The safe stress for a hollow steel column which carries an axial load of 2100 kN is 125 MN/m². if the external diameter of the column is 30cm, what will be the internal diameter? A)25 cm B)26.19cm C)30.14 cm D)27.9 cm

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Quizplus · Accepted Answer

The safe stress ($\sigma$) is given by the formula $\sigma = \frac{P}{A}$, where $P$ is the axial load and $A$ is the cross-sectional area. For a hollow column, $A = \pi/4 \times (D^2 - d^2)$, where $D$ is the external diameter and $d$ is the internal diameter. Given $\sigma = 125 \, \text{MN/m}^2 = 125 \times 10^6 \, \text{N/m}^2$, $P = 2100 \, \text{kN} = 2100 \times 10^3 \, \text{N}$, and $D = 30 \, \text{cm} = 0.3 \, \text{m}$, we can solve for $d$.Rearranging the formula for $\sigma$, we get $A = \frac{P}{\sigma} = \frac{2100 \times 10^3}{125 \times 10^6} = 0.0168 \, \text{m}^2$.Substituting $A$ and $D$ into the area formula, $0.0168 = \frac{\pi}{4} \times (0.3^2 - d^2)$.Solving for $d$, we find $d \approx 0.2619 \, \text{m} = 26.19 \, \text{cm}$.

The Safe Stress for a Hollow Steel Column Which Carries