Question F12.1: If the beam is subjected to a shear force of V = 100 kN, det......

Consider two vertical rectangles and a horizontal rectangle.

\begin{aligned} I & =2\left[\frac{1}{12}(0.02)\left(0.2^{3}\right)\right]+\frac{1}{12}(0.26)\left(0.02^{3}\right) \\ & =26.84\left(10^{-6}\right) \mathrm{m}^{4} \end{aligned}

Take two rectangles above A.

\begin{aligned} Q_{A} & =2[0.055(0.09)(0.02)]=198\left(10^{-6}\right) \mathrm{m}^{3} \\ \tau_{A} & =\frac{V Q_{A}}{I t}=\frac{100\left(10^{3}\right)\left[198\left(10^{-6}\right)\right]}{\left[26.84\left(10^{-6}\right)\right] 2(0.02)} \\ & =18.4  \mathrm{MPa} \end{aligned}