Area moment of inertia

For a plane area $A$ lying in the $xy$-plane, the second moments about the $x$- and $y$-axes are defined by integrating area times the square of its distance from the axis:

where: $y$ and $x$ are the perpendicular distances of an area element $dA$ from the $x$-axis and $y$-axis respectively. Both integrals are always positive (distances are squared) and carry units of length$^4$ (mm$^4$ or m$^4$). Because the integrand grows with the square of distance, area placed far from the axis contributes disproportionately — this single fact explains why I-sections, with material pushed to the flanges, are so much stiffer in bending than a solid rectangle of the same area. Related to $I_x$ and $I_y$ is the polar moment of inertia, taken about a point (the origin $O$) rather than a line:

where: $r^2 = x^2 + y^2$ is the squared distance from $O$ to the element $dA$. $J_O$ measures resistance associated with radial distance and reappears later when circular shafts are analysed for torsion. For composite shapes — a bracket built from rectangles, a channel with a circular hole — you never re-integrate from scratch. Instead you split the shape into simple pieces whose centroidal moments of inertia are already known (from a standard-shapes table), shift each piece's moment onto the shared reference axis using the parallel-axis theorem below, and sum the contributions (subtracting for any cut-out). The figure shows this decomposition for an I-shaped section.

The radius of gyration,

where: $I$ is a moment of inertia and $A$ is the area, is the distance at which the entire area could be concentrated as a single point without changing $I$. It is used later for slenderness ratios in columns. Standard results worth memorising: rectangle ($b \times h$) about its centroidal axis parallel to $b$, $\bar I_x = bh^3/12$; circle of radius $r$, $\bar I_x = \pi r^4/4$ and $\bar J = \pi r^4/2$; triangle (base $b$, height $h$) about its centroidal axis parallel to the base, $\bar I_x = bh^3/36$.

Problem. An I-shaped cross-section is built from three rectangles: a top flange 120 mm wide by 20 mm deep, a web 20 mm wide by 160 mm deep, and a bottom flange 120 mm by 20 mm, stacked symmetrically about a common vertical centreline, for a total depth of 200 mm. Find the moment of inertia $\bar I_x$ about the horizontal centroidal axis, then find the radius of gyration and the moment of inertia about the bottom edge of the section.

Step 1. Locate the centroid. By symmetry (identical flanges, centred web) the centroid lies at mid-height, 100 mm from the bottom edge.

Step 2. Moment of inertia of the web, which is already centred on the centroidal axis ($d=0$): $\bar I_{web} = \dfrac{20\times160^3}{12} = \dfrac{20\times4.096\times10^6}{12} = 6.827\times10^6\ \text{mm}^4$.

Step 3. Moment of inertia of each flange, transferred to the centroidal axis. A flange's own centroid sits at $d = \tfrac{160}{2}+\tfrac{20}{2} = 90$ mm from the section centroid, with area $A_{flange}=120\times20=2400\ \text{mm}^2$ and own centroidal moment $\bar I_{flange,own} = \dfrac{120\times20^3}{12} = 8\times10^4\ \text{mm}^4$. By the parallel-axis theorem, $I_{flange,transferred} = 8\times10^4 + 2400\times90^2 = 8\times10^4+1.944\times10^7 = 1.952\times10^7\ \text{mm}^4$. Two identical flanges give $2\times1.952\times10^7 = 3.904\times10^7\ \text{mm}^4$.

Step 4. Add the pieces: $\bar I_x = 6.827\times10^6+3.904\times10^7 = \mathbf{4.587\times10^7\ \text{mm}^4}$. Most of the stiffness comes from the flanges, far from the axis — the design reason I-sections are efficient in bending.

Step 5. Radius of gyration. Total area $A = 2400+2400+(20\times160) = 8000\ \text{mm}^2$, so $k=\sqrt{4.587\times10^7/8000} = \sqrt{5734}= \mathbf{75.7\ \text{mm}}$.

Competitive stretch

Find $I$ about the bottom edge of the section instead. The distance from the section centroid to the bottom edge is $d=100$ mm, so by the parallel-axis theorem $I_{bottom} = \bar I_x + Ad^2 = 4.587\times10^7 + 8000\times100^2 = 4.587\times10^7+8\times10^7 = \mathbf{1.259\times10^8\ \text{mm}^4}$, noticeably larger than $\bar I_x$, confirming the centroidal axis gives the minimum.

• Confusing the area moment of inertia ($I=\int y^2\,dA$, units length$^4$) with the mass moment of inertia ($I=\int r^2\,dm$, units mass·length$^2$) from dynamics — these are entirely different quantities describing different physics.
• Applying the parallel-axis theorem between two non-centroidal axes directly — it only transfers to or from a centroidal axis; to go between two arbitrary parallel axes you must route through the centroid in two steps.
• Forgetting to subtract a cut-out's moment of inertia in a composite section, or using the hole's own centroid distance instead of its distance from the section's overall centroidal axis.
• Using the wrong $d$ in $Ad^2$ — it must be the perpendicular distance between the two parallel axes, not a diagonal length or an unrelated dimension from the figure.

Ask the tutor: if splitting an oddly shaped bracket into simple pieces for a composite moment-of-inertia calculation feels confusing, ask the AI tutor to walk through the decomposition and the parallel-axis bookkeeping step by step.