University:
Massachusetts Institute of Technology
Course:
12.005 | Applications of Continuum Mechanics to Earth, Atmospheric, and Planetary Sciences
Academic year:

2024
Views:

332

Pages:

3
Author:

Gloria R.

Problem Set #4 1 Problem 2 Just NE of Los Angeles, the San Andreas fault trends approximately N65◦W–S65◦E. To within observational error, the displacement gradient there is observed to be (each year): � 0.15 0.24 0.00 −0.15 � where x1 is East and x2 is North, and the units are 10−6 strain. a) Write the (two dimensional) strain tensor, the rotation tensor, and the areal dilation. b) What are the directions of maximal principal compression and extension? c) Is this what you would expect, if the San Andreas is a strikeslip fault? Solution a) The strain and rotation tensor are found directly from the displacement gradient tensor dui = dxj � du1 dx1 du2 dx1 du1 dx2 du2 dx2 1 = 2 � � � = 0.15 0.24 0.00 −0.15 � × 10−6 /year. The strain tensor is 1 εij = 2 � dui duj + dxj dxi � 1 2 du dx1 du2 1 + du dx1 dx2 du1 dx2 + du1 dx2 − 0 du2 dx1 � � 0.15 0.12 ×10−6 /year. 0.12 −0.15 � 0.00 +0.12 ×10−6 /year. −0.12 0.00 = 2 2 du dx2 � The rotation tensor is 1 ωij = 2 � dui duj − dxj dxi The area dilation, � δA , A 1 = 2 � du2 dx1 0 − du1 dx2 du2 dx1 � = � is the trace of the strain tensor δA = εii = ε11 + ε22 = (0.15 − 0.15)10−6 /year = 0/year A and the area is constant. b) To ﬁnd the principal strains and directions of the strain tensor, we follow the same procedure that we used to ﬁnd the principal stresses and directions of the stress tensor (see the solutions for problem set 3 for details). Problem Set #4 2 In brief, the principal strains, per year, are the solutions to the characteristic equation � � � � 0.15 − λ 0.12 � � � × 10−6 det(εij − λδij ) = � � 0.12 −0.15 − λ � � ⇒ λ = ± (0.12)2 + (0.15)2 10−6 ≈ ±0.19 × 10−6 where the positive and negative signs correspond to extension and contrac ˆ associated with each λ are tion, respectively. The principal directions, n, found as the solutions to � 0.15 � 0.19 0.12 0.12 −0.15 � 0.19 � � −6 × 10 n1 n2 � � = 0 0 � where the principal directions must form a right handed coordinate system. The solutions can be expressed as � principal σnn = 0.19 0 0 −0.19 � � −6 × 10 and n ˆ j(n) = +0.94 −0.33 +0.33 +0.94 � , where the ﬁrst matrix is the principal strain tensor, and the columns of the second matrix deﬁne the principal frame, given in terms of the original coordinate system. Note that the ﬁnal solution to this problem is nonunique in terms of the signs of the eigenvectors, which is to say it is nonunique in terms of π2 rotations about the axes. Problem Set #4 3 c) Since the San Andreas is strike slip, the strain tensor will be simple shear in a coordinate system with one of the coordinates aligned along the fault trace, and the other normal to the trace. The principal directions of strain for a simple shear are 45◦ from the original coordinates, since rotat ing simple shear 45◦ results in pure shear (i.e. no oﬀ diagonal elements in the rotated strain tensor). Therefore, one would expect that the principal directions of the strain tensor for the San Andreas fault would be aligned 65◦ − 45◦ = 20◦ from the coordinates given above. For the observed strain tensor, the principal directions are about 19.33◦ from the coordinates, taking care that the sense of slip on the San Andreas is right lateral, we see that the principal directions and strains for the strain tensor above are essentially coincident with what we would expect.

12.005 Problem Set #4 with Solutions

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