# Simply supported beam under uniformly distributed load

Property Distribution Maximum
Shear $$\displaystyle wx - \frac{wL}{2}$$ $$\displaystyle \frac{wL}{2}$$
Moment $$\displaystyle \frac{wx^{2}}{2} - \frac{wLx}{2}$$ $$\displaystyle \frac{wL^{2}}{8}$$
Deflection $$\displaystyle \frac{w}{12EI}\bigg( \frac{x^{4}}{2} - Lx^{3} + \frac{L^{3}x}{2} \bigg)$$ $$\displaystyle \frac{5wL^{4}}{384EI}$$
Symbol Description
$$x$$ Distance from left support
$$w$$ Uniformly distributed load (force per unit length)
$$L$$ Span
$$EI$$ Bending stiffness of beam
$$v$$ Deflection
$$\phi$$ Rotation
$$\kappa$$ Curvature
$$M$$ Moment
$$V$$ Shear

Given an applied force per unit length, $$w$$. The shear force at a position, $$x$$ is given by the expression:

\begin{align} V(x) & = \int w\ dx \\ & = wx + C \end{align}

from symmetry:

\begin{align} V(\frac{L}{2}) & = 0 \\ w\frac{L}{2} + C & = 0 \\ C & = -\frac{wL}{2} \end{align}

therefore:

$V(x) = wx - \frac{wL}{2}$

The moment at a position, $$x$$ is given by the expression:

\begin{align} M(x) & = \int V(x)\ dx \\ & = \frac{wx^{2}}{2} - \frac{wLx}{2} + D \end{align}

from boundary conditions:

\begin{align} M(0) & = 0 \\ \implies D & = 0 \end{align}

therefore:

$M(x) = \frac{wx^{2}}{2} - \frac{wLx}{2}$

Moment is at a maximum where $$x = \frac{L}{2}$$

\begin{align} M(\frac{L}{2}) & = \frac{w(\frac{L}{2})^{2}}{2} - \frac{wL(\frac{L}{2})}{2} \\ & = wL^{2} \Bigg( \frac{1}{8} - \frac{1}{4} \Bigg) \\ & = -\frac{wL^{2}}{8} \end{align}

Curvature is given by the expression

\begin{align} M(x) & = \kappa (x) EI \\ \kappa(x) & = \frac{w}{2EI} \Bigg( x^{2} - Lx \Bigg) \end{align}

The rotation of the beam at a position, $$x$$, assuming small displacements, is given by the expresion:

\begin{align} \phi (x) & = \int \kappa (x) \ dx \\ & = \frac{w}{2EI}\Bigg( \frac{x^{3}}{3} - \frac{Lx^{2}}{2} + F \Bigg) \end{align}

By symmetry,

\begin{align} \phi (\frac{L}{2}) & = 0 \\ \\ 0 & = \Bigg( \frac{ (\frac{L}{2}) ^ 2}{3} - \frac{L(\frac{L}{2})^{2}}{2} + F \Bigg) \\ F & = L^{3} \Bigg( \frac{1}{8} - \frac{1}{24} \Bigg) \\ F & = \frac{1}{12}L^{3} \end{align}

Therefore:

$\phi(x) = \frac{w}{2EI}\Bigg( \frac{x^{3}}{3} - \frac{Lx^{2}}{2} + \frac{L^{3}}{12} \Bigg)$

The defelction of a beam at a position, $$x$$, assuming small displacements, is given by the expression:

\begin{align} v(x) &= \int \phi (x) \ dx \\ & = \frac{w}{2EI}\Bigg( \frac{x^{4}}{12} - \frac{Lx^{3}}{6} + \frac{L^{3}x}{12} + G \Bigg) \end{align}

from boundary conditions:

\begin{align} v(0) & = 0 \\ \implies G & = 0 \end{align}

Therefore:

\begin{align} v(x) = \frac{w}{12EI}\Bigg( \frac{x^{4}}{2} - Lx^{3} + \frac{L^{3}x}{2} \Bigg) \end{align}

The maximum deflection occurs where $$x = \frac{L}{2}$$

\begin{align} v(\frac{L}{2}) & = \frac{w}{12EI}\Bigg( \frac{(\frac{L}{2})^{4}}{2} - L(\frac{L}{2})^{3} + \frac{L^{3}(\frac{L}{2})}{2} \Bigg) \\ & = \frac{wL^{4}}{12EI}\Bigg( \frac{1}{2 \times 16} - \frac{1}{8} + \frac{1}{2 \times 2} \Bigg) \\ & = \frac{5wL^{4}}{384EI} \end{align}