Faults and stress – Transcript

00:02an ambition in structural geology is to try and deduce the stress regime under which specific structures developed there's also questions that we might want to ask are what conditions could faults slip under can we deduce the regional stress regime under which thoughts developed in regards to specific questions such as why are normal fault sleep and thrust falls gently dipping we can pose the problem is normal faulting

00:29easier than thrusting and are there particular ways in which the conditions in the earth might promote faulting the tool we're going to use for this is something called the more circle it's a graphical display developed by the german engineer otto moore at the end of the 19th century so let's think a little bit about stress in the earth so let's imagine a column of rock and a

00:56cube of rock at the bottom of it the stress is acting on this unit cube primarily coming from the vertical load of rocks on top and you might expect those rocks at the bottom to squeeze out sideways under that load and indeed they would except for the fact there are adjacent columns there doing the same thing to their cubes of rock at their bases so that this particular cube in here is

01:22being compressed in all directions we can describe the stress state in terms of three orthogonal axes and the rock is being loaded by entirely compressive stresses the maximum compressive stress we give the donation sigma 1 the minimum compressive stress sigma 3 and an intermediate stress axis to create our three-dimensional pattern

01:48sigma 2. but let's keep this simple in a 2d world and so we'll neglect sigma 2 for now and just consider the situation sigma 1 sigma 3. and in this particular situation sigma 1 the maximum compressive stress is vertical and it's trying to push the rocks out against sigma 3 in this direction here so it's the difference between sigma 1 and sigma 3

02:13that will permit deformation so one question we might want to ask is what sort what orientations of faults might develop under this loading condition so let's think of this materials containing micro floors they might be micro fractures grain boundaries and so forth which are yet to slip but will become the seeds for faults and we can think about this principle stresses acting

02:40on one of these surfaces the principal stresses shown by the blue arrows will resolve onto that surface into two components a normal stress acting perpendicular to the fault or the potential fault and a shear stress acting along it so as we can see from the zoom in that what we've done is set up a vector triangle relating sigma 1

03:06to the orientation of the normal stress sigma n and the shear stress on the potential fault sigma s through these equations here so let's plot the relationship of sigma n the normal stress to the shear stress sigma s so here we have a graph which plots sigma s the shear stress against normal stress and let's plot the situation

03:32when the principal stresses lie parallel to the normal stress sigma 1 is a higher value than sigma 3 so plots further along the normal stress axis halfway between sigma 3 and sigma 1 is the mean stress so in this situation we have the normal stress equal to sigma 3 there's no resolved

03:58shear stress in this orientation let's look at a new orientation of a fault slightly around here so now there's an obliquity between the principal stress field and sigma n we get a new relationship now and there's a shear stress exerted on the fault move a bit further around the shear stress increases

04:22and increases some more as we change these angular relationships as we get to this position here the normal stress acting on our four plane is halfway between sigma one and sigma three so it's the equal to the value of the mean stress so sigma n is equal to the mean stress and we can read off what the shift stress would be in that orientation let's keep going around plot more

04:50orientations of the potential slip surfaces and the result shear stress and normal stress as we come down eventually to hear where the normal stress is parallel to sigma 1.

05:03so we plot the relationship between normal stress and shear stress for all the orientations of faults for a given pair of principal stress magnitudes sigma 1 and single 3. these relationships plotted the circle it's called the mohr circle after automotive let's look at the angular relationships and we can see that the relationship between sigma 1 and the normal stress is given this

05:29angle theta plots as 2 theta on the mohr circle which is an angle measured around from sigma 1 to the radius drawn from the mean stress out to our fault orientation at the edge and from that we can read off a shear stress and a normal stress from the graph so that's the mohr circle and it provides a relationship for a given sigma 1 sigma 3 pair for the relationships between shear

05:57stress and normal stress for a variety of fault orientations or potential fault orientations but do they slip in order to answer that we need a failure criteria criteria which says under which faults will slip what's plotted on the graph now is the so-called coulomb failure criteria which is relates to frictional sliding on the surface the orientations of fractures below that

06:24surface are stable beyond it are unstable and they're critical for slipping if thought orientations lie on the failure line or failure criterion so that's the point here on our particular mohr circle for the values of sigma one and sigma three we've used to construct it we can read off the relationship between

06:49that point on the failure envelope and find off the orientation there there's two theta look at its relationship to our block diagram on the side and you can see the relationship between sigma 1 sigma 3 the normal stress and that fault is primed to move it which it's slipping so what can we do with this information or we can do it to establish which

07:16orientations of faults are likely to move given a particular stress rating let's look at this situation in here we've got a fault it will slip in that sense so it is a normal fault in order for the radius line from mean stress out to the margin of the circle to intersect the failure criterion 2 theta there has to be greater than 90 degrees it has to be an open angle to

07:42get us to that side of the graph therefore the single angle between sigma 1 and sigma n the normal stress must be greater than 45 degrees in turn if we look at the vector triangle in here the angle made therefore between sigma 1 and the false surface must be less than 45 degrees to work

08:07that triangle through so let's zoom that in to that situation the angle between sigma 1 and the normal fork plate must be less than 45 degrees in order for that to slip if we now put a conjugate fault in so put the mirror image four plane surface in here we can see how this works sigma one must bisect

08:33the acute angle between the two fault surfaces sigma three bisects the open angle between these two conjugate fault planes so here we have a set of faults seen in outcrop from new zealand we can draw in the faults see the offsets and we can therefore deduce the orientation of sigma one bisecting

08:57the acute angle between these conjugate force we can flip our diagram over in this case the bisector of the acute angle is horizontal we're generating thrust faults in this situation let's look at an outcrop again from new zealand we can draw on the faults show the offsets in there

09:23the bisector of the acute angle sigma 1 is horizontal so horizontal sigma 1 is the stress state that characterizes thrust voltage so let's step this into three dimensions and this was worked out by anderson in the early part of the 20th century showing how conjugate faults relate to the orientation in three dimensions to

09:50the three principle stresses sigma one and sigma three we've looked at and sigma two the intermediate stress axis as we can see in this diagram for normal false sigma one is vertical sigma three is horizontal and that situation flips for thrust faulting so sigma 3 is vertical sigma 1 is horizontal for thrusts and for strike slip both sigma 1 and sigma 3 are horizontal

10:15sigma 2 is vertical in this case an important rule is that sigma 2 runs parallel to the intersection between our two conjugate faults this is andersonian faulting so let's look at this on a map from part of the isle of mull a series of faults have been mapped out let's just pick them out in here and they're making quasi

10:40conjugate series in here we can identify the bisector of the acute angle through there that's sigma one the faults presumably very steep they're cutting from coast to coast across a hillside without changing trend very much and the sigma 3 will be bisecting the open angle and sigma 2 will be vertical so we can use the pattern of faults on the map to deduce the

11:05stress situation under which those faults formed anderson's theory applies to false at initiation so it's the important rule sigma 1 by 6 the acute angle normal faults must therefore dip at angles of greater than 45 degrees commonly 60 or 70 degrees at initiation conversely thrust must dip at angles

11:30significantly less than 45 degrees in order for sigma 1 to bisect the acute angle and in general thrusts initiate angles of 20 or 30 degrees or thereabouts so let's go back to the more circle here we've got the situation that we developed earlier of sigma 1 vertical sigma 3 horizontal and the mohr circle just are butting the failure criteria

11:57so normal faults are about to move they're able to do so because the circle is large enough to intersect the failure envelope so let's look at a situation now where sigma 1 and sigma 3 are almost the same as the difference the differential stress is very small you can see that we're a long way from the failure envelope so the faults aren't going to be slipping here let's release sigma 3 a little so sigma

12:273 is falling the mean stress is reducing but the circle is enlarging because the difference between sigma 1 and sigma 3 is increasing until it intersects the failure criterion and a fault will slip so the higher the differential stress the greater the chance of fault slip reducing horizontal stress increases the