Moment of Inertia — I-Beam or Built-Up Section

The fastest way to the strong-axis $I$ of a doubly symmetric I-shape is not to add flanges and web, it is to subtract. Take the full bounding rectangle $BH^3/12$ and remove the two rectangular voids that flank the web, each of width $(B-t_w)/2$ and depth $h = H - 2t_f$. A 200 × 400 welded girder with 20 mm flanges and a 10 mm web gives $(200 \times 400^3 - 190 \times 360^3)/12 = 327{,}946{,}667$ mm⁴. Check it the long way — web $10 \times 360^3/12 = 38.88$ million, plus two flanges each contributing $133{,}333 + 4000 \times 190^2 = 144.53$ million — and you land on the same number. The box-minus-voids route is one expression instead of five and it never asks you to find a centroid.

What the arithmetic shows is where the stiffness lives. In that girder the two flanges supply 88% of $I$ and the web supplies 12%, even though the web is nearly half the steel. The web is not there for bending at all; it is there to carry shear and to hold the flanges apart. Change the depth from 400 to 500 mm with the same plates and $I$ rises by roughly 60% for no extra weight, which is the reason plate girders get deep and thin rather than short and fat.

The limits worth knowing: this is the strong axis only, it assumes sharp corners rather than the rolled fillets that make a real W-shape a percent or two stiffer than the formula, and it says nothing about whether the section can actually reach that stiffness. A deep thin web buckles, and an unbraced compression flange rolls sideways long before the fibres reach yield. Compactness and lateral bracing are separate checks that $I$ cannot see.