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Solutions (8th ed structural analysis) chapter 9

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9–1. Determine the vertical displacement of joint A. Each
bar is made of steel and has a cross-sectional area of
600 mm2. Take E = 200 GPa. Use the method of virtual
work.

C

B

2m

A
D
1.5 m

The virtual forces and real forces in each member are shown in Fig. a and b,
respectively.


#

Member

n(kN)

N(kN)

L(m)

nNL (kN2 m)

AB
AD
BD
BC

1.25
-0.75
-1.25
1.50

6.25
-3.75
-6.25
7.50

2.50
3
2.50
1.50

19.531
8.437
19.531
16.875

©

64.375



64.375 kN2 # m
nNL
1 kN # ¢ Av = a
=
AE
AE
¢ Av =

64.375 kN # m
AE
64.375(103) N # m

=

c0.6(10 - 3) m2 d c200(109) N>m2 d

= 0.53646 (10 - 3) m
= 0.536 mm T

Ans.

308

5 kN

1.5 m


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–2.

Solve Prob. 9–1 using Castigliano’s theorem.

C

B

2m

A

Member
AB
AD
BD
BC

0N
0P

N(kN)
1.25 P
-0.750 P
-1.25 P
1.50 P

Na

L(m)

6.25
-3.75
-6.25
7.50

2.5
3
2.5
1.5

19.531
8.437
19.531
16.875

©

64.375

1.25
-0.75
-1.25
1.50

D

0N
bL(kN # m)
0P

N1P = 5kN2

1.5 m

1.5 m

5 kN

0N L
b
¢ Av = a Na
0P AE
=

64.375 kN # m
AE
64 # 375(103) N # m

=

30.6(10 - 3) m243200(109) N>m24

= 0.53646 (10 - 3) m
= 0.536 mm T

Ans.

*9–3. Determine the vertical displacement of joint B. For
each member A ϭ 400 mm2, E ϭ 200 GPa. Use the method
of virtual work.

F

E

D

1.5 m

Member

n

N

L

nNL

AF
AE
AB
EF
EB
ED
BC
BD
CD

0
-0.8333
0.6667
0
0.50
-0.6667
0
0.8333
-0.5

0
-37.5
30.0
0
22.5
-30.0
0
37.5
-22.5

1.5
2.5
2.0
2.0
1.5
2.0
2.0
2.5
1.5

0
78.125
40.00
0
16.875
40.00
0
78.125
16.875

A

2m

© = 270
nNL
1 # ¢ Bv = a
AE
270(103)
¢ Bv =

400(10 - 6)(200)(109)

= 3.375(10 - 3) m = 3.38 mm T

Ans.

309

C

B
45 kN
2m


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*9–4.

Solve Prob. 9–3 using Castigliano’s theorem.

F

E

D

1.5 m

A

C

B
45 kN
2m

2m

Member
AF
AE
AB
BE
BD
BC
CD
DE
EF

0N
0P

N
0
-0.8333P
0.6667P
0.5P
0.8333P
0
-0.5P
0.6667P
0

N(P = 45)

0
-0.8333
0.6667
0.5
0.8333
0
-0.5
-0.6667
0

L

0
-37.5
30.0
22.5
37.5
0
-22.5
-30.0
0

1.5
2.5
2.0
1.5
2.5
2.0
1.5
2.0
2.0

Na

0N
bL
0P

0
78.125
40.00
16.875
78.125
0
16.875
40.00
0
© = 270

0N L
270
dBv = a N a
b
=
0P AE
AE
270(103)
=

400(10 - 6)(200)(109)

= 3.375(10 - 3) m = 3.38 mm

Ans.

9–5. Determine the vertical displacement of joint E. For
each member A ϭ 400 mm2, E ϭ 200 GPa. Use the method
of virtual work.

F

E

D

1.5 m

A

C

B
45 kN

Member

n

N

L

nNL

AF
AE
AB
EF
EB
ED
BC
BD
CD

0
-0.8333
0.6667
0
0.50
-0.6667
0
0.8333
-0.5

0
-37.5
30.0
0
22.5
30.0
0
37.5
-22.5

1.5
2.5
2.0
2.0
1.5
2.0
2.0
2.5
1.5

0
78.125
40.00
0
-16.875
40.00
0
78.125
16.875

2m

© = 236.25
nNL
1 # ¢Ev = a
AE
236.25(103)
¢Ev =

400(10 - 6)(200)(109)

= 2.95(10 - 3) m = 2.95 mm T

Ans.

310

2m


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–6.

Solve Prob. 9–5 using Castigliano’s theorem.

F

E

D

1.5 m

A

C

B
45 kN
2m

Member
AF
AE
AB
BE
BD
BC
CD
DE
EF

0N
0P

N
0
-(0.8333P + 37.5)
0.6667P + 30
22.5 - 0.5P
0.8333P + 37.5
0
-(0.5P + 22.5)
-(0.6667P + 30)
0

N(P = 45)

0
-0.8333
0.6667
-0.5
0.8333
0
-0.5
-0.6667
0

0
-37.5
30.0
22.5
37.5
0
-22.5
-30.0
0

L

Na

1.5
2.5
2.0
1.5
2.5
2.0
1.5
2.0
2.0

0
78.125
40.00
-16.875
78.125
0
16.875
40.00
0

2m

0N
bL
0P

© = 236.25
0N
¢ Ev = a N
0P

236.25
L
=
AE
AE

236.25(103)
=

400(10 - 6)(200)(109)

= 2.95(10 - 3) m = 2.95 mm T

Ans.

9–7. Determine the vertical displacement of joint D. Use
the method of virtual work. AE is constant. Assume the
members are pin connected at their ends.

D

E

3m

A
B
4m

The virtual and real forces in each member are shown in Fig. a and b,
respectively

#

n(kN)

N(kN)

L(m)

nNL(kN2 m)

AB
BC
AD
BD
CD
CE
DE

0.6667
0.6667
-0.8333
0
-0.8333
0.500
0

10.0
10.0
-12.5
15.0
-12.5
27.5
0

4
4
5
3
5
3
4

26.667
26.667
52.083
0
52.083
41.25
0

©

198.75

nNL
198.75 kN2 # m
=
1 kN # ¢ Dv = a
AE
AE
¢ Dv =

198.75 kN # m
199 kN # m
=
AE
AE

Ans.

T

311

4m
15 kN

Member

C

20 kN


© 2012 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.

9–7.

Continued

*9–8.

Solve Prob. 9–7 using Castigliano’s theorem.

D

E

3m

A
B
4m

4m
15 kN

Member
AB
BC
AD
BD
CD
CE
DE

0N
0P

N (kN)
0.6667P + 10.0
0.6667P + 10.0
-(0.8333P + 12.5)
15.0
-(0.8333P + 12.5)
0.5P + 27.5
0

0.6667
0.6667
-0.8333
0
-0.8333
0.5
0

N(P = 0) kN
10.0
10.0
-12.5
15.0
-12.5
27.5
0

L(m)

198.75 kN # m
199 kN # m
=
AE
AE

T

0N
b L (kN m)
0P

#

4
4
5
3
5
3
4

26.667
26.667
52.083
0
52.083
41.25
0

©

198.75

0N L
b
¢ Dv = a N a
0P AE

=

Na

Ans.

312

C

20 kN


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–9. Determine the vertical displacement of the truss at
joint F. Assume all members are pin connected at their end
points. Take A ϭ 0.5 in2 and E ϭ 29(103) ksi for each
member. Use the method of virtual work.

500 lb
300 lb
3 ft

3 ft

F

E

D

3 ft

C
A

B

600 lb

L
nNL
=
[( -1.00)( - 600)(3) + (1.414)(848.5)(4.243) + ( - 1.00)(0)(3)
¢ Fv = a
AE
AE
+ (- 1.00)(- 1100)(3) + (1.414)(1555.6)(4.243) + ( -2.00)( - 1700)(3)
+ (- 1.00)(- 1400)(3) + (- 1.00)(- 1100)(3) + (- 2.00)( -1700)(3)](12)
47425.0(12)
=

9–10.

0.5(29)(106)

= 0.0392 in. T

Ans.

Solve Prob. 9–9 using Castigliano’s theorem.

500 lb
300 lb
3 ft
F

3 ft
E

D

3 ft

C
A

B

600 lb

0N L
1
b
=
¢ Fv = a Na
[- (P + 600)]( - 1)(3) + (1.414P + 848.5)(1.414)(4.243)
0P AE
AE
+ ( - P)( - 1)(3) + ( -(P + 1100))( -1)(3)
+ (1.414P + 1555.6)(1.414)(4.243) + (- (2P + 1700)) ( - 2)(3)
+ ( - (P + 1400)( -1)(3) + ( -(P + 1100))( -1)(3)
(55.97P + 47.425.0)(12)
+ (- (2P + 1700))(- 2)(3)](12) =
(0.5(29(10)6)
Set P = 0 and evaluate
¢ Fv = 0.0392 in. T

Ans.

313


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–11. Determine the vertical displacement of joint A. The
cross-sectional area of each member is indicated in the
figure. Assume the members are pin connected at their end
points. E ϭ 29 (10)3 ksi. Use the method of virtual work.

D

2 in2

2 in2

2 in2
2

3 in

3 in2
4 ft
7k

The virtual force and real force in each member are shown in Fig. a and b,
respectively.

n(k)

AB
BC
AD
BD
CD
CE
DE

-1.00
-1.00
22
-2.00
22
-1.00
0

#

N(k)

L(ft)

nNL(k2 ft)

-7.00
-7.00
7 22
-14.00
7 22
-4.00
0

4
4
4 22
4
4 22
4
4

28
28
5622
112
5622
16
0

nNL
1 k # ¢ Av = a
AE
1 k # ¢ Av =

(56 22 + 5622) k2 # ft

(29 + 28 + 112 + 16)k2 # ft
(3in2)[29(103) k>in2]

¢ Av = 0.004846 ft a

+

(2in2)[29(103) k>in2]

12 in
b = 0.0582 in. T
1 ft

Ans.

314

3 in2 4 ft

B

A

Member

E

3 in2

C

4 ft
3k


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*9–12.

Solve Prob. 9–11 using Castigliano’s theorem.

D

2 in2

2 in2

2 in2
2

3 in

3 in

2

4 ft
7k

AB
BC
AD
BD
CD
CE
DE

0N
0P

N(k)
-P
-P
22P
-2P
22P
-(P - 3)
0

N1P = 7k2

-1
-1
22
-2
22
-1
0

-7
-7
7 22
-14
7 22
-4
0

Na

L(ft)
4
4
422
4
422
4
4

0N
b L(k # ft)
0P

28
28
56 22
112
56 22
16
0

dN L
b
¢ Av = a Na
dP AE
(28 + 28 + 112 + 16) k # ft
=

(3 in2)[29(103)k>m2]

= 0.004846 ft a

+

5622 + 5622 k2 # ft
(2 in2)[29(103)k>in2]

12 in
b = 0.0582 in T
1 ft

Ans.

315

3 in2 4 ft

B

A

Member

E

3 in2

C

4 ft
3k


© 2012 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.

9–13. Determine the horizontal displacement of joint D.
Assume the members are pin connected at their end points.
AE is constant. Use the method of virtual work.

8m
2k

D

6 ft

3k

C

6 ft

The virtual force and real force in each member are shown in Fig. a and b,
respectively.
A

Member

n(k)

N(k)

AC
BC
BD
CD

1.50
-1.25
-0.75
1.25

5.25
-6.25
-1.50
2.50

L(ft)

nNL(k2

# ft)

6
10
12
10

47.25
78.125
13.50
31.25

©

170.125

nNL
1k # ¢ Dh = a
AE
1k # ¢ Dh =

¢ Dh =

170.125 k2 # ft
AE
170 k # ft
:
AE

Ans.

316

B


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–14.

Solve Prob. 9–13 using Castigliano’s theorem.

8m
2k

D

6 ft

3k

C

6 ft

A

Member

N(k)

0N
0P

N 1P = 2k2

L(ft)

AC
BC
BD
CD

1.50P + 2.25
-(1.25P + 3.75)
-0.750P
1.25P

1.50
-1.25
-0.750
1.25

5.25
-6.25
-1.50
2.50

6
10
12
10

47.25
78.125
13.5
31.25

©

170.125

Na

0N
b L(k # ft)
0P

dN L
b
¢ Dh = a Na
dP AE
=

170.125 k # ft
AE

=

170 k # ft
:
AE

Ans.

317

B


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–15. Determine the vertical displacement of joint C of the
truss. Each member has a cross-sectional area of A = 300 mm2.
E = 200 GPa. Use the method of virtual work.

H

B
4m

The virtual and real forces in each member are shown in Fig. a and b respectively.

AB
DE
BC
CD
AH
EF
BH
DF
BG
DG
GH
FG
CG

0.6667
0.6667
1.333
1.333
-0.8333
-0.8333
0.5
0.5
-0.8333
-0.8333
-0.6667
-0.6667
1

N(kN)
6.667
6.667
9.333
9.333
-8.333
-8.333
5
5
-3.333
-3.333
-6.6667
-6.6667
4

#

L(m)

nNL(kN2 m)

4
4
4
4
5
5
3
3
5
5
4
4
3

17.78
17.78
49.78
49.78
34.72
34.72
7.50
7.50
13.89
13.89
17.78
17.78
12.00
© = 294.89

nNL
294.89 kN2 # m
1kN # ¢ Cv = a
=
AE
AE
¢ Cv =

294.89 kN # m
AE
294.89(103) N # m

=

[0.3(10 - 3) m2][200(109) N>m2]

= 0.004914 m = 4.91 mm

Ans.

T

318

C
4m

3 kN

n(kN)

F
3m

A

Member

G

4m
4 kN

E

D
4m
3 kN


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*9–16.

Solve Prob. 9–15 using Castigliano’s theorem.

H

G

F
3m

A
4m

4m
3 kN

Member

N(kN)

0N
0P

N1P = 4 kN2

AB
DE
BC
CD
AH
EF
BH
DF
BG
DG
GH
FG
CG

0.6667P + 4
0.6667P + 4
1.333P + 4
1.333P + 4
-(0.8333P + 5)
-(0.8333P + 5)
0.5P + 3
0.5P + 3
-0.8333P
-0.8333P
-(0.6667P + 4)
-(0.6667P + 4)
P

0.6667
0.6667
1.333
1.333
-0.8333
-0.8333
0.5
0.5
-0.8333
-0.8333
-0.6667
-0.6667
1

6.667
6.667
9.333
9.333
-8.333
-8.333
5
5
-3.333
-3.333
-6.667
-6.667
4

L(m)

Na

0N
bL (k # m)
0P

4
4
4
4
5
5
3
3
5
5
4
4
3

17.78
17.78
49.78
49.78
34.72
34.72
7.50
7.50
13.89
13.89
17.78
17.78
12.00

©

294.89

0N L
294.89 kN # m
¢ Cv = a Na
b
=
0P AE
AE
294.89(103) N # m
=

[0.3(10-3) m2][200(109) N>m2]

= 0.004914 m
= 4.91 mm

Ans.

T

319

C

B

4m
4 kN

E

D
4m
3 kN


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–17. Determine the vertical displacement of joint A.
Assume the members are pin connected at their end points.
Take A = 2 in2 and E = 29 (103) for each member. Use the
method of virtual work.

D

E
8 ft

A
B

8 ft
1000 lb

8 ft

C

500 lb

1
nNL
¢ Av = a
=
[2(-2.00)(-2.00)(8) + (2.236)(2.236)(8.944) + (2.236)(2.795)(8.944)]
AE
AE
164.62(12)
=

9–18.

(2)(29)(103)

= 0.0341 in. T

Ans.

D

Solve Prob. 9–17 using Castigliano’s theorem.

E
8 ft

0N L
1
¢ Av = a Na
b
=
[-2P(-2)(8) + (2.236P)(2.236)(8.944)
0P AE
AE

A

+ (-2P)(-2)(8) + (2.236P + 0.5590)(2.236)(8.944)](12)

8 ft
1000 lb

Set P = 1 and evaluate
164.62(12)
¢ Av =

(2)(29)(103)

= 0.0341 in. T

Ans.

320

B
500 lb

8 ft

C


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–19. Determine the vertical displacement of joint A if
members AB and BC experience a temperature increase of
¢T = 200°F. Take A = 2 in2 and E = 29(103) ksi. Also,
a = 6.60 (10 - 6)>°F.

D

E
8 ft

A
B

8 ft

8 ft

C

¢ Av = a na¢TL = (-2)(6.60)(10-6)(200)(8)(12) + (-2)(6.60)(10-6)(200)(8)(12)
= -0.507 in. = 0.507 in. c

Ans.

D

*9–20. Determine the vertical displacement of joint A if
member AE is fabricated 0.5 in. too short.
E

8 ft

A
8 ft

¢ Av = a n¢L = (2.236)(-0.5)
= -1.12 in = 1.12 in. c

Ans.

321

B

8 ft

C


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–21. Determine the displacement of point C and the
slope at point B. EI is constant. Use the principle of virtual
work.

P

B

C
L
2

Real Moment function M(x): As shown on figure (a).

L
2

Virtual Moment Functions m(x) and mu(x): As shown on figure (b) and (c).
Virtual Work Equation: For the displacement at point C,
L

1#¢ =

mM
dx
L0 EI

1 # ¢C = 2 c
¢C =

L

1
x1 P
1
a b a x1 b dx1 d
EI L0
2
2

PL3
48 EI

Ans.

T

For the slope at B,
L

1#u =

muM
dx
L0 EI
L

1 # uB

uB =

9–22.

L

1
2
x1 P
x2 P
1
=
c
a b a x1 bdx1 +
a1 b a x2 bdx2 d
EI L0 L
2
L
2
L0

PL2
16EI

Ans.

Solve Prob. 9–21 using Castigliano’s theorem.

P

B

C

Internal Moment Function M(x): The internal moment function in terms of
the load P' and couple moment M' and externally applied load are shown on
figures (a) and (b), respectively.
Castigliano’s Second Theorem: The displacement at C can be determined
0M(x)
x
= =
and set P¿ = P.
0P¿
2
L
0M dx
¢ =
b
Ma
0P¿ EI
L0

with

L

¢C

1
1
P
x
= 2c
a xb a b dx d
EI L0 2
2

=

PL3
48EI

Ans.

T

To determine the slope at B, with

0M(x1)
x1 0M(x2)
x2
=
= 1 ,
and
0M¿
L
0M¿
L

setting M¿ = 0.
L

u =

L0

Ma
L

uB =

=

0M dx
b
0M¿ EI
L

1
2
x1
x2
1
P
1
P
a x1 b a bdx1 +
a x2 b a1 b dx2
EI L0 2
L
EI L0 2
L

PL2
16EI

Ans.
322

L
2

L
2


© 2012 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.

9–23. Determine the displacement at point C. EI is
constant. Use the method of virtual work.

P

A

C
B
a

a

L

mM
dx
L0 EI

1 # ¢C =

a

2Pa3
T
3EI

=

*9–24.

a

1
(x )(Px1)dx1 +
(x2)(Px2)dx2 d
c
EI Lo 1
L0

¢C =

Ans.

P

Solve Prob. 9–23 using Castigliano’s theorem.

A

C
B
a

0M1
= x1
0P¿

0M2
= x2
0P¿

Set P = P¿
M1 = Px1
L

¢C =

=

L0

Ma

M2 = Px2
a

a

0M
1
bdx =
c
(Px1)(x1)dx1 +
(Px2)(x2)dx2 d
0P¿
EI L0
L0

2Pa3
3EI

Ans.

323

a


© 2012 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.

9–25. Determine the slope at point C. EI is constant. Use
the method of virtual work.

P

A

C
B
a

L

1 # uC =

L0

a

muMdx
EI

a
(x1>a)Px1dx1
(1)Px2dx2
+
EI
EI
L0
L0
a

uC =

=

Pa2
Pa2
5Pa2
+
=
3EI
2EI
6EI

9–26.

Ans.

Solve Prob. 9–25 using Castigliano’s theorem.

P

A

Set M¿ = 0

C
B

L

uC =

L0
a

=

Ma

(Px1)(a1x1)dx1

L0

a

dM dx
b
dM¿ EI

EI

a

+

2

=

Pa
Pa2
5Pa2
+
=
3EI
2EI
6EI

(Px2)(1)dx2
EI
L0
Ans.

324

a


© 2012 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.

9–27. Determine the slope at point A. EI is constant. Use
the method of virtual work.

P

A

C
B
a

L

1 # uA =
uA =

L0

muM
dx
EI

a
a
x1
1
Pa2
a1 b(Px1)dx1 +
(0)(Px2)dx2 d =
c
a
EI L0
6EI
L0

*9–28.

Ans.

P

Solve Prob. 9–27 using Castigliano’s theorem.

0M1
x1
= 1 a
0M¿

A

0M2
= 0
0M¿

M1 = -Px1
L

L0

Ma

=

-Pa2
6EI

=

Pa2
6EI

C
B
a

Set M¿ = 0

uA =

a

M2 = Px2
a
a
x1
0M dx
1
b
=
c
( -Px1)a 1 b dx1 +
(Px2)(0)dx2 d
a
0M¿ EI
EI L0
L0

Ans.

325

a


© 2012 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.

9–29. Determine the slope and displacement at point C.
Use the method of virtual work. E = 29(103) ksi,
I = 800 in4.

6k
A

B

6 ft

Referring to the virtual moment functions indicated in Fig. a and b and the real moment
function in Fig. c, we have
L

1k # ft # uc =

6 ft
6 ft ( -1)3-(6x + 12)]
(-1)(-12)
muM
2
dx =
dx1 +
dx2
EI
EI
L0 EI
L0
L0

=

uc =

252 k2 # ft3
EI

252(122) k # in2
252 k # ft2
= 0.00156 rad
=
EI
[29(103) k>in2](800 in4)

Ans.

and
L

1k # ¢ C =

mM
dx =
L0
L0 EI
=

¢C =

6 ft

6 ft ([-(x + 6)]3-(6x + 12)]
(-x1)(-12)
2
2
dx1 +
dx2
EI
EI
L0

1944 k2 # ft3
EI

1944(123) k # in3
1944 k # ft3
= 0.415 in
=
EI
[29(103) k>in2](800 in4)

Ans.

T

326

C

6 ft

12 kиft


© 2012 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.

9–30.

Solve Prob. 9–29 using Castigliano’s theorem.

6k
A

B

6 ft

0M1
= -1 and
For the slope, the moment functions are shown in Fig. a. Here,
0M¿
0M2
= -1. Also, set M¿ = 12 kft, then M1 = -12 k # ft and
0M¿
M2 = -(6x2 + 12) k # ft. Thus,
L

uc =

L0

uc =

Ma

0M dx
b
=
0M¿ EI
L0

6 ft

6 ft
(-12)(-1)
-(6x2 + 12(-1)
dx2 +
dx2
EI
EI
L0

252(122) k # in2
252 k # ft2
= 0.00156 rad
=
EI
[29(103 k>in2](800 in4)

Ans.

0M1
For the displacement, the moment functions are shown in Fig. b. Here,
= -x1
0P
0M2
and
= -(x2 + 6). Also set, P = 0, then M1 = -12 k # ft and
0P
M2 = -(6x2 + 12) k # ft. Thus,
L

¢C =

L0

a

0M dx
b
=
0P EI
L0
=

6 ft

6 ft
(-12)(-x1)
-(6x2 + 12)[-(x2 + 6)]
dx1 +
dx2
EI
EI
L0

1944 k # ft3
EI
1944(123) k # in3

=

[29(103) k>in2](800 in4)

= 0.145 in T

327

Ans.

C

6 ft

12 kиft


© 2012 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.

9–31. Determine the displacement and slope at point C of
the cantilever beam. The moment of inertia of each segment
is indicated in the figure. Take E = 29(103) ksi. Use the
principle of virtual work.

A

B
IAB ϭ 500 in.4
6 ft

Referring to the virtual moment functions indicated in Fig. a and b and
the real moment function in Fig. c, we have
L

1k # ft # uc =

3 ft
6 ft
(-1)(-50)
(-1)( -50)
m0M
dx1 +
dx2
dx =
EIBC
EIAB
L0 EI
L0
L0

1k # ft # uc =

150 k2 # ft3
300 k2 # ft3
+
EIBC
EIAB

uc =

150 k # ft2
300 k # ft2
+
EIBC
EIAB

150(122) k # in2
=

300(122) k # in2

[29(103) k>in2](200 in4)

+

[29(103) k>in2](500 in4)
Ans.

= 0.00670 rad
And
L

mM
dx =
L0 EI
L0

1 k # ¢C =
1 k # ¢C =

¢C =

3 ft

225 k2 # ft3
1800 k2 # ft3
+
EIBC
EIAB
1800 k2 # ft3
225 k # ft3
+
EIBC
EIAB

225(123) k # in3
=

3

6 ft
-x1(-50)
-(x2 + 3)(-50)
dx1 +
dx2
EIBC
EIAB
L0

2

1800(123) k # in3
4

[29(10 ) k>in ](200 in )

+

[29(103) k>in2](500 in4)

= 0.282 in T

328

Ans.

C
IBC ϭ 200 in.4
3 ft

50 kиft


© 2012 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.

*9–32.

Solve Prob. 9–31 using Castigliano’s theorem.
A

B
IAB ϭ 500 in.4
6 ft

For the slope, the moment functions are shown in Fig. a. Here,

0M2
0M1
=
= -1.
0M¿
0M¿

Also, set M¿ = 50 k # ft, then M1 = M2 = -50 k # ft.
Thus,
L

uC =

L0

Ma

0M dx
b
=
0M¿ EI
L0

3 ft

6 ft
-50(-1)dx
-50(-1)dx
+
EIBC
EIAB
L0

150 k # ft2
300 k # ft2
+
EIBC
EIAB

=

150(122) k # in2
=

300(122) k # in2

[29(103 k>in2)](200 in4)

+

[29(103) k>in2](500 in4)

= 0.00670

Ans.

For the displacement, the moment functions are shown in Fig, b. Here,
and

0M1
= -x1
0P

0M2
= -(x2 + 3). Also, set P = 0, then M1 = M2 = -50 k # ft. Thus,
0P
L

¢C =

L0

Ma

0M dx
b
=
0P EI L0
=

3 ft

6 ft
( -50)(-x)dx
(-50)[-(x2 + 3)]dx
+
EIBC
EIAB
L0

225 k # ft3
1800 k # ft3
+
EIBC
EIAB
225(123) k # in3

=

3

2

1800(123) k # in3
4

[29(10 ) k>in ](200 in )

+

[29(103) k>in2](500 in4)

= 0.282 in T

Ans.

329

C
IBC ϭ 200 in.4
3 ft

50 kиft


© 2012 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.

9–33. Determine the slope and displacement at point B.
EI is constant. Use the method of virtual work.

400 N
300 N/m

B

A
3m

Referring to the virtual moment function indicated in Fig. a and b, and real moment
function in Fig. c, we have
L

1 N # m # uB =

3m
(-1)[-(150x2 + 400x)]
m0M
dx =
dx
EI
L0 EI
L0

3150 N2 # m3
EI

1 N # m # uB =
uB =

3150 N # m2
EI

Ans.

And
L

1 N # ¢B =

mM
dx =
L0
L0 EI

3 m ( -x)3-(150x2

1 N # ¢B =

6637.5 N2 # m3
EI

¢B =

6637.5 N # m3
EI

+ 400x)4

EI

Ans.

T

330


© 2012 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.

9–34.

Solve Prob. 9–33 using Castigliano’s theorem.

400 N
300 N/m

B

A
3m

For the slope, the moment function is shown in Fig. a. Here,

0M
= -1.
0M¿

Also, set M¿ = 0, then M = -(150x2 + 400x) N # m. Thus,
L

uB =

=

L0

Ma

0M dx
b
=
0M¿ EI
L0

3m

-(150x2 + 400x)(-1)
dx
EI

3150 N # m2
EI

Ans.

For the displacement, the moment function is shown in Fig. b. Here,

0M
= -x.
0P

Also, set P = 400 N, then M = (400x + 150x2) N # m. Thus,
L

¢B =

=

L0

Ma

0M dx
b
=
0P EI
L0

3m

-(400x + 150x2)(-x)
dx
EI

6637.5 N # m3
T
EI

Ans.

331


© 2012 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.

9–35. Determine the slope and displacement at point B.
Assume the support at A is a pin and C is a roller. Take
E = 29(103) ksi, I = 300 in4. Use the method of virtual work.

4 k/ft

A
10 ft

Referring to the virtual moment functions shown in Fig. a and b and the
real moment function shown in Fig. c,
L

1 k # ft # uB =

10 ft
(0.06667x1)(30x1 - 2x21)dx1
muM
dx =
EI
L0 EI
L0
5 ft

+
2

1 k # ft # uB =
uB =

270.83 k
EI

L0

(-0.06667x2)(30x2 - 2x22)
dx2
EI

# ft3

270.83(122) k # in2
270.83 k # ft2
= 0.00448 rad
=
EI
329(103) k>in24(300 in4)

Ans.

And
L

1 k # ¢B =

mM
dx =
L0 EI
L0

10 ft

(0.3333x1)(30x1 - 2x21)dx1
EI
5 ft

+
1 k # ¢B =

¢B =

L0

(0.6667x2)(30x2 - 2x22)dx2
EI

2291.67 k # ft3
EI

2291.67(123) k # in3
2291.67 k # ft3
= 0 # 455 in T
=
EI
329(103) k>in24(300 in4)

Ans.

332

C

B
5 ft


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