2.23 Numerical Examples
47
Differentiating the stress components with respective axes, we get
∂σ x
∂ x
= 3y
2 z + 2,
∂τ xy
∂ y
= 0,
∂τ xz
∂z
= 3x y
2
Substituting in (a), 3y
2 z + 2 + 0 + 3x y
2
At point (1, −1, 2), we get 3 × 1 × 2 + 2 + 3 × 1 × 1 = 11 which is not equal to
zero.
Similarly,
∂σ y
∂ y
= 5xz + 3,
∂τ yz
∂z
= 3x y
2
+ 0
∴ (ii) becomes 0 + 5xz + 3 + 3x y
2
At point (1, −1, 2), we get 5 × 1 × 2 + 3 + 3 × 1 × 1 = 16 which is not equal to
zero.
And
∂σ z
∂z
= y
2
,
∂τ yz
∂ y
= 6x yz + 2x,
∂τ xz
∂ x
= 3y
2 z + 2y
Therefore, (iii) becomes 3y
2 z + 2y + 6x yz + 2x + y
2
At the point (1, −1, 2), we get 3 × 1 × 2 + 2 × (−1) + 6 × 1 × (−1) × 2 + 2 ×
1 + (−1)
2
= −5 which is not equal to zero.
Hence, the given stress components do not satisfy the equilibrium equations.
Recalling (a), (b) and (c) with body forces, the equations can be modified as below.
∂σ x
∂ x
+
∂τ xy
∂ y
+
∂τ xz
∂z
+ F x = 0
( d )
∂τ xy
∂ x
+
∂σ y
∂ y
+
∂τ yz
∂z
+ F y = 0
( e )
∂τ xz
∂ x
+
∂τ yz
∂ y
+
∂σ z
∂z
+ F z = 0
( f )
where F x , F y and F z are the body forces.
Substituting the values in (d), (e) and (f), we get body forces so that the stress
components become under equilibrium.
Therefore,
3 × 1 × 2 + 2 + 3 × 1 × 1 + F x = 0
∴ F x = −11
Also,