Question

A compound reverted spur gear train is to be designed to reduce the speed of a 180HP engine's rotating shaft from 3500rpm to 100rpm exactly. (a) (5 points) Specify appropriate numbers of teeth (N2, N3, N4 and N5) to minimize the gearbox size. A pressure angle of 20° is used for all the gears. Interference of gears is not acceptable. (b) (5 points) If all the gears are cut with 20° full-depth system. Assume a diametral pitch of 6 teeth/in, and allowable stress of 60kpsi. Find a suitable face width for gear 5 (see the following figure). (c) (2 points) Provide at least two (2) solutions (by modifying some parameters above, while keeping the same gear train value) to further reduce the size of the gear box. 2 5

Answer 1

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

• N2 = 126 teeth
• N3 = 18 teeth
• N4 = 120 teeth
• N5 = 24 teeth

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.

1. Consider gear materials that have been heat treated or case hardened to increase the surface strength. Adjustments can be made to the diametral pitch if necessary to achieve a good balance of size, material, and cost.
2. If the stresses are all much lower than the material strengths, a larger diametral pitch is in order, which will reduce the size of the gears and the gearbox.

Comments:

• References to the equations used are from Shigley.
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