1. The Concept — Why Beams Bend

When a beam is subjected to transverse loads, it develops internal bending moments (as shown in the BMD). These bending moments cause the beam to curve. The fibres on the concave side are compressed (shortened), and the fibres on the convex side are stretched (lengthened). Somewhere between them — at the neutral axis — the fibres neither stretch nor compress.

The bending stress at any point within the cross-section depends on how far that point is from the neutral axis. The farther from the NA, the greater the stress. This linear variation is the foundation of the flexure formula.

2. Assumptions of Simple Bending Theory

• The beam is subjected to pure bending (moment only, no shear — or shear effects are negligible).
• The material is linear elastic (obeys Hooke’s law: σ = Eε).
• The material is homogeneous and isotropic.
• Plane sections remain plane after bending (Bernoulli’s hypothesis) — cross-sections rotate but do not warp.
• The beam is initially straight and has a constant cross-section.
• The beam bends in the plane of symmetry of the cross-section.
• Deformations are small (small angle approximation applies).

3. The Flexure Formula — Derivation

From the assumption that plane sections remain plane, the strain at a distance y from the neutral axis is:

ε = y/R, where R = radius of curvature of the bent beam.

Using Hooke’s law: σ = Eε = Ey/R

The bending moment is the integral of stress × distance over the cross-section:

M = ∫σ·y·dA = (E/R)∫y²dA = EI/R

Where I = ∫y²dA = second moment of area (moment of inertia).

4. Neutral Axis & Stress Distribution

The neutral axis (NA) is the line within the cross-section where the bending stress is zero. For a beam made of a single material, the NA passes through the centroid of the cross-section.

Stress distribution across the cross-section:

• Stress varies linearly from zero at the NA to maximum at the outermost fibres.
• In sagging (positive moment): fibres above the NA are in compression, fibres below are in tension.
• In hogging (negative moment): fibres above the NA are in tension, fibres below are in compression.
• For symmetric sections (rectangle, circle, I-beam symmetric about horizontal axis): the maximum tensile and compressive stresses are equal in magnitude.
• For unsymmetric sections (T-section, L-section): the NA is not at mid-height, so maximum tensile and compressive stresses differ.

5. Moment of Inertia — Standard Sections

For composite sections (I-beams, T-beams, built-up sections), use the parallel axis theorem:

6. Section Modulus

Section modulus is the single most important parameter for beam design. A larger Z means the beam can carry a larger bending moment at the same allowable stress. This is why I-beams (which concentrate material far from the NA) are far more efficient in bending than solid rectangular sections of the same weight.

Design equation: For a given allowable stress σ_allow and maximum bending moment M_max:

7. Worked Numerical Examples

Example 1: Rectangular Beam — Maximum Bending Stress

Problem: A simply supported beam of rectangular cross-section (100 mm wide × 200 mm deep) has a span of 4 m and carries a central point load of 20 kN. Find the maximum bending stress.

Example 2: Circular Beam

Problem: A cantilever beam of circular cross-section (diameter 80 mm) is 2 m long and carries a point load of 5 kN at the free end. Find the maximum bending stress at the fixed end.

Example 3: Beam Design — Required Section Modulus

Problem: A simply supported beam carries a UDL of 8 kN/m over a span of 6 m. Allowable bending stress is 150 MPa. Find the required section modulus and suggest a rectangular section (width = depth/2).

8. Common Mistakes Students Make

• Using the wrong moment of inertia formula: I = bd³/12 for a rectangle (NOT bd³/3, which is I about the base, not the centroid). Always use the centroidal I unless applying the parallel axis theorem.
• Confusing I and Z: I = second moment of area (mm⁴). Z = section modulus = I/y_max (mm³). They have different units and different physical meanings.
• Mixing mm and m units: If M is in N·mm and I is in mm⁴, the result σ is in N/mm² = MPa. If M is in N·m and I is in m⁴, σ is in Pa. Never mix: e.g., M in N·m with I in mm⁴ gives nonsense.
• Forgetting that NA passes through the centroid: For unsymmetric sections (T-beam, L-section), you must first find the centroid position, then calculate I about the centroidal axis. Using the geometric midpoint instead of the centroid is a common and serious error.
• Applying the flexure formula where shear is significant: The formula σ = My/I assumes pure bending. Near supports and concentrated loads, shear stress can be significant. For short, deep beams, shear effects may dominate over bending.

9. Frequently Asked Questions

What is the flexure formula?

The flexure formula σ = My/I gives the bending stress at any point within a beam’s cross-section. M is the bending moment, y is the distance from the neutral axis, and I is the moment of inertia. The maximum stress occurs at the outermost fibre (y = y_max) and can be simplified to σ_max = M/Z, where Z = I/y_max is the section modulus.

What is the neutral axis?

The neutral axis is the line within a beam’s cross-section where bending stress is zero. Fibres on one side are in tension, fibres on the other side are in compression. For a beam of uniform material, the neutral axis passes through the centroid of the cross-section. All bending stress calculations are measured as distances from this axis.

What is section modulus?

Section modulus Z = I/y_max is the property that directly relates the maximum bending moment a section can carry to the maximum bending stress: σ_max = M/Z. A larger Z means the beam is stronger in bending. I-beams have high section modulus because their material is concentrated far from the neutral axis, making them the most efficient shapes for resisting bending.