Holpzaphel_-_Nonlinear-Solid-Mechanics-a-Contin
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2.3 l\llaterial, Spatial Derivatives |
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v1 = v-i (:1;.1> :1;2 , t), v2 = v2 ( x 1 , :r2 , t ).,-.v:1 = 0 the associated |
mot.ion is caned plane. |
A motion is potential if there exists a spatial velocity field |
v = grad lJ>, where the |
spatial scalar fie]d <I> is called the veloc-ity potential. A motion satisfying c;urlv = o is irrotational. By recalling relation (1.274) we conclude that if a motion is potential then it is automatic.ally an .irrotational motion.
EXERCISES
:1. |
For a one-dimens.ional problem the displacement field U = U (.X, t) is given by |
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the equation |
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U =ct)( , |
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with c denoting a constant. The relation between the spatial and the material |
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coordinates is given by :r = (1 + ct) .X. |
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Derive the first and second derivatives of U with respect to time t in both the |
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material and spatial descriptions. |
2. |
Recall Example 2.2 on p. 63. The spatial components a. 1 = ;r 1, a 2 = J:2 , a.~i = 0, |
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i.e. eq. (2.14), are derived directly from Lhe acceleration in the material descrip- |
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tion, Le. eq. {2.12)~ by means of given motion (2.10). |
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Show that the spatial components a.a, a = 1., 2, 3, may also be derived from |
expression {2..27) using eq. (2.13 ), i.e~ ·v 1 = :1:2 , v2 = :r 1, v:i = 0.
3. ln a certain region the spatial veloc.ity components of v = v(x, t) are given as
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;(··.a |
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;.2·) _l-/Jt |
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'l1'• |
= (\i ( '1~2.,r'' |
+ ..,.a) Ll -{Jl |
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l.. l |
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(l .I.. |
+ .J, |
t •l'2 (:, |
2 |
-4; • ' 1ol.I 2 |
• •2 L- |
where n, /3 > 0 are given constants. Find the components of the spatial acceler- ation field a ;::: a(x.~ t) at point (1, 0, 0) and time t = 0.
4. Consider a plane motion defined by the velocity components
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{J(T) ( •-r•. |
'l"• ·f··) |
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V··1 = |
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I. ~ ' ' 2 '.I |
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[h:1 |
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... |
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where cl> is a harmonic scalar field. Show that the plane motion is .irro.tational and find the spatial acceleration field.
70 |
2 Kinematics |
2.4 Deformation Gradient
One goal of th.is section is to study the deformation (i.e. the changes of size and shape) of a continuum body occurring when moved from the reference configuration Q0 to some current configuration n. Fo.r simplicity, we often omit the arguments of the tensor quantities in subsequent considerations.
Deformation gradient. As we know from Section 2.1, a typical point X E 0 0 identified by the position vector X maps into the point x E n with position vector x.. Now we want to know how curves and tangent vectors deform.
Consider a material (or undeformed) curve x = r(~) c no, (..YA = rA(~)), where f, denotes a parametrization (see Figure 2.4_). The material curve is associated with the reference -configuration n0 of the continuum body. Hence, the material curve is not .a function of time. During a certain motion x the material curve deforms into a spatial (or deformed) curve x = 1(~, t) c n, (:z;a = 1'n(~, t) ), at ti.me'/:.
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r.~·..i'(~, t) |
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.. . . |
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time t" = 0 |
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"Figure 2.4 Defonnation of a material curve r c no into a spatial curve I C n.
The spatial curve at a fixed time t is then defined by the parametric equation
x = 1'(Ct) = x(r(€)J) or
We denote the spatial tangent vector to the spatial curve as dx and the material tangent vector to the material curve as dX. They are de-fined by
dx == ,../(~ |
·t)d c |
dX = I''(c)d.C |
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(2.37) |
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with the abbreviation (e )' = D( •) / D~. In the literature the tangent vectors dx and dX, vihich are infinitesimal vector elements in the current and reference configuration (see
2.4 Deformation Gradient |
71 |
:...figure 2.4), are often referred to as the spatial (or deformed) line element and the
:,·material (or undeformed) line element, respectively.
~.{~·:=.:·::·By using (2.36) and the chain rule we find that 1'(~, t;) = (Bx(X, t)/DX)r'(~).
.. Hence, from eq. (2.37_) we deduce the fundamental relation
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.. : .
··:·:.= ..
dx = F(X, t)dX |
or |
(2.38) |
\vhere the definition
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"f:. . ·· .
F (X, t_) = |
·Bx(X, t) |
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t |
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· DX |
= Grad.x X1 |
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oxa |
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G I |
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-(2.39) |
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r:t |
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r aA = D~,. |
:rl"af ..\"~<\:fa |
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..--\A
"_is to be used. The quantity Fis crucial in nonlinear continuum mechanics and is a pri-
·inary measure of deformation, called the deformation ·gradient. In general_, F has nine
·,~.o~ponents for aU t,, and it characterizes the behavior of motion in the neighborhood
.of a. poi.nt.
Expression (2.38) clearly defines a linear transfonnation which .generates .a vector Clx·by the action of the second-order .tensor F on the vector dX. Hence, eq. (2.3.8) ·:~-e-rves as a transformation rule. Therefore, Fis said to be a two-point tensor involving points in two distinct configurations. One index describes spatial coordinates, :i:a, and _the other material coordinates., ..YA. In sum.ma.ry: material tangent vectors map (i.e.
:~r~~sform) into spatial tangent vectors via the deformation gradient.
.·.·... _We suppose that the -derivative of the inverse motion x- 1 with respect to the current _·position x of a (material) point exists so that
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ax-1 (x, t) |
=-= .gradX(x, t;) |
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F |
1(x, t) = |
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ax |
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(2AO) |
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DxA |
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or |
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,,.. |
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F 4a |
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gt cl(1a~ |
....-\ A |
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·xa |
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the tensor r-L is the inverse of the deformation g-radicnt. It carries the spatial line element dx into the material line element dX according to the (linear) transformu-
.tion rule dX = .F- 1(x, ·t)dx, or, in index notation, cLY;t = ~:;~-cb:a.
Generally, the nonsingular (invertible, i.e,. detF ¥= 0) tensor F depends on X ·which d~notes a so-called inhomogeneous de-lorm.ation. A deformation of a body in ques-
.t~.()~ is said to be homogeneous if F does not depend on the space coordinates. The co-mponents f:1A depend only on time. Every part of a specimen deforms as the whole ·does. The associated motion is called affine~ For a rigid-body translation for which the
.diSplacement field .is independent of X we have F = I, f;1..i = c5aA· However, if there
~s no .motion we have F = I and x = X.
·72 |
2 Kinematics |
EXAMPLE 2.5 For a .two-dimensional problem the deformation is given by the explicit equations
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1.v. |
(2.41) |
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·= 2 + ·-...{\. J - |
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..i\.') |
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Determine the matrix representation ..of the deformation gradient and its inverse and study the deformation of a unit square with reg.ion n0 (see Figure 2.5). Consider a material line element a 0 and a spatial line element b with unit lengths. Show how a0 deforms, and carry .out the :inverse ope.ration with b. For the components of a 0 and b take (1/ J21 1/ J2) and (1_, 0), respectively.
Solution.. |
By recalling definitions (2.39) and (2.40) we find |
after some simple |
algebra with the given defonnation (2.4 I) that |
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(2..42) |
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[FJ = .[ 3/; |
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The given deformation x carries the unit vector :a0 into its new position character-
.ized by a (see Figure 2..5). By means of (2.42) 1 and the given components of a0 we may specify the deformed .position of a0 , i.e.
-2 |
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] [ |
[a]= [F][ao] = · 3/2 |
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1/v'2]
1//2
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~'2 :'( -31 ] |
·1;•) |
(2.43) |
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.v ,c, |
la!= 5. - . |
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The inverse deformation x-1 carries the unit vector b into |
its undeformed position |
characterized by b0 • By analogy with the above, using (2.42.h and the given compo- |
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nents of b, we may specify .the undeformed position of b, Le. |
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[ho] = [F |
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Jib] |
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-1/5 |
2/5 |
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1 ] |
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-:3/5 |
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( 2 ) /'2 |
(2.44) |
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-;- |
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Iho I = . -=- |
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Observe that during deformation x the length of a 0 |
increases from l |
tip to 5 1/ 2 ., while |
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during the inverse map x-1 the length of b decreases from 1 up to (2/5) 112 (see Fig...
ure 2.5). Iii
.................................. ,.... ,,.. :.. , ...... ,, ........... ···:·:···:···,.. , .. :····'·''····:··' |
.................·,:.··:···=·····:···:'·". , ...·.··:·':······.·:··.·:··:·:··,. ···':······'."•:'·."···'··'··'·:,.... , ·············=··:'''''·"·."'·':·'····::··:··'·······:··::·..."·'····'."··':··:··'·"··'·"··'··:··...,.. ,....... ,.............................. . |
2.4 Deformation Grad·ient |
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4
• ':" ." , • ~ • • ~ • • , •, , ... , , : . • ( •• , : . , .. , • , , I "• "'. , "• :
3 ,., ..,.., ........... ,., ... .
I • ~ |
lbl = 1 |
2 ··------ .. :~·::·::·:Jr |
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1.-'----....
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laol |
= 1 |
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... . .-!-:-.·~. |
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-1 |
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C1 ... J. ·'... |
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bar·z .•·..: ·· . • . •••
Figure 2.5 Deformation of a unit square showing the map x of a material line element a0 into a and the inverse map x-1. of a spatial line element b into b0 •
Displacement gradient tensor. |
To combine the |
deformation gradient |
with the |
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displacement vector we deduce from (2.4) and definition (2.39) that |
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GradU = Gradx(X, t) - GradX |
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= F(X, t) - I |
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= FaA - c5aA · |
(2.45) |
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a-··\. .4 |
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The second-order tensor GraclU is called the d·isplacement gradient tensor in the mate.rial descr~ption. From (2.5) we find using definition (2.40) that
gradu = gradx - gTadX(x, t)
=I - F- 1(x, t) |
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(2.46) |
where the second-order tensor gradu is called the displacement ·gradient tensor in
the spatial description.,.
74 2 Kinematics ·
Note the following relationships between the material gradient and ·mate.rial divergence, and the spatial gradient and spatial divergence of the smooth scalar, vector and tensor fields, <I>, u .and A~ respectively. The fields are defined on the current configura-
tion of a continuum body. By the chain rule we have the useful properties |
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or |
D<I> |
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- =F |
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a• 'a |
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...c\. .."t |
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gradu = Gradu F- |
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av· |
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u. ,,, |
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Nanson's -formula. |
We already know that points, curves,· tangent vectors, for -ex- |
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.ample, X, r, dX, map onto points, curves, tangent vectors x, I~ .dx~ respective-ly. An arbitrary differential vector maps via the deformation gradient F (see the linear tra.ns.formation(2.3
However, .a unit vector N normal to an infinitesimal material (or undeformed) sur~ face e:Jcment dS does not map to a unit vector n normal to the associated infinitesimal spatial (or deformed) surt·ace element els via F, as shown in the following.
We perform the change in volume between the reference and the current configuration at
dv = .J(X, t)dlf |
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(2.50) |
J(X, t) == detF(Xl t) |
> 0 , |
(2.5 l) |
in which J is the detenninant of the deformation gradient F, known as the volume ratio (or Jacobian determinant). In (2.50), dl/ and dv denote infinites·imal vo]ume ele.ments defined in the reference and current configurations called material (or un.. deformed) and spatial (or deformed) volume elements, respect.ive.ly. Further, we assume that the volume is a continuous (or at least .a p·iecewise continuous) function of continuum particles so that dV' = d....Y1d..\"'2d.Y:1 and dv = fh1d:r:2d:r.:_1 (continuum idealization).
Since :F is invertible we have J(X, t) = det.F(X,.t) # 0. Because of the im-
penetrabiJity of matter, i.e. volume elements cannot have· negative volumes, we re- ject J (X., t) < 0 which mathematically is possible. Consequently, the volume ratio
J (X, t) > 0 must be greater than zero for all X E fl0 and for all times t. The inverse
of relation (2.5]) follows with identity (l.109) as J- 1 =
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introduced in eq. (2..40)..
If there is no motion (F = land x = X), we obtain the consistency condition J = 1~ since det.F = detl = .L However, a motion (or a defonnation) with J = 1 (at every
2.4 Deformation Gradient |
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p.artide in every configuration and time t) is called isocho,ric or volume-preserving. It keeps the volume constant.
In order to compute the relationship between the unit vectors n and N -consider an arbitrary material line element dX, which maps to dx during a certain motion X· We now ·express the infinitesimal volume element in the current con.figuration dv as a dot
product. By means of (2.50) we have the following relation |
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dv = ds ·dx = ,JdS · dX , |
(2.52) |
with ds = dsn and dS = dSN denoting vector elements of infinitesimally sma]) areas
~ie~ned in the current and reference configurations~ respectively. |
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With transformation (2.38) and identity ( 1.81) we ·may rewrite eq. (2.52) as |
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JdS) ·dX= 0 . |
(2.53) |
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Since (2.53) holds for arbitrary material line e·lements dX_, we find that |
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ds .= JF -~rdS. , |
(2.54) |
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Wmch shows how the vector elements of the infinitesimally small areas ds and d~ on the current and .reference configurations are related. Relationship (2.54) is well-known as Nanson's formula.
EXERCISES.
..- 1. Consider a parametric curve in space of the form X = r(~), with .
de-fining a ·helix. Find the length of the helix for 0 < ~ < II Cf lr'(c;)Id~).
.: :· 2. In a deformation of a three-dimensional problem, the displac-ement components
of u are found to be
'll - r1· |
1·r·1 |
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·l..._·,l-4~ |
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Compute the matrix repres.entations of F_,_1 and F and deduce that the defonnation is isochoric.
3. Let a -continuum body undergo a homogeneous de.fonnation whic.h is defined by where the components of the vector c and the tensor
A a.re constants or time9'.dependent functions ..
76 |
2 Kinematics |
(a) |
Show that [F] ="{A] and interpret this result. |
(b) |
Consider the components of vector c .and tensor A given ·by |
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(A] = |
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where a, f3, |
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Show that a particle that lies on a spherical |
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surface in the current configuration initially lies on the surface of an e] lipsoid.
(c)Consider (b) and detennine how a unit nonnal N of an infinitesimal small area dS in the reference configuration deforms. Take N = 1/v'3 (el - e2+ eJ).
2.5Strain Tensors
We have l~arnt from the ·preceding section that the deformation _gradient is the fundamental kinematic (second-order) tensor in finite deformation kinematics that characterizes changes of .material elements during motion. The aim of this section is to determine these changes in the form ·of (second-order) strain tensors related to either the reference or the current configuration.
Note that unlike displacements, which are measurable quantities, strains are based on a concept that is introduced to simplify analyses. Therefore, numerous definitions and names of strain tensors have been proposed in the literature. We discuss (and compare) the most com1non definitions of strain tensors established in nonlinear continuum
mechanics.
Material strain tensors.. We compute the change in length between two neighbor- ing points X and Y, located in region n0 , occurring during a motion (see Figure 2.6).
By neighboring we mean that X is 'close' to Y.
The geometry in the referen.ce co11figuratio11 is given by
y = y + (X - |
Y-X |
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XI = x + dX ' |
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de-= IY - XI , |
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(.2.56) |
clX .=dean |
ao = IY -·XI . |
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We denote the (material) length of the material line element dX = Y - X by de. It is
the distance between the neighboring points x |
E no and y E no, i.e. de = :f y - XI |
with ds /IXI << 1 (JXI # 0). The unit vector a0, |
laoI = 1, at the referential position X |
2.5 Strai·n Tensors |
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d~'~::.~tw;::::::::.':):r::,•
:<~~.,"!,,!i:::r:::::::i~;:I:::
time t
', .~:."'..·. :.:. :_....·...·..:_:·.. ,.......
'.., .....:_-.:_.:_:·.:··.'..::·,-,:.:.:·
.,.... ~:\;:,'/-:·..:....... ·
Au0 = ·Fao
fAl)ol =A
time ·t = 0
laol = 1
Figure 2.6 Deformation of a material line element with Jeng.th de into a spatial line element with length Ade.
describes the direction of the material line element (which may be imagined as a fiber), as illustrated .in Fi_gure 2.6. Hence., additional.ly, we find using (2.56).1 that
dX · dX = dc:a0 • dea0 = dc:2 • |
(2.57) |
Note that the vector quantities dX and n0 are naturally associated with the reference
~on.figuration of the body.
Certain motions transform the two neighboring points X and Y into their displaced positions x = x(X, t) and y = of region .Q, respectively. We now ask how close is x to y. Using Taylor's expansion according to (1.238}, {.l.23.9), y may be expressed by means of (2.55), (2.56) and the deformation gradient (2.39), as
y = x(Y, t) = x(X + dcao, t) |
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= x(X_, t) + dcF(X, t_)ao + o.(Y - X) , |
(2.58_) |
78 2 Kinematks
where the Landau order symbol o(Y-X) refers to a small error that tends to zero faster than Y - X ~ o. With motion x = x(X, t) and (2.56)2, (2.56)3 it follows subsequently from (2.58) that
y - x = dcF(X, t).a0 + o(Y - X)
= F(X, t)(Y - X) + o(Y - X) , |
(2.59) |
which clearly shows that the term F(Y - X) linearly approximates the relative motion
.y - x. The ·more Y approaches X the better is the approximation.$ the smaller is de- =
1v-x1.
Next, we define the stretch vector .Aa0 .in the direction of lhe unit vector a0 at
X E 0.o_, i.e. |
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An0 (X, t) = F(X, 't)ao , |
(2.60) |
with length ,.\ = l.A110 Icalled stretch ratio or simply the stretch (see Figure 2.6). Then, the length of a spatial line element (originally in the direction of a0 ), Le. the. distance
between the two neighboring places x and y., is obtained from (2.59) . by neglecting
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terms of order dc:2 • Using definitions (1.15) 1 and (2.60) we find with y - x:;::: dc-Fa0 that
(2.61)
.In summary: a material line element dX at .X with length de at time ·t = -0 becomes the length Ade at time "t. The stretch ,\ is a m·easure of how much the unit vector a0 has stretched. We say that a line element is extended, unstretched or com-pressed according to A > 1, ,.\ :;:::; 1 or A < L. respectively.
With definitions (l..15h, (2.·60) and property (1..8.1), the square of,.\ .is computed according to
,\2 = Ano · A110 == Fao · Fao
= a0 · Frr Fao = a0 · Ca0 , |
(2.62) |
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(2.63) |
where we have introduced the right Cauchy-Green tensor C as an important strain measure in material coordinates (F is on the right). Frequently in the literature C is referred to as the Green deformation tensor. From (2.63) we learn that to determine the stretch of a fiber one only needs the direction a0 at a point X E 0 0 and the se-cond- order tensor C.
Note that C is symmetric and positive definite at each X E f20 . Thus,
and u · Cu > 0 for all u =!= o . (2.64)
