Question 20.2: The beam section shown in Fig. P.20.2 has been idealized int......
The beam section shown in Fig. P.20.2 has been idealized into an arrangement of direct stress carrying booms and shear stress only carrying panels. If the beam section is subjected to a vertical shear load of 1,495 N through its shear center, each of booms 1, 4, 5, and 8 has an area of 200 mm², and each of booms 2, 3, 6, and 7 has an area of 250 mm², determine the shear flow distribution and the position of the shear center.
Answer: Wall 12, 1.86 N/mm; 43, 1.49 N/mm; 32, 5.21 N/mm; 27, 10.79 N/mm; remaining
distribution follows from symmetry. 122 mm to the left of the web 27
From Eq. (20.6), and referring to Fig. S.20.2(a)
q_{s}=-\left({\frac{S_{x}I_{x x}-S_{y}I_{x y}}{I_{x x}I_{y y}-I_{x y}}}\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) (20.6)
q_{s}=-{\frac{S_{y}}{I_{x x}}}\sum\limits_{r=1}^{n}B_{r}y_{r}where
I_{x x}=4\times250\times80^{2}+2\times200\times50^{2}+2\times200\times40^{2}i.e.,
I_{x x}=8.04\times10^{6}\,{\mathrm{mm}}^{4}Then
q_{s}=-1.86\times10^{-4}\sum\limits_{r=1}^{n}B_{r}y_{r}from which
The remaining shear flow distribution follows from symmetry; the complete distribution is shown in Fig. S.20.2(b).
Taking moments about the mid-point of web 27,
which gives
x_{S}=-122\operatorname*{min}\left(\mathrm{i.e.,to the}\operatorname{left of web\,}27\right)
