Gear Project Design

BICOL UNIVERSITY C O LL E GE O F E N GI N EE R IN G Legazpi City

S U B M I T T E D

T O:

ENGR. MANUEL RUSTRIA,MME Professor

S U B M I T T E D

B Y:

RUSSEL JAMES ALAMER BSME4B

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L ET TE R

O F

T RA NS MI TT AL

February 11 ,2010

Engr. Manuel Rustria Professor in Charge College of Engineering Legazpi City

I hereby submit this design as a partial fulfillment to the completion of the subject Machine Design 2. Given this day of February 11 ,2010.

Sincerely yours, Russel James O. Alamer









ACKNOWLEDGEMENT

The designer expresses his deep gratitude and sincere appreciation to those who inspired and helped him in the realization of this design project. To His professor Engr. Manuel Rustria for his valuable advices and suggestions in the designing and calculating process. To his classmates who shared their knowledge and resources in the fulfillment of the project: Fernado Pillejera Jr., Andrew Ferreras, Christian Gongona, Mark Eljun Vibar and Joseph Grimaldo. To his parents Rodolfo and Jonibeth Alamer for their encouragement and financial support. Above all to the Lord Jesus Christ, his God and Savior Who enables the designer to do everything according to His will.









TABLE

OF

CONTENTS

TITLE PAGE…………………………………………………… PAGE………………………………………………………………………………i …………………………i

LETTER OF TRANSMITTAL……………………………………………………………ii

ACKNOWLEDGEMENT ………………………………………………………………...iii

TABLE OF CONTENTS………………………………………………………………....iv

STATEMENT OF THE PROBLEM……………………………………………………..1

COMPUTATIONS………………………………………………………………………..2

SUMMARY OF THE COMPUTED DIMENSIONS……………………………………11

SKETCHES……………………………………………………………………………….

BIBLIOGRAPHY………………………………………………………………………….13

APPENDICES……………………………………………………………………………..14









S TA TE ME NT O F T HE P RO BL EM

PROBLEM 676] A gear reduction unit is to be designed according according to the data in the table and the following specifications. specifications. Hp

Mg

80

2

Max. ctr. Dist., in. 5¼

Rpm, Pinion

Kind of load Mon. N sf

720

Minor pulsations, 1.2

The Velocity ratio may be varied by an amount necessary to have whole tooth numbers. The given center distance is the permissible maximum. The teeth are to be 20˚FD with Np≥ 18 teeth, with 17 as the minimum acceptable. acceptable. The service is continuous, with indefinite life. Use Buckingham’s dynamic-load for average gears. a) Decide upon the material with its treatment, pitch, and face width. Start out being orderly with your calculations calculations so that you do not need to copy all of them for your report. The report should show calculations for the final decision first, but all significant calculations should be in the appendix. These latter calculations calculations should show: that a cheap material cannot be used; that through hardened- steel (minimum permissible tempering temperature is 800˚F), flame or induction-hardened induction-hardened teeth have all been considered in detail. b) To complete the design of the gears. A shaft size is needed. At the option of the instructor (i) compute shaft diameters for pure torsion only using a conservative design stress as S s = 6ksi (to cover stress concentration factor, minor bending, on the assumption that the bearings will be quite close to the gears, etc.); or (ii) make a tentative assumption assumption of the distance between bearings, and design the shafts by a rotational procedure. It would be logical for the input and output to be via flexible couplings. Let the shaft material be









d) Make a sketch of each gear gear (on separate separate sheets of paper) including including on it all dimensions and information for its manufacture. e) At the instructor’s instructor’s option (i) (i) choose choose rolling type bearings, bearings, or (ii) design design sleeve bearings. f) Decide Decide upon upon all details details of of the housin housing g to enclose enclose the the gears, gears, with with sketches sketches depicting them. g) Your final final report report should be be arranged arranged (1) Title Page; (2) a summary summary of the the final design decisions, computed dimensions, and material specifications; (3) sketches; (4) final calculations; calculations; (5) other calculations. calculations.

1











COMPUTATIONS] GIVEN:

Power = P = 80 hp Gear Ratio=M g=2 Center distance= Cd=5 ¼ Pinion Revolution Revolution = N = 720 rpm Kind of Loading = Minor Pulsations Nsf = 1.2 Pressure Angle = Ф = 20 ˚FD Design the gears based on endurance strength SOLUTION: Mw = Mg = 2, since the pinion is the driver. Solving for the diameters: diameters: Cd= (Dg+ Dp) ½ 2(5.25)= Dg+ Dp ◘ Dg=(10.5)-Dp ……..eq.1 ……..eq.1 Since Mw = Dg/Dp =2, Substitute eq. 1: 2= (10.5-Dp)/ Dp 2 Dp + Dp =10.5 Dp=10.5/3 Dp=3.5 in. ◘ Dg=10.5 - 3.5 Dg=7 in. Assuming no. of teeth for pinion, pinion , Np = 17 Pd = Np/Dp = 17/3.5 = 4.85, use 5 with Pd = 5 and, For N For Ng: Ng= Pd(Dg) = (7)(5) =35









= [π(3.4 in.)(720 rpm)]/(12 in/ ft.) = 640.8849013 fpm For transmitted Load F t t : Ft = 80hp/Vm = [(80hp)(33000 ft.-lb/min-hp)]/ 646.5397681 fpm = 4119.304409 lb. assume face width, width , b =1.75, [8 < bPd <12.5] bPd = 1.75(5) = 8.75 …satisfied! By Buckingham’s average Dynamic Load, Fd = Ft + I = Ft + [o.o5Vm(bc + Ft) ]/[0.05Vm + (bc + F t)1/2]….eq. ]….eq. 2 Solving for c, assume steel and steel, at AT25: c = 1660e 1660e permissible error, e , from AF20 e = .0005 (for precision cut ) therefore, c = (.0005)(1660)(1000) = 830 substitute in eq. 2, Fd = 4119.304409 + [0.05(640.8849013)(1.75x830 + 4119.304409)]/ 4119.304409) ]/ [0.05(640.8849013) + (1.75x830 (1.75x830 + 4119.304409) 4119.304409)1/2] = 5792.810194 lb taking limited load, Fw = DpbQK g Solving for Q, Q = (2 Mg) / (Mg + 1) = (2x2.055)/(2.055+1) = 1.346 since Fw = Fd= 5792.810194, 5792.810194, K g = Fw / (DpbQ) = 5792.810194/(1.346x3.4x 5792.810194/(1.346x3.4x1.75) 1.75)









For gear, At BHN = 269 Choose C1137, at table AT9, Design of Machine Elements by V.M. Faires Solve for the corresponding tempering temperature, ˚F 700 t˚F 1000

BHN 277 269 229

t˚F =tempering tempaerature tempaerature = 750˚F solving for S n’y: for Gear, Sn’ = 250(269) psi =67,250psi assuming Load near at middle, with 35 teeth, y= 0.633 ,from table AT24 therefore, Sn’y = (.633)(67,250) = 42,569.25 psi for pinion, Sn’ = 250(331) psi = 82,750 psi assuming Load near at middle, with 17 teeth, y = .512, from table AT24 therefore, therefore, Sn’y = (.512)(82,750) = 42,368 psi With Sn’y of pinion < S n’y of Gear Therefore, Gear is stronger. Use k f f as 1.5, Fs = Sn’yp (b)/[(K f f)(P ) (Pd) = (1.75)(42,368)/(1.5 x 5) = 9885.8667 lb. ≤ Fs ≥ Fd, 9,885.8667 ≥ 5,759.986, satisfied









• Basic Clearance = 0.25/Pd = 0.25/5 = 0.05 in. • Clearance = 0.35/Pd = 0.35/5 = 0.07 in. For Gear: • Dedendum Diameter = Dg – 2d = 7 – 2(0.25) = 6.5 in. • Addendum Diameter = Dg + 2a = 7 + 2(0.2) = 7.4 in. • Rim Diameter = Dedendum Diameter – 2(rimT) = 6.5 – 2(0.3518) = 5.7694 in. For Pinion: • Dedendum Diameter = DP – 2d = 3.4 – 2(0.25) = 3.9 in. • Addendum Diameter = DP + 2a = 3.4 + 2(.2) = 3 in. For shaft of pinion: P = 2πTN T = P/ 2πN = [80 hp x (33000 ft. lb x 12 in)/ (min-hp-ft)]/ (2π)(720 rpm) = 7002.817 lb.-in.









Solve for the Hub Diameter. D h, for pinion, use 2 ½ in. since D p = 3.43 Solve for Hub Length, L hub, Lhub = 1.625(Ds) =1.625(1 11/16 in.) =2.74 in. For Key: With suggested material C1118 (cold drawn), Su = 80ksi, Sy = 75 ksi, F.S. = 2.125 (based in S y) Sds = (0.6)(75ksi)/ 2.125 = 21.176 ksi Sdc = (75ksi)/2.125 = 35.29ksi since Sds /S /Sdc ≠ 0.5, use flat key, At AT19, Machine Design by Faires, At Shaft Diameter = 1 11/16 in., b = 3/8 t=¼ base tolerance: 0.0025 in.

Solving for length of key, Using Sdc, L = 4T / (Sdc)(t)(Ds) = 4(7002.817496 lb. in.)/[(35,291 psi)(1 11/16 in.)(1/4 in.)] = 1.88 in. Using Sds, L = 2T/ (S ds)(b)(Ds) = (2 x 7002.817496 lb. in.)/ [(1 11/16 in.)(3/8 in.)(21,176.47psi)] = 1.045 in Assume the length of the key = length of hub, Lkey = Lhub = 2.74 in.











At given Sd = 6ksi, with suggested material material AISI 1137 (cold finished) Solve for D s, Ds = 3√[(6x14,417.32 lb. in. x 1.25 in.)/(πx6000lb. in2)] = 1.7901 in., use 1 15/16 in ,from Design of Machine Elements, by Faires For tolerance, tolerance , Assuming class RC 1, from table 3.1,Design of Machine Elements, by Faires At given Shaft diameter: -0.004 in. -0.007 in. For hub of gear, Solve for the Hub Diameter. D h, Dh = 1.8(Ds) = 1.8(1 15/16 in.) = 3.4875., use 3 ½ in. Solve for Hub Length, L hub, Lhub = 1.625(Ds) =1.625(1 15/16 in.) =3.15 in. For Key: With suggested material C1118 (cold drawn), Su = 80ksi, Sy = 75 ksi, F.S. = 2.125 (based in S y)









Solving for Rim thickness, tr, tr = 0.56(Pc) = 0.56(π/Pd) = 0.56(π/5) = .3518 in. Solving for Web thickness, tw, tw= 0.55(Pc) = 0.55(π/Pd) = 0.55(π/5) = .3455 in. For the arms, The designer has decided not to make use of arms, since the diameter of both pinion and gear is small, it would just add cost on the machining. For the bead, Since there is no arm designed, it is not advisable to use a bead. for the Bearing of pinion, with Ds = 1 11/16 in., use BRG. 308 with Bore Dia. = 40mm or 1.5748 in. Outside Dia. = 90mm or 3.5433 in. for the Bearing of gear, with Ds = 1 11/16 in., use BRG. 304 with Bore Dia. = 20mm or .7874 in. Outside Dia. Dia. = 52mm 52mm or 2.04 in. As widely used material for bearing, choose Copper-base alloys Since the selected bearing for the pinion and gear is not exactly suited to their shaft









K g = Fw / (DpbQ) = 5646.75 / (3x1.75 x1.346) = 799.087 From table AT 26 in the 20 deg. column p.603 With K g = 799 Since it is out of range, choose the Choose K g = 400, ∑BHN=600, From p. 57 T.B. Cast Iron in general sense includes white cast iron, malleable iron, and nodular iron, but when cast iron is used without a qualifying adjective, gray cast iron, spoken of as gray iron is meant. From that definition we can use table AT 6 p. 570 T.B., and we can notice from that given type, if we select two materials and get their ∑BHN it is smaller compare to our computed ∑BHN. For example if we take the two material with the highest BHN, ASTM60 with ∑BHN= 302 and nodular cast iron 100-70-03(heat treated) (ASTM A396-58) with BHN= 270, their ∑BHN= 572, since 600>572 therefore we cannot use that materials. ~END









D ES IG N BEARING ~GEAR

Rolling Bearing Material: B e oDf icaum Tyopre t:eter O uN ts:ide Diameter BH No. of Teeth Gear Diameter

S UM MA RY

BRG. 30temperature 4 AISI C1137 with tempering 750˚F 20mPm r i.o7n8c7u4t in. recois 52mm o2r692.04 in. 35 7 in.

HUB

Material GEAR AND Hub Diameter Center distance Hub length

Velocity ratio KEY Material thickness base Diametral pitch Length

Gear ratio

Face width RIM

Addendum: Material Thickness: Dedendum WEB

Working Materialdepth Thickdepth ne s s whole ARM

Fillet radius SHAFT

clearance Material

AISI C1137 with tempering temperature 750˚F 3 ½ in. 5.2 in. 3.15 in. 2.0558 AISI C1118 cold drawn 2.0558 3/8 in. 51/2 in. (tolerance: -0.0025 in.) 3.15 in. 1.75in

PINION DIMENSIONS

AISI C1/5 113in 7 with tempering temperature 750˚F .3518in. 1.25/5 in. AISI C2/5 113in. 7 with tempering temperature 750˚F .3455in. 2.25/5 in. N.A. 0.3/5 0.35/5AISI 1137 cold finished









B I B L I O G R A P H Y 1. V. M. FAIRES, DESIGN OF MACHINE ELEMENTS 4th EDITION, THE

MACMILLAN COMPANY, NEW YORK, 1969 2. PSME PSME CODE, CODE, THE PSME PSME CODE COMMIT COMMITTEE, TEE, 1984 1984











APPENDICES

Addendum. The radial distance between the Pitch Circle and the top of the teeth.

Backlash. Play between mating teeth.

Base Circle . The circle from which is generated the involute curve upon which the

tooth profile is based.









Flank . The working surface of a gear tooth, located between the pitch diameter and

the bottom of the teeth

Gear. The larger of two meshed gears. If both gears are the same size, they are both

called "gears".

Land. The top surface of the tooth.









Ratio. Ratio of the numbers of teeth on mating gears.

Root Circle. The circle that passes through the bottom of the tooth spaces.

Root Diameter. The diameter of the Root Circle.

Working Depth. The depth to which a tooth extends into the space between teeth

on the mating gear.









Elasticity. It is the property of a material to regain its original shape after

deformation when the external forces are removed. This property is desirable for materials used in tools and machines. It may be noted that steel is more elastic than rubber.

Plasticity. It is property of a material which retains the deformation produced under

load permanently. This property of the material is necessary for forgings, in stamping









Involute. the curve formed by path of a point on a straight line, called the

generatrix, as it rolls along a convex base curve. (The base curve is usually a circle.) This curve is generally used as the profile of gear teeth.

Gear Project Design

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