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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

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  • $\begingroup$ Concrete does not behave well in tension. $\endgroup$
    – Solar Mike
    Commented Jun 5 at 10:01
  • $\begingroup$ Yes, I'm aware, but we've made some coasters that can snap and some that don't. I want to distinguish them quantitatively. $\endgroup$
    – Hendrix13
    Commented Jun 5 at 11:32
  • $\begingroup$ Perhaps you should check a better reference than wiki... see engineeringtoolbox.com/young-modulus-d_417.html $\endgroup$
    – Solar Mike
    Commented Jun 5 at 12:57
  • $\begingroup$ The reference is to a general well known formula. Regardless, I could not find anything on your link that helps answer this question. $\endgroup$
    – Hendrix13
    Commented Jun 5 at 14:42
  • $\begingroup$ I am confused about the dimensions of the sample in question - 8 mm thick with a 90 mm diameter. A cylindrical beam should have a diameter and length, what does "thick" denote? Also, 8 mm is too short for a beam, is it the height of a cylindrical pilaster? $\endgroup$
    – r13
    Commented Jun 5 at 17:58

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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.

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  • $\begingroup$ Thanks for the clarification and help here with my understanding of the concept but it doesn't define it for an upright cylinder so my question is still unanswered. $\endgroup$
    – Hendrix13
    Commented Jun 6 at 2:36
  • $\begingroup$ @Hendrix13 please edit your question and change the figure to show what you mean. then I will edit my answer to address that. $\endgroup$
    – kamran
    Commented Jun 6 at 7:34
  • $\begingroup$ Thanks. Is my question title "What is the formula to define the flexural strength of an upright cylinder?" not clear enough? If not I can make the edits. $\endgroup$
    – Hendrix13
    Commented Jun 6 at 9:08
  • $\begingroup$ @Hendrix13, No, it's not clear. An upright cylinder will behave like a beam if it is subjected to lateral force in the middle and is supported from top to bottom. It will act like a cantilever beam if supported only from one end. If the ratio of length to diameter is less than 10, it becomes a short column. A scaled diagram showing dimensions, supports, and forces is needed to let me see what you intend to do. $\endgroup$
    – kamran
    Commented Jun 6 at 16:11
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    $\begingroup$ Let's say your concrete tensile strength is 100 psi, ignoring fo the moment that concrete performs poorly in tension and cracks suddenly, when the height of your sample is in the denominator of the stress equation, it means the thicker your sample, the smaller the stress. In other words, if you double the thickness of a sample that was barely stressed near 100 psi, you can load it up 4 times as much as before. $\endgroup$
    – kamran
    Commented Jun 24 at 17:14

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