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

Special cases. Two extreme situations are of interest. First, if the drop height h for the
block is much greater than the maximum spring deflection D max , then the work term WD max
in Equation (a) can be neglected; thus,
and the maximum spring deflection is
Wh
1
= 2
k∆ 2
max
2Wh
∆ max = = 2∆sth
k
For the other extreme, if the drop height h of the block is zero, then
⎡ 2(0) ⎤
∆ max = D st ⎢1+ 1+ ⎥ = 2D
⎣ D st ⎦
In other words, when the block is dropped from the top of the spring as a dynamic load,
the spring deflection is twice as large as it would have been if the block were slowly and
gradually placed on top of the spring. When a load is applied so gradually that the maximum
deflection is the same as the static deflection, the impact factor is 1.0. However, if the load
is applied suddenly, the effect produced in the elastic system is significantly amplified.
impact from a Weight Moving Horizontally: By a procedure similar to that used for a
freely falling weight, the impact load of a horizontally moving weight can be investigated.
Suppose a block having mass m = W/g slides on a smooth (i.e., frictionless) horizontal
surface with a velocity v, as shown in Figure 17.18. The kinetic energy of the block before
1
it contacts the spring is 2
mv
2 . If the mass of the spring is neglected and the spring responds
elastically, then the principle of conservation of energy requires that the kinetic energy of
the block before it contacts the spring will be transformed into stored energy in the spring
at its fully compressed position:
1 ⎛W
⎞
2 g v 2
1
2
k 2
⎜ ⎟ = Dmax
⎝ ⎠
Thus,
st
v
m
731
IMPACT LOAdINg
Δ max
k
FIGURE 17.18 Horizontally
moving weight and spring.
D =
max
If we again define the static deflection of the spring caused by the weight of the block
as D st = W/k (note that this is horizontal deflection, which would occur in the spring from
the application of a horizontal force equal in magnitude to the weight of the block), then
the maximum spring deflection can be expressed as
Wv
gk
2
D
max
= D
st
2
v
gD
st
(17.24)
and the impact force F max of the block on the spring can be stated as
F W v 2
max =
gD
st
(17.25)
where the impact factor
n =
2
v
gD
st
(17.26)