Moment of Inertia & Section Modulus

Calculate second moment of area (I) and section modulus (Z) for rectangular and circular cross-sections, plus the parallel axis theorem

Inputs

b

Width

h

Height (bending direction)

Formula Interpretation

Moment of Inertia — Definition

I = \sum y^2 \Delta a \quad (\text{mm}^4)

The second moment of area sums the product of each area element Δa and the square of its distance y from the neutral axis. Larger I means the cross-section resists bending more strongly. I is a purely geometric property of the cross-section.

Section Modulus — General

Z_c = \frac{I}{y_c}, \quad Z_t = \frac{I}{y_t}

For unsymmetric sections, the compression fibre (y_c) and tension fibre (y_t) may be at different distances from the neutral axis, giving different moduli. For symmetric sections Z_c = Z_t = Z = I/y, where y is the distance to either extreme fibre.

Rectangle

I = \frac{bh^3}{12}, \quad Z = \frac{bh^2}{6}

For a rectangular section b (width) × h (height in bending direction). Note the cubic dependence on h: doubling h multiplies I by 8 but Z by only 4. This is why beams are oriented with their larger dimension vertical.

Circle

I = \frac{\pi d^4}{64}, \quad Z = \frac{\pi d^3}{32}

For a solid circular section of diameter d. A circle is symmetric about every axis, so I and Z are identical in all bending directions. Hollow circular sections (tubes) achieve high I with less material mass.

Parallel Axis Theorem

I_s = I + Al^2

Shifts the moment of inertia from the centroidal axis to a parallel axis at distance l. A is the full cross-sectional area. This theorem is essential for composite sections: compute I for each part about its own centroid, then shift each to the common reference axis and sum.

Knowledge Points

Orientation Matters for Rectangles

A 100 × 210 mm rectangle has Z = 735 000 mm³ when h=210 mm is vertical, but only 350 000 mm³ when rotated 90°. The taller orientation is 2.1× stronger in bending. This is why floor joists and I-beams are always placed with their web vertical.

Z Directly Governs Bending Stress

The maximum bending stress σ_max = M/Z, where M is the bending moment. A larger section modulus Z means a lower stress for the same load. Selecting the right cross-section shape and orientation is therefore the primary lever for reducing bending stress.

I vs Z: Different Design Objectives

Use I when calculating deflection (stiffness: δ ∝ 1/(EI)). Use Z when calculating bending stress (strength: σ = M/Z). A section can have a large I but a small Z if the extreme fibres are far from the neutral axis — e.g. a very deep, narrow T-beam.

Parallel Axis — Composite Sections

To find I of a composed shape (e.g. an I-beam made of flanges + web), split into rectangles, compute each I₀ about its own centroid, then shift all to the common centroid using I_s = I₀ + Ae². Sum all I_s values for the total I.

Worked Example

A simply supported beam has a rectangular cross-section 100 mm × 210 mm. Compare the bending strength when the beam is placed in orientation (a) with h=210 mm vertical versus orientation (b) with h=100 mm vertical.

Knowns

• Cross-section: b × h = 100 mm × 210 mm
• Orientation (a): h = 210 mm vertical (strong axis)
• Orientation (b): h = 100 mm vertical (weak axis, rotated 90°)

Solution

Step 1 — Section Modulus, Orientation (a) [h=210 mm vertical]

Z_{(a)} = \frac{bh^2}{6} = \frac{100 \times 210^2}{6} = 735{,}000\,\text{mm}^3

Step 2 — Section Modulus, Orientation (b) [h=100 mm vertical]

Z_{(b)} = \frac{bh^2}{6} = \frac{210 \times 100^2}{6} = 350{,}000\,\text{mm}^3

Step 3 — Bending Stress Comparison

\sigma_{(a)} = \frac{M}{Z_{(a)}} = \frac{M}{735{,}000} \qquad \sigma_{(b)} = \frac{M}{Z_{(b)}} = \frac{M}{350{,}000}

Step 4 — Stress Ratio

\frac{\sigma_{(b)}}{\sigma_{(a)}} = \frac{735{,}000}{350{,}000} \approx 2.1

Orientation (b) produces 2.1× more bending stress under the same moment. Therefore, beam orientation (a) — with the larger dimension vertical — is 2.1× stronger than orientation (b).

Extended Knowledge

•The I-beam (H-section) concentrates material far from the neutral axis through the flanges, greatly increasing $I$ and $Z$ compared to a solid rectangle of equal mass. The web connects the flanges and carries shear. This shape achieves the highest bending strength per unit weight.
•For a hollow tube (outer diameter $D$ , inner diameter $d$ ): $I = \pi(D^4 - d^4)/64$ , $Z = \pi(D^4 - d^4)/(32D)$ . Removing the low-contribution core material near the neutral axis reduces weight while preserving most of the bending resistance.
•A T-beam has different distances $y_c$ (to compression flange) and $y_t$ (to tension bottom), giving $Z_c \neq Z_t$ . Concrete is weak in tension, so engineers place reinforcing steel near the tension fibre — and the cross-section is designed so $Z_t$ governs (tension limits first).
•Structural steel codes publish tables of $S$ (elastic section modulus) and $Z_p$ (plastic section modulus) for standard profiles. Designers select sections by comparing required $Z = M_{design} / \sigma_{allow}$ against catalogue values, without computing integrals.
•When two materials (e.g. steel + concrete) act together in bending, the weaker material is ''transformed'' to an equivalent area of the stronger, scaled by the modular ratio $n = E_1/E_2$ . The parallel axis theorem then gives the composite section''s $I$ .