Warping Function and Torsion of Non-Circular Cylinders
edit
The displacements are given by
u
1
=
−
α
x
2
x
3
;
u
2
=
α
x
1
x
3
;
u
3
=
α
ψ
(
x
1
,
x
2
)
{\displaystyle u_{1}=-\alpha x_{2}x_{3}~;~~u_{2}=\alpha x_{1}x_{3}~;~~u_{3}=\alpha \psi (x_{1},x_{2})}
where
α
{\displaystyle \alpha }
ψ
{\displaystyle \psi }
σ
13
=
μ
α
(
ψ
,
1
−
x
2
)
;
σ
23
=
μ
α
(
ψ
,
2
+
x
1
)
{\displaystyle \sigma _{13}=\mu \alpha (\psi _{,1}-x_{2})~;~~\sigma _{23}=\mu \alpha (\psi _{,2}+x_{1})}
where
μ
{\displaystyle \mu }
τ
=
(
σ
13
2
+
σ
23
2
)
{\displaystyle \tau ={\sqrt {(\sigma _{13}^{2}+\sigma _{23}^{2})}}}
Equilibrium is satisfied if
∇
2
ψ
=
0
∀
(
x
1
,
x
2
)
∈
S
{\displaystyle \nabla ^{2}{\psi }=0~~~~\forall (x_{1},x_{2})\in {\text{S}}}
Traction-free lateral BCs are satisfied if
(
ψ
,
1
−
x
2
)
d
x
2
d
s
−
(
ψ
,
2
+
x
1
)
d
x
1
d
s
=
0
∀
(
x
1
,
x
2
)
∈
∂
S
{\displaystyle (\psi _{,1}-x_{2}){\frac {dx_{2}}{ds}}-(\psi _{,2}+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}
or,
(
ψ
,
1
−
x
2
)
n
^
1
+
(
ψ
,
2
+
x
1
)
n
^
2
=
0
∀
(
x
1
,
x
2
)
∈
∂
S
{\displaystyle (\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}
The twist per unit length is given by
α
=
T
μ
J
~
{\displaystyle \alpha ={\frac {T}{\mu {\tilde {J}}}}}
where the torsion constant
J
~
=
∫
S
(
x
1
2
+
x
2
2
+
x
1
ψ
,
2
−
x
2
ψ
,
1
)
d
A
{\displaystyle {\tilde {J}}=\int _{S}(x_{1}^{2}+x_{2}^{2}+x_{1}\psi _{,2}-x_{2}\psi _{,1})dA}
Example 1: Circular Cylinder
edit
Choose warping function
ψ
(
x
1
,
x
2
)
=
0
{\displaystyle \psi (x_{1},x_{2})=0}
Equilibrium (
∇
2
ψ
=
0
{\displaystyle \nabla ^{2}{\psi }=0}
(
0
−
x
2
)
d
x
2
d
s
−
(
0
+
x
1
)
d
x
1
d
s
=
0
∀
(
x
1
,
x
2
)
∈
∂
S
{\displaystyle (0-x_{2}){\frac {dx_{2}}{ds}}-(0+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}
Integrating,
x
2
2
+
x
1
2
=
c
2
∀
(
x
1
,
x
2
)
∈
∂
S
{\displaystyle x_{2}^{2}+x_{1}^{2}=c^{2}~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}
where
c
{\displaystyle c}
There is no warping of cross sections for circular cylinders
J
~
=
∫
S
(
x
1
2
+
x
2
2
)
d
A
=
∫
S
r
2
d
A
=
J
{\displaystyle {\tilde {J}}=\int _{S}(x_{1}^{2}+x_{2}^{2})dA=\int _{S}r^{2}dA=J}
The twist per unit length is
α
=
T
μ
J
{\displaystyle \alpha ={\frac {T}{\mu J}}}
The non-zero stresses are
σ
13
=
−
μ
α
x
2
;
σ
23
=
μ
α
x
1
{\displaystyle \sigma _{13}=-\mu \alpha x_{2}~;~~\sigma _{23}=\mu \alpha x_{1}}
The projected shear traction is
τ
=
μ
α
(
x
1
2
+
x
2
2
)
=
μ
α
r
{\displaystyle \tau =\mu \alpha {\sqrt {(x_{1}^{2}+x_{2}^{2})}}=\mu \alpha r}
Compare results from Mechanics of Materials solution
ϕ
=
T
L
G
J
⇒
α
=
T
G
J
{\displaystyle \phi ={\frac {TL}{GJ}}~~\Rightarrow ~~\alpha ={\frac {T}{GJ}}}
and
τ
=
T
r
J
⇒
τ
=
G
α
r
{\displaystyle \tau ={\frac {Tr}{J}}~~\Rightarrow ~~\tau =G\alpha r}
Example 2: Elliptical Cylinder
edit
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}
∇
2
ψ
=
0
{\displaystyle \nabla ^{2}{\psi }=0}
(
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}
a
{\displaystyle a}
b
{\displaystyle b}
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}}}
J
{\displaystyle J}
J
~
<
J
{\displaystyle {\tilde {J}}<J}
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
∂
{\displaystyle \partial }
∂
{\displaystyle \partial }
u
3
{\displaystyle u_{3}}
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}}
Example 3: Rectangular Cylinder
edit
In this case, the form of
ψ
{\displaystyle \psi }
(
ψ
,
1
−
x
2
)
n
^
1
+
(
ψ
,
2
+
x
1
)
n
^
2
=
0
∀
(
x
1
,
x
2
)
∈
∂
S
{\displaystyle (\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}}
Suppose that
2
a
{\displaystyle 2a}
2
b
{\displaystyle 2b}
a
>
b
{\displaystyle a>b}
a
{\displaystyle a}
x
1
{\displaystyle x_{1}}
b
{\displaystyle b}
x
2
{\displaystyle x_{2}}
ψ
,
1
=
x
2
on
x
1
=
±
a
,
and
ψ
,
2
=
−
x
1
on
x
2
=
±
b
{\displaystyle \psi _{,1}=x_{2}~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~\psi _{,2}=-x_{1}~~{\text{on}}~~x_{2}=\pm b}
A suitable
ψ
{\displaystyle \psi }
∇
2
ψ
=
0
{\displaystyle \nabla ^{2}{\psi }=0}
ψ
¯
=
x
1
x
2
−
ψ
{\displaystyle {\bar {\psi }}=x_{1}x_{2}-\psi }
Then the equilibrium condition becomes
∇
2
ψ
¯
=
0
{\displaystyle \nabla ^{2}{\bar {\psi }}=0}
The traction-free BCs become
ψ
¯
,
1
=
0
on
x
1
=
±
a
,
and
ψ
¯
,
2
=
2
x
1
on
x
2
=
±
b
{\displaystyle {\bar {\psi }}_{,1}=0~~{\text{on}}~~x_{1}=\pm a~~,{\text{and}}~~{\bar {\psi }}_{,2}=2x_{1}~~{\text{on}}~~x_{2}=\pm b}
Let us assume that
ψ
¯
(
x
1
,
x
2
)
=
f
(
x
1
)
g
(
x
2
)
{\displaystyle {\bar {\psi }}(x_{1},x_{2})=f(x_{1})g(x_{2})}
Then,
∇
2
ψ
¯
=
ψ
¯
,
11
+
ψ
¯
,
22
=
f
″
(
x
1
)
g
(
x
2
)
+
g
″
(
x
2
)
f
(
x
1
)
=
0
{\displaystyle \nabla ^{2}{\bar {\psi }}={\bar {\psi }}_{,11}+{\bar {\psi }}_{,22}=f^{''}(x_{1})g(x_{2})+g^{''}(x_{2})f(x_{1})=0}
or,
f
″
(
x
1
)
f
(
x
1
)
=
−
g
″
(
x
2
)
g
(
x
2
)
=
η
{\displaystyle {\frac {f^{''}(x_{1})}{f(x_{1})}}=-{\frac {g^{''}(x_{2})}{g(x_{2})}}=\eta }
edit
In both these cases, we get trivial values of
C
1
=
C
2
=
0
{\displaystyle C_{1}=C_{2}=0}
Let
η
=
−
k
2
;
k
>
0
{\displaystyle \eta =-k^{2}~~~;~~k>0}
Then,
f
″
(
x
1
)
+
k
2
f
(
x
1
)
=
0
⇒
f
(
x
1
)
=
C
1
cos
(
k
x
1
)
+
C
2
sin
(
k
x
1
)
g
″
(
x
2
)
−
k
2
g
(
x
2
)
=
0
⇒
g
(
x
2
)
=
C
3
cosh
(
k
x
2
)
+
C
4
sinh
(
k
x
2
)
{\displaystyle {\begin{aligned}f^{''}(x_{1})+k^{2}f(x_{1})=0~~\Rightarrow &~~f(x_{1})=C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\\g^{''}(x_{2})-k^{2}g(x_{2})=0~~\Rightarrow &~~g(x_{2})=C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\end{aligned}}}
Therefore,
ψ
¯
(
x
1
,
x
2
)
=
[
C
1
cos
(
k
x
1
)
+
C
2
sin
(
k
x
1
)
]
[
C
3
cosh
(
k
x
2
)
+
C
4
sinh
(
k
x
2
)
]
{\displaystyle {\bar {\psi }}(x_{1},x_{2})=\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\right]}
Apply the BCs at
x
2
=
±
b
{\displaystyle x_{2}=\pm b}
ψ
¯
,
2
=
2
x
1
{\displaystyle {\bar {\psi }}_{,2}=2x_{1}}
[
C
1
cos
(
k
x
1
)
+
C
2
sin
(
k
x
1
)
]
[
C
3
sinh
(
k
b
)
+
C
4
cosh
(
k
b
)
]
=
2
x
1
[
C
1
cos
(
k
x
1
)
+
C
2
sin
(
k
x
1
)
]
[
−
C
3
sinh
(
k
b
)
+
C
4
cosh
(
k
b
)
]
=
2
x
1
{\displaystyle {\begin{aligned}\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\\\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[-C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\end{aligned}}}
or,
F
(
x
1
)
G
′
(
b
)
=
2
x
1
;
F
(
x
1
)
G
′
(
−
b
)
=
2
x
1
{\displaystyle F(x_{1})G^{'}(b)=2x_{1}~~;~~~F(x_{1})G^{'}(-b)=2x_{1}}
The RHS of both equations are odd. Therefore,
F
(
x
1
)
{\displaystyle F(x_{1})}
cos
(
k
x
1
)
{\displaystyle \cos(kx_{1})}
C
1
=
0
{\displaystyle C_{1}=0}
F
(
x
1
)
[
G
′
(
b
)
−
G
′
(
−
b
)
]
=
0
{\displaystyle F(x_{1})\left[G^{'}(b)-G^{'}(-b)\right]=0}
Hence,
G
′
(
b
)
{\displaystyle G'(b)}
sinh
(
k
b
)
{\displaystyle \sinh(kb)}
C
3
=
0
{\displaystyle C_{3}=0}
ψ
¯
(
x
1
,
x
2
)
=
C
2
C
4
sin
(
k
x
1
)
sinh
(
k
x
2
)
=
A
sin
(
k
x
1
)
sinh
(
k
x
2
)
{\displaystyle {\bar {\psi }}(x_{1},x_{2})=C_{2}C_{4}\sin(kx_{1})\sinh(kx_{2})=A\sin(kx_{1})\sinh(kx_{2})}
Apply BCs at
x
1
=
±
a
{\displaystyle x_{1}=\pm a}
ψ
¯
,
1
=
0
{\displaystyle {\bar {\psi }}_{,1}=0}
A
k
cos
(
k
a
)
sinh
(
k
x
2
)
=
0
{\displaystyle Ak\cos(ka)\sinh(kx_{2})=0}
The only nontrivial solution is obtained when
cos
(
k
a
)
=
0
{\displaystyle \cos(ka)=0}
k
n
=
(
2
n
+
1
)
π
2
a
,
n
=
0
,
1
,
2
,
.
.
.
{\displaystyle k_{n}={\frac {(2n+1)\pi }{2a}}~~,~~~n=0,1,2,...}
The BCs at
x
1
=
±
a
{\displaystyle x_{1}=\pm a}
ψ
¯
(
x
1
,
x
2
)
=
∑
n
=
0
∞
A
n
sin
(
k
n
x
1
)
sinh
(
k
n
x
2
)
{\displaystyle {\bar {\psi }}(x_{1},x_{2})=\sum _{n=0}^{\infty }A_{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}
Applying the BCs at
x
1
=
±
b
{\displaystyle x_{1}=\pm b}
∑
n
=
0
∞
A
n
k
n
sin
(
k
n
x
1
)
cosh
(
k
n
b
)
=
2
x
1
⇒
∑
n
=
0
∞
B
n
sin
(
k
n
x
1
)
=
2
x
1
{\displaystyle \sum _{n=0}^{\infty }A_{n}k_{n}\sin(k_{n}x_{1})\cosh(k_{n}b)=2x_{1}~~\Rightarrow ~~~\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})=2x_{1}}
Using the orthogonality of terms of the sine series,
∫
−
a
a
sin
(
k
n
x
1
)
sin
(
k
m
x
1
)
d
x
1
=
{
0
i
f
m
≠
n
a
i
f
m
=
n
{\displaystyle \int _{-a}^{a}\sin(k_{n}x_{1})\sin(k_{m}x_{1})dx_{1}={\begin{cases}0&{\rm {if}}~m\neq n\\a&{\rm {if}}~m=n\end{cases}}}
we have
∫
−
a
a
[
∑
n
=
0
∞
B
n
sin
(
k
n
x
1
)
]
sin
(
k
m
x
1
)
d
x
1
=
∫
−
a
a
[
2
x
1
]
sin
(
k
m
x
1
)
d
x
1
{\displaystyle \int _{-a}^{a}\left[\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})\right]\sin(k_{m}x_{1})dx_{1}=\int _{-a}^{a}\left[2x_{1}\right]\sin(k_{m}x_{1})dx_{1}}
or,
B
m
a
=
4
k
m
2
sin
(
k
m
a
)
{\displaystyle B_{m}a={\frac {4}{k_{m}^{2}}}\sin(k_{m}a)}
Now,
sin
(
k
m
a
)
=
sin
(
(
2
m
+
1
)
π
2
)
=
(
−
1
)
m
{\displaystyle \sin(k_{m}a)=\sin \left({\frac {(2m+1)\pi }{2}}\right)=(-1)^{m}}
Therefore,
A
m
=
B
m
k
m
cosh
(
k
m
b
)
=
(
−
1
)
m
32
a
2
(
2
m
+
1
)
3
π
3
cosh
(
k
m
b
)
{\displaystyle A_{m}={\frac {B_{m}}{k_{m}\cosh(k_{m}b)}}={\frac {(-1)^{m}32a^{2}}{(2m+1)^{3}\pi ^{3}\cosh(k_{m}b)}}}
The warping function is
ψ
=
x
1
x
2
−
32
a
2
π
3
∑
n
=
0
∞
(
−
1
)
n
sin
(
k
n
x
1
)
sinh
(
k
n
x
2
)
(
2
n
+
1
)
3
cosh
(
k
n
b
)
{\displaystyle \psi =x_{1}x_{2}-{\frac {32a^{2}}{\pi ^{3}}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}\sin(k_{n}x_{1})\sinh(k_{n}x_{2})}{(2n+1)^{3}\cosh(k_{n}b)}}}
The torsion constant and the stresses can be calculated from
ψ
{\displaystyle \psi }
edit
The traction free BC is obviously difficult to satisfy if the cross-section
is not a circle or an ellipse.
ϕ
(
x
1
,
x
2
)
{\displaystyle \phi (x_{1},x_{2})}
σ
13
=
ϕ
,
2
;
σ
23
=
−
ϕ
,
1
{\displaystyle {\sigma _{13}=\phi _{,2}~~;~~\sigma _{23}=-\phi _{,1}}}
You can easily check that this definition satisfies equilibrium.
d
ϕ
d
s
=
0
∀
(
x
1
,
x
2
)
∈
∂
S
{\displaystyle {{\frac {d\phi }{ds}}=0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}}
where
s
{\displaystyle s}
ϕ
=
0
∀
(
x
1
,
x
2
)
∈
∂
S
{\displaystyle {\phi =0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}}
From the compatibility condition, we get a restriction on
ϕ
{\displaystyle \phi }
∇
2
ϕ
=
C
∀
(
x
1
,
x
2
)
∈
S
{\displaystyle {\nabla ^{2}{\phi }=C~~\forall ~(x_{1},x_{2})\in {\text{S}}}}
where
C
{\displaystyle C}
Using relations for stress in terms of the warping function
ψ
{\displaystyle \psi }
∇
2
ϕ
=
−
2
μ
α
∀
(
x
1
,
x
2
)
∈
S
{\displaystyle {\nabla ^{2}{\phi }=-2\mu \alpha ~~\forall ~(x_{1},x_{2})\in {\text{S}}}}
Therefore, the twist per unit length is
α
=
−
1
2
μ
∇
2
ϕ
{\displaystyle {\alpha =-{\frac {1}{2\mu }}\nabla ^{2}{\phi }}}
The applied torque is given by
T
=
−
∫
S
(
x
1
ϕ
,
1
+
x
2
ϕ
,
2
)
d
A
{\displaystyle {T=-\int _{S}(x_{1}\phi _{,1}+x_{2}\phi _{,2})dA}}
For a simply connected cylinder,
T
=
2
∫
S
ϕ
d
A
{\displaystyle {T=2\int _{S}\phi dA}}
The projected shear traction is given by
τ
=
(
ϕ
,
1
)
2
+
(
ϕ
,
2
)
2
{\displaystyle {\tau ={\sqrt {(\phi _{,1})^{2}+(\phi _{,2})^{2}}}}}
The projected shear traction at any point on the cross-section is tangent to the contour of constant
ϕ
{\displaystyle \phi }
at that point.
ψ
{\displaystyle \psi }
ϕ
{\displaystyle \phi }
ψ
,
1
=
1
μ
α
ϕ
,
2
+
x
2
;
ψ
,
2
=
−
1
μ
α
ϕ
,
1
−
x
1
{\displaystyle {\psi _{,1}={\frac {1}{\mu \alpha }}\phi _{,2}+x2~;~~\psi _{,2}=-{\frac {1}{\mu \alpha }}\phi _{,1}-x1}}
The equations
∇
2
ϕ
=
−
2
μ
α
∀
(
x
1
,
x
2
)
∈
S
;
ϕ
=
0
∀
(
x
1
,
x
2
)
∈
∂
S
{\displaystyle \nabla ^{2}{\phi }=-2\mu \alpha ~~\forall ~(x_{1},x_{2})\in {\text{S}}~~;~~~\phi =0~~\forall ~(x_{1},x_{2})\in \partial {\text{S}}}
are similar to the equations that govern the displacement of a membrane
that is stretched between the boundaries of the cross-sectional curve
and loaded by an uniform normal pressure.
The stress function is proportional to the displacement of the membrane from the plane of the cross-section.
The stiffest cross-sections are those that allow the maximum volume to be developed between the deformed membrane and the plane of the cross-section for a given pressure.
The shear stress is proportional to the slope of the membrane.
The equation
∇
2
ϕ
=
−
2
μ
α
{\displaystyle \nabla ^{2}{\phi }=-2\mu \alpha }
ϕ
=
ϕ
p
+
ϕ
h
{\displaystyle \phi =\phi _{p}+\phi _{h}}
where
ϕ
p
{\displaystyle \phi _{p}}
ϕ
h
{\displaystyle \phi _{h}}
ϕ
p
=
−
μ
α
x
1
2
;
ϕ
p
=
−
μ
α
x
2
2
{\displaystyle \phi _{p}=-\mu \alpha x_{1}^{2}~~;~~\phi _{p}=-\mu \alpha x_{2}^{2}}
and, in cylindrical co-ordinates,
ϕ
p
=
−
μ
α
r
2
2
{\displaystyle \phi _{p}=-{\frac {\mu \alpha r^{2}}{2}}}
The homogeneous equation is the Laplace equation
∇
2
ϕ
=
0
{\displaystyle \nabla ^{2}{\phi }=0}
f
(
z
)
{\displaystyle f(z)}
z
=
x
1
+
i
x
2
=
r
e
i
θ
{\displaystyle z=x_{1}+ix_{2}=re^{i\theta }}
Thus,
ϕ
h
=
Re
(
f
(
z
)
)
or
ϕ
h
=
Im
(
f
(
z
)
)
{\displaystyle \phi _{h}={\text{Re}}(f(z))~~{\text{or}}~~\phi _{h}={\text{Im}}(f(z))}
Suppose
f
(
z
)
=
z
n
{\displaystyle f(z)=z^{n}}
Then, examples of
ϕ
h
{\displaystyle \phi _{h}}
ϕ
h
=
C
1
r
n
cos
(
n
θ
)
;
ϕ
h
=
C
2
r
n
sin
(
n
θ
)
;
ϕ
h
=
C
3
r
−
n
cos
(
n
θ
)
;
ϕ
h
=
C
4
r
−
n
sin
(
n
θ
)
{\displaystyle \phi _{h}=C_{1}r^{n}\cos(n\theta )~~;~~\phi _{h}=C_{2}r^{n}\sin(n\theta )~~;~~\phi _{h}=C_{3}r^{-n}\cos(n\theta )~~;~~\phi _{h}=C_{4}r^{-n}\sin(n\theta )}
where
C
1
{\displaystyle C_{1}}
C
2
{\displaystyle C_{2}}
C
3
{\displaystyle C_{3}}
C
4
{\displaystyle C_{4}}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
Circular cross-section.
Elliptical cross-section.
Circle with semicircular groove.
Equilateral triangle. There are a few other papers which propose closed-form or semi-closed-form solutions to the torsion problem for cross-sections with irregular shapes [ 1] [ 2] [ 3]
Example: Equilateral Triangle
edit
Torsion of a cylinder with a triangular cross section
The equations of the three sides are
side
∂
S
(
1
)
:
f
1
(
x
1
,
x
2
)
=
x
1
−
3
x
2
+
2
a
=
0
side
∂
S
(
2
)
:
f
2
(
x
1
,
x
2
)
=
x
1
+
3
x
2
+
2
a
=
0
side
∂
S
(
3
)
:
f
3
(
x
1
,
x
2
)
=
x
1
−
a
=
0
{\displaystyle {\begin{aligned}{\text{side}}~\partial S^{(1)}~:~~&f_{1}(x_{1},x_{2})=x_{1}-{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(2)}~:~~&f_{2}(x_{1},x_{2})=x_{1}+{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(3)}~:~~&f_{3}(x_{1},x_{2})=x_{1}-a=0\end{aligned}}}
Let the Prandtl stress function be
ϕ
=
C
f
1
f
2
f
3
{\displaystyle \phi =Cf_{1}f_{2}f_{3}}
Clearly,
ϕ
=
0
{\displaystyle \phi =0}
ϕ
{\displaystyle \phi }
C
{\displaystyle C}
C
{\displaystyle C}
ϕ
{\displaystyle \phi }
ϕ
{\displaystyle \phi }
ϕ
=
C
(
x
1
−
3
x
2
+
2
a
)
(
x
1
+
3
x
2
+
2
a
)
(
x
1
−
a
)
{\displaystyle \phi =C(x_{1}-{\sqrt {3}}x_{2}+2a)(x_{1}+{\sqrt {3}}x_{2}+2a)(x_{1}-a)}
Plugging into the compatibility condition
∇
2
ϕ
=
12
C
a
=
−
2
μ
α
{\displaystyle \nabla ^{2}{\phi }=12Ca=-2\mu \alpha }
Therefore,
C
=
−
μ
α
6
a
{\displaystyle C=-{\frac {\mu \alpha }{6a}}}
and the Prandtl stress function can be written as
ϕ
=
−
μ
α
6
a
(
x
1
3
+
3
a
x
1
2
+
3
a
x
2
2
−
3
x
1
x
2
2
−
4
a
3
)
{\displaystyle \phi =-{\frac {\mu \alpha }{6a}}(x_{1}^{3}+3ax_{1}^{2}+3ax_{2}^{2}-3x_{1}x_{2}^{2}-4a^{3})}
The torque is given by
T
=
2
∫
S
ϕ
d
A
=
2
∫
−
2
a
a
∫
−
(
x
1
+
2
a
)
/
3
(
x
1
+
2
a
)
/
3
ϕ
d
x
2
d
x
1
=
27
5
3
μ
α
a
4
{\displaystyle T=2\int _{S}\phi dA=2\int _{-2a}^{a}\int _{-(x_{1}+2a)/{\sqrt {3}}}^{(x_{1}+2a)/{\sqrt {3}}}\phi dx_{2}dx_{1}={\frac {27}{5{\sqrt {3}}}}\mu \alpha a^{4}}
Therefore, the torsion constant is
J
~
=
27
a
4
5
3
{\displaystyle {\tilde {J}}={\frac {27a^{4}}{5{\sqrt {3}}}}}
The non-zero components of stress are
σ
13
=
ϕ
,
2
=
μ
α
a
(
x
1
−
a
)
x
2
σ
23
=
−
ϕ
,
1
=
μ
α
2
a
(
x
1
2
+
2
a
x
1
−
x
2
2
)
{\displaystyle {\begin{aligned}\sigma _{13}=\phi _{,2}&={\frac {\mu \alpha }{a}}(x_{1}-a)x_{2}\\\sigma _{23}=-\phi _{,1}&={\frac {\mu \alpha }{2a}}(x_{1}^{2}+2ax_{1}-x_{2}^{2})\end{aligned}}}
The projected shear stress
τ
=
σ
13
2
+
σ
23
2
{\displaystyle \tau ={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}}
is plotted below
Stresses in a cylinder with a triangular cross section under torsion
The maximum value occurs at the middle of the sides. For example,
at
(
a
,
0
)
{\displaystyle (a,0)}
τ
max
=
3
μ
α
a
2
{\displaystyle \tau _{\text{max}}={\frac {3\mu \alpha a}{2}}}
The out-of-plane displacements can be obtained by solving for the
warping function
ψ
{\displaystyle \psi }
u
3
=
α
x
2
6
a
(
3
x
1
2
−
x
2
2
)
{\displaystyle u_{3}={\frac {\alpha x_{2}}{6a}}(3x_{1}^{2}-x_{2}^{2})}
The displacement field is plotted below
Displacements
u
3
{\displaystyle u_{3}}
Thin-walled Open Sections
edit
Examples are I-beams, channel sections and turbine blades.
We assume that the length
b
{\displaystyle b}
t
{\displaystyle t}
t
{\displaystyle t}
ξ
{\displaystyle \xi }
Using the membrane analogy, we can neglect the curvature of the membrane
in the
ξ
{\displaystyle \xi }
d
ϕ
d
η
2
=
−
2
μ
α
{\displaystyle {\frac {d^{\phi }}{d\eta ^{2}}}=-2\mu \alpha }
which has the solution
ϕ
=
μ
α
(
t
2
4
−
η
2
)
{\displaystyle \phi =\mu \alpha \left({\frac {t^{2}}{4}}-\eta ^{2}\right)}
where
η
{\displaystyle \eta }
The stress field is
σ
3
ξ
=
∂
∂
ϕ
η
=
−
e
μ
β
η
;
σ
3
η
=
0
{\displaystyle \sigma _{3\xi }={\frac {\partial }{\partial }}{\phi }{\eta }=-e\mu \beta \eta ~~;~~~\sigma _{3\eta }=0}
Thus, the maximum shear stress is
τ
max
=
μ
β
t
max
{\displaystyle \tau _{\text{max}}=\mu \beta t_{\text{max}}}
Thin-walled Closed Sections
edit
The Prandtl stress function
ϕ
{\displaystyle \phi }
ϕ
1
{\displaystyle \phi _{1}}
0
{\displaystyle 0}
σ
3
s
=
ϕ
1
t
{\displaystyle \sigma _{3s}={\frac {\phi _{1}}{t}}}
where
s
{\displaystyle s}
t
{\displaystyle t}
The value of
ϕ
1
{\displaystyle \phi _{1}}
ϕ
1
=
2
μ
α
A
∮
S
d
S
t
{\displaystyle \phi _{1}={\frac {2\mu \alpha A}{\oint _{S}{\frac {dS}{t}}}}}
where
A
{\displaystyle A}
The torque is approximately
T
=
2
A
ϕ
1
{\displaystyle T=2A\phi _{1}}
Introduction to Elasticity
↑ Approximate Torsional Analysis of Multi-layered Tubes with Non-circular Cross-Sections, Gholami Bazehhour, Benyamin, Rezaeepazhand, Jalil, Journal: Applied composite materials
ISSN: 0929-189X
Date: 12/2011
Volume: 18
Issue: 6
Page: 485-497
DOI: 10.1007/s10443-011-9213-z
↑ Simplified approach for torsional analysis of non-homogenous tubes with non-circular cross-sections, B Golami Bazehhour, J Rezaeepazhand, Date: 2012: Journal: International Journal of Engineering, Volume: 25, Issue: 3, ↑ Torsion of tubes with quasi-polygonal holes using complex variable method, Gholami Bazehhour, Benyamin, Rezaeepazhand, J.
Journal: Mathematics and mechanics of solids
ISSN: 1081-2865
Date: 05/2014
Volume: 19
Issue: 3
Page: 260-276
DOI: 10.1177/1081286512462836
![{\displaystyle {\bar {\psi }}(x_{1},x_{2})=\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\cosh(kx_{2})+C_{4}\sinh(kx_{2})\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bcb2287f2c30081bbf32884dfdc0a1b041a4235)
![{\displaystyle {\begin{aligned}\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\\\left[C_{1}\cos(kx_{1})+C_{2}\sin(kx_{1})\right]\left[-C_{3}\sinh(kb)+C_{4}\cosh(kb)\right]&=2x_{1}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8f3546d1357cc29b7f3b8e541db9f4809ba2b02)
![{\displaystyle F(x_{1})\left[G^{'}(b)-G^{'}(-b)\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7d9023ae4253379b8f94da24302aff5143a00b4)
![{\displaystyle \int _{-a}^{a}\left[\sum _{n=0}^{\infty }B_{n}\sin(k_{n}x_{1})\right]\sin(k_{m}x_{1})dx_{1}=\int _{-a}^{a}\left[2x_{1}\right]\sin(k_{m}x_{1})dx_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a7507d4b2ef54e1539e8202728c76e9fef29c6e)
