Answer in Mechanical Engineering for Zeeshan #214685

The following particulars relate to a symmetrical tangent cam operating a roller follower :-

Least radius = 36 mm, nose radius = 30 mm, roller radius = 21 mm, distance between cam

shaft and nose centre = 28 mm, angle of action of cam = 150°, cam shaft speed = 750 r.p.m.

Assuming that there is no dwell between ascent and descent, determine the lift of the valve

and the acceleration of the follower at a point where straight flank merges into the circular

nose.

“BT=OT-OB\nBT=OP+PT-OT\n=d+r_2-r_1\n=25+3.2-16=12.2”

“R= frac{r^2_1-r_2^2+d^2-2r_1 d cos alpha}{2(r_1-r_2-d cos alpha)}\nR= frac{16^2-3.2^2+25^2-2*1.6*25 d cos75}{2(16-3.2-25 cos 75)}\nR=52.82 mm”

Flank angle “phi”

“frac{PO}{sin phi}= frac{PQ}{sin(180- phi)}\nsin phi = frac{sin(180- 75)}{52.82-3.2} implies phi = 29.6 ^0”

Acceleration at the end of the contact with the flank when “theta = phi=29.6^0”

“a=omega ^2(R-r_1) cos phi \na=(frac{2 pi *600}{60}) ^2 (52.82-16)*10^{-3} cos 29.6\na=129.39 m/s^2”

Reterdation at the beginning of the contact with the nose

“a=-omega ^2(R-r_1) cos phi \na=-(frac{2 pi *600}{60}) ^2 (25)*10^{-3} cos 29.6\na=-25.21 m/s^2”

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