Question 13.14: The cantilever beam shown in Fig. 13.16 tapers along its len...

Choosing the origin of axes at the free end B we have, from Eq. (13.10)

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_B\left(\frac{dv}{dx} \right) _B – (v_A-v_B) = \int_{B}^{A}{\frac{M}{EI_X}x } \ dx                                                         (i)

in which I_x , the second moment of area at any section X, is given by

I_X = I_0\left(1+\frac{x}{L} \right)

Also (dv/dx)_A = 0, x_B = 0, v_A = 0 so that Eq. (i) reduces to

v_B = \int_{0}^{L}{\frac{M_x}{EI_0 (1+\frac{x}{L} )} } dx                                                          (ii)

The geometry of the M/EI diagram in this case will be complicated so that the analytical approach is most suitable. Therefore since M = −Wx, Eq. (ii) becomes

v_B = – \int_{0}^{L}{\frac{Wx^2}{EI_0 (1+\frac{x}{L} )} } dx

or

v_B = – \frac{WL}{EI_0} \int_{0}^{L}{\frac{x^2}{L+x} } dx                                                          (iii)

Rearranging Eq. (iii) we have

v_B = – \frac{WL}{EI_0} \left[\int_{0}^{L}{(x-L) \ dx } + \int_{0}^{L}{\frac{L^2}{L+x} \ dx } \right]

Hence

v_B = – \frac{WL}{EI_0} \left[\left(\frac{x^2}{2} – Lx \right) + L^2 \log_e(L-x) \right] _0^L

so that

v_B = – \frac{WL^3}{EI_0} \left(-\frac{1}{2} + \log_e 2 \right)

i.e.

v_B = -\frac{0.19WL^3}{EI_0}