Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

For segment CD, draw a free-body diagram that cuts through segment CD around
support E of the beam. From the free-body diagram, derive the following equation for
the real internal moment M in segment CD of the beam:
M
45 kN/m
45 kN/m
2
M =− ( x3
− 3 m) + (150 kN) x
2
3m≤
x ≤ 4.5 m
3
3
V
D
x 3
3 m
E
150 kN
Finally, derive the following equation for the real internal moment M in segment DE
of the beam:
M
M
= ( 150 kN)
x 4
0 ≤ x ≤ 3 m
4
V
x 4
E
150 kN
The equations for m and M, along with the appropriate limits of integration, are summarized
in the table that follows. The results for the integration in each segment are also given.
x Coordinate
Beam
Segment
origin
limits
(m)
m
(kN ⋅ m)
M
(kN ⋅ m)
∫
⎛ ⎞
m⎜
M ⎟
⎝ EI ⎠
dx
1
45
AB A 0–3 x
2 1 - x + 300x
2
1 2 1
1
BC A 3–4.5 x
45
2 2 - x -180( 2
x - 3) + 300 x
2 2 2 2
1
CD E 3–4.5 x
45
2 3 x 2
- ( 3 - 3) + 150
2 x
1
DE E 0–3 x
2 4 150x
4
3
1,122.188 kN 2 -m
3
EI
1,875.762 kN 2 -m
3
EI
1,550.918 kN 2 -m
3
EI
675.0 kN 2 -m
3
EI
5,223.868 kN 2 -m
3
EI
pRoBLEmS
p17.34 Determine the vertical displacement of joint B for the
truss shown in Figure P17.34/35. Assume that each member has
a cross-sectional area A = 1.25 in. 2 and an elastic modulus
E = 10,000 ksi. The loads acting on the truss are P = 21 kips and
Q = 7 kips.
p17.35 Determine the horizontal displacement of joint B for
the truss shown in Figure P17.34/35. Assume that each member
has a cross-sectional area A = 1.25 in. 2 and an elastic modulus
E = 10,000 ksi. The loads acting on the truss are P = 21 kips and
Q = 7 kips.
y
A
x
Q
6.5 ft
P
B
5 ft
C
4.5 ft
FIGURE p17.34/35
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