\[\begin{align} v(x) & = \begin{cases} \displaystyle \frac{(x_{0} - L)}{EI}P \bigg( \frac{x^{3}}{6L} + \frac{x_{0}^{2}x}{6L} - \frac{x_{0}x}{3} \bigg), & \text{if}\ x \leq x_{0} \\ \displaystyle \frac{x_{0}P}{EI} \bigg( \frac{x^{3}}{6L} - \frac{x^{2}}{2} + \frac{x_{0}^{2}x}{6L} + \frac{Lx}{3} - \frac{x_{0}^{2}}{6} \bigg), & \text{if}\ x \geq x_{0} \end{cases} \end{align}\]
| Symbol | Description |
|---|---|
| \(v(x)\) | Displacement |
| \(x\) | Distance from left hand support |
| \(x_{0}\) | Distance from left hand support to point load |
| \(L\) | Span |
| \(P\) | Load |
Moment balance
\[\begin{align} P x_{0} & = L R_{2} \\ R_{2} & = \frac{P}{L} x_{0} \end{align}\]
Force balance
\[\begin{align} R_{1} + R_{2} & = P \\ R_{1} = P - \frac{P}{L} x_{0} \\ R_{1} = P (1 - \frac{x_{0}}{L}) \end{align}\]
\[\begin{align} V(x) & = \begin{cases} -R_{1}, & \text{if}\ x < x_{0} \\ R_{2}, & \text{if}\ x > x_{0} \end{cases} \\ & = \begin{cases} -P (1 - \frac{x_{0}}{L}), & \text{if}\ x < x_{0} \\ \frac{P}{L} x_{0}, & \text{if}\ x > x_{0} \end{cases} \end{align}\]
\[\begin{align} M(x) &= \int V(x)\ dx \\ & = \begin{cases} \displaystyle \int -P \bigg(1 - \frac{x_{0}}{L}\bigg)\ dx, & \text{if}\ x < x_{0} \\ \displaystyle \int \frac{P}{L} x_{0}\ dx, & \text{if}\ x > x_{0} \end{cases} \\ & = \begin{cases} \displaystyle -Px \bigg(1 - \frac{x_{0}}{L}\bigg) + A, & \text{if}\ x \leq x_{0} \\ \displaystyle \frac{Px}{L} x_{0} + B, & \text{if}\ x \geq x_{0} \end{cases} \end{align}\]
Impose boundary conditions to resolve \(A\) and \(B\):
\[\begin{align} M(0) & = 0 \\ \implies A & = 0 \end{align}\]
\[\begin{align} M(L) & = 0 \\ \frac{PL}{L} x_{0} + B & = 0 \\ B & = -P x_{0} \end{align}\]
Therefore the bending moment distribution can be expressed as follows:
\[\begin{align} M(x) & = \begin{cases} \displaystyle -Px \bigg(1 - \frac{x_{0}}{L}\bigg), & \text{if}\ x \leq x_{0} \\ \displaystyle (x - L)\bigg(\frac{P}{L} x_{0}\bigg), & \text{if}\ x \geq x_{0} \end{cases} \end{align}\]
\[\begin{align} \kappa(x) = \frac{M(x)}{EI} & = \begin{cases} \displaystyle -\frac{Px}{EI} \bigg(1 - \frac{x_{0}}{L}\bigg), & \text{if}\ x \leq x_{0} \\ \displaystyle \bigg(\frac{x_{0}P}{EI}\bigg)\bigg(\frac{x}{L} - 1\bigg), & \text{if}\ x \geq x_{0} \end{cases} \\ & = \begin{cases} \displaystyle \frac{P(x_{0} - L)}{EI} \bigg(\frac{x}{L}\bigg), & \text{if}\ x \leq x_{0} \\ \displaystyle \bigg(\frac{x_{0}P}{EI}\bigg)\bigg(\frac{x}{L} - 1\bigg), & \text{if}\ x \geq x_{0} \end{cases} \end{align}\]
\[\begin{align} \phi(x) & = \int \kappa(x)\ dx \\ & = \begin{cases} \displaystyle \frac{P(x_{0} - L)}{EI} \int \frac{x}{L}\ dx, & \text{if}\ x \leq x_{0} \\ \displaystyle \frac{x_{0}P}{EI} \int \frac{x}{L} - 1\ dx, & \text{if}\ x \geq x_{0} \end{cases} \\ & = \begin{cases} \displaystyle \frac{P(x_{0} - L)}{EI} \bigg(\frac{x^{2}}{2L}+D\bigg), & \text{if}\ x \leq x_{0} \\ \displaystyle \frac{x_{0}P}{EI} \bigg(\frac{x^{2}}{2L} - x + F\bigg), & \text{if}\ x \geq x_{0} \end{cases} \end{align}\]
Ensure continuity where \(x = x_{0}\):
\[\begin{align} \frac{x_{0}-L}{EI}P\Bigg(\frac{x_{0}^{2}}{2L}+D\Bigg) & = \frac{x_{0}P}{EI}\Bigg(\frac{x_{0}^{2}}{2L}-x_{0}+F\Bigg) \\ \frac{x_{0}^{3}}{2L} + x_{0}D - \frac{1}{2}x_{0}^{2} - LD & = \frac{x_{0}^{3}}{2L} - x_{0}^{2} + Fx_{0} \\ x_{0}D + \frac{1}{2}x_{0}^{2} - LD & = Fx_{0} \\ F & = D + \frac{1}{2}x_{0} - \frac{LD}{x_{0}} \end{align}\]
\[\begin{align} v(x) & = \int \phi(x)\ dx \\ & = \begin{cases} \displaystyle \frac{P(x_{0} - L)}{EI} \int \frac{x^{2}}{2L}+D\ dx, & \text{if}\ x \leq x_{0} \\ \displaystyle \frac{x_{0}P}{EI} \int \frac{x^{2}}{2L} - x + D + \frac{1}{2}x_{0} - \frac{LD}{x_{0}}\ dx, & \text{if}\ x \geq x_{0} \end{cases} \\ & = \begin{cases} \displaystyle \frac{P(x_{0} - L)}{EI} \bigg( \frac{x^{3}}{6L} + Dx + G \bigg), & \text{if}\ x \leq x_{0} \\ \displaystyle \frac{x_{0}P}{EI} \bigg( \frac{x^{3}}{6L} - \frac{x^{2}}{2} + Dx + \frac{x_{0} x}{2} - \frac{LDx}{x_{0}} + H \bigg), & \text{if}\ x \geq x_{0} \end{cases} \end{align}\]
Impose boundary condition, \(v = 0\) where \(x = 0\).
\[\begin{align} v(0) & = 0 \\ \implies G & = 0 \end{align}\]
Impose boundary condition, \(v = 0\) where \(x = L\).
\[\begin{align} v(L) & = 0 \\ 0 & = \frac{x_{0}P}{EI} \Bigg( \frac{L^{3}}{6L} - \frac{L^{2}}{2} + DL + \frac{x_{0} L}{2} - \frac{L^{3}D}{x_{0}} + H \Bigg) \\ H & = \frac{L^{2}}{3} - DL - \frac{x_{0}L}{2} + \frac{DL^{2}}{x_{0}} \end{align}\]
Ensure continuity at \(x = x_{0}\)
\[\begin{align} \frac{x_{0} - L}{EI} P \Bigg( \frac{x_{0}^{3}}{6L} + Dx_{0} \Bigg) & = \frac{x_{0}P}{EI}\Bigg (\frac{x_{0}^3}{6L} - \frac{x_{0}^{2}}{2} + Dx_{0} + \frac{x_{0}^2}{2} - LD + \frac{L^{2}}{3} - DL - \frac{x_{0}L}{2} + \frac{DL^{2}}{x_{0}} \Bigg) \\ \frac{x_{0}^4}{6L} + Dx_{0}^{2} - \frac{x_{0}^{3}}{6} - LDx_{0} & = \frac{x_{0}^{4}}{6L} - \frac{x_{0}^{3}}{2} + Dx_{0}^{2} + \frac{x_{0}^{3}}{2} - DLx_{0} + \frac{L^{2}}{3} x_{0} - DLx_{0} - \frac{L}{2} x_{0}^{2} + DL^{2} \\ -\frac{x_{0}^{3}}{6} & = \frac{L^{2}}{3} x_{0} - DLx_{0} - \frac{L}{2} x_{0}^{2} + DL^{2} \\ DLx_{0} - DL^{2} & = \frac{L^{2}}{3} x_{0} - \frac{L}{2}x_{0}^{2} + \frac{x_{0}^{3}}{3} \\ DL(x_{0} - L) & = (x_{0} - L)\Bigg( \frac{x_{0}^{2}}{6} - \frac{Lx_{0}}{3} \Bigg) \\ D & = \frac{x_{0}^2}{6L} - \frac{x_{0}}{3} \end{align}\]
Substitute expression for \(D\) back into expression for \(H\).
\[\begin{align} H & = \frac{L^{2}}{3} - \Bigg( \frac{x_{0}^{2}}{6L} - \frac{x_{0}}{3} \Bigg) L - \frac{x_{0} L}{2} + \Bigg( \frac{x_{0}^2}{6L} - \frac{x_{0}}{3} \Bigg) \frac{L^{2}}{x_{0}} \\ H & = \frac{L^{2}}{3} - \frac{x_{0}^{2}}{6} + \frac{x_{0}L}{3} - \frac{x_{0}L}{2} + \frac{x_{0}L}{6} - \frac{L^{2}}{3} \\ H & = -\frac{x_{0}^2}{6} \end{align}\]
Therefore we can write an expression for the displacement:
\[\begin{align} v(x) & = \begin{cases} \displaystyle \frac{(x_{0} - L)}{EI}P \bigg( \frac{x^{3}}{6L} + \frac{x_{0}^{2}x}{6L} - \frac{x_{0}x}{3} \bigg), & \text{if}\ x \leq x_{0} \\ \displaystyle \frac{x_{0}P}{EI} \bigg( \frac{x^{3}}{6L} - \frac{x^{2}}{2} + \frac{x_{0}^{2}x}{6L} - \frac{x_{0}x}{3} + \frac{x_{0} x}{2} - \frac{Lx}{x_{0}}\bigg(\frac{x_{0}^{2}}{6L} - \frac{x_{0}}{3} \bigg) - \frac{x_{0}^2}{6} \bigg), & \text{if}\ x \geq x_{0} \end{cases} \\ & = \begin{cases} \displaystyle \frac{(x_{0} - L)}{EI}P \bigg( \frac{x^{3}}{6L} + \frac{x_{0}^{2}x}{6L} - \frac{x_{0}x}{3} \bigg), & \text{if}\ x \leq x_{0} \\ \displaystyle \frac{x_{0}P}{EI} \bigg( \frac{x^{3}}{6L} - \frac{x^{2}}{2} + \frac{x_{0}^{2}x}{6L} + \frac{Lx}{3} - \frac{x_{0}^{2}}{6} \bigg), & \text{if}\ x \geq x_{0} \end{cases} \end{align}\]
where \(x = \frac{L}{2}\) and \(\frac{L}{2} \geq x_{0}\)
\[\begin{align} v\bigg(\frac{L}{2}\bigg) &= \frac{x_{0}P}{EI} \Bigg( \frac{(\frac{L}{2})^{3}}{6L} - \frac{(\frac{L}{2})^{2}}{2} + \frac{x_{0}^{2}(\frac{L}{2})}{6L} + \frac{L(\frac{L}{2})}{3} - \frac{x_{0}^{2}}{6} \Bigg) \\ & = \frac{x_{0}P}{EI} \bigg( \frac{L^{2}}{6 \times 8} - \frac{L^{2}}{8} + \frac{x_{0}^{2}}{12} + \frac{L^{2}}{6} - \frac{x_{0}^{2}}{4 \times 6} \bigg) \\ & = \frac{x_{0}P}{EI} \bigg( L^{2} \bigg[ \frac{1}{48} - \frac{6}{48} + \frac{8}{48} \bigg] - \frac{x_{0}^{2}}{12} \bigg) \\ & = \frac{x_{0}P}{EI} \bigg( \frac{3L^{2}}{48} - \frac{x_{0}^{2}}{12} \bigg) \\ & = \frac{PL^{3}}{EI} \bigg( \frac{3x_{0}}{48L} - \frac{x_{0}^{3}}{12L^{3}} \bigg) \\ & = \frac{PL^{3}}{48EI} \bigg( \frac{3x_{0}}{L} - 4\frac{x_{0}^{3}}{L^{3}} \bigg) \end{align}\]