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

Multiplying the two factors of the second term on the left gives
2
v − 2v v − 2v h = 0
max
st max st
Now solve for v max , using the quadratic formula:
v
max
=
2
2 v ± (-2 v ) - 4(1)( -2 v h)
st
st
2
st
2
= v ± v + 2v h (a)
st
st
st
The maximum deflection of the beam at B if the load is dropped from the height of
5 in. can now be computed:
2
v = 0.080757 in. + (0.080757 in.) + 2(0.080757 in.)(5 in.)
max
= 0.080757 in. + 0.902270 in.
= 0.983027 in. = 0.983 in.
Ans.
largest Bending Stress: The maximum dynamic force P max exerted on the beam is calculated
from the maximum dynamic deflection v max . If the beam behaves elastically and
the stress–strain curve applicable to P max is the same as the stress–strain curve for P, then
the dynamic load is calculated as follows:
v
max
Pmaxa b
=
2 2
3LEI
3LEI
P
ab
v
6 2 4
3(240 in.)(29×
10 lb/in. )(340 in. )
∴ max = (0.983027 in.)
2 2 max =
2 2
(144 in.) (96 in.)
= 36,518.0 lb = 36,500 lb
The maximum dynamic bending moment in the beam is
M
max
Pmaxb
L a (36,518.0 lb)(96 in.)
6
= = (144 in.) = 2.103437 × 10 lb⋅in.
240 in.
and the largest bending stress in the beam is
Mmaxc
σ max = =
I
6
(2.103437 × 10 lb⋅in.)(14in./2)
340 in.
4
= 43,306.1 psi = 43,300 psi Ans.
Note that the impact factor is
vmax
0.983027 in.
n = = = 12.173
v 0.080757 in.
st
SiMPliFiED SolutioN
An equation similar to Equation (17.23) can be derived for the beam’s impact factor n.
Recall Equation (a):
2
v = v ± v + 2v h
max st st
st
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