Deflection of Beams — Double Integration and Macaulay's Method with Examples

A beam can be strong enough yet still fail in service if it sags too much — cracking plaster, jamming doors and alarming occupants. Predicting deflection is therefore as important as checking stress. This article covers the double integration method and the more powerful Macaulay's bracket method, with worked examples.

The Governing Equation

From the flexure relationship, the curvature of the elastic curve is M/EI. For small deflections:

EI · d²y/dx² = M(x)

Integrating once gives the slope EI·(dy/dx); integrating again gives the deflection EI·y. The two constants of integration are found from boundary conditions (zero deflection at supports, zero slope at a fixed end, etc.).

Standard Deflection Formulas (Memorise These)

(SS = simply supported.) Note deflection is inversely proportional to EI — the flexural rigidity — so doubling I halves the sag.

Macaulay's Method — The Idea

Instead of writing a separate moment equation for each beam segment, Macaulay's method writes one expression using angle brackets ⟨x − a⟩. The rule: a bracketed term is included only when its contents are positive; when negative, the whole term is dropped (set to zero). Brackets are integrated as a unit:

∫⟨x − a⟩ⁿ dx = ⟨x − a⟩ⁿ⁺¹ / (n+1)

This makes point loads, UDLs and applied moments fall out automatically across the whole span.

Worked Example 1 — Cantilever (Double Integration)

A cantilever of length 3 m carries a 10 kN point load at the free end. E = 200 GPa, I = 4.5 × 10⁷ mm⁴. Find the free-end deflection and slope.

1. δ = WL³/3EI = (10 000 × 3000³) / (3 × 200 000 × 4.5×10⁷) = (10 000 × 2.7×10¹⁰)/(2.7×10¹³) = 10 mm
2. θ = WL²/2EI = (10 000 × 3000²)/(2 × 200 000 × 4.5×10⁷) = 9×10¹⁰/1.8×10¹³ = 0.005 rad (0.29°)

Worked Example 2 — Simply Supported Beam (Macaulay)

A simply supported beam span 6 m carries a 40 kN point load 2 m from the left support A. Set up the Macaulay expression and find the deflection under the load.

1. Reactions: R_A = 40×4/6 = 26.67 kN; R_B = 40×2/6 = 13.33 kN.
2. Moment (Macaulay): M = 26.67x − 40⟨x − 2⟩ (kN, m).
3. EI y'' = 26.67x − 40⟨x − 2⟩
4. EI y' = 13.335x² − 20⟨x − 2⟩² + C₁
5. EI y = 4.445x³ − 6.667⟨x − 2⟩³ + C₁x + C₂
6. Boundary conditions: y = 0 at x = 0 → C₂ = 0; y = 0 at x = 6 → 4.445(216) − 6.667(64) + 6C₁ = 0 → 960.1 − 426.7 + 6C₁ = 0 → C₁ = −88.9.
7. Deflection at x = 2 m: EI y = 4.445(8) − 0 − 88.9(2) = 35.56 − 177.8 = −142.2 kN·m³
8. With EI in consistent units, y = −142.2/EI (downward). For EI = 200×10⁶ kN·mm² converted appropriately, substitute to get the numerical sag.

The negative sign confirms downward deflection. The single bracketed expression handled the discontinuity at the load with no segment-by-segment algebra.

Why Deflection Control Matters

• IS 456 limits final deflection to span/250 (and span/350 or 20 mm after partitions are built, whichever is less).
• Steel floors are often deflection-governed, not strength-governed.
• Pre-camber is built into long-span beams and bridges to offset dead-load deflection.

Common Mistakes

• Integrating bracketed terms with respect to x as if they were ordinary terms — they must stay bracketed.
• Applying a UDL that does not run to the end of the beam without adding a compensating upward UDL in the brackets.
• Forgetting that constants C₁, C₂ are found from boundary conditions, not assumed zero.