The deformation of the bar of rectangular section under its own weight is equal to ____ the deformation due to a direct load equal to the weight of the body applied at the lower end.

Deformation under own weight

Consider a small strip ‘dx’ at a distance ‘x’ as shown in figure. We shall find change in length for ‘dx’ and then integrate for whole length. Force  exerted by weight below strip ‘dx’

Force (P) = Volume below ‘dx’ x specific weight = A × ρ (A = cross – sectional area)

Elongation of the strip

\({\left( {\delta I} \right)_{dx}} = \frac{{\left( {Axl} \right)dx}}{{A \times E}}\left[ {\begin{array}{*{20}{c}} {load = Ax\rho }\\ {length = dx} \end{array}} \right]\)

\({\left( {\delta l} \right)_{dx}} = \frac{{\rho xdx}}{E}\)

For total elongation integrate and take limits

\(\delta l = \mathop \smallint \limits_o^l \frac{\rho }{E}dxd = \frac{\rho }{E}\left( {\frac{{{x^2}}}{2}} \right)_o^l\)

\(\delta l = \frac{{\rho {l^2}}}{{2E}}\)                                …..i)

Now elongation due to load (W)

\(\delta l = \frac{{Wl}}{{2E}}\)

Load = own weight (given)

W = ρlA

\(\therefore A = \frac{W}{{\rho l}}\)

\(\therefore {\left( {\delta l} \right)_2} = \frac{{\rho {l^2}}}{E}\)                     ……ii)

Now form equation (i) and (ii)

\(\frac{{\delta l}}{{\delta {l_2}}} = \frac{1}{2}\)

\(\therefore \delta l = \frac{{\delta {l_2}}}{2}\)