A steel wire is stretched with a stress of \(70 \mathrm{MPa}(10,000 \mathrm{psi})\) at \(20^{\circ} \mathrm{C}\left(68^{\circ} \mathrm{F}\right) .\) If the length is held constant, to what temperature must the wire be heated to reduce the stress to \(17 \mathrm{MPa}(2500 \mathrm{psi}) ?\) v

Short Answer

Expert verified
Answer: The steel wire needs to be heated to \(42.08^{\circ} \mathrm{C}\) to reduce the stress to \(17 \, \mathrm{MPa}\).

Step by step solution

01

Identify the given information and required temperature change

We have the following information: Initial stress: \(\sigma_1 = 70 \, \mathrm{MPa}\) Final stress: \(\sigma_2 = 17 \, \mathrm{MPa}\) Initial temperature: \(T_1 = 20^{\circ} \mathrm{C}\) We need to find the final temperature \(T_2\).
02

Calculate the change in stress

To find the change in stress, we subtract the final stress from the initial stress. $$\Delta \sigma = \sigma_1 - \sigma_2 = 70 \, \mathrm{MPa} - 17 \, \mathrm{MPa} = 53 \, \mathrm{MPa}$$.
03

Calculate the change in temperature

We'll use the stress-strain relationship: $$\frac{\Delta \sigma}{E} = \alpha \Delta T$$. Rearrange this equation to get: $$\Delta T = \frac{\Delta \sigma}{\alpha E}$$ We need the values for the coefficient of linear expansion, \(\alpha\), and Young's modulus, \(E\), for steel. You can find these values from reference materials or your textbook: \(\alpha = 12 \times 10^{-6} \, \mathrm{K}^{-1}\) \(E = 200 \, \mathrm{GPa}\) Plug the values into the equation: $$\Delta T = \frac{53 \times 10^6 \, \mathrm{Pa}}{(12 \times 10^{-6} \, \mathrm{K}^{-1})(200 \times 10^9 \, \mathrm{Pa})}$$ $$\Delta T = 22.08 \, \mathrm{K}$$
04

Calculate the final temperature

Finally, add \(\Delta T\) to the initial temperature to find the final temperature: $$T_2 = T_1 + \Delta T = 20^{\circ} \mathrm{C} + 22.08 \, \mathrm{K}$$ $$T_2 = 42.08^{\circ} \mathrm{C}$$ So, the steel wire needs to be heated to \(42.08^{\circ} \mathrm{C}\) to reduce the stress to \(17 \, \mathrm{MPa}\).

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Key Concepts

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

Thermal Expansion Coefficient
The thermal expansion coefficient is a fundamental property that describes how the size of a material changes with temperature. Different materials will expand or contract at different rates when heated or cooled, which is quantified by this coefficient, usually denoted as \( \alpha \).

For many solids, the expansion is linear, meaning it's directly proportional to the change in temperature. It is expressed in units of inverse temperature (usually per degree Celsius \( ^\circ\text{C}^{-1} \) or per Kelvin \( \text{K}^{-1} \) ). When a material is heated and its length is restrained, thermal stress can develop because the material cannot expand freely. This property helps us understand and calculate such stress under varying thermal conditions, as was necessary in the exercise with the steel wire.
Young's Modulus
Young's modulus, often represented by \( E \), is a measure of a material's stiffness and is a fundamental characteristic in the field of materials science and engineering. It describes the stress-strain relationship for materials exhibiting elastic behavior—at least within certain limits.

Defined as the ratio of stress (force per unit area) to strain (proportional deformation), Young's modulus has the units of pressure, usually expressed in pascals \( \text{Pa} \). In simple terms, a high value of \( E \), which steel possesses, signifies that a lot of force is needed to change its shape; thus, it is considered a \textbf{stiff} material. This concept is critical when calculating the necessary changes in temperature to alter the stress within a material, as it influences how much force is transferred into dimensional changes rather than deformation.
Stress-Strain Relationship
Stress is defined as the force applied over a cross-sectional area, and the resulting distortion or strain is how much a material deforms under that force. The relationship between stress and strain is often linear for small deformations, allowing for the calculation of changes in material length, area, or volume with formulas that apply Young's modulus and the thermal expansion coefficient.

An outstanding point to remember is that the relationship assumes the deformation is elastic—meaning the material will return to its original shape once the force is removed. If the stress exceeds a material's elastic limit, permanent deformation, or even failure, can occur. The exercise demonstrated this relationship by showing the calculations needed to determine the stress reduction in a wire through heating, under the assumption that the wire's deformation remained within its elastic range.

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Most popular questions from this chapter

The constant \(A\) in Equation 19.2 is \(12 \pi^{4} R / 5 \theta_{\mathrm{D}}^{3},\) where \(R\) is the gas constant and \(\theta_{\mathrm{D}}\) is the Debye temperature (K). Estimate \(\theta_{\mathrm{D}}\) for aluminum, given that the specific heat is \(4.60 \mathrm{J} / \mathrm{kg}-\mathrm{K}\) at \(15 \mathrm{K}\).

(a) Briefly explain why thermal stresses may be introduced into a structure by rapid heating or cooling. (b) For cooling, what is the nature of the surface stresses? (c) For heating, what is the nature of the surface stresses?

(a) Briefly explain why porosity decreases the thermal conductivity of ceramic and polymeric materials, rendering them more thermally insulative. (b) Briefly explain how the degree of crystallinity affects the thermal conductivity of polymeric materials and why.

A bimetallic strip is constructed from strips of two different metals that are bonded along their lengths. Explain how such a device may be used in a thermostat to regulate temperature.

The difference between the specific heats at constant pressure and volume is described by the expression $$c_{p}-c_{v}=\frac{\alpha_{v}^{2} v_{0} T}{\beta}\quad\quad\quad\quad\quad(19.10)$$ where \(\alpha_{v}\) is the volume coefficient of thermal expansion, \(v_{0}\) is the specific volume (i.e., volume per unit mass, or the reciprocal of density \(), \beta\) is the compressibility, and \(T\) is the absolute temperature. Compute the values of \(c_{v}\) at room temperature \((293 \mathrm{K})\) for aluminum and iron using the data in Table 19.1, assuming that \(\alpha_{v}=3 \alpha_{l}\) and given that the values of \(\beta\) for \(A l\) and \(F e\) are \(1.77 \times 10^{-11}\) and \(2.65 \times 10^{-12}(\mathrm{Pa})^{-1},\) respectively.

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