Calculating stress as a result of an impact?

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #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
20,118
4,204
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
20,118
4,204
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
20,118
4,204
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.
 

Related Threads on Calculating stress as a result of an impact?

  • Last Post
Replies
1
Views
2K
  • Last Post
3
Replies
56
Views
12K
Replies
1
Views
9K
Replies
6
Views
6K
Replies
3
Views
9K
  • Last Post
Replies
5
Views
118K
  • Last Post
Replies
1
Views
5K
Replies
7
Views
5K
Replies
2
Views
558
Replies
6
Views
22K
Top