Question 20.1: Idealize the box section shown in Fig. P.20.1 into an arrang......
Idealize the box section shown in Fig. P.20.1 into an arrangement of direct stress carrying booms positioned at the four corners and panels that are assumed to carry only shear stresses. Hence, determine the distance of the shear center from the left-hand web.
Answer: 225 mm
From either Eq. (20.1) or (20.2), and referring to Fig. S.20.1(a)
B_{1}=\frac{t_{\mathrm{D}}b}{6}\left(2+\frac{\sigma_{2}}{\sigma_{1}}\right) (20.1)
B_{2}=\frac{t_{\mathrm{D}}b}{6}\left(2+\frac{\sigma_{1}}{\sigma_{2}}\right) (20.2)
B_{1}=60\times10+40\times10+\frac{500\times10}{6}(2+1)+\frac{300\times10}{6}(2-1)
i.e.,
i.e.,
B_{2}=3540\,\mathrm{mm}^{2}=B_{3}Since the section is now idealized, the shear flow distribution due to an arbitrary shear load S_{y} applied through the shear center is, from Eq. (20.11), given by
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^s_0 t_Dx ds+\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^s_0 t_Dy ds+\sum\limits_{r=1}^{n}B_{r}y_{r}\right)+q_{s,0} (20.11)
q_{s}=-{\frac{S_{y}}{I_{x x}}}\sum\limits_{r=1}^{n}B_{r}y_{r}+q_{s,0} (i)
in which
I_{x x}=2\times4000\times150^{2}+2\times3540\times150^{2}=339\times10^{6}\,{\mathrm{mm}}^{4}.‘Cut’ the section in the wall 12. Then
q_{\mathrm{b,12}}=q_{\mathrm{b,43}}=0q_{\mathrm{b,}41}=-{\frac{S_{y}}{I_{x x}}}\times4000\times(-150)=1.77\times10^{-3}S_{y}
q_{\mathrm{b,32}}=-{\frac{S_{y}}{I_{x x}}}\times3540\times(-150)=1.57\times10^{-3}S_{y}
Since the shear load is applied through the shear center, the rate of twist is zero and q_{s,0} is given by Eq. (17.28) in which
q_{s,0}=-{\frac{\oint q_{b}\,\mathrm{d}s}{\oint\mathrm{d}s}} (17.28)
{\oint{\frac{\mathrm{d}s}{t}}}=2\times{\frac{500}{10}}+{\frac{300}{10}}+{\frac{300}{8}}=167.5
Then
q_{s,0}=-{\frac{1}{167.5}}S_{y}\Biggl(1.57\times10^{-3}\times{\frac{300}{8}}-1.77\times10^{-3}\times{\frac{300}{10}}\Biggr)which gives
q_{s,0}=-0.034\times10^{3}S_{y}The complete shear flow distribution is then as shown in Fig. S.20.1(b).
Taking moments about the intersection of the horizontal axis of symmetry and the left-hand web,
from which
x_{S}=225\,{\mathrm{mm}}
