14 x 0 i also the gear a is rotating at 1200 rpm

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: revolutions Total motion + Gear C TA TB + 0 –y + x× –x –y – y–x TA TB –y + x × + TA TC +x × –y TA TB TA TD × TB TE TA TC + x× –y Gear E TA TB × TB TC =+ –y Add – y revolutions to all elements Compound gear B-D –y + x × TA TD × TB TE –y TA TC –y + x × TA TD × TB TE Since the annular gear C is fixed, therefore from the fourth row of the table, – y +x× ∴ TA =0 TC –y + x × or 14 =0 100 – y + 0.14 x = 0 ...(i) Also, the gear A is rotating at 1200 r.p.m., therefore – x – y = 1200 ...(ii) From equations (i) and (ii), x = – 1052.6, and y = – 147.4 1. Speed and direction of rotation of gear E From the fourth row of the table, speed of gear E, NE = – y + x × 14 41 TA TD × = 147.4 – 1052.6 × × 43 98 TB TE = 147.4 – 143.4 = 4 r.p.m. = 4 r.p.m. (anticlockwise) Ans. 2. Fixing torque required at C We know that torque on A = PA × 60 1850 × 60 = = 14.7 N-m 2π N A 2π × 1200 Since the efficiency is 100 per cent throughout, therefore the power available at E (PE) will be equal to power supplied at A (PA). 468 l Theory of Machines = ∴ Torque on E PA × 60 1850 × 60 = = 4416 N-m 2π × N E 2π × 4 ∴ Fixing torque required at C = 4416 – 14.7 = 4401.3 N-m Ans. Example 13.22. An over drive for a vehicle consists of an epicyclic gear train, as shown in Fig. 13.29, with compound planets B-C. B has 15 teeth and meshes with an annulus A which has 60 teeth. C has 20 teeth and meshes with the sunwheel D which is fixed. The annulus is keyed to the propeller shaft Y which rotates at 740 rad /s. The spider which carries the pins upon which the planets revolve, is driven directly from main gear box by shaft X, this shaft being relatively free to rotate with respect to wheel D. Find the speed of shaft X, when all the teeth have the same module. When the engine develops 130 kW, what is the holding torque on the wheel D ? Assume 100 per cent efficiency throughout. Fig. 13.29 Solution. Given : TB = 15 ; TA = 60 ; TC = 20 ; ωY = ωA = 740 rad /s ; P = 130 kW =...
View Full Document

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.

Ask a homework question - tutors are online