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Chapter 6: Problem 28

A bar of a steel alloy that exhibits the stress-strain behavior shown in Figure \(6.21\) is subjected to a tensile load; the specimen is 300 mm (12 in.) long and has a square cross section \(4.5 \mathrm{~mm}(0.175 \mathrm{in}\).) on a side. (a) Compute the magnitude of the load necessary to produce an elongation of \(0.45 \mathrm{~mm}\) \((0.018\) in.). (b) What will be the deformation after the load has been released?

Short Answer

Expert verified
Answer: The magnitude of the load necessary to produce an elongation of 0.45 mm in the steel bar is approximately 6075 N. After the load has been released, the deformation is approximately 0, implying that the steel bar would regain its original length.

Step by step solution

01

a) Calculating the Tensile Load:

Hooke's Law defines the relationship between stress (蟽) and strain (蔚) as 蟽 = E蔚, where E is the Modulus of Elasticity. We need to determine the stress at which the elongation of 0.45 mm occurs. First, we need to calculate the strain from the elongation: 蔚 = (elongation) / (original_length) = (0.45 mm) / (300 mm) = 0.0015 Now, we can find the stress by using the relationship between stress and strain given in Figure 6.21, which is approximately linear: 蟽 = E蔚 Let's say the Modulus of Elasticity of the steel alloy (E) as given in Figure 6.21 is 200 GPa. So, converting GPa to N/mm虏, E = 200 x 10^3 N/mm虏. 蟽 = 200 x 10^3 N/mm虏 x 0.0015 = 300 N/mm虏 Now, we can calculate the force: Area of the cross-section (A) = Side^2 = (4.5 mm)^2 = 20.25 mm虏 Force (F) = Area (A) x stress (蟽) = 20.25 mm虏 x 300 N/mm虏 = 6075 N So, the magnitude of the load necessary to produce an elongation of 0.45 mm is 6075 N (approximately).
02

b) Deformation after Load Release:

When the load is released, the steel bar will partially recover its shape due to its elastic properties. However, it may not return to its original length completely. To determine the permanent deformation, we need to find the strain after unloading. This can again be using the data provided in Figure 6.21 regarding the relationship between stress and strain. Let's say that after unloading, the stress (蟽') is reduced to 0 N/mm虏 since the load is removed. The relationship between the 蟽' and 蔚' can also be used from Figure 6.21, and assuming it's linear: 蟽' = E蔚' Now, as 蟽' = 0, 蔚' will also be 0. This implies that the deformation after the load has been released is also 0. In reality, there might be a little permanent deformation, and finding its value would require more information than provided in the exercise. However, for answering this part, the deformation after the load has been released is approximately 0, implying that the steel bar would regain its original length after unloading.

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Key Concepts

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

Stress-Strain Behavior
When a material like steel is subjected to a tensile load, it experiences stress and strain. **Stress** is the internal force per unit area exerted within a material, while **strain** represents the deformation of the material, or how much it changes in length compared to its original state.

The relationship between stress and strain can be visualized using a stress-strain curve. This curve shows how a material reacts as loads are applied or removed. Typically, for small deformations, the behavior of many materials, including steel, can be considered linear. This is why Hooke's Law, which states that stress is directly proportional to strain, comes into play for such linear parts of the curve.

It's crucial to understand that this linear relationship only holds until the yield point of the material, which is where it begins to deform plastically and loses its elasticity. Past this point, the material will not return to its original shape once the load is removed.
Modulus of Elasticity
The Modulus of Elasticity, denoted by 'E', is a fundamental property of materials that measures their ability to deform elastically. Specifically, it's the ratio of stress to strain in the linear portion of the stress-strain curve.

A higher Modulus of Elasticity means the material is stiffer and requires more stress to produce a given amount of strain. In simpler terms, materials with high values of 'E', like steel, are resistant to being deformed elastically.

For example, within the context of the problem at hand, we used a typical value of 200 GPa for steel's Modulus of Elasticity. This was applied during calculations using the formula \( \sigma = E\epsilon \), allowing us to determine the required tensile load to produce a specific elongation.
Permanent Deformation
Permanent deformation occurs when a material has been stressed beyond its elastic limit and does not return to its original shape after the load is removed. This is typically seen when the load surpasses the yield strength of the material.

In the stress-strain curve, the portion beyond the yield point represents permanent deformation. Here, any increase in strain results in large inelastic deformations with little increase in stress. After such scenarios, materials do not completely recover, showing a permanent change in length or shape.

In our exercise, though it is stated that the deformation after release is nearly zero, in real-world scenarios some minor permanent deformation might occur if the material was pushed beyond its elastic range. Without specific data or actual testing details, the precise value of such permanent deformation cannot be determined.

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Most popular questions from this chapter

In Section \(2.6\) it was noted that the net bonding energy \(E_{N}\) between two isolated positive and negative ions is a function of interionic distance \(r\) as follows: $$ E_{N}=-\frac{A}{r}+\frac{B}{r^{n}} $$ where \(A, B\), and \(n\) are constants for the particular ion pair. Equation \(6.25\) is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity \(E\) is proportional to the slope of the interionic force-separation curve at the equilibrium interionic separation; that is, $$ E \propto\left(\frac{d F}{d r}\right)_{r_{0}} $$ Derive an expression for the dependence of the modulus of elasticity on these \(A, B\), and \(n\) parameters (for the two-ion system) using the following procedure: 1\. Establish a relationship for the force \(F\) as a function of \(r\), realizing that $$ F=\frac{d E_{N}}{d r} $$ 2\. Now take the derivative \(d F / d r\). 3\. Develop an expression for \(r_{0}\), the equilibrium separation. Because \(r_{0}\) corresponds to the value of \(r\) at the minimum of the \(E_{N}\)-versus-r curve (Figure \(2.8 b\) ), take the derivative \(d E_{N} / d r\), set it equal to zero, and solve for \(r\), which corresponds to \(r_{0}\). 4\. Finally, substitute this expression for \(r_{0}\) into the relationship obtained by taking \(d F / d r\).

The following true stresses produce the corresponding true plastic strains for a brass alloy: $$ \begin{array}{cc} \hline \begin{array}{c} \text { True Stress } \\ \text { (psi) } \end{array} & \text { True Strain } \\ \hline 50,000 & 0.10 \\ 60,000 & 0.20 \\ \hline \end{array} $$ What true stress is necessary to produce a true plastic strain of \(0.25\) ?

Using the solution to Problem \(6.13\), rank the magnitudes of the moduli of elasticity for the following hypothetical X, Y, and Z materials from the greatest to the least. The appropriate \(A, B\), and \(n\) parameters (Equation \(6.25)\), for these three materials are shown in the following table; they yield \(E_{N}\) in units of electron volts and \(r\) in nanometers: $$ \begin{array}{cccc} \hline \text { Material } & \boldsymbol{A} & \boldsymbol{B} & \boldsymbol{n} \\\ \hline \mathrm{X} & 2.5 & 2.0 \times 10^{-5} & 8 \\ \mathrm{Y} & 2.3 & 8.0 \times 10^{-6} & 10.5 \\ \mathrm{Z} & 3.0 & 1.5 \times 10^{-5} & 9 \\ \hline \end{array} $$

The following table gives a number of Rockwell B hardness values that were measured on a single steel specimen. Compute average and standard deviation hardness values. $$ \begin{array}{lll} 83.3 & 80.7 & 86.4 \\ 88.3 & 84.7 & 85.2 \\ 82.8 & 87.8 & 86.9 \\ 86.2 & 83.5 & 84.4 \\ 87.2 & 85.5 & 86.3 \end{array} $$

A cylindrical rod of copper \((E=110 \mathrm{GPa}\), \(16 \times 10^{6}\) psi) having a yield strength of 240 MPa ( \(35,000 \mathrm{psi})\) is to be subjected to a load of \(6660 \mathrm{~N}\left(1500 \mathrm{lb}_{\mathrm{f}}\right)\). If the length of the rod is \(380 \mathrm{~mm}\) (15.0 in.), what must be the diameter to allow an elongation of \(0.50 \mathrm{~mm}\) \((0.020 \mathrm{in} .)\) ?
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