Engineering Mechanics - Statics Episode 2 Part 8 pdf

Engineering Mechanics - Statics Chapter 7
Support Reactions: From FBD (b),
wL
2
L
4
⎛
⎜
⎝
⎞
⎟
⎠
B
y
L
2
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
B
y
wL
4
=
From FBD (a),
A
y

wL
2
−
B
y
− 0=
A
y
3
wL
4
=
B
y
−
L
2
w
L
2
⎛
⎜
⎝
⎞
⎟
⎠
L
4
⎛
⎜

⎝
⎞
⎟
⎠
− M
A
+ 0=
M
A
w
L
2
4
⎛
⎜
⎝
⎞
⎟
⎠
=
Shear and Moment Functions:
From FBD (c) For
0 x≤ L≤ A
y
wx− V− 0=
V
w
4
3L 4x−()=
M

A
A
y
x− wx
x
2
⎛
⎜
⎝
⎞
⎟
⎠
+ M+ 0= M
w
4
3Lx 2x
2
− L
2
−
()
=
681
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
The beam has depth a and is subjected to a uniform distributed loading w which acts at an
angle

θ
from the vertical as shown. Determine the internal normal force, shear force, and
moment in the beam as a function of x. Hint:The moment loading is to be determined from a
point along the centerline of the beam (x axis).
Given:
a 2 ft=
L 10 ft=
θ
30 deg=
w 50
lb
ft
=
Solution:
0 x≤ L≤
Σ
F
x
= 0;
Nwsin
θ
()
x+ 0=
Nx() w− sin
θ
()
x=
Σ
F
y

= 0;
V− wcos
θ
()
x− 0=
Vw− cos
θ
()
x=
Σ
M = 0;
wcos
θ
()
x
x
2
⎛
⎜
⎝
⎞
⎟
⎠
wsin
θ
()
x
a
2
⎛

⎜
⎝
⎞
⎟
⎠
− M+ 0=
Mx() w− cos
θ
()
x
2
2
⎛
⎜
⎝
⎞
⎟
⎠
wsin
θ
()
xa
2
⎛
⎜
⎝
⎞
⎟
⎠
+=

Problem 7-56
Draw the shear and moment
diagrams for the beam.
Given:
w 250
lb
ft
= L 12 ft=
682
Problem 7-55
© 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−
1
2
x
wx
L
− 0= Vx()
wx
2
2L
−
1

lb
=
Σ
M = 0;
M
x
2
wx
L
⎛
⎜
⎝
⎞
⎟
⎠
1
3
x
⎛
⎜
⎝
⎞
⎟
⎠
+ 0= Mx()
w− x
3
6L
1
lb ft⋅

=
024681012
2000
1000
0
Distance (ft)
Force (lb)
Vx()
x
ft
024681012
1
.
10
4
5000
0
Distance (ft)
Moment (lb-ft)
Mx()
x
ft
Problem 7-57
The beam will fail when the maximum shear force is V
max
or the maximum moment is M
max
.
Determine the largest intensity w of the distributed loading it will support.
683

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
Given:
L 18 ft=
V
max
800
lb=
M
max
1200
lb ft⋅=
Solution:
For
0 x≤ L≤
V
w− x
2
2L
= M
w− x
3
6L
=
V
max
wL

2
=
w
1
2
V
max
L
⎛
⎜
⎝
⎞
⎟
⎠
= w
1
88.9
lb
ft
=
M
max
w
2
L
2
6
=
w
2

6
M
max
L
2
⎛
⎜
⎝
⎞
⎟
⎠
= w
2
22.2
lb
ft
=
Now choose the critical case
w min w
1
w
2
,
()
= w 22.22
lb
ft
=
Problem 7-58
The beam will fail when the maximum internal moment is M

max
. Determine the position x of the
concentrated force
P
and its smallest magnitude that will cause failure.
684
© 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:
For
ξ

< x,
M
1
P
ξ
Lx−()
L
=
For
ξ
> x,
M
2
Px L
ξ
−

()
L
=
Note that M
1
= M
2
when x =
ξ
M
max
M
1
= M
2
=
Px L x−()
L
=
P
L
Lx x
2
−
()
=
x
Lx x
2
−

()
d
d
L
2x−= x
L
2
=
Thus,
M
max
P
L
L
2
⎛
⎜
⎝
⎞
⎟
⎠
L
L
2
−
⎛
⎜
⎝
⎞
⎟

⎠
=
P
2
L
2
⎛
⎜
⎝
⎞
⎟
⎠
=
P
4M
max
L
=
Problem 7-59
Draw the shear and moment diagrams
for the beam.
Given:
w 30
lb
ft
= M
C
180
lb ft⋅=
a 9 ft= b 4.5 ft=

Solution:
A
y
− a
1
2
wa
a
3
⎛
⎜
⎝
⎞
⎟
⎠
+ M
C
− 0=
685
© 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
wa
6
M
C
a

−=
A
y
B
y
+
1
2
wa− 0=
B
y
wa
2
A
y
−=
x
1
0 0.01
a, a =
A
y
1
2
w
x
a
⎛
⎜
⎝

⎞
⎟
⎠
x− V
1
x()− 0= V
1
x() A
y
wx
2
2a
−
⎛
⎜
⎝
⎞
⎟
⎠
1
lb
=
A
y
− x
1
2
w
x
a

⎛
⎜
⎝
⎞
⎟
⎠
x
x
3
⎛
⎜
⎝
⎞
⎟
⎠
+ M
1
x()+ 0= M
1
x() A
y
x
wx
3
6a
−
⎛
⎜
⎝
⎞

⎟
⎠
1
lb ft⋅
=
x
2
a 1.01a, ab+ =
A
y
1
2
wa− B
y
+ V
2
x()− 0= V
2
x() A
y
B
y
+
wa
2
−
⎛
⎜
⎝
⎞

⎟
⎠
1
lb
=
A
y
− x
1
2
wa x
2a
3
−
⎛
⎜
⎝
⎞
⎟
⎠
+ B
y
xa−()− M
2
x()+ 0=
M
2
x() A
y
xB

y
xa−()+
wa
2
x
2a
3
−
⎛
⎜
⎝
⎞
⎟
⎠
⋅−
⎡
⎢
⎣
⎤
⎥
⎦
1
lb ft⋅
=
024681012
200
100
0
100
Distance (ft)

Force (lb)
V
1
x
1
()
V
2
x
2
()
x
1
ft
x
2
ft
,
686
© 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
024681012
200
100
0
100
Diastance (ft)
Moment (lb-ft)

M
1
x
1
()
M
2
x
2
()
x
1
ft
x
2
ft
,
Problem 7-60
The cantilevered beam is made of material having a specific weight
γ
.
Determine the shear and
moment in the beam as a function of x.
Solution:
By similar triangles
y
x
h
d
= y

h
d
x=
W
γ
V=
γ
1
2
yxt
⎛
⎜
⎝
⎞
⎟
⎠
=
γ
1
2
h
d
x
⎛
⎜
⎝
⎞
⎟
⎠
xt

⎡
⎢
⎣
⎤
⎥
⎦
=
γ
ht
2d
⎛
⎜
⎝
⎞
⎟
⎠
x
2
=
Σ
F
y
= 0;
V
γ
ht
2d
⎛
⎜
⎝

⎞
⎟
⎠
x
2
− 0=
V
γ
ht
2d
⎛
⎜
⎝
⎞
⎟
⎠
x
2
=
Σ
M = 0;
M−
γ
ht
2d
⎛
⎜
⎝
⎞
⎟

⎠
x
2
x
3
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
687
© 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
γ
ht
6d
− x
3
=
Problem 7-61
Draw the shear and moment
diagrams for the beam.
Given:
kip 10
3

lb=
w
1
30
lb
ft
= w
2
120
lb
ft
=
L 12 ft=
Solution:
w
1
L
L
2
⎛
⎜
⎝
⎞
⎟
⎠
1
2
w
2
w

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

1
−
2
⎛
⎜
⎝
⎞
⎟
⎠
L
3
⎛
⎜
⎝
⎞
⎟
⎠
+=
A
y
w
1
x−
1
2
w
2
w
1
−

()
x
L
⎛
⎜
⎝
⎞
⎟
⎠
x− Vx()− 0=
Vx() A
y
w
1
x−
1
2
w
2
w
1
−
()
x
2
L
⎛
⎜
⎝
⎞

⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦
1
lb
=
A
y
− xw
1
x
x
2
⎛
⎜
⎝
⎞
⎟
⎠
+
1
2
w
2

w
1
−
()
x
L
⎛
⎜
⎝
⎞
⎟
⎠
x
x
3
⎛
⎜
⎝
⎞
⎟
⎠
+ Mx()+ 0=
Mx() A
y
xw
1
x
2
2
⎛

⎜
⎝
⎞
⎟
⎠
− w
2
w
1
−
()
x
3
6L
⎛
⎜
⎝
⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kip ft⋅
=

688
© 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
024681012
500
0
500
Distance (ft)
Force (lb)
Vx()
x
ft
024681012
0
1
2
Distance (ft)
Moment (kip-ft)
Mx()
x
ft
Problem 7-62
Draw the shear and moment
diagrams for the beam.
Units Used:
kip 10
3
lb=

Given:
F
1
20
kip= F
2
20
kip= w 4
kip
ft
= a 15 ft= b 30 ft= c 15 ft=
Solution:
F
1
ab+()wb
b
2
⎛
⎜
⎝
⎞
⎟
⎠
+ F
2
c− A
y
b− 0= A
y
F

1
ab+
b
⎛
⎜
⎝
⎞
⎟
⎠
wb
2
+ F
2
c
b
⎛
⎜
⎝
⎞
⎟
⎠
−=
A
y
B
y
+ F
1
− F
2

− wb− 0= B
y
F
1
F
2
+ wb+ A
y
−=
689
© 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
x
1
0 0.01
a, a =
F
1
− V
1
x()− 0= V
1
x() F
1
−
1
kip
=

F
1
xM
1
x()+ 0= M
1
x() F
1
− x
1
kip ft⋅
=
x
2
a 1.01a, ab+ =
F
1
− wx a−()− A
y
+ V
2
x()− 0= V
2
x() F
1
− wx a−()− A
y
+
⎡
⎣

⎤
⎦
1
kip
=
F
1
xA
y
xa−()− wx a−()
xa−
2
⎛
⎜
⎝
⎞
⎟
⎠
+ M
2
x()+ 0=
M
2
x() F
1
− xA
y
xa−()+ w
xa−()
2

2
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kip ft⋅
=
x
3
ab+ 1.01 ab+(), ab+ c+ =
V
3
x() F
2
− 0= V
3
x() F
2
1
kip
=
M
3
x()− F
2
ab+ c+ x−()− 0= M

3
x() F
2
− ab+ c+ x−()
1
kip ft⋅
=
0 102030405060
100
50
0
50
100
Distance (ft)
Force (kip)
V
1
x
1
()
V
2
x
2
()
V
3
x
3
()

x
1
ft
x
2
ft
,
x
3
ft
,
690
© 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
0 102030405060
400
200
0
200
Distance (ft)
Moment (kip-ft)
M
1
x
1
()
M
2

x
2
()
M
3
x
3
()
x
1
ft
x
2
ft
,
x
3
ft
,
Problem 7-63
Express the x, y, z components of internal loading in the rod at the specific value for y, where
0 y< a<

.
Units Used:
kip 10
3
lb=
Given:
y 2.5 ft=

w 800
lb
ft
= F 1500 lb=
a 4 ft= b 2 ft=
Solution: In general we have
V
F
0
wa y−()
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
=
691
© 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
wa y−()
ay−
2
⎛
⎜

⎝
⎞
⎟
⎠
F− b
F− ay−()
⎡
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎦
=
For these values we have
V
1500.00
0.00
1200.00
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
lb= M
900.00
3000.00
−
2250.00−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb ft⋅=
Problem 7-64
Determine the normal
force, shear force, and
moment in the curved rod
as a function of
θ
.
Given:
c 3=
d 4=
Solution:
For
0
θ

≤
π
≤
Σ
F
x
= 0;
N
d
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
P cos
θ
()
−
c
c
2
d
2
+

⎛
⎜
⎝
⎞
⎟
⎠
P sin
θ
()
− 0=
N
P
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
dcos
θ
()
csin
θ
()
+

()
=
Σ
F
y
= 0;
V
d
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
P sin
θ
()
−
c
c
2
d
2
+
⎛

⎜
⎝
⎞
⎟
⎠
P cos
θ
()
+ 0=
V
P
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
dsin
θ
()
ccos
θ
()
−
()

=
692
© 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 = 0;
d−
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
Pr rcos
θ
()
−
()
c
c
2
d
2

+
⎛
⎜
⎝
⎞
⎟
⎠
Prsin
θ
()
+ M+ 0=
M
Pr
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
ddcos
θ
()
− csin
θ
()

−
()
=
Problem 7-65
The quarter circular rod lies in the horizontal
plane and supports a vertical force
P
at its
end. Determine the magnitudes of the
components of the internal shear force,
moment, and torque acting in the rod as a
function of the angle
θ
.
Solution:
Σ
F
z
= 0;
VP=
Σ
M
x
= 0;
MPrcos
θ
()
+ 0=
MP− rcos
θ

()
=
MPrcos
θ
()
=
Σ
M
y
= 0;
TPrl sin
θ
()
−
()
+ 0=
TP− r l sin
θ
()
−
(
=
TPr1 sin
θ
()
−
(
=
693
© 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
Draw the shear and moment
diagrams for the beam.
Given:
M
B
800
lb ft⋅= a 5 ft=
F 100 lb= b 5 ft=
Solution:
x
1
0 0.01
a, a = V
1
x() F
1
lb
= M
1
x() F− ab+ x−()M
B
−
⎡
⎣
⎤
⎦
1

lb ft⋅
=
x
2
a 1.01a, ab+ = V
2
x() F
1
lb
= M
2
x() F− ab+ x−()
1
lb ft⋅
=
0246810
99.8
99.9
100
Distance (ft)
Force (lb)
V
1
x
1
()
V
2
x
2

()
x
1
ft
x
2
ft
,
0246810
2000
1000
0
Distane (ft)
Moment (lb-ft)
M
1
x
1
()
M
2
x
2
()
x
1
ft
x
2
ft

,
694
Problem 7-66
© 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
Draw the shear and moment diagrams
for the beam.
Units Used:
kN 10
3
N=
Given:
w 3
kN
m
= F 10 kN=
L 6 m=
Solution:
Vx() wL x−()F+[]
1
kN
=
Mx() w−
Lx−()
2
2
FL x−()−
⎡

⎢
⎣
⎤
⎥
⎦
1
kN m⋅
=
0123456
0
20
40
Distance (m)
Force (kN)
Vx()
x
0123456
150
100
50
0
Distance (m)
Moment (kN-m)
Mx()
x
695
Problem 7-67
© 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 7-68
Draw the shear and moment
diagrams for the beam.
Units Used:
kN 10
3
N=
Given:
F 7 kN= M 12 kN m⋅= a 2 m= b 2 m= c 4 m=
Solution:
Guesses A 1 N= B 1 N=
Given AB+ F− 0= F− aM− Ba b+ c+()+ 0=
A
B
⎛
⎜
⎝
⎞
⎟
⎠
Find AB,()=
x
1
0 0.01
a, a = x
2
a 1.01a, ab+ = x
3
ab+ 1.01 ab+(), ab+ c+ =

V
1
x
1
()
A
1
kN
= V
2
x
2
()
AF−()
1
kN
= V
3
x
3
()
B−
1
kN
=
M
1
x
1
()

Ax
1
kN m⋅
= M
2
x
2
()
Ax
2
Fx
2
a−
()
−
kN m⋅
= M
3
x
3
()
Ba b+ c+ x
3
−
()
1
kN m⋅
=
012345678
5

0
5
Distance (m)
Force (kN)
V
1
x
1
()
V
2
x
2
()
V
3
x
3
()
x
1
x
2
, x
3
,
696
© 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
012345678
5
0
5
10
15
Distance (m)
Moment (kN-m)
M
1
x
1
()
M
2
x
2
()
M
3
x
3
()
x
1
x
2
, x
3

,
Problem 7-69
Draw the shear and moment
diagrams for the beam.
Units Used:
kN 10
3
N=
Given:
a 2 m= b 4 m= w 1.5
kN
m
=
Solution:
wb a−()
ba−
2
⎛
⎜
⎝
⎞
⎟
⎠
A
y
b− 0= A
y
wb a−()
2
2b

= A
y
0.75
kN=
x
1
0 0.01
a, a = V
1
x() A
y
1
kN
= M
1
x() A
y
x
1
kN m⋅
=
x
2
ba− 1.01 ba−(), b = V
2
x() A
y
wx a−()−
⎡
⎣

⎤
⎦
1
kN
=
M
2
x() A
y
xw
xa−()
2
2
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kN m⋅
=
697
© 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
0 0.5 1 1.5 2 2.5 3 3.5 4
4

2
0
2
Distance (m)
Force (kN)
V
1
x
1
()
V
2
x
2
()
x
1
x
2
,
0 0.5 1 1.5 2 2.5 3 3.5 4
1
0
1
2
Distance (m)
Moment (kN-m)
M
1
x

1
()
M
2
x
2
()
x
1
x
2
,
Problem 7-70
Draw the shear and moment diagrams
for the beam.
Given:
w 30
lb
ft
= M
C
180
lb ft⋅=
a 9 ft= b 4.5 ft=
Solution:
A
y
− a
1
2

wa
a
3
⎛
⎜
⎝
⎞
⎟
⎠
+ M
C
− 0=
A
y
wa
6
M
C
a
−=
698
© 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
B
y
+

1
2
wa− 0=
B
y
wa
2
A
y
−=
x
1
0 0.01
a, a =
V
1
x() A
y
wx
2
2a
−
⎛
⎜
⎝
⎞
⎟
⎠
1
lb

= M
1
x() A
y
x
wx
3
6a
−
⎛
⎜
⎝
⎞
⎟
⎠
1
lb ft⋅
=
x
2
a 1.01a, ab+ =
V
2
x() A
y
B
y
+
wa
2

−
⎛
⎜
⎝
⎞
⎟
⎠
1
lb
=
M
2
x() A
y
xB
y
xa−()+
wa
2
x
2a
3
−
⎛
⎜
⎝
⎞
⎟
⎠
−

⎡
⎢
⎣
⎤
⎥
⎦
1
lb ft⋅
=
024681012
200
100
0
100
Distance (ft)
Force (lb)
V
1
x
1
()
V
2
x
2
()
x
1
ft
x

2
ft
,
699
© 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
024681012
200
100
0
100
Diastance (ft)
Moment (lb-ft)
M
1
x
1
()
M
2
x
2
()
x
1
ft
x
2

ft
,
Problem 7-71
Draw the shear and moment
diagrams for the beam.
Given:
kip 10
3
lb=
w
1
30
lb
ft
= w
2
120
lb
ft
=
L 12 ft=
Solution:
w
1
L
L
2
⎛
⎜
⎝

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

⎜
⎝
⎞
⎟
⎠
w
2
w
1
−
2
⎛
⎜
⎝
⎞
⎟
⎠
L
3
⎛
⎜
⎝
⎞
⎟
⎠
+=
Vx() A
y
w
1

x−
1
2
w
2
w
1
−
()
x
2
L
−
⎡
⎢
⎣
⎤
⎥
⎦
1
lb
=
Mx() A
y
xw
1
x
2
2
− w

2
w
1
−
()
x
3
6L
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kip ft⋅
=
700
© 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
024681012
500
0
500
Distance (ft)
Force (lb)
Vx()

x
ft
024681012
0
1
2
Distance (ft)
Moment (kip-ft)
Mx()
x
ft
Problem 7-72
Draw the shear and moment diagrams for
the shaft. The support at A is a journal
bearing and at B it is a thrust bearing.
Given:
F
1
400
lb= F
2
800
lb=
w 100
lb
in
= a 4 in= b 12 in= c 4 in=
Solution:
F
1

awb
b
2
⎛
⎜
⎝
⎞
⎟
⎠
− F
2
b− Bb c+()+ 0= B
w
b
2
2
⎛
⎜
⎝
⎞
⎟
⎠
F
2
b+ F
1
a−
bc+
= B 950.00 lb=
701

© 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
x
1
0 0.01
a, a = V
1
x() F
1
−
1
lb
= M
1
x() F
1
− x
1
lb in⋅
=
x
2
a 1.01a, ab+ = V
2
x
2
()
B− F

2
+ wa b+ x
2
−
()
+
⎡
⎣
⎤
⎦
1
lb
=
M
2
x
2
()
Ba b+ c+ x
2
−
()
F
2
ab+ x
2
−
()
− w
ab+ x

2
−
()
2
2
⋅−
⎡
⎢
⎣
⎤
⎥
⎦
1
lb in⋅
=
x
3
ab+ 1.01 ab+(), ab+ c+ = V
3
x
3
()
B−
lb
= M
3
x
3
()
Ba b+ c+ x

3
−
()
1
lb in⋅
=
0 5 10 15 20
1000
0
1000
2000
Distance (in)
Force (lb)
V
1
x
1
()
V
2
x
2
()
V
3
x
3
()
x
1

in
x
2
in
,
x
3
in
,
0 5 10 15 20
2000
0
2000
4000
Distance (ini)
Moment (lb-in)
M
1
x
1
()
M
2
x
2
()
M
3
x
3

()
x
1
in
x
2
in
,
x
3
in
,
702
© 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 7-73
Draw the shear and moment
diagrams for the beam.
Units Used:
kN 10
3
N=
Given:
F
1
10
kN= F
2

10
kN= M
0
12
kN m⋅=
a 2 m= b 2 m= c 2 m= d 2 m=
Solution:
F
1
bc+ d+()M
0
− F
2
d+ A
y
ab+ c+ d+()− 0=
A
y
F
1
bc+ d+()M
0
− F
2
d+
ab+ c+ d+
= A
y
8.50
kN=

A
y
B
y
+ F
1
− F
2
− 0= B
y
F
1
F
2
+ A
y
−= B
y
11.50
kN=
x
1
0 0.01
a, a =
V
1
x() A
y
1
kN

= M
1
x() A
y
x
1
kN m⋅
=
x
2
a 1.01a, ab+ =
V
2
x() A
y
F
1
−
()
1
kN
= M
2
x() A
y
xF
1
xa−()−
⎡
⎣

⎤
⎦
1
kN m⋅
=
x
3
ab+ 1.01 ab+(), ab+ c+ =
V
3
x() A
y
F
1
−
()
1
kN
= M
3
x() A
y
xF
1
xa−()− M
0
+
⎡
⎣
⎤

⎦
1
kN m⋅
=
x
4
ab+ c+ 1.01 ab+ c+(), ab+ c+ d+ =
V
4
x() B
y
−
1
kN
= M
4
x() B
y
ab+ c+ d+ x−()
1
kN m⋅
=
703
© 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
012345678
15
10

5
0
5
10
Distance (m)
Force (kN)
V
1
x
1
()
V
2
x
2
()
V
3
x
3
()
V
4
x
4
()
x
1
x
2

, x
3
, x
4
,
012345678
0
10
20
30
Distance (m)
Moment (kN-m)
M
1
x
1
()
M
2
x
2
()
M
3
x
3
()
M
4
x

4
()
x
1
x
2
, x
3
, x
4
,
Problem 7-74
Draw the shear and moment
diagrams for the shaft. The support
at A is a journal bearing and at B it
is a thrust bearing.
704
© 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 200 lb= w 100
lb
ft
= M 300 lb ft⋅= a 1 ft= b 4 ft= c 1 ft=
Solution:
Fa b+()Ab− wb
b
2
⎛

⎜
⎝
⎞
⎟
⎠
+ M− 0= A
Fa b+()
wb
2
2
⎛
⎜
⎝
⎞
⎟
⎠
+ M−
b
= A 375.00lb=
x
1
0 0.01
a, a = V
1
x
1
()
F−
1
lb

= M
1
x
1
()
F− x
1
1
lb ft⋅
=
x
2
a 1.01a, ab+ = V
2
x
2
()
F− A+ wx
2
a−
()
−
⎡
⎣
⎤
⎦
1
lb
=
M

2
x
2
()
F− x
2
Ax
2
a−
()
+ w
x
2
a−
()
2
2
−
⎡
⎢
⎣
⎤
⎥
⎦
1
lb ft⋅
=
x
3
ab+ 1.01 ab+(), ab+ c+ = V

3
x
3
()
0
1
lb
= M
3
x
3
()
M−
1
lb ft⋅
=
0123456
400
200
0
200
Distance (ft)
Force (lb)
V
1
x
1
()
V
2

x
2
()
V
3
x
3
()
x
1
ft
x
2
ft
,
x
3
ft
,
705
Given:
© 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.