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1. What is deformation?Let o denote the initial configuration of a physical body (e.g., an airplane part, a blob of putty, etc.). As illustrated in Fig. 1.1, let X denote the initial position of a point in the body. Disallowing fracture (where particles break into two separated particles), each point in a body can have only one initial (pre-deformation) location. Consequently, the initial position vector, X , can be regarded as a unique name of any particular particle of interest. When we say particle X we are referring to the particle that was initially located at position X . Let denote the configuration of the body after deformation. The goal of this chapter is to define the standard methods and lexicon used to quantify the relationship between and o . Because each material particle X moves to a new location, x and because no two distinct particles are allowed to deform into the same location, the key relationship between and o is the mapping function, x = ( X ) that relates the initial and deformed position vectors. Often, a mapping function does not need to be found per se. Instead, kinematics is founded on the notion that a mapping function exists. Figure 1.1 shows a pseudo 2D deformation in which a square grid has been conceptually painted onto the material. We have filled in a few of the material elements (grid cells) with solid color assist with some upcoming discussions. Of course, since the out-ofplane thickness has changed (by about 50% from the looks of it), this deformation is actually three-dimensional. A pseudo 2D deformation has out-of-plane thickness changes, but no out-of-plane shearing; motion is characterized principally with respect to the two inplane coordinates. Most of the concepts in this document will be illustrated in the context of pseudo 2D deformations, but they apply equally well to general deformations in which material elements can shear and rotate in arbitrary directions.
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Lo L
u x
Figure 1.1. A general 2D deformation. A slice of putty is deformed into a new configuration in pseudo two-dimensional manner. The cross-section and the out-of plane length change arbitrarily, but no out-of-plane shearing is allowed. In this sketch, a grid has been conceptually painted onto the material to facilitate visualizing the deformation.
DeformationSince each particle occupies its own unique location in space in both the initial and deformed configurations, we may assert that there exists a one-to-one function (called a mapping function) such thatMaterial interpenetration? Nonphysical!
x = (X) (1.1) Examples of mapping functions are given in Figure 1.2. A locally admissible, but Section 5. Naturally, no single material element globally inadmissible deformation. (i.e., no grid cell in Fig. 1.1) should be permitted to invert. This constraint is called local admissibility of the deformation, and it is ensured if the matrix [ x i X j ] has a positive determinant. In finite element simulations, a technique called hourglass control is typically used to ensure local admissibility. However, local admissibility is not sufficient to guarantee global admissibility, which requires that the mapping function in Eq. (1.1) be invertible and therefore prohibits material inter-penetration. Figure 1.2, for example, shows a globally inadmissible deformation that is everywhere locally admissible. In computational mechanics, contact algorithms must be used to avoid the globally inadmissible phenomenon of material interpenetration. We will soon discuss in great detail the deformation gradient tensor, F , 2Copyright is reserved. Individual copies may be made for personal use. No part of this document may be reproduced for profit.
whose ij components are given by [ x i X j ] . Occasionally, problems in mechanics specify only how the F tensor varies in space without making explicit reference to the mapping function, x = ( X ) . In these problems, another admissibility condition (called ) ensures that the mapping function exists even though it is the compatibility condition not being used explicitly compatibility ensures that the F tensor field is integrable even if we choose not to integrate it.
DisplacementThe displacement of a particle is a vector extending from a particles initial location to its deformed location (see Fig. 1.1): uxX (1.2)
The initial and final position vectors, x and X , are origin-dependent (they depend on where you place the origin). The displacement vector u is a free vector it wont change displacement can be generalized if you move the origin. The definitions of positions and to permit different origins to be used in the initial and deformed configurations. If the deformed origin is located by extending a vector c from the initial origin, then x in the above expressions must be replaced by c + x . The origin-to-origin vector c does not vary with position it is spatially constant.
The sketch of deformation (Fig. 1.1) shows a square grid conceptually painted onto the undeformed body, and some of the grid cells have been filled in with solid color to help us better visualize the deformation. Note that each little material square deforms to a shape that is close to a parallelogram in shape. This approximation (squares-to-parallelograms) becomes exact in the limit of an infinitely refined grid. In three dimensions, cubes deform into parallelepipeds. On surfaces embedded in 3D space, infinitesimal squares deform into infinitesimal parallelograms. The deformation gradient is a tensor that quantifies both the 3D and 2D shape change as well as overall material rotation, making it superior to strain as an all-encompassing measure of deformation of material elements. If you have the deformation gradient tensor, then you can compute the strain; the converse is not true. The purpose of this section is to describe how to graphically approximate the deformation gradient tensor by simply looking at the deformation. We will postpone giving an exact mathematical definition of the deformation gradient tensor.
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Any cube can be characterized by the three orthonormal vectors that form its edges. Likewise, the deformed parallelepiped can be characterized by the three vectors that form its edges. The deformation gradient tensor F quantifies the changes in these edge vectors into columns of a 3 3 matrix of compoby assembling the three deformed edge vectors nents. For rectangular Cartesian coordinate systems, the columns of the deformation gradients component matrix contain the deformed parallelepiped edge vectors, expressed relative to the undeformed edge vectors that defined the initial cube. By relative, we mean that all length changes are expressed as multiples of the initial cubes edge lengths, and all directions are expressed relative to the initial cubes edge directions. Consequently (by appropriately defining our unit of length) we may pretend that the initial infinitesimal cube is a unit cube whose three edge vectors are aligned with the orthonormal laboratory basis, { E1, E2, E3 } . Upon deformation, these edge vectors deform into a new set of vectors, { g 1, g 2, g 3 } , that define the edges of the deformed parallelepiped. If you find the of the i th deformed edge vector and assemble those components into the lab components i th column of F , then the result will be the component matrix for F with respect to the basis. Mathematically, this means that { E 1, E 2, E 3 } (lab) gk = F Ek or, equivalently, g k = F jk E j , (2.2) (2.1)
where F jk are the jk components of F with respect to the lab basis.* The k th column of [ F jk ] contains the lab components of g k . These equations provide us with a method for approximating the local deformation gradient at a point whenever we have accurate pictures of the deformed cube (or, for 2D, the deformed square). The deformed g k edge vectors are not generally orthogonal or of unit length. They are called material vectors because they travel along with the same set of material points (rotating with and stretching in proportion to the grid lines). In general, if M is a material flow with the vector in the initial configuration (not necessarily of unit length), then it will material to become a new (stretched and rotated) vector m in the deformed configuration, given by m = F M , and the fiber stretch is defined = m M .(2.3)
* Comparing Eqs. (2.1) and (2.2), you might (wrongly) think that we should have written Fkj instead of F jk in Eq. (2.2). To see why Eq. Eq. (2.2) is correct, note that any vector v may be expressed in terms of the lab basis as v = v k E k , where v k = v E k . Stated differently, any vector v may be writ ten v = ( v Ek ) E k , or, since the dot product is commutative, v = ( E k v ) E k . This identity applies for any vector v . Applying it to one of the stretched edge vectors, g k , gives g k = ( E i g k ) E i , or, substituting Eq. (2.1), g k = Fik E i , where F ik = E i F E k .
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Thus, if the F tensor is constructed from knowledge of how the unit lab base vectors material at a point, then you can immediately compute how any material would flow with vector (not necessarily aligned with the lab basis and not necessarily of unit length) would flow with the material. This is the fundamental characteristic of a linear transformation. We will explain later that F ij = x i X j . The mapping function x = ( X ) is generally nonlinear, but its derivative at a point is linear with respect to small changes in position. This statement is analogous to what a freshman calculus student encounters when first introduced to the concept of differentiation. A function y = f ( x ) might be nonlinear, but one can always (for smooth functions) define a straight (i.e., linear) line that is tangent to the curve at any point and the slope of the line is given by dy dx . The deformation gradient tensor, F ij = x i X j , plays a role in three dimensions that is analogous to the tangent slope dy dx in one dimension. The difficulty lies in visualization. In one dimension, we can always plot the function y = f ( x ) and then show the tangent line on the graph. In three dimensions, we must understand the mapping function x = ( X ) in a more abstract manner by talking about what it does (locally) to little cubesof material. Our perspective here would be analogous to describing the 1D mapping y = f ( x ) by stating that a little line segment dx deforms into a new line segment dy that differs in length by a ratio s = dy dx so that dy = sdx . For generalization into 3D mapping, this last equation becomes d x = F d X . Before refining this loose mathematical discussion, lets spend some time constructing some deformation gradient tensors graphically. What we will be doing here is analogous to taking out a ruler and estimating the local slope of a 1D curve y = f ( x ) by measuring local rise over run.
EXAMPLESConsider the top material element sketched in Fig. 2.1. Zooming in on the undeformed and deformed shapes of these elements, and introducing a ruler for which the initial cube has unit length gives the following drawing:
E2 Undeformed
g = 1.78 E 1 + 1.27 E 2 2
g 1 = 1.3 E 1 0.05 E 2
Figure 2.1. A graphical determination of the deformation gradient This sketch shows a zoomed in picture of the small material element at the top of Figure 1.1 on page 2. A ruler has been drawn on this figure so that the undeformed square is a unit square.
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Using the rulers as a guide, we can read off the components of each of the deformed edge vectors to obtain: 1.3 { g1 } = 0.05 and 1.78 { g2 } = 1.27 (2.4)
Assembling these components into columns of a 2 2 matrix gives [ F ] = 1.3 1.78 0.05 1.27 The fiber stretches are 1 = g 1 E 1 = 1.3 and 2 = g 2 E 2 = 2.2 (2.6) which mean that an infinitesimal fiber originally aligned with E 1 will, after deformation, have a length 1.3 times as large. A fiber originally aligned with E 2 will change length by a factor of 2.2. The deformation in Fig. 1.1 is actually three dimensional the out-of-plane thickness has increased by about 50%, which corresponds to an out-of-plane stretch ( L Lo ) equal to 1.5 . Consequently, the E 3 base vector has deformed to become g 3 = 1.5 E 3 . Conse quently, the pseudo 2D deformation gradient in Eq. (2.5) can be upgraded to 3D as 1.3 1.78 0 [ F ] = 0.05 1.27 0 0 0 1.5(2.5)
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If we zoom in on the middle element in our original sketch and graphically measure the deformed material element edge vectors, then the deformation gradient is found as follows:
g = 1.5 E 2 2 E2 Deformed Undeformed
E1 Component arrays: 1 { g1 } = 0 0 { g2 } = 1.5
g = E1 1
Figure 2.2. A graphical approximation of the local deformation gradient This figure shows an enlarged picture of the middle material element shown in Figure 1.1 on page 2. This material element happens to be in a state of uniaxial strain in the 2-direction.
Finally, the material element in the lower left of Figure 1.1 on page 2 is deformed according to a deformation gradient approximated graphically as follows:
g = 0.4 E 1 + E 2 2Deformed
E1 Component arrays: 1 { g1 } = 0 Figure 2.3.
0.4 { g2 } = 1
Graphical determination of the deformation gradient for the bottom element in Fig. 1.1
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The first column contains g 1 and the second column contains g 2 . Specifically, 1 2 2 and { g 2 } = { g1 } = 1 1 4
--E g = 2 E1 1 42 2
rmed Defo
g =0.5 E 1 E 2 1
Figure 2.4. Sketching the deformation when the deformation gradient is known This deformation corresponds to the deformation gradient in Eq. (2.8).
Being only a collection of four numbers, the deformation gradient matrix in Eq. (2.9) does not have much obvious physical meaning by itself. But, by drawing the associated deformation, we can now see that this deformation is essentially a movement of the material consisting of stretching it in the 2-direction, then shearing it in the 1-direction, and finally rotating it clockwise.
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Note that the tensors are multiplied in reverse order of application. Changing the order of application will result in a different material deformation (See Fig. 2.5).
stretch horizontally
rotate 45 degrees
Figure 2.5.
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Therefore, the deformation gradient tensor must be invertible. This is the mathematical statement of local admissibility, discussed on page 2.
Tracking surfacesThe gk vectors move with the material. They always lie along the same set of points (more correctly, theyre always tangent to and stretch with the grid lines). In three dimensions, the triad { g 1, g 2, g 3 } forms the edges of a parallelepiped that defines the deformed element. shape of the material Upon deformation, a face of the parallelepiped is defined by the two material edge vectors that form that face. If, for example, we consider the face formed by g 2 and g 3 , then the normal to this face is proportional to g 2 g 3 10Copyright is reserved. Individual copies may be made for personal use. No part of this document may be reproduced for profit.
Even though the edge vectors move with the material, the outward normals to planes do not move with the material. Consider the deformation shown in Fig. 2.4, for which the deformation gradient is 2 [F] = 1 1 -4 1 -2
To extend this example to three dimensions (so that we can talk about planes), lets suppose that the deformation is actually that of a thick plate (no stretching occurs in the outof-plane direction), so the 3D deformation gradient is1 -2
2 0(2.21)
We have already explained that g k comes from the k th column of F , which means gk = F Ek (2.22) Now we introduce complementary vectors k defined by i = F T E i (2.23) Note that i g k = ik (2.24) Incidentally, because the vector i obeys Eq. (2.24), it is called the dual or contravari in publications that employ convected coordinates, the ant vector associated with g k and, g i (which is distinguished from g because it uses a supervector i is typically denoted i will continue to use the symbol for the dual vector. script rather than a subscript). We i Suppose, for example, 1 is perpendicular to both g 2 and g 3 ; hence, it must be per as the normal goes, the pendicular to the material surface formed by g 2 and g . As far 3 for computational convenience. magnitude of 1 is inconsequential and may be chosen Specifically, our definition in Eq. (2.23) corresponds to ( g2 g3 ) 1 = -------------------- - , where g o g 1 ( g 2 g 3 ) go In three dimensions [ FT ] 1 4 0 2 = ----15 8 2 0 0 01(2.25)
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The columns of this matrix contain the lab components of the v k vectors, sketched in Fig. 2.6. This picture shows quite clearly that normals to material planes are not material vectors the dont flow with the material.
N2 E2 Undeformed
N1 E1 g 1
g 2
Figure 2.6. Tracking material surfaces This deformation corresponds to the deformation gradient in Eq. (2.8). The deformed edges are coincident with g i = F E i outward normals to the material surfaces are parallel to i = F T Ei .
In general, if a material plane (i.e., one that moves with the material) has a normal N (not the required to be of unit length) in the undeformed configuration, then after deformation, vector = FT N (2.27) (again not generally of unit length) will be normal to the plane in the deformed configuration. If the unit normal is desired, then this result can simply be normalized.
rmed Defo 1
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Area vectorsLet M 1 and M 2 be any two material vectors in the undeformed configuration. The parallelogram area of the formed by these two vectors is given by Ao = M1 M2 (2.28)
If the material vectors are infinitesimal, then this area vector will be infinitesimal. Upon deformation, the M k vectors deform into new stretched and rotated vectors, m 1 and m 2 , vector that define a new area A = m1 m2 Substituting Eq. (2.3) into this expression gives A = ( F M 1 ) ( F M2 ) Using an identity from tensor analysis, this result can be written as A = F c ( M1 M 2 ) or A = F c Ao (2.32) (2.30) (2.29)
where F c is the cofactor of F . Since the deformation gradient is invertible, the cofactor may be alternatively written F c = ( det F ) F T (2.33)
Recalling that J = det F and also recalling that M 1 M 2 is just the undeformed area vec tor, A o , Eq. (2.32) maybe written A = J F T A o This famous result is called Nansons relation.(2.34)
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This conclusion holds because (recall) the deformation can be described through a mapping: x = ( X) (2.37) In a rectangular Cartesian coordinate system, the components of the deformation gradient tensor are xi F ij = ------ Xj In matrix form, x1 x1 x1 -------- -------- ------- X1 X2 X 3 x2 x2 x2 [ F ] = -------- -------- ------- X1 X2 X 3 x3 x3 x3 -------- -------- ------- X1 X2 X 3
Homogenous deformationsWe have described the deformation gradient tensor in the context of inhomogenous deformations. Referring to the definition in Eq. (2.38), note that the deformation gradient tensor itself generally varies with position. Consequently, the deformation of one infinitesimal material cubes generally will be different from that of another cube located elsewhere in the body.
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A homogenous deformation is a very special kind of deformation in which the deformation gradient has the same value everywhere in the body. In this case, Eq. (2.36) can be integrated to give the mapping function, x = F X + c, (2.40) where c is a constant of integration representing material translation. For homogenous deformations, straight lines deform to straight lines, planes deform to planes, cubes (no matter how large) deform to parallelepipeds, and spheres deform to ellipsoids. Contrast this with an inhomogenous deformation such as the one illustrated in Fig. 1.1, where originally straight grid lines become curved and only infinitesimal cubes deform to parallelepipeds. These distinctions are illustrated in Fig. 2.7.
Initial Configuration
Pure Rotation
Homogeneous Deformation
Figure 2.7. Distinctions between pure rotation, homogenous deformation and inhomogenous deformation.
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Displacements for Homogenous deformations. Recall that, for a homogenous deformation, x = F X + c, (2.41) where F is spatially uniform and c (which is also spatially uniform) represents the dis of the particle originally located at the origin in the reference frame. placement The displacement field, u = x X , is therefore u = H X + c , where H F I Of course, Eq. (2.41) can be alternatively written
X = F 1 x C , where C F 1 c (2.43) Since neither F nor c vary spatially, the new alternative translation vector C is also spa The vector tially constant. C is the displacement of the particle currently at the origin. Using Eq. (2.43), the displacement, u = x X , can be written 1 u = h X + C , where h I F (2.44) Both Eq. (2.42) and Eq. (2.44) will give the same result for the displacement field. EXAMPLE: Consider purely circular motion about the 3-direction. Since the particle at the origin does not move, we have c = C = 0 . For circular motion about the x 3 axis, the deformation gradient is given by cos sin 0 [ F ] = sin cos 0 0 0 1 Therefore x 1 = X 1 cos X2 sin x 2 = X 1 sin + X 2 cos x3 = X3 We can introduce cylindrical coordinate systems such that X 1 = R cos , X 2 = R sin x 1 = r cos , x 2 = r sin Then Eq. (2.46) becomes r cos = R cos cos R sin sin r sin = R cos sin + R sin cos , or, using trig identities on the right hand sides, r cos = R cos ( + ) r sin = R sin ( + ) , 16Copyright is reserved. Individual copies may be made for personal use. No part of this document may be reproduced for profit.
(2.47) (2.48)
In other words, under circular motion, the radial coordinate remains unchanged while the angular coordinate increases by the rotation angle. Now lets look at the far less obvious results for the displacement vector. It would be easy to fall into the trap of thinking that the radial component of displacement must be zero. However, even though the radius does not change, the radial displacement is nonzero. To prove this methodically, lets go back to the RCS result of Eq. (2.46). The displacements are found by simply subtracting the initial Cartesian coordinate X k to give u 1 = X 1 ( cos 1 ) X 2 sin u 2 = X 1 sin + X 2 ( cos 1 ) u3 = 0
From here, we can introduce a cylindrical coordinate system. The base vectors for the system are different in the reference and current systems. Namely, the reference cylindrical basis is E R = cos E 1 + sin E 2 E = sin E 1 + cos E 2 and therefore, with respect to this system, u R = u E R = u 1 cos + u 2 sin = cos 1 u = sin
Mk = f Ek . (2.56) o If these vectors become { m 1, m 2, m 3 } in the deformed configuration, then you can into a second matrix f so that likewise assemble their lab components mk = f Ek . (2.57) We seek the deformation gradient tensor that connects the initial and final configuration directly. In other words, we seek a tensor F such that mk = F Mk . (2.58) Solving Eq. (2.56) for E k gives E k = f 1 M k . Substituting this result into Eq. (2.57) o gives m k = f f 1 M k o Comparing Eq. (2.58) with Eq. (2.59) tells us that F = f f 1 o(2.59)
Method 2: Another way to achieve the same result is to employ dyads and dual bases. Again, let { M 1, M 2, M 3 } denote three linearly independent infinitesimal material vectors 3D configuration. Let { , , } be the associated dual (or contrain the undeformed 1 2 3 variant) vectors. That is, M2 M3 M3 M 1 M1 M2 , , , 1 = -------------------- 2 = -------------------- 3 = --------------------Mo Mo Mo (2.61) where M o ( M 1 M 2 ) M 3 If the material line segments { M 1, M 2, M 3 } deform to become new line segments tensor may be constructed by { m 1, m 2, m 3 } , then the deformationgradient F = m k k3
Here, the symbol denotes dyadic multiplication. For any two vectors a and b , the dyad, a b , is a second order tensor whose ij component equals a i b j . Eq. (2.60) and Eq. (2.62) will both give the same result. Example: Suppose that a mounting bracket for the landing gear on an airplane was photographed during the most recent inspection of the plane. Following an unusually rough landing, a new photograph revealed that the bracket had undergone some severe deformation as shown in Fig. 2.8. We will use the lines formed by the ACME landers company logo to determine the deformation gradient for the bracket. For the purpose of this exam-
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ple, we will presume that the bracket underwent homogenous deformation. Of course, in real life, the deformation would likely be inhomogenous and the technique illustrated below would have to be applied using smaller line segments (such pairs of intersecting microscopic scratches) to find the deformation at various points in the bracket.
Afterd lan er s
ers d n a l
Figure 2.8.
Introducing rulers as shown, the straight lines forming the company logo can be regarded as vectors. The line below the word ACME will be taken to be M 1 , pointing up and to the left; the line below landers will be M 2 , pointing up and to the right. Before deformation, the rulers show these vectors to have lab components as follows: M 1 = 0.875 E 1 + 0.7 E 2 M 2 = 0.45 E 1 + 0.125 E 2 After deformation, the logo vectors deformed into m 1 = 1.1 E 1 + 0.1 E 2 m 2 = 0.325 E 1 + 0.175 E 2 We will presume that the bracket did not deform at all out of plane so that(2.63) (2.64)
(2.65) (2.66)
M 3 = E 3 and m 3 = E 3 (2.67) Using method #1 for finding the deformation gradient tensor, the f is constructed by o assembling the lab components of the undeformed vectors into columns: [f ] = o 0.875 0.45 0 0.7 0.125 0 0 0 1
[f] =
Applying Eq. (2.60) gives [ F ] = [ f ] [ f ] 1 o 1.1 0.325 0 0.875 0.45 0 = 0.1 0.175 0 0.7 0.125 0 0 0 1 0 0 11
1.1 0.325 0 0.2945 1.060 0 0.860 0.496 0 = 0.1 0.175 0 1.6495 2.062 0 = 0.259 0.467 0 0 0 1 0 0 1 0 0 1
Solving the same problem using method #2, we must first construct the dual vectors associated with { M 1, M 2, M 3 } . To start, note that M 1 M 2 = 0.4244 E 3 (2.71) (2.72) M 2 M 3 = 0.125 E 1 0.45 E 2 (2.73) M 3 M 1 = 0.7 E 1 0.875 E 2 Therefore, referring to Eq. (2.61), M o ( M 1 M 2 ) M 3 = 0.4244 1 = 0.29455 E 1 + 1.0604 E 2 2 = 1.6495 E 1 + 2.06186 E 2 3 = E3 , Recall that the deformed vectors were measured to be(2.74)
m 1 = 1.1 E 1 + 0.1 E 2 (2.76) (2.77) m 2 = 0.325 E 1 + 0.175 E 2 Referring to Eq. (2.62), we need to compute the dyads, m 1 1 , m 2 2 , and m 3 3 : 1.1 0.324005 1.16644 0 m 1 1 = 0.1 0.29455 1.0604 0 = 0.029455 0.10604 0 0 0 0 0 0.325 0.5360875 0.6701045 0 m 2 2 = 0.175 1.6495 2.06186 0 = 0.2886625 0.3608255 0 0 0 0 0(2.78)
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0 0 0 0 [ m2 2 ] = 0 0 0 1 = 0 0 0 1 0 0 1 Now, applying Eq. (2.62) adding up the above tensors gives 0.860 0.496 0 [ F ] = 0.259 0.467 0 0 0 1
which agrees with the result obtained using method #1. Obviously, for this particular problem, method #1 is superior in its computational simplicity. The fundamental concepts associated with method #2 are nonetheless important to understand and often easier to work with in theoretical discussions. Strain rosette example: a strain rosette orients three strain gages at equal ( 120 ) angles. To ensure measurement at a single point, the gages are centered at the same location; however, to analyze the rosette, you can imagine that the gages are actually placed along edges of an equilateral triangle and then presume that the material in the triangle deforms homogeneously by the same amount as the original central point. Strain gages return information about length changes but not orientation changes. In principle, if you could accurately measure the orientation changes of the first two gages, then the third one would not be necessary. The redundant third strain gage directly measures the relative orientation change between the first two gages. Thus, strain rosettes really measure only stretch they do not characterize rotation. The full deformation gradient (which includes rotation) can always be constructed by first analyzing the rosette with the fiducial assumption that the first gage does not rotate. This will produce a fiducial deformation gradient and the actual deformation gradient can be constructed by then superimposing the actual rotation of the first gage onto the fiducial deformation. In three dimensions, constructing a deformation gradient is more complicated because the reorientations of two gages must be tracked.
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3. Polar decompositionA rotation is a special kind of deformation in which material vectors permissibly change orientation, but they dont change length. In this case, it can be shown [see gobag rotation document] that the associated deformation gradient will be orthogonal (its inverse will equal its transpose). Furthermore, since element inversions are prohibited, rotations will be proper orthogonal (determinant will equal +1 ) Rotations are discussed in great detail in the gobag Rotation document. A stretch is a completely different special kind of deformation in which there exist three material vector orientations in the 3D initial configuration that will change in length, but not in orientation. In this case, the deformation gradient tensor will be both symmetric and positive definite. The proviso about positive definiteness is important. If a deformation gradient tensor is symmetric, that does not necessarily mean it is a stretch. For example, a pure rotation of 180 will result in a symmetric deformation gradient, but its not a stretch because it possesses a negative eigenvalue. Being symmetric, a stretch is diagonal in its principal basis. The principal values, called principal stretches, equal the ratio of deformed to undeformed lengths of the three non-rotating material fibers. Material fibers that are not aligned with the principal directions of a stretch will change orientation, but for every fiber rotating one way, there will be another rotating in the opposite direction, making the net rotation of material fibers zero for a pure stretch deformation. The polar decomposition theorem is a mathematical statement that the (local) deformation of a material element may be regarded as a combination of a pure stretch and a pure rotation. You can stretch a material and then rotate it or vice versa. Recall that order of application of sequentially applied deformations matters. Therefore, if you want to end up with the same deformation, then you must conclude that the stretch used in the stretchthen-rotate scenario would have to be different from the stretch used in the rotate-thenstretch scenario. You might also suspect that the rotations would also have to be different in the two scenarios. While it is true that the stretches are different, it can be proved that the rotation is the same in both cases. The polar decomposition states that there exists a rotation tensor R , a stretch-first tensor U and a stretch last tensor V such that Stretch then rotate:
Note that the Jacobian is J = det F = det U = det V . The following page summarizes key polar decomposition formulas (proof omitted).
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POLAR DECOMPOSITIONF = RU First STRETCH then ROTATE
F = VR First ROTATE then STRETCH
This figure shows a deformation for which principal Hencky (logarithmic) strains are 40% and the material rotation is 150 . This deformation can be accomplished via two stages: stretch and rotation. The polar decomposition theorem says that you will get the same final result regardless of whether you rotate first or rotate last. The key governing equations are: U = ( FT F ) 1 / 2 V = (F FT)1 / 2 = R U RT 1 1 R = FU R = V F For 2D deformations, the rotation tensor can be directly computed by F + F 22 F 12 F 21 1 - 11 [ R ] = --------------------------------------------------------------------- ( F 11 + F 22 ) 2 + ( F 12 F 21 ) 2 F 21 F 12 F 22 + F 11 Then U = R T F and V = F R T
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Recall that the deformation of an infinitesimal line element d X is given by dx = F dX In terms of the polar rotation and stretch, this becomes
dx = R U dX (3.4) Broken down in this manner, we see that U d X quantifies the stretching of the material that results strictly from the material dis fiber d X , as well as the part of the fiber rotation tortion (i.e., its shape change only). Under the pure stretch, U d X , fibers that are aligned with the principal direc with the principal directions of U wont rotate. Fibers not aligned tions will rotate, but for any fiber that rotates one way, there will exist a different fiber rotating in the opposite direction. Consequently, the overall material rotation under a pure stretch will be zero. The total deformation is obtained after the pure stretch by the application of the rotation tensor R , which rotates all fibers about the same axis by the same angle of rotation. Hence, R is really is an excellent measure of overall material rotation. EXAMPLE: Consider the deformation in Eq. (2.81): 0.860 0.496 0 [ F ] = 0.259 0.467 0 0 0 1 This is a two-dimensional deformation, so the rotation tensor is found by F + F22 F12 F 21 1 - 11 [ R ] = --------------------------------------------------------------------- ( F 11 + F 22 ) 2 + ( F 12 F 21 ) 2 F 21 F 12 F 22 + F 11 1 - 1.327 0.755 = -----------------------------------------------------2 2 ( 1.327 ) + ( 0.755 ) 0.755 1.327 = 0.8692 0.4945 0.4945 0.8692(3.6)
Since, for 2D deformations, R 11 = cos and R21 = sin , where is the rotation angle, we note that = 30 . The reference stretch is[ U ] = [ R ] T [ F ] = 0.8692 0.4945 0.860 0.496 = 0.876 0.200 0.4945 0.8692 0.259 0.467 0.200 0.651
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Note that the off-diagonals ( V 12 and V 21 ) are nearly equal to zero, which means the principal directions of the spatial stretch just happened to turn out to coincide with the lab basis for this particular example. Also note that the 11 component of the stretch is nearly equal to 1. For this particular problem, if it werent for computational round-off, V 12 and V 21 have turned out to be identically zero and V 11 would have been identically equal to 1. We are able to make this assertion because we know how we created Fig. 2.8 (which corresponds to this deformation) in the first place. That figure was created by using the computers drawing tools to first rotate the undeformed configuration by an angle = 30 and then to reduce the height of the rotated drawing by approximately 50% (note that V 22 is nearly equal to 50% of unity). Since the drawing was obtained by first rotating and then stretching, we know that the rotation angle computed from the polar decomposition had to come out equal to our initial rotation angle, and the spatial stretch had to come out to be a 50% vertical compression, as it has! We could have alternatively constructed the drawing by using a stretch then rotate sequence involving a 50% compression of the initial configuration along a line oriented at 30 followed by a rotation of +30 . In general, the principal directions of stretch are neither coincident with the lab directions nor related in any way to the rotation angle. Incidentally, upgrading the above stretches and rotations to 3D gives 0.8692 0.4945 0 [ R ] = 0.4945 0.8692 0 0 0 1 0.876 0.200 0 [ U ] = 0.200 0.651 0 0 0 1 0.993 0.006 0 [ V ] = 0.006 0.534 0 0 0 1
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4. StrainA infinite number of strain definitionsThe deformation gradient is truly the best measure of deformation. It is especially attractive because it contains information about both stretch and material rotation. For one-dimensional homogenous uniaxial deformations, the axial strain is often defined asL Lo - = 1, = -------------Lo
L = ---Lo
The parameter is called the axial stretch. Axial strain is often defined alternatively as L Lo - = [ 1 1 ] = -------------L Many advanced materials models define the strain as = ln (4.3) Note that all you really need to describe the amount of deformation is the stretch . All of these strains definitions are of the general Seth-Hill form,* 1 - [ k 1 ] = -k The parameter k characterizes the strain definition. By appropriate choices for k , any of the above strain definitions can be achieved. For example, Eq. (4.1) is obtained by choosing k =1 , while k =2 Eq. (4.2) corresponds to k = 1 . The log strain in Eq. (4.3), also known as Hencky strain, is obtained in the limit as k 0 . All of these various strain definitions are equivalent to first order when length k0 changes are small (i.e., when is close to Figure 4.1. 1). Thats because a two-term Taylor sures. series expansion of Eq. (4.4) centered at =1 gives 1 , regardless of the value of k .* Referred to as Erikson-Doyle in the rheology literature. (4.4) (4.2)
k =2 k0 k = 2
k = 2
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The variation of strain with stretch is shown in Fig. 4.1 for various choices of the SethHill parameter k . All of the strain measures are negative in compression ( < 1 ) and positive in tension ( > 1 ) . For positive choices of k , extreme compression toward the limit as 0 results in the strain measure approaching a finite limit. Physically, one would expect that an infinite load would be required to compress a material down to nothing. Consequently, if a choice k > 0 were to be used in a linear stress-strain relationship, = E , this behavior would imply that complete compression (a physical impossibility) could be obtained via a finite load. Such a result is nonphysical and therefore a nonlinear constitutive model must be used when using k > 0 in a highly compressed material. Otherwise numerical solutions will exhibit compression instabilities. When choosing a negative Seth-Hill parameter ( k < 0 ) , the opposite happens: positive choices for k become unstable if a linear stress-strain relation is used in a highly tensile (large stretch) application.* The logarithmic strain ( k 0 ) is the only strain choice that avoids both of these (compression and tension) pitfalls. This does not mean that a materials stress-strain relationship will necessarily be linear with respect to a logarithmic strain measure; it only means that a first-order Taylor series expansion of the actual (generally nonlinear) stress-strain function is likely to be more accurate than other strain measures. Other strain measures can be used, and the same level of accuracy can be obtained if more terms in a Taylor series expansion are employed. Recall that a stretch is the ratio of deformed to undeformed lengths. For general deformations, we have already shown that the Jacobian ( J = det F = det U = det V ) equals analog of the linthe ratio of deformed to undeformed volumes. Thus, J is the volumetric ear stretch and we may define volume strains in a manner consistent with the linear strain definition given in Eq. (4.4). Namely, the consistent volumetric strain is defined 1 - [Jk 1] v = -k(4.5)
This volumetric strain is well posed for any 3D deformation. Soon, we will define strain tensors and then prove that the consistent volumetric strain is not equal to the trace of the strain tensor except when k =0 (i.e., log strain).
* This is less likely to be troublesome in practice because usually some other inelastic failure mechanism (e.g. fracture) usually initiates well before the unstable tensile regime is reached.
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Strain tensorsFor 3D deformations, linear strain measures can still be defined in terms of the ratio of the deformed to undeformed lengths of an individual material fiber. However, for 3D, differently oriented fibers will experience different amounts of stretch. Furthermore, two material fibers can change their relative orientations under 3D deformations, so we need to quantify that phenomenon as well. We have already emphasized that the deformation gradient tensor already does an excellent job in this respect. The deformation gradient is, by far, the superior measure of material deformation. Nonetheless, a lot of people are more comfortable with using strain instead of the deformation gradient, so we will describe now how to compute strain from a deformation gradient tensor. Keep in mind that strain tensors lose information (about rotation). You can construct any of the strain tensors if you have the deformation gradient, but you cannot do the reverse. The 1D definition of strain given in Eq. (4.3) was phrased in terms of the ratio of deformed to undeformed lengths. We can extend this concept to 3D by using the polar decomposition theorem. The stretch tensor from the polar decomposition is positive definite and symmetric and positive definite. Consequently, it is diagonal in its principal basis and all of the eigenvalues are positive. In the absence of material rotation (i.e., if R = I ) any material fiber oriented in one the i th principal stretch directions will change length by a factor equal to the eigenvalue, i , but it will not change orientation. This suggests that we can use the Seth-Hill generalized strain definition in each of these directions, and then assemble the resulting three principal strains into a tensor that is diagonal with respect to the stretch directions. When there is material rotation (i.e., if R I ) then two strain defini one that uses the spatial tions are possible, one that uses the reference stretch U and we can upgrade the Seth-Hill stretch V . Hence, to define a 3D generalized strain tensor, strain measure in Eq. (4.4) to either a reference strain, 1 - [ Uk I ] = -k or a spatial strain, 1 - [ Vk I ] = -k Of course, if k =0 is used, these expressions become, in the limit, = ln ( U ) and = ln ( V ) (4.8) (4.7) (4.6)
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In general, for non-even powers of k ,* computing U k [or ln ( U ) ] requires expressing U the k th power in its principal basis, raising each principal stretch to [or taking logs of the principal stretches if log strain is desired], and then transforming back to the lab basis. Specifically if [ U ] is the matrix of lab components of U and if a matrix [ Q ] is con eigenvectors of U , then structed to contain the lab components of the orthonormalized [ U ] = [ Q ]T [ ] [ Q ] (4.9) where [ ] is a diagonal matrix containing the principal stretches associated with the principal basis. The lab components of the general Seth-Hill strain are found by 1 - ( [ Q ][ k ][ Q ]T [ I ] ) [ ] = -k (4.10)
Similarly, components of the logarithmic strain with respect to the lab basis are found by [ ] = [ Q ] [ ln ] [ Q ] T (4.11) where [ ln ] is a diagonal matrix containing the natural logs of the principal stretches. For the logarithmic strain, note that tr = ln ( 1 ) + ln ( 2 ) + ln ( 3 ) = ln ( 1 2 3 ) = ln J = v (4.12) The logarithmic strain is the only large-deformation strain measure whose trace equals the consistent volumetric strain. Some sketches of pure stretch deformations and their associated logarithmic principal strains are given on the following page.
* We will explain later why an eigenvalue decomposition is not necessary for even Seth-Hill parameters. Thats the key reason why even powers are used at all -- they are simpler computationally.
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Undeformed configuration
Hencky strain: 0%
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Rotation: 60 deg
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For pure stretch deformations, there is no difference between and defined in spatial and referEqs. (4.6) and (4.7). However, whenever there is material rotation, the ence strains will differ from each other by = R RT ,(4.13)
where R is the rotation from the polar decomposition theorem. This distinction (illus the previous page) becomes of paramount importance when constructing matetrated on rial constitutive laws for anisotropic materials. Modeling an anisotropic material requires specification of the material orientation. If a constitutive model is constructed using the spatial strain, then that model must transform the initial material orientation into the new (rotated) orientation prior to applying the model. Additionally, the principle of material frame indifference demands that models that use spatial strain must introduce special objective rates when ever the constitutive model is applied in rate form. When using the reference strain, however, neither rotation of the material orientation nor special rates are required. The price paid for using reference strain measures is the spatial strain (which is whats usually available from structural mechanics host codes) has to be transformed back into the unrotated configuration prior to application of the constitutive model, and then the output of the model must be transformed back to the spatial configuration. Computationally, the reference strain option is more efficient and guaranteed to satisfy the principle of material frame indifference. We already mentioned that an advantage of the logarithmic strain is that its trace equals the volumetric strain. Furthermore, the log strain tends to be the choice that makes the stress-strain function most linear in uniaxial tests. However, the logarithmic strain requires a polar decomposition in order to obtain a stretch tensor, and the polar decomposition is labor-intensive because it requires an eigenvalue analysis to take the square root of a tensor. Moreover, computing rates of the logarithmic strain is extraordinarily complicated because, in addition to the principal stretches being a function of time, the principal directions also change in time. For many people, these disadvantages far outweigh the advantages. Recall (from the polar decomposition summary sheet) that U = ( F T F ) 1 / 2 . Conse be written quently, the generalized Seth-Hill reference strain in Eq. (4.6) may 1 - [ ( FT F )k / 2 I ] = -(4.14) k All of the disadvantages we cited for the logarithmic strain will apply whenever the SethHill parameter not an even (and nonzero) integer. Even values for the parameter k are computationally attractive because there is no need to compute the square root of a tensor. In particular, the choice k =2 corresponds to the what is called the Lagrangean strain tensor. With this choice, the square root in Eq. (4.14) goes away, giving 1 T -[F F I] E = -2 (4.15)
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The Lagrange strain is a reference strain. Consequently, it may be regarded as a strain that results from the pure reference (right) stretch prior to the rotation part of the deformation. Superimposing even more rotation will have no effect on the Lagrange strain. Anisotropic constitutive models that use the Lagrange strain in conjunction with a reference stress measure do not require rotation of the material orientation into the current state. These models will automatically satisfy the principle of material frame indifference without needing to introduce any special objective rates. The Lagrange strain can be readily computed directly from the deformation gradient tensor (no polar decomposition is required), and its rate is also easily found by 1 - [ FT F + FT F ] E = -2 (4.16)
A fundamental identity from continuum mechanics states that the rate of the deformation gradient is related to the materials spatial velocity gradient, L v x , by F = LF (4.17) Therefore, Eq. (4.16) can be written E = FT D F , where 1 - [ L + LT ] D -2 (4.18)
The tensor D is often called the rate of deformation, which is an unfortunate and mis because, despite having some rate-like properties, it is not the proper rate of leading name any tensor. By this, we mean that it is possible to construct closed deformation paths* for which the time integral of D is not zero. Frequently in mechanics, it is useful to introduce a general unrotation operation, denoted by an overbar, and defined for any spatial second-order tensor Y by T YR YR (4.19) Using the overbar notation, Fig. 4.18 can be written E = VDV where D RT D R and V = FT R = RT V R = U * i.e., deformations that begin at one configuration, move through other configurations, and then return to the starting configuration. (4.22) (4.21)
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Recalling that u = x X , note that du d(x X) dx dX H = ------ = ------------------- - = ------ ------ = F I dX dX dX dX Solving for F in Eq. (4.24) and substituting the result into Eq. (4.15) gives 1 - [ H + H T + H T H ] This is eq. (3.19) in the book. E = -2
This is the formula one typically finds in an elasticity textbook, whereas continuum mechanics textbooks typically favor the more elegant and computationally simple Eq. (4.15).
1 - [ U2 I ] = R U I -(4.29) 2 In the complete absence of material stretch (i.e., when U I ) the tensor + I does finite stretches, for such indeed approximately equal the polar rotation R . However, an interpretation does not apply and the WHR tensor is essentially useless. Under a pure stretch (i.e., a non-rotational deformation where F = U ), the WHR tensor becomes 34Copyright is reserved. Individual copies may be made for personal use. No part of this document may be reproduced for profit.
= e E , where e is the engineering ( k = 1 ) strain measure. Thus, since engineering equal to Lagrange strain, the WHR tensor will not be zero or the identity even not strain is under pure stretches. It is a lousy measure of material rotation. For finite stretch problems, forget about the WHR, and use the polar rotation tensor instead. Incidentally, noting that the last term in Eq. (4.28) is the Lagrange strain, the WHR tensor may be written = FIE (4.30)
Therefore, once again, choosing a Seth-Hill parameter that is an even integer has the computational advantage that no polar decomposition is required. The Eulerian strain e is a spatial strain measure corresponding to a Seth-Hill parameter k = 2 : 1 - [ I ( F FT )1 ] = 1 -- [ I ( F T F 1 ) ] = -2 2 35Copyright is reserved. Individual copies may be made for personal use. No part of this document may be reproduced for profit.
Recalling that u = x X . Differentiating this with respect to the spatial position vector gives du d(x X) dx dX - = -------------------- -----h = ---- - = ---- - = I F 1 dx dx dx dx Solving for F 1 in Eq. (4.38) and substituting the result into Eq. (4.36) gives 1 - [ h + hT hT h ] = -2 (4.38)
SS = V I (4.41) Here SS means small stretch (i.e., each principal stretch is nearly equal to 1). For small stretch deformations, any identity that was true only for a particular Seth-Hill parameter becomes true (approximately) for the small stretch tensor as well. For example, the volumetric strain is in general equal to the trace of the strain tensor only for logarithmic strain definitions. However, for small stretches, this statement holds approximately regardless of the strain definition. Even if stretches are small, however, there might still be large rotations, so SS SS (4.42)
* The phrase small stretch does not mean the principal stretches are close to zero; it means they are close to 1.
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In order to assert that all strain definitions are approximately equivalent, we must add the proviso that the rotation is small as well. This means that the deformation gradient tensor is approximately equal to the identity and therefore the displacement gradient tensor is approximately zero. If the displacement gradients are small, then the last term in Eq. (4.25) is negligible, and Eq. (4.25) becomes 1 - (H + HT ) SDG = -2 (4.43)
Here, SDG stands for small displacement gradients. For small displacement gradients, not only is the stretch small, but the rotation is small too. Therefore SDG = U I , or 1 - (H + HT ) + I U = -(4.45) 2 For small displacement gradients, the infinitesimal rotation tensor is approximately 1 - (H HT) R SDG = -2 (4.46) (4.44)
This approximation must be handled carefully since this expression does not correspond to an orthogonal tensor.
Volume strainIn the section on the polar decomposition, we noted that the Jacobian J equals the determinant of stretch, which is simply the product of the principal stretches: J = 1 2 3(4.47)
Furthermore, in the section on the deformation gradient, we showed that the ratio of deformed to initial volumes equals the Jacobian VJ = ----Vo
Thus, the Jacobian is the volumetric analog of the linear stretch, = L L o . When a strain is defined using a particular choice k for the Seth-Hill strain parameter, then 1 - ( Uk I ) = -k 37Copyright is reserved. Individual copies may be made for personal use. No part of this document may be reproduced for profit.
The consistent definition of volumetric strain uses the same value for the Seth-Hill parameter so that 1 - (Jk 1) v = -k(4.50)
In this section, we will show how the consistent volumetric strain is related to invariants of the strain tensor we will prove that it is not equal to the trace of the strain tensor except under conditions of small strain or when k 0 . From Eq. (4.49), we know Uk = I + k Therefore, taking the determinant,(4.51)
J k = det ( I + k ) (4.52) Using a fundamental theorem from 3D matrix analysis, the determinant on the right hand side can be expanded to give J k = det ( I ) + I C : ( k ) + I : ( k ) C + det ( k ) (4.53) where the superscript C denotes the cofactor. The cofactor of the identity is just the identity itself, so the second term on the right hand side simplifies to kI 1 , where I 1 tr . Also, ( k ) C = k 2 ( ) C , so the third term becomes k 2 I 2 , where I 2 is the second invariant of strain, given by I 2 = tr ( ) C . Finally, the last term is just the third invariant I 3 times may be written k 3 . Thus, the above equation J k = 1 + kI1 + k 2 I 2 + k 3 I 3 and the consistent volumetric strain becomes v = I 1 + kI 2 + k 2 I 3(4.55) (4.54)
For arbitrary deformations, note that the volume strain will be identically equal to the trace of the strain only if k =0 . In other words, the logarithmic (Hencky) strain measure is the only one whose consistent volume strain measure is given by the trace of the strain tensor. Of course, for small deformations, the strain measures become indistinguishable because I 3 I 2 I 1 , so approximating the volumetric strain by the trace of the strain is acceptable in that case.
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These deformations have different stretches and different velocity gradients but they both have the same rate of deformation tensor. This alone should tell you that theres something strange about the rate of deformation tensor. It fails to comprise the same information as the true rate of a stretch tensor.
The right hand side is linear with respect to F , so it follows that there must exist a fourth order tensor Y , which depends on F , but is independent of F , such that ) (4.61) ( D ij = Y ijmn F D = Y :F mn Specifically, 1 1 ) - ( F 1 + jm F in Y ijmn = -2 im jn(4.62)
Lets suppose (for the purpose of contradiction) that there does exist some strain tensor . If so, then the chain rule would require that that is a function of F such that D = d E = ------ :F (4.63) dF , then comparing with Eq. (4.61), it would have to be true that If D = d - = Y -----(this will be contradicted) (4.64) dF or, using Eq. (4.62), ij 1 1 ) ------------ ( F 1 + jm F in = -2 im jn Fmn(this will be contradicted) (4.65)
On the surface, this looks like a fairly straightforward set of partial differential equations (one equation for each of the ways to select values for the free indices). You might throw caution to the wind and simply launch into trying to integrate this system (if you do, keep 1 equals ( F 1 ) , not 1 F ). However, you will find that no matter how in mind that F rs rs rs a solution. hard you try, you cant find This system has no solution. There are two ways to prove it. The first way looks at an integrability condition. At the root, we need to prove that Y :F is not an exact differential If it were, then Eq. (4.64) would be valid, which would imply that 2 E ij 2 Eij Y ijpq Y ijmn -------------- = ------------------------ = ------------------------ = ------------ F pq F mn F pq F pq F mn Fmn(this will be contradicted) (4.66)
Conversely, if this condition were true, then Eq. (4.64) would be integrable. Hence, we need to show that Eq. (4.67) is false. First we need a lemma:1 Frs 1 F 1 ----------- = F rj ks F jk
Y ijmn 1 1 F 1 ) -------------- = ----- ( F 1 F 1 + jm F ip qn 2 im jp qn F pq Swapping roles of mn with pq gives Y ijpq 1 1 F 1 ) ------------- = ----- ( F 1 F 1 + jp F im nq F mn 2 ip jm nq The integrability condition requires that Y ijmn Y ijpq --------------- = ------------- Fpq Fmn
for all F , or (this will be contradicted) (4.71)
1 F 1 + F 1 F 1 = F 1 F 1 + F 1 F 1 im F jp qn jm ip qn ip jm nq jp im nq
This relationship must hold for all F for Eq.(4.65) to be integrable. This relationship does not hold for all F , but its not real obvious why. To prove that its false for at least one F , we need to contrive a counterexample, but the right counterexample is not very evident. This leads us to an alternative way to prove that there does not exist any tensor such that D = t2 t2
and therefore
dt 0 , then
Below, we prove that neither D nor D are true rates, which is why careful writers dont often call D the stretching instead of the rate typeset them as rates. Careful speakers be confused with stretch. of deformation, but we worry that such a term might To demonstrate that D is not the time rate of any proper function of F , we need a path that is closed in F for which D dt 0 . For this purpose, consider the following four-stage deformation (illustrated in the figure):[ F] = 1 0 (1 t) + 1 0 t 01 0 2 [ F] = 1 0 (2 t) + 1 1 ( t 1 ) 02 0 2 [ F] = 1 1 (3 t) + 1 1 ( t 2 ) 02 0 1 [ F] = 1 1 (4 t) + 1 0 ( t 3 ) 01 0 1
for 0 < t < 1 for 1 < t < 2 for 2 < t < 3 for 3 < t < 4
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= 0.
The corresponding symmetric part of the velocity gradient can be computed using (4.57) to give[D] = 0 0 1 0 ---------1+t 0 14 14 0 0 0 1 0 --------t4 0 12 0
t = 0
t = 1
t = 2
dt D 0
dt + D dt + D dt + D dt D 0 1 2 3
= =
0 0 + 0 1 4 + 0 0 + 0 1 2 14 0 0 ln 2 0 ln 2 1 2 0 0 1 4 1 4 0
This result is non-zero, which [applying Eq. (4.73)], completes the proof that D is not a true rate of any quantity that can be expressed as a function of the deformation gradient. The preceding derivation used a four-stage deformation sequence: 1. elongate vertically, 2. shear to the right, 3. compress vertically, and 4. shear to the left. This path was selected to simplify the equations as much as possible for the counterexample. Now that we know that D is not a true rate, a natural follow subclass of deformations, if any, up activity is to determine what will permit D to be a true rate. In other words, we are now exploring thetime integral of D associated with closed paths in restricted to belong to some parF for which all values of F are ticular subclass. It can be shown that D is a true rate for any deformation that involves no rotation of the principal stretch directions. t = 3
t = 4
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In the preceding counterexample, the nonzero contributions to the time integral of D might have appeared (at first glance) to arise solely from the simple shearing stages, so its natural to question whether the time integral of D might turn out to be zero whenever we limit ourselves to deformations that are always pure stretches (i.e. deformations where the polar rotation tensor R always equals the identity I . For this class of deforma D and D . Hence, this tions, there is no distinction between question is the same as asking whether or not D , defined in Eq. (4.59), is a true rate. The answer again turns out no and the following three-stage loading is a counterexample:[ F] = 1 0 (1 t) + 1 0 t 01 0 2 [ F] = 1 0 (2 t) + 3 2 1 2 (t 1 ) 12 32 02 [ F] =32 12 1 2 (3 t) + 1 0 (t 2) 32 01
This problem can be analyzed in a manner similar to what was done in the four-stage loading counterexample, but the math is messy (so we recommend using a symbolic math program such as mathematica). The time integral of D again turns out the be nonzero even though the deformation consists entirely of pure stretches.
t = 3
The reason why neither D nor D are true rates is linked to of stretch. The above example illustrates that the the changing of the principaldirections principal directions of stretch can change without having any rotation (i.e. the polar rotation can remain R = I ).
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The last two of these equations state that no motion occurs transverse to the 1-direction. All motion is characterized by the 1D deformation mapping function f . Here, the function f depends on the nature of the loading. The deformation gradient is given by x1 x 1 x1 -------- -------- -------X1 X2 X3
f ( x1 )
Note that the local stretch varies only along the axial direction. When the stretch is constant, the deformation is said to be homogeneous uniaxial strain. Suppose, for example, that a piston contains a gas. If the piston is slowly compressed from a length L o to L , then the deformation will satisfy x 1 X 1 = L L o , or x 1 = ( L Lo ) X 1 , and therefore = L L o(5.7)
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If, on the other hand, the piston is suddenly compressed from a length L o to L , then the material displacements at a given instant in time will be highly inhomogenous because of wave motion back and forth. Even if a deformation is equilibrated, it might not be homogenous. Consider, for example, a long cylinder of a compressible gel sitting on a table so that the axis of the cylinder points up. Then gravity will make material more compressed at the bottom of the cylinder than at the top.
X Inhomogenous uniaxial strain
X Homogenous uniaxial strain
u X x X
u x
The left hand side of the above figure shows inhomogenous uniaxial strain. Note that the material elements (i.e., the little squares of material) dont deform very much on the right side of the bar, but they are stretched by more than 400% on the left hand side. You might expect a deformation somewhat like this if you hang a long cylinder of cork* from the ceiling. For homogenous uniaxial strain, each little material element stretches by the same amount everywhere.
* Cork is a material whose Poissons ratio is approximately zero. Consequently, it would have no offaxis (transverse) contractions. This constraint is needed to ensure uniaxial strain conditions.
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For uniaxial strain, the left-hand side is not required to remain motionless you might additionally have rigid translation as well, resulting a displacement vector that is longer by the amount of translation. The key is that the displacement vector always connects the initial and deformed locations of the same particle. Inhomogenous uniaxial strain
o X
X Figure 5.1.
Mapping function for uniaxial strain. For uniaxial strain, all motion is one-dimensional. Specifically, the mapping function is of the form x1 = f ( X1 ) x1 = X2 x1 = X1 which corresponds to the following deformation gradient matrix J 0 0 df ( X 1 ) [ F ] = 0 1 0 , where J = --------------dX 1 0 0 1(5.8a) (5.8b) (5.8b)
Since J is the Jacobian, it equals the ratio of deformed volume to initial volume. For this 1D uniaxial strain deformation (in which the cross-sectional area is constant), the volume ratio is identically equal to the length ratio.
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The function f quantifies the entire deformation x1 except for rigid translation. It is the one-dimensional analog of the deformation gradient tensor. Since f is a single function of a single variable, uniaxial strain is one of the few deformations that allows you to actually plot a graph of the mapping function. For example, the graph of the mapping corresponding to Fig. 5.1 can be approximated by picking up the picture of the deformed state to make it become the vertical axis in a plot of x 1 vs. X1 , as illustrated in Fig. 5.2. Note that the graph of the mapping function has a steeper slope at the left side of the bar than at the right; this indicates that material at the left side is stretched more than that at the right. A similar plot for homogeneous uniaxial strain would be a straight line, indicating that all elements are stretched by the same amount. The intensity of stretching is quantified X1 by the actual value of the slope. If the slope equals unity at a point, then the material element there has not stretched at all. Slopes larger than unity correspond to stretching. Slopes Figure 5.2. Graph of a 1D uniaxial smaller than unity correspond to compression. strain deformation mapping function. The vertical intercept of the mapping graph is the amount of rigid translation that is superimposed on the deformation. The slope of the mapping graph equals the material stretch (deformed length divided by initial length). Therefore, since length is always positive, the slope must be everywhere positive. In other words, a physically realizable uniaxial strain deformation must have a monotonically increasing mapping graph it will never dip down.
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Triaxial strainTriaxial strain is a homogenous axially symmetric deformation. It differs from uniaxial strain only in that lateral deformations are permitted. Specifically, the deformation gradient is of the form F = A PA + T P T (5.13) Here, the subscripts A and T stand for axial and transverse respectively, and the coefficient tensors are projection operations defined PA n n and PT I n n (5.15) The vector n is a unit vector in the direction of the symmetry axis, and n n denotes the ij compodyadic multiplication of n with itself (the result is a second-order tensor with P acts on an arbitrary vector v , the result is the part of that nents given by n i n j . When A vector in the direction of the symmetry axis. When P T acts on the vector, the result is the Stated mathematically, for any vecpart transverse (perpendicular) to the symmetry axis. tor, v PA v = ( v n )n (5.16) PA v = v ( v n ) n (5.17) We have stated that triaxial strain is a being here described as a homogenous deformation, which means that the deformation gradient tensor F is spatially constant and the defini therefore be integrated to give tion of the deformation gradient, d x = F d X may x = F X + C, (5.18) where C is an integration constant representing rigid translation. In light of Eqs. (5.13), (5.17), the deformation mapping function can be written (5.16) and x = A ( X n ) n + T [ X ( X n ) n ] + C (5.19) In terms of RCS coordinates with the 1-direction aligned with n , the deformation gradient and the mapping function may be written(5.14)
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A 0 0 [F] = and x1 = A X1 + C 1 x2 = T X2 + C2 x3 = T X3 + C3(5.21) (5.22) (5.23)
Here, X T denotes the distance of X from the symmetry axis. Note that uniform spherical deformation (discussed on page 47) is a special case of triaxial deformation in which A = L = and C = 0 . Also, homogenous uniaxial strain and = 1 . is a special case of triaxial deformation in which A = T
6. Axisymmetric deformationsAn axisymmetric deformation is one for which all motion is axial and/or radial with respect to a given unit vector N (the axis of symmetry). No angular motion is permitted. Cylindrical coordinates are a natural choice for this type of motion. Given an initial position vector X = R ER + ZEZ The displacement vector for axisymmetric deformations will be of the form(6.1)
u = u R ( R, Z ) E R + u Z ( R, Z ) E Z (6.2) Note that the displacement has no angular component. The radial and axial components of displacement depend only on the radial and axial coordinates. When the displacement field is of this form, the displacement gradient becomes uR, R 0 u R, Z [H] = 0 u Z, R uR ----0 with respect to cylindrical basis R 0 u Z, Z(6.3)
For axially symmetric deformations, the deformed position vector will have the same radial base vector as the initial position vector, so it will be of the form x = x R ( R, Z ) E R + x Z ( R, Z ) E Z , where x R = R + u R ( R, Z ) and x Z = Z + u Z ( R, Z ) Consequently, the deformation gradient will be of the form
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where x R, R = 1 + u R, R x Z, R = u Z, R
A cylindrical basis can be bypassed through the use of projection operations, as illustrated below for the constricted rod problem.
Constricted rodConsider a deformation in which a rod expands and/or contracts about an axis of symmetry, while planes that were initially perpendicular to the axis of symmetry remain planar. Then, taking the base of the rod to be fixed, we may say that xZ = Z and xR = R(6.7) (6.6)
Here, and are scalar-valued functions of Z. They do not vary with R. To avoid material inversion (i.e., to ensure local admissibility), both and must be positive. Applying Eq. (6.5), 0 sh 0 0 0 0 h
[F] = where
<cylindrical basis>
h ( Z ) = + Z ( Z ) R ( Z ) and s ( Z ) = -------------------h(Z) Note that the deformation gradient may be written 1 0 s 0 0 [F] = 0 1 0 0 0 0 0 1 0 0 h
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In this form, it is clear that this deformation can be viewed as a triaxial stretch followed by a simple shear. The function s ( Z ) is the deformed slope of the deformed shape of a line initially defined by R = 1 Initial configuration Special case: linear axisymmetry. Recall that the general deformation analyzed above corresponded to the mapping xZ = Z xR = R(6.11) (6.12)
Consider the special case of uniform axial stretch and a linearly varying radial profile = constant and x z --- R + xR = c k L(6.14) (6.13)
Deformed configuration
Axisymmetric deformation. where c and k are constants and L is the unde- Figure 6.1. In this figure, formed length of the bar. Note that cR is the X 1 4 - X and h ( X 1 ) = 1 3 -- ----deformed radial coordinate for a point whose ini- g ( X 1 ) = -5 1 5 25 tial cylindrical coordinates are ( R, 0 ) , and kR L is the slope of the line passing through this point. Substituting Eq. (6.11) into Eq. (6.14) gives
Reb: what are the g and h functions referenced in Figure? Did I change notation and forget to update that caption? My notes.nb now has this deformation as an example.k Z ---------xR = c + L R from which it follows that kZ = c + ---------L Then h = and Rk s = ----L(6.17) (6.16) (6.15)
Note that s is the slope of the line originally defined by R = 1 . Fig. 6.1 shows this type of deformation for = 4 5 , c = 1 , and k = 1 8 . With these choices, h = 4 5 , Zs = R 8 L , and = 1 -------. Therefore the deformation gradient is 10 L
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1 1 0 s 0 0 [F] = 0 1 0 0 0 = 0 0 1 0 0 h <cylindrical basis> For any point on the center line, Z1 -------10 L [F] = 0 0 0 Z1 -------10 L 0 0 0 4 -5 0 0
0 0 4 -5(6.18)
which shows that points on the centerline undergo a triaxial stretching that becomes increasingly compressive as Z increases.
( x 1, x 2 )
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For this deformation, we seek: 1. The mapping function. 2. The deformation gradient. 3. The polar rotation tensor R and reference (right) stretch U . 4. The Lagrange strain Let X = X 1 E 1 + X 2 E 2 denote the initial position of a point in the body, and let the deformed location of that same point. x = x 1 E 1 + x2 E 2 denote First lets find the mapping function. In other words, we seek formulas for the deformed coordinates ( x 1, x 2 ) of a point expressed in terms of the initial coordinates ( X 1, X 2 ) of that same point. Any initially vertical line can be identified by its initial abscissa value, X 1 . Each initially vertical line bends into the shape of an arc of radius X 1 , with an arc length given by X 1 , where is the arc angle. The arc angle increases in proportion to X 2 . Therefore, a constant exists such that = X2 The arc length of a point originally located at ( X1, X 2 ) must be given by s = X1 X2 Applying this equation to the center line gives R max = RH Thereforec max - = --- = ---------H R
For example, if you are located 2/3 of the way up the bar in the undeformed configuration, then you will be 2/3 of the way along the arc in the deformed configuration. Consider a point ( X 1, X 2 ) , not necessarily on the center line. After deformation, this point deforms to a new point ( x 1, x 2 ) located a distance r = X 1 from the origin. Consequently, X2 - x 1 = r cos = X1 cos ---- H max X2 ----x 2 = r sin = X 1 sin H max(7.7)
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X2 x1 - = cos ----- F 11 = ------- H max X1 X2 x2 - = sin ----F 21 = ------- H max X1 Assembled into matrix form, [ F ] = c s , s c
x1 X1 X2 - = c ----- sin ----- F 12 = ------- H max X2 R x2 X1 X2 - = c ----- cos ----F 22 = ------- H max X2 R(7.9)
The Jacobian equals to the ratio of the deformed to undeformed volumes and can be computed by J = det F = The polar decomposition gives [ R ] = c s s c and [U] = 1 0 0 (7.14)
Note that this deformation consists of a stretch in the 2-direction, followed by a rotation into the final deformed orientation. The Lagrange strain is 0 0 1 1 T 2 --[E] = {[F] [F] [I]} = {[U] [I]} = 1 2 2 - (2 1) 0 -2 The reference logarithmic strain is [ ] = ln [ U ] = 0 0 0 ln The logarithmic volume strain is given by v = tr = ln The displacement gradient is given by(7.18) (7.17)
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[] =
Unlike the polar rotation, there is not really any physical interpretation for this tensor. For pure stretch deformations, it equals the difference between the engineering strain (SethHill parameter k =1 ) and the Lagrange strain (Seth-Hill parameter k =2 ). Under sufficiently large stretches, this difference will be neither zero nor the identity even in the absence of rotation.
Simple ShearSimple shear involves two orthonormal unit vectors, m and n , often taken to be in the 1 and 2 directions, respectively. Displacement u is in the direction of m . The magnitude of displacement varies as a function of n X .Since u = x X , this means that simple shear has a mapping of the form x = X + g( n X )m where the function g is specified. Sinusoidal g-function2 1 1 2 1 2
linear g-function
n m m
Figure 7.1. Simple shear deformation of a unit square. When the g-function is nonlinear, the mapping is nonlinear if the g-function is nonlinear. The local slope of formerly vertical lines is quantified by the variable, which is a constant for simple linear shear.
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g(n X) = 2(n X) (7.22) where is a constant (and the factor of 2 is introduced for convenience). Then the deformation gradient is dx dX F = -----F 1 = -----(7.23) also - = I + 2 m n - = I 2 m n dX dx In the general case, where the g-function is nonlinear, the deformation gradient is of this form except that is 1 - g( n X ) = ( X ) = -2 (7.24)
Note the structure of the deformation gradient. A deformation is a shear whenever F is the identity I plus a dyad a b for which a b = 0 , in which case a defines the shearing direction, b defines the shearing plane, and = ( ab ) 2 is the shearing strain. Introducing a basis for which the 1 direction is parallel to m and the 2-direction is parallel to n , 1 2 0 [F] = 0 1 0 0 0 1 and [ F ] 1 1 2 0 = 0 1 0 0 0 1(7.25)
Then the polar rotation is given by c s 0 [R] = s c 0 , 0 0 1 where tan = In terms of , s - 0 1 2 c [F] = 0 1 0 0 0 1 and s = sin = -----------------1 + 2 and 1 c = cos = ----------------(7.27) 1 + 2
Note that a positive corresponds to a negative (clockwise) polar rotation angle, which should be evident from Fig. 7.1. The unit rotation axis is a = m n . Letting A denote form as the associated skew tensor, the rotation may be written in symbolic R = ( cos ) ( I a a ) + a a + ( sin ) A The right stretch is(7.29)
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s s2
+ [ U ] = [ R ]T [ F ] = s 1 ------------0 c 0 0 1 The left stretch is 1 + s2 ------------c T [ V] = [ F][ R] = s 0 In symbolic form, 2 c( c + 1) U = I s ( m n + n m ) + ---------------------------- n n c 2 c(c + 1) V = I s ( m n + n m ) + ---------------------------- m m c For small distortions (i.e., small so that c 1 and s ), these become U I (mn + nm) V I (mn + nm ) For large distortions (i.e., 2 ) 4 U = I ( m n + n m ) + -------------nn ()nn 2 s 1 0 0
0 1
(7.34) (7.35)
4 (7.37) V = I ( m n + n m ) + -------------mm ()mm 2 This result makes sense. As shear increases, fibers that were originally aligned with n in the reference configuration become stretched without bound and approach alignment with m . The n n component in the reference stretch U increases without bound because n is original orientation of the severely stretched fibers. the The m m component in the spatial orientation approached by stretch V is increases without bound because m is the final these severely stretched fibers. Time effects in shear. Suppose that m and n are constant in time, but that varies with time. Then F = 2mn 0 2 0 [F] = 0 0 0 0 0 0
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L = F F 1 = 2 m n
0 2 0 [L] = 0 0 0 0 0 0 0 10 [D] = 1 0 0 0 00 0 1 0 [ W ] = 1 0 0 0 0 0 0 1 0 [ ] = c2 1 0 0 0 0 0
D = sym L = ( m n + n m )
W = skw L = ( m n n m )
= R RT = c2(mn nm )
Note that, for constant , the vorticity, w = , remains constant for all time. However, the polar angular velocity, = c 2 , approaches zero because c 0 . Thus, the polar rotation gives a better description of material rotation since material fibers do not tumble and tumble indefinitely. The horizontal fibers dont rotate at all, and the initially vertical fibers rotate no more than 90 .
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TorsionAn informative way to get started with torsion is to first recall some elementary concepts about rotation. The mapping of an initial vector X to become a rotated vector x is given by x = QX (7.43) in which the rotation tensor Q is proper orthogonal. For time-varying problems, the rota and a skew-symmetric tensor is introduced such that tion tensor Q varies with time, velocity vector . If Q = Q . The axial vector associated with is the angular is angular velocity is constant in a time-varying rotation problem, the angle of rotation = t . For time-varying rigid rotation, Eq. (7.43) is spatially linear because Q depends only on time, not on position. Torsion, on the other hand, is not rigid it involves material distortion. Nevertheless, as suggested in Fig. 7.2, torsion still involves rotation of position vectors about the cylinder axis N = n . Therefore, Eq. (7.43) still applies, but it is no longer spatially linear rotation because the angle and therefore Q , vary with distance along the cylinder axis. N n
max = H
Figure 7.2. Torsion of a right-circular cylinder of radius R=1 and height H=4 by a twist parameter = 0.2 radians per unit length.
where the torsion parameter is the twist angle per unit length along the axis of the cylinder, and z is the axial coordinate, given in symbolic form by z = NX = nx 59Copyright is reserved. Individual copies may be made for personal use. No part of this document may be reproduced for profit.
In equation Eq. 7.44, the variable z is playing a role analogous to time t for time-varying rotations with a constant angular velocity. Analogous to the introduction of an angular velocity tensor such that Q = Q in time varying problems, we introduce an angu such that lar twist tensor dQ - = Q -----(7.46) dz For the uniform twist problem, the axial vector associated with is = n (7.47) For a constant twist parameter , the twist tensor is also constant even though Q var ies with position. It is useful to define a vector v by v = x = x Equivalently, v = r = r where r = ( I nn ) x (7.50) The vector r is the part of the position vector perpendicu of rotation. The vector v plays a role analolar to the axis gous to velocity for time-varying rotations, and it points in the angular direction. Therefore vN = 0 In a cylindrical system, r = r er and v = r e (7.53) To find the deformation gradient, we must differentiate Eq. (7.43) with respect to X , using the product rule because Q itself varies with position by virtue of its dependence on the axial coordinate z: d(Q X) dQT dQT dx d X dz - = Q + X -------- - ------ = Q -----F = ------ = --------------------- + X --------d X d X d X d X dz d X From Eq. (7.46) and then Eq. (7.43) and ultimately Eq. (7.48), note that(7.54) (7.51) Figure 7.3. (7.52) Helper vectors.
ez (7.48)
e er
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dQ dQT - X = Q X = x = v - = -----X --------dz dz Also, from Eq. (7.45), dz-----= N dX Therefore, Eq. (7.54) becomes F = Q + vN Incidentally, note that F = ( I + vn ) Q
F = Q (I + uN) where u Q T v (7.59) Recall from the previous section on shear that a deformation is a shear whenever its deformation gradient is the identity I plus a dyad a b for which a b = 0 , in which case a = ( ab ) 2 is the shear defines the shearing direction, b defines the shearing plane, and therefore note that the deformation gradient for torsion ing strain. From Eq. (7.58), we may be regarded as a rigid rotation Q followed by a tangent shear of shear strain magni -- r . Alternatively, from Eq. (7.59), tude 1 torsion may be seen as a tangent shear of the 2 same magnitude followed by rotation Q . In terms of an orthonormal cylindrical basis { e r, e , e z } having the 3-direction aligned with N , [F] = cos ( z ) sin ( z ) 0 sin ( z ) cos ( z ) r 0 0 1 w.r.t. cylindrical basis { e r, e , e z } (7.60)
The left Cauchy-Green tensor is given by B = F FT = ( Q + v N ) ( QT + Nv ) Multiplying out gives B = Q QT + vN QT + Q Nv + vN Nv Recalling that Q represents rotation about the unit vector N , note that T QQ = I Q N = N QT = n NN = 1 Therefore Eq. (7.62) becomes(7.61)
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where (recall)
v = x = x
The right Cauchy-Green tensor is C = FT F = I + Nu + uN + u2NN The component matrix is [C] = 1 0 0 0 1 r 0 r 1 + 2 r2 where u Q T v (7.68)
Here, the E-basis is the un-rotation of the e-basis by Q T . Application of a large deformation constitutive equation to torsion. Suppose that you wish to know how much torque is required to achieve a given twist . Then you must introduce a constitutive equation that relates deformation to stress. To illustrate the process, consider the following very simplistic constitutive model for an incompressible rubber-like material = 2GE + q( x )I (7.70) Here, q ( x ) is an undetermined function (discussed below), is the Cauchy stress tensor, and E is the Signorini strain defined in Eq. (4.32), 1 2 1 - [ V I ] = --(B I) , E = -(7.71) 2 2 or, using Eq. (7.66) 1 - (vn + nv + vv ) where (recall) v = x = x . E = -(7.72) 2 The undetermined function q ( x ) is present because, for an incompressible material, the pressure cannot be determined exclusively from knowledge of the deformation. Instead, q ( x ) must be determined to satisfy the equations of equilibrium and the boundary condi Assuming no body forces, equilibrium requires tions. ij, j = 0 Substituting Eq. (7.72) into (7.70), and then into (7.73) gives G ( v i, j n j + n i v j, j + v i, j v j + v i v j, j ) + q , i = 0(7.74) (7.73)
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Noting that v m, p = mp and recalling that is skew symmetric, it follows that G ( ij n j + ij v j ) + q , i = 0 Note that ij n j is the indicial form of n = n = n n = 0 Therefore, Eq. (7.74) becomes, in symbolic form, dq - = 0 G v + ---- dx or, recalling that v = x , dq ----- = G x dx Integrating gives 1 -Gx x q ( x ) = c -2 or
1 -G r 2 q ( x ) = c -(7.80) 2 where c is an integration constant and r is the radial vector defined in Eq. (7.50). The above result may be alternatively written 1 -G r 2 q ( x ) = c -2 or, since and r are perpendicular to each other, 1 - G2r2 , q ( x ) = c -2 Therefore, putting Eqs. (7.82) and (7.72) back into (7.70) gives 1 - G 2 r 2 I where (recall) v = r . = G ( v n + n v + v v ) + c -(7.83) 2 This solution is complete except for the undetermined integration constant c, which must be found by the boundary condition that the lateral surface of the cylinder must be traction free. The outward normal to the lateral surface is radial and therefore perpendicular to both v and n . Hence, the only way to satisfy the traction-free condition over the lateral is if boundary 1 - G 2 R2 c = -2 so that the final solution for the stress field is(7.84) (7.81)
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1 - G 2 ( R 2 r 2 ) I where (recall) v = r . = G ( v n + n v + v v ) + -(7.85) 2 We are now in a position to determine the net force and torque required to achieve a specified twist . The traction at the top surface of the cylinder is t = n or, using Eq. (7.85) and recalling that v is perpendicular to n , 1 - G2( R2 r2 ) n t = G v + -2 or, introducing a cylindrical basis, 1 - G 2 ( R 2 r 2 ) ez t = [ Gr ] e + -2 (7.88) (7.86)
The first term is the tangential part of traction and the second term is the normal part of traction. Note that the normal traction is nonzero. In fact, the net axial force is given byR
F axial =
tz dAA
1 -- G 2 ( R 2 r 2 ) 2 r dr = 1 -- G2R4 2 4
This force is positive and therefore tension is required to achieve the specified deformation. The torque is given byR
rt dAA
r [ Gr ] 2 r dr0
Keep in mind that these results are inextricably tied to the choice of constitutive model. If a different model is used, the results will be different. However, it is to be expected that the trends will be similar. For example, increasing the radius significantly increases the force an torque (in this case by a power of 4).
Vortex<need to add this section>
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We will denote these known physical position vectors by { x LL, x HH, x LH, x HL } . Here, the is the location of the point subscripts H and L denote high and low. The vector x LL whose logical coordinates are ( L, L ) . The vector x HL is the location of the point whose logical coordinates are ( H, L ) , and so on. x HH
x LH
x HL
2 1 ( L, L ) ( H, L ) x LL
If ( 1, 2 ) are the logical coordinates of a point in the reference square, then the physical coordinates are given by H 1 H 2 H 1 L 2 - ------------------ x LL + ------------------ ------------------ x x ( 1, 2 ) = ----------------- H L H L H L L H LH L 1 H 2 L 1 L 2 - ------------------ x HL + ------------------ ------------------ x + ----------------- L H H L L H L H HH For the case that ranges from L = 0 to H = 1 , this becomes x ( 1, 2 ) = ( 1 1 ) ( 1 2 ) x LL + + ( 1 ) ( 1 2 ) x HL + In this case, define ( 1 1 ) ( 2 ) x LH ( 1 ) ( 2 ) xHH (7.92)
x g 1 = -------- = ( x HL x LL ) ( 1 2 ) + ( x HH x LH ) 2 1 x g 2 = -------- = ( x LH x LL ) ( 1 1 ) + ( x HH x HL ) 1 2 and the logical deformation gradient associated with this mapping is dx x d 1 x d 2 ------ + -------- = g1 e1 + g2 e2 = -------- -------- -------d 1 d 2 d 65
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where e 1 and e 2 are the orthonormal reference base vectors. This mapping can be used to describe either the spatial or reference configuration for an actual deformation. In that case, the actual reference configuration has vectors, G 1 and G 2 , defined similarly to the g-vectors above. Then the actual deformation gradientcan be evaluated by F = g1 G 1 + g2 G 2 (7.96) where G 1 and G 2 , are the duals of G 1 and G2 . As a practical matter, it is usually easier the deformation to evaluate gradient by [ F ] = [ g ] [ G ] 1 , (7.97) where the matrix [ g ] contains g 1 and g 2 in its columns and [ G ] contains G 1 and G 2 in its columns.
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Lawn care is a time consuming and tedious task for most people. A person needs to ensure that the lawn looks perfect and that is the reason why lawn care is a tedious and time consuming task. You need to keep your lawn clean all the times since many advantages come along with having a beautiful lawn. You can increase the aesthetic value of your home when you have a clean lawn. Staying at home is fun for most people who have beautiful lawns in their backyards since they can spend your whole day lying on well kept grass. If you are planning to sell your house, you need to have a beautiful lawn since it can increase the value of the house. You need to contact a lawn service when you need to have your lawn taken care of for you to get the best services.
It can prove to be a hard task to find the perfect lawn services for most people. In the market today, many companies offer lawn care services, and that is the reason why it can be a confusing task to hire the right lawn service. Another thing that makes it harder for people to choose the right lawn care service is the price that they charge for their services. For you to find the right lawn care services, you need to consider a few factors before you choose one. Making a hiring choice without considering these factors may lead you to a poor lawn service company that will not give you the right services that you require. To read more about this page, click the
The first thing you need to consider before you choose a lawn service company is certification and insurance. Ensuring that a lawn service has the necessary credentials that allow them to carry out the task in your house is necessary before you hire any lawn service. You need to find a company that has been in the business for a long time for you to get the right service. If you find a company that has experience in the field, you may get the best service since they can have the necessary expertise to carry out the lawn care services. Go to the reference of this
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Before you choose a lawn service, the other thing that you need to consider is your budget. Before you choose a lawn service, you need to determine the right amount of financing that you will need to complete the task. Finding a company that will offer you their services at a price that suits your budget is necessary after you make a budget. You need to avoid a lawn service that charges a low cost for their services when hiring a lawn care service since low costs can translate to poor services. Take a look at the information about lawn care at
Lawn Care