Question

A 16-tooth right-handed helical pinion turns at 2000 rpm and transfers 5kw to a 38-tooth gear with 30 helix-angle 25 pressure angle. Normal module is 2 mm. Determine pitch diameters; tangential module; axial and tangential pitch; normal pressure angle; transmitted load.

Answer 1

Given Data:

No. of tooth on pinion(Z₁) = 16;

Rotational speed of pinion(N) = 2000 rpm;

Power (P) = 5 kw =5000 w;

No. of tooth on gear(Z₂) =38;

Helix angle(ß) = 30°;

Pressure angel(α)=25°;

Normal module = 2mm;

Solution:

1. Pitch diameters:

According to formula for pitch diameters,

Pitch diameter (d) = No. Of tooth * Transverse module= z*m

Since z is known let’s find transverse module(m) first

Transverse module (m) =

= 2.3094mm cos (30) cos (

Hence,

Pinion pitch diameter :

dp = Zp * m = 16 * 2.3094 = 36.95 = 37mm

Gear pitch diameter :

dg-Z*m - 38* 2.3094 -87.75 mm

2)Axial Pitch :

$P_a = P/ tan(\beta ) = \pi *m/tan(\beta) = 2.3094*\pi/tan(30°) = 12.557 mm$

3) Normal Pressure Angle:

$\alpha _n = tan^-1(tan\alpha*cos\beta) \alpha_n = \tan^-1(tan(25°)*cos(30°)) = 21.98°= 22°$

Therefore, normal pressure angle = 22 degrees.

4) Transmitted load :

$Torque(T) = P*60/2\pi N = 5*10^3*60/2\pi*2000 = 23.88 N-m$

$Radial force(f_r) = f_t*tan(\alpha_n)/cos(\beta) = 1290*tan(22)/cos(30) = 601.82 N$

$Axial Force(f_a) = f_t*tan(\beta) = 1290*tan(30°) = 744.78 N$

$Normal force (f_n) = \sqrt (f_t^2+f_a^2+f_r^2 )= \sqrt (1290)^2+(744.78)^2+(601.82^2) = 1606.54 N$

Hence, Total transmitted load = 1606.54 N.