Stress and strain from a screw dislocation19

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

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
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
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern