Structural Geology
Lecture 27
Buckle Folds 
(Mechanics of Folding)

The mechanics of folding of layered rocks has evolved from the model for the folding of an elastic plate.  Figure 27-1 shows an elastic plate with a thickness of 2y.  The plate is folded upon the application of a force P distributed on the area DA at the ends of the plate.  As the plate is folded the upper portion of the plate is stretched (stippled area) whereas the lower portion of the plate is compressed.  A plane between the compressed and stretched portion of the plate is folded but neither stretched nor compressed.  This plane is referred to as the neutral surface.  A plane normal to the fold axis cuts the neutral surface to form a line called the neutral fiber.  In Figure 27-1 the neutral fiber has the length l.   A line a distance of y above the neutral fiber has the length l+ Dl.  The stretch Dl is proportional to
	Dl/l  =  y/R.

(Fig. 27-1)

An elastic fold such as is shown in Figure 27-1 is known as a buckle.  The force P necessary to cause the buckle is proportional to the distance from the neutral fiber.  For the force per unit area on a small strip at y from the neutral fiber

	DP/ DA  =  E.y/R       

	s  = E e    
	
	e = Dl / l                                                   

where E is the Young's modulus.  For elastic plates the bending moment (M) is defined as the torque about the neutral fiber.
	        
	M  =  ∫ ydP
	         cross section                                                            

 or let
	
	dP  =  E.y/RdA                                                                 

then 
	               ˙
	M  =  E/R∫ y2dA
	                cross section                                                      

Now the moment of inertia (I) about a horizontal axis through its "unit mass" is

	I  =  ∫ y2dA                                                                               

 so

	M  =  E/RI.                                                                     

Curvature is defined as the rate of change of a curve (i.e. the neutral fiber) in the vertical direction (y) as a function of distance in the horizontal direction (Fig. 26-2).

	1/R  »  - (d2y / d2x). 

(Fig. 26-2)

Folds can also be defined in terms of a bending moment at x where  the equation for a beam is

	M(x)  =  EI (d2y / d2x)  =  Py    
                                   
Rearranging this equation we can construct a homogeneous 2nd order differential equation

	                         Py
	d2y / d2x  =  - ___
	                         EI                                                                  
 The same equation for the buckling of a plate is

	M(x)  =  EI/(1- n) . d2y / d2x.

where n is the Poisson's ratio for that plate.  For small deflections the shape of a buckled beam can be represented by the equation
	
	y  = yosin πx/L                                                              

where L is the half-wave length of the buckle fold.  If we take the derivative of this equation

and then take the second derivative

Replacing terms on the right side of this equation, we derive

	,                                                            

the second order differential equation for the bending moment can be written in terms of the force necessary to cause the buckle.  Taking the homogeneous 2nd order differential equation from above and rewriting it in terms of P, we derive

Substituting for 

	.                                                                   

If the end of the buckle fold is fixed 

The physical significance of this force P is that if P is less than a certain value there will be no bending but if the force is slightly greater, the beam will bend suddenly where

	 =  the Euler force at which a beam will buckle.

Figure 26-2 shows a first and second mode fold.  The second mode is modeled as

	y  =   yo sin ( 2π/L) 2                                                          

The Euler buckling formula for the second mode fold is 

	P2  =  EI(2π/L) 2                                                              

This equation is the load necessary to hold a member of length 2L in a 2nd mode of buckling.
Rocks layers differ from the single member which is constrained neither on top of bottom.  Rather anytime that a rock layer folds, that layer has to contend with moving aside the rock layers above and below.  In modeling folds these adjacent layers are handled in several manners.  We will consider two approaches to the "confined member" fold.  First we will consider the situation where both the layer of interest and those above and below are elastic.  The layer in question is considered competent and located in an incompetent elastic media.
Johnson (1970) considers the problem of a confined member loaded axially (Fig. 26-3).  The elastic member is responding to axial loads, P, shear forces, Q; bending moments, M; and a distributed load, q.  The distributed load is due to the elastic medium having an elastic constant k.  The load on the confining medium is then proportional to the deflection, y, of the folded layer.  Consider a small element of length dx and thickness equal to the confined member.  The axial load on both ends are P but they are not colinear so that they bend the member where bending moment on one end is M and the moment on the other end is M + dM.
   This element can be examined by summing the moments around the right end (Fig. 26-3).  The moments exerted by P and Q + dQ on the right-hand end vanish, because their lever arms are zero.  Summing moments

	SM = 0

	- M + (M + dM) - Q dx + P( dy) = 0

	dM - Q dx + P dy = O 

 Now divide by dx and take the limits as dx approaches zero.

	dM/dx  -  Q  +  pdy/dx  =  0

	Q  =  p  dy/dx  +  dM/dx.

 By summing the forces in the y direction we find that

	Q - (Q + dQ) - k(y) dx  =  0

	dQ/dx =  -  ky                                                                      

This is equivalent to the relation between the shear force and the distributed load derived earlier for a member that was not supporting an axial load. 

Combining the equation for forces in y direction and the moments we have
	
	d2M / d2x  +  p (d2y / d2x)  =  -ky                                                
 We learned earlier that

	M  =  EI (d2y / d2x)                                                                      

So now

	EI (d4y / d4x) +  P (d2y / d2x)  +  ky  =  0                                 

The general solution to this equation has the form of a sinusoidal deflection of the rod.

	y  =  y0cos lx                                                                    

(Fig. 26-3)

Where the wave length of the deformation is L the wave number is l

	L  =  2π/l                                                                              

The force with which the rod is pushing against the lateral constrain is

	-ky  =  Pl2y  -  EIl4y                                                          

where Pl2y is the driving force generated by compression P and EIl4y is the restraint against deformation by bending rigidity.  There is a wave length for which -ky = 0 and the two terms are equal.  This value for l is the Euler Wave Length.

	Le  =  2πÖ(EI/P).