Question 16.14: A uniform cantilever of arbitrary cross-section and length l......
A uniform cantilever of arbitrary cross-section and length l has section properties I_{x x},I_{y y},\operatorname{and} I_{xy} with respect to the centroidal axes shown in Fig. P.16.14. It is loaded in the vertical (yz) plane with a uniformly distributed load of intensity w/unit length. The tip of the beam is hinged to a horizontal link which constrains it to move in the vertical direction only (provided that the actual deflections are small). Assuming that the link is rigid and that there are no twisting effects, calculate
(a) The force in the link;
(b) The deflection of the tip of the beam.
Answer: (a)\ 3wlI_{x y}/8I_{x x};\;({\bf b})\;w l^{4}/8E I_{x x}.
(a) From Eqs (16.28),
\left\{\begin{matrix} u^{\prime \prime} \\ν^{\prime \prime}\end{matrix} \right\} =\frac{-1}{E\left(I_{xx}I_{yy}-I^{2}_{xy}\right) } \begin{bmatrix} -I_{xy} & I_{xx} \\ I_{yy} & -I_{xy} \end{bmatrix} \left\{\begin{matrix} M_{x} \\ M_{y} \end{matrix} \right\} (16.28)
u^{\prime\prime}={\frac{M_{x}I_{x y}-M_{y}I_{x x}}{E\Big(I_{x x}I_{y y}-I_{x y}^{2}\Big)}} (i)
Referring to Fig. P.16.14,
M_{x}=-{\frac{w}{2}}(l-z)^{2} (ii)
and
M_{y}=-T(l-z) (iii)
in which T is the tension in the link. Substituting for M_{\mathrm{x}}\operatorname{and}M_{y} from Eqs (ii) and (iii) in Eq. (i),
u^{\prime\prime}=-\frac{1}{E\left(I_{xx}I_{y y}-I_{x y}^{2}\right)}\left[w\frac{I_{x y}}{2}(l-z)^{2}-T I_{x x}(l-z)\right]Then
u^{\prime}=-\frac{1}{E\Big(I_{xx}I_{y y}-I_{x y}^{2}\Big)}\Big[w\frac{I_{x y}}{2}\Big(l^{2}z-l z^{2}+\frac{z^{3}}{3}\Big)-T I_{x x}\Big(l z-\frac{z^{2}}{2}\Big)+A\Big]When z = 0, u^{\prime}=0 so that A = 0. Hence,
When z = 0, u = 0 so that B = 0. Hence,
u=-\frac{1}{E\Big(I_{xx}I_{y y}-I_{x y}^{2}\Big)}\bigg[w\frac{I_{x y}}{2}\bigg(l^{2}\frac{z^{2}}{2}-l\frac{z^{3}}{3}+\frac{z^{4}}{12}\bigg)-T I_{x x}\bigg(l\frac{z^{2}}{2}-\frac{z^{3}}{6}\bigg)\bigg] (iv)
Since the link prevents horizontal movement of the free end of the beam, u = 0 when z=l. Hence, from Eq. (iv),
w\frac{I_{x y}}2\biggl(\frac{l^{4}}2-\frac{l^{4}}3+\frac{l^{4}}{12}\biggr)-T I_{x x}\biggl(\frac{l^{3}}2-\frac{l^{3}}6\biggr)=0whence,
T={\frac{3w l I_{x y}}{8I_{x x}}}(b) From Eqs (16.28),
v^{\prime\prime}={\frac{M_{x}I_{y y}-M_{y}I_{x y}}{E\left(I_{x x}I_{y y}-I_{x y}^{2}\right)}} (v)
The equation for υ may be deduced from Eq. (iv) by comparing Eqs (v) and (i). Thus,
\upsilon=\frac{1}{E\Big(I_{xx}I_{y y}-I_{x y}^{2}\Big)}\Bigg[w\frac{I_{x y}}{2}\bigg(l^{2}\frac{z^{2}}{2}-l\frac{z^{3}}{3}+\frac{z^{4}}{12}\bigg)-T I_{x y}\bigg(l\frac{z^{2}}{2}-\frac{z^{3}}{6}\bigg)\Bigg] (vi)
At the free end of the beam where z=l,
v_{\mathrm{FE}}={\frac{1}{E\left(I_{x x}I_{y y}-I_{x y}^{2}\right)}}\left({\frac{w I_{yy}l^{4}}{8}}-T I_{x y}{\frac{l^{3}}{3}}\right)which becomes, since T=3~w lI_{x y}/8I_{x x},
v_{\mathrm{FE}}={\frac{w l^{4}}{8E I_{x x}}}