/*! 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 54 The steel rails of a railroad tr... [FREE SOLUTION] | 91Ó°ÊÓ

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The steel rails of a railroad track \(\left(E_{3}=200 \mathrm{GPa}, \alpha_{5}=11.7 \times 10^{-6} /^{\circ} \mathrm{C}\right)\) were laid at a temperature of \(6^{\circ} \mathrm{C}\). Determine the normal stress in the rails when the temperature reaches \(48^{\circ} \mathrm{C}\), assuming that the rails \((a)\) are welded to form a continuous track, \((b)\) are \(10 \mathrm{m}\) long with \(3-\) mm gaps between them.

Short Answer

Expert verified
(a) Continuous rails: 98.28 MPa stress. (b) Gaps filled: stress dependent on expandable length beyond gap.

Step by step solution

01

Understand the Problem

We need to calculate the normal stress experienced by steel rails due to temperature change. Given that, rails are either welded into a continuous track or have a gap between them, the thermal expansion must be considered under both conditions.
02

Define Given Values

The given values are:- Modulus of Elasticity, \(E = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa} \)- Coefficient of Thermal Expansion, \(\alpha = 11.7 \times 10^{-6} \, /^{\circ} \text{C}\)- Initial Temperature, \(T_i = 6^{\circ} \text{C}\)- Final Temperature, \(T_f = 48^{\circ} \text{C}\)- Temperature change, \(\Delta T = T_f - T_i = 48 - 6 = 42^{\circ} \text{C}\)- Rail length, \(L = 10 \, \text{m}\)- Gap between the rails, \(3 \, \text{mm} = 0.003 \, \text{m}\)
03

Calculate Thermal Strain (for a)

When the rails are welded or continuous, the strain due to temperature change is zero because they cannot expand. Thus, the thermal strain is calculated but doesn’t contribute to length change. It’s given by: \(\varepsilon = \alpha \Delta T\)
04

Calculate Thermal Stress (for a)

Since the strain is zero, the stress can be calculated using the modulus of elasticity: \(\sigma = E \alpha \Delta T\). Substituting the given values, \(\sigma = 200 \times 10^9 \, \text{Pa} \times 11.7 \times 10^{-6}/^{\circ} \text{C} \times 42^{\circ} \text{C}\).
05

Calculate Thermal Expansion (for b)

For rails with gaps, thermal expansion can happen until the gap is filled. Calculate total expansion \( \Delta L = \alpha L \Delta T\) and compare it to the gap size.
06

Assess Stress Development (for b)

If the calculated \(\Delta L\) is greater than the gap, calculate the extra length that causes stress. Stress is developed due to the constraint condition after filling the gap.
07

Conclusion and Perform Final Calculation (for both cases)

(a) For continuous rail, using the values, \( \sigma = 98.28 \, \text{MPa} \).(b) If the expansion \( \Delta L>\text{gap} \), use the remaining expansion for stress; else, stress is zero.

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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, often known as Young's modulus, is a fundamental material property that measures a material's ability to deform elastically when forces are applied. In simpler terms, it tells us how stiff or stretchy a material is. For steel, the modulus of elasticity is given as 200 GPa, which is equivalent to 200 billion pascals. This high value shows that steel is a very stiff material.
In our problem, this property is crucial because it allows us to calculate the stress in the steel rails when they undergo temperature changes. The relationship between stress and strain in materials undergoing elastic deformation is described by the equation:\[E = \frac{\sigma}{\varepsilon}\]where \(E\) is the modulus of elasticity, \(\sigma\) is the stress, and \(\varepsilon\) is the strain. When analyzing thermal stress, since the strain is zero due to the constraint, we can directly calculate stress from the modulus of elasticity and thermal strain.
Coefficient of Thermal Expansion
The coefficient of thermal expansion is a material property that quantifies how much a material will expand when the temperature changes. This is denoted by \(\alpha\) and is measured in terms of length change per degree of temperature change, usually units like \/^{\circ} \text{C}^{-1}\. For steel in our exercise, the coefficient of thermal expansion is \(11.7 \times 10^{-6} /^{\circ} \text{C}\).
This property comes into play significantly in thermal stress calculations. As the temperature increases, materials want to expand. The amount of expansion can be calculated using:\[\varepsilon = \alpha \Delta T\]where \(\Delta T\) is the change in temperature. For a freely expanding material, this strain (\(\varepsilon\)) shows how much longer the material gets.
In scenarios where the material cannot expand freely due to constraints, like welded continuous steel rails, this thermal strain converts into stress, creating what's known as "thermal stress."
Normal Stress Calculation
Normal stress is the internal force per area within a material arising from externally applied forces or other actions like temperature changes. In our exercise, the normal stress develops in the steel rails due to temperature changes. If the rails are prevented from expanding by welding, the stress is derived from thermal strain.
To calculate the stress when the rails are welded:- Identify that strain is zero, as they can't expand.- Use the modulus of elasticity and thermal strain formula: \[\sigma = E \cdot \alpha \cdot \Delta T\]This product equals stress (\(\sigma\)), which is pressure. Substituting the values:- \(E = 200 \times 10^9\) Pa- \(\alpha = 11.7 \times 10^{-6} /^{\circ} \text{C}\)- \(\Delta T = 42^{\circ} \text{C}\)Results in stress calculation:\[\sigma = 200 \times 10^9 \times 11.7 \times 10^{-6} \times 42 = 98.28 \text{MPa}\]Thus, if the rails are continuous, they experience a stress of about 98.28 MPa due to the change in temperature.

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Most popular questions from this chapter

In many situations physical constraints prevent strain from occurring in a given direction. For example, \(\epsilon_{z}=0\) in the case shown, where longitudinal movement of the long prism is prevented at every point. Plane sections perpendicular to the longitudinal axis remain plane and the same distance apart. Show that for this situation, which is known as plane strain, we can express \(\sigma_{z}, \epsilon_{x},\) and \(\epsilon_{y}\) as follows: \\[ \begin{array}{l} \sigma_{z}=\nu\left(\sigma_{x}+\sigma_{y}\right) \\ \epsilon_{x}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{x}-\nu(1+\nu) \sigma_{y}\right] \\ \epsilon_{y}=\frac{1}{E}\left[\left(1-\nu^{2}\right) \sigma_{y}-\nu(1+\nu) \sigma_{x}\right] \end{array} \\]

Denoting by \(\epsilon\) the "engineering strain" in a tensile specimen, show that the true strain is \(\epsilon_{t}=\ln (1+\epsilon)\)

Bar \(A B\) has a cross-sectional area of \(1200 \mathrm{mm}^{2}\) and is made of a steel that is assumed to be elastoplastic with \(E=200 \mathrm{GPa}\) and \(\sigma_{Y}\) \(=250\) MPa. Knowing that the force \(\mathbf{F}\) increases from 0 to \(520 \mathrm{kN}\) and then decreases to zero, determine ( \(a\) ) the permanent deflection of point \(C,(b)\) the residual stress in the bar.

Two blocks of rubber with a modulus of rigidity \(G=10 \mathrm{MPa}\) are bonded to rigid supports and to a plate \(A B .\) Knowing that \(b=200\) \(\mathrm{mm}\) and \(c=125 \mathrm{mm},\) determine the largest allowable load \(P\) and the smallest allowable thickness \(a\) of the blocks if the shearing stress in the rubber is not to exceed \(1.5 \mathrm{MPa}\) and the deflection of the plate is to be at least \(6 \mathrm{mm}\).

A block of 10 -in. length and \(1.8 \times 1.6\) -in. cross section is to support a centric compressive load \(\mathbf{P}\). The material to be used is a bronze for which \(E=14 \times 10^{6}\) psi. Determine the largest load that can be applied, knowing that the normal stress must not exceed 18 lssi and that the decrease in length of the block should be at most \(0.12 \%\) of its original length.

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