The Finite Element Method for the Analysis of Non-Linear and ...

The Finite Element Method for the Analysis of 

Non-Linear and Dynamic Systems 

Prof. Dr. Eleni Chatzi 

Lecture 1 - 17 September, 2013 

Institute of Structural Engineering Method of Finite Elements II 1

Course Information 

Instructor 

Prof. Dr. Eleni Chatzi, email: chatzi@ibk.baug.ethz.ch 

Office Hours: HIL E14.3, Wednesday 10:00-12:00 or by email 

Assistant 

Juan Escalln Osorio, HIL E10.2, email: ejuan@student.ethz.ch 

Course Website 

Lecture Notes and Homeworks will be posted at: 

http://www.ibk.ethz.ch/ch/education 

Suggested Reading 

Nonlinear Finite Elements for Continua and Structures by T. 

Belytschko, W. K. Liu, and B. Moran, John Wiley and Sons, 2000 

The Finite Element Method: Linear Static and Dynamic Finite 

Element Analysis by T. J. R. Hughes, Dover Publications, 2000 

The Finite Element Method Vol. 2 Solid Mechanics by O.C. 

Zienkiewicz and R.L. Taylor, Oxford : Butterworth Heinemann, 2000 


Course Outline 

Review of the Finite Element method - Introduction to 

Non-Linear Analysis 

Non-Linear Finite Elements in solids and Structural Mechanics 

- Overview of Solution Methods 

- Continuum Mechanics & Finite Deformations 

- Lagrangian Formulation 

- Structural Elements 

Dynamic Finite Element Calculations 

- Integration Methods 

- Mode Superposition 

Eigenvalue Problems 

Special Topics 

- Extended Finite Elements, Multigrid Methods, Meshless Methods 


Grading Policy 

Performance Evaluation - Homeworks (100%) 

Homework 

Homeworks are due in class 2-3 weeks after assignment 

Computer Assignments may be done using any coding language 

(MATLAB, Fortran, C, MAPLE) - example code will be 

provided in MATLAB 

Commercial software such as CUBUS, ABAQUS and SAP will 

also be used for certain Assignments 

Homework Sessions will be pre-announced and it is advised to bring 

a laptop along for those sessions 


Review of the Finite Element Method (FEM) 

Classification of Engineering Systems 

Discrete 

Continuous 

q|y+dy 

h 1 

q|x 

dy 

q|x+dx 

dx 

q|y 

h 2 

L 

Flow 

of water 

Permeable Soil 

F = KX 

Direct Stiffness Method 

( ) 

k ∂ 2 φ 

∂ 2 x + ∂2 φ 

∂ 2 y 

Impermeable Rock 

= 0 

Laplace Equation 

FEM: Numerical Technique for approximating the solution of continuous 

systems. We will use a displacement based formulation and a stiffness 

based solution (direct stiffness method). 



How is the Physical Problem formulated? 

The formulation of the equations governing the response of a system under 

specific loads and constraints at its boundaries is usually provided in the 

form of a differential equation. The differential equation also known as the 

strong form of the problem is typically extracted using the following sets 

of equations: 

1 Equilibrium Equations 

aL + ax 

ex. f (x) = R + (L − x) 

2 

2 Constitutive Requirements 

Equations 

ex. σ = Eɛ 

3 Kinematics Relationships 

ex. ɛ = du 

dx 

Axial bar Example 

q(x)=ax 

x 

ax aL 

f(x) 

L-x 

R 

R 



How is the Physical Problem formulated? 

Differential Formulation (Strong Form) in 2 Dimensions 

Quite commonly, in engineering systems, the governing equations are of a 

second order (derivatives up to u ′′ or ∂2 u 

∂ 2 x 

) and they are formulated in terms 

of variable u, i.e. displacement: 

Governing Differential Equation ex: general 2nd order PDE 

A(x, y) ∂2 u 

∂ 2 x + 2B(x, y) ∂2 u 

∂x∂y + C(x, y) ∂2 u 

∂u 

∂ 2 y 

= φ(x, y, u, 

∂y , ∂u 

∂y ) 

Problem Classification 

B 2 − AC < 0 ⇒ elliptic 

B 2 − AC = 0 ⇒ parabolic 

B 2 − AC > 0 ⇒ hyperbolic 

Boundary Condition Classification 

Essential (Dirichlet): u(x 0 , y 0 ) = u 0 

order m − 1 at most for C m−1 

Natural (Neumann): 

∂u 

∂y (x 0, y 0 ) = ˙u 0 

order m to 2m − 1 for C m−1 



Differential Formulation (Strong Form) in 2 Dimensions 

The previous classification corresponds to certain characteristics for each 

class of methods. More specifically, 

Elliptic equations are most commonly associated with a diffusive or 

dispersive process in which the state variable u is in an equilibrium 

condition. 

Parabolic equations most often arise in transient flow problems where 

the flow is down gradient of some state variable u. Often met in the 

heat flow context. 

Hyperbolic equations refer to a wide range of areas including 

elasticity, acoustics, atmospheric science and hydraulics. 


Strong Form - 1D FEM 

Reference Problem 

Consider the following 1 Dimensional (1D) strong form (parabolic) 

d 

dx (c(x)du dx ) + f(x) = 0 

− c(0) d dx u(0) = C 1 

u(L) = 0 

(Neumann BC) 

(Dirichlet BC) 

Physical Problem (1D) Diff. Equation Quantities 

One dimensional Heat 

flow 

Axially Loaded Bar 

d dT 

Ak 

dx dx + Q = 0 

d du 

AE 

dx dx + b = 0 

T=temperature 

A=area 

k=thermal 

conductivity 

Q=heat supply 

u=displacement 

A=area 

E=Young’s 

modulus 

B=axial loading 

Constitutive 

Law 

Fourier 

q = −k dT/dx 

q = heat flux 

Hooke 

σ = Edu/dx 

σ = stress 


Weak Form - 1D FEM 

Approximating the Strong Form 

The strong form requires strong continuity on the dependent field 

variables (usually displacements). Whatever functions define these 

variables have to be differentiable up to the order of the PDE that 

exist in the strong form of the system equations. Obtaining the 

exact solution for a strong form of the system equation is a quite 

difficult task for practical engineering problems. 

The finite difference method can be used to solve the system 

equations of the strong form and obtain an approximate solution. 

However, this method usually works well for problems with simple 

and regular geometry and boundary conditions. 

Alternatively we can use the finite element method on a weak 

form of the system. This weak form is usually obtained through 

energy principles which is why it is also known as variational form. 



From Strong Form to Weak form 

Three are the approaches commonly used to go from strong to weak 

form: 

Principle of Virtual Work 

Principle of Minimum Potential Energy 

Methods of weighted residuals (Galerkin, Collocation, Least 

Squares methods, etc) 

*We will mainly focus on the third approach. 



From Strong Form to Weak form - Approach #1 

Principle of Virtual Work 

For any set of compatible small virtual displacements imposed on the body 

in its state of equilibrium, the total internal virtual work is equal to the 

total external virtual work. 

∫ 

W int = 

where 

Ω 

∫ 

¯ɛ T τ dΩ = W ext = 

Ω 

∫ 

ū T bdΩ + ū ST T S dΓ + ∑ 

Γ 

i 

ū iT R C 

i 

T S : surface traction (along boundary Γ) 

b: body force per unit area 

R C : nodal loads 

ū: virtual displacement 

¯ɛ: virtual strain 

τ : stresses 




Principle of Minimum Potential Energy 

Applies to elastic problems where the elasticity matrix is positive definite, 

hence the energy functional Π has a minimum (stable equilibrium). 

Approach #1 applies in general. 

The potential energy Π is defined as the strain energy U minus the work of 

the external loads W 

Π = U − W 

U = 1 ∫ 

ɛ T CɛdΩ 

2 Ω 

∫ 

∫ 

W = ū T bdΩ + ū ST T s dΓ T + ∑ 

Ω 

Γ T i 

ū T i R C 

i 

(b T s , R C as defined previously) 




Galerkin’s Method 

Given an arbitrary weight function w, where 

S = {u|u ∈ C 0 , u(l) = 0}, S 0 = {w|w ∈ C 0 , w(l) = 0} 

C 0 is the collection of all continuous functions. 

Multiplying by w and integrating over Ω 

∫ l 

0 

w(x)[(c(x)u ′ (x)) ′ + f (x)]dx = 0 

[w(0)(c(0)u ′ (0) + C 1 ] = 0 



Using the divergence theorem (integration by parts) we reduce the 

order of the differential: 

∫ l 

0 

∫ l 

wg ′ dx = [wg] l 0 − gw ′ dx 

0 

The weak form is then reduced to the following problem. Also, in 

what follows we assume constant properties c(x) = c = const. 

Find u(x) ∈ S such that: 

∫ l 

0 

w ′ cu ′ dx = 

∫ l 

0 

wfdx + w(0)C 1 

S = {u|u ∈ C 0 , u(l) = 0} 

S 0 = {w|w ∈ C 0 , w(l) = 0} 


Weak Form 

Notes: 

1 Natural (Neumann) boundary conditions, are imposed on the 

secondary variables like forces and tractions. 

For example, ∂u 

∂y (x 0, y 0 ) = ˙u 0 . 

2 Essential (Dirichlet) or geometric boundary conditions, are imposed 

on the primary variables like displacements. 

For example, u(x 0 , y 0 ) = u 0 . 

3 A solution to the strong form will also satisfy the weak form, but not 

vice versa.Since the weak form uses a lower order of derivatives it can 

be satisfied by a larger set of functions. 

4 For the derivation of the weak form we can choose any weighting 

function w, since it is arbitrary, so we usually choose one that satisfies 

homogeneous boundary conditions wherever the actual solution 

satisfies essential boundary conditions. Note that this does not hold 

for natural boundary conditions! 


FE formulation: Discretization 

How to derive a solution to the weak form? 

Step #1:Follow the FE approach: 

Divide the body into finite elements, e, connected to each other 

through nodes. 

e 

x 1 

e 

x 2 

e 

Then break the overall integral into a summation over the finite 

elements: 

[ 

∑ ∫ x e ∫ ] 

2 

x e 

w ′ cu ′ 2 

dx − wfdx − w(0)C 1 = 0 

x1 

e x1 

e 

e 


1D FE formulation: Galerkin’s Method 

Step #2: Approximate the continuous displacement using a discrete 

equivalent: 

Galerkin’s method assumes that the approximate (or trial) solution, u, can 

be expressed as a linear combination of the nodal point displacements u i , 

where i refers to the corresponding node number. 

u(x) ≈ u h (x) = ∑ i 

N i (x)u i = N(x)u 

where bold notation signifies a vector and N i (x) are the shape functions. 

In fact, the shape function can be any mathematical formula that helps us 

interpolate what happens at points that lie within the nodes of the mesh. 

In the 1-D case that we are using as a reference, N i (x) are defined as 1st 

degree polynomials indicating a linear interpolation. 

As will be shown in the application presented in the end of this lecture, for the 

case of a truss element the linear polynomials also satisfy the homogeneous 

equation related to the bar problem. 



Shape function Properties: 

Bounded and Continuous 

One for each node 

Ni e(x j e) = δ ij, where 

{ 1 if i = j 

δ ij = 

0 if i ≠ j 

The shape functions can be written as piecewise functions of the x 

coordinate: 

This is not a convenient notation. 

⎧ 

x − x i−1 

Instead of using the global coordinate 

, x i−1 ≤ x < x i 

⎪⎨ x i − x i−1 x, things become simplified when 

N i (x) = xi + 1 − x 

, x i ≤ x < x 

using coordinate ξ referring to the 

i+1 

⎪⎩ 

x i+1 − x i local system of the element (see page 

0, otherwise 25). 



Step #2: Approximate w(x) using a discrete equivalent: 

The weighting function, w is usually (although not necessarily) chosen to 

be of the same form as u 

w(x) ≈ w h (x) = ∑ i 

N i (x)w i = N(x)w 

i.e. for 2 nodes: 

N = [N 1 N 2 ] u = [u 1 u 2 ] T w = [w 1 w 2 ] T 

Alternatively we could have a Petrov-Galerkin formulation, where w(x) is 

obtained through the following relationships: 

w(x) = ∑ i 

(N i + δ he 

σ 

δ = coth( Pee 

2 ) − 2 

Pe e 

dN i 

dx )w i 

coth = ex + e −x 

e x − e −x 



Note: Matrix vs. Einstein’s notation: 

In the derivations that follow it is convenient to introduce the 

equivalence between the Matrix and Einstein’s notation. So far we 

have approximated: 

u(x) = ∑ i 

N i (x)u i (Einstein ′ s notation) = N(x)u (Matrix notation) 

w(x) = ∑ i 

N i (x)w i = N(x)w 

(similarly) 

As an example, if we consider an element of 3 nodes: 

3∑ 

u(x) = N i (x)u i = N 1 u 1 + N 2 u 2 + N 3 u 3 ⇒ 

i 

⎡ 

u(x) = [N 1 N 2 N 3 ] ⎣ 

⎤ 

u 1 

u 2 

⎦ = N(x)u 

u 3 



Step #3: Substituting into the weak formulation and rearranging 

terms we obtain the following in matrix notation: 

∫ l 

0 

∫ l 

0 

w ′ cu ′ dx − 

∫ l 

(w T N T ) ′ c(Nu) ′ dx − 

0 

wfdx − w(0)C 1 = 0 ⇒ 

∫ l 

0 

w T N T fdx − w T N(0) T C 1 = 0 

Since w, u are vectors, each one containing a set of discrete values 

corresponding at the nodes i, it follows that the above set of equations can 

be rewritten in the following form, i.e. as a summation over the w i , u i 

components (Einstein notation): 

∫ l 

0 

( ∑ 

i 

) 

dN i (x) 

u i c 

dx 

⎛ 

⎝ ∑ j 

⎞ 

dN j (x) 

w j 

⎠ dx 

dx 

∫ l 

− f ∑ 

0 

j 

w j N j (x)dx − ∑ j 

w j N j (x)C 1 

∣ ∣∣∣∣∣x=0 

= 0 



This is rewritten as, 

[ 

∑ ∫ ( 

l ∑ 

w j 

j 

0 

i 

cu i 

dN i (x) 

dx 

) ] 

dN j (x) 

− fN j (x)dx + (N j (x)C 1 )| 

dx 

x=0 

= 0 

The above equation has to hold ∀w j since the weighting function w(x) is 

an arbitrary one. Therefore the following system of equations has to hold: 

∫ ( ) 

l ∑ dN i (x) dN j (x) 

cu i − fN j (x)dx + (N j (x)C 1 )| 

0 

dx dx 

x=0 

= 0 j = 1, ..., n 

i 

After reorganizing and moving the summation outside the integral, this 

becomes: 

[ 

∑ ∫ ] 

l 

c dN ∫ 

i(x) dN j (x) 

l 

u i = fN j (x)dx + (N j (x)C 1 )| 

dx dx 

x=0 

= 0 j = 1, ..., n 

0 

i 

0 



We finally obtain the following discrete system in matrix notation: 

Ku = f 

where writing the intagral from 0 to l as a summation over the 

subelements we obtain: 

K = A e K e −→ K e = 

∫ x e 

2 

x1 

e 

N Ț xcN ,x dx = 

∫ x e 

2 

x1 

e 

B T cBdx 

f = A e f e −→ f e = 

∫ x e 

2 

x e 1 

N T fdx + N T h| x=0 

where A is not a sum but an assembly (see page and, x denotes 

differentiation with respect to x. 

In addition, B = N ,x = dN(x) is known as the strain displacement 

dx 

matrix. 


1D FE formulation: Iso-Parametric Formulation 

Iso-Parametric Mapping 

This is a way to move from the use of global coordinates (i.e.in 

(x, y)) into normalized coordinates (usually (ξ, η)) so that the finally 

derived stiffness expressions are uniform for elements of the same 

type. 

x 

x 1 

e 

x 2 

e 

−1 1 

ξ 

Shape Functions in Natural Coordinates 

x(ξ) = ∑ 

N i (ξ)xi e = N 1 (ξ)x1 e + N 2 (ξ)x2 

e 

i=1,2 

N 1 (ξ) = 1 2 (1 − ξ), N 2(ξ) = 1 (1 + ξ) 

2 



Map the integrals to the natural domain −→ element stiffness matrix. 

Using the chain rule of differentiation for N(ξ(x)) we obtain: 

K e = 

∫ x 

e 

2 

x e 1 

where N ,ξ = d dξ 

and x ,ξ = dx 

dξ = x 2 e − x1 

e 

2 

N Ț xcN ,xdx = 

∫ 1 

−1 

(N ,ξ ξ ,x) T c(N ,ξ ξ ,x)x ,ξ dξ 

[ 1 (1 − ξ) 1 

(1 + ξ) ] = [ −1 

2 2 2 

= h = J (Jacobian) and h is the element length 

2 

ξ ,x = dξ 

dx = J−1 = 2/h 

From all the above, 

[ ] 

K e c 1 −1 

= 

x2 e − x 1 

e −1 1 

Similary, we obtain the element load vector: 

f e = 

∫ x 

e 

2 

x e 1 

N T fdx + N T h| x=0 = 

∫ 1 

−1 

1 

2 

N T (ξ)fx ,ξ dξ + N T (x)h| x=0 

Note: the iso-parametric mapping is only done for the integral. 

Institute of Structural Engineering Method of Finite Elements II 26 

]


So what is meant by assembly? (A e ) 

It implies adding the components of the stiffness matrix that correspond to 

the same degrees of freedom (dof). 

In the case of a simple bar, it is trivial as the degrees of freedom (axial 

displacement) are as many as the nodes: 

1:K 1 2:K 2 

1 2 3 

Red indicates the node each 

component corresponds to 

1 2 

2 3 

Element Stiffness Matrices (2x2): K 1 = c 1 −1 

h −1 1 1 

, K2 = c 1 −1 

2 h −1 1 

2 

3 

1 2 3 

1 −1 0 

Total Stiffness Matrix (4x4): K = c −1 1 + 1 −1 

h 

0 −1 1 

1 

2 

3 



In the case of a frame with beam elements, the stiffness matrix of the elements is 

typically of 4x4 size, corresponding to 2 dofs on each end (a displacement and a 

rotation): 

φ1 

φ2 

*The process will be shown explicitly 

during the HW sessions 

u1 

u2y 

φ2 

1 1:K 1 2 

2:K 2 

u2x 

Green indicates the dof each 

component corresponds to 

φ3 

u3 

u1 φ1 

3 

⎡ K 11 

1 1 K 12 

Element Stiffness Matrices (4x4): K 1 1 1 

⎢K = 12 K 22 

⎢ 1 1 

⎢ 

K 13 K 23 

1 1 ⎣K 14 K 24 

u2y φ2 

1 K 13 

1 K 14 

1 K 23 

1 K 24 

1 K 33 

1 K 34 

1 K 34 K 44 

u2x 

φ2 

⎡ K 11 

2 2 

u1 

K 12 

, K 2 2 2 

φ1 ⎢K = 12 K 22 

1 ⎦ ⎥⎥⎥⎤ u2y ⎢ 2 2 

⎢ 

K 13 K 23 

φ2 

2 2 ⎣K 14 K 24 

u3 φ3 

2 2 K 13 K 14 

2 2 K 23 K 24 

2 2 K 33 K 34 

2 K 34 K 2 44 ⎦ ⎥⎥⎥⎤ 

u2x 

φ2 

u3 

φ3 

φ2 u2y u2x 

K 2 

44 + K 22 

1 K 34 0 φ2 

Total Stiffness Matrix (2x2): K = 1 K 34 

1 K 33 

2 K 24 u2y 

0 2 K 24 

2 K 11 

u2x 

Fixed dofs are 

not included in 

the total 

stiffness matrix 


Axially Loaded Bar Example 

A. Constant End Load 

Given: Length L, Section Area A, Young’s modulus E 

Find: stresses and deformations. 

Assumptions: 

The cross-section of the bar does not change after loading. 

The material is linear elastic, isotropic, and homogeneous. 

The load is centric. 

End-effects are not of interest to us. 



A. Constant End Load 

Strength of Materials Approach 

From the equilibrium equation, the axial force at a random point x 

along the bar is: 

f(x) = R(= const) ⇒ σ(x) = R A 

From the constitutive equation (Hooke’s Law): 

ɛ(x) = σ(x) 

E 

Hence, the deformation is obtained as: 

δ(x) = ɛ(x) 

x 

= R 

AE 

⇒ δ(x) = Rx 

AE 

Note: The stress & strain is independent of x for this case of 

loading. 



B. Linearly Distributed Axial + Constant End Load 

From the equilibrium equation, the axial force at random point x 

along the bar is: 

f(x) = R + 

aL + ax 

2 

(L − x) = R + a(L2 − x 2 ) 

( depends on x) 

2 

In order to now find stresses & deformations (which depend on x) 

we have to repeat the process for every point in the bar. This is 

computationally inefficient. 



From the equilibrium equation, for an infinitesimal element: 

∆σ 

dσ 

Aσ = q(x)∆x + A(σ + ∆σ) ⇒ A }{{} 

lim + q(x) = 0 ⇒ A 

∆x dx + q(x) = 0 

∆x→0 

Also, ɛ = du 

dx , σ = Eɛ, q(x) = ax ⇒ AE d 2 u 

dx 2 + ax = 0 

Strong Form 

AE d 2 u 

dx + ax = 0 

2 

u(0) = 0 essential BC 

f(L) = R ⇒ AE du 

dx ∣ = R natural BC 

x=L 

Analytical Solution 

u(x) = u hom + u p ⇒ u(x) = C 1x + C 2 − ax 3 

6AE 

C 1, C 2 are determined from the BC 



An analytical solution cannot always be found 

Approximate Solution - The Galerkin Approach (#3): Multiply by the weight function 

w and integrate over the domain 

Apply integration by parts 

∫ L 

0 

∫ L 

0 

[ 

AE d2 u 

dx 2 wdx = AE du ] l 

dx w − 

0 

AE d2 u 

dx 2 wdx = [ 

∫ L 

∫ 

AE d2 u 

L 

0 dx 2 wdx + axwdx = 0 

0 

AE du 

dx 

∫ L 

0 

AE du dw 

dx dx dx ⇒ 

(L)w(L) − AE 

du 

dx (0)w(0) ] 

− 

∫ L 

0 

AE du dw 

dx dx dx 

But from BC we have u(0) = 0, AE du (L)w(L) = Rw(L), therefore the approximate 

dx 

weak form can be written as 

∫ L 

AE du 

∫ 

dw 

L 

0 dx dx dx = Rw(L) + axwdx 

0 



Variational Approach (#1) 

Let us signify displacement by u and a small (variation of the) displacement by δu. Then 

the various works on this structure are listed below: 

In addition, σ = E du 

dx 

∫ L 

δW int = A σδεdx 

0 

δW ext = Rδu| x=L 

∫ L 

δW body = qδudx 

0 

Then, from equilibrium: δW int = δW ext + δW body 

→ 

∫ L 

A 

0 

E du 

dx 

d(δu) 

dx 

∫ L 

dx = qδudx + Rδu| x=L 

0 

This is the same form as earlier via another path. 



In Galerkin’s method we assume that the approximate solution, u can be expressed as 

u(x) = 

n∑ 

u j N j (x) 

w is chosen to be of the same form as the approximate solution (but with arbitrary 

coefficients w i ), 

w(x) = 

j=1 

n∑ 

w i N i (x) 

i=1 

Plug u(x),w(x) into the approximate weak form: 

∫ L n∑ dN j (x) 

AE u j 

0 

dx 

j=1 

n∑ 

i=1 

w i 

dN i (x) 

dx 

w i is arbitrary, so the above has to hold ∀ w i : 

n∑ 

∫ L n∑ 

dx = R w i N i (L) + ax w i N i (x)dx 

i=1 

0 

i=1 

n∑ 

[∫ L dN j (x) 

AE dN ] 

∫ 

i (x) 

L 

dx u j = RN i (L) + axN i (x)dx 

j=1 

0 dx dx 

0 

i = 1 . . . n 

which is a system of n equations that can be solved for the unknown coefficients u j . 



The matrix form of the previous system can be expressed as 

K ij u j = f i where K ij = 

and f i = RN i (L) + 

∫ L 

0 

∫ L 

0 

dN j (x) 

dx 

axN i (x)dx 

AE dN i(x) 

dx 

dx 

Finite Element Solution - using 2 discrete elements, of length h (3 nodes) 

From the[ iso-parametric ] formulation we know the element stiffness matrix 

1 −1 

K e = AE . Assembling the element stiffness matrices we get: 

h −1 1 

⎡ 

K tot = ⎣ 

K11 e K12 1 0 

K12 1 K22 1 + K11 2 K12 

2 

0 K12 2 K22 

2 

⎤ 

⎦ ⇒ 

K tot = AE h 

⎡ 

⎣ 

1 −1 0 

−1 2 −1 

0 −1 1 

⎤ 

⎦ 



We also have that the element load vector is 

f i = RN i (L) + 

∫ L 

0 

axN i (x)dx 

Expressing the integral in iso-parametric coordinates N i (ξ) we have: 

dξ 

dx = 2 h , x = N1(ξ)x 1 e + N 2(ξ)x2 e , ⇒ 

f i = R| i=4 + 

∫ L 

0 

a(N 1(ξ)x e 1 + N 2(ξ)x e 2 )N i (ξ) 2 h dξ 


Strong Form - 2D Linear Elasticity FEM 

Governing Equations 

Equilibrium Eq: ∇ s σ + b = 0 ∈ Ω 

Kinematic Eq: ɛ = ∇ s u ∈ Ω 

Constitutive Eq: σ = D · ɛ ∈ Ω 

Traction B.C.: τ · n = T s ∈ Γ t 

Displacement B.C: u = u Γ ∈ Γ u 

Hooke’s Law - Constitutive Equation 

Plane Stress 

Plane Strain 

τ zz = τ xz = τ yz = 0, ɛ zz ≠ 0 ɛ zz = γ xz = γ yz = 0, σ zz ≠ 0 

⎡ 

⎤ 

⎡ 

D = 

E 1 ν 0 

1 − ν ν 0 

⎣ ν 1 0 ⎦ 

E 

D = 

⎣ ν 1 − ν 0 

1 − ν 2 (1 − ν)(1 + ν) 

0 0 

0 0 

1−ν 

2 

1−2ν 

2 

⎤ 

⎦ 


2D FE formulation: Discretization 

Divide the body into finite elements connected to each other through 

nodes 



Shape Functions in Natural Coordinates 

N 1 (ξ, η) = 1 4 (1 − ξ)(1 − η), N 2(ξ, η) = 1 (1 + ξ)(1 − η) 

4 

N 3 (ξ, η) = 1 4 (1 + ξ)(1 + η), N 4(ξ, η) = 1 (1 − ξ)(1 + η) 

4 

Iso-parametric Mapping 

x = 

y = 

4∑ 

N i (ξ, η)xi 

e 

i=1 

4∑ 

i=1 

N i (ξ, η)y e 

i 


Bilinear Shape Functions 


2D FE formulation: Matrices 

from the Principle of Minimum Potential Energy (see slide #9) 

∂Π 

∂d = 0 ⇒ K · d = f 

where 

∫ 

∫ 

∫ 

K e = B T DBdΩ, f e = N T BdΩ + 

Ω e Ω e 

Γ e T 

Gauss Quadrature 

I = 

∫ 1 ∫ 1 

−1 −1 

Ngp Ngp 

N T t s dΓ 

f (ξ, η)dξdη 

∑ ∑ 

= W i W j f (ξ i , η j ) 

i=1 i=1 

where W i , W j are the weights and 

(ξ i , η j ) are the integration points.

The Finite Element Method for the Analysis of Non-Linear and ...

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