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I need to calculate $K$ or equivalent stiffness of a beam that has a varying $I$ across its length. $I$ has been given to me as a function of $x$, or length.

I have been unable to find a proper equation or set of equations describing how to obtain $K$ for this situation. I've found equations for constant $I$ values across the length of beams. Is there an equation I am simply unaware of, or some general procedure for deriving $K$ for variable $I$ values?

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You could insert the variable $I(x)$ into the integral equation for the rotation and the deflection.

First determine your model. Then determine the equation of the moment $M(x)$. Then enter this in the equation of rotation.

rotation: $\theta = \int \frac{M(x)}{E*I(x)}dx $

solve this equation (or let wolfram alpha do it for you) , add the relevant boundary conditions (such as $\theta(0) = 0$ for a clamped beam) and then solve

deflection: $\nu = \int \theta dx$

and add the relevant boundary conditions.

Good luck!

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The short answer is you can't.

The slightly longer answer is you can, but the solution is specific to the adopted cross-section.

The much longer answer will now demonstrate why this is the case by means of an example. In this example, we're just going to take a simple cantilever beam (fixed at $x=0$) of span $L$. The beam will have a variable stiffness $E \cdot I(x)$, since it is a simple rectangular cross-section, with a height that varies (linearly) from $h_0$ at the fixed end to $h_1$ at the free end. The applied load will be a simple concentrated load $P$ at the free end. What is the deflection at the free end?

As @jos already mentioned in their answer, we use the beam equations.

This is an isostatic beam, so the bending moment equation can be trivially obtained as

$$M = P(L-x)$$

this must then be divided by the beam's stiffness and the result must be integrated to obtain the beam's tangent.

$$\theta = \int\limits_0^L\frac{P(L-x)}{E \cdot I(x)}\text{d}x$$

Here we can already see the problem. In ordinary beams we can just remove $EI$ from the integral and go on our merry way. Here, however, $I(x)$ will also need to be integrated.

So, what's the equation for $I(x)$? In this case, it will be

$$I(x) = \frac{b \cdot h(x)^3}{12} = \frac{b \cdot \left(h_0 + \left(h_1-h_0\right)\dfrac{x}{L}\right)^3}{12} = \frac{b \cdot \left(h_0 + \Delta h\dfrac{x}{L}\right)^3}{12}$$

So now we "just" have to find $\theta$.

$$\begin{gather} \theta = \frac{1}{E}\int\frac{12P(L-x)}{b \cdot \left(h_0 + \Delta h\dfrac{x}{L}\right)^3}\text{d}x \\ \theta = \frac{6PL^3}{Eb\Delta h^2}\left(\frac{-\Delta hL+2\Delta hx+h_0L}{(\Delta hx+h_0L)^2} + C_1\right) \end{gather}$$

Already getting ugly, right? Now, we just have to integrate this once more to get the deflection.

$$\begin{gather} \delta = \frac{6PL^3}{Eb\Delta h^2}\int\limits_0^L\left(\frac{-\Delta hL+2\Delta hx+h_0L}{(\Delta hx+h_0L)^2} + C_1\right)\text{d}x \\ \delta = \frac{6P}{Eb\Delta h^2}\left(\dfrac{L(\Delta h+h_0)}{\Delta h(\Delta hx+h_0L)}+\dfrac{2\ln(\Delta hx+h_0L)}{\Delta h}+C_1x+C_2\right) \end{gather}$$

Now, to find $C_1$ and $C_2$:

$$\begin{gather} \theta(0) = 0 = \frac{-\Delta hL+h_0L}{(h_0L)^2} + C_1 \\ \therefore C_1 = \frac{\Delta hL-h_0L}{(h_0L)^2} \\ \delta(0) = 0 = \dfrac{L(\Delta h+h_0)}{\Delta hh_0L}+\dfrac{2\ln(h_0L)}{\Delta h}+C_2 \\ \therefore C_2 = -\dfrac{L(\Delta h+h_0)}{\Delta hh_0L}-\dfrac{2\ln(h_0L)}{\Delta h} \\ \end{gather}$$

$$\begin{align} \therefore \delta = \frac{6P}{Eb\Delta h^2}&\left(\dfrac{L(\Delta h+h_0)}{\Delta h(\Delta hx+h_0L)}+\dfrac{2\ln(\Delta hx+h_0L)}{\Delta h}+ \\ \frac{\Delta hL-h_0L}{(h_0L)^2}x-\dfrac{L(\Delta h+h_0)}{\Delta hh_0L}-\dfrac{2\ln(h_0L)}{\Delta h}\right) \end{align} $$

And there you have it. To check my work, I created a beam with $h_0 = 200\ \text{mm}$, $h_1 = 100\ \text{mm}$, $b = 100\ \text{mm}$, $L = 10\ \text{m}$, and $E = 10000\ \text{MPa}$, and applied a load $P = 1\ \text{kN}$ at the end.

I simulated the beam by cutting it into discrete segments, each with a different uniform cross-section. To test the sensitivity, I made two models: one with ten segments and one with twenty. Here are the results:

enter image description here

Using the equation obtained above, I got 81.78 mm, for an error of 1% compared to the models. Given those are approximations, that looks pretty good.

From this, we can then obviously obtain the stiffness of the beam: $1\ \text{kN}/81.78\ \text{mm} = 12.2\ \text{kN/m}$

So, now that we've gone through all this, what can we learn? Basically, that this is really hard. And the solutions aren't elegant. This was just about the most trivial cross-section ever and things already got messy. I mean, those are logarithms, for crying out loud. What if the section varied both in height and width? What if the section didn't vary linearly, but polynomially (or something else)? What if we were dealing with an asymmetrical I-section, with six varying variables ($t_{f,top}$, $b_{f,top}$, $t_{f,bot}$, $b_{f,bot}$, $t_{w}$, $h_{w}$)?

This is why the most common solution when dealing with variable sections is to do as I did: split your beam into discrete segments, each with a different interpolated cross-section. Some professional FEA programs make this easy and some even have variable sections built in, but I don't know how they're implemented.

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  • $\begingroup$ Just to alleviate any confusion, should the deflection value be 80.78 mm and not 81.78 mm. I am basing this on your diagram which plots the deflections. Thanks for the work and the explanation. $\endgroup$ – Perry H. Feb 26 '20 at 8:49
  • $\begingroup$ @PerryH. The models got 80.78 and 80.86 mm. However, the analytical equation I derived via integration of the fundamental beam equation gave 81.78 mm. So the "right" answer is 81.78 mm, and the diagrams are just to show my derivation is correct (the error is due to the fact that my analytical solution perfectly describes the changing cross-section, while the models use constant sections in each segment). $\endgroup$ – Wasabi Feb 26 '20 at 14:13

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