Hong Life CalculatorsUnderstanding the beam deflection and bending calculator — concept and principle
This calculator gives the maximum deflection, maximum bending moment and maximum bending stress of a straight elastic beam for four textbook cases: a simply supported beam with a centre point load or a uniform load, and a cantilever with an end point load or a uniform load. Use it for quick checks of shelves, joists and steel beams.
How it is calculated
- Deflection: simply supported, centre load PL³/(48EI); simply supported, uniform load 5wL⁴/(384EI); cantilever, end load PL³/(3EI); cantilever, uniform load wL⁴/(8EI).
- Moment: PL/4, wL²/8, PL and wL²/2 respectively. For uniform cases, enter w in kN/m in the load field.
- Stress: σ = M ÷ Z, with M in kN·m and Z in cm³, giving MPa.
- Span/deflection: L ÷ δ, the usual way to state stiffness limits.
E is entered in GPa and I in cm⁴; the calculator converts EI to N·m² (E × 10⁹ × I × 10⁻⁸).
The principle
Euler–Bernoulli beam theory says curvature equals M/(EI): the bending moment bends the beam, and its flexural rigidity EI resists. Integrating curvature twice along the span, with the support conditions (zero deflection at supports, zero slope at a fixed end), gives the deflection formulas. Deflection grows with the cube of span for point loads and the fourth power for uniform loads, which is why a small increase in span makes a beam much more flexible.
Worked example
- Inputs: simply supported, centre point load P = 10 kN, L = 4 m, E = 200 GPa, I = 2,140 cm⁴, Z = 214 cm³ (values typical of a steel I-section about 200 mm deep).
- EI = 200 × 10⁹ × 2,140 × 10⁻⁸ = 4.28 × 10⁶ N·m².
- Deflection: 10,000 × 4³ ÷ (48 × 4.28 × 10⁶) = 0.003115 m = 3.115265 mm.
- Moment: 10 × 4 ÷ 4 = 10 kN·m.
- Stress: 10 × 10⁶ N·mm ÷ 214,000 mm³ = 46.72897 MPa.
- Span/deflection: 4 ÷ 0.003115 ≈ 1,284, so δ ≈ L/1,284.
| Span (m) | Max deflection (mm) |
|---|---|
| 1 | 0.0486760125 |
| 2 | 0.3894080997 |
| 3 | 1.314252336 |
| 4 | 3.115264798 |
| 5 | 6.084501558 |
| 6 | 10.51401869 |
| 8 | 24.92211838 |
How to read the result
Building codes commonly limit deflection to values such as L/250 to L/360 depending on use; L/1,284 is comfortably stiff. Compare the stress with the allowable or design strength of the material (for S235 steel the yield strength is 235 MPa), applying the safety factors your code requires.
Common mistakes
- Entering a total uniform load in kN instead of w in kN/m.
- Using the weak-axis I of an I-beam when it bends about its strong axis.
- Forgetting the beam's own weight and other loads on the same span.
Frequently asked questions
How do I combine two load cases?
In the linear range you can add deflections and moments from separate cases (superposition), as long as the maximums occur at the same point.
Why does a uniform load use 5/384?
It comes from integrating the parabolic moment diagram wx(L − x)/2 twice with zero deflection at both supports.
Which value matters more, stress or deflection?
Both must pass. Long, light beams are usually governed by deflection, short heavily loaded beams by strength.
Sources: W. C. Young and R. G. Budynas, Roark's Formulas for Stress and Strain, McGraw-Hill · AISC, Steel Construction Manual, Part 3 (beam diagrams and formulas)