Simply supported beam under uniformly distributed load

Figure 1: Simply supported beam
Figure 1: Simply supported beam
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 \(x = \frac{L}{2}\)

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