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Maximum Deflection (at center)
1.3333
mm
Deflection (metres) 0.00133333 m
Formula δ = P·L³ / (48·E·I)

What this calculator does

This tool computes the maximum vertical deflection of a simply supported beam (pinned at one end, roller at the other) carrying a single point load applied at mid-span. For this classic load case the largest deflection occurs directly under the load, at the center of the span, and is given by \(\delta = \frac{P L^{3}}{48 E I}\). The result is a universal structural-mechanics formula and applies regardless of country or building code — codes only affect the allowable limits you compare against.

Simply supported beam with a central point load and mid-span deflection
A simply supported beam with a central point load P, showing the maximum deflection at mid-span.

How to use it

Enter the central point load P in newtons, the beam span L in metres, the Young's modulus E in gigapascals (steel ≈ 200 GPa, aluminium ≈ 69 GPa), and the second moment of area (moment of inertia) I in mm⁴. The calculator converts everything to consistent SI units (E to pascals, I to m⁴), applies the formula, and reports the deflection in millimetres and metres.

The formula explained

$$\delta = \frac{P \cdot L^{3}}{48 \cdot E \cdot I}$$ The deflection grows with the cube of the span, so doubling the length increases sag eightfold. Stiffer materials (higher E) and deeper sections (higher I) reduce deflection proportionally. The factor 48 in the denominator is specific to a central point load on a simply supported span.

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Cross section showing the second moment of area of a rectangular beam
The flexural rigidity EI combines material stiffness E with the beam cross-section's second moment of area I.

Worked example

A steel beam with \(P = 10{,}000 \ \text{N}\), \(L = 4 \ \text{m}\), \(E = 200 \ \text{GPa} \ (2\times10^{11} \ \text{Pa})\) and \(I = 50{,}000{,}000 \ \text{mm}^4 \ (5\times10^{-5} \ \text{m}^4)\): $$\delta = \frac{10000 \times 4^{3}}{48 \times 2\times10^{11} \times 5\times10^{-5}} = \frac{640{,}000}{480{,}000{,}000} = 0.001333 \ \text{m} \approx 1.33 \ \text{mm}.$$

FAQ

Is this the maximum deflection? Yes — for a central point load on a simply supported beam the maximum deflection is at mid-span.

Does it include the beam's own weight? No, only the applied point load. Add the self-weight (a uniformly distributed load, \(\frac{5 w L^{4}}{384 E I}\)) separately if needed.

What units should I use? P in newtons, L in metres, E in GPa, I in mm⁴. The tool handles unit conversion internally and returns mm.

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