Stress and Strain: Basic Terms and Concepts
Stress and Strain: Important terms
In
traditional geology the unit of pressure is the bar, which is
about equal to atmospheric pressure. It is also about equal to the
pressure under 10 meters of water.
For pressures deep in the earth we use the kilobar, equal to 1000 bars. The pressure beneath 10 km of water, or at the bottom of the deepest oceanic trenches, is about 1 kilobar. Beneath the Antarctic ice cap (maximum thickness about 5 km) the pressure is about half a kilobar at greatest.
For pressures deep in the earth we use the kilobar, equal to 1000 bars. The pressure beneath 10 km of water, or at the bottom of the deepest oceanic trenches, is about 1 kilobar. Beneath the Antarctic ice cap (maximum thickness about 5 km) the pressure is about half a kilobar at greatest.
In
the SI System, the fundamental unit of length is the meter and mass
is the kilogram.
Important units used in geology include:
Important units used in geology include:
 Energy: Joule: kgm^{2}/sec^{2}. Five grams moving at 20 meters per second have an energy of one joule. This is about equal to a sheet of paper wadded up into a ball and thrown hard.
 Force: Newton: kgm/sec^{2}. On the surface of the Earth, with a gravitational acceleration of 9.8 m/sec^{2}, a newton is the force exerted by a weight of 102 grams or 3.6 ounces.
 Pressure: Pascal = Newton/m^{2} or kg/msec^{2}.
 A newton spread out over a square meter is a pretty feeble force.
 Atmospheric pressure is about 100,000 pascals. A manila file folder (35 g, 700 cm^{2} area) exerts a pressure of about 5 pascals.
By
comparison with traditional pressure units, one bar = 100,000
pascals. One megapascal (Mpa) equals 10 bars, one Gigapascal (Gpa)
equals 10 kilobars.
Using Units in Calculations
The
fundamental rule in using units in calculations is that units obey
the same algebraic rules as other quantities
Example: Converting Traditional Density to SI density
Density
is conventionally represented as grams per cubic centimetre. How do we represent density in the SI system?
1
gram/cm^{3} = (0.001 kg)/(.01 m)^{3} = 10^{3}
kg/10^{6} m^{3} =
1000 kg/m^{3}
1000 kg/m^{3}
Thus,
to convert traditional to SI density, multiply by 1000. Thus,
2.7 gm/cm^{3} = 2700 kg/m^{3}, etc.
Example: Pressure Beneath a Stone Block
What's
the pressure beneath a granite block 20 meters long, 15 meters wide
and 10 meters high, with density 2.7 gm/cm^{3}?
First,
we find the mass of the block. Mass is volume times density
or 20 x 15 x 10 m^{3} x 2700 kg/m^{3} = 8.1 x 10^{6} kg.
Note that we have m^{3} times kg/m^{3}, and the m^{3} terms cancel out to leave the correct unit, kilograms.
or 20 x 15 x 10 m^{3} x 2700 kg/m^{3} = 8.1 x 10^{6} kg.
Note that we have m^{3} times kg/m^{3}, and the m^{3} terms cancel out to leave the correct unit, kilograms.
Now
the force the block exerts is given by mass times acceleration, in
this case the acceleration of gravity, or 9.8 m/sec^{2}.
Thus the force the block exerts is 8.1 x 10^{6} kg x 9.8 m/sec^{2}, or 7.9 x 10^{7} kgm/sec^{2}.
Referring to the SI units listed above, we see that these are indeed the correct units for force. The block exerts 7.9 x 10^{7} newtons of force on the ground beneath it.
Thus the force the block exerts is 8.1 x 10^{6} kg x 9.8 m/sec^{2}, or 7.9 x 10^{7} kgm/sec^{2}.
Referring to the SI units listed above, we see that these are indeed the correct units for force. The block exerts 7.9 x 10^{7} newtons of force on the ground beneath it.
The
pressure the block exerts is force divided by area, or 7.9 x 10^{7}
newtons/(20 m x 15 m) = 265,000 pascals (verify that the units are
correct). This is only 2.65 bars, the pressure beneath 27 meters of
water. Scuba divers can stand that pressure easily, but nobody would
want to lie under a tenmeter thick slab of rock. This should bother
you.
It
should be intuitively obvious that the pressure will be the same
regardless of the area of the block. Can you show why this is so?
Conversion Factors
Often
students find it hard to decide whether to multiply or divide by a
conversion factor. For example, one meter = 3.28 feet. To convert 150
feet to meters, do you multiply or divide by 3.28?
If
you think of the conversion factor as merely a number, it can be a
puzzle. But consider:
1
meter = 3.28 feet. Therefore 1 m/3.28 feet = 1 and 3.28 feet/1 m = 1
Conversion
factors are not just numbers, but units too. Every conversion
factor, with units included, equals unity. That part about
including units is allimportant. So, given a conversion problem, use
the conversion factor to eliminate unwanted units, produce desired
units, or both.
To
convert 150 feet to meters, we want to get rid of feet and obtain
meters. The conversion factor is 3.28 feet/1 m. Multiplying gives us
492 feet^{2}/m ^{2}. It's perfectly correct  it
might be a valid part of some other calculation  but not what we
need here. We need to get rid of feet and obtain meters, which means
we need meters in the numerator (upstairs) and feet in the
denominator (downstairs).
150
feet x 1m/2.38 feet = 45.7 meters. Feet cancel out, leaving us with
only meters.
A
more complex example: convert 10 miles per hour to meters per second.
Here, none of the units we want in the final answer are
present in the initial quantity. But we know:
 1 mile = 5280 feet
 1 meter = 3.28 feet
 1 hour = 60 minutes
 1 minute = 60 seconds
We
want to get rid of miles and hours and get meters and seconds. So we
want our conversion factors to eliminate miles and hours:
10
mi/hr x (5280 feet/1 mi) x (1 hr/60 min)
Also,
we want our end result to be in meters/second so at some point we
will have to have
Something
x (1m/3.28 feet) x (1 min/60 sec) This is the only way to get
m/sec using the conversion factors given. We will, of course, have to
get rid of the feet and minutes somehow.
Putting
it all together we get
10
mi/hr x (5280 feet/1 mi) x (1 hr/60 min) x (1m/3.28 feet) x (1
min/60 sec) = 4.47 m/sec
Miles
cancel, hours cancel, feet cancel, minutes cancel, and we end up with
m/sec, just what we needed.
Some
people prefer to use a grid arrangement as shown below:
10 miles  5280 feet  1 m  1 hour  1 min  =  4.47 m 
1 hour  1 mile  3.28 feet  60 min  60 sec  1 sec 
In
this example we get rid of miles and feet to get meters first, then
we get rid of hours and minutes to get seconds.
Stress Terms
Stress
is defined as force per unit area. It has the same units as pressure,
and in fact pressure is one special variety of stress. However,
stress is a much more complex quantity than pressure because it
varies both with direction and with the surface it acts on.
Compression
Stress
that acts to shorten an object.
Tension
Stress
that acts to lengthen an object.
Normal Stress
Stress
that acts perpendicular to a surface. Can be either Congressional or tensional.
Shear
Stress
that acts parallel to a surface. It can cause one object to slide
over another. It also tends to deform originally rectangular objects
into parallelograms. The most general definition is that shear acts
to change the angles in an object.
Hydrostatic
Stress
(usually compressional) that is uniform in all directions. A scuba
diver experiences hydrostatic stress. Stress in the earth is nearly
hydrostatic. The term for uniform stress in the earth is lithostatic.
Directed Stress
Stress
that varies with direction. Stress under a stone slab is directed;
there is a force in one direction but no counteracting forces
perpendicular to it. This is why a person under a thick slab gets
squashed but a scuba diver under the same pressure doesn't. The scuba
diver feels the same force in all directions.
In
geology we never see stress. We only see the results of stress as
it deforms materials. Even if we were to use a strain gauge to
measure insitu stress in the rocks, we would not measure the stress
itself. We would measure the deformation of the strain gauge (that's
why it's called a "strain gauge") and use that to
infer the stress.
Strain Terms
Strain
is defined as the amount of deformation an object experiences
compared to its original size and shape. For example, if a block 10
cm on a side is deformed so that it becomes 9 cm long, the strain is
(109)/10 or 0.1 (sometimes expressed in percent, in this case 10
percent.)
Note that strain is dimensionless.
Note that strain is dimensionless.
Longitudinal or Linear
Strain
Strain
that changes the length of a line without changing its direction. Can
be either compressional or tensional.
Compression
Longitudinal
strain that shortens an object.
Tension
Longitudinal
strain that lengthens an object.
Shear
Strain
that changes the angles of an object.
Shear causes lines to rotate.
Shear causes lines to rotate.
Infinitesimal Strain
Strain
that is tiny, a few percent or less. Allows a number of useful
mathematical simplifications and approximations.
Finite Strain
Strain
larger than a few percent. Requires a more complicated mathematical
treatment than infinitesimal strain.
Homogeneous Strain
Uniform
strain. Straight lines in the original object remain straight.
Parallel lines remain parallel. Circles deform to ellipses. Note that
this definition rules out folding, since an originally straight layer
has to remain straight.
Inhomogeneous Strain
How
real geology behaves. Deformation varies from place to place. Lines
may bend and do not necessarily remain parallel.
Terms for Behavior of Materials
Elastic
Material
deforms under stress but returns to its original size and shape when
the stress is released. There is no permanent deformation. Some
elastic strain, like in a rubber band, can be large, but in rocks it
is usually small enough to be considered infinitesimal.
Brittle
Material
deforms by fracturing. Glass is brittle. Rocks are typically brittle
at low temperatures and pressures.
Ductile
Material
deforms without breaking. Metals are ductile. Many materials show
both types of behavior. They may deform in a ductile manner if
deformed slowly, but fracture if deformed too quickly or too much.
Rocks are typically ductile at high temperatures or pressures.
Viscous
Materials
that deform steadily under stress. Purely viscous materials like
liquids deform under even the smallest stress. Rocks may behave like
viscous materials under high temperature and pressure.
Plastic
Material
does not flow until a threshhold stress has been exceeded.
Viscoelastic
Combines
elastic and viscous behavior.
Models of glacioisostasy frequently assume a viscoelastic earth: the crust flexes elastically and the underlying mantle flows viscously.
Models of glacioisostasy frequently assume a viscoelastic earth: the crust flexes elastically and the underlying mantle flows viscously.
Stress and Young's Modulus
Stress is force per area  strain is deformation of a solid due to stress
Stress
Stress is the
ratio of applied force F and cross section A, defined
as "force per area".
Direct Stress or Normal Stress
Stress normal to
the plane is usually denoted "normal stress" and can
be expressed as
σ = F_{n}
/ A (1)
where
σ = normal
stress ((Pa) N/m^{2}, psi)
F_{n}
= normal component force (N, lb_{f }(alt.
kips))
A = area (m^{2},
in^{2})
 a kip is a nonSI unit of force  it equals 1,000 poundsforce
 1 kip = 4448.2216 Newtons (N) = 4.4482216 kilonewtons (kN)
Example  Tensile Force acting on a Rod
A force of 10 kN
is acting on a circular rod with diameter 10 mm. The stress in the
rod can be calculated as
σ = 10 10^{3}
(N) / (π (10 10^{3} (m) / 2)^{2})
=
127388535 (N/m^{2})
=
127 (MPa)
Shear Stress
Stress parallel
to the plane is usually denoted "shear stress" and
can be expressed as
τ = F_{p}
/ A (2)
where
τ = shear
stress ((Pa) N/m^{2}, psi)
F_{p}
= parallel component force (N, lb_{f})
A = area (m^{2},
in^{2})
Strain
Strain is
defined as "deformation of a solid due to stress" and can
be expressed as
ε = dl / l_{o}
= σ / E (3)
where
dl = change
of length (m, in)
l_{o}
= initial length (m, in)
ε = unit less measure of engineering strain
E = Young's
modulus (Modulus of Elasticity) (Pa, psi)
Example  Stress and Change of Length
The rod in the
example above is 2 m long and made of steel with Modulus of
Elasticity 200 GPa. The change of length can be calculated by
transforming (3) as
dl = σ
l_{o }/ E
=
127 10^{6} (Pa) 2 (m) / 200 10^{9}
(Pa)
=
0.00127 (m)
=
1.27 (mm)
Young's Modulus  Modulus of Elasticity (or Tensile Modulus)  Hooke's Law
Most metals have
deformations that are proportional with the imposed loads over a
range of loads. Stress is proportional to load and strain is
proportional to deformation expressed by the Hooke's law like
E = stress /
strain = (F_{n} / A) / (dl / l_{o})
(4)
where
E = Young's
modulus (N/m^{2}) (lb/in^{2},
psi)
Modulus of
Elasticity or Young's Modulus are commonly used for metals and metal
alloys and expressed in terms 10^{6}
lb_{f}/in^{2}, N/m^{2}
or Pa. Tensile modulus are often used for plastics and expressed
in terms 10^{5} lb_{f}/in^{2}
or GPa.
Shear Modulus
S = stress /
strain = (F_{p} / A) / (s / d)
(5)
where
S = shear
modulus (N/m^{2}) (lb/in^{2},
psi)
F_{p}
= force parallel to the faces which they act
A = area
(m^{2}, in^{2})
s =
displacement of the faces (m, in)
d = distance
between the faces displaced (m, in)
Elastic Moduli
Material  Young's Modulus  Shear Modulus  Bulk Modulus  
10^{10} N/m^{2}  10^{6} lb/in^{2}  10^{10} N/m^{2}  10^{6} lb/in^{2}  10^{10} N/m^{2}  10^{6} lb/in^{2}  
Aluminum  7.0  10  2.4  3.4  7.0  10 
Brass  9.1  13  3.6  5.1  6.1  8.5 
Copper  11  16  4.2  6.0  14  20 
Glass  5.5  7.8  2.3  3.3  3.7  5.2 
Iron  9.1  13  7.0  10  10  14 
Lead  1.6  2.3  0.56  0.8  0.77  1.1 
Steel  20  29  8.4  12  16  23 
Stress and Strain
Up
to now we have been studying the dynamics of rigid
bodies, that is, idealised objects that have a definite size and shape, but one in
which the particles making up the object are constrained so that the
relative positions of the particles never changes. In other words,
the rigid body does not ever stretch, squeeze or twist.
However, we know that in reality this does occur, and we need to find a way to describe it. This is done by the concepts of stress, strain and elastic modulus.
Stress is a measurement of the strength of a material,
strain is a measure of the change in the shape of the object that is undergoing stress and
elastic modulus is a measurement of the amount of stress needed to change the shape of the object.
There are three main types of stress.
If we stretch or compress an object, we are subjecting it to a tensile stress.
If an object is subjected to a force along an entire surface, changing its volume, then it is said to be
experiencing a bulk stress.
Finally, if the force is acting tangentially to the surface, causing it to twist, then we are subjecting it to a shear stress.
However, we know that in reality this does occur, and we need to find a way to describe it. This is done by the concepts of stress, strain and elastic modulus.
Stress is a measurement of the strength of a material,
strain is a measure of the change in the shape of the object that is undergoing stress and
elastic modulus is a measurement of the amount of stress needed to change the shape of the object.
There are three main types of stress.
If we stretch or compress an object, we are subjecting it to a tensile stress.
If an object is subjected to a force along an entire surface, changing its volume, then it is said to be
experiencing a bulk stress.
Finally, if the force is acting tangentially to the surface, causing it to twist, then we are subjecting it to a shear stress.
Tensile Stress
Consider
a bar of cross sectional area A
being subjected to equal and opposite forces F
pulling at the ends. If this were a rope, we would say that it is
experiencing a tension force. Taking this concept over, we say that
the bar is under tension,
and is experiencing a stress that we define to be the ratio of the force to the cross sectional area
and is experiencing a stress that we define to be the ratio of the force to the cross sectional area
Stress
= F/A

(62)

This stress is called the tensile stress because every part of the object is subjected to a tension.
The SI unit of stress is the Newton per square meter, which is called the Pascal
1
Pascal = 1 Pa = 1 N/m^{2}
Example:
A 250 kg bob is attached to a steel cable with a diameter of 0.05 m. If we take the cable to be essentially massless, what is the tensile stress experienced by the cable?
A 250 kg bob is attached to a steel cable with a diameter of 0.05 m. If we take the cable to be essentially massless, what is the tensile stress experienced by the cable?
The stress is just the force
divided by the area
If
the bar is being pressed instead of pulled, then we say that it is
undergoing compressive
stress instead of
tensile stress.
Tensile Strain
The
fractional amount that an object stretches when it is subjected to a
tensile stress is called the tensile strain. Mathematically, we write
this as
(63)

where
l_{0}
is the original unstressed length of the bar.
Elastic Modulus
Robert
Hooke found that, when the forces are not too large, the amount of
strain experience by an object was directly proportional to the
stress. This is another example of Hooke's
law.
Define the elastic modulus to be
Define the elastic modulus to be
(64)

Using
the definitions of stress and strain, this can be rearranged to yield
For
tensile stress, the elastic modulus is called the Young's modulus and
is denoted by Y.
When a material is stressed, the dimensions perpendicular to the direction of the stress become smaller by an amount proportional to the fractional change in length. This can be written as
When a material is stressed, the dimensions perpendicular to the direction of the stress become smaller by an amount proportional to the fractional change in length. This can be written as
(65)

where
is a dimensionless constant called Poison's ratio. Like Young's
modulus, it is a property of the material and can be used to characterise it.
Example:
A 10000 kg box hangs by a 20 m long cable which has a cross sectional area of 0.15 m^{2}. When an additional 250 kg is added to the box, the cable is seen to stretch 0.001 mm. What is the stress, strain and Young's modulus for the cable? What is the material used in the cable?
A 10000 kg box hangs by a 20 m long cable which has a cross sectional area of 0.15 m^{2}. When an additional 250 kg is added to the box, the cable is seen to stretch 0.001 mm. What is the stress, strain and Young's modulus for the cable? What is the material used in the cable?
Comparing
this with a standard chart of material characteristics, we see that
the cable was probably made of tungsten.
Shear Stress and Strain
Now
consider a force that is applied tangentially to an object
The
ratio of the shearing force to the area A
is called the shear
stress
If
the object is twisted through an angle ,
then the strain is
Shear
Strain = tan
Finally,
we can define the shear modulus, M_{S},
as
The
shear modulus is also known as the torsion
modulus.
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