Introduction to Finite Element Method


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1 Introduction to Finite Element Method Dr. Rakesh K Kapania Aerospace and Ocean Engineering Department Virginia Polytechnic Institute and State University, Blacksburg, VA AOE 524, Vehicle Structures Summer, 216 c 216 Rakesh K. Kapania, Mitchell Professor, Aerospace and Ocean Engineering, Virginia Polytechnic Institute and State University, Blacksburg, VA,
2 Introduction to Finite Element Method Complex structures under different types of loads Solution is often governed by ordinary or partial differential equations Simple domains or simple forcing functions analytical techniques closed form solution Complicated problems approximate solutions c Rakesh K. Kapania AOE 524, Vehicle Structures 2
3 Approximate methods Approximate methods 1. SemiAnalytical Methods 2. Numerical Methods SemiAnalytical Methods Series Expansion using Fourier, Bessel and other orthogonal functions RayleighRitz Method based on a variational principle Methods of Weighted Residual (MWR) (Galerkin Method and the Collocation Method) The Green s Functions approach (Integral Equations Approach) c Rakesh K. Kapania AOE 524, Vehicle Structures 3
4 Approximate methods (contd...) Numerical Methods Finite difference method Finite element method Spectral method Boundary element method Numerical methods (except the finite difference method) are essentially derived from one or the other semianalytical methods. For example Finite element method discrete implementation of the RayleighRitz Method or the Method of Weighted Residuals Spectral method Series expansion method with use of FFT (coefficients) Boundary element method Integral equation approach. c Rakesh K. Kapania AOE 524, Vehicle Structures 4
5 Approximate methods (contd...) A key feature of both of these categories is to convert an ordinary or partial differential equations into a set of linear (or nonlinear depending on the problem at hand) algebraic equations, i.e. Ka = Q. K is called the stiffness matrix. a is the vector of generalized coordinates Q is a vector of generalized forces. This set of equations can be solved numerically. c Rakesh K. Kapania AOE 524, Vehicle Structures 5
6 Semi Analytical Methods In the semianalytic methods, the solution u(x, y) is generally expressed as u(x, y) = φ o (x, y) + N n=1 a n φ n (x, y) where a n are called the generalized coordinates and φ n (x, y) are called the trial functions and are defined over the entire domain Ω. The function φ o (x, y) is chosen such that it satisfies the nonhomogeneous part of the boundary conditions. Note that this function is completely known and there is no undetermined coefficient in front of this function. c Rakesh K. Kapania AOE 524, Vehicle Structures 6
7 Trial Functions The trial functions i.e. functions φ n (x, y) have global support. φ n (x, y) satisfy the homogeneous part of the boundary conditions and must satisfy certain requirements of completeness and linear independence (the trial functions should be linearly independent) Selection of the trial functions is non trivial Accuracy of the solution generally increases as the number of terms N in the expansion are increased Semianalytical methods differ from each other depending upon the manner in which the coefficients a n are determined c Rakesh K. Kapania AOE 524, Vehicle Structures 7
8 Methods of Weighted Residuals In the Methods of Weighted Residuals the above approximation to the solution is substituted in the governing equations and the resulting error or the residual R(x, y) is minimized in various ways. In general, the following condition is used to obtain the values a n. R(x, y)ψ j (x, y)dω =, j = 1, 2, 3,...N Ω where the ψ j (x, y) are called the weight (hence the name Methods of Weighted Residuals) or the test functions, Ω, represents the domain. N is the number of terms used in the original expansion. c Rakesh K. Kapania AOE 524, Vehicle Structures 8
9 Methods of Weighted Residuals (contd...) The above condition is thus applied N different times and leads to a set of N algebraic equations in the coefficients, a n. The weight functions may or may not be same as the trial functions. For example, in the collocation scheme, the weight functions are the Dirac s Delta functions and in the Galerkin Method the weight functions and the trial functions are the same. c Rakesh K. Kapania AOE 524, Vehicle Structures 9
10 Ritz Method (Principle of Stationary Potential Energy) The unknown coefficients a n s are determined by using the underlying variational principle. In structures, for example, the coefficients are determined from the condition that at equilibrium the total potential energy Π of a structure under given loading must be minimum, i.e. the first variation of the total potential energy be equal to zero. In this method the N algebraic equations are obtained by satisfying the following N conditions Π a n = ; 1 n N Note that the total Potential energy is defined as Π = U + V, where U is the strain energy and V is the potential of the applied loads. c Rakesh K. Kapania AOE 524, Vehicle Structures 1
11 Principle of Virtual Work The principle of minimum potential energy on which the Ritz method is based, is restricted to conservative external forces i.e. to forces that can be derived from a potential. For non conservative forces, one has to resort to the more general Principle of Virtual Work which can be stated as: If a structure is in equilibrium and remains in equilibrium while it is subjected to a virtual distortion, the external virtual work δw vir done by the external forces acting on the structure is equal to the internal virtual work δw vir done by the internal stresses (δw vir = δw int ). c Rakesh K. Kapania AOE 524, Vehicle Structures 11
12 Principle of Virtual Work (contd..) The term virtual displacement implies an imaginary infinitesimal displacement that is superimposed on the deformed configuration while keeping the external loads unchanged and which is consistent with the boundary conditions (i.e. it is zero on that part of the boundary where the essential boundary conditions are specified). The virtual displacements, which should be piecewise continuous, to satisfy the displacement compatibility inside the domain, can be considered as the delta (δ) operator (a variation of the displacement field) or they may be considered as the test functions that are used in the method of weighted residuals (e.g. in Galerkin Method). c Rakesh K. Kapania AOE 524, Vehicle Structures 12
13 Principle of Virtual Work (contd...) Approximate methods using the principle of virtual work can be as easily derived as those using the principle of minimum potential energy. Also note that the principle of virtual work is both a sufficient and necessary condition for satisfying the equilibrium of a structure. Indeed, you can derive the equations of equilibrium and the associated boundary conditions from the principle of virtual work. c Rakesh K. Kapania AOE 524, Vehicle Structures 13
14 Principle of Virtual Work (contd...) For a general three dimensional structure, the expressions for external and internal virtual work can be written as : δw vir = B i δu i dvol + T (v) i δu i ds Vol δw int = Vol S 1 τ ij δɛ ij dvol where δu i is the virtual displacement field, B i is the body force field, T (v) i is the external surface traction acting on S 1 part of the boundary, v is the unit normal to the body. τ i j are the nine internal stress components (i, j = 1,2,3) and δɛ i j are the virtual strain components. c Rakesh K. Kapania AOE 524, Vehicle Structures 14
15 Principle of Virtual Work (contd...) Also note that a repeated index indicates a summation over that index (i.e. u i v i = u 1 v 1 + u 2 v 2 + u 3 v 3 ). Following this convention, the right hand side in the second equation is a sum of nine components. c Rakesh K. Kapania AOE 524, Vehicle Structures 15
16 Ritz Method For conservative forces, the principle of virtual work and the principle of stationary potential energy lead to the same equation, namely δ(u + V ) =. Here U is the strain energy and V is the potential of the applied loads. For a general 3D structure the Strain Energy is given as U = dudvol Where du is the strain energy density and is given as du = Vol ɛij τ ij dɛ ij For a linear elastic material, du = 1 2 τ ijɛ ij Also recall that following the summation convention, the right hand side is a sum of nine terms. c Rakesh K. Kapania AOE 524, Vehicle Structures 16
17 Ritz Method (contd...) The potential V of the applied loads for a general 3D structure is given as V = B i u i dvol + u i ds Vol Note that δv = δw vir S 1 T (v) i For a discrete load, Q, with a corresponding displacement, q, V can be written as: V = Qq c Rakesh K. Kapania AOE 524, Vehicle Structures 17
18 Example A Cantilever Beam Restrained By a Spring. The expression for the strain energy for a beam supported by a spring at the free end (x = L) is given as U = 1 2 EI ( d 2 ) 2 w dx 2 dx + 1 ( ) 2 2 k w (x=l) c Rakesh K. Kapania AOE 524, Vehicle Structures 18
19 Example (contd..) The expression for the potential of the applied load V is given as V = p(x)w(x)dx where w(x) is the transverse deflection of the beam. The approximate solution is obtained by using the principle of minimum total potential energy (Π): where Π = U + V. δπ = c Rakesh K. Kapania AOE 524, Vehicle Structures 19
20 Example (contd...) To obtain the approximate solution, we assume the transverse deflection as: N w(x) = C i φ i (x) i=1 where φ i (x) are the trial functions. Since there are no prescribed non homogeneous displacements and slope boundary conditions, we do not need φ (x). c Rakesh K. Kapania AOE 524, Vehicle Structures 2
21 Example (contd...) The trial functions must satisfy some requirements: (i) functions should be linearly independent. This means that the following equation will hold if and only if all C i s are zero. N C i φ i (x) = i=1 (ii) Functions must satisfy the essential boundary conditions. For the problem at hand, it means that both φ i (x) = and φ i (x) = at x =. Note that in the Ritz method the force (natural) boundary conditions are not explicitly satisfied. As the number of terms in the expansion for the dependent variable is increased, the natural boundary conditions tend to be satisfied naturally (Hence the name Natural Boundary Conditions). c Rakesh K. Kapania AOE 524, Vehicle Structures 21
22 Example (contd...) (iii) The trial functions must form a complete set. This means that the solution of the problem can be approximated arbitrarily close by a linear combination of the given trial functions as the number of terms in the expansion are increased. If we are using sine (or cosine) series, we must use sin(ix)(cosine(ix)) before we use sin(i + 1)x(cosine(i + 1)x). Similarly as we use (x) i+1, we must use (x) i. c Rakesh K. Kapania AOE 524, Vehicle Structures 22
23 Example (contd...) Substituting the expansion for w(x) in the expression for total potential energy, and using the principle of minimum potential energy, we get: N δπ δπ = δc i = i=1 δc i Since the variation in each of the generalized displacements C i s arbitrary and independent from the variation in the rest of the generalized displacements C i s, it means that Π C i = ; 1 i N c Rakesh K. Kapania AOE 524, Vehicle Structures 23
24 Example (contd...) This will yield a set of N equations. For the present beam problem, the expression for Π can be written as: C i Π L ( = 2 ) w 2 w EI C i x 2 C i x 2 dx + kw w L L L p(x) w dx C i C i [ N ] [ N ] = C j φ j (x) φ i (x)dx + k C j φ j (L) φ i (L) j=1 j=1 p(x)φ i (x)dx ] = N [ EI (x)φ i (x)φ j (x)dx + kφ i (L)φ j (L) j=1 p(x)φ i (x)dx = N j=1 C j K ij C j Q i c Rakesh K. Kapania AOE 524, Vehicle Structures 24
25 Example (contd...) Where the stiffness coefficient K ij and the generalized force Q i are given as: K ij = EI (x)φ i (x)φ j (x)dx + kφ i (L)φ j (L) Q i = p(x)φ i (x)dx For the present problem, we can choose where i varies from 1 to N. ( x ) i+1 φ i (x) = L c Rakesh K. Kapania AOE 524, Vehicle Structures 25
26 Example (contd...) Note that these trial functions satisfy all the essential boundary conditions, namely, φ() = and φ () =. Also the φ s are linearly independent and note that to satisfy the essential boundary conditions, we are starting our expansion from ( x ) 2 ( x ), (for i = l) and not from i.e. a constant term as L L required by the completeness requirement. The solution to our problem does not contain any constant or linear terms in the expansion. The present expansion will thus satisfy the completeness requirement. c Rakesh K. Kapania AOE 524, Vehicle Structures 26
27 Example (contd...) Substituting the chosen trial functions in the expressions for the stiffness matrix and the generalized forces (note (φ i (L) = 1), we obtain: ( ) x i+j 2 K ij = EI (x)(i)(j)(i + 1)(j + 1) dx + k = Q i = = ( (i)(j)(i + 1)(j + 1) (i + j 1) ( x ) ( x ) i+1 p dx L L p L (i + 3) ) ( EI L 3 L i+j+2 ) + α ( EI L 3 Using the above expressions we can obtain the approximate solutions using various values of N. c Rakesh K. Kapania AOE 524, Vehicle Structures 27 )
28 Example (contd...) Case (i) (n = 1) The set of linear equations for this case can be written as: EI L 3 (4 + α)c 1 = p L 4 p L C 1 = 4(4 + α)ei c Rakesh K. Kapania AOE 524, Vehicle Structures 28
29 Example (contd...) Case (ii) (n = 2) The set of linear equations for this case can be written as: [ EI 4 + α 6 + α L α 12 + α C 2 ] { C1 C 2 } = p L { Solution of this set of two equations will give us the values of the undetermined constants C 1 and C 2. { } C1 p L = 4 { } 36 + α 2EI (12 + 4α) (14 + α) } c Rakesh K. Kapania AOE 524, Vehicle Structures 29
30 Example (contd...) Case (iii) N = 3 The set of three linear equations is: EI C 1 L 3 C 2 = p L C α 6 + α 8 + α 6 + α 12 + α 18 + α 8 + α 18 + α α Using Gauss elimination, the solution is C 1 = (8α + 123)p L 4 24EI (α + 3) C 2 = (13α + 72)p L 4 24EI (α + 3) C 3 = p L 4 48EI c Rakesh K. Kapania AOE 524, Vehicle Structures 3
31 Example (contd...) Case (iv) N = 4 The set of four equations is: EI L α 6 + α 8 + α 1 + α 6 + α 12 + α 18 + α 24 + α 8 + α 18 + α α 4 + α α 24 + α 4 + α 7 + α C 1 C 2 C 3 C 4 = p L Solution of these four equations using Gauss elimination yields: C 1 = (7α + 2)p L 4 24EI (α + 3) C 3 = C 2 = (3α + 2)p L 4 8EI (α + 3) C 4 = p L 4 12EI It can be shown that this is also the exact solution to this problem. As a result, we do not have to proceed any further. c Rakesh K. Kapania AOE 524, Vehicle Structures
32 Example (contd...) For Complex problem exact solution is not known We have to decide how many terms in the expansion will yield an accurate solution We stop when we see that addition of more terms does not make any significant change in the desired response quantity (generally stresses) For a given number of terms, the accuracy of the stresses will be less than that for the displacements Differentiation of an approximation is almost always less accurate than the function c Rakesh K. Kapania AOE 524, Vehicle Structures 32
33 Penalty Approach Solution for various values of the spring constant can be easily determined by using appropriate value of α. As a special case, consider the case of a simplysupported end at x = L. This case is achieved by letting the value of the spring constant to be infinite (α infinity). The limiting value of the solution is: w(x) = p L 4 [ 7 ( x ) 2 3 ( x ) 3 1 ( x ) ] 5 + EI 24 L 8 L 12 L c Rakesh K. Kapania AOE 524, Vehicle Structures 33
34 Penalty Approach (contd...) Note that in obtaining this solution we did not satisfy the essential boundary condition at x = L explicitly. This suggests a way around satisfying the essential boundary conditions explicitly, by placing a spring with very large stiffness at the point where the displacement is zero. This approach is called the penalty approach. Instead of looking for trial functions that are zero at x =, and, x = L and have slope = at x =, we only satisfied the essential boundary conditions at x = exactly and will satisfy the essential boundary condition at x = L only approximately. c Rakesh K. Kapania AOE 524, Vehicle Structures 34
35 Limitations of the Ritz Method In the above approach, difficulty is that the selection of trial functions is very time consuming and for complicated domains with complex set of restraints, it may well be impossible to come up with a set of trial functions that will satisfy all the essential boundary conditions. c Rakesh K. Kapania AOE 524, Vehicle Structures 35
36 Finite Element Method In finite element method, in which the domain is divided ( discretized ) into a large number of smaller domains ( called finite elements) and simple polynomials that are defined only locally (i.e. only over a given finite element) are used as trial functions. Using either a weighted residual method or Ritz method, a set of algebraic equations are derived for every finite element and are then assembled (in a procedure called assemblage ) together to form a global set of equations. Another advantage the finite element method has over the global methods is that the matrix K is banded (sparse). This is due to the fact that a given trial function is defined only locally. It can easily solve linear and nonlinear problems and static and dynamic problems. c Rakesh K. Kapania AOE 524, Vehicle Structures 36
37 Finite Element Method (contd...) c Rakesh K. Kapania AOE 524, Vehicle Structures 37
38 Finite Element Method (contd...) Consider the solution of the ordinary differential equation governing the axial response of a bar with axial rigidity EA(x), and subjected to an axial force p(x): d dx ( EA(x) du dx ) = p(x) Let the boundary conditions be : u() = and u(l) = 1. An approximate solution of this equation can be found by satisfying the stationary condition for the total potential potential energy Π, which for an axial bar is given as: Π = 1 ( ) du 2 EA(x) dx p(x)u(x)dx 2 dx In the finite element method, we assume u(x) to be : N u(x) = a i φ i (x) i=1 c Rakesh K. Kapania AOE 524, Vehicle Structures 38
39 Finite Element Method (contd...) In the finite element method, the generalized coefficients are the values of the dependent variables (the axial displacement u in the present case) at the nodes (i.e a i = u i ). This is an important difference between the finite element method and the Ritz method. In the Ritz method, the generalized coordinates a i s, in general, do not have any physical significance. c Rakesh K. Kapania AOE 524, Vehicle Structures 39
40 Finite Element Method (contd...) By taking the first variation of the potential energy with respect to the nodal displacement values, we obtain: Π δπ = u i = u i N i=1 Since the variation of each of the generalized coordinates (nodal displacements u i s) is arbitrary and independent from the variation of all the other generalized coordinates, we get a set of N equations: Π =, i = 1, 2...N u i c Rakesh K. Kapania AOE 524, Vehicle Structures 4
41 Finite Element Method (contd...) The left hand side in the above equation can also be written as: Π L ( ) ( ) u u L = EA(x) dx p(x) (u(x))dx u i x u i x) u i = = N j=1 [ EA(x) [ ( N ) u j φ j (x) j=1 φ i (x) p(x)φ i (x) ] dx ] EA(x)φ i (x)φ j (x)dx u j p(x)φ i (x)dx where a prime indicates the derivative with respect to x. c Rakesh K. Kapania AOE 524, Vehicle Structures 41
42 Finite Element Method (contd...) For a linear problem, the resulting set of equations will be linear and can be written as: Ka = Q. Here, K the stiffness matrix a the vector of nodal displacements Q the nodal load vector. The elements of the stiffness matrix K, and the load vector Q are given as: K ij = Q i = EA(x)φ i (x)φ j (x)dx p(x)φ i (x)dx c Rakesh K. Kapania AOE 524, Vehicle Structures 42
43 Finite Element Method (contd...) The stiffness coefficient K ij have a physical meaning: it is the force required at node i due to a unit displacement at node j provided all other displacements are zero. Note that the stiffness matrix is symmetric but sparse. For example, the stiffness coefficients K 13, K 14,... =. To illustrate this, let us examine the stiffness coefficient K 13. K 13 = φ 1(x)φ 3(x)dx Product is zero because the trial function φ 1 (x) is zero on that part of the domain on which the trial function φ 3 (x) is nonzero and viceversa. Similarly we can see that for the secondorder ordinary differential equations, the stiffness matrix will have a bandwidth of 3. c Rakesh K. Kapania AOE 524, Vehicle Structures 43
44 Finite Element Method (contd...) For example, in the third row, the three nonzero coefficients will be : K 32, K 33, and K 34, and in the i th row, the three nonzero coefficients will be K i,i 1, K i,i, and K i,i+1. Let P i be an axial force acting at node i. In that case, the ith generalized force will be modified as Q i = P i + Q i c Rakesh K. Kapania AOE 524, Vehicle Structures 44
45 Finite Element Method (contd...) This can be easily seen from the fact that if a point force P k is acting at the k th node, the expression for the total potential energy will be modified to contain an additional term, P k u k. When the variation of this term is taken, it leads to the additional force term P i in the expression for the generalized force Q i. If for the given problem, we divide the domain (, L) using four elements (or five nodes), the resulting set of equations will be as follows: K 11 K 12 K 21 K 22 K 23 K 32 K 33 K 34 K 43 K 44 K 45 K 54 K 55 u 1 = u 2 u 3 u 4 u 5 = 1 = Q 1 Q 2 Q 3 Q 4 Q 5 c Rakesh K. Kapania AOE 524, Vehicle Structures 45
46 Finite Element Method (contd...) Note that in the above equation, there are five unknowns including: the two generalized forces Q 1 and Q 5 and the three generalized (nodal) displacements u 2, u 3, and u 4. Also note that the 5 by 5 global stiffness matrix K in the above set of linear equations is singular. This is due to the fact that the axial bar is not restrained from moving freely in the space. If we determine the eigenvalues of this global stiffness matrix, one of the eigenvalue will be found to be zero and all the elements in the corresponding eigenvector will be unity (i.e. a rigid body motion in the x direction). c Rakesh K. Kapania AOE 524, Vehicle Structures 46
47 Finite Element Method (contd...) For numerical results to converge, it is very important that the global stiffness matrix be singular and contain appropriate rigid body modes. Lack of this singularity will imply that a rigid body motion (in our case, all u i = 1) will result in a nonzero force vector, i.e. a free body motion will result in a nonzero axial strain which is of course not true. To obtain a unique solution for a given problem we remove this singularity by using the essential boundary conditions. In the present problem, this will be accomplished by substituting u 1 = and u 5 = 1. c Rakesh K. Kapania AOE 524, Vehicle Structures 47
48 Finite Element Method (contd...) The three displacements can be found by solving the following reduced set of equations: K 22 K 23 u 2 Q 2 K 32 K 33 K 34 u 3 = Q 3 u 5 K 43 K 44 u 4 Q 4 K 45 The first vector on right hand side can be obtained by performing the required integrals in the expressions for generalized forces and the second vector contains products of some stiffness coefficients with the generalized displacement u 5 = 1. c Rakesh K. Kapania AOE 524, Vehicle Structures 48
49 Finite Element Method (contd...) Once the three unknown displacements are determined from the above equations, we can obtain the unknown generalized forces Q 1 and Q 5 by substituting the values of all the nodal displacements in the full stiffness matrix. Note that Q 1 represents the reaction at the node 1 and the Q 5 represents the force required to produce u 5 = 1. We can easily change the boundary conditions in the finite element method. For example, suppose that we have applied a force P at the free edge (x = L). In that case, the displacement at the free edge (u 5 if we divide the domain in four elements) will be unknown but the force Q 5 = Q 5 + P 5 will be known. c Rakesh K. Kapania AOE 524, Vehicle Structures 49
50 Finite Element Method (contd...) As another example, suppose we have a spring of stiffness k at the free end (x = L). This is equivalent to applying a force P 5 = ku 5 at the free end. Since this term contains the unknown u 5 we can take it to the left hand side in the fifth (last) equation. This will change the stiffness coefficient K 55 to K 55 + k. c Rakesh K. Kapania AOE 524, Vehicle Structures 5
51 Example EI 4 w x 4 = p (p is uniform) I (w) = w() = w () = EIw x=l = M EI (w ) = [ ] 1 2 EI (w ) 2 pw dx M w x x=l c Rakesh K. Kapania AOE 524, Vehicle Structures 51
52 Example (contd...) Assume w(x) = N C i φ i (x) (Nparameter Ritz method) j=1 φ i must satisfy essential boundary conditions. i.e. φ(x) = at x = ; φ () = Furthermore, the φ i s should be linearly i.e. independent a 1 φ 1 + a 2 φ a n φ n = if a 1 = a 2 = = a n = c Rakesh K. Kapania AOE 524, Vehicle Structures 52
53 Example (contd...) i.e. if we chose φ 1 = x 2, we cannot choose another φ j = ax 2, where a is a constant. Let φ j (x) = x j+1, j = 1, 2, 3,, N since x j+1 = at x = at x j+1 x = (j + 1)x j = x = c Rakesh K. Kapania AOE 524, Vehicle Structures 53
54 Example (contd...) Furthermore, x j+1 is linearly independent of x j+2, x j+3, w = C 1 x 2 + C 2 x 3 + C 3 x 4 + C 4 x C N x N+1 if N = 2, we stop at x 3 if N = 3, we stop at x 4.. w = w = N C j x j+1 j=1 N C j (j + 1)(j)x j 1 j=1 c Rakesh K. Kapania AOE 524, Vehicle Structures 54
55 Example (contd...) I = 1 2 EI ( N C j φ j j=1 M [ N j=1 ) 2 p C j φ j x=l ( N j=1 C j φ j ) ] dx = I [C 1, C 2, C 3,, C N ] I C 1 = I I = = C 2 C N which are a set of linear equations. c Rakesh K. Kapania AOE 524, Vehicle Structures 55
56 Example (contd...) I C i = 2 EI 2 [ N C j φ j j=1 ] φ i dx l pφ i dx M φ i x=l = for i = 1, 2,, N N [ ] EI φ i φ j dx C j = j=1 for i=1,2,,n l pφ i dx + M φ i x=l c Rakesh K. Kapania AOE 524, Vehicle Structures 56
57 Example (contd...) Let K ij = f i = EI φ i φ j dx = B(φ i, φ j ) pφ i dx + M φ i x=l = l(φ i ) N K ij C j j=1 for i=1,2,3,,n K 11 K 12 K 1N K 21 K 22 K 2N K N1 K N2 K 1N = f i c Rakesh K. Kapania AOE 524, Vehicle Structures 57 C 1 C 2. C N = f 1 f 2. f N
58 Example (contd...) Note K ij = K ji [K ]{C } = {f } where [K] is the stiffness matrix, { C } is the generalized displacement vector, and { f } is the generalized load vector. Also it can be shown that [K] is positive definite. So, {C } = [K ] 1 {f } c Rakesh K. Kapania AOE 524, Vehicle Structures 58
59 Example (contd...) Let N=2: w = C 1 x 2 + C 2 x 3 i.e. φ 1 = x 2, φ 2 = x 3, φ 1 = 2, φ 2 = 6x K 11 = = = 4EIL EI φ 1 φ 1 dx EI 2 2 dx K 12 = K 21 = = = 6EIL 2 EI 2 6x dx EI φ 1 φ 2 dx K 22 = = 12EIL 3 EI φ 2 φ 2 dx = EI 6 6 dx c Rakesh K. Kapania AOE 524, Vehicle Structures 59
60 Example (contd...) f 1 = = pφ 1 dx + M (φ 1) x=l px 2 dx + M (2x) x=l = pl3 + 2M L 3 [ ] { } 4L 6L 2 C1 EI 6L 2 12L 3 C 2 f 2 = = = pφ 2 dx + M (φ 2) x=l px 3 dx + M (3x 2 ) x=l = pl4 + 3M L 2 4 { } pl M L pl M L 2 c Rakesh K. Kapania AOE 524, Vehicle Structures 6
61 Example (contd...) { C1 C 2 } = 1 EI [ 4L 6L 2 6L 2 12L 3 ] { 1 pl M L pl M L 2 } = [ 1 EI 1 12L 3 6L 2 12L 4 6L 2 4L ] { pl M L pl M L 2 } = { 1 4pL M L EIL 4 4 pl6 18M L 4 2pL 5 12M L 3 + pl 4 L + 12M L 3 } = 1 { 5 2 pl6 + 6M L 4 12EIL 4 pl 5 c Rakesh K. Kapania AOE 524, Vehicle Structures 61 }
62 Example (contd...) w(x) = C 1 x 2 + C 2 x 3 ( 5pL = 2 ) + 12M x EI ( ) pl x 3 12EI Now, for N = 3 w(x) = C 1 x 2 + C 2 x 3 + C 3 x 4 A better approximation Objective : To determine C 1, C 2, C 3 φ 1 = x 2 ; φ 1 = 2x; φ 1 = 2 φ 2 = x 3 ; φ 2 = 3x 2 ; φ 2 = 6x φ 3 = x 4 ; φ 3 = 4x 3 ; φ 3 = 12x 2 c Rakesh K. Kapania AOE 524, Vehicle Structures 62
63 Example (contd...) K 13 = K 31 K 23 = K 32 = EI φ 1 φ 3 dx = EI φ 2 φ 3 dx = EI 2 12x 2 dx = EI 6x 12x 2 dx K 13 = 8EIL 3 K 23 = 18EIL 4 c Rakesh K. Kapania AOE 524, Vehicle Structures 63
64 Example (contd...) K 22 = = EI φ 3 φ 3 dx EI (12x 2 ) 2 dx f 3 = = pφ 3 dx + M (φ 1) x=l p x 4 dx + M 4x 3 x=l K 22 = 144EIL5 5 f 3 = pl M L 3 c Rakesh K. Kapania AOE 524, Vehicle Structures 64
65 Example (contd...) EIL 4 6L 8l 2 6L 12L 2 18L 3 8L 2 18L L4 SOLVING GIVES: C 1 C 2 C 3 = PL M L PL M L 2 PL M L 3 C 1 = pl2 + 2M 4EI C 2 = pl 6EI p C 3 = 24EI c Rakesh K. Kapania AOE 524, Vehicle Structures 65
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