# Dynamic Fracture of rod

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1. Apr 16, 2015

### Trevorman

1. The problem statement, all variables and given/known data

Brittle materials as glass and ceramics have generally a much lower tensile strength than compressive strength. A rigid mass impacts the end of a bar (free-free) of such a material and a piece of the other end drops off. Is there a relation between the length of the piece and the dynamic tensile strength of the material?

Clarification
A pressure wave is propagating through the rod, when the wave is reaching the free end at x = L, the wave reflects and becomes a tensile wave.

2. Relevant equations

$\sigma =E \frac{\partial u}{\partial x}$
E - Youngs modulus
$\sigma$ - stress
u - displacement

3. The attempt at a solution
Newtons second law

$\sigma_0 A - M * \frac{\partial^2 u }{\partial t^2}=0$
M - Mass of rod
$\frac{\partial^2 u }{\partial t^2}$ -acceleration of the particles
$\sigma_0$ - Initial stress

Don't really know how to tackle this problem and where to begin.

2. Apr 16, 2015

### Staff: Mentor

Your Newton's 2nd law equation is not really the correct version for describing the propagation of a wave along the rod. Do a differential force balance on the section of the rod between x and x + Δx, taking into account the mass times acceleration of the material within this segment of the rod, and the tensile forces acting on the two ends of the segment (which differ from one another). The tensile force at each end is equal to the stress on that end times the cross sectional area. Then express each stress as Edu/dx at that location. Let's see what you come up with.

Chet

3. Apr 17, 2015

### Trevorman

Hi thank you, this is my result from your suggestion. Is this correct?

$\sigma A - (\sigma + \Delta \sigma )A = A \Delta x \rho \frac{\partial^2 u}{\partial t^2}$
$-E\frac{\partial u}{\partial x}A =A\Delta x \rho \frac{\partial^2 u}{\partial t^2}$

So the piece that is flying off $\Delta x$?

4. Apr 17, 2015

### Staff: Mentor

This is not quite right. The final differential equation should read:

$$\frac{E}{\rho}\frac{\partial ^2 u}{\partial x^2}=\frac{\partial ^2 u}{\partial t^2}$$

This equation is also satisfied by the strain ε.

Now for your problem. I have no idea how to get the answer fundamentally. Just can't see through how to do it. But, the above differential equation must be the starting point.

However, given the nature of the question, we can certainly guess the answer using dimensional analysis. The only length scale in the problem is the length of the rod L. So the length that breaks off, l, must be proportional to L. Also the length that breaks off must be proportional to the ultimate strength. So, dimensionally, the only equation that makes sense to me is:

$$l=L\frac{σ_u}{E}$$

But, how this can be obtained from the fundamental analysis of the problem evades me.

Chet

5. Apr 20, 2015

### Trevorman

Thank you for helping me Chet!

I am familiar with the Wave equation but I cannot figure out how to setup the problem to get out the length of the piece, it is a tricky question. I will come back when I have more information of how the problem can be solved, been scratching my head over this for a while.

6. May 4, 2015

### Trevorman

Okey so I have some additional information now.

When a "stiff" mass hits a rod a exponentially decaying pressure-wave is formed. The pressure wave has a front with the size -zV. The wave propagates forward through the bar and when it is reflected at the free end it replace the signs and become a tension wave which propagates backwards. The rod is then both the reflection and the "tail" of the pressure wave. Net drag force is the difference between the two and if fracture occur in a particular position, you can calculate the tension there (ie, failure stress). It will depend on how long the broken part is.

where z is the impedance and V is the velocity of the stiff mass hitting the rod.
$z = \frac{AE}{c}$
c is the velocity of the propagation speed of the wave, and depends on which way the wave is travelling. $c=\sqrt{\frac{\rho}{E}}$, where E is $E_p$ when it is a pressure wave and $E_t$ when there is a tension wave.

The solution for the wave equation is the sum of forward and backward waves.