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ABAQUS Lecture Notes

By:Mohammad Javad Kazemzadeh-Parsi

Assistant professor of Mechanical Engineering

Islamic Azad University, Shiraz Branch

Shiraz, Iran



ABAQUS Lecture Notes M.J. Kazemzadeh-Parsi

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Chapter 1

IntroductionThe finite element method is a numerical method that can be used for the accurate solution

of complex engineering problems. Although the origins of the method can be traced to severalcenturies back, most of the computational details have been developed in mid 1950s,

primarily in the context of the analysis of aircraft structures. Thereafter, within a decade, the

potential of the method for the solution of different types of applied science and engineering

problems was recognized. Over the years, the finite element technique has been so well

established that today, it is considered to be one of the best methods for solving a wide

variety of practical problems efficiently. In addition, the method has become one of the active

research areas not only for engineers but also for applied mathematicians. One of the main

reasons for the popularity of the method in different fields of engineering is that once a

general computer program is written, it can be used for the solution of a variety of problems

simply by changing the input data.The ABAQUS finite element software has strong capabilities for solving, specifically, nonlinear

problems and was developed by Hibbitt, Karlsson&Sorenson, Inc. The solution of a general

problem by ABAQUS involves three stages: ABAQUS Preprocessor, ABAQUS Solver, and

ABAQUS Postprocessor. ABAQUS/CAE or another suitable pre-processor provides a

compatible input file to ABAQUS. ABAQUS/Standard or ABAQUS/Explicit can be used as

ABAQUS/Solver to solve the problem. The ABAQUS/Standard, based on implicit algorithm, is

good for static, strongly nonlinear problems. ABAQUS/Explicit, based on explicit algorithm, is

intended for dynamic problems. Both ABAQUS/Standard and ABAQUS/Explicit can be

executed under ABAQUS/CAE. The ABAQUS/CAE or another suitable postprocessor can be

used for displaying the output (results) of the problem.ABAQUS/CAE provides a complete ABAQUS environment that provides a simple, consistent

interface for creating, submitting, monitoring, and evaluating results from ABAQUS/Standard

and ABAQUS/Explicit simulations. ABAQUS/CAE is divided into modules, where each module

defines a logical aspect of the modeling process; for example, for defining the geometry,

defining the material properties, and generating a mesh. As we move from one module to

another module, we build the model from which ABAQUS/CAE generates an input file that

we can submit to the ABAQUS/Standard or ABAQUS/Explicit for carrying the analysis. After

completing the analysis, the unit (ABAQUS/Standard or ABAQUS/Explicit) sends the

information to ABAQUS/CAE to allow us to monitor the progress of the job, and generates an

output database. Finally, we use the visualization module of ABAQUS/CAE (also licensedseparately as ABAQUS/Viewer) to read the output database and view the results of analysis.

The ABAQUS/Viewer provides graphical displays of ABAQUS finite element models and

results. It obtains the model and results information from the output database. We can

control the output information displayed. For example, we can obtain plots such as

undeformed shape, deformed shape, contours,x-ydata, and time history animation from

ABAQUS/Viewer.

Basic Concepts of the Finite Element MethodThe basic idea in the finite element method is to find the solution of a complicated problem

by replacing it by a simpler one. Since the actual problem is replaced by a simpler one infinding the solution, we will be able to find only an approximate solution rather than the exact




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solution. The existing mathematical tools will not be sufficient to find the exact solution (and

sometimes, even an approximate solution) of most of the practical problems. Thus, in the

absence of any other convenient method to find even the approximate solution of a given

problem, we have to prefer the finite element method. Moreover, in the finite element

method, it will often be possible to improve or refine the approximate solution by spending

more computational effort.

In the finite element method, the solution region is considered as built up of many small,

interconnected subregions called finite elements. As an example of how a finite element

model might be used to represent a complex geometrical shape, consider the milling machine

structure shown in Figure 1-1(a). Since it is very difficult to find the exact response (like

stresses and displacements) of the machine under any specified cutting (loading) condition,

this structure is approximated as composed of several pieces as shown in Figure 1-1(b) in the

finite element method. In each piece or element, a convenient approximate solution is

assumed and the conditions of overall equilibrium of the structure are derived. The

satisfaction of these conditions will yield an approximate solution for the displacements and

stresses. Figure 1-2 shows the finite element idealization of a fighter aircraft.

Figure 1-1

Figure 1-2




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Historical BackgroundAlthough the name of the finite element method was given recently, the concept dates back

for several centuries. For example, ancient mathematicians found the circumference of a

circle by approximating it by the perimeter of a polygon as shown in figure 1-3. In terms of

the present-day notation, each side of the polygon can be called a “finite element”. By

considering the approximating polygon inscribed or circumscribed, one can obtain a lower

bound S(l) or an upper bound S(u) for the true circumference S. Furthermore, as the number of

sides of the polygon is increased, the approximate values converge to the true value. These

characteristics, as will be seen later, will hold true in any general finite element application.

Figure 1-3

To find the differential equation of a surface of minimum area bounded by a specified closedcurve, Schellback discretized the surface into several triangles and used a finite difference

expression to find the total discretized area in 1851. In the current finite element method, a

differential equation is solved by replacing it by a set of algebraic equations. In 1943, Courant

presented a method of determining the torsional rigidity of a hollow shaft by dividing the

cross section into several triangles and using a linear variation of the stress function ϕ over

each triangle in terms of the values of ϕ at net points (called nodes in the present day finite

element terminology). This work is considered by some to be the origin of the present-day

finite element method. Since mid-1950s, engineers in aircraft industry have worked on

developing approximate methods for the prediction of stresses induced in aircraft wings. In

1956, Turner, Cough, Martin, and Topp presented a method for modeling the wing skin usingthree-node triangles. At about the same time, Argyris and Kelsey presented several papers

outlining matrix procedures, which contained some of the finite element ideas, for the

solution of structural analysis problems.

The name finite element was coined, for the first time, by Clough in 1960. Although the finite

element method was originally developed mostly based on intuition and physical argument,

the method was recognized as a form of the classical Rayleigh-Ritz method in the early 1960s.

Once the mathematical basis of the method was recognized, the developments of new finite

elements for different types of problems and the popularity of the method started to grow

almost exponentially. The digital computer provided a rapid means of performing the many

calculations involved in the finite element analysis and made the method practically viable.

Along with the development of high-speed digital computers, the application of the finite




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element method also progressed at a very impressive rate. Zienkiewicz and Cheung presented

the broad interpretation of the method and its applicability to any general field problem.

With this broad interpretation of the finite element method, it has been found that the finite

element equations can also be derived by using a weighted residual method such as Galerkin

method or the least squares approach. This led to widespread interest among applied

mathematicians in applying the finite element method for the solution of linear and nonlinear

differential equations. It is to be noted that traditionally, mathematicians developed

techniques such as matrix theory and solution methods for differential equations, and

engineers used those methods to solve engineering analysis problems. Only in the case of

finite element method, engineers developed and perfected the technique and applied

mathematicians use the method for the solution of complex ordinary and partial differential

equations. Today, it has become an industry standard to solve practical engineering problems

using the finite element method. Millions of degrees of freedom (dof) are being used in the

solution of some important practical problems.

A brief history of the beginning of the finite element method was presented by Gupta and

Meek. The rapid progress of the finite element method can be seen by noting that, annuallyabout 3800 papers were being published with a total of about 56,000 papers and 380 books

and 400 conference proceedings published as estimated in 1995. With all the progress, today

the finite element method is considered one of the well-established and convenient analysis

tools by engineers and applied scientists.

Engineering Applications of the Finite Element MethodAs stated earlier, the finite element method was developed originally for the analysis of

aircraft structures. However, the general nature of its theory makes it applicable to a wide

variety of boundary value problems in engineering. A boundary value problem is one in which

a solution is sought in the domain (or region) of a body subject to the satisfaction of

prescribed boundary (edge) conditions on the dependent variables or their derivatives. Table

1-1 gives specific applications of the finite element in the three major categories of boundary

value problems, namely (1) equilibrium or steady-state or time-independent problems, (2)

eigenvalue problems, and (3) propagation or transient problems.

In an equilibrium problem, we need to find the steady-state displacement or stress

distribution if it is a solid mechanics problem, temperature or heat flux distribution if it is a

heat transfer problem, and pressure or velocity distribution if it is a fluid mechanics problem.

In eigenvalue problems also, time will not appear explicitly. They may be considered as

extensions of equilibrium problems in which critical values of certain parameters are to be

determined in addition to the corresponding steady-state configurations. In these problems,we need to find the natural frequencies or buckling loads and mode shapes if it is a solid

mechanics or structures problem, stability of laminar flows if it is a fluid mechanics problem,

and resonance characteristics if it is an electrical circuit problem.

The propagation or transient problems are time-dependent problems. This type of problem

arises, for example, whenever we are interested in finding the response of a body under time-

varying force in the area of a solid mechanics and under sudden heating or cooling in the field

of heat transfer.




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Commercial Finite Element Program PackagesThe general applicability of the finite element method makes it a powerful and versatile toolfor a wide range of problems. Hence, a number of computer program packages have been

developed for the solution of a variety of structural and solid mechanics problems. Some of

the programs have been developed in such a general manner that the same program can be

used for the solution of problems belonging to different branches of engineering with little or

no modification.

Many of these packages represent large programs that can be used for solving real complex

problems. For example, the NASTRAN (National Aeronautics and Space Administration

Structural Analysis) program package contains approximately 150,000 FORTRAN statements

and can be used to analyze physical problems of practically any size, including a complete

aircraft, an automobile, and a space shuttle.




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The availability of supercomputers has made a strong impact on the finite element

technology. In order to realize the full potential of these supercomputers in finite element

computation, special parallel numerical algorithms, programming strategies, and

programming languages are being developed. The use of personal computers and

workstations in engineering analysis and design is becoming increasingly popular as the price

of hardware is decreasing dramatically. Many finite element programs, especially suitable for

the personal computer and workstation environment, have been developed. Among the main

advantages are a user-friendly environment and inexpensive graphics.

Solutions Using Finite Element SoftwareThree Steps of Finite Element Solution

The solution of any engineering analysis problem using commercial FEA software involves the

following three steps:

Preprocessing:In this step, the geometry, material properties, loads (actions) and boundary conditions are

given as input data. In-built automatic mesh generation modules develop the finite element

mesh with minimal input from the analyst on the type of elements and mesh density to be

used. The analyst can display the data as well as the finite element mesh generated for visual

inspection and verification for correctness.

Numerical analysis:

The software automatically generates the element characteristics (stiffness) matrices and

characteristic (load) vectors, assembles them to generate the system equations, implements

the specified boundary conditions and solves the equations to find the nodal values of the

field variable (displacements) and computes the element resultants (stresses and strains).

Post processing:

The solution of the problem, such as nodal displacements and element stresses, can be

displayed either numerically in tabular form or graphically (two- or three-dimensional plots

of deformed shape or stress variation. The analyst can choose the mode of display for the

results.

Checking the Results of FEA

It is extremely important to check the results given by the FEA software. Usually a simplerversion of the actual problem is to be solved using the software so that the results can be

compared with known solutions (obtained by other methods such as a simplified analysis

technique). In addition, the analyst must ensure that the results agree with engineering

intuition and behavior. Also, one needs to verify whether the solution satisfies the specified

boundary and symmetry conditions. If necessary, the problem needs to be solved by changing

the boundary conditions, loads or materials to find whether the resulting FEA solutions

behave as per engineering intuition and expectations.

Basic Element Shapes

In most engineering problems, we need to find the values of a field variable such asdisplacement, stress, temperature, pressure, and velocity as a function of spatial coordinates




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(x, y, z). In the case of transient or unsteady-state problems, the field variable has to be found

as a function of not only the spatial coordinates (x, y, z) but also time (t). The geometry

(domain or solution region) of the problem is often irregular. The first step of the finite

element analysis involves the discretization of the irregular domain into smaller and regular

subdomains, known as finite elements. This is equivalent to replacing the domain having an

infinite number of degrees of freedom (dof) by a system having a finite number of dof.

A variety of methods can be used to model a domain with finite elements. Different methods

of dividing the domain into finite elements involve varying amounts of computational time

and often lead to different approximations to the solution of the physical problem. The

process of discretization is essentially an exercise of engineering judgment. Efficient methods

of finite element idealization require some experience and knowledge of simple guidelines.

For large problems involving complex geometries, finite element idealization based on

manual procedures requires considerable effort and time on the part of the analyst. Some

automatic mesh generation programs have been developed for the efficient idealization of

complex domains requiring minimal interface with the analyst.

The shapes, sizes, number, and configurations of the elements have to be chosen carefullysuch that the original body or domain is simulated as closely as possible without increasing

the computational effort needed for the solution. Mostly the choice of the type of element is

dictated by the geometry of the body and the number of independent coordinates necessary

to describe the system. If the geometry, material properties, and the field variable of the

problem can be described in terms of a single spatial coordinate, we can use the one-

dimensional or line elements shown in Figure 1-4(a). The temperature distribution in a rod

(or fin), the pressure distribution in a pipe flow, and the deformation of a bar under axial load,

for example, can be determined using these elements. Although these elements have a cross-

sectional area, they are generally shown schematically as a line element (Figure 1-4(b)).In

some cases, the cross-sectional area of the element may be non-uniform.

Figure 1-4

For a simple analysis, one-dimensional elements are assumed to have two nodes, one at each

end, with the corresponding value of the field variable chosen as the unknown (degree of

freedom). However, for the analysis of beams, the values of the field variable (transverse

displacement) and its derivative (slope) are chosen as the unknowns (dof) at each node as

shown in Figure 1-4(c).




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When the configuration and other details of the problem can be described in terms of two

independent spatial coordinates, we can use the two-dimensional elements shown in Figure

1-5. The basic element useful for two-dimensional analysis is the triangular element. Although

a quadrilateral element (or its special forms, the rectangle and parallelogram) can be obtained

by assembling two or four triangular elements, as shown in Figure 1-6, in some cases the use

of quadrilateral (or rectangle or parallelogram) elements proves to be advantageous. For the

bending analysis of plates, multiple dof (transverse displacement and its derivatives) are used

at each node.

Figure 1-5

Figure 1-6

If the geometry, material properties, and other parameters of the body can be described by

three independent spatial coordinates, we can idealize the body by using the three-dimensional elements shown in Figure 1-7. The basic three-dimensional element, analogous




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to the triangular element in the case of two-dimensional problems, is the tetrahedron

element. In some cases the hexahedron element, which can be obtained by assembling five

tetrahedrons as indicated in Figure 1-7, can be used advantageously. Some problems, which

are actually three-dimensional, can be described by only one or two independent

coordinates. Such problems can be idealized by using an axisymmetric or ring type of

elements shown in Figure 1-8. The problems that possess axial symmetry, such as pistons,

storage tanks, valves, rocket nozzles, and reentry vehicle heat shields, fall into this category.

Figure 1-7

Figure 1-8

For the discretization of problems involving curved geometries, finite elements with curved

sides are useful. Typical elements having curved boundaries are shown in Figure 1-9. Theability to model curved boundaries has been made possible by the addition of mid-side nodes.

Finite elements with straight sides are known as linear elements, whereas those with curved

sides are called higher-order elements.




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Figure 1-9

Discretization ProcessVarious considerations to be taken in the discretization process are discussed in the following

sections.

Type of Elements

Often, the type of elements to be used will be evident from the physical problem. For

example, if the problem involves the analysis of a truss structure under a given set of load

conditions (Figure 1-10(a)), the type of elements to be used for idealization is obviously the

“bar or line elements” as shown in Figure 1-10(b). Similarly, in the case of stress analysis of

the short beam shown in Figure 1-11(a), the finite element idealization can be done using

three-dimensional solid elements as shown in Figure 1-11(b). However, the type of elements

to be used for idealization may not be apparent, and in such cases one has to choose the type

of elements judicially. As an example, consider the problem of analysis of the thin-walled shell

shown in Figure 1-12(a). In this case, the shell can be idealized by several types of elements

as shown in Figure 1-12(b). Here, the number of dof needed, the expected accuracy, the ease

with which the necessary equations can be derived, and the degree to which the physical

structure can be modeled without approximation will dictate the choice of the element type

to be used for idealization. In certain problems, the given body cannot be represented as an

assemblage of only one type of elements. In such cases, we may have to use two or more

types of elements for idealization. An example of this would be the analysis of an aircraft

wing. Since the wing consists of top and bottom covers, stiffening webs, and flanges, three

types of elements—namely, triangular plate elements (for covers), rectangular shear panels

(for webs), and frame elements (for flanges)—have been used in the idealization shown in

Figure 1-13.




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Figure 1-10

Figure 1-11

Figure 1-12




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Figure 1-13

Size of Elements

The size of elements influences the convergence of the solution directly, and hence it has to

be chosen with care. If the size of the elements is small, the final solution is expected to be

more accurate. However, we have to remember that the use of smaller-sized elements will

also mean more computation time. Sometimes, we may have to use elements of differentsizes in the same body. For example, in the case of stress analysis of the box beam shown in

Figure 1-14(a), the size of all the elements can be approximately the same, as shown in Figure

1-14(b). However, in the case of stress analysis of a plate with a hole shown in Figure 1-15(a),

elements of different sizes have to be used, as shown in Figure 1-15(b). The size of elements

has to be very small near the hole (where stress concentration is expected) compared to

distant places. In general, whenever steep gradients of the field variable are expected, we

have to use a finer mesh in those regions. Another characteristic related to the size of

elements that affects the finite element solution is the aspect ratio of the elements. The

aspect ratio describes the shape of the element in the assemblage of elements. For two-

dimensional elements, the aspect ratio is taken as the ratio of the largest dimension of theelement to the smallest dimension. Elements with an aspect ratio of nearly unity generally

yield best results.




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Figure 1-14

Figure 1-15

Location of Nodes

If the body has no abrupt changes in geometry, material properties, and external conditions

(e.g., load and temperature), the body can be divided into equal subdivisions and hence thespacing of the nodes can be uniform. On the other hand, if there are any discontinuities in the

problem, nodes have to be introduced at these discontinuities, as shown in Figure 1-16.

Figure 1-16




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Number of Elements

The number of elements to be chosen for idealization is related to the accuracy desired, size

of elements, and the number of dof involved. Although an increase in the number of elements

generally means more accurate results, for any given problem, there will be a certain number

of elements beyond which the accuracy cannot be significantly improved. This behavior is

shown graphically in Figure 1-17. Moreover, since the use of a large number of elements

involves a large number of dof, we may not be able to store the resulting matrices in the

available computer memory.

Figure 1-17

Simplifications Afforded by the Physical Configuration of the Body

If the configuration of the body as well as the external conditions are symmetric, we may

consider only half of the body for finite element idealization. The symmetry conditions,

however, have to be incorporated in the solution procedure. This is illustrated in Figure 1-18,

where only half of the plate with a hole, having symmetry in both geometry and loading, isconsidered for analysis. 1 Since there cannot be a horizontal displacement along the line of

symmetry AA, the condition that u=0 has to be incorporated while finding the solution.




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Figure 1-18

Finite Representation of Infinite Bodies

In most of the problems, like in the analysis of beams, plates, and shells, the boundaries of

the body or continuum are clearly defined. Hence, the entire body can be considered for

element idealization. However, in some cases, as in the analysis of dams, foundations, and

semi-infinite bodies, the boundaries are not clearly defined. In the case of dams (Figure 1-19),since the geometry is uniform and the loading does not change in the length direction, a unit

slice of the dam can be considered for idealization and analyzed as a plane strain problem.

However, in the case of the foundation problem shown in Figure 1-20(a), we cannot idealize

the complete semi-infinite soil by finite elements. Fortunately, it is not really necessary to

idealize the infinite body. Since the effect of loading decreases gradually with increasing

distance from the point of loading, we can consider only that much of the continuum in which

the loading is expected to have a significant effect as shown in Figure 1-20(b). Once the

significant extent of the infinite body is identified as shown in Figure 1-20(b), the boundary

conditions for this finite body have to be incorporated in the solution. For example, if the

horizontal movement only has to be restrained for sides AB and CD (i.e., u=0), these sides aresupposed to be on rollers as shown in Figure 1-20(b). In this case, the bottom boundary can

be either completely fixed (u=v=0) or constrained only against vertical movement (v=0). The

fixed conditions (u=v=0alongBC) are often used if the lower boundary is taken at the known

location of a bedrock surface.




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Figure 1-19

Figure 1-20

NODE NUMBERING SCHEME

The finite element analysis of practical problems often leads to matrix equations in which the

matrices involved will be banded. The advances in the finite element analysis of large practical

systems have been made possible largely due to the banded nature of the matrices.

Furthermore, since most of the matrices involved (e.g., stiffness matrices) are symmetric, the

demands on the computer storage can be substantially reduced by storing only the elements

involved in half bandwidth instead of storing the entire matrix.

The bandwidth of the overall or global characteristic matrix depends on the node numbering

scheme and the number of dof considered per node. If we can minimize the bandwidth, the

storage requirements as well as solution time can also be minimized. Since the number of dof

per node is generally fixed for any given type of problem, the bandwidth can be minimized by

using a proper node numbering scheme. As an example, consider a three-bay frame with rigid

joints, 20 stories high, shown in Figure 1-21. Assuming that there are 3 dof per node, there

are 252 unknowns in the final equations (including the dof corresponding to the fixed nodes),

and if the entire stiffness matrix is stored in the computer, it will require 2522 =63504

locations. The bandwidth (strictly speaking, half-bandwidth) of the overall stiffness matrix can

be shown to be 15, and thus the storage required for the upper half-band is only 15×252=3780

locations.




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Figure 1-21

Before we attempt to minimize the bandwidth, we discuss the method of calculating the

bandwidth. For this, we consider again the rigid jointed frame shown in Figure 1.21. By

applying constraints to all the nodal dof except number 1 at node 1 (joint A), it is clear that an

imposed unit displacement in the direction of 1 will require constraining forces at the nodes

directly connected to node A—that is, B and C. These constraining forces are nothing but the

cross-stiffnesses appearing in the stiffness matrix, and these forces are confined to the nodes

B and C. Thus, the nonzero terms in the first row of the global stiffness matrix (Figure 1.22)will be confined to the first 15 positions. This defines the bandwidth (B)as the maximum

difference between the numbered dof at the ends of any member + 1.

This definition can be generalized so as to be applicable for any type of finite element as

Bandwidth (B)=(D+1)*f. Where D is the maximum largest difference in the node numbers

occurring for all elements of the assemblage, and f is the number of dof at each node.

Figure 1-21




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The previous equation indicates that D has to be minimized in order to minimize the

bandwidth. Thus, a shorter bandwidth can be obtained simply by numbering the nodes across

the shortest dimension of the body. This is clear from Figure 1-23 also, where the numbering

of nodes along the shorter dimension produces a bandwidth of B=15 (D=4), whereas the

numbering along the longer dimension produces a bandwidth of B=66 (D=21).

As observed previously, the bandwidth of the overall system matrix depends on the manner

in which the nodes are numbered. For simple systems or regions, it is easy to label the nodes

so as to minimize the bandwidth. But for large systems, the procedure becomes nearly

impossible. Hence, automatic mesh generation algorithms, capable of discretizing any

geometry into an efficient finite element mesh without user intervention, have been

developed. Most commercial finite element software has built-in automatic mesh generation

codes. An automatic mesh generation program generates the locations of the node points

and elements, labels the nodes and elements, and provides the element –node connectivity

relationships.

Figure 1-23

Automatic Mesh GenerationMesh generation is the process of dividing a physical domain into smaller subdomains (called

elements) to facilitate an approximate solution of the governing ordinary or partial

differential equation. For this, one-dimensional domains (straight or curved lines) are

subdivided into smaller line segments, two-dimensional domains (planes or surfaces) are

subdivided into triangle or quadrilateral shapes, and three-dimensional domains (volumes)

are subdivided into tetrahedron and hexahedron shapes. If the physical domain is simple and

the number of elements used is small, mesh generation can be done manually. However, most

practical problems, such as those encountered in aerospace, automobile, and construction

industries have complex geometries that require the use of thousands and sometimes

millions of elements. In such cases, the manual process of mesh generation is impossible and

we have to use automatic mesh generation schemes based on the use of a CAD or solid

modeling package.




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Automatic mesh generation involves the subdivision of a given domain, which maybe in the

form of a curve, surface, or solid (described by a CAD or solid modeling package) into a set of

nodes (or vertices) and elements (subdomains) to represent the domain as closely as possible

subject to the specified element shape and size restrictions. Many automatic mesh generation

schemes use a “bottom-up” approach in that nodes (or vertices or corners of the domain) are

meshed first, followed by curves (boundaries), then surfaces, and finally solids. Thus, for a

given geometric domain of the problem, nodes are first placed at the corner points of the

domain, and then nodes are distributed along the geometric curves that define the

boundaries. Next, the boundary nodes are used to develop nodes in the surface(s), and finally

the nodes on the various surfaces are used to develop nodes within the given volume (or

domain). The nodes or mesh points are used to define line elements if the domain is one-

dimensional, triangular, or quadrilateral elements if the domain is two-dimensional, and

tetrahedral or hexahedral elements if the domain is three-dimensional.

The automatic mesh generation schemes are usually tied to solid modeling and computer-

aided design schemes. When the user supplies information on the surfaces and volumes of

the material domains that make up the object or system, an automatic mesh generatorgenerates the nodes and elements in the object. The user can also specify minimum

permissible element sizes for different regions of the object. Many mesh generation schemes

first create all the nodes and then produce a mesh of triangles by connecting the nodes to

form triangles (in a plane region). In a particular scheme, known as Delaunay triangulation,

the triangular elements are generated by maximizing the sum of the smallest angles of the

triangles; thus the procedure avoids generation of thin elements.

The most common methods used in the development of automatic mesh generators are the

tesselation and octree methods. In the tesselation method, the user gives a collection of node

points and also an arbitrary starting node. The method then creates the first simplex element

using the neighboring nodes. Then a subsequent or neighboring element is generated byselectingthe node point that gives the least distorted element shape. The procedure is

continued until all the elements are generated. The step-by-step procedure involved in this

method is illustrated in Figure 1-24 for a two-dimensional example. Alternately, the user can

define the boundary of the object by a series of nodes. Then the tesselation method connects

selected boundary nodes to generate simplex elements. The stepwise procedure used 3 in

this approach is shown in Figure 1-25.

Figure 1-24




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Figure 1-25

The octree methods belong to a class of mesh generation schemes known as tree structure

methods, which are extensively used in solid modeling and computer graphics display

methods. In the octree method, the object is first considered enclosed in a three-dimensional

cube. If the object does not completely (uniformly) cover the cube, the cube is subdivided

into eight equal parts. In the two-dimensional analog of the octree method, known as the

quadtree method, the object is first considered enclosed in a square region. If the object does

not completely cover the square, the square is subdivided into four equal quadrants. If any

one of the resulting quadrants is full (completely occupied by the object) or empty (not

occupied by the object), then it is not subdivided further. On the other hand, if any one of the

resulting quadrants is partially full (partially occupied by the object), it is subdivided into four

quadrants. This procedure of subdividing partially full quadrants is continued until all the

resulting regions are either full or empty, or until some predetermined level of resolution is

achieved. At the final stage, the partially full quadrants are assumed to be either full or empty

arbitrarily based on a pre-specified criterion.

UnitsBefore starting to define any model, you need to decide which system of units you will use.

Abaqus has no built-in system of units. Do not include unit names or labels when entering

data in Abaqus. All input data must be specified in consistent units. Some common systems

of consistent units are shown in table 1-2.

Table 1-2




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Chapter 2

Geometric ModelingIn this chapter geometric modeling in the part module of the ABAQUS is investigated.

Example 1

Two and three dimensional wire parts




22

Example 2

Two dimensional shell parts

Part (1)

Part (2)

Part (3)




23

Part (4)

Part (5)




24

Part (6)

Part (7)




25

Example 3

Three dimensional shell parts




26

Example 4

Three dimensional solid parts

Part (1)

Part (2)

Part (3)




27

Part (4)

Part (5)

Part (6)




28

Part (7)

Part (8)

Part (9)




29

Part (10)

Part (11)

Part (12)




30

Part (13)

Part (14)

Part (15)




31

Part (16)

Part (17)

Part (18)




32

Chapter 3

Truss Structures

Example 1The structure is a simple, pin-jointed truss that is constrained at the left end and mounted on

rollers at the right end. The members can rotate freely at the joints. The frame is prevented

from moving out of plane. A simulation is required to de termine the structure’s static

deflection and the peak stress in its members when a 10 kN load is applied as shown in figure.

All members are circular steel rods 5 mm in diameter. Elastic properties are E=200 GPa and

v=0.29.




33

Example 2

E =10e4 ksi, A=2 in2, P4=P8= 100 kip




34

Example 3

E=2×10e7 N/cm2, A=2cm2 for all members

All dimensions in centimeters; All base nodes fixed




35

Example 4Finite element to be used: T3D2 = 3D two-node truss element

Dimensions in figure are in mm; Area of cross section of each bar: 3225.8 mm2

Material: Aluminum; E= 69 GPa

Load applied: Vertical load of 10,000 N at node 1




36

Example 5In this example the 52 bar space truss (dome structure) with configuration shown in the

following figures is considered. At each free node (1 –13) it is attached a non-structural mass

of 50 kg. The material is steel with Young’s modulus equal to 210 GPa, Poisson ration equal

to 0.29 and specific mass of 7800 kg/m3. The 52 bars are divided into eight groups, as shown

in the table.




37




38

Other Examples




39




40

Chapter 4

Two Dimensional Elasticity

Example 1

Hole in PlateA Rectangular plate with central circular hole is subjected to a uniformly distributed axial force

as shown. If the plate was made of carbon steel with 210 GPa Young modulus and 0.29 Poisson

ratio determine the maximum deflection and Von-Mises stress due to the specified loading.

The plate thickness is 0.002 and the resultant of the axial loading is 2 kN. Use any symmetry

in the model if it is appropriate. Also do a grid study analysis and determine the optimal mesh

size.

0

5

10

15

20

25

30

35

40

45

50

0 1000 2000 3000 4000 5000 6000 7000

M a x i m

u m S

t r e s s ( M p a )

Number of Nodes

S (Mpa) Q4

S (Mpa) T3




41

Example 2

L Shaped BracketAn L shaped plate of thickness 0.05 inch is subjected to a distributed force as shown. If the

plate was made of aluminum with 10.4 Mpsi Young modulus and 0.333 Poisson ratiodetermine the maximum deflection and Von-Mises stress due to the specified loading.




42

Example 3

Tensile BracketA bracket plate with an internal hole is loaded axially. The plate was made of carbon steel

with 210 GPa Young modulus and 0.29 Poisson ratio. Determine the maximum deflection andVon-Mises stress due to an axial tensile loading of 10 kN distributed along the right and left

edges.




43

Example 4

Connecting RodConsider the following connecting rod (All dimensions are in mm). Assume the left hole

constrained completely in all directions and 1 KN vertical force is distributed over the internalsurface of the right hole. If it was 10mm in thickness, determine the maximum Von-mises

stress and maximum displacement for the plane-stress case. The rod is made of ST37

structural steel with 200GPa Young modulus and 0.3 Poisson ratio.

The solution is as follows. All deformations are in mm and all stresses are in MPa.




44




45

Example 5

Curved SliderConsider the following slider (all dimensions are in inch). The left hole is constrained in all

directions and the right hole constrained only in y direction. Also assume that 250 lb force isdistributed in the semicircular arc at the upper end of the slit with an angle of 30 degree with

respect to the horizon. If the thickness of the plate was 1/8 inch and it was made of structural

steel with 30Gpsi Young modulus and 0.3 Poisson ratio determine the maximum Von-mises

stress and also the maximum deflection in the object.

The solution is as follows. All deformations are in inch and all stresses are in psi.




46




47

Example 6

Concentrated ForceConsider a 2D square object supported on a flat frictionless surface as shown in the following

figure. A concentrated force is applied at the midpoint of its upper edge. Use linear andquadratic quadrilateral elements to evaluate the maximum Von-Misses stress under the load.

Examine different mesh size and draw convergence curve. The plate is made from structural

steel with E=200GPa and v=0.3. Its thickness is 1mm and assume plane stress.

0

20

40

60

80

100

120

140

160

180

0 500 1000 1500 2000 2500 3000 3500 4000 4500

M a x i m u m S

t r e s s ( M p a )

Number of elements

Element Q4

Element Q8




48

Example 7

Sharp Corner

Consider a 2D object with sharp corners which is clamped at left and is under a uniform tensilestress at right as shown in the following figure. Use linear and quadratic quadrilateral

elements to evaluate the maximum Von-Misses stress in the object. Examine different mesh

size and draw convergence curve. The plate is made from structural steel with E=200GPa and

v=0.3. Its thickness is 1mm and assume plane stress.

Q4 Elements Q8 Elements




49

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

0 2000 4000 6000 8000

M a x i m u m S

t r e s s ( M

p a )

Number of elements

Element Q4

(Mpa)




50

Example 8

Rounded CornerConsider a 2D object with round corners which is clamped at left and is under a uniform

tensile stress at right as shown in the following figure. Use linear and quadratic quadrilateralelements to evaluate the maximum Von-Misses stress in the object. Examine different mesh

size and draw convergence curve. The plate is made from structural steel with E=200GPa and

v=0.3. Its thickness is 1mm and assume plane stress.

Q4 Elements Q8 Elements




51

1

1.05

1.1

1.15

1.2

1.25

1.3

0 2000 4000 6000 8000 10000

M a x i m u m S

t r e s s ( M p a )

Number of elements

Element Q4 (Mpa)

Element Q8 (Mpa)




52

Example 9

Solid Slab under Variable Load DistributionA long rectangular solid slab is clamped completely along two opposite sides as shown in the

figure. The slab is subjected to a transvers triangular distributed force with maximum of 100tonne/m for unit depth. The slab was made of concrete with 25 GPa Young modulus, 0.2

Poisson ratio and 2320 kg/m3 mass density. Determine the maximum deflection and

maximum normal stress in the slab due to the specified loading.




53

Example 10

Solid Slab under Variable Load Distribution




54

Example 11

Pipe Made of Two Different Materials




55

Chapter 5

Three Dimensional Elasticity

Example 1Cantilever Beam




56

Example 2

Cantilever Beam under Variable Loading




57

Example 3

L Shaped Bracket 1

Consider the following 3D bracket (all dimensions are in mm). Assume that the left end of theobject is clamped completely and a 1KN shearing force is distributed over the internal surface

of the hole and is pointing to the right. The bracket is made of structural steel with 200GPa

Young modulus and 0.3 Poisson ratio. In this circumstance, determine the maximum Von-

mises stress and maximum displacement produced in the object.

The solution is as follows. All deformations are in mm and all stresses are in MPa.




58




59

Example 4

L Shaped Bracket 2

Consider the following 3D bracket (all dimensions are in mm). Assume that the internalsurfaces of the two holes in the right leg are clamped completely and a 3KN shearing force is

distributed over the internal surface of the upper hole and is pointing to the right. The bracket

is made of structural steel with 200GPa Young modulus and 0.3 Poisson ratio. In this

circumstance, determine the maximum Von-Mises stress and maximum displacement

produced in the object.




60

Example 5

Pin Loading




61

Chapter 6

Heat Transfer

Example 1Conductive Heat Transfer in a Square

Three sides of a square plate are maintained at constant temperature of 0 °C and the fourth

one is kept at 100°C. The plate is made of carbon steel with average thermal conductivity of

52 W/mK. Under this conditions, determine the temperature distribution in the plate.

K=(53.2+50.7)/2=52 W/mK




62




63

0.00

10.00

20.00

30.00

40.00

50.00

60.00

70.00

80.00

90.00

100.00

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

Mesh 1

Mesh 2

Mesh 3

Mesh 4




64

Example 2

Three Dimensional Needle Fin

Consider a needle fin with circular cross section made of carbon steel which is under naturalconvection in the quiet air. The base of the fin is maintained at 100°C while the ambient

temperature is 25°C. Under this condition, determine the temperature distribution along the

fin.

The thermal conductivity of the fin can be approximated for 50°C as fillows.

k=(53.2+50.7)/2=52 W/mK

The coefficient of convective heat transfer can be approximated for horizontal cylinders as

the follows:25.0

25.0)179.079.1(

od

t t h W/m2K for laminar conditions

3/125.0 )32.016.2( t t h W/m2K for turbulent conditions




65

In this problem we consider 25

t , 7525100 t and 01.0o

d therefore one can obtain

9.12h

0

10

20

30

40

50

60

70

80

90

100

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2




66

Chapter 7

Thermal Stress

Example 1Thermal Stress in a Rod

A steel link, with no internal stresses, is pinned between two solid structures at a reference

temperature of 0°C (273 K). One of the solid structures is heated to a temperature of 75°C

(348 K). As heat is transferred from the solid structure into the link, the link will attempt to

expand. However, since it is pinned, this cannot occur, and as such, stress is created in the

link. A steady-state solution of the resulting stress will be found to simplify the analysis. Loads

will not be applied to the link, only a temperature change of 75°C. The link is steel with a

modulus of elasticity of 200 GPa, a thermal conductivity of 60.5 W/mK, and a thermal

expansion coefficient of 12e-6/K.




67

Example 2

Thermal Stress in a Square

Consider a square 2D elastic object which is under thermal stress condition. Three sides ofthe square are maintained in constant temperature of 0°C and the fourth one is kept in 100°C

as shown in the following figure. Under this condition, determine the temperature

distribution in the body and also obtain the maximum displacement and Von-Mises stress

induced in it for both cases of plane stress and plane strain. The square is made of carbon

steel with the following material properties.

Coefficient of thermal conductivity: 52 W/mK

Coefficient of thermal expansion: 10.8e-6 1/°C

Young modulus: 200e9 Pa

Poisson ratio: 0.3.




68

Plane stress case




69

Plane strain case




70

Example 3

Thermal Stress in a Pipe

Consider a circular cross section long pipe with internal diameter of 60mm and externaldiameter of 100mm. The internal surface of the pipe is maintained in 100 °C and its external

surface is kept in 0°C. Under this condition, determine the temperature distribution,

maximum displacement and Von-Mises stress induced in the pipe due to thermal stress. The

pipe is made of carbon steel with the following material properties.

Coefficient of thermal conductivity: 52 W/mK

Coefficient of thermal expansion: 10.8e-6 1/°C

Young modulus: 200e9 Pa

Poisson ratio: 0.3.




71




72

Chapter 8

Beam Structures

Example 1

Beam Bridge

Young’smodulus: E=70e3

Poisson’s ratio: v=0.0 it is not required for beams

THE FOLLOWING TABLE IS PRINTED FOR NODES BELONGING TO NODES




73

Example 2

Beam Bridge

The two-dimensional bridge structure is simply supported at its lower corners. The structureis composed of steel T-sections (E=210 GPa, ν=0.25) oriented as shown below. The detail of

the cross section is also shown. A uniform distributed load of 1000 N/m is applied to the lower

horizontal members in the vertical downward direction. Determine the stresses and the

vertical displacements.




74




75

Example 3

Table

The following table frame structure is supported by a frictionless surface under its legs. Thestructure is composed of steel box sections (E=210 GPa, ν=0.25). A uniform distributed load

of 1000 N/m is applied to the upper horizontal members in the vertical downward direction.

Determine the stresses and the displacements of the structure.




76




77

Example 4

Crane

The following structure is a part of a crane which is clamped at four left joints. The main beamsof the structure are composed of steel circular cross section pipes with 80mm outer diameter

and 4mm in thickness. The braces are also made of steel pipes of 60mm outer diameter and

3mm in thickness. This structure, In addition to the weights of its own members, supports a

concentrated force of 6KN at the mid span of the last member on the right. Under this

circumstance, determine the stresses and displacements of the structure. For steel, consider

E=210 GPa, ν=0.25 and mass density as 7800 kg/m3. The gravitational acceleration is

9.81m/s2.




78

Example 5

BridgeTo deal with a more realistic example, the following discussion considers the arch structure

shown in the figure, a simplified model of a pedestrian bridge with the arches in a basket-handle configuration. The span L=1200 in (30.5 m); the width W=96 in (2.44 m). Cross sections

are selected either from a list of 32 AISC HSS round sections or a list of 75 AISC HSS rectangular

tube sections. The lists are compiled by selecting the lightest section for each shape variety.

The model includes one geometric decision variable, the span-to-depth ratio of the arch (γ),

which ranges from 4 to 12, plus the five section variables indicated in Fig.5as follows:

Rib: The main arch rib, a round HSS section.

Brace: Members that connect the two arches near the crown,

a round HSS section.

Hanger: The suspenders which transfer load from the deck to

the arch, a round HSS section.Tie beam: The longitudinal beams at the deck level, a rectan-gular HSS section.

Transverse beam: The beams which span between the tie, a

rectangular HSS section.

Concerning structural modeling, the structure uses a pin sup-port for each arch at one end of

the bridge and roller supports at

the other. The hanger members are pin-ended, and the connections

between the arches and the tie beams are also pinned. Concerning

materials, the rectangular tube sections use a yield stress of 46 ksi

(317 MPa), and the round sections use a yield stress of 42 ksi

(290 MPa). Dead load includes the self weight of the model plus

a superimposed dead load on the deck area of 0:08 kip=ft

2

(3.83 kPa). The live load is 0:85 k=ft

2

(4.07 kPa) distributed on

the deck area. Superimposed dead and live loads are applied to

the nodes of each transverse beam, at the beam ends, and a midspan

node according to tributary area. The vertical deflection of the

nodes of the tie beam are limited toL=1;000 for live load only. The

analysis includes four load combinations: two to check stiffness

and two to check strength and stability. The combinations for stiff-ness include one with full

live load and one with live load on half

the span; these combinations use linear analysis. The combinations

for strength and stability include one with dead plus live and an-other with dead plus live load

on half the span; these combinations

are factored according to the AISC LRFD code (AISC 2001) and

use large displacement analysis. Note that these load combinations

are unrealistically simple, in particular because they do not account

for lateral loads. Concerning stability criteria, the algorithm checks

each member to account for member-level compression stability

according to the AISC LRFD requirements discussed previously

by using an effective length factorK¼1:0 and an unsupported




79

length equal to the member length. System-level stability was

considered as described in the discussion of stability constraints




80

Chapter 9

Shell and Plate Structures

Example 1

Rectangular PlatePlate is fixed at one edge and supported by rollers at the opposite edge. Other two edges are

free (no support). A concentrated transverse force of 100 N applied at center. Material:

E=10e3 N/mm2, Poisson ratio = 0.3




81

Example 2

L Shaped Beam StructureA frame structure which is composed of two I beams is shown in the following. The

dimensions of the cross section are also shown. Assume the structure is completely fixed atpoint A and a vertical force of 1KN is applied at the end at B. It is also needed to consider the

weight of the own structure. The structure is made of structural steel with 200GPa Young

modulus, 0.3 Poisson ratio and 7800kg/m3 mass density. In these conditions determine stress

distribution and displacements of the structure.




82




83

Example 3

Pressure Vessel

A pressure vessel is composed of a cylindrical body with hemispherical caps. The diameter ofthe vessel is 1m and its total length is 2m. Two pipes of 0.2m diameter are attached to the

central part of the vessel with relative angle of 90 degrees. Each pipe is 0.3m in length and

the distance between centerlines is 0.6m. The connection point of the pipes and the vessel is

filleted with radius of 0.05m. The thickness of the vessel and pipes is 0.005m and are made of

structural steel (E=200GPa, v=0.3). Assume a constant pressure of 2atm is exerted on the

internal surfaces and the free ends of the pipes are clamped completely. In these conditions

determine the stress and displacement distribution in the vessel.




84




85

Example 4

PipeE=200GPa

Niu=0.3Thichness=0.01 m

Internal Pressure=1e6 Pa

BC: onlt the translational DOFs of the edges of the bolt holes are restrained




86




87




88

Chapter 10

Composite Structures

Composite is a macroscopic mixture of at least two materials. One of the materials is thematrix in which the other materials called reinforcements are embedded.

The following figure shows a schematic of an Airbus 380 airplane (the largest airplane in the

world as of 2008). This airplane has more than 50% of its structure made of composite

materials. These components include the flaps, ailerons, rudder, radome etc. Most of these

components are flat in shape and they are usually made using hand-lay-up (HLU) and

autoclave molding techniques. The next figure shows a schematic of the hand-lay-up

fabrication technique and a representative lay-up sequence. Autoclave molding is a well-

established method for composites used in the aero-space industry with certified resins and

fibers. A photograph of an auto-clave is also shown.




89




90

The following figure shows a pressure vessel made of composite materials using the

combination of hand-lay-up and filament winding processes. Composite pressure vessels are

light weight and can contain pressures higher than those contained by metallic vessels. These

components are made using the filament winding process.

The following figure shows a photograph of a filament winding machine.




91

Figure 1.3(a) shows a component made using pultrusion. Pultrusion is used to make many

structures for civil engineering applications. Figure 1.3(b) shows the schematic of the

pultrusion process, and Figure 1.3(c) shows a photograph of a lab scale pultrusion machine.




92




93

Figure 1.4(a) shows a composite component made using the liquid composite molding (LCM)

method (5 piece). LCM has been used to make automobile composite components. Figure

1.4(b) shows a schematic of the liquid composite molding process.

Depending on the purpose of the analysis, different modeling techniques for composites can

be used:

Microscopic modeling Matrix and reinforcements are separately modeled as deformable continua.

Each element is composed of a single homogeneous material.

Layered modeling

Each element is composed of several layers of different materials.

Smeared modeling

The composite is modeled as an equivalent homogeneous material with stacked or single

layer element configuration




94




95

Example 1

Glass/Epoxy

E1=38.6 GPa

E2=8.27 GPa

G12=4.14 GPa

Viu12=0.26

Layer thickness =0.002 m

Layup: [90, 45, 0, 45, 90]




96

Chapter 11

Free Vibration

Example 1




97

Chapter 12

Linear Buckling

Example 1




98

Chapter 13

Contact Stress

Example 1




99

Chapter 14

Plastic Deformation

Example 1




100

Chapter 15

Flow in Porous Media

Example 1




101

Chapter 16

Flow of Viscose Fluids

Example 1




102

Appendix 1

Material Properties




103




104




105




106




107

Appendix 2

Stress Concentration Factors




108

Appendix 3

Meshing Techniques

Top-down meshingTop-down meshing relies on the geometry of a part to define the outer bounds of the mesh.

The top-down mesh matches the geometry; you may need to simplify and/or partition

complex geometry so that Abaqus/CAE recognizes basic shapes that it can use to generate a

high-quality mesh. In some cases top-down methods may not allow you to mesh portions of

a complex part with the desired type of elements. The top-down techniques—structured,

swept, and free meshing—and their geometry requirements are well-defined, and loads and

boundary conditions applied to a part are associated automatically with the resulting mesh.

Structured meshing

Structured meshing is the top-down technique that gives you the most control over yourmesh because it applies preestablished mesh patterns to particular model topologies. Most

unpartitioned solid models are too complex to be meshed using preestablished mesh

patterns. However, you can often partition complex models into simple regions with

topologies for which structured meshing patterns exist. Figure 17 –3 shows an example of a

structured mesh.

Swept meshing

Abaqus/CAE creates swept meshes by internally generating the mesh on an edge or face and

then sweeping that mesh along a sweep path. The result can be either a two-dimensional

mesh created from an edge or a three-dimensional mesh created from a face. Like structuredmeshing, swept meshing is a top-down technique limited to models with specific topologies

and geometries. Figure 17 –4 shows an example of a swept mesh.




109

Free meshing

The free meshing technique is the most flexible top-down meshing technique. It uses no

preestablished mesh patterns and can be applied to almost any model shape. However, free

meshing provides you with the least control over the mesh since there is no way to predict

the mesh pattern. Figure 17 –5 shows an example of a free mesh.

Bottom-up meshing

Bottom-up meshing uses the part geometry as a guideline for the outer bounds of the mesh,

but the mesh is not required to conform to the geometry. Removing this restriction gives you

greater control over the mesh and allows you to create a hexahedral or hexahedral-dominated mesh on geometry that is too complex for the structured or swept meshing

techniques. Bottom-up meshing can be applied to any solid model shape. It provides you with

the most control over the mesh, since you select the method and the parameters that drive

the mesh. However, you must also decide whether the resulting mesh is a suitable

approximation of the geometry. If it is not, you can delete the mesh and try a different

bottom-up meshing method or partition the region and mesh the resulting smaller regions

with either bottom-up or top-down meshing techniques.

To mesh a single bottom-up region, you may have to apply several successive bottom-up

meshes. For example, you may use an extruded bottom-up mesh to generate part of a region,




110

then use the element faces of the extruded mesh as a starting point to generate a swept mesh

for features that the extruded mesh did not include.

Loads and boundary conditions are applied to geometry. Unlike a top-down mesh, a bottom-

up mesh may not be fully associated with geometry. Therefore, you should check that the

mesh is correctly associated with the geometry in areas where loads or boundary conditions

are applied. Proper mesh-geometry association will ensure that the loads and boundary

conditions are correctly transferred to the mesh during the analysis. (For more information,

see “Mesh-geometry association,” Section 17.11.4.) Because of the extra effort required by

the user to create a satisfactory mesh compared to the automated top-down meshing

processes, bottom-up meshing is recommended for use only when top-down meshing cannot

generate a suitable mesh.

Figure 17 –6 shows an example of a bottom-up meshed part. Although this part is relatively

simple, it requires two regions and four bottom-up meshes to completely mesh the part.

Abaqus/CAE displays bottom-up meshed regions using a mixture of the region geometry color(light tan) and the mesh color (light blue) to emphasize that the geometry and mesh may not

be associated. Displaying both the geometry and the mesh allows you to view and edit the

mesh-geometry associativity.

What is structured meshing?

The structured meshing technique generates structured meshes using simple predefined

mesh topologies. Abaqus/CAE transforms the mesh of a regularly shaped region, such as a

square or a cube, onto the geometry of the region you want to mesh. You can apply the

structured meshing technique to simple two-dimensional regions (planar or curved) or to

simple three-dimensional regions that have been assigned the Hex or Hex-dominated

element shape option. For example, Figure 17 –42 illustrates how simple mesh patterns for

triangles, squares, and pentagons are applied to more complex shapes.




111

Three-dimensional structured meshingFigure 17 –49 illustrates examples of simple three-dimensional regions that can be meshed

using the structured meshing technique.

Meshing more complex regions with this technique may require manual partitioning. If you

do not partition a complex region, your only meshing option may be the free meshingtechnique with tetrahedral elements. Meshes constructed using the structured meshing

technique consist of hexahedral elements, which are preferred over tetrahedral elements.

You can eliminate holes (whether they pass all the way through the part instance or just part

way through) by partitioning their circumferences into halves, quarters, etc. For example, the

four partitions in Figure 17 –51 convert the part instance from one region with a hole to four

regions without holes.




112

You should limit arcs to 90° or less to avoid concavities along sides and at edges. For example,

the part instance in Figure 17 –52 has been partitioned so that the single region with 180° arcs

becomes two regions with 90° arcs.

All the faces of the region must have geometries that could be meshed using the two-

dimensional structured meshing technique. For example, without partitioning, the semicircles

at either end of the part in Figure 17 –53 have only two sides each. (A face must have at least

three sides to be meshed using the structured meshing technique.) If you partition the part

into two halves, each semicircle is divided into two faces with three sides each.

Exactly three edges of the region must meet at each vertex. For example, the vertex at the

top of an unpartitioned pyramid in Figure 17 –54 is connected to four edges. However, if you

partition the pyramid into two tetrahedral regions, the vertex is connected to only three

edges for each individual region.




113

What is swept meshing?

Abaqus/CAE uses swept meshing to mesh complex solid and surface regions. The swept

meshing technique involves two phases:

Abaqus/CAE creates a mesh on one side of the region, known as the source side.

Abaqus/CAE copies the nodes of that mesh, one element layer at a time, until the final side,

known as the target side, is reached. Abaqus/CAE copies the nodes along an edge, and this

edge is called the sweep path. The sweep path can be any type of edge —a straight edge, a

circular edge, or a spline. If the sweep path is a straight edge or a spline, the resulting mesh

is called an extruded swept mesh. If the sweep path is a circular edge, the resulting mesh is

called a revolved swept mesh.

For example, Figure 17 –65 shows an extruded swept mesh. To mesh this model, Abaqus/CAE

first creates a two-dimensional mesh on the source side of the model. Next, each of the nodesin the two-dimensional mesh is copied along a straight edge to every layer until the target

side is reached.

To determine if a region is swept meshable, Abaqus/CAE tests if the region can be replicatedby sweeping a source side along a sweep path to a target side. In general, Abaqus/CAE selects

the most complex side (for example, the side that has an isolated edge or vertex) to be the

source side. In some cases you can use the mesh controls to select the sweep path. If some

regions of a model are too complex to be swept meshed, Abaqus/CAE asks if you want to

remove these regions from your selection before it generates a swept mesh on the remaining

regions. You can use the free meshing technique to mesh the complex regions, or you can

partition the regions into simplified geometry that can be structured or swept meshed.

When you assign mesh controls to a region, Abaqus/CAE indicates the direction of the sweep

path and allows you to control the direction. If the region can be swept in more than onedirection, Abaqus/CAE may generate a very different two-dimensional mesh on the faces that




it can select as the source side. As a result, the direction of the sweep path can influence the

uniformity of the resulting three-dimensional swept mesh, as shown in Figure 17 –66.

abaqus lecture notes

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