Lecture 9 - Spur gear design CONTENTS
1. Problem 1 Analysis 2. Problem 2 Analysis 3. Problem 3 Spur gear design
PROBLEM 1 – SPUR GEAR DESIGN
In a conveyor system a step-down gear drive d rive is used. The input pinion is made of 18 teeth, 2.5 o mm module, 20 full depth teeth of hardness 330 Bhn and runs at 1720 rpm. The driven gear is of hardness 280 Bhn 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 are 420 and a nd 385 MPa respectively. The gears are made by hobbing ho bbing 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 figure. Conveyor drive Layout diagram
Solution: The bending fatigue stress is found from AGMA eq uation: σ
Ft K v K o Km bmJ
(1)
Z2= Z1 x (N1/N2) = 18 X (1720/860) = 36
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
V = π dn/60000 = π x 45 x 1720/60000 = 4.051m/s
Kv
50 (200V) 200V)0.5
(2)
50 Z
J (sharing) K v
K o
K m
Pinion 18 0.338
1.569 1.25 1.6
Gear
1.569 1.25 1.6
36 0.385
The J value is obtained from Graph 2 for sharing teeth as in practice. K o and K m values are obtained from Tables 3 and 4 respectively for the given conditions.
Graph 2 - Geometric Factor J:
SPUR GEAR –TOOTH BENDING STRESS (AGMA)
Table 3 -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 4. Load distribution factor K m 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
Over 2.2
Over 2.2
Over 2.2
Over 2.2
Accuracy and mounting such that less than full-face contact exists
For pinion:
Ft K v Ko Km bmJ
σ
Ft 35x2.5x0.338
x 1.569x1.25x1.6
(3)
= 0.1061 Ft For Gear: σ
Ft K v K o Km bmJ Ft = x 1.569 x 1.25 x1.6 35x2.5x0.385 = 0.0932 Ft =
Fatigue strength of the material is given by:
σe = σe’ k L k v k s k r k T k f k m Prop.
(5)
σut MPa σ’e=0.5σut MPa k L K v
k s
Pinion
420
210
1
1
0.8
Gear
385
187.5
1
1
0.8
(4)
SPUR GEAR – PERMISSIBLE TOOTH BENDING STRESS (AGMA)
Endurance limit of the material is given by:
σe = σe’ k L k v k s k r k T k f k m
(27)
Where, σe’ endurance limit of rotating-beam specimen k L = load factor , = 1.0 for bending loads k v = size factor, = 1.0 for m < 5 mm and = 0.85 for m > 5 mm k s = surface factor, is taken from Graph 4 based on the ultimate tensile strength of the material for cut, shaved, and ground gears. k r = reliability factor given in Table 5. o k T = temperature factor, = 1 for T≤ 350 C o = 0.5 for 350 < T ≤ 500 C
Graph 4 Surface factor K s
o
Reliability of 90%, working temperature <150 C and reversible is assumed. k f = 1.0 since it is taken in J factor. k m = 1.0 for reverse bending assumed here. Prop. k r k T k f k m Pinion 0.897 1.0 1.0 1.0 Gear
0.897 1.0
1.0 1.0
Fatigue strength of the material is:
σe = σe’ k L k v k s k r k T k f k m
(6)
Permissible bending stress
[σ ]
σe
(7)
n
Hence the design equation from bending consideration is :
σ ≤ [σ]
(8)
Factor of safety required is : s = 1.5 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
The table shows that the pinion is weaker than gear. Maximum tangential force that can be transmitted is: Ft= 947 N Maximum power that can be transmitted is : W = Ft v / 1000 = 947 x 4.051 /1000 = 3.84 kW
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 o mm module, 20 full depth teeth of hardness 340 Bhn and runs at 1720 rpm. The driven gear is of hardness 280 Bhn 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 are 420 and 385 MPa respectively. The gears are made by hobbing process. From surface durability consideration, find 8 the maximum power that can transmitted by the drive with a factor of safety 1.2 for a life of 10 cycles. Drive layout is shown in the figure. Conveyor drive Layout diagram
Given Data: i = n1/n2 = 1720/860 = 2 Z2= Z1 x i = 18 X 2 = 36 n
Z
m
d = mZ
b
Pinion 1720rpm 18 2.5 mm 45 mm
35 mm
Gear
35 mm
860 rpm
36 2.5 mm 90 mm
Bhn
Ø
Pinion
340
20
Gear
280
20
Reliability Life 8
Temp
o
99 %
10
<120 C
o
o
99 %
-
<120 C
o
Solution: The induced dynamic contact stress is given by equation (9)
σH
Cp
Ft b d1 I
K V Ko Km
(10)
For steel vs steel from Table 1 C p = 191
MPa
SPUR GEAR – CONTACT STRESS
I
sin φ cos φ i 2 i 1
(11)
I
sin 20o cos 20o 2 2 2 1
0.1071
SPUR GEAR – SURFACE DURABILITY
Table 2 -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 3. Load distribution factor K m Face width b ( 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
Over 2.2
Over 2.2
Over 2.2
Over 2.2
Accuracy and mounting such that less than full-face contact exists
V = π dn/60000 = π x 45 x 1720/60000 = 4.051m/s For hobbed gear
Kv
50
(200V)0.5 50
(12)
Z
K v
K o
K m
Pinion 18 1.569 1.25 1.6
Gear
σH
36 1.569 1.25
Cp 191
Ft b d1 I
1.6
K V Ko Km Ft
35x45x0.1071
1.569x1.25x1.6
26.051 Ft MPa SPUR GEAR – SURFACE DURABILITY
Surface fatigue strength of the material is
σsf = σsf ‘ K L K r K T
(13)
7
For steel 10 cycles life & 99% reliability from Table 4 σsf ’ = 28(Bhn) - 69 = 2.8x340 – 69 = 954 MPa 8 K L = 0.9 for 10 cycles from Fig.1 K R = 1.0. for 99% reliability from Table 5 SPUR GEAR – SURFACE FATIGUE STRENGTH
SPUR GEAR – ENDURANCE LIMIT
Table 5. Reliability factor K R Reliability (%) K R 50
1.25
99
1.00
99.9
0.80
SPUR GEAR – ALLOWABLE SURFACE FATIGUE STRESS (AGMA)
[ σH ] = σSf / f s = 954/1.2 = 795 MPa For factor of safety f s = 1.2 Design equation is : σH ≤ [ σH ] 26.051 √ Ft = 795 Ft = 931 N Maximum Power that can be transmitted W = Ft V/1000 = 931x4.051/1000 = 3.51 kW
Problem 3 - Design of Spur gear
A pair of gears is to be designed to transmit 30 kW power from a pinion running at 960 rpm to a 8 o gear running at 320 rpm. Design the gears so that they can last for 10 cycles. Assume 20 full depth involute spur gear for the system. Motor shaft diameter is 30 mm. 8
o
Data: W = 30 kW ; n1= 960 rpm; n2 = 320 rpm; Life = 10 cycles ; 20 full depth involute spur gear . Solution: i = n1 / n2 = 960 / 320 = 3 In order to keep the size small and meet the centre distance Z1 = 17 chosen Z2 = i Z1 = 3 x 17 = 51
Torque:
2 n1
2 x x 960
60
60
T1
w
30x1000 100.48
100.48rad / s
298.57Nm
From Lewis equation we have for pinion
σp
Ft
2T1
bYm
bYZ1m2
[ σ ]p
(1)
Steel C50 WQT with 223 Bhn hardness of and tensile strength of 660 MPa a nd steel C45 OQT with hardness 205 Bhn and tensile strength of 600 MPa are assumed for the pinion and the gear. For C50 form the data book is permissible static bending stress [σ] p = 542 MPa and for C45 [σ]g = 487 MPa and face width b= 10m is chosen for both wheels.
Problem 3 - Design of Spur gear- Buckingham approach
Equation (1) we have σ p
T1 5YZ1m3
[σ ]p
(2 )
For the pinion Y = 0.25808 for Z1= 17, from the Table 1 For the gear, Y = 0.39872, for Z2 = 51 from the Table 1; For gear, Y[σ]g = 0.39872x 487 = 194.17 For pinion, Y[σ] p = 0.25808 x 542 = 139.87
SPUR GEAR - TABLE 1 FOR MODIFIED LEWIS FORM FACTOR
For the same face width hence pinion will be weaker and considered for the design. σp
T 5YZ1m
298.57 x 1000 3
542MPa
5x0.25808 x 17m
13610 3
m3
(3 )
m = 2.93 mm, . Since motor shaft diameter is 30 mm, to get sufficiently large pinion m = 4 mm is taken Wheel
Z
m
b=10m
d
V =wrv
40 mm
68mm
3.42 m/s
C 50
223
51 4mm 40 mm 204mm 3.42 m/s
C 45
205
Pinion 17 4mm Gear
Material Hardness
We will now use Buckingham dynamic load approach for the design Ft = T1/r 1 = 298.57/0.034 = 8781 N Buckingham dynamic load:
Fi
9.84V(Cb +Ft ) 9.84V + 0.4696 Cb+Ft
(4)
For V=3.42 m/s permissible error is e= 0.088 mm from graph6. From table 7, if we choose I class commercial cut gears, expected error is 0.050 for m=4mm. In order to keep the dynamic load low precision cut gears are chosen. e = 0.0125 SPUR GEARS – PERMISSIBLE ERROR – GRAPH 6
SPUR GEARS – EXPECTED ERROR IN TOOTH PROFILE –TABLE 7
SPUR GEARS – VALUE OF C –TABLE 6
C = 11860e = 11860 x 0.0125 = 148.25 Substituting the values Ft = 8781 N , C = 148.25, V=3.42 m/s, b= 40mm in eqn (4) Buckingham dynamic load:
9.84x3.42(148.25x40 + 8781)
Fi
9.84x3.42 + 0.4696 148.25x40+ 8781
5464N
(5)
Fd = Ft + Fi = 8781 + 5464 = 14245 N Beam strength of the pinion Ftp = bYm [σ] p = 40x0.25808 x4x542 = 22381 N Since Ftp(22381)> Fd(14245) the design is safe from tooth bending failure consideration. Wear strength of the pinion is:
Ft s
b d1 I
C p = 191 MPa I
0.5
[ σH ]
2
Cp
(6)
from the table for steel vs steel
sin φ cos φ i 2 i 1
sin 20o c os 20o 3 0.1205 2 3 1
Substituting i =3, Ф=20 we get I= 0.1205 0
SPUR GEAR – BUCKINGHAM CONTACT STRESS EQUATION
Surface fatigue strength of the pinion material is σsf = σsf ’ K L K R K T σsf ’ = 2.8(Bhn) – 69 MPa = 2.8 x 223-69 = 555.4 MPa 8 K L= 0.9 for 10 cycles life from graph1 K R = 1.0 taken for 99 reliability o K T = 1.0 for operating temperature <120 C Assumed.
SPUR GEAR – SURFACE FATIGUE STRENGTH (AGMA)
Table 5 Reliability factor K R Reliability (%) K R 50
1.25
99
1.00
99.9
0.80
Surface fatigue strength of the pinion material is σsf = σsf ’ K L K R K T = 555.4x0.9x1x1 = 500 MPa Assuming a factor of safety s = 1.1 [σH] = σsf /s = 500/1.1 = 455 MPa Wear strength of the pinion is:
Ft s =b d1 I
[ σH ]
Cp
2
455 = 40x68x0.1205 191
2
=1860 N
Since Fts (1860) << Fd (14245) Hence the design is not safe. Revision is necessary. Increase the surface hardness of the pinion material to 475 Bhn and also increase the b to 13m = 13 x 4 = 52 mm.
Surface fatigue strength of the pinion material is σsf = σsf ’ K L K R K T σsf ’ = 2.8(Bhn) – 69 MPa = 2.8 x 475-69 = 1261 MPa 8 K L= 0.9 for 10 cycles life from graph1 K R = 1.0 taken for 99 reliability o K T = 1.0 for operating temperature <120 C Assumed. Surface fatigue strength of the pinion material is σsf = σsf ’ K L K R K T = 1261x0.9x1x1 = 1135 MPa Assuming a factor of safety s = 1.1 [σH] = σsf /s = 1135 /1.1 = 1032 MPa
Ft s =b d1 I
[σH ]
2
=52x68x0.1205
Cp
1032 191
2
=12439N
Since Fts (12439) < Fd (14245) still Not safe. Hence increase the module to 5mm. Wheel
Z
m
b=13m
d
V =wrv
65 mm
85mm
4.27 m/s
C 50
475
51 5mm 65 mm 255mm 4.27 m/s
C 45
450
Pinion 17 5mm Gear
Material Hardness
With new dimensions Fd = 16098 N Fts 19436 N. Since Fts > Fd , the revised design is safe from surface fatigue (pitting) considerations.
Problem 3 - Design of Spur gear-AGMA Approach
AGMA equation for tooth bending stress σ
Ft K v K o Km b m J
Ft = 8781 N, b= 10m, K v =1.15 assumed
Kv
78 (200V) 78
0.5
78 (200x3.42) 0.5 78
0.5
1.15
SPUR GEAR –TOOTH BENDING STRESS (AGMA)
K o = 1.25 is taken assuming uniform power source and moderate shock load from the table3 K m = 1.3 taken assuming accurate mounting and precision cut gears for face width of about 50mm. J = 0.34404 for pinion Z1= 17 mating with gear Z2=51 For gear J = 0.40808
Table 3 -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 4. Load distribution factor K m 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
Over 2.2
Over 2.2
Over 2.2
Over 2.2
Accuracy and mounting such that less than full-face contact exists
σe = σe’ k L k v k s k r k T k f k m The pinion is of steel C50 OQT with 223Bhn hardness and tensile strength of 660 MPa and the gear is of C45 OQT 30 with hardness 210 Bhn and tensile strength of 465 MPa. For pinion σe’ = 0.5 σut = 0.5 x 660 = 330 MPa k L = 1 for bending, k V = 1 assumed expecting m to be <5mm; k S = 0.73 from the graph 4 for σut= 660MPa, k f = 1.- and k m = 1.33 for σut= 660MPa
σe = σe’ k L k v k s k r k T k f k m = 330x1x1x0.73x1x1x1x1.33 = 320.4 MPa Factor of safety on bending of 1.5 assumed [σ] = σe / s = 320.4 / 1.5 =213.6 MPa
SPUR GEAR – PERMISSIBLE TOOTH BENDING STRESS (AGMA)
Graph 4 Surface factor k s
Substituting the value in the equation
2T1 K v Ko Km b Z1 m2 J
σ
9539
m
3
298540 x 1.15 x 1.25 x 1.3 10m x 17xm2 x 0.34404
213.6
m = 3.55 mm take m=4 mm. From this module the dimensions calculated are given in Table form Wheel
Z
m
b=10m
d
V =wrv
40 mm
68mm
3.42 m/s
C 50
223
51 4mm 40 mm 204mm 3.42 m/s
C 45
205
Pinion 17 4mm Gear
Material Hardness
Ft = T1 / 2 = 29854/34 = 8781N The tooth has to be checked from surface durability considerations now. The contact stress equation of AGMA is given below:
σH
Cp
Ft b d1 I
K V K o Km
[σ ]
0.5
C p = 191 MPa from the table for steel vs steel 0 Substituting i =3, Ф=20 we get I= 0.1205
I Kv
sin20o c os 20 o 3 0.1205 2 3 1
sin φ cos φ i 2 i 1 78
(200V)
0.5
78
(200x3.42)
78
0.5
0.5
1.15
78
SPUR GEAR – SURFACE DURABILITY
Table 2 -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 3. Load distribution factor K m Face width b ( 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
Over 2.2
Over 2.2
Over 2.2
Over 2.2
Accuracy and mounting such that less than full-face contact exists
K o = 1.25 and K m = 1.3 assumed as in the case of bending stress calculation
σH
Cp
Ft b d1 I
K V Ko Km
191
8781x1.15x1.25x1.3 40x68x0.1205
σH = 1209 MPa Surface fatigue strength of the pinion material is σsf = σsf ’ K L K R K T σsf ’ = 2.8(Bhn) – 69 MPa = 2.8 x 223-69 = 555.4 MPa 8 K L= 0.9 for 10 cycles life from graph1 K R = 1.0 taken for 99 reliability o K T = 1.0 for operating temperature <120 C Assumed. Surface fatigue strength of the pinion material is σsf = σsf ’ K L K R K T = 555.4x0.9x1x1 = 500 MPa Assuming a factor of safety s = 1.1 [σH] = σsf /s = 500/1.1 = 455 MPa
σH = 1209 MPa Since σH (1209) > [σH ] (455) The design is not safe and surface fatigue failure will occur. Solutions: Increase the surface hardness of the material to 475 Bhn and also increase the b to 13m = 13 x 4 = 52 mm
Surface fatigue strength of the pinion material is σsf = σsf ’ K L K R K T σsf ’ = 2.8(Bhn) – 69 MPa = 2.8 x 475-69 = 1261 MPa 8 K L= 0.9 for 10 cycles life from graph1 K R = 1.0 taken for 99 reliability o K T = 1.0 for operating temperature <120 C Assumed.
Surface fatigue strength of the pinion material is σsf = σsf ’ K L K R K T = 1261x0.9x1x1 = 1135 MPa Assuming a factor of safety s = 1.1 [σH] = σsf /s = 1135 /1.1 = 1032 MPa
σH
Cp
Ft b d1 I
K V K o Km
8781x1.15x1.25x1.3 52x68x0.1205
191
As σH (1185) >[σH ] (1032) the design is not safe from surface durability considerations. Hence increase the module to 5mm and take b=13m σH
Cp
Ft b d1 I
K V K o Km
191
8781x1.15x1.25x1.3 65x85x0.1205
σH =948 MPa < [σH ] (1032 MPa) . Hence the design is safe now from surface durability considerations.
Final specification of the pinion and gear are given in the table
Wheel
Z
m
Pinion 17 5mm Gear
b=13m
d
65 mm
85mm
51 5mm 65 mm 255mm
Wheel Material Steel Hardness Manufacturing quality Pinion
C50 OQT
475 Bhn
Precision cut
Gear
C45 OQT
450 Bhn
Precision cut
