# Stiffness Matrices Spring and Bar Elements Lecture Notes

Stiffness Matrices, Spring and Bar Elements Lecture Notes Dr. Rakhmad Arief Siregar Universiti Malaysia Perlis 1

Introduction n The primary characteristics of a finite element are embodied in the element stiffness matrix The stiffness matrix contains geometric and material behaviour information for structural finite element that indicates the resistance of the element to deformation when subjected to loading. In this chapter we will discuses the finite element characteristics of two relatively simple onedimensional structural elements, a linearly elastic spring. 2

Linear spring as a finite element n n n A linear elastic spring is a mechanical device capable of supporting axial loading only, over a reasonable operating range Here the elongation or contraction of spring is directly proportional to the applied axial loading The constant of proportional between deformation and load is referred to as the spring constant, spring rate, or spring stiffness, (k), the unit is force per unit length 3

Linear spring as a finite element n n n Elastic spring supports axial loading only Local coordinate system as an x-axis oriented along the length of the spring u 1 and u 2 are nodal displacement for node 1 and node 2, respectively 4

Linear spring as a finite element n n n The total elongation of the spring is known as the net force of the spring. Assuming both nodal displacements are zero the spring is undeformed the net spring is given by: = u 2 - u 1 The resultant axial force in the spring is: f = k =k (u 2 - u 1) 5

Linear spring as a finite element n For equilibrium f 1 + f 2 = 0 or f 1 = -f 2 then; f 1 =- k (u 2 - u 1) f 2 = k ( u 2 - u 1 ) n or in the form of matrix: 6

Linear spring as a finite element n or in the form of [ ke ] { u } = { f } n where: n ke is defined as element stiffness matrix 7

Linear spring as a finite element n n In general case, the forces are prescribed and the objective is to solve the unknown nodal displacements. The solution is represented by: 8

System Assembly in Global Coordinates n n n Derivation of the element stiffness matrix for a spring element was based on equilibrium condition The same procedure can be applied to a connected system of spring elements by writing the equilibrium equation for EACH NODE The process as described as “assembly” as we take individual stiffness components and “put them together” to obtain system equations 9

System Assembly in Global Coordinates System of two springs with node numbers, element numbers, nodal displacements and nodal forces n n n It is assumed that the springs have different spring constants k 1 and k 2 The global nodes number are node 1, 2, 3 The global nodal displacements are U 1, U 2, and U 3 10

Free body diagrams 11

Free body diagrams n The equilibrium condition for each spring: 12

System Assembly in Global Coordinates n To assemble, use displacement compatibility conditions which relate element displacements to system displacements: 13

System Assembly in Global Coordinates n n The compatibility conditions state the physical fact that the springs are connected at node 2, Remain connected at node 2 after deformation and must have the same nodal displacement 14

System Assembly in Global Coordinates n We expand both matrix equation to 3 x 3 so that the displacement vector can be in the same condition 15

System Assembly in Global Coordinates n The addition of both 3 x 3 matrix will yields as: 16

System Assembly in Global Coordinates n Referring to above condition, we can simplify the load as: 17

System Assembly in Global Coordinates n which is in the form of [K ] { U } = { F } 18

System Assembly in Global Coordinates n Where the system stiffness matrix [ K ] is: 19

Example 2. 1 Consider the two element system depicted in fig below given that Node 1 is attached to a fixed support, yielding the displacement constraint U 1 = 0, k 1 = 50 lb/in, k 2 = 75 lb/in F 2 = F 3 = 75 lb. Determine U 2 and U 3 F 2 = - 50 lb 20

2 spring bar elements n Referring to above condition, we can simplify the load as: 21

Solution n System assembly in global coordinates 22

Solution n The first algebraic equation: -50 = F 1 n The second and third equations: 23

Steps n n n Formulate the individual element stiffness matrices Write the element to global displacement relations Assemble the global equilibrium equation in matrix form Reduce the matrix equation for the unknown nodel displacements (primary variables) Solve for the reaction forces (secondary variable) by back substitution 24

Problem 32. 11 Perform the same computation as in Sec. 32. 4 but change the force to 1. 5 and spring constants to Spring 1 2 3 4 k 0. 75 1. 5 0. 5 2 25

4 spring bar elements U 4 3 U 5 4 4 5 26

Solution 32. 11 Spring 1 2 3 4 k 0. 75 1. 5 0. 5 2 27

Solution 32. 11 n Answer x 1 = 2, x 2 = 3, x 3 = 6, and x 4 = 6. 75. 28

Problem 32. 12 Perform the same computation as in Sec. 32. 4 but change the force to 2 and spring constants to Spring 1 2 3 4 5 k 0. 25 0. 5 1. 5 0. 75 1 29

32. 9 Spring 1 2 3 4 5 k 0. 25 0. 5 1. 5 0. 75 1 30

Solution n for x 1 = 8, x 2 = 12, x 3 = 13. 33333, x 4 = 16, and x 5 = 18. 31

Elastic Bar Element n n n Applications of spring are used in many machineries. Spring element is also often used to represent the elastic nature of supports for more complicated systems. A more generally applicable element is an elastic bar subjected to axial forces only. 32

Elastic Bar Element n The formulation of the finite element characteristics of an elastic bar element is based on the following assumptions: n n The bar is geometrically straight The material obeys Hooke’s law Forces are applied only at the ends of the bar The bar supports axial loading only: bending, torsion and shear are not transmitted to the element via nature of its connections to other element 33

Spring element n Do you still remember the equilibrium condition for spring element? 34

Spring & Bar element n Do you still remember the equilibrium condition for spring element? 35

Bar element n By utilizing Hook’s law: Spring constant 36

Problem 2. 11 n Figure below depicts an assembly of two bar elements made of different materials. Determine the nodal displacements, elements stresses and the reaction force. 37

Solution 2. 11 n Element equations are 38

Solution 2. 11 n Global stiffness matrix 39

Solution 2. 11 n n Global stiffness matrix You may found the nodal displacement and reaction force by solving this matrix 40

Solution 2. 11 n Answer: U 2= 6. 7 x 10 -3 in U 3= 24. 5 x 103 in F 1= -20 x 103 lb n How to calculate the stresses? 41

Solution 2. 11 n Calculation of stress by Hook’s law 42

Solution 2. 11 U 2= 6. 7 x 10 -3 in U 3= 24. 5 x 10 -3 in F 1= 20 x 103 lb 43

Solution 2. 11 n Answer 44

n End 45

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