Math Problem Statement

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.

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.

Solution

The problem involves a vertical load PP 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 PypP_{\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=0.10sq inA = 0.10 \, \text{sq in}
  • Yield point stress, σyp=40,000psi\sigma_{\text{yp}} = 40,000 \, \text{psi}
  • Angles: α=45\alpha = 45^\circ, β=30\beta = 30^\circ

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

Steps to calculate PypP_{\text{yp}}:

  1. Force in each wire at yield: The force FypF_{\text{yp}} that each wire can support is given by: Fyp=σyp×AF_{\text{yp}} = \sigma_{\text{yp}} \times A where σyp\sigma_{\text{yp}} is the yield stress and AA is the cross-sectional area.

    Substituting the values: Fyp=40,000×0.10=4,000lb per wireF_{\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 α=45\alpha = 45^\circ with the vertical, so the vertical component of the force in each wire is: Fvertical=Fyp×cos(α)F_{\text{vertical}} = F_{\text{yp}} \times \cos(\alpha)

    Substituting α=45\alpha = 45^\circ: Fvertical=4,000×cos(45)=4,000×12=2,828.43lb per wireF_{\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: Pyp=5×Fvertical=5×2,828.43=14,142.15lbP_{\text{yp}} = 5 \times F_{\text{vertical}} = 5 \times 2,828.43 = 14,142.15 \, \text{lb}

However, the expected answer is Pyp=16,600lbP_{\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 α\alpha and β\beta, which may require incorporating the angle β=30\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 α\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 β\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.

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Math Problem Analysis

Mathematical Concepts

Statics
Forces in Equilibrium
Trigonometry

Formulas

F_yp = sigma_yp * A
F_vertical = F_yp * cos(alpha)
P_yp = n * F_vertical

Theorems

Force Decomposition in Components

Suitable Grade Level

Grades 11-12 or University Engineering