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Introduction to the Stiffness (Displacement) Method

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Introduction to the Stiffness (Displacement) Method MANE 4240 & CIVL 4240 Reading assignment: Introduction to Finite Elements Chapter 2: Sections 2.1-2.5 + Lecture notes Summary: Introduction to the Stiffness • Developing the finite element equations for a system of (Displacement) Method: springs using the “direct stiffness” approach • Application of boundary conditions Analysis of a system of springs • Physical significance of the stiffness matrix • Direct assembly of the global stiffness matrix Prof. Suvranu De • Problems FEM analysis scheme F1x F F 2x 3x x Step 1: Divide the problem domain into non overlapping regions (“elements”) connected to each other through special points k1 k2 (“nodes”) Step 2: Describe the behavior of each element Problem Analyze the behavior of the system composed of the two springs Step 3: Describe the behavior of the entire body by putting loaded by external forces as shown above together the behavior of each of the elements (this is a process known as “assembly”) 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 1 F1x k F2x k2 F3x F F F 1 x 1x 2x 3x x 1 2 3 Element 2 k k Element 1 1 2 d1x d2x d3x Solution Node 1 Step 1: In order to analyze the system we break it up into smaller Solution parts, i.e., “elements” connected to each other through “nodes” Step 2: Analyze the behavior of a single element (spring) F1x k F2x k2 F3x 1 x 1 2 3 Element 2 Element 1 © 2002 Brooks/Cole Publishing / Thomson Learning™ d1x d2x d3x Node 1 Two nodes: 1, 2 ˆ ˆ Unknowns: nodal displacements d , d ,d , Nodal displacements: d1x d2x 1x 2x 3x ˆ ˆ Nodal forces: f1x f2x Spring constant: k Behavior of a linear spring (recap) F x k k 1 F d k d F = Force in the spring © 2002 Brooks/Cole Publishing / Thomson Learning™ d = deflection of the spring Local (xˆ , yˆ , zˆ ) and global (x,y,z) coordinate systems k = “stiffness” of the spring Hooke’s Law F = kd 2 Note T fˆ ˆ 1. The element stiffness matrix is “symmetric”, i.e. kˆ = kˆ 1x f2x 2. The element stiffness matrix is singular, i.e., © 2002 Brooks/Cole Publishing / Thomson Learning™ ˆ 2 2 Hooke’s law for our spring element det (k ) = k − k = 0 ˆ ˆ ˆ f2x = k (d2x − d1x ) Eq (1) The consequence is that the matrix is NOT invertible. It is not Force equilibrium for our spring element (recap free body diagrams) possible to invert it to obtain the displacements. Why? ˆ ˆ The spring is not constrained in space and hence it can attain f1x + f2x = 0 ˆ ˆ ˆ ˆ Eq (2) multiple positions in space for the same nodal forces ⇒ f1x = −f2x = −k (d2x − d1x ) e.g., ⎧ˆ ⎫ Collect Eq (1) and (2) in matrix form ⎪f1x ⎪ ⎡ 2 - 2⎤⎧1⎫ ⎧- 2⎫ ⎧fˆ ⎫ ⎡ k - k⎤⎧dˆ ⎫ ⎨ ⎬ = ⎢ ⎥⎨ ⎬ = ⎨ ⎬ ˆ ˆ ˆ ⎪ 1x ⎪ ⎪ 1x ⎪ ⎪ˆ ⎪ - 2 2 2 2 f = k d ⎨ ⎬ = ⎢ ⎥⎨ ⎬ ⎩f2x ⎭ ⎣ ⎦⎩ ⎭ ⎩ ⎭ ⎪fˆ ⎪ - k k ⎪dˆ ⎪ ⎩ 2x ⎭ ⎣ ⎦⎩ 2x ⎭ Element force Element nodal ⎪⎧fˆ ⎪⎫ ⎡ 2 - 2⎤⎧3⎫ ⎧- 2⎫ Element ˆ kˆ ˆ 1x vector displacement f d ⎨ ⎬ = ⎢ ⎥⎨ ⎬ = ⎨ ⎬ stiffness ˆ - 2 2 4 2 vector ⎩⎪f2x ⎭⎪ ⎣ ⎦⎩ ⎭ ⎩ ⎭ matrix Solution To assemble these two results into a single description of the Step 3: Now that we have been able to describe the behavior of response of the entire structure we need to link between the local each spring element, lets try to obtain the behavior of the original and global variables. structure by assembly Question 1: How do we relate the local (element) displacements Split the original structure into component elements back to the global (structure) displacements? Element 2 F1x k F2x k2 F3x Element 1 1 x 12k1 23k 2 1 2 3 Element 1 Element 2 ˆ (1) (1) ˆ (1) ˆ (1) (2) (2) ˆ (2) ˆ (2) f ˆ f d ˆ ˆ f d d1x d2x d3x 1x d1x 2x 2x f1x d1x 2x 2x Node 1 (1) (1) ⎧fˆ ⎫ ⎡ k - k ⎤⎧dˆ ⎫ ˆ (2) ˆ (2) (1) ⎪ 1x ⎪ 1 1 ⎪ 1x ⎪ ⎪⎧f ⎪⎫ ⎡ k 2 - k 2 ⎤⎪⎧d ⎪⎫ dˆ = d ⎨ ⎬ = ⎢ ⎥⎨ ⎬ 1x = 1x 1x 1x ˆ (1) ˆ (1) ⎨ (2) ⎬ ⎢ ⎥⎨ (2) ⎬ Eq (5) ⎪f ⎪ ⎣- k1 k1 ⎦⎪d ⎪ ˆ - k k ˆ (1) (2) ⎩ 2x ⎭ ⎩ 2x ⎭ ⎩⎪f2x ⎭⎪ ⎣ 2 2 ⎦⎩⎪d2x ⎭⎪ ˆ ˆ d2x = d1x = d2x (1) (1) (1) ˆ kˆ ˆ ( 2) ˆ ( 2) ( 2) f d fˆ k dˆ ˆ (2) d2x = d3x Eq (3) Eq (4) 3 Hence, equations (3) and (4) may be rewritten as Question 2: How do we relate the local (element) nodal forces back to the global (structure) forces? Draw 5 FBDs (1) F1x k F k2 F ⎧ˆ ⎫ ⎡ k - k ⎤⎧d ⎫ ˆ (2) 1 2x 3x ⎪f1x ⎪ 1 1 1x ⎪⎧f ⎪⎫ ⎡ k 2 - k 2 ⎤⎧d2x ⎫ x ⎨ ⎬ = ⎨ ⎬ 1x = ˆ (1) ⎢ ⎥ ⎨ (2) ⎬ ⎢ ⎥⎨ ⎬ ⎪f ⎪ - k1 k1 d2x ˆ - k k d D 3 ⎩ 2x ⎭ ⎣ ⎦⎩ ⎭ ⎩⎪f2x ⎭⎪ ⎣ 2 2 ⎦⎩ 3x ⎭ 1 A BC2 d d d3x Or, we may expand the matrices and vectors to obtain 1x 2x 2 3 ˆ (1) ⎧ 0 ⎫ ⎡0 0 0 ⎤ ⎧f ⎫ ⎡ k1 − k1 0⎤⎧d ⎫ ⎧d1x ⎫ ⎪ 1x ⎪ ⎪ 1x ⎪ ⎪ˆ (2) ⎪ ⎢ ⎥⎪ ⎪ (1) ⎢ ⎥ f1x d (1) (2) ˆ = - k k 0 d2x ⎨ ⎬ = 0 k 2 − k 2 ⎨ 2x ⎬ fˆ ˆ (1) fˆ ˆ (2) ⎨f2x ⎬ 1 1 ⎨ ⎬ ⎢ ⎥ F1x 1x f2x 1x f 2x F3x ⎢ ⎥ F2x ⎪ ⎪ ⎪ ⎪ ⎪fˆ (2) ⎪ ⎢0 - k k ⎥⎪d ⎪ © 2002 Brooks/Cole Publishing / Thomson Learning™ 0 ⎢ 0 0 0⎥⎩d3x ⎭ ⎩ 2x ⎭ ⎣ 2 2 ⎦⎩ 3 x ⎭ ⎩ ⎭ ⎣ ⎦ ˆ (1) ( 2) e ( 2)e d At node1: F - f = 0 (1)e (1)e d fˆ ˆ 1x 1x fˆ kˆ k ˆ (1) ˆ (2) Eq (6) Eq (7) At node 2 : F2x - f2x − f1x = 0 (1)e kˆ Expanded element stiffness matrix of element 1 (local) At node 3: F - fˆ (2) = 0 (1)e 3x 2x fˆ Expanded nodal force vector for element 1 (local) d Nodal load vector for the entire structure (global) In vector form, the nodal force vector (global) But we know the expressions for the expanded local force vectors ⎧F ⎫ ⎧ fˆ (1) ⎫ from Eqs (6) and (7) 1x ⎪ 1x ⎪ ⎪ ⎪ ˆ (1) ˆ (2) (1)e (1)e (2)e (2)e F = ⎨F2x ⎬ = ⎨f2x + f1x ⎬ fˆ = kˆ d and fˆ = kˆ d ⎪F ⎪ ⎪ fˆ (2) ⎪ ⎩ 3x ⎭ ⎩ 2x ⎭ Hence (1)e (2)e (1)e (2)e (1)e (2)e Recall that the expanded element force vectors were F = fˆ + fˆ = kˆ d + kˆ d = ⎜⎛kˆ + kˆ ⎟⎞d ⎝ ⎠ ⎧ˆ (1) ⎫ f1x ⎧ 0 ⎫ (1)e (2)e (2) ⎪ (1) ⎪ ⎪ˆ ⎪ ˆ ˆ ˆ f1x F = K d f = ⎨f2x ⎬ and f = ⎨ ⎬ ⎪ ⎪ ⎪fˆ (2) ⎪ ⎩0 ⎭ ⎩ 2x ⎭ F = Global nodal force vector Hence, the global force vector is simply the sum of the expanded element nodal force vectors d = Global nodal displacement vector ⎧F1x ⎫ ⎪ ⎪ ˆ (1)e ˆ (2)e K = Global stiffness matrix F = ⎨F2x ⎬ = f + f ⎪ ⎪ = sum of expanded element stiffness matrices ⎩F3x ⎭ 4 For our original structure with two springs, the global stiffness The system equations F = K d imply matrix is ⎧F1x ⎫ ⎡ k1 − k1 0 ⎤ ⎧d1x ⎫ ⎡ k1 − k1 0⎤ ⎡0 0 0 ⎤ ⎪ ⎪ ⎪ ⎪ F = ⎢- k k + k − k ⎥ d K = ⎢- k k 0⎥ + ⎢0 k − k ⎥ ⎨ 2x ⎬ ⎢ 1 1 2 2 ⎥ ⎨ 2x ⎬ ⎢ 1 1 ⎥ ⎢ 2 2 ⎥ ⎪F ⎪ ⎢ 0 - k k ⎥ ⎪d ⎪ ⎢ 0 0 0⎥ ⎢0 - k k ⎥ ⎩ 3x ⎭ ⎣ 2 2 ⎦ ⎩ 3x ⎭ ⎣ ⎦ ⎣ 2 2 ⎦ (1)e (2)e kˆ kˆ F = k d − k d ⎡ k − k 0 ⎤ 1x 1 1x 1 2x 1 1 ⇒ F = -k d + (k + k )d − k d = ⎢- k k + k − k ⎥ 2x 1 1x 1 2 2x 2 3x ⎢ 1 1 2 2 ⎥ F3x = -k2d2x + k 2d3x ⎣⎢ 0 - k 2 k 2 ⎦⎥ NOTE These are the 3 equilibrium equations at the 3 nodes. 1. The global stiffness matrix is symmetric 2. The global stiffness matrix is singular F1x k F k2 F 1 2x 3x x 3 1 A BC2 D Notice that the sum of the forces equal zero, i.e., the structure is in d d1x d2x 3x static equilibrium. 2 3 F1x + F2x+ F3x =0 fˆ (1) ˆ (1) ˆ (2) ˆ (2) F1x 1x f2x f1x f 2x F3x F2x Given the nodal forces, can we solve for the displacements? © 2002 Brooks/Cole Publishing / Thomson Learning™ ˆ (1) ˆ (1) F1x = k1()d1x − d2x = f1x To obtain unique values of the displacements, at least one of the At node1: F1x - f1x = 0 F = -k d + (k + k )d − k d nodal displacements must be specified. ˆ (1) ˆ (2) 2x 1 1x 1 2 2x 2 3x At node 2 : F2x - f2x − f1x = 0 = − k1()()d1x − d2x + k 2 d2x − d3x At node 3: F - fˆ (2) = 0 3x 2x ˆ (1) ˆ (2) = f2x + f1x ˆ (2) F3x = -k2 ()d2x − d3x = f2x 5 Direct assembly of the global stiffness matrix Node element connectivity chart : Specifies the global node number corresponding to the local (element) node numbers Global F1x k F2x k2 F3x ELEMENT Node 1 Node 2 Local node number 1 x 1 1 2 1 2 3 Global node number Element 1 Element 2 d d d 1x 2x 3x 2 2 3 Local Element 2 Element 1 12k1 23k2 (1) fˆ (1) ˆ (1) fˆ (1)dˆ ˆ (2) ˆ (2) fˆ (2)dˆ (2) 1x d1x 2x 2x f1x d1x 2x 2x Stiffness matrix of element 1 Stiffness matrix of element 2 Example 2.1 d d1x d2x d2x 3x (1) k - k d (2) ⎡ k - k ⎤ d ˆ ⎡ 1 1 ⎤ 1x ˆ 2 2 2x k = ⎢ ⎥ k = ⎢ ⎥ 22 3 4 - k k - k 2 k 2 ⎣ 1 1 ⎦ d2x ⎣ ⎦ d3x © 2002 Brooks/Cole Publishing / Thomson Learning™ Global stiffness matrix Compute the global stiffness matrix of the assemblage of d d d1x 2x 3x springs shown above d d d d ⎡ k1 - k1 0 ⎤ d1x 1x 2x 3x 4x ⎢ ⎥ ⎡ 1000− 1000 0 0 ⎤ d1x K = - k1 k1 + k 2 - k 2 d2x ⎢ ⎥ ⎢−+−1000() 1000 2000 2000 0 ⎥ d ⎢ 0 - k k ⎥ d K=⎢ ⎥ 2x ⎣ 2 2 ⎦ 3x ⎢ 0−+− 2000() 2000 3000 3000⎥ d ⎢ ⎥ 3x Examples: Problems 2.1 and 2.3 of Logan 0 0− 3000 3000 ⎣ ⎦ d4x 6 Imposition of boundary conditions Example 2.3 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) 3 Homogeneous boundary condition at node 1 © 2002 Brooks/Cole Publishing / Thomson Learning™ k =100N/m F =5N k1=500N/m 2 3x Compute the global stiffness matrix of the assemblage of 1 x springs shown above 2 3 Element 1 Element 2 d =0 d d ⎡⎤k-k011 1x 2x 3x ⎢⎥ K-kkkk-kk =+++⎢⎥1123() 23 ⎢⎥ ⎣⎦0 -kk()()23++ kk 23 System equations Eq(2) and (3) are used to find d2x and d3x by solving 0 ⎡ 600− 100⎤⎡⎤⎡⎤d2x 0 ⎡⎤⎡⎤⎡⎤500 -500 0 dF11xx ⎢ ⎥⎢⎥⎢⎥= ⎢⎥⎢⎥⎢⎥ ⎣−100 100⎦⎣⎦⎣⎦d3x 5 -500 600 -100d2x = 0 ⎢⎥⎢⎥⎢⎥ ⎡⎤d ⎡⎤0.01 m ⎢⎥⎢⎥⎢⎥ 2x ⎣⎦⎣⎦⎣⎦0 -100 100d3x 5 ⇒=⎢⎥⎢⎥ ⎣⎦d3x ⎣⎦0.06 m Global Stiffness Nodal Nodal disp load matrix NOTICE: The matrix in the above equation may be obtained from vector vector the global stiffness matrix by deleting the first row and column Note that F1x is the wall reaction which is to be computed as part 500 -500 0 of the solution and hence is an unknown in the above equation ⎡ ⎤ ⎡⎤600− 100 ⎢-500 600 -100⎥ ⎢⎥ Writing out the equations explicitly ⎢ ⎥ ⎣⎦−100 100 ⎢ 0 -100 100 ⎥ -500d2x= F 1x Eq(1) ⎣ ⎦ 600dd23xx−= 100 0 Eq(2) Note use Eq(1) to compute F12xx =-500d=− 5N −+=100dd23xx 100 5 Eq(3) 7 Imposition of boundary conditions…contd.
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