# Uniqueness theorem

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```					Uniqueness theorem

Let us consider now whether our equations can have more than one correct solution

corresponding to given surface and body forces.

The equations of equilibrium expressed in terms of stresses have been established as

 ij ,i  f j  0                                                                         (a)

In the expanded form, one has

 x  xy  xz                             
           Xx  0                    
x    y    z                              
 xy  y  yz                             

           Xy  0                                                             (2-12)
x    y    z                              
 xz  yz  z                             
           Xz  0                    
x    y    z                              


The kinematic relationships are

 ij 
1
2
ui, j  u j ,i                                                               (b)

In the expanded form, one has

u        v        u v
x         ,  y  ,  xy                                                            (2-2)
x        y        y x

and

w          w u        w v
z         ,  xz     ,  yz                                                       (2-3)
z          x z        y z

With the compatibility constrains of

 ij ,kl   kl ,ij   ik , jl   jl ,ik                                                (c)

Again, in the expanded form

1
 2 x   y   xy                           
2     2

 2                                    
y 2   x    xy                             
 2 y  2 z   yz
2                              
 2                                    
z 2  y     yz                            

 2 z  2 x  2 xz                         
 2 
x 2    z     xz                           

                                     (2-13)
  x 1    yz  xz  xy 
2

                        
yz 2 x  x          y     z            

 2 y 1    yz  xz  xy               
                                   
xz 2 y  x         y    z 

 2 z 1    yz  xz  xy               
                                   
xy 2 z  x         y    z              


and the constitutive law is given by

1
 ij      1     ij   ij kk                                                 (d)
E                           

In the expanded form

1
11    1     11   11  11   22   33  
E                                               

 xx  1     xx   1  xx   yy   zz     xx    yy   zz 
1                                                  1
E                                             E                        

Likewise,

1
 yy       yy    xx   zz  
E                         

 zz    xx   yy 
1
 zz 
E                        

1        1                              1                      1
12   12         1     12  12 kk   1     12  0 
                          E                   2G  12
2        E

1          1
 xy   xy           xy
2         2G

Likewise,

2
1            1
 xz   xz             xz
2           2G

1            1
 yz   yz             yz
2           2G

It is intended to prove that a solution to equations synthesized from Equations (1) through (d) is

unique for a given region with reasonable boundary conditions. A counterargument is followed

whereby it is assumed that two different solutions, say ui1 and ui 2 , exist. It is designated that the

difference between the solutions by asterisk (*) superscripts, that is,

ui*  ui   ui
1         2
(e)

and to show that ui*  0 .

The corresponding differences for strains and stresses are

        
 ij   ij1   ij2
*
(f)

and

        
 ij   ij1   ij2
*
(g)

Since the body forces are identical in the two systems, f j*  0 , and since both systems are in

equilibrium

 ij , j  0
*
(h)

Consider now the energy difference

1
U*        V  ij ij dV
* *
(i)
2

where U is the strain energy. The strain energy is non-negative (strain energy gained is stored in

the deformed elastic body). Replacing  ij by equation (b), yields
*

3
V  ij ij dV  2 V 2 ui, j  u j ,i  ij dV  2 V ui, j ij dV
1      * *         1 1 *            *        *       1     *     *
(k)
2

Note the i and j in equation (j) become repeated indices (dummy indices). Recalling the product

differentiation rule  ui* ij    ij ui*, j  ui* ij , j , gives
*        *                *
,j

V  ij ij dV  2 V ui  ij , j dV  2 V ui  ij , j dV  2 V ui  ij , j dV  0  2 S ui  ij n j dA
1      * *         1      * *              1     * *             1      * *                  1     * *
(l)
2

In equation (l), an application of the divergence theorem is evident. Note the integral involving

 ij , j vanishes by virtue of equation (h). Also recognizing  ij n j  T * (surface traction vector or
*                                                             *

applied loads), equation (l) can now be rewritten as

1
2
* *         1     * *             1     * *       1

1   2 1

 2
V  ij ij dV  2 S ui  ij n j dV  2 S ui Ti dA  2 S ui  ui Ti  Ti dA                                  (m)

The right hand side of equation (m) vanishes since the applied loads must be identical in the two

systems. Hence, the energy difference between the two solutions also vanishes and they are

identical. Therefore, ui1  ui 2  0 . QED

Alternately, Gustave Robert Kichhoff (1824-1887) presented the uniqueness theorem in 1876

long before the invention if the indicial notation. He published at a tender age of 26 the first

satisfactory theory of plate bending.

Let’s consider now whether the four priori equations (a) through (d) can have more than one

solution corresponding to given surface and body forces.

Let  x . . . . . . ,  xy . . . . represent a solution for loads Ti and body forces f i , and let

 x . . . . . . ,  xy . . . . represent a second solution for the same loads Ti and body forces f i .

Then for the first solution one has the following equations:

4
 x  xy  xz                             
              Xx  0                    
x    y     z                                

 xy         y         yz               
                         Xy  0                                                (2-12)
x            y           z                  

 xz  yz  z                             
             Xz  0
x     y     z                               


and

Tx   xl   xym   xzn

Ty   xyl   ym   yzn

Tz   xzl   yzm   zn

where l, m, n are the direction cosines of the outward normal to the surface of the body at the

point under consideration.

For the second solution, one has

 x  xy  xz                          
                Xx  0                 
x     y      z                              

 xy        y        yz              
                          Xy  0   
x           y            z                 

 xz   yz  z                         
               Xz  0
x       y      z                            


and

Tx   xl   xym   xzn

Ty   xyl   ym   yzn

Tz   xzl   yzm   zn

and also the conditions of compatibility.

5
By subtraction one finds that the stress distribution given by the differences  x   x , . . . ,

 xy   xy , . . . , satisfies the equations


  x   x           xy
   xy            0
xz         xz


x                    y                        z                      


  xy   xy              
y           y            yz
   yz    0       


x                    y                        z                      

  xz   xz         yz
   yz             0
z         z



x                    y                        z                      


and

                                           
Tx   x   x l   xy   xy m   xz   xz n

T        l        m        n
y           xy   xy                y          y             yz        yz

T        l        m        n
z            xz   xz               yz         yz             z         z

where all external forces and body forces vanish. The conditions of compatibility will also be

satisfied by the corresponding strain differences,  x   x , . . . . .,  xy   xy , . . .

Thus this differential stress and strain distribution is one which corresponds to zero surface and

body forces. The work done by these differential stresses and strains is zero, and it follows that

 u dV vanishes.
V       0                    The strain energy density function u 0 is non-negative. Therefore the integral

can vanish only if u 0 vanishes at all points of the body. This requires that each of the differential

strain components  x   x , . . . . .,  xy   xy , . . . should be zero.                    The two sets of

strains  x , . . . . .,  xy , . . ., and  x , . . . . .,  xy , . . ., and consequently the two sets of stresses are

identical. That is, the equations can yield only one solution corresponding to given loads. QED

6

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 views: 0 posted: 9/15/2012 language: English pages: 6