3 Engineering Example 2

3.1 Torsion of a mild-steel bar

Introduction

For materials such as mild-steel, the relationship between applied shear stress and shear strain (deformation) can be described as follows.

• For small values of the shear strain, the shear stress ( $\tau$ ) and shear strain ( $\omega$ ) are proportional to one another, i.e.

  $\omega =\frac{1}{G}\times \tau$ (1)

  (where $G$ is the shear modulus ). This is known as elastic behaviour .

• There is a maximum shear stress that the material is capable of supporting. If the shear strain is increased further, the shear stress remains roughly constant. This is known as plastic behaviour .

Figure 3 summarises the relationship between shear stress and shear strain; the point $\left({\omega }_{{}_{{}_Y}},{\tau }_{{}_Y}\right)$ is known as the yield point .

Figure 3

Now suppose that one end of a bar of circular cross section is twisted through an angle $\theta$ , then the shear strain on the surface is given by

${\omega }_{{}_{{}_S}}=\frac{R\theta }{L}$ (2)

(where $R$ and $L$ are the radius and length of the bar respectively), while the shear strain, at a distance $r$ from the central core, is given by

$\omega =\frac{r\theta }{L}$ (3)

The torque transmitted by a bar is given by the integral

$T={\int \; <!--nolimits\; -->}_0^{R}2\pi r^2\tau \left(r\right)dr$ (4)

As the shear strain is a function of distance from the central axis of the bar, it may be that the shear strain on the surface is greater than the critical shear strain ${\omega }_{{}_{{}_Y}}$ . In this scenario the shear stress is given by

$\tau =\left\{\begin{array}{cc}\frac{{\tau }_{{}_Y}}{{\omega }_{{}_{{}_Y}}}\omega \hfill & \omega \le {\omega }_{{}_{{}_Y}}\hfill \\ & \\ {\tau }_{{}_Y}\hfill & \omega >{\omega }_{{}_{{}_Y}}\hfill \end{array}\right.$ (5)

i.e. the regions near the central axis exhibit elasticity, but in those regions near the surface the elastic limit has been exceeded and the metal exhibits plasticity (see Figure 4).

Figure 4

Problem in words

Find an expression for the torque transmitted by a bar as a function of the angle $\theta$ through which one end is turned.

Mathematical statement of problem

Using Equations (3) to (5), find a formula for $T$ in terms of the variable $\theta$ .

Mathematical analysis

Substituting (3) into (5)

For small values of $\theta$ , $r_e\ge R$ so that the whole of the bar will be in the elastic region, i.e.

$\tau =\frac{{\tau }_{{}_Y}}{{\omega }_{{}_{{}_Y}}}\frac{r\theta }{L}$

Now (4) becomes

$T={\int \; <!--nolimits\; -->}_0^{R}2\pi r^2\frac{{\tau }_{{}_Y}}{{\omega }_{{}_{{}_Y}}}\frac{r\theta }{L}dr=2\pi \frac{{\tau }_{{}_Y}}{{\omega }_{{}_{{}_Y}}}\frac{\theta }{L}{\int \; <!--nolimits\; -->}_0^R{r}^{3}dr=2\pi \frac{{\tau }_{{}_Y}}{{\omega }_{{}_{{}_Y}}}\frac{\theta }{L}{\left[\frac{r^4}{4}\right]}_0^R=\frac{\pi }{2}\frac{{\tau }_{{}_Y}}{{\omega }_{{}_{{}_Y}}}\frac{\theta }{L}{R}^4$ (6)

i.e. the torque is directly proportional to the twist, $\theta$ .

For larger $\theta$ , $r_e<R$ , so that (4) becomes

But $r_e=L{\omega }_{{}_{{}_Y}}∕\theta$ , so

Equation (6) will apply when $r_e\ge R$ , i.e. $\left(L{\omega }_{{}_{{}_Y}}∕\theta \right)\ge R$ or $\theta \le \left(L{\omega }_{{}_{{}_Y}}∕R\right)$ , so that combining (6) and (7) gives overall

$T=\left\{\begin{array}{cc}\frac{\pi }{2}\frac{{\tau }_{{}_Y}}{{\omega }_{{}_{{}_Y}}}\frac{\theta }{L}{R}^4\hfill & \theta \le \frac{L{\omega }_{{}_{{}_Y}}}{R}\hfill \\ & \\ \frac{2\pi }{3}{\tau }_{{}_Y}{R}^3-\frac{\pi }{6}{\tau }_{{}_Y}\frac{L^3{\omega }_{{}_{{}_Y}}^3}{{\theta }^3}\hfill & \theta >\frac{L{\omega }_{{}_{{}_Y}}}{R}\hfill \end{array}\right.$ (8)

Interpretation and further comment

At the critical value of $\theta$ , i.e. when the outer edge begins to exhibit plasticity, both formulae in (8) give

$T_{crit}=\frac{\pi }{2}{\tau }_{{}_Y}{R}^3$

Furthermore, the first derivatives are both

$\frac{dT}{d\theta }=\frac{\pi }{2}\frac{{\tau }_{{}_Y}}{{\omega }_{{}_{{}_Y}}}\frac{R^4}{L}$

i.e. the curves join smoothly.

The second derivatives, though, are not equal (zero in one case). In the theoretical limit as $\theta \to \infty$

$T=\frac{2\pi }{3}{\tau }_{{}_Y}{R}^3$

so this is the total torsional torque which can be carried by the bar. (The critical torque above is three-quarters of this value.) However, clearly $\theta \to \infty$ is merely a theoretical limit since the bar would, in fact, shear at a finite value of $\theta$ .