Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 VARIOUS ISSUES IN THE LARGE STRAIN THEORY OF TRUSSES STAMM 2006 Symposium on Trends in Applications of Mathematics to Mechanics Vienna, Austria, 10–14.

Similar presentations


Presentation on theme: "1 VARIOUS ISSUES IN THE LARGE STRAIN THEORY OF TRUSSES STAMM 2006 Symposium on Trends in Applications of Mathematics to Mechanics Vienna, Austria, 10–14."— Presentation transcript:

1 1 VARIOUS ISSUES IN THE LARGE STRAIN THEORY OF TRUSSES STAMM 2006 Symposium on Trends in Applications of Mathematics to Mechanics Vienna, Austria, 10–14 July 2006 C. A. Kaklamanis and K.V. Spiliopoulos Department of Civil Engineering National Technical University of Athens

2 2 OUTLINE INTRODUCTION  Manifold View of the Truss and Embeddings KINEMATICS  Notation and Point Transformation  The Deformation Gradient and its (Diagonal) Polar Decomposition  Various Strain Measures  The Velocity Gradient and the Deformation and Spin Tensors  Various Strain Rates CONSTITUTIVE MODELING  Strain Energy Rate and Strain Energy Density Rate  Conjugate Stress and Strain Measures  A Linear, Hyperelastic Model and Constitutive Transformations  The Strain Energy and the Finite Element Methodology AN EXAMPLE CONCLUDING REMARKS

3 3 INTRODUCTION (Manifold View of the Truss and Embeddings) Mathematically, view the truss as a 3D differentiable manifold. Use three parameters for particle identification, (s 1, s 2, s 3 ). The truss manifold consists of the s 1 and the s 2 – s 3 submanifolds. To make observations, we consider embeddings of the truss submanifolds. An acceptable embedding is denoted by m C, with m being some possible configuration labeling, such that m = 0 is the natural reference configuration. All quantities associated with a given configuration obtain the corresponding left superscript. We postulate that:  Acceptable embeddings of the s 1 submanifold correspond to straight line segments with length m L.  Acceptable embeddings of the s 2 – s 3 submanifold correspond to bounded plane areas being normal to the embedded s 1 line.  To describe this area, we need two linearly independent vectors (not necessarily orthogonal) that span it, while being normal to the unit vector that lies along the s 1 line.

4 4 KINEMATICS (Notation and Point Transformation) To develop the truss kinematics, consider the truss in two different configurations as shown in the Figure: Note that and are not necessarily orthogonal pairs. We will use n y or m z for these pairs when they are not orthogonal and x otherwise. F represents the transformation relating m C with n C. This is the transformation that we want to find. X1X1 X3X3 X 2 Global Frame mCmC m X (i) mX(j)mX(j) mXs(nx1)mXs(nx1) i j m x 3 m x 2 m x 1 Local Frames j i nXs(nx1)nXs(nx1) n X (j) n X (i) n x 1 n x 3 n x 2 F nCnC

5 5 KINEMATICS (Notation and Point Transformation) Consider a free vector a. Its representation in some frame ψ is denoted as a│ ψ. We have that: a│ ψ = R ψ · a│ X where R ψ is the transformation relating the ψ and X components of a. With reference to the previous figure, we find that: (1) where n L and m L are the truss lengths in each configuration. n λ y and m λ z are the norms of the n y, m z vector pairs used to span the truss’s cross section in each configuration. Bold subscripts denote the vector pairs used to describe the truss’s cross section.

6 6 KINEMATICS (The Deformation Gradient) Equation (1) is a point mapping, telling us how points in n C transform to point in m C. To describe truss deformations, we need to know how line elements transform. A line element is a vector, bound to some point of the embedded manifold, that connects this point with some point that is infinitesimally close to it. Mathematically, a line element is a vector belonging to the manifold’s tangent space. By (1), we get by differentiation, that: (2) where and are line elements of the truss in n C and m C respectively. is known as the deformation gradient. By (2) and (1), we see that points and line elements transform from one configuration to another in exactly the same way, i.e. via the deformation gradient.

7 7 KINEMATICS (The Polar Decomposition of the Deformation Gradient) The deformation gradient contains all the kinematic information necessary to describe the motion of the truss. In order to uncover the building blocks of this motion, we polar – decompose it, getting: (3) where R is an orthogonal matrix representing a 3D rotation of the truss and U and V are symmetric, positive definite matrices. There are three types of second order tensors: Eulerian, Lagrangian and mixed – type (or two – point) for which the m C only, the n C only, or both configurations are related. Clearly, F and R are two – point tensors, whereas U is Lagrangian and V Eulerian. We would like to find the conditions under which, U or V are intrinsically diagonal, since this is a great simplification. These conditions have to do with the vector pairs used to describe the truss’s cross section, as well as with the frames used in the various representations.

8 8 KINEMATICS (Conditions for Intrinsically Diagonal U and V Matrices) We find that: where (Note two eigenvalues only) Note that the z – y subscript, indicates that the non – orthogonal m z pair used to span the truss’s cross section in m C, is related to the n y pair used to span the truss’s cross section in n C, via the following: Note that n y o and m z o used in the representations, are the orthogonal frames corresponding to the orthogonal rotation part of the polar decomposition of the non – orthogonal matrix relating the n y and m z frames with the X frame.

9 9 KINEMATICS (Conditions for Intrinsically Diagonal U and V Matrices) Similarly, when orthogonal pairs are used for the description of the truss’s cross section (denoted by use of x subscript), as well as the corresponding frames for the representations, we get: where Note that here we have three distinct eigenvalues as opposed to the non – orthogonal pairs case where we had only two. It can be shown that the conditions for diagonal U or V amount to succeeding in having the Eulerian counterparts of U equaling the Lagrangian counterparts of V where the Eulerian counterpart of U is given by: By these relations, we find that the truss’s motion consists of a series of transformations that can take place in any order: a rigid body rotation given by R and a deformation involving stretching in three directions, described by Λ.

10 10 KINEMATICS (Various Strain Measures) Our concern lies in describing the deformation portion of the truss’s motion. Hence, we must use Λ to do that. (In the following we assume orthogonal vector pairs only). A quite general way of doing this, is by means of Seth’s strain measures: (4) Where k = …, -2, -1, 0, 1, 2, …. E is a Lagrangian tensor, whereas ε is the Eulerian counterpart of E. The strain measures in (4) (all equally plausible), measure the deformation experienced by the truss in going from the reference n C to the current m C configuration. For k = 1, we get the Biot strain, or corrotated engineering. For k = 2, we get the Green – Lagrange strain, for k = -2 we get the Almansi strain, whereas for k = -1, we get the Hyperbolic strain. For k = 0, we get, known as the Lagrangian and Eulerian Logarithmic (or Hencky) strains respectively.

11 11 KINEMATICS (The Velocity Gradient and the Deformation and Spin Tensor) The rate of change of change of the truss’s velocity with respect to position, is the velocity gradient m L. We let the following: where We find that: where are the deformation (symmetric part of m L) and spin tensors (antisymmetric part of m L) The Lagrangian counterpart of m d is We are now able to calculate the strain rates:

12 12 KINEMATICS (Various Strain Rates) The material rates of the various strain measures are found to be: For k = 0, note that we get the material rates of the Logarithmic strains to be identical with the corresponding rates of deformation, a result that is not true in general, but only in cases where the principal directions of U and V remain fixed. The material rates are not objective. An objective rate is Jaumann’s rate, given by: When k = 0 and a is the corrotating or the reference local frame, we get that: This result is a special case of a more general one given by Xiao in 1997, where it is shown that the Logarithmic rate of the Eulerian Logarithmic strain is identical to the Eulerian rate of deformation. For the truss, we find that it is the Jaumann’s rate is identical to Xiao’s Logarithmic rate.

13 13 CONSTITUTIVE MODELING (Strain Energy & Strain Energy Density Rate )


Download ppt "1 VARIOUS ISSUES IN THE LARGE STRAIN THEORY OF TRUSSES STAMM 2006 Symposium on Trends in Applications of Mathematics to Mechanics Vienna, Austria, 10–14."

Similar presentations


Ads by Google