# Lecture 14 â€“ BEVEL GEARS PROBLEMS

Document Sample

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

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, n 1 = 600 rpm, n 2 = 200 rpm
Shaft angle:∑ = γ 1 + γ 2 =60o , The semi pitch cone angles are shown in Fig.14.1.
d 1 = 200 mm, φ = 20o and b = 40mm.

n1 600
Solution: (a)       i         3
n2 200

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

d 2 = i d 1 = 3 x 200 = 600 mm
r 1 = 0.5 d 1 = 0.5x200 = 100 mm
r 2 = 0.5 d 2 = 0.5x600 = 300 mm
sin           sin60o
tan γ 2 =            =               =1.0392
1               1
+cos          +cos60o
i              3
γ 2 = 46.1o

γ 1 = ∑ - γ 2 = 60 – 46.1 = 13.9o
r 2av = r 2 – 0.5bsin γ 2 = 300 - 0.5x40x sin46.1 = 285.59 mm
r 1av = r 1 – 0.5bsin γ 1 = 100 - 0.5x40x sin13.9 = 95.2 mm

Solution: (b)
V 1 = πd 1av n 1 /60000 = π x (2x95.2)x600 /60000
= 5.98 m/s

1000W 1000x10
Ft                   1673N
V1av   5.979

Ft    1673
Fn                 1780N
cos  cos20o

F 2a = F n sin φsinγ 2 = 1780x sin20o sin46.1o = 439 N
F 2r = F n sin φcosγ 2 = 1780x sin20o cos46.1o = 422 N

Solution: (c)

Torque = F t x (0.5d 2av ) x 10-3
= 1673 x (0.5x 285.59) x 10-3
= 238.9 Nm

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

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

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:
n 1 = 960 rpm, W = 5 kW, Z 1 = 15, Z 2 = 45, m = 5 mm
d 1 = 75 mm, d 2 = 225 mm

Solution:
The pitch angles are

d            75 
 1  tan1  1   tan1        18.43
o

 d2          225 

d            225 
 2  tan1  2   tan1        71.57
o

 d1          75 
d 1av = d 1 - b sinγ 1 = 75 – 30sin18.43o = 65.52 mm

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

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

d1 n1  x 65.52x960
Vav =                        3.29 m / s
60000       60000

Transmitted tangential force:
1000W 1000 x 5
Ft =        =         =1519 N
v     3.29

(This acts in the +ve z direction as shown in Fig.3.)

F r = F t tan φ cosγ 2 =1519 tan 20ocos71.57o= 175 N
F a = F t tan φ cosγ 2 =1519tan 20o sin71.57o= 525 N
d 2av = d 2 - b cos γ 2 = 225-30sin71.57o=196.54mm.
r 2av = 0.5 d 2av = 0.5x196.54 = 98.27mm.

Where F r is acting - x direction and F a 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

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

Torque: T = F t x r 2av =1519 x 98.27x10-3= 149.27 Nm
As per the given problem the bearing at C takes the entire thrust load. Hence, F c y = F a
= 525 N.
Taking moment about horizontal axis through D,
-F c zx 150 + F t x 92.76 = 0,
i.e, -F c zx 150 +1519 x 92.76 = 0,           F c z = 959.3 N
∑ Fz = 0, from which F D z = 1519 – 959.3 = 559.7 N

Taking moment about vertical axis through D,
F c x x 150 – F r x 92.76 – F a x 98.27 = 0
i.e, F c x x150 – 175 x 92.76 – 525x98.27 = 0
F c x = 452.2 N

Taking moment about vertical axis through C,
F D x x 150 + F r x 90 - F a x 98.27 = 0
F D x x 150 + 175 x (90-32.76) - 525 x 98.27 = 0
F D x = 277.2 N

Fig.14.4 Calculated forces on bevel gear shaft

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

Torque: T 1 = F t x r 1av = 1519x32.76 x 10-3=49.76 Nm
As per the given problem the bearing at A takes the entire thrust load. Hence, F A x = F a
= 175 N.

Taking moment about horizontal axis through B,
-F A zx 75 + F t x(75+ 61.73) = 0,
i.e, -F A zx 75 +1519 x 136.73 = 0,          F A z = 2769 N
∑ Fz = 0, from which F B z = 1519 - 2769 = 1250 N

Taking moment about vertical axis through B,
F A y x 75 – F r x 136.73 + F a x 32.76 = 0
i.e, F A y x75 –525x 136.73 +175 x 32.76 = 0
F A y = 881 N

∑ Fy = 0, from which F B y = 881– 525 = 356 N

Fig.14.5 Forces acting on bevel pinion shaft

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

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

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, n 1 = 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
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
σ ut = 1240 MPa. The gear is made of ductile iron grade 120/90/02 of hardness 331 Bhn
and tensile strength σ ut = 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, σ ut = 1240 MPa,
Hardness=380 Bhn
σ sf ’ = 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

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

σ e ’ = 0.5σ ut =.0.5x1240 =620 MPa
k L = 1.0 for bending
k V = 1.0 for bending for m ≤ 5 module,
k s = 0.645 for σ ut = 1240 MPa from Fig.14.6
k r = 0.897 for 90% reliability from the Table 14.1
k T = 1.0 with Temp. < 120oC, k f = 1.0
k m = 1.33 for σ ut = 1240 MPa from the Fig.14.7
σ e = 620x1x1x0.645x1x1x0.897x1.33 = 477 MPa

Fig.14.6 Surface factor, k S

Table 14.1 Reliability factor k r

k f = fatigue stress concentration factor. Since this factor is included in J factor, its value
is taken as 1.

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

k m = 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 k m = 1.33 for σ ut less than 1.4 GPa.

Fig.14.7 Miscellaneous effects factor, k m

Corrected fatigue strength of the gear material:
σe = σe’ kL kv ks kr kT kf km
σ e ’ = 0.35σ ut =.0.35x974 =340.9 MPa
k L = 1.0 for bending
k V = 1.0 for bending for m ≤ 5 module,
k s = 0.673 for σ ut = 974 MPa from Fig.14.6
k r = 0.897 for 90% reliability from the Table 14.1
k T = 1.0 with Temp. < 120oC, k f = 1.0
k m = 1.33 for σ ut = 974 MPa from Fig. 14.7
σ e = 340.9x1x1x0.673x0.897x1x1x1.33 = 273.7MPa

Surface fatigue strength of pinion is:

σ sf = σ sf ’ K L K H K R K T

σ sf ’ = surface fatigue strength of the material = 2.8 (Bhn) – 69       From Table 14.2
= 2.8 x 380 -69 = 995 MPa

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

Table 14.2 Surface fatigue strength σ sf ’ (MPa) for metallic spur gears
(107 cycle life, 99% reliability and temperature <120oC)

K L = 0.9       for 108 cycles from Fig.14.8
K H = 1.005       for K = 380/331 = 1.14 & i=4 from Fig.14.9
K R = 1.0      for 99% reliability from Table 14.3
K T = 1.0      assuming temp. < 1200C

For the pinion material,
σ sf1 = σ sf ’ K L K H K R K T = 995 x 0.9 x 1 x1.005 x 1
= 900 MPa

Fig. 14.8 Life factor, K L

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

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

Table 14.3 Reliability factor K R

Reliability (%)           50           99           99.9K R

KR              1.25          1.0           0.80

K T = 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: σ sf ’ = 0.95[2.8(Bhn)-69]= 0.95[2.8x331-69] = 815 MPa
K L = 0.97       for 0 .39 x 108 cycles from Fig.14.8
K H = 1.005       for K = 380/331 = 1.14 & i=4 from Fig.14.9
K R = 1.0      for 99% reliability from Table 14.3
K T = 1.0      assuming temp. < 1200C
σ sf 2 = σ sf ’ K L K H K R K T = 815 x 0.97 x 1.005 x1 x 1 = 795 MPa

Permissible stresses in bending fatigue:
Pinion material: [σ 1b ] = σ e / s = 477 /1.5 =218 MPa
Gear material: [σ 2b ] = σ e / s = 273.7 /1.5 =182.5 MPa

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

Permissible stresses in contact fatigue:
Pinion material: [σ 1H ] = σ sf1 / 1.2= 900/1.2=750 MPa
Gear material: [σ 2H ] = σ sf 2 /1.2 = 795/1.2=663 MPa

Z 1 = 20 assumed for 20o pressure angle gears.
z 2 = i z 1 = 2.5 x 20 = 50. To have hunting tooth action, the value of z 2 is taken to be 51.
Hence i = z 2 / z 1 = 51 / 20 = 2.55
n 2 = 750 /2.55 = 294 rpm
tanγ 1 = z 2 / z 1 20 / 51= 0.3922, Hence γ 1 = 21.4o
γ 2 = ∑ - γ 1 = 90o - 21.4o = 69.6o

2n1 2x750
1                78.5rad / s
60    60

Torque:           1000W 1000x6
T1                 76.43 Nm
1   78.5
Bending stress in pinion is given by:

F               2T1
σb1  t K v K o Km          K v K o Km
bmJ             8m3 Z1J

Assuming b = 8m and putting F t = 2T 1 /d 1 where d 1 = m Z 1

Z1         20
Z v1                        21.5
cos  1   cos 21.4o

Z2       51
Z v2                      139.2
cos  2 cos 68.5o

J = 0.37 for Z v1 = 21.4 mating against Z v2 = 139.2 from Fig.14.10
K v = 1.25 assumed expecting the V to be 8 m/s from Fig.14.11
K o = 1.75 for induction motor and heavy shocks from Table 14.4.
K m = 1.25 from Table 14.5 for straddle mounted gears.

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

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, K v

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

Table 14.4 -Overload factor K o

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 K m

2T1                 2x(76.43x103 )
σb1 =           K v K o Km =                x1.25x1.75x1.25
8m3 Z1J              8m3 x20x0.37

7060.4
σb1 =
m3

7060.4
σb1 =           [ σb1 ]  218MPa
m3

m = 3.2 mm

Similarly for the gear: J =0.375 for Z v2 = 139.2 mating with Z v1 =21.5 from Fig. 14.10

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

2T2                 2x(2.55x76.43x103 )
σb2 =            K v Ko Km =                     x1.25x1.75x1.25
8m3 Z2 J               8m3 x51x0.375

6966.3
σb2 =            [σb2 ]  182.5
m3

m = 3.37 mm, Take a standard module of 4 mm
b = 8 m = 8 x 4 = 32 mm,
L = d 1 / sinγ 1 = 0.5x80/sin21.5o =109 mm

Bevel Gear Contact stress

b < L / 3 = 109/3 = 36.33mm. b= 32mm satisfies this requirement.
F t = T 1 / 0.5d 1 = 76.43 x 103 / 0.5x80 =1911 N
V 1 = ω 1 r 1 = 78.5 x (0.5x80) x 10-3= 3.14 m/s
Bevel gear contact stress is given by:
Ft
σH =Cp           K V Ko Km
b dI

C p = 0.93x 166 = 154.38 from Table 14.6.
C p 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

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

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

Ft                              1911
σH =Cp              K V K o K m =154.38                1.11x1.75x1.25
b dI                          32x80x0.107

[σ H1 ] = 750 MPa , [σ H2 ] = 663 MPa

σ H = 635 MPa < [σ H1 ] or [σ H2 ] , Hence the design is safe.

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