Lecture 5 ch - ETH Zürich · Remember that is ... The Timoshenko Beam Element ... dx Title:...

Institute of Structural Engineering Page 1

Method of Finite Elements I

Chapter 5

Structural Elements:The truss & beam elements

Method of Finite Elements I30-Apr-10

Today’s Lecture Contents• The truss element

• The Euler/Bernoulli beam element

FE Classification

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 Element

i. 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)

Strain-Displacement relation

define The 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

uu N N

1 1 1 1N x N

B Jx L

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:

Elastic material:

From the Continuous to the Discrete form

Therefore the relation turns to

Rearranging terms:

Therefore the relation turns to

Rearranging terms:

Finally 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

Example 2: Truss Element under uniform distributed tensile force

Strong Form Solution

Boundary Conditions

Solution

Finite Element Solution

But now we need to account for the distributed loading

Boundary Conditions

Solution

0 0.5 1 1.5 2Length (m)

other two Prismatic Element

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

perpendicular to the beam axis

Assumptions

The 2−node Euler/Bernoulli beam element

other two Prismatic Element

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

perpendicular to the beam axis

Finite Element Idealization

We are looking for a 2-node finite element formulation

Kinematic Field

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

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

: normal stress: bending moment

R : curvature

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

: dwslopedx

1: d wcurvatureR dx

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

d w Mdx EI

Assume an infinitesimal element:

Force Equilibrium:

V x p x x V x x

V x x V xp x

Moment Equilibrium:

xM x V x x M x x p x x

M x x M x xV x p xx

dM V x

dV p xdx

Basic Beam Theory

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

Interpolation Scheme

Beam homogeneous differential equation

4 ( ) 0d wEI p xdx

uDirichlet w wdwdx

d wNeumann EI Mdxd wEI Sdx

Boundary Conditions

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:

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

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

So 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

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

Interpolation Scheme in isoparametric coordinates

1 1 221 1 121 1 221 1 12

Hermite Polynomials

2 1 3 1 5 2 6 2

2 3 5 6 1 1 2 2

with 2 2 2

andl l dx Lw N w N N w N J

dl lN N N N N d w w

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

Since the axial and bending displacement fields are uncoupled

Shape Function Matrix [N]Shape Function Matrix [N]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 TheoryEuler/ Bernoulli Theory

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

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

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

The evaluation of the Beam Element force vector

The beam element force vector is readily derived as:

T Te eTe e

d N d Nf N p x dx M S

Assuming constant distributed load p:

Te e e

LpLf N x pdx p N Jd

The Timoshenko Beam Element

The transverse shear strain now also comes into play:dwdx

Lecture 5 ch - ETH Zürich · Remember that is ... The Timoshenko Beam Element ... dx Title:...

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