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\,}
at the boundary of the cross-section (which is what we need for solid cross sections).
Since, the traction-free boundary conditions are satisfied by
ϕ
{\displaystyle \phi \,}
, all we have to do is satisfy the compatibility condition to get the value of
C
{\displaystyle C\,}
. If we can get a closed for solution for
C
{\displaystyle C\,}
, then the stresses derived from
ϕ
{\displaystyle \phi \,}
will satisfy equilibrium.
Expanding
ϕ
{\displaystyle \phi \,}
out,
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 }
. For the equilateral triangle, after some
algebra, we get
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}\,}
in a cylinder with a triangular cross section.