If, for now, we look at the “classical worst case,” or the outer fiber stress, y is 5. Then we can plot the actual stress as a function of position. Here we will look at only one half of the length of the beam. The other side is symmetrical (Fig. 2).
We can see that the stress is zero directly above the support at the end. The stress is highest directly under the load. Assuming there is no yielding, there will be a linear increase of stress with distance as we approach the position in dead center. Note also that if the location of interest is above the neutral axis, the stress will be compressive. If it is below the neutral axis, it will be tensile. This is commonly addressed by assigning one of the stresses (usually compressive) a negative sign. We won’t worry about this any more than to say that you have to look at the absolute values of these normal stresses when comparing to the strength values, to determine if the stress has exceeded the strength.
Now, if we assume that the yield strength of the material is over 72,000 psi, we will have the case where there is no yielding. But what if the yield strength is only 50,000 psi? See Figure 3. Now we can see that the outer fiber stresses exceed the strength for the material at all positions between approximately 83 and 120 inches. That goes for the far side of the center of the length as well. So, 120-83=37 inches, times 2 (for both ends of the beam), or 74 of the 240 inches of length have been at least yielded at the surface layers – top and bottom.
Will the beam totally collapse in this case? How can we determine if this is the case? Well, for starters, we can look at how deep the yielded condition goes, at the “worst” position (directly under the load). For this, we go back to our equation, Stress = 6Pxy/h4, but now we are going to keep x at 120 inches, and let y, the distance from the center of the height of the beam, vary instead (Fig. 4).
Here we can easily see that if we have the 50 ksi yield strength material, the 1.5 inches of thickness nearest the top and bottom surfaces (again, we have symmetry, this time upper and lower around the horizontal divider plane) have exceeded the yield strength of the material (Fig. 5).
So, there is the inner core, 7 inches thick that has NOT yielded! Therefore, as a gross oversimplification that you should use conceptually to understand what is happening, and not assume you can do a similar calculation to make a safely designed object (because this would be a bad idea), the beam is deflected, but since there is an unyielded core, if the load were removed, it would partially return toward its original shape.
Next time, we will bring case hardening more fully into the discussion.