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| {{#Wiki_filter:SOLVIA Verification Manual Nonlinear Examples EXAMPLE B68 UNCONFINED SEEPAGE FLOW THROUGH A GRAVITY DAM Objective To perform an unconfined seepage flow analysis using PLANE conduction elements. | | {{#Wiki_filter:}} |
| Physical Problem The steady-state free surface seepage through the dam shown in the figure below is considered. The dam material is isotropic with permeability constants ky = k, = 2.5 ft/hr.
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| Finite Element Model The figures on page B68.2 show the finite element mesh employed. The permeability matrix is evaluated using (4 x 4) Gauss integration. The bottom figure shows a contour plot of the distortion of the elements expressed in degrees.
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| Solution Results The input data on page B68.4 has been used for the solution. A contour plot of the pressure is shown in the top figure on page B68.3. A good comparison with the free surface solution in [1], page 223 (full line in figure above) can be observed. A vector plot of the flux is shown in the bottom figure on page B68.3.
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| Reference
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| [1] Harr, M.E., Groundwater and seepage, McGraw Hill, New York, 1962.
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| Version 99.0 B68.1
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| SOLVIA Verification Manual Nonlinear Examples B68 UNCONFINED SEEPAGE FLOW THROUGH A GRAVITY DAM ORIGINAL *
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| * Z SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB B68 UNCONFINED SEEPAGE FLOW THROUGH A GRAVITY DAM ORIGJNAL 1i. L r, V 6' '8 I k Ii S~DISTOR7TON
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| 'lAX 4S*O
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| / I 42 273 S36. 8 ' 8 S3'.363 25 908
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| '-2453 S9.55,129
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| .0880 IN 1 .360S SCLVIA-PRE 99.0 CLVA ENGINEERING A3 Version 99.0 B68.2
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| SOLVIA Verification Manual Nonlinear Examples B68 NNC3NFINED SEEPAGI -LOW THROUGH A GRAVITY DAY OR<ETNAL z T-"E I PRESSURE MAX 0.C0003 J .37,174 S0.32416 0.27359 0o.2230:
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| S.17243 S0.12:86 0 .071279
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| ., C. 20702 MIN-4.5864E-3 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B68 UNCONFINED SEEPAGE FLOW THROUGH A GRAVITY DAM ORIGINAL I I TIME 1 z
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| I-SEEPACE F UX L.A486, ECL.','A-POST 99 SOLVI/A ENGINEERING AB:,
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| I Version 99.0 B68.3
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B68 UNCONFINED SEEPAGE FLOW THROUGH A GRAVITY DAM' DATABASE CREATE COORDINATES / ENTRIES NODE Y Z 1 / 2 12.64 / 3 16. / 4 12. 4. / 5 4. 4.
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| LINE STRAIGHT 1 2 EL*f6 MIDNODES=I LINE STRAIGHT 2 3 EL=4 MIDNODES=1 LINE COMBINED 1 3 2 LINE STRAIGHT 1 5 EL-5 MIDNODES=1 RATIO-i.
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| T-MATERIAL 1 SEEPAGE PERMEABILITY=2.5 GAMMA-.1 EGROUP 1 PLANE STRAIN INT=4 GSURFACE 1 3 4 5 EL1=20 EL2=5 NODES=8 T-LOADS SEEPAGEHEADS INPUT=LINES 1 5 4.
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| 2 3 0.
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| T-LOADS SEEPAGEHEADS INPUT=NODES DELETE 2 MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES MESH NSYMBOLS=MYNODES CONTOUR-DISTORTION SOLVIA-TEMP END SOLVIA-POST input
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| * B68 UNCONFINED SEEPAGE FLOW THROUGH A GRAVITY DAM T-DATABASE CREATE WRITE FILENAME='b68.1is' SET OUTLINE=YES NSYMBOLS=MYNODES MESH CONTOUR=TPRESSURE MESH VECTOR=TFLUX EMAX NUMBER=3 END Version 99.0 1368.4
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B69 SEMI-INFINITE REGION UNDERGOING TWO PHASE CHANGES Objective To verify the TRUSS conduction element for analysis of phase changes.
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| Physical Problem The figure below shows the semi-infinite region considered. The initial temperature within the domain is To = 100' and the surface temperature is prescribed to be T, = 0'. The first and second phase change temperature interval is 0'. The thermal conductivity and specific heat are constant between the temperature 0' and 100'.
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| A00 k = 1.0 thermal conductivity C = 1.0 specific heat per Ts = 0° T0 =1000- 0 unit volume y00 Finite Element Model The figure on page B69.2 shows the used finite element model which consists of forty TRUSS conduction elements. The temperature degree of freedom at node 1 is deleted to impose the zero surface temperature. The element conduction matrix is evaluated using one point integration and a lumped specific heat matrix is employed. The time integration is performed using the Euler backward method in 67 time steps.
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| Version 99.0 B69.1
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| SOLVIA Verification Manual Nonlinear Examples Solution Results The numerical solution results are obtained by using the input data on pages B69.4 and B69.5. The temperature distributions within the domain are given in the top figure on page B69.3 for the solution times 0.2, 0.4 and 1.0 and are compared with the corresponding analytical solutions ([1] page 290).
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| The heat flux in the element at time 1.0 is shown in the bottom figure on page B69.3.
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| User Hints
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| * The material conductivity and specific heat are assumed to remain constant in the entire range of temperatures from 0' to 1000. As a result, the semi-infinite domain is discretized using linear TRUSS conduction elements with constant material properties.
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| Reference
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| [1] Carslaw, H.S., and Jaeger, J.C., Conduction of Heat in Solids, 2nd Ed., Oxford University Press, Inc., N.Y., 1959.
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| B69 SE,'I-iNFTNITE REG-LON UNDERGOING T.40 PHASE -HANGES ORIG:NAL .-- 0.S z i
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| i I---- 1-11 --- I ---- I I - I -- - - - - I --- --
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| ORIG NAL -- 0 z
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| -X SCL"/A-PRE 99.0 SGLV:A \N' ThORNAB Version 99.0 B69.2
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| SOLVIA Verification Manual Nonlinear Examples B69 SE>- INF!N 9EGION UNDERGO 3 0>0G O PHASE CHANGz ST.............
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| .. . .: .. ( .. ...... . ..... . .............
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| 'D 1 2* K 7 /*
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| (% x 2W (01,,>
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| 0 6 7 3 TAN' SOLV A-POST 99.0 SOLVEA ENG:NEERING AB I
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| 369 SEMd-INFINITE UEGION UNDERGOING TW; P-AS CHANGES f.
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| 2 SOLVIA-POST 99 CL/IA ENGC7 EEI\NG AB Version 99.0 B69.3
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B69 SEMI-INFINITE REGION UNDERGOING TWO PHASE CHANGES' DATABASE CREATE T-MASTER NSTEP=30 DT=0.004 20 0.01 17 0.04 T-ANALYSIS TRANSIENT HEATMATRIX=LUMPED METHOD=BACKWARD-EULER T-ITERATION TTOL=i.E-6 ITEMAX=i00 COORDINATES 1 0. TO 2 8.
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| T-MATERIAL 1 CONDUCTION K=i. SPECIFICHEAT=i. LH1=80. LH2=30.
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| LINE STRAIGHT 1 2 EL=-00 RATIO=20 EGROUP 1 TRUSS GLINE 1 2 EL-f00 EDATA / 1 1.
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| T-INITIAL TEMPERATURE INPUT=LINE 1 2 100.
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| T-PHASETRANSITIONS
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| : 50. 0. 0
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| : 90. 0. 0 FIXBOUNDARIES TEMPERATURE / 1 SET NSYMBOL=YES VIEW=-Y MESH NNUMBERS=MYNODES SUBFRAME=12 MESH BCODE=TEMPERATURE SOLVIA-TEMP END SOLVIA-POST input
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| * B69 SEMI-INFINITE REGION UNDERGOING TWO PHASE CHANGES T-DATABASE CREATE WRITE FILENAME='b69.1is' AXIS ID=i VMIN-0. VMAX=8. LABEL='DISTANCE' AXIS ID-2 VMIN=0. VMAX=i00. LABEL='TEMPERATURE' USERCURVE 1 / READ 'B69TI02.DAT' USERCURVE 2 / READ 'B69TI04.DAT' USERCURVE 3 / READ 'B69TI10.DAT' NPLINE NAME=NODES / 1 101 TO 199 2 SET DIAGRAM=GRID NLINE LINENAME=NODES TIME=0.2 XAXIS=1 YAXIS=2 SYMBOL=i SET SUBFRAME=OLD NLINE LINENAME=NODES TIME=0.4 XAXIS=-i YAXIS=-2 SYMBOL=2 NLINE LINENAME=NODES TIME=i.0 XAXIS=-i YAXIS=-2 SYMBOL=4 PLOT USERCURVE 1 XAXIS=-i YAXIS=-2 PLOT USERCURVE 2 XAXIS=-i YAXIS=-2 PLOT USERCURVE 3 XAXIS=-I YAXIS=-2 EPLINE NAME=TRUSS / 1 1 TO 100 1 ELINE LINENAME=TRUSS KIND:TFLUX OUTPUT=ALL SUBFRAME=NEXT END Version 99.0 1369.4
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B70 USER-SUPPLIED MATERIAL MODEL (RAMBERG-OSGOOD)
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| Objective To demonstrate the implementation of the Ramberg-Osgood constitutive law as a user-supplied material model for SOLID elements.
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| Physical Problem A cantilever structure subjected to uniaxial compression is considered as shown in figure below.
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| I .- y Po Mi)
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| The chosen Ramberg-Osgood constitutive law constants:
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| 'ID E=3x 107 v=0.30 4
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| a0 =6x10 S=0.05 n =10.0 Tolerance for effective stress iterations = 101° Iteration limit for effective stress iterations = 25 In the uniaxial load case the stress-strain relation can be described by
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| - GO i (0
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| =-+p Lgo ) "
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| where so = yield strain O = yield stress material constant n = power hardening exponent Material Model The FORTRAN source subroutine used to implement the constitutive law is given on pages B170.5 to B170.9.
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| Finite Element Model The cantilever structure is discretized using 4 twenty-node SOLID elements, see figure on page B70.2. The solution is obtained using the full Newton iteration with line searches.
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| Solution Results Using the input data given on page B170.4 the obtained numerical solution is shown in the figures on page B70.3. The obtained solution agrees well with the corresponding analytical solution.
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| Version 99.0 B70.1
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| SOLVIA Verification Manual Nonlinear Examples B70 USER-SUPP.IED MATER AL M0DEL RAMBEoG-OSSOOC CRI3NA TIME I PRESSURE 30000 MASTER 000 11 H 111111 SOLV:A-PRE 99.0 SOLViA ENGINEERING AB 370 USER-SUPPLIED MATERIAL MODEL RAMBERG-CSGOOD ORIGINAL - - I.
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| z MAX D:$PL. 0-
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| : 0. 4037 TIME A K Y Y-D-R DISPL.
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| MlAX 0
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| -8. 766SE-3
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| -0.02630C
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| "-0.06'.366 0 043833
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| -. 0378899
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| -0.3964329
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| -. 1 1396
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| )
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| MrNAKC2 0 . IA-IOST 9 CLIA ENGINEERING AS Version 99.0 B70.2
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| SOLVIA Verification Manual Nonlinear Examples B7C LSE-S-J PI - 77 ATR A~IL'ICFE ýANBEPG-OSGOOD
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| ;701
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| -Ij 0
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| 0 20030 400330 60000 8C003 3. -8.0 - . -4, -2.0 0 -1 VA~j C- 77NFF Nýý -C I E STPATN-YY SOLVIA-POST 99.0 SLVIA ENCINEE.NR7 Version 99.0B0. B70.3
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B70 USER-SUPPLIED MATERIAL MODEL ( RAMBERG-OSGOOD DATABASE CREATE MASTER IDOF=000111 NSTEP=4 TOLERANCES TYPE=F RNORM=0.6 ITERATION METHOD=FULL-NEWTON LINE-SEARCH=YES TIMEFUNCTION 1 0 0 / I 3E4 / 2 6E4 / 4 9E4 COORDINATES 1 0. 16. / 2 / 3 1. / 4 1, 16.
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| NGENERATION NSTEP=4 ZSTEP=-- / 1 TO 4 MATERIAL 1 USER-SUPPLIED CTIT=3.E7 CTI2-0.3 CTI3=6.E4, CTI4=0.05 CTIS=10 SCPT=1.E-10 SCP2=25 INTER=10
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| -200 0 0 0 0 0 0 200 0 0 0 0 0 0 EGROUP 1 SOLID GVOLUME 1 2 3 4 5 6 7 8 EL1=4 EL2=1 EL3=1 NODES=20 LOADS ELEMENT INPUT=SURFACE 1 4 5 8 1.
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| FIXBOUNDARIES 1 INPUT=LINES / 5 6 / 1 2 / 2 6 FIXBOUNDARIES 2 INPUT=SURFACE / 2 3 6 7 FIXBOUNDARIES 3 INPUT=LINES / 3 4 / 1 2 / 2 3 LOADS TEMPERATURE TREF=0.
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| VECTOR LENGTH=0.5 MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES BCODE=ALL VECTOR=LOAD SOLVIA END SOLVIA-POST input
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| * B70 USER-SUPPLIED MATERIAL MODEL ( RAMBERG-OSGOOD DATABASE CREATE WRITE FILENAME='b70.lis' MESH ORIGINAL=DASH DMAX=1 CONTOUR=DY NHISTORY NODE=1 DIRECTION=2 OUTPUT=ALL XVARIABLE=1 SUBFRAME=21 EXYPLOT ELEMENT=i POINT=1 XKIND=EYY YKIND=SYY OUTPUT=ALL END Version 99.0 B70.4
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| SOLVIA Verification Manual Nonlinear Examples SUBROUTINE CUSER3 ( KTR, STRESS, STRAIN, SIG, EPS, DEPS, DEPST, 1 THSTRi, THSTR2, SCP, ICP, ARRAY, IARRAY, D, 2 ALFA, CTD, ALFAA, CTDD, CTI, CTJ, PROP, 3 TMPI, TMP2, TIME, DT, INTER, KEY )
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| C* --- ---------------------------------------------------- SOLVIA CUSER3 C*I C*I USER-SUPPLIED MATERIAL MODEL FOR SOLID ELEMENT C*I C*I This subroutine is to be supplied by the USER to calculate C*I the STRESSES and CONSTITUTIVE MATRIX of a special material.
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| C*I C*I This subroutine is called in USER3 for each INTEGRATION POINT C*I for each SOLID element to perform the following operations C*I C*I KEY = 1 Initialize the working arrays during INPUT phase C*I KEY = 2 Calculate element stresses C*I KEY = 3 Calculate the stress-strain matrix C*I KEY = 4 Print element stresses and other desired variables C*I C*I VARIABLES C*I C*I NUMBEL Element number C*I IPT Integration point number C*I STRESS(6) Stress components, received at time TAU and C*I updated by USER-SUPPLIED coding to correspond C*I to time TAU + DTAU C*I STRAIN(6) Total strain components at time T + DT C*I EPS(6) Total strain components at time T C*I DEPS(6) Subdivided incremental strain components C*I DEPS(I) = ( STRAIN(I) - EPS(I) )/INTER - DEPST(I)
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| C*I ( Passed to subroutine CUSER3 by SOLVIA program )
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| C*I DEPST(6) Components of subincremental thermal strain C*I ( Passed to subroutine CUSER3 by SOLVIA program )
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| C*I THSTRI(6) Total thermal strain at time TAU C*I THSTR2(6) Total thermal strain at time TAU + DTAU C*I INTER Number of subdivisions for the strain increments C*I KTR Current subdivision number C*I EQ. 1, First subdivision C*I EQ. INTER, Last subdivision C*I SCP(5) User-defined solution control parameters C*I ICP(5) User-defined solution control parameters C*I ARRAY(23) Working array ( REAL ), received at time TAU and C*I updated by USER-SUPPLIED coding to correspond to C*I time TAU - DTAU C*I IARRAY(2) Working array ( INTEGER ), received at time TAU C*I and updated by USER-SUPPLIED coding to correspond C*I to time TAU + DTAU C*I D(6,6) Stress-strain matrix, to be calculated by USER C*I SUPPLIED coding C*I ALFA Coefficient of thermal expansion at time TAU C*I CTD(5) Temperature-dependent material constants at C*I time TAU C*I ALFAA Coefficient of thermal expansion at time C*I TAU + DTAU C*I CTDD(5) Temperature-dependent material constants at C*I time TAU + DTAU C*I CTI(8) User-defined constant material properties C*I CTJ(8) User-defined constant material properties C*I PROP(16,*) User-defined material property table C*I TMPI Temperature at integration point at time TAU C*I TMP2 Temperature at integration point at time C*I TAU + DTAU C*I TIME Time at current step, T + DT C*I DT Time step increment, DT C*I Version 99.0 B70.5
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| SOLVIA Verification Manual Nonlinear Examples C*I Note that the variables passed to the subroutine CUSER3 C*I when KEY = 3, 4 are these calculated last; i.e., calculated C*I in the last subdivision when KEY = 2. Hence, these variables C*I are not calculated when KEY = 3, 4.
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| C*I C*I Note also that when a mixed formulation is used, C*I PRESSURE-INTERPOLATION = YES in EGROUP command, the stress C*I strain matrix D is used in the stress calculation, KEY = 2, C*I in the SOLVIA program. The user should keep the D-matrix C*I updated during the integration of stresses for a correct C*I pressure interpolation.
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| C*I C*I There are no controls or error checking of user input data C*I for the User-Supplied material model in SOLVIA-PRE.
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| C*I IMPLICIT DOUBLE PRECISION ( A-H, O-Z COMMON /COMTEM/ RTEMP(100), JTEMP(200)
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| COMMON /EL/ IND,ICOUNT,NPAR(30),NUMEG,NEGL,NEGNL,IMASS,IDAMP, 1 ISTATNDOF,KLINIEIG,IMASSNIDAMPN COMMON /VAR/ NG,MODEX,IUPDTKSTEP,ITEMAX, IEQREF,ITE,KPRI, 1 IREF,IEQUIT, IPRI,KPLOTN,KPLOTE COMMON /ELDATA/ NUMBEL,NEL,IPT,IELD,ND,IPS,ISVE,IPRNT COMMON /DISDR/ DISD(9)
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| COMMON /THREED/ DE,NPT,IDW,NND9,ISOCOR COMMON /TAPES/ IIN,IOUT COMMON /INCOMP/ CUDIS(6,6),CUPRE(6),PRESEL(4),PPROJ,PKAPPA,PREDI, 1 IDINCONPRESS,NPRACC,NPROLD,N3C C
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| DIMENSION STRESS(*), STRAIN(*), SIG(*), EPS(*), DEPS(*),
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| 1 DEPST(*), THSTRI(*), THSTR2(*), SCP(5), ICP(5),
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| 2 ARRAY(23), IARRAY(2), D(6,*), CTD(5), CTDD(5),
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| 3 CTI(8), CTJ(8), PROP(16,*)
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| GOTO ( 100, 200, 300, 400 ), KEY C*I C*I C*I K EY 1 C*I C*I Initialize components of real and integer working arrays, C*I ARRAY(23) and IARRAY(2)
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| C*I 100 CONTINUE C*I C*I I N S E R T U S E R - S U P P L I E D C O D I N G C*I GOTO 900 C*I C*I C*I K E Y = 2 C*I C*I Integration of element stresses, STRESS C*I 200 CONTINUE C*I C*I **I N S E R T U S E R -S U P P L I E D C O D I N G C*I C*I Ramberg-Osgood material C*I C*I CTI( 1 ) Youngs modulus C*I CTI( 2 ) Poissons ratio C*I CTI( 3 ) Yield stress C*I CTI( 4 ) Beta, material parameter C*I CTI( 5 ) N, power hardening constant C*I SCP( 1 ) Tolerance for effective stress iterations C*I SCP( 2 ) Iteration limit for effective stress iterations Version 99.0 1370.6
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| SOLVIA Verification Manual Nonlinear Examples C*I ARRAY( 1 ) Ratio of effective stress to yield stress C*I ARRAY( 2 ) Total effective strain C*I ARRAY( 3 ) Constant in stress-strain matrix C*I ARRAY( 4 ) Constant in stress-strain matrix C*I ARRAY( 5 Constant in stress-strain matrix C*I TOLER = I.D-6 C*I YM = CTI( 1 )
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| PR = CTI( 2 YS = CTI( 3 BB = CTI( 4 )
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| XN = CTI( 5 TOLS = SCP( 1 IMAX = INT( SCP( 2 ) )
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| C*I Xl = 2.D0 / 9.D0 X2 = 2.D0 * ( 1.D0 - PR ) / 3 .00 XNl = XN - 1.D0 YE = YS / YM C*I C*I Effective strain C*I E12 = STRAIN( 1 ) - STRAIN( 2 ))
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| E23 = STRAIN( 2 ) - STRAIN( 3 )
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| E31 = STRAIN( 3 ) STRAIN( 1 E4 = STRAIN( 4 E5 = STRAIN( 5 E6 = STRAIN( 6 C*I EQE ( E12
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| * E12 + E23
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| * E23 E31
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| * E31 )
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| * X1 1 (E4 *E4 + E5 *E5 + E6 *E6 * *15.D0 IF ( EQE .LT. 1.D-20 ) GOTO 230 EQE = SQRT( EQE )
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| RE = EQE / YE RS ARRAY( 1 IF ( RS .LT. TOLER ) RS = TOLER C *I C *I Newton-Raphson iteration to obtain equivalent stress C2*I ITR = 0 210 F BB
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| * RS ** XN + X2
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| * RS - RE IF ( ABS( F ) .LT. TOLS ) GOTO 220 ITR ITR + i IF ( ITR GT. IMAX ) GOTO 800 FP - XN
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| * BB
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| * RS ** INI - X2 IF ( ABS( FP ) LT. TOLS ) GOTO 810 RS RS - F / FP IF ( RS .LT. TOLER ) RS = TOLER GOTO 210 C;*1 C 2*1 Element stresses C *I 220 ET - RS
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| * YS / EQE ARRAY( 1 ) - RS GOTO 240 C*I 230 ET = YM / X2 240 DVOL = YM/( 3.D* (l.D0 - 2.D0
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| * PR DDEV = ET
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| * X1 DD = DVOL + 2.D0
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| * DDEV DV = 1.5D0
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| * DDEV DF = DVOL - DDEV C*I Version 99.0 B70.7
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| SOLVIA Verification Manual Nonlinear Examples ARRAY( 2 ) == EQE DD ARRAY( 3 )
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| ARRAY( 4 ) = DF ARRAY( 5 = DV C*I DO 250 I = 1, 6 DO 250 J = 1, 6 D( I, J ) 0.D0 250 CONTINUE C*I DO 260 I 1 3 1,
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| D( I, I ) = DD II = I + 3 D( II, II ) - DV 260 CONTINUE D( 1, 2 ) = DF D( 1, 3 ) = DF D( 2, 3 ) DF D( 2, 1 ) - DF D( 3, 1 ) = DF D( 3, 2 ) = DF C*I STRESS 1 = DD
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| * STRAIN( 1 DF * (STRAIN(2) + STRAIN(3)
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| STRESS 2 = DD
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| * STRAIN( 2 (STRAIN(l)
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| DF * + STRAIN(3) )
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| STRESS 3 = DD
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| * STRAIN( 3 DF * (STRAIN(l) + STRAIN(2)
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| STRESS( 4 = DV
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| * STRAIN( 4 STRESS 5 = DV
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| * STRAIN( 5 STRESS 6 = DV
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| * STRAIN( 6 C*I GOTO 900 C*I C*I Convergence failed C*I 800 WRITE IOUT, 8000 NPAR( 30 ), NUMBEL, IPT WRITE IOUT, 8100 ITR STOP C*I 810 WRITE IOUT, 8000 NPAR( 30 ), NUMBEL, IPT WRITE IOUT, 8200 STOP C*I C*I C*I K E Y = 3 C*I C*I Form constitutive law, stress-strain matrix D C*I 300 CONTINUE C*I C*I *
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| * INSERT U S E R - S U P P L I E D COD ING C*I 00 = ARRAY( 3 )
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| DF = ARRAY( 4 )
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| DV = ARRAY( 5 )
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| C*I DO 310 I = 1, 6 DO 310 J = 1, 6 D( I, J 0.D0 310 CONTINUE C*I DO 320 I 1, 3 D( I, I ) DD II = I + 3 D( II, II ) DV Version 99.0 B70.8
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| SOLVIA Verification Manual Nonlinear Examples 320 CONTINUE D( 1, 2 ) DF D( 1, 3 ) DF D( 2, 3 ) DF D( 2, 1 = DF D( 3, 1 = DF D( 3, 2 ) DF C*I GOTO 900 C*I C*I C*I K E Y = 4 C*I C*I Print of element results, STRESS and STRAIN, etc.
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| C*I 400 CONTINUE C*I C*I
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| * I N S E R T U S E R - S U P P L I E D C O D I N G C*I C*I C*I Current effective stress C*I EQS ARRAY( )*CTI( 3)
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| C*I C*I Print heading and element number C*I IF ( IPT .EQ. 1 ) WRITE ( IOUT, 2000 ) NUMBEL WRITE ( IOUT, 2100 ) IPT, ( STRESS( I ), I = 1, 6 ), EQS C*I GOTO 900 C*I 2000 FORMAT ( /, ' ELEMENT POINT STRESS-XX STRESS-YY', 5X, 1 'STRESS-ZZ STRESS-XY STRESS-XZ STRESS-YZ', 3X, 2 'EFF. STRESS', //, lX, 15 )
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| 2100 FORMAT C 12X, 13, 1X, 7( IX, IPEl3.5 8000 FORMAT ( /, ' ***ERROR: SOLVIA program STOPS', / 14X, 1 'EGROUP 4, 1', ' ( SOLID User-Supplied model )', / 14X, 2 'Element I, 5', ' Integration point = ', 13 8100 FORMAT ( 14X, 'Number of iterations = ', I5, / 14X, 1 'Max number of iterations reached in the effective stress 2 'iterations', / 14X, 'Subroutine CUSER3', / )
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| 8200 FORMAT ( 14X, 'Newton-Raphson iterations failed due to zero', /,
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| 1 14X, 'derivative in effective stress calculations', / 14X, 2 'Subroutine CUSER3', /
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| 900 RETURN C
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| END Version 99.0 B70.9
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B71 NATURAL FREQUENCIES OF A PRE-LOADED BEAM Objective To verify the change in natural frequencies of a beam due to pre-load.
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| Physical Problem A simply supported beam as shown in the figure below is considered. The beam is pre-loaded with an axial compressive load of 600 KN.
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| FF Fb L-= 5.Om p =7800kg,'m3 b=0.1m F=-600.103 N 2
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| E =2.0.10"' N/m Finite Element Model The model is shown in the top figure on page B71.3. It consists of twenty 2-node beam elements.
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| Solution Results The axial load is applied in the first run using the input data on page 1371.3. Then, the input data on page B71.4 is employed in a restart analysis to calculate the frequencies. The restart information on file SOLVIA08.DAT was saved in the first run and then used as starting conditions in the second run.
| |
| For comparison the same analyses were made using the same BEAM element model but with the GENERAL section. The section parameters were the same as for the rectangular section except that no shear deformation was included.
| |
| The theoretical value of the natural frequency, fl, for a beam without shear deformation is given in
| |
| [I], page 144.
| |
| -7I Ei2
| |
| .i 22-FC" where L = axial length I = area moment of inertia m = mass per unit length F = axial load, negative if the load is compressive Version 99.0 B71.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples Calculated frequencies (Hz):
| |
| Theory SQOLVIA SOLVIA with shear def. without shear def.
| |
| Frequency 1 2.726 2.709 2.722 Frequency 2 32.28 32.21 32.27 Frequency 3 78.36 78.03 78.35 The mode shapes are shown in the figure below and in the figure on page B71.3.
| |
| An analysis with no pre-load gave the following frequencies (Hz):
| |
| Theory SOLVIA SOLVIA with shear def. without shear def.
| |
| Frequency 1 9.185 9.181 9.185 Frequency 2 36.74 36.68 36.74 Frequency 3 82.66 82.36 82.66 Reference
| |
| [1] Blevins, Robert D., Formulas for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company, 1979.
| |
| 37' NATURAL AREQUENCIES OF A PRELCADED BEAN
| |
| 'lAX DISPL. ! SE-3 z TIIE I LOAD
| |
| £ES 3 'D'O11 C 0O11 RE.FERELNrc i O. Z CAX DS25L. - 3.37394 MO0DE ' FREO 2.7994 SOL!IA-CSOS 9. -LVIA _NG NEER !F G A,(
| |
| Version 99.0 B71.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples B/' NA-URAL -RE'zUENCIES CF A DRE< Aý BE AM' R EERENCE I 1< . z "MAX D CSPL C,07161s MCDE 2 FREC 32.207 z
| |
| IAX DCSP. P -0 010716!1 M1DE 3 FRED 78.033 SCLVIA-PGST 99.0 SOLVIA ENGINE-RING AB SOLVIA-PRE input, first run HEADING 'B71 NATURAL FREQUENCIES OF A PRELOADED BEAM'
| |
| * First run DATABASE CREATE MASTER IDOF=i00011 KINEMATICS DISPLACEMENT=LARGE ITERATION METHOD=BFGS COORDINATES 1 / 2 0. 5.
| |
| MATERIAL 1 ELASTIC E=2.Ell DENSITY=7800 EGROUP 1 BEAM SECTION 1 RECTANGULAR WTOP 0.1 D=0.1 BEAMVECTOR / 1 0. 0. 1.
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| GLINE 1 2 AUX=-1 EL=20 FIXBOUNDARIES 23 / !
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| FIXBOUNDARIES 3 / 2 LOADS CONCENTRATED 2 2 -600.E3 SOLVIA END Version 99.0 B71.3
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input, first run
| |
| * B71 NATURAL FREQUENCIES OF A PRELOAD BEAM
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| * First run DATABASE CREATE END SOLVIA-PRE input, second run
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| * B71 NATURAL FREQUENCIES OF A PRELOADED BEAM
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| * Second run DATABASE OPEN MASTER MODEX=RESTART TSTART=l.0 ANALYSIS TYPE-DYNAMIC MASSMATRIX=LUMPED FREQUENCIES SUBSPACE-ITERATION NEIG-3 SSTOL=i.E-10 SOLVIA END SOLVIA-POST input, second run
| |
| * B71 NATURAL FREQUENCIES OF A PRELOADED BEAM
| |
| * Second run DATABASE RESTART WRITE FILENAME='b7la.lis' FREQUENCIES EMAX SET NSYMBOLS-YES VIEW=X MESH NNUMBER-MYNODES BCODE=ALL VECTOR=LOAD SUBFRAME=12 SET RESPONSETYPE=VIBRATIONMODE ORIGINAL=YES MESH TIME=i MESH TIME=2 SUBFRAME=12 MESH TIME=3 END Version 99.0 1371.4
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| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B72 PLASTIC COLLAPSE OF A BEAM STRUCTURE, PIPE-SECTION Objective To verify the PIPE element in a collapse analysis using the automatic iteration method.
| |
| Physical Problem A plane frame with hinged support is subjected to a horizontal load, see figure below. The frame has a pipe section and the material is elastic-perfectly plastic.
| |
| F h = 1200 mm D = 120 mm t=2.5 mm 17% I D 5 E =2.0.10 N/mm 2
| |
| v=0.3 h 2 74l 7, Gy = 250 N/mm Finite Element Model Seven 2-node PIPE elements are used in the model. At the junctions of the horizontal and vertical members short elements are used to describe the plastic mechanism. A small displacement analysis is used. The model is shown in the left figure on page B72.2.
| |
| Solution Results The input data on page B72.3 is used to calculate the collapse load with eight integration points along the circumferential direction (TINT=8). In addition, results for TINT=12 and TINT=24 have been obtained for comparison.
| |
| Case Collapse load, KN Theory 15.00 SOLVIA TINT=8 14.23 SOLVIA TINT= 12 14.67 SOLVIA TINT=24 14.92 Deflection of node 6 as function of the load multiplier and a contour plot of the t-moment are shown in the right figure on page B72.2.
| |
| Version 99.0 B72.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples User Hints
| |
| * Note that the model used is intended for calculation of the limit load. For a more detailed analysis of the stress distribution a finer mesh needs to be used also in areas away from the locations of plastic mechanism.
| |
| 872 P5_ASTC CCLLAPSE OF A 0EAM STRUCTURE0 P'PE-SE5T'ON 872 PLAST/0C COL5APS OF A BEAM STRUCTURE. PIPE-SECTEON 3R8GINAL - 20. Z 1J @"
| |
| Y 7
| |
| ' 'S MAST-R 200 1000 6000 800 2000 12000 11000 i IC01 a -CAD MULTOI ER 0MBD1A l Y.l R3GINAL - 200 MAX D:SPL - 35.532 2 TiME TIME i0 41 6
| |
| -1 SMONMENT-T7 NO AVERAGTNG MAX 8.5374E6 7 .4703E6 3359E6
| |
| , S312O:SE6
| |
| : 3. 0672E
| |
| -t 0672E5
| |
| -3 20. SE 5.3359E5 7 4703E6 FORCE 71N-5.8374E5 SOLV0A-0RE 99 0 SOLV'A 08 ANOONERN SOLVIA-POST 99.0 SOLVIA ENGONEE'RNG A8 Version 99.0 B72.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B72 PLASTIC COLLAPSE OF A BEAM STRUCTURE, PIPE-SECTION' DATABASE CREATE MASTER IDOF=100011 NSTEP=10 AUTOMATIC-ITERATION NODE=3 DIRECTION=2 DISPLACEMENT=-0, CONTINUATION=YES DISPMAX=100 COORDINATES / ENTRIES NODE Y Z 1 / 2 0. 1100. / 3 0. 1200. / 4 100. 1200.
| |
| 5 1100. 1200. / 6 1200. 1200. / 7 1200. 1100.
| |
| 8 1200. / 9 600. 600.
| |
| MATERIAL 1 PLASTIC E=2.E5 NU=0.3 YIELD=250. ET=0.
| |
| EGROUP 1 PIPE TINT=8 RESULTS=FORCES SECTION 1 DIAMETER=120 THICKNESS=2.5 ENODES / 1 9 1 2 TO 7 9 7 8 FIXBOUNDARIES 23 / 1 8 FIXBOUNDARIES 234 / 9 LOADS CONCENTRATED / 3 2 1.
| |
| SET VIEW=X NSYMBOLS=YES PLOTORIENTATION-PORTRAIT MESH NNUMBERS=YES BCODE=ALL SUBFRAME-12 MESH ENUMBERS=YES VECTOR=LOAD SOLVIA END SOLVIA-POST input
| |
| * B72 PLASTIC COLLAPSE OF A BEAM STRUCTURE, PIPE-SECTION DATABASE CREATE WRITE FILENAME='b72.1is' SET PLOTORIENTATION=PORTRAIT VIEW=X SUBFRAME 12 NHISTORY NODE=6 DIRECTION=2 SYMBOL:1 OUTPUT=ALL CONTOUR AVERAGE=NO MESH CONTOUR:MT DMAX=TRUE END Version 99.0 B 72. 3
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B73 PLASTIC COLLAPSE OF A BEAM STRUCTURE, BOX-SECTION Objective To verify the ISOBEAM element and RIGIDLINK in a collapse analysis using the AUTOMATIC ITERATION method.
| |
| Physical Problem A plane frame with hinged support is subjected to a horizontal load, see figure below. The frame has a uniform box cross-section and the material is elastic-perfectly plastic.
| |
| F
| |
| - Rigid link Beam section 1 Center node h " Beam section 2 I74 h 74 Geometry Material 5 2 h = 2000 mm E = 2.0-10 N/mm b = 100 mm v =0.3 2
| |
| t=2 mm = 250 N/mm
| |
| -y Finite Element Model The 2-node ISOBEAM element and RIGIDLINK are used in the model. Each side of the box section is modeled using an ISOBEAM and connected to the center of the section with a RIGIDLINK. At the junctions of the horizontal and vertical members short elements are used to describe the plastic mechanism. A small displacement analysis is used. The model is shown in the figures on page B73.2.
| |
| Solution Results The input data on pages B73.3 and B73.4 is used. The obtained SOLVIA collapse load is compared with the analytical solution.
| |
| Collapse load, N SOLVIA 7366 Theory 7204 The deflection at node 17 as a function of the load multiplier and the deformed mesh with the applied load are shown in the figure on page B73.3.
| |
| Version 99.0 1373.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples User Hints
| |
| "* As in Example B72 short elements are used only at the locations of plastic mechanism. For a detailed analysis of the stress distribution a fine mesh needs to be used for the entire model.
| |
| "* Note that some other common cross-sections may be modeled using isobeam elements and rigid links.
| |
| B73 PLASTIC COLLAPSE OF A BEAM TRJUCTURE. 30X-SE:77ON 373 PLASTIC COLLAPSE OF A BEAM STRUCTURE, BOX SEC-ION 0RIGINAL - - *00. Z 0R7GINAL - 00 Z E'
| |
| C. L ZONE A2 V U!
| |
| 0ROGINAL - C00 Z ORIGINAL - oo.
| |
| ZONE A. LK ZONE A3
| |
| ,ý7 90 V 9-V 327 SOLVWA,-RE 990 SOL.AUNOINEI E.RNG SOL'¢ZA SNGfN.A NG AS AB 0LV.A-RE 99.0 Version 99.0 1373.2
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples B73 ASTIC C3L I APSE OF A BEAM' SRUCTURE, BOX-SET-3N
| |
| ,AX SPL. p-6.71 6-- Z 0T TIE IL
| |
| 'I _ 0 y
| |
| -L T -il I* I L CAD OAD N1,TV PIP. _A'BDA 76 .
| |
| 1 OLVIA-PCST 99.0 SLV/IA ENG-INEERING AB SOLVIA-PRE input HEADING B73 PLASTIC COLLAPSE OF A BEAM STRUCTURE, BOX-SECTION' DATABASE CREATE MASTER IDOF=100011 NSTEP-10 AUTOMATIC-ITERATION NODE-li DIRECTION=2 DISPLACEMENT=10, CONTINUATION-YES DISPMAX-500 COORD INAT ES 1 / 2 0. -49. /3 -4. 4 0. 49. 5 49.
| |
| 16 0. 0. 2000. /17 0. 0. 2049. /18 -49. 0. 2000.
| |
| 19 0. 0. 1951. 20 49. 0. 2000.
| |
| NGENERATION NSTEP=5 ZSTEP--1900 / 1 To 5 NGENERATION NSTEP-10 ZSTEP-2000 / 1 TO 5 NOENERATION NSTEP=5 YSTEP-2000 /16 TO 20 NGENERATION NSTEP=25 YSTEP=2000 ZSTEP=-2000 /1 TO 5 N CENERATIONNSTEP230 YSTEP=2000 ZSTEP01900 1 TO 5 NGENERATION NSTEP--35 YSTEP=2000 / 1 To 5 MATERIAL 1 PLASTIC E9S2.EA NUG0.3 YIELDE250. ET0.
| |
| EGROUP 1 ISOBEAM SECTION 1 SDIMD2. DDIMN100.
| |
| SECTION 2 SDIM TDIMN96.
| |
| CO2.
| |
| Version 99.0 B73.3
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
| |
| ENODES 1 1 2 7 TO 4 1 5 10 5 6 7 12 TO 8 6 10 15 9 16 17 22 TO 12 16 20 25 13 26 27 32 TO 16 26 30 35 17 31 32 37 TO 20 31 35 40 EDATA / ENTRIES EL SECTION 1 1 STEP 2 TO 19 1 / 2 2 STEP 2 TO 20 2 RIGIDLINK 2 1 TO 5 1/ 7 6 TO 10 6 / 12 11 TO 20 11 22 21 TO 30 21 / 32 31 TO 35 31 / 37 36 TO 40 36 FIXBOUNDARIES 23 / 1 36 LOADS CONCENTRATED / 11 2 1.
| |
| SET VIEW=X NSYMBOL=YES PLOTORIENTATION=PORTRAIT ZONE NAME=A1 INPUT=ELEMENTS / 1 5 9 15 19 ZONE NAME=A2 INPUT=ELEMENTS / 2 6 9 14 18 ZONE NAME=A3 INPUT=ELEMENTS / 3 7 9 13 17 MESH SUBFRAME=12 MESH ZONENAME=A1 NNUMBER=YES MESH MESH ZONENAME=A2 NNUMBERS=YES SUBFRAME=12 ZONENAME=A3 NNUMBERS=YES SOLVIA END SOLVIA-POST input
| |
| * B73 PLASTIC COLLAPSE OF A BEAM STRUCTURE, BOX-SECTION DATABASE CREATE WRITE FILENAME='b73.1is' SUBFRAME 21 NHISTORY NODE=17 DIRECTION=2 SYMBOL=1 OUTPUT=ALL MESH VIEW=X VECTOR-LOAD ELIST ZONENAME=EL14 SELECT=S-EFFECTIVE END Version 99.0 B73.4
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B74 ANALYSIS OF A RUBBER BUSHING Objective To verify the PLANE STRAIN element and the use of rigid links in large strain rubber analysis.
| |
| Physical Problem The figure below shows the rubber bushing considered. The shaft and the frame are assumed to be perfectly bonded to the rubber. The problem considered is to evaluate the radial stiffness of the long bushing [1].
| |
| Fram e D = 60 mm d=30 mm Rubber material 2
| |
| Shaft CI = 0.177 N/mm C2 = 0.045 N/mm2 2
| |
| S= 666 N/mm
| |
| [ L .L Finite Element Model Due to symmetry only one half of the model is considered. The shaft is modeled using rigid links and the rubber is modeled using 32 9-node PLANE STRAIN elements. The frame is assumed to be rigid so the nodes on the outer surface of the rubber are fixed in Y- and Z-directions. Prescribed dis placements in the negative Z-direction of the shaft is used as applied loading. The figures on page B174.3 show the finite element model used in the analysis.
| |
| Solution Results Theoretical results under small displacement assumptions can be found in [ 1], page 21. This problem is also analyzed in [2],
| |
| An approximate value for the radial (z-direction in model) stiffness for long bushings in [1] is k = 3LLG = 59.9.L N/mm (using G = 2(Ci +C 2 )= 0.444N/mm2 )
| |
| Using the input data as shown on pages B74.6 and B74.7 we get a radial bushing stiffness of k = 58.9 -L N/mm for a displacement of 1 mm.
| |
| As shown in the SOLVIA-POST figures on page B74.4 the bushing has an almost linear radial stiffness characteristic up to 5 mm shaft displacement.
| |
| Version 99.0 1374.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples Contour plots of the shear strain and the hydrostatic pressure and a vector plot of principal stretches for a shaft displacement of 3 mm are also shown on page B74.4.
| |
| An approximate value for the radial stiffness for short bushings in [1] is k =f3LG = 93.2 N/mm for L=10mm Using a PLANE STRESS finite element model we get a radial bushing stiffness of k = 94.1 N/mm The corresponding contour and vector plots from the PLANE STRESS analysis can be seen on page B74.5.
| |
| User Hints
| |
| "* Note that in rubber analysis the elements may become highly distorted giving bad performance. It is important to have a fine mesh in regions with high strain concentrations. A finer mesh is neces sary in this example if the response should be traced further up to 10 mm shaft displacement.
| |
| "* Note that the 8- and 9-node PLANE Strain/Axisymmetric elements have a linear pressure assump tion in the mixed (displacement-pressure) formulation. The 4-node element has a constant pressure assumption.
| |
| References
| |
| [1] Lindley, P.B., "The Stiffness of Rubber Springs" in Allen, P.W., Lindley, P.B., and Payne, A.R. (eds), "Use of Rubber in Engineering", Proceedings, Conference, Imperial College, London, 1966.
| |
| [2] Zdunek, A.B., and Bercovier, M., "Numerical Evaluation of Finite Element Methods for Rubber Parts", Proceedings of the Sixth International Conference on Vehicle Structural Mechanics, Society of Automotive Engineers, Detroit, 1986.
| |
| Version 99.0 B74.2
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples 374 ýV'NALYSS OF A UBEBLSH:N
| |
| !RIG: \A iI SOLVTA-PRE 9 9 .0 SO\/T14 ENGINEER-1NG AB 374 ANALYSTS OF A RUBBER BLSH NG ORTG:NAL i - . z LY EAXEq-v-.4EN ~ IEE .(' I G 4 SOLVIA-PRE Version 99.0B7. 1374.3
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples B74 ANALYT OF A P SBB 3USH-N IG r-AX D:P ___- 4.9501 L 'IAX : ýL - 2>.SC TIME 3 Ly 7- ll 3 ST-RA N-RS lEA N I; 7 NO AIERAAOINC
| |
| 'lAX 13011 MIAX 2.2987 0.63110 0 :8445 0.6290S O0.2622C.
| |
| -I.3745
| |
| : 2. 0123
| |
| -271,02
| |
| ,'-7N-2.2721 MIN-3. 044 1 j SOLVIA-PCST 99.0 SOLVIA ENGINEERING AS 371 ANAL. -ýSOF A R~UBBER BUSHING M'AX D15PI -H 495C(2 TI,-IE 3
| |
| .7-P, 7 6357 T
| |
| )L--DCS 92 D Sol ' A .. NC-INE--&i\ ASý Version 99.0B7. B74.4
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples B74A ,ANALYSIS OF A RUBBER BUSH NG, PLANE TRESS MAX DISP' Z MAX DISP. 3 TiM- 3 L, TIMIF 3 STRA, N-RS NO AVERAG:NG MAX 0.020277 j MEAN NO AVERAGrTNG MAX 0 .7834 STRESS M,*-7Cý1273 1
| |
| ý,3 0.078747 3. "0668
| |
| " "'-C 4476 .0.57238
| |
| -0.2L378 8.7978E- 3
| |
| -0. 27680 -0. 039642
| |
| - 34281 A72 5 - 088082 S
| |
| ' -0.40883
| |
| -0 . 4748S -0. 13652
| |
| -0 18496 M:N-O. 50785 MIN-. 039'8 SOLVIA-POST 99.0 SCL\/IA ENGINEERING AB B74A ANALYSIS OF A R U BBER BUSHING, PLANE STRESS MAX DISPL. - 3 Ly T-ME 3 LY
| |
| * . _* _, _ _
| |
| * _, * .* .* _
| |
| * _ _,. _, -., -.J N.
| |
| LL U0 N.
| |
| N p
| |
| S RET CH
| |
| :3 I .279 1'.IA ,UE O L- A --NCTi -R C.71586 C01,IA -RCST 99. 0 S~' I-L'/A ENGINN Version 99.0 1374.5
| |
| | |
| SOLVLk Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B74 ANALYSIS OF A RUBBER BUSHING' DATABASE CREATE MASTER IDOF-I00111 NSTEP=-0 DT=0.5 KINEMATICS DISPLACEMENTS=LARGE STRAINS=LARGE TOLERANCES TYPE=F RNORM=1.
| |
| ITERATION FULL-NEWTON LINE-SEARCH=YES TIMEFUNCTION 1 o o / 10 10 SET NODES-9 SYSTEM 1 CYLINDRICAL COORDINATES / ENTRIES NODE R THETA 1 30 -90 / 2 15 -90 / 3 15 90 4 30 90 / 5 MATERIAL 1 RUBBER CG=0.177 C2=0.045 KAPPA-666 EGROUP 1 PLANE STRAIN PRESSURE-INTERPOLATION=YES GSURFACE 1 4 3 2 ELI=8 EL2=4 SYSTEM=1 FIXBOUNDARIES 23 INPUT=LINES 1 4 FIXBOUNDARIES 2 INPUT=LINES / 1 2 3 4 FIXBOUNDARIES 2 INPUT=NODES / 5 RIGIDLINKS INPUT=LINES 2 3 5 LOADS DISPLACEMENTS 5 3 -1.
| |
| MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES MESH EAXES=STRESS-RST BCODE=ALL SOLVIA END Version 99.0 B74.6
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
| |
| * B74 ANALYSIS OF A RUBBER BUSHING DATABASE CREATE STRESSREFERENCE=ELEMENT WRITE FILENAME='b74.1is' TOLERANCES SET NSYMBOLS-MYNODES CONTOUR AVERAGE=NO MESH CONTOUR-ERS TIME=3 SUBFRAME=21 MESH CONTOUR=SMEAN TIME=3 MESH VECTOR-STRETCH TIME=3 SUBFRAME=21 NHISTORY NODE=5 DIRECTION=3 KIND=REACTION XVARIABLE=1 OUTPUT=ALL EMAX SELECT=STRETCH NUMBER=3 TYPE=MAXIMUM EMAX SELECT=STRETCH NUMBER=3 TYPE-MINIMUM END Version 99.0 1374.7
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| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B75 LARGE DEFLECTION OF A CURVED ELASTIC CANTILEVER Objective To verify the large displacement behaviour of the BEAM element in a curved structure.
| |
| Physical Problem A curved elastic cantilever, 45-degree bend, is subjected to a concentrated vertical tip load as shown in the figure below. The bend has a radius of 100 in. The large displacement 3D-response of the cantilever shall be calculated for a tip load of 600 lb.
| |
| z Beam cross-section a= 1 in.
| |
| Y aL Elastic material x Ez= 107 psi 45*
| |
| v=0.
| |
| F
| |
| -- R=100 in.
| |
| Finite Element Model The cantilever bend is modeled using eight BEAM elements. A concentrated load is applied at the tip end in the Z-direction and the other end of the bend is fixed, see bottom figure on page B75.2. The load is incremented with twenty equal increments up to 600 lb. and the Full-Newton method is used in the equilibrium iterations. The element meshing of the model is made as described in [jI].
| |
| Solution Results The input data shown on page B75.5 is used in the SOLVIA analysis.
| |
| The history of the tip node displacements can be seen in the figures on page B75.3. The history of the axial force at the fixed end is also shown on page B75.3.
| |
| The following results can be compared with results presented in [1].
| |
| Tip displacements Load level SOLVIA Crisfield [1,2]
| |
| X Y Z X Y Z 150 -2.6304 -4.5051 25.399 300 -7.0864 -12.220 40.464 -7.14 -12.18 40.53 450 -10.784 -18.801 48.698 -10.86 -18.78 48.79 600 -13.559 -23.895 53.609 -13.68 -23.87 53.71 Version 99.0 B75.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples A solution has also been made using thirty-two 27-node SOLID elements.
| |
| The following results were obtained:
| |
| Load level Tip displacements, SOLID case X Y z 150 -2.5951 -4.3464 25.109 300 -7.0470 -11.935 40.195 450 -10.784 -18.477 48.506 600 -13.607 -23.567 53.477 The graphical output from SOLVIA-POST can be seen on page B75.4. The axial stress arr is shown along the line of integration points closest to the inner lower comer of the beam.
| |
| References
| |
| [1] NAFEMS, Ref. R0024, "A Review of Benchmark Problems for Geometric Non-Linear Behaviour of 3-D Beams and Shells" (
| |
| | |
| ==SUMMARY==
| |
| ), Prinja, N.K., and Clegg, R.A.
| |
| [2] Crisfield, M.A., "A Consistent Co-Rotational Formulation for Non-Linear, Three-Dimensional Beam Elements", J. Computer Methods in Applied Mechanics and Engineering, Vol. 81 (1990), pp. 13 1- 15 0 .
| |
| B75 LARGE DEFLECTION OF A CLRVED ELASTIU CANTILEVER ORIS7NAL Z TI'1E C ZONE EG3 X -_ v
| |
| ,AXES\ RST I SCL I/A-DRE 90 SCL',iZA ENGINEERING AB I Version 99.0 B75.2
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples B75 -RE CN 37A CURVED *LASTIC CAN -, EV E iL (r.
| |
| '4
| |
| -J z
| |
| 9-00 200 i0o 0 VALUF TT'1EFNC7-2N 7'L Cý: T'1-MEJN -CNI SOLVIA 22 - 99.0 SC -VA ENCTNLERING AB B7 ARGE DEF-E TON OF A CURVWC ELAST7C CAN-- 1 0 0
| |
| -'O 1
| |
| "2, 0 201 Gd V4ýLUE F ft ''F- J C1 -N I T
| |
| /ZI-A-CST 99.0 SOL/IA \ >7EERING AB Version 99.0B5. B75.3
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples j37SA L4F<- EFLE O 30F A 'JRVE AST Iý A,\ 1ER ,SL I T DI E N
| |
| Crý (3 ~0 L.
| |
| N z
| |
| 00, 3 4C 60 86
| |
| ý'EFJNCT1N C UR SOL V7A -POST 99.0 SCLVI'\ ENTNý -RING AB B7SA LARO_ DEFLETION CF A CURVE~ ELASTIr CAN\TI EER, S M'AX DI:SPL. 60.212 Z TIME 'C
| |
| '- AD 26 .6 C-. V -N(- 11 SOLK/A \C \NEERI\K. AB Version 99.0B74 B75.4
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B75 LARGE DEFLECTION OF A CURVED ELASTIC CANTILEVER' DATABASE CREATE MASTER NSTEP=20 DT=0.5 KINEMATICS DISPLACEMENTS=LARGE ITERATION FULL-NEWTON TOLERANCES ETOL=I.E-6 TIMEFUNCTION 1
| |
| : 0. 0. / 10. 600.
| |
| SYSTEM 1 SPHERICAL X=i00.
| |
| COORDINATES ENTRIES NODE R THETA PHI 1 / 2 100. 135. 90. / 3 100. 180. 90.
| |
| MATERIAL 1 ELASTIC E=-.E7 NU 0.
| |
| EGROUP 1 BEAM RESULTS-FORCES SECTION 1 RECTANGULAR WTOP:1. D-i.
| |
| GLINE Ni=2 N2=3 AUX=1 EL-8 SYSTEM=I FIXBOUNDARIES / 1 3 LOADS CONCENTRATED 2 3 1.
| |
| VIEW ID=i XVIEW=i.5 YVIEW=I. ZVIEW=0.5 SET NSYMBOLS=YES NNUMBERS=MYNODES VIEW-i MESH ZONENAME=EGi VECTOR=LOAD EAXES=RST SOLVIA END SOLVIA-POST input
| |
| * B75 LARGE DEFLECTION OF A CURVED ELASTIC CANTILEVER DATABASE CREATE WRITE FILENAME='b75.1is' SUBFRAME 21 NHISTORY NODE=2 DIRECTION=1 XVARIABLE=i OUTPUT-ALL NHISTORY NODE=2 DIRECTION=2 XVARIABLE=i OUTPUT=ALL SUBFRAME 21 NHISTORY NODE=2 DIRECTION=3 XVARIABLE=i OUTPUT=ALL EHISTORY ELEMENT=8 POINT=2 KIND=FR XVARIABLE=! OUTPUT=ALL ELIST ZONENAME=EL8 NLIST KIND-REACTIONS END Version 99.0 B75.5
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B76 INITIAL STRAIN IN A CURVED BEAM Objective To verify the BEAM, ISOBEAM and PIPE elements when using the initial strain option. Note that when the initial strain is nonzero for any of these elements then the corresponding element group is considered nonlinear by the program.
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| Physical Problem A curved beam as shown in the figure below has an initial axial strain of 0.001. The radial displacement due to the initial strain is to be calculated.
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| R=2 m E=2.l10[ N/im 2 einit = 0.001 u = radial displacement u
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| Z Finite Element Model Three separate models of the same structure as shown on page B76.2 are considered.
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| Model 1: 2-node BEAM element Model 2: 4-node ISOBEAM element Model 3: 4-node PIPE element Solution Results The theoretical radial displacement is u = -R -eirtit A minus sign is needed because a positive initial strain means that the beam shrinks when it is released from the initial configuration. The initial strain is the strain occurring in the configuration with zero displacements and with zero thermal strain.
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| The input data shown on pages B76.3 and B76.4 gives the following results for the radial displacement in [mm]:
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| Version 99.0 B76.1
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| SOLVIA Verification Manual Nonlinear Examples B76 IN'TAL STRA-N -N! A UR"/WEDBEAM ORIGfNAL. S z 1 0 S C22 2:1t 2ý 0.
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| 209
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| ý100 208 C32 207 4)3 3!3 2062C 02 308 30 "2203
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| '0
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| \ 2
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| '?305
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| ,20 t
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| \30 4 B12 1
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| )-33
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| \,303 302 MAS T ER
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| '30!
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| B '0:1~
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| 3''
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| C E 0 AS SOLV7A ENIGINEER\ING AB SOLVIA 295 99,C 376 INITIAL STRAIN IN A CURVED 3EAM ORIGINAL -2E-3 z MAX DISP!
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| TIME I X
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| S/>?
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| I<\
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| A- '
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| -t A
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| -A
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| -4
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| -- 4 D-7SPLACEMEN7 2E-3
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| ý0LV,/A ENGiNEEý,R7,'G 4B
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| _/:A-POST 99.0 Version 99.0 B76.2
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEAD 'B76 INITIAL STRAIN IN A CURVED BEAM' DATABASE CREATE MASTER IDOF=-00011 SYSTEM 1 CYLINDRICAL COORDINATES SYSTEM=i ENTRIES NODE R THETA XL 1 2. 0. 0.
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| 2 2. 90. 0.
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| 3 0.
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| 21 2. 0. 2.
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| 22 2. 90. 2.
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| 23 0. 0. 2.
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| 31 2. 0. 4.
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| 32 2. 90. 4.
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| 33 0. 0. 4.
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| MATERIAL 1 ELASTIC E=2.Eii NU=0.3 EGROUP 1 BEAM MATERIAL=1 RESULT=FORCES SECTION 1 RECTANGULAR WTOP=0.02 D=0.02 SET SYSTEM=1 GLINE 1 2 3 EL=12 EDATA / ENTRIES EL INIT-STRAIN 1 0.001 TO 12 0.001 EGROUP 2 ISOBEAM MATERIAL=i RESULT=FORCES SECTION 1 SDIM=0.02 TDIM=0.02 GLINE 21 22 23 EL=4 NODES=4 NFIRST=201 EDATA / ENTRIES EL INIT-STRAIN 1 0.001 TO 4 0.001 EGROUP 3 PIPE MATERIAL=1 RESULT=FORCES SECTION 1 DIAMETER=0.02 THICKNESS=0.002 GLINE 31 32 33 EL=4 NODES=4 NFIRST=301 EDATA / ENTRIES EL INIT-STRAIN 1 0.001 TO 4 0.001 FIXBOUNDARIES / 3 23 33 FIXBOUNDARIES 24 / 2 22 32 FIXBOUNDARIES 34 / 1 21 31 SET HEIGHT=0.25 MESH NNUMBER=YES NSYMBOL=YES BCODE=ALL SOLVIA END Version 99.0 B76.3
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B76 INITIAL STRAIN IN A CURVED BEAM DATABASE CREATE WRITE 'b76.1is' MESH ORIGINAL=DASHED VECTOR=DISPLACEMENT NSYMBOL=YES NPLINE LINE-EGI / 2 111 STEP -1 TO 101 1 NPLINE LINE-EG2 / 22 211 STEP -1 TO 201 21 NPLINE LINE-EG3 / 32 311 STEP -1 TO 301 31 NVARIABLE DISPY DIRECTION=2 NVARIABLE DISPZ DIRECTION=3 NVARIABLE Y DIRECTION=2 KIND=COORDINATE NVARIABLE Z DIRECTION=3 KIND:COORDINATE RESULTANT DISP-RAD STRING=' (Y*DISPY+Z*DISPZ)/(SQRT(Y*Y+Z*Z))'
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| RLINE LINE-EGi DISP-RAD OUTPUT=LIST RLINE LINE-EG2 DISP-RAD OUTPUT=LIST RLINE LINE-EG3 DISP-RAD OUTPUT=LIST END Version 99.0 B76.4
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B77 TEMPERATURE GRADIENTS IN A CURVED BEAM Objective To verify the BEAM, ISOBEAM and PIPE elements when subjected to temperature gradient loadings.
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| Physical Problem A curved beam forming a ring which is fixed at one point as shown in the figure below has tempera ture gradients in both the radial and the X directions. The mean temperature of the beam is zero. The X-displacement and the bending moment due to the radial temperature gradient are to be calculated.
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| R-2 m The temperature gradients are 1 2 E=2.10 N/m
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| - = 50oC/m (X-direction) a=l-10-1 1/°C as
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| -= 100 °C / m (radial direction)
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| Pipe section: at H Fixed Diameter = 30 mm Thickness = 1 mm where 0 is the temperature and s and t are local element axes.
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| Rectangular cross-section of 20 x 20 mm in BEAM and ISOBEAM elements.
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| Finite Element Model Three separate models of the same structure as shown on page B77.2 are considered.
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| Model 1: 2-node BEAM element Model 2: 4-node ISOBEAM element Model 3: 4-node PIPE element Solution Results The theoretical X-displacement at point A is DO u.X=-2R.R a at The theoretical bending moment in the beam is Mt = E It a. where It is the moment of inertia for bending about the t-axis.
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| as The input data shown on pages B77.5 and B77.6 gives the following results:
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| Version 99.0 B77.1
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| SOLVIA Verification Manual Nonlinear Examples X-displacement in [mm] at point A:
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| The t-axis bending moment in Nm for a rectangular cross-section:
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| The t-axis bending moment in Nm for a pipe section:
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| The t-axis bending moment along the elements are shown on pages B77.3 and B77.4.
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| B77 'P -,ATURE GRADIENTS ',N A CURVE'D BEAN 3RIINAL ý- 3.5 z X.. __.
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| 34 V
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| 33ý6 EAXES=RST B A A 10 S L'IA- 99. LVOA EANEARDIO AS Version 99.0 B77.2
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| SOLVIA Verification Manual Nonlinear Examples 377 -"7vPERATURE GRADIENTS '\1A CURVED BEAM OR: CNA MAX - SPI 8E-3 TIME X DISPLACEMENT 8E-3 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B77 '7\PERATURE SRAD EN7S IN A CURVED BEAM TiME I ORGTNAL i-I z X
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| m t po r...
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| BEAM L A 9S' SCL/IA 7\GCNEERIN( AB Version 99.0 B77.3
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| SOLVIA Verification Manual Nonlinear Examples 377ýPýRAT RE SRA DiE\,q IN A :JR~V. SEAM ITil 71E T I."'
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| 01
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| - <1/2 2W.( A -
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| -o Ti LRA' I CB P IPE 3CLVIA POST 99.C SO' \IA -INC-NEER>j'C- AB Version 99.0B'7. B77.4
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEAD 'B77 TEMPERATURE GRADIENTS IN A CURVED BEAM' DATABASE CREATE SYSTEM 1 CYLINDRIC COORDINATES SYSTEM=i ENTRIES NODE X R 1 0 2 2 0 0 3 3 2 4 3 0 5 6 2 6 6 0 MATERIAL 1 THERMO-ELASTIC TREF=0
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| -i00 2.Eii 0.3 i.E-5 100 2.Eli 0.3 i.E-5 EGROUP 1 BEAM MATERIAL-i RESULT=FORCES SECTION 1 RECTANGULAR WTOP=0.02 D=0.02 GLINE 1 1 AUX=2 EL-72 SYSTEM=i LOADS ELEMENTS TGRADIENTS 1 S 50. 50. TO 72 s 50. 50.
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| 1 t 100. 100. TO 72 t 100. 100.
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| EGROUP 2 ISOBEAM MATERIAL=i RESULT=FORCES SECTION 1 SDIM=0.02 TDIM=0.02 GLINE 3 3 AUX=4 EL=24 NODES=4 SYSTEM-I LOADS ELEMENTS TGRADIENTS 1 s 50. 50. TO 24 s 50. 50.
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| 1 t 100. 100. TO 24 t 100. 100.
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| EGROUP 3 PIPE MATERIAL=i RESULThFORCES SECTION 1 DIAMETER=3E-2 THICKNESS=i.E-3 GLINE 5 5 AUX=6 EL 24 NODES=4 SYSTEM=i LOADS ELEMENTS TGRADIENTS 1 s 50. 50. TO 24 s 50. 50.
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| 1 t 100. 100. TO 24 t 100. 100.
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| LOAD TEMPERATURE TREFE0.
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| FIXBOUNDARIES INPUT=ZONE ZONE=MYNODES SET HEIGHT=0.25 MESH NNUMBER=MYNODES NSYMBOL-MYNODES BCODE=ALL EAXES=RST SOLVIA END Version 99.0 1377.5
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B77 TEMPERATURE GRADIENTS IN A CURVED BEAM DATABASE CREATE WRITE 'b77.iis' NMAX EGI DIRECTION=i NMAX EG2 DIRECTION=i NMAX EG3 DIRECTION=i EMAX SET SMOOTHNESS=YES MESH ORIGINAL=DASHED VECTOR=DISP EGROUP 1 EPLINE BEAM 1 1 2 TO 72 1 2 EGROUP 2 EPLINE ISOBEAM 1 1 2 TO 24 1 2 EGROUP 3 EPLINE PIPE 1 1 2 TO 24 1 2 MESH PLINES=ALL SUBFRAME=21 ELINE BEAM KIND=MT ELINE ISOBEAM KIND=MT SYMBOL=i SUBFRAME=21 ELINE PIPE KIND=MT SYMBOL=i END Version 99.0 B77.6
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B78 TEMPERATURE GRADIENTS IN A PLATE Objective To verify the SHELL elements when using the temperature gradient option and the SOLID element when subjected to a linear temperature field.
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| Physical Problem A circular plate with a concentric circular hole is exposed to a uniform temperature gradient through its thickness.
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| R,=2m R 2 = 0.5 m h=0.2 m (thickness)
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| E=2.10" N/m 2 A y Fixed v = 0.3 oa= 1.10-5 1/°C a =100 °C/m at where 0 is the temperature and t is the element t-direction (positive thickness direction and positive X-direction).
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| Finite Element Model Four separate models of the same structure as shown on page B78.2 are considered.
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| Model 1: 4-node SHELL element Model 2: 9-node SHELL element Model 3: 16-node SHELL element Model 4: 27-node SOLID element The temperature load is applied as element temperature gradients in the SHELL elements and as nodal temperatures for the SOLID elements.
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| Solution Results The theoretical X-displacement at point A is at The input data shown on pages B78.4 to B78.6 gives the following results for the displacement in mm at point A:
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| Version 99.0 B78.1
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| SOLVIA Verification Manual Nonlinear Examples Theory 4-node SHELL 9-node SHELL 16-node SHELL 27-node SOLID
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| -8.000 -8.000 -8.000 -8.000 -8.000 Contour plots of the von Mises effective stress are shown on pages B78.3 and B78.4. The plate should be stress free in this example. The stress free condition corresponds to a spherical shape of the plate which cannot exactly be modelled by the isoparametric elements. The von Mises stresses are, however, very small which can be seen by a comparison with the temperature stress at zero displace ments:
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| Ea= =-28.57.106 N / m 2 (0=o10C) 1-V B78 TEMPERATURE GRAD7ENTS IN A PLATE ORIGINAL I z TIME I x Y MASTER7 P 88 2000 3 0 0!00 CC 0 0O TGRADIENT SOLV A-PRE 99.0 SOLVIA ENGINEERING AB Version 99.0 B78.2
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| SOLVIA Verification Manual Nonlinear Examples B78 TEMPERATURE GRADIENTS iN A PLATE ORIGINAL ý-- --4 I. z
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| ý.AX DISPL.; 8.00 E-3, TIME X *'
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| DISPLACEMENT 8.00SE-3 SOLVIA-POST 99.0 SOLVIA ENGTNEERING AB B78 TEMPERATURE GRADIENTS IN A PLATE B78 TEMPERATURE GRADIENTS :N A PLATE MAX DISPL. -8E-3 MAX DISPL. -SXE-3 TIME , zZ TIME I ZONE ED: ZONE -r2 2 Y MISES MISES GASE ON2ZXNE BASED ON ZONE SHEL. 'TOP SHELL TOP MAX 3,3307E MAY 962 27 991 .06 3.1243E-4
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| : 2. 7jSE-4 808.63
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| :: 706 .20 2 29871E 1[.8860E 6503.7 1 4732E-4
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| ,* .0604E-41 4-node SHELL eiements 9-node SHELL elemens
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| .4763E-5 2.348SE-I tS
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| ,94,06 S296.58 MIN 2.8462E-6 TIN 142.84 SOLVIA ENGINEERING AG SOLVIA-GOST 99 0 SOLVTA ENGINEERING 3 SOL/AA-POST 99 0 Version 99.0 B78.3
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| SOLVIA Verification Manual Nonlinear Examples 878 TEMPERATURE GRADIENTS 7N A PLATE B78 TEMPERATURE GRADIENTS 7N A PLATE MAX DSPL R8E-3 Z MAX D1APL. -
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| * S OSE-3 TIME 1 ZONE EG3 ZONE EG4 MISES BASED ON ZONE SHELL TOP BASED ON ZONE MAX 5419.3 MAX 81446
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| !5143.6 S4Sg1.1 S690 773.01 09 4038.6 3486.2 524,27 2933.7 ?* 358.44
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| !i 441.3S 238t.2 27S 23 16-node SHELL elements 1928.8 27-node SOLED elements S1278.3 MIN 1000. 1 807MG8RI A[
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| SOLVIA-POST 990 SOLVIA ENGINEERING AB SOLVIA-POST 99.0 SOLVIA ENGINEERING AB SOLVIA-PRE input HEAD 'B78 TEMPERATURE GRADIENTS IN A PLATE' DATABASE CREATE SYSTEM 1 CYLINDRIC COORDINATES SYSTEM=i ENTRIES NODE X R 1 0 2 2 0 0.5 11 -3.0 2 12 -3.0 0.5 21 -6.0 2 22 -6.0 0.5 31 -9. 0 2 32 -9 .0 0.5 33 -9 .2 2 34 -9. 2 0.5 35 -9.1 2 36 -9.1 0.5 Version 99.0 B78.4
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
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| MATERIAL 1 THERMO-ELASTIC TREF=0.
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| -1000 2.E11 0.3 i.E-5 1000 2.E11 0.3 i.E-5 EGROUP 1 SHELL MATERIAL=i SET SYSTEM=1 GSURFACE 2 1 1 2 EL1=4 EL2=36 NODES=4 THICKNESS 1 2.E-1 LOAD ELEMENTS TYPE=TGRADIENT 1 100. TO 144 100.
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| FIXBOUNDARIES / 1 FIXBOUNDARIES 3 / 2 EGROUP 2 SHELL MATERIAL=1 GSURFACE 12 11 ii 12 EL1=4 EL2=36 NODES=9 THICKNESS 1 2.E-1 LOAD ELEMENTS TYPE-TGRADIENT 1 100. TO 144 100.
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| FIXBOUNDARIES / 11 FIXBOUNDARIES 3 / 12 EGROUP 3 SHELL MATERIAL=i GSURFACE 22 21 21 22 EL1=4 EL2=36 NODES=-6 THICKNESS 1 2.E-1 LOAD ELEMENTS TYPE=TGRADIENT 1 100. TO 144 100.
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| FIXBOUNDARIES / 21 FIXBOUNDARIES 3 / 22 LINE CYLINDRIC 31 31 EL=36 MIDNODES=*
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| LINE CYLINDRIC 32 32 EL=36 MIDNODES=i LINE CYLINDRIC 33 33 EL=36 MIDNODES=1 LINE CYLINDRIC 34 34 EL=36 MIDNODES=1 EGROUP 4 SOLID MATERIAL=1 GVOLUME 32 31 31 32 34 33 33 34 EL1=4 EL2=36 NODES=27 SYSTEM=i LOAD TEMPERATURE TREFE0. INPUT=N6DES INTERPOLATION=X COORDi=-9.6 V1=-50. COORD2=-8.6 V2=50. SYSTEM=0 31 TO 36 2327 TO 4264 FIXBOUNDARIES 35 FIXBOUNDARIES 23/ 31 FIXBOUNDARIES 3 / 36 MESH NNUMBER=MYNODES NSYMBOL=MYNODES BCODE=ALL CONTOUR=T-GRADIENT SOLVIA END Version 99.0 B78.5
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B78 TEMPERATURE GRADIENTS IN A PLATE DATABASE CREATE write 'b78.lis' N*AX EGI DIRECTION=I23 NMAX EG2 DIRECTION=123 NMAX EG3 DIRECTION=123 NMAX EG4 DIRECTION=123 MESH ORIGINAL=DASHED VECTOR=DISPLACEMENT SET PLOTORIENTATION=PORTRAIT CONTOUR AVERAGING=ZONE MESH EGi CONTOUR=MISES TEXT '4-node SHELL elements' YPT=2 MESH EG2 CONTOUR=MISES TEXT '9-node SHELL elements' YPT=2 MESH EG3 CONTOUR=MISES TEXT '16-node SHELL elements' YPT=2 MESH EG4 CONTOUR=MISES TEXT '27-node SOLID elements' YPT=2 END Version 99.0 B78.6
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B79 REINFORCED CONCRETE BEAM, ISOBEAM Objective To verify the rebar section option in a reinforced concrete beam analysis using ISOBEAM elements.
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| Physical Problem A simply supported reinforced concrete beam subjected to two symmetrically applied concentrated loads as shown in the figure below is considered. The steel reinforcement area Ast = 0.62 in2 is to be used in this analysis and the steel reinforcement is located 2.06 in. from the bottom surface of the beam. This problem is also analyzed in Example B 19 using PLANE STRESS elements.
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| P/2 P/2 PLANE OF PLANE OF CONTRAFLEXURE CONTRAFLEXURE 1
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| 2, 62 in"2 P/2 I Ast= in2 a2"OO BEAM DIMENSIONS Material parameters CONCRETE:
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| 4 Density 0.2172. 10- 3 lbf s 2 /in Initial tangent modulus 3060 ksi Poisson's ratio 0.2 Uniaxial cut-off tensile strength 0.458 ksi Uniaxial maximum compressive stress (SIGMAC) -3.74 ksi Compressive strain at SIGMAC -0.002 in/in Uniaxial ultimate compressive stress -3.225 ksi Uniaxial ultimate compressive strain -0.003 in/in STEEL: 4 2
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| Density 0.7339 .10-3 lbf. s /in Young's modulus 30000 ksi Initial yield stress 44 ksi Strain hardening modulus 300 ksi Finite Element Model Using symmetry only one-half of the structure need to be considered. Ten parabolic ISOBEAM elements are used in the finite element model as shown in the left figure on page B79.2. REBAR sections with an elastic-plastic material model are used to model the reinforcement. The solution is obtained using the AUTOMATIC-ITERATION method.
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| Version 99.0 1379.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples Solution Results The input data on pages B79.4 and B79.5 is used in the finite element analysis.
| |
| The right figure below shows the mid-span deflection of the beam as a function of the load. The response curve for the ISOBEAM element model with symbols included is compared with the corresponding result curve from Example B 19 which is shown as a solid line without symbols.
| |
| The top figure on page B79.3 shows the axial stress-strain history at the four integration point levels closest to the midspan. The bottom figure shows the variation of the axial stress along the four integration point levels for the last solution time.
| |
| User Hints
| |
| "* The ISOBEAM element is constrained so that plane sections remain plane during deformation.
| |
| Therefore, the ISOBEAM model shows a somewhat larger load capacity than the PLANE STRESS model in Example B 19.
| |
| " The plastic strain concentration observed in the PLANE STRESS model is not present in the ISOBEAM model because of the constraint. No variation of the stresses can be observed in the axial direction for the constant moment portion of the beam.
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| " Listing and scanning of REBAR and CONCRETE results can be obtained in SOLVIA-POST by using the SELECT parameter in the ELIST, EEXCEED and EMAX commands.
| |
| 879 REINFORCED CONCRETE BEAM, 10BEAM B79 REINFORCED CONCRETE BEAM. :SOBEAM OR:GINAL - tO. .
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| TINE I L0 "F0RCE 0.5 MASTE 1O001: z 9 :0y E_
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| m ORIGINAL - " Z Y
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| 01 P
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| 0 t 2 3 4 5 6 7 9 9 10 t
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| 2 4 6 4 '
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| LAMBDA 6AXES=RST SOLVIA-PRE 99.0 SOLV0A ENGINEERING AB SOL/IA-POST 99.0 SOLVfA ENGINEER:NG AB Version 99.0 B79.2
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| F'6LS 0-66 UO1SJ10A 0 66!06V/CO
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| -1 / 02 02z 0 Q,t
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| .n... ....
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| .N....' . ......
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| r7 (2, ~c7,
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| ..... I2 I-
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| ýLliýlr 09 oi 71
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| 'F-A' -
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| 7 (2
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| "0 ]-JK.
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| WV39OSI 'L4V]9 31--)ZDNC3 a3dOJ0NI2d 6L5
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| ý]V SNIdDýNION] VIAOOOS 0'66 iSOc VIAOýOS
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| ~N~NI~ -c'Z d 0; J e 03 9-01 0 05os 0 00;ý-
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| At')
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| 0 k 7v]('CsI -Fv2g 1-:ýZ2NG3 CDKAONT3ýý 6C6 soldLuuxj jrouquoN s~idtxwx~ uOIILC0)UJQA VIA"IOS fl~UiIUuNItfUNAT
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B79 REINFORCED CONCRETE BEAM, ISOBEAM' DATABASE CREATE MASTER IDOF100011 NSTEP=30 AUTOMATIC-ITERATION NODE=3 DIRECTION=3 DISPLACEMENT=-0.01, DISPMAX=0.6 CONTINUATION=YES TOLERANCES TYPE=F RNORM=10 RMNORM=50 RTOL=0.01 COORDINATES / ENTRIES NODE Y Z 1 / 2 50. / 3 68. / 4 0. 10.
| |
| MATERIAL 1 CONCRETE EO=3.06E3 NU=0.2 SIGMAT=0.458, SIGMAC=-3.74 EPSC=-0.002 SIGMAU=-3.225 EPSU=-0.003, BETA=0.75 KAPPA=15 STIFAC=0.0001 SHEFAC=0.5 MATERIAL 2 PLASTIC E=3.E4 YIELD=44 ET=300.
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| EGROUP 1 ISOBEAM RESULT=TABLES MATERIAL=i REBARMATERIAL=2 GLINE 1 2 4 EL=6 NODES=3 GLINE 2 3 4 EL=4 NODES=3 SECTION 1 SDIM=12. TDIM=6.
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| REBARSECTION 1 1 -3.94 0. 0.62 STRESSTABLE 1 112 122 132 142 212 222 232 242 FIXBOUNDARIES 3 1 FIXBOUNDARIES 24 / 3 FIXBOUNDARIES 234 / 4 LOADS CONCENTRATED / 2 3 -0.5 SET VIEW=X NSYMBOLS=YES PLOTORIENTATION=PORTRAIT MESH NNUMBERS=MYNODES VECTOR=LOAD BCODE=ALL SUBFRAME=d2 MESH EAXES=RST ENUMBERS=YES GSCALE=OLD SOLVIA END Version 99.0 1379.4
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
| |
| * B79 REINFORCED CONCRETE BEAM, ISOBEAM DATABASE CREATE WRITE FILENAME='b79.1is' AXIS 1 VMIN=-0.6 VMAX=0. LABEL='DISPLACEMENT' AXIS 2 VMIN=0. VMAX=i2. LABEL-'LAMBDA' USERCURVE 1 SORT=NO READ FILENAME='b79disp.dat' SET VIEW=X DIAGRAM=GRID SET PLOTORIENTATION=PORTRAIT NHISTORY NODE=3 DIRECTION=3 OUTPUT=ALL XAXIS-2 YAXIS= SYMBOL=I PLOT USERCURVE 1 XAXIS=-2 YAXIS=-- SUBFRAME=OLD SET PLOTORIENTATION-LANDSCAPE SUBFRAME 22 EXYPLOT ELEMENT=10 POINT=212 ](KIND=ERR YKIND=SRR SYMBOL=i EXYPLOT ELEMENT=-0 POINT=222 (KIND=ERR YKIND=SRR SYMBOL=i EXYPLOT ELEMENT=-0 POINT=232 (KIND=ERR YKIND=SRR SYMBOL-i EXYPLOT ELEMENT=-0 POINT=242 (KIND-ERR YKIND=SRR SYMBOL=i EPLINE LEVEL1 / 1 112 212 TO L0 112 212 EPLINE LEVEL2 / 1 122 222 TO L0 122 "222 EPLINE LEVEL3 / 1 132 232 TO L0 132 232 EPLINE LEVEL4 / 1 142 242 TO L0 142 242 SUBFRAME 22 ELINE LEVEL1 KIND=SRR SYMBOL=i ELINE LEVEL2 KIND=SRR SYMBOL=1 ELINE LEVEL3 KIND=SRR SYMBOL-i ELINE LEVEL4 KIND=SRR SYMBOL=1 ELIST SELECT=CONCRETE ELIST SELECT=REBAR END Version 99.0 B79.5
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B80 REINFORCED CONCRETE BEAM, NONLINEAR-ELASTIC ISOBEAM Objective To verify the NONLINEAR-ELASTIC material model in a concrete beam structure analysis using ISOBEAM elements.
| |
| Physical Problem The same simply supported reinforced concrete beam as analyzed in Example B79 is to be analyzed, but using the NONLINEAR-ELASTIC material to model the uniaxial behaviour of the concrete material model.
| |
| P/2 P/2
| |
| : 36) - 50 - - -
| |
| 5 01 PLANE OF CONTRAFLEXURE d PLANE OF CONTRAFLEXURE BEAM DIMENSIONS Finite Element Model Using symmetry only one-half of the structure need to be considered. Ten parabolic ISOBEAM elements are used in the finite element model as shown in the left bottom figure on page B180.2. The concrete material is modeled using the NONLINEAR-ELASTIC material model and the steel reinforcement is modeled using the REBAR section option. The solution is obtained using the AUTOMATIC-ITERATION method.
| |
| Solution Results The input data on pages B80.4 and B80.5 is used in the finite element analysis.
| |
| The right figure on page B180.2 shows the mid-span deflection of the beam as a function of the load.
| |
| The response curve for the nonlinear-elastic model with symbols included is compared with the corresponding results from the concrete model in Example B79 which are shown as a solid line without symbols. The two curves are almost coinciding.
| |
| The top figure on page B80.3 shows the axial stress-strain history at the four integration point levels closest to the midspan. The bottom figure shows the variation of the axial stress along the four integration point levels for the last solution time.
| |
| Version 99.0 1380.1
| |
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| SOLVIA Verification Manual Nonlinear Examples User Hints
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| " The NONLINEAR-ELASTIC material model can be used to simulate concrete materials in simple beam structures as in this example. Two reasons for the good agreement between the two material models are the one-dimensional action of the concrete in the beam structure and that the principal stresses do not change directions significantly during the response.
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| " The uni-axial stress-strain curve for the concrete model can be obtained by performing a one element analysis as shown in the input data on page 180.6. A 4-node PLANE STRESS2 element with dimensions 1 x 1 is subjected to uni-axial compression using prescribed displacements. The Poisson's ratio is set to zero and the load is applied in small increments up to the uni-axial ultimate strain, EPSU. The corresponding stress-strain curve can be found in SOLVIA-POST by using the EXYPLOT command and the stress-strain data can be included in a NONLINEAR-ELASTIC material model. The uni-axial stress-strain curve can be seen in the figure on page B80.4.
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| B80 RE'NFORCED CONCRETE BEAM. NONLINEAR-ELASTIC TSOBEAM 990 REINFORCED CONCRETE BEAM, NONL:NEAR-ELASTIC ISOBEAM ORIGINAL F-------- 10. Z ° TIME I L FORCE 0.$
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| MA3TER iOCoit C h01
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| -N 0 ORIGINAL 10. Z L*
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| 7 i 2 3 4 5 89 iG L i 0 2 4 6 8 12 ILAM9DA
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| .AXES=RST SOLVIA-PRE 99 SOLVIA ENGINEERING AS SOLVIA-POST 99.0 SOLVIA ENGINEERING AS Version 99.0 1380.2
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| SOLVIA Verification Manual Nonlinear Examples B80 REINFCRCED CONCRETE BEAM, NONL:NEAR-ELASTTC 1SOBEAM "22
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| .3 o0 00 0. 1.0 1 5 2.0 2.5 G00 2 A S E !0 P 212 STRAL*N[-RR El . P 2 .2 2 ST.R A .N RR 0....
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| 0 (NJ
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| ýE 2**l
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| -S0.0 -[00.0 S-0.0 0.n S0.0o -1.0 -0.8 -o.6 -0.4 -0.2
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| -Ai0- 3 ES E 10 P 232 STRA N-RR EG 10 P 242 STRA7N-RR SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B80 REINFORCED CONCRETE BEAM, NONLINEAR-ELASTIC ISOBEAM TIME 25 TIME 25 0*
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| 0* .........
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| *i...... ..........;:,....... * -,* .................... ...........................
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| 0*
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| (.1 0*
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| 9)
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| -J .0)
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| '2* ,LEVE i 2) 0 20 40 60 80 0 20 40 60 80 0 20 40 F025 LEVEL2 TME 2S* TIME 2S
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| -Y
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| : 2. .0 0y*
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| '2" 91 ?... .................... ..
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| L0*
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| 0*
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| 0 20 10 60 80 0 '0 40 60 SC LEVE-L3 OLEEL SOLV A-POST 99.0 SOLVIA ENGINEERTNG ,
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| Version 99.0 1380.3
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| SOLVIA Verification Manual Nonlinear Examples OjUNu r Ni-aXlt SC STREBS-SrRAINCUVE ý9 N*0*, E
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| .0 SOtt JSSEER!NG 5OLS1A-POST99 0 SOLVIA-PRE input HEADING 'B80 REINFORCED CONCRETE BEAM, NONLINEAR-ELASTIC ISOBEAM' DATABASE CREATE MASTER IDOF=100011 NSTEP=30 AUTOMATIC-ITERATION NODE=3 DIRECTION=3 DISPLACEMENT=-0.01, ALFA=1 DISPMAX=0.5 CONTINUATION=YES TOLERANCES TYPE=F RNORM=5 RMNORM=20 RTOL=0.01 COORDINATES / ENTRIES NODE Y Z 1 / 2 50. / 3 68. / 4 0. 10.
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| MATERIAL 1 NONLINEAR-ELASTIC
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| -0.003 -3.225
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| -0.00285 -3.34155
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| -0.0027 -3.45087
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| -0.00255 -3.54928
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| -0.0024 -3.63242
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| -0.00225 -3.69530
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| -0.0021 -3.73244
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| -0 .002 -3.74
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| -0.00195 -3.73802
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| -0.0018 -3.70629
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| -0.00165 -3.63190
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| -0.0015 -3.51049
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| -0.00135 -3.33912
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| -0.0012 -3.11675
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| -0.00105 -2.84450
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| -0.0009 -2.52571
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| -0.00075 -2.16570
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| -0.0006 -1.77127
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| -0.00045 -1.35007
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| -0.0003 -0.90985
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| -0.00015 -0.45781
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| -0.00003 -0.0918
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| : 0. 0.
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| 1.496732E-4 0.458 2 .2451E-3 0.
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| : 1. 0.
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| MATERIAL 2 PLASTIC E=3.E4 YIIELD=44 ET=300.
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| Version 99.0 B 80.4
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
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| EGROUP 1 ISOBEAM RESULTS=TABLES MATERIAL=i REBARMATERIAL=2 GLINE 1 2 4 EL=6 NODES=3 GLINE 2 3 4 EL=4 NODES=3 SECTION 1 SDIM=12. TDIM=6.
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| REBARSECTION 1 1 -3.94 0. 0.62 STRESSTABLE 1 112 122 132 142 212 222 232 242 FIXBOUNDARIES 3 / 1 FIXBOUNDARIES 24 / 3 FIXBOUNDARIES 234 / 4 LOADS CONCENTRATED / 2 3 -0.5 SET VIEW=X NSYMBOLS=YES PLOTORIENTATION=PORTRAIT MESH NNUMBERS=MYNODES VECTOR=LOAD BCODE=ALL SUBFRAME=12 MESH EAXES=RST ENUMBERS=YES GSCALE=OLD SOLVIA END SOLVIA-POST input
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| * B80 REINFORCED CONCRETE BEAM, NON-LINEAR-ELASTIC ISOBEAM DATABASE CREATE WRITE FILENAME='b80.1is' AXIS 1 VMIN=-0.6 VMAX=0. LABEL='DISPLACEMENT' AXIS 2 VMIN=0. VMAX=12. LABEL='LAMBDA' USERCURVE 1 SORT=NO READ FILENAME='b80disp.dat' SET VIEW=X DIAGRAM=GRID SET PLOTORIENTATION=PORTRAIT NHISTORY NODE=3 DIRECTION=3 OUTPUT=ALL XAXIS=2 YAXIS=1 SYMBOL=1 PLOT USERCURVE 1 XAXIS=-2 YAXIS=-1 SUBFRAME=OLD SET PLOTORIENTATION=LANDSCAPE SUBFRAME 22 EXYPLOT ELEMENT=10 POINT=212 XKIND=ERR YKIND=SRR SYMBOL=1 EXYPLOT ELEMENT=10 POINT=222 XKIND=ERR YKIND=SRR SYMBOL=1 EXYPLOT ELEMENT=10 POINT=232 XKIND=ERR YKIND=SRR SYMBOL=1 EXYPLOT ELEMENT=10 POINT=242 XKIND=ERR YKIND=SRR SYMBOL=1 EPLINE LEVEL1 / 1 112 212 TO 10 112 212 EPLINE LEVEL2 / 1 122 222 TO 10 122 222 EPLINE LEVEL3 / 1 132 232 TO 10 132 232 EPLINE LEVEL4 / 1 142 242 TO 10 142 242 SUBFRAME 22 ELINE LEVELI KIND=SRR SYMBOL=1 ELINE LEVEL2 KIND=SRR SYMBOL=1 ELINE LEVEL3 KIND=SRR SYMBOL=i ELINE LEVEL4 KIND=SRR SYMBOL=i ELIST ELIST SELECT=REBAR SUMMATION KIND=LOAD SUMMATION KIND=REACTION DETAILS=YES END Version 99.0 1380.5
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B80UNI UNI-AXIAL STRESS-STRAIN CURVE FOR CONCRETE, DATABASE CREATE MASTER IDOF=100111 NSTEP=100 TIMEFUNCTION 1
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| : 0. 0. / 100. 1.
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| COORDINATES / ENTRIES NODE Y Z 1 1. 1. / 2 0. 1. 3 / 4 1.
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| MATERIAL 1 CONCRETE E0=3.06E3 NU=0. SIGMAT=0.458, SIGMAC=-3.74 EPSC=-0.002 SIGMAU=-3.225 EPSU=-0.003, BETA=0.75 KAPPA=15 STIFAC=0.0001 SHEFAC=0.5 EGROUP 1 PLANE STRESS2 ENODES / 1 1 2 3 4 EDATA / 1 0.1 FIXBOUNDARIES 2 / 1 2 3 4 FIXBOUNDARIES 3 / 3 4 LOADS DISPLACEMENTS 1 3 -0.003 / 2 3 -0.003 SOLVIA END SOLVIA-POST input
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| * B80UNI UNI-AXIAL STRESS-STRAIN CURVE FOR CONCRETE DATABASE CREATE WRITE FILENAME='bBOuni.lis' SET PLOTORIENTATION=PORTRAIT DIAGRAM=GRID EXYPLOT ELEMENT=1 POINT=1 XKIND=EZZ YKIND=SZZ OUTPUT=ALL END Version 99.0 B80.6
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B81 BRESLER-SCORDELIS CONCRETE BEAM Objective To verify the PLANE concrete material model and demonstrate the use of the AUTOSTEP method in a concrete analysis.
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| Physical Problem Bresler and Scordelis investigated experimentally the shear strength in reinforced concrete beams [1].
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| A total of 12 simply supported beams were tested. One of the beams, called OA-2 in [1], is to be analyzed in this example. The figure below shows the concrete beam and the reinforcement.
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| . P L/2 L/2 i4 pi 4 Concrete Steel L = 4.57 m E0 = 28700MPa E = 218000 MPa hd h = 0.56 m v =0.2 CY= 555 MfPa b = 0.305 m a, = -23.7 MPa ET-= 10000 MPa d = 0.465 m -9 = -0.0022 Ast = 3290.10-'62 a, = 4.33MPa Finite Element Model The finite element model consists of fifty 8-node PLANE STRESS2 elements using the concrete ma terial model. The steel reinforcement is modeled with ten parabolic TRUSS elements using an elastic plastic material model- The AUTOSTEP method is used to control the load incrementation up to the collapse load of the beam. Force tolerances are used with a force norm of 80 kN and a force tolerance of one percent. The BFGS iteration method is employed. Due to symmetry only half of the beam needs to be considered as shown in the finite element model on page B81.3.
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| Solution Results The input data used in the SOLVIA analysis is shown on pages B81.8 and B81.9.
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| Bresler and Scordelis reported an ultimate load of 80 kips (appr. 356 kIN) with a maximum mid-span deflection of 0.46 in (appr. 11.7 mm) for the OA-2 beam.
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| Version 99.0 B81.1
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| SOLVIA Verification Manual Nonlinear Examples In the experimental study the OA-2 beam showed typical initial flexural cracks in the lower part of the beam section followed by diagonal tension cracks. The failure mode of the OA-2 beam was denoted a "diagonal tension failure" and it occurred as a result of longitudinal splitting in the com pression zone near the point load and also by horizontal splitting along the tensile reinforcement near the end of the beam [1].
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| In the first analysis the value for SIGMAT, the tensile strength, is set equal to the modulus of rupture for the concrete material. The results from this analysis can be seen on pages B81.3 and B81.4. The mid-span deflection of the beam as a function of the load can be seen in the bottom figure on page B81.3. A vector plot of crack normals and a contour plot of the state of the concrete in the last load step can be seen on page B81.4. Note that the concrete has crushed in the compressed region under the point load and that a large number of flexural cracks have opened in the concrete but we see no longitudinal splitting in the compressed zone or horizontal splitting along the reinforcement. SOLVIA calculates the ultimate load as 360.8 kN with a maximum mid-span deflection of 11.7 mm.
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| In the second analysis, B81A, SIGMAT is set to half the value of the modulus of rupture. The KAPPA-value is increased to keep the tension stiffening effect constant. The SHEFAC value is increased to 0.75 to keep the shear strength in the crack planes. The results from the analysis can be seen on pages B81.5 and B81.6. The mid-span deflection as a function of the load can be seen in the top figure on page B81.5. A vector plot of crack normals in the bottom figure on page B81.5 shows that a horizontal crack has started in the compressed zone and that diagonal cracks have developed along the reinforcement near the end of the beam. A zone under the point load is crushed as seen in the top figure on page B81.6.
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| SOLVIA calculated the ultimate load as 351.2 kN with a maximum mid-span deflection of 12.9 mm.
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| In the third analysis, B81B, a nonlinear-elastic material model is used to simulate the concrete behaviour. In the nonlinear-elastic model the directions of the principal stresses and principal strains coincide so it is a simple "rotating crack plane" approach. The results from the analysis can be seen on pages B81.6 and B81.7. The ultimate load is found to be 339.6 kN with a mid-span deflection of 10.4 mm.
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| User Hints
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| "° Note that the shear stress in an active crack plane cannot be larger than SIGMAT*SHEFAC in the SOLVIA concrete model. This restriction is included in the model to prevent non-physical large shear stresses in crack planes.
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| ° Note that the value of SIGMAT, the maximum tensile strength in the concrete, is critical in this example. Decreasing the value of SIGMAT by fifty percent, the "diagonal tension" failure mode found in the experiments is indicated in analysis B81A.
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| Reference
| |
| [1] Bresler, B., and Scordelis, A., "Shear Strength of Reinforced Concrete Beams", ACI Journal, Proceedings V. 60, No. 1, January, 1963.
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| Version 99.0 B81.2
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| SOLVIA Verification Manual Nonlinear Examples B81 BRESLER-SCORDEL7S CONCRETE BEAM ORIGINAL - 0.2 TIME 1 FORCE i00100 M1ASTER B 101 C i10111 ORIGINAL -- 0.2 z
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| -- y EAXES=RST SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB B81 BRESLER-SCORDELIS CONCRETE BEAM 1-C Cz C ',K
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| ,NI tS-0 200 25 0 300 VALUE OF TINEFJNC -LCN i SOLVTA-POST 99 0 SOLVIA ENG1 NEERING AB Version 99.0 B81.3
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| SOLVIA Verification Manual Nonlinear Examples B81 BRESLER-SCORDEL>S CONCRETE BEAM
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| ý'AX DSPL. H-- 0. 01186 TIME 18.04 -Y CRACK NORMA SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B81 BRESLER-SCORDELIS CONCRETE BEAM z
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| MAX DISPL. i-- 0.0:186 TIME 18.04 Ly LOAD I .804E5 CONCRETE U CRUSHED MNORMAL
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| .:CLOSED I CRACK 2 CRACKS 3 CRACKS SOLViA-POST 99.0 SOLVIA ENG:NEERING AB Version 99.0 B81.4
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| SOLVIA Verification Manual Nonlinear Examples B8IA BRESLER-SCORDELiS CONCRETE BEAM, SIGMAT=2.15 MP1 O
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| ,<1 J.
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| 2:
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| 2:
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| \.
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| C so [0s 200 250 300 350 qo06 VALUE 0F TZMEFJNCCTON 1 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B81A BRESLER-SCORDELDS CONCRETE BEAM, SIGMAT=2.,1S MPa MAX DISPL. -- 0.012959 z TIME t7.575 Ly CRACK NORMAL SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 B 81.5
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| SOLVIA Verification Manual Nonlinear Examples B81A BRESLER-SCORDELIS CONCRETE BEAM, SI*MAT=2.!5 MP MAX DISPL. i 0.012959 Z TIME 17575 Y
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| LOAD 1.7S7SES CONCRETE SCRUSHED NCRMAL CLOSED I CRACK 2 CRACKS 3 CRACKS SOLVIA-POST 99.0 SOLVIA ENGINEERING AB I
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| B81B BRESLER-SCORDELIS CONCRETE BEAM, NONLINEAR-ELASTiC PLANE CD.
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| 0J O
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| C)
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| --SO T- --O I---
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| 0 200 250 300 VALUE OF FNC Ti"ft "INI SOLVIA-POST 99.0 SOLV7A ENGINEERING AB Version 99.0 B81.6
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| SOLVIA Verification Manual Nonlinear Examples B81B 3RESLER-SCORDEL.S CONCRETE BEAM, NONLINEAR-ELASTIC PLANE MAX DISPL.' 0.01056 TIME !6.981 Y SPRINCIPAL 2.3594E7
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| -2.3594E7 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB 381B BRESLER-SCORDELTS CONCRETE BEAM, NONLINEAR-ELASTIC PLANE MAX D7SPL. 0.01056
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| . Z TIME 16.981 1 REACTION
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| : 8. $748E5 TRESCA MAX 2. 4105E7!
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| S2.2599E7
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| ' 1....046 2E 7 3559E7
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| : 7. 5329E6
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| : 4. 5197E6
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| :.5066E6 MIN 0 SOL!IA-POST 99.0 SOLVIA ENGINEERTNG AB Version 99.0 1381.7
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B81 BRESLER-SCORDELIS CONCRETE BEAM' DATABASE CREATE MASTER IDOF=I00111 NSTEP=50 ITERATION METHOD=BFGS AUTO-STEP DTMAX=1 ITELOW=7 ITEHIGH=15 FDECREASE=0.2 TMAX=20.
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| TOLERANCES TYPE=F RNORM=80.E3 RTOL=0.01 ITEMAX=30 STOL=0.3 TIMEFUNCTION 1
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| : 0. 0. / 20. 400.
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| COORDINATES / ENTRIES NODE Y Z 1 0. 0.56 / 2 2.285 0.56 3 0. 0.095 / 4 2.285 0.095 5 0. 0. / 6 2.285 MATERIAL 1 CONCRETE EO=28700.E6 NU=0.2 SIGMAT=4.33E6, SIGMAC=-23.7E6 EPSC=-0.0022 SIGMAU=-20.2E6, EPSU=-0.0031 BETA=0.75 KAPPA=iS. STIFAC=0.0001, SHEFAC=0.5 MATERIAL 2 PLASTIC E=218000.E6 YIELD=555.E6 ET-10000.E6 EGROUP 1 PLANE STRESS2 MATERIAL=1 GSURFACE 2 1 3 4 EL1=10 EL2=4 NODES=8 GSURFACE 4 3 5 6 EL1=10 EL2=1 NODES=8 EDATA / 1 0.305 EGROUP 2 TRUSS MATERIAL=2 GLINE 3 4 EL=-0 NODES=3 EDATA / 1 0.00329 FIXBOUNDARIES 3 INPUT=NODE / 5 FIXBOUNDARIES 2 INPUT-LINE / 2 4 / 4 6 LOADS CONCENTRATED / 2 3 -500.
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| SET VIEW=X NSYMBOLS=MYNODES MESH NNUMBERS=MYNODES VECTOR=LOAD BCODE=ALL SUBFRAME=12 MESH EAXES=RST GSCALE=OLD SOLVIA END Version 99.0 B81.8
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B81 BRESLER-SCORDELIS CONCRETE BEAM DATABASE CREATE WRITE FILENAME='b8l.1is' SET VIEW=X DIAGRAM=GRID NHISTORY NODE=4 DIRECTION=3 XVARIABLE=1 OUTPUT=ALL MESH VECTOR=CRACK-NORMAL MESH VECTOR=LOAD CONTOUR=CONCRETE SUMMATION KIND=LOAD SUMMATION KIND=REACTION DETAILS=YES EGROUP 1 ZONE NAME=MID INPUT=ELEMENT / 1 STEP 10 TO 41 ELIST ZONENAME=MID SELECT=CONCRETE ELIST ZONENAME=TRUSS END Version 99.0 B 81.9
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B82 COMPRESSION OF A RUBBER O-RING, PLANE STRAIN Objective To verify the rubber material model in a PLANE STRAIN and PLANE AXISYMMETRIC contact analysis.
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| Physical Problem The compression characteristics of O-rings have been investigated by Lindley [1]. One of these 0 rings, denoted A2, will be analyzed using a plane strain approach. The geometry of the O-ring section can be seen in the figure below. Lindley has used a constant Young's modulus, E, in his investigation as defined in the figure below. For comparison, a PLANE AXISYMMETRIC model with and without friction is also analyzed.
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| P D = 9.73 cm d= 1.2cm A C, = 2.75 kgf./cm 2 d C2 = 0.50 kgf./cm 2 2
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| K= 29250 kgf./cm 2
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| "E= 19.5 kgf./cm L~D Finite Element Model In the first model, the toroidal rubber ring is idealized as a rubber cylinder pressed between two plates. Due to symmetry only a quarter of the cylinder needs to be considered. The model consists of 150 4-node PLAIN STRAIN elements with 10 elements in the radial direction and 15 elements in the circumferential direction. The applied loading is described using prescribed displacements. The con tact conditions are modeled with the contactor surface defined on the rubber cylinder and a rigid frictionless target surface defined as the lower plate. The Mooney-Rivlin constants C, and C, and the bulk modulus Kare defined in the figure above. Ten equal load steps are applied for the prescribed displacement using Full-Newton equilibrium iterations. The finite element model can be seen in the top figure on page B82.3.
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| Solution Results The input data used in the SOLVIA analysis can be seen on pages B82.9 and B82.10.
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| In his experimental study, Lindley found an empirical force-displacement relationship P A)3/2 S-=Ed( 1.25(-d +50 d where Young's modulus is held constant.
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| Version 99.0 1382.1
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| SOLVIA Verification Manual Nonlinear Examples The bottom figure on page B82.3 shows the force-displacement curve with symbols as calculated in the SOLVIA analysis. The solid line without symbols is the empirical formula described by Lindley.
| |
| The top figure on page B82.4 shows the effective stress distribution in the rubber in the last load step and the bottom figure on the same page shows the hydrostatic pressure distribution in the last load step. The figures on page B82.5 show the distribution of maximum and minimum principal stretches.
| |
| For comparison a PLANE AXISYMMETRIC model has also been used to analyze the rubber O-ring as shown in the top figure on page B82.6. In example B82A the friction coefficient between the rubber and the steel plate is 0.7 which corresponds to a dry O-ring [I]. The SOLVIA-PRE input is given on page B82.11.
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| The bottom figure on page B82.6 shows the force-displacements curve with symbols as calculated by SOLVIA together with the empirical formula described by Lindley drawn as a solid line. The figures on page B82.7 show the von Mises stress distribution and the hydrostatic pressure distribution in the last load step.
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| In example B82B the friction coefficient between the rubber and the steel plate is 0. The top figure on page B82.8 shows the force-displacement curve with symbols as calculated by SOLVIA together with the curve from Lindley. The bottom figure on page B82.8 shows the distribution of the von Mises stress distribution in the last load steps. Note that with no frictional resistance between the rubber and the steel plate the radial displacement of the O-ring is large and the force needed to compress the rubber is significantly reduced.
| |
| User Hints
| |
| "* Note that the friction between the rubber and the steel is important in the PLANE AXISYMMETRIC analysis as the friction stiffens the O-ring.
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| " The 4-node PLANE STRAIN/AXISYMMETRIC element with the u/p-formulation has a constant pressure assumption.
| |
| " Note that the axisymmetric element extends one radian in the circumferential direction.
| |
| Reference
| |
| [1] Lindley, P.B., "Compression Characteristics of Laterally-unrestrained Rubber O-rings", J. Instn Rubb. Ind. 4, pp. 209-213, 1967.
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| Version 99.0 B82.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples B82 COMPRESS7ON OF A RUBBER O-RING, PLANE STRAIN ORIGINAL
| |
| * H 0.i z y
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| t3 SOLVIA-PRE 99.0 SOLVIA ENGINEER!NG AB B82 COMPRESSION OF A RUBBER O-RING, PLANE STRAIN
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| ,Li L-1 C,
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| ... 3..
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| . i. ..
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| 0.00 O.OS 0.20 0.5 0.30 DISPLACEL-IENI1 SOLVA -POST 99.0 SOLViA ENGINEER NG AB Version 99.0 B 82.3
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| SOLVIA Verification Manual Nonlinear Examples B82 COMPRESSION OF A RUBBER O-RING PLANE STRAIN CRIGINAL - 0.1 z I-MAX DISPL. 0.49575 7I7E 10
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| . . .7..
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| I
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| /
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| /
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| /
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| REACTION 14.6So MISES MAX 30. 532
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| : 28. 657 24.907 21 . 158 17.408
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| : 13. 658 9.9077
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| : 6. 1-578
| |
| : 2. 4079 MIN 0.53294 SOLVWA-POST 99.0 SOLVIA ENGINEERING AB I
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| B82 COMPRESSION OF A RUBBER O-RING PLANE STRAIN ORIGINAL i-- 0.1 z MAX DISPL. 0 .4957-5 L y TIME 10
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| /
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| /
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| /
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| /
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| CONTACT 9.38ai MEAN STRESS MAX 0.46248 0.7S980
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| ý-3.2041
| |
| -S. 6489
| |
| -8. 0935 S-'0. 538
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| :-1.2.983
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| -15. 427
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| -:7.87?
| |
| MIN-9. 094 SOLVIA-POST 99 SOLVTA ENGINEERING AB Version 99.0 1382.4
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| SOLVIA Verification Manual Nonlinear Examples B82 COMPRESSION OF A RUBBER r-RING PLANE STRAIN ORI 0 \NAL -- 0.1 z I
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| MAX DISPL. 0. 49575 Y
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| TIME 13
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| /
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| PRINCIPAL STRETCH MAX MAX 2.3262 S2. 2450 9 99 2.082
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| .75741 1-iii 1 .5949
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| .4324 1 .2699 1.1074 MIN 1.0261 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B82 COMPRESSION OF A RUBBER O-RING PLANE STRAIN ORIGINAL z MAX DISPL.
| |
| , - 0.1 0.49575 Ly TIME 10
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| /
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| /
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| /
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| PRINCIPAL STRETCH MTN MAX 0.97973 0o.94536
| |
| ::0. 87661 0.80786
| |
| : 0. 7391.
| |
| : 0. 67036
| |
| : 0. 6016.
| |
| "0 53286 MIN 0 42973 SOLV A-POST 99.0 COLVIA ENGINEERING AB Version 99.0 B82.5
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| SOLVIA Verification Manual Nonlinear Examples B82A COMPRESS-7N OF A RUBBER C-RING AXIS(MMETRIC ORIGINAL H - ; 0.2 z R
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| 2221 20,19 18 17 18 15 t4 1311 10 9 8 7 6 5 4 3 2 1
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| ,25 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB B82A COMPRESSTON OF A RUBBER O-RING, AXISYMMETRIC
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| - -10.
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| -:1 2'o ... . ... .... .... ...... . ... .....
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| :2 O.OC .. .. . .........
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| 3.,O3 0 t0 0 .5 0. 20 0.25 J.30 Di£SPLACEIE N1-SOLVIA-P1ST 99.0 SOLVIA ENGINEERING AB Version 99.0 B82.6
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| SOLVIA Verification Manual Nonlinear Examples B82A COMPR! 3N OF A RUBBER O-RING, AXISY1MMETRIC ORIGINAL ,-- 39.2 MAX DISPI 1 0. 52894 Y TIME 10 R
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| ,/
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| REACTION 188. 13 MISES MAX 44.721 S42. 029
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| : 36. 44 311. 259 25.874
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| : 20. 188
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| ' S. 103 9.7183 4.3332 MIN 1.6406 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B82A COMPRESSION OF A RUBBER 0-RING, AXISYMMETRTC ORIGINAL 0.2 z MAX DISPL.
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| TIME 10 iC0.2894 LyR
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| /'-7-/
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| CONTACT 97.585 MEAN STRESS MAX 3.0303 r7 .07 i 16 9 3 iii-3. 9098
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| ;(-8.8-366
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| *<-13.1 -63 ii~i-
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| - 7. 790 i* -22.4i7
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| !iiii-27.0o41 iii-3 ' .67 MIN-33.984 SOLV'A- POST 99.0 SOLVIA ENGINEER NG A8 Version 99.0 B82.7
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| SOLVIA Verification Manual Nonlinear Examples B82B COMPRESSION OF A RUBBER O-RING AXISYMMETRIC
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| "'r*)
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| 0*
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| 0 -
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| C .30 i1 0.00 .05 0. t O. 15 0.20 0.2S DISPLACEMENT SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B82B COMPRESSION OF A RUBBER C-RING, AXISYMMETRIC ORIGINAL , 0 . z MAX DISPL.
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| TIME 10 1.6838 LYR REACTION
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| / 85. 332 MISES MAX 18.241 S7. 41S 1 5.762 1;:
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| -2.4S7 i_ 805
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| :::9. 5.S24
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| : 7. 4998
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| . 8473 MIN 5 0211 SOLVIA-POST 99.I SOLVIA ENGINEERING AB Version 99.0 1382.8
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B82 COMPRESSION OF A RUBBER O-RING, PLANE STRAIN' DATABASE CREATE MASTER IDOF=100111 NSTEP=10 DT=-.
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| ITERATION METHOD=FULL-NEWTON TOLERANCES TYPE F RNORM-10. RTOL=0.01 TIMEFUNCTION 1
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| : 0. 0. / 10. 0.3 SYSTEM 1 CYLINDRICAL COORDINATES / ENTRIES NODE R THETA 1 0.6 0. TO 11 / 12 0.6 -90. / 13 0.6 -12.
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| COORDINATES SYSTEM=- / ENTRIES NODE Y Z 14 0. -0.6 / 15 0.9 -0.6 MATERIAL 1 RUBBER C1=2.75 C2=0.5 KAPPA=29250.
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| EGROUP 1 PLANE STRAIN PRESSURE-INTERPOLATION:YES GSURFACE 11 12 13 11 ELi=i0 EL2=13 NODES=4 SYSTEM=i GSURFACE 11 13 1 11 EL1=10 EL2=2 NODES-4 SYSTEM=i CGROUP 1 CONTACT2 SUBTYPE-STRAIN CONTACTSURFACE 1 INPUT=LINE 12 13 CONTACTSURFACE 2 INPUT=NODE 15 14 CONTACTPAIR 1 CONTACTOR=1 TARGET=2 FIXBOUNDARIES / 14 15 FIXBOUNDARIES 2 LINE / 11 12 CONSTRAINTS 1 3 11 3 1. TO 10 3 11 3 1.
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| LOADS DISPLACEMENTS / 11 3 -1.
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| MESH NNUMBERS=MYNODES NSYMBOLS=MYNODES SOLVIA END Version 99.0 B82.9
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B82 COMPRESSION OF A RUBBER O-RING, PLANE STRAIN DATABASE CREATE WRITE FILENAME='b82.1is' AXIS 1 VMIN-0. VMAX=0.3 LABELSTRING='DISPLACEMENT' AXIS 2 VMIN=--5. VMAX=O. LABELSTRING='FORCE' USERCURVE 1 READ B82DISP.DAT SET VIEW=X DIAGRAM=GRID OUTLINE=YES ORIGINAL=YES NHISTORY NODE=1 DIRECTIONW3 KIND-REACTION XAXIS=- YAXIS=2, SYMBOL=l XVARIABLE=l OUTPUT=ALL PLOT USERCURVE 1 XAXIS=-l YAXIS=-2 SUBFRAME=OLD MESH CONTOUR=MISES VECTOR=REACTION MESH CONTOUR=SMEAN VECTOR=CONTACT MESH CONTOUR=LPMAX MESH CONTOUR=LPMIN CLIST GLIST ZONENAME=Cl EMAX SELECT=S-EFFECTIVE NUMBER=5 EMAX SELECT=STRETCHES TYPE-MAXIMUM NUMBER=5 EMAX SELECT=STRETCHES TYPE=MINIMUM NUMBER=5 END Version 99.0 B82.10
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input B82A HEADING 'B82A COMPRESSION OF A RUBBER O-RING, AXISYMMETRIC' DATABASE CREATE MASTER IDOF=100111 NSTEP=10 DT=-.
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| ITERATION METHOD-FULL-NEWTON TOLERANCES TYPE=F RNORM=10. RTOL=0.01 TIMEFUNCTION 1
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| : 0. 0. / 10. 0.3 SYSTEM 1 CYLINDRICAL Y=4.865 Z=0.6 COORDINATES / ENTRIES NODE R THETA 1 0.6 0. TO 11 / 12 0. 180. TO 22 0.6 180 23 0.6 -168. / 24 0.6 -12.
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| DELETE 12 COORDINATES SYSTEM=0 / ENTRIES NODE Y Z 25 3.865 / 26 5.865 MATERIAL 1 RUBBER C1=2.75 C2=0.5 KAPPA=29250.
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| EGROUP 1 PLANE AXISYMMETRIC PRESSURE-INTERPOLATION=YES GSURFACE 11 22 23 11 EL1=10 EL2=2 NODES=4 SYSTEM=I GSURFACE 11 23 24 11 EL1=10 EL2=26 NODES=4 SYSTEM=1 GSURFACE 11 24 1 11 EL1=10 EL2=2 NODES-4 SYSTEM=i CGROUP 1 CONTACT2 SUBTYPE=AXISYMMETRIC CONTACTSURFACE 1 INPUT=LINE 23 24 CONTACTSURFACE 2 INPUT=NODE 26 25 CONTACTPAIR 1 CONTACTOR=1 TARGET=2 FRICTIONCOEFFICIENT=0.7 FIXBOUNDARIES / 25 26 CONSTRAINTS 1 3 ii 3 1. TO 10 3 11 3 1.
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| 13 3 11 3 1. TO 22 3 11 3 1.
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| LOADS DISPLACEMENTS / 11 3 -1.
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| MESH NNUMBERS=MYNODES NSYMBOLS=MYNODES SOLVIA END Version 99.0 B82.11
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B83 COMPLEX-HARMONIC ANALYSIS OF A PRE-LOADED BEAM Objective To verify the the COMPLEX-HARMONIC method on a pre-loaded structure.
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| Physical Problem A simply supported beam as shown in the figure below is considered. A large axial compressive pre load and a small mid-span sinusoidal force are applied.
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| yPin F l b P=10N E=2.0.10" N/m 2 3
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| L=5m p = 7800 kg/m 3
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| b = 0.1 m F = -600.10 N 1=8.3333.l0- m A=b.b=0.01 m 2 Finite Element Model The model is shown on page B83.4 and consists of twenty 2-node BEAM elements. Mass propor tional Rayleigh damping which corresponds to a modal damping ratio of 0.02 at the first natural frequency is considered. Since co, = 27c. 2.7212 radians per second the mass proportional damping coefficient a to be used in Rayleigh damping is u. = 2. - (o = 0.6839. Transverse shear deformation effects are excluded.
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| Solution Results The analytical steady state solution is based on the Euler-Bemoulli beam theory. If the displacement in the Y-direction is denoted v(x,t) then the equilibrium equation is 2
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| aD v D'v a v pA +apA-+ EI x -+na-xxx=- 2 Psin t at Separation of variables by v = p(x). Y(t) gives the undamped mode shapes and circular frequencies as
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| % n(x)= sin~n1JL n= 1,2,3,....
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| _n 2 T2 EI1+ L2F 2 2 L ýpA n 7g2EI We may note that setting w,, = 0 gives the Euler buckling loads Fcr F n 1-2E 7cEl Version 99.0 B83.1
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| SOLVIA Verification Manual Nonlinear Examples The steady state solution is obtained in complex representation as V=Imtveiot J= l*-" F P.D (cos0n +isinO,).sin n-rx .eint where
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| -[*) + (2(p p2 )21 -1/2
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| =(
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| COn 20)n, I n 1,5,9,...
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| I n 3,7,11,....
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| { n even 0 =tan -' 2 _.P n Using the amplitude I-V and the phase angle 0 the steady state response can be written v = Im{ Iv [e where
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| + 8 .sin s. sn--
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| pA=c C02 o s O si n L
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| I n~ CO 2 21 sin 0 sin n -tx
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| [D oW*=2 n- si n L 0 =tan - '
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| l8. . cos 8, nrcx 2 L)
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| The axial load is applied in the first run using the input data on page B83.5. Then, the input data for the second run on pages B83.5 and B83.6 is employed in a restart analysis to calculate the complex harmonic response. The maximum displacement amplitude of node 3 is 4.385 mm and occurs at the frequency 2.720 Hz. The theoretical solution gives a maximum amplitude of 4.377 mm at the fre quency 2.726 Hz. The amplitude and phase angle response calculated by SOLVIA as well as the theoretical solution are shown on page B83.4. The SOLVIA amplitude solution is shown using both linear and logarithmic axes.
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| The vector plot in the lower right figure is an SRSS case combination of the real and the imaginary response at the frequency 2.720 Hz, where the maximum amplitude response occurs.
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| User Hints
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| * Since the damping values are rather small the command NHARMONIC can also be used.
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| "* The Rayleigh damping adds a mass and a stiffness proportional part to the total consistent damp ing matrix.
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| C = O* + +/-KLIN + CCONC + CELEMENT where Version 99.0 B83.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples M Total system mass matrix.
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| KLIN Stiffness matrix based on the elements in all linear element groups.
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| a, [3 Rayleigh damping factors specified by user.
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| CcoNc Damping matrix contributed by the concentrated nodal dampers.
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| CELEMENT Damping matrix contributed by the GENERAL and SPRING elements with viscous damping and PLANE, SOLID and SHELL with material damping.
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| The relation between the modal damping ratio and the Rayleigh damping factors 2wo 2 The relation is shown graphically for a = 0.6839 and [3= 5. 10 3 in the figure below.
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| * Stiffness proportional damp proportional tfMass damping:
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| * r ' ' 20 ' 350 4 0 0o CIRCULARFRECUENCY
| |
| *We can only use mass proportional Rayleigh damping in this example, since the BEAM elements are considered nonlinear due to the calculation of the geometric stiffness.
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| * For PLANE, SOLID and SHELL elements viscous and hysteretic material damping can be used in COMPLEX-HARMONIC analysis.
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| * The same model is used in example B71 where the natural frequency with a lumped mass matrix assumption was calculated to 2.722 0z.
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| "* The BEAM element with a consistent mass matrix includes rotary inertia which lowers the natural frequencies somewhat. The theoretical frequencies for the Euler-Beoulli beam do not consider rotary inertia.
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| "* The complex harmonic method in SOLVIA includes the effect of damping also on the distribution "ofthe amplitude response. Hence, the amplitude distribution at a resonant frequency may not correspond exactly to the corresponding mode shape calculated for the undamped model.
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| References
| |
| "[1] Clough. R.W., and Penzien, J., Dynamics of Structures, Second Edition, McGraw-Hill, 1993.
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| [2] Blevins, Robert D., Formulas for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company, 1979.
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| Version 99.0 B83.3
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| SOLVIA Verification Manual Nonlinear Examples 383A COMPLEX-HARMONIC ANALYSIS OF A PRE-LOADED BEAM, SECOND RUN 883A COMP1CX-KARMONIC ANALYSIS OF A PRE-LOADEO BEAM. SECOND RUN
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| :i........ ..... .. .... ..... ...... ........
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| IN.
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| C !
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| *.!..I . . . .... . . ... . ..! ........ . . . . . . ......... ... .. ........... ... .. . ... . . .. . .. .
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| I 6 2 3 4 S 2
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| FRECBELOCY ar
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| : 99. 5 OLVIA ENGINEER:NG AB SOLVIA -OST 99.0 SOLVIA ENGINEERING AB SOLVIA-POST 2 30 S B83A COMPLEX-HARMONIC ANALYSIS OF A PRE-.OADED BEAM. SECOND RUN T'eoretical so!ion SLVIFENGIEERIG A SOL1ý-POT 993 CL3 -CPE-AMC NA O RELAEDBA, SIOO U C'C
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| 'NI' I on CO N .. N.....
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| .CO .. .............. C C83A COPCECARC0NI AALOS:S. A PRE-LOADED BEAC.SECCND RUN ORIGINAL -
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| . . .. . . . . . . . . .. . . . . ~~~~~~~~~IA. VALUE*
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| .. CGINAL- CL
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| , ZSýE-l- _,
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| OXSLACE-I.T SOL,/A-00ST 99A SOLVIA ENGINE01RBNG AB A P C ANC L C L C T_
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| E.,
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| 93.4 EN;INE_ýING Ab Version 99.0 83.4
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input, first run HEADING 'B83 COMPLEX-HARMONIC ANALYSIS OF A PRE-LOADED BEAM, FIRST RUN' DATABASE CREATE MASTER IDOF=001110 KINEMATICS DISPLACEMENT=LARGE ITERATION METHOD=BFGS COORDINATES 1 / 2 5. / 3 2.5 MATERIAL 1 ELASTIC E=2.EI1 DENSITY=7800 EGROUP 1 BEAM SECTION 1 GENERAL RINERTIA=1.40577000E-05 SINERTIA=8.33333333E-06 TINERTIA=8.33333333E-06 AREA=1.E-02 BEAMVECTOR / 1 0. 0. 1.
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| GLINE 1 2 AUX=-- EL-20 FIXBOUNDARIES 12 / 1 FIXBOUNDARIES 2 /2 LOADS CONCENTRATED / 2 1 -600.E3 SOLVIA END SOLVIA-PRE input, second run DATABASE OPEN HEADING 'B83A COMPLEX-HARMONIC ANALYSIS OF A PRE-LOADED BEAM, SECOND RUN' MASTER IDOF=001110 MODEX=RESTART TSTART=-.
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| RAYLEIGH-DAMPING ALPHA=0.6839 BETA-0.
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| COMPLEX-HARMONIC 1.0 8 / 2.6 200 / 2.8 20 / 4.0 6 6.0 DELETE LOADS CONCENTRATED LOADS CONCENTRATED / 3 2 i0.
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| SOLVIA END Version 99.0 1383.5
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input, second run
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| * B83A COMPLEX-HARMONIC ANALYSIS OF A PRE-LOADED BEAM, SECOND RUN DATABASE CREATE WRITE 'b83a.lis' SET PLOTORIENTATION PORTRAIT DIAGRAM=GRID NCOMPLEX-HARMONIC NODE=3 DIRECTION=2 KIND=DISPLACEMENT OUTPUT=ALL AXES-SCALE=3 NCOMPLEX-HARMONIC NODE=3 DIRECTION=2 KIND=DISPLACEMENT AXES-SCALE=l USERCURVE 1 XLABEL='FREQUENCY' / READ B83DISP.DAT USERCURVE 2 XLABEL='FREQUENCY' YLABEL='PHASE ANGLE' READ B83PHASE.DAT AXIS 1 VMIN=- VMAX=5.OE-3 LABEL='DISPLACEMENT' PLOT USERCURVE 1 YAXIS=l TEXT STRING='Theoretical solution' X=2 Y=23.5 SET PLOTORIENTATION=LANDSCAPE PLOT USERCURVE 2 TEXT STRING='Theoretical solution' X=3 Y=15.75 MESH VIEW-Z BCODE=ALL NNUMBER=MYNODES NSYMBOL=YES ORIGINAL=YES DEFORMED=NO SUBFRAME=I2 CASECOMBINATION 1 STANDARD FUNCTION=SRSS 2.720 1. REAL 2.720 1. IMAGINARY SET RESPONSETYPE=CASECOMBINATION MESH ORIGINAL=YES NSYMBOL=YES VECTOR=DISPLACEMENT NLIST N3 END Version 99.0 B83.6
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B84 IMMERSION OF A SHELL STRUCTURE INTO WATER, BUOYANCY EFFECTS Objective To verify the SHELL element load option with interpolation of intensity along a global coordinate axis considering large displacements.
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| Physical Problem The figure below shows a shell structure to be immersed into water. The buoyancy force as a function of the immersion depth is to be calculated.
| |
| z R=l m vLength L =4 m Thickness h = 0.020 m 2
| |
| E =2.1.10" N/m v=0.3 3
| |
| Pwater =
| |
| 10 0 0 kg / in
| |
| -- =9.81 n/s 2 Finite Element Model The finite element model is shown on page B84.2 and consists of 4-node SHELL elements. Pre scribed displacements at node 2 and constraint equations are used to move the structure in the negative Z-direction. The prescribed displacements are applied in ten time steps. Full-Newton iteration and force tolerances are used.
| |
| The water pressure load is applied by the element load option of a linear pressure variation fixed in space. When the model is immersed into the water the pressure is automatically calculated based on the current large displacement configuration. The element load option of an element fixed pressure is also available. Both options of pressure load are deformation dependent.
| |
| Solution Results If the shell structure is assumed to be rigid the total vertical reaction F can be expressed as a function of the immersion depth 8 as F=-gp rLR2(0 -sin )/2 where 0=2arccos- RR---) for 3Ž0 The input data on pages B84.4 and B84.5 gives the following results:
| |
| Vertical displacement Total vertical reaction force (N) 8 Theory SOLVIA m B84 B84A 0.1 -2304 -2273 -2304 0.2 -6416 -6352 -6407 0.5 -24101 -24027 -24081 1.0 -61638 -61378 -61571 Version 99.0 B84.1
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| | |
| SOLVIA Verification Manual Nonlinear Examples Case B84A has the same input as B84 except that 40 elements instead of 20 are used along the circumference.
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| The top figure on page B84.3 shows the force-displacement diagram calculated by SOLVIA for case B84 together with the theoretical solution. The two curves almost coincide. The configuration in the last solution step together with the reaction force are shown in the bottom figure.
| |
| The nodal loads on the structure at time 10 are shown in the top figure on page B84.4.
| |
| User Hints
| |
| "* With 4-node SHELL elements the geometry of the half cylinder is approximated by an assembly of flat elements with all nodes lying on the midsurface of the cylinder. Hence, the volume of the model is smaller than the volume of the perfect half cylinder.
| |
| The pressure variation along the edge of the flat 4-node element is linear, which agrees with the water pressure variation. Hence, the calculated buoyancy force is smaller than the theoretical value for a perfect half cylinder, since the only error is due to the smaller volume.
| |
| " A better description of the circular geometry is achieved by the 9-node and 16-node SHELL elements. However, the pressure variation along the edges of these elements is still linear. Since the actual pressure along the circular boundary varies in a sinusoidal manner there will be an approximation in the pressure load application, which counteracts the better geometry of these SHELL elements.
| |
| "* The change in volume due to elastic deformation of the SHELL structure could be significant if the SHELL thickness is small.
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| B84 IMMERSION OF A SHELL STRUCTURE INTO WATER, BUOYANCY EFFECTS 0 . Z ORIGINAL ;
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| ZONE SHELL x -y E.
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| MASTER B 01O00O 0:0 !3u_0 C iO00G SHELL TOP SSHEL BOTTON NON-SHELL SOLV A ENGINEERING AB SOLVTA-PRE 9 9.0 Version 99.0 B84.2
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| SOLVIA Verification Manual Nonlinear Examples B84 :MMERSTCN OF A SHELL STRUCTURE INTO WATER, 3UOYANCY EFFECTS N
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| w 0
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| 0 -I 0.0 0 .0) 0.2 0.3 0.4 0.s 0.6 0 7 0.8 0.9 DISPLACEIENT SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B84 IMMERSICN OF A SHELL STRUCTURE INTO WATER, BUOYANCY EFFECTS ORIGINAL -- -ý 0 .S z MAX DISPL. i t N TIME 10 X- I-Y ZONE SHELL REACTION 61378 SOLVTA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 B84.3
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| SOLVIA Verification Manual Nonlinear Examples B84 IMMERSICN OF A SHELL STRUCTURE INTO WATER, BUOYANCY EFFECTS ORIGINAL 0-.2 ziZ 7TME 10 LOAD 1018.9 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB SOLVIA-PRE input DATABASE CREATE HEADING 'B84 IMMERSION OF A SHELL STRUCTURE INTO WATER, BUOYANCY EFFECTS' PARAMETER $CIRC=20 $LONG=6 $SHORT=4 SET NODES=4 MASTER NSTEP=10 DT=1 MEMORY SOLVIA=20 KINEMATICS DISPLACEMENT=LARGE ITERATION METHOD=FULL-NEWTON TOLERANCES TYPE=F RTOL=1.E-3 RNORM=I.E5 RMNORM=1.E5 PORTHOLE SAVEDEFAULT=NO TIMEFUNCTION 1 / 0. 1. / 10 1.0 TIMEFUNCTION 2 / 0. 0. / 10 1.0 COORDINATES ENTRIES NODE X Y Z 1 4. 1. TO 3 4. -1. / 4 0. 1. TO 6 0. -1.
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| MATERIAL 1 ELASTIC E=2.0E11 NU=0.3 DENSITY=-.
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| Version 99.0 B84.4
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
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| EGROUP 1 SHELL MATERIAL=i SYSTEM 1 CYLINDRIC LINE CYLINDRIC 1 3 EL=$CIRC ROTATION=NEGATIVE SYSTEM=i LINE CYLINDRIC 4 6 EL=$CIRC ROTATION=NEGATIVE SYSTEM=i GSURFACE 1 3 6 4 EL2=$LONG GSURFACE 2 3 1 2 ELI=$SHORT GSURFACE 5 4 6 5 EL1=$SHORT THICKNESS 1 0.020 LOADS ELEMENT INPUT=SURFACE INTERPOLATION=Z SYSTEM=GLOBAL-SPACEFIXED COORD1=-I. Vi=0. COORD2=-2.
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| V2=9810. VMIN=0.
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| 1 3 6 4 t. . . . . 1 2 3 1 2 t. . . . . 1 5 4 6 5 t. . . . . 1 FIXBOUNDARIES 1 INPUT=LINE / 1 2 / 2 3 / 4 5 / 5 6 FIXBOUNDARIES 2 INPUT-LINE / 1 4 / 3 6 LOAD DISPLACEMENT 2 3 -1. 2 CONSTRAINTS 1323/3323/4323/5323/6323 COLOR EFILL=SHELL VIEW ABC X=2 Y-i Z=0.5 SET BCODE=ALL NNUMBER=MYNODES NSYMBOL=MYNODES MESH ZONE=SHELL VIEW=ABC SOLVIA END SOLVIA-POST input
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| * B84 IMMERSION OF A SHELL STRUCTURE INTO WATER,
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| * BUOYANCY EFFECTS DATABASE CREATE WRITE 'b84.1is' AXIS 1 VMIN=.0 VMAX=1.0 LABELSTRING='DISPLACEMENT' AXIS 2 VMIN=-7.5E4 VMAX=0.0 LABELSTRING='FORCE' USERCURVE 1 / READ B84.DAT SET DIAGRAM=GRID NHIST NODE=2 DIRECTION=3 KIND=REACTION XAXIS=i YAXIS=2, SYMBOL=i XVAR=2 OUTPUT=ALL PLOT USERCURVE 1 XAXIS=-i YAXIS=-2 SUBFRAME=OLD VIEW ABC X=2 Y=1 Z=0.7 MESH ZONE-SHELL ORIGINAL=DASHED VECTOR=REACTION VIEW=ABC MESH VECTOR=LOAD VIEW=X ORIGINAL=YES DEFORMED=NO END Version 99.0 B84.5
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B85 PURE SHEAR OF A RUBBER SHEET Objective To verify the rubber material model in plane pure shear deformation using PLANE STRESS2 and PLANE STRAIN elements.
| |
| Physical Problem A quadratic rubber sheet of unit side length deforms with displacement u in pure shear.
| |
| Initial configuration at time 0 Deformed configuration at time t Mooney-Rivlin constants:
| |
| C 1 =0.125MPa SC 2 = 0.025 MPa I0, = 10 3 MPa (bulk modulus) 0 0 0 1 0 1 1+u Finite Element Model In the first analysis one PLANE STRESS2 element has been used and in the second analysis one PLANE STRAIN element. The deformation is prescribed using LOADS DISPLACEMENT in ten increments with a total displacement of u = 2. Full Newton equilibrium iterations are used.
| |
| Solution Results The plane deformation in pure shear is described by 0 0 ttX= x 1 +U X2 tX 2 = 0X, 0 0 where X and 'x 2 are the deformed coordinates of the point with the original coordinates x, and x2.
| |
| The displacement u is defined in the figure. The following deformation measures can be formed:
| |
| 0 = [0x'] [1 ?] (Deformation gradient tensor)
| |
| St x
| |
| 00=[I 1 j] (Cauchy - Green deformation tensor)
| |
| ( c-I)= 0 u/2] (Green- Larangestrain tensor) 0 FkCI
| |
| ý uI 2 2/2] Larng, The left indices t and 0 indicate that the tensor quantity is occurring in the deformed configuration at solution time t and referenced to the original undeformed configuration at time 0. In the following description of this example the left indices are not written out but are understood to be there. Hence, for example, the following understanding regarding the notation is used:
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| X= X , 'C=C and gE=e Version 99.0 B85.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples The principal stretches X are obtained from det(C- _2I)= 0 which gives u2 +4]
| |
| The principal directions On referenced to the initial configuration are obtained from (C- k22)0n = 0 which gives
| |
| °n =[=2 1]
| |
| n2 =L 1]
| |
| The principal stretch directions 'n in the deformed configuration are obtained from tn = X On which gives 1
| |
| n, = [2 +u, 1]
| |
| n2=-I +u, 1]
| |
| Extending this plane problem to three dimensions we can write xX=Io u 1 u C = lIUu 2 01 0]L0 Cu=/U u/2 0]
| |
| +u 0 00 The term x33 is unity for pure shear in the plane stress case because the material is incompressible.
| |
| Since the volume of a differential element is constant during deformation we have that det X= I which gives x33 = 1.
| |
| The term x33 is also unity in the plane strain case because, by definition, there is no deformation in the x 3 - direction.
| |
| In the calculation of the analytical expressions for the stresses we need to distinguish between the plane stress and the plane strain case.
| |
| Plane Stress In the plane stress case the incompressibility of the rubber material is enforced by 1- I C 33 = (Cl C 2 1 -C 1 2 C 2 since incompressibility means that also det C = 1.
| |
| Version 99.0 B85.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples The invariants of the Cauchy-Green deformation tensor are It = C 1 1 +C22 +C 33 = C11 +C 22 +(Cl 1 C 22 -_C 2 C21) 12 = Cl C22 +C C +C3 3 CH -C12 C21 = C11 C 22 +-(cH+C22 XCII C 22 -C1 2 Cz 1 )-Y -C 2 C21 22 33 13 =detC=l The Mooney-Rivlin material model is determined by the potential W= C, (i, -3)+C 2 (02 -3) where C1 and C2 are the Mooney-Rivlin material constants. The second Piola-Kirchhoff stresses are calculated from the potential as Sij = -W = 2 -W =23W a Il +2 DW a3I ac i C1/2j all aCij DI2 aCij which gives U2)
| |
| S1 =-2Clu2-2 C 2 u2(2+
| |
| 2 S 2 2 =*-2C 2u S t2 =2C0u+2C 2(l+u2)u S 33 = 0 The transformation to Cauchy stresses is performed by 1
| |
| I*XSXT det X which gives 2 2 "t7 +u S 22 = 2C 1 u 11 = SIl
| |
| +2uS 12 2
| |
| T 22 =S 22 =-2 C2 u
| |
| 'r2 =St 2 +uS2 2 =2(C 1 +C 2 )u 1133 = 0 Plane Strain In the plane strain case the rubber material is compressible and a bulk modulus is specified. A sepa rately interpolated pressure field is employed to subtract in an integration over each element the hydrostatic pressure calculated by the interpolated displacements. This procedure allows the material to be almost incompressible.
| |
| The invariants of the Cauchy-Green deformation tensor are I1 =C 11 +C 2 2 +/-C 33 12 -C 1l C 22 +C 22 C 3 3 +C3 3 CH -C 12 C 21 13 = C 33 (C 1 1 C 22 -C12 C 2 1)
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| The Mooney-Rivlin material model is determined by the potential w = c, (J, - 3)+ c2 0( 2 - 3)+ - K(J 3 - 1)2 Version 99.0 B85.3
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| | |
| SOLVIA Verification Manual Nonlinear Examples where J-1"3!3 U J 5"3-2/3 J?=[1/2 Ji = 11 *13 J2 1* 3 3 3 and where K is the bulk modulus. The isochoric (volume preserving) deformation (in J1 and J2 ) and the dilatational (volume changing) deformation (in J ) are then separated in the potential, see [I ]. The second Piola-Kirchhoff stress components based on the displacement interpolation are calculated as aw aw where
| |
| _aw D1 1 a Waw DJ 2 .W DJ 3 aC i i lO aJcii DJ2 aC ij WJ3aCij and aCij D I r a il
| |
| _ aCij I 3i3 a 3 _]
| |
| aCij 3
| |
| -1 /3
| |
| =(
| |
| -J2 _3 212 . I3 )I -213 i ac 313 aC-)
| |
| DJ 3 - I__L aC ij 2 3 a C ii Using the prescribed displacement field for pure shear we obtain 2 U2 S 1 u 2u5 1 + 2C 2
| |
| )-2 S 22
| |
| =-(C 3
| |
| u2) 2 Cu S33= (C 2+
| |
| 33 Transformation to Cauchy stresses gives C2 ) -
| |
| 'r I = (2 C, +
| |
| 3
| |
| =2 I +
| |
| =-(C 2 C2)- 2u5 2u5 1133 = (C2 -Cl)-2 3
| |
| T 12 = (Cl -C 2 )2u We note that the hydrostatic pressure based on the interpolated displacements is zero since Irl + T 2 2 + T33 = 0. Hence, the separately interpolated pressure, in addition to the interpolation of displacements, is not needed in this case. The interpolated pressure amplitudes are consequently zero.
| |
| Version 99.0 B85.4
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples Inserting u = 1 and u = 2 in the analytical expressions for the Cauchy stress components and the principal stretches the values in the table below are obtained. Included in the table are also the corresponding values computed by SOLVIA using the input data on pages B85.7 and B85.8. The SOLVIA values and the analytical values are identical within the listed number of digits.
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| u=l u=2 Theory SOLVIA Theory SOLVIA Stretches:
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| Xt 1.6180 1.6180 2.4142 2.4142 X2 0.61803 0.61803 0.41421 0.41421 k.3 1.0000 1.0000 1.0000 1.0000 Plane stress:
| |
| ""i1 0.250000 0.250000 1.000000 1.000000 T22 -0.050000 -0.050000 -0.200000 -0.200000 "733 0 0 0 0
| |
| ""12 0.300000 0.300000 0.600000 0.600000 Plane strain:
| |
| "0.183333 0.183333 0.733333 0.733333 t 22 -0.116667 -0.116667 -0.466667 -0.466667
| |
| -0.0666667 -0.0666667 -0.266667 -0.266667
| |
| 'r2 0.300000 0.300000 0.600000 0.600000 Pressure 0 0 0 0 The figures on the next page show the variation of the stress components with increasing shear displacement u, and also the principle stretches and stresses corresponding to u = 2.
| |
| User Hints
| |
| ° Note that the vector version of Green-Lagrange strain, which is output by SOLVIA, has shear strain components which are twice the corresponding tensor component values. Hence, 2-0F12 = u is output by SOLVIA in this example. This is in analogy with the engineering strain
| |
| = 2. e (i j) where ej is the tensor component of the shear strain.
| |
| -ij
| |
| " The shear stress component in this example of plane pure shear varies linearly with the shear displacement amplitude u. The direct stress components are initially zero but they grow in magnitude with increasing shear.
| |
| " The directions of principal stretch are orthogonal.
| |
| Reference
| |
| [1] Ogden. R.W.. "Volume Changes Associated with the Deformation of Rubber-Like Solids",
| |
| J. Mech. Phys. Solids, 1976, Vol. 24, pp. 323-338.
| |
| Version 99.0 B85.5
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| | |
| SOLVIA Verification Manual Nonlinear Examples PLANE STRESS 85 PURE SHEAR OF A RUBBER SHEET, PLANE STRESS BBS PURE SHEAR OF A RUBBER SHEET, 2 Z MAX OISPL.
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| TIME 2 A
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| STRETCH 2.4ý 2 LJ - 7 H 7 91 91< MAX DISPL. 2 TIME 2 - y AA.
| |
| ---ý_--A_
| |
| . 1 SPRINCIPAL o I o L.5 2.
| |
| i.2485 DEFORMATION -1.248S SOLVIA ENGINEERING AB SOLVA -POST 99 0 SOLVEA ENGINEERING AB SOLVIA-POST 99.0 PLANE STRAIN BSSA PURE SHEAR OF A RUBBER SHEET, PLANE STRAIN BASA PURE SHEAR OF A RUBBER SHEET.
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| MAX DISPL. 2 z TIME 2 LKy i..I....
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| STRETCH
| |
| ,2ý
| |
| . 41421 0.z2 MAX DISPL. 2 L T!ME 2 Al
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| :iZT----x-
| |
| -- I SPRINCIPAL
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| .1 .s 2.0 0 98 86 DE LAORMAAi AON -09 8186 SOLVIA ENGINEERING AB SOLVIA ENGINEERING AB OLVIA-POST 990 SOLVIA-POST 99 0 Version 99.0 B85.6
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B85 PURE SHEAR OF A RUBBER SHEET, PLANE STRESS' DATABASE CREATE MASTER IDOF=i00111 NSTEP=10 DT=0.2 ITERATION FULL-NEWTON TIMEFUNCTION 1
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| : 0. 0./ 2. 2.
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| COORDINATES / ENTRIES NODES Y Z 1 1. 1. / 2 0. 1. / 3 / 4 1.
| |
| MATERIAL 1 RUBBER C1=0.125 C2=0.0250 KAPPA=1000.
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| EGROUP 1 PLANE STRESS2 ENODES 1 1234 EDATA / 1 0.1 FIXBOUNDARIES 3 1 TO 4 FIXBOUNDARIES 2 / 1 4 LOADS DISPLACEMENTS 1 2 1. / 2 21.
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| SOLVIA END SOLVIA-POST input
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| * B85 PURE SHEAR OF A RUBBER SHEET, PLANE STRESS DATABASE CREATE WRITE FILENAME='b85.1iS' AXIS 1 VMIN=0. VMAX=2. LABELSTRING='DEFORMATION' AXIS 2 VMIN--0.5 VMAX=i. LABELSTRING='STRESS' SET DIAGRAM=GRID PLOTORIENTATION=PORTRAIT EHISTORY ELEMENT=4 POINT=i KIND=SYY XAXIS:1 YAXIS=2 SYMBOL=-,
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| OUTPUT=ALL EHISTORY ELEMENT=1 POINT=i KIND=SZZ XAXIS=-I YAXIS=-2 SYMBOL=2, OUTPUT=ALL SUBFRAME=OLD EHISTORY ELEMENT=i POINT=1 KIND=SYZ XAXIS=-I YAXIS=-2 SYMBOL:3, OUTPUT=ALL SUBFRAME=OLD ELIST TSTART=I ELIST SELECT=STRETCH TSTART=
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| ELIST TSTART=2 ELIST SELECT=STRETCH TSTART=2 MESH VECTOR=STRETCH SUBFRAME=i2 MESH VECTOR=SPRINCIPAL END Version 99.0 B 85.7
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B85A PURE SHEAR OF A RUBBER SHEET, PLANE STRAIN' DATABASE CREATE MASTER IDOF=-00111 NSTEP=10 DT=0.2 ITERATION FULL-NEWTON TIMEFUNCTION 1
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| : 0. 0. / 2. 2.
| |
| COORDINATES / ENTRIES NODES Y Z 1 1. 1. / 2 0. 1. / 3 / 4 1.
| |
| MATERIAL 1 RUBBER Ci=0.125 C2=0.0250 KAPPA-O000.
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| EGROUP 1 PLANE STRAIN PRESSURE-INTERPOLATION=YES ENODES 1 1 2 3 4 FIXBOUNDARIES 3 1 TO 4 FIXBOUNDARIES 2 / 1 4 LOADS DISPLACEMENTS 1 2 1. / 2 21.
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| SOLVIA END SOLVIA-POST input
| |
| * B85A PURE SHEAR OF A RUBBER SHEET, PLANE STRAIN DATABASE CREATE WRITE FILENAME='b85a.lis' AXIS 1 VMIN=C. VMAX=2. LABELSTRING='DEFORMATION' AXIS 2 VMIN=-0.5 VMAX=1. LABELSTRING='STRESS' SET DIAGRAM=GRID PLOTORIENTATION=PORTRAIT EHISTORY ELEMENT=i POINT=i KIND=SYY XAXIS=i YAXIS=2 SYMBOL=-,
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| OUTPUT=ALL EHISTORY ELEMENT=i POINT=i KIND=SZZ XAXISý-i YAXIS=-2 SYMBOL=2, OUTPUT=ALL SUBFRAME=OLD EHISTORY ELEMENT=i POINT=I1 KIND=SYZ XAXIS=-I YAXIS--2 SYMBOL=3, OUTPUT=ALL SUBFRAIE=OLD EHISTORY ELEMENT=i POINT=i KIND=SXX XAXIS=-i YAXIS=-2 SYMBOL=4, OUTPUT=ALL SUBFRAME=OLD ELIST STRAIN=YES TSTART=i ELIST SELECT=STRETCH TSTART=1 ELIST STRAIN=YES TSTART=2 ELIST SELECT=STRETCH TSTART=2 MESH VECTOR=STRETCH SUBFRAME=i2 MESH VECTOR=SPRINCIPAL END Version 99.0 B85.8
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| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B86 IN-PLANE BUCKLING OF AN UNSYMMETRIC DEEP ARCH Objective To verify the AUTOMATIC-ITERATION method for post collapse analysis with PLANE STRESS2 elements.
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| Physical Problem A deep elastic arch as shown in the figure below is to be analyzed. Only deformation in the arch plane is considered. The boundary conditions are unsymmetric.
| |
| Geometry data Material properties Reference load 9 2 El r = 100.0 m E=210.10 N/m Pref = - = 1750-10 3 N a=1.0 m v=0.0 ct =215' Finite Element Model The arch is modeled using 64 8-node PLANE STRESS2 elements. The model is shown in the figure on page B86.2. The reduced integration order of 2 x 2 and the force convergence criterion are used.
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| Skew system is employed at the right boundary to achieve plane stress conditions. The vertical load is distributed in a consistent way along the nodes of the load application section. Details of the model at the boundaries and at the load application are shown on page B86.3.
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| Solution Results The input data is shown on pages B86.5 and B86.6. The deformed mesh at four solution steps are shown in the bottom figure on page B86.3. The horizontal and vertical displacements at one of the load application nodes (node 2) are drawn using symbols in the top figures on page B86.4. The curves without symbols in the same figures are the corresponding solution from [1]. The two curves are almost coinciding.
| |
| Version 99.0 B86.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples The bottom figure on page B86.4 shows the vertical reaction at the hinged support.
| |
| The peak load and the corresponding horizontal and vertical displacements are compared with the results reported in [1]:
| |
| Peak load factor Y-displacement (in) Z-displacement (in)
| |
| SOLVIA 8.967 -61.2 -114.0 Reference [11 9.027 -60.4 -112.7 Reference
| |
| [1] Lyons, P. and Holsgrove, S., "Finite Element Benchmarks for 2D Beams and Axisymmetric Shells Involving Geometric Nonlinearity (Summary), NAFEMS, Ref. ROO 10 B86 IN-PLANE BUCKLrNG OF AN UNSYMMETRIC DEEP ARCH CRIGINAL 20. Z TIME I Ly ZONE EGI A ERCA 16667T-6 SOLVIA ENGINEERING AB SOLV_,A-PRE 99-C Version 99.0 B86.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples 386 !N-PLANE BUCKLING OF AN UNSYMMETRIC DEEP ARCH OR:GINAL Z ORIGINAL 1. Z TME I TIME 1 L
| |
| -y ZONE E33 ZONE E32 FORCE FORCE
| |
| : 1. 1667E6 1. 16667E6 i2 11 C
| |
| NAXES=SKE: NAXES=SKEW MASTER MASTER 1001I, 100111 ORTGINA; - 2. Z ORIGINAL
| |
| * 2. Z TIME 1 IL, TIME .
| |
| Ly ZONE N3+I ZONE Ni+1 _OAD LOAD 0.0
| |
| /
| |
| 0.0 C
| |
| a b NAXES=SKEW H-NAXES=SKE MASTER MASTER 100B1'I B 1011!
| |
| S Ii~i11 C l:1i!l SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB B86 IN-PLANE BUCKLING OF AN UNSYMMETRIC DEEP ARCH ORIGINAL - 20. Z MAX DISPL. i.63.22 Ly TIME 21 LOAD I.4648E5
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| ;OLI A-POST 99.0 3CV[A ENGINEERING A3 Version 99.0 B86.3
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples B86 IN-PLANE 8UC<LNG OF AN UNSYMMETRIC DEEP ARCH B86 IN-PLANE BUCKLING OF AN UNSYMMETREC DEEP ARCH
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| 'N
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| '-4 7
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| 1' 26 4 a2 10 4 A 4 Io COADMULTIPLIER LAMBDA COADMTULTIPLIER LAMBDA SOLVHA-POST 99.0 SOLV:A ENGINEERING AB SCLVIA-POST 99.0 SOLVIA ENGINEERING AB 386 IN-PLANE BUCKLING OF AN UNSYMIIETRIC DEEP ARCH 4
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| SID S* 'N 2o LDALD NUCfIRL ER LArBDA SOLVIA-TOST 99.0 SOLVIA ENGDNEERING AB Version 99.0 B86.4
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B86 IN-PLANE BUCKLING OF AN UNSYMMETRIC DEEP ARCH' DATABASE CREATE MASTER IDOF=100111 NSTEP=100 KINEMATICS DISPLACEMENTS=LARGE AUTOMATIC-ITERATION NODE=2 DIRECTION=3 DISPLACEMENT=-1. ,
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| DISPMAX=120.5 CONTINUATION=YES TOLERANCES TYPE=F RNORM=1.E5 SYSTEM 1 CYLINDRIC COORDINATES ENTRIES NODE R THETA 1 100.5 -17.5 2 100.5 90.0 3 100.5 197.5 4 100.0 -17.5 5 99.5 -17.5 6 99.5 90.0 7 99.5 197. 5 8 100.0 197.5 9 100.0 90.0 SKEWSYSTEM EULERANGLES 1 -17.5 NSKEWS 11 / 51 MATERIAL 1 ELASTIC E=210.E9 NU=0.
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| EGROUP 1 PLANE STRESS2 INT=2 EDATA / 1 1.
| |
| GSURFACE 1 2 6 5 EL1=32 EL2=1 NODES=8 SYSTEM=1 GSURFACE 2 3 7 6 EL1=32 EL2=1 NODES=8 SYSTEM=I FIXBOUNDARIES 23 / 4 8 FIXBOUNDARIES 3 / 1 5 LOADS CONCENTRATED 2 3 -291.666E3 6 3 -291.666E3 9 3 -1166.666E3 SET VIEW=X NSYMBOLS=MY VECTOR=LOAD MESH EGi SUBFRAME 22 SET NNUMBERS=MY BCODE=ALL NAXES=SKEW MESH E33 MESH E32 MESH N3+1 MESH N1+1 SOLVIA END Version 99.0 B86.5
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
| |
| * B86 IN-PLANE BUCKLING OF AN UNSYMMETRIC DEEP ARCH DATABASE CREATE WRITE FILE='b86.1is' SET VIEW=X SMOOTHNESS=YES MESH ORIGINAL=YES TIME=21 SET GSCALE=OLD TEXT=NO AXES=NO MESH TIME=40 SUBFRAME=OLD MESH TIME=-0 SUBFRAME=OLD MESH VECTOR=LOAD SUBFRAME=OLD USERCURVE 1 SORT=NO READ B86Y.DAT USERCURVE 2 SORT=NO READ B86Z.DAT AXIS 1 VMIN=O VMAX=-0 LABEL='LOAD MULTIPLIER LAMBDA' AXIS 2 VMIN=-80 VMAX=O. LABEL='NODE 9 Y-DIR DISPL.'
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| AXIS 3 VMIN=-140 VMAX=-. LABEL='NODE 9 Z-DIR DISPL.'
| |
| SET PLOTORIENTATION=PORTRAIT NHISTORY NODE=9 DIRECTION=2 XAXIS=i YAXIS=2 SYMBOL=i OUTPUT=ALL PLOT USERCURVE 1 SUBFRAME=OLD XAXIS=-I YAXIS=-2 NHISTORY NODE=9 DIRECTION=3 XAXIS=1 YAXIS=3 SYMBOL=i OUTPUT=ALL PLOT USERCURVE 2 SUBFRAME=OLD XAXIS=-1 YAXIS=-3 NHISTORY NODE=4 DIRECTION=3 KIND=REACTION SYMBOL=1 END Version 99.0 1386.6
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| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B87 LATERAL BUCKLING OF A RIGHT-ANGLE FRAME Objective To verify the AUTOMATIC-ITERATION method and the BEAM element in buckling and postbuckling analysis of a three dimensional beam structure.
| |
| Physical Problem A very slender right-angle frame shown in the figure below is analyzed. No shear effects in the beams are considered.
| |
| L Section data:
| |
| b=30mm b - -- t= 0.6 mm (thickness) 2 A = 18.0 mm 1, = 2.1329 ram4 I = 1350.0 mrnm 4 It =0.54mMn Perturbation load in Z-direction at tip:
| |
| 4 1P0i-Pz = Pref Reference load at tip:
| |
| pr L4 L 0.820035 N b
| |
| 2 Length L 240 mm Material, E = 71240 N/mrm "v= 0.31 Finite Element Model The frame is modeled using eight 2-node BEAM elements. The model is shown in the left figure on page B87.2. The AUTOMATIC-ITERATION method with force/moment convergence criterion is used.
| |
| Solution Results The input data on pages B87.4 and B87.5 is used. The deformed mesh and the displacements of nodes 3 and 2 as function of the load factor are shown in figures on pages B87.2 and B87.3. The load-lateral displacement curve shows a good agreement with the results reported in reference [1].
| |
| The analysis case B87A is a linearized buckling analysis using the input data on page B87.6. The ap plied load is 10% of Pref and the critical load factor obtained is 10.0212 giving a critical buckling load of 0.822, Pref" This value is in good agreement with the buckling load of 0.838. Pref reported in [1].
| |
| Version 99.0 B87.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples User Hints
| |
| "* This structure is quite extreme since the ratio b/t=50. The buckling is rather sudden and the buckling displacement is predominantly in the lateral (Z) direction. The buckling also includes torsion.
| |
| " The post-buckling behaviour is first associated with a very small tangent stiffness but the load carrying capacity in the negative Z-direction is then gradually increasing.
| |
| " The convergence rate for this example is quite slow and many load steps are needed in the automatic-iteration analysis.
| |
| " The linearized buckling analysis in this example is straight-forward and presents no difficulties.
| |
| The moment caused by the load gives a rotation that is practically zero due to the section dimensions of the force and the direction of the load. The moment does then not jeopardize the application of linearized buckling. For a further discussion of the limits for application of linearized buckling, see Example B88.
| |
| Reference
| |
| [1] Kouhia, R. "On Kinematical Relations of Spatial Framed Structures", J. Computers &
| |
| Structures, Vol. 40, No. 5, pp. 1185-1191, 1991.
| |
| B87 LATERAL BUCKLING OF A RIGHT-ANGLE FRAME B87 LATERAL BUCKLING OF A RIGHT-ANGLE FRAME ORIGINAL -- 50. Y TIME I L,
| |
| _OAO TEN IADA AULTP
| |
| ¢o g
| |
| FORCE 0.82003S i 2,3 IASTER 000R0 OAD '1ULTTPL:EQ LAMEDA I 11N1N SOLVIA-PRE 99.0 SOL'I!A ENGTNEERING AB SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 B87.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples 887 LATERAL BUCKLING OF A RIGHT-ANGLE FRAME 887 LATERAL BUCKLING OF A RTGHT-ANGLE FRAME o
| |
| S/// ?
| |
| x oa g
| |
| , S 1.0 20 LOAD MULTIPLOER .A90 LOAD MULT:PLIER LAMBDA t2 00 l. 13 'S 2 0.
| |
| O's 1.0 t.5 2.0 LOAD MULTIP9IER LAMBOA LOAD MULTIP.IER LAMBDA SOLVIA-POST 99.0 0 SOLVIA ENGRIEERING AB SOLVIA-POST 99.0 SOLVIA ENGINEERING AS 987 LATERAL BUCKLING OF A RIGHT-ANGLE FRAME B87 LATERAL BUCKLING OF A RIGHT-ANGLE FRAME ORIGINAL - ' so. I ORIGINAL I o O. Y MAX DISPL. 364.88 TlME 1500 LKx MAX DISPL. 364.88 TIME ISS0 LK LOAD LOD 1 .3901 t.3901 ORIG:NAL - - 00 MAX DISPL 364.88 LX TIME !SO0 CAD E.3901 SOLVIA ENGINEERING AB SOLVIA/0OST 99 0 SOL/IA-POST 99.0 SOLVIA ENGINEERING AS Version 99.0 B 87.3
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| | |
| SOLVIA Verification Manual Nonlinear Examples B87A LINEARIZED LATERAL BUCKLING OF A RIGHT-ANGLE FRAME 387A LIIIEAR:ZZD LATERAL BUCKL:NG OF A RIGHT-ANGLE -RAME REFERENCE - So. Y REFERENCE - -50.
| |
| MAX DISPL.- 17. 461 MAX DISPL - 17 461 LLA LAMBDA 10 .32 MODE I LAMBDA 10.021 MODE 1 II REFERENCE - - 20. Z MAX DISPL *- 17 .61 L-x MODE I LAMBDA 10.021 SOLVIA ENGINEERING AB SOLVIA-POST 99.0 SOLVIA ENGINEERING AB I SOLV:A-DOST 99.0 SOLVIA-PRE input DATABASE CREATE HEADING 'B87 LATERAL BUCKLING OF A RIGHT-ANGLE FRAME' MASTER NSTEP=1500 KINEMATICS DISPLACEMENTS=LARGE AUTOMATIC-ITERATION NODE=3 DIRECTION=i DISPLACEMENTS=-0.05, DISPMAX=400 CONTINUATION=YES TOLERANCES TYPE=F RNORM=10. RMNORM=1000 RTOL=0.001 COORDINATES 1 0 240 2 240 240 3 240 0 MATERIAL 1 ELASTIC E=71.24E3 NU=0.31 EGROUP 1 BEAM RESULT=FORCE BEAMVECTOR 1 0 0 1 GLINE 1 2 AUX=-I EL=4 GLINE 2 3 AUX=-I EL=4 Version 99.0 B87.4
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
| |
| SECTION 1 GENERAL RINERTIA=2.1329 SINERTIA=i350.0 /
| |
| TINERTIA=0.54 AREA=18.0 FIXBOUNDARIES / 1 LOADS CONCENTRATED 3 1 -. 820035 3 3 .820035E-4 SET PLOTORIENTATION=PORTRAIT VIEW=Z MESH NSYMBOL=YES NNUMBERS=MYNODES BCODE=ALL VECTOR-LOAD SOLVIA END SOLVIA-POST input
| |
| * B87 LATERAL BUCKLING OF A RIGHT-ANGLE FRAME DATABASE CREATE WRITE FILENAME= 'b87. lis' SET DIAGRAM=GRID PLOTORIENTATION=PORTRAIT SUBFRAME 12 NHISTORY NODE=3 DIRECTION=3 NHISTORY NODE=3 DIRECTION=1 SUBFRAME 12 NHISTORY NODE=3 DIRECTION=2 NHISTORY NODE=2 DIRECTION=3 SUBFRAME 12 NHISTORY NODE=2 DIRECTION=i NHISTORY NODE=2 DIRECTION=2 SUBFRAME 12 SET ORIGINAL=DASHED NSYMBOL=YES VECTOR=LOAD MESH VIEW=Z MESH VIEW=-Y SUBFRAME 12 VIEW 1 -1 0 0 ROTATE=-90 MESH VIEW-i END Version 99.0 B87.5
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input, linearized buckling DATABASE CREATE HEADING 'B87A LINEARIZED LATERAL BUCKLING OF A RIGHT-ANGLE FRAME' MASTER NSTEP=1500 KINEMATICS DISPLACEMENTS=LARGE BUCKLINGLOADS NEIG=1 TOLERANCES TYPE=F RNORM=10. RMNORM=1000 RTOL=0.001 COORDINATES 1 0 240 2 240 240 3 240 0 MATERIAL 1 ELASTIC E=71.24E3 NU=0.31 EGROUP 1 BEAM RESULT=FORCE BEAMVECTOR 1 0 0 3.
| |
| GLINE 1 2 AUX=-- EL=4 GLINE 2 3 AUX=-- EL=4 SECTION 1 GENERAL RINERTIA=2.1329 SINERTIA=1350.0 TINERTIA=0.54 AREA=18.0 FIXBOUNDARIES / 1 LOADS CONCENTRATED 3 1 -. 820035E-1 SET PLOTORIENTATION=PORTRAIT VIEW=Z MESH NSYMBOL-YES NNUMBERS=MYNODES BCODE-ALL VECTOR=LOAD SOLVIA END SOLVIA-POST input, linearized buckling
| |
| * B87A LINEARIZED LATERAL BUCKLING OF A RIGHT-ANGLE FRAME DATABASE CREATE WRITE FILENAME-'b87a.lis' BUCKLINGLOADS SET DIAGRAM=GRID PLOTORIENTATION=PORTRAIT SET RESPONESTYPE=BUCKLING SET ORIGINAL=DASHED NSYMBOL=YES SUBFRAME 12 MESH VIEW-Z MESH VIEW=-Y SUBFRAME 12 VIEW 1 -1 0 0 ROTATE=-90 MESH VIEW=1 END Version 99.0 B 87.6
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B88 LEE'S FRAME BUCKLING PROBLEM Objective To verify the AUTOMATIC-ITERATION method and the PLANE STRESS2 element in a large displacement analysis with snap-through and snap-back behaviour.
| |
| Physical Problem A plane frame as shown in the figure below is to be analyzed. Only deformation in the plane of the frame is considered.
| |
| I0.2 L 0.8 L Geometry data P4 L= 1.2 m 4l F. F/, d = 0.02 m (in-plane thickness) t= 0.03 m (out-of-plane thickness)
| |
| Material properties L E = 71.74.10 9 N/m 2 v=0.0 Reference load El P --ef = 996.389 N Finite Element Model The frame is modeled using 8-node PLANE STRESS2 elements. The model is shown in the figure on page B88.3. The force convergence criterion is used. Details of the model at the boundaries and at the load application are shown in the top figure on page B88.4.
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| Solution Results The input data is shown on pages B88.6 and B88.7. The deformed meshes at 8 solution steps are shown in the bottom figure on page B88.4. The vertical displacement at the load application node (node 2) is shown as a function of the applied load factor X using symbols in the top figure on page B88.5. The curve without symbols is the corresponding solution from [1]. The two curves are very close.
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| The bottom figure on page B88.5 shows the vertical reaction force at the bottom of the column (node 11) as a function of the load factor k.
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| The peak load at the instability point is compared with the results reported in [1]:
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| SOLVIA Reference [1]
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| Peak load factor 18.51 18.55 Version 99.0 B88.1
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| | |
| SOLVIA Verification Manual Nonlinear Examples User Hints
| |
| " As can be seen from the load-displacement diagram on page B88.5 this example is a large dis placement problem which exhibits a snap-through and snap-back behaviour. The last portion of the loading shows a significant stiffening since most of the load is taken by axial stress in the nearly vertical beam, which was initially horizontal. The changes in the tangent stiffness can be seen to be gradual without any sudden instability.
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| "* Also the first portion of the load-displacement curve up to the first instability point is nonlinear with large displacements since the vertical load introduces a moment in the column. This moment causes the column to deform laterally and the axial load in the column amplifies this lateral deformation.
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| " This example is not suitable to be analyzed by the linearized buckling method in the program. In linearized buckling it is assumed that the geometric stiffness matrix changes in a linear manner with the load. It is further assumed that the prebuckling displacements are small so that the shape of the stress distribution is the same during the loading and only the levels of stress are changed.
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| Linearization based on the configurations at times t and t+At gives then the equilibrium equation at an arbitrary prebuckling load level after time t as (tK-XAK)u= AR (1) where tK, `+AtK are the tangent stiffness matrices at times t and t+At, respectively, tR, t+AtR are the load vectors at times t and t+At, respectively, AR = t+AtR - tR is the load increment vector from the configuration at time t, AK = tK -t+AtK is the geometric stiffness matrix for the load increment AR, u is the displacement increment vector from the configuration at time t,
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| : k. is the load factor.
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| In practice, the stress-free configuration, usually at t=0, is often selected to be the reference configuration. When the assumptions for linearized buckling are satisfied tK is then the linear tangent stiffness matrix and AK is the geometric stiffness matrix corresponding to the load level AR as in classical linearized buckling.
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| At the buckling load level an infinite number of displacement increment vectors satisfy equilibrium. Taking two such displacement increment vectors and subtracting the corresponding equilibrium equations we obtain the eigenvalue problem (t K - *cAK) (p = 0 (2) where (pis the buckling mode shape (the difference between the two displacement increment vectors) and kc, is the load factor at buckling. Since the right hand side of (2) is a zero force vector, the mode shape (pscaled by a factor c to any selected magnitude can be added to the solution of (1) at the critical buckling load level.
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| XAR In schematic terms the load-displacement relation X.oAR (tK- ocrAK)(U+IcP) from the reference configuration is shown in the fig ure for the lowest buckling load and for positive X.
| |
| Upon loading the load-displacement curve is gov erned by (1). At the buckling load level an infinite XAR = (tK -AK)u number of displacement solutions is possible since the eigenmode equation (2) is satisfied. The lowest Subuckling load is 'R + Xcr AR.
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| Version 99.0 B88.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples
| |
| " In this frame buckling example the basic assumptions for application of linearized buckling are not satisfied. Even if a very small load is used for linearization, comparatively large moments are introduced in the frame causing rotations. A consequence is that the matrix AK is not the geometric stiffness matrix which in classical linearized buckling could be formed for the column and beam under axial membrane load.
| |
| Examples of linearized buckling analysis are given in B43, B45, B46, B50 and B87.
| |
| "* If the vertical load is moved to the corner of the frame the problem can meaningfully be analyzed by the linearized buckling method.
| |
| " If linearized buckling analysis is applied to a structure it is recommended to always check the buckling behaviour by successively increasing the load using small steps in a large displacement analysis in which the model is disturbed from its perfect geometrical form. The buckling mode shapes can often be used to create such a disturbance when scaled to a magnitude that safely corresponds to the actual physical conditions.
| |
| Reference
| |
| [1] Lyons, P. and Holsgrove, S., "Finite Element Benchmarks for 2D Beams and Axisymmetric Shells Involving Geometric Nonlinearity (Summary), NAFEMS, Ref. ROO10 B88 LEES FRAME BUCKLING PROBLEM z
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| ORIGINAL TIME i
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| --- 0.2 L~y I
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| FORCE 996.389 SOLVIA-PRE 99 0 SOLVIA ENGINEERING AB Version 99.0 B88.3
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| SOLVIA Verification Manual Nonlinear Examples B88 LEES FRAME BUCKLING PROBLEM ORIGINAL 0 CO.S Z ORIGINAL 0.02 Z ZONE CCRNER L ZONE NS+2 Ly 3
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| MAST .R MASTER 1001 100111 B 11111 Z OR7GINAL 0.02 Z ORIGINAL i-- 0.02 --
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| 7nNF N 1+2 L TiME I ZONE N2+2
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| ~FORCE 996.389 MASI ER 10011 ,"'ASTER
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| ,t B t1 t111 1~001[I1 SOLV!A-PRE 99.0 SOLVIA ENGINEERING AB B88 LEES FRAME BUCKLING PROBLEM ORIGINAL v-- - 0. 2 Z MAX DTSPL. i 0.28244 TIME S LOAD 36090 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 B88.4
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| SOLVIA Verification Manual Nonlinear Examples B88 LEES FRAME BUCKLING PROBLEM a
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| r 4
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| ',O I
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| L I
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| t=:e
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| ' * =*-- --- :_I .Th I
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| -10 -S 1l0 Is 20 25 30 35 1 01 LOAD MULTIPLIER LAMBDA SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B88 LEES FRAME BUCKLING PROBLEM I
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| -'44-C-
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| -CAD, -Li'PLIEP L,-\MBDA SCLV A-POST 99.0 OLVIA ENDGNEERING AB Version 99.0 B88.5
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B88 LEES FRAME BUCKLING PROBLEM' DATABASE CREATE MASTER IDOF=-00111 NSTEP=50 KINEMATICS DISPLACEMENTS=LARGE TOLERANCES TYPE=F RNORM=1000. RTOL=d.E-4 AUTOMATIC-ITERATION NODE=2 DIRECTION=3 DISPLACEMENT=-0.04 DISPMAX=0.94 CONTINUATION=YES COORDINATES ENTRIES NODE Y Z 1 1.2 1.21 2 0.24 1.21 3 0.01 1.21 4 -0.01 1.21 5 1.2 1.2 6 1.2 1.19 7 0.24 1.19 8 0.01 1.19 9 -0.01 1.19 10 0.01 11 12 -0.01 MATERIAL 1 ELASTIC E=71.74E9 NU=0.3 EGROUP 1 PLANE STRESS2 EDATA / 1 0.03 SET NODES=9 GSURFACE 1 2 7 6 EL1=24 EL2=2 GSURFACE 2 3 8 7 EL1=8 EL2=2 GSURFACE 3 4 9 8 EL1=2 EL2=2 ADDZONE=CORNER GSURFACE 9 12 10 8 EL1=32 EL2=2 FIXBOUNDARIES 23 / 5 11 LOADS CONCENTRATED 2 3 -996.389 MESH VECTOR=LOAD NSYMBOLS-MYNODS SET NSYMBOL=YES NNUMBER-MYNODES BCODE=ALL MESH CORNER SUBFRAME=22 MESH N5+2 MESH N11+2 MESH N2+2 VECTOR=LOAD SOLVIA END Version 99.0 B88.6
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B88 LEES FRAME BUCKLING PROBLEM DATABASE CREATE WRITE b88.1is COLOR LINE MESH ORIGINAL=DASHED TIME=5 SET SUBFRAME=OLD TEXT=NO AXES=NO SET GSCALE=OLD MESH TIME=9 MESH TIME=-I MESH TIME=20 MESH TIME-25 MESH TIME=32 MESH TIME=38 MESH VECTOR=LOAD SET SUBFRAME=NEXT USERCURVE 1 SORT=NO / READ B88.DAT AXIS 1 VMIN=--0 VMAX=40 LABEL='LOAD MULTIPLIER LAMBDA' AXIS 2 VMIN=-i VMAX=0 LABEL='NODE 2 Z-DIR DISPL.'
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| AXIS 3 VMIN=O VMAX=16.E3 LABEL='NODE ii FORCE-ZZ' NHISTORY NODE=2 DIRECTION=3 XAXIS=I YAXIS=2 SYMBOL=i OUTPUT=ALL PLOT USERCURVE 1 SUBFRAME-OLD XAXIS=-i YAXIS=-2 NHISTORY NODE=-i DIRECTION=3 YAXIS=3 KIND=REACTION SYMBOL=I END Version 99.0 B88.7
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B89 BRESLER-SCORDELIS CONCRETE BEAM, A-2 WITH STIRRUPS Objective To demonstrate the SOLID concrete material model in a reinforced concrete beam analysis.
| |
| Physical Problem The experimental study by Bresler and Scordelis [1] included twelve reinforced concrete beams. One such beam, denoted OA-2, is analyzed in Example B81 using PLANE STRESS2 elements. In this example the beam denoted A-2 will be analyzed. The figure below shows the concrete beam and the steel reinforcement.
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| a P
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| \\\ \
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| [ LU2 [12 s-I I
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| L = 4.57 m Stirrup spacing, a = 0.21 m h =0.56 m A,, (bottom) = 3290- 10-6 m2 b =0.305 m ASc (top) = 252- 10-6 m2 2
| |
| d =0.465 m A,, (stirrup) = 645.10-6 m b c =0.05 m Concrete Steel (bottom) Steel (top) Steel (stirrup)
| |
| E0 = 28700 MPa E= 218000 MPa E = 201400 MIPa E = 189700 MPa v =0.2 Gy = 555MPa -y
| |
| = 345 MPa Gy = 325MPa a, = -24.3 MPa ET = 10000 MPa ET =10000 MPa ET = 10000 MPa ZC = -0.0021 at = 3.72 MPa u = 0.0031 6, = -21.6 MPa The main difference between the OA-2 and the A-2 beam is the stirrup reinforcement including two longitudinal bars at the top of the beam section. The spacing of the stirrup reinforcement along the beam is 0.21 m in the experiment. The values of the uniaxial cut-off tensile stress a, and the minimum compressive stress a, have been given the values of f and f, respectively, listed in [].
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| Version 99.0 1389.1
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| SOLVIA Verification Manual Nonlinear Examples Finite Element Model The finite element model is shown in the figures on page B89.4 and the top figure on page B89.5.
| |
| Due to symmetry only one quarter of the beam need to be considered. The concrete is modelled using 27-node SOLID elements and the steel reinforcement is modelled using 3-node TRUSS elements. The area of the TRUSS elements is distributed in a consistent way. The procedure is similar to calculation of consistent nodal forces due to pressure loading and line loading.
| |
| The beam load is applied as a triangular pressure load over the mid-section cover layer element. Re inforcement is used in the lateral direction of the cover layer element. This can be compared with the steel plate used in the experimental study.
| |
| One elastic solid element is used at the support to avoid excessive nonlinear effects due to stress concentrations.
| |
| Advantage of the repetitive nature of the structure is taken in the model generation so that the basic building block of elements shown at the top of page B89.4 is copied 9 times.
| |
| The coefficient KAPPA in the concrete material model is set to 30 which gives a slow decrease of the tensile stress following a tensile failure.
| |
| The AUTOSTEP method is used to trace the incremental response and the equilibrium iterations are performed using the BFGS method.
| |
| Solution Results The input data to the SOLVIA analysis can be found on pages B89.12 to B89.14.
| |
| In the experimental study the A-2 beam showed similar diagonal tension cracking as the OA-2 beam but there was no sudden failure. Instead the A-2 beam with web reinforcement could withstand larger deflections prior to failure and local plastic yielding may have developed in the stirrup reinforcement
| |
| [I ]. The collapse load was measured to 110 kips (489 kN) and the failure mechanism was denoted as shear-compression failure.
| |
| The results from the SOLVIA analysis can be seen in the figures on pages B89.5 to B89.8.
| |
| The bottom figure on page B89.5 shows the calculated mid-span deflection as a function of the load together with the experimental results from ref. [1]. The experimental curve is drawn without symbols. The top figure on page B89.6 shows a vector plot of the crack normal pattern and the bottom figure shows the state of concrete in the last load step. The top figure on page B89.7 shows a contour plot of the force distribution in the stirrup reinforcement in the last load step. The stress along the top and bottom reinforcement and in the concrete along 4 lines SECT1 to SECT4 in the vertical direction are shown on page B89.8. The location of the lines are plotted in the bottom figure of page B89.7.
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| History plots of the highest stressed truss element in the stirrup and top reinforcement are shown in the bottom figures on page B89.8.
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| Version 99.0 B89.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples The analysis stops when the top reinforcement starts to yield and the concrete begins to crush at the top surface. No local yielding in the stirrup or the bottom horizontal reinforcement has occurred in the analysis.
| |
| The results for a case B89A with a reduced cut-off tensile stress (Q = 2.0.106 N/mm2 ) and an increased shear stiffness factor (SHEFAC = 0.75) are shown on pages B89.9 to B89.1 1. The calcu lated mid-span deflection curve is closer to the experimental curve than in the previous analysis. At the load level of 380 kN an increased transfer of stress to a stirrup element occurs which gives local stirrup yielding, see the left bottom figure on page B89. 11. The corresponding crack pattern in the middle of the beam with inclined cracks due to the beam section shear is shown in the bottom figure on page B89.9.
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| User Hints
| |
| " The complexities in a reinforced concrete analysis are already evident in the earlier concrete examples, see B 19, B79, B80 and B81. With an increased amount of reinforcement as in this example it is particularly important to realize that the stress transfer between the concrete and the reinforcement is difficult to analyze in all details. A coarser model may remove some of these difficulties but the price is a higher degree of averaging.
| |
| The mid-span deflection curve shows a stiffer behaviour in case B89 than in B89A. The larger value for the tensile cut-off stress in case B89 in combination with the value of 30 for KAPPA results in larger tensile stresses in the cracked portion of the beam than in case B89A and hence a stiffer behaviour.
| |
| " The value of SHEFAC governs the capacity to transfer shear stresses across crack planes. A larger value of SHEFAC decreases in general the numerical difficulties that may be encountered during the analysis. In case B89A a larger value of SHEFAC was used to balance the effect of the lower cut-off tensile stress on the shear transfer capacity across crack planes.
| |
| " Local difficulties due to concentrated load applications and point type of supports may often occur in reinforced concrete analysis. It is important to spread out the loadings and support reactions over a realistic area. In this example the central load is modeled as a pressure load and the element closest to the hinged support has an elastic material. Local difficulties are also relieved by dis tributing the areas of the TRUSS element reinforcement in a consistent way.
| |
| "* Details of the SOLVIA concrete material model are given in [2].
| |
| Reference
| |
| [1] Bresler, B., and Scordelis, A., "Shear Strength of Reinforced Concrete Beams", ACI Journal, Proceedings V. 60, No. 1, January, 1963.
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| [2] "The Concrete Material Model in SOLVIA", SOLVIA Engineering AB report, rev. August, 1995.
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| Version 99.0 B89.3
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| SOLVIA Verification Manual Nonlinear Examples 889 BRESLER-SCORDELIS CONCRETE BEAM, A-2 WITH STIRRUPS ORIGINAL P -. r 1 Z ORIGINAL C C0.S Z 1 05 X '.. y ZONE TRUSS y.7 I
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| SOLVTA-PRE 99.0 SOLVIA ENGINEERING AB B89 BRESLER-SCORDELIS CONCRETE BEAM, A-2 WITH STIRRUPS ORIGINAL --- 0.2 Z TIME 0.S x y PRESSURE 43L80 MASTER 0111
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| [) 100 1:
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| E 101 :1 F 1 101 !
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| SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB Version 99.0 B89.4
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| SOLVIA Verification Manual Nonlinear Examples B89 BRESLER-SC RDELIS CONCRETE BEAM, A-2 WIT. STIRRUPS ORIGINAL
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| * 0.2 ZONE TRUSS X -YY SOLVIA-PRE 99.0 SOLVIA ENGINEE-RING AS B89 BRESLERISCORDELIS CONCRETE BEAM, A-2 WITH STIRRUPS IL 9K C
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| S,~
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| 7
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| *2 (I)
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| >2 0 tO0 O 200 300 350 100 GO SOLVIA-COST 99.0 \OLV/A ENGINEERING AB Version 99.0 B89.5
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| SOLVIA Verification Manual Nonlinear Examples B89 BRESLER-SCORDELS CONCRETE BEAM, A-2 UITH STIRRUPS MAX DISPL. - 0.013218 Z TIME 21.2S ZONE SOLID X/ y I CRACK NORMAL SOLVIA-POST 99.0 SOLVIA ENGINEERTNG AB B89 BRESLER-SCORDELIS CONCRETE BEAM, A-2 1-TH STIRRUPS MAX D7SPL.i 0.0132[8 Z TIME 21.25 X 1,Y REACTION
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| : 3. 3685E5 CONCRETE 7 CRUSHED NORMAL CLCSED i CRACK 2 CRACKS 3 CRACKS SOLVIA-POST 99.0 SOLV A ENGINEER7NG AB Version 99.0 B89.6
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| SOLVIA Verification Manual Nonlinear Examples B89 BRESLER-SCORDELIS CONCRETE BEAK A-2 WITH STIRRUPS MAX DISPL. - C 013209 Z TIME 2t.25 ZONE EG4 X y FORCE-R NO AVERAGING P-> MAX 1474.7 4-00
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| !132S. 1 14 .... t 02S. 8 S726. 6C S427.36 1t28. 13 MIN-320.72 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B89 BRESLER-SCORDELIS CONCRETE BEAM, A-2 W4ITH STIRRUPS ORIGINAL 0.2 G X Y SOLVIA-POST 99.0 SOLV]A EN( GINC>RINC AB Version 99.0 B89.7
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| 00 00 cyll f-0 V)
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| SOLVIA Verification Manual Nonlinear Examples B89A BRESLER-SCORDELIS CONCRETE BEAM, A-2 WITH STIRRUPS, S-GT=2.0E6 22.. . . . ....
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| <¢-
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| I0o i,"0 200 250 300 350 400 450 500 FORCE SOLVIA-POST 99.C SOLVITA ENGINEERING AB B89A BRESLER-SCORDELIS CONCRETE BEAM, A-2 WITH STIRRUPS, SIGT=2.OE6 MAX DTSPL.i 0.014S21 Z TIME 19.209 ZONE SOLID X y L CRACK NORMA SOLVIA-DOST 99.0 CL',!TA ENGTNEE.RNG AE Version 99.0 B89.9
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| SOLVIA Verification Manual Nonlinear Examples B89A BRESLER-SCORDELIS CONCRETE 3EAM, A-2 A!TH STIRRUPS, SIGT=2.0E6 MAX DISPL. 0.014521 Z I7ME 19.209 xA O RE ACTION 3.4345SE5 CONCRETE CRUSHED NORMAL CLOSED I CRACK 2 CRACKS 3 CRACKS SOLVIA POST 99.0 SOLVIA ENGINEERING AB B89A BRESLER-SCORDELIS CONCRETE BEAM, A-2 WITH STIRRUPS, SIGT:2.CE6 MAX DISPL. 0.014521 Z TIME 19.209 ZONE EG4 X
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| !29 j-FORCE-R NO AVERAG NG MAX 3816.9
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| [] 374.8 2790 6 "2106.3
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| £ 22 .1 14..
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| 737.85 i"53.6!4 MIN-288.51 SOLVIA-POS 99. 0 OLV-A ENGINEERING AB Version 99.0 B89.10
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B89 BRESLER-SCORDELIS CONCRETE BEAM, A-2 WITH STIRRUPS' DATABASE CREATE MASTER IDOF=000111 NSTEP=50 DT=0.5 AUTO-STEP DTMAX=1 DTMIN=IE-5 FDECREASE=0.2 ITELOW=5 ITEHIGH=15 TMAX=25 TOLERANCES TYPE=F RNORM=80.E3 RTOL=0.05 ITEMAX=30 STOL=0.3 SET MYNODES=200 PRINTOUT MIN TIMEFUNCTION 1
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| : 0. 0. / 25. 500.
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| COORDINATES ENTRIES NODE Z 1 0.56 TO 3 0.51 TO 11 0.095 TO 13 NGENERATION TIME=2 NSTEP=13 XSTEP=-0.07625 1 TO 13 NGENERATION TIME=2 NSTEP=39 YSTEP=-0.11425 1 TO 39 MATERIAL 1 CONCRETE EQ 28700.E6 NU=0.2 SIGMAT=3.72E6, SIGMAC=-24.6E6 EPSC=-0.0021 SIGMAU=-21.6E6, EPSU=-0.0031 BETA=0.75 KAPPA=30. STIFAC=0.0001, SHEFAC=0.5 MATERIAL 2 PLASTIC E=218000.E6 YIELD=555.E6 ET=*0000.E6 MATERIAL 3 PLASTIC E=201400.E6 YIELD=345.E6 ET=10000.E6 MATERIAL 4 PLASTIC E=189700.E6 YIELD=325.E6 ET=10000.E6 MATERIAL 5 ELASTIC E=28700.E6 NU=0.2 SET NODES-27 EGROUP 1 SOLID MATERIAL=1 GVOLUME 79 1 27 105 81 3 29 107 EL1=1 EL2=1 EL3=1 GVOLUME 81 3 29 107 89 11 37 115 EL1=1 EL2=1 EL3=4 GVOLUME 89 11 37 115 91 13 39 117 EL1=1 EL2=1 EL3=1 SET NODES=3 EGROUP 2 TRUSS MATERIAL=2 GLINE 11 89 / GLINE 24 102 / GLINE 37 115 EDATA 1 274.2E-6 / 2 1096.4E-6 / 3 274.2E-6 EGROUP 3 TRUSS MATERIAL=3 ENODES GLINE 3 81 / GLINE 16 94 / GLINE 29 107 EDATA 1 21.E-6 / 2 84.E-6 / 3 21.E-6 EGROUP 4 TRUSS MATERIAL-4 GLINE 3 29 / GLINE 42 68 / GLINE 81 107 GLINE 37 11 / GLINE 76 50 / GLINE 115 89 GLINE 29 37 EL=4 / GLINE 68 76 EL=4 / GLINE 107 115 EL=4 GLINE 16 24 EL=4 / GLINE 55 63 EL=4 / GLINE 94 102 EL=4 GLINE 3 11 EL=4 / GLINE 42 50 EL=4 / GLINE 81 89 EL=4 Version 99.0 B 89.1I2
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
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| EDATA 1 5.85E-6 / 2 23.39E-6 / 3 5.85E-6 4 5.85E-6 / 5 23.39E-6 / 6 5.85E-6 7 0,97E-6 TO 10 0.97E-6 / 11 3.90E-6 TO 14 3.90E-6 15 0.97E-6 TO 18 0.97E-6 / 19 3.90E-6 TO 22 3.90E-6 23 15.60E-6 TO 26 15.60E-6 / 27 3.90E-6 TO 30 3.90E-6 31 0.97E-6 TO 34 0.97E-6 / 35 3.90E-6 TO 38 3.90E-6 39 0.97E-6 TO 42 0.97E-6 MESH NNUMBERES=YES NSYMBOL=YES SUBFRAME=21 MESH TRUSS NNUMBERES=YES NSYMBOL=YES TRANSLATE ZONENAME=ELEMENTS Y=-0.2285 COPIES-9 EGROUP 5 TRUSS MATERIAL=4 GLINE 80 1i 06 EL=i / GLINE 41 67 EL=1 / GLINE 2 28 EL-i GLINE 79 1 05 EL-I / GLINE 40 66 EL=i / GLINE 1 27 EL=I GLINE 876 8"74 EL=i / GLINE 8 89 887 EL=i / GLINE 902 900 EL=I EDATA 1 23.39E-6 / 2 93. 6E-6 / 3 23 . 39E-6 4 5.85E-6 / 5 23. 39E- 6 / 6 5. 85E-6 7 5.85E-6 / 8 23. 39E- 6 / 9 5.85E-6 SET NODES=27 EGROUP 6 SOLID MATERIAL=5 GVOLUME 874 796 822 900 876 798 824 902 EL1=1 EL2=1 EL3=1 ZONE XSYM GLOBAL-LIMITS XMIN=-0.153 XMAX=-0.152 ZONE YSYM GLOBAL-LIMITS YMIN=-1.E-3 YMAX=1.E-3 FIXBOUNDARIES 1 INPUT=ZONE XSYM FIXBOUNDARIES 2 INPUT-ZONE YSYM FIXBOUNDARIES 3 INPUT=LINE / 876 902 EGROUP 1 LOADS ELEMENT 1 t 0. 14348. 14348. 0.
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| ENODES DELETE 60 MESH VECTOR=LOAD BCODE=ALL MESH TRUSS NSYMBOL=MYNODES SOLVIA END Version 99.0 B89.13
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B89 BRESLER-SCORDELIS CONCRETE BEAM, A-2 WITH STIRRUPS DATABASE CREATE WRITE FILENAME='b89.1is' USERCURVE 1 / READ B89.DAT AXIS 1 VMIN=0 VMAX=500 LABEL='FORCE' AXIS 2 VMIN=-23E-3 VMAX=0 LABEL='MID-SPAN DEFLECTION' SET DIAGRAM=GRID NHISTORY NODE=29 DIRECTION=3 XVAR=1 XAXIS=1 YAXIS=2 SYMBOL=-,
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| OUTPUT=ALL PLOT USERCURVE 1 XAXIS=-1 YAXIS=-2 SUBFRAME=OLD MESH SOLID VECTOR=CRACKNORMAL MESH VECTOR=REACTION CONTOUR=CONCRETE ZONE REINF INPUT=EGROUP / 2 3 CONTOUR LEVEL=6 AVERAGING=NO MESH ZONENAME=EG4 CONTOUR=FR EGROUP 4 MESH EL114 ENUMB=YES NSYMBOL=YES GSCALE=OLD TEXT=NO SUBFRAME=OLD EGROUP 2 EPLINE BOTTOM 28 2 1 STEP 3 TO 1 2 1 EGROUP 3 EPLINE TOP 28 2 1 STEP 3 TO 1 2 1 EGROUP 1 EPLINE SECT1 1 9 8 7 TO 6 9 8 7 EPLINE SECT2 13 9 8 7 TO 18 9 8 7 EPLINE SECT3 25 9 8 7 TO 30 9 8 7 EPLINE SECT4 37 9 8 7 TO 42 9 8 7 MESH PLINE=ALL ELINE BOTTOM KIND=SRR SYMBOL=i ELINE TOP KIND=SRR SYMBOL=i AXIS 3 VMIN=-30E6 VMAX=5E6 LABEL='STRESS-YY' ELINE SECT1 KIND=SYY SYMB=* YAXIS=3 ELINE SECT2 KIND=SYY SYMB=1 YAXIS=3 ELINE SECT3 KIND=SYY SYMB=1 YAXIS=3 ELINE SECT4 KIND=SYY SYMB=1 YAXIS=3 EGROUP 4 EHISTORY EL=114 POINT=2 KIND=SRR XVAR=1 XAXIS=1 SYMBOL=d EGROUP 3 EHISTORY EL=3 POINT=i KIND=SRR XVAR=- XAXIS=1 SYMBOL=I EMAX EMAX EG4 NUMBER=5 END Version 99.0 B89.14
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| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B90 COMPOUND CYLINDER Objective To verify the PLANE AXISYMMETRIC element with a thermo-orthotropic and a thermo-elastic material under pressure and temperature loading.
| |
| Physical Problem A long compound cylinder with free and open ends are subjected to an internal pressure and a uni form temperature. Away from the ends the plane cross-section of the cylinder remain plane under the deformation, i.e., uniform axial strain. The inner material is isotropic and the outer material is or thotropic.
| |
| Geometry Material r, = 23 mm orientation C a r, = 25 mm r = 27 mm r3 t C Inner material, isotropic 2
| |
| E =2.1.10' N/mm v=0.3 a*=2.10-' 1/°C Outer material, orthotropic 2
| |
| Ea =Eb =5. 103 N/mm 2
| |
| Load case 1: E, = 1.3_105 N/mm 2
| |
| Internal pressure p, = 200 N/mm Vac = 'Vbc = Vab - 0.25 Load case 2: f2a =ab =2.10-5 1/°C Internal pressure p, = 200 N/mm 2 and ac, =3.10-6 1/C 2
| |
| temperature rise of 130 0 C Gab = 0.5.104 N/mm 2
| |
| Gac = Gbc = 1. 104 N/mm Finite Element Model The top figure on page B90.4 shows the finite element model. It consists of ten PLANE AXISYMMETRIC elements. Constraint equations are used to obtain a uniform axial strain in the elements.
| |
| Solution Results The hoop, radial and axial directions are denoted as 1, 2 and 3, respectively, in the index notation below for Young's modulus, Poisson's ratio and the thermal expansion coefficient. The material relation is then Version 99.0 B90.1
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| | |
| SOLVIA Verification Manual Nonlinear Examples E-I -(X, W -vu31lFhl E2 (du /dr-c 2 T) = 12 1 -V 32 " r E3 (eý - 3T) - 13 - V23 I G'j
| |
| [
| |
| Since directions 2 (radial) and 3 (axial) have the same properties we can simplify to E(u/
| |
| E2 (du / dr-r- aT) ct 2T) H
| |
| = 1
| |
| -V12
| |
| -V2 [ -V21 .l(Yhl 1 -V22 LE2 (ý- ,T) L-V 1 2 -V 22 I The relations to the parameters used in the SOLVIA solution are Ea = E2 Eb== E 2 Ec = El Vab =22 "Vbc = V'2 1 Vac = V 2 1 The equilibrium equation is Chy = r + Ur dr Since there are no edge effects the axial strain is constant so that de, / dr = 0 giving d&F = dub dor dr dr + dr The following expressions for the stress components and the radial displacement are then obtained if PI, P2 and p3 are the pressures at radii r1, r2 and r3, respectively:
| |
| Outer orthotropic cylinder, r2 _ r <r 3 uh = K, 3PrP-1l + K 2 32 r P2 +K 3 3
| |
| S= K, r '- + K 2 r -1 +K3 a = K,rI'-I (v 22 + vI 2 1 ) + K 2 r0- 1 (V22 + v1 2 P32 ) + K 3 (v 22 + v 12 )+ E2 (e - (X2 T) u(r)= r{ I[Ki(P -v 21 (1+v 22 +v 12 P3))-r"-' +K 2(0 2 -V2 2,(1 +v 22 +Vs2 3)).rPz' +
| |
| K3 (l-V2, (1+,,22 +V12 ))]+ <T--V,2 (e7 - a2T)I where K r3- ' (P2 + K3 )+ r,-2 (-3 +K 3 )
| |
| r2.,0-1 r3 r2 r3 P 1 1 K3 r3 -' (P2 1+ K 3 )- r. 3-I '1 (p31 +K 3 )
| |
| - r2 13- r3 1,-
| |
| 3 -I _r 2 3-1 r3 13 K3 = El(a 2 -a )T + EIv12 I -_ }e, - aT /(A, - A 3)
| |
| A, = -V1 2 V21 A 3 = V[2(1- V 2) 031 =A /A, 3 /[ 2 -VA; Al Version 99.0 B90.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples Inner isotropic cylinder, r, < r _ r, :
| |
| h = rl~, ,
| |
| p -r"P 2 r (2 -p_ )/rI)/(r22 r.)
| |
| Gr2(r pl-r~p2 +r~r2(p 2 2 2p)/(r r- )
| |
| ur)= 2~r~{ 1 r P 2 (r
| |
| - '-p 2) +E(e - c2 p ) PT u(r) =r irPr 2P(
| |
| [r', -. v~)rr2 ,2r2 _2-r1 Pi )(l+V' Il+ctT-v(ecxTd The total axial load must be zero: 2j radr+ frazdr =0 Sr, r )
| |
| A further equation is obtained since the radial displacement at the interface radius r = r, must be the same for the outer and inner cylinders. Hence, we have two equations which can be solved for the unknowns P 2 , the pressure at the interface, and e, the axial strain.
| |
| The analytical solution for the stresses at the three radii is given in the table below together with the solution obtained by SOLVIA using the input data on pages B90.7 and B90.8:
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| 2 Stresses in (N/mm )
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| Inner cylinder Outer cylinder r` 1r2 r2 r3 Load case 1 ah Theory 1565 1430 875 759 SOLVIA 1565 1430 875 759 c7 Theory -200 -64 -64 0 SOLVIA -200 -64 -64 0
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| *7 Theory 10 10 -17 -2 SOLVIA 10 10 -17 -2 Load case 2 Gh Theory 1381 1260 1056 935 SOLVIA 1381 1260 1056 935 Gr Theory -200 -78.5 -78.5 0 SOLVIA -200 -78.5 -78.5 0 oy Theory 9 9 -18 1 SOLVIA 9 9 -18 1 As can be seen in the table the SOLVIA results are in excellent agreement with the analytical results.
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| The SOLVIA results have been obtained from the contour stress figures on pages B90.4 and B90.5.
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| The variation of the stress components in the radial direction is shown on page B90.6.
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| Version 99.0 B90.3
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| SOLVIA Verification Manual Nonlinear Examples 390 COMPOUND CYLINDER OR7GINAL - 0.2 2 TIME 2 Ly 78E PRESSURE 200 MASTER 2S 8 01"[
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| ORIGINAL H-- 02 L L Y R
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| EAXES ORTHOTROPIC SOLVIA-DRE 99.0 SOLVIA ENGINEERING AB I
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| 890 COMPOUND CYLONDER 890 COMPOUND CYLINDER MAX 0ISPL. - 0 .17769 7L Y MAXD:SPL, 0 2!749 Z TIME I ZONE -G! R T75E 2 ZONE E51 LYR STRESS-XX STRESS-XX BASED ON ZONE BASED ON ZONE MAX 1565,34 MAX :381. 40 1555.86 L539. 92 .!373.81 1
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| 1358,62 1522 97 . 343.44 556. 02 1489 07 S328.
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| 1313,08 2
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| 1472.1 .1297 90 1t45. 18 1282 72 1438.23 526753 M1N 1429.76 MIN 1259.94 MAX DISPL S0.17216 Z Z MAX DISPL. s- 0.21749 TIME 1 LY TOME 2 LY ZONE EG2 R ZONE E32 R STRESS-XX STRESS-XX 3ASED ON ZONE 3ASED ON ZONE "MAX 874.64 MAX -'.06ý02 I ____ S
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| &52.92 838. 45 !1018 24 1:003. t2 S 8509.49 795.02 S988.009 780,58 :;972.856 957 783 766.06 MIN 758.3-MTN 935.11'3 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB SOLVIA-POST 990 SOLVWA ENGINEERING A8 Version 99.0 B90.4
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| | |
| SOLVIA Verification Manual Nonlinear Examples 590 COMPOUND CYL:NDER MAX DISPL. - 0.17769 TIME t ZONE E'5l S, STRESS-YY BASED ON ZONE STRESS-YY BASED ON ZONE MAX- 64 .222 MAX-78. 459 1-56.0 48
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| -72. 793
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| .- 89 736 101 23
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| -106.68 -116.40
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| -131 .58
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| -1476 6
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| 161 .93
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| -1744 5 -177. ,1.
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| ,-192.29
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| -191 .39 MIN-199.88 Z MAX DISPL. - 0.21749 Z MAX DISPL -- 0 .172i6 TIME , T-ME 2 Ly ZONE ES2 R ZONE EG2 R S3TRESS-YY - 20 37RESS-YY BASED 3N ZONE BASED IN ZONE MAX 0.032095 MAX 04036023
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| -160120111
| |
| *-3. 9924
| |
| : -4.68S
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| -20 090
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| -28 139 -34.3*3
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| -36 188 -44 127
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| -44 237 -53.941
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| -52 286 -63. 755
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| -60 335 ,,-73.569 MIN-64.359 M11N-78. 76 SOLVIA-POST 99.0 8OLVIA ENGINEERING A9 SOLVIA-POST 99.0 SOLVIA ENGINEERING A0 B90 COMPOUND CYLINDER MAX OISPL. 0---.21749 Z TIME 2 LY ZONE EG1 R STRESS-ZZ BASED ON ZONE MAX 8 75740 8 75299 0 '4117 8 73535 8 72653 8 7177 8 70808 8.70006 8.6912' 0N0 8. 68683 MAX DOSPL. 0 21749 Z ZIME 2 ZONE EG2 R SRESS-ZZ BASED ON 2ONE
| |
| 'lAX 0.76957 S90,38453
| |
| -2.627
| |
| -50009
| |
| "-9.6173
| |
| - 925
| |
| -16.542 MIN-t7.696 SOLV/!A-OST 99.0 SOLVIA ENGINEERING AS Version 99.0 B90.5
| |
| | |
| 7i 0
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| ý7_
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| 7m cz
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input DATABASE CREATE HEAD 'B90 COMPOUND CYLINDER' MASTER IDOF=-00111 COORDINATES ENTRIES NODE Y Z 1 23. 0. TO 13 25. 0. TO 25 27. 0.
| |
| 51 23. 1. TO 63 25. 1. TO 75 27. 1.
| |
| MATERIAL 1 THERMO-ELASTIC
| |
| -1000 2.1E5 0.3 2E-5 1000 2.1E5 0.3 2E-5 MATERIAL 2 THERMO-ORTHOTROPIC, EA=5E3 EB=5E3 EC=1.3E5, NUAC=0.25 NUBC=0.25 NUABE0.25 ,
| |
| GAB=0.5E4 GAC=d.0E4 GBC=1.0E4, ALPHAA=2E-5 ALPHAB=2E-5 ALPHAC=3E-6 EGROUP 1 PLANE AXISYMMETRIC MATERIAL=d GSURFACE 1 13 63 51 ELI=6 NODES-9 EGROUP 2 PLANE AXISYMMETRIC MATERIAL=2 GSURFACE 13 25 75 63 EL1=6 NODES=9 FIXBOUNDARIES 3 INPUT=LINE / 1 25 CONSTRAINT 52 3 51 3 TO 75 3 51 3 EGROUP 1 LCASE 1 LOADS ELEMENT INPUT=LINE 1 51 200.
| |
| LOADS TEMPERATURE 1 0. TO 125 0.
| |
| LCASE 2 LOADS ELEMENT INPUT=LINE 1 51 200.
| |
| LOADS TEMPERATURE 1 130. TO 125 130.
| |
| SET HEIGHT=0.25 MESH NSYMBOL=MYNODES NNUMBER=MYNODES BCODE=ALL ENUMBERS=GROUP VECTOR=LOAD SUBF=12 MESH EAXES=ORTHOTROPIC SOLVIA END Version 99.0 B90.7
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
| |
| * B90 COMPOUND CYLINDER DATABASE CREATE WRITE 'b90.1is' CONTOUR AVERAGING=ZONE SET PLOTORIENTATION=PORTRAIT CONTOUR=SXX MESH EGi TIME=1 SUBFRAME=12 MESH EG2 TIME=I MESH EGi TIME=2 SUBFRAME=12 MESH EG2 TIME=2 SET CONTOUR=SYY MESH EG1 TIME=i SUBFRAME=I2 MESH EG2 TIME=i MESH EGi TIME=2 SUBFRAME=i2 MESH EG2 TIME=2 SET CONTOUR=SZZ MESH EGL TIME=i SUBFRAME=12 MESH EG2 TIME=i MESH EGi TIME=2 SUBFRAME=i2 MESH EG2 TIME=2 SET PLOTORIENTATION=LANDSCAPE EGROUP 1 EPLINE THICK 1 9 6 3 TO 6 9 6 3 EGROUP 2 EPLINE THICK 1 9 6 3 TO 6 9 6 3
| |
| *e ELINE THICK KIND=SXX TIME-i SYMBOL=i OUTPUT=ALL ELINE THICK KIND=SYY TIME=I SYMBOL=i ELINE THICK KIND=SZZ TIME=1 SYMBOL=i
| |
| *E ELINE THICK KIND=SXX TIME=2 SYMBOL=i OUTPUT=ALL ELINE THICK KIND=SYY TIME-2 SYMBOL=i ELINE THICK KIND=SZZ TIME=2 SYMBOL=I END Version 99.0 B90.8
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| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B91 CONCRETE SLAB UNDER UNIAXIAL MOMENT, JAIN AND KENNEDY Objective To verify the SHELL element under uniaxial moment loading when used with the CONCRETE material model and with rebar layers.
| |
| Physical Problem A concrete slab simply supported along two lines, see figure. The slab is singly reinforced with two layers of steel. Four cases are to be analyzed having the first layer of steel located with the [3-angles 0, 15, 30 and 45 degrees, respectively. The second layer of steel is orthogonal to the first layer and placed directly on top of the first layer. The uniaxial moment is generated by two line loads, each located at a distance of 0.150 m to the nearest support line. Experimental results for this problem have been reported by Jain and Kennedy [I].
| |
| 0.860,m rebar layer 2 rebar I
| |
| 'ayerI
| |
| /2T 0.460 ,
| |
| Pt- -- -- -
| |
| P12 P/2 0.150l 0.460 0.150r
| |
| ,0050 ,m[" 't0050,,
| |
| o ash "'r' 0.760, .,
| |
| Concrete data: Steel data:
| |
| E0 = 29.10 9 N/m 2 (7= 1-5 MPa E = 200.10 9 N/i 2
| |
| = -34 MPa (Y,=-32 MPa Gy = 220 MPa
| |
| = -0.002 F- -0.003 Layer 1: Diameter = 0.0048 m v=0 K= 10 cc1 = 0.065m T =0.038 m t= 0.0119 in (0.625 of half thickness)
| |
| Layer 2: Diameter = 0.0048 m cc 2 = 0.055 m t2 = 0.0071 m (0.376 of half thickness)
| |
| Finite Element Model The finite element model is shown on page B91.3. The entire slab has been modeled since the cases with inclined reinforcement have no symmetry planes. A separate element group is used for the constant moment part of the slab. The force is applied as element line load along two of the boundaries of this element group. The concrete material model has been assumed to have a tensile strength of 1.5 MPa and a tension stiffening coefficient of K= 10.
| |
| Version 99.0 1391.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples Solution Results The input data for the first run B91A, where the orientation of the reinforcement is aligned with the sides of the slab (13 = 0), is shown on pages B91.9 - B91.10. The input data for the following three runs B91B, B91C and B9lD are the same except that the value for 13is set to 15, 30 and 45 degrees, respectively, in the EDATA command of the two element groups.
| |
| The calculated curve for the moment load as a function of the center deflection of the slab is shown for each of the four cases on pages B91.4 - B91.8 and can be compared with the corresponding experimental curve from [1 ] drawn as a solid line without symbols. The ultimate moment loads obtained in the experiment are practically the same for the four cases and the calculated values agree well with the experimental values. It can be noted that the calculated cases 3 = 15 and 13= 30 show an almost constant moment load level already when the first steel layer yields. The ultimate moment load level, when also the second steel layer yields, occur for a significant additional deflection of the slab. Such two moment load levels are not so apparent in the corresponding experimental curves.
| |
| Calculated principal moments, crack normal directions at the lower slab face, deformations and reactions, rebar stress histories for element 2, rebar plastic strains, concrete stress-strain history at the lower slab face of element 2 and the concrete bending stress distribution in the thickness direction at element 2 are shown for each of the four cases on pages B91.4 - B91.8. The integration point in element 2 closest to the center of the slab has been selected.
| |
| Only rebar layer 1 is stressed in the case 13= 0 which gives a symmetric response of the slab. Both rebar layers are stressed about the same amount in the case 13= 45. A slight asymmetry in the response must occur since the rebar layers cannot act isotropically at all load levels due to the varying distance to the uncracked portion of the concrete. In the cases 13= 15 and 3 = 30 the rebar layer 1 is stressed much before rebar layer 2 so the response is clearly asymmetric with reaction forces that cause some torsion moment loading of the slab. The asymmetry of the response is however gradually decreasing with the plastic elongation of rebar layer 1.
| |
| With the selected material data for the concrete the calculated moment-deflection curve agrees well with the experimental results also during the loading portion up to the ultimate moment for the cases with 0 and 15 degrees orientation of the reinforcement. The cases with 30 and 45 degrees rebar directions show, however, larger deflections for the calculated response compared to the experimentally obtained deflections.
| |
| Kinks in the calculated response curves occur when the concrete starts to crack at an integration point.
| |
| In the cases with 30 and 45 degrees rebar directions cracking at the third integration point in the thickness direction results in a rather sudden increase in deflection. A larger number of integration points in the thickness direction should give a more continuous advancement of the cracking. In the analysis 6 integration points were used in the thickness direction of the slab. When the ultimate moment load is approached only the upper integration point layer of the slab shows significant compressive stresses in the bending direction while the lower integration point layers have practically no stresses in the bending direction.
| |
| User Hints
| |
| * The value of the concrete tensile strength and the amount of tension stiffening affect the calculated moment-deflection curves significantly. The values assumed in the analyses have been selected based on the response for the case 3 = 0, which is reasonably close to the experimental curve.
| |
| "* It is important to have more integration points in the thickness direction of the slab than the default number. Six Gauss integration points were used in this example.
| |
| Version 99.0 B91.2
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples Note that a specified line load is applied to all element edges of the current element group which belong to the line. If the line is internal of the element group the line load would then be applied at least twice, since each portion of an internal line is part of at least two element edges by definition.
| |
| In this example two element groups are used so that the line load can be applied along the boundary of one of the element groups.
| |
| Reference
| |
| [1] Jain, S., and Kennedy, J., Yield Criterion for Reinforced Concrete Slabs, Journal of the Structural Division, Proceedings of the American Society of Civil Engineers, Vol. 100, No ST3, March, 1974.
| |
| B91A CONCRETE SLAB JNDER JNIAXIAL MOMENT. JAIN AND KENNEDY, BETA=O ORIGINAL - 0.
| |
| TIME So 391A CONCRETE SLAB JNDER UNIAXIAL MOMENT. JAIN AND KENNEDY. BETA=O ORi:SNAL TIME 50
| |
| -- 0.
| |
| yz EFOSRC 333,333 MASTSR oO O0I A 00100, C tOOO0t D ICIOOA SCtLVA -NG:NEERING AB SOLVTA-PRE 99.0 Version 99.0 1391.3
| |
| | |
| a: 2 2 Eu I.)
| |
| o I I' I
| |
| SI Vi K A 0 - --- -'-
| |
| Eu, a*
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| ~~~~~~~~
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| . .. . . . . .. . . I. . .
| |
| 0 ON, ON
| |
| | |
| cv C, I r : ý I ý. - I.; " , ý11-0 1/4ZO*..
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| N 6 cv 1
| |
| 0 N 0 LU 0
| |
| cv V
| |
| 1/4 1/41/4 1/41/4 1/4 1/4 1/4 1/41/4 1/4 1/4*4L 1/4 1/4 -
| |
| IU!
| |
| 1/4 1/4
| |
| -t eli S
| |
| 1/41/431/41/41/431/4 1/4 1/41/4 1/4 1/4 1/4 1/4 1/4 1/4 1/41/41/41/41/41/4 - 1/4 1/4' CI
| |
| | |
| 0) 0 C/)
| |
| 6::
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| V 0 A 2 T Li to
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| /
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| 0t A' AUA AA A A
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| I
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| /
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| .4/
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| N (4
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| 4 4 I,.
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| U (4
| |
| - - N (S
| |
| .4 0
| |
| ON ON 0
| |
| t.14 ..4 0
| |
| .04 3-"
| |
| | |
| -51e:3 r-14 :3-:3%
| |
| 0 49 z
| |
| 5S 4 ~6 Z 91 i' 31 00 3 '5
| |
| :3 N :3 * :3
| |
| :3
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| '5 I-1l
| |
| :3 41/4 T
| |
| sA CZ 0
| |
| 0 V) -0 U0
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input
| |
| : HEADING,
| |
| 'B91A CONCRETE SLAB UNDER UNIAXIAL MOMENT, JAIN AND KENNEDY, BETA=0 DATABASE CREATE MASTER IDOF=000001 NSTEP=100 DT=50 AUTO-STEP DTMAX=50 TMAX=10000 ITELOW=5 ITEHIGH=15 DISPMAX=0.008 TOLERANCES TYPE=F RNORM=1000 RMNORM=1000 RTOL=0.001 ITEMAX=50 SET MYNODES=200 TIMEFUNCTION 1 0 0 / 10000 10000 COORDINATES ENTRIES NODE X Y Z 1 0.430 0.230 0.
| |
| 2 0.380 0.230 0.
| |
| 3 0.230 0.230 0.
| |
| 4 0. 0.230 0.
| |
| 5 -0.230 0.230 0.
| |
| 6 -0.380 0.230 0.
| |
| 7 -0.430 0.230 0.
| |
| 11 0.430 0. 0.
| |
| 12 0.380 0. 0.
| |
| 13 0.230 0. 0.
| |
| 14 0. 0. 0.
| |
| 15 -0.230 0. 0.
| |
| 16 -0.380 0. 0.
| |
| 17 -0.430 0. 0.
| |
| 21 0.430 -0.230 0.
| |
| 22 0.380 -0.230 0.
| |
| 23 0.230 -0.230 0.
| |
| 24 0. -0 .230 0.
| |
| 25 -0.230 -0.230 0.
| |
| 26 -0.380 -0.230 0.
| |
| 27 -0.430 -0.230 0.
| |
| REBAR-LAYER 1 D=0.0048 CC=0.065 TCO0R=0.625 PSI=0.
| |
| REBAR-LAYER 2 D=0.0048 CC=0.055 TCOOR=0.376 PSI=90.
| |
| MATERIAL 1 CONCRETE EO=29.E9 NU=0.0 SIGMAT=1.5E6, SIGMAC=-34.E6 EPSC=-0.002 SIGMAU=-32.E6 EPSU=-0.003, BETA=0.75 KAPPA=10 STIFAC=0.0001 SHEFAC=0.75 MATERIAL 2 PLASTIC E=200.E9 YIELD=220.E6 ET=0.
| |
| EGROUP 1 SHELL STRESSREFERENCE=ELENENT TINT=6 MATERIAL=1 REBARMATERIAL=2 THICKNESS 1 0.038 GSURFACE 5 3 23 25 ELi=4 EL2=2 NODES=4 LAYER-SECTION 1 TREF=0. 1 2 LOADS ELEMENT TYPE=FORCE INPUT=LINE 3 23 thickness 6.66667 6.66667 5 25 thickness 6.66667 6.66667 EDATA ENTRIES EL NTH BETA 1 1 0. TO 8 1 0.
| |
| EGROUP 2 SHELL STRESSREFERENCE=ELEMENT TINT=6 MATERIAL=i REBARMATERIAL=2 THICKNESS 1 0.038 GSURFACE 2 1 21 22 EL1=1 EL2=2 NODES=4 GSURFACE 3 2 22 23 EL1=2 EL2=2 NODES=4 GSURFACE 6 5 25 26 EL1=2 EL2=2 NODES=4 GSURFACE 7 6 26 27 ELi=1 EL2=2 NODES=4 LAYER-SECTION 1 TREF=0. 1 2 Version 99.0 B91.9
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
| |
| EDATA ENTRIES EL NTH BETA 1 1 0. TO 12 1 0.
| |
| FIXBOUNDARIES 3 INPUT=LINES / 2 22 / 6 26 FIXBOUNDARIES 12 / 14 FIXBOUNDARIES 1 / 4 SET NNUMBERS=MYNODES NSYMBOL=MYNODES BCODE=ALL VECTOR=LOAD MESH EAXES=RST MESH ENUMBERS=GROUP VIEW=-I SOLVIA END SOLVIA-POST input
| |
| * B91A Concrete slab under uniaxial moment, Jain and Kennedy , beta=0 deg DATABASE CREATE TOLERANCES WRITE 'b9la.lis' SYSTEM 1 CARTESIAN PHI=180 AXIS I VMIN=0. VMAX=0.009 LABEL='Center deflection (m)'
| |
| AXIS 2 VMIN=0. VMAX=2200 LABEL='Moment load (Nm/m)'
| |
| SET DIAG=GRID NXYPLOT XNODE=14 XDIR=3 YKIND=TIME YDIR=l OUTPUT=ALL XAXIS=1 YAXIS=2, SYSTEM=1 SYMBOL=
| |
| * read test result USERCURVE 1 / READ 'B91A.DAT' PLOT USERCURVE 1 XAXIS=1 YAXIS=2 SUBFRAME=OLD CONTOUR AVERAGING=NO MESH CONTOUR=MPMAX VECTOR=MPRINCIPAL SHELLSURFACE TOP MESH CONTOUR=CONCRETE VECTOR=CRACKNORMALS MESH WHOLE CONTOUR=D3 SYSTEM=l VECTOR=REACTIONS VIEW=-I EGROUP 1 EHIST E=2 P=i KIND=SAA REBAR=1 EHIST E=2 P=1 KIND=SAA REBAR=2 SHELLSURFACE REBAR=1 MESH CONTOUR=EPACC EAXES=REBAR VIEW=I SHELLSURFACE REBAR=2 MESH CONTOUR=EPACC EAXES=REBAR EXYPLOT E=2 P=6 XKIND=ERR YKIND=SRR EPLINE THICK / 2 6 5 4 3 2 1 ELINE THICK KIND=SRR SYMBOL=l ELIST E2 END Version 99.0 B91.10
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B92 TILTING-PAD JOURNAL BEARING, RIGID PADS Objective To demonstrate the use of FILM elements in a tilting-pad journal bearing with the assumption that the pad and shaft surfaces are rigid and do not conduct any heat.
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| Physical Problem The geometry of the pivoted-pad journal bearing is shown in the figures below. Three pivoted pads are used. Each pad is free to rotate about the pivot point. This bearing is analyzed with elastic pads including thermal deformations in Example B93. Elasticity and heat conduction in both the shaft and the pads are treated in Example B96.
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| AZ t I I Oil properties:
| |
| Geometry: Bearing characteristics:
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| Radii Machined-in clearance Density 3
| |
| rshat = 140- 10-3 m c = rpad - rsbaft = 210.10-- m p = 880 kg/m "rpad= 140.210-10-3 m Pivot circle clearance Conductivity k=0.145W/m°C atl 0 °C Angles c'= 126.10.4 m k =0.133 W/m°C at 150 'C 0A =35' OB = 7.5' Preload factor Specific heat c - c' m:=
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| c
| |
| -=:0.4 c, =1.71.106 j/m 3 °C Distance pad surface pivot point The absolute viscosity of the oil as d =100.10- 3 m Rotational speed (o = 314.16 rad / s function of temperature is shown in top figures on page B92.2 Bearing axial width w=210 3 m Vertical load Reservoir oil temperature T, = 45 °C F= 100-103 N B92.l Version 99.0 99.0 B92.1
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| SOLVIA Verification Manual Nonlinear Examples I B2 IDl O 17 AS CJ6001- or S -C , : . IC S io o o *6 60 6o 6 1 6iz *0/ .6 SO6 IA N[NGICEEDI61 AB
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| *01VIA-pOST 99.0 262[ASNGACEeR6G 2 SO'IIIA-POST91 0 SOL'IIA-PDST99 0 Finite Element Model Due to the symmetry about the plane X=0 only one half of the bearing is modelled. An 8x3 mesh of 18-node FILM elements is used at each pad. The generation is performed in a cylindrical system using the comer node numbers of the three film segments as shown below. The surfaces of each segment with FILM elements have different radii of curvature, one radius for the pad and one for the shaft surface. A preloaded bearing is then obtained when the pad surface of each segment is translated into the correct radial position. The resulting film thicknesses are shown as a contour plot in the right top figure of page B92.6.
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| 3 3 film film 1 99 reservoir oil node user defined nodes at X=0 user defined nodes atX=-0.105 m Each pad surface is kept rigid by rigid links to the pivot node. Similarly, each shaft surface is kept rigid by rigid links to the center node of the shaft. The pad and shaft surface nodes with their rigid links are shown in the bottom figures of page B92.5.
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| Each shaft surface is given a rotational velocity of 314.16 rad/s (3000 rpm) about the negative X direction, thus co, =-314.16.
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| Version 99.0 B92.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples The film pressure along three of the segment edges is fixed to zero, leaving the film pressure free along the remaining nodes of the edge in the symmetry plane, see the two bottom figures of page B92.6. Note that views from both below and above are necessary to show the pressure boundary codes since the FILM element hides some of the nodes due to its thickness, even though the thickness is very small.
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| The pivot points of the pads are free to rotate about the X-axis, while the shaft center node is fixed for rotation about the X-axis so that the frictional moment due to the oil viscosity can be taken as a reaction moment.
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| The command T-FILMCONSTRAINTS FACTOR=RESERVOIRFLOW is used to prescribe that the oil temperature at each inlet node is determined as a mix of 30% reservoir oil flow and the remaining 70% as hot oil carry over flow from the neighboring outlet node corresponding to the inlet node, see the schematic figure below. The values are adjusted by the program if the available outlet oil flow is less than 70%, see below. The inlet node temperature is thus determined as cout QoverToul +cR QvTR Cin Qin where oil flow at the inlet node = QOVr +Q V oil flow at the corresponding outlet node reservoir oil flow to the inlet node Q) >0.3-Qi reservoir oil temperature TR = 45 °C hot oil carry over flow from the outlet node to the inlet node Qover -<QOut The thermal film constraints are shown graphically in the left top figure on page B92.6.
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| The alternative option in T-FILMCONSTRAINTS namely FACTOR=CARRYOVERFLOW is used in Example B97.
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| Solution Results The input data on pages B92.10 to B92.13 is used in the analysis.
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| The force/moment tolerances govem the accuracy of the analysis. The used force/moment tolerances Version 99.0 B92.3
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| | |
| SOLVIA Verification Manual Nonlinear Examples govern also the accuracy of the pressure and temperature solutions since a small change in pressure or in temperature causes a large change in the film surface loads compared to the force/moment tolerance values. The prescribed pressure and temperature tolerances are therefore rather loose which also has a positive effect on the initial convergence rate.
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| 3 The artificial springs of the two lower film segments have the default value of 1.0- 10N1/ m for the spring stiffness. This means that a pressure of 10 MPa causes a spring deformation of 100 gtm during the iterations, which stabilizes the iteration path to convergence. However, the oil pressure for the top film segment will be small in the converged configuration due to the increase in film thickness when the shaft is moving downwards under the action of the vertical load. The top film segment needs, therefore, a lower value for the artificial spring stiffness when the film pressure becomes small. This is achieved by specifying the spring stiffness variation with pressure, parameter DKDP in command FILMVELOCITY, to be 1.0 104 N / (m 3Pa). The artificial spring stiffness varies then with the 3
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| pressure and is for example 1.0.1011 N / m 3 when the pressure is 10 MPa and 1.0 1010N / m when the pressure is 1 MPa. A too large artificial spring stiffness will cause slow convergence, since the out-of balance forces can only achieve very small displacement corrections during the iterations.
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| The used values of tolerances and artificial spring stiffnesses give good convergence characteristics and the solution can be obtained in one step.
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| The displacement vector plot on page B92.7 shows the essentially vertical displacement of the shaft and the rotations of the pads. The distributions of film temperature, film flow, film power loss, film thickness and Reynolds number are shown as contour plots on pages B92.7 and B92.8. Diagram results of film temperature, film pressure and film thickness are shown on pages B92.8 and B92.9. In these diagrams the element result lines are defined using result points nearest the symmetry plane and nearest the side edge. The edge results have symbols in the diagrams.
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| The numerical results are summarized in the tables below. Note that the results apply to half the bearing.
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| Film element results, Pad I Pad 2 Pad 3 Total half bearing model:
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| Summation of pressure load in -25.9- 10' 1.9-10, -25.9- 10' -50.1- 103 vertical Z-direction, N 3 2.1. 4.6 10-4 2.3"10-4 9.0.10-4 Inlet oil flow m /s Power loss W 4.6.10' 2.4- 103 5.1"103 12.1- 103 Film thickness, max m 132- 10-6 209. 10 140 10' Film thickness, min m 60-10-6 180.10- 67-10 NI/m 2 4.8. 106 0.2. 106 4.8.106 Film pressure, max 0 64.0 57.3 55.9 Inlet film temperature') C 0 79.7 60.3 71.7 Outlet film temperature') °C
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| : 1) At the element result point nearest the symmetry plane.
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| Version 99.0 B92.4
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| | |
| SOLVIA Verification Manual Nonlinear Examples Finite element results at the center of the shaft (node 7), half bearing model:
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| Vertical displacement e'z , m -65.1.10 Horizontal displacement e'y , m -6.1. 10.6 Attitude angle 0 = tanlf(e'Y/e'z) , degrees 5.4 Eccentricity ratio iF= e'/c' 0.52 Frictional torque, reaction moment Tx , Nm -38.5 Frictional power cox Tx, W 12.1-10' User Hints 0 Note that the summation of the film pressure in a film element is performed over the top surface (t=+ 1) of the element with the positive pressure forces acting in the positive t-axis direction.
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| The thickness of the film element at an integration point is calculated as the projection onto the t axis of the line between the interpolated top and bottom surface points corresponding to the location of the integration point. The t-axis is always normal to the midsurface of the FILM element.
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| B92 TILTING-PAD JOURNAL SEARANG. R7GID PADS B92 TILTING-PAD JOURNAL BEARING, R7IGD PADS ORIGINAL - 0 05 ORIGINAL - -- .OS Z Z
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| ZONE R1G!LL:NKS ZONE SHAFT Y
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| X X 0112 k! 3 SOLV0A-PRE 99.0 SOLVIA ENGINEERING AS SOLVIA-PRE 99 3 SOLlIA ENGWNEERING AS Version 99.0 B92.5
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| | |
| , LULJL'JL'*L0LU' L T - CL'"
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| CL LU LU
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| 'o>
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| zo I '2",,
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| ° a 2' LU, CL C o
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| 00..
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| >';L r'
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| SOLVIA Verification Manual Nonlinear Examples 592 T-LTING-PAD JOURNAL BEARING, RIGID PAD5 392 TILT NG-PAD JOURNAL BEARING, R-GID PADS MAX D:SRL. -. 5355E-S Z MAX DISPL 6.535SE-S TIME L L -'ME I ZONE F_.L
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| -Y X
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| ILLM TEMPERATURE MAX 79.9t8
| |
| ! 78.322 75.[29 7* 936
| |
| : 58. 743 65 SSO 52.357 59.164 DISPLACEMENT 55 97' 6.5S3SSE-S MIN 54.374 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB SOLVIA-POST 99.0 SOLVIA ENGINEERING AB
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| ý B92 TILTING-PAD JOURNAL BEARING, RI1ID PADS B92 TILTING-PAD JOURNAL BEARING, RIG7D PADS MAX DISPL 65355E-5 MAX DISPL. 6.5355E-5 I 7ME 1 OTME I ZONE FILM Z ZONE F:LM X2 y F-A1
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| !'FLPRESSUREZM'42-FO S *!i*d *' - FILM POWERLOSS MAX 1.864tE6 'lAX 3.3373E.
| |
| S4.5587E6
| |
| ...1 2. 8'.16!E5 3.948IE5 2.94767-5
| |
| ; 3.33 745E6 2.6879ES i 2.72685E6 10 2I.16126 INEER1.5055E6 Si.9085E 2 .583ES ii 8.9-82E5 I .6488E5 i 2.841655 MI .290E5 38 MINI21163 M[IN G A2BgE SOLVIA ENGINEERING AB SCLV.A-POST 99.0 SOLVCA ENGZNE5PIN AB SOLVIA-POST 99.0 Version 99.0 B92.7
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| SOLVIA Verification Manual Nonlinear Examples 892 0 LTING-PAD DDURNAL BEARING. RIGID PADS 892 71-TING-PAD 'OURNWL BEARA'NG RIGID -ACD OR0G:NA -
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| ZONE F 0.'
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| 0 K7 ZONE FILM Z
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| X y
| |
| F',LI12 oYM MI-E 7 FILM M3SYF THICKESS MAX 2.0958E S2.0023E-4 FIL.'11-S'TM I St52E-I
| |
| .6282E-1
| |
| .4412E-4, 12S41E-4~
| |
| 1087!0 4 8.8005E-51 6.302E S' 8
| |
| MIN S 99506-0 SOLVIA POST 99.0 SOLVIA ENGIGNEERING AB 89Z T.ICINTPA JOUORNR N0 8 0 TN. RIGIDPADS
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| 'IýE i..
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| I 01 ;ýLIIA-S7 SS D CLIIAEýIGIII-Eý111G1,3 L'ý-IýST1ý G 0 OUT;I E..GI1EER01G0 392 -ILITIIG-PAOJ0JR11L --1RI11G,RIGID -1G1 392 TI 00 GPIT TORNA 00 SE00NG GI 0 PADS
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| --. 1-1 ..... --.-
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| . ....... Lýk -!- I I : I.- !
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| ,VIRAE90GI0E1090A Version 99.0B9S B92.8
| |
| | |
| 0 z
| |
| ON ci ON 2 1 CI
| |
| 'ol' 0
| |
| ON ON 0
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| 0 C'S C1
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| :73-
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B92 TILTING-PAD JOURNAL BEARING, RIGID PADS' DATABASE CREATE MASTER IDOF=100011 TEMPERATURE=COUPLED TOLERANCES RNORM=1E3 RMNORM=1E3 RTOL=0.01 TYPE=F ITEMAX=100 T-ITERATION TTOL=0.1 PARAMETER $RI=140E-3 SR2=140.21E-3 $R3=240E-3 PARAMETER $MAT=1 SET EPS=1.E-8 MYNODES=200 SET PLOTORIENTATION=PORTRAIT NSYMBOL=MYNODES NNUMBERS=MYNODES VIEW 1 1 0 1 SYSTEM 1 CYLINDRICAL COORDINATES SYSTEM=1 ENTRIES NODE R THETA 1 $R1 -82.5 2 $R1 -12.5 3 $R1 55.0 4 SRi 125.0 5 $R1 -167.5 6 SRi -97.5 7 0. 0.
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| 11 $R2 -82.5 12 $R2 -12.5 13 $R3 -47.5 99 400E-3 -45 NGENERATION TIMES=2 SYSTEM=1 YSTEP=137.5 NSTEP=10 11 TO 13 NGENERATION ZSTEP=-0.105 NSTEP=i00 SYSTEM=i 1 TO 6 / 11 12 / 21 22 / 31 32 READ B92.DAT SET NODES=i8 EGROUP 1 FILM MATERIAL=1 GVOLUME 32 31 131 132 6 5 105 106 EL1=8 EL2=3 SYSTEM=i ADDZONE=FILMi FILMVELOCITY INPUT=SURFACE OMEGA=314.16 AX=0. AY=C. AZ=C. BX=-1. BY=0. BZ=0.0 5 6 106 105 EGROUP 2 FILM MATERIAL=1 GVOLUME 22 21 121 122 4 3 103 104 ELi=8 EL2=3 SYSTEM=1 ADDZONE=FILM2 FILMVELOCITY INPUT=SURFACE OMEGA=314.16 AX=C. AY=0. AZ=C. BX=-1. BY=-. BZ=0.0, DKDP=1E4 KMIN=1E9 3 4 104 103 EGROUP 3 FILM MATERIAL=i GVOLUME 12 11 111 112 2 1 101 102 EL1=8 EL2=3 SYSTEM=1 ADDZONE=FILM3 FILMVELOCITY INPUT=SURFACE OMEGA=314.16 AX=C. AY=C. AZ=C. BX=-1. BY=-. BZ=0.0 1 2 102 101 RIGIDLINK INPUT=SURFACE ADDZONE=PAD1 31 32 132 131 33 RIGIDLINK INPUT=SURFACE ADDZONE=PAD2 22 21 121 122 23 RIGIDLINK INPUT=SURFACE ADDZONE=PAD3 12 11 I11 112 13 MESH RIGIDLINKS NSYMBOLS=YES Version 99.0 1392.10
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
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| RIGIDLINK INPUT=SURFACE ADDZONE=SHAFT1 5 6 106 105 7 RIGIDLINK INPUT=SURFACE ADDZONE=SHAFT2 3 4 104 103 7 RIGIDLINK INPUT=SURFACE ADDZONE=SHAFT3 1 2 102 101 7 ZONE SHAFT ZONE OPERATION=ADD ZONEl=SHAFTI SHAFT2 SHAFT3 MESH SHAFT NSYMBOLS=YES TRANSLATE ZONE=PAD1 Y=56.75E-6 Z=61.93E-6 SYSTEM=0 COPIES=
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| TRANSLATE ZONE=PAD2 Z=-84E-6 SYSTEM=0 COPIES=0 TRANSLATE ZONE=PAD3 Y=-56.75E-6 Z=61.93E-6 SYSTEM=0 COPIES=0 FIXBOUNDARIES DIRECTION=FILM-PRESSURE INPUT=LINE 1 101 / 101 102 / 102 2 3 103 / 103 104 / 104 4 5 105 / 105 106 / 106 6 FIXBOUNDARIES DIRECTION=23 13 23 33 FIXBOUNDARIES 4 / 7 LOAD CONCENTRATED 7 3 -50E3 T-LOAD TEMPERATURE 99 45.0 T-FILMCONSTRAINT INPUT=LINE MINPUT=LINE FACTOR=RESERVOIRFLOW, RESERVOIRNODE=99 12 112 21 121 .3 32 132 11 111 .3 22 122 31 131 .3 MESH TFILMCONSTRAINTS NSYMBOLS=YES T-INITIAL TEMP TREF=45 SOLVIA MESH FILM CONTOUR=THICKNESS MESH FILM BCODE=FILMPRESSURE VIEW 2 1 0 -1 MESH FILM BCODE=FILMPRESSURE END Version 99.0 B92.11
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE read file B92.DAT
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| * File B92.DAT MATERIAL $MAT OIL DENSITY=880 PRESSURE-FACTOR=2.7E-8
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| * temperature viscosity pressure
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| : 10. 0.1210
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| : 15. 0.0890
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| : 20. 0.0660
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| : 25. 0.0510
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| : 30. 0.0400
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| : 35. 0.0330
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| : 40. 0.0260
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| : 45. 0.0210
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| : 50. 0.0175
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| : 55. 0.0145
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| : 60. 0.0123
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| : 65. 0.0105
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| : 70. 0.0090
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| : 75. 0.0080
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| : 80. 0.0070
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| : 85. 0.0062
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| : 90. 0.0054
| |
| : 95. 0.0049 100. 0.0044 105. 0.0041 110. 0.0038 115. 0.0035 120. 0.0032 125. 0.00295 130. 0 .0027 135. 0 .0025 140. 0.0023 145. 0 .00215 150. 0.0020 END DATA T-MATERIAL $MAT TEMP-CONDUCTION
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| * temperature conductivity specific heat
| |
| : 10. 0.145 1.71E6 150. 0.133 1.71E6 END DATA SOLVIA-POST input
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| * B92 TILTING-PAD JOURNAL BEARING, RIGID PADS SET PLOTORIENTATION=PORTRAIT DIAGRAM=GRID DATABASE CREATE WRITE 'b92.iis' EGROUP 1 EPLINE FILM1-SYM 1 9 6 3 TO 8 9 6 3 EPLINE FILMI-EDGE 17 7 4 1 TO 24 7 4 1 EGROUP 2 EPLINE FILM2-SYM 1 9 6 3 TO 8 9 6 3 EPLINE FILM2-EDGE 17 7 4 1 TO 24 7 4 1 Version 99.0 B92.12
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| ffZ69 0-66 UOTS-TOA CNE MOqEWqIl=GNIN MqIa NOIIVWWnS MOqJWqI2=GNlN ZW=2 NOIIVKWnS MOqdWqIZ=QNIN TWqIa NOIIVNKnS MOqaNqIJ=GNIN WqIa NOIIVWWnS SEnSS2ýIdWqIJ=GXIN WqIa NOIIVWNnS SSOqHSMOdWqI3=GNIN WqIa NOIIVKWnS NOIlDVEýl=GNIN ý-N ISIrIN LN ISIqN E=SIXV2ý ýT=SIXVX SSaNN3IHIWqIZ=QNIN N2ýS- ZKUI,ý RNIrI:4 E=SI=i ZT=SIXVX SSRNN3IHIWrlId=CINIA W71S - Z=E HNIrIE TT=SIXVX SS2NNDIHIWqIJ=GNI]q N,7ýS-T=Z SNIrI2 GqO=HNVUJ2nS T=709H7,S ýT-SIXVX SýIOSS9HdWrIIa=QNIY SDC3 - ENrII2 HNI72 ET=SIXVX 2ýInSS9'ddN7Ia=CININ N7ýS - HNI72 070=9NVýIZSJOS T='109N2ýS E-SIXV-Tý ZT-=SIXVX SdfISSaUdNrIIZ=GNIN S9GH- ýWrI I Z SNI72 EýT=SIXVX SýIfISSHddNrlIa=GNIX NTS-EWrIIJ SNI72 CqO=ENVUa9nS T=rIOEN7,S TT-=SIXVX HýMSSSUdNHa=GNIN H9G2-TN'III SNI72 Z=SIXV2ý TT=SIXVX HýMSSaUdNrHa=(INIM N,ýS-TWr= aNIrIS T=rIORW7.S T-SIXV2ý ET-=SIXVX R9GR-EWqIZ ENT=
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| T=SIXVK CT=SIXVX E=VUHdWHIWqIJ=GNIN W2ýS-EW=a aNIrIE GrIo=aNvýIjarls T=qOgK,ýS T-=SIX)fA ZT-=SIXVX S=VE2dWSlWqIJ=GNIM SE)G2-ZwrIIj UNIrIE T=SIXV7, ZT=SIXVX W2ýS-ZWIIZ HNIrIH T=qOGN7,S T-SIXV2ý TT-=SIXVX 2DGS-TWrIIJ SNYIE T=SIXV.7, TT=SIXVX HHnIVýlHdKSIWqIJ=GNIN W2ýS-TWUIZ RNIrIS 2dvDsGNvq=NOIIVIN2IUoloqd 12S ST'O=XVWA O=NINA ET SIX'V ZNUII, =rl2gVrI BT
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| * O=XVWA O=NIKA ET sixv
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| (-Iuoz)) Indut JSOI-VIA'IOS soldwuxa mouquoN junuel/V uOIIL'Z)UIJ3A VIA-10S
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B93 TILTING-PAD JOURNAL BEARING, ELASTIC PADS Objective To demonstrate the use of FILM elements in a tilting-pad journal bearing with thermo-elastic pads that are cooled on the external surfaces.
| |
| Physical Problem The geometry of the bearing and the elastic pads are shown in the figures below. The pads have a thermo-elastic material and have convection surfaces on the outside faces. A rigid shaft with axisymmetric thermal conditions is assumed. Each pad is free to rotate about the X-direction at the pivot point. This bearing is analyzed in Example B92 with rigid pads and no heat conduction.
| |
| Elasticity and heat conduction in both the shaft and the pads are treated in Example B96.
| |
| 3 The values of rahaft, rpadI c', F, woand the properties of the oil are the same as in Example B92.
| |
| Pad material and geometry:
| |
| Young's modulus E = 2.10" N / m2 Conductivity k = 51.9 W / m°C Poisson's ratio v = 0.3 Convection coefficient h = 200 W I m C Thermal expansion coefficient Ambient temperature Ta = 45 0C a=1.1.10-5 m/m°C Angles = 35° c = 42.5' Pad thickness h -5010- 3 m Distance pad surface - pivot point Reference temperature giving zero thermal d = 100- 10-3 m stresses TREF = 25°C Version 99.0 B93.1
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| | |
| SOLVIA Verification Manual Nonlinear Examples Finite Element Model Due to the symmetry about the plane X=0 only one half of the bearing is considered. The key nodes of the model are shown on page B93.4. Pad 3 is first generated using a sequence of GVOLUME commands. A single pivot point is used but the pivot force on the pad is distributed to a line load in the symmetry plane. The line is kept rigid by rigid links to the pivot point. The external faces of the pad is cooled using thermal convection surfaces. Pad 3 is then copied by rotation to create pads 1 and 2, and all three pads are moved to the correct radial position.
| |
| The film segment at each pad consists of 8x 3 16-node FILM elements.
| |
| Each circumferential line of shaft nodes is constrained to be thermally axisymmetric in order to model the averaging effect due to the shaft rotation. The shaft surface is made rigid by rigid links to the center node.
| |
| The film pressure boundary conditions, the applied vertical shaft load and the thermal constraints for the oil film inlet and outlet flows are defined as in Example B92.
| |
| Solution Results The input data on pages B93.8 - B93.12 is used in the analysis.
| |
| The vertical shaft force of 100 kN and the rotational speed of 314.16 rad/s (3000 rpm) are applied in a single load step.
| |
| The force/moment tolerance governs the accuracy of the analysis as described in Example B92. The reference force and moment are, however, set to larger values since the norms of the unbalance force and moment vectors have now many more components due to the much increased number of displacement degrees of freedom. The achieved accuracy can be checked in SOLVIA-POST using the command TOLERANCES.
| |
| The artificial spring stiffnesses used by the FILM elements are the same as in Example B92.
| |
| The displacement boundary conditions for the pads are shown on page B93.4. The convection surfaces of the pads are indicated by yellow color in the left top plot of page B93.5. The film thickness is displayed in a contour plot on the same page as well as the film temperature constraints and the axisymmetric temperature constraints for the shaft surface.
| |
| The displacement vector plot on page B93.5 shows the essentially vertical displacement of the shaft and the rotations of the pads. The distributions of pad temperature, film temperature, film flow, film pressure, film power loss, film thickness and Reynolds number are shown as contour plots on pages B93.5 and B93.6.
| |
| Diagram results of film temperature, film pressure and film thickness are shown on pages B93.6 and B93.7. In these diagrams the element result lines are defined using result points nearest the symmetry plane and nearest the side edge. The edge results have symbols in the diagrams.
| |
| The thermal expansion of the pads in the radial direction gives somewhat smaller film thicknesses and larger film pressures, in particular for the upper pad, than calculated in Example B92. The film thickness is also affected by the elasticity and the thermal gradients in the pads.
| |
| The numerical results are summarized in the tables below. Note that the results apply to half the bearing.
| |
| Version 99.0 B93 .2
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples Film element results, Pad 1 Pad 2 Pad 3 Total half bearing model:
| |
| Summation of pressure load in -28.9- 103 7.8-103 -28.9.-10' -50.0 .103 vertical Z-direction, N m3 / s 2.3.10-4 4.1-10-4 25-10-4 8.9.10-4 Inlet oil flow Power loss W 4.7- 103 2.8. 10' 5.1 - 10' 12.6- 103 Film thickness, max m 170 220. 10' 172-10-.6 Film thickness, min m 49.10' 133-10-6 58-10' Film pressure, max N / Mi 6.4- 106 1.0-10 6 6.1.10' Inlet film temperature' °C 67.1 58.9 58.3 Outlet film temperature') °C 83.7 63.6 76.1
| |
| : 1) At the element result point nearest the symmetry plane.
| |
| Finite element results at the center of the shaft (node 7), half bearing model:
| |
| Vertical displacement e' ,m -55.2- 10-6 Horizontal displacement e'y , m -25.-10-6 Attitude angle = tan-(e',/e'z) , degrees 2.6 Eccentricity ratio F'= e/c' 0.44 Frictional torque, reaction moment Tx , Nm -40.1 Frictional power cox -.T, W 12.6. 103 User Hints The solution is based on the assumption of laminar oil flow which prevails when Reynolds number is smaller than about 500 - 1500, depending on a number of detail flow conditions for the oil. The maximum Reynolds number is calculated to be 683, see the contour plot on page B93.6.
| |
| The oil flow portion at a film inlet node that comes from the reservoir is defined to be minimum 30%, see command T-FILMCONSTRAINT FACTOR=RESERVOIRFLOW in the input data on page B93.10. However, a larger portion is taken from the reservoir when the outlet oil flow from the neighboring oil film node is not sufficient to make up the remaining 70%. In this case the inlet oil temperature will be proportionally lower. The alternative option T-FILMCONSTRAINTS FACTOR=CARRYOVERFLOW is used in Example B97.
| |
| Version 99.0 B93.3
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples CZ 11 film film 99 reservoir oil node user defined nodes at X=O user defined nodes at X =-0.105 m B93 TILTING-PAO JOURNAL BEARING. ELASTIC PADS 393 TTLTING-PAO JOURNAL BEARINS, ELASTIC PADS ORIGINAL *0.05 Z ORIGINAL - 0.05 Z ZONE SOLID ZONE SOLID Z Y 'SX
| |
| '1ASTiR 'AOS7F OOO0A 5B N I 1AS S i!OCLO A 222222 A 22?22A SOL/:A ENGINEER!NG AB SOLVIA-PRE 99.0 SOLVSA ENGINEERING AB SOLV'A-PRE 99 Version 99.0 B93.4
| |
| | |
| cn
| |
| -ci a
| |
| CA 0 * .
| |
| 6 p *°I -ci V
| |
| tS, law :5 U-)
| |
| a' N
| |
| ŽA'77.7y
| |
| /
| |
| 0 47 `
| |
| 0p a*
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| 82
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| | |
| Cl)
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| A-Himff aim 1 man*
| |
| V a
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| z aLa] a, '.1 Jý h xu=mg 1 is pass, ph NtW7 0
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| 0 0 Vill q
| |
| 03 0
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B93 TILTING-PAD JOURNAL BEARING, ELASTIC PADS' DATABASE CREATE MASTER TEMPERATURE=COUPLED TOLERANCES TYPE=F RNORM=1E5 RMNORM=lE5 RTOL=0.01 ITEMAX=100 T-ITERATION TTOL=0.1 PARAMETER $RI=140E-3 $R2=140.21E-3 $R3=190E-3 $R4=240E-3 SET EPS=1.E-6 MYNODES=l000 SYSTEM 1 CYLINDRICAL COORDINATES SYSTEM=l ENTRIES NODE R THETA 1 $Rl -82.5 2 $Rl -12.5 3 SRI 55.0 4 $Rl 125.0 5 $RT -167.5 6 $R1 -97.5 11 $R3 -12.5 12 $R2 -12.5 13 $R2 -38.75 14 $R2 -56 .25 15 $R2 -82 .5 16 $R3 -82.5 17 $R3 -51.875 18 $R3 -43.125 19 $R4 -51.0 TO 21 $R4 -44.0 99 400E-3 -45 7 0 0 NGENERATION TIMES=2 SYSTEM=l YSTEP=137.5 NSTEP=10 12 15 NGENERATION ZSTEP=-0.105 NSTEP=100 SYSTEM=1 1 TO 6 / 11 TO 35 SET MIDNODES=l LINE CYLINDRIC 12 13 EL=3 / LINE CYLINDRIC 112 113 EL=3 LINE CYLINDRIC 13 14 EL=2 / LINE CYLINDRIC 113 114 EL=2 LINE CYLINDRIC 14 15 EL=3 / LINE CYLINDRIC 114 115 EL=3 LINE COMBINED 12 15 13 14 / LINE COMBINED 112 115 113 114
| |
| * Oil material PARAMETER $MAT=1 READ B92.DAT
| |
| * Pad material MATERIAL 2 THERMO-ELASTIC TREF=25
| |
| -10000 2.Eli 0.3 1.IE-5 10000 2.ElT 0.3 1.IE-5 T-MATERIAL 2 TEMP-CONDUCTION
| |
| -10000 51.9 0.
| |
| 10000 51.9 0.
| |
| * Pad convection material T-MATERIAL 4 CONVECTION H=200.
| |
| * Film elements, each pad film is an element group SET NODES=16 EGROUP 1 FILM MATERIAL=l GVOLUME 32 35 135 132 6 5 105 106 ELl=8 EL2=3 SYSTEM=l ADDZONE=FILM1 Version 99.0 B93.8
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
| |
| FILPVELOCITY INPUT=SURFACE OMEGA=314.16 ,
| |
| AX=C. AY=0. AZ=C. BX=--. BY=C. BZ=0.0 5 6 106 105 EGROUP 2 FILM MATERIAL=1 GVOLUME 22 25 125 122 4 3 103 104 ELi=8 EL2=3 SYSTEM=i ADDZONE=FILM2 FILMVELOCITY INPUT=SURFACE OMEGA=314.16 ,
| |
| AX=C. AY=C. AZ=-. BX=-1. BY=C. BZ=0.0 DKDP=-E4 KMIN=-E9 3 4 104 103 EGROUP 3 FILM MATERIAL=i GVOLUME 12 15 115 112 2 1 101 102 ELi=8 EL2=3 SYSTEM=i ADDZONE=FILM3 FILMVELOCITY INPUT=SURFACE OMEGA=314.16 ,
| |
| AX=-. AY=C. AZ=C. BX=-1. BY=C. BZ=0.0 1 2 102 101
| |
| * Generate pad 3 SET NODES=20 EGROUP 4 SOLID MATERIAL=2 GVOLUME ii 12 13 18 111 112 113 118 ELi=2 SYSTEM=1 ADDZONE=PAD3 GVOLUME 18 13 14 17 118 113 114 117 SYSTEM=1 ADDZONE=PAD3 GVOLUME 17 14 15 16 117 114 115 116 SYSTEM=1 ADDZONE=PAD3 GVOLUME 21 18 17 19 121 118 117 119 ELI=2 EL3=3 SYSTEM=1 ADDZONE=PAD3
| |
| * Boundaries for pad 3 FIXBOUNDARIES 12356 / 20
| |
| * Rigid links at pivot point RIGIDLINK INPUT=LINE ALPHA=l.1E-5 TREF=25 ADDZONE=PAD3
| |
| * nl n2 masternode 19 21 20
| |
| * Convection surfaces for pad 3 T-BOUNDARIES SEGMENTS CON-MATERIAL=4 INPUT=SURFACE 11 18 118 111 / 18 21 121 118 / 21 19 119 121 19 17 117 119 / 17 16 116 117 / 111 112 113 118 113 114 117 118 / 114 115 116 117 / 118 117 119 121 12 11 111 112 / 16 15 115 116 T-LOADS ENVIRONMENT CONVECTION=YES INPUT=MATERIAL 4 45.
| |
| * Generate pad 2 and pad 1 ROTATE PAD3 Ni=7 N2=15 N3=25 ADDZONE=PAD2 ROTATE PAD3 Ni=7 N2=15 N3=35 ADDZONE=PAD1
| |
| * Move all pads to correct radial position TRANSLATE ZONE=PADI Y=56.75E-6 Z=61.93E-6 SYSTEM=C COPIES=C TRANSLATE ZONE=PAD2 Z=-84E-6 SYSTEM=C COPIES=C TRANSLATE ZONE=PAD3 Y=-56.75E-6 Z=61.93E-6 SYSTEM=C COPIES=C
| |
| * Axisymmetric temperature condition for shaft surfaces LINE COMBINED 1 6 2 3 4 5 T-CONSTRAINT INPUT=LINE ADDZONE=T-SHAFT 1 6 1 1.
| |
| TRANSLATE ZONE=T-SHAFT X=-0.0175 SYSTEM=0 COPIES=6 ADDZONE=ABCD
| |
| * Make shaft surfaces rigid RIGIDLINK INPUT=SURFACE ADDZONE=SHAFT 1 2 102 101 7 3 4 104 103 7 5 6 106 105 7 Version 99.0 B93.9
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
| |
| * Film pressure boundary conditions FIXBOUNDARIES DIRECTION=FILM-PRESSURE INPUT=LINE 1 101 / 101 102 / 102 2 3 103 / 103 104 / 104 4 5 105 / 105 106 / 106 6
| |
| * Load and boundary conditions, shaft center node LOADS CONCENTRATED 7 3 -50E3 FIXBOUNDARIES 1456 / 7
| |
| * Symmetric boundary conditions for pad plane X=0 ZONE XSYM INPUT=CYLINDRICAL XLMIN=0 SYSTEM=l FIXBOUNDARIES 1 INPUT=ZONE ZONEl=XSYM
| |
| * Temperature and boundary condition for reservoir oil node FIXBOUNDARIES / 99 T-LOAD TEMPERATURE 99 45.0
| |
| * Define the oil inlet flow portion from reservoir T-FILMCONSTRAINT INPUT=LINE MINPUT=LINE FACTOR=RESERVOIRFLOW, RESERVOIRNODE=99 2 102 3 103 0.3 4 104 5 105 0.3 6 106 1 101 0.3 T-INITIAL TEMP TREF=25 SOLVIA SET PLOTORIENTATION=PORTRAIT NSYMBOL=MYNODES COLOR EFILL=ORANGE MESH ZONE=SOLID VIEW=I BCODE=ALL MESH ZONE=SOLID VIEW=-I BCODE=ALL SET PLOTORIENTATION=LANDSCAPE COLOR EFILL=CON-MATERIAL MESH ZONE=SOLID VIEW=I NNUMBERS=MY SUBF=21 MESH ZONE=SOLID VIEW=-I NNUMBERS=MY VIEW 1 1 0 1 MESH ZONE=FILM CONTOUR=THICKNESS SET NNUMBERS=MY MESH TCONSTRAINTS NSYMBOLS=YES MESH TFILMCONSTRAINTS NSYMBOLS=YES END Version 99.0 B93.10
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
| |
| * B93 TILTING-PAD JOURNAL BEARING, ELASTIC PADS SET MODE=BATCH PLOTORIENTATION=LANDSCAPE DIAGRAM=GRID DATABASE CREATE WRITE 'b93.1is' VIEW ABOVE 1 0 1 VIEW BELOW 1 0 -1 EGROUP 1 EPLINE FILMi-SYM 1 9 6 3 TO 8 9 6 3 EPLINE FILMi-EDGE 17 7 4 1 TO 24 7 4 1 EGROUP 2 EPLINE FILM2-SYM 1 9 6 3 TO 8 9 6 3 EPLINE FILM2-EDGE 17 7 4 1 TO 24 7 4 1 EGROUP 3 EPLINE FILM3-SYM 1 9 6 3 TO 8 9 6 3 EPLINE FILM3-EDGE 17 7 4 1 TO 24 7 4 1 MESH VIEW=X VECTOR=DISP MESH SOLID CONTOUR=TEMP VECTOR=REACTION VIEW=ABOVE MESH SOLID CONTOUR=TEMP VECTOR=REACTION VIEW=BELOW MESH FILM CONTOUR=FILMTEMP VIEW=ABOVE MESH FILM CONTOUR=TEMP VIEW=ABOVE MESH FILM CONTOUR=TEMP VIEW=BELOW MESH FILM CONTOUR=FILMPRESSURE VECTOR=FILMFLOW VIEW=ABOVE MESH FILM CONTOUR:FILMPOWERLOSS MESH FILM CONTOUR=FILMTHICKNESS PLINE=ALL MESH FILM CONTOUR=FILMREYNOLDS AXIS 1 VMIN=50 VMAX=90 LABEL='FILM-TEMP' AXIS 2 VMIN=0 VMAX=8E6 LABEL='FILM-PRESSURE' AXIS 3 VMIN=0 VMAX=216E-6 LABEL='FILM-THICKNESS' AXIS 11 VMIN=0 VMAX=0.18 LABEL='FILMI' AXIS 12 VMIN=0 VMAX=0.18 LABEL='FILM2' AXIS 13 VMIN=0 VMAX=0.18 LABEL='FILM3' ELINE FILMi-SYM KIND=FILMTEMPERATURE XAXIS=11 YAXIS= 1 OUTPUT=ALL ELINE FILMi-EDGE KIND=FILMTEMPERATURE XAXIS=-11 YAXIS=
| |
| SUBFRAME=OLD SYMBOL=1 ELINE FILM2-SYM KIND=FILMTEMPERATURE ELINE FILM2-EDGE KIND=FILMTEMPERATURE XAXIS=12 YAXIS=1 OUTPUT=ALL XAXIS=-12 YAXIS=-l ,
| |
| SUBFRAME=OLD SYMBOL:1 ELINE FILM3-SY* KIND=FILMTEMPERATURE XAXIS=13 YAXIS=1 OUTPUT=ALL ELINE FILM3-EDGE KIND=FILMTEMPERATURE XAXIS:-13 YAXIS=l ,
| |
| SUBFRAME=OLD SYMBOL=I ELINE FILMI-SYM KIND=FILMPRESSURE XAXIS=11 YAXIS=2 ELINE FILMI-EDGE KIND:FILMPRESSURE XAXIS=-Il YAXIS=2 SUBFRAME=OLD SYMBOL=I Version 99.0 B93.11
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input ELINE FILM2-SYM KIND=FILMPRESSURE XAXIS=12 YAXIS=2 ELINE FILM2-EDGE KIND=FILMPRESSURE XAXIS=-12 YAXIS=2 SUBFRAME=OLD SYMBOL=1 ELINE FILM3-SYM KIND=FILMPRESSURE XAXIS=13 YAXIS=2 ELINE FILM3-EDGE KIND=FILMPRESSURE XAXIS=-13 YAXIS=2 SUBFRAME=OLD SYMBOL=1 ELINE FILM1-SYM KIND=FILMTHICKNESS XAXIS=11 YAXIS=3 ELINE FILMI-EDGE KIND=FILMTHICKNESS XAXIS=-ll YAXIS=3 SUBFRAME=OLD SYMBOL=1 ELINE FILM2-SYM KIND=FILMTHICKNESS XAXIS=12 YAXIS=3 ELINE FILM2-EDGE KIND=FILMTHICKNESS XAXIS=-12 YAXIS=3 SUBFRAME=OLD SYMBOL=1 ELINE FILM3-SYM KIND=FILMTHICKNESS XAXIS=13 YAXIS=3 ELINE FILM3-EDGE KIND=FILMTHICKNESS XAXIS=-13 YAXIS=3 SUBFRAME=OLD SYMBOL=1 NLIST N7 NLIST N7 KIND=REACTIONS SUMMATION FILM KIND=FILMPOWERLOSS SUMMATION FILM KIND=FILMPRESSURE SUMMATION FILM KIND=FILMFLOW SUMMATION FILM1 KIND=FILMFLOW SUMMATION FILM2 KIND=FILMFLOW SUMMATION FILM3 KIND=FILMFLOW SUMMATION PAD1 KIND=REACTION SUMMATION PAD2 KIND=REACTION SUMMATION PAD3 KIND=REACTION SUMMATION KIND=REACTION TOLERANCES END Version 99.0 B93.12
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| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B94 JOURNAL BEARING, FULL SOMMERFELD SOLUTION Objective To verify the pressure solution produced by the FILM element when the film temperature is constant and the oil material model is specified as independent of the oil pressure.
| |
| Physical Problem The journal bearing shown in figure below is to be analyzed when the rotational speed is 3000 rpm.
| |
| No side leakage of oil shall occur. Shaft and bearing axes are assumed to be parallel. The full Sommerfeld solution is sought.
| |
| A z
| |
| Bearing radius r, = 0.1001 m Shaft radius rb = 0.1000 m Eccentricity e = 20-10-6 m e Y Bearing width B = 0.05 m Absolute viscosity - = 0.02 Pa s Rotational speed Ob = 314.16 rad/s Pressure at maximum film thickness (0=0) Po"= Pa Finite Element Model The finite element model consists of 60X2 16-node FILM elements. are The top left figure on page B94.3 shows a contour plot of the film thickness. Two cylindrical systems used to generate the film elements. All nodes of the FILM elements have zero displacements. The command FIXBOUNDARIES is not used for all nodes because a displacement solution requires at least one active displacement degree of freedom. Therefore, the shaft nodes were selected to have zero Z displacements using the LOADS DISPLACEMENTS command.
| |
| Solution Results The theoretical solution for this problem can be found for example in reference [I], page 221.
| |
| P(-rP0 69Lco n (2+)+ cosO)
| |
| Pressure:
| |
| where c =ra rb F= e/ c e is defined in the figure above.
| |
| Version 99.0 B94.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples Maximum pressure: Pma, - PO = O)b Pm 3F(4-E 2 )J(4-5 2
| |
| +e4 )
| |
| where p =m 2(2+ 2)(1- 2)
| |
| Load capacity:
| |
| F,=0 Film oil flux: q -frbmo hm where h.- 2 c (1 -2 (the film thickness at maximum pressure)
| |
| The input data used in the SOLVIA analysis can be seen on pages B94.4 and B94.5.
| |
| A comparison with the analytical solution is given in the table:
| |
| Analytical SOLVIA Maximum film pressure (MPa) 7.74 7.75 Load capacity in Y-direction (kN) -119 -119 Load capacity in Z-direction (N) 0 5.3.10-*
| |
| Oil film flux in 0 direction (in 2 / s) 0.00148 0.00148 A contour plot of the film pressure is shown in top right figure on page B94.3. The film pressure around the shaft is shown in a diagram on page B94.3 together with the analytical solution. A very good agreement with the analytical solution is observed. The distributions of film power loss, film thickness and oil film flux are also shown.
| |
| User Hints
| |
| " The oil film flux is defined as the volume oil flow per unit film width and is, therefore, constant around the shaft. It is analytically evaluated at the points having maximum (or minimum) pressure, where it is equal to the Couette flow per unit length, since the oil flow due to the pressure gradient is zero at these points.
| |
| " An oil pressure much below the vapor pressure of the oil cannot occur in practice since it would result in cavitation. The FILM element is currently formulated for incompressible oil flow without any cavitation. The solution of this example corresponds to the case that a hydrostatic pressure is applied to the oil film so that both positive and negative pressures with respect to the reference hydrostatic pressure is possible.
| |
| Version 99.0 B94.2
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples The midsurface of the FILM element is used as the reference surface by SOLVIA when calculating the oil film pressure (and temperature). The midsurface is obtained by taking the average deformed coordinates of corresponding top and bottom surface nodes. The analytical reference surface is positioned at the shaft surface. The applied surface velocity and the film thickness are thus the same in SOLVIA as in the analytical solution but the circumferential length of the two reference surfaces are slightly different.
| |
| Reference
| |
| [1] Hamrock, Bernard, J., Fundamentals of fluid film lubrication, McGraw-Hill, 1994 B94 JOURNALSEARING.FULLSoGMERFELO SCLUTION ORGNI 00 702 1 0 00 000L 000 OUlRA F02 MIX D0S0L U0SS6C-24 z 00 o o.79, z. r ezcNEFI FL
| |
| ** OSE-,
| |
| 090 O000S0V0-I MIN90750E6 SOLIA A -
| |
| SNEýEýI 90 000 00oo-0C00 S ; - 1,0SL: TI 18OZ
| |
| *Eý31 0 L11lA 0099 0 i 394 JýURNA-BEAR!NG*ULSMM*ENFELD SOLUTIOýN 094 OURNAL 0ULL BEARI0NG. SOLUTION 00MMERFEL0
| |
| : jML__
| |
| so o os o j C 00. . 0 11100
| |
| .AB.. . . . . . . . ... ... ~~~~~~~~~~G
| |
| . . .%!. EN .INER.I SOL .'A -100N<ERING 0 0 L,'iA-POS799 0 0 L_ 0nE0FEL _ OUI0 394 JOURNAL0EAR0 0 NG, ELD SOLUTIO ULLSO 0MER0 N 790 0CU'NA7E0N.
| |
| IT
| |
| ;0URN*L -J00\,,. 010 53L-IA-VS 3 10 0 0. 1.*NE*N
| |
| : 1. 7 SOL',0-0SI 99 Version 99.0 B94.3
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE:
| |
| HEADING 'B94 JOURNAL BEARING, FULL SOMMERFELD SOLUTION' DATABASE CREATE MASTER IDOF=100111 NSTEP=1 TEMPERATURE=LOAD SET EPS=1.E-6 PARAMETER sX=0.05 $RB=0.i $RA=0.1001 $ANG=90 $EL-60 SYSTEM 1 CYLINDRICAL Z=0.
| |
| SYSTEM 2 CYLINDRICAL Z=-20.E-6 COORDINATES SYSTEM=i ENTRIES NODE XL R THETA 1 0. $RA SANG 2 $X $RA $ANG COORDINATES SYSTEM=2 ENTRIES NODE XL R THETA 3 0. $RB SANG 4 $X $RB $ANG SET MIDNODES=i LINE CYLINDRICAL Ni=i N2=1 EL=$EL SYSTEM=i ROTATION=LONG LINE CYLINDRICAL Ni-2 N2=2 EL=$EL SYSTEM-i ROTATION-LONG LINE CYLINDRICAL Ni=3 N2=3 EL=$EL SYSTEM=2 ROTATION=LONG LINE CYLINDRICAL Ni=4 N2=4 EL=SEL SYSTEM=2 ROTATION=LONG MATERIAL 1 OIL
| |
| -1000. 0.02 1000. 0.02 END DATA SET NODES=d6 EGROUP 1 FILM MATERIAL=i RSINT=2 GVOLUME 1 1 2 2 3 3 4 4 EL2=2 FILMVELOCITY INPUT=SURFACE OMEGA=314.16 MOTION=ROTATION, AX=0. AY-0. AZ=-20.E-6 BX=i. BY=0. BZ=-20.E-6 3344 FIXBOUNDARIES DIRECTION=23 INPUT=SURFACE 3344 FIXBOUNDARIES DIRECTION=2 INPUT=SURFACE 1122 FIXBOUNDARIES DIRECTION FILMPRESSURE INPUT=LINE 12 LOADS DISPLACEMENTS INPUT=SURFACE 1 1 2 2 3 0.
| |
| INITIAL TEMPERATURE TREF=30.
| |
| LOADS TEMPERATURE TREF=30.
| |
| MESH FILM CONTOUR-THICKNESS ZONE OUTER CYLINDRICAL SYSTEM*I RMIN=0.10005 ZONE INNER CYLINDRICAL SYSTEM=2 RMAX=0.i0005 SOLVIA END Version 99.0 B94.4
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
| |
| * B94 JOURNAL BEARING, FULL SOMMERFELD SOLUTION DATABASE CREATE SET ZONE=FILM DIAGRAM-GRID WRITE 'b94.1is' MESH CONTOUR=FILMPRESSURE AXIS 1 VMIN:0 VMAX=0.7 AXIS 2 VMIN=-1OE6 VMAX=1OE6 EPLINE JOURNAL 1 3 1 TO 60 3 1 ELINE JOURNAL KIND=FILMPRESSURE XAXIS=1 YAXIS=2 USERCURVE 1 READ B94.DAT PLOT USERCURVE 1 XAXIS=-I YAXIS=-2 SUBFRAME=OLD ELINE JOURNAL KIND=FILMPOWER ELINE JOURNAL KIND=FILMTHICKNESS ELINE JOURNAL KIND=FILMFLUX EMAX SUMMATION KIND=FILMPRESSURE SUMMATION KIND=FILMPOWERLOSS SUMMATION KIND=FILMFLOW SUMMATION OUTER KIND=REACTION SUMMATION INNER KIND=REACTION END Version 99.0 B94.5
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B95 HEAT GENERATION IN A PARALLEL-SURFACE SLIDER BEARING Objective To verify the temperature and pressure solution for the FILM element when an isoviscous oil material is used.
| |
| Physical Problem A parallel-surface slider bearing with constant film thickness is to be analyzed. No heat transfer to the mating surfaces is assumed, i.e. adiabatic conditions. The oil film is assumed to be infinitely wide so that no side flow can occur. The inlet pressure is 10 MPa which causes a linear variation of the pressure to the outlet pressure of 0 Pa.
| |
| z T- inlet - outlet Oil properties: Speed ti = 10 m/s Geometry:
| |
| Bearing length 1= 0.1 m Absolute viscosity pt = 0.02 Pa s 3 w = 0.1 m Density p = 800 kg/iM Bearing width Film thickness h = 100.10-6 m Thermal conductivity k = 0.15 W/m 0 C Specific heat per unit volume c= 1.106 J/m 3 °C Inlet conditions (X = 0.0): Outlet conditions ( X = 0.1):
| |
| P =10-106 Pa P = 0.0 Pa T=0 0C Finite Element Model to The model is shown in the top left plot on page B95.3. Twenty 8-node FILM elements are used model the oil film. The fixed boundary conditions for displacements are specified by the command film LOADS DISPLACEMENT so that loads and reactions will be available for the top and bottom surfaces. The temperature boundary conditions are given by the command T-FILMCONSTRAINTS and all inlet nodes are dependent of node 9, which is designated to be the oil reservoir.
| |
| Solution Results The input data on pages B95.4 and B95.5 is used in the analysis.
| |
| Version 99.0 1395.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples The energy equation in one dimension with adiabatic conditions along the length is pV-2(cT)--I d d bk- aT =
| |
| dhk dx dx( dx) where c = specific heat per unit mass (= c, / p)
| |
| V =ih h3 dp = volume oil flow per unit width (film flux) 2 12,g dx pV = mass flow of oil per unit width p c V = heat flow of oil per unit width and unit temperature dp) + 'u62 = generated heat per unit surface area (film power loss) q =h 3 dpL q 1 ý Ydx) h 1 In this example hk << pVc so the solution to the energy equation is T= qqX pcV where X is the distance from the inlet boundary with zero temperature.
| |
| The shear stress is
| |
| ('~x'P=2h d-dp+ th h (at the top surface) 2h dxp= ýth (lz')bt - -2dx-+ h (at the bottom surface)
| |
| The shear forces in the positive X-direction acting on the surfaces with area A are (F,) top =--(Zuzx) top .A (FX)bot = (zx) bot *A The following analytical values are then obtained and the values agree excellently with the corresponding SOLVIA results:
| |
| Result type Unit Analytical SOLVIA Film flux m2 / s 0.91667.10.3 0.91667-10.3 2 61667 61667 Film power loss W/ m Outlet oil temperature C 6.7273 6.7273 Shear force on top surface N 30.000 30.000 Shear force on bottom surface N 70.000 70.000 Pressure force on top surface N 50000 50000 Pressure force on bottom surface N -50000 -50000 The SOLVIA reactions are the same as the corresponding forces but with reversed signs. Contour plots of film temperature, film heat flux, film pressure, film flux, film power loss and the Reynolds number are shown on pages B95.3 and B95.4. The heat flux in the film thickness direction at the top surface (film t-top flux) is also shown to verify that the boundary is indeed adiabatic.
| |
| Version 99.0 B95.2
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples User hints i The command SUMMIATION KIND=FILMPRESSURE gives the total force due to pressure alone (no shear force from the viscosity) acting at the top surface bounding the film. However, both the pressure and the shear forces are included when summation of applied forces and calculated reactions are requested.
| |
| i Note that SUMMATION KIND=TFLOW does not consider heat flow in FILM elements. The heat flow in the thickness direction of FILM elements through the top and bottom surfaces can be obtained by SUMMATION KIND=FILMTFLUX.
| |
| In the summation of applied forces over both surfaces the force components in the Z-direction (due to pressure) cancel each other. However, the X-direction force components (due to viscosity) are added since the shear stresses due to the pressure flow act in the same direction at both surfaces.
| |
| 50 -1 RATIOS NARRRALCEL
| |
| -N AEN SURFRCE SLIDER3EARING IO'RIGINALrM- o 31z ioNEý'ILI xY SDLIA99SR 1 NE9IS A S9 ' EC sE EZ EIIA R i A9 6N7273 16,1 2.9432 2 993Z 42915 IIL'IIAEICIINEIRlii AS CR[SIN-L 0011E 9S HEAT ZENERATION IN A PAZRAALE-SURACESLIDER CEARENG 945 MEAT, NE9ATION 11A ARALL -AURFACESLIER NEAR:NG ZONC F: 9 ZONE F1-9A 9S A EATM EE~zNI A'E-SR'C LAE6ERNOIEFI R
| |
| > L NIrc TIX E1& .
| |
| 9A E C ERAITCN:N A AT TRARLLEL-SURACIOE BEARIG BT E'EATESRATION TNA RALHE-SURFRCE SF:D 49E 9
| |
| IIX ISIL I 125E-21 I ýC[SýL 2,325E-21 MAX TI1.
| |
| 201E FILN
| |
| <f*. ~ ~ ~ , ýSL*
| |
| L, 0 :14 FLUX 1
| |
| 6R10
| |
| :121C
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| * 3 A2SOEC SOLVIAECO E.RING AB SO0VIA-o0Sl99 1 Version 99.0 B95.3
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| SOLVIA Verification Manual Nonlinear Examples
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| $95 97N1A 162$
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| N A 9E L-LURFACN 2L:DEý 9G 49$ -$ N A PARL9EL-4U$ E S9q 7,ENA:0NN :7 4E24N0 6, 9'5 2 92$-2 4 2SPL 2 D25-22 9r.E IlLtl 701E I ZONEF:Lý IIX 16-6 667 666. 6 8
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| j1667.$86
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| , ; 396670207 jii $9 6969501 EYNOLDS RLIA 49664.$67 7(6 9 49978 61664 287 5 17 916$
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| 667 SOVAPS 9aSOLVIA NGNEERIN1GAI $S-31_ 6$-D7 99.0 SOLVIA-PRE input HEADING 'B95 HEAT GENERATION IN A PARALLEL-SURFACE SLIDER BEARING' DATABASE CREATE MASTER IDOF=000111 NSTEP=1 TEMPERATURE=COUPLED SET EPS=1.E-6 TIMEFUNCTION 1
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| : 0. 0. / 1. 1.
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| COORDINATES 1 0.1 0.1 50.E-6 // 2 0.0 0 .1 50 .E-6 4 0.1 0.0 50 .E-6 3 0.0 0.0 50.E-6
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| / 6 0.0 0.1 -50 .E-6 5 0.1 0.1 -50.E-6 7 0.0 0.0 -50.E-6
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| / 8 0.1 0.0 -50 .E-6 9 -0.05 0.05 MATERIAL 1 OIL DENSITY=800
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| * temp viscosity
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| -1000. 0.02 1000. 0.02 END DATA T-MATERIAL 1 TEMP-CONDUCTION
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| * temp conductivity specific heat
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| -1000. 0.15 1.E6 1000. 0.15 1.E6 END DATA T-INITIAL TEMPERATURE TREF=0.
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| SET NODES=8 EGROUP 1 FILM MATERIAL=1 RSINT=2 GVOLUME 1 2 3 4 5 6 7 8 EL1=10 EL2=2 EL3=1 FILMVELOCITY INPUT=SURFACE OMEGA=10 MOTION=TRANSLATION TIMEFUNCTION=1, AX-0. AY=0. AZ=0. BX=-. BY=0. BZ:0.
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| 1234 LOADS FILMPRESSURE 2 INPUT=LINE
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| * nl n2 p1 p timefunction 2 3 1.E7 1.E7 1 1 4 0. 0. 1 LOADS DISPLACEMENTS INPUT=SURFACE
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| * nl n2 n3 n4 dir value 1 2 3 4 1 0.
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| 1 2 3 4 2 0.
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| 1 2 3 4 3 0.
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| 5 6 7 8 1 0.
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| 5 6 7 8 2 0.
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| 5 6 7 8 3 0.
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| Version 99.0 B95.4
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
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| T-LOADS TEMPERATURE
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| * node value timefunction 9 0. 1 T-FILMCONSTRAINTS INPUT=LINE M-INPUT=NODE
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| * snl sn2 master factor 2 3 9 1.
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| MESH CONTOUR=THICKNESS NNUMBER=MYNODES NSYMBOL=MYNODES SOLVIA END SOLVIA-POST input
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| * B95 HEAT GENERATION IN A PARALLEL-SURFACE SLIDER BEARING
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| * Results from temperature analysis T-DATABASE CREATE SET ZONE=FILM WRITE 'b95.lis' SET LINESPERPAGE=9999 NEWPAGE=NO MESH CONTOUR=FILMTEMPERATURE MESH CONTOUR=FILMTFLUX MESH CONTOUR=FILMTTOPFLUX ELIST ZONENAME=ELI SUMMATION KIND=TFLOW SUMMATION KIND=FILMTFLUX
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| * Results from stress analysis DATABASE CREATE MESH CONTOUR=FILMPRESSURE VECTOR=FILMFLOW MESH CONTOUR=FILMFLUX VECTOR=FILMFLUX MESH CONTOUR=FILMPOWERLOSS MESH CONTOUR=FILMREYNOLDS ZONE UPPER GLOBAL ZMIN=0.
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| ZONE LOWER GLOBAL ZMAX=0.
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| SUMMATION UPPER KIND=LOADS SUMMATION LOWER KIND=LOADS SUMMATION KIND=LOADS SUMMATION UPPER KIND=REACTIONS SUMMATION LOWER KIND=REACTIONS SUMMATION KIND=REACTIONS SUMMATION KIND=FILMPRESSURE SUMMATION KIND=FILMPOWERLOSS SUMMATION KIND=FILMFLOW END Version 99.0 B95.5
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B96 TILTING-PAD JOURNAL BEARING, ELASTIC PADS AND SHAFT Objective To demonstrate the use of FILM elements in a tilting-pad journal bearing with thermo-elastic pads and shaft. The pads are cooled on the external surfaces and the shaft on the end surface.
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| Physical Problem The same bearing as in Examples B92 and B93 is selected to be analyzed. Example B92 treats the case with rigid shaft and pads and with adiabatic film conditions. Elastic pads including thermal de formation and heat conduction are analyzed in Example B93. In this example both the shaft and the pads are elastic and can deform due to the temperature. The pads have convection surfaces on the out side faces and the shaft has a convection surface at the end face. Axisymmetric thermal conditions for the shaft are considered. The pivoted pads are supported with radial elastic springs.
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| An overall picture of the geometry is shown below. The geometry and material properties of the pads are shown in the figures on page B 93.1. The values of rshaft, c', F, co and the properties of the oil are the same as in Examples B93 and B92. The radial pad spring stiffness is krad = 1200 N / .m.
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| 1°
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| °Z Z
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| Section A-A X=O shaft film pad krdi2 Shaft data:
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| Young's modulus E = 2-10" N/ m2 Convection coefficient h = 200 W / m 2 °C Poisson's ratio v = 0.3 Ambient temperature of the shaft Ta = 45 'C Thermal expansion coefficient ca = 1.1 .10-5 1/'C Zero thermal stresses for TREF = 25 ° C Bearing width w = 210. 10-3 m Conductivity k = 51.9 W / m°C Version 99.0 B96.1
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| SOLVIA Verification Manual Nonlinear Examples Finite Element Model Due to the symmetry about the plane X=0 only one half of the bearing is considered. The pads and the film elements are generated as in example B93. The shaft consists of 20-node SOLID elements. Each circumferential line of shaft nodes at the the oil film surface is constrained to be thermally axisymmetric in order to model the averaging effect due to the shaft rotation. A convection surface is applied at the end of the shaft.
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| Each pad is supported by a radial spring to allow radial expansion of the pad pivot due to an assumed elasticity in the support structure.
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| The key nodes of the model and the displacement boundary conditions are shown on page B96.4. The pad convection surfaces are colored yellow in the plot of convection surfaces on page B96.5. A con tour plot of the initial film thickness is also shown on page B96.5 together with two plots of thermal constraint couplings.
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| Solution Results The input data on pages B96.9 - B96.14 is used in the analysis.
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| Default values of the FILM element artificial spring stiffness are used for the bottom pads. The top pad uses the artificial stiffness KNORM=3. 10".
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| The displacement vector plot on page B96.6 shows the essential displacements. The shaft has a verti cal displacement due to the external load and a radial displacement due to thermal expansion. The pads rotate around the pivot node and translate in the radial direction.
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| The distribution of temperature and film temperature are shown as contour plots on page B96.6. The distribution of film pressure, film power loss, film thickness and Reynolds number are shown on page B96.7. Diagram results of film temperature, film pressure and film thickness are shown on page B96.7 - B96.8. In these diagrams element result lines are defined using result points nearest the sym metry plane and nearest the edge. The edge results have symbols in the diagrams.
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| The numerical results are summarized in the tables below. Note that the results apply to half the bearing.
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| FILM element results Pad 1 Pad 2 Pad 3 Total Summation of pressure load in -31.4.103 12.7- 103 -31.4-103 -50.1 _103 vertical Z-direction, N m3 / s 2.0-10-4 3.2 10-- 2.2- 10-4 7.4.l0-4 Inlet oil flow Power loss W 4.6.10' 3.2. 10' 4.9.03 12.7.03 Film thickness, max m 157- 10 181 6 155.10 43 96-10-6 49. 10-6 Film thickness, min m 7.2.106 1.7.106 6.8.106 Film pressure, max Pa Version 99.0 B96.2
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| SOLVIA Verification Manual Nonlinear Examples FILM element results, cont. Pad 1 Pad 2 Pad 3 Total 0 72.2 63.5 63.4 Inlet film temperature" °C Outlet film temperature' ) °C 89.9 70.8 83.3 Radial displacement at pad 71. 10. 21-10' 71.10.
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| pivot node, m Heat flow from the oil film to 0.22- 10' 0.13- 103 0.18-10 3 0.53.103 the pads, W 0.50. 103 -0.29- 103 0.0- 10i 0.2 t- 103 Heat flow from the oil film to the shaft W
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| ') At the element result points nearest the symmetric line.
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| Finite elements results at the center of the shaft (node 7 ), half bearing model:
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| Vertical displacement e'z , m 5.10-6 Horizontal displacement e'y , m -1.1-10-6 Attitude angle q = tan-'(e'y/e'z) degrees 0.9 Eccentricity ratio s' = e'/c' 0.58 Frictional torque, reaction moment Tx , Nm -40.5 Frictional power, ox .Tx, W 12.7-10' User Hint It is important in this example that the pivot nodes have a radial stiffness that models the elasticity in the supporting structure. The temperature of the supporting structure is assumed to be the same as the reference temperature, thus no thermal expansion. If no elasticity in the supporting structure is used in the analysis then the thermal expansion of the shaft and the pads would cause the oil film thickness to approach zero. If this models an actual physical situation it could indicate shaft seizure.
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| In this example the thermal constraint for the inlet oil flow in command T-FILMCONSTRAINTS is set to the option FACTOR=RESERVOIRFLOW. The alternative option FACTOR=CARRYOVERFLOW is used in Example B97. Note that the two options may not always be each others inverse. For example, when FACTOR=RESERVOIRELOW the available carry over flow may be less than the portion (inlet flow) - (specified minimum reservoir flow), in which case the reservoir flow portion is increased by the program.
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| Version 99.0 B96.3
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| SOLVIA Verification Manual Nonlinear Examples 11 film film 99 reservoir oil node user defined nodes at X=-0.105 m user defined nodes at X=0 B96 TILTING-PAD JOURNAL BEARING. ELASTIC PADS AND SHAFT 396 TI-T/NG-PAD JOURNAL BEARING. ELASTIC PADS AND SHAFT ORIGINAL - S-.5 Z ORIGINAL "--0.05 Z ZONE ELE MENT7SI ZONE ELEMENTS x y MASTER MASTER 90 A
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| C 1000" C 1000!I D 1010![
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| O 01011 F 2 F20 U1, G 222222 A22222 SLVTA-PRA F 99 0 SOLVIA ENGINEERING AB SOL/IA-PRE 99.0 'OLVIA ENGINEERING AB Version 99.0 1396.4
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| CD 0
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| (A 0
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| 0 "0 (1 1
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| 0 0
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| 0 0
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| 0 P
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| "0 ON ON 2
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| 0 0
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| 0 0
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| 0 4E 2n'T (A
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| L.
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| ;,q - p-rW1 04 Cl e
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| C?
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| F -t ON
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| -Fl-V vF7 F -F ro C5
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| -F-F -F ON ON 0
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| ... 0 1
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| II)
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| p
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| . 7 0 A
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| 'I
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| -i 0
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| 0 0
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| JI, 0
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| 0 00 z0 0
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| 0 (V
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| 1 01 0
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B96 TILTING-PAD JOURNAL BEARING, ELASTIC PADS AND SHAFT' DATABASE CREATE MASTER TEMPERATURE=COUPLED TOLERANCES TYPE=F RNORM=5E4 RMNORM=5E4 RTOL=-E-2 ITEMAX=100 ITERATION FULL-NEWTON T-ITERATION TTOL=0.1 PRINTOUT MIN
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| $R3=190E-3 $R4=240E-3 SR5=270E-3 PARAMETER $RI=140E-3 $R2=140.21E-3 SET EPS=1.E-6 MYNODES=1000 SYSTEM 1 CYLINDRICAL COORDINATES SYSTEM=1 ENTRIES NODE R THETA 1 $Ri -82.5 2 $Ri -12.5 3 $RI 55.0 4 $Ri 125.0 5 $Ri -167.5 6 $Ri -97.5
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| $R3 -12.5 12 $R2 -12.5 13 $R2 -38.75 14 $R2 -56.25 15 $R2 -82.5 16 $R3 -82.5 17 $R3 -51.875 18 $R3 -43.125 19 $R4 -51.0 TO 21 $R4 -44.0 90 $R5 -47.5 91 $Ri 90.0 99 400E-3 -45 7 0 0 NGENERATION TIMES=2 SYSTEM=1 YSTEP=137.5 NSTEP=10 12 15 NGENERATION ZSTEP=-0.105 NSTEP=100 SYSTEM=i I TO 7 / 11 TO 35 NGENERATION ZSTEP=-0.210 NSTEP=200 SYSTEM=1 1 TO 7 / 11 TO 35 SET MIDNODES=1 LINE CYLINDRICAL 12 13 EL=3 / LINE CYLINDRICAL 112 113 EL=3
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| / LINE CYLINDRICAL 113 114 EL=2 LINE CYLINDRICAL 13 14 EL=2 115 EL=3
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| / LINE CYLINDRICAL 114 LINE CYLINDRICAL 14 15 EL=3 115 113 114
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| / LINE COMBINED 112 LINE COMBINED 12 15 13 14
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| / LINE CYLINDRICAL 4 5 EL-3 ROTATION=SHORT LINE CYLINDRICAL 2 3 EL=3 LINE CYLINDRICAL 6 1 EL=i
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| * Oil material PARAMETER $MAT=1 READ B92.DAT
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| * Pad and shaft material MATERIAL 2 THERMO-ELASTIC TREF=25
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| -10000 2.Eli 0.3 1.1E-5 10000 2.El 0.3 1.1E-5 T-MATERIAL 2 TEMP-CONDUCTION
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| -10000 51.9 0.
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| 10000 51.9 0.
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| Version 99.0 1396.9
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
| |
| * Pad convection material T-MATERIAL 4 CONVECTION H=200.
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| * Film elements, each pad film is an element group SET NODES=16 EGROUP 1 FILM MATERIAL=l GVOLUME 32 35 135 132 6 5 105 106 ELi=8 EL2=3 SYSTEM=I ADDZONE=FILMI FILMVELOCITY INPUT=SURFACE OMEGA=314.16 NA=7 NB=107 5 6 106 105 EGROUP 2 FILM MATERIAL=1 GVOLUME 22 25 125 122 4 3 103 104 ELI=8 EL2=3 SYSTEM=l ADDZONE=FILM2 NA=7 NB=107 KNORM=3E10 FILMVELOCITY INPUT=SURFACE OMEGA=314.16 3 4 104 103 EGROUP 3 FILM MATERIAL=l EL2=3 SYSTEM=i ADDZONE=FILM3 GVOLUME 12 15 115 112 2 1 101 102 EL1=8 NA=7 NB=107 FILMVELOCITY INPUT=SURFACE OMEGA=314.16 1 2 102 101
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| * Generate pad 3 SET NODES=20 EGROUP 4 SOLID MATERIAL=2 GVOLUME 11 12 13 18 111 112 113 118 ELi=2 SYSTEM=1 ADDZONE=PAD3 GVOLUME 18 13 14 17 118 113 114 117 SYSTEM=l ADDZONE=PAD3 GVOLUME 17 14 15 16 117 114 115 116 SYSTEM=l ADDZONE=PAD3 GVOLUME 21 18 17 19 121 118 117 119 EL1=2 EL3=3 SYSTEM=1 ADDZONE=PAD3
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| * Symmetric boundary condition for the pads at X=0 ZONE XSYM-PADS INPUT=CYLINDRICAL XLMIN=0 SYSTEM=1 FIXBOUNDARIES 1 INPUT=ZONE ZONE1=XSYM-PADS
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| * Boundaries for pad 3 FIXBOUNDARIES / 90 BOUNDARIES 101011 / 20 SKEWSYSTEM EULERANGLES 1 -47.5 0 0 NSKEW 20 1
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| * Spring element to pivoted pad node EGROUP 5 SPRING DIRECTION=AXIALTRANSLATION PROPERTYSET 1 K=600.E6 ENODES ADDZONE=PAD3 1 20 90
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| * Rigid links at pivot point to distribute pivot force RIGIDLINK INPUT=LINE ALPHA=.-IE-5 TREF=25 ADDZONE=PAD3
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| * ni n2 masternode 19 21 20
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| * Convection surface s for pad 3 INPUT=SURFACE T-BOUNDARIES SEGMENT S CON-MATERIAL=4 11 18 118 ll / 18 21 121 118 / 21 19 11 9 121 3 118 19 17 117 119 / 17 16 116 117 / 111 112 11 9 121 113 114 117 118 / 114 115 116 117 / 118 117 11 12 11 i1 112 / 16 15 115 116 T-LOADS ENVIRONMENT CONVECTION=YES INPUT=MATERIAL 4 45.
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| * Generate pad 2 and pad 1 ROTATE PAD3 N1=7 N2=15 N3=25 ADDZONE=PAD2 ROTATE PAD3 Ni=7 N2=15 N3=35 ADDZONE=PADl Version 99.0 B96.10
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
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| * Move all pads to correct radial position TRANSLATE ZONE=PAD1 Y=56.75E-6 Z=61.93E-6 SYSTEM=- COPIES=
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| TRANSLATE ZONE=PAD2 Z=-84E-6 SYSTEM=- COPIES=
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| TRANSLATE ZONE=PAD3 Y=-56.75E-6 Z=61.93E-6 SYSTEM=C COPIES=C
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| * Axisymmetric temperature condition for the shaft surface LINE COMBINED 1 1 2 3 4 5 6 T-CONSTRAINT INPUT=LINE ADDZONE=T-SHAFT
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| * snl sn2 master factor 1 1 1 1.
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| COPYSETTINGS FIXBOUNDARY=NO TRANSLATE ZONE=T-SHAFT X=-0.0175 SYSTEM=- COPIES=6 ADDZONE=ABCD
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| * The elastic shaft EGROUP & SOLID MATERIAL=2 GVOLUME 2 1 i01 102 7 7 107 107 SYSTEM=1 ADDZONE=SHAFT GVOLUME 3 2 102 103 7 7 107 107 ELI=3 SYSTEM=I ADDZONE=SHAFT GVOLUME 4 3 103 104 7 7 107 107 SYSTEM=l ADDZONE=SHAFT GVOLUME 5 4 104 105 7 7 107 107 EL1=3 SYSTEM=1 ADDZONE=SHAFT GVOLUME 6 5 105 106 i01 7 7 107 107 SYSTEM=l ADDZONE=SHAFT GVOLUME 1 6 106 7 7 107 107 SYSTEM=l ADDZONE=SHAFT GVOLUME 102 i01 201 202 107 107 207 207 EL2=2 SYSTEM=i ADDZONE=SHAFT GVOLUME 103 102 202 203 107 107 207 207 EL1=3 SYSTEM=l ADDZONE=SHAFT GVOLUME 104 103 203 204 107 107 207 207 SYSTEM=l ADDZONE=SHAFT GVOLUME 105 104 204 205 107 107 207 207 ELl=3 SYSTEM=1 ADDZONE=SHAFT GVOLUME 106 105 205 206 107 107 207 207 SYSTEM=i ADDZONE=SHAFT GVOLUME i01 106 206 201 107 107 207 207 SYSTEM=l ADDZONE=SHAFT
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| * Shaft convection material T-MATERIAL 5 CONVECTION H=200.
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| * Convection at shaft end surface T-BOUNDARIES SEGMENTS CON-MATERIAL = INPUT=SURFACE 201 202 207 207 / 202 203 207 207 / 203 204 207 207 204 205 207 207 / 205 206 207 207 / 206 201 207 207 T-LOADS ENVIRONMENT CONVECTION=YES INPUT=MATERIAL 5 45.
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| * Constrain the rotation of the shaft as a rigid body CONSTRAINT
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| * slave dir master dir factor 91 2 7 2 1.
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| 7 4 -140E-3
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| * Film pressure boundary conditions FIXBOUNDARIES DIRECTION=FILM-PRESSURE INPUT=LINE 1 i01 / i01 102 / 102 2 3 103 / 103 104 / 104 4 5 105 / 105 106 / 106 6
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| * Load and boundary conditions for shaft center node LOADS CONCENTRATED 7 3 -5CE3 FIXBOUNDARIES 156 / 7
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| * Use prescribed X-rotation to get reaction torque LOADS DISPLACEMENT 7 4 0.
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| * Symmetric boundary condition for the shaft at X=C FIXBOUNDARIES 1 INPUT=SURFACE 12 77 / 2 3 77 /3 4 7 7 4 5 77 / 5 6 77 /6 1 7 7 Version 99.0 B96.11
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
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| * Temperature and boundary conditions for the reservoir oil node FIXBOUNDARIES / 99 T-LOAD TEMPERATURE 99 45.0
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| * Define the oil inlet flow portion from reservoir T-FILMCONSTRAINT INPUT=LINE MINPUT=LINE FACTOR=RESERVOIRFLOW, RESERVOIRNODE=99
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| * snl sn2 mnl mn2 factor 2 102 3 103 0.3 4 104 5 105 0.3 6 106 1 101 0.3 T-INITIAL TEMP TREF=25 ZONE PADS INPUT=EGROUP / 4 SOLVIA SET PLOTORIENTATION=PORTRAIT NSYMBOL=MYNODES NNUMBER-MYNODES COLOR EFILL=ORANGE MESH ZONE=ELEMENTS VIEW=I BCODE=ALL MESH ZONE=ELEMENTS VIEW=-I BCODE=ALL SET PLOTORIENTATION=LANDSCAPE MESH ZONE=PADS VIEW=I MESH ZONE=SHAFT NNUMBER=MYNODES MESH ZONE=SPRING NNUMBER=YES NSYMBOL=YES COLOR EFILL=CON-MATERIAL MESH ZONE=ELEMENT VIEW=I SUBF=21 MESH ZONE=ELEMENT VIEW=-I MESH ZONE=SOLID HIDDEN=NO OUTLINE=YES NNUMBER=NO VIEW=I MESH ZONE=FILM CONTOUR=THICKNESS MESH ZONE=TCONSTRAINTS MESH ZONE=TFILMCONSTRAINTS END Version 99.0 B96.12
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B96 TILTING-PAD JOURNAL BEARING, ELASTIC PADS AND SHAFT SET MODE=BATCH PLOTORIENTATION=LANDSCAPE DIAGRAM=GRID DATABASE CREATE WRITE 'b96.1is' VIEW ABC 1 0 1 VIEW DEF 1 0 -1 EGROUP 1 EPLINE FILMI-SYM 1 9 6 3 TO 8 9 6 3 EPLINE FILMI-EDGE 17 7 4 1 TO 24 7 4 1 EGROUP 2 EPLINE FILM2-SYM 1 9 6 3 TO 8 9 6 3 EPLINE FILM2-EDGE 17 7 4 1 TO 24 7 4 1 EGROUP 3 EPLINE FILM3-SYM 1 9 6 3 TO 8 9 6 3 EPLINE FILM3-EDGE 17 7 4 1 TO 24 7 4 1 ZONE PADS INPUT=EGROUP 4
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| MESH VIEW=X VECTOR=DISP ORIGINAL=YES DEFORMED=NO CONTOUR VMIN=50.
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| MESH ELEMENT CONTOUR=TEMP VIEW=I VECTOR=REACTION MESH ELEMENT CONTOUR=TEMP VIEW=-I VECTOR=REACTION CONTOUR VMIN=AUTO MESH PADS CONTOUR=TEMP VIEW=I MESH SHAFT CONTOUR=TEMP MESH FILM CONTOUR=FILMTEMP VIEW=ABC MESH FILM CONTOUR=TEMP VIEW=ABC MESH FILM CONTOUR=TEMP VIEW=DEF MESH FILM CONTOUR=FILMPRESSURE VECTOR=FILMFLOW VIEW=ABC MESH FILM CONTOUR=FILMPOWERLOSS MESH FILM CONTOUR=FILMTHICKNESS PLINE=ALL MESH FILM CONTOUR=FILMREYNOLDS AXIS I VMIN=50 VMAX=90 LABEL='FILM-TEMP' AXIS 2 VMIN=0 VMAX=8E6 LABEL='FILM-PRESSURE' AXIS 3 VMIN=0 VMAX=216E-6 LABEL='FILM-THICKNESS' AXIS 11 VMIN=0 VMAX=0.18 LABEL='FILM1' AXIS 12 VMIN=0 VMAX=0.18 LABEL='FILM2' AXIS 13 VMIN=0 VMAX=0.18 LABEL='FILM3' ELINE FILMI-SYM KIND=FILMTEMPERATURE XAXIS=11 YAXIS=1 OUTPUT=ALL ELINE FILMT-EDGE KIND=FILMTEMPERATURE XAXIS=-11 YAXIS=-1 SUBFRAME=OLD SYMBOL=i ELINE FILM2-SYM KIND=FILMTEMPERATURE XAXIS=12 YAXIS=1 OUTPUT=ALL ELINE FILM2-EDGE KIND=FILMTEMPERATURE XAXIS=-12 YAXIS=-
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| SUBFRAME=OLD SYMBOL=1 Version 99.0 B96.13
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| tl'969 0-66 UOISJOA ONE moqai=GNIN 99a NOIIVKNns MOqdl=QNI'A VSH NOIIVKWnS Moaal=GNIN Eava NOIIVNWnS moqjl=GNIN ýGvd NOIIVWNnS MOqal=GNIY TGVd NOIIVKWnS xnqalwaIz=QNIm ý:Ds NOIJVKNnS Xn'IZIW'IIZ=GNIN ZSR NOIIVKNnS TOS NOIIVKWnS HIVS'dD asvavlval NOII3VH'd=GNIN DNIddS JSIqN DNI'ddS IS=
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| DNI'ddS ISIUN waij xvwa NIN=Hd7,1 K7IJ=I3ZqHS wqis XVKH XVW=3d7,1 W719=IDEUSS wqia NOIIZ)VsU=GNI)l NOIIVwxfls NOIIDVS'd=GNIY EGVd NOIIVNNDS NOII3V2'd=GNIH ZGVd NOIIVWwns NOIIDVSE=GNIý4 TC[Vd NOIIVKWfIS MOqJW'IIZ=GNIN ý:WUIZ Noijvwwns MOqlWaIZ=GNIN ZNUIJ NOIIVWNnS MOq3WrII3=GNIN TNqIg NOIIVKwns MOqJWrIIZ=(JNIN WaIZ Noilvillwns H'dnSSSýIdWrIIZ=CINIX WrIId NOIIVKNnS SSOrDIEMOdNrIIZ=GNIN N'IIq NOIIVKWnS SNOII3VHE=GNIM LN ISIIN LN ISFIN
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| ýf.
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| T='109NTS GqO=HNVElEnS E=SIXV.7 ET-=SIXVX SSSNADIHIN'IIZ=GNIN aDCH-ENqIa aNIqS
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| ý:=SIXVA ET=SIXVX SSHNHDIHINqId=GNI)3 W2ýS-MqIl RNIIE T=rlOgK-TS ClJO=ENVýIZEnS E=SIXV.7 ZT-=SIXVX SSSNNDIHIWqIa=GNIN HOGE-ZNqIJ HNII2 E=SIXV;ý ZT=SIXVX SSRNN3IHIW'IIE=GND3 W2ýS-ýWqId SNI712 T='lOgK7,S GUO=RKV8ZEnS E=SIXV,'ý TT-=SIXVX SSSNNZ)IHIWrIIZ=(ININ RDG2-TWqIa HNII2 E=SIXVT, TT=SIXVX SS2NYZ)IHIWqIZ=CNIN W-TS-TNqIJ aNIqZ T='IOgN.KS CIUO=2KVEZEJ2S ET-=SIXVX RUnSS2EdW'IIZ=CININ 9D(J2-EWqIJ UNYI2 Z=SIXVT ýT=SIXVX HdnSSHHdNIIa=CININ W2ýS-MaIZ HNIIS T=rlOgK7,S C1qO=HNV8ZEDS Z=SIXV,ý ZT-=SIXVX HdfISSEýldW7IIE=CININ RDCIR-ZNqIJ aNII2 Z:=SIXV-T ZT=SIXVX HýH'ISSSEdW'Ha=(ININ W7ýS-ZW=l 2NIq2 T=rIOgK-TS TIO=HNVýIZEnS Z=SIXV7ý TT-=SIXVX SdfISSSddWrIIZ=CININ R9C[2-TWqI3 HNIIS Z=SIXVJý TT=SIXVX SUnSS2ý1dWrII2=CNIN W2ýS-TWrIIJ 2NIrIE T=rIOgKIS GaO=HNVEZ9j2S T=SIXV)ý ýT-SIXVX 2ýInJVUSdW2lNrIIZ=CNIN aDGE-EN7Ia RNIq2 q'IV=JndJnO T=SIXV2ý ýýT=SIXVX SdnIVE2dNHlWrIIa=CININ K7,S-MaIa HNIrIS
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| (-Iuozo) Indut JLSO(I-VIA-IOS sojdwuxg.mouijuoN InnurIN UOTILIDUPOA VIA-IOS
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B97 TILTING-PAD THRUST BEARING, SPRING SUPPORTED Objective To verify the use of the FILM element in an axial thrust bearing with spring supported thermo-elastic pads and temperature dependent oil viscosity.
| |
| Physical Problem The thrust bearing selected to be analyzed in the steady-state condition is represented in the figure below by one of its 12 identical pads. The pads are supported by springs and circumferential motion is fixed at two support points for each pad. The external surfaces of the pads are cooled by the surrounding oil. The response is cyclicly symmetric so it is sufficient to consider one of the pads.
| |
| Geometry r, = 0.3125 m ojr I r(igid ru nn er
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| *h0 r2 = 0.475 m film ZA t/2 t = 0.060 m pad Ta Tsupport point -
| |
| I
| |
| 't/2 I X t = 0.050m elastic bed t,1 Initial film thickness ho = 120.10-6 m Pad material E=2.06.10" N/M2 v = 0.3 (x=12.10-6 1/°C TR, = 20 'C (giving G= 0) k = 45 W/ m°C Elastic bed 6 2 Ez =70_10 N/iM 2
| |
| E. =Ey =I_10-3 N/M 3
| |
| G=1.103 N/ M 3
| |
| Reservoir oil temperature TR = 450 C Oil density p= 880 kg/ M Pad oil temperature T, = 35°C k=0.145W/m°C 0 atl °C Oil conductivity Pad convection coefficient h = 500 W / m2 °C k=0.133W/rm°C at 150 'C Oil specific heat cv=1.71.10 6 j/m 3 OC Hot oil carry over factor = 0.95 Rotational speed (o, = -62.832 rad / s (600 rpm) The absolute viscosity of the oil as function of temperature is shown on page B97.2.
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| Vertical load on each pad F = 100 .10 N Version 99.0 1397.1i
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples
| |
| [ S91 T N - T.3-9 9R'0F'3G
| |
| !IS -1N a91 T T 3LG:,PTT,
| |
| --THUDSI StNI P -ED S"P I ,~
| |
| EýIT CC!
| |
| -C E3%I Finite Element Model The finite element model is built in the same way as in Examples B92, B93 and B96, where radial tilting pad bearings were modeled. The model here consists of 10 x 10 8-node FILM elements and 8 node SOLID elements for the pad and for the spring bed. The pad has a thermo-elastic material. The uniaxial spring bed is modeled with an orthotropic material. Rigid links are used for the runner which is assumed to be rigid and the runner surface is considered thermally isolated (adiabatic).
| |
| 95 % of the outlet oil flow at the boundary to the neighboring pad is considered to be carry over flow, is thus to be part of the inlet flow to the neighboring pad. The temperature of the oil that cools the pad flow is set to 45 set to 35 'C. The temperature of the reservoir oil that is mixing, with the carry over
| |
| 'C in order to model a somewhat higher oil temperature for the inlet flow from the reservoir.
| |
| film The model is shown on pages B97.4 and B97.5. The 4" and 5"' plot on page B97.5 shows the oil surface of the pad colored blue and the cooled surfaces colored yellow.
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| Solution Results The input data on pages B97.9 - B97.14 is used and the results are summarized in the table below.
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| Shaft vertical displacement, mm -2.2228 Shaft frictional torque, T, Nm -73.73 Frictional power, co,, -Ti, kW 4.633 Vertical pressure load on runner surface, N 1.00.105 3
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| Inlet oil flow, M /.S 1.491.10.
| |
| Power loss, kW 4.636 Verstion 99.0ults.
| |
| Maximum film thickness, ýtm 124 Minimum film thickness, atm 26 Maximum film pressure, MPa 11.7 Max inlet film temperature. line INLET, 'C 78 Min inlet film temperature, line INLET, 'C 66 Max outlet film temperature, line OUTLET, aC 105 Min outlet film temperature. line OUTLET, 'C 86 Version 99.0 B97.2
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| | |
| SOLVIA Verification Manual Nonlinear Examples Contour plots of the oil film pressure, temperature, thickness, power loss per unit area and in-plane heat flux are shown on page B97.6.
| |
| Contour plots of the pad displacements and the pad temperature are shown in the bottom plots of page B97.6 and the top plots of page B97.7. Vector plots of the pad oil pressure load and the pad reactions are also shown on page B97.7.
| |
| The variations of the film thickness and the pressure along the circumferential line MID are shown in the bottom part of page B97.7. Inlet and outlet line variations of oil film temperature, thickness and volume oil flux are shown on page B97.8.
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| The temperature solution plots for the pad include heat fluxes (vector and contour) and oil film in plane heat fluxes in the circumferential r-direction and in the axial t-direction. The top t-direction oil film flux is zero since the runner surface is modeled as isolated.
| |
| The calculated pressure and oil film thickness agree well with measured values at 9 points in the oil film. The temperature of the pad surface neighboring the film agrees well with the measured temperatures towards the outlet but the measured temperatures are cooler towards the inlet. The difference is believed to be due to lamination of the hot oil carry over flow and the cooler reservoir oil flow before thermal mixing is accomplished, see the user hints below.
| |
| User Hints
| |
| "* The vertical displacement of the spring supported pad is very large compared to the oil film thickness. The artificial springs in the FILM element make it still possible to compute the load response in one load step.
| |
| 3
| |
| " The artificial stiffness of the FILM element at the pressure 1 MPa is set to 5-1011 N / M , which is 3
| |
| considerably larger than the spring bed modulus 70.106 /0.05 = 1.4. 10 N / M . The value of 3
| |
| DKDP=-2 -104 gives an artificial stiffness of 3.2 .101 N / m at the pressure of 10 MPa. The larger stiffness at 1 MPa provides initial stability in the iterations of the oil film response. The lower stiffness at 10 MPa provides a good convergence rate when the iterated film pressure is close to the solution and the out-of balance forces are small
| |
| " The calculated temperature of the hot oil carry over flow at the inlet is about 95 'C, which is considerably higher than the temperature of the reservoir oil surrounding the pad. It seems reasonable to assume that the thermal mixing of the hot oil carry over flow and the cooler reservoir oil flow is not instantaneous but is gradually achieved along the pad surface. This would imply that the pad surface temperature at the inlet is cooler than calculated due to effects of the thermal lamination.
| |
| "* This case of assumed effective cooling of the pad external surface is modeled by using the 2
| |
| convection coefficient h = 500 W / m °C. The heat flow from the oil film to the pad is then calculated to be 0.85 kW which can be compared with the total calculated power loss of 4.6 kW.
| |
| " The calculated thermal and elastic deformations of the pad are significant as can be seen from the contour plot of the oil film thickness and the thickness variation along the lines MID, INLET and OUTLET. The temperature gradient through the pad causes the pad surface to be of convex shape which affects the solution.
| |
| Version 99.0 B97.3
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples bed x
| |
| B97 TILTING-PAD THRUST BEAP:NG SPRING SUPPORTED ORIGINAL z x -, y 22 32
| |
| !.42 OLVIA-PRE 99.0 SOLVIA ENG-NI RiNG AB Version 99.0 B97.4
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B97 TILTING-PAD THRUST BEARING, SPRING SUPPORTED' DATABASE CREATE MASTER TEMPERATURE=COUPLED TOLERANCES TYPE=F RTOL=lE-3 RNORM=1.E5 ITEMAX=100 T-ITERATION TTOL=10.
| |
| SET EPS=1.E-6 PARAMETERS $TRES=45 $TPAD=35 $FLOW=0.95 PARAMETERS $KNORM=5E11 $PNORM=IE6 $DKDP=-2E4 PARAMETER $MAT=2 $TINIT=70 $H=500 PRINTOUT MIN SYSTEM 1 CYLINDRICAL THETA=-90.0 COORDINATES ENTRIES NODE R THETA XL 1 0.475 -10.4 0.06012 2 0.475 13.5 0.06012 3 0.3125 13.5 0.06012 4 0.3125 -10.4 0.06012 5 0. 0. 0.06012 11 0.475 -10.4 0.060 12 0.475 13.5 0.060 13 0.3125 13.5 0.060 14 0.3125 -10.4 0.060 21 0.475 -10.4 0.030 22 0.475 14.6 0.030 23 0.3125 14.6 0.030 24 0.3125 -10.4 0.030 25 0.475 -14.6 0.030 26 0.3125 -14.6 0.030 27 0.475 0.0 0.030 28 0.3125 0.0 0.030 31 0.475 -10.4 0.0 32 0.475 14.6 0.0 33 0.3125 14.6 0.0 34 0.3125 -10.4 0.0 35 0.475 -14.6 0.0 36 0.3125 -14.6 0.0 41 0.475 -10.4 -0.050 42 0.475 14.6 -0.050 43 0.3125 14.6 -0.050 44 0.3125 -10.4 -0.050 45 0.475 -14.6 -0.050 46 0.3125 -14.6 -0.050 51 0.4 20.0 0.062
| |
| * Pad material MATERIAL 1 THERMO-ELASTIC TREF=20
| |
| -1000 2.06EII 0.3 12.E-6 5000 2.06E11 0.3 12.E-6 T-MATERIAL 1 TEMP-CONDUCTION
| |
| -1000 45 3.588E6 5000 45 3.588E6
| |
| * Spring bed material MATERIAL 3 ORTHOTROPIC EA=IE3 EB=1E3 EC=70E6 GAB=IE3 GAC=1E3 GBC=1E3 Version 99.0 1397.9
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
| |
| T-MATERIAL 3 TEMP-CONDUCTION
| |
| -1000 1. I.E6 5000 1. l.E6 T-MATERIAL 4 CONVECTION H=$H
| |
| * Oil material READ B97ISOVG68.DAT
| |
| * Oil film EGROUP 1 FILM MATERIAL=$MAT SET SYSTEM=l NODES=8 GVOLUME 1 2 3 4 11 12 13 14 EL1=10 EL2-10 EL3=1 FILMVELOCITY INPUT=SURFACE OMEGA=62.831853, AX=- AY=C AZ=- BX=- BY=C BZ=-1.
| |
| KNORM=$KNORM PNORM=$PNORM DKDP=$DKDP 1 2 3 4 LINE CYLINDRICAL 22 27 EL=6 LINE CYLINDRICAL 27 21 EL=4 LINE COMBINE 22 21 27 LINE CYLINDRICAL 23 28 EL=6 LINE CYLINDRICAL 28 24 EL=4 LINE COMBINE 23 24 28
| |
| * Pad EGROUP 2 SOLID MATERIAL=l GVOLUME 11 12 13 14 21 22 23 24 EL2=10 EL3=2 GVOLUME 21 22 23 24 31 32 33 34 EL3=2 GVOLUME 21 24 26 25 31 34 36 35 EL2=2
| |
| * Spring bed EGROUP 3 SOLID MATERIAL=3 GVOLUME 31 32 33 34 41 42 43 44 GVOLUME 35 31 34 36 45 41 44 46 AXES-ORTHOTROPIC 1 NODI=31 NODJ=32 NODK=33 T-INITIAL TEMPERATURE TREF=$TINIT
| |
| * Reservoir oil temperature at node 51 T-LOADS TEMPERATURE 51 $TRES
| |
| * Constrained inlet film temperature T-FILMCONSTRAINT INPUT=LINE M-INPUT=LINE RESERVOIRNODE=51
| |
| .FACTOR=CARRYOVERFLOW
| |
| * snl sn2 mnl mn2 factor 2 3 1 4 SFLOW
| |
| * Spring bed boundaries FIXBOUNDARIES DIR=123 INPUT=SURFACE 41 42 43 44 45 41 44 46
| |
| * Boundaries for pad pivot nodes FIXBOUNDARIES DIR=i / 28 27 FIXBOUNDARIES DIR=2 / 28
| |
| * Film element pressure boundaries FIXBOUNDARIES DIR=FILMPRESSURE INPUT=LINE 12 / 23 / 34 / 41 Version 99.0 B97.10
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
| |
| * Cooling of external pad surfaces T-BOUNDARIES SEGMENTS CON-MATERIAL=4 INPUT=SURFACE 31 32 33 34 35 31 34 36 25 35 36 26 21 25 26 24 11 21 24 14 22 12 13 23 32 22 23 33 14 13 23 24 24 23 33 34 26 24 34 36 12 11 21 22 22 21 31 32 21 25 35 31 T-LOADS ENVIRONMENT CONVECTION=YES INPUT=SURFACE 31 32 33 34 $TPAD 35 31 34 36 $TPAD 25 35 36 26 $TPAD 21 25 26 24 $TPAD 11 21 24 14 $TPAD 22 12 13 23 $TPAD 32 22 23 33 STPAD 14 13 23 24 $TPAD 24 23 33 34 $TPAD 26 24 34 36 $TPAD 12 11 21 22 $TPAD 22 21 31 32 STPAD 21 25 35 31 $TPAD
| |
| * Rotor load LOADS CONCENTRATED
| |
| * node dir fac 5 3 -100.E3 RIGIDLINK INPUT=SURFACE
| |
| * nl n2 n3 n4 master 1 2 3 4 5 FIXBOUNDARIES 12456 INPUT=NODE / 5 SOLVIA SET NNUMBER=MYNODES NSYMBOL=MYNODES COLOR EFILL=GROUP MESH MESH EGi CONTOUR=THICKNESS MESH EG3 BCODE=ALL COLOR EFILL=CON-MATERIAL MESH EG2 VIEW=I BCODE=ALL MESH EG2 VIEW=-I BCODE=ALL MESH RIGIDLINK BCODE=ALL VIEW=I COLOR LINE MESH FILM VIEW=Z EAXES=RST MESH FILM ENUMBER=YES MESH FILM NNUMBER=Y NSYMBOL=YES END Version 99.0 B97.11
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE read file B971SOVG68.DAT
| |
| "* ISOVG68 OIL
| |
| "* Discretization from cSt-curve assuming constant density=880 Kg/m3
| |
| "* Extrapolation above 110 degrees C MATERIAL $MAT OIL DENSITY=880 PRESSURE-FACTOR=2.7E-8
| |
| * temperature viscosity pressure
| |
| : 10. 0.4000
| |
| : 15. 0.2700
| |
| : 20. 0.1900
| |
| : 25. 0.1400
| |
| : 30. 0.1000
| |
| : 35. 0.0790
| |
| : 40. 0.0610
| |
| : 45. 0.0460
| |
| : 50. 0.0380
| |
| : 55. 0.0310
| |
| : 60. 0.0260
| |
| : 65. 0.0210
| |
| : 70. 0.0180
| |
| : 75. 0.0150
| |
| : 80. 0.0130
| |
| : 85. 0.0115
| |
| : 90. 0.0100
| |
| : 95. 0.0088 100. 0.0077 105. 0 .0070 110. 0 .0063 120. 0.0051 130. 0.0044 140. 0.0037 150. 0.0032 END DATA T-MATERIAL $MAT TEMP-CONDUCTION
| |
| * temperature conductivity specific heat
| |
| : 10. 0.145 1.71E6 150. 0.133 1.71E6 END DATA Version 99.0 B97.12
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
| |
| * B97 TILTING-PAD THRUST BEARING, SPRING SUPPORTED DATABASE CREATE WRITE 'b97.1is' COLOR EFILL GROUP SET ZONE=FILM OUTLINE=YES MESH CONTOUR=FILMPRESSURE VECTOR=FILMFLOW MESH CONTOUR=FILMTEMPERATURE MESH CONTOUR=FILMTHICKNESS MESH CONTOUR=FILMPOWERLOSS MESH CONTOUR=FILMFLUX VECTOR=FILMFLUX SET.ZONE=SOLID MESH CONTOUR=DISP VIEW=I SET ZONE=EG2 MESH CONTOUR=DX MESH CONTOUR=DY MESH CONTOUR=DZ MESH CONTOUR=TEMPERATURE SET ZONE=SOLID MESH VECTOR=LOAD MESH VECTOR=REACTION VIEW=-I SET ZONE=EGi EPLINE MID 50 2 4 TO 41 2 4 MESH PLINES=MID VIEW=Z ELINE MID KIND=FILMTHICKNESS SYMBOL=1 ELINE MID KIND=FILMPRESSURE SYMBOL=1 EPLINE INLET 100 1 2 STEP 10 TO 10 1 2 EPLINE OUTLET 91 3 4 STEP 10 TO 1 3 4 DELETE EPLINE MID MESH PLINES=ALL AXIS 1 VMIN=48 VMAX=110 LABEL='FILMTEMPERATURE' AXIS 2 VMIN=0 VMAX=0.18 LABEL='INLET (symbols) and OUTLET' 3 VMIN=0 VMAX=2E-3 LABEL='FILM FLUX IN CIRCUMFERENTIAL DIRECTION' AXIS AXIS 4 VMIN=0 VMAX=130E-6 LABEL='FILM THICKNESS' ELINE INLET KIND=FILMTEMPERATURE YAXIS=* XAXIS=2 SYMBOL=1 ELINE OUTLET KIND=FILMTEMPERATURE YAXIS=1 XAXIS=2 SUBFRAME=OLD ELINE INLET KIND=FILMTHICK YAXIST 4 XAXIS=2 SYMBOL=X ELINE OUTLET KIND=FILMTHICK YAXIS=4 XAXIS*2 SUBFRAME=OLD ELINE INLET KIND=FILMFLUXR YAXIS=3 XAXIS=2 SYMBOL=U ELINE OUTLET KIND=FILMFLUXR YAXIS=3 XAXIS=2 SUBFRAME=OLD Version 99.0 B97.13
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| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input (cont.)
| |
| SUMMATION ZONENAME=FILM KIND=) FILMPOWERLOSS SUMMATION ZONENAME=FILM KIND= FILMPRESSURE ZONE INLET INPUT=NODES 2 110 TO 118 3 ZONE OUTLET INPUT=NODES 1 128 TO 136 4 SUMMATION ZONENAME=FILM KIN]D=FILMFLOW SUMMATION ZONENAME=INLET KIN]D=FILMFLOW DETAILS=YES DETAILS=YES SUMMATION ZONENAME=OUTLET KIN]D=FILMFLOW ZONENAME=SOLID KIN]D=LOAD SUMMATION SUMMATION ZONENAME=SOLID KINID=REACTION NLIST N5 NLIST N5 KIND=REACTION EMAX FILM EMAX FILM SELECT=FILM EMAX FILM SELECT=FILM TYPE=MIN TOLERANCES INTERMEDIATE=YES T-DATABASE CREATE MESH EG2 CONTOUR=TFLUX VECTOR=TFLUX VIEW=I MESH EG2 CONTOUR=TFLUX VECTOR=TFLUX VIEW=-I SET ZONE=FILM VIEW=Z MESH CONTOUR=FILMTRFLUX MESH CONTOUR=FILMTSFLUX MESH CONTOUR=FILMTTOPFLUX MESH CONTOUR=FILMTBOTFLUX SUMMATION FILM KIND=FILMTFLUX ELIST E1+1 SELECT=BASICS END Version 99.0 B97.14
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| | |
| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B98 SECTOR-SHAPED THRUST BEARING, ISOVISCOUS OIL AND RIGID PAD Objective To verify the pressure, oil flow and power loss calculations by the FILM element when modeling a sector-shaped thrust bearing using fixed bearing surfaces and a constant oil viscosity.
| |
| Physical Problem A thrust bearing from ref. [1I is selected to be analyzed, see the figure and data below. The pad and the runner are assumed to be rigid. The sector angle is 5/6 radians since the total number of pads is 6 and they together extend 5 radians. One of the pads is considered in this analysis.
| |
| Radii rj = 0.070 m, r2 = 0.130 m Tu 5 T1 Anles0 2 B =- + - radians 12 2 12 2 Absolute viscosity p. = 0.049 Pa s Rotation speed co = 157.0796 rad / s (1500 rpm)
| |
| Film thickness h(r,0)= h +h0k r sin(B 0 - 0) r/rmsin(oB - 0A) where k = 3.0 h 0 = 50-10-6 m 1r r 1 +r 2 2
| |
| Finite Element Model The described finite element model consists of 16-node film elements in a 20 x 20 mesh. The circular edge lines are defined with RATIO=5 which gives a finer mesh towards the outlet at 0 = OB. Rigid links are used to connect the nodes of the two surfaces to the corresponding center nodes 10 and 11 so that forces and reactions at the center nodes will be calculated. The bottom (pad) film surface is tilted with respect to its radial line 0 = B using the commands MSURFACE and LINE PROJECTION so 0
| |
| that the film thickness at node 9 is 200 4tm. The variation of the film thickness is then as stated in the expression above, which was used in ref. [1]. The top figures on page B98.3 show the film thickness distribution and the orientations of the element axes.
| |
| Solution Results The total pressure load in the Z-direction for the considered part of thrust bearing is given by 0a, rý3 F, J fprdrdO =Po h 2
| |
| ° OAr, Version 99.0 B98.1
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples where b = r2 - r1 and P0 = 0.02732 is the dimensionless pressure from reference [I].
| |
| The inlet (and outlet ) oil flow for the considered part is Q = q 0 brO)h0 where q 0 = 1.719 is the dimensionless oil flow from reference [1].
| |
| The power loss for the considered part is E = Eo b ho r3 where E0 = 0.4448 is the dimensionless power loss from reference [11].
| |
| The resultant force F, due to pressure is calculated in reference [1] to act at the point with the dimensionless coordinates x0 = 0.2579 and y0 = 0.9631. A transformation is made to the X-Y system shown in the figure above and these point coordinates are here denoted Xp and Yp.
| |
| The input data used in the SOLVIA analysis for 16 node FILM elements in the 20 x 20 mesh can be seen on page B98.5 and B98.6. Further analyses were made using 10 x 10 and 40 x 40 meshes with the 16 node FILM element and also a 20 x 20 mesh of 8 node FILM elements.
| |
| The resultant pressure load, oil flow and power loss are directly obtained using the SUMMATION command. The coordinates of the point where the resultant pressure load acts have been calculated from static equilibrium using the moment at the center node listed by command SUMMATION.
| |
| The SOLVIA results as well as the results from reference [I] are listed in the table below.
| |
| SOLVIA FILM elements Units Reference 16 node 8 node
| |
| [1] 40x40 20x20 1OxlO 20x20 Pressure load F. N 5047 5083.2 5083.1 5082.0 5050.3 Oil flow Q 10- 4m 3 / S 0.8101 0.80757 0.80727 0.80608 0.80158 Power loss E W 645.3 645.95 645.95 645.93 645.09 Coordinate Xp m -0.01539 -0.015466 -0.015465 -0.015459 -0.015386 Coordinate YP m 0.09851 0.098507 0.098507 0.098506 0.098483 A good agreement with reference [1] can be observed. The distribution of pressure, power loss and radial (r) and tangential (s) oil flux are shown in contour plots on pages B98.3 and B98.4. The variations of pressure and power loss per unit area are also shown along a radial and a circumferential line on page B98.4.
| |
| Reference
| |
| [1] Floberg, L., On the Optimum Design of Sector-Shaped Tilting-Pad Thrust Bearings, Acta Polytechnica Scandinavica, Mechanical Engineering Series No 45, Stockholm 1969 Version 99.0 B98.2
| |
| | |
| C/)
| |
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| |
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| CD 0n 1
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| |
| 'C-44 / 0 0 0 0 0 0 0 000000 0 0 (000 o0 [0
| |
| | |
| 0 C) td
| |
| ýo FILI
| |
| | |
| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B98 SECTOR-SHAPED THRUST BEARING, ISOVISCOUS OIL AND RIGID PAD' DATABASE CREATE PARAMETER $Ri=0.070 $R2= 0.130 $RM=0.100 $FIA=66.1268 PARAMETER $FIB=I13.8732 $HO=-50E-6 $Hi=-200E-6 SET EPS=i.E-6 SYSTEM 1 CYLINDRICAL AZ=I. BX=1.
| |
| COORDINATES / ENTRIES NODE R THETA XL 1 $R1 SFIB $HO / 2 $R1 $FIA $HO 3 $R2 $FIA SHO / 4 $R2 $FIB $HO 5 $RI $FIB 0. / 6 $R1 $FIA 0.
| |
| 7 $R2 $FIA 0. / 8 $R2 $FIB 0.
| |
| 9 $RM $FIA $Hi / 10 0. 0. 0.
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| 11 0. 0. SHO SET MIDNODES=I LINE CYLINDRICAL N1=i N2=2 EL=20 RATIO=5 LINE CYLINDRICAL N1=5 N2=6 EL=20 RATIO=5 LINE CYLINDRICAL N1=4 N2=3 EL=20 RATIO=5 LINE CYLINDRICAL Ni=8 N2 =7 EL=20 RATIO=5 MSURFACE 1 PLANE NI=i N2=4 N3-9 LINE PROJECTION NI=1 N2=2 TARGETSURFACE=i ZDIR=I.
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| LINE PROJECTION NI=3 N2=4 TARGETSURFACE=i ZDIR=I.
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| MATERIAL 1 OIL DENSITY=900.
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| -1000. 0.049 1000. 0.049 SET NODES=16 EGROUP 1 FILM MATERIAL=1 RSINT=2 GVOLUME 8 5 6 7 4 1 2 3 ELI=20 EL2=20 EL3=1 FILMVELOCITY INPUT=SURFACE OMEGA=-57.0796 BZ=I.0 8567 INITIAL TEMPERATURE TREF=60.
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| LOADS TEMPERATURE TREF=60.
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| RIGIDLINKS INPUT=SURFACE M-INPUT=NODE 4 1 2 3 1i 8 5 6 7 10 LOADS DISPLACEMENTS i0 i 0. /10 2 0. /10 3 0. /i0 4 0. /i0 5 0. /10 6 0.
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| ii 0. /11 2 0. /11 3 0. /i1 4 0. /i1 5 0. / ii6 0.
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| FIXBOUNDARIES DIRECTION=FILM-PRESSURE INPUT=LINE 12 / 23 / 34 / 41 SOLVIA SET PLOTORIENTATION=PORTRAIT NNUMBERS=MY NSYMBOLS=MY VIEW=Z MESH CONTOUR=THICKNESS MESH FILM EAXES=RST END Version 99,0 B98.5
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B98 SECTORSHAPED THRUST BEARING, ISOVISCOUS OIL AND RIGID PAD DATABASE CREATE WRITE 'b98.lis' SET DIAGRAM=GRID OUTLINE=YES VIEW=Z ZONE=FILM MESH CONTOUR=FILMPRESSURE MESH CONTOUR=FILMRFLUX VECTOR=FILMFLOW MESH CONTOUR=FILMSFLUX VECTOR=FILMFLOW MESH CONTOUR=FILMPOWERLOSS EPLINE CIRCLE 9 2 1 STEP 20 TO 389 2 1 EPLINE RADIUS 260 1 3 TO 241 1 3 MESH PLINE=ALL ELINE RADIUS KIND=FILMPRESSURE ELINE RADIUS KIND=FILMPOWERLOSS ELINE CIRCLE KIND=FILMPRESSURE ELINE CIRCLE KIND=FILMPOWERLOSS SUMMATION KIND=FILMPRESSURE SUMMATION KIND=FILMFLOW SUMMATION KIND=FILMPOWERLOSS ZONE UPPER GLOBAL ZMIN=-20E-6 ZONE LOWER GLOBAL ZMAX=-20E-6 SUMMATION UPPER KIND=LOAD SUMMATION UPPER KIND=REACTION SUMMATION LOWER KIND=LOAD SUMMATION LOWER KIND=REACTION END Version 99.0 1398.6
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| SOLVIA Verification Manual Nonlinear Examples EXAMPLE B99 SIMPLY SLPPORTED CONCRETE SLAB, JAIN AND KENNEDY Objective To verify the SHELL element in an analysis with elastic-plastic rebars and with the concrete material model.
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| Physical Problem The experimental study by Jain [1 I includes twenty-nine reinforced concrete slabs. In this example the slab denoted El will be analyzed. The experimental study of these slabs are also reported by Jain and Kennedy in [2].
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| Dimension of the plate a, = 910 mm 2 b1 "- pb, a1 ==610 mm 1010mm b 2 = 710mm T=38mm laye-r 1-- - -
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| _ _ Rebar layer 1 a1 , = 4.8 mm a2 cc 1= 51 mm t, = 12 mm TCOORD , = -2t =-0.6316 T
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| Rebar layer 2 0 2 = 4.8 mm tA cc 2 = 42 mm T t, =7mm TCOORD --2t
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| --- 2 -0.3684 Direction of the rebars P=0c Steel Concrete E=200-10 9 N/rM E 0 =29-109 N/I M2 F =-34MPa c7 = 220 MPa v=0 . = -32 MPa ET = 0 K = 20 d, = 2.75 MPa SHEFAC=0.5 'E = -0.002 SHELL3D=YES su = -0.003 Version 99.0 B99.1
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| SOLVIA Verification Manual Nonlinear Examples Finite Element Model The finite element model is shown on page B99.3. The 4-node SHELL element is used. Due to symmetry when P3 = 0' only one quarter of the plate needs to be considered. The simply supported boundary conditions are modeled by nonlinear spring elements to allow comer uplifting of the plate.
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| The spring elements have only stiffness in compression. The AUTO-STEP method and BFGS iteration method are used.
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| Solution Results The input data to the SOLVIA analysis can be found on pages B99.6 - B99.8. The calculated load deflection curve is shown on page B99.3 together with the experimental results. The experimental curves are drawn without symbols. A good agreement can be observed.
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| Contour plots of the results at the last load step are shown on pages B99.3 - B99.5. The principal moments, maximum and minimum, are shown on page B99.3. Contour plots of the maximum principal stress at the bottom and top surfaces are shown on pages B99.3 - B99.4. The concrete crack state at the bottom, mid and top surfaces are shown on page B99.4. The contour plots of stress, strain and accumulated plastic strain in the rebars are shown on pages B99.4 and B99.5. The rebar results are displayed without nodal averaging.
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| History plots of the stress in the rebars of the center element 144, point 1, show that rebar layer 2 reaches the plastic state before rebar layer 1.
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| In case B99A the only input data change compared to B99 is that the material parameter SHELL3D=NO. The calculated load - deflection curve for case B99A can be seen on page B99.5 to be very close to the curve for case B99.
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| In case B99B the only input data change compared to B99 is that the material parameter cut-off tensile stress is reduced from 2.75 MPa to 1.5 MPa and the calculated load - deflection curve is shown on page B99.5. The ultimate load when dt = 1.5 MPa is close to the ultimate load when a, = 2.75 MPa but larger deflections occur during the loading phase.
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| Reference
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| [1] Jain, S. C., "Ultimate Strength and Behavior in Reinforced Concrete Slabs," thesis presented to the University of Windsor, at Windsor, Canada, in 1971, in partial fulfillment of the requirements for the degree of Doctor of Philosophy.
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| [2] Jain, S. C. and Kennedy, J.B., "Yield Criterion for Reinforced Concrete Slabs", Journal of the Structural Division, ASCE, Vol. 100, No. ST3, March 1974.
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| User hint When SHELL3D=YES is specified for the concrete material of SHELL elements with midsurface nodes then no strain occurs in the thickness direction except for possible thermal strain. This is equivalent to rigid stirrups so direct stresses 7, in the thickness direction of the concrete can develop. The capability to accommodate inclined principal compressive stresses in the concrete due to out-of-plane shear improves. In this particular example with v = 0 and mainly plate bending the influence of the value of the SHELL3D parameter on the load - deflection curve can be seen to be small.
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| Version 99.0 B99.2
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| UN C) A:
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| ON
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| ,-,UI Cr j3 i f-I-I 0
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| C)
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| C) 0 C) 0 C*
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| C) K, Ii 2 j C)
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| V) K 2 A6 2<
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| '9 ri -;-
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| A A 0
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| -t
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| -K 22 2 2 2 2 2 2 r W7 A '2 NC N
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| N N'N N N.
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| N N NA N, N N C N N N NN N cC7 N' NN N\
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| N N' N \
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| N N 3/4 ,'
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| 'N N' NC,/
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| N N N4 N 2 N
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| CIO '<'C V 0 cn
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| ON 0
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| 0 z
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| 1 II)
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| ON ON 0
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| rJ)
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING 'B99 SIMPLY SUPPORTED CONCRETE SLAB, JAIN AND KENNEDY DATABASE CREATE MASTER NSTEP=100 AUTO-STEP DTMAX=1 TMAX=20 ITELOW=5 ITEHIGH=i5 DISPMAX=0.010 TOLERANCES TYPE=F RNORM=1000 RMNORM=1000 RTOL=0.1 ITEMAX=50 SET MYNODES=200 TIMEFUNCTION 1 0 0 / 20 200000 COORDINATES ENTRIES NODE X Y Z 1 0.455 0.305 0. TO 13 0. 0.305 0. TO 25 0. 0. 0. TO 37 0.455 0. 0. TO 49 0.455 0.305 0.
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| 51 0. 0.355 0.
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| 52 0.455 0.355 0.
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| 53 0.505 0.355 0.
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| 54 0.505 0.305 0.
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| 55 0.505 0. 0.
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| 101 0.455 0.305 -0.020 TO 113 0. 0.305 -0.020 137 0.455 0. -0.020 TO 149 0.455 0.305 -0.020 DELETE 49 149 1
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| REBAR-LAYER 1 D=0.0048 CC=0.051 TCOOR=-0.6316 PSI=0.
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| REBAR-LAYER 2 D-0.0048 CC-0.042 TCOOR=-0-3684 PSI=90.
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| MATERIAL 1 CONCRETE EQ=29.E9 NU=0.0 SIGMAT=2.75E6, SIGMAC=-34.E6 EPSC=-0.002 SIGMAU=-32.E6 EPSU=-0.003, BETA=0.75 KAPPA=20 STIFAC=0.0001 SHEFAC=0.5 SHELL3D=YES MATERIAL 2 PLASTIC E=200.E9 YIELD=220.E6 ET=0.
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| EGROUP 1 SHELL STRESSREFERENCE=ELEMENT TINT=6 MATERIAL=1 REBARMATERIAL=2 THICKNESS 1 0.038 GSURFACE 1 13 25 37 EL1=12 EL2=12 NODES=4 GSURFACE 52 51 13 1 EL1=12 EL2=2 NODES=4 GSURFACE 53 52 1 54 EL1=2 EL2=2 NODES=4 GSURFACE 54 1 37 55 ELi=2 EL2=12 NODES=4 LAYER-SECTION 1 TREF=0. 1 2 LOADS ELEMENT INPUT=SURFACE 1 13 25 37 t 1.0 EDATA ENTRIES EL NTH BETA 1 1 0. TO 196 1 0.
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| Version 99.0 B99.6
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input (cont.)
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| FIXBOUNDARIES 156 INPUT=LINE / 25 13 / 13 51 FIXBOUNDARIES 246 INPUT=LINE / 25 37 / 37 55 FIXBOUNDARIES / 101 TO 148 EGROUP 2 SPRING DIRECTION=AXIALTRANSLATION ENODES 1 1 101 TO 13 13 113 14 37 137 TO 25 48 148 PROPERTYSET 1 K=TABLE DISP-FORCE / 1 1 / IE-20 1 / IE9 PROPERTYSET 2 K=TABLE DISP-FORCE / 1 3 / 1E-20 3 / 2E9 EDATA ENTRIES EL PROPERTYSET 13 1 / 14 1 1 2 TO 12 2 15 2 TO 25 2 MESH SHELL NNUMBERS=MYNODES NSYMBOL=MYNODES BCODE=ALL MESH SHELL CONTOUR=PRESSURE VECTOR=LOAD MESH SHELL COLOR EFILL=GROUP EGROUP 1 MESH E144 ENUMBER=YES GSCALE=OLD SUBFRAME=OLD MESH SPRING NNUMBERS=MYNODES NSYMBOL=MYNODES BCODE=ALL SOLVIA END Version 99.0 B99.7
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| SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input
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| * B99 Simply supported concrete slab, Jain and Kennedy DATABASE CREATE WRITE 'b99.lis' TOLERANCES SYSTEM 1 CARTESIAN PHI=180 AXIS 1 VMIN=0. VMAX=0.02 LABEL=-center deflection (m)'
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| AXIS 2 VMIN=0. VMAX=l50000 LABEL='pressure load (Pa)'
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| SET DIAG:GRID NXYPLOT XNODE=25 XDIRECTION=3 YKIND=TIME YDIR=l OUTPUT=ALL, XAXIS=1 YAXIS=2 SYSTEM=1 SYMBOL=1
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| * read test results USERCURVE 1 / READ 'B99B0.DAT' PLOT USERCUVE 1 XAXIS=- YAXIS=2 SUBFRAME=OLD SET ZONE=EGI MESH CONTOUR=MPMAX MESH CONTOUR=MPMIN SHELLSURFACE BOT MESH CONTOUR=SPMAX SHELLSURFACE TOP MESH CONTOUR=SPMIN SHELLSURFACE BOT MESH CONTOUR=CONCRETE VECTOR=CRACK SHELLSURFACE MID MESH CONTOUR=CONCRETE SHELLSURFACE TOP MESH CONTOUR=CONCRETE VECTOR=CRACK MESH WHOLE CONTOUR=DZ VECTOR=REACTIONS VIEW=-I SHELLSURFACE REBAR=1 SET EAXES=REBAR CONTOUR AVERAG=NO MESH CONTOUR=SAA VIEW=I MESH CONTOUR=EAA SHELLSURFACE REBAR=2 MESH CONTOUR=SAA MESH CONTOUR=EAA EHIST E=144 P=* KIND=SAA REBAR=1 EHIST E=144 P=1 KIND=SAA REBAR=2 SHELLSURFACE REBAR=1 MESH CONTOUR=EPACC EAXES=REBAR SHELLSURFACE REBAR=2 MESH CONTOUR=EPACC EAXES=REBAR END Version 99.0 B99.8
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| SOLVIA Verification Manual Nonlinear Examples INDEX Arch Hertz plane strain B57 circular B45 rubber B82 deep B86 3-D B59, B62 shallow B21 Coupled displacement-temperature analysis, TRUSS B40 see FILM element Beam, see also Cantilever Creep building frame B22 beam B5, B39 compressive load effect on frequencies Creep law no. 1 (Bailey-Norton) B63 B71 Creep law no. 2 (primary and secondary) curved B75, B76, B77 B64 reinforced concrete B19, B79, B80, B81 cylinder B 15 simply supported B47, B83 PLANE STRESS2 element B5, B39 temperature gradient B77 thick-walled cylinder B 15 Bearings, see FILM element Cyclic loading Birth/death option B32, B37 creep B15 Buckling isotropic hardening B8 imperfections B4, B44 kinematic hardening B9 large displacement B3, B87, B88 Cylinder linearized B43, B45, B46, B50, B87, B88 compound B90 snap-back B41, B86 concrete B35 snap-through B40, B41, B42, B86 panel B3 thermal B50 thick-walled B13, B14, B15, B35, B49 Buoyancy B84 Damping Building, impact load B22 Rayleigh B83 Cable B1O,B51,B58 Dam seepage B67, B68 failure modeling B33 Deformation dependent loading B48, B84 Cantilever Excavation B32 creep B5, B39 FILM element curved B75 journal bearing B92, B93, B94, B96 deformation-dependent load B48 thrust bearing B97, B98 end moment B52 tilting pad B92, B93, B96, B97 plastic B26, B27, B34 slider bearing B95 PLATE bending B25, B27 Sommerfeld B94 SHELL bending B24, B26 rigid pads B92 thermal stress B38 elastic pads B93, B96, B97 torsion B53 elastic rotor B97 Column B46 Frame Complex-harmonic analysis B83 buckling B87, B88 Concrete building B22, B33 beam B19, B79, B80, B81, B91 plastic collapse B72, B73 cylinder B35 Frequency analysis B7, B71 material curves B20 Freezing, melting B30, B31, B69 slab B91, B99 Friction B58, B59, B82 Contact Gap, TRUSS element B6, B7 dynamic B6, B7, B59, B60, B61, B62 Geologic material friction B58, B59, B82 curve-description B18 Hertz axisymmetric B56 Drucker-Prager B23 Version 99.0 1.1
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| SOLVIA Verification Manual Nonlinear Examples Hardening Rubber isotropic B8, B54 bushing B74 kinematic B9, B55 o-ring B82 Impact load B22, B61 pure shear B85 Imperfections B4, B44 sheet BIl Initial strain B10, B51, B76 Seepage flow Latent heat, see Freezing, melting free surface B67, B68 Linearized buckling, see Buckling one-dimensional B66 Mode superposition analysis B7, B47 Shell bending Mooney-Rivlin material, see Rubber PLATE element B25, B27 Nonlinear-elastic material B33, B80 SHELL element B24, B26 Oil film bearing, see FILM element Soil specimen B23 O-ring B82 Solidification B30, B31 Orthotropic cylinder B90 Space shuttle temperatures B29 Pendulum B2 Spherical shell B12, B36, B42, B65 Phase transition, thermal B30, B31, B69 Spurious modes B4 Pipe whip Stiffened plate, see Plate direct time integration B6 Substructures B22 mode superposition B7 Temperature analysis Plastic analysis stationary B38 cantilever bending B26, B27, B34 transient B28, B29, B30, B31, B69 collapse, limit load B72, B73 variable time-step B28 multi-linear model B54, B55 Temperature boundary conditions spherical cap B36 convection B28 thick-walled cylinder B 13, B 14, B49 heat flow B29 torsion B53 radiation B28, B29 Plate Temperature gradient load buckling B4, B43, B44, B50 beam B77 large displacements B 16 plate B78 stiffened B43, B44 Thermal stress B14, B17, B38, B50, B90 temperature gradient B78 Tilting pad bearing, see FILM element Pre-load Torsion B53 axial compressive in beam B71, B83 TRUSS structures B1, B40, B41 Prestressing, see Initial strain Tunnel B18, B32 Ramberg-Osgood material B70 ULJ (Updated-Lagrangian-Jaumann)
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| Reinforced concrete, see Concrete analysis B48 Removal, addition of elements B32, B37 Unloading/reloading B 13 Residual stresses B14, B34 User-supplied material B70 Restart Water pressure B84 dynamic B47, B71, B83 Wave propagation B61 static B21, B39 Version 99.0 1.2
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| 10 State Street, Wohwrn, M'1assachusetts 01801 Stevenson & Associates Tel 617 932 9580 Fax 617 933 4428 A structural-mechanicalconsulting engineeringfirm BOSTON-CLL 7E-AND July 2, 2002 American Electric Power Nuclear Generation Group ..
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| 500 Circle Drive Buchanan, MI 49107
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| | |
| ==Subject:==
| |
| License for SOLVIA Computer Code Let this letter serve as a license for American Electric Power Company to use SOLVIA copyrighted verification documentation 1 for the purposes of regulatory compliance in answering of questions as posed by the USNRC. This grants American Electric Power the right have a copy of this documentation and to send a copy of the same to the USNRC for their use in the review of the SOLVIA code.
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| The SOLVIA verification documents transmitted herewith each bear a SOLVIA copyright notice. The NRC is permitted to make the number of copies of the information contained in these reports which are necessary for its internal use in connection with generic and plant specific reviews and approvals as well as the issuance, denial, amendment, transfer, renewal, modification, suspension, revocation or violation of a license, permit, order, or regulation, copyright protection notwithstanding. The NRC is permitted to make the number of copies beyond those necessary for its internal use which are necessary in order to have one copy available for public viewing in the appropriate docket files in the public document room in Washington, DC and in local public document rooms as may be required by NRC regulations if the number of copies submitted is insufficient for this purpose. Copies made by the NRC must include the copyright notice in all instances.
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| Stevenson & Associates hereby grants this license as US distributor of the SOLVIA code.
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| Very truly yoLrs, W lterDj rdjevic P re '- en a) SOLVIA Finite Element Program, Version 99.0, SOLVIA Verification Manual Linear Examples, Report SE 99-4, SOLVIA Engineering AB, 2000.
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| b) SOLVIA Finite Element Program, Version 99.0, SOLVIA Verification Manual Nonlinear Examples, Report SE 99-5, SOLVIA Engineering AB, 2000.
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| | |
| ATTACHMENT 7 TO AEP:NRC:2520 COMMITMENTS The following table identifies those actions committed to by Indiana Michigan Power Company (I&M) in this document. Any other actions discussed in this submittal represent intended or planned actions by I&M. They are described to the Nuclear Regulatory Commission (NRC) for the NRC's information and are not regulatory commitments.
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| Commitment Date I&M will revise the ice condenser endwall structural calculation and August 23, 2002 provide the NRC the results of the revised calculation with respect to required rebar yield strength.}}
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