pressure effect on a pipe

7/29/2019 pressure effect on a pipe

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Pressure Effect on a Pipe

Introduction

Pressure effect is a zero external force resultant acting on a pipe. For this reasonit is treated as the temperature effect, the load equivalent to the stress inducedby ts effects is applied as external load by changing its negative/positive sign.Two possible condition can be considered for the pipe end:

a) the pipe is capped at the ends; the internal and external pressureact on the relevant cross section

b) the pipe is open at the ends and no actions are present

Pipe Stress Fields

The stress state is self-equilibrated and can be analyzed into axial, radial andcircumferential (hoop) stress, considering the following symbols,[1]:

P i internal pressure

P o external pressure

Di inner steel diameter

Do outer steel diameter

Axial Stress The axial stress is present only in the end cap condition and itis equal to the difference between the internal and the external pressure(the fluid pressure is taken as a modulus):

σx =P iD

2

i − P oD2

o

D2o −D2

i

(1)

σx = 0

Radial Stress The radial stress can be deduced using the Lamé stress dis-

tribution using as boundary condition the internal and external pressureacting respectively on the inner and outer radius, and it is equal in botcondition capped and free ends:

σr(r) =P iD

2

i − P oD2

o

D2o −D2

i

−

D2

iD2

o

D2(r)

(P i − P o)

D2o −D2

i

(2)

1



Circumferential Stress The hoop stress can be determined using the samestress distribution used in the radial case and result:

σh(r) =P iD

2

i − P oD2

o

D2o −D2

i

+D2

iD2

o

D2(r)·

(P i − P o)

D2o −D2

i

(3)

As above highlighted the stress field is self-equilibrated and no external forceis applied to the structure like considering the temperature loads. In FiniteElement code the force rising from the volume integration of the deformationinduced by self-equilibrated load (F s−e) are applied “as is” to the structure tocompute the displacement (U sol):

F ext + F s−e = K · U sol (4)

where the F s−e is applied as any other external force in a standard Finite Ele-

ment Analysis and is calculated according to the linear elastic Hooke law:

F s−e = E ·Asteel

α∆T +

1

E (σx − ν (σr + σh))

(5)

Then the self-equilibrated force induced deformation (εs−e) are subtracted tothe deformation obtained from the solution (εsol) to calculated internal elementactions (the effective external force resultant on steel pipe):

F int = K · U sol − F s−e (6)

Test Example

To verify how FEM codes implement the pipe behavior some simple run havebeen performed on ANSYS1 and Code_Aster. Two run has been performed fora 1 meter long linear stretch of a pipe subject to a temperature and pressureload considering the two different case of restrained and free axial elongation.The test data are reported in table1

Fully Restrained Pipe

The axial force that rises to restrain the pipe can be calculated using equations4 and 5.

1. First applying the pressure and temperature effect of equation 5 as anexternal force to the global system of equilibrium equation whose solution

will result in a zero value for the U sol.1ANSYS, and any and all ANSYS, Inc. brand, product, service and feature names, logos

and slogans are registered trademarks or trademarks of ANSYS, Inc. or its subsidiaries in the

United States or other countries, service and feature names or trademarks are the property of

their respective owners.

2



Do 1t 0.1

tins 0.1ρins 3000H 0P in 1.0 · 109

ρsteel 7850E 2.1 · 1011

α 1.16 · 10−5

∆T 40

Table 1: Data for the FE analysis

2. Then calculating the internal force at the element level using equation 6results:

F int = −F s−e

The data used to run the model are in the table 1. To solve the model theANSYS FE code has been used with the mesh showed in figure 1.

The pipe elongation, considering fully restrained ends, has to be absorbed bythe constrain resulting in the force R along the pipe axis corresponding to anelongation (considering pipe length of 1m):

R

Asteel · E

Reaction that can be determined by equations 6 and 5 where the substitution

of the stress relationships 1, 2 and 3 leads to:

R

Asteel · E = α∆T +

1 − 2ν

E

P iD2

i − P oD2

o

D2o −D2

i

The constraint reaction has two contribution one form the temperature and theother from the pressure:

R = AsteelE (α∆T ) + Asteel(1 − 2ν )P iD

2

i − P oD2

o

D2o −D2

i

the stress induced by the reaction is in the present example decomposed in thetable 2

ANSYS Results

Using the script file testpipe59.mac one with fully restrained ends and theother with free elongation condition.

3



Reaction Self-Equilibrated Total

E (α∆T ) (1 − 2ν )P iD

2

i−P oD

2

o

D2o−D2

i

P iD2

i−P oD

2

o

D2o−D2

i

−9.744 · 107 −7.111 · 108

−8.0855 · 108 1.7778 · 109 9.6923 · 108

Table 2: Composition in the axial direction of the stress

Figure 1: Mesh

4



σx 1.7778E+09σr 0.0000E+00

σh 3.5556E+09εth 4.6400E-04ε pr 3.3862E-03

Table 3: Stress in the case end cap

Figure 2: Axial stress with KEYOPT(8)=0 end cap

Capped Ends In this case considering not present the outer pressure thedeveloped stresses are showed in table 3 with the resulting deformation for thethermal and pressure components. The pipe is stretched due to the effects of the end cap internal pressure and the temperature. The reaction force of therestraint is a compressive force and its value is:

Theory ANSYS

R = −2.2861 · 108 -2.2861E+08

The resulting axial stress on the pipe summing up the pressure, temperatureand reaction force effects is

Theory ANSYS

σTot

x = 9.9623·

108

9.69E+008

The result of the ANSYS simulation is in complete agreement with the abovecalculations as is showed in figure 2 where the steel axial stress is reported

5



σx 0.0000E+00σr 0.0000E+00

σh 3.5556E+09εth 4.6400E-04ε pr -5.0794E-03

Table 4: Stress in the case “free-end-cap”

Figure 3: Axial stress in the free cap ends

Ends without Cap In this case the pressure deformation effect is a contrac-tion of the pipe while the temperature produce an elongation. The net effect isa contraction resulting in a pulling reaction force whose value is:

Theory ANSYS

R = 2.740 · 108 2.7404E+08

The only action, along the x coordinate direction, is the tensile reaction forcewhich induces in the steel the following total stress:

Theory ANSYS

σTotx = 9.9623 · 108 9.6923E+08

The result of the ANSYS simulation is in complete agreement with the above

calculations as is showed in figure 3 where the steel axial stress is reported.

Code_Aster

Code_Aster does not have a specialized Beam Element for pipe simulation thatcan manage external pressure load. The pressure effect has been simulated as an

6



initial strain and a temperature load using the scripting feature of Code_Asterthat enable the user to use python language in the .comm file.

Capped Ends In the first example an initial strain, simulating the pressureeffect, has been applied plus the temperature load. the reactions are listed below

CHAMP AUX NOEUDS DE NOM SYMBOLIQUE REAC_NODA

NUMERO D’ORDRE: 1 INST: 0.00000E+00

NOEUD DX DY DZ DRX DRY DRZ

N1 2.28612E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N2 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N3 -2.28612E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

while the internal action are:

CHAMP PAR ELEMENT AUX NOEUDS DE NOM SYMBOLIQUE SIEF_ELNO_ELGA


M1 N VY VZ MT MFY MFZ

N1 -2.75505E+07 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00N2 -2.75505E+07 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00


N2 -2.75505E+07 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00

N3 -2.75505E+07 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

and the nodal force, coincident with the internal actions are:

CHAMP AUX NOEUDS DE NOM SYMBOLIQUE FORC_NODA



N1 2.75505E+07 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N2 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N3 -2.75505E+07 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

The initial strains, εx = 3.386 · 10−3, are considered as an external load

applied to the ends of the structure whose amount areP

= 2.01

·

10

8N exactlythe difference between the reaction, R = 2.28612 · 108, and the nodal force

F = 2.75505 · 107. These considerations are supported by the strain valuesreported below:

CHAMP PAR ELEMENT AUX NOEUDS DE NOM SYMBOLIQUE DEGE_ELNO_DEPL


M1 EPXX GAXY GAXZ GAT KY KZ

N1 -4.64000E-04 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N2 -4.64000E-04 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

M2 EPXX GAXY GAXZ GAT KY KZ

N2 -4.64000E-04 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N3 -4.64000E-04 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

Simulating the pressure effect as an additional temperature load can be doneusing the following formula:

∆T P =1

α

1 − 2ν

E

P iD2

i − P oD2

o

D2o −D2

i

The reactions are as expected identical to the previous results:

7





NOEUD DX DY DZ DRX DRY DRZN1 2.28612E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N2 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N3 -2.28612E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

while the nodal forces in this case are coincident with the reaction due tothe absence of the external forces leading to the following internal actions:




N1 -2.28612E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00

N2 -2.28612E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00


N2 -2.28612E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00

N3 -2.28612E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

The resulting stresses are drawn from the internal action:

CHAMP PAR ELEMENT AUX NOEUDS DE NOM SYMBOLIQUE SIPO_ELNO_DEPL


M1 SN SVY SVZ SMT SMFY SMFZ

N1 -8.08551E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N2 -8.08551E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00


N2 -8.08551E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N3 -8.08551E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00

It is plain that it does not take into account the self-equilibrated stress field

induce by the internal and external pressure that has to be summed to the stressinduced by the reactions, stated in other form the axial force is a compressiveforce not the “effective pulling force” that takes into account the pressure action.

Open Ends In this case the pressure effect is simulated by the followingequation:

R

Asteel · E = α∆T +

−2ν

E

P iD2

i − P oD2

o

D2o −D2

i

The reactions are:




N1 -2.74042E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N2 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N3 2.74042E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

that results in a traction inside the pipe as results from the list of the internalaction that follows:




N1 2.74042E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00

N2 2.74042E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

8



Figure 4: Axial stress in the cap ends case for simply supported conditions


N2 2.74042E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00

N3 2.74042E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

leading to the correct value for the internal stress:




N1 9.69227E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N2 9.69227E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00


N2 9.69227E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N3 9.69227E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00

Simply Supported Pipe

The same calculations, of the previous section, have been performed let the pipeto move along its axis.

ANSYS Results

ANSYS confirm to handle the pipe stress in the correct way as the figures 4 and5 show respectively for the cap ends and open ends conditions. The axial stress

is calculated in both cases accordingly to the theory.

Code_Aster Case

Due to the workarounds needed in Code_Aster Beam Element, to deal withinternal/external pressure different results are obtained considering cap or open

9



Figure 5: Axial stress in the open ends case for simply supported conditions

ends.

Cap Ends Both workarounds have been analyzed for simulating the pressureeffect

Initial Strain Considering the pressure as an initial strain load it gener-ates an external load that, in the isostatic conditions as the one in this case,

transfers the load to the structure. The cap ends condition lead to the followingnull reaction:




N1 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N2 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N3 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

while the internal actions are:




N1 2.01062E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00

N2 2.01062E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

M2 N VY VZ MT MFY MFZN2 2.01062E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00

N3 2.01062E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

the initial strain equivalent load induces, as expected, a pulling force insidethe pipe leading to the following stresses, equivalent to the whole pressure effect,as reported in table 2, and not only to the axial component as expected:

10





M1 SN SVY SVZ SMT SMFY SMFZN1 7.11111E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N2 7.11111E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00


N2 7.11111E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N3 7.11111E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00

Thermal Load The simulation of pressure effects as thermal expansiongive rise to a null stress state as it is expected:




N1 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N2 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00


N2 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00N3 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00

Open Ends For open ends too both workarounds have been tested

Initial Strain In this case, simulating the pressure effect with initial strainload, lead to a compression effect on the isostatic structure, leading to a nullreactions with the following internal actions:




N1 -3.01593E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00

N2 -3.01593E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00


N2 -3.01593E+08 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00

N3 -3.01593E+08 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

and stresses:




N1 -1.06667E+09 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N2 -1.06667E+09 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00


N2 -1.06667E+09 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N3 -1.06667E+09 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00

Thermal Load The simulation of pressure as thermal load give more sat-isfactory results. In fact the displacements shows a contraction of the pipe due

to the pressure effect:CHAMP AUX NOEUDS DE NOM SYMBOLIQUE DEPL



N1 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N2 -2.30768E-03 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

N3 -4.61537E-03 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00

11



and the resulting stress state is null as in the following:




N1 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N2 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00


N2 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 -0.00000E+00 0.00000E+00

N3 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 0.00000E+00 -0.00000E+00

Flexural Response

Further tests have been performed to study the response, of the FE model, toloads generating bending in the pipe. To this aim a 120m long straight pipehas been considered, with the same mechanical characteristics of the previous

section, with fully restrained condition in the vertical plane and two differentconditions in the axial direction

1. free end displacements in the axial direction

2. restrained end displacements in the axial direction

The loads considered was the gravity, the internal pressure as 1 · 109Pa and notemperature load.

The following example would simulate the response of a straight pipe to avertical load as gravity. In some condition the contribution of the axial stressin carrying the vertical loads is significant and must be taken into account,this result in considering the equilibrium of the pipe referred to the deformedconfiguration and not to the undeformed one. In condition under examination

the resulting strains are usually still in the range of small deformation so theanalysis should consider ” large rotation and small deformation ”.

Code_Aster

In Code_Aster considering the material as linear elastic, “COMP_ELAS” op-tion of the solution phase, only the beam element “POU_D_T_GD” can handle ge-ometric non-linearities, both large rotation and large deformation, [2], althoughin the present examples the large deformation option is not essential.

The vertical loads considered are the pipe steel weight and the insulationweight as listed in table 1. Code_Aster has no specific element to simulatesubmerged pipe and the pipe steel weight is simulated as gravity load using thefollowing command:

GRAVITY=AFFE_CHAR_MECA(MODELE=PipeFEM,

PESANTEUR=_F(GRAVITE=-9.81,

DIRECTION=(0.,0.,1.,),),);

while the insulation weight is simulated by linear distributed pressure using thefollowing command:

12



INSUWE=AFFE_CHAR_MECA(MODELE=PipeFEM,

FORCE_POUTRE=_F(TOUT=’OUI’,

FX=0.0,FY=0.0,FZ=-Wins,),);

Free Ends

The deflection of the pipe for the present case is listed below:

CHAMP AUX NOEUDS DE NOM SYMBOLIQUE DEPL

NUMERO D’ORDRE: 1 INST: 1.00000E-01


N1 1.96639E-18 0.00000E+00 -1.41270E-19 0.00000E+00 2.18563E-18 0.00000E+00

N2 -2.05766E-02 0.00000E+00 -4.55424E-02 0.00000E+00 2.25819E-02 0.00000E+00

N3 -4.28868E-02 0.00000E+00 -1.71584E-01 0.00000E+00 4.04861E-02 0.00000E+00

N4 -6.76689E-02 0.00000E+00 -3.60148E-01 0.00000E+00 5.40491E-02 0.00000E+00

N5 -9.48932E-02 0.00000E+00 -5.94612E-01 0.00000E+00 6.36072E-02 0.00000E+00

N6 -1.24044E-01 0.00000E+00 -8.59710E-01 0.00000E+00 6.94962E-02 0.00000E+00

N7 -1.54347E-01 0.00000E+00 -1.14153E+00 0.00000E+00 7.20510E-02 0.00000E+00

N8 -1.84946E-01 0.00000E+00 -1.42749E+00 0.00000E+00 7.16055E-02 0.00000E+00

N9 -2.15041E-01 0.00000E+00 -1.70635E+00 0.00000E+00 6.84931E-02 0.00000E+00

N10 -2.43977E-01 0.00000E+00 -1.96818E+00 0.00000E+00 6.30467E-02 0.00000E+00

N11 -2.71306E-01 0.00000E+00 -2.20436E+00 0.00000E+00 5.55989E-02 0.00000E+00

N12 -2.96814E-01 0.00000E+00 -2.40758E+00 0.00000E+00 4.64827E-02 0.00000E+00

N13 -3.20523E-01 0.00000E+00 -2.57187E+00 0.00000E+00 3.60311E-02 0.00000E+00

N14 -3.42670E-01 0.00000E+00 -2.69256E+00 0.00000E+00 2.45776E-02 0.00000E+00

N15 -3.63670E-01 0.00000E+00 -2.76631E+00 0.00000E+00 1.24559E-02 0.00000E+00

N16 -3.84065E-01 0.00000E+00 -2.79112E+00 0.00000E+00 2.43678E-17 0.00000E+00

N17 -4.04460E-01 0.00000E+00 -2.76631E+00 0.00000E+00 -1.24559E-02 0.00000E+00

N18 -4.25461E-01 0.00000E+00 -2.69256E+00 0.00000E+00 -2.45776E-02 0.00000E+00

N19 -4.47608E-01 0.00000E+00 -2.57187E+00 0.00000E+00 -3.60311E-02 0.00000E+00

N20 -4.71317E-01 0.00000E+00 -2.40758E+00 0.00000E+00 -4.64827E-02 0.00000E+00

N21 -4.96824E-01 0.00000E+00 -2.20436E+00 0.00000E+00 -5.55989E-02 0.00000E+00

N22 -5.24153E-01 0.00000E+00 -1.96818E+00 0.00000E+00 -6.30467E-02 0.00000E+00

N23 -5.53089E-01 0.00000E+00 -1.70635E+00 0.00000E+00 -6.84931E-02 0.00000E+00

N24 -5.83184E-01 0.00000E+00 -1.42749E+00 0.00000E+00 -7.16055E-02 0.00000E+00

N25 -6.13784E-01 0.00000E+00 -1.14153E+00 0.00000E+00 -7.20510E-02 0.00000E+00

N26 -6.44086E-01 0.00000E+00 -8.59710E-01 0.00000E+00 -6.94962E-02 0.00000E+00

N27 -6.73237E-01 0.00000E+00 -5.94612E-01 0.00000E+00 -6.36072E-02 0.00000E+00

N28 -7.00461E-01 0.00000E+00 -3.60148E-01 0.00000E+00 -5.40491E-02 0.00000E+00

N29 -7.25244E-01 0.00000E+00 -1.71584E-01 0.00000E+00 -4.04861E-02 0.00000E+00

N30 -7.47554E-01 0.00000E+00 -4.55424E-02 0.00000E+00 -2.25819E-02 0.00000E+00

N31 -7.68130E-01 0.00000E+00 -9.93900E-20 0.00000E+00 4.06535E-21 0.00000E+00

the vertical reaction is:




N1 5.48147E-06 0.00000E+00 1.91664E+06 0.00000E+00 -3.80451E+07 0.00000E+00

N2 -2.77790E-05 0.00000E+00 2.18036E-06 0.00000E+00 -7.78586E-07 0.00000E+00

.........N30 4.76188E-05 0.00000E+00 6.54851E-07 0.00000E+00 5.76675E-06 0.00000E+00

N31 -1.04353E-04 0.00000E+00 1.91664E+06 0.00000E+00 3.80451E+07 0.00000E+00

while the nodal forces are:

13





NOEUD DX DY DZ DRX DRY DRZN1 5.48147E-06 0.00000E+00 1.85275E+06 0.00000E+00 -3.80451E+07 0.00000E+00

N2 -2.77790E-05 0.00000E+00 -1.27776E+05 0.00000E+00 -7.78586E-07 0.00000E+00

N3 -6.77833E-06 0.00000E+00 -1.27776E+05 0.00000E+00 -3.25963E-06 0.00000E+00

N4 -6.78358E-05 0.00000E+00 -1.27776E+05 0.00000E+00 -8.35210E-06 0.00000E+00

N5 6.86479E-05 0.00000E+00 -1.27776E+05 0.00000E+00 -1.30422E-05 0.00000E+00

.........

The internal action are listed in table 5 were due to results symmetry onlythe first half of the pipeline has been reported.

As a further comparison the Euler type Finite Element has been used, andthe deflection is listed here




N1 0.00000E+00 0.00000E+00 4.25790E-36 0.00000E+00 1.83789E-35 0.00000E+00

N2 -2.03175E-02 0.00000E+00 -4.70844E-02 0.00000E+00 2.27304E-02 0.00000E+00

N3 -4.06349E-02 0.00000E+00 -1.75573E-01 0.00000E+00 4.07580E-02 0.00000E+00

N4 -6.09524E-02 0.00000E+00 -3.67326E-01 0.00000E+00 5.44186E-02 0.00000E+00

N5 -8.12698E-02 0.00000E+00 -6.05547E-01 0.00000E+00 6.40483E-02 0.00000E+00

N6 -1.01587E-01 0.00000E+00 -8.74785E-01 0.00000E+00 6.99828E-02 0.00000E+00

N7 -1.21905E-01 0.00000E+00 -1.16093E+00 0.00000E+00 7.25582E-02 0.00000E+00

N8 -1.42222E-01 0.00000E+00 -1.45122E+00 0.00000E+00 7.21103E-02 0.00000E+00

N9 -1.62540E-01 0.00000E+00 -1.73423E+00 0.00000E+00 6.89751E-02 0.00000E+00

N10 -1.82857E-01 0.00000E+00 -1.99988E+00 0.00000E+00 6.34884E-02 0.00000E+00

N11 -2.03175E-01 0.00000E+00 -2.23945E+00 0.00000E+00 5.59862E-02 0.00000E+00

N12 -2.23492E-01 0.00000E+00 -2.44554E+00 0.00000E+00 4.68045E-02 0.00000E+00

N13 -2.43810E-01 0.00000E+00 -2.61209E+00 0.00000E+00 3.62791E-02 0.00000E+00

N14 -2.64127E-01 0.00000E+00 -2.73442E+00 0.00000E+00 2.47459E-02 0.00000E+00

N15 -2.84444E-01 0.00000E+00 -2.80917E+00 0.00000E+00 1.25409E-02 0.00000E+00

N16 -3.04762E-01 0.00000E+00 -2.83430E+00 0.00000E+00 1.61447E-15 0.00000E+00

N17 -3.25079E-01 0.00000E+00 -2.80917E+00 0.00000E+00 -1.25409E-02 0.00000E+00

N18 -3.45397E-01 0.00000E+00 -2.73442E+00 0.00000E+00 -2.47459E-02 0.00000E+00

N19 -3.65714E-01 0.00000E+00 -2.61209E+00 0.00000E+00 -3.62791E-02 0.00000E+00

N20 -3.86032E-01 0.00000E+00 -2.44554E+00 0.00000E+00 -4.68045E-02 0.00000E+00

N21 -4.06349E-01 0.00000E+00 -2.23945E+00 0.00000E+00 -5.59862E-02 0.00000E+00

N22 -4.26667E-01 0.00000E+00 -1.99988E+00 0.00000E+00 -6.34884E-02 0.00000E+00

N23 -4.46984E-01 0.00000E+00 -1.73423E+00 0.00000E+00 -6.89751E-02 0.00000E+00

N24 -4.67302E-01 0.00000E+00 -1.45122E+00 0.00000E+00 -7.21103E-02 0.00000E+00

N25 -4.87619E-01 0.00000E+00 -1.16093E+00 0.00000E+00 -7.25582E-02 0.00000E+00

N26 -5.07937E-01 0.00000E+00 -8.74785E-01 0.00000E+00 -6.99828E-02 0.00000E+00

N27 -5.28254E-01 0.00000E+00 -6.05547E-01 0.00000E+00 -6.40483E-02 0.00000E+00

N28 -5.48571E-01 0.00000E+00 -3.67326E-01 0.00000E+00 -5.44186E-02 0.00000E+00

N29 -5.68889E-01 0.00000E+00 -1.75573E-01 0.00000E+00 -4.07580E-02 0.00000E+00

N30 -5.89206E-01 0.00000E+00 -4.70844E-02 0.00000E+00 -2.27304E-02 0.00000E+00

N31 -6.09524E-01 0.00000E+00 -4.39371E-36 0.00000E+00 -4.81482E-35 0.00000E+00

The data shows no interaction between axial and vertical force in fact thedisplacement in the x direction of the node N31 is exactly the contraction in-duced by the fictious temperature that simulates the pressure effect.

The results can be summarized in terms of maximum deflection at the pipemiddle vertical and moment reaction at ends in the following table:

14






N1 2.09188E+04 0.00000E+00 -1.85263E+06 0.00000E+00 3.43587E+07 0.00000E+00

N2 2.09188E+04 0.00000E+00 -1.85263E+06 0.00000E+00 3.43587E+07 0.00000E+00


N2 5.43863E+04 0.00000E+00 -1.72411E+06 0.00000E+00 2.72415E+07 0.00000E+00

N3 5.43863E+04 0.00000E+00 -1.72411E+06 0.00000E+00 2.72415E+07 0.00000E+00


N3 7.54675E+04 0.00000E+00 -1.59541E+06 0.00000E+00 2.06362E+07 0.00000E+00

N4 7.54675E+04 0.00000E+00 -1.59541E+06 0.00000E+00 2.06362E+07 0.00000E+00


N4 8.63934E+04 0.00000E+00 -1.46688E+06 0.00000E+00 1.45428E+07 0.00000E+00

N5 8.63934E+04 0.00000E+00 -1.46688E+06 0.00000E+00 1.45428E+07 0.00000E+00


N5 8.92228E+04 0.00000E+00 -1.33867E+06 0.00000E+00 8.96021E+06 0.00000E+00

N6 8.92228E+04 0.00000E+00 -1.33867E+06 0.00000E+00 8.96021E+06 0.00000E+00


N6 8.58382E+04 0.00000E+00 -1.21083E+06 0.00000E+00 3.88712E+06 0.00000E+00

N7 8.58382E+04 0.00000E+00 -1.21083E+06 0.00000E+00 3.88712E+06 0.00000E+00


N7 7.79451E+04 0.00000E+00 -1.08329E+06 0.00000E+00 -6.77791E+05 0.00000E+00

N8 7.79451E+04 0.00000E+00 -1.08329E+06 0.00000E+00 -6.77791E+05 0.00000E+00


N8 6.70746E+04 0.00000E+00 -9.55967E+05 0.00000E+00 -4.73558E+06 0.00000E+00

N9 6.70746E+04 0.00000E+00 -9.55967E+05 0.00000E+00 -4.73558E+06 0.00000E+00


N9 5.45853E+04 0.00000E+00 -8.28746E+05 0.00000E+00 -8.28686E+06 0.00000E+00

N10 5.45853E+04 0.00000E+00 -8.28746E+05 0.00000E+00 -8.28686E+06 0.00000E+00


N10 4.16656E+04 0.00000E+00 -7.01530E+05 0.00000E+00 -1.13319E+07 0.00000E+00

N11 4.16656E+04 0.00000E+00 -7.01530E+05 0.00000E+00 -1.13319E+07 0.00000E+00


N11 2.93352E+04 0.00000E+00 -5.74242E+05 0.00000E+00 -1.38704E+07 0.00000E+00

N12 2.93352E+04 0.00000E+00 -5.74242E+05 0.00000E+00 -1.38704E+07 0.00000E+00


N12 1.84455E+04 0.00000E+00 -4.46834E+05 0.00000E+00 -1.59022E+07 0.00000E+00

N13 1.84455E+04 0.00000E+00 -4.46834E+05 0.00000E+00 -1.59022E+07 0.00000E+00


N13 9.67891E+03 0.00000E+00 -3.19293E+05 0.00000E+00 -1.74267E+07 0.00000E+00

N14 9.67891E+03 0.00000E+00 -3.19293E+05 0.00000E+00 -1.74267E+07 0.00000E+00


N14 3.54878E+03 0.00000E+00 -1.91631E+05 0.00000E+00 -1.84433E+07 0.00000E+00

N15 3.54878E+03 0.00000E+00 -1.91631E+05 0.00000E+00 -1.84433E+07 0.00000E+00


N15 3.97887E+02 0.00000E+00 -6.38866E+04 0.00000E+00 -1.89518E+07 0.00000E+00

N16 3.97887E+02 0.00000E+00 -6.38866E+04 0.00000E+00 -1.89518E+07 0.00000E+00

.......

Table 5: Free ends internal action

15



Code_Aster Theory

Middle deflection PUO_D_T_GD -2.79112E+00Middle deflection PUO_D_D -2.83430E+00

End Vertical reaction 1.91664E+06End Moment reaction -3.80451E+07Maximum Axial Force 8.92228E+04

Restrained Axial Ends

For the present case the center line beam deflection is listed below:




N1 2.79796E-32 0.00000E+00 2.51703E-35 0.00000E+00 0.00000E+00 0.00000E+00

N2 1.20714E-05 0.00000E+00 -7.34947E-03 0.00000E+00 3.45595E-03 0.00000E+00

N3 -1.67498E-06 0.00000E+00 -2.35151E-02 0.00000E+00 4.54176E-03 0.00000E+00

N4 -2.58435E-05 0.00000E+00 -4.20760E-02 0.00000E+00 4.70555E-03 0.00000E+00

N5 -4.95773E-05 0.00000E+00 -6.05342E-02 0.00000E+00 4.51065E-03 0.00000E+00

N6 -6.85372E-05 0.00000E+00 -7.79180E-02 0.00000E+00 4.17622E-03 0.00000E+00

N7 -8.14849E-05 0.00000E+00 -9.38494E-02 0.00000E+00 3.78751E-03 0.00000E+00

N8 -8.84148E-05 0.00000E+00 -1.08181E-01 0.00000E+00 3.37768E-03 0.00000E+00

N9 -8.97811E-05 0.00000E+00 -1.20857E-01 0.00000E+00 2.95963E-03 0.00000E+00

N10 -8.62049E-05 0.00000E+00 -1.31853E-01 0.00000E+00 2.53839E-03 0.00000E+00

N11 -7.83673E-05 0.00000E+00 -1.41162E-01 0.00000E+00 2.11590E-03 0.00000E+00

N12 -6.69701E-05 0.00000E+00 -1.48779E-01 0.00000E+00 1.69293E-03 0.00000E+00

N13 -5.27227E-05 0.00000E+00 -1.54705E-01 0.00000E+00 1.26977E-03 0.00000E+00

N14 -3.63367E-05 0.00000E+00 -1.58937E-01 0.00000E+00 8.46534E-04 0.00000E+00

N15 -1.85248E-05 0.00000E+00 -1.61477E-01 0.00000E+00 4.23272E-04 0.00000E+00

N16 8.39977E-18 0.00000E+00 -1.62323E-01 0.00000E+00 -2.85359E-18 0.00000E+00

N17 1.85248E-05 0.00000E+00 -1.61477E-01 0.00000E+00 -4.23272E-04 0.00000E+00

N18 3.63367E-05 0.00000E+00 -1.58937E-01 0.00000E+00 -8.46534E-04 0.00000E+00

N19 5.27227E-05 0.00000E+00 -1.54705E-01 0.00000E+00 -1.26977E-03 0.00000E+00

N20 6.69701E-05 0.00000E+00 -1.48779E-01 0.00000E+00 -1.69293E-03 0.00000E+00

N21 7.83673E-05 0.00000E+00 -1.41162E-01 0.00000E+00 -2.11590E-03 0.00000E+00

N22 8.62049E-05 0.00000E+00 -1.31853E-01 0.00000E+00 -2.53839E-03 0.00000E+00

N23 8.97811E-05 0.00000E+00 -1.20857E-01 0.00000E+00 -2.95963E-03 0.00000E+00

N24 8.84148E-05 0.00000E+00 -1.08181E-01 0.00000E+00 -3.37768E-03 0.00000E+00

N25 8.14849E-05 0.00000E+00 -9.38494E-02 0.00000E+00 -3.78751E-03 0.00000E+00

N26 6.85372E-05 0.00000E+00 -7.79180E-02 0.00000E+00 -4.17622E-03 0.00000E+00

N27 4.95773E-05 0.00000E+00 -6.05342E-02 0.00000E+00 -4.51065E-03 0.00000E+00

N28 2.58435E-05 0.00000E+00 -4.20760E-02 0.00000E+00 -4.70555E-03 0.00000E+00

N29 1.67498E-06 0.00000E+00 -2.35151E-02 0.00000E+00 -4.54176E-03 0.00000E+00

N30 -1.20714E-05 0.00000E+00 -7.34947E-03 0.00000E+00 -3.45595E-03 0.00000E+00

N31 1.11918E-31 0.00000E+00 -3.92288E-33 0.00000E+00 0.00000E+00 0.00000E+00

The ends reaction now show a considerable higher horizontal force due tothe axial restrain:

CHAMP AUX NOEUDS DE NOM SYMBOLIQUE REAC_NODANUMERO D’ORDRE: 10 INST: 1.00000E+00


N1 -3.01869E+08 0.00000E+00 1.91664E+06 0.00000E+00 -7.85448E+06 0.00000E+00

N2 -1.46627E-05 0.00000E+00 7.09115E-08 0.00000E+00 -2.70084E-08 0.00000E+00

........

N29 -2.25306E-05 0.00000E+00 -1.41757E-06 0.00000E+00 -2.60537E-07 0.00000E+00

16



N30 -1.17421E-05 0.00000E+00 6.56946E-07 0.00000E+00 7.60891E-07 0.00000E+00

N31 3.01869E+08 0.00000E+00 1.91664E+06 0.00000E+00 7.85448E+06 0.00000E+00

The same is for the nodal force:




N1 -3.01869E+08 0.00000E+00 1.85275E+06 0.00000E+00 -7.85448E+06 0.00000E+00

N2 -1.46627E-05 0.00000E+00 -1.27776E+05 0.00000E+00 -2.70084E-08 0.00000E+00

N3 9.35793E-06 0.00000E+00 -1.27776E+05 0.00000E+00 -2.25846E-08 0.00000E+00

N4 -6.67572E-06 0.00000E+00 -1.27776E+05 0.00000E+00 4.83997E-08 0.00000E+00

while the internal action are reported in table 6 where only the first half pipe internal action are reported due to results symmetry. As in the previoussection the same analysis has been performed using the POU_D_E elementgetting the following displacements:

------>




N1 -2.16840E-19 0.00000E+00 -2.68544E-36 0.00000E+00 3.58186E-36 0.00000E+00

N2 1.62289E-18 0.00000E+00 -4.70844E-02 0.00000E+00 2.27304E-02 0.00000E+00

N3 2.24193E-18 0.00000E+00 -1.75573E-01 0.00000E+00 4.07580E-02 0.00000E+00

N4 -1.15443E-18 0.00000E+00 -3.67326E-01 0.00000E+00 5.44186E-02 0.00000E+00

N5 2.47616E-18 0.00000E+00 -6.05547E-01 0.00000E+00 6.40483E-02 0.00000E+00

N6 3.09520E-18 0.00000E+00 -8.74785E-01 0.00000E+00 6.99828E-02 0.00000E+00

N7 2.71039E-18 0.00000E+00 -1.16093E+00 0.00000E+00 7.25582E-02 0.00000E+00

N8 3.32943E-18 0.00000E+00 -1.45122E+00 0.00000E+00 7.21103E-02 0.00000E+00

N9 3.94847E-18 0.00000E+00 -1.73423E+00 0.00000E+00 6.89751E-02 0.00000E+00

N10 4.56751E-18 0.00000E+00 -1.99988E+00 0.00000E+00 6.34884E-02 0.00000E+00

N11 5.18655E-18 0.00000E+00 -2.23945E+00 0.00000E+00 5.59862E-02 0.00000E+00

N12 5.80559E-18 0.00000E+00 -2.44554E+00 0.00000E+00 4.68045E-02 0.00000E+00N13 5.42079E-18 0.00000E+00 -2.61209E+00 0.00000E+00 3.62791E-02 0.00000E+00

N14 -9.87118E-19 0.00000E+00 -2.73442E+00 0.00000E+00 2.47459E-02 0.00000E+00

N15 -1.37193E-18 0.00000E+00 -2.80917E+00 0.00000E+00 1.25409E-02 0.00000E+00

N16 -7.77983E-18 0.00000E+00 -2.83430E+00 0.00000E+00 3.26499E-16 0.00000E+00

N17 -7.16079E-18 0.00000E+00 -2.80917E+00 0.00000E+00 -1.25409E-02 0.00000E+00

N18 -6.54175E-18 0.00000E+00 -2.73442E+00 0.00000E+00 -2.47459E-02 0.00000E+00

N19 -5.92271E-18 0.00000E+00 -2.61209E+00 0.00000E+00 -3.62791E-02 0.00000E+00

N20 -5.30367E-18 0.00000E+00 -2.44554E+00 0.00000E+00 -4.68045E-02 0.00000E+00

N21 -4.68463E-18 0.00000E+00 -2.23945E+00 0.00000E+00 -5.59862E-02 0.00000E+00

N22 -4.06559E-18 0.00000E+00 -1.99988E+00 0.00000E+00 -6.34884E-02 0.00000E+00

N23 -4.45040E-18 0.00000E+00 -1.73423E+00 0.00000E+00 -6.89751E-02 0.00000E+00

N24 -3.83136E-18 0.00000E+00 -1.45122E+00 0.00000E+00 -7.21103E-02 0.00000E+00

N25 -3.21232E-18 0.00000E+00 -1.16093E+00 0.00000E+00 -7.25582E-02 0.00000E+00

N26 -2.59328E-18 0.00000E+00 -8.74785E-01 0.00000E+00 -6.99828E-02 0.00000E+00

N27 -1.97424E-18 0.00000E+00 -6.05547E-01 0.00000E+00 -6.40483E-02 0.00000E+00

N28 -1.35520E-18 0.00000E+00 -3.67326E-01 0.00000E+00 -5.44186E-02 0.00000E+00N29 -1.74001E-18 0.00000E+00 -1.75573E-01 0.00000E+00 -4.07580E-02 0.00000E+00

N30 -1.12096E-18 0.00000E+00 -4.70844E-02 0.00000E+00 -2.27304E-02 0.00000E+00

N31 -2.16840E-19 0.00000E+00 -1.05118E-35 0.00000E+00 -1.48809E-35 0.00000E+00

and as it would be expected no difference with the free ebds case in the

17



vertical plane has been evidenced. The whole data can be summarized in thefollowing table:

Code_Aster Theory

Middle deflection POU_D_T_GD -1.62323E-01Middle deflection POU_D_T_GD -2.83430E+00

End Vertical reaction 1.91664E+06End Moment reaction -7.85448E+06Maximum Axial Force 3.01872E+08

ANSYS

For the present case the command SSTIFF,ON NLGEOM,ON is used to con-sider the large rotation and small strain case using the submerged dedicated

pipe element PIPE59.

Free Ends

The results are reported in terms of axial force, bending moment and verticaldisplacements respectively in figures (6), (7) and (8) while the reactions at endsare listed below:

***** POST1 TOTAL REACTION SOLUTION LISTING *****

LOAD STEP= 1 SUBSTEP= 8

TIME= 1.0000 LOAD CASE= 0

THE FOLLOWING X,Y,Z SOLUTIONS ARE IN THE GLOBAL COORDINATE SYSTEM

NODE FX FY FZ MX MY MZ

1 0.0000 0.0000 0.17584E+07 0.0000 -0.35128E+08 0.0000

121 0.0000 0.17584E+07 0.35128E+08 0.0000

TOTAL VALUES

VALUE 0.0000 0.0000 0.35168E+07 0.0000 0.48429E-06 0.0000

The following summary table can be written:

ANSYS Theory

Middle deflection -2.586End Vertical reaction 0.17584E+07End Moment reaction 0.35128E+08Maximum Axial Force 7.6905E+04

Axially Restrained Ends

In the case of the restrained ends it is possible to compare the numerical solutionwith an analytical one. The governing eqution are:

EJ · yIV −N · yII = q (x)

18






N1 3.01872E+08 0.00000E+00 -1.33112E+06 0.00000E+00 5.25827E+06 0.00000E+00

N2 3.01872E+08 0.00000E+00 -1.33112E+06 0.00000E+00 5.25827E+06 0.00000E+00


N2 3.01874E+08 0.00000E+00 -5.17830E+05 0.00000E+00 1.65208E+06 0.00000E+00

N3 3.01874E+08 0.00000E+00 -5.17830E+05 0.00000E+00 1.65208E+06 0.00000E+00


N3 3.01873E+08 0.00000E+00 -2.01445E+05 0.00000E+00 2.49206E+05 0.00000E+00

N4 3.01873E+08 0.00000E+00 -2.01445E+05 0.00000E+00 2.49206E+05 0.00000E+00


N4 3.01873E+08 0.00000E+00 -7.83660E+04 0.00000E+00 -2.96536E+05 0.00000E+00

N5 3.01873E+08 0.00000E+00 -7.83660E+04 0.00000E+00 -2.96536E+05 0.00000E+00


N5 3.01872E+08 0.00000E+00 -3.04861E+04 0.00000E+00 -5.08841E+05 0.00000E+00

N6 3.01872E+08 0.00000E+00 -3.04861E+04 0.00000E+00 -5.08841E+05 0.00000E+00


N6 3.01872E+08 0.00000E+00 -1.18600E+04 0.00000E+00 -5.91432E+05 0.00000E+00

N7 3.01872E+08 0.00000E+00 -1.18600E+04 0.00000E+00 -5.91432E+05 0.00000E+00


N7 3.01871E+08 0.00000E+00 -4.61403E+03 0.00000E+00 -6.23563E+05 0.00000E+00

N8 3.01871E+08 0.00000E+00 -4.61403E+03 0.00000E+00 -6.23563E+05 0.00000E+00


N8 3.01871E+08 0.00000E+00 -1.79519E+03 0.00000E+00 -6.36063E+05 0.00000E+00

N9 3.01871E+08 0.00000E+00 -1.79519E+03 0.00000E+00 -6.36063E+05 0.00000E+00


N9 3.01870E+08 0.00000E+00 -6.98573E+02 0.00000E+00 -6.40927E+05 0.00000E+00

N10 3.01870E+08 0.00000E+00 -6.98573E+02 0.00000E+00 -6.40927E+05 0.00000E+00


N10 3.01870E+08 0.00000E+00 -2.71927E+02 0.00000E+00 -6.42820E+05 0.00000E+00

N11 3.01870E+08 0.00000E+00 -2.71927E+02 0.00000E+00 -6.42820E+05 0.00000E+00


N11 3.01870E+08 0.00000E+00 -1.05906E+02 0.00000E+00 -6.43557E+05 0.00000E+00

N12 3.01870E+08 0.00000E+00 -1.05906E+02 0.00000E+00 -6.43557E+05 0.00000E+00


N12 3.01870E+08 0.00000E+00 -4.12561E+01 0.00000E+00 -6.43844E+05 0.00000E+00

N13 3.01870E+08 0.00000E+00 -4.12561E+01 0.00000E+00 -6.43844E+05 0.00000E+00


N13 3.01869E+08 0.00000E+00 -1.59959E+01 0.00000E+00 -6.43956E+05 0.00000E+00

N14 3.01869E+08 0.00000E+00 -1.59959E+01 0.00000E+00 -6.43956E+05 0.00000E+00


N14 3.01869E+08 0.00000E+00 -5.94313E+00 0.00000E+00 -6.43999E+05 0.00000E+00

N15 3.01869E+08 0.00000E+00 -5.94313E+00 0.00000E+00 -6.43999E+05 0.00000E+00


N15 3.01869E+08 0.00000E+00 -1.50812E+00 0.00000E+00 -6.44013E+05 0.00000E+00

N16 3.01869E+08 0.00000E+00 -1.50812E+00 0.00000E+00 -6.44013E+05 0.00000E+00

.....

Table 6: Restrained ends internal action

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Figure 6: Free end axial internal force graph


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whose solution, in the case of constant distribuited load, is:

y(x) = A + B · x + C · sinh(

N

EJ · x) + D · cosh(

N

EJ · x) +

q · x

2N

where N is the axial force.The restrained axial ends results are reported in the figures respectively for

the axial force, bending moment and vertical displacements. The reaction forcesfor the ends are listed below:

PRINT REACTION SOLUTIONS PER NODE

***** POST1 TOTAL REACTION SOLUTION LISTING *****

LOAD STEP= 1 SUBSTEP= 7

TIME= 1.0000 LOAD CASE= 0

THE FOLLOWING X,Y,Z SOLUTIONS ARE IN THE GLOBAL COORDINATE SYSTEM

NODE FX FY FZ MX MY MZ

1 -0.30183E+09 0.0000 0.17584E+07 0.0000 -0.73074E+07 0.0000

121 0.30183E+09 0.0000 0.17584E+07 0.73074E+07 0.0000

TOTAL VALUES

VALUE 0.0000 0.0000 0.35168E+07 0.0000 -0.55879E-08 0.0000

The following summary table can be written:

ANSYS Theory

Middle deflection -.149413 -.16196761End Vertical reaction 0.17584E+07 1916635.2End Moment reaction 0.734E+07 7961868.Maximum Axial Force 3.02E+08

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Conclusion

At the conclusion of the simple test example the following consideration can bedrawn:

1. ANSYS is able to correctly handle the effect of internal and external pres-sure giving rise to the correct stress distribution on the axial, radial andcircumferential directions in both, cap and open ends;

2. The two different condition of with and without cap have no effect on thestress on the pipe steel section on the case of restrained ends, the axialtotal stress is the same in the two conditions what is different it is thereaction absorbed by the constrains

3. Code_Aster can simulate the condition of open ends leading to a cor-rect stress state for both, fully restrained and simply supported boundaryconditions if the pressure effects are induced by an equivalent thermal

expansion load

4. Code_Aster, in the condition of cap ends, cannot represent correctly thestress state and some pulling force have to be fictitiously applied to gen-

erate the correct stress state5. Code_Aster, in condition of isostatic structure, as in the simply supported

one, obtains the correct stress state if the pressure effect is represented bya thermal load in the open ends situation.

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Free EndCode_Aster ANSYS

Middle deflection -2.79112E+00 -2.586End Vertical reaction 1.91664E+06 0.17584E+07End Moment reaction -3.80451E+07 0.35128E+08Maximum Axial Force 8.92228E+04 7.6905E+04

Axial RestrainCode_Aster ANSYS Theory

Middle deflection -1.62323E-01 -.149413 -.16196761End Vertical reaction 1.91664E+06 0.17584E+07 1916635.2End Moment reaction -7.85448E+06 0.734E+07 7961868.Maximum Axial Force 3.01872E+08 3.02E+08 3.02E+08

Table 7: Final comparing table

6. For the flexural examples the results are compared in table (7) where itis possible to highlight how ANSYS results are stiffer then the ones of Code_Aster.

References

[1] ANSYS® Academic Research, Release 12.1, Help System, Coupled FieldAnalysis Guide, ANSYS, Inc.

[2] Code_Aster Documentation, Modélisation statique et dy-

namique des poutres en grandes rotations, http://www.code-aster.org/V2/doc/default/man_r/r5/r5.03.40.pdf

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pressure effect on a pipe

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