Chapter - 3 Design of bevel gear3.1 IntroductionThe bevel gears are used for transmitting power at a constant velocity ratio between two shafts whose axes intersect at a certain angle. The pitch cones of bevel gears are analogous to the pitch cylinders of spur and helical gears. The basic shape of a bevel gear tooth is almost the same as that of the spur gear. The tooth tapers off as it approaches the apex. The contour of the tooth also varies along its entire length. The angle between the shafts will depend on the conditions of drive. It is usually 9o, but can have other angles also.Classification of Bevel GearsThe bevel gears may be classified into the following types, depending upon the angles between the shafts and the pitch surfaces.Mitre gears. When equal bevel gears (having equal teeth and equal pitch angles) connect two shafts whose axes intersect at right angle, then they are known as mitre gears.Angular bevel gears. When the bevel gears connect two shafts whose axes intersect at an angle other than a right angle, then they are known as angular bevel gears.Crown bevel gears. When the bevel gears connect two shafts whose axes intersect at an angle greater than a right angle and one of the bevel gears has a pitch angle of 90º, then it is known as a crown gear.Internal bevel gears. When the teeth on the bevel gear are cut on the inside of the pitch cone, then they are known as internal bevel gears.

3.2 Nomenclature of bevel Gear 1. Pitch cone. It is a cone containing the pitch elements of the teeth.2. Cone centre. It is the apex of the pitch cone. It may be defined as that point where the axes of two mating gears intersect each other.3. Pitch angle. It is the angle made by the pitch line with the axis of the shaft. It is denoted by ‘θP’ or δ

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7. Face angle. It is the angle subtended by the face of the tooth at the cone centre. It is denoted by ‘φ’. The face angle is equal to the pitch angle plus addendum angle.8. Root angle. It is the angle subtended by the root of the tooth at the cone centre. It is denoted by ‘θR’. It is equal to the pitch angle minus dedendum angle.9. Back (or normal) cone. It is an imaginary cone, perpendicular to the pitch cone at the end of the tooth.10. Back cone distance. It is the length of the back cone. It is denoted by ‘RB’. It is also called back cone radius.11. Backing. It is the distance of the pitch point (P) from the back of the boss, parallel to the pitch point of the gear. It is denoted by ‘B’.12. Crown height. It is the distance of the crown point (C) from the cone centre (O), parallel to the axis of the gear. It is denoted by ‘HC’.13. Mounting height. It is the distance of the back of the boss from the cone centre. It is denoted by ‘HM’.14. Pitch diameter. It is the diameter of the largest pitch circle.15. Outside or addendum cone diameter. It is the maximum diameter of the teeth of the gear. It is equal to the diameter of the blank from which the gear can be cut. Mathematically, outside diameter, DO = DP + 2 a cos θP where DP = Pitch circle diameter, a = Addendum, and θP = Pitch angle.16. Inside or dedendum cone diameter. The inside or the dedendum cone diameter is given by Dd = DP – 2d cos θP Where Dd = Inside diameter, and d = Dedendum.3.3 Material Selection for the spur gear`

3.4 Geometry of the bevel gearLet, θp1 = pitch angle for the pinion, θs = Angle b/n the two shaft axis θp2 = pitch angle for the gear Dp = pitch diameter of the pinionDp = pitch diameter of the geartan θp1 = , θp1 = tan -1 () = tan -1(18/51) = 19.40tan θp2 = , θp2 = tan -1 () = tan -1(51/18) = 70.60θs = 900 To determine the module, we can use the following formula;- M = Where , T1 = Pinion torque Z1 = Number of teeth of the pinion

= pitch angle of the pinion δbp = The allowable stress

δbp = , δe = the endurance limits of the material

From the selected material δe = 370Mpa and δbp = 180N/mm2T1 = 52.52 from the previous calculation

Z1 = 18, δ1 = 19.40

M = = 2.17 implies m= 2.5

M= 2.5

S.no Description For Pinion For Gear

1 Number of teeth Z1 = 18 Z2 = 51

       

2 Pitch Circle Diameter d1 = m1*Z1 d2 = m1*Z2

    d1 = 2.5*18 = 45mm d2 = 2.5*51 = 127.5mm

3 Transmission Ratio i = n1/n2 = Z2/Z1 = 52/18 = 2.833

4 Pitch Cone angle θp1= tan -1( Z1/Z2) θp2= tan -1( Z2/Z1)

  ( Shaft angle is equal to 900) θp1= tan -1( 18/51) θp2= tan -1( 51/18)

    θp1= 19.40 θp2= 70.6.90

5 Shaft angle (£) θs = θp1 + θp2

    θs = 17.10 + 72.90 = 900

6 Tip Circle Diameter da1 = d1 + 2*m cos θp1 da2 = d2 + 2*m cos θp2

    da1 = 45 + 2*2.5 cos 19.40 da2 = 127.5 + 2*2.5cos70.60

    da1 = 49.7mm da2 = 129.16mm

7 Cone Distance R = d1/ 2*sin θp1 R = d2/ 2*sin θp2

    R = 45/ 2*sin 19.4 R = 260/ 2*sin70.6

    R = 68mm R = 68mm

8 Face Width bmax < R/3

    b = R/3 = 68/3 = 23 mm

9 Middle Circle Diameter dm1 = d1 - b sin θp1 dm2 = d2 - b sin θp2

    dm1 = 45 - 23 sin 19.4 dm2 = 127.5- 23sin 70.6

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    dm1 = 37.36m dm2 = 105.8mm

    dm1≈ 38mm dm2 ≈ 106mm

s.no Description For Pinion For Gear

10 Virtual number of teeth ( ZV1) ZV1 = Z1 / COS θp1 ZV2 = Z2 / COS ð2

    ZV1 = 18 / COS 19.4 ZV2 = 51 / COS 70.6

    ZV1 = 21.2 ZV2 = 83.78

11 Middle Module mm = dm1/Z1 = dm2/Z2

    mm = 37.36/18 = 105.8/51 = 2.07

12 Top Clearance ( C ) C = 0.2 * M = 0.2 * 2.5= 0.5mm

13 Whole Depth ( h ) h = 2.2 * M = 2.2 * 2.5 = 5.5 mm

14 Addendum ha1 = ha2 = m = 2.5mm

15 Dedendum hf1 = hf2 = 1.2 * m

    hf1 = hf2 = 1.2 * 2.5 = 3 mm

16 Addendum angle øa1= øa2 == tan -1( M/R)

    øa1= øa2 == tan -1( 2.5/68)

    øa1= øa2 == 2.10

17 Dedendum angle øf1= øf2 == tan -1( 1.2M/R)

    øf1= øf2 == tan -1( 1.2*2.5/68)

    øf1= øf2 == 2.530

18 Blank Cone angle (Face angle) da1 = θp1+ øa1 da2 = θp2+ øa2

    da1 = 19.4 + 2.1 = 21.50 da2 = 70.6+ 2.53 = 73.130

19 Crown height (CH1) CH1 = d2/2 - msinθp1 CH2 = d1/2 - msinθp2

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Circumferential force at the middle is give by

Ftm = For calculating the bending and other stress consider the force is taken toact at the tip corner of the gear tooth

qk1 qe1 qk2 qe2From, the chart qk1 = 3 qe1 = 1 qk2 = 2.4 qe2 = 1The allowable bending stress can be taken from the table which is

There for Ftm = = =2764.2N b = 23mm

mm = 0.8m = 0.8(2.5) = 2.07 = 174.175Mpa = 139.342Mpa Hence, δb1 = 174.175mpa < δbp = 180Mpa the design is safe

    CH1 = 127.5/2 – 2.5*sin 19.4 CH2 = 45/2 – 2.5*sin70.6

    CH1= 62.9mm CH2 = 20.14mm

 20  Back cone distance  Ra1 = Rtanθp1  Ra2 = Rtan θp2

= 68*tan19.4 = 23.95 = 68tan70.6 = 190.97

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