How does the surface area of a piece of glass (or any general material; sapphire isn't technically glass) correlate with strength? Strength as in resistance to pressure concentrated at a small point?

Intuitively, something tells me that the larger a piece of glass is, the less resistant it will be to pressure at a small point. Take the extreme example. A "piece" of glass that consists of one molecule. Applying pressure to this one molecule won't fracture the molecule. (But then perhaps this example is too extreme).

I ask because of this comparison, done with a sheet of sapphire "glass" that is clearly much larger than the other, competing piece of glass. Would the size of the sapphire give it an inherent disadvantage in this test?

enter image description here

Here is a video of the complete test: https://www.youtube.com/watch?v=u6B-2jTvh1w


Yes, you're right that sample sizes need to be standardized in strength tests. In the picture you included, we see a ring-on-ring strength test. A small ring on top presses against a sample of uniform thickness resting on a larger ring. The loading is (up to second order effects) localized within the disc-shaped area of the sample located above the lower ring. By applying standardized ring diameters and standardized sample thicknesses (here 1 mm), unbiased comparisons are achieved.

  • $\begingroup$ I am not sure that this test is as unbiased as you make it out to be. The larger size of the sample on the left gives the edges some additional stiffness - which will actually reduce the curvature of the sample somewhat, and thus reduce the tensile stresses that would give rise to crack propagation. $\endgroup$ – Floris Aug 19 '14 at 2:14
  • $\begingroup$ @Floris - that is the "second order effects" I was referring to. Note that these go against OP's suspicion that the larger sample is disadvantaged in the test. $\endgroup$ – Johannes Aug 19 '14 at 2:49
  • $\begingroup$ I agree with everything except the assertion these are second order effects - see the mathematics in my answer. Otherwise I agree with you (didn't see your comment before posting my answer - but still, it seems we came to the same conclusion independently). $\endgroup$ – Floris Aug 19 '14 at 3:06

The caption in your image is "strength after abrasion damage test".

Typically, the strength of a material under stress strongly depends on surface defects. A defect (crack, groove) leads to a stress concentration at the tip: this stress concentration allows a material to break (a crack to propagate) with a relatively low applied force. You experience this when you try to open a packet of ketchup: the little notch that has been cut for you makes it easy to tear the packet open. Example of what I'm talking about (from http://www.notchcommunications.com/uploads/files/photo-1.jpg)

enter image description here

The geometry of a crack in the surface has a MAJOR impact on the strength is a subsequent ring on ring test - see some fundamental descriptions at http://www.engineersedge.com/material_science/stress_concentration_fundamentals_9902.htm . In essence, the length of the crack (perpendicular to the direction of stress) and the radius of the crack tip (sharper = weaker) determine the stress concentration. This is a reason that etching a material can sometimes make it stronger: the acid (or other etching fluid) makes the crack tips blunter, and so reduces the stress concentration.

All that is just preliminary to a discussion of the picture drawn. From http://www.nasa-otm.eu/lw_resource/datapool/Publications/J._Europ._Ceram._Soc._Malzbender.pdf , equation 1 is $$\sigma_b = \frac{3F(1+v)}{2\pi\cdot t^2}\cdot \left(\ln\frac{r_2}{r_1}+ \frac{1-v}{1+v}\cdot \frac{r_2^2- r_1^2}{2\cdot r_3^2}\right)$$

This shows that the maximum bending stress $\sigma_b$ for a plate in a dual ring test geometry depends on the radii of the two rings stressing the sample, $r_1$ and $r_2$, but also (weakly) on the radius of the sample itself, $r_3$.

This confirms your suspicion - that final term can increase the stress when the radius of the sample is not much bigger than the outer ring.

If we take $r_1 = 3$, $r_2 = 4$, and $r_3 = 4$, with a Poisson ratio of 0.25, then the last term in brackets evaluates to 0.320 . As you make $r_3$ larger, the maximum stress does in fact get smaller. Here is a simple table:

\begin{array}{|c|c|c|}\hline r_3 & \textrm{factor}&\textrm{fraction}\\ \hline 4& 0.320& 100\%\\ \hline 5 & 0.304 & 95\%\\ \hline 6 & 0.297 & 93\%\\ \hline 7& 0.294 & 92\%\\ \hline 8 & 0.292 & 91\%\\ \hline 9 & 0.291 & 91\%\\ \hline 10 & 0.290 & 90\%\\ \hline \end{array}

My point is that the radius of the disk beyond the testing geometry does have a (small) effect - which is exactly what you were asking about. In the testing geometry as shown, you have a different dimension along one axis than another - in other words, no longer a axis symmetric setup. This makes the analysis a bit more complex, but to first order you can say that the stress along one dimension (where the sample is longer) will be lower than along the other dimension, and the total effect is no more than 5%.

The difference is not very large - and it biases the result such that a larger sample will appear stronger. Given that in the image you showed, the smaller sample is supposed to demonstrate its superior strength, it seems that the bias is against the hypothesis they would like to prove. This makes the conclusion more compelling.

But again I stress - the details of the abrasion test are actually the most important: how the samples were scratched is far more important than the sample dimensions (given they have the same thickness and the same ring geometry). Gorilla glass is much harder than "ordinary" glasss, with a Mohr's hardness of around 9 - close to that of sapphire. This means it scratches less easily - a prerequisite for performing well in this test.

But the real secret behind Gorilla glass is in the residual stresses. As described in http://chemistry.about.com/od/howthingswork/f/What-Is-Gorilla-Glass.htm , after the glass is made it is dipped into a bath of molten potassium. The potassium (K) ions displace the smaller sodium (Na) ions in the surface of the glass, and when the glass cools down, these larger ions give rise to a compressive stress on the surface. This compression help protect the surface from scratching; if also offsets tension during the stress test, and helps prevent crack propagation. And therein lies the secret…

Incidentally, the original Gorilla glass was invented in the '60s and didn't go anywhere - except on race cars which cared about the lightest, strongest glass possible. This changed when Apple needed something like that for the iPhone. The rest is history.

  • $\begingroup$ Statistically, wouldn't a larger pane of material be more prone to abrasion? The two sheets of material were abraded by putting them in a container with other things (car keys, etc.) and the container was rotated for 45 minutes. $\endgroup$ – Dissenter Aug 19 '14 at 3:24
  • 1
    $\begingroup$ You are right - without knowing the details of the abrasion test, I had assumed it was "done properly", with each part of the surface equally likely to be damaged. I would imagine that in the geometry you describe, the edges of the sample are much more likely to be abraded than the center. Since the ring-on-ring test actually stresses the center of the sample more, this opens the test methodology to even more criticism. $\endgroup$ – Floris Aug 19 '14 at 3:31
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    $\begingroup$ youtube.com/watch?v=u6B-2jTvh1w $\endgroup$ – Dissenter Aug 19 '14 at 4:01

The fact that the picture is describing "Strength after abrasion damage" would seem to imply that the discussion falls under the rubric of fracture mechanics. "Abrasion damage" would imply that small cracks already exist IMO. In that case, the major influences would be the characteristics of the material and the geometry of the crack, and only then would the effect of the sample geometry come into play.

Starting with the Griffith equation for fracturing stress $\sigma_f$

$\sigma_f = \sqrt{\dfrac{G_cE}{\pi a}}$

we see that both the material characteristics(the critical strain energy release rate $G_c$ and Young's modulus $E$) and the crack length $a$ come into play.

That would apply if the plate is large enough so that edge effects can be neglected (i.e. the plate dimensions are large compared to the crack length). I don't know if that's the case or not for the ring stress test depicted, and if the sample size was chosen so the test is properly standardized as @Johannes mention in his answer.

If that's not the case, things get complicated. It depends on the crack geometry relative to the sample as well as the size of the sample itself.Then it's a matter of computing the effects of the geometry on the stress intensity factor, and that seems to be done with numerical methods like finite element analysis. I can't comment further on that ,it's outside my expertise.


You don't appear to be asking for strength here, but rather asking for hardness - a wholly different concept. Resistance to penetration or distortion by a small focused force?

Neither strength (which is an inherent property of the material) nor hardness (also an inherent property) can be changed by altering dimensions. RESISTANCE TO BREAKAGE may be, if you're talking about bending your hypothetical sheet around a fulcrum... but then only the distance between your distant pressure points and the fulcrum would make a difference.

If you're speaking strictly about penetration resistance, AND if the material in question is "plastic" in nature, then larger dimensions will tend to diminish the effect of the penetrating object. If the material is "crystalline" in nature, that doesn't hold true except in a very very small scale wherein a crystalline lattice may be very slightly disturbed by a pentrating object.


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