For a Solid rectangular section, show that the maximum shear stress= (1.5xAverage Shear Stress).

(4 marks) May-16, May-17

A symmetrical I-section beam has flange width=220 mm, flange thickness=20mm, web thickness=15 mm and web depth=300mm. It carries a shear force of 110kN. Sketch shear stress distribution across the section.

Let F be the shear force in N

Average shear stress $q_{avg}=\frac{total shear force}{total area}$

$q_{avg}=\frac{F}{b*d}$

Now, shear stress at (1)-(1) section is $q_{1-1}$,

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

$=\frac{F*(b*(\frac{h}{2}-y))*[y+\frac{1}{2}(\frac{h}{2}-y)]}{\frac{b*h^3*b}{12}}$

$=\frac{12F*(\frac{h}{2}-y)*[y+\frac{h}{4}-\frac{y}{2})]}{b*h^3}$

$=\frac{12F}{bh^3}[(\frac{h}{2}-y)(\frac{h}{4}+\frac{y}{2})]$

$=\frac{12F}{bh^3}(\frac{h}{2}-y)*\frac{1}{2}(\frac{h}{2}+y)$

$q_{1-1}=\frac{6F}{bh^3}[(\frac{h}{2}^2-y^2]$ (parabolic equation)

q will be max when y=0 i.e at neautral axis (NA)

$q_{max}=\frac{6F}{bh^3}*\frac{h^2}{4}=\frac{3}{2}*\frac{F}{bh}$

And also, q will be …

For a Solid rectangular section the maximum shear stress= (1.5xAverage Shear Stress). Numerical on Symmetrical I Section

Both solutions are wrong for the t-section. In the flange there is significant horizontal stress that both solutions are disregarding