The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.
We call the outer side ,a and inner side b.
$$ I = \frac {(a^4-b^4)}{12} $$
And $$ S = \frac {(a^4-b^4)}{6a} $$
So roughly, you need three of the smaller size beams for each 40x40 mm.
But since the deflection of the smaller beam will be threeTHREE times more under the same load, we need to check on that not to become too excessive.
The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.
We call the outer side ,a and inner side b.
$$ I = \frac {(a^4-b^4)}{12} $$
And $$ S = \frac {(a^4-b^4)}{6a} $$
So roughly, you need three of the smaller size beams for each 40x40 mm. But since the deflection of the smaller beam will be three times more under the same load, we need to check on that not to become too excessive.
The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.
We call the outer side ,a and inner side b.
$$ I = \frac {(a^4-b^4)}{12} $$
And $$ S = \frac {(a^4-b^4)}{6a} $$
So roughly, you need three of the smaller size beams for each 40x40 mm.
But since the deflection of the smaller beam will be THREE times more under the same load, we need to check on that not to become too excessive.
The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.
We call the outer side ,a and inner side b.
$$ I = \frac {(a^4-b^4)}{12} $$
And $$ S = \frac {(a^4-b^4)}{6a} $$
So roughly, you need three of the smaller size beams for each 40x40 mm. But since the deflection of the smaller beam will be three times more under the same load, we need to check on that not to become too excessive.
