# The wheel b has 20 teeth and gears with a and also

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Unformatted text preview: ame direction, then the input and output torques will be in opposite directions. Similarly, when the input and output shafts rotate in opposite directions, then the input and output torques will be in the same direction. Example 13.19. Fig. 13.26 shows an epicyclic gear train. Pinion A has 15 teeth and is rigidly fixed to the motor shaft. The wheel B has 20 teeth and gears with A and also with the annular fixed wheel E. Pinion C has 15 teeth and is integral with B (B, C being a compound gear wheel). Gear C meshes with annular wheel D, which is keyed to the machine shaft. The arm rotates about the same shaft on which A is fixed and carries the compound wheel B, C. If the motor runs at 1000 r.p.m., find the speed of the machine shaft. Find the torque exerted on the machine shaft, if the motor develops a torque of 100 N-m. Fig. 13.26 Solution. Given : TA = 15 ; TB = 20 ; TC = 15 ; NA = 1000 r.p.m.; Torque developed by motor (or pinion A ) = 100 N-m First of all, let us find the number of teeth on wheels D and E. Let T D and T E be the number of teeth on wheels D and E respectively. Let dA, dB, dC, dD and dE be the pitch circle diameters of wheels A , B , C, D and E respectively. From the geometry of the figure, dE = dA + 2 dB and dD = dE – (dB – dC) Since the number of teeth are proportional to their pitch circle diameters, therefore, T E = T A + 2 T B = 15 + 2 × 20 = 55 and T D = T E – (T B – T C) = 55 – (20 – 15) = 50 Speed of the machine shaft The table of motions is given below : 464 l Theory of Machines Table 13.21. Table of motions. Revolutions of elements Step No. Conditions of motion Arm Pinion A Compound wheel B-C Wheel D 1. Arm fixed-pinion A rotated through + 1 revolution (anticlockwise) 0 +1 – TA TB – 2. Arm fixed-pinion A rotated through + x revolutions 0 +x – x× TA TB –x × 3. Add + y revolutions to all elements +y +y 4. Total motion +y x +y +y y – x× TA TC × TB TD TA TC × TB TD +y TA TB y –x× TA TC × TB TD Wheel E − TA TB T × =− A TB TE TE − x× TA TE +y y − x× TA TE We know...
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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.

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