# The Basics of FEA Procedure

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1 CHAPTER 2 The Basics of FEA Procedure 2.1 Introduction This chapter discusses the spring element, especially for the purpose of introducing various concepts involved in use of the FEA technique. A spring element is not very useful in the analysis of real engineering structures; however, it represents a structure in an ideal form for an FEA analysis. Spring element doesn t require discretization (division into smaller elements) and follows the basic equation F = ku. We will use it solely for the purpose of developing an understanding of FEA concepts and procedure. 2.2 Overview Finite Element Analysis (FEA), also known as finite element method (FEM) is based on the concept that a structure can be simulated by the mechanical behavior of a spring in which the applied force is proportional to the displacement of the spring and the relationship F = ku is satisfied. In FEA, structures are modeled by a CAD program and represented by nodes and elements. The mechanical behavior of each of these elements is similar to a mechanical spring, obeying the equation, F = ku. Generally, a structure is divided into several hundred elements, generating a very large number of equations that can only be solved with the help of a computer. The term finite element stems from the procedure in which a structure is divided into small but finite size elements (as opposed to an infinite size, generally used in mathematical integration). The endpoints or corner points of the element are called nodes. Each element possesses its own geometric and elastic properties. Spring, Truss, and Beams elements, called line elements, are usually divided into small sections with nodes at each end. The cross-section shape doesn t affect the behavior of a line element; only the cross-sectional constants are relevant and used in calculations. Thus, a square or a circular cross-section of a truss member will yield exactly the same results as long as the cross-sectional area is the same. Plane and solid elements require more than two nodes and can have over 8 nodes for a 3 dimensional element. A line element has an exact theoretical solution, e.g., truss and beam elements are governed by their respective theories of deflection and the equations of deflection can be found in an engineering text or handbook. However, engineering structures that have stress concentration points e.g., structures with holes and other discontinuities do not have a theoretical solution, and the exact stress distribution can only be found by an

2 experimental method. However, the finite element method can provide an acceptable solution more efficiently. Problems of this type call for use of elements other than the line elements mentioned earlier, and the real power of the finite element is manifested. In order to develop an understanding of the FEA procedure, we will first deal with the spring element. In this chapter, spring structures will be used as building blocks for developing an understanding of the finite element analysis procedure. Both spring and truss elements give an easier modeling overview of the finite element analysis procedure, due to the fact that each spring and truss element, regardless of length, is an ideally sized element and does not need any further division. 2.3 Understanding Computer and FEA software interaction - Using the Spring Element as an example In the following example, a two-element structure is analyzed by finite element method. The analysis procedure presented here will be exactly the same as that used for a complex structural problem, except, in the following example, all calculations will be carried out by hand so that each step of the analysis can be clearly understood. All derivations and equations are written in a form, which can be handled by a computer, since all finite element analyses are done on a computer. The finite element equations are derived using Direct Equilibrium method. Example 2.1 Two springs are connected in series with spring constant k 1, and k 2 (lb./in) and a force F (lb.) is applied. Find the deflection at nodes 2, and 3. Solution: k 1 k 2 o o o F Figure 2.1 For finite element analysis of this structure, the following steps are necessary: Step 1: Derive the element equation for each spring element. Step 2: Assemble the element equations into a common equation, knows as the global or Master equation. Step 3: Solve the global equation for deflection at nodes 1 through 3. FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-2

3 Detailed description of these steps follows. Step 1: Derive the element equation for each spring element. First, a general equation is derived for an element e that can be used for any spring element and expressed in terms of its own forces, spring constant, and node deflections, as illustrated in figure 2.2. u i u j f i f j e Figure 2.2 Element e can be thought of as any element in the structure with nodes i and j, forces f i and f j, deflections u i and u j, and the spring constant k e. Node forces f i and f j are internal forces and are generated by the deflections u i and u j at nodes i and j, respectively. For a linear spring f = ku, and f i = k e (u j u i ) = -k e (u i -u j ) = - k e u i + k e u j For equilibrium, f j = -f i = k e (u i -u j ) = k e u i - k e u j Or -f i = k e u i - k e u j - f j = - k e u i + k e u j Writing these equations in a matrix form, we get f k k u i e e i = f j k u e k e j The above matrix equation is a general form of an equation of a spring elements, and can be used to derive element equations for any spring element in this example, and in general, it is valid for any linear spring element. Thus, equations for each elements can be written as follows: FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-3

4 Element 1: k 1 f 1 f f1 f 2 (1) k1 = k 1 k k 1 1 u u 1 2 Where, the superscript on the force matrix indicates the corresponding element. Element 2: k 2 f 2 f f 2 f3 (2) k2 = k 2 k k 2 2 u u 2 3 Thus, f (1) 1 = -k 1 (u 1 u 2 ) (1) f 2 = k 1 (u 1 -u 2 ) f (2) 2 = -k 2 (u 2 u 3 ) (2) f 3 = k 2 (u 2 -u 3 ) This completes the procedure for step 1. Note that f 3 = F (lb.). This will be substituted in step 2. The above equations represent individual elements only and not the entire structure. Step 2 : Assemble the element equations into a global equation. The basis for combining or assembling the element equation into a global equation is the equilibrium condition at each node. When the equilibrium condition is satisfied by summing all forces at each node, a set of linear equations is created which links each element force, spring constant, and deflections. In general, let the external forces at each node be F 1, F 2, and F 3, as shown in figure 2.3. Using the equilibrium equation, we can find the element equations, as follows. Node1: F = 0 = f + F 1 1 or F = f = k ( u u ) = k u k u F 1 f 1 Node 1 FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-4

5 Node 2: F = 0 = f 2 (1) + f 2 (2) + F 2 F 2 or F 2 = -f (1) 2 f (2) 2 = (1) = -k 1 (u 1 u 2 ) + k 2 (u 2 u 3 ) f 2 f 2 (2) = -k 1 u 1 + k 1 u 2 + k 2 u 2 k 2 u 3 Node 2 Node 3: F = 0, f 3 (2) + F 3 = 0 Or F 3 = -f 3 (2) = -k 2 (u 2 - u 3 ) F 3 Node 3 f 3 (2) The superscript e in force f n (e) indicates the contribution made by the element number e, and the subscript n indicates the node n at which forces are summed. Rewriting the equations, we get, k 1 u 1 k 1 u 2 = F1 - k 1 u 1 + k 1 u 2 + k 2 u 2 k 2 u 3 = F2 (2.1) - k 2 u 2 + k 2 u 3 = F3 These equations can now be written in a matrix form, giving k 1 - k 1 0 u 1 F 1 - k 1 k 1 + k 2 - k 2 u 2 = F k 2 k 2 u 3 F 3 This completes step 2 for assembling the element equations into a global equation. At this stage, some important conceptual points should be emphasized and will be discussed below Procedure for Assembling Element stiffness matrices The first term on the left hand side in the above equation represents the stiffness constant for the entire structure and can be thought of as an equivalent stiffness constant, given as FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-5

6 k 1 - k 1 0 [K eq ] = - k 1 k 1 + k 2 - k k 2 k 2 A single spring element with a value K eq will have an identical mechanical property as the structural stiffness in the above example. The assembled matrix equation represents the deflection equation of a structure without any constraints, and cannot be solved for deflections without modifying it to incorporate the boundary conditions. At this stage, the stiffness matrix is always symmetric with corresponding rows and columns interchangeable. The global equation was derived by applying equilibrium conditions at each node. In actual finite element analysis, this procedure is skipped and a much simpler procedure is used. The simpler procedure is based on the fact that the equilibrium condition at each node must always be satisfied, and in doing so, it leads to an orderly placement of individual element stiffness constant according to the node numbers of that element. The procedure involves numbering the rows and columns of each element, according to the node numbers of the elements, and then, placing the stiffness constant in its corresponding position in the global stiffness matrix. Following is an illustration of this procedure, applied to the example problem. Element 1: 1 2 k 1 K (1) = k 1 -k 1 1 -k 1 k Element 2: 2 3 k 2 K (2) = k 2 -k 2 2 -k 2 k Assembling it according with the above-described procedure, we get, k 1 -k 1 0 [ K g ] = 2 -k 1 k 1 + k 2 -k k 2 k 2 FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-6

7 Note that the first constant k 1 in row 1 and column 1 for element 1 occupies the row 1 and column 1 in the global matrix. Similarly, for element 2, the constant k 2 in row 2 and column 2 occupies exactly the same position (row 2 and column 2) in the global matrix, etc. In a large model, the node numbers can occur randomly, but the assembly procedure remains the same. It s important to place the row and column elements from an element into the global matrix at exactly the same position corresponding to the respective row and column Force matrix At this stage, the force matrix is represented in a general form, with unknown forces F 1, F 2, and F 3 F 1 F 2 F 3 Representing the external forces at nodes 1, 2, and 3, in general terms, and not in terms of the actual known value of the forces. In the example problem, F 1 = F 2 = 0 and F 3 = F. the actual force matrix is then 0 0 F Generally, the assembled structural matrix equation is written in short as {F}=[k]{u}, or simply, F = k u, with the understanding that each term is an m x n matrix where m is the number of rows and n is the number of columns. Step 3: Solve the global equation for deflections at nodes. There are two steps for obtaining the deflection values. In the first step, all the boundary conditions are applied, which will result in reducing the size of the global structural matrix. In the second step, a numerical matrix solution scheme is used to find deflection values by using a computer. Among the most popular numerical schemes are the Gauss elimination and the Gauss-Sedel iteration method. For further reading, refer to any numerical analysis book on this topic. In the following examples and chapters, all the matrix solutions will be limited to a hand calculation even though the actual matrix in a finite element solution will always use one of the two numerical solution schemes mentioned above. FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-7

8 2.3.3 Boundary conditions In the example problem, node 1 is fixed and therefore u 1 = 0. Without going into a mathematical proof, it can be stated that this condition is effected by deleting row 1 and column 1 of the structural matrix, thereby reducing the size of the matrix from 3 x 3 to 2 x 2. In general, any boundary condition is satisfied by deleting the rows and columns corresponding to the node that has zero deflection. In general, a node has six degrees of freedom (DOF), which include three translations and three rotations in x, y and z directions. In the example problem, there is only one degree of freedom at each node. The node deflects only along the axis of the spring. In this section, the finite element analysis procedure for a spring structure has been established. The following numerical example will utilize the derivation and concepts developed above. Example 2.2 In the given spring structure, k 1 = 20 lb./in., k 2 = 25 lb./in., k 3 = 30 lb./in., F = 5 lb. Determine deflection at all the nodes. K 1 k 2 P k 3 F o o o o Figure 2.3 Solution We would apply the three steps discussed earlier. Step 1: Derive the Element Equations As derived earlier, the stiffness matrix equations for an element e is, K (e) = k e -k e -k e k e Therefore, stiffness matrix of elements 1, 2, and 3 are, 1 2 Element 1: K (1) = Element 2: 2 3 K (2) = FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-8

9 Element 3: 3 4 K (3) = Step 2: Assemble element equations into a global equation Assembling the terms according to their row and column position, we get [K g ] = Or, by simplifying [K g ] = The global structural equation is, F u 1 F 2 = u 2 F u 3 F u 4 Step 3: Solve for deflections First, applying the boundary conditions u 1 =0, the first row and first column will drop out. Next, F 1 = F 2 = F 3 = 0, and F 4 = 5 lb. The final form of the equation becomes, u 2 0 = u u 4 FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-9

10 This is the final structural matrix with all the boundary conditions being applied. Since the size of the final matrices is small, deflections can be calculated by hand. It should be noted that in a real structure the size of a stiffness matrix is rather large and can only be solved with the help of a computer. Solving the above matrix equation by hand we get, 0 = 45 u 2 25 u 3 u = -25 u u 3 30 u 4 Or u 3 = u = -30 u u 4 Example 2.3 In the spring structure shown k 1 = 10 lb./in., k 2 = 15 lb./in., k 3 = 20 lb./in., P= 5 lb. Determine the deflection at nodes 2 and 3. Solution: k 1 k 2 k 3 o o o o Figure 2.4 Again apply the three steps outlined previously. Step 1: Find the Element Stiffness Equations Element 1: 1 2 [K (1) ] = Element 2: 2 3 [K (2) ] = Element 3: 3 4 [K (3) ] = FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-10

11 Step 2: Find the Global stiffness matrix = Now the global structural equation can be written as, F u 1 F 2 = u 2 F u 3 F u 4 Step 3: Solve for Deflections The known boundary conditions are: u 1 = u 4 = 0, F 3 = P = 3lb. Thus, rows and columns 1 and 4 will drop out, resulting in the following matrix equation, 0 = u u 3 Solving, we get u 2 = & u 3 = Example 2.4 In the spring structure shown, k 1 = 10 N/mm, k 2 = 15 N/mm, k 3 = 20 N/mm, k 4 = 25 N/mm, k 5 = 30 N/mm, k 6 = 35 N/mm. F2 = 100 N. Find the deflections in all springs. k 1 k 3 k 2 F 2 k 6 Fig. 2.5 k FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-11 k 5

12 Solution: Here again, we follow the three-step approach described earlier, without specifically mentioning at each step. Element 1: 1 4 [K (1) ] = Element 2: 1 2 [K (2) ] = Element 3: 2 3 [K (3) ] = Element 4: 2 3 [K (4) ] = Element 5: 2 4 [K (5) ] = Element 6: 3 4 [K (6) ] = The global stiffness matrix is, [K g ] = And simplifying, we get [K g ] = FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-12

13 And the structural equation is, F u 1 F 2 = u 2 F u 3 F u 4 Now, apply the boundary conditions, u 1 = u 4 = 0, F 2 = 100 N. This is carried out by deleting the rows 1 and 4, columns 1 and 4, and replacing F 2 by 100N. The final matrix equation is, u 2 0 = u 3 Which gives u 2 = u Deflections: Spring 1: u 4 u 1 = 0 Spring 2: u 2 u 1 = Spring 3: u 3 u 2 = Spring 4: u 3 u 2 = Spring 5: u 4 u 2 = Spring 6: u 4 u 3 = Boundary Conditions with Known Values Up to now we have considered problems that have known applied forces, and no known values of deflection. Now we will consider the procedure for applying the boundary conditions where, deflections on some nodes are known. Solutions of these problems are found by going through some additional steps. As discussed earlier, after obtaining the FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-13

14 structural global matrix equation, deflections are found by solving the equation by applying a numerical scheme in a computer solution. However, when there are known nodal values and unknown nodal forces, the method is not directly applicable. In this situation, the structural equation is first modified by incorporating all boundary conditions and then the final matrix equation is solved by a computer using a numerical method, as mentioned earlier. The following procedure traces the necessary steps for solving problems that involve known nodal values Procedure for incorporating the known Nodal Values in the Final Structural Equation There are two methods that are frequently used for applying boundary conditions to a structural matrix equation. In one method, the matrices are partitioned into two parts with known and unknown terms. In the second method, the known nodal values are applied directly in the structural matrix. Both methods can be used with equal effectiveness. The first method will not be discussed here. Details of the second method follow. Consider the following linear equations, k 11 u 1 + k 12 u 2 + k 13 u 3 + k 14 u 4 = F 1 (2.2) k 21 u 1 + k 22 u 2 + k 23 u 3 + k 24 u 4 = F 2 (2.3) k 31 u 1 + k 32 u 2 + k 33 u 3 + k 34 u 4 = F 3 (2.4) k41u 1 + k 42 u 2 + k 43 u 3 + k 44 u 4 = F 4 (2.5) These linear algebraic equations can be written in matrix form as follows. k 11 k 12 k 13 k 14 u 1 F 1 k 21 k 22 k 23 k 24 u 2 = F 2 k 31 k 32 k 33 k 34 u 3 F 3 k 41 k 42 k 43 k 44 u 4 F 4 Let the known nodal value at node 2 be u 2 = U 2 (a constant), then by the linear spring equation F 2 = k 22 U 2 Therefore, equation ( )) above can be reduced to k 22 u 2 = k 22 U 2 = F 2 and the matrix with this boundary condition can be written as k 11 k 12 k 13 k 14 u 1 F 1 0 k u 2 = F 2 k 31 k 32 k 33 k 34 u 3 F 3 k 41 k 42 k 43 k 44 u 4 F 4 FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-14

15 Now, equations 2.2, 2.4, 2.5 also contain the u 2 term and therefore these equations must also be modified. We can modify equation 1 by transferring the term k 12 u 2 to the right hand side and replacing u 2 by U 2. The modified equation can be written as K 11 u k 13 u 3 + k 14 u 4 = F 1 k 12 U 2 Similarly, equations 3 and 4 can be written as K 31 u k 33 u 3 + k 34 u 4 = F 3 k 32 U 2 K 41 u k 43 u 3 + k 44 u 4 = F 4 k 42 U 2 The final matrix equation is k 11 0 k 13 k 14 u 1 F 1 k 1 U 2 0 k u 2 = k 22 U 2 k 31 0 k 33 k 34 u 3 F 3 k 32 U 2 k 41 0 k 43 k 44 u 4 F 4 k 42 U 2 The dotted line indicates changes made in the enclosed terms. The final matrix remains symmetric and has the same size. The boundary conditions for forces can now be incorporated and a numerical solution scheme can be used to solve this equation. This procedure is summarized in the following simple, step-by-step approach. Given the known boundary conditions at node 2: u i = u 2 = U 2, follow these steps to incorporate the known nodal values. Note that, here, i = 2 and j = 1,2,3,4. Step 1: Set all terms in row 2 to zero, except the term in column 2 (kij = 0, k ii = k 22 0) Step 2: Replace F 2 with the term k 22 U 2 (F i = k ii u i ) Step 3: Subtract the value k i2 U2 from all the forces, except F 2 ( subtract k ji from the existing values of f j ), where i = 1, 3, and 4 Step 4: Set all the elements in column 2 to zero, except, row2 (all k ji = 0, k ii # 0) The above procedure now will be applied in the following example problem. FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-15

16 Example 2.5 In example problem 2.4 replace the force F by a nodal deflection of 1.5 mm on node 2 and rework the problem. Solution Rewriting the final structural matrix equation in example 2.4, we have F u 1 F 2 = u 2 F u 3 F u 4 Boundary condition are: u 1 = u 4 = 0, and u 2 = U 2 = 1.5mm. Applying the 4 steps described above in sequence, Step 1: Set all terms in row 2 to zero, except the term in column 2 (kij = 0, k ii = k 22 0) F u 1 F 2 = u 2 F u 3 F u 4 Step 2: Replace F 2 with the term k 22 U 2 = (90)(1.5) = 135, (F i = k ii u i ) F u = u 2 F u 3 F u 4 Step 3: Subtract the value k 22 U 2 from all the forces, except F 2 (subtract k ji from the existing values of f j ) F 1 F 1 (15)(1.5) = 22.5 Row 1: k j2 = k 12 = -15 F 3 F 3 (-45)(1.5) = 67.5 Row 2: k j2 = k 32 = -45 F 4 F 4 (-30)(1.5) = 45 Row 2: k j2 = k 42 = - 30 Note: F 1 = F 3 = F 4 = 0. FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-16

17 The new force equation now is, Step 4: Set all the elements in column 2 to zero, except, row2 (all k ji = 0, k ii 0) Or, k 12 = k 32 = k 42 = 0, and the new equation is, u = u u u 4 This is the final equation after the nodal value u 2 = 1.5 mm is incorporated into the structural equation. The same procedure can be followed for the boundary conditions u 1 = u 4 = 0. It can be stated that for zero nodal values, the procedure will always lead to elimination of rows and columns corresponding to these nodes, that is, the first and fourth rows as well as columns will drop out. The reader is encouraged to verify this statement. Thus, the final equation is, Solving for u 2 and u 3, we get Spring deflection is: 90 0 u 2 = u u 2 = 1.5 u Spring 1: u 2 u 1 = Spring 2: u 3 u 1 = Spring 3: u 3 u 2 = Spring 4: u 3 u 2 = Spring 5: u 4 u 2 = Spring 6: u 4 u 3 = FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-17

18 2.3.6 Structures that can be Modeled Using a Spring Elements As mentioned earlier, almost all engineering structures (linear structures) are similar to a linear spring, satisfying the relation F = ku. Therefore, any structure that deflects only along its axial direction (with one degree of freedom) can be modeled as a spring element. The following example illustrates this concept. Example 2.6 A circular concrete beam structure is loaded as shown. Find the deflection of points at 8, 16, and the end of the beam. E = 4 x 106 psi y 12 in 3 in lb x 24 in Figure 2.6 Solution The beam structure looks very different from a spring. However, its behavior is very similar. Deflection occurs along the x-axis only. The only significant difference between the beam and a spring is that the beam has a variable cross-sectional area. An exact solution can be found if the beam is divided into an infinite number of elements, then, each element can be considered as a constant cross-section spring element, obeying the relation F = ku, where k is the stiffness constant of a beam element and is given by k = AE/L. In order to keep size of the matrices small (for hand- calculations), let us divide the beam into only three elements. For engineering accuracy, the answer obtained will be in an acceptable range. If needed, accuracy can be improved by increasing the number of elements. As mentioned earlier in this chapter, spring, truss, and beam elements are line-elements and the shape of the cross section of an element is irrelevant. Only the cross-sectional area is needed (also, moment of inertia for a beam element undergoing a bending load need to be defined). The beam elements and their computer models are shown in figure 2.8. FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-18

19 Here, the question of which cross-sectional area to be used for each beam section arises. A good approximation would be to take the diameter of the mid-section and use that to approximate the area of the element. k 1 k 2 k k 1 k 2 k Beam sections Figure 2.7 Equivalent spring elements Cross-sectional area The average diameters are: d 1 = 10.5 in., d 2 = 7.5 in., d 3 = 4.5. (diameters are taken at the mid sections and the values are found from the height and length ratio of the triangles shown in figure 2.10), which is given as 12/L = 3/(L-24), L = 32 Average areas are: A 1 = in2 A 2 = in2 A 3 = 15.9 in2 24 in 12 in d 1 d 2 d 3 Original Averaged L- 24 L Figure 2.8 Figure in Stiffness k 1 = A 1 E/L 1 = (86.59)( /8) = lb./in., similarly, FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-19

20 k 2 = A 2 E/L 2 = lb./in. k 3 = A 3 E/L 3 = lb./in. Element Stiffness Equations Similarly, [K (1) ] = [K (2) ] = Global stiffness matrix is [K (3) ] = [K g ] = Now the global structural equations can be written as, u 1 F u 2 = F u 3 F u 4 F4 Applying the boundary conditions: u 1 = 0, and F 1 = F 2 = F 3 = 0, F 4 = 5000 lb., results in the reduced matrix, u u 3 = u FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-20

21 Solving we get, u u 3 = in. u The deflections u 2, u 3, and u 4 are only the approximate values, which can be improved by dividing the beam into more elements. As the number of elements increases, the accuracy will improve. FEA Lecture Notes by R. B. Agarwal Finite Element Analysis 2-21

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