Moments of inertia of areas

Why are steel beams shaped like the letter I, floor joists set on edge rather than laid flat, and bicycle frames and flagpoles made of hollow tubes? The answer is a geometric property of a cross-section called its second moment of area, or area moment of inertia. Lesson 6 found the first moment of an area, ∫y dA, which locates the centroid. The second moment, ∫y² dA, measures how far an area is spread from an axis, and it controls how strongly a beam resists bending, how much it deflects, how much load a column can carry before it buckles, and where the resultant of fluid pressure acts on a submerged gate. This lesson defines the second moment, finds it for standard shapes by integration, introduces the parallel-axis theorem for composite sections, and shows why distributing material far from the centroid makes efficient structural members. It completes the statics toolkit that mechanics of materials builds on.

Definitions

For an area A in the x-y plane, the second moments about the x and y axes are:

Each element of area is weighted by the square of its distance from the axis, so area far from the axis counts much more than area near it. Second moments are always positive, have units of length to the fourth power (mm⁴ or m⁴), and depend on the axis chosen. Engineers most often need them about axes through the centroid.

Two related quantities are useful. The polar moment of inertia about an axis perpendicular to the plane through the origin is \(J_O = \int r^2\,dA = I_x + I_y\); it governs the twisting of circular shafts. The radius of gyration, \(k = \sqrt{I/A}\), is the distance at which the whole area could be concentrated to give the same second moment; it appears in column buckling formulas.

Do not confuse the area moment of inertia with the mass moment of inertia, ∫r² dm, which measures a body's resistance to angular acceleration in dynamics. They share a name and a form, but one is a property of a shape and the other of a mass distribution.

Second moments by integration

The parallel-axis theorem

Composite sections are built from parts whose own centroids do not lie on the section's centroidal axis. The parallel-axis theorem relates the second moment about any axis to that about a parallel axis through the area's own centroid, a distance d away:

The Ad² term is often much larger than the part's own Ī, which is why material far from the centroid is so effective. The theorem also shows that the second moment about the centroidal axis is the smallest possible about any parallel axis.

For a composite section, the procedure extends the centroid table of Lesson 6: find the section's centroid, then for each part add its own centroidal second moment and its area times the square of its distance from the section's centroid.

# Second moments of area of sections built from rectangles: (width b, height h, centroid height y), mm
def props(parts):
    A = sum(b * h for b, h, y in parts)
    yb = sum(b * h * y for b, h, y in parts) / A
    I = sum(b * h ** 3 / 12 + b * h * (y - yb) ** 2 for b, h, y in parts)   # parallel-axis theorem
    return A, yb, I

tee = [(200, 20, 190), (20, 180, 90)]
A, yb, I = props(tee)
top = 200 - yb
print(f"T-section: A = {A:.0f} mm2, centroid {yb:.1f} mm up, I = {I / 1e6:.2f} x 10^6 mm4")
print(f"  section modulus: top {I / top / 1e3:.0f} x 10^3 mm3, bottom {I / yb / 1e3:.0f} x 10^3 mm3")

d, bf, tf, tw = 310.0, 165.0, 9.7, 5.8          # wide-flange beam, about W310x39
ibeam = [(bf, tf, d - tf / 2), (tw, d - 2 * tf, d / 2), (bf, tf, tf / 2)]
A2, yb2, I2 = props(ibeam)
side = A2 ** 0.5                                  # solid square bar with the same area
Isq = side ** 4 / 12
print(f"I-beam: A = {A2:.0f} mm2, I = {I2 / 1e6:.1f} x 10^6 mm4, S = {I2 / (d / 2) / 1e3:.0f} x 10^3 mm3")
print(f"Square bar of equal area ({side:.1f} mm): I = {Isq / 1e6:.2f} x 10^6 mm4, ratio {I2 / Isq:.0f}")

M = 142e6                                         # service moment from Lesson 8, N mm
print(f"Bending stress in the I-beam under 142 kN m: {M * (d / 2) / I2:.0f} MPa")

T-section: A = 7600 mm2, centroid 142.6 mm up, I = 28.80 x 10^6 mm4
  section modulus: top 502 x 10^3 mm3, bottom 202 x 10^3 mm3
I-beam: A = 4886 mm2, I = 84.1 x 10^6 mm4, S = 542 x 10^3 mm3
Square bar of equal area (69.9 mm): I = 1.99 x 10^6 mm4, ratio 42
Bending stress in the I-beam under 142 kN m: 262 MPa

The comparison is striking. The I-beam and a solid square bar contain the same amount of steel, yet the I-beam's second moment is about 42 times larger, because most of its area is in flanges 150 mm from the centroid, while the bar's material is clustered around it. The web holds the flanges apart and carries most of the shear (Lesson 8); the flanges do most of the work in bending.

Why the second moment matters

Mechanics of materials, the course that follows statics, uses the second moment in nearly every formula. Three results show why it matters so much:

• Bending stress. In a beam carrying bending moment M, the normal stress at distance y from the neutral axis, which passes through the centroid (Lesson 6), is \(\sigma = My/I\). The largest stress is at the extreme fibre, distance c from the axis, so \(\sigma_{\max} = Mc/I = M/S\), where \(S = I/c\) is the section modulus. Steel section tables list S for every shape.
• Deflection. A beam's deflection is inversely proportional to EI, the product of the material's stiffness (modulus of elasticity E) and the second moment. A simply supported beam with a uniform load deflects \(5wL^4/(384EI)\) at midspan.
• Buckling. A slender column buckles sideways at the Euler load \(P_{cr} = \pi^2EI/L^2\) (for pinned ends), and it buckles about the axis with the smaller I. This is why the compression members of Lesson 5's trusses are made from shapes with large I in every direction, such as tubes and wide-flange sections.

Centre of pressure revisited

Lesson 7 found the force of water on a submerged plane surface, \(F = \rho g h_c A\), and noted that it acts below the centroid. The second moment gives the location directly. For a vertical surface whose centroid is at depth \(h_c\), the centre of pressure is at depth:

For Lesson 7's gate, 1.0 m wide and 1.5 m high with its centroid 2.75 m deep: \(\bar{I} = 1.0 \times 1.5^3/12 = 0.281\) m⁴, so \(h_p = 2.75 + 0.281/(2.75 \times 1.5) = 2.82\) m, matching the trapezoid calculation. The formula works for any shape: circular inspection hatches, triangular gates and inclined panels (using distances measured along the incline).

Principal axes

For unsymmetric shapes such as angles (L-sections) and Z-sections, the second moment changes with the orientation of the axis, and a third property, the product of inertia \(I_{xy} = \int xy\,dA\), is needed. There is always one pair of perpendicular axes through the centroid, the principal axes, about which I takes its maximum and minimum values and the product of inertia is zero; for any shape with an axis of symmetry, that axis is a principal axis. A single angle loaded vertically bends about its principal axes, not about its legs, so it deflects sideways as well as down, a behaviour engineers must account for when angles are used as lintels or bracing. Steel tables list the principal second moments and the least radius of gyration of every angle, because a slender angle in compression buckles about its weakest axis.