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Chapter 17: Problem 52

a) Suppose a bimetallic strip is constructed of copper and steel strips of thickness \(1.0 \mathrm{~mm}\) and length \(25 \mathrm{~mm},\) and the temperature of the strip is reduced by \(5.0 \mathrm{~K}\). Determine the radius of curvature of the cooled strip (the radius of curvature of the interface between the two strips). b) If the strip is \(25 \mathrm{~mm}\) long, how far is the maximum deviation of the strip from the straight orientation?

Short Answer

Expert verified
Answer: The radius of the curvature of the bimetallic strip is 2.5 m, and the maximum deviation from the straight orientation is approximately 12.5 mm.

Step by step solution

01

Finding the difference in length of the two materials

To calculate the radius of curvature of the cooled strip, we first need to find the difference in the length of copper and steel after being cooled by 5.0 K. The change in length is given by the formula: \(\Delta L = L_0 \cdot \alpha \cdot \Delta T\), where \(L_0\) is the initial length, \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the change in temperature. For copper, the coefficient of linear expansion, \(\alpha_{Cu} = 17 \times 10^{-6} \mathrm{K^{-1}}\). So, the change in length of copper, \(\Delta L_{Cu} = L_0 \cdot \alpha_{Cu} \cdot \Delta T = 25 \times 10^{-3} \mathrm{m} \cdot 17 \times 10^{-6} \mathrm{K^{-1}} \cdot (-5) \mathrm{K} = -2.125 \times 10^{-6} \mathrm{m}\). For steel, the coefficient of linear expansion, \(\alpha_{Fe} = 12 \times 10^{-6} \mathrm{K^{-1}}\). So, the change in length of steel, \(\Delta L_{Fe} = L_0 \cdot \alpha_{Fe} \cdot \Delta T = 25 \times 10^{-3} \mathrm{m} \cdot 12 \times 10^{-6} \mathrm{K^{-1}} \cdot (-5) \mathrm{K} = -1.5 \times 10^{-6} \mathrm{m}\). Now, we need to find the difference in length, \(\Delta L = |\Delta L_{Cu} - \Delta L_{Fe}| = |(-2.125 - (-1.5)) \times 10^{-6} \mathrm{m}| = 6.25 \times 10^{-7} \mathrm{m}\).
02

Calculate the radius of curvature

Now, we can calculate the radius of curvature using the formula: \(R = \frac{L^2}{2 \cdot \Delta L}\). Plugging in the values, we get: \(R = \frac{(25 \times 10^{-3} \mathrm{m})^2}{2 \cdot 6.25 \times 10^{-7} \mathrm{m}} = 2500 \times 10^{-6} \mathrm{m} = 2.5 \mathrm{m}\). So, the radius of curvature of the cooled strip is 2.5 m.
03

Calculate the maximum deviation (sagitta)

The maximum deviation or sagitta can be found using the formula: \(s = R - \sqrt{R^2 - (\frac{L}{2})^2}\). Plugging in the values, we get: \(s = 2.5 \mathrm{m} - \sqrt{(2.5 \mathrm{m})^2 - (\frac{25 \times 10^{-3} \mathrm{m}}{2})^2} = 2.5 \mathrm{m} - \sqrt{6.244 \mathrm{m^2}} \approx 0.0125 \mathrm{m}\). So, the maximum deviation of the strip from the straight orientation is approximately 0.0125 m or 12.5 mm.

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Key Concepts

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

Coefficient of Linear Expansion
Understanding the coefficient of linear expansion is crucial when studying materials that are subject to temperature changes. It is a thermophysical property indicating how much a material's length changes per degree change in temperature. Represented by the symbol \( \alpha \), it's typically expressed in the unit \( \mathrm{K^{-1}} \) (inverse kelvins).

For example, as mentioned in the exercise, copper has an \( \alpha \) value of \( 17 \times 10^{-6} \mathrm{K^{-1}} \) and steel has \( 12 \times 10^{-6} \mathrm{K^{-1}} \). When they are bonded together to form a bimetallic strip and the temperature changes, each metal will expand or contract at its characteristic rate. The difference in their expansion rates will cause the strip to bend, with the curvature depending on the temperature change and the physical properties of both materials. By using the formula \( \Delta L = L_0 \cdot \alpha \cdot \Delta T \), we can determine the change in length (\( \Delta L \) ) of each material when the temperature changes by \( \Delta T \) kelvins.
Radius of Curvature
The radius of curvature, \( R \), is a measure that describes the degree of bending of a curve. In the context of a bimetallic strip, it refers to the radius of the imaginary circle that the curved strip would form. The larger the radius, the less pronounced the curve is. Conversely, a smaller radius indicates a tighter curve.

As seen from the solution to our problem, the radius of curvature of the bimetallic strip after being subjected to a temperature change can be calculated using \( R = \frac{L^2}{2 \cdot \Delta L} \), where \( L \) is the length of the strip and \( \Delta L \) denotes the difference in contraction or expansion between the two attached materials. This value is essential in engineering applications where precise knowledge of a material's behavior under thermal stress is required for design and safety.
Thermal Expansion
Thermal expansion is the tendency of matter to change its shape, area, and volume in response to a change in temperature. This physical principle is behind the predictable bending of a bimetallic strip. When heated or cooled, materials expand or contract based on their individual coefficients of linear expansion.

Temperature fluctuations cause the physical dimensions of materials to change - an essential consideration in construction, manufacturing, and material selection. For instance, bridges are built with expansion joints that accommodate the thermal expansion and contraction of materials. Without accounting for this effect, structures may fail due to thermal stress. The exercise demonstrates how understanding thermal expansion is not just about the theory but has practical implications for solving real-world engineering challenges.

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Most popular questions from this chapter

A medical device used for handling tissue samples has two metal screws, one \(20.0 \mathrm{~cm}\) long and made from brass \(\left(\alpha_{\mathrm{b}}=18.9 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right)\) and the other \(30.0 \mathrm{~cm}\) long and made from aluminum \(\left(\alpha_{\mathrm{a}}=23.0 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right)\). A gap of \(1.00 \mathrm{~mm}\) exists between the ends of the screws at \(22.0^{\circ} \mathrm{C}\). At what temperature will the two screws touch?

In order to create a tight fit between two metal parts, machinists sometimes make the interior part larger than the hole into which it will fit and then either cool the interior part or heat the exterior part until they fogether. Suppose an aluminum rod with diameter \(D_{1}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right)\) is to be fit into a hole in a brass plate that has a diameter \(D_{2}=10.000 \mathrm{~mm}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right) .\) The machinists can cool the rod to \(77.0 \mathrm{~K}\) by immersing it in liquid nitrogen. What is the largest possible diameter that the rod can have at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) and just fit into the hole if the rod is cooled to \(77.0 \mathrm{~K}\) and the brass plate is left at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} ?\) The linear expansion coefficients for aluminum and brass are \(22 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) and \(19 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\), respectively.

For a class demonstration, your physics instructor uniformly heats a bimetallic strip that is held in a horizontal orientation. As a result, the bimetallic strip bends upward. This tells you that the coefficient of linear thermal expansion for metal T, on the top is _____ that of metal B, on the bottom. a) smaller than b) larger than c) equal to

A copper cube of side length \(40 . \mathrm{cm}\) is heated from \(20 .{ }^{\circ} \mathrm{C}\) to \(120{ }^{\circ} \mathrm{C}\). What is the change in the volume of the cube? The linear expansion coefficient of copper is \(17 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

Express each of the following temperatures in degrees Celsius and in kelvins. a) \(-19^{\circ} \mathrm{F}\) b) \(98.6^{\circ} \mathrm{F}\) c) \(52^{\circ} \mathrm{F}\)
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