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

(b) If the rod has a constant diameter of 18 mm, the strain energy in the rod is
U
total
2 2
F L (61, 072.6 N) (1,500 mm)
= =
2 2
2AE 2(70,000 N/mm )(254.4690 mm )
= 157,043.9 Nmm ⋅
Equate the strain energy stored in the prismatic rod to the work done by the falling
collar to calculate the maximum deformation of the rod:
1
P
2
max
δ
δ
max
max
= 157,043.9 Nmm ⋅
As before, compute the static deformation:
2(157,043.9 Nmm) ⋅
=
= 5.1429 mm
61,072.6 N
δ
st
=
2
δmax
2( h + δ )
max
=
2
(5.1429 mm)
2(180 mm + 5.1429 mm)
= 0.071430 mm
Now, from the relationship
compute the static load:
δ =
st
FL st
AE
F
st
2 2
(0.071430 mm)(254.4690 mm )(70,000 N/mm )
= = 848.25 N
1,500 mm
Consequently, the largest mass that can be dropped if the entire rod has a diameter
of 18 mm is
m
F st 848.25 N
= = = 86.5 kg
Ans.
2
g 9.807 m/s
Note that the allowable mass for case (b) is about 40 percent larger than the allowable
mass for case (a).
Comments: The results for cases (a) and (b) seem to be paradoxical because a larger mass
can be dropped when some of the material in the rod is removed. This apparent discrepancy
is probably best explained by considering strain-energy densities. The strain-energy
density of the 18 mm diameter segment when it is subjected to the dynamic load is
u
1
σ
2
(240 MPa)
= = = 0.4114 MPa
2E 2(70,000 MPa)
1 2 1
This strain-energy density is represented by area OCD on the stress–strain diagram
shown. The strain-energy density in the 25 mm diameter segment when it is subjected to
the maximum dynamic load is
u
2
σ
2
(124.416 MPa)
= = = 0.1106 MPa
2E 2(70,000 MPa)
2 2 2
where
61,072.6 N
σ 2 = = 124.416 MPa
490.8739 mm
740