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

Next, consider the overhang span between A and B. No bending moments act in this
portion of the beam; thus, the beam does not bend, but it does rotate because it is attached
to the center span. The overhang portion AB rotates by an angle equal to the rotation angle
θ B , which occurs at the left end of the center span. The deflection of overhang AB is due
exclusively to this rotation, and accordingly, the beam deflection at A can be calculated
from the rotation angle θ B of the center span.
Finally, consider the overhang span between D and E. The deflection at E is a combination
of two effects. The more obvious effect is the deflection at the free end of a
cantilever beam subjected to a concentrated load. This deflection, however, does not account
for all of the deflection at E. The standard cantilever beam cases found in Appendix C
assume that the beam does not rotate at the fixed support; in other words, the cantilever
cases assume that the support is rigid. Overhang DE, however, is not connected to a rigid
support. It is connected to center span BD, which is flexible. As the center span flexes, the
overhang rotates downward, and this is the second effect that causes deflection at E. To
calculate the beam deflection at E, we must consider both cases: the cantilever and the
simply supported beam.
SolutioN
For this beam, the elastic modulus is E = 29,000 ksi and the moment of inertia is
I = 2,100 in. 4 . The term EI, which appears in all the calculations, has the value
4 6 2
EI = (29,000 ksi)(2,100 in. ) = 60.9 × 10 kip⋅in.
The bending moment produced at D by the 40 kip load is M = (40 kips)(8 ft) = 320 kip ⋅ ft =
3,840 kip ⋅ in.
Case 1—upward Deflection of Center Span
The upward deflection of point C in the center span is computed from the elastic curve
equation for a simply supported beam subjected to a concentrated moment at D:
A
8 ft
v
B
v C
C
D
320 kip.ft
8 ft 8 ft 8 ft
40 kips
E
x
Mx
v
6LEI x 2
( 3 Lx 2 L 2
=− − + )
(a)
Beam deflection at C: Substitute the following values into
Equation (a):
M
x = 8ft = 96in.
L = 16 ft = 192 in.
EI
=−320 kip⋅ ft = −3,840 kip⋅in.
= 60.9 × 10 kip⋅in.
6 2
Use these values to compute the beam deflection at C:
v
C
Mx
6LEI x 2
( 3 Lx 2 L 2
=− − + )
( −3,840 kip⋅in.)(96 in.)
2 2
=−
[(96 in.) − 3(192 in.)(96 in.) + 2(192 in.) ] = 0.1453 in. Ans.
6 2
6(192 in.)(60.9 × 10 kip⋅in. )
Case 2—Downward Deflection of overhang AB
The downward deflection of point A on the overhang span is computed from the rotation
angle produced at support B of the center span by the concentrated moment, which acts at
D. In the beam table, the magnitude of the rotation angle at the end of the span opposite
the concentrated moment is given by
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