the deformation gradient
TRANSCRIPT
Course: Continuum Mechanics II; Topic: Kinematics
Instructors: OA Fakinlede, O Adewumi & OA George
The Deformation Gradient
Deformations & Motions
โข We are primarily concerned here with deformations and motions in material entities.
โข The first important quantity we shall encounter is the deformation gradient. We shall see that is is a tensor. It will also become clear that all the information we need concerning the deformation or motion are contained in this tensor.
โข We will try to compute it for important deformations & motions. It will lead us to other geometric quantities of importance. The first lecture will focus on the Deformation Gradient.
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Topics in this Lecture
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Referential (Material) & Spatial (Current) Configurations
The Deformation Gradient Tensor
Polar Decomposition Theorem
Simple Deformations & Motions
Displacement, Stretch, Strain & Other Measures of Deformation
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Insight Versus Math & Computation
โข This and subsequent chapters call for two achievements: Insight and Computations. Experience shows that you may feel intimidated by the mathematics and computations, thinking, wrongly, that they are difficult. I have news for you, THEY ARE NOT as difficult as they appear to be!
โข Mathematica will be used in many cases to demonstrate the (Symbolic) computations. Simulation software will solve, numerically, the difficult differential equations. Every difficult step will be further explained if you ask for help.
โข Insight is the key. It is possible to be able to work the Math and compute without gaining Insight! This is a trap you should avoid!
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Simulations Input & Output
โข Century 21, Engineers do not have to manually solve the differential equations that arise from their analyses. Software is usually available to do that.
โข They need to understand the inputs and results of the software they use in order to supply relevant boundary conditions, obtain correct results and make useful judgments.
โข When, for example, you perform a simulation in Fusion 360, outputs include the symmetrical strain tensor and the displacement vector.
โข Fusion 360 also gives you the Equivalent Strain.
โข Weeks Two and Three will supply what you need to โฆ
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Two Week Goals
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Understand why strain, not deformation gradient, is the choice for quantifying deformation
2
Understand the meaning of each element in the strain tensor; and know why the strain tensor is necessarily symmetrical
3
Know that the elementary definition of strain is only approximate: understand when it cannot properly measure strain
4
See that rotations, translations and other rigid body motions produce zero strain but non-zero deformation gradient.
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Understand why rigid body motions create ambiguity in strain measurement. How they are removed to make strains unique.
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Understand various deformations and motions such as Uniaxial, biaxial or triaxial extension, bending, shear, dilatation, torsion, etc.
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Pointwise Transformation
โข We look at the body we are concerned about in the ambient environment of a 3-Dimensional Euclidean Point Space, โฐ: there is a subset of โฐ between which is in a one to one correspondence with each point in the body. A deformation is a mapping from this subset to another subset of the same space:
๐ฑ = ๐(๐)
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โข Where ๐ represents an arbitrary point in the material region of concern, and ๐ฑ, its image in the spatial. A deformation that changes over time, is a motion. If at specific ๐ก =1,2,โฆ , ๐ in time, we have the discrete functions,
๐ฑ1 = ๐1 ๐ , ๐ฑ2 = ๐2 ๐ ,โฆ , ๐ฑ๐ = ๐๐ ๐ ,โฆ
โข Motion can also be described by the single, continuous, time dependent function, ๐ฑ = ๐ ๐, ๐ก or ๐ฑ = ๐๐ก(๐)
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โข Motion is defined as set of mappings, ๐ฑ = ๐ . , ๐ก , ๐ก โ โ.
โข We assume that our subset of โฐ is connected.
โข Each member of the set of mappings, that is, each specific deformation in the set, is a known as a configuration or description at a point in time. We can take the configurations as photographs of the body as it undergoes its motion.
โข If we take that view, even though we can have several photographs, at most one of them, represents the current state of the body. This configuration is called the Spatial Configuration.
Pointwise Transformation
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Back to elementary calculus:
Connected region, simply
connected region, multiply
connected region.
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Referential & Spatial Configurations
โข Consider, for simplicity, the 2-D region undergoing a deformation as shown by what was originally a circle and a straight line.
โข As a result of deforming the region, we have shape changes that we observe.
โข Notice that at the time the second picture is seen, the first no longer exists. However, we keep a photo of it, and refer to it as a โReferential or Material Configurationโ.
โข To distinguish the one we are presently observing, we call the latter the โSpatial or Current Configurationโ.
โข Mathematically, we can look at this as a vector ๐๐transformed to vector ๐๐ฑ as shown by the purple arrows.
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Referential & Spatial Configurations
We are looking at the two vectors, ๐๐ in the referential configuration, and ๐๐ฑ in the spatial.
โข Note that these do not exist physically together. We bring them up, in the same diagram, artificially, for the purpose of gaining analytical insight.
โข Using this artifice, we can see the entire deformation as the transformation of a vector ๐๐ to the vector ๐๐ฑ. The former can represent any material vector in the referential (original) configuration, while the latter represents its image in the spatial (or current) configuration.
โข We have removed the translation of the vectors, bringing their origins together to show that elongation (or contraction) with the rotation that have been caused by the deformation.
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Spatial and Referential Vectors
โข It is obvious that there is a relationship between the infinitesimal spatial and material vector that transformed to it. Each spatial vector, ๐๐ฑ being a vector, consists of three scalars functions, ๐๐ฅ1, ๐๐ฅ2, ๐๐ฅ3 , or ๐๐ฅ๐ผ, ๐ผ = 1,2,3.
โข By the vector equation, ๐ฑ = ๐(๐), we mean ๐ฅ๐ผ = ๐๐ผ ๐1, ๐2, ๐3 , and, from elementary multivariate calculus, we have,
๐๐ฅ๐ผ =๐๐๐ผ๐๐๐
๐๐๐
โข In vector form, assuming the referential system is spanned by ๐๐ , ๐ = 1โฆ3 and that the spatial system is spanned by ๐๐ผ, ๐ผ =1,2,3. Adding the relevant bases, we have the vector equation,
๐๐ฅ๐ผ๐๐ผ =๐๐๐ผ๐๐๐
๐๐ผ โ๐๐ ๐๐๐๐๐
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Several ways to
explain this dyad
Observe that there are
nine components here:
three for each ๐ผ
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The Deformation Gradient
More compactly, we write, ๐๐ฑ = ๐ ๐๐
โข where transformation tensor field
๐ ๐, ๐ก = Grad ๐ ๐, ๐ก =๐๐ฅ๐ผ๐๐๐
๐๐ผ โ๐๐ =๐๐๐ผ๐๐๐
๐๐ผ โ๐๐
โข the material (referential) gradient of the deformation or motion function, ๐ ๐, ๐ก . This deformation gradient is the tensor that transforms material vectors to spatial vectors in the region of interest.
โข It contains ALL information about the deformation or motion.
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Textbooks write these equations as equivalent.
However, the rightmost is the correct expression
because it differentiates the function ๐ฑ = ๐ ๐, ๐ก while
the other differentiates the value.
When is it a
deformation? Motion?
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The Deformation Gradient
โข Is the fundamental tensor describing deformation and motion. Its second base vector, ๐๐, is a reciprocal base, as is obvious from the fact that the variable it represents is below.
โข Under Cartesian coordinates, there is NO difference between a base vector and its reciprocal base vector.
โข Once we are in curvilinear systems such as cylindrical or spherical polar, there will be differences as we shall illustrate.
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The Deformation Gradient
โข It is conventional to use the capital letter Grad for the gradient in the deformation gradient to emphasize the fact that the gradient is taken of the deformation (or motion) function with respect to the Referential system. When a gradient is taken with respect to the Spatial system, we shall write is as grad.
โข Observe that Grad ๐ ๐, ๐ก โ grad ๐ ๐, ๐ก = ๐
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The Deformation Gradient
Easily proved because
โข The deformation gradient is the material gradient of the deformation function;
grad ๐ ๐, ๐ก =๐๐ฅ๐ผ๐๐ฅ๐
๐๐ผ โ๐๐
= ๐ฟ๐ผ๐๐๐ผ โ๐๐= ๐
โข ๐ฑ = ๐ฅ๐ผ๐๐ผ is, in Cartesian Coordinates, the fundamental spatial variable. It is dependent on the referential vector variable, ๐ = ๐๐๐๐ .
โข This relationship is called โthe deformationโ or โthe motionโ, ๐ฑ = ๐ ๐, ๐ก
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Volume Ratio
โข The change in the material vectors implies there are also changes in areas and volumes. To obtain the change in volume as a result of the deformation, consider an infinitesimal tetrahedron in the referential state.
โข The volume of the tetrahedron,1
6๐๐1, ๐๐2, ๐๐3 โ 0
โข i.e. the volume will not vanish if the three vectors are neither colinear nor all coplanar. As a result of the motion, the corresponding spatial vectors will form a deformed tetrahedron.
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Volume Ratio
โข Each side will be a transformed referential vector into the spatial: ๐๐ฑ1, ๐๐ฑ2, ๐๐ฑ3 will be related to the material vectors in such a way that,
๐๐ฑ๐ = ๐ ๐๐๐
โข The volume ratio between the spatial and material configurations,
๐ฝ =๐๐ฑ1, ๐๐ฑ2, ๐๐ฑ3๐๐1, ๐๐2, ๐๐3
=๐ ๐๐1, ๐ ๐๐2, ๐ ๐๐3๐๐1, ๐๐2, ๐๐3
= det ๐ .
โข The linear independence of vectors ๐๐1, ๐๐2, ๐๐3 is guaranteed by the non-vanishing of the tetrahedron or we shall have chosen a trivial volume. However, what guarantee do we have for the spatial tetrahedron?
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Consider the Possibility:
๐๐ฑ = ๐ ๐๐ = ๐จ
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the zero vector. What can this mean physically? The linear independence of the denominator in the determinant expression guarantees non-vanishing of the numerator provided the deformation gradient is an invertible tensor. Mathematically, the Jacobian (determinant of ๐ ) of the transformation is zero.
โข We were able to find a non-trivial (not a zero tensor) transformation tensor that transforms a real vector into nothingness! We, by a deformation transformation destroyed matter!
โข Our physical considerations preclude this possibility. We exclude from consideration such a possibility. And since we cannot have ๐ฝ = 0, we can therefore conclude that
๐ฝ > 0
โข Since continuity forces it to pass through zero to negative; if it cannot be zero, it cannot be negative. The only allowable transformations have a positive determinant.
Q2.2519
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The Reference Map
โข The set of mappings that gives each deformation, and consequently, the entire motion is a set of one-to-one mappings. Such mappings are invertible. It follows that, at each time ๐ก, we have,
๐ = ๐โ1 ๐ฑ, ๐ก
โข From which we can find the reference configuration that resulted in each spatial configuration at a time ๐ก. The material point that occupied the spatial position ๐ฑ at time ๐ก can be computed by the reference map.
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Simple Deformation
โข Consider a rectangular 2-D region and draw a circle of a unit radius and a diagonal line as shown.
โข We can plot the line and the circle using parametric expressions for each as shown in the picture.
This is generated by the Mathematica code:
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Simple Deformation
โข Consider a simple 2-D deformation function,
๐ฑ = ๐ ๐, ๐ผ = ๐1 + ๐ผ๐2 ๐1 +๐2๐ผ
4๐1 +1
๐ผ๐2
For any ๐ผ โ 0, the prescribed parameter signifies a specific deformation. Let ๐ผ = 0.2 then, the above deformation can be plotted as shown. The picture here is generated by the Mathematica code:
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Simple Motion, Reference Map
Consider, for example, the motion,๐ฑ = ๐ ๐, ๐ก = ๐ก๐1 + ๐1๐2 ๐1 + ๐2๐1 + ๐ก๐2 ๐2 + ๐ก๐3
โข Where ๐1 and ๐2 are constants, and ๐ก is the time variable. To obtain the reference map, we can invert this function and obtain,
๐ = ๐โ๐ ๐ฑ, ๐ก =๐ก๐ฅ1 โ ๐1๐ฅ2๐ก2 โ ๐1๐2
๐1 +๐ก๐ฅ2 โ ๐2๐ฅ1๐ก2 โ ๐1๐2
๐2 +๐ฅ3๐ก๐3
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Examples of Simple Motions
The following examples of simple motions have been named:
โข Pure translation, ๐ ๐, ๐ก = ๐ + ๐(๐ก), where ๐ is a differentiable vector-valued function of time.
โข Pure rotation, ๐ ๐, ๐ก = ๐(๐ก)๐, where ๐ is a proper orthogonal function. (A way of saying that it is a rotation function of time).
โข Simple Shear. ๐ ๐, ๐ก = ๐ + ๐ผ ๐ก ๐1 โ๐2 ๐, where ๐ผ is a differentiable, scalar valued function of time. Q: Transpose the dyad and what do you get? Compare to the original shear motion.
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Simple Shear
โข The Mathematica graphic is about Uniform Shear.The Deformation gradient here is easily calculated by hand. Do this to ensure you donโt get lost in the mechanical computation and lose the context:
๐ฑ = ๐ ๐ = 0.5 + ๐1 + 0.5๐2 ๐1 + ๐2๐2
โข for the element occupying ๐1๐1 + ๐2๐2 initially.
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Deformation Gradient of Simple Shear
๐๐ฅ1
๐๐1= 1,
๐๐ฅ1
๐๐2= 0.5,
๐๐ฅ1
๐๐3= 0 and
๐๐ฅ2
๐๐2=
๐๐ฅ3
๐๐3with all other
components of the deformation gradient vanishing.1 0.5 00 1 00 0 1
โข is the matrix of the deformation gradient components.
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Uniform Extension
โข Consider the unit cube with the transformation vector:
๐ฑ = ๐ ๐ = ๐ผ1๐1๐1 + ๐ผ2๐2๐2 + ๐ผ3๐3๐3
โข Note that uniaxial extension can be obtained by allowing two of the constants to be unity while biaxial will be ensured by one of the constants becoming one as follows:
โข Uniaxial: ๐ฑ = ๐ ๐ = ๐ผ1๐1๐1 + ๐2๐2 + ๐3๐3
โข Biaxial: ๐ฑ = ๐ ๐ = ๐1๐1 + ๐ผ2๐2๐2 + ๐ผ3๐3๐3
โข Pure Dilatation: ๐ฑ = ๐ ๐ = ๐ผ(๐1๐1 + ๐2๐2 + ๐3๐3)
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The 1 โ ๐ฅ โ1 RiddleTo keep you awake in class!
Primary 5: Long
Division
Long Division
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Foolish Crammer
1 โ ๐ฅ โ1 = 1 + ๐ฅ + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 +โฏ
โข No thought, no sense just memorize and give back to the teacher when asked!
โข Something a Primary 5 pupil can achieve! Simple Long division!
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Senior Secondary School 3: Binomial Theorem
๐ + ๐ ๐ = ๐๐๐0 + ๐๐๐โ1๐1 +๐ ๐ โ 1
2!๐๐โ2๐2 +
๐ ๐ โ 1 (๐ โ 2)
3!๐๐โ3๐3 +โฏ
โข For example
๐ + ๐ 3 = ๐3๐0 + 3๐3โ1๐1 +3 3 โ 1
2!๐3โ2๐2 +
3 3 โ 1 (3 โ 2)
3!๐3โ3๐3 +โฏ
= ๐3 + 3๐2๐ + 3๐๐2 + ๐3
1 โ ๐ฅ โ1 = 1โ1(โ๐ฅ)0+ โ1 1โ2 โ๐ฅ 1 +โ1 โ2
2!1โ3 โ๐ฅ 2 +
โ1 โ2 โ3
3!1โ4 โ๐ฅ 3
+โ1 โ2 โ3 (โ4)
4!1โ5 โ๐ฅ 4 +
โ1 โ2 โ3 (โ4)(โ5)
5!1โ6 โ๐ฅ 5โฆ
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GEG 30x Engineering Mathematics
โข Taylor Series near the point ๐ฅ = ๐,
๐ ๐ฅ = ๐ ๐ + ๐ฅ โ ๐ ๐โฒ ๐ +๐ฅ โ ๐ 2
2!๐โฒโฒ ๐ + โฏ+
๐ฅ โ ๐ ๐
๐!๐๐ ๐ + โฏ
Here we are evaluating the function, ๐ ๐ฅ =1
1โ๐ฅnear the point, ๐ฅ = 0.
๐โฒ ๐ฅ =1
1 โ ๐ฅ= โ1 1 โ ๐ฅ โ2 โ1 โ ๐โฒ 0 = 1
๐โฒโฒ ๐ฅ =๐
๐๐ฅ
1
1 โ ๐ฅ
2
= โ2 1 โ ๐ฅ โ3 โ1 โ ๐โฒโฒ 0 = 2 ร 1 = 2!
It is not difficult to show that ๐๐ 0 = ๐! so that, as before,
1 โ ๐ฅ โ1 = 1 + ๐ฅ + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 +โฏ
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Mathematica Compliant Engineer
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But is the expansion correct?
โข For example, let us try ๐ฅ = 2 and find the answer given by the expansion:
1 โ ๐ฅ โ1 = 1 + ๐ฅ + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 + ๐ฅ5 +โฏ
โข So that,1 โ 2 โ1 = 1 + 2 + 22 + 23 + 24 + 25 +โฏ
= 1 + 2 + 4 + 8 + 16 + 24 +โฏ
โข If at this point, you do not know what is happening and cannot solve this riddle, you are a foolish crammer! You have learned nothing here! Matters not what marks you got!
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Solution to Riddle
Parenthetical Issues:
โข Convergence properties of Infinite Series
โข Remainder estimate of Taylor expansion
โข You did not bother cramming those.
It turns out that they are far more important than cramming the formula.
See how easy it is for software to spit out the result. If I need the result in
my business, I will rather buy the software than hire you! I need someone
that understands how to interpret the results and knows when they are
valid! Crammers are more than useless to any serious business!
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The Girl & Boy
โข There are several skills that can be picked up this semester:
1. Solid Modeling
2. Animation
3. Simulation
4. Computation
5. Design & Analysis
Each of these, on their own, if you were to master them, not only can fetch your daily bread, can also make you an asset to your family, your people and your country. I am afraid not many seem to realize that!
Continuum Mechanics is to give you insight to the theoretical
underpinnings of these. Memorization is OK. Insight and
understanding are more important. When you choose
cramming over insight, you are self-immolating!
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Polar Decomposition
Theorem
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โข In this section, we are looking at a multiplicative decomposition, motivated by the reality that it is NOT the entire transformation wrought by the deformation gradient that concerns us in the study of geometrical changes resulting from the application of loads.
โข It successfully separates portions of the deformation gradient that do not cause shape changes from the parts that are relevant in geometric modifications resulting from the transformation.
For a given deformation gradient ๐ญ, there isa unique rotation tensor ๐น, and unique, positive definite, symmetric tensors ๐ผ and ๐ฝ for which, ๐ญ = ๐น๐ผ = ๐ฝ๐น
โข ๐ผ is called the Right Stretch Tensor, and ๐ฝ the Left Stretch Tensor.
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โข The proof of this important theorem is given subsequently. More important though is to know the meaning:
โข Beginning from any material configuration, the transformation given by the deformation gradient leads to the spatial configuration. However, this transformation can be achieved in two two-stage processes.
โข A stretch in the material configuration through the Right Stretch Tensor ๐; followed by a rotation by the rotation tensor ๐ to the spatial configuration. Note that the rotation tensor is neither a material nor a spatial tensor. It is, like the deformation gradient, a two-toe tensor; operating on a material vector and producing a spatial tensor.
โข A transformation to the spatial configuration by the rotation tensor ๐, followed by a stretch to the final state in that configuration by the left stretch tensor. The latter is a spatial tensor as it takes a spatial vector (output of the rotation tensor), and returns a spatial vector.
Meaning of Polar Decomposition
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Polar Decomposition: Proof
There are two stages to establish the Polar Decomposition of the Deformation Gradient:
1. We show that the Right Cauchy-Green Tensor, ๐ = ๐ ๐๐
is symmetric and positive definite.
2. Use this fact to find ๐, such that ๐ = ๐ ๐๐ = ๐๐.๐ = ๐ ๐โ1
3. Then find ๐ in terms of ๐.
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Symmetry & Positive
Definiteness of ๐
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โข It is obvious that ๐ = ๐ T๐ is symmetric because its transpose is
๐T = ๐ T๐ = ๐
โข Now select ANY real non-zero vector ๐ฎ. We can find a vector ๐ = ๐ ๐ฎ.
The quadratic form, ๐ฎ โ ๐ T๐ ๐ฎ = ๐ โ ๐ ๐ฎ = ๐ 2 > 0
Since we selected ๐ฎ arbitrarily, and we have reached the conclusion that any quadratic form with ๐ is always greater than zero, we have proved that ๐ is positive definite.
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Square Root of ๐
โข Every Positive Definite Tensor has a positive definite square root: The right Cauchy-Green Tensor
๐๐ = ๐ = ๐ T๐ = ๐T ๐T๐ ๐ = ๐T๐ ๐ = ๐2
โข Shows that ๐ = ๐ T๐ = (๐๐)T๐๐ so that ๐ = ๐๐. To complete the proof, write,
๐ = ๐๐ = ๐๐
and we immediately find that ๐ = ๐๐๐T by simply post-multiplying the above by ๐T. Since ๐ is positive definite, so must ๐. Obvious, if we consider arbitrary real vectors ๐ฎ, ๐ฏ such that ๐ฏ โก ๐T๐ฎ,
๐ฎ โ ๐๐ฎ = ๐ฎ โ ๐๐๐T๐ฎ = ๐ฎ โ ๐๐๐ฏ = ๐ฏ โ ๐๐T๐ฎ = ๐ฏ โ ๐๐ฏ > 0
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Right Cauchy Green Tensor
โข In the proof of Polar Decomposition Theorem, we encounter another important tensor: The Right Cauchy-Green Tensor
๐ = ๐ T๐ = ๐T ๐T๐ ๐ = ๐T๐ ๐ = ๐2
โข It is said to be โRightโ because there is a โLeftโ Cauchy-Green Tensor:๐ ๐ T that can be obtained by the product of the deformation gradient and its transpose.
โข In the former, the deformation gradient is at the right hand; in the latter, it is at the left side โ hence the distinguishing names. It is easily shown that,
๐๐ = ๐ ๐ ๐
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โข Remember that referential and spatial configurations DO NOT coexist. Bringing vectors from either state together is artificial, not real.
โข Tensors that come to our attention are classified by what kinds of arguments they can take and what kind of vectors they produce.
โข On the other hand, vectors are classified by where they reside. For example, the material vector is so called because it is made up of elements in the referential (material) configuration. Spatial tensors are similarly defined.
Referential (Material) & Spatial Vectors, Tensors
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Material, Spatial & Two-Towed Tensors
โข A tensor that takes a material argument and produces a material result is defined as a material tensor.
โข Right Cauchy Green Tensor, Right Stretch Tensor
โข A tensor that takes a spatial argument and produces a spatial result is defined as a spatial tensor.
โข Left Cauchy-Green Tensor, Left Stretch Tensor
โข A Tensor that takes a material argument and produces a spatial result transforms a vector from the referential state to an image in the spatial configuration.
โข Examples: Deformation Gradient, Rotation Tensor
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Determine the Type of Tensor
โข Consider a spatial vector ๐ฌ. The dot product ๐ฌ โ ๐๐ฑ has physical significance while ๐ฌ โ ๐๐ does not as the two operands do not exist at the same time so an operation between them makes no physical sense. Clearly,
๐ฌ โ ๐๐ฑ = ๐ฌ โ ๐ ๐๐ = ๐๐ โ ๐ T๐ฌ
โข meaning that ๐ T๐ฌ is a material vector so that ๐ T transforms spatial vectors to material. Beginning with a material vector ๐ญ. The physically meaningful product,
๐ญ โ ๐๐ = ๐ญ โ ๐ โ1๐๐ฑ = ๐๐ฑ โ ๐ โ๐๐ญ
โข Showing that ๐ โ๐, just like ๐ , transforms material to spatial while ๐ โ1 like ๐ T transforms spatial vectors to material. These tensors are two-toed.
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Area Ratio
โข For an element of area ๐๐ in the deformed body with a vector ๐๐ฑprojecting out of its plane (does not have to be normal to it). For the elemental volume, we have the following relationship:
๐๐ฏ = ๐ฝ๐๐ = ๐๐ โ ๐๐ฑ = ๐ฝ๐๐ โ ๐๐
โข where ๐๐ is the element of area that transformed to ๐๐ and ๐๐ is the image of ๐๐ฑ in the undeformed material. Noting that, ๐๐ฑ =๐ ๐๐ we have,
๐๐ โ ๐ ๐๐ โ ๐ฝ๐๐ โ ๐๐ = 0= ๐ T๐๐ โ ๐ฝ๐๐ โ ๐๐
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Area Ratio
โข For an arbitrary vector ๐๐, we have:๐ T๐๐ โ ๐ฝ๐๐ = ๐จ
โข so that,๐๐ = ๐ฝ๐ โT๐๐ = ๐ ๐๐๐
โข where ๐ ๐ is the cofactor tensor of the deformation gradient.
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Deriving Reciprocal Bases
โข Reciprocal bases are easily derived. The dyad product of a reciprocal bases with the natural bases yield the identity tensor.
โข In Cartesian systems, the natural and the reciprocal bases coincide. In curvilinear coordinates such as Cylindrical and Spherical Polar, this is not so. For example, for spherical polar, the reciprocal basis can be derived from the natural basis (obtained by differentiating the position vector), using superscript to represent the reciprocal bases, as follows:
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Spherical Reciprocal Bases
๐1
๐2
๐3=
1
๐1 โ ๐10 0
01
๐2 โ ๐20
0 01
๐3 โ ๐3
๐๐๐๐๐
๐ sin ๐ ๐๐
=
1 0 0
01
๐20
0 01
๐2 sin2 ๐
๐๐๐๐๐
๐ sin ๐ ๐๐=
๐๐๐๐๐๐2
๐ sin ๐ ๐๐
๐2 sin2 ๐
=
๐๐๐๐๐๐๐
๐ sin ๐
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Natural & Reciprocal Bases
โข This can be repeated for other coordinate systems, e.g. the cylindrical. The results can be summarized as follows:
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Coordinate System Natural Basis Vectors Reciprocal Base Vectors
Cartesian๐๐ซ
๐๐ฅ1= ๐1;
๐๐ซ
๐๐ฅ2= ๐2;
๐๐ซ
๐๐ฅ3= ๐3
{๐1, ๐2, ๐3}
Cylindrical Polar๐๐ซ
๐๐= ๐๐;
๐๐ซ
๐๐= ๐๐๐;
๐๐ซ
๐๐ง= ๐๐ง
๐๐;๐๐
๐; ๐๐ง
Spherical Polar๐๐ซ
๐๐= ๐๐;
๐๐ซ
๐๐= ๐๐๐;
๐๐ซ
๐๐= ๐ sin๐๐๐ ๐๐;
๐๐๐;
๐๐
๐ sin ๐
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Deformation Gradient: Cartesian to Cartesian
โข If the Referential State is based on Cartesian unit vectors, ๐๐ , ๐ = 1,2,3 and we have ๐๐ผ , ๐ผ = 1,2,3 for the spatial state. Deformation takes the form, ๐ฑ =๐ ๐, ๐ก = ๐ ๐1, ๐2, ๐3, ๐ก in this case, the Deformation Gradient,
๐ =๐๐๐ผ๐๐๐
๐๐ผ โ๐๐
= ๐1 ๐2 ๐3
๐๐1๐๐1
๐๐1๐๐2
๐๐1๐๐3
๐๐2๐๐1
๐๐2๐๐2
๐๐2๐๐3
๐๐3๐๐1
๐๐3๐๐2
๐๐3๐๐3
โ๐1๐2๐3
.
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Deformation Gradient: Cylindrical to Cylindrical
โข In the case of orthogonal systems, linear or curvilinear, this relationship becomes simply dividing by the magnitude of the respective natural base vector. The deformation gradient from a material configuration in cylindrical Polar coordinates ๐ , ฮ, ๐ to a spatial configuration ๐, ๐, ๐ง in the same coordinate system is,
๐ = ๐๐ ๐๐๐ ๐๐ง
๐๐
๐๐
๐๐
๐ฮ
๐๐
๐๐๐๐
๐๐
๐๐
๐ฮ
๐๐
๐๐๐๐ง
๐๐
๐๐ง
๐ฮ
๐๐ง
๐๐
โ
๐๐น๐ฮ๐ ๐๐
= ๐๐ ๐๐ ๐๐ง
๐๐
๐๐
1
๐
๐๐
๐ฮ
๐๐
๐๐
๐๐๐
๐๐
๐
๐
๐๐
๐ฮ๐๐๐
๐๐๐๐ง
๐๐
1
๐
๐๐ง
๐ฮ
๐๐ง
๐๐
โ
๐๐น๐ฮ๐๐
.
โข We used upper case to depict the Material system. It is the reciprocal system.
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Deformation Gradient: Spherical to Spherical
โข If both were spherical, ๐, ฮ,ฮฆ โ ๐, ๐, ๐ the deformation gradient becomes,
โข
๐ = ๐๐ ๐๐๐ ๐ sin ๐ ๐๐
๐๐
๐๐
๐๐
๐ฮ
๐๐
๐ฮฆ๐๐
๐๐
๐๐
๐ฮ
๐๐
๐ฮฆ๐๐
๐๐
๐๐
๐ฮ
๐๐
๐ฮฆ
โ
๐๐๐ฮ๐๐ฮฆ
๐ sinฮ
= ๐๐ ๐๐ ๐๐
๐๐
๐๐
1
๐
๐๐
๐ฮ
1
๐ sinฮ
๐๐
๐ฮฆ
๐๐๐
๐๐
๐
๐
๐๐
๐ฮ
๐
๐ sinฮ
๐๐
๐ฮฆ
๐ sin ๐๐๐
๐๐
๐ sin ๐
๐
๐๐
๐ฮ
๐ sin ๐
๐ sinฮ
๐๐
๐ฮฆ
โ
๐๐๐ฮ๐ฮฆ
.
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Coordinate System Pairings
โข We are completely free to represent the initial coordinate system any way we like.
โข We can therefore have Cartesian to Spherical Polar or Spherical Polar to Cylindrical Polar transformations.
โข It is a matter of which system best describes the deformation transformations as will be shown in examples.
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Illustrative Example
โข Consider the length 2๐ฟ of a thin rod uniformly bent into a semicircle as shown.
โข Referential configuration is the straight rod, Spatial, after the bending, is the semi-circular rod. If the rodโs length does not increase as a result of shape change, then ๐๐ = 2๐ฟ. Clearly, radius ๐ =2๐ฟ/๐
โข A point previously located at the distance ๐ฅ from the origin is now at angle ๐. The relationship between the two is linear:
๐ฅ
2๐ฟ=๐
๐โ ๐ =
๐๐ฅ
2๐ฟ
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๐ฅ 0 ๐ฟ
2
๐ฟ โ๐ฟ
๐ 0 ๐
4
๐
2โ๐
2
How else can you
obtain this formula?
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Bar to Semicircular
Region
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Bar to Semicircular Region
โข Imagine that we bent the bar shown into a semicircular region. Transformation function can be found by the following consideration: Note that each horizontal filament in the original bar becomes a circular filament in the spatial configuration. The vertical undeformed sections become radial sections in the spatial state. Let the centerline be a semicircle at a distance ๐ and let the thickness contract uniformly with a factor ๐ผ
โ ๐ฅ1 = ๐ = ๐1 ๐1, ๐2, ๐3, ๐ก = ๐ + ๐ผ๐2, and
๐ฅ2 = ๐ = ๐2 ๐1, ๐2, ๐3, ๐ก =๐๐12๐ฟ
โข If the bar contracts uniformly in ๐3 direction, ๐ฅ3 = ๐ง = ๐3 ๐1, ๐2, ๐3, ๐ก = ๐ฝ๐3
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Take ๐1 axis along the bar, ๐2 vertically, and ๐3along the axis of bending.
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Referential & Spatial Configurations
โข Clearly, the referential configuration here is the bar; Spatial is the semicircular bar.
โข Deformation is such that the spatial is in cylindrical coordinates, the referential is in Cartesian.
โข Deformation gradient requires the reciprocal Cartesian bases which are the same as the Cartesian. In the spatial, we use the cylindrical. The full computation given in Q4.7, is repeated here:
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Deformation Gradient
๐ = ๐๐ ๐๐๐ ๐๐ง
๐๐
๐๐1
๐๐
๐๐2
๐๐
๐๐3๐๐
๐๐1
๐๐
๐๐2
๐๐
๐๐3๐๐ง
๐๐1
๐๐ง
๐๐2
๐๐ง
๐๐3
โ๐1๐2๐3
= ๐๐ ๐๐ ๐๐ง ๐
0๐๐
๐๐20
๐๐
๐๐10 0
0 0๐๐ง
๐๐3
โ๐1๐2๐3
= ๐๐ ๐๐ ๐๐ง
0 ๐ผ 0๐๐
2๐ฟ0 0
0 0 ๐ฝ
โ๐1๐2๐3
=๐๐
2๐ฟ๐๐ ๐ผ๐๐ ๐ฝ๐๐ง โ
๐1๐2๐3
=๐๐
2๐ฟ๐๐ โ๐1 + ๐ผ๐๐ โ๐2 + ๐ฝ๐๐ง โ๐3
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Right Cauchy-Green/Stretch Tensors
โข Clearly,
๐ = ๐ T๐ =๐๐
2๐ฟ๐1 โ๐๐ + ๐ผ๐2 โ๐๐ + ๐ฝ๐3 โ๐๐ง
๐๐
2๐ฟ๐๐ โ๐1 + ๐ผ๐๐ โ๐2 + ๐ฝ๐๐ง โ๐3
=๐๐
2๐ฟ๐1 โ๐๐
๐๐
2๐ฟ๐๐ โ๐1 +โฏ+ ๐ฝ๐3 โ๐๐ง ๐ฝ๐๐ง โ๐3
=๐๐
2๐ฟ
2
๐1 โ๐1 + ๐ผ2๐2 โ๐2 + ๐ฝ2๐3 โ๐3
โข since each set of basis vectors is orthonormal, and the Right Stretch Tensor,
๐ =๐๐
2๐ฟ๐1 โ๐1 + ๐ผ๐2 โ๐2 + ๐ฝ๐3 โ๐3
โข Is the square root of the Right Cauchy Green tensor. The positive square roots are taken since
both ๐ as well as ๐ are necessarily positive definite and can only have positive eigenvalues.
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Computing Functions in Cylindrical Systems
๐ = ๐ ๐ , ๐ฉ, ๐ = ๐๐ ๐ , ๐ฉ, ๐ ; ๐ = ๐ ๐ , ๐ฉ, ๐ = ๐๐ ๐ ,๐ฉ, ๐ ; ๐ง = ๐ง ๐ , ๐ฉ, ๐ = ๐๐ง ๐ ,๐ฉ, ๐
๐๐ฑ =๐๐ฑ
๐๐๐๐ =
๐๐
๐๐๐๐ = ๐ ๐๐ฑ
The spatial position vector, ๐ฑ = ๐๐๐ ๐, ๐ + ๐ง๐๐ง โ
๐๐ฑ =๐๐ฑ
๐๐๐๐ +
๐๐ฑ
๐๐๐๐ +
๐๐ฑ
๐๐ง๐๐ง = ๐๐๐๐ + ๐
๐๐๐ ๐, ๐
๐๐๐๐ + ๐๐ง๐๐ง
= ๐๐๐๐ + ๐๐๐๐๐ + ๐๐ง๐๐ง
Similarly, in the Referential,
๐๐ =๐๐
๐๐ ๐๐ +
๐๐
๐๐ฉ๐๐ฉ +
๐๐
๐๐๐๐ = ๐๐ ๐๐ + ๐ ๐๐ฉ๐๐ฉ + ๐๐๐๐
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Cylindrical Deformation Gradient
๐๐ฑ =๐๐
๐๐๐๐ = ๐ ๐๐ = ๐๐ ๐๐๐ ๐๐ง
๐๐๐๐๐
๐๐๐๐๐ฉ
๐๐๐๐๐
๐๐๐๐๐
๐๐๐๐๐ฉ
๐๐๐๐๐
๐๐๐ง๐๐
๐๐๐ง๐๐ฉ
๐๐๐ง๐๐
โ
๐๐ ๐๐ฉ๐ ๐๐
๐๐ ๐ ๐๐ฉ๐๐
So that the deformation gradient, in terms of unit vector sets {๐๐ , ๐๐ , ๐๐ง} and ๐๐ , ๐๐ฉ , ๐๐ , ๐ can be written as,
๐ =
๐๐๐๐๐
1
๐
๐๐๐๐๐ฉ
๐๐๐๐๐
๐๐๐๐๐๐
๐
๐
๐๐๐๐๐ฉ
๐๐๐๐๐๐
๐๐๐ง๐๐
1
๐
๐๐๐ง๐๐ฉ
๐๐๐ง๐๐
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