Reduction of Displacement Field to Normal Form

REDUCTION OF THE DISPLACEMENT FIELD TO A NORMAL FORM
V.V. Kibitkin1, A.I. Solodushkin1, A.P. Zykova1,
1
V.E. Rubtsov , N.L. Savchenko
UDC 539.214, 539.374, 004.932
1
Simple algorithms were given to eliminate the rotation and constant vector for a planar vector field. The algorithm of
compensation of rotation was suggested on a basis of functional minimization. A formula of the functional was found
too. Similar approach has been used also to find the components of a constant vector. It was made possible to
diminish the error for deformation, and eliminate the constant vector in order to analyze the displacement field. It was
proposed to characterize a vector field by its mean length of components (algebraic or modulus) and related standard
deviation. This approach has been demonstrated on the example of model vector field.
Keywords: displacement vector field, error, compensation, rotation, constant vector, ceramics, fracture stages, correlation.
INTRODUCTION
Digital image correlation (DIC) technique is effective for studying deformation behaviors of
materials due to its high precision, versatile space resolution and relatively simple procedure of data
processing [1-3]. When studying hard materials, in particular, ceramics, the specimens may be of
arbitrary sizes so that region of interest (ROI) practically coincides with the specimen’s lateral face
and displacement vector fields (DVF) obtained from that face using the DIC method can be referred to
as the response of the whole specimen. The ROI images may be always selected so that a series of
high-precision DVFs will be obtained after mathematical processing. In particular, this is the reason
why no high-speed camera is needed for the DIC if the ROI surface coincided with the camera focal
plane.
During the DIC procedure, the specimen can however rotate in-plane as a whole and thus increase
the margin of error. For example, specimen is usually rotated at the very beginning of the compression
test due to sliding between its end faces and the testing machine platens.
On the other hand, any displacement field has a constant vector component uc being independent
of the coordinates, so these fields can vary at each new measurement [4]. From the view point of
mechanics of deformed solid, some constant vector can be added to the displacements field, because
deformation components are not dependent on a constant. However, this constant does not allow us to
estimate accurately the mean length of a vector field.
A vector field, measured with DIC in coordinate system XOY during span t , can be written as
follows:
(1)
u ( x, y)  u0 ( x, y)  u ( x, y)  uc  uer ( x, y) ,
where u0 ( x, y) is the ideal displacement field which shows the materials flow in ROI  , u ( x, y) is
the vector field formed by rotation of this region at angle θ, uer ( x, y) is the vector referred to as
measurement error or a residual field.
The objective of this work is to transform a displacement vector field to its normal form, evaluate
the mean length of vectors of the DVF and the related standard deviation (SD). In addition, this work
aims to finding a correlation between these amplitude characteristics and current mechanical state of a
material.
COMPENSATION OF ROTATION
1
Institute of Strength Physics and Materials Science of the Siberian Branch of the Russian Academy of Sciences,
Tomsk, Russia, e-mail: vvk@ispms.ru, s.ai@ispms.ru, zykovaap@ispms.ru, rvy@ispms.ru, savnick@ispms.ru
Rotation is a typical source of error in DIC measurements and many algorithms have been
developed to diminish the error for each subset [5-7]. It is a problem for conventional DIC algorithms
because they are not capable of identifying new positions when rotation has already occured. In the
general case the compensation of 3D rotation is a challenge so that some, special algorithms were
developed [8].
Let us dwell on how to eliminate the rotation error for planar displacement field. To compensate
the error, the first step could be finding angle of rotation θ, then adding other vectors related to the
rotation to each vector of u ( x, y ) , but with inverse direction (-θ). It is clear that if some field turned at
some angle then the summation vector length would increase. However, this is not the case with the
rotation field where total vector length depends on the rotation direction.
Let there exists in a global coordinate system XOY a discrete vector field u ( x, y ) with
dimensions M  N , where N and M are the numbers of rows and columns, respectively (Fig. 1).
Here the directions of ordinates axes are downward, because this is more convenient for computation
procedure.
Fig. 1. The scheme of calculating the vector field of rotation.
Let us introduce a new coordinate system XOY whose center coincides with that of the vector
field. Then pass a radius-vector r0 from the origin O to a position A( x0 , y0 ) and turn it clockwise to
form point B. Pass another radius-vector r1 from O to position B( x1 , y1 ), so that their difference will
give us the rotation vector u :
u  r1  r0 .
(2)
The lengths of these two vectors are equal, so that r0  x  y  x  y , and it is seen from Fig. 1
2
2
0
2
0
2
1
2
1
that x1  r0 sin(   ) , y1  r0 cos(   ) ,   arctg ( y0 / x0 ) , and r0  x02  y02 .
Taking into account equation (2), the projections of rotation vector can be written as
ux  x02  y02 cos(   )  x0 , u y  x02  y02 sin(   )  y0 ,
(3)
(4)
for M / 2  x  M / 2 ,  N / 2  y  N / 2 , and x0  0 , y0  0 .
A change to global coordinate system XOY needs to be replaced by
(5)
x  xM / 2, y  y  N / 2 ,
under conditions 0  x  M / 2 and 0  y  N / 2 . These conditions ensure the points location into
displacement field.
Taking into consideration (3), (5), the relations (4) can be rewritten using the coordinate system
XOY as
ux  y0 sin   2 x0 sin 2 ( / 2) ,
uy  (1)( x0 sin   2 y0 sin 2 ( / 2)) .
(6)
Equation (1) shows that the initial vector field must be rotated in the inverse direction. Because the
field uer ( x, y) , as a rule, is not equal to zero, the task of finding the rotation angle  can be reduced to
locating a minimal value of some functional. The functional is understood as a result of applying
certain actions to one or several functions and finally obtaining a function or a number. If the required
vector field was found to be the rotation field, then the compensation would produce the error vector
field. Otherwise, any other field would give some contribution to the functional. Here we use square
measure
(7)
SS ( )   (u  u )2  min .

Digital camera in DIC gives us digital data, so the functional must be rewritten as:
N
M
SS ( )   (ux i, j  uxi , j )2  (uyi, j  uyi , j )2  / M  N  min .


i 1 j 1
(8)
This functional, which is a sum of all vector components, is normalized to the dimensions of array
ux(i, j) or uy(i, j) , where ux(i, j) is array of longitudinal displacements, uy(i, j) is array of transversal
displacements, and i, j can be viewed as discrete coordinates in global coordinate system. This
normalization removes the dependence of the functional on the pitch T , i.e. is the space period
between partial vectors of the field.
The compensation of the rotation should be made with respect to the point of origin O . It is clear
that if the compensation is valid, then the total length would be minimal.
To save the calculation time, another functional can be applied as follows:
Sm ( )   u  u  min ,

or in its discrete form it is:
N
M
Sm ( )    ux i, j  uxi , j  uyi, j  uyi , j  / M  N  min .
(9)


i 1 j 1
These functionals produce the only solution, because several different rotations result in the same
turn. Since the rotation transformation should be applied for all discrete points of the vector field, then
the replacements must be made as follows:
x0  x j  j ,
y0  yi  i .
(10)
Here i is the row number, and j is the column number. The vector length is measured in pixels,
which is later recalculated into micrometers (
).
The relations (6) were rewritten in discrete form to obtain:
uxi, j  (i  M / 2)sin   2( j  N / 2)sin 2 ( / 2) ,
uyi, j  (1)( j  M / 2)sin   2(i  N / 2)sin 2 ( / 2)) .
(11)
Formulae (11) set the rotation of the field where   0 indicates a clockwise turn. To find the
rotation angle numerically, one can use a zero-order method, where the only information related to a
functional, should be taken into consideration [8].
An algorithm of uniform search is outlined the next steps. The interval min    max is divided
onto n  1 equal parts. Then one needs to calculate the functional SS (n ) n times and in order to find
its minimal value. So,   min(SS ) . Here k  min  k   where   (max  min ) / (n+1 ) for
0  k  n  1. The interval  is also absolute error of the rotation angle  .
Finally, to eliminate the rotation error, we must add the rotation vector field u ( x, y) of inverse
angle (  ) to the measured field u ( x, y ) .
ucomp ( x, y )  u ( x, y )  u ( x, y )  u ( x, y )  u ( x, y ) .
(12)
Thus, after this angle compensation for initial field we arrive to
ucomp ( x, y)  u0 ( x, y)  uc  uer ( x, y) .
(13)
Next task is to subtract the vector uc  ucx  ex  ucy  ey which is constant for each point of the field.
Because the components of vectors are independent, they can be found separately.
COMPENSATION OF THE CONSTANT VECTOR
It is easy to prove that the sum of all projections is minimal for the two vector components if a
number t x (t y ) has been subtracted from each vector where t x  ucx and t y  ucy . Therefore, the main
goal is to find the minima of the difference functionals Fx (tx ) and Fy (t y ) :
Fx (tx )   (u x ( x, y )  t x )  min ,
Fy (t y )   (u y ( x, y)  t y )  min

(14)

In this case u x ( x, y ) and u y ( x, y) are longitudinal and transverse components, derived after
angle compensation. Equations (14) must be rewritten in discrete forms to obtain
N
M
N
M
Fx (t x )   ( uxi , j  t x ) / M  N  min , Fy (t y )   ( uyi , j  t y ) / M  N  min .
i 1 j 1
(15)
i 1 j 1
Such a functional can be treated as a function of a single variable. The problem has a solution if
the function is unimodal in some range of t x (or t y ) that can be chosen as that of extremum values of
x(y)-component displacements. Experimental vector fields meet this requirement, because total vector
length is limited and the sum of constant vectors is a single vector.
Therefore, any displacement vector field can be reduced to a normal form by compensation of the
rotation and constant vectors. Finally, we obtain a combination of an ideal vector and error fields
u0 ( x, y) and uer ( x, y) respectively.
comp
ucomp
( x, y )  uc  u0 ( x, y )  uer ( x, y ) .
,u ( x, y )  u
(17)
VECTOR FIELD LENGTH CHARACTERISTICS
The absolute error for the length of a particular vector uermean equals approximately to the DIC
measurement error uerDIC , which is usually a known value.
N
M
uermean   ( uxer (i, j)  u ery (i, j ) ) / M  N  uerDIC .
(18)
i 1 j 1
For most precise measurements, the direction of the vector is not important while the vector length
is. The modulo mean quantities and related dispersion are obtained using expressions as follows:
N
M
uxm   ( uxi , j ) / M  N ,
i 1 j 1
N
N
i 1 j 1
M
N
Dxm   ( uxi , j  u mx )2 / M  N ,
 my  Dmy ,
u0m  (u xm ) 2  (u ym ) 2 .
(19)
M
Dmy   ( uyi , j  u my )2 / M  N ,
i 1 j 1
 mx  D xm ,
M
u ym   ( uyi , j ) / M  N ,
i 1 j 1
m 
Dxm  D my 
Dm ,
(20)
where Dm is the variance,  m is the standard deviation.
In addition to the modular mean values, the algebraic ones can also be discussed. In mechanical
testing, the normalized vector fields are usually symmetrical relative to the line of applied force.
Therefore, their algebraic mean values tend to approach a minimum while their modulo values tend to
approach some maximum.
To illustrate these algebraic values, a modulus sign needs to be eliminated from eqs. (19), (20):
N
M
u   (uxi , j ) / M  N ,
a
x
i 1 j 1
N
M
u   (uyi , j ) / M  N ,
a
y
i 1 j 1
u0a  (u xa ) 2  (u ya ) 2
(19 )
N
M
N
Dxa   (uxi , j  u ax )2 / M  N ,
i 1 j 1
 ax  D xa ,
M
Dya   (uyi , j  u ay )2 / M  N ,
i 1 j 1
 ay  Day ,
 a  Dxa  Dya .
(20 )
A displacement field is referred to two consequent time moments t1 and t 2 , where t  t2  t1 .
Normalization of the algebraic or modulo characteristic on this interval gives us their rate equivalents.
Variations of a displacement field m and  a can be also used:
m   m / u0m ,
a   a / u0a .
In this case m can be considered as a field characteristic.
(21)
MODEL VECTOR FIELD
Let us demonstrate how a vector field can be viewed after compensation on constant vector. The
simpliest field u ( x, y )  u x ( x, y )  ex  u y ( x, y )  e y of elastic displacements on the specimen surface
(Fig.2,a) is given by two its components [9,10]:
u y ( x, y )  b  y
ux ( x, y)  a  x ,
,
(22)
where a and b are constants. These equations in digital format are given in (23):
ux ( j )  a  j , u y (i )  b  i .
Here 0  i  M , 0  j  M and M is even.
a
(23)
b
Fig. 2. Initial model field of elastic displacements (a) and its view after compensation on constant vector (b).
Now let found the equation for sum of all vector components. The sum of 0-th string is equal
L0  aS0  0  b  M  aS0 ,
M
where
S0   i  M  ( M  1) / 2.
The
sum
of
first
string
is
i 0
L1  aS0  1 bM  aS0  bM , sum for the k-index the sum is L1  aS0  1 bM  aS0  bM .
Summation over all M rows gives us formula (24) for total components length L :
L  aMS0  bMS0  (a  b)M 2 (M  1) / 2 .
Taking into account the constant vector, one can rewrite:
M
M
i 0
j 0
(24)
L   (b  i  c y )   (a  j  cx )  (bM / 2  bM / 2  c y M )  (aM / 2  aM / 2  cx M ) .
c
2
2
(25)
It is clear that sum is minimal if cx  aM / 2 and cy  bM / 2 . Usually M  1 . The vector
field will have symmetrical view owing to the constant vector uc  (M / 2)  (aex  bey ) (Fig.2,b).
Using (24) and (25), we get
Lc  (a  b)M / 2 .
(27)
2
It is observable from (24) and (27) that the sum of vectors component decreases by a factor M
after the compensation. Mean vector length of a field u0m (or u0mC for compensated field) is defined by
normalizing by dimensions:
m
u0m  L / M 2  (a  b) M / 2 ,
u0C
 LC / M 2  (a  b) / 2M .
(28)
Longitudinal vector component ux linearly changes from zero to maximum value uxmax  aM ,
then its mean length is uxm  aM / 2 , that is in accodance with eq. (27). The variance can be estimated
from (29):
M
m
2
2
Dx   (ux  a  j ) / M .
(29)
j 0
Obvious transformations give us
M
M
2
2
2
Dx  a 2  (M / 2  j )  a 2  (M / 4  M  j  j ) .
M j 0
M j 1
M
Taking into account that  i 2  M  ( M  1)(2M  1) / 6 , we obtain Dx  a  M /12 . In a similar
i 0
manner Dy  bM /12 . Finally we obtain the variance D :
D  Dx  Dy  (a  b)M /12 .
(30)
On the other hand, eq. (30) can be rewritten as
D  (u x
max
 u y ) /12 .
max
(31)
One can see (31) that the variance is proportional to dimensional change. If the vector field covers
all space of a specimen then this change, normalized on initial length shows us strain.
This example demonstrates that compensation by constant vector allows more clear viewing of the
vector field as well as that the variance of vectors components is related to particular vector field thus
showing its mechanical reaction.
CONCLUSIONS
The mathematical processing allows reconstructing a displacement field to normal form where
rotation has been eliminated and the constant vector found. This work discloses a procedure for
compensating the detrimental contribution of these two factors and reducing a vector field to a
normalized form. Such a vector field gives us the mean absolute and algebraic lengths as well as
related standard deviations. These characteristics are considered to be mechanical response of a
material.
ACKNOWLEDGMENT
The work was performed according to the Government research assignment for ISPMS SB RAS,
project FWRW-2021-0012.
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