BICOL UNIVERSITYC O L L E G E O F E N G I N E E R I N G

Legazpi City

S U B M I T T E D B Y :

RUSSEL JAMES ALAMERBSME4B

S U B M I T T E D T O :

ENGR. MANUEL RUSTRIA,MMEProfessor

L E T T E R O F T R A N S M I T T A L

February 11 ,2010

Engr. Manuel RustriaProfessor in ChargeCollege of EngineeringLegazpi 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

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A C K N O W L E D G E M E N T

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.

T A B L E O F C O N T E N T S

TITLE PAGE………………………………………………………………………………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

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S T A T E M E N T O F T H E P R O B L E M

PROBLEM 676]

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

Hp Mg Max. ctr. Dist., in. Rpm, Pinion Kind of load Mon. Nsf

80 2 5 ¼ 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. 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 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

should show: that a cheap material cannot be used; that through hardened- steel (minimum permissible

tempering temperature is 800˚F), flame or 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 Ss = 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 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 cold finished

AISI 1137. Design the keys for cold-drawn AISI C1118. Use better materials than these only for good reason.

c) Determine the dimensions of the hub, arms or webs, and rims and beads of both gears.

d) Make a sketch of each gear (on separate sheets of paper) including on it all dimensions and information for

its manufacture.

e) At the instructor’s option (i) choose rolling type bearings, or (ii) design sleeve bearings.

f) Decide upon all details of the housing to enclose the gears, with sketches depicting them.

g) Your final report should be arranged (1) Title Page; (2) a summary of the final design decisions, computed

dimensions, and material specifications; (3) sketches; (4) final calculations; (5) other calculations.

1

COMPUTATIONS]

GIVEN: Power = P = 80 hpGear Ratio=Mg=2Center distance= Cd=5 ¼ Pinion Revolution = N = 720 rpmKind of Loading = Minor PulsationsNsf = 1.2Pressure Angle = Ф = 20 ˚FDDesign the gears based on endurance strength

SOLUTION:Mw = Mg = 2, since the pinion is the driver.

Solving for the diameters:Cd= (Dg+ Dp) ½

2(5.25)= Dg+ Dp

◘ Dg=(10.5)-Dp ……..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, Np = 17

Pd = Np/Dp = 17/3.5 = 4.85, use 5

with Pd = 5 and, For Ng:

Ng= Pd(Dg) = (7)(5) =35

with Pd = 5, Np = 17, solve for the new diameters,

Dp= Np/Pd

= 17/5 = 3.4

with Pd=5, Ng= 35,Dg= Ng/Pd

= 35/5 = 7

the new Gear ratio,since it can be adjusted, Mg’,Mw’=Mg’= Ng/Np=35/17 =2.0588

the new center distance Cd’,Cd’ = (Dg + Dp)/2 = (7 + 3.5)/2 = 5.25 in.

Solving for the Pitch line Speed, Vm,

Vm= πDpn = [π(3.4 in.)(720 rpm)]/(12 in/ ft.) = 640.8849013 fpm

For transmitted Load Ft:Ft = 80hp/Vm

= [(80hp)(33000 ft.-lb/min-hp)]/ 646.5397681 fpm2

= 4119.304409 lb.assume face 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 + Ft)1/2]….eq. 2

Solving for c, assume steel and steel, at AT25:c = 1660e

permissible error, e , from AF20e = .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) ]/[0.05(640.8849013) + (1.75x830 + 4119.304409)1/2] = 5792.810194 lb

taking limited load,Fw = DpbQKg

Solving for Q,

Q = (2 Mg) / (Mg + 1) = (2x2.055)/(2.055+1) = 1.346

since Fw = Fd= 5792.810194,

Kg = Fw / (DpbQ) = 5792.810194/(1.346x3.4x1.75) = 726.5533919

From AT26,With Kg = 726.5533919

…out of range!Choose Kg = 400, ∑BHN=600,

BHNp = ∑BHN -BHNg

Assuming BHNg = 269BHNp = 600-269 = 331

For Pinion, at BHN= 331Choose 4150 at 1200˚F, at table AT9, Design of Machine Elements by V.M. Faires

For gear, At BHN = 269Choose C1137, at table AT9, Design of Machine Elements by V.M. Faires

Solve for the corresponding tempering temperature,

˚F BHN700 277t˚F 2691000 229

t˚F =tempering tempaerature = 750˚F

3

solving for Sn’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 psifor pinion,

Sn’ = 250(331) psi = 82,750 psi

assuming Load near at middle, with 17 teeth,y = .512, from table AT24

therefore,Sn’y = (.512)(82,750) = 42,368 psi

With Sn’y of pinion < Sn’y of GearTherefore, Gear is stronger.

Use kf as 1.5,Fs = Sn’yp (b)/[(Kf)(Pd) = (1.75)(42,368)/(1.5 x 5) = 9885.8667 lb. ≤

Fs ≥ Fd,9,885.8667 ≥ 5,759.986, satisfied

FsNsf ≥ Fd, Nsf = 1.29,885.8667 ≥ (1.2)(5,759.986)9,885.8667 ≥ 6,911.98, satisfied

Computing Other dimensions for the Gear and pinion,Using AGMA Tooth System• Addendum a = 1/Pd = 1/5 = 0.2 in.• Dedendum d = 1.25/Pd = 1.25/5 = 0.25 in.• Working Depth = 2/Pd = 2/5 = 0.4 in.• Whole Depth = 2.25/Pd = 2.25/5 = 0.45 in.• Circular Tooth thickness = π/2Pd = π/ 2(5) = π/10 = 0.314 in.• Fillet radius = 0.3/ Pd = 0.3/5 = 0.06 in.• 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.

at given loading, minor pulsations,Ks = 1.0 – 1.5Use 1.25 (average)

At given Sd = 6ksi, with suggested material AISI 1137 (cold finished)Solve for Ds,

Ds = 3√[(6x7002.817 lb. in. x 1.25 in.)/(πx6000lb. in2)]

= 1.669in., use 1 11/16 in. ,from Design of Machine Elements, by Faires

For 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, Solve for the Hub Diameter. Dh,for pinion,

use 2 ½ in. since Dp = 3.43

Solve for Hub Length, Lhub,

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 Sy)

Sds = (0.6)(75ksi)/ 2.125 = 21.176 ksi

Sdc = (75ksi)/2.125 = 35.29ksi

since Sds/Sdc ≠ 0.5, use flat key,

At AT19, Machine Design by Faires,At Shaft Diameter = 1 11/16 in.,

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b = 3/8t = ¼base 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/ (Sds)(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.

…..

For shaft of gear:

With Mg’ = 2.0558= np/ ng

ng = 720rpm/2.0588 = 349.72 rpm

P = 2πTN

T = P/ 2πN = [80 hp x (33000 ft. lb x 12 in)/ (min-hp-ft)]/ (2π)( 349.72 rpm) = 14,417.32 lb.-in.

at given loading, minor pulsations,Ks = 1.0 – 1.5Use 1.25

At given Sd = 6ksi, with suggested material AISI 1137 (cold finished)Solve for Ds,

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, 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. Dh,

Dh = 1.8(Ds) = 1.8(1 15/16 in.) = 3.4875., use 3 ½ in.

Solve for Hub Length, Lhub,

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 Sy)

Sds = (0.6)(75ksi)/ 2.125 = 21.176 ksi

Sdc = (75ksi)/2.125 = 35.29ksi

since Sds/Sdc ≠ 0.5, use flat key,

At AT19, Machine Design by Faires,At Shaft Diameter = in.,

b = ½ t = 3/8 base tolerance: 0.0025 in.

Solving for length of key,

Using Sdc,L = 4T / (Sdc)(t)(Ds) = 4(14,417.32lb. in.)/[(35,291 psi)(1 15/16 in.)(3/8 in.)] = 2.25 in.

Using Sds,L = 2T/ (Sds)(b)(Ds) = (2 x 14417.32 lb. in.)/ [(1 15/16 in.)(1/2 in.)(21,176.47psi)] = 1.41 in

Assume the length of the key = length of hub,Lkey = Lhub = 3.15 in.

For the rim,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. 308with 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. 304with Bore Dia. = 20mm or .7874 in. Outside Dia. = 52mm or 2.04 in.

As widely used material for bearing, choose Copper-base alloysSince the selected bearing for the pinion and gear is not exactly suited to their shaft diameters, therefore, lathe machining is suggested, in order to get the exact fit..

Testing if cast iron can be utilized in the design…

by BUCKINGHAM’S EQ.

Fd = Ft + I

= Ft + [o.o5Vm(bc + Ft) ]/[0.05Vm + (bc + Ft)1/2] Solving for c, assume steel and steel, at table AT 25 p. 601 T.B.

c = 1660e (1000)Permissible error, e, from Figure AF 20

e = .0005(for precision cut)Therefore,

c = (.0005) (830) (1000) = 415

Fd = 4119.304+[0.05(640.88)(1.75 x415+4119.304)]/[0.05(640.88)+(1.75x415+ 4119.304)1/2] = 5646.75 lbf

Limiting Load for Wear (Fw) Fw = DpbQKg

Solving for Q,

Q = (2 Mg) / (Mg + 1) = (2x2.055)/(2.055+1) = 1.346

Taking Fw = Fd= 5646.75 lbf

Kg = Fw / (DpbQ) = 5646.75 / (3x1.75 x1.346) = 799.087

From table AT 26 in the 20 deg. column p.603With Kg = 799

Since it is out of range, choose the Choose Kg = 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 E S I G N S U M M A R Y

BEARING Rolling Bearing BRG. 304

Bore Diameter 20mm or .7874 in.Outside Diameter 52mm or 2.04 in.

HOUSINGMaterial SAE 1045

Height 10. 2 inchesLength 10 inchesDepth 14 inches

```For Detail dimensions, see at sketches

~GEARMaterial: AISI C1137 with tempering temperature 750˚FType of cut: Precision cutBHN: 269No. of Teeth 35Gear Diameter 7 in.

HUBMaterial AISI C1137 with tempering temperature 750˚FHub Diameter 3 ½ in.Hub length 3.15 in.

KEYMaterial AISI C1118 cold drawnthickness 3/8 in.base 1/2 in. (tolerance: -0.0025 in.)Length 3.15 in.

RIMMaterial AISI C1137 with tempering temperature 750˚FThickness: .3518in.

WEBMaterial AISI C1137 with tempering temperature 750˚FThickness .3455in.

ARM N.A.

SHAFTMaterial AISI 1137 cold finishedShaft Diameter 1 15/16 in (Tolerance: -0.0004 in. -0.0007 in.)

~PINIONMaterial: AISI 4150 with tempering temperature 1200˚FBHN: 331Type of Cut Precision cutNo. of Teeth 17Pinion Diameter 3.4 in.

HUBMaterial AISI 4150 with tempering temperature 1200FHub Diameter 2 ½ in.Hub length 2.74 in.

KEYMaterial AISI C1118Thickness 1/4 in.Base 3/8 in. (Tolerance: -0.004 in. -0.007 in.)Length 2.74 in

SHAFTMaterial AISI 1137 cold finishedShaft Diameter 1 11/16 in. (Tolerance: -0.003 in. -0.016 in.)

BEARING Rolling Bearing BRG. 308

Bore Diameter 40mm or 1.5748 in.Outside Diameter 90mm or 3.5433 in.

GEAR AND PINION DIMENSIONS

Center distance 5.2 in.

Velocity ratio 2.0558

Gear ratio 2.0558

Diametral pitch 5

Face width 1.75in

Addendum: 1/5 in

Dedendum 1.25/5 in.

Working depth 2/5 in.

whole depth 2.25/5 in.

Fillet radius 0.3/5

clearance 0.35/5

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 CODE, THE PSME CODE COMMITTEE, 1984

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A P P E N D I C E S

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.

Center Distance. The distance between centers of two gears.

Circular Pitch. Millimeter of Pitch Circle circumference per tooth.

Circular Thickness. The thickness of the tooth measured along an arc following the Pitch Circle

Clearance. The distance between the top of a tooth and the bottom of the space into which it fits on the meshing gear.

Contact Ratio. The ratio of the length of the Arc of Action to the Circular Pitch.

Dedendum. The radial distance between the bottom of the tooth to pitch circle.

Diametral Pitch. Teeth per mm of diameter.

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

Face Width. The width of the tooth measured parallel to the gear axis.

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.

10

Line of Action. That line along which the point of contact between gear teeth travels, between the first point of contact

and the last.

Module. Millimeter of Pitch Diameter to Teeth.

Pinion. The smaller of two meshed gears.

Pitch Circle. The circle, the radius of which is equal to the distance from the center of the gear to the pitch point.

Diametral pitch. Teeth per millimeter of pitch diameter.

Pitch Point. The point of tangency of the pitch circles of two meshing gears, where the Line of Centers crosses the

pitch circles.

Pressure Angle. Angle between the Line of Action and a line perpendicular to the Line of Centers.

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.

Kilogram. The kilogram is defined as the mass of international prototype (standard block of platinumiridium

alloy) of the kilogram, kept at the International Bureau of Weights and Measures at Sevres near Paris.

Second. The second is defined as the duration of 9 192 631 770 periods of the radiation corresponding

to the transition between the two hyperfine levels of the ground state of the caesium – 133 atom.

Strength. It is the ability of a material to resist the externally applied forces without breaking

or yielding. The internal resistance offered by a part to an externally applied force is called *stress.

Stiffness. It is the ability of a material to resist deformation under stress. The modulus of

elasticity is the measure of stiffness.

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 images on coins and

in ornamental work.

Ductility. It is the property of a material enabling it to be drawn into wire with the application

of a tensile force. A ductile material must be both strong and plastic. The ductility is usually

measured by the terms, percentage elongation and percentage reduction in area. The ductile material

commonly used in engineering practice (in order of diminishing ductility) are mild steel, copper,

aluminium, nickel, zinc, tin and lead.

cast iron. is obtained by re-melting pig iron with coke and limestone in a furnace known as cupola.

It is primarily an alloy of iron and carbon. The carbon contents in cast iron varies from 1.7 per cent to 4.5 per cent. It

also contains small amounts of silicon, manganese, phosphorous and sulphur. The carbon in a

cast iron is present in either of the following two forms: 1. Free carbon or graphite, and 2. Combined carbon or

cementite.

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.

11