Stress and strain from a screw dislocation19

# Stress and strain from a screw dislocation19 - 12.005...

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12.005 Lecture Notes 19 Stress and strain from a screw dislocation Figure 19.1 Need to get traction = 0 at surface. First consider medium. Assume: u 1 = u 3 = 0 u 2 = S θ 2 π ∇ ⋅ u = 0 no compression, only shear Symmetry cylindrical coordinates, r , θ , z with z parallel to x 2 axis; u r = u θ = 0 Solution is σ θ z = µ r u z θ = µ S 2 π r We can get by coordinate transformation. σ ij D Dislocation axis S/2 S/2 Fault displacement S Surface fault x 3 x 2 x 1 Figure by MIT OCW.

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How to get traction on surface? Trick – image dislocation Figure 19.2 Solution for matched image dislocation is whole space gives = 0 on surface of ½ space! σ i 3 Shear strain at surface: Figure 19.3 x 3 x 1 Image dislocation 2D D Actual dislocation Surface θ D Surface r 2D x 1 x 3 Figure by MIT OCW. Figure by MIT OCW.
r 2 = x 1 2 + x 3 2 From each dislocation ε z θ = S 2 π r Rotating strain tensor ε 12 = ε z θ cos θ = ε z θ x 3 r At surface ε 12 D = S 2 π x 3 x 1 2 + x 3 2 + 2 D x 3 2 D x 3 ( ) 2 + x 1 2

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