1. Stress, Strain, Pressure, Deformation, Strength, etc.
Maurice Dusseault

2. Stresses (I)
Stresses in a solid sediment arise because of gravity and geological history
Stresses are different in diffferent directions
Three principal stresses are orthogonal, and the vertical direction is usually one of them
Overburden weight is = v (+/- 5%)
The lateral stresses, hmin and HMAX (or h and H) are at 90 degrees to one another

3. Stresses (II)
Normal stresses () act orthogonal to a plane and cause the material to compress
Shear stresses () act parallel to a plane and cause the material to distort
y-
yx
Static equilibrium:
xx
yy
xy = - yx
xy
x+
x -
xy
yx
y+

4. Pressures Pressures refer to the fluid potential (p)
Pressures can be hydrostatic, less than hydrostatic (rare) or greater (common). Called underpressured or overpresssured
Pressures at a point are the same in all directions because they are within the fluid
We assume that capillary effects are not important for large stresses and pressures
Differences in pressures lead to flow

5. Normally pressured range:
Pressures at Depth
pressure (MPa)
Fresh water: ~10 MPa/km
Sat. NaCl brine: ~12 MPa/km
~10 MPa
Hydrostatic pressure distribution: p(z) = rwgz
1 km
Underpressured case: underpressure ratio = p/(rwgz), a value less than 1
Overpressured case: overpressure ratio = p/(rwgz), a value greater than 1.2
underpressure
overpressure
Normally pressured range:
0.95 < p(norm) < 1.2
depth

6. Effective Stress () Principle
The famous Terzaghi concept (1921)
Only “effective” stresses [’ ] affect strength and deformation behaviour
Effective stress is the stress component transmitted through the solid rock matrix
Total stresses are the sum of the effective stresses and the pressures:  =  + p, or:
ij = []ij + [p]

7. Effective Stresses Pressure is the same in all directions (a fluid)po f2 f1 f3 f4
sv + po = sv (or Sv)
sh + po = sh (or Sh)
Pressure is the same in all directions (a fluid)
Effective stress is the sum of the grain-to-grain (matrix) forces
The sum of p and s gives total stresses, s
Usually, sv = r(z)dz
shmin , sHMAX must be measured or estimated

8. Rock Strength (I)
Strength is the resistance to shear stress (shear strength), compressive normal stress (crushing strength), tensile stress (tensile strength), or bending stress (beam strength).
All of these depend on effective stresses (), therefore we must know the pore pressure (p)
Rock specimen strength is usually very different than rock mass strength because of joints, bedding planes, fissures, etc.
Which to use? Depends on the problem scale.

9. Prepared rock specimen
Rock Strength (II)
Tensile strength (To) is extremely difficult to measure: it is direction-dependent, flaw-dependent, sample size-dependent, ...
To is used in fracture models (HF, thermal fracture, tripping or surge fractures)
For a large reservoir, To may be assumed to be zero because of joints, bedding planes, etc. F
Prepared rock specimen A F
To = F/A

10. Rock Strength (III)
Shear strength is a vital geomechanics strength aspect, often critical for design
Shearing is associated with:
Borehole instabilities, including breakouts, failure
Reservoir shear and induced seismicity
Casing shear and well collapse
Reactiviation of old faults, creation of new ones
Hydraulic fracture in soft, weak reservoirs
Loss of cohesion and sand production
Bit penetration, particularly PCD bits
n
slip plane
sn is normal effective stress
t is the shear stress  plane 
rock

11. Rock Strength (IV) a
slip planes
Shear strength depends on the frictional behaviour and the cohesion of the rock
Carry out a series of triaxial shearing tests at different 3, plot each as a stress-strain curve, determine peak strengths r r a
1 - 3
peak
strength
stress difference a
axial strain

12. Curves 1 - 3 a stress difference axial strain peak strength
1 - 3
Curves
peak
strength
massive damage,
shear plane develops
damage
starts
sudden stress drop (brittle)
cohesion
breaking
stress difference
continued damage
ultimate or
residual
strength
“elastic” part of  curve a
seating, microcrack closure
axial strain

13. Rock Strength (V)
To plot a yield criterion from triaxial tests, plot 1, 3 at failure on equally scaled n axes, join with a semicircle, then sketch tangent (= Y)  Y
cohesion c
n To
3
1

14. 1  3 Plotting Method
1
3
Uniaxial
compressive
strength, C0
tan2
Curved or
linear fit
Plot 1, 3 values at peak strength on axes
Fit a curve or a straight line to the data points
The y-intercept is the unconfined strength
Y = (1 - 3tan2 - C0 = 0)
(straight line approximation)

15. Rock Strength (VI)
Area - A
Strength of joints or faults require shear box tests
Specimen must be available and aligned properly in a shear box
Different stress values (N) are used
N - normal force S
Shear box
S - shear
force N S A
Linear “fit”
Curvilinear “fit”
data point N c
cohesion A

16. Yield Criterion
This type of plot is called a Mohr-Coulomb plot. Y is usually called a Mohr-Coulomb “yield” or “failure” criterion
It represents the shear strength of the rock (S/A) at various normal stresses (N/A), A is area of plane
For simplicity, a straight line fit is often used S A N
Linear “fit”
Curvilinear “fit”
data point
cohesion c
= t
= sn Y

17. Cohesion Bonded grains 1 - 3 Crystal strength Interlocking grains
Cohesive strength builds up rapidly with strain
But! Permanently lost with fabric damage and debonding of grains
1 - 3
complete s-e curve
cohesion
mobilization
stress difference
friction
mobilization a

18. mob = cohesion + friction f = c + n
Frictional resistance to slip between surfaces
Must have movement () to mobilize it
Slip of microfissures can contribute
Slip of grains at their contacts develops
Friction is not destroyed by strain and damage
Friction is affected by normal effective stress
Friction builds up more slowly with strain
mob = cohesion + friction f = c + n

19. Estimating Rock Strength
Laboratory tests OK in some cases (salt, clay), and are useful as indicators in all cases
Problems of fissures and discontinuities
Problems of anisotropy (eg: fissility planes)
Often, a reasonable guess, tempered with data, is adequate, but not always
Size of the structure (eg: well or reservoir) is a factor, particularly in jointed strata
Strength is a vital factor, but often it is difficult to choose the “right” strength value

20. Strength Anisotropy  UCS UCS Vertical core 0° 30° 60° 90°  
Bedding inclination

21. Crushing Strength (I)
Some materials (North Sea Chalk, coal, diatomite, high porosity UCSS) can crush
Crushing is collapse of pores, crushing of grains, under isotropic stress (minimal )
Tests involve increasing all-around effective stress (’) equally, measuring V/’
Tests can involve reducing p in a highly stresses specimen (ie: ’ increases as p drops)
UCSS = unconsolidated sandstone

22. Crushing Strength (II)
Apply p,  ( > p), allow to equilibrate ( =  - p)
Increase  by increasing  or dropping p  =  - p
Record volumetric strain, plot versus effective stress
The curve is the crushing behavior with + V  p    LE
crushing
material V

23. Rock Stiffness
To solve any ’-problem, we have to know how the rock deforms in response to a stress change
This is often referred to as the “stiffness”
For linear elastic rock, only two parameters are needed: Young’s modulus, E, and Poisson’s ratio, (see example)
For more complicated cases, more parameters are required

24. Rock Stiffness Determination
Stiffness controls stress changes
Estimate stiffness using correlations based on geology, density, porosity, lithology, ....
Use seismic velocities (vP, vS) for an upper-bound limit (invariably an overestimate)
Use measurements on laboratory specimens (But, there are problems of scale and joints)
In situ measurements (THE tool, others ...)
Back-analysis using monitoring data

25. What are E and ?  deformation L Young’s modulus (E): radial
E is how much the material
compresses under a change
in effective stress
radial
dilation r L L L
strain () =
Poisson’s ratio 
 is how much rock expands
laterally when compressed.
If = 0, no expansion (eg: sponge)
In  = 0.5, complete expansion,
therefore volume change is zero

E =  r
 = L

26. Rock Properties from Correlations
A sufficient data base must exist
The GMU must be properly matched to the data base; for example, using these criteria:
Similar lithology
Similar depth of burial and geological age
Similar granulometry and porosity
Estimate of anisotropy (eg: shales and laminates)
Correlation based on geophysical properties (KBES)
Use of a matched analogue advised in cases where core cannot be obtained economically
GMU = geomechanical unit

27. What is a Matched Analogue?
Rocks too difficult to sample without damage, or too expensive to obtain
Study logs, mineralogy, even estimate the basic properties (f, E, n…)
Find an analogue that is closely matched, but easy to sample for laboratory specimens
Use the analogue material as the basis of the test program
Don’t push the analogue too far!!

28. Seismic Wave Stiffness
vP, vS are dynamic responses affected by rock density and elastic properties
Because seismic strains are tiny, they do not compress microcracks, pores, or contacts
Thus, ED and D are always higher than the static test moduli, ES and S
The more microfissures, pores, point contacts, the more ED > ES, x 1.3, even to x 10
If porosity ~ 0,  very high, ES approaches ED

29. Laboratory Stiffness
Cores and samples are microfissured; these open when stress is relieved, E may be underestimated
In microfissured or porous rock, crack closure, slip, and contact deformation dominate stiffness
ES and S under confining stress are best values
Joints are a problem: if joints are important in situ, their stiffness may dominate rock response, but it is difficult to test in the laboratory

30. Cracks and Grain Contacts (I)E1 E2
Microflaws can
close, open, or
slip as  changes E1 E3
Flaws govern rock stiffness
The nature of the grain-to-grain
contacts and the overall porosity
govern the stiffness of porous SS

31. Cracks and Grain Contacts (II)
Point-to-point contacts are much more compliant than long (diagenetic) contacts
Large open microfissures are compliant
Oriented contacts or microfissures give rise to anisotropy of mechanical properties
Rocks with depositional structure or exposed to differential stress fields over geological time develop anisotropy through diagenesis

32. How Do We “Test” This Rock Mass?
Joints and fractures can be at scales of mm to several meters
Large  core: 115 mm
Core plugs: mm
If joints dominate, small-scale core tests are “indicators” only
This issue of “scale” enters into all Petroleum Geomechanics analyses
A large core specimen
A core “plug”
1 m
Machu Picchu, Peru, Inca Stonecraft

33. Induced Anisotropy
UCSS# subjected to large stress differences develops anisotropy (contacts form in 1 direction and break in 3 direction)
A brittle isotropic rock develops microcracks mainly parallel to 1 direction
Now, these rocks have developed anisotropy because of their -history (i.e.: damage)
This is a challenging area of analysis
# UCSS = unconsolidated sandstone

34. In Situ Stiffness Measurements
Pressurization of a packer-isolated zone, with measurement of radial deformation
Direct borehole jack methods (mining)
Geotechnical pressuremeter modified for high pressures (membrane inflated at high pressure, radial deformation measured)
Correlation methods (penetration, indentation, others?)
These are not widely used in Petroleum Eng.

35. Back-Analysis for Stiffness
Apply a known effective stress change, measure deformations (eg: uplift, compaction)
Use an analysis model to back-calculate the rock properties (best-fit approach)
Includes all large-scale effects
Can be confounded by heterogeneity, anisotropy, poor choice of GMU, ...
Often used as a check of assumptions

36. Direct Borehole Stability Problems
Stuck drill pipe
differential pressure sticking
wedging in the borehole
Stuck casing during installation
Lost circulation (in many cases)
Mudrings, cuttings build-up in washouts
Borehole squeeze

37. Indirect Stability Problems
Slow advance rates in drilling
Longer hole exposure = greater costs
Longer exposure = greater chance of instability
Solids build-up and loss of mud control
Blowouts
Washouts, sloughing cause tripping and drilling difficulties, swabbing
A blowout eventually develops as control is lost