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