1 Hibbeler - Resistências dos Materiais - 7ªed - Soluções

Prévia do material em texto

ISM for 
Problem 1-1
Determine the resultant internal normal force acting on the cross section through point A in each
column. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the column
has a mass of 200 kg/m.
a( ) Given: g 9.81
m
s
2
� wBC 300
kg
m
� 
LBC 3m� wCA 400
kg
m
� 
FB 5kN� LCA 1.2m� 
FC 3kN� 
Solution:
�n6 Fy = 0; FA wBC g˜� � LBC˜� wCA g˜� � LCA˜� FB� 2FC� 0=
FA wBC g˜� � LBC˜ wCA g˜� � LCA˜� FB� 2FC�� 
FA 24.5 kN Ans
b( ) Given: g 9.81
m
s
2
� w 200 kg
m
� 
L 3m� F1 6kN� 
FB 8kN� F2 4.5kN� 
Solution:
�n6 Fy = 0; FA w L˜( ) g˜� FB� 2F1� 2F2� 0=
FA w L˜( ) g˜ FB� 2F1� 2F2�� 
FA 34.89 kN Ans
Problem 1-2
Determine the resultant internal torque acting on the cross sections through points C and D of the
shaft. The shaft is fixed at B.
Given: TA 250N m˜� 
TCD 400N m˜� 
TDB 300N m˜� 
Solution:
Equations of equilibrium:
+ TA TC� 0=
TC TA� 
TC 250 N m˜ Ans
+ TA TCD� TD� 0=
TD TCD TA�� 
TD 150 N m˜ Ans
Problem 1-3
Determine the resultant internal torque acting on the cross sections through points B and C.
Given: TD 500N m˜� 
TBC 350N m˜� 
TAB 600N m˜� 
Solution:
Equations of equilibrium:
6 Mx = 0; TB TBC TD�� 0=
TB TBC� TD�� 
TB 150 N m˜ Ans
6 Mx = 0; TC TD� 0=
TC TD� 
TC 500 N m˜ Ans
Problem 1-4
A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting
on the section through point A.
Given: P 80N� 
T 30deg� I 45deg� 
a 0.3m� b 0.1m� 
Solution:
Equations of equilibrium:
 + 6Fx'=0; NA P cos I T�� �˜� 0=
NA P cos I T�� �˜� 
NA 77.27 N Ans
+ 6Fy'=0; VA P sin I T�� �˜� 0=
VA P sin I T�� �˜� 
VA 20.71 N Ans
+ 60A=0; MA P cos I� �˜ a˜ cos T� �� P sin I� �˜ b a sin T� ��� �˜� 0=
MA P� cos I� �˜ a˜ cos T� � P sin I� �˜ b a sin T� ��� �˜�� 
MA 0.555� N m˜ Ans
Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Problem 1-5
Determine the resultant internal loadings acting on the cross section through point D of member AB.
Given: ME 70N m˜� 
a 0.05m� b 0.3m� 
Solution:
Segment AB: Support Reactions
+ 60A=0; �ME By 2 a˜ b�( )˜� 0=
By
ME�
2a b�� By 175� N 
+At B: Bx By
150
200
§¨
©
·
¹˜� Bx 131.25� N 
Segment DB: NB Bx�� VB By�� 
+ 6Fx=0; ND NB� 0=
ND NB�� ND 131.25� N Ans
+ 6Fy=0; VD VB� 0=
VD VB�� VD 175� N Ans
+ 60D=0; MD� ME� By a b�( )˜� 0=
MD ME� By a b�( )˜�� 
MD 8.75� N m˜ Ans
Problem 1-6
The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internal
loadings acting on the cross section at point D.
Given: P 5000N� 
a 0.8m� b 1.2m� c 0.6m� d 1.6m� 
e 0.6m� 
Solution:
T atan b
d
§¨
©
·
¹� T 36.87 deg 
I atan a b�
d
§¨
©
·
¹ T�� I 14.47 deg 
Member AB: 
+ 60A=0; FBC sin I� �˜ a b�( )˜ P b( )˜� 0=
FBC
P b( )˜
sin I� � a b�( )˜� 
FBC 12.01 kN 
Segment BD: 
 + 6Fx=0; ND� FBC cos I� �˜� P cos T� �˜� 0=
ND FBC� cos I� �˜ P cos T� �˜�� 
ND 15.63� kN Ans
+ 6Fy=0; VD FBC sin I� �˜� P sin T� �˜� 0=
VD FBC� sin I� �˜ P sin T� �˜�� 
VD 0 kN Ans
+ 60D=0; FBC sin I� �˜ P sin T� �˜�� � d c�
sin T� �˜ MD� 0=
MD FBC sin I� �˜ P sin T� �˜�� � d c�
sin T� �˜� 
MD 0 kN m˜ Ans
Note: Member AB is the two-force member. Therefore the shear force and moment are zero.
Problem 1-7
Solve Prob. 1-6 for the resultant internal loadings acting at point E.
Given: P 5000N� 
a 0.8m� b 1.2m� c 0.6m� d 1.6m� 
e 0.6m� 
Solution:
T atan b
d
§¨
©
·
¹� T 36.87 deg 
I atan a b�
d
§¨
©
·
¹ T�� I 14.47 deg 
Member AB: 
+ 60A=0; FBC sin I� �˜ a b�( )˜ P b( )˜� 0=
FBC
P b( )˜
sin I� � a b�( )˜� 
FBC 12.01 kN 
Segment BE: 
 + 6Fx=0; NE� FBC cos I� �˜� P cos T� �˜� 0=
NE FBC� cos I� �˜ P cos T� �˜�� 
NE 15.63� kN Ans
+ 6Fy=0; VE FBC sin I� �˜� P sin T� �˜� 0=
VE FBC� sin I� �˜ P sin T� �˜�� 
VE 0 kN Ans
+ 60E=0; FBC sin I� �˜ P sin T� �˜�� � e˜ ME� 0=
ME FBC sin I� �˜ P sin T� �˜�� � e˜� 
ME 0 kN m˜ Ans
Note: Member AB is the two-force member. Therefore the shear force and moment are zero.
Problem 1-8
The boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist and
load weigh 1500 N, determine the resultant internal loadings in the crane on cross sections through
points A, B, and C.
Given: P 1500N� w 750 N
m
� 
a 2.1m� b 1.5m� 
c 0.6m� d 2.4m� e 0.9m� 
Solution:
Equations of Equilibrium: For point A
+ 6Fx=0; NA 0� Ans
+
VA w e˜� P� 0=6Fy=0;
VA w e˜ P�� VA 2.17 kN Ans
+ 60A=0; MA� w e˜( ) 0.5 e˜( )˜� P e( )˜� 0=
MA w e˜( )� 0.5 e˜( )˜ P e( )˜�� MA 1.654� kN m˜ Ans
Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point B
+ 6Fx=0; NB 0� Ans
+
VB w d e�( )˜� P� 0=6Fy=0;
VB w d e�( )˜ P�� VB 3.98 kN Ans
+ 60B=0; �MB w d e�( )˜[ ] 0.5 d e�( )˜[ ]˜� P d e�( )˜� 0=
MB w d e�( )˜[ ]� 0.5 d e�( )˜[ ]˜ P d e�( )˜�� 
MB 9.034� kN m˜ Ans
Note: Negative sign indicates that MB acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point C
+ 6Fx=0; VC 0� Ans
+
NC� w b c� d� e�( )˜� P� 0=6Fy=0;
NC w� b c� d� e�( )˜ P�� 
NC 5.55� kN Ans
+ 60B=0; MC� w c d� e�( )˜[ ] 0.5 c d� e�( )˜[ ]˜� P c d� e�( )˜� 0=
MC w c d� e�( )˜[ ]� 0.5 c d� e�( )˜[ ]˜ P c d� e�( )˜�� 
MC 11.554� kN m˜ Ans
Note: Negative sign indicates that NC and MC acts in the opposite direction to that shown
on FBD.
Problem 1-9
The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of the
tooth, i.e., at the centroid point A of section a-a.
Given: P 400N� 
T 30deg� I 45deg� 
a 4mm� b 5.75mm� 
Solution: D I T�� 
Equations of equilibrium: For section a -a
 + 6Fx'=0; VA P cos D� �˜� 0=
VA P cos D� �˜� 
VA 386.37 N Ans
+ 6Fy'=0; NA sin D� �� 0=
NA P sin D� �˜� 
NA 103.53 N Ans
+ 60A=0; MA� P sin D� �˜ a˜� P cos D� �˜ b˜� 0=
MA P� sin D� �˜ a˜ P cos D� �˜ b˜�� 
MA 1.808 N m˜ Ans
Problem 1-10
The beam supports the distributed load shown. Determine the resultant internal loadings on the cross
section through point C. Assume the reactions at the supports A and B are vertical.
Given: w1 4.5
kN
m
� w2 6.0
kN
m
� 
a 1.8m� b 1.8m� c 2.4m� 
d 1.35m� e 1.35m� 
Solution: L1 a b� c�� 
L2 d e�� 
Support Reactions: 
+ 60A=0; By L1˜ w1 L1˜� � 0.5 L1˜� �� 0.5w2 L2˜� � L1 L23�§¨©
·
¹˜� 0=
By w1 L1˜� � 0.5( )˜ 0.5w2 L2˜� � 1 L23 L1˜�
§¨
©
·
¹
˜�� By 22.82 kN 
+ 6Fy=0; Ay By� w1 L1˜� 0.5w2 L2˜� 0=
Ay By� w1 L1˜� 0.5w2 L2˜�� Ay 12.29 kN 
Equations of Equilibrium: For point C 
+ 6Fx=0; NC 0� Ans
+ 6Fy=0; Ay w1 a b�( )˜�ª¬ º¼ VC� 0=
VC Ay w1 a b�( )˜�ª¬ º¼�� VC 3.92 kN Ans
+ 60C=0; MC w1 a b�( )˜ª¬ º¼ 0.5˜ a b�( )˜� Ay a b�( )˜� 0=
MC w1 a b�( )˜ª¬ º¼� 0.5˜ a b�( )˜ Ay a b�( )˜�� 
MC 15.07 kN m˜ Ans
Note: Negative sign indicates that VC acts in the opposite direction to that shown on FBD.
Problem 1-11
The beam supports the distributed load shown. Determine the resultant internal loadings on the cross
sections through points D and E. Assume the reactions at the supports A and B are vertical.
Given: w1 4.5
kN
m
� w2 6.0
kN
m
� 
a 1.8m� b 1.8m� c 2.4m� 
d 1.35m� e 1.35m� 
Solution: L1 a b� c�� 
L2 d e�� 
Support Reactions: 
+ 60A=0; By L1˜ w1 L1˜� � 0.5 L1˜� �� 0.5w2 L2˜� � L1 L23�§¨©
·
¹˜� 0=
By w1 L1˜� � 0.5( )˜ 0.5w2 L2˜� � 1 L23 L1˜�
§¨
©
·
¹
˜�� By 22.82 kN 
+ 6Fy=0; Ay By� w1 L1˜� 0.5w2 L2˜� 0=
Ay By� w1 L1˜� 0.5w2 L2˜�� Ay 12.29 kN 
Equations of Equilibrium: For point D 
+ 6Fx=0; ND 0� Ans
+ 6Fy=0; Ay w1 a( )˜�ª¬ º¼ VD� 0=
VD Ay w1 a( )˜�� VD 4.18 kN Ans
+ 60D=0; MD w1 a( )˜ª¬ º¼ 0.5˜ a( )˜� Ay a( )˜� 0=
MD w1 a( )˜ª¬ º¼� 0.5˜ a()˜ Ay a( )˜�� MD 14.823 kN m˜ Ans
Equations of Equilibrium: For point E 
+ 6Fx=0; NE 0� Ans
+ 6Fy=0; VE 0.5w2 0.5 e˜( )˜� 0=
VE 0.5w2 0.5 e˜( )˜� VE 2.03 kN Ans
+ 60D=0; ME� 0.5w2 0.5 e˜( )˜ª¬ º¼ e3
§¨
©
·
¹˜� 0=
ME 0.5� w2 0.5 e˜( )˜ª¬ º¼ e3
§¨
©
·
¹˜� ME 0.911� kN m˜ Ans
Note: Negative sign indicates that ME acts in the opposite direction to that shown on FBD.
Problem 1-12
Determine the resultant internal loadings acting on (a) section a-a and (b) section b-b. Each section is
located through the centroid, point C.
Given: w 9
kN
m
� T 45deg� 
a 1.2m� b 2.4m� 
Solution: L a b�� 
Support Reactions: 
+ 60A=0; Bx� L˜ sin T� �˜ w L˜( ) 0.5 L˜( )� 0=
Bx
w L˜( ) 0.5 L˜( )˜
L sin T� �˜� Bx 22.91 kN 
+ 6Fy=0; Ay w L˜ sin T� �˜� 0=
Ay w L˜ sin T� �˜� Ay 22.91 kN 
+ 6Fx=0; Bx w L˜ cos T� �˜� Ax� 0=
Ax w L˜ cos T� �˜ Bx�� Ax 0� kN 
(a) Equations of equilibrium: For Section a - a : 
+ 6Fx=0; NC Ay sin T� �˜� 0=
NC Ay� sin T� �˜� 
 + 6Fy=0; VC Ay cos T� �˜� w a˜� 0=
VC Ay� cos T� �˜ w a˜�� VC 5.4� kN Ans
+ 60A=0; MC� w a˜( ) 0.5 a˜( )˜� Ay cos T� � a˜� 0=
MC w a˜( ) 0.5 a˜( )˜ Ay cos T� � a˜�� MC 12.96� kN m˜ Ans
(b) Equations of equilibrium: For Section b - b : 
+ 6Fx=0; NC w a˜ cos T� �˜� 0=
NC w� a˜ cos T� �˜� NC 7.64� kN Ans
+ 6Fy=0; VC w a˜ sin T� �˜� Ay� 0=
VC w a˜ sin T� �˜ Ay�� VC 15.27� kN Ans
+ 60A=0; MC� w a˜( ) 0.5 a˜( )˜� Ay cos T� � a˜� 0=
MC w a˜( ) 0.5 a˜( )˜ Ay cos T� � a˜�� MC 12.96� kN m˜ Ans
NC 16.2� kN Ans
Problem 1-13
Determine the resultant internal normal and shear forces in the member at (a) section a-a and (b)
section b-b, each of which passes through point A. Take T�= 60 degree. The 650-N load is applied
along the centroidal axis of the member.
Given: P 650N� T 60deg� 
(a) Equations of equilibrium: For Section a - a : 
+ 6Fy=0; P Na_a� 0=
Na_a P� 
Na_a 650 N Ans
+ 6Fx=0; Va_a 0� Ans
(b) Equations of equilibrium: For Section b - b : 
 + 6Fy=0; Vb_b� P cos 90deg T�� �˜� 0=
Vb_b P cos 90deg T�� �˜� 
Vb_b 562.92 N Ans
+ 6Fx=0; Nb_b P sin 90deg T�� �˜� 0=
Nb_b P sin 90deg T�� �˜� 
Nb_b 325 N Ans
Problem 1-14
Determine the resultant internal normal and shear forces in the member at section b-b, each as a
function of T. Plot these results for 0o Td 90od . The 650-N load is applied along the centroidal axis
of the member.
Given: P 650N� T 0� 
 Equations of equilibrium: For Section b - b : 
+ 6Fx0; Nb_b P cos T� �˜� 0=
Nb_b P cos T� �˜� Ans
 + 6Fy=0; Vb_b� P cos T� �˜� 0=
Vb_b P� cos T� �˜� Ans
Problem 1-15
The 4000-N load is being hoisted at a constant speed using the motor M, which has a weight of 450 N.
Determine the resultant internal loadings acting on the cross section through point B in the beam. The
beam has a weight of 600 N/m and is fixed to the wall at A.
Given:
W1 4000N� w 600
N
m
� 
W2 450N� 
a 1.2m� b 1.2m� c 0.9m� 
d 0.9m� e 1.2m� 
f 0.45m� r 0.075m� 
Solution:
Tension in rope: T
W1
2
� 
T 2.00 kN 
Equations of Equilibrium: For point B 
+ 6Fx=0; NB� T�� � 0=
NB T�� NB 2� kN Ans
+ 6Fy=0; VB w e( )˜� W1� 0=
VB w e( )˜ W1�� VB 4.72 kN Ans
+ 60B=0; MB� w e( )˜[ ] 0.5˜ e( )˜� W1 e r�( )˜� T f( )˜� 0=
MB w e( )˜[ ]� 0.5˜ e( )˜ W1 e r�( )˜� T f( )˜�� 
MB 4.632� kN m˜ Ans
Problem 1-16
Determine the resultant internal loadings acting on the cross section through points C and D of the
beam in Prob. 1-15.
Given: W1 4000N� 
w 600
N
m
� 
W2 450N� 
a 1.2m� b 1.2m� c 0.9m� 
d 0.9m� e 1.2m� 
f 0.45m� r 0.075m� 
Solution:
Tension in rope: T
W1
2
� T 2.00 kN 
Equations of Equilibrium: For point C LC d e�� 
+ 6Fx=0; NC� T�� � 0=
NC T�� NC 2� kN Ans
+ 6Fy=0; VC w LC� �˜� W1� 0=
VC w LC� �˜ W1�� VC 5.26 kN Ans
+ 60C=0; MC� w LC� �˜ª¬ º¼ 0.5˜ LC� �˜� W1 LC r�� �˜� T f( )˜� 0=
MC w LC� �˜ª¬ º¼� 0.5˜ LC� �˜ W1 LC r�� �˜� T f( )˜�� 
MC 9.123� kN m˜ Ans
Equations of Equilibrium: For point D LD b c� d� e�� 
+ 6Fx=0; ND 0� ND 0 kN Ans
+ 6Fy=0; VD w LD� �˜� W1� W2� 0=
VD w LD� �˜ W1� W2�� VD 6.97 kN Ans
+ 60C=0; MD� w LD� �˜ª¬ º¼ 0.5˜ LD� �˜� W1 LD r�� �˜� W2 b( )˜� 0=
MD w LD� �˜ª¬ º¼� 0.5˜ LD� �˜ W1 LD r�� �˜� W2 b( )˜�� 
MD 22.932� kN m˜ Ans
Problem 1-17
Determine the resultant internal loadings acting on the cross section at point B.
Given: w 900
kN
m
� 
a 1m� b 4m� 
Solution: L a b�� 
Equations of Equilibrium: For point B 
+ 6Fx=0; NB 0� 
NB 0 kN Ans
+ 6Fy=0; VB 0.5 w
b
L
§¨
©
·
¹˜
ª«¬
º»¼˜ b( )˜� 0=
VB 0.5 w
b
L
§¨
©
·
¹˜
ª«¬
º»¼˜ b( )˜� 
VB 1440 kN Ans
+ 60B=0; MB� 0.5 w
b
L
§¨
©
·
¹˜
ª«¬
º»¼˜ b( )˜
b
3
§¨
©
·
¹˜� 0=
MB 0.5� w
b
L
§¨
©
·
¹˜
ª«¬
º»¼˜ b( )˜
b
3
˜� 
MB 1920� kN m˜ Ans
Problem 1-18
The beam supports the distributed load shown. Determine the resultant internal loadings acting on the
cross section through point C. Assume the reactions at the supports A and B are vertical.
Given: w1 0.5
kN
m
� a 3m� 
w2 1.5
kN
m
� 
Solution: L 3 a˜� w w2 w1�� 
Support Reactions: 
+ 60A=0; By L˜ w1 L˜� � 0.5 L˜( )� 0.5 w( ) L˜[ ] 2L3§¨© ·¹˜� 0=
By w1 L˜� � 0.5( )˜ 0.5 w( ) L˜[ ] 23§¨© ·¹˜�� 
By 5.25 kN 
+ 6Fy=0; Ay By� w1 L˜� 0.5 w( ) L˜� 0=
Ay By� w1 L˜� 0.5 w( ) L˜�� 
Ay 3.75 kN 
Equations of Equilibrium: For point C 
+ 6Fx=0; NC 0� NC 0 kN Ans
+ 6Fy=0; VC w1 a˜� 0.5 w
a
L
§¨
©
·
¹˜
ª«¬
º»¼˜ a( )˜� Ay� 0=
VC w1� a˜ 0.5 w
a
L
§¨
©
·
¹˜
ª«¬
º»¼˜ a( )˜� Ay�� 
VC 1.75 kN Ans
+ 60C=0; MC w1 a˜� � 0.5 a˜( )� 0.5 w aL§¨© ·¹˜ª«¬ º»¼˜ a( )˜ a3§¨© ·¹˜� Ay a˜� 0=
MC w1� a˜� � 0.5 a˜( ) 0.5 w aL§¨© ·¹˜ª«¬ º»¼˜ a( )˜ a3§¨© ·¹˜� Ay a˜�� 
MC 8.5 kN m˜ Ans
Problem 1-19
Determine the resultant internal loadings acting on the cross section through point D in Prob. 1-18.
Given: w1 0.5
kN
m
� a 3m� 
w2 1.5
kN
m
� 
Solution: L 3 a˜� w w2 w1�� 
Support Reactions: 
+ 60A=0; By L˜ w1 L˜� � 0.5 L˜( )� 0.5 w( ) L˜[ ] 2L3§¨© ·¹˜� 0=
By w1 L˜� � 0.5( )˜ 0.5 w( ) L˜[ ] 23§¨© ·¹˜�� 
By 5.25 kN 
+ 6Fy=0; Ay By� w1 L˜� 0.5 w( ) L˜� 0=
Ay By� w1 L˜� 0.5 w( ) L˜�� 
Ay 3.75 kN 
Equations of Equilibrium: For point D 
+ 6Fx=0; ND 0� ND 0 kN Ans
+ 6Fy=0; VD w1 2a( )˜� 0.5 w
2a
L
§¨
©
·
¹˜
ª«¬
º»¼˜ 2a( )˜� Ay� 0=
VD w1� 2a( )˜ 0.5 w
2 a˜
L
§¨
©
·
¹˜
ª«¬
º»¼˜ 2a( )˜� Ay�� 
VD 1.25� kN Ans
+ 60D=0; MD w1 2a( )˜ª¬ º¼ a( )� 0.5 w 2 a˜L
§¨
©
·
¹˜
ª«¬
º»¼˜ 2a( )˜
2a
3
§¨
©
·
¹˜� Ay 2a( )˜� 0=
MD w1� 2a( )˜ª¬ º¼ a( ) 0.5 w 2 a˜L
§¨
©
·
¹˜
ª«¬
º»¼˜ 2a( )˜
2a
3
§¨
©
·
¹˜� Ay 2a( )˜�� 
MD 9.5 kN m˜ Ans
Problem 1-20
The wishbone construction of the power pole supports the three lines, each exerting a force of 4 kN
on the bracing struts. If the struts are pin connected at A, B, and C, determine the resultant internal
loadings at cross sections through points D, E, and F.
Given: P 4kN� a 1.2m� b 1.8m� 
Solution:
Support Reactions: FBD (a) and (b).
Given
+ 60A=0; By a( )˜ Bx 0.5 b˜( )˜� P a( )˜� 0= [1]
+ 60C=0; Bx 0.5 b˜( )˜ P a( )˜� By a( )˜� P a( )˜� 0= [2]
Solving [1] and [2]: Initial guess: Bx 1kN� By 2kN� 
Bx
By
§¨
©¨
·
¹
Find Bx By�� �� Bx
By
§¨
©¨
·
¹
2.67
2
§¨
©
·
¹ kN 
From FBD (a):
+ 6Fx=0; Bx Ax� 0=
Ax Bx� Ax 2.67 kN 
+ 6Fy=0; Ay P� By� 0=
Ay P By�� Ay 6 kN 
From FBD (b):
+ 6Fx=0; Cx Bx� 0=
Cx Bx� Cx 2.67 kN 
+ 6Fy=0; Cy By� P� P� 0=
Cy 2P By�� Cy 6 kN 
Equations of Equilibrium: For point D [FBD (c)].
+ 6Fx=0; VD 0� VD 0 kN Ans
+ 6Fy=0; ND 0� ND 0 kN Ans
+ 60D=0; MD 0� MD 0 kN m˜ Ans
For point E [FBD (d)].
+ 6Fx=0; VF Ax Cx�� 0= VF Ax� Cx�� VF 0 kN Ans
+ 6Fy=0; NF Ay Cy�˜� 0= NF Ay Cy�� NF 12 kN Ans
+ 60F=0; MF Ax Cx�� � 0.5 b˜( )˜� 0= MF Ax Cx�� �0.5 b˜( )˜� MF 4.8 kN m˜ Ans
+ 6Fx=0; Ax VE� 0= VE Ax� 
+ 6Fy=0; NE Ay� 0= NE Ay� 
+ 60E=0; ME Ax 0.5 b˜( )˜� 0= ME Ax 0.5 b˜( )˜� ME 2.4 kN m˜ Ans
For point F [FBD (e)].
VE 2.67 kN Ans
NE 6 kN Ans
Problem 1-21
The drum lifter suspends the 2.5-kN drum. The linkage is pin connected to the plate at A and B. The
gripping action on the drum chime is such that only horizontal and vertical forces are exerted on the
drum at G and H. Determine the resultant internal loadings on the cross section through point I.
Given: P 2.5 kN� T 60deg� 
a 200mm� b 125mm� c 75mm� 
d 125mm� e 125mm� f 50mm� 
Solution:
Equations of Equilibrium: Memeber Ac and BD are
two-force members.
6Fy=0; P 2 F˜ sin T� �� 0= [1]
F
P
2 sin T� �˜� [2]
F 1.443 kN 
Equations of Equilibrium: For point I.
+ 6Fx=0; VI F cos T� �˜� 0= VI F cos T� �˜� VI 0.722 kN Ans
+ 6Fy=0; NI� F sin T� �˜� 0= NI F sin T� �˜� NI 1.25 kN Ans
+ 60I=0; MI� F cos T� �˜ a( )˜� 0=
MI F cos T� �˜ a( )˜� MI 0.144 kN m˜ Ans
Problem 1-22
Determine the resultant internal loadings on the cross sections through points K and J on the drum lifter
in Prob. 1-21.
Given: P 2.5 kN� T 60deg� 
a 200mm� b 125mm� c 75mm� 
d 125mm� e 125mm� f 50mm� 
Solution:
Equations of Equilibrium: Memeber Ac and BD are
two-force members.
6Fy=0; P 2 F˜ sin T� �� 0= [1]
F
P
2 sin T� �˜� [2]
F 1.443 kN 
Equations of Equilibrium: For point J.
+ 6Fy'=0; VI 0� VI 0 kN Ans
 + 6Fx'=0; NI F� 0= NI F�� NI 1.443� kN Ans
+ 60J=0; MJ 0� MJ 0 kN m˜ Ans
Note: Negative sign indicates that NJ acts in the opposite direction to that shown on FBD.
Support Reactions: For Member DFH : 
+ 60H=0; FEF c( )˜ F cos T� � a b� c�( )˜� F sin T� �˜ f( )˜� 0=
FEF F cos T� �˜ a b� c�c§¨©
·
¹˜ F sin T� �˜
f
c
§¨
©
·
¹˜�� 
FEF 3.016 kN 
Equations of Equilibrium: For point K.
+ 6Fx=0; NK FEF� 0= NK FEF� NK 3.016 kN Ans
+ 6Fy=0; VK 0� VK 0 kN Ans
+ 60K=0; MK 0� MK 0 kN m˜ Ans
Problem 1-23
The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings
acting on the cross section at B. Neglect the weight of the wrench CD.
Given: g 9.81
m
s
2
� U 12 kg
m
� P 60N� 
a 0.150m� b 0.400m� 
c 0.200m� d 0.300m� 
Solution: w U g˜� 
6Fx=0; NBx 0N� Ans
6Fy=0; VBy 0N� Ans
6Fz=0; VBz P� P� w b c�( )˜� 0=
VBz P P� w b c�( )˜�� 
VBz 70.6 N Ans
60x=0; TBx P b( )˜� P b( )˜� w b˜( ) 0.5 b˜( )˜� 0=
TBx P� b( )˜ P b( )˜� w b˜( ) 0.5 b˜( )˜�� TBx 9.42 N m˜ Ans
60y=0; MBy P 2a( )˜� w b˜( ) c( )˜� w c˜( ) 0.5 c˜( )˜� 0=
MBy P 2 a˜( )˜ w b˜( ) c( )˜� w c˜( ) 0.5 c˜( )˜�� MBy 6.23 N m˜ Ans
60z=0; MBz 0N m˜� Ans
Problem 1-24
The main beam AB supports the load on the wing of the airplane. The loads consist of the wheel
reaction of 175 kN at C, the 6-kN weight of fuel in the tank of the wing, having a center of gravity at
D, and the 2-kN weight of the wing, having a center of gravity at E. If it is fixed to the fuselage at A,
determine the resultant internal loadings on the beam at this point. Assume that the wing does not
transfer any of the loads to the fuselage, except through the beam.
Given: PC 175kN� PE 2kN� PD 6kN� 
a 1.8m� b 1.2m� e 0.3m� 
c 0.6m� d 0.45m� 
Solution:
6Fx=0; VAx 0kN� Ans
6Fy=0; NAy 0kN� Ans
6Fz=0; VAz PD� PE� PC� 0=
VAz PD PE PC��� 
VAz 167� kN Ans
60x=0;
MAx PD a( )˜ PE a b� c�( )˜� PC a b�( )˜�� MAx 507� kN m˜ Ans
60y=0; TAy PD d( )˜� PE e( )˜� 0=
TAy PD� d( )˜ PE e( )˜�� TAy 2.1� kN m˜ Ans
60z=0; MAz 0kN m˜� Ans
MAx PD a( )˜� PE a b� c�( )˜� PC a b�( )˜� 0=
Problem 1-25
Determine the resultant internal loadings acting on the cross section through point B of the signpost.
The post is fixed to the ground and a uniform pressure of 50 N/m2 acts perpendicular to the face of the
sign.
Given: a 4m� d 2m� 
p 50
N
m
2
� 
b 6m� e 3m� 
c 3m� 
Solution: P p c( )˜ d e�( )˜� 
6Fx=0; VBx P� 0=
VBx P� 
VBx 750 N Ans
6Fy=0; VBy 0N� Ans
6Fz=0; NBz 0N� Ans
60x=0; MBx 0N m˜� Ans
60y=0; MBy P b 0.5 c˜�( )˜� 0=
MBy P b 0.5 c˜�( )˜� 
MBy 5625 N m˜ Ans
60z=0; TBz P e 0.5 d e�( )˜�[ ]˜� 0=
TBz P e 0.5 d e�( )˜�[ ]˜� 
TBz 375 N m˜ Ans
Problem 1-26
The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the
pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through
point D. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y
direction. The journal bearings at A and B exert only y and z components of force on the shaft.
Given: P1z 400N� P2y 200N� P3y 80N� 
a 0.3m� b 0.4m� c 0.3m� d 0.4m� 
Solution: L a b� c� d�� 
Support Reactions:
60z=0; 2P3y d( )˜ 2P2y c d�( )˜� Ay L( )˜� 0=
Ay 2P3y
d
L
˜ 2P2y
c d�
L
˜�� 
Ay 245.71 N 
6Fy=0; Ay� By� 2 P2y˜� 2 P3y˜� 0=
By Ay� 2 P2y˜� 2 P3y˜�� By 314.29 N 
60y=0; 2P1z b c� d�( )˜ Az L( )˜� 0=
Az 2P1z
b c� d�
L
˜� Az 628.57 N 
6Fz=0; Bz Az� 2 P1z˜� 0=
Bz Az� 2 P1z˜�� Bz 171.43 N 
Equations of Equilibrium: For point D.
6Fx=0; NDx 0N� Ans
6Fy=0; VDy By� 2 P3y˜� 0=
VDy By 2 P3y˜�� 
VDy 154.3 N Ans
6Fz=0; VDz Bz� 0=
VDz Bz�� 
VDz 171.4� N Ans
60x=0; TDx 0N m˜� Ans
60y=0; MDy Bz d 0.5 c˜�( )˜� 0=
MDy Bz d 0.5 c˜�( )˜ª¬ º¼�� MDy 94.29� N m˜ Ans
60z=0; MDz By d 0.5 c˜�( )˜� 2 P3y˜ 0.5 c˜( )˜� 0=
MDz By� d 0.5 c˜�( )˜ 2 P3y˜ 0.5 c˜( )˜�� MDz 148.86� N m˜ Ans
Problem 1-27
The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the
pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through
point C. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y
direction. The journal bearings at A and B exert only y and z components of force on the shaft.
Given: P1z 400N� P2y 200N� P3y 80N� 
a 0.3m� b 0.4m� c 0.3m� d 0.4m� 
Solution: L a b� c� d�� 
Support Reactions:
60z=0; 2P3y d( )˜ 2P2y c d�( )˜� Ay L( )˜� 0=
Ay 2P3y
d
L
˜ 2P2y
c d�
L
˜�� Ay 245.71 N 
6Fy=0; Ay� By� 2 P2y˜� 2 P3y˜� 0=
By Ay� 2 P2y˜� 2 P3y˜�� By 314.29 N 
60y=0; 2P1z b c� d�( )˜ Az L( )˜� 0=
Az 2P1z
b c� d�
L
˜� Az 628.57 N 
6Fz=0; Bz Az� 2 P1z˜� 0=
Bz Az� 2 P1z˜�� Bz 171.43 N 
Equations of Equilibrium: For point C.
6Fx=0; NCx 0N� Ans
6Fy=0; VCy Ay� 0=
VCy Ay� 
VCy 245.7 N Ans
6Fz=0; VCz Az� 2P1z� 0=
VCz Az� 2P1z�� 
VCz 171.4 N Ans
60x=0; TCx 0N m˜� Ans
60y=0; MCy Az a 0.5 b˜�( )˜� 2 P1z˜ 0.5 b˜( )˜� 0=
MCy Az a 0.5 b˜�( )˜ 2 P1z˜ 0.5 b˜( )˜�� MCy 154.29 N m˜ Ans
60z=0; MCz Ay a 0.5 b˜�( )˜� 0=
MCz Ay� a 0.5 b˜�( )˜� MCz 122.86� N m˜ Ans
Problem 1-28
Determine the resultant internal loadings acting on the cross section of the frame at points F and G.
The contact at E is smooth.
Given: a 1.2m� b 1.5m� c 0.9m� P 400N� 
d 0.9m� e 1.2m� T 30deg� 
Solution: L d
2
e
2�� 
Member DEF : 
+ 60D=0; NE b( )˜ P a b�( )˜� 0=
NE P
a b�
b
˜� NE 720 N 
Member BCE : 
+ 60B=0; FAC
e
L
§¨
©
·
¹˜ d( )˜ NE sin T� �˜ c d�( )˜� 0=
FAC
L
e d˜
§¨
©
·
¹ NE sin T� �˜ c d�( )˜ª¬ º¼˜� 
FAC 900 N 
+ 6Fx=0; Bx FAC
d
L
§¨
©
·
¹˜� NE cos T� �˜� 0=
Bx FAC�
d
L
§¨
©
·
¹˜ NE cos T� �˜�� 
Bx 83.54 N 
+ 6Fy=0; By� FAC
e
L
§¨
©
·
¹˜� NE sin T� �˜� 0=
By FAC
e
L
§¨
©
·
¹˜ NE sin T� �˜�� 
By 360 N 
Equations of Equilibrium: For point F.
+ 6Fy'=0; NF 0� NF 0 N Ans
 + 6Fx'=0; VF P� 0= VF P� VF 400 N Ans
+ 60F=0; MF P 0.5 a˜( )˜� 0= MF P 0.5 a˜( )˜� MF 240 N m˜ Ans
Equations of Equilibrium: For point G.
+ 6Fx=0; Bx NG� 0= NG Bx� NG 83.54 N Ans
+ 6Fy=0; VG By� 0= VG By� VG 360 N Ans
+ 60G=0; MG� By 0.5 d˜( )˜� 0= MG By 0.5 d˜( )˜� MG 162 N m˜ Ans
Problem 1-29
The bolt shankis subjected to a tension of 400 N. Determine the resultant internal loadings acting on
the cross section at point C. 
Given:
P 400N� 
r 150mm� 
T 90deg� 
Solution:
Equations of Equilibrium: For segment AC.
+ 6Fx=0; NC P� 0= NC P� NC 400 N Ans
+ 6Fy=0; VC 0� VC 0 N Ans
+ 60G=0; MC P r( )˜� 0= MC P� r( )˜� MC 60� N m˜ Ans
Problem 1-30
The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings
acting on the cross section through B.
Given: P 750N� MC 800N m˜� 
U 12 kg
m
� g 9.81 m
s
2
� 
a 1m� b 2m� c 2m� 
Solution:
Py
4
5
§¨
©
·
¹ P˜� Pz
3
5
� P˜� 
Equations of Equilibrium: For point B.
6Fx=0; VBx 0kip� Ans
6Fy=0; NBy Py� 0=
NBy Py�� Ans
NBy 600� N Ans
6Fz=0; VBz Pz� U g˜ c˜� U g˜ b˜� 0=
VBz Pz� U g˜ c˜� U g˜ b˜�� 
VBz 920.9 N Ans
60x=0; MBx Pz b( )˜� U g˜ c˜ b( )˜� U g˜ b˜ 0.5 b˜( )˜� 0=
MBx Pz� b( )˜ U g˜ c˜ b( )˜� U g˜ b˜ 0.5 b˜( )˜�� 
MBx 1606.3 N m˜ Ans
60y=0; TBy 0N m˜� Ans
60z=0; MBy Mc� 0=
MBy MC�� 
MBy 800� N m˜ Ans
Problem 1-31
The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadings
acting on the cross section through A which is located at an angle T from the horizontal.
Solution: P kN� T deg� 
Equations of Equilibrium: For point A.
+ 6Fx=0; NA� P cos T� �˜� 0=
NA P cos T� �˜� Ans
 + 6Fy=0; VA P sin T� �˜� 0=
VA P sin T� �˜� Ans
+ 60A=0; MA P r˜ 1 cos T� ��� �˜� 0=
MA P r˜ 1 cos T� ��� �˜� r Ans
Problem 1-32
The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determine
the resultant internal loadings acting on the cross section through point B. Hint: The distance from the
centroid C of segment AB to point O is CO = 0.9745r.
Given: T 22.5deg� r m� a 0.9745r� w kN
m
2
� 
Solution:
Equations of Equilibrium: For point B.
6Fz=0; VB
S
4
r˜ w˜� 0= VB 0.785 w˜ r˜� Ans
6Fx=0; NB 0� Ans
60x=0; TB
S
4
r˜ w˜ 0.09968r( )˜� 0= TB 0.0783w r2˜� Ans
60y=0; MB
S
4
r˜ w˜ 0.37293r( )˜� 0= MB 0.293� w r2˜� Ans
Problem 1-33
A differential element taken from a curved bar is shown in the figure. Show that dN/dT = V, 
dV/dT = -N, dM/dT = -T, and dT/dT = M, 
Solution:
Problem 1-34
The column is subjected to an axial force of 8 kN, which is applied through the centroid of the
cross-sectional area. Determine the average normal stress acting at section a–a. Show this distribution
of stress acting over the area’s cross section.
Given: P 8kN� 
b 150mm� d 140mm� t 10mm� 
Solution:
A 2 b t˜( ) d t˜�� A 4400.00 mm2 
V P
A
� V 1.82 MPa Ans
Problem 1-35
The anchor shackle supports a cable force of 3.0 kN. If the pin has a diameter of 6 mm, determine the
average shear stress in the pin.
Given: P 3.0kN� 
d 6mm� 
Solution:
+ 6Fy=0; 2 V˜ P� 0=
V 0.5P� 
V 1500 N 
A
S d2˜
4
� A 28.2743 mm2 
Wavg
V
A
� Wavg 53.05 MPa Ans
Problem 1-36
While running the foot of a 75-kg man is momentarily subjected to a force which is 5 times his
weight. Determine the average normal stress developed in the tibia T of his leg at the mid section a-a.
The cross section can be assumed circular, having an outer diameter of 45 mm and an inner diameter
of 25 mm. Assume the fibula F does not support a load.
Given: g 9.81
m
s
2
 
M 75kg� 
do 45mm� di 25mm� 
Solution:
A
S
4
do
2
di
2�§© ·¹˜� A 1099.5574 mm2 
V 5M g˜
A
� V 3.345 MPa Ans
Problem 1-37
The thrust bearing is subjected to the loads shown. Determine the average normal stress developed on
cross sections through points B, C, and D. Sketch the results on a differential volume element located
at each section.
Units Used: kPa 10
3
Pa� 
Given: P 500N� Q 200N� 
dB 65mm� dC 140mm� dD 100mm� 
Solution:
AB
S dB2˜
4
� AB 3318.3 mm2 
VB
P
AB
� VB 150.7 kPa Ans
AC
S dC2˜
4
� AC 15393.8 mm2 
VC
P
AC
� VC 32.5 kPa Ans
AD
S dD2˜
4
� AD 7854.0 mm2 
VD
Q
AD
� VD 25.5 kPa Ans
Problem 1-38
The small block has a thickness of 5 mm. If the stress distribution at the support developed by the load
varies as shown, determine the force F applied to the block, and the distance d to where it is applied.
Given: a 60mm� b 120mm� t 5mm� 
V1 0MPa� V2 40MPa� V3 60MPa� 
Solution:
F AVµ´¶ d=
F 0.5 V2˜ a t˜( )˜ V2 b t˜( )˜� 0.5 V3 V2�� �˜ b t˜( )˜�� 
F 36.00 kN Ans
Require:
F d˜ Ax V˜µ´¶ d=
d
0.5 V2˜ a t˜( )˜ª¬ º¼ 2a3˜ V2 b t˜( )˜ª¬ º¼ a 0.5 b˜�( )˜� 0.5 V3 V2�� �˜ b t˜( )˜ª¬ º¼ a 2 b˜3�§¨© ·¹˜�
F
� 
d 110 mm Ans
Problem 1-39
The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a
couple is applied to the lever, determine the average shear stress in the pin between the pin and lever.
Given: a 250mm� b 12mm� 
d 6mm� P 20N� 
Solution:
+ 60O=0; V b˜ P 2a( )˜� 0=
V P
2a
b
§¨
©
·
¹˜� 
V 833.33 N 
A
S d2˜
4
� A 28.2743 mm2 
Wavg
V
A
� Wavg 29.47 MPa Ans
Problem 1-40
The cinder block has the dimensions shown. If the material fails when the average normal stress
reaches 0.840 MPa, determine the largest centrally applied vertical load P it can support.
Given: Vallow 0.840MPa� 
ao 150mm� ai 100mm� 
bo 2 1 2� 3�( )˜ 2�[ ] mm˜� 
bi 2 1 3�( )˜[ ] mm˜� 
Solution:
A ao bo˜ ai bi˜�� A 1300 mm2 
Pallow Vallow A( )˜� 
Pallow 1.092 kN Ans
Problem 1-41
The cinder block has the dimensions shown. If it is subjected to a centrally applied force of P = 4 kN,
determine the average normal stress in the material. Show the result acting on a differential volume
element of the material.
Given: P 4kN� 
ao 150mm� ai 100mm� 
bo 2 1 2� 3�( )˜ 2�[ ] mm˜� 
bi 2 1 3�( )˜[ ] mm˜� 
Solution:
A ao bo˜ ai bi˜�� A 1300 mm2 
V P
A
� 
V 3.08 MPa Ans
Problem 1-42
The 250-N lamp is supported by three steel rods connected by a ring at A. Determine which rod is
subjected to the greater average normal stress and compute its value. Take T = 30°. The diameter of
each rod is given in the figure.
Given: W 250N� T 30deg� I 45deg� 
dB 9mm� dC 6mm� dD 7.5mm� 
Solution: Initial guess: FAC 1N� FAD 1N� 
Given
+ 6Fx=0; FAC cos T� �˜ FAD cos I� �˜� 0= [1]
+ 6Fy=0; FAC sin T� �˜ FAD sin I� �˜� W� 0= [2]
Solving [1] and [2]:
FAC
FAD
§¨
©¨
·
¹
Find FAC FAD�� �� 
FAC
FAD
§¨
©¨
·
¹
183.01
224.14
§¨
©
·
¹ N 
Rod AB:
AAB
S dB2˜
4
� AAB 63.61725 mm2 
VAB
W
AAB
� VAB 3.93 MPa 
Rod AD :
AAD
S dD2˜
4
� AAD 44.17865 mm2 
VAD
FAD
AAD
� VAD 5.074 MPa 
Rod AC:
AAC
S dC2˜
4
� AAC 28.27433 mm2 
VAC
FAC
AAC
� VAC 6.473 MPa Ans
Problem 1-43
Solve Prob. 1-42 for T = 45°.
Given: W 250N� T 45deg� I 45deg� 
dB 9mm� dC 6mm� dD 7.5mm� 
Solution: Initial guess: FAC 1N� FAD 1N� 
Given
+ 6Fx=0; FAC cos T� �˜ FAD cos I� �˜� 0= [1]
+ 6Fy=0; FAC sin T� �˜ FAD sin I� �˜� W� 0= [2]
Solving [1] and [2]:
FAC
FAD
§¨
©¨
·
¹
Find FAC FAD�� �� 
FAC
FAD
§¨
©¨
·
¹
176.78
176.78
§¨
©
·
¹ N 
Rod AB:
AAB
S dB2˜
4
� AAB 63.61725 mm2 
VAB
W
AAB
� VAB 3.93 MPa 
Rod AD :
AAD
S dD2˜
4
� AAD 44.17865 mm2 
VAD
FAD
AAD
� VAD 4.001 MPa 
Rod AC:
AAC
S dC2˜
4
� AAC 28.27433 mm2 
VAC
FAC
AAC
� VAC 6.252 MPa Ans
Problem 1-44
The 250-N lamp is supported by three steel rods connected by a ring at A. Determine the angle of
orientation T of AC such that the average normal stress in rod AC is twice the average normal stress in
rod AD. What is the magnitude of stress in each rod? The diameter of each rod is given in the figure.
Given: W 250N� I 45deg� 
dB 9mm� dC 6mm� dD 7.5mm� 
Solution:Rod AB: AAB
S dB2˜
4
� AAB 63.61725 mm2 
Rod AD : AAD
S dD2˜
4
� AAD 44.17865 mm2 
Rod AC: AAC
S dC2˜
4
� AAC 28.27433 mm2 
Since VAC 2VAD= Therefore
FAC
AAC
2
FAD
AAD
=
Initial guess: FAC 1N� FAD 2N� T 30deg� 
Given
FAC
AAC
2
FAD
AAD
= [1]
+ 6Fx=0; FAC cos T� �˜ FAD cos I� �˜� 0= [2]
+ 6Fy=0; FAC sin T� �˜ FAD sin I� �˜� W� 0= [3]
Solving [1], [2] and [3]:
FAC
FAD
T
§¨
¨¨
©
·
¸
¹
Find FAC FAD� T�� �� FAC
FAD
§¨
©¨
·
¹
180.38
140.92
§¨
©
·
¹ N 
T 56.47 deg 
VAB
W
AAB
� VAB 3.93 MPa Ans
VAD
FAD
AAD
� VAD 3.19 MPa Ans
VAC
FAC
AAC
� VAC 6.38 MPa Ans
Problem 1-45
The shaft is subjected to the axial force of 30 kN. If the shaft passes through the 53-mm diameter hole
in the fixed support A, determine the bearing stress acting on the collar C. Also, what is the average
shear stress acting along the inside surface of the collar where it is fixed connected to
the 52-mm diameter shaft?
Given: P 30kN� 
dhole 53mm� dshaft 52mm� 
dcollar 60mm� hcollar 10mm� 
Solution:
Bearing Stress:
Ab
S
4
dcollar
2
dhole
2�§© ·¹˜� 
Vb
P
Ab
� Vb 48.3 MPa Ans
Average Shear Stress:
As S dshaft� �˜ hcollar� �˜� 
Wavg
P
As
� Wavg 18.4 MPa Ans
Problem 1-46
The two steel members are joined together using a 60° scarf weld. Determine the average normal and
average shear stress resisted in the plane of the weld.
Given:
P 8kN� T 60deg� 
b 25mm� h 30mm� 
Solution:
Equations of Equilibrium: 
 + 6Fx=0; N P sin T� �˜� 0=
N P sin T� �˜� N 6.928 kN 
+ 6Fy=0; V P cos T� �˜� 0=
V P cos T� �˜� V 4 kN 
A
h b˜
sin T� �� 
V N
A
� V 8 MPa Ans
Wavg
V
A
� Wavg 4.62 MPa Ans
Problem 1-47
The J hanger is used to support the pipe such that the force on the vertical bolt is 775 N. Determine the
average normal stress developed in the bolt BC if the bolt has a diameter of 8 mm. Assume A is a pin.
Given: P 775N� 
a 40mm� b 30mm� 
d 8mm� T 20deg� 
Solution:
Support Reaction: 
 6FA=0; P a( )˜ FBC cos T� �˜ a b�( )˜� 0=
FBC
P a˜
a b�( ) cos T� �˜� 
FBC 471.28 N 
Average Normal Stress: 
ABC
S d2˜
4
� 
V
FBC
ABC
� 
V 9.38 MPa Ans
Problem 1-48
The board is subjected to a tensile force of 425 N. Determine the average normal and average shear
stress developed in the wood fibers that are oriented along section a-a at 15° with the axis of the board.
Given: P 425N� T 15deg� 
b 25mm� h 75mm� 
Solution:
Equations of Equilibrium: 
 + 6Fx=0; V P cos T� �˜� 0=
V P cos T� �˜� V 410.518 N 
+ 6Fy=0; N P sin T� �˜� 0=
N P sin T� �˜� N 1 N 
Average Normal Stress: 
A
h b˜
sin T� �� 
V N
A
� V 0.0152 MPa Ans
Wavg
V
A
� Wavg 0.0567 MPa Ans
Problem 1-49
The open square butt joint is used to transmit a force of 250 kN from one plate to the other. Determine
the average normal and average shear stress components that this loading creates on the face of the
weld, section AB.
Given: P 250kN� T 30deg� 
b 150mm� h 50mm� 
Solution:
Equations of Equilibrium: 
 + 6Fx=0; V� P sin T� �˜� 0=
V P sin T� �˜� V 125 kN 
+ 6Fy=0; N P cos T� �˜� 0=
N P cos T� �˜� N 216.506 kN 
Average Normal and Shear Stress: 
A
h b˜
sin 2T� �� 
V N
A
� V 25 MPa Ans
Wavg
V
A
� Wavg 14.434 MPa Ans
Problem 1-50
The specimen failed in a tension test at an angle of 52° when the axial load was 100 kN. If the diameter
of the specimen is 12 mm, determine the average normal and average shear stress acting on the area of
the inclined failure plane. Also, what is the average normal stress acting on the cross section when
failure occurs?
Given: P 100kN� 
d 12mm� T 52deg� 
Solution:
Equations of Equilibrium: 
+ 6Fx=0; V P cos T� �˜� 0=
V P cos T� �˜� V 61.566 kN 
+ 6Fy=0; N P sin T� �˜� 0=
N P sin T� �˜� N 78.801 kN 
Inclined plane:
A
S
4
d
2
sin T� �
§¨
©
·
¹� 
V N
A
� V 549.05 MPa Ans
Wavg
V
A
� Wavg 428.96 MPa Ans
Cross section: 
A
Sd2
4
� 
V P
A
� V 884.19 MPa Ans
Wavg 0� Wavg 0 MPa Ans
Problem 1-51
A tension specimen having a cross-sectional area A is subjected to an axial force P. Determine the
maximum average shear stress in the specimen and indicate the orientation T of a section on which it
occurs. 
Solution:
Equations of Equilibrium: 
+ 6Fy=0; V P cos T� �˜� 0=
V P cos T� �˜=
Inclined plane:
Aincl
A
sin T� �=
W V
Aincl
= W P cos T� �˜ sin T� �˜
A
= W P sin 2T� �˜
2A
=
dW
dT
P cos 2T� �˜
A
=
dW
dT 0=
cos 2T� � 0=
2T 90deg=
T 45deg� Ans
Wmax
P sin 90q� �˜
2A
= Wmax
P
2A
=
Ans
Problem 1-52
The joint is subjected to the axial member force of 5 kN. Determine the average normal stress acting
on sections AB and BC. Assume the member is smooth and is 50-mm thick.
Given: P 5kN� T 45deg� I 60deg� 
dAB 40mm� dBC 50mm� t 50mm� 
Solution: D 90deg I�� D 30.00 deg 
AAB t dAB˜� 
ABC t dBC˜� 
+ 6Fx=0; NAB cos D� �˜ P cos T� �˜� 0=
NAB
P cos T� �˜
cos D� �� 
NAB 4.082 kN 
+ 6Fy=0; NAB� sin D� �˜ P sin T� �˜� NBC� 0=
NBC NAB� sin D� �˜ P sin T� �˜�� 
NBC 1.494 kN 
VAB
NAB
AAB
� VAB 2.041 MPa Ans
VBC
NBC
ABC
� VBC 0.598 MPa Ans
Problem 1-53
The yoke is subjected to the force and couple moment. Determine the average shear stress in the bolt
acting on the cross sections through A and B. The bolt has a diameter of 6 mm. Hint: The couple
moment is resisted by a set of couple forces developed in the shank of the bolt.
Given: P 2.5kN� M 120N m˜� 
ho 62mm� hi 50mm� 
d 6mm� T 60deg� 
Solution:
As a force on bolt shank is zero, then
WA 0� Ans
Equations od Equilibrium:
6Fz=0; P 2Fz� 0=
Fz 0.5P� Fz 1.25 kN 
6Mz=0; M Fx hi� �˜� 0=
Fx
M
hi
� Fx 2.4 kN 
Average Shear Stress: 
A
Sd2
4
� 
The bolt shank subjected to a shear force of VB Fx
2
Fz
2�� 
WB
VB
A
� WB 95.71 MPa Ans
Problem 1-54
The two members used in the construction of an aircraft fuselage are joined together using a 30°
fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld.
Assume each inclined plane supports a horizontal force of 2 kN.
Given:
P 4.0kN� 
b 37.5mm� hhalf 25m� 
T 30deg� 
Solution:
Equations of Equilibrium: 
 + 6Fx=0; V� 0.5P cos T� �˜� 0=
V 0.5P cos T� �˜� V 1.732 kN 
+ 6Fy=0; N 0.5P sin T� �˜� 0=
N 0.5P sin T� �˜� N 1 kN 
Average Normal and Shear Stress: 
A
hhalf� � b˜
sin T� �� 
V N
A
� V 533.33 Pa Ans
Wavg
V
A
� Wavg 923.76 Pa Ans
Problem 1-55
The row of staples AB contained in the stapler is glued together so that the maximum shear stress the
glue can withstand is W max = 84 kPa. Determine the minimum force F that must be placed on the
plunger in order to shear off a staple from its row and allow it to exit undeformed through the groove
at C. The outer dimensions of the staple are shown in the figure. It has a thickness of 1.25 mm
Assume all the other parts are rigid and neglect friction.
Given: Wmax 0.084MPa� 
a 12.5mm� b 7.5mm� 
t 1.25mm� 
Solution:
Average Shear Stress:
A a b˜ a 2t�( ) b t�( )˜[ ]�� 
Wmax
V
A
= V Wmax� � A˜� 
V 2.63 N 
Fmin V� 
Fmin 2.63 N Ans
Problem 1-56
Rods AB and BC have diameters of 4mm and 6 mm, respectively. If the load of 8 kN is applied to the
ring at B, determine the average normal stress in each rod if T = 60°.
Given: W 8kN� T 60deg� 
dA 4mm� dC 6mm� 
Solution:
Rod AB: AAB
S dA2˜
4
� 
Rod BC : ABC
S dC2˜
4
� 
+ 6Fy=0; FBC sin T� �˜ W� 0=
FBC
W
sin T� �� 
FBC 9.238 kN 
+ 6Fx=0; FBC cos T� �˜ FAB� 0=
FAB FBC cos T� �˜� 
FAB 4.619 kN 
VAB
FAB
AAB
� VAB 367.6 MPa Ans
VBCFBC
ABC
� VBC 326.7 MPa Ans
Problem 1-57
Rods AB and BC have diameters of 4 mm and 6 mm, respectively. If the vertical load of 8 kN is
applied to the ring at B, determine the angle T�of rod BC so that the average normal stress in each rod
is equivalent. What is this stress?
Given: W 8kN� 
dA 4mm� dC 6mm� 
Solution:
Rod AB: AAB
S dA2˜
4
� 
Rod BC : ABC
S dC2˜
4
� 
+ 6Fy=0; FBC sin T� �˜ W� 0=
+ 6Fx=0; FBC cos T� �˜ FAB� 0=
Since FAB V AAB˜=
FBC V ABC˜=
Initial guess: V 100MPa� T 50deg� 
Given
V ABC˜ sin T� �˜ W� 0= [1]
V ABC˜ cos T� �˜ V AAB˜� 0= [2]
Solving [1] and [2]:
V
T
§¨
©
·
¹ Find V T�� �� 
T 63.61 deg Ans
V 315.85 MPa Ans
Problem 1-58
The bars of the truss each have a cross-sectional area of 780 mm2. Determine the average normal
stress in each member due to the loading P = 40 kN. State whether the stress is tensile or compressive.
Given: P 40kN� 
a 0.9m� b 1.2m� A 780mm2� 
Solution: c a
2
b
2�� c 1.5 m 
h
b
c
� v a
c
� 
Joint A:
+ 6Fy=0; v( ) FAB˜ P� 0= FAB
P
v
� FAB 66.667 kN 
+ 6Fx=0; h( ) FAB˜ FAE� 0= FAE h( ) FAB˜� FAE 53.333 kN 
VAB
FAB
A
� VAB 85.47 MPa (T) Ans
VAE
FAE
A
� VAE 68.376 MPa (C) Ans
Joint E:
+ 6Fy=0; FEB 0.75P� 0= FEB 0.75P� FEB 30 kN 
+ 6Fx=0; FED FAE� 0= FED FAE� FED 53.333 kN 
VEB
FEB
A
� VEB 38.462 MPa (T) Ans
VED
FED
A
� VED 68.376 MPa (C) Ans
Joint B:
+ 6Fy=0; v( ) FBD˜ v( ) FAB˜� FEB� 0= FBD FAB
FEB
v
§¨
©
·
¹�� FBD 116.667 kN 
+ 6Fx=0; FBC h( )FAB� h( )FBD� 0= FBC h( )FAB h( )FBD�� FBC 146.667 kN 
VBC
FBC
A
� VBC 188.034 MPa (T) Ans
VBD
FBD
A
� VBD 149.573 MPa (C) Ans
Problem 1-59
The bars of the truss each have a cross-sectional area of 780 mm2. If the maximum average normal
stress in any bar is not to exceed 140 MPa, determine the maximum magnitude P of the loads that can
be applied to the truss.
Vallow 140MPa� Given:
a 0.9m� b 1.2m� A 780mm2� 
Solution: c a
2
b
2�� c 1.5 m 
h
b
c
� v a
c
� 
For comparison purpose, set P 1kN� 
Joint A:
+ 6Fy=0; v( ) FAB˜ P� 0= FAB
P
v
� FAB 1.667 kN 
+ 6Fx=0; h( ) FAB˜ FAE� 0= FAE h( ) FAB˜� FAE 1.333 kN 
VAB
FAB
A
� VAB 2.137 MPa (T)
VAE
FAE
A
� VAE 1.709 MPa (C)
Joint E:
+ 6Fy=0; FEB 0.75P� 0= FEB 0.75P� FEB 0.75 kN 
+ 6Fx=0; FED FAE� 0= FED FAE� FED 1.333 kN 
VEB
FEB
A
� VEB 0.962 MPa (T)
VED
FED
A
� VED 1.709 MPa (C)
Joint B:
+ 6Fy=0; v( ) FBD˜ v( ) FAB˜� FEB� 0= FBD FAB
FEB
v
§¨
©
·
¹�� FBD 2.917 kN 
+ 6Fx=0; FBC h( )FAB� h( )FBD� 0= FBC h( )FAB h( )FBD�� FBC 3.667 kN 
VBC
FBC
A
� VBC 4.701 MPa (T)
VBD
FBD
A
� VBD 3.739 MPa (C)
Since the cross-sectional areas are the same, the highest stress occurs in the member BC,
which has the greatest force
Fmax max FAB FAE� FEB� FED� FBD� FBC�� �� Fmax 3.667 kN 
Pallow
P
Fmax
§¨
©
·
¹
Vallow A˜� �˜� Pallow 29.78 kN Ans
Problem 1-60
The plug is used to close the end of the cylindrical tube that is subjected to an internal pressure of p =
650 Pa. Determine the average shear stress which the glue exerts on the sides of the tube needed to
hold the cap in place.
Given: p 650Pa� a 25mm� 
di 35mm� do 40mm� 
Solution:
Ap
S di2˜
4
� As S do˜� � a( )� 
P a( )˜ FBC cos T� �˜ a b�( )˜� 0=
P p Ap� �˜� P 0.625 N 
Average Shear Stress: 
Wavg
P
As
� 
Wavg 199.1 Pa Ans
Problem 1-61
The crimping tool is used to crimp the end of the wire E. If a force of 100 N is applied to the handles,
determine the average shear stress in the pin at A. The pin is subjected to double shear and has a
diameter of 5 mm. Only a vertical force is exerted on the wire.
Given: P 100N� 
a 37.5mm� b 50mm� c 25mm� 
d 125mm� dpin 5mm� 
Solution:
From FBD (a):
+ 6Fx=0; Bx 0� 
Bx 0 N 
+ 60D=0; P d( )˜ By c( )˜� 0=
By P
d
c
˜� 
By 500 N 
From FBD (b):
+ 6Fx=0; Ax 0� 
Ax 0 N 
+ 60E=0; Ay a( )˜ By a b�( )˜� 0=
Ay By
a b�
a
˜� 
Ay 1166.67 N 
Average Shear Stress: 
Apin
S dpin2˜
4
� 
VA 0.5 Ay� �˜� VA 583.333 N 
Wavg
VA
Apin
� Wavg 29.709 MPa Ans
Problem 1-62
Solve Prob. 1-61 for pin B. The pin is subjected to double shear and has a diameter of 5 mm.
Given: a 37.5mm� b 50mm� c 25mm� 
d 125mm� dpin 5mm� P 100N� 
Solution:
From FBD (a):
+ 6Fx=0; Bx 0� 
Bx 0 N 
+ 60D=0; P d( )˜ By c( )˜� 0=
By P
d
c
˜� 
By 500 N 
Average Shear Stress: Pin B is subjected to doule shear
Apin
S dpin2˜
4
� 
VB 0.5 By� �˜� VB 250 N 
Wavg
VB
Apin
� Wavg 12.732 MPa Ans
Problem 1-63
The railcar docklight is supported by the 3-mm-diameter pin at A. If the lamp weighs 20 N, and the
extension arm AB has a weight of 8 N/m, determine the average shear stress in the pin needed to
support the lamp. Hint: The shear force in the pin is caused by the couple moment required for
equilibrium at A.
Given: w 8
N
m
� P 20N� 
a 900mm� h 32mm� 
dpin 3mm� 
Solution:
From FBD (a):
+ 6Fx=0; Bx 0� 
+ 60A=0; V h( )˜ w a˜( ) 0.5a( )˜� P a( )˜� 0=
V w a˜( ) 0.5 a
h
§¨
©
·
¹˜ P
a
h
˜�� V 663.75 N 
Average Shear Stress: 
Apin
S dpin2˜
4
� 
Wavg
V
Apin
� Wavg 93.901 MPa Ans
Problem 1-64
The two-member frame is subjected to the distributed loading shown. Determine the average normal
stress and average shear stress acting at sections a-a and b-b. Member CB has a square cross section
of 35 mm on each side. Take w = 8 kN/m.
Given: w 8
kN
m
� 
a 3m� b 4m� A 0.0352� �m2� 
Solution: c a
2
b
2�� c 5 m 
h
a
c
� v b
c
� 
Member AB:
 6MA=0; By a( )˜ w a˜( ) 0.5a( )˜� 0=
By 0.5w a˜� By 12 kN 
+ 6Fy=0; v( ) FAB˜ By� 0=
FAB
By
v
� FAB 15 kN 
Section a-a: 
Va_a
FAB
A
� Va_a 12.24 MPa Ans
Wa_a 0� Wa_a 0 MPa Ans
Section b-b: 
+ 6Fx=0; N FAB h( )˜� 0= N FAB h( )˜� N 9 kN 
+ 6Fy=0; V FAB v( )˜� 0= V FAB v( )˜� V 12 kN 
Ab_b
A
h
� 
Vb_b
N
Ab_b
� Vb_b 4.41 MPa Ans
Wb_b
V
Ab_b
� Wb_b 5.88 MPa Ans
Problem 1-65
Member A of the timber step joint for a truss is subjected to a compressive force of 5 kN. Determine
the average normal stress acting in the hanger rod C which has a diameter of 10 mm and in member B
which has a thickness of 30 mm.
Given: P 5kN� T 60deg� I 30deg� 
drod 10mm� h 40mm� t 30mm� 
Solution:
AB t h˜� 
Arod
S
4
drod
2˜� 
+ 6Fx=0; P cos T� �˜ FB� 0=
FB P cos T� �˜� FB 2.5 kN 
+ 6Fy=0; Fc P sin T� �˜� 0=
FC P sin T� �˜� FC 4.33 kN 
Average Normal Stress:
VB
FB
AB
� VB 2.083 MPa Ans
VC
FC
Arod
� VC 55.133 MPa Ans
Problem 1-66
Consider the general problem of a bar made from m segments, each having a constant cross-sectional
area Am and length Lm. If there are n loads on the bar as shown, write a computer program that can be
used to determine the average normal stress at any specified location x. Show an application of the
program using the values L1 = 1.2 m, d1 = 0.6 m, P1 = 2 kN, A1 = 1875 mm
2, L2 = 0.6 m, d2 = 1.8 m,
P2 = -1.5 kN, A2 = 625 mm
2.
Problem 1-67
The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear
stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a
diameter of 18 mm.
Given: P 15kN� 
a 0.5m� b 1m� c 1.5m� 
d 1.5m� e 0.5m� 
T 30deg� dpin 18mm� 
Solution: L a b� c� d� e�� 
Support Reactions:
 60A=0; By� L( )˜ P L a�( )˜� 4P c d� e�( )˜� 4P d e�( )˜� 2P e( )˜� 0=
By P
L a�
L
˜ 4 P˜ c d� e�
L
˜� 4 P˜ d e�
L
˜� 2 P˜ e
L
˜�� 
By 82.5 kN 
 + 6Fy=0; By� P� 4 P˜� 4 P˜� 2 P˜� Ay� 0=
Ay By� P� 4 P˜� 4P� 2 P˜�� 
Ay 82.5 kN 
FBC
By
sin T� �� FBC 165 kN 
Ax FBC cos T� �˜� Ax 142.89 kN 
AverageShear Stress: 
Apin
S dpin2˜
4
� 
For Pins B and C: 
WB_and_C
0.5FBC
Apin
� WB_and_C 324.2 MPa Ans
For Pin A: 
FA Ax
2
Ay
2�� FA 165 kN 
WA
0.5FA
Apin
� WA 324.2 MPa Ans
Problem 1-68
The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of the
loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins are
in double shear as shown, and each has a diameter of 18 mm.
Given: Wallow 80MPa� 
a 0.5m� b 1m� c 1.5m� 
d 1.5m� e 0.5m� 
T 30deg� dpin 18mm� 
Solution: L a b� c� d� e�� 
For comparison purpose, set P 1kN� 
Support Reactions:
 60A=0; By� L( )˜ P L a�( )˜� 4P c d� e�( )˜� 4P d e�( )˜� 2P e( )˜� 0=
By P
L a�
L
˜ 4 P˜ c d� e�
L
˜� 4 P˜ d e�
L
˜� 2 P˜ e
L
˜�� By 5.5 kN 
 + 6Fy=0; By� P� 4 P˜� 4 P˜� 2 P˜� Ay� 0=
Ay By� P� 4 P˜� 4P� 2 P˜�� Ay 5.5 kN 
FBC
By
sin T� �� FBC 11 kN 
Ax FBC cos T� �˜� Ax 9.53 kN 
FA Ax
2
Ay
2�� FA 11 kN 
Require: 
Fmax max FBC FA�� �� Fmax 11 kN 
Apin
S dpin2˜
4
� 
Pallow
P
Fmax
§¨
©
·
¹
Wallow 2Apin� �˜ª¬ º¼˜� Pallow 3.70 kN Ans
Problem 1-69
The frame is subjected to the load of 1 kN. Determine the average shear stress in the bolt at A as a
function of the bar angle T. Plot this function, 0 Td 90od , and indicate the values of T for which this
stress is a minimum. The bolt has a diameter of 6 mm and is subjected to single shear.
Given: P 1kN� dbolt 6mm� 
a 0.6m� b 0.45m� c 0.15m� 
Solution:
Support Reactions:
 60C=0; FAB cos T� �˜ c( )˜ FAB sin T� �˜ a( )˜� P a b�( )˜� 0=
FAB
P a b�( )˜
cos T� � c( )˜ sin T� � a( )˜�=
Average Shear Stress: Pin B is subjected to doule shear
W
FAB
Abolt
= Abolt
S dbolt2˜
4
� 
W 4P a b�( )˜
cos T� � c( )˜ sin T� � a( )˜�ª¬ º¼ S dbolt2˜§© ·¹˜
=
dW
dT
4P a b�( )˜
S dbolt2˜
sin T� � c( )˜ cos T� � a( )˜�
cos T� � c( )˜ sin T� � a( )˜�ª¬ º¼2
˜=
dW
dT 0= sin T� � c( )˜ cos T� � a( )˜� 0= tan T� �
a
c
=
T atan a
c
§¨
©
·
¹� 
T 75.96 deg Ans
Problem 1-70
The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of the
beam, 1ft xd 12ftd . If the hoist is rated to support a maximum of 7.5 kN, determine the maximum
average normal stress in the 18-mm-diameter tie rod BC and the maximum average shear stress in the
16-mm-diameter pin at B.
Given: P 7.5kN� xmax 3.6m� 
a 3m� T 30deg� 
drod 18mm� dpin 16mm� 
Solution:
Support Reactions:
 
60C=0; FBC sin T� � a( )˜˜ P x( )˜� 0=
FBC
P x( )˜
sin T� � a( )˜=
Maximum FBC occurs when x= xmax . Therefore, 
FBC
P xmax� �˜
sin T� � a( )˜� FBC 18.00 kN 
Arod
S drod2˜
4
� Apin
S dpin2˜
4
� 
Wpin
0.5 FBC˜
Apin
� Wpin 44.762 MPa Ans
Vrod
FBC
Arod
� Vrod 70.736 MPa Ans
Problem 1-71
The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal
and average shear stresses acting over the shaded section, which is oriented at T from the horizontal.
Plot the variation of these stresses as a function of T (0o Td 90od ).
Solution:
Equations of Equilibrium: 
 + 6Fx=0; V P cos T� �˜� 0=
V P cos T� �˜=
 + 6Fy=0; N P sin T� �˜� 0=
N P sin T� �˜=
Inclined plane:
AT
A
sin T� �=
V N
A
= V P
A
sin T� �2˜= Ans
Wavg
V
A
= Wavg
P
2A
sin 2T� �˜= Ans
Problem 1-72
The boom has a uniform weight of 3 kN and is hoisted into position using the cable BC. If the cable has
a diameter of 15 mm, plot the average normal stress in the cable as a function of the boom position T for
0
o Td 90od .
Given: W 3kN� 
a 1m� 
do 15mm� 
Solution: Angle B: IB 0.5 90deg T�� �=
IB 45deg 0.5T�=
Support Reactions:
 60A=0; FBC sin IB� �˜ a( )˜ W 0.5a( )˜ cos T� �� 0=
FBC
0.5W cos T� �˜
sin 45deg 0.5T�� �=
Average Normal Stress: 
VBC
FAB
ABC
= ABC
S do2˜
4
� 
VBC
2W
S do2˜
§¨
©
·
¹
cos T� �
sin 45deg 0.5T�� �˜= Ans
Problem 1-73
The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributed
loading along its length and to two concentrated loads as shown, determine the average normal stress in
the bar as a function of for 0 x� 0.5md .
Given: P1 3kN� P2 6kN� 
w 8
kN
m
� A 400 10 6�� �˜ m2� 
a 0.5m� b 0.75m� 
Solution: L a b�� 
+ 6Fx=0; N� P1� P2� w L x�( )˜� 0=
N P1 P2� w L x�( )˜�=
Average Normal Stress:
V N
A
=
V
P1 P2� w L x�( )˜�
A
= Ans
Problem 1-74
The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributed
loading along its length and to two concentrated loads as shown, determine the average normal stress in
the bar as a function of for 0.5m x� 1.25md .
Given: P1 3kN� P2 6kN� 
w 8
kN
m
� A 400 10 6�� �˜ m2� 
a 0.5m� b 0.75m� 
Solution: L a b�� 
+ 6Fx=0; N� P1� w L x�( )˜� 0=
N P1 w L x�( )˜�=
Average Normal Stress:
V N
A
=
V
P1 w L x�( )˜�
A
= Ans
Problem 1-75
The column is made of concrete having a density of 2.30 Mg/m3. At its top B it is subjected to an axial
compressive force of 15 kN. Determine the average normal stress in the column as a function of the
distance z measured from its base. Note: The result will be useful only for finding the average normal
stress at a section removed from the ends of the column, because of localized deformation at the ends.
Given: P 3kN� U 2.3 103� � kg
m
3
� g 9.81 m
s
2
� 
r 180mm� h 0.75m� 
Solution:
A S r2˜� w U g˜ A˜� 
+ 6Fz=0; N P� w h z�( )˜� 0=
N P w h z�( )˜�=
Average Normal Stress:
V N
A
=
V P w h z�( )˜�
A
= Ans
Problem 1-76
The two-member frame is subjected to the distributed loading shown. Determine the largest
intensity of the uniform loading that can be applied to the frame without causing either the average
normal stress or the average shear stress at section b-b to exceed V = 15 MPa, and W = 16 MPa
respectively. Member CB has a square cross-section of 30 mm on each side.
Given: Vallow 15MPa� Wallow 16MPa� 
a 4m� b 3m� A 0.0302� �m2� 
Solution: c a
2
b
2�� c 5 m 
v
b
c
� 
h
a
c
� 
Set wo 1
kN
m
� 
Member AB:
 6MA=0; By� b( )˜ wo b˜� � 0.5b( )˜� 0=
By 0.5wo b˜� 
By 1.5 kN 
Section b-b: 
By FBC h( )˜= FBC
By
h
� 
FBC 1.88 kN 
+ 6Fx=0; FBC h( )˜ Vb_b� 0= Vb_b FBC h( )˜� 
Vb_b 1.5 kN 
+ 6Fy=0; Nb_b� FBC v( )˜� 0= Nb_b FBC v( )˜� 
Nb_b 1.125 kN 
Ab_b
A
v
� 
Vb_b
Nb_b
Ab_b
� Vb_b 0.75 MPa 
Wb_b
Vb_b
Ab_b
� Wb_b 1 MPa 
Assume failure due to normal stress: wallow wo
Vallow
Vb_b
§¨
©
·
¹
˜� wallow 20.00
kN
m
 
Assume failure due to shear stress: wallow wo
Wallow
Wb_b
§¨
©
·
¹
˜� wallow 16.00
kN
m
 Ans
Controls ! 
Problem 1-77
The pedestal supports a load P at its center. If the material has a mass density U, determine the radial
dimension r as a function of z so that the average normal stress in the pedestal remains constant. The
cross section is circular.
Solution:
Require: V P w1�
A
= V P w1� dw�
A dA�=
P dA˜ w1 dA˜� A dw˜=
dw
dA
P w1�
A
=
dw
dA
V= [1]
dA S r dr�( )2 Sr2�=
dA 2Sr dr˜=
dw Sr2 U g˜� �˜ dz˜=
From Eq.[1],
Sr2 U g˜� �˜ dz˜
2Sr dr˜ V=
r U g˜� �˜ dz˜
2dr
V=
U g˜
2V 0
z
z1( )µ´¶ d
r
1
r
r
1
r
µ´
µ¶
d=
U g˜ z˜
2V ln
r
r1
§¨
©
·
¹= r r1 e
U g˜
2V
§¨
©
·
¹ z˜˜=
However, 
V P
S r12˜
=
Ans
r r1 e
S r12˜ U˜ g˜
2P
§¨
©¨
·
¹ z˜˜=
Problem 1-78
The radius of the pedestal is defined by r = (0.5e-0.08y2) m, where y is given in meters. If the material
has a density of 2.5 Mg/m3, determine the average normal stress at the support.
Given:ro 0.5m� h 3m� g 9.81
m
s
2
 
r ro e
0.08� y2
m˜= U 2.5 103� �˜ kg
m
3
� 
yunit 1m� 
Solution:
dr S e 0.08� y
2§© ·¹ dy˜=
Ao Sro2� Ao 0.7854 m2 
dV S r2� � dy˜=
dV Sro2 e 0.08� y
2§© ·¹
2
dy˜=
V
0
3
ySro2 e 0.08� y
2§© ·¹
2
yunit� �˜ª«¬
º»¼
µ´
µ¶ d� 
V 1.584 m
3 
W U g˜ V˜� 
W 38.835 kN 
V W
Ao
� 
V 0.04945 MPa Ans
Problem 1-79
The uniform bar, having a cross-sectional area of A and mass per unit length of m, is pinned at its
center. If it is rotating in the horizontal plane at a constant angular rate of Z, determine the average
normal stress in the bar as a function of x.
Solution:
Equation of Motion :
+ 6Fx=MaN=MZ�r2 ;
N m
L
2
x�§¨©
·
¹˜ Z x
1
2
L
2
x�§¨©
·
¹�
ª«¬
º»¼˜=
N
m Z˜
8
L
2
4 x
2˜�� �˜=
Average Normal Stress:
V N
A
=
V m Z˜
8A
L
2
4 x
2˜�� �˜= Ans
Problem 1-80
Member B is subjected to a compressive force of 4 kN. If A and B are both made of wood and are
10mm. thick, determine to the nearest multiples of 5mm the smallest dimension h of the support so that
the average shear stress does not exceed Wallow = 2.1 MPa.
Given: P 4kN� 
t 10mm� Wallow 2.1MPa� 
a 300mm� b 125mm� 
Solution: c a
2
b
2�� c 325 mm 
h
a
c
� v b
c
� 
V P v( )˜� V 1.54 kN 
Wallow
V
t h˜=
h
V
t Wallow˜
� h 73.26 mm 
Use h 75mm� h 75 mm Ans
Problem 1-81
The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure
shear stress for the bolts is Wfail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5.
Given: P 80kN� Wfail 350MPa� J 2.5� 
Solution:
Wallow
Wfail
J� Wallow 140 MPa 
Vbolt 0.5
P
2
§¨
©
·
¹˜� Vbolt 20 kN 
Abolt
Vbolt
Wallow
=
S
4
§¨
©
·
¹ d
2˜
Vbolt
Wallow
=
d
4
S
Vbolt
Wallow
§¨
©
·
¹
˜� 
d 13.49 mm Ans
Problem 1-82
The rods AB and CD are made of steel having a failure tensile stress of Vfail = 510 MPa. Using a factor
of safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the load
shown. The beam is assumed to be pin connected at A and C.
Given: P1 4kN� P2 6kN� P3 5kN� 
a 2m� b 2m� c 3m� d 3m� 
J 1.75� Vfail 510MPa� 
Solution: L a b� c� d�� 
Support Reactions:
 60A=0; FCD L( )˜ P1 a( )˜� P2 a b�( )˜� P3 a b� c�( )˜� 0=
FCD P1
a
L
§¨
©
·
¹˜ P2
a b�
L
˜� P3
a b� c�
L
˜�� FCD 6.70 kN 
 60C=0; FAB� L( )˜ P1 b c� d�( )˜� P2 c d�( )˜� P3 d( )˜� 0=
FAB P1
b c� d�
L
˜ P2
c d�
L
˜� P3
d
L
§¨
©
·
¹˜�� FAB 8.30 kN 
Average Normal Stress: Design of rod sizes
Vallow
Vfail
J� Vallow 291.43 MPa 
For Rod AB
Abolt
FAB
Vallow
=
S
4
§¨
©
·
¹ dAB
2˜
FAB
Vallow
=
dAB
4
S
FAB
Vallow
§¨
©
·
¹
˜� dAB 6.02 mm Ans
For Rod CD
Abolt
FCD
Vallow
=
S
4
§¨
©
·
¹ dCD
2˜
FCD
Vallow
=
dCD
4
S
FCD
Vallow
§¨
©
·
¹
˜� dCD 5.41 mm Ans
Problem 1-83
The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is
fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if
the allowable shear stress for the key is Wallow = 35 MPa.
Given: P 200N� Wallow 35MPa� 
L 500mm� a 20mm� 
b 25mm� 
Solution:
 60A=0; Fa_a a( )˜ P L( )˜� 0=
Fa_a P
L
a
˜� 
Fa_a 5000 N 
For the key
Aa_a
Fa_a
Wallow
= b d˜
Fa_a
Wallow
=
d
1
b
Fa_a
Wallow
§¨
©
·
¹
˜� 
d 5.71 mm Ans
Problem 1-84
The fillet weld size a is determined by computing the average shear stress along the shaded plane,
which has the smallest cross section. Determine the smallest size a of the two welds if the force
applied to the plate is P = 100 kN. The allowable shear stress for the weld material is Wallow = 100 MPa.
Given: P 100kN� Wallow 100MPa� 
L 100mm� T 45deg� 
Solution:
Shear Plane in the Weld: Aweld L a˜ sin T� �˜=
Aweld
0.5P
Wallow
=
L a˜ sin T� �˜ 0.5PWallow=
a
1
L sin T� �˜
0.5P
Wallow
§¨
©
·
¹
˜� 
a 7.071 mm Ans
Problem 1-85
The fillet weld size a = 8 mm. If the joint is assumed to fail by shear on both sides of the block along
the shaded plane, which is the smallest cross section, determine the largest force P that can be applied
to the plate. The allowable shear stress for the weld material is Wallow = 100 MPa.
Given: a 8mm� L 100mm� 
T 45deg� Wallow 100MPa� 
Solution:
Shear Plane in the Weld: Aweld L a˜ sin T� �˜=
P Wallow 2Aweld� �˜=
P Wallow 2 L a˜ sin T� �˜� �ª¬ º¼˜� 
P 113.14 kN Ans
Problem 1-86
The eye bolt is used to support the load of 25 kN. Determine its diameter d to the nearest multiples of
5mm and the required thickness h to the nearest multiples of 5mm of the support so that the washer
will not penetrate or shear through it. The allowable normal stress for the bolt is Vallow = 150 MPa and
the allowable shear stress for the supporting material is Wallow = 35 MPa.
Given: P 25kN� dwasher 25mm� 
Vallow 150MPa� Wallow 35MPa� 
Solution:
Allowable Normal Stress: Design of bolt size
Abolt
P
Vallow
=
S
4
§¨
©
·
¹ d
2˜ PVallow
=
d
4
S
P
Vallow
§¨
©
·
¹
˜� 
d 14.567 mm 
Use d 15mm� d 15 mm Ans
Allowable Shear Stress: Design of support thickness
Asupport
P
Wallow
= S dwasher� �˜ h˜ PWallow=
h
1
S dwasher� �˜
P
Wallow
§¨
©
·
¹
˜� 
h 9.095 mm 
Use d 10mm� d 10 mm Ans
Problem 1-87
The frame is subjected to the load of 8 kN. Determine the required diameter of the pins at A and B if
the allowable shear stress for the material is Wallow = 42 MPa. Pin A is subjected to double shear,
whereas pin B is subjected to single shear.
Given: P 8kN� Wallow 42MPa� 
a 1.5m� b 1.5m� c 1.5m� d 0.6m� 
Solution: TBC 45deg� 
Support Reactions: From FBD (a),
 60D=0; FBC sin TBC� �˜ c( )˜ P c d�( )˜� 0=
FBC P
c d�
sin TBC� � c( )˜˜� 
FBC 15.839 kN 
From FBD (b),
 60A=0; Dy a b�( )˜ P c d�( )˜� 0=
Dy P
c d�
a b�˜� Dy 5.6 kN 
+ 6Fx=0; Ax P� 0= Ax P� Ax 8 kN 
 + 6Fy=0; Dy Ay� 0= Ay Dy� 
FA Ax
2
Ay
2�� FA 9.77 kN 
Apin
0.5FA
Wallow
=
S
4
§¨
©
·
¹ d
2˜
0.5FA
Wallow
=
d
4
S
0.5FA
Wallow
§¨
©
·
¹
˜� d 12.166 mm Ans
For pin B: Pin A is subjected to single shear, and FB FBC� 
Apin
FB
Wallow
=
S
4
§¨
©
·
¹ d
2˜
FB
Wallow
=
d
4
S
FB
Wallow
§¨
©
·
¹
˜� d 21.913 mm Ans
Problem 1-88
The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile
stress of Vallow = 200 MPa, determine the required diameter of each wire if the applied load is P = 5
kN.
Given: P 5kN� Vallow 200MPa� 
a 4m� b 3m� T 60deg� 
Solution:
c a
2
b
2�� h a
c
� v b
c
� 
At joint A:
Initial guess: FAB 1kN� FAC 2kN� 
Given
+ 6Fx=0; FAC h( )˜ FAB sin T� �˜� 0= [1]
 + 6Fy=0; FAC v( )˜ FAB cos T� �˜� P� 0= [2]
Solving [1] and [2]:
FAB
FAC
§¨
©¨
·
¹
Find FAB FAC�� �� 
FAB
FAC
§¨
©¨
·
¹
4.3496
4.7086
§¨
©
·
¹ kN 
For wire AB
AAB
FAB
Vallow
=
S
4
§¨
©
·
¹ dAB
2˜
FAB
Vallow
=
dAB
4
S
FAB
Vallow
§¨
©
·
¹
˜� dAB 5.26 mm Ans
For wire AC
AAC
FAC
Vallow
=
S
4
§¨
©
·
¹ dAC
2˜
FAC
Vallow
=
dAC
4
S
FAC
Vallow
§¨
©
·
¹
˜� dAC 5.48 mm Ans
Problem 1-89
The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile
stress of Vallow = 180 MPa, and wire AB has a diameter of 6 mm and AC has a diameter of 4 mm,
determine the greatest force P that can be applied to the chainbefore one of the wires fails.
Given: Vallow 180MPa� 
a 4m� b 3m� T 60deg� 
dAB 6mm� dAC 4mm� 
Solution:
c a
2
b
2�� h a
c
� v b
c
� 
Assume failure of AB: FAB AAB� � Vallow˜=
FAB
S
4
§¨
©
·
¹ dAB
2˜ Vallow˜� FAB 5.09 kN 
At joint A:
Initial guess: P1 1kN� FAC 2kN� 
Given
+ 6Fx=0; FAC h( )˜ FAB sin T� �˜� 0= [1]
 + 6Fy=0; FAC v( )˜ FAB cos T� �˜� P1� 0= [2]
Solving [1] and [2]:
P1
FAC
§¨
©¨
·
¹
Find P1 FAC�� �� P1
FAC
§¨
©¨
·
¹
5.8503
5.5094
§¨
©
·
¹ kN 
Assume failure of AC: FAC AAC� � Vallow˜=
FAC
S
4
§¨
©
·
¹ dAC
2˜ Vallow˜� FAC 2.26 kN 
At joint A:
Initial guess: P2 1kN� FAB 2kN� 
Given
+ 6Fx=0; FAC h( )˜ FAB sin T� �˜� 0= [1]
 + 6Fy=0; FAC v( )˜ FAB cos T� �˜� P2� 0= [2]
Solving [1] and [2]:
FAB
P2
§¨
©¨
·
¹
Find FAB P2�� �� FAB
P2
§¨
©¨
·
¹
2.0895
2.4019
§¨
©
·
¹ kN 
Chosoe the smallest value: P min P1 P2�� �� P 2.40 kN Ans
Problem 1-90
The boom is supported by the winch cable that has a diameter of 6 mm and an allowable normal stress
of Vallow = 168 MPa. Determine the greatest load that can be supported without causing the cable to fail
when T = 30° and I = 45°. Neglect the size of the winch.
Given: Vallow 168MPa� do 6mm� 
T 30deg� I 45deg� 
Solution:
For the cable:
Tcable Acable� � Vallow˜=
Tcable
S
4
§¨
©
·
¹ do
2˜ Vallow˜� 
Tcable 4.7501 kN 
At joint B:
Initial guess: FAB 1 kN� W 2 kN� 
Given
+ 6Fx=0; Tcable� cos T� � FAB cos I� �˜� 0= [1]
 + 6Fy=0; W� FAB sin I� �˜� Tcable sin T� �˜� 0= [2]
Solving [1] and [2]:
FAB
W
§¨
©
·
¹ Find FAB W�� �� 
FAB
W
§¨
©
·
¹
5.818
1.739
§¨
©
·
¹ kN Ans
Problem 1-91
The boom is supported by the winch cable that has an allowable normal stress of Vallow = 168 MPa. If
it is required that it be able to slowly lift 25 kN, from T = 20° to T = 50°, determine the smallest
diameter of the cable to the nearest multiples of 5mm. The boom AB has a length of 6 m. Neglect the
size of the winch. Set d = 3.6 m.
Given: Vallow 168MPa� W 25 kN� 
d 3.6m� a 6m� 
Solution:
Maximum tension in canle occurs when T 20deg� 
sin T� �
a
sin \� �
d
= \ asin d
a
§¨
©
·
¹ sin T� �˜
ª«¬
º»¼� \ 11.842 deg 
At joint B:
Initial guess: FAB 1 kN� Tcable 2 kN� 
Given I T \�� 
+ 6Fx=0; Tcable� cos T� � FAB cos I� �˜� 0= [1]
 + 6Fy=0; W� FAB sin I� �˜� Tcable sin T� �˜� 0= [2]
Solving [1] and [2]:
FAB
Tcable
§¨
©¨
·
¹
Find FAB Tcable�� �� 
FAB
Tcable
§¨
©¨
·
¹
114.478
103.491
§¨
©
·
¹ kN 
For the cable:
Acable
P
Vallow
=
S
4
§¨
©
·
¹ do
2˜
Tcable
Vallow
=
do
4
S
Tcable
Vallow
§¨
©
·
¹
˜� 
do 28.006 mm 
Use do 30mm� do 30 mm Ans
Problem 1-92
The frame is subjected to the distributed loading of 2 kN/m. Determine the required diameter of the
pins at A and B if the allowable shear stress for the material is Wallow = 100 MPa. Both pins are
subjected to double shear.
Given: w 2
kN
m
� Wallow 100MPa� 
r 3m� 
Solution: Member AB is atwo-force member 
T 45deg� 
Support Reactions:
 60A=0; FBC sin T� �˜ r( )˜ w r˜( ) 0.5r( )˜� 0=
FBC
0.5w r˜
sin T� �� FBC 4.243 kN 
 + 6Fy=0; Ay FBC sin T� �˜� w r˜� 0=
Ay FBC� sin T� �˜ w r˜�� Ay 3 kN 
+ 6Fx=0; Ax FBC cos T� �˜� 0=
Ax FBC cos T� �˜� Ax 3 kN 
Average Shear Stress: Pin A and pin B are subjected to double shear
FA Ax
2
Ay
2�� FA 4.243 kN 
FB FBC� FB 4.243 kN 
Since both subjected to the same shear force V = 0.5 FA and V 0.5FB� 
Apin
V
Wallow
=
S
4
§¨
©
·
¹ dpin
2˜ VWallow
=
dpin
4
S
V
Wallow
§¨
©
·
¹
˜� 
dpin 5.20 mm Ans
Problem 1-93
Determine the smallest dimensions of the circular shaft and circular end cap if the load it is required to
support is P = 150 kN. The allowable tensile stress, bearing stress, and shear stress is (Vt)allow = 175
MPa, (Vb)allow = 275 MPa, and Wallow = 115 MPa.
Given: P 150kN� Vt_allow 175MPa� 
Wallow 115MPa� Vb_allow 275MPa� 
d2 30mm� 
Solution:
Allowable Normal Stress: Design of end cap outer diameter
A
P
Vt_allow
=
S
4
§¨
©
·
¹ d1
2
d2
2�§© ·¹˜ PVt_allow=
d1
4
S
P
Vt_allow
§¨
©
·
¹
˜ d22�� d1 44.62 mm Ans
Allowable Bearing Stress: Design of circular shaft diameter
A
P
Vb_allow
=
S
4
§¨
©
·
¹ d3
2§© ·¹˜ PVb_allow=
d3
4
S
P
Vb_allow
§¨
©
·
¹
˜� d3 26.35 mm Ans
Allowable Shear Stress: Design of end cap thickness
A
P
Wallow
= S d3˜� � t˜ PWallow=
t
1
S d3˜
P
Wallow
§¨
©
·
¹
˜� t 15.75 mm Ans
Problem 1-94
If the allowable bearing stress for the material under the supports at A and B is (Vb)allow = 2.8 MPa,
determine the size of square bearing plates A' and B' required to support the loading. Take P = 7.5 kN.
Dimension the plates to the nearest multiples of 10mm. The reactions at the supports are vertical.
Given: Vb_allow 2.8MPa� P 7.5 kN� 
P1 10 kN� P2 10 kN� 
P3 15 kN� P4 10 kN� 
a 1.5m� b 2.5m� 
Solution: L 3 a˜ b�� 
Support Reactions:
 60A=0; By 3a( ) P2 a( )˜� P3 2a( )˜� P4 3a( )˜� P L( )˜� 0=
By P2
a
3a
§¨
©
·
¹˜ P3
2a
3a
§¨
©
·
¹˜� P4
3a
3a
§¨
©
·
¹˜� P
L
3a
§¨
©
·
¹˜�� By 35 kN 
 60B=0; Ay� 3 a˜( )˜ P1 3 a˜( )˜� P2 2 a˜( )˜� P3 a( )˜� P b( )˜� 0=
Ay P1
3a
3a
§¨
©
·
¹˜ P2
2a
3a
§¨
©
·
¹˜� P3
a
3a
§¨
©
·
¹˜� P
b
3a
§¨
©
·
¹˜�� Ay 17.5 kN 
For Plate A:
Aplate_A
Ay
Vb_allow
= aA
2
Ay
Vb_allow
=
aA
Ay
Vb_allow
� 
aA 79.057 mm 
Use aA x aA plate: aA 80mm= Ans
For Plate B
Aplate_B
By
Vb_allow
= aB
2
By
Vb_allow
=
aB
By
Vb_allow
� 
aB 111.803 mm 
Use aB x aB plate: aB 120mm= Ans
Rb = 35kN 
 
(b is in subscript)
Problem 1-95
If the allowable bearing stress for the material under the supports at A and B is (Vb)allow = 2.8 MPa,
determine the maximum load P that can be applied to the beam. The bearing plates A' and B' have
square cross sections of 50mm x 50mm and 100mm x 100mm, respectively.
Given: Vb_allow 2.8MPa� 
P1 10 kN� P2 10 kN� 
P3 15 kN� P4 10 kN� 
a 1.5m� b 2.5m� 
aA 50mm� aB 100mm� 
Solution: L 3 a˜ b�� 
Support Reactions:
 60A=0; By 3a( ) P2 a( )˜� P3 2a( )˜� P4 3 a˜( )˜� P L( )˜� 0=
By P2
a
3a
§¨
©
·
¹˜ P3 2
a
3a
˜§¨©
·
¹˜� P4
3 a˜
3a
§¨
©
·
¹˜� P
L
3a
§¨
©
·
¹˜�=
 60B=0; Ay� 3 a˜( )˜ P1 3 a˜( )˜� P2 2 a˜( )˜� P3 a( )˜� P b( )˜� 0=
Ay P1
3a
3a
§¨
©
·
¹˜ P2
2a
3a
§¨
©
·
¹˜� P3
a
3a
§¨
©
·
¹˜� P
b
3a
§¨
©
·
¹˜�=
For Plate A: Ay aA
2§© ·¹ Vb_allow˜� 
aA
2§© ·¹ Vb_allow˜ P1 3a3a
§¨
©
·
¹˜ P2
2a
3a
§¨
©
·
¹˜� P3
a
3a
§¨
©
·
¹˜� P
b
3a
§¨
©
·
¹˜�=
P P1
3a
b
§¨
©
·
¹˜ P2
2a
b
§¨
©
·
¹˜� P3
a
b
§¨
©
·
¹˜� aA
2§© ·¹ Vb_allow˜ 3ab
§¨
©
·
¹˜�� P 26.400 kN 
Pcase_1 P� 
For Plate B: By aB
2§© ·¹ Vb_allow˜� 
aB
2§© ·¹ Vb_allow˜ P2 a3a
§¨
©
·
¹˜ P3 2
a
3a
˜§¨©
·
¹˜� P4
3 a˜
3a
§¨
©
·
¹˜� P
L
3a
§¨
©
·
¹˜�=
P P2�
a
L
§¨
©
·
¹˜ P3 2
a
L
˜§¨©
·
¹˜� P4
3 a˜
L
§¨
©
·
¹˜� aB
2§© ·¹ Vb_allow˜ 3aL˜�� P 3.000 kN 
Pcase_2 P� 
Pallow min Pcase_1 Pcase_2�� �� Pallow 3 kN Ans
Problem 1-96
Determine the required cross-sectional area of member BC and the diameter of the pins at A and B if
the allowable normal stress is Vallow = 21 MPa and the allowable shear stress is Wallow = 28 MPa.
Given: Vallow 21MPa� Wallow 28MPa� 
P 7.5kip� T 60deg� 
a 0.6m� b 1.2m� c 0.6m� 
Solution: L a b� c��Support Reactions:
 60A=0; By L( )˜ P a( )˜� P a b�( )˜� 0=
By P
a
L
˜ P a b�
L
˜�� 
FBC
By
sin T� �� FBC 38.523 kN By 33.362 kN 
Bx FBC cos T� �˜� Bx 19.261 kN 
 + 6Fy=0; By� P� P� Ay� 0= Ay By� P� P�� Ay 33.362 kN 
+ 6Fx=0; Bx Ax� 0= Ax Bx� Ax 19.261 kN 
FA Ax
2
Ay
2�� FA 38.523 kN 
Member BC: 
ABC
FBC
Vallow
� ABC 1834.416 mm2 Ans
Pin A:
AA
FA
Wallow
=
S
4
§¨
©
·
¹ dA
2˜
FA
Wallow
=
dA
4
S
FA
Wallow
§¨
©
·
¹
˜� dA 41.854 mm Ans
Pin B:
AB
0.5FBC
Wallow
=
S
4
§¨
©
·
¹ dB
2˜
0.5FBC
Wallow
=
dB
4
S
0.5FBC
Wallow
§¨
©
·
¹
˜� dB 29.595 mm Ans
Problem 1-97
The assembly consists of three disks A, B, and C that are used to support the load of 140 kN.
Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and the
diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is (Wallow)b =
350 MPa and allowable shear stress is Wallow = 125 MPa.
Given: P 140kN� 
Wallow 125MPa� Vb_allow 350MPa� 
hB 20mm� hC 10mm� 
Solution:
Allowable Shear Stress: Assume shear failure dor disk C
A
P
Wallow
= S d2˜� � hC˜ PWallow=
d2
1
S hC˜
P
Wallow
§¨
©
·
¹
˜� d2 35.65 mm Ans
Allowable Bearing Stress: Assume bearing failure dor disk C
A
P
Vb_allow
=
S
4
§¨
©
·
¹ d2
2
d3
2�§© ·¹˜ PVb_allow=
d3 d2
2 4
S
P
Vb_allow
§¨
©
·
¹
˜�� d3 27.60 mm Ans
Allowable Bearing Stress: Assume bearing failure dor disk B
A
P
Vb_allow
=
S
4
§¨
©
·
¹ d1
2§© ·¹˜ PVb_allow=
d1
4
S
P
Vb_allow
§¨
©
·
¹
˜� d1 22.57 mm 
Since d3 > d1, disk B might fail due to shear.
W P
A
= W PS d1˜ hB˜
� W 98.73 MPa < Wallow (O.K.!)
Therefore d1 22.57 mm Ans
Problem 1-98
Strips A and B are to be glued together using the two strips C and D. Determine the required thickness
t of C and D so that all strips will fail simultaneously. The width of strips A and B is 1.5 times that of
strips C and D.
Given: P 40N� t 30mm� 
bA 1.5m� bB 1.5m� 
bC 1m� bD 1m� 
Solution:
Average Normal Stress: Requires,
VA VB= VB VC= VC VD=
N
bA� � t˜
0.5N
bC� � tC� �˜=
tC
0.5 bA� � t˜
bC
� tC 22.5 mm Ans
Problem 1-99
If the allowable bearing stress for the material under the supports at A and B is (Vb)allow = 2.8 MPa,
determine the size of square bearing plates A' and B' required to support the loading. Dimension the
plates to the nearest multiples of 10mm. The reactions at the supports are vertical. Take P = 7.5 kN.
Given: Vb_allow 2.8MPa� P 7.5kN� 
w 10
kN
m
� a 4.5m� b 2.25m� 
Solution: L a b�� 
Support Reactions:
 60A=0; By a( ) w a( )˜ 0.5 a˜( )˜� P L( )˜� 0=
By w 0.5 a˜( )˜ P
L
a
§¨
©
·
¹˜�� By 33.75 kN 
 60B=0; Ay� a( )˜ w a( )˜ 0.5 a˜( )˜� P L( )˜� 0=
Ay w 0.5 a˜( )˜ P
b
a
§¨
©
·
¹˜�� Ay 18.75 kN 
Allowable Bearing Stress: Design of bearing plates
For Plate A:
Area
Ay
Vb_allow
= aA
2
Ay
Vb_allow
=
aA
Ay
Vb_allow
� 
aA 81.832 mm 
Use aA x aA plate: aA 90mm� aA 90 mm Ans
For Plate B
Area
By
Vb_allow
= aB
2
By
Vb_allow
=
aB
By
Vb_allow
� 
aB 109.789 mm 
Use aB x aB plate: aB 110mm� aB 110.00 mm Ans
Problem 1-100
If the allowable bearing stress for the material under the supports at A and B is (Vb)allow = 2.8 MPa,
determine the maximum load P that can be applied to the beam. The bearing plates A' and B' have
square cross sections of 50mm x 50mm and 100mm x 100mm, respectively.
Given: Vb_allow 2.8MPa� 
w 10
kN
m
� a 4.5m� b 2.25m� 
aA 50mm� aB 100mm� 
Solution: L a b�� 
Support Reactions:
 60A=0;
By a( )˜ w a( )˜ 0.5 a˜( )˜� P L( )˜� 0=
P By
a
L
§¨
©
·
¹˜ w
a
L
§¨
©
·
¹˜ 0.5 a˜( )˜�=
 60B=0; Ay� a( )˜ w a( )˜ 0.5 a˜( )˜� P b( )˜� 0=
P Ay�
a
b
§¨
©
·
¹˜ w
a
b
§¨
©
·
¹˜ 0.5 a˜( )˜�=
Allowable Bearing Stress:
Assume failure of material occurs under plate A. Ay aA
2§© ·¹ Vb_allow˜� 
P aA
2§© ·¹ Vb_allow˜ª¬ º¼� ab
§¨
©
·
¹˜ w a˜( )
0.5a
b
˜�� P 31 kN 
Pcase_1 P� 
Assume failure of material occurs under plate B. By aB
2§© ·¹ Vb_allow˜� 
P By
a
L
§¨
©
·
¹˜ w
a
L
§¨
©
·
¹˜ 0.5 a˜( )˜�� P 3.67 kN 
Pcase_2 P� 
Pallow min Pcase_1 Pcase_2�� �� Pallow 3.67 kN Ans
Problem 1-101
The hanger assembly is used to support a distributed loading of w = 12 kN/m. Determine the average
shear stress in the 10-mm-diameter bolt at A and the average tensile stress in rod AB, which has a
diameter of 12 mm. If the yield shear stress for the bolt is Wy = 175 MPa, and the yield tensile stress for
the rod is Vy = 266 MPa, determine the factor of safety with respect to yielding in each case.
Given: Wy 175MPa� w 12
kN
m
� 
Vy 266MPa� 
a 1.2m� b 0.6m� e 0.9m� 
do 10mm� drod 12mm� 
Solution:
c a
2
e
2�� h a
c
� v e
c
� 
Support Reactions: L a b�� 
 60C=0; FAB v( )˜ª¬ º¼ a( )˜ w L( )˜ 0.5 L˜( )˜� 0=
FAB w
L
a v˜
§¨
©
·
¹˜ 0.5 L˜( )˜� 
FAB 27 kN 
For bolt A: Bolt A is subjected to double shear, and V 0.5FAB� V 13.5 kN 
A
S
4
§¨
©
·
¹ do
2˜� W V
A
� W 171.89 MPa Ans
FS
Wy
W� FS 1.02 Ans
For rod AB: N FAB� N 27 kN 
A
S
4
§¨
©
·
¹ drod
2˜� V N
A
� V 238.73 MPa Ans
FS
Vy
V� FS 1.11 Ans
Problem 1-102
Determine the intensity w of the maximum distributed load that can be supported by the hanger
assembly so that an allowable shear stress of Wallow = 95 MPa is not exceeded in the 10-mm-diameter
bolts at A and B, and an allowable tensile stress of Vallow = 155 MPa is not exceeded in the
12-mm-diameter rod AB.
Given: Wallow 95MPa� Vallow 155MPa� 
a 1.2m� b 0.6m� e 0.9m� 
do 10mm� drod 12mm� 
Solution: c a
2
e
2�� h a
c
� v e
c
� 
Support Reactions: L a b�� 
 60C=0; FAB v( )˜ª¬ º¼ a( )˜ w L( )˜ 0.5 L˜( )˜� 0=
FAB w
L
a v˜
§¨
©
·
¹˜ 0.5 L˜( )˜=
Assume failure of pin A or B:
V 0.5FAB= V Wallow A˜= A
S
4
§¨
©
·
¹ do
2˜� 
0.5 w˜ L
a v˜
§¨
©
·
¹˜ 0.5 L˜( )˜ Wallow
S
4
§¨
©
·
¹ do
2˜ª«¬
º»¼˜=
w
a v˜
0.5L( )
2
Wallow˜
S
4
§¨
©
·
¹ do
2˜ª«¬
º»¼˜� 
w 6.632
kN
m
 (controls!) Ans
Assuming failure of rod AB:
N FAB= N Vallow A˜= A
S
4
§¨
©
·
¹ drod
2˜� 
w
L
a v˜
§¨
©
·
¹˜ 0.5 L˜( )˜ Vallow
S
4
§¨
©
·
¹ drod
2˜ª«¬
º»¼˜=
w
a v˜
0.5L
2
Vallow˜
S
4
§¨
©
·
¹ drod
2˜ª«¬
º»¼˜� 
w 7.791
kN
m
 
Problem 1-103
The bar is supported by the pin. If the allowable tensile stress for the bar is (Vt)allow = 150 MPa, and
the allowable shear stress for the pin is Wallow = 85 MPa, determine the diameter of the pin for which
the load P will be a maximum. What is this maximum load? Assume the hole in the bar has the same
diameter d as the pin. Take t =6 mm and w = 50 mm. 
Given: Wallow 85MPa� 
Vt_allow 150MPa� 
t 6mm� w 50mm� 
Solution: Given
Allowable Normal Stress: The effective cross-sectional
area Ae for the bar must be considered here by taking
into account the reduction in cross-sectional area
introduced by the hole. Here, effective area Ae is equal
to (w - d) t, and Vallow equals to P /Ae .
Vt_allow
P
w d�( ) t˜= [1]
Allowable Shear Stress: The pin is subjected to double
shear and therefore the allowable W equals to 0.5P /Apin,
and the area Apin is equal to ( S/4) d2. 
Wallow
2
S
§¨
©
·
¹
P
d
2
§¨
©
·
¹
˜=
[2]
Solving [1] and [2]: Initial guess: d 20mm� P 10kN� 
P
d
§¨
©
·
¹ Find P d�( )� P 31.23 kN Ans
d 15.29 mm Ans
Problem 1-104
The bar is connected to the support using a pin having a diameter