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

20 kips
b = 4 ft a=
16 ft
v
x
v C
x=
10 ft
Beam deflection at C: At point C, x = 10 ft = 120 in. and the
beam deflection there for this case is
v
C
Pbx
=− − −
6LEI L 2 b 2 x 2
( )
(20 kips)(48 in.)(120 in.)
2 2 2
=−
[(240 in.) − (48 in.) − (120 in.) ]
6 2
6(240 in.)(15.022 × 10 kip⋅in. )
=−0.2178 in.
Combine the two Cases
The deflection at C is the sum of the separate deflections in
cases 1 and 2.
vC =−0.5053 in. − 0.2178 in. = −0.723 in. Ans.
A
v
B
C
L=
20 ft
20 kips 30 kips
v C
4 ft 6 ft 3 ft 7 ft
D
E
x
ExAmpLE 10.11
The simply supported beam shown consists of a W24 × 76 structural
steel wide-flange shape [E = 29,000 ksi; I = 2,100 in. 4 ]. For the
loading shown, determine
(a) the beam deflection at point C.
(b) the beam deflection at point A.
(c) the beam deflection at point E.
Plan the Solution
Before starting to solve this problem, sketch the deflected shape
of the elastic curve. The 40 kip load causing the beam to bend
downward at E, in turn causing the beam to bend upward between
v
the simple supports. Since B is a pin support, the deflection of the
beam at B will be zero, but the slope will not be zero.
Let us consider the beam span between B and C in more
detail. What is it exactly that causes the beam to bend upward in
A
B
this region? Certainly, the 40 kip load is involved, but more precisely,
8 ft
the 40 kip load creates a bending moment, and it is this
bending moment that causes the beam to bend upward. For that reason, the effect of a
concentrated moment applied at one end of a simply supported span is the only consideration
required in order to compute the beam deflection at C.
A
8 ft
v
B
C
D
8 ft 8 ft 8 ft
C
D
8 ft 8 ft 8 ft
40 kips
E
40 kips
E
x
x
431