2 Stress Strain 3425

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```					   Stress and strain
Can cause you pain,
Take this course
To save remorse.

Kutna Hora,
Czechoslovakia - building
damaged by subsidence
due to mining (photo from
Legget, 1973)
Stress and Strain
•stress and strain

•compressive, tensile and shear

•balance of force

•behavior of rock material under stress

•analysis of stress - Mohr’s circle

•strength of rock and soil - angle of internal friction
and cohesion

•testing of rock and soil
Stress = force / area
Strain = deformation (percent of original)
Types of Stress ( Monroe and Wicander, 1998)

Compressive Stress

Tensile Stress

Shear Stress
Compression shown in road cut, Palmdale, California
(Tarbuck and Lutgens, 1999)
Marble slab broken in tension, Rock Creek Cemetery,
Washington, D.C. ( Monroe and Wicander, 1998)
Elastic
Plastic
Elastic Limit
Rupture
Brittle
Strength

Monroe and Wicander, 1998
Stress
Force = m*a = m*g
Stress = force/area
(kg m/sec2 = 1 Newton which is force;
Stress, Pa
F/Area = kg/m sec2 = 1 Pa = pressure)
g= 9.8 m/sec2
γ = unit weight = ρr g
Depth, m

For homogeneous, dry rock:
Lithostatic stress = σc = ρrgz = γz
Example: Suppose rock has a density of ρr =
3.0 g/cm3. What is the stress at 10 m depth?
Typical values of ρ (from Rahn, p. 76)

Material           Density (g/cm3)
Water               1.0
Tuff                2.1
Sandstone           2.2
Limestone           2.5
Granite             2.6
Basalt              2.7
Gneiss              2.7
A generic value of ρ = 2.7 is often used for dry bulk soil
density [density of the soil mass (solids plus voids)] if more
specific values are not available.
Hydrostatic pressure
Example: The density of the soil is 2.5
g/cm3 and the density of water is 1.0
20 m
g/cm3. Assume that the water table is at
the ground surface. What is the dry
lithostatic stress at a depth of 20 m?
What is the hydrostatic pressure?

Total lithostatic stress – stress due to              Stress, Pa
rock and water
What is the total lithostatic stress?
Depth,
m

Note: if the water table is at some depth,
d, then the hydrostatic pressure is
σw = ρwg (z-d)
Effects of strata – the stress due to the weight of each
geologic layer must be calculated separately, then added together
from the top down

Example: The following geology is present at a site:
Stress, Pa

10 m, ρr = 2.0 g/cm3

20 m, ρr = 3.0 g/cm3
Depth,
m
10 m, ρr = 4.0 g/cm3
•Balance of force - if the forces on an object are not balanced in
each direction, the object will accelerate
•Balance of compressive, tensile, and shear forces are
necessary
•Balance of moments – if the moments around a point are not
balanced, the object will rotate
beam

Symbols:
- the support is unable to move vertically or horizontally but
is able to rotate
- the support is unable to move vertically or horizontally,
unable to rotate
- support is unable to move vertically, may move horizontally
and is able to rotate
F1
Balance of forces: forces in
up to zero in order to
prevent motion

Example: Vertical forces
Σ Fy = 0
F1 + F2 = 0
F1 = - F2
F2
To prevent motion:                                 P
•forces in any one direction must add
up to zero
P/2                P/2
Fx = 0, Fy = 0, Fz = 0

y
x
z
force vectors must be resolved and                         Use right hand
their horizontal and vertical                               rule for
coordinate
components must add up to zero                              directions
Pv = P sinθ                          P
Pv
Ph = P cosθ                          θ

Ph
Example: Resolve the force of 20 N
acting at angle of 35o from the horizontal.

F = 20N        Fv

35o

Fh
Example : Suppose a bar is supported at two ends. There is a force
of 50 N downward at the center of the bar (length =L). What are the
forces at either end of the bar supporting it?
Fy = 0;    F1 + F2 + P = 0; F1 + F2 = -P
by symmetry    F1= F2 = -P/2

Fx = 0; there are no forces in the x direction
Fz = 0; there are no forces in the z direction

P

F1                          F2
W =N*L

L
Fa                                 Fb

Example: Find the unknown support forces in the beam .
Total force on the beam w= N*L     (could be the weight of the beam)
 Fx = 0 - there are no horizontal forces
 Fy = 0 ; Fa + Fb = total downward force
Fa + Fb = NL

note: assume that the distributed weight acts at the centroid of the beam - halfway
Fb = NL
2
using  Fy= 0, Fa = NL
2
Moments about a point in a beam are calculated by multiplying
the distance times the perpendicular component of force

L1               L2             L3

F1         point A         F2                F3

Around point A, the bending moment is:
MA = F1 * L1 + F2* L2 - F3 * (L2 + L3)

Note that the direction of rotation is important in the equation
To prevent rotation around a point, moments about the point have to

Example : 10 Pa is applied in the center of a beam of length, L. Then
Fy = 0; F1 + F2 + 10 Pa = 0; F1 + F2 = -10 Pa

•Instead of finding the individual forces by symmetry, use the concept of
•The support around point 1 is free to rotate so the support cannot add
any bending moment. Therefore the sum of the moments about point
one are the bending moment from the force, 10 Pa, and the bending
moment from the force, F2.

10 Pa * L/2 – F2 * L = 0
10 Pa
F2 = 10 Pa / 2 = 5 Pa
and F1 = 5 Pa
F2
F1
Unconfined compression test
(uniaxial compression test)
- No lateral pressure
- Vertical force (stress) is applied until
failure
- Vertical strains measured
Shear and Normal Stress
In compression, the failure from an applied
force may be due to compressive stress but
is often due to shear stress

Normal stress - applied force normal to a     Normal
hypothetical or real surface of the rock      stress

core divided by the area of the end
Shear
stress
Shear stress – component of force parallel
to hypothetical or real surface divided by
the area of that surface
Consider a rock core under a compressive stress, F and area of
end, Ao.
FS
The forces must balance. The stresses do not necessarily balance.
Now consider area of the section at angle θ
θ
F
Hyp
Hyp * cos θ = adj

A =Ao / cos θ                                                                FN

Now consider the force:                                              F
FN = cosθ;    FN = F cosθ         - normal force                         FS
F
FS = sin θ;   FS = F sinθ         - shear force
F                                                                    θ

N = FN = F cosθ;           N = F cosθ cos θ
A     A                   Ao
S = FS = F sin θ;          τ = S = F sinθ cosθ
A      A                       Ao
τ = F sin2θ
2Ao
Example: Suppose a stress of 500 kN is applied
evenly over the ends of a rock core sample with the
ends having an area of 0.5 m2.
What is the shear and normal stress on a plane 30o
500 kN
from the horizontal?
30o

σN = 500 kN cos2 30o = 750 kPa
0.5 m2

τ = 500 kN sin (2* 30o ) = 433 kPa
2 * 0.5 m2
Triaxial Compression Test
-Soil encased in membrane and subjected to
lateral stress from water or gas

-Vertical load is applied until failure

-Test repeated for several confining stresses

-Results plotted and analyzed for Cohesion
and angle of friction
3-D and 2-D states of stress
Principle stress - the stress state at any point in the earth can be resolved into
stress vectors in any three perpendicular directions. There is one orientation
where there is zero shear stress. The stress state in this direction consists of the
principal stresses:
Maximum, minimum, and intermediate stresses are all perpendicular to
each other
σ1 = maximum stress (compression)
σ2 = intermediate stress
σ3 = minimum stress
The orientation of the maximum stress is called the principal stress
direction
Common tangent
υ, angle of              to all circles
internal friction
Shear stress, τ
Shear stress at failure

c, cohesion
σ2                    σ1   normal stress

three different stress states

c = unit cohesion
values can be 0 with clean sand; 1000 - 6000
N/m2 are common

υ = angle of internal friction
= angle of repose - 30o - 40o is typical
τ                                   τ
τ, at failure
τ, at failure

σ                                  σ

No cohesion                             No friction

τ
τ, at failure          Examples
σ

Cohesion and friction
Coulomb's Law or the Mohr-Coulomb Law – a
description of the shear strength of the material

τfriction = c + σN tan υ
More Mohr’s Circle:

τ       υ

c

σ2      σ1      σN

Note that the radius of the circle (maximum
shear stress) is
r = 1 - 2
2
and the normal stress at the center of the circle is
center = 1 + 2
2
B
τ

         2
A σ2         C        D       σ1       σN

What is the shear and normal stress at some angle, , from the maximum
principal stress direction?

Point C is at the center of the circle and CB = r = 1 - 2
2

CD = CB* cos 2 = 1 - 2 cos 2               N,B = 1 + 2 + 1 - 2 * cos 2
2                                    2          2
= 2 + (1 - 2) cos2 

 = DB = CB* sin 2 = 1 - 2 sin 2
2
The shear stress at any angle, , is  = 1 - 2 sin 2
2

Conjugate sets - pairs of discontinuity sets perpendicular to each
other and 45o from the maximum compressive stress direction

Example: Suppose σ1 = 20 MPa and σ3 = 10 MPa on a rock core.
What is the maximum shear stress and what is its orientation?
What is the shear stress on a surface inclined 40o to σ3?
What is the normal stress on this surface?
Stress in saturated soil

total vertical stress = vertical solid particle stress plus water
pressure

total horizontal stress = horizontal solid particle stress plus
water pressure

effective stress = total vertical stress minus buoyancy of
water

buoyancy of water = hydrostatic stress
Effective stress σ΄ = σTotal - w

water pressure underground supports part of
the weight of the rock and of water

•Deformation of the rock or soil is only dependent
on the stress in the rock skeleton, the effective
stress

•Water cannot support shear stress; shear
strength of rock and soil is based on the effective
stress
Example: Suppose a vertical cliff of sand is to be
constructed. The sand has the following properties:
Φ = 32o
c=0
ρ 1 = 1.9 g/cm3 (above water table- moist sand)
ρ 2 = 2.9 g/cm3 (below water table – saturated sand
including water)

7m   Water table

sand           20 m

τ resistance

What is the effective stress and the shear resistance in
the sand 20 m below the top of the cliff?
Example: Consider stress across a fracture within the
ground. Assume saturated conditions. The shear strength
within the fracture is:
τf = c + ( σN - σW)tan υ

τf= c + σN ΄tan υ
σTotal

σW

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