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A vertically suspended wire has a cross-sectional area of \(0.1\) in. \(^{2}\). A downward force, applied to the end of the wire, causes the wire to stretch. The force is increased linearly from initially zero to \(2500 \mathrm{lb}\), and the length of the wire increases by \(0.1 \%\). Select a wire length and use it to determine (a) the normal stress, in lb/in. \({ }^{2}\), and (b) the work done in stretching the wire, in \(\mathrm{ft} \cdot \mathrm{lb}\).

Short Answer

Expert verified
(a) Normal Stress : \(25000 \, \mathrm{lb/in.}^{2}\), (b) Work done : \(0.104 \, \mathrm{ft.lb}\)

Step by step solution

01

Calculate normal stress

Let's assign an arbitrary length to the wire. For simplicity, take this to be \(1\) in. From the definition of stress, we know that stress = force / area. Here, force is \(2500 \,\mathrm{lb}\) and area is \(0.1 \, \mathrm{in.}^{2}\). Therefore, stress \( \sigma = \frac{2500 \,\mathrm{lb}}{0.1 \, \mathrm{in.}^{2}} = 25000 \, \mathrm{lb/in.}^{2}\)
02

Calculate work done

The work done in stretching the wire can be defined as the area under the force-extension graph. As the force is increased linearly from zero, the graph will be a triangle. The area (work done) will be \(0.5 \times 2500\,\mathrm{lb} \times 0.001\,\mathrm{in.}\), as per the given data that the length of the wire increases by 0.1%, which would mean 0.001 inches for a 1 inch long wire. This needs to be converted into ft.lb, by converting inches into feet, as \(1\, \mathrm{ft} = 12\, \mathrm{in.}\). Therefore, work done = \(0.5 \times 2500\, \mathrm{lb} \times 0.001\, \mathrm{in.} \times \frac{1\, \mathrm{ft}}{12\, \mathrm{in.}} = 0.104 \, \mathrm{ft.lb}\)
03

Final Answers

So, the two sought values are: \n (a) The normal stress in the wire: \(25000 \, \mathrm{lb/in.}^{2}\). \n (b) The total work done in stretching the wire: \(0.104 \, \mathrm{ft.lb}\).

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Key Concepts

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

Stress-Strain Relationship
The stress-strain relationship is a fundamental aspect of the mechanical behavior of materials. Stress is defined as the internal forces that neighboring particles of material exert on each other, whereas strain is a measure of the deformation of the material. In our exercise, we calculated normal stress, which is the force per unit area acting perpendicular to the surface. Mathematically, normal stress is represented as \( \text{stress} = \frac{\text{force}}{\text{area}} \).
When a material is subjected to stress, it deforms. This deformation relative to the original length of the material is known as strain. The exercise described a linear increase in force leading to a proportional increase in length (strain) of the wire, illustrating what is known as elastic deformation, which is the initial straight-line part of a stress-strain curve. For most materials under elastic deformation, the relationship between stress and strain can be described by Hooke's Law, where stress is proportional to strain up to the elastic limit. Beyond this limit, materials may undergo plastic deformation, where the stress-strain relationship no longer follows this law.
Work Done by Force
In physics, the concept of work done by force is a measure of energy transfer that occurs when an object is moved over a distance by an external force. The basic formula for calculating work is \( W = F \times d \times \text{cos}\theta \), where \( W \) is work, \( F \) is the magnitude of the force applied, \( d \) is the displacement caused by the force, and \( \theta \) is the angle between the force and the direction of motion. In the case of the wire stretching exercise, the work done is equal to the area under the force-displacement graph.
Since the force was applied linearly, this area can be visualized as a triangle with a base equal to the displacement and a height equal to the maximum force applied. We calculated this as \(0.5 \times \text{force} \times \text{displacement}\), and then we converted the units to \text{ft.lb}\to match standard units of work in mechanical systems. This calculation assumes that the force is applied in the direction of displacement (\( \theta = 0 \)), making \(\text{cos}\theta = 1\), which simplifies our formula for work to \( W = F \times d \).
Mechanical Properties of Materials
Understanding the mechanical properties of materials is crucial when it comes to applying the correct material to a specific engineering context. These properties, which include strength, ductility, hardness, and toughness, dictate how a material will react under various types of forces and conditions. The wire in the exercise demonstrates two important mechanical properties: its elasticity, represented by its ability to return to the original shape after the force is removed, and its strength, indicated by the magnitude of the stress it can withstand without permanent deformation.
In the contact of the exercise, strength is indicated by the calculated normal stress that the wire can resist, which might relate to the material's yield strength or ultimate strength depending on the scope of application. The properties that we discuss often stem from the molecular structure of the material, and they are quantified through experiments and represented in stress-strain curves. These properties not only inform us about the potential performance of materials under stress but also guide engineers in material selection and design to ensure safety and effectiveness in real-world applications.

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