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1 1D MODELS Logan; chapter 2. 2 What is the relation between forces acting on spring ends and displacements of these ends? Or, using FEA terminology,

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Presentation on theme: "1 1D MODELS Logan; chapter 2. 2 What is the relation between forces acting on spring ends and displacements of these ends? Or, using FEA terminology,"— Presentation transcript:

1 1 1D MODELS Logan; chapter 2

2 2 What is the relation between forces acting on spring ends and displacements of these ends? Or, using FEA terminology, what is the relation between nodal displacements and forces acting on these nodes? To formulate this relation we must first assume displacement pattern in-between nodes. Let’s assume linear displacement: k The displacement function has as many coefficients as the element has degrees of freedom, in this case 2 DEFINITION OF SPRING ELEMENT We assume linear displacement along the length of element. This is our displacement interpolation function. 0 ≤ x ≤ L

3 3 We now want to express displacement u along the element as a function of nodal displacements This is the relation between nodal displacements and displacements along the length of elements (i.e. in-between nodes) DEFINITION OF SPRING ELEMENT

4 4 The same can be written in matrix form: where are called shape functions DEFINITION OF SPRING ELEMENT

5 5 Let’s say that element experiences force T This force causes nodal displacements and Relation between force and nodal displacements Force acting on node 1 Force acting on node 2 This matrix relates nodal displacements of spring element to forces acting on nodes of that element. This matrix is called element stiffness matrix DEFINITION OF SPRING ELEMENT k

6 6 Element stiffness matrix describes relation between nodal loads and nodal displacement of an element. This is matrix is: Square Symmetric Singular (determinant equals zero) Positive Diagonal Terms

7 7 Find displacements of point 2; find reactions in point 1 when force F 2 is acting on node 2 This problem lends itself well toward discretization with one spring element: 1. 2.2. F2F2 ONE SPRING 1. 2.2. F2F2 0 F2F2 F Reaction THIS IS CALLED A BAR ELEMENT

8 8 k1k1 k2k2 Find displacements of point 3; find reactions in points 1 and 2 This problem lends itself well toward discretization with two spring elements: 1.. 3 2. F3F3 Solution steps 1.Formulate stiffness matrix for individual elements 2.Expand individual stiffness matrices so that they are associated with all degrees of freedom in the system 3.Assemble global stiffness matrix 4.Solve for displacements and reaction forces TWO SPRINGS IN SERIES

9 9 Relation between nodal displacements and forces in element 1 Relation between nodal displacements and forces in element 2 TWO SPRINGS IN SERIES or

10 10 Relation between nodal displacements and forces in element 1int he expanded form Relation between nodal displacements and forces in element 2 in the expanded form TWO SPRINGS IN SERIES expanded

11 11 Expanded matrices of elements 1 and 2 can now be added to form the global stiffness matrix Relation between nodal displacements and nodal forces has been formed: Vector of nodal displacements Vector of nodal loads Global stiffness matrix TWO SPRINGS IN SERIES These matrixes been expanded before global stiffness matrix could be assembled. The process in which individual matrices are expanded and then added to form global stiffness matrix is called Direct Stiffness Method also known as the displacement method or matrix stiffness method

12 12 TWO SPRINGS IN SERIES The process in which individual matrices are expanded and then added to form global stiffness matrix is called Direct Stiffness Method also known as the displacement method or matrix stiffness method

13 13 In our case node 1 and node 2 are fixed: This is A set of three equations with three unknowns: d 3, F 1, F 2 TWO SPRINGS IN SERIES

14 14 k1k1 k2k2 k3k3 Find displacements of points 3 and 4; find reactions in points 1 and 2. This problem lends itself well toward discretization with three spring elements. 1..3.34. 2. F1F1 F2F2 THREE SPRINGS IN SERIES Solution steps 1.Formulate stiffness matrix for individual elements 2.Expand individual stiffness matrices so that they are associated with all degrees of freedom in the system 3.Assemble global stiffness matrix 4.Solve for displacements and reaction forces

15 15 Therefore these matrixes must be expanded before global stiffness matrix can be assembled. Note that: THREE SPRINGS IN SERIES

16 16 Stiffness matrix of element 1 before expansion And after expansion THREE SPRINGS IN SERIES k1k1 k2k2 k3k3 1..3.34. 2. F1F1 F2F2

17 17 Stiffness matrix of element 2 before expansion And after expansion THREE SPRINGS IN SERIES k1k1 k2k2 k3k3 1..3.34. 2. F1F1 F2F2

18 18 Stiffness matrix of element 3 before expansion And after expansion THREE SPRINGS IN SERIES k1k1 k2k2 k3k3 1..3.34. 2. F1F1 F2F2

19 19 Stiffness matrices can now be added to assemble global stiffness matrix Therefore, the relation between all nodal displacements and all nodal forces is: Global stiffness matrix Vector of nodal displacements Vector of nodal loads THREE SPRINGS IN SERIES

20 20 This is a set of four linear algebraic equations with fours unknowns: d 3, d 4 nodal displacements f 1, f 3 reaction forces THREE SPRINGS IN SERIES Known nodal displacements (displacements boundary conditions): d 1 =0d 2 =0 Known nodal loads (force boundary conditions): f 3 = - F 1 f 4 = F 2

21 21 THREE SPRINGS IN SERIES/PARALLEL Before expansion: Element 1 Element 2 Element 3

22 22 THREE SPRINGS IN SERIES/PARALLEL And after expansion

23 23 Symmetric This means that k ij = k ji. This is always the case when the displacements are directly proportional to the applied loads. Square The number of rows are equal to the number of columns in the matrix. Singular The element stiffness matrix is singular (the determinate of the matrix is equal to zero) since no constraints (prescribed displacements and/or rotation) have been applied. Positive Diagonal Terms All the terms in the main diagonal (upper left to lower right) must be positive. If k ii is negative then the force and it's corresponding displacement would be oppositely directed, which is physically unreasonable. If k ii = 0, then the displacement would produce no reaction force resisting it, which would imply that the structure is unstable. STIFFNESS MATRIX CHARACTERISTICS

24 24 BETTI’S THEOREM* (RECIPROCITY THEOREM) Betti’s theorem, discovered by Enrico Betti in 1872 states that for all linear elastic structures subject to two sets of forces P i and Q i, the work done by the set P though the displacement produced by set Q is equal to the work done be the set Q through displacements produced by set P. WHY IS STIFFNESS MATRIX SYMMETRIC? *A theorem is a statement which has been proved on the basis of previously established statements, such as other theorems, and previously accepted statements, such as axioms

25 25 1 2 P Q ΔQ1ΔQ1 ΔP1ΔP1 Betti’s theorem example Consider a beam on which two points 1 and 2 have been defined. First we apply force P at point 1 and measure the vertical displacement of point 2. Then we remove force P and apply force Q at point 2. Betti’s reciprocity theorem states that: WHY IS STIFFNESS MATRIX SYMMETRIC?

26 26 WHY IS STIFFNESS MATRIX SQUARE? Number of unknowns equals the number of equations

27 27 WHY IS STIFFNESS MATRIX SINGULAR? Displacement boundary conditions have not yet been defined

28 28 WHY ARE ALL TERMS ON THE MAIN DIAGONAL MUST BE POSITIVE? Negative terms would indicate negative stiffness. Zero terms would indicate no stiffness in given direction.

29 29 12345 Sequential numbering 1243 Element stiffness k=10N/mm Bandwidth = 3

30 30 15342 Out of sequence numbering 1243 Element stiffness k=10N/mm Bandwidth = 5


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