Force Analysis Spur Gears

Wt = 33000 hp / V

Power, hp

hp= T n / 63000

in-lb rpm

Pitch line velocity, ft/min.

V = d n / 12

Applied Torque, in-lb.

T = d2 Wt

Transmitted load, lb.

Wt = F t

32

The equation in SI units

The tangential force Wt, this is not the entire force that acts between the gear and teeth. Tooth inaccuracies and deflections, misalignments,and the like produce dynamic effects that also act on the teeth. The dynamic load Fd or total gear-tooth load, in U.S. customary units, is estimated using one of the following formulas:

DYNAMIC EFFECTSDYNAMIC EFFECTS

V is the pitch line velocity in fpm. To convert to m/s divide the given values in these equations by 196.8.

Force Analysis Helical Gears

n = normal pressure angle

= helix angle

t = tangential pressure angle

tann = tan t cos

Wr = W sinnWt = W cosn cos W = W cos sin

t = pressure angle (20o or 25o) = helix angle (10, 20, 30, or 40o)

Wa = W cosn sin Where W= total force

Wr = radial component

Wt = tangential component (transmitted load)

Wa = axial component (thrust load)

Wr = Wt tant

Wa = Wt tan

Force Analysis Bevel Gears

= Pressure angle (20o)

Force Analysis Worm Gear Sets

Three orthogonal components of W without considering friction

considering friction

Relations between forces acting on the worm and the gear

Force Analysis Worm Gear Sets

Two useful relations, friction force and transmitted loads.

Efficiency of worm gear setsEfficiency of worm gear sets

(pressure angle)= 20o, 25o, 30o (max. pressure angle)= 25o, 35o, 45o

f is dependent on the relative or sliding velocity

VG and VW are the pitch-line velocity of the gear and the worm rsp and Vs the sliding velocity

Stress Analysis Spur Gears

Bending Strength

Lewis equation (1892 Wilfred Lewis, Philadelphia Engineers Club)

Bending stress is maximum at the root of the tooth,

Substituting for x and introducing p (circular pitch),

Using similar triangles,

F

W t

Lewis equation, where

Substituting P = / p and Y = y

Substituting for x and introducing p (circular pitch),

The form factor y is called Lewis form factor.

Modification of Lewis Equation

Assumptions made in deriving Lewis equation

1. The load is applied to the tip of a single tooth.

2. The radial component of the load, Wr , is negligible.

3. The load is distributed uniformly across the full face width.

4. Stress concentration in the tooth fillet is negligible.

Lewis equation, where

Let bending stress be designated by the allowable static bending stress 0 and so tangential load Ft, by the allowable bending load Fb. Then, this equation becomes

For satisfactory gear peiformance, it is necessary that:

Modification of Lewis EquationModifications according to AGMA standards(American Gear Manufacturers Association)

W t tangential transmitted loadKo overload factor

Lewis equation

Ko overload factorKV dynamic factorKS size factorPd transverse diameteral pitchF face width of the narrower memberKm load-distribution factorKB rim-thickness factorJ geometry factor for bending strength

which includes root fillet stress concentration factor Kf

Bending Stress Modifying FactorsGeometry factor J

For 250

The size factor Ks = 1 for helical gears

Bending Stress Modifying FactorsDynamic factor Kv

Application factor,(Overload factor) Ka , (Ko )

Bending Stress Modifying Factors

AGMA has not established standards for size factor and recommends that Ks be set to 1.

Size factor Ks

Load Distribution factor Km

Rim thickness factor KB

Backup ratio

KB = -2mB + 3.4 0.5 mB 1.2KB = 1.0 mB > 1.2

Km = 1.6 may be used as a conservative value for face width less than 2 in.

AGMA Bending Strength Equation

Sfb is the allowable fatigue bending stress, psi

KR is the reliability factor

KT is the temperature factor

KL is life factor

Allowable Stress

KR is the reliability factor

Temperature factor KT

AGMA recommends using temperature factor of 1 for operating temperatures up to 250 oF.

Reliability factor KR

KT = (460 + T)/620.

for higher temp.

The The life Factor Klife Factor KLL rectifies the allowable stress for the rectifies the allowable stress for the required number of stress cycles required number of stress cycles other than 10other than 1077

AGMA Bending Strength Equation

Stress cycle factor KL

AGMA Bending-Fatigue Strength, Sfb

THE WEAR STRENGTH OF A GEAR TOOTH: THE WEAR STRENGTH OF A GEAR TOOTH: THE BUCKNGHAM FORMULATHE BUCKNGHAM FORMULA

The maximum contact pressure Po between thetwo cylinders, for v = 0.3, may be computed from:

PPo o ~ S~ Se e

Multiplying both sides by b, and replacing the total load Fab denoted by Fw:

the allowable wear load Fw

For helical gearFor helical gear

For For satisfactory gear performance, the usual satisfactory gear performance, the usual requirement is that:requirement is that:

FFww > > FFdd

AGMA Surface Stress Equation

Cp elastic coefficient, (lb/in2)0.5

Wt transmitted tangential loadCa overload factor (same as Ka)Cv dynamic factor (same as Kv)Cs size factor (same as Ks)Cm load-distribution factor (same as Km)

OR

Cm load-distribution factor (same as Km)Cf surface condition factord pitch diameter of the pinionF face width of the narrowest memberI geometry factor

mN = 1 for spur gears

mG = speed ratio = NG /NP

Geometry factorcost sint

2mN

cost sint2mN

mG

mG

mG + 1

mG - 1

I =

external gears

internal gears

AGMA Surface Stress Equation

AGMA Elastic coefficient CP

Surface finish factor Cf

The surface condition factor Cf is used to account for such considerations as surfacesurfacefinish, residual stress, and plasticity effectsfinish, residual stress, and plasticity effects. The Cf =1 for a smooth surface finish. When rough finishes are present , 1.25 is reasonable. If both rough finish and residual stress exist, 1.5 is the suggested value.

AGMA Surface Strength Equation

Sfc is the allowable contact stress, psi

CR is the reliability factor (same as KR)

CT is the temperature factor (same as KT)

CL is the surface-life factor

CR is the reliability factor (same as KR)

CH is the hardness ratio factor

Hardness ratio factor, CHCH should only be used for the gear design, for pinion design set CH equal to 1.0.

AGMA Surface Strength Equation

Pitting resistance stress cycle factor ZN

AGMA Surface Strength EquationAGMA allowable surface fatigue strength

Design specification, the contact stress must not exceed the design stress value:

c,all c,all c c

AGMA Spur Gear Design Equations

Surface strength design equation, Stress = strength

Design steps

Select material, start with grade 1 with low hardness.

Calculate the transmitted load

Select material, start with grade 1 with low hardness.

Choose a face width, dP F dP

Select standard full depth gears with pressure angle of 20o or 25o.

Decide on: load type (uniform, non-uniform), mounting accuracy, # of cycles to failure, gear quality and reliability.

Solve the design equation for the diameter.Diameter and materialOutcome is

Or, select diameter and solve for material.

AGMA Spur Gear Design Equations

Bending strength design equation

Design steps

Solve the design equation to obtain the diametral pitch, P

Assume a value for J (geometry factor), 0.35 to 0.45

Solve the design equation to obtain the diametral pitch, Pd

Calculate the number of teeth for pinion and gear, determine the actual J and check against the assumed one, iterate if needed.

For power transmission, 2 < Pd (diametral pitch) < 16

Iterate until optimum design is achieved.

The number of teeth on the pinion should be over 18 to avoid interference.

Diametral pitch, POutcome is

Design ExampleDesign a pair of spur gears to transfer power from a 25 hp motor to a fan. The motor turns at 1000 rpm and the desired fan speed is 500 rpm. Specify the material, diameter, and number of teeth for both gears.

Selections and assumptions

1. Standard full depth gears with pressure angle, = 20o, will be used.

3. Select face width F = dP (pinion diameter) , dP/2 F dP

2. Assume good quality gear with, Qv= 10, will be used.

4. Assume uniform load K = 1

5. Assume operating temp. less than 250o (KT=CT = 1), new gear (Cf = 1), and a small size gear (Ks=Cs = 1).

4. Assume uniform load Ka= 1

KL = CL = 1

Design for

107 life cycle

KR = CR= 1Select 99% reliability

Design ExampleSurface failure

From assumptions and selections list;

Ca= 1 CR = 1 CL = 1 Cs = 1 Cf = 1 CH = 1 (pinion), , , , , ,

Select material: use A-3 steel (hardness HB = 300), surface strength range from 120,000 to 135,000 psi. Use Sc = 125,000 psi in calculation.

, t = 20omN = 1 for spur gears

mG = speed ratio = P / G = 1000/500 = 2

= 0.107cost sint

2mN

mGmG + 1

I =

Look up CP (elastic coefficient) = 2300