Study Companion: S24 — Area Moments of Inertia

Centroids came from the first moment of an area, \int y\,dA. Whenever a distributed force grows in proportion to the distance from an axis, its moment about that axis brings in one more factor of distance: \int y^2\,dA, the second moment of the area, or moment of inertia. It appeared in S23's centre of pressure, and it is the reason beams are made deep, I-shaped and hollow. This lesson defines it, its polar and radius-of-gyration forms, how to move it between parallel axes, and how to add it up for a composite section.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, Appendix A, articles A/1 ("Introduction"), A/2 ("Definitions") and A/3 ("Composite Areas"), with Sample Problems A/1–A/8. Results for common shapes are Meriam's Table D/3. Axes: x to the right, y up, z out of the plane of the area.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

A steel ruler, 30\,\mathrm{mm} wide and 3\,\mathrm{mm} thick, is bent about a horizontal axis, first lying flat and then standing on its edge. The cross-section is the same; only its orientation changes.

For the same stress at its outermost fibres, how much more bending moment does it carry on edge? The plausible guess is "the same: same steel, same area". The answer is in Nugget 02.

Nugget 01: When Intensity Grows With Distance, the Moment Is a Second Moment of Area

Claim. If a force distributed over an area has an intensity proportional to the distance from an axis, each element's force is proportional to distance times area, and its moment to distance squared times area. The total moment is then a constant times \int y^2\,dA, a property of the area's shape alone.

The picture. Meriam's Fig. A/1: three places the integral appears.

(a) A plate under pressure proportional to the distance y from an axis AB in its plane, as in S23. (b) The stress across a beam's section bent by end couples, \sigma = ky, tension on one side of the axis O–O and compression on the other. (c) A twisted circular shaft, whose shear stress grows with the radius, \tau = kr.

Check with numbers. A vertical gate 1\,\mathrm m wide and 2\,\mathrm m deep, its top edge in the water surface, under p = \rho gy. Its moment about the surface line is M = \int y(\rho gy\,dA) = \rho g\int y^2\,dA = \rho g\,bh^3/3 = 9810(1)(2)^3/3 = 26.16\,\mathrm{kN\,m}. The force is R = \rho g\bar hA = 9810(1)(2) = 19.62\,\mathrm{kN}, so its line is M/R = 1.333\,\mathrm m down: the two-thirds depth of S23, now as a ratio of second to first moment.

The idea. Meriam (A/1): when forces are distributed over an area with an intensity proportional to the distance from a moment axis, the elemental force is proportional to distance times area and the elemental moment to distance squared times area; the total moment involves \int(\text{distance})^2\,d(\text{area}), the moment of inertia or second moment of the area. In (a), the moment about AB of p = ky on dA is py\,dA = ky^2\,dA; in (b), dM = y(\sigma\,dA) = ky^2\,dA; in (c) the area is normal to the moment axis and M = \int r(\tau\,dA) = k\int r^2\,dA with a radial coordinate. "Second moment" is the more fitting name: the first moment y\,dA is multiplied once more by the arm y. "Inertia" comes only from the formal likeness to the inertia of rotating bodies; the area's moment of inertia is a purely mathematical property with no physical meaning by itself.

The equation (definition: second moment; with the physical law of moments when the intensity is ky):

Worked number. The same gate, 3\,\mathrm m deep: I = bh^3/3 = 9\,\mathrm{m^4}, M = 9810(9) = 88.3\,\mathrm{kN\,m}; the moment grows with the cube of the depth.

Real example (one you can see). A bookshelf sags much less with its boards on edge than flat; bridge girders, joists and rafters are all deep and thin for the same reason: their resisting moment is set by \int y^2\,dA.

Plausible wrong answer. "The moment is the force times the centroid's distance, so \int y\,dA is enough." That holds for a uniform intensity; when the intensity itself grows with y, each element's arm is counted twice.

Check. Of the three examples, which has the area normal to the moment axis? (The shaft, (c).)

Narration spine. A plate tilts into water; arrows grow on it with depth, and each arrow's moment arm lights up, the same distance again: y times ky. A beam section's stress arrows flip from compression to tension across its axis; a shaft's arrows spin, growing outward. Each time the same integral writes itself.

Bridge. The integral needs a name and notation for the different axes.

Nugget 02: Rectangular and Polar Moments of Inertia Are Never Negative

Claim. About the x- and y-axes, I_x = \int y^2\,dA and I_y = \int x^2\,dA; about the z-axis through O, the polar moment I_z = \int r^2\,dA = I_x + I_y. Distance is squared, so every element adds a positive amount: a moment of inertia is never negative or zero, unlike a first moment.

The picture. Meriam's Fig. A/2.

Check with numbers (Meriam's Sample Problem A/1). A rectangle b wide and h high, horizontal strips b\,dy at y from its centroidal axis x_0:

The puzzle. The ruler flat (b = 30, h = 3\,\mathrm{mm}): I = 30(3)^3/12 = 67.5\,\mathrm{mm^4}. On edge (b = 3, h = 30\,\mathrm{mm}): I = 3(30)^3/12 = 6750\,\mathrm{mm^4}, a hundred times more. With stress \sigma = ky (Nugget 01), the moment is M = kI; for the same stress \sigma_{\max} at the outer fibres, k = \sigma_{\max}/c with c the distance to them (1.5\,\mathrm{mm} flat, 15\,\mathrm{mm} on edge), so M = \sigma_{\max}I/c: 45\sigma_{\max} flat against 450\sigma_{\max} on edge. On edge the ruler carries ten times the moment for the same stress, and for the same stress gradient k a hundred times.

The idea. Meriam (A/2): by definition dI_x = y^2\,dA and dI_y = x^2\,dA, so

the first two rectangular and the third the polar moment of inertia (sometimes J), using x^2 + y^2 = r^2; for areas with rectangular boundaries, Eq. A/3 is the easy way to I_z. An element at -y adds as much as one at +y, so I is always positive, while a first moment may be positive, negative or zero. Dimensions are length⁴: m⁴ or mm⁴. As for centroids, rectangular coordinates suit straight boundaries and polar coordinates circular ones, and the element should make the integration as simple as possible.

The equation (definition: rectangular and polar moments of inertia):

Worked number. A 120 \times 40\,\mathrm{mm} rectangle about its centroidal axes: \bar I_x = 120(40)^3/12 = 0.640(10^6)\,\mathrm{mm^4}, \bar I_y = 40(120)^3/12 = 5.76(10^6)\,\mathrm{mm^4}, \bar I_z = 6.40(10^6)\,\mathrm{mm^4}.

Real example (one you can see). Floor joists are always installed on edge, never flat, and a sheet of paper that flops when flat stands up when folded into a corrugation: both put area far from the bending axis.

Plausible wrong answer. "Same area, same strength." The ruler's area is 90\,\mathrm{mm^2} both ways, but its second moment differs a hundredfold.

Check. Can a hole make a composite area's I negative? (No: the remaining area is still real area, each element contributing y^2\,dA > 0; a hole only reduces I.)

Narration spine. The ruler lies flat; its section's thin strips sit close to the axis, their y^2 bars tiny. It rotates onto its edge: the strips spread far above and below the axis and their y^2 bars grow, the total climbing a hundredfold.

Bridge. I has awkward units, length⁴. A single length can describe how far out the area sits.

Nugget 03: The Radius of Gyration Is a Root-Mean-Square Distance

Claim. The radius of gyration k_x = \sqrt{I_x/A} is the distance at which the whole area, squeezed into a thin strip, would have the same moment of inertia. It is the root-mean-square distance of the area from the axis, and it is always larger than the centroid's distance.

The picture. Meriam's Fig. A/3, for a rectangle about its base.

Check with numbers. A rectangle of height h about its base: I_x = bh^3/3 (SP A/1), A = bh, so k_x^2 = h^2/3 and k_x = h/\sqrt3 = 0.577h. Its centroid is at \bar y = h/2 = 0.5h. For h = 40\,\mathrm{mm}: k_x = 23.1\,\mathrm{mm}, the centroid 20\,\mathrm{mm}; k_x^2/\bar y^2 = (1/3)/(1/4) = 4/3.

The idea. Meriam (A/2): picture the area A concentrated in a narrow strip at distance k_x from the x-axis; it has the same moment of inertia if k_x^2A = I_x. Similarly for y, and for a narrow ring of radius k_z about O:

The radius of gyration measures how the area is distributed about the axis, and a moment of inertia can be specified by k and A. Meriam's caution: do not confuse k with the centroid's coordinate. \bar y^2 is the square of the mean distance of the elements from the axis; k_x^2 is the mean of the squares, and the square of the mean is less than the mean of the squares, so I_x \ne A\bar y^2.

The equation (definition: radius of gyration):

Worked number. A circle of radius r (SP A/3, Nugget 05): k_z = r/\sqrt2 = 0.707r and k_x = r/2; check k_x^2 + k_y^2 = r^2/4 + r^2/4 = r^2/2 = k_z^2.

Real example. Steel section tables list the radius of gyration of every rolled shape beside its I: column-buckling checks use the slenderness ratio, length over k, directly.

Plausible wrong answer. "I_x = A\bar y^2." That puts all the area at its mean distance; the spread of the area about that distance adds more. For the rectangle about its base, A\bar y^2 = bh^3/4, short of bh^3/3.

Check. Which is larger for any area about a given axis, k or the centroid's distance \bar y? (k, unless all the area is at one distance from the axis.)

Narration spine. The rectangle slides together, row by row, into a single thin strip; the strip stops at h/\sqrt3, above the centroid's marker at h/2, and the two dashed lines are labelled "root mean square" and "mean".

Bridge. Tables give I about centroidal axes. Real problems need it about other axes.

Nugget 04: The Parallel-Axis Theorem Adds Area Times Distance Squared

Claim. The moment of inertia about any axis equals that about the parallel axis through the centroid plus the area times the square of the distance between them: I = \bar I + Ad^2. The centroidal value is therefore the smallest for any direction.

The picture. Meriam's Fig. A/4, and a triangle about three parallel axes.

Check with numbers (Meriam's Sample Problem A/2). A triangle of base b and height h; strips parallel to the base, dA = [(h - y)b/h]\,dy:

The centroid is h/3 up: \bar I = bh^3/12 - (bh/2)(h/3)^2 = bh^3/36. Up to the vertex, 2h/3 from the centroid: I_{\text{vertex}} = bh^3/36 + (bh/2)(2h/3)^2 = bh^3/4. For b = 60\,\mathrm{mm}, h = 90\,\mathrm{mm}: 3.645, 1.215 and 10.935(10^6)\,\mathrm{mm^4}.

The idea. Meriam (A/2), transfer of axes: with x_0–y_0 through the centroid C and x at distance d_x, an element's moment about x is (y_0 + d_x)^2\,dA; integrating, I_x = \int y_0^2\,dA + 2d_x\int y_0\,dA + d_x^2\int dA. The first term is \bar I_x; the second is zero because \int y_0\,dA = A\bar y_0 and \bar y_0 = 0 for centroidal axes (S18's first moment about a centroidal axis); the third is Ad_x^2:

These are the parallel-axis theorems. The axes must be parallel, and one must pass through the centroid. The axes may be in the plane or normal to it. SP A/1 uses them: the rectangle about its base, \frac1{12}bh^3 + bh(h/2)^2 = \frac13bh^3; about a corner O, \bar I_z + A[(b/2)^2 + (h/2)^2] = \frac13A(b^2 + h^2).

The equation (mathematical identity: the parallel-axis theorem, from the zero first moment about centroidal axes):

Worked number. The 120 \times 40\,\mathrm{mm} rectangle about its long edge: 0.640(10^6) + 4800(20)^2 = 2.56(10^6)\,\mathrm{mm^4}, four times its centroidal value.

Real example. An I-beam's flanges contribute almost all its stiffness through their Ad^2 terms: their own centroidal \bar I is tiny, but their area sits far from the beam's axis.

Plausible wrong answer. "The parallel-axis theorem lets me subtract Ad^2 to go to any axis." Subtracting is allowed only when moving to the centroidal axis.

Check. For a given direction, about which axis is I smallest? (The centroidal one: every other parallel axis adds a positive Ad^2.)

Narration spine. An area with its centroidal axis; a parallel axis slides away, and a bar labelled Ad^2 grows on top of the fixed bar \bar I. Then a triangle with three axes, base, centroid, vertex, and three numbers, 1/12, 1/36, 1/4, dropping into place.

Bridge. Integrating needs an element; the best ones have known moments of their own.

Nugget 05: Pick an Element Whose Own Second Moment Is Known

Claim. Choose an element whose every part is at the same distance from the axis (a strip parallel to it, a ring about a pole), or one whose own moment of inertia is known (a thin rectangle about its base, y^3\,dx/3). Then one integration covers the area.

The picture. Meriam's Sample Problems A/4 and A/3.

Check with numbers. SP A/4: the area under x = ky^2 from x = 0 to 4, through (4, 3), so k = 4/9. (a) Horizontal strip, all at distance y: dA = (4 - x)\,dy = 4(1 - y^2/9)\,dy, I_x = \int_0^3 4y^2(1 - y^2/9)\,dy = 72/5 = 14.4 (units)⁴. (b) Vertical strip of height y = \frac32\sqrt x, whose own moment about its base (SP A/1) is \frac13y^3\,dx: I_x = \frac13\int_0^4(\frac32\sqrt x)^3\,dx = 14.4. SP A/3, a circle of radius r: a ring has all its area at r_0, dA = 2\pi r_0\,dr_0, so

by symmetry and Eq. A/3; directly, \int_0^{2\pi}\int_0^r(r_0\sin\theta)^2r_0\,dr_0\,d\theta = \pi r^4/4, the same, with more work.

The idea. Meriam (A/2): the guidelines for elements are those of S18, now for squared distances. SP A/1's hint: starting from dx\,dy and integrating over x first just multiplies by b, giving the strip y^2b\,dy. SP A/4: the two strips are equally good, but the vertical one needs the rectangle's moment about its base. SP A/3: polar coordinates and the lowest-order element, the ring, whose polar moment is obviously its area times r_0^2. SP A/6, the area between the y-axis and two circular arcs of radius a centred at O and A(a, 0): a vertical strip covers it in one integration (a horizontal strip would need two because of the corner), and its moment is that of a strip of height y_2 minus one of height y_1, dI_x = \frac13(y_2^3 - y_1^3)\,dx, with y_2 = \sqrt{a^2 - x^2}, y_1 = \sqrt{a^2 - (x - a)^2} and limits 0 to a/2, where the arcs meet: I_x = (a^4/96)(9\sqrt3 - 2\pi) = 0.0969a^4.

The equation (geometric relationship: second moments of simple elements):

Worked number. A solid circular shaft 50\,\mathrm{mm} in diameter: I_x = \pi(25)^4/4 = 3.07(10^5)\,\mathrm{mm^4}, J = I_z = 6.14(10^5)\,\mathrm{mm^4}, k_z = 17.7\,\mathrm{mm}.

Real example. Drive shafts are tubes rather than solid bars: the polar moment is dominated by the rings near the outside, so the core adds weight and little J.

Plausible wrong answer. "For a vertical strip, dI_x = y^2\,dA with y the strip's height." Its parts are at all distances from 0 to y; its moment is \frac13y^3\,dx, not y^2(y\,dx).

Check. Why is the ring the natural element for the polar moment of a circle? (All its area is at the same distance r_0 from the centre.)

Narration spine. The parabolic area; a horizontal strip sweeps up with one distance label; a vertical strip sweeps across, its moment drawn as a little y^3/3 stack. Then a disc fills ring by ring from the centre, the outer rings carrying most of the total.

Bridge. When neither axis passes through the centroid, the transfer must be done in two steps.

Nugget 06: Transfer Through the Centroid, Never Directly Between Two Off-Centroid Axes

Claim. To move a moment of inertia between two parallel axes neither of which is centroidal, go first to the centroidal axis (subtracting Ad_1^2), then to the target (adding Ad_2^2). A direct transfer by the difference of the distances is wrong.

The picture. Meriam's Sample Problem A/5: a semicircle of radius 20\,\mathrm{mm} whose diameter is 15\,\mathrm{mm} above the x-axis.

Check with numbers. About its diameter, half a circle's: I = \frac12(\pi r^4/4) = \pi(20)^4/8 = 2\pi(10^4)\,\mathrm{mm^4}. The centroid is \bar r = 4r/3\pi = 8.49\,\mathrm{mm} above the diameter; A = \pi(20)^2/2 = 628.3\,\mathrm{mm^2}.

• Step 1, to the centroid: \bar I = 2\pi(10^4) - 628.3(8.49)^2 = 1.756(10^4)\,\mathrm{mm^4}.
• Step 2, to x, 15 + 8.49 = 23.49\,\mathrm{mm} below the centroid: I_x = 1.756(10^4) + 628.3(23.49)^2 = 36.4(10^4)\,\mathrm{mm^4}.

A direct "transfer" of 15\,\mathrm{mm} from the diameter would give 2\pi(10^4) + 628.3(15)^2 = 20.4(10^4)\,\mathrm{mm^4}, 44 % too small.

The idea. Meriam (A/2): if a transfer is needed between two parallel axes neither of which passes through the centroid, first transfer to the parallel centroidal axis, then from it to the second axis. The cross term 2d\int y_0\,dA vanishes only when the reference axis is centroidal. SP A/5's hint: here neither axis passes through C, so two transfers are needed; with the centroid on one of them, one would do.

The equation (mathematical identity: two applications of Eq. A/6):

Worked number. The triangle of Nugget 04: from its base (d_1 = h/3) to its vertex (d_2 = 2h/3): bh^3/12 - (bh/2)(h^2/9) + (bh/2)(4h^2/9) = bh^3/12 + bh^3/6 = bh^3/4. Shifting by h directly would give bh^3/12 + bh^3/2, wrong.

Real example. Engineers checking a bolted plate about a bolt line, or a crane-rail cap about the girder's axis, always route the calculation through each part's own centroid, exactly as section tables are arranged.

Plausible wrong answer. "Shift by the distance between the two axes." Only if one of them is centroidal.

Check. A semicircle's I about its diameter is known. Is that diameter a centroidal axis? (No: the centroid is 4r/3\pi above it.)

Narration spine. Three parallel lines under the semicircle: the known one, the centroid's and the target. An arrow tries to jump straight from the first to the third and turns red; it reroutes up to the centroid's line, then down to the target, and the number 36.4(10^4) appears.

Bridge. Real sections are several shapes, some removed. Their moments of inertia add.

Nugget 07: Composite Areas: Add Moments of Inertia, Never Radii of Gyration

Claim. The moment of inertia of a composite area about an axis is the sum of its parts' moments about that same axis, holes counted negative: I = \sum(\bar I + Ad^2). Radii of gyration do not add; the composite's is \sqrt{I/A} from the totals.

The picture. Meriam's Sample Problems A/7 and A/8: an 80 \times 60\,\mathrm{mm} rectangle less a quarter circle and a triangle.

Check with numbers (SP A/7). Part (1), the rectangle: A = 4800\,\mathrm{mm^2}, centroid 30\,\mathrm{mm} above x and 40\,\mathrm{mm} from y. Part (2), the quarter circle (r = 30, \bar r = 12.73\,\mathrm{mm}), negative: A = -\frac14\pi(30)^2, d_x = 60 - 12.73, d_y = 12.73, \bar I = -(\pi/16 - 4/9\pi)30^4. Part (3), the triangle, negative: A = -\frac12(40)(30), d_x = 10, d_y = 80 - 40/3, \bar I_x = -\frac1{36}(40)(30)^3, \bar I_y = -\frac1{36}(30)(40)^3.

SP A/8 gets I_x from moments about each part's own base: rectangle \frac13(80)(60)^3 = 5.76(10^6); quarter circle about its base -\pi(30)^4/16 = -0.1590(10^6), to its centroid -0.0445(10^6), to x -1.624(10^6); triangle about its base, on x, -\frac1{12}(40)(30)^3 = -0.090(10^6). Total 4.05(10^6)\,\mathrm{mm^4} again, and k_x = \sqrt{4.05(10^6)/3490} = 34.0\,\mathrm{mm}. (Meriam prints the triangle's term as -0.90(10^6); its own total uses the correct -0.09(10^6).)

The idea. Meriam (A/3): since I is a sum of distance squared times area, the moment of inertia of a composite area about an axis is simply the sum of those of its parts about the same axis; negative parts (holes) have negative I. With many parts, tabulate A, \bar I, d and Ad^2 for each, and I = \sum\bar I + \sum Ad^2. SP A/8's hints: transfer the quarter circle to its own centroid before moving it to x; watch the signs, since a negative area has negative \bar I and A; and use common sense: the holes reduce the rectangle's I. Radii of gyration may not be added: the composite's k = \sqrt{I/A} uses the total I and total A, and its polar k = \sqrt{I_z/A} with I_z = I_x + I_y about axes through the chosen point.

The equation (mathematical identity: additivity of the integral, with Eq. A/6 per part):

Worked number (an I-beam against a solid square of the same area). Flanges 100 \times 10\,\mathrm{mm}, web 10 \times 180\,\mathrm{mm}, overall depth 200\,\mathrm{mm}, area 3800\,\mathrm{mm^2}: I = 2[100(10)^3/12 + 1000(95)^2] + 10(180)^3/12 = 22.9(10^6)\,\mathrm{mm^4}. A square of the same area, 61.6\,\mathrm{mm} on a side: 3800^2/12 = 1.20(10^6)\,\mathrm{mm^4}. The I-section has 19 times the moment of inertia for the same steel.

Real example (one you can see). Look at any steel-framed building or rail bridge: the beams are I-sections, nearly all their area in the two flanges far from the axis, joined by a thin web.

Plausible wrong answer. "Add the parts' radii of gyration." k is a square root of I/A; the parts' k's combine neither by addition nor by any simple rule; recompute from totals.

Check. In SP A/7, why are the Ad² terms of parts 2 and 3 negative? (Their areas are negative: they are holes.)

Narration spine. The plate assembles from a rectangle with two pieces lifting out; a table fills row by row, the negative entries red. The two column totals add to 4.05(10^6). Then an I-beam and a square of equal area stand side by side, with I bars of 22.9 and 1.2.

Bridge. All of these were about axes parallel to the coordinate axes. Rotate the axes and a new quantity, the product of inertia, appears. (S25.)

Synthesis

On edge, the ruler carries ten times the bending moment for the same outer-fibre stress, and its moment of inertia is a hundred times larger, though nothing changed but its orientation. The second moment weighs each piece of area by the square of its distance, so area far from the axis counts for far more. That is why gates feel their lowest metre most, why beams are deep, and why I-sections and tubes exist.

The tools are few. The definitions I_x = \int y^2\,dA, I_z = I_x + I_y; the radius of gyration k = \sqrt{I/A}; the parallel-axis theorem I = \bar I + Ad^2, always through the centroid; and additivity over composite parts, holes negative. The moment of inertia of an area weighs each element by the square of its distance from the axis: it is smallest about the centroid, grows by Ad^2 elsewhere, and adds part by part.

• Central idea: I = \int y^2\,dA, the moment of a distributed intensity proportional to distance.
• Key relationship: I = \bar I + Ad^2 between parallel axes, one centroidal.
• Most important assumption: the axes are parallel and one passes through the centroid; composite parts are summed about one common axis.
• Open question: about inclined axes, I depends on the angle. Which orientation gives the largest and smallest values? (S25.)

Computed views

Neither MATLAB reference covers area moments of inertia. These views are our own, built on the lesson's Meriam examples, in both languages so you can work in whichever you prefer. Both print the same numbers.

Python (numpy, scipy). S24_area_moments.py integrates \int y^2\,dA with quad for the ruler, the rectangle, the triangle and the shaft. It builds the I-beam from three rectangles with the shared sections.py, which applies the parallel-axis theorem part by part, and compares the beam with a square of the same area.

Part 1: by integration (mm^4)
  ruler flat: I = 67.5, M per unit stress = I / c = 45 mm^3
  ruler on edge: I = 6750.0, M per unit stress = I / c = 450 mm^3
  120 x 40 rectangle: I_x = 6.4e+05, I_y = 5.76e+06, J = 6.4e+06
    about its long edge: 2.56e+06
  triangle about its base 0.08333 b h^3 (1/12), about its vertex 0.2500 (1/4)
  50 mm shaft: I_x = 3.068e+05, J = 6.136e+05, k_z = 17.7 mm
  gate 3 m deep: I = b h^3 / 3 = 9.0 m^4, M = 88.3 kN m
Part 2: composites and the parallel-axis theorem (mm)
  I-beam: A = 3800 mm^2, I_x = 2.293e+07 mm^4, I_y = 1.682e+06 mm^4
  square of the same area (61.6 mm): I = 1.203e+06, ratio 19.1

MATLAB (Octave-compatible). S24_area_moments.m uses integral, and section_props.m for the composites.

ruler: flat I = 67.5, on edge 6750.0 mm^4; I/c = 45 and 450 mm^3
120 x 40: I_x = 6.4e+05, I_y = 5.76e+06, J = 6.4e+06 mm^4
50 mm shaft: I_x = 3.068e+05, J = 6.136e+05 mm^4
I-beam: A = 3800 mm^2, I_x = 2.293e+07, I_y = 1.682e+06; square 1.203e+06, ratio 19.1

Traps

• Using the first moment when the intensity grows with distance. The moment needs \int y^2\,dA. (From Nugget 01.)
• Expecting a negative moment of inertia. Squared distances make it positive. (From Nugget 02.)
• Taking I = A\bar y^2. That ignores the spread about the centroid; k > \bar y. (From Nugget 03.)
• Subtracting Ad^2 to reach a non-centroidal axis. Transfers go to or from the centroid. (From Nugget 04.)
• Taking y^2\,dA for a strip perpendicular to the axis. Use its own moment, \frac13y^3\,dx. (From Nugget 05.)
• Transferring directly between two off-centroid axes. Go through the centroid. (From Nugget 06.)
• Adding radii of gyration. Add I's and A's, then take \sqrt{I/A}. (From Nugget 07.)
• Dropping a hole's sign in one column only. Both A and \bar I are negative. (From Nugget 07.)

Self-check

1. A 120 \times 40\,\mathrm{mm} rectangle: find \bar I_x, \bar I_y, \bar I_z, I about its long edge, and the three centroidal radii of gyration.
2. A triangle of base 60\,\mathrm{mm} and height 90\,\mathrm{mm}: find I about its base, its centroidal axis parallel to the base, and the parallel axis through its vertex.
3. A circle of diameter 50\,\mathrm{mm}: find I_x, I_z, k_x and k_z.
4. A tube of outer diameter 60\,\mathrm{mm} and inner 50\,\mathrm{mm}: find J by treating the bore as a negative area, then compare with a solid shaft of the same area.
5. For SP A/4's area (under y = \frac32\sqrt x, 0 \le x \le 4), find I_y.
6. Find the moment of inertia of SP A/5's semicircle about its own axis of symmetry.
7. A student moves SP A/5's moment of inertia from the diameter straight to the x-axis by adding A(15)^2. What answer do they get, and what is wrong?
8. For a rectangle about its base, compare k_x^2 with \bar y^2, and explain the difference.
9. A T-section: flange 100 \times 20\,\mathrm{mm} on top of a web 20 \times 80\,\mathrm{mm} (total depth 100\,\mathrm{mm}). Find its centroid and its moment of inertia about the horizontal centroidal axis.
10. In question 9, why may the flange's and web's radii of gyration not be added?
11. A vertical gate 1\,\mathrm m wide and 2\,\mathrm m deep with its top at the water surface: find the moment of the water pressure about the surface line, and the depth of the resultant.
12. A ruler 25 \times 2\,\mathrm{mm}: by what factor does I grow from flat to on edge?

Fully worked solutions

1. \bar I_x = 120(40)^3/12 = 0.640(10^6), \bar I_y = 40(120)^3/12 = 5.76(10^6), \bar I_z = 6.40(10^6)\,\mathrm{mm^4}; about the long edge 0.640(10^6) + 4800(20)^2 = 2.56(10^6)\,\mathrm{mm^4}. k_x = \sqrt{0.640(10^6)/4800} = 11.55\,\mathrm{mm}, k_y = 34.64\,\mathrm{mm}, k_z = 36.51\,\mathrm{mm} (and 11.55^2 + 34.64^2 = 36.51^2).

2. Base: bh^3/12 = 60(90)^3/12 = 3.645(10^6); centroid: bh^3/36 = 1.215(10^6); vertex: bh^3/4 = 10.935(10^6)\,\mathrm{mm^4}.

3. r = 25\,\mathrm{mm}: I_x = \pi r^4/4 = 3.07(10^5)\,\mathrm{mm^4}, I_z = \pi r^4/2 = 6.14(10^5)\,\mathrm{mm^4}, k_x = r/2 = 12.5\,\mathrm{mm}, k_z = r/\sqrt2 = 17.7\,\mathrm{mm}.

4. J = \frac\pi{32}(60^4 - 50^4) = 0.659(10^6)\,\mathrm{mm^4}, area \frac\pi4(60^2 - 50^2) = 864\,\mathrm{mm^2}. A solid shaft of that area has r = \sqrt{864/\pi} = 16.6\,\mathrm{mm} and J = \pi r^4/2 = 0.119(10^6)\,\mathrm{mm^4}: the tube has 5.5 times the polar moment.

5. Vertical strip at x, height \frac32\sqrt x, every part at distance x from the y-axis: I_y = \int_0^4 x^2\cdot\frac32\sqrt x\,dx = \frac32\cdot\frac{4^{3.5}}{3.5} = 54.9 (units)⁴.

6. About the axis of symmetry the semicircle is half a circle's area about a diameter: \pi r^4/8 = \pi(20)^4/8 = 6.28(10^4)\,\mathrm{mm^4} (that axis is centroidal for the semicircle).

7. 2\pi(10^4) + 628.3(15)^2 = 20.4(10^4)\,\mathrm{mm^4}. The diameter is not a centroidal axis, so the cross term does not vanish; the correct route through the centroid gives 36.4(10^4)\,\mathrm{mm^4}.

8. k_x^2 = h^2/3 and \bar y^2 = h^2/4. \bar y^2 is the square of the mean distance; k_x^2 the mean of the squared distances; the difference, h^2/12, is the spread of the area about its centroid, \bar I_x/A.

9. A = 2000 + 1600 = 3600\,\mathrm{mm^2}; from the bottom, \bar y = [2000(90) + 1600(40)]/3600 = 67.78\,\mathrm{mm}. \bar I = 100(20)^3/12 + 2000(22.22)^2 + 20(80)^3/12 + 1600(27.78)^2 = 3.14(10^6)\,\mathrm{mm^4}.

10. Each part's k refers to its own centroidal axis, and k^2 = I/A is a ratio: the composite's k = \sqrt{3.14(10^6)/3600} = 29.5\,\mathrm{mm} comes only from the totals.

11. M = \rho g\int y^2\,dA = \rho gbh^3/3 = 9810(1)(8)/3 = 26.16\,\mathrm{kN\,m}; R = \rho g\bar hA = 19.62\,\mathrm{kN}; depth M/R = 1.333\,\mathrm m, two thirds of the depth.

12. I_{\text{edge}}/I_{\text{flat}} = [2(25)^3/12]/[25(2)^3/12] = (25/2)^2 = 156.