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

or in terms of the deformation δ as
U
AEδ = 2
(17.13)
2L
721
ELASTIC STRAIN ENERgy FOR
AxIAL dEFORMATION
The total strain energy of a bar that consists of several segments (each having constant
force, area, and elastic modulus) is equal to the sum of the strain energies in all the segments.
For example, the strain energy in the multisegment bar shown in Figure 17.11 is
equal to the sum of the strain energies in segment AB and segment BC. In general terms,
the strain energy of a bar with n segments can be expressed as
U =
n 2
∑
i=
1
Fi
Li
2AE
i
i
(17.14)
where F i is the internal force in segment i and L i , A i , and E i are, respectively, the length,
area, and elastic modulus of segment i.
For a nonprismatic bar having a slightly tapered, variable cross section and a continuously
varying axial force (Figure 17.12), then the total strain energy can be derived by integrating
the strain energy in a differential element dx over the total length of the bar:
L
2
[ Fx ( )]
U = ∫
2 AxE ( )
dx
0
(17.15)
Here, F(x) is the internal force and A(x) is the cross-sectional area at a distance x from the
origin of the bar.
A
B
P 1
C
FIGURE 17.11 Bar with
multiple prismatic segments.
A
x
L
dx
A(x)
B
P 2
F(x)
FIGURE 17.12 Nonprismatic
bar subjected to varying axial
loading.
ExAmpLE 17.1
Segmented rod ABC is made of a brass that has a yield strength σ Y = 124 MPa
and a modulus of elasticity E = 115 GPa. The diameter of segment (1) is 25 mm,
and the diameter of segment (2) is 15 mm. For the loading shown, determine the
maximum strain energy that can be absorbed by the rod if no permanent deformation
is caused.
Plan the Solution
The maximum force that can be applied to the segmented rod will be dictated by
the capacity of segment (2). From the yield strength and the cross-sectional area
of segment (2), determine the maximum force P. The internal force in each segment will
equal the external load. The strain energy in each segment can be calculated from the internal
force along with the length, area, and elastic modulus of each segment. The total
strain energy U is simply the sum of the strain energies in segments (1) and (2), as indicated
by Equation (17.14).
A
(1)
900 mm
B
(2)
1,100 mm
C
P
SolutioN
Compute the cross-sectional areas of segments (1) and (2):
A
A
π π
= d = (25 mm)
4 4
= 490.874 mm
π π
= d = (15mm)
4 4
= 176.715 mm
1 1 2 2 2
2 2 2 2 2