Page 216 - Analysis and Design of Machine Elements

Analysis and Design of Machine Elements
194
8.4.3 Tooth Surface Fatigue Strength Analysis
A helical gearing has an additional margin for increasing load carrying capacity. In a
helical gear drive, the inclined contact line due to helix angle progress from the tip of
teeth across the pitch line to the lower flank of tooth. The gradually changed contact
length leads to a smooth engagement. Moreover, a large contact ratio in a helical gear
drive implies more teeth are engaged simultaneously and share the applied loads. It is
the gradual teeth engagement and lower average load per tooth that gives helical gears
the ability to transmit heavy loads at high speeds.

8.4.3.1 Contact Stress Calculation
Pitting resistance for helical gear teeth is evaluated by the same approach as discussed
for spur gears, with minor adjustment to account for geometrical differences due to helix
angle. The Hertz formula is employed in contact stress calculations.
Similar to spur gears, incorporating Eqs. (8.11) and (8.46), the design load is
calculated by
KF t
F = KF = (8.47)
n
ca
cos cos b
t
When helical gears mesh, engagement is gradual and load is propagated diagonally
across the tooth surface. The load is shared by all the teeth in engagement. The total
length of inclined contact lines is affected by transverse and face contact ratio, and con-
tact ratio factor Z is introduced. The total length of contact line is calculated by [5]

b
L = 2 (8.48)
Z cos b

The relationship between normal radius of curvature and transverse radius of cur-
n
vature in a helical gear can be expressed as [12]
t
t
= (8.49)
n
cos
b
Similar to a spur gear, the transverse radius of curvature is calculated in the trans-
t
verse plane by
1
= d sin (8.50)
t 1 t
2
Since strength analysis is performed in the normal plane, therefore,
1 1 2cos b 2cos b 2cos b u ± 1
± = ± = ⋅ (8.51)
d sin ud sin d sin u
n1 n2 1 t 1 t 1 t
Incorporating Eqs. (8.47), (8.48) and (8.51) and Hertz formula Eq. (2.51), we have
√
( )
√ 1 1 √
√ ±
√ F KF 2cos
√ n ⋅ 1 2 t ⋅ b ⋅ u ± 1
E
H
= √ ( ) = ⋅ Z Z
L 1− 2 1− 2 b cos d sin u
1 + 2 t 1 t
E 1 E 2
√ √
KF t u ± 1 2cos b
= ⋅ ⋅ ⋅ Z Z
E
bd 1 u sin cos t
t