Math Problem Statement

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Determine the required diameter if the stress is not to exceed 140 MPa.

Solution

We are tasked with determining the required diameter dd of the steel wire so that the stress does not exceed 140 MPa, given a tensile load of 2000 N and other conditions from the image. Let's work through the problem step by step.

Given Data:

  • Length of the wire L=15mL = 15 \, \text{m}
  • Load P=2000NP = 2000 \, \text{N}
  • Maximum allowable stress σ=140MPa=140×106Pa\sigma = 140 \, \text{MPa} = 140 \times 10^6 \, \text{Pa}
  • Maximum allowable elongation ΔL=6mm=0.006m\Delta L = 6 \, \text{mm} = 0.006 \, \text{m}
  • Young’s modulus E=200GPa=200×109PaE = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa}

Step 1: Relationship for Stress

Stress σ\sigma is related to the force PP and the cross-sectional area AA by the formula: σ=PA\sigma = \frac{P}{A} For a circular cross-section, the area AA is: A=πd24A = \frac{\pi d^2}{4} where dd is the diameter.

Substitute AA into the stress equation: σ=4Pπd2\sigma = \frac{4P}{\pi d^2} Rearranging for dd: d=4Pπσd = \sqrt{\frac{4P}{\pi \sigma}} Now, let's plug in the values: d=4×2000π×140×106d = \sqrt{\frac{4 \times 2000}{\pi \times 140 \times 10^6}}

Step 2: Elongation Condition

The elongation ΔL\Delta L of the wire can be calculated using Hooke's Law: ΔL=PLAE\Delta L = \frac{PL}{AE} Again, using the area A=πd24A = \frac{\pi d^2}{4}, we substitute it into the elongation equation: ΔL=4PLπd2E\Delta L = \frac{4PL}{\pi d^2 E} Rearranging to solve for dd using the maximum elongation: d=4PLπEΔLd = \sqrt{\frac{4PL}{\pi E \Delta L}} Substitute the values: d=4×2000×15π×200×109×0.006d = \sqrt{\frac{4 \times 2000 \times 15}{\pi \times 200 \times 10^9 \times 0.006}}

Conclusion:

I'll compute both values for the diameter based on the stress and elongation conditions. Let me calculate it for you.The required diameters based on both conditions are:

  1. Stress condition: d=4.26mmd = 4.26 \, \text{mm}
  2. Elongation condition: d=5.64mmd = 5.64 \, \text{mm}

Since the problem specifies that the total elongation must not exceed 6 mm, we use the larger diameter from the elongation condition, which is 5.64 mm. Therefore, the required diameter is approximately 5.64 mm to satisfy both conditions.

Would you like further details or clarifications? Here are some related questions you might find useful:

  1. How does Young's modulus affect material elongation?
  2. What happens if the diameter chosen is below the calculated value for elongation?
  3. Can you explain how stress and strain are related in materials?
  4. How does the length of the wire influence its elongation?
  5. What safety factors should be considered in such stress-strain analysis?

Tip: Always check both stress and strain limits to ensure material safety and functionality under load.

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Math Problem Analysis

Mathematical Concepts

Stress-Strain Analysis
Hooke's Law
Circular Cross-Section
Young's Modulus

Formulas

Stress σ = P / A
Elongation ΔL = PL / AE
Area of a circle A = πd² / 4
d = √(4P / (πσ)) for stress condition
d = √(4PL / (πEΔL)) for elongation condition

Theorems

Hooke's Law for linear elasticity
Stress-Strain Relationship in materials

Suitable Grade Level

Undergraduate Level - Engineering (Strength of Materials)