To find the minimum external diameter \( D_O \) of a cast iron column that can carry a load of \( 1.6 \, \text{MN} \) without exceeding a stress of \( 90 \, \text{N/mm}^2 \), we can follow these steps:
### Step 1: Understand the given values
- Internal diameter \( D_i = 200 \, \text{mm} \)
- Load \( F = 1.6 \, \text{MN} = 1.6 \times 10^6 \, \text{N} \)
- Maximum allowable stress \( \sigma_m = 90 \, \text{N/mm}^2 \)
### Step 2: Write the formula for stress
The stress \( \sigma \) in the column can be expressed as:
\[
\sigma = \frac{F}{A}
\]
Where \( A \) is the cross-sectional area of the column. For a hollow circular section, the area \( A \) can be calculated as:
\[
A = \frac{\pi}{4} (D_O^2 - D_i^2)
\]
### Step 3: Set up the equation for maximum stress
We want the stress not to exceed \( \sigma_m \):
\[
\sigma_m = \frac{F}{A}
\]
Substituting the expression for \( A \):
\[
\sigma_m = \frac{F}{\frac{\pi}{4} (D_O^2 - D_i^2)}
\]
### Step 4: Rearrange the equation to find \( D_O \)
Rearranging gives:
\[
D_O^2 - D_i^2 = \frac{4F}{\pi \sigma_m}
\]
Thus,
\[
D_O^2 = D_i^2 + \frac{4F}{\pi \sigma_m}
\]
Taking the square root:
\[
D_O = \sqrt{D_i^2 + \frac{4F}{\pi \sigma_m}}
\]
### Step 5: Substitute the known values
Now we can substitute the known values into the equation:
- Convert \( D_i \) to meters for consistency:
\[
D_i = 200 \, \text{mm} = 0.2 \, \text{m} = 200 \, \text{mm}
\]
- Substitute \( F \), \( \sigma_m \), and \( D_i \):
\[
D_O = \sqrt{(200 \, \text{mm})^2 + \frac{4 \times (1.6 \times 10^6 \, \text{N})}{\pi \times (90 \, \text{N/mm}^2)}}
\]
### Step 6: Calculate the values
Calculating the area:
\[
D_O = \sqrt{(200)^2 + \frac{4 \times 1.6 \times 10^6}{\pi \times 90}}
\]
Calculating \( \frac{4 \times 1.6 \times 10^6}{\pi \times 90} \):
\[
\frac{4 \times 1.6 \times 10^6}{\pi \times 90} \approx 22400.4 \, \text{mm}^2
\]
Now substituting back:
\[
D_O = \sqrt{40000 + 22400.4} = \sqrt{62400.4} \approx 249.8 \, \text{mm}
\]
### Step 7: Final result
Thus, the minimum external diameter \( D_O \) should be approximately:
\[
D_O \approx 250.2 \, \text{mm}
\]
