Question 13.2: Determine the deflection curve and the deflection of the fre...
Determine the deflection curve and the deflection of the free end of the cantilever shown in Fig. 13.3(a).
The bending moment, M, at any section X is given by
M=-\frac{w}{2} (L-x)^2 (i)
Substituting for M in Eq. (13.3) and rearranging we have
\frac{d^2v}{dx^2} =\frac{M}{EI} (13.3)
EI\frac{d^2v}{dx^2} =-\frac{w}{2} (L-x)^2 = -\frac{w}{2} (L^2 – 2Lx + x^2) (ii)
Integration of Eq. (ii) yields
EI\frac{dv}{dx} =-\frac{w}{2} \left(L^2x – Lx^2 + \frac{x^3}{3} \right) +C_1
When x = 0 at the built-in end, v = 0 so that C_1 = 0 and
EI\frac{dv}{dx} =-\frac{w}{2} \left(L^2x – Lx^2 + \frac{x^3}{3} \right) (iii)
Integrating Eq. (iii) we have
EIv = -\frac{w}{2} \left(L^2\frac{x^2}{2} – \frac{Lx^3}{3} +\frac{x^4}{12} \right) +C_2
and since v = 0 when x = 0, C_2 = 0. The deflection curve of the beam therefore has the equation
v= – \frac{w}{24 \ EI} (6L^2x^2 – 4Lx^3 + x^4) (iv)
and the deflection at the free end where x = L is
v_{tip} = -\frac{wL^4}{8EI} (v)
which is again negative and downwards. The applied loading in this case may be easily expressed in mathematical form so that a solution can be obtained using Eq. (13.5), i.e.
\frac{d^4v}{dx^4} = -\frac{w}{EI} (vi)
in which w = constant. Integrating Eq. (vi) we obtain
EI\frac{d^3v}{dx^3} =-wx + C_1
We note from Eq. (13.4) that
\frac{d^3v}{dx^3} = -\frac{S}{EI} (i.e. −S = − wx + C_1 )
When x = 0, S = −wL so that
C_1 = wL
Alternatively we could have determined C_1 from the boundary condition that when x = L, S = 0.
Hence
EI\frac{d^3v}{dx^3} =-w(x-L) (vii)
Integrating Eq. (vii) gives
EI\frac{d^2v}{dx^2} =-w\left(\frac{x^2}{2} – Lx \right) + C_2
From Eq. (13.3) we see that
\frac{d^2v}{dx^2} =\frac{M}{EI}
and when x = 0, M = −wL²/2 (or when x = L, M = 0) so that
C_2 = -\frac{wL^2}{2}
and
EI\frac{d^2v}{dx^2} = -\frac{w}{2}(x^2 – 2Lx + L^2)
which is identical to Eq. (ii). The solution then proceeds as before.