7.5 Torsion of Circular Cross-Section
233
i.e.
x
2
+ y
2
= constant
where (x, y) are the co-ordinates of any point on the boundary. Hence, the boundary
is a circle.
From Equation (c), we can write
w = θψ(x, y)
i.e.
w = θC.
The polar moment of inertia for the section is
J =
(x
2
+ y
2
)dxdy = I P
But
M t = G I P θ
or
θ =
M t
G I P
.
Therefore,
w =
M t C
G I P
which is a constant. Since the fixed end has zero w at least at one point, w is zero at
every cross-section (other than the rigid body displacement). Thus the cross-section
does not warp.
Further, the shear stresses are given by the Equations (d) and (e) as
τ xz = G
∂w
∂ x
− yθ
= Gθ
∂ψ
∂ x
− y
τ yz = G
∂w
∂ y
+ xθ
= Gθ
∂ψ
∂ y
+ x
233
i.e.
x
2
+ y
2
= constant
where (x, y) are the co-ordinates of any point on the boundary. Hence, the boundary
is a circle.
From Equation (c), we can write
w = θψ(x, y)
i.e.
w = θC.
The polar moment of inertia for the section is
J =
(x
2
+ y
2
)dxdy = I P
But
M t = G I P θ
or
θ =
M t
G I P
.
Therefore,
w =
M t C
G I P
which is a constant. Since the fixed end has zero w at least at one point, w is zero at
every cross-section (other than the rigid body displacement). Thus the cross-section
does not warp.
Further, the shear stresses are given by the Equations (d) and (e) as
τ xz = G
∂w
∂ x
− yθ
= Gθ
∂ψ
∂ x
− y
τ yz = G
∂w
∂ y
+ xθ
= Gθ
∂ψ
∂ y
+ x