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

y
103.7 MPa
40.2 MPa
18.5°
x
63.5 MPa
143.9 MPa
An appropriate sketch of the in-plane principal stresses, the maximum in-plane shear
stress, and the orientation of these planes is shown.
(d) Absolute Maximum Shear Stress
The absolute maximum shear stress τ abs max can be calculated from the absolute maximum
shear strain:
τ
absmax
− 6
= G γ = (44,000 MPa)(1,636 × 10 ) = 72.0 MPa Ans.
absmax
Alternatively, τ abs max can be calculated from the principal stresses if we note that σ p3 =
σ z = 0 on the free surface of the copper alloy component:
τ
absmax
σ
=
− σ
p1 p3
2
143.9 − 0
= = 72.0 MPa
Ans.
2
574
Including Temperature Effects in the Generalized Hooke’s Law
Temperature can be included in the three-dimensional stress–strain relations as follows:
εx 1
= [ σx E
− νσ ( y + σz
)] + α∆T
1
εy = [ σ y
E
− νσ ( x + σz
)] + α∆T
(13.28)
1
εz = [ σz E
− νσ ( x + σ y )] + α∆T
Here, α is the coefficient of thermal expansion and ∆T is the change in temperature. Note
that temperature change does not have any effect on shear stresses or shear strains.
In terms of strains, the three-dimensional stress–strain relations that include temperature
effects can be written as follows:
E
E
σ x =
T
(1 )(1 2 ) [(1 − νε ) x + νε ( x + εz
)] −
ν ν
1 2ν α ∆
+ −
−
E
E
σ y =
T
(1 )(1 2 ) [(1 − νε ) y + νε ( x + εz
)] −
ν ν
1 2ν α ∆
+ −
−
E
E
σ z =
T
(1 )(1 2 ) [(1 − νε ) z + νε ( x + ε y
ν ν
)] −
+ −
1 2ν α ∆ −
(13.29)
For the case of plane stress, the generalized Hooke’s law relationships that include temperature
can be expressed in terms of stresses as follows:
1
εx = ( σ x − νσ y ) + α ∆T
E
1
εy = ( σ y − νσ x ) + α ∆T
E
v
εz =− ( σ x + σ y)
+ α∆T
E
(13.30)
In terms of strains, the Hooke’s law relationships can be expressed as the following two
equations:
E
E
σ x = ( εx + νε y)
− T
2
1 ν
1 ν α ∆
− − (13.31)
E
E
σ y = ( εy + νε x ) − T
2
1 − ν
1 ν α ∆ −