Answer to Question #223619 in Mechanical Engineering for Thabang

Question

QUESTION 1 GEARS [40]

1.1 Write a problem statement to verify spur gear set material selected from Table 

6.9, page 127 of the prescribed textbook.

(3)

1.2 Elaborate on the product design specification for the design of the spur gear 

set (Figure 1 above) referring to the performance aspect, consider the 

transmission of power, the gear ratio, and efficiency.

(6)

1.3 Verify the suitable material for pinion and gear (Figure 1 above) based on 

Lewis’ formula if both the pinion and gear were selected from Table 6.9, page 

127.

(31)

QUESTION 2 MINIMUM DIAMETER OF THE OUTPUT SHAFT [10]

2.1 The output solid steel shaft is supported by the ball bearings, as shown in 

Figures 1 & 2 above. The shaft operates at room temperature. Determine the 

minimum allowable diameter of the output shaft based on ASME CODE and 

select the standard size of the shaft from the Macsteel catalogue, provided 

with the question paper.

Answer 1

Spur gears can be manufactured of metals like steel or brass or polymers like nylon or polycarbonate. Plastic gears are quieter but at the price of strength and loading capacity.

1.1- In this problem, we have to design the single set up spur gearbox, and for that, 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 first have to consider the gear tooth type, 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 must first 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 using the theory of elasticity or by some numerical methods such 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 gear module is more as compare to the pinion. But by the material specifications, gear material is allowable stress is less as compare to pinion material.