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

SolutioN
(a) The areas of the two rod segments are as follows:
A
A
1
2
π
= (18mm)
4
= 254.4690 mm
π
= (25 mm)
4
= 490.8739 mm
2 2
2 2
The maximum stress occurs in segment (1). The maximum dynamic load that can be
applied to this segment without exceeding the 240 MPa limit is
2 2
P = σ A = (240 N/mm )(254.4690 mm ) = 61, 072.6 N
max max 1
For this dynamic load, the strain energy in compound rod ABC can be determined
from Equation (17.14):
U
total
F1 2 L1
F2 2 L2
= +
2AE
2AE
1 1
2 2
2
(61, 072.6 N) ⎡ 600 mm 900 mm ⎤
= +
2
2(70,000 N/mm ) ⎣⎢
2 2
254.4690 mm 490.8739 mm ⎦⎥
= 111,664.5 Nmm ⋅
Equate the strain energy stored in the compound rod to the work done by the falling
collar to determine the maximum deformation of the entire rod due to the impact load:
1
P
2
δ
max max
δ
max
= 111,664.5 Nmm ⋅
2(111,664.5 Nmm) ⋅
=
61,072.6 N
= 3.6568 mm
The static deformation of the entire rod can be related to the dynamic deformation by
2
δ - 2 δ ( h + δ ) = 0
max
which was derived in Example 17.5. The static deformation is thus
st
max
δ
st
=
2
δmax
2( h + δ )
max
=
2
(3.6568 mm)
2(180 mm + 3.6568 mm)
= 0.036405 mm
The static deformation of this compound rod can be expressed as
FL 1 1 FL 2 2 Fst L1
L2
δ st = + = ⎡ ⎤
⎢ + ⎥
AE AE E ⎣ A A ⎦
1 1
2 2
2
(0.036405 mm)(70,000 N/mm )
∴ Fst
=
600 mm 900 mm
+
2 2
254.4690 mm 490.8739 mm
1
2
= 608.00 N
Consequently, the largest mass that can be dropped is
m
F st 608.00 N
= = = 62.0 kg
Ans.
2
g 9.807 m/s
739