The Concept of Force
Units
Load
The Concept of Stress
 Defined—That which changes, or tends to change, the state of rest or state of motion of a body
.
 Newton’s First Law of Motion—An object at rest will remain at rest and an object in motion will continue in motion with a constant velocity unless it experiences a net force, in which case it is caused to accelerate. A net force arises when the forces are not balanced.
 Newton’s Second Law of Motion—Acceleration of an object is directly proportional to the net force on it, and inversely proportional to the mass of the object.
 The definition of force is based on mass (m) and acceleration (a): force = mass x acceleration or F = ma
 Force units must be mass times acceleration units
 Mass is the amount of material a body contains
 Units
 Kilogram (MKS and SI)
 Gram (cgs)
 Slug (1 kg = 0.068521 kilograms) [English Standard]
 Volume (V) is the amount of space occupied by a mass
 Density (ρ) is the mass per unit volume
 M = ρV or ρ = M/V
 Weight is the magnitude of the force of gravity acting on a body
 Units
 Newton (SI)—the force required to impart an acceleration of one meter per second per second to a body of 1 kg mass
 Dyne (cgs)—force required to impart an acceleration of one centimeter per second per second to a body whose mass is 1 gram

 Poundal—the unit of force that will accelerate a 1 pound mass to a velocity of 1 ft/sec in 1 sec
 Pound—the unit of force that will accelerate a 1 pound mass to a velocity of 32ft/sec in 1 sec
 Forces are vector quantities
 Magnitude
 Direction
 Resolve a force acting at an angle to a plane into components perpendicular and parallel to the plane
 F
Y
/F = sin 
 F
Y
= F sin 
 F
X
/F = cos 
 F
X
= F cos 
 Torque is the tendency of a force to rotate a body about some axis
 It is the product of the force and the perpendicular distance (moment arm) from the line of action of the force to the axis of rotation.
 Static equilibrium—All forces and torques are balanced
 Dynamic equilibrium—All torques are balanced, but not all forces are balanced
 Body forces act on the mass of a body in a way that depends on the amount of material in the body but is independent of the forces created by adjacent surround materials
 Gravity is the most important body force
 Electromagnetic forces are also significant in some applications
 Contact forces are pushes or pulls across real or imaginary surfaces of contact such as a fault between adjacent parts of a rock body
 Produced by gravitational loading, thermal loading, or displacement loading
 Gravitational loading is the weight of the rock column above the point you are interested in
 Thermal loading involves expansion and contraction, and is generated when a rock wants to expand or contract, but cannot
 Displacement loading refers to large-scale mechanical disturbance of rocks—collision of plates, regional bending, etc.

 Forces can be thought of in terms of how much acceleration they impart to a body in a given amount of time
 Forces can also be described by how much weight they will support; the weight is called load.
 Since force = mass times acceleration, weight equals mass times the acceleration of gravity
 To support a mass of 2.3 kg (2.3 x 9.87 = 22.7 newtons) requires an opposing force of 22.7 newtons to support it.
 That is an opposing force of about 50 pounds.
 Usually we would say this was 22.7 kg of force (here is confusion) which is about
50 lbs (22.7 x 2.2 = 49.94)
 Stress is defined as force per unit area
 Units
 Newtons/m 2
 Dynes per cm 2
 Pounds per in 2 (Sounds like pressure doesn’t it. It is!)
 Pascals (1 newton per m 2 )
 1 atmosphere  1bar  10 6 dynes/cm 2  10 5 Newtons/m 2  10 5 pascal
 1 bar is 100 kilopascales
 1 bar is 0.1 megapascales
 Since stresses in the earth are measured in kilobars, we usually use megapascals or gigapascals
 Volume (V) = width (W) x breadth (B) x height (H)
 V = 2m X 2m X 2m = 8 m 3
 Density (ρ) = 2.7 g/cm 3 = 2700 kg/m 3
 Mass (M) = ρ V = 2700 kg/m 3 X 8 m 3 = 21,600 kg
 The acceleration due to gravity (g) is 9.8 m/s 2

 Force (F) = mass (M) X acceleration (g)
F = Mg =21,600 kg X 9.8m/s 2 = 211,680 kg-m/s 2
 F = 211,680 N

 F = 211,680 N
 Stress (  ) = force (F)/area (A)

 Area (A) =  r 2 = 3.14 X (0.5 m) 2 = 0.79 m 2
 = F/A = 211,680 N/0.79 m 2 = 267,949 N/m 2
  = 268 kPa
•ρ = 2700 kg/m3
•V = 800 m X 800 m X 800 m
•V = 512,000,000 m3
•M = V ρ
•F = Mg = Vρg
•F = 512,000,000 m3 X 2700 kg/m3 X 9.8 m/s2
•F = 13,547,520,000,000 kg-m/s2
•A = 640,000 m2
•σ = F/A
•σ = 13,547,520,000,000 kg-m/s2 / 640,000 m2
•σ = 21,168,000 Pa
•σ = 21.17 MPa
=

XY
cos a F
4
= σ
XX
cos a F
5
= -
σ
X’X’
(1) F
1
=

YX
sin a F
2
= σ
YY
F
6
=

X’Y’
(1)
sin a F
3
F
1X’
= F
1 cosα = 
YX sinα cosαF
2X’
= F
2 sinα = σ
YY sinα sinαF
3X’
= F
3 sinα = 
XY cosα sinαF
4X’
= F
4 cosα = σ
XX cosα cosαF
5X’
= σ
X’X’
σ
FX’
= 2

XY sinα cosα + σ
YY sin
2 α + σ
XX
cos
2 α - σ
X’X’ = 0
σ
X’X
= σ
XX cos
2 α + 2 
XY sinα cosα + σ
YY
sin
2 α
F
1Y’
= F
1
sin

= – 
YX
sin

sin

F
2Y’
= F
2
cos

= σ
YY
sin

cos

F
3Y’
= F
3
cos

=

XY cos

cos

F
4Y’
=
–F
4
– 
XX
cos

sin

+

sin

=
– 
XX
cos

sin

F
6Y’
YY
sin

cos

+

X’Y’
= 0
= t
X’Y’

FY’
=
– 
XY
sin
2 
+

XY
cos
2 
= – 
XY
(sin
2  – cos 2 
) – 
XX
– 
YY
) sin

cos

+

X’Y’
= 0

X’Y’
= (

XX
– 
YY
) sin

cos

+

XY
(sin
2  – cos 2 
)
To find when the normal stress (

X’X’ ) is at a minimum or a maximum we take the derivative of the normal stress and set it equal to zero.

d

X’X’ = d (

XX cos 2 
+ 2

XY
sin

cos

+

YY
sin 2
=

XX
(2cos2

)(
– sin

) + 2

XY
=
–
2

XX
sin

cos

+ 2

YY
– d

X’X’ = 2

XX

) = 0
[(cos

)(cos

) + (sin

)(
– sin

)] +

YY
(2sin

)(cos

)= 0
(sin

cos

) + 2

XY
sin

cos
 –
2

YY
= 2(

XX
– 
YY
) sin

cos

+ 2

XY
(sin
2
(cos2
 –
sin2

) = 0
(sin

cos

)
–
2

XY
(cos2
 –
sin2

) = 0
 –
cos
2 
) = 0

X ’ Y ’ = (

XX
– 
YY
) sin

cos

+

XY
(sin
2  –
cos
2 
)
Note: d

X’X’ =
–
2

X ’ Y ’
This means that when d

X’X’ = 0 ,

X ’ Y
= 0 or when

X’X’ is at a maximum or a minimum, there is no shear stress.
Let’s now try to figure out how often shear stress is 0 and normal stress is at a maximum or a minimum d

X’X’
= –2

X’Y’
and when d

X’X’
= 0, the shear stress is also 0
Remember two other trig identities
2 sin

cos

= sin 2

and cos
2 
– sin
2 
= cos 2

2(

XX
–

YY
) sin

cos

+ 2

XY
(sin
2 
– cos
2
(

XX

) = 0
–

YY
)sin 2

= 2
 any situation, we know
XY
cos 2

2

XY
/ (

XX
,

YY

XX
–

YY
) = sin 2

/ cos 2

= tan 2

Since in
, and

XY
are known, we can calculate a or the angle that defines the planes for which normal stress is at a maximum or a minimum and shear stress is 0
2

XY
/ (

XX
– 
YY
) = tan 2

From our example
2(3)/(5-3) = 6/2 = 3 = tan 2

2

= Tan
-1
(3)
2

= 71.565

= 35.782
Since we already proved that principle stresses act at right angles to each other, we know that there must be another plane with 0 shear stress at 90°.
So 35.782° + 90° = 125.782°
Let us calculate the normal stress acting on a plane oriented such that

= 35.782°

X’X’
=

XX
cos 2 
+ 2

XY sin

cos

+

YY
sin 2 
= 5(.811)
2
+ 2(3) (.585) (.811) + 3(.585)
2
= 3.289 + 2.847 + 1.027


X’X’
= 7.172
For the other plane on which

= 125.782°

X’X’
=

XX
cos 2 
+ 2

XY sin

cos

+

YY
sin 2 
= 5(-.585)
2
+ 2(3) (.811) (-.585) + 3(.811)
2
= 1.711 - 2.847 +1.973

X’X’
= 0.837
To check, let’s solve for the shear stress ( 
X’Y’
) when

= 35.782°

X’Y’
= (

XX
–

YY
) sin

cos

+

XY
= (5-3)(.585)(.811) + 3[(.585)
2
(sin 2 
– cos 2 
)
– (.811) 2
]
= (2)(0.474435) + 3(0.342225 - 0.657721)
= (2)(0.474435) + 3(-0.315496)
= 0.94887- 0.946488

X’Y’
= ~0
We now know that the maximum normal stress is 7.172 Kb and that it acts on a plane making a 35.782° angle with the Y axis
We now know that the minimum normal stress is 0.837 Kb and that it acts on a plane making a 125.782 ° angle with the Y axis
Principle stresses act on planes of no shear stress
The maximum principle stress is

1
The minimum principle stress is

3
They are at right angles
We can designate new axes parallel to

1 and

3
Planes perpendicular to these axes would be principal planes
Since principal planes have no shear stress, the equations for the normal and shear stress on any plane an be reduced to:
 n
=

1
cos

= (

1
–

3
2 
+

3 sin 2 
and
)sin

cos

Since we know some trig identities: sin
2 cos
2

= –((cos2

– 1)/2)

= (cos2

+ 1)/2 sin

cos

= sin2

/2
We can rewrite the previous equations as
 n
=

1
=

1
cos
2 
+

3 sin
2 
(cos2

+ 1)/2 –

3
(cos2

– 1)/2)
=

1
cos2

/2 +

1
/2 –

3 cos2

/2 +

3
/2
= (

1
+

3
)/2 + ((

1
–

3
 and since

= (

1
= ((

1
–

3
–

3
)/2) cos2

)sin

cos

)/2) sin2


Sliding on a fault plane will occur when the shear stress (

) exceeds the resistance to shear
The resistance to shear is partly due to friction
Frictional resistance depends on the normal stress (
 n of friction (

)
) on the plane and on the coefficient
Frictional resistance is
  n
In rocks, the other part of the resistance to shear is called the cohesion (C)
Faulting will occur when
   n
+ C
Let’s do a simple example with sand, which has no cohesion
C = 0;

= 0.6;

1
= 6 bars;

3
= 2 bars
(

1
+

3
)/2 = 4 and (

1
–

3
)/2 = 2
Let’s do another example with real rock
In this case
C = 1.05;

= 0.53;

1
= 5.5 Kb;

(

1
+

3
)/2 = 3.075 and (

1
–

3
3
= 0.65 Kb
)/2 =2.425

Principal Stresses act on planes of no shear stress

Principal stresses are the maximum and minimum stresses

Principal stresses are at right angles

Failure will happen on planes making a 30° angle with the direction of 
1
It is also possible to take our two equations (one for
 n equations for a circle where (

1 and one for

) and rewrite them as
+

3
)/2 is the center of the circle and
(

1
–

3
)/2 is the radius
Such a circle would plot in

-

space
Every point on the circle represents the normal and shear stress on some plane in the material
Let’s go back to our original example where 
1 makes a 35.782° angle with the Y axis and

= 7.172 Kb and acts on a plane that
3
= 0.837 Kb and acts on a plane that makes a 125.782° angle with the Y axis

Let’s use Mohr Circles to look at the results of a laboratory test on rocks.
The table below gives the stress conditions at which failure by faulting occurred:

1

3
1100 bars 100 bars
1400 bars 200 bars
2000 bars 400 bars
3200 bars 800 bars
5600 bars 1600 bars
6800 bars 2000 bars