Chapter 6 - Lecture 10

Chapter 6 - Lecture 10 - [hkl] Critical resolved shear...

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Chapter 6 Mechanical Behavior Issues to address… Elastic behavior Plastic behavior Slip system Solution hardening Resolved shear stress
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Elastic Deformation Figure 6-18
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Elastic Deformation In the absence of stress the center-to-center separation distance of two Fe atoms is 0.2480nm. Under a tensile stress of 1,000MPa along this direction. The atomic separation distance increases to 0.2489 nm. Calculate the modulus of elasticity along this direction
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Plastic Deformation Figure 6-19
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Plastic Deformation Figure 6-20
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Figure 6-21 Plastic Deformation
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Caterpillar vs Dislocation Motion Figure 6-22
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Slip System Figure 6-23
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Slip System Figure 6-24
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Table 6.8 Major Slip Systems in the Common Metal Structures
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Figure 6-26 Solution Hardening Ceramics, intermetallic compounds
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Resolved Sheared Stress Figure 6-27 Slip plane (hkl) [hkl] λ cos F φ cos / A σ τ cos cos cos cos cos / cos = = = A F A F s s A P =
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Resolved Sheared Stress Figure 6-27 Slip plane (hkl)
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Unformatted text preview: [hkl] Critical resolved shear stress, c cos cos c c = cos / A cos cos cos cos cos / cos = = = A F A F Example A zinc single crystal is being pulled in tension, with the normal to its basal plan (0001) at 60 degree to the tensile axis and with the slip direction at 40 degree to the tensile axis. (a)What is the resolved shear stress acting in the slip direction when a tensile stress of 0.690 MPa is applied. (b) What tensile stress is necessary to reach the critical resolved shear stress of 0.94MPa [11 2 0] Practice A crystalline grain of iron in a metal plate is situated so that a tensile load is oriented along the [110] crystal direction. If the applied stress is 50 Mpa, what will be the resolved shear stress along the direction within the (101) plane? [11 1 ]...
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This note was uploaded on 10/03/2011 for the course EMCH 371 taught by Professor Dai-hattrick during the Spring '11 term at South Carolina.

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Chapter 6 - Lecture 10 - [hkl] Critical resolved shear...

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