---
title: "Simple band brake absorbing power"
subject: "Machine Design II"
chapter: 7
key: "semester_7/machine_design_ii:band-brake-power-absorbed"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/band-brake-power-absorbed
site: BCT Engineering
---

# Simple band brake absorbing power

Machine Design II · Chapter 7: Clutches and brakes

- Asked in: year not given, 8 marks
- Video: [BAND BRAKE II L-6 NUMERICAL-SIMPLE BAND BRAKE II #theory_of_machine #theoryofmachine #machinedesign](https://www.youtube.com/watch?v=rVVh3wUsphg) — SCA
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/band-brake-power-absorbed
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-7

## Question

A simple band brake acts on a drum of diameter 500 mm that rotates at 300 rpm and absorbs 15 kW. The coefficient of friction is 0.25 and the angle of contact is 270 degrees. One end of the band is attached to the fulcrum of the lever and the other end to a pin on the lever at 60 mm from the fulcrum; the operating force acts at 700 mm from the fulcrum. Find (a) the braking torque and the tensions in the band, (b) the force required at the end of the lever when the slack end is attached to the lever pin and when the tight end is attached to it, and (c) the width of the steel band if its thickness is 3 mm, the allowable stress is 60 MPa and the joint efficiency is 80%.

## Short answer

Braking torque is 477.5 N m. Tensions: T1 = 2759 N, T2 = 850 N. Lever force: 72.8 N (slack on pin) or 236.5 N (tight on pin). Band width: 20 mm.

## Answer

#### Data
$D = 500$ mm ($r = 0.25$ m), $N = 300$ rpm, $P = 15$ kW, $\mu = 0.25$, $\theta = 270^\circ = 4.712$ rad, $a = 60$ mm, $l = 700$ mm.

#### (a) Torque and tensions
$$
\omega = \frac{2\pi\times300}{60} = 31.42\ \text{rad/s}, \qquad T_B = \frac{P}{\omega} = \frac{15000}{31.42} = 477.5\ \text{N m}
$$

Net braking force on the band: $T_1 - T_2 = \dfrac{T_B}{r} = \dfrac{477.5}{0.25} = 1910\ \text{N}$.

Ratio of tensions (belt friction): $\dfrac{T_1}{T_2} = e^{\mu\theta} = e^{0.25\times4.712} = e^{1.178} = 3.248$.

$$
T_2 = \frac{1910}{3.248-1} = 849.5\ \text{N}, \qquad T_1 = 3.248\times849.5 = 2759.4\ \text{N}
$$

```text
  fulcrum O    pin A (a=60)         force F
     |-----------|------------------->| l=700
      \band ends/
        drum with tight T1, slack T2
```

#### (b) Lever force
Taking moments about the fulcrum, $F\,l = T_{pin}\,a$.

- Slack end on the pin ($T_{pin} = T_2$): 

$$
F = \frac{T_2\,a}{l} = \frac{849.5\times60}{700} = 72.8\ \text{N}
$$

- Tight end on the pin ($T_{pin} = T_1$):

$$
F = \frac{T_1\,a}{l} = \frac{2759.4\times60}{700} = 236.5\ \text{N}
$$

The slack-end arrangement needs about one third of the force, so it is used in the direction of main rotation.

#### (c) Band width
Maximum tension is $T_1$. With joint efficiency $\eta$:

$$
T_1 = \sigma\,b\,t\,\eta \;\Rightarrow\; b = \frac{2759.4}{60\times3\times0.8} = 19.2\ \text{mm} \;\Rightarrow\; 20\ \text{mm}
$$

**Answer:** $T_B = 477.5$ N m, $T_1 = 2759$ N, $T_2 = 850$ N; lever force 72.8 N (slack end on pin) or 236.5 N (tight end on pin); band width 20 mm.

## Check yourself

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

1. How is the braking torque found?
   A. T_B = P/omega = 15000/31.42 = 477.5 N m
   B. T_B = P x omega
   C. T_B = (T1 + T2) r
   D. T_B = P/N
   Answer: A. Torque equals power over angular speed. (Key point: Braking torque)

2. How are the band tensions T1 and T2 found?
   A. T1 + T2 = T_B/r and T1/T2 = mu theta
   B. T1 - T2 = T_B/r and T1/T2 = e^(mu theta)
   C. T1 - T2 = T_B and T1/T2 = mu theta
   D. T1 x T2 = T_B/r and T1/T2 = e^theta
   Answer: B. Net force is T_B/r = 1910 N and the ratio is e^(0.25 x 4.712) = 3.248, giving T2 = 849.5 N, T1 = 2759 N. (Key point: Belt tension ratio)

3. Why is the slack end attached to the pin in the direction of main rotation?
   A. It needs about one third of the lever force (72.8 N vs 236.5 N)
   B. It gives a higher braking torque
   C. It lowers the band tension
   D. It gives a smaller band width
   Answer: A. F = T_pin a / l, so with T2 instead of T1 the force is much smaller. (Key point: Lever force comparison)

4. Which tension and efficiency are used for the band width?
   A. T2 with joint efficiency
   B. T1 with joint efficiency: T1 = sigma b t eta
   C. T1 - T2 with no efficiency
   D. Average tension without efficiency
   Answer: B. The maximum tension T1 governs, so b = 2759.4/(60 x 3 x 0.8) = 19.2 mm, use 20 mm. (Key point: Band width)

## 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
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