Design a spur gear drive to transmit 25 H.P at 850 r.p.m speed reduction is 2.5. Materials for pinion and wheel are C15 steel and CI grade 30 respectively. Take pressure angle of $20^0$. Design bending stress for pinion material is $85M/mm^2$ and surface endurance limit for pinion material is $620N/mm^2$.

Mumbai University > Mechanical Engineering > Sem 7 > Machine Design 2

Marks: 10M

Year: Dec 2016

Given

P = 25H.P

N = 850 rpm

i = 2.5

C15

C30

$\alpha = 20^0$

$[\sigma_b] = 85N/mm^2$

$[\sigma_c] = 620 N/mm^2$

Solution:

No. of teeth on pinion $z_1 = \frac{2}{sin^2 \alpha} = 18$

No. of teeth on wheel $z_2 = i z_1 = 46$

corrected i = 2.55

$Y_1 = \pi \bigg[0.154 - \frac{0.912}{z_1}\bigg] = 0.3246$

$Y_2 = \pi \bigg[0.154 - \frac{0.912}{z_2} \bigg] = 0.4215$

Strength of pinion = $85 x 0.3246 = 27.591 N/mm^2$

Strength of wheel = $35.829 N/mm^2$

Calculate of module based on beam strength

$m \geq 1.26 \sqrt[3]{\frac{[]M_t}{[\sigma_b] \psi Y_1 z_1}}$

$[M_t] = 1.5 \times \frac{18650 \times 60}{2 \pi 850} = 314.284 \times 10^3 N-mm$

$m \geq 5.021$

m = 6 (PSG 8.2)

Checking for contact stress

$\sigma_c = 0.74 \frac{i + 1}{a} \sqrt{\frac{i+1}{ib} E [M_t]}$

$E = 2.15 \times 10^5 N/mm^2$ (PSG 814) $a = m(z_1 + z_2)/2 = 192 mm$ $\sigma_c = 541 N/mm^2$ $\sigma_c \gt [\sigma_c]$ safe

Principal dimensions

module m = 6 mm

centre distance a = 192 m

Bottom clearance c = 0.25m = 1.5mm

tooth depth h = 2.25m m = 13.5 mm

PCD $d_1 = mz_1 = 108 \hspace{1cm} d_2 = mz_2 = 276$

ACD $d_{a1} = (z_1 + 2f_0)m = 120 \hspace{1cm} d_{a2} = (z_2 + 2f_0)m = 288mm$

DCD $d_{f_1} = (z - 2f_0)m-2c = 93\hspace{1cm} d_{f_2} = 261 mm$