0
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

Assuming the shaft and sprocket masses are small and performing static analysis:

enter image description here

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 $$R_A+R_B=R+R=2R=F_\text{chain}\hspace{0.5cm}\Rightarrow\hspace{0.5cm}R=\frac{T_\text{out}}{d}$$

$\endgroup$
4
  • $\begingroup$ So basically the entire 3200N force will also be applied radially on the bearings? I don't think this is correct. I think the confusion is perhaps caused from my explanation, trying to simplify things I actually complicated them more. The bearings in question are not carrying the weight of the system (which is a vehicle, 200kg), they are only carrying the weight of the drive unit, which is negligible $\endgroup$
    – php_nub_qq
    Jun 9, 2022 at 10:06
  • $\begingroup$ You should include a drawing of the system $\endgroup$
    – drC1Ron
    Jun 9, 2022 at 10:17
  • $\begingroup$ It is exactly as you have drawn it, but those bearings bear only the weight of the shaft and sprocket, not the entire 200kg. The sprocket drive does however move the 200kg which is the total weight of the vehicle, and it is driven by a motor $\endgroup$
    – php_nub_qq
    Jun 9, 2022 at 12:58
  • $\begingroup$ The force from the chain must be supported somewhere $\endgroup$
    – drC1Ron
    Jun 10, 2022 at 6:18
0
$\begingroup$

Depending on where the torque was measured, the force "$F$" equals the torque divided by the "diameter" of the sprocket.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.