Calculating stress as a result of an impact?

In summary, the maximum stress on an object upon impact can be calculated by considering the impulse force and the speed of sound in the material. This can be applied to a simple example of a solid linearly elastic cylinder hitting a rigid wall, where a compression wave and a recovery wave travel along the cylinder at the speed of sound in the metal. By applying Hooke's law and accounting for kinematics, the stress in the compressed region, as well as the total time of contact and force magnitude, can be determined analytically. The stress and strain return to zero after the object loses contact with the wall.
  • #1
I'm wondering if there is a calculation that can be used to work out the maximum stress on an object upon impact.

Say for example an object strikes a rigid wall, how would one go about calculating the stress on that object?
 
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  • #2
If you had a ball hit a rigid wall, like in your example, there is an impulse force associated with that. If you can calculate the impulse force then you should be able to calculate your stress just like any other scenario.
 
  • #3
tina-duncan said:
I'm wondering if there is a calculation that can be used to work out the maximum stress on an object upon impact.

Say for example an object strikes a rigid wall, how would one go about calculating the stress on that object?

This is a complicated stress analysis problem involving a moving boundary and time-dependent deformations. You can get a feel for how this might work by considering a solid linearly elastic cylinder hitting a rigid wall head-on. When the front end of the cylinder hits the wall, it stops, but the back end of the cylinder continues moving. A compression wave starts out at the contact surface, and travels backward along the cylinder. Within the compression region, the velocity is zero, but further back, the material is still moving at the original velocity. Eventually, the entire cylinder is stopped but is under compressive stress. Then, the compressive stress begins to be relieved at the trailing end of the cylinder first. The front end of the cylinder remains compressed (and motionless), but a new wave starts traveling forward along the cylinder from the rear. In the rear region where the stress is relieved, the material velocity is now opposite to the original velocity and of the same magnitude. Eventually the new dilatation wave reaches the leading end of the cylinder, and the entire cylinder is now traveling backward at the magnitude of the original velocity. At this point, the cylinder loses contact with the wall, and the rebound is complete. You can solve this problem analytically because of the simple geometry if you apply Hooke's law properly and are careful about accounting for the kinematics. This will tell you the stress in the compressed region, as well at the total time of contact and the magnitude of the force during the contact interval.

Chet
 
  • #4
Thanks for your response Chet. I would greatly appreciate it if you would show how to derive the analytical solution for the situation you described. Thanks!
 
  • #5
TryingToThink said:
Thanks for your response Chet. I would greatly appreciate it if you would show how to derive the analytical solution for the situation you described. Thanks!
Incidentally, Welcome to Physics Forums.

I'm not going to go through all the details since that would be too lengthy for this venue. But, I will give you an abridged analysis. The compression wave and recovery waves travel along the cylinder at the speed of sound in the metal during the time that the cylinder is in contact with the wall. From the analysis of the dynamics in conjunction with hooke's law, the speed of sound in the metal is [itex]\sqrt{\frac{E}{ρ}}[/itex], where E is the young's modulus and ρ is the density. If the cylinder is of length L, the amount of time that it is in contact with the wall is [itex]\frac{2L}{\sqrt{\frac{E}{ρ}}}[/itex]. During the contact, the analysis shows that the force is constant. The impulse of the force is equal to the change in momentum. Therefore,
[tex]F\frac{2L}{\sqrt{\frac{E}{ρ}}}=2mv[/tex]
where m is the mass of the cylinder, and v is its approach and rebound speed. The mass m is equal to the cross sectional area A times the length times the density:

m=ALρ

If we substitute this into the previous momentum balance equation, we get:
[tex]\frac{F}{A}=\sqrt{Eρ}v=E\frac{v}{\sqrt{\frac{E}{ρ}}}[/tex]
The left hand side of this equation is the compressive stress at the contact surface (as well as throughout the compression region). The compressive strain in this region is, according to the above equation, given by [itex]\frac{v}{\sqrt{\frac{E}{ρ}}}[/itex]. This is the ratio of the approach speed to the speed of sound in the metal.

I hope this helps.

Chet
 
  • #6
Excellent, thanks Chet. So does this mean that, during the release wave, the rod is in tension with stress of equal magnitude to that during the compression?

Thanks
 
  • #7
TryingToThink said:
Excellent, thanks Chet. So does this mean that, during the release wave, the rod is in tension with stress of equal magnitude to that during the compression?

Thanks
No. Actually, in the release wave region, the stress and strain returns to zero. After the rod loses contact, the stress throughout the rod is zero.
 

1. How is stress calculated as a result of an impact?

Stress is calculated using the formula: stress = force / area. The force is the impact force applied to the object, and the area is the cross-sectional area of the object that is experiencing the impact. This calculation gives the stress in units of force per area, such as N/m^2 or Pa.

2. What factors affect the stress calculation in an impact?

The main factors that affect the stress calculation in an impact are the magnitude of the impact force, the duration of the impact, and the area over which the force is applied. Additionally, the material properties of the object, such as its elasticity and strength, can also affect the stress calculation.

3. How is stress different from strain in an impact?

Stress and strain are related concepts in the field of mechanics. Stress is the force applied to an object per unit area, while strain is the resulting deformation of the object. In an impact, stress is the force of the impact on the object, while strain is the resulting change in shape or size of the object.

4. Can stress as a result of an impact be predicted?

Yes, stress as a result of an impact can be predicted using mathematical models and simulations. These models take into account factors such as the impact force, duration, and object material to estimate the resulting stress. However, experimental testing is still necessary to validate these predictions.

5. What are the potential consequences of high stress as a result of an impact?

High stress as a result of an impact can lead to deformation, damage, or failure of the object. This can have significant consequences, especially in critical structures such as buildings, bridges, or vehicles. Therefore, understanding and accurately calculating stress as a result of an impact is crucial for ensuring the safety and reliability of these structures.

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