gcarmich
Electrical
- Sep 4, 2007
- 5
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
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