Introduction to Finite Elements in Engineering, 3rd Ed, echecs16.infoupatla - Ebook download as PDF File .pdf) or read book online. Introduction to Finite Elements in Engineering, 3rd Ed, echecs16.infoupatla_2 - Ebook download as PDF File .pdf), Text File .txt) or read book online. Introduction to. Introduction to Finite. Elements in Engineering. F O U R T H E D I T I O N. TIRUPATHI R. CHANDRUPATLA. Rowan University. Glassboro, New Jersey. ASHOK.

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The Passive Voice The third present tense form uses a conjugation of to do ( do, does) with the verb.. Vera was su Introduction to Finite Element. Introduction-to-Finite-Elements-in-Engineering-3rd-Ed-T-R-chandrupatla. sergiu Baetu. Loading Preview. Sorry, preview is currently unavailable. You can. Download Introduction to Finite Elements in Engineering By Tirupathi R. Chandrupatla, Ashok D. Belegundu – Introduction to Finite Engineering is ideal for.

The development of finite element theory is combined with examples and exercises involving engineering applications. The steps used in the development of the theory are implemented in complete, self-contained computer programs. While the strategy and philosophy of the previous editions has been ret This book provides an integrated approach to finite element methodologies. While the strategy and philosophy of the previous editions has been retained, the Third Edition has been updated and improved to include new material on additional topics. Chapter topics cover fundamental concepts, matrix algebra and gaussian elimination, one-dimensional problems, trusses, two-dimensional problems using constant strain triangles, axisymmetric solids subjected to axisymmetric loading, two-dimensional isoparametric elements and numerical integration, beams and frames, three-dimensional problems in stress analysis, scalar field problems, dynamic considerations, and preprocessing and postprocessing. For practicing engineers as a valuable learning resource.

Algebraic equation sets that arise in the steady state problems are solved using numerical linear algebra methods, while ordinary differential equation sets that arise in the transient problems are solved by numerical integration using standard techniques such as Euler's method or the Runge-Kutta method. In step 2 above, a global system of equations is generated from the element equations through a transformation of coordinates from the subdomains' local nodes to the domain's global nodes.

This spatial transformation includes appropriate orientation adjustments as applied in relation to the reference coordinate system. The process is often carried out by FEM software using coordinate data generated from the subdomains.

FEA as applied in engineering is a computational tool for performing engineering analysis. It includes the use of mesh generation techniques for dividing a complex problem into small elements, as well as the use of software program coded with FEM algorithm. In applying FEA, the complex problem is usually a physical system with the underlying physics such as the Euler-Bernoulli beam equation , the heat equation , or the Navier-Stokes equations expressed in either PDE or integral equations , while the divided small elements of the complex problem represent different areas in the physical system.

FEA is a good choice for analyzing problems over complicated domains like cars and oil pipelines , when the domain changes as during a solid state reaction with a moving boundary , when the desired precision varies over the entire domain, or when the solution lacks smoothness.

FEA simulations provide a valuable resource as they remove multiple instances of creation and testing of hard prototypes for various high fidelity situations. Another example would be in numerical weather prediction , where it is more important to have accurate predictions over developing highly nonlinear phenomena such as tropical cyclones in the atmosphere, or eddies in the ocean rather than relatively calm areas.

Colours indicate that the analyst has set material properties for each zone, in this case a conducting wire coil in orange; a ferromagnetic component perhaps iron in light blue; and air in grey. Although the geometry may seem simple, it would be very challenging to calculate the magnetic field for this setup without FEM software, using equations alone. FEM solution to the problem at left, involving a cylindrically shaped magnetic shield.

The ferromagnetic cylindrical part is shielding the area inside the cylinder by diverting the magnetic field created by the coil rectangular area on the right. The color represents the amplitude of the magnetic flux density , as indicated by the scale in the inset legend, red being high amplitude.

The area inside the cylinder is low amplitude dark blue, with widely spaced lines of magnetic flux , which suggests that the shield is performing as it was designed to.

History[ edit ] While it is difficult to quote a date of the invention of the finite element method, the method originated from the need to solve complex elasticity and structural analysis problems in civil and aeronautical engineering. Its development can be traced back to the work by A. Hrennikoff [4] and R. For each extract the element displacement vector q from the Q vector, uSing element connectiVity, and determine the element stresses.

Step 3. Evaluate the reaction force at each support from R" -C O,. L Section 3.

The accuracy of the solution. Choice of C. Let us expand the first equation in Eq. Note that is a load applied at the support if any , and that FtfC is general- ly of small magnitude. The reader may wish to choose a sample problem and experiment with this using, say, lOS or 10 6 to check whether the reaction forces differ by much.

Using the penalty approach for handling boundary conditions, do the following: When using the penalty approach, therefore, a large num- ber C is added to the first aDd third diagonal elements of K. Choosing C based on Eq, 3. Determine the displacement field, stress, and support reactions in the body.

To do this, assume that the wall does not exist. From this result, we see that contact does occur.

The problem has to be re-solved, since the boundary conditions are now different The displacement 74 Chapter 3 One-Dimensional Problems at B' is specified to be 1. Consider the two-element finite element model in Fig. Also, the number C x 1. Such boundary conditions are referred to as multipoint constraints in the literature. The penalty approach will now be shown to apply to this type of boundary condition.

These modifications are given as 3. A nonphysical argument is used here to arrive at the modified X tential energy in Eq. Multipoint constraints are the most eral types of boundary conditions, from which other types can be treated as special cases.

A rigid bar of negligible mass, pinned at one end, is supported by a steel rod and an aluminum rod. What are the boundary conditions for your model? Solve the equations for Q. Then determine element stresses. Next, mUltipoint constraints given in part a are considered. For the first constraint,Qt - 0. The addition to the stiffness matrix is obtained from Eqs. Similarly, the consideration of the second multipoint constraint Q2 - 0. Thus, the results become. Double-precision arithmetic on the computer is rec- ommended when there are several multipoint constraints.

In some problems, however, use of quadratic interpolation leads to far more accurate results. In this section, quadratic shape functions will be introduced, and the corresponding element stiffness matrix and load vectors will be derived. The reader should note that the basic procedure is the same as that used in the linear one-dimensional element earlier. Consider a typical three-node quadratic element, as shown in Fig. In the local numbering scheme, the left node will be numbeted 1, the right node 2, and the midpoint 3.

Node 3 has been introduced for the purposes of passing a quadratic fit and is called an internal node. Similarly, N2 equals umty at node 2 and equals zero at the other two nodes: N, equals unity al node 3 and equals zero at nodes 1 and 2. F Th. These shape functions are called Lagrange shape functions. Now the displacement field within the element is written in terms of the nodal displacements as 3.

Thus, u in Eq. Using Eqs. Recall that when using lin- ear shape functions, the strain and stress came out to be constant within the element. We now have expressions for u, E, and j in Eqs. Again, in the finite element model considered here, it will be assumed that crOSS- sectional area A e , body force F, and traction force T are constant within the element. QTKQ - QTF, where the structural stiffness matrix K and nodal load vector F are assembled from element stiff- ness matrices and load vectors, respectively.

EDlllple 3. Determine the axial stress distribution in the rod. Consider only the centrifugal force. Ignore bending of the rod. The model has a total of five degrees of freedom. The element stiffness matrices are from Eq.

Global dol -8 16 2 , [7 kl "'" 10 x 0. Thking average values of f over each element, we have f - 0. The element connectivity table is as follows: I Global Node Nos. X - 2i L - x The exact stress distribution based on this equation is also shown in Fig. That is, the thennal stress problem will be considered. Note that a positive I: T implies a rise in temperature. The stress--strain law in the presence of! From this figure, we see that the stress-strain relation is given by IT E.

The second term yields the desired element load vector a f , as a result of the temperature change: T is the average change in temperature within the element. The temperature load vector in Eq.

The tem- perature IS then raised 10 60 0 e. The element temperature forces due to. Since dofs 1 and 3 are fixed. Consider the bar in Fig. Q2S in. Find the bandwidth NBW for the one-dimensional model whose nodes are numbered as shown in Fig. A finite element solution using one-dimensional.

Consider a finite element with shape functions NIC, and N2 t used to interpolate the displacement field within the element Fig. It is desired to attach a spring to node 22 of a structure modeled using as shown in Fig.

Consider the 1-D model of the structure shown in Fig. With the help of a sketch. What is the strain energy in the structure? Determine the nodal displacements, element stresses, and support reactions. Solve this problem by hand calculation, adopting the elim. Solve by hand calculation. Problems 91 3. Determine the stress in each material. You may name the nodes for both the elements.

Determine the DOdai displacements, element stresses, and support reactions. Complete Example 3. Plot the stress distributions on Fig. Show all work such as element matrices, assembly, boundary condItions, and solutIOn. Compare finite element and Rayleigh-Ritz solutions by providing plots of u x vs.

Consider the multipoint constraint 3Qp - Qq: Indicate what modifications need to be made to the banded stiffness ma- trix S to implement this constraint.

Also, if the bandwidth of the structure is n 1 , what will be the new bandwidth when the constraint is introduced? The rigid beam in Fig. Find the stress in each vertical member. The boundary condition is of the mUltipoint constraint type. IS t1g tened. This problem reinforces the fact that once the shape functions are assumed. Certain arbitrary shape functions are given. Consider the one.. That is, develop the B matrix. You need not evaluate the integrals.

Derive the element stiffness matrix II; for the one.. Introduce the linearity of width for part a and diameter for part b using the shape functions used for the displacement interpolation. For plotting and extrapolation purposes see Chapter 12 , it is sometimes necessary to obtain nodal stress values from element stress values that are obtained from a computer run. Specifically,consider the element stresses, TI, Uz, and U3.

Obtain Si from the least-squares criterion. Plot the distribution of stress from the nodal values. Determine the stresses in the 4 in.

For the vertical rod shown in Fig. Use E Introduce weight contribution to the nodal loads into the program and solve using two elements and four elements. Comment on the stress distribution. For Fig. Then plot number of elements vs. The structure in Fig. Deter- mine the displacements, stresses, and support reactions.

Solve this problem by hand cal- culation, using the eIimination method for handling boundary conditions.

As String. Enabled - True crndStart. IzzputBcur "Izzput F,tl. Input il, Title Line Input '1, Dummy: Dr N Next I , Df.. M Line Input tl, Dummy For I.. NOC N, 1: PM N3, AREA N.. EL I X2I , r. AL S Nl, SIN, HPC I, 1: S 12, Abs I2 - II.. NU I React I.. File d: Visible - False txtView. Visible a True txtView. Two- dimensional trusses or plane trusses are treated in Section 4. In Section 4. A typical plane truss is shown in Fig.

A truss structure consists only of two-force members. That is, every truss element is in direct tension or compression Fig. In a truss, it is required that all loads and reactions are applied only at the joints and that all members are connected togeth- er at their ends by frictionless pin joints. Every engineering student has, in a course on statics.

These methods. Further,joint displacements are not readily obtainable. The finite element method on the other hand is applicable to stati- cally detenninate or indeterminate structures alike. The finite element method also pr0- vides joint deflections. Effects of temperature changes and support settlements can also be routinely bandied. Q, Q, Q. The steps involved are discussed here.

Local and Global Coordinate Systems The main difference between the one-dimensional structures considered in Chapter 3 and trusses is that the elements of a truss have various orientations. To account for these different orientations, local and global coordinate systems are introduced as follows: A typical plane-truss element is shown in local and global coordinate systems in Fig. In the local numbering scheme, the two nodes of the element are numbered 1 and 2.

The local coordinate system consists of the x' -axis, which runs along the element from node 1 toward node 2. All quantities in the local coordinate system will be denot- ed by a prime '. The global system is fixed and does not depend on the orientation of the element.

Note that x,y, and z form a right -handed coordinate sys- tem with the z-axis coming straight out of the paper. A systematic numbering scheme is adopt- ed here: A node whose global node number is j has associated with it dofs 2j - 1 and 2j. Further, the global displacements associated with node j are Q2i-l and Q2i' as shown in Fig. Let qi and q2 be the displacements of nodes 1 and 2, respectively, in the local coordinate system.

Thus, the element displacement vector in the local coordinate system is denoted by 4. Thus, 4. Similarly, 4.

These direction cosines are the cosines of the angles that the local x'-axis makes with the globaIx-,y-axes, respectively. Equations 4.

Referring to Fig. We then have 2. Element Stiffness Matrix An important observation will now be made: The truss element is a one-dimensional element when viewed in the local coordinate system. Consequently, from Eq. The problem at hand is to develop an expression for the element stiffness matrix in the global coor- dinate system.

This is obtainable by considering the strain energy in the element. From the previous equa- tion, we obtam the element stiffness matrix in global coordinates as 4. Note that a positive stress im- plies that the element is in tension and a negative stress implies compression. Example 4. Complete the following: The nodal coordinate data are as follows: For example, the con- nectivity of element 2 can be defined as instead of as in the previous table.

How- ever, ca1culations of the direction cosines will be consistent with the adopted connectivity scheme. Using formulas in Eqs.

Now, using Eq. By adding the element stiffness contributions, noting the element connectivity, we get 1 2 3 4 5 6 7 8 The elimination approach discussed in Chapter 3 will be used here. The rows and columns corresponding to dofs 1,2,4,7, and 8, which corre- spond to fixed supports, are deleted from the K matrix. The reduced finite element equations are given as Q6 The connectIVIty 1 is 1 - 2. Section 4. We need to determine the reac- tion forces along dois 1, 2,4, 7,and 8, which correspond to fixed supports.

Thus, we have 0 T is the average change in tem- perature in the element. It may be noted that the initial strain lEo can also be induced by forcing members into places that are either too long or too short, due to fabrication errors. We will now express the load vector in Eq.

Since the potential energy associated with this load is the same in magnitude whether measured in the local or global roordinate systems. Once the displacements are obtained by solving the finite element equations, the stress in each truss element is obtained from see Eq.

There are no other loads on the structure. Determine the nodal displacements and element stresses as a result of this temperature increase. Use the elimination approach. Node 2 settles bv 0. Wnte down. Use the penalty approach. Onlv the oad vector needs to be assembl d d h. Usmg Eg. I ments 2 and 3 are. Using the elimination approach, we can delete all roW! The resulting finite element equations are UJ psi b Support 2 settles by 0.

In the penalty approach for handling boundary conditions Chapter 3 , recall that a large spring constant C is added to the diagonal elements in the struc- tural stiffness matrix at those dofs where the displacements are specified. Typically, C may be chosen 10 4 times the largest diagonal element of the unmodified stiffness matrix see Eq. Further, a force Ca is added to the force vector, where a is the specified displacement.

Conse- quently, the modified finite element equations are given by In the program TRUSS, that is provided, these equations are automatically generated and solved from the user's input data. The output from the program is. The local and global coo d' t f 3 D lruss.

Note that the local coord' I I. The nodal displacement vector in global coordinates is now Fig. The element stiffness matrix in global coordinates is given by Eq. Two methods, the banded approach and the sky- line approach, are discussed in Chapter 2.

In the banded approach, the elements of each element stiffness matrix k e are directly placed in a banded matrix S. In the skyline approach, the elements of k e are placed in a vector form with certain identification pointers.

The bookkeeping aspects of this assembly procedure for banded and skyline solution are discussed in the sections that follow. Assembly for Banded Solution The assembly of elements of Ie" into a banded global stiffness matrix S is now discussed for a two-dimensional truss element. Consider an element e whose connectivity is indi- cated as follows: Element 1 Global Node Nos. Consider a truss element e connected to, say, nodes 4 and 6. The degrees of freedom for the element are 7, 8, 11, and Thus, the entries in the global stiffness matrix for this element will be 1 Skyline Assembly As discussed in Chapter 2, the first step in skyline assembly involves the evaluation of the skyline height or the colunm height for each cation.

Consider the element e with the end nodes j andj shown in Fig. Then, starting with a tor of identifiers, ID, we look at the four degrees of freedom 2i - 1, 2i, 2j - 1, and 2j. This is precisely represented in the table given in Fig. At this stage all the skyline heights have been determined and placed in I - 1. The next step involves assembling the element stiffness values into the column vector A. The correspondence of the global locations of the square stiffness matrix com- ing from an element shown in Fig.

Other programs provided may be similarly modified for skyline solution instead of banded solution. Consider the truss element shown in Fig, P4. The x-,y-coordinates of the two nodes are indicated in the figure. A truss element, with local node numbers 1 and 2, is shown in Fig. For the truss in Fig.

For the two-bar truss shown in Fig. For the three-bar truss shown in Fig. For the two-dimensional truss configuration shown in Fig. S, determine the bandwidth for stiffness storage in a banded form. Choose an alternative numbering scheme and determine the corresponding bandwidth. Comment on the strategy that you use for decreasing the bandwidth. II, Problems 4.

A small railroad bridge is constructed of steel members, all of which have a cross-sectional area of mm 2. A train stops on the bridge.

Estimate how much the point R moves horizontally because of this loading. Also determine the nodal displacements and element stresses. Consider the truss in Fig. Cross-sectional areas in square inches are shown in parentheses. Consider symmetry and model only one-baH of the truss shoWD- Determine displacements and element stresses. Cross-sectional areas for each element are as follows: Element Area in. Member is mm long. Member was manufactured to be mm long instead of nun.

However, it was forced into place. Determine a the stresses in the members assuming that member was manufactured to its correct length of 5 1 mm and b the stresses in the members as a result of member being forced into place and the load P, of course.

Treat this as an initial strain problem and use the temperature load vector expression in the text. Expressions for the element stress Eq. Generalize these expressions for a three-dimensional truss element. Fmd deflections at nodes, stresses in members, and reactions at supports for the truss shown in Fig. Find the deflections at the nodes for the truss configuration shown in Fig.

I sides of the 3-D truss. If the members in the truss in Problem 4. If f7 c is the com- pressive stress in a member, then the factor of safety for buckling may be taken as Perl Ao: Introduce this into the computer program TRUSS2D to calculate the factors of safety 10 compression members and print them in the output file.

Identify the tetrahedral pat- terns in the truss. Chandrupatla ana A.

SEI3, 2. L SEI3, 4. SN" SN.. PMII3, SN -,.. SN Stress I.. Dr,l Next I End Sub '. The displacements, traction components, and distrib- uted body force values are functions of the position indicated by X, y. Section 5. Stiffness and load concepts are then developed using energy and Galerkin approaches.

Figure 5. The points where the comers of the triangles meet are ca1led nodes, and each triangle formed by three nodes and three sides is called an element. The ele- ments fill the entire region except a small region at the boundary.

For the triangulation shown in Fig. In the two-dimensional problem discussed here, each node is permitted to displace in the two directions x andy. Thus, each node has two degrees of freedom dofs. As seen from the numbering scheme used in trusses, the displacement components of node j are taken as Q2j-l in the x direction and Q2j in the y direction.

We denote the global dis- placement vector as 5. Computationally, the information on the triangulation is to be represented in the form of nodal coordinates and connectivity. The nodal coordinates are stored in a two- dimensional array represented by the total nwnber of nodes and the two coordinates per node. The connectivity may be clearly seen by isolating a typical element, as shown in Fig.

For the three nodes designated locally as 1,2, and 3, the corresponding globaJ node numbers are defined in Fig. This element connectivity information becomes an array of the size and number of elements and three nodes per element. A typical con- nectivity representation is shown in Table 5.

Most standard fmite element codes use the convention of going around the element in a counterclockwise direction to avoid cal- culating a negative area. However, in the program that accompanies this chapter, ordering is not necessary. Table 5. The displacement components of a local node j in Fig. We denote the element displacement vector as 5. Yt X2. Y2 and X3' Y3 have the global corre- spondence established through Thble 5.

The local representation of nodal coordinates and degrees of freedom provides a setting for a simple and clear representation of ele- ment characteristics. As discussed earlier, the finite element method uses the concept of shape functions in systematically developing these interpolations. For the constant strain triangle. The three shape functions Nt.

N 2 , and N3 corresponding to nodes I, 2, and 3, respectively. Shape function Nt is 1 at node 1 and linearly reduces to 0 at nodes 2 and 3. The values of shape function Nt thus define a plane surface shown shaded in Fig. Any linear combination of these shape functions also represents a plane surface. N 2 , and N3 are therefore not linearly independent; only two of these are indepen- dent. The independent shape functions are conveniently represented by the pair t.

At this stage, the similarity with the one- dimensional element Chapter 3 should be noted: The shape functions can be physicaUy represented by area coonIinates. A point x, y in a triangle divides it into three areas, AI. Isoparametric Representation The displacements inside the element are now written using the shape functions and the nodal values of the unknown displacement field.

We have or, using Eq. This is isoparametric representation. This ap- proach lends to simplicity of development and retains the uniformity with other com- plex elements. Equation 5. Example 5. Solution Using the isoparametric representation Eqs. From Eqs. On taking the derivative ofx andy, J [X" y,,] X23 Y. H the points 1,2, and 3 are ordered in a counter- clockwise manner,det J is positive in sign.

We have A IldetJI 5. Most computer codes use a counterclockwise order for the nodes and use det J for evaluating the area. He was on the faculty at GMI from through He received the Ph.

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