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Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Module 2- GEARS
Lecture 9 - SPUR GEAR DESIGN
Contents
9.1 Problem 1 Analysis
9.2 Problem 2 Spur gear
9.1 PROBLEM 1 – SPUR GEAR DESIGN
In a conveyor system a step-down gear drive is used. The input pinion is made of 18
teeth, 2.5 mm module, 20o full depth teeth of hardness 330Bhn and runs at 1720 rpm.
The driven gear is of hardness 280Bhn and runs with moderate shock at 860 rpm. Face
width of wheels is 35 mm. The gears are supported on less rigid mountings, less
accurate gears and contact across full face may be assumed. The ultimate tensile
strength of pinion and gear materials is 420 and 385MPa respectively. The gears are
made by hobbing process. Find the tooth bending strength of both wheels and the
maximum power that can be transmitted by the drive with a factor of safety 1.5. The
layout diagram is shown in the Fig 9.1.
Fig 9.1 Conveyor drive layout
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Solution:
The bending fatigue stress is found from AGMA equation as,
tv o mK K
FK
b m J (9.1)
We know that, Z2= Z1 x (N1/N2)
Substituting values from table 1,
Z2= 18 X (1720/860) = 36
Table 9.1 Data given for gear and pinion
N Z m d b
Pinion 1720rpm 18 2.5 mm 45 mm 35 mm
Gear 860 rpm 36 2.5 mm 90 mm 35 mm
Using the values from Table 9.1,
V = π dn/60000 = π x 45 x 1720/60000
= 4.051m/s
We know that 0.5
v
50 (200V)K
50
(9.2)
Table 9.2 J values for pinion and gear
Z J (sharing) Kv Ko Km
Pinion 18 0.338 1.569 1.25 1.6
Gear 36 0.385 1.569 1.25 1.6
The J value is obtained from Fig. 9.2 for sharing teeth as in practice. Ko and Km values
are obtained from Tables 9.3 and 9.4 for the given conditions.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Fig.9.2 - Geometric Factor J
SPUR GEAR –TOOTH BENDING STRESS (AGMA)
Table 9.3 - 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
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Table 9.4 - Load distribution factor Km
Face width ( mm)
Characteristics of Support 0 - 50 150 225 400 up
Accurate mountings, small bearing
clearances, minimum deflection, precision
gears
1.3
1.4
1.5
1.8
Less rigid mountings, less accurate gears,
contact across the full face
1.6
1.7
1.8
2.2
Accuracy and mounting such that less than
full-face contact exists
Over
2.2
Over
2.2
Over
2.2
Over
2.2
For pinion:
(9. 3)
tv o m
t
t
K K
F1.569x1.25x1.6
35x 2.5x0.338
= 0.1061 F
FK
b m J
x
And for Gear:
(9.4)
tv o m
t
t
σ = K K
F1.569x1.25x1.6
35= x
x 2.5x0.385
= 0.0932 F
FK
b m J
Fatigue strength of the material is given by,
σe = σe’ kL kv ks kr kT kf km (9.5)
Table 9.5 Properties of pinion and gear
Prop. σut MPa σ’e=0.5σut MPa kL Kv ks
Pinion 420 210 1 1 0.8
Gear 385 187.5 1 1 0.8
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
SPUR GEAR – PERMISSIBLE TOOTH BENDING STRESS (AGMA)
Endurance limit of the material is given by:
σe = σe’ kL kv ks kr kT kf km (9.6)
Where, σe’ is the endurance limit of rotating-beam specimen
From table 9.5,
kL = load factor = 1.0 for bending loads
kv = size factor = 1.0 for m < 5 mm and
= 0.85 for m > 5 mm
ks = surface factor, is taken from Fig.9.3 based on the ultimate tensile strength of the
material
for cut, shaved, and ground gears.
kr = reliability factor given in Table 9.5.
kT = temperature factor = 1 for T≤ 350oC
= 0.5 for 350 < T ≤ 500oC
Fig.9.3 Surface factor Ks
Reliability of 90%, working temperature <150o C and reversible is assumed.
kf = 1.0 since it is taken in J factor.
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
km = 1.0 for reverse bending assumed here
Table 9.6 K terms of pinion and gear
Prop. kr kT kf km
Pinion 0.897 1.0 1.0 1.0
Gear 0.897 1.0 1.0 1.0
Table 9.7 Reliability factor R
Reliability factor R 0.50 0.90 0.95 0.99 0.999 0.9999
Factor Kr 1.000 0.897 0.868 0.814 0.753 0.702
Permissible bending stress
(9.7) e[ ]
Hence the design equation from bending consideration is,
(n
σ ≤ [ σ ] (9.8)
Factor of safety required = 1.5
```````````````````````Table 9.8 Strength values of pinion and gear
Prop. σe MPa [σ]= σe / s MPa σ MPa FT N
Pinion 150.7 100.5 0.1061 Ft 947
Gear 134.6 89.7 0.0932 Ft 962
Table 9.8 shows that the pinion is weaker than gear. And maximum tangential force that
can be transmitted is: Ft= 947 N
So, the maximum power that can be transmitted is:
W = Ft v / 1000 = 947 x 4.051 /1000
= 3.84 kW
-----------------
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
9.2 PROBLEM 2 – SPUR GEAR DESIGN
In a conveyor system a step-down gear drive is used. The input pinion is made of 18
teeth, 2.5 mm module, 20o full depth teeth of hardness 340Bhn and runs at 1720rpm.
The driven gear is of hardness 280Bhn and runs with moderate shock at 860 rpm. Face
width of wheels is 35mm. The gears are supported on less rigid mountings, less
accurate gears and contact across full face may be assumed. The ultimate tensile
strength of pinion and gear materials is 420 and 385MPa respectively. The gears are
made by hobbing process. From surface durability consideration, find the maximum
power that can be transmitted by the drive with a factor of safety 1.2 for a life of 108
cycles. Drive layout is shown in the Fig 9.4.
Fig. 9.4 Conveyor drive Layout diagram
Data given:
i = n1/n2 = 1720/860 = 2
Z2= Z1 x i = 18 X 2 = 36
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
Table 9.9 Data given for pinion and gear
n Z m d = mZ b
Pinion 1720rpm 18 2.5 mm 45 mm 35 mm
Gear 860 rpm 36 2.5 mm 90 mm 35 mm
Table 9.10 Properties of gear and pinion
Bhn Ø Reliability Life Temp
Pinion 340 20o 99 % 108 <120oC
Gear 280 20o 99 % - <120oC
Solution:
The induced dynamic contact stress is given by equation below,
tH p V o m
1
FC K K K
bd I (9.9)
When both pinion and gear material are made up of steel, from Table 9.11,
Cp = 191 (9.10) MPa
SPUR GEAR – CONTACT STRESS
Table 9.11 Elastic coefficient Cp for spur gears in MPa
Gear material Pinion Material(μ=0.3 in all cases)
Steel Cast iron Al Bronze
Tin Bronze
Steel, E=207Gpa 191 166 162 158
Cast iron, E=131Gpa 166 149 149 145
Al Bronze, E=121Gpa 162 149 145 141
Tin Bronze, E=110Gpa 158 145 141 137
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
(9.11) sin cos iI
2 i 1
Substituting the values from table 10, o osin 20 cos20 2
I 02 2 1
.1071
SPUR GEAR – SURFACE DURABILITY
From table 3 and 4,
V = π dn/60000 = π x 45 x 1720/60000
= 4.051m/s
For hobbed gear,
(9.12) 0.5)
K v
50 (200V
50
Table 9.14 K Values of pinion and gear
Z Kv Ko Km
Pinion 18 1.569 1.25 1.6
Gear 36 1.569 1.25 1.6
Substituting values from Table 14, we have,
t
H p V o m1
t
t
FC K K K
bd I
F191 1.569x1.25x1.6
35x45x0.1071
26.051 F MPa
Surface fatigue strength of the material is given by,
σsf = σsf‘ KL Kr KT (9.13)
From table 10, for steel life is 107 cycles & reliability 99% and from Table 9.15,
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
σsf’ = 28(Bhn) – 69 = 2.8x340 – 69 = 954MPa
KL = 0.9 for 108 cycles from Fig.9.2
KR = 1.0. for 99% reliability from Table 9.10
SPUR GEAR – SURFACE FATIGUE STRENGTH
Table 9.15 Surafce fatigue strength σsf for metallic spur gears (107 cycle life 99%
reliability and temperature <120 0 C)
Material σsf’(MPa)
Steel 2.8 (Bhn)-69MPa
Nodular iron 0.95 (2.8(Bhn)-69MPa)
Cast iron, grade 20 379
Cast iron, grade 30 482
Cast iron, grade 40 551
Tin Bronze, AGMA 2C (11% Sn) 207
Aluminium Bronze (ASTM 148 – 52)
(Alloy 9C – H.T.)
448
Fig. 9.5 Life factor Kl
Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Madras
SPUR GEAR – ENDURANCE LIMIT
Table 9.16 Reliability factor KR
Reliability (%) KR
50 1.25
99 1.00
99.9 0.80
SPUR GEAR – ALLOWABLE SURFACE FATIGUE STRESS (AGMA)
We know that,
[ σH ] = σSf / fs = 954/1.2 = 795MPa
For factor of safety fs = 1.2
Design equation is, σH ≤ [ σH ]
26.051 √ Ft = 795 Ft = 931 N
Maximum Power that can be transmitted is,
W = Ft V/1000 = 931x4.051/1000 = 3.51kW
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