A T-section beam has a top flange of (120mm x 20mm) and the web of (20mm x 100mm). The overall depth is 120mm. It is subjected to a shear force of 60kN. Draw the shear stress distribution diagram.

Shear force $F=60*10^3 N$

1. Calculation of moment of inertia and $\bar y_{base}$

$\bar y_{base}$ = $\frac{20*100*\frac{100}{2}+20*120*(100+\frac{20}{2})}{20*100+20*120}$

= 82.72mm

$A =20*100+20*120=4400mm$

$I_{base}=\frac{20*100^3}{3}+\frac{120*20^3}{12}+20*120*(100+\frac{20}{2})^2$

$I_{base}=35.7867*10^6 mm^4$

Now,

$I_{NA}$ = $I_{base}-A*(\bar y_{base})^2$

$=35.7867*10^6-4400*(82.72)^2$

$=5.679*10^6 mm^4$

2. Calculation of shear stress:

$q_{1-1}=\frac{F*A \bar y}{Ib}$ at section (1)-(1)

$q_{1-1}=\frac{60*10^3*20*120*(32.28-\frac{20}{2})}{5.679*10^6*b}$

When b=120mm, $q_{1-1}=4.70 N/mm^2$

When b=20mm, $q_{1-1}= 28.24 N/mm^2$

At neautral axis (N.A) :

$q_{NA}=\frac{60*10^3*82.72*20*\frac{82.72}{2}}{5.679*10^6*10}$

$q_{NA}=36.14 N/mm^2$ (considered area below neutral axis)

Note: For calculating shear stress at NA i.e Neutral Axis always go for below Area neutral axis to save time.