section%209.3%20-%20Property%20Mechanisms

section%209.3%20-%20Property%20Mechanisms - 9.3 Strength of...

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9.3 Strength of Metals Objectives Describe the atomic basis for mechanical properties. Calculate resolved shear stress and yield strength. As we learned in the previous section, the yield stress of a material is the stress at which permanent deformation begins. Below the yield stress, in the elastic region, deformation occurs by stretching of bonds. When the force is released the bonds can return to their original length, and thus the deformation is recoverable. In contrast, above the yield stress bonds are broken, and thus the deformation is not recoverable. In this section we begin by figuring out what actually happens to a crystal when the bonds break. Then we will see how the yield stress can be calculated. Figure 9.3.1: Illustration of one possible bond-breaking mechanism that leads to plastic deformation. GI 9.3.1 Describe the bond-breaking process that needs to occur for the deformation in Figure 9.3.1 to happen. GI 9.3.2 How many bonds total need to break in the process shown in Figure 9.3.1? GI 9.3.3 How many bonds need to break simultaneously in the process shown in Figure 9.3.1?
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Figure 9.3.2: Bond-breaking process when there is a dislocation present. GI 9.3.4 Describe the bond-breaking process that needs to occur for the deformation in Figure 9.3.2 to happen. GI 9.3.5 How many bonds total need to break in the process shown in Figure 9.3.2? GI 9.3.6 How many bonds need to break simultaneously in the process shown in Figure
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This note was uploaded on 03/30/2010 for the course EMA 3010 taught by Professor Unknown during the Spring '08 term at University of Florida.

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section%209.3%20-%20Property%20Mechanisms - 9.3 Strength of...

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