32
2 Analysis of Stress
sin
dθ
2
=
dθ
2
and cos
dθ
2
= 1
Neglecting higher order terms and simplifying, we get
r
∂σ r
∂r
dr dθ + σ r dr dθ − σ θ dr dθ +
∂τ r θ
∂θ
dr dθ = 0
on dividing throughout by rdθ dr, we have
∂σ r
∂r
+
1
r
∂τ r θ
∂θ
+
σ r − σ θ
r
+ F r = 0
(2.45)
Similarly resolving all the forces in the θ-direction at right angles to r-direction,
we have
− σ θ dr cos
dθ
2
+
σ θ +
∂σ θ
∂θ
dθ
dr cos
dθ
2
+ τ r θ dr sin
dθ
2
+
τ r θ +
∂τ r θ
∂θ
dθ
dr
sin
dθ
2
− τ r θ r dθ + (r + dr ) dθ
τ r θ +
∂τ r θ
∂r
dr
+ F θ = 0
On simplification, we get
∂σ θ
∂θ
+ τ r θ + τ r θ + r
∂τ r θ
∂r
dθ dr = 0
Dividing throughout by rdθ dr, we get
1
r
·
∂σ θ
∂θ
+
∂τ r θ
dr
+
2τ r θ
r
+ F θ = 0
(2.46)
In the absence of body forces, the equilibrium equations can be represented as:
∂σ r
∂r
+
1
r
∂τ r θ
∂θ
+
σ r − σ θ
r
= 0
1
r
∂σ θ
∂θ
+
∂τ r θ
∂r
+
2τ r θ
r
= 0
(2.47)
2.22 General State of Stress in Three Dimensions
in Cylindrical Co-ordinate System
See Fig. 2.17.
The equilibrium equations for three-dimensional state are given by