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Lecture 7: Dynamic Analysis - Force and Stress in the Subsurface

Lecture Topics: Force Stress Lithostatic stress

McMaster University

McMaster University

Todays lecture: SGRR p. 98-119

Structural Analysis University

Analysis of geologic structures involves three successive steps:

1 . Descriptive or geometric analysis - quantitatively describe geometry of structures

2. Kinematic analysis - determine movements, changes in shape or strain

3. Dynamic analysis - determine direction, magnitude of forces and stresses

Dynamic Analysis - Stress McMaster University

Deformation (strain response) depends upon: magnitude and orientation of stresses imposed on rock time (strain rate) rock strength (next week)

e = (lfJ0)/lo

f t

Basic Definitions McMaster University

Stress - intensity of forces acting on rock body Strain - change in size or shape of a rock body resulting from applied forces

Stress

Strain

Force: Newton's First Law McMaster University

Force - 'push or pull' required to change the state of rest motion of a body object at rest is state of 'static equilibrium' all forces are balanced

Fi+F2+F3 = Fi + F2+F3

AF=o

Force: Newton's Second Law University

Force = mass x acceleration

F = ma

Units of Force Dimensions of force:

F = ma [M*LT-2]

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Basic unit of force is the Newton (N): force required to impart acceleration of 1 ms-2 to a body of 1 kilogram mass

1 N = 1 kgms-2 (SI units)

Types of Forces Body Forces - result from field acting on the entire mass

independent of surface forces gravity magnetic field

Surface Forces - forces produced by action of one body across surface of contact tectonic forces transmitted across a fault plane

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F

Forces in Geology McMaster University

Forces generated bJ plate tectonism - ridge spreading, subduction igneous intrusions meterorite impacts

Asihcnospfwre

Levin: Fig. 5-45 Ulhosphorc

Force: Vector Quantity McMaster University

Force is a vector quantity having both magnitude and direction obeys laws of vector addition/subtraction vector algebra

F

DIRECTION

Vector Addition/Subtraction University

Resultant vector can be found by adding and subtracting vector quantities

40 N 1 «

10 N

30 N

NET FORCE = F1 - F2 = 30 N

Parallelogram Rule McMaster University

Forces conform to rules of vector algebra resultant of any two vectors can be found graphically draw vectors tail to tail and find diagonal

FR is resultant vector of forces acting on point P

Vector Addition in 3-D McMaster University

Any force vector FR can be resolved into 3 principal components acting at right angles in a Cartesian co-ordinate system

FD = R Fx2+Fy

2+Fz2

Stress is the concentration of force 'intensity' of the applied force also known as 'traction' also a vector quantity

a = F_

per unit area

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Units of Stress Stress in Earth Science is usually measured in Pascals

• 1 pascal (Pa) = force of 1N acting on an area of 1 m2

• 1 Pa = 1 Nnr2 = 1 kgms_2*m"2 = 1 kgs_2rrr1

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F=1 N

1 m2

1 Pa= 1 N / m 2

Units of Stress university

1 kilopascal (kPa) = 1000 Pa (103 Pa)

1 megapascal (MPa) = 106 Pa

1 gigapascal (GPa) = 109 Pa

1 bar= 105Pa

Lithostatic Stress Vertical stress produced by mass of overlying crust

upper crust gradient is about 26.5 MPa/km mantle gradient is approx 35 MPa/km

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1 kbar=10 8 Pa 10kbar= 1 GPa

Lithostatic Stress Problem Calculate the lithostatic stress at base granite cube 1000 m on a

side with a density of 2700 kgm-3

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1000 m < >

Density = 2700 kgnr3

1000 m

A

GRANITE 1000 m

Lithostatic stress?

Mass of granite cube

m = p x V = 2700 kgrrr3x (1000 m)3

= 2.7x1012kg

Force at base of cube

F = m xa = 2.7x 1012 kg x 9.8 ms-2

= 2.65x1013kgms-2

= 2.65x1013N

000 m

1000 m

26.5 MPa

Stress at base of cube: a = F / A

= 2.65 x 1013 HI (1000 m x 1000Tn) = 2.65x107 Pa(Nm-2) = 26.5 MPa

The granite block exerts a force of 26.5 MPa

000 m

1000 m

I 26.5 MPa

ithostatic Stress Gradient McMaster University

Note that we can also write the lithostatic stress in terms of depth, z:

a F/A mg/A

Vpg/A Azpg/A PQZ

(m = Vp)

where p = density g = gravitational acceleration z = depth

Example McMaster University

Calculate lithostatic stress at the base of the continental crust at depth of 40 km?

a = pgh 2700 kgnr3 x 9.8 ms"2 x 40,000 m 1.05 x10 9 Pa

= 1.0GPa

! c

Temperature in C 400 600

Stress Components A stress acting on any surface (arbitrarily oriented plane) can be

resolved into two components normal stress, on (sigma) - stress acting normal to plane shear stress, x (tau) - stress acting tangential to plane a.k.a. traction

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O" n

v. T = 0

Stress Components McMaster University

In case of static equilibrium, all stresses balanced by equal opposite forces case for most structural problems

+ a + a n

- a n

n + T

Stress Sign Convention compressive stress is positive tensile stress is negative

a n

v *

O" n

Compressive Stress - Positive

a n

Tensile Stress - Negative

Stress Sign Conventions Sign of shear stresses indicated by 'sense' of motion

clockwise or 'right-handed' shear is negative counterclockwise or 'left-handed' shear positive

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Clockwise Shear - Negative (Right-handed)

Counterclockwise Shear - Positive (Left-handed )

What is Sign of Shear? McMaster University

Coordinate System McMaster University

Orientation of stress referenced to 3-dimensional Cartesian coordinate system x, y in horizontal plane z is vertical by convention

Z(Up)

Y (North) X (East)

Principle Stress Axes McMaster University

Principle axes are directions along which shear components are zero normal stress at a maximum, no shear

Z(Up)

Y (North) X (East)

Stress Acting on a Plane Stress acting at an angle to 2-D plane will generate two sh

stress components and one normal component

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C7

Stress Ellipsoid The total stress field acting on a point can be represented by the

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stress ellipsoid:

<x

a.| - greatest principal stress

a 2 - intermediate principal stress direction

a 3 - least principal stress

triaxial stress is general case where <JA> <52> a3

Ellipsoid

2-D Stress Ellipse McMaster University

Generally we simplify problems by dealing with stress within a single plane ♦ planes containing ♦ a1 and a3

♦ a1 and a2

♦ a2 and a3

Calculation of Stress Components Calculate the normal (an) and shear stress (x) components for a stress of 50 MPa inclined at 60° to the XY plane

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(Txz=50MPa

^^>- -x T=?

McMaster University Calculation of Stress Components

Normal and shear stress components can be calculated by solving for the lengths of the vectors

an = sin 6 • GX2

= sin 60 ■ 50 Mpa

= 43.3 MPa

T = cos 9 ■ a xz

= cos 60 ■ 50 Mpa

= 25 MPa

n a =43.4 MPa

T =25 MPa