Question 20.4: Find the position of the shear center of the rectangular fou......
Find the position of the shear center of the rectangular four boom beam section shown in Fig. P.20.4. The booms carry only direct stresses, but the skin is fully effective in carrying both shear and direct stress. The area of each boom is 100 mm².
The x axis is an axis of symmetry so that I_{xy}=0, also the shear center, S, lies on this axis. Apply an arbitrary shear load, S_{y}, through S. The internal shear flow distribution is then given by Eq. (20.11) in which S_{x}=0 and I_{xy}=0. Thus
q_{s}=-\left({\frac{S_{x}I_{x x}-S_{y}I_{x y}}{I_{x x}I_{y y}-I_{x y}^{2}}}\right) \left(\int_{0}^{s}t_{\mathrm{D}}x\,\mathrm{d}s+\sum\limits_{r=1}^{n}B_{r}x_{r}\right) -\left(\frac{S_{y}I_{y y}-S_{x}I_{x y}}{I_{x x}I_{y y}-I_{x y}^{2}}\right) \left(\int_{0}^{s}t_{\mathrm{DJ}}\,\mathrm{d}s+\sum\limits_{r=1}^{n}B_{r}y_{r}\right)+q_{s,0} (20.11)
q_{s}=-{\frac{S_{y}}{I_{xx}}}\left(\int_{0}^{s}t_{\mathrm{D}}y\,\mathrm{d}s+\sum\limits_{r=1}^{n}B_{r}y_{r}\right)+q_{s,0} (i)
in which, from Fig. S.20.4,
I_{x x}=4\times100\times40^{2}+2\times0.64\times240\times40^{2}+{\frac{0.36\times80^{3}}{12}}+{\frac{0.64\times80^{3}}{12}}i.e.,
I_{x x}=1.17\times10^{5}\,\mathrm{mm}^{4}‘Cut’ the section at O. Then, from the first two terms on the right-hand side of Eq. (i),
q_{\mathrm{b,Ol}}=-{\frac{S_{y}}{I_{x x}}}\int_{0}^{\mathrm{o_{1}}}0.64s_{1}\,\mathrm{d}s_{1}i.e.,
q_{\mathrm{b,Ol}}=-0.27\times10^{-6}S_{y}s_{1}^{2} (ii)
and
q_{b,1}=-4.32\times10^{-4}S_{y}Also,
q_{b,12}=-\frac{S_{y}}{I_{x x}}\left(\int_{0}^{s_{2}}0.64\times40\,\mathrm{d}s_{2}+100\times40\right)-4.32\times10^{-4}S_{y}i.e.,
q_{\mathrm{b,12}}=-10^{-4}S_{y}(0.22s_{2}+38.52)whence,
q_{b,2}=-91.32\times10^{-4}S_{y}Finally,
i.e.,
q_{b.23}=-10^{-4}S_{y}\big(0.12s_{3}-0.15\times10^{-2}s_{3}^{2}+125.52\big) (iv)
The remaining q_{b} distribution follows from symmetry. From Eq. (17.27),
q_{s,0}=-{\frac{\oint(q_{\mathrm{b}}/G t)\mathrm{d}s}{\oint\mathrm{d}s/G t}} (17.27)
q_{s,0}=-{\frac{\oint(q_{\mathrm{b}}/t)\mathrm{d}s}{\oint\mathrm{d}s/t}} (v)
in which
\textstyle\oint{\frac{\mathrm{d}s}{t}}=\frac{80}{0.64}+\frac{2\times240}{0.64}+\frac{80}{0.36}=1097.2Now substituting in Eq. (v) for q_{\mathrm{b,O1}},q_{\mathrm{b,12}},\mathrm{and}\;q_{\mathrm{b,23}} from Eqs (ii)-(iv), respectively,
from which
q_{s,0}=70.3\times10^{-4}S_{y}The complete shear flow distribution is then
q_{O1}=-10^{-4}S_{y}(0.27\times10^{-2}s_{1}^{2}-70.3) (vi)
q_{12}=q_{34}=-10^{-4}S_{y}(0.22s_{2}-31.78) (vii)
q_{23}=-10^{-4}S_{y}(0.12s_{3}-0.15\times10^{-2}s_{3}^{2}-55.22) (viii)
Taking moments about the mid-point of the wall 23,
Substituting for q_{\mathrm{Ol}}{\mathrm{~and~}}q_{12} from Eqs (vi) and (vii) in Eq. (ix),
from which
\xi_{S}=142.5\,\mathrm{mm}