Question

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

• A. The elongation of the rod is $\Delta l = \frac{1}{2} \frac{\rho g l^2}{E}$. The relative increment of its volume is $\frac{\Delta V}{V} = \frac{1}{2} (1-2\nu) \frac{\rho g l}{E}$.
• B. The elongation of the rod is $\Delta l = \frac{\rho g l^2}{E}$. The relative increment of its volume is $\frac{\Delta V}{V} = (1-2\nu) \frac{\rho g l}{E}$.
• C. The elongation of the rod is $\Delta l = \frac{1}{2} \frac{\rho g l}{E}$. The relative increment of its volume is $\frac{\Delta V}{V} = \frac{1}{2} \frac{\rho g l}{E}$.
• D. The elongation of the rod is $\Delta l = \frac{\rho g l^2}{2E}$. The relative increment of its volume is $\frac{\Delta V}{V} = \frac{1}{2} \frac{\rho g l^2}{E}$.

Answer 1

Answer: (A)

The elongation of the rod is $\Delta l = \frac{1}{2} \frac{\rho g l^2}{E}$. The relative increment of its volume is $\frac{\Delta V}{V} = \frac{1}{2} (1-2\nu) \frac{\rho g l}{E}$.

Answer 2

a) The elongation is calculated by integrating the strain along the length of the rod, where the strain is proportional to the stress due to the rod's own weight. The stress varies linearly with depth. $\Delta l = \int_{0}^{l} \frac{\rho g (l-x)}{E} dx = \frac{\rho g}{E} \left[ lx - \frac{x^2}{2} \right]_{0}^{l} = \frac{\rho g}{E} \left( l^2 - \frac{l^2}{2} \right) = \frac{1}{2} \frac{\rho g l^2}{E}$ b) The relative volume change is found by integrating the volumetric strain of each element. Volumetric strain is $(1-2\nu)$ times the longitudinal strain. The total volume change is $(1-2\nu)$ times the initial cross-sectional area times the total elongation. Dividing by the initial volume $V=Al$ gives the relative change. $\frac{\Delta V}{V} = (1-2\nu) \frac{\Delta l}{l} = (1-2\nu) \frac{1}{l} \left( \frac{1}{2} \frac{\rho g l^2}{E} \right) = \frac{1}{2} (1-2\nu) \frac{\rho g l}{E}$