4 ms·

>Also "compressive strength" is not really a thing. This is true, but neither is "tensile strength" really a thing for the same reason. A simple uni-axial ten

>Also "compressive strength" is not really a thing. This is true, but neither is "tensile strength" really a thing for the same reason. A simple uni-axial tensile test is not really uni-axial, but a combination of orthogonal normal stresses that ultimately results in shear failure. I've heard it said that "all failure is shear failure", and I think that's true. When you look closely at the ductile fracture surface of a ruptured tensile specimen, the characteristic "cupping"[0] appearance consists of various surfaces at 45 degrees from the direction of the applied load. Principle shear stresses are always oriented 45 degrees from the principle normal stresses. [0]https://upload.wikimedia.org/wikipedia/commons/1/1b/DuctileFailure.jpg https://upload.wikimedia.org/wikipedia/commons/1/1b/DuctileF...

hwillis 1y ago
> This is true, but neither is "tensile strength" really a thing for the same reason. Well, not the same reason. Shear strength is in many ways more important because failure usually propagates from an origin like a rip- by shearing. But fundamentally, tensile strength and shear strength are both much more empirical than compressive strength. Tensile strength can be used with ~99% accuracy on a wire that is microns long or miles long. If you double the relative width of a testing sample it will dramatically change the measured compressive strength even without buckling.
throw7383753 1y ago
You could take that further and say that shear does not exist. It's a construct we have created to deal with tension failures in other planes. Everything basically comes down to a truss model. The failure you idealize as a shear stress is really a failure of a tension element in a miniature truss. This is true in bolt shear, scissors, etc. This is why the code forces you to use strut and tie model deep concrete beams. As much as you may want to idealize a shear stress, really what's happening is the beam is arching over the span.
- hydrogen7800 1y ago
  Ha, yes! I've even heard some argue that stress isn't even real! Whatever "real" means. Nothing is real; it's models all the way down.
  - bozhark 1y ago
    All models are wrong… Some are useful
rda2 1y ago
For those who are interested in learning more: "All failure is shear failure" - this is a simple explanation of Tresca's Yield Criterion. For materials with higher compressive than tensile strength, the equivalent is the Mohr-Coulomb failure criterion. https://en.wikipedia.org/wiki/Material_failure_theory https://en.wikipedia.org/wiki/Material_failure_theory
- kurthr 1y ago
  For a simple example try twisting a ductile aluminum bolt (or clay), twisting a brittle piece of chalk (or concrete), or twisting a composite (twig). They all fail differently (and the first two at 45deg to each other. Mohr's circle is interesting, and fatigue failure more interesting still. https://youtu.be/1YTKedLQOa0?t=533 https://youtu.be/1YTKedLQOa0?t=533

hwillis 1y ago

> This is true, but neither is "tensile strength" really a thing for the same reason. Well, not the same reason. Shear strength is in many ways more important because failure usually propagates from an origin like a rip- by shearing. But fundamentally, tensile strength and shear strength are both much more empirical than compressive strength. Tensile strength can be used with ~99% accuracy on a wire that is microns long or miles long. If you double the relative width of a testing sample it will dramatically change the measured compressive strength even without buckling.

throw7383753 1y ago

You could take that further and say that shear does not exist. It's a construct we have created to deal with tension failures in other planes. Everything basically comes down to a truss model. The failure you idealize as a shear stress is really a failure of a tension element in a miniature truss. This is true in bolt shear, scissors, etc. This is why the code forces you to use strut and tie model deep concrete beams. As much as you may want to idealize a shear stress, really what's happening is the beam is arching over the span.

hydrogen7800 1y ago
Ha, yes! I've even heard some argue that stress isn't even real! Whatever "real" means. Nothing is real; it's models all the way down.
- bozhark 1y ago
  All models are wrong… Some are useful

rda2 1y ago

For those who are interested in learning more: "All failure is shear failure" - this is a simple explanation of Tresca's Yield Criterion. For materials with higher compressive than tensile strength, the equivalent is the Mohr-Coulomb failure criterion. https://en.wikipedia.org/wiki/Material_failure_theory https://en.wikipedia.org/wiki/Material_failure_theory

kurthr 1y ago
For a simple example try twisting a ductile aluminum bolt (or clay), twisting a brittle piece of chalk (or concrete), or twisting a composite (twig). They all fail differently (and the first two at 45deg to each other. Mohr's circle is interesting, and fatigue failure more interesting still. https://youtu.be/1YTKedLQOa0?t=533 https://youtu.be/1YTKedLQOa0?t=533