CIVL4332 L1 Introduction to Finite Element Methodjoegattas.com/wp-content/uploads/2015/10/4332-Lectures-S2-2015.pdf · CIVL4332 L1 Introduction to Finite Element Method ... Introduction

SCHOOL OF CIVIL ENGINEERING

CIVL4332 L1Introduction to Finite Element Method


by Joe Gattas, Faris Albermani

Introduction

The FEM is a numerical technique for solving physical problems such as stress analysis. Using FE formu-lation, the equilibrium equation

[@]T [D]{@}{u}+ {B} = 0 (1.1)

is transformed into a system of simultaneous algebraic equations of the form

[K]{r} = {R} (1.2)

This transformation operates at the scale of individual domain subdivisions (finite elements), which are usedas piecewise patches to build a complete 1D, 2D, or 3D domain. The stiffness method seen in CIVL3340similarly used an element discretization, with each element having an analytical stiffness and discontinuousnodal displacements. The FEM method has a numerical stiffness and a continuous displacement field.

• Advantages of FEM: versatility and accuracy control

• Disadvantages of FEM: numerical (rather than analytical) solution and large computer output.

CIVL4332 lectures on the finite element method are as follows.

L1: Introduction to Finite Element Method

L2: Continuum Mechanics: Stress and Equilibrium

L3: Continuum Mechanics: Strain and Constitutive Laws

L4: FEM: Formulation

L5: FEM: Shape Functions

L6: FEM: Isoparametric Elements

page 1 of 52



FEM Analysis Types

There are numerous types of finite element analysis. Different solutions can be obtained by using differentforms of the general finite element stiffness equation:

• Linear Static Analysis: Kr = R.

• Non-Linear Static Analysis: [K + KG

]r = R, where KG

is a geometric stiffness matrix that is afunction of stresses.

• Linear Buckling (Eigenvalue) Analysis: |K + �cr

KG

| = 0, where �cr

is the load factor for the criticalbuckling load. Does not solve deformation path but finds the point where displacement is possiblewithout applied load (buckling).

• Dynamic Analysis: Mr + Cr +Kr = R(t), where M is the mass matrix, C is the damping matrix,and r and r first and second derivatives of displacement with respect to time.

• Natural Frequency (Vibration) Analysis: |K+!2M | = 0, where !2 is natural frequency of the system.

This lecture series shall primarily focus on linear static analysis.

FEM Procedure Overview

The broad procedure for the finite element method is as follows:

Pre-processing

1. Select element type and discretization.

2. Displacement (shape) functions [N ] are used to approximate the displacement field {u} within theelement from using nodal values {r}, with {u} = [N ]{r}.

3. Establish strain-displacement relations {"} = [@]{u} and constitutive laws {�} = [D]{"}.

4. The element stiffness relation is obtained, usually with an energy approach used to derive the stiff-ness [k

e

]{re

} = {fe

}.

Solution

5. Assemble the structural stiffness matrix [K] and nodal force vector {R} from element stiffnesses [ke

]and element forces {f

e

}.

6. Impose boundary conditions and solve the global equation for nodal displacements [K]{r} = {R}.

Post-processing

7. Stress recovery: element stresses are calculated by back-substitution into the element equation.

page 2 of 52



Steps 2-4 are full finite element procedures that are simplified in the direct stiffness method seen inCIVL3340. The full FEM process is effective for elements of any complexity.

Example 1.1: Direct Stiffness vs Finite Element

Formulate the element stiffness of a 1D truss element with area A, length L, and Young’s modulus E usingthe Direct Stiffness and Finite Element methods.

Direct Stiffness Method

Truss elements have a single axial translational displacementand single corresponding axial force. Basic stiffness can be de-rived from basic linear elastic material elongation: " = �/L =�/E = P/(EA).

[k] =AE

L

Δ

P

Transform basic stiffness to local x0 � y0 system.

[k0] = T TkT =AE

L

2

664

1 0 �1 00 0 0

1 0Sym. 0

3

775

1D case: global x�y axes align with local x0�y0 axes, considerx axis (u displacements) only.

[ke

] =AE

L

1 �1�1 1

� ui’vi’

uj’

vj’

Finite Element Method

Step 2: Displacement field and shape functions

A truss element has two nodes, each with a nodal coordinate x,nodal displacement r, and nodal force f . It also has constantcross-section area A and length L.

{u} =

u1

u2

�, {r} =

r1r2

�, {f} =

f1f2

�

A,L1 2

r2 f2,x2r1 f1,x1

page 3 of 52



A continuous linear displacement u field is assumed to exist be-tween end nodes. Shape functions N are used to obtain inter-mediate displacement values from end node displacements r.

u = [N ]{r}

= [ N1 N2 ]

r1r2

�

= N1r1 +N2r2

1 2

r2r1u

Shape functions for a two-node element with a linear displace-ment field are:

[N1] = 1� x

L

[N2] =x

L

1 2

N1

1 2

N2

1

1

Step 3: Strain and stress fields

The strain field for a 1D element is:

["] = "xx

Strain is the differential of displacement:

"xx

= [@/@x]{u}= [@/@x][N ]{r} = [B]{r}

The B matrix is defined as the differential of N

[B] = [@][N ] = [�1/L 1/L]

The stress field for a 1D element is:

{�} = {�xx

}

Material constitutive laws relate stress andstrain:

{�} = [D]{"}

For 1D case, �xx

= E"xx

so [D] = [E].

{�xx

} = [E]{"} = [E][B]{r}

Step 4: Energy approach to obtain element stiffness

Using an energy formulation if can be shown that element stiffness is equal to the integral of BTDB overthe element volume. For a 1D element, volume V = AL, which gives:

[ke

] =

Z

V

BTDBdV = A

Z

L

BTDBdx

= A

Z

L

�1/L1/L

�[E] [�1/L 1/L] dx =

AE

L

1 �1�1 1

�

Note: this integral is solved numerically, not analytically, in a typical FEM solver.

page 4 of 52


CIVL4332 L2Continuum Mechanics: Stress and Equilibrium



Continuum mechanics is an important part in the finite element method as stress and strain relationshipsare used to solve the stiffness integral.

Coordinate system and Applied Forces

Most problems are formulated in 3D in FEM so we shall adopt a 3D Cartesian right-handed coordinatesystem. This is referred to as the x

i

system, where i = 1, 2, 3. Axes x1, x2, and x3 are used instead ofx, y, and z, respectively.

x3x1

x2

Within the a 3D Cartesian system we can locate a continuum, which is a continuous body. This continuumis general and can be broadly applied, for example as a fluid, magnetic field, traffic flow, etcetera. Thepresent notes will focus on a structural type analysis for stresses of an elastic continuum.

An elastic continuum can be subjected to some forces that induce stresses, strains, and displacementsthrough the continuum. The forces at some point P within the body can be classified as:

page 5 of 52



• Body force Bi

for i = 1, 2, 3. These are forces per unit volume or unit mass of the body. Examplebody forces include gravity loads and inertial forces.

• Surface traction T n

i

for i = 1, 2, 3. These are forces per unit area (in 3D) or per unit length (in 2D).The traction is defined with reference to a plane with normal unit vector {n}.

x3x1

x2

B1

B2

B3

n Tin

3D Stress Definition

The application of forces with induce a stress in the body. The stress at any point is defined with the Cauchystress tensor [�

ij

]:

[�ij

] =

2

4�11 �12 �13

�21 �22 �23

�31 �32 �33

3

5 (2.3)

For a 3D problem, the stress tensor has nine stress components at any point:

• Direct (Normal) Stress Components: �11, �22, �33

• Shear Stress Components: �12, �13, �23, �21, �31, �32

It is essential to understand where these stresses are acting on the body and the sign convention used. Apoint P within a body can be enlarged to give the following infinitesimal volume dx⇥ dy ⇥ dz.

page 6 of 52



x3x1

x2

P

x3x1

x2

P

Stress components �ij

act on the interfaces of this volume. The first index i denotes the interface andthe second index j denotes the direction of the component, with positive directions corresponding to theoutward normal vector of each face.

Direct stresses are positive if both indices havedirection vectors of the same sign, e.g. �11 ispositive if it acts on an interface with a normalin the direction of x1 and with a positive force inthe direction of x2. This gives tension as pos-itive and compression as negative. Positive di-rect stresses are shown below.

x3x1

x2

P σ11

σ22

σ33

Shear stresses are positive if both indices havedirection vectors of the same sign, e.g. �12 ispositive if it acts on an interface with a normalin the direction of x1 and with a positive force inthe direction of x2. Positive shear stresses areshown below.

x3x1

x2

Pσ12

σ23

σ31 σ13

σ32

σ21

Example 2.1: Stress Tensor Symmetry

Prove that �ij

= �ji

for all i 6= j at any point in a continuum by using the three equations of momentequilibrium.

page 7 of 52



x3

x1

x2

Oσ12

σ23

σ31

σ13

σ32

σ21

σ11

σ22

σ33

σ23

σ32

O

σ32

σ23

x3

x2

Traction stress �23 acts on an area dx1dx3. Traction stress �32 acts on an area dx1dx2. Considering thesum of moments about axis x1 at point O:

XM

x1 = 0 = (�23dx1dx3)dx2 � (�32dx1dx2)dx3

�23dx1dx2dx3 = �32dx1dx2dx3

�23 = �32

Similar summations of Mx2 and M

x3 show �13 = �31 and �12 = �21, respectively.

The moment equations of equilibrium can thus be used to prove �ij

= �ji

. The stress tensor is thereforesymmetric with six unique stress components, written as stress vector {�}T = {�11, �22, �33, �12, �13, �23}.

Stress Transformation

A cut can be taken through point P to generate a plane with unit normal vector n. There will be stresses onthe new interface, represented by traction vector {T n

i

}. This stress vector can be obtained from the stresstensor as:

{T n

i

} = [�ij

]{n} (2.4)

The traction stress vector is composed of normal (direct) stress vector {�n

} and shear stress vector {⌧}.The magnitude of the normal stress can be found be taking the dot product of {n} and {T n

i

}.

�n

= {n}T {T n

i

} (2.5)

The normal stress vector is then{�

n

} = �n

{n} (2.6)

and {⌧} can be found by subtraction of {T n

i

} and {�n

}{⌧} = {T n

i

}� {�n

} (2.7)

page 8 of 52



x3x1

x2

n

Tin

σnτ

3D Case

x1

x2

Tin

σnτ

n

2D Case

Example 2.2: Octahedral Traction Stress

The octahedral plane is the plane whose normal vector makes equal angles with coordinate system axes,that is with equal direction cosines of 1/

p3. Calculate the normal (direct) and shear stress components on

the octahedral plane through point P given it has the following stress tensor:

�ij

=

2

410 3 123 8 412 4 5

3

5 MPa

The octahedral plane has the following nor-mal vector:

{n} =

2

4n1

n2

n3

3

5 =

2

41/p3

1/p3

1/p3

3

5

x3x1

x2

n

The traction vector is given as:

{T n

i

} = [�ij

]{n}= [14.4, 8.7, 12.1]T MPa

The normal component is:

{�n

} = {n}T{T n

i

}{n} = [11.7, 11.7, 11.7]T MPa

page 9 of 52



The shear stress component is:

{⌧} = {T n

i

}� {�n

} = [2.7,�3.1, 0.4]T MPa

In addition to finding stress on target plane, a complete [�ij

] defined in a new coordinate system ui

can beobtained with rotation of stress tensor[�

ij

] defined in old coordinate system xi

. Old and new stress tensorsare related by:

[�ij

] = [L][�ij

][L]T (2.8)

where [L] is a rotation matrix containing direction cosines a between old and new system axes.

[L] =

2

4au1x1 a

u1x2 au1x3

au2x1 a

u2x2 au2x3

au3x1 a

u3x2 au3x3

3

5 (2.9)

x3 x1

x2

u1

u2

u3

x2u1acos 1- u

x1u1acos 1-

x31acos

11-u

Stress Invariants, Principal Stresses, and Stress Decomposition

A stress tensor will change depending on a chosen coordinate system orientation, however it possessesthree stress invariants that are constant across all orientations. As stress invariants are the same in anyorientation, they can be used to check stress transformations, i.e. I1 = I1, I2 = I2, and I3 = I3, where I1,I2, and I3 are the first, second, and third stress invariants, respectively.

The first stress invariant is the sum of direct stresses:

I1 = �11 + �22 + �33 (2.10)

The second stress invariant is the sum of the three minors of the stress tensor:

I2 =

�11 �12

�21 �22

�+

�11 �13

�31 �33

�+

�22 �23

�32 �33

�(2.11)

page 10 of 52



The third stress tensor is the determinant of the stress tensor:

I3 = |�ij

| (2.12)

If a cut is made on an appropriate plane through a point in a continuum, the traction stress vector willcorrespond to the normal stress vector, i.e. shear stress on that plane is zero. This corresponds to theprincipal stress state. It is the 3D version of the 2D principal stresses seen in CIVL2330.

x1

x2

Tin σn

n= σ

nτ

σ1σ2σ3

τ1τ2

τ3

Mathematically this is represented as:[�

ij

]{n} = �{n} (2.13)

The solution to this equation in terms of unknown � is given as:2

4(�11 � �) �12 �13

�21 (�22 � �) �23

�31 �32 (�33 � �)

3

5 {n} = 0 (2.14)

A non-trivial solution to this equation occurs when Det|�ij

� �| = 0, which is a cubic polynomial in termsof �. The three cubic roots �1, �2, �3 are the three principal stresses at that point, such that �1 > �2 > �3.The characteristic cubic equation for principal stresses can be written as

�3 � I1�2 + I2� � I3 = 0 (2.15)

Three principal stress directions are {n1}, {n2}, and {n3} associated with principal stress directions �1,�2, and �3, respectively. They can be found by back-substitution of the principal stresses into (2.14).

For a principal stress state, the stress invariants simplify to:

I1 = �1 + �2 + �3 (2.16)I2 = �1�2 + �1�3 + �2�3 (2.17)I3 = �1�2�3 (2.18)

page 11 of 52



Any stress state [�ij

] can be decomposed into the deviatoric stress tensor [sij

] and the hydrostatic pressurep:

[�ij

] = [sij

] + p[�ij

] (2.19)

where p = I1/3 and [�ij

] = 1 if i = j but otherwise is zero, i.e. it is a 3⇥3 identity matrix. Deviatoric stressis thus given by:

[sij

] = [�ij

]� p[�ij

] =

2

4(�11 � p) �12 �13

�21 (�22 � p) �23

�31 �32 (�33 � p)

3

5 (2.20)

These two values represent important phyical aspects of the point under consideration. The deviatoricstress corresponds to deformation. It represents a pure shear stress state which has �

n

= 0, ⌧ 6= 0, andI1 = 0 and is usually responsible for failure of the material; it is used to derive the von-Mises yield criteriondiscussed below. The hydrostatic stress is responsible for dilation.

Hydrostatic Deviatoric

There are six stress components in a general stress state or three stress components in a principal stressstate. When assessing material failure is is convenient to express the stress as a single value. For ductilematerials such as metals, the von-Mises (effective) stress is most appropriate and is given by:

�VM

=

r(�11 � �22)2 + (�22 � �33)2 + (�33 � �11)2 + 6(�2

12 + �223 + �2

13)

2(2.21)

For a principal stress state this simplifies to:

�VM

=

r(�1 � �2)2 + (�2 � �3)2 + (�3 � �1)2

2(2.22)

von-Mises stress is always positive so cannot be used to indicate load direction and it can be higher thanthe maximum principal stress �1. Different effective failure stress relationships should be used for othermaterials, for example the Drucker-Prager yield criterion can be used for concrete.

The principal shear stresses can be obtained from principal direct stresses from Mohr’s circle manipulationsas:

⌧1 = |�1 � �2|/2 (2.23)⌧2 = |�1 � �3|/2 (2.24)⌧3 = |�2 � �3|/2 (2.25)

page 12 of 52



Equilibrium Equations

The three moment equilibrium equations were used to generate a symmetric stress tensor. Three forceequilibrium equations remain. Consider forces acting on an infinitesimal element in the x1 direction:

σ11σ11

σ21

σ21

+σ11∂

∂ x1∂ x1

+σ21∂

∂ x2∂ x2

dx1

dx2B1

x1x2 x3 x1

x2

σ11

σ31

σ31+σ31∂

∂ x3∂ x3

dx3

In addition to the traction forces acting on element interfaces there is a body force B1.

⌃Fx1 = 0 =

✓@�11

@x1@x1

◆@x2@x3+

✓@�21

@x2@x2

◆@x1@x3+

✓@�31

@x3@x3

◆@x2@x1+B1@x1@x2@x3 (2.26)

which reduces to@�11

@x1+

@�21

@x2+

@�31

@x3+B1 = 0 (2.27)

This can be repeated for directions x2 and x3, which can be written in shorthand as:

�ji,j

+Bi

= 0 (2.28)

where �ji,j

is the differential of �ji

with respect to j.

In matrix form this becomes:[@]T{�}+ {B} = 0 (2.29)

where

[B]T = {B1, B2, B3} (2.30)

[�]T = {�11, �22, �33, �12, �23, �13} (2.31)

[@]T =

2

4@

@x10 0 @

@x20 @

@x3

0 @

@x20 @

@x1

@

@x30

0 0 @

@x30 @

@x2

@

@x1

3

5 (2.32)

The differential operator [@]T is extensively used in FEM.

page 13 of 52



Example 2.3: Principal Stresses

Calculate the principal stresses and associated principal stress directions, given that:

�ij

=

2

4�10 9 5

0 0Sym. 8

3

5 MPa

and one of the principal stresses is 4MPa.

The invariants of the given stress tensor are:

I1 = �11 + �22 + �33 = �2

I2 =

�11 �12

�21 �22

�+

�11 �13

�31 �33

�+

�22 �23

�32 �33

�= �81� 0� 105 = �186

I3 = |�ij

| = 0� 9⇥ 9⇥ 8 + 0 = �648

Given a principal stress of 4MPa, (��4) is a factor of the characteristic equation �3�I1�2+I2��I3 = 0:

�3 � I1�2 + I2� � I3 = 0 = �3 + 2�2 � 186� + 648

(� � 4)(a�2 + b� + c) = a�3 + (b� 4a)�2 + (c� 4b)� � 4c

which gives a = 1, b = 6, and c = �162. The roots of (a�2 + b� + c) are 10.08MPa and -16.08MPa,which gives principal stresses as:

�1 = 10.08 MPa , �2 = 4 MPa , �3 = �16.08 MPa

Substitution of a principal stress into the princpal stress solution gives:2

4(�10� �1) 9 5

(0� �1) 0Sym. (8� �1)

3

5

2

4n1x1

n1x2

n1x3

3

5 =

2

4000

3

5 (2.33)

This equation cannot be solved directly as it has already been used to generate the principal stresses. Usethe additional condition that {n1}T{n1} = 1:

2

4n1x1

n1x2

n1x3

(0� �1) 0Sym. (8� �1)

3

5

2

4n1x1

n1x2

n1x3

3

5 =

2

4100

3

5 (2.34)

which gives {n1}T = ±{0.363, 0.324, 0.873}. A similar substitution gives {n2}T =±{�0.362, �0.815, 0.453} and {n3}T = ±{�.859, 0.481, 0.178}.

page 14 of 52


CIVL4332 L3Continuum Mechanics: Strain and Constitutive Laws

CIVL4332 L3Continuum Mechanics: Strain and ConstitutiveLaws


Analysis of Strain

The strain tensor ["ij

] is similar to the stress tensor, i.e. it defines the strain field at a point in a continuum:

["ij

] =

2

4"11 "12 "13"21 "22 "23"31 "32 "33

3

5 (3.35)

The tensor is symmetric, "ij

= "ji

, and so has six unique strain components, written as stress vector{"}T = {"11, "22, "33, �12, �13, �23}. The first three components are direct strains and the second threecomponents are the engineering shear strain components where �

ij

= 2"ij

, discussed further below.

Direct strain components represent the length change between two points in a continuum, with positivetaken as extension and negative as contraction. Finite-element uses Green-Lagrange strain, defined as:

" =1

2

(Lo

+�L)2 � L2o

L2o

=�L

Lo

+1

2

✓�L

Lo

◆2

⇡ �L

Lo

for small strain ( < 1%) (3.36)

where Lo

is the initial length between two points and Lo

+ �L is the length of the same two points afterdeformation. If �L = 0 after load is applied to a body than there has been a displacement without changein length and the body has undergone a rigid body motion.

This small displacement strain " = �L/Lo

is called engineer’s strain and is as previously used in structuralmechanics courses. For large displacements, logarithmic or true strain e is used instead of ":

e =

ZL

L

o

dL

L= ln

dL

L(3.37)

page 15 of 52



Shear strain components represent the change in angle between three points in a continuum:

� = ✓ � ✓o

(3.38)

where ✓o

is the initial angle between three points and ✓ is the angle of the same three points after deforma-tion.

AB

A

BLo Lo+Δ LA

B

A

Boθ

CCθ

Direct Strain Shear Strain

For all strain components, the first subscript denotes the direction of the fibre under strain and the secondsubscript denotes the direction of the fibre’s displacement. All transformations derived for the stress tensorare also valid for the strain tensor, e.g. a traction vector for strain on an interface with unit normal vector nis �n = ["]{n}. How the strains are occurring within the body and can be understood as follows.

For direct strains, consider a 1D fibre with initiallength dx.

x1dx

dx +∂

∂ x dxu

1

11

11

After load is applied, it has length:

dx+@u1

@x1dx

The engineer’s strain is then:

"11 =�L

Lo

=@u1

@x1(3.39)

For shear strains, consider two fibres with sidelengths dx1 and dx2:

x1

dx1

x2

dx2

∂ u1

∂ u2ε1221

x1

dx1

x2

dx2 ε 12γ

After load is applied, these are distorted to gen-erate two angles. Using a small angle trigono-metric approximation (tan x = x), there are:

"12 =@u2

@x1, "21 =

@u1

@x2(3.40)

"12, "23, and "13 are referred to as tensorialshear strains as they correspond to strain ten-sor components. It is more common to the alter-nate definition referred to as engineering shearstrains where all shear displacement acts on asingle edge: �

ij

= 2"ij

.

page 16 of 52



Example 3.1: Strain Transformations

The is a interface cut ABC within a continuum. Point A, B, and C are located a unit distance along axesx1, x2, and x3 respectively. Point D is located halfway between points A and B.

During an experiment, the strain ["] is measured at D as:

["] =

2

40.01 �0.005 0

0.02 0.01Sym. �0.03

3

5 (3.41)

Calculate the change in the right angle ADC.

x3x1

x2

D

B

CA

The points have the following coordinates:

A = [1, 0, 0]

B = [0, 1, 0]

C = [0, 0, 1]

D = [0.5, 0.5, 0]

The change in angle is a shear strain �.

Material direction of two fibres

��!DA = A�D = [0.5,�0.5, 0]��!DC = C �D = [�0.5,�0.5, 1]

page 17 of 52



Convert these to a unit vector:

nDA

=

��!DA

|DA| =[0.5,�0.5, 0]

1/p2

=1p2

8<

:

1�10

9=

;

nDC

=

��!DC

|DA| =[�0.5,�0.5, 1]p

3/p2

=1p6

8<

:

�1�12

9=

;

We can check if unit vectors are correct as they are orthogonal nT

DA

nDC

= 0.

Strain vector of fibre DC

{�nDC

} = ["]{nDC

} =1p6

8<

:

�0.0020.002�0.028

9=

; (3.42)

�nDC

can be projected onto {nDC

} to find the direct strain component. As {nDC

} and {nDA

} are orthogonal,�nDC

can be projected onto {nDA

} to find the shear strain component.

"ADC

= {nDA

}T{�nDC

} = �0.0029 rad. (3.43)

The change in angle ADC is �ADC

= 2"ADC

= �0.0058 rad.

x3x1

x2

δnDC

εADC

page 18 of 52



Strain-Displacement Relations

At any point P in a continuum, there are three displacements ui

, where i = 1, 2, 3. Strain is the partialderivative of displacement and so given the displacement vector {u}T = {u1 u2 u3}, strain-displacementrelations can be written as:

{"} = [@]{u} (3.44)2

6666664

"11"22"33�12�13�23

3

7777775=

2

66666664

@

@x10 0

0 @

@x20

0 0 @

@x3@

@x2

@

@x10

0 @

@x3

@

@x2@

@x30 @

@x1

3

77777775

2

4u1

u2

u3

3

5 =

2

66666664

@u1@x1@u2@x2@u3@x3

@u2@x1

+ @u1@x2

@u3@x2

+ @u2@x3

@u1@x3

+ @u3@x1

3

77777775

(3.45)

where [@] is defined for stress equilibrium relations and engineering shear strains are given in the strainvector, not tensorial shear strains.

This relation shows that three displacements generate six components of strain. Going in the oppositedirection however, six strain components cannot determine a unique continuous displacement field, asthe system is overdeteriminant. The strain components therefore have to be constrained by compatibilityequations. Six compatibility equations similar to the one below need to be satisfied in a 3D system.

@2"11@x2

2

+@2"22@x2

1

� 2@2"12@x1@x2

= 0 (3.46)

Of the six 3D compatibility equations, three are independent. All but the first equation shown above isautomatically constrained in a plane strain or plane stress field.

The compatibility equations are generated as follows. Shear strain is related to displacement with:

�12 =

✓@u1

@x2+

@u2

@x1

◆= 2"12

Deformation u must be continuous and there must be some gradient in the strain field, i.e. @"12/@x1 6= 0and @"12/@x2 6= 0. The strain gradient can therefore be obtained by taking the derivative of the previousequation:

@3u1

@x1@x22

+@3u2

@x21@x2

= 2@2"12@x1@x2

Direct strain relations "11 =@u1@x1

and "22 =@u2@y12

can be substituted to simplify:

@2"11@x2

2

+@2"22@x2

1

= 2@2"12@x1@x2

which gives the compatibility equation (3.46).

page 19 of 52



Constitutive Laws

Thus far, we have related forces to stress and displacements to strain. The final relations required are thoseto link stress and strain. The relations are termed constitutive laws and are to do with the system material.It is therefore more complicated that prior relations which only relate to system geomety.

The general form of the constitutive law is written the matrix [D] that relates stress and strain as follows.

{�} = [D]{"} (3.47)

The simplest material is an isotropic, linear elastic material. The linear elastic material requires only twomaterial properties to define: Young’s Modulus E and Poisson’s ratio v which are obtained from materialtests. The isotropic linear elastic material is the simplest case of the constitutive law. Orthotropic materialssuch as composite laminates require 9 material properties to define, and completely anisotropic materialsrequire 21 material properties to define.

For direct stresses in an isotropic material, there is coupling from the Poisson’s effect. An axial stress �11

will induce an axial strain of E"11 and transverse strains of "22 = �v"11 and "33 = �v"11.

σ11

ε11

σ12

γ12

EGσ11

ε11

ε22 ε11= -v

σ11

For shear stresses in an isotropic material there is no coupling. A shear stress �12 will induce an axial strainof G�12, where G is the shear modulus related as:

G =E

2(1 + v)(3.48)

This gives the following stress-strain relation:

2

6666664

"11"22"33�12�13�23

3

7777775=

1

E

2

6666664

1 �v �v 0 0 01 �v 0 0 0

1 0 0 02(1 + v) 0 0

2(1 + v) 0Sym. 2(1 + v)

3

7777775

2

6666664

�11

�22

�33

�12

�13

�23

3

7777775(3.49)

page 20 of 52



the inverse of this gives the constitutive matrix as:

[D] =E

(1 + v)(1� 2v)

2

6666664

1� v v v 0 0 01� v v 0 0 0

1� v 0 0 01�2v2 0 0

1�2v2 0

Sym. 1�2v2

3

7777775(3.50)

All natural materials have 0 < v < 0.5. As v ! 0.5 the material becomes incompressible as the 1/(1�2v)causes [D] ! 1. Meta-materials are those that have been engineered to possess properties not foundin nature, including negative Poisson’s ratios, i.e. �1 < v < 0. There are termed auxetic materials andinclude origami-inspired meta-materials, applied for example as a deployable stent1:

Recalling the equilibrium relation of this previous lecture. This can now be written with the constitutive lawand using the strain-displacement relation as:

[@]T{�}+ {B} = 0

[@]T [D]{"}+ {B} = 0

[@]T [D]{@}{u}+ {B} = 0 (3.51)

1Figure taken from You, Z. and Kaori K. ”Expandable tubes with negative Poissons ratio and their application in medicine.”Origami4: Fourth International Meeting of Origami Science, Mathematics, and Education. 2009.

page 21 of 52



Continuum Mechanics Summary

The relations in 3D continuum mechanics can be summarised as follows.

ForcesB

i

, T n

i

Displacements ⇥3{u

i

}Stiffness

[@]T [D]{@}{u}+ {B} = 0

EquilibriumEquations ⇥3

[@]T{�} + {B} = 0

Strain-DisplacementRelations ⇥6{"} = [@]{u}

Stress components ⇥6{�}

Constitutive Laws ⇥6{�} = [D]{"}

Strain components ⇥6{"}

There are 15 unknown quantities: 6 stress components, 6 strain components, and 3 displacement com-ponents. These are found with 15 relations: 3 equations of equilibrium, 6 strain-displacement relations, 6constitutive laws. The stiffness relation can be used to solve displacements directly, but element stressesand strains must then be calculated back-substitution into other relations.

Simplified (1D and 2D) Mechanics

The above derivations are for complete 3D continuum systems. All lower-order structural systems can beobtained from simplified continuum conditions as follows.

1D Truss Element with x1 longitudinal axis and u1 longitudinal displacements only.

{u} = u1

{"} = "11 = @u1/@x1

[@] = @/@x1

{�} = �11 = E"11

[D] = E

x1F

,u11x

page 22 of 52



1D Beam Element with x1 longitudinal axis, x2 transverse axis, and u1 and u2 displacements only.

{u} = u1 = �x2@u2

@x1

{"} = "11 = @u1/@x1 = �x2@2u2/@x

21

[@] = �x2@2/@x2

1

{�} = �11 = E"11

[D] = E

x1F ,u11x

x2 ,u2 F2x

2u 1xd d/θ =

2D Plane Stress A 2D continuum located in the x1�x2 plane with all stresses in-plane and an out-of-planelength much smaller than in-plane lengths x3 << x1 and x3 << x2, for example as in a cantilever with anend point load.

Two in-plane displacement components:

{u}T = {u1 u2}

All stress is in plane, i.e. there is no out-of-planestress �33 = �13 = �23 = 0, leaving three stresscomponents:

{�} = {�11 �22 �12}

There is no coupling in shear strain so �13 = �23 =0. There is coupling for direct strain so there willbe out-of-plane (Poisson) expansion or contraction"33 6= 0:

{"}T = {"11 "22 �12}"33 = �v("11 + "22)

x1

F

,u1

1x

x2,u2

F2x

x3

σ11

O

O

σ12

σ12

σ11

σ22

σ22

Stress-displacement relations and constitutive laws are then:

[@] =

2

4@

@x10

0 @

@x2@

@x2

@

@x1

3

5 and [D] =E

1� v2

2

41 v 0v 1 00 0 1�v

2

3

5 (3.52)

page 23 of 52



2D Plane Strain A 2D continuum located in the x1 � x2 plane with all strains in-plane and an out-of-planelength much larger than in-plane lengths x3 >> x1 and x3 >> x3, for example as in a dam with uniformhydrostatic pressure along length.

Two in-plane displacement components:

{u}T = {u1 u2}

All strain is in plane, i.e. there is no out-of-planestrain "33 = "13 = "23 = 0, leaving three straincomponents:

{"}T = {"11 "22 �12}

There is no coupling in shear stress so �13 = �23 =0. There is coupling for direct stress so there will beout-of-plane stress �33 6= 0:

{�}T = {�11 �22 �12}�33 = v(�11 + �22)

x1 ,u1

x2,u2

x3 ε

11

O

O

ε12

ε12

ε11

ε22

ε22

Stress-displacement relations and constitutive laws are then:

[@] =

2

4@

@x10

0 @

@x2@

@x2

@

@x1

3

5 and [D] =E

(1 + v)(1� 2v)

2

41� v v 0v 1� v 00 0 1�2v

2

3

5 (3.53)

2D Thin Plate Element A 2D element with x1 longitudinal axis, x2 transverse axis, and u1, u2, and u3

displacement components. The element has out-of-plane stress and strain, but is assumed to be thinx3 << x1 and x3 << x2 such that out-of-plane shear can be neglected �13 = �23 = 0.

{u} =

⇢u1

u2

�= �x3

⇢@

@x1@

@x2

�u3

{"}T = {"11 "22 �12}

[@]T =

@2

@x21

@2

@x22

2@2

@x1@x2

�

{�}T = {�11 �22 �12}

and constitutive matrix [D] the same as for theplane stress case.

x1 ,u1

x3,u3

x2 ,u2

3u 1xd d/θ =13

3u 2xd d/θ =23

F3x

page 24 of 52



Example 3.2: 2D Continuum Mechanics

Establish an equivalent 2D system for the following 3D bodies:

• a) A long, thick-walled cylinder under constant internal pressure.

• b) A fin web connection plate.

a) The cylinder has no longitudinal stress, so a plane strain condition can be assumed.

σr

σθ

b) The fin plate has no out-of-plane stress, so a plane stress condition can be assumed:

σ11

σ22

page 25 of 52


CIVL4332 L4FEM Formulation



As stated in earlier lectures, the Finite Element Method transforms the continuum mechanics equilibriumequation [@]T [D]{@}{u} + {B} = 0 into a system of numerically-solvable simultaneous algebraic equa-tions of the form [K]{r} = {R}. The following lecture describes the necessary steps to achieve thistransformation and thus generate a complete FEM formulation.

Discretization

The first step in the Finite Element Method is to break up the continuum to be solved into discrete elementsin a process that is called discretization. The need for a discretized form arises because the equilibriumequation is written in terms of an unknown displacement field {u}. For a general continuum, it is difficultto know the form of this displacement field. However it is easy know the displacement field for a smallelement, for example a 2-node bar element or 4-node square element. The continuous displacement fieldfor the whole continuum is therefore easier to approximate as a stepwise assembly of small elements. It isessential to choose an element suitable for the particular problem, e.g. a disc cannot be approximated with1D truss elements, but could be approximated with 2D plate elements.

The process is analogous to curve-fitting with a straight-line approximation, i.e. it is difficult to determine afunction to fit a complex curve, but it is easy to assume a straight line between discrete points on the curveto generate approximately the same form.

page 26 of 52



Exact Coarse Appx. Fine Appx.

As with any approximation, there is some error between the approximate and ideal system. The necessarylevel of discretization varies substantially between FEM models. It is therefore always necessary to under-take a mesh refinement or mesh sensitivity study. Such studies increase the number of elements (e.g. from80 to 160 to 320 elements) or the change element type (e.g. quad or tri plate elements) systematically toestablish a mesh type and density that gives a stable solution.

Displacement (Shape) Functions

Shape functions are used to transform nodal displacements {r} into a continuous displacement field {u}.

{u} = [N ]{r} (4.54)

where [N ] is the shape function matrix. This consists of individual shape functions N to relate each nodaldisplacement to the continuous displacement in a particular element:

{u} = [N1N2N3N4]

8>><

>>:

r1r2r3r4

9>>=

>>;

{u} = N1r1 +N2r2 +N3r3 +N4r4

r4

x3x1

x2

1

2

3

4r1

r2

r3

u

page 27 of 52



The above relation allows stress and strain fields to be expressed directly in terms of nodal displacements:

{"} = [@]{u} = [@][N ]{r} = [B]{r} (4.55){�} = [D]{"} = [D][B]{r} (4.56)

where [B] is the shape function derivative matrix:

[B] = [@][N ] (4.57)

The next lecture wll discuss how to generate suitable shape functions.

Energy Formulation for Element Stiffness

Previously, stiffness has been derived for elements directly as the inverse of flexibility coefficients obtainedfrom structural mechanics solutions. This is feasible for simple elements such as 1D truss, beam, or frameelements, however for more complex elements it is challenging, e.g. an 8 or 20-node brick element.

Instead, an energy approach is used in FEM to calculate the element stiffness matrix as follows.

Internal (Strain) Energy

The internal (strain) energy U of a system is found by integrating the stress times strain over the volume Vof material in the body:

U =1

2

Z

V

{�}T{"}dV (4.58)

This can be expanded in terms of stress and strain components. 1D case with uniform cross-section:

U =A

2

Z�11"11dx1

2D case with uniform thickness:

U =t

2

Z Z�11"11 + �22"22 + �12"12dx1dx2

Full 3D case:

U =1

2

Z Z Z�11"11 + �22"22 + �33"33 + �12"12 + �23"23 + �13"13dx1dx2dx3

Internal energy (4.58) can also be expressed in terms of nodal displacements by substituting (4.55) and(4.56). This gives:

U =1

2

Z

V

{�}T{"}dV

=1

2

Z

V

([D][B]{r})T ([B]{r})dV

=1

2{r}T

✓Z[B]T [D][B]dV

◆{r} (4.59)

page 28 of 52



External Energy

The external energy is generated as the sum of work WS

from traction forces {T}, WB

from body forces{B}, and W

N

from nodal forces {fn

} applied to the body:

W = WS

+WB

+WN

=

Z{u}T{T}dS +

Z{u}T{B}dV + {r}T{f

N

} (4.60)

where S is the surface over which the traction force {T} is applied.

WS

and WB

can also be written directly in terms of nodal displacement vector. From (4.54), it is known that{u}T = {r}T [N ]T . This gives:

WS

=

Z{u}T{T}dS = {r}T

Z[N ]T{T}dS = {r}T{f

S

} (4.61)

WB

=

Z{u}T{B}dV = {r}T

Z[N ]T{B}dV = {r}T{f

B

} (4.62)

W = {r}T ({fB

}+ {fS

}+ {fN

}) (4.63)

= {r}T{R} (4.64)

where {fS

}, {fB

}, and {R} are energy-equivalent nodal load vector for surface, body, and all loads,respectively.

Energy Equilibrium

Of all displacements that might satisfy the boundary conditions of a structure, those that satisfy the equa-tions of equilibrium occur when there is a stationary value of potential energy, i.e. the difference betweeninternal work and external work is at a minimum.

uStable Equilibrium

Neutral Equilibrium

Unstable EquilibriumΠ

The total potential energy ⇧ is related to internal work U and external work W as:

⇧ = U �W (4.65)

The stationary condition is then�⇧ = �U � �W = 0 (4.66)

where � is a small admissible variation. In other words, when a structure is subjected to a load, it will find adeformed configuration that uses the minimum amount of energy to achieve a certain displacement.

page 29 of 52



Equilibrium therefore is given by:

@⇧

@{r} =@U

@{r} � @W

@{r} = 0

@U

@{r} =@W

@{r}

Substituting (4.58) and (4.64) then gives:

@(12{r}T

�R[B]T [D][B]dV

�{r})

@{r} =@({r}T{R})

@{r}✓Z[B]T [D][B]dV

◆{r} = {R}

[K]{r} = {R}

where stiffness matrix [K] is:

[K] =

Z[B]T [D][B]dV (4.67)

Example 4.1: Spring Element

Use the energy approach to derive the stiffness matrix of 1D spring element:

kere = f e

k1 2

r2 f2,r1 f1,A two-node spring element has nodal displacements {r}T = {r1 r2} and nodal forces {f}T = {f1 f2}.Internal energy is:

U =1

2

Z

V

{�}T{"}dV =A

2

Z�11"11dx1

=AE

2

Z"211dx1 =

AE

2

ZL

0

✓r2 � r1

L

◆2

dx1

=AE

L

(r2 � r1)2

2= k

(r2 � r1)2

2

External work is:W = r1f1 + r2f2

page 30 of 52



Potential energy is:

⇧ = U �W = k(r2 � r1)2

2� r1f1 � r2f2

which is stationary when:

�⇧ = 0 =@⇧

@r1�r1 +

@⇧

@r2�r2

The non-trivial solutions to this are when @⇧@r1

= 0 and @⇧@r2

= 0:

@⇧

@r1= 0 gives k(r1 � r2) = f1

@⇧

@r2= 0 gives k(r2 � r1) = f2

which is matrix form becomes

k

1 �1�1 1

� r1r2

�=

f1f2

�

FEM Formulation Summary

The FEM formulation is therefore primarily concerned with expressing the main unknowns in a systemstiffness equation in terms of nodal displacements. It can be summarised as follows:

Displacement Field {u} = [N ]{r}Strain Field {"} = [@]{u} = [@][N ]{r} = [B]{r}Stress Field {�} = [D]{"} = [D][B]{r}

Element Strain Energy U =1

2

Z{�}T{"}dV =

1

2{r}T

✓Z[B]T [D][B]dV

◆{r}

External Energy W = WN

+WB

+WS

= {r}T{R}Stationary Potential �⇧ = �U � �W = 0 , which gives

Stiffness Relation {R} = [K]{r} =

✓Z[B]T [D][B]dV

◆{r}

Stiffness Matrix [K] =

Z[B]T [D][B]dV

Example 4.2: FEM Truss Stiffness

Revisit the Finite Element formulation of a 1D truss element shown in Example 1.1.

page 31 of 52


CIVL4332 L5FEM Shape Functions



Displacement Fields

Consider the following element with an axial load, length L, cross-section area A, and Young’s Modulus E.In a finite element formulation it can be discretise into five 1D elements, each with length l.

L,A,E

l

21 3 4 5

A general form of the displacement field for axial (1D) loading is as a polynomial function over the beamlength with degree n� 1:

u(x) = a1 + a2x+ a3x2 + ...+ a

n

xn�1 (5.68)

which has n coefficients and 0 x L. Coefficient a1 does not contribute to the strain field and so itcorresponds to rigid body motion. The remaining coefficients correspond to displacement gradients.

The degree of polynomial necessary to approximate the field depends on the complexity of the problem. Anelement with a uniform axial load has a uniform stress and strain. Strain is the derivative of displacement,so a uniform strain gives a linear displacement field, i.e. a first-degree polynomial. An element with anon-uniform axial load will have a non-uniform strain and so a higher-order polynomial is necessary forapproximation.

The number of derivatives of displacement that are continuous between elements is referred to as thedegree of continuity of an element. A C0 element has zero continuous derivatives i.e. it is continuous in

page 32 of 52



terms of displacement only. A C1 element has a continuous first derivative of displacement (strain) as wellas continuous displacement.

u

ε

u

ε ε

uC 0 C 0 C 1

Displacement Fields with Shape Functions

The 1D element in the previous section with a uniform axial load has a linear displacement field. This canbe expressed as:

u = a1 + a2x

{u} =< 1 x >

⇢a1a2

�

{u} =< x > {a} (5.69)

where {a} is the vector of polynomial coefficients.

Nodal displacements values u(x = 0) = r1 and u(x = l) = r2 can be used to relate polynomial coefficientsto nodal displacements. In matrix form this becomes:

⇢r1r2

�=

1 01 l

�⇢a1a2

�

{r} = [A]{a} (5.70){a} = [A]�1{r} (5.71)

where [A]�1 =1

l

l 0�1 1

�(5.72)

Substituting (5.71) into (5.69) give:

{u} =< x > [A]�1{r}{u} = [N ]{r} (5.73)

where [N ] is a matrix containing the shape or interpolation functions:

[N ] =< x > [A]�1 (5.74)

page 33 of 52



which for this example is:

[N ] =< 1 x >1

l

l 0�1 1

�

=< 1� x

l

x

l>

N1 = 1� x

l(5.75)

N2 =x

l(5.76)

This is expressed in the physical coordinate system however it can also be expressed in the natural coor-dinate system by replacing the normalised x/L term with ⇠:

N1 = 1� ⇠

N2 = ⇠

where 0 ⇠ 1.

All shape functions have some common properties:

• they are always between 0 and 1;

• they have a value of 1 at their own node and 0 at every other node;

• the sum of shape functions is equal to 1 at any point in the element; and

• the superposition of shape functions times nodal displacements gives the whole displacement field.

r2r1u

1 2N1 1 2N2

1 1N2N1

1+

N1 N2r1 r2 N1 N2r1 r2+

Lagrange Interpolation

The above uses a polynomial function to generate shape functions. Another approach is to use Lagrangeinterpolation to generate shape functions directly. Lagrange interpolation represents a field for an elementwith a polynomial field of degree p is represented as:

� =nX

1

Lx

i

�i

(5.77)

page 34 of 52



where Lx

i

is the Lagrange function, � is some field (displacement, stress, strain, etc.), �i

is the field valueat node i, and the number of nodes n is one greater than the polynomial degree, i.e. n = p+ 1.

For a Lagrange displacement field � = u, �i

= ri

, and so Lx

i

= Ni

.

� =nX

1

Lx

i

�i

= Lx1�1 + L

x2�2 + ...+ Lx

n

�n

+

u = N1r1 +N2r2 + ...+Nn

rn

As Lagrange interpolation is for displacement only, Lagrange elements are C0 elements.

The Lagrange function is a chain product (⇧) along all element axes:

Lx

i

=⇧n

j+1(xj

� x)

⇧n

j+1(xj

� xi

), for i 6= j (5.78)

Example 5.1: Lagrange Shape Functions

Use Lagrange interpolation to find shape functions for a 1D element:

i) with a linear (first-order) displacement field.

ii) with a quadratic (second-order) displacement field.

Plot the quadratic shape functions.

i) A 1D linear displacement field requires p+ 1 = 2 nodes.

Lx1 = N1 =

(x2 � x)

(x2 � x1)=

(l � x)

l= 1� x

l

Lx2 = N2 =

(x1 � x)

(x1 � x2)=

(0� x)

�l=

x

ll

1 2

x2x1

ii) A 2D quadratic displacement field requires p+1 = 3 nodes, with coordinates x1 = 0, x2 = 0.5l, x3 = l.

Lx1 = N1 =

(x2 � x)(x3 � x)

(x2 � x1)(x3 � x1)= 1� 3

x

l+ 2

⇣xl

⌘2

Lx2 = N2 =

(x1 � x)(x3 � x)

(x1 � x2)(x3 � x2)= 4

x

l� 4

⇣xl

⌘2

Lx3 = N3 =

(x1 � x)(x2 � x)

(x1 � x3)(x2 � x3)= �x

l+ 2

⇣xl

⌘2

1 3

x3x12

x2

page 35 of 52



Quadratic shape functions are:

N1 N2

1 1

N3

1

2D Shape functions

Tri Element

A triangular element has 3 nodes, each with two displacements (rx

, ry

) per node.

2

1

3

r1

r2

xy

r3

r4r5

r6

The element has a displacement field that can be represented as a two-variable polynomial:

u = a1 + a2x+ a3y + a4x2 + a5xy + a6y

2 + ... (5.79)

A bilinear triangular has a linear displacement field in both x and y direction:

u = a1 + a2x+ a3y

{u} =< x > {a}

{u} =< 1 x y >

8<

:

a1a2a3

9=

; (5.80)

Considering first x-displacements, nodal coordinates (x1, y1), (x2, y2), and (x3, y3) are used to relate poly-

page 36 of 52



nomial coefficients to nodal displacements:

{r} = [A]{a} (5.81)8<

:

r1r3r5

9=

; =

2

41 x1 y11 x2 y21 x3 y3

3

5

8<

:

a1a2a3

9=

; (5.82)

The displacement field {u} in the x directions is then:

{u} =< x > [A]�1{r}= [N ]{r}

= [N1 N2 N3]

8<

:

r1r3r5

9=

; (5.83)

r3r1

N1 N2

111

1

2

3

1

N2N1 + N3+ N1 N2r1 r3+ N3 r5+

r5u

1

1

N3

The displacement field {v} in the y directions is also linear, so can be represented by the same displace-ment functions:

{v} = [N1 N2 N3]

8<

:

r2r4r6

9=

; (5.84)

and a complete displacement field description is then:

{u} =

⇢uv

�=

N1 0 N2 0 N3 00 N1 0 N2 0 N3

�

8>>>>>><

>>>>>>:

r1r2r3r4r5r6

9>>>>>>=

>>>>>>;

(5.85)

page 37 of 52



Each shape function is associated with a single element node and shares the same properties as the 1Dexample shape functions, i.e. N1 +N2 +N3 = 1, etc.

The general form of the shape functions for a triangle C0 element with a linear displacement field is:

Ni

=xb

i

+ yci

+ fiP3

1 fi=

xbi

+ yci

+ fi

2A(5.86)

where:

bi

= yj

� yk

(5.87)ci

= xk

� xj

(5.88)fi

= xj

yk

� xk

yj

(5.89)

where if i = 1, j = 2 and k = 3; if i = 2, j = 3 and k = 1; and if i = 3, j = 1 and k = 2.

Example 5.2: Tri Element Displacement Field

A three-node constant strain triangular finite element has nodal coordinates (x, y) of (2,2), (6,4), and (3,8)and nodal displacements (u, v) of (1,1), (-1,1), and (-2,-1). For a plane stress case, calculate the element:

i) shape functions; and

ii) stress components,

assuming the element has a Young’s Modulus of 100MPa and a Poisson’s ratio of 0.1.

i) A constant strain element has a linear displacement field, so Lagrange interpolation can be used to findshape functions:

b1 = y2 � y3 = �4 c1 = x3 � x2 = �3 f1 = x2y3 � x3y2 = 36

b2 = y3 � y1 = 6 c2 = x1 � x3 = �1 f2 = x3y1 � x1y3 = �10

b3 = y1 � y2 = �2 c3 = x2 � x1 = 4 f3 = x1y2 � x2y1 = �4

2A = f1 + f2 + f3 = 22

Shape functions are then as per (5.86)

N1 =�4x+�3y + 36

22

N2 =6x+�1y � 10

22

N3 =�2x+ 4y +�4

22

page 38 of 52



ii) The shape function derivative matrix is given as:

[Bi

] = [@][Ni

] =

2

4@

@x

00 @

@y

@

@y

@

@x

3

5N1 00 N1

�=

1

2A

2

4bi

00 c

i

ci

bi

3

5

Stress and strain can be generated from shape function derivatives:

{�} = [D]{"} = [D][B]{r}

=E

1� v2

2

41 v 0v 1 00 0 1�v

2

3

5 [B1|B2|B3]

8>>>>>><

>>>>>>:

u1

v1u2

v2u3

v3

9>>>>>>=

>>>>>>;

=100

1� 0.12

2

41 0.1 00.1 1 00 0 1�0.1

2

3

5 1

22

2

4�4 0 6 0 �2 00 �3 0 �1 0 4�3 �4 �1 6 4 �2

3

5

8>>>>>><

>>>>>>:

11�11�2�1

9>>>>>>=

>>>>>>;

=

8<

:

�31.2�39.5�12.4

9=

; MPa

Quad Element

A bilinear quadrilateral plate element has linear displacement field in both directions. A linear first-orderpolynomial in requires n = p + 1 = 2 nodes, per direction, so the quad element requires (p + 1)2 = 4nodes. Nodes should be numbered 1 to 4 either anti-clockwise or clockwise. Either is suitable as long as itis consistent across elements.

If expressed in the physical (x, y) system, the quad plate has width 2a and height 2b. In FEM formulationsit is easier to work in the natural (⇠, ⌘) system, where x = ⇠a, and y = ⌘b.

The Lagrange chain product generates C0 element shape functions when multiplied along all coordinate

page 39 of 52



axes:

Ni

= L⇠

L⌘

N1 =⇠2 � ⇠

⇠2 � ⇠1

⌘4 � ⌘

⌘4 � ⌘1=

(1� ⇠)(1� ⌘)

4

N2 =⇠1 � ⇠

⇠1 � ⇠3

⌘3 � ⌘

⌘3 � ⌘2=

(1 + ⇠)(1� ⌘)

4

N3 =⇠4 � ⇠

⇠4 � ⇠3

⌘2 � ⌘

⌘2 � ⌘3=

(1 + ⇠)(1 + ⌘)

4

N4 =⇠3 � ⇠

⇠3 � ⇠4

⌘1 � ⌘

⌘1 � ⌘4=

(1� ⇠)(1 + ⌘)

4

The general form of this is:

Ni

=(1 + ⇠⇠

i

)(1 + ⌘⌘i

)

4(5.90)

x

y

ξ

η

(1,1)

(1,-1)

(-1,1)

(-1,-1)

v3u3

v2u2

v1u1

v4u4

2a

2b

1 2

4 3

Higher Continuity and Hermite Interpolation

Increasing the degree of polynomial for a particular element will increase the complexity of the displacementfield, however it will not increase the continuity of the field between elements. For example, a beam elementhas a cubic displacement field so has n = p + 1 = 4 nodes per element. A series of cubic C0 beamelements will have continuous displacement v but not a continuous displacement gradient ✓.

{u} =< 1 x x2 x3 >

8>><

>>:

a1a2a3a4

9>>=

>>;(5.91)

page 40 of 52



Lagrange interpolation always produces a C0 element so a different interpolation is used for higher-continuity:the Hermite interpolation. It uses the same polynomial but increases the DOF per node instead of the num-ber of nodes per element. A Hermite C1 beam element will therefore interpolate both deflection v anddeflection gradient ✓.

v v v

v v v

C 0

C 1v xd d/θ =θ

θ

Deflection v and rotation ✓ can be written as polynomials as:

v = a1 + a2x+ a3x2 + a4x

3

✓ =@v

@x= a2 + 2a3x+ 3a4x

2

θjθivi

vj

Four boundary conditions are then found with substitution of nodal values:

v(x = 0) = v1 = a1

✓(x = 0) = ✓1 = a2

v(x = l) = v2 = a1 + a2l + a3l2 + a4l

3

✓(x = l) = ✓2 = a2 + 2a3l + 3a4l2

The polynomial coefficient matrix is therefore a 4⇥ 4 matrix:

{r} = [A]{a}8>><

>>:

v1✓1v2✓2

9>>=

>>;=

2

664

1 0 0 00 1 0 01 l l2 l3

0 1 2l 3l2

3

775

8>><

>>:

a1a2a2a3

9>>=

>>;(5.92)

Shape functions can then be obtained from [N ] =< x > [A]�1 as

N1 = 1� 3⇠2 + 2⇠3 for v1 (5.93)N2 = ⇠L(⇠ � 1)2 for ✓1 (5.94)N3 = 3⇠2 � 2⇠3 for v2 (5.95)N4 = ⇠2L(⇠ � 1) for ✓2 (5.96)

page 41 of 52



N1

1

1

N

N2N

N3

1

1

N

N4N

page 42 of 52


CIVL4332 L6FEM Isoparametric Elements



Introduction

Development of finite elements in terms of global coordinate system is only possible for simple types of ele-ments. For more versatile elements (e.g. non-rectangular elements, elements with curved sides, elementsa different orientations in space, etc.) isoparametric formulation is used.

31 2

43

1

2

1

24

86

5

7

Isoparametric elements are elements for which the same shape functions are used to interpolate elementdisplacement field as well as element geometry. The element can then be formulated in natural coordinatesystem defined by the element geometry rather than its orientation in the physical coordinates system.Mapping between the natural and physical systems is used in the element formulation as:

Displacement field: {u} = [N ]{r}Element geometry : {X} = [N ]{C}

Where {r} and {C} are element nodal displacement and element nodal coordinates, respectively. Forisoparametric elements [N ] and [N ] are identical. If [N ] is of a higher degree than [N ] the element issubparametric (the reverse is superparametric).

page 43 of 52



Example 6.1: Isoparametric Geometry

A three-node triangular finite element has the following interpolation for element geometry and displacementfield:

� =nX

1

Lx

i

�i

Nodal coordinates (x, y) are (1,1), (8,2), and (4,6) and corresponding nodal horizontal displacements are1, -3, and 3 mm. Calculate the horizonal displacement within the element at location (2,2).

Lagrange interpolation is used for element geometry and displacement field. The element therefore has alinear interpolation and shape functions can be evaluated from polynomial u = a1 + a2x + a3y as seen inprevious examples.

However, as displacement is only required at point P = (2, 2), an alternate method can be used to onlyevaluate shape function values at point P:

uP

= N1,Pu1 +N2,Pu2 +N3,Pu3

As it is an isoparametric element, geometry can be used to solve these shape function values:

xP

= N1,Px1 +N2,Px2 +N3,Px3 = 2

yP

= N1,Py1 +N2,Py2 +N3,Py3 = 2

One additional equation is obtained fromP

Ni

= 1:

N1,P +N2,P +N3,P = 1

Solving this system of equations gives:

N1,P =3

4, N2,P =

1

16, N3,P =

3

16

Substitution gives horizonal displacement as:

uP

= 1⇥ 3

4� 3⇥ 1

16� 3⇥ 3

16= 0

Lagrange interpolation is used to map the physical coordinates (x, y) of an irregular element to a regulartemplate element expressed in natural coordinates (⇠, ⌘).

The coordinate from any point P within the template element can be mapped to the physical element usingthe same shape functions obtained for displacement field mapping, multiplied by the nodal coordinatesrather than nodal displacements.

page 44 of 52



x,uy,v

ξη

(1,1)

(1,-1)

(-1,1)

(-1,-1)

y3x3

12

43

( , )

y2x2( , )

y4x4( , )

y1x1( , )1 2

4 3yx( , )PP(ξ,η)

Interpolation of displacement field:

{u} =

⇢uv

�= [N ]{r} =

N1 0 N2 0 N3 0 N4 00 N1 0 N2 0 N3 0 N4

�

8>>>>>>>>>><

>>>>>>>>>>:

u1

v1u2

v2u3

v3u4

v4

9>>>>>>>>>>=

>>>>>>>>>>;

Interpolation of element geometry:

{X} =

⇢xy

�= [N ]{C} =

N1 0 N2 0 N3 0 N4 00 N1 0 N2 0 N3 0 N4

�

8>>>>>>>>>><

>>>>>>>>>>:

x1

y1x2

y2x3

y3x4

y4

9>>>>>>>>>>=

>>>>>>>>>>;

(6.97)

where the shape function is as described in the previous lecture for a bilinear quad element:

Ni

=(1 + ⇠⇠1)(1 + ⌘⌘

i

)

4

page 45 of 52



Jacobian Matrix

The shape functions Ni

are written in the natural system (⇠, ⌘), however the strain matrix [Bi

] = [@][Ni

]requires differentiation with respect to the physical system (x, y). The Jacobian matrix is therefore used tomap differentials w.r.t. the natural system to differentials w.r.t. the physical system:

⇢@

@⇠

@

@⌘

�= [J ]

⇢@

@x

@

@y

�(6.98)

where the Jacobian [J ], is:

[J ] =

"@x

@⇠

@y

@⇠

@x

@⌘

@y

@⌘

#(6.99)

Computationally, this is evaluated as:

[J ] =

"@N1@⇠

@N2@⇠

@N3@⇠

@N4@⇠

@N1@⌘

@N2@⌘

@N3@⌘

@N4@⌘

#2

664

x1 y1x2 y2x3 y3x4 y4

3

775 (6.100)

After calculation of [J ], the inverse is found as:

⇢@

@x

@

@y

�= [J ]�1

⇢@

@⇠

@

@⌘

�=

j11 j12j21 j22

�⇢@

@⇠

@

@⌘

�=

⇢j11

@

@⇠

+ j12@

@⌘

j21@

@⇠

+ j22@

@⌘

�(6.101)

where j11, j12, j21, j22 are elements of the Jacobian inverse. The shape function derivative matrix can thenbe expressed in natural coordinates as:

[B(x, y)] =

2

4@N

i

@x

00 @N

i

@y

@N

i

@y

@N

i

@x

3

5 =

2

64j11

@N

i

@⇠

+ j12@N

i

@⌘

0

0 j21@N

i

@⇠

+ j22@N

i

@⌘

j21@N

i

@⇠

+ j22@N

i

@⌘

j11@N

i

@⇠

+ j12@N

i

@⌘

3

75 = [B(⇠, ⌘)] (6.102)

Example 6.2: Jacobian Matrix

For the Quad-4 rectangular element shown below, the nodal displacements are:

{r}T = {0, 0, 0, 0, 0.25, 0.5, 0, 0}

Calculate the shear strain �xy

at the centre of the element.

page 46 of 52



x

y

ξ

η

(1,1)

(1,-1)

(-1,1)

(-1,-1)

v3u3

v2u2

v1u1

v4u4

2a

2b

1 2

4 3

Strain field is related to nodal displacements as:

{"} =

2

4"xx

"yy

�xy

3

5 = [B]{r} = [B1|B2|B3|B4]

8>>>>>>>>>><

>>>>>>>>>>:

0000

0.250.500

9>>>>>>>>>>=

>>>>>>>>>>;

Only node three has displacement so only matrix [B3] is required for interpolation.

[B3] = [@][N3] =

2

4@N3@x

00 @N3

@y

@N3@y

@N3@x

3

5

where N3 is given by the bilinear Quad-4 shape function for i = 3 so (⇠3, ⌘3) = (1, 1):

N3 =(1 + ⇠)(1 + ⌘)

4

Mapping from physical to natural coordinates gives x = a⇠ and y = b⌘. This is used to populate theJacobian matrix:

[J ] =

"@x

@⇠

@y

@⇠

@x

@⌘

@y

@⌘

#=

a 00 b

�

[J ]�1 =1

|J |

b 00 a

�=

1

ab

b 00 a

�=

1/a 00 1/b

�

page 47 of 52



which can be used to swap from physical to natural partial derivatives:⇢

@

@x

@

@y

�= [J ]�1

⇢@

@⇠

@

@⌘

�=

1/a 00 1/b

�⇢@

@⇠

@

@⌘

�=

⇢ 1a

@

@⇠

1b

@

@⌘

�

The strain components are therefore given by:

2

4"xx

"yy

�xy

3

5 = [B3(⇠, ⌘)]

⇢0.250.5

�=

2

64

1a

@N3@⇠

0

0 1b

@N3@⌘

1b

@N3@⌘

1a

@N3@⇠

3

75⇢

0.250.5

�

and the shear component as:

�xy

=1

b

@N3

@⌘(0.25) +

1

a

@N3

@⇠(0.5)

=1 + ⇠

4b(0.25)

1 + ⌘

4a(0.5)

At the centre of the element, (⇠, ⌘) = (0, 0), which gives shear strain of:

�xy

=1

16a+

1

8b

The stiffness matrix integral can be re-formulated in the (2D) natural system as:

[K] =

Z[B]T [D][B]dV

= t

Z Z[B]T [D][B]dxdy

= t

Z 1

�1

Z 1

�1

[B(⇠, ⌘)]T [D][B(⇠, ⌘)]|J |d⇠d⌘ (6.103)

where |J | is the determinant of the Jacobian matrix which gives an element area mapping between physicaland natural systems. The physical finite element must be geometrically well-proportioned to avoid a singular[J ] matrix. Ill-proportioning could include large side length aspect ratios and internal element angles.

page 48 of 52



Example 6.3: Jacobian Stiffness Formulation

For a 1D quadratic (three-node) element with length L, Young’s modulus E, and cross-section area A:

i) Formulate the shape functions of Example 5.1 in the natural system (�1 ⇠ 1).

ii) Find the element stiffness matrix.

i) Physical coordinates are x1 = 0, x2 = 0.5l, x3 = L and corresponding natural coordinates are coordi-nates ⇠1 = �1, ⇠2 = 0, ⇠3 = 1.

The mapping function is therefore given as:

⇠ = 2x

L� 1

1 3

x3x12

x2

Shape functions from Example 5.1 can then be reformulated in the natural system as:

N1 = 1� 3x

l+ 2

⇣xl

⌘2

=1

2⇠(⇠ � 1)

N2 = 4x

l� 4

⇣xl

⌘2

= (1� ⇠2)

N3 = �x

l+ 2

⇣xl

⌘2

=1

2⇠(⇠ + 1)

ii) The inverse Jacobian matrix is:

[J ]�1 =d⇠

dx=

2

L

and the shape function differential matrix with respect to the natural system is:

B = [B1 B2 B3] =

dN1

dx

dN2

dx

dN3

dx

�= [J ]�1

dN1

d⇠

dN2

d⇠

dN3

d⇠

�

=2

L

⇠ � 1

2, � 2⇠ , ⇠ +

1

2

�

page 49 of 52



The stiffness matrix is then formulated from:

[K] =

Z[B]T [D][B]dV = A

Z[B]T [D][B]dx = A

Z +1

�1

[B]T [D][B]|J |d⇠

= A

Z +1

�1

2

L

2

4⇠ � 1

2�2⇠⇠ + 1

2

3

5 [E]2

L

⇠ � 1

2, � 2⇠ , ⇠ +

1

2

�L

2d⇠

=2AE

L

Z +1

�1

2

4(⇠ � 1

2)2 �2⇠(⇠ � 1

2) (⇠ � 12)(⇠ +

12)

4⇠2 �2⇠(⇠ + 12)

Sym. (⇠ + 12)

2

3

5 d⇠

=2AE

L

2

4(⇠ � 1

2)3/3 �(⇠2(4⇠ � 3))/6 (⇠(4⇠2 � 3))/12

4⇠3/3 �(⇠2(4⇠ + 3))/6Sym. (⇠ + 1

2)3/3

3

5+1

�1

=AE

6L

2

414 �16 2

32 �16Sym. 14

3

5

Numerical Integration (Gauss Quadrature) and Stress Recovery

Integration in the natural system is much easier, as bounds are between -1 and +1. There is still however alarge (and computationally demanding) number of integrals that need to be evaluated to determine elementstiffnesses. An efficient numerical integral process is therefore adopted termed Gauss quadrature. Thisintegral is written as:

[K] =X

i

X

j

wi

wj

f(⇠i

⌘j

) (6.104)

where

f(⇠i

⌘j

) = t[B(⇠i

, ⌘j

)]T [D][B(⇠i

, ⌘j

)]|J(⇠i

, ⌘j

)| (6.105)

and wi

and wj

are weight functions, i and j are the sampling (Gauss) points upto the number of Gausspoints N

GP

, i.e. 1 i NGP

and 1 j NGP

.

Gauss quadrature can be understood as a more accurate and efficient version of common numerical inte-gral methods such as the midpoint, trapezium, or Simpon’s rules. It yields exact results for polynomials ofdegree 2n� 1 with n function evaluation points.

Gauss points are selected at a certain location with the element and given a pre-determined weighting,listed in the below table. N

GP

as listed in the table is a 1D case. A 2D case will use N2GP

and a 3D casewill N3

GP

, with the same listed weightings.

page 50 of 52



ξ+1-1

f( )ξ

-1/√3 ξ

f( )ξ

+1/√3ξ

f( )ξ

0 +1-1 0

Midpoint Trapezoidal Gauss

NGP ⇠i

or ⌘i

wi

or wj

1 0 2.02 +1/

p3, �1/

p3 1.0, 1.0

3 +p0.6, 0, �

p0.6 5/9, 8/9, 5/9

ξη

+1-1+1/√3-1/√3

ξ

η

+1-1+√0.6-√0.6

ξη

+1-1

NGP= 1 NGP= 2 NGP= 32 22

Once nodal displacements are obtained, element stresses can be calculated or each element from

{�} = [D][B]{r}

Stresses are most accurate at Gauss points of a quadrature rule one order less than that required for fullintegration of the element stiffness. This is because FEM typically generates an over-stiff response from thedisplacement field mapping assumption and reduced Gauss point integration counteracts this somewhat.To give satisfactory numerical values for a bilinear element, two gauss points is sufficient. For quadraticelements, two to three gauss points is sufficient. Stresses at any location within the element can be obtainedusing interpolation/extrapolation from Gauss points.

Example 6.3: Numerical Integration

Compare explicit and numerical (gauss quadrature) integration of:

I =

Z 1

�1

f(⇠)d⇠ =

Z 1

�1

(1 + ⇠2)d⇠

page 51 of 52



Explicit: Z 1

�1

(1 + ⇠2)d⇠ = [⇠ + ⇠3/3]1�1 =8

3= 2.6667

Numerical:

Gauss is exact if p 2(NGP

� 1). Above function is second-order polynomial, gauss quadrature is exactwith N

GP

= 2. For NGP

= 2, (⇠i

= ±1/p3, w

i

= 1):

I =X

wi

f(⇠i

) = 1(1 + (�1/p3)2) + 1(1 + (1/

p3)2) = 2.6667

which has 0% error.

For NGP

= 1, (⇠i

= 0, wi

= 2):

I =X

wi

f(⇠i

) = 2(1 + 02) = 2.0

which has 33% error.

page 52 of 52

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