Torsion

Table of contents

Stress
Angle of twist

Check out the Efficient Engineer video on torsion here.

Stress

1: Consider a thin walled circular shaft under torsion. We want to find the relation between the applied Moment

M

and the shear stress

\tau

The definition of thin wall is formally that $t << R$ . Also, it is assumed that $\tau$ is constant in the radial direction. Taking a look at a small section of the cross-section we have

dM=TR

where $dM$ is a infinitesimal moment and $T$ denotes the radial force, which is given by the shear stress and area

T=\tau dA

If we take a closer look at the area, we can realize that it can be defined by the infinitesimal ange $d\varphi$ as we "roll out" the small strip (valid if t << R)

dA = R d \varphi t

Then we have

dM = TR = dA \tau R=\tau tR^2 d\varphi

the total moment is the sum of all infinitesimal moments

\def\arraystretch{2.0} \begin{array}{l} M=\int dM =\int_0^{2\pi } \tau R^2 d\varphi \\ M=2\pi R^2 t\tau \end{array}

\Rightarrow \tau =\dfrac{M}{2\pi R^2 t} \;\left\lbrack \dfrac{N}{ {\mathrm{m} }^2 }\right\rbrack

The denominator is known as the torsional resistance, $W$ .

\boxed{ \tau_{\max } =\dfrac{M}{W} }

For a solid section we have

W=\dfrac{\pi R^3 }{2}=\dfrac{\pi d^3 }{16}

For a thick tube we have

W=\dfrac{\pi \left(R^4 -r^4 \right)}{32R}=\dfrac{\pi \left(D^4 -d^4 \right)}{16D}

and since earlier, for a thin tube with a thickness of $t$ we have

W=2\pi R^2 t

The shear stress $\tau$ will vary linearly from zero at the center of the shaft to $\tau_{\max }$ at its outer surface, this can be expressed as

\tau =\dfrac{\rho }{R}\tau_{\max }

The infinitesimal moment can also be given as the product of the moment-arm $\rho$ and the force

dM = \rho \tau dA

Thus the total moment is

M=\int \rho \tau dA=\int \rho \frac{\rho }{R}\tau_{\max } dA

Note that $\frac{\tau_{\max } }{R}$ is constant, so we can pull it out of the integral, and we get

M=\dfrac{\tau_{\max } }{R}\int \rho^2 dA

This integral is known as the polar moment of inertia.

\boxed{J=\int \rho^2 dA}

Which we can also use to compute the maximum torsional shear stress

\boxed{ \tau =\dfrac{M \rho }{J}\Rightarrow \tau_{\max } =\dfrac{MR}{J} }

So we see the relation between the polar moment of inertia $J$ and the torsional resistance $W$ is given by

W=\dfrac{J}{R}

Angle of twist

2: Consider a circular shaft under torsion. We want to find the relation between the applied Moment

M

and the angle of twist

\theta

For small angles we have the geometric relation

\gamma dx = Rd \theta

If we assume that the material is linear elastic, Hooke's law holds and

\tau =G\gamma

where $G$ is the shear modulus, and is either determined experimentally, or computed from the Young's modulus and Poisson's ratio. It can be shown that

G=\dfrac{E}{2\left(1+\nu \right)}

Thus we can get the relation between the twist angle and applied moment by adding up all small contributions

d\theta =\dfrac{\gamma }{R} dx = \dfrac{\tau }{GR } dx

we know that for a solid rod $\tau =\frac{2M}{\pi R^3 }$ , so

d\theta =\dfrac{2M}{\pi R^4 }\dfrac{1}{G} dx

\Rightarrow \int_0^{\theta } d\theta =\int_0^L \dfrac{2M}{\pi R^4 }\dfrac{1}{G} dx

= \boxed{ \theta =\dfrac{LM}{GJ} }

$GJ$ is known as the torsional rigidity. This assumes that everything is constant along the length of the rod. If something varies, then we need to integrate.

\boxed{ \theta =\int_0^L \dfrac{M\left(x\right)}{G\left(x\right)J\left(x\right)} dx }

⚠ Note

Note that this relation is valid under small deformations and for circular cross sections. There are approximate formulas for torsional resistance (

W

) in various sources (check the formulary for some examples). A far more accurate method is to utilize Finite Element Analysis for more complex cross sections to determine stresses and displacements.

Applied Mechanics

Stress

Angle of twist