What is the required load to bend a steel plate in a radius of 508mm?

  • Thread starter Thread starter swapnasis
  • Start date Start date
  • Tags Tags
    Bending Load
Click For Summary

Discussion Overview

The discussion revolves around calculating the load required to bend a steel plate into a specific radius of 508mm. Participants explore the application of beam bending formulas and the implications of large deflections in the context of bending mechanics.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant outlines the project details, including the dimensions of the steel plate and the setup with support rollers, seeking assistance with calculations.
  • Another participant agrees with the initial approach but notes that the beam deflection formula is only accurate for small deflections, suggesting that large deflections may lead to inaccuracies.
  • A later reply emphasizes the need for assumptions regarding flat sections and the influence of shear and normal stresses when calculating stress in large bending scenarios.
  • Multiple participants express confusion about the distinction between calculating stress and the required load to achieve the desired bending radius, indicating a need for clarity in the calculations.
  • One participant suggests looking into the bending of simply supported thin plates under point load for further guidance and provides links to resources.
  • Additional participants introduce their own related projects, seeking formulas for bending plates and further assistance.

Areas of Agreement / Disagreement

Participants generally agree on the complexity of the problem and the limitations of the beam bending formulas for large deflections. However, there is no consensus on the exact calculations or methods to determine the required load, as confusion persists regarding the distinction between stress and load calculations.

Contextual Notes

Limitations include assumptions about material properties, the effects of large deformations, and the specific conditions under which the formulas apply. The discussion does not resolve these complexities.

Who May Find This Useful

Individuals working on projects involving the bending of steel plates, particularly in engineering or materials science contexts, may find the discussion relevant.

swapnasis
Messages
4
Reaction score
0
Hi...

I am doing a project to calculate the load required to bend a plate.. Here are the details..

There is steel plate of 2000mm width and 25mm thcikness placed over 02 support steel rollers of diameter 180mm. Another steel roller of same diameter press from the top of the plate at the center of the bottom support rollers. How much load is required to bend the plate in a radius of 508mm?

I was calculating the above with the help of beam bending formula. Means one bean placed on top of 02 support beams and load is given at the center of the beam for bending. But I didn't get the answer perfectly.

Can somebody help me to find out the solution?

Swap..
 
Engineering news on Phys.org
Hi swapnasis.
Your approach is quite right. But the beam deflection formula is applied for only small deflections. But in your case deflection is quite large and hence you cannot rely on the accuracy of the results. Moreover if the steel used is M.S. then local plastic deformation of the plate at the center roller will lead to more inaccuracy of the result.
 
koolraj09 said:
Hi swapnasis.
Your approach is quite right. But the beam deflection formula is applied for only small deflections. But in your case deflection is quite large and hence you cannot rely on the accuracy of the results. Moreover if the steel used is M.S. then local plastic deformation of the plate at the center roller will lead to more inaccuracy of the result.

Thanks KOOLRAJ for your reply. But still I am confused about the formula and answer.. Can you please help me showing me the calculations with the beam bending formula? Atleast I can reach near to the exact solution..
 
Hi again.

Stress in large bending deformation:
Big bending asymptote stress.png

For large deformations of the body, the stress in the cross-section is calculated using an extended version of this formula. First the following assumptions must be made:

1. Assumption of flat sections - before and after deformation the considered section of body remains flat (i.e. is not swirled).
2. Shear and normal stresses in this section that are perpendicular to the normal vector of cross section have no influence on normal stresses that are parallel to this section.

Large bending considerations should be implemented when the bending radius ρ is smaller than ten section heights h:

ρ < 10h

With those assumptions the stress in large bending is calculated as:



F is the normal force
A is the section area
M is the bending moment
ρ is the local bending radius (the radius of bending at the current section)
Ix' is the area moment of inertia along the x axis, at the y place (see Steiner's theorem)
y is the position along y-axis on the section area in which the stress σ is calculated

For more info visit http://en.wikipedia.org/wiki/Bending
 
Last edited:
Hi...

Thanks again. I got your reply. But still it is confusing. I don't need to calculate the stress in the cross section. I need to calculate the load(Bending force) to bend the cross section on such condition. How much exact load(bending force) we require to bend the cross section of the said dimension as I have mentioned earlier?

Hope to get the reply soon..

koolraj09 said:
Hi again.

Stress in large bending deformation:
Big bending asymptote stress.png

For large deformations of the body, the stress in the cross-section is calculated using an extended version of this formula. First the following assumptions must be made:

1. Assumption of flat sections - before and after deformation the considered section of body remains flat (i.e. is not swirled).
2. Shear and normal stresses in this section that are perpendicular to the normal vector of cross section have no influence on normal stresses that are parallel to this section.

Large bending considerations should be implemented when the bending radius ρ is smaller than ten section heights h:

ρ < 10h

With those assumptions the stress in large bending is calculated as:



F is the normal force
A is the section area
M is the bending moment
ρ is the local bending radius (the radius of bending at the current section)
Ix' is the area moment of inertia along the x axis, at the y place (see Steiner's theorem)
y is the position along y-axis on the section area in which the stress σ is calculated

For more info visit http://en.wikipedia.org/wiki/Bending
 
Hi...

Thanks again for the reply. But still it is confusing. I am not calculating the stress in the cross section. I want to know the required load(Force) to bend the cross section. I am again indicating the details..


There is a steel plate(YS= 80,000Psi) of 2000mm width, 2000mm long and 25mm thickness. The plate is supported at both end from the bottom with 02 steel rolls of 180mm diameter. To bend the steel plate with a radius of 508mm, another steel roll of 180mm diameter is presses from the top of the plate at the center. Now the question is how much load(force) is required to bend the steel plate to achieve the required radius? It is really not a beam but a steel plate which is to be bent for a vessel.

Please give me the reply if you can solve the problem.

Thanks again for giving me the time..

koolraj09 said:
Hi again.

Stress in large bending deformation:
Big bending asymptote stress.png

For large deformations of the body, the stress in the cross-section is calculated using an extended version of this formula. First the following assumptions must be made:

1. Assumption of flat sections - before and after deformation the considered section of body remains flat (i.e. is not swirled).
2. Shear and normal stresses in this section that are perpendicular to the normal vector of cross section have no influence on normal stresses that are parallel to this section.

Large bending considerations should be implemented when the bending radius ρ is smaller than ten section heights h:

ρ < 10h

With those assumptions the stress in large bending is calculated as:



F is the normal force
A is the section area
M is the bending moment
ρ is the local bending radius (the radius of bending at the current section)
Ix' is the area moment of inertia along the x axis, at the y place (see Steiner's theorem)
y is the position along y-axis on the section area in which the stress σ is calculated

For more info visit http://en.wikipedia.org/wiki/Bending
 
hey swap .. I read thru ur problem very fast .. I got not much time .. As far as I understood u have to find the load which causes a displacement of 508mm. I would suggest go thru the bending of Simply supported thin plates under point load. I am sending a link as well. I am sure this will help u. Sorry can't solve the prob for u as i am kinda bz .. but this shud help u for sure ... Njoi and best o luck :)

http://www.efunda.com/formulae/solid_mechanics/plates/casestudy_list.cfm

http://www.efunda.com/formulae/solid_mechanics/plates/calculators/SSSS_PPoint.cfm

or els u can read Theory of thin plates by timoshenko .. Find the equation for displacement in the plate due to applied load. u know the end conditions ... U know the displacement and u can find the Load required .. Best o luck ..
 
hiii...swapnasis

I am also doing a project to calculate the load required to bend a plate.. Here are the details..

There is steel plate of 1600mm width and 12mm thcikness for three roller bending machine now i want to calculate rolls diameter and bending radius of plate can u help me?
I want formulas to calculate it..
 
hii swap i am also doing the same project can u pls tell me the formulas for bending plate
 

Similar threads

Replies
1
Views
2K
Replies
13
Views
2K
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
5K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
33
Views
5K
  • · Replies 14 ·
Replies
14
Views
5K