Answer to Question #227542 in Mechanical Engineering for Trilok

Question

The number of teeth on the gear and pinion of two spur gears in mesh are 40 and 24 respectively. Both the gears have a module of 8 mm and pressure angle of 20 degree. If the pinion rotates at 500 rpm, what will be the sliding velocity at the moment the tip of the tooth of the pinion has contact with gear flank? Take addendum equal to one module. Also, find the maximum velocity of sliding.

Answer 1

Given T=40, t=24, N₁=500rpm, addendum=8mm, m=8mm, pressure angle= 20 degrees

w₁=$\frac{2πN_1}{60}=\frac{2π*500}{60}=52.36rad/s$

w₂=$w_1*\frac{t}{T}=52.36*\frac{24}{40}=31.42rad/s$

r=m*$\frac{t}{2}=8*\frac{24}{2}=96mm$

R=$m*\frac{T}{2}=8*\frac{40}{2}=160mm$

r_A=(96+8)mm=104mm

R_A=(160+8)mm=168mm

KP=$\sqrt{(R_A)^{2}-R^{2}cos^{2}\phi}-RSin\phi$

$=\sqrt{168^{2}-160^{2}cos^{2}20}-160sin20\newline =(74.96-54.72)mm=20.24mm\newline PL=\sqrt{(r_A)^{2}-r^{2}cos^{2}\phi}-rSin\phi\newline =\sqrt{104^{2}-96^{2}cos20}-96sin20\newline =(51.75-32.83)mm=18.92mm\newline 1)sliding\ velocity\ at\ engagement;\newline v_s=(w_1+w_2)KP\newline =(52.36+31.42)20.24\newline =1695.71mm/s\newline 2)velocity\ at \ disenchantment;\newline =(w_1+w_2)PL\newline=(52.36+31.42)18.92=1585.12mm/s.\newline \therefore max \ sliding\ velocity =1695.71mm/s$