Beam Deflection Calculator
Calculate the maximum deflection of a simply-supported beam under uniform or point load.
About the Beam Deflection Calculator
This calculator computes the maximum mid-span deflection of a simply-supported beam under either a Uniformly Distributed Load (UDL) or a Central Point Load. Beam deflection is a critical parameter in structural engineering, ensuring that structural members meet code serviceability limits.
Uniform Load (UDL) Formula
$$ \delta = \frac{5 w L^4}{384 E I} $$
Central Point Load Formula
$$ \delta = \frac{P L^3}{48 E I} $$
How to use this calculator
- Select your **Load Type** using the tabs above (Uniform UDL or Central Point Load).
- For UDL, enter the load per unit length \(w\) (e.g. \(\text{N/m}\)). For a Point Load, enter the concentrated central force \(P\) (e.g. \(\text{N}\)).
- Enter the span length of the beam \(L\) in meters.
- Enter the material's modulus of elasticity \(E\) in GPa (e.g., steel is typically \(200\,\text{GPa}\), concrete is around \(25\text{–}35\,\text{GPa}\)).
- Enter the cross-section's second moment of area (Moment of Inertia) \(I\) in \(\text{mm}^4\).
- Click **Calculate** to see the step-by-step conversion, final deflection in millimeters, and the span-to-deflection ratio (\(L/\delta\)).
The formulas explained
Beam deflection behaves differently depending on how the load is distributed across the span:
- Uniform Distributed Load (UDL): The formula is \(\delta = \frac{5 w L^4}{384 E I}\). Deflection under uniform load scales with the **fourth power** of the span length (\(L^4\)). A small change in beam length has a massive impact on the bending.
- Central Point Load: The formula is \(\delta = \frac{P L^3}{48 E I}\). Concentrating the load at a single point at the center of the span causes larger localized bending compared to a distributed load of the same total magnitude, scaling with the **third power** (\(L^3\)).
Where the variables are:
- \(\delta\) = maximum deflection of the beam at midspan (m)
- \(w\) = uniform distributed load per unit length (N/m)
- \(P\) = central point load (N)
- \(L\) = span length of the beam (m)
- \(E\) = modulus of elasticity, or material stiffness (Pa)
- \(I\) = area moment of inertia of the beam cross-section (\(\text{m}^4\))
Worked Examples
Example 1: Uniformly Distributed Load (UDL)
Problem: Find the maximum deflection of a simply-supported beam with a UDL \(w = 2{,}000\,\text{N/m}\), a span \(L = 3\,\text{m}\), a modulus \(E = 200\,\text{GPa}\), and an inertia \(I = 8.0 \times 10^{-6}\,\text{m}^4\).
- Apply the UDL formula: \(\delta = \frac{5 w L^4}{384 E I}\).
- Convert and substitute: \(\delta = \frac{5 \times 2{,}000 \times 3^4}{384 \times (200 \times 10^9) \times (8.0 \times 10^{-6})}\).
- Compute the values: \(\delta = \frac{810{,}000}{614{,}400{,}000} \approx 0.00132\,\text{m}\).
Answer: Deflection is \(1.32\,\text{mm}\) (a deflection ratio of \(L/2275\)).
Example 2: Central Point Load
Problem: Find the maximum deflection of the same beam under a central point load of \(P = 500\,\text{N}\), a span \(L = 2\,\text{m}\), and the same properties.
- Apply the Point Load formula: \(\delta = \frac{P L^3}{48 E I}\).
- Substitute the values: \(\delta = \frac{500 \times 2^3}{48 \times (200 \times 10^9) \times (8.0 \times 10^{-6})}\).
- Compute the values: \(\delta = \frac{4{,}000}{76{,}800{,}000} \approx 0.000052\,\text{m}\).
Answer: Deflection is \(0.052\,\text{mm}\) (a deflection ratio of \(L/38400\)).
Frequently asked questions
What is a UDL?+
A Uniformly Distributed Load (UDL) is a load spread evenly along the entire length of the beam, like a floor's self-weight.
What is moment of inertia?+
A cross-sectional property describing resistance to bending; larger values indicate a stiffer beam shape.
What limit of deflection is acceptable?+
Deflection limits are defined by building codes. Standard limits are often span/360 for floors under live loads (preventing bouncy floors or plaster cracks) and span/240 for roof structures.
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