SEM327 – Engineering Assignment - Dynamics of Machines Assignment - Assessment Answer

November 20, 2018
Author : Sara Lanning

Solution Code: 1FII

Question: Engineering Assignment

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Engineering Assignment

Case Scenario/ Task

Part 1) Gear design

For the planetary gear train shown in the figure below, the arm connected to gear 6 and the input motor rotates around the input axis. Gear 1 is fixed to the input axle. Gears 2 and 3, 4 and 5, 6 and 6, 7 and 8 are fixed together respectively. Gears 2 and 3, and 9 are fixed in space but allowed to rotate. If the tooth numbers are N1 = 15, N2 = 40, N3 = 20, N5 = 48, N6 = 12, N8 = 35, N9 = 11.


A. Recognize and point out what type each gear is and finish the teeth counting for all gears. B. Determine the speed and direction of the output shaft if the input speed is 80 rpm CCW.

Gear design

Part 2) Rotating Shafts

Determine the mass to be added on the rotors below in the plane A at the radius 65mm considering both the static and dynamic balances using the tabular method

Rotating ShaftsPart 3) Belts and Pulleys

Two pulleys, one 320 mm diameter and the other 500 mm diameter, are on parallel shafts 2.05 m apart, and are connected by a cross-belt.

Determine: a) The length of the belt required and the angle of contact between the belt and each pulley; and b) What power can be transmitted by the belt when the larger pulley rotates at 500 r.p.m., if the maximum permissible tension in the belt is 1.5 kN, and the coefficient of friction between the belt and the pulley is 0.20?

Belts and Pulleys

Part 4) Clutches (Marks 7.5)

A flywheel system (overrunning clutch) is depicted in the figure below.

Two coaxial shafts (A and B) are connected by a single-plate clutch of internal radius 45 mm and external radius 135 mm, with both sides of the plate being used. The coefficient of friction is assumed as 0.3. Assume the pressure is (a) uniform, and (b) inversely proportional to radius.

Determine what the required spring force is to enable the maximum power transmission of 5.5 kW at an angular speed of 900 revs/min?


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Solution: Engineering Assignment

Solution 1)

We have T1 = 15, T2 = 40, T3 = 20, T5 = 48, T6 = 12, T8 = 35, T9 = 11.. further we have input rpm as 80 revolutions per minute

Tooth and speed is connected with the following formula.

N1T1 = N2T2 , which for a simple gear connection.

Gear 1,2,3,4 ,5 8and 9  are simple spur gears. Gear 6 acts like an idler to the epicyclic gear train consisting of gear 5, 6 and 7.

Number of teeth on each gear is as follows

It is known that driver and driven gears only matters irrespective of number of idlers introduced in the system. Hence the speed of output shaft can be calculated on the basis of input gear.

It implies N9 = N1T1/T9 = 1200/11 = 109.09 rpm in clock wise direction.

So the speed of gear 8 becomes N8 = 1200 /35 = 240/7 = 34.285

As gear 7 and 8 are joined together their speed remains same, so N7 = 34.385 or 240/7 rpm

As gear 5, 6 and 7 forms a epicyclic combination, it follows following procedure

If arm a is remain fixed and assuming speed of gear 6 as x then speed of gear 7 will be T6*X/T7

Now as speed of arm a is 80 it implies the speed of gear 6 and 7 is increased by 80.

Thus the equation becomes , à 12 x /T7 = 320/7                                                             (a)

Further for a compound epicyclic gears arm speed can be written in term s of the two gears as

From above calculation putting the values we get à  à x/T7 = 80/21 = 3.89

Solution 2)

Tabulur calculation is used to calculate components of forces and momentum and balance it with the unknown quantity. Here the unknown is fourth mass. The radius of the fourth mass is known to be 65 mm. The tabular method is followed as below.

m r L ? mrcos? mrsin? mrlcos? mrlsin?
1.5 20 200 30 25.980 15 5196.1524 3000
2.5 35 100 225 -61.871 -61.871 -6187.1845 -6187.1845
5 50 40 105 -64.70 241.481 -2588.19 9659.2583
Total -100.591 194.61 -3579.2221 6472.0738


From Static Balancing

Resultant product of required mass mArA =  = 219.061 kgmm

As mArA  = 219.061 kgmm, we have rA = 65mm à mA = 219.061/65 = 3.37 kg.

From Dynamic balancing

Deriving angle of the fourth mass ,  = tan-1    = -61.058  degrees.

Solution 3)

For a Crossed belt Length of belt is given by

L = ?(R+r) +  + 2C,                                                                                                                                 (i)

Where R is radius of bigger pulley and r is radius of smaller pulley. C represents centre to centre distance of pulley.

And angle of contact ? =                                                                                          (ii)

  1. For length of belt using equation (i) we get L = ?(0.5+0.32) + + 2x2.05 = 7 m

Angle of contact =  =   = 227.156 degrees

  1. In a belt system there is a slack and tight side whose ratio depends on the coefficient of friction by the following equation. More involving equation are also

  • ,                                                                                                                     (iii)

where ? = coefficient of friction, ? = angle of contact, T1 = tension on tight side and T2 = tension on the slack side.

  • ,                                                                                                                            (iv)

D = diameter of pulley, N = revolution per minute

  • , gives the equation for power.                                                (v)

From equation (iii) we can get T1 =   à T2 = 1500/2.20984 = 678.7821N

also from equation (iv) we get v = 13.08 m/s

Hence the power can be calculated from the equation

, P = (2000-678.7821)x13.08  = 17281.53 W = 17.28 kW

Hence the power required is 17.28 kW.

Solution 4

We are provided with inner and outer diameter of an over-running clutch. Further the power transmission and angular speed has been provided. Lets summarize the knowns and unknowns.

Power (kW) = 5.5, N = 900 rpm , D = 135mm, d = 45mm, ? = 0.3

Formulas involved. (Kett, 2012)

For constant pressure (p) ,

  • Torque ,                                                                                                                (i)

For pressure inversely propotional to (r)

  • Torque ,                                                                                                                 (ii)

Again Power,                                                                                                                             (iii)

Using Equation (ii) we get 5.5 =  à T = 58356.8124 Nmm

  1. Now using the value of torque in equation (i) we get ,= 3844.1816 N =3.84kN
  2. Now using the value of torque in equation (ii) we get , = 4322.7268 N = 4.32 kN

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