/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 A copper bar \(A B\) of length \... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 13

A copper bar \(A B\) of length \(0.635 \mathrm{m}\) and diameter \(50 \mathrm{mm}\) is placed in position at room temperature with a gap of \(0.2 \mathrm{mm}\) between end \(A\) and a rigid restraint (see figure). The bar is supported at end \(B\) by an elastic spring with spring constant \(k=210 \mathrm{MN} / \mathrm{m}\) (a) Calculate the axial compressive stress \(\sigma_{c}\) in the bar if the temperature of the bar only rises \(27^{\circ} \mathrm{C}\). (For copper, use \(\alpha=17.5 \times 10^{-6 /^{\circ}} \mathrm{C}\) and \(E=110\) GPa. (b) What is the force in the spring? (Neglect gravity effects. (c) Repeat part (a) if \(k \rightarrow \infty\)

Short Answer

Expert verified
(a) The axial compressive stress is 90.5 MPa. (b) The force in the spring is 63.0 kN. (c) With infinite stiffness, the stress is 90.5 MPa.

Step by step solution

01

Understanding the Problem

We have a copper bar that expands due to temperature rise. Initially, the bar has a gap of 0.2 mm, and it's supported by a spring. We need to calculate the compressive stress when the bar expands and touches a rigid restraint. We also need to find the force in the spring and reevaluate under the condition that the spring has an infinite stiffness.
02

Calculating Thermal Expansion

First, calculate the change in length ( \( \Delta L \) ) of the copper bar due to the temperature increase using the formula \( \Delta L = \alpha L \Delta T \) where \( \alpha = 17.5 \times 10^{-6} /^{\circ} \text{C} \) and \( \Delta T = 27^{\circ} \text{C} \). For \( L = 0.635 \text{ m} \), compute \( \Delta L \).
03

Calculating Constraints of Movement

The total possible expansion ( \( \Delta L \) ) due to temperature increase of the bar will first close the initial gap. Calculate the true expansion causing compressive stress, \( \Delta L_{eff} = \Delta L - 0.2 \times 10^{-3} \text{ m} \).
04

Applying Hooke's Law to Calculate Stress

Use Hooke's Law to find the axial compressive stress \( \sigma_c = E \epsilon \) where \( \epsilon = \frac{\Delta L_{eff}}{L} \). Multiply \( E = 110 \text{ GPa} = 110 \times 10^9 \text{ N/m}^2 \) with the strain \( \epsilon \) calculated from step 3.
05

Calculating Force in the Spring

The force in the spring ( \( F_s \) ) can be calculated using \( F_s = k \Delta L_{eff} \). Substituting \( k = 210 \times 10^6 \text{ N/m} \), compute the force in the spring.
06

Considering Infinite Stiffness

When \( k \rightarrow \infty \), the support is perfectly rigid. This means the bar does not compress under the given constraints. The entire \( \Delta L \) is prevented by the gap closure, meaning the entire thermal expansion transfers into stress as calculated in step 4. Hence, \( \Delta L_{eff} = \Delta L \).

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Key Concepts

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

Thermal Expansion
When materials are subjected to a change in temperature, they expand or contract. This change in dimension is known as thermal expansion. The extent of expansion can be calculated using the formula:\[ \Delta L = \alpha L \Delta T \]where:
  • \( \Delta L \) is the change in length,
  • \( \alpha \) is the coefficient of thermal expansion, specific to the material,
  • \( L \) is the original length,
  • \( \Delta T \) is the change in temperature.
For copper in the given problem, \( \alpha = 17.5 \times 10^{-6} /^{\circ} \text{C} \). As the temperature of the copper bar increases, it expands. The total change in length needs to be considered when calculating other physical effects, like stress.
Hooke's Law
Hooke's Law is a principle that relates the force needed to extend or compress a spring to the distance it is stretched or compressed. It is expressed by the formula:\[ \sigma = E \epsilon \]where:
  • \( \sigma \) is the stress,
  • \( E \) is the Young's Modulus of the material, measuring its stiffness,
  • \( \epsilon \) is the strain, defined as the change in length relative to the original length.
For the exercise, stress on the copper bar after its thermal expansion can be determined using its Young's Modulus, \( E = 110 \text{ GPa} = 110 \times 10^9 \text{ N/m}^2 \). When the bar expands and touches the rigid restraint, the compressive stress is calculated through Hooke's Law, translating the expansion into stress.
Spring Force
The spring in this case helps support the copper bar. It responds to changes in the position of the bar caused by thermal expansion. The force exerted by the spring can be found using the formula:\[ F_s = k \Delta L_{eff} \]where:
  • \( F_s \) is the spring force,
  • \( k \) is the spring constant, given as \( 210 \text{ MN/m} = 210 \times 10^6 \text{ N/m} \),
  • \( \Delta L_{eff} \) is the effective change in length after allowing the initial gap.
In this setup, the spring compresses slightly until the initial gap is closed by the thermal expansion of the bar. This tension or compression within the spring directly correlates to the applied force and the potential energy stored within the spring. Neglecting gravity makes the calculations simpler.
Material Properties
The inherent characteristics of materials play a crucial role in how they react under physical changes. Key material properties include:
  • Coefficient of thermal expansion (\( \alpha \)): Indicates how much a material will expand per degree change in temperature.
  • Young's Modulus (\( E \)): Represents the material's ability to withstand changes in length when under lengthwise tension or compression.
  • Spring constant (\( k \)): Defines the stiffness of the spring, impacting how much force is exhibited when it is compressed or stretched.
Understanding these properties ensures the correct application of physical laws to solve real-world engineering problems. Copper, for example, is known for its high thermal conductivity and moderate thermal expansion, which is pertinent when it is used in thermal applications like in the exercise.

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Most popular questions from this chapter

A rigid steel plate is supported by three posts of high-strength concrete each having an effective crosssectional area \(A=40,000 \mathrm{mm}^{2}\) and length \(L=2 \mathrm{m}\) (see figure). Before the load \(P\) is applied, the middle post is shorter than the others by an amount \(s=1.0 \mathrm{mm}\) Determine the maximum allowable load \(P_{\text {allow if }}\) if the allowable compressive stress in the concrete is \(\sigma_{\text {allow }}=20\) MPa. (Use \(E=30\) GPa for concrete.)

A hollow circular tube \(T\) of length \(L=380 \mathrm{mm}\) is uniformly compressed by a force \(P\) acting through a rigid plate (see figure). The outside and inside diameters of the tube are 76 and \(70 \mathrm{mm}\), respectively. A concentric solid circular bar \(B\) of \(38 \mathrm{mm}\) diameter is mounted inside the tube. When no load is present, there is a clearance \(c=\) \(0.26 \mathrm{mm}\) between the bar \(B\) and the rigid plate. Both bar and tube are made of steel having an elastoplastic stress-strain diagram with \(E=200 \mathrm{GPa}\) and \(\sigma_{Y}=250 \mathrm{MPa}\) (a) Determine the yield load \(P_{Y}\) and the corresponding shortening \(\delta_{Y}\) of the tube. (b) Determine the plastic load \(P_{P}\) and the corresponding shortening \(\delta_{P}\) of the tube. (c) Construct a load-displacement diagram showing the load \(P\) as ordinate and the shortening \(\delta\) of the tube as abscissa. (Hint: The load-displacement diagram is not a single straight line in the region \(0 \leq P \leq P_{Y}\).)

A flat bar of width \(b\) and thickness \(t\) has a hole of diameter \(d\) drilled through it (see figure). The hole may have any diameter that will fit within the bar. What is the maximum permissible tensile load \(P_{\max }\) if the allowable tensile stress in the material is \(\sigma_{t} ?\)

A brass wire of diameter \(d=1.6 \mathrm{mm}\) is stretched between rigid supports with an initial tension \(T\) of \(200 \mathrm{N}\) (see figure). (Assume that the coefficient of thermal expansion is \(21.2 \times 10^{-6 / 0} \mathrm{C}\) and the modulus of elasticity is 110 GPa. (a) If the temperature is lowered by \(30^{\circ} \mathrm{C}\), what is the maximum shear stress \(\tau_{\max }\) in the wire? (b) If the allowable shear stress is \(70 \mathrm{MPa}\), what is the maximum permissible temperature drop? (c) At what temperature change \(\Delta T\) does the wire go slack?

Wires \(B\) and \(C\) are attached to a support at the lefthand end and to a pin- supported rigid bar at the righthand end (see figure). Each wire has cross- sectional area \(A=19.3 \mathrm{mm}^{2}\) and modulus of elasticity \(E=210 \mathrm{GPa}\) When the bar is in a vertical position, the length of each wire is \(L=2.032 \mathrm{m} .\) However, before being attached to the bar, the length of wire \(B\) was \(2.031 \mathrm{m}\) and wire \(C\) was \(2.030 \mathrm{m}\) Find the tensile forces \(T_{B}\) and \(T_{C}\) in the wires under the action of a force \(P=3.115 \mathrm{kN}\) acting at the upper end of the bar.
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