I have to derive two equations in order to get a Merit Index.

so the question is how you derive the

m=b^2*l*r , r=density

and Fy = (b^4/12 /(b/2) * sy/(m/b^2*r)

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- Nov 4th 2010, 10:01 AM #1

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- Nov 4th 2010, 10:29 AM #2
Uh... I guess that b is the breadth of a square and l the length of a cuboid with a square face...

Then from $\displaystyle \rho = \dfrac{m}{V}$

We know that the volume is $\displaystyle V = b \times b \times l$

Hence, we get:

$\displaystyle \rho = \dfrac{m}{b^2l}$

$\displaystyle m = b^2l\rho$

I don't know what you mean by Fy and sy...

Is that it?

$\displaystyle F_y = \left(\dfrac{\frac{b^4}{12}}{\frac{b}{2}}\right)\l eft(\dfrac{s_y}{\frac{m}{b^2r}}\right)$

- Nov 4th 2010, 10:37 AM #3

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well I am in that point , Fy = failure load of a beam equation , sy=yield strees .

the final Merit index is = sy^2/3 / ρ . I am in the point Fy=b^4/12 / b/2 * σy/(m/b^2*ρ) .

So the thing is , how you get to that σy^2/3 / ρ ?

To make it a bit clearer it will be in the end Fy = b^3 *6 * m σy/2^3/ρ . So this σy/2^3/ρ will be in the final equation as it is now . without any mass or length in its function.

I hope I am clear now .

- Nov 4th 2010, 10:48 AM #4
So, this is right then and you use this to find the Merit index?

$\displaystyle F_y = \left(\dfrac{b^4/12}{b/2}\right)\left(\dfrac{\sigma y}{m/(b^2\rho)}\right)$

This is strange because some terms cancel out... to become:

$\displaystyle F_y = \dfrac{b^5 \sigma y \rho}{6m}$

- Nov 4th 2010, 11:06 AM #5

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- Nov 4th 2010, 11:17 AM #6

- Nov 4th 2010, 11:32 AM #7