Answer to Question #210763 in Mechanical Engineering for Dimpho Thupaemang

Question

An input shaft connected to the motor rotates at 1200 rpm, while the output speed must be 570 rpm. The contact force between the gears is transmitted through the pressure angle of 20°. The radial

force is 321,90 N. The material for both gear and pinion is 817M40.

Use table 6.8 on page 126 to select the suitable gear and pinion (make use of the (35) Lewis formula). Then, do one set of calculations and tabulate the answers.

Answer 1

Given:

 N₁ = 1200 rpm

N₂ = 570 rpm

$\phi = 20^0$

$F_r = 321.90 kN$

The below diagram shows the force analysis used for gear tooth:

$F_r = F_t *tan \phi \implies F_t = \frac{321.90 }{ tan 20 }= 884.41 kN$

$torque, T_1 = F_t R_1$

$Gear \space ratio, G = \frac{ N_1 }{ N_2 } = 2.1$

Assume that the teeth of the pinion are 20,i.e., z₂ = 20

then teeth on gear, z₁ = 20/2.1 ~ 10

Assume that the width of the gear = 200 mm

From the databook of machine design,

    For z = 20, Y = 0.320

    and S_b = 850 MPa

Applying Lewis eqaution,

 $F_t = b_mY*(S_b)$

  884.41 *1000 =$200 * (\frac{D_1}{10})*0.320 * 850$

   D₁ = 162.57 mm

Pinion diameter = 162.57 mm   

Gear diameter =$2.1 * 162.57 =341.4 mm$