CHAPTER 4: Isoparametric Elements and Solution Techniques · 1 CHAPTER 4: Isoparametric Elements and Solution Techniques NODE NUMBERING How global stiffness coefficients Kij are stored

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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

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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.)

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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.

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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.

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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.

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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.

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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.

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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.

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Transformation of Nodal dof

■ For example,

■ k’ operates on u1’ and u2’.

φφ sincos 11'1 vuu +=

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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,

=== '

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'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.

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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.

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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 +=+=

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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=

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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.

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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 :

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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

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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,

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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 !

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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 )