Question 16.13: A uniform thin-walled beam ABD of open cross-section (Fig. P......
A uniform thin-walled beam ABD of open cross-section (Fig. P.16.13) is simply supported at points B and D with its web vertical. It carries a downward vertical force W at the end A in the plane of the web. Derive expressions for the vertical and horizontal components of the deflection of the beam midway between the supports B and D. The wall thickness t and Young’s modulus E are constant throughout.
Answer: u=0.186W l^{3}/E a^{3}t,\quad ν=0.177W l^{3}/E a^{3}t.
From Eqs (16.28), the horizontal component of deflection, u, is given by
\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)
\displaystyle u^{\prime\prime}=\frac{M_{x}I_{x y}-M_{y}I_{x x}}{E\left(I_{x x}I_{y y}-I_{x y}^{2}\right)} (i)
in which, for the span BD, referring to Fig. P.16.13, M_{x}=-R_{\mathrm{D}}z,\,M_{y}=0, where R_{\mathrm{{D}}} is the vertical reaction at the support at D. Taking moments about B,
R_{\mathrm{D}}2l+W l=0so that
R_{\mathrm{D}}=-W/2\ \ (\mathrm{downward})Eq. (i) then becomes
u^{\prime\prime}={\frac{W I_{x y}}{2E\left(I_{xx}I_{y y}-I_{x y}^{2}\right)}}z (ii)
From Fig. P.16.13,
I_{x x}={\frac{t(2a)^{3}}{12}}+2a t(a)^{2}+2\left[{\frac{t(a/2)^{3}}{12}}+t{\frac{a}{2}}\left({\frac{3a}{4}}\right)^{2}\right]={\frac{13a^{3}t}{4}}I_{y y}={\frac{t(2a)^{3}}{12}}+2{\frac{a}{2}}t(a)^{2}={\frac{5a^{3}t}{3}}
I_{x y}={\frac{a}{2}}t(-a)\left({\frac{3a}{4}}\right)+a t\Bigl(-{\frac{a}{2}}\Bigr)(a)+{\frac{a}{2}}t(a)\biggl(-{\frac{3a}{4}}\biggr)+a t\Bigl({\frac{a}{2}}\Bigr)(-a\Bigr)=-{\frac{7a^{3}t}{4}}
Equation (ii) then becomes
u^{\prime\prime}=-{\frac{42W z}{113E a^{3}t}} (iii)
Integrating Eq. (iii) with respect to z
u^{\prime}=-{\frac{21W}{113E a^{3}t}}z^{2}+Aand
u=-{\frac{7W}{113E a^{3}t}}z^{3}+A z+B (iv)
When z = 0, u = 0 so that B = 0. Also u = 0 when z = 2l, which gives
A=-{\frac{28W l}{113E a^{3}t}}Then
u={\frac{7W}{113E a^{3}t}}{\bigl(}-z^{3}+4l^{2}z{\bigr)} (v)
At the mid-span point where z=l, Eq. (v) gives
u={\frac{0.186Wl^{3}}{E a^{3}t}}Similarly,
v={\frac{0.177W l^{3}}{E a^{3}t}}