Transcript VARIOUS ISSUES IN THE LARGE STRAIN THEORY OF TRUSSES

VARIOUS ISSUES IN THE
LARGE STRAIN THEORY OF TRUSSES
C. A. Kaklamanis and K.V. Spiliopoulos
Department of Civil Engineering
National Technical University of Athens
STAMM 2006
Symposium on Trends in Applications of
Mathematics to Mechanics
Vienna, Austria, 10–14 July 2006
1
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
2
INTRODUCTION (Manifold View of the Truss and Embeddings)






Mathematically, view the truss as a 3D differentiable manifold.
Use three parameters for particle identification, (s1, s2, s3).
The truss manifold consists of the s1 and the s2 – s3 submanifolds.
To make observations, we consider embeddings of the truss submanifolds.
An acceptable embedding is denoted by mC, 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 s1 submanifold correspond to straight line segments
with length mL.
 Acceptable embeddings of the s2 – s3 submanifold correspond to bounded plane
areas being normal to the embedded s1 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 s1 line.
3
KINEMATICS (Notation and Point Transformation)

To develop the truss kinematics, consider the truss in two different configurations as
shown in the Figure:
F
nX
2
nx
n
3
n
xˆ 2
xˆ 3
mx
(j)
mx
2
3
n
i
nX
1
j
nC
nx
nx
X3
xˆ 1
m
m
mX
(i)
nXs(nx
1)
xˆ 3
(i)
m
i
mC
xˆ 2
j
mx
1
Local
Frames
xˆ 1
mXs(nx
mX
1)
(j)
X2
Global
Frame
X1



Note that n xˆ 2 , n xˆ 3 and m xˆ 2 , m xˆ 3 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 mC with nC. This is the transformation that we
want to find.
4
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 = Rψ · a│X
where Rψ is the transformation relating the ψ and X components of a.

With reference to the previous figure, we find that:

where

X zˆ  m X  i   m X zˆA c 
m
n Fzˆ yˆ
m
n  zˆ yˆ

m

 mn 

 0
 0

 R  mn Fzˆyˆ
0
m
n  zy 2
0

X



m

n  zy 3 
0
0
m
n Fzˆ yˆ

 R 1


n
m
n Fzˆ yˆ
m
L
m


n
n
L
X yˆ  n X  i   n X yAˆ  c 
X

(1)

 R z1 mn  zˆ yˆ  R y
m
m
n  zy 2

n
 z2
 y2
m
m
n  zy 3

n
 z3
 y3
n
L and mL are the truss lengths in each configuration. nλy and mλz 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.
5
KINEMATICS (The Deformation Gradient)

Equation (1) is a point mapping, telling us how points in nC transform to point in mC.

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:


d mX zˆ
where d nX yˆ
m
n Fzˆ yˆ 




m
n Fzˆ yˆ

 d nX yˆ

(2)
and d mX zˆ are line elements of the truss in nC and mC 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.
6
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:
m
(3)
F
 mR
 mU
 mV
 mR
n
zˆ yˆ

n
zˆ yˆ

n
zˆ yˆ

n
zˆ yˆ

n
zˆ yˆ

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 mC only, the nC 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.
7
KINEMATICS (Conditions for Intrinsically Diagonal U and V Matrices)

We find that:
m
n Fzˆ yˆ n
yo
 mn R zˆ yˆ mn zˆ yˆ 
m
 mn R zˆ yˆ
n Fzˆ yˆ m
zo
m
 mn R zˆ yˆ
n Vzˆ yˆ n
yo

 mn U zˆ yˆ
 mn Vzˆ yˆ
m
mn R zˆ yˆ
n Vzˆ yˆ n
yo
m
m
zo
zo
 mn zˆ yˆ mn R zˆ yˆ where
mn R Tzˆ yˆ
m
n U zˆ yˆ n
yo
m
zo
mn R Tzˆ yˆ
 mn  0
0 


m
m
m
m
V

U



0

0
n zˆ  yˆ m
n
zˆ  yˆ n
n zˆ  yˆ
n zy 2

 (Note two eigenvalues only)
zo
yo
m
 0

0
n  zy 2 

Note that the z – y subscript, indicates that the non – orthogonal mz pair used to span
the truss’s cross section in mC, is related to the ny pair used to span the truss’s cross
m
section in nC, via the following:
 z 2 m  z3
 n
m
n
zˆ 2 m zˆ 3  n yˆ 2 n yˆ 3
y
y
2

 mn R zˆ yˆ  mn U zˆ yˆ
3
Note that nyo and mzo 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 ny and mz frames with the X frame.
8
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:
m
n Fxˆ n
x



 mn R xˆ mn  xˆ 
m
m
n Vxˆ n  n R xˆ
x
m
n Fxˆ m
x
 mn R xˆ 
m
n U xˆ m
x
 mn xˆ  mn R xˆ
 mn  0
0 


m
m
m
m
V

U



0

0
where
n xˆ m
n
xˆ n
n xˆ
n x2


z
x
m
 0

0

n
x

3 
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: mn Rmn Umn R T
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.
9
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:
m k



1
0
0
n

m k 
m k 
1
1
k
m
k
m
 0

0
n  xˆ n  n  xˆ m 
n  xˆ  I 
n  x 1
x
x
2

(4)
k
k
k
m
 0

0
n  x 1


3
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 nC to the current mC 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
m 0 
n  xˆ n
x
 mn  xˆ0 

m
x
 ln

m
n  xˆ
, known as the Lagrangian and
Eulerian Logarithmic (or Hencky) strains respectively.
10
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 mL.
m
m
n 1
We let the following: n  xˆ t   x t   x

m

n
0
0
  t 

 
m
 n x   0
where m  x t    0
 x t 
0
2



m
 0
0
0
 x t 



3
m
m
m
We find that: L xˆ t  m  d xˆ t  m  w xˆ t  m

x
x
0
n

0
x2
n
 n
L
0   L
0 
m
 m
L
n
 x  
L
3 
x
m
m
m 1
m
w xˆ t  m  0
where d xˆ t  m x  
x t   x t 
x
m
are the deformation (symmetric part of L) and spin tensors (antisymmetric part of
m
L)
m
The Lagrangian counterpart of md is

D mn R T m dmn R
We are now able to calculate the strain rates:
11
KINEMATICS (Various Strain Rates)

The material rates of the various strain measures are found to be:
m k 
 xˆ m
n
x
 m d xˆ
m
x
m  k 
n  xˆ n
x
m kx n xk
 m D xˆ
n
x
m kx n xk

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:
m
  k   mn   k   mn   k   m w  m w  mn   k 
n
a

a
a
a
a
a
When k = 0 and a is the corrotating or the reference local frame, we get that:
0 
m


n
a

m
d
a
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.
12
CONSTITUTIVE MODELING (Strain Energy & Strain Energy Density Rate )
13