OrthogonalizedStrained Boxes
Strain
It was argued above that strain in a medium may be assessedby examining the manner in which a cube of the unstrained material is distortedby forces within and outside the medium. It has also been argued that strain takes two basic forms called volumestrain and vector
The vector triple product
It was noted that if three vectors {a, b, g} form three edges of a parallelepiped, then thescalar part of their product is equal to thevolume of the parallelepiped.
The cube with edges a,b, and g is strained into a parallelepiped in which the edgesare rotated relative to each other and possibly compressed or lengthened, butthey still enclose the same matrix, after the strain.
When the edge vectors are mutually orthogonal, the vectortriple product is always a scalar. If we change the order of the vectors, the scalar may be either 1 or–1. The order used here is aright-handed coordinate system in which the earlier listed vectors act upon thelater listed vectors. In ,
is multiplied by
and the productis multiplied by
. It is equallyvalid to choose the cyclic permutations
or
and there aresituations when these permutations are more appropriate choices (seebelow). All these triple vectorproducts are equal to 1.0. Thecomplementary set of permutations in which the order is reversed,
,
, and
, will yield triple vector products of -1.
Since the choice of the vectors is arbitrary, except thatthey are mutually perpendicular unit vectors, this result is applicable to anythree mutually perpendicular unit vectors. We will generally choose orderings that reflects a righthand coordinate system, so that the volume will be positive.
Definition of Box
When we place a hypothetical unit cube in a medium and allowit to be strained along with the medium, we are effectively placing a frame ata point in the medium. The edge vectors extending from one corner of the cubeform a unit frame. Such a frame iscalled a box. A box is actually a set of extension vectors, because it haslocation and the lengths and directions of the vectors may change with theanatomical motion of the strain. Unlike a frame of reference, the edge vectors may move relative to eachother. The box does haveorientation and it makes sense to consider changes in its orientation, but abox is not a frame of reference.
The central concept of this chapter will be is strainframes. A strain frame is a way offinding a set of mutually orthogonal vectors that meaningfully represent theorientation of a strained box. Italso turns out that they also form the elements of a frame that may serve asthe basis of an anatomical description of the internal rotations between theedge vectors of a strained box.
When a box is placed at a point in a medium to measurestrain it is said to be a test box.
Volume Strain and Diagonal Strain
It is readily seen that if the orthogonal edge vectors arenot unit vectors, then the scalar of the product is the product of thelengths. This is what one wouldexpect of an index of volume, . If a unit boxis distorted so that the edges remain perpendicular, but change in length, thenthe box experiences a strain. Thisis intuitive if we consider a box that is stretched so that one edge vector,say
, doubles in length while the others,
, remain unit vectors.
The volume ofthe box doubles which is certainly a strain. Such a strain will be called a volumetric strain. Volume isclearly a scalar quantity.
The strain in this scenario is actually more complex than avolumetric strain. To see this,consider the following scenario. Suppose that one edge increases by a factor of 2.0, but the other twoedges are reduced by a factor of .
The volume does not change, therefore there is no volumetricstrain, but the box is clearly strained, because it is distorted relative tothe original cubic box. In orderto capture this strain we look at another relation between the three edgevectors, a feature reflected in the diagonal vector.
The diagonal of the vector of the elongated box rotates28.171¡ about an axis that is in the direction of relative to thediagonal vector of the original cubic box. This strain that changes the shape of the cube will becalled diagonal strain. There was alsodiagonal strain in the example where one edge doubled in length without theothers changing, but it was combined with a volumetric strain.
These three boxes have the same volume, therefore thereis no volumetric strain in moving from one to the other.
There are many ways in which a bit of a medium may bestrained, such as changing its location, its shape or its orientation.
A third example illustrates that volumetric strain can alsooccur in isolation. One can have avolumetric strain without changing the shape of the box.
The volume increases eight-fold and the diagonal becomestwice as long, but the shape of the box is not changed because the diagonal ofthe enlarged box is parallel with the diagonal of the original, unstrained,box. This type of strain is notcommon and it is less interesting than the more common types where therelationships between the edge vectors change.
Rotational or Vector Strain
When the edge vectors rotate relative to each other, so thatthey are no longer mutually orthogonal the strain quaternion has a vectorcomponent and the strain is a rotationalor vector strain.
Unlike volume strain, which does not depend on the orientationof the test box, vector strain is highly dependent upon the orientation of thebox relative to the strain. Thelast chapter addressed some of the properties of vector strain by computing thedistortion in a small spherical bubble of material centered upon a givenlocation and submitted to a particular strain. It gave some insights into the relationships between thestrain and the distortions experienced by test boxes with differentorientations. In this chapter, wewill examine another side of the strained test box and the strain quaternion.
Uniform and Directional Strain
The first example above was an instance of a directionalstrain and the last example is an instanceof uniform or isomorphicstrain. Fundamentally, the difference between the uniform expansionand directional expansion is that with uniform expansion or contraction, nomatter how we choose the box edges, they remain orthogonal after thestrain. If we are following aspherical bubble of material, it remains spherical although it may increase ordecrease in diameter. As aconsequence, any three orthogonal points remain orthogonal.
With directional expansion, the box edges remain orthogonalonly if we choose them to lie parallel with the directions of expansion orcontraction. A spherical bubble ofmaterial becomes an ellipsoidal bubble and it is only if we chose mutuallyorthogonal points on the initial sphere that become major or minor axes of theellipse that the axes remain mutually perpendicular.
If the triple vector product, that is, the strainquaternion, is a scalar quaternion after the strain, then the axes are alignedwith the directions of expansion and contraction. Any other choice of box edges will experience a rotationalor vector strain. That is, theirtriple vector product will have a vector component, therefore, it will be afull blown quaternion, with both scalar and vector components.
In the first example considered, we got a nil rotationcomponent because we happened to choose an arrangement of edge vectors that didnot change direction with the strain. The relationships between the directions of the edge vectors wereunchanged by the strain, they remained mutually orthogonal.
Multiplying the icomponents by 2 and leaving the other components alone and then multiplying outin the vector triple product yields the following.
Clearly, the first and last components of the productare not mutually perpendicular after the strain and that is reflected in thevector component of the strain quaternion, which indicates that they haverotated about an axis of rotation in the negative j
We can find a box that experiences only volumetric strainwhen the strain is compressive or tensile, by choosing one edge so that it liesparallel with the direction of expansion or contraction and a second edgeparallel with any other possible orthogonal strain(s). Such a box isconceptually straightforward to construct. In the situation that we have just been considering, thereis only the one strain, the stretching along the i
Because of the symmetry of the a and gaxes relative to the axis of strain their distortions happen to cancel out inthis particular instance and the diagonal is the same as for the unstrainedbox, namely {1, 1, 1}. However, ifwe choose a unstrained box that is the original cubic box rotated 45¡ about thediagonal of the box, then the diagonal of the strained box is not the same asfor the unstrained box.
The unstrained box is given by the following edge vectors.
The strained box is given by the following edge vectors,in which the i components are doubled.
The diagonal of the unstrained box is {1, 1, 1} and thediagonal of the strained box is {2, 1, 1}. The difference between the diagonals is in the i
Boxes With Non-Orthogonal Edge Vectors
If the edge vectors are not mutually orthogonal, then thevector triple product is always a quaternion, the scalar of that quaternion isthe volume of the parallelepiped, S(
Example 1.
Let ,
and
. The third edgevector is the direction of the diagonal of the unit cubic box.
The strain quaternion is as follows.
The new volume, , is 1.0 and the vertical vector,
, is tilted forward 54.7356¡ relative to the perpendicular tothe
plane about aunit vector axis of rotation that is in the direction of
.
The Strain Frame
It is possible to construct a set of mutually orthogonalunit vectors that are associated with the strained box, which give a definiteorientation to the box. Theconstruction is as follows.
The non-orthogonal strained vector set {
The unit vector perpendicular to the plane is designatedby the symbol
and the axis ofrotation for the
componentrelative to
is the unitvector designated by the symbol
. The vectorsare perpendicular to each other because
is perpendicularto the
and
is perpendicularto the plane determined by
and
. Consequently,
lies in the
plane.
We can complete the frame by computing the perpendicularto the plane determined by .
, is the unit vector parallel to the axis of rotation thatturns
into
.
, is parallel to the axis of rotation that turns
into
and it is alwaysin the plane determined by
and
. Consequently,these two vectors and their right-handed mutual perpendicular,
, form an orientation frame,
, for the vector triplet,
. If we take thetwo vectors
and
in that order,then the right hand mutually orthogonal vector is the ratio of
to
.
The strain frame.
We can readily compute , for the present situation, where the edge vectors are
,
and
.
The strain frame is not symmetric in that it gives a specialvalue to the and its perpendicular or normal vector,
. Still, it canbe viewed as an orientation frame in that it generally gives a uniqueorthogonal frame to a set of strained vectors.
The strain frame of an unstrained box is not uniquelydefined by this protocol, since and therefore
and
may be any twomutually orthogonal vectors in the
plane. It willbe convenient to leave the strain frame for an unstrained frame undetermineduntil it is compared with a strained frame at which time it takes on thenearest value to the strained frame. In the example just considered the ratio of the strained to theunstrained frame is 1.0 because we chose the value of the unstrained frame inwhich the
and
axes are equalto those of the strained frame and the
axis is equal to k inboth the strained and unstrained frames. There is not a change in orientation. That result conforms with our intuition that pure shear doesnot change the orientation of the medium.
The case of ![]()
Let us consider the general situation where the that turns
into
, that is r.
.
The angle is the anglebetween the
vectors.
In the formalism of vector analysis and with arbitrarylength vectors, is given by thefollowing expression.
. The unitvector
is perpendicularto both g and
is the anglebetween g and
The vector sis the unit vector of the axis of rotation for the shear that rotates theperpendicular to the base () relative to the base (the
). Therefore, inthe instance of shear in one plane, the vector component of the triple vectorproduct quaternion,
, gives one the axis of rotation of the shear,
If we return to the example above and retain , then the strain quaternion is readily computed.
The perpendicular to the . One can tellby inspection that these are the correct values for the two vectors.
We find that the unitary strain quaternion in this scenario()is composed of a volumetric strain and a shear strain.
The case of ![]()
The case where , but
is perpendicularto their plane is another interesting case to consider.
and
, therefore
. The principalfactor in the shear is that the perpendicular is reduced by a factor of
, where
and
when
isturned into
.
The product of and
is easilywritten.
The product is a quaternion with its vector in the directionof the perpendicular to the first two axes. The magnitude of the quaternion is the negative of the ratioof the length of to the length of
and the angle ofthe quaternion is the angle between them.
The third edge vector, , is aligned with the perpendicular,
, therefore the triple vector product may be written down.
The volumetric strain is proportional to the sine of theangle between and
and the shearstrain is proportional to the cosine of the angle times the negativeperpendicular to the
.
It is notable that the strain quaternion is the same if weuse the form or the
form for thetriple vector product. That meansthat when we use the first triple vector product in the next section, it willreduce to each of the results that have already been considered when one or theother rotation is reduced to zero.
We can illustrate this strain by allowing the strained boxto have the edge vectors ,
, and
. The volumequaternion is readily calculated.
The vector of the rotation is directed in the negative k and
axes is reducedby 45¡ from its angular excursion in the unstrained test box.
Note that the and
vectors areundefined in this situation, because there are an infinity of possible candidates.
is valid
since
is aligned with
and thereforethere is no specific rotation that rotates one into the other.
is the nullvector,
is alsoundefined. Consequently, we cannotdefine a strain frame in the usual way. However, if we reassign the vectors, so that the
edge is viewedas tilted relative to the
plane, then itis possible to construct a frame using the same analysis as in the previoussection. In that scenario,
. Changing theangle between the first and second axes does not change the orientation of thebox.
No mutually orthogonal edges
In each of these first two situations, there is a singleaxis of rotation and the strain quaternion could be expressed in terms of thataxis of rotation. We now considerthe situation in which none of the edge vectors is perpendicular to any of theother edge vectors.
It has already been established that can be writtenin terms of
.
The third edge, , can be written in terms of
.
These expressions can be combined to give the triplevector product.
The vector was defined tobe a unit vector that completed the right-handed strain frame with
and
, in that order, so that
.
This expression for the strain quaternion reduces to theprevious two expressions for the strain quaternion when or
. That is whatwe would expect since they are extreme examples of this situation.
If we are primarily interested in the form of the solution,then it is more convenient to use the unit strain quaternion.
We can use trigonometric identities to rewrite the strainquaternion in terms of the internal angles. Since
it follows that
The strain quaternion has been expressed as a function ofthe basis vectors of the strain frame and the angular excursions of twostrains. The volumetric strain isthe product of the two component strains, which is what one would expect.and
vectors changesthe volume in proportion to the sine of the angle between them and the tiltingof the vertical vector changes the volume in proportion to the cosine of theangle it forms with the vertical unit vector.
The axis of rotation for the vector component is dependentupon the relative magnitudes of the angular excursions. and
are near a rightangle
, that is the test box is only slightly distorted by thestrain, the vector of the strain quaternion lies near the
. As eitherangle moves away from a right angle the strain vector shifts towards the
axis. This canbe more easily appreciated if we normalize on the
component andreplace the tilt of
relative to thevertical with the interior angle,
. The more theedges converge, the proportionately greater the
componentbecomes. For small amounts ofconvergence (q and
approximately at right angles), the
component isrelatively small.
The volumetricstrain is a function of both interior angles.
Put in other words, as the test box becomes more distortedthe volume shrinks and the component in the direction of the vector becomeslonger. The components in thedirections of the
and
vectors becomeshorter.
Given the strain rotations of a cubic box , one can write down the strain quaternion,
. The strainquaternion allows one to compute the relations between the edges of thedistorted box.
The Inversion of the Generalized Strain Quaternion
We have explored the calculation of the strain quaternion inthe case where 1.) all the edge vectors are orthogonal, 2.) the casewhen the first and second edge vectors are not orthogonal,
3.) the casewhere the third component is not orthogonal to the first two,
and 4.) the casewhere none of the vectors are orthogonal to any other edge vectors
. The strainvector for the first case is the null vector. The second and third cases give
and
, respectively, as their strain vectors.
The expression for the strain quaternion in the fourth caseis more difficult in that there is interaction between the angular excursionsbetween the vectors, so that all components of the vector component of thestrain quaternion are functions of both angles and it is necessary to introducea third basis vector to the frame.
When the strain quaternion is computed for the fourth casethe expression is going to be in terms of , instead of
. However, oncewe have computed the three orientation frame vectors, it is simply a matter ofcomputing the projection of the strain vector upon each frame vector.
Then, we can write the strain quaternion as follows.
Given this quaternion, we can write down four equations thatallow one to determine the values of and
.
These equations lead directly to the values of and
.
Consequently, the angular excursion about the axis thatcarries
into
is
and the angularexcursion about the
axis thatcarries
into the
is
. The anglebetween
and
is
. Therefore,there is a fairly direct calculation that allows one to extract the angularexcursions for both distortions, given the strain quaternion for thegeneralized distortion.
An Example
Let us consider an example that utilizes theseobservations. The box is distorted into the box
. The strainquaternion is readily computed.
We loose no generality in substituting i
The vector isobviously
in thissituation.
The vector is theunit vector of the ratio of
to
.
The component is the ratio of
to
.
We can compute the projections of the vector component ofthe strain quaternion upon these frame vectors. We start with the equations from above.
Again, we write the equations from above for the angularexcursions of the rotations and substitute into the equations.
It is easily confirmed that the volume of the unit cube isreduced to , that the
edge vector isrotated –45¡ relative to the
edge vectorabout the
axis, and thatthe
edge vector isat an angle of 35.2644¡ to the
.
Another Example
In the last example all the edge vectors remained unitvectors after the strain. If thematrix is incompressible, then the unit vectors will become longer.. Then thestrain quaternion is the product of the three edge vectors.
This is very like the result that was obtained with the unitvectors, differing only in that there is not a term.
Summary:
We began this section on strain with the consideration of aninteresting mathematical relationship, namely, that the scalar of three vectorsis the volume occupied by the parallelepiped that has those vectors as its edgevectors. For a given strain, thechange in volume of a test box is independent of the orientation of the testbox, even though the box may be distorted in very different ways depending uponits initial orientation.
The same strain may create rather different appearingstrained boxes depending upon the orientation of the test box.
In the last chapter, we saw how the dependence of the vectorstrain upon the test box orientation is a function of the set of mutuallyorthogonal test box axes that one chooses from a unit sphere surrounding thelocation. The distortions of suchspherical bubbles into strain ellipsoids explain the changes in the test boxaxes with strain.
In this chapter, we returned to a consideration of strainedboxes and defined a new concept, the strain frame. It was shown that the strainquaternion can be expressed a function of the components of the strain frameand that one can invert the strain quaternion to obtain the rotations of thetest box axes. If no two edgevectors are mutually orthogonal, then the vector component of the strainquaternion is not obviously indicative of the internal rotations. and
axes.
axis.
projection becomessmaller.