Question 13.13: Figure 13.15(a) shows a cantilever beam of length L carrying...

In this problem the bending moment and M/EI diagrams have different geometrical shapes. Choosing the origin of axes at C, Eq. (13.10) becomes

x_B\left(\frac{dv}{dx} \right) _B – x_A\left(\frac{dv}{dx} \right) _A -(v_B – v_A) = \int_{A}^{B}{\frac{M}{EI} } x \ dx                                                (13.10)

x_A\left(\frac{dv}{dx} \right) _A – x_C\left(\frac{dv}{dx} \right) _C -(v_A – v_C) = \int_{C}^{A}{\frac{M}{EI} } x \ dx                                                         (i)

in which (dv/dx)_A = 0, x_C = 0, v_A = 0 . Hence

v_C = \int_{0}^{L}{\frac{M}{EI}x } \ dx                                                   (ii)

From the geometry of the M/EI diagram (Fig. 13.15(c)) and taking moments of areas about C we have

v_C = \left[\left(\frac{-WL}{2EI} \right) \frac{L}{2} \frac{3L}{4} + \frac{1}{2} \left(\frac{-WL}{2EI} \right) \frac{L}{2}\frac{5L}{6} + \frac{1}{2} \left(\frac{-WL}{EI} \right) \frac{L}{2} \frac{2}{3} \frac{L}{2} \right]

which gives

v_C = -\frac{3WL^3}{8EI}

Analytically we have

v_C = \left[\int_{0}^{L/2}{\frac{-Wx^2}{EI/2} } \ dx + \int_{L/2}^{L}{\frac{-Wx^2}{EI} } \ dx \right]

or

v_C = -\frac{W}{EI} \left\{\left[\frac{2x^3}{3} \right]_0^{L/2} + \left[\frac{x^3}{3} \right]_{L/2}^L \right\}

Hence

v_C = -\frac{3WL^3}{8EI}

as before.