Simple band brake absorbing power

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): $\dfrac{T_1}{T_2} = e^{\mu\theta} = e^{0.25\times4.712} = e^{1.178} = 3.248$.

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

  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.