Math Problem Statement

A steel wire 15 m long hanging vertically supports a tensile load of 2000 N. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 140 MPa and the total elongation is not to exceed 6 mm. Assume E = 200 GPa.

Solution

We are given a steel wire with the following parameters:

  • Length of the wire L=15mL = 15 \, \text{m}
  • Tensile load F=2000NF = 2000 \, \text{N}
  • Maximum stress σmax=140MPa=140×106Pa\sigma_{\text{max}} = 140 \, \text{MPa} = 140 \times 10^6 \, \text{Pa}
  • Maximum elongation ΔLmax=6mm=6×103m\Delta L_{\text{max}} = 6 \, \text{mm} = 6 \times 10^{-3} \, \text{m}
  • Young's modulus for steel E=200GPa=200×109PaE = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa}

We need to determine the required diameter of the steel wire such that:

  1. The stress does not exceed 140 MPa.
  2. The total elongation does not exceed 6 mm.

Step 1: Calculate the required cross-sectional area based on maximum stress

The tensile stress σ\sigma is defined as: σ=FA\sigma = \frac{F}{A} where:

  • σ\sigma is the tensile stress,
  • FF is the applied force,
  • AA is the cross-sectional area.

We are given that the maximum stress is 140 MPa. Therefore, the required cross-sectional area AA can be calculated as: A=FσmaxA = \frac{F}{\sigma_{\text{max}}}

Step 2: Calculate the elongation

The elongation ΔL\Delta L of the wire is given by Hooke's Law for materials in tension: ΔL=FLAE\Delta L = \frac{F L}{A E} We are given that the maximum elongation ΔLmax=6mm\Delta L_{\text{max}} = 6 \, \text{mm}, and we can use this to calculate the area AA based on the elongation limit.

We will use these two conditions (stress limit and elongation limit) to find the required diameter.

Let me perform these calculations now.### Results:

  1. The required cross-sectional area based on the stress limit is 1.43×105m21.43 \times 10^{-5} \, \text{m}^2.
  2. The required cross-sectional area based on the elongation limit is 2.5×105m22.5 \times 10^{-5} \, \text{m}^2.

Since the elongation limit provides a more stringent requirement, we use that for the calculation.

Therefore, the required diameter of the steel wire is approximately 5.64mm5.64 \, \text{mm}.

Would you like further clarification or details on any part of this calculation?

Here are 5 related questions you might want to explore:

  1. How would the required diameter change if the load were increased?
  2. How does the Young's modulus affect the elongation and diameter requirements?
  3. What would happen to the wire if the stress exceeded 140 MPa?
  4. How can we account for the weight of the wire in similar calculations?
  5. What is the impact of different materials on wire design?

Tip: Always ensure the material's stress and strain limits are within the safe range to avoid failure in structural applications.

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Math Problem Analysis

Mathematical Concepts

Stress and Strain
Hooke's Law
Material Properties

Formulas

Stress: σ = F / A
Elongation: ΔL = (F * L) / (A * E)
Area of circular cross-section: A = π * d^2 / 4

Theorems

Hooke's Law for Linear Elasticity

Suitable Grade Level

Undergraduate Engineering