sm12 7

________________________________________________________________________
PROBLEM (*12.7) A solid tapered bar AB of length L is fixed at one end and subjected to .force
P at the other end (Fig. Pl 2.7). Verify that the strain energy of the bar is
U = 2
2
PL
Ed
π
(P12.7)
where d is the diameter at the free end.
*SOLUTION
2
2d
x
dV Adx dx
L
ρπ ===
dx
2d O
d
L
A
B
P
ρ
L
x
ρ
or
22
22
2
() 4
dd
dV x dx dx
LL
π
π
==
x
Thus,
2
22 22
4
(4)
PP L
AdxL dx
σππ == = P
We have
24
0244
8
2LP
2
EdxE
σ
π
==
U Let 22
2
2PL
C
d
π
=
2
2
021L
L
VL L
dx
UUdVC C
x
x
===−
∫∫
2
2
11
2PL
CLL Ed
π
⎡⎤
=−+=
⎢⎥
⎣⎦ Q.E.D
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