Bevel Gears.ppt (revised).pdf

23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Bevel Gears • Bevel gears are used to transmit power between two intersecting shafts, angle between the axis can be less than, equal to or more than 900. • The pitch surfaces of the bevel gears are rolling cones. The teeth of bevel gears are tapered, the outer larger part is called ‘heel’ & the inner part is called the ‘toe’. • The bevel gear has many diverse applications such as locomotives, marine applications, automobiles, printing presses, cooling towers, power plants, steel plants, railway track inspection machines, etc. Dr P R Venkatesh Mech Dept RVCE Bangalore

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Bevel gear lifts floodgate by means of central screw. 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Bevel gear on roller shutter door

23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Terminology Of Bevel Gears Pitch Cone: It is an imaginary cone, the surface of which contains the pitch lines of all teeth in the bevel gear. Cone distance: It is the length of the pitch cone element. It is denoted by ‘L’. Pitch angle: It is the angle which the pitch line makes with the axis of the gear. It is denoted by ‘d’. Back cone: It is an imaginary cone whose elements are perpendicular to the elements of the pitch cone. Back cone distance: It is the length of the back cone element. It is also called back cone radius and is denoted by ‘rb’.

23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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hf

Co ne dis ta

nce L

ha

Axis of Gear O d

Face width b 23-Feb-18 8:37 PM

B

d

rb s u i d a er n co k c Ba

Terminology of Bevel Gears

Dr P R Venkatesh Mech Dept RVCE Bangalore

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d d 

Two bevel gears in mesh showing Shaft angle & pitch angle

23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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 

Right Angled Bevel gear 

Obtuse angled bevel gear

Acute angled bevel gear

23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Formative number of teeth in Bevel Gears • Formative number of teeth or equivalent number of teeth in a bevel gear is equal to the number of teeth in an equivalent spur gear having pitch radius equal to back cone radius. Let z= number of teeth on bevel gear ze= Equivalent number of teeth rb=Back cone radius r=Pitch radius at the larger end m= module, d=Pitch cone angle 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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d

rb

r d

d

Virtual Or Equivalent number of teeth on Bevel gear 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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d 2r Number of teeth on pitch circle z= = m m 2rb Number of teeth on back cone radius z e = m ze 2rb m r From the above equations, =  = z m 2r cos d  r ze 1 r    =  From the geometry of bevel gear,rb =  z cos d  cos d  z  ze = Eqn 12.35 Page 218 cosδ Note : The equivalent number of teeth must be used in computation of Lewis form factor. 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Terms used in Bevel Gears l = Cone distance (Slant height of cone) b= Face width

d =Semi pitch cone angle or Pitch angle d=Pitch diameter gear/pinion=mz h a = Addendum height, h f = Dedendum height

 a &  d =Addendum & dedendum angle  c = cutting angle = (d   d )  f = Face angle = (d   d ) 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Pitch Cone angle ( d ) For Acute angle between shaft axes, sin  Eqn12.29(a) Page 215 i  cos  sin  tand 2 = Eqn12.29(b) Page 215 1    cos  i For right angle bevel gears, 1 tand1 = Eqn 12.32 (a) Page 217, tand 2 = i i For Obtuse angle between shaft axes, tand1 =

Eqn 12.32 (b) Page 217

sin(180   ) tand1 = Eqn 12.31 (a ) Page 217 i  cos(180   ) sin(180   ) tand 2 = Eqn 12.31 (b) Page 217 1    cos(180   ) Dr P R Venkatesh Mech Dept RVCE i Bangalore

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Design Procedure for Bevel Gears : Step 1 : To identify the weaker member Find  d 1 y1 for Pinion &  d 2 y2 for Gear where y1 & y2 are the Lewis form factors for pinion & gear calculated on the basis of equivalent number of teeth z e Step 2 : Beam strength or Load carrying capacity (a) Tangential tooth load 1000  P  Cs Ft = 12.7(a), Page 205, Cs from T 12.8, P 235 v (b) Lewis equation  Lb  Ft =  d Cv bYm   Eqn 12.37, Page 218  L  where  d = allowable stress Table 12.7, Page 234 Dr P R Venkatesh Mech Dept RVCE C from equations 12.38 a & 12.38 b, Page 219

23-Feb-18 v 8:37 PM

Bangalore

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Face width b = 10m Eqn 12.36a, Page 218 L Also b  Eqn 12.36b, Page 218. Take whichever is least. 3 1 Cone distance L= d12  d 2 2 Eqn 12.33, Page 217 2 m  2 2  z1  z2   As d = mz, L may be written as = 2   Equating the values of Ft from (a) & (b), module m can be obtained. For standard value of module, refer T12.24, Page 242 Dr P R Venkatesh Mech Dept RVCE Bangalore

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Step 3 : Dimensions of Gears : (1) Pitch diameter of pinion d1 = mz1 , of gear d 2 = mz2 (2) Face width of gear b=6m to 10 m

Eqn 12.36(a), page 218

(3) Cone distance L=3b 12.36 b, P 218 (4) Addendum h a , Dedendum h f from Table 12.23, page 242. (5) Outer dia of pinion d o1 = d1  2ha cos d1 Outer dia of gear d o 2 = d 2  2ha cos d 2

Eqn12.30a, P 215 Eqn12.30b, P 215

 2h sin d1  (6) Addendum angle  a = tan 1  a  Eqn12.29c, Page215 d1   2h f sin d1  1  (7) Dedendum angle  d = tan   Eqn12.29d , Page215 d 1   (8) Lateral or Radial load: Fr = Fte tan  cos d1 Eqn 12.45 d , Page 219, where Fte = Effective tooth load=

Ft L L  0.5b

12.45 ( a), page 219

(9) Axial load or thrust: Dr P R Venkatesh Mech Dept RVCE = F tan  sin d Eqn 12.45(Bangalore e), Page 219 23-Feb-18F8:37 PM a te 1

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Step 4: Check for dynamic & wear load : K 3v(Cb  Ft ) (a) Dynamic load Fd = Ft  K 3v  Cb  Ft

Eq12.40, Page 219

where C = Dynamic load factor from T12.12 Page 236 K 3 = 20.67  d1bQe K  (b) Wear load Fw =  Eq12.41 Page 219   cos d1  K = load stress factor from T12.16, Page 239 2 ze 2 Qe = Ratio factor = ze 2  ze1

Eq12.41 Page 219

If Fw  Fd , the design is safe. 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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PROBLEM 1 Type 1 (Where number of teeth are given/Assumed)

A cast steel (Untreated) pinion having 24 teeth transmits a power of 9 KW through a high grade CI (heat treated) gear at a velocity ratio 3. Pinion rotates at 720 rpm. The angle between the shaft axes is 600. The teeth are 141/20 system . Design the gears completely. 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Pitch Cone angle( δ) For Acute angle between shaft axes, sin  tand1 = Eqn12.67 Page 171 i  cos  0 sin 60 o Here  = 600 , i = 3  tand1 =  d = 13.9 1 3  cos 600 sin  tand 2 = Eqn12.68 Page 171 1    cos  i sin 600 tan d 2 =  d 2 = 46.10 1 0  cos 60   3 Here z1 = 24, z2 = iz1 = 3  24 = 72 teeth z1 24 Equivalent number of teeth on pinion z e1 = = = 24.73 cos d1 cos13.9 z2 72 = = 103.86 Dr P R Venkatesh Mech Dept RVCE cos d 2 cos 46.1

Equivalent number of teeth on gear z e 2 = 23-Feb-18 8:37 PM

Bangalore

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Step 1 : To identify the weaker member : From T 12.7/ P 234, for cast steel pinion (untreated),

 d 1 = 138.3Mpa For high grade Cast iron (grade 35) (heat treated),  d 2 = 78.5Mpa Lewis form factor for pinion & gear based on equivalent teeth 0.684 y1 = 0.124  ze1

0

1 Eqn12.5 (c), page 204 ( For 14 system) 2

0.684 0.684 y1 = 0.124  = 0.0963, Also y 2 = 0.124  = 0.1174 24.73 103.83  d 1 y1 (for Pinion)=78.5  0.0963=13.318 &  d 2 y2 (for Gear)=138.3  0.1174=9.22 (< d 1 y1 ) Gear is weaker. Hence design must be based on gear Dr P R Venkatesh Mech Dept RVCE 23-Feb-18 8:37 PM

Bangalore

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Step 2 : Beam strength or Load carrying capacity (a) Tangential tooth load 1000  P  Cs Ft = 12.7(a), Page 205 v  d 2 N 2   m  z2  N 2 Here, P=9 KW, v= = 60000 60000   m  72  240 v = = 0.905m 60000 C s = 1.5, (Assuming medium shocks, 8-10 hrs service/day) 1000  9 1.5 14917 Substituting the above, Ft = = 0.905m m 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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(b) Lewis equation :  Lb  Ft =  d Cv bYm   Eqn 12.37, Page 218  L  Here  d =  d 2 =78.5 Mpa, Y= y 2 =   0.1174 = 0.3688 1 m m 2 2 2 2 Cone distance L= d1  d 2 = z1  z2 = 242  722 2 2 2

L 37.94m L = 37.94m , Face width b   12.65m(> 10m) 3 3 Taking the least value b = 10m L  b 37.94m  10m = = 0.7364. L 37.94m Substituting ,the above values, Also

Ft = 78.5  Cv  10m  0.3688  m  0.7634 2 23-Feb-18 Ft = 8:37 221.15 m Cv PM

(2)Dr P R Venkatesh Mech Dept RVCE Bangalore

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Equating (1) & (2), 14917 = 221m 2Cv  m 3Cv = 67.49 (3) m As the value of Cv is around 0.5, initial trial value of module is m = 3 2  RHS = 3 2  67.49 = 5.13  5mm (standard value) Now, velocity v=0.905  m =0.905  5=4.525 m/sec 6.1 Velocity factor Cv = 6.1  v (Assuming generated teeth)

Eqn12.38b, Page219

6.1 Cv = = 0.574 6.1  4.525  m 3Cv = 53  0.574 = 71.76 > 67.49 (required value)  Selected module of 5Drmm is satisfactory. P R Venkatesh Mech Dept RVCE 23-Feb-18 8:37 PM

Bangalore

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Step 3 : Dimensions of Gears : (1) Pitch diameter of pinion d1 = mz1 = 5  24 = 120mm (2) Pitch diameter of gear d 2 = mz2 = 5  72 = 360mm (2) Face width of gear b=10  m = 10  5 = 50mm (3) Cone distance L=37.94m =37.94  5=189.7 mm 14917 (4) Tangential tooth load Ft = = 2983.4 N 5 (4) Addendum h a , Dedendum h f from Table 12.23, page 242 For 14.50 system, h a = 1m = 1 5 = 5mm, h f = 1.157 m = 1.157  5 = 5.785mm (5) Outer dia of pinion d o1 = d1  2ha cos d1

Eqn12.30a, P 218

d o1 = 120  (2  5  cos13.9) = 129.7 mm Outer dia of gear d o 2 = d 2  2ha cos d 2 d

= 360  (2  5 Drcos 46.1)Mech = 366.93 P R Venkatesh Dept RVCE mm

o 2PM 23-Feb-18 8:37

Bangalore

Eqn12.30b, P 218 24

Continued  2ha sin d1  (6) Addendum angle  a = tan   d 1    2  5  sin13.9  0  a = tan 1  = 1.15  120   1

Eqn12.29c, Page 218

 2h f sin d1  (7) Dedendum angle  d = tan   Eqn12.29d , Page 218 d1    2  5.785  sin13.9  0  d = tan 1  = 1.33  120   (8) Radial load on gear & end thrust (axial load) on pinion Fr = Fte tan  cos d1 Eqn 12.45d , Page 219, where 1

Ft L 2983.42 189.7 = = 3580 N L  0.5b 189.7  0.5  63.2  Fr = 3580  tan14.5  cos13.9 = 899 N Fte = Effective tooth load=

(9) Axial thrust on gear & radial load on pinion Fa = Fte tan  sin d1

Eqn12.45e, Page 219

P R Venkatesh Mech Dept F = 3580  tan14.5 Drsin13.9 = 222.4 NRVCE

a 23-Feb-18 8:37 PM

Bangalore

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Step 4: Check for dynamic & wear load : K 3v(Cb  Ft ) (a) Dynamic load Fd = Ft  K 3v  Cb  Ft

Eq12.40, Page 219

where C = Dynamic load factor from T12.12 Page 236 For v=4.525m/sec, From Table 12.14, page 237, tooth error e  0.0675 (Average value for v=4 & 5 m/sec) From T 12.12, Page 236, for 14.50 system CI & Steel combination, 454.8 Dynamic load factor C=  0.0675 = 511.65 N / mm 0.06 K 3 = 20.67 for SI system of units (always), Here,b=63.2 mm 20.67  4.525  (511.65  50  2983.4)  Fd = 2983.4  20.67  4.525  (511.65  50  2983.4)

Dynamic load Fd = 2983.4 +10176.6 = 13160N 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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 d1bQe K  (b) Wear load Fw =  Eq12.41 Page 219   cos d1  2 ze 2 Qe = Ratio factor = Eq12.41 Page 219 ze 2  ze1 2 103.83 Q = = 1.61, Also b=50 mm, d1 = 13.90 103.83  24.73  120  50 1.61 K  Fw =   = 9951.4K Newton cos13.9   For safe design, Fw  Fd i.e 9951.4K 13160  K 1.322 From T 12.16, Page 239, for 14.5 system & K=1.383(  1.17), 0

BHN for Pinion = 350, BHN for gear = 350 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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PROBLEM 2 Type 1 (Where number of teeth are given/Assumed)

Design a pair of bevel gears to transmit 12 KW at 300 rpm of the gear & 1470 rpm of the pinion. The angle between the shaft axes is 900. The pinion has 20 teeth and the material for gears is cast steel (d=183.33 N/mm2), BHN 320. Take service factor as 1.25 and check the gears for wear & dynamic load. suggest suitable surface hardness for the gear pair. 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Pitch Cone angle ( d ) For right angle between shaft axes, 1 tand1 = Eqn12.32a Page 217, i 1470 Here i= = 4.9  z2 = iz1 = 4.9  20 = 98 teeth 300 1  1  1  1  0  Pitch angle of pinion δ1 = tan   = tan  = 11.53  i 4. 9     Also tand 2 = i Eqn12.32b Page 217 i.e. Pitch angle of gear=δ 2 = tan 1 (4.9) = 78.46 0 Equivalent number of teeth on pinion z e1 Equivalent number of teeth on gear z e2 23-Feb-18 8:37 PM

z1 20 = = = 20.4 cos d1 cos11.53

z2 98 = = = 490 cos d 2 cos 78.46

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Step 1 : To identify the weaker member : As both the pinion & gear are made of same material, the pinion is the weaker member.

 d 1 =  d 2 = 183.33Mpa Lewis form factor for pinion & gear based on equivalent teeth 0.912 y1 = 0.154  ze1

Eqn12.5(d ), page 205 ( For 200 FDI system)

0.912 0.912 y1 = 0.154  = 0.1093 , Also y 2 = 0.154  = 0.152 20.4 490  d 1 y1 (for Pinion)=183.33  0.1093=20.03 &  d 2 y2 (for Gear)=183.33  0.152=27.89 As  d 1 y1   d 2 y2 , Pinion is weaker. Hence Drdesign must be based on pinion. P R Venkatesh Mech Dept RVCE 23-Feb-18 8:37 PM

Bangalore

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Step 2 : Beam strength or Load carrying capacity (a) Tangential tooth load 1000  P  Cs Ft = 12.7(a), Page 205 v  d1 N1   m  z1  N1 Here, P=12 KW, v= = 60000 60000   m  20 1470 v = = 1.5394m 60000 C s = 1.25, (given in the problem statement) 1000 12 1.25 9744 Substituting the above, Ft = = 1.5394m m 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

(1) 31

(b) Lewis equation :  Lb  Ft =  d Cv bYm   Eqn 12.37, Page 218  L  Here  d =  d 1 =183.33 Mpa, Y= y1 =   0.1094 = 0.3437 1 m m 2 2 2 2 Cone distance L= d1  d 2 = z1  z2 = 202  982 2 2 2

L 50m L = 50m , Face width b   16.67m 3 3 From eqn 12.3 6(d)page 173, b = 10m if L > 30m Take b = 10m L  b 50m  10m Also = = 0.8. L 50m Substituting ,the above values, Ft = 183.33  Cv 10m  0.3437  m  0.8 2  F = 504.08 m Cv 23-Feb-18 8:37t PM

(2)

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Equating (1) & (2), 9744 = 504.08m 2Cv  m 3Cv =19.33 (3) m As the value of Cv is around 0.5, initial trial value of module is m = 3 2  RHS = 3 2  19.33 = 3.38mm From Table 12.24, page 242 (OR Table 12.2, page 229), the preferred values are 3 mm & 4mm. Taking m = 3mm , velocity v=1.5394  m =1.5394  3 =4.62 m/sec Velocity factor Cv =

6.1 6.1  v

Eqn12.38(b), Page 219 (Assuming generated teeth)

6.1 = 0.57  m 3Cv = 33  0.57 =15. 36 < 19.33 (required value) 6.1  4.62  module of 3 mm is not satisfactory. For m=4mm, v=1.5394  4=6.16m/sec, Cv = 0.498 Cv =

 m 3Cv = 43  0.498 = 31.87 > 19.33 (required value) Hence Selected module of 4 mm is satisfactory. 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Step 3 : Specification of Gears : (1) Pitch diameter of pinion d1 = mz1 = 4  20 = 80mm (2) Pitch diameter of gear d 2 = mz2 = 4  98 = 392mm (2) Face width of gear b=10  m = 10  4 = 40mm (3) Cone distance L=50m =50  4=200 mm 9744 (4) Tangential tooth load Ft = = 2436N 4 (4) Addendum h a , Dedendum h f For 200 involute system, h a = 1m = 1 4 = 4 mm, h f = 1.25 m = 1.25  4 = 5mm (5) Outer dia of pinion d o1 = d1  2ha cos d1 d o1 = 80  (2  4  cos11.53) = 87.84mm Outer dia of gear d o 2 = d 2  2ha cos d 2 d

23-Feb-18 8:37 PM o 2

= 392  (2 Dr4PR Venkatesh cos 78.46) 393.6 mm Mech Dept = RVCE Bangalore

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Step 3 Continued  2h sin d1  (6) Addendum angle  a = tan 1  a  d 1   1  2  4  sin11.53  0  a = tan  = 1.145  80   2h f sin d1  1  (7) Dedendum angle  d = tan   d 1  

Eqn12.29(c), Page 215

Eqn12.29( d ), Page 215

 2  5  sin11.53  0 = 1.43  80   (8) Radial load on gear & end thrust (axial load) on pinion Fr = Fte tan  cos d1 Eqn12.45( d ), Page 219, where

 d = tan 1 

Ft L 2436  200 = = 2707 N L  0.5b 200  0.5  40  Fr = 2707  tan 200  cos11.530 = 965.3 N Fte = Effective tooth load=

(9) Axial thrust on gear & radial load on pinion Fa = Fte tan  sin d1 Eqn12.45(e), Page 219. 0 Mech Dept RVCE Dr P R Venkatesh F = 2707  tan 200  sin11.53 = 197 N Bangalore

23-Feb-18 8:37a PM

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Step 4: Check for dynamic & wear load : (a) Dynamic load Fd = Ft 

K 3v(Cb  Ft ) K 3v  Cb  Ft

Eq12.40, Page 219

where C = Dynamic load factor from T 12.12 Page 236 For v = 6.16 m / sec , From Table 12.14, page 237, tooth error e  0.059 (As 6.16m/sec  6m/sec) From T 12.12, Page 236, for 200 FDI system Steel & Steel combination, Dynamic load factor C= 686.7 N / mm (for e=0.06) K 3 = 20.67 for SI system of units (always), Here,b=40 mm  Fd = 2436 

20.67  6.16  (686.7  40  2436) 20.67  6.16  (686.7  40  2436)

Dynamic load Fd = 2436 +12681 = 15117N

23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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 d bQ K  (b) Wear load Fw =  1 e  Eq 12.41 Page 219  cos d1  2 ze 2 Qe = Ratio factor = Eq12.41 Page 219 ze 2  ze1

 es 2 sin  2  490 0 Q = = 1.92, Also b=40 mm, d1 = 11.53 . K = 490  20.4 1.4  es = 2.75( BHN )  70

1 1      E1 E2 

12.15b, page208

12.15 d , page 208, Given BHN=320,

Surface endurance limit  es = 2.75(320)  70 = 810Mpa, Elastic modulus E1 = E2 = 207Gpa = 207  103 Mpa (As both pinion & gear are of Steel)

810  Load stress factor K=

2

sin 200  1 1   = 1.55 3   1.4 10  207 207   80  40 1.92  1.55  Fw =   = 9719 Newton
For safe design, Fw  Fd i.e 6270.5K 15 117  K  2.4 From T 12.16, Page 239, for 200 system & K=3.226(  2.4), BHN for Pinion = 450, BHN for gear 450 Mech Dept RVCE Dr P R=Venkatesh 23-Feb-18 8:37 PM

Bangalore

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PROBLEM 3 Type 2 (Where pitch diameter/velocity are given)

A pair of straight tooth bevel gears at right angles is to transmit 5 KW at 1200 rpm of the pinion. The diameter of the pinion is 80 mm & the velocity ratio is 3.5. The teeth are 14.50 involute form. Both the pinion & gear are made of Cast iron with an allowable stress of 55 Mpa. Determine the module, face width from the standpoint of strength & also check the design for dynamic & wear load. 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Pitch Cone angle ( d ) For right angle between shaft axes, 1 Eqn12.32a Page 217, i Here i=3.5, d1 = 80mm  d 2 = i  d1 = 3.5  80 = 280mm tand1 =

 1  0  Pitch angle of pinion δ1 = tan 1  = 15.95   3.5  Also tand 2 = i Eqn12.32b Page 217, i.e. Pitch angle of gear=δ 2 = tan 1 (3.5) = 74.05 0 Equivalent number of teeth on pinion z e1 = Equivalent number of teeth on gear z e2 = d   z=  23-Feb-18  8:37 PM m 

z1 80 83.2 = = cos d1 m cos15.95 m

z2 280 1399.6 = = cos d 2 m cos 78.46 m

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Step 1 : To identify the weaker member : As both the pinion & gear are made of same material, the pinion is the weaker member. Hence design must be based on pinion. For 14.50 involute system, the Lewis form factor for pinion based on equivalent number of teeth is 0.684 y1 = 0.124  83.2 m y1 = (0.124  8.22 103 m)



23-Feb-18 8:37 PM



Dr P R Venkatesh Mech Dept RVCE Bangalore

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Step 2 : Beam strength or Load carrying capacity (a) Tangential tooth load 1000  P  Cs Ft = v Here, P=5 KW, v=

12.7( a), Page 205

 d1 N1   80 1200 =

60000 60000  Pitch line velocity v = 5.027 m / sec C s = 1.0, (Assuming steady loads, 8-10 hrs/day service) 1000  5 1 Substituting the above, Ft = = 995 N 5.027 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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(b) Lewis equation :  Lb  Ft =  d Cv bYm   Eqn 12.37, Page 218  L  Here  d =  d 1 =55 Mpa, Y= y1 =  y1 =  (0.124  8.22  103 m)  Y=(0.3896-0.026m ) 1 1 2 2 Cone distance L= d1  d 2 = 802  2802 = 145.6mm 2 2 L 145.6 Let face width b    48.3mm,say 48mm. 3 3  L  b   145.6  48   =  = 0.67  L   145.6  6.1 6.1 Velocity factor C v = = = 0.548 6.1  v 6.1  5.027 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Substituting ,the above values, Ft = 55  0.548  48  (0.3896-0.026m )  m  0.67  Ft = 377.64m  25.2m 2

(2)

Equating (1) & (2), 995 = 377.64m  25.2m 2 Solving, module m =3.41mm From Table 12.24, page 242 (OR Table 12.2, page 229), the preferred value is 4 mm. Take m = 4mm. 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Step 3 : Specification of Gears : (1) Pitch diameter of pinion d1 = mz1 80  4  z1  Number of teeth on pinion= 20 teeth (2) Pitch diameter of gear d 2 = mz2  280 = 4  z2  Number of teeth on gear=70 teeth (3) Addendum h a , Dedendum h f from Table 12.23, page 242 For 14.50 involute system, h a = 1m = 1 4 = 4 mm, h f = 1.157 m = 1.157  4 = 4.63mm (5) Outer dia of pinion d o1 = d1  2ha cos d1 d o1 = 80  (2  4  cos15.93) = 87.7 mm Outer dia of gear d o 2 = d 2  2ha cos d 2 d o 2 = 280  (2  4  cos 74.05) = 282.2mm 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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Step 3 Continued  2h sin d1  (6) Addendum angle  a = tan 1  a  d 1   1  2  4  sin15.95  0  a = tan  = 1.57  80   2h f sin d1  1  (7) Dedendum angle  d = tan   d 1  

Eqn 12.29 (c), Page 215

Eqn 12.29 ( d ), Page 215

 2  4.63  sin15.95  0 = 1.82  80   (8) Radial load on gear & end thrust (axial load) on pinion Fr = Fte tan  cos d1 Eqn 12.45( d ), Page 219, where

 d = tan 1 

Ft L 995 145.6 = = 1191.4 N L  0.5b 145.6  0.5  48  Fr = 1191.4  tan14.50  cos15.950 = 296.25 N Fte = Effective tooth load=

(9) Axial thrust on gear & radial load on pinion Fa = Fte tan  sin d1 Eqn 12.45(e), Page 219 P R Venkatesh0Mech Dept RVCE F = 1191.4  tan14.50 Drsin15.95 = 84.67 N Bangalore

23-Feb-18 8:37 a PM

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Step 4: Check for dynamic & wear load : K 3v(Cb  Ft ) (a) Dynamic load Fd = Ft  K 3v  Cb  Ft

Eq 12.40, Page 219

where C = Dynamic load factor from T12.12 Page 236 For v=5.027  5m/sec, From Table 12.14, page 235, tooth error e  0.064 From T 12.12, Page 236, for 14.50 system CI & CI combination, 331.3 Dynamic load factor C=  0.064 = 353.4 N / mm 0.06 K 3 = 20.67 for SI system of units (always), Here,b=48 mm 20.67  5.027  (353.4  48  995)  Fd = 995  20.67  5.027  (353.4  48  995)

Dynamic load Fd = 995 + 7843.1 = 8838.1N 23-Feb-18 8:37 PM

Dr P R Venkatesh Mech Dept RVCE Bangalore

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 d1bQe K  (b) Wear load Fw =  Eq12.41 Page 219   cos d1  2 ze 2 Qe = Ratio factor = Eq12.41 Page 219 ze 2  ze1 z1 z2 20 70 z e1 = = = 20.8, z e 2 = = = 254.73 0 0 cos d1 cos15.95 cos d 2 cos 74.05 2  254.73 = 1.85, Also b=48 mm, d1 = 15.950. 254.73  20.8  80  48 1.85  K  Fw =   = 7388 K Newton 0 cos15.95   Hence design will be unsatisfactory from stand point of wear. Hence find 'K'

Q =

For safe design, Fw  Fd i.e 7388 K  8838.1  K 1.2 From T12.16, Page 239, for 14.50 system & K=1.275(  1.2), BHN for Pinion = 400, BHN for gear = 300 Dr P R Venkatesh Mech Dept RVCE 23-Feb-18 8:37 PM

Bangalore

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