Can you please show me a detailed solution to this problem.?
Solution:
1)
For e = (N2 / N3) (N4 / N5) = 35. For an exact ratio, we will
choose to factor the train value into integers, such that
N2 / N3 = 7 (1)
N4 / N5 = 5 (2)
Assuming a constant diametral pitch in both stages, the geometry
condition to satisfy the in-line requirement of the compound
reverted configuration is
N2 + N3 = N4 + N5 (3)
With three equations and four unknowns, one free choice is available. It is necessary that all of the unknowns be integers. We will use a normalized approach to find the minimum free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity, thus N3 = 1. From (1), N2 = 9. From (3),
N2 + N3 = 7 + 1 = 8 = N4 + N5
Substituting N4 = 5 N5 from (2) gives
8 = 5 N5 + N5 = 6 N5
N5 = 8 / 6 = 4 / 3
To eliminate this fraction, we need to multiply the original
free choice by a multiple of 3. In addition, the smallest gear
needs to have sufficient teeth to avoid interference.
From k = 1, phi= 20°, and m = 9, the minimum number of
teeth on the pinion to avoid interference is 17. Therefore, the
smallest multiple of 3 greater than 17 is 18.
Setting N3 = 18 and repeating the solution of equations (1), (2),
and (3) yields
2)
d= N/P = 24/6 = 4.0 in
Table 14-2: Y = 0.296
V = (pi)dn/12 = pi(4.0)(100)/12 = 104.7 ft/min
Eq. (14-4b): Kv = (6000 + 104.7)/6000 = 1.0175
Eq. (13-35) : Wt = 33000 H/V = 33 000 (180/104.7) = 56733.52 lbf
Eq. (14-7): F=(Kv.Wt.P)/sigma.Y = (1.0175 x 56733.2 x 6 ) / (60000 x 0.296) = 19.50 in
Use Face Width F = 19.5 in
3) Please consider the following two solutions to further reduce the size of the gear box keeping the same gear train ratio.
Comments:
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