Bevel Gear Problem

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram

Indian Institute of Technology Madras

Module 2- GEARS

Lecture 14 – BEVEL GEARS PROBLEMS Contents 14.1 Bevel gear analysis

14.2 Bevel gear analysis

14.3 Bevel gear design

14.1 BEVEL GEARS – PROBLEM 1

A pair of bevel gears is transmitting 10 kW from a pinion rotating at 600 rpm to gear

mounted on a shaft which intersects the pinion shaft at an angle of 60o. The pinion has

an outside pitch diameter of 200 mm, a pressure angle of 20o and a face width of 40

mm, and the gear shaft is rotating at 200 rpm. Determine ( a ) the pitch angles for the

gears, ( b ) the forces on the gear, and ( c ) the torque produced about the shaft axis

Fig.14.1 Intersecting shafts and semi pitch cone angles Data: W = 10 kW, n1 = 600 rpm, n2 = 200 rpm

Shaft angle:∑ = γ1+ γ2=60o , The semi pitch cone angles are shown in Fig.14.1.

d1 = 200 mm, φ = 20o and b = 40mm.

1

2

n 600i 3

n 200 Solution: (a)



d2 = i d1 = 3 x 200 = 600 mm

r1 = 0.5 d1 = 0.5x200 = 100 mm

r2 = 0.5 d2 = 0.5x600 = 300 mm

o

2o

o2

sin sin60tanγ = = =1.0392

1 1+cos +cos60

i 3

γ =46.1

γ1 = ∑ - γ2 = 60 – 46.1 = 13.9o

r2av = r2 – 0.5bsin γ2 = 300 - 0.5x40x sin46.1 = 285.59 mm

r1av = r1 – 0.5bsin γ1 = 100 - 0.5x40x sin13.9 = 95.2 mm

Solution: (b)

V1 = πd1avn1 /60000 = π x (2x95.2)x600 /60000

= 5.98 m/s

t

1av

1000W 1000x10F 1673N

V 5.979

F2a = Fnsin φsinγ2 = 1780x sin20o sin46.1o = 439 N

F2r = Fnsin φcosγ2 = 1780x sin20o cos46.1o = 422 N

Solution: (c) Torque = Ft x (0.5d2av ) x 10-3

= 1673 x (0.5x 285.59) x 10-3

= 238.9 Nm

----------------------

tn o

F 16731780N

cos cos20

F



14.2 BEVEL GEARS – PROBLEM 2 The bevel pinion shown in Fig.14.2 rotates at 960 rev/min in the clockwise direction,

viewing from the right side and transmits 5 kW to the gear. The mounting distances, the

location of all bearings, and the radii of the pitch circles of the pinion and gear are

shown in pitch cones in the figure. Bearings A and C should take the thrust loads. Find

the bearing forces on the gear shaft.

Fig.14.2 Bevel gear arrangement, (All dimensions are in mm)

Data:

n1 = 960 rpm, W = 5 kW, Z1 = 15, Z2 = 45, m = 5 mm

d1 = 75 mm, d2 = 225 mm

Solution:

The pitch angles are

od

d1 11

12

75tan tan 18.43

225

d1av = d1- b sinγ1 = 75 – 30sin18.43o = 65.52 mm

od

d1 12

21

225tan tan 71.57

75



The pitch-line velocity corresponding to the average pitch radius is

n x xm s1 1

av

d 65.52 960V = 3

60000 60000

.29 /

Transmitted tangential force:

t

1000W 1000 x 5F = = =1519 N

v 3.29

(This acts in the +ve z direction as shown in Fig.3.) Fr = Ft tan φ cosγ2 =1519 tan 20ocos71.57o= 175 N

Fa = Ft tan φ cosγ2 =1519tan 20o sin71.57o= 525 N

d2av = d2 - b cos γ2 = 225-30sin71.57o=196.54mm.

r2av = 0.5 d2av = 0.5x196.54 = 98.27mm.

Where Fr is acting - x direction and Fa is in the –y direction. All forces are acting at a

distance of 98.27 mm from the shaft centre line and 32.76 mm from the apex of the

pitch cones as in Fig.14.3.

Fig.14.3 Various forces acting on the bevel gear and the shaft reactions



Torque: T = Ft x r2av =1519 x 98.27x10-3= 149.27 Nm

As per the given problem the bearing at C takes the entire thrust load. Hence, Fcy = Fa

= 525 N.

Taking moment about horizontal axis through D,

-Fczx 150 + Ft x 92.76 = 0,

i.e, -Fczx 150 +1519 x 92.76 = 0, Fc

z = 959.3 N

∑ Fz = 0, from which FDz = 1519 – 959.3 = 559.7 N

Taking moment about vertical axis through D,

Fcx x 150 – Fr x 92.76 – Fa x 98.27 = 0

i.e, Fcx x150 – 175 x 92.76 – 525x98.27 = 0

Fcx = 452.2 N

Taking moment about vertical axis through C,

FDx x 150 + Frx 90 - Fa x 98.27 = 0

FDx x 150 + 175 x (90-32.76) - 525 x 98.27 = 0

FDx = 277.2 N

Fig.14.4 Calculated forces on bevel gear shaft



Torque: T1 = Ft x r1av = 1519x32.76 x 10-3=49.76 Nm

As per the given problem the bearing at A takes the entire thrust load. Hence, FAx = Fa

= 175 N.

Taking moment about horizontal axis through B,

-FAzx 75 + Ft x(75+ 61.73) = 0,

i.e, -FAzx 75 +1519 x 136.73 = 0, FA

z = 2769 N

∑ Fz = 0, from which FBz = 1519 - 2769 = 1250 N

Taking moment about vertical axis through B,

FAy x 75 – Fr x 136.73 + Fa x 32.76 = 0

i.e, FAy x75 –525x 136.73 +175 x 32.76 = 0

FAy = 881 N

∑ Fy = 0, from which FBy = 881– 525 = 356 N

Fig.14.5 Forces acting on bevel pinion shaft

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14.3 BEVEL GEARS – PROBLEM 3 A bevel gear pair has to be designed to transmit 6 kW power at 750 rpm. The shaft

angle is 90o. Speed ratio desired is about 2.5. The prime mover is induction motor and

the driven side is connected to a belt conveyor.

Data: W = 6 kW, n1= 750 rpm, i ≈ 2.5 and ∑ = 90o

Prime mover is electric motor. Out put is linked to a conveyor.

Solution:

To solve the problem the following assumptions are made.

1. The gears are to be mounted on anti-friction bearings in a gear box and are subjected

to extensive shock due to sudden loading of the belt conveyor.

2. The conveyor gearbox has to last for 20 years for which hardened gears are

selected.

3. The gears are of continuous duty and are straddle mounted on antifriction ball

bearings.

4. The pinion material is made of C45 steel of hardness 380 Bhn and tensile strength

σut= 1240 MPa. The gear is made of ductile iron grade 120/90/02 of hardness 331 Bhn

and tensile strength σut= 974 MPa. Both gears are hobbed, HT and OQ&T and ground.

5. A factor of safety of 1.5 and 1.2 on bending and contact fatigue strengths of the

materials was assumed.

Solution:

We will first determine the allowable stresses for the pinion and gear materials.

For pinion material, σut = 1240 MPa,

Hardness=380 Bhn

σsf’ = 2.8 (Bhn) – 69 = 2.8x380-69=995 MPa

Corrected bending fatigue strength of the pinion material:

σe = σe’ kL kv ks kr kT kf km



σe’ = 0.5σut =.0.5x1240 =620 MPa

kL = 1.0 for bending

kV = 1.0 for bending for m ≤ 5 module,

ks = 0.645 for σut = 1240 MPa from Fig.14.6

kr = 0.897 for 90% reliability from the Table 14.1

kT = 1.0 with Temp. < 120oC, kf = 1.0

km = 1.33 for σut = 1240 MPa from the Fig.14.7

σe = 620x1x1x0.645x1x1x0.897x1.33 = 477 MPa

Fig.14.6 Surface factor, kS

Table 14.1 Reliability factor kr

kf = fatigue stress concentration factor. Since this factor is included in J factor, its value

is taken as 1.



km = Factor for miscellaneous effects. For idler gears subjected to two way bending,

= 1. For other gears subjected to one way bending, the value is taken from the Fig.14.7.

Use km = 1.33 for σut less than 1.4 GPa.

Fig.14.7 Miscellaneous effects factor, km Corrected fatigue strength of the gear material:

σe = σe’ kL kv ks kr kT kf km

σe’ = 0.35σut =.0.35x974 =340.9 MPa

kL = 1.0 for bending

kV = 1.0 for bending for m ≤ 5 module,

ks = 0.673 for σut = 974 MPa from Fig.14.6

kr = 0.897 for 90% reliability from the Table 14.1

kT = 1.0 with Temp. < 120oC, kf = 1.0

km = 1.33 for σut = 974 MPa from Fig. 14.7

σe = 340.9x1x1x0.673x0.897x1x1x1.33 = 273.7MPa

Surface fatigue strength of pinion is:

σsf = σsf’ KL KH KR KT

σsf’ = surface fatigue strength of the material = 2.8 (Bhn) – 69 From Table 14.2

= 2.8 x 380 -69 = 995 MPa



Table 14.2 Surface fatigue strength σsf’ (MPa) for metallic spur gears

(107 cycle life, 99% reliability and temperature <120oC)

KL = 0.9 for 108 cycles from Fig.14.8

KH = 1.005 for K = 380/331 = 1.14 & i=4 from Fig.14.9

KR = 1.0 for 99% reliability from Table 14.3

KT = 1.0 assuming temp. < 1200C

For the pinion material,

σsf1 = σsf’ KL KH KR KT = 995 x 0.9 x 1 x1.005 x 1

= 900 MPa

Fig. 14.8 Life factor, KL



Fig. 14.9 Hardness ratio factor, KH K = Brinell hardness ratio of pinion and gear, KH = 1.0 for values of K below 1.2

Table 14.3 Reliability factor KR

Reliability (%) 50 99 99.9KR

KR 1.25 1.0 0.80

KT = temperature factor, = 1 for T≤ 120oC, based on Lubricant temperature. Above

120oC, it is less than 1 to be taken from AGMA standards.

For gear: σsf’ = 0.95[2.8(Bhn)-69]= 0.95[2.8x331-69] = 815 MPa

KL = 0.97 for 0 .39 x 108 cycles from Fig.14.8

KH = 1.005 for K = 380/331 = 1.14 & i=4 from Fig.14.9

KR = 1.0 for 99% reliability from Table 14.3

KT = 1.0 assuming temp. < 1200C

σsf 2= σsf’ KL KH KR KT = 815 x 0.97 x 1.005 x1 x 1 = 795 MPa

Permissible stresses in bending fatigue:

Pinion material: [σ1b] = σe/ s = 477 /1.5 =218 MPa

Gear material: [σ2b] = σe/ s = 273.7 /1.5 =182.5 MPa



Permissible stresses in contact fatigue:

Pinion material: [σ1H] = σsf1 / 1.2= 900/1.2=750 MPa

Gear material: [σ2H] = σsf 2 /1.2 = 795/1.2=663 MPa

Z1= 20 assumed for 20o pressure angle gears.

z2 = i z1 = 2.5 x 20 = 50. To have hunting tooth action, the value of z2 is taken to be 51.

Hence i = z2 / z1 = 51 / 20 = 2.55

n2 = 750 /2.55 = 294 rpm

tanγ1 = z2 / z1 20 / 51= 0.3922, Hence γ1 = 21.4o

γ2 = ∑ - γ1= 90o - 21.4o = 69.6o

1

1

2 n 2 x75078.5rad/ s

60 60

Torque:

11

1000W 1000x6T 7

78.5

6.43 Nm

Bending stress in pinion is given by:

t 1b1 v o m v o m3

1

F 2Tσ K K K K K K

bmJ 8m Z J

Assuming b = 8m and putting Ft = 2T1/d1 where d1 = m Z1

1v1 o

1

Z 20Z 21.5

cos cos21.4

2v2 o

2

Z 51Z 139.2

cos cos68.5

J = 0.37 for Zv1 = 21.4 mating against Zv2 = 139.2 from Fig.14.10

Kv = 1.25 assumed expecting the V to be 8 m/s from Fig.14.11

Ko = 1.75 for induction motor and heavy shocks from Table 14.4.

Km = 1.25 from Table 14.5 for straddle mounted gears.



Fig. 14.10 Number of teeth in gear for which geometry factor

J is desired, pressure angle 20o and shaft angle 90o

Fig. 14.11 Dynamic load factor, Kv



Table 14.4 -Overload factor Ko

Driven Machinery

Source of power

Uniform

Moderate Shock

Heavy Shock

Uniform

1.00

1.25

1.75

Light shock

1.25

1.50

2.00

Medium shock

1.50

1.75

2.25

Table 14.5 BEVEL GEARS – MOUNTING FACTOR Km

3

1v o mb1 3 3

1

2T 2x(76.43x10 )σ = K K K = x1.25x1.75x1.25

8m Z J 8m x20x0.37

b1 3

7060.4σ =

m

b1 b13] 218MPa

7060.4σ = [σ

m

m = 3.2 mm Similarly for the gear: J =0.375 for Zv2 = 139.2 mating with Zv1 =21.5 from Fig. 14.10



3

2v o mb2 3 3

2

2T 2x(2.55x76.43x10 )σ = K K K = x1.25x1.75x1.25

8m Z J 8m x51x0.375

3b2 b2

6966.3[ ] 182

mσ = σ .5

m = 3.37 mm, Take a standard module of 4 mm

b = 8 m = 8 x 4 = 32 mm,

L = d1 / sinγ1 = 0.5x80/sin21.5o =109 mm

Bevel Gear Contact stress

b < L / 3 = 109/3 = 36.33mm. b= 32mm satisfies this requirement.

Ft = T1 / 0.5d1= 76.43 x 103 / 0.5x80 =1911 N

V1= ω1r1= 78.5 x (0.5x80) x 10-3= 3.14 m/s

Bevel gear contact stress is given by:

t

H p V o

Fσ =C K K K

bdI m

Cp = 0.93x 166 = 154.38 from Table 14.6.

Cp values of 1.23 times the values given in the table are taken to account for a

somewhat more localized contact area than spur gears.

Table 14.6 Elastic Coefficient Cp for spur gears, in MPa0.5



Fig. 14.12 Geometry factor I for straight bevel gear pressure angle 20o and shaft angle 90o I = 0.107 from Fig.14.12. Other factors are same in bending fatigue stress equation.

Kv = 1.11 for V = 3.14 m/s from Fig. 14.11, for quality 10 gears

t

H p V o m

F 1911σ =C K K K =154.38 1.11x1.75x1.25

bdI 32x80x0.107

[σH1] = 750 MPa , [σH2] = 663 MPa

σH = 635 MPa < [σH1] or [σH2] , Hence the design is safe.

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Bevel Gear Problem

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