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Force on gear tooth calculation based on HP and RPM

Force on gear tooth calculation based on HP and RPM

Force on gear tooth calculation based on HP and RPM

(OP)
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


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