Answer :
Answer:
To determine the stresses in the wires and the elongation due to the loading on the beam, we follow these steps:
### Given Data
- Length of beam (\( L \)): 1.5 m
- Uniform load (\( W \)): 4 kN (4000 N)
- Number of support wires (\( n \)): 3
- Initial length of each wire (\( L_0 \)): 2.4 m
- Diameter of blast wires (\( d_{\text{blast}} \)): 4.0 mm
- Diameter of steel wire (\( d_{\text{steel}} \)): 1.8 mm
- Young's modulus of elasticity for steel (\( E_s \)): 205 GPa = 205 kN/mm²
- Young's modulus of elasticity for blast wires (\( E_b \)): 85 GPa = 85 kN/mm²
### Step-by-Step Solution:
#### 1. Calculate the Vertical Reactions at the Supports
The total load applied to the beam is 4000 N. Since the beam is simply supported by three wires, each wire carries an equal load due to symmetry:
\[ R = \frac{4000 \text{ N}}{3} = 1333.33 \text{ N} \]
#### 2. Determine the Elongation in the Wires
The elongation (\( \Delta L \)) in each wire due to the load can be calculated using the formula:
\[ \Delta L = \frac{WL_0}{AE} \]
where:
- \( W \) is the load (1333.33 N),
- \( L_0 \) is the initial length of the wire (2.4 m),
- \( A \) is the cross-sectional area of the wire,
- \( E \) is the Young's modulus of elasticity.
#### 3. Calculate the Cross-sectional Areas (\( A \)) of the Wires
The cross-sectional area \( A \) of a wire with diameter \( d \) is given by:
\[ A = \frac{\pi d^2}{4} \]
For blast wires (\( d_{\text{blast}} = 4.0 \text{ mm} \)):
\[ A_{\text{blast}} = \frac{\pi (4.0 \text{ mm})^2}{4} = 12.57 \text{ mm}^2 \]
For steel wire (\( d_{\text{steel}} = 1.8 \text{ mm} \)):
\[ A_{\text{steel}} = \frac{\pi (1.8 \text{ mm})^2}{4} = 2.55 \text{ mm}^2 \]
#### 4. Calculate Elongation in Each Wire
##### For Blast Wires:
\[ \Delta L_{\text{blast}} = \frac{4000 \times 2.4}{12.57 \times 85} \text{ mm} \]
\[ \Delta L_{\text{blast}} \approx 2.26 \text{ mm} \]
##### For Steel Wire (at midpoint):
\[ \Delta L_{\text{steel}} = \frac{4000 \times 2.4}{2.55 \times 205} \text{ mm} \]
\[ \Delta L_{\text{steel}} \approx 4.65 \text{ mm} \]
#### 5. Calculate Stresses in the Wires
The stress (\( \sigma \)) in each wire can be calculated using the formula:
\[ \sigma = \frac{W}{A} \]
##### For Blast Wires:
\[ \sigma_{\text{blast}} = \frac{4000}{12.57} \text{ N/mm}^2 \]
\[ \sigma_{\text{blast}} \approx 318.51 \text{ N/mm}^2 \]
##### For Steel Wire:
\[ \sigma_{\text{steel}} = \frac{4000}{2.55} \text{ N/mm}^2 \]
\[ \sigma_{\text{steel}} \approx 1568.63 \text{ N/mm}^2 \]
### Summary
- **Elongation**: Blast wires elongate approximately 2.26 mm, while the steel wire elongates about 4.65 mm.
- **Stresses**: Blast wires experience a stress of approximately 318.51 N/mm², and the steel wire experiences a stress of about 1568.63 N/mm².
These calculations provide insights into how the load affects the wires' elongation and stresses based on their respective material properties and dimensions.
Explanation:
