118
4 Stress–Strain Relations
giving all the stress components (or strain components) zero, for this case of zero
body and surface forces.
Therefore,
σ
x − σ
x
=
σ
y σ
y
=
σ
z − σ
z
= 0
and
(τ
xy − τ
xy ) = (τ
yz − τ
yz ) = (τ
zx − τ
zx ) = 0
This shows that the set σ
x , σ
y , σ
z , . . . .τ σ
zx is identical to the set
σ
x , σ
y , σ
z , . . . .τ
zx and there is one and only one solution for the elastic problem.
4.9 Numerical Examples
Example 4.1 The following are the principal stresses at a point in a stressed material.
Taking E = 210 kN/mm
2 and ν = 0.3, calculate the volumetric strain and Lame’s
constants.
σ x = 200 N/mm
2
, σ y = 150 N/mm
2
, σ z = 120 N/mm
2
Solution We have
ε x =
1
E
σ x − ν
σ y + σ z
=
1
210 × 10 3 [200 − 0.3(150 + 120)]
∴ ε x = 5.67 × 10
−4
ε y =
1
E
σ y − ν(σ z + σ x )
=
1
210 × 10 3 [150 − 0.3(120 + 200)]
∴ ε y = 2.57 × 10
−4
ε z =
1
E
σ z − ν
σ x + σ y
=
1
210 × 10 3 [120 − 0.3(200 + 150)]
∴ ε z = 7.14 × 10
−5
4 Stress–Strain Relations
giving all the stress components (or strain components) zero, for this case of zero
body and surface forces.
Therefore,
σ
x − σ
x
=
σ
y σ
y
=
σ
z − σ
z
= 0
and
(τ
xy − τ
xy ) = (τ
yz − τ
yz ) = (τ
zx − τ
zx ) = 0
This shows that the set σ
x , σ
y , σ
z , . . . .τ σ
zx is identical to the set
σ
x , σ
y , σ
z , . . . .τ
zx and there is one and only one solution for the elastic problem.
4.9 Numerical Examples
Example 4.1 The following are the principal stresses at a point in a stressed material.
Taking E = 210 kN/mm
2 and ν = 0.3, calculate the volumetric strain and Lame’s
constants.
σ x = 200 N/mm
2
, σ y = 150 N/mm
2
, σ z = 120 N/mm
2
Solution We have
ε x =
1
E
σ x − ν
σ y + σ z
=
1
210 × 10 3 [200 − 0.3(150 + 120)]
∴ ε x = 5.67 × 10
−4
ε y =
1
E
σ y − ν(σ z + σ x )
=
1
210 × 10 3 [150 − 0.3(120 + 200)]
∴ ε y = 2.57 × 10
−4
ε z =
1
E
σ z − ν
σ x + σ y
=
1
210 × 10 3 [120 − 0.3(200 + 150)]
∴ ε z = 7.14 × 10
−5