“exact” Elements of beams (right and curved) · Code_Aster Version default Titre : Éléments "exacts" de poutres (droites et courbes) Date : 02/04/2013 Page : 1/67 Responsable

Code_Aster Version default

Titre : Éléments "exacts" de poutres (droites et courbes) Date : 02/04/2013 Page : 1/67Responsable : Jean-Luc FLÉJOU Clé : R3.08.01 Révision : 10779

“Exact” elements of beams (right and curved)

Summarized :

This document presents the beam elements of Code_Aster based on an exact resolution of the equations of the continuous model carried out for each element of the mesh.

The beams can be right (Elements POU_D_T and POU_D_E) or curves (Elements POU_C_T). The section, constant or variable over the length, can be of an unspecified form. The material is homogeneous, isotropic, elastic linear.

The assumptions selected are the following ones:• Assumption of Eulerian: the transverse shears are neglected, as well as the inertia of rotation. This

assumption is checked for strong slenderness (element POU_D_E).• Assumption of Timoshenko: the transverse shears and all the terms of inertia are taken into account. This

assumption is to be used for weak slenderness (elements POU_D_T and POU_C_T).• Assumption of Saint-Coming: torsion is free.

The processing of the various loadings and the quantities expected as a result (forced - forces) is also presented.

Warning : The translation process used on this website is a "Machine Translation". It may be imprecise and inaccurate in whole or in part and is provided as a convenience.

Licensed under the terms of the GNU FDL (http://www.gnu.org/copyleft/fdl.html)



Contents1 Notations4 .....................................................................................................................................................

2 Introduction5 .................................................................................................................................................

3 equations of the mouvement6 ......................................................................................................................

3.1 tension-compression6

3.1.1 Balance equation local6 ...........................................................................................................

3.1.2 Method of Lagrangien7 .............................................................................................................

3.2 pure torsion (torsion of Saint-Coming) ............................................................................................ 7

3.2.1 Balance equation local7 ...........................................................................................................

3.2.1.1 Beam of section circulaire8 ............................................................................................

3.2.1.2 Beam of section quelconque8 ........................................................................................

3.2.2 Method of Lagrangien9 .............................................................................................................

3.3 bending simple9 .................................................................................................................................

3.3.1 Balance equation local10 .........................................................................................................

3.3.2 Method of Lagrangien12 ...........................................................................................................

4 Beam element droite14 ................................................................................................................................

4.1 longitudinal Motion of tension - compression14 .................................................................................

4.1.1 Determination of the matrix of rigidité14 ..................................................................................

4.1.2 Determination of the second membre15 ..................................................................................

4.1.3 Computation of the forces to the nodes of the poutre15 ..........................................................

4.1.4 Determination of the matrix of masse16 ..................................................................................

4.2 Motion of free torsion around the axis longitudinal17 .........................................................................

4.3 Motion of flexion18 .............................................................................................................................

4.3.1 Bending in the plane (xOz) .................................................................................................. 18

4.3.2 Bending in the plane (xOy) .................................................................................................. 20

4.3.3 Determination of the coherent mass matrix with the matrix of rigidité20 .................................

4.3.3.1 Bending in the plane (xOz) ........................................................................................ 20

4.3.3.2 Motion of bending around axis (OZ) .......................................................................... 22

4.4 Mass matrix reduced by the technique of the masses concentrées23 ...............................................

4.5 Stiffness matrix centrifuge25 ..............................................................................................................

5 Straight beams particulières28 .....................................................................................................................

5.1 Eccentricity of the axis of torsion compared to the axis neutre28 ......................................................

5.2 Sections variables29 ..........................................................................................................................

5.2.1 Computation of the matrix of rigidité30 ....................................................................................

5.2.1.1 Determination of the equivalent section () ................................................................ 30

5.2.1.2 Determination of a constant of equivalent torsion (Ceq) ........................................... 32

5.2.1.3 Determination of the geometrical moments équivalents34 ............................................

5.2.2 Computation of the matrix of masse36 ....................................................................................

5.2.2.1 By the method masses équivalentes36 ..........................................................................





5.2.2.2 By the method of the masses concentrated (diagonal matrix) ................................. 37

6 geometrical Stiffness - Structure précontrainte39 ........................................................................................

7 Beam courbe44 ............................................................................................................................................

7.1 Flexibility matrix for bending in the plane of the beam [C1] .......................................................... 49

7.2 Flexibility matrix for bending out of plane of the beam [C2] .......................................................... 50

8 Chargements53 ............................................................................................................................................

8.1 Loading by déformation53 ..................................................................................................................

8.1.1 For the straight beam of Eulerian and the straight beam of Timoshenko53 ............................

8.1.2 For the curved beam of Timoshenko54 ...................................................................................

8.2 Loading due to the pesanteur54 .........................................................................................................

8.3 Loadings répartis56 ............................................................................................................................

8.3.1 Straight beam with section constante56 ...................................................................................

8.3.2 Straight beams with section variable57 ....................................................................................

8.3.3 Beam courbe57 ........................................................................................................................

8.4 Loading thermique57 ..........................................................................................................................

8.5 Loading électrique58 ..........................................................................................................................

8.5.1 secondary Driver finished right or infini58 ................................................................................

8.5.2 secondary Driver described by part of mesh ASTER59 ...........................................................

9 Torsor of the forces - Torsor of the stresses (or generalized forces) - nodal Forces and réactions60 .........

9.1 the torsor of the efforts60 ...................................................................................................................

9.2 the tensor of the contraintes61 ...........................................................................................................

9.3 Computation of the nodal forces and réactions63 ..............................................................................

10 Element of barre64 .....................................................................................................................................

11 Bibliographie65 ...........................................................................................................................................

12 Description of the versions of the document65 ..........................................................................................

ForewordThis documentation of reference of the beam elements was carried out from a work completed per m.t. Bourdeix, P. Hemon, O. Wilk of the Institute Aerotechnics of the National Academy of Arts and Trades, in the frame of an External Contract of Research and Development with this laboratory.

The volume of this document is due at the same time to the required accuracy and the didactic character of the talk, which is voluntarily preserved.





1 Notationsthe notations used here are not all identical to those used in [U4.42.01], for reasons of compactness and homogeneity with [R3.08.03].One gives the correspondence between this notation and that of the documentation of use.

DX , DY ,DZ and DRX , DRY , DRZ are the names of the degrees of freedom associated with

the components with displacement u , v ,w , x ,y ,z .

C constant of torsion JX

ey , ez eccentricity of the center of torsion/shears EY ,EZ

E modulus Young E

Poisson's ratio NU

G modulates of Coulomb = E

21

I y , I z geometrical moments of bending compared to the axes y , z IY , IZ

I p geometrical moment polar

I x

polar main moment of inertia around the longitudinal axis x

k y , kz shear coefficients1

AY,

1AZ

K stiffness matrix

M mass matrix

M x ,M y , M z moments around the axes x , y , z MT ,MFY , MFZ

N normal force with the section N

S area of the section A

u , v ,w translations on the axes x , y , z DX , DY , DZ

V y ,V z shears along the axes y , z VY , VZ

density RHO

cT transverse shearing stress

x , y , z rotations around the axes x , y , z DRX ,DRY , DRZ





2 Introductiona beam is a solid generated by a surface of area S whose geometrical center of inertia G follows a

curve C called the average fiber or neutral fiber. The area S is the cross-section (cross section) or profile, and it is supposed that if it is evolutionary, its evolutions (size, form) continuous and progressive when G are described the line average one.

For the study of the beams in general, one makes the following assumptions:• the cross-section of the beam is indeformable,• transverse displacement is uniform on the cross-section.

These assumptions make it possible to express displacements of an unspecified point of the section, according to displacements of the point corresponding located on the line average one, and according to an increase in displacement due to the rotation of the section around the transverse axes. The latter can be neglected (POU_D_E) or be the object of a modelization (POU_D_T and POU_C_T).

The discretization in “exact” elements of beam is carried out on a linear element with two nodes and six degrees of freedom by nodes. These degrees of freedom are the three translations u , v ,w and

the three rotations x , y , z .

Appear 2-abeam element and degrees of freedom.

Waited until the strains are local, it is built in each top of the mesh a local base depending on the element on which one works. The continuity of the fields of displacements is ensured by a basic change, bringing back the data in the global database.

In the case of the straight beams, one traditionally places the line average one on the axis x of the

local base, transverse displacements being thus carried out in the plane y , z .

Finally when we arrange quantities related to the degrees of freedom of an element in a vector or an elementary matrix (thus of dimension 12 or 122 ), one arranges initially the variables for top 1 then those of top 2. For each node, one stores initially the quantities related to the three translations, then those related to three rotations. For example, a vector displacement will be structured in the following way:

u1 , v1 ,w1 , x1

, y1,z 1

sommet 1

u2 , v2 , w2 ,x2, y2

, z2

sommet 2





3 The equations of motionIn this chapter one presents the equations of the motion of the beams in tension - compression, torsion and bending in the elastic domain. In each case, these equations are deduced by application from the Lagrange's equations, resulting from the principle of Hamilton, or by writing the local equilibrium of a segment of beam. We chose to point out the two methods, the reader will be able to refer to that which is most familiar for him. One limits oneself here to the cases where the only loadings are distributed loadings (not concentrated forces).

3.1 The traction and compressionthe tension - compression is the translatory movement on the longitudinal axis of the beam.

3.1.1 Balance equation room

One considers a segment length dx subjected to an axial load N [Figure .1.11-a] intern and an

external force f ext per unit of length.

Appear 3.1.1-a : Segment of beam charged axially.

The beam has a section S x and consists of a material of density x and Young modulus E x . The basic principle of the mechanics makes it possible to write:

−N(x )+N(x+dx)+ ∫x

x+dx

f ext(s)ds=∫x

x+dx

ρ(s)S (s)∂2 u (s)∂ t 2 ds

where U is displacement on the axis x of the segment

Thus:

N(x+dx)−N(x )

dx+

1dx ∫x

x+dx

fext (s)ds=1dx ∫x

x+dx

ρS∂2 u∂ t2 ds

While passing in extreme cases when dx 0 , one obtains:

d N x

dxf ext

x =S

∂2 u∂ t 2 x [3.1.1-1]

only the first order terms are preserved and one replaces in [éq3.1.1-1], then one uses the Hooke's law and the assumption that the beam consists of longitudinal fibers working only in traction and compression to express the axial load by:

N x =ES∂u∂ x

[3.1.1-2]





One obtains thus after simplification by dx :

∂

∂ x ES∂u∂ x f ext=S

∂2 u∂ t 2

[3.1.1-3]

which represents the local equilibrium with the first order of a beam, for a motion of tension - compression.

3.1.2 Method of Lagrangian

Taking again the segment of beam of the figure3.1.1-a total kinetic energy of the beam length l is written:

E c=12∫0

lS ∂u

∂ t 2

dx

One will note for the continuation E c e=

12 S ∂u

∂ t 2

elementary kinetic energy.

The internal energy of strain, thanks to the Hooke's law is written:

E p int=

12∫0

lES ∂u

∂ x 2

dx

One will note in the same way E pinte

=12

ES ∂u∂ x

2

.

One also has the job of the external force given by:

E pext=∫o

lf ext u dx

and at the elementary level E p exte

=f ext .u

the Lagrangian one is given by:L=E c−E pint

−E pext

and Lagrangian density:L=Ec e

−E pinte

−E p exte

For the monodimensional continuous system, the Lagrange's equation is written in this case:

∂L∂u−∂

∂ x ∂L∂u ' − ∂

∂ t ∂ L∂ u =0 [3.1.2-1]

where u' and u respectively indicate derivative compared to x and compared to time. Its application brings back for us obviously to the equation of the motion of a beam in tension - compression [éq3.1.1-3].

3.2 Pure torsion (torsion of Saint-Coming)torsion is the rotation movement around the longitudinal axis of the beam. It is supposed here that the center of gravity is confused with the center of rotation (of torsion) [R3.03.03], and the warping of the section is neglected. The case of the eccentring of the center of torsion compared to the center of gravity is treated with [§2727].






One considers a segment length dx put in rotation due to one moment M x [Figure3.2.1-a] intern

and an external couple x per unit of length.

Appear 3.2.1-a : Segment of beam in rotation around Ox

the segment is turned of an angle x compared to the not deformed position. We have as follows:

3.2.1.1 Circular beam of section

−M x x M x xdx ∫x

xdxx s ds=∫x

xdxI qx

∂2 x

∂ t 2ds

with I q x=∫s

r 2 ds is the main moment of inertia plane of the section S around the rotational axis

0, x .

As for the tension, one obtains after division by dx and transition in extreme cases:

d M x

dxx=I

x

∂2x

∂ t 2

The constitutive law is introduced:

M x=G I p

∂x

∂ x

where G is the modulus of Coulomb (or shear modulus) and I p the polar geometrical moment

compared to the center of gravity of the section. (One has besides: I x= I p for a material of

density homogeneous).We obtain the statement then:

∂

∂ x G I p

∂x

∂ x + x=Ix

∂2 x

∂ t2 [3.2.1.1-1]

which represents the local equilibrium with the first order of a segment of beam for a motion of torsion.

3.2.1.2 Unspecified beam of section

to take account of warping while remaining on the free assumption of torsion, in the case of the noncircular sections one is led to replace the moment I p by a constant of torsion C (lower than I p )

in the equation of torsion ([R3.03.03] for the computation of C ).





By definition M x=G C∂ x

∂ x. One obtains then:

∂

∂ x G Cx

∂ x +x=C∂

2 x

∂ t2[3.2.1.2-1]

When the center of gravity of the section is not the center of rotation, this statement is not valid and the motions of torsion and of bending are coupled.


We have same way as with [the §77 kinetic energy (for example for a circular beam of section):

Ec=∫0

l 12

Ix ∂x

∂ t 2

dx ,

internal potential energy

E p int=∫0

l 12

GI p ∂x

∂ t 2

dx ,

and the work of the external couple

E p ext=∫0

l x x dx

By applying the Lagrange's equation [éq3.1.2-1] to the variable x , one leads naturally

to [éq3.2.1.1-1] giving the motion of a beam in pure torsion.

3.3 The pure bendingbending is the rotation and translatory movement around an axis perpendicular to the longitudinal axis of the beam. One speaks here about pure bending (around Oy or Oz ). One limits oneself to the case of the straight beams. The curved beams are treated with [§4445].

One describes the equation of bending in the plane O , x , z , the extension to the plane O , x , y is immediate [Figure 3.3-a ].

Appear 3.3-a : Bending of a beam in the plane O , x , z .

The translation along the axis O , z is noted w and rotation around O , y is noted y .






One considers a segment length dx subjected to the shears V z , the bending moment M y , an

external force t zext distributed uniformly per unit of length, and an external couple m yext

distributed

uniformly per unit of length [Figure3.3.1-a].

Appear 3.3.1-a : Segment of beam in bending in the plane O , x , z

the local equilibrium of the forces and the moments (on the section of X-coordinate xdx ) gives for the forces:

−V z x V z xdx ∫x

xdxt z ext

ds=∫x

xdxS ds

∂2 w

∂ t 2

and for the moment:

−M y x M y xdx ∫x

xdxm yext

ds−∫x

xdxV z x ds=∫x

xdx I y ds

∂2y

∂ t2

One neglects the terms in dx² . While passing in extreme cases when dx tends towards 0, one obtains:

∂V z

∂ xt zext

=S ∂2 w∂ t2

∂M y

∂ x−V zm yext

= I y

∂2 y

∂ t 2

It is noted that the force uniformly distributed t zext produces a term which is of the second order in the

equilibrium of the moments and is thus neglected. One introduces then the behavior models of strength of materials.

M y=EI y

∂y

∂ x and V z=k z SG ∂w

∂ xy [3.3.1-1]

the statement [éq3.3.1-1] of V z is due to Timoshenko [feeding-bottle4] where k z is the shear

coefficient in the direction z . It characterizes beam of Timoshenko the model; it will be seen thereafter that the model beam of Eulerian corresponds to a simplification of the model of Timoshenko. I z is the geometrical moment of the section compared to the axis O , y .





Consequently, one leads to the two equations coupled in w and y for bending in the plane

O , x , z .

∂

∂ x [ k z SG ∂w∂ x y ]t z ext

= S∂

2 w∂ t 2 [3.3.1-2]

∂

∂ x [ EI y

∂ y

∂ x ]−k z SG ∂w∂ x

q ym y ext= I y

∂2 y

∂ t 2 [3.3.1-3]

When the beam is uniform, i.e. the section and the material are constant on the longitudinal axis, the equations [éq3.3.1-2] and [éq3.3.1-3] are reduced to only one equation in w . For that, one only once derives compared to X-coordinate X the balance equation from the moments [éq3.3.1-3].

EI y

∂3 y

∂ x3 −k z SG ∂2 w∂ x2

∂ y

∂ x = I y

∂3 y

∂ x∂ t2

It will be noted that this handling eliminates the presence from the term resulting from an external couple uniformly distributed. Then, the equation [éq3.3.1-2] can be put in the form:

∂ y

∂ x=∂

2 w∂ x2−

1k z SG

t z ext

k zG∂

2 w∂ t 2

EI y∂

4 w∂ x4 S

∂2 w∂ t 2 − I y[ 1 E

k z G ] ∂4 w

∂ x2∂ t 2

2 I y

k zG∂

4 w∂ t 4 −t zext

=0 [3.3.1-4]

There remains useful for this kind of equation to point out the physical meaning of the various terms, so during simplifications to be aware of the neglected effects.

EIy

∂4 w

∂ x4 balance the charging density in the direction of the translation due to the bending moment.

S∂

2 w

∂ t2 is the term of inertia of translation.

Iy

∂4 w

∂ x2∂ t2

represent the inertia of rotation of bending.

Iy

E

k z G

∂4 w

∂ x2∂ t2 is an additional term of the inertia of rotation due to the taking into account of the

transverse shears (assumption of Timoshenko).

2 Iy

kzG

∂4 w

∂ t4 result from the coupling between the inertia of rotation and the inertia of translation

coming from the shears.

The model of beam of Timoshenko (POU_D_T or POU_C_T), takes into account the set of these terms, in particular those which are relating to the shears. One can thus model beams of weak slenderness.





The model of beam of Eulerian (POU_D_E) is a simplification since the strains in shears are neglected as well as the inertia of rotation (what is justified because it intervenes in the dynamic studies only for the high modes). These assumptions are justified in the case of a beam of sufficiently large slenderness. So for the model of Eulerian, the equation of the motion of bending, in the general case of the beams with variable section is written:

∂

∂ x2 EI y∂

2 w∂ x2 S ∂

2 w∂ t 2

−t z ext=0 [3.3.1-5]

In addition, it is indeed the shears which cause the rotation of the cross-sections compared to the neutral axis. To neglect this effect thus amounts writing that V z=0 what brings to [éq3.3.1-1].

y=−∂w∂ x

[3.3.1-6]

which is the translation of the assumption of Eulerian.

Concerning bending in the plane O , x , y , the same approach leads to [éq3.3.1-7] for the beam of Timoshenko with:

{∂

∂ x [k ySG ∂ v

∂ x−

z ]ty

ext

=S∂2 v

∂ t2

∂

∂ x [EIz

∂z

∂ x ]−kySG ∂v

∂ x−

zmz ext

= Iz

∂2

z

∂ t2

[3.3.1-7]

and when the section is constant:

EI z∂

4 v∂ t 4− S

∂2 v∂ t 2− I z 1

Ek y G ∂

4 v∂ x2

∂ t 2

2 I z

k y G∂

4 v∂ t 4t yext

=0 [3.3.1-8]

the use of the assumption of Eulerian in the plane O , x , y z=∂ v

∂ x makes it possible to lead

to the equation of the motion of bending for a beam of Eulerian according to [éq3.3.1-9]:

∂

∂ x2 [ EI z∂

2 v∂ x2 ]−S ∂

2 v∂ t 2

t yext=0 [3.3.1-9]


kinetic energy is expressed by:

E c=∫0

l 12 I y ∂y

∂ t 2

dx∫0

l 12S ∂w

∂ t 2

dx

according to displacements in rotation and translation.





Internal potential energy is worth:

E p int=∫0

l 12

EI ∂ y

∂ x 2

dx∫0

l 12 [∬s

cT ∂w∂ x y dS ] dx

where cT is transverse shear stress and the term

∂w∂ x

y the shear strain. The model of beam of

Eulerian this term while the model of Timoshenko neglects puts forth a hypothesis on the distribution

of the stresses cT in the section, compatible with the statement [éq3.3.1-1]. In the general case of

the model of Timoshenko, internal potential energy is written:

E p int=∫0

l 12

EI ∂ y

∂ x 2

dx∫0

l 12

k z SG ∂w∂ x y

2

dx

The potential of the external loads is expressed as for him by:

E p ext=−∫0

lt zext

dx−∫0

lm y ext

y dx

The use of the Lagrange's equation [éq3.1.2-1] applied once to the variable w then with the variable

y brings back for us to the two equations [éq3.3.1-2] and [éq3.3.1-3] describing motion in bending of

a segment of beam.





4 Beam element rightOne describes in this chapter obtaining the elementary matrixes of stiffness and mass for the beam element right, according to the model of Eulerian ( POU_D_E ) or Timoshenko ( POU_D_T ). The stiffness matrixes are calculated with option “RIGI_MECA”, and the mass matrixes with option “MASS_MECA” for the coherent matrix, and option “MASS_MECA_DIAG” for the diagonalized mass matrix.

4.1 Longitudinal motion of traction and compressiona difficulty to write the variational formulation comes owing to the fact that there can be in structures made up of beams of the concentrated loadings (assimilable to the Dirac). The balance equation [éq3.1.1-1] must be replaced by:

d Ndx

x f ext x ∑i=1

N

f ici x =0

One omitted by simplicity the inertia forces which would undergo the same processing as the external

forces f ext .

i represent the function of Dirac located at the point i , are f ic

to them the concentrated forces

applied to the beam.

For the application of the finite element method, the balance equation must be written in the form of the principle of the virtual wors which is in this case:

∫G

N d vdx

dx=∫G

f ext⋅vdx∑i=1

N

f ici v [4.1-1]

Any confusion being excluded, i indicates the measurement of Dirac associated with the point i ,

y is a field of kinematically admissible longitudinal displacement unspecified.

In practice, it is supposed that there is no concentrated force inside the beam elements, but only with the nodes ends.

4.1.1 Determination of the stiffness matrix

It corresponds to the statement of the virtual wor of the internal forces according to a given displacement. I.e.:

∫0

LN

d vdx

dx for an element length L .

The elastic behavior model is introduced:

N x =ESd udx

While choosing for function-test:

v x =1 x =1−xL

and v x =2 x =xL

corresponding respectively to the degrees of freedom u1 and u2 of the two nodes of the element,

one obtains directly:

∫0

LN

d 1

dxdx=∫0

L−

ESL

dudx

dx=−ESL

[ u L −u 0 ]





and

∫0

LN

d 2

dxdx=∫0

L ESL

dudx

dx=ESL

[u L −u 0 ]

the stiffness matrix of the element is thus:

K=ESL [ 1 −1−1 1 ]

Note:In the statement of the virtual wor of the internal forces, u intervenes only for u 0 and

u Lu was not discretized inside the element. This is why the element is described as “exact”: one

obtains the exact solution with the nodes, but only with the nodes.

4.1.2 Determination of the second member

the second member is the statement of the virtual wor of the forces applied. The second associated member with the distributed loading and the functions tests previously introduced is:

[ f 1

f 2 ] with f 1=∫0

1f ext x 1− x

L dx and f 2=∫0

1f ext x

xL

dx

Note:In AFFE_CHAR_MECA_F , one can introduce f ext like an unspecified function of x . On the level

of the computation of f 1 and f 2 , on the other hand, integration is made by supposing that

f ext varies linearly between the values taken with the nodes ends. If one must model a radial force distributed nonlinear, it is then necessary to discretize more finely.But let us insist on the fact that whatever the form of f ext x (polynomial or different), if one

can calculate the integrals exactly f 1 and f 2 , the solution of the static problem will be exact with the nodes of the problem.

The virtual wor of the concentrated forces (given by assumption to the nodes of the elements) does not intervene directly on the level of the element.

One introduces these concentrated forces in the form of nodal forces, directly in the assembled vector of the second member.

4.1.3 Computation of the forces to the nodes of the beam

the complete Principle of the Virtual wors is written indeed on the assembled system. In addition, by

writing the formula of integration per part on all the structure (beam [ x0 , x1 ] ):

∫

N v, x

dx = [N x1 v x1−N xo v xo ]∑i=1

N

[∣N∣]iiv

−∑j=1

M

∫

j

N, x

v dx

[4.1.3-1]

j representing all the intervals without discontinuity of normal force, therefore without concentrated

force, and the [∣N∣]i jumps of N between these intervals.





Indeed, by bringing this statement closer to the Principle of the Virtual wors, one finds, for each concentrated loading (by choosing suitable functions v test):

i=1, N [∣N∣]i= f ic

Each finite element of beam is by assumption an interval without discontinuity. There can thus be discontinuity of the internal forces N from one element to another if there exists a concentrated force on the node connecting the two elements.

The internal forces for an element are determined in the following way:The balance equation inside an element is:

N , x f rep=0

The formula of integration by parts [éq4.1.3-1] on the element gives:

∫0

LN x v , x dx=[ N L v L −N 0 v 0 ]∫0

Lf ext x v x dx

While considering N L and N 0 like data, one could have obtained this formula directly Principle of the Virtual wors [éq4.1-1]. By still taking the functions test:

v x =1 x =1−xL

and v x =2 x =xL

One obtains:

−ESL

[u L −u 0 ]=−N 0 f 1

ESL

[ u L −u 0 ]=N L f 2

soit [−N 0 N L ]=[ K ] [ u 0

u L ]−[ f 1

f 2]

I.e. the internal forces are obtained by cutting off with the product K⋅U the nodal forces equivalent

to the distributed loads f ext .It is also observed that they are of opposite sign. So that the sign is the same one from one element to another, it is thus necessary to change the sign of N 0 calculated by this method. It is what is made by the computation of option EFGE_ELNO.

4.1.4 Determination of the mass matrix

the mass matrix to be coherent with the stiffness matrix is given starting from the same functions test. However, it is not possible any more to calculate exactly the associated nodal forces without making assumption on the form of the solution. The computation from the mass matrix will involve an error of discretization.

A dynamic computation will thus require a discretization of structure of beam in small elements, which is not the case for a static computation. It goes without saying that in the case of a dynamic computation, the computation of the forces which one will lead as with [§1515] by cutting off the nodal forces from inertia is also approximate. The solution u is selected in the space generated by the functions tests (i.e. polynomials of degree to most equal to 1):

u=u 0 1 x u L 2 x





The mass matrix appears in the statement of the virtual wor due to the inertia forces:

W=VTM U , U= u1

u2

Work is also written:

W=∫0

LV x m u x ,t dx

with m=∫sdS=S in the case of a homogeneous material.

While taking u x ,t =1 x u1 t 2 x u2 t , one a:

W=∫0

LVTS 1 x

2 x 1 x 2 x U dx

W = VT S∫0

L 1 x 2 x 1 x 2 x dx U

the mass matrix is thus written:

M= S ∫0

L1

2 dx ∫0

L12 dx

∫0

L12 dx ∫0

L2

2 dx

and done calculations:

M= SL

6 2 11 2

corresponding respectively to the degrees of freedom u1 and u2 of the two nodes of the element.

4.2 Free motion of torsion around the longitudinal axisthe problem is similar to that of the tension compression. For a beam , charged by concentrated

twisting moments x x distributed and moments ic

, the principle of the virtual wors is written:

∫

M xddx

dx=∫ x⋅ dx∑

i=1

N

ic⋅i ,∀

The constitutive law is:

M x x =G⋅C⋅d x

dx

Except for the variables, this equation has the same form as that of the motion of tension - compression. By means of the same reasoning, one obtains the same statements for the mass matrixes and of stiffness elementary is:

K=G C

L 1 −1−1 1 et M=C L

6 2 11 2

corresponding respectively to the degrees of freedom x1 and x2 of the two nodes of the element,

The computation of the mass matrix like having previously required to discretize the field solution.





The second member, due to the couple x distributed, is obtained in the same way that for the

motion of tension - compression:

∫0

L1 x x dx

∫0

L2 x x dx

1 x = 1− xL

2 x = xL

4.3 Motion of bendingWe place ourselves here in the frame of a straight beam at constant section of Timoshenko type. We take account of the effects of transverse shears. The beam of Eulerian - Bernoulli will be then treated by simplification of the equations of Timoshenko.

The description of bending is more complex than preceding motions, but a wise choice of functions tests will enable us to get of the same results forms.

4.3.1 Bending in the plane (xOz)

With obvious notations and while not being interested initially, as in the preceding cases, with the forces of inertia, the principle of the virtual wors is written for the motion of bending in the plane x0z :

∫

V z ' M y '=∫ t zext

m yext ∑

i=1

N

t ici mi

ci , [4.3.1-1]

for very , kinematically admissible.

The stiffness matrix results from the statement of the virtual wor of the internal forces which one will clarify by means of the behavior model then while integrating by parts:

∫

V z 'M y ' =∫

k z SG w 'y 'E I yy' '

=k z SG [w L ' L L −w 0 ' 0 0 ]−∫

k z SG w '' ' ∫

k z SG y '

EI y [y L ' L − y 0 ' 0 ]−∫ EI y y ''

The functions tests which one will choose “will check the balance equations without second member”, i.e. [éq3.3.1-2] and [éq3.3.1-3]:

{ '' '=0EI y ''−k z SG =0

[4.3.1-2]

Under these conditions, the nodal forces, statement of the work of the internal forces in these virtual displacements given are expressed exactly, without assumption on the form of the solution, according to displacements at the end of beam as in the preceding cases:

∫

Vz ' M

y ' = k

zSG [wL ' L L −w 0 ' 00 ]

EIy [y

L ' L −y0 ' 0 ]

[4.3.1-3]

Note::It is clear that the condition [éq 4.3.1-2] led to functions tests depending explicitly on the geometrical and material characteristics of the beam, but that does not have laugh at awkward.





The couples of functions selected tests are: , =i,

i4 i= [1, 4 ]

where, while having noted y=12 EI y

k z SGL2 , the functions i are defined by:

Displacements:

1 x =

1

1y[2 xL

3

−3xL 2

−y

x

L1 y]

2 x =

L

1y[−xL

3

4

y

2 x

L 2

−2

y

2 xL ]

3 x =

1

1y

[−2 xL 3

3xL 2

y x

L ]

4 x =

L

1y[−xL

3

2−

y

2 x

L 2

y

2 xL ]

[]

Rotations:

5x =6

L 1 yxL [ 1−

xL ]

6 x =1

1 y

[ 3 xL

2

−4 y xL 1 y ]

7 x =−6

L 1 yxL [ 1−

xL ]

8x =1

1 y

[ 3 xL

2

−2 y xL ]

[4.3.1-5]

corresponding each one respectively to the degrees of freedom w1 y1 w2 , and y2

of the two

nodes of the element,

One checks without difficulty that the couples i , i4 check well [éq4.3.1-2]. Moreover:

1, 5 0 =1 , 0 1 , 5 L =0 , 0

2, 6 0 = 0, 1 2 , 6 L = 0 , 0

3, 7 0 =0 , 0 3 , x7 L = 1 , 0

4, 8 0 =0 , 0 4 , 8 L = 0 , 1

[4.3.1-6]

the stiffness matrix results easily from [éq4.3.1-3] (by ordering the columns according to

w 0 , y 0 , w L , y L .





K=12 EI y

L3 1y 1 −

L2

−1 −L2

4 y L2

12L2

2− y L2

12

Sym 1 L2

4 y L2

12

It is clear that the computation of the forces acts in the same way that to [§1515].

4.3.2 Bending in the plane (xOy)

the stiffness matrix for a motion of bending in the plane xOy is in the same way obtained that in the preceding case. The functions tests which lead to an exact statement of the nodal forces must this time check (equation similar to [éq4.3.1-2]):

{ '' '=0EI z ''−k y SG '=0

[4.3.2-1]

the couples of functions selected tests are:

1 , −5 ; −2 , 6 ; 3 , −7 ; −4 , 8

corresponding each one respectively to the degrees of freedom v1 z1 v2 , and z2 of the two

nodes of the element,

Them i being given by [éq4.3.1-5] while having replaced y par. z=12EI z

k y SGL 2 the stiffness

matrix obtained is:

K=12 EI

z

L3 1z 1

L

2−1

L

2

4 z L2

12−

L

2

2−z L2

12

1 −L

2

sym 4z L

2

12

4.3.3 Determination of the coherent mass matrix with the stiffness matrix

Option “MASS_MECA” of operator CALC_MATR_ELEM.

4.3.3.1 Bending in the plane (xOz)

Let us consider the motion of bending in the plane xO z , the work of the inertia forces is written:

W=∫o

L

wm wy I zy dx with m=∫S

dS and I z=∫S

y 2dS





In the case of a homogeneous material, we have: m=S and I z= I z

w x , t y x , t are discretized on the basis of function tests introduced for the computation of the

stiffness matrix, that is to say:

w x, t =1x w

1t

2x

y1

t + 3 x w

2t

4 x

y2

t

yx,t =

5 x w

1 t

6 x

y1

t 7x w

2 t

8 x

y2

t

in other words: w=w t xw

with w~=

w1

y1

w2

y2

and w = 1

2

3

4

y=w

~

t

ywith y

=5

6

7

8

By integrating these notations in the statement of the work of the inertia forces, one a:

W=∫0

L

mw~

twwt w

~

I z

w~

t

y

y

t w~

dx

One from of deduced the form of the mass matrix:

M= mij mij=∫o

L S x i x j x I z x i+ 4 x j+ 4 x dx

for i 1 to 4 and j from 1 to 4. That is to say:

M = S

1 y 2 13 L357 L y

10L y

2

3−11 L2

210−11 L2 y

120−

L2 y2

249 L703 L y

10L y

2

613L2

4203 L2 y

40

L2 y2

24

L3

105

L3 y

60

L3 y2

120−13 L2

420−3 L2 y

40−

L2 y2

24−L3

140−

L3 y

60−L3 y

2

120

13L35

7 L y

10

L y2

3−11 L2

21011 L2 y

120L2 y

2

24

L3

105L3y

60

L3 y2

120

I z

1 y 2 65 L

−110 y

2−65L

−110y

2

2 L15

Ly

6L y

2

3110− y

2−L30−Ly

6

L y2

6

65 L

110− y

2

2 L15

L y

6L y

2

3

It is necessary well to note, as in [§1616], that in the dynamic case, one is not ensured to have an exact solution the nodes, as it is the case in static.





4.3.3.2 Motion of bending around axis (OZ)

In the same way, for the motion of bending around the axis O z , in the plane xO y , the work of the inertia forces is written:

∫O

L

vmv

z

I y

z dx with I y

=∫S z 2 dS= I y

This time v x , t and z x , t are discretized in accordance with [§2020] by:

v x, t =1 x v1 t −2 x z1

t + 3 x v2 t −4 x z2t

z x, t =−5 x v1 t 6 x z1t −7 x v2 t 8 x z 2

t

We obtain the following mass matrix then:

M =S

1 z 2 13 L357 Lz

10Lz

2

311 L2

21011 L2z

120

L2z2

249 L703 L z

10

Lz2

6−13 L2

420−3 L2z

40−

L2z2

24

L3

105L3 z

60L3 z

2

12013L2

4203 L2z

40

L2 z2

24−L3

140−

L3z

60−

L3z2

120

13 L35

7 L z

10

Lz2

3−11 L2

210−11 L2z

120−L2z

2

24

L3

105

L3 z

60

L3 z2

120

I y

1z 2 65 L

110−z

2−65 L

110−z

2

2 L15

L z

6

Lz2

3−110 z

2−L30−Lz

6Lz

2

6

65 L

−110 z

2

2 L15

L z

6L z

2

3

In the model of beam of Eulerian-Bernoulli, the effects of transverse shears are neglected. It is thus enough, to obtain the mass matrixes and of stiffness associated with this model, to cancel the

variables y and z contained in the mass matrixes and of stiffness of the model of Timoshenko. (

y and z the coefficients of form utilize k y and k z , opposite of the shear coefficients).

It will be noted that in the model Eulerian-Bernoulli programmed in Aster, the inertia of rotation is also

neglected. It is thus necessary, for this model, to cancel the terms in "" I z and "" I y in the mass

matrix of the model of Timoshenko.

4.4 Mass matrix reduced by the technique of the lumped massesthe mass matrix is thus reduced to a diagonal matrix and is obtained by option “MASS_MÉGA_DIAG” of operator CALC_MATR_ELEM.

The element beam is considered with constant section S and constant density .The technique known as of “Lumping” consists in adding on the diagonal all the terms of line of the coherent matrix and to cancel all the extra-diagonal terms.





With regard to the diagonal component related to the motion of tension - compression M 11 and that

related to the motion of torsion M 44 , we have:

M 11= S L2

M 44= I y+I z L2

with I y , I z : geometrical moments

One can consider that these components were obtained by dividing the beam element in two equal

shares length L2

then by associating the mass and inertia obtained with the node of the half -

element. For, M 44 the preceding statement corresponds to a choice: one could also have written:

M 44=CL2

.

Note: Comparison with the integration methods numerical.One can note that if one carries out a in the following way approached integration:

∫ef=∑

i=1,n

mese n

f aie

( aie : nodes i of the element e n : many nodes of the element)

one obtain result identical (for one beam: mes e=L and n=2 ).

The diagonal components related to motions of bending which are programmed are:

M 22= S L2

,

M 33= S L2

,

M 55=Min S L3

105 , S

L2

48 I y2L15

,

M 66=Min S L3

105 , S

L2

48 I z2L15

One finds well the components M 22 and M 33 related to the translations of motions of bending by the technique of the lumped masses to the nodes. On the other hand, the origin of the formulas used for the components M 55 and M 66 dependant on rotations, is unknown. One can simply notice that one finds the values:

S L3

105 I

z

2 L

15

S L3

105 I

y

2 L

15

for the diagonal components of the equivalent mass matrix [§2020]. But this matrix is not diagonal. Nevertheless, the results got by this method remain correct.





4.5 Stiffness matrix centrifugesthe centrifugal stiffness matrix results from a loading rotation characterized by the Flight Path Vector

= x

y

z t

. u= uxu

yu

z t

The vector displacement is posed.

Its statement is Ku=−

12∫V

∧u⋅∧u dV

One can write: Ku=−

12∫V

ut

t⋅u dV with =

0 −z y

z 0 −x

−y x 0 While breaking up displacements according to the shape functions N i , i=1,2 ,i

y ,iz , i=1,4

defined previously, it comes: u=Nq where q all the degrees of freedom of the 2 nodes of the

beam indicate. On an element Ve

the elementary matrix Ke

is written:

Ke=−

12∫

V e N t

t⋅N dV

e

To have a coherent centrifugal stiffness matrix with the stiffness matrix and the coherent mass matrix, U is discretized on the basis of function tests introduced for the computation of the stiffness matrix, that is to say:

uxx, t =N

1x u

1 t + N

2x u

2 t

uyx, t =

1z x v

1 t −

2z x

z1

t + 3z x v

2t −

4z x

z2

t

uz x, t =1y x w1 t 2

y x y1

t + 3y x w2 t 4

y x y2

t

One from of deduced the matrix easily N :

N=N1 0 0 0 0 0 N2 0 0 0 0 0

0 1z 0 0 0 −2

z 0 3z 0 0 0 −4

z

0 0 1y 0 2

y 0 0 0 3y 0 4

y 0

The matrixes being hollow, one can easily clarify the product N t

t⋅N

It is a matrix [12,12] symmetric constituted by the following blocks:

k e=k11 symk12 k22

k11=y

2z

2N1

2sym sym sym sym sym

−xyN11z x

2z2 1

z 2 sym sym sym sym

−xzN11y −yz1

y1z x

2y2 1

y 2 sym sym sym

0 0 0 x2N1

2sym sym

−xzN12y −yz1

z2y x

2y2 1

y 2y 0 x

2y2 2

y2 sym

xyN12z

−x2z

21

z2z

yz1y2z 0 yz 2

y2z

x2z

22

z2





k22=y

2z

2N2

2sym sym sym sym sym

−xyN23z x

2z2 3

z 2 sym sym sym sym

−xzN23y −yz3

y3z x

2y2 3

y 2 sym sym sym

0 0 0 x2N2

2sym sym

−xzN24y −yz3

z4y x

2y2 3

y 4y 0 x

2y2 4

y2 sym

xyN24z

−x2z

23

z4z

yz3y4z 0 yz 4

y4z

x2z

24

z2

k12= y

2z

2N 1N 2 −xy N 21

z−xz N 21

y 0 −xz N 22y

xz N 22z

−x yN 13z

x2z

21

z3

z−yz1

y3

z 0 −yz2y3

z−x

2z

22

z3

z

−xz N 13y

− yz3y1

zx

2 y

21

y3

y 0 x2 y

22

y3

yyz3

y2z

0 0 0 x2 N 1N 2 0 0

−xz N 14y

− yz1z4

yx

2 y

21

y4

y 0 x2 y

22

y4

yyz2

z4

y

x yN 14z

−x2z

21

z4

z yz1

y4

z 0 yz2y4

zx

2z

22

z4

z

It remains L to integrate this matrix on an element beam length.

the coefficients y etétant z defined into 4.3.1 and 4.3.2, for k=y , z , one a:

∫0

LN11

kdx=L60 9 k

1k ∫0

LN21

kdx=L60 21− k

1k

∫0

LN12

kdx=−L2

120 4 k

1k ∫0

LN22

kdx=−L2

120 6− k

1k

∫0

LN13

kdx=L60 20 1

1k ∫0

LN23

kdx=L60 10− 1

1k

∫0

LN14çkdx=

L2

120 5 11k ∫0

LN24

kdx=L2

120 5− 11k

One notices that the integrals ∫0

Li

k2dx , k=y , z result from the intégralesen ∫0

L iyizdx making

y=z .

It thus remains to calculate the 16 integrals ∫0

Liy jzdx , who are not a priori symmetric in there and

Z (in fact, some are it).

∫0

L

1y1zdx=

L1y 1z

1335 720

yzyz

3

∫0

L

2y2zdx=

L2

1y 1z 1105

1

120yz

yz

120





∫0

L3y3zdx=∫0

L1y1zdx ∫0

L4y4zdx=∫0

L2y2zdx

∫0

L

1y2zdx=

−L2

1y 1z 11210

10y12z

240yz

24

∫0

L

2y1zdx=

−L2

1y 1z 11210

12y10z

240yz

24

∫0

L

1y3zdx=

L1y 1z

970 320

yzyz

6

∫0

L3y1zdx=∫0

L1y3zdx ∫0

L1y4zdx=−∫0

L2y3zdx ∫0

L4y1zdx=−∫0

L3y2zdx

∫0

L

2y3zdx=

−L2

1y 1z 13420

5y4z

120yz

24

∫0

L

3y2zdx=

−L2

1y 1z 13420

4y5z

120yz

24

∫0

L

2y4zdx=

−L3

1y 1z 1140

yzyz

120

∫0

L4y2zdx=∫0

L2y4zdx ∫0

L3y4zdx=−∫0

L1y2zdx ∫0

L4y3zdx=−∫0

L2y1zdx





5 Particular straight beamsIt is a question in this chapter of taking into account straight beams whose section has properties which were ignored until now, in particular the beams having a excentré center of torsion compared to the neutral axis (the section does not have 2 axes of symmetry), and those whose section evolves continuously on their axis.

5.1 Eccentring of the axis of torsion compared to the neutral axisthe center of torsion is the point which remains fixed when the section is subjected to the only twisting moment. It is also called shear center because a force applied in this point does not produce rotation

x

Appears 5.1-a : Section with center of torsion.

At the point C , the effects of bending and torsion are uncoupled, one can thus use the results established in the preceding chapter. One finds the components of displacement at the point O by considering the relation of rigid body:

u O =u C OC∧

with =

x

00 vector rotation and OC=

0e

y

ez the position of the center of torsion in the local

coordinate system of the beam. In fact, one obtains:

{u

x=u

xc

uy=u

yc

ez

x

uz=uzc

−eyx

[5.1-1]





the change of variables given by [éq 5.1-1 ] is written matriciellement:

uxc1

uyc1

uz c1

xc1

yc1

zc1

uxc2

uyc2

uz c2

xc2

yc2

z c2

=1 0 0 0 0 00 1 0 −ez 0 00 0 1 ey 0 0 01 0 0 1 0 01 0 0 0 1 01 0 0 0 0 1

1 0 0 0 0 00 1 0 −ez 0 0

0 0 0 1 ey 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

ux

1

u y1

uz1

x1

y1

z1

ux2

uy2

uz2

x2

y2

z2

It is thus enough to determine the elementary matrixes of mass Mc and stiffness Kc in the

reference C , x , y , z where motions of bending and torsion are decoupled then to be transported in the reference related to the neutral axis O , x , y , z by the following transformations:

K=PT K c P

et M=PT Mc P

The values of ey and ez are with being provided to Code_Aster via operand SECTION: “GENERALE” of operator AFFE_CARA_ELEM, default values being obviously zero values.

5.2 Variable sectionsIt is possible to take into account evolutionary sections in a continuous way for the straight beams of Timoshenko and Eulerian (POU_D_E and POU_D_T only). One distinguishes two types of variation of section:• linear or refines,• quadratic or homothetic.

The distinction between the two types is conceived easily by taking the example of a rectangular beam:• if only one of side dimensions varies, in a linear way is supposed, then the area of the cross-

section varies linearly, and is given by:

S x =S1 [1 S 2

S1

−1 x

L ]

• when two side dimensions vary (in a linear way), the area of section will evolve in a quadratic way.

S x =S1 [1 S2

S1

−1 x

L ]2





Code_Aster makes it possible to treat sections “RINGS”, “RECTANGLE” and “GENERALE”, but for obvious reasons of geometry, all these types of section cannot admit the two types of variation. The following table summarizes the existing possibilities.

Constant section Linear Quadraticcircle yes not yesright-angled yes yes

according to thereyes

general yes not yes

For section “RECTANGLE”, it is the user who chooses the type of variation, by specifying “AFFINE” or “HOMOTHETIC” in AFFE_CARA_ELEM. It should well be noted that in case “AFFINE”, dimensions cannot vary that according to Y.

We consider generally that the section varies according to the formula [éq5.2-1]:

S x =S11+cxL

m

[5.2-1]

S1 is the initial section x=0c is fixed by it by the knowledge of the final section S2 in x=L .

m give the degree of variation: m=1 linear variation, m=2 quadratic variation.

The section varying, it goes from there in the same way inertias I y x , I z x and I px .We will thus have:

I y x =I y11+cxL

m+ 2

[5.2-2]

I z x =I z11+cxL

m+ 2

[5.2-3]

I p x =I p11+cxL

m+ 2

[5.2-4]

c is given for each formula from the value for x = L : I y2, I z2

, I p 2.

The Coulomb and Young E moduli G are supposed to be constant.The principle adopted by Code_Aster consists in calculating equivalent, constant characteristics of section on the beam, starting from the real characteristics data at the two ends. These equivalent characteristics thus depend on the phenomenon to which they contribute, in particular, are distinct for the effects of stiffness or inertia.

5.2.1 Computation of the stiffness matrix

5.2.1.1 Determination of the equivalent section ( Seq )

the determination of the equivalent section uses neither the method taken with [§1414] to obtain the exact stiffness matrix nor an approximation of the solution by a polynomial function as described with [§1616]. In fact, the method employed deviates from the finite element method and even from the method of Galerkin, it consists in carrying out a resolution of the problem of the beam with variable section without forces distributed imposed, which makes it possible to clarify the forces at the ends according to displacements. This method is “coherent” with that of [§1414] because the functions tests defined in [§1414] are worth 1 or 0 on the ends of the beam, therefore [éq4.1-1] the nodal forces can be “comparable” with forces.





In addition, this method makes it possible to get exact results for the static problem without distributed force and conduit as we will see it with a value Seq understood enters S1 and S2 which, in the general case, guarantees the convergence of the solution approximated towards the exact solution (without however knowing the order of convergence).

The cross-section of the beam being variable, the equation of traction and compression in static effortlessly distributed imposed is written:

∂∂ x [E S x

∂u∂ x ]=0 [5.2.1.1-1]

with N x =E S x ∂u∂ x

We let us determine the stiffness matrix in the general case [éq5.2-1], we deduce thereafter the values from them from the equivalent sections for the cases m=1 (progression linear) and m=2 (quadratic progression).

While integrating [éq5.2.1.1-1], we have:

E S x ∂ u∂ x=C1

or, by taking account of the statement of S x :

E S 1 1+c xL

m∂u∂ x=C1

The constant of integration is given starting from the values of thrust loads to the nodes.We integrate once again in order to obtain the forces with the nodes according to displacements u 0=u1 and u L =u2 :

∂u∂ x=

C1

E S1 1+c x

L −m

from where u x =C1

E S 1

Lc

ln1+c xL +C 2 if m=1

and u x=

C1

E S1

Lc

1

l−m 1cxL

m−1C2

if m≥2

It is noted that the statement of u x is far from being polynomial.By taking account owing to the fact that:

C1=−N1 for x=0C1=N2 x=L

and that:

C2=1+c

−m1 u 0 −u L

1+c −m1

−1

We obtain:

for m=1 , {N 1=E S 1 c

L ln 1c u1−u2

N 2=E S 1 c

L ln 1c u2−u1





and for m=2 , {N 1=E S 1 c 1−m

L [−u1 1c

m−1

1−1c m−1 u2 1c

m−1

1−1c m−1 ]

N 2=E S 1 c 1−m

L [u1 1c

m−1

1−1c m−1 −u2 1c

m−1

1−1c m−1 ]

While replacing c by its value, that is to say:

• if m=1 ,

c=S2

S1

−1 ,

N 1=EL

S2−S1 ln S2−ln S1

u1−u2

N 2=EL

S2−S1 ln S2−ln S1

u2−u1

• if m=2 ,

c= S2

S1

−1

N1=EL S1 S2 u1−u2

N 2=EL

S1 S2 u2−u1

We note that the stiffness matrixes, in the two treated cases, will have the same form as for a constant section if one takes as equivalent section:

Seq=S2−S1

ln S2−ln S1

for a section varying linearly

Seq=S1 S2 for a section varying in a quadratic way

5.2.1.2 Determination of a constant of equivalent torsion (Ceq)

the equation of pure torsion of a beam with variable section, is written:

∂

∂ x G C x ∂ qx

∂ x =0 [5.2.1.2-1]

with C x =C11c xL

m2

( m=1 or 2 )

the method is the same one as for the computation of the equivalent section: it is a question of

integrating the preceding equation in order to obtain the forces (torques x1, x2

) according to

displacements with the nodes x1,x2

and to deduce some, by comparison with the formulas with

constant section, statement the one polar geometrical moment are equivalent.





By integration of [éq5.2.1.2-1], we have:

G C11cxL

m2∂ x

∂ x=D1 ,

D1 the constant of integration is determined by the torques applied to the nodes.Of: ∂ x

∂ x=

D1

G C 11c

xL

−m2

,

We deduce:x x =

D1

G C1

−Lc m+1 1c

xL

− m1

D2

We determine D2 starting from the system:

{x1

=−D

1L

GC1 c m1 D2

x2

=−D1 L

GC1c m1

1c −m1 D2

that is to say: D2= x1

1c − m1c x2

1c − m1−1

By taking account owing to the fact that:

D1=−1 for x=0D1=2 x=L

we let us have finally:for m=1 ,

{1=

G

L

2 C2 C1

23−C1C2

23

C2

23−C

1

23

x1

−x2

2=

G

L

2 C2C1

23−C1C2

23

C2

23−C1

23

x2

−x1

We thus take in the case of a section varying linearly, a polar geometrical moment are equivalent C eq of the following form:

Ceq=

2 C 2C1

23−C1C2

23

C2

23−C1

23

linear variation





for m=2 ,

{1=

G

L

3C2 C1

34−C1C2

34

C2

34−C

1

34

x1

−x2

2=G

L

3C2C

1

34−C

1C

2

34

C2

34−C

1

34

x2

−x1

In the case of a section varying in a quadratic way, the polar geometrical moment is written:

Ceq=

3C 2 C1

34−C1C2

34

C 2

34−C1

34

quadratic variation

5.2.1.3 Determination of the equivalent geometrical moments

In fact, it does not seem possible to find, as we did for the section or the polar geometrical moment, of

equivalent geometrical moments I y eqet I zeq which would come to replace the geometrical moments

I y et I z in the statement of the terms of the stiffness matrix.

We expose here the method suggested by J.R. BANERJEE and F.W. WILLIAMS [feeding-bottle 3] which clarifies the stiffness matrix in the case of a motion of bending of a beam Eulerian - Bernoulli with variable section (linear or quadratic).

The results correspond to those programmed in Code_Aster for a beam of the type Eulerian - Bernoulli with variable section (linear or quadratic). By extension, the same approach is applied there for the beams of the Timoshenko type. The results are not here detailed.

Let us consider bending in the plane x o y .On the basis of the static equation of the motion of bending of a beam of the type Eulerian - Bernoulli:

∂

2

∂ x2 [EI z x ∂

2 v

∂ x 2 ]=0 , and z=∂ v∂ x

v x is expressed according to four constants of integration C1 , C2 , C3 , C4 . These constants

are determined by the values of displacements to the nodes:

v o =v1 v L =v2

z o =z1z L =z2

that is to say: v1

z1

v2

z2

=B C1

C2

C3

C4

, B matrix 4×4 .





Forces: V y x =∂

∂ x [EI z x ∂

2 v

∂ x2 ]and the moments: M z x =EI z x

∂2 v

∂ x2 ,

also express themselves according to these constants of integration, and one can write:

T y1

M z1

T y2

M z2

=DC 1

C 2

C 3

C4 , D matrix 4×4 .

The stiffness matrix corresponds to the product DB−1 . The terms of this matrix are clarified in the

following tables.

Let us recall that I z x =I z11c xL

m2

, m=1 or 2 ,

Pose W 1=EI z1

cL

, W 2=EI z 1 cL

2

, W 3=EI z1 cL

3

.

quadratic Variation afiine Variationm=1 m=2

c= I 2

I1

13−1 c= I 2

I1

14−1

1 c2 4 c²3 c3 2 c 2c c3

3 c c1 2 c c12 c3

4 2 c c12 c3 4 c²

5=c2−4 2c² c1

6=c3−5 4 c² c1

c2 ln c1−2 c c3

c1

the matrix K is written then:

K= 1

W 31 W 22 −W 31 W 23

W 14 −W 22 W 15

Sym W 31 −W 23

W 16

Now let us consider bending in the plane xO z .For the sections with quadratic variation, the approach is identical. But it differs for the sections with linear variation (according to y only).





One calculates the terms of the stiffness matrix corresponding to bending in the plane 0, x , z by the values given in the following table.

Variation closely connected: bending in the plane 0, x , z

c= I 2

I1−1

1 2ln c1

c

2 2−1

3 c1 1−2

4 1c−2

5=c2−4

6=c3−5

c2 ln c1−2 c

In the case of the beams of Timoshenko, for the shear coefficients, one applies to the reduced section k⋅S the relations used for the section, namely:

k y S eq=

k y2

S2−k

y1

S1

ln k y2

S2 −ln k y1

S1

k z S eq=

k z2

S2−k z1

S1 ln k z

2

S2 −ln k z1

S1 } if the variation is closely connected,

ky S eq=S1 S2 k y1

k y2

kz S eq=S1 S2 k z1 kz 2} if the variation is quadratic,

and one introduces the additional terms into K in the same way that for a constant section.

Computations are not here detailed. One obtains of an the same K matrix forms than previously with

for principal modification the value of :• variation closely connected:

=c2 ln c1 −2c

2

12c3 c2

• quadratic variation:

=c3

c1

2

3c3 c23c3





5.2.2 Computation of the mass matrix

5.2.2.1 By the method of the equivalent masses

Of the “average” values are calculated for the section, the reduced section, and the moments, namely:

S=S1+S2

2 if the variation is closely connected

S=1

L∫0

LS x dx=

S1+S

2S

1 S

2

3 if the variation is quadratic

I y=I y

1

I y2

2

Iz=

I z1

I z2

2

I

x

=I

y1

Iy2

Iz1

Iz 2

2

} whatever the variation

the mass matrix is then calculated like that of a beam having these characteristics.

5.2.2.2 By the method of the masses concentrated (diagonal matrix)

• If the section varies in a way closely connected, the programmed matrixes correspond, with regard to motions of traction and compression and torsion, with those of the beams prismatic, by means of the sections and equivalent inertias of torsion:◦pour the tension compression:

L 3S1S2

80

03S2S1

8

◦pour torsion:

L I y1

Iz 1

Iy2

Iz2

4 1 00 1x1

x2





◦pour motions of bending:

S1 L

2

M5,5 M6,6 0

S2 L

20

M11,11 M 12,12

v1 w1

z1y

1

v2 w2

z2y

2

with M5,5=M

11.11=min S

eqL 3

105,S

eqL2

48 L

15 I y1

Iy

1

M6,6=M

12.12=min S

eqL3

105,S

eqL 2

48 L

15 I z1

Iz

1 Seq= S1S2

2 • If the section varies in a homothetic way, the matrixes are programmed, for various motions, in

the following way:◦pour the traction and compression:

5 S1S2

12L 0

0 S15 S2

12L

u1

u2

◦pour torsion:

LI y1

Iy2

Iz1

Iz2

4 1 00 1x1

x2

◦pour bending in each of the two planes:





5 S1S2

12L

M5,5 M 6,6 0

S 15S 2

12L

M11,11 M12,12

v1 w1

z1y

1

v2 w2

z2y

2

with M 5,5=M 11 , 11=min [ S1S2 2

L3

105 , S1S2

2L3

48 ] I y1I y2

L15

and M 6,6=M 12 , 12=min [ S1S 2 2

L3

105 , S 1S2

2L3

48 ] I z1I z 2

L15





6 geometrical Stiffness - prestressed StructureOption: “RIGI_MECA_GE”In the case of a prestressed structure, therefore subjected to initial forces (known and independent of time), one cannot neglect in the balance equation the terms introduced by the change of geometry of the virgin state of stress in a prestressed state [feeding-bottle2].

Appear 6-a : Various states.

This change of geometry does not modify the balance equation, in the frame of the assumption of the small disturbances (HP) around V o (and of V * ), only by the addition of a linear term in displacements with which the matrix associated is called geometrical stiffness matrix and who expresses himself by:

W G=∫V o

∂uk3D

∂ x i

ijo ∂vk

3D

∂ x j

dV

where u3D is displacement (resp. v3D the virtual displacement kinematically admissible) taken from

V * (but compared to V o in the frame of the assumption of the small disturbances) and o

prestressing (of Cauchy if one wants) since one is in the frame of the assumption of the small disturbances.

W G being a symmetric bilinear form in u3D and v3D , it can be interpreted like the variation of a

potential UG .

W G=UG

One a:

2U G=∫V o

∂ uk3D

∂ x iij

o ∂ uk3D

∂ x j

For a model of beam 3D, the stress tensor is reduced in the local axes of the beam to the components

xx , xy et xx , from where:

2U G=∫V o

xxo ∂ui

3D

∂ x∂ ui

3D

∂ x2 xy

o ∂u i3D

∂ x∂u i

3D

∂ y2 xz

o ∂ ui3D

∂ x∂ ui

3D

∂ z

The terms ∂ux3D

∂ x 2

, ∂ux

3D

∂ x

∂ux3D

∂ y and

∂ux3D

∂ x

∂ ux3D

∂ zare neglected [feeding-bottle5]. Moreover, in the

local coordinate system of the center of torsion of beam:

{u

x

3D x , y , z =u x z

y x −y

z x

uy

3D x , y , z =v x −z

x x

uz

3D x , y , z =w x y

x x





and ∇ u=u'z

y

'−y

z

'−

z

y

v '−zx

' 0 −x

w 'yx

'

x0

from where one draws, according to the preceding assumption:

2U G=∫V o

xxo [ v '−z x

' 2w ' y x

' 2]2 xy

o w ' y x' x2 xz

o v '−z x' − x

However, the generalized forces are connected to the stresses by the statements:

N o=∫

S

xx

o Vy=∫

S

xy

o Vz=∫

S

xz

o

My=∫

S

zxx

o Mz=∫

S

−yxx

o

One from of deduced:

2 UG=∫o

LN ° v '

2w '

2 −2 My

o v ' x

'−2 M

z

o w ' x

'2V

y

o w ' x−2 V

z

o v ' x

∫V o

xx

o y2z2 x

' 22 xy

o yxz

o z x

x

'

By supposing, moreover, that xxo is constant in the discretized element (what inaccurate for example

for a vertical beam is subjected to its inertia loading) and that x linearly varies compared to x

x=1− x

L 1

x

L

2, from where x

' =1L1

1L2

, it comes:

∫V o

xxo y2

z 2 x'

2= 1 , 2 [

N o

LI yI z

S −

N o

L

I y I z

S

−N o

L

I yI z

S

N o

L

I yI z

S]1

2

By neglecting in particular the terms which had with the influence of the shears on the mode of buckling or vibration, and by supposing that the distributed loads are null on an element, one a:

N x = constant, My=M y2

−My1

x

LM

y1

V y x = constant,

V z x = constant, Mz=M z2

−Mz1

x

LM

z1

Under this assumption and for the model of Eulerian-Bernoulli (for the model of Timoshenko, one uses the same matrix), one obtains the following matrix:

A=A1 A2

0 A3





Higher triangular part of the geometrical stiffness matrix with:

A1 =

1u1

2v1

3w1

4

x1 5

y1 6

z1

1

2 1.2 No

L −

M y1o

2L−

M y2o

2LV z

o

2

N o

10

3 1.2 No

L −

M z1o

2L−

M z2o

2L−V y

o

2 −

N o

10

4 I y I z N

o

S L

−M z1oM z2

o

12−

LV yo

12

M y1o−M y2

o

12−LV z

o

12

52 L N o

15

62 L N o

15

A2 =

7u2

8v 2

9w2

10

x2 11

y2 12

z2

1

2 −1.2 No

L

M y1o

2 L

M y2o

2 L

V z

o

2

N o

10

3 −1.2 No

L

M z1o

2 L

M z2o

2 L

−V y

o

2

−N o

10

4

M y1o

2 LM y2

o

2 L

−V z

o

2

M z1o

2 L

M z2o

2 L

V y

o

2

− I yI z N

o

S L

M z1o−M z2

o 12

−LV y

o

12

−M y1oM y2

o 12

−LV z

o

12

5 −N o

10

M z1o−M z2

o 12

LV y

o

12

−L N o

30

6 −N o

10

−M y1oM y2

o 12

LV y

o

12

−L N o

30





A3 =

the 7u2

8v 2

9w2

10

x2 11

y2 12

z2

7

8 1.2 No

L

−M y1

o

2 L−M y2

o

2 L

−V z

o

2

−N o

10

9 1.2 No

L

−M z1

o

2L−M z2

o

2 L

V y

o

2

N o

10

10 I y I z N

o

S L

−M z1oM z2

o 12

−LV y

o

12

M y1o−M y2

o 12

−LV z

o

12

112 L N o

15

122 L N o

15

By means of equalities ∂M y

∂ x−V z=0 and

∂M z

∂ x−V y=0 , one finds the programmed matrix.

Moreover, to be able to deal with the problems of discharge of thin beams, requested primarily by bending moments and normal force, it is necessary to add the assumption of rotations moderated in torsion [feeding-bottle4], [feeding-bottle5].

This results in the following shape of the field of displacements:

U x , y , z =u x z y x x x z x − y z x −x x y x

In addition, if the center of torsion C is not confused with the center of gravity, it is necessary to write:

{V x , y , z =v x ,C − z−zC x

W x , y , z =w x , C y− yC x

These two modifications bring additional terms in the geometrical stiffness matrix:The assumption of moderate rotations results in adding in 2UG the term:

2UG1=∫0

L−M z

o x y ,xM yo x z ,xV y

o x yV z

o x y

The terms of the matrix A to be added are:





4−5 : M z1

o

2 10−11 : −

M z2o

2

4−6 : −M y1

o

2 10−12 :

M y2o

2

With regard to the eccentricity of the center of torsion, it is necessary to add the terms corresponding to:

U G

2=N o z c∫0

Lv ' x

'−N o yc∫0

Lw ' x

'− yc V y

ozc V z

o ∫0

Lx x

'

yc M zo−zc M y

o ∫0

L

x'

2

Moreover, it is necessary to carry out a change of reference as with [§2727].





7 Large displacements the processing of large displacements is available for the straight beams in Code_Aster . It is activated by the key word DEFORMATION=' GROT_GDEP'. This option makes it possible to take into account the reactualization of the geometry with each iteration. With each iteration the geometrical stiffness matrix is brought up to date, chapter 6, and is added to the tangent matrix. That allows a better convergence of computations as that is explained in [R3.08.09].





8 Curved beamTo compute: the stiffness matrix for a curved beam element, we calculate via various stages.

We leave the balance equations which integrated will give us a matrix (noted J

) making it possible

to determine the forces in a point of the beam knowing the forces in another point. This matrix will take into account the local basic change.Then by writing the potential energy of the element and by noticing the decoupling of bending in the plane of the element of bending out of this plane, one determines the two flexibility matrixes.Finally the flexibility matrixes being calculated, one obtains the local stiffness matrix by means of the Castigliano's principle, which must be recomputed in the global database to be assembled.

Appear 8-a : Local coordinate system of the curved beam.

Cf: Instruction manual of Code_Aster (booklet [U4.2]: modelization index C p26/30).To attach the forces applied in a point P of structure to the forces obtained in another point Q of structure, one integrates the balance equations static of a curved beam (effortlessly distributed).

We here will limit we to study the curved beam with constant section (with taking into account of the transverse shears) and with constant radius of curvature.





Appear 8-b : Locate average (local coordinate system)

the balance equations static are:

{N , s−V 1=0

V 1, sNR=0

V 2, s=0

{M T , s−

M 1

R=0

M 1, sM T

R−V 2=0

M 2, sV 1=0

this for:

Appear 8-c : Curvilinear reference.

To integrate, one uses the conditions in P :

{N=F y

V1=− F x

V 2=F z

{M T=M y

M 1=− M x

M 2=M z





While integrating and while passing in the system of following axis:

Appear 8-d : System of axis chosen by integration.

One obtains:

[N

V1

V2

MT

M 1

M 2

]=cos −sin 0 0 0 0sin cos 0 0 0 0

0 0 1 0 0 0

0 0 R cos−1 cos −sin 0

0 0 R sin sin cos 0

R cos−1 −Rsin 0 0 0 1

J

[F x

F y

F z

M x

My

Mz

]

We now will take into account the mechanical characteristics by means of potential energy:

E p=1

2∫s1

s2

N2

E S

V 12

k 1S G

V 22

k 2 S GMT

2

E IM1

2

E I 1

M 2

2

E I2ds

(the sign "" ~ means that we use the internal forces) (note: f =− f by the principle of action-reaction).

Recall of the behavior models in the frame of the model of Timoshenko:

{N=ES

∂ u

∂ s

V 1=k1 SG ∂w1

∂ s−2

V 2=k2 SG ∂w2

∂ s1 {M

T=EI

∂

∂ s

M1=−EI 1

∂1

∂ s

M2=−EI

2

∂2

∂ s





Torsor of the internal forces: Kinematical torsor:

{T } ∣N s TM T s M

{C} ∣s

1n

1

2n

2

u sw1n

1w

2n

2

From where:

E p=1

2∫s

1

s2 N ,V 1, V 2 [1

ES0 0

0 1

k 1 SG0

0 01

k2 SG] N

V 1

V 2ds

12∫s1

s2 M T , M1, M 2 [

1

EI0 0

01

EI1

0

0 0 1

EI 2

]M T

M 1

M 2ds

or:

E p=12∫0

[F x

F y

F z

M x

M y

M z

]P

T

JT [

1ES

1k1 SG

1k2 SG

0

1GC

01

EI 1

1EI 2

]J [F x

F y

F z

M x

M y

M z

]P

Rd

J being the matrix obtained previously.





One can thus calculate the flexibility matrix [C ] :

[C ]=∫0

JT [

1ES

1k1 SG

1k2 SG

0

1GC

01

EI 1

1EI 2

]J Rd

We can notice that the matrix J can break up into two pennies independent matrixes, a part concerning bending in the plane of the element, the other concerning bending out of plane of the element.

Note: This decomposition will make it possible to reverse the matrixes more easily than we will obtain a little further.

J J1 , J2

Bending in the plane of the element:

[N

V1

M 2]=

cos −sin 0sin cos 0

Rcos−1 −Rsin 1J

1

[F

x

Fy

M z]

Bending out of plane of the element:

[M T

M 1

V2]=cos −sin Rcos−1

sin cos Rsin 0 0 1

J2

[M

x

My

F z]





8.1 Flexibility matrix for bending in the plane of the beam [C1]

[C1 ]=∫0

[ cos sin R cos−1 −sin cos −R sin

0 0 1 ] [1

ES0 0

01

k 1 SG0

0 01

EI 2

] J1R d

=∫0

[x11 x12

R cos −1 EI 2

x22 −R sin

EI 2

sym.1

EI 2

] R d

with:

x11=cos2ES

sin2k1 SG

R2 cos −1 2

EI 2

x12=−cos sin ES

cos sin k 1SG

−R2 cos −1 sin

EI 2

x22=sin2ES

cos2k1 SG

R2 sin2

EI 2Appendix:

∫o

cos2 d∫o

sin2 d=

∫o

cos2 d=

1

2sin cos = 1

42−sin 2

∫o

sin2d =

1

2sin cos = 1

42−sin2

∫o

sincos d =sin2

2

from where:

C111=

R4ES

2 sin 2 R

4k1 SG2−sin2

R3

4 EI 2

6 −8sinsin 2 ,

C221=

R4 ES

2sin 2 R4 k1 SG

2 −sin 2 R3

4 EI 2

2−sin 2 ,





C331=RE I 2

C121 =−

R2 E S

sin2R3

2 k1S Gsin 2−

R3

2 E I 2sin2−

R3

2E I 2[2 cos−2 ]

C131=

R2

E I 2sin−

C231=

R2

E I 2cos −1

8.2 Flexibility matrix for bending out of plan of the beam [C2]

[C2 ]=∫0

[cos sin 0−sin cos 0

R cos −1 Rsin 1] [1EI

0 0

01

EI 1

0

0 01

k 2 SG] J 2 R d

=∫0

[cos2

EI

sin2

EI 1

−cos sin

EI

cos sinEI 1

R cos −1 cosEI

R sin2

EI 1

sin2

EI

cos2

EI 1

−R cos−1 sin

EI

R cossinEI 1

sym.R2 cos−1

2

EI

R2sin2

EI 1

1

k 2 SG] R d

C112=−

R4 EI

2sin 2 R4 EI 1

2sin 2

C222=

R4 EI

2−sin 2 R4 EI 1

2sin 2

C332=−

R3

2 EI[6 −8sin sin 2 ] R3

4 EI 1 2sin 2 Rb

k 2 SG

C122=−

R2 EI

sin2

R2 EI 1

sin2

C132=

R2

4 EI 2sin 2 −4 sin R2

4 EI 1 2−sin 2

C232 =−

R2

2 EI sin22 cos−2 R2

2 EI 1sin2

having determined the flexibility matrix, we will be able to calculate the stiffness matrix.One thus has:

E p=12

TPeT

[C ]TPe

, this ∀TPe

( e : for outside)





Like one a: TQe=J

TP

e (Recall: TP

e et TQe

are not described in the same base).

One also has:

E p=12

TQeT

J

−1T

[C ] J−1 TQ

e , ∀TQe

Using the theorem of Castigliano, one can obtain displacements associated with the external forces

T e .

Up=[C ]T

Pe in the local base of the point P s , n

1,n

2 Pand U

Q=J

−1T

[C ] J−1 T

Qe the local base of the point Q s ,n

1, n

2 Q

∀TPe and ∀T

Qe

By breaking up the problem into two subproblems, and by means of the principle of superposition, one can write:

TQtotale

= TPPe

effort provenantdu point P

TPQe

effort provenantdu point Q

=−TPPi−T

PQi

TQtotale

= TQPe

effort provenantdu point Q

TQQe

effort provenantdu point P

=−TQPi−T

QQi

(one applies two load vector forces (independent one of the other) to the point P and the point Q ).

Action coming on a side: TPPi=−T

Pe=−K Up K=[C ]

1 Action coming on other side: TPQ

i =+ J−1TQ

e=J

−1 [J K JT ]UQ=K J

T UQ

We obtain as follows:

TP total

e=K UP−K J

T UQ and the same TQ total

e=−J

K UPJ

K J

T UQ

the stiffness matrix for “displacements” UP (given in its local base) and for “displacements” UQ

(given in its local base) is:

K=[K −K JT

−JK J

KJ

T ]

Moreover, in the case of beams with hollow circular section (elbows of pipework), one divides I y and

I z by coefficients of compliance given by the user, to take into account the variation of stiffness due to ovalization (cf [§6062]).

For the mass matrix, one considers only the reduced matrix (lumped masses), and one makes the simplifying assumption that the statement given for the straight beams [§2222] remains valid by considering a right element length R . .





One obtains:

M 11=M 22=M 33=M 77=M88=M 99=SR .

2

M 44=M10 10= I yI z R .

2

M 55=M 11 11=2 I y R .

15minS R

3

105,S R

2

48 M 66=M 12 12=

2 I z R .15

min S R 3

105,S R

2

48

This assumption is restrictive and does not allow to take well into account the inertia of bending or torsion due to the curvature. It is thus better in this case modelize a curved beam by several elements POU_C_T.





9 Loadingsthe various types of loading developed for beam elements “MECA_POU_D_E”, “MECA_POU_D_T”, “MECA_POU_C_T” in linear elasticity are:

Types or options MECA_POU_D_E MECA_POU_D_T MECA_POU_C_TCHAR_MECA_EPSI_R :loading by imposition of a strain (of standard thermal stratification)

developed developed developed

CHAR_MECA_PESA_R :loading due to gravity


CHAR_MECA_FR1D1D :distributed loading by actual values


CHAR_MECA_FF1D1D :distributed loading by function


CHAR_MECA_TEMP_R :developed “thermal”

loading developed developed

CHAR_MECA_FRELEC :loading “forces electric” created by a secondary driver developed

right developed not

CHAR_MECA_FRLAPL :loading “forces electric” created by a developed unspecified secondary

driver developed not

9.1 Loading by strainOPTION: “CHAR_MECA_EPSI_R”One calculates the loading from a strain state (this option was developed to take into account the thermal stratification in the pipework). The strain is given by the user using key word PRE_EPSI in AFFE_CHAR_MECA.

9.1.1 For the straight beam of Eulerian and the straight beam of Timoshenko

The model takes into account only one work in traction and compression and pure bending (not of shears, not of twisting moment).The following behavior models are used:

N=ES∂u

∂ x, M

y=E I

y

∂y

∂ x, M

z=E I

z

∂z

∂ x

Thus, one obtains the second elementary member associated with this loading:

• at item 1: Fx

1

=E S1

∂u

∂ x, M

y1

=E Iy

1

∂ y

∂ x, M

z1

=E Iz

1

∂z

∂ x

• at item 2: Fx

2

=E S2

∂u

∂ x, M

y2

=E Iy

2

∂ y

∂ x, M

z2

=E Iz

2

∂z

∂ x

While being given ∂ u∂ x

, ∂ y

∂ x and

∂z

∂ x for a beam, one can affect a loading to him.





9.1.2 For the curved beam of Timoshenko

9.1.2-a 9.1.2-acurved beam.

The method used can be compared with that previously presented. One uses to obtain the loading with the nodes, the local stiffness matrix K which multiplied by displacements with the nodes u

gives the forces F applied to the nodes: F=K u

The method takes into account only the strain related to the length of the beam. But one takes into account only the length projected on x , i.e. the shortest distance connecting the two extreme points

of the curved beam: 2R sin ( R : radius of curvature). This length is multiplied by a strain rate,

which gives the strain state of the beam. Then one projects on x and on y .

One has thus for u :

u=−2 R sin EPX cos2 R sin EPX sin 0000

2 R sin EPX cos2 R sin EPX sin

}point 1

}point 2

with EPX=∂ u∂ x

9.2 Loading due to gravityOPTION: “CHAR_MECA_PESA_R”the force of gravity is given by the modulus of acceleration g and a normalized vector indicating the

direction n .The principle to distribute the loading on the two nodes of the beam is to take account of the shape functions x associated with each degree of freedom with the element [§1414]. We thus have for a

loading in gravity an equivalent nodal force Q :

Q=∫0

L x S g n dx

Note:The shape functions used are (simplifying assumption) those of the model Eulerian - Bernoulli.





It is necessary, of course, to be placed in the local coordinate system of the beam to do this calculation.

Axial load (in x ) :

F x i=∫0

LiS g⋅xdx

1=1− xL

, 2=xL

from where: F x1=g⋅x L S1

3

S2

6 at item 1,

F x2=g⋅x L S1

6

S2

3 at item 2,

for a section varying linearly S=S1S2−S1 xL

and

F x1=g⋅x L 3S12S1 S2S2

12 ,F x2=g⋅x L S22S1 S23S2

12 a section varying homothétiquement S [ S1S2−S1

xL ]

2

Twisting moment:Without taking into account of warping, it does not exist.• In the plane xO z :

F z 1=g⋅z∫0

L1 S dx

M y1=g⋅z∫0

L2 S dx

F z 2=g⋅z∫0

L3 S dx

M y2=g⋅z∫0

L4 S dx

1=2 xL

3

−3 xL

2

1 , 2=L [− xL

3

2 xL

2

−xL ] ,

3=−2 xL

3

3 xL

2

,4=L [− xL

3

xL

2

]





from where, for a section varying linearly:

Fy1

=g⋅y L 7S1

20

3 S2

20 , Mz1

=g⋅y L 2 S1

20

S 2

30 F

y2

=g⋅y L 3 S1

20

7 S2

20 , Mz2

=−g⋅y L2 S1

30

S2

20

and, for a section varying homothétiquement:

F z1

=g⋅z L 8 S15S 1 S 22 S2

30 , M y1

=g⋅z L2 −2 S1−2S1 S 2−S 2

60 F

z2

=g⋅z L 2 S15S

1S

28 S

2

30 , My2

=g⋅z L2 S12S

1S

22S

2

60

• In the plane x o z :

F y

1

=g⋅y∫o

L1S dx , M z

1

=g⋅y∫o

L−2 S dx

F y2

=g⋅y∫o

L3S dx , M z

2

=g⋅y∫o

L−4 S dx

from where, for a section varying linearly:

Fy1

=g⋅y L 7S1

20

3 S2

20 , Mz1

=g⋅y L 2 S1

20

S 2

30 F

y2

=g⋅y L 3 S1

20

7 S2

20 , Mz2

=−g⋅y L2 S1

30

S2

20

and, for a section varying homothétiquement:

F y1=g⋅y L8 S 15S 1 S 22 S 230 , M z1

=g⋅y L2 2 S 12 S1 S 2S 260

F y2=g⋅y L 2 S15S 1S 28S 230 , M z 2

=g⋅y L2 −S1−2 S1 S 2−2 S 260

9.3 Distributed loadingsOPTIONS: “CHAR_MECA_FR1D1D”, “CHAR_MECA_FF1D1D”,

the loadings are given under key word FORCE_POUTRE, either by actual values in AFFE_CHAR_MECA (option CHAR_MECA_FR1D1D), or by functions in AFFE_CHAR_MECA (option CHAR_MECA_FF1D1D).The various possibilities are:

loading constant loading linearly varyingstraight beam with constant section developed developedstraight beam with section varying linearly developed notstraight beam with section varying homothétiquement

developed not

curved beam developed not





For the straight beams with constant section, the distributed loading is composed of forces and moments constant or varying linearly along the axis of the beam. In all the other cases the distributed load is only made up of constant force.The method used to compute: the loading to impose on the nodes is that presented to [§1515].

9.3.1 Straight beam with constant section

For a loading constant or varying linearly, one obtains:

F x1=L n1

3

n2

6 , F x2=L n1

6

n2

3 , M x1=L m t1

3

mt 2

6 , M x2=L mt 1

6

mt 2

3

n1 and n2 are the components of the axial loading as in points 1 and 2, m t1 m t2

are the

components of the twisting moment as in points 1 and 2, coming from the data of the user replaced in the local coordinate system.

t y1 t y2

, t z1 and t z2

are the components of the shears, m y1, m y2

, m z1 and m z2

are the

components of the moments of bending, at items 1 and 2, coming from the user data replaced in the local coordinate system.

F y1=L 7t y1

20

3 t y2

20 −m z1

mz 2

2M y1

=−L2 t z 1

20

t z2

30 −m y2

−m y1L

12

F y2=L 3 t y1

20

7 t y2

20 m z1

m z2

2M y2

=L2 t z1

30

t z2

20 m y2

−m y1L

12

F z1=L 7 t z1

20

3 t z2

20 m y1

m y2

2M z1

=L2 t y1

20

t y2

30 −m z2

−m z1L

12

F z2=L 3 t z1

20

7 t z2

20 −my1

m y2

2M z2

=−L2 t y1

30

t y2

20 m z2

−m z1L

12

9.3.2 Straight beams with variable section

the provided loading must be constant. One uses a method similar to that used by the inertia loading [§5455].

9.3.3 Curved beam

the provided loading must be constant along the element. The loading reduced to the nodes is equivalent to that which one can obtain by taking again the results of [§5758]. With a constant load, that becomes:

Fx1

=Ln

2F

x2

=Ln

2F

y1

=Lt

y

2F

y2

=Lt

y

2

Mz 1

=L 2t y

12M

z 2

=−L 2t y

12F

z1

=Lt z

2F

z2

=Lt z

2

My

1

=−L2t

z

12M

y2

=L 2t

z

12

9.4 Thermal loadingOPTION: “CHAR_MECA_TEMP_R”to obtain this loading, it is necessary to calculate the strain induced by the difference in temperature T−T référence :





u1=−L T−T référence u2=L T−T référence

( : thermal coefficient of thermal expansion)

Then, one calculates the induced forces: F=K u . As K is the local stiffness matrix with the element, one must then carry out a basic change to obtain the values of the components loading in the total reference.

9.5 Electric loadingOPTIONS: “FORCE_ELEC”, “INTE_ELEC”.This kind of loading makes it possible to take into account the force of Laplace acting on a principal driver, due to the presence of a secondary driver.The secondary driver « 2 » is right and it is not based on part of the Aster mesh if one uses option “FORCE_ELEC”.The secondary driver is not necessarily right and it can lean on part of the Aster mesh if one uses option “INTE_ELEC”.The linear density of the force of Laplace exerted in a point M of the principal driver by the secondary driver is written:

f M =12

e1∧∫2

e2∧r∥r∥3 ds with ∥e1∥

2=∥e2∥

2=1

Note::To obtain the force of Laplace, the vector turned over following the computation carried out by one of these two options must be multiplied by the temporal function of intensity specified by the operator “DEFI_FONC_ELEC” . [U4.MK.10]

9.5.1 Conducting secondary right finished or infiniteOPTION: “FORCE_ELEC” [U4.44.01]For a finished secondary driver, we have:

f M =e1

2∧

nd sin 1−sin2

with n=e2∧d

dd=∥d∥

For an infinite secondary driver, we have:

f M =e1∧nd

Three types of loading are possible:• driver infinite parallel right,• multiple infinite parallel drivers right,• right driver in unspecified position finished or infinite.





In the case of a driver infinite parallel right, its position can be given of two ways:• maybe by vector translation of the principal driver to the secondary driver,• maybe by the distance enters the two drivers and by a point of the secondary driver.

One thus obtains a constant loading on all the principal element. One can thus use the technique presented to [§5657] [§5758] to compute: the loading with the nodes.

F1=f M L

2F

2=f M

L

2

M1=e

1∧f M

L 2

12M

2=−e

1∧f M

L2

12

f M =cste∀M

For the case of the parallel drivers right infinite multiple, one must give the vector directly “forces of Laplace” normalized. This one being usually given in the total reference, the vector should be determined “forces of Laplace” in the local coordinate system of the principal element. Thus of the same way that previously, one calculates the loading with the nodes.

In the case of a right driver in unspecified position finished or infinite, its position is given by two points P1 and P2 such as flow circulates of P1 worms P2 . The loading is not obligatorily constant along the principal element.

The method chosen to compute: the loading reduced to the nodes is obviously the same one as before. But here, integration is done numerically by discretizing the element in a certain number (in practice: 100 points enters P1 and P2 ).

One integrates as follows:

F1=∫0

Lf M [2 x

L 3

−3 xL

2

1] dx

F2=∫0

Lf M [−2 x

L 3

3 xL

2

] dx

M 1=∫0

L−e1∧f M [ L− x

L 3

2 xL

2

−xL ] dx

M 2=∫0

L−e1∧

f M [ L− xL

3

xL

2

] dx

Note:

As f M ⋅x = 0 and e1∧f M ⋅x=0 ,x = e

1 one uses only the shape functions associated with the problem with bending.

9.5.2 Secondary driver describes by part of Aster meshOPTION: “INTE_ELEC”, [U4.44.01]the secondary driver in its group is not necessarily right. But it is described only by right elements. Its length is obligatorily finished.Its position can be supplemented by the use of one vector translation or a symmetry plane (by a point and a normalized normal vector of the symmetry plane) compared to the principal driver.





Except the fact that it is necessary to add the interaction of the various elements of the secondary driver on the principal driver, the method is the same one as previously (case iii ).

Note:

For numerical integration, one does not use 5 points understood enters P1

and P2 .





10 Torsor of the forces - Torsor of the stresses (or generalized forces) - nodal Forces and reactionsOPTIONS: “EFGE_ELNO”, “SIEF_ELGA”, “SIPM_ELNO”, “SIPO_ELNO”, “FORC_NODA”,

“REAC_NODA”.Option “EFGE_ELNO” makes it possible to calculate the torsor of the forces to the 2 nodes of each element of “beam”.Options “SIPM_ELNO” and “SIPO_ELNO” make it possible to calculate the maximum values of the components tensor of the stresses intervening in a model of “beam”.Option “SIEF_ELGA” allows the computation of the nodal forces (option “FORC_NODA” and of reactions “REAC_NODA”).

Note:When one of these options end in “_C”, that means that the values of displacements (and thus of the forces) are complex.

10.1 The torsor of forcesOPTION: “EFGE_ELNO”One seeks to calculate with the two nodes of each element “beam” constituting the mesh of studied structure, the forces exerted on the element “beam” by the rest of structure. The values are given in the local base of each element.By integrating the balance equations, one obtains [§1515] the forces in the local coordinate system of the element:

RLOC=K LOCe uLOCMLOC

e uLOC−f LOCe

where: RLOC = −N 1 ,−V Y1 ,−V Z

1 ,−M T1 ,−M Y

1 ,−M Z1 , N 2 ,V Y

2 ,V Z2 ,M T

2 , M Y2 ,M Z

2 K LOC

e elementary matrix of stiffness of the “exact” element of beam,

M LOCe

elementary matrix of mass of the element beam,

f LOCe

vector of the forces “distributed” on the element beam,

u LOC vector “degree of freedom” limited to the element beam,

u LOC vector “acceleration” limited to the element beam.

One changes then the signs of nodal efforts 1 [§1515].

OPTION: “SIEF_ELGA”option “SIEF_ELGA” is established for reasons of compatibility with other options. It is used only for

computation of the nodal forces and the reactions. It is calculated by: RLOC=K LOCe uLOC .

10.2 The tensor of the stressesOPTIONS: “SIPM_ELNO”, “SIPO_ELNO”.

One seeks of the components to calculate the maximum values xx , xy and xz which are

connected to the forces by:

N = ∫s xx dS ,

V y = ∫s xy dS ,

V z = ∫s xz dS ,

M T = ∫s y xz−z xy dS ,

M y = ∫sz xx dS ,

M z = ∫s− y xx dS .





With option “SIPM_ELNO”, one calculates the maximum effect of all the forces on xx , xy and xz

.

• For xy , one a:

xzi Max

=V Y

i

AYi

M Ti

J XiRT

i for the node i , i=1,2

with: AY=k y A A : area of the section

with: k y : constant of shears

with: J X : constant of torsion

with: RT : radius of torsion

• For xz , one a:

xzi Max

=V Z

i

AZi

M Ti

J Xi

RTi ,

• For xx , one a:

◦Pour a rectangular section:

xz

iMax

=∣N i

Ai∣∣MY

i

I Yi

HZ

i

2 ∣∣MZ

i

I Zi

HZ

i

2 ∣ si N i≥0,

xz

i Max

=−∣N i

Ai ∣∣MY

i

I Yi

HZ

i

2 ∣∣MZ

i

I Zi

HZ

i

2 ∣ si N i0

with I , I Z : geometrical moments

with H , HZ : sides of the rectangle

◦Pour a circular section:

xx

iMax

=∣N i

Ai∣MY

i 2M Z

i 2 Ri

IY

i si N i≥0,

xxi Max

=−∣N i

Ai ∣ M Yi

2

M Zi

2 Ri

IY

i si N i0

with R : radius of the section

◦Pour a general section:

xxi Max

=N i

AiM Y

i

I Yi RZ

i−

M Zi

I Zi RY

i

In this last case, one calculates at the point RYi ,RZ

i . RYi et RZ

i are the distances to neutral fiber.

To find these formulas, one uses the relations between the stresses and the strains then those

between the internal forces and u , x ; xy ; xz ; x , x ; y , x ; z , x





{ xx=E xx

xy=2G xy

xz=2G xz ,

{xx=u , xz y , x−y z , x

2xy= xy−z x , x xy=V , x−z 2xy= xz yx , x xz=W , xy ,

NV Y

V Z

M T

M Y

M Z

=EA

k Y GA 0kZ GA

GJ0 EI Y

EI Z

u , x xy

xz

x , x

y , x

z , x

(the various types of motion are decoupled when one works with the principal axes (by definition)). One obtains:

xx x , y , z =E [ NEAz

M Y

EI Y

− yM Z

EI Z ] ,

xy x , y , z =G [ V Y

k Y GA−z

M T

G J X] ,

xz x , y , z =G [ V Z

k Z GA y

M T

G J X ]

With option “SIPO_ELNO”, computation is somewhat different. One searches the maxima effects of

each force N , VY ,V Z , M T , M X , M Y on the components xx , xy and xz . One finds the

preceding results in broken up form. The vector result is written:

11 ,21 , 31, 41, 51 , 61

12 ,22 , 32 , 42 , 52 ,

62

with 1i=

N i

Ai for the node i , i=1,2

1i=

N i

Ai

3i=

V Z i

AZ i

,

4i=

MT i

JX i

RT i ,





the rectangular sections:

5i=

M Y i

I Y i

H Z i

2 ,

6i=−

MZ i

IZ i

HY i

2,

5i=M Y i

I Y i

Ri ,

6i=−M

Z i

IY i

Ri ,

for the unspecified sections with point: Ri , RZ i

5i=M

Y i

I Y i

RZ i ,

6i=−M

Z i

I Z i

RY i

In the case of beam with hollow circular section (pipes), the compliance of the thin shells not being well represented, certain quantities should be corrected. Two coefficients are used for that:• a coefficient of compliance cflex (also used by the stiffness [§4445]),• a coefficient of intensification of the stresses isigm .

In particular, one can take as a starting point the rules RCC_M.

I classique=I classique

cflex cflex≥1

5=M Y

I Y corrigéRZ

isigmcflex

,

6=M Z

I Z corrigéRY

isigmcflex

isigm≥1

only in bending

10.3 Computation of the nodal forces and reactionsOPTION: “FORC_NODA”This option calculates a force vector nodal on all structure. It produces a field at nodes in command CALC_CHAMP by assembly of the elementary terms.For this computation, one usually uses in 3D [R5.03.01] for example the principle of the virtual wors and one writes:

F=QT

where QT the matrix represents symbolically associated with the operator divergence. For an

element, one writes the work of the strain field virtual:

QT h=∫

u wh

For the beam elements, one calculates simply the nodal forces by assembly of the elementary nodal forces calculated by option SIEF_ELGA, which are expressed by:

[FLOC ]=[KLOC ] [ULOC ] [§7 .1 ]

OPTION: “REAC_NODA”This option, called by CALC_CHAMP, makes it possible to obtain the reactions R to the bearings

starting from the nodal forces F by:

R=F−FcharF iner

F char et F iner being nodal forces associated with the loadings given (specific and distributed) and with

the forces of inertia.





11 Element of barKey word “BARS”a bar is a straight beam of constant section comprising only the degrees of freedom of traction and compression. The equation of motion, the stiffness matrix and the forces are thus those of the beams (right of constant section) relating to the traction and compression.However the mass matrix must take into account the degrees of freedom in the 3 directions of space, mainly so that in computations of dynamics the mass is taken into account in each direction (see

[V2.02.146]). Thus, if one notes L the length of the element, A the area of his section and his

density, the elementary mass matrix is the following one (with the components in the order DX 1 DY 1 DZ 1 DX 2 DY 2 DZ 2 ) with m= A L :

M elem=m /3 0 0 m /6 0 00 m /3 0 0 m /6 00 0 m /3 0 0 m /6m/6 0 0 m /3 0 00 m /6 0 0 m /3 00 0 m /6 0 0 m /3

[11-1]

the only geometrical characteristic is the area of the cross section [U4.42.01 §6].





12 Bibliography[1] J.S. PRZEMIENIECKI. Theory of Structural Matrix Analysis, Mc Graw Hill, New - York. 1968.

[2] MR. GERADIN, D. RIXEN. Theory of vibrations, application to the dynamics of structures, Masson, Paris. 1993.

[3] J.R.BANERJEE, F.W. WILLIAMS. Exact Bernoulli-Eulerian static stiffness matrix for has arranges of tappered beam-columns. Interm. J. for Numerical Methods in Engineering, vol. 23, pp 1615 - 1628. 1986. [4] J.L. BATOZ

4 G. DHATT. Modelization of structures by finite elements – HERMES. [5] V. OF TOWN

5 OF GOYET. Nonlinear static analysis by the finite element method of formed spatial structures by beams with asymmetric sections. Thesis of the University of Liege. 1989. [6] C. MASSONNET

6 , S. CESCOTTO. Mechanics of the materials. Cd. Boeck-Wesmael s.a. 1994. [7] P. DESTUYNDER

7 Introduction to the computation of structures. Course C. Scientific computation. 1990. Description of

13 the versions of the document Version AsterAuthor (S), Organization

(S) Description ofthe modifications 3 J.M.PROIX, P.MIALON,

3EDF/IMA/MMN, IAT St CYR) initial Text 10.2 F.VOLDOIRE, J. - L

. FLÉJOU EDF

AMA Various

corrections of form . Addition

of a synthesis of the degrees of freedom and shape functions. 11.0 X.DESROCHES Addition of the stiffness matrix

centrifuges12.0 J-L.FLÉJOU Moments distributed

on prismatic

beam.



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