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1717. . Shaft DesignShaft Design

ObjectivesObjectives Compute forces acting on shafts from gears, pulleys, and Compute forces acting on shafts from gears, pulleys, and

sprockets.sprockets.

Find bending moments from gears, pulleys, or sprockets that Find bending moments from gears, pulleys, or sprockets that are transmitting loads to or from other devices.are transmitting loads to or from other devices.

Determine torque in shafts from gears, pulleys, sprockets, Determine torque in shafts from gears, pulleys, sprockets, clutches, and couplings.clutches, and couplings.

Compare combined stresses to suitable allowable stresses, Compare combined stresses to suitable allowable stresses, including any required stress reduction factors such as stress including any required stress reduction factors such as stress concentration factors and factors of safety.concentration factors and factors of safety.

Determine suitability of shaft design and/or necessary size of Determine suitability of shaft design and/or necessary size of shafting.shafting.

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IntroductionIntroduction

Shaft must have adequate Shaft must have adequate torsionaltorsional strength to strength to transmit torque and not be over stressed.transmit torque and not be over stressed.Shafts are mounted in bearings and transmit power Shafts are mounted in bearings and transmit power through devices such as gears, pulleys, cams and through devices such as gears, pulleys, cams and clutches.clutches.Components such as gears are mounted on shafts Components such as gears are mounted on shafts using keys.using keys.Shaft must sustain a combination of bending and Shaft must sustain a combination of bending and torsionaltorsional loads.loads.

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Standard diameters of shaftsStandard diameters of shafts

1/41/45 to 85 to 8

1/81/83 to 53 to 5

1/161/16UptoUpto 33

Diameter Diameter increments (in.)increments (in.)

Diameter (in.)Diameter (in.)

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Torsion of circular shaftsTorsion of circular shafts

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Torsion of circular shaftsTorsion of circular shafts

= the angle of twist (radians)= the angle of twist (radians)T = the applied torque (inT = the applied torque (in--lb.)lb.)L = shaft length (in.)L = shaft length (in.)J = polar moment on inertia of the shaft cross section J = polar moment on inertia of the shaft cross section (in(in44))G = shear modulus of elasticity of the shaft material G = shear modulus of elasticity of the shaft material (lb/in(lb/in22))

JGLT twist,of Angle =

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2

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TorsionalTorsional Shear StressesShear Stresses

TorsionalTorsional shear stress, Sshear stress, SSS ==

J = Polar moment of inertia = J = Polar moment of inertia = c = radius of the shaft c = radius of the shaft T = TorqueT = Torqued = diameter of shaftd = diameter of shaft

Torque

JcT

32d 4

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Shear Stress in a shaftShear Stress in a shaft

Shear stress, SShear stress, SSS ==WhereWhere

T = torqueT = torqueD = diameter of the shaft = D = diameter of the shaft =

Torque

3T16D

3

SST16

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Forces on spur gear teethForces on spur gear teeth

FFtt = Transmitted force= Transmitted forceFFnn = Normal force or separating = Normal force or separating forceforceFFrr = Resultant force = Resultant force = pressure angle= pressure angleFFnn = F= Ftt tan tan

cosFF tr =

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Forces on spur gear teethForces on spur gear teeth

Power, orPower, or

Torque, T = FTorque, T = Ftt r and r = r and r = DDpp /2 /2 Combining the above we can writeCombining the above we can write

63,000nTP =

nD63,000P2

DT2F

ppt

==

n P63,000T =

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Loads from Bevel gearsLoads from Bevel gears

An additional axial force will be acting on the shaft An additional axial force will be acting on the shaft because of the bevel anglebecause of the bevel angleFor the pinion it is relatively small, and can be For the pinion it is relatively small, and can be neglected.neglected.For the larger gear it will be significant and will be For the larger gear it will be significant and will be larger than the radial separating force.larger than the radial separating force.

Loads from Bevel gearsLoads from Bevel gears

Force transmitted, FForce transmitted, Fn n = F= Ftt tan tan coscos = Pressure angle= Pressure angle = Cone angle= Cone angleAxial Force, Axial Force, FFaa = F= Ftt tan tan sin sin Resultant Force, FResultant Force, Frr = = F = FF = Fnn or or FFaa depending on whichever is largerdepending on whichever is larger

22t FF +

3

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Loads from Worm gearsLoads from Worm gears

Axial

Driving

Separating

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Loads from Worm gearsLoads from Worm gears

Driving force on the worm gear, FDriving force on the worm gear, Ftt = = TToo = Output torque= Output torqueSeparating force, FSeparating force, Fss ==

wherewhere = lead angle= lead angle = normal pressure angle= normal pressure anglef = coefficient of frictionf = coefficient of friction

wg

o

rT

sinf-coscossinFt

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Loads from Worm gearsLoads from Worm gears

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Loads from Worm gearsLoads from Worm gears

Axial force on the worm gearAxial force on the worm gear

wherewhere = lead angle= lead angle = normal pressure angle= normal pressure anglef = coefficient of frictionf = coefficient of friction

+=

sinf-coscoscosf sincosFF t(gear)a(gear)

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Loads from Belts and ChainsLoads from Belts and Chains

For a belt, Total load, FFor a belt, Total load, Ftt = F= Fff + + FFbbNet driving force, Net driving force, FFdd = F= Fff FFbbDriving torque, TDriving torque, T = = FFdd rrr = effective radius of pulley or sprocketr = effective radius of pulley or sprocketFor a chain For a chain FFbb = 0= 0

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Bending of circular shaftsBending of circular shafts

Shafts transmit power through gears and Shafts transmit power through gears and pulleyspulleysThese produce bending load in addition to These produce bending load in addition to torsiontorsionUse strength of material approach to calculate Use strength of material approach to calculate the reaction forces and bending momentsthe reaction forces and bending moments

4

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Bending of circular shaftsBending of circular shafts

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Bending of circular shaftsBending of circular shafts

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Shaft Design ProblemsShaft Design Problems

Step 1:Step 1: Calculate the torque on the shaft from powerCalculate the torque on the shaft from powerStep 2:Step 2: Find the Find the torsionaltorsional stress in the shaftstress in the shaftStep 3:Step 3: Calculate the loads coming from gears, belts Calculate the loads coming from gears, belts or chainsor chainsStep 4:Step 4: Calculate the bending moment due to the Calculate the bending moment due to the acting forces. If necessary combine the forces.acting forces. If necessary combine the forces.Step 5:Step 5: Calculate the bending stress in the shaftCalculate the bending stress in the shaftStep 6:Step 6: Combine the bending stress and the Combine the bending stress and the torsionaltorsionalstress using the theories discussed in chapter 4stress using the theories discussed in chapter 4

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Shaft shown drives a gear set that is transmitting 5 hp at 1750 rpm.

Shaft is supported in self-aligning ball bearings and gears are both 10 pitch, 40 tooth, 20 spur gears.

Find torsional and bending stresses in shaft.

Example Problem 17-1: Design Stresses in Shafts

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Find the torsion in the shaft:

(2-6)

hp = Tn

63,000

then:

(17-1)

T = 63,000 hp

n

T = 63,000 (5)

1750

T = 180 in-lb

Example Problem 17-1: Design Stresses in Shafts (contd.)

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Find the torsional stress in the shaft.

First find Z':

(Appendix 3)

Z' = D3

16

Z' = (.75 in)3

16

Z' = .083 in3

(3-6)

Ss = TZ'

Ss = 180 in-lb.083 in3

Ss = 2170 lb/in2

Example Problem 17-1: Design Stresses in Shafts (contd.)

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Find the load at the gear pitch circle:

(11-4)

Dp = NTPd

Dp = 4010

Dp = 4 inches

(12-3)

Ft = 2 TDP

Ft = 2 (180 in-lb)

4 in

Ft = 90 lb

Example Problem 17-1: Design Stresses in Shafts (contd.)

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Find the resultant force on the shaft:

(12-2)

Fr = Ft

cos

Fr = 90 lb

cos 20

Fr = 96 lb

Find the maximum moment:

(Appendix 2)

Mm = FL4

Mm = 96 lb (15 in)

4

Mm = 360 in-lb

Example Problem 17-1: Design Stresses in Shafts (contd.)

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Find the stress:

S = MZ

(Appendix 3)

Z = D3

32

Z = (.75 in)3

32

Z = .041in3

S = MZ

S = 360 in-lb.041 in3

S = 8780 lb/in2

Example Problem 17-1: Design Stresses in Shafts (contd.)

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Combined Stresses in ShaftsCombined Stresses in Shafts

As seen in Chap 4As seen in Chap 4

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Combined maximum shear stressCombined maximum shear stress

= Maximum combined shear stress= Maximum combined shear stressS = normal stress S = normal stress SSSS = shear stress= shear stressThis can be rewritten asThis can be rewritten as

T = Torque in the shaftT = Torque in the shaftM = Maximum momentM = Maximum moment

1/222

S 2SS

+=

( )1/2223 MTD5.1 +=

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(4-5)

=

Ss2 +

S2

2

Substituting stresses from previous example problem:

=

(2170 lb/in2)

2 +

87802 lb/in

2 2

= 4900 lb/in2

This should be compared to shear stress allowables.

Example Problem 17-2: Combined Stresses in Shafts

From previous example problem, find the combined stress using the maximum shear stress theorem:

6

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Maximum Normal Stress TheoryMaximum Normal Stress Theory

= equivalent combined normal stress= equivalent combined normal stressS = normal stress from bending or axial loadsS = normal stress from bending or axial loadsSSSS = shear or = shear or torsionaltorsional stressstress

This can be written asThis can be written as

1/222

S 2SS

2S

+=

[ ]1/2223 )M(TMD5.1 ++=

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Example Problem 17-3: Combined Stresses in Shafts

From Example Problem 17-1, find the combined stress using the maximum normal stress theory:

2

21

2222

2

2122

/9300

28780)/2170(

2/8780

22

inlb

ininlbinlb

SSS s

=

++=

+=

Substituting stresses from Example Problem 17-1:

This should be compared to the normal stress allowable.

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Solid Circular shaftSolid Circular shaft

( )3 1/222 MT

5.1D +=

[ ]3 1/222 )M(TM

5.1D ++=

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Critical speeds of shaftsCritical speeds of shafts

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Critical speeds of shaftsCritical speeds of shaftsOperating speed should be 20% away from the critical speed.Operating speed should be 20% away from the critical speed.Vibration frequency, f is given byVibration frequency, f is given by

f = frequency in cycles per second, Hzf = frequency in cycles per second, Hzk = force constant, force per inch of deflectionk = force constant, force per inch of deflectiong = acceleration due to gravity, 386.4 in./sg = acceleration due to gravity, 386.4 in./s22

W = weight in pounds, lbW = weight in pounds, lb

Wgk

21f =

7

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Change the frequency to rpmChange the frequency to rpmCritical speed, Critical speed, NNcc = 60 = 60 ffAlso k is weight divided by deflectionAlso k is weight divided by deflection

Wk =

WgW

260Nc =

1187.7Nc =

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Shaft with n concentrated loadsShaft with n concentrated loads

RayleighRayleighs equation is used.s equation is used.

2nn

233

222

211

nn332211c W...WWW

W...WWW187.7N

++++

++++=

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First, find deflection:(Appendix 2)

= FL3

48 EI

(Appendix 3)

I = D4

64

I = (.75 in)4

64

I = .016 in4

= 96 lb (15 in)4

48 (30 x 106 lb/in2) (.016 in4)

= .21 inch

Find the estimated critical speed for the shaft in Example Problem 17-1 (assume the entire shaft diameter is inch).

Example Problem 17-5: Critical Speed

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Example Problem 17-5: Critical Speed (contd.)

This is approximate, and additional multiples would exist at 820, 1230, and 1640 rpm.

rpmN

N

N

c

c

c

41021.

188

188

=

=

=

(17-14)

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