Force on gear tooth calculation based on HP and RPM

[gcarmich]

I've recently been learning about gear design.

I can calculate the strength of my gear tooth using formula:

F(max)=((Bending strength of material (N/mm2) * gear width (mm) * Lewis Factor)/diametral pitch (mm))*Barth speed factor

where bending strength = tensile strength (N/mm2)/3

but I need to determine how much force the drive shaft will put on the drive gear (1:1 gear ratio) to ensure the gear tooth is strong enough.

If I have a 15hp engine with a max rpm of 5500 and a pitch diameter of 35mm (1.41732in) am I correct to solve this as:

safe tooth load (psi)=2(63025*HP)/RPM*pitch diameter?

487.235 psi=2(63025*15)/5500*1.41732 = 3.359N/mm2

tooth face area = 70.14mm2

force on tooth=3.359N/mm2*70.14mm2=235N=52.83lbf?

Thank you,
Gil

 

[mfgenggear]

see this

https://www.facebook.com/groups/389731063332700