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

In summary: According to the information provided, the required load to bend the plate in a radius of 508mm is approximately 285N.
  • #1
swapnasis
4
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..
 
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  • #2
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.
 
  • #3
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..
 
  • #4
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:
  • #5
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
 
  • #6
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
 
  • #7
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 ..
 
  • #8
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..
 
  • #9
hii swap i am also doing the same project can u pls tell me the formulas for bending plate
 

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

What is bending load calculation?

Bending load calculation is the process of determining the amount of stress and strain placed on a structural element, such as a beam or column, when subjected to an external force or load. It is an important aspect of structural design to ensure that the element can safely withstand the applied load without failure.

Why is bending load calculation important?

Bending load calculation is important because it helps engineers and designers determine the appropriate size and strength of structural elements, which is crucial for ensuring the safety and stability of a structure. It also helps in identifying potential weak points and determining the most efficient and cost-effective design.

What factors are considered in bending load calculation?

Several factors are taken into account in bending load calculation, including the type of material, the cross-sectional shape and size of the element, the magnitude and direction of the load, and the support conditions. These factors influence the amount of stress and strain experienced by the element and ultimately determine its ability to resist bending forces.

What methods are used for bending load calculation?

There are several methods for calculating bending loads, including hand calculations, analytical methods, and computer-aided analysis using finite element analysis (FEA) software. The choice of method depends on the complexity of the structure and the level of accuracy required.

What are some common applications of bending load calculation?

Bending load calculation is used in various industries, including construction, aerospace, automotive, and manufacturing, to design and analyze structures and components that will be subjected to bending forces. Some common applications include bridges, buildings, aircraft wings, and machine parts.

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