CHAP 2 BAR TRUSS FINITE ELEMENT Direct Stiffness

CHAP 2 BAR & TRUSS FINITE ELEMENT Direct Stiffness Method FINITE ELEMENT ANALYSIS AND DESIGN Nam-Ho Kim 1

INTRODUCTION TO FINITE ELEMENT METHOD • What is the finite element method (FEM)? – A technique for obtaining approximate solutions of differential equations. – Partition of the domain into a set of simple shapes (element) – Approximate the solution using piecewise polynomials within the element F Structure Fi xe d u Piecewise-Linear Approximation Element x 2

INTRODUCTION TO FEM cont. • How to discretize the domain? – Using simple shapes (element) – All elements are connected using “nodes”. 5 6 1 1 7 2 2 8 3 3 Nodes Elements 4 – Solution at Element 1 is described using the values at Nodes 1, 2, 6, and 5 (Interpolation). – Elements 1 and 2 share the solution at Nodes 2 and 6. 3

INTRODUCTION TO FEM cont. • Methods – Direct method: Easy to understand, limited to 1 D problems – Variational method – Weighted residual method • Objectives – Determine displacements, forces, and supporting reactions – Will consider only static problem 4

1 -D SYSTEM OF SPRINGS F 2 F 4 2 1 2 6 3 1 5 3 4 4 5 F 3 • Bodies move only in horizontal direction • External forces, F 2, F 3, and F 4, are applied • No need to discretize the system (it is already discretized!) • Rigid body (including walls) • Spring NODE ELEMENT 5

SPRING ELEMENT • Element e – Consist of Nodes i and j – Spring constant k(e) e i j – Force applied to the nodes: – Displacement ui and uj – Elongation: – Force in the spring: – Relation b/w spring force and nodal forces: – Equilibrium: 6

SPRING ELEMENT cont. • Spring Element e – Relation between nodal forces and displacements – Matrix notation: – k: stiffness matrix – q: vector of DOFs – f: vector of element forces 7

SPRING ELEMENT cont. • Stiffness matrix – It is square as it relates to the same number of forces as the displacements. – It is symmetric. – It is singular, i. e. , determinant is equal to zero and it cannot be inverted. – It is positive semi-definite • Observation – For given nodal displacements, nodal forces can be calculated by – For given nodal forces, nodal displacements cannot be determined uniquely 8

SYSTEM OF SPRINGS cont. F 2 • Element equation and assembly F 4 2 1 2 6 3 1 5 3 4 4 5 F 3 9

SYSTEM OF SPRINGS cont. 10

SYSTEM OF SPRINGS cont. F 2 F 4 2 1 2 6 3 1 5 3 4 4 5 F 3 11

SYSTEM OF SPRINGS cont. • Relation b/w element forces and external force • Force equilibrium F 2 F 4 2 1 2 6 3 1 5 3 4 4 5 F 3 3 F 3 • At each node, the summation of element forces is equal to the applied, external force 12

SYSTEM OF SPRINGS cont. • Assembled System of Matrix Equation: • [Ks] is square, symmetric, singular and positive semi-definite. • When displacement is known, force is unknown R 1 and R 5 are unknown reaction forces 13

SYSTEM OF SPRINGS cont. • Imposing Boundary Conditions – Ignore the equations for which the RHS forces are unknown and strike out the corresponding rows in [Ks]. – Eliminate the columns in [Ks] that multiply into zero values of displacements of the boundary nodes. 14

SYSTEM OF SPRINGS cont. • Global Matrix Equation • Global Stiffness Matrix [K] – square, symmetric and positive definite and hence non-singular • Solution • Once nodal displacements are obtained, spring forces can be calculated from 15

UNIAXIAL BAR • For general uniaxial bar, we need to divide the bar into a set of elements and nodes • Elements are connected by sharing a node • Forces are applied at the nodes (distributed load must be converted to the equivalent nodal forces) • Assemble all elements in the same way with the system of springs p(x) F • Solve the matrix equation x for nodal displacements Statically indeterminate • Calculate stress and strain p(x) using nodal displacements F Statically determinate 16

1 D BAR ELEMENT • Two-force member • Only constant cross-section • Element force is proportional to relative displ • First node: i second code: j • Force-displacement relation Similar to the spring element 17

1 D BAR ELEMENT cont. – Matrix notation – Either force or displacement (not both) must be given at each node. – Example: ui = 0 and fj = 100 N. – What happens when fi and fj are given? • Nodal equilibrium – Equilibrium of forces acting on Node I – In general Fi Element e fi(e) fi(e+1) Node i Element e+1 18

1 D BAR ELEMENT cont. • Assembly – Similar process as spring elements – Replace all internal nodal forces with External Applied Nodal Force – Obtain system of equations [Ks]: Structural stiffness matrix {Qs} Vector of nodal DOFs {Fs}: Vector of applied forces • Property of [Ks] – Square, symmetric, positive semi-definite, singular, non-negative diagonal terms • Applying boundary conditions – Remove rigid-body motion be fixing DOFs – Striking-the-nodes and striking-the-columns (Refer to sprint elements) [K]: Global stiffness matrix {Q} Vector of unknown nodal DOFs {F}: Vector of known applied forces 19

1 D BAR ELEMENT cont. • Applying boundary conditions cont. – [K] is square, symmetric, positive definite, non-singular, invertible, and positive diagonal terms – Can obtain unique {Q} • Element forces – After solving nodal displacements, the element force can be calculated – Element stress • Reaction Forces Note Pi = Pj – Use [Ks]{Qs} = {Fs}: the rows that have been deleted (strike-the-rows) – Or, use 20

EXAMPLE • 3 elements and 4 nodes • At node 2: • Equation for each element: 21

EXAMPLE cont. • How can we combine different element equations? (Assembly) – First, prepare global matrix equation: Displacement vector Stiffness matrix Applied force vector – Write the equation of element 1 in the corresponding location 22

EXAMPLE cont. – Write the equation of element 2: – Combine two equations of elements 1 and 2 23

EXAMPLE cont. – Write the equation of element 3 – Combine with other two elements Structural Stiffness Matrix 24

EXAMPLE cont. • Substitute boundary conditions and solve for the unknown displacements. – Let K 1 = 50 N/cm, K 2 = 30 N/cm, K 3 = 70 N/cm and f 1 = 40 N. – Knowns: F 1, F 2, u 3, and u 4 – Unknowns: F 3, F 4, u 1, and u 2 25

EXAMPLE cont. – Remove zero-displacement columns: u 3 and u 4. – Remove unknown force rows: F 3 and F 4. – Now, the matrix should not be singular. Solve for u 1 and u 2. – Using u 1 and u 2, Solve for F 3 and F 4. 26

EXAMPLE cont. • Recover element data Element force 27

EXAMPLE • • • Statically indeterminate bars E = 100 GPa RL F = 10, 000 N A 1 = 10− 4 m 2, A 2 = 2× 10− 4 m 2 Element stiffness matrices: A B C F 0. 25 m RR 0. 4 m • Assembly 28

EXAMPLE cont. • Applying BC • Element forces or Element stresses • Reaction forces 29

PLANE TRUSS ELEMENT • What is the difference between 1 D and 2 D finite elements? – 2 D element can move x- and y-direction (2 DOFs per node). – However, the stiffness can be applied only axial direction. • Local Coordinate System – 1 D FE formulation can be used if a body-fixed local coordinate system y is constructed along the length of the element Y – The global coordinate system (X and Y axes) is chosen to X represent the entire structure – The local coordinate system (x and y axes) is selected to align the x-axis along the length of the element 50 N x 8 cm 12 cm 30

PLANE TRUSS ELEMENT cont. • Element Equation (Local Coordinate System) – Axial direction is the local x-axis. – 2 D element equation y Local coordinates f 1 2 Global coordinates x – is square, symmetric, positive semi-definite, and non-negative diagonal components. • How to connect to the neighboring elements? – Cannot connect to other elements because LCS is different – Use coordinate transformation 31

COORDINATE TRANSFORMATION • Transform to the global coord. and assemble • Transformation matrix 32

COORDINATE TRANSFORMATION cont. • The same transformation force vector • Property of transformation matrix 33

ELEMENT STIFFNESS IN GLOBAL COORD. • Element 1 y K f N 2 N 1 x • Transform to the global coordinates 34

ELEMENT STIFFNESS IN GLOBAL COORD. cont. • Element stiffness matrix in global coordinates – Depends on Young’s modulus (E), cross-sectional area (A), length (L), and angle of rotation (f) – Axial rigidity = EA – Square, symmetric, positive semi-definite, singular, and non-negative diagonal terms 35

EXAMPLE • Two-bar truss – In local coordinate 50 N Element 1 – Diameter = 0. 25 cm – E = 30 106 N/cm 2 • Element 1 N 2 8 cm Element 2 N 1 N 3 12 cm y K N 1 N 2 f 1 f = 33. 7 o 1 E = 30 x 106 N/cm 2 A = pr 2 = 0. 049 cm 2 L = 14. 4 cm x 36

EXAMPLE cont. • Element 1 cont. – Element equation in the global coordinates y • Element 2 N 2 f 2 = – 90 o E = 30 x 106 N/cm 2 A = pr 2 = 0. 049 cm 2 K L = 8 cm N 3 f 2 x 37

EXAMPLE cont. • Assembly – After transforming to the global coordinates Element 1 Element 2 • Boundary Conditions – Nodes 1 and 3 are fixed. – Node 2 has known applied forces: F 2 x = 50 N, F 2 y = 0 N 38

EXAMPLE cont. • Boundary conditions (striking-the-columns) – Striking-the-rows • Solve the global matrix equation 39

EXAMPLE cont. • Support reactions – The reaction force is parallel to the element length (two-force member) • Element force and stress (Element 1) – Need to transform to the element local coordinates 40

EXAMPLE cont. • Element force and stress (Element 1) cont. – Element force can only be calculated using local element equation – – There is no force components in the local y-direction In x-direction, two forces are equal and opposite The force in the second node is equal to the element force Normal stress = 60. 2 / 0. 049 = 1228 N/cm 2. – 60. 2 N 1 2 41

OTHER WAY OF ELEMENT FORCE CALCULATION • Element force for plane truss – Write in terms of global displacements 42

EXAMPLE 2 • Directly assembling global matrix equation (applying BC in the element level) • Element property & direction cosine table Elem AE/L i -> j 1 206× 105 1 -> 3 2 206× 105 1 -> 2 3 206× 105 1 -> 4 f -30 90 210 l = cosf 0. 866 0 0. 866 m = sinf 0. 5 1 0. 5 F 2 45 3 1 4 1 3 • Since u 3 and v 3 will be deleted after assembly, it is not necessary to keep them 43

SPACE TRUSS ELEMENT • A similar extension using coordinate transformation Y • 3 DOF per node – u, v, and w – fx, fy, and fz f. Y K f. X N 2 f. Z • Element stiffness matrix is 6 x 6 N 1 X Z • FE equation in the local coord. 44

SPACE TRUSS ELEMENT cont. • Relation between local and global displacements – Each node has 3 DOFs (ui, vi, wi) – Direction cosines 45

SPACE TRUSS ELEMENT cont. • Relation between local and global force vectors • Stiffness matrix 46

THERMAL STRESSES • Temperature change causes thermal strain L No stress, no strain (a) at T = Tref L DL No stress, thermal strain Thermal stress, no strain (b) at T = Tref + DT • Constraints cause thermal stresses • Thermo-elastic stress-strain relationship Thermal expansion coefficient 47

THERMAL STRESSES cont. • Force-displacement relation • Finite element equation Thermal force vector • For plane truss, transform to the global coord. 48

EXAMPLE 3 z • Space truss Node 1 2 3 4 Elem x 0 0 0 1 EA/L y 0 – 1 1 0 i -> j l z 0 1 1 1 m 2 E 3 E 2 1 y 4 E 1 10, 000 N n 1 1 -> 4 0 2 2 -> 4 0 3 3 -> 4 0 x 49

Project 1 • Fully-stressed design of a ten-bar truss – Most efficient usage of material = all members are at yield stress – Modify cross-sectional areas (individually) to minimize weight b Parameters Dimension, b Safety factor, SF Load, P 1 Load, P 2 Density, Young’s modulus, E Allowable stress, Initial area Min. CS area Values 360 inches 1. 5 66. 67 kips 0. 1 lb/in 3 104 ksi 25 ksi* 1. 0 in 2 0. 1 in 2 *for Element 9, allowable stress is 75 ksi 5 b 3 1 8 y 1 0 7 x 6 5 3 6 1 2 9 b 4 4 2 P 1 P 2 50

PROJECT 1 cont. • Schedule: Project due: 2/23 (Wed) – Submission: Submit report (Word or PDF file, Max 10 pages), CAE file and/or programs by 1: 55 PM in Sakai (penalty for late submission) • Report – Formal report, including title, summary, introduction, approach, results, discussion, appendix (input file, programs), and/or references • What to include in the report – Use the same definition of nodes and elements in the problem definition. (You need to interpret your Abaqus results accordingly) – Plot of overlapped undeformed and deformed geometry with stress contour – FE analysis results (tables) for initial design (nodal displacements and element stresses, S 11) – Your methodology for design updates (algorithms or flowchart, etc) – Fully-stressed design (cross-sectional areas and stress of each element, use tables) 51 – Discussions (what you learned through the project)

PROJECT 1 • Analysis and Design of Plane Truss – E = 100 GPA, A = 1. 0 cm 2, L = 0. 3 m or 2 y 7 13 4 14 20 x 1 8 1 6 15 21 3 9 2 8 10 16 22 5 m 3 10 17 23 7 11 4 12 18 24 9 12 5 14 19 25 11 6 13 – Apply axial loading, transverse shear, and bending loads – Calculate equivalent axial rigidity, flexural rigidity, and shear rigidity – Use average deflection of Nodes 13 and 14 – Verify the equivalent properties by adding two more bays 52