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Unformatted text preview: on Arm Gear A Gear C * TA 1. Arm B fixed, gear A rotated
through + 1 revolution (i.e. 1
revolution anticlockwise) 0 +1 ± 2. Arm B fixed, gear A rotated
through + x revolutions
Add + y r evolutions to all
elements 0 +x ± x× +y +y +y +y x +y y ± x× 3.
4. * Total motion Gear B – TC
TC TA TC
– x× TA
TB +y TA
TC y – x× TA
TB The ± sign is given to the motion of the wheel C because it is in a different plane. So we cannot indicate the
direction of its motion specifically, i.e. either clockwise or anticlockwise. 460 l Theory of Machines Since the speed of the arm is 100 r.p.m. anticlockwise, therefore from the fourth row of the
y = + 100
Also, the speed of the driving shaft X or gear A is 100 r.p.m. clockwise.
x + y = – 100
x = – y – 100 = – 100 – 100 = – 200
∴ Speed of the driven shaft i.e. shaft Y ,
N Y = Speed of gear B = y – x × 40 TA = 100 – – 200 × 30 TB = + 366.7 r.p.m. = 366.7 r.p.m. (anticlockwise) Ans.
Example 13.17. In a gear train, as
shown in Fig. 13.23, gear B is connected to the
input shaft and gear F is connected to the output
shaft. The arm A carrying the compound wheels
D and E, turns freely on the output shaft. If the
input speed is 1000 r.p.m. counter- clockwise
when seen from the right, determine the speed of
the output shaft under the following conditions :
1. When gear C is fixed, and 2. when
gear C is rotated at 10 r.p.m . counter clockwise.
Solution. Given : T B = 20 ; T C = 80 ;
T D = 60 ; T E = 30 ; T F = 32 ; N B = 1000 r.p.m.
(counter-clockwise) Fig. 13.23 The table of motions is given below :
Table 13.19. Table of motions.
Revolutions of elements
No. Conditions of motion Arm A Gear B
shaft) 1. Arm fixed, gear B rotated
through + 1 revolution (i.e.
1 revolution anticlockwise) 0 +1 2. Arm fixed, gear B rotated
through + x revolutions 0 +x 3. Add + y revolutions to all
elements +y +y 4. Total motion +y x +y Compound
wheel D-E + TB
TD Gear C – TB TD
=– + x× TB
TD +y y + x× Gear F (or
output shaft) – TB TE
TD TF TB
TC – x...
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- Fall '08