Page 245 - Analysis and Design of Machine Elements

Gear Drives
223
Steps Computation Results Units
(2) Decide values 1) Calculate the load factor K = 1.737
to be used in the K = K K K K H =1.0×1.10 × 1.2×1.316
v
H
A
calculation =1.737
Z = 189.8 MPa 1/2
2) For the match of steel pinion and steel gear, the E
1/2
elastic coefficient is Z = 189.8 MPa .
E
Z = 2.392
3) For a standard helical gear, the pressure angle is H
∘
20 , zone factor is calculated by
√
Z = 2cos b
H
sin t cos t
From Eq. (8.36)
( ) ( )
tan n ∘
= arctan = arctan tan 20 ∘
t
cos cos 18.19
∘
= 20.96
And from Eq. (8.37)
∘
= arctan(tan cos ) = arctan(tan18.19
b
t
∘
cos 20.96 )=17.06 ∘
Therefore, the zone factor is
√
√ ∘
Z = 2cos b = 2 cos 17.06 ∘ = 2.392
∘
H
sin t cos t sin 20.96 cos 20.96
= 1.586
4) From Eq. (8.39), the transverse contact ratio is
[ ( )]
1 1 = 1.988

= 1.88 − 3.2 + cos
z z
[ ( 1 1 1 2 )] ∘
= 1.88 − 3.2 + cos 18.19 = 1.586
19 76
The face contact ratio is calculated from
Eq. (8.40)
=0.318 z tan =0.318 × 1.0

d 1
∘
×19 tan 18.19 =1.988
Z = 0.794
5) Since > 1, substitute = 1 in Eq. (8.52), the

contact ratio factor is
√
4
Z = (1 − )+

3
√
1
Z = = 0.794
1.586
Z = 0.975
6) From Eq. (8.53), the helix angle factor is
√
√ ∘
Z = cos = cos 18.19 = 0.975

7) Compute tangential force F t
2T 1 2 × 9.948 × 10 4
F = = = 4974.12N
t
d 1 39.999
Contact strength
8) Check the tooth surface fatigue strength
√ is enough
= Z Z Z Z KF t ⋅ u±1
H H E
bd 1 u
= 2.392 × 189.8 × 0.794 × 0.975 ×
√
1.737×4974.12 ⋅ 4±1 = 913.1 ≤ [ ]
40×39.999 4 H
7. Gear drawings (omitted)