Steven Dutch, Professor Emeritus, Natural and Applied Sciences,University of Wisconsin - Green Bay

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.

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-m^{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: kg-m/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. A Fig Newton weighs about 15 grams; therefore one SI Newton equals approximately 7 Fig Newtons.**Pressure**: Pascal = Newton/m^{2}or kg/m-sec^{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.

The fundamental rule in using units in calculations is that **units obey
the same algebraic rules as other quantities**

Density is conventionally represented as grams per cubic centimeter. 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}

Thus, to convert traditional to SI density, **multiply by 1000**. Thus,
2.7 gm/cm^{3} = 2700 kg/m^{3}, etc.

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.

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} kg-m/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 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?

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 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 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 compressional 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 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.

**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.

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23 February 1999, Last Update 24 February 1999*