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

Case 3—uniformly Distributed load on Right overhang
Both simply supported and cantilever beam equations will be required to compute deflections
at E; only simply supported beam equations will be necessary to compute the beam
deflections at A and C.
Beam deflection at E: Consider the cantilever beam deflection at E of the 2 m long overhang.
From Appendix C, the maximum deflection of a cantilever beam with a uniformly
distributed load is given as
v
v
80 kN/m
v
max
4
wL
=−
8EI
(g)
A
B C D E
3 m 3 m 3 m 2 m
v E
x
Let
and
w =−80 kN/m
L = 2m
3 2
EI = 43.2 × 10 kN⋅m
and use Equation (g) to compute one portion of the beam deflection
at E:
v
E
4 4
wL (80 kN/m)(2 m)
=− =−
3 2
8EI 8(43.2 × 10 kN⋅m)
−3
=− 3.704 × 10 m = −3.074 mm
v A
A
v
B
θ B
C
v C
θD
3 m 3 m 3 m 2 m
D
80 kN/m
160 kN.m
E
v E
x
This calculation implicitly assumes that the beam is fixed to a rigid
support at D. However, the overhang is not attached to a rigid support
at D; rather, it is attached to a flexible beam that rotates in
response to the moment produced by the 80 kN uniformly distributed
load. The rotation of the overhang at D must be accounted for
in determining the deflection at E.
The moment at D due to the 80 kN distributed load is
M = (0.5)(80 kN/m)(2 m) 2 = 160 kN · m, which acts clockwise as
shown. The rotation angles at the ends of the span of a simply supported
beam subjected to a concentrated moment are given by
Equations (b) and (c). Let
M
=−160 kN⋅m
L
= 6m(i.e., thelength ofthe centerspan)
and
3 2
EI = 43.2 × 10 kN⋅m
and use Equation (b) to compute the rotation angle at D:
ML (
θ = = − − 160 kN ⋅ m)(6 m)
D 3 2
3EI 3(43.2 × 10 kN⋅m)
−3
=− 7.407 × 10 rad
The beam deflection at E is computed from the rotation angle θ D and the overhang length:
= θ = − × − 3 −3
v x ( 7.407 10 rad)(2 m) =− 14.814 × 10 m = −14.814 mm
E D DE
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