A symmetrical I-section with flange 250mm x 20mm has a web 160mm x 10mm, if the shear force acting on the section is 80 kN. Find the maximum shear stress developed in the section and draw shear stress distribution diagram.

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

$I_{NA}=\frac{250*(20+160+20)^3}{12}-\frac{(250-10)*160^3}{12}$

$I_{NA}=84.75*10^6 mm^4$

$\bar y_{base}=\frac{250*20*\frac{20}{2}+10*160*(20+\frac{160}{2})+250*20*(20+160+\frac{20}{2})}{250*20+10*160+250*20}$

$\bar y_{base}=100mm$

$F=80KN=80*10^3 N$ (given)

1. Calculation of shear stress:

$q_{1-1}=\frac{F*A \bar y}{Ib}$

$=\frac{80*10^3*20*250*(100-\frac{20}{2})}{84.75*10^6*b}$

When b=10mm, $q_{1-1}=42.47 N/mm^2$

When b=250mm, $q_{1-1}= 1.69 N/mm^2$

$q_{NA}=\frac{80*10^3*[250*20*(100-\frac{20}{2})+(100-20)*10*(\frac{100-20}{2})]}{84.75*10^6*10}$

$q_{NA}=45.49 N/mm^2$ (maximum at neutral axis always for symmetrical section)