Page 243 - Analysis and Design of Machine Elements

221
Gear Drives
Steps Computation Results Units
10) Compute and compare the values of Y Y Sa
Fa
Y Y /[ ] for the pinion and gear.

F
Y Y
Y
Y Fa1 Sa1 2.771 × 1.556 × 0.689 × 0.883
=
[ ] 444.4
F1
= 0.00590
Y Y Y Y
Fa2 Sa2 2.214 × 1.773 × 0.689 × 0.883
=
[ ] 454.9
F2
= 0.00525
(3) Design 1) Calculate trial modules m = 2.0 mm
n
calculation
√
m ≥ 3 2KT 1 cos 2 ⋅ Y Fa Y Sa Y Y
n 2
d z 1 [ F ]
√ ∘
4
3 2×1.3×9.948×10 cos 2 14 2.771×1.556×0.689×0.883
= ⋅
1.0×19 2 444.4
= 1.58mm
Select m = 2.0.
n
4. Geometrical (1) Centre distance a = 100 mm
calculation (z + z )m n (19 + 76)× 2
2
1
a = = ∘ = 97.91mm
2cos 2 × cos 14
Select a = 100 mm.
(2) Modified helix angle = 18.19 ∘
(z + z )m (19 + 76)× 2 mm
= arccos 1 2 n = arccos d = 39.999
1
2a 2 × 100
∘ d = 159.996 mm
= 18.19 2
∘
Select = 18.19 .Since calculated is close to the
∘
initially selected helix angle 14 , Y , Y , Y and

Sa
Fa
Y do not need to be modified.

m z 2 × 19
d = n 1 = ∘ = 39.999mm
1
cos cos 18.19
m z 2 × 76
n 2
d = = ∘ = 159.996mm
2
cos cos 18.19
(3) Face width of the gear b = 45 mm
b = d =1 × 39.999mm =39.999mm 1 mm
d 1 b = 40
Select b = 40 mm, b = 45 mm. 2
2 1
5. Modify the load 1) Calculate pitch line velocity v
factor d n × 39.999 × 960
1 1
v = = = 2.01m∕s
60 × 1000 60 × 1000
2) Calculate the ratio of face width and tooth height
b/h.
Tooth height h = 2.25m = 2.25 × 2 = 4.5 mm
n
The ratio of b/h = 40/4.5 = 8.88
(continued)