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Furthermore the computer implementation becomes more transparent because the sequence of computer operations can be placed in close correspondence with the DSM steps.
Subsequent Chapters incorporate ingredients of the mathematical interpretation when it is felt convenient to do so. However, the exposition avoids excessive entanglement with the mathematical theory when it may obfuscate the physics. A historical outline of the evolution of Matrix Structural Analysis into the Finite Element Method is given in Appendix H, which provides appropriate references.
In Chapters 2 through 6 the time is frozen at about , and the DSM presented as an aerospace engineer of that time would have understood it. This is not done for sentimental reasons, although that happens to be the year in which the writer began his thesis work on FEM under Ray Clough.
Elements based on equilibrium, mixed and hybrid variational formulations. Flexibility and mixed solution methods of solution. Kirchhoff-based plate and shell elements. Continuum-based plate and shell elements. The behavior of the total system is that of the individual elements plus their interaction. A key factor in the initial acceptance of the FEM was that the element interaction can be physically interpreted and understood in terms that were eminently familiar to structural engineers.
The unknown function or functions is locally approximated over each element by an interpolation formula expressed in terms of values taken by the function s , and possibly their derivatives, at a set of node points generally located on the element boundaries.
The states of the assumed unknown function s determined by unit node values are called shape functions. The trial function space may be inserted into the governing equations and the unknown node values determined by the Ritz method if the solution extremizes a variational principle or by the Galerkin, least-squares or other weighted-residual minimization methods if the problem cannot be expressed in a standard variational form.
The concept of element-by-element breakdown and assembly, while convenient in the computer implementation, is not theoretically necessary. The mathematical interpretation permits a general approach to the questions of convergence, error bounds, trial and shape function requirements, etc.
It also facilitates the application of FEM to classes of problems that are not so readily amenable to physical visualization as structures; for example electromagnetics and thermal conduction. These three methods appeared as ad-hoc computational devices created by engineers and physicists to deal with problems posed by new science and technology electricity, quantum mechanics, and delta-wing aircraft, respectively with little help from the mathematical establishment.
Only some time after the success of the new techniques became apparent were new branches of mathematics operational calculus, distribution theory and piecewise-approximation theory, respectively constructed to justify that success. That versatility makes those structures a good choice to illustrate the transition from the hand-calculation methods taught in undergraduate courses. An actual plane truss structure. That shown is typical of a roof truss used in residential building construction.
The idealized model of the example truss as a pin-jointed assemblage of bars is shown in Figure 2. It has three members and three joints. They can also be analyzed by the computer-oriented FEM.
The example plane truss structure. Hence the engineer turns to the pin-jointed assemblage of axial force elements and uses it to carry out the structural analysis. The axially-carrying members and frictionless pins of this structure are only an approximation of a real truss.
But with the example truss we can go over the basic DSM steps without getting mired into too many members. Both E and A are assumed to be constant along each member. Experience has shown. It should be noted that as a practical structure the example truss is not particularly useful — the one depicted in Figure 2. The member lengths are denoted by L 1. Pin-jointed idealization of example truss: These members connect joints 1—2.
Consequently members will carry some bending as well as direct axial loading. Members are generically identified by index e because of their close relation to finite elements.
The term state variables is used more often in nonlinear analysis. Joints are generically identified by indices such as i. The set of all joint forces can be arranged as a 6-component column vector: Classical structural mechanics tells us that the displacements of the truss are completely defined by the displacements of the joints.
The dual nomenclature is used in the initial Chapters to stress the physical interpretation of the FEM. In a idealized pin-jointed truss. The values of u x and u y at joint i will be called u xi and u yi and.
In the general FEM. Other names for it in the literature are structure coordinate system and overall coordinate system. The components at joint i will be denoted as f xi and f yi. This data will tell us which components of f and u are true unknowns and which ones are known a priori. The x and y displacement components will be denoted by u x and u y.
All member axial forces can be characterized by the x and y components of these forces. The key ingredients of the stiffness method of analysis are the forces and displacements at the joints. They are also called the degrees of freedom or state variables of the system.
In structural analysis procedures of the pre-computer era such information was used immediately by the analyst to discard unnecessary variables and thus reduce the amount of bookkeeping that had to be carried along by hand. This statement is a particular case of the more general finite element theory. Then f is simply the i th column of K.
This may seem strange. Furthermore it will be assumed that if all displacements vanish. For instance if in 2. Such effects are produced by actions such as temperature changes or lack-of-fit fabrication.
The following interpretation of the entries of K is highly valuable for visualization and checking. And indeed there is a member level or element level. Because the assumed behavior of the truss is linear. The entries of the stiffness matrix are often called stiffness coefficients and have a physical interpretation discussed below. Choose a displacement vector u such that all components are zero except the i th one. The axes origin is arbitrary and may be placed at the member midpoint or at one of the end joints for convenience.
Localization Next. These are collectively called breakdown steps and are described below. These systems are called local coordinate systems or member-attached coordinate systems.
See Figure 2. As shown in Figure 2. In structural mechanics the property just noted is called interpretation of stiffness coefficients as displacement influence coefficients.
Disconnection To carry out the first step of the DSM we proceed to disconnect or disassemble the structure into its components. The member properties include the length L. Breakdown of example truss into individual members 1. This step is illustrated in Figure 2. Thus K y1x2. The two end joints are called i and j. In the general finite element method they receive the name element coordinate systems. Think of the truss member in Figure 2. E and A. The most straightforward technique relies on the Mechanics of Materials approach covered in undergraduate courses.
Beer and E. Mechanics of Materials. Chapter 2 of F. Combining 2. When these relations are interpreted from the standpoint of the FEM.
If the member properties are uniform along its length. Two other methods for obtaining the local force-displacement relation 2. In the following Chapter we will complete the main DSM steps by putting the truss back together and solving for the unknown forces and displacements. This is a general result. Or in words: Use the displacement interpolation E2.
Can you interpret this result physically?
Interpretation hint: Assembly and Solution 3—1. Hand Assembly by Augmentation and Merge. Recovery of Reaction Forces. Applying Displacement BCs by Reduction. Chapter 3: Solving for Displacements. Assembly Rules. Recovery of Internal Forces and Stresses. Coordinate Transformations. This process also embodies two substeps: Upon deriving the stiffness relations at the element level in terms of the local coordinate system. On some equation solvers. This process is called assembly.
On the computer these steps are done concurrently. The solution step completes the DSM proper. Coordinate Transformations Before describing the globalization step. The necessary transformations are easily obtained by inspection of Figure 3. Postprocessing steps may follow. Then the modified equations are submitted to a linear equation solver. Next comes the solution.. Assembly involves two substeps: After all members are processed we have the free-free master stiffness equations.
The expressions 3. The general proof relies on the Principle of Virtual Work. Furthermore in the general case T is not even a square matrix. But this property does not extend to more general elements. This relation is not accidental and can be proved to hold generally. A comparison of 3. However this invariance was only checked by explicit computation for a truss member in Exercise 2.
The counterpart of the transposition relation is the adjointness property. However the congruential transformation relations 3. Globalization From now on we reintroduce the member index.
K e is called a member stiffness matrix in global coordinates. The process is technically called merging of individual members. The merge operation can be physically interpreted as reconnecting that member in the process of fabricating the complete structure. If the angle is zero we recover 2. This is mathematically governed by two rules of structural mechanics: The proof of 3. The globalized member stiffness matrices for the example truss can now be easily obtained by inserting appropriate values into 3.
For a truss structure. For member 1. Force equilibrium: The sum of forces exerted by all members that meet at a joint balances the external force applied to that joint. This translates into 2 3 2 3 the two component equations: Application of the foregoing rules at this particular joint gives Rule 1: Here f3 is the known external joint force applied on the joint.
Hand Assembly by Augmentation and Merge To directly visualize how the two assembly rules translate to member merging rules. These forces would act in the directions shown if both members 2 and 3 were in tension. Notational conventions to this effect are explained in Figure 3. Compatibility of displacements: The joint displacements of all mem- bers meeting at a joint are the same. Joint forces f 2 3 3 and f3 are applied by the joint on the members. The second one can be visualized by considering a joint as a free body.
But this augmentation enables us to write the matrix relation: The force equilibrium of joint 3 of the example truss. The first rule is physically obvious: To prepare the equations for an equation solver we need to separate known and unknown components of f and u.
In this Section a technique suitable for hand computation is described. It is virtually foolproof for hand computations.
The physical interpretation of singularity is that there are still unsuppressed rigid body motions: This explanation of the assembly process is conceptually the easiest to follow and understand. The mathematical interpretation of this behavior is that rows and columns of K are linear combinations of each other see Remark 3.
Solving for Displacements Solving the reduced system by hand for example. From Figure 2. Subsequent Chapters discuss how more general constraint forms.
To apply 3. This solution vector is expanded to six components by including the specified values 3—9. This space is spanned by three independent rigid body motions: The coefficient matrix of this system is no longer singular. Recovery of Reaction Forces Premultiplying the complete displacement solution 3. In trusses the only internal forces are the axial member forces. Two such steps are described below. It is easy to check that the complete force system is in equilibrium. Such quantities are said to be derived because they are recovered from the displacement solution.
Recovery of Internal Forces and Stresses Frequently the structural engineer is not primarily interested in displacements but in internal forces and stresses. Extract the displacements of member e from the displacement solution u to form u e.
The recovery of derived quantities is part of the so-called postprocessing steps of the DSM. The axial force p e in member e can be obtained as follows. These are in fact the most important quantities for preliminary design.
But often the analyst needs information on other mechanical quantities. Compute the deformation d relative displacement and recover the axial force from the equivalent spring constitutive relation: Furthermore we get three reaction forces: These forces are denoted by p 1.
There are two major differences: I Member stiffness matrices are not expanded. For simplicity we shall assume here that K is stored as a full square matrix. An alternative interpretation of 3. This is more in tune with the Theory of Elasticity viewpoint discussed in Exercise 2.
Signs shown for these forces correspond to tension. Difference II is a more advanced topic that is deferred to the last part of the book. For the example truss the freedom-pointer technique expresses the entries of K as the sum. The internal forces in the example truss are the axial forces p 1. II The master stiffness matrix K is stored using a special format that takes advantage of symmetry and sparseness.
Exercise 3. Note that 3. Entries K pq that do not get any contributions from the right hand side remain zero. Advantage may be taken of the symmetry of K e and K to roughly halve the number of additions. The ordering of the last two is irrelevant. It is sufficient to mark the equations that correspond to displacement BCs. Reduction is convenient for hand computations because it cuts down on the number of equations to solve. To apply support conditions without rearranging the equations we clear set to zero rows and columns corresponding to prescribed zero displacements as well as the corresponding force components.
But it has a serious flaw for computer implementation: Rearrangement can be as or more expensive than solving the equations.
Solving this modified system yields the complete displacement solution 3. The resulting system is called the modified set of master stiffness equations. It was previously noted that on the computer the number of equations is not the only important consideration. The solver is then programmed to skip those equations. Truss structure for Exercise 3. Then check by hand that using that formula you get 3. Verify that this system cannot be solved for the joint displacements u y2.
Partial answer: Rederive the reduced system 3. Verify that overall force equilibrium x forces. Partial result: Any method: Using the DSM carry out the following steps: Additional Topics 4—1. Worked-Out Example 2. Worked-Out Example 1. Thermomechanical Behavior. Application of DBCs by Reduction. Chapter 4: Thermomechanical Stiffness Equations.
These include: This Chapter covers some topics that were left out from Chapters 2—3 for clarity. Recall the master stiffness equations 3. Inserting the known data into 4. This happens. There are cases. Mathematically these are called non-homogenous boundary conditions.
These steps were illustrated with the hand analysis of a plane truss structure. The treatment of this generalization of the FEM equations is studied in the following subsections.
Columns 1, 2 and 4 are removed by transferring all known terms from the left to the right hand side: Note that its coefficient matrix is exactly the same as in the reduced system 3. The right hand side, however, is different. It consists of the applied joint forces modified by the effect of known nonzero displacements.
This is a consequence of the fact that the example truss is statically determinate. The force systems internal and external in such structures are insensitive to movements such as foundation settlements. As there, the main objective is to avoid rearranging the master stiffness equations. To understand the process it is useful to think of being done in two stages.
First equations 1, 2 and 4 are modified so that they become trivial equations, as illustrated for the example truss and the support conditions 4.
The solution of this system recovers 4. In the next stage, columns 1, 2 and 4 of the coefficient matrix are cleared by transferring all known terms to the right hand side, following the same procedure explained in 4.
As before, this is called the modified master stiffness system. Observe that the equations retain the original order. Solving this system yields the complete displacement solution 4. Note that if all prescribed displacements are zero, forces on the right hand side are not modified, and one would recover 3.
First the applied forces in the right-hand side are modified for the effect of nonzero prescribed displacements, and the prescribed displacements stored in the reaction-force slots. This is called the force modification procedure. Second, rows and columns of the stiffness matrix are cleared as appropriate and ones stored in the diagonal positions.
This is called the stiffness modification procedure. It is essential that the procedures be executed in the indicated order, because stiffness terms must be used to modify forces before they are cleared. The reduction and modification techniques for applying DBCs can be presented in compact matrix form. In this matrix equation, subvectors u2 and f1 collect displacement and force components, respectively, that are known, given or prescribed. On the other hand, subvectors u1 and f2 collect force and displacement components, respectively, that are unknown.
The force components in f2 are reactions on supports; consequently f2 is called the reaction force vector. This is the reduced master equation system. If the displacement B. Examples that illustrate 4. The computer-oriented modification technique retains the same joint displacement vector as in 4. This system is often denoted as.
For the computer implemenmtation it is important to note that the partitioned form 4.
The equations are not explicitly rearranged and retain their original numbers. The example shows that u1 and u2 are generally interspersed throughout u. Thus, matrix operations such as K12 u2 required indirect pointer addressing to avoid explicit array rearrangements.
If u vanishes, so does f. This behavior does not apply, however, if there are initial force effects. A common source of initial force effects are temperature changes. Imagine that a plane truss structure is unloaded that is, not subjected to external forces and is held at a uniform reference temperature.
External displacements are measured from this environment, which is technically called a reference state. Now suppose that the temperature of some members changes with respect to the reference temperature while the applied external forces remain zero. Because the length of members changes on account of thermal expansion or contraction, the joints will displace. If the structure is statically indeterminate those displacements will induce strains and stresses and thus internal forces.
These names reflect what is viewed as the physical source of initial force effects at the continuum level. The treatment results in an initial force vector that has to be added to the applied mechanical forces. The member is prismatic and uniform.
Therefore we begin with the derivation of the matrix stiffness equations of a generic truss member. It is assumed that the disconnection and localization steps of the DSM have been carried out.
This is conventionally chosen to be the temperature throughout the structure at which the displacements. For many structures. Generic truss member subjected to mechanical and thermal effects: For linear structural models all such sources may be algebraically treated in the same way as thermal effects. There are other physical sources of initial force effects. The temperature T is also uniform. Thermomechanical Behavior Consider the generic plane-truss member shown in Figure 4.
We introduce the concept of reference temperature Tr e f. In this Section we go over the analysis of a plane truss structure whose members undergo temperature changes from a reference state. For clarity the member identification subscript will be omitted in the following development until the transformation-assembly steps.
If the member is disassembled or disconnected. It may. This thermal stress is further discussed in Remark 4.
Equilibrium of truss member under thermomechanical forces. This may be positive or negative. This coefficient will be assumed to be uniform over the generic member. It is physically obvious for an unconstrained member such as that depicted in Figure 4. In response to this stress the length changes by d M. In structures such as buildings and bridges Tr e f is often taken to be the mean temperature during the construction period. At the other extreme.
Those zero displacements. This motivates the use of expansion joints in pavements. It is an instance of an initial force vector at the element level.
This stress can cause buckling or cracking in severely heated structural members that are not allowed to expand or contract. Passing to matrix form: It follows that fT contains the negated joint forces internal forces that develop in a heated or cooled bar if joint motions are precluded.
The last relation in 4. Displacement boundary conditions can be applied by reduction or modification of these equations. Effective forces by themselves are of little use in design. Globalization At this point we restore the member superscript so that the member stiffness equations 4. The only difference is that the effective joint force vector f contains a superposition of mechanical and thermal forces. Processing the reduced or modified system by a linear equation solver yields the displacement solution u.
Solution The master system 4. Two truss members are connected in series as shown and fixed at the ends. Find the stress in the members. Structure for worked-out Example 1. The member lengths are 4 and 6. On removing the first and third equations. Worked-Out Example 1 The first problem is defined in Figure 4.
Member 1 is in tension and member 2 in compression. To recover mechanical internal forces in member e. The thermal force vectors. To reduce clutter note that all y motions are suppressed so only the x freedoms are kept: The thermal forces for each member in global coordinates are obtained by using 4.
The truss is mecanically unloaded. Worked-Out Example 2 The second example concerns the example truss of Chapters Since the prescribed displacements are zero. Completing u with the prescribed zero displacements and premultiplying by K gives the complete effective force vector: This nice property extends to the general Finite Element Method.
Suppose that a prestress force FP is present in a bar. This is a consequence of the example frame being statically determinate. As can be seen there is a wide variety of physical effects.
The following example indicate that this is done for a bar element. The sum of the two: Local effects that lead to initial forces at the member level are: A physical interpretation of 4. Equate to a thermal elongation: Engineers also call these prestresses. The good news is that once the member equations 4. Since in linear analysis all such effects can be treated as initial forces. But all of them can handle temperature variation inputs. These translate into internal forces and stresses.
This is input to the program as a fictitious temperature change. If this device is used. The trick is not necessary for personal or open-source codes over which you have full control. The disconnected member was supposed to have length L. Prove that statically determinate truss structures are free of thermal stresses. Use a reduction scheme to apply the displacement BCs.
Everything else is the same. Programming Remarks. Testing the Member Stiffness Module. Class Demo Scripts.
Module Description. Case Sensitivity. Chapter 5: Programming Style and Prerequisites. Why Mathematica? Good documentation and abundance of application books at all levels.
Availability on a wide range of platforms that range from PCs and Macs through Unix workstations. But there is an excellent tutorial available: