# Modification of Beam Equation for Thin Sheet

I am trying to determine the modulus of elasticity (Young's Modulus) for a rigid paper sheet. The issue I am having is that my results for the Young's Modulus change depending on what orientation (length extending out or width extending out) I use to measure it.

The material of the paper is not trivial, and as a result I am not able to find a datasheet on the material itself. Instead, I did some simple tests to calculate the Young's Modulus, especially since it is a thick sheet that behaves more like a sheet of metal than floppy paper.

I took the sheet, fixed one end, and measured the deflection at the other end (and other relevant parameters). I did this with both the length and the width extending freely over the fixed end. I accounted for curvature in the sheet when measuring the deflection in order to make the measurement as representative of an ideal beam as possible (there were some assumptions I made about the curvature, but I will not go into details about how I eliminated the effect of curvature in my deflection measurement unless I have to, as I believe the problem lies elsewhere). I then used the Euler-Bernoulli beam equations for distributed load, $$q$$, (in this case, the distributed load was due to gravity) to arrive at the Young's Modulus, $$E$$. $$E = \frac{qL^4}{8I\delta_{gravity}}$$ $$q=$$ distributed load, $$L=$$ the length of the paper extending over the fixed end, $$I=$$ second moment of area, which changed depending on orientation, and $$\delta_{gravity}=$$ deflection after removing effects of curvature.

The results I achieved were consistent ( within 0.3 GPa of each other), but changed drastically, by almost an order of magnitude or more, when changing whether it was the width of the paper that was modeled as the "length" of my beam, or the length of the paper.

I am wondering what caused this big change in the Young's Modulus, since I don´t know if it is typical for Young's Modulus to vary by such a large amount between orientations (10-90 times difference). Paper is an anisotropic material (thank you, alephzero), so it could be related to this fact. I believe I am missing some factor in the equation that is depenedent on orientation, such as a correction factor for structural stiffness due to sheet curvature, or something with the Poisson ratio, but of course, this belief can be misleading me.

Things to note:

The material is foamboard, only the outside of the material is made of paper in the strictest sense. The inside is made of foam.

I changed the distributed load, $$q$$, depending on the orientation, since the weight/length changes depending on whether the width or the length. I did the same thing for the second moment of area, $$I$$.

For the second moment of area, $$I$$, I modified the equation to more precisely model the shape of the curvature, but it did not change the value of $$I$$ significantly.

• Paper is not an isotropic material. At a microscopic scale, it is a fibre-reinforced composite material and the fibre orientations are not random. At a macroscopic scale, a reasonable approximation is that it is orthotropic. Oct 4 '19 at 12:17
• Thank you, I will investigate this further and correct the question to reflect this. Oct 4 '19 at 13:33
• "I believe I am missing some factor in the equation" - because of the way paper is made, the principal material axes are along the "grain" of the paper (i.e. along the length of the roll before it is cut into sheets) and transverse to it. For an orthotropic material the elastic modulus in the two directions, and the shear modulus, are three independent quantities, The value of Possion's ratio is also different in the two directions. The isotropic formula $E/G = 2(1+\nu)$ doesn't apply. For example, woven cloth may have $E/G$ of the order of $10^6$ or bigger. Oct 4 '19 at 22:04
• @alephzero I'd suggest adding these comments as an answer. They give a good explanation (more investigation by the OP will be necessary to see if they are correct for their particular case) which may very well be useful for future readers. Hell, it was useful for me. I read the question and immediately thought of anisotropy, but would've assumed paper behaved isotropically on a macro scale (that the fibre distributions were random). Learn something new every day.
– Wasabi
Oct 5 '19 at 17:46

## 1 Answer

I would calculated the I being allocated to two different materials, foam and paper. one way of getting a sense of the difference between the two is to let the paper beam on its side flatwise, and measure the E of the paper by neglecting the contribution of foam, if it is not paper on both sides glue 2 back to back to make it so. Because in that configuration the paper is the farthest from neutral axis and most stressed. Then you can use this E to compare to the vertical beam position as the papers contribution. Because of the embedded errors in measurements of the weight and deflections extra care is needed. such as for cantilever just use a continuous length of the paper 3 lengths and leave 2 length in a frame or bracket, secured over a table or something like that.

Use 10 sheets of paper bound together by paper tape at intervals to make the beam 10 times thicker, not to add to stiffness of the beam. Then divide you I by 10. this will average out small inconsistencies in fabrication.

If practical add a concentrated load several times bigger than the weight of the paper at the end and calculate deflection due to both distributed and concentrated loads this takes care of possible uneven thickness of the paper. $$\ E= \frac{Pl^3}{3\delta_yI }$$.

Try to look for shear warping or lateral buckling especially near the support, even if very small and hard to see, the out of plane secondary forces can produce twisting and torque which has to be counted for.