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Unformatted text preview: 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 = ⇒ no compression, only shear Symmetry cylindrical coordinates, ⇒ r , θ , z with z parallel to x 2 axis; u r = u θ = Solution is σ θ z = µ r ∂ u z ∂ θ = µ S 2 π r We can get by coordinate transformation. σ ij D D i s l o c a t i o n a x i s S/2 S/2 Fault displacement S S u r f a c e f a u l t x 3 x 2 x 1 Figure by MIT OCW. 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...
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This note was uploaded on 10/25/2010 for the course MIT Geodynamic taught by Professor Ywn during the Fall '10 term at MIT.
 Fall '10
 ywn

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