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?


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.