In case of beams with circular cross-section, what is the ratio o

SOLUTION

Explanation:

As we know, Shear stress is given:

\(\tau = F \times \frac{{A\bar y}}{{Ib}} = \frac{F}{{3I}}\left( {{R^2} - {y^2}} \right)\)

So, shear stress distribution across a circular section is parabolic.

With the increase of y, the shear stress decreases.

At y = R, τ = 0

At y = 0, τ = τmax

So at the neutral axis, the shear stress is maximum.

\(\begin{array}{l} {\tau _{max}} = \frac{F}{{3I}}\times{R^2}\\ I = \frac{\pi }{{64}}{D^4} = \frac{\pi }{4}{R^4}\\ {\tau _{max}} = \frac{F}{3}\times{R^2}\times\frac{4}{{\pi {R^4}}} = \frac{4}{3} \times \frac{F}{{\pi {R^2}}} \end{array}\)

Average shear stress is:

\({\tau _{avg}} = \frac{{shear\;force}}{{Area}} = \frac{F}{{\pi {R^2}}}\)

So, Maximum shear stress becomes,

\({\tau _{max}} = \frac{4}{3} \times {\tau _{avg}}\)

\(\frac{{{\tau _{max}}}}{{{\tau _{avg}}}} = \frac{4}{3}\)