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

By inspection, the rotation angle at D must be negative; that is, the beam slopes downward
to the right at the roller support. The magnitude of the beam deflection at E due to the
center span rotation at D is computed from the beam slope and the length of overhang DE:
v = θ L = (0.0040355 rad)(96 in.) = 0.3874 in.
E D DE
By inspection, the overhang will deflect downward at E; consequently, this deflection
component is
The total deflection at E is the sum of deflections (d) and (f):
vE =− 0.3874 in.
(f)
vE =−0.1937 in. − 0.3874 in. = −0.581 in.
Ans.
mecmovies
ExAmpLE
m10.10 Determine expressions for the slope θ C and the deflection
v C at end C of the beam shown. Assume that EI is
constant for the beam.
ExAmpLE 10.12
70 kN
v
80 kN/m
The simply supported beam shown consists of a W410 × 60 structural
steel wide-flange shape [E = 200 GPa; I = 216 × 10 6 mm 4 ].
For the loading shown, determine
A
B C D E
3 m 3 m 3 m 2 m
x
(a) the beam deflection at point A.
(b) the beam deflection at point C.
(c) the beam deflection at point E.
Plan the Solution
Although the loading in this example is more complicated, the
same general approach used to solve Example 10.8 will be used
for this beam. The loading will be separated into three cases:
70 kN
v
v
80 kN/m
x
x
A
B
C
D
E
A
B
C
D
E
3 m 3 m 3 m 2 m
3 m 3 m 3 m 2 m
Case 1—Concentrated load on left overhang.
Case 2—uniformly distributed load on center span.
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