Data: B = D = 120 mm, t = 20mm, F = 125 kN

$Find \tau_{max}$

$\mathrm{b}=\mathrm{d}=120-2 \mathrm{t}=120-2 \mathrm{x} 20=80 \mathrm{mm}$

Consider the area above the N.A.

$\left.\text { Shear stress }\left(\tau_{1}\right) \text { at the bottom of flange by taking width ( } \mathrm{b}=120 \mathrm{mm}\right)$

$\tau_{1}=\frac{F A \overline{Y}}{I b}$

$\overline{Y}=60-\frac{20}{2}=50 \mathrm{mm}$

$\mathrm{I}=\frac{1}{12}\left(B D^{3}-b d^{3}\right)=\frac{1}{12}\left(120 \times 120^{3}-80 \times 80^{3}\right)=13.866 \times 10^{6} \mathrm{mm}^{4}$

$\therefore \tau_{1}=\frac{\left(125 \times 10^{3}\right) \times(120 \times 20) \times 50}{13.866 \times 10^{6} \times 120}=9.015 \mathrm{N} / \mathrm{mm}^{2}$

Shear stress $\left(\tau_{2}\right)$ at the bottom of flange by taking width

$(\mathrm{b}=20+20=40 \mathrm{mm})$

$\therefore \tau_{2}=\tau_{1} \times \frac{120}{40}=9.015 \times \frac{120}{40} \quad 27.045 \mathrm{N} / \mathrm{mm}^{2}$

width at N.A. = 20 + 20 = 40 mm

Web area above the N.A.

$\overline{Y}=\frac{40}{2}=20 mm$

$\therefore$ Additional Shear Stress due to web area above the N.A. is given by

$\tau_{\text {additional }}=\frac{F A \overline{Y}}{I b}=\frac{\left(125 \times 10^{3}\right)(1600)(20)}{13.866 \times 10^{6} \times 40}=7.212 N / m m^{2}$

$$ \begin{aligned} \tau_{\max }=\tau_{\mathrm{NA}} &=\tau_{2}+\tau_{\text {additional }} \\ &=27.045+7.212 \\ &=34.256 \mathrm{N} / \mathrm{mm}^{2} \end{aligned} $$