/*! 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 33 A rod has a radius of 10 \(\math... [FREE SOLUTION] | 91Ó°ÊÓ

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A rod has a radius of 10 \(\mathrm{mm}\). If it is subjected to an axial load of \(15 \mathrm{N}\) such that the axial strain in the rod is \(\epsilon_{x}=2.75\left(10^{-6}\right),\) determine the modulus of elasticity \(E\) and the change in its diameter. \(\nu=0.23\).

Short Answer

Expert verified
The modulus of elasticity 'E' of the rod is \(17.37 GPa\) and the diameter of the rod decreases by \(0.013mm\)

Step by step solution

01

Calculate the Modulus of Elasticity

Let's start by calculating the modulus of elasticity 'E', making use of Hooke's law. According to Hooke's law, stress=\(σ\)=strain=\(ε\) multiplied by modulus of elasticity=\(E\). Therefore, we can rearrange the formula to find 'E': \(E=σ/ε\). The stress σ can be found using the formula: σ=P/A, where P is the Force applied (15N) and A is the area of the rod. Since the rod is circular, A=\(π r^{2}\), where r is the radius (0.01m).
02

Calculate Stress and Modulus of Elasticity

First calculate the area: A=\(π (0.01)^{2}\) = \(3.142 × 10^{-4} m^{2}\). Next compute the stress: σ=P/A = \(15/(3.142 × 10^{-4}) = 4.77×10^{4} \mathrm{N/m^{2}}\). We then substitute this and the given strain into the formula for 'E': \(E = σ/ε = (4.77×10^{4})/(2.75 × 10^{-6}) = 1.737 ×10^{10} \mathrm{N/m^{2}}\) or \(17.37 GPa\)
03

Calculate the change in diameter

Axial strain leads to a decrease in diameter or lateral strain. The lateral strain=\(ε_{l}\) is equal to \(-νε_{x}\), where \(ν\) is Poisson's ratio(0.23) and \(ε_{x}\) is the axial strain \(2.75 × 10^{-6}\). Substituting these into the formula, we get \(ε_{l} = -νε_{x} = -(0.23)(2.75×10^{-6}) = -0.63 × 10^{-6}\). The change in diameter can be calculated using the formula \(\Delta D = ε_{l} × original diameter = -0.63 × 10^{-6} × 0.02m = -0.013 × 10^{-3}m\) or \(-0.013mm\).
04

Conclusion

So, the modulus of elasticity 'E' of the rod is \( 17.37 GPa\) and the diameter of the rod decreases by \(0.013mm\).

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Key Concepts

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

Axial Strain
Axial strain is a fundamental concept in mechanics that describes how much a material stretches or compresses along its axis when subjected to an external force.
It is a dimensionless measure and is calculated as the change in length divided by the original length of the material.
In our exercise, the axial strain of the rod is given as \(\epsilon_{x} = 2.75 \times 10^{-6}\). This indicates a very tiny extension relative to the rod's original length.

Axial strain is important for understanding how different materials behave under load.
  • It helps engineers predict the durability of materials.
  • The measurement aids in designing structures that can withstand specific amounts of load.
  • It is crucial for determining other related mechanical properties.
Overall, axial strain is a key property to assess structural integrity and ensure safety in mechanical components.
Poisson's Ratio
Poisson's ratio \((u)\) is a measure of the contraction or expansion that occurs perpendicular to the direction of applied stress.
When a material is stretched or compressed in one direction, it tends to contract or expand in the other directions. This tendency is captured by Poisson's ratio.
In this exercise, the Poisson's ratio for the rod is given as \(u = 0.23\).

Poisson's ratio helps in understanding the lateral deformation of the material when axial strain is applied.
  • If \(u\) is positive, the material contracts laterally when stretched axially, and expands laterally when compressed.
  • A higher \(u\) suggests more lateral contraction or expansion, while a lower \(u\) indicates less lateral change.
In our example, the negative sign in the lateral strain calculation \(\epsilon_{l} = -u \epsilon_{x}\) reflects the reduction in diameter as a result of axial strain. This concept is crucial for applications where precise changes in dimensions are necessary, such as designing mechanical joints or thin-walled structures.
Stress and Strain Calculations
Stress and strain calculations are foundational to understanding the mechanical behavior of materials under load.
In this exercise, stress \((\sigma)\) is calculated using the formula \(\sigma = P/A\), where \(P\) is the force applied, and \(A\) is the cross-sectional area.
For our circular rod with a radius of 10 mm, the formula for the area \(A\) is \(\pi r^2\), which gives us \(A = 3.142 \times 10^{-4}\ \text{m}^{2}\).

Once the stress is found as \(4.77 \times 10^{4}\ \text{N/m}^{2}\), it is used to calculate the modulus of elasticity \((E)\) through \(E = \sigma/\epsilon\).
These calculations link stress, strain, and material properties, providing insights into how materials will respond under different conditions.
  • Stress describes internal forces within a material.
  • Strain measures deformation resulting from those forces.
  • The modulus of elasticity indicates material stiffness, showing how much a material will deform under a given stress.
Understanding these calculations is crucial for engineers to select suitable materials for construction and to predict the longevity and safety of structures.

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Most popular questions from this chapter

Derive an expression for an equivalent bending moment \(M_{e}\) that, if applied alone to a solid bar with a circular cross section, would cause the same energy of distortion as the combination of an applied bending moment \(M\) and torque \(T\).

A bar with a circular cross-sectional area is made of SAE 1045 carbon steel having a yield stress of \(\sigma_{Y}=150 \mathrm{ksi}\). If the bar is subjected to a torque of 30 kip \(\cdot\) in. and a bending moment of 56 kip \(\cdot\) in., determine the required diameter of the bar according to the maximum-distortion-energy theory. Use a factor of safety of 2 with respect to yielding.

The state of strain on an element has components \(\boldsymbol{\epsilon}_{x}=-400\left(10^{-6}\right), \boldsymbol{\epsilon}_{y}=0, \gamma_{x y}=150\left(10^{-6}\right) .\) Determine the equivalent state of strain on an element at the same point oriented \(30^{\circ}\) clockwise with respect to the original element. Sketch the results on this element.

Consider the general case of plane strain where \(\epsilon_{x}, \epsilon_{y},\) and \(\gamma_{x y}\) are known. Write a computer program that can be used to determine the normal and shear strain, \(\epsilon_{x^{\prime}}\) and \(\gamma_{x^{\prime} y^{\prime}},\) on the plane of an element oriented \(\theta\) from the horizontal. Also, include the principal strains and the element's orientation, and the maximum in-plane shear strain, the average normal strain, and the element's orientation.

A soft material is placed within the confines of a rigid cylinder which rests on a rigid support. Assuming that \(\boldsymbol{\epsilon}_{x}=0\) and \(\boldsymbol{\epsilon}_{y}=0,\) determine the factor by which the modulus of elasticity will be increased when a load is applied if \(\nu=0.3\) for the material.

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