We have a rolling machine with 3 rollers; the bottom has two rollers with diameters of 168 mm and one top roller with a diameter of 219 mm. I want to calculate the required force to bend a 10 mm thick plate which is 2,500 mm wide. The rolling machine is connected to a 10 HP motor with a gear box and the rollers have an rpm of 8.

  • $\begingroup$ What's the spacing between the two bottom rollers? $\endgroup$
    – Mark
    Jun 25, 2015 at 11:10

2 Answers 2


The deflection of a beam, $\delta$ mm in terms of the force is:

$$\delta = \frac{WL^3}{48EI}$$

Where $I = \frac{1}{12}b*h^3$ – in your case h is 10 mm, b = 2500 mm. E is the modulus, and based on your material – for steel it is 200 GPa. Using this, and the knowledge of the unspecified L (length between the bottom rollers), you can calculate how much you will bend the plate.

Now, to keep the plate permanently bent, you will need to exceed the elastic limit of the material – so you need to bend it so the stresses exceed the yield strength at the temperature you're working the material. For your case, that's when:

$$ \sigma = \frac{WLh}{8I}$$

Exceeds your yield strength – which for steel runs at 250 MPa. The fun about permanent bending – you had to exceed some sort of $\delta_{min}$ to meet this yield strength, then your sheet will continue bending as you apply more force, to a full $\delta$. However, when you remove the piece from the machine, it will "spring back" – the worst it will spring back will be $\delta – \delta_{min}$ – so you usually have to overshoot by a bit, check to see if it's bent enough, and bend it a second time.

  • 4
    $\begingroup$ This answer is mostly correct, but its worth mentioning that the yield strength of the steel is highly dependent on the steel itself. And the deflection you should aim for is equal to $\delta_y + \delta$, where $\delta_y$ is the deflection at yield and $\delta$, the desired final deflection. This way, the steel will elastically lose all of $\delta_y$ but keep $\delta$. Also, in this case the given equation for $\delta$ is no longer valid since it assumes elastic deformations. The force to be applied should be equal to the yield force, since the piece will keep deforming under constant load. $\endgroup$
    – Wasabi
    Jun 25, 2015 at 13:31
  • $\begingroup$ what is the value of W in above formula $\endgroup$
    – user5079
    Feb 17, 2016 at 11:55

W = P/L. The load is assumed to be uniform along the top rol length. Another L is the spacing of bottom rollers, called the bending span. This L only before the bending starts. Say at the top 12 o'clock at each bottom roller. Once it does, the L becomes shorter as the support of the plate moves to 1 o'clock at the left and 11 o'clock at the right roller. The shorter the span L the more power is needed.


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.