Page 235 - Analysis and Design of Machine Elements

213
Steps Computation Results Gear Drives Units
2) Since the numbers of load cycles of the pinion
6
and gear are greater than 3 × 10 cycles, select
the life factor for bending stress as K = 0.85,
FN1
K = 0.87.
FN2
[ ] = 371.6 MPa
3) Calculate allowable bending stress F1
Select safety factor S = 1.4, from Eq. (8.86), [ ] = 264.7 MPa
F F2
allowable bending stress is
K 0.85 × 306
[ ]= FN1 F lim 1 Y = × 2.0
F 1 S F ST 1.4
= 371.6 MPa
K 0.87 × 213
[ ]= FN2 F lim 2 Y = × 2.0
F 2 S F ST 1.4
= 264.7MPa
(2) Decide values 1) Compute the load factor K K = 1.397
to be used in the K = K K K K =1 × 1.10 × 1.0×1.27
calculation A v F F
=1.397
2) Specify tooth form factor Y
Fa
From Table8.4,byinterpolating,wehave
Y = 2.85; Y = 2.232
Fa1 Fa2
3) Specify stress correction factor Y
sa
From Table 8.4, by interpolation, we have
Y = 1.54; Y = 1.762
sa1 sa2
4) Compute and compare the values of Y Y /[ ]
Fa Sa F
for the pinion and gear
Y Fa1 Sa1 = 2.85 × 1.54 = 0.0118
Y
[ ] 371.6
F1
Y Y 2.232 × 1.762
Fa2 Sa2 = = 0.0148
[ ] 264.7
F2
√
√ 3 2×1.397×9.948×10 4
(3) Design m ≥ 3 2KT 1 2 ⋅ Y Fa Y Sa = 1×19 2 × 0.0148 = m = 2.5 mm
[ F ]
d z
calculation 2.25mm 1 z = 26
1
The module obtained by contact strength analysis z = 104
2
is greater than that obtained by bending strength
analysis. Since module relates closely to the tooth
bending strength, while the pitch diameter of a
gear is relevant to contact strength. Therefore, the
module obtained from bending strength analysis is
selected and round down to a standard value of
2.5 mm. The pitch diameter obtained from contact
strength analysis d = 66.671 mm is selected to
1
calculated the number of teeth of the pinion,
d 66.671
z = 1 = = 26.67
1 m 2.5
Select z = 26.
1
The number of teeth of the gear
z = uz = 4.0 × 26 = 104, select z = 104.
2 1 2
Thus, the designed gear can meet both contact and
bending strength requirements.
(continued)