CodeCogs

Example 1 [imperial]

Problem

Two gear wheels $A$ and $B$ are mounted on parallel shafts so that they may revolve separately or may be meshed together externally.

The wheels were originally turning freely in the same direction.

Find:

a) The speed and direction of rotation of wheel $A$ , if the gears are suddenly meshed, assuming that there is no back-lash of the teeth.

b) The loss of energy in ft. lb. due to impact.

Workings

a) Let $n$ r.p.m. be the final speed of $A$ so that the final speed of $B$ is $2n$ in the opposite direction.

The impulse is equal and opposite on the two wheels at the moment of "meshing".

By moments about their respective axes, it follows that: The change of angular momentum of $A$ / The change of angular momentum of $B$ = The radius of $A$ / The radius of $B$

i.e. $\displaystyle\frac{10\times 12^2(300-n)}{12\times 10^2(2n+100)}=\displaystyle\frac{100}{50}$ . Or $360-1.2n=4n+200$
$\frac{160}{5.2}=30.75\;r.p.m.$

b) The loss of Kinetic Energy

The loss of K.E. $=\left [\frac{1}{2}\times \frac{10}{32.2}\times \left ( \frac{12}{12} \right )^2\left ( 300^2-30.75^2 \right )+$
$+\frac{1}{2}\times \frac{12}{32.2}\times \left ( \frac{10}{12} \right )^2\left ( 100^2-61.5^2 \right ) \right ]\times\left ( \frac{2\pi}{60} \right )^2=$

$=\frac{\pi^2}{64.4\times 900}\times \left ( 890,500+51,800 \right ) =161\;ft.lb.$ `

Solution

The speed is $30.75\;r.p.m.$
The loss of Kinetic Energy is $161\;ft.lb.$

Example 2 [imperial]

Problem

A car weighs 1650 lb., road wheels 26.4 in. diameter. The moment of inertia , $I$ , of the rotating parts is $7\displaystyle\frac{1}{2}\;lb.ft.^2$ and for each road wheel $\displaystyle 42\;lb.ft.^2$ .

The clutch is disengaged; the engine is idling at 500 r.p.m., the car is coasting at 18 m.p.h., the bottom gear of $19\displaystyle\frac{1}{2}$ to $1$ is engaged and the clutch pedal is suddenly released.

Find the change in speed of the car and the change in kinetic energy.

Workings

Initial Engine Speed

$=500\times \frac{2\pi}{60}=52.4\;rad./sec.$

(1)

Initial Wheel Speed

$=18\times \frac{88}{60}\times \frac{12}{26.4\times \pi}\times 2\pi=24\;rad./sec.$

(2)

Let $\omega$ rad./sec. be the final speed of the engine. The final speed of the road wheels will then be $\displaystyle\frac{\omega}{19.5}$ .

Since the speed of the car is that of the wheel circumference, the effective inertia of the wheels and car is:

$4\times 42+1650\left (\frac{ \pi\times 26.4}{12\times 2\;\pi} \right )^2=2165\;lb.ft.^2$

(3)

If the impulsive torque on the engine (i.e. at the clutch) is $\tau$ then that on the wheels is $19.5\tau$ , neglecting losses.

Hence:

19.5 $X$ Gain of angular momentum of engine = Loss of angular momentum of wheels and car.

From equations (#1) (#2) and (#3)

$19.5\times 7.5(\omega-52.4)=2165\left ( 24-\frac{\omega}{19.5} \right )$ Or, $2840\omega-149,000=1,011,000-2165\omega$

Giving $\omega=\displaystyle\frac{1,160,000}{5055}=232\;rad./sec.$

This corresponds to a final car speed of: $\left ( \displaystyle\frac{232}{19.5} \right )\left ( \displaystyle\frac{26.4}{24} \right )\left ( \displaystyle\frac{60}{88} \right )=8.9\;m.p.h.$

i.e. A reduction in speed of $9.1 m.p.h.$

The loss of Kinetic Energy is:

$\frac{1}{2\times 32.2}\times 2165\times \left [ 24^2-\left ( \frac{232}{19.5} \right )^2-7.5\left ( 232^2-52.4^2 \right ) \right ]$ $=\frac{941,000-378,000}{2\times 32.2}=8750\;ft.lb.$

Solution

The change in speed is $9.1 m.p.h.$
The change in kinetic energy is $8750\;ft.lb.$

Example 3 [imperial]

Problem

Whilst accelerating, the driving torque at the input shaft $A$ of a single-stage reduction gear has a constant value $\tau_A$ and the resisting torque at the output shaft $B$ is also constant with a value of $T_B$ . The respective moments of inertia of the two shafts systems are $I_A$ and $I_B$ and the speed of $A$ is $n$ times that of $B$ .

Show that the acceleration of $B$ is given by:

$\alpha_B=\frac{nT_A-T_B}{n^2I_A+I_B}$

and find the gear ratio for which this has a maximum value.

If $I_A$ is equivalent to 150 lb. at 7 in. radius and $I_B$ is equivalent to 750 lb. at 9 in. radius and if $T_A-25\;lb.ft.,\;\;T_B=45\;lb.ft.$ show that the required value of the ratio is about 5.19 and find the requirement for the output shaft to attain a speed of 240 r.p.m. if is initially at rest.

Workings

The torque to accelerate A = $I_A\alpha_A=I_A\;n\,\alpha_B$

The torque transmitted = $T_A-n\,I_A\;\alpha_B\;\;$ (At shaft $A$ )

And the torque transmitted = $n(T_A-n\,I_A\;\alpha_B)\;\;\;$ (At shaft $B$ )

The torque to accelerate $B$ = $I_B\alpha_B$

The output torque = $n(T_A-n\,I_A\;\alpha_B)-I_B\alpha_B=T_B$

From which:

$\alpha_B=\frac{nT_A-T_B}{n^2I_A+I_B}$

(4)

To find the maximum value for the gear ratio $n$ the above equation is differentiate w.r.t.t. and hence:

$\frac{(n^2I_A+I_B)T_A-(nT_A-T_B)\;2n\;I_A}{(n^2I_A+I_B)^2}=0$ $\therefore \;\;\;\;\;n^2I_AT_A-2nI_AT_A-I_BT_A=0$

Solving the quadratic for $n$ :

$n=\frac{I_AT_B+\sqrt{I_A^2T_B^2+I_AI_BT_A^2}}{I_AT_A}=\frac{T_B}{T_A}+\sqrt{\left ( \frac{T_B}{T_A} \right )^2+\frac{I_B}{I_A}}$

$=\frac{45}{25}+\sqrt{\left ( \frac{45}{25} \right )^2+\frac{750\times 9^2}{150\times 7^2}}$

Thus,

$n=1.8+3.39=5.19$

(5)

Using equations (#1) and (#2)

$\alpha_B=\frac{(5.19\times 25-45)\times 12^2\times 32.2}{5.19^2\times 150\times 7^2+750\times 9^2}=1.52\;rad./sec.^2$

The time to reach a speed of 250 r.p.m. from rest $=\displaystyle\frac{240\times 2\pi}{60\times 1.52}=16.6\; sec.$

Solution

The gear ratio is $n=\displaystyle\frac{45}{25}+\sqrt{\left ( \displaystyle\frac{45}{25} \right )^2+\displaystyle\frac{750\times 9^2}{150\times 7^2}}$
The requirement is $16.6\; sec.$

Example 4 [imperial]

Problem

A motor shaft $A$ rotates $G$ times as fast as a shaft $B$ geared to $A$ .

If the torque on $A$ is constant and $I_A$ and $I_B$ are the moments of inertia of $A$ and $B$ respectively, find an expression for $G$ when $B$ has maximum angular acceleration. Evaluate the value of $G$ when $I_A=20\;lb.ft.^2$ and $I_B=180\;lb.ft.^2$ .

If a steady driving torque of 45 lb.ft. is applied to $A$ and a variable resisting torque is applied to $B$ , find the time to increase the speed of $B$ from 150 r.p.m. to 240 r,p,m, when the resisting torque varies directly with the speed and has a value of 75 lb.ft. at 150 r.p.m.

Workings

If the torque on $A$ is $\tau_A$ then the equivalent torque on $B$ is:

$G\tau _a$ =(The equivalent inertia on $B$ ) $\times \alpha_B$

Thus, $G\tau_A=(G^2I_V+I_B)\;\alpha_B$ . Therefore, $\;\;\;\alpha_B=\displaystyle\frac{G}{G^2\;I_A+I_B}\times \tau_A$

Differentiating with respect to $G$ for a maximum: $(G^2\;I_A+I_B)1=G\times 2GI_A$ $\therefore \;\;\;\;\;\;G=\sqrt{\frac{I_B}{I_A}}$

Substituting in values, $G=\sqrt{\displaystyle\frac{180}{20}}=3$

Since the speed of $B$ is $\omega\;rad./sec.$ , the equivalent torque on $B$ is:

$G\tau_A-\tau_B=3\times 45-\frac{75\omega\times 30}{150\pi}=135-4.78\;\omega$

(6)

The equivalent moment of inertia of the system referred to shaft $B$ is: $\left ( G^2\;I_A+I_B \right )$

And therefore as the system is accelerating, the equivalent torque on $B$ $=\left ( G^2\;I_A+I_B \right )\;\frac{d\omega}{dt}$

Substituting in values, $=\left [ \displaystyle\frac{9\times 20+180}{32.2} \right ]\;\displaystyle\frac{d\omega}{dt}=11.2\;\displaystyle\frac{d\omega}{dt}$

Combining with equation (#1), $11.2\;\displaystyle\frac{d\omega}{dt}+4.78\;\omega=135$

Separating the variables and integrating, $t=11.2\displaystyle\int_{15.7}^{25.1}\displaystyle\frac{d\omega}{135-4.78\;\omega}$

To increase the speed of $B$ from 150 to 240 r.p.m.

$t=\frac{11.2}{4.78}\;\ln\;\frac{135-4.78\times 25.2}{135-4.78\times 15.7}=2.35\;\ln4=3.25\;sec.$

Solution

$G=3$
The time to increase the speed is $3.25\;sec.$

Example 5 [imperial]

Problem

In a gear unit the input and output shafts are parallel and directly geared to rotate in opposite directions with a speed reduction of 4. Friction losses can be neglected.

If the input to the unit is 30 h.p. at 1400 r.p.m., find the externally applied fixing torque which must be exerted on the unit casing. The direction of the applied torque at the input shaft may be considered positive.

If the inertias of the input and output shafts and connected parts are equivalent to 350 lb. at 8 in. radius and 450 lb. at 12 in. radius respectively, and the resisting torque at output is constant, find the additional constant torque which must be applied at the input to raise its speed from 1400 r.p.m. to 1500 r.p.m. in 1 sec. and also the associated fixing torque on the casing.

Workings

The input torque = $\displaystyle\frac{30\times 550\times 60}{2\pi\times 1400}=113\;lb.ft.$

The output torque = $4\times 113=452\;lb.ft.\;\;$ (Neglecting Losses)

Torque on the casing = $113+452=565\;lb.ft.$

(See "Gear Trains" equation (5))

The equivalent inertia on the input shaft = $350\left ( \displaystyle\frac{8}{12} \right )^2+\left ( \displaystyle\frac{459}{4^2} \right )\left ( \displaystyle\frac{12}{12} \right )^2$ $=155+28.2=183.2\;lb.ft.^2$

The acceleration = $(1500-1400)\times \displaystyle\frac{2\pi}{60}=10.47\;rad,/sec.^2$

The additional torque at input = $183.2\times \displaystyle\frac{10.47}{32.2}=59.6\;lb.ft.$

Input torque transmitted, The additional torque at input = $183.2\times \displaystyle\frac{10.47}{32.2}=59.6\;lb.ft.$

= $59.6$ - The torque needed to accelerate the input shaft= $59.6-\frac{350}{32.2}\times \left ( \frac{8}{12} \right )^2\times 10.47=9.02\;lb.ft.$

The torque to the output shaft = $4\times 9.02=36.08\;lb.ft.$

The extra torque on the casing = $9.02+36.8=45.82\;lb.ft.$

Solution

The additional torque $= 183.2\times \displaystyle\frac{10.47}{32.2}=59.6\;lb.ft$
The extra torque on the casing $=9.02+36.8=45.82\;lb.ft.$

Example 6 [imperial]

Problem

In the epicyclic train shown in the diagram all the teeth are of 4 d.p. Wheel $F(40 t)$ which is mounted on shaft $A$ is stationary.

$D(50 t)$ and $E(30 t)$ are made of one piece of metal and can rotate freely on a pin $C$ which is attached to plate $B$ . A similar pair of pinions $D^1$ and $E^1$ are mounted on a pin $C^1$ . $B$ can rotate freely on shaft $A$ . Wheel $G$ is mounted on the driving shaft $H$ . The polar moment of inertia of the disc $B$ including the two pins $C$ and $C^1$ is $1200\;lb.in^2$ . $D$ and $E$ together weigh 7 lb with polar radius of gyration about their own centre-line of 3 in. $G$ has a polar moment of inertia of $200\;lb.in^2.$

Find the torque on shaft $H$ to accelerate the system so that $B$ has an angular acceleration of $5\;rad./sec.^2$

Workings

The number of teeth on $G$ = $t_F+t_P-t_E =40+50-30=60\;teeth$

The method of finding the speed and hence the acceleration ratios for an epicyclic gear chain can be found in Gear Trains. As shown there, a table is drawn up as follows:

The radius arm of pins $C$ and $C^1$ = $\displaystyle\frac{t_F+t_P}{2\times \text{diametral pitch}}$

$=\frac{40+50}{2\times 4}=11.25\;in.$

The moment of inertia of the disc $B$ and the mass of wheels $D-E$ about the axis of $B$ .

$1200+2\times 7\times 11.25^2=2970\;lb.in^2$

The equivalent inertia referred to shaft $H$ calculated from the speed ratio which have been determined.

$=200+2\times 3^2(7\times 3^2)+\left ( \frac{5}{3} \right )^2\times 2970=9530\;lb.in^2$

Also from the speed ratios:

Acceleration of shaft $H$ = $3\;rad./sec^2$ $\therefore \;\;\;\;\;\;\tau_H=\left ( \frac{9530}{32.2\times 12} \right )\times 3=74\;lb.in.$

Solution

The torque is $74\;lb.in.$

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Example 8 [imperial]

Problem

The engine of a motor-car runs at 3420 r.p.m. when the road speed is 60 m.p.h. The weight of the car is 2400 lb. The inertia of the rotating parts of the engine corresponds to 24 lb. at a radius of gyration of 0.48 ft. and that of the road wheels is 240 lbs at 0.8 ft.

The efficiency of the engine and transmission is 0.9 and the wind resistance is 200 lb. The road wheel diameter is 2.5 ft.

Estimate the horse-\b{power} developed by the engine when the car travels at 60 m.p.h. with an acceleration of 3 ft./sec.

Workings

The speed ratio between the engine and the wheels is: