A S/S beam has a span of 9m and carries a UDL of 1600 N/m over the entire span. It is T section with flange dimensions (120mm x 20mm) and web dimensions (25mm x 100mm). Overall depth of section=120mm. Calculate the maximum shear stress. Draw the shear stress distribution diagram.

Note

Web dimension is printed as (100mm x 25mm) instead of (25mm x 100mm) in the question paper as the Overall depth of the section is 120 which is the sum of the depth of flange + depth of web=20+100=120mm.

1.Calculation of moment of inertia $I_{NA}$ and $\bar y$:

$A=120*20+100*25=4900 mm^2$

$\bar y_{base} = \frac{100*25*\frac{100}{2} + 20*120 * \frac{20}{2}+100}{100*25+20*120}$

$\bar y_{top}=120-\bar y_{base}=40.61mm$

$I_{base}=\frac{25*100^3}{3} + \frac{120*20^3}{12} + 120*20*(100+\frac{20}{2})^2$

$=37.453*10^6 mm^4$

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

$=37.453*10^6-4900*79.39^2$

$6.56*10^6 mm^4$

2.A s/s simply supported beam has a span of 9m and carried a UDL of 1600 N/m over entire span.

Max shear force F=$\frac{1600*9}{2}$

$F=7200N$

3.Calculation of maximum shear stress:

Let shear stress at section 1-1 be $q_{1-1}$

$q=\frac{F*A\bar y}{I_{NA}b}$

$q_{1-1}- \frac{7200*(120*20* (\bar y_{top} - \frac{20}{2}}{6.56*10^6*b}$

when b=120mm and $y_{top}=40.61mm$

$q_{1-1}=0.67 N/mm^2$ (1)

when b=25mm and $\bar y_{top}=40.61mm$

$q_{1-1}=3.23 N/mm^2$ (2)

Let shear stress as neutral axis be $q_{NA}$

$q_{NA}= \frac{7200*79.39*25*\frac{79.39}{2}}{6.56*10^6*25}$

$q_{NA}=3.46 N/mm^2$

On comparing 1,2,3

$q_{NA}=3.46 N/mm^2$ is maximum