________________________________________________________________________
PROBLEM (12.5) A tubular aluminum alloy shaft AB (G = 4 x 106 psi) is fixed at the end B and
carries two torques at A and C as shown in Fig. P12.5. What is the strain energy stored in the shaft?
SOLUTION
Using the method of sections:
400 .
AB
Tlb 100 .
BC
Tlbin in =⋅
=
⋅
The polar moment of inertia is
44
(3 2 ) 6.381 .
32
Ji
4
n
π
=−=
The strain energy is then
222
1()
22AB AB BC BC
TL
UT LTL
GJ GJ
== +
∑
22
6
1[(400) (25) (100) (10)]
2(4 10 )(6.381)
=+
×
nlb
−
=−
Ui
3
80.3(10 ) .
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