The shear stress distribution over a beam cross - section is show
The shear stress distribution over a beam cross - section is shown in figure. The beam is of
A. Equal flange I - section
B. Unequal flange I - section
C. Circular cross-section
D. T - section
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Explanation:
Shear stress distribution:
The value of shear stress at any level is given by
\({\bf{\tau }} = \frac{{{\bf{FA\bar Y}}}}{{{\bf{Ib}}}}\)
where F = Applied/Resisting shear force, I = Moment of Inertia of the complete cross-section about NA, b = width of the required level where shear stress is required,
A = Area either above or below the required level, Y = centroidal distance of considered area from NA.
Now, as we can see from the equation width 'b' is inversely proportional to the shear stress of the section.
So from the top of the beam as the width decreases the shear stress will go on increase up to the neutral axis and beyond the neutral axis as with the increase of width down the shear stress starts decreasing.
Shear stress will be maximum at the neutral axis.
