Problem 3.8.9^∗

Solution

In the statement, the tensile strength $\sigma^*$ has been relabeled. In this problem, tensile strength means strength under tensile stress.

Diagrams

We begin by looking at the stress diagrams (changes in pressure) after the forward traveling wave (see the figure for the problem) touches the free boundary of the rod, toward which the positive half-wave of stress was initially moving. In the diagrams, this free boundary coincides with the vertical axis, on which we plot the values of the stress $\sigma$.

Suppose the forward traveling wave has entered the boundary by a distance less than $L/2$. Here is the diagram (we use the method from problem $3.8.6$ in terms of stress).

Analysis of Fig. D1 shows that when the forward wave has entered the boundary by a distance less than $L/2$, there is no tension in the rod and no spalling can occur (spalling is fracture at a point where the tensile strength is exceeded).

Clearly, if the forward traveling wave has entered the boundary by a distance $L/2$, there is no stress in the rod (draw the diagram to check).

Now suppose the forward traveling wave has entered the boundary by a distance greater(!) than $L/2$. Here is the diagram.

Analysis of Fig. D2 shows that when the forward wave has entered the boundary by a distance greater than $L/2$, the rod contains only points under tension (negative stress), and at some point the maximum tensile stress $\sigma^*$ may be exceeded in magnitude.

Spalling point

The cases corresponding to diagram D2 make it clear that the maximum tensile stress is always below the leftmost point of the forward traveling wave, below point $A$ in diagram D2. Spalling will occur at this point.

Someone will figure out that after the moment when the forward wave has entered the boundary by $L/2$, the stress below point $A$ follows a reversed sine wave: point $A$ corresponds to the stress value in the backward traveling wave below it.

If the rest of the solution is tied to point $A$, let us switch to a reference frame tied to point $A$. In this case, the stress below it follows a reversed sine wave (after tension begins to appear in the rod). Then the backward wave passes point $A$ at a speed of $2c$, and if we virtually extend the sinusoidal profile of the backward wave, the value below point $A$ varies periodically with period $T = 2L/(2c)$.

Thus, the law of variation of the tensile stress after tension appears

$$\sigma = -\sigma_0\sin\left(\frac{2\pi}{T}t\right).$$

Hence, taking into account $\sigma^* < 0$, the time $\tau$ after tension appears at which the spalling point appears

$$\tau = \frac{L}{2\pi c}\arcsin\left(\frac{\sigma^*}{\sigma_0}\right).$$

The distance from the boundary at which the maximum tension will be reached (see D2)

$$l = \frac{L}{2} - c\tau$$

and

$$\boxed{l = \frac{L}{2}\left(1 - \frac{1}{\pi}\arcsin\frac{\sigma^*}{\sigma_0}\right).}$$

For $\sigma_0\gg\sigma^*$:

$$l \approx \frac{L}{2}\left(1 - \frac{1}{\pi}\frac{\sigma^*}{\sigma_0}\right) \approx \frac{L}{2}.$$

For $\sigma_0\approx\sigma^*$:

$$l \approx \frac{L}{2}\left(1 - \frac{1}{\pi}\frac{\pi}{2}\right) \approx \frac{L}{4}.$$