Torsion

Semi-inverse method

In the semi-inverse method some features of the solution are assumed (from the displacements), and then it is checked that all equations of elasticity and boundary conditions are satisfied. The solution is then unique.

Assumptions (Saint-Venant):

1. Each cross-section rotates as a rigid body in its plane through angle $\theta z$, with $\theta$ = twist per unit length.
2. The section also warps (out-of-plane displacement), the same for all sections: $w = \theta\,\psi(x,y)$, where $\psi$ is the warping function.

For a circular shaft $\psi = 0$ (no warping), which is why plane sections stay plane there.

Displacements and strains

$$u = -\theta z\,y,\qquad v = \theta z\,x,\qquad w = \theta\,\psi(x,y)$$ $$\gamma_{xz} = \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} = \theta\left(\frac{\partial \psi}{\partial x} - y\right),\qquad \gamma_{yz} = \theta\left(\frac{\partial \psi}{\partial y} + x\right)$$

All other strains are zero, so $\tau_{xz} = G\gamma_{xz}$, $\tau_{yz} = G\gamma_{yz}$ are the only stresses.

Prandtl stress function

With no body force, the only non-trivial equilibrium equation is $\dfrac{\partial\tau_{xz}}{\partial x} + \dfrac{\partial\tau_{yz}}{\partial y} = 0$. This is identically satisfied by

$$\tau_{xz} = \frac{\partial \phi}{\partial y},\qquad \tau_{yz} = -\frac{\partial \phi}{\partial x}$$

Compatibility: eliminate $\psi$ from the strains:

$$\frac{\partial\gamma_{xz}}{\partial y} - \frac{\partial\gamma_{yz}}{\partial x} = -2\theta \ \Rightarrow\ \frac{\partial\tau_{xz}}{\partial y} - \frac{\partial\tau_{yz}}{\partial x} = -2G\theta$$ $$\boxed{\nabla^2\phi = \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} = -2G\theta}$$

Boundary condition: the lateral surface is stress free, so the resultant shear stress must be tangent to the boundary. This requires $d\phi/ds = 0$ along the boundary, i.e. $\phi =$ constant, taken as zero for a solid (simply connected) section.

Torque

$$\begin{aligned} T &= \iint (x\,\tau_{yz} - y\,\tau_{xz})\,dA = -\iint\left(x\frac{\partial\phi}{\partial x} + y\frac{\partial\phi}{\partial y}\right)dA \end{aligned}$$

Integrating by parts and using $\phi = 0$ on the boundary gives each term $= -\iint \phi\,dA$, so

$$\boxed{T = 2\iint \phi\,dA}$$

The torsional constant follows from $T = GJ\theta$.

Check: solid circular shaft of radius $R$

Try $\phi = \dfrac{G\theta}{2}\left(R^2 - x^2 - y^2\right)$. Then $\nabla^2\phi = \dfrac{G\theta}{2}(-2-2) = -2G\theta$ and $\phi = 0$ at $r = R$. $\checkmark$

$$\tau_{xz} = -G\theta y,\quad \tau_{yz} = G\theta x \Rightarrow \tau = G\theta r$$ $$T = 2\int_0^R \frac{G\theta}{2}(R^2 - r^2)\,2\pi r\,dr = 2\pi G\theta\left(\frac{R^4}{2} - \frac{R^4}{4}\right) = G\theta\,\frac{\pi R^4}{2} = GJ\theta$$

which is the familiar result $J = \pi R^4/2$.

Membrane analogy

Prandtl observed that the equation for the torsion stress function, $\nabla^2\phi = -2G\theta$, has the same form as the equation for the small deflection $z$ of a uniformly stretched membrane under pressure $p$ and tension $S$:

$$\nabla^2 z = -\frac{p}{S}$$

So if a thin membrane (soap film) is stretched over a hole of the same shape as the cross-section and inflated slightly, and $p/S$ is made equal to $2G\theta$, the deflection $z$ is equal to $\phi$.

   hole shaped like the section
   +--------------------+
   |   .--------.       |   membrane bulges by z
   |  /  contours \     |   (contour lines = lines of
   |  \          /      |    shear stress direction)
   |   '--------'       |
   +--------------------+

Analogy rules

The analogy gives the stress picture without solving the equation, and explains why the stress is zero at the sharp convex corner of a square and highest at the mid-point of the sides.

Narrow rectangle ($b \gg t$)

The membrane is nearly cylindrical, with a parabolic profile across the thickness $t$. Then

$$\tau_{max} = \frac{3T}{b\,t^2},\qquad \theta = \frac{3T}{G\,b\,t^3},\qquad J = \frac{b\,t^3}{3}$$

The shear stress varies linearly across the thickness, from zero at the mid-plane to $\tau_{max}$ at the long edges.

Open and closed sections

An open thin-walled section (angle, channel, I) behaves as a set of narrow rectangles: $J = \tfrac13\sum b_it_i^3$. The torsional stiffness depends on $t^3$, and the shear flow circulates only within each thin strip.

A closed tube carries a shear flow around the whole wall, enclosing area $A$, with $J = 4A^2/\oint ds/t$, which depends on $t$ only to the first power and on the enclosed area squared. A closed tube is therefore typically tens to hundreds of times stiffer and gives much lower stress than an open section of the same material, which is why box or tubular sections are used where torsion is important.

Bredt-Batho theory

For a thin-walled closed section, the shear flow $q = \tau t$ is constant around the wall, and the torque is $T = 2Aq$, where $A$ is the area enclosed by the mid-line of the wall.

        100 mm (mean)
    +-----------------+  t = 4 mm
    |                 |
    |                 | 60 mm   sides t = 6 mm
    |                 |
    +-----------------+  t = 4 mm

Data

$A = 100\times60 = 6000$ mm$^2$, $T = 3\ \text{kN m} = 3\times10^6$ N mm, $G = 80\times10^3$ MPa.

Shear flow and stress

$$q = \frac{T}{2A} = \frac{3\times10^6}{2\times6000} = 250\ \text{N/mm}$$ $$\begin{aligned} \tau_{top,bottom} &= \frac{q}{t} = \frac{250}{4} = 62.5\ \text{MPa} \\ \tau_{sides} &= \frac{250}{6} = 41.67\ \text{MPa} \end{aligned}$$

The maximum shear stress occurs in the thinnest wall (4 mm): $62.5$ MPa.

Angle of twist per unit length

Equating the strain energy to the work done by the torque gives

$$\theta = \frac{T}{4A^2G}\oint\frac{ds}{t}$$ $$\oint\frac{ds}{t} = 2\left(\frac{100}{4}\right) + 2\left(\frac{60}{6}\right) = 50 + 20 = 70$$ $$\theta = \frac{3\times10^6\times70}{4\times(6000)^2\times80\times10^3} = \frac{2.1\times10^8}{1.152\times10^{13}} = 1.823\times10^{-5}\ \text{rad/mm}$$ $$\theta = 0.01823\ \text{rad/m} = 1.045^\circ\text{/m}$$

Equivalent torsional constant

$$J_{eq} = \frac{4A^2}{\oint ds/t} = \frac{4\times 6000^2}{70} = 2.057e+06\ \text{mm}^4$$

Check: $\theta = T/(GJ_{eq}) = 3\times10^6/(80\times10^3\times2.057\times10^6) = 1.823\times10^{-5}$ rad/mm. $\checkmark$

Total twist for 1.5 m

$$\phi = \theta L = 0.01823\times1.5 = 0.0273\ \text{rad} = 1.57^\circ$$

Answer: $q = 250$ N/mm; $\tau = 62.5$ MPa (top and bottom), $41.7$ MPa (sides); $\theta = 0.0182$ rad/m; $J_{eq} = 2.06\times10^6$ mm$^4$; twist over 1.5 m $= 0.0273$ rad ($1.57^\circ$).

Method

For a thin open section, the section behaves like a set of narrow rectangles. Each strip has the same twist per unit length $\theta$, and carries a torque in proportion to its $J_i = b_it_i^3/3$ (membrane analogy).

$$J = \sum \frac{b_it_i^3}{3},\qquad \theta = \frac{T}{GJ},\qquad \tau_i = \frac{T\,t_i}{J},\qquad T_i = T\,\frac{J_i}{J}$$

Torsion constant

$$\begin{aligned} J_{web} &= \frac{100\times6^3}{3} = \frac{21600}{3} = 7200\ \text{mm}^4 \\ J_{flange} &= \frac{60\times8^3}{3} = \frac{30720}{3} = 10240\ \text{mm}^4\ \text{(each)} \\ J &= 7200 + 2(10240) = 27680\ \text{mm}^4 \end{aligned}$$

Shear stress

$T = 200\ \text{N m} = 2\times10^5$ N mm.

$$\begin{aligned} \tau_{web} &= \frac{Tt_w}{J} = \frac{2\times10^5\times6}{27680} = 43.35\ \text{MPa} \\ \tau_{flange} &= \frac{2\times10^5\times8}{27680} = 57.80\ \text{MPa} \end{aligned}$$

The maximum shear stress, $57.8$ MPa, occurs at the long edges of the thicker flanges.

Twist

$$\theta = \frac{T}{GJ} = \frac{2\times10^5}{80\times10^3\times27680} = 9.03\times10^{-5}\ \text{rad/mm} = 0.0903\ \text{rad/m} = 5.17^\circ\text{/m}$$

Torque shared by strips

(The two flanges together carry 74.0 percent.)

The calculation neglects the stress concentration at the re-entrant corners and the extra stiffness at junctions; a correction factor (about 1.1 to 1.3 for rolled sections) is used for rolled channels and angles.

Answer: $\tau_{web} = 43.4$ MPa, $\tau_{flange,max} = 57.8$ MPa, $\theta = 0.0903$ rad/m ($5.17^\circ$/m); torque shared: web $52.0$ N m, each flange $74.0$ N m.