Dear Rizwan,

You have almost solved your own problem.

For the load analysis of each shaft, which seem to be OK, solve the 6 equilibrium equations:

horizontal forces in x-direction = 0

horizontal forces in z-direction = 0

vertical forces in y-direction = 0

moments around x-axis = 0

moments around z- axis = 0

moments around y- axis = 0

Torque = Power/omega = Tangential force * Radius

omega = radial speed = 2 * 3.141593 * rpm / 60

Do not forget to make a vector diagram of the forces on the gear teeth.

If you make no errors then the axial load on a bearing is 0, because the axial forces stay within the gears themselves.

success

Teus ■

Dear Rizwan,

Power in kW is the same for each gear, disregarded losses. (Law of conservation of energy)

The gearbox, which you describe, is an rpm step up gearbox.

The torque on shaft 2 equals the torque on shaft 1 divided by the gear ratio.

Power = Torque 1 x omega 1

Power = Torque 2 x omega 2

The tangential forces, radial forces and axial forces on the meshing teeth are equal for both gears. (Newton: action equals reaction)

Tangential force = Torque 1 / Radius 1

Tangential force = Torque 2 / Radius 2

radial force = tangential force / tan(alpha)

axial force = tangential force / tan(beta)

(if I did this correctly)

Because this is a double gear wheel, each gear teeth set carries half the power.

The axial force of one gear set (beta = +) compensates the axial gear force of the other set (beta = -).

The axial resultant force is therefore 0

From the equilibrium of forces (in space), the resulting bearing reactions (in space) is then determined.

I know it is quite some effort, which needs the upmost accuracy.

Are you sure that you apply sleeve bearings?

Success

Teus ■

there will be no axial load as the gears are double helical in action

it is not specified whether the pinion is the input or output?

is it a speed increaser?

by the looks of the arrows it seems that the 520 gear is the input?

please clarify ■