The Finite Element Method for the Analysis of Non-Linear and Dynamic Systems Prof. Dr. Eleni Chatzi Lecture 6-5 November, 015 Institute of Structural Engineering Method of Finite Elements II 1
Introduction to Dynamic Analysis Introduction Swiss Federal Institute of Technology Page 3 The very basics - Newton s nd law of motion: The Introduction very basics ti to Dynamic Analysis k Unstretched position Newtons nd law of motion: kδ mx = w k( Δ+ x) Δ Introduction m m Stretched position w= mg mx The w + very k( Δ+ basics x) = 0 k Unstretched (static equilibrium) position Newtons nd law of motion: kδ x xx, mx = w k( x) Δ m m mx kδ+ k( Δ+ x) = mx + kx = 0 Stretched position k w= mg x+ mx x = w x+ + k( ωδ+ n x= x) 0= 0 (static equilibrium) π m ωτ n = π τ = = π m ωn k x xx, 1 1 k fn = = mx k k x mx kx τ π m 0 0 x(t) x= Asin k ω cos t nt+ B ωnt x= x sinωnt+ xcosω t x = + Asinω x= x+ n t ω + n x= Bcosω 0 n t x(t) = ẋ0 sinω π n mt + x 0 cosω n t ωτ n = π ωτ = = π m n ωn k 1 1 k fn = = where ω n = π k τ π m T = m 0 0 x= Asinω t+ Bcosω t t x= x sinω t+ xcosω t thod of Finite Elements II n n n n Method of Finite Elements II Institute of Structural Engineering Method of Finite Elements II
Introduction to Dynamic Analysis We have previously considered the equilibrium equations governing the linear dynamic response of a system of finite elements: MÜ + C U + KU = R F I (t) + F D (t) + F E (t) = R(t) where: M : Mass Matrix K : Stiffness Matrix U : Displacements U : Velocities Ü : Accelerations where: F I (t) = MÜ F D (t) = C U F E (t) = KU Inertial Force Damping Force Internal Force Institute of Structural Engineering Method of Finite Elements II 3
Introduction to Dynamic Analysis When is Dynamic Analysis required? Whether a dynamic analysis is needed or not is generally up to engineering judgment requires understanding of the interaction between loading and structural response! In general if the loading varies over time with frequencies higher than the Eigen-frequencies of the structure dynamic analysis will be required Institute of Structural Engineering Method of Finite Elements II 4
Introduction to Dynamic Analysis Objective Solve the dynamic Equation of motion (numerically) MÜ + C U + KU = R In principle the equilibrium equations may be solved by any standard numerical integration scheme BUT! Efficiency numerical efforts must be considered and it is worthwhile to look at special techniques of integration which are especially suited for the analysis of finite element assemblies. *The section on Direct Integration Methods is based on Prof. M. Faber s notes of the FEM II course - Fall 009 Institute of Structural Engineering Method of Finite Elements II 5
Direct Integration Methods These methods rely in discretizing the continuous problem. The original problem formulation is essentially the dynamic equation of motion which is for structural problems is a second order Ordinary Differential Equation (ODE). This involves, after the solution is defined at time zero, the attempt to satisfy dynamic equilibrium at discrete points in time. Most methods use equal time intervals at t, t, 3 t...n t. However this is not mandatory; in some cases a variable time step might be employed. This is most commonly the case for special classes of problems such as Impact Problems. Institute of Structural Engineering Method of Finite Elements II 6
Direct Integration Methods Direct means: The equations are solved in their original form Two ideas are utilized 1. The equilibrium equations are satisfied only at time steps, i.e. at discrete times with intervals t. A particular variation of displacements, velocities and accelerations within each time interval is assumed The accuracy depends on these assumptions as well as the choice of time intervals! Institute of Structural Engineering Method of Finite Elements II 7
Direct Integration Methods Assumptions The displacements, velocities and accelerations U 0 : U 0 : Ü 0 : Displacement vector at time t Velocity vector at time t Acceleration vector at time t are assumed to be known and we aim to establish the solution of the equilibrium equations for the period 0 - T. For this purpose we sub-divide T into n intervals of length t = T/n and establish solutions for the times t, t, 3 t,..., T Institute of Structural Engineering Method of Finite Elements II 8
Direct Integration Methods We distinguish principally between Implicit and Explicit methods Explicit methods: Solution is based on the equilibrium equations at time t 1st order Example: The Forward Euler Method Assuming we want to approximate the solution of the initial value problem ẏ = f(t, y(t)), y(t 0 ) = y 0 by using the first two terms of the Taylor expansion of y, which represents the linear approximation around the point (t 0, y(t 0 )) we obtain the forward Euler integration rule: y n+1 = y n + hf(t n, y n ) Institute of Structural Engineering Method of Finite Elements II 9
Direct Integration Methods Implicit methods: Solution is based on the equilibrium equations at time t + t 1st order Example: The Backward Euler Method Assuming we want to approximate the solution of the same initial value problem the backward Euler integration rule is obtained as: y n+1 = y n + hf(t n+1, y n+1 ) Since f(t, y) is, in general, a non-linear function of y, iteration is required to solve this last equation for y n+1. Institute of Structural Engineering Method of Finite Elements II 10
Direct Integration Methods Explicit methods calculate the state of a system at a later time from the state of the system at the current time, while implicit methods find a solution by solving an equation involving both the current state of the system and the later one. Note that: Implicit integration is not necessarily more accurate than explicit. It can be less accurate! The major benefit of implicit integration is stability. Many of these methods are able to run with any arbitrarily large time step for any input unless we are lying at the limits of floating point math (unconditionally stable). Obviously a large time step implies throwing away accuracy. For most real structures, which contain stiff elements, a very small time step is required in order to obtain a stable solution. Therefore, all explicit methods are conditionally stable. The condition is the size of the selected time step. Institute of Structural Engineering Method of Finite Elements II 11
Direct Integration Methods Most Commonly used Direct Integration Methods (for the case of the Dynamic Equation of Motion) 1 The Central Difference Method (CDF) The Houbolt method 3 The Newmark method 4 The Wilson θ method 5 Coupling of integration operators The difference in items 1-4 lies in the way we choose a discretized equivalent of the derivatives. The overall setup of the solution is very much similar for all methods. Additionally depending on the resulting equations some schemes are explicit (CDF) and others implicit (Houbolt, Newmark, Wilson θ) Institute of Structural Engineering Method of Finite Elements II 1
Direct Integration Methods Most Commonly used Direct Integration Methods Velocity Central Difference Method Acceleration U(t) t - Δt t t + Δt U(t) t - Δt t t + Δt t - Δt t t + Δt Reminder: The difference lies in the way we choose a discretized equivalent of the derivatives. Institute of Structural Engineering Method of Finite Elements II 13
Direct Integration Methods Most Commonly used Direct Integration Methods Displacement The Houbolt Method Velocity U(t) Acceleration t - Δt t - Δt t t + Δt These approximations are derived via the displacement approximation. Houbolt s method uses a third-order interpolation of displacements extending two steps back in time. Reminder: The difference lies in the way we choose a discretized equivalent of the derivatives. Institute of Structural Engineering Method of Finite Elements II 14
Direct Integration Methods Most Commonly used Direct Integration Methods The Newmark Method Displacement Velocity The Newmark method uses a second order Taylor expansion for approximating Velocities and U! t+δt = U! t + Δt U!! V Accelerations: U t+δt = U t + Δt U!! V + Δt U!!, where U!! V, U!! D are approximations of the Acceleration D if if t t + Δt t t + Δt Reminder: The difference lies in the way we choose a discretized equivalent of the derivatives. Institute of Structural Engineering Method of Finite Elements II 15
The Central Difference Method Approximate the velocity (first derivative) as: t U = 1 t (t+ t U t- t U) (1) Approximate the acceleration (second derivative) as: tü = 1 t (t+ t U t U + t- t U) () Substitute Eqns (1),() into MÜ + C U + KU = R: ( 1 t M + 1 ) ( t C t+ t U = t R K ) ( t M t 1 U t M 1 ) t C t- t U (a 0 M + a 1 C) t+ t U = t R (K a M) t U (a 0 M a 1 C) t- t U Institute of Structural Engineering Method of Finite Elements II 16
The Central Difference Method We see that we do not need to factorize the stiffness matrix, i.e, we only rely on info from the current step t to go to the next one t + t (explicit method) We also see that in order to calculate the displacements at time t we need to know the displacements at time 0 and t In general, 0 U, 0 U, 0 Ü are known and we can use Eqns (1),() to obtain t U: - t U = 0 U t 0 U + t 0Ü Institute of Structural Engineering Method of Finite Elements II 17
The Central Difference Method Solution Procedure A. Initial Calculations Institute of Structural Engineering Method of Finite Elements II 18
The Direct Central Integration Difference ti Methods Method Solution Procedure The Central difference method: Solution procedure: B. B: For For each each time time step step 1) Calculate effective loads at time t: ( ) ( ) ˆ Δ R = R K a M U a M a C U t t t t t 0 1 ) Solve for the displacements U at time t + Δt LDL Tt +Δ t U = t Rˆ 3) If required, solve for the corresponding velocities and accelerations U ( Δ U U+ +Δ = a U) 0 t t t t t t U Δ +Δ = a 1 ( U+ U ) t t t t t thod of Finite Elements II Institute of Structural Engineering Method of Finite Elements II 19
Direct Integration ti Methods The Central Difference Method The effectiveness Central difference of the central methoddifference method depends on the efficiency of the time step solution since generally a small discretization The effectiveness is required. of the central difference method depends on the efficiency of the time step solution because we need a lot of For them this reason the method is usually only applied when a lumped (diagonal) For this reason massthe matrix method can is beusually assumed only and applied when the when velocity a lumped dependent mass matrix damping can be (C) assumed can beand neglected, when the i.e.: velocity dependent damping can be neglected, i.e.: 1 t+δt t M U= Rˆ Δt t+δt t ˆ Δt 1 U i = Ri ( ), mii > 0 t t t t ΔΔ t RˆR = R K M U M U m ii Δt Δt () () KU t = K i t U= t F i t () ˆ t t i 1 t Δ t t i i R = R F M ( U U ) Δt ethod of Finite Elements II i Institute of Structural Engineering Method of Finite Elements II 0
The Central Difference Method Direct Integration ti Methods Example - DOF system k 1 = 4 U, U, U 1 1 1 m = R = 0 1 k = 1 0 m = 1 R = 10 0 U 6 1 U1 0 + = 0 1 U 4 U 10 U, U, U k 3 = Method of Finite Elements II For this system the natural periods are T 1 = 4.45, T =.8 Institute of Structural Engineering Method of Finite Elements II 1
The Central Difference Method Example - DOF system Institute of Structural Engineering Method of Finite Elements II
Direct Integration Methods The Central Difference Method Example - DOF system The Central difference method: Example: 1 1 1 a = 0, a 1, a a0, a3 Δ t = Δ t = = a 1 1 a0 = = 1.8, a 1 = = 1.79, (0.8) 0.8 For Δt = 0.8 1 1 a = = 5.5, a 3 = = 0.039 (0.8) 5.5 Δt 0 0 0 0 Ui = 0.8 0.039039 0 0 + 10 = 0.039 ˆ 0 0 0 5.5 0 M = 1.8 1.79 0 1 + 0 0 = 0 1.8 t ˆ 0 45.0 t 5.5 0 t Δt R = 10 + 1.5 U 0 1.8 U 0 U 6 1 0 1 + U Method of Finite Elements II Institute of Structural Engineering Method of Finite Elements II 3
The Central Difference Method Swiss Federal Institute of Technology Page 18 Direct Integration ti Methods Example - DOF system U The Central difference method: Example: The equation which must be solved for each time step is: 0 U 6 1 U1 0 0 1 + = 4 U 10 The equation which must be solved for each time step is: t+δt 5.5 0 U1 t ˆ = t+δ t 0 1.8 R U t t Δt t 450 55 ˆ 0 45.0 U 5.5 0 1 U 1 R = t t t 10 + 1.5 U 0 1.8 Δ U Method of Finite Elements II Institute of Structural Engineering Method of Finite Elements II 4
The Central Difference Method Example - DOF system The results are: 6 5 4 CentralDifference_U1 CentralDifference_U 3 1 0 0 0.5 1 1.5.5 3 3.5 4 Institute of Structural Engineering Method of Finite Elements II 5
The Houbolt Method Swiss Federal Institute of Technology Page 0 Houbolt Direct Integration Method tiderivative Methods Approximation MU + CU + KU = R c t+δ t U 1 ( t+δt 5 t 4 t Δt t Δt = U U+ U U ) Δtt 1 (11 18 9 U = U U+ U U ) 6Δt t+δ t t+δt t t Δt t Δt a b a and b inserted in c 1 t+δt + + = Δt M 6Δt C K U t+δt R 5 3 t 4 3 t Δt 1 1 t Δt + + + + + Δt M Δt C U Δt M Δt C U Δt M 3Δt C U ethod of Finite Elements II Institute of Structural Engineering Method of Finite Elements II 6
The Houbolt Method We will not consider the Houbolt in more detail however it is noted that it is necessary to factorize the stiffness matrix (implicit method) Furthermore, if the mass and damping terms are neglected, the Houbolt method results in the static analysis equations: K t+ t U = t+ t R Notice how this is not true for the CDM! Institute of Structural Engineering Method of Finite Elements II 7
The Direct Newmark Integration Method ti Methods In 1959 Newmark presented a family of single-step integration The Newmark method methods for the solution of structural dynamic problems for both blast and seismic loading. This method may be seen as an extension of the Wilson θ method U = U + (1 δ) U + δ U Δt t+δ t t t t+δt 1 U= U+ U Δ t+ ( α) U + α U Δt t +Δ t t t t t +Δ t δ and α are parameters which may be adjusted to achieve accuracy and stability The terms essentially result from the use of a Taylor series expansion: δ =0.5, α = 1/6 corresponds to the linear acceleration method which also correspond t U = t- t U to + the t t- t Wilson U + t θ method t- tü t 3 + with θ =... t- t 1 Newmark originally proposed δ =0.5, 05 α = 1/4 which results in an t unconditionally U = t- t stable U + tscheme t- t Ü + t (the trapetzoidal rule) thod of Finite Elements II!! t- t... U +... 3! U +... Institute of Structural Engineering Method of Finite Elements II 8
The Newmark Method Notes: δ and a are parameters, effectively acting as weights for calculating the approximation of the acceleration, and may be adjusted to achieve accuracy and stability δ = 0.5, a = 1/6 is known as the linear acceleration method, which also correspond to the Wilson θ method with θ = 1 Newmark originally proposed δ=0.5, a = 1/4, which results in an unconditionally stable scheme (the trapezoidal rule) Institute of Structural Engineering Method of Finite Elements II 9
The Newmark Method We now solve for the displacements, velocities and accelerations by inserting the above into the dynamic equilibrium equation: where Institute of Structural Engineering Method of Finite Elements II 30
The Newmark Method Solution Procedure A. For each time step Institute of Structural Engineering Method of Finite Elements II 31
The Direct Newmark Integration ti Method Methods Solution Procedure The Newmark method : Solution procedure: B. For each time step B: For each time step 1) Calculate effective loads at time t: t +Δ t ˆ t +Δ t t t ( t R = R+ M a 0 U+ a U+ a3 U ) t t t + C( a U+ a U + a U ) 1 4 5 ) Solve for the displacements U at time t + Δt Tt t t t LDL +Δ U = +Δ Rˆ 3) Solve for the corresponding velocities and accelerations t+δ t t+δt t t t U = a0( U U) a U a3 U t+δ t t t+δ t t U= U+ a U+ a U thod of Finite Elements II 7 6 Institute of Structural Engineering Method of Finite Elements II 3
The Newmark Method Stability of the Newmark Method For zero damping the Newmark method is conditionally stable if δ 1, α 1 and t 1 ω max δ α where ω max is the maximum natural frequency. The Newmark method is unconditionally stable if α δ 1 Institute of Structural Engineering Method of Finite Elements II 33
The Newmark Method Stability of the Newmark Method However, if δ 1, errors are introduced. These errors are associated with numerical damping and period elongation, i.e. a seemingly larger damping and period of oscillation than in reality. Because of the unconditional stability of the average acceleration method, it is the most robust method to be used for the step-by-step dynamic analysis of large complex structural systems in which a large number of high frequencies, short periods, are present. The only problem with the method is that the short periods, which are smaller than the time step, oscillate indefinitely after they are excited. The higher mode oscillation can however be reduced by the addition of stiffness proportional (artificial) damping. source: csiberkeley.com Institute of Structural Engineering Method of Finite Elements II 34
The Newmark Method dof system example t = 0.8s 6 5 4 Discrete Time Step, Δt=1 sec 1st Floor CMD nd Floor CMD 1st Floor NM nd Floor NM 1st Floor true nd Floor true Displacement 3 1 0 1 0 1 3 4 5 6 7 8 9 10 time Institute of Structural Engineering Method of Finite Elements II 35
The Newmark Method dof system example t = 1s Displacement 0 15 10 5 0 Discrete Time Step, Δt=1 sec 1st Floor CMD nd Floor CMD 1st Floor NM nd Floor NM 1st Floor true nd Floor true 5 10 0 4 6 8 10 time Institute of Structural Engineering Method of Finite Elements II 36
The Wilson θ Method APPENDIX The Wilson θ Method In 1973, the general Newmark method was made unconditionally stable by the introduction of a θ factor. The introduction of this factor is motivated by the observation that an unstable solution tends to oscillate about the true solution. Therefore, if the numerical solution is evaluated within the time increment the spurious oscillations are minimized. This can be accomplished by a simple modification to the Newmark method by using a time step defined by: t = θ t, θ 1.0 source: csiberkeley.com Reminder: For θ = 1 the Wilson θ method is equivalent to the Newmark linear acceleration method with δ = 0.5, a = 1/6. Institute of Structural Engineering Method of Finite Elements II 37
Direct Integration ti Methods The Wilson θ Method The Wilson θ method In this method the acceleration is assumed to vary linearly from time t to t + Δt τ U = U + U U θδt θ ( ) t + τ t t + Δ t t By integration we obtain τ U = U + U τ + U U θδt θ ( ) t+ τ t t t+ Δt t U 1 1 = U + U τ + τ + τ U θ U U 6θΔt θ ( ) t+ τ t t t 3 t+ Δt t thod of Finite Elements II Institute of Structural Engineering Method of Finite Elements II 38
The Wilson θ Method The Wilson θ method Setting τ = θ Δt we get θδt U = U + U + U θ ( ) t+δ θ t t t+δt t 1 U= U+ U θδ t+ Δ U + U 6 θ ( θ t) ( ) t +Δ θ t t t t +Δ t t from which we can solve Swiss Federal Institute of Technology Page 4 t+δ θ t 6 t+δt t 6 t t Direct U = Integration U U U U ti Methods θδt θ ( ) ( ) θδt t+δ θ t 3 t+δ θ t t t θδt t U = ( U U ) U U The Wilson θδt θ method Method of Finite Elements II 6 6 U = U U U U θδt +Δ θ t t t+δt t t t ( ) ( θ ) θδt 3 θδ t U = ( U U) U U θδt We now solve for the displacements, velocities and accelerations by inserting into the dynamic equilibrium equation t+δ θ t t+ θδ t t+ θδ t t+ θδt M U + C U + K U= R R = R + θ ( R R ) t + θδ t t t +Δ t t Institute of Structural Engineering Method of Finite Elements II 39
The Wilson θ Method Solution Procedure A. Initial Calculations Implicit Procedure! Institute of Structural Engineering Method of Finite Elements II 40
The Direct Wilson Integration θ Method ti Methods Solution Procedure The Wilson θ method : Solution procedure: B. For each time step B: For each time step 1) Calculate effective loads at time t+ t: t +Δ θ t ˆ t t +Δ θ t t t t t R = R+ θ ( R R) + M( a 0 U+ a U+ U ) t t t + C( a1 U+ U + a3 U ) ) Solve for the displacements U at time t + Δt Tt θ t t θ t LDL +Δ U = +Δ Rˆ 3) Solve for the corresponding velocities and accelerations t+δ t t+ θδt t t t U = a4( U U) + a5 U + a6 U t+δ t t t+δt t U = U+ a ( U + U ) t+δ t 7 U= t U+ Δ t t U + a ( t+δt t ) 8 U + U thod of Finite Elements II Institute of Structural Engineering Method of Finite Elements II 41
Direct Integration Methods Coupling of integration operators For some problems it may be an advantage to combine the different types of integration schemes e.g. if a structure is subjected to dynamic load effect from hydrodynamic loading then the analysis of the hydrodynamic forces may be assessed using an explicit scheme and the structural response by using an implicit scheme. The best choice of strategy will depend on the problem at hand in regard with stability and accuracy! Institute of Structural Engineering Method of Finite Elements II 4