A shear force of V = 300 kN is applied to the box girder. Determine the shear flow at points A and B.

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Step 1Find the moment of inertia of the entire section about neutral axis:\begin{array}{l}I=\frac{B D^{3}}{12}-\frac{b d^{3}}{12} \\ I=\frac{(200 \mathrm{~mm})(400 \mathrm{~mm})^{3}}{12}-\frac{(180 \mathrm{~mm})(380 \mathrm{~mm})^{3}}{12} \\ I=2.43586 \times 10^{8} \mathrm{~mm}^{4}\end{array} Step 2Shear flow at point A:q_{A}=\frac{V A \bar{y}}{I}Since the point A lies at the top fibre therefore no area lies above it. So A = 0.\begin{array}{l}q_{A}=\frac{V(0) \bar{y}}{I}=0 \\ q_{A}=0 \mathrm{kN} / \mathrm{mm}^{2}\end{array} Step 3Consider the point B.Area of the section lying above B is given by:\begin{array}{l}A=10 \mathrm{~mm} \times 200 \mathrm{~mm} \\ A=2000 \mathrm{~mm}^{2}\end{array}Distance of centroid of this area from  neutral axis is given by:\bar{y}=\frac{200 \mathrm{~mm}}{2}=100 \mathrm{~mm}Now shear flow at point B:\begin{array}{l}q_{B}=\frac{V A \bar{y}}{I} \\ q_{s}=\frac{(300 \mathrm{kN})\left(2000 \mathrm{~mm}^{2}\right)(100 \mathrm{~mm})}{2.43586 \times 10^{8} \mathrm{~mm}^{4}} \\ q_{s}=0.246 \mathrm{kN} / \mathrm{mm}^{2}\end{array}  ...