$$ \text { Data: } \mathrm{W}=60 \mathrm{kN}, \text { Flange: } 75 \mathrm{mm} \times 12 \mathrm{mm}, \mathrm{Web} ; 100 \mathrm{mm} \times 15 \mathrm{mm} $$ ![enter image description here][1] $$ \begin{array}{l}{\mathrm{a}_{1}=15 \times 88=1320 \mathrm{mm}^{2}} \ {\mathrm{a}_{2}=72 \times 12=900 \mathrm{mm}^{2}} \ {\mathrm{a}_{1}+\mathrm{a}_{2}=1320+900=2220 \mathrm{mm}^{2}} \ {y_{1}=\frac{88}{2}=44 \mathrm{mm}} \ {y_{2}=88+\frac{12}{2}=94 \mathrm{mm}}\end{array} $$ $$ \begin{array}{l}{\overline{Y}_{\text {base }}=\frac{a_{1} y_{1}+a_{2} y_{2}}{a_{1}+a_{2}}=\frac{(1320 \times 44)+(900 \times 94)}{2220}=64.27 \mathrm{mm} \text { from base }} \ {\mathrm{h}_{1}=64.27-\frac{88}{2}=20.27 \mathrm{mm}}\end{array} $$ $$ \begin{array}{l}{\mathrm{h}_{2}=23.73+\frac{12}{2}=29.73 \mathrm{mm}} \ {\mathrm{I}_{\mathrm{NA}}=(\mathrm{MI})_{1}+(\mathrm{MI})_{\mathrm{II}}} \ {\mathrm{I}_{\mathrm{NA}}=\left(\mathrm{I}_{\mathrm{C}}+\mathrm{Ah}^{2}\right)_{1}+\left(\mathrm{I}_{\mathrm{C}}+\mathrm{Ah}^{2}\right)_{\mathrm{II}}}\end{array} $$ $$ \begin{array}{l}{\mathrm{I}_{\mathrm{NA}}=\left(\frac{\mathrm{bd}^{3}}{12}+(\mathrm{b} \times \mathrm{d}) \times \mathrm{h}^{2}\right)_{1}+\left(\frac{\mathrm{bd}^{3}}{12}+(\mathrm{b} \times \mathrm{d}) \times \mathrm{h}^{2}\right)_{\mathrm{II}}} \ {\mathrm{I}_{\mathrm{NA}}=\left(\frac{15 \times 88^{3}}{12}+(1320) \times(20.27)^{2}\right)_{\mathrm{I}}+\left(\frac{75 \times 12^{3}}{12}+(900) \times(29.73)^{2}\right)_{\mathrm{II}}}\end{array} $$ $$ \begin{array}{l}{\mathrm{I}_{\mathrm{NA}}=(1394192.228)_{\mathrm{I}}+(806285.67)_{\mathrm{II}}} \ {\mathrm{I}_{\mathrm{NA}}=2200477.838 \mathrm{mm}^{4}} \ {\mathrm{q}_{0}=0 \quad \text { At top and bottom of section. }} \ {q=\frac{S A \overline{Y}}{b I}}\end{array} $$ $$ \begin{array}{l}{q_{1}=\frac{60 \times 10^{3} \times 75 \times 12 \times 29.73}{75 \times 2200477.838}=9.73 \mathrm{N} / \mathrm{mm}^{2}} \ {q_{2}=\frac{60 \times 10^{3} \times 75 \times 12 \times 29.73}{15 \times 2200477.838}=48.64 \mathrm{N} / \mathrm{mm}^{2}}\end{array} $$ $$ \begin{array}{l}{q_{(\max )}=\frac{60 \times 10^{3} \times(75 \times 12 \times 29.73+15 \times 23.73 \times 11.815)}{15 \times 2200477.838}=56.32 \mathrm{N} / \mathrm{mm}^{2}} \ {q_{(\mathrm{max})}=\frac{60 \times 10^{3} \times 15 \times(64.27 \times 32.135)}{15 \times 2200477.838}=56.32 \mathrm{N} / \mathrm{mm}^{2}}\end{array} $$