Elasticity — Question 5

Question 5

A hanging rod carries a nonuniform stress from its own weight

A uniform vertical rod hangs from its top under its own weight. Its length is $L=100\,\mathrm m$, density $\rho=8000\,\mathrm{kg/m^3}$ and modulus $E=200\,\mathrm{GPa}$. Use $g=10\,\mathrm{m/s^2}$ and constant area $A$; neglect geometric changes.

1. Using $x$ measured downward from the top, find the tension profile and total elongation.

2. Find the largest stress and explain why doubling area does not change the self-weight elongation.

Question 5 - Solution

A hanging rod carries a nonuniform stress from its own weight

(a) Derivation and calculation

A section at depth $x$ supports the rod below it: $$\boxed{N(x)=\rho Ag(L-x)},\qquad\sigma(x)=\rho g(L-x).$$ Integrating strain over length gives $$\delta=\int_0^L\frac{N(x)}{EA}\,dx=\frac{\rho gL^2}{2E}.$$ $$\boxed{\delta=\frac{8000(10)(100)^2}{2(200\times10^9)}=.00200\ \mathrm m=2.00\ \mathrm{mm}}.$$

(b) Result and interpretation

Maximum stress occurs at the top: $$\boxed{\sigma_{\max}=\rho gL=8000(10)(100)=8.00\ \mathrm{MPa}}.$$ Doubling $A$ doubles both $N(x)$ and $EA$: $$\frac{2N(x)}{E(2A)}=\frac{N(x)}{EA}.$$ The strain profile and total self-weight extension therefore remain unchanged.

Physical check

Tension decreases linearly from the full rod weight at the top to zero at the free end.