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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.

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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.

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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.

(a) 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

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

(1)

Thus in the limit

$\mu R=\delta T$

(2)

Resolving Radially

Drawing of pulley showing tensions and radial T

$T_r'=T\sin\frac{1}{2}\delta \theta$

(3)

$T_r''=(T+\delta T)\sin\frac{1}{2}\delta \theta$

(4)

Adding

$R=(T+\delta T)\sin\frac{1}{2}\delta \theta +T\sin\frac{1}{2}\delta \theta$

(5)

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

(6)

Hence in the limit

$R=T\delta \theta$

(7)

Combining equations (2) and (7) and eliminating R

$\frac{\delta T}{T}=\mu\, \delta \theta$

(8)

This equation can now be integrated over the whole arc $\theta$ .

$\therefore \;\;\;\;\;\;\ln\left ( \frac{T_1}{T_2} \right )=\mu \,\theta$

(9)

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

(10)

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.

(b) 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 and 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, the the force on an element

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

(11)

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

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

(12)

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

(13)

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

(14)

The Total Tensions on the two sides are then -

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

(15)

where $T_1\;\;and\;\;T_2$ are the tensions transmitting the power and are calculated from equation (7)

Equation (7) can now be written as -

$\frac{T_T-T_C}{T_S-T_C}=e^{\mu \;\theta }$

(16)

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

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

$R=2N\sin\beta$

(17)

$\text{But Friction}=\mu \times 2N=\frac{\mu \,R}{\sin\beta }\;\;\;\;\text{using the above equation}$

(18)

$=\mu '\,R\;\;\;\;\;\;\;Where\;\;\;\mu '=\frac{\mu }{\sin\beta }$

(19)

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

(a) Imperial

$\text{The horse-power transmitted}=(T_1-T_2)\times\frac{v}{550}$

(20)

Using equation (10) this can be rewritten to give the maximum power transmitted i e.

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

(21)

And using equation (15)

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

(22)

(b) S.I.Units

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

(23)

Combining with equation (10)

$=T_1\left ( 1-e^{-\mu \,\theta } \right )\times\frac{v}{1,000}$

(24)

And using equation (15)

$=\left ( T_T-T_C \right )\left ( 1-e^{-\mu \,\theta } \right )\times\frac{v}{1,000}$

(25)

For a limiting value of $T_T$ It can be shown that the above equations for the power transmitted have a maximum value when -

$v=\sqrt{\left ( \frac{g\,T_T}{3w} \right )}\;\;\;\;\;i.e.\;\;\;\;T_C=\frac{1}{3}T_T$

(26)

It is usual to assume that the mean tension $\frac{1}{2}(T_T-T_S)$ 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 (10) that:

$T_1=\frac{2\,T_0}{1+e^{-\mu \,\theta }}$

(27)

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

Band Brakes

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.

Shoe Brakes

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)

Worked Examples

To see the workings associated with the following questions, "Click" on the button.

Example 1

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 $2\tfrac{1}{2}$ turns of rope and it is being wound off at 100 ft/min.? (U.L.)

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Example 2

A flexible elastic band passes over a pulley 20 in. in diameter which is supported in fixed horizontal bearings

The end A is anchored to the ground and the end B is held by a turnbuckle whose lower end is also anchored to the ground at C..

The turnbuckle, which has right-and left-hand square-threaded screws of $1\tfrac{1}{4}\;in.$ external diameter and $\frac{1}{4}\;in.$ pitch is used to tighten the band on the pulley until the couple on the turnbuckle is 100 lb.in.,the tension on the band being then assumed to be uniform.

The pulley is now rotated clockwise against the frictional resistance of the band. Assuming that the total length of the band now remains unchanged, find the couple required to produce this rotation.

The coefficient of friction between the screws and nuts = 0.15 and between the band and the pulley = 0.25. Neglect friction at the bearings and any extension between B and C (U.L.)

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Example 3

In the torque amplifier shown in the diagram, the input and output shafts rotate at the same speed and in the same direction. They are connected by a flexible band, attached to radius arms, which make 1.25 turns around a coaxial drum. The drum is driven at high speed by an electric motor. If the coefficient of friction between the drum and the band is 0.21, find the torque required at the input shaft to overcome a torque of 1.5 lb.ft. at the output shaft. If the drum speed is 5 r.p.s. and the shaft speeds are 2 r.p.s. find for the above conditions, the energy lost in drum friction expressed in ft.lb./sec.

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Example 4

An elastic belt makes contact with a pair of flat pulleys over arcs of $180^0$ , the coefficient of friction being 0.45. The initial tension is 90 lb. and it may be assumed that the sum of the tensions on the two sides remains constant. Find the difference in the tensions at which the belt slips on the pulleys.

If the belt is 1 inch wide and $\frac{1}{4}$ thick and the elastic modulus for the material is 35,000 lb./sq.in.estimate the effect on the maximum tension difference which w2ould be caused if the pulley centre moved together by $\frac{1}{8}$ in., if they were initially 3 ft. apart.

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Example 5

An open belt connects two flat pulleys. The smaller pulley is 1 ft. in diameter and runs at 200 r.p.m. The angle of lap on this pulley is $160^0$ and the coefficient of friction between belt and pulley face is 0.25. The belt is on the point of slipping when $3\tfrac{1}{2}$ h.p. is being transmitted.

Which of the following alternatives would be more effective in increasing the power which could be transmitted -

Increase the initial tension in the belt by 10%
Increase the coefficient of friction by 10% by the application of a suitable dressing onto the belt.

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Example 6

A leather belt $\tfrac{1}{4}$ in. thick connects two pulleys of 3 ft. 2 in. and 1 ft. 9 in. diameters, carried on parallel shafts at 11 ft. 6 in. centres; the speed of the larger pulley is 240 r.p.m.. The coefficient of friction is 0.25 and the maximum permissible load on the belt is 260 lb. Find the maximum horse-power and the necessary initial tension.

Plot on a base of initial mean tension, curves showing the horse-power which can be transmitted

Maximum loading but neglecting the possibility of slip.
Limitations due to slip but neglecting the possibility of overloading the belt.

Hence determine the permissible range of initial tension if

$7\tfrac{1}{2}$

(80)

is to be transmitted. (U.L.)

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Example 7

A small pulley of radius $r_1$ on a line shaft drives a large pulley of radius $r_2$ on a machine vertically below it, the centre distance being a distance d.

Show that slipping will be equally likely to occur at either pulley, if the tension in the belt where it runs over the larger pulley is given by -

$T=\frac{e^{\mu (\pi -2\theta )}-1}{e^{\pi +2\theta }-e^{\mu -2\theta }}\times w\;d$

(81)

Where w is the weigth of a unit length of belt and $\sin\theta=\frac{r_2-r_1}{d}$ (U.L.)

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Example 8

A leather belt transmits 25 h.p. from a pulley 3 ft. in diameter which runs at 300 r.p.m.; $\theta=165^0,\;\;\;\mu=0.27,$ the density of the belt is 0.035 lb/cu.in. and the maximum stress is not to exceed 350 lb/sq.in.

In the belt is $\frac{1}{4}$ in. thick, find the least possible width required.

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Example 9

An open belt drive connects two pulleys of 48 in. and 20 in.diameters, on parallel shafts 12 ft. apart. The belt weighs 0.6 lb/ft length and the maximum tension in it is not to exceed 4r500 lb. The coefficient of friction is 0.3.

The 48 in pulley, which is the driver, runs at 200 r.p.m.. Due to belt slip on one of the pulleys, the velocity of the driven shaft is only 450 r.p.m.. Calculate the torque on each of the two shafts, the horse-power transmitted and the horsepower lost in friction.

What is the efficiency of the drive ? (U.L.)

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Example 10

A pulley of 12 in. diameter running at 1,500 r.p.m. transmits 20 h.p. to an elastic belt driving a similar pulley. Find the required width of belt and the uniform tension when stationary, if the maximum tension under load is 100 lb./in.width. It may be assumed that the sum of the tensions in the two sides remains constant, $e^{\mu \,\theta }=2.5$ and the weight of the belt is 0.12 lb./in. width per foot length.

If the belt is 0.2 in. thick and E = 35,000 lb./sq.in., find also the frictional loss of speed of the driven pulley.

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Example 11

A rope drive transmits 800 h.p. from a pulley of effective diameter 12.5 ft., which runs at a speed of 90 r.p.m. The angle of lap is $160^{0}$ , the groove angle $45^{0}$ , the coefficient of friction is 0.28, the weight of the rope 0.9 lb./ft. and the allowable tension in the rope is 500 lb.

Find the smallest number of ropes that will be required.

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Example 12

Derive the expression:-

$\frac{T_1-\frac{wv^2}{g}}{T_2-\frac{wv^2}{g}}=e^{\mu \,\theta }$

(126)

Which connects the pulls on the two sides of a belt subjected to centripetal action, $T_1\;\;and\;\;T_2$ being the total pulls.

A V-belt having a lap of $180^0$ , has a cross-section of 1 sq.in. and runs in a groove of included angle $2\beta=45^0$ . The density of the belt is 0.05 lb./cu.in. and the maximum stress is limited to 600 lb./sq.in, the coefficient of friction being 0.15.

Find the maximum horse-power that can be transmitted if the wheel has a mean diameter of 12 in. and runs at 1000 r.p.m. (U.L.)

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Example 13

In a belt drive the radius of the driving pulley is r ft., the angle of lap of the belt is $\theta$ , the coefficient of friction between the belt and the pulley is $\mu$ and the weight of the belt is $\rho$ lb.per foot length.

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The drive is initially stationary and a torque is then applied to the driving pulley giving it an angular acceleration os $\alpha\; rad./sec^2$ . If the applied torque is such that slipping is about to commence when the belt speed has reached v ft./sec., determine the difference between the tight side tension $T_1$ and the slack-side tension $T_2$ in terms of $T_1,\;r\;v,\;\theta,\;\alpha, \;\rho,\;and\;\mu$ .

From this relationship discuss the extent to which the difference of the tensions when slipping is about to commence is affected by the acceleration of the belt as compared to the speed of the belt (U.L.)

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Example 14

In the band-and block brake shown in the diagram, the band is lined with 12 blocks each of which subtends an angle of $15^0$ at the centre of the wheel. If, when the brake is in action, the greatest and least tensions in the brake strap ate P and Q show that :-

$\frac{P}{Q}=\left ( \frac{1+\mu \,\tan7\tfrac{1}{2}^0}{1-\mu \,\tan7\tfrac{1}{2}^0} \right )^{12}$

(147)

Where $\mu$ is the coefficient of friction for the blocks.

With the lever arrangement illustrated, find the least force required at A for the brake to absorb 300 h.p. at 240 r.p.m. Take $\mu$ as 0.4 (U.L.)

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Example 15

The diagram shows a brake drum of 11 in. in diameter, acted on by two brake shoes which are mounted on a pin A and which are pushed apart by two hydraulically pistons at B each of which exerts a force of P lb. on the shoe with which it makes contact. The brake lining extends $69^0$ above and $60^0$ below the horizontal centre line. The coefficient is 0.2. The radial pressure between the lining and the drum is proportional to the rate of wear of the lining.