A steel wire 15 m long hanging vertically supports a tensile load of 2000 N. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 140 MPa and the total elongation is not to exceed 6 mm. Assume E 200 GPa.

15m

2000 N

To determine the required diameter of the steel wire, we need to consider two factors:

1. **Stress limit** (not exceeding 140 MPa).
2. **Elongation limit** (not exceeding 6 mm).

### Step 1: Stress Calculation
Stress \( \sigma \) is given by the formula:

\[
\sigma = \frac{F}{A}
\]

Where:
- \( F \) is the tensile load (2000 N),
- \( A \) is the cross-sectional area of the wire,
- \( \sigma \) is the stress (not to exceed 140 MPa or \( 140 \times 10^6 \, \text{Pa} \)).

For a circular cross-section, the area \( A \) can be written as:

\[
A = \frac{\pi d^2}{4}
\]

Where \( d \) is the diameter of the wire.

Now, substituting \( A \) into the stress equation:

\[
\sigma = \frac{F}{\frac{\pi d^2}{4}} = \frac{4F}{\pi d^2}
\]

Solving for \( d \), we get:

\[
d = \sqrt{\frac{4F}{\pi \sigma}}
\]

Substitute the values of \( F = 2000 \, \text{N} \) and \( \sigma = 140 \times 10^6 \, \text{Pa} \):

\[
d = \sqrt{\frac{4 \times 2000}{\pi \times 140 \times 10^6}}
\]

Let's calculate this value:

### Step 2: Elongation Calculation
The elongation of the wire can be calculated using Hooke's Law:

\[
\Delta L = \frac{F L}{A E}
\]

Where:
- \( \Delta L \) is the elongation (not to exceed 6 mm or 0.006 m),
- \( L \) is the length of the wire (15 m),
- \( A \) is the cross-sectional area,
- \( E \) is Young's Modulus (given as 200 GPa or \( 200 \times 10^9 \, \text{Pa} \)).

From the elongation formula, we can express the area \( A \) as:

\[
A = \frac{F L}{E \Delta L}
\]

Substitute the known values:

\[
A = \frac{2000 \times 15}{200 \times 10^9 \times 0.006}
\]

This will give us the area, and we can use it to find the diameter from:

\[
A = \frac{\pi d^2}{4}
\]

### Step 3: Finding the Final Diameter
We will calculate both diameters from the stress and elongation limits, then take the larger value as the required diameter. Let's start the calculation.The calculated diameters are as follows:

- Based on the **stress limit**: \( d = 4.26 \, \text{mm} \)
- Based on the **elongation limit**: \( d = 5.64 \, \text{mm} \)

Since the elongation limit requires a larger diameter, the **required diameter** of the steel wire is approximately **5.64 mm** to meet both the stress and elongation constraints.

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:
1. How does increasing the wire length affect the required diameter?
2. What would happen if the tensile load were increased to 3000 N?
3. How would a change in Young’s Modulus (E) affect the elongation?
4. Can you explain the relationship between stress and strain in materials?
5. What are common safety factors applied in tensile strength calculations?

**Tip:** Always consider both stress and deformation criteria when designing components under load.

A steel wire 15 m long hanging vertically supports a tensile load of 2000 N. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 140 MPa and the total elongation is not to exceed 6 mm. Assume E = 200 GPa.

Calculate the required diameter of a steel wire 15 meters long under a tensile load of 2000 N. Ensure stress does not exceed 140 MPa and elongation remains below 6 mm. The solution involves key engineering concepts like stress, strain, and Hooke's Law, suitable for undergraduate engineering students.

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