Question 13.15: Determine the horizontal and vertical components of the defl...
Determine the horizontal and vertical components of the deflection of the free end of the cantilever shown in Fig. 13.18. The second moments of area of its unsymmetrical section are I_z,I_y and I_{zy} .
The bending moments at any section of the beam due to the applied load W are
M_z = −W(L − x), \ \ \ M_y = 0
Then Eq. (13.14) reduces to
\frac{d^2u}{dx^2} = \frac{M_yI_z – M_zI_{zy}}{E(I_zI_y – I_{zy}^2)} (13.14)
\frac{d^2u}{dx^2} =\frac{W(L-x)I_{zy}}{E(I_zI_y-I_{zy}^2)} (i)
Integrating with respect to x
\frac{du}{dx} =\frac{WI_{zy}}{E(I_zI_y-I_{zy}^2)} \left(Lx-\frac{x^2}{2}+C_1 \right)
When x = 0, (du/dx) = 0 so that C_1 = 0 and
\frac{du}{dx} =\frac{WI_{zy}}{E(I_zI_y-I_{zy}^2)} \left(Lx-\frac{x^2}{2} \right) (ii)
Integrating Eq. (ii) with respect to x
u=\frac{WI_{zy}}{E(I_zI_y-I_{zy}^2)} \left(\frac{Lx^2}{2} – \frac{x^3}{6} + C_2 \right)
When x = 0, u = 0 so that C_2 = 0. Therefore
u=\frac{WI_{zy}}{6E(I_zI_y-I_{zy}^2)} \left(3Lx^2 – x^3 \right) (iii)
At the free end of the cantilever where x = L
u_{fe}=\frac{WI_{zy}L^3}{3E(I_zI_y-I_{zy}^2)} (iv)
The deflected shape of the beam in the xy plane is found in an identical manner from Eq. (13.15) and is
\frac{d^2v}{dx^2} = \frac{M_zI_y – M_yI_{zy}}{E(I_zI_y – I_{zy}^2)} (13.15)
v=-\frac{WI_y}{6E(I_zI_y-I_{zy}^2)} (3Lx^2-x^3) (v)
from which the deflection at the free end is
v_{fe}=-\frac{WI_yL^3}{3E(I_zI_y-I_{zy}^2)} (vi)
The absolute deflection, δ_{fe} , at the free end is given by
δ_{fe} = (u^2_{fe} + v^2_{fe})^{\frac{1}{2} } (vii)
and its direction is at tan^{−1}(u_{fe}/v_{fe}) to the vertical.
Note that if either Gz or Gy is an axis of symmetry I_{zy} = 0 and Eqs. (iv) and (vi) reduce
to
u_{fe}=0 \ \ \ v_{fe}= – \frac{WL^3}{3EI_z} (compare with Eq. (v) of Ex. 13.1)