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I have a thick dowel (43mm diameter) of a certain timber (Tasmanian Oak), I want to use it for a chin-up bar (of all things), but I'm unsure of its strength.

The datasheet for this timber supplies a Modulus of rupture, $\sigma$ (117 MPa).

However, I see that Modulus of rupture is calculated with rectangular beams.

$$\sigma = \frac{3FL}{2d^2b}$$

$F$ - Downwards force; $L$ - length; $d$ - depth; $b$ - breadth. Assuming a 3-point model.

How can I translate that rectangular index of flexural strength into something useful for a cylinder?

I tried being clever, replacing a rectangular area $A$ with something circular, i.e. in the denominator $2d^2b$ -> $2d .d.w$ -> $2dA$; $A = \pi r^2$.

$$\sigma = \frac{3FL}{2\pi (\frac{d}{2}) ^2 d}$$

But thinking about it, that's probably quite false.

  • The force goes from a "line" across the top of the usual rectangular beam to a point on the circumference of the dowel
  • Translating down through the circular section, that load will be resisted/supported by adjacent fibers of the wood,
  • that's lateral or radial, so shear forces are much higher through the cross section (surely?)
  • most reinforced at the "equator"
  • decreasing down to its narrowest point (the other "pole" underneath), exactly at the place tensile stress is greatest.

So a quick substitution of cross-sectional area of a cylinder will not do.

Quite stumped. But not a physicist. Can you help?

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Looks like the modulus of rupture (flexural strength) for a circular cross-section is $\frac{F L}{\pi R^3}$ (https://en.wikipedia.org/wiki/Three-point_flexural_test)

EDIT(1/28/2021): So if the length of the bar is 0.5 m, it can withstand approximately 7300 N, so it can withstand a hockey team of 6:-)

EDIT(1/28/2020): Of course, dynamic load also needs to be taken into account.

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  • $\begingroup$ Ah, that formula wasn't on the Modulus of rupture Wikipedia page. I didn't think to look at the test. Thanks! [goes off to make strange proposal to a hockey team...] $\endgroup$ Commented Jan 30, 2021 at 1:04

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