KIT Workshop
Dr. Pritam ChakrabortyAsst. Prof., AE, IIT Kanpur
16th January, 2020Outreach Audi., IIT Kanpur
Outline
▪ NonlinearityDefinition, Examples.
▪ Newton Raphson
▪ Nonlinear 1D Bar ProblemODEs, weak Form, solution strategy, MATLAB code
▪ Incremental Method
▪ Postprocessing
▪ Other Direct Solvers
▪ Problems to Solve
References
▪ Nonlinear Finite Elements for Continua and StructuresBy Ted Belytschko, Wing Kam Liu, Brian Moran, Khalil Elkhodary
▪ Finite element proceduresBy Klaus-Jürgen Bathe
▪ Non‐Linear Finite Element Analysis of Solids and Structures, Vol 1By M. A. Crisfield
▪ Nonlinear Finite Element MethodsBy Peter Wriggers
Nonlinearity
A physical system (a body or assembly, etc.) is defined as nonlinear if the output(s) or response(s), y, of the system due to input(s), x, do not possess a linear relationship, or,
𝑦 ≠ 𝑚𝑥 + 𝑐
In structural mechanics, the outputs and inputs are forces, stresses, displacements, strains, etc.
Nonlinearity implies that force do not follow a linear relationship with displacement, or vice-versa.
Types of Nonlinearity
▪ GeometricInvolves large displacement and rotation, and/or deformation of structural elements
Examples: cables, membranes, frames, tyres, metal forming
Hyperelastic seal CableBuckling
Types of Nonlinearity
▪ Physical or MaterialNonlinear response function between stress and strainExamples: metal plasticity, viscoelasticity, soil, damage
Stress-strain Damage of Porous Material
Types of Nonlinearity
▪ Boundary ConditionsDeformation dependent boundary condition, contact
Contact Rolling
Geometry, material and boundary nonlinearity
Geometric Nonlinearity
Example: Large displacement of a rigid beam [Wriggers, 2008]
Initial Configuration
Equilibrated Configuration
From Moment Balance 𝐹𝑙 cos ∅ = 𝑐 ∅
Hence, spring rotation, f, output, is a nonlinear function of applied force, F, input.
Geometric Nonlinearity
Example: Large displacements of elastic springs [Wriggers, 2008]
Initial Configuration
Equilibrated Configuration
Kinematics
Elongation
Deflection
Force Balance
𝑓 = 𝑙 1 +𝑤
𝑙
2
− 1
𝐹 = 2𝑇𝑠𝑖𝑛 𝜃 = 2𝑇𝑤
𝑙 + 𝑓
For a linear spring 𝑇 = 𝑐𝑓
𝑤
𝑙1 −
1
1 +𝑤𝑙
2=
𝐹
2𝑐𝑙
Hence,
Or, deflection of spring mid-point, w, output, is a nonlinear function of applied force, F, input.
𝐹
2𝑐𝑙
Geometric Nonlinearity
Example: Snap through [Wriggers, 2008]
Initial Configuration Equilibrated Configuration
Truss members
Kinematics
Original length - L0 ; Current Length - LDeflection of A - w
ℎ − 𝑤 2 + 𝑙2 = 𝐿2
ℎ2 + 𝑙2 = 𝐿02
𝑓 = 𝑙 1 +ℎ − 𝑤
𝑙
2
− 1 +ℎ
𝑙
2
Force Balance
𝐹 = 2𝑇𝑠𝑖𝑛 𝛼 − 𝜑 = 2𝑇ℎ − 𝑤
𝐿
For a linear spring 𝑇 = 𝑐𝑓
Hence, 𝑤 − ℎ
𝑙1 −
𝐿0
𝑙 1 +𝑤 − ℎ𝑙
2
=𝐹
2𝑐𝑙
Loss of equilibrium at D and restored at ESnap through –equilibrium at D or E for a F and different w.
𝐹
2𝑐𝑙
𝐿0𝑙= 1.25
Material Nonlinearity
Example: Response of elasto-plastic bar assembly [Wriggers, 2008]
Bars 1 and 2 experience same strain, e, as that of assembly
Force Equilibrium: F = F1 + F2 = s1 A + s2 A
Stress of the assembly: S = F/2A
Elasto-Perfectly Plastic When bars 1 and 2 are elastic
𝜎1 = 2𝐸𝜀 𝜎2 = 𝐸𝜀 𝜀 =𝐹
3𝐸𝐴𝜎1 =
2𝐹
3𝐴𝜎2 =
𝐹
3𝐴
Σ =3
2𝐸𝜀 = ሖ𝐸𝜀
Onset of plasticity in bar 2
𝜀∗ =𝜎𝑌𝐸
𝜎1 = 2𝐸𝜀∗ 𝜎2 = 𝜎𝑌 Σ∗ =3
2𝜎𝑌
Bar 1 elastic and bar 2 plastic
𝐹 = 𝜎𝑌𝐴 + 2𝐸𝜀𝐴 Ƹ𝜀 = 𝜀 − 𝜀∗ 𝐹 = 3𝜎𝑌𝐴 + 2𝐸𝐴 Ƹ𝜀
Σ = Σ − Σ∗ = 𝐸 Ƹ𝜀
Material Nonlinearity
Onset of plasticity in bar 1
𝜎1 = 3𝜎𝑌 𝜎2 = 𝜎𝑌
𝜀∗∗ =3𝜎𝑌2𝐸
=3
2𝜀∗
𝐹 = 𝜎𝑌𝐴 + 3𝜎𝑌𝐴
Σ∗∗ = 2𝜎𝑌
Bar 1 and 2 plastic
Σ = 2𝜎𝑌
Newton Raphson
In the examples, nonlinear algebraic equations describe the relationship between forces (stresses) and displacements (deflections, strains)
𝐹𝑙
𝑐=
∅
cos∅1st example:
Evaluating𝐹𝑙
𝑐∅for a is trivial. However the reverse requires solving of roots.
Newton-Raphson is a widely used method for finding out roots.
Iteratively solves by linearizing a nonlinear equation.
Algorithm
Problem: f(u) is some function of u. Find u0 such that f(u0) = 0.
Solution: Let us guess a u = u1. No that lucky, so, f(u1) ≠ 0
But at u0 = u1 + Δu1, which we don’t know, f(u0) = 0 or f(u1 + Δu1) = 0Taylor series expansion of f(u) about u1:
𝑓 𝑢1 + ∆𝑢1 = 𝑓 𝑢1 + ቚ𝑑𝑓
𝑑𝑢 𝑢1∆𝑢1 +
1
2ቚ
𝑑2𝑓
𝑑𝑢2 𝑢1∆𝑢1
2 +⋯ = 0
Truncate the series after linear term and equate to zero (Linearization)
𝑓𝐿 𝑢1 + ∆෦𝑢1 = 𝑓 𝑢1 + ቚ𝑑𝑓
𝑑𝑢 𝑢1∆෦𝑢1 = 0 ∆෦𝑢1 = −
𝑓 𝑢1
ฬ𝑑𝑓𝑑𝑢 𝑢1
Update: 𝑢2 = 𝑢1 + ∆෦𝑢1
However, u2 need not be u0 or f(u2) ≠ 0Repeat step 1 about u2 to obtain an updated value
Step 1:
Flow chart
Initial Guess Is
Update
Y
N
i = Iteration number
Example
𝐹𝑙
𝑐=
𝑢
cos(𝑢)
𝑢0cos(𝑢0)
= 2Find u0 such that Or𝑓 𝑢 = 2 cos 𝑢 − 𝑢
And find u0 such that f(u0) = 0
Initial guess of u = 0 and df/du = -2sin(u) - 1𝑓 𝑢1
𝑓𝐿 𝑢2 = 0
𝑓 𝑢2𝑓𝐿 𝑢3 = 0
𝑓𝐿 𝑢4 = 0
𝑓 𝑢3
𝑓 𝑢4
𝑓𝐿 𝑢5 = 0
u 0 2 0.99514 1.030107 1.029867
f(u) 2 -2.83229 0.093631 -0.00065 -2.98E-08
MATLAB Code
Boundary Value Problems
Mechanical behaviour of continuous bodies or their assembly can be suitably represented by boundary value or initio-boundary value problems
Example: Elastic Cantilever Beam
xy
z𝜕𝜎𝑖𝑗
𝜕𝑥𝑗= 0
ሶ𝜎𝑖𝑗 = 𝜎𝑖𝑗(𝐷𝑘𝑙)
𝐿𝑖𝑗 = 𝐷𝑖𝑗 +𝑊𝑖𝑗 =𝜕𝑣𝑖𝜕𝑥𝑗
Governing
Constitutive
Kinematic
• At x = y = z = 0u = v = w = 0
• At x = 400 y = -60 z = 40Fz = -2500
• On all the faces (exclude the fixed face)tx = ty = tz = 0
Boundary Conditions
Problem can be nonlinear• geometrically if cantilever beam undergoes large bending • physically (material) if the stress or stress rate is a nonlinear function of displacement
or velocity gradients
Hence, solve nonlinear PDE(s) to model deformation behaviour of continuous bodies
Nonlinear Differential Equations
Let L is a differential operator and operates on some scalar or vector or tensor valued
function, u. (e.g. 𝐿 =𝜕
𝜕𝑥)
The operator is said to be linear if for w = u + v, Lw = Lu + Lv
If the above doesn’t hold then the ordinary or partial differential equation is nonlinear
Example: Linear Homogeneous ODE
𝑑2𝑢
𝑑𝑥2+ 𝑢 = 0 for some u(x), and
𝑑2𝑣
𝑑𝑥2+ 𝑣 = 0 for some v(x). Then if w(x) = u(x) + v(x), then
𝑑2𝑤
𝑑𝑥2+ 𝑤 = 0
Example: Nonlinear Homogeneous ODE
𝑑2𝑢
𝑑𝑥2+ 𝑢2 = 0 for some u(x), and for some v(x). Then if w(x) = u(x) + v(x), then
𝑑2𝑣
𝑑𝑥2+ 𝑣2 = 0
𝑑2𝑤
𝑑𝑥2+𝑤2 =
𝑑2𝑢
𝑑𝑥2+ 𝑢2 +
𝑑2𝑣
𝑑𝑥2+ 𝑣2 + 2𝑢𝑣 ≠ 0
FEM for Nonlinear Differential Equations
▪ A numerical technique for solving ordinary (ODEs) and partial differential equations (PDEs)
▪ Based on decomposition of domain into finite regions or elements
▪ Interpolative approximation of the primary variable(s) in every element
▪ Weak form of ODE or PDE in every element gives algebraic equations
▪ Linear: Stiffness matrix and force vector (RHS vector) derivable from weak form▪ Nonlinear: Linearization to obtain the tangent stiffness (Jacobian) matrix and RHS vector (Residual)
▪ Assembly of tangent stiffness matrix and RHS vector for all elements
▪ Apply boundary conditions
▪ Linear: Solve to obtain unknown nodal values▪ Nonlinear: Solve to obtain update on initial guess of nodal values
Calculate RHS vector and check for convergence (Newton Raphson)
Example: Nonlinear bar in 1D
L0
L
A0(X)
A(x)
u0
A tapered bar fixed on left face and displaced by u0 on right face.
The bar is incompressible elastic and undergoes large deformation - nonlinear
Assumption: Stress and strain components, other than sxx and exx , are zero
Initial or Reference Configuration (t = 0)
Deformed or Current Configuration (t)
Ƹ𝑒1
X: Position vector of a particle, A, at t = 0.x: Position vector of A at t
x(X,t)
Kinematics
Motion: Maps any particle from the initial position to its current position or x = x( X, t)For a continuous body (without discontinuities such as cracks), the mapping is uniqueThus, inverse exists, i.e. X = x-1( x, t)
Displacement of particle A: u = x – XIn terms of initial position, u = ො𝑢 𝑋, 𝑡 = 𝑥 𝑋, 𝑡 − 𝑋
current position, u = 𝑢 𝑥, 𝑡 = 𝑥 − 𝑥−1(𝑥, 𝑡)
Deformation Gradient (F) of A and its small neighbourhood
A A’(X) (X+DX)
t=0
A A’(x) (x+Dx)
t
∆𝑥 = 𝑥 𝑋 + ∆𝑋, 𝑡 − 𝑥 𝑋, 𝑡 ∆𝑥 ≈ 𝑥 𝑋, 𝑡 +𝑑𝑥
𝑑𝑋(𝑋, 𝑡)∆𝑋 − 𝑥 𝑋, 𝑡
Taylor series till linear term
∆𝑥 =𝑑𝑥
𝑑𝑋𝑋, 𝑡 ∆𝑋 = 𝐹∆𝑋 𝐹 =
𝑑𝑥
𝑑𝑋= 1 +
𝑑ො𝑢
𝑑𝑋
Initial Current
Deformation Gradient
Kinematics
Adopting the Green-Lagrange strain definition to account for large stretching of material filament
This strain is defined w.r.t. to the infinitesimal length of filament in initial position or DX.and strain is defined in the initial configuration of the body (t = 0).
In terms of u:
Under small strain assumption, x ~ X, giving
Thus the Green-Lagrange strain introduces a nonlinear term to account for large deformation
Force Balance
Deformed body under force equilibrium
PP
Cutting Planex
P P(x)
In current configuration: P(x) = P = constant Or P(x(X,t)) = P’(X,t) = constant
Condition of force equilibrium in initial configuration
In terms of stress
sN – Nominal stress component sxx; s – True stress component sxx
From incompressibility: A0 DX = A Dx or A0 = AF
Constitutive Equation
Define a linear nominal stress vs Green Lagrange strain relation
Both the stress and strain are defined in the initial configuration
Nonlinear ODE
Governing Equation
Strain - Displacement
Stress - Strain
Boundary Conditions: u(x = 0) = 0 and u(x = L0) = u0
Lagrangian and Eulerian Descriptions
Lagrangian DescriptionBalance Laws, Kinematics and Constitutive equations are described in the initial configuration (X)
Eulerian DescriptionBalance Laws, Kinematics and Constitutive equations are described in the current configuration (x)
Equilibrium
The constitutive and kinematics are described in the rate form
Hypoelastic
Lagrangian FEM
In structural/solid mechanics, the interest is the deforming body and its state of stress, etc.Hence, the solving methodology must track the motion of the domain or the initial mesh (discretized domain)
Two widely used approaches:
Total Lagrangian :
▪The Lagrangian description of mass and momentum balance, kinematics and constitutive equations are solved▪Interpolation and weak form are defined over the initial configuration ▪Mesh remains undistorted
Lagrangian FEM
Updated Lagrangian :
▪The Eulerian description of mass and momentum balance, kinematics and constitutive equations are solved▪The deformation gradient is retained to connect the configurations▪Interpolation and weak form are defined over the current configuration ▪Mesh distorts with deformation
▪Used in FEM softwares
Demonstration of solving the 1-D problem using the Total Lagrangian Method and related derivations
▪Derive the residual vector and tangent stiffness matrix for 1 element▪Assembly for n elements▪Application of BCs▪MATLAB CODE
▪Use of Gauss Quadrature▪General representation of residual vector and tangent stiffness matrix▪MATLAB CODE
Assembly – Pseudo Code
Assembly – Pseudo Code
Incremental Method
In structural problems, the loads and/or displacements are applied over a finite interval of time.
P
t
PF
d
t
dF
Applied
d
P
PF, dF
Response
When solving, the interest is only at certain times or state In previous examples, stress, strain, etc. were obtained at (PF, dF)
For nonlinear problems, achieving the final state from initial state can lead to convergence difficulties
One reason – Due to an initial guess far from the actual solution
OR
Incremental Method
Divide the applied loads and/or displacements into finite number of intervals
P
t
PF
t1 t2 t3 t4 t5 tf
Time points - t1, t2, ...
Time Increment – Dtn = tn+1- tn , n – discrete points
For static problems with rate independent constitutive models, time points and increments are markers
The applied load and/or displacement increments determine convergence of N-R iterations
Example: Using the tapered bar problem subjected to load at the right end
P
t
PF
t1 t2 t3 t4 t5 tf
P1
P2
The load is assumed to vary linearly with time followed by equal divisions of the total
The final state is reached in the following manner:
Knowing the initial state, first solve for applied load of P1
using NR (initial guess t = 0)
Knowing the state at t1, solve for applied load of P2 using NR (initial guess t = t1)
So on and so forth till PF
Example: Incremental Method
Demonstration of solving the 1-D problem using the Incremental method
▪MATLAB code
Flow chart
Flow chart
IN
OUT
Postprocessing
▪ Nodal values of only the primary variables are accurate without any further manipulation
▪ Stress, strain, etc. are accurate only at the Gauss points
▪ Nodal stresses, etc. are extrapolation or obtained through other specialized techniques (superconvergent patch recovery [Zienkiewicz, 1992])
▪ Nodal forces (reactions) can be obtained as follows for the 2 element 2 DirichletBC problem - once the NR iterations have converged.
Some Other Direct Solvers
▪ NR method has quadratic convergence
▪ However, requires ▪The exact tangent stiffness matrix to be calculated in each iteration▪Assembly of the local matrices▪Inversion (Factorization) of the tangent stiffness matrix in every iteration –n3 operation
▪ While still the most effective and widely used, other modified versions of Newton’s update can help
▪Reducing number of factorization▪Calculation of exact tangent stiffness
Modified Newton Raphson
▪The tangent stiffness matrix is derived only in the first iteration of the iterative update of a time point▪ The choice of time points depends on the nonlinearity (iterations to converge)
Quasi Newton Method
▪The iterations in a time point starts with an approximate stiffness matrix▪ Updated iteratively by the enforcing the secant condition
Secant condition
▪The Broyden-Fletcher-Goldfarb-Shanno (BFGS) method is the most popular for the secant update of the matrix – rank two update
BFGS Update
Problem: 1d-linear bar with nonlinear body force
▪ Derive the residual vector and tangent stiffness matrix for an element using linear Lagrange shape functions.
▪ For the bar with 2 elements, show the equation(s) to solve.
Boundary Conditions: u(x = 0) = 0 and u(x = L0) = u0
Problem 2: 1d nonlinear bar with C2 elements
▪ Use quadratic shape functions to derive the residual vector and tangent stiffness matrix.
▪ For a single element obtain the nonlinear equation(s) to solve.▪ Find the suitable n-point rule if quadratic shape functions are used in
the master element.
Governing Equation
Strain - Displacement
Stress - Strain
Boundary Conditions: u(x = 0) = 0 and u(x = L0) = u0