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

strain energies in all the segments. The strain energy of a shaft with n segments can be
expressed as
723
ELASTIC STRAIN ENERgy FOR
TORSIONAL dEFORMATION
U
=
n 2
∑
i=
1
Ti
Li
2J G
i
i
(17.18)
where T i is the internal force in segment i and L i , J i , and G i are, respectively, the length,
polar moment of inertia, and shear modulus of segment i.
For a nonprismatic shaft having a slightly tapered, variable cross section and a continuously
varying internal torque, the strain energy can be derived by integrating the strain
energy in a differential element dx over the total length of the shaft:
U
L
2
[ Tx ( )]
= ∫
2 JxG ( )
dx
0
(17.19)
Here, T(x) is the internal torque and J(x) is the polar moment of inertia of the crosssectional
area at a distance x from the origin of the shaft.
ExAmpLE 17.2
Three identical shafts of identical torsional rigidity JG and length L are subjected
to torques T as shown. What is the elastic strain energy stored in each shaft?
Plan the Solution
The elastic strain energy for cases (a) and (b) can be determined from Equation
(17.16). The strain energy for case (c) can be found from Equation (17.18).
(a)
A
T
SolutioN
From Equation (17.16), the strain energy for case (a) is
L
C
U
a
2
T L
= Ans.
2JG
(b)
T
In case (b), strain energy is created in only one-third of the shaft, from A
to B:
U
b
2 2
T ( L/3)
T L
= = Ans.
2JG
6JG
In case (c), the internal torque in segment AB is 2T and the internal
torque in segment BC is T. From Equation (17.18), the total strain energy in
shaft ABC is then
2 2 2 2
(2 T)( L/3)
T (2 L/3)
2T L T L
Uc
= + = +
2JG
2JG
3JG
3JG
Ans.
2
T L
=
JG
Notice that the sum of the strain energies for cases (a) and (b) does not equal the
strain energy for case (c); that is, U c ≠ U a + U b . The torque term in Equations (17.16) and
(17.18) is squared; thus, superposition is not valid for strain energies.
(c)
A
A
L
3
L
3
B
T
B
2L
3
2L
3
C
C
T