Page 233 - Analysis and Design of Machine Elements

211
Steps Computation Results Gear Drives Units

(1) Decide the 1) Select the endurance limit of the pinion and gear
allowable contact from Table 8.6
stresses Contact endurance limit of the pinion
Hlim1 = 740 MPa
Contact endurance limit of the gear
Hlim2 = 550
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 is greater than 5×10 ,selectthe life factor
for contact stress slightly less than 1.0, as K HN1 =
0.90; K HN2 = 0.95. More precise data can be found
in design handbooks or standards.
[ ] = 666 MPa
5) Calculate allowable contact stresses H1
Select reliability of 99% and safety factor S = 1.0, [ ] = 522.5 MPa
H H2
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
1
calculation 2) The torque transmitted by the pinion is T = 9.948 × 10 Nmm
6
6
T =9.55 × 10 P ∕n =9.55 × 10 × 10∕960
1
1
1
4
=9.948 × 10 N ⋅ mm
= 1.0
3) For symmetric supporting, select the face d
width factor as 1.0, as recommended in
d
Section 8.6.6.
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 . For a Z = 2.5
H
E
∘
standard spur gear, the pressure angle is 20 , zone
factor is Z = 2.5.
H
(3) Design 1) Calculate trial pitch diameter of the pin-
calculation
ion using the smaller allowable stress
√
( ) 2
3 2KT 1 u+1 Z E Z H
d ≥ ⋅ ⋅
1t u [ H ]
d
√
( ) 2
3 2×1.3×9.948×10 4 4+1 189.8×2.5
= ⋅ ⋅ mm
1 4 522.5
= 64.363mm
2) Calculate pitch line velocity, v
d n × 64.363 × 960
1t 1
v = = = 3.23m∕s
60 × 1000 60 × 1000
3) Calculate face width
b = ⋅ d =1 × 64.363mm =64.363mm
d 1t
(continued)