© 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.

9–1. 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

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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.

9–7.

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

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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.

9–13. 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

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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.

9–23. 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

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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.

9–25. 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

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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.

9–27. 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

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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.

9–29. 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

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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.

9–30.

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

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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.

9–31. 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

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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.

*9–32.

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

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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.

9–33. 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

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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.

9–34.

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

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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.

9–35. 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

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

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

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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.

9–7.

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

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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.

9–13. 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

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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.

9–23. 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

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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.

9–25. 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

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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.

9–27. 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

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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.

9–29. 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

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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.

9–30.

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

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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.

9–31. 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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