7.2 General Solution of the Torsion Problem
227
γ yz = θ
∂ψ
∂ y
+ x
Also, by Hooke’s law, the stress–strain relationships are given by
σ x = 2Gε x + λe, τ xy = Gγ xy
σ y = 2Gε y + λe, τ yz = Gγ yz
σ z = 2Gε 2 + λe, τ xz = Gγ xz
where e = ε x + ε y + ε z .
and
λ =
ν E
(1 + ν)(1 − 2ν)
.
Substituting (a), (b) and (c) in the above equations, we obtain
σ x = σ y = σ z = τ xy = 0
τ xz = G
∂w
∂ x
− yθ
= Gθ
∂ψ
∂ x
− y
(d)
τ yz = G
∂w
∂ y
+ xθ
= Gθ
∂ψ
∂ y
+ x
(e)
It can be observed that with the assumptions (a), (b) and (c) regarding deformation,
there will be no normal stresses acting between the longitudinal fibres of the shaft
or in the longitudinal direction of those fibres. Also, there will be no distortion in
the planes of cross-sections, since ε x , ε y and γ xy vanish. We have at each point, pure
shear defined by the components τ xz and τ yz .
However, the stress components should satisfy the equations of equilibrium given
by:
∂σ x
∂ x
+
∂τ xy
∂ y
+
∂τ xz
∂z
+ F x = 0
∂σ y
∂ y
+
∂τ xy
∂ x
+
∂τ yz
∂z
+ F y = 0
∂σ z
∂z
+
∂τ xz
∂ x
+
∂τ yz
∂ y
+ F z = 0
Assuming negligible body forces, and substituting the stress components into
equilibrium equations, we obtain
∂τ xz
∂z
= 0,
∂τ zy
∂z
= 0,
∂τ xz
∂ x
+
∂τ zy
∂ y
= 0
(7.3)