Institute of Structural Engineering Page 1 Chapter 2 The Direct Stiffness Method
Institute of Structural Engineering Page 2 Direct Stiffness Method (DSM) Computational method for structural analysis Matrix method for computing the member forces and displacements in structures DSM implementation is the basis of most commercial and open-source finite element software Based on the displacement method (classical hand method for structural analysis) Formulated in the 1950s by Turner at Boeing and started a revolution in structural engineering
Institute of Structural Engineering Page 3 Goals of this Chapter DSM formulation DSM software workflow for linear static analysis (1 st order) 2 nd order linear static analysis linear stability analysis
Institute of Structural Engineering Page 4 Computational Structural Analysis Y X Physical problem Continuous mathematical model strong form Discrete computational model weak form Modelling is the most important step in the process of a structural analysis!
Institute of Structural Engineering Page 5 System Identification (Modelling) Y 3 4 6 4 1 3 5 2 1 2 X 5 6 Global Coordinate System Nodes Elements Boundary conditions Loads Node numbers Element numbers and orientation
Institute of Structural Engineering Page 6 Deformations System Deformations Nodal Displacements System identification nodes, elements, loads and supports deformed shape (deformational, nodal) degrees of freedom = dofs
Institute of Structural Engineering Page 7 Degrees of Freedom Truss Structure Frame Structure u i u i u i = ( u dx, u dy ) dof per node u i = ( u dx, u dy, u rz ) 7 * 2 = 14 dof dof of structure 8 * 3 = 24 dof
Institute of Structural Engineering Page 8 Elements: Truss 1 dof per node u x L, E, F DX P 1 P 2 N X/Y = local coordinate system u x = displacement in direction of local axis X DX = displacement of truss end P 1 DX = (u 2 u 1 ) P 2 P1 = EF L (u 1 u 2 ) P 2 = EF L ( u 1 + u 2 ) compatibility const. equation equilibrum e = DX L s = E e P 2 = P 1 = N N = ʃ E s = EF s = EF L DX p = k u p : (element) stiffness matrix k : (element) nodal forces u : (element) displacement vector
Institute of Structural Engineering Page 9 Elements: Beam 3 dof per node L, E, F u x u y DX DY u y RZ u x = displacement in direction of local axis X u y = displacement in direction of local axis Y k u
Institute of Structural Engineering Page 10 Elements: Global Orientation local θ R θ = cos θ sin θ 0 0 0 sin θ cos θ 0 0 0 0 0 1 0 0 0 0 cos θ sin θ 0 0 0 sin θ cos θ 0 0 0 0 0 1 global u glob = u = R u loc k glob = k = R T k loc R
Institute of Structural Engineering Page 11 Beam Stiffness Matrix FX S = FY S = MZ S = FX S = FY S = MZ E = UX S UY S UZ S UX E UY E UZ E k 11 k 12 k 13 p p k 22 k 23 symm. is ie k 33 k 14 k 15 k 16 k 24 k 25 k 26 k 34 k 35 k 36 k 44 k 45 k 46 k 55 k 56 k 66 k iss k ise uis k k u ies iee p = k u ie E UX E =1 S FY S e.g. k 24 = reaction in global direction Y at start node S due to a Element stiffness matrix in global orientation unit displacement in global direction X at end node E
Institute of Structural Engineering Page 12 Nodal Equilibrum 3 6 f 4 r4: Vector of all forces acting at node 4 4 5 2 r4 = - k 6ES u 3 + contribution of element 6 due to start node displacement u 3 - k 6EE u 4 + contribution of element 6 due to end node displacement u 4 - k 5EE u 4 + contribution of element 5 due to start node displacement u 4 - k 5ES u 2 + contribution of element 5 due to start node displacement u 2 external load f 4 Equilibrum at node 4: r 4 = - k 5SE u 2 -k 6ES u 3 - k 5EE u 4 - k 6EE u 4 + f 4 = 0
Institute of Structural Engineering Page 13 Global System of Equations r 1 = - u 1 k 1EE + k 3SS + k 4SS u 2 u 3 u 4 k 3SE k 4SE + f 1 = 0 3 6 4 4 5 1 3 2 r 2 = - k 3ES k 2EE + k 3EE + k 5SS k 5SE + f 2 = 0 1 2 r 3 = - k 4ES k 4EE + k 6SS k 6SE + f 3 = 0 r 4 = - k 5ES k 6ES k 5EE + k 6EE + f 4 = 0 - K U + F = 0 F = K U
Institute of Structural Engineering Page 14 Global System of Equations F = global load vector = Assembly of all fe K = global stiffness matrix = Assembly of all ke U = global displacement vector = unknown F = K U = equilibrium at every node of the structure
Institute of Structural Engineering Page 15 Solving the Equation System What are the nodal displacements for a given structure (= stiffness matrix K ) due to a given load (= load vector F )? K U = F left multiply K -1 K U = K -1 F K -1 U = K -1 F Inversion possible only if K is non-singular (i.e. the structure is sufficiently supported = stable)
Institute of Structural Engineering Page 16 Beam Element Results 1. Element nodal displacements Disassemble u from resulting global displacements U 2. Element end forces Calculate element end forces = p = k u 3. Element stress and strain along axis Calculate moment/shear from end forces (equilibrium equation) Calculate curvature/axial strain from moments/axial force 4. Element deformations along axis Calculate displacements from strain (direct integration)
Institute of Structural Engineering Page 17 1. Adjust global load vector Lateral Load 2. Adjust element stresses f = local load vector => add to global load vector F e.g. bending moment M: M due to u M due to f M diagram
Institute of Structural Engineering Page 18 Linear Static Analysis (1 st order) Workflow of computer program 1. System identification: Elements, nodes, support and loads 2. Build element stiffness matrices and load vectors 3. Assemble global stiffness matrix and load vector 4. Solve global system of equations (=> displacements) 5. Calculate element results Exact solution for displacements and stresses
Institute of Structural Engineering Page 19 2 nd Order Effects or the influence of the axial normal force Normal forces change the stiffness of the structure!
Institute of Structural Engineering Page 20 Geometrical Stiffness Matrix Truss Very small element rotation => Member end forces (=nodal forces p ) perpendicular to axis due to initial N k G = geometrical stiffness matrix of a truss element NOTE: It s only a approximation p = ( k + k G ) u
Institute of Structural Engineering Page 21 Beams: Geometrical Stiffness k G = geometrical stiffness matrix of a beam element k G =
Institute of Structural Engineering Page 22 Linear Static Analysis (2 nd order) What are the 2 nd order nodal displacements for a given structure due to a given load? Global system of equations ( K + K G ) U = F U = ( K + K G ) -1 F Inversion possible only if K + K G is non-singular, i.e. - the structure is sufficiently supported (= stable) - initial normal forces are not too big
Institute of Structural Engineering Page 23 Linear Static Analysis (2 nd order) Workflow of computer program 1. Perform 1 st order analysis 2. Calculate resulting axial forces in elements (=N e ) 3. Build element geometrical stiffness matrices due to N e 4. Add geometrical stiffness to global stiffness matrix 5. Solve global system of equations (=> displacements) 6. Calculate element results NOTE: Only approximate solution!
Institute of Structural Engineering Page 24 Stability Analysis How much can a given load be increased until a given structure becomes unstable? K G = f(n max ) N max = λ max N 0 K G (N max ) = λ max K G (N 0 ) = λ max K G0 2 nd order analysis No additional load possible (K + λ max K G0 ) U = F (K + λ max K G0 ) ΔU = ΔF = 0 linear algebra (A - λ B) x = 0 Eigenvalue problem
Institute of Structural Engineering Page 25 Stability Analysis Eigenvalue problem (A - λ B) x = 0 Solution λ = eigenvalue x = eigenvector e.g. Buckling of a column (K - λ K G0 ) x = 0 λ = critical load factor x = buckling mode λ F x λ N 0
Institute of Structural Engineering Page 26 Stability Analysis Workflow of computer program 1. Perform 1 st order analysis 2. Calculate resulting axial forces in elements (=N 0 ) 3. Build element geometrical stiffness matrices due to N 0 4. Add geometrical stiffness to global stiffness matrix 5. Solve eigenvalue problem NOTE: Only approximate solution!