ME 4244 Course Project by Jeffrey A. Kornuta Mechanical Engineering Louisiana State University Spring 2007 Abstract This paper will investigate a simple machine system which consists of several components. These components include a rotating shaft complete with two gears, two bearings, and keys to mate the gears with the shaft. First, a statics analysis will be performed in order to determine all forces acting on the system. Next, the bearings will be selected based on a 90% reliability rating, followed by the shaft materials selection and analysis (both statics and fatigue). Finally, the gears will be analyzed for both bending and contact stresses, including an adequate key selection to mate the gears to the shaft.

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Contents 1 Introduction 2 System Analysis 2.1 Statics Analysis . 2.2 Bearing Analysis 2.3 Shaft Analysis . . 2.4 Gear Analysis . . 2.5 Key Analysis . .

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3 Conclusions

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List of Tables 1

Summary of results from the statics analysis. . . . . . . . . . . . . .

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List of Figures 1 2 3 4

The system under consideration [2]. . . . . . . . Static force configuration on the system. . . . . Example of an angular contact ball bearing [1]. Example of a spur gear [3]. . . . . . . . . . . . .

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Introduction

Figure 1: The system under consideration [2].

To design is either to formulate a plan for the satisfaction of a specified need or to solve a problem. If the plan results in the creation of something having a physical reality, then the product must be functional, safe, reliable, competitive, usable, manufacturable, and marketable [2]. The design problem under consideration is given in Figure 1. This gear reduction system must reduce the rpm of another adjoining system while maintaining its specified dimensions and exerted forces. The system consists of a 42” long shaft, 24” diameter gear, 12” diameter gear, two bearings, and two keys. The shaft diameter is given as 2” while its given rotational speed is n = 500 rpm. Also, the life of the system must be designed for a life of 12,000 hours. The force configuration is given in Figure 1, with the axial load Ox = 150 lb applied from other components outside the system. From this stage, a statics analysis must be completed in order to begin with the specific materials selection and safety factor calculations.

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32 305 and cup 32 305.

Ans.

11-13 z R OzO

y

Ox

z

√ R = 0.95

T = 2706.3 lb-in

82.1

O

T = 240(12

16" y R Oy O

T A

226

210 451

14"

T

B

z R CzC

12" C

2706 F= 6 cos 2

x y

RC Cy

In xy-plane:

Figure 2: Static force configuration on the system.

2 2.1

#

System Analysis Statics Analysis

M O = −82.1(16) − 210( y

RC = 181 lbf

y analysis. The The principle step in designing this system is to perform the statics statics analysis relies on the six well-known equations O

In xz-plane: X ~

F = 0

X

~ = 0. M

R = 82 + 210 − 181 =

#

(1)

(2) − 452(3 M O = 226(16)

After performing the necessary calculations [Appendix 1-2], the following rez sults are obtained for the reaction forces: C After obtaining these values, the system design analysis may continue; however, z first the power rating is specified since the torque T has been calculated. HP =

Tn = 21.47 hp. 63025

R = −237 lbf R O = 226 − 451 + 237 R O = (111(3)2 + 122 ) 1/2

RC = (1812 + 2372 ) 1/2 The adequate bearings must be chosen for a certain reliability to obtain the reFeOtakes = the 1.2(112) = 134. quired system performance. At O, it is assumed that the bearing entire Ox = 150 lb load. Also, the combined radial load is found to be Or = 111.4 lb. FeC = = 357. Thus, for the bearing at O, a proper bearing size must be chosen. For 1.2(298) a design 40 000(200)(60) 4 xD = 106 2.2

Bearing Analysis

Table 1: Summary of results from the statics analysis. Value 497.68 lb 2706.3 lb-in -150 lb 110.9 lb 10.75 lb 181.51 lb -236.27 lb

Variable F T Ox Oy Oz Cy Cz

life of 12,000 hours, xD = LD /106 = 360. For a combined reliability of 90%, each bearing must be 95% reliable. For an application factor of one and an 02-series angular contact bearing,  C10 = Fe

xD x0 + (θ − x0 )(ln(1/R))1/b

1/a .

(4)

Because Fe = X(Or) + Y (Ox), Fe must first have an assumed value so that a trial and error solution may be found1 . Thus, a value of 700 N is chosen. This equivalent force requires a 12 mm bore bearing with C10 = 5.84 kN. Using the corresponding value for C0 , Ox/C0 is calculated in order to find X and Y . This process is repeated twice in order to obtain a final bore size of 20 mm for an angular contact ball bearing [Appendix 3]. Next, the correct bearing is chosen for C. Cr = 297.94 lb, while Cx = 0 because it is assumed that O takes the entire axial load. Again, assuming a reliability of 95% with an application factor of one, C10 = 11.06 kN. Thus, choosing an angular contact ball bearing, a bore size of 20 mm is chosen. The following presents a short summary of the results: • At O, 02-Series Angular Contact Ball Bearing, 20 mm bore (95%) • At C, 02-Series Angular Contact Ball Bearing, 20 mm bore (95%) • Combined reliability: R = 90% • System design life: 12,000 hours. Because the shaft for this system is 2”, it would be assumed that the shaft must be reduced in diameter in order for the bearings to be installed. However, 1

X and Y depend on the ratio Ox/C0 ; thus, their values vary for each bearing bore size.

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in practice, larger bearings would simply be chosen to fit the shaft. Thus, if implemented on this system, the combined reliability would obviously be much greater than 90%.

Figure 3: Example of an angular contact ball bearing [1].

2.3

Shaft Analysis

After constructing the shear force, torsion, and bending moment diagrams [Appendix 4], the point of greatest interest (ie., the most likely point to fail that must be designed against) is at B. The maximum moment M = 3.575 kip-in, while the max stress is σmax = 4.52 ksi and the shear stress τ = 1.72 ksi. Thus, the von Mises stress is calculated to be σ0 =

p

2 σmax + 3τ 2 = 5.42 ksi.

(5)

Note that since the shaft has an axial load of only 150 lb, it may be considered negligible for this analysis. Thus, if a safety factor of two was desired, the yield strength of the material must be Sy = σ 0 nd ≥ 10.83 ksi. Since this is the case, a cheap steal will be chosen: AISI 1030 CD. With this shaft material, the safety factor against static failure is 11.77 [Appendix 4]. The fatigue analysis shall be completed next. First Se must be computed; only taking into account ka , Se = ka (Sut /2) = 32.6 ksi, (6) where kb = kc = kd = ke = kf = 1 (assumed). Also, it is assumed that there exists no stress concentrations of any sort on the shaft surface, thus Kt = Kf = Kts = Kf s = 1. Modeling the system under pure rotation, it is obtained

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σa0 = σa = 4.52 ksi p 0 2 = 2.98 ksi. σm = 3τm

(7) (8)

Thus, using the Goodman criteria for the analysis [Appendix 5], the safety factor n = 5.62, which is a reasonable result. Also, since the safety factor against static failure is so high, damage due to shaft deflection shall be ignored. Here is a summary of the results: • Shaft material is AISI 1030 CD steel. Again, the diameter is 2”. • Safety factor against static failure is 11.77. • Safety factor against fatigue failure is 5.62.

2.4

Gear Analysis

Gear A and gear B have diameters of 24” and 12”, respectively. Because the outer surface of the gears must be hard in order to reduce wear, a high carbon steel is chosen. The specific steel chosen is AISI 4140 (Grade 1) steel from which the gears have been milled. The gears are spur gears, and for simplicity it is assumed that each gear’s mate is of the same material and same dimensions. Additionally, the face width is 1.5” for each gear, and the pitch is 6 T/in for each gear. The pressure angle φ for gear A and B is 20◦ and 25◦ , respectively. First, gear A is analyzed. The allowable bending stress for A is given as σall = Kv

W tp , FY

(9)

where Kv is the velocity factor, W t is the tangential force, p is the pitch, F is the face width, and Y is a tabulated correction factor. Thus, for bending, σall = 7.217 ksi [Appendix 5]. Utilizing the same type of procedure for fatigue as seen in the shaft analysis, Se is first calculated in order to design against an infinite life. Since the endurance limit is found to be Se = 53.13 ksi, the safety factor against bending is n = Se /σall = 7.36. For contact stresses, the simplified contact stress equation is utilized:  1/2  1 1 Kv W t + , (10) σc = −Cp F cos φ r1 r2 where r1 = 0.5(dG sin φ), and r2 = 0.5(d √ P sin φ). Since this gear is a steel gear driving another steel gear, Cp = 2300 psi from Table 14-8 [2]. Thus, σc = 38.96 ksi. 7

In order too calculate the safety factor, the value for the contact strength Sc must be found. From Figure 14-5 [2] and using the Brinell hardness for 4140 HB = 197, Sc = 92.53 ksi. Therefore, the safety factor against contact stress is calculated to be  2 Sc n= = 5.64. (11) σc

Figure 4: Example of a spur gear [3]. For the analysis of gear B, the procedure follows the strategy ensued for gear A. The allowable stress is calculated to be σall = 9.67 ksi, while the endurance limit Se is found to be Se = 53.13 ksi; thus, the safety factor against bending stress can be expressed as n = Se /σall = 5.5. For contact stresses, σc = 56.54 ksi and Sc = 92.53 ksi from Figure 14-5 [2]. Consequently, n = (Sc /σc )2 = 2.68. For each gear, the power rating HP = W t V /33000 matched the power rating found during the statics analysis. The following is a summary of the results for the gears: • Each gear is milled from high-carbon AISI 4140, Grade 1 steel. • Gear A had a safety factor of 7.36 under bending stresses and 5.64 under contact stresses. • Gear B had a safety factor of 5.5 under bending stresses and 2.68 under contact stresses.

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• Each gear’s power rating matched what had been previously defined for the system: 21.47 hp.

2.5

Key Analysis

The purpose of the keys is to mate the gears with the shaft. Because the torque T for each gear is identical, the dimensions of each key will also be identical. The force at the surface of the shaft P = T /r = 2.706 kips. Since the key material is assumed to be the material of the shaft (AISI 1030), the shear strength Ssy = 0.577Sy = 36.81 ksi. Using the following expression, the safety factor n may be obtained [Appendix 6]: P Ssy = , (12) n tF where t is the width of the key and F is the length of the key. The width t = 0.5” from Table 7.6 [2], and the length F is assumed to be the face width F = 1.5” in order to be conservative. Thus, the safety factor n = 10.2, which is more than adequate to prevent crushing. The following is a summary of the results for the keys: • The torque on each gear is identical, thus the key dimensions will be identical. • The key material is AISI 1030 CD steel. • The width of the key is 1/2” while the length is simply equal to the face width, F = 1.5”. • The safety factor for the key is 10.2, a very safe value to prevent crushing.

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Conclusions

Overall, the design of the system meets the required specifications. By following a logical design procedure, one may conceive the details of any system with ease. In the case of the system analyzed in this paper, the details of the system most assuredly meet the defined requirements. The following is a short summary of the pertinent system specifications. Bearings: • At O, 02-Series Angular Contact Ball Bearing, 20 mm bore (95%). • At C, 02-Series Angular Contact Ball Bearing, 20 mm bore (95%). • Combined bearing reliability: R = 90%. 9

• System design life: 12,000 hours. • For these bearings, the shaft would need to be cut; however, in practice larger bearings would be utilized, increasing the reliability. Shaft: • Shaft material is AISI 1030 CD steel. • 2” diameter, 500 rpm. • Safety factor against static failure is 11.77. • Safety factor against fatigue failure is 5.62. Gears: • Each gear is milled from high-carbon AISI 4140, Grade 1 steel. • Both gears have a pitch 6 T/in and face width 1.5”. • Gear A has a safety factor of 7.36 under bending stresses and 5.64 under contact stresses. • Gear B has a safety factor of 5.5 under bending stresses and 2.68 under contact stresses. • Each gear’s power rating matches what had been previously defined for the system: 21.47 hp. Keys: • The key material is AISI 1030 CD steel. • The width of each key is 1/2” while the length is simply equal to the face width 1.5”. • The safety factor for the key is 10.2, a safe value to prevent crushing. As for the overall system, this gear reduction system operates at 500 rpm with a horsepower rating of 21.47 hp. The design life is for 12,000 hours, and the overall factor of safety is equal to the lowest component factor of safety, 2.68. Some parts of the system might be thought of as over-engineered; however, the materials chosen are very common and (mostly) inexpensive. Moreover, this analysis of a gear reduction system attempts to portray Mechanical Engineering design in a practical light. 10

References [1] Bearing. www.reliance.com/mtr/cmbemn.htm. [2] R. G. Budynas, J. K. Nisbett. Shigley’s Mechanical Engineering Design. McGraw Hill, Eighth edition, 2008. [3] Spur gear. www.empirehobby.co.kr/shop/data/2/xray 5784.jpg.

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