Question 13.1: Determine the deflection curve and the deflection of the fre...

The load W causes the cantilever to deflect such that its neutral plane takes up the curved shape shown Fig. 13.2(b); the deflection at any section X is then v while that at its free end is v_{tip} . The axis system is chosen so that the origin coincides with the built-in end where the deflection is clearly zero.

The bending moment, M, at the section X is, from Fig. 13.2(a)

M = −W(L − x)      (i.e. hogging)                                                 (i)

Substituting for M in Eq. (13.3) we obtain

\frac{d^2v}{dx^2} =\frac{M}{EI}                                       (13.3)

\frac{d^2v}{dx^2} =-\frac{W}{EI} (L-x)

or in more convenient form

EI \frac{d^2v}{dx^2} =- W(L-x)                               (ii)

Integrating Eq. (ii) with respect to x gives

EI\frac{dv}{dx} = -W\left(Lx – \frac{x^2}{2} \right) +C_1

where C_1 is a constant of integration which is obtained from the boundary condition that dv/dx = 0 at the built-in end where x = 0. Hence C_1 = 0 and

EI\frac{dv}{dx} = -W\left(Lx – \frac{x^2}{2} \right)                                              (iii)

Integrating Eq. (iii) we obtain

EIv = -W\left(\frac{Lx^2}{2} – \frac{x^3}{6} \right) +C_2

in which C_2 is again a constant of integration. At the built-in end v = 0 when x = 0 so that C_2 = 0. Hence the equation of the deflection curve of the cantilever is

v= – \frac{W}{6EI} (3Lx^2 – x^3)                                                 (iv)

The deflection, v_{tip} , at the free end is obtained by setting x = L in Eq. (iv). Thus

v_{tip} = – \frac{WL^3}{3EI}                                        (v)

and is clearly negative and downwards.