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Chapter 7: Problem 27

(a) Show, for a tensile test, that $$ \% \mathrm{CW}=\left(\frac{\epsilon}{\epsilon+1}\right) \times 100 $$ if there is no change in specimen volume during the deformation process (i.e., \(A_{0} l_{0}=A_{d} l_{d}\) ). (b) Using the result of part (a), compute the percent cold work experienced by naval brass (the stress-strain behavior of which is shown in Figure 6.12) when a stress of 400 MPa \((58,000\) psi) is applied.

Short Answer

Expert verified
Question: Derive the formula for percentage cold work (% CW) during a tensile test, considering no change in the specimen volume, and calculate the % CW for naval brass when a stress of 415 MPa is applied. Answer: The formula for % CW during a tensile test is % CW = (ε/(ε + 1)) × 100, where ε is the strain. To calculate the % CW for naval brass when the stress is 415 MPa, find the corresponding strain value (ε_x) from the stress-strain curve and substitute it into the formula: % CW = (ε_x/(ε_x + 1)) × 100.

Step by step solution

01

Part (a): Deriving the % CW formula

Let's start by writing the expression for the initial volume (V0) and the deformed volume (Vd), and then equate them since there is no volume change. Initial volume, \(V_0 = A_0 l_0\) Deformed volume, \(V_d = A_d l_d\) As V0 = Vd, \(A_0 l_0 = A_d l_d\) ...(1) Now we know that, Strain, \(\epsilon = \frac{l_d - l_0}{l_0}\) Rearrange the expression for \(l_d\), \(l_d = l_0(1+\epsilon)\) ...(2) Now, let's substitute equation (2) into equation (1), \(A_0 l_0 = A_d l_0(1+\epsilon)\) Rearrange the equation to solve for \(\frac{A_d}{A_0}\), \(\frac{A_d}{A_0} = \frac{1}{1+\epsilon}\) Now we can calculate the % CW from the ratio of deformed and initial cross-sectional areas as follows: % CW \(=\frac{A_0 - A_d}{A_0} \times 100\) Substitute the expression for \(\frac{A_d}{A_0}\), \(\% \mathrm{CW}=\left(\frac{\epsilon}{\epsilon + 1}\right) \times 100\) So, this is the formula for % CW during a tensile test, considering no change in volume.
02

Part (b): Calculate % CW for Naval Brass

Now, let's use the given stress value of 415 MPa and the stress-strain behavior of naval brass to find the strain (\(\epsilon\)) when the stress is 415 MPa. From the Figure 6.12 (not provided), we can approximate the strain value by finding the point on the curve that corresponds to a stress of 415 MPa. Suppose, we find the strain value to be \(\epsilon_x\). Now, using the % CW formula derived in part (a), we can calculate the % CW experienced by naval brass as: \(\% \mathrm{CW} =\left(\frac{\epsilon_x}{\epsilon_x + 1}\right) \times 100\) After solving this equation with the obtained strain value, we will have the percent cold work experienced by naval brass when a stress of 415 MPa is applied.

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Key Concepts

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

Tensile Test
Imagine pulling on a candy bar until it stretches and eventually breaks—that's essentially what a tensile test does with materials such as metals. In a tensile test, we clamp a specimen of the material into a machine and apply a force that stretches it until it deforms or fractures. The goal is to measure the material's strength and ductility. This test is crucial for understanding how materials will behave under different types of forces.

During the test, we record the force applied and the amount the material stretches. The data collected helps us plot a stress-strain curve, which shows the material's response to the applied stress. It's an essential method for engineers to select the right material for the right job, ensuring safety and performance.
Strain
Strain is a fundamental concept in materials science that measures how much a material deforms under stress. Put simply, it's the amount of stretch or compression a material experiences compared to its original length. Mathematically, it's expressed as a ratio, where the change in length is divided by the original length (\( \frac{\text{change in length}}{\text{original length}} \)).

Strain isn't just about pulling something apart; it can also be about squeezing it together, which is called compressive strain. The tensile strain is what's most commonly associated with the tensile test, and it's a critical value for understanding a material's ductility or how much it can be stretched before breaking.
Stress-Strain Behavior
The stress-strain behavior of a material is akin to a personality profile for how it reacts under pressure. When we plot the applied stress against the resulting strain during a tensile test, we get a curve typically starting with a straight line where the material behaves elastically—meaning it'll go back to its original shape once the stress is removed.

However, keep applying stress, and the material will start to yield and deform permanently; this is the plastic region. The curve's peak represents the maximum strength of the material. This visual representation—the stress-strain curve—is an incredibly powerful tool for materials engineering. It helps predict how materials will behave in real-world applications and influences the design choices of everything from bridges to medical implants.
Material Deformation
Material deformation is the technical term for when stuff bends, stretches, or changes shape under a force. This can take place suddenly or over time. Returning to our earlier mention of strain, it's the measure of this deformation.

There are two main types of deformation: elastic and plastic. Elastic deformation is like a rubber band snapping back after being stretched—that's temporary change. Plastic deformation is like molding clay—it stays where you leave it. This is a permanent change. Understanding how and why materials deform is essential in designing safer structures and creating new materials that can withstand extreme environments.
Naval Brass Properties
Naval brass is a specific type of brass that's prized for its resilience in marine environments—it doesn't corrode easily in seawater, thanks to its unique ingredients. This brass alloy mixes copper, zinc, and sometimes a little tin to create a material that's tough and resistant to wear and tear, which is why it's a favorite for ship components and, historically, naval artillery shells.

When we perform a tensile test on naval brass, we can determine its mechanical properties, like ultimate tensile strength, yield strength, and percent cold work—the latter being the amount of deformation it can endure without heating it up (cold working it). These properties help engineers decide where and how to use naval brass most effectively, ensuring its qualities are utilized to the fullest in applications where corrosion-resistance and durability are paramount.

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Most popular questions from this chapter

An undeformed specimen of some alloy has an average grain diameter of \(0.040 \mathrm{~mm}\). You are asked to reduce its average grain diameter to \(0.010 \mathrm{~mm}\). Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.

The lower yield point for an iron that has an average grain diameter of \(5 \times 10^{-2} \mathrm{~mm}\) is 135 MPa (19,500 psi). At a grain diameter of \(8 \times\) \(10^{-3} \mathrm{~mm}\), the yield point increases to \(260 \mathrm{MPa}\) \((37,500 \mathrm{psi})\). At what grain diameter will the lower yield point be \(205 \mathrm{MPa}(30,000 \mathrm{psi})\) ?

Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows: $$ \begin{array}{lcr} \hline & \begin{array}{c} \text { Circular } \\ \text { (diameter, } \mathbf{m m}) \end{array} & \begin{array}{c} \text { Rectangular } \\ (\mathbf{m m}) \end{array} \\ \hline \text { Original dimensions } & 15.2 & 125 \times 175 \\ \text { Deformed dimensions } & 11.4 & 75 \times 200 \\ \hline \end{array} $$ Which of these specimens will be the hardest after plastic deformation, and why?

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 \(16 \mathrm{~mm}\) and \(11 \mathrm{~mm}\), respectively. The second specimen, with an initial radius of \(12 \mathrm{~mm}\), must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation.

(a) From the plot of yield strength versus (grain diameter) \(^{-1 / 2}\) for a \(70 \mathrm{Cu}-30 \mathrm{Zn}\) cartridge brass, Figure \(7.15\), determine values for the constants \(\sigma_{0}\) and \(k_{y}\) in Equation \(7.7\). (b) Now predict the yield strength of this alloy when the average grain diameter is \(1.0 \times 10^{-3} \mathrm{~mm}\)
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