The following procedure is here.

Consider a cantilever fixed at one end and loaded at the other one. In cartesian coordinates (if $y$ is horizontal and $x$ vertical, meaning that the load acts parallel to the $x$ axis) the equation of the curvature is: $$\frac{\frac{d^2x}{dy^2}}{\left[1+\left(\frac{dx}{dy}\right)^2\right]^{3/2}}=\frac{M(y)}{EI},$$

where $M(y)$ is the moment of bending, $I$ the moment of inertia of the cross-sectional area of the beam, and $E$ is the Young's modulus of the beam's material.

Considering the variable $z=dx/dy$ and the lenght of the beam: $$s(l)=\int_0^l\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy,$$

where $l$ is the projection of the beam onto the $y$ axis. Then we can write:

$$\frac{z}{\sqrt{1+z^2}}=\int_0^y\frac{M(y)}{EI}dy = G(y).$$

The bending moment is $M(y)=P(l-y)$, where $P$ is the applied force parallel to the $x$ axis. Therefore


After this introduction I present my question:

I carried out an experiment of bending spaghettis by placing them horizontally with a fixed end, and loaded at the other end. The aim was to find the spaghettis' Young's modulus.

Unfortunately the theory developed for these deflections is mainly broad for small deflections (they take some approximations in the first equation I wrote). However I measured $l,P,I$ to be $0.17, 0.1, 5.75\times 10^{-12}$, in mks.

My idea is to integrate numerically the last equation so that $s(l,E)=L$, where $L$ is the length of the spaghetti: $0.2$m. This means that the integration is made over $y$, from $0$ to $l$. Then, since $l$ is known, $s$ really depends on $E$. So there must be some $E$ for which the integral is $L$. However I need a really large $E$ (~$1\times 10^9$ Pa) so that the expression in the square root is positive, and this doesn't make sense because the spaghettis shouldn't have such a big $E$.

What could be wrong with this approach?

  • $\begingroup$ Have you looked at many tables of the Young's modulus? Note the units typically used to tabulate these moduli. How sure are you that this value is actually large? $\endgroup$ Commented May 16, 2015 at 19:44
  • $\begingroup$ @dmckee, because I used the "small deflection" theory to solve for $E$ and found it to be around $3\times 10^8 Pa$ $\endgroup$ Commented May 16, 2015 at 19:49
  • 1
    $\begingroup$ Let me rephrase. What is a typical value of $E$ for a modest strength, day-to-day material. How does that compare to what you have computed for pasta? $\endgroup$ Commented May 16, 2015 at 19:51
  • $\begingroup$ It is cool how you experimentally tested a recent scientific result (2010) at home - @VladimirVargas $\endgroup$
    – Cicero
    Commented May 16, 2015 at 20:20
  • $\begingroup$ @dmckee I checked some tables (none of them have spaghettis as expected haha), but, from an intuitive point of view I expect spaghettis to have a very small $E$, like tenths of MPa or something like that. However, with my data, $E>4\times 10^8$ Pa is good enough to have a positive argument in the square root. So thanks for making me think of the possibility of having a greater $E$ than the one I expected without any real reason whatsoever :) $\endgroup$ Commented May 16, 2015 at 21:06

1 Answer 1


Generally speaking, Young's modulus is measured in terms of Gigapascals, or $10^9$ Pascals, or at least Megapascals, or $10^6$ Pascals. Hence, your answer should be fine. Here is a table with Young's modulus values for day to day materials. Notice how $E$ is measured in Gigapascals.


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