Short question:

Is it possible (and if so, how) to transform a curve of the form

$$l\left(\frac{a}{b}\right) = \frac{1-\frac{a}{b}}{\frac{K}{3}\left(2+\left(1-\frac{a}{b}\right)^3\right)}$$

into the straight line

$$u\left(\frac{a}{b}\right)=1-\frac{a}{b} ,$$

where $K$ is a positive, real constant, $a,b\in[0,\infty)$ (see below for more info on this upper limit), $\frac{a}{b}\in[0,1]$ and $l,u\in[0,1]$ by making that when $a\to0$, the curve becomes the straight line and, when $a\to\infty$, it becomes the first curve? I know it doesn't make much sense mathematically speaking, but allow me to explain further below.

Detailed explanation of the problem:

I am dealing with composite materials models, and the theory says there is a lower limit for fracture given by the curve $l\left(\frac{a}{b}\right)$, and an upper one described by the straight line $u\left(\frac{a}{b}\right)$. $a$ and $b$ are geometric parameters (hole diameter and width of the test specimen), so, theoretically, they can take any values from $0$ to $\infty$ as long as $\frac{a}{b}\in[0,1]$.

Now, in my experiments I have found that when $a$ grows for a fixed value of $\frac{a}{b}$, the results go from being closer to the straight line to being closer to the curve. Also, when $\frac{a}{b}$ is varied for a given value of $a$, the experimental curves seem to have a form that is in between the straight line and the curve $l\left(\frac{a}{b}\right)$. Here is a graph with slightly different notation. I think that $K=3$ in that graph.

I wonder if it's possible to find a general curve $c=f\left(\frac{a}{b}\right)$ that transform smoothly from $l\left(\frac{a}{b}\right)$ to $u\left(\frac{a}{b}\right)$ when $a$ diminishes (or viceversa when $a$ grows).

Conceptually there would be a problem when approaching either $u$ or $l$ because these equations are defined for $\frac{a}{b}\in[0,1]$ with finite values of $a$ and $b$ (because they are geometric parameters), but the experiments seem to indicate they would be the tendency of the desired curve $c$ when $a\to0$ and $a\to\infty$. I don't really care about these extreme cases because they are ideal models that don't actually occur, so even though this is a mathematical question, it's important to keep the "real" characteristics of the problem in mind.

Sorry for any inaccuracies in my math syntax and for my lack of knowledge to properly tag this question.

  • $\begingroup$ I forgot to mention the obvious fact that, for the case $\frac{a}{b}=1$, there is no test specimen, and the strength is zero. Thus, $c\left(\frac{a}{b}=1\right)=l\left(\frac{a}{b}=1\right)=u\left(\frac{a}{b}=1\right)=0$. $\endgroup$
    – yuyu2809
    Apr 6, 2017 at 17:44

1 Answer 1


After some thinking, the answer seems obvious to me now:

If we assume that $\frac{a}{b}\in[0,1]$ no matter what value $a$ takes, it is evident that the desired curve $c\left(\frac{a}{b}\right)$, or maybe more accurately, $c\left(a,b\right)$, should be of the form

$$c\left(a,b\right) = \frac{1-\frac{a}{b}}{\left[\frac{K}{3}\left(2+\left(1-\frac{a}{b}\right)^3\right)\right]^{m\arctan(a)}},$$

with $m$ a positive parameter to be determined experimentally for every value of $a$. That way, when $a\to0$, we get only the numerator (which is the upper limit straight line), and when $a\to\infty$, taking $m=\frac{2}{\pi}$ would make $c\left(a,b\right) = l\left(\frac{a}{b}\right)$.

I have still some questions to resolve, but it seems that $m$ should be lower than $\frac{2}{\pi}$ for the curve to adjust to the experiments. I would not reach the lower limit curve with this value, but since $a$ represents a diameter and it will never be actually infinity, it does not seem to be a problem; if anything, I would get a more precise lower limit for the composite material strength. I will also have to consider the units of $a$ when adjusting $m$, and maybe I will find that this parameter is also a function of $a$ in some way.

Sometimes asking a question makes you see the answer more easily, so thanks to this platform for allowing me to do so.


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