Introduction to the Stiffness (Displacement) Method: Analysis of a system of spr

Introduction to the Stiffness (Displacement) Method: Analysis of a system of spr www.phwiki.com

Introduction to the Stiffness (Displacement) Method: Analysis of a system of spr

Spoon, Doug, Faculty Advisor has reference to this Academic Journal, PHwiki organized this Journal Introduction to the Stiffness (Displacement) Method: Analysis of a system of springs Prof. Suvranu De MANE 4240 & CIVL 4240 Introduction to Finite Elements Reading assignment: Chapter 2: Sections 2.1-2.5 + Lecture notes Summary: Developing the finite element equations as long as 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”)

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Problem Analyze the behavior of the system composed of the two springs loaded by external as long as ces as shown above Given F1x , F2x ,F3x are external loads. Positive directions of the as long as ces are along the positive x-axis k1 in addition to k2 are the stiffnesses of the two springs 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” Unknowns: nodal displacements d1x, d2x, d3x, Solution Step 2: Analyze the behavior of a single element (spring) Two nodes: 1, 2 Nodal displacements: Nodal as long as ces: Spring constant: k

Local ( , , ) in addition to global (x,y,z) coordinate systems 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 Hooke’s law as long as our spring element Eq (1) Force equilibrium as long as our spring element (recap free body diagrams) Eq (2) Collect Eq (1) in addition to (2) in matrix as long as m Element as long as ce vector Element nodal displacement vector Element stiffness matrix

Note The element stiffness matrix is “symmetric”, i.e. The element stiffness matrix is singular, i.e., The consequence is that the matrix is NOT invertible. It is not possible to invert it to obtain the displacements. Why The spring is not constrained in space in addition to hence it can attain multiple positions in space as long as the same nodal as long as ces e.g., Solution Step 3: Now that we have been able to describe the behavior of each spring element, lets try to obtain the behavior of the original structure by assembly Split the original structure into component elements Eq (3) Eq (4) Element 1 Element 2 To assemble these two results into a single description of the response of the entire structure we need to link between the local in addition to global variables. Question 1: How do we relate the local (element) displacements back to the global (structure) displacements Eq (5)

Hence, equations (3) in addition to (4) may be rewritten as Eq (6) Eq (7) Or, we may exp in addition to the matrices in addition to vectors to obtain Exp in addition to ed element stiffness matrix of element 1 (local) Exp in addition to ed nodal as long as ce vector as long as element 1 (local) Nodal load vector as long as the entire structure (global) Question 2: How do we relate the local (element) nodal as long as ces back to the global (structure) as long as ces Draw 5 FBDs In vector as long as m, the nodal as long as ce vector (global) Recall that the exp in addition to ed element as long as ce vectors were Hence, the global as long as ce vector is simply the sum of the exp in addition to ed element nodal as long as ce vectors

But we know the expressions as long as the exp in addition to ed local as long as ce vectors from Eqs (6) in addition to (7) Hence For our original structure with two springs, the global stiffness matrix is NOTE The global stiffness matrix is symmetric The global stiffness matrix is singular The system equations imply These are the 3 equilibrium equations at the 3 nodes.

Notice that the sum of the as long as ces equal zero, i.e., the structure is in static equilibrium. F1x + F2x+ F3x =0 Given the nodal as long as ces, can we solve as long as the displacements To obtain unique values of the displacements, at least one of the nodal displacements must be specified. Direct assembly of the global stiffness matrix Global Element 1 Element 2 Local

Node element connectivity chart : Specifies the global node number corresponding to the local (element) node numbers Global node number Local node number Stiffness matrix of element 1 d1x d2x d2x d1x Stiffness matrix of element 2 Global stiffness matrix d2x d3x d3x d2x d1x d1x Examples: Problems 2.1 in addition to 2.3 of Logan Example 2.1 Compute the global stiffness matrix of the assemblage of springs shown above

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Example 2.3 Compute the global stiffness matrix of the assemblage of springs shown above 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 System equations Note that F1x is the wall reaction which is to be computed as part of the solution in addition to hence is an unknown in the above equation Writing out the equations explicitly 0 Eq(1) Eq(2) Eq(3) Global Stiffness matrix Nodal disp vector Nodal load vector

Eq(2) in addition to (3) are used to find d2x in addition to d3x by solving NOTICE: The matrix in the above equation may be obtained from the global stiffness matrix by deleting the first row in addition to column NOTICE: 1. Take care of homogeneous boundary conditions by deleting the appropriate rows in addition to columns from the global stiffness matrix in addition to solving the reduced set of equations as long as the unknown nodal displacements. 2. Both displacements in addition to as long as ces CANNOT be known at the same node. If the displacement at a node is known, the reaction as long as ce at that node is unknown ( in addition to vice versa) Imposition of boundary conditions contd. Nonhomogeneous boundary condition: spring 2 is pulled at node 3 by 0.06 m) k1=500N/m k2=100N/m x 1 2 3 Element 1 Element 2 d1x=0 d2x d3x=0.06m

Steps in solving a problem Step 1: Write down the node-element connectivity table linking local in addition to global displacements Step 2: Write down the stiffness matrix of each element Step 3: Assemble the element stiffness matrices to as long as m the global stiffness matrix as long as the entire structure using the node element connectivity table Step 4: Incorporate appropriate boundary conditions Step 5: Solve resulting set of reduced equations as long as the unknown displacements Step 6: Compute the unknown nodal as long as ces

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