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. Part 1. 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. Part 2. 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). Part 3. 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.

Answer 1

The intermediate shaft RPM is assumed as 700 RPM. The minimum number of teeth required on gear 2 for the desired speed ratio so as to avoid the interference can be calculated. The number of teeth on gear 3 can then be calculated. The angular speed of gear 4 will also be 700RPM and thus number of teeth on gear 4 and gear 5 can then be calculated. For the given diametral pitch, module is obtained and then diameter of gear is obtained. Once diameter is known, the tangential force can be calculated as power and angular speed are known (torque can be calculated). Once tangential force on gear is known, using the Lewis Equation for bending strength on gear, the width of gear can be obtained. The Lewis Form factor is taken as 0.4 from the data book.

In order to further reduce the sizer of gear box, it would be suggestable to go for helical gear drive or bevel gear drive instead of spur gear. The transmission ratio will be the same and it will consume less space.

Intenmediate Shaft TPM to be Foo Assuming aft {{ 1 + 2 (2+2) Siuže]"-1 | 11 Tomiu to (Minimum Number of teeths avoid Interference) LE 700 3500 x = 20° 2x1x1") min = 15.7 = 16 teetha {1+ W5 ('*2) %2016-1 = 16 Zg Zz 23 = 16x (3500) = Z2 = 80 teetha 11 w2 = 700 RPM = W Wslon ws - 100 (min) - 2 = 16. 4 2xxx/7) { [ + (17) + ('7+2) Su%20 24 7= = 1/2 teeth 25 1 P=6 teeth lin (16) in / teeth P 100x 745= 134.1 KW TE ER BY Power = 180 HP = (Power) = (Torque) xw 1600 X (134-1) = T 21/100) Akkuning "Lewrit form foeter Y=0.4 60 Pa = 6 inches / teeth (Diametral Pitch) = 60 kpku = 6oooo pri T = 12805.6 Nom = 113339.11 lb-inch T +(128056) = (F)(9,20) - F = (upam) (113359-11) -) = 4047.8 16% (60000) = (11047.3) x 6 6 (0.4) (!()*** b = 1.0119 inches 2