Question

2,2 kW to a gear Question 3 A pinion having 26 involute teeth of 6 module rotates at 350 r/min and transm at 350 r/min and transmits wheel having 62 teeth; the addendum on both wheels is 14 of the circular obliquity is 200 Determine: 3.1 the lengths and arc of the path of approach, and 3.2 the normal force between the teeth at an instant when there is only one pair of teeth in contact.

Answer 1

Given] No. of teeth on pinion: Zp= 26 m=6mm (module) Speed of pinion = Np= 350 rpm power transmitted = p = 2.2 kW wheel teeth = ZG-62 addendum : a = Circular pitch ) :¢(im) A-90- { A + addendum ratio? 0 = 20 solution | [3.11 Length & Are of path of approach (POA, AOA): POA = JRG² R²205²6 - R sind Ra - addendum radii of cancel = Rta = m ZG + A . Лm 2 -( 6X62 (5x62 + T K ) mm I Ra= 190.71 mm - - ma R= pitch radii of wheela me - | R=186 mm . . :: Poda 190.77?1862 052 (20) - 186 Sin (20) . -

| POA = 12-681 mm) Arc of apprach. POA 12.68 AJA = cosa = cos (25) Ao A = 13.495 mm 3.2 Normal force when there is between teeth at instant only one pair in contact (Fo] FR -- FY - FREE FLIN - (FBP of pinion ) Fo= Normal force on tooth component of Fo Fr. Fra vertical of horizontal where Fur tangential load on tooth responsible for power transmission Radial load an troth, For Now. Torque on pinon: FEXr=T= Sop To= m2 p - Zane GOX(2.2 X133w) 25x 350 Fr X78X10-3. -3 8= 6X26mm r=78x10²m I Fe=769.SAN

Now; from figure above! Fina Florin e :. The 769,54 .. in Cos (20) [In=818.93N