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I have been making cylindrical concrete coasters and, for fun, have been seeing if my friend's can snap them. Every subsequent mix design I'll change the mix itself or the dimensions of the geometry to make the coaster stronger so that, hopefully, my friend's can't snap them.

I know that my mixes have high compressive strength and that they can't get much better in terms of maximising strength from that parameter (i.e. they are good mix designs), but I'm unsure about the flexural strength. I want to look at other ways of increasing the flexural strength other than changing the mix design (because iterating mix design testing can take a while and a lot of resources); logical and quick ways.

It makes sense to me that by increasing the thickness or height of the coaster that I will be increasing the flexural strength (i.e. because I'm imagining that it'd be easier to snap a thinner coaster than a thicker one and have experienced this), however, when I look at the formula I see that the reverse is true. The formula I've seen online (i.e from Wikipedia) is as below for a rectangular prism:

https://en.wikipedia.org/wiki/Flexural_strength#/media/File:Flexural_strength.svg

I believe this formula if it were to be adjusted for cylindrical samples would still be for the sample on its side being exposed to flexural stress rather than it sitting upright and being flexed (i.e. imagine breaking a thin cylindrical coaster). You can see from the formula that the thickness parameter is in the denominator, so increasing its value would decrease the theoretical flexural strength... this seems counterintuitive to me.

So, my question is, what would the correct formula be here? My latest coasters are 8 mm thick with a 90 mm diameter. How could I adjust the dimensions to make the flexural strength higher so that my friend can't snap them? He is quite strong but I'm determined to get him on this (without breaking his fingers!). For full clarity, I've added a labelled diagram below where h = height (or thickness or length), F = Force and d = diameter.

enter image description here

Hendrix13
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1 Answers1

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You are reading the symbol $ \sigma \ $ wrong.

It means stress, not strength. Stress means tension/compression force per unit of area. And it makes sense a thicker member experiences less stress versus a thinner member which experiences more stress and will break.

The correct strength formula for a rectangular beam like what you show on your diagram is: $$M_{max}=FL/4$$ $$ \sigma_{max}=\frac{Mc}{I}, \ c=D/2, \ and \ \ I=\frac{BD^3}{12}$$ So $$\sigma_{max}=\frac{3FL}{2BD^2}$$

And the strength of that beam is $$ \frac{\sigma_{max \ of\ conc\ mix}I}{D/2}=\sigma_{max \ of\ conc\ mix}.S $$

  • S section modulus= 2I/D
  • $\sigma_{max}$ stress under this loading
  • $\sigma_{max \ of\ conc\ mix}$ Max tensile/compressive strength of conc mix.

Edit

In this case, your beam's "I" varies proportional to its secants perpendicular to an axis passing through the two supports, say the (D) axis. Because all the other parameters are the same. So the variation in the I is a semi-circle graph, with the greatest I located in the center of the D and the smallest at the two ends. The Moment varies linearly as the famous triangle with the maximum M happening under the center of the beam. $ M_{max}= FD/4$

This means the maximum stress is the same equation as before except we plug in 90 mm for B and we are careful to replace the D in my original answer with "h= 8mm" as per your annotation.

The equation of Maximum stress is the same but the deflection is much less than a prismatic beam and requires a bit of calculation.

kamran
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