For a strongest rectangular beam cut from a circular log, the rat

SOLUTION

For the given timer section of diameter ‘D’:

Let a rectangular section of dimension ‘b × d’ is to be cut out:

The beam is strongest if the section modulus is maximum.

\({\rm{z}} = \frac{{\rm{I}}}{{{{\rm{y}}_{{\rm{max}}}}}} = \frac{{\frac{{{\rm{b}}{{\rm{d}}^3}}}{{12}}}}{{\frac{{\rm{d}}}{2}}} = \frac{{{\rm{b}}{{\rm{d}}^2}}}{{\rm{b}}}\)

\(\therefore {\rm{z}} = \frac{{{\rm{b}}{{\rm{d}}^2}}}{6}\)

\({\rm{z}} = \frac{{{\rm{b}}\left( {{{\rm{D}}^2} - {{\rm{b}}^2}} \right)}}{6} = \frac{{{\rm{b}}{{\rm{D}}^2} - {{\rm{b}}^3}}}{6}\)

For maximum section modulus

\(\frac{{dz}}{{db}} = 0\;{\rm{or\;}}\frac{{{\rm{dz}}}}{{{\rm{dd}}}} = 0\)

We are differentiating section modulus w.r.t width for easier calculation

\(\Rightarrow \frac{{\rm{d}}}{{{\rm{db}}}}\left( {\frac{{{\rm{d}}{{\rm{D}}^2} - {{\rm{b}}^3}}}{6}} \right) = 0\)

⇒ D2 - 3b2 = 0

\(\Rightarrow {\rm{b}} = \frac{{\rm{D}}}{{\sqrt 3 }}\)

\({\rm{d}} = \sqrt {{{\rm{D}}^2} - {{\rm{b}}^2}} = \sqrt {{{\rm{D}}^2} - \frac{{{{\rm{D}}^2}}}{3}} = \sqrt {\frac{2}{3}} \times {\rm{D}}\)

Thus, the cross-sectional area for a rectangular beam is given by

\({A} = {\rm{b}} \times {\rm{d}} = \frac{{\rm{D}}}{{\sqrt 3 }} \times \sqrt {\frac{2}{3}} \times {\rm{D}} = \frac{{\sqrt 2 {{\rm{D}}^2}}}{3}\)

\(\therefore {\rm{A}} = \frac{{\sqrt 2 }}{3}{{\rm{D}}^2}\)

\(\therefore \frac{{\rm{b}}}{{\rm{d}}} = \frac{1}{{\sqrt 2 }} = 0.707\)