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Maximum holding torque of a band brake

This exercise is about the appliance of the belt friction formula (here is an online belt friction calculation tool).

Task

The band brake shown should be operated right-turning and left-turning. What is the maximum holding torque in right-turning and left-turning rotation without the brake drum rotating?

Band brake for clockwise and counterclockwise rotation with operating lever
Band brake for clockwise and counterclockwise rotation with operating lever

Solution

A solution video will be published soon on this channel.

The first step is the free-cut of the operator lever. The force FS doesn’t have a real impact here, but it is included for formal reasons.

Forces on the operating lever
Forces on the operating lever
\[ \require{cancel} \]

The balance of forces in x-direction is

\[ \tag{1} \sum F_x = 0 = F_{Ax} -- F_S \cdot \cos 45° \]

The balance of forces in y-direction is

\[ \tag{2} \sum F_y = 0 = F_{Ay} + F_S \cdot \sin 45° + F_B -- F \]

Both balances don’t yield useful informations, but the balance of moments around A gives

\[ \tag{3} \sum M(A) = 0 = F_B \cdot r -- F \cdot l \]

\[ \tag{4} F_B = \frac{F \cdot l}{r} \]

Next, the relationship between the rope force FS and the braking force FB is noted using the rope friction formula (Euler-Eytelwein). This is done separately for the two different directions of rotation.

Right-turning

Brake drum with counterclockwise moment
Right turning drum, the braking moment acts left turning
\[ \require{cancel} \]

In the case of the left-turning moment, F B lies in the strained strand, F S in the loose strand. The following applies:

\[ \tag{5} F_B \leq F_S \cdot e^{µ \cdot \alpha} \]

\[ \tag{6} F_S \geq F_B \cdot e^{-µ \cdot \alpha} \]

In union with the equilibrium of moments around the drum pivot, it follows:

\[ \tag{7} \sum M = 0 = F_S \cdot r + M_L -- F_B \cdot r \]

\[ \tag{8} M_L = F_B \cdot r -- F_S \cdot r \]

\[ \tag{9} M_L = \frac{F \cdot l}{\bcancel{r}} \cdot \bcancel{r} -- \frac{F \cdot l}{\bcancel{r}} \cdot e^{-µ \cdot \alpha} \cdot \bcancel{r}\]

\[ \tag{10} M_L = F \cdot l \cdot \left( 1 -- e^{-µ \cdot \alpha} \right) \]

Left-turning

Brake drum with clockwise turning moment
Left turning drum, the braking moment acts right turning
\[ \require{cancel} \]

At the right-turning moment, FB lies in the loose strand, FS in the strained strand. The following applies:

\[ \tag{11} F_S \leq F_B \cdot e^{µ \cdot \alpha} \]

In union with the equilibrium of moments around the drum pivot, it follows:

\[ \tag{12} \sum M = 0 = F_S \cdot r -- M_R -- F_B \cdot r \]

\[ \tag{13} M_R = F_S \cdot r -- F_B \cdot r \]

\[ \tag{14} M_R = \frac{F \cdot l}{\bcancel{r}} \cdot e^{µ \cdot \alpha} \cdot \bcancel{r} -- \frac{F \cdot l}{\bcancel{r}} \cdot \bcancel{r} \]

\[ \tag{15} M_R = F \cdot l \cdot \left(e^{µ \cdot \alpha} -- 1 \right) \]