Structural Analysis (Solutions) Chapter 9 by Wajahat

308
The virtual forces and real forces in each member are shown in Fig. a and b,
respectively.
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 GPa. Use the method of virtual
work.
E = 200
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D
CB
2 m
A
1.5 m
5 kN
1.5 m
Member n(kN) N(kN) L(m) nNL (kN2 m)
AB 1.25 6.25 2.50 19.531
AD -0.75 -3.75 3 8.437
BD -1.25 -6.25 2.50 19.531
BC 1.50 7.50 1.50 16.875
64.375©
#
Ans.= 0.536 mm T
= 0.53646 (10-3
) m
=
64.375(103
) N # m
c0.6(10-3
) m2
d c200(109
)N>m2
d
¢Av
=
64.375 kN # m
AE
1 kN # ¢Av
= a
nNL
AE
=
64.375 kN2 # m
AE

309
Ans.= 0.536 mm T
= 0.53646 (10-3
) m
=
64 # 375(103
) N # m
30.6(10-3
)m2
43200(109
)N>m2
4
=
64.375 kN # m
AE
¢Av
= aNa
0N
0P
b
L
AE
9–2. Solve Prob. 9–1 using Castigliano’s theorem.
D
CB
2 m
A
1.5 m
5 kN
1.5 mMember N(kN) L(m)
AB 1.25 P 1.25 6.25 2.5 19.531
AD -0.750 P -0.75 -3.75 3 8.437
BD -1.25 P -1.25 -6.25 2.5 19.531
BC 1.50 P 1.50 7.50 1.5 16.875
64.375©
Na
0N
0P
bL(kN # m)N1P = 5kN2
0N
0P
*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.
Member n N L nNL
AF 0 0 1.5 0
AE -0.8333 -37.5 2.5 78.125
AB 0.6667 30.0 2.0 40.00
EF 0 0 2.0 0
EB 0.50 22.5 1.5 16.875
ED -0.6667 -30.0 2.0 40.00
BC 0 0 2.0 0
BD 0.8333 37.5 2.5 78.125
CD -0.5 -22.5 1.5 16.875
© = 270
Ans.¢Bv
=
270(103
)
400(10-6
)(200)(109
)
= 3.375(10-3
)m = 3.38 mm T
1 # ¢Bv
= a
nNL
AE
C
1.5 m
A
DEF
45 kN
2 m
B
2 m

310
Ans.=
270(103
)
400(10-6
)(200)(109
)
= 3.375(10-3
)m = 3.38 mm
dBv
= aN a
0N
0P
b
L
AE
=
270
AE
*9–4. Solve Prob. 9–3 using Castigliano’s theorem.
C
1.5 m
A
DEF
45 kN
2 m
B
2 m
C
1.5 m
A
DEF
45 kN
2 m
B
2 m
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.
Member N N(P = 45) L
AF 0 0 0 1.5 0
AE -0.8333P -0.8333 -37.5 2.5 78.125
AB 0.6667P 0.6667 30.0 2.0 40.00
BE 0.5P 0.5 22.5 1.5 16.875
BD 0.8333P 0.8333 37.5 2.5 78.125
BC 0 0 0 2.0 0
CD -0.5P -0.5 -22.5 1.5 16.875
DE 0.6667P -0.6667 -30.0 2.0 40.00
EF 0 0 0 2.0 0
© = 270
Na
0N
0P
bL
0N
0P
Member n N L nNL
AF 0 0 1.5 0
AE -0.8333 -37.5 2.5 78.125
AB 0.6667 30.0 2.0 40.00
EF 0 0 2.0 0
EB 0.50 22.5 1.5 -16.875
ED -0.6667 30.0 2.0 40.00
BC 0 0 2.0 0
BD 0.8333 37.5 2.5 78.125
CD -0.5 -22.5 1.5 16.875
© = 236.25
Ans.¢Ev =
236.25(103
)
400(10-6
)(200)(109
)
= 2.95(10-3
)m = 2.95 mm T
1 # ¢Ev = a
nNL
AE

311
Ans.¢Dv
=
198.75 kN # m
AE
=
199 kN # m
AE
T
1 kN # ¢Dv
= a
nNL
AE
=
198.75 kN2 # m
AE
C
1.5 m
A
DEF
45 kN
2 m
B
2 m
Member N N(P = 45) L
AF 0 0 0 1.5 0
AE -(0.8333P + 37.5) -0.8333 -37.5 2.5 78.125
AB 0.6667P + 30 0.6667 30.0 2.0 40.00
BE 22.5 - 0.5P -0.5 22.5 1.5 -16.875
BD 0.8333P + 37.5 0.8333 37.5 2.5 78.125
BC 0 0 0 2.0 0
CD -(0.5P + 22.5) -0.5 -22.5 1.5 16.875
DE -(0.6667P + 30) -0.6667 -30.0 2.0 40.00
EF 0 0 0 2.0 0
© = 236.25
Na
0N
0P
bL
0N
0P
Ans.=
236.25(103
)
400(10-6
)(200)(109
)
= 2.95(10-3
)m = 2.95 mm T
¢Ev
= aN
0N
0P
L
AE
=
236.25
AE
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.
20 kN
15 kN
A
B C
ED
4 m
3 m
4 m
Member n(kN) N(kN) L(m) nNL(kN2 m)
AB 0.6667 10.0 4 26.667
BC 0.6667 10.0 4 26.667
AD -0.8333 -12.5 5 52.083
BD 0 15.0 3 0
CD -0.8333 -12.5 5 52.083
CE 0.500 27.5 3 41.25
DE 0 0 4 0
198.75©
#
The virtual and real forces in each member are shown in Fig. a and b,
respectively

312
9–7. Continued
Ans.=
198.75 kN # m
AE
=
199 kN # m
AE
T
¢Dv
= aN a
0N
0P
b
L
AE
20 kN
15 kN
A
B C
ED
4 m
3 m
4 m
Member N(kN) N(P = 0)kN L(m) (kN m)
AB 0.6667P + 10.0 0.6667 10.0 4 26.667
BC 0.6667P + 10.0 0.6667 10.0 4 26.667
AD -(0.8333P + 12.5) -0.8333 -12.5 5 52.083
BD 15.0 0 15.0 3 0
CD -(0.8333P + 12.5) -0.8333 -12.5 5 52.083
CE 0.5P + 27.5 0.5 27.5 3 41.25
DE 0 0 0 4 0
198.75©
#Na
0N
0P
bL
0N
0P

313
Ans.=
47425.0(12)
0.5(29)(106
)
= 0.0392 in. T
+ (-1.00)(-1400)(3) + (-1.00)(-1100)(3) + (-2.00)(-1700)(3)](12)
+ (-1.00)(-1100)(3) + (1.414)(1555.6)(4.243) + (-2.00)(-1700)(3)
¢Fv
= a
nNL
AE
=
L
AE
[(-1.00)(-600)(3) + (1.414)(848.5)(4.243) + (-1.00)(0)(3)
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.
300 lb
500 lb
600 lb
C
BA
F E D
3 ft
3 ft 3 ft
300 lb
500 lb
600 lb
C
BA
F E D
3 ft
3 ft 3 ft
Set P = 0 and evaluate
Ans.¢Fv
= 0.0392 in. T
+ (-(2P + 1700))(-2)(3)](12) =
(55.97P + 47.425.0)(12)
(0.5(29(10)6
)
+ (-(P + 1400)(-1)(3) + (-(P + 1100))(-1)(3)
+ (1.414P + 1555.6)(1.414)(4.243) + (-(2P + 1700)) (-2)(3)
+ (-P)(-1)(3) + (-(P + 1100))(-1)(3)
[-(P + 600)](-1)(3) + (1.414P + 848.5)(1.414)(4.243)¢Fv
= aNa
0N
0P
b
L
AE
=
1
AE

314
The virtual force and real force in each member are shown in Fig. a and b,
respectively.
Ans.¢Av
= 0.004846 ft a
12 in
1 ft
b = 0.0582 in. T
1 k # ¢Av
=
(29 + 28 + 112 + 16)k2 # ft
(3in2
)[29(103
)k>in2
]
+
(5622 + 5622) k2 # ft
(2in2
)[29(103
)k>in2
]
1 k # ¢Av
= a
nNL
AE
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.
7 k
3 k
A
B
C
ED
4 ft
4 ft
4 ft
3 in2
2 in2
2 in22 in2
3 in2
3 in2
3 in2
Member n(k) N(k) L(ft) nNL(k2 ft)
AB -1.00 -7.00 4 28
BC -1.00 -7.00 4 28
AD
BD -2.00 -14.00 4 112
CD
CE -1.00 -4.00 4 16
DE 0 0 4 0
562242272222
562242272222
#

315
Ans.= 0.004846 ft a
12 in
1 ft
b = 0.0582 in T
=
(28 + 28 + 112 + 16) k # ft
(3 in2
)[29(103
)k>m2
]
+
5622 + 5622 k2 # ft
(2 in2
)[29(103
)k>in2
]
¢Av
= aNa
dN
dP
b
L
AE
7 k
3 k
A
B
C
ED
4 ft
4 ft
4 ft
3 in2
2 in2
2 in22 in2
3 in2
3 in2
3 in2
Member N(k) L(ft)
AB -P -1 -7 4 28
BC -P -1 -7 4 28
AD
BD -2P -2 -14 4 112
CD
CE -(P - 3) -1 -4 4 16
DE 0 0 0 4 0
56224227222222P
56224227222222P
Na
0N
0P
bL(k # ft)N1P = 7k2
0N
0P

316
The virtual force and real force in each member are shown in Fig. a and b,
respectively.
Ans.¢Dh
=
170 k # ft
AE
:
1k # ¢Dh
=
170.125 k2 # ft
AE
1k # ¢Dh
= a
nNL
AE
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.
8 m
6 ft
6 ft
2 k
3 k
BA
D
C
Member n(k) N(k) L(ft) nNL(k2 ft)
AC 1.50 5.25 6 47.25
BC -1.25 -6.25 10 78.125
BD -0.75 -1.50 12 13.50
CD 1.25 2.50 10 31.25
170.125©
#

317
Ans.=
170 k # ft
AE
:
=
170.125 k # ft
AE
¢Dh
= aNa
dN
dP
b
L
AE
8 m
6 ft
6 ft
2 k
3 k
BA
D
C
Member N(k) L(ft)
AC 1.50P + 2.25 1.50 5.25 6 47.25
BC -(1.25P + 3.75) -1.25 -6.25 10 78.125
BD -0.750P -0.750 -1.50 12 13.5
CD 1.25P 1.25 2.50 10 31.25
170.125©
Na
0N
0P
bL(k # ft)N 1P = 2k2
0N
0P

318
Ans.= 0.004914m = 4.91mm T
=
294.89(103
) N # m
[0.3(10-3
) m2
][200(109
) N>m2
]
¢Cv
=
294.89 kN # m
AE
1kN # ¢Cv
= a
nNL
AE
=
294.89 kN2 # m
AE
9–15. Determine the vertical displacement of joint C of the
truss.Each member has a cross-sectional area of
Use the method of virtual work.E = 200 GPa.
A = 300 mm2
.
E
3 m
A
B
4 m
H
C
G
D
3 kN
4 m4 m4 m
4 kN
3 kN
F
Member n(kN) N(kN) L(m) nNL(kN2 m)
AB 0.6667 6.667 4 17.78
DE 0.6667 6.667 4 17.78
BC 1.333 9.333 4 49.78
CD 1.333 9.333 4 49.78
AH -0.8333 -8.333 5 34.72
EF -0.8333 -8.333 5 34.72
BH 0.5 5 3 7.50
DF 0.5 5 3 7.50
BG -0.8333 -3.333 5 13.89
DG -0.8333 -3.333 5 13.89
GH -0.6667 -6.6667 4 17.78
FG -0.6667 -6.6667 4 17.78
CG 1 4 3 12.00
#
The virtual and real forces in each member are shown in Fig. a and b respectively.
© = 294.89

319
Ans.= 4.91 mm T
= 0.004914 m
=
294.89(103
) N # m
[0.3(10-3
) m2
][200(109
) N>m2
]
¢Cv
= aNa
0N
0P
b
L
AE
=
294.89 kN # m
AE
E
3 m
A
B
4 m
H
C
G
D
3 kN
4 m4 m4 m
4 kN
3 kN
F
Member N(kN) L(m)
AB 0.6667P + 4 0.6667 6.667 4 17.78
DE 0.6667P + 4 0.6667 6.667 4 17.78
BC 1.333P + 4 1.333 9.333 4 49.78
CD 1.333P + 4 1.333 9.333 4 49.78
AH -(0.8333P + 5) -0.8333 -8.333 5 34.72
EF -(0.8333P + 5) -0.8333 -8.333 5 34.72
BH 0.5P + 3 0.5 5 3 7.50
DF 0.5P + 3 0.5 5 3 7.50
BG -0.8333P -0.8333 -3.333 5 13.89
DG -0.8333P -0.8333 -3.333 5 13.89
GH -(0.6667P + 4) -0.6667 -6.667 4 17.78
FG -(0.6667P + 4) -0.6667 -6.667 4 17.78
CG P 1 4 3 12.00
294.89©
Na
0N
0P
bL(k # m)N1P = 4 kN2
0N
0P

320
Ans.=
164.62(12)
(2)(29)(103
)
= 0.0341 in. T
¢Av
= a
nNL
AE
=
1
AE
[2(-2.00)(-2.00)(8) + (2.236)(2.236)(8.944) + (2.236)(2.795)(8.944)]
9–17. Determine the vertical displacement of joint A.
Assume the members are pin connected at their end points.
Take and for each member. Use the
method of virtual work.
E = 29 (103
)A = 2 in2
A
B C
E
8 ft
1000 lb 500 lb
8 ft
8 ft
D
A
B C
E
8 ft
1000 lb 500 lb
8 ft
8 ft
D
Set P = 1 and evaluate
Ans.¢Av
=
164.62(12)
(2)(29)(103
)
= 0.0341 in. T
+ (-2P)(-2)(8) + (2.236P + 0.5590)(2.236)(8.944)](12)
¢Av
= aNa
0N
0P
b
L
AE
=
1
AE
[-2P(-2)(8) + (2.236P)(2.236)(8.944)

321
Ans.= -0.507 in. = 0.507 in. c
¢Av
= ana¢TL = (-2)(6.60)(10-6
)(200)(8)(12) + (-2)(6.60)(10-6
)(200)(8)(12)
9–19. Determine the vertical displacement of joint A if
members AB and BC experience a temperature increase of
Take and ksi. Also,
a = 6.60 (10-6
)>°F.
E = 29(103
)A = 2 in2
¢T = 200°F.
A
B C
E
8 ft 8 ft
8 ft
D
A
B C
E
8 ft 8 ft
8 ft
D*9–20. Determine the vertical displacement of joint A if
member AE is fabricated 0.5 in. too short.
Ans.= -1.12 in = 1.12 in. c
¢Av
= an¢L = (2.236)(-0.5)

322
Real Moment function M(x): As shown on figure (a).
Virtual Moment Functions m(x) and : As shown on figure (b) and (c).
Virtual Work Equation: For the displacement at point C,
Ans.
For the slope at B,
Ans.uB =
PL2
16EI
1 # uB =
1
EI
c
L
L
1
0
a
x1
L
b a
P
2
x1bdx1 +
L
L
2
0
a1 -
x2
L
b a
P
2
x2bdx2 d
1 # u =
L
L
0
muM
EI
dx
¢C =
PL3
48 EI
T
1 # ¢C = 2c
1
EI L
L
1
0
a
x1
2
b a
P
2
x1bdx1 d
1 # ¢ =
L
L
0
mM
EI
dx
mu(x)
9–21. Determine the displacement of point C and the
slope at point B. EI is constant. Use the principle of virtual
work.
BC
P
L
2
L
2
BC
P
L
2
L
2
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
with
Ans.
To determine the slope at B, with , and
setting .
Ans.=
PL2
16EI
uB =
1
EI L
L
1
0
a
P
2
x1b a
x1
L
bdx1 +
1
EI L
L
2
0
a
P
2
x2b a1 -
x2
L
bdx2
u =
L
L
0
Ma
0M
0M¿
b
dx
EI
M¿ = 0
0M(x2)
0M¿
= 1 -
x2
L
0M(x1)
0M¿
=
x1
L
=
PL3
48EI
T
¢C = 2c
1
EI L
L
1
0
a
P
2
xb a
x
2
bdxd
¢ =
L
L
0
Ma
0M
0P¿
b
dx
EI
0M(x)
0P¿
= =
x
2
and set P¿ = P.

323
Ans.=
2Pa3
3EI
T
¢C =
1
EI
c
L
a
o
(x1)(Px1)dx1 +
L
a
0
(x2)(Px2)dx2 d
1 # ¢C =
L
L
0
mM
EI
dx
9–23. Determine the displacement at point C. EI is
constant. Use the method of virtual work.
A C
B
aa
P
A C
B
aa
P*9–24. Solve Prob. 9–23 using Castigliano’s theorem.
Set
Ans.=
2Pa3
3EI
¢C =
L
L
0
Ma
0M
0P¿
bdx =
1
EI
c
L
a
0
(Px1)(x1)dx1 +
L
a
0
(Px2)(x2)dx2 d
M1 = Px1 M2 = Px2
P = P¿
0M1
0P¿
= x1
0M2
0P¿
= x2

324
Ans.=
Pa2
3EI
+
Pa2
2EI
=
5Pa2
6EI
uC =
L
a
0
(x1>a)Px1dx1
EI
+
L
a
0
(1)Px2dx2
EI
1 # uC =
L
L
0
muMdx
EI
9–25. Determine the slope at point C. EI is constant. Use
the method of virtual work.
A C
B
aa
P
A C
B
aa
P9–26. Solve Prob. 9–25 using Castigliano’s theorem.
Set
Ans.=
Pa2
3EI
+
Pa2
2EI
=
5Pa2
6EI
=
L
a
0
(Px1)(1
ax1)dx1
EI
+
L
a
0
(Px2)(1)dx2
EI
uC =
L
L
0
Ma
dM
dM¿
b
dx
EI
M¿ = 0

325
Ans.
Set
Ans.=
Pa2
6EI
=
-Pa2
6EI
uA =
L
L
0
Ma
0M
0M¿
b
dx
EI
=
1
EI
c
L
a
0
(-Px1)a1 -
x1
a
bdx1 +
L
a
0
(Px2)(0)dx2 d
M1 = -Px1 M2 = Px2
M¿ = 0
0M2
0M¿
= 0
0M1
0M¿
= 1 -
x1
a
uA =
1
EI
c
L
a
0
a1 -
x1
a
b(Px1)dx1 +
L
a
0
(0)(Px2)dx2 d =
Pa2
6EI
1 # uA =
L
L
0
muM
EI
dx
9–27. Determine the slope at point A. EI is constant. Use
the method of virtual work.
A C
B
aa
P
A C
B
aa
P

326
Referring to the virtual moment functions indicated in Fig. a and b and the real moment
function in Fig. c, we have
Ans.
and
Ans.¢C =
1944 k # ft3
EI
=
1944(123
) k # in3
[29(103
) k>in2
](800 in4
)
= 0.415 in T
=
1944 k2 # ft3
EI
1k # ¢C =
L
L
0
mM
EI
dx =
L
6 ft
0
(-x1)(-12)
EI
dx1 +
L
6 ft
0
([-(x2 + 6)]3-(6x2 + 12)]
EI
dx2
uc =
252 k # ft2
EI
=
252(122
) k # in2
[29(103
) k>in2
](800 in4
)
= 0.00156 rad
=
252 k2 # ft3
EI
1k # ft # uc =
L
L
0
muM
EI
dx =
L
6 ft
0
(-1)(-12)
EI
dx1 +
L
6 ft
0
(-1)3-(6x2 + 12)]
EI
dx2
9–29. Determine the slope and displacement at point C.
Use the method of virtual work. ksi,
I = 800 in4
.
E = 29(103
)
6 ft 6 ft
A B C 12 kиft
6 k

327
For the slope, the moment functions are shown in Fig. a. Here, and
.Also, set , then and
.Thus,
Ans.
For the displacement, the moment functions are shown in Fig. b. Here,
and .Also set, P = 0, then and
.Thus,
Ans.=
1944(123
) k # in3
[29(103
) k>in2
](800 in4
)
= 0.145 in T
=
1944 k # ft3
EI
¢C =
L
L
0
a
0M
0P
b
dx
EI
=
L
6 ft
0
(-12)(-x1)
EI
dx1 +
L
6 ft
0
-(6x2 + 12)[-(x2 + 6)]
EI
dx2
M2 = -(6x2 + 12) k # ft
M1 = -12 k # ft
0M2
0P
= -(x2 + 6)
0M1
0P
= -x1
uc =
252 k # ft2
EI
=
252(122
) k # in2
[29(103
k>in2
](800 in4
)
= 0.00156 rad
uc =
L
L
0
Ma
0M
0M¿
b
dx
EI
=
L
6 ft
0
(-12)(-1)
EI
dx2 +
L
6 ft
0
-(6x2 + 12(-1)
EI
dx2
M2 = -(6x2 + 12) k # ft
M1 = -12 k # ftM¿ = 12 kft
0M2
0M¿
= -1
0M1
0M¿
= -1
6 ft 6 ft
A B C 12 kиft
6 k

328
Referring to the virtual moment functions indicated in Fig. a and b and
the real moment function in Fig. c, we have
Ans.
And
Ans.=
225(123
) k # in3
[29(103
) k>in2
](200 in4
)
+
1800(123
) k # in3
[29(103
) k>in2
](500 in4
)
= 0.282 in T
¢C =
225 k # ft3
EIBC
+
1800 k2 # ft3
EIAB
1 k # ¢C =
225 k2 # ft3
EIBC
+
1800 k2 # ft3
EIAB
1 k # ¢C =
L
L
0
mM
EI
dx =
L
3 ft
0
-x1(-50)
EIBC
dx1 +
L
6 ft
0
-(x2 + 3)(-50)
EIAB
dx2
= 0.00670 rad
=
150(122
) k # in2
[29(103
) k>in2
](200 in4
)
+
300(122
) k # in2
[29(103
) k>in2
](500 in4
)
uc =
150 k # ft2
EIBC
+
300 k # ft2
EIAB
1k # ft # uc =
150 k2 # ft3
EIBC
+
300 k2 # ft3
EIAB
1k # ft # uc =
L
L
0
m0M
EI
dx =
L
3 ft
0
(-1)(-50)
EIBC
dx1 +
L
6 ft
0
(-1)(-50)
EIAB
dx2
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 . Use the
principle of virtual work.
E = 29(103
) ksi
A B C
6 ft
IAB ϭ 500 in.4 IBC ϭ 200 in.4
3 ft
50 kиft

329
For the slope, the moment functions are shown in Fig. a. Here, .
Also, set , then .
Thus,
Ans.
For the displacement, the moment functions are shown in Fig, b. Here,
and .Also, set P = 0, then .Thus,
Ans.= 0.282 in T
=
225(123
) k # in3
[29(103
) k>in2
](200 in4
)
+
1800(123
) k # in3
[29(103
) k>in2
](500 in4
)
=
225 k # ft3
EIBC
+
1800 k # ft3
EIAB
¢C =
L
L
0
Ma
0M
0P
b
dx
EI
=
L
3 ft
0
(-50)(-x)dx
EIBC
+
L
6 ft
0
(-50)[-(x2 + 3)]dx
EIAB
M1 = M2 = -50 k # ft
0M2
0P
= -(x2 + 3)
0M1
0P
= -x1
= 0.00670
=
150(122
) k # in2
[29(103
k>in2
)](200 in4
)
+
300(122
) k # in2
[29(103
) k>in2
](500 in4
)
=
150 k # ft2
EIBC
+
300 k # ft2
EIAB
uC =
L
L
0
Ma
0M
0M¿
b
dx
EI
=
L
3 ft
0
-50(-1)dx
EIBC
+
L
6 ft
0
-50(-1)dx
EIAB
M1 = M2 = -50 k # ftM¿ = 50 k # ft
0M1
0M¿
=
0M2
0M¿
= -1
A B C
6 ft
IAB ϭ 500 in.4 IBC ϭ 200 in.4
3 ft
50 kиft

330
Referring to the virtual moment function indicated in Fig. a and b, and real moment
function in Fig. c, we have
Ans.
And
Ans.¢B =
6637.5 N # m3
EI
T
1 N # ¢B =
6637.5 N2 # m3
EI
1 N # ¢B =
L
L
0
mM
EI
dx =
L
3 m
0
(-x)3-(150x2
+ 400x)4
EI
uB =
3150 N # m2
EI
1 N # m # uB =
3150 N2 # m3
EI
1 N # m # uB =
L
L
0
m0M
EI
dx =
L
3 m
0
(-1)[-(150x2
+ 400x)]
EI
dx
9–33. Determine the slope and displacement at point B.
EI is constant. Use the method of virtual work.
3 m
A
B
400 N
300 N/m

331
For the slope, the moment function is shown in Fig. a. Here, .
Also, set , then .Thus,
Ans.
For the displacement, the moment function is shown in Fig. b. Here, .
Also, set , then .Thus,
Ans.=
6637.5 N # m3
EI
T
¢B =
L
L
0
Ma
0M
0P
b
dx
EI
=
L
3 m
0
-(400x + 150x2
)(-x)
EI
dx
M = (400x + 150x2
) N # mP = 400 N
0M
0P
= -x
=
3150 N # m2
EI
uB =
L
L
0
Ma
0M
0M¿
b
dx
EI
=
L
3 m
0
-(150x2
+ 400x)(-1)
EI
dx
M = -(150x2
+ 400x) N # mM¿ = 0
0M
0M¿
= -1
3 m
A
B
400 N
300 N/m

332
Referring to the virtual moment functions shown in Fig. a and b and the
real moment function shown in Fig. c,
Ans.
And
Ans.¢B =
2291.67 k # ft3
EI
=
2291.67(123
) k # in3
329(103
) k>in2
4(300 in4
)
= 0 # 455 in T
1 k # ¢B =
2291.67 k # ft3
EI
+
L
5 ft
0
(0.6667x2)(30x2 - 2x2
2
)dx2
EI
1 k # ¢B =
L
L
0
mM
EI
dx =
L
10 ft
0
(0.3333x1)(30x1 - 2x1
2
)dx1
EI
uB =
270.83 k # ft2
EI
=
270.83(122
) k # in2
329(103
) k>in2
4(300 in4
)
= 0.00448 rad
1 k # ft # uB =
270.83 k2 # ft3
EI
+
L
5 ft
0
(-0.06667x2)(30x2 - 2x2
2
)
EI
dx2
1 k # ft # uB =
L
L
0
muM
EI
dx =
L
10 ft
0
(0.06667x1)(30x1 - 2x1
2
)dx1
EI
Assume the support at A is a pin and C is a roller. Take
Use the method of virtual work.I = 300 in4
.E = 29(103
) ksi,
BA C
4 k/ft
10 ft 5 ft

333
For the slope, the moment functions are shown in Fig. a. Here,
and .Also, set , then
and .Thus,
Ans.
For the displacement, the moment fractions are shown in Fig. b. Here,
and .Also, set , then
and .Thus
Ans.=
2291.67 k # ft3
EI
=
2291.67(123
) k # in3
[29(103
)k>in2
](300 in4
)
= 0.455 in T
+
L
5 ft
0
(30x2 - 2x2
2
)(0.6667x2)dx2
EI
¢B =
L
L
0
Ma
0M
0P
b
dx
EI
=
L
10 ft
0
30x1 - 2x1
2
(0.3333x1)dx1
EI
M2 = (30x2 - 2x2
2
) k # ftM1 = (30x1 - 2x1
2
) k # ft
P = 0
0M2
0P
= 0.6667x2
0M1
0p
= 0.3333x1
=
270.83 k # ft2
EI
=
270.83(122
) k # in2
[29(103
)k>in2
](300 in4
)
= 0.00448 rad
+
L
5 ft
0
(30x2 - 2x2
2
)(0.06667x2)dx2
EI
uB =
L
L
0
Ma
∂M
∂M¿
b
dx
EI
=
L
10 ft
0
(30x1 - 2x1
2
)(0.06667x1)dx1
EI
M2 = (30x2 - 2x2
2
) k # ftM1 = (30x1 - 2x1
2
) k # ft
M¿ = 0
0M2
0M¿
= 0.06667x2
0M1
0M¿
= 0.06667x1
BA C
4 k/ft
10 ft 5 ft

334
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.
Ans.
And
Ans.= 0.469 in T
=
2291.67(123
)k # in3
[29(103
) k>in2
](300 in4
)
+
100(12) k # in
[12(103
) k>in2
](7.50 in2
)
¢B =
2291.67 k # ft3
EI
+
100 k # ft
GA
=
2291.67 k2 # ft3
EI
+
100 k2 # ft
GA
+
L
5 ft
0
(0.6667x2)(30x2 - 2x2
2
)
EI
dx2 +
L
5 ft
0
1c
(- 0.6667)(4x2 - 30)
GA
ddx2
=
L
10 ft
0
(0.3333x1)(30x1 - 2x1
2
)
EI
dx1 +
L
10 ft
0
1c
0.3333(30 - 4x1)
GA
ddx1
1 k # ¢B =
L
L
0
mM
EI
dx +
L
L
0
ka
nV
GA
bdx
uB =
270.83 k # ft2
EI
=
270.83(122
) k # in2
[29(103
)k>in2
(300 in4
)]
= 0.00448 rad
=
270.83 k2 # ft3
EI
+ 0
+
L
5 ft
0
(-0.06667x2(30x2 - 2x2
2
)
EI
dx2 +
L
5 ft
0
1c
0.06667(4x2 - 30)
GA
ddx2
=
L
10 ft
0
0.06667x1(30x1 - 2x1
2
)
EI
dx1 +
L
10 ft
0
1c
0.06667(30 - 4x1)
GA
ddx1
1 k # ft # uB =
L
L
0
muM
EI
dx +
L
L
0
ka
nV
GA
bdx
Assume the support at A is a pin and C is a roller. Account
for the additional strain energy due to shear. Take
, , and assume
AB has a cross-sectional area of . Use the
A = 7.50 in2
G = 12(103
) ksi,I = 300 in4
E = 29(103
) ksi
BA C
4 k/ft
10 ft 5 ft

335
9–37. Continued

336
Referring to the virtual and real moment functions shown in Fig. a and b,
respectively,
Ans.¢C =
w0L4
120EI
T
1 # ¢C =
L
L
0
mM
EI
dx = 2
L
L
2
0
a
1
2
b a
w0L
4
x -
w0
3L
x3
b
EI
dx
9–38. Determine the displacement of point C. Use the
method of virtual work. EI is constant.
BA
C
w0
L__
2
L__
2

337
The moment function is shown in Fig. a. Here . Also, set , then
.Thus
Ans.=
w0L4
120EI
T
¢C =
L
L
0
Ma
0M
0P
b
dx
EI
= 2
L
L
2
0
a
w0L
4
x -
w0
3L
x3
b a
1
2
xb
EI
dx
M =
w0L
4
x -
w0
3L
x3
P = 0
0M
0P
=
1
2
x
BA
C
w0
L__
2
L__
2

338
Referring to the virtual moment functions shown in Fig. a and b and the real
moment functions in Fig. c, we have
Ans.
And
Ans.¢A =
22.95 kN # m3
EI
T
1 kN # ¢A =
22.95 kN2 # m3
EI
1 kN # ¢A =
L
L
0
mM
EI
dx =
L
3 m
0
(-x1)(-0.3333x1
3
)
EI
dx1 +
L
3 m
0
(-x2)(6x2 - 3x2
2
)
EI
dx2
uA =
9 kN # m2
EI
1 kN # m # uA =
9 kN2 # m3
EI
+
L
3 m
0
(0.3333x2)(6x2 - 3x2
2
)
EI
dx2
1 kN # m # uA =
L
L
0
muM
EI
dx =
L
3 m
0
(-1)(-0.3333x1
3
)
EI
dx1
*9–40. Determine the slope and displacement at point A.
Assume C is pinned. Use the principle of virtual work. EI is
constant.
B C
6 kN/m
3 m 3 m
A

339
The slope, the moment functions are shown in Fig. a. Here,
and .Also, set , then and . Thus
Ans.
The displacement, the moment functions are shown in Fig. b. Here, and
.Also, set , then and . Thus
Ans.=
22.95 kN # m3
EI
T
¢A =
L
L
0
Ma
0M
0P
b
dx
EI
=
L
3 m
0
(-0.3333x1
3
)(-x1)
EI
dx1 +
L
3 m
0
(6x2 - 3x2
2
)(-x2)
EI
dx2
M2 = 6x2 - 3x2
2
M1 = -0.3333x1
3
P = 0
0M2
0P
= -x2
0M1
0P
= -x1
uA =
9 kN # m2
EI
uA =
L
L
0
Ma
0M
0M¿
b
dx
EI
=
L
3 m
0
(-0.3333x1
3
)(-1)
EI
dx1 +
L
3 m
0
(6x2 - 3x2
2
)(0.3333x2)
EI
dx2
M2 = 6x2 – 3x2
2
M1 = – 0.3333x1
3
M¿ = 0
0M2
0M¿
= -0.3333x2
0M1
0M¿
= -1
B C
6 kN/m
3 m 3 m
A

340
Referring to the virtual and real moment functions shown in Fig. a and b, respectively,
Ans.¢D =
1397 k # ft3
EI
T
1 k # ¢D =
1397.33 k2 # ft3
EI
+ 2
L
4 ft
0
(- 0.5x3)(12x3 - 1.50x3
2
)
EI
dx3
1 k # ¢D =
L
L
0
mM
EI
dx =
L
4 ft
0
(- 0.5x1)(-12x1)
EI
dx1 +
L
4 ft
0
[- 0.5(x2 + 4)][- (20x2 + 48)]
EI
dx2
9–42. Determine the displacement at point D. Use the
principle of virtual work. EI is constant.
8 k
4 ft4 ft
3 k/ft
BA C
4 ft 4 ft
D

341
The moment functions are shown in Fig. a. Here,
and .Also set ,
and .Thus,
Ans.=
1397.33 k # ft3
EI
=
1397 k # ft3
EI
T
+ 2
L
4 ft
0
(12x3 - 1.50x3
2
)(0.5x3)
EI
dx3
+
L
4 ft
0
[-(20x2 + 48)][-(0.5x2 + 2)]
EI
dx2
¢D =
L
L
0
Ma
0M
0P
b
dx
EI
=
L
4 ft
0
(-12x1)(-0.5x1)
EI
dx1
M3 = 12x3 - 1.50x3
2
M1 = -12x1, M2 = -(20x2 + 48)
P = 0
0M3
0P
= 0.5x3
0M2
0P
= - (0.5x2 + 2)
0M1
0P
= -0.5x,
9–43. Determine the displacement at point D. Use
Castigliano’s theorem. EI is constant.
8 k
4 ft4 ft
3 k/ft
BA C
4 ft 4 ft
D

342
Ans.=
440 k # ft3
EI
T
(¢D)x =
L
L
0
mM
EI
dx =
L
Lo
0
(x)(600x)dx
EI
+
L
L
0
(10)(750x)dx
EI
+ 0
*9–44. Use the method of virtual work and determine the
vertical deflection at the rocker support D. EI is constant.
10 ft
8 ft
A
D
C
B
600 lb
10 ft
8 ft
A
D
C
B
600 lb
Set ,
Ans.=
440 k # ft3
EI
T
(¢D)v =
L
L
0
M
EI
a
0M
0P
bdx =
L
10
0
(600x)(x)dx
EI
+
L
1
0
(750x)(10)dx
EI
+ 0
P = 0

343
Ans.=
5wL4
8EI
¢Ch
=
l
EI
c
L
L
0
(1x1)a
wx1
2
2
bdx1 +
L
L
0
(1L)a
wL2
2
bdx2 d
1 # ¢Ch
=
L
L
0
mM
EI
dx
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.
L
L
A
B
C
w
L
L
A
B
C
w
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.
Ans.=
wL4
4EI
¢Bv
=
l
EI
c
L
L
0
(0)a
wx1
2
2
bdx1 +
L
L
0
(L - x2)a
wL2
2
bdx2 d
l # ¢Bv
=
L
L
0
mM
EI
dx

344
P does not influence moment within vertical segment.
Set
Ans.¢B =
L
L
0
Ma
0M
0P
b
dx
EI
=
L
L
0
a-
wL2
2
b(x)
dx
EI
=
wL4
4EI
P = 0
0M
0P
= x
M = Px -
wL2
2
L
L
A
B
C
w
B
C
A
10 ft
8 ft
200 lb/ft
400 lb/ft9–49. Determine the horizontal displacement of point C.
EI is constant. Use the method of virtual work.
Referring to the virtual and real moment functions shown in Fig. a and b,
respectively,
Ans.¢Ch
=
1148 k # ft3
EI
;
1 k # ¢Ch
=
1147.73 k2 # ft3
EI
1 k # ¢Ch
=
L
L
0
mM
EI
dx =
L
8 ft
0
(-x1)(–0.1x1
2
)
EI
dx1 +
L
10 ft
0
(-8)[-(0.2x2
2
+ 6.40)]
EI
dx2

345
B
C
A
10 ft
8 ft
200 lb/ft
400 lb/ft
9–49. Continued
The moment functions are shown in Fig. a. Here, and .Also,
set , then and .
Thus,
Ans.=
1147.73 k # ft3
EI
=
1148 k # ft3
EI
;
¢Ch
=
L
L
0
Ma
0M
0P
b
dx
EI
=
L
8 ft
0
(-0.1x1
2
)(-x1)
EI
dx1 +
L
10 ft
0
[-(0.2x2
2
+ 6.40)](-8)
EI
dx2
M2 = -(0.2x2
2
+ 6.40)M1 = -0.1x1
2
P = 0
0M2
0P
= -8
0M1
0P
= -x1

346
Ans.= 2.81 mm T
=
225(103
)
200(109
)(400)(106
)(10-12
)
=
3150(103
)x - 25(103
)x2
43
0
EIAB
(¢C)v =
L
L
0
mM
EI
dx =
L
3
0
(3 - x)(50)(103
)dx
EIAB
+ 0
9–51. Determine the vertical deflection at C. The cross-
sectional area and moment of inertia of each segment is
shown in the figure. Take Assume A is a
fixed support. Use the method of virtual work.
E = 200 GPa.
C
50 kN
A B
3 m
1 mAAB ϭ 18(103
) mm2
IAB ϭ 400(106
) mm4
ABC ϭ 6.5(103
) mm2
IBC ϭ 100(106
) mm4
C
50 kN
A B
3 m
1 mAAB ϭ 18(103
) mm2
IAB ϭ 400(106
) mm4
ABC ϭ 6.5(103
) mm2
IBC ϭ 100(106
) mm4
*9–52. Solve Prob. 9–51, including the effect of shear and
axial strain energy.
See Prob. 9–51 for the effect of bending.
Note that each term is zero since n and N or v and V do not occur simultaneously
in each member. Hence,
Ans(¢C)v
= 2.81 mm T
U = a
nNL
AE
+
L
L
0
Ka
nV
GA
bdx

347
Ans.= 2.81 mm T
=
225(103
)
200(109
)(400)(106
)(10-12
)
=
3150(103
)x - 25(103
)x2
43
0
EIAB
(¢C)v
=
L
L
0
M
EI
a
0M
0P
bdx =
L
3
0
(50)(103
)(3 - x)dx
EIAB
+ 0
C
50 kN
A B
3 m
1 mAAB ϭ 18(103
) mm2
IAB ϭ 400(106
) mm4
ABC ϭ 6.5(103
) mm2
IBC ϭ 100(106
) mm4
B C
A D
12 ft
IAB ϭ 600 in.4
IBC ϭ 900 in.4
ICD ϭ 600 in.4
5 ft 5 ft
12 k
9–54. Determine the slope at A. Take
The moment of inertia of each segment of the frame is
indicated in the figure. Assume D is a pin support. Use the
E = 29(103
) ksi.
Ans.=
(75 - 25 + 25)
EIBC
=
75(144)
29(103
)(900)
= 0.414(10-3
) rad
uA =
L
L
0
muM
EI
dx =
L
5
0
(1 - 0.1x)(6x)dx
EIBC
+
L
5
0
(0.1x)(6x)dx
EIBC
+ 0 + 0

348
Ans.=
(75 - 25 + 25)
EI0C
=
75(144)
29(103
)(900)
= 0.414(10-3
) rad
uA =
L
L
0
M
EI
a
0M
0M¿
bdx =
L
5
0
(6x)(1 - 0.1x)dx
EIBC
+
L
5
0
(6x)(0.1x)dx
EIBC
+ 0 + 0
Set M¿ = 0,
B C
A D
12 ft
IAB ϭ 600 in.4
IBC ϭ 900 in.4
ICD ϭ 600 in.4
5 ft 5 ft
12 k
3 ft
A D
5 k
CB
2 k
2 in.2
2 in.2
4 ft
1 in.2
1 in.2
*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.
Ans.= 0.0401 in. :
(¢c)h
= a
nNL
AE
=
1.33(4.667)(4)(12)
2(29)(103
)
+
(1)(5)(3)(12)
(1)(29)(103
)
+ 0 +
(-8.33)(-1.667)(5)(12)
(1)(29)(103
)

349
Ans.= 0.0401 in. :
+
(-8.33)(-1.667)(5)(12)
(1)(29)(103
)
(¢c)h
= Na
0N
0P
b
L
AE
=
(4.667)(1.33)(4)(12)
2(29)(103
)
+
(5)(1)(3)(12)
(1)(29)(103
)
+ 0
Set P = 0,
3 ft
A D
5 k
CB
2 k
2 in.2
2 in.2
4 ft
1 in.2
1 in.2
A
B
C
6 ft
10 ft
45Њ
400 lb/ ft9–58. 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.
Ans.=
79.1k.ft3
EI
:
=
1
EI
c(333.47x3
- 27.05x4
)|6
0 + (42.15x3
)|10
0 d
(¢c)h
=
L
L
0
mM
EI
dx =
L
4
0
(0.541x)(1849.17x - 200x2
)dx
EI
+
L
10
0
(0.325x)(389.5x)dx
EI
Member N force
AB 1.33P + 4.667 1.33
BC P + 5 1
BD -1.667P - 8.33 -1.667
CD 0 0
0N
0P

350
Ans.=
79.1k.ft3
EI
:
=
1
EI
c(333.47x3
- 27.5x4
)|6
0 + (42.15x3
)|10
0 d
(¢c)h
=
L
L
0
M
EI
a
0M
0P
bdx =
L
4
0
(1849.17x - 200x2
)(0.541x)dx
EI
+
L
10
0
(389.5x)(0.325x)dx
EI
Set P = 0.
A
B
C
6 ft
10 ft
45Њ
400 lb/ ft
A
B
D
E
C
6 ft
10 ft
5 k
8 ft 8 ft
*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.
Ans.=
1.25(103
)
3EI
=
4.17 k # ft3
EI
T
(¢c)v
=
L
L
0
mM
EI
dx = 2c
L
10
0
(0.25x)(1.25x)dx
EI
+
L
10
0
(-0.25x)(-1.25x)dx
EI
d

351
Ans.=
1.25(103
)
3EI
=
4.17 k # ft3
EI
T
(¢c)v
=
L
L
0
M
EI
a
0M
0P
bdx = 2c
L
10
0
(1.25x)(0.25x) dx
EI
+
L
10
0
(-1.25x)(-0.25x) dx
EI
d
Set P = 5 k.
A
B
D
E
C
6 ft
10 ft
5 k
8 ft 8 ft

Structural Analysis (Solutions) Chapter 9 by Wajahat

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Structural Analysis (Solutions) Chapter 9 by Wajahat