A shear force of V = 450 kN is applied to the box girder. Determine the shear flow at points C and D.

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Step 1enven thatmoment of mertia of cross-section\begin{array}{l}I=\frac{1}{12}(0.2)(0.4)^{3}-\frac{1}{12}(0.18)(0.38) \\\left.I=0.2436 \times 10 \mathrm{~m}^{4}\right) \\A_{c}^{\prime}=0 ; \\\text { Sheer fiow }\left(Q_{e}\right)=A^{\prime} y_{c}=0\end{array}\begin{aligned}\theta_{0} & =\bar{y}_{1}^{\prime} A_{1}^{\prime}+y_{2}^{\prime} A_{2}^{\prime} \\& =0.195(0.01)(0.09)+0.15(0.10)(0.01) \\\left(Q_{D}\right) & =0.3255 \times 10^{-3} \mathrm{~m}^{3}\end{aligned}Step 2 ...