20 Jun

Parabolas, Rectangles, Triangles & Snakes (Part 5)

Part 5 – Centroids, More Integrals….

In the last post we went through a process to find a general force integral which involved knowing the width and height of the stress block in terms of how far from the neutral axis depth you are. Then by using numerical integration methods we would be able to calculate the concrete force by evaluating this integral. Note that except for certain cases like a constant stress or constant width I don’t think you can even solve the integral by hand, so your best chance of success is solving it numerically unless of course you’re a masochist and love integration that much. Good thing Python makes it easy to solve integrals numerically.

In this post we’ll look at extending the previous post (and the result we had magicked up for the force integral) by going on to find the centroid, or in other words the location where the resultant force acts within our concrete compression block.

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14 Jun

Parabolas, Rectangles, Triangles & Snakes (Part 4)

Part 4 – Integrals, its Business Time

In the first post in this series I posted the following integral for the x direction centroid of a part of the compression block for EC2 parabolic-rectangular derivation :-

\overline{x}=\frac{\displaystyle{\int_{y_1}^{y_2}}\;\bigg[\frac{(x_1-x_2)(y_2-y)}{(y_2-y_1)}+x_2\bigg]\;\bigg[(1-(1-\displaystyle{\frac{y}{k_1c}})^n\bigg]\;\bigg[\frac{(x_{b\,mid}-x_{t\,mid})(y_2-y)}{(y_2-y_1)}+x_{t\,mid}\bigg]\;\bigg[dy\bigg]}{\displaystyle{\int_{y_1}^{y_2}}\;\bigg[\frac{(x_1-x_2)(y_2-y)}{(y_2-y_1)}+x_2\bigg]\;\bigg[(1-(1-\displaystyle{\frac{y}{k_1c}})^n\bigg]\;\bigg[dy\bigg]}

Which at first glance looks a bit intimidating, but it breaks down to quite a simple formulation which I’ll try my best to explain in this post and the next post.

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