CIV-05

UDL midspan deflection

Simply supported beam: δ = 5 w L⁴ / (384 EI).

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DeflectionEuler–BernoulliEN 1992-1-1

Governing equation

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

where

w: Uniform load (kN/m)
L: Span (m)
E: Modulus (GPa)
I: Second moment (cm⁴)
\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-05 — UDL midspan deflection) is the form associated with Euler–Bernoulli · EN 1992-1-1. Working symbols: $w$ , $L$ , $E$ , $I$ $\rightarrow$ $\delta$ . Four integrations of EI y'' = M(x) with y(0)=y(L)=0 give the midspan 5/384 factor for uniform load.

Purpose

Purpose: compute

\delta

from

w

L

E

I

in Structural & civil via

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

Simply supported beam: δ = 5 w L⁴ / (384 EI). 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 = 15.000\,\mathrm{kN/m}

L = 6.000\,\mathrm{m}

E = 30.000\,\mathrm{GPa}

I = 12000.000\,\mathrm{cm⁴}

, the governing relation

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

yields

\delta = 70.313\,\mathrm{mm}

. Prismatic, Euler–Bernoulli, small deflection, simple supports. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Uniform load w15 kN/mSpan L6 mModulus E30 GPaSecond moment I12000 cm⁴

Outputs

Midspan deflection \delta70.313 mm

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Narration of this film

Prismatic, Euler–Bernoulli, small deflection, simple supports.

Four integrations of EI y'' = M(x) with y(0)=y(L)=0 give the midspan 5/384 factor for uniform load.

Reading speed

Watch on YouTube