Deformation/fracture of material on impact

In summary, the conversation discusses calculating the deformation of a material upon impact by a mass moving along its normal. The given information includes the mass, velocity, contact area, thickness, and material constants. Equations for stress, strain, and Hooke's law are provided, along with an example calculation. However, some clarifications are needed, such as taking into account the thickness of the plate in the stress equation and using the correct symbol for original length in the strain equation. The increasing deformation with increasing thickness is expected behavior, and determining the breaking point of the plate requires additional information about the material's properties.
  • #1
ITAmember
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I want to calculate the deformation of a plate of a material upon being struck by a mass moving along the plate's normal, and if the impact force is sufficient, if the plate will break or not. The given information can be assumed to be: mass, velocity, and contact area of the mass, and the thickness of the plate along with necessary material constants (e.g. Young's modulus).

For the sake of an example, let's assume a 100 kg hammer moving at 10 [itex]\frac{m}{s}[/itex] with a contact area of 0.01 [itex]m^{2}[/itex] strikes a 0.1 m thick steel plate, of which Young's modulus is equal to 200 GPa.

Some equations:

Stress
[itex]\sigma=\frac{F}{A}[/itex]

Strain
[itex]\tau=\frac{\Delta{L}}{L_{0}}[/itex]

Hooke's law
[itex]E=\frac{stress}{strain}=\frac{\sigma}{\tau}=\frac{{F}L_{0}}{{A}\Delta{L}}[/itex]

To calculate [itex]\Delta{L}[/itex] I need force, which AFAIK is non-trivial to calculate. I found this equation

[itex]F=m\frac{v^{2}}{2\Delta{L}}[/itex]

Solving Hooke's law for F, substituting, and solving for [itex]\Delta{L}[/itex] yields the following

[itex]F=\frac{EA\Delta{L}}{L_{0}}[/itex]
[itex]\Delta{L}=mL_{0}\frac{v^{2}}{2EA}[/itex]

This finds the deformation of the material using the given information, and with substituting the example the deformation is 0.0005m, which sounds reasonable for a 0.1m steel plate. However, I noticed that as the thickness of the plate increased, so did the deformation. I suppose this makes sense given the equation for strain, but seems counter-intuitive and leads me to wonder if I am doing this wrong. In addition, I have no idea where to start with the breaking point of the plate and how much force that requires.

All help is appreciated.
 
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  • #2




Thank you for your post. It seems like you have made some good progress in your calculations so far. However, there are a few things that I would like to clarify and suggest for your calculations.

Firstly, in your equation for stress, you have correctly used the force and contact area of the mass, but you have not taken into account the thickness of the plate. The correct equation for stress would be \sigma=\frac{F}{A\Delta{L}}. This takes into account the fact that the force is distributed over the entire area of the plate, not just the contact area.

Secondly, in your equation for strain, you have used the wrong symbol for the original length of the plate. It should be L_{0}, not \Delta{L}. This is the original length of the plate before it is deformed by the impact force.

Now, to address your concern about the increasing deformation with increasing thickness of the plate - this is actually expected behavior. As you mentioned, the equation for strain shows that the deformation is directly proportional to the thickness of the plate. This means that a thicker plate will deform more under the same impact force compared to a thinner plate. This is because the thicker plate has more material to stretch and deform.

As for the breaking point of the plate, this is a more complex issue and cannot be determined solely based on the given information. The breaking point of a material depends on various factors such as its strength, toughness, and the type of loading it is subjected to. In this case, we would need to know the material's ultimate strength and other properties to determine if the impact force is sufficient to cause the plate to break.

I would suggest consulting a materials science textbook or a structural engineer for more information on determining the breaking point of a material. I hope this helps in your calculations. Good luck with your project!
 

FAQ: Deformation/fracture of material on impact

What is deformation and fracture of material on impact?

Deformation and fracture of material on impact refers to the changes that occur in a material when it is subjected to an external force or impact. This force can cause the material to change its shape, size, or position, and in extreme cases, it can lead to the material breaking or fracturing.

What causes deformation and fracture of material on impact?

The main cause of deformation and fracture of material on impact is the application of a force that is greater than the material's strength or ability to withstand. This can be due to sudden impact, high-speed collisions, or other external forces such as explosions or extreme temperatures.

How does the type of material affect its deformation and fracture on impact?

The type of material plays a crucial role in its deformation and fracture on impact. Some materials, such as metals, have high strength and can withstand greater forces before deforming or fracturing. Other materials, like glass or ceramics, are more brittle and can fracture easily under impact.

What factors influence the severity of deformation and fracture on impact?

Several factors can influence the severity of deformation and fracture on impact, including the type and strength of the material, the velocity and direction of the impact, and the temperature and environment in which the impact occurs. These factors can also interact with each other, making it challenging to predict the exact outcome of an impact on a material.

How can we prevent or minimize deformation and fracture of material on impact?

Preventing or minimizing deformation and fracture of material on impact can be achieved through various methods, such as using stronger materials, designing structures to absorb or distribute impact forces, and applying protective coatings or materials. It is also essential to consider the potential impacts and forces a material may encounter in its intended use during the design and testing phases.

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