Torsion

## Contents

* 1 Torsion of Circular Cylinders
* 1.1 About the problem:
* 1.1.1 Assumptions:
* 1.1.2 Find:
* 1.2 Solution:
* 2 Torsion of Non-Circular Cylinders
* 2.1 About the problem
* 2.1.1 Assumptions:
* 2.1.2 Find:
* 2.2 Solution:
* 2.2.1 Displacements
* 2.2.2 Strains
* 2.2.3 Stresses
* 2.2.4 Equilibrium
* 2.2.5 Internal Tractions
* 2.2.6 Boundary Conditions on Lateral Surfaces
* 2.2.7 Boundary Conditions on End Surfaces
* 2.2.8 The Green-Riemann Theorem
* 3 The Torsion Problem Summarized
* 4 Related Content

## Torsion of Circular Cylinders

[edit | edit source]

Torsion of a cylinder with a circular cross section

### About the problem:

[edit | edit source]

* Circular Cylinder.
* Centroidal axis thru the center of each c.s.
* Length  L {\displaystyle L} , Outer radius  c {\displaystyle c} .
* Applied torque  T {\displaystyle T} .
* Angle of twist  ϕ {\displaystyle \phi } .

#### Assumptions:

[edit | edit source]

* Each c.s. remains plane and undistorted.
* Each c.s. rotates through the same angle.
* No warping or change in shape.
* Amount of displacement of each c.s. is proportional to distance from end.

#### Find:

[edit | edit source]

* Shear strains in the cylinder ( γ {\displaystyle \gamma } ).
* Shear stress in the cylinder ( τ {\displaystyle \tau } ).
* Relation between torque ( T {\displaystyle T} ) and angle of twist ( ϕ {\displaystyle \phi } ).
* Relation between torque ( T {\displaystyle T} ) and shear stress ( τ {\displaystyle \tau } ).

### Solution:

[edit | edit source]

If  γ {\displaystyle \gamma } is small, then

(1) L γ = r ϕ ⇒ γ = r ϕ L {\displaystyle {\text{(1)}}\qquad L\gamma =r\phi ~~\Rightarrow ~~{\gamma ={\frac {r\phi }{L}}}}

Therefore,

(2) γ max = c ϕ L ⇒ γ = r γ max {\displaystyle {\text{(2)}}\qquad \gamma _{\text{max}}={\frac {c\phi }{L}}~~\Rightarrow ~~\gamma ={\frac {r}{\gamma }}_{\text{max}}}

If the deformation is elastic,

(3) τ = G γ ⇒ τ = r ϕ G L {\displaystyle {\text{(3)}}\qquad \tau =G\gamma ~~\Rightarrow ~~{\tau ={\frac {r\phi G}{L}}}}

Therefore,

(4) τ max = r ϕ G L ⇒ τ = r τ max {\displaystyle {\text{(4)}}\qquad \tau _{\text{max}}={\frac {r\phi G}{L}}~~\Rightarrow ~~\tau ={\frac {r}{\tau }}_{\text{max}}}

The torque on each c.s. is given by

(5) T = ∫ A τ r d A = ϕ G L ∫ A r 2 d A = G ϕ J L {\displaystyle {\text{(5)}}\qquad T=\int _{A}\tau rdA={\frac {\phi G}{L}}\int _{A}r^{2}dA={\frac {G\phi J}{L}}}

where  J {\displaystyle J} is the polar moment of inertia of the c.s.

(6) J = { 1 2 π c 4 solid circular c.s. 1 2 π ( c 2 4 − c 1 4 ) annular circular c.s. {\displaystyle {\text{(6)}}\qquad J={\begin{cases}{\frac {1}{2}}\pi c^{4}&{\text{solid circular c.s.}}\\\\{\frac {1}{2}}\pi (c_{2}^{~4}-c_{1}^{~4})&{\text{ annular circular c.s.}}\end{cases}}}

Therefore,

(7) τ = T r J ⇒ τ max = T c J {\displaystyle {\text{(7)}}\qquad {\tau ={\frac {Tr}{J}}}~~\Rightarrow \tau _{\text{max}}={\frac {Tc}{J}}}

and

(8) ϕ = T L J G {\displaystyle {\text{(8)}}\qquad {\phi ={\frac {TL}{JG}}}}

## Torsion of Non-Circular Cylinders

[edit | edit source]

Torsion of a noncircular cylinder

### About the problem

[edit | edit source]

* Solution first found by St. Venant.
* Tractions at the ends are statically equivalent to equal and opposite torques  ± T = ± T e ^ 3 {\displaystyle \pm \mathbf {T} =\pm T{\widehat {\mathbf {e} }}{3}} .
* Lateral surfaces are traction-free.

#### Assumptions:

[edit | edit source]

* An axis passes through the center of twist ( x 3 {\displaystyle x_{3}} axis).
* Each c.s. projection on to the  x 1 − x 2 {\displaystyle x_{1}-x_{2}} plane rotates,but remains undistorted.
* The rotation of each c.s. ( ϕ {\displaystyle \phi } ) is proportional to  x 3 {\displaystyle x_{3}} .  : (9) ϕ = α x 3 {\displaystyle {\text{(9)}}\qquad \phi =\alpha x_{3}} where  α {\displaystyle \alpha } is the  twist per unit length.
* The out-of-plane distortion (warping) is the same for each c.s. and is proportional to  α {\displaystyle \alpha } .

#### Find:

[edit | edit source]

* Torsional rigidity ( T / α {\displaystyle T/\alpha } ).
* Maximum shear stress.

### Solution:

[edit | edit source]

#### Displacements

[edit | edit source]

u 1 = r cos ⁡ ( ϕ + θ ) − r cos ⁡ θ = x 1 ( cos ⁡ ϕ − 1 ) − x 2 sin ⁡ ϕ u 2 = r sin ⁡ ( ϕ + θ ) − r sin ⁡ θ = x 1 sin ⁡ ϕ + x 2 ( cos ⁡ ϕ − 1 ) u 3 = α ψ ( x 1 , x 2 ) {\displaystyle {\begin{aligned}u_{1}&=r\cos(\phi +\theta )-r\cos \theta =x_{1}(\cos \phi -1)-x_{2}\sin \phi \\\u_{2}&=r\sin(\phi +\theta )-r\sin \theta =x_{1}\sin \phi +x_{2}(\cos \phi -1)\\\u_{3}&=\alpha \psi (x_{1},x_{2})\end{aligned}}}

where  ψ ( x 1 , x 2 ) {\displaystyle \psi (x_{1},x_{2})} is the { warping function}.\\\ If  ϕ = α x 3 << 1 {\displaystyle \phi =\alpha x_{3}<<1} (small strain),

(10) u 1 ≈ − α x 2 x 3 ; u 2 ≈ α x 1 x 3 ; u 3 = α ψ ( x 1 , x 2 ) {\displaystyle {\text{(10)}}\qquad {u_{1}\approx -\alpha x_{2}x_{3}~;~~u_{2}\approx \alpha x_{1}x_{3}~;~~u_{3}=\alpha \psi (x_{1},x_{2})}}

#### Strains

[edit | edit source]

ε i j = 1 2 ( u i , j + u j , i ) {\displaystyle \varepsilon _{ij}={\frac {1}{2}}\left(u_{i,j}+u_{j,i}\right)}

Therefore,

ε 11 = 1 2 ( 0 + 0 ) = 0 ε 22 = 1 2 ( 0 + 0 ) = 0 ε 33 = 1 2 ( 0 + 0 ) = 0 ε k k = ε 11 + ε 22 + ε 33 = 0 ε 12 = 1 2 ( − α x 3 + α x 3 ) = 0 ε 23 = 1 2 ( α ψ , 2 + α x 1 ) (11) ε 31 = 1 2 ( α ψ , 1 − α x 2 ) (12) {\displaystyle {\begin{aligned}\varepsilon _{11}&={\frac {1}{2}}\left(0+0\right)=0\\\\\varepsilon _{22}&={\frac {1}{2}}\left(0+0\right)=0\\\\\varepsilon _{33}&={\frac {1}{2}}\left(0+0\right)=0\\\\\varepsilon _{kk}&=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=0\\\\\varepsilon _{12}&={\frac {1}{2}}\left(-\alpha x_{3}+\alpha x_{3}\right)=0\\\\\varepsilon _{23}&={\frac {1}{2}}\left(\alpha \psi _{,2}+\alpha x_{1}\right){\text{(11)}}\qquad \\\\\varepsilon _{31}&={\frac {1}{2}}\left(\alpha \psi _{,1}-\alpha x_{2}\right){\text{(12)}}\qquad \end{aligned}}}

#### Stresses

[edit | edit source]

σ i j = 2 μ ε i j + λ ε k k δ i j {\displaystyle \sigma _{ij}=2\mu \varepsilon _{ij}+\lambda \varepsilon _{kk}\delta _{ij}}

Therefore,

σ 11 = 0 σ 22 = 0 σ 33 = 0 σ k k = 0 σ 12 = 0 σ 23 = μ α ( ψ , 2 + x 1 ) (13) σ 31 = μ α ( ψ , 1 − x 1 ) (14) {\displaystyle {\begin{aligned}\sigma _{11}&=0\\\\\sigma _{22}&=0\\\\\sigma _{33}&=0\\\\\sigma _{kk}&=0\\\\\sigma _{12}&=0\\\\\sigma _{23}&=\mu \alpha (\psi _{,2}+x_{1}){\text{(13)}}\qquad \\\\\sigma _{31}&=\mu \alpha (\psi _{,1}-x_{1}){\text{(14)}}\qquad \end{aligned}}}

#### Equilibrium

[edit | edit source]

σ j i , j = 0 no body forces. {\displaystyle \sigma _{ji,j}=0~~~~{\text{no body forces.}}}

Therefore,

σ 11 , 1 + σ 21 , 2 + σ 31 , 3 = 0 ⇒ 0 = 0 σ 12 , 1 + σ 22 , 2 + σ 32 , 3 = 0 ⇒ 0 = 0 σ 13 , 1 + σ 23 , 2 + σ 33 , 3 = 0 ⇒ μ α ( ψ , 11 + ψ , 22 ) = μ α ∇ 2 ψ = 0 (15) {\displaystyle {\begin{aligned}\sigma _{11,1}+\sigma _{21,2}+\sigma _{31,3}=0&\Rightarrow ~~0=0\\\\\sigma _{12,1}+\sigma _{22,2}+\sigma _{32,3}=0&\Rightarrow ~~0=0\\\\\sigma _{13,1}+\sigma _{23,2}+\sigma _{33,3}=0&\Rightarrow ~~\mu \alpha (\psi _{,11}+\psi _{,22})=\mu \alpha \nabla ^{2}{\psi }=0{\text{(15)}}\qquad \end{aligned}}}

#### Internal Tractions

[edit | edit source]

* Normal to cross sections is  n ^ = e ^ 3 {\displaystyle {\widehat {\mathbf {n} }}{}={\widehat {\mathbf {e} }}{3}} .
* Normal traction  t n = t ∙ n ^ = 0 {\displaystyle t^{n}=\mathbf {t} \bullet {\widehat {\mathbf {n} }}{}=0} .
* Projected shear traction is  t s = σ 13 2 + σ 23 2 {\displaystyle t^{s}={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}} .
* Traction vector at a point in the cross section is { tangent} to the cross section.

#### Boundary Conditions on Lateral Surfaces

[edit | edit source]

* Lateral surface traction-free.
* Unit normal to lateral surface appears as an in-plane unit normal to the boundary  ∂ S {\displaystyle \partial S} .

We parameterize the boundary curve  ∂ S {\displaystyle \partial S} using

x = x ~ ( s ) , 0 ≤ s ≤ l ; x ~ ( 0 ) = x ~ ( l ) {\displaystyle \mathbf {x} ={\tilde {\mathbf {x} }}(s)~,~~0\leq s\leq l~~;~~~{\tilde {\mathbf {x} }}(0)={\tilde {\mathbf {x} }}(l)}

The tangent vector to  s {\displaystyle s} is

ν ^ = d x d s and n ^ = ν ^ × e ^ 3 ⇒ n ^ = d x 2 d s e ^ 1 − d x 1 d s e ^ 2 {\displaystyle {\widehat {\boldsymbol {\nu }}}={\frac {d\mathbf {x} }{ds}}~{\text{and}}~~{\widehat {\mathbf {n} }}{}={\widehat {\boldsymbol {\nu }}}\times {\widehat {\mathbf {e} }}_{3}~~\Rightarrow ~~~{\widehat {\mathbf {n} }}{}={\frac {dx_{2}}{ds}}{\widehat {\mathbf {e} }}{1}-{\frac {dx_{1}}{ds}}{\widehat {\mathbf {e} }}_{2}}

The tractions  t 1 {\displaystyle t_{1}} and  t 2 {\displaystyle t_{2}} on the lateral surface are identically zero. However, to satisfy the BC  t 3 = 0 {\displaystyle t_{3}=0} , we need

t 3 = n 1 σ 13 + n 2 σ 23 = 0 ⇒ ( ψ , 1 − x 2 ) n 1 + ( ψ , 2 + x 1 ) n 2 = 0 {\displaystyle t_{3}=n_{1}\sigma _{13}+n_{2}\sigma _{23}=0~~\Rightarrow ~~~\left(\psi _{,1}-x_{2}\right)n_{1}+\left(\psi _{,2}+x_{1}\right)n_{2}=0}

or,

(16) ( ψ , 1 − x 2 ) d x 2 d s + ( ψ , 2 + x 1 ) d x 1 d s = 0 {\displaystyle {\text{(16)}}\qquad \left(\psi _{,1}-x_{2}\right){\frac {dx_{2}}{ds}}+\left(\psi _{,2}+x_{1}\right){\frac {dx_{1}}{ds}}=0}

#### Boundary Conditions on End Surfaces

[edit | edit source]

The traction distribution is statically equivalent to the torque  T {\displaystyle \mathbf {T} } . At  x 3 = L {\displaystyle x_{3}=L} ,

t 1 = σ 13 ; t 2 = σ 23 ; t 3 = σ 33 = 0 {\displaystyle t_{1}=\sigma _{13}~;~~t_{2}=\sigma _{23}~;~~t_{3}=\sigma _{33}=0}

Therefore,

F 1 = ∫ S σ 13 d S = μ α ∫ S ( ψ , 1 − x 2 ) d S {\displaystyle F_{1}=\int _{S}\sigma _{13}~dS=\mu \alpha \int _{S}(\psi _{,1}-x_{2})~dS}

From equilibrium,

∇ 2 ψ = 0 ⇒ ψ , 1 − x 2 = ( ψ , 1 − x 2 ) + x 1 ( ψ , 11 + ψ , 22 ) = ψ , 1 + x 1 ψ , 11 − x 2 + x 1 ψ , 22 = ( x 1 ψ , 1 − x 1 x 2 ) , 1 + ( x 1 ψ , 2 + x 1 x 1 ) , 2 = [ x 1 ( ψ , 1 − x 2 ) ] , 1 + [ x 1 ( ψ , 2 + x 1 ) ] , 2 {\displaystyle {\begin{aligned}\nabla ^{2}{\psi }=0~~\Rightarrow ~~~\psi _{,1}-x_{2}&=(\psi _{,1}-x_{2})+x_{1}(\psi _{,11}+\psi _{,22})\\\&=\psi _{,1}+x_{1}\psi _{,11}-x_{2}+x_{1}\psi _{,22}\\\&=(x_{1}\psi _{,1}-x_{1}x_{2})_{,1}+(x_{1}\psi _{,2}+x_{1}x_{1})_{,2}\\\&=\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}\end{aligned}}}

Hence,

(17) F 1 = μ α ∫ S [ x 1 ( ψ , 1 − x 2 ) ] , 1 + [ x 1 ( ψ , 2 + x 1 ) ] , 2 d S {\displaystyle {\text{(17)}}\qquad F_{1}=\mu \alpha \int _{S}\left[x_{1}(\psi _{,1}-x_{2})\right]_{,1}+\left[x_{1}(\psi _{,2}+x_{1})\right]_{,2}dS}

#### The Green-Riemann Theorem

[edit | edit source]

If  P = f ( x 1 , x 2 ) {\displaystyle P=f(x_{1},x_{2})} and  Q = q ( x 1 , x 2 ) {\displaystyle Q=q(x_{1},x_{2})} then

∫ S ( Q , 1 − P , 2 ) d S = ∮ ∂ S ( P d x 1 + Q d x 2 ) {\displaystyle \int _{S}(Q_{,1}-P_{,2})dS=\oint _{\partial S}(Pdx_{1}+Qdx_{2})}

with the integration direction such that  S {\displaystyle S} is to the left.

Applying the Green-Riemann theorem to equation (17), and using equation (16)

(18) F 1 = μ α ∮ ∂ S − x 1 ( ψ , 2 + x 1 ) d x 1 + x 1 ( ψ , 1 − x 2 ) d x 2 = 0 {\displaystyle {\text{(18)}}\qquad F_{1}=\mu \alpha \oint _{\partial S}-x_{1}(\psi _{,2}+x_{1})dx_{1}+x_{1}(\psi _{,1}-x_{2})dx_{2}=0}

Similarly, we can show that  F 2 = 0 {\displaystyle F_{2}=0} .  F 3 = 0 {\displaystyle F_{3}=0} since  t 3 = 0 {\displaystyle t_{3}=0} .

The moments about the  x 1 {\displaystyle x_{1}} and  x 2 {\displaystyle x_{2}} axes are also zero.

The moment about the  x 3 {\displaystyle x_{3}} axis is

M 3 = ∫ S ( x 1 σ 23 − x 2 σ 13 ) d S = μ α ∫ S ( x 1 ψ , 2 + x 1 2 − x 2 ψ 1 + x 2 2 ) d S = μ α J ~ {\displaystyle M_{3}=\int _{S}(x_{1}\sigma _{23}-x_{2}\sigma _{13})dS=\mu \alpha \int _{S}(x_{1}\psi _{,2}+x_{1}^{2}-x_{2}\psi _{1}+x_{2}^{2})dS=\mu \alpha {\tilde {J}}}

where  J {\displaystyle J} is the torsion constant. Since  M 3 = T {\displaystyle M_{3}=T} , we have

α = T μ J ~ {\displaystyle \alpha ={\frac {T}{\mu {\tilde {J}}}}}

If  ψ = 0 {\displaystyle \psi =0} , then  J ~ = J {\displaystyle {\tilde {J}}=J} , the polar moment of inertia.

## The Torsion Problem Summarized

[edit | edit source]

* Find a warping function  ψ ( x 1 , x 2 ) {\displaystyle \psi (x_{1},x_{2})\,} that is harmonic. and satisfies the traction BCs.
* Compatibility is not an issue since we start with displacements.
* The problem is independent of applied torque and the material properties of the cylinder.
* So it is just a geometrical problem. Once  ψ {\displaystyle \psi } is known, we can calculate
* The displacement field.
* The stress field.
* The twist per unit length.

## Related Content

[edit | edit source]

Introduction to Elasticity