/*! 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 20 A brass alloy is known to have a... [FREE SOLUTION] | 91Ó°ÊÓ

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A brass alloy is known to have a yield strength of \(240 \mathrm{MPa}(35,000\) psi), a tensile strength of \(310 \mathrm{MPa}(45,000 \mathrm{psi})\), and an elastic modulus of 110 GPa \(\left(16.0 \times 10^{6}\right.\) psi). A cylindrical specimen of this alloy \(15.2 \mathrm{~mm}(0.60 \mathrm{in}\).) in diameter and \(380 \mathrm{~mm}(15.0\) in.) long is stressed in tension and found to elongate \(1.9 \mathrm{~mm}(0.075\) in.). On the basis of the information given, is it possible to compute the magnitude of the load necessary to produce this change in length? If so, calculate the load; if not, explain why.

Short Answer

Expert verified
If so, determine the magnitude. Answer: No, it is not possible to compute the magnitude of the load necessary to produce the elongation of 1.9 mm in the given brass alloy cylindrical specimen due to the calculated stress (550 MPa) being outside of the acceptable range between the yield strength (240 MPa) and tensile strength (310 MPa).

Step by step solution

01

Calculate the cross-sectional area

The cross-sectional area of the cylindrical specimen (A) can be calculated using the formula for the area of a circle: \(A = \pi r^{2}\) where r is the radius of the cylinder. In this case, the diameter of the cylinder is given as 15.2 mm, so the radius (r) is: \(r = \frac{15.2}{2} = 7.6 \; \text{mm}\) Now we can calculate the cross-sectional area (A): \(A = \pi(7.6^2) = 181.6 \; \text{mm}^2\)
02

Calculate the elongation ratio

To find the elongation ratio (ER), we need to divide the elongation of the cylinder (1.9 mm) by its original length (380 mm): \(\text{ER} = \frac{1.9}{380} = 0.005\)
03

Calculate the stress

Now we will use the relationship between stress, elongation ratio, and elastic modulus: \(\text{Stress (}\sigma\text{)} = \text{Elongation Ratio (ER)} \cdot \text{Elastic Modulus (E)}\) Plugging in the given values, we get: \(\sigma = 0.005 \cdot 110 \cdot 10^{3} = 550\; \text{MPa}\)
04

Check if the stress is within the acceptable range

We need to check if the calculated stress (550 MPa) falls within the acceptable range, which is between the yield strength (240 MPa) and the tensile strength (310 MPa) of the brass alloy. In this case, the calculated stress (550 MPa) is not within the acceptable range, so it is not possible to compute the magnitude of the load necessary to produce this change in length. If the stress had been within the acceptable range, we would proceed to step 5 to calculate the load, but in this case, the problem can't be solved as the calculated stress is higher than the tensile strength of the alloy.

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Key Concepts

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

Yield Strength
Yield strength is an essential concept in materials science that refers to the stress level at which a material begins to deform plastically. This means that beyond this point, the material will not return to its original shape when the stress is removed.
For the brass alloy in the exercise, the yield strength is given as 240 MPa. This value indicates the limit of elastic behavior in the material. Understanding yield strength is crucial for design safety, ensuring that materials can withstand applied forces without undergoing permanent deformation.
It is important to note that yield strength is determined through experimental testing and varies between different materials and alloys. Designers use this information to select appropriate materials for specific applications, ensuring they remain in the elastic deformation region during use.
Tensile Strength
Tensile strength, also often known as ultimate tensile strength (UTS), represents the maximum stress a material can withstand while being stretched or pulled before breaking. For the brass alloy mentioned, this value is 310 MPa. It is a crucial property that indicates the material's ability to handle maximum tensile forces before failure.
Comparing tensile strength with yield strength gives insights into a material's ductility and toughness. If the stress exceeds the tensile strength, the material will fail, often catastrophically. Hence, tensile strength serves as an upper safety limit for design and is considered when determining the load-carrying capacity of a material.
It is important for engineers to consider both tensile and yield strengths when designing components to ensure they operate safely under applied forces.
Elastic Modulus
Elastic modulus, also known as Young's modulus, is a measure of a material's stiffness or rigidity. It is defined as the ratio of stress to strain when deformation is entirely elastic. For the given brass alloy, the elastic modulus is 110 GPa.
The elastic modulus provides insight into how much a material will deform under a specific load within its elastic range. Higher elastic modulus means a stiffer material that deforms less under stress. This is critical when precision and minimal deformation are required.
Elastic modulus is fundamental in design and engineering, influencing choices for applications involving bending, flexibility, and vibration. Understanding this concept is key to predicting how materials will behave under various loading conditions.
Stress Calculation
Stress calculation is a foundational skill in engineering and materials science, essential for evaluating how materials respond under forces.
In this context, stress (\(\sigma\)) is calculated using the formula \(\sigma = \frac{F}{A}\), where \(F\) is the applied force and \(A\) is the cross-sectional area of the material. Additionally, the relationship between stress, elastic modulus, and elongation ratio is expressed as \(\sigma = \text{ER} \times E\).
This exercise specifically showed how to calculate stress based on the elongation of a cylindrical specimen and its material properties. In this case, the calculated stress exceeded the tensile strength of the brass, indicating potential failure under the given load.
Stress analysis helps in designing safe and effective structures by predicting how materials will behave under load, ensuring reliability and integrity in engineering applications.

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Most popular questions from this chapter

Briefly cite the differences between the recovery and recrystallization processes.

6.24 A cylindrical rod \(120 \mathrm{~mm}\) long and having a (- diameter of \(15.0 \mathrm{~mm}\) is to be deformed using a tensile load of \(35,000 \mathrm{~N}\). It must not experience either plastic deformation or a diameter reduction of more than \(1.2 \times 10^{-2} \mathrm{~mm}\). Of the following materials listed, which are possible candidates? Justify your choice(s). \begin{tabular}{lccc} \hline & Modulus of Material & \begin{tabular}{c} \mathrm{ Yield } \(\\\ {\text { Elasticity }} \\ {\text { (GPa) }}\) & Strength (MPa) & Poisson's Ratio \\ \hline Aluminum alloy & 70 & 250 & \(0.33\) \\ \hline Titanium alloy & 105 & 850 & \(0.36\) \\ \hline Steel alloy & 205 & 550 & \(0.27\) \\ \hline Magnesium alloy & 45 & 170 & \(0.35\) \\ \hline \end{tabular} \end{tabular}

Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 15 and \(12 \mathrm{~mm}\), respectively. The second specimen, with an initial radius of \(11 \mathrm{~mm}\), must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation.

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 \(6.31\) 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,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 2.10b), 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 \(d F / d r\).

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