Introduction to Stiffness Method - Introduction to Finite Elements - Lecture Slides, Slides of Mathematical Methods for Numerical Analysis and Optimization

Download Introduction to Stiffness Method - Introduction to Finite Elements - Lecture Slides and more Slides Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! 1 Introduction to the Stiffness (Displacement) Method: Analysis of a system of springs Reading assignment: Chapter 2: Sections 2.1-2.5 + Lecture notes Summary: • Developing the finite element equations for a system of springs using the “direct stiffness” approach • Application of boundary conditions • Physical significance of the stiffness matrix • Direct assembly of the global stiffness matrix • Problems FEM analysis scheme Step 1: Divide the problem domain into non overlapping regions (“elements”) connected to each other through special points (“nodes”) Step 2: Describe the behavior of each element Step 3: Describe the behavior of the entire body by putting together the behavior of each of the elements (this is a process known as “assembly”) Problem Analyze the behavior of the system composed of the two springs loaded by external forces as shown above k1 k2 F1x F2x F3x x Given F1x , F2x ,F3x are external loads. Positive directions of the forces are along the positive x-axis k1 and k2 are the stiffnesses of the two springs Docsity.com 2 Solution Step 1: In order to analyze the system we break it up into smaller parts, i.e., “elements” connected to each other through “nodes” k1 k2 F1x F2x F3x x k1 k2F1x F2x F3x x 1 2 3 Element 1 Element 2 Node 1 d1x d2x d3x Unknowns: nodal displacements d1x, d2x, d3x, © 2002 Brooks/Cole Publishing / Thomson Learning™ Solution Step 2: Analyze the behavior of a single element (spring) k1 k2F1x F2x F3x x 1 2 3 Element 1 Element 2 Node 1 d1x d2x d3x Two nodes: 1, 2 Nodal displacements: Nodal forces: Spring constant: k 1xd̂ 2xd̂ 1xf̂ 2xf̂ © 2002 Brooks/Cole Publishing / Thomson Learning™ Local ( , , ) and global (x,y,z) coordinate systemsx̂ ŷ ẑ F d F x k k d k 1 Hooke’s Law F = kd Behavior of a linear spring (recap) F = Force in the spring d = deflection of the spring k = “stiffness” of the spring Docsity.com 5 For our original structure with two springs, the global stiffness matrix is ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −+ − = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −+ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = 22 2211 11 k̂ 22 22 k̂ 11 11 kk-0 kkkk- 0kk kk-0 kk0 000 000 0kk- 0kk K )2()1( ee NOTE 1. The global stiffness matrix is symmetric 2. The global stiffness matrix is singular The system equations imply 3x22x23x 3x22x211x12x 2x11x11x 3x 2x 1x 22 2211 11 3x 2x 1x dkd-kF dkd)kk(d-kF dkdkF d d d kk-0 kkkk- 0kk F F F += −++= −= ⇒ ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −+ − = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ These are the 3 equilibrium equations at the 3 nodes. dKF = 0f̂-F:3nodeAt 0f̂f̂-F:2nodeAt 0f̂-F:1nodeAt (2) 2x3x (2) 1x (1) 2x2x (1) 1x1x = =− = k1 k2F1x F2x F3x x 1 2 3 d1x d2x d3x A B C D © 2002 Brooks/Cole Publishing / Thomson Learning™ (1) 1xf̂ (1)2xf̂ (2) 1xf̂ (2)2xf̂2xF1xF 3x F 2 3 ( ) (1)1x2x1x11x f̂ddkF =−= ( ) ( ) (2) 1x (1) 2x 3x2x22x1x1 3x22x211x12x f̂f̂ ddkddk dkd)kk(d-kF += −+−−= −++= ( ) (2)2x3x2x23x f̂dd-kF =−= Notice that the sum of the forces equal zero, i.e., the structure is in static equilibrium. F1x + F2x+ F3x =0 Given the nodal forces, can we solve for the displacements? To obtain unique values of the displacements, at least one of the nodal displacements must be specified. Docsity.com 6 Direct assembly of the global stiffness matrix k1 k2F1x F2x F3x x 1 2 3 Element 1 Element 2d1x d2x d3x Global Element 1 k11 2 (1) 1xd̂ (1) 1xf̂ (1) 2xf̂ (1) 2xd̂ Element 2 k22 3 (2) 1xd̂(2)1xf̂ (2) 2xf̂ (2) 2xd̂ Local Node element connectivity chart : Specifies the global node number corresponding to the local (element) node numbers 322 211 Node 2Node 1ELEMENT Global node number Local node number ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = 11 11)1( kk- k-k k̂ Stiffness matrix of element 1 d1x d2x d2xd1x Stiffness matrix of element 2 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = 22 22)2( kk- k-k k̂ d2x d3x d3xd2x Global stiffness matrix ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ += 22 2211 11 kk-0 k-kkk- 0k-k K d2x d3x d3xd2x d1x d1x Examples: Problems 2.1 and 2.3 of Logan Example 2.1 Compute the global stiffness matrix of the assemblage of springs shown above ( ) ( ) 1000 1000 0 0 1000 1000 2000 2000 0 K 0 2000 2000 3000 3000 0 0 3000 3000 −⎡ ⎤ ⎢ ⎥− + −⎢ ⎥= ⎢ ⎥− + − ⎢ ⎥−⎣ ⎦ d3xd2xd1x d4x d2x d3x d1x d4x © 2002 Brooks/Cole Publishing / Thomson Learning™ 22 3 4 Docsity.com 7 Example 2.3 Compute the global stiffness matrix of the assemblage of springs shown above ( ) ( ) ( ) 1 1 1 1 2 3 2 3 2 3 2 3 k -k 0 K -k k k k - k k 0 - k k k k ⎡ ⎤ ⎢ ⎥= + + +⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ © 2002 Brooks/Cole Publishing / Thomson Learning™ 3 Imposition of boundary conditions Consider 2 cases Case 1: Homogeneous boundary conditions (e.g., d1x=0) Case 2: Nonhomogeneous boundary conditions (e.g., one of the nodal displacements is known to be different from zero) Homogeneous boundary condition at node 1 k1=500N/m k2=100N/m F3x=5N x1 2 3 Element 1 Element 2d1x=0 d2x d3x System equations 1 1 2 3 500 -500 0 -500 600 -100 0 0 -100 100 5 x x x x d F d d ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Note that F1x is the wall reaction which is to be computed as part of the solution and hence is an unknown in the above equation Writing out the equations explicitly 2x 1 2 3 2 3 -500d 600 100 0 100 100 5 x x x x x F d d d d = − = − + = 0 Eq(1) Eq(2) Eq(3) Global Stiffness matrix Nodal disp vector Nodal load vector Eq(2) and (3) are used to find d2x and d3x by solving Note use Eq(1) to compute 1 2x=-500d 5xF N= − 2 3 2 3 600 100 0 100 100 5 0.01 0.06 x x x x d d d m d m − ⎡ ⎤⎡ ⎤ ⎡ ⎤ =⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⇒ =⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ NOTICE: The matrix in the above equation may be obtained from the global stiffness matrix by deleting the first row and column 500 -500 0 -500 600 -100 0 -100 100 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ 600 100 100 100 −⎡ ⎤ ⎢ ⎥−⎣ ⎦ Docsity.com 10 Physical significance of the stiffness matrix The first equation is 1313212111 Fdkdkdk =++ Force equilibrium equation at node 1 What if d1=1, d2=0, d3=0 ? 313 212 111 kF kF kF = = = Force along node 1 due to unit displacement at node 1 Force along node 2 due to unit displacement at node 1 Force along node 3 due to unit displacement at node 1 While nodes 2 and 3 are held fixed Similarly we obtain the physical significance of the other entries of the global stiffness matrix Columns of the global stiffness matrix Physical significance of the stiffness matrix ijk = Force at node ‘i’ due to unit displacement at node ‘j’keeping all the other nodes fixed In general This is an alternate route to generating the global stiffness matrix e.g., to determine the first column of the stiffness matrix Set d1=1, d2=0, d3=0 k1 k2F1 F2 F3 x 1 2 3 Element 1 Element 2d1 d2 d3 Find F1=?, F2=?, F3=? Physical significance of the stiffness matrix For this special case, Element #2 does not have any contribution. Look at the free body diagram of Element #1 x k1 (1) 1xd̂ (1) 1xf̂ (1) 2xf̂ (1) 2xd̂ (1) (1) (1) 2x 1 2x 1x 1 1 ˆ ˆ ˆf (d d ) (0 1)k k k= − = − = − (1) (1) 1x 2x 1 ˆ ˆf f k= − = Physical significance of the stiffness matrix F1 F1 = k1d1 = k1=k11 F2 = -F1 = -k1=k21 F3 = 0 =k31 (1) 1xf̂ Force equilibrium at node 1 (1) 1 1x 1 ˆF =f k= Force equilibrium at node 2 (1) 2xf̂ F2 (1) 2 2x 1 ˆF =f k= − Of course, F3=0 Docsity.com 11 Physical significance of the stiffness matrix Hence the first column of the stiffness matrix is 1 1 2 1 3 0 F k F k F ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪= −⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ To obtain the second column of the stiffness matrix, calculate the nodal reactions at nodes 1, 2 and 3 when d1=0, d2=1, d3=0 1 1 2 1 2 3 2 F k F k k F k −⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪= +⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭ Check that Physical significance of the stiffness matrix To obtain the third column of the stiffness matrix, calculate the nodal reactions at nodes 1, 2 and 3 when d1=0, d2=0, d3=1 1 2 2 3 2 0F F k F k ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪= −⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ Check that Steps in solving a problem Step 1: Write down the node-element connectivity table linking local and global displacements Step 2: Write down the stiffness matrix of each element Step 3: Assemble the element stiffness matrices to form the global stiffness matrix for the entire structure using the node element connectivity table Step 4: Incorporate appropriate boundary conditions Step 5: Solve resulting set of reduced equations for the unknown displacements Step 6: Compute the unknown nodal forces Docsity.com