Page 241 - Analysis and Design of Machine Elements

Example Problem 8.3 Gear Drives 219
Design a pair of hard tooth surface helical gears for the reducer in Example Problem 8.1,
using the same design data.

Solution

Steps Computation Results Units
1. Selection of (1) Select helical gears according to the layout and Helical gears
gear type, gear requirement of the reducer.
accuracy grades, (2) Since the conveyor is ordinary machinery whose Gear accuracy
materials and heat speed is not very high, select gear accuracy grade grade 7
treatments 7 from Table 8.5.
(3) Material selection 40Cr for pinion
Select alloy steel 40Cr (AISI 5140) with hardness and gear
550HV for both the pinion and gear, heat treated
by flame hardening.
2. Select initial (1) Select the number of teeth of the pinion z 1 z = 19
1
variables Select z = 19, thenumberofgearteeth is z = 76
1
z = uz = 4.0 × 19 = 76. So select z = 76. 2
2 1 2
∘
(2) Select helical angle = 14 .
3. Design by gear Design formula by Eq. (8.63)
teeth bending √ 2KT cos Y Y Y Y
2
Fa Sa
strength m ≥ 3 1 ⋅
n
z 2 [ ]
d 1 F
(1) Decide the 1) Select the endurance limit of the pinion and gear
allowable bending from Table 8.6
stresses Bending endurance limit of the pinion and gear
Flim1 = Flim2 = 366 MPa
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
h
1
1
× 15) = 4.15 × 10 9
9
N =4.15 × 10 ∕4.0=1.04 × 10 9
2
3) 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
[ ] = 444.4 MPa
4) Calculate allowable bending stress F1
Select safety factor S = 1.4, from Eq. (8.86), [ ] = 454.9 MPa
F F2
allowable bending stress is
K FN1 F lim 1 0.85 × 366

[ ]= Y ST = × 2.0
F1
S F 1.4
= 444.4 MPa
K 0.87 × 366
[ ]= FN2 F lim 2 Y = × 2.0
F2 S ST 1.4
= 454.9MPa F
(continued)