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