What is the shape of the shear stress distribution across a recta
A. Triangular
B. Parabolic only
C. Rectangular only
D. A combination of rectangular and parabolic shape
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
The average shear stress is given by
\(\tau = \frac{{F\left( {A\bar y} \right)}}{{I \times b}}\)
where F = shear force, Ay̅ = moment of the area taken into consideration, I = moment of inertia, b = width of the section.
For rectangular cross-section as shown in the figure below,
\(\bar y = \frac{{\left( {\frac{h}{2} - y} \right)}}{2} + y = \frac{{\left( {\frac{h}{2} + y} \right)}}{2}\)
\(A\bar y = b\left( {\frac{h}{2} - y} \right) \times \frac{1}{2}\left( {\frac{h}{2} + y} \right) = \frac{b}{2}\left( {\frac{{{h^2}}}{4} - {y^2}} \right)\)
\(I = \frac{{b{h^3}}}{{12}}\)
Therefore,
\(\tau = \frac{F}{{\left( {\frac{{b{h^3}}}{{12}}} \right) \times b}} \times \frac{b}{2}\left( {\frac{{{h^2}}}{4} - {y^2}} \right) = \frac{{6F}}{{b{h^3}}} \times \left( {\frac{{{h^2}}}{4} - {y^2}} \right)\)
∴ Shear stress is Parabolic in nature.
For the shear stress to be maximum y should be equal to zero
\(\tau = \frac{{6F}}{{b{h^3}}} \times \left( {\frac{{{h^2}}}{4} - 0} \right) = \frac{{6F}}{{b{h^3}}} \times \frac{{{h^2}}}{4} = \frac{{3F}}{{2\left( {bh} \right)}} = \frac{{3F}}{{2A}}\)
