- $I_y=\frac{b\times h^3}{12}=\frac{10 cm\times20 cm^3}{12}=6,666.67 cm^4$
- $S_y=h_1\times b\times((h-e_z)-\frac{h_2}2)=10 cm\times10 cm\times((20 cm-10 cm)-\frac{10 cm}2)=500 cm^3$
- $\tau=V_L=\frac{V_z\times S_y}{I_y\times b}=\frac{5.53 kN\times500 cm^3}{6,666.67 cm^4}=0.415 kN/cm=41.5 kN/m$
- Iy is the second moment of area,
- Sy is the statical moment,
- h1 is the height of the upper cross-section part,
- h2 is the height of the lower cross-section part,
- ez is the centroidal distance,
- h is the total height.