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

2024
Views:

434

Pages:

12
Author:

basteishx23y

Problem Set #2 1 Problem 1 Write out symbolically the relationship between the traction vector acting on a surface, (T_{i}) the normal vector describing the orientation of that surface, (n_{j}) and the local stress tensor, (\sigma_{ji}) using: a) vector-matrix notation b) the summation symbol Σ c) Einstein summation notation d) full explicit evaluation writing out each term Solution a) (NOTE: Since symbol notation is not convenient to write in TEX I will use (\vec{T}), (\hat{n}) and σ for a vector, unit length vector and tensor respectively.) [\vec{T} = \sigma^{T}\hat{n}] and since (\sigma^{T} = \sigma), (\vec{T} = \sigma\hat{n}). b) [T_{i} = \sum_{j=1}^{3}\sigma_{ji}<0>hat{n}_{j}, i = 1, 2, 3] where I retained the hat on the unit length vector, this is optional but I find it to be convenient. c) [T_{i} = \sigma_{ji}\hat{n}_{j}] where it is assumed that you sum over the dummy index j, and again it is assumed that (i = 1, 2, 3). d) [T_{1} = \sigma_{11}\hat{n}{1} + \sigma{21}\hat{n}{2} + \sigma{31}\hat{n}{3}] [T{2} = \sigma_{12}\hat{n}{1} + \sigma{22}\hat{n}{2} + \sigma{32}\hat{n}{3}] [T{3} = \sigma_{13}\hat{n}{1} + \sigma{23}\hat{n}{2} + \sigma{33}\hat{n}_{3}] note that the order of the i or j are not so important because the stress tensor is symmetric, but in general it is important to keep the i's and j's straight. Problem Set #2 2 Problem 2 [Modified from Turcotte and Schubert, 2-12 (2-24 in version 2)]. The state of stress at a point on a fault plane is σyy = 150 MPa, σxx = 200 MPa, and σxy = 0 MPa. What is the normal traction and shear traction on the fault if the fault strikes N-S and dips 35 degrees to the west? Do you expect the fault to slip, given your expectation of the value of the coefficient of friction? Solution [Diagram showing a fault plane with normal and shear stress components, strike and dip angles] Following the convention that compression is negative (BEWARE, T&S uses the opposite convention! I will use negative for compression.) the stress tensor is: [\sigma_{ij} = \begin{pmatrix} -200 & 0 \ 0 & -150 \end{pmatrix} MPa] This problem can be solved in a couple of ways. With Mohr's Circles, with rotation matrices and by using the formulas learned in class. The Mohr's circle is the simplest so I will present that first. Mohr's Circle Solution Normal and shear stress are given by Mohr's circle construction by [\sigma_{n} = \sigma_{c} + \sigma_{r}cos(2\theta)] [\tau = \sigma_{r}sin(2\theta)] [\sigma_{c} = \frac{\sigma_{yy} + \sigma_{xx}}{2} = \frac{-150 - 200}{2} MPa = -175 MPa] [\sigma_{r} = \frac{\sigma_{yy} - \sigma_{xx}}{2} = \frac{-150 + 200}{2} MPa = 25 MPa] [Mohr's circle diagram with center at -175 MPa and radius 25 MPa] τ σ

12.005 Problem Set #2 with Solutions

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