CHAPTER 4: Isoparametric Elements and Solution Techniques · 1 CHAPTER 4: Isoparametric Elements...
Transcript of CHAPTER 4: Isoparametric Elements and Solution Techniques · 1 CHAPTER 4: Isoparametric Elements...
1
CHAPTER 4: Isoparametric Elements and Solution Techniques
NODE NUMBERING
■ How global stiffness coefficients Kij are stored strongly influences computer storage requirement and program execution speed.
■ Storage format depends on how nodes are numbered.
■ Shown is a six-node structure. Say, each node has one dof only.
■ Element stiffness matrices look like
2
Assembly ■ Element stiffness matrices are assembled by placing their entries in the proper rows and columns of the global stiffness matrix K.
■ Element 2-4, for example, is connected to nodes 2 and 4. Hence, its stiffness coefficients are placed in rows 2 and 4 of K.
■ Coefficients c and d are thus added because elements 1-2 and 2-4 collectively resist a force applied to node 2. (Think that there are two parallel springs at node 2, one from each neighboring element.)
3
Bandwidth ■ The assembled K has non-zero areas as shown in figure
■ K is banded, i.e., all coefficients beyond a certain distance from the diagonal are zero.
■ With good node numbering, nonzero coefficients cluster in a narrow band along the diagonal. The bandwidth is the width of this band
■ There may be zeros within the band; there are only zeros outside. ■ Here “half bandwidth” is 3.
4
Poor Node-Numbering
■ There are many zeros within the skyline. These zeros would have to be stored in the computer with most storage schemes. ■ When the equilibrium equation Kd=R is solved by a direct solver such as Gaussian elimination, such zeros become nonzero.. Undesirable: round-off error increases!
■The “half bandwidth” is 6 in this case.
5
Sparse Matrices and storage ■ In practice global stiffness matrices contain many zeros. Hence, they are sparse matrices. (In addition, they are almost always symmetrical.)
■ An array of 100x100 Q4 elements will generate10,000 element matrices, each with 8x8=64 coefficients. With 202x202=40804 nodes, stiffness matrix has 408042=1.67 billion coefficients with less than 640,000 non-zero coefficients!
■ Only those coefficients between the diagonal and the skyline need be stored in the computer (only the half band).
■ Software usually revises node and element numbering for efficiency.
6
Example
■ Prob. 4.1-a: One dof per node. Two-node elements. Shown is the node numbering that results in as few coefficients as possible between the skyline and the diagonal.
7
Equation Solving ■ Solution of Kd=R (*) :
◆ Direct methods: Gaussian elimination✦ forward reduction: orderly elimination of unknowns from lower
equations in (*). Has to be done only once when there are multiple load cases (i.e., more than one R vector).
✦ back substitution: substitution of solved unknowns into upper equations.
◆ Iterative methods: start with an initial guess for the solution vector and stop when convergence is achieved. Each load case has to be treated as a new problem.
■ Computational time for a direct method: ∝ nb2 ; n=order of K , b=bandwidth of K
■ Direct methods are better for small b or multiple load cases.
8
Transformation Between Coordinate Systems
■ A structural element may be arbitrarily oriented in a structure, that is, w.r.t. the structure axes (global axes).
■ Finite element stiffness matrices are much easier to formulate in an element coordinate system (local coordinates).
■ Element matrices are, therefore, calculated in local coordinates and transformed to global coordinates. Software does this.
9
Transformation of Nodal dof
■ For example,
■ k’ operates on u1’ and u2’.
φφ sincos 11'1 vuu +=
10
Element Stiffness Matrix in Global Coordinate System ■ It can also be shown that r=TTr’ where r and r’ are the element nodal load vectors expressed in global and local axes, respectively.
■ Then,
=== '
2
'1' example,for ; ''''uudTdkdkr
kdTdkTrT == ' ,by sides both gMultiplyin TT
■ Hence,
■ T would be 2x6 and k would be 6x6 for a 3-D finite element.
11
Offsets
■ Sometimes need to connect elements that don’t share common nodes.
■Or, the elements may physically be in contact but some software won’t allow direct connection of different types of elements to the same node.
■ In either case offset connection is used.
■ In this connection one node is made “slave” to a “master” node.
■ The slave node either undergoes the same displacement as the master or moves as dictated by the master.
12
Offset Example
■ Below, nodes 3 and 4 are two nodes of a plane element and nodes 1 and 2 are those of a beam. The corresponding nodes are connected by rigid links 3-1 and 4-2.
■ For example,
■ Hence, beam nodes 1 and 2 are made slaves to 3 and 4.
24421331 , avvbuu zz ϑϑ +=+=
13
Transformation of Stiffness Matrices for Offset ■ Expressing the relation between slave and master dof as a matrix equation,
■ If the beam element stiffness matrix operating on dof at nodes 1 and 2 is k’ and that operating on dof at nodes 3 and 4 is k, the two are related by
■ Hence nodes 1 and 2 are eliminated and don’t appear in KD=R. TkTk 'T=
14
Isoparametric Elements ■ Isoparametric formulation: to model nonrectangular elements, elements with curved sides, “infinite” elements for unbounded media, etc.
■ Four-node plane isoparametric quadrilateral:
■ Natural coordinate system ξη is used in addition to global XY.
■ Origin of ξη system is at the average of corner coordinates.
■ Element sides always defined by ξ=±1, η= ±1 regardless of the size and shape.
15
Coordinates and Displacement of a Point in an Isoparametric Element ■ A point within an isoparametric element has two sets of coordinates, (ξ, η) and (X, Y) related by same (iso) shape (interpolation) functions as displacements
(Recall the shape functions for a bilinear quadrilateral, Q4, Eq. 3.4-3)
■ Displacements of a point within the element :
16
Gradients in the Two Coordinate Systems
■ u and v displacements are parallel to X and Y, not to ξ and η !!!
■ “Isoparametric” means the same shape functions are used to interpolate both coordinates and displacements.
■ Strains:
Using the chain rule:
),(but etc., , ηξε fuXuX =∂∂=
awayright evaluatet can' weso Xu ∂∂
ξξξ ∂∂
∂∂
+∂∂
∂∂
=∂∂ Y
YuX
Xuu
17
The Jacobian Matrix ■ Doing the same thing for derivative w.r.t. η and rewriting
J: the Jacobian matrix
■ From the interpolation equation for the coordinates,
18
Strains ■ Inverting the matrix in the above equation,
■ Strains:
etc. , *12
*11 ηξ
ε∂∂
+∂∂
=∂∂
=uJuJ
Xu
X
1-*12
*11 ofrow first thein terms theare and where JJJ
Exercise: Express the other strain components similarly !
19
Stiffness Matrix ■ Hence, to compute strains, derivatives of u and v w.r.t. the natural coordinates are needed. For example,
■ The 3x8 strain-displacement matrix B relating strains to nodal dof can then be written.
■The 8x8 element stiffness matrix is given by
where we used
( |J| is like a scale factor between areas, equal to A/4 for a rectangle or parallelogram )