Practical structural geology

Quick refresher

Each source has its own scale. 3D seismic (a survey that covers a whole block of ground) shows the shape of horizons (the surfaces between rock layers) and faults that are larger than its resolution (the smallest detail it can tell apart). Wells give exact formation tops (the depths where each formation begins), but only at points. The dipmeter (a tool that measures the dip of the beds) and core show details a few centimetres across, right beside the borehole. A map built from a sparse exploration network cannot show folds that are shorter than the spacing of the data. Yet these short folds give the curvature from which we predict fractures.
Faults, salt and seals (rock that stops fluids from moving) divide the section into pressure compartments. Inside one compartment, an isolated dipping bed evens out its pressure along the water gradient (the way water pressure rises with depth). The shales around it do not. So at the crest of the bed the pressure is higher than in the shales at the same depth, and at its base it is lower.
The widest fractures are those perpendicular to the least horizontal stress. So the permeability of a fractured reservoir is highest along $\sigma_H$ (the largest horizontal stress). Depletion (the fall of reservoir pressure as fluid is produced) closes fractures. Injection pressure, as it approaches $\sigma_h$ (the smallest horizontal stress), opens fractures along the same direction.
Balancing rests on conservation of volume. In a section drawn along the direction of transport (the direction in which the rocks moved), the area of each unit stays the same. In flexural-slip folding (layers bend by sliding over one another, like the pages of a book) the lengths of the beds stay the same too. The restored section (the section put back into its state before deformation) must look geologically plausible.
The excess area of a horizon above the regional level equals the shortening multiplied by the height of the horizon above the detachment (the surface along which the beds slid). We build an area–depth line from several horizons. Its slope is the shortening, and the depth where the area is zero is the depth of the detachment.
A balanced section is not necessarily a correct one. Balancing rejects impossible versions, but it does not pick out a single one.

Key relationships

s = \left(\dfrac{\sin(\pi a/\lambda)}{\pi a/\lambda}\right)^{2}

the fraction of the curvature of a fold of wavelength $\lambda$ that the second difference of depths shows (the second difference is a way to measure bending from three points), when the points are a distance $a$ apart

\Delta p_{top} = (G_{sh} - G_w)\,(z_c - z_{top})

how much the pressure at the crest of an isolated dipping bed exceeds the pressure of the shales; $z_c$ is the depth of the centroid (the depth where the bed and the shales have equal pressure)

\sigma_n = \sigma_H\sin^2\theta + \sigma_h\cos^2\theta

the normal stress (the push at right angles to the fracture wall) on a vertical fracture whose strike (its compass direction) makes an angle $\theta$ with the direction of $\sigma_H$

S = \sum_i l_i\,(1 - \cos\delta_i),\qquad A = S\,h_d

the shortening found from the length of the bed, and the excess area above the regional level; $\delta_i$ is the dip of the $i$ -th segment of the bed, $h_d$ the height of the horizon above the detachment

Common pitfalls

Predicting fractures from the curvature of a map built from 2D lines (single seismic lines, not a 3D block) or from sparse wells.
Taking the pressure in a sand bed to be equal to the pressure in the surrounding shales at the same depth.
Expecting the same permeability in all directions in a fractured reservoir, and forgetting that depletion and injection change it.
Balancing a section drawn in time instead of depth, with a distorted vertical scale, or cut at a slant to the direction of transport.
Treating a balanced section as proven.

After this lesson you can

compare how structures look on seismic data, on maps, in wells and in core, and explain which task a map is good for, given the network of observations it was built from
estimate what fraction of a fold's curvature a map shows when its points are a given distance apart, and what this means for predicting fractures
read from the dip plot of a well (a graph of bed dips against depth) the structural dip, unconformities, fold hinges and the bending of layers near faults
calculate how much the pressure at the crest of an isolated dipping bed exceeds the pressure in the shales around it (the centroid effect), and how far the seal is from hydraulic fracturing (being split open by pressure)
estimate the opening (aperture) and the permeability of fracture sets with different orientations from the stresses, and how they change when the reservoir is depleted and when fluid is injected
check a section by balancing it: find the shortening from the length of a bed, the excess area, the depth of the detachment and the area–depth line

Structure in the data: what each source sees

Nobody sees the structure of a field as a whole. We assemble it from data of different kinds and scales, and every source has its own resolution and its own blind spots. Seismic data show the shape of horizons and the faults between wells. But the depths depend on the velocities of sound in the rock, and vertically seismic data can tell apart only units tens of metres thick (resolution). Wells give exact formation depths, but only at points. The dipmeter and the image logs (pictures of the borehole wall) show details a few centimetres across, but only beside the borehole. The task is not to find the “best” source. The task is to bring them together into one consistent picture.

What each source of structural data shows

Source: 3D seismic
Scale: bin (grid cell) 12–50 m; vertically tens of metres
Seen well: shape of horizons, faults whose throw (vertical offset) is greater than the resolution, salt, unconformities, curvature
Seen poorly: small faults, steep limbs, structure beneath salt

Source: 2D seismic
Scale: lines hundreds of metres to kilometres apart
Seen well: large folds and faults
Seen poorly: structures smaller than the line spacing, how faults connect from one line to the next

Source: Formation tops in wells
Scale: a point; vertically decimetres
Seen well: absolute depths, thicknesses, missing or repeated parts of the rock sequence
Seen poorly: position of a fault between wells

Source: Dipmeter and image logs
Scale: centimetres beside the borehole
Seen well: dips of layers, fold hinges, bending beside faults, fractures, breakouts (places where the borehole wall caves in)
Seen poorly: anything farther than a metre from the borehole

Source: Core
Scale: millimetres to centimetres
Seen well: slickensides (polished, scratched fault surfaces), breccias (rock broken into angular pieces), fracture fills (minerals that fill cracks)
Seen poorly: orientation, if the core is not oriented

Source: Drilling and tests
Scale: an interval of the borehole
Seen well: drilling-fluid losses and pressure jumps at faults, inflow from fractures
Seen poorly: geometry

Source: Analogue outcrops (similar rocks exposed at the surface)
Scale: metres to kilometres
Seen well: shape of folds and faults, how one passes into another, fracture pattern
Seen poorly: reservoir properties at depth

Source	Scale	Seen well	Seen poorly
3D seismic	bin (grid cell) 12–50 m; vertically tens of metres	shape of horizons, faults whose throw (vertical offset) is greater than the resolution, salt, unconformities, curvature	small faults, steep limbs, structure beneath salt
2D seismic	lines hundreds of metres to kilometres apart	large folds and faults	structures smaller than the line spacing, how faults connect from one line to the next
Formation tops in wells	a point; vertically decimetres	absolute depths, thicknesses, missing or repeated parts of the rock sequence	position of a fault between wells
Dipmeter and image logs	centimetres beside the borehole	dips of layers, fold hinges, bending beside faults, fractures, breakouts (places where the borehole wall caves in)	anything farther than a metre from the borehole
Core	millimetres to centimetres	slickensides (polished, scratched fault surfaces), breccias (rock broken into angular pieces), fracture fills (minerals that fill cracks)	orientation, if the core is not oriented
Drilling and tests	an interval of the borehole	drilling-fluid losses and pressure jumps at faults, inflow from fractures	geometry
Analogue outcrops (similar rocks exposed at the surface)	metres to kilometres	shape of folds and faults, how one passes into another, fracture pattern	reservoir properties at depth

Formation depths in wells are exact only if the directional survey (the measured path of the well) and the wellhead elevation are known (map data). We look for faults in wells as missing or repeated parts of the section (fault in a well), and on seismic data as breaks and offsets of reflections (the lines on a seismic image that mark rock boundaries) (tracking faults).

The questions we ask about the structure change with the stage of the project. In exploration, the type and size of the trap matter, and so does the time when it formed and when oil and gas filled it (time of the structure and time of charge). In appraisal (the work after a discovery to find out how big and how good it is), three things matter: the closure (the height of the trap from its crest down to its spill point), the faults that bound the accumulation, and whether they seal (sealing). In development, the details matter: blocks, small faults, fracturing and changes of thickness. A map that is good enough at one stage is often unsuitable at the next. Every map carries the mark of its purpose. These marks are the network of data it was built from, the velocity model (how fast sound travels in the rocks, used to turn time into depth) and the method of interpolation (how values are filled in between the data points) (gridding). So when you get someone else's map, first learn its “passport”: what data it was built from, for what task and by what method, and when it was last checked against new wells.

Outcrops (places where rock is exposed at the surface) show what neither seismic data nor a well can show: how a fault turns into a fold, how many fractures there are per metre, and how they stop at the boundaries between layers. But near the surface the rock has lost its load (it is no longer buried under other rocks) and is weathered, and a structure as big as a field does not fit in one outcrop. From an analogue we take the geometry and the way structures relate to one another. We check the numbers against data from the reservoir itself (fractures in outcrops).

Procedure for the structural interpretation of a field

1
Data
Check the wellhead coordinates, the elevations, the directional surveys, the formation tops and how the seismic data are tied to the wells (so that each seismic reflection is matched with a known formation top).
2
Framework
Follow the horizons and faults through the seismic data. Build time structure maps (maps of the travel time of sound to each horizon) and convert them to depth (depth conversion).
3
Wells
Correlate the well sections (match the same layers from well to well), find the missing and repeated intervals and read the dip plot.
4
Style
Decide which kind of deformation acted on the area (stretching, squeezing or sideways sliding) and check that all the structures agree with it (structural style).
5
Check
Build sections along the direction of transport and balance them. Make the maps, sections and wells agree with one another (reconciliation).
6
Consequences
Assess how well the faults seal, the pressure compartments, the fracturing and the stresses. Pass the results on to the volumetrics (the calculation of volumes), the reservoir model and the well plans.

A map is built for a task: curvature and data spacing

We predict fractures in the crests of folds from curvature: the more sharply a layer is bent, the more its outer part is stretched (extension in bending). We find the curvature from the second difference of depths at three points at equal distances $a$ (formula of the lesson on folds). But the second difference sees only folds that are clearly longer than the spacing. The easiest way to show this is to write the profile of the top as a sum of waves, that is, cosines of different lengths. Any profile can be split up in this way.

What fraction of the curvature of a fold the second difference shows

\frac{z_1 - 2z_0 + z_2}{a^2} = \sum_k A_k\left(\frac{2\pi}{\lambda_k}\right)^{2} s_k,\qquad s_k = \left(\frac{\sin(\pi a/\lambda_k)}{\pi a/\lambda_k}\right)^{2}

where:

$z_1,\ z_0,\ z_2$ depths of the top at three points a distance $a$ apart; $z_0$ is at the crest, m
$a$ spacing of the points from which the curvature is calculated, m
$A_k$ amplitude of the $k$ -th wave — half the height difference between its crest and its trough, m
$\lambda_k$ wavelength of the $k$ -th wave — the distance between neighbouring crests, m
$A_k(2\pi/\lambda_k)^2$ true curvature of the $k$ -th wave at its crest, m⁻¹
$s_k$ fraction of the curvature of the $k$ -th wave that the second difference shows, —

Units: lengths in metres, curvature in m⁻¹

Holds when

We write the profile of the top as a sum of cosines. The formula is for a point where all the waves have a crest. At other points each wave enters the sum with its own factor, but the fraction $s_k$ is the same.
The slopes are small, as in the formula for the crest in the lesson on folds.
The depths have no errors. A random error in depth is divided by $a^2$ , so at a very small spacing the noise distorts the curvature more than the loss of short waves does.

When the spacing is a quarter of the wavelength, the second difference shows 81% of the curvature. At half the wavelength it shows 41%. When the spacing equals the wavelength, it shows zero: all three points fall on the same point of the wave cycle (the same phase), so the fold disappears. The curvature of a wave grows as $1/\lambda^2$ , so the short waves give most of the curvature. These are exactly the waves that a sparse network loses. To predict fractures, we calculate the curvature from detailed 3D seismic data. The spacing must be smaller than a quarter of the length of the shortest folds we need to see, but not so small that depth noise distorts the result.

Worked exampleDoes a map from 2D lines see the folds that fracture the layer?

Given (assumed data)

Top of a dolomite layer (shape assumed): a gentle anticline (an arch-shaped fold) — amplitude 80 m, wavelength 10 km; on it second-order folds (small folds on top of the big one) — amplitude 8 m, wavelength 800 m; the crest at a depth of 2400 m
Mechanical layer: dolomite 6 m thick; it bends on its own as one unit and slips on thin shale partings
Detailed map (3D seismic): curvature is calculated from points 100 m apart
Map from 2D lines: lines 1 km apart
Strain (relative stretching) at which the rock breaks in extension: about 0.02% — the order of magnitude given in the lesson on folds

Solution

True curvature at the crest is the sum of the curvatures of the two waves: $80\cdot(2\pi/10{,}000)^2 + 8\cdot(2\pi/800)^2 = 3.1583\cdot10^{-5} + 4.9348\cdot10^{-4} = 5.2506\cdot10^{-4}$ m⁻¹, radius $R = 1905$ m. The second-order folds give 94% of the curvature.
Detailed map: 100 m on either side of the crest the top lies at a depth of 2402.501 m; $z'' = 2\cdot(2402.501 - 2400)/100^2 = 5.002\cdot10^{-4}$ m⁻¹, $R = 1999$ m. Extension at the top of the layer: $e = h/(2R) = 3/1999 = 0.150$ %.
Map from 2D lines: 1 km from the crest the top lies at a depth of 2423.279 m; $z'' = 2\cdot(2423.279 - 2400)/1000^2 = 4.656\cdot10^{-5}$ m⁻¹, $R \approx 21{,}480$ m; $e = 3/21{,}480 = 0.0140$ %.
Comparison with the limit: on the detailed map the extension is 7.5 times greater than 0.02%, on the 2D map it is 0.70 of the limit.

Answer: The detailed map shows 0.15% extension in the crests of the second-order folds. That is 7.5 times above the limit, so we expect fractures there. The map from 2D lines gives 0.014%. That is below the limit, so from this map we would not expect fractures.

What it means. The map from 2D lines shows the gentle anticline correctly, and for finding a trap that is enough. For predicting fractures it is unsuitable. The second-order folds, which are what fracture the layer, lie between the lines. The second difference from points 1 km apart keeps only $s = (\sin 1.25\pi/1.25\pi)^2 = 0.032$ of their curvature, and at a spacing of 100 m it keeps $s = 0.95$ . So even the detailed map slightly underestimates the curvature: it gives 0.15% extension, while the true value is 0.16%. The example is simplified. Real second-order folds do not repeat strictly periodically, and depths from seismic data contain noise, which the second difference makes larger when the spacing is small. The data in this example are assumed for illustration.

CalculatedProfile of the top in the example. The solid line is the true surface; the detailed map with points 100 m apart repeats it almost exactly. The dashed line is a smooth curve through the points of the 2D lines, which are 1 km apart. The second-order folds, 800 m long, vanish between the lines. The limbs of the gentle anticline seem steeper, because 2 km from the crest the point has fallen in the trough of a small fold.

The same question — do the data suit the task? — faces every map. A map built from sparse wells knows nothing about faults whose throw is smaller than the error of the picks (the depths chosen for the formation tops). An isochore map (a map of the thickness of a layer) made from old 2D lines will not show small changes of thickness. Faults smaller than the seismic resolution cannot be seen at all, and we estimate their number statistically (faults below resolution).

Bed dips in a well: what to read from the dip plot

The dipmeter and borehole image logs give the dip of the layers (the angle at which a bed slopes) every few centimetres along the borehole (image logs and sinusoids). On a dip plot, each measurement is a circle placed at the dip angle for its depth. The tail of the circle points along the dip azimuth (the compass direction in which the bed slopes down), so the symbol looks like a “tadpole”. We read such a plot not point by point but interval by interval, looking for stretches that behave in a similar way.

Constant dip (green pattern): The angle and the azimuth hardly change. This is a dip domain: the structural dip of the unit. In an undisturbed sequence, the boundary between two domains is a fold hinge (the line of sharpest bending) that the well has crossed. If the azimuths of the two domains are almost opposite, the well has crossed the axial surface (the surface that runs through all the hinges of the fold) (fold axis from the dipmeter).
Dip increasing with depth (red pattern): The azimuth stays constant, while the angle increases downwards. The beds bend towards something that lies below: a fault, the edge of an incised channel (a channel cut down into older beds), a salt body, or a buried high or reef that the layers drape over.
Dip decreasing with depth (blue pattern): The bending dies out downwards. Examples are layers below a fault, and cross-laminae (thin inclined layers inside a bed) that flatten out towards the base of a set (sedimentary structures).
A jump without a transition zone: The angle and the azimuth change at once. This means an unconformity (a surface where part of the rock record is missing) or a fault. An unconformity is confirmed by missing section that grows towards the crest of the uplift. A fault is confirmed by the fault plane itself on the image log (a steep sinusoid that cuts the layers), by a zone of crushing and fractures, and by drilling-fluid losses.

The bending of layers beside a fault shows the strike of the fault (its compass direction along the surface): the azimuth of the bent layers is roughly perpendicular to the fault. One bend cannot tell us which way the fault dips. At a normal fault (the block above the fault plane has moved down), the layers are dragged by friction against the movement, and they dip in the same direction as the fault plane. At a reverse fault they dip towards it. The hanging wall (the block above the fault plane) of a listric fault (a fault that flattens out with depth) sags towards the fault (rollover), so its layers dip towards the fault plane (listric faults). The direction comes from the fault plane itself on the image log, from correlation with neighbouring wells and from seismic data. Dips in the well are also needed for correlation. We use them to convert thicknesses measured along the borehole into true vertical thickness and true stratigraphic thickness (TVT and TST). Without this, faults in dipping beds and in deviated wells are missed, or found where they do not exist.

SchematicDip plot of a vertical well (schematic). Above the unconformity there is an almost horizontal cover. Below it is the limb of a fold, dipping at 24° to the north-east. Above the fault the dip increases to 40°, because the hanging wall sags towards the fault. The fault plane dips to the south-west at 63°, towards the layers. Below the fault the dip is again 24°. Correlation with a neighbouring well shows that about 60 m of section is missing at the fault.

Structure and abnormal pressure

Structure controls pore pressure in two ways. First, faults, salt and impermeable seals divide the section into pressure compartments. These are blocks in which the pressure evens out inside the block but not with the neighbouring blocks (isolated compartments). Second, the relief of permeable bodies inside a compartment (how much higher one end of the body is than the other) redistributes the pressure. A bed evens out the pressure over its own volume, while the surrounding shales do not.

Blocks on either side of a sealing fault often lie on different pressure lines (pressure plotted against depth). In neighbouring wells the same bed has a different pressure, and on a pressure–depth plot the water lines are parallel but shifted apart (several hydrodynamic systems). A pressure jump where a well crosses a fault is a reliable sign of a seal. The reverse is not true. Equal initial pressure on both sides does not prove that the fault will not hold back flow during development (faults in development). So we predict pressure block by block, not with one curve for the whole area. In the subsalt carbonates of the Pre-Caspian Basin (carbonate rocks below the salt) the regional seal is salt, and the pressure in them is often abnormally high (example from Kazakhstan).

Inside a compartment, relief matters. Take a sand body — a lens, a dipping bed or the limb of an anticline — with shales on all sides. In these shales, overpressure (pressure above the normal water pressure) arose through undercompaction: the shales were buried so fast that their water had no time to escape, so the water carries part of the weight of the rocks above (undercompaction). Below the top of the overpressured zone, the pressure in the shales rises almost in parallel with the overburden pressure (the weight of all the rock above). So the effective stress in the shales (the part of the load carried by the grains) stays about what it was at the top of the zone. The sandstone is permeable, and over geological time the pressure in it evens out along the water gradient, which is much gentler. The pressure lines of the sandstone and the shales cross. Above the crossing point, the pressure in the sandstone is higher than in the shales at the same depth. Below it, the pressure is lower. At the top of the body, water leaves it for the shales. At the base, water enters it from the shales. In the steady state the outflow and the inflow are equal. The depth where the pressures of the sandstone and the shales are equal is called the centroid. The redistribution of pressure by a dipping body is called lateral transfer.

Pressure in an isolated dipping bed and the excess at its top

p_r(z) = p_{sh}(z_c) + G_w\,(z - z_c),\qquad \Delta p_{top} = p_r(z_{top}) - p_{sh}(z_{top}) = (G_{sh} - G_w)\,(z_c - z_{top})

where:

$p_r(z)$ water pressure in the sand body at depth $z$ , MPa
$p_{sh}(z)$ pore pressure in the surrounding shales, MPa
$z_c$ depth of the centroid, where the pressures of the body and the shales are equal; for a body whose area is spread evenly over its depth range, the middle of its height, m
$z_{top}$ depth of the top (crest) of the body, m
$G_w$ pressure gradient of the water in the bed, MPa/m
$G_{sh}$ pore pressure gradient of the shales over the depth interval of the body; with undercompaction at a constant effective stress, it equals the overburden pressure gradient, MPa/m
$\Delta p_{top}$ pressure excess at the crest of the body over the pressure of the shales at the same depth, MPa

Units: pressures in MPa, gradients in MPa/m, depths in metres

Holds when

The body is connected and filled with water. Hydrocarbons at the top add their buoyancy (the extra pressure that comes from being lighter than water) to the pressure.
The flows through the shales are steady, and the shales let water through equally well over the whole surface of the body. Then the centroid is the mean depth of the surface of the body, weighted by area (the derivation is in the advanced section).
The pressure gradient of the shales is constant over the depth interval of the body. If the shales deeper down are more compacted and let less water through, the centroid moves up, and the excess at the crest is less than the calculated value.

The excess grows with the height of the body and with the difference between the gradients. The greater the relief of an isolated bed (the steeper its dip, or the higher the fold), the more the pressure at its crest differs from the shale pressure at the same depth. At its base it differs by the same amount in the opposite direction. Some methods estimate pressure from the properties of the shales (d-exponent, a number worked out from the drilling rate that shows how compacted the rock is). They give the pressure of the shales, not of the sandstone, so they underestimate the pressure at the crest.

Worked examplePressure at the crest of a dipping sand body and the limit of a gas accumulation

Given (assumed data)

Gradients: water in the bed 0.0104 MPa/m (density 1060 kg/m³), overburden pressure 0.0226 MPa/m
Shales: overpressure from undercompaction begins at 1500 m; below it the effective stress in the shales is constant
Sand body: crest at 2800 m, deepest point at 4000 m; area spread evenly with depth; shales on all sides
Fracture pressure of the shales of the seal at the crest: 56.7 MPa (from a leak-off test in a neighbouring well, a test that finds the pressure at which the rock starts to take fluid)
Densities in the bed: gas 230 kg/m³, oil 750 kg/m³

Solution

Effective stress in the shales below the top of the overpressure: $(0.0226 - 0.0104)\cdot1500 = 18.3$ MPa, so the shale pressure is $p_{sh}(z) = 0.0226\,z - 18.3$ .
The centroid is the middle of the height of the body: $z_c = (2800 + 4000)/2 = 3400$ m; the shale pressure there is $0.0226\cdot3400 - 18.3 = 58.54$ MPa.
Crest: in the shales $0.0226\cdot2800 - 18.3 = 44.98$ MPa (equivalent mud weight 1.64 g/cm³, the density of a drilling fluid whose column gives this pressure), in the sandstone $58.54 - 0.0104\cdot600 = 52.30$ MPa (1.90 g/cm³). That is 7.32 MPa more, exactly $(0.0226 - 0.0104)\cdot600$ .
Base of the body, 4000 m: in the shales 72.10 MPa, in the sandstone $58.54 + 6.24 = 64.78$ MPa — 7.32 MPa less.
Margin to hydraulic fracturing of the seal at the crest: $56.7 - 52.30 = 4.40$ MPa. A gas accumulation brings the pressure at its top up to the fracture pressure when it reaches a height $H = 4.40/((1060 - 230)\cdot9.81\cdot10^{-6}) = 4.40/0.008142 = 540$ m (hydraulic-fracturing limit of the seal).
If we took the sandstone pressure to be equal to the shale pressure, the margin would be $56.7 - 44.98 = 11.72$ MPa and the limit would be 1439 m. That is more than the whole height of the body, 1200 m.

Answer: The pressure at the crest of the sand body is about 52.3 MPa, which is 7.3 MPa higher than in the shales at the same depth. The margin to hydraulic fracturing of the seal is only 4.4 MPa, so a gas accumulation cannot be taller than about 540 m.

What it means. For drilling, this is the difference between equivalent mud weights of 1.64 and 1.90 g/cm³. A well at the crest, with its drilling fluid chosen from the shale pressure, will meet a bed at a pressure much higher than expected. The gap between the pressure in the sandstone and the fracture pressure of the seal (2.06 g/cm³) is narrow, so there is little room to choose the drilling-fluid density. A well near the base is the opposite case. It will open a sandstone at a pressure lower than in the shales above it, with a risk of losses (drilling fluid draining into the rock) and stuck pipe (the drill string jams in the hole). For exploration the conclusion is that hydraulic fracturing of the seal, not the spill point, limits a gas accumulation in such a body. The same seal will hold an oil accumulation much higher, because the density difference between water and oil is smaller. The limit is about 1447 m, which is more than the height of the body. The estimate is simplified. We took the area of the body to be spread evenly with depth, and the shales to let water through equally well everywhere. For a dome whose area grows with depth, the centroid lies deeper than the middle, and the excess at the crest is greater. The data in this example are assumed for illustration.

CalculatedPressure against depth for the example: hydrostatic pressure (the pressure of a water column; dashed), overburden pressure (thin line), pressure in the shales with overpressure from 1500 m (solid), and the water pressure in the sand body from the crest (2800 m) to the base (4000 m). The lines of the shales and the sandstone cross at the centroid, at 3400 m. The diamond is the fracture pressure of the seal at the crest.

Pressure also creates structures. Shales with overpressure are weak layers along which thrusts (faults where one block is pushed over another) break away, and the belts of folds above them are wide and gentle (critical wedge). In deltas such shales feed growth faults (faults that move while the sediment is still being deposited) and shale diapirs (columns of shale that push upwards). A rise of pore pressure lowers the effective stresses, and it can reactivate a fault or open fractures in a seal (slip on a fracture and pore pressure, natural hydraulic fracturing). Then the pressure is “bled off” along the fault into the beds above, and “charged” sands (sands filled with high-pressure fluid) appear in them without signs of overpressure in the shales.

Stresses, fractures and the direction of flow

A reservoir whose permeability changes noticeably with effective stress is called stress-sensitive. This shows most strongly in fractured rocks and in weak, high-porosity rocks such as loose sands and chalk (a soft, fine-grained limestone). For the weak rocks, the main concerns are compaction as the pressure falls and subsidence of the surface (compaction of the reservoir). For fractured rocks, the main concern is direction.

Fractures are opened or squeezed shut not by the stress regime in which they formed millions of years ago, but by today's regime. So we begin with the direction of the present-day stresses. These sources give it:

borehole breakouts (places where the borehole wall caves in) and drilling-induced fractures (cracks made by the drilling itself), seen on image logs and with a multi-arm caliper (a tool that measures the width of the hole in several directions): breakouts are elongated along $\sigma_h$ , induced fractures along $\sigma_H$ (breakouts and induced fractures)
the azimuth of hydraulic fractures, found by microseismic monitoring (listening for the tiny tremors that fracturing causes), and mini-frac injections (small test injections) that give the magnitude of $\sigma_h$ (closure pressure)
permeability that differs with direction, found from interference tests between wells (a pressure change in one well is watched in another) and from early breakthrough of injected water, which is itself a trace of open fractures with one orientation
shear-wave splitting in dipole sonic logs (a shear wave splits into a fast and a slow wave in rock that has a preferred direction) and azimuthal anisotropy of seismic data (seismic velocity that changes with the compass direction)
regional stress data, with the warning that near faults, near salt and in the crests of folds the direction may rotate

The fractures that are most open are those squeezed less than the rest: the ones perpendicular to $\sigma_h$ (fractures under load). The normal stress on a vertical fracture depends on the angle between its strike and the direction of $\sigma_H$ :

Normal stress on a vertical fracture

\sigma_n = \sigma_H\sin^2\theta + \sigma_h\cos^2\theta,\qquad \sigma_n' = \sigma_n - p_p

where:

$\sigma_n$ total normal stress on the plane of the fracture, MPa
$\sigma_H,\ \sigma_h$ greatest and least horizontal stresses, MPa
$\theta$ angle between the strike of the fracture and the direction of $\sigma_H$ , °
$\sigma_n'$ effective normal stress that squeezes the fracture shut, MPa
$p_p$ pore (reservoir) pressure, MPa

Units: stresses and pressure in MPa, angle in degrees

Holds when

The fracture is vertical and one of the principal stresses (the three stresses that act at right angles to one another) is vertical. For an inclined fracture, the contribution of $\sigma_v$ is added (advanced section).
We measure the angle $\theta$ from the strike of the fracture. In stresses on a plane the angle is measured from the normal to the fracture (the line at right angles to it), and from that line $\sigma_H$ lies at $90° - \theta$ .
The effective stress on the fracture is the total stress minus the pore pressure, as in the lesson on fractures.
The stresses are present-day, at a given point; near faults and salt they may differ from the regional ones.

At $\theta = 0$ the fracture strikes along $\sigma_H$ , and only $\sigma_h$ squeezes it. This is the most open direction. At $\theta = 90°$ , $\sigma_H$ squeezes the fracture. A fracture at 30° to $\sigma_H$ is squeezed harder than one parallel to $\sigma_H$ , by a quarter of the stress difference. The closure law from the lesson on fractures gives the aperture under load, and the permeability of the system grows as the cube of the aperture.

Worked examplePermeability of two fracture sets now, after depletion and near an injector

Given (assumed data)

Stresses and pressure at a depth of 2900 m: $\sigma_H = 62$ MPa, $\sigma_h = 48$ MPa, initial reservoir pressure 30 MPa
Fractures: two vertical sets: A — striking along $\sigma_H$ , B — along $\sigma_h$ ; the spacing is the same; the hydraulic aperture (the opening that counts for flow) without load is 250 µm in both; the matrix is almost impermeable
Closure: hyperbolic law (how the opening shrinks as the stress grows) with the parameters from the lesson on fractures: $V_m = 180$ µm, $K_{ni} = 0.08$ MPa/µm
Depletion: pressure reduced by 15 MPa over the whole accumulation; the horizontal stresses fall by $A = 2/3$ of this amount
Injector (injection well): bottomhole pressure 44 MPa; the stresses near it are taken as unchanged

Solution

Effective normal stresses: on set A $48 - 30 = 18$ MPa, on set B $62 - 30 = 32$ MPa.
Closure and aperture: A — $\dfrac{18\cdot180}{0.08\cdot180 + 18} = 100.00$ µm, aperture $250 - 100.00 = 150.00$ µm; B — $\dfrac{32\cdot180}{14.4 + 32} = 124.14$ µm, aperture 125.86 µm.
With equal spacing, the permeabilities of the sets are in the ratio of the cubes of the apertures. Along $\sigma_H$ (set A) the reservoir conducts $(150.00/125.86)^3 = 1.69$ times better than along $\sigma_h$ .
Depletion: the horizontal stresses fall by $\tfrac23\cdot15 = 10$ MPa and the pressure falls by 15 MPa, so the effective stresses rise by 5 MPa, to 23 and 37 MPa. The apertures become 139.30 and 120.43 µm. The permeability along $\sigma_H$ will be $(139.30/150.00)^3 = 0.801$ of the initial value, and along $\sigma_h$ it will be $(120.43/125.86)^3 = 0.876$ . The anisotropy (the ratio of the two permeabilities) is 1.55.
At the injector: $48 - 44 = 4$ MPa on set A and $62 - 44 = 18$ MPa on set B; the apertures are 210.87 and 150.00 µm. The permeability along $\sigma_H$ is $(210.87/150.00)^3 = 2.78$ times the initial, the anisotropy 2.78.

Answer: At first the reservoir conducts 1.7 times better along

\sigma_H

than across it. Depletion reduces the permeability of both sets, to 0.80 and 0.88 of the initial value, and it slightly smooths out the difference. Near an injector the permeability along

\sigma_H

grows almost threefold.

What it means. Set A, which is open wider, is softer (it closes more easily), so it loses more than set B in depletion. Near injectors the picture is the reverse. As the pressure approaches $\sigma_h$ , it opens fractures along $\sigma_H$ . At a pressure equal to $\sigma_h$ they open to their aperture without load, which means 4.6 times the initial permeability. Above $\sigma_h$ they grow as hydraulic fractures in the same direction (fracture orientation). This leads to practical rules. We place rows of injectors along $\sigma_H$ , so that the water moves towards the producers across the open fractures. We do not place producers on the continuation of an injector line along $\sigma_H$ . We drill horizontal wells along $\sigma_h$ , so that they cross the open fractures (Horizontal wells). We keep the injection pressure below $\sigma_h$ if water breakthrough (water reaching the producers too early) is unacceptable. The model is simplified. In reality the stresses near an injector rise a little with the pressure, the apertures of fractures differ, and shear on critically stressed fractures can keep them open. The data in this example are assumed for illustration.

CalculatedPermeability of a fracture set as a fraction of its value at 18 MPa, against effective normal stress — from the closure law of the lesson on fractures (aperture without load 250 µm). The points are the states of the sets in the example: A₀ and B₀ at the start, A₁ and B₁ after depletion by 15 MPa, A₂ and B₂ near the injector (A — fractures along

\sigma_H

, B — along

\sigma_h

). The ratio of the heights of the points of one state is the anisotropy.

Depletion changes the stresses themselves, too. The horizontal stresses in a depleted block fall (poroelastic model). At the boundaries of the block — near sealing faults, at the top and at the base — the rock is unloaded unevenly. A fault can slip when the ratio of the shear stress (the stress that makes one side slide past the other) to the effective normal stress on it comes close to the coefficient of friction. Then faults come to life, casing strings are sheared (cut across by the movement), and sometimes a fault starts to conduct fluid (critically stressed fractures). In weak rocks the compaction of the reservoir reaches the surface as subsidence (compaction and subsidence).

Section balancing: what it rests on and how a section is restored

Most of the structure underground cannot be seen. Seismic data image steep limbs and the zones beneath thrusts poorly, and wells give only points. The same set of data allows several interpretations, and some of them are geometrically impossible. Section balancing checks an interpretation. We “restore” the deformation backwards, to the state before it, and look at whether a plausible original shape results. A section that can be restored in this way is called balanced. The method was developed in the second half of the twentieth century in the study of thrust belts (belts where rock was pushed over its neighbours). Today it is used to check interpretations in compressional belts (squeezed regions) and in extensional basins (stretched regions) alike.

At its base is the conservation of the volume of rock. Suppose the section is drawn along the direction of transport, the rocks do not move out of its plane (plane strain), and their volume does not change. Then the area of each unit stays the same in the section. If the sequence bends with slip between the layers, like a deck of cards (flexural slip), the lengths and thicknesses of the beds stay the same as well. This gives the two main checks: by bed length (line-length balance) and by area (area balance).

Pin line: A line across which the layers did not slip. It lies in the undeformed part of the section or on the axial surface of a symmetric fold. In the restoration, we measure bed lengths from it.
Loose line: A line at the other end of the section. In the restored section it must come out straight: vertical, or smoothly inclined if the beds slipped over one another. Steps and kinks in it mean that the bed lengths do not agree.
Regional level: The position a horizon would have without the fold or fault. We extend it from the undeformed parts. The area raised above the regional level is the excess area.
Detachment (décollement): A gently dipping surface of slip along a weak layer such as salt or shale. Above it the sequence is crumpled; below it the rock is hardly deformed (detachment).

How a section is restored

1
Section
Build the section in depth, at true scale (the same scale vertically and horizontally) and along the direction of transport, across the strike of the folds and thrusts (section from a map).
2
Lines
Draw the pin and loose lines and the regional levels of the horizons.
3
Restoration
Straighten the beds (by length for flexural slip, or by area) and remove the displacements on faults, beginning with the youngest. Before that, decompact the units deposited while the structures were growing, that is, restore their original thickness (restoring thicknesses).
4
Assessment
Check that the restored section is plausible. The beds must be continuous, the thicknesses must agree with the stratigraphy of the area (the known order of the rock layers), the faults must have possible dips, and the loose line must be straight.
5
Correction
If the section is not plausible, change the interpretation and repeat. Change the position and displacements of faults, the depth of the detachment, or the shape of the folds at depth.

CalculatedThe section of the example below, at true scale. The upper part shows an anticline made of straight segments above the detachment, with horizons 1–3 (regional levels dashed, the excess area of horizon 1 highlighted). A ductile layer (soft rock that flows) lies on the detachment and is thickened in the core of the fold. The pin and loose lines are marked. The detachment is drawn at the depth found from the area, and the deep boundary suggested by seismic data is shown too. The lower part shows the beds before deformation (the distances between them are arbitrary). Their lengths are equal, and the loose line stands farther away than now, by the shortening

S

Shortening, excess area and the depth of the detachment

Shortening from the bed length

S = l_0 - W = \sum_i l_i\,(1 - \cos\delta_i),\qquad \varepsilon = \frac{S}{l_0}

where:

$S$ shortening of the section — a length; the stretch ratio $l/l_0$ from the lesson on strain here equals $1 - \varepsilon$ , m
$l_0$ length of the bed between the pin and loose lines, measured along the bed — its length before deformation, m
$W$ present horizontal distance between the pin and loose lines, m
$l_i,\ \delta_i$ length and dip angle of the $i$ -th straight segment of the bed, m, °
$\varepsilon$ relative shortening, —

Units: lengths in metres, angles in degrees

Holds when

Flexural slip: the length and thickness of the bed do not change.
Before the deformation the bed lay horizontally; the section is drawn in depth, at true scale, along the direction of transport.
We account for displacements on faults within the section; they are added to the bed length.

Each dipping segment is shortened by $l(1 - \cos\delta)$ . At a dip of 10° this is 1.5% of its length, at 30° it is 13%, and at 50° it is 36%. So the steep limbs give most of the shortening, and an error in their dip is costly. All the beds between the pin and loose lines must give the same shortening. If they do not, then a fault has been missed in the interpretation, a bed has been traced wrongly, or the sequence has been deformed by something other than bending.

Excess area and the depth of the detachment

A = S\,h_d,\qquad h_d = \frac{A}{S},\qquad A(z) = S\,(z_d - z)

where:

$A$ excess area of the horizon above its regional level, m²
$S$ shortening, m
$h_d$ height of the regional level of the horizon above the detachment, m
$z$ regional depth of the horizon, m
$z_d$ depth of the detachment, m

Units: lengths in metres, areas in m²

Holds when

The area of the unit between the horizon and the detachment does not change: plane strain, no pressure solution or compaction.
The detachment is horizontal and there is no deformation beneath it.
All horizons are shortened equally. Then the points of the area–depth graph lie on a straight line.

Before deformation, the unit between a horizon and the detachment was a rectangle $l_0 \times h_d$ . After shortening, its width decreased by $S$ while its area stayed the same. So the “extra” area is raised above the regional level: $A = S\,h_d$ . The deeper the detachment, the larger the area that the same shortening raises. We build an area–depth graph from several horizons. The slope of the line is the shortening, and the depth where the area goes to zero is the depth of the detachment. If the points do not lie on a line, then either the shortening varies across the section or part of the area has not been accounted for. The idea of finding the depth of a detachment from the area of a fold appeared as early as the beginning of the twentieth century.

Worked exampleShortening of a fold and depth of the detachment: is the deep boundary compatible with the section?

Given (assumed data)

Horizon 1 (top of the carbonates) between the pin and loose lines: 1000 m horizontal; back limb (the gentle side) 1200 m at a dip of 30°; crest 600 m horizontal; forelimb (the steep side) 783.24 m at a dip of 50°; 1000 m horizontal
Height of the fold and regional depth of horizon 1: the crest is raised 600 m above the regional level of 1500 m
Horizon 3: 600 m deeper; the same limbs, a narrower crest — 159.45 m
Deep boundary on seismic data, proposed as the detachment: 4100 m

Solution

Bed length: $l_0 = 1000 + 1200 + 600 + 783.24 + 1000 = 4583.24$ m. Width: $W = 1000 + 1200\cos30° + 600 + 783.24\cos50° + 1000 = 1000 + 1039.23 + 600 + 503.46 + 1000 = 4142.69$ m.
Shortening $S = 4583.24 - 4142.69 = 440.55$ m. That is 9.6% of the bed length.
The excess area is a trapezoid 600 m high with parallel sides of 600 m (the crest) and $600 + 1039.23 + 503.46 = 2142.69$ m: $A_1 = 600\cdot(600 + 771.35) = 0.8228$ km².
Depth of the detachment below the regional level of horizon 1: $h_d = 0.8228\cdot10^6/440.55 = 1868$ m. So the detachment is at a depth of about 3368 m.
Check on horizon 3: $A_3 = 600\cdot(159.45 + 771.35) = 0.5585$ km². The area has decreased by $0.8228 - 0.5585 = 0.2643$ km². This is exactly $S\cdot600 = 0.2643$ km², as the area–depth line requires. Both horizons give the same depth of the detachment.
If a boundary at 4100 m were the detachment, then at the same shortening the area would have to be $440.55\cdot(4100 - 1500) = 1.145$ km². That is 1.39 times more than has been drawn. And from the drawn area the shortening would be only $0.8228\cdot10^6/2600 = 316.5$ m instead of 440.55 m.

Answer: The fold is shortened by 441 m (9.6%). The area drawn, at such a shortening, requires a detachment at a depth of about 3370 m. A boundary at 4100 m is incompatible with this section.

What it means. A section with the deep boundary as the detachment is not balanced. There are two ways out. Either the fold is larger at depth than drawn — for example, its core is thickened by thrust slices or by ductile rocks that seismic data cannot tell apart. Or the detachment lies higher, in a shale unit at about 3370 m, and the deep boundary has nothing to do with the fold. We choose using other data: whether wells in the area show a weak layer at about 3370 m, and how neighbouring folds behave. For prediction this is not a small matter. The shape of the fold at depth and the position of the crest of the lower horizons depend on the depth of the detachment (shift of the crest with depth). Note that the crest narrows downwards by 0.734 m for each metre of depth and vanishes at a depth of 2317 m, whereas the detachment is deeper. So between these depths the fold cannot be parallel (a fold in which every bed keeps its thickness). Its core is filled by ductile rocks injected into the crest, or it is cut by faults (advanced section). The model is simplified. Real folds do not consist of perfectly straight segments, and the ductile layer of the core can leave the plane of the section. The data in this example are assumed for illustration.

CalculatedArea–depth graph for the example. The excess areas of horizons 1–3 (points) lie on a line whose slope equals the shortening of 440.55 m. The line goes to zero at the depth of the detachment, about 3370 m. The dashed line is the one that a detachment at 4100 m would need: the areas drawn are too small for it.

The same principle works in extension. The hanging wall of a listric fault slides down the curved fault surface and sags below the regional level (listric faults). The area of the sag of the pre-rift horizon (a horizon that existed before the rifting) equals the horizontal displacement $d$ (the heave) multiplied by the depth of the detachment beneath that horizon: $h_d = A/d$ . We also use the shape of the sag of the hanging wall to construct the shape of the fault surface at depth. The methods for this state how the hanging wall is deformed: by vertical or inclined simple shear (vertical shear, inclined shear). First we decompact the units deposited while the displacement was going on. In the hanging wall they are thicker, and their weight compacts the rocks under the horizon more and deepens its sag. Without a correction, the area of the sag is overestimated.

Terms and symbols

Dip domain: An interval of a well, or a part of a section, in which the beds dip in the same way. The boundaries of dip domains are fold hinges, unconformities and faults.
Pressure compartment: A block of rock bounded by barriers. The pressure evens out inside the block but not with the neighbouring blocks.
Lateral transfer and centroid: Lateral transfer is the redistribution of pressure by an isolated dipping bed. The centroid is the depth at which the pressure of the bed equals the pressure of the surrounding shales.
Margin to hydraulic fracturing of the seal: The difference between the fracture pressure of the seal and the fluid pressure beneath it. It limits the height of the accumulation and the window of drilling-fluid densities (the range of densities that can be used safely).
Stress-sensitive reservoir: A reservoir whose permeability changes noticeably with effective stress and also depends on the direction of that stress.
Balanced section: A section that can be restored to a plausible shape before the deformation.
Pin and loose lines: Lines at the ends of a section that is being restored. We measure bed lengths from the first line. In the restored section the second line must be straight.
Regional level and excess area: The regional level is the position a horizon would have without the fold. The excess area is the area raised above it.
Shortening $S$: The difference between the bed length and the present width of the section between the pin and loose lines.
Depth to detachment $h_d$: The height of a horizon above the detachment, which is the same as the depth of the detachment below the horizon. From the area, $h_d = A/S$ .
Area–depth graph: A graph of the excess areas of several horizons against their regional depths. The slope of the line is the shortening, and the depth where it reaches zero is the depth of the detachment.

Common mistakes and misconceptions

Wrong

“The map from 2D lines shows a gentle crest — so there are few fractures in it.”

In fact

Such a map cannot see folds that are shorter than the line spacing, and these folds give the curvature. To predict fractures we need curvature from 3D seismic data, calculated with a spacing smaller than a quarter of the wavelength of the folds.

Wrong

“The pressure in the sandstone equals the pressure in the shales at the same depth, so a prediction from the shales is good for the whole section.”

In fact

An isolated dipping bed evens out its pressure along the water gradient. At its crest the pressure is higher than in the shales at the same depth, and at its base it is lower. In the example of the lesson the difference is 7.3 MPa.

Wrong

“Fractures in all directions pass fluid equally well, and their permeability does not change during development.”

In fact

Fractures along $\sigma_H$ are open wider. Depletion squeezes them shut. Injection pressure, as it approaches $\sigma_h$ , opens them and can turn them into hydraulic fractures in the same direction.

Wrong

“The section is balanced, so the interpretation is correct.”

In fact

Balancing only rejects impossible versions. Usually several different sections can be balanced, and we choose between them using wells, seismic data and analogues.

Wrong

“Any section along a seismic line is suitable for balancing.”

In fact

We need a section in depth, at true scale and along the direction of transport. If the section is in time, has a distorted vertical scale, or is drawn at a slant to the transport, lengths and areas are distorted.

In depth: advanced detail

Why the area of a parallel fold decreases linearly with depth

A fold made of straight segments (a kink fold) with constant layer thickness has a simple structure. Each dipping segment is a band between two parallel axial surfaces, and these surfaces bisect the angles at the hinges (cut them in half). The axial surface of a hinge with a kink (a sharp bend) of angle $\delta$ deviates from the vertical by $\delta/2$ . So with depth, the band of the back limb shifts towards the crest by $\Delta z\,\mathrm{tan}(\delta_1/2)$ , and the band of the forelimb moves to meet it by $\Delta z\,\mathrm{tan}(\delta_2/2)$ . The lengths and dips of the limbs do not change. The height of the uplift $H$ is the same for all horizons, but the crest narrows: $c(z) = c_1 - \Delta z\,(\mathrm{tan}(\delta_1/2) + \mathrm{tan}(\delta_2/2))$ .

The excess area of a horizon is a trapezoid: $A = H\,(c + (b_1 + b_2)/2)$ , where $b = H/\mathrm{tan}\,\delta$ are the horizontal projections of the limbs. So the area decreases with depth at the rate $H\,(\mathrm{tan}(\delta_1/2) + \mathrm{tan}(\delta_2/2))$ . And this rate is exactly the shortening: for each limb $H\,\mathrm{tan}(\delta/2) = l\sin\delta\cdot\dfrac{1 - \cos\delta}{\sin\delta} = l\,(1 - \cos\delta)$ . So in a parallel fold the area and the bed lengths give the same shortening, and the points of the area–depth graph lie on the line $A = S\,(z_d - z)$ .

The crest narrows to zero at a depth of $c_1/(\mathrm{tan}(\delta_1/2) + \mathrm{tan}(\delta_2/2))$ below horizon 1, which is about 817 m in the example. The detachment found from the area lies deeper. The area goes to zero only when the “crest” becomes a negative width $-(b_1 + b_2)/2$ . Below the pinch-out point (where the crest narrows to nothing) a parallel fold cannot continue. The area that the shortening requires is taken up by the ductile rocks of the core — salt or shale — that were injected into the crest (or faults cut the core). This is how detachment folds are built, and that is why the check by area is more reliable for them than the check by the length of individual beds.

Where the centroid comes from

Water flows through the shales around the sand body according to Darcy's law, which says that the flow is proportional to the pressure difference. Through each square metre of the surface of the body, the flow is proportional to the pressure difference $p_{sh} - p_r$ and to the conductivity of the shales (Rock permeability). The pressure in the sandstone rises along the water gradient, and in the shales it rises more steeply. So the difference is linear in depth: $p_{sh}(z) - p_r(z) = (G_{sh} - G_w)(z - z_c)$ . If the body neither loses nor gains water (the steady state), the total flow through the whole surface is zero: $\int (z - z_c)\,\dd S = 0$ when the conductivity of the shales is the same everywhere. Hence $z_c = \int z\,\dd S/\int \dd S$ , the mean depth of the surface of the body, weighted by area.

For a dipping bed of constant width, the area is spread evenly with depth, and the centroid is the middle of its height. For a dome or an anticline, the area grows with depth, so the centroid moves down and the pressure excess at the crest grows. If the shales deeper down let less water through, the lower part of the surface contributes less, and the centroid moves up. In practice, equilibrium has not always been reached. Young, rapidly subsiding bodies may not yet have had time to even out the pressure, and the inflow of water from the shales depends on their own compaction.

Normal stress on an inclined fracture

Let the principal stresses be $\sigma_v$ , $\sigma_H$ and $\sigma_h$ . Let the normal to the fracture make angles with their directions whose cosines are $n_v$ , $n_H$ and $n_h$ ( $n_v^2 + n_H^2 + n_h^2 = 1$ ). Then the normal stress on the fracture is $\sigma_n = \sigma_v n_v^2 + \sigma_H n_H^2 + \sigma_h n_h^2$ . For a vertical fracture $n_v = 0$ , $n_H = \sin\theta$ and $n_h = \cos\theta$ , which gives the formula of the lesson. A gently dipping fracture in a normal-faulting regime (where the vertical stress is the greatest) is squeezed almost by the overburden pressure, so it is less open than vertical fractures. In a thrust regime, where the least stress is vertical, it is the gently dipping fractures that are open.

The shear stress on the same plane is $\tau^2 = \sigma_v^2 n_v^2 + \sigma_H^2 n_H^2 + \sigma_h^2 n_h^2 - \sigma_n^2$ . We compare the ratio $\tau/\sigma_n'$ with the coefficient of friction. It shows how close a fracture or fault is to slipping (stresses on a plane). In depletion and injection this ratio changes, and so does the risk of reactivating faults.

Connections to other topics

Structural styles and tectonic history — structural styles, the domino model, inversion and the timing of structures.
Faults — faults in wells, on maps and on seismic data, sealing by faults.
Folds, salt diapirs and their relationship to faults — curvature and extension of beds in bending, folds and salt.
Fractures and naturally fractured reservoirs — the origin of fractures, their closure under load and data on fracturing.
Seismic interpretation — tracking horizons and faults and converting time to depth.
Cross-sections from maps, map combination and faults on maps — sections from maps and the reconciliation of maps, sections and wells.
Abnormal formation pressures: origin, prediction and recognition — the mechanisms of overpressure and its recognition while drilling.
Reservoir and seal — the height of an accumulation limited by hydraulic fracturing of the seal.
Rock compressibility and the effect of stress on core properties — permeability under load, compaction of the reservoir and surface subsidence.
Spatial correlation and the variogram — how to describe the variability of reservoir properties between wells.

Check yourself

1
Second-order folds have a wavelength of about 600 m. We calculate the curvature from the second difference of depths at three points. At what point spacing will the curvature of these folds practically vanish?
1. A.150 m
2. B.600 m
3. C.The spacing does not matter, only the accuracy of the depths does
4. D.50 m
Show the answer
Correct answer: B. 600 m
When the spacing equals the wavelength, all three points fall on the same phase of the wave: $s = (\sin\pi/\pi)^2 = 0$ . At 150 m, a quarter of the wavelength, the second difference shows about 81% of the curvature. At 50 m it shows almost all of it. The accuracy of the depths matters too. Their error is divided by $a^2$ , so too small a spacing makes the noise larger.
2
An isolated bed full of water dips from 3100 to 3700 m, and its area is spread evenly with depth. The pressure in the surrounding shales rises with a gradient of 0.0225 MPa/m, and the water gradient in the bed is 0.0105 MPa/m. By how much is the pressure at the crest of the bed higher than in the shales at the same depth? And where is it more dangerous to drill?
Show the answer
The centroid is in the middle, at 3400 m, which is 300 m below the crest. The excess is $(0.0225 - 0.0105)\cdot300 = 3.6$ MPa. A well at the crest will meet a pressure 3.6 MPa above the prediction from the shales, so there is a risk of a kick. At the base of the bed the pressure is lower by the same amount than in the shales above it. That brings a risk of losses and stuck pipe.
3
The greatest horizontal stress is directed along azimuth 40°, the regime is normal-faulting ( $\sigma_v > \sigma_H > \sigma_h$ ). Which natural fractures are most likely to be open wider than the rest?
1. A.Horizontal
2. B.All equally
3. C.Vertical, striking 40°
4. D.Vertical, striking 130°
Show the answer
Correct answer: C. Vertical, striking 40°
A vertical fracture striking along $\sigma_H$ is squeezed only by the least stress, $\sigma_h$ . A fracture striking 130° is squeezed by $\sigma_H$ , and a horizontal fracture is squeezed by the greatest stress, $\sigma_v$ . So the permeability is highest along azimuth 40°, and injected water will go first of all in that direction.
4
Between the pin and loose lines, the top of the limestones is 7350 m long and the base of the same unit is 7050 m long. The width of the section is 6600 m. What does this mean?
Show the answer
From the top the shortening is 750 m, and from the base it is 450 m. In flexural slip the beds of one unit must give the same shortening. In the restored section the loose line would come out with a step of 300 m. So there is an error in the interpretation. A fault that repeats the top has been missed, or the base has been traced wrongly, or the unit has been deformed by something other than bending (it has been thickened or shortened along the layers). The section must be rebuilt.
5
A fold raises a horizon above the detachment. The excess area is 1.2 km², and the shortening from the bed length is 800 m. How far below the regional level of the horizon does the detachment lie? Can a strong seismic reflection 3 km below the horizon be the detachment?
Show the answer
$h_d = A/S = 1.2\cdot10^6/800 = 1500$ m. A detachment 3 km beneath the horizon would need, at the same shortening, an area of $800\cdot3000 = 2.4$ km². That is twice what has been drawn. So one of three things is true. The detachment is higher, at about 1.5 km. Or the fold is larger at depth than drawn. Or the shortening from the bed length is overestimated — for example, a bed has been wrongly lengthened in a limb.
6
Which section cannot be checked by balancing in its own plane?
1. A.A section through the tilted blocks of a rift across the strike of the normal faults
2. B.A section across a fold-and-thrust belt along the direction of transport of the thrust sheets
3. C.A section across the strike of a strike-slip zone with a flower structure
4. D.A section through an anticline above a detachment, built in depth
Show the answer
Correct answer: C. A section across the strike of a strike-slip zone with a flower structure
In a strike-slip zone the blocks move along the fault, which means across the plane of the section, so area is not conserved in it. The other sections are drawn along the direction of transport and, if the other conditions are met, can be balanced.

Updated: 28 October 2026