379
8–21. Determine the slope and displacement at point C. Use
the principle of virtual work. EI is constant.
A
2 m1 m
B
C
3 m
12 kN
>
m
SOLUTION
Referring to the virtual moment functions shown in Figs. a
and b, and the real moment functions in Fig. c,
1
kN
#
m
#
uC=
LL
0
m
u
M
EI
dx =
L2 m
0
(–0.1667x
1
)(16x
1
)dx
1
EI
+
L1 m
0
–0.1667(x
2
+2)(–6x
2
2+16x
2
+32)dx
2
EI
+
L3 m
0
(0.1667x
3
)(–6x
3
2+32x
3
)dx3
EI
1
kN
#
m
#
uC=
4.6667 kN2#m3
EI
u
C=
4.67 kN #m2
EI
Ans.
And
1
kN
#
∆C=
LL
0
mM
EI
dx =
L2 m
0
(0.5x
1
)(16x
1
)dx
1
EI
+
L1 m
0
0.5(x
2
+2)(–6x
2
2+16x
2
+32)dx2
EI
+
L3 m
0
(–6x
3
2+32x
3
)(0.5x
3
)dx3
EI
1
kN
#
∆C=
152.5 kN2#m3
EI
∆C=
T Ans.
Ans.
uC=
4.67 kN #m2
EI
∆C=
152.5 kN #m3
EI
T
380
EI
EI
381
3 m 4
m4
m
BD
A
C
16 kN
>
m
8–23. Determine the displacement at point D. Use the
principle of virtual work. EI is constant.
SOLUTION
3EI
382
3EI
3[29(10
3
) k>in
2
](170 in
4
)
Ans.
uA=0.0219 rad
∆A=2.34 in.T
383
B
A
C
10 ft 10 ft
10 k
10 ft
2 k>ft
8–25. Solve Prob. 8–24 using Castigliano’s theorem.
3EI
3[29(10
3
) k>in
2
](170 in
4
)
Ans.
uA=0.0219 rad
∆A=2.34 in.T
AB
C
2 m
IAB 5 600 (106) mm4IBC 5 150 (106) mm
4
1 m
12 kN>m
8 kN>m
8–27. Solve Prob. 8–26 using Castigliano’s theorem.
SOLUTION
386
EI
24EI
Ans.
u
A=
wL3
24EI
A
w
A
L
B
C
L
u
w
A
L
B
C
L
8–29. Solve Prob. 8–28 using Castigliano’s theorem.
w
EI
EI
B
C
12 ft 6 ft
A
2 k>ft
8–31. Solve Prob. 8–30 using Castigliano’s theorem.
SOLUTION
SOLUTION
*8–32. Bar ABC has a rectangular cross section of 300mm by
100 mm. Attached rod DB has a diameter of 20 mm. Determine
the vertical displacement of point C due to the loading.
Consider only the effect of bending in ABC and axial force in
DB.
E=200
GPa.
3 m
20 kN
AB
C
4 m
D
100 mm
300 mm
3 m
3 m
20 kN
AB
C
4 m
D
100 mm
300 mm
3 m
8–33. Bar ABC has a rectangular cross section of 300 mm by
100 mm. Attached rod DB has a diameter of 20 mm. Determine
the slope at A due to the loading. Consider only the effect of
bending in ABC and axial force in DB.
E=200
GPa.
SOLUTION
392
SOLUTION
8–34. Determine the slope and displacement at point B.
Assume the support at A is a pin and C is a roller. Take
E=200
GPa,
I=150(106)
mm4.
Use the method of virtual
work.
AC
4 m2 m
16 kN>m
B
Ans.
uB=0.00231 rad
∆B=7.82 mmT
393
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8–35. Solve Prob. 8–34 using Castigliano’s theorem.
AC
4 m2 m
16 kN>m
B
3[200(10
9
) N>m
2
][150(10
–6
) m
4
]
For the displacement, the moment functions are shown in Fig. b.
Here,
0M
1
0P
=
2
3
x
1
and
0M
2
0P
=
x
2
3
.
Also, set
P=0.
Then
M1=(48x1–8x1
2) kN #m
and
M2=(48x2–8x2
2) kN #m.
Thus,
∆
B=
L
L
0
M
a
0M
0Pb
dx
EI
=
L
2 m
0
(48x1–8x1
2)
a2
3 x1
b
dx1
EI
+
L
4 m
0
(48x2–8x2
2)
ax
2
3
b
dx
2
EI
∆
B=
704 kN #m3
3EI
=
704(103) N #m3
3[200(10
9
) N>m
2
][150(10
–6
) m
4
]
=0.007822 m =7.82 mmT
Ans.
uB=0.00231 rad
∆B=7.82 mmT
0M
6
0M
x
M1=(48x1–8x1
2) kN #m
2) kN #m.
208(103) N #m2
SOLUTION
AC
4 m2 m
16 kN>m
B
*8–36. 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 if the cross section is
a wide flange. Take
E=200
GPa,
I=150(106)
mm4,
G=75
GPa,
and assume AC has a cross-sectional area of
A=2.50(103)
mm2.
Use the method of virtual work.
8–36 (Continued)
SOLUTION
5EI
8–37. Determine the displacement of point C. Use the
method of virtual work. EI is constant.
BA
C
3 k>ft
12 ft 12 ft
397
BA
C
3 k>ft
12 ft 12 ft
8–38. Solve Prob. 8–37 using Castigliano’s theorem.
SOLUTION
0M
1
5EI
Ans.
∆C=
41472 k #ft3
5EI
T
29(10
3
)(245)