To calculate the maximum load that the cast iron column can carry, we need to use the following formula for compressive load:

$\text{Max Load (P)} = \sigma \times A$

Where:

• $\sigma$ is the compressive stress (in $N/m^2$),
• $A$ is the cross-sectional area of the column.

Step 1: Convert diameters from cm to meters

• Outside diameter $D_{\text{outside}} = 15 \, \text{cm} = 0.15 \, \text{m}$
• Inside diameter $D_{\text{inside}} = 12 \, \text{cm} = 0.12 \, \text{m}$

Step 2: Calculate the cross-sectional area of the hollow column

The cross-sectional area of a hollow cylinder is given by the difference between the areas of the outer and inner circles:

$A = \frac{\pi}{4} \left(D_{\text{outside}}^2 - D_{\text{inside}}^2\right)$

Substitute the values:

$A = \frac{\pi}{4} \left(0.15^2 - 0.12^2\right)$ $A = \frac{\pi}{4} \left(0.0225 - 0.0144\right)$ $A = \frac{\pi}{4} \times 0.0081 \, \text{m}^2$ $A = 0.0020 \, \text{m}^2$

Step 3: Calculate the maximum load

Given that the compressive stress $\sigma = 80 \, \text{MN/m}^2 = 80 \times 10^6 \, \text{N/m}^2$:

$P = \sigma \times A$ $P = 80 \times 10^6 \times 0.0020$ $P = 160,000 \, \text{N}$

Final Answer:

The maximum load the column can carry is $160 \, \text{kN}$.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. How would the load change if the inside diameter increased to 13 cm?
2. What would be the effect on the load if the compressive stress limit was increased to 100 MN/m²?
3. How can we calculate the safety factor for the column?
4. What is the difference in maximum load between a solid column and a hollow one of the same outer diameter?
5. How would the material properties affect the load-bearing capacity?

Tip: Always check the units before performing calculations to ensure consistency and avoid errors.