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Dynamic Analysis
Stress 1
Geol
Reading: Chapter 3
Outline
- Units
- Normal, shear components
- Pressure
- Lithostatic Stress
- Stress Tensor
- Resolving stress on a plane
Strain = deformation
Stress = cause of deformation
Force
F= Mass. acceleration
kg .m/s^2
Newtons
A normal person can throw a bowling ball at 90 N
Resolving a Force Vector
Fy= F. SinA
F
A
Fx= F. CosA
y
x
Normal
Shear
Does applied force alone control
deformation?
F
F
Land’O Lakes
2
Stress
Stress = force / area
Units = Pascals
= N/m^2
=kg/(m. s^2 )
Stress vs. Pressure
- Units of pressure = force/area
- 33 psi = 33 pounds per square inch
What is the difference?
- Pressure - Scalar quantity
- Stress - Tensor quantity
How much is 1 Pascal?
- 1 Pa= 0.00014 psi
- Car tire ~200,000 Pa
- 1 Megapascal (1 MPa) = 1 million Pa
= 1 x 10 6 Pa
Calculate State of
Stress at 1000 m depth
Lithostatic Stress
- Similar to hydrostatic pressure
- Magnitude of stress components is the same
in all directions
4
Resolving stress on a plane
Must first correct for the change in the area
x
z
σxx
σzz
a
σzz Sin a
σxx Cos a
Then, you can resolve the vectors into normal and shear components
Example
σxx= 45 Mpa
σzz = 26 Mpa
x- north
z- up
What are the normal and shear stresses acting
on a fault oriented 090/40S?
Step 1: Understand the geometry
up
σxx
σzz
N
Dip=
Cross Sectional View
S
fault
Map View
North
fault
Step 2: Find Magnitude of stress components along x and z
z
x
a= 90-dip = 50 deg.
σfx = σxx Cos a
= 45 MPa Cos 50
= 28.9 Mpa
σfz = σzz Sin a
= 26 MPa Sin 50
= 19.9 Mpa
σxx
σzz
a
F
σfx
σfz
Step 3: Add the two vectors
z
x
σf^2 = σfx^2 + σfz^2
σf = sqrt{28.9 2 +19.9 2 } Mpa
σf = 35.1 Mpa
b= arctan (σfz /σfx )
b= 34.5 deg.
σfx
a
F
σfz
σf
b
Step 4: Resolve σf into normal and
shear components
z
x
c = a-b = 50- 34.5 = 15.
σn = σf Cos c
= 35.1 Mpa Cos 15.
= 33.8 Mpa
σs = σf Sin c
= 35.1 Mpa Sin 15.
= 9.32 MPa
σn
a
F
σf
b
σs
c
5
Magnitude of
Normal Force
and
Normal Stress
As a function of
Angle Θ
σn Max
σn = σ Cos 2 Θ
Magnitude of
Shear Force
and
Shear Stress
As a function of
Angle Θ
σs Min σs Min
σs = σ ½ Sin 2 Θ
Stress Components
x
z
σxx σxx
σzz
σzz
σxz σxz
σzx
σzx
Principal Stress Components You always can find an orientation of the reference cube where there won’t be any resolved shear stresses
x
z σ^1
σ 1
σ 3
σ 3
Most compressive
Least compressive
Stress
components
Stress Ellipse- Normal Stress
σ 3
σ 3
σ 1
σ 1
Maximum Normal Stress
Minimum Normal Stress
