A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at points A and B.

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Step 1Section properties for given box are\begin{array}{l}I_{N \cdot A}=\left\{\begin{array}{l}\frac{(0.145)(0.3)^{3}}{12} \mathrm{~m}^{4}-\frac{(0.125)(0.28)^{3}}{12} \mathrm{~m}^{4} \\ +2\left[\frac{(0.125)(0.01)^{3}}{12}+0.125(0.01)(0.105)^{2}\right] \mathrm{m}^{4}\end{array}\right\}=125.17 \times 10^{-6} \mathrm{~m}^{4} \\ Q_{A}=\bar{y}_{2}^{\prime} A^{\prime}=0.145 \times(0.125 \times 0.01) \\ Q_{A}=0.18125 \times 10^{3} \mathrm{~m}^{3} \\ Q_{B}=\bar{y}_{1}^{\prime} A^{\prime}=0.105 \times(0.125 \times 0.01) \\ Q_{B}=0.13125 \times 10^{3} \mathrm{~m}^{3}\end{array} Step 2Shear flow at A and B can be determined as\begin{array}{l}q_{A}=\frac{1}{2}\left(\frac{V Q_{A}}{I}\right) \\ q_{A}=\frac{1}{2}\left(\frac{\left(18 \times 10^{3} \mathrm{~N}\right)\left(0.18125 \times 10^{3} \mathrm{~m}^{3}\right)}{125.17 \times 10^{-6} \mathrm{~m}^{4}}\right) \\ q_{A}=13033 \mathrm{~N} / \mathrm{m}=13.033 \mathrm{kN} / \mathrm{m} \\ q_{B}=\frac{1}{2}\left(\frac{V Q_{B}}{I}\right) \\ q_{B}=\frac{1}{2}\left(\frac{\left(18 \times 10^{3} \mathrm{~N}\right)\left(0.13125 \times 10^{3} \mathrm{~m}^{3}\right)}{125.17 \times 10^{-6} \mathrm{~m}^{4}}\right) \\ q_{B}=9437 \mathrm{~N} / \mathrm{m}=9.437 \mathrm{kN} / \mathrm{m}\end{array} ...