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Torque

Publisher Summary

This chapter provides an overview on torque. When two equal forces act on a body they cause the body to rotate, and the system of forces is called a couple. The turning moment of a couple is called a torque, T. Torque is a gauge that measures how much force is acting on an object that causes it to rotate. The unit of torque is the Newton meter, Nm. A common and simple method of transmitting power from one shaft to another is by means of a belt passing over pulley wheels that are keyed to the shafts. Typical applications include an electric motor driving a line of shafting and an engine driving a rotating saw.

1. When two equal forces act on a body as shown in Figure 38.1, they cause the body to rotate, and the system of forces is called a couple.

2. The turning moment of a couple is called a torque, T. In Figure 38.1, torque = magnitude of either force × perpendicular distance between the forces, i.e.

$T = F d$ $T = F d$

The unit of torque is the newton metre, Nm.

3. When a force F newtons is applied at a radius r metres from the axis of, say, a nut to be turned by a spanner, the torque T applied to the nut is given by:

$T = F r N m$ $T = F r N m$

4. Figure 38.2(a) shows a pulley wheel of radius r metres attached to a shaft and a force F newtons applied to the rim at point P. Figure 38.2(b) shows the pulley wheel having turned through an angle θ radians as a result of the force F being applied.

The force moves through a distance s, where arc length s = rθ.

$\begin{array}{l} Work done = force x distance moved by force \\ = F \times r θ = Fr θ Nm = Fr θ J . \end{array}$ $\begin{array}{l} Work done = force x distance moved by force \\ = F \times r θ = Fr θ Nm = Fr θ J . \end{array}$

However, Fr is the torque T,

Hence, work done = Tθ joules.

$\begin{array}{l} Power = \frac{work done}{time taken} \\ = \frac{T θ}{time taken}, for a constant torque, T . \\ However, = \frac{angle θ}{time taken} = angular velocity, ω rad/s . \end{array}$ $\begin{array}{l} Power = \frac{work done}{time taken} \\ = \frac{T θ}{time taken}, for a constant torque, T . \\ However, = \frac{angle θ}{time taken} = angular velocity, ω rad/s . \end{array}$

Hence, power, P = T ω watts.

Angular velocity, ω = 2πn rad/s, where n is the speed in rev/s.

Hence, power, P = 2πn T watts.

Thus the torque developed by a motor whose spindle is rotating at 1000 rev/min and developing a power of 2.50 kW is given by:

$\begin{array}{l} P =2 π n T, from which, \\ torque T = \frac{P}{2 π n} = \frac{2500}{2 π (\frac{1000}{60})} \\ = 23.87 Nm \end{array}$ $\begin{array}{l} P =2 π n T, from which, \\ torque T = \frac{P}{2 π n} = \frac{2500}{2 π (\frac{1000}{60})} \\ = 23.87 Nm \end{array}$

6. From para. 4, work done = Tθ, and if this work is available to increase the kinetic energy of a rotating body of moment of inertia I, then

$T θ = I (\frac{ω_{2}^{2} - ω_{1}^{2}}{2})$ $T θ = I (\frac{ω_{2}^{2} - ω_{1}^{2}}{2})$

where ω₁ and ω₂ are the initial and final angular velocities,

$\begin{array}{l} i . e . T θ = I (\frac{ω_{2} + ω_{1}}{2}) (ω_{2} - ω_{1}) \\ However, (\frac{ω_{2} + ω_{1}}{2}) is the mean angular velocity, i .e . \frac{θ}{t}, \end{array}$ $\begin{array}{l} i . e . T θ = I (\frac{ω_{2} + ω_{1}}{2}) (ω_{2} - ω_{1}) \\ However, (\frac{ω_{2} + ω_{1}}{2}) is the mean angular velocity, i .e . \frac{θ}{t}, \end{array}$

where t is the time and (ω₂ − ω₁) is the change in angular velocity, i.e. αt, where α is the angular acceleration.

$Hence, T θ = I (\frac{θ}{t}) (α t)$ $Hence, T θ = I (\frac{θ}{t}) (α t)$

from which,

$torque T = I α,$ $torque T = I α,$

where

I is the moment of inertia in kg m²,

α is the angular acceleration in rad/s² and

T is the torque in Nm.

Thus if a shaft system has a moment of inertia of 37.5 kg m², the torque required to give it an angular acceleration of 5.0 rad/s² is given in:

$torque T = I α = (37 .5) (5 .0) = 187 .5 Nm .$ $torque T = I α = (37 .5) (5 .0) = 187 .5 Nm .$

Power transmission by belt drives

(i) A common, and simple method of transmitting power from one shaft to another is by means of a belt passing over pulley wheels which are keyed to the shafts, as shown in Figure 38.3. Typical applications include an electric motor driving a line of shafting and an engine driving a rotating saw.

(ii) For a belt to transmit power between two pulleys there must be a difference in tensions in the belt on either side of the driving and driven pulleys. For the direction of rotation shown in Figure 38.3, F₂ > F₁.

The torque T available at the driving wheel to do work is given by:

$T = (F_{2} - F_{1}) r_{x} Nm,$ $T = (F_{2} - F_{1}) r_{x} Nm,$

and the available power P is given by:

$P = T ω = (F_{2} - F_{1}) r_{x} ω_{x} watts$ $P = T ω = (F_{2} - F_{1}) r_{x} ω_{x} watts$

(iii) From para. 12(i), page 291, the linear velocity of a point on the driver wheel, v_x = r_xω_x. Similarly, the linear velocity of a point on the driven wheel, v_y = r_yω_y.

Assuming no slipping, v_x = v_y

i.e. r_x ω_x = r_y ω_y

Hence, r_x (2πn_x) = r_y (2πn_y)

$from which, \frac{r_{x}}{r_{y}} = \frac{n_{y}}{n_{x}}$ $from which, \frac{r_{x}}{r_{y}} = \frac{n_{y}}{n_{x}}$

(iv)

$\begin{array}{l} Efficiency = \frac{useful work output}{energy input} \times 100 % o r \\ efficiency = \frac{power output}{power intput} \times 100 % . \end{array}$ $\begin{array}{l} Efficiency = \frac{useful work output}{energy input} \times 100 % o r \\ efficiency = \frac{power output}{power intput} \times 100 % . \end{array}$

For example, a 15 kW motor is driving a shaft at 1150 rev/min by means of pulley wheels and a belt. The tensions in the belt on each side of the driver pulley are 400 N and 50 N and the diameters of the driver and driven pulley wheel are 500 mm and 750 mm respectively. The power output from the motor is given by:

$\begin{array}{l} Power output = (F_{2} - F_{1}) r_{x} ω_{x} \\ = (400 - 50) (\frac{500}{2} \times 10^{- 3}) (\frac{1150 \times 2 π}{60}) \\ = 10.54 k W . \end{array}$ $\begin{array}{l} Power output = (F_{2} - F_{1}) r_{x} ω_{x} \\ = (400 - 50) (\frac{500}{2} \times 10^{- 3}) (\frac{1150 \times 2 π}{60}) \\ = 10.54 k W . \end{array}$

$\begin{array}{l} Hence, the efficiency of the motor = \frac{power output}{power input} \\ = \frac{10.54}{15} \times 10 % \\ = 70.27 % . \end{array}$ $\begin{array}{l} Hence, the efficiency of the motor = \frac{power output}{power input} \\ = \frac{10.54}{15} \times 10 % \\ = 70.27 % . \end{array}$

The speed of the driven pulley is obtained from

$\frac{r_{x}}{r_{y}} = \frac{n_{y}}{n_{x}}$ $\frac{r_{x}}{r_{y}} = \frac{n_{y}}{n_{x}}$

i.e. speed of driven pulley wheel,

$n_{y} = \frac{n_{x} r_{x}}{r_{y}} = \frac{(1500) (0.25)}{(\frac{0.75}{2})} = 767 rev/ \min$ $n_{y} = \frac{n_{x} r_{x}}{r_{y}} = \frac{(1500) (0.25)}{(\frac{0.75}{2})} = 767 rev/ \min$

(i) The ratio of the tensions for a flat belt when the belt is on the point of slipping is

$\frac{T_{1}}{T_{2}} = e^{u θ}$ $\frac{T_{1}}{T_{2}} = e^{u θ}$

where

µ is the coefficient of friction between belt and pulley,

θ is the angle of lap, in radians (see Figure 38.4), and

e is the exponent 2.718.

(ii) For a vee belt as in Figure 38.5, the ratio is

$\frac{T_{1}}{T_{2}} = e^{^{u θ} / \sin α}$ $\frac{T_{1}}{T_{2}} = e^{^{u θ} / \sin α}$

where α is the half angle of the groove and of the belt.

This gives a much larger ratio than for the flat belt. The V-belt is jammed into its groove and is less likely to slip. Referring to Figure 38.5, the force of fraction on each side is µR_N where R_N is the normal (perpendicular) reaction on each side. The triangle of forces shows that $R_{N} = \frac{R / 2}{\sin α}$ $R_{N} = \frac{R / 2}{\sin α}$ where R is the resultant reaction.

The force of friction giving rise to the difference between T₁ and T₂ is therefore

$μ R_{N} \times 2 = (\frac{μ}{\sin α}) R$ $μ R_{N} \times 2 = (\frac{μ}{\sin α}) R$

The corresponding force of friction for a flat belt is µR. Comparing the forces of friction for flat and V-belts it can be said that the V-belt is equivalent to a flat belt with a coefficient of friction given by µ/ (sin α).

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Table of Contents for Chapter 38: Torque

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Publisher Summary

Power transmission by belt drives

Table of Contents for
Chapter 38: Torque