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

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. Questions: 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.

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 m_Ar_A = = 219.061 kgmm As m_Ar_A = 219.061 kgmm, we have r_A = 65mm à m_A = 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

3844. Now using the value of torque in equation (i) we get ,= 3844.1816 N =3.84kN
3845. Now using the value of torque in equation (ii) we get , = 4322.7268 N = 4.32 kN