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How would one calculate the amount of weight a steel bar could hold before breaking?

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Apologies for the terrible diagram. So if I had a steel bar of the length 18in, and the cross-section with a diameter of 1 inch, how much weight could it take before breaking? (By break, I mean assume one end was stuck to the ground and the other balancing a large weight on the top, "breaking" means bending or snapping in such a way the weight falls.)

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There are two possible modes of failure here.

  1. Compressive yield, which occurs when the stress in the rod exceeds the yield limit. The average stress in the rod would be $$\left. \sigma = \frac{F}{A} \;\right\} F \leq \sigma_{\rm yield\,} A$$ Where $F$ is the axial load applied, and $A=\frac{\pi}{4} d^2$ is the cross-sectional area.

    The yield limit of the bar is a material property that you have to look up.

  2. Buckling failure, where the rod bows due to the axial load. The typically happens at a lower force level than compressive yield. $$ F \leq \frac{\pi^2 E I}{(K \ell)^2} $$ Where $E$ is the modulus of elasticity, $I = \frac{\pi}{64}d^4$ is the area moment of the section, $\ell$ is the free length of the rod, and $K$ is a constant that depends on the end-supports. See wikipedia.

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  • $\begingroup$ Would yield necessarily cause the weight to fall, as the original poster specifies? Yield implies only that the bar wouldn't return to its original length if the weight were removed. I think a more appropriate stress level would be the ultimate strength under compression. At this point, the bar would shear apart. $\endgroup$ – Chemomechanics Nov 17 '18 at 1:47
  • $\begingroup$ @Chemomechanics - yielding means plastic deformation, and I think this will lead to an unstable situation where the rod is starting to bend more and more until the weight falls off. After yielding there is no way to keep the rod straight anymore. $\endgroup$ – ja72 Nov 17 '18 at 2:46
  • $\begingroup$ Yielding doesn't imply instability, especially for a ductile metal that hardens upon yielding. What you're describing seems to be the end of the stress-strain curve (when the material comes apart), not the yield point. (One can verify via an online search that compressive stress-strain curves for metals generally extend far past the yield point.) $\endgroup$ – Chemomechanics Nov 17 '18 at 5:41
  • $\begingroup$ What I was implying (but not saying) is that it is highly unlikely the load remains exactly axial after yielding, adding to the stresses due to a slight bending moment, which then causes more yield and amplifies the eccentricity causing a runaway situation. But you are right, if the material is subject to work hardening it will tend to stabilize things. $\endgroup$ – ja72 Nov 17 '18 at 17:14
  • $\begingroup$ And I agree that without work hardening, eccentricity is much more likely to be a runaway process, as you describe. $\endgroup$ – Chemomechanics Nov 17 '18 at 17:16

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