Internal expanding shoe brake analysis

Method

For a pivoted shoe, the normal pressure is proportional to $\sin\theta$ ($\theta$ from the pivot line): $p = p_a\sin\theta/\sin\theta_a$, where $\theta_a$ is the angle of maximum pressure. As $\theta_2 = 120^\circ > 90^\circ$, $\theta_a = 90^\circ$ and $\sin\theta_a = 1$.

Data: $r = 0.15$ m, $a = 0.11$ m, $b = 0.04$ m, $\theta_1 = 10^\circ$, $\theta_2 = 120^\circ$, $p_a = 0.8\times10^6$ Pa, $\mu = 0.28$, $c = 0.2$ m.

         drum centre O
            |  a = 110
   pivot A -+--- shoe from 10 deg to 120 deg
            |
        F at c = 200 from A

Braking torque

$$T = \mu p_a b\,r^2(\cos\theta_1 - \cos\theta_2)$$ $$T = 0.28\times0.8\times10^6\times0.04\times0.15^2\times(0.9848 + 0.5) = 299.3\ \text{N m}$$

Moment of the normal forces about the pivot

$$M_N = p_a b\,r\,a\int_{\theta_1}^{\theta_2}\sin^2\theta\,d\theta = p_a b\,r\,a\left[\frac{\theta_2-\theta_1}{2} - \frac{\sin2\theta_2-\sin2\theta_1}{4}\right]$$

The bracket $= \dfrac{1.9199}{2} - \dfrac{(-0.8660) - 0.3420}{4} = 0.9599 + 0.3020 = 1.2619$ (angles in radians: $\theta_2-\theta_1 = 1.9199$ rad).

$$M_N = 0.8\times10^6\times0.04\times0.15\times0.11\times1.2619 = 666.3\ \text{N m}$$

Moment of friction forces about the pivot

$$M_f = \mu p_a b\,r\int_{\theta_1}^{\theta_2}\sin\theta\,(r - a\cos\theta)\,d\theta = \mu p_a b\,r\left[r(\cos\theta_1-\cos\theta_2) - \frac{a}{2}(\sin^2\theta_2 - \sin^2\theta_1)\right]$$ $$M_f = 0.28\times0.8\times10^6\times0.04\times0.15\,[\,0.15\times1.4848 - 0.055\times(0.75-0.0302)\,] = 246.1\ \text{N m}$$

Actuating force

• Self-energising (friction moment assists the actuating force): $F\,c = M_N - M_f$

$$F = \frac{666.3 - 246.1}{0.2} = 2101\ \text{N}$$

• Non-self-energising (friction moment opposes): $F\,c = M_N + M_f$

$$F = \frac{666.3 + 246.1}{0.2} = 4562\ \text{N}$$

The self-energising (leading) shoe needs less than half the force for the same maximum pressure.

Self-locking

Self-locking occurs when $M_f \ge M_N$ ($F \le 0$). $M_f$ is proportional to $\mu$:

$$\mu_{lock} = 0.28\times\frac{666.3}{246.1} = 0.76$$

Answer: $T = 299$ N m per shoe; $F = 2.10$ kN (self-energising), $4.56$ kN (non-self-energising); self-locking at $\mu \ge 0.76$, so the brake is not self-locking at $\mu = 0.28$.