Method of Finite Elements I - Homepage | ETH Zürich · 2020. 3. 29. · Method of Finite Elements I. 30-Apr-10 Uniaxial Element i. The longitudinal direction is sufficiently larger

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Method of Finite Elements I

Chapter 5

The Euler Bernoulli Beam

Book Chapters[O] V2/Ch1[F] Ch1



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



Reminder: 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



The Weak form of the problem can be derived either via use of the Principle of Virtual Work (energy methods) or via use of the Galerkin approximation.

Form Strong to Weak Form(reminder from Lecture 3)

( )

2

2 ( )

(0) 0

0

x L

Strong Form

Boundary Conditions (BC)Essential BC

Natural BC

uduL AE

d

Rd

d uA x

x

E qx

σ=

=

=

=

⇒ =

x

R

q(x)=qx



Variational Formulation – Weak Form

Prismatic Element:

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

Reminder: The 2-node truss element



Finite Element Idealization – Global Coordinates

Truss Element Shape Functions

in global coordinates

( ) [ ] 11 22

uu x N N

u

=



Strain-Displacement relation

defineThe truss element

stress displacement matrix

Constant Variation of strains along the element’s length



Finite Element Idealization – Isoparametric Coordinates (ξ)

Truss Element Shape Functions

in isoparametric coordinates

, 2A l =ξ

1ξ = − 1ξ =0ξ =

( ) ( )11 12

N ξ ξ= −

-1 1

( ) ( )21 12

N ξ ξ= +

-1 1

( ) ( ) ( ) 11 22

uu N N

uξ ξ ξ

=

( )2

1 21

1 1(1 ) (1 )2 2 i ii

x x x N xξ ξ ξ=

= − + + =∑

We introduce the following coordinate transformation:



Isoparametric formulation:

Strain Displacement Matrix

Constant Variation of strains along the element’s length

( ) ( ) ( ) 11 22

uu N N

uξ ξ ξ

=

( ) ( )11 12

N ξ ξ= − ( ) ( )21 12

N ξ ξ= +

( ) ( ) ( ) [ ]1 1 1 1N x N NB Jx x L

ξ ξξξ ξ

−∂ ∂ ∂∂= = = = −∂ ∂ ∂ ∂

, 2A l =ξ

1ξ = − 1ξ =0ξ =



Rewriting the Weak Form using the Basis Function approximationIt 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

{ } ( ) ( )0

or L

TF N x q x dx = ∫for distributed loads q(x)

virtual strains

virtual disp.

actualstress

actual load



or more conveniently

where:

The FEM EquationThe discrete form of the truss element equilibrium equation reduces to

11

220

11 1

11 2

2 1 2 12 1

L

x

x

LL L

L

u fE Adx

u f×

× ××

=

−−∫



The FEM EquationThe discrete form of the truss element equilibrium equation reduces to

Performing the integration, the following expression is derived

Finally, the FEM Equation to solve is:

11

220

11 1

11 2

2 1 2 12 1

L

x

x

LL L

L

u fE Adx

u f×

× ××

=

−−∫

[ ] [ ]

{ }

{ }

11

22 2 2 2 1 2 1

x

xx

u fK K

u fK u f× × ×

= ⇒ =



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


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


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

beam axis

Assumptions



Finite Element Idealization

We are looking for a 2-node finite element formulation



Finite Element Idealization

We are looking for a 2-node finite element formulation

*DOFs relating to bending



Kinematic Field

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

dw wdx

θ ′= =

neutral axis



Kinematic FieldPlane sections remain plane AND perpendicular to the beam axis



Plane 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)

Kinematic Field



Plane sections remain plane AND perpendicular to the beam axis

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

Kinematic Field



Basic Beam TheoryMechanics Principles , ,M Ey E y

I Rσ σ ε σ= − = = −

: normal stress : bending moment

1/R : curvature

where

Mσ

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

: dwslopedx

θ =2

2

1: d wcurvatureR dx

κ = =

Therefore: M Ey yI R

σ = − ⇒ − M yI

= −2

2

d w Mdx EI

⇒ =

p q

bax

Δx

y



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

=

2 4

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

= ⇒ =

Basic Beam TheoryAssume an infinitesimal element:



The Strong Form Equation (continuous)

Beam homogeneous differential equation

4

4 ( )d wEI p xdx

=

: on

on

uDirichlet w wdwdx θ

θ θ

= Γ

= = Γ

2*

2

3

: on

on

M

S

d wNeumann EI Mdx

d wEI Sdx

= Γ

= Γ

Boundary Conditions

*we assume here for simplicity only the effect of distributed loads p(x) and no distributed moments m(x)

*in case of prescribed concentrated loads (S) or moments (M) acting o the boundary Γ



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:

In matrix form:

Therefore the rotation would be a second order polynomial expression:

(I)



Interpolation SchemeWe demand that the relation holds 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:

(shape function matrix)

Interpolation Scheme



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



Interpolation Scheme in global coordinates

The resulting shape function must be C1 continuous, i.e., both the deflection, w(x), and its derivative the slope θ, must be continuous over the entire member, but also: between adjacent beam elements.



In order to meet the C1 continuity requirement between adjacent elements, the Hermitian cubic shape functions are used.

These shape functions are conveniently expressed in terms of the same dimensionless “natural” coordinate that was used for the bar (truss) element.

Interpolation Scheme in isoparametric coordinates

( )2

1 21

1 1(1 ) (1 ) where Jacobian 2 2 2i ii

dx lx x x N x Jd

ξ ξ ξξ=

= − + + = = =∑

ξ-1 0 1



Interpolation Scheme in isoparametric coordinates

©C

arlo

s Fe

lippa

1 1w =

2 1w =

( ) ( ) ( ) ( ) ( )2 1 3 1 5 2 6 2 with 2dx lw N w N N w N J Jd

ξ ξ ξ θ ξ ξ θξ

= + + + = = =

( )2N ξ=

( )3N ξ=

( )5N ξ=

( )6N ξ=

http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch12.d/IFEM.Ch12.Slides.d/IFEM.Ch12.Slides.pdf



Interpolation Scheme

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

Since the axial and bending displacement fields are uncoupled

Shape Function Matrix [N]



For a 2-dimensional beam element, the relevant strains are

Strain-Displacement compatibility relations

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 formWe can re-write the strain displacement relation in the following vector multiplication form

But u0, w can be expressed in terms of the shape function interpolation

strain-displacement matrix

[ ]Bε



Strain-Displacement relations in Global CoordinatesSo the strain-displacement matrix assumes the following form

where: and:

By substituting the expressions of the shape functions in global coordinates:

[ ] 2 3 2 3 2 3 2 32 3 2 2 3 2

1 0 0 0 0 3 2 2 3 20 1 0

x xL L

N x x x x x x x xxL L L L L L L L

−

= − + − + − − +

[ ] 2 21 6 12 6 6 12 61 4 1 2x x x xB y y y yL L L L L L Lε = − − − + −

The strain-displacement matrix [Bε] in global coordinates is:



Strain-Displacement relations in Isoparametric CoordinatesSo the strain-displacement matrix assumes the following form

In isoparametric coordinates this differentiation is achieved via the chain rule:

( ) ( )( ) ( ) ( )Chain Rule 1, ,of Differentiation

2 for 1, 4ii i ii x i xdN xdN x dN dNdN N J i

dx d dx d L dξ ξ ξξ

ξ ξ ξ−= → = = = =

[ ] 1, 2, 3, 4, 5, 6,2 2 2 22 4 4 2 4 4y y y yB N N N N N NL L L L L Lε ξ ξξ ξξ ξ ξξ ξξ = − − − −

Which yields the following expression for [Bε] in terms of ξ:

( ) ( )( ) ( )( )

( ) ( )

2Chain Rule

, ,of Differentiation2

21

, 2 2

2 2

2 4 for 2,3,5,6

i iii xx i xx

i ii xx

dN x dN xd N x d d dN Ndx dx L d d L d dx

dN d NdN J id L d L d

ξ ξ ξξ ξ ξ

ξ ξξ ξ ξ

−

= = → = =

⇒ = = =



Strain-Displacement relations in Isoparametric CoordinatesSo the strain-displacement matrix assumes the following form

Let us rewrite the expressions of the shape functions in isoparametric coordinates Ν(ξ):

[ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2

1 10 0 0 02 2

1 10 1 2 1 1 0 1 2 1 14 8 4 8

Nl l

ξ ξ

ξ ξ ξ ξ ξ ξ ξ ξ

− +

= − + − + + − − + −

( ) ( )1 6 6[ ] 1 3 1 1 3 1 with dw dwB y y y y JL L L d dxε

ξ ξξ ξξ

= − − − − − + =

[ ] 1, 2, 3, 4, 5, 6,2 2 2 22 4 4 2 4 4y y y yB N N N N N NL L L L L Lε ξ ξξ ξξ ξ ξξ ξξ = − − − −

The strain-displacement matrix [Bε] in isoparametric coordinates then is:



Principle of Virtual Work for the Beam

( ) ( )0

L

xx xx i yi jVi j j

dw dwdV wp x m x dx w F Mdx dx

δε σ δ δ δ δ = + + + ∑ ∑∫ ∫ ∫ ∫

Assume an imposed virtual displacement field on the beam: , dwwdx

δ δ

Internal Virtual Work External Virtual Work

( )p x

( )m x

Ω



Principle of Virtual Work for the Beam

2 2

2 2xx xxV V

d w d wdV y E y dVdx dx

δε σ δ

= − −

∫ ∫ ∫ ∫ ∫ ∫

( ) ( )0

L

xx xx i yi jVi j j

dw dwdV wp x m x dx w F Mdx dx

δε σ δ δ δ δ = + + + ∑ ∑∫ ∫ ∫ ∫

Internal Virtual Work External Virtual Work

The internal work is calculated as:

Assume an imposed virtual displacement field on the beam: , dwwdx

δ δ

2 22

2 20

L

xx xxVA

d w d wdV y dA dxdx dx

δε σ δ

⇒ =

∫ ∫ ∫ ∫ ∫ ∫2 2

2 20

L

xx xxV

d w d wdV EI dxdx dx

δε σ δ

⇒ =

∫ ∫ ∫ ∫



The evaluation of the Beam Element Stiffness Matrix

The beam element stiffness matrix is then readily derived as:

where the curvature displacement matrix [B] is the matrix that links curvature to nodal deformations:

( )e TL

K B EI B dx= ∫

[ ] 1 6 63 1 3 1BL L L

ξ ξξ ξ = − − +

[ ]2

2

i

izj

jz

wd w Bdx w

θ

θ

=

[B] is obtained from [Bε] by retailing only the bending degrees of freedom and therefore elements B2, B3, B5, B6 divided by –y :





Performing the integration with respect to dx yields the following stiffness for the bending DOFs:

where: is the cross-sectional moment of inertia and

( )e TL

K B EI B dx= ∫

3 3 2

2

3 2 3 2

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

ebending

EI EI EI EIL L L LEI EI EI EIL L L LK

EI EI EI EIL L L LEI EI EI EIL L L L

− −

= − − −

−





By adding the terms of the axial stiffness we obtain the complete beam element stiffness matrix

( )e TL

K B EI B dx= ∫



The evaluation of the Beam Element force vectorThe beam element force vector is readily derived as:

( ) ( ) ( ) ( ) ( )TF

Me

e

T Te eTe e e

yL L

ff

d N d Nf N p x dx m x dx N F M

dx dx

Ω

Γ

Γ

Γ

= + + +

∫ ∫

Assuming constant distributed load p:

( )

( )( )( )( )

2

3

5

6

1

612

6

Te e

L L

N LN pLf N x pdx p JdNN L

ξξ

ξξξ

Ω

= = = −

∫ ∫ Lp

distributed forces concentrated forces onthe boundary

Slide Number 1Slide Number 2Slide Number 3Slide Number 4Slide Number 5Form Strong to Weak Form� (reminder from Lecture 3)Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Slide Number 27Slide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Slide Number 40Slide Number 41Slide Number 42Slide Number 43Slide Number 44

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Method of Finite Elements I - Homepage | ETH Zürich · 2020. 3. 29. · Method of Finite Elements I. 30-Apr-10 Uniaxial Element i. The longitudinal direction is sufficiently larger