Question

The client requires a single step-up spur gearbox to transmit a 10 kW power from a shaft rotating at 1000 rpm, while the output speed must be 1700 rpm. The material for gear and pinion is 817M40 and 655M13, respectively. Use table 6.10 on page 128 to select the suitable gear and pinion. 1.1 Briefly state the problem.2 State the design specification considering the following design factors: 1. Functionality 2. Stress.3 State formula and standards relevant to the single step-up spur gearbox design.4 If the pinion has 20 teeth, use the Lewis formula to determine the face width of the pinion and gear. Then, do one set of calculations and tabulate the answers.5 Conclusion

Answer 1

1.1- In this problem, we have to design the single set up spur gearbox, and for that there some of the data is given, but some of the data is not given, and that data that is not given is taken from the databook. In this problem, 10kw of power is transmitted from pinion to gear, and with the help of material specifications, we can design the gear.

1.2- Functionality- The spur gear is mostly used in industrial equipment for transferring power, torque, and speed. This also facilitates the constant speed and +ve drive. If we are talking about the question, we have first to consider the type of gear tooth, and for the best single set up spur gearbox, we used 20⁰ full depth involute type teeth. And we also know that for good functionality, the face width varies between 8m to 12m.

Stress- To ensure gear safety, we have first to calculate the maximum tooth stresses because if we don't know this, it can cause tooth damaged. The bending stress mostly occurs in the case of spur gear. And the maximum stress can be calculated by using the theory of elasticity or by some numerical methods that as FEA. Then before taking any material for gear design, we have to know about their strengths, and it comes under the requirements.

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 $b_p=19.32mm=20 mm$

Conclusion- Under this problem, after doing one set of calculations we get the face width and the module of the gear is more as compare to the pinion. but by the material specifications, gear material is allowable stress is less as compare to pinion material.