Question

A steel spur pinion has a pitch of 6 teeth/in, 22 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of 1200 rev/min and transmits 15 hp to a 60-tooth gear. If the face width is 2 in, estimate the bending stress.

Answer 1

27.7Mpa

Step-by-step explanation:

d = Nm = 22(6) = 132mm

From values of Lewis from factor Y.

Y = 0.331

V = πdn = π(0.132)(1200/60)

= 8.297 m/s

Using Dynamic Effect equation (cut/milled profile)

Kv = (6.1 + 8.297)/6.1

= 2.36

W¹ = Hp/πdn

Converting Hp to W

= 15hp ≈ 11190W

= 11190/π(0.132)(1200/60)

= 11190/8.29= 1349.82N

∆ = KvW¹/mFY

converting inches to millimeter 1in ≈ 2.54mm

= 2.36(1349.82)/ 0.006 (0.058)(0.331)

= 3185.5752/0.000115188

= 27655443.2753

=27.7Mpa.

Answer 2

7.63 ksi

Step-by-step explanation:

Given that:

pitch(p) = 6 teeth/in

pressure angle
`(\phi) = 20^0`

Pinion speed
`(n_p)` = 1200 rev/min

Power(H) = 15 hp

Teeth on gear
`(N_(\sigma))` = 60

Teeth on pinion
`(N_p)` = 22

Face width (b) = 2 in

To find the diameter from the parameters above ; we have:

(d) =
`(N)/(p)`

=
`(22)/(6)`

= 3.667 in

Using values of the Lewis factor Y for
`(N_p)` = 22

Y = 0.331

Then finding the velocity; we have the formula;

`V = (\pi d n_p)/(12)`

`V = (\pi *3.667*1200)/(12)`

V = 1152 ft/min

For the cut or mi;;ed profile; the velocity factor can be determined as

`K_v = (1200+V)/(1200)`

`K_v = (1200+1152)/(1200)`

`K_v = 1.96`

Then we proceed to determine the value of our tangential load also as follows:

`W^t = (T)/((d)/(2) )`

`W^t = (63025H)/((n_pd)/(2) )`

`W^t = (63025*15)/(1200*(3.667)/(2) )`

= 429.79 lbf

Finally ; the bending stress is calculated as :

`(\sigma) = (K_ vW^tp)/(FY)`

`(\sigma) = (1.96*429.79*6)/(2*0.331)`

`(\sigma) = 7634.94 psi`

`(\sigma) =7.63 ksi`

Thus, the bending stress is 7.63 ksi