Problem: Bevel Gear
Design a bevel drive with 90o shaft angle having following particulars:
Motor Capacity = 10kW
Motor Speed = 1460 rpm
Transmission Ratio = 3.5
Service Factor = 1.5
Material: C60 steel for both the wheels.
Solution:
1.Teeth no. Z
Take Zp =18
Therefore, Zg = i . Zp = 3.5 x 18 = 63
2. Pitch cone angle, θ
θp = tan -1 (1/i) = tan -1 (1/3.5) = 15.945o
θg = tan -1 i = tan -1 (3.5) = 74.055o
Check: θp + θg = 90o
3. Virtual Number of Teeth Zv
Zvp = Zp/(cos θp) = 18.72
Zvg = Zg/(cos θg) = 229.33
4. Form Factor, Y = 0.484 – (3.28/Zv)
Yp = 0.309
Yg = 0.47
5. Basic Allowable Stress, So
Taking a factor of safety of 3.5 over the ultimate strength for C60 steel,
So = Su / f.s. = 800/3.5 =230MPa (approx.)
6. Weaker wheel
For pinion: [So.Y]p = 71.07
For gear: [So.Y]g = 108.1
As [So.Y]p < [So.Y]g , PINION is the weaker wheel and hence the Pinion is to be designed.
7. Design Torque, Tp
𝑇𝑝 = 𝐶𝑠 ×
= 1.5 ×
60 ×106
2𝜋
60 ×106
2𝜋
𝐾𝑊
×𝑛
𝑝
10
× 1460 = 98110 N.mm
8. Module, m
3
𝑚= √
2𝑇𝑝 ∙ 𝐶𝑤
𝑆𝑜 ∙𝐶𝑣 ∙𝑋∙𝑌𝑝 ∙𝑍𝑝 ∙𝛽
Putting, Tp = 98110 N.mm
So = 230 MPa
X = 8 (assumed)
Zp = 18
Yp = 0.309
β=1−
𝑏
𝐿
We have, 𝑚 =
25.3681
𝑚= √
𝐶𝑣
3
=1−
3
√
𝑋.𝑚
𝑚𝑍𝑝
(
)
2 sin 𝜃𝑝
=1−
2 sin 𝜃𝑝 . 𝑋
𝑍𝑝
= 0.7558
25.3681
𝐶𝑣
𝜋.𝑚.𝑍 .𝑛
𝑝 𝑝
𝑉=
60000
= 1.3760176 × 𝑚
…
3.686
3.2802
3.2621
3.2613
3.5 (std.)
𝐶𝑣 =
…
5.0721
4.5137
4.4888
…
…
5.5
5.5 + √𝑉
0.5
0.7095
0.7214
0.7219
…
…
We take standard module m= 3.5 mm
9. Important Dimensions
Item
Number of Teeth, Z
Module, m
Face Width, b = X.m
PCD , D= m . Z
Pitch Cone Angle, θ
Small end Dia, S = D- 2b sin θ
Axial face width, ba = b cos θ
Back Cone rad., r = D/(2 cos θ)
Crown height, H= D/(2 tan θ)
Pitch Cone rad., L = D/(2 sin θ)
Pinion
18
3.5
28
63
15.945
47.616 mm
26.923 mm
32.76 mm
110.253 mm
114.66 mm
Gear
63
220.5
74.055
166.655 mm
7.692 mm
401.326 mm
31.499 mm
114.66 mm
Also, Pitch line Velocity, 𝑉 =
𝜋.𝑚.𝑍𝑝 .𝑛𝑝
60000
5.5
= 4.816 𝑚/𝑠
𝐶𝑣 = 5.5+√𝑉 = 0.7148
10. Dynamic Load Fd
Fd = F t + F i
Ft = 2Tp/Dp = 2 x 98110/63 = 3115 N
𝜋 𝐷𝑝 + 𝑆𝑝 2
Pitch circular mass for pinion = Mp = 8 (
) . 𝑏𝑎𝑝 . 𝜌
2
𝜋 63 + 47.616 2
) × 26.923 × 7.83 × 10−6 kg
=8(
2
= 0.2532 kg
𝜋 𝐷𝑔 + 𝑆𝑔 2
Pitch circular mass for gear = Mg = 8 (
𝜋 220.5 + 166.655 2
= (
8
2
) . 𝑏𝑎𝑔 . 𝜌
) × 7.692 × 7.83 × 10−6 kg
2
= 0.8863 kg
𝑀𝑝 .𝑀𝑔
Equivalent Mass = M = 𝑀 +𝑀 = 0.1969 kg
𝑝
G=
tan 20°(1−cos 20°)
[𝑟
𝜋 2
𝑔
( )
1
𝑔
1
𝑏𝑝
+ 𝑟 ]
𝑏𝑔
= 5.948 x 10-3 mm-1
Rigid Body Acceleration Force = F1 = 2GMV2 KN
= 2 x 5.948 x 10-3 x 0.1969 x 4.8162 KN
= 0.05432 KN = 54.32 N
Asymptotic Load = F2 = k . ε + Ft
𝑏
28
Here, k = 1 1 =
N/mm = 329778 N/mm
1
1
9[
+
𝐸𝑝 𝐸𝑔
9[
]
+
2.12×105 2.12×105
]
For class 7 gear,
εp = εg = 8 + 3.55 √𝜋. 𝑚 , micron
= 8 + 3.55 √𝜋. 3.5 , micron = 19.77 micron
Hence, ε = εp + εg = 39.54 micron = 0.03954 mm
Therefore, F2 = k . ε + Ft = 329778 x 0.03954 + 3115 = 16154.4 N
𝐹 .𝐹
Equivalent Acceleration Force , Fa = 𝐹 1+𝐹2 = 54.138 N
1
2
So, Fi = √𝐹𝑎 (2𝐹2 − 𝐹𝑎 ) = 1321 N
Dynamic Load, Fd = Ft + Fi = 4436 N
Note : Ft / Cv = 3115 / 0.7148 = 4357 N
11. Check Wear Strength : Fw
𝐹𝑤 =
𝐷𝑝 . 𝑏 . 𝐾 . 𝑄
𝑄= 𝑍
cos 𝜃𝑝
2𝑍𝑣𝑔
𝑣𝑝 + 𝑍𝑣𝑔
2 ×229.33
= 18.72+229.33 = 1.849
Therefore, 𝐾 ≥
But, 𝐾 =
≥ 1.25 𝐹𝑑
1.25 𝐹𝑑 cos 𝜃𝑝
𝐷𝑝 . 𝑏 . 𝑄
𝑆𝑒𝑛,𝑐 2 .sin 𝜙
1.4
1.25 ×4436 × cos 15.945
=
63 × 28 × 1.849
1
1
𝑝
𝑔
[𝐸 + 𝐸 ]
= 1.6347
1.9616 =
𝑆𝑒𝑛,𝑐 2 .sin 20°
1.4
[
1
2.12×105
+
1
2.12×105
]
Sen,c = 842 MPa = 2.8 BHN – 70
Therefore, Required BHN = 326
12. Forces on Pinion Tooth
Tangential Forces in pinion (& gear),
Ft = 2T/D = 3115 N
2𝑇
2𝑇
Effective Tangential tooth load, F = 𝐷 = 1
𝑚
𝐷
𝐹𝑡
𝑚
1−
Therefore, 𝐹 = 𝐹𝑡 . 𝐷 =
𝑏
2𝐿
2
(𝐷+𝑆)
= 3548.24 N
Crushing load on tooth Fc = F tan φ
= 3548.24 tan 20o = 1291.45 N (Common in gear and pinion)
Radial force on pinion Frp = Fc cos θp = 1241.766 N = Fag
Axial force on pinion Fa = Fc sin θp = 354.78 N = Frg