(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 \(1.0 \times 10^{-3} \mathrm{~mm}\)

Short Answer

Expert verified
In order to determine the constants \(\sigma_{0}\) and \(k_{y}\) from the plot of yield strength versus grain diameter \(^{-1/2}\) for a 70 Cu-30 Zn cartridge brass (Figure 7.15), you must first choose two points on the graph and use them to form a system of linear equations. Upon solving this system, the obtained \(\sigma_{0}\) and \(k_{y}\) constants can be combined with the provided average grain diameter of \(2.0 \times 10^{-3} \mathrm{mm}\) to predict the yield strength of the alloy. The final prediction will be a numerical value expressed in the relevant units (e.g., MPa or psi).

Step by step solution

01

Determining the constants \(\sigma_{0}\) and \(k_{y}\)

Inspect Figure 7.15 and determine two points on the graph, preferably far apart. Points (1) and (2) are described using the coordinates \((x_1, y_1)\) and \((x_2, y_2)\), respectively. Then, \(x_1 = d_1^{-1/2}\), \(y_1 = \sigma_{Y1}\), \(x_2 = d_2^{-1/2}\), \(y_2 = \sigma_{Y2}\) Solve these equations for \(\sigma_{0}\) and \(k_{y}\) by setting up the following two equations: \(\sigma_{Y1} = \sigma_{0} + k_{y} d_1^{-1/2}\) and \(\sigma_{Y2} = \sigma_{0} + k_{y} d_2^{-1/2}\) These can be written as a system of linear equations: \(\begin{bmatrix} 1 & d_1^{-1/2} \\ 1 & d_2^{-1/2} \end{bmatrix}\begin{bmatrix} \sigma_0 \\ k_{y} \end{bmatrix}=\begin{bmatrix} \sigma_{Y1} \\ \sigma_{Y2} \end{bmatrix}\) Using the method of your choice, solve this system for \(\sigma_0\) and \(k_{y}\).
02

Predicting the yield strength

Now that you have values for the constants \(\sigma_{0}\) and \(k_{y}\), use Equation 7.7 to predict the yield strength for a grain diameter of \(2.0 \times 10^{-3} \mathrm{mm}\): \(\sigma_{Y} = \sigma_{0} + k_{y} (2.0 \times 10^{-3})^{-1/2}\) Compute the yield strength and report your result.

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Key Concepts

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

Hall-Petch Relationship
Understanding the Hall-Petch relationship is crucial for your grasp of how the microscopic structure of materials can influence their macroscopic properties. This relationship describes how the strength of a polycrystalline material varies with its grain size. Smaller grains can obstruct the movement of dislocations, which are defects within the crystal structure that contribute to plastic deformation. As grain size decreases, the yield strength, or the stress at which a material begins to deform plastically, increases.

The Hall-Petch equation, which can be represented as \( \sigma_Y = \sigma_0 + k_y d^{-1/2} \), connects yield strength \(\sigma_Y\) to the grain diameter \(d\) in a quantifiable manner. Here, \(\sigma_0\) is the material's intrinsic yield strength without the influence of grain boundaries, and \(k_y\) is the Hall-Petch slope, a constant that quantifies the grain boundary strengthening effect. So, smaller grain diameters lead to higher yield strengths because of the \(d^{-1/2}\) term.
Grain Size Effects on Yield Strength
When technical jargon is stripped away, the influence of grain size on yield strength is quite straight-forward. The material is composed of grains, much like a mosaic is made of tiles. The boundaries between these grains act as barriers to dislocation motion, akin to a wall halting the progress of an intruder. Dislocations can be thought of as carriers of plastic deformation, and when their path is blocked, deformation — and thus material failure — take more effort to occur.

In more technical terms, as the average grain diameter decreases, more grain boundaries are present per unit area, leading to an enhanced 'hallmark' effect of preventing dislocation motion. This results in improved yield strength, making the material harder to permanently deform. This phenomenon explains why processes like cold working, which refine grain structure, are used to enhance the strength of materials. Educators often use a simplified visualization or analogy of a wall and intruders (dislocations) to help students visualize why smaller grains enhance yield strength.
Yield Strength Calculation
Let's dive into how to actually calculate yield strength based on grain size. Referring back to the Hall-Petch equation, once we gather empirical data for yield strength at different grain sizes, we can plot this data and determine the values for \(\sigma_0\) and \(k_y\). In the exercise, you find two points on the yield strength vs. grain size graph. Once you have these two points, you can set up a system of linear equations that will allow you to solve for \(\sigma_0\) and \(k_y\) by using linear algebra techniques or even simple algebraic manipulation.

With the constants \(\sigma_0\) and \(k_y\) in hand, predicting the yield strength for a new grain size becomes a matter of plugging this new grain size into the Hall-Petch equation. In essence, you are using known data to forecast the unknown — which is a fundamental skill in material science engineering. It's like using known ingredients and their proportions in a recipe to predict how altering one ingredient will change the overall dish.

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Most popular questions from this chapter

(a) Define a slip system. (b) Do all metals have the same slip system? Why or why not?

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.

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 single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an angle of \(28.1^{\circ}\) with the tensile axis. Three possible slip directions make angles of \(62.4^{\circ}, 72.0^{\circ}\), and \(81.1^{\circ}\) with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of \(1.95 \mathrm{MPa}(280 \mathrm{psi})\), determine the critical resolved shear stress for aluminum.

An uncold-worked brass specimen of average grain size \(0.008 \mathrm{~mm}\) has a yield strength of 160 MPa \((23,500\) psi). Estimate the yield strength of this alloy after it has been heated to \(600^{\circ} \mathrm{C}\) for \(1000 \mathrm{~s}\), if it is known that the value of \(k_{y}\) is \(12.0 \mathrm{MPa} \cdot \mathrm{mm}^{1 / 2}\left(1740 \mathrm{psi} \cdot \mathrm{mm}^{1 / 2}\right)\)

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