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Given a distribution of shear force on a uniform rod, how do we determine whether the rod will break? And if it does break, at which point is it most likely to break? Is it at the point where shear force is at maximum?

These questions come from my thinking of the following question, so it would be helpful if anyone can contribute to that as well. A piece of chalk (idealised as a 1D object) falls on the ground. Assuming proper parameters are known, how to determine if the chalk will break or not?

Yu Lu
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Let's take the example of a mill-rolled steel beam. We load it in some well-defined way and then ask the question, for a uniform stress distribution, where will it fail along its length?

From an engineering standpoint the question has no answer because our usual stress-distribution models do not contain any material or manufacturing flaws.

In fact, the reason there is an entire field of engineering called nondestructive testing is that invisible flaws are what trigger early failures, which process is unpredictable without knowing in advance where the flaws are located in the part.

So we perform an ultrasound, x-ray or eddy current scan of the beam's interior and/or a magnaflux or dye-penetrant inspection of the beam surface, and find the slag inclusions or surface cracks which will trigger failure first.

niels nielsen
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It is certainly possible for a rod to fail in simple shear—that’s what a shear pin is designed to do, for example. But I’d say the more common failure is likely due to bending—when you want to break a stick without tools, you grip it at the ends and bend it over your knee, you don’t grab it in the middle with both hands and try to dislocate it laterally.

This type of failure will occur where the bending moment (the integral along the length of the rod of the shear stress) is highest. If you know the moment of inertia of the cross-section of the rod and its material properties, you can determine the bending moment which will cause failure.

Ben51
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