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A 0.5 -m-long steel rod with a 1 -cm diameter is stretched in a tensile test. What is the work required to obtain a relative strain of \(0.1 \% ?\) The modulus of elasticity of steel is \(2 \times 10^{8} \mathrm{kPa}\).

Short Answer

Expert verified
The work required is approximately 3.925 Joules.

Step by step solution

01

Calculate the Original Cross-Sectional Area

To calculate the cross-sectional area, use the formula for the area of a circle: \[ A = \pi r^2 \]The diameter of the rod is 1 cm, so the radius \( r \) is 0.5 cm or 0.005 m. Substituting, we have:\[ A = \pi (0.005)^2 \approx 7.85 \times 10^{-5} \text{ m}^2 \]
02

Determine Change in Length

Relative strain is given by the formula:\[ \text{Strain} = \frac{\Delta L}{L_0} \]where \( \Delta L \) is the change in length and \( L_0 \) is the original length. Given that \( \text{Strain} = 0.001 \) (or \(0.1\%\)) and \( L_0 = 0.5 \text{ m}\), we find:\[ \Delta L = \text{Strain} \times L_0 = 0.001 \times 0.5 = 0.0005 \text{ m} \]
03

Calculate the Stress

Stress \( \sigma \) is related to strain and modulus of elasticity \( E \) by:\[ \sigma = E \times \text{Strain} \]The modulus of elasticity \( E \) of steel is \(2 \times 10^{8} \text{ kPa} = 2 \times 10^{11} \text{ Pa} \). Thus:\[ \sigma = 2 \times 10^{11} \times 0.001 = 2 \times 10^{8} \text{ Pa} \]
04

Calculate the Force Applied

Force \( F \) can be determined using stress \( \sigma \) and area \( A \):\[ F = \sigma \times A \]Plug in the values:\[ F = (2 \times 10^{8} \text{ Pa}) \times (7.85 \times 10^{-5} \text{ m}^2) \approx 15700 \text{ N} \]
05

Compute Work Done

The work done \( W \) to stretch the rod can be calculated by:\[ W = \frac{1}{2} F \Delta L \]Substitute \( F \) and \( \Delta L \) from previous steps:\[ W = \frac{1}{2} \times 15700 \text{ N} \times 0.0005 \text{ m} \approx 3.925 \text{ J} \]

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Key Concepts

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

Modulus of Elasticity
The modulus of elasticity, also known as Young's modulus, is a fundamental property that characterizes the stiffness of a material. It is denoted by the symbol "E" and is defined as the ratio of stress (\( \sigma \)) to strain (\( \epsilon \)) within the elastic limit of a material. In simpler terms, it describes how much a material will deform under a given load.
  • High modulus of elasticity: Indicates a stiff material that does not easily deform.
  • Low modulus of elasticity: Reflects a more flexible material that will deform more easily.
In this exercise, the modulus of elasticity for steel is given as \(2 \times 10^{11} \text{ Pa}\). This value implies that steel is a relatively stiff material, which means it will not stretch much under small loads. However, understanding this concept is crucial when predicting how materials behave in real-world applications.
Tensile Test
A tensile test is a crucial experimental procedure used to determine various mechanical properties of materials, especially focusing on how they behave when subjected to tensile (pulling) forces. This involves deforming a material sample to understand how it responds to the applied force. Some important aspects of tensile tests include:
  • Setup: The material, such as a rod or wire, is elongated under controlled conditions.
  • Outcome: Results provide stress-strain data, helping to identify material properties like tensile strength, ductility, and modulus of elasticity.
  • Application: Widely used in engineering fields to ensure material reliability and design safety.
In our problem, the tensile test involves stretching a steel rod with a specified diameter and length to achieve a specific strain of 0.1%. Such tests help in understanding the material's strength and elasticity before being used in structural applications.
Stress and Strain
Stress and strain are two essential concepts in materials science that help us understand material deformation.
  • Stress: Defined as the force applied per unit area (\( \sigma = \frac{F}{A} \)). It describes how internal forces in the material resist the applied load. In this exercise, we computed the stress as \(2 \times 10^{8} \text{ Pa}\).
  • Strain: The deformation experienced by the material when stress is applied (\( \epsilon = \frac{\Delta L}{L_0} \)). It is a dimensionless quantity representing the relative change in shape or size. Here, a strain of 0.1% or 0.001 was used.
Understanding the relationship between stress and strain is crucial in predicting material performance, especially when designing structures and systems to ensure they do not fail under load. The modulus of elasticity links these quantities, providing a measure of material stiffness.
Work Done
Work done in the context of stretching materials involves calculating the energy required to deform a material under certain conditions. It reflects the energy transferred when a force is applied to move an object. For elasticity, work done is given by:\[ W = \frac{1}{2} F \Delta L \]This formula comes from the area under the stress-strain curve in linear elastic deformation phases. In our exercise, we calculated the work needed to achieve 0.1% strain on the steel rod, which amounted to approximately 3.925 Joules.Key points about work done in this case:
  • Predicts necessary energy for a material change due to stress.
  • Helps assess a material's energy efficiency and resilience.
Understanding work done is vital for ensuring that materials can withstand required loads without exceeding their energy capacities, ensuring longevity and safety of structures.

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Most popular questions from this chapter

A hydraulic hoist raises a \(1750-\mathrm{kg}\) car \(1.8 \mathrm{~m}\) in an auto repair shop. The hydraulic pump has a constant pressure of \(800 \mathrm{kPa}\) on its piston. What is the increase in potential energy of the car and how much volume should the pump displace to deliver that amount of work?

A piston/cylinder \((0.5 \mathrm{~kg}\) steel altogether) maintaining a constant pressure has \(0.2 \mathrm{~kg} \mathrm{R}-134 \mathrm{a}\) as saturated vapor at \(150 \mathrm{kPa}\). It is heated to \(40^{\circ} \mathrm{C}\), and the steel is at the same temperature as the R-134a at any time. Find the work and heat transfer for the process.

Helium gas expands from \(125 \mathrm{kPa}, 350 \mathrm{~K}\) and \(0.25 \mathrm{~m}^{3}\) to \(100 \mathrm{kPa}\) in a polytropic process with \(n=1.667 .\) How much work does it give out?

A copper block of volume \(1 \mathrm{~L}\) is heat treated at \(500^{\circ} \mathrm{C}\) and now cooled in a \(200-\mathrm{L}\) oil bath initially at \(20^{\circ} \mathrm{C}\), as shown in Fig. \(\mathrm{P} 3.118\). Assuming no heat transfer with the surroundings, what is the final temperature?

A cylinder fitted with a frictionless piston contains \(2 \mathrm{~kg}\) of superheated refrigerant \(\mathrm{R}-134 \mathrm{a}\) vapor at \(350 \mathrm{kPa}, 100^{\circ} \mathrm{C}\). The cylinder is now cooled so that the \(\mathrm{R}-134 \mathrm{a}\) remains at constant pressure until it reaches a quality of \(75 \% .\) Calculate the heat transfer in the process.

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