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Chapter 2: Problem 2

A 4.8 -ft-long steel wire of \(\frac{1}{4}\) -in.- diameter is subjected to a \(750-\) lb tensile load. Knowing that \(E=29 \times 10^{6}\) psi, determine (a) the elongation of the wire, \((b)\) the corresponding normal stress.

Short Answer

Expert verified
Elongation: 0.0302 inches; Stress: 15275.05 psi.

Step by step solution

01

Identify Given Values

First, let's list all the known values from the problem: - Length of the steel wire, \( L = 4.8 \text{ ft} = 4.8 \times 12 = 57.6 \text{ in.} \)- Diameter of the wire, \( d = \frac{1}{4} \text{ in.} \)- Tensile load, \( F = 750 \text{ lb} \)- Modulus of Elasticity \( E = 29 \times 10^6 \text{ psi} \)We need to find (a) the elongation of the wire and (b) the corresponding normal stress.
02

Calculate the Cross-sectional Area

The cross-sectional area \( A \) of the wire is calculated using the formula for the area of a circle:\[A = \pi \left(\frac{d}{2}\right)^2\]Substitute the diameter:\[A = \pi \left(\frac{1}{4}\right)^2 = \pi \left(\frac{1}{16}\right) = \frac{\pi}{16} \approx 0.0491 \text{ in}^2\]
03

Calculate Elongation Using Hooke's Law

The elongation \( \Delta L \) can be calculated using the formula:\[\Delta L = \frac{FL}{AE}\]Substitute the values:\[\Delta L = \frac{750 \times 57.6}{0.0491 \times 29 \times 10^6} \approx 0.0302 \text{ in.}\]
04

Calculate the Normal Stress

Normal stress \( \sigma \) is calculated using the formula:\[\sigma = \frac{F}{A}\]Substitute the known values:\[\sigma = \frac{750}{0.0491} \approx 15275.05 \text{ psi}\]
05

Summary of Results

The elongation of the wire is approximately \( 0.0302 \) inches, and the corresponding normal stress is approximately \( 15275.05 \) psi.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tensile Load
When we talk about a tensile load, we're discussing a force that attempts to stretch or elongate an object. Imagine pulling on both ends of a rope; that's a tensile force at work.
In our example, the steel wire is subjected to a tensile load of 750 pounds. This force is trying to pull the wire apart, stretching the steel to a certain extent.
If you don't apply a load properly, it can lead to failure or breaking of the material. When designing materials to carry loads, understanding tensile load is crucial.
  • A tensile load is applied along the length of the material.
  • It is measured in units such as pounds (lb) or Newtons (N).
  • Tensile loads are essential for ensuring materials perform as desired under stress.
Being mindful of tensile loads helps engineers design safer and more efficient structures.
Normal Stress
Normal stress is a measure of how much a force is squeezing or stretching an object per unit area. It's like the pressure the force exerts on the material.
In our steel wire scenario, the normal stress is calculated based on the tensile load and the wire's cross-sectional area. We use the formula:\[\sigma = \frac{F}{A}\]
Here, \(\sigma\) represents normal stress, \(F\) is the force or load applied, and \(A\) is the cross-sectional area.
  • Normal stress is expressed in units like pounds per square inch (psi) or Pascals (Pa).
  • High normal stress can lead to material deformation or failure.
  • Understanding normal stress helps in determining the material's ability to withstand forces without failure.
Being aware of the normal stress ensures the structural integrity of materials under load.
Hooke's Law
Hooke's Law is a principle that states that the amount an elastic material stretches is directly proportional to the force applied, as long as the deformation is not permanent. This relation helps predict how materials behave under different loads.
The elongation or stretch of a material can be calculated using:\[\Delta L = \frac{FL}{AE}\]
Here, \(\Delta L\) is the change in length (elongation), \(F\) is the applied force, \(L\) is the original length, \(A\) is the cross-sectional area, and \(E\) is the modulus of elasticity.
  • Hooke's Law is applicable only within the elastic limit of the material.
  • It provides insights into designing materials that need to return to their original shape after deformation.
  • Understanding this law is essential for fields such as mechanical and civil engineering.
By using Hooke's Law, we can efficiently predict and manage material behavior under tension.
Modulus of Elasticity
The modulus of elasticity, often denoted as \(E\), quantifies a material's ability to resist deformation when a force is applied. It's like a material's stiffness indicator.
For the steel wire in our problem, \(E\) is given as 29 million psi, showcasing how strong the material is in resisting changes to its shape.
This modulus plays a vital role in determining how much a material stretches under a specific tensile load.
  • Materials with higher modulus values are stiffer and can withstand more stress without deformation.
  • It is measured in units such as psi or GPa (gigapascals).
  • A crucial factor in engineering, helping in selecting materials for different applications.
Understanding the modulus of elasticity aids in predicting and managing how materials respond to applied forces.

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Most popular questions from this chapter

In a standard tensile test a steel rod of 22 -mm diameter is subjected to a tension force of 75 kN. Knowing that \(\nu=0.30\) and \(E=200 \mathrm{GPa},\) determine \((a)\) the elongation of the rod in a \(200-\mathrm{mm}\) gage length, ( \(b\) ) the change in diameter of the rod.

A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2 -kip tensile load is applied to it. Knowing that \(E=29 \times 10^{6} \mathrm{psi},\) determine \((a)\) the smallest diameter rod that should be used, ( \(b\) ) the corresponding normal stress caused by the load.

An 18 -m-long steel wire of 5 -mm diameter is to be used in the manufacture of a prestressed concrete beam. It is observed that the wire stretches \(45 \mathrm{mm}\) when a tensile force \(\mathbf{P}\) is applied. Knowing that \(E=200 \mathrm{GPa}\), determine ( \(a\) ) the magnitude of the force \(\mathbf{P}\), (b) the corresponding normal stress in the wire.

Anylon thread is to be subjected to a 10 -N tension. Knowing that \(E=3.2 \mathrm{GPa},\) that the maximum allowable normal stress is \(40 \mathrm{MPa},\) and that the length of the thread must not increase by more than \(1 \%\), determine the required diameter of the thread.

In many situations physical constraints prevent strain from occurring in a given direction. For example, \(\epsilon_{z}=0\) in the case shown, where longitudinal movement of the long prism is prevented at every point. Plane sections perpendicular to the longitudinal axis remain plane and the same distance apart. Show that for this situation, which is known as plane strain, we can express \(\sigma_{z}, \epsilon_{x},\) and \(\epsilon_{y}\) as follows: \\[ \begin{array}{l} \sigma_{z}=\nu\left(\sigma_{x}+\sigma_{y}\right) \\ \epsilon_{x}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{x}-\nu(1+\nu) \sigma_{y}\right] \\ \epsilon_{y}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{y}-\nu(1+\nu) \sigma_{x}\right] \end{array} \\]
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