A steel wire hangs from a high-altitude balloon. The steel has unit weight 77 kN/m3 and yield stress of 280 MPa. The required factor of
Safety against yield is 2.0. The maximum permissible length of the wire is
Approximately:

Question

Quizplus · Accepted Answer

The maximum permissible length of the wire can be calculated using the formula for stress due to the self-weight of the wire, which is $\sigma = \rho g h$, where $\sigma$ is the stress, $\rho$ is the density (which can be converted from unit weight), $g$ is the acceleration due to gravity (approximately $9.81 m/s^2$), and $h$ is the height or length of the wire. Given a factor of safety of 2, the allowable stress is $280 \, \text{MPa} / 2 = 140 \, \text{MPa}$. Converting unit weight to density, $\rho = 77 \, \text{kN/m}^3 = 77000 \, \text{N/m}^3$. Rearranging the formula to solve for $h$, we get $h = \sigma / (\rho g)$. Substituting the given values, $h = 140 \times 10^6 \, \text{N/m}^2 / (77000 \, \text{N/m}^3 \times 9.81 \, \text{m/s}^2)$, which calculates to approximately 1800 meters.

A Steel Wire Hangs from a High-Altitude Balloon