A homogeneous cable of length L and uniform cross section is

A homogeneous cable of length L and uniform cross section is suspended from one end. Denoting by p the density (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of the cable due to its own weight, Show that the same elongation would be obtained if the cable were horizontal and if a force equal to half of its weight were applied at each end.

Solution

Let\'s assume y is the element point in cable

Weight of cable W = m*g = density*volume*g = pho*g*L*A

weight of portion below element point P = rho*g*A*(L-y)

x is elongation in cable

dx = P*dy/(EA) = rho*g*A*(L-y)*dy/(EA) = rho*g*(L-y)*dy/E

integration from 0 to L of x

x = (rho*g/E)*[Ly - 0.5y^2]^L_{0 =}

x = rho*g*L^2/(2E)

B. F = EAx/L = EA*rho*g*L^/(2EL) = rho*g*A*L/2 = W/2