Finite Element Method 2D frame systems - Lublinakropolis.pol.lublin.pl/users/jpkmb/FEM_4.pdfThe element of a 2D frame is more general than a truss element presented in Chapter 2 because

Finite Element Method 2D frame systems

Jerzy Podgórski

Introduction

The correct choice of model for a structure is very important for quality and exactness of the results obtained. The choice of frame or truss (for example, a truss with fixed nodes) is often subjective and it depends on experience and intuition of the analyst.

In this chapter, we will present the following model of a bar structure - a 2D frame which gives more possibilities.

Introduction

The element of a 2D frame is more general than a truss element presented in Chapter 2 because with help of this element we can also model ideal truss structures (articulated connection of elements at nodes). We can simply say that a frame is a structure whose bars can be bent while truss elements can be only compressed and stretched.

Introduction

It has the following consequences:

• bar (an element) of a frame can be loaded between nodes,

• modelling of different types of loads is possible,

• connection of an element with a node can be a fixed or an articulated connection,

• node of a 2D frame has 3 degrees of freedom,

Introduction

In the case of plane frames, we will neglect index Z of rotation angles in our notation because all rotation angles on the plane XY (which we will use to describe the structure) are rotations with respect to the Z axis. Let us assume that a frame element is straight and homogeneous which means that it is made from a homogeneous material and has a constant cross section.

Introduction

the global coordinate system

Introduction

the local coordinate system

The element stiffness matrix for a 2D frame

We group nodal displacements and forces in column matrices just as we did previously. They are called vectors:

• displacement and nodal forces vector of the first node i and the last node j in the local system

u'i

ix

iy

i

u

u

u' j

jx

jy

j

u

u

f 'i

ix

iy

i

F

F

M

f ' j

jx

jy

j

F

F

M


• element displacement vector in the local coordinate system

element forces vector in the local coordinate system

uu

u'

'

'e i

j

ix

iy

i

jx

jy

j

u

u

u

u

ff

f'

'

'e i

j

ix

iy

i

jx

jy

j

F

F

M

F

F

M


We can also describe all the vectors formulated above in the global system:

u i

iX

iY

i

u

u

u j

jX

jY

j

u

u

f i

iX

iY

i

F

F

M

f j

jX

jY

j

F

F

M

uu

ue i

j

iX

iY

i

jX

jY

j

u

u

u

u

ff

fe i

j

iX

iY

i

jX

jY

j

F

F

M

F

F

M


As in the previous chapters, the relationship between nodal forces and nodal displacements will be of great importance. This relation (analogous to a truss) in the local coordinate system has the form:

and in the global system

K u f' ' 'e e e

K u fe e e


At the moment, we will concentrate on searching for the stiffness matrix K’e in the local coordinate system and next its transformation to the global system.

Equilibrium equations of the element lead to the following relations between nodal forces:

F F F F Fx ix jx ix jx 0

F F Fy iy jy 0 M M M F Li i j jy 0


It has been shown that three equations are unable to calculate six components of the vector . The discussion concerning element strains will provide these missing equations. The deformation caused by the axial forces Fix and Fjx is identical to the deformation of a truss element, hence we take advantage of previously determined dependence:

NEA

Lu ujx ix F

EA

Lu uix ix jx F

EA

Lu ujx ix jx


We will obtain the remaining equations when we consider the flexural deformation of an element and the relationship between shearing forces and bending moments. The well-known relationship between curvature and bending moment is:

1

1

2

2

23

2

d y

dx

dy

dx

M x

EJ z


ρ - the radius of a curvature,

E - Young’s modulus of a material,

Jz - the moment of inertia of an element cross section

1

1

2

2

23

2

d y

dx

dy

dx

M x

EJ z


The equilibrium of one section of a bar in bending gives the equation:

T xdM x

dx( )

( )

d y

dx

M x

EJ z

2

2

( )

Since we are dealing with linear structures with small deflections, we assume dy/dx<< 1, which simplifies the relationship between curvature and bending moment to the well-known form:


The opposite sign of the right side of to the one that we have usually assumed, comes from the sense of the y axis of the local coordinate system which is orientated anticlockwise in our assumptions.

Differentiating this equation twice, we obtain the relationship:

d y

dx

M x

EJ z

2

2

( )

d y

dx

q x

EJ

y

z

4

4

( )


qy(x) denotes the distributed load which is perpendicular to the axis of an element. Here the element is free from nodal loads, thus

Finally, we obtain the set of differential equations:

d y

dx

q x

EJ

y

z

4

4

( )

0yq

04

4

dx

yd

zEJ

xM

dx

yd )(2

2

zEJ

xT

dx

yd )(3

3


After integrating relations we obtain the following equations:

• bending line of the frame element:

• bending moment:

• shearing force:

04

4

dx

yd

y x Cx

Cx

C x C( ) 1

3

2

2

3 46 2

M x EJ C x Cz( ) 1 2

T x EJ Cz( ) 1

C1 ... C4 are integration constants which should

be determined on the basis of boundary

conditions.


We have four boundary conditions:

• at node i , x=0:

• at node j , x=L:

y uiy( )0 dy

dxx

i

0

y L u jy( ) dy

dxx L

j


After inserting these conditions into

, we obtain the

following values of the integration constants:

y x Cx

Cx

C x C( ) 1

3

2

2

3 46 2

CL

u u

Li j

jy iy

1 2

62

CL

u u

Li j

jy iy

2

14 2 6

C i3

C uiy4


Hence after putting the above equations into

and we obtain:

M x EJ C x Cz( ) 1 2 T x EJ Cz( ) 1

M MEJ

L

u u

Liz

i j

jy iy

( )0 4 2 6

M M LEJ

L

u u

Ljz

i j

jy iy

( ) 2 4 6

F TEJ

L

u u

Liyz

i j

jy iy

( )0 6 6 12

2

F T LEJ

L

u u

Ljyz

i j

jy iy

( )

26 6 12


Finally, tabulating equations ,

and from the previous slide

in a suitable sequence we obtain the stiffness matrix:

FEA

Lu uix ix jx

FEA

Lu ujx ix jx


The relationships described by equations

are called transformation formulae of the displacement method in structural mechanics (in some other form).

M MEJ

L

u u

Liz

i j

jy iy

( )0 4 2 6

M M LEJ

L

u u

Ljz

i j

jy iy

( ) 2 4 6

F TEJ

L

u u

Liyz

i j

jy iy

( )0 6 6 12

2

F T LEJ

L

u u

Ljyz

i j

jy iy

( )

26 6 12

The stiffness matrix in the local coordinate system

The transfer of the matrix to the global coordinate system is done according to rules analogous to the rules described for 2D truss element. In order to obtain the transformation matrix of an element, we need Ri that is, the transformation matrix from the local system to the global one for the node i.


Since the third degree of freedom is a rotation with respect to the z axis which does not change its location because it is always perpendicular to the plane xy, the rotation will be the same as for a truss element:

or

u u uiX ix iy cos sin

u u uiY ix iy sin cos

iZ iz i

i

iy

ix

i

iY

iX

u

u

cs

sc

u

u

100

0

0

iii uRu


In accordance that the frame element is straight, the transformation matrix of the node j is identical to Ri which leads to the final form of the element stiffness matrix:

Re

c s

s c

c s

s c

0 0 0 0

0 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0 1

After multiplying matrices we obtain the stiffness matrix of a frame element in the global coordinate system. Unfortunately, its form is rather complex.

K R K Re e e e '

T


2

2

AL

J z cosc sins

Static reduction of the stiffness matrix

Frame elements are not always joined at a node ensuring the agreement of all displacements of nodes and in the bar section at this node. Articulated joints shown in next figure are examples of such incomplete connections.

At this joint, the angle of the nodal rotation does not influence the rotation of the element section of a node. The latter e2 can rotate independently of the node.


The element joint scheme with one element able to rotate (an articulated joint).


We determine the unknown angle of the rotation of such an element using an additional equation which is given by the equilibrium condition of moments in a joint. Hence we can reduce the number of degrees of freedom of the element because the additional equilibrium condition allows us to eliminate one displacement from the set of equations.


Example 1 - articulated connection.

Additional equilibrium condition of a section at the node i : Mi = 0 leads, after considering

equations , ,

to conditions:

f 'i

ix

iy

i

F

F

M

f ' j

jx

jy

j

F

F

M

K u f' ' 'e e e

EJ

L

u

L

u

L

z iy

i

jy

j6 4 6 2 0


we calculate the required value of the rotation angle of the section at the node i:

EJ

L

u

L

u

L

z iy

i

jy

j6 4 6 2 0

i

iy jy

j

u

L

u

L

3

2

3

2

1

2


After putting this result into and taking into consideration matrix K’e we obtain:

K u f' ' 'e e e

,3336122

1

2

3

2

3612

22

j

jyiyzj

jy

j

jyiyiyziy

L

u

L

u

L

EJ

L

u

L

u

L

u

L

u

L

EJF

j

jyiyzj

jy

j

jyiyiyzjy

L

u

L

u

L

EJ

L

u

L

u

L

u

L

u

L

EJF 333612

2

1

2

3

2

3612

22

j

jyiyzj

jy

j

jyiyiyzj

L

u

L

u

L

EJ

L

u

L

u

L

u

L

u

L

EJM 33346

2

1

2

3

2

326


The new stiffness matrix of an element with the joint at the node i:

Superscripts (3,i) indicate that the third degree of freedom is eliminated at the first node.

Example 2 - moveable connection.

Additional equilibrium condition of a section at the node j: Fjx = 0 leads to condition:

Fix = 0

it does not change the relations for the remaining nodal forces.



The stiffness matrix of such an element takes the following form:

Superscripts (1,j) indicate that the first degree of freedom is eliminated at the last node.


The above process is called the static reduction of a stiffness matrix. Now we will give the matrix notation of an operation leading to a reduced stiffness matrix. For the sake of simplicity, we assume that the last degree of freedom of an element is the eliminated degree of freedom.


Nodal forces, nodal displacements vectors and the stiffness matrix are divided into blocks:

where according to the symmetry of the matrix we have:

K K11 11 T

K K01 10 T


is the matrix 1x1 and thus it is a scalar, the blocks and are also scalars. The results of the multiplication of previous matrix blocks are:

From f0 equation we calculate:

and after inserting into f1 equation we obtain:

or

K’’ - the condensed element stiffness matrix.

00K

0101111 uKuKf 0001010 0 uKuKf

u K K u0 001

01 1

f K u K K K u1 11 1 10 001

01 1 f K u1 1 ' '


Vector of an element load still remains to be determined. We obtain it by composing both the load vector of an element with rigid connections with nodes and the vector of the load caused by displacements of nodes free from constraints:

f f f

f

f

f

f

f

f

o uo

o

u

u

1

0

1

0

1

0


Since then:

and hence

because other displacements contained in are equal to zero.

f f f0 0 0 0 o u

f f K u K u0 0 01 1 00 0u o o o

u K f0 00

1

0o o

1u


Finally, we obtain

In this way, we can eliminate any degree of freedom but it requires some more complex transformations. We leave this problem to be solved by the reader.

f f K K f1 1 10 00

1

0 o o o o

Boundary conditions of plane frame structures

Supports for plane frames include articulated and fixed supports all listed in presentation 2. The latter ones prevent the rotation of a support node. Symbolic notation of these supports and the boundary conditions describing them are shown in next slides.

For non-typical supports, we propose to consider the use of suitable boundary elements instead of these supports.


Y

X

Mr

RrY

b) rigid-movable support (a displacement in the direction of the global axis)X

r

u rY = 0r = 0

RrX

Y

Y

X

X

Mr

Mr

RrY

RrX

a) rigid support

c) rigid movable support (a displacement in the direction of the global axis)Y

r

r

u rX = 0

u rY = 0

u rX = 0

r = 0

r = 0


Y

X

Mr

Rr Y


r

u r Y

= 0

r = 0

Rr X

Y

Y

X

X

Mr

Mr

Rr Y

Rr X

a) rigid support


r

r

u r X

= 0

u r Y

= 0

u r X

= 0

r = 0

r = 0


Y

X

Mr

Rr Y


r

u r Y

= 0

r = 0

Rr X

Y

Y

X

X

Mr

Mr

Rr Y

Rr X

a) rigid support


r

r

u r X

= 0

u r Y

= 0

u r X

= 0

r = 0

r = 0


Considering boundary conditions requires the modification of a global stiffness matrix of a structure and it is done identically as for a plane truss, thus, we will not describe the way of modifying this matrix here. A whole range of other supports such as moveable skew supports and elastic supports considered analogously to supports of trusses described for plane truss is also possible.

Boundary elements of 2D frames

Introducing a boundary element is a convenient way to avoid problems connected with the consideration of different, non-typical boundary conditions. It allows, in fact, us to model fixed and fixed-movable supports with approximate exactness and to substitute elastic supports.


Now we will present a single elastic support inclined at some angle. The scheme of this element and notations used are shown here:


Stiffness of springs: hrx and hry are forces which should be applied to their ends in order to induce unitary extensions. Rotation stiffness of a support gr is a moment necessary to induce the rotation of the node r equal to one radian.


The stiffness matrix of such an element in the local coordinate system has the form:

Kb

rx

ry

r

h

h

g

0 0

0 0

0 0


Its transformation to the global system is done analogously to the case of normal frame or truss elements except that it concerns one node only . The rotation matrix is

given by . Hence we can write the

equation transforming the matrix to the global system:

K R K Rb

rb

r ' T

d y

dx

q x

EJ

y

z

4

4

( )

K R K Re e e e '

T


After taking into consideration and

we obtain:

d y

dx

q x

EJ

y

z

4

4

( )

Kb

rx

ry

r

h

h

g

0 0

0 0

0 0

K

b

rx ry rx ry

rx ry rx ry

r

c h s h sc h h

sc h h c h s h

g

2 2

2 2

0

0

0 0

sins cosc


If we model flexible supports we ought to assume high stiffness of a suitable spring. In most cases, stiffness of the order of 11030 assures similarity between results obtained with this method and the results obtained with the exact method.

Internal forces due to a static load

The variety of loads which can act on a frame structure is greater than it was in the case of a truss. Frame elements can be affected by concentrated (forces, moments), distributed (pressure, moment loads) and temperature loads. The formulation of equilibrium equations requires substitution of internode loads for an equivalent set of concentrated forces and moments acting on nodes.


Equation define

displacements of an element bending in the direction of the y axis of the global system. After adding the equations describing the displacements in an axial direction, we obtain relations defining the displacements vector for any point between nodes

y x Cx

Cx

C x C( ) 1

3

2

2

3 46 2

u Nux

u x

u x

x

x

ye


N - the rectangular matrix of shape functions. It contains two blocks:

Ni(x) - matrix of the shape functions for the first node,

Nj(x) - matrix of the shape functions for the last node.

u Nux

u x

u x

x

x

ye

N N Nx x xi j


We can obtain both matrices from equations

and :

y x Cx

Cx

C x C( ) 1

3

2

2

3 46 2L u ujx ix

53

53

1

10

0

00

L

LxiN

64

64

2

10

0

00

L

LxjN


The convenient non-dimensional coordinate

is introduced here. Non-dimensional displacement functions (i = 1,2 ... 6)) and their derivatives , are surveyed on the following slides.

53

53

1

10

0

00

L

LxiN

64

64

2

10

0

00

L

LxjN

i

i

i

Lx /

0 0.2 0.4 0.6 0.8 1

0.5

1

1 ( )


ω1 = 1 – ξ

Shape function of frame element with parallel displacement in node i.

ω2 = ξ

Shape function of frame element with parallel displacement in node j.

0 0.2 0.4 0.6 0.8 1

0.5

1

2 ( )


ω3 = 1 - 3ξ2 + 2ξ3

Shape function of frame element with perpendicular displacement in node i.

0 0.2 0.4 0.6 0.8 1

0.5

1

3 ( )


ω4 = ξ2(3 - 2ξ)

Shape function of frame element with perpendicular displacement in node j.

0 0.2 0.4 0.6 0.8 1

0.5

1

4 ( )


ω5 = ξ(1 - 2ξ + ξ3)

Shape function of frame element with rotation in node i.

0 0.2 0.4 0.6 0.8 1

0.1

0.2

5 ( )


1

ω6 = - ξ2(1 - ξ)

Shape function of frame element with rotation in node j.

0 0.2 0.4 0.6 0.8 1

0.2

0.16 ( )


1

ω’1 = - 1

The rotation function of frame element with parallel displacement in node i.

0 0.2 0.4 0.6 0.8 1

1

0.5' 1 ( )


ω’2 = 1

The rotation function of frame element with parallel displacement in node j.

0 0.2 0.4 0.6 0.8 1

0.5

1

' 2 ( )


ω’3 = - 6ξ(1 - ξ)

The rotation function of frame element with perpendicular displacement in node i.

0 0.2 0.4 0.6 0.8 1

2

1' 3 ( )


ω’4 = 6ξ(1 - ξ)

The rotation function of frame element with perpendicular displacement in node j.

0 0.2 0.4 0.6 0.8 1

1

2

' 4 ( )


ω’5 = 1 - 4ξ + 3ξ2

The rotation function of frame element with rotation in node i.

0 0.2 0.4 0.6 0.8 1

1

1

' 5 ( )


ω’6 = - ξ(2 - 3ξ)

The rotation function of frame element with rotation in node j.

0 0.2 0.4 0.6 0.8 1

1

1

' 6 ( )


ω’’1 = 0

The bending moment function of frame element with parallel displacement in node i.

0 0.2 0.4 0.6 0.8 1

1

1

'' 1 ( )


ω’’2 = 0

The bending moment function of frame element with parallel displacement in node j.

0 0.2 0.4 0.6 0.8 1

1

1

'' 2 ( )


ω’’3 = - 6 + 12ξ

The bending moment function of frame element with perpendicular displacement in node i.

0 0.2 0.4 0.6 0.8 1

6

6

'' 3 ( )


ω’’4 = 6 - 12ξ

The bending moment function of frame element with perpendicular displacement in node j.

0 0.2 0.4 0.6 0.8 1

6

6

'' 4 ( )


ω’’5 = - 4 + 6ξ

The bending moment function of frame element with rotation in node i.

0 0.2 0.4 0.6 0.8 1

4

1

2

'' 5 ( )


ω’’6 = - 2 + 6ξ

The bending moment function of frame element with rotation in node j.

0 0.2 0.4 0.6 0.8 12

1

4

'' 6 ( )



Let us consider now the bar (an element) of a plane frame loaded with static loads


We will find nodal forces by making use of conditions of element equilibrium. We will use the principle of virtual work here:

- the work of nodal forces,

- the work of external forces (static loads).

Lne e f u'

T

ef

L q x u x q x u x m x x dxz y y x x o

L

0


Concentrated forces and moments can also be analysed by describing them in the following way:

where δ(xo) is Dirac’s delta.

Pxxxq o )()(

ooo MxxM )(


The Dirac’s delta δ(xo) is defined as:

,0)( oxx

,)( oxx

,0)( oxx

while x<x0;

while x=x0;

while x>x0;


The element equilibrium is maintain when

Ln+Lz = 0, which means:

where q(x) is the vector of external loads:

L

ee dxxx0

TTuquf

xm

xq

xq

x

o

y

x

q


Putting the expression describing the element displacements vector

into

we obtain relations:

u Nux

u x

u x

x

x

ye

L

ee dxxx0

TTuquf

f u q Nu'e e e

L

dxT T

0

f Nq'eL

dxT

0


These relations enables us to replace loads acting on elements by loads acting on nodes. It should be noted here that there are forces acting on the nodes in the equilibrium equations and that these forces act against those acting on the element thus, they should be subtracted from the nodal forces vector of the structure.


We check the effectiveness of equation

for three simple examples when:

• the load with a concentrated force is applied to the centre of an element,

• the load with a concentrated moment,

• the distributed load which is constant for the whole element.

f Nq'eL

dxT

0


Example No 1.

The frame element loaded with a concentrated force.


We introduce a non-dimensional coordinate to make the calculations more convenient and write the concentrated force as follows:

and after putting it into Eqn. we obtain:

q

0

05

0

P .

f Nq'eL

dxT

0


1

0

6

4

5

3

1

0

T

T

2

5.0

5.0

0

5.0

5.0

0

0

5.0

0

d

L

LPdPL

j

ie

N

Nf

8/

5.0

0

8/

5.0

0

5.0

5.0

0

5.0

5.0

0

6

4

5

3

L

LP

L

LP

which means that

0ixF PFiy2

1 PLM i

8

1

0jxF PFjy2

1 PLM j

8

1


Example No 2.

The frame element loaded with a concentrated moment.


We write the concentrated moment applied to the centre of an element by using Dirac’s delta:


f Nq'eL

dxT

0

q

0

0

0 5M

L

o.


which means that

d

L

L

L

Md

L

M o

j

ioe

1

0

6

4

5

3

1

0

T

T

5.0

/5.0

0

5.0

/5.0

0

5.0

0

0

N

Nf

4

12

304

12

30

L

L

M o

0ixF

0jxF

oiy ML

F2

3

ojy ML

F2

3

oi MM4

1

oj MM4

1


Example No 3.

The frame element loaded with a uniformly distributed load.


The continuous load uniformly distributed on the whole length of an element gives a load vector:


q

qo

0

1

0

f Nq'eL

dxT

0


which means that

f '

/

/

/

/

eo oq L

L

L

d q LL

L

0

0

0

1 2

12

0

1 2

12

3

5

4

6

0

1

0ixF

0jxF

LqF oiy2

1

LqF ojy2

1

LqM oi12

1

LqM oj12

1

Forces caused by a temperature load

The action of a temperature on frame elements can cause flexion. This happens when the temperature field is not homogeneous in the cross section. In the case of a truss, the flexion of bars did not cause increasing nodal forces because truss elements are connected by means of jointed nodes.


Bars of frame structures can make a node rotate, hence we have to determine forces at the node in the element undergoing the action of the non-uniform temperature field.

Let us consider an element of which the upper fibres are affected by an increase in a temperature Δtg, and the lower fibres are affected by an increase in a temperature Δtd.


The temperature distribution in the element cross section.


The temperature field can be written as follows:

- the increase in the

temperature of the middle fibres,

- the difference of temperatures between extreme fibres,

h - is the height of the cross section,

t x y t xy

ht xo h,

dggdo ytyth

t 1

dgh ttt


yd - the distance between the centre of gravity and the lower fibres,

yg - is the distance between the centre of gravity and the upper fibres.

t x y t xy

ht xo h,


Strains of the element fibres induced by the temperature field are equal to

where αt is the expansion coefficient of the material.

t t t o hy t y t ty

h


If bars cannot deform freely, then stresses rise inside them:

which the internal forces result from:

x t t o hE E t ty

h

N dA E t dAt

hydAx

A

t o

A

h

A


Since the second integral occurring in previous equation is the static moment with regard to the z axis which crosses the centre of gravity, this moment has to be equal to zero. Thus, we obtain

like in the case of a truss element.

N x t x EAt t o


The second internal force caused by temperature stresses is the bending moment:

The first integral has to be equal to zero similarly

to and the second one

is the moment of inertia of the cross section calculated with regard to the middle axis.

M x x ydA E t x ydAt

hy dAt x t o

A

h

AA

2

N dA E t dAt

hydAx

A

t o

A

h

A


Thus, we can write an equation describing the bending moment due to temperature stresses as

where is the moment of inertia of

the element section with regard to the z axis crossing the centre of gravity of the section.

M xt x

hEJt

t hz

A

z dAyJ 2


We calculate forces at nodes making use of the principle of virtual work just as we did before in this presentation:

where

u f te et

t

L

x dxT T

' 0

t t

t

t

N x

M x

0


tt is the vector of the internal forces induced by a temperature. The zero value of the expression in the second row of the vector comes from the fact that the temperature does not cause shearing forces in the elements, ε(x) is the vector of displacements gradients:

ey

x

dx

d

dy

dudx

du

x uBε

)(


B is the matrix of derivatives of shape functions:

.

On the basis of

we calculate:

ex uBε )(

B B B i j

53

53

1

10

0

00

L

LxiN

64

64

2

10

0

00

L

LxjN


, , (i = 1,2 ... 6)) are nondimensional functions given before in this presentation.

Bi x

L

L

L L

10 0

01

01 1

1

3 5

2 3 5

'

' '

' ' ' '

B j x

L

L

L L

10 0

01

01 1

2

4 6

2 4 6

'

' '

' ' ' '

i

i

i


On the basis of we calculate

components of the nodal forces vector:

After inserting matrices Bi(x) and Bj(x) into

equation , we obtain

u f te et

t

L

x dxT T

' 0

f B t'ett

L

dx T

0

f B t'ett

L

dx T

0


where and are non-dimensional coordinates at both the beginning and end of the action interval of the temperature load (following figure).

f '

'

' '

' '

'

' '

' '

ett

o

zh

zh

o

zh

zh

E

A t d

J

hLt d

J

ht d

A t d

J

hLt d

J

ht d

1

3

5

2

4

6

1

2

1

2

1

2

1

2

1

2

1

2

1 2


The temperature loaded frame element.


In the case when the temperature load is constant and occurs along the whole length of the element, we obtain the following equation from equations of Ni(x) and Nj(x):

f 'ett

o

z h

o

z h

E

A t

J t

hA t

J t

h

0

0


,

and

These equations describe internal forces acting on the element. Forming the load vector we should subtract components of the vector from suitable components of the global vector.

53

53

1

10

0

00

L

LxiN

64

64

2

10

0

00

L

LxjN

L q x u x q x u x m x x dxz y y x x o

L

0

Lne e f u'

T

Documents

Finite Element Method 2D frame systems - Lublinakropolis.pol.lublin.pl/users/jpkmb/FEM_4.pdfThe element of a 2D frame is more general than a truss element presented in Chapter 2 because