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. NOTICE: Nonhomogeneous boundary condition: spring 2 is pulled at 1. Take care of homogeneous boundary conditions node 3 by 0.06 m) k =500N/m k2=100N/m by deleting the appropriate rows and columns from the 1 x global stiffness matrix and solving the reduced set of 1 equations for the unknown nodal displacements. 2 3 Element 1 Element 2 d =0 d d =0.06m 2. Both displacements and forces CANNOT be known at 1x 2x 3x the same node. If the displacement at a node is known, the reaction force at that node is unknown (and vice versa)

System equations Now use only equation (2) to compute d2x 0

⎡⎤⎡⎤⎡⎤500 -500 0 dF11xx 600d2x = 100(0.06) ⎢⎥⎢⎥⎢⎥-500 600 -100d = 0 ⇒=dm0.01 ⎢⎥⎢⎥⎢⎥2x 2x ⎣⎦⎣⎦⎣⎦⎢⎥⎢⎥⎢⎥0 -100 100 dF33xx 0.06 Note that now F1x and F3x are not known. Now use Eq(1) and (3) to compute F1x =-5N and F3x=5N

Writing out the equations explicitly

-500d2x= F 1x Eq(1) 600d −= 100(0.06) 0 2x Eq(2) −+100dF 100(0.06) = 23xxEq(3)

8 Recap of what we did…contd. Recap of what we did 2. Relate the local forces of each element to the global forces at Step 1: Divide the problem domain into non overlapping regions the nodes (use FBDs and force equilibrium). e (“elements”) connected to each other through special points F = ∑fˆ (“nodes”) Element Finally obtain nodal Step 2: Describe the behavior of each element ( f ˆ = kˆ d ˆ ) displacement F = K d vector

Step 3: Describe the behavior of the entire body (by “assembly”). Where the global stiffness matrix This consists of the following steps K = ∑ ke 1. Write the force-displacement relations of each spring in expanded form Global e ˆ ˆ e ˆ nodal f = k d displacement vector

Recap of what we did…contd. Physical significance of the stiffness matrix Apply boundary conditions by partitioning the matrix and vectors F1x k F k2 F 1 2x 3x x ⎡K K ⎤⎧d ⎫ ⎧F ⎫ 1 2 3 11 12 1 = 1 Element 2 ⎢ ⎥⎨ ⎬ ⎨ ⎬ Element 1 d ⎣K21 K22⎦⎩d2⎭ ⎩F2⎭ d1x d2x 3x In general, we will have a ⎡k11 k12 k13 ⎤ stiffness matrix of the form Solve for unknown nodal displacements K = ⎢k k k ⎥ (assume for now that we do not ⎢ 21 22 23 ⎥ know k , k , etc) ⎢k k k ⎥ K22d2 =F2 −K21d1 11 12 ⎣ 31 32 33 ⎦ Compute unknown nodal forces The finite element ⎡k11 k12 k13 ⎤⎧d1 ⎫ ⎧F1 ⎫ F =K d +K d ⎪ ⎪ ⎪ ⎪ 1 11 1 12 2 ⎢k k k ⎥ d = F force-displacement ⎢ 21 22 23 ⎥⎨ 2 ⎬ ⎨ 2 ⎬ relations: ⎪ ⎪ ⎪ ⎪ ⎣⎢k31 k32 k33 ⎦⎥⎩d3 ⎭ ⎩F3 ⎭

9 Physical significance of the stiffness matrix Physical significance of the stiffness matrix

The first equation is In general Force equilibrium k d + k d + k d = F 11 1 12 2 13 3 1 equation at node 1 k = Force at node ‘i’ due to unit displacement at node ‘j’ ij keeping all the other nodes fixed Columns of the global stiffness matrix This is an alternate route to generating the global stiffness matrix e.g., to determine the first column of the stiffness matrix What if d1=1, d2=0, d3=0 ? While nodes 2 and 3 are held fixed Set d1=1, d2=0, d3=0 F1 = k11 Force along node 1 due to unit displacement at node 1 F1 k F2 k2 F3 1 x F2 = k 21 Force along node 2 due to unit displacement at node 1 1 2 3 F3 = k31 Force along node 3 due to unit displacement at node 1 Element 1 Element 2 d1 d2 d3 Similarly we obtain the physical significance of the other Find F =?, F =?, F =? entries of the global stiffness matrix 1 2 3

Physical significance of the stiffness matrix Physical significance of the stiffness matrix Force equilibrium at node 1 For this special case, Element #2 does not have any contribution. F1 ˆ (1) Look at the free body diagram of Element #1 F=f11x1= k (1) (1) ˆ (1) fˆ dˆ d2x 1x 1x x Force equilibrium at node 2

(1) (1) F ˆ k1 ˆ 2 f1x f2x ˆ (1) F=f22x= −k 1 (1) (1) (1) (1) ˆ ˆˆˆ f2x F = k d = k =k f(dd)(01)2x=−=−=−kkk 1 2x 1x 1 1 1 1 1 1 11

Of course, F3=0 F2 = -F1 = -k1=k21 ˆˆ(1) (1) ff1x=− 2x =k 1 F3 = 0 =k31

10 Physical significance of the stiffness matrix Physical significance of the stiffness matrix Hence the first column of the stiffness matrix is To obtain the third column of the stiffness matrix, calculate the ⎧⎫⎧F11k ⎫ nodal reactions at nodes 1, 2 and 3 when d =0, d =0, d =1 ⎪⎪⎪ ⎪ 1 2 3 ⎨⎬⎨F21=−k ⎬ ⎪⎪⎪ ⎪ Check that ⎩⎭⎩F3 0 ⎭ To obtain the second column of the stiffness matrix, calculate the ⎧F1 ⎫⎧0 ⎫ ⎪ ⎪⎪ ⎪ nodal reactions at nodes 1, 2 and 3 when d1=0, d2=1, d3=0 ⎨F22⎬⎨=−k ⎬ ⎪ ⎪⎪ ⎪ Check that ⎩⎭⎩F32k ⎭

⎧⎫⎧Fk11− ⎫ ⎪⎪⎪ ⎪ ⎨⎬⎨F212=+kk ⎬ ⎪⎪⎪ ⎪ ⎩⎭⎩Fk32− ⎭

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

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