# Number of layers in a laminated steel transplant, using Young's Modulus

I'm not sure if my calculations are correct for the above question. $$E = \frac{\sigma}{\epsilon} = \frac{\frac{My}{I}}{\frac{\delta}{L}} = k. \frac{1}{I}$$ Where k is constant for both beams, therefore: $$E_1I_1 = E_2I_2$$

$$\frac{I_1}{I_2} = \frac{1}{10}$$

$$\frac{\frac{W(H/n)^3}{12}}{\frac{WH^3}{12}} = \frac{1}{n^3} = \frac{1}{10}$$

$$n = 10^{\frac{1}{3}}$$ Is this correct? If so should I round up or down for the number of layers? Also any comments on part (b) are welcome :)

• Instead of rounding, compute the modulus for each neighboring integer value and choose the closer one. At the very least explain why you made the choice that you did. Dec 20 '16 at 17:13

This is quite a poorly-worded question. After all, how do the layers interact with one another? Can they be considered to be fully bonded to one another? I assume not, because then they've behave as one element, which kind of beats the purpose. Should we assume that the load is evenly distributed between each of the layers (so, for a given load $q$, if you have three layers then they'll behave like three beams each under $q/3$)? Something else?

I'm going to assume that the load is evenly distributed between each of the layers.

So, you have some sort of load $q$ being supported by $N$ layers of material. This means that the load being resisted by each layer is equal to $\dfrac{q}{N}$.

Now, the inertia of a rectangular cross-section is

$$I = \dfrac{bh^3}{12}$$

So, if we have $N$ layers sharing the available height $h$, the inertia of each layer is equal to

$$I = \dfrac{b\left(\dfrac{h}{N}\right)^3}{12} = \dfrac{bh^3}{12N^3}$$

Now, depending on the support conditions (simply supported, cantilever, fixed-and-pinned, etc), the deflection will be equal to (for simple loading conditions):

$$\delta =K\dfrac{qL^x}{EI}$$

where $K$ depends on the boundary conditions and loading and $x=3$ for concentrated loads and $x=4$ for distributed loads. For example, a simply-supported beam under uniformly distributed loads has

$$\delta =\dfrac{5}{384}\dfrac{qL^4}{EI}$$

So, lets call the deflection of the single beam $\delta_1$ (with $E_1=20\text{ GPa}$) and the deflection of one of the layers $\delta_2$ (with $E_2=200\text{ GPa}$).

\begin{align} \delta_1 &= \delta_2 \\ K\dfrac{qL^x}{E_1I_1} &= K\dfrac{\dfrac{q}{N}L^x}{E_2I_2} \\ \dfrac{1}{E_1I_1} &= \dfrac{1}{NE_2I_2} \\ E_1I_1 &= NE_2I_2 \\ E_1I_1 &= 10NE_1I_2 \\ I_1 &= 10NI_2 \\ \dfrac{bh^3}{12} &= 10N\dfrac{bh^3}{12N^3} \\ 1 &= 10\dfrac{1}{N^2} \\ \therefore N &= \lceil\sqrt{10}\rfloor = 3 \end{align}

With three layers, each layer has an inertia $\frac{1}{27}$ that of the solid beam. But they each only need to withstand a third of the load, meaning the they will deflect 9 times more than a solid beam of equal elastic modulus.

• Thank you! Yes the questions are deliberately vague. We're supposed to outline every assumption or inference we make throughout our solutions. Dec 20 '16 at 18:31
• I really like when professors make questions intentionally vague like this (when they make it clear beforehand they expect assumptions to be made). You rarely get a real problem with everything laid out for you.
– JMac
Dec 21 '16 at 13:19