Cantilever deflection and slope by Castigliano

Data

$L = 2$ m, $w = 10$ kN/m, $P = 8$ kN, $EI = 200\times10^{9}\times 8\times10^{-5} = 16\times10^{6}$ N m$^2$ $= 16000$ kN m$^2$.

Bending moment

Take $x$ from the free end. Add a fictitious couple $M_0$ at the free end for the slope (to be set to zero after differentiation). Let $P$ be treated as a variable load for the deflection.

$$M(x) = -\left(Px + \frac{w x^2}{2} + M_0\right)$$

(hogging; the sign cancels in $M\,\partial M/\partial P$.)

  P = 8 kN   w = 10 kN/m (all along)
   |  v v v v v v v v v v v v
   v
   o==============================# fixed end
   x = 0  <---------- 2 m ------>

Deflection at the free end

$$\frac{\partial M}{\partial P} = -x$$ $$\begin{aligned} \delta &= \frac{\partial U}{\partial P} = \frac{1}{EI}\int_0^L M\frac{\partial M}{\partial P}\,dx = \frac{1}{EI}\int_0^L \left(Px + \frac{wx^2}{2}\right)x\,dx \\ &= \frac{1}{EI}\left[\frac{PL^3}{3} + \frac{wL^4}{8}\right] \end{aligned}$$ $$\delta = \frac{1}{16000}\left[\frac{8\times2^3}{3} + \frac{10\times 2^4}{8}\right] = \frac{21.333 + 20}{16000} = 2.583\times10^{-3}\ \text{m}$$

Here $PL^3/3EI = 1.333$ mm (point load) and $wL^4/8EI = 1.250$ mm (UDL).

Slope at the free end

$$\frac{\partial M}{\partial M_0} = -1$$ $$\theta = \frac{\partial U}{\partial M_0}\Big|_{M_0=0} = \frac{1}{EI}\int_0^L\left(Px + \frac{wx^2}{2}\right)dx = \frac{1}{EI}\left[\frac{PL^2}{2} + \frac{wL^3}{6}\right]$$ $$\theta = \frac{16 + 13.333}{16000} = 0.00183\ \text{rad}$$

The positive signs mean that the free end moves in the direction of the load (downward) and rotates in the same sense as the fictitious couple $M_0$, which is the sense of the rotation caused by the loads.

Answer: Deflection at the free end $= 2.58$ mm downward; slope $= 0.00183$ rad ($0.105^\circ$).