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Consider a homogeneous beam with a rectangular c/s, its neutral axis will be at equal distances from the top and bottom edges.

enter image description here

The first moment of the entire area about the NA will be zero, since we have as many $y\,dA$ terms as $-y\,dA$ terms.

I was investigating whether this is true for other c/s shapes too, which are not doubly symmetric, for example,

enter image description here

Will the first moment of area $\int_A y\,dA$ about the NA be equal to zero for all c/s shapes? Even for the one that I show right above?

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Yes, the sum of the first moment of inertia will be zero. That is actually how we find the neutral axis.

Let's call the are on top of the neutral axis area T and the bellow axis area B.

Then the location of the neutral axis with respect to a line on the bottom of the C/S will be. $$ \overline{y} = \frac{areaT*\overline{y_t}+areaB* \overline{y_b}}{areaT+areaB}$$

It will be closer to the larger end sort of like seeking the line that if you put the C/S on it it will balance itself.

he is an example of a T beam.

neutral x

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  • $\begingroup$ Oh yes, I so forgot that the very basis of determining the location of neutral axis, is the equation $\int_A ydA=0$. Thanks $\endgroup$ Feb 25, 2022 at 19:13
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Short answer is yes it will be zero. That is one of the main points for selecting the neutral axis because it simplifies the calculations.

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