Example 2 : Elliptical cylinder
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Choose warping function
ψ ( x 1 , x 2 ) = k x 1 x 2 {\displaystyle \psi (x_{1},x_{2})=kx_{1}x_{2}\,} where k {\displaystyle k\,} is a constant.
Equilibrium (∇ 2 ψ = 0 {\displaystyle \nabla ^{2}{\psi }=0} ) is satisfied.
The traction free BC is
( k x 2 − x 2 ) d x 2 d s − ( k x 1 + x 1 ) d x 1 d s = 0 ∀ ( x 1 , x 2 ) ∈ ∂ S {\displaystyle (kx_{2}-x_{2}){\frac {dx_{2}}{ds}}-(kx_{1}+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}} Integrating,
x 1 2 + 1 − k 1 + k x 2 2 = a 2 ∀ ( x 1 , x 2 ) ∈ ∂ S {\displaystyle x_{1}^{2}+{\frac {1-k}{1+k}}x_{2}^{2}=a^{2}~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}} where a {\displaystyle a\,} is a constant.
This is the equation for an ellipse with major and minor axes a {\displaystyle a\,} and b {\displaystyle b\,} ,
where
b 2 = ( 1 + k 1 − k ) a 2 {\displaystyle b^{2}=\left({\frac {1+k}{1-k}}\right)a^{2}} The warping function is
ψ = − ( a 2 − b 2 a 2 + b 2 ) x 1 x 2 {\displaystyle \psi =-\left({\frac {a^{2}-b^{2}}{a^{2}+b^{2}}}\right)x_{1}x_{2}} The torsion constant is
J ~ = 2 b 2 a 2 + b 2 I 2 + 2 a 2 a 2 + b 2 I 1 = π a 3 b 3 a 2 + b 2 {\displaystyle {\tilde {J}}={\frac {2b^{2}}{a^{2}+b^{2}}}I_{2}+{\frac {2a^{2}}{a^{2}+b^{2}}}I_{1}={\frac {\pi a^{3}b^{3}}{a^{2}+b^{2}}}} where
I 1 = ∫ S x 1 2 d A = π a b 3 4 ; I 2 = ∫ S x 2 2 d A = π a 3 b 4 {\displaystyle I_{1}=\int _{S}x_{1}^{2}dA={\frac {\pi ab^{3}}{4}}~~;~~I_{2}=\int _{S}x_{2}^{2}dA={\frac {\pi a^{3}b}{4}}} If you compare J ~ {\displaystyle {\tilde {J}}} and J {\displaystyle J\,} for the ellipse, you will find that J ~ < J {\displaystyle {\tilde {J}}<J} . This implies that the torsional rigidity is less than that predicted with the assumption that plane sections remain plane.
The twist per unit length is
α = ( a 2 + b 2 ) T μ π a 3 b 3 {\displaystyle \alpha ={\frac {(a^{2}+b^{2})T}{\mu \pi a^{3}b^{3}}}} The non-zero stresses are
σ 13 = − 2 μ α a 2 x 2 a 2 + b 2 ; σ 23 = − 2 μ α b 2 x 1 a 2 + b 2 {\displaystyle \sigma _{13}=-{\frac {2\mu \alpha a^{2}x_{2}}{a^{2}+b^{2}}}~~;~~\sigma _{23}=-{\frac {2\mu \alpha b^{2}x_{1}}{a^{2}+b^{2}}}} The projected shear traction is
τ = 2 μ α a 2 + b 2 b 4 x 1 2 + a 4 x 2 2 ⇒ τ max = 2 μ α a 2 b a 2 + b 2 ( b < a ) {\displaystyle \tau ={\frac {2\mu \alpha }{a^{2}+b^{2}}}{\sqrt {b^{4}x_{1}^{2}+a^{4}x_{2}^{2}}}~~\Rightarrow ~~\tau _{\text{max}}={\frac {2\mu \alpha a^{2}b}{a^{2}+b^{2}}}~~(b<a)} Shear stresses in the cross section of an elliptical cylinder under torsion
For any torsion problem where ∂ S {\displaystyle \partial S\,} is convex, the maximum projected shear traction occurs at the point on ∂ S {\displaystyle \partial S\,} that is nearest the centroid of S {\displaystyle S\,} .
The displacement u 3 {\displaystyle u_{3}\,} is
u 3 = − ( a 2 − b 2 ) T x 1 x 2 μ π a 3 b 3 {\displaystyle u_{3}=-{\frac {(a^{2}-b^{2})Tx_{1}x_{2}}{\mu \pi a^{3}b^{3}}}} Displacements (u 3 {\displaystyle u_{3}\,} ) in the cross section of an elliptical cylinder under torsion