How Much a Beam Sags

A loaded beam bends, and its bending obeys a fourth-order equation: the load fixes the fourth derivative of the sag. Four integrations, four constants, four conditions at the ends. Out comes the exact shape.

How it works

Higher-order linear equations

In one sentence

An $n$ th-order linear equation needs $n$ conditions to pin down a solution; $EIy'''' = w$ on $[0, L]$ with $y = y'' = 0$ at both ends has the unique solution $y = \frac{wx(L^3 - 2Lx^2 + x^3)}{24EI}$ .

General formula

{\htmlData{sym=0}{E}}{\htmlData{sym=1}{I}}\,{\htmlData{sym=2}{\frac{d^4y}{dx^4}}} = {\htmlData{sym=3}{w}} \;\Rightarrow\; {\htmlData{sym=5}{y}}(x) = \frac{{\htmlData{sym=3}{w}}\,x\left({\htmlData{sym=4}{L}}^3 - 2{\htmlData{sym=4}{L}}x^2 + x^3\right)}{24\,{\htmlData{sym=0}{E}}{\htmlData{sym=1}{I}}}

Successive integrals of the load give the shear force, the bending moment, the slope and finally the deflection: four steps, four constants, four end conditions.

What the symbols mean

$E$
Young's modulus
How stiff the material is.
$I$
Second moment of area
How the cross-section's shape resists bending.
$\frac{d^4y}{dx^4}$
Fourth derivative
Equals load divided by stiffness.
$w$
Load
Weight per meter along the beam.
$L$
Span
Distance between the supports.
$y$
Deflection
How far the beam sags at position x.

A beam's bending stiffness $EI$ times its fourth derivative equals the load per length $w$ : a fourth-order linear equation.
Integrate four times; each integration brings a constant, so four conditions are needed.
A simply supported beam can't move or resist bending at its ends: $y = y'' = 0$ at $x = 0$ and $x = L$ .
The sag is largest at the middle, $\frac{5wL^4}{384EI}$ : double the span and the sag grows sixteen-fold.

Learn the math

Higher Order Linear EquationsDiff Eq · Lesson 39, Higher Order Differential Equations Higher Order Nonhomogeneous EquationsDiff Eq · Lesson 40, Higher Order Differential Equations