Question 9.5: Design a pair of spur gears to be used as a part of the driv...

For this problem, P = 3.0 hp and $n_p = 1750$ rpm, $K_o= 1.75$ (uniform driver heavy shock driven machine). Then $P_{des} = (1.75) (3.0 hp) = 5.25$ hp. Try $P_d = 12$ for the initial design.

Step 2. Specify the number of teeth in the pinion. For small size, use 17 to 20 teeth as a start.

For this problem, let’s specify $N_P = 18$.

Step 3. Compute the nominal velocity ratio from $VR = n_p/n_G$

For this problem, use $n_G=462.5$ rpm at the middle of the acceptable range.

$VR = n_p/n_G = 1750/462.5 = 3.78$

Step 4. Compute the approximate number of teeth in the gear from $N_G = N_P(VR).$

For this problem, $N_G= N_P(VR) = 18(3.78) = 68.04$. Specify $N_G = 68.$

Step 5. Compute the actual velocity ratio from $VR = N_G/N_P$

For this problem. $VR = N_G/N_P = 68/18 = 3.778$.

Step 6. Compute the actual output speed from $n_G= n_P (N_P/N_G).$
For this problem, $n_G=n_P (N_P/N_G) = (1750 rpm)( 18/68) = 463.2$ rpm. OK.

Step 7. Compute the pitch diameters, center distance, pitch line speed, and transmitted load and judge the general acceptability of the results.

For this problem, the pitch diameters are:

$D_P = N_P/P_d= 18/12 = 1.500$ in
$D_G=N_G/P_d = 68/12 = 5.667$ in

Center distance:

$C= (N_P + N_G)/(2P_d) = (18 + 68)/(24) = 3.583$ in

Pitch line speed: $v_t = πD_Pn_P/12 = [ π( 1500 ) (1.750 ) ] /12 = 687$ ft/min
Transmitted load: $W_t = 33 000(P)/v_t = 33 000(3.0)/687 = 144$ lb

These values .seem to be acceptable.

Step 8. Specify the face width of the pinion and the gear using Equation (9-28) as a guide.

Nominal value of $F = 12/P_d$                  (9-28)

For this problem: Lower limit $= 8/P_d = 8/12 = 0.667$  in.
Upper limit $= 16/P_d = 16/12 = 1.333$ in
Nominal value $= 12/P_d = 12/12 = 1.00$  in. Use this value.

Step 9. Specify the type of material for the gears and determine $C_P$ from Table 9-9.
For this problem, specify two steel gears. $C_P = 2300$.

Step 10. Specify the quality number, $Q_v$ using Table 9-2 as a guide. Determine the dynamic factor from Figure 9-21.

For this problem, specify $Q_v = 6$ ,for a wood chipper. $K_v = 1.35.$

Step 11. Specify the tooth form, the bending geometry factors for the pinion and the gear from Figure 9-17 and the pitting geometry factor from Figure 9-23.

For this problem, specify 20° full depth teeth.$J_P = 0.325, J_G = 0.410, I = 0.104$

Step 12. Determine the load distribution factor ,$K_m,$ from Equation (9-16) and Figures 9-18 and 9-19. The precision class of the gear system design must be specified.
Values may be computed from equations in the figures or read from the graphs.

For this problem: $F = 1.00 in. D_P = 1.500. F/D_P = 0.667. Then C_{pf} = 0.042.$
Specify open gearing for the wood chipper, mounted to the frame. $C_{ma} = 0.264.$
Compute: $K_m = 1.0 + C_{pf} + C_{ma} + 0.042 + 0.264 = 1.3$

Step 13. Specify the size factor, $K_s$ from Table 9-6.

For this problem, $K_s= 1.00$ for $P_d = 12$

Step 14. Specify the rim thickness factor, $K_B,$ from Figure 9-20.
For this problem, specify a solid gear blank.$K_B, = 1.00.$
Step 15. Specify a service factor. SF. typically from 1.00 to 1.50. based on uncertainty of data.
For this problem, there is no unusual uncertainty. Let SF = 1.00.

Step 16. Specify a hardness ratio factor, $C_H,$ for the gear, if any. Use $C_H = 1.00$ for
the early trials until materials have been specified. Then adjust $C_H$ if significant differences exist in the hardness of the pinion and the gear.

Step 17. Specify a reliability factor using Table 9-8 as a guideline.

For this problem, specify a reliability of 0.99. $K_R= 1.00.$

Step 18. Specify a design life. Compute the number of loading cycles for the pinion and the gear. Determine the stress cycle factors for bending $(Y_N)$ and pitting $(Z_N)$ for the pinion and the gear.

For this problem, intermittent use is expected. Specify the design life to be 3000 hours, similar to agricultural machinery. The numbers of loading cycles are:

$N_{CP} = (60)(3000 \ hr)(1750rpm)(1) = 3.15 × 10^8$ cycles
$N_{eG}  = (60)(3000 \ hr)(462.5rpm)(1) = 8.33 × 10^7$ cycles

Then, from Figure 9-22, $Y_{NP} = 0.96. Y_{NG} = 0.98.$ From Figure 9-24, $Z_{NP} = 0.92. Z_{NG}= 0.95$

Step 19. Compute the expected bending stresses in the pinion and the gear using Equation (9-15).

$s_{tP}=\frac{W_{t}P_d}{FJ_{P}}K_oK_sK_mK_BK_v=\frac{(144)(12)}{(100)(0.325)} (1.75)(1.0)(1.31)(1.0)(1.35)=16400 \ psi$
$s_{tG}= s_{tP}(J_p/J_G) = (16 400)(0.325/0.410) = 13 000 \ psi$

Step 20. Adjust the bending stresses using Equation 9-20.
For this problem, for the pinion:

$S_{atP}\gt S_{tP}\frac{K_R(SF)}{Y_{NP}} =(16 400)\frac{(1.00)(1.00)}{0.96}= 17 100$ psi

For the gear:

$S_{atG}\gt S_{tG}\frac{K_R(SF)}{Y_{NG}} =(13000)\frac{(1.00)(1.00)}{0.98}= 13 300$psi

Step 21. Compute the expected contact stress in the pinion and the gear from Equation (9-25). Note that this value will be the same for both the pinion and the gear.

$s_{c}=C_P\sqrt{\frac{W_{t}K_oK_sK_mK_v}{FD_{P}I}} =2300\sqrt{\frac{(144)(1.75)(1.0)(1.31)(1.35)}{(1.00)(1.50)(0.104)} }=122900$ psi

Step 22. Adjust the contact stresses for the pinion and the gear using Equation (9-27).

$S_{acP}\gt S_{cP}\frac{K_R(SF)}{Z_{NP}}=(122 900)\frac{(1.00) (1.00)}{(0.92)} =133 500$ psi

For the gear:

$S_{acG}\gt S_{cG}\frac{K_R(SF)}{Z_{NG}C_H}=(122 900)\frac{(1.00) (1.00)}{(0.95)(1.00)} =129300$ psi

Step 23. Specify materials for the pinion and the gear that will have suitable through hardening or case hardening to provide allowable bending and contact stresses greater than those required from Steps 20 and 22. Typically the contact stress is the controlling factor. Refer to Figures 9-10 and 9-11 and Tables 9-3 and 9-4 for data on required hardness. Refer to Appendices 3 to 5 for properties of steel to specify a
particular alloy and heat treatment.

For this problem contact stress is the controlling factor. Figure 9-11 shows that through hardening of steel with a hardness of HB 320 is required for both the pinion and the gear. From Figure A 4-4 , we can specify AISI 4140 OQT 1000  steel that has a hardness of HB 341, giving a value of $s_{ac} = 140 000$ psi. Ductility is adequate as indicated by the 18% elongation value. Other materials could be specified.