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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 1 Lecture 6; The Finite Element Method 1-dimensional spring systems

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 2 With a 1-dimensional spring system we here mean springs in series where the individual springs are linear, i.e. expose linear force-elongation relations Note that the identity of a spring is indicated by a ring-symbol, while its stiffness is referred to as k

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 3 Why study (linear) spring systems? You are familiar with springs, since you have met them in a number of courses By using what you already know (e.g. equilibrium), we may find a matrix formulation which has the same structure as an FE-formulation for a general elastostatic problem of Solid Mechanics

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 4 Example problem Let us first, in order to have something to compare with, in the "traditional" way determine the spring forces and spring elongations in the example below. We will later on solve the same problem by a structured FE-treatment.

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 5 Example problem; cont. In order to solve our problem, we need to look at equilibrium, constitutive relations and compatibility. First, make free body diagrams of the nodes/connection points 2 and 3, resp., and study equilibrium (T denotes a spring force) A statically indeterminate problem (2 eq. & 3 unknowns)

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 6 Example problem; cont. The constitution tells us (linear springs) where δ denotes the spring elongation. We now have 5 eq. for 6 unknowns, which we fix by compatibility Let us now do some calculations (see the next page)

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 7 Example problem; cont. By equilibrium With the left expressions inserted into the right one we get By constitution and compatibility The elongations are finally found by inserting the obtained spring forces into the constitutive relations (only a scaling with k)

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 8 Structured analysis of spring systems Let us now focus on a general structured FE approach for spring systems, which contains the following steps I.Find, for each spring, a relation between the spring loading and the spring displacements II.Find, for the spring structure, a relation between the structural loading and the structural displacements III.Solve the structural problem for the given loading and restraints, and calculate entities such as spring forces and spring elongations Let us now see how this is done!

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 9 Step I Find, for each spring, a relation between spring loading and the spring displacements

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 10 Spring equilibrium Let us now study equilibrium for an arbitrary spring in a spring system as illustrated below, where the index e refers to the element number, where the spring nodes/end points are labeled n1e and n2e, resp., and where the spring loading is described by the forces f1e and f2e, resp. Note that the forces are defined positive in the positive x-direction!

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 11 Spring equilibrium; cont. Let us make an imaginary cut in the spring, insert the spring force Te, and study equilibrium for each part of the spring On matrix form we thus have

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 12 Spring equilibrium; cont. Let us now proceed and look at the spring deformation, where the elongation (as before) is labeled δe, while the displacement of its end nodes are called d1e and d2e; resp. Obviously or, on matrix form

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 13 Spring equilibrium; cont. Since the springs are linear, we have

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 14 Transformation diagram We thus have got the following, so called, transformation diagram spring loading or element forces spring displacements or element displacements

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 15 Transformation diagram; cont. A counter-clockwise trip in the transf. diag. will give us the so called spring stiffness [k]e, which relates the element displacements to the element forces

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 16 Transformation diagram; cont. The complete transf. diag. for the spring thus takes the form

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 17 Step II Find, for the spring structure, the relation between the structual loading and the structural displacements

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 18 Structural loading and structural displacements For the spring structure, we introduce structural nodes, with associated structural loads and structural displacements, according to the example below

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 19 Structural loading and structural displacements; cont. We introduce the structural load matrix {F} and the structural displacement matrix {D} as illustrated below

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 20 The structural problem What we now seek is a relation between the structural load matrix {F} and the structural displacement matrix {D}, which will take the form Note that we here use lower case letters for spring related quantities, and capital letters for structural quantities.

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 21 The structural problem; cont. In the transformation diagram we thus have to find

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 22 The relation between spring disp. and structural disp. We fix this by so called connectivity matrices (unique for each spring). For spring 1 we get

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 23 The relation between spring disp. and structural disp.; cont. For spring 2 we get

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 24 The relation between spring disp. and structural disp.; cont. For spring 3 we get

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 25 The relation between spring disp. and structural disp.; cont. Thus,

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 26 The relation between spring loads and structural loads Let us now continue, and study the relation between spring loads and structural loads. If looking closely at the structural nodes, and using Newtons 3:rd law and the previously defined spring loads, we get the following situation Equilibrium implies

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 27 The relation between spring loads and structural loads; cont. Thus or or, actually

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 28 The relation between spring loads and structural loads; cont. Thus

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 29 The structural problem By an anti-clockwise trip in the transf. diag. we get

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 30 The structural problem; cont. external loads internal loads

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 31 Step III Solve the structural problem for given loading and restraints, and calculate entities such as spring forces and spring elongations Let us see what this looks like in our example!

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 32 Example problem Find the spring forces and spring elongations in the example below With the connectivity matrices previously found, we get (see the next page)

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 33 Example problem; cont.

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 34 We thus have to solve Example problem; cont. = F = 0 = - F = ?

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 35 The unknown displacements D2 and D3 are now found by solving the equation system one gets when eliminating all rows and columns associated with locked displacements Example problem; cont. = F = 0 = - F = ? implying

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 36 The spring elongations are then found by (c.f. the transformation diagram) Example problem; cont. which for instance gives Finally, the spring forces are found by (c.f. the transformation diagram) Implying e.g. that As before :)

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 37 The way used for finding the structural problem described above (assembling the spring stiffnesses by using connectivity matrices) constitutes the general structured way to describe FEM. In a hand calculation situation, it is often easier to use the procedure described below (not using connectivity matrices). Also when considering implementation in a computer code it is generally too expensive to directly work with connectivity matrices (both with respect to computational time and memory allocation). Instead one saves the positions of the non-zero elements in the connectivity matrices. Example problem; cont.

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 38 The contribution to [K] from element 1, which we here call [K]1, is Example problem; cont. = D1 = D2 The “direct identification” method

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 39 The contribution to [K] from element 2, which we here call [K]2, is Example problem; cont. = D2 = D3 The “direct identification” method; cont.

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 40 The contribution to [K] from element 3, which we here call [K]3, is Example problem; cont. = D3 = D4 The “direct identification” method; cont.

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 41 The resulting structural stiffness is found by adding all contributions (giving the same result as before) Example problem; cont. The “direct identification” method; cont.

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 42 After having solved the structural problem, we proceed in the following way when calculating the spring forces (note that we in the “direct identification” procedure have not set up any connectivity matrices). Example problem; cont. The “direct identification” method; cont.

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 43 As an example, we get for element 1 Example problem; cont. The “direct identification” method; cont. = k = D2 = D1 T1 As before :)

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 44 An alternative treatment of our example problem by taking advantage of the symmetry Decreasing the length by two, increases the stiffness by two!

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 45 The contribution to [K] from element 1, which we here call [K]1, is Example problem; cont. = D1 = D2

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 46 The contribution to [K] from element 2, which we here call [K]2, is Example problem; cont. = D2 = D3

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 47 We thus have to solve Example problem; cont. = F = 0 = ? = 0 and gets = F As before :)

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Lecture 6; The Finite Element Method 1-dimensional spring systems (modified ) 48 The spring stiffness matrix [k]e and the structural stiffness matrix [K] are symmetric and singular (can not be inverted). However, if we have prevented rigid body motions, we may solve the structural problem (the reduced stiffness matrix obtained by removing rows and columns associated with locked displacements is not singular). One may finally also note that the sum of all components in a row or column of [k]e or [K] is always zero. Some final comments

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