Page 238 - Analysis and Design of Machine Elements

Analysis and Design of Machine Elements
216
Steps Computation Results Units
Z = 0.786
7) Since > 1, substitute by = 1 in Eq. (8.52), the

contact ratio factor is
√
4
Z = (1 − )+
3

√
1
= = 0.786
1.62
Z = 0.985
1) From Eq. (8.53), the helix angle factor is
√
√ ∘
Z = cos = cos 14 = 0.985

(3) Design 2) Calculate pitch diameter of the pinion d using the
calculation 1
smaller allowable stress
√
( ) 2
3 2KT 1 u±1 Z H Z E Z Z
d ≥ ⋅
1t u [ H ]
d
√
3 2 × 1.3 × 9.948 × 10 4 4 + 1 ( 2.433 × 189.8 × 0.786 × 0.985 ) 2
= ⋅ ⋅
1 4 522.5
= 53.29 mm
3) Calculate pitch line velocity, v
d n × 53.29 × 960
v = 1t 1 = = 2.67m∕s
60 × 1000 60 × 1000
4) Calculate face width
b = d =1 × 53.29mm =53.29mm
d 1t
5) Calculate the ratio of face width to tooth height b/h
∘
d cos 53.29 × cos 14
1t
Module m = = = 2.72mm
nt
z 1 19
Tooth height h = 2.25 m = 2.25 × 2.72 mm = 6.12 mm
nt
The ratio of b/h = 53.29/6.12 = 8.71
K = 1.588
6) Calculate the load factor.
From Eq. (8.12), the load factor is
K = K K K K
A
v

Determine the application factor from Table 2.1 as
K = 1.0.
A
−1
According to v = 2.67 m s , gear accuracy grade 7,
from Eqs. (8.13)–(8.16), the dynamic factor is found as
K = l.10.
v
From Table 8.2, select transverse load factor K = K
Ha Fa
= 1.1.
Determine the face load factor K from Table 8.3. For

accuracy grade 7 and a symmetrically supported pinion,
2
2
K H = a + a [1 + a ] + a b
1
3
2
4
d
d
2
2
=1.12 + 0.18 × [1 + 0 × 1 ]×1 +2.3
×10 −4 ×53.29 = 1.312
From Eq. (8.18)
(b∕h) 2 8.71 2
N = = = 0.887
0 1 +(b∕h)+(b∕h) 2 1 + 8.71 + 8.71 2
From Eq. (8.17)
K =(K ) N 0 = 1.312 0.887 = 1.27
F H
The load factor is then
K = K K K K H =1.0×1.10 × 1.1×1.312
A
H
v
=1.588
(continued)