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Unformatted text preview: el R. The
numbers of teeth on the wheels are :
P = 114, Q = 120, R = 120, X = 36, Y = 24 and Z = 30. Fig. 13.30. The driving shaft I makes 1500 r.p.m.clockwise looking from our right and the input at I is
7.5 kW.
1. Find the speed and direction of rotation of the driven shaft O and the wheel P.
2. If the mechanical efficiency of the drive is 80%, find the torque tending to rotate the fixed
wheel R.
Solution. Given : T P =144 ; T Q = 120 ; T R = 120 ; T X = 36 ; T Y = 24 ; T Z = 30 ; N I = 1500
r.p.m. (clockwise) ; P = 7.5 kW = 7500 W ; η = 80% = 0.8
First of all, consider the train of wheels Z ,R and Q (arm). The revolutions of various wheels
are shown in the following table. 470 l Theory of Machines
Table 13.25. Table of motions.
Revolutions of elements Step No. Conditions of motion Q (Arm) Z (also I) R (Fixed) 1. Arm fixedwheel Z rotates through + 1
revolution (anticlockwise) 0 +1 – TZ
TR 2. Arm fixedwheel Z rotates through + x revolutions 0 +x – x× TZ
TR 3. Add + y revolutions to all elements +y +y 4. Total motion +y x +y +y y –x× TZ
TR Since the driving shaft I as well as wheel Z rotates at 1500 r.p.m. clockwise, therefore
x + y = – 1500 ...(i) Also, the wheel R is fixed. Therefore y –x× TZ
=0
TR y=x× or 30
TZ
= x×
= 0.25 x
120
TR ...(ii) From equations (i) and (ii),
x = – 1200, and y = – 300 Now consider the train of wheels Y , Q, arm A , wheels P and X . The revolutions of various
elements are shown in the following table.
Table 13.26. Table of motions.
Revolutions of elements
Step
No. Conditions of motion Arm A, B
and Shaft O Wheel Y Compound
wheel QX
TY
TQ 1. Arm A fixedwheel Y
rotates through + 1
revolution (anticlockwise) 0 +1 – 2. Arm A fixedwheel Y rotates
through + x 1 revolutions 0 + x1 – x1 × 3. Add + y 1 revolutions to all
elements + y1 + y1 4. Total motion + y1 x1 + y 1 + TY
TQ + y1
y1 – x1 × or TY
= y = – 300
TQ y1 – x1 × 24
= – 300
120 TY TX
×
TQ TP + x1 × TY TX
×
TQ TP + y1
TY
TQ Since the speed of compound wheel QX is same as...
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 Fall '08
 CELGHORN

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