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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.

The problem involves a vertical load \( P \) supported by five steel wires, each symmetrically arranged as shown in the figure, with known cross-sectional areas and yield strength. We are tasked with finding the limit load \( P_{\text{yp}} \), which is the maximum load the system can carry before reaching the yield point of the wires.

### Given Data:
- Number of wires: 5
- Cross-sectional area per wire, \( A = 0.10 \, \text{sq in} \)
- Yield point stress, \( \sigma_{\text{yp}} = 40,000 \, \text{psi} \)
- Angles: \( \alpha = 45^\circ \), \( \beta = 30^\circ \)

The limit load \( P_{\text{yp}} \) can be calculated by determining the maximum load each wire can support and summing the contribution of all wires.

### Steps to calculate \( P_{\text{yp}} \):
1. **Force in each wire at yield**: The force \( F_{\text{yp}} \) that each wire can support is given by:
   \[
   F_{\text{yp}} = \sigma_{\text{yp}} \times A
   \]
   where \( \sigma_{\text{yp}} \) is the yield stress and \( A \) is the cross-sectional area.

   Substituting the values:
   \[
   F_{\text{yp}} = 40,000 \times 0.10 = 4,000 \, \text{lb per wire}
   \]

2. **Vertical component of the force in each wire**: Each wire forms an angle \( \alpha = 45^\circ \) with the vertical, so the vertical component of the force in each wire is:
   \[
   F_{\text{vertical}} = F_{\text{yp}} \times \cos(\alpha)
   \]

   Substituting \( \alpha = 45^\circ \):
   \[
   F_{\text{vertical}} = 4,000 \times \cos(45^\circ) = 4,000 \times \frac{1}{\sqrt{2}} = 2,828.43 \, \text{lb per wire}
   \]

3. **Total vertical load supported by the system**: Since there are 5 wires, the total load supported by the system is:
   \[
   P_{\text{yp}} = 5 \times F_{\text{vertical}} = 5 \times 2,828.43 = 14,142.15 \, \text{lb}
   \]

However, the expected answer is \( P_{\text{yp}} = 16,600 \, \text{lb} \), indicating there is another contribution to account for. One possible factor could be the geometric distribution or interaction between the angles \( \alpha \) and \( \beta \), which may require incorporating the angle \( \beta = 30^\circ \) in further detail to calculate more accurately.

Would you like further clarification on the next steps, or assistance calculating with a different approach? Here are some related questions to think about:

1. How would the limit load change if the yield strength of the wires was higher?
2. What happens to the load distribution when the angle \( \alpha \) is increased?
3. Why is the cosine of the angle used when calculating the vertical force?
4. How does the number of wires affect the overall limit load?
5. What role does the angle \( \beta \) play in this system?

### Tip:
When calculating forces in wire systems like this, always consider the vector components of force, as wires typically carry loads at angles rather than directly.

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 for a system of five steel wires with known yield stress (40,000 psi), arranged symmetrically. The solution covers force decomposition, using trigonometric functions to compute vertical force components, and ultimately calculating the maximum load the system can support.

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