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Section modulus: how strong is this beam section in bending?

Calculate the second moment of area, the elastic section modulus and the radius of gyration of rectangles, tubes, round bars and I-beams, and the bending stress for a given moment.

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Results

Elastic section modulus S

21.39

Second moment of area I
98.93
Cross-section area
13.87
Radius of gyration
2.67
Bending stress at the extreme fiber
1,683

What section properties tell you

How much a beam bends, and how much stress builds up in it, depends not only on the material but on the shape of its cross-section. Three properties describe that shape:

  • Second moment of area I (also called area moment of inertia), in in⁴ or cm⁴: governs stiffness and deflection.
  • Elastic section modulus S = I ÷ c, where c is the distance from the neutral axis to the outer fiber, in in³ or cm³: governs bending stress.
  • Radius of gyration r = √(I ÷ A): used in column buckling.

The bending stress at the outer fiber is σ = M ÷ S, where M is the bending moment.

Formulas (bending about the strong axis)

Shape I S
Solid rectangle b × h b h³ ÷ 12 b h² ÷ 6
Hollow rectangle, wall t [B H³ − (B − 2t)(H − 2t)³] ÷ 12 2I ÷ H
Solid circle, diameter d π d⁴ ÷ 64 π d³ ÷ 32
Round tube, outside D, wall t π [D⁴ − (D − 2t)⁴] ÷ 64 2I ÷ D
Symmetric I-beam [b h³ − (b − t_w)(h − 2t_f)³] ÷ 12 2I ÷ h

Example

A 2×10 joist has actual dimensions of 1.5 × 9.25 in. Its section modulus is 1.5 × 9.25² ÷ 6 = 21.39 in³ and its moment of inertia 98.9 in⁴. Under a bending moment of 3,000 lb·ft (36,000 lb·in), the bending stress is 36,000 ÷ 21.39 = about 1,680 psi.

Depth matters most

Because the depth is squared in S and cubed in I, doubling the depth of a rectangular beam makes it four times stronger and eight times stiffer, while doubling its width only doubles both. That is why joists are placed on edge and why I-beams put most of their material in the flanges, far from the neutral axis.

Checking a design

  • Compare the bending stress with the allowable bending stress of the material: for lumber, the adjusted Fb from the NDS design values; for steel, the limits of the AISC specification.
  • Steel design often uses the plastic section modulus Z, which is larger than S (1.5 times for a rectangle, about 1.1 to 1.15 for an I-beam).
  • Use actual, not nominal, lumber sizes: a 2×10 is 1.5 × 9.25 in.
  • Check deflection, shear and bearing as well; the beam deflection calculator covers deflection.

This calculator is a learning and estimating tool. Structural members in buildings must be designed or checked by a qualified professional under the local code.

Frequently asked questions

What is the section modulus of a rectangle?

S = b × h² ÷ 6, where b is the width and h the depth in the direction of bending.

What is the difference between moment of inertia and section modulus?

The second moment of area I governs stiffness and deflection. The section modulus S = I ÷ c governs bending stress, where c is the distance to the outer fiber.

How do you calculate bending stress?

Divide the bending moment by the section modulus: σ = M ÷ S. 36,000 lb·in on a 2×10 (S = 21.39 in³) gives about 1,680 psi.

What is the section modulus of a 2×10?

About 21.39 in³ using the actual size of 1.5 × 9.25 in, with a moment of inertia of about 98.9 in⁴.

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Results are estimates provided for information purposes. They do not replace professional advice.