Question

Two-stage gearbox design with input and output shafts on the same axis (a_ı = a_II) will be made. The gearbox will be driven by an electric motor of P= 7kW 1500 RPM and it is foreseen to operate under all conditions. Step cycle rates are i1=4.55, i2=3.67.

First stage: module m_I=1.75 mm, z₁= 13, ß₀₁=19^o and gears are made of C15 material

Second stage: z₃=12, ß₀₂= 19^o and gears are made of 16MnCr5 material

a)Find the module of the second stage (m₂)?

b)Since the axle spacing of the first stage will be the axle spacing of the gearbox, take all the necessary measures ( by making the most appropriate decisions) to achieve this "the number of teeth of the wheels will be rounded to the nearest integer."

c)Size the second stage pinion and gear.

Z4 iz z 23 Z2

P.S: You don't have to solve this problem numerically. It would be great too if you show the way I should follow.

Answer 1

1. Find the Torque to be transmitted from the formula P = T * (omega) [ T = P / (OMEGA) = 7000 W * 60 / (1500 * 2* 3.14) = 44.5 Nm ]

2. Able to work under all conditions, means power train should be able to absorb the shock which is present during the start up high load. Meaning actual torque capacity will be 5-10 % more than that calculated in point 1.

3. Find the no of teeth Z2, using the formula i1 = Z2 / Z1. ----> Z2 will come 59

4. Find Dia of gear 2 by formula m1 = m2 = Dz2 / Z2 ----- > it will come 103.25 mm ie r2 = 51.625 mm

5. Use formula : m = 2 (r1 + r2) / (z1 + z2) ----- > find r1 .

6. Find Z4 from i2 = Z4 / Z3 -----> 44.

7. Find M2 from formula : M = 2(r3 + r4) / Z3 + Z4 ------ > 2.25 mm

8. Dimensions of the second stage Gear and Pinion can be found from the material strength corresponding to the given matrial grade, and the stresses generated from the Torque to be transmitted.