# 5 table 135 table of motions revolutions of elements

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Unformatted text preview: 5. Table 13.5. Table of motions. Revolutions of elements Step No. Conditions of motion Arm EF Gear C Gear B 1. Arm fixed-gear C rotates through + 1 revolution ( i . e . 1 rev. anticlockwise) 0 +1 – TC TB 2. Arm fixed-gear C rotates through + x revolutions 0 +x –x× TC TB 3. Add + y r evolutions to all elements +y +y +y 4. Total motion +y x +y y –x× Gear A – TC TB T × =– C TB TA TA – x× TC TA +y TC TB y –x× TC TA 444 l Theory of Machines Speed of gear C We know that the speed of the arm is 18 r.p.m. therefore, y = 18 r.p.m. and the gear A is fixed, therefore y –x× ∴ TC =0 TA 18 – x × or 32 =0 72 x = 18 × 72 / 32 = 40.5 ∴ Speed of gear C = x + y = 40.5 + 18 = + 58.5 r.p.m. = 58.5 r.p.m. in the direction of arm. Ans. Fig. 13.10 Speed of gear B Let dA, dB and dC be the pitch circle diameters of gears A , B and C respectively. Therefore, from the geometry of Fig. 13.10, dB + dC d =A 2 2 or 2 dB + dC = dA Since the number of teeth are proportional to their pitch circle diameters, therefore 2 TB + T C = TA ∴ Speed of gear B or 2 T B + 32 = 72 or T B = 20 TC 32 = 18 – 40.5 × = – 46.8 r.p.m. 20 TB = 46.8 r.p.m. in the opposite direction of arm. Ans. = y –x× Example 13.7. An epicyclic train of gears is arranged as shown in Fig.13.11. How many revolutions does the arm, to which the pinions B and C are attached, make : 1. when A makes one revolution clockwise and D makes half a revolution anticlockwise, and 2. when A makes one revolution clockwise and D is stationary ? The number of teeth on the gears A and D are 40 and 90 respectively. Fig. 13.11 Solution. Given : T A = 40 ; T D = 90 First of all, let us find the number of teeth on gears B and C (i.e. TB and T C). Let dA, dB, dC and dD be the pitch circle diameters of gears A , B , C and D respectively. Therefore from the geometry of the figure, dA + dB + dC = dD or dA + 2 dB = dD ...(3 dB = dC) Since the number of teeth are proportional to their pitch circle diameters, therefore, TA + 2 T B = T D ∴ T B = 25, or and 40 + 2 T B = 90 T C = 2...
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## This note was uploaded on 02/13/2014 for the course MIE 301 taught by Professor Celghorn during the Fall '08 term at University of Toronto- Toronto.

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