Trusses 1112020 1 Element Formulation by Virtual Work

Trusses 11/1/2020 1

Element Formulation by Virtual Work u Use virtual work to derive element stiffness matrix based on assumed displacements – Principle of virtual work states that if a general structure that is in equilibrium with its applied forces deforms due to a set of small compatible virtual displacements, the virtual work done is equal to its virtual strain energy of internal stresses. 11/1/2020 2

u At element level, d. Ue = d. We – d. Ue = virtual strain energy of internal stresses – d. We = virtual work of external forces acting through virtual displacements 11/1/2020 3

u We now assume a simple displacement function to define the displacement of every material point in the element. u Usually use low order polynomials u Here u = a 1 + a 2 x – u is axial displacement – a 1, a 2 are constants to be determined – x is local coordinate along member 11/1/2020 4

u The constants are found by imposing the known nodal displacements ui, uj at nodes i and j ui = a 1 + a 2 xi uj = a 1 + a 2 xj u ui, uj are nodal displacements u xi, xj are nodal coordinates 11/1/2020 5

u letting xi = 0, xj = L, we get – a 1 = ui – a 2 = (uj-ui)/L u We can write – [N] = matrix of element shape functions or interpolation functions – {d} = nodal displacements 11/1/2020 6

N 1=1 Variation of N 1 N 2=1 Variation of N 2 11/1/2020 7

u Strain is given by u where [B] is a matrix relating strain to nodal displacement (matrix of derivatives of shape function) 11/1/2020 8

u Now s = E(e- eo )= E[B]{d}-E eo – Stress and strain are constant in a member u Define internal virtual strain energy for a set of virtual displacements {dd} to be 11/1/2020 9

- de = virtual strain - s = stress level at equilibrium – d. V = volume u Virtual work of nodal forces is T d. We = {dd} {f} u Then, virtual work is given by 11/1/2020 10

u Substituting u Canceling and rearranging gives T {dd} gives [k]{d}={F} where For thermal problem 11/1/2020 11

u for a truss we get u this formulation method also applies to 2 -d and 3 -d elements 11/1/2020 12

Procedure for Direct Stiffness Method (Displacement Method) 1. Discretize into finite elements, Identify nodes, elements and number them in order. 2. Develop element stiffness matrices [Ke] for all the elements. 3. Assemble element stiffness matrices to get the global stiffness matrix ([KG] =S [Ke]). The size of of global stiffness matrix = total d. o. f of the structure including at boundary nodes. Assembly is done by matching element displacement with global displacements. Also develop appropriate force vector (by adding element force vectors) such that equation of the type [KG] {u}={F} is obtained. 11/1/2020 13

Procedure for Direct Stiffness Method 4. Apply kinematic boundary conditions. Without applying boundary conditions, [KG] will be singular. (minimum number of boundary conditions required is to arrest ‘Rigid Body’ displacements). 5. Solve for unknown displacements {u} ( {u}= [KG] – 1{F}). 6. Once displacements are determined find (a) reactions by picking up appropriate rows from the equation {F}=[KG] {u}, (b) Find element forces {f}=[Ke] {ue}, (c) Element stresses given by {se}= [D][B]{ue}. 11/1/2020 14

11/1/2020 15

2 A, L, E 11/1/2020 16

11/1/2020 17

Reactions { 11/1/2020 18

Element Forces f 2 f 1 A, L, E 2 A, L, E 11/1/2020 19

11/1/2020 20

11/1/2020 21

11/1/2020 22

u 1 u 2 11/1/2020 u 3 23

11/1/2020 24

11/1/2020 25

11/1/2020 26

11/1/2020 27

Direct Element Formulation u truss element acts like 1 -d spring – l >> transverse dimensions – pinned connection to other members (only axial loading). – usually constant cross section and modulus of elasticity 11/1/2020 28

» A = cross section area » E = modulus of elasticity » L = length 11/1/2020 29

u Assume displacements are much smaller than overall geometry – vertical displacements of horizontal member produce no vertical force u Stiffness matrix is written in local element coordinates aligned along element axis u want stiffness matrix for arbitrary orientation 11/1/2020 30

u rotate coordinate systems using rotation matrix [R] u displacement components in global coordinates are related to displacement components in local coordinates by {d’}=[R]{d} – {d} = displacement in global coordinates – {d’} = displacement in local element coordinates 11/1/2020 31

11/1/2020 32

11/1/2020 33

u start with member on x axis, element equations are or {k’}{d’}={f’} u Note that y equations are all zero 11/1/2020 34

11/1/2020 35

At node i A similar matrix can be obtained at node j 11/1/2020 36

u Matrix [R] is: 11/1/2020 37

u Similarly , force components are related by {f’} = [R]{f} u Local force displacement relation is [k’]{d’} = {f’} u global force displacement relation is [k][R]{d} = [R]{f} -1 T u using fact that [R] = [R] , we get T [R] [k][R]{d} = {f} 11/1/2020 38

u then [k] = stiffness matrix in global T coordinates is [R] [k’][R] 11/1/2020 39

u Structure equation is [k] {D} = {F} – [k] = structure stiffness matrix – {D} = nodal displacement vector – {F} = applied load vector 11/1/2020 40

11/1/2020 41

2 E le m e n t C o o r d in a t e i- n o d e y x 1 1 2 0 3 0 C o o r d in a te 0 2 L s in 4 5 L e n g th j- n o d e 2 2 x L c o s 4 5 L s in 4 5 11/1/2020 C y s L c o s 4 5 s in 4 5 L c os 4 5 -sin 4 5 42

11/1/2020 43

11/1/2020 44

11/1/2020 45

11/1/2020 46

11/1/2020 47

11/1/2020 48

11/1/2020 49

11/1/2020 50

v’ 3 11/1/2020 v 3 u’ 3 q u 3 51

11/1/2020 52

11/1/2020 53

11/1/2020 54

Finite Element Model u usually use existing codes to solve problems u user responsible for – creating the model – executing the program – interpreting the results 11/1/2020 55

u arrangement of nodes and elements is known as the mesh u plan to make the mesh model the structure as accurately as possible 11/1/2020 56

u for a truss – each member is modeled as 1 truss element – truss members or elements are connected at nodes – node connections behave like pin joints – truss element behaves in exact agreement with assumptions – no need to divide a member into more than 1 element 11/1/2020 57

– such subdivision will cause execution to fail » due to zero stiffness against lateral force at the node connection where 2 members are in axial alignment 11/1/2020 58

11/1/2020 59

u there is geometric symmetry – often possible to reduce the size of problem by using symmetry – need loading symmetry as well 11/1/2020 60

11/1/2020 61

u Fig. 3 -5 and 3 -6 show symmetric loads and the reduced model – need to impose extra conditions along the line of symmetry » displacement constraints: nodes along the line of symmetry must always move along that line » changed loads: the load at the line of symmetry is split in two 11/1/2020 62

Computer input assistance ua preprocessor is used to assist user input u required inputs are – data to locate nodes in space – definition of elements by node numbers – type of analysis to be done – material properties – displacement conditions – applied loads 11/1/2020 63

u interactive preprocessors are preferable – you can see each node as it is created – elements are displayed as they are created – symbols are given for displacement and load conditions – usually allow mesh generation by replication or interpolation of an existing mesh – allow inserting nodes along lines – allow entering a grid by minimum and maximum positions plus a grid spacing 11/1/2020 64

u truss element consists of 2 node numbers that connect to form element u other information for truss is – modulus of elasticity – cross sectional area u data can form a material table u assign element data by reference to the table 11/1/2020 65

u boundary or displacement conditions are set by selecting a node and setting its displacement U do not over constrain a structure by prescribing zero displacements where there is no physical support 11/1/2020 66

u loading conditions are set by selecting nodes and specifying force or moment components U check model carefully at this point 11/1/2020 67

Analysis Step u mostly transparent to user u small truss models have enough accuracy and performance for an accurate solution u a large model has a large number of elements and nodes 11/1/2020 68

u numerical solution may not be accurate if there are full matrices u get better accuracy if the nonzero terms are close to the diagonal – reduces the number of operations and round off error (banded matrix) 11/1/2020 69

u in FE model, element or node numbering can affect bandwidth – good numbering pattern can minimize bandwidth – different methods based on node or element numbering – to minimize, plan numbering pattern so nodes that connect through an element have their equations assembled close together 11/1/2020 70

u In Fig. 3 -7, node numbers are considered, X’s show nonzero terms 11/1/2020 71

u In Fig. 3 -8, node numbers are considered 11/1/2020 72

u many programs have bandwidth or wavefront minimizers available u most programs will keep original numbering for display but use the minimized number scheme 11/1/2020 73

u numerical algorithms, numerical range of the computer affect solution u relative stiffness of members can influence results – problems when members of high and low stiffness connect – can exceed precision of computer – physical situation is usually undesirable 11/1/2020 74

u Approximation error for truss is zero u Most common error messages (errors) come from – incorrect definition of elements – incorrect application of displacement boundary conditions 11/1/2020 75

– may get non-positive definite structure stiffness matrix from not enough boundary conditions to prevent rigid body motion » two elements connect in-line Þ zero lateral stiffness » truss structure not kinematically stable (linkage) 11/1/2020 76

u next look at stress components – in continua, stress components are related to averaged quantities at the nodes – trusses have a stress in each member (not easy to plot) u truss model is exact so it does not usually need refinement 11/1/2020 77

Output Processing and Evaluation u Get numerical results with input data followed by all nodal displacements and element stresses u first graphic to look at is the deformed shape of the structure – nodal displacements are exaggerated to show structure deformation – check to ensure model behaves as expected 11/1/2020 78

u linear elastic analysis, failure is by – overstressing – buckling (have to find members with significant compression and use Euler's buckling equation) 11/1/2020 79

Final Remarks u few situations where a truss element is the right element for modeling behavior 11/1/2020 80