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

ExAmpLE 13.10
A 2014-T6 aluminum alloy [E = 73 GPa; ν = 0.33; α = 23.0 × 10 −6 /°C]
block with dimensions of 250 mm by 250 mm by 90 mm is placed in
the smooth rigid die fixture shown. Deformation of the block in the y
and z directions is unrestricted; however, the deformation of the block
is restricted to 0.20 mm in the x direction. The block is subjected to a
uniformly distributed compressive stress of σ y = −360 MPa, and the
temperature is increased by 45°C. Determine
(a) The normal stress σ x .
(b) The strains in the block.
(c) The change in volume of the block.
Plan the Solution
We know that the stress applied to the block in the y direction is σ y = −360 MPa. Since the
block is unrestrained in the z direction, we also know that the stress in the z direction is
σ z = 0, making this situation a plane stress problem. The first question to explore is
whether the combination of the Poisson effect and the thermal expansion will cause the
block to contact the rigid die in the x direction. If so, then we will know the value of the
normal strain in the x direction.
SolutioN
(a) Normal Stress σ x
Let us assume that the combination of the Poisson effect and the thermal expansion will
cause the block to expand so that it contacts the rigid die in the x direction. If so, then the
normal strain in the x direction will be
0.20 mm
−6
e x = = 800 × 10 mm/mm
250 mm
We calculate σ x from Equation (13.30), where e x is assumed to be 800 × 10 −6 mm/mm:
z
y
Rigid
die
x
0.20 mm
360 MPa
250 mm
250 mm
90 mm
1
ex = ( σ x − νσ y ) + α ∆T
E
∴ σ = E( e −α∆ T)
+ νσ
x x y
3 −6 −6
= (73× 10 MPa)[800 × 10 − (23× 10 / ° C)(45° C)] + (0.33)( −360 MPa)
=−135.96 MPa
The fact that σ x is a compressive stress justifies our assumption that the block will contact
the rigid die in the x direction.
(b) Strains in the Block
All three stresses are now known:
σ =− 135.96 MPa σ =− 360 MPa σ = 0 MPa
x y z
Ans.
We have determined that the normal strain in the x direction is e x = 800 × 10 −6 mm/mm.
Next, we will calculate e y and e z from Equations (13.28):
1
ey = [ σ y − v ( σ x + σ z )] + α ∆T
E
1
−6
=
[ −360 MPa − (0.33)( − 135.96 MPa + 0)] + (23× 10 / ° C)(45°
C)
3
73 × 10 MPa
−6
=− 3,282 × 10 mm/mm
Ans.
575