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

The right-hand side of this equation can be manipulated algebraically to give
v
max
⎡ 2h
⎤
= vst
⎢1+ 1+
⎥
⎣ vst
⎦
so that the impact factor can be written as
vmax
2h
n = = 1+ 1+
v
v
st
The static deflection was calculated previously as v st = 0.080757 in. On the basis of
this static deflection, the impact factor for the 5 in. drop height is
st
2(5 in.)
n = 1+ 1+ = 12.173
Ans.
0.080757 in.
The static results can now be multiplied by the impact factor to give the dynamic
deformation and bending stress:
σ
v = nv = 12.173(0.080757 in.) = 0.983 in.
Ans.
max
max
st
= nσ
= 12.173(3,557.65 psi) = 43,300 psi
Ans.
st
ExAmpLE 17.7
(2)
C
Collar D shown is released from rest and slides without friction downward a distance of
180 mm, where it strikes a head fixed to the end of compound rod ABC. The compound
rod is made of aluminum [E = 70 GPa], and the diameters of rod segments (1) and (2) are
18 mm and 25 mm, respectively.
900 mm
600 mm
(1)
B
Collar D
A
180 mm
(a) Determine the largest mass of the collar for which the maximum normal stress in the
rod is 240 MPa.
(b) If the diameter of rod segment (2) is reduced to 18 mm, what is the largest mass of the
collar for which the maximum normal stress in the rod is 240 MPa?
Plan the Solution
From the maximum normal stress, calculate the maximum dynamic force allowed in
the segment that has the smaller cross-sectional area, segment (1). Next, compute the
total strain energy stored in the compound rod for this maximum load. Then, equate the
total strain energy to the work performed by the maximum force on the rod to calculate
the maximum deformation of the compound rod. Now use the dynamic deformation
and the drop height to determine first the static deformation and then the static load.
Then, determine the allowable mass from the static load. Finally, repeat the process for
a rod that has a constant diameter of 18 mm, and compare the allowable masses for the
two cases.
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