Chapter 7 - FEM

Practical Considerations in Modeling; Interpreting Results

Introduction

In this section we will discuss some modeling considerations and guidelines, including mesh size, natural subdivisions, and the use of symmetry and associated boundary conditions, We will also introduce the concept of static condensation, which enables us to apply the basis of the CST stiffness matrix to a quadrilateral quadrilateral element.

Finite Element Modeling

Finite element modeling is partly an art guided by visualizing physical interactions taking place within a body , In modeling the user is confronted with the difficult task of understanding understanding physical behavior taking place and understanding understanding the physical behavior of various elements available for use, Matching the appropriate finite element to the physical behavior being modeling is one of many decisions that must be made by the modeler. Understanding the boundary conditions can be one of the most difficult task a modeler must face in construction a useable finite element model. Aspect Ratio and Element Shape

The aspect ratio is define as the ratio of the longest dimension to the shortest dimension of a quadrilateral element , In general, as the aspect ratio increases, the inaccuracy of the finite element solution increases, Consider the five different finite element model shown in the figure below. A plot of the resulting error in the displacement displacement at point A of the beam verse aspect ratio is given. In addition, the numerical answers are given in the following table.

CIVL 7117 Finite Elements Methods in Structural Mechanics

Page 286


Page 287


Page 287

Case

Aspect Ratio

Number of Nodes

Number of Elements

Point A

Point B

% Error

1

1.1

84

60

-1.093

-0.346

5.2

2

1.5

85

64

-1.078

-0.339

6.4

3

3.6

77

60

-1.014

-0.238

11.9

4

6.0

81

64

-0.886

-0.280

23.0

5

24.0

85

64

-0.500

-0.158

56.0

Exac Exactt Solu Soluti tion on

-1.1 -1.152 52

-0.3 -0.360 60

In general, elements that yield the best results are compact and regular in shape will: (1) aspect ratios near one; and (2) corner angles of quadrilaterals near 90°. CIVL 7117 Finite Elements Methods in Structural Mechanics

Page 288

Use of Symmetry

The use of symmetry will often expedite the modeling of a problem. Symmetry allows us to consider a reduced problem instead of the actual problem. This will allow us to use a finer discretization of element with less computational cost. Natural Subdivisions at Discontinuities

There are a variety of natural subdivisions for finite element discretizations. For example, natural locations of nodes occur at concentrated loads or discontinuities in loading, other types of boundary conditions, and abrupt changes in geometry of materials. Sizing of Elements and Mesh Refinement

A discretization depends on the geometry of the structure, the loading, and the boundary conditions. For example, areas of high, rapidity changing stresses require a finer mesh then regions where the stress is constant.


Page 289

Here the use of symmetry is applied to a soil mass subjected to a foundation loading (66 nodes and 50 elements). Note that at the place of symmetry the displacements in the direction perpendicular to the plane muse be zero. This is modeled by rollers at nodes 2 - 6.


Page 290

The figure above illustrates the use of triangular elements for transitions from smaller quadrilaterals to larger quadrilaterals. The transitions are required since CST elements do not have immediate nodes along their edges. If an element had an intermediate node, the resulting equations would be inconsistent with the energy formulation for the CST equations.


Page 291


Page 292


Page 293

Infinite Medium

A typical example of infinite medium is a soil foundation problem. The guideline for the finite element model is that enough material must be included such that the displacements are nodes and stresses within the elements becomes negligibly small at locations far from the foundation load. The level of discretization can be determined by a trail-and-error procedure in which the horizontal and vertical distances from the load are varied and the resulting effects on the displacements and stresses are observed. For a homogeneous soil mass, experience has shown the influence of a footing becomes insignificant if the horizontal distance of the model is taken as approximately four and six times the width of the footing and the vertical distance is taken as approximately four to ten times the width of the footing. Checking the Model

The discretized finite element model should be checked carefully before results are computed. Ideally, a model should be checked by an analyst not involved in the preparation of the model, who is then more likely to be objective. Preprocessors with their detailed graphical display capabilities now make it comparatively easy to find errors, particularly with a misplaced node or missing element or a misplaced load or boundary condition. Preprocessors include the ability to color, shrink, rotate, and section a model mesh. Checking the Results and Typical Postprocessor Results

An analyst should probability spend as mush time processing, checking, and analyzing results as spent in data preparation.


Page 294

Plate with a hole in the center.

Deformed shape of plate superimposed over an unreformed shape


Page 295

The wrench in this example is modeled by 307 constraint strain triangular elements (plane stress assumption). Below is a plot of the deformed shape of the wrench over the original mesh.


Page 296

On the right is a plot of the stress in the wrench and the plot on the left is a plot of the defected shape. This analysis was preformed using WinFElt finite element structural analysis system.


Page 297

Equilibrium and Compatibility of Finite Element Results

An approximate solution for a stress analysis problem using the finite element method based on assumed displacement fields does not generally satisfy all the requirements for equilibrium and compatibility that an exact theory-of-elasticity solution satisfies. However, remember that relatively few exact solutions exist. Hence, the finite element method is a very practical one for obtaining reasonable, but approximate, numerical solutions. We now describe some of the approximations generally inherent with finite element solutions. 1.

Equilibrium of nodal forces and moments is satisfied. This is true because the global equation F = Kd is a nodal equilibrium equation whose solution for d is such that the sums of all forces and moments applied to each node are zero. Equilibrium of the whole structure is also satisfied because the structure reactions are included in the global forces, and hence, in the nodal equilibrium equations.

2.

Equilibrium within an element is not always satisfied. However, for the constant-strain bar and the constant-strain triangle, element equilibrium is satisfied. Also the cubic displacement function is shown to satisfy the basic beam equilibrium differential equation, and hence, to satisfy element force and moment equilibrium.

3.

Equilibrium is not usually satisfied between elements. A differential element including parts of two adjacent finite elements is usually not in equilibrium (see the figure below). For line elements, such as used for truss and frame analysis, interelement equilibrium is satisfied. However, for two- and three-dimensional elements, interelement equilibrium is not usually satisfied. Also, the coarseness of the mesh causes this lack of interelement equilibrium to be even more pronounced.


Page 298

4.

Compatibility is satisfied within an element as long as the element displacement field is continuous; hence, individual elements do not tear apart.

5.

In the formulation of the element equations, compatibility is invoked at the nodes. Hence, elements remain connected at their common nodes. Similarly, the structure remains connected to its support nodes because boundary conditions are invoked at these nodes.

6.

Compatibility may or may not be satisfied along interelement boundaries. For line elements such as bars and beams, interelement boundaries are merely nodes. The constant-strain triangle remain straight sided when deformed and therefore, interelement compatibility exists for these elements. Incompatible elements, those that allow gaps or overlaps between elements, can be acceptable and even desirable.


Page 299

Convergence of Solution

When the mesh size is reduced - that is the number of elements is increased we are ensured of monotonic convergence of the solution when compatible and complete displacement functions are sued.

Case

Number of Elements

Number of Nodes

Aspect Ratio

Point A

1

12

21

2

-0.740

2

24

39

1

-0.980

3

32

45

3

-0.875

4

64

85

1.5

-1.078

5

80

105

1.2

-1.100

Exact Solution CIVL 7117 Finite Elements Methods in Structural Mechanics

-1.152 Page 300

Interpretation of Stresses

In the stiffness or displacement formulation of the finite element method, used in this course, the primary quantities determined are the interelement model displacements of the assemblage. Secondary quantities, such as stress and strain, are computed based on these nodal displacements. In the case of the bar and constant-strain triangles, stresses are constant over the element. For these elements, it is common practice to assign the stress to the centroid of the element with acceptable results. An alternative procedure sometimes is to use an average (possibly weighted) value of the stresses evaluated at each node of the element. This averaging method is often based interpolating the element nodal values using the element shape functions. The averaging method is called smoothing . While the results from smoothing may be pleasing to the eye, they may not indicate potential problems with the model and the results. You should always view the unsmoothed contour plots as well. Highly discontinuous contours between elements in a region of an unsmoothed plot indicate modeling problems and typically require additional refinement of the element mesh in the suspect region.

Static Condensation

Let’s consider the concept of static condensation and used it to develop the stiffness matrix of a quadrilateral element. Consider a general quadrilateral element as shown below.


Page 301

An imaginary node 5 is temporary introduced at the intersection of the diagonals of the quadrilateral to create four triangles. We can superimposed the stiffness matrices of the four triangles to create the stiffness matrix of the quadrilateral element, where the internal imaginary node 5 degrees of freedom are said to be condensed out so that they never inter the final equations. Let’s start by partitioning the equilibrium equations:

K11 K12  de  Fe   =  K   21 K 22   di   Fi 

where d i is the vector of displacements corresponding to the imaginary internal node, F i is the vector of loads at the internal node, and d e and F e are the actual displacements and loads, respectively. Rewriting the above equations we gives: K11d e + K12d i = Fa K 21de + K 22d i = Fi

Solving for d i gives: d i = −K 22 −1K21de + K 22 −1Fi

Substituting the above equation, we obtain the condensed equilibrium equation: kc dc = F c

where kc = k11 − k12k22 −1k 21 Fc = Fe − k12k22 −1Fi

where k c and called the condensed stiffness matrix and the condensed F c are load vector , respectively. An advantage of the four-CST quadrilaterals is that the solution becomes less dependent on the skew of the subdivision mesh. The skew means a directional stiffness bias that is built into a model trough certain discretization patterns.


Page 302

The stiffness matrix of a typical triangular element, call it element 1, labeled with nodes 1, 2, and 5 is given as: (1) (1) (1) k11  k12 k 15  (1) (1) (1)  k (1)  = k21 k22 k25  (1) (1) (1) k51  k52 k 55  

where k ij (1) is a 2 x 2 matrix. The assembled stiffness matrix for the quadrilateral is:


Page 303

Example Problem

Consider the quadrilateral with internal node 5 and dimensions as shown below. Apply the static condensation technique.

Using the CST stiffness matrix for plain strain, we get: 1

2

5

3

4

5

 1.5 1.0 0.1 0.2 −1.6  1.0 3.0 −0.2 2.6 −0.8  1.5 −1.0 −1.6 E  0.1 −0.2 [k (1) ] = [k (3) ] =  4.16  0.2 2.6 −1.0 3.0 0.8  −1.6 −0.8 −1.6 0.8 3.2   −1.2 −5.6 1.2 −5.6 0.0

−1.2  −5.6  1.2   −5.6  0.0   11.2 

4

1

5

2

3

5

 1.5 1.0 0.1 0.2 −1.6  1.0 3.0 −0.2 2.6 −0.8  1.5 −1.0 −1.6 E  0.1 −0.2 (2) (4) [k ] = [k ] =  4.16  0.2 2.6 −1.0 3.0 0.8  −1.6 −0.8 −1.6 0.8 3.2   −1.2 −5.6 1.2 −5.6 0.0


−1.2  −5.6  1.2   −5.6  0.0   11.2 

Page 304

The resulting assembled matrix before static condensation is:

After partitioning, the condensed stiffness matrix is:


Page 305

Flowchart for the Solution of Place Stress/Strain Problems

The following flowchart is typical for a finite element process used for the analysis of plane stress and plane strain problems.


Page 306

Example Problem

Consider the thin plate subjected to the surface traction shown in the figure below (as discussed in Chapter 6). Use WinFElt to model this problem. Assume plane stress conditions.

Assume plane stress conditions. Let E = 30 x 10 6 psi, ν = 0.30, and t = 1 in. Determine the nodal displacements and the element stresses. Let’s discretize the plate into two elements as shown below:

The general WinFElt input file format is: problem description title = " " type = static : transient : static-substitution : modal : spectral : static-thermal : transient-thermal nodes = nnn elements = nnn


Page 307

analysis parameters start = xxx stop = xxx step = xxx beta = 0.25 gamme = 0.5 alpha = 0.0 nodes = [nnn, nnn, ...] dofs = [Tx, Ty, Tz, Rx, Ry, Rz] mass-mode = lumped : consistent nodes x_name_x elements material properties forces distributed loads constraints end

For this problem the WinFElt file is: problem description title = "example_p291" nodes = 4 elements = 2 analysis = static analysis parameters dofs = [Tx, Ty] nodes 1 x 2 x 3 x 4 x

= 0 = 0 = 20 = 20

y y y y

= 0 = 10 = 10 = 0

z z z z

= = = =

0 0 0 0

constraint = pin constraint = pin constraint = free force = x_force force = x_force

CSTPlaneStress elements 1 nodes = [1,3,2] material = plate 2 nodes = [1,4,3] material = plate

material properties plate t = 1.0 E = 30e6 forces x_force

nu = 0.30

Fx = 5000.0

distributed loads


Page 308

constraints pin Tx = c free Tx = u

Ty = c Ty = u

end

In this input file the keywords: title, nodes, elements , and analysis define the type of problem. Under analysis parameters section the keyword dofs define the degrees of freedom for each node. Next the nodes are defined using the format: nnn

x = xxx

y = xxx

z = xxx

force = x_name_x

constraint = x_name_x

where nnn is the node number, x, y, and z are the coordinates of the node, force is the name of the point force, and constraint is the name of the cons taint definition. Next the elements are defined using the following format: nnn

nodes = [nnn, nnn, ...]

material = x_name_x

load = x_name_x

where nnn is the element number, the nodes parameter defines the nodes that form he element, material defines the element material properties, and load define any distributed loads on the element. Next the element material properties are defined using the following format: x_name_x

A = xxx Ix = xxx Rk = xxx Kx = xxx

t = xxx E = xxx G = xxx nu = xxx Iy = xxx Iz = xxx J = xxx Rm = xxx Ky = xxx Kz = xxx c = xxx

kappa = xxx

where x_name_x is the name of the material and the remaining keyword define the section properties. Next the nodal forces are defined using the following format: x_name_x

Fx = xxx

Fy = xxx

Fz = xxx

Mx = xxx

My = xxx

Mz = xxx

where x_name_x is the name of the nodal force and the remaining keyword define the magnitude and direction of the concentrated force. Next the elemental distributed loads are defined using the following format: x_name_x

direction = x_name_x

values = (1, xxx) (2, xxx)

where x_name_x is the name of the element distributed load the remaining keyword define the magnitude, shape, and direction of the distributed load. CIVL 7117 Finite Elements Methods in Structural Mechanics

Page 309

Next the constraint conditions are defined using the following format: x_name_x

Tx = c:u

Ty = c:u

Tz = c:u

Rx = c:u

Ry = c:u

Rz = c:u

where x_name_x is the name of the constraint condition, the keyword u and c state whether the degree of freedom is unknown or constrained. The input file is terminated with an end statement. After opening the input file in WinFElt, click on the solve button and an new window will open with displaying the results: ** example_p219 ** Nodal Displacements ----------------------------------------------------------------------------Node # DOF 1 DOF 2 DOF 3 DOF 4 DOF 5 DOF 6 ----------------------------------------------------------------------------1 0 0 0 0 0 0 2 0 0 0 0 0 0 3 0.00060958 4.1633e-06 0 0 0 0 4 0.0006637 0.00010408 0 0 0 0 Element Stresses ------------------------------------------------------------------------------1: 1004.8 301.44 2.4019 1004.8 301.43 0.19566 2: 995.2 -1.201 -2.4019 995.2 -1.2068 -0.13812

Reaction Forces ----------------------------------Node # DOF Reaction Force ----------------------------------1 Tx -5000 1 Ty -3002.4 2 Tx -5000 2 Ty 3002.4


Page 310

The results are similar to those presented in the textbook on page 298. In addition, WinFElt has some graphical visualization capabilities. By selecting the appropriate options on the Controls menu, a color contour of the displacements can be plotted.

If more elements are utilized it may be possible to visualize the smoothed stress contours.

Note that the stress contours are not correct and most likely due to the fact the CST elements do not provide continuous inter-element stress values.


Page 311

WinFElt help file for basic keywords :Problem description The problem description section is used to define the problem title and the number of nodes and elements in the problem. The problem description section is the only section which you cannot repeat within a given input file. :nodes = integer :elements = integer These numbers will be used for error checking so the specifications given here must match the actual number of nodes and elements given in the definition sections. Note that the definitions for nodes and elements do not have to be given in numerical order, as long as nodes 1 ... m and elements 1 ... n (where m is the number of nodes and n is the number of elements) all get defined in one of the element and node definition sections in the file. :analysis = problem type defines the type of problem that you wish to solve. Currently it can either be static, transient, static-substitution, modal, static-thermal, transient-thermal, or spectral. If you do not specify anything, static analysis will be assumed. :Nodes The nodes section(s) must define all of the nodes given in the problem. :x = constant expression :y = constant expression :z = constant expression Each node must be located with an x, y, and z coordinate using x=, y=, and z= assignments. Coordinates are taken as 0.0 if they are not otherwise defined. If a coordinate is left unspecified for a given node, it takes the value for that coordinate from the previous node. :constraint = string A node must have a constraint assigned to it by a constraint=statement. The constraint name specified by string must be a valid constraint defined in the constraints section of the input file. The default constraint leaves the node completely free in all six degrees of freedom. Like coordinates, if a constraint is left unspecified, the node will be assigned the same constraint as the previous node. :force = string


Page 312

Forces (applied point loads and moments) on a node are optional and are applied using the force= statement; if a force is not specified there will be no force applied to that node. :mass = constant expression

specifies an optional lumped mass at this node.

:Elements An element definition section begins with the keywords xxxxx elements where xxxxx is the symbolic name of a type of element. All elements under this section heading will be taken to be the given element type. There could be multiple sections that defined beam elements, but each must begin with the keywords beam elements. If there were truss elements in the same problem, they would have to be defined in sections which began with the keywords truss elements. Currently available types are spring, truss, beam, beam3d, timoshenko, CSTPlaneStrain, CSTPlaneStress, iso2d_PlaneStrain, iso2d_PlaneStress, quad_PlaneStrain, quad_PlaneStress, htk, brick, rod, and ctg. A FElt problem is not limited to one element type; the routines for assembling the global stiffness matrix takes care of getting the right parts of the right element stiffness matrices into the global matrix. :nodes = [n1, n2, ...] Each element must have a list of nodes to which it is attached. The node list is defined with the nodes=[ ... ] statement. The length of the list inside the square brackets varies with element type, but must always contain the full number of nodes which the element type definition requires. :material = string A material property must be assigned to every element with a material= statement. If the material is never specified, an element will take the same material property as the previous element. If nothing ever gets assigned to an element, the default material property will have zeros for all of its characteristics. :load = string An element can have up to three optional distributed loads. Each load is assigned with a separate load= assignment. The named load in each assignment must be a valid object defined in the distributed loads section. Each element type may treat a distributed load differently so you should be careful that the names given for the loads on a given element match the names of distributed loads which are defined in a manner conformant with what that element type is expecting. :Material properties The material properties section(s) is quite simple. Each material has a name followed by a list of characteristics, such as steel E = 30e6 A = 1.0 ... CIVL 7117 Finite Elements Methods in Structural Mechanics

Page 313

Note that no element types require a material property with every characteristic defined. In fact, most only use three or four characteristics. Density (rho) is always necessary for the materials in a transient, modal, or spectral analysis problem if you actually want your elements to have any inertia. Different element types can certainly use the same material as long as that material definition contains the right characteristics for each type of element that uses it. :E = constant expression

Specifies the material's Young's (elastic) modulus.

:A = constant expression

Specifies the material's cross-sectional area.

:t = constant expression

Specifies the material's thickness.

:rho = constant expression

Specifies the material's density.

:nu = constant expression

Specifies the material's Poisson's ratio.

:G = constant expression

Specifies the material's bulk or shear modulus.

:J = constant expression

Specifies the material's torsional stiffness.

:Ix = constant expression

Specifies the material's I_xx moment of inertia.

:Iy = constant expression

Specifies the material's I_yy moment of inertia.

:Iz = constant expression

Specifies the material's I_zz moment of inertia.

:kappa = constant expression

Specifies the material's shear force correction factor.

:Rk = constant expression

Specifies the material's Rayleigh stiffness damping coefficient.

:Rm = constant expression

Specifies the material's Rayleigh mass damping coefficient.

:Kx = constant expression

Specifies the material's thermal conductivity along the x-direction.

:Ky = constant expression

Specifies the material's thermal conductivity along the y-direction.

:Kz = constant expression

Specifies the material's thermal conductivity along the z-direction.

:c = constant expression

Specifies the material's heat capacitance.


Page 314

:Constraints :Tx = constraint specification :Ty = constraint specification :Tz = constraint specification :Rx = constraint specification :Ry = constraint specification :Rz = constraint specification The constraints section(s) must define all of the named constraints within the problem. Each constraint is defined by a name followed by a list of DOF specifications of the form Tx=? Ty=? Tz=? Rx=? Ry=? Rz=? where the ? can either be c for constrained, u for unconstrained or a valid, possibly time-dependent, expression for cases where a displacement boundary condition is required (e.g., settlement of support, or time-varying temperature along a boundary in a transient thermal analysis problem). Note that specifying Tx=c is equivalent to Tx=0.0. The T and R refer to translation and rotation, respectively and the subscripted axis letter indicates that the specification refers to translation along, or rotation about, that axis. An additional specification of h or hinged is allowed for the rotational DOFs for cases where it is necessary to model a nodal (momentless) hinge. Note that a hinge specification currently has meaning only on beam, beam3d and timoshenko elements. If a specification is never made for a DOF, then the problem is assumed to be unconstrained in that degree of freedom. Getting the constraints right is an important part of getting a reasonable solution out of a finite element problem so you should be aware of what DOF are active in a given problem (this will depend on which types of elements are being used ... even in a 2-d problem, the global stiffness matrix will take displacement in the z direction into account if there are 3-d elements in use, consequently, those displacements should be constrained). :ITx = constant expression :ITy = constant expression :ITz = constant expression :IRx = constant expression :IRy = constant expression :IRz = constant expression In order to account for initial conditions in transient analysis problems, a constraint specification may also include the initial (at time t=0) displacements, and velocity and acceleration in the translational DOFs. Initial displacements are given with ITx=, IRz=, etc. :Vx = constant expression :Vy = constant expression :Vz = constant expression :Ax = constant expression :Ay = constant expression :Az = constant expression You can specify initial accelerations with Ax=, Ay=, and Az=. Initial velocities are given by Vx=, Vy=, and Vz=. Unspecified velocities and displacements will be taken as 0.0. If there are no initial accelerations specified (i.e., none of the nodes have a constraint with an acceleration CIVL 7117 Finite Elements Methods in Structural Mechanics

Page 315

assigned) then the initial acceleration vector will be solved for by the mathematical routines based on the initial force and velocity vectors. If any of the nodes have a constraint which has an acceleration component defined (even if that component is assigned to 0.0) then the mathematical routines will not solve for an initial acceleration vector; they will build one based on the constraint information, assigning 0.0 to any component that was not specified. What this means is that you cannot specify the initial acceleration for only a few nodes and expect the mathematical routines to simply solve for the rest of them. If you specify any of the initial accelerations then you are effectively specifying all of them. :Forces :Fx = variable expression :Fy = variable expression :Fz = variable expression :Mx = variable expression :My = variable expression :Mz = variable expression The forces section(s) defines all of the point loads used in the problem and actually looks a lot like the constraints section. The magnitudes of the forces in the six directions are specified by Fx=? Fy=? Fz=? Mx=? My=? Mz=?. If the value for a given force component is not given it is assumed to be zero. The directions for both forces and the constraints as defined above should be given in the global coordinate system (right-hand Cartesian). If you are doing transient analysis, the force definitions can be more complicated than a simple numerical assignment or expression. FElt allows you to specify a transient force as either a series of time-magnitude pairs (a discrete function in time) or as an actual continuous function of time. This latter fact means that you can define a force as Fx=sin(t) rather than having to discretize the sine function. These continuous forcing functions are a special case of expressions as discussed above. :Sfx = variable expression :Sfy = variable expression :Sfz = variable expression :Smx = variable expression :Smy = variable expression :Smz = variable expression For spectral analysis problems, you can explicitly specify input spectra using Sfx=, Smy=, etc. if you want to to compute the actual power spectrum of the output. These spectra can be analytic functions of w or discrete frequency, power pairs. :Distributed loads The distributed loads section(s) must contain a definition for each distributed load that was assigned in the element definition section(s). A valid definition for a distributed load is a symbolic name followed by the keywords direction=xxx and values=(n,x) (m,y) ... The direction assignment must be set to one of parallel, perpendicular, LocalX, LocalY, LocalZ, GlobalX, GlobalY, GlobalZ, radial, axial. The valid directions for a given element type, and what those directions CIVL 7117 Finite Elements Methods in Structural Mechanics

Page 316

refer to for that element type are described in the individual element descriptions in chapter~\ref{elements. The values assignment is used to assign a list of load pairs to the named load. A load pair is given in the form (n, x) where n is the local node number to which the magnitude given by x applies. Generally, two pairs will be required after the values= token. The imaginary line between the two magnitudes at the two nodes defines an arbitrarily sloping linearly distributed load. This allows you to specify many common load shapes: a constant distributed load of magnitude y, including cases of self-weight, a load which slopes from zero at one node to x at a second node, or a linear superposition of these two cases in which the load has magnitude y at node 1 and magnitude $x+y$ at node 2. The definition of this load would be: distributed loads load_case_1

direction=perpendicular

values=(1,-2000) (2,-6000)

:Analysis parameters The analysis parameters section is required only if you are doing some type of transient, modal, or spectral analysis (e.g., analysis=transient, analysis=spectral in the problem description section). For modal analysis it is simply used to set the type of element mass matrices that will be formed, but for transient and spectral analyses it contains information that further defines the problem and the parameters for the numerical integration in time. :start = constant expression the start of the frequency range of interest in spectral analysis :stop = constant expression the end of the time (transient analysis) or frequency (spectral analysis) range of interest. :duration = constant expression an alias for stop= in transient analysis problems. :step = constant expression the time or frequency step to be used between the start and stop points :dt = constant expression an alias for step= in transient analysis problems. :beta = constant expression :gamma = constant expression :alpha = constant expression integration parameters in the structural and thermal dynamic integration schemes :mass-mode = for the types of element mass matrices that should be formed, either lumped CIVL 7117 Finite Elements Methods in Structural Mechanics

Page 317

or consistent (note again that this is the only assignment that is required in this section for a modal analysis problem); :nodes =[integer, integer, ... ] defines a list of nodes which you are interested in seeing output for. The list of nodes should just be a comma delimited list of valid global node numbers. :dofs = [ ... ] defines the list of local DOF that you are interested in for the nodes that you are interested in. The list of DOF should be a list of symbolic DOF names (Tx, Ty, Tz, Rx, Ry, Rz). You will get solution output for each of these DOF at each of the nodes that you specified in the node list. :Rk = constant expression :Rm = constant expression global Rayleigh damping parameters with Rk= and Rm=. If either of these parameters is nonzero then the global damping matrix will be formed using these two parameters and the global mass and stiffness matrices as opposed to being formed from elemental Rayleigh parameters and elemental mass and stiffness matrices.


Page 318

Chapter 7 - FEM

Recommend Documents