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

(b) The lateral strain is
0.004 in.
ε lat = − =−0.002025 in./in.
1.975 in.
From Equation (3.6), Poisson’s ratio can be computed as
εlat
0.002025 in./in.
ν =− =− − = 0.352
Ans.
ε 0.005750 in./in.
long
(c) The shear modulus is then computed from Equation (3.7) as
G =
E 1, 408,916 psi
=
2(1 + ν)
2(1 + 0.352)
= 521,049 psi = 521,000 psi Ans.
ExAmpLE 3.2
1.3 m
(1)
100 mm
Strain
gage
A B C
1.5 m 1.0 m
P
Rigid bar ABC is supported by a pin at A and a 100 mm wide
by 6 mm thick aluminum [E = 70 GPa; α = 22.5 × 10 −6 /°C;
ν = 0.33] alloy bar at B. A strain gage affixed to the surface of
the aluminum bar is used to measure its longitudinal strain.
Before load P is applied to the rigid bar at C, the strain gage
measures zero longitudinal strain at an ambient temperature
of 20°C. After load P is applied to the rigid bar at C and the
temperature drops to −10°C, a longitudinal strain of 2,400 µε
is measured in the aluminum bar. Determine
(a) the stress in member (1).
(b) the magnitude of load P.
(c) the change in the width of the aluminum bar (i.e., the
100 mm dimension).
Plan the Solution
This problem illustrates some misconceptions that are common in applying Hooke’s law
and Poisson’s ratio, particularly when temperature change is a factor in the analysis.
SOLUTION
(a) Since the elastic modulus E and the longitudinal strain ε are given in the problem,
one might be tempted to compute the normal stress in aluminum bar (1) from
Hooke’s law [Equation (3.4)]:
σ
= ε = µε ⎡ 1, 000 MPa
⎣ ⎢ ⎤ ⎡ 1 mm/mm ⎤
E (70 GPa)(2,400 )
1GPa
⎢
⎦⎥
⎣1, 000,000 µε
⎥
⎦
= 168 MPa
1 1 1
This calculation is not correct for the normal stress in member (1). Why is it incorrect?
From Equation (2.7), the total strain ε total in an object includes a portion ε σ due to
stress and a portion ε T due to temperature change. The strain gage affixed to member
(1) has measured the total strain in the aluminum bar as ε total = 2,400 µε = 0.002400
mm/mm. In this problem, however, the temperature of member (1) has dropped 30°C
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