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Representation theorem equivalent body force Moment tensor

Representation theorem equivalent body force Moment tensor

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Representation theorem equivalent body force Moment tensor

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  1. Representation theoremequivalent body forceMoment tensor QUANTITATIVE SEISMOLOGY CHAPTER 2. and 3.

  2. Reciprocity theorems: Bettis theorem

  3. Bettis theorem • no initial conditions involved for u or v • It remains true if u, utt, T(u,n), f are evaluated at t1 but v, vtt, T(v,n), g are evaluated at a different time t2. • Choose t1=t and t2=τ-t and integrate from 0 to τ acceleration terms reduced to terms that depend only on the initial and final values.

  4. For u and v with a quiescent past

  5. Introducing Green function … ith component displacement at (x,t) due to an nth unit impulse force acting at x=ξ and t=ι

  6. Bettis theorem

  7. Representation theorem Interchange space and time symbols

  8. Representation theorem

  9. Representation theorems for an internal surface; body-force equivalents for discontinuities in Traction and displacement S Σ+ Σ- v

  10. (Spatial reciprocity of G is applied) S Σ+ Σ- Σ++Σ-+S Assume both G and u satisfy homogeneous boundary conditions on S  S itself is no longer our concern

  11. body-force equivalents What are body-force equivalents mean ?

  12. Traction discontinuity contributes to displacement field (Delta function is used to “localized” T on Σ within V)

  13. body-force equivalent of a traction discontinuity on Σ

  14. displacement discontinuity contributes to displacement How to localize [u] on Σ within V ?

  15. body-force equivalent of a displacement discontinuity on Σ The body-force equivalents hold for a general inhomogeneous anisotropic medium, and they are remarkable in their dependence on properties of the elastic medium only at the fault surface itself.

  16. ξ3 Slip jump as a buried faulting [u]=uΣ+ - uΣ- ξ2 ξ1 Σ • Σ lies in the ξ3plane • Isotropic media • Let ξ1 the direction of slip

  17. In isotropic medium, other C13pq vanish, except C1313=C1331=μ Point forces distributed over the plane η3=0- and η3=0-

  18. Total moment about the η2-axis due to f1.

  19. Total moment about the η2-axis due to f1.

  20. Similarly, we may find the moment due to f3 • total moment from f1 and f3 =0 • body-force equivalents from slip discontinuity across the fault plane  double couple

  21. η3 η3 -δ(η3) η1 η1 What does this mean ? See figure 3.3 for the distribution of body-force over the Σ plane

  22. Take a Break

  23. CHAPTER 3. P.46,48

  24. CHAPTER 3. P.48,49,51

  25. CHAPTER 3. P.51

  26. Chapter 2. P.27,28