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12.005 Lecture Notes 19 Stress and strain from a screw dislocation x3 ce rfa t u S ul fa Fault displacement S D o ati c o sl s Di axi x2 n S/2 S/2 x1 Figure 19.1 Figure by MIT OCW. Need to get traction = 0 at surface. First consider ∞ medium. Assume: u1 = u3 = 0 u2 = Sθ 2π ∇ ⋅ u = 0 ⇒ no compression, only shear Symmetry ⇒ cylindrical coordinates, r, θ , z with z parallel to x2 axis; Solution is σ θ z = ur = uθ = 0 µ ∂uz µ S = r ∂θ 2π r We can get σ ij by coordinate transformation. How to get traction on surface? Trick – image dislocation x3 2D Image dislocation Surface D Actual dislocation x1 Figure 19.2 Figure by MIT OCW. Solution for matched image dislocation is whole space gives σ i3 = 0 on surface of ½ space! Shear strain at surface: x3 2D Surface D θ Figure 19.3 Figure by MIT OCW. r x1 r 2 = x12 + x32 S From each dislocation ε zθ = 2π r Rotating strain tensor ε12 = ε zθ cosθ = ε zθ x3 r At surface ⎡ ⎤ 2D − x3 SD S ⎢ x3 ⎥= ε = + 2 2 2 2 2 ⎥ 2π ⎢ x1 + x3 2D − x3 + x12 ⎦ π (D + x1 ) ⎣ D 12 ( ) where x3 is actual dislocation x + x32 2 1 2D − x3 (2D − x ) + x 2 3 2 1 is image dislocation Displacement u2 = x1 ∫ε D 12 dx1 = ∞ x ⎞ S⎛ 2 ⎜1− tan −1 1 ⎟ D⎠ 2⎝ π Aside – slip discontinuity objectionable? σ 12 → ∞ along x1 = 0 as x3 → 0 (stress singularity at tip of fault) Alternative model 0 Apply uniform σ 12 Cut 0 ≤ x3 ≤ D, set σ 12 = 0 1/ 2 ⎛ ⎞ 2 x1 ⎟ S ⎜⎛ x1 ⎞ u2 = ⎜⎜1+ 2 ⎟⎟ − 2 ⎜⎝ D ⎠ D⎟ ⎝ ⎠ Displacement (meters) 2 6 1 3.5 0 0 2 4 6 8 10 Distance from fault (km) NE Figure 19.4 Figure by MIT OCW. Virtually indistinguishable! St. Venant’s principle Elastostatics – if boundary tractions on a part S1 of the boundary of S are replaced by statically equivalent traction dist, effects on stress dist are negligible at pts whose distance from S1 is large compare to size of S. Usual context – long beam under end load (non-uniform) See's average end load Figure 19.5 Figure by MIT OCW. Apply to loading 1/2 space Figure 19.6 Pt source approximates in seismology.