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12.520 Lecture Notes 17 Stress and strain from a screw dislocation x3 ce rfa Su ault f Fault displacement S x2 ion D sl Di at oc s i ax S/2 S/2 x1 Figure 17.1 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; ur = uθ = 0 µ ∂uz µ S = r ∂θ 2π r We can get σ ij by coordinate transformation. Solution is σ θ z = How to get traction on surface? Trick – image dislocation x3 2D Image dislocation Surface D Actual dislocation x1 Figure 17.2 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 � r Figure 17.3 Figure by MIT OCW. r 2 = x12 + x3 2 S 2π r From each dislocation ε zθ = Rotating strain tensor ε12 = ε zθ cosθ = ε zθ x3 r At surface ⎡ ⎤ 2 D − x3 SD S ⎢ x3 ⎥= ε = + 2 2 2 2 2 ⎥ 2π ⎢ x1 + x3 2 D − x3 + x12 ⎦ π ( D + x1 ) ⎣ D 12 ( ) where x3 is actual dislocation x + x32 2 1 2D − x3 (2D − x ) + x 2 3 is image dislocation 2 1 Displacement u2 = x1 ∫ε D 12 ∞ dx1 = x ⎞ S⎛ 2 ⎜1− tan −1 1 ⎟ 2⎝ π D⎠ x1 Aside – slip discontinuity objectionable? σ 12 → ∞ along x1 = 0 as x3 → 0 (stress singularity at tip of fault) Alternative model Apply uniform σ 120 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 Distance from fault (km) NE Figure 17.4 Figure by MIT OCW. Virtually indistinguishable! 8 10 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 17.5 Figure by MIT OCW. Apply to loading 1/2 space Figure 17.6 Pt source approximates in seismology.