Lecture 05 Force and Stress S05
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Transcript of Lecture 05 Force and Stress S05
1
Structural Geology
GLY 4400 - Lecture 5
Force and Stress –
Normal and Shear Stress
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Rocks and Force
• Rocks constantly experience the force of gravity• They may also experience a variety of other
forces, including tectonic forces and forces associated with impact
• Previously, we saw that force is defined by the following equation:– F = mA
– where F is the force vector, m is mass, and A is the acceleration vector
3
Responses to Force
• Rocks respond to applied forces in one of two ways:– A. Movement – Newtonian mechanics
– B. Distortion – continuum mechanics
Figure 3.2 in text
4
Types of Force
• Body forces– Fb ∝ m
• Surface forces– Fs ∝ area
5
Definition of Stress
• We have previously seen that stress is an internal force set up as the result of external forces acting on a body
• Stress is usually represented by the Greek letter sigma, σ– σ = F/Area
• Subscripts are often attached to σ to add additional information
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Stress in Different Dimensions
• In two dimensional problems, stress is a vector quantity, and is sometimes called traction
• In three dimensions, stress is a second-order tensor, which will be discussed shortly
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Traction
• Figure 3.3 in text
• Stress in an arbitrary direction may be resolved into components
• A. Normal stress, denoted σn
• B. Shear stress, denoted σs or τ (tau)
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Resolution of the Stress Vector
• Figure 3_4a illustrates the principle of stress resolution
• A plane face is ABCD in the drawing
• Note: The section through a cube implies a plane, i.e. two dimensions
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Force and Stress
• A force, F, is applied along rib AB• Line EF in the drawing is the trace of a plane which makes an angle θ with the
top and bottom surfaces of ABCD • The force can be resolved into components Fn perpendicular to the plane, and Fs
parallel to the plane
Figure 3.4a & b in text
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Fn and Fs
• σ = F/AB (Note: F = σAB)
• Fn = F cos θ = σAB cos θ
• Since AB = EF cos θ,
• Fn = σEF cos2θ
• Fs = F sin θ = σAB sin θ = EF sin θ cos θ
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Trigonometry Identity
• We can use the following trigonometric identity to simplify Fs
– sin θ cos θ = ½ (sin 2θ)
• Fs = σEF ½ (sin 2θ)
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Normal and Shear Stress
• σn = Fn/EF = σ cos2θ
• σs = Fs/EF = σ ½ (sin 2θ)
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Stress Vector Resolution
• Thus, the stress vector acting on a plane can be resolved into vector components normal and parallel to the plane
• Their magnitudes vary as a function of the orientation of the plane
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Normal Force and Stress vs. θ
• Plot of the normalized values of normal force and the normal stress versus theta
• The curves have a slightly different shape, but in both cases the normalized values decrease and go to zero at θ = 90º
Figure 3.4c in text
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Shear Force and Stress vs. θ
• The curves in this case are nearly identical until θ = 25º, then the shear force increases faster than the shear stress
• After 45º, the shear force continues to increase, but the shear stress again goes to zero at 90 º
Figure 3.4d in text
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Stress Ellipse
• Figure 3. 5a shows a plane cut by four other planes (a through d)
• The stresses on each plane are plotted, and are perpendicular to their respective planes
• Since the body is at rest, every stress is opposed by an equal an opposite stress
• We can connect the endpoints of the two dimensional stress vectors with a smooth curve, generating the ellipse shown
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Stress Ellipsoid
• If we were to draw similar ellipses in the two additional, mutually perpendicular, planes, we could then combine the data to generate a three dimensional ellipsoid, as shown in figure 3.5b
• This is known as the stress ellipsoid
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Principal Stresses
• Ellipsoids are characterized by three principal axes
• In the stress ellipsoid, these axes are known as the principal stresses
• Each principal stress is a vector
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Stress Using Cartesian Coordinates
• Stress can be visualized in another manner
• Using a standard three dimensional Cartesian coordinate system, and we picture a point cube, as shown in figure 3.6
• We can resolve the stress acting on each face of the cube into three components Figure 3.6 in text
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Stress Notation
• The face normal to the x-axis has a component σxx
• First subscript refers to the plane, in this case the one normal to the x-axis
• Second subscript refers to the component along axis x• In addition, we have two shear stresses, σxy and σxz, which lie
along the y and z axes within the plane under consideration
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Table of Stress Components
Stress on face normal to:
In the direction of:
x y z
Xx σxx σxy σxz
Yy σyx σyy σyz
Zz σzx σzy σzz
Doing the same for the other principal stress axes, we generate a table
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Stress Components
• The normal stress components are σxx, σyy, and
σzz
• The shear stress components are σxy, σxz, σyx, σyz, σzx, and σzz
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Equivalence of Shear Stress Components
• Since the object is at rest, three of the six shear stress components must be equivalent to the other three (otherwise the object would move) – σxy = σyx, σxz = σzx, and σyz = σzy
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Independent Stress Components
• This leaves six independent components:
Stress on face normal to:
In the direction of:
x y z
Xx σxx σxy σxz
Yy σxy σyy σyz
Zz σxz σyz σzz
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Principal Stress TableStress on face normal to:
In the direction of:
x y z
Xx σxx 0 0
Yy 0 σyy 0
Zz 0 0 σzz
Thus oriented, the axes are known as the principal axes of stress, and the planes perpendicular are the principal planes of stress
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Isotropic Stress
• It is possible that the three principal stresses will be equal in magnitude
• If this condition is met, the stress is said to be isotropic
• The stress ellipsoid becomes a sphere
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Anisotropic Stress
• When the principal stresses are unequal, they are said to be anisotropic
• We then introduce another convention:– σ1 σ2 σ3
• σ1 is called the maximum principal stress
• σ2 is the intermediate principal stress
• σ3 is the minimum principal stress
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Types of Stress
• General Triaxial Stress – σ1 > σ2 > σ3 … 0
• Biaxial stress, with one axis at zero – σ1 > 0 > σ3 or σ1 > σ2 > 0
• Uniaxial tension – σ1 = σ2 = 0; σ3 < 0
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Uniaxial Stress
• Uniaxial compression– σ2 = σ3 = 0; σ1 > 0
• Hydrostatic or lithostatic pressure– σ1 = σ2 = σ3
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Gabriel Auguste Daubrée
• Daubrée (1814-1896) was an early experimenter in many aspects of the geological sciences
• He taught mineralogy at the French School of Mines
• He introduced synthesis techniques and extended these to general experimental work
31
Daubrée Experiment
• For example, Figure 3-7a shows a picture of wax placed between two wooden plates, an experiment first performed by Daubrée
• He reported some of his results at the first International Geological Conference in 1878 (Paris)
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Diagram of Daubrée Experiment
• Plane AB is arbitrary, and it makes an angle θ with σ3
• We can make two other simplifying assumptions:– Line AB has unit length
– The plane represented by AB within the block has unit area
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Forces in Balance
• The forces parallel and perpendicular to AB must balance
• We resolve force AB into the component
of the force BC (parallel to σ1) along CD
plus the component of the force AC along CD (parallel to σ3)
• Note that ACD = θ
34
Resolving Forces
– forceBC = σ1cosθ (Force = stress area)
– forceAC = σ3sinθ
– AreaBC = 1 (cos θ)
– AreaAC = 1 (sin θ)
• On the AB surface, there is a normal stress, σn and a shear stress, σs
35
Normal Stress
• The normal stress is the same as the stress along CD:– σn = σ1cosθcosθ + σ3sinθsinθ = σ1cos2θ +
σ3sin2θ
• Since cos2θ = ½ (1 + cos2θ) and sin2θ = ½ (1 - cos2θ) we get– σn = ½ (σ1 + σ3) + ½ (σ1 - σ3) cos2θ
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Shear Stress
• We resolve force AB into the component
of the force BC along AB plus the
component of the force AC along AB– σs = σ1cosθsinθ - σ3sinθcosθ =
(σ1 - σ3) sinθcosθ
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Simplification of Shear Stress
• Substituting sinθcosθ = ½ sin2θ gives σs = ½(σ1 - σ3)sin2θ
• The planes of maximum normal stress are at θ = 0 relative to σ3, because cos2θ = 1 at θ = 0
• The planes of maximum shear stress are at 45 relative to σ3, because sin2θ = 1 at 45