CONTINUUM MECHANICS for ENGINEERS

(6.7-8b)
(6.7-8c)
in which φ = φ (r,θ). To qualify as an Airy stress function φ must once again
satisfy the condition 4 φ = 0 which in polar form is obtained from Eq 6.7-6 as
For stress fields symmetrical to the polar axis Eq 6.7-9 reduces to
or
4
φ = ∂ ⎛
⎜
⎝ ∂
σ
τ
θ
rθ
2
∂ φ
= 2
∂r
∂φ
φ
φ
=
r ∂θ r r θ r r θ
−
2
1 1 ∂ ∂ ⎛ 1 ∂ ⎞
=− 2
∂∂ ∂
⎜
⎝ ∂
⎟
⎠
2
2 2
2
1 ∂ 1 φ 1 φ 1 φ
+ 2 2 2 2 2 2
∂ θ
θ
+
∂
∂
⎟ ∂
+
∂
∂
∂ +
∂
r r r r
⎜
r r r r ∂
4
r
4
φ = ∂ ⎛
⎜
⎝ ∂
⎞
⎟ 0
⎠
=
(6.7-9)
(6.7-10a)
(6.7-10b)
It may be shown that the general solution to this differential equation is
given by
φ = A ln r + Br 2 ln r + Cr 2 + D (6.7-11)
so that for the symmetrical case the stress components take the form
⎞ ⎛
⎠ ⎝
1 ∂ ⎞ φ 1 φ
⎟ 0
⎠
∂
+ ∂ ⎛ ⎞
⎟
⎝ ⎠
=
2
2
+ 2
r r ∂r
⎜ 2
∂r
r ∂r
4
3
2
∂ 3 ∂ 2 ∂
+ 2r−r+ r 0
4
3
2
∂r
∂r
∂r
r
∂
∂ =
φ φ φ φ
φ A
σ r = B r C
r r r
∂ 1
= + ( 1+ 2ln )+ 2
2
∂
σ
θ
2
∂ φ A
= =− + B( 3+ 2ln r 2 2 )+ 2C
∂r
r
τ r θ = 0
(6.7-12a)
(6.7-12b)
(6.7-12c)
When there is no hole at the origin in the elastic body under consideration,
A and B must be zero since otherwise infinite stresses would result at that