Determine the shear stress at points A and B if the beam is subjected to a shear force of V = 600 kN. Represent the state of stress on a volume element of these points.
Consider a vertical rectangle and two squares.
\begin{aligned} I= & \frac{1}{12}(0.1)\left(0.3^{3}\right)+(2) \frac{1}{12}(0.1)\left(0.1^{3}\right) \\ & =0.24167\left(10^{-3}\right) \mathrm{m}^{4} \end{aligned}
Take top half of area (above A ).
\begin{aligned} Q_{A} & =y_{1}^{\prime} A_{1}^{\prime}+y_{2}^{\prime} A_{2}^{\prime} \\ & =\left[\frac{1}{2}(0.05)\right](0.05)(0.3)+0.1(0.1)(0.1) \\ & =1.375\left(10^{-3}\right) \mathrm{m}^{3} \end{aligned}
\tau_{A}=\frac{V Q}{I t}=\frac{600\left(10^{3}\right)\left[1.375\left(10^{-3}\right)\right]}{\left[0.24167\left(10^{-3}\right)\right](0.3)}=11.4 \mathrm{MPa}
Take top square (above B ).
Q_{B}=y_{2}^{\prime} A_{2}^{\prime}=0.1(0.1)(0.1)=1\left(10^{-3}\right) \mathrm{m}^{3}
\tau_{B}=\frac{V Q}{I t}=\frac{600\left(10^{3}\right)\left[1\left(10^{-3}\right)\right]}{\left[0.24167\left(10^{-3}\right)\right](0.1)}=24.8 \mathrm{MPa}