232
7 Torsion of Prismatic Bars
As φ = constant at the boundary and x 1 , x 2 , y 1 and y 2 represent points on the
lateral surface, then
M t =
¨
φdxdy +
¨
φdxdy = 2
¨
φdxdy
(7.15)
Hence, it is observed that each of the integrals in Eq. (7.15) contributing one half
of the torque due to τ xz and the other half due to τ yz .
Thus all the differential equations and boundary conditions are satisfied if the
stress function φ, obeys Eq. (7.8) and is used to obtain M t , and the solution obtained
in this manner is the exact solution of the torsion problem.
7.5 Torsion of Circular Cross-Section
The Laplace equation is given by
∂
2
ψ
∂ x 2 +
∂
2
ψ
∂ y 2 = 0
where ψ = warping function.
The simplest solution to the above equation is
ψ = constant = C
But the boundary condition is given by Eq. (7.6) is
∂ψ
∂ x
− y
dy
dS
−
∂ψ
∂ y
+ x
dx
dS
= 0
Therefore, with ψ = C, the above boundary condition becomes
(0 − y)(dy/dS) − (0 + x)(dx/dS) = 0
−y
dy
dS
− x
dx
dS
= 0
or
d
dS
x
2
+ y
2
2
= 0
7 Torsion of Prismatic Bars
As φ = constant at the boundary and x 1 , x 2 , y 1 and y 2 represent points on the
lateral surface, then
M t =
¨
φdxdy +
¨
φdxdy = 2
¨
φdxdy
(7.15)
Hence, it is observed that each of the integrals in Eq. (7.15) contributing one half
of the torque due to τ xz and the other half due to τ yz .
Thus all the differential equations and boundary conditions are satisfied if the
stress function φ, obeys Eq. (7.8) and is used to obtain M t , and the solution obtained
in this manner is the exact solution of the torsion problem.
7.5 Torsion of Circular Cross-Section
The Laplace equation is given by
∂
2
ψ
∂ x 2 +
∂
2
ψ
∂ y 2 = 0
where ψ = warping function.
The simplest solution to the above equation is
ψ = constant = C
But the boundary condition is given by Eq. (7.6) is
∂ψ
∂ x
− y
dy
dS
−
∂ψ
∂ y
+ x
dx
dS
= 0
Therefore, with ψ = C, the above boundary condition becomes
(0 − y)(dy/dS) − (0 + x)(dx/dS) = 0
−y
dy
dS
− x
dx
dS
= 0
or
d
dS
x
2
+ y
2
2
= 0