Question

State the product design specification considering output shaft design factor as 1. Performance. 

Answer 1

Spur gears are commonly employed in industrial machinery to transmit power, torque, and speed. This also makes the constant speed and +ve drive possible. If we are talking about the question, we must first evaluate the type of gear tooth, and we utilized 200 full depth involute type teeth for the best single set up spur gearbox. And we also know that it is for the best.

Force on the spur gear teeth $F_t=\frac{100*P*C_s}{v}$

Torque $T=F_t*r$

Velocity $v=\frac{\pi d N}{60}$

Lewi's equation, $F_t= \sigma _b*\pi*y*m*b*c_v$

Dynamic force transmitted $F_d=F_t+F_i$

Check for wear $F_w=d*b*Q*k$ where $Q=\frac{2zg}{zp+zg}$

Face width$\implies 8m \eqslantless b\eqslantless12m$

Given data

P=10KW; N_g =1000 rpm; N_p =1700 rpm; Z_p =20

For material (1) 817M40 , the allowable stress $\sigma _g=221 MPa$

(2) 655M13 , $\sigma _p=345 MPa$

For pinion face width we know that Tangential force, $F_t=\frac{1000*P*C_s}{v}; C_s=1$ for steady load

Velocity pitch line $v=\frac{\pi d N}{60} \implies \frac{\pi*m*Z_p* N_p}{60}=\frac{\pi*m*20* 1700}{60}=1780.23m$

$F_t=\frac{1000*10*1}{1780.23m}=\frac{5.61}{m}$

Use lewi's equation

$F_t= \sigma _b*\pi*y*m*b*c_v$ ......(1)

$y=0.154-\frac{0.912}{Z_p}=0.1084$

$C_v=\frac{6.1}{6.1+562.2m}$

Put all the values in equation 1

$\implies \frac{5.61}{m}=34.5* \pi*0.1084*m*12m*10^6* \frac{6.1}{6.1+562.2m}$

b=12m because $9m \eqslantless b\eqslantless12m$

$\implies 31.11+3153.942m=8600215447m^3$

$m=1.61*10^{-3}meter (m)=1.61(mm)$

Face width for pinion $b=12*1.61=19.32mm$

Now for face width of gear $F_t=\frac{1000*P*C_S}{V}; C_S=1$

$v=\frac{\pi d_gN_g}{60} \implies \frac{\pi *m*Z_g*N_g}{60}$

$\frac{N_p}{N_g}=i=\frac{Z_g}{Z_p} \implies \frac{1700}{1000}=\frac{Z_g}{20} \implies Z_g=34$ No. of teeth on gear

$v=\frac{\pi *m*34*1000}{60} \implies 1780.23 m$

$F_t=\frac{1000*10*1}{1780.23 m} \implies \frac{5.61}{m}$

Use lewi's equation

$F_t= \sigma _b*\pi*y*m*b*c_v$ ......(11)

$y=0.154-\frac{0.912}{Z_g}=0.127$

$C_v=\frac{6.1}{6.1+562.2m}$

Put all the values in equation 11

$\implies \frac{5.61}{m}=\pi*0.127*m*221*10^6* \frac{6.1}{6.1+562.2m}$

$\implies 31.11+3153.942m=6454415930m^3$

$m=1.78*10^{-3}meter (m)=1.78(mm)$

Face width for pinion $b_g=12m=21.36mm=22mm$

Face width for pinion $b_p=19.32mm=20 mm$