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2 Analysis of Stress
Mean stress = σ m =
1
3
σ x + σ y + σ z
=
1
3
(40 + 30 + 20)
= 30 k Pa
Deviator stress tensor =
⎡
⎣
(σ x − σ m ) τ xy
τ xz
τ xy
σ y − σ m
τ yz
τ xz
τ yz (σ z − σ m )
⎤
⎦
=
⎡
⎣
(40 − 30)
20
30
20
(30 − 30)
40
30
40
(20 − 30)
⎤
⎦
=
⎡
⎣
10 20 30
20 0 40
30 40 −10
⎤
⎦ kPa
Spherical stress tensor =
⎡
⎣
σ m 0 0
0 σ m 0
0 0 σ m
⎤
⎦
=
⎡
⎣
30 0 0
0 30 0
0 0 30
⎤
⎦ kPa
Example 2.11 The stress components at a point in a body are given by
σ x = 3x y
2 z + 2x, τ xy = 0
σ y = 5x yz + 3y τ yz = τ xz = 3x y
2 z + 2x y
σ z = x
2 y + y
2 z
Determine whether these components of stress satisfy the equilibrium equations
or not as the point (1, −1, 2). If not, then determine the suitable body force required
at this point so that these stress components become under equilibrium.
Solution The equations of equilibrium are given by
∂σ x
∂ x
+
∂τ xy
∂ y
+
∂τ xz
∂z
= 0
( a )
∂τ xy
∂ x
+
∂σ y
∂ y
+
∂τ yz
∂z
= 0
( b )
∂τ xz
∂ x
+
∂τ yz
∂ y
+
∂σ z
∂z
= 0
( c )