# Introduction to Finite Element Method

Save this PDF as:

Size: px
Start display at page:

## Transcription

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. Semi-Analytical Methods 2. Numerical Methods Semi-Analytical Methods Series Expansion using Fourier, Bessel and other orthogonal functions Rayleigh-Ritz 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 semi-analytical methods. For example Finite element method discrete implementation of the Rayleigh-Ritz 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 co-ordinates 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 semi-analytic 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 co-ordinates 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 non-homogeneous 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 Semi-analytical 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 3-D 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 3-D 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 simply-supported 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 co-ordinates 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 co-ordinates, 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 co-efficient 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 non-zero and vice-versa. Similarly we can see that for the second-order 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 non-zero 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 non-zero 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) (N-parameter 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

### Stiffness Matrices, Spring and Bar Elements

CHAPTER Stiffness Matrices, Spring and Bar Elements. INTRODUCTION The primary characteristics of a finite element are embodied in the element stiffness matrix. For a structural finite element, the stiffness

### Introduction to Continuous Systems. Continuous Systems. Strings, Torsional Rods and Beams.

Outline of Continuous Systems. Introduction to Continuous Systems. Continuous Systems. Strings, Torsional Rods and Beams. Vibrations of Flexible Strings. Torsional Vibration of Rods. Bernoulli-Euler Beams.

### Lecture 27: Structural Dynamics - Beams.

Chapter #16: Structural Dynamics and Time Dependent Heat Transfer. Lectures #1-6 have discussed only steady systems. There has been no time dependence in any problems. We will investigate beam dynamics

### International Journal of Advanced Engineering Technology E-ISSN

Research Article INTEGRATED FORCE METHOD FOR FIBER REINFORCED COMPOSITE PLATE BENDING PROBLEMS Doiphode G. S., Patodi S. C.* Address for Correspondence Assistant Professor, Applied Mechanics Department,

### Bending of Simply Supported Isotropic and Composite Laminate Plates

Bending of Simply Supported Isotropic and Composite Laminate Plates Ernesto Gutierrez-Miravete 1 Isotropic Plates Consider simply a supported rectangular plate of isotropic material (length a, width b,

### Nonlinear FEM. Critical Points. NFEM Ch 5 Slide 1

5 Critical Points NFEM Ch 5 Slide Assumptions for this Chapter System is conservative: total residual is the gradient of a total potential energy function r(u,λ) = (u,λ) u Consequence: the tangent stiffness

### ACCURATE FREE VIBRATION ANALYSIS OF POINT SUPPORTED MINDLIN PLATES BY THE SUPERPOSITION METHOD

Journal of Sound and Vibration (1999) 219(2), 265 277 Article No. jsvi.1998.1874, available online at http://www.idealibrary.com.on ACCURATE FREE VIBRATION ANALYSIS OF POINT SUPPORTED MINDLIN PLATES BY

### COORDINATE TRANSFORMATIONS

COORDINAE RANSFORMAIONS Members of a structural system are typically oriented in differing directions, e.g., Fig. 17.1. In order to perform an analysis, the element stiffness equations need to be expressed

### L e c t u r e. D r. S a s s a n M o h a s s e b

The Scaled smteam@gmx.ch Boundary Finite www.erdbebenschutz.ch Element Method Lecture A L e c t u r e A1 D r. S a s s a n M o h a s s e b V i s i t i n g P r o f e s s o r M. I. T. C a m b r i d g e December

### Resources. Introduction to the finite element method. History. Topics

Resources Introduction to the finite element method M. M. Sussman sussmanm@math.pitt.edu Office Hours: 11:1AM-12:1PM, Thack 622 May 12 June 19, 214 Strang, G., Fix, G., An Analysis of the Finite Element

### Mathematics FINITE ELEMENT ANALYSIS AS COMPUTATION. What the textbooks don't teach you about finite element analysis. Chapter 3

Mathematics FINITE ELEMENT ANALYSIS AS COMPUTATION What the textbooks don't teach you about finite element analysis Chapter 3 Completeness and continuity: How to choose shape functions? Gangan Prathap

### Professor Terje Haukaas University of British Columbia, Vancouver Notation

Notation This document establishes the notation that is employed throughout these notes. It is intended as a look-up source during the study of other documents and software on this website. As a general

### FEAP - - A Finite Element Analysis Program

FEAP - - A Finite Element Analysis Program Version 8.5 Theory Manual Robert L. Taylor Department of Civil and Environmental Engineering University of California at Berkeley Berkeley, California 94720-1710

### Finite element modelling of structural mechanics problems

1 Finite element modelling of structural mechanics problems Kjell Magne Mathisen Department of Structural Engineering Norwegian University of Science and Technology Lecture 10: Geilo Winter School - January,

Conjugate Gradient (CG) Method by K. Ozawa 1 Introduction In the series of this lecture, I will introduce the conjugate gradient method, which solves efficiently large scale sparse linear simultaneous

### EFFECT OF A CRACK ON THE DYNAMIC STABILITY OF A FREE}FREE BEAM SUBJECTED TO A FOLLOWER FORCE

Journal of Sound and

### A RATIONAL BUCKLING MODEL FOR THROUGH GIRDERS

A RATIONAL BUCKLING MODEL FOR THROUGH GIRDERS (Hasan Santoso) A RATIONAL BUCKLING MODEL FOR THROUGH GIRDERS Hasan Santoso Lecturer, Civil Engineering Department, Petra Christian University ABSTRACT Buckling

### Dispersion relation for transverse waves in a linear chain of particles

Dispersion relation for transverse waves in a linear chain of particles V. I. Repchenkov* It is difficult to overestimate the importance that have for the development of science the simplest physical and

### Module 3. Analysis of Statically Indeterminate Structures by the Displacement Method

odule 3 Analysis of Statically Indeterminate Structures by the Displacement ethod Lesson 14 The Slope-Deflection ethod: An Introduction Introduction As pointed out earlier, there are two distinct methods

### BACKGROUNDS. Two Models of Deformable Body. Distinct Element Method (DEM)

BACKGROUNDS Two Models of Deformable Body continuum rigid-body spring deformation expressed in terms of field variables assembly of rigid-bodies connected by spring Distinct Element Method (DEM) simple

### THE USE OF DYNAMIC RELAXATION TO SOLVE THE DIFFERENTIAL EQUATION DESCRIBING THE SHAPE OF THE TALLEST POSSIBLE BUILDING

VII International Conference on Textile Composites and Inflatable Structures STRUCTURAL MEMBRANES 2015 E. Oñate, K.-U.Bletzinger and B. Kröplin (Eds) THE USE OF DYNAMIC RELAXATION TO SOLVE THE DIFFERENTIAL

### Geometry-dependent MITC method for a 2-node iso-beam element

Structural Engineering and Mechanics, Vol. 9, No. (8) 3-3 Geometry-dependent MITC method for a -node iso-beam element Phill-Seung Lee Samsung Heavy Industries, Seocho, Seoul 37-857, Korea Hyu-Chun Noh

### Unit 15 Shearing and Torsion (and Bending) of Shell Beams

Unit 15 Shearing and Torsion (and Bending) of Shell Beams Readings: Rivello Ch. 9, section 8.7 (again), section 7.6 T & G 126, 127 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering

### Fracture Mechanics of Composites with Residual Thermal Stresses

J. A. Nairn Material Science & Engineering, University of Utah, Salt Lake City, Utah 84 Fracture Mechanics of Composites with Residual Thermal Stresses The problem of calculating the energy release rate

### Lucio Demeio Dipartimento di Ingegneria Industriale e delle Scienze Matematiche

Scuola di Dottorato THE WAVE EQUATION Lucio Demeio Dipartimento di Ingegneria Industriale e delle Scienze Matematiche Lucio Demeio - DIISM wave equation 1 / 44 1 The Vibrating String Equation 2 Second

### Practice Problems For Test 3

Practice Problems For Test 3 Power Series Preliminary Material. Find the interval of convergence of the following. Be sure to determine the convergence at the endpoints. (a) ( ) k (x ) k (x 3) k= k (b)

### RECURSIVE DIFFERENTIATION METHOD FOR BOUNDARY VALUE PROBLEMS: APPLICATION TO ANALYSIS OF A BEAM-COLUMN ON AN ELASTIC FOUNDATION

Journal of Theoretical and Applied Mechanics, Sofia, 2014, vol. 44, No. 2, pp. 57 70 RECURSIVE DIFFERENTIATION METHOD FOR BOUNDARY VALUE PROBLEMS: APPLICATION TO ANALYSIS OF A BEAM-COLUMN ON AN ELASTIC

### Stress, Strain, Mohr s Circle

Stress, Strain, Mohr s Circle The fundamental quantities in solid mechanics are stresses and strains. In accordance with the continuum mechanics assumption, the molecular structure of materials is neglected

### Part C: Beam-Columns. Beams Basic Formulation The Temporal Solution The Spatial Solution

Part C: Beam-Columns Beams Basic Formulation The Temporal Solution The Spatial Solution Rayleigh-Ritz Analysis Rayleigh s Quotient The Role of Initial Imperfections An Alternative Approach Higher Modes

### FINITE VOLUME METHOD: BASIC PRINCIPLES AND EXAMPLES

FINITE VOLUME METHOD: BASIC PRINCIPLES AND EXAMPLES SHRUTI JAIN B.Tech III Year, Electronics and Communication IIT Roorkee Tutors: Professor G. Biswas Professor S. Chakraborty ACKNOWLEDGMENTS I would like

### 4. SHAFTS. A shaft is an element used to transmit power and torque, and it can support

4. SHAFTS A shaft is an element used to transmit power and torque, and it can support reverse bending (fatigue). Most shafts have circular cross sections, either solid or tubular. The difference between

### Linear Algebra in Hilbert Space

Physics 342 Lecture 16 Linear Algebra in Hilbert Space Lecture 16 Physics 342 Quantum Mechanics I Monday, March 1st, 2010 We have seen the importance of the plane wave solutions to the potentialfree Schrödinger

### Structural Analysis III Compatibility of Displacements & Principle of Superposition

Structural Analysis III Compatibility of Displacements & Principle of Superposition 2007/8 Dr. Colin Caprani, Chartered Engineer 1 1. Introduction 1.1 Background In the case of 2-dimensional structures

### Lecture 8: Flexibility Method. Example

ecture 8: lexibility Method Example The plane frame shown at the left has fixed supports at A and C. The frame is acted upon by the vertical load P as shown. In the analysis account for both flexural and

### THE MECHANICAL BEHAVIOR OF ORIENTED 3D FIBER STRUCTURES

Lappeenranta University of Technology School of Engineering Science Degree Program in Computational Engineering and Technical Physics Master s Thesis Alla Kliuzheva THE MECHANICAL BEHAVIOR OF ORIENTED

### Lecture 2: Finite Elements

Materials Science & Metallurgy Master of Philosophy, Materials Modelling, Course MP7, Finite Element Analysis, H. K. D. H. Bhadeshia Lecture 2: Finite Elements In finite element analysis, functions of

### Statics and Influence Functions From a Modern Perspective

Statics and Influence Functions From a Modern Perspective Friedel Hartmann Peter Jahn Statics and Influence Functions From a Modern Perspective 123 Friedel Hartmann Department of Civil Engineering University

### Elastic-Plastic Fracture Mechanics. Professor S. Suresh

Elastic-Plastic Fracture Mechanics Professor S. Suresh Elastic Plastic Fracture Previously, we have analyzed problems in which the plastic zone was small compared to the specimen dimensions (small scale

### Iterative Methods for Linear Systems

Iterative Methods for Linear Systems 1. Introduction: Direct solvers versus iterative solvers In many applications we have to solve a linear system Ax = b with A R n n and b R n given. If n is large the

### Effect of Mass Matrix Formulation Schemes on Dynamics of Structures

Effect of Mass Matrix Formulation Schemes on Dynamics of Structures Swapan Kumar Nandi Tata Consultancy Services GEDC, 185 LR, Chennai 600086, India Sudeep Bosu Tata Consultancy Services GEDC, 185 LR,

### LINEAR AND NONLINEAR SHELL THEORY. Contents

LINEAR AND NONLINEAR SHELL THEORY Contents Strain-displacement relations for nonlinear shell theory Approximate strain-displacement relations: Linear theory Small strain theory Small strains & moderate

### Advanced Structural Analysis Prof. Devdas Menon Department of Civil Engineering Indian Institute of Technology, Madras

Advanced Structural Analysis Prof. Devdas Menon Department of Civil Engineering Indian Institute of Technology, Madras Module - 5.2 Lecture - 28 Matrix Analysis of Beams and Grids (Refer Slide Time: 00:23)

### Applications with Laplace

Week #8 : Applications with Laplace Goals: Solving application ODEs using Laplace transforms Forced Spring/mass system Deformation of beams under load Transport and diffusion of contaminants in groundwater

### Generic Strategies to Implement Material Grading in Finite Element Methods for Isotropic and Anisotropic Materials

University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Engineering Mechanics Dissertations & Theses Mechanical & Materials Engineering, Department of Winter 12-9-2011 Generic

### Nonlinear analysis in ADINA Structures

Nonlinear analysis in ADINA Structures Theodore Sussman, Ph.D. ADINA R&D, Inc, 2016 1 Topics presented Types of nonlinearities Materially nonlinear only Geometrically nonlinear analysis Deformation-dependent

### HOMEWORK PROBLEMS FROM STRANG S LINEAR ALGEBRA AND ITS APPLICATIONS (4TH EDITION)

HOMEWORK PROBLEMS FROM STRANG S LINEAR ALGEBRA AND ITS APPLICATIONS (4TH EDITION) PROFESSOR STEVEN MILLER: BROWN UNIVERSITY: SPRING 2007 1. CHAPTER 1: MATRICES AND GAUSSIAN ELIMINATION Page 9, # 3: Describe

### Chapter 3 LAMINATED MODEL DERIVATION

17 Chapter 3 LAMINATED MODEL DERIVATION 3.1 Fundamental Poisson Equation The simplest version of the frictionless laminated model was originally introduced in 1961 by Salamon, and more recently explored

### A Constant Displacement Iteration Algorithm for Nonlinear Static Push-Over Analyses

120 A Constant Displacement Iteration Algorithm for Nonlinear Static Push-Over Analyses ABSTRACT Graham C. Archer Assistant Professor, School of Civil Engineering, University of Purdue 1284 Civil Engineering

### Partial Differential Equations

Partial Differential Equations Xu Chen Assistant Professor United Technologies Engineering Build, Rm. 382 Department of Mechanical Engineering University of Connecticut xchen@engr.uconn.edu Contents 1

### I = i 0,

Special Types of Matrices Certain matrices, such as the identity matrix 0 0 0 0 0 0 I = 0 0 0, 0 0 0 have a special shape, which endows the matrix with helpful properties The identity matrix is an example

### Research Article Element for Beam Dynamic Analysis Based on Analytical Deflection Trial Function

Mathematical Problems in Engineering Volume 215 Article ID 582326 1 pages http://dx.doi.org/1.1155/215/582326 Research Article Element for Beam Dynamic Analysis Based on Analytical Deflection Trial Function

### Y. M. TEMIS AND V. V. KARABAN 1

J. KSIAM Vol.5, No.2, 39-51, 2001 BOUNDARY ELEMENT TECHNIQUE IN TORSION PROBLEMS OF BEAMS WITH MULTIPLY CONNECTED CROSS-SECTIONS Y. M. TEMIS AND V. V. KARABAN 1 Abstract. This paper shows how boundary

### CAAM 336 DIFFERENTIAL EQUATIONS IN SCI AND ENG Examination 1

CAAM 6 DIFFERENTIAL EQUATIONS IN SCI AND ENG Examination Instructions: Time limit: uninterrupted hours There are four questions worth a total of 5 points Please do not look at the questions until you begin

### AN IMPROVED NUMERICAL MODEL FOR CALCULATING SHIP HULL FRAME TRANSVERSAL STRUCTURE

COMPUTATIONAL MECHANICS New Trends and Applications E. Oñate and S. R. Idelsohn (Eds.) CIMNE, Barcelona, Spain 1998 AN IMPROVED NUMERICAL MODEL FOR CALCULATING SHIP HULL FRAME TRANSVERSAL STRUCTURE Oscar

### Chapter 2 Vector-matrix Differential Equation and Numerical Inversion of Laplace Transform

Chapter 2 Vector-matrix Differential Equation and Numerical Inversion of Laplace Transform 2.1 Vector-matrix Differential Equation A differential equation and a set of differential (simultaneous linear

### Stability of flow past a confined cylinder

Stability of flow past a confined cylinder Andrew Cliffe and Simon Tavener University of Nottingham and Colorado State University Stability of flow past a confined cylinder p. 1/60 Flow past a cylinder

### Problem 1A. Calculus. Problem 3A. Real analysis. f(x) = 0 x = 0.

Problem A. Calculus Find the length of the spiral given in polar coordinates by r = e θ, < θ 0. Solution: The length is 0 θ= ds where by Pythagoras ds = dr 2 + (rdθ) 2 = dθ 2e θ, so the length is 0 2e

### By drawing Mohr s circle, the stress transformation in 2-D can be done graphically. + σ x σ y. cos 2θ + τ xy sin 2θ, (1) sin 2θ + τ xy cos 2θ.

Mohr s Circle By drawing Mohr s circle, the stress transformation in -D can be done graphically. σ = σ x + σ y τ = σ x σ y + σ x σ y cos θ + τ xy sin θ, 1 sin θ + τ xy cos θ. Note that the angle of rotation,

### Flow-Induced Vibration Analysis of Supported Pipes with a Crack

Flow-Induced Vibration Analysis of Supported Pipes with a Crack Jin-Hyuk Lee 1 *, Samer Masoud Al-Said,3 1 Department of Mechanical Engineering, American University of Sharjah, Sharjah, UAE, Department

### Adaptive Analysis of Bifurcation Points of Shell Structures

First published in: Adaptive Analysis of Bifurcation Points of Shell Structures E. Ewert and K. Schweizerhof Institut für Mechanik, Universität Karlsruhe (TH), Kaiserstraße 12, D-76131 Karlsruhe, Germany

### CHAPTER -6- BENDING Part -1-

Ishik University / Sulaimani Civil Engineering Department Mechanics of Materials CE 211 CHAPTER -6- BENDING Part -1-1 CHAPTER -6- Bending Outlines of this chapter: 6.1. Chapter Objectives 6.2. Shear and

### 8.1 Bifurcations of Equilibria

1 81 Bifurcations of Equilibria Bifurcation theory studies qualitative changes in solutions as a parameter varies In general one could study the bifurcation theory of ODEs PDEs integro-differential equations

### Lecture 12: Detailed balance and Eigenfunction methods

Miranda Holmes-Cerfon Applied Stochastic Analysis, Spring 2015 Lecture 12: Detailed balance and Eigenfunction methods Readings Recommended: Pavliotis [2014] 4.5-4.7 (eigenfunction methods and reversibility),

### Damping of materials and members in structures

Journal of Physics: Conference Series Damping of materials and members in structures To cite this article: F Orban 0 J. Phys.: Conf. Ser. 68 00 View the article online for updates and enhancements. Related

### Dynamic Model of a Badminton Stroke

ISEA 28 CONFERENCE Dynamic Model of a Badminton Stroke M. Kwan* and J. Rasmussen Department of Mechanical Engineering, Aalborg University, 922 Aalborg East, Denmark Phone: +45 994 9317 / Fax: +45 9815

### 2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at A and supported at B by rod (1). What is the axial force in rod (1)?

IDE 110 S08 Test 1 Name: 1. Determine the internal axial forces in segments (1), (2) and (3). (a) N 1 = kn (b) N 2 = kn (c) N 3 = kn 2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at

### MA 201: Method of Separation of Variables Finite Vibrating String Problem Lecture - 11 MA201(2016): PDE

MA 201: Method of Separation of Variables Finite Vibrating String Problem ecture - 11 IBVP for Vibrating string with no external forces We consider the problem in a computational domain (x,t) [0,] [0,

### Chapter 3 Conforming Finite Element Methods 3.1 Foundations Ritz-Galerkin Method

Chapter 3 Conforming Finite Element Methods 3.1 Foundations 3.1.1 Ritz-Galerkin Method Let V be a Hilbert space, a(, ) : V V lr a bounded, V-elliptic bilinear form and l : V lr a bounded linear functional.

### The Direct Stiffness Method II

The Direct Stiffness Method II Chapter : THE DIRECT STIFFNESS METHOD II TABLE OF CONTENTS Page. The Remaining DSM Steps.................2 Assembly: Merge....................2. Governing Rules.................2.2

### MEI solutions to exercise 4 1

MEI-55 solutions to exercise Problem Solve the stationary two-dimensional heat transfer problem shown in the figure below by using linear elements Use symmetry to reduce the problem size The material is

### (a) If A is a 3 by 4 matrix, what does this tell us about its nullspace? Solution: dim N(A) 1, since rank(a) 3. Ax =

. (5 points) (a) If A is a 3 by 4 matrix, what does this tell us about its nullspace? dim N(A), since rank(a) 3. (b) If we also know that Ax = has no solution, what do we know about the rank of A? C(A)

### Flexural-Torsional Buckling of General Cold-Formed Steel Columns with Unequal Unbraced Lengths

Proceedings of the Annual Stability Conference Structural Stability Research Council San Antonio, Texas, March 21-24, 2017 Flexural-Torsional Buckling of General Cold-Formed Steel Columns with Unequal

### Topic 2 Quiz 2. choice C implies B and B implies C. correct-choice C implies B, but B does not imply C

Topic 1 Quiz 1 text A reduced row-echelon form of a 3 by 4 matrix can have how many leading one s? choice must have 3 choice may have 1, 2, or 3 correct-choice may have 0, 1, 2, or 3 choice may have 0,

### Thermomechanical Effects

3 hermomechanical Effects 3 Chapter 3: HERMOMECHANICAL EFFECS ABLE OF CONENS Page 3. Introduction..................... 3 3 3.2 hermomechanical Behavior............... 3 3 3.2. hermomechanical Stiffness

### Advanced Structural Analysis Prof. Devdas Menon Department of Civil Engineering Indian Institute of Technology, Madras

Advanced Structural Analysis Prof. Devdas Menon Department of Civil Engineering Indian Institute of Technology, Madras Module No. # 5.4 Lecture No. # 30 Matrix Analysis of Beams and Grids (Refer Slide

### Introduction to Timoshenko Beam Theory

Introduction to Timoshenko Beam Theory Aamer Haque Abstract Timoshenko beam theory includes the effect of shear deformation which is ignored in Euler-Bernoulli beam theory. An elementary derivation is

### UNIT-II MOVING LOADS AND INFLUENCE LINES

UNIT-II MOVING LOADS AND INFLUENCE LINES Influence lines for reactions in statically determinate structures influence lines for member forces in pin-jointed frames Influence lines for shear force and bending

### arxiv: v1 [math.co] 3 Nov 2014

SPARSE MATRICES DESCRIBING ITERATIONS OF INTEGER-VALUED FUNCTIONS BERND C. KELLNER arxiv:1411.0590v1 [math.co] 3 Nov 014 Abstract. We consider iterations of integer-valued functions φ, which have no fixed

### 18.303: Introduction to Green s functions and operator inverses

8.33: Introduction to Green s functions and operator inverses S. G. Johnson October 9, 2 Abstract In analogy with the inverse A of a matri A, we try to construct an analogous inverse Â of differential

### Differential Quadrature Method for Solving Hyperbolic Heat Conduction Problems

Tamkang Journal of Science and Engineering, Vol. 12, No. 3, pp. 331 338 (2009) 331 Differential Quadrature Method for Solving Hyperbolic Heat Conduction Problems Ming-Hung Hsu Department of Electrical

### ADI iterations for. general elliptic problems. John Strain Mathematics Department UC Berkeley July 2013

ADI iterations for general elliptic problems John Strain Mathematics Department UC Berkeley July 2013 1 OVERVIEW Classical alternating direction implicit (ADI) iteration Essentially optimal in simple domains

### An introduction to the mathematical theory of finite elements

Master in Seismic Engineering E.T.S.I. Industriales (U.P.M.) Discretization Methods in Engineering An introduction to the mathematical theory of finite elements Ignacio Romero ignacio.romero@upm.es October

### Damage detection of damaged beam by constrained displacement curvature

Journal of Mechanical Science and Technology Journal of Mechanical Science and Technology 22 (2008) 1111~1120 www.springerlink.com/content/1738-494x Damage detection of damaged beam by constrained displacement

### Stochastic Collocation Methods for Polynomial Chaos: Analysis and Applications

Stochastic Collocation Methods for Polynomial Chaos: Analysis and Applications Dongbin Xiu Department of Mathematics, Purdue University Support: AFOSR FA955-8-1-353 (Computational Math) SF CAREER DMS-64535

### [POLS 8500] Review of Linear Algebra, Probability and Information Theory

[POLS 8500] Review of Linear Algebra, Probability and Information Theory Professor Jason Anastasopoulos ljanastas@uga.edu January 12, 2017 For today... Basic linear algebra. Basic probability. Programming

### 1.050 Content overview Engineering Mechanics I Content overview. Selection of boundary conditions: Euler buckling.

.050 Content overview.050 Engineering Mechanics I Lecture 34 How things fail and how to avoid it Additional notes energy approach I. Dimensional analysis. On monsters, mice and mushrooms Lectures -3. Similarity

### 2. Review of Linear Algebra

2. Review of Linear Algebra ECE 83, Spring 217 In this course we will represent signals as vectors and operators (e.g., filters, transforms, etc) as matrices. This lecture reviews basic concepts from linear

### NONLINEAR DYNAMIC ANALYSIS OF COMPLEX STRUCTURES

EARTHQUAKE ENGINEERING AND STRUCTURAL DYNAMICS, VOL. 1, 241-252 (1973) NONLINEAR DYNAMIC ANALYSIS OF COMPLEX STRUCTURES E. L. WILSON~ University of California, Berkeley, California I. FARHOOMAND\$ K. J.

### det(ka) = k n det A.

Properties of determinants Theorem. If A is n n, then for any k, det(ka) = k n det A. Multiplying one row of A by k multiplies the determinant by k. But ka has every row multiplied by k, so the determinant

### u(o, t) = ZX(x, t) = O, u(l, t) = 62 STUDIES ON GEODESICS IN VIBRATIONS OF ELASTIC BEAMS

38 MA THEMA TICS: A. D. MICHAL PROC. N. A. S. dicated in the passage in question. As thus measured, it does not appear at all constant (for constant amplitude) as I should expect it to be. This seems to

### 3 QR factorization revisited

LINEAR ALGEBRA: NUMERICAL METHODS. Version: August 2, 2000 30 3 QR factorization revisited Now we can explain why A = QR factorization is much better when using it to solve Ax = b than the A = LU factorization

### MAT 242 CHAPTER 4: SUBSPACES OF R n

MAT 242 CHAPTER 4: SUBSPACES OF R n JOHN QUIGG 1. Subspaces Recall that R n is the set of n 1 matrices, also called vectors, and satisfies the following properties: x + y = y + x x + (y + z) = (x + y)

### V( x) = V( 0) + dv. V( x) = 1 2

Spectroscopy 1: rotational and vibrational spectra The vibrations of diatomic molecules Molecular vibrations Consider a typical potential energy curve for a diatomic molecule. In regions close to R e (at

### A Moving Least Squares weighting function for the Element-free Galerkin Method which almost fulfills essential boundary conditions

Structural Engineering and Mechanics, Vol. 21, No. 3 (2005) 315-332 315 A Moving Least Squares weighting function for the Element-free Galerkin Method which almost fulfills essential boundary conditions

### 2. Mechanics of Materials: Strain. 3. Hookes's Law

Mechanics of Materials Course: WB3413, Dredging Processes 1 Fundamental Theory Required for Sand, Clay and Rock Cutting 1. Mechanics of Materials: Stress 1. Introduction 2. Plane Stress and Coordinate

### THEME A. Analysis of the elastic behaviour of La Aceña arch-gravity dam

THEME A Analysis of the elastic behaviour of La Aceña arch-gravity dam Gjorgi KOKALANOV, Professor, Faculty of Civil Eng., Skopje, Republic of Macedonia Ljubomir TANČEV, Professor, Faculty of Civil Eng.,

### Chapter 3: Stress and Equilibrium of Deformable Bodies

Ch3-Stress-Equilibrium Page 1 Chapter 3: Stress and Equilibrium of Deformable Bodies When structures / deformable bodies are acted upon by loads, they build up internal forces (stresses) within them to