Advanced Structural Geology
Advanced Structural Geology GEOL 542
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This 2 page Class Notes was uploaded by June Gottlieb on Friday October 23, 2015. The Class Notes belongs to GEOL 542 at University of Idaho taught by Simon Kattenhorn in Fall. Since its upload, it has received 18 views. For similar materials see /class/227866/geol-542-university-of-idaho in Geology at University of Idaho.
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Date Created: 10/23/15
and the chain rule For example from the sham rule Substituting we get 913 Jib ax 7 a case as Mne F l 31 away Adding the two Second derivatives ioge her reduces to the following a compannimy F coordinates HecaH 1mm the de mtion of the Awry stress function mm c 31 9339IBu ala greal 39 away was Um Geomechauics GeoEGeol 442542 Putting the F quot quot39 v anatinn into Practice Example 1 Incompatible Strains Suppose that someone indicates that the following functions define the variations in strain throughout a deforming body 2 7 2 2 2 2 exxxy exy xy eyy xy We can test these functions against the compatibility equation to determine if they are acceptable for an elastically deforming body The compatibility equation gives us 2x2 2y2 8xy Obviously this cannot be true except where x and y are both zero These strains are thus not compatible We can continue to demonstrate this by attempting to find the displacement functions for this strain field We accomplish this by integrating the strains Bu 8 M a auzlsxxdx x From this we get u l3X3y2 f1y C1 Similarly v l eyy dy 13x2y3 f2x C2 In these solutions fI is a function of y only and f2 is a function of x only C1 and C2 are arbitrary constants We know that the shear strain is defined by displacement gradients through the kinematic relation 8 1 auav xy 2 8y 8x If we differentiate our above functions for u and v with respect to y and x respectively we get alszsyM and 3122xy3 3y 3 dy 8x 3 dX The shear strain is half the sum of these two gradients 1 3 3 1df1Y 1df2x 8 y 3CXy Ky 2 dy 2 dx This cannot possibly equal xzy2 as we originally defined sxy These strains are thus incompatible and cannot be integrated to find a legitimate displacement field
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