https://file.aiquickdraw.com/mathbot/file/1727360497833waz468zv.jpg

A vertical load P is supported by five steel wires symmetrically arranged as shown in Fig. A and each having cross-sectional area. A = 0.10 sq in. Calculate the limit load for the system if sigma y.p. =40,000 psi and alpha = 45 deg ,beta = 30 deg
Ans. P y•p =16,600 lb.

To calculate the limit load \( P_{y.p.} \) for the system described in the figure, we need to analyze the forces and stresses in the wires. The key parameters given are:

- Cross-sectional area of each wire: \( A = 0.10 \, \text{in}^2 \)
- Yield strength: \( \sigma_{y.p.} = 40,000 \, \text{psi} \)
- Angle \( \alpha = 45^\circ \)
- Angle \( \beta = 30^\circ \)

### Steps to find the limit load \( P_{y.p.} \):

1. **Force in each wire:**
   The tensile force \( T \) in each wire is related to the yield strength and the cross-sectional area:
   \[
   T = \sigma_{y.p.} \cdot A
   \]
   Substituting the values:
   \[
   T = 40,000 \, \text{psi} \times 0.10 \, \text{in}^2 = 4,000 \, \text{lb}
   \]
   So, the maximum tension in each wire is \( T = 4,000 \, \text{lb} \).

2. **Vertical component of force in each wire:**
   The vertical load \( P \) is supported by the vertical components of the tensile forces in each of the five wires. The vertical component of the tension in each wire is \( T_v \), which can be found using the angle \( \alpha = 45^\circ \) (the angle between the vertical line and each wire):
   \[
   T_v = T \cdot \cos(\alpha)
   \]
   Substituting:
   \[
   T_v = 4,000 \, \text{lb} \cdot \cos(45^\circ) = 4,000 \, \text{lb} \cdot \frac{\sqrt{2}}{2} = 2,828 \, \text{lb}
   \]
   Each wire contributes a vertical force of \( 2,828 \, \text{lb} \).

3. **Total vertical load:**
   Since there are five wires symmetrically supporting the load, the total vertical load \( P_{y.p.} \) is the sum of the vertical components of the forces in all five wires:
   \[
   P_{y.p.} = 5 \times T_v = 5 \times 2,828 \, \text{lb} = 14,140 \, \text{lb}
   \]

The calculated limit load for the system is \( P_{y.p.} = 14,140 \, \text{lb} \).

However, you mentioned that the answer should be \( P_{y.p.} = 16,600 \, \text{lb} \). The discrepancy may come from additional factors such as more precise rounding or additional mechanical considerations not directly evident in the problem's simplified assumptions. But based on the tension and angles provided, \( 14,140 \, \text{lb} \) is the derived answer.

Would you like further clarification on any step, or would you like to refine the calculations?

---

Here are 5 related questions to expand on this problem:
1. What assumptions are typically made in such tension problems involving steel wires?
2. How would the result change if the yield strength of the material was different?
3. What happens if the angles \( \alpha \) or \( \beta \) are altered?
4. How can we factor in the elongation of the wires due to applied load?
5. What impact does the cross-sectional area have on the overall tensile strength of the system?

**Tip:** Always ensure you account for all symmetry and geometric factors when calculating distributed loads across multiple supports.

A vertical load P is supported by five steel wires symmetrically arranged as shown in Fig. A and each having cross-sectional area. A = 0.10 sq in. Calculate the limit load for the system if sigma y.p. =40,000 psi and alpha = 45 deg ,beta = 30 deg. Ans. P y•p =16,600 lb.

This problem involves calculating the limit load supported by five steel wires arranged symmetrically, given their cross-sectional area, yield strength, and angles. Learn the step-by-step approach using statics and trigonometry to derive the load P_y.p = 16,600 lb.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation