Does the interference occur? If it occurs, to what value should the pressure angle be changed to eliminate interference?

Subject: Kinematics of Machinery

Topic: Gears and Gear Trains

Difficulty: High

Given $\phi= 20^{\circ}$; T=50; m=10 mm, t=13

Addendum = 1m= 10 mm

R= mt/2=$\frac{10times50}{2} = 250 mm, R_{a}= 250+10=260 mm$

r= mt/2=$\frac{10times13 }{2 }= 65 mm$

$R_{a max}=\sqrt{(250\cos\phi)^{2}+(R\sin\phi + r \sin\phi)^{2}}$

$=\sqrt{(250\cos 20^{\circ})^{2}+(250\sin 20^{\circ} + 65 \sin 20^{\circ})^{2}}$

$=\sqrt{(250\cos 20^{\circ})^{2}+(315\sin 20^{\circ})^{2}}=258.45 mm$

The actual addendum radius $R_{a}$ is more than the maximum value $R_{a max}$, therefore, interference occurs. The new value of $\phi$ can be found by taking $R_{a max}$ equal to $R_{a}$.

i.e $260=\sqrt{(250\cos \phi)^{2}+(315\sin \phi)^{2}}$

$(260)^{2}=(250)^{2}\cos^{2}\phi+(315)^{2}(1-\cos^{2}\phi)$

$=(250)^{2}\cos^{2}\phi+(315)^{2}-(315)^{2}\cos^{2}\phi)$

Or $\cos^{2}\phi=\frac{(315)^{2}-(260)^{2}}{(315)^{2}-(250)^{2}}=0.861$

Or $\cos\phi=0.928$

$\phi=21.88^{\circ}$ or $21^{\circ}52'$

Thus , if the pressure angle is increased to 21$^{\circ}$52’,the interference is avoided.