Indiacon Civil

Subjects › Strength of Materials › Intermediate

Deflection of beams

Intermediate Strength of Materials Reviewed 24 Oct 2026 2 min read

Strength is not the only requirement for a beam; it must also not sag too much. Excess deflection cracks partitions and finishes, ponds water on roofs, makes floors feel bouncy and jams doors and windows. Codes therefore set limits such as span divided by a number, and designers calculate deflections or use span-to-depth rules.

The elastic curve follows from bending theory: E I d²y/dx² = M(x). Integrate the bending moment once to get slope and twice to get deflection, using the support conditions to fix the constants. Macaulay's method handles several point loads in one expression. The moment-area method reads slope change as the area of the M/EI diagram and deflection as its first moment, convenient for cantilevers. Conjugate beam and unit load methods give the same answers by other routes.

PR_AR_B
Diagram: Indiacon original, not to scale.

A handful of standard results cover most hand checks. A simply supported beam under a uniform load w deflects 5 w L⁴ / (384 E I) at mid-span; under a central point load P, P L³ / (48 E I). A cantilever with an end load deflects P L³ / (3 E I) and with a uniform load w L⁴ / (8 E I). Deflection rises with the fourth power of span for distributed loads, so doubling span multiplies sag sixteen times. Reinforced concrete adds cracking, creep and shrinkage, which can make long-term deflection two to three times the short-term value.

Key formulas
  • E I y″ = M(x)
  • Simply supported, UDL: δ = 5 w L⁴ / (384 E I)
  • Simply supported, central load: δ = P L³ / (48 E I)
  • Cantilever: end load P L³/(3EI); UDL w L⁴/(8EI)
Assumptions and limits
  • Formulas assume linear elastic, prismatic beams with small deflections and negligible shear deformation (fine for span/depth above about 10).
  • For reinforced concrete use cracked-section stiffness and add long-term creep and shrinkage effects.
Solved example

Problem. A steel beam of 6 m span carries 15 kN/m (service). I = 8 × 10⁷ mm⁴, E = 200 GPa. Find mid-span deflection and compare with span/300.

  1. w = 15 N/mm, L = 6 000 mm.
  2. δ = 5 × 15 × 6 000⁴ / (384 × 200 000 × 8 × 10⁷).
  3. 6 000⁴ = 1.296 × 10¹⁵; numerator = 9.72 × 10¹⁶; denominator = 6.144 × 10¹⁵.
  4. δ = 15.8 mm. Limit 6 000/300 = 20 mm, so it passes.

Answer. About 15.8 mm, within span/300 = 20 mm

Quick quiz

1How does UDL deflection vary with span?

With the fourth power of span.

2Name two hand methods for beam deflection.

Double integration (Macaulay) and moment-area; also conjugate beam or unit load.

3Why is long-term RC deflection larger than short-term?

Cracking, creep and shrinkage.

Code note
  • IS 456 (India): Plain and reinforced concrete: general rules for materials, durability, design and construction.
  • IS 800 (India): General construction in steel: limit state design of steel members and connections.

Clauses and documents to read

  • IS 456:2000 cl. 23.2 (control of deflection)
  • IS 800:2007 cl. 5.6.1 (deflection limits)

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • J. M. Gere, B. J. Goodno, Mechanics of Materials, Cengage, latest edition.
  • F. P. Beer, E. R. Johnston, J. T. DeWolf, Mechanics of Materials, McGraw Hill, latest edition.
  • E. P. Popov, Engineering Mechanics of Solids, Prentice Hall, 2nd edition.

Free and official sources

All references for this subject →

Found an error? Write to indiacongroupofengineers [at] gmail [dot] com. For corrections and feedback. We never ask for payment or personal details.