Stress, strain and structural analysis

Quick refresher

Stress is force per unit area (MPa). Strain is a change of lengths, angles and volume, and it has no units. The difference between the principal stresses (the largest and smallest stresses acting on the rock) changes the shape of a rock. The mean stress only compacts it.
The shear stress on a plane is greatest at 45° to $\sigma_1$ (the largest principal stress). But the plane closest to slipping is the one whose normal (the line at right angles to it) makes about 60° with $\sigma_1$ , at a friction coefficient of 0.6. Pore pressure reduces the effective normal stress and makes slip easier.
Brittle rocks break, forming fractures and faults. Ductile rocks flow and crumple. Rocks become more ductile with depth, with rising temperature and when deformed slowly. Salt and undercompacted shales (shales that are less compacted than their depth would give) are ductile even near the surface.
Attitude is made of the strike azimuth (the compass direction of a horizontal line on the bed), the dip azimuth (the compass direction in which the bed slopes down) and the dip angle (the steepness of the slope). Russian-language notation gives the dip azimuth and the dip angle (“dip azimuth 130° ∠ 35°”). English-language notation more often gives the strike by the right-hand rule and the dip angle (040/35).
The apparent dip on a section is less than the true dip: $\mathrm{tan}\,\delta' = \mathrm{tan}\,\delta\,\sin\beta$ . We see the true dip only on a section across the strike, and vertical exaggeration makes the dip look steeper.
Structure contours (lines of equal depth) run along the strike, and closer spacing means a steeper dip: $\mathrm{tan}\,\delta = \Delta z/w$ . From the depths of a horizon in three wells we find its dip, and then we predict its depth in a new well.

Key relationships

\sigma_n = \frac{\sigma_1 + \sigma_3}{2} + \frac{\sigma_1 - \sigma_3}{2}\cos 2\theta,\quad \tau = \frac{\sigma_1 - \sigma_3}{2}\sin 2\theta

stresses on a plane whose normal makes an angle $\theta$ with $\sigma_1$

e = \frac{l - l_0}{l_0},\qquad \gamma = \mathrm{tan}\,\psi

relative extension and shear strain

\mathrm{tan}\,\delta' = \mathrm{tan}\,\delta\,\sin\beta,\qquad \mathrm{tan}\,\delta_V = V\,\mathrm{tan}\,\delta'

apparent dip on a section at an angle $\beta$ to the strike, and the same angle on a section with vertical exaggeration $V$

\mathrm{tan}\,\delta = \frac{\Delta z}{w},\qquad h = h_\text{v}\cos\delta

dip angle from structure contours; true thickness of a bed from the thickness in a vertical well

Common pitfalls

Mixing up shear and normal stress, and total and effective stress. The shear stress and the effective normal stress $\sigma_n - p$ decide slip on a plane.
Assuming that a rock that is brittle in a laboratory test stays brittle over millions of years. At strain rates of about $10^{-15}$ s⁻¹ and temperatures of a few hundred degrees, even strong rocks flow.
Writing a strike without saying which rule gives the dip side. By the right-hand rule, 040/35 and 220/35 are different planes.
Taking an angle measured on a section that is not across the strike, or on a section with vertical exaggeration, for the true dip.
Taking the thickness of a dipping bed in a vertical well for the true thickness, and even more so in a deviated well (one that is not vertical).

After this lesson you can

tell stress from strain, find the normal and shear stresses on a tilted plane, and estimate how close pore pressure (the pressure of the fluid in the pores) brings a plane to slipping
describe strain by extension (stretching), shear and the strain ellipse, and compare strain rates in nature and in the laboratory
explain what decides whether a rock behaves in a brittle way (it breaks) or a ductile way (it flows), and estimate the depth of the brittle–ductile transition in the Earth's crust
write the attitudes of planes and lines in the Russian-language and English-language forms, and convert one notation into the other
calculate the apparent dip on a section (the dip seen on a cross-section that is not across the strike), the angle on a section with vertical exaggeration (a stretched vertical scale), and the true thickness of a bed from a well
find the dip from the structure contours of a map and from the depths of a horizon in three wells

What structural analysis studies

Layers of sedimentary rocks are laid down almost horizontally. But in wells and on seismic sections they are tilted, folded, broken by faults and cut by fractures. All this is the result of the deformation of the rocks under forces that acted on them for millions of years. Structural geology studies the forms that deformation created, which are called structures: a fold (a bend in the layers), a fault (a break along which rock has moved), a dome (a rounded uplift), a fracture set (a group of parallel cracks). It also studies the processes that created them.

Structural analysis answers questions at three levels. Geometry: what shape and position the layers, faults and fractures have now, and where and at what angle they dip. Kinematics (the study of movement): what displacements and changes of shape led to this geometry, that is, what moved, rotated, shortened or stretched, and in what order. Dynamics, or geomechanics: what stresses caused these displacements, and why the rock responded with a fracture or a bend. The engineer needs all three. Geometry defines the trap and the positions of wells. Kinematics gives the order of events, for example whether the trap formed before the oil arrived (timing of trap formation). Geomechanics describes how the formation behaves in drilling, hydraulic fracturing (cracking the rock with fluid under high pressure) and development.

Where structural data come from

Source: Seismic
What it provides: the shape of horizons (rock surfaces) and large faults in three dimensions
Scale: from hundreds of metres to tens of kilometres

Source: Wells: depths of beds, dipmeter (a tool that measures bed dip), borehole image logs
What it provides: accurate depths, dip angles and directions of beds and fractures at the wellbore
Scale: from centimetres to metres at the wellbore

Source: Core
What it provides: bed dip, fractures, slickensides (scratches left by slip), dissolution seams (stylolites)
Scale: from millimetres to metres

Source: Outcrops (rock exposed at the surface) and geological maps
What it provides: attitudes and relationships of the rocks at the surface
Scale: from metres to kilometres

Source	What it provides	Scale
Seismic	the shape of horizons (rock surfaces) and large faults in three dimensions	from hundreds of metres to tens of kilometres
Wells: depths of beds, dipmeter (a tool that measures bed dip), borehole image logs	accurate depths, dip angles and directions of beds and fractures at the wellbore	from centimetres to metres at the wellbore
Core	bed dip, fractures, slickensides (scratches left by slip), dissolution seams (stylolites)	from millimetres to metres
Outcrops (rock exposed at the surface) and geological maps	attitudes and relationships of the rocks at the surface	from metres to kilometres

Data of different scales complement each other. Faults whose throw (vertical offset) is too small for seismic to resolve are seen only in wells and core. Seismic shows the shape of the structure between wells.

Stress in rocks

Inside a rock, the neighbouring particles push on any imaginary plane that we draw through it. Stress is this force divided by the area of the plane. The unit is the pascal, and in practice the megapascal (1 MPa = 10⁶ Pa ≈ 10 atm). We split the force into two parts. The part at right angles to the plane is the normal stress $\sigma_n$ , which presses one side of the plane against the other. The part parallel to the plane is the shear stress $\tau$ , which tries to slide one side along the other. In geology and geomechanics we count compression as positive, because almost all normal stresses in the subsurface are compressive. (In the mechanics of materials the sign is usually the opposite.)

At every point there are three planes, at right angles to one another, on which there is no shear stress. The normal stresses on them are the principal stresses. In structural geology we number them by size: $\sigma_1 \ge \sigma_2 \ge \sigma_3$ . In a quiet sedimentary basin one of them is vertical. This is the weight of the rocks, $\sigma_v$ . The other two are horizontal: $\sigma_H$ (the maximum) and $\sigma_h$ (the minimum). Which of the three is the largest decides the tectonic regime (stress regimes, how stresses are estimated and measured).

For deformation, the difference between the stresses matters more than their size. The mean stress $(\sigma_1 + \sigma_2 + \sigma_3)/3$ squeezes the rock equally from all sides and reduces its volume. This is how sediments are compacted. The differential stress $\sigma_1 - \sigma_3$ changes the shape. It creates shear stresses that slide grains, open fractures and crumple layers. Pore pressure $p$ carries part of the load, so the effective stress $\sigma - p$ acts on the rock skeleton, the solid framework of grains (in more detail). The effective stress reduces the compression across any plane, but it does not change the differential stress. In the friction law and in failure criteria we subtract the whole pore pressure. For the elastic compression of the skeleton we subtract only a fraction of it, $\alpha p$ , where $\alpha \le 1$ is the Biot coefficient (poroelastic model).

Stresses on an inclined plane (two-dimensional case)

\sigma_n = \frac{\sigma_1 + \sigma_3}{2} + \frac{\sigma_1 - \sigma_3}{2}\cos 2\theta, \qquad \tau = \frac{\sigma_1 - \sigma_3}{2}\sin 2\theta

where:

$\sigma_n$ normal stress on the plane, MPa
$\tau$ shear stress on the plane, MPa
$\sigma_1,\ \sigma_3$ greatest and least principal stresses, MPa
$\theta$ angle between the direction of $\sigma_1$ and the normal (perpendicular) to the plane, °

Units: any identical units of stress; the angle in degrees or radians

Holds when

The plane contains the direction of the intermediate principal stress $\sigma_2$ , so it does not enter the formulas.
The stresses are uniform near the plane. The formulas follow from the balance of forces on a triangular prism (advanced section).
For effective stresses, subtract the pore pressure from $\sigma_1$ , $\sigma_3$ and $\sigma_n$ . The shear stress $\tau$ does not depend on it.

At $\theta = 0$ the plane is perpendicular to $\sigma_1$ and carries only the normal stress $\sigma_1$ . At $\theta = 90°$ it carries only $\sigma_3$ . The shear stress is greatest, $(\sigma_1 - \sigma_3)/2$ , at $\theta = 45°$ . If we plot the pairs $(\sigma_n, \tau)$ for all $\theta$ , we get a circle, the Mohr circle. Its centre is at $(\sigma_1 + \sigma_3)/2$ and its radius is $(\sigma_1 - \sigma_3)/2$ . The plane with angle $\theta$ corresponds to the point turned from $\sigma_1$ by an angle $2\theta$ .

A ready-made plane of weakness, such as a fracture or a fault without cohesion (without any bonding across it), begins to slide when the shear stress reaches the friction force: $\tau = \mu\,(\sigma_n - p)$ . For most rocks the coefficient of friction $\mu$ lies between 0.6 and 0.85, almost whatever the composition (Byerlee's law). For faults filled with clay material it is markedly lower, about 0.2–0.4. A similar criterion describes the strength of intact rock that has not yet fractured, but it also includes cohesion (Mohr–Coulomb criterion).

CalculatedMohr circle in effective stresses for the example below, before the pore pressure rises (solid circle) and after it has risen by 5.6 MPa (dashed). After the rise, the point of the fracture reaches the friction line

\tau = 0.6\,(\sigma_n - p)

. The size of the circle does not change.

Worked exampleHow far a dipping fracture is from slipping

Given (assumed data)

Depth: 2600 m
Overburden pressure gradient: 23.0 kPa/m; the vertical $\sigma_v$ is the greatest principal stress $\sigma_1$
Minimum horizontal stress $\sigma_h = \sigma_3$: 41.0 MPa
Pore pressure: hydrostatic (the weight of a water column), 10.2 kPa/m
Plane: a fracture without cohesion, dipping at 60°, striking along $\sigma_H = \sigma_2$
Coefficient of friction $\mu$: 0.6

Solution

Principal stresses and pressure: $\sigma_1 = 0.023\cdot 2600 = 59.8$ MPa, $\sigma_3 = 41.0$ MPa, $p = 0.0102\cdot 2600 = 26.52$ MPa. The centre of the Mohr circle is $(59.8 + 41.0)/2 = 50.40$ MPa, the radius is $9.40$ MPa.
$\sigma_1$ is vertical, so the normal to a plane dipping at 60° makes an angle $\theta = 60°$ with it, and $2\theta = 120°$ : $\sigma_n = 50.40 + 9.40\cos 120° = 45.70$ MPa, $\tau = 9.40\sin 120° = 8.14$ MPa.
Effective normal stress $\sigma_n - p = 45.70 - 26.52 = 19.18$ MPa; the ratio $\tau/(\sigma_n - p) = 8.14/19.18 = 0.42$ is less than the coefficient of friction, 0.6.
Slip begins when $\tau = 0.6\,(\sigma_n - p)$ , that is, at $p = \sigma_n - \tau/0.6 = 45.70 - 13.57 = 32.13$ MPa. This is 5.6 MPa above the present value.

Answer: At present the fracture is stable: the shear stress is 42% of the effective normal stress, against a friction threshold of 60%. Slip will begin if the pore pressure rises by about 5.6 MPa.

What it means. We do not need to change the tectonic forces for this. It is enough to raise the pore pressure, for example by injecting water near a fault at a pressure above the reservoir pressure (waterflooding). This is how we explain induced earthquakes (quakes caused by human activity) and leaks along faults during injection. So we limit the injection pressure near large faults, using a geomechanical calculation. The dip of 60° is not chosen by chance. At $\mu = 0.6$ , the planes closest to slipping are those with $\theta \approx 60°$ . And where the crust is being stretched, normal faults most often dip at about 60° (Faults). The simplifications push the estimate to the safe side. First, we ignore the cohesion of the fracture, but with cohesion slip would come later. Second, we ignore that a rise of pressure in the reservoir partly raises the horizontal stress. With a vertical $\sigma_1$ this makes the Mohr circle smaller and also moves slip further away. The data in this example are assumed for illustration.

Strain: extension, shear and the strain ellipse

Strain says how much the size and shape of a body have changed, using numbers without units. Any movement of rocks has four parts: moving as a whole (translation), rotation, a change of shape, and a change of volume. The first two change nothing inside the body. Strain in the narrow sense means the last two.

Extension, stretch and shear strain

e = \frac{l - l_0}{l_0}, \qquad S = \frac{l}{l_0} = 1 + e, \qquad \gamma = \mathrm{tan}\,\psi

where:

$e$ relative elongation of a line (extension); a negative value is shortening, —
$S$ stretch, —
$l_0,\ l$ length of the line before and after deformation, m
$\gamma$ shear strain, —
$\psi$ angular shear: the angle by which a line that was perpendicular to a given direction has turned away from the right angle, °

Units: dimensionless quantities; lengths in the same units

Holds when

The quantities describe the finite (total, accumulated) strain, not the path by which it developed.
In homogeneous strain (the same everywhere) they are the same throughout the volume. In inhomogeneous strain they refer to a small region.

In the lesson on basins, the lithosphere (the rigid outer shell of the Earth) is stretched by a factor of $\beta$ . That is the stretch $S = \beta$ . Shortening by a quarter is $e = -0.25$ . Compaction of sediments is also strain. A clay that loses porosity from 60 to 15% at a constant grain volume shortens vertically by more than half, $e \approx -0.53$ (calculating thicknesses with compaction).

Draw a circle on a rock before deformation. After homogeneous deformation it becomes an ellipse, the strain ellipse. In three dimensions a sphere becomes an ellipsoid with semi-axes $S_1 \ge S_2 \ge S_3$ . The axial ratio $R = S_1/S_3$ shows how much the shape is distorted. The directions of the axes are the directions of greatest extension and greatest shortening. Natural “circles” are objects that were close to spheres at the start: ooids (small round limestone grains), reduction spots in red beds (round pale spots in red rock), and some fossils.

Worked exampleStrain of a limestone from flattened ooids

Given (assumed data)

Mean long axis of the ooids in a thin section (a thin slice of rock viewed under a microscope) $a$: 1.32 mm
Mean short axis $b$: 0.75 mm
Assumptions: before deformation the ooids were spheres; their cross-sectional area did not change

Solution

Original diameter, if the cross-sectional area stays the same: $\pi a b/4 = \pi d_0^2/4$ , so $d_0 = \sqrt{ab} = \sqrt{1.32\cdot 0.75} = 0.995$ mm.
Stretches: $S_1 = 1.32/0.995 = 1.327$ , $S_3 = 0.75/0.995 = 0.754$ .
Extensions: $e_1 = 32.7$ %, $e_3 = -24.6$ %. Axial ratio: $R = 1.32/0.75 = 1.76$ .

Answer: The rock is shortened by about 25% across the long axes of the ooids and extended by 33% along them.

What it means. The direction of shortening shows where the pressure came from during deformation. Its size shows how much the sequence has been crumpled. In fold belts, such estimates help to restore the original width of the basin. The result underestimates the shortening in two cases. In the first, part of the carbonate has dissolved under pressure. Then the cross-sectional area has decreased, $\sqrt{ab}$ is less than the original diameter, and the extension is, on the contrary, overstated. In the second, the ooids are stronger than the cement and deformed less than the rock as a whole. Then both the extension and the axial ratio are understated. Real ooids are not perfectly round, so we measure tens of grains and allow for the scatter of their original shape. The data in this example are assumed for illustration.

It matters not only how much a rock deforms, but also how fast. The strain rate $\dot e = e/t$ is measured in s⁻¹. In a laboratory strength test, a sample is shortened by a few per cent over tens of minutes, which is a rate of about $10^{-5}$ s⁻¹. In nature, deformation is billions of times slower, and this changes the behaviour of rocks (next section).

Worked exampleHow fast a mountain belt deforms

Given (assumed data)

Shortening rate of the belt from satellite geodesy (precise measurement of ground position by satellite): 15 mm/yr
Width of the deforming belt: 300 km
Strain rate in a laboratory test: $10^{-5}$ s⁻¹

Solution

Strain rate: $\dot e = v/W = 0.015/(3\cdot 10^{5}) = 5.0\cdot 10^{-8}$ yr⁻¹; there are $3.156\cdot 10^{7}$ s in a year, so $\dot e = 1.58\cdot 10^{-15}$ s⁻¹.
Shortening by 1% builds up in $0.01/\dot e = 200$ thousand years, and in 1 million years it reaches $5.0$ %.
The laboratory test is faster than nature by a factor of $10^{-5}/\dot e \approx 6.3\cdot 10^{9}$ .

Answer: The belt deforms at a rate of about 1.58·10⁻¹⁵ s⁻¹, which is almost ten orders of magnitude (powers of ten) slower than a sample in the laboratory.

What it means. The shortening rate and the width are rough values for the Tian Shan (tectonics of Kazakhstan). Deformation is spread unevenly across the belt. It is concentrated on faults and in foothill folds, so local rates may be several times higher than the average. The main conclusion does not depend on the details. Over geological time a rock goes through a path that takes tens of minutes in the laboratory. But it goes through it at rates slow enough for creep (slow flow of solid rock) and pressure solution to act. So we cannot carry strength measured on core over directly to tectonic processes. The data in this example are assumed for illustration.

Brittle and ductile behaviour of rocks

Compress a rock sample and slowly increase the differential stress. At first the rock deforms elastically: after the load is removed, it returns to its original shape (Rock compressibility and the effect of stress on core properties). Then the yield point is reached, and the deformation can no longer be undone. Two paths are possible after that. In brittle behaviour, microfractures build up in the rock, and it suddenly breaks along one or several surfaces. The stress drops sharply. In ductile behaviour, the deformation spreads through the whole volume. The rock flows without losing continuity, at an almost constant or even rising stress.

SchematicStress–strain curves of one rock at different confining pressures (the pressure that squeezes the rock from all sides), as a schematic. At low pressure the rock fails in a brittle way, with a drop in stress. At medium pressure the behaviour is transitional. At high pressure the rock flows in a ductile way, without loss of continuity.

Brittle behaviour creates fractures, faults and crush zones. Ductile behaviour creates folds, the flow of salt and shales, flattened grains and cleavage (closely spaced parallel surfaces along which the rock splits). The same rock can behave in either way. The conditions in which it deforms decide.

What makes a rock more ductile

Factor: Effective confining pressure
Shifts towards ductile behaviour: increase
Why: friction on fractures that are just starting grows, so they cannot open and link up

Factor: Pore pressure
Shifts towards ductile behaviour: decrease (a rise in pressure makes the rock more brittle)
Why: fluid pressure takes load off the skeleton and reduces the effective confinement

Factor: Temperature
Shifts towards ductile behaviour: increase
Why: defects of the crystal lattice (faults in the regular pattern of atoms) move more easily, and dissolution and redeposition speed up

Factor: Strain rate
Shifts towards ductile behaviour: decrease
Why: slow mechanisms, creep and pressure solution, have time to relieve the stresses

Factor: Rock composition
Shifts towards ductile behaviour: salt, gypsum, shales; on heating — calcite, then quartz, then feldspars
Why: in these minerals, flow starts at lower stresses and temperatures

Factor	Shifts towards ductile behaviour	Why
Effective confining pressure	increase	friction on fractures that are just starting grows, so they cannot open and link up
Pore pressure	decrease (a rise in pressure makes the rock more brittle)	fluid pressure takes load off the skeleton and reduces the effective confinement
Temperature	increase	defects of the crystal lattice (faults in the regular pattern of atoms) move more easily, and dissolution and redeposition speed up
Strain rate	decrease	slow mechanisms, creep and pressure solution, have time to relieve the stresses
Rock composition	salt, gypsum, shales; on heating — calcite, then quartz, then feldspars	in these minerals, flow starts at lower stresses and temperatures

Changing each factor in the opposite direction makes the rock more brittle. Water in the rock and a small grain size also help flow.

Together these factors set the brittle–ductile transition in the Earth's crust. Above it, faults slip in jerks, and almost all earthquakes within continents happen here. Below it, the rocks flow. The main control is temperature. At geological strain rates, quartz starts to flow at about 300 °C and feldspars at about 450–500 °C. So the depth of the transition depends on the heat flow (the heat coming up from inside the Earth). In hot rifts (zones where the crust is stretching) the transition is shallower. Beneath cold shields (the old, stable cores of continents) it is deeper.

Worked exampleAt what depth the crust stops being brittle

Given (assumed data)

Temperature distribution: the model from the lesson on the structure of the Earth: heat flow 65 mW/m², radiogenic heat (heat from radioactive decay) in the upper 15 km of the crust
Onset of ductile flow: quartz — about 300 °C, feldspars — about 450 °C

Solution

In the upper 15 km, $T(z) = 10 + 26 z - 0.36 z^2$ (°C, $z$ in km; $q_0/\lambda = 26$ °C/km, $A/(2\lambda) = 0.36$ °C/km²). The equation $T = 300$ °C gives $z = 13.8$ km.
At a depth of 15 km the temperature is 319 °C. Deeper, we ignore radiogenic heat, the gradient is 15.2 °C/km, and 450 °C is reached at $15 + (450 - 319)/15.2 = 23.6$ km.

Answer: In this model the crust is brittle down to about 14 km. Deeper, quartz-rich rocks flow, and from a depth of about 24 km feldspar-rich rocks flow too.

What it means. This agrees with the fact that earthquakes within continents usually happen no deeper than 15–20 km. The sedimentary cover with oil and gas accumulations (the first kilometres, at 50–150 °C) lies entirely in the brittle region. That is why fractures and faults matter so much in reservoirs (Fractures and naturally fractured reservoirs, Faults). The exceptions are salt and undercompacted shales, which flow even at low temperatures. The boundary is not sharp: the transition spans kilometres, and 300 and 450 °C are rounded values. The water content of the rocks, the grain size and the strain rate can shift the boundary by kilometres. In a hot basin with a heat flow of 90 mW/m², 300 °C is reached at a depth of only about 9 km. The data in this example are assumed for illustration.

Salt and shales flow even near the surface

Rock salt flows under the weight of the overlying sediments at the usual temperatures of a basin, over thousands and millions of years. In the Pre-Caspian Basin, the salt of the Kungurian stage of the Lower Permian has been squeezed into domes and ridges. More than a thousand have been identified, the deformation of the salt continues, and some domes reach the surface (basins of Kazakhstan). Sandstones and carbonates at the same depths break in a brittle way. Undercompacted shales with abnormally high pore pressure also behave in a ductile way, and they too form diapirs (columns of rising material; Folds, salt diapirs and their relationship to faults). Engineers take salt creep into account when drilling and casing wells: flowing salt narrows the hole and loads the casing string, the steel pipe that lines the well (Casing: purpose, types, properties and standards).

Attitude: strike, dip and how they are recorded

We give the position of a plane (the top of a bed, a fracture, a fault) by its attitude. The strike line is a horizontal line on the plane. Its direction is the strike azimuth. The dip line is at right angles to the strike and runs down the plane in the steepest direction. Its azimuth is the dip azimuth (dip direction). The angle between the dip line and the horizontal is the dip angle $\delta$ (dip): from 0° for a horizontal bed to 90° for a vertical one. We measure azimuths from north, clockwise.

SchematicA block diagram of a bed with the attitude dip azimuth 130° ∠ 35° (schematic). On the horizontal top face, the top of the bed crops out (reaches the surface) along the strike line. On the vertical west–east and north–south faces we see apparent dips of about 28 and 24°. The true dip

\delta = 35°

appears only in the vertical plane in the dip direction, which is the dashed triangle inside the block.

The strike azimuth is ambiguous: a horizontal line points in two opposite directions, for example 40° and 220°. The dip azimuth has only one value. So Russian-language practice records the dip azimuth and the dip angle, and finds the strike when needed as the dip azimuth minus or plus 90°. English-language practice more often writes the strike and the dip angle, and it agrees on which side of the strike the plane dips. By the right-hand rule, a bed dips to the right when you look along the recorded strike. That is, the strike azimuth is 90° less than the dip azimuth.

The same plane in different notations

Notation: dip azimuth 130° ∠ 35°
How to read it: Russian-language: dip azimuth 130°, dip angle 35°; the form “dip azimuth SE 130, dip angle 35” also occurs

Notation: 130/35
How to read it: dip azimuth and dip angle (dip direction / dip), the same as in the Russian-language notation

Notation: 040/35
How to read it: strike by the right-hand rule and dip angle: we look along 40°, and the bed dips to the right, towards 130°

Notation: N40°E, 35°SE
How to read it: quadrant notation (angles measured from north or south towards east or west): strike 40° east of north, dip 35° to the south-east

Notation	How to read it
dip azimuth 130° ∠ 35°	Russian-language: dip azimuth 130°, dip angle 35°; the form “dip azimuth SE 130, dip angle 35” also occurs
130/35	dip azimuth and dip angle (dip direction / dip), the same as in the Russian-language notation
040/35	strike by the right-hand rule and dip angle: we look along 40°, and the bed dips to the right, towards 130°
N40°E, 35°SE	quadrant notation (angles measured from north or south towards east or west): strike 40° east of north, dip 35° to the south-east

The notations “130/35” and “040/35” look alike. Without an explanation, “130/35” could be read as a strike by the right-hand rule, that is, a dip towards 220°. And “220/35” by that rule means a dip towards 310°. These are completely different planes. So reports state which notation they use, and in case of doubt we recalculate everything into the dip azimuth.

We give the position of a line (a fold axis, the line where a bed meets a fault, the scratches called slickensides on a fault surface) by its trend and its plunge. The trend is the azimuth of the vertical plane that contains the line, taken in the direction in which the line goes down. The plunge is the angle between the line and the horizontal. For example, a line plunging at 25° to the south has a trend of 180° and a plunge of 25°.

In outcrops we measure attitudes with a geological compass that has a clinometer (a built-in angle gauge). In wells, the dipmeter and borehole image logs give them. Borehole image logs are unwrapped images of the borehole wall, made from electrical resistivity or from the reflection of ultrasound. A tilted plane cuts a cylindrical borehole along an ellipse. On the unwrapped image, the ellipse looks like a sinusoid (a wavy line). Its lowest point shows the dip azimuth. The height of the sinusoid $H$ , for a hole of diameter $d$ , gives the angle between the plane and the plane perpendicular to the hole: $\mathrm{tan}\,\delta = H/d$ . For a vertical well this is the dip angle itself. For a deviated well we recalculate the result, taking into account the inclination and azimuth of the hole (advanced section). In this way, image logs give the dip of beds and fractures every few centimetres along the hole.

In most petroleum basins the layers are not overturned, so the question of which way up a layer is does not arise. In fold belts, some limbs (sides) of folds are overturned. Then we find the top of a layer from primary structures (features formed when the sediment was laid down): cross-bedding, graded bedding, and scour marks at the base (Primary sedimentary structures).

True and apparent dip; the thickness of a bed in a well

On a vertical section drawn across the strike (along the dip line), the slope of the bed equals the true dip angle. On any other vertical section the slope is smaller. This smaller angle is the apparent dip $\delta'$ . On a section along the strike, a bed looks horizontal.

Apparent dip on a vertical section

\mathrm{tan}\,\delta' = \mathrm{tan}\,\delta\cdot\sin\beta = \mathrm{tan}\,\delta\cdot\cos\omega

where:

$\delta'$ apparent dip on the section, °
$\delta$ true dip, °
$\beta$ angle between the line of the section and the strike, °
$\omega$ angle between the line of the section and the dip direction, $\omega = 90° - \beta$ , °

Units: angles in degrees; $\tan$ is the tangent (written tg in Russian-language notation)

Holds when

The bed is flat (planar) within the area. For a curved surface, we apply the formula to the tangent plane (the flat plane that just touches the surface).
The section is vertical. For a tilted section, for example along the hole of a deviated well, we need a vector calculation (advanced section).

Walk a distance $L$ along the line of the section. Along the dip direction we move only $L\cos\omega$ , and movement along the strike does not change the depth. So the bed goes down by $L\cos\omega\,\mathrm{tan}\,\delta$ . This gives the formula. The apparent dip is always less than the true dip. For a gentle dip it is almost proportional to $\sin\beta$ . Steep beds look steep on almost any section, except those close to the strike.

CalculatedApparent dip against the angle between the section and the strike, for a true dip of 10°, 35° and 70°, from the formula of the lesson. The point is the example below: a profile at 50° to the strike of a bed dipping at 35°.

Worked exampleDip of a bed on a seismic profile

Given (assumed data)

Attitude of the top of the bed (from wells): dip azimuth 130° ∠ 35°
Direction of the seismic profile: west–east (azimuth 90°)
Vertical scale of the section: 5 times larger than the horizontal

Solution

Strike by the right-hand rule: $130° - 90° = 40°$ (and the opposite, 220°). The angle between the profile and the dip direction is $\omega = 130° - 90° = 40°$ , and between the profile and the strike $\beta = 50°$ .
$\mathrm{tan}\,\delta' = \mathrm{tan}\,35°\cdot\cos 40° = 0.7002\cdot 0.7660 = 0.5364$ , hence $\delta' = 28.2°$ .
On a section with a fivefold vertical exaggeration this bed will be drawn at an angle of $\mathrm{arctan}(5\cdot 0.5364) = 69.6°$ .

Answer: On the west–east profile the bed dips at 28.2° instead of the true 35°, and on a section with a fivefold vertical exaggeration it looks almost vertical, at 69.6°.

What it means. We cannot use the angle taken from such a profile as the true dip, for example to convert thicknesses or to predict the depth of the bed in a neighbouring well. One profile cannot give the true dip. We need a second profile in another direction, or well data (advanced section). Strong vertical exaggeration is common on seismic sections and correlation panels. It helps us see thin layers, but it turns gentle folds into steep ones and makes faults almost vertical. The data in this example are assumed for illustration.

The dip of a bed also changes its thickness in a well. A vertical well crosses a dipping bed in the vertical direction. The distance between the top and the base along the hole is the true vertical thickness $h_\text{v}$ (TVT). It is greater than the true stratigraphic thickness $h$ (TST), which is measured at right angles to the bedding. In Russian-language geological texts the thickness of a layer is more often called by a word that literally means “power”.

True thickness of a bed from the vertical thickness

h = h_\text{v}\cos\delta

where:

$h$ true (stratigraphic) thickness — measured along the normal to the bedding, m
$h_\text{v}$ vertical thickness — between the top and the base in a vertical well, m
$\delta$ dip of the bed, °

Units: any identical units of length

Holds when

The top and base are parallel, and the dip is constant over the thickness of the bed.
The well is vertical. In a deviated well, the length of the hole in the bed is related to the true thickness through the angle between the hole and the normal to the bed (advanced section).

We draw thickness maps either from true thicknesses (isopach maps) or from vertical thicknesses (isochore maps). To find the volume of rock between two structure surfaces, we need the second kind (Gross rock volume from maps). At a dip of 10° the difference is about 1.5%, and at 35° it is already about 18%.

Worked exampleThickness of a sandstone in a vertical well

Given (assumed data)

Top and base of the bed in a vertical well: 2412.0 and 2436.5 m
Dip of the bed (from the image log): 35°

Solution

Vertical thickness: $h_\text{v} = 2436.5 - 2412.0 = 24.5$ m.
True thickness: $h = 24.5\cdot\cos 35° = 24.5\cdot 0.8192 = 20.07$ m, which is 18% less.

Answer: The true thickness of the bed is about 20.07 m, against 24.5 m measured vertically.

What it means. Suppose we compare this well with a well on a gentle limb (the flatter side of a fold), where the bed dips at 5°. There the vertical thickness will be 18% smaller than here. But this “thinning” is only apparent: the true thickness is the same. So before correlation and before drawing thickness maps, we bring all thicknesses to one kind. The data in this example are assumed for illustration.

Structure contours, structure maps and sections

A structure contour map shows the shape of a geological surface in plan view (seen from above). It is a map of structure contours: lines of equal elevation of the top or base of a bed (or of equal depth below sea level). Each contour is a horizontal line on this surface. So contours run along the strike, and the dip is at right angles to them, towards greater depths. The closer the contours, the steeper the dip.

Dip angle from structure contours

\mathrm{tan}\,\delta = \frac{\Delta z}{w}

where:

$\delta$ dip angle, °
$\Delta z$ contour interval — the difference in depth between neighbouring contours, m
$w$ horizontal distance between neighbouring contours, measured perpendicular to them, m

Units: $\Delta z$ and $w$ in the same units; $w$ is the distance on the ground, that is, the distance on the map multiplied by the second number of the scale (25,000 in 1:25,000)

Holds when

The surface between neighbouring contours is taken to be planar.
We take the distance at right angles to the contours. In any other direction, we would get an apparent dip.

On a map at a scale of 1:25,000 with a contour interval of 25 m, a distance of 8 mm between contours is 200 m on the ground and a dip of about 7°, while a distance of 2 mm is 50 m and a dip of about 27°.

We draw contours from points of known depth: wells, and seismic horizons converted from time to depth. The topic Structure maps and contouring covers how to choose the contour interval, how to interpolate (fill in values) between points, and what errors to avoid. The simplest case is three wells that have penetrated one horizon. If we can treat the surface between them as flat, we find the dip from the three depths and predict the depth at any point. This is the three-point problem.

Worked exampleDip of the top from three wells, and the depth in a planned well

Given (assumed data)

Well A: origin; top at a depth of 2380 m
Well B: 1800 m east and 600 m north of A; top at 2512 m
Well C: 700 m east and 2000 m north of A; top at 2340 m
Planned well D: 1200 m east and 1200 m north of A
Depths: vertical, below sea level

Solution

The elevation of the top is a plane $e = -z = -2380 + g_x x + g_y y$ , where $g_x$ and $g_y$ are the slopes of the elevation to the east and to the north. Differences relative to well A: for B $1800\,g_x + 600\,g_y = -132$ , for C $700\,g_x + 2000\,g_y = 40$ .
The determinant of the system is $1800\cdot 2000 - 600\cdot 700 = 3.18\cdot 10^{6}$ ; $g_x = \dfrac{-132\cdot 2000 - 600\cdot 40}{3.18\cdot 10^{6}} = -0.09057$ , $g_y = \dfrac{1800\cdot 40 + 132\cdot 700}{3.18\cdot 10^{6}} = 0.05170$ .
Dip angle: $\mathrm{tan}\,\delta = \sqrt{g_x^2 + g_y^2} = 0.1043$ , $\delta = 6.0°$ . The elevation falls fastest in the direction opposite to the vector $(g_x;\ g_y)$ , which is towards the south-east. The dip azimuth is 120° and the strike is 30°.
Depth of the top in well D: $z = 2380 - (-0.09057\cdot 1200 + 0.05170\cdot 1200) = 2426.6$ m.

Answer: The top dips at about 6.0° with a dip azimuth of 120°. In the planned well D we expect it at a depth of about 2427 m.

What it means. The calculation is valid only if the surface between the wells is flat. A fold, a fault or a change in thickness between them will distort the prediction. So we give the depth in the planned well as a range and check it against seismic data. A check on the arithmetic: the plane must reproduce the depth of each of the three wells. Computer programs that build surfaces use the same calculation: they fit planes to triangles between many wells. The data in this example are assumed for illustration.

A geological cross-section shows the structure of the rocks in a vertical plane along a chosen line. We build it from maps, wells and seismic data. Several sections in different directions, drawn in perspective, make a block diagram or a fence diagram. We see the true dip only on a section across the strike. On the others we see an apparent dip. Sections are also often drawn with vertical exaggeration: the vertical scale is larger than the horizontal one by a factor of $V$ , so that thin beds can be seen. All slopes then become steeper.

Angle on a section with vertical exaggeration

\mathrm{tan}\,\delta_V = V\cdot\mathrm{tan}\,\delta'

where:

$\delta_V$ the angle at which the bed is drawn on the section, °
$V$ exaggeration factor: how many times the vertical scale is larger than the horizontal one, —
$\delta'$ apparent dip on this section (the true dip on a section across the strike), °

Units: angles in degrees

Holds when

The exaggeration is the same throughout the section. On a seismic section in time (where the vertical axis is time, not depth), the vertical scale also changes with the wave velocity, so angles are distorted unevenly (Seismic interpretation).

A gentle dip of 3° at $V = 10$ looks like 28°. A slope of 25 m in 200 m (about 7°, as in the figure below) at $V = 4$ looks like 27°. Thicknesses on such a section are also stretched by a factor of $V$ . So we state both scales on every section, and we estimate angles on a section without exaggeration ( $V = 1$ ). The topic Cross-sections from maps, map combination and faults on maps covers how to draw a section from a structure map and how to show faults on maps.

SchematicA structure map of the sloping top of a bed (contours every 25 m, dip of about 7° to the south-east) and a section along the line A–A′ across the strike, with the vertical scale exaggerated 4 times. On the section the bed is drawn at an angle of about 27°. Schematic.

Traces of deformation inside the rock: cleavage, boudinage, stylolites

Deformation leaves traces not only in the shape of layers but also in the fabric of the rock itself (the way its grains and structures are arranged). Oriented elements form an oriented fabric, or deformation fabric. These elements are platy clay minerals laid parallel, planes of splitting, dissolution seams and fractures. The fabric matters to the engineer because it makes the properties of the rock depend on direction.

Compaction fabric of shales: As shale is buried, it loses water and porosity, and its platy clay particles rotate and lie parallel to the bedding. In compacted shales and mudstones, the velocity of elastic waves and the electrical conductivity are higher along the bedding than across it. This is called anisotropy (properties that depend on direction), and we take it into account when we process seismic data and well logs. Permeability across the layers is several times lower than along them, and in thinly layered sequences it is orders of magnitude lower. Layered sequences of different hardness push the bit sideways, and the hole deviates by itself (Deflection tools and downhole motors).
Cleavage: a system of closely spaced parallel planes along which the rock splits into thin plates. It forms in shaly rocks under strong compression, when the clay minerals begin to recrystallise (grow into new crystals). It occurs in fold belts and in the deepest parts of basins. Cleavage planes lie almost at right angles to the direction of greatest shortening, and they may cross the bedding at a high angle. Under still stronger compression and heating, shale turns into slate (a metamorphic rock: one changed by heat and pressure), which splits along the cleavage into even thin plates.
Boudinage: the stretching of a sequence in which strong layers alternate with soft, yielding ones. A strong layer is torn into elongated blocks called boudins, and the soft layer flows into the gaps. Sequences lying on ductile salt may also be torn into separate blocks when they are stretched, and the salt flows in between them (Folds, salt diapirs and their relationship to faults).
Stylolites: jagged (toothed) dissolution seams, most often in limestones and dolomites, less often in sandstones. Under load, the mineral dissolves where grains touch. The insoluble residue (clay particles, organic matter, iron oxides) builds up as a thin film, and the grains on both sides of the seam move closer together. The height of the “teeth” shows the minimum amount of rock that has dissolved. Stylolites parallel to the bedding come from the weight of the rocks. Tilted and vertical ones come from tectonic compression.

Stylolites are a frequent cause of vertical heterogeneity (uneven properties from layer to layer) in carbonate reservoirs. A seam lined with clay residue may be poorly permeable, and it may divide a bed into layers that connect weakly. The dissolved carbonate is redeposited nearby in the pores as cement and reduces porosity. Small fractures sometimes develop along stylolites, and they do the opposite: they raise permeability. So we describe stylolites separately on core and compare how often they occur with the inflow profile (the contribution of each interval to the flow into the well; Reservoir heterogeneity and permeability averaging). Fractures without displacement are also a deformation fabric. The topic Fractures and naturally fractured reservoirs covers their origin and their influence on development, and the topic Faults covers faults.

Terms and symbols

Stress: The force per unit area of a plane inside a rock, in MPa. The normal stress $\sigma_n$ presses the sides of the plane together, and the shear stress $\tau$ slides them.
Principal stresses $\sigma_1 \ge \sigma_2 \ge \sigma_3$: The normal stresses on three planes at right angles to one another, which carry no shear stress.
Differential stress: The difference $\sigma_1 - \sigma_3$ . It changes the shape of the rock, while the mean stress changes its volume.
Mohr circle: A drawing of the stresses on all planes that contain the direction of $\sigma_2$ : a circle with its centre at $(\sigma_1 + \sigma_3)/2$ and radius $(\sigma_1 - \sigma_3)/2$ .
Strain: The change of size and shape of a body: extension $e$ , stretch $S$ , shear strain $\gamma$ , change of volume.
Strain ellipse: The ellipse into which a circle turns under homogeneous strain. The axial ratio $R = S_1/S_3$ measures the distortion.
Strain rate $\dot e$: Strain per unit time, in s⁻¹. In tectonics it is about $10^{-15}$ – $10^{-13}$ s⁻¹, and in laboratory tests about $10^{-5}$ s⁻¹.
Brittle and ductile behaviour: Breaking with the formation of fractures and faults, or flowing without loss of continuity. The boundary between them in the crust is the brittle–ductile transition.
Attitude: The strike azimuth, dip azimuth and dip angle of a plane; for a line, the trend and plunge.
Right-hand rule: A convention for recording the strike. The bed dips to the right of the recorded strike direction, so the dip azimuth equals the strike azimuth + 90°.
Apparent dip $\delta'$: The slope of a bed on a vertical section that is not perpendicular to the strike; always less than the true dip.
Structure contour: A line of equal elevation (depth) of a geological surface; a structure map is a map of such contours.
Vertical exaggeration factor $V$: How many times the vertical scale of a section is larger than the horizontal one; it makes all slopes steeper: $\mathrm{tan}\,\delta_V = V\,\mathrm{tan}\,\delta'$ .
True thickness and vertical thickness: The thickness of a bed along the normal to the bedding ( $h$ , TST) and along the vertical ( $h_\text{v}$ , TVT): $h = h_\text{v}\cos\delta$ .
Stereographic projection: A representation of the orientations of planes and lines on a circle: a line by a point, a plane by an arc or a pole.
Cleavage, boudinage, stylolites: Deformation fabrics: planes of splitting in shaly rocks, strong layers torn apart in a soft matrix, jagged dissolution seams.

Common mistakes and misconceptions

Wrong

“There is a shear stress on the fracture, so it slips.”

In fact

Friction resists slip, and friction is proportional to the effective normal stress. The fracture stays still as long as $\tau < \mu\,(\sigma_n - p)$ . A rise in pore pressure can upset this balance without any change in the tectonic forces.

Wrong

“Salt and shale at a depth of 3 km are as brittle as sandstone.”

In fact

Salt flows at any temperature found in a basin, and undercompacted shales flow when the pore pressure is high. Next to brittle reservoirs, they behave like a very thick (viscous) liquid: they form domes and diapirs, narrow the hole and load the casing strings.

Wrong

“On a seismic section the bed dips at 40°, so its dip is 40°.”

In fact

The angle on a section is only an apparent dip if the section does not run across the strike, and it is exaggerated if the vertical scale is exaggerated. We find the true dip from two directions, from wells or from a structure map.

Wrong

“Strike 220°, dip 35° is an unambiguous record.”

In fact

Without the rule or the dip side, these are two different planes: one dipping towards 130° and one towards 310°. It is safer to record the dip azimuth: “dip azimuth 310° ∠ 35°”.

Wrong

“A deviated well penetrated the bed over 60 m, so its thickness is 60 m.”

In fact

The length of hole in the bed equals the true thickness only if the hole is perpendicular to the bedding. In general $h = \Delta l\cos\gamma$ , where $\gamma$ is the angle between the hole and the normal to the bed.

In depth: advanced detail

Vector calculation: a bed in a deviated well and the true dip from two sections

It is convenient to solve attitude problems with vectors in an “east–north–up” coordinate system. The normal to a plane with dip azimuth $a$ and dip angle $\delta$ , pointing upwards, is $\mathbf n = (\sin\delta\sin a;\ \sin\delta\cos a;\ \cos\delta)$ . The axis of a well with inclination $\theta$ (its angle from the vertical) and azimuth $\varphi$ (the notation of the lesson Directional survey calculations) points downwards along the vector $\mathbf w = (\sin\theta\sin\varphi;\ \sin\theta\cos\varphi;\ -\cos\theta)$ . Here $\theta$ is the inclination of the hole, not the angle of the plane from the section on stresses. And $\gamma$ below is an angle, not the shear strain.

Angle between the hole and the normal to the bed, and the length of hole in the bed

\cos\gamma = \left|\sin\delta\,\sin\theta\,\cos(a - \varphi) - \cos\delta\,\cos\theta\right|, \qquad h = \Delta l\,\cos\gamma

where:

$\gamma$ angle between the axis of the hole and the normal to the bedding, °
$\delta,\ a$ dip angle and dip azimuth of the bed, °
$\theta,\ \varphi$ inclination and azimuth of the hole, °
$\Delta l$ length of hole between the top and the base of the bed, m
$h$ true thickness of the bed, m

Units: angles in degrees; lengths in the same units

Holds when

The bed is flat, and its top and base are parallel. The hole is straight over the interval of the bed.
The azimuths of the bed and of the hole are measured from the same north.

The formula gives the absolute value of the dot product $\mathbf n\cdot\mathbf w$ (a way to multiply two vectors to get a number). For a vertical well ( $\theta = 0$ ) it gives $\cos\gamma = \cos\delta$ , that is, $h = h_\text{v}\cos\delta$ . If the hole goes against the dip, up the dip ( $\varphi = a + 180°$ ), then $\gamma = |\theta - \delta|$ . If it goes down the dip ( $\varphi = a$ ), then $\gamma = \theta + \delta$ (for $\theta + \delta \le 90°$ ).

Worked exampleOne bed in three wells

Given (assumed data)

Bed: dip azimuth 130° ∠ 35°, true thickness 20.07 m (as in the example of the lesson)
Well 1: vertical
Well 2: inclination 40°, azimuth 310°: going against the dip
Well 3: inclination 40°, azimuth 130°: going down the dip

Solution

Well 1: $\cos\gamma = \cos 35° = 0.81915$ ; the length of hole in the bed is $20.07/0.81915 = 24.5$ m.
Well 2: $\gamma = |40° - 35°| = 5°$ , $\cos\gamma = 0.99619$ ; the length is $20.07/0.99619 = 20.1$ m.
Well 3: $\gamma = 40° + 35° = 75°$ , $\cos\gamma = 0.25882$ ; the length is $20.07/0.25882 = 77.5$ m.

Answer: The same bed, 20 m thick, takes up 24.5 m of hole in the vertical well, 20.1 m in the well going against the dip, and 77.5 m in the well going down the dip.

What it means. The length of hole in the bed is not the thickness: in well 3 it is almost four times the true thickness. To get the net thickness (the thickness of reservoir rock) for hydrocarbons in place, and to correlate wells, we recalculate the intervals with $\cos\gamma$ . For horizontal wells that run along the bed, $\gamma$ is close to 90°, and hundreds of metres of hole lie in the bed (Horizontal wells). We take the inclination and azimuth of the hole from the directional survey, and the dip of the bed from image logs or a structure map. The data in this example are assumed for illustration.

The same approach solves the inverse problem. Suppose we have measured apparent dips on two sections in different directions. Each of them defines a line that lies in the plane of the bed. The normal to the bed is at right angles to both lines, so it equals their cross product (another way to combine two vectors). We find the dip azimuth and dip angle from its components. For a sinusoid on the image of a deviated well, the formula $\mathrm{tan}\,\delta = H/d$ gives the angle between the bed and the plane perpendicular to the hole. We get the attitude by rotating this plane together with the hole.

Stereographic projection

When there are many orientations, such as hundreds of fractures from an image log or the bedding along the whole hole, we show them on a stereographic projection (stereonet). Imagine a sphere whose centre lies on a plane or a line. A line pierces the lower half of the sphere at one point, and a plane cuts it along an arc of a great circle (a circle as big as the sphere). We project the lower half of the sphere onto a flat horizontal circle. Horizontal lines land at the edge of the circle, and a vertical line lands at the centre. In structural geology the equal-area projection (the Schmidt net) is used more often. It preserves areas, so it is good for counting the density of points. The equal-angle projection (the Wulff net) preserves angles.

Equal-area projection of a line onto the lower hemisphere

r = R\sqrt{2}\,\sin\frac{90° - \rho}{2}, \qquad x = r\sin\varphi,\quad y = r\cos\varphi

where:

$r$ distance of the point from the centre of the projection circle, mm on the drawing
$R$ radius of the projection circle, mm
$\rho$ plunge of the line, °
$\varphi$ trend of the line, °
$x,\ y$ coordinates of the point: to the east and to the north of the centre, mm

Units: $r$ , $R$ , $x$ , $y$ in the same units of length; angles in degrees

Holds when

We use the lower-hemisphere projection, which is the standard one in structural geology.
We show a plane either by an arc (the projections of all the lines lying in it) or by a single point, its pole. The pole is the projection of the normal, which plunges at an angle $90° - \delta$ in the direction opposite to the dip.

A horizontal line ( $\rho = 0$ ) falls on the edge of the circle, $r = R$ . A vertical one falls at the centre. It is easier to show a group of fractures by their poles: fractures of one set give a cluster of points, and the centre of the cluster is the mean orientation of the set.

CalculatedEqual-area projection of the lower hemisphere. It shows the bedding with dip azimuth 130° ∠ 35° and a fracture with dip azimuth 250° ∠ 54° (their arcs and poles), and the line where they intersect, drawn with the formulas of the lesson.

Worked exampleLine of intersection of the bedding and a fracture

Given (assumed data)

Bedding: dip azimuth 130° ∠ 35°
Fracture (from the image log): dip azimuth 250° ∠ 54°

Solution

Normals (east; north; up): $\mathbf n_1 = (0.4394;\ -0.3687;\ 0.8192)$ , $\mathbf n_2 = (-0.7602;\ -0.2767;\ 0.5878)$ .
The line of intersection lies in both planes, so it is at right angles to both normals. Its direction is the cross product $\mathbf l = \mathbf n_1\times\mathbf n_2 = (0.0100;\ -0.8810;\ -0.4019)$ , of length $|\mathbf l| = 0.9684$ . The vertical component is negative, so the vector already points downwards.
The horizontal projection $(0.0100;\ -0.8810)$ points almost exactly south: the trend is $179.4°$ . The plunge is $\arcsin(0.4019/0.9684) = 24.5°$ .

Answer: The fractures of this set intersect the layers along lines plunging at 24.5° to the south (trend 179.4°).

What it means. Such lines show up on core and on bedding surfaces as ribbing (fine ridges). The rock blocks between the fractures are elongated along these lines, and often so are channels of higher permeability. The same calculation finds the axis of a cylindrical fold (a fold shaped like part of a cylinder) from the dip of the beds on its limbs, and the line along which a fault cuts a bed. On a stereographic projection, the line of intersection is the point where two arcs meet. The data in this example are assumed for illustration.

Where the formulas for stresses on a plane come from

Cut a triangular prism of unit thickness out of the rock. Its sloping face has length 1, and the normal to it makes an angle $\theta$ with $\sigma_1$ . The other two faces are perpendicular to $\sigma_1$ and $\sigma_3$ , and they have lengths $\cos\theta$ and $\sin\theta$ . The forces on them are $\sigma_1\cos\theta$ along $\sigma_1$ and $\sigma_3\sin\theta$ along $\sigma_3$ . The prism is in equilibrium (the forces balance), so their sum balances the force on the sloping face. We project this force onto the normal and onto the face itself, and we get $\sigma_n = \sigma_1\cos^2\theta + \sigma_3\sin^2\theta$ and $\tau = (\sigma_1 - \sigma_3)\sin\theta\cos\theta$ .

The double-angle formulas $\cos^2\theta = (1 + \cos 2\theta)/2$ , $\sin^2\theta = (1 - \cos 2\theta)/2$ and $\sin\theta\cos\theta = \tfrac12\sin 2\theta$ bring them to the form given in the lesson. This is the parametric equation of a circle with its centre at $(\sigma_1 + \sigma_3)/2$ and radius $(\sigma_1 - \sigma_3)/2$ : the Mohr circle. Planes tilted with respect to $\sigma_2$ lie inside the region between three circles built on the pairs of principal stresses. For faults and fractures striking along $\sigma_2$ , one large circle is enough.

Connections to other topics

Rock compressibility and the effect of stress on core properties — the elasticity of rocks and effective stress in the reservoir.
Rock mechanics for hydraulic fracturing — principal stresses in the reservoir, their magnitudes and how they are measured.
Fracture gradient and leak-off testing — tectonic stress regimes and the orientation of fractures.
Sand production: types, prediction and cost of the problem — the strength of intact rock and the Mohr–Coulomb criterion.
Faults — types of faults, their geometry and their role in traps and reservoir communication.
Folds, salt diapirs and their relationship to faults — folds and salt diapirs.
Fractures and naturally fractured reservoirs — fractures, fracture sets and fractured reservoirs.
Structure maps and contouring — how we build structure maps from wells and seismic data.
Cross-sections from maps, map combination and faults on maps — how to draw sections from maps and how to show faults on maps.
Traps and timing — structural traps: crest, closure, time of formation.
Stratigraphy and geologic time — how we reconstruct the sequence of events, including deformations.

Check yourself

1
The vertical stress is $\sigma_1 = 60$ MPa, the minimum horizontal stress $\sigma_3 = 40$ MPa, the pore pressure 25 MPa. On which plane, striking along $\sigma_2$ , is the shear stress greatest?
1. A.On one inclined at 45°
2. B.On the horizontal one
3. C.It is the same on all planes
4. D.On the vertical one
Show the answer
Correct answer: A. On one inclined at 45°
$\tau = \tfrac12(\sigma_1 - \sigma_3)\sin 2\theta$ is greatest at $\theta = 45°$ , and it equals 10 MPa. The horizontal and vertical planes are principal planes, so there is no shear stress on them. Pore pressure does not change the shear stress. It reduces only the effective normal stress. But the plane closest to slipping is not the plane with the greatest $\tau$ . Friction grows with the normal stress, and at $\mu = 0.6$ a plane steeper than 45° reaches the threshold first. Its normal makes an angle of about 60° with $\sigma_1$ .
2
A bed has the attitude “dip azimuth 225° ∠ 20°”. What dip will be seen on a north–south seismic profile, and in which direction?
Show the answer
The angle between the profile (azimuth 180°) and the dip direction, 225°, is 45°. $\mathrm{tan}\,\delta' = \mathrm{tan}\,20°\cdot\cos 45° = 0.364\cdot 0.707 = 0.257$ , $\delta' \approx 14.4°$ . On the profile the bed dips to the south. The strike of the bed by the right-hand rule is 135°, and the profile runs at 45° to it.
3
On a structure map at a scale of 1:10,000, contours with an interval of 10 m are 5 mm apart. What is the dip angle?
1. A.about 1°
2. B.about 2°
3. C.about 11°
4. D.about 45°
Show the answer
Correct answer: C. about 11°
5 mm on a map at 1:10,000 is 50 m on the ground. $\mathrm{tan}\,\delta = 10/50 = 0.2$ , $\delta \approx 11.3°$ . A common mistake is to use the distance on the map and forget the scale.
4
On the image log of a vertical well of 216 mm diameter, a bed boundary appears as a sinusoid 0.30 m high with its lowest point at an azimuth of 250°. Write the attitude in two ways.
Show the answer
$\mathrm{tan}\,\delta = H/d = 0.30/0.216 = 1.39$ , $\delta \approx 54°$ . The lowest point of the sinusoid shows the dip direction. The Russian-language notation is “dip azimuth 250° ∠ 54°”. By the right-hand rule it is 160/54: strike 160°, and the bed dips to the right, towards 250°. This same plane is used in the section on the stereographic projection.
5
Why does rock salt in a sedimentary basin flow, while sandstone at the same depth fails by fracturing?
1. A.The salt is hotter than the sandstone
2. B.The sandstone is under a higher pore pressure
3. C.Salt is always younger than sandstone
4. D.Salt deforms by creep and by dissolution and redeposition even at low temperatures and stresses
Show the answer
Correct answer: D. Salt deforms by creep and by dissolution and redeposition even at low temperatures and stresses
Halite (rock salt) is soft, the defects of its crystal lattice move easily, and it dissolves in pore water. So at geological strain rates it flows even at surface temperature. Quartz starts to flow only at about 300 °C, so sandstones at the depths of accumulations deform in a brittle way.
6
A rock deforms at a rate of $10^{-14}$ s⁻¹. How long will it take to shorten by 10%? Compare with a test on core at $10^{-5}$ s⁻¹.
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$t = e/\dot e = 0.1/10^{-14} = 10^{13}$ s, which is about 317 thousand years. In the laboratory it is $0.1/10^{-5} = 10^{4}$ s, less than three hours. At such different rates the rock behaves differently: in nature, creep and pressure solution have time to act.

Updated: 26 October 2026