# What is the moment of inertia of an I-beam? [closed]

A thin rod of length L and mass m, perpendicular to the axis of rotation, rotating about its center. Its moment of inertia would be $$I=(1/12)mL^2$$ (see picture). If I want to use an I-beam instead of a circular rod, what will be the formula for the moment of inertia?

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• I am guessing it is the same because the distribution of mass in the direction of rotation is not important here. There is a slight effect due to the width of the beam, but it can be safely ignored for long beams. Dec 5, 2023 at 17:33

Just use the thin rod formula if the pieces that form the I cross section are small relative to the length.

Wikipedia has a nice list of moments of inertia. For a flat plate (the top and bottom of the I cross section), the $$I =\frac 1 {12}(w^2 + l^2)$$ where $$w$$ is the width of the beam and $$l$$ is the length. Consider $$w=6$$ and $$l=100$$ then $$6^2 + 100^2$$ is 10,000 plus unimportant amount from $$w$$.

Happy that you ask! Here's my solution.

• Assume the width and length of the rectangular surface of the upper part of the I-beam are $$w\ and\ L$$ respectively, and the thickness of the whole I-beam is $$h$$(which turns out not to influence the total moment of inertia)
1. Simple Proof of the Stretch rule(the-thickness-doesn't-matter-rule): If the chosen axis's direction is $$\hat z$$, then the stretches in the $$\hat z$$-direction do not affect the overall moment of inertia, provided that the distribution of mass remains unchanged except in the direction parallel to the axis. By choosing z as the rotating axis, we assume that if we shift the whole I-beam's coordinate from $$z$$ to $$z+dz$$, its moment of inertia stays the same. This means if we shift a cross-section(while the x and y coordinates stay the same) to a different z position, the overall moment of inertia remains. Therefore, because of the space symmetry(or translational invariance), the stretch rule is valid.
2. Moment of inertia of a rectangle of $$w$$ and $$h$$ with arbitrary thickness rotating with the center of mass of the rectangle:
$$I=\int r^2 dm=\int (\sqrt{x^2+y^2})^2 dm=\int x^2 dm+\int y^2 dm=I_y+I_x\$$where $$I_y,I_x$$ is the moment of inertia while choosing y,x as the principle axis. So,$$I_{rectangle}=\frac{1}{12}mw^2+\frac{1}{12}mL^2=\frac{1}{12}m(L^2+w^2)$$
3. The moment of inertia of the I-beam:
$$I_{I-beam}=2I_{rectangle}+I_{middle}=\frac{1}{6}m(L^2+w^2)+\frac{1}{12}mL^2=\frac{1}{12}m(3L^2+2w^2)$$

Therefore,$$I_{I-beam}=\frac{1}{12}m(3L^2+2w^2)$$

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