Design a pair of spur gears to be used as a part of the drive for a chipper to prepare pulpwood
for u.se in a paper mill. Intermittent use is expected. An electric motor transmits 3.0 horsepower to the pinion at 1750 rpm and the gear must rotate between 460 and 465 rpm. A compact design is desired.

Question

Accepted Answer

General Design Procedure

Step 1. Considering the transmitted power. P. the pinion speed, [latex]n_p[/latex], and the application, refer to Figure 9-27 to determine a trial value for the diametral pitch, [latex]P_d.[/latex] The overload factor. [latex]K_o[/latex] can be determined from Table 9-5, considering both the power source and the driven machine.

TABLE 9-5 Suggested overload factors, [latex]K_o[/latex]
	Driven Machine
Power source	Uniform	Light shock	Moderate shock	Heavy shock
Uniform	1	1.25	1.5	1.75
Light shock	1.2	1.4	1.75	2.25
Moderate shock	1.3	1.7	2	2.75

For this problem, P = 3.0 hp and [latex]n_p = 1750[/latex] rpm, [latex]K_o= 1.75[/latex] (uniform driver heavy shock driven machine). Then [latex]P_{des} = (1.75) (3.0 hp) = 5.25[/latex] hp. Try [latex]P_d = 12[/latex] 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 [latex]N_P = 18[/latex].

Step 3. Compute the nominal velocity ratio from [latex]VR = n_p/n_G[/latex]

For this problem, use [latex]n_G=462.5[/latex] rpm at the middle of the acceptable range.

[latex]VR = n_p/n_G = 1750/462.5 = 3.78[/latex]

Step 4. Compute the approximate number of teeth in the gear from [latex]N_G = N_P(VR).[/latex]

For this problem, [latex]N_G= N_P(VR) = 18(3.78) = 68.04[/latex]. Specify [latex]N_G = 68.[/latex]

Step 5. Compute the actual velocity ratio from [latex]VR = N_G/N_P[/latex]

For this problem. [latex]VR = N_G/N_P = 68/18 = 3.778[/latex].

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

[latex]D_P = N_P/P_d= 18/12 = 1.500[/latex] in
[latex]D_G=N_G/P_d = 68/12 = 5.667[/latex] in

Center distance:

[latex]C= (N_P + N_G)/(2P_d) = (18 + 68)/(24) = 3.583[/latex] in

Pitch line speed: [latex]v_t = πD_Pn_P/12 = [ π( 1500 ) (1.750 ) ] /12 = 687[/latex] ft/min
Transmitted load: [latex]W_t = 33 000(P)/v_t = 33 000(3.0)/687 = 144[/latex] 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 [latex]F = 12/P_d [/latex] (9-28)

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

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

TA BLE 9-9 Elastic coefficient. [latex]C_P [/latex]
		Gear material and modulus of elasticity. [latxe]E_G, lb/in^2(MPa)[/latex]

	Modulus of elasticity, [latex]E_P, lb/in^2 (MPa)[/latex]	Steel [latex]30×10^6[/latex] [latex](2 ×10^5)[/latex]	Malleable iron [latex]25 ×10^6[/latex] [latex](1.7 ×10^5)[/latex]	Nodular iron [latex]24×10^6[/latex] [latex](1.7 ×10^5)[/latex]	Cast iron [latex]22×10^6[/latex] [latex](1.5 ×10^5)[/latex]	Aluminum bronze [latex]17.5×10^6[/latex] [latex](1.2 ×10^5)[/latex]	Tin bronze [latex]16×10^6[/latex] [latex](1.1 ×10^5)[/latex]
Pin ion material	Modulus of elasticity, [latex]E_P, lb/in^2 (MPa)[/latex]	Steel [latex]30×10^6[/latex] [latex](2 ×10^5)[/latex]			Cast iron [latex]22×10^6[/latex] [latex](1.5 ×10^5)[/latex]
Steel	[latex]30×10^6[/latex]	2180	2180	2160	2100	1950	1900
	[latex](2 ×10^5)[/latex]	(191)	(181)	( 179)	(174)	(162)	(158)
Mal l. i ron	[latex]25 ×10^6[/latex]	2 180	2090	2070	2020	1900	1850
	[latex](1.7 ×10^5)[/latex]	(181)	(174)	(172)	(168)	(158)	(154)
Nod . iron	[latex]24×10^6[/latex]	2160	2070	2050	2000	1880	1830
	[latex](1.7 ×10^5)[/latex]	(179)	(172)	(170)	(166)	(156)	(152)
Cast iron	[latex]22×10^6[/latex]	2100	2020	2000	1960	1850	1800
	[latex](1.5×10^5)[/latex]	(174)	(168)	(166)	(163 )	(154)	(149)
Al. bronze	[latex]17.5×10^6[/latex]	1950	1900	1880	1850	1750	1700
	[latex](1.2×10^5)[/latex]	(162)	(158)	(156)	(154)	(145)	(141)
Tin bronze	[latex]16 ×10^6[/latex]	1900	1850	1830	1800	1700	1650
	[latex](1.1×10^5)[/latex]	(158)	(154)	(152)	(149)	(141)	(137)
Source: Extracted from AGMA Standard 2001-C95. Fundamental Rating Factors and Calculation Methods for Involute Spur and Helical Gear Teeth, with the permission of the publisher. American Gear Manufacturers Association. 1500 King Street, Suite 201, Alexandria, VA 22314. Note: Poisson's ratio = 0.30; units for [latex]C_P[/latex] are [latex](lb/in^2)^{0.5}[/latex] or [latex](MPa)^{0.5}[/latex].

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

TABLE 9-2 Recommended AGMA quality numbers
Application	Quality number	Application	Quality number
Cement mixer drum drive	3-5	Small power drill	7-9
Cement kiln	5-6	Clothes washing machine	8-10
Steel mill drives	5-6	Printing press	9-11
Grain harvester	5-7	Computing mechanism	10-11
Cranes	5-7	Automotive transmission	10-11
Punch press	5-7	Radar antenna drive	10-12
Mining conveyor	5-7	Marine propulsion drive	10-12
Paper-box-making machine	6-8	Aircraft engine drive	10-13
Gas meter mechanism	7-9	Gyroscope	12-14
Machine tool drives and drives for other high-quality mechanical systems
Pitch line speed (fpm)	Quality number		Pitch line speed (m/s)
0-800	6-8		0-4
800-2000	8-10		4-11
2000-1000	10-12		11-22
Over 4000	12-14		Over 22

For this problem, specify [latex]Q_v = 6[/latex] ,for a wood chipper. [latex] K_v = 1.35.[/latex]

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.[latex] J_P = 0.325, J_G = 0.410, I = 0.104[/latex]

Step 12. Determine the load distribution factor ,[latex]K_m,[/latex] 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: [latex]F = 1.00 in. D_P = 1.500. F/D_P = 0.667. Then C_{pf} = 0.042.[/latex]
Specify open gearing for the wood chipper, mounted to the frame. [latex]C_{ma} = 0.264.[/latex]
Compute: [latex]K_m = 1.0 + C_{pf} + C_{ma} + 0.042 + 0.264 = 1.3[/latex]

Step 13. Specify the size factor, [latex]K_s[/latex] from Table 9-6.

TABLE 9-6 Suggested size factors. [latex]K_s[/latex]
Diametral pitch. P_d	Metric module, m	Size factor, [latex]K_s[/latex]
≥5	≤5	1
4	6	1.05
3	8	1.15
2	12	1.25
1.25	20	1.4

For this problem, [latex]K_s= 1.00[/latex] for [latex]P_d = 12[/latex]

Step 14. Specify the rim thickness factor, [latex]K_B,[/latex] from Figure 9-20.
For this problem, specify a solid gear blank.[latex]K_B, = 1.00.[/latex]
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, [latex]C_H,[/latex] for the gear, if any. Use [latex]C_H = 1.00[/latex] for
the early trials until materials have been specified. Then adjust [latex]C_H[/latex] 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.

TABLE 9-8 Reliability factor, [latex]K_R[/latex]
Reliability	[latex]K_R[/latex]
0.90, one failure in 10	0.85
0.99, one failure in 100	1
0.999. one failure in 1000	1.25
0.9999, one failure in 10 000	1.5

For this problem, specify a reliability of 0.99. [latex]K_R= 1.00.[/latex]

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 [latex](Y_N)[/latex] and pitting [latex](Z_N)[/latex] 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:

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

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

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

[latex]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[/latex]
[latex]s_{tG}= s_{tP}(J_p/J_G) = (16 400)(0.325/0.410) = 13 000 \ psi[/latex]

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

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

For the gear:

[latex]S_{atG}\gt S_{tG}\frac{K_R(SF)}{Y_{NG}} =(13000)\frac{(1.00)(1.00)}{0.98}= 13 300[/latex]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.

[latex]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 [/latex] psi

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

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

For the gear:

[latex]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[/latex] 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.

TABLE 9-3 Allowable stress numbers for case-hardened steel gear materials
	Allowable bending stress number,[latex] s_{at}[/latex] (ksi)			Allowable contact stress number, [latex] s_{ac}[/latex] (ksi)
Hardness at surface	Grade 1	Grade 2	Grade 3	Grade 1	Grade 2	Grade 3
Flame- or induction-hardened:
50 HRC	45	55		1 70	190
54 HRC	45	55		175	195
Carhurized and case-hardened :
55-64 HRC 55				180
58-64 HRC 55		65	75	180	225	275
Nitrided through-hardened		steel :
83.5 HR 15N		See Figure 9-14.		150	163	175
84.5 HR 15N		See Figure 9-14.		155	168	180
Nitrided nitralloy [latex]135M:^a[/latex]
87.5 HR 15N		See Figure 9-15.
90.0 HR 15N		See Figure 9-15.		170	183	195
Nitrided nitralloy [latex]N :^a[/latex]
87.5 HR 15N 90.0 HR 15N	See Figure 9-15. See Figure 9-15.			172	188	205
Nitrided 2.5% chrome( no aluminum ): 87.5 HR 15N	See Figure 9-15.			155	172	189
90.0 HR 15N	See Figure 9-15.			176	196	216
Source: Extracted from AGMA Standard 2001 -C9.S. Fundamental Ratins.; Factors and Calculation Methods for Involute Spur and Helical Gear Teeth, with the permission of the publisher. American Gear Manufacturers Association, 1500 King Street. Suite 201. Alexandria. VA 22314, Nitralloy is a proprietary family of steels containing approximately 1.0% aluminum which enhances the formation of hard nitrides.

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 [latex]s_{ac} = 140 000[/latex] psi. Ductility is adequate as indicated by the 18% elongation value. Other materials could be specified.

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

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