CIV-33

UDL midspan deflection

δ = 5 w L⁴ / (384 E I). Simply supported, uniform load.

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DeflectionEuler–Bernoulli

Governing equation

\delta=\dfrac{5 w L^4}{384 EI}

where

w: UDL (kN/m)
L: Span (m)
E: Modulus (GPa)
I: Second moment (cm^4)
\delta: Midspan deflection (mm)

Lecture brief

Historical brief

From Euler’s 1744 elastica and Navier’s beam theory to Mohr’s circle and transformed-section RC, structural mechanics grew as a closed-form craft before finite elements. The lab keeps those governing lines for buckling, flexure, joints and influence. This sheet (CIV-33 — UDL midspan deflection) is the form associated with Euler–Bernoulli. Working symbols: $w$ , $L$ , $E$ , $I$ $\rightarrow$ $\delta$ . Integrate EI y'' = M(x) = w x (L−x)/2 twice. Max at midspan.

Purpose

Purpose: compute

\delta

from

w

L

E

I

in Structural & civil via

\delta=\dfrac{5 w L^4}{384 EI}

δ = 5 w L⁴ / (384 E I). Simply supported, uniform load. Use it when a real structural & civil question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

w = 12.000\,\mathrm{kN/m}

L = 6.000\,\mathrm{m}

E = 200.000\,\mathrm{GPa}

I = 8000.000\,\mathrm{cm^4}

, the governing relation

\delta=\dfrac{5 w L^4}{384 EI}

yields

\delta = 12.66\,\mathrm{mm}

. A sagging beam, a UDL, a midspan tick. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

UDL w12 kN/mSpan L6 mModulus E200 GPaSecond moment I8000 cm^4

Outputs

Midspan deflection \delta12.66 mm

Reading speed

Watch on YouTube

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Narration of this film

A sagging beam, a UDL, a midspan tick.

Integrate EI y'' = M(x) = w x (L−x)/2 twice. Max at midspan.

Reading speed

Watch on YouTube