The elastic limit of brass is 379 MPa . What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit ?

Options

A. 1.00 mm
B. 0.90 mm
C. 1.36 mm
D. 1.16 mm

Explanation

To find the minimum diameter of a brass rod that can support a 400 N load without exceeding its elastic limit of 379 MPa, we can follow these steps:

### Step 1: Understand the relationship between stress, force, and area
The stress \( \sigma \) on the rod can be defined as:
\[ \sigma = \frac{F}{A} \]
where:
- \( F \) is the force applied (400 N),
- \( A \) is the cross-sectional area of the rod.

### Step 2: Convert the elastic limit to standard units
The elastic limit is given as 379 MPa, which can be converted to Newtons per square meter (N/m²):
\[ 379 \text{ MPa} = 379 \times 10^6 \text{ N/m}^2 \]

### Step 3: Express the area in terms of diameter
The area \( A \) of a circular cross-section can be expressed in terms of the diameter \( D \):
\[ A = \frac{\pi D^2}{4} \]

### Step 4: Set up the equation for stress
Substituting the expression for area into the stress formula gives:
\[ \sigma = \frac{F}{\frac{\pi D^2}{4}} = \frac{4F}{\pi D^2} \]

### Step 5: Equate the stress to the elastic limit
To ensure that the stress does not exceed the elastic limit, we set:
\[ \frac{4F}{\pi D^2} = 379 \times 10^6 \]

### Step 6: Solve for \( D^2 \)
Rearranging the equation to solve for \( D^2 \):
\[ D^2 = \frac{4F}{\pi \cdot 379 \times 10^6} \]
Substituting \( F = 400 \, \text{N} \):
\[ D^2 = \frac{4 \times 400}{\pi \cdot 379 \times 10^6} \]
\[ D^2 = \frac{1600}{\pi \cdot 379 \times 10^6} \]

### Step 7: Calculate \( D^2 \)
Using \( \pi \approx 3.14 \):
\[ D^2 = \frac{1600}{3.14 \cdot 379 \times 10^6} \]
Calculating the denominator:
\[ 3.14 \cdot 379 \approx 1186.06 \]
Thus,
\[ D^2 \approx \frac{1600}{1186.06 \times 10^6} \]
\[ D^2 \approx \frac{1600}{1186060000} \approx 1.349 \times 10^{-6} \]

### Step 8: Take the square root to find \( D \)
\[ D \approx \sqrt{1.349 \times 10^{-6}} \approx 0.00116 \, \text{m} = 1.16 \, \text{mm} \]

### Conclusion
The minimum diameter of the brass rod should be approximately **1.16 mm**.