- #1

- 13

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter tina-duncan
- Start date

- #1

- 13

- 0

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

- #2

- 36

- 5

- #3

Chestermiller

Mentor

- 21,062

- 4,643

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

- 3

- 0

- #5

Chestermiller

Mentor

- 21,062

- 4,643

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

- 3

- 0

Thanks

- #7

Chestermiller

Mentor

- 21,062

- 4,643

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.

Thanks

Share: