Structural Geology
Lecture 11
Force-Balance and Mohr's Equations
(Mohr Stress)

The Force-Balance Analysis

In high school physics we were taught that pressure is defined as force per unit area.   Stress may have the same definition.  The distinction between pressure and stress in structural geology is based on the nature of the material on which the force is acting.  The distinction is made depending on whether the material in question has a shear strength.  Materials such as rock are said to have a shear strength because they maintain their shape when placed unsupported on a table.  If we are dealing with a rock that has a shear strength (fluids such as water and gases do not), then we say that it exerts a stress on its surroundings.  Materials which have a shear strength can exert different stresses in different directions.  In contrast, water, without a shear strength would proceed to run over the table top seeking the lowest spot.  If we wish to describe the force per unit area that a liquid or gas is exerting on its container, we use the term pressure.  Water in the pores in rocks exerts a pressure on the grains surrounding the pore.  

The most useful equations for teaching the concept of stress are the equations for normal (sn) and shear (t) stress in terms of principal stresses (s1,s3).   The derivation of these equations is based on a force-balance problem which assumes that a body subject to forces is in equilibrium which means that all forces in any direction add to zero.  

(Fig. 11-1)

Consider the forces acting parallel to the front face of the triangular solid shown in Figure 11-1.  These forces are acting in a plane so that this force-balance problem is two dimensional.  If the area of the hypothenuse face is unity (i.e. 1), then the area of the left face is unity times sin q whereas the area of the bottom face is unity times cos q.  Forces acting on the three faces are stresses multiplied by the area of the face [Force =  (force/area) ´ (area)].  Summing the forces in the horizontal and vertical directions (the forces add to zero in both directions), we obtain

	S Fh   =  0  =  Fx - S sinq - N cosq  	(11-1a)

	S Fv  =  0  =   Fy - N sinq + S cosq	(11-1b)

Rewriting the force terms using components of stress, we obtain the normal and shear stresses acting on a plane an angle of q to s1 in terms of the principal stresses.  

	s1 cos q  -  sn cosq  -  t sinq   =  0	(11-2a)

	s3 sin q  -  sn sinq  +  t cosq   =  0	(11-2b)

In many geological applications the plane of interest is a fault plane.  Now we can solve these two equations for sn and t

	 sn   =  s3 cos2q  +  s1  sin2q 	(11-3a)

	t   =   (s1 -  s3) cosq sinq    	(11-3b)

 Remember that  

	sin2q  =  2 sinq cosq   	(11-4)

and also    
	cos2q   =  1/2(1  +  cos2q )   and  sin2q   =  1/2(1  -  cos2q )	(11-5)

We derive
	sn  =   1/2(s1  +  s3)  +  1/2(s1  -  s3)cos2q            	(11-6a)            

and 
	t  =    1/2(s1  -  s3)sin2q .	(11-6b)

Note that in these equations q is the angle between s1 and the normal to the fault plane.  This convention leads to a positive sign in the middle of equation 11-6.  Chapter 10 of Marshak and Mitra (1988) uses the convention that q is the angle between the normal to the fault plane and s1.  If the q is the angle between s1 and the fault plane the sign convention equation 1-29a becomes

	sn  =   1/2(s1  +  s3)  -  1/2(s1  -  s3)cos2q 	(11-7).

Unless each convention is memorized the reader will be confused when using these equations in the future.

The general force-balance problem is shown in figure 11-2 where the back and side faces of the unit triangle are subject to both normal and shear stresses.  In other words the coordinate system x-y is not parallel to the principal stress directions.  The general case is given to illustrate a property of stress called invariance with respect to coordinate system.  In two dimensions we define stress as a force per unit of line length, in contrast to the three dimensional situation where stress is a force per unit area (Fig. 11-1).  Consider a coordinate system Oxy with an arbitrary line AB cutting the x and y axes such that the normal to the line AB makes an angle q with the x-axis.  This gives a right triangle AOB with sides OA (parallel to Ox) and OB (parallel to Oy) and a hypothenuse AB.    Across the line AB a stress vector p can be applied making an angle q with the x-axis.  Remember that p = df/dA when dA Þ 0, so a stress vector can represent stress at a point.  Otherwise stress is defined on a line (2-D) or a surface (3-D).  The stress vector p can be resolved into components parallel to the x and y axes:  p = px + py.  Even though P is called a vector it still has units of stress (force/length in two dimensions).  Because the triangle ABO is in equilibrium the sum of the force-vectors on all sides must balance.  In 2-D stress multiplied by line length will give a force vector.  So
	
	pxAB  =  sxOB  +  tyxOA. 	(11-8)

 
 
(Fig. 11-2)

If length a = AB, then a´cosq = OB and a´sinq = OA.  If we divide by length a, then
	
	px  =  sx cosq  +  tyx sinq.          	(11-9a)                                          

Likewise	

	py  =    sy sinq  +  txy cosq                          	(11-9b)                         

Now consider a normal stress sn and shear stress t  across AB in terms of the components of the stress vector

	sn  =  px cosq  +  py sinq                    	(11-9a)                                  

The equality holds for t
	
	t  =  py cosq  -  px sinq.                                            	(11-9b)

Substituting for px and py and remembering that  sin2q  =  2 sinq cosq   and also    
cos 2q  =  cos2q  -  sin2q

	sn  =  sx cos2q +  2txy sinq cosq  +  sy sin2q	(11-10a)

and
	
	t  =  1/2(sy  -  sx) sin2q  +  txy cos2q.                           	(11-10b)

These are the general equations for a stress system in which the orientation of the principal stresses are unknown.  The following exercise illustrates the case for the principal stresses being parallel to the coordinate axes.

(Fig. 11-3)

Now we consider the effect of rotating the coordinate system on the values of stress (Fig. 11-3).  To accomplish this coordinate rotation let sn and t  be sx' and  t x'y' where Ox'y' is rotated q from Oxy.  From a previous equation we have
	
	sx'  =  sx cos2q  +  2txy sinq cosq  +  sy sin2q	(11-11a)

Now we must find sy' in the new coordinate system.  This is accomplished by replacing q by  q  +  1/2π and we get

	sy'  =  sx sin2q  -  2t xy sinq cosq  +  sy cos2q 	(11-11b)

If we add sx' and sy' and remember that  sin2q  +  cos2q  =  1, we get
	
sx'  +  sy'  =  sx  +  sy.     	(11-12)                                            	
This shows that the sum of the normal stresses is invariant or unchanged by rotation of the coordinate system.  Likewise

	t x'y'  =  0.5(sy  -  sx) sin2q  +  t xy cos2q.              	(11-13)          

Principal stresses are found in planes containing no shear stress.  If in these planes, tx"y"  =  0, then from the previous equation	                 
                                     
				2txy
	                 tan2Q  =     ____	(11-14)
	                                               sx - sy                                                      

where Q is the one angle between the coordinate system Ox"y"  and Oxy where the shear stresses vanish along the directions Ox" and Oy".  In this coordinate system the only stresses are the normal stresses sx" and sy". This coordinate system contains the principal stress axes and the components, sx" and sy" are known as the principal stresses.

(Fig. 11-4)
Let us now call these principal stresses s1 and s2 (Fig. 10-4) and then we shall choose the coordinate system such that the x and y axes are in the direction of the principal stresses, s1 and s2.  Now the normal sn and shear t stresses across a line whose normal is inclined at q to s1 is
	
	sn  =  s1.cos2q   +   s2.sin2q             	(11-15)                                   

This is the convention used in Chapter 10 of Marshak and Mitra (1988).  Again remember that 
	
	cos2q   =  1/2(1  +  cos2q )   and  sin2q   =  1/2(1  -  cos2q )  

we can rearrange the equation for the normal stress
	
	sn  =   1/2(s1  +  s2)  +  1/2(s1  -  s2)cos2q    	(repeat of 11-6a)                    

and shear stress
	
	t  =    1/2(s1  -  s2)sin2q .  	(repeat of 11-6b)                                           

From this last equation we see that shear stress is greatest when q  =  π/4 and 3π/4.
Using this same coordinate system where the axes are parallel to the principal stresses s1 and s2, we can look at the stress vectors px and py.  They become

	px  =  s1.cosq      and      py  =  s2.sin q.               	(11-16)

(Fig. 11-5)

Substituting into the equation  sin2q  +  cos2q  =  1,  we can generate the equation for an ellipse where the px and py are on the ellipse called the stress ellipse.

	px2              py2
	___   +        ___   =     1
	s12              s22                                               	(11-17)                  

The semi axes of the ellipse are s1 and s2.
The Mohr's circle is a graphical method of representing the state of stress of a rock in two dimensions (Fig. 11-5).  The equations used for the Mohr's circle representation are those derived above for the coordinate system with axes parallel to the principal stresses.  The Mohr's circle may be used to derive the normal sn and shear t  stresses on any plane whose normal is oriented at q from s1.   The coordinate system for the Mohr's circle representation is sn along the horizontal axes with increasing compression to the right and t  along the vertical axes.  Critical points along the sn-axes are OP = s1, OQ = s2, and  C  =  1/2(s1  +  s2).  The angle measured as PCA counter clockwise from OP is 2q.  Now we have
	
	sn  = OB  =  OC + CB   =  1/2(s1 +  s2)  +  1/2(s1 -  s2)cos2q  	(11-18a)

and
	
t  =  AB   =  1/2(s1  -  s2)sin2q.  	(11-18b)