# How to calculate forces on bearings in a chain drive

I have a chain drive which is moving a load of 200kg. I have 95Nm on the input shaft which has 15 teeth, which results in 152Nm on the output with 24 teeth.

The shafts are supported on bearings on both sides and the sprockets sit in the middle. I'm trying to calculate the lateral forces that will be generated on the bearings and I'm not really sure how to do that.

Taking for example the output shaft, its sprocket is 0.095m in diameter.

$$152/0.0475=3200N$$

That would mean there is about 300kg of force acting on the sprocket's teeth. This is already a little weird to me as I'd imagine forces wouldn't exceed the total weight of the load, but okay.

I'm having trouble figuring out how much of that will get transferred to the bearings in lateral forces?

If you have an output torque $$T_\text{out}$$ on the output shaft, and the sprocket diameter is $$d$$ then chain force $$F_\text{chain}$$ is found from $$T_\text{out}=\frac{d}{2}F_\text{chain}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}F_\text{chain}=\frac{2T_\text{out}}{d}$$ The reaction forces in the bearings are equal, i.e. $$R_A=R_B=R$$ such that $$R_A+R_B=R+R=2R=F_\text{chain}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}R=\frac{T_\text{out}}{d}$$
Depending on where the torque was measured, the force "$$F$$" equals the torque divided by the "diameter" of the sprocket.