, moment of inertia about axis b-b is
I b = I y + Ad 2 = 3.86 + (3.83)(2.030)2 = 19.643 in 4
For both sections,
I z = 2 I b = 39.286 in 4
c = width = 4.060 in.
M all
29
U
58
= 19.333 ksi
F .S . 3.0
 I (19.333)(39.286)
= all =
c
4.060
 all =
=
=
Mc
I
M all = 187.1 kip  in. 
PROBLEM 4.10
Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compressive stresses in portion BC
of the beam.
SOLUTION
A
y0
A y0

8
7.5
60

6
4
24

4
0.5

18
Yo =
2
86
86
= 4.778 in.
18
Neutral axis lies 4.778 in. above the base.
1
1
b1h13 + A1d12 = (8)(1)3 + (8)(2.772) 2 = 59.94 in 4
12
12
1
1
3
2
I 2 = b2 h2 + A2 d 2 = (1)(6)3 + (6)(0.778) 2 = 21.63 in 4
12
12
1
1
I 3 = b3 h33 + A3 d32 = (4)(1)3 + (4)(4.278) 2 = 73.54 in 4
12
12
I = I1 + I 2 + I 3 = 59.94 + 21.63 + 73.57 = 155.16 in 4
ytop = 3.222 in. ybot = −4.778 in.
I1 =
M − Pa = 0
M = Pa = (25)(20) = 500 kip  in.
 top = −
 bot = −
30
Mytop
I
=−
(500)(3.222)
155.16
Mybot
(500)(−4.778)
=−
I
155.16
 top = −10.38 ksi (compression) 
 bot = 15.40 ksi (tension) 
PROBLEM 4.39
A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa)
are bonded together to form the composite beam shown. Knowing that
the beam is bent about a horizontal axis by a couple of moment M =
35 N  m, determine the maximum stress in (a) the aluminum strip, (b)
the copper strip.
SOLUTION
Use aluminum as the reference material.
n = 1.0 in aluminum
n = Ec /Ea = 105/75 = 1.4 in copper
Transformed section:
A, mm2
nA, mm2
y0 , mm
nA y0 , mm 3

144
144
9
1296

144
201.6
3
Σ
Y0 =
345.6
604.8
1900.8
1900.8
= 5.50 mm
345.6
The neutral axis lies 5.50 mm above the bottom.
n1
1.0
b1h13 + n1 A1d12 =
(24)(6)3 + (1.0)(24)(6)(3.5) 2 = 2196 mm 4
12
12
n2
1.4
3
2
I 2 = b2 h2 + n2 A2 d 2 =
(24)(6)3 + (1.4)(24)(6)(2.5) 2 = 1864.8 mm 4
12
12
I = I1 + I 2 = 4060.8 mm 4 = 4.0608  10−9 m 4
I1 =
(a)
Aluminum:
n = 1.0,
 =−
y = 12 − 5.5 = 6.5 mm = 0.0065 m
nMy
(1.0)(35)(0.0065)
=−
= −56.0  106 Pa = −56.0 MPa
−9
I
4.0608  10
 = −56.0 MPa 
(b)
Copper:
n = 1.4,
 =−
y = −5.5 mm = −0.0055 m
nMy
(1.4)(35)(−0.0055)
=−
= 66.4  106 Pa = 66.4 MPa
I
4.0608  10−9
 = 66.4 MPa 
31
PROBLEM 4.42
The 6  12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8  106 psi and for
steel is 29  106 psi. Knowing that the beam is bent about a horizontal axis by a
couple of moment M = 450 kipin., determine the maximum stress in (a) the wood,
(b) the steel.
SOLUTION
Use wood as the reference material.
For wood,
n =1
For steel,
n=
Es 29  106
=
= 16.1111
Ew 1.8  106
For C8  11.5 channel section (see the screenshot taken from
Appendix E on the next page),
A = 3.38 in 2 , tw = 0.220 in.,
x = 0.571 in., I y = 1.32 in 4
For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section.
The centroid of the wood (part 2) lies 0.220 + 6.00 = 6.22 in. above the bottom.
Transformed section:
A, in2
3.38
Part
1
72
2

Y0 =
nA, in2
54.456
y , in.
0.571
72
6.22
31.091
447.84
d, in.
3.216
2.433
478.93
126.456
478.93 in 3
= 3.787 in.
126.456 in 2
nAy , in 3
d = y0 − Y0
The neutral axis lies 3.787 in. above the bottom of the section.
I1 = n1 I1 + n1 A1d12 = (16.1111)(1.32) + (54.456)(3.216)2 = 584.49 in 4
n2
1
b2 h23 + n2 A2 d 22 = (6)(12)3 + (72)(2.433)2 = 1290.20 in 4
12
12
4
I = I1 + I 2 = 1874.69 in
I2 =
M = 450 kip  in
(a)
Wood:
n = 1,
w = −
(b)
Steel:
(1)(450)(8.433)
= −2.02 ksi
1874.69
n = 16.1111,
s = −
32
n My
I
y = 12 + 0.220 − 3.787 = 8.433 in.
 =−
 w = −2.02 ksi 
y = −3.787 in.
(16.1111)(450)(−3.787)
= 14.65 ksi
1874.67
 s = 14.65 ksi 
33
PROBLEM 4.51
Knowing that the bending moment in the reinforced concrete beam is +100
kip  ft and that the modulus of elasticity is 3.625  106 psi for the concrete
and 29  106 psi for the steel, determine (a) the stress in the steel, (b) the
maximum stress in the concrete.
SOLUTION
n=
Es
29  106
=
= 8.0
Ec
3.625  106

 
As = (4)   (1)2 = 3.1416 in 2
4
nAs = 25.133 in
2
Locate the neutral axis.
x
(24)(4)( x + 2) + (12 x)   − (25.133)(17.5 − 4 − x) = 0
2
96 x + 192 + 6 x 2 − 339.3 + 25.133x = 0
x=
Solve for x.
or
6 x 2 + 121.133x − 147.3 = 0
−121.133 + (121.133)2 + (4)(6)(147.3)
= 1.150 in.
(2)(6)
d3 = 17.5 − 4 − x = 12.350 in.
1
1
b1h13 + A1d12 =
(24)(4)3 + (24)(4)(3.150) 2 = 1080.6 in 4
12
12
1
1
I 2 = b2 x3 = (12)(1.150)3 = 6.1 in 4
3
3
I1 =
I3 = nA3d32 = (25.133)(12.350)2 = 3833.3 in 4
I = I1 + I 2 + I 3 = 4920 in 4
 =−
(a)
Steel:
nMy
I
where M = 100 kip  ft = 1200 kip  in.
y = −12.350 in.
n = 8.0
s = −
(b)
Concrete:
 s = 24.1 ksi 
n = 1.0, y = 4 + 1.150 = 5.150 in.
c = −
34
(8.0)(1200)(−12.350)
4920
(1.0)(1200)(5.150)
4920
 c = −1.256 ksi 
PROBLEM 4.55
Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip  in., determine (a) the maximum stress in each of the three
metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
n=
Es 30  106
=
= 3.0 in steel
Ea 10  106
Eb 15  106
=
= 1.5 in brass
Ea 10  106
n = 1.0 in aluminum
n=
For the transformed section,
n1
1
b1h13 + n1 A1d12 = (1.5)(0.5)3 + (0.75)(1.0)2
12
12
4
= 0.7656 in
I1 =
n2
1.5
b2 h23 + n2 A2 d 22 =
(1.5)(0.5)3 + (1.5)(0.75)(0.5)2 = 0.3047 in 4
12
12
n3
3.0
3
I 3 = b3 h3 =
(1.5)(0.5)3 = 0.0469 in 4
12
12
I 4 = I 2 = 0.3047 in 4
I 5 = I1 = 0.7656 in 4
I2 =
I=
5
I
i
= 2.1875 in 4
1
(a)
Aluminum:
=
nMy (1.0)(12)(1.25)
=
= 6.86 ksi
I
2.1875

Brass:
=
nMy (1.5)(12)(0.75)
=
= 6.17 ksi
I
2.1875

Steel:
=
nMy (3.0)(12)(0.25)
=
= 4.11 ksi
I
2.1875

1
M
12  103
=
= 548.57  10−6 in.−1
6
Ea I (10  10 )(2.1875)
(b)

35


=
 = 1823 in. = 151.9 ft

PROBLEM 4.64
Semicircular grooves of radius r must be milled as shown in the top and
bottom portions of a steel member. Using an allowable stress of 10 ksi,
determine the largest bending moment that can be applied to the member
when (a) r = 0.25 in., (b) r = 0.5 in.
SOLUTION
(a)
d = D − 2r = 3 − 0.5 = 2.50 in.
D
3
=
= 1.20
d
2.50
From Fig. 4.25,
I =
r
0.25
=
= 0.10
d
2.55
K = 2.07
1 3
1 ( )
bh =
1 (2.5)3 = 1.30208 in 4
12
12
 =K
Mc
I
 M =
I
Kc
=
(10 ksi)(1.30208 in 4 )
(2.07)(1.25 in.)
 = 5.03 kip  in. 
(b)
From Fig. 4.25,
K = 1.60
M =
36
D
3
=
= 1.50
d
2.0
d = D − 2r = 3 − 2 ( 0.5 ) = 2.0 in.
I
Kc
I =
=
r
0.5
=
= 0.25
d
2.0
1 3
1( )
bh =
1 (2.0)3 = 0.66667 in 4
12
12
(10 ksi)(0.66667 in 4 )
(1.60)(1 in.)
 = 4.17 kip  in. 
PROBLEM 4.134
The couple M is applied to a beam of the cross section shown in a plane
forming an angle  with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
For W310  38.7 rolled steel shape,
I z = 84.9  106 mm 4 = 84.9  10−6 m 4
I y = 7.20  106 mm 4 = 7.20  10−6 m 4
1
1
yD = − yE =   (310) = 155 mm
2
2
1
z A = z E = − z B = − z D =   (165) = 82.5 mm
2
y A = yB =
M z = (16  103 ) cos 15 = 15.455  103 N  m
M y = (16  103 ) sin 15 = 4.1411  103 N  m
(a)
tan  =
Iz
84.9  10−6
tan  =
tan 15 = 3.1596
I
7.20  10−6
 = 72.4
 − 72.4 − 15 = 57.4
(b)
Maximum tensile stress occurs at point E.
E = −
M z yE M y z E
(15.455  103 )(−155  10−3 ) (4.1411  103 )(82.5  10−3 )
+
=−
+
Iz
Iy
84.9  10−6
7.20  10−6
= 75.7  106 Pa = 75.7 MPa
37


PROBLEM 4.141
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine the stress at point A.
SOLUTION
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (1.894, 0.800)  106 mm 4
Z : (0.614, 0.800)  106 mm 4
E : (1.254, 0)  106 mm 4
2
2
R = EF + FZ = 0.6402 + 0.8002  10−6 = 1.0245  106 mm 4
I v = (1.254 − 1.0245)  106 mm 4 = 0.2295  106 mm 4 = 0.2295  10−6 m 4
I u = (1.254 + 1.0245)  106 mm 4 = 2.2785  106 mm 4 = 2.2785  10−6 m4
FZ 0.800  106
=
= 1.25
 m = 25.67
FE 0.640  106
M v = M cos  m = (1.2  103 ) cos 25.67 = 1.0816  103 N  m
tan 2 m =
M u = − M sin  m = −(1.2  103 ) sin 25.67 = −0.5198  103 N  m
u A = y A cos  m − z A sin  m = 45 cos 25.67 − 45 sin 25.67 = 21.07 mm
v A = z A cos  m + y A sin  m = 45 cos 25.67 + 45 sin 25.67 = 60.05 mm
A = −
M v u A M u vA
(1.0816  103 )(21.07  10−3 ) (−0.5198  103 )(60.05  10−3 )
+
=−
+
Iv
Iu
0.2295  10−6
2.2785  10−6
= 113.0  106 Pa
38
 A = 113.0 MPa 
PROBLEM 4.144
The tube shown has a uniform wall thickness of 12 mm. For the
loading given, determine (a) the stress at points A and B, (b) the point
where the neutral axis intersects line ABD.
SOLUTION
Add y- and z-axes as shown. Cross section is a 75 mm  125-mm rectangle with a 51 mm  101-mm rectangular
cutout.
1
1
(75)(125)3 − (51)(101)3 = 7.8283  106 mm 4 = 7.8283  10−6 m 4
12
12
1
1
I y = (125)(75)3 − (101)(51)3 = 3.2781  103 mm 4 = 3.2781  10−6 m 4
12
12
A = (75)(125) − (51)(101) = 4.224  103 mm 2 = 4.224  10−3 m 2
Iz =
Resultant force and bending couples:
P = 14 + 28 + 28 = 70 kN = 70  103 N
M z = −(62.5 mm)(14 kN) + (62.5 mm)(28kN) + (62.5 mm)(28 kN) = 2625 N  m
M y = −(37.5 mm)(14 kN) + (37.5 mm)(28 kN) + (37.5 mm)(28 kN) = −525 N  m
(a)
A =
P M z yA M y zA
70  103
(2625)(−0.0625) (−525)(0.0375)
−
+
=
−
+
−3
A
Iz
Iy
4.224  10
7.8283  10−6
3.2781  10−6
 A = 31.5 MPa 
= 31.524  106 Pa
B =
P M z yB M y z B
70  103
(2625)(0.0625) (−525)(0.0375)
−
+
=
−
+
−3
A
Iz
Iy
4.224  10
7.8283 10−6
3.2781 10−6
 B = −10.39 MPa 
= −10.39  106 Pa
(b)
Let point H be the point where the neutral axis intersects AB.
z H = 0.0375 m,
0=
yH = ?,  H = 0
P M z yH M y z H
−
+
A
Iz
Iy
 P Mz H  7.8283  10−6  70  103
(−525)(0.0375) 
+
 +
=


−
3
A
I y 
2625
3.2781  10−6 
 4.224  10

= 0.03151 m = 31.51 mm
yH =
Iz
Mz
31.51 + 62.5 = 94.0 mm
39
Answer: 94.0 mm above point A. 
PROBLEM 4.197
The vertical portion of the press shown consists of a rectangular
tube of wall thickness t = 10 mm. Knowing that the press has been
tightened on wooden planks being glued together until P = 20 kN,
determine the stress at (a) point A, (b) point B.
SOLUTION
Rectangular cutout is 60 mm  40 mm.
A = (80)(60) − (60)(40) = 2.4  103 mm 2 = 2.4  10−3 m 2
I =
1
1
(60)(80)3 − (40)(60)3 = 1.84  106 mm 4
12
12
= 1.84  10−6 m 4
c = 40 mm = 0.040 m e = 200 + 40 = 240 mm = 0.240 m
P = 20  103 N
M = Pe = (20  103 )(0.240) = 4.8  103 N  m
(a)
A =
P Mc
20  103
(4.8  103 )(0.040)
+
=
+
= 112.7  106 Pa
−3
A
I
2.4  10
1.84  10−6
 A = 112.7 MPa 
(b)
B =
P Mc
20  103
(4.8  103 )(0.040)
−
=
−
= −96.0  106 Pa
A
I
2.4  10−3
1.84  10−6
 B = −96.0 MPa 
40