Page 237 - Analysis and Design of Machine Elements

215
Steps Computation Results Gear Drives Units

N = 4.15 × 10 9
2) The number of load cycles is calculated by 1
Eq. (8.84) as N = 1.04 × 10 9
2
N =60n jL = 60 × 960 × 1 × (16 × 300
1
1
h
× 15) = 4.15 × 10 9
9
N =4.15 × 10 ∕4.0=1.04 × 10 9
2
3) Since the number of load cycles of the pinion and
7
gear are greater than 5 × 10 ,selectlifefactor
for contact stress slightly less than 1.0, as K =
HN1
0.90; K = 0.95. More precise data can be found
HN2
in design handbooks or standards.
[ ] = 666 MPa
4) Calculate allowable contact stresses H1
Select reliability of 99% and safety factor S = 1.0, [ ] = 522.5 MPa
H2
H
the allowable contact stress is calculated by
Eq. (8.85), as
K
[ ]= HN1 H lim 1 = 0.90 × 740 = 666MPa
H1
S H
K
[ ]= HN2 H lim 2 = 0.95 × 550 = 522.5MPa
H2
S H
(2) Decide values 1) Select a trial load factor K = 1.3
to be used in the t 4
calculation 2) The torque transmitted by the pinion is T = 9.948 × 10 N⋅mm
1
6
T =9.55 × 10 P ∕n =9.55 × 10 × 10∕960
6
1 1 1
4
=9.948 × 10 N ⋅ mm
= 1.0
3) For symmetric supporting, select face width fac- d
tor as 1.0.
d
Z = 189.8 MPa 1/2
4) For the match of steel pinion and steel gear, the E
1/2
elastic coefficient is Z = 189.8 MPa .
E
Z = 2.433
5) 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 tan 20 ∘
= arctan = arctan ∘
t
cos cos 14
∘
= 20.56
and from Eq. (8.37)
= arctan(tan cos ) = arctan(tan14 ∘
b
t
∘
cos 20.56 )=13.13 ∘
Therefore, the zone factor is
√
√ ∘
Z = 2cos b = 2 cos 13.13 ∘ = 2.433
∘
H
sin t cos t sin 20.56 cos 20.56
= 1.62
6) From Eq. (8.39), the transverse contact ratio is
estimated as = 1.509
[ ( )]
1 1
= 1.88 − 3.2 + cos
z z
[ ( 1 1 1 2 )] ∘
= 1.88 − 3.2 + cos 14 = 1.62
19 76
The face contact ratio is calculated by Eq. (8.40) as
=0.318 z tan =0.318 × 1.0
d 1
∘
×19 tan 14 =1.509
(continued)