Method of Finite Elements I - Homepage | ETH Zürich · 2019-03-31 · Method of Finite Elements I...

Institute of Structural Engineering

Method of Finite Elements I

Chapter 5

Structural Prismatic Elements:The truss & beam elements


Method of Finite Elements I

Chapter Goals

Learn how to formulate the Finite Element Equations for 1D elements, and specifically The bar element (review)

The Euler/Bernoulli beam element

What is the Weak form?

What order of elements do we use?

What is the isoparametric formulation?

How is the stiffness matrix formulated?

How are the external loads approximated?


Method of Finite Elements I30-Apr-10

Today’s Lecture Contents• The truss element

• The Euler/Bernoulli beam element



FE Classification



The 2-node truss element

Uniaxial Elementi. The longitudinal direction is sufficiently larger than the other two ii. The bar resists an applied force by stresses developed only along its

longitudinal direction

Prismatic Elementi. The cross-section of the element does not change along the element’s length

Assumptions



Variational Formulation

Prismatic Element:

Uniaxial Element: Only the longitudinal stress and strain components are taken into account. The traction loads degenerate to distributed line loads



Finite Element Idealization

Truss Element Shape Functions

(in global coordinates)

( ) [ ] 11 2

2

uu x N N

u

=



Strain-Displacement relation

defineThe truss element

stress displacement matrix

Constant Variation of strains along the element’s length



Or we may use the isoparametric formulation:

Once again:The truss element

stress displacement matrix

Constant Variation of strains along the element’s length

Isoparametric Shape Functions

( ) ( ) ( ) 11 2

2

uu N N

uξ ξ ξ

=

( ) ( )11 12

N ξ ξ= −

( ) ( )21 12

N ξ ξ= +

( ) ( ) ( ) [ ]1 1 1 1N x N N

B Jx x L

ξ ξξξ ξ

−∂ ∂ ∂∂= = = = −

∂ ∂ ∂ ∂

-11

-1 1



From the Continuous to the Discrete form – from Strong to Weak form

It is convenient to re-write the Principle of Virtual Work in matrix form

where:

and:

Elastic material:

and for concentrated loads on the element end nodes 1, 2, or:

( ) ( )0

LTF N x q x dx = ∫

for distributed loads q(x)



From the Continuous to the Discrete form

Therefore the relation turns to

Rearranging terms:




Therefore the relation turns to

Rearranging terms:

This has to hold xuδ∀ which implies that:




the discrete form of the truss element equilibrium equation reduces to

or more conveniently

where:

Performing the integration, the following expression is derived



Example 1: Truss Element under concentrated tensile force

Strong Form Solution Finite Element Solution

Boundary Conditions

Solution

Boundary Conditions

Solution

11

22

1 11 1

x

x

uf EAuLf

− = −

1

2

01 11 1100

xf EAuL

− = −

( )

( ) ( ) ( )2

1 1 2 2

100 , Lu u L thenEA

u x N x u N x u

= =

= +



Example 2: Truss Element under uniform distributed tensile force

Strong Form Solution

Boundary Conditions

Solution



1

20

1 112

Lx x

xx

xf q LL q dx

xfL

− − = =

∫


Finite Element Solution

But now we need to account for the distributed loading

Boundary Conditions

Solution

Displacement at node 2@ x=L

11

22

1 11 1

x

x

uf EAuLf

− = −

21

22

1 1 11 1 12 2

x xuq L q LEA uuL EA

− − = ⇒ = −





Displacement at node 2@ x=L

2

2 2xq L

uEA

=

( ) ( ) ( )1 1 2 2 2xq L

u x N x u N x u xEA

= + =





2

2 2xq L

uEA

=

( ) ( ) ( )1 1 2 2 2xq L

u x N x u N x u xEA

= + =




0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2Length (m)

( ) ( ) ( )1 1 2 2 2xq L

u x N x u N x u xEA

= + =




0

0.5

1

1.5

2

0 0.5 1 1.5 2Length (m)

( ) ( ) ( )1 21 2 2

xdN x dN x q Lu x u u

dx dx EA= + =



Uniaxial Elementi. The longitudinal direction is sufficiently larger than the

other two Prismatic Element

i. The cross-section of the element does not change along the element’s length

Euler/ Bernoulli assumptioni. Upon deformation, plane sections remain plane AND

perpendicular to the beam axis

Assumptions

The 2-node Euler/Bernoulli beam element



The 2-node Euler/Bernoulli beam element

Uniaxial Elementi. The longitudinal direction is sufficiently larger than the other two

Prismatic Elementi. The cross-section of the element does not change along the element’s length

Euler/ Bernoulli assumptioni. Upon deformation, plane sections remain plane AND perpendicular to the

beam axis




We are looking for a 2-node finite element formulation




©Carlos Felippa

DOFs relating to pure bending

1w2w

2w

1w

http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch06.d/IFEM.Ch06.Slides.d/IFEM.Ch06.Slides.pdf



Kinematic Field

Two deformation components are considered in the 2-dimensional case1. The axial displacement2. The vertical displacement

dw wdx

θ ′= =



Kinematic FieldPlane sections remain plane AND perpendicular to the beam axis




Point A displacement(that’s because the section remains plane)

(that’s because the plane remains perpendicular to the neutral axis)




Therefore, the Euler/ Bernoulli assumptions lead to the following kinematic relation



Basic Beam Theory

, ,M Ey yI E R

σσ ε σ= = = −

: normal stress : bending moment

R : curvature

where

Mσ

Reminder: from the geometry of the beam, it holds that:

w

: dwslopedx

θ =

θ2

2

1: d wcurvatureR dx

κ = =

(that’s because the plane remains perpendicular to the neutral axis)Therefore:

2

2

d w Mdx EI

= −

p q

bax

Δx

y



Assume an infinitesimal element:

Force Equilibrium:

( ) ( ) ( )( ) ( ) ( ) 0

0

x

V x p x x V x x

V x x V xp x

x∆ →

+ ∆ − + ∆ = ⇒

+∆ −= →

∆

Moment Equilibrium:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 0

02

2x

xM x V x x M x x p x x

M x x M x xV x p xx

∆ →

∆− − ∆ + + ∆ − ∆ = ⇒

+∆ − ∆= + →

∆( )dM V x

dx=

( )dV p xdx

=

Basic Beam Theory

2 4

2 4, ( ) ( )d M d wFinally p x EI p xdx dx

= ⇒ − =



Interpolation Scheme

Beam homogeneous differential equation

4

4 ( ) 0d wEI p xdx

+ =

: on

on

uDirichlet w wdwdx θθ θ

= Γ

= = Γ

2

2

3

: on

on

M

S

d wNeumann EI Mdx

d wEI Sdx

− = Γ

− = Γ

Boundary Conditions



Interpolation Scheme – the Galerkin Method

Beam homogeneous differential equation

That’s a fourth order differential equation, therefore a reasonable assumption for the interpolation field would be at least a third order polynomial expression:

or in matrix form:

Therefore the rotation would be a second order polynomial expression:

(I)




The relation must hold for the arbitrary displacements at the nodal points:

Therefore, by solving w.r.t the polynomial coefficients :

Now we can derive the 2-dimensional Euler/Bernoulli finite element interpolation scheme (i.e. a relation between the continuous displacement field and the beam nodal values):




Therefore by substituting in (I):

Or more conveniently:

where:



0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0N5

0.2 0.4 0.6 0.8 1.0

0.14

0.12

0.10

0.08

0.06

0.04

0.02

N6

Interpolation Scheme in global coordinates

After some algebraic manipulation the following expressions are derived for the shape functions

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0N2

0.2 0.4 0.6 0.8 1.0

0.02

0.04

0.06

0.08

0.10

0.12

0.14

N3

Institute of Structural Engineering Page 37


Interpolation Scheme in isoparametric coordinates

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

21 2

21 3

22 5

22 6

1 1 241 1 141 1 241 1 14

Hermite Polynomials

w N

N

w N

N

ξ ξ ξ

θ ξ ξ ξ

ξ ξ ξ

θ ξ ξ ξ

→ = − +

→ = − +

→ = + −

→ = + −

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) [ ]

2 1 3 1 5 2 6 2

2 3 5 6 1 1 2 2

with 2 2 2

, 2 2

e e

andl l dx lw N w N N w N J

dl lN N N N N d w w

ξ ξ ξ θ ξ ξ θξ

ξ ξ ξ θ θ

= + + + = =

= =




What about the axial displacement component ??????

Since the axial and bending displacement fields are uncoupled

Shape Function Matrix [N]



Remember that is the beam’s curvature

Strain-Displacement compatibility relations

2-dimensional beam element

The normal strain :

The shear strain :

The normal strain : Euler/ Bernoulli Theory

The Euler/ Bernoulli assumptions predict zero variation of both the shear strain and the vertical component of the normal strain



Strain-Displacement relations in matrix form

We can re-write the strain displacement relation in the following form

and we can easily substitute the matrix interpolation form



Strain-Displacement relations in matrix form

So the strain-displacement matrix assumes the following form

where: and:

and by substituting the expressions of the shape functions, the strain-displacement matrix Bassumes the following form

or in isoparametric coordinates:

2

2

1 6 61 3 1 1 3 1 with & e e edw dw d wB J B dL L L d dx d

ξ ξξ ξξ ξ

= − − + = =



The evaluation of the Beam Element Stiffness Matrix

The beam element stiffness matrix is readily derived as:

Performing the integration with respect to x:

where: is the cross-sectional moment of inertia and is

( )e TK B EI Bd= Ω∫



The evaluation of the Beam Element force vector

The beam element force vector is readily derived as:

( ) ( ) ( ) T

S

Me

e

TeTe e e

ff

d Nf N p x dx M N S

dx

Ω

Γ

ΓΩ

Γ

= + +

∫

Assuming constant distributed load p:

( ) ( )

1

612

6

Te e e

LpLf N x pdx p N Jd

Lξ

ξ ξΩΩ

= = = −

∫ ∫L

p

Method of Finite Elements I - Homepage | ETH Zürich · 2019-03-31 · Method of Finite Elements I...

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