A copper rod of length l is suspended from the ceiling by one of its ends. Find:
(a) the elongation `Deltal` of the rod due to its own weight ,
(b) the relative increment of its volume `DeltaV//V`.

Explanation

(a) As free end has zero tension, thus the tension in the rod at a vestical distance y from its lower end
`T=m/lgy` (1)
Let `dell` be the elongation of the element of length `dy`, then
`dell=(sigma(y))/(E)dy`
`=(T)/(SE)dy=(mgydy)/(SlE)=rhogydy//E` (where `rho` is the density of the copper)
Thus the sought elongation
`Deltal=intdell=rhogunderset(0)overset(l)int(ydy)/(E)=1/2rhogl^2//E` (2)
(b) If the longitudinal (tensile) strain is `epsilon=(Deltal)/(l)`, the accompanying lateral (compressive) strain is given by
`epsilon^'=(Deltar)/(r)=-muepsilon` (3)
Then since `V=pir^2l` we have
`(DeltaV)/(V)=(2Deltar)/(r)+(Deltal)/(l)`
`=(1-2mu)(Deltal)/(l)` [Using(3)]
where `(Deltal)/(l)` is given in part (a), `mu` is the Poisson ratio for copper.