Engineering Mechanics - Statics Episode 2 Part 7 pdf

Engineering Mechanics - Statics Chapter 7
Problem 7-19
Determine the normal force, shear force, and moment at a section passing through point C.
Units Used:
kN 10
3
N=
Given:
P 8 kN= c 0.75 m=
a 0.75m= d 0.5 m=
b 0.75m= r 0.1 m=
Solution:
Σ
M
A
= 0;
T− dr+()Pa b+ c+()+ 0=
TP
ab+ c+
dr+
⎛
⎜
⎝
⎞
⎟
⎠
= T 30kN=
+
→
A
x

T= A
x
30
kN=
Σ
F
x
= 0;
+
↑
Σ
F
y
= 0;
A
y
P= A
y
8
kN=
+
→
Σ
F
x
= 0;
N
C
− T− 0=
N

C
T−= N
C
30
− kN=
+
↑
Σ
F
y
= 0;
V
C
P+ 0=
V
C
P−= V
C
8
− kN=
Σ
M
C
= 0;
M
C
− Pc+ 0=
M
C
Pc= M

C
6
kN m⋅=
Problem 7-20
The cable will fail when subjected to a tension T
max
. Determine the largest vertical load P the frame
will support and calculate the internal normal force, shear force, and moment at a section passing
through point C for this loading.
Units Used:
kN 10
3
N=
641
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Given:
T
max
2
kN=
a 0.75 m=
b 0.75 m=
c 0.75 m=
d 0.5 m=
r 0.1 m=
Solution:
Σ

M
A
= 0;
T
max
− rd+()Pa b+ c+()+ 0=
PT
max
dr+
ab+ c+
⎛
⎜
⎝
⎞
⎟
⎠
= P 0.533 kN=
+
→
Σ
F
x
= 0;
T
max
A
x
− 0= A
x
T

max
= A
x
2
kN=
+
↑
A
y
P− 0= A
y
P= A
y
0.533
kN=
Σ
F
y
= 0;
+
→
N
C
A
x
−= N
C
2
− kN=
Σ

F
x
= 0;
N
C
− A
x
− 0=
+
↑
Σ
F
y
= 0;
V
C
− A
y
+ 0= V
C
A
y
= V
C
0.533
kN=
M
C
A
y

c= M
C
0.400
kN m⋅=
Σ
M
C
= 0;
M
C
− A
y
c+ 0=
Problem 7-21
Determine the internal shear force and moment acting at point C of the beam.
Units Used:
kip 10
3
lb=
Given:
w 2
kip
ft
=
a 9 ft=
642
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7

Solution:
Σ
F
x
= 0;
N
C
0
= N
C
0
=
Σ
F
y
= 0;
wa
2
wa
2
− V
C
− 0= V
C
0
=
Σ
M
C
= 0;

M
C
wa
2
⎛
⎜
⎝
⎞
⎟
⎠
a−
wa
2
⎛
⎜
⎝
⎞
⎟
⎠
a
3
⎛
⎜
⎝
⎞
⎟
⎠
+ 0=
M
C

wa
2
3
= M
C
54.00
kip ft⋅=
Problem 7-22
Determine the internal shear force and moment acting at point D of the beam.
Units Used:
kip 10
3
lb=
Given:
w 2
kip
ft
=
a 6 ft=
b 9 ft=
Solution:
N
D
0
=
Σ
F
x
= 0;
Σ

F
y
= 0;
wb
2
w
a
b
⎛
⎜
⎝
⎞
⎟
⎠
a
2
⎛
⎜
⎝
⎞
⎟
⎠
− V
D
− 0=
V
D
wb
2
w

a
b
⎛
⎜
⎝
⎞
⎟
⎠
a
2
⎛
⎜
⎝
⎞
⎟
⎠
−=
V
D
5.00
kip=
Σ
M
D
= 0;
M
D
wb
2
⎛

⎜
⎝
⎞
⎟
⎠
a−
wa
b
⎛
⎜
⎝
⎞
⎟
⎠
a
2
⎛
⎜
⎝
⎞
⎟
⎠
a
3
⎛
⎜
⎝
⎞
⎟
⎠

+ 0=
M
D
wb
2
⎛
⎜
⎝
⎞
⎟
⎠
a
wa
3
6b
⎛
⎜
⎝
⎞
⎟
⎠
−=
M
D
46.00
kip ft⋅=
643
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics Chapter 7
Problem 7-23
The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the internal
normal force, shear force, and moment at (a) point C, which is just to the right of the bearing at
A, and (b) point D, which is just to the left of the
F
2
force.
Units Used:
kip 10
3
lb=
Given:
F
1
2500
lb= a 6 ft=
F
2
3000
lb= b 12 ft=
w 75
lb
ft
= c 2ft=
Solution:
Σ
M
B
= 0;

A
y
− bc+()F
1
ab+ c+()+ wb
b
2
c+
⎛
⎜
⎝
⎞
⎟
⎠
+ F
2
c+ 0=
A
y
1
2
2F
1
ab+ c+()wb
2
2cb+
()
+ 2F
2
c+

bc+
= A
y
4514
lb=
Σ
F
x
= 0;
B
x
0
lb=
Σ
F
y
= 0;
A
y
F
1
− wb− F
2
− B
y
+ 0=
B
y
A
y

− F
1
+ wb+ F
2
+= B
y
1886
lb=
Segment AC :
F
1
aM
C
+ 0=
Σ
M
C
= 0;
M
C
F
1
− a= M
C
15
− kip ft⋅=
Σ
F
x
= 0;

N
C
0
= N
C
0
=
F
1
− A
y
+ V
C
− 0=
Σ
F
y
= 0;
V
C
A
y
F
1
−= V
C
2.01
kip=
Segment BD:
Σ

M
D
= 0;
M
D
− B
y
c+ 0=
M
D
B
y
c= M
D
3.77
kip ft⋅=
Σ
F
x
= 0;
N
D
0
= N
D
0
=
Σ
F
y

= 0;
V
D
F
2
− B
y
+ 0=
V
D
F
2
B
y
−= V
D
1.11
kip=
644
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Problem 7-24
The jack AB is used to straighten the bent beam DE using the arrangement shown. If the axial
compressive force in the jack is P, determine the internal moment developed at point C of the
top beam. Neglect the weight of the beams.
Units Used:
kip 10
3

lb=
Given:
P 5000 lb=
a 2 ft=
b 10 ft=
Solution:
Segment:
Σ
M
C
= 0;
M
C
P
2
⎛
⎜
⎝
⎞
⎟
⎠
b+ 0=
M
C
P
2
− b=
M
C
25.00

− kip ft⋅=
Problem 7-25
The jack AB is used to straighten the bent
beam DE using the arrangement shown. If
the axial compressive force in the jack is P,
determine the internal moment developed at
point C of the top beam. Assume that each
beam has a uniform weight density
γ
.
Units Used:
kip 10
3
lb=
Given:
P 5000 lb=
645
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
γ
150
lb
ft
=
a 2 ft=
b 10 ft=
Solution:
Beam:

+
↑
Σ
F
y
= 0;
P 2
γ
ab+()− 2R− 0=
R
P
2
γ
ab+()−=
R 700 lb=
Segment:
Σ
M
C
= 0;
M
C
Rb+
γ
ab+()
ab+
2
⎛
⎜
⎝

⎞
⎟
⎠
+ 0=
M
C
R− b
γ
ab+()
2
2
−=
M
C
17.8
− kip ft⋅=
Problem 7-26
Determine the normal force, shear force, and moment in the beam at sections passing through
points D and E. Point E is just to the right of the F load.
Units Used:
kip 10
3
lb=
Given:
a 6 ft=
w 1.5
kip
ft
= b 6 ft=
c 4 ft=

F 3 kip=
d 4 ft=
646
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Σ
M
B
= 0;
1
2
wa b+()
ab+
3
⎛
⎜
⎝
⎞
⎟
⎠
A
y
ab+()− 0=
A
y
1
2
wa b+()

ab+
3
⎛
⎜
⎝
⎞
⎟
⎠
ab+
=
A
y
3
kip=
+
→
Σ
F
x
= 0;
B
x
0
=
+
↑
Σ
F
y
= 0;

B
y
A
y
+
1
2
wa b+()− 0=
B
y
A
y
−
1
2
wa b+()+=
B
y
6
kip=
+
→
Σ
F
x
= 0;
N
D
0
=

+
↑
A
y
1
2
aw
ab+
⎛
⎜
⎝
⎞
⎟
⎠
a− V
D
− 0=
Σ
F
y
= 0;
V
D
A
y
1
2
aw
ab+
⎛

⎜
⎝
⎞
⎟
⎠
a−= V
D
0.75
kip=
Σ
M
D
= 0;
M
D
1
2
aw
ab+
⎛
⎜
⎝
⎞
⎟
⎠
a
a
3
⎛
⎜

⎝
⎞
⎟
⎠
+ A
y
a− 0=
M
D
1
−
2
aw
ab+
⎛
⎜
⎝
⎞
⎟
⎠
a
a
3
⎛
⎜
⎝
⎞
⎟
⎠
A

y
a+= M
D
13.5
kip ft⋅=
+
→
N
E
0
=
Σ
F
x
= 0;
+
↑
Σ
F
y
= 0;
V
E
− F− B
y
− 0=
V
E
F− B
y

−= V
E
9
− kip=
Σ
M
E
= 0;
M
E
B
y
c+ 0=
M
E
B
y
− c= M
E
24.0
− kip ft⋅=
647
Solution:
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Determine the normal force, shear force, and moment at a section passing through point D of the
two-member frame.
Units Used:

kN 10
3
N=
Given:
w
1
200
N
m
=
w
2
400
N
m
=
a 2.5 m=
b 3 m=
c 6 m=
Solution:
Σ
M
A
= 0;
w
1
− c
c
2
⎛

⎜
⎝
⎞
⎟
⎠
1
2
w
2
w
1
−
()
c
2c
3
⎛
⎜
⎝
⎞
⎟
⎠
−
a
a
2
c
2
+
F

BC
c
()
+ 0=
F
BC
w
1
c
2
2
w
2
w
1
−
()
c
2
3
+
⎡
⎢
⎣
⎤
⎥
⎦
a
2
c

2
+
ac
= F
BC
2600
N=
+
→
Σ
F
x
= 0;
A
x
c
a
2
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
BC
= A

x
2400
N=
+
↑
Σ
F
y
= 0;
A
y
w
1
c−
1
2
w
2
w
1
−
()
c−
a
a
2
c
2
+
⎛

⎜
⎝
⎞
⎟
⎠
F
BC
+ 0=
A
y
w
1
c
1
2
w
2
w
1
−
()
c+
a
a
2
c
2
+
⎛
⎜

⎝
⎞
⎟
⎠
F
BC
−= A
y
800
N=
+
→
Σ
F
x
= 0;
A
x
− N
D
+ 0= N
D
A
x
= N
D
2.40
kN=
+
↑

Σ
F
y
= 0;
A
y
w
1
b−
1
2
w
2
w
1
−
()
b
c
⎛
⎜
⎝
⎞
⎟
⎠
b− V
D
− 0=
V
D

A
y
w
1
b−
1
2
w
2
w
1
−
()
b
2
c
⎛
⎜
⎝
⎞
⎟
⎠
−= V
D
50
N=
Σ
M
D
= 0;

A
y
− bw
1
b
b
2
⎛
⎜
⎝
⎞
⎟
⎠
+
1
2
w
2
w
1
−
()
b
c
⎛
⎜
⎝
⎞
⎟
⎠

b
b
3
⎛
⎜
⎝
⎞
⎟
⎠
+ M
D
+ 0=
M
D
A
y
b() w
1
b
2
2
⎛
⎜
⎝
⎞
⎟
⎠
−
1
2

w
2
w
1
−
()
b
3
3c
⎛
⎜
⎝
⎞
⎟
⎠
−= M
D
1.35
kN m⋅=
648
Problem 7-27
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Problem 7-28
Determine the normal force, shear force, and moment at sections passing through points E and F.
Member BC is pinned at B and there is a smooth slot in it at C. The pin at C is fixed to member CD.
Units Used:
kip 10

3
lb=
Given:
M 350 lb ft⋅=
w 80
lb
ft
= c 2 ft=
F 500 lb= d 3 ft=
θ
60 deg= e 2 ft=
a 2 ft= f 4 ft=
b 1 ft= g 2 ft=
Solution:
Σ
M
B
= 0;
1−
2
wd
2d
3
⎛
⎜
⎝
⎞
⎟
⎠
F sin

θ
()
d− C
y
de+()+ 0=
C
y
wd
2
3
⎛
⎜
⎝
⎞
⎟
⎠
F sin
θ
()
d+
de+
= C
y
307.8
lb=
+
→
Σ
F
x

= 0;
B
x
F cos
θ
()
− 0= B
x
F cos
θ
()
= B
x
250
lb=
+
↑
Σ
F
y
= 0;
B
y
1
2
wd− F sin
θ
()
− C
y

+ 0=
B
y
1
2
wd Fsin
θ
()
C
y
−+= B
y
245.2
lb=
+
→
Σ
F
x
= 0;
N
E
− B
x
− 0= N
E
B
x
−= N
E

250
− lb=
649
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
+
↑
Σ
F
y
= 0;
V
E
B
y
− 0= V
E
B
y
= V
E
245
lb=
Σ
M
E
= 0
M

E
− B
y
c− 0= M
E
B
y
− c= M
E
490
− lb ft⋅=
+
→
Σ
F
x
= 0;
N
F
0
= N
F
0
lb= N
F
0.00
lb=
+
↑
Σ

F
y
= 0;
C
y
− V
F
− 0= V
F
C
y
−= V
F
308
− lb=
Σ
M
F
= 0;
C
y
f() M
F
+ 0= M
F
f− C
y
= M
F
1.23

− kip ft⋅=
Problem 7-29
The bolt shank is subjected to a tension F. Determine the internal normal force, shear force,
and moment at point C.
Given:
F 80 lb=
a 6 in=
Solution:
Σ
F
x
= 0;
N
C
F+ 0=
N
C
F−=
N
C
80.00
− lb=
Σ
F
y
= 0;
V
C
0
=

Σ
M
C
= 0;
M
C
Fa+ 0=
M
C
F− a=
M
C
480.00
− lb in⋅=
Problem 7-30
Determine the normal force, shear force, and moment acting at sections passing through points B
and C on the curved rod.
650
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Units Used:
kip 10
3
lb=
Given:
F
1
300

lb=
θ
30 deg= r 2 ft=
φ
45 deg=
F
2
400
lb=
Solution:
Σ
F
x
= 0;
F
2
sin
θ
()
F
1
cos
θ
()
− N
B
+ 0=
N
B
F

2
− sin
θ
()
F
1
cos
θ
()
+=
N
B
59.8
lb=
Σ
F
y
= 0;
V
B
F
2
cos
θ
()
+ F
1
sin
θ
()

+ 0=
V
B
F
2
− cos
θ
()
F
1
sin
θ
()
−=
V
B
496
− lb=
Σ
M
B
= 0;
M
B
F
2
rsin
θ
()
+ F

1
rrcos
θ
()
−
()
+ 0=
M
B
F
2
− rsin
θ
()
F
1
r 1 cos
θ
()
−
()
−=
M
B
480
− lb ft⋅=
+
→
Σ
F

x
= 0;
F
2
A
x
− 0= A
x
F
2
= A
x
400
lb=
+
↑
Σ
F
y
= 0;
A
y
F
1
− 0= A
y
F
1
= A
y

300
lb=
Σ
M
A
= 0;
M
A
− F
1
2
r+ 0= M
A
2
F
1
r= M
A
1200
lb ft⋅=
Σ
F
x
= 0;
N
C
A
x
sin
φ

()
+ A
y
cos
φ
()
+ 0=
N
C
A
x
− sin
φ
()
A
y
cos
φ
()
−= N
C
495
− lb=
Σ
F
y
= 0;
V
C
A

x
cos
φ
()
− A
y
sin
φ
()
+ 0=
V
C
A
x
cos
φ
()
A
y
sin
φ
()
−= V
C
70.7
lb=
Σ
M
C
= 0;

M
C
− M
A
− A
x
rsin
φ
()
− A
y
rrcos
φ
()
−
()
+ 0=
651
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
M
C
M
A
− A
x
rsin
φ

(
)
− A
y
r 1 cos
φ
(
)
−
(
)
+= M
C
1.59
− kip ft⋅=
Problem 7-31
The cantilevered rack is used to support each end of a smooth pipe that has total weight W.
Determine the normal force, shear force, and moment that act in the arm at its fixed support A along
a vertical section.
Units Used:
kip 10
3
lb=
Given:
W 300 lb=
r 6 in=
θ
30 deg=
Solution:
Pipe:

+
↑
Σ
F
y
= 0;
N
B
cos
θ
()
W
2
− 0=
N
B
1
2
W
cos
θ
()
⎛
⎜
⎝
⎞
⎟
⎠
= N
B

173.205
lb=
Rack:
+
→
Σ
F
x
= 0;
N
A
− N
B
sin
θ
()
+ 0=
N
A
N
B
sin
θ
()
= N
A
86.6
lb=
+
↑

Σ
F
y
= 0;
V
A
N
B
cos
θ
()
− 0=
V
A
N
B
cos
θ
()
= V
A
150
lb=
Σ
M
A
= 0;
M
A
N

B
rrsin
θ
()
+
cos
θ
()
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
M
A
N
B
rrsin
θ
()
+
cos
θ
()
⎛
⎜
⎝
⎞

⎟
⎠
= M
A
1.800
kip in⋅=
652
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Problem 7-32
Determine the normal force, shear force, and moment at a section passing through point D of the
two-member frame.
Units Used:
kN 10
3
N=
Given:
w 0.75
kN
m
=
F 4 kN=
a 1.5 m= d 1.5 m=
b 1.5 m= e 3=
c 2.5 m= f 4=
Solution:
Σ
M

C
= 0;
B
x
− cd+()
f
e
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
Fd+ 0=
B
x
fdF
e
2
f
2
+ cd+()
= B
x
1.2
kN=

Σ
M
A
= 0;
w− cd+()
cd+
2
⎛
⎜
⎝
⎞
⎟
⎠
B
y
ab+()+ B
x
cd+()+ 0=
B
y
w
cd+()
2
2
⎡
⎢
⎣
⎤
⎥
⎦

B
x
cd+()−
ab+
= B
y
0.40
kN=
+
→
Σ
F
x
= 0;
N
D
− B
x
− 0= N
D
B
x
−= N
D
1.2
− kN=
+
↑
Σ
F

y
= 0;
V
D
B
y
+ 0= V
D
B
y
−= V
D
0.4
− kN=
653
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Σ
M
D
= 0;
M
D
− B
y
b+ 0= M
D
B

y
b= M
D
0.6
kN m⋅=
Problem 7-33
Determine the internal normal force, shear force, and moment acting at point A of the smooth
hook.
Given:
θ
45 deg=
a 2 in=
F 20 lb=
Solution:
Σ
F
x
= 0;
N
A
F cos
θ
()
− 0=
N
A
F cos
θ
()
= N

A
14.1
lb=
Σ
F
y
= 0;
V
A
F sin
θ
()
− 0=
V
A
F sin
θ
()
= V
A
14.1
lb=
Σ
M
B
= 0;
M
A
N
A

a− 0=
M
A
N
A
a= M
A
28.3
lb in⋅=
Problem 7-34
Determine the internal normal force, shear force, and moment acting at points B and C on the
curved rod.
654
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Units Used:
kip 10
3
lb=
Given:
F 500 lb=
θ
2
30
deg=
r 2 ft=
a 3=
θ

1
45
deg=
b 4=
Solution:
Fb
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
sin
θ
2
()
Fa
a
2
b
2
+
⎛
⎜
⎝

⎞
⎟
⎠
cos
θ
2
()
− N
B
+ 0=
Σ
F
N
= 0;
N
B
F
a( )cos
θ
2
()
b sin
θ
2
()
−
a
2
b
2

+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
= N
B
59.8
lb=
Σ
F
V
= 0;
V
B
Fb
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟

⎠
cos
θ
2
()
+
Fa
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
sin
θ
2
()
+ 0=
V
B
F−
bcos
θ
2
()

a( )sin
θ
2
()
+
a
2
b
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
= V
B
496
− lb=
Σ
M
B
= 0;
M
B
Fb
a

2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
rsin
θ
2
()
+ F
a
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
rrcos
θ
2

()
−
()
+ 0=
M
B
Fr
b− sin
θ
2
()
a− a( )cos
θ
2
()
+
a
2
b
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
= M

B
480
− lb ft⋅=
Also,
Σ
F
x
= 0;
A
x
−
Fb
a
2
b
2
+
+ 0=
A
x
F
b
a
2
b
2
+
⎛
⎜
⎝

⎞
⎟
⎠
= A
x
400.00
lb=
Σ
F
y
= 0;
A
y
Fa
a
2
b
2
+
− 0=
655
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
A
y
F
a
a

2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
= A
y
300.00
lb=
Σ
M
A
= 0;
M
A
−
Fa
a
2
b
2
+
⎛
⎜
⎝

⎞
⎟
⎠
2r+ 0=
M
A
2Fra
a
2
b
2
+
= M
A
1200
lb ft⋅=
Σ
F
x
= 0;
N
C
A
x
sin
θ
1
()
+ A
y

cos
θ
1
()
+ 0=
N
C
A
x
− sin
θ
1
()
A
y
cos
θ
1
()
−= N
C
495
− lb=
Σ
F
y
= 0;
V
C
A

x
cos
θ
1
()
− A
y
sin
θ
1
()
+ 0=
V
C
A
x
cos
θ
1
()
A
y
sin
θ
1
()
−= V
C
70.7
lb=

Σ
M
C
= 0;
M
C
− M
A
− A
y
rrcos
θ
1
()
−
()
+ A
x
rsin
θ
1
()
− 0=
M
C
M
A
− A
y
rrcos

θ
1
()
−
()
+ A
x
rsin
θ
1
()
−= M
C
1.59
− kip ft⋅=
Problem 7-35
Determine the ratio a/b for which the shear force will be zero at the midpoint C of the beam.
Solution:
Find A
y
:
Σ
M
B
= 0;
656
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7

1
2
2ab+()w
1
3
ba−()
⎡
⎢
⎣
⎤
⎥
⎦
A
y
b− 0=
A
y
w
6b
2ab+()ba−()=
This problem requires
V
C
0.
=
Summing forces vertically for the section, we have
+
↑
Σ
F

y
= 0;
w
6b
2ab+()ba−()
1
2
a
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
w
2
− 0=
w
6b
2ab+()ba−()
w
8
2ab+()=
4 ba−()3b=
b 4a=
a
b

1
4
=
Problem 7-36
The semicircular arch is subjected to a uniform distributed load along its axis of w
0
per unit length.
Determine the internal normal force, shear force, and moment in the arch at angle
θ
.
Given:
θ
45 deg=
Solution:
Resultants of distributed load:
657
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
F
Rx
0
θ
θ
w
0
r
⌠
⎮

⌡
d sin
θ
()
= rw
0
1
cos
θ
()
−
()
=
F
Rx
rw
0
1
cos
θ
()
−
()
=
F
Ry
0
θ
θ
w

0
r
⌠
⎮
⌡
d cos
θ
()
= rw
0
sin
θ
()
=
F
Rx
rw
0
sin
θ
()()
=
M
Ro
0
θ
θ
w
0
r

⌠
⎮
⌡
d r= r
2
w
0
θ
=
Σ
F
x
= 0;
V− F
Rx
cos
θ
()
+ F
Ry
sin
θ
()
− 0=
Vrw
0
1
cos
θ
()

−
()
⎡
⎣
⎤
⎦
cos
θ
()
rw
0
sin
θ
()()
⎡
⎣
⎤
⎦
sin
θ
()
−=
Vw
0
r cos
θ
()
1−
()
=

a cos
θ
()
1−=
a 0.293−= Varw
0
=
Σ
F
y
= 0;
NF
Ry
cos
θ
()
+ F
Rx
sin
θ
()
+ 0=
Nrw
0
1
cos
θ
()
−
()

⎡
⎣
⎤
⎦
− sin
θ
()
rw
0
sin
θ
()()
⎡
⎣
⎤
⎦
cos
θ
()
−=
Nw
0
− rsin
θ
()
=
b sin
θ
()
−=

b 0.707−= Nw
0
rb=
Σ
M
o
= 0;
M− r
2
w
0
θ
()
+ brw
0
r+ 0=
Mw
0
r
2
θ
b+
()
=
c
θ
b+=
c 0.0783= Mcr
2
w

0
=
658
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
The semicircular arch is subjected to a uniform distributed load along its axis of w
0
per unit length.
Determine the internal normal force, shear force, and moment in the arch at angle
θ
.
Given:
θ
120 deg=
Solution:
Resultants of distributed load:
F
Rx
0
θ
θ
w
0
r
⌠
⎮
⌡
d sin

θ
()
= rw
0
1
cos
θ
()
−
()
=
F
Rx
rw
0
1
cos
θ
()
−
()
=
F
Ry
0
θ
θ
w
0
r

⌠
⎮
⌡
d cos
θ
()
= rw
0
sin
θ
()
=
F
Rx
rw
0
sin
θ
()()
=
M
Ro
0
θ
θ
w
0
r
⌠
⎮

⌡
d r= r
2
w
0
θ
=
Σ
F
x
= 0;
V− F
Rx
cos
θ
()
+ F
Ry
sin
θ
()
− 0=
Vrw
0
1
cos
θ
()
−
()

⎡
⎣
⎤
⎦
cos
θ
()
rw
0
sin
θ
()()
⎡
⎣
⎤
⎦
sin
θ
()
−=
Vw
0
r cos
θ
()
1−
()
=
a cos
θ

()
1−=
a 1.500−= Varw
0
=
Σ
F
y
= 0;
NF
Ry
cos
θ
()
+ F
Rx
sin
θ
()
+ 0=
Nrw
0
1
cos
θ
()
−
()
⎡
⎣

⎤
⎦
− sin
θ
()
rw
0
sin
θ
()()
⎡
⎣
⎤
⎦
cos
θ
()
−=
Nw
0
− rsin
θ
()
=
b sin
θ
()
−=
659
Problem 7-37

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
b 0.866−= Nw
0
rb⋅⋅=
Σ
M
o
= 0;
M− r
2
w
0
θ
()
+ brw
0
r+ 0=
Mw
0
r
2
θ
b+
()
=
c
θ

b+=
c 1.2284= Mcr
2
w
0
=
Problem 7-38
Determine the x, y, z components of internal loading at a section passing through point C in the pipe
assembly. Neglect the weight of the pipe.
Units Used:
kip 10
3
lb=
Given:
F
1
0
350
400
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
F

2
150
0
300
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
a 1.5 ft= b 2 ft= c 3 ft=
Solution:
r
1
c
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r

2
0
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
F
C
F
1
− F
2
−= F
C
150.00−
350.00−
700.00
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
lb=
660
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
M
C
r
1
− F
1
× r
2
F
2
×−= M
C
1400.00
1200.00
−
750.00−
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
lb ft⋅=
Problem 7-39
Determine the x, y, z components of internal loading at a section passing through point C in the pipe
assembly. Neglect the weight of the pipe.
Units Used:
kip 10
3
lb=
Given:
F
1
80−
200
300
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
F
2
250
150

−
200−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
a 1.5 ft= b 2 ft= c 3 ft=
Solution:
r
1
c
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
2
0
b

0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
F
C
F
1
− F
2
−= F
C
170.00−
50.00−
500.00
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=

M
C
r
1
− F
1
× r
2
F
2
×−= M
C
1000.00
900.00
−
260.00−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb ft⋅=
Problem 7-40
Determine the x, y, z components of internal loading in the rod at point D.
661
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may

be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Units Used:
kN 10
3
N=
Given:
M 3 kN m⋅=
F
7
12
−
5−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
a 0.75 m=
b 0.2 m=
c 0.2 m=
d 0.6 m=
e 1 m=
Solution:
Guesses
C

x
1
N= C
y
1
N= B
x
1
N=
B
z
1
N= A
y
1
N= A
z
1
N=
Given
0
A
y
A
z
⎛
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎠
B
x
0
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
C
x
C
y
0
⎛
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎠
+ F+ 0=
e−
bc+ d+
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
×
0
bc+
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
B
x
0
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

×+
0
0
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
C
x
C
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+
0
bc+ d+

0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F×+
0
0
M−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
662
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
A
y
A

z
B
x
B
z
C
x
C
y
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A

y
A
z
, B
x
, B
z
, C
x
, C
y
,
()
=
A
y
A
z
B
x
B
z
C
x
C
y
⎛
⎜
⎜
⎜

⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
53.60−
87.00
109.00
82.00
−
116.00−
65.60
⎛
⎜
⎜
⎜
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
kN=
Guesses
V
Dx
1
N= N
Dy
1
N=
V
Dz
1
N= M
Dx
1
Nm⋅=
M
Dy
1
Nm⋅= M
Dz

1
Nm⋅=
Given
C
x
C
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
V
Dx
N
Dy
V
Dz
⎛
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎠
+ 0=
0
b−
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
C
x
C
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎠
×
M
Dx
M
Dy
M
Dz
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
0
0
M−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

+ 0=
V
Dx
N
Dy
V
Dz
M
Dx
M
Dy
M
Dz
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎠
Find V
Dx
N
Dy
, V
Dz
, M
Dx
, M
Dy
, M
Dz
,
()
=
V
Dx
N
Dy
V
Dz
⎛
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎠
116.00
65.60
−
0.00
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
M
Dx
M
Dy
M
Dz
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
49.20
87.00
26.20
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN m⋅=
663
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
Determine the x, y, z components of internal loading in the rod at point E.
Units Used:
kN 10
3
N=
Given:
M 3 kN m⋅=
F
7
12

−
5−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
a 0.75 m=
b 0.4 m=
c 0.6 m=
d 0.5 m=
e 0.5 m=
Solution:
Guesses
C
x
1
N= C
y
1
N= B
x
1
N=
B
z

1
N= A
y
1
N= A
z
1
N=
Given
0
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
B
x
0
B
z
⎛

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
C
x
C
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ F+ 0=
d− e−
bc+
0
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×
0
b
0
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
B
x
0
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+
0
0
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
C
x
C
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+
0
bc+
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F×+
0

0
M−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
664
Problem 7-41
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 7
A
y
A
z
B
x
B
z
C
x
C
y
⎛

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
y
A
z
, B
x
, B
z
, C
x
, C

y
,
()
=
A
y
A
z
B
x
B
z
C
x
C
y
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎠
53.60−
87.00
109.00
82.00
−
116.00−
65.60
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
kN=

Guesses
N
Ex
1
N= V
Ey
1
N=
V
Ez
1
N= M
Ex
1
Nm⋅=
M
Ey
1
Nm⋅= M
Ez
1
Nm⋅=
Given
0
A
y
A
z
⎛
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
N
Ex
V
Ey
V
Ez
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
e−
0
0
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
0
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×
M
Ex
M
Ey
M
Ez
⎛
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
N
Ex
V
Ey
V
Ez
M
Ex
M
Ey
M
Ez
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find N
Ex
V
Ey
, V
Ez
, M
Ex
, M
Ey
, M
Ez
,
()
=
N
Ex
V
Ey

V
Ez
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0.00
53.60
87.00
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
M
Ex
M
Ey
M

Ez
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0.00
43.50
−
26.80−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN m⋅=
Problem 7-42
Draw the shear and moment dia
g
rams for the shaft in terms of the
p
arameters shown; There is

665
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.