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The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initial diameter of the specimen is \(d_{1},\) show that when the diameter is \(d,\) the true strain is \(\boldsymbol{\epsilon}_{t}=2 \ln \left(d_{1} / d\right)\)

Short Answer

Expert verified
The true strain is \(\epsilon_t = 2 \ln \left(\frac{d_1}{d}\right)\).

Step by step solution

01

Understanding the Problem

The problem involves plastic deformation of a tensile specimen, where the volume remains constant during deformation. We are asked to relate the true strain with initial and current diameters of the specimen.
02

Initial Setup and Assumptions

Let's assume the initial length and diameter of the specimen are \(L_1\) and \(d_1\), respectively. The initial volume \(V_1\) can be expressed as \(V_1 = \pi \left(\frac{d_1}{2}\right)^2 L_1\). Similarly, when the diameter changes to \(d\), the length becomes \(L\), and the volume \(V\) is \(V = \pi \left(\frac{d}{2}\right)^2 L\).
03

Constant Volume Condition

Since the volume is constant, we have \(V_1 = V\). Thus, we equate the two volume expressions: \[\pi \left(\frac{d_1}{2}\right)^2 L_1 = \pi \left(\frac{d}{2}\right)^2 L\].
04

Simplify the Equation

Cancel out the common terms, \(\pi\) and the factor \(\frac{1}{4}\) from both sides: \[ (d_1)^2 L_1 = (d)^2 L \].
05

Expressing Length Ratio

Rearrange the equation to express the ratio of the lengths in terms of the diameters: \[\frac{L}{L_1} = \left(\frac{d_1}{d}\right)^2\].
06

Definition of True Strain

True strain \(\epsilon_t\) is given by \(\epsilon_t = \ln \left(\frac{L}{L_1}\right)\). Substitute the length ratio from the previous step: \[\epsilon_t = \ln \left(\frac{d_1}{d}\right)^2\].
07

Applying Logarithmic Properties

Apply the logarithmic property \(\ln (x^y) = y \ln (x)\): \[\epsilon_t = 2 \ln \left(\frac{d_1}{d}\right)\].
08

Conclusion

Thus, we have shown that when the diameter changes from \(d_1\) to \(d\), the true strain is \(\epsilon_t = 2 \ln \left(\frac{d_1}{d}\right)\).

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Key Concepts

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

Plastic Deformation
In the realm of material science, **plastic deformation** is a key concept that refers to the permanent change in shape of a material under an applied force. When materials undergo plastic deformation, they do not return to their original shape after the stress is removed. This is in contrast to elastic deformation, where materials return to their original form once the force is no longer applied.

For tensile specimens, plastic deformation occurs after the yield point is passed. Initially, the specimen might stretch elastically, but given enough force, it will begin to thin out and extend further in a process of irreversible deformation.

Understanding plastic deformation is crucial for many engineering applications. It tells us how much a material can be deformed before it breaks or changes in undesirable ways. This understanding helps in selecting the right materials for specific applications, ensuring safety and efficiency in design.
Constant Volume
The principle of **constant volume** during plastic deformation asserts that as a material undergoes plastic change, the volume of the tensile specimen remains unchanged. This might seem counterintuitive, as the specimen visibly deforms and changes shape.

The constant volume condition is based on the preservation of mass concept. Since the material isn't adding or losing mass during deformation, the volume stays constant even though the dimensions change. This phenomenon can be explained using volume conservation formulas.

In practice, during plastic deformation of a tensile specimen, although length increases, the cross-sectional area decreases proportionally to maintain constant volume. Thus, understanding this principle is essential for accurate calculations in engineering problems, like finding true strain in deforming materials.
Tensile Specimen
A **tensile specimen** is a sample of material used in tensile testing to measure its mechanical properties. It's primarily used to assess how a material reacts to forces that attempt to pull it apart. Typically, a tensile specimen is a cylindrical or rectangular rod of material.

Tensile testing is a fundamental test conducted to determine various properties, such as ultimate tensile strength, yield strength, and elongation. During the test, a specimen is subjected to a controlled tension until failure occurs.

This process allows engineers to assess whether the material's properties meet required standards. It helps in identifying performance characteristics, which are critical in applications where failure could be catastrophic.

Through tensile tests, engineers can also plot stress-strain curves, which provide vital insights into material behavior under stress.
Logarithmic Properties
**Logarithmic properties** play a significant role in the mathematical characterization of materials science concepts such as true strain. True strain takes into account the natural logarithm of the ratio of initial to final dimensions, providing a more precise measure of deformation for engineering analyses.

Using logarithmic properties, specifically \( \ln(x^y) = y \ln(x) \), allows us to simplify expressions and compute values conveniently. In the context of the exercise, the true strain \(\epsilon_t\) becomes calculable using the formula: \(\epsilon_t = 2 \ln(d_1/d)\).

The use of logarithms in strain calculations has the advantage of accumulating continuous changes rather than treating deformation increments as linear. This results in a more realistic representation of how materials stretch, compress, or shear over time and produces accurate analysis in material testing and design.

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Most popular questions from this chapter

A 4 -ft concrete post is reinforced with four steel bars, each with a \(\frac{3}{4}\) -in. diameter. Knowing that \(E_{s}=29 \times 10^{6}\) psi and \(E_{c}=3.6 \times 10^{6} \mathrm{psi}\), determine the normal stresses in the steel and in the concrete when a 150 -kip axial centric force \(\mathbf{P}\) is applied to the post.

The concrete post \(\left(E_{c}=3.6 \times 10^{6} \mathrm{psi} \text { and } \alpha_{c}=5.5 \times 10^{-6} /^{\circ} \mathrm{F}\right)\) is reinforced with six steel bars, each of \(\frac{z}{8}\) -in. diameter \(\left(E_{x}=29 \times 10^{6} \mathrm{psi} \text { and } \alpha_{x}=6.5 \times 10^{-6} /^{\circ} \mathrm{F}\right) .\) Determine the normal stresses induced in the steel and in the concrete by a temperature rise of \(65^{\circ} \mathrm{F}\)

Anylon thread is to be subjected to a 10 -N tension. Knowing that \(E=3.2 \mathrm{GPa},\) that the maximum allowable normal stress is \(40 \mathrm{MPa},\) and that the length of the thread must not increase by more than \(1 \%\), determine the required diameter of the thread.

A 250 -mm-long aluminum tube \((E=70 \mathrm{GPa})\) of 36 -mm outer diameter and 28 -mm inner diameter can be closed at both ends by means of single- threaded screw-on covers of 1.5 -mm pitch. With one cover screwed on tight, a solid brass rod \((E=105 \mathrm{GPa})\) of 25 -mm diameter is placed inside the tube and the second cover is screwed on. Since the rod is slightly longer than the tube, it is observed that the cover must be forced against the rod by rotating it one-quarter of a turn before it can be tightly closed. Determine (a) the average normal stress in the tube and in the rod, (b) the deformations of the tube and of the rod.

A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2 -kip tensile load is applied to it. Knowing that \(E=29 \times 10^{6} \mathrm{psi},\) determine \((a)\) the smallest diameter rod that should be used, ( \(b\) ) the corresponding normal stress caused by the load.

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