---
title: "Internal expanding shoe brake analysis"
subject: "Machine Design II"
chapter: 7
key: "semester_7/machine_design_ii:internal-expanding-shoe-brake"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/internal-expanding-shoe-brake
site: BCT Engineering
---

# Internal expanding shoe brake analysis

Machine Design II · Chapter 7: Clutches and brakes

- Asked in: year not given, 8 marks
- Video: [Design of Machine Elements - Internal Expanding Shoe Brake](https://www.youtube.com/watch?v=duHGuGWsQgE) — Dr. M. Raja Roy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/internal-expanding-shoe-brake
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-7

## Question

An internal expanding shoe brake has a drum of radius 150 mm. The shoe is pivoted at a point 110 mm from the drum centre. The friction lining has a width of 40 mm, extends from 10 degrees to 120 degrees measured from the line joining the pivot to the drum centre, and the allowed maximum pressure is 0.8 MPa. The coefficient of friction is 0.28. The actuating force acts perpendicular to the pivot line at a distance of 200 mm from the pivot. Find the braking torque of the shoe, the actuating force when the rotation makes the shoe self-energising, the actuating force when the shoe is non-self-energising, and the value of the coefficient of friction for which the shoe would self-lock.

## Short answer

Using pressure distribution p ∝ sinθ, braking torque per shoe is 299 N m. Self-energising actuating force is 2.10 kN, non-self-energising is 4.56 kN. The brake self-locks at coefficient of friction 0.76.

## Answer

#### 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.

```text
         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 $\mu = 0.28$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the pivoted shoe with theta_2 = 120 degrees, where is the pressure maximum?
   A. At theta = 90 degrees, since p is proportional to sin(theta)
   B. At theta = 10 degrees
   C. At theta = 120 degrees
   D. At the pivot
   Answer: A. p = p_a sin(theta)/sin(theta_a), and since theta_2 > 90 degrees, theta_a = 90 degrees. (Key point: Pressure distribution)

2. The braking torque of the shoe is given by
   A. T = mu p_a b r^2 (sin(theta_2) - sin(theta_1))
   B. T = mu p_a b r (cos(theta_1) - cos(theta_2))
   C. T = mu p_a b r^2 (cos(theta_1) - cos(theta_2))
   D. T = p_a b r^2 (cos(theta_1) - cos(theta_2))
   Answer: C. Integrating mu p b r^2 sin(theta) dtheta gives this, 299.3 N m. (Key point: Braking torque)

3. Which moment equations apply?
   A. Self-energising: F c = M_N - M_f; non-self-energising: F c = M_N + M_f
   B. Self-energising: F c = M_N + M_f; non-self-energising: F c = M_N - M_f
   C. Both: F c = M_N
   D. Self-energising: F c = M_f - M_N
   Answer: A. Friction moment helps the actuating force in the self-energising case, so less force (2101 N) is needed than 4562 N. (Key point: Actuating force cases)

4. When does the shoe self-lock and what is mu_lock?
   A. When M_f >= M_N; mu_lock = 0.28 x 666.3/246.1 = 0.76
   B. When M_f <= M_N; mu_lock = 0.28
   C. When F is maximum; mu_lock = 0.5
   D. When M_N = 0; mu_lock = 0.76
   Answer: A. M_f is proportional to mu, so locking occurs at mu = 0.28 x M_N/M_f = 0.76, hence not at 0.28. (Key point: Self-locking)

## Related in Machine Design II

- [Brake energy and drum temperature rise](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Abrake-energy-absorbed-temperature-rise) — Machine Design II, chapter 7
- [Simple band brake absorbing power](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aband-brake-power-absorbed) — Machine Design II, chapter 7
- [Clutch versus brake and self-energising](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aclutch-brake-differences-self-energizing) — Machine Design II, chapter 7

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