Page 273 - Analysis and Design of Machine Elements

Steps Computation Results Wormgear Drives 251
Units
2. Design From Eq. (9.16), the design formula by contact
by surface strength is
contact ( Z E ) 2
2
strength m d ≥ 9.4KT 2
1
z [ ]
2
H
a = 200 mm
1) Determine the torque acting on the wormgear T 2 mm
Select z = 2, efficiency = 0.8, then m = 8
1
P P d = 80 mm
1
T = 9.55 × 10 6 2 = 9.55 × 10 6 1 =
2 n n ∕i
2
9 × 0.8 1
6
9.55 × 10 × = 0.94
1460∕20
6
× 10 N •mm
2) Specify load factor K
For a stable work load, select application factor
as K = 1.15 from Table 2.1. Select the face load factor
A
as K = 1.0; dynamic factor is K = 1.05 for low speed
v
and minor impact. The load factor is then
K = K K K =1.15 × 1.0×1.05 = 1.21
v
A

3) Determine elastic factor Z
E
Since hardened steel worms mating with a cast tin
1/2
bronze wormgear, select Z = 155 MPa .
E
4) Specify allowable contact stress [ ]
H
Since the wormgear material is cast tin bronze

ZCuSnl0P1, metal mould casting and the hardness
of worm surface is greater than 45 HRC, from
Table9.2,selectthe basicallowablecontact stress
′
as [ ] = 268 MPa.
H
The number of cycles
1460
N = 60jn L = 60 × 1 × × 10000 = 4.38 × 10 7
2 h 20
Contact life factor
√ √
8 10 7 10 7
K = = 8 = 0.8314
HN N 4.38 × 10 7
Then the allowable contact stress is
′
[ ]= K [ ] =0.8314 × 268 = 222.8MPa
H HN H
2
5) Calculate the value of m d 1
( ) 2
Z
2
m d ≥ 9.4KT 2 E = 9.4 × 1.21 × 0.94 ×
1
z [ ]
2
H
( 155 ) 2
10 6 = 3234
2 × 20 × 222.8
From Table 9.1, select m = 8 mm, d = 80 mm,
1
a = 200 mm.
3. Parameters 1) Parameters and dimensions of the worm q = 10
and dimensions From Eq. (9.4), the worm diameter factor is = 11.31
of the worm d ∘
and wormgear q = 1 = 10
m
From Eq. (9.5), lead angle is
( )
z 2 ∘
′
= arctan 1 = arctan = 11 18 36 ′′
q 10
∘
= 11.31