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

before the strain measurement. From Equation (2.6), the strain caused by the temperature
change in the aluminum bar is
ε
T
−6
= α∆ T = (22.5 × 10 /C)( ° − 30° C) = −0.000675 mm/mm
Hence, the strain caused by normal stress in member (1) is
ε = ε + ε
total
∴ εσ = εtotal
− εT
= 0.002400 mm/mm − ( −0.000675 mm/mm)
= 0.003075 mm/mm
σ
Using this strain value, we can now compute the normal stress in member (1) from
Hooke’s law:
σ1 = E ε = (70 GPa)(0.003075 mm/mm) = 215.25 MPa = 215 MPa Ans.
(b) The axial force in member (1) is computed from the normal stress and the bar area:
2
F = σ A = (215.25 N/mm )(100 mm)(6 mm) = 129,150 N
1 1 1
Now write an equilibrium equation for the sum of moments about joint A, and solve
for load P:
Σ M = (1.5 m)(129,150 N) − (2.5 m) P = 0
A
T
∴ P = 77,490 N = 77.5 kN Ans.
(c) The change in the width of the bar is computed by multiplying the lateral (i.e., transverse)
strain ε lat by the 100 mm initial width. To determine ε lat , the definition of
Poisson’s ratio [Equation (3.6)] is used:
εlat
ν =− ∴ εlat
= −νε
ε
long
Using the given value of Poisson’s ratio and the measured strain, we could calculate
ε lat as
ε =− νε =− (0.33)(2,400 µε ) =− 792 µε
lat
long
This calculation is not correct for the lateral strain in member (1). Why is it
incorrect?
The Poisson effect applies only to strains caused by stresses (i.e., mechanical effects).
When they are unrestrained, homogeneous, isotropic materials expand uniformly in
all directions as they are heated (and contract uniformly as they cool). Consequently,
thermal strains should not be included in the calculation of Poisson’s ratio. For this
problem, the lateral strain should be calculated as
ε lat =− (0.33)(0.003075 mm/mm) + ( − 0.000675 mm/mm) =−0.0016898 mm/mm
The change in the width of the aluminum bar is, therefore,
δ width = ( − 0.0016898 mm/mm)(100 mm) =−0.1690 mm Ans.
long
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