Problem on Shear Stress

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.

3 Answers

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Let F be the shear force in N

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


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

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





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

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


And also, q will be minimum or zero when y=\frac{h}{2} i.e at extreme fibre.

Now, $\frac{q_{max}}{q_{avg}}=\frac{\frac{3}{2}(F/bh)}{F/bh}=\frac{3}{2}$

$q_{max}=\frac{3}{2} q_{avg}$ for solid rectangular section

F=110KN=$110*10^3 N$

$\bar y_{base}=\frac{220*20*\frac{20}{2}+300*15*(20+\frac{300}{2})+220*20*(20+300+\frac{20}{2})}{220*20+300*15+220*20}$

$\bar y_{base}=170mm$




$I_{NA}=2.59*10^8 mm^4$

Now, shear stress:

enter image description here

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


When b=15mm, $q_{1-1}=19.93 N/mm^2$ (top and bottom flange will be same; due to symmetrical I section)

When b=220mm, $q_{1-1}= 1.35 N/mm^2$

Now, shear syress at N-A i.e q_{N-A}


$q_{NA}=24.71 N/mm^2$ (maximum)


For a Solid rectangular section the maximum shear stress= (1.5xAverage Shear Stress). part 1  of 2 part 2 of 2 Numerical on Symmetrical I Section part 1 part 2


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

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