Answer to Question #206281 in Mechanical Engineering for Md nazmul hassan

Question

Question No. 18: (a) In an epicyclic gear train, an arm carries two gears A and B having 36 and 45 teeth respectively. If the arm rotates at 150 r.p.m in the anticlockwise direction about the centre of the gear A which is fixed, determine the speed of gear B. If the gear A instead of being fixed, makes 300 r.p.m in the clockwise direction, what will be the speed of gear B?

(b) Prove that the resultant unbalanced force is minimum when half of the reciprocating masses are balanced by rotating masses.

 

Answer 1

Speed of gear B when gear A is fixed

Since the speed of the arm is 150 r.p.m. anticlockwise, therefore from the third row of the table

The velocity of gear A = 0 (fixed)

$a + b = 0$

$a = – b = – 150 r.p.m$

Speed of gear B, $N_B =a – b \frac{T_A}{T_B}= 150 – 150 ×\frac{36}{45} = + 270 r.p.m$

= 270 r.p.m. (anticlockwise)

Speed of gear B when gear A makes 300 r.p.m. clockwise

Since gear A makes 300 r.p.m. clockwise, therefore from the third row of the table

$a + b = – 300$

$b = –300 – a = –300 – 150 = – 450 r.p.m$

$N_B = a–b \frac{T_A}{T_B} = 150 + 450 × \frac{36}{45}= + 510 r.p.m$

= 510 r.p.m. (anticlockwise)

Part b

To minimize the effect of the unbalance force a compromise is, usually made, is of the $\frac{2}{3}$ reciprocating mass is balanced or a value between to $\frac{1}{2}$ to $\frac{3}{4}$

If ‘c’ is the fraction of the reciprocating mass, then the primary force balanced by the mass  cmr ω² cos θ and the primary force unbalanced by the mass = (1-c) mr ω² cos θ

The vertical component of centrifugal force which remains unbalanced = c mr ω ² sin θ

In reciprocating engines, unbalance forces in the direction of the line of stroke are more

dangerous than the forces perpendicular to the line of stroke.