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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

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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

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Types of Force

• Body forces– Fb ∝ m

• Surface forces– Fs ∝ area

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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

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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

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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

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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