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 60 o. The pinion has an outside pitch diameter of 200 mm, a pressure angle of 20 o 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:∑ 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.
Solution: (a)
i
n1 n2
600 200
3
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
tan γ 2 =
1 +cos i o γ2 =46.1
sin60o
=
1 +cos60o 3
=1.0392
γ 1 = ∑ - γ 2 = 60 – 46.1 = 13.9 o 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 Ft
Fn
1000 W 1000x10 V1av
5.979
Ft
1673
cos
cos20o
1673N
1780N
F 2a = F n sin φsinγ 2 = 1780x sin20 o sin46.1o = 439 N F 2r = F n sin φcosγ 2 = 1780x sin20 o cos46.1 o = 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
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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
tan
1
1
tan
1
2
d1 d2
d2 d1
tan
tan
1
1
75 225
225 75
18.43o
71.57o
d 1av = d 1 - b sinγ 1 = 75 – 30sin18.43 o = 65.52 mm
The pitch-line velocity corresponding to the average pitch radius is
Vav =
d1 n1 60000
x 65.52x960
60000
3.29 m / s
Transmitted tangential force: Ft =
1000W 1000 x 5 = =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 20 ocos71.57 o= 175 N F a = F t tan φ cosγ 2 =1519tan 20o sin71.57 o= 525 N d 2av = d 2 - b cos γ 2 = 225-30sin71.57 o=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
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 = 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
c
y
= Fa
Torque: T 1 = Ft 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 = 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 FB 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 FB y = 881– 525 = 356 N
Fig.14.5 Forces acting on bevel pinion shaft
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A
x
= Fa
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 90 o. 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 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 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: σ e = σ e ’ k L k v k s k r k T k f k m
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. < 120 oC, 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.
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: σ e = σ e ’ k L k v k s k r k T k f k m 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. < 120 oC, 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 = 2.8 x 380 -69 = 995 MPa
From Table 14.2
Table 14.2 Surface fatigue strength
’
σ sf
(MPa) for metallic spur gears
(107 cycle life, 99% reliability and temperature <120 oC)
K L = 0.9 K H = 1.005
for 108 cycles from Fig.14.8 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. < 120 0C
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
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 (%) KR
50
99
99.9K R
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 10 8 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. < 120 0C
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
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 20 o pressure angle gears. z 2 = i z 1 = 2.5 x 20 = 50. To have hunting tooth action, the value of z 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 1
2 x750 78.5rad/ s 60
60
Torque:
1000 W 1000x6 78.5 1
T1
76.43 Nm
Bending stress in pinion is given by:
σb1
Ft bmJ
K v Ko Km
2T1 8m3 Z1J
K v K o Km
Assuming b = 8m and putting F t = 2T 1 /d 1 where d 1 = m Z1
Z v1
Z v2
Z1 cos Z2 cos
1
2
20 cos 21.4o 51 cos 68.5 o
21.5
139.2
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.
2
is taken to be 51.
Fig. 14.10 Number of teeth in gear for which geometry factor J is desired, pressure angle 20 o and shaft angle 90 o
Fig. 14.11 Dynamic load factor, K v
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 = Kv Ko Km = x1.25x1.75x1.25 8m3 Z1J 8m3 x20x0.37
σb1 =
σb1 =
7060.4 m3
7060.4 m3
[ σb1 ] 218MPa
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
σb2 =
σb2 =
2T2 2x(2.55x76.43x10 3 ) K K K = x1.25x1.75x1.25 v o m 8m3 Z2 J 8m3 x51x0.375
6966.3
[σb2 ]
m3
182.5
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/sin21.5 o =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 10 3 / 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: σH =Cp
Ft b dI
K V K o Km
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 MPa 0.5
Fig. 14.12 Geometry factor I for straight bevel gear pressure angle 20 angle 90o
o
and shaft
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
σH
=Cp
Ft b dI
K V K o K m =154.38
1911 32x80x0.107
1.11x1.75x1.25
[σ H1 ] = 750 MPa , [σ H2 ] = 663 MPa
σ H = 635 MPa < [σ H1 ] or [σ H2 ] , Hence the design is safe.
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