Example 133 the speed ratio of the reverted gear

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: ar train, as shown in Fig. 13.5, is to be 12. The module pitch of gears A and B is 3.125 mm and of gears C and D is 2.5 mm. Calculate the suitable numbers of teeth for the gears. No gear is to have less than 24 teeth. Solution. G iven : Speed ratio, N A/ N D = 1 2 ; m A = m B = 3.125 mm ; m C = m D = 2.5 mm Let Fig. 13.5 N A = Speed of gear A , T A = Number of teeth on gear A , rA = Pitch circle radius of gear A , N B, N C , N D = Speed of respective gears, T B, T C , T D = Number of teeth on respective gears, and rB, rC , rD = Pitch circle radii of respective gears. * We know that circular pitch, pc = ∴ 2πr = πm T r1 = m.T1 m.T2 m.T3 m.T4 ; r2 = ; r3 = ; r4 = 2 2 2 2 Now from equation (i), m.T1 m.T2 m.T3 m.T4 + = + 2 2 2 2 T 1 + T 2 = T 3 + T4 or r= m.T , where m is the module. 2 436 l Theory of Machines Since the speed ratio between the gears A and B and between the gears C and D are to be same, therefore N * NA = C = 12 = 3.464 NB ND Also the speed ratio of any pair of gears in mesh is the inverse of their number of teeth, therefore TB TD = = 3.464 TA TC ...(i) We know that the distance between the shafts x = rA + rB = rC + rD = 200 mm or ∴ and and and m .T ... 3 r = 2 m .T mA .TA m .T m .T + B B = C C + D D = 200 2 2 2 2 3.125 (T A + T B) = 2.5 (T C + T D) = 400 ...(∵ mA = m B, and m C = m D) T A + T B = 400 / 3.125 = 128 ...(ii) T C + T D = 400 / 2.5 = 160 ...(iii) From equation (i), T B = 3.464 T A. Substituting this value of T B in equation (ii), T A + 3.464 T A = 128 or T A = 128 / 4.464 = 28.67 say 28 Ans. T B = 128 – 28 = 100 Ans. Again from equation (i), T D = 3.464 T C. Substituting this value of T D in equation (iii), T C + 3.464 T C = 160 or T C = 160 / 4.464 = 35.84 say 36 Ans. T D = 160 – 36 = 124 Ans. Note : The speed ratio of the reverted gear train with the calculated values of number of teeth on each gear is NA TB × TD 100 × 124 = = = 12.3 N D TA × TC 28 × 36 13.7. Epicyclic Gear Train We have already discussed that in an epicyclic gear train, the axes of the shafts, over which the gears are mounted, may move relative to a fixed axis. A simple epicyclic gear train is shown in Fig. 13.6, where...
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