Belt and Rope Drives Brakes

Power is the time rate at which work is done or energy is transferred. In calculus terms, power is the derivative of work with respect to time.

An engine or motor is a machine designed to convert energy into useful mechanical motion.

Centrifugal force represents the effects of inertia that arise in connection with rotation and which are experienced as an outward force away from the center of rotation.

A band brake is a brake in which the frictional force is applied by increasing the tension in a flexible band to tighten it around the drum.

The Theory behind Belt and Rope drives using both Flat and Grooved pulleys. Included in this section are drum brakes

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Introduction

Where power has to be transmitted between two shafts which are at a distance apart, a belt or rope drive is frequently used. In most cases the power transmitted relies upon the friction between the rope or belt and the rim of the pulley.

However in the case of toothed belts and chains, friction does not play a beneficial part, no slip is allowable and a precise relationship between the movement of the pulleys is obtained. Examples of this sort of drive are the cam belt used on many modern engines and the simple bicycle chain.

Rope drives, as such, are not now much used, but the principle is alive and well in shipping where capstans and windlasses are used to tighten ropes and chains.

Flat belts were used extensively in both agriculture to connect both steam and early tractor to static machinery and also in early factories where all the machinery in a building was powered by one engine driving a series of shafts, pulley wheels and belts. These belts could be long enough to transmit power from one floor of a building to the one above. A selection of arrangements of belt drives is shown in the diagram below.

Today the majority of belts in use are Vee often used in sets of three,side by side. A modern development seen on some cars, is a flat belt with a series of mini Vees on the inner surface.

Flat Pulley Drives

Although described as Flat many pulleys used with flat belts are actually slightly curved so that the diameter at the middle is slightly larger than that at the ends.

This helps to keep the belt on the pulley.

It is also normal on long belts to twist them as shown in the diagram. This can be seen in the above diagram and increases the angle of contact between the belt and the pulley. Another technique to increase the angle is the use of idlers which are small pulley wheels which help to control slack in the belt.

At Low Speed

The maximum torque that can be transmitted between the belt and the pulley occurs when limiting friction is developed around the arc of contact.

The forces acting on a short length of the belt which subtends an angle of $\delta \theta$ at the centre of the pulley, are shown on the above diagram.

Resolving Tangentially: $\mu R=(T+\delta T)\cos\frac{1}{2}\delta \theta -T\cos\frac{1}{2}\delta \theta$

Thus in the limit:

$\mu R=\delta T$

(1)

Resolving Radially:

Drawing of pulley showing tensions and radial $T$ : $T_r'=T\sin\displaystyle\frac{1}{2}\delta \theta$ therefore, $T_r''=(T+\delta T)\sin\displaystyle\frac{1}{2}\delta \theta$

Adding $R=(T+\delta T)\sin\frac{1}{2}\delta \theta +T\sin\frac{1}{2}\delta \theta$ $\text{As}\;\;\delta \theta \rightarrow 0\;\;\;\sin\frac{1}{2}\delta \theta \rightarrow \theta \;\;\text{and}\;\;\delta T\rightarrow 0$

Hence in the limit:

$R=T\delta \theta$

(2)

Combining equations (#1) and (#2) and eliminating $R$ : $\frac{\delta T}{T}=\mu\, \delta \theta$

This equation can now be integrated over the whole arc $\theta$ . $\therefore \;\;\;\;\;\;\ln\left ( \frac{T_1}{T_2} \right )=\mu \,\theta$ Or,

$\frac{T_1}{T_2}=e^{\mu \theta }$

(3)

This is the maximum possible ratio of tensions between the "tight" and "slack" sides of the belt and consequently for a drive between pulleys of unequal diameter, the smaller angle of lap should be used.

It should be noted that in some applications such as Capstans and Windlasses the rope is passed completely round the pulley, sometimes more than once.

Allowance for Centrifugal Force

The analysis above assumes, in effect, that the belt and pulley is at rest. In practice this is not often the case when the belt is subjected to a centrifugal force as if passes round the pulley.

If $w$ is the weight of a unit length of belt and $v$ is the linear belt speed, then the force on an element:

$=\left ( \frac{w\,r\,\delta \theta }{g} \right )\times \frac{v^2}{r}=\frac{w\,v^2}{g}\times\delta \theta$

This force has to be balanced by a tension $T_C$ in the belt. The value of this tension is obtained by resolving radially.

Hence $2\,T_c\times \sin\displaystyle\frac{1}{2}\delta \theta =\left ( \displaystyle\frac{wv^2}{g} \right )\delta \theta$

$\text{As}\;\;\;\;\;\delta \theta \rightarrow 0\;\;\;\sin\frac{1}{2}\delta \theta \rightarrow \frac{\delta \theta }{2}$

$\therefore \;\;\;\;\;T_C=\frac{w\,v^2}{g}$

(4)

The Total Tensions on the two sides are then:

$T_{tight}=T_1+T_C\;\;\,\;\;\;T_{slack}=T_2+T_C$

(5)

where $T_1$ and $T_2$ are the tensions transmitting the power and are calculated from equation (#2).

Equation (#2) can now be written as: $\displaystyle\frac{T_T-T_C}{T_S-T_C}=e^{\mu \;\theta }$

Grooved Pulley Drives

Although rope can be used for such drives, the vast majority rely on V-belts. In many applications the pulley has more than one groove and there are obviously an appropriate number of belts.

Assuming that the included angle of the groove is $2\beta$ , the Normal Reaction $R$ is given by :

$R=2N\sin\beta$ But Friction

$=\mu \times 2N=\displaystyle\frac{\mu \,R}{\sin\beta }= mu '\,R$

(6)

(using the above equation)
Where $mu '=\displaystyle\frac{\mu }{\sin\beta }$

The analysis now becomes similar to a flat pulley where $\mu '$ is the equivalent coefficient of friction.

The Power Transmitted

Two sets of equations are given below to cater for both measurement systems:

Imperial

The horse-power transmitted

$=(T_1-T_2)\times\frac{v}{550}$

(7)

Using equation (#3) this can be rewritten to give the maximum power transmitted i e.

$h.p.=T_1(1-e^{-\mu\, \theta })\times \frac{v}{550}$

(8)

And using equation (#5)

$h.p.=(T_T-T_C)(1-e^{-\mu\, \theta })\times \frac{v}{550}$

(9)

S.I.Units

Power transmitted in Kilo Watts $=(T_1-T_2)\times \displaystyle\frac{v}{1,000}$

Combining with equation (#3), $=T_1\left ( 1-e^{-\mu \,\theta } \right )\times\displaystyle\frac{v}{1,000}$

And using equation (#5) $=\left ( T_T-T_C \right )\left ( 1-e^{-\mu \,\theta } \right )\times\displaystyle\frac{v}{1,000}$

For a limiting value of $T_T$ , it can be shown that the above equations for the power transmitted have a maximum value when:
$v=\sqrt{\left ( \displaystyle\frac{g\,T_T}{3w} \right )}$ i.e. $T_C=\displaystyle\frac{1}{3}T_T$

It is usual to assume that the mean tension $\displaystyle\frac{1}{2}(T_T-T_S)$ is constant at all speeds and is therefore equal to the initial tension $T_0$ .
If $T_C$ is neglected it then follows from equation (#3) that: $T_1=\displaystyle\frac{2\,T_0}{1+e^{-\mu \,\theta }}$

And hence the power transmitted at a given speed is proportional to the initial tension.

The principles of Band Brakes are similar to those of the preceding paragraphs, giving, in this case the braking torque which can be applied to a rotating drum.

There has been a slow but steady movement away from the use of Shoe and Drum Brakes by the automotive industry. The method of tackling problems associated with this type of brake can be seen in Example (15)

Example 1 [imperial]

Problem

A number of trucks are to be hauled by a rope passed round a hydraulic capstan; assuming a coefficient of friction of 0.25 between the rope and the capstan and a constant manual pull of 30 lb., derive an expression for the maximum pull on the trucks in terms of complete turns of rope round the capstan.

What will be the horse power exerted by the capstan under these conditions when there are $2\displaystyle\frac{1}{2}$ turns of rope and it is being wound off at 100 ft/min.?

Workings

From equation (3) $\frac{T_1}{T_2}=e^{\mu \,\theta }$

From the question:
$T_2=30\,lb.$ , $\mu =0.25$ and $\theta =2\,\pi\, n\;radians$

From which: $T_1=30\;e^{0.25\times2\,\pi \,n}=30(4.82)^n$

When $n = 2.5$ , $T_1=1520\;lb.$

Using equation (7) the horse-power of the capstan is given by: $\frac{(T_1-T_2)\times v}{550} = \frac{(1520-30)\times 100}{550\times 60}=4.52 \;h.p.$

To which must be added the power supplied manually which is: $\frac{30\times 100}{550\times 60}=0.091\;h.p.$ So the horse-power taken by the trucks $= 4.52+0.091=4.611\;h.p.$