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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.
9–1. 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
1 kN # ¢ Av = a
¢ Av =
64.375 kN2 # m
nNL
=
AE
AE
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
N(kN)
0N
0P
N1P = 5kN2
L(m)
AB
AD
BD
BC
1.25 P
-0.750 P
-1.25 P
1.50 P
1.25
-0.75
-1.25
1.50
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
¢ Av = a Na
=
Na
D
0N
bL(kN # m)
0P
Member
1.5 m
1.5 m
5 kN
0N L
b
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 ⫽ 400 mm2, E ⫽ 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
1 # ¢ Bv = a
nNL
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
N
0N
0P
N(P = 45)
L
AF
AE
AB
BE
BD
BC
CD
DE
EF
0
-0.8333P
0.6667P
0.5P
0.8333P
0
-0.5P
0.6667P
0
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
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 ⫽ 400 mm2, E ⫽ 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
1 # ¢Ev = a
nNL
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
N
0N
0P
N(P = 45)
L
Na
AF
AE
AB
BE
BD
BC
CD
DE
EF
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
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
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
¢ Ev = a N
0N
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
1 kN # ¢ Dv = a
¢ Dv =
nNL
198.75 kN2 # m
=
AE
AE
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
0N
b L (kN m)
0P
Member
N (kN)
0N
0P
N(P = 0) kN
L(m)
AB
BC
AD
BD
CD
CE
DE
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
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
¢ Dv = a N a
=
0N L
b
0P AE
198.75 kN # m
199 kN # m
=
AE
AE
T
Ans.
312
Na
#
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 ⫽ 0.5 in2 and E ⫽ 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
¢ Fv = a
L
nNL
=
[( -1.00)( - 600)(3) + (1.414)(848.5)(4.243) + ( - 1.00)(0)(3)
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
=
[- (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)
¢ Fv = a Na
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 ⫽ 29 (10)3 ksi. Use the method of virtual work.
D
2 in2
2 in2
2 in2
2
3 in
7k
The virtual force and real force in each member are shown in Fig. a and b,
respectively.
#
n(k)
N(k)
L(ft)
nNL(k2 ft)
AB
BC
AD
BD
CD
CE
DE
-1.00
-1.00
22
-2.00
22
-1.00
0
-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
3 in2
4 ft
4 ft
C
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
nNL(k2
# ft)
Member
n(k)
N(k)
L(ft)
AC
BC
BD
CD
1.50
-1.25
-0.75
1.25
5.25
-6.25
-1.50
2.50
6
10
12
10
47.25
78.125
13.50
31.25
©
170.125
nNL
1k # ¢ Dh = a
AE
1k # ¢ Dh =
170.125 k2 # ft
AE
¢ Dh =
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
¢ Dh = a Na
Na
0N
b L(k # ft)
0P
dN L
b
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.
#
Member
n(kN)
N(kN)
L(m)
nNL(kN2 m)
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
6.667
6.667
9.333
9.333
-8.333
-8.333
5
5
-3.333
-3.333
-6.6667
-6.6667
4
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
=
AE
AE
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
¢ Cv =
F
3m
A
1kN # ¢ Cv = a
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
0N
bL (k # m)
0P
Member
N(kN)
0N
0P
N1P = 4 kN2
L(m)
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
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
Na
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
¢ Av = a
C
500 lb
1
nNL
=
[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.
8 ft
(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 =
L0
muM
dx
EI
a
uA =
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 k⭈ft
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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 k⭈ft
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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
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
3 ft
6 ft
-x1(-50)
-(x2 + 3)(-50)
dx1 +
dx2
EIBC
EIAB
L0
1 k # ¢C =
mM
dx =
L0 EI
L0
1 k # ¢C =
225 k2 # ft3
1800 k2 # ft3
+
EIBC
EIAB
¢C =
1800 k2 # ft3
225 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
328
Ans.
C
IAB ⫽ 500 in.4
IBC ⫽ 200 in.4
6 ft
3 ft
50 k⭈ft
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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
For the slope, the moment functions are shown in Fig. a. Here,
B
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
IAB ⫽ 500 in.4
IBC ⫽ 200 in.4
6 ft
3 ft
50 k⭈ft
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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
1 N # m # uB =
3150 N2 # m3
EI
uB =
3150 N # m2
EI
Ans.
And
L
1 N # ¢B =
mM
dx =
L0
L0 EI
3 m ( -x)3-(150x2 + 400x)4
1 N # ¢B =
6637.5 N2 # m3
EI
¢B =
6637.5 N # m3
EI
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
+
L0
(-0.06667x2)(30x2 - 2x22)
dx2
EI
2
1 k # ft # uB =
270.83 k # ft3
EI
270.83(122) k # in2
2
uB =
270.83 k # ft
EI
=
329(103) k>in24(300 in4)
Ans.
= 0.00448 rad
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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*9–36.
Solve Prob. 9–35 using Castigliano’s theorem.
4 k/ft
A
10 ft
For the slope, the moment functions are shown in Fig. a. Here,
0M1
0M2
= 0.06667x1 and
= 0.06667x2. Also, set M¿ = 0, then
0M¿
0M¿
M1 = (30x1 - 2x21) k # ft and M2 = (30x2 - 2x22) k # ft. Thus,
L
uB =
L0
Ma
∂M dx
b
=
∂M¿ EI L0
10 ft
(30x1 - 2x21)(0.06667x1)dx1
EI
5 ft
+
270.83(122) k # in2
270.83 k # ft2
=
=
EI
L0
(30x2 - 2x22)(0.06667x2)dx2
EI
[29(103)k>in2](300 in4)
Ans.
= 0.00448 rad
For the displacement, the moment fractions are shown in Fig. b. Here,
0M1
0M2
= 0.3333x1 and
= 0.6667x2. Also, set P = 0, then
0p
0P
M1 = (30x1 - 2x21) k # ft and M2 = (30x2 - 2x22) k # ft. Thus
L
¢B =
L0
Ma
0M dx
b
=
0P EI L0
10 ft
30x1 - 2x21(0.3333x1)dx1
EI
5 ft
+
2291.67(123) k # in3
2291.67 k # ft3
=
EI
L0
(30x2 - 2x22)(0.6667x2)dx2
EI
=
[29(103)k>in2](300 in4)
Ans.
= 0.455 in T
333
C
B
5 ft
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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.
9–37. Determine the slope and displacement at point B.
Assume the support at A is a pin and C is a roller. Account
for the additional strain energy due to shear. Take
E = 29(103) ksi, I = 300 in4, G = 12(103) ksi, and assume
AB has a cross-sectional area of A = 7.50 in2. Use the
method of virtual work.
4 k/ft
A
10 ft
The virtual shear and moment functions are shown in Fig. a and b and the real shear
and moment functions are shown in Fig. c.
L
L
muM
nV
ka
dx +
bdx
GA
L0 EI
L0
1 k # ft # uB =
10 ft
=
L0
10 ft
0.06667x1(30x1 - 2x21)
0.06667(30 - 4x1)
1c
dx1 +
ddx1
EI
GA
L0
5 ft
+
270.83 k2 # ft3
+ 0
EI
=
uB =
L0
5 ft
(-0.06667x2(30x2 - 2x22)
0.06667(4x2 - 30)
dx2 +
1c
ddx2
EI
GA
L0
270.83(122) k # in2
270.83 k # ft2
= 0.00448 rad
=
EI
[29(103)k>in2(300 in4)]
Ans.
And
L
1 k # ¢B =
10 ft
=
L
mM
nV
dx +
ka
bdx
EI
GA
L0
L0
L0
10 ft
(0.3333x1)(30x1 - 2x21)
0.3333(30 - 4x1)
dx1 +
1c
ddx1
EI
GA
L0
5 ft
+
L0
5 ft
(0.6667x2)(30x2 - 2x22)
(- 0.6667)(4x2 - 30)
dx2 +
1c
ddx2
EI
GA
L0
=
100 k2 # ft
2291.67 k2 # ft3
+
EI
GA
¢B =
2291.67 k # ft3
100 k # ft
+
EI
GA
2291.67(123)k # in3
=
3
2
100(12) k # in
4
[29(10 ) k>in ](300 in )
+
[12(103) k>in2](7.50 in2)
= 0.469 in T
Ans.
334
C
B
5 ft
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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.
9–37.
Continued
335
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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.
9–38. Determine the displacement of point C. Use the
method of virtual work. EI is constant.
w0
A
B
L
__
2
Referring to the virtual and real moment functions shown in Fig. a and b,
respectively,
w0 3
1 w0L
a ba
x x b
2
4
3L
mM
1 # ¢C =
dx = 2
dx
EI
L0 EI
L0
L
2
L
¢C =
w0L4
120EI
Ans.
T
336
C
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.
9–39.
w0
Solve Prob. 9–38 using Castigliano’s theorem.
A
B
L
__
2
0M
1
The moment function is shown in Fig. a. Here
= x. Also, set P = 0, then
0P
2
w0L
w0 3
M =
xx . Thus
4
3L
L
¢C =
=
L0
Ma
w0L4
120EI
0M dx
b
= 2
0P EI
L0
L
2
a
w0 3 1
w0L
x x b a xb
4
3L
2
dx
EI
Ans.
T
337
C
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.
*9–40. Determine the slope and displacement at point A.
Assume C is pinned. Use the principle of virtual work. EI is
constant.
6 kN/ m
A
B
3m
Referring to the virtual moment functions shown in Fig. a and b and the real
moment functions in Fig. c, we have
L
3m
(-1)( -0.3333x31)
muM
dx =
dx1
EI
L0
L0 EI
1 kN # m # uA =
3m
+
L0
(0.3333x2)(6x2 - 3x22)
dx2
EI
2
1 kN # m # uA =
uA =
9 kN # m3
EI
9 kN # m2
EI
Ans.
And
L
1 kN # ¢ A =
mM
dx =
L0 EI
L0
1 kN # ¢ A =
22.95 kN2 # m3
EI
¢A =
3m
3m
( -x1)(-0.3333x31)
(-x2)(6x2 - 3x22)
dx1 +
dx2
EI
EI
L0
22.95 kN # m3
T
EI
Ans.
338
C
3m
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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.
9–41.
Solve Prob. 9–40 using Castigliano’s theorem.
6 kN/ m
A
B
3m
0M1
= -1
The slope, the moment functions are shown in Fig. a. Here,
0M¿
0M2
= -0.3333x2. Also, set M¿ = 0, then M1 = – 0.3333x31 and M2 = 6x2 – 3x22. Thus
and
0M¿
L
3m
3m
(-0.3333x31)(-1)
(6x2 - 3x22)(0.3333x2)
0M dx
Ma
uA =
b
=
dx1 +
dx2
0M¿ EI
EI
EI
L0
L0
L0
uA =
9 kN # m2
EI
Ans.
0M1
= -x1 and
The displacement, the moment functions are shown in Fig. b. Here,
0P
0M2
= -x2. Also, set P = 0, then M1 = -0.3333x31 and M2 = 6x2 - 3x22. Thus
0P
L
3m
3m
(-0.3333x31)(-x1)
(6x2 - 3x22)(-x2)
0M dx
¢A =
Ma
b
=
dx1 +
dx2
0P EI
EI
EI
L0
L0
L0
=
22.95 kN # m3
T
EI
Ans.
339
C
3m
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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.
9–42. Determine the displacement at point D. Use the
principle of virtual work. EI is constant.
8k
3 k/ft
A
B
4 ft
4 ft
Referring to the virtual and real moment functions shown in Fig. a and b, respectively,
L
1 k # ¢D =
mM
dx =
L0 EI
L0
4 ft
4 ft
(- 0.5x1)( -12x1)
[- 0.5(x2 + 4)][- (20x2 + 48)]
dx1 +
dx2
EI
EI
L0
4 ft
+ 2
1 k # ¢D =
¢D =
L0
(- 0.5x3)(12x3 - 1.50x23)
dx3
EI
1397.33 k2 # ft3
EI
1397 k # ft3
T
EI
Ans.
340
C
D
4 ft
4 ft
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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.
9–43. Determine the displacement at point D. Use
Castigliano’s theorem. EI is constant.
8k
3 k/ft
A
B
4 ft
The moment functions are shown in Fig. a. Here,
0M1
= -0.5x,
0P
0M3
0M2
= - (0.5x2 + 2) and
= 0.5x3. Also set P = 0,
0P
0P
M1 = -12x1, M2 = -(20x2 + 48) and M3 = 12x3 - 1.50x23. Thus,
L
¢D =
L0
Ma
0M dx
b
=
0P EI
L0
4 ft
(-12x1)(-0.5x1)
dx1
EI
4 ft
+
L0
[-(20x2 + 48)][-(0.5x2 + 2)]
dx2
EI
4 ft
+ 2
=
L0
(12x3 - 1.50x23)(0.5x3)
dx3
EI
1397 k # ft3
1397.33 k # ft3
=
T
EI
EI
Ans.
341
4 ft
C
D
4 ft
4 ft
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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.
*9–44. Use the method of virtual work and determine the
vertical deflection at the rocker support D. EI is constant.
600 lb
10 ft
B
A
8 ft
L
(¢ D)x =
mM
dx =
L0 EI
L0
=
9–45.
Lo
L (10)(750x)dx
(x)(600x)dx
+
+ 0
EI
EI
L0
440 k # ft3
T
EI
D
C
Ans.
600 lb
Solve Prob. 9–44 using Castigliano’s theorem.
10 ft
B
A
8 ft
Set P = 0,
L
(¢ D)v =
10
1
(600x)(x)dx
(750x)(10)dx
M 0M
a
bdx =
+
+ 0
EI
EI
L0
L0
L0 EI 0P
440 k # ft3
T
=
EI
D
C
Ans.
342
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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.
9–46. The L-shaped frame is made from two segments,
each of length L and flexural stiffness EI. If it is subjected
to the uniform distributed load, determine the horizontal
displacement of the end C. Use the method of virtual work.
C
w
L
L
1 # ¢ Ch =
mM
dx
L0 EI
A
L
L
wx21
l
wL2
¢ Ch =
c
(1x1) a
bdx1 +
(1L)a
bdx2 d
EI L0
2
2
L0
=
B
5wL4
8EI
L
Ans.
9–47. The L-shaped frame is made from two segments,
each of length L and flexural stiffness EI. If it is subjected
to the uniform distributed load, determine the vertical
displacement of point B. Use the method of virtual work.
C
w
L
A
B
L
l # ¢ Bv =
mM
dx
L0 EI
L
¢ Bv =
=
L
wx21
l
wL2
c
(0) a
bdx1 +
(L - x2)a
bdx2 d
EI L0
2
2
L0
wL4
4EI
Ans.
343
L
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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.
*9–48.
Solve Prob. 9–47 using Castigliano’s theorem.
C
w
L
A
B
L
P does not influence moment within vertical segment.
M = Px -
wL2
2
0M
= x
0P
Set P = 0
L
¢B =
L0
L
Ma
0M dx
wL4
wL2
dx
b
=
ab (x)
=
0P EI
2
EI
4EI
L0
Ans.
9–49. Determine the horizontal displacement of point C.
EI is constant. Use the method of virtual work.
400 lb/ft
B
A
10 ft
8 ft
Referring to the virtual and real moment functions shown in Fig. a and b,
respectively,
L
1 k # ¢ Ch =
mM
dx =
L0 EI
L0
1 k # ¢ Ch =
1147.73 k2 # ft3
EI
¢ Ch =
1148 k # ft3
;
EI
8 ft
(-x1)(–0.1x21)
EI
10 ft
dx1 +
L0
(-8)[-(0.2x22 + 6.40)]
EI
C
dx2
Ans.
344
200 lb/ ft
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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.
9–49.
Continued
9–50.
Solve Prob. 9–49 using Castigliano’s theorem.
400 lb/ft
B
A
10 ft
8 ft
0M1
0M2
= -x1 and
= -8. Also,
0P
0P
2
2
set P = 0, then M1 = -0.1x1 and M2 = -(0.2x2 + 6.40).
The moment functions are shown in Fig. a. Here,
Thus,
L
¢ Ch =
L0
Ma
0M dx
b
=
0P EI L0
=
8 ft
10 ft
(-0.1x21)(-x1)
[-(0.2x22 + 6.40)]( -8)
dx1 +
dx2
EI
EI
L0
1147.73 k # ft3
1148 k # ft3
=
;
EI
EI
Ans.
345
C
200 lb/ ft
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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.
9–51. Determine the vertical deflection at C. The crosssectional area and moment of inertia of each segment is
shown in the figure. Take E = 200 GPa. Assume A is a
fixed support. Use the method of virtual work.
ABC ⫽ 6.5(103) mm2
IBC ⫽ 100(106) mm4
C
AAB ⫽ 18(10 ) mm
IAB ⫽ 400(106) mm4
3
2
A
50 kN
1m
B
3m
L
(¢ C)v =
3
(3 - x)(50)(103)dx
mM
+ 0
dx =
EIAB
L0 EI
L0
=
=
3150(103)x - 25(103)x2430
EIAB
225(103)
200(109)(400)(106)(10 - 12)
= 2.81 mm T
Ans.
*9–52. Solve Prob. 9–51, including the effect of shear and
axial strain energy.
ABC ⫽ 6.5(103) mm2
IBC ⫽ 100(106) mm4
C
AAB ⫽ 18(10 ) mm
IAB ⫽ 400(106) mm4
3
2
A
See Prob. 9–51 for the effect of bending.
L
U = a
nV
nNL
+
Ka
bdx
AE
GA
L0
Note that each term is zero since n and N or v and V do not occur simultaneously
in each member. Hence,
(¢ C)v = 2.81 mm T
Ans
346
1m
B
3m
50 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.
9–53.
Solve Prob. 9–51 using Castigliano’s theorem.
ABC ⫽ 6.5(103) mm2
IBC ⫽ 100(106) mm4
C
AAB ⫽ 18(10 ) mm
IAB ⫽ 400(106) mm4
3
2
50 kN
1m
A
B
3m
L
(¢ C)v =
=
3
(50)(103)(3 - x)dx
M 0M
+ 0
a
b dx =
EIAB
L0 EI 0P
L0
3150(103)x - 25(103)x2430
EIAB
225(103)
=
200(109)(400)(106)(10 - 12)
= 2.81 mm T
Ans.
12 k
9–54. Determine the slope at A. Take E = 29(103) ksi.
The moment of inertia of each segment of the frame is
indicated in the figure. Assume D is a pin support. Use the
method of virtual work.
5 ft
5 ft
B
C
IBC ⫽ 900 in.
4
IAB ⫽ 600 in.4
12 ft
ICD ⫽ 600 in.4
A
L
uA =
=
5
5
(1 - 0.1x)(6x)dx
(0.1x)(6x)dx
muM
dx =
+
+ 0 + 0
EIBC
EIBC
L0 EI
L0
L0
(75 - 25 + 25)
75(144)
= 0.414(10 - 3) rad
=
EIBC
29(103)(900)
Ans.
347
D
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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.
9–55.
Solve Prob. 9–54 using Castigliano’s theorem.
12 k
5 ft
5 ft
B
C
IBC ⫽ 900 in.4
IAB ⫽ 600 in.4
12 ft
ICD ⫽ 600 in.4
A
D
Set M¿ = 0,
L
uA =
5
5
(6x)(1 - 0.1x)dx
(6x)(0.1x)dx
M 0M
+
+ 0 + 0
a
b dx =
EI
0M¿
EI
EIBC
L0
L0
L0
BC
=
(75 - 25 + 25)
75(144)
= 0.414(10 - 3) rad Ans.
=
EI 0 C
29(103)(900)
*9–56. Use the method of virtual work and determine the
horizontal deflection at C. The cross-sectional area of each
member is indicated in the figure. Assume the members are
pin connected at their end points. E = 29(103) ksi.
2k
1 in.2
B
1 in.2
4 ft
2 in.2
2 in.2
A
D
3 ft
(¢ c)h = a
(1)(5)(3)(12)
(-8.33)(-1.667)(5)(12)
1.33(4.667)(4)(12)
nNL
+
+ 0 +
=
AE
2(29)(103)
(1)(29)(103)
(1)(29)(103)
= 0.0401 in. :
Ans.
348
C
5k
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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.
9–57.
Solve Prob. 9–56 using Castigliano’s theorem.
2k
1 in.2
B
Member
N force
0N
0P
AB
BC
BD
CD
1.33P + 4.667
P+5
-1.667P - 8.33
0
1.33
1
-1.667
0
(¢ c)h = N a
2 in.2
2 in.2
A
D
(5)(1)(3)(12)
(4.667)(1.33)(4)(12)
0N L
+
+ 0
b
=
3
0P AE
2(29)(10 )
(1)(29)(103)
(-8.33)(-1.667)(5)(12)
+
(1)(29)(103)
= 0.0401 in. :
Ans.
3 ft
9–58. Use the method of virtual work and determine the
horizontal deflection at C. E is constant. There is a pin at A,
and assume C is a roller and B is a fixed joint.
400 lb/ ft
B
C
6 ft
10 ft
45⬚
A
L
(¢ c)h =
5k
1 in.2
4 ft
Set P = 0,
C
4
10
(0.541x)(1849.17x - 200x2)dx
(0.325x)(389.5x)dx
mM
dx =
+
EI
EI
EI
L0
L0
L0
=
1
c(333.47x3 - 27.05x4)|60 + (42.15x3)|10
0 d
EI
=
79.1k.ft3
:
EI
Ans.
349
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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.
9–59.
Solve Prob. 9–58 using Castigliano’s theorem.
400 lb/ ft
B
C
6 ft
10 ft
45⬚
A
Set P = 0.
L
(¢ c)h =
4
10
(1849.17x - 200x2)(0.541x)dx
(389.5x)(0.325x)dx
M 0M
a
bdx =
+
EI
0P
EI
EI
L0
L0
L0
=
1
c(333.47x3 - 27.5x4)|60 + (42.15x3)|10
0 d
EI
=
79.1k.ft3
:
EI
Ans.
*9–60. The frame is subjected to the load of 5 k. Determine
the vertical displacement at C. Assume that the members
are pin connected at A, C, and E, and fixed connected at
the knee joints B and D. EI is constant. Use the method of
virtual work.
C
6 ft
D
B
5k
10 ft
E
A
8 ft
L
(¢ c)v =
10
10
(0.25x)(1.25x)dx
(-0.25x)(-1.25x)dx
mM
dx = 2c
+
d
EI
EI
EI
L0
L0
L0
=
1.25(103)
4.17 k # ft3
=
T
3EI
EI
Ans.
350
8 ft
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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.
9–61. Solve Prob. 9–60 using Castigliano’s theorem.
C
6 ft
D
B
5k
10 ft
E
A
8 ft
Set P = 5 k.
L
(¢ c)v =
=
10
10
(1.25x)(0.25x) dx
(-1.25x)(-0.25x) dx
M 0M
a
bdx = 2 c
+
d
EI
EI
L0 EI 0P
L0
L0
1.25(103)
4.17 k # ft3
=
T
3EI
EI
Ans.
351
8 ft