# Need help with a force calculation (deformation of a cylinder)

• las59036
In summary: I'm not sure if you are looking for that. There are other books that might be more applicable to your problem.
las59036
TL;DR Summary
Trying to calculate the force required to deform a cross section of a solid cylinder. To create an image, imagine a cylinder laying on a flat surface, and then a section of it defined as a length of X, being subjected to a compressive load.
Trying to calculate the force required to deform a cross section of a solid cylinder. To create an image, imagine a cylinder laying on a flat surface, and then a section of it defined as a length of X, being subjected to a compressive load. I'm trying to find the correct formula for this. Any help is greatly appreciated.

Deformation under an axial load is PL/AE.
P is the applied compressive force at the end in a direction perpendicular to the end, A is the Area of the cross section, E is the modulus of Elasticity of the material, and L is the total length of the cylinder. Assumed to be an elastic material with a linear stress/strain relationship. The cylinder will deform under any load, so I’m not sure what deformation or load you are looking for. The formula gives the total compressive deformation of the full length cylinder.

PhanthomJay said:
Deformation under an axial load is PL/AE.
P is the applied compressive force at the end in a direction perpendicular to the end, A is the Area of the cross section, E is the modulus of Elasticity of the material, and L is the total length of the cylinder. Assumed to be an elastic material with a linear stress/strain relationship. The cylinder will deform under any load, so I’m not sure what deformation or load you are looking for. The formula gives the total compressive deformation of the full length cylinder.
That formula would work for this application?

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Try this. It's from Formulas for Stress and Strain, 5th Edition, by Roark and Young. That book, or any of the older or newer editions, is THE source for these types of calculations.

las59036
@las59036 Are you talking about just an applied load or an impacting load?

las59036 said:
That formula would work for this application?
Oh, no, based on your sketch, I misread your problem. Looks like you are providing a clamping force like a vice load, a compressive loading all around the cylinder of exposed length X? If the clamping force is F, the clamping bearing pressure on the cylinder is F/(X(d)), where d is the diameter of the cylinder. The cylinder will deform (squeeze)under any load, but it takes a large load to deform it significantly (crushing, beyond yield stress). You need experimental data or a Finite element analysis to determine how much deforming under what load. It depends heavily on the material type. Electric connections of cable are often formed this way by squeezing the splice joint over the wire (with die castings), and depending on the diameter, if steel, you are talking in some cases 60 tons of force American for an inch diameter cable, using a hydraulic press. Much less for small diameter stuff.

las59036 and Dr.D
This is a contact stress problem, and as such will depend upon the material properties of the support (Young's Modulus, Poisson's Ratio) as well as that of the cylinder.

las59036 and PhanthomJay
PhanthomJay said:
Oh, no, based on your sketch, I misread your problem. Looks like you are providing a clamping force like a vice load, a compressive loading all around the cylinder of exposed length X? If the clamping force is F, the clamping bearing pressure on the cylinder is F/(X(d)), where d is the diameter of the cylinder. The cylinder will deform (squeeze)under any load, but it takes a large load to deform it significantly (crushing, beyond yield stress). You need experimental data or a Finite element analysis to determine how much deforming under what load. It depends heavily on the material type. Electric connections of cable are often formed this way by squeezing the splice joint over the wire (with die castings), and depending on the diameter, if steel, you are talking in some cases 60 tons of force American for an inch diameter cable, using a hydraulic press. Much less for small diameter stuff.
Yeah, the diameter is about 0.1 inches and the material being nickel. Would the formula that you listed give me an idea of the load required?

JBA said:
@las59036 Are you talking about just an applied load or an impacting load?
An applied load that will cause a deformation.

Elastic (comes back if you remove the force) or plastic (stays squashed) deformation? Do you want to know how much load to make it yield, do you want to calculate a spring constant, do you want to squash it flat, or do you want to form it around something? Do you have accurate mechanical properties for the material?

las59036 said:
An applied load that will cause a deformation.
An infinitely small load will cause an infinitely small elastic deformation, so the simple answer is "an infinitely small load".

The formulas in Roark apply to elastic deformation. There are techniques for calculating plastic deformation, but they are more difficult than elastic deformation. If you want to form it around something, hand calculations are not practical. For that, you need FEA. And that FEA is not the simple linear static FEA that most people are familiar with, it is an advanced package with large deformation, plasticity, and contact.

Is it a "knife-edge" loading, or is the load distributed over a small well-defined angle of the circumference?

jrmichler said:
Elastic (comes back if you remove the force) or plastic (stays squashed) deformation? Do you want to know how much load to make it yield, do you want to calculate a spring constant, do you want to squash it flat, or do you want to form it around something? Do you have accurate mechanical properties for the material? An infinitely small load will cause an infinitely small elastic deformation, so the simple answer is "an infinitely small load".

The formulas in Roark apply to elastic deformation. There are techniques for calculating plastic deformation, but they are more difficult than elastic deformation. If you want to form it around something, hand calculations are not practical. For that, you need FEA. And that FEA is not the simple linear static FEA that most people are familiar with, it is an advanced package with large deformation, plasticity, and contact.
Plastic deformation, I want to squish it flat to a thickness of about .05 inches. The original diameter is 0.1 inches. Basically, I want to form it by compressing the cylinder from the top and bottom with a flat surface of maybe a steel square block. I have the mechanical properties of the steel and nickel as well. I was hoping to find a formula that could at least give me an estimate of the amount of force I would need to do this, since I do not have any FEA resources at my disposal.

In my judgment, a contact problem like (with varying contact region) is pretty difficult and would require FEA.

Chestermiller said:
In my judgment, a contact problem like (with varying contact region) is pretty difficult and would require FEA.
Yeah, a very small section of the cylinder would be under a compressive load. FEA software is not at my disposal, I was hoping to find a formula to estimate the required force to flatten the section of the cylinder.

Dr.D said:
This is a contact stress problem, and as such will depend upon the material properties of the support (Young's Modulus, Poisson's Ratio) as well as that of the cylinder.
Would you happen to know of a formula that could give me an estimate of the force needed for the deformation?

For flattening a wire, I think you will have more results with an impact loading delivered via a hammer blow. I've flattened lots of wires that way without worrying for a moment how much force with involved. If you hit it once and it is where you want it, you're done. If not, hit it again.

jrmichler, JBA and PhanthomJay
You can analyze forever (paralysis by analysis), or you can do like @Dr.D suggested and just hit it with a hammer.

There are procedures for hand calculating plastic deformation, but they only work for simple shapes. And the calculation is not simple. Even with FEA, you would need to run an experiment to verify the assumptions in the FEA calculation. If you need to run the experiment anyway, just skip the analysis and do the experiment first. Then you will know.

If you want to use a press instead of a hammer, go to Harbor Freight and get one of these:

It is powerful enough to squash a short cylinder 0.1" diameter. And cheaper than spending the time to try to calculate a solution.

JBA and PhanthomJay
Would you be willing to accept the result for a square cross section as an upper bound to the force for a circular cross section?

Since you want the equation, here goes. These are from Manufacturing Processes for Engineering Materials, by Serope Kalpakjian. First, we take a simplified look at the process. The view is down the length of the cylinder, and they simplify by assuming a rectangular wire. Notice how friction between the die and the object causes a rectangular object to get rounded sides. Also note that a cylinder compressed to half diameter will look very much like a rectangle similarly compressed.

Next, we look at the pressure distribution between the die and the object. The pressure distribution is an exponential curve if there is sliding friction between the die and the object, and triangular if there is no sliding (the object sticks to the die).

Finally, they make things easier by reducing the equations to a pair of charts. A long (length several times the diameter) wire is the plane strain case on the left.

The left chart is for rectangular specimens, but by the time your wire is squashed to half diameter, this chart will be close enough. You only need to calculate the pressure at the point of maximum squish because that will the maximum pressure and force. You will still need to do the experiment...

JBA
jrmichler said:
Since you want the equation, here goes. These are from Manufacturing Processes for Engineering Materials, by Serope Kalpakjian. First, we take a simplified look at the process. The view is down the length of the cylinder, and they simplify by assuming a rectangular wire. Notice how friction between the die and the object causes a rectangular object to get rounded sides. Also note that a cylinder compressed to half diameter will look very much like a rectangle similarly compressed.
View attachment 249178

Next, we look at the pressure distribution between the die and the object. The pressure distribution is an exponential curve if there is sliding friction between the die and the object, and triangular if there is no sliding (the object sticks to the die).
View attachment 249179

Finally, they make things easier by reducing the equations to a pair of charts. A long (length several times the diameter) wire is the plane strain case on the left.
View attachment 249180
The left chart is for rectangular specimens, but by the time your wire is squashed to half diameter, this chart will be close enough. You only need to calculate the pressure at the point of maximum squish because that will the maximum pressure and force. You will still need to do the experiment...
What would you think about neglecting friction with the platens? It would always be possible to put a low viscosity oil in between. In the contact problem with a horizontal circular cylinder, the friction effect might be less because of the geometry. Doing a horizontal rectangular cylinder without friction would be much simpler analytically.

Lubrication in a forging process such as this is almost always in the boundary lubrication regime, although an extremely viscous lubricant might get into the mixed regime. In that case, the viscosity would be so high that viscous drag would need to be considered. An excellent book on the subject is Tribology in Metalworking: Friction, Lubrication and Wear by John Schey. It's out of print, but available used from Amazon.

Lower pressure metal forming processes, such as sheet metal forming, can actually switch between mixed and boundary lubrication regimes during the forming process. Check out SAE Paper 950698 for some experimental data showing exactly that.

jrmichler said:
Try this. It's from Formulas for Stress and Strain, 5th Edition, by Roark and Young. That book, or any of the older or newer editions, is THE source for these types of calculations.

These formulas are for thin curved beams (or otherwise circular rings). The OP asks about solid cylinder. In this case different equations from Roark's apply (chapter 14, table 14.1, case 2a. Cylinder on a flat plate): $$\sigma_{max}=0.591 \sqrt{\frac{pE}{D}}$$ and $$\Delta D=\frac{4p(1-\nu^{2})}{\pi E} \left( \frac{1}{3}+ ln \frac{2D}{2.15 \sqrt{\frac{pD}{E}}} \right)$$ where: ##p=\frac{P}{L}## - load per unit length, ##P## - force, ##L## - cylinder length, ##E## - Young's modulus, ##D## - cylinder diameter, ##\nu## - Poisson's ratio.

Of course these formulas are for elastic region only. If you want to include nonlinearity you will have to use FEA.

## 1. What is a force calculation?

A force calculation is the process of determining the amount of force required to cause a specific deformation or change in an object. This calculation is based on the physical properties of the object, such as its size, shape, and material composition.

## 2. How is force calculated?

Force is calculated using the formula F = ma, where F is the force, m is the mass of the object, and a is the acceleration. In the context of deformation of a cylinder, the force calculation would also take into account the material's elasticity and the amount of deformation desired.

## 3. What factors can affect the force calculation for a cylinder?

The force calculation for a cylinder can be affected by various factors, including the material of the cylinder, its dimensions, the magnitude and direction of the applied force, and the environment in which the cylinder is placed.

## 4. How can I determine the amount of deformation in a cylinder?

The amount of deformation in a cylinder can be determined by using the formula d = FL/AE, where d is the deformation, F is the force applied, L is the length of the cylinder, A is the cross-sectional area, and E is the modulus of elasticity of the material. This formula takes into account the force applied and the physical properties of the cylinder.

## 5. What are some applications of force calculations for cylinder deformation?

Force calculations for cylinder deformation have various applications, including engineering and design, material testing, and structural analysis. It can also be used in industries such as automotive, aerospace, and construction to ensure the durability and safety of cylindrical structures and components.

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