Distortions in Media: Shear Strain

The compressive/tensile strain that was considered in thelast chapter was uniformly distributed within the medium.  When lateral movement was notconstrained, the strain led to a flow that was not uniform.  In this chapter, we will considerstrain that is, as it were, across the medium, without compression orexpansion. It also causes adifferential flow, which shears the medium. We will consider two types of shear, linear shear androtational shear. Linear shear isin a single direction throughout the medium. Rotational shear is generated when one parallel end platerotates relative to the other.

Linear Strain

Linear Strain in Crystalline Materials

In a crystalline material a shearing force is distributedevenly over the layers of the crystal, so the displacement is a linear functionof the distance from the fixed plane. Consequently, if the two surfaces are separated by a distance ‘w’ andthe moving layer is displaced a distance , then the shift at a location, , is proportional to its relative distance from the fixedplate to the moving plate, measured perpendicular to the fixed plate.

We might expect substances like bone to shear in thismanner. It is a very simpleanatomy.

Linear Strain in Amorphous Materials

Amorphous media will tend to flow when sheared and thedistribution of the displacement will not be uniform. Material near a fixed surface will not move as readily asmaterial at some distance from the endplates. We will encounter a situation approximating laminarflow. In laminar flow, thevelocity of flow is proportional to the square of the distance from the fixedsurface.

The shear of one platerelative to another while preserving the distance between them. The two parallel plates are separatedby w and the shear is s.

It is computationally easiest to assume flat upper and lowerboundaries for the matrix, which will be assumed parallel to the direction ofshear. For definiteness, assumethat the plates are separated by a distance w, the bottom plate is fixed andupper plate moves a distance in the directionof the shear, , and that the two plates are parallel and horizontal.  The amount of shear in any horizontalplane, between the two plates is given by a second order polynomial.

We specify that the displacement is zero adjacent to theplates. Consequently, we canevaluate the expression for the local displacement at those two locations andobtain the value of k as a function of the maximal displacement and thedistance between the two plates.

We do not know the value of the maximal local displacement, , but we do know the magnitude of the shear displacement andthe shear is the summation of all the local displacements.  From that, it is possible to computethe value of maximal local displacement.

The maximal local displacement is readily computed fromthis result by noting that the total displacement is .

We can now write the expression for the local displacementand the expression for the shear.

and

For any location , the shear will move it to a new location that is the samedistance from the plates, but moved parallel with the shear vector an amountthat depends upon its location relative to the fixed plate.

The shear profile is not a linear function of the distanceof the location from the plates, but there is not a strong curvature to theleading edge of the displaced matrix. The shear profiles are the same throughout the matrix between theplates.

The shear is symmetrical about the middle horizontal planemidway between the plates. Consequently, it will look the same whether the upper plate alone or thelower plate alone is moving or both are moving as long as the relative movementbetween the plates is the same.

Extension Frame

Let us construct an infinitesimal extension frame at thelocation . Also, let thestrain be in the direction of the yaxis. The change in the frame asthe medium experiences the shear strain described above will be given by thechanges in all the infinitesimal extension vectors

We subtract the strained location vector from each ofthe stained combination location and extension vectors to obtain the strainedextension frame.

Since is aninfinitesimal, we can set all terms with higher powers of equal to zero.

If we compute the strain quaternion, it is much as expected.

There is no change in volume, because the scalar of thestrain quaternion is unity, but the vertical axis is rotated about the axisthat is mutually perpendicular to the vertical axis and the axis in thedirection of the shear. Therotation of the vertical axis is nil when the vertical component is at theupper or lower boundary of the slab of material, meaning that there is no shearat the boundaries of the slab. Theangle of the strain quaternion () is plotted versus the depth () in a 2 unit thick slab in the following figure.

The assumed shears range from modest () to very large () in steps of 0.1 times the thickness of the slab.  Most biological shears would be near tothe low end of this range. However, between the surfaces in a joint, the shears might become quitelarge.

The orientation frame

As with the compressive/tensile strain, orthogonalizing theframe leads to a frame that is aligned with the original orientationframe. Consequently, theorientation is not changed by linear shear.

Rotational Strain

Another way that a shear strain may occur is if the mediumis subjected to a rotational force applied to its upper face that causes itsupper surface to rotate relative to its lower face. This type of force causes a rotational strain.

Let us assume a circular slab that lies between two platesand let the lower plate be fixed while the upper plate rotates about an axisthrough the center of the slab. Alocation (r) is designated by itsdistance from the center of the slab (r) and its height above the lower plate(h). The locus r is rotated about the axis of rotation () through an angle when the upper plateis rotated through the angle . We have todetermine the magnitude of from the anatomyof the system. As before, theoriginal locus is labeled by and the newlocus is . The change inlocation is . The rotationof the upper rim of the slab is given by a simple quaternion expression.

It is assumed that all r move in the same manner, tracing a circular trajectory in theslab. Upon that assumption we canreduce the problem to two dimensions in which the movement occurs in circularband concentric with the axis of rotation. We essentially peel away that bandand lay it flat for many of the calculations.

The upper (green) plateis rotated through an angular excursion of q about an axis ofrotation, r. Producing a rotational shear between the two plates.  For locations between the plates, l₀, the flow of the matrix is in a ring concentric withthe axis of rotation to an extent that depends upon the distance from the locationto the fixed plate (blue) and the separation between the two plates, 2c.

The initial location vector () is resolved into two component vectors, one parallel to theperpendicular, h, and one perpendicularto it, r.

We will also need to designate a tangential unit vector,, that is perpendicular to r and h.

The maximal shear, the movement at the interface between theslab of material and the moving plate is the angular excursion, in radians,times the radial distance, concentric with the axis of rotation.

However, the shear is along a circular trajectory,therefore net distance traveled is the final location minus the initiallocation.

Since , there is a tendency for the sheared matrix to movecentrally, however, that space is already occupied by the more central rings,so the matrix must flow in its own concentric ring. Still, there is a centrally directed tendency to compressthe matrix. If the matrix containsvertical fibers that link the two bounding surfaces, then the fibers will tendto pass directly between their attachment sites. The surface generated by a circle of such fibers will beconcave in its middle, giving the matrix a more hourglass-like shape.  This anatomy has been developed in somedetail elsewhere (Langer 2005n; Langer2005o),where the strains in the vertebral artery are computed from its anatomy.

Since a cylindrical ring is topologically a flat surface wecan, as it were, peel it away to treat the flow as we did in linear shear andlay it back into its original position to see how that flow, f, appears in situ.

We have already derived the expression for , given , when the shear is linear.

With several simple changes, a similar expression describesrotational strain. The total shearis an angular excursion, , and the perpendicular offset is . The change inlocation, , in radians, is also expressed as angular excursion.

The distance is the distancewithin the ring, rather than the chord that connects the initial and finallocations. Consequently, theexpression relating the initial and final locations may be written as follows.

With rotational shear, the final location is more conciselyexpressed as a quaternion product with the initial location.  Note that the angular excursion can bereduced to a scalar factor times the total angular shear.  This means that the angular excursionof the flow is independent of the distance from the axis of rotation, which isin accord with our expectations. This is of physical importance, because it means that the strain is uniformthroughout the radius of the matrix.

The distribution of shear is basically taking the linearshear and wrapping it on to a cylindrical surface. The general shape of the distribution was sketched in theintroductory figure for this section and it is plotted from detailedcalculations in the following figure.

Torsional Shear

There is a type of torsion in which the strain is notuniform radially. If the torqueforce is applied to the outer surface of the matrix, as in unscrewing a jarlid, then the strain differential is between the outer surface and the axis ofrotation. If the tangential forceis assumed to be uniform over the outer surface, then the strain is greatest atthe outer surface and it is attenuated as one moves centrally.  However, unless the center is fixed,the mass will begin to rotate as a unit and the strain will be resolved.  A variant of this scenario that ispotentially more interesting is when the mass is fixed or retarded by otherforces and a torque force is applied to a portion of the outer surface.  For instance, when the mass is a boneshaft and the torque is due to the pull at a muscle attachment.  That problem is far more difficult andtherefore will be deferred until we have more experience with strain.

Extension frames for rotatory shear

A convenient and logical frame for extension is a unitvector in the direction of r, the unittangent at , and a vertical unit vector. The manner in which the radial and tangent vectors have beendefined ensures that if they form the first and second extension vectorsrespectively, then the extension frame will be right-handed if the verticalvector points in the direction of the axis of rotation.

The original location moves to a new location given by theexpression given above.

The transformation of the first component of theextension frame is given by the following expression.

If , then –

The second component, the tangent vector, transforms inmuch the same manner, but we must use the rotation for conical rotation,because the differential vector is not in the same direction as the locationvector.

In both cases the infinitesimal extension axis is rotatedaround a vertical axis to the same extent as the location.  That is essentially what one wouldexpect from the anatomy of the situation. If we orthogonalize by constructing the third axis of the extensionframe as the ratio of the second axis to the first axis, then the orientationchanges as if multiplied by a unit quaternion with its vector in the verticaldirection and its angle equal to the angular excursion of the location vector.

The last component of the extension frame transforms much asthe vertical component of the linear shear transformed.  The calculations are a bit morecomplex. We must use theexpression for conical rotations, but we may use or as the locationvector. The angular excursion ofthe rotation of the compound vector is a function of the infinitesimalextension vector.

The first term is the expression for in , so may be expressedas , the excursion of the location plus anadditional excursion . Theratio of to is the tilt ofthe vertical axis in the tangent plane to the ring at the new location.  The tangent plane rotates with thelocation, so, the actual tilt is the tilt rotated through the angular excursionof the location, .

The vector rotateshorizontally, so the vertical magnitude of the vertical component does notchange with the rotation so the strain quaternion has a scalar of unity.  The volume does not change.  That is what one expects with a shearstrain.

Orientation Frame

The first two components of the strain frame remain mutuallyorthogonal and they rotate in the horizontal plane through an angle equal tothe change in location. If weorthogonalize the strain frame by setting the third component equal to theratio of the first two components, then the resulting orientation frame rotatesthrough the angle about the axisof rotation for the slab,. Consequently,in this instance, the orientation does change with the strain, but the changeis relatively simple, being the same for all points in the matrix.

Summary

In many ways shear strain is simpler thancompressive/tensile strain. Thereis a uniformity and simplicity in the distributions of strain that was notpresent with the unconstrained compressive/tensile strain.

As with the compressive/tensile strain it is not necessarilythe case that the distribution of strain is a second order function of thedistance from a moving plate. However, it is impossible to gain much insight ifwe do not assume some distribution and the second order distribution yieldsreasonable outcomes that accord, at least qualitatively, with commonobservation. It might beinteresting to assume distributions of strain other than linear or second orderfunctions and explore their implications for the overall distribution of shear.

While the rotational and linear shear cases are verysimilar, it was interesting to explore the rotational case because of thegreater use of quaternion analysis in the rotational model.  Quaternions make the analysis verystraightforward, basically as easy as with the linear shear anatomy.  The rotational shear anatomy allowed usto generate a model where there is a change in orientation in the medium, allbe it very simple.

We have not explored all the possibilities for shearanatomies. As alluded to earlierin the chapter, we might explore the consequences of a shear applied to thelateral wall of the slab or we might consider a matrix in which there arefibrous elements that make the strain anisotropic or force the strain to be inthe most direct line between two moving points.

We might also combine compressive/tensile strain with shearstrain as with a wrenching movement that screws surfaces together whilerotating them with respect to each other. We might consider situations in which one end plate tilts relative theother as when a vertebra tilts forward and backward upon another.  We might consider the combination ofstrained elements as with the gelatinous nucleus pulposus confined within thefibrous annulus fibrosus in a lumbar intervertebral disc.  However, the intent here was to illustratethe application of the tools that have been developed to a few simpleanatomies, therefore, these more complex situations will be deferred to anothertime and place.