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

Plan the Solution
The axial deformation and the normal stress in the rod under static conditions are determined
with the weight of the collar applied to the lower end of the rod. Work and energy
principles can be used to equate the work done by the 1,200 N weight as the collar falls
30 mm and elongates the rod to the strain energy stored in the rod at the instant of maximum
deformation. From this energy balance, the maximum rod deformation can be
calculated. The maximum deformation can then be used to determine the maximum
dynamic force, the corresponding normal stress, and the impact factor n.
SolutioN
(a) The static force applied to the rod is simply the weight W of the collar; thus,
The cross-sectional area of the rod is
Fst = W = 1, 200 N
π π
A = d = (15mm) = 176.7146 mm
4 4
2 2 2
The axial deformation in the 15 mm diameter rod due to the 1,200 N weight of the
collar is
FL st
(1,200 N)(750 mm)
δ st = = = 0.025465 mm = 0.0255 mm Ans.
2 2
AE (176.7146 mm )(200,000 N/mm )
and the static normal stress is
Fst
1, 200 N
σ st = = = 6.79061 MPa = 6.79 MPa Ans.
2
A 176.7146 mm
(b) The maximum rod deformation when the collar is dropped can be determined from
work and energy principles. The external work done by the 1,200 N weight as the collar
is dropped from height h must equal the strain energy stored by the rod at its maximum
deformation. Recall from Section 17.3 that the strain energy stored in an axial member
can be expressed in terms of the member deformation by Equation (17.13); thus,
External work = Internal strain energy
2
AEδmax
Fst
( h + δmax
) =
2L
AE
2
δmax
- Fst
( h + δmax
) = 0
2L
2
FL st
δmax
- 2 (
AE h + δmax
) = 0
Recognizing that the term F st L/AE is the static deformation δ st , rewrite the last
equation as
2
δ - 2 δ ( h + δ ) = 0
max
Now multiply out the two factors of the middle term to get
st
max
2
δ - 2δ δ - 2δ
h = 0
max
st max st
and solve for δ max , using the quadratic formula:
δ
max
=
2
2 δ ± (-2 δ ) - 4(1)( -2 δ h)
st
st
2
st
= δ ± δ + 2δ
h
st st 2 st
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