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Tuesday, 30 June 2015

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 kilo-bar, equal to 1000 bars. The pressure beneath 10 km of water, or at the bottom of the deepest oceanic trenches, is about 1 kilo-bar. Beneath the Antarctic ice cap (maximum thickness about 5 km) the pressure is about half a kilo-bar 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:
  • Energy: Joule: kg-m2/sec2. 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: kg-m/sec2. On the surface of the Earth, with a gravitational acceleration of 9.8 m/sec2, a newton is the force exerted by a weight of 102 grams or 3.6 ounces.

  • Pressure: Pascal = Newton/m2 or kg/m-sec2
  • 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 cm2 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/cm3(0.001 kg)/(.01 m)310-3 kg/10-6 m3 =
1000 kg/m3
Thus, to convert traditional to SI density, multiply by 1000. Thus, 2.7 gm/cm3 = 2700 kg/m3, 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/cm3?
First, we find the mass of the block. Mass is volume times density
or 20 x 15 x 10 m3 x 2700 kg/m3 = 8.1 x 106 kg.
Note that we have m3 times kg/m3, and the m3 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/sec2.
Thus the force the block exerts is 8.1 x 106 kg x 9.8 m/sec2, or 7.9 x 107 kg-m/sec2.
Referring to the SI units listed above, we see that these are indeed the correct units for force. The block exerts 7.9 x 107 newtons of force on the ground beneath it.
The pressure the block exerts is force divided by area, or 7.9 x 107 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 ten-meter 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 all-important. 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 feet2/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 in-situ 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 (10-9)/10 or 0.1 (sometimes expressed in percent, in this case 10 percent.) 
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.
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 glacio-isostasy 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 = Fn / A         (1)
where
σ = normal stress ((Pa) N/m2, psi)
Fn = normal component force (N, lbf (alt. kips))
A = area (m2, in2)
  • a kip is a non-SI unit of force - it equals 1,000 pounds-force
  • 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 = 10 103 (N) / (π (10 10-3 (m) / 2)2)
   = 127388535 (N/m2
   = 127 (MPa)

Shear Stress

Stress parallel to the plane is usually denoted "shear stress" and can be expressed as
τ = Fp / A         (2)
where
τ = shear stress ((Pa) N/m2, psi)
Fp = parallel component force (N, lbf)
A = area (m2, in2)

Strain

Strain is defined as "deformation of a solid due to stress" and can be expressed as
ε = dl / lo = σ / E         (3)
where
dl = change of length (m, in)
lo = 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/ E
     = 127 106 (Pa) 2 (m) / 200 109 (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 = (Fn / A) / (dl / lo)         (4)
where
E = Young's modulus (N/m2) (lb/in2, psi)
Modulus of Elasticity or Young's Modulus are commonly used for metals and metal alloys and expressed in terms 106 lbf/in2, N/m2 or Pa. Tensile modulus are often used for plastics and expressed in terms 105 lbf/in2 or  GPa.

Shear Modulus

S = stress / strain = (Fp / A) / (s / d)         (5)
where
S = shear modulus (N/m2) (lb/in2, psi)
Fp = force parallel  to the faces which they act
A = area (m2, in2)
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
1010 N/m2 106 lb/in2 1010 N/m2 106 lb/in2 1010 N/m2 106 lb/in2
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.

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
 


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/m2
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 mass-less, 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 l0 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
 


(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
 


(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 m2. 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, MS, as
The shear modulus is also known as the torsion modulus.



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