Question 5 of 6: Long-shoe external drum brake - maximum pressure, torque and power

Question 5: Long-shoe external drum brake - maximum pressure, torque and power

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) the maximum lining pressure $p_{a}$; (b) the braking torque and the power the brake can absorb at 250 rpm.

Approach. Treat the shoe as a long shoe: the lining pressure varies as the sine of the angle measured from the hinge-pin line, so the normal and friction tractions are integrated over the arc, moments are taken about the pin to find the peak pressure from the actuating force, and the friction moment about the drum axis then gives the torque.

1. Locate the hinge pin relative to the drum centre. The pin sits 155 mm to the left of the drum axis and above it; scaling the drawing gives the pin line at $\alpha = 55^\circ$ from the vertical, so$$a = \frac{155}{\sin 55^\circ} = 189.2\ \text{mm}$$This distance $a$ is the moment arm of the whole traction system about the pin, and it is the one quantity that must be extracted from the figure rather than read off a dimension.
2. Convert the lining arc to angles measured from the pin line. The lining runs from 45° on one side of the vertical to 45° on the other, and the pin line is 55° from the vertical, so$$\theta_{1} = \alpha - 45^\circ = 10^\circ, \qquad \theta_{2} = \alpha + 45^\circ = 100^\circ$$Since $\theta_{2} > 90^\circ$, the maximum of $\sin\theta$ falls inside the lining, so the pressure peaks at $\theta_{a} = 90^\circ$ and $\sin\theta_{a} = 1$. Taking $\theta_{a} = \theta_{2}$ here would be the standard error.
3. Pressure distribution. The shoe is rigid and pivots about A, so the normal approach of the lining to the drum, and hence the pressure, varies as$$p(\theta) = p_{a}\,\frac{\sin\theta}{\sin\theta_{a}} = p_{a}\sin\theta$$with $\theta$ measured from OA. The pressure is a tenth of its peak at the leading edge and falls only to 98 % of the peak at the trailing edge.
4. Moment of the normal tractions about the pin. Integrating $dN = p\,b\,r\,d\theta$ with its moment arm $a\sin\theta$ about A,$$M_{N} = \frac{p_{a}br a}{\sin\theta_{a}}\int_{\theta_{1}}^{\theta_{2}}\sin^{2}\theta\,d\theta = p_{a}(75)(150)(189.2)(0.9564) = 2.036\times10^{6}\,p_{a}$$in N·mm with $p_{a}$ in MPa.
5. Moment of the friction tractions about the pin. Each friction element $dF = f\,dN$ acts tangentially, with moment arm $(r - a\cos\theta)$ about A:$$M_{F} = \frac{f p_{a}br}{\sin\theta_{a}}\int_{\theta_{1}}^{\theta_{2}} \sin\theta\,(r - a\cos\theta)\,d\theta = f p_{a}br\left[r(1.1585) - a(0.4698)\right]$$$$M_{F} = 0.4\,p_{a}(75)(150)\left[150(1.1585) - 189.2(0.4698)\right] = 0.382\times10^{6}\,p_{a}$$
6. Decide the sense of the friction moment. The drum turns counter-clockwise, so at the lining - which lies above the drum axis - the drum surface sweeps towards the pin. Taking moments about A shows that this friction moment acts in the same sense as the normal-traction moment, i.e. it resists the applied moment: the shoe is de-energising. The equilibrium of the shoe is therefore$$F_{a}c = M_{N} + M_{F}$$Had the drum turned the other way the friction moment would subtract, the shoe would be self-energising, and the same actuating force would produce a much higher pressure.
7. Maximum pressure. Substituting $F_{a}c = 5000(405) = 2.025\times10^{6}$ N·mm,$$p_{a} = \frac{F_{a}c}{M_{N}+M_{F}} = \frac{2.025\times10^{6}}{(2.036+0.382)\times10^{6}} = \boxed{0.838\ \text{MPa}}$$which is comfortably within the 1.0-1.7 MPa allowable of a moulded or woven asbestos-free lining.
8. Braking torque. Every friction element acts at the constant radius $r$ about the drum axis, so the integral carries $\sin\theta$ rather than $\sin^{2}\theta$:$$T = \frac{f p_{a} b r^{2}(\cos\theta_{1}-\cos\theta_{2})}{\sin\theta_{a}} = 0.4(0.838)(75)(150)^{2}\left[\cos 10^\circ - \cos 100^\circ\right]$$$$T = 0.4(0.838)(75)(22\,500)(0.9848+0.1736) = 654.9\times10^{3}\ \text{N}\cdot\text{mm} = \boxed{655\ \text{N}\cdot\text{m}}$$
9. Power capacity. With $\omega = 2\pi(250)/60 = 26.18$ rad/s,$$\dot{W} = T\omega = 654.9(26.18) = 17.15\times10^{3}\ \text{W} = \boxed{17.1\ \text{kW}}$$This is the instantaneous rate at which the brake converts mechanical energy to heat at rated speed; continuous operation at this level would require the drum surface to dissipate about 0.5 MW/m² and would not be sustainable without forced cooling.