ML022030446

Attachment to Response to NRC Request for Additional Information Re Containment Structure Conformance to Design Basis Requirements. Solvia Verification Manual, Pages B30.1 - B67.6

ML022030446
Person / Time
Site:	Cook
Issue date:	07/16/2002
From:	Greenlee S Indiana Michigan Power Co
To:	Document Control Desk, Office of Nuclear Reactor Regulation
References
AEP:NRC:2520, TAC MB3603, TAC MB3604
Download: ML022030446 (192)

Text

SOLVIA Verification Manual EXAMPLE B30 SOLIDIFICATION OF A SEMI-INFINITE SLAB OF LIQUID Objective To verify the TRUSS conduction element in SOLVIA-TEMP including phase changes.

Physical Problem A uniform infinite slab of liquid is considered to be initially at zero temperature as shown in the figure below. Suddenly, the temperature of the surface of the liquid is reduced to -450F and main tained constant. The freezing front position and the temperature in the slab is to be predicted.

y ins)

MATERIAL PROPERTIES k = 1.08 Btu/in sec 'F c = 1.0 Btu/in3 'F L = 70.26 Btu/in 3

Freezing point 0

X (ins)

Of -- 0.1 '17 Finite Element Model The finite element model consists of 32 equally spaced TRUSS elements. The conductivity, specific heat and density of the liquid and solid phases are assumed equal and constant. The material prop erties are shown in the figure above. A lumped heat capacity matrix must be employed. The Euler backward method of time integration with a time step of 0.05 sec is used in the finite element analy sis. Equilibrium iterations are employed at every time step and the tolerance is 1.0. 10.

Solution Results The input data shown on page B30.4 has been used in the analysis. The calculated front positions at four solution times are shown as circles in the figure on page B30.2 together with the analytical solution from [1], page 285, drawn as a solid line. The calculated distributions of the latent heat at the four solution times are shown in the left top figure on page B30.3.

The calculated temperature history of the innermost node 33 and of node 9 (1 inch from surface) are shown in the right top figure of page B30.3. The analytical solution [1] of the node 9 temperature history is drawn as a solid line for comparison.

The calculated solution agrees well with the analytical solution.

Version 99.0 Nonlinear Examples B30.1

SOLVIA Verification Manual Nonlinear Examples User Hints In a general problem where heat is absorbed or liberated in connection with phase changes, the specific heat may vary significantly over the phase change temperature interval, see bottom figure on page B30.3 for an example. It may then be convenient to idealize the specific heat curve according to the dashed line of the figure and to model the area between the peak and the dashed line as latent heat.

Reference

[1]

Carslaw, H.S., and Jaeger, J.C., Conduction of Heat in Solids, Clarendon Press, Oxford, 1959.

0 SOLVIA solution

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0r 0I 0.0 0.5

.0 i.S 2.0 Time (sec)

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0 2.5 3.0 3.5 4.C B30.2 1

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B30 SOLIDIFICATION OF A SEMI-INFINITE SLAB OF LIQUID' DATABASE CREATE T-MASTER NSTEP=80 DT=0.05 T-ANALYSIS TRANSIENT HEATMATRIX=LUMPED METHOD=BACKWARD-EULER T-ITERATION TTOL=1.E-6 METHOD=MODIFIED-NEWTON ITEMAX=i00 COORDINATES 1

TO 33

4.

T-MATERIAL 1

CONDUCTION K=1.08 SPECIFICHEAT=1.

LHi=70.26 T-MATERIAL 2

CONVECTION H=1.E6 EGROUP 1

TRUSS ENODES

/

1 1 2 TO 32 32 33 EDATA

/

1

1.

T-PHASETRANSITIONS

-0.1 0.0 0 T-BOUNDARIES SEGMENTS CON-MATERIAL-2 INPUT=NODES 111 T-LOADS ENVIRONMENT CONVECTION=YES INPUT-NODES 1 -45.

SOLVIA-TEMP END SOLVIA-POST input B30 SOLIDIFICATION OF A SEMI-INFINITE SLAB OF LIQUID T-DATABASE CREATE WRITE FILENAME='b30.1is' NPLINE NAME=TRUSS

/

1 TO 33 SET PLOTORIENTATION-PORTRAIT SUBFRAME 22 NLINE LINENAME=TRUSS KIND=LHl TIME=1 SYMBOL=i OUTPUT=ALL NLINE LINENAME=TRUSS KIND=LH1 TIME=2 SYMBOL=i OUTPUT-ALL NLINE LINENAME=TRUSS KIND=LHi TIME=3 SYMBOL=i OUTPUT-ALL NLINE LINENAME=TRUSS KIND=LH1 TIME=4 SYMBOL=i OUTPUT=ALL AXIS ID=I VMIN=-26.0 VMAX=0 LABEL='NODE 9 TEMPERATURE' USERCURVE 1 /

READ

'B30N9.DAT' SUBFRAME 12 NHISTORY NODE=33 OUTPUT=ALL NHISTORY NODE=9 OUTPUT=ALL YAXIS=I SYMBOL=i PLOT USERCURVE 1 YAXIS=-i SUBFRAME=OLD END Version 99.0 Nonlinear Examples B30.4

SOLVIA Verification Manual EXAMPLE B31 SOLIDIFICATION OF A CORNER REGION Objective To verify the PLANE conduction element in SOLVIA-TEMP when employed in the analysis of transient heat conduction and phase changes.

Physical Problem The corner of a uniform infinitely long container carrying a liquid with initial temperature Oi and freezing temperature O, is considered as shown in figure below. At time t=0+, the temperature of the container surface 0 ' is reduced to a temperature lower than Of and is maintained constant. The solution for the temperature and the position of the phase transition region is to be determined.

Initial temperature 6i = 0.3 'C Fluid Freezing temperature Of = 0.0

• C Freezing temperature increment AOf = 0.0 'C s

Temperature of container surface O0 =1 -

C Conductivity k = 1.0 W/m °C Specific heat c 1.0 J/m 3 °C "Latent heat L = 1.0 J/m 3 Container

.=

L

- 0.25 fc(@r-@s)

Finite Element Model The finite element model considered is shown in the figure on page B31.2. A 20x20 mesh of 4-node PLANE conduction elements are used to model the corner of the liquid with side length of two meters. A lumped heat capacity assumption must be used in the model since phase change effects are included. The Euler backward method of time integration with a time step of 0.020 sec. is employed in the step-by-step solution. Equilibrium iteration is performed at each time step using the modified Newton-Raphson method and the tolerance 1.0-10-8 is used.

Solution Results The input data on pages B31.4 and B31.5 is used in the finite element analysis. Contour plots of the temperature distribution at time 0.1 and 0.2 are shown in the left top figure on page B31.3. The element flux at time 0.1 and 0.2 is shown in vector plots in the right top figure on page B3 1.3.

Contour diagrams of the element heat flux, the boundary nodal heat flow and the deviation of element heat flux are shown in the bottom diagrams of page B31.3. Note that the latent heat effect introduces a discontinuity in the heat flux between elements at the zone of freezing.

The distribution of temperature and latent heat along the diagonal is shown in the top diagrams of page B31.4.

Version 99.0 Nonlinear Examples B3 1.1

SOLVIA Verification Manual The solution agrees well with the theoretical solution presented in [ 1 ]. This example has also been analyzed in [21 and [3].

User Hints The 4-node element is chosen in this case because a lumped heat capacity matrix must be used when phase changes are analyzed. A higher order element would give a wavy solution due to the used lumping of nodal heats.

References

[1]

Budhia, H., and Kreith, F., "Heat Transfer with Melting or Freezing in a Wedge", Int. J. Heat Mass Trans., 16, pp. 195-211, 1973.

[2]

Comini, G., Del Guidice, S., Lewis, R.W., and Zienkiewicz, O.C., "Finite Element Solution of Non-Linear Heat Conduction Problems with Special Reference to Phase Change", Int. J. Num.

Meth, Eng., Vol. 8, pp. 613-624, 1974.

[3]

Morgan, K., Lewis, R.W., and Zienkiewicz, O.C., "An Improved Algorithm for Heat Conduction Problems with Phase Change", Int. J. Num. Meth. Eng., Vol. 12, pp. 1191-1195, 1978.

ORIGINAL 0.2 831 SOLIDIFICATICN OF A CORNER REGION I

ORIGINAL 0.2 LK, SOLVIA-PRE 99 Q Version 99.0 z L, SOLVIA *NGANEURANG AB Nonlinear Examples B31.2

SOLVIA Verification Manual Nonlinear Examples 331 SOLIOIF -AT:

N OF A CORNER REGION ORIGINAL O.S TIME 0 i

A 331 SCLIDIFICATION OF A CORNER REGION ORIGINAL 0.5 TIME 0-'

ORIGINAL 0.5 TIME 0.2 831 SOLIDIFICAT0ON OF A CORNER RE3ION OLV'A-POST 99.0 50LVIA EN(

z L, HEAT FLOW 0,2j103 HEAT FLUX MAX 2.3104 S2.6347 S2.2831, 1

0.932 S!. 808 229S A

7825 0

o5269S 0.17566 MIN 8.3135E-6 z L_

DEVIATION OF HEAT FLUX MAX 0 75630 I 0.70903 061449 J 0.6!199 1

0.12 42

0. 33088 0.23634

o. t4181

0. 047269 "TIN 0

1NEtR-NG A8 831 SOLIDIFiCA7iON OF A CORNER REGION 2

HEAT FLOW C0. ! 9 8-3 HEAT FLUX MAX 1.9089 7896 I 3124 3.0738 0.59655 0.35793 MIN 3.3088E-6

-y OR 1GINAL TIME 0,2 SOLVIA-ROST 99.3 DSI/AT.ON OF HEAT FLUX MAX 0.8588 0.54027 0.47603 C 40280 0.32956

0. 25632 z

L, HEAT FLUX 2.8[04 z

ORIGINAL 0

TiME 3

0 Version 99.0

3. 336618 M1.N 3 SOLVIA ENGI.,EERING. AR B31.3

SOLVIA Verification Manual Nonlinear Examples B31 SOLDIF:CATION OF A CORNER REGION 71r'E C I 0o o0s IA 1

a 3

2 A.

DIAGONAL 'IDE I -

21 I E0, 2

6.

0. 9

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.CRNER REGION

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q

-0 3

i 3

1 2

3.0 DIAGONAL NODE D

2:

SOLVIA-POAT 99.0 SOLVIA ENGINEERING AS SOLVIA-PRE input HEADING

'B31 SOLIDIFICATION OF A CORNER REGION' DATABASE CREATE T-MASTER NSTEP-10 DT=0.02 T-ANALYSIS TRANSIENT HEATMATRIX=LUMPED METHOD=BACKWARD-EULER T-ITERATION TTOL=1.E-8 METHOD=MODIFIED-NEWTON ITEMAX=-00 COORDINATES

/

ENTRIES NODE Y

Z 1

TO 21 2.0 2.0

/

22 2.0

/

23

0.

2.0 T-INITIAL TEMPERATURE TREF=0.3 T-MATERIAL 1

CONDUCTION K=1.

SPECIFICHEAT=1.

LHL=0.25 T-MATERIAL 2

CONVECTION H=1.E6 EGROUP 1

PLANE STRAIN GSURFACE 1 22 21 23 EL1=20 EL2=20 NODES=4 T-PHASETRANSITIONS

0.

0.

T-BOUNDARIES SEGMENTS CON-MATERIAL=2 INPUT=LINES 1 22

/

1 23 T-LOADS ENVIROMENT CONVECTION=YES INPUT=LINES 1 22 -1. -1.

/

1 23 -1. -1.

Version 99.0 7

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2" 3 ;; "T...........

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B31.4

Nonlinear Examples SOLVIA Verification Manual SOLVIA-PRE input (cont.)

SET NSYMBOLSýMYNODES HEIGHT=0.2 MESH NNUMBERS=MYNODES SUBFRAME=21 SET HEIGHT=0.15 MESH ENUMBERS=YES SOLVIA-TEMP END SOLVIA-POST input B31 SOLIDIFICATION OF A CORNER REGION T-DATABASE CREATE WRITE FILENAME='b31.1iS' SET MESH MESH PLOTORIENTATION=PORTRAIT OUTLINE=YES CONTOUR=TEMPERATURE TIME=0.1 SUBFRAME=12 CONTOUR=TEMPERATURE TIME=0.2 MESH VECTOR=TFLUX TIME=0.1 SUBFRAME=12 MESH VECTOR=TFLUX TIME=0.2 MESH CONTOUR=TFLUX VECTOR=TFLOW TIME=0.1 SUBFRAME=12 MESH CONTOUR=FDEVIATION TIME=0.1 MESH CONTOUR=TFLUX VECTOR=TFLOW TIME=0.2 SUBFRAME=12 MESH CONTOUR=FDEVIATION TIME=0.2 1

TO 21 NPLINE NAME=DIAGONAL SUBFRAME 12 NLINE LINENAME=DIAGONAL NLINE LINENAME=DIAGONAL KIND=TEMPERATURE TIME=0.1 OUTPUT=ALL KIND=TEMPERATURE TIME=0.2 OUTPUTýALL SUBFRAME 12 NLINE LINENAME=

NLINE LINENAME=

END KIND=LHI TIME=0.1 OUTPUT=ALL KIND=LH1 TIME=0.2 OUTPUT=ALL Version 99.0 B31.5

SOLVIA Verification Manual EXAMPLE B32 EXCAVATION OF A TUNNEL Objective To demonstrate the element birth/death option when using the PLANE STRAIN and TRUSS elements in analyzing the construction of a tunnel.

Physical Problem The construction of a tunnel shown in the figure below requires the creation of the tunnel opening by removal of material and the erection of props at the center line. The displacements and stresses during different phases of the construction are to be determined.

a = 20.0 ft 60 tt below ground Level r = 24.0 ft Rock/Soil:

Prop a

E =4.32.10 4 ksf V 0.0 p =1 20 lb/ft3 r

Prop:

Opening E= 4.32.10 6 ksf S2r v=0.0 Finite Element Model The finite element model considered is shown in the top figure on page B32.4. The model is very coarse and is intended for demonstration of element birth/death analysis procedures only. Eight 4-node PLANE STRAIN elements are used to model one-half of the underground region. The element death option is active for the element modeling the tunnel region, and the element birth option is active for the TRUSS element modeling the prop. The following step-by-step solution is performed to simulate the different phases of construction:

Step 1:

All elements except the TRUSS element are active and a uniform loading of q = 80 kips/ft2 is applied at the top surface.

Step 2:

The element modeling the tunnel region becomes inactive to simulate the creation of the opening. The TRUSS element remains inactive and the applied loading is the same as in step 1.

Step 3:

The TRUSS element becomes active to simulate the addition of the prop. The applied loading is the same as in step I.

The applied loading is increased to 120kips / ft2 and the response of the Step 4:

completed tunnel is obtained.

Version 99.0 Nonlinear Examples B32.1

SOLVIA Verification Manual Solution Results The input data on pages B32.6 and B32.7 is used for the finite element solution. The deformed con fliurations of the underground region predicted by the finite element model for the four load steps are shown in the bottom figures on page B32.4. The vector plots of the principal stresses and the contour plots of the principal stresses in the plane strain elements for the four load steps are given on page B32.5. The distribution of stresses during the different phases of construction can be observed.

User Hints

" No comparison of these finite element results with experimental or other computed results are given, but the solution reported here satisfies some qualities that should be observed; namely, a redistribution of stresses due to the excavation of the tunnel (step 2). no change in stresses due to the addition of the prop only (step 3), and the tensile stress developed in the roof due to the additional loading is relatively small, when the prop is active (step 4).

" Note that the birth/death conditions at the solution time t+At determine whether an element is active or inactive also during the formation of a new stiffness matrix at the beginning of the time step (time t). However, all other nonlinearities affecting the reformed stiffness matrix are based on the conditions at time t.

"* For the evaluation of the nodal equivalent forces and the final stresses, the conditions at the solution time t+At always govern, also for the elements with birth/death option.

If TBIRTH and TDEATH denote the time of birth and death, respectively, of an element, the following table shows whether the element is active in the stiffness matrix reformed at time t and in the final stress calculation at time t+At:

Element birth option:

Element death option:

Element active if t + At Ž TBIRTH Element inactive if t + At< TBIRTH Element active if t + At_* TDEATH Element inactive if t + At > TDEATH Element birth-then-death option: (This option is possible but is not used in this example)

Element active if t + At Ž TBIRTH and t + At

• TDEATH Element inactive if t + At < TBIRTH or t + At > TDEATH Note that the table can also be represented pictorially as shown below:

Element birth:

TBIRTH in the shaded range corresponds to the element included in the stiffness matrix at time t and in the force vector at time t+At; the element is stress free at time t.

time t+At First solution time the element is active.

Version 99.0 t

"t Nonlinear Examples B32.2

SOLVIA Verification Manual TDEATH in the shaded range corresponds to the element Element death:

not included in the stiffness matrix at time t and in the force vector at time t+At.

-~time t

t+At First solution time the element is inactive.

The above shows that for the options used it makes no difference what value of TBIRTHITDEATH is specified as long as this value lies in the shaded regions.

Hence, consider that in the birth option the element is to be active at time t+At, then we can input for TB IRTH the value t+(At/1000) and this means:

The element stiffness matrix is included in the calculation of the stiffness matrix, if reformed at time t, and the force vector at time t+At.

The element is stress free at time t, i.e. the stresses accumulated in the element are only due to the displacements from time t onwards.

Instead of the value t+(At/1000) for TBIRTH we could employ any value for TBIRTH satis fying t+(At!1000) _< TBIRTH < t+At without any change in the solution results.

Also, consider that in the death option the element is to be inactive at time t+At, then we can input for TDEATH the value of (t+At) - At/1000 and this means that the element stiffness matrix and force vectors are not included for the solution at time t+At. Here, instead of the value (t+At) - At/1000 for TDEATH we could use any value satisfying the condition t < TDEATH < (t+At) -At/1000 without any change in the solution results.

"Note that the only nonlinearity in this solution is due to the removal/addition of the elements representing the tunnel and the prop. Hence, since in this example a new stiffness matrix is calculated at each step corresponding to the conditions at the end of each such step, this matrix does not change during the solution step and a linear system is, therefore, considered in each step.

Considering this tunnel analysis, convergence in the equilibrium iteration is, therefore, immediate in all the steps.

" Note that the TRUSS element representing the prop has as its initial length the distance between the element end nodes in the deformed configuration at time 2.0. Hence, the element is born stress free. This holds for all elements that are born in a solution.

" Further examples of the use of the birth and death options are given in Example B37.

Version 99.0 Nonlinear Examples B32.3

SOLVIA Verification Manual Nonlinear Examples ORGINA:I-L 11 TI'E B32 EXCAVAT>ON OF A TUNNEL Z

ORIGINAL o

10.

PRESSURE 80 SOLVIA-PRE 99.0 MAS5TER I100111 C

1 101!

SOLVIA ENGINE-RTNG AB ORIGINAL 10 MlAX DISPL.

0.2061 TIME 4 SOLVIA-POST 99n SOLVIA ENGINEERING AB Version 99.0 332 EXCAVATION OF A TUNNEL ORIGNAL

• - 10.

MAX DISP.

0,1037 TiME I B32.4

SOLVIA Verification Manual Nonlinear Examples B32 EXCAVATION OF A TUNNE, ORIGINAL 10 TIME I

ZONE EG2-3 SPR9NCIPAL 80

-80 ORIGINAL 10 Z

TIME I

ZONE EG2-3 PRINC:PAL STRESS IN MAX-8 0.30030 979,996913~

-79.998677

-79.999559

-80.000441

-80.301323

-80

.002205

.- 80.033087 MIN-80.300003 SOLVOA-POST 99.0 SOLVIA ENGINEERING AB 932 EXCAVATION OF A TUNNEL ORIGINAL -

10.

TIM.E 2 ZONE EG2 ORIG.NAL 10 TIME 2 ZONE 6G2 SOLVIA-POST 99.0 SOLVIA SPRINCIPAL 161 5S Z

Ly PRINCIPAL STRESS 1AX MAX 93. 326

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72 "63 069 50 966 38.864 26.761 14 668 S2.55cS S

MIN-3. 4959 ENGINEERING AB 832 EXCAVATION CF A TUNNEL ORIGINAL 1

!0.

70ME 4 ZONE EGI-2 ORIG:NAL -

SO.

ZTNE 3 ZONE CE 1-2 SOLVIA-OST 99 3

SOLVIA SPRI.NC ýýL 161.0 Z 1-y PRINCIPAL STRESS MAX TAX 93 326 S87.274 S75.!72 63.069 S.966 38.864 2R 761 iii 14,658 2.5555

'IN-495 0B ENGNEER.NG NA ORIGINAL 10 7ME 4

ZONE EGI-2 SOLIIA-POST 99.0 SOLVIA EN B32.5 Z

-7 215 2NC 3AL 215 23

-210 23 Z

L~y PRINCIPAL STRESS MAX.

MAX 47. 53 4'

0521 38.436 32.351 26.266

20. 181 S8.0ll!

1.9262 TOIN-!.163 GINEERING AR Version 99,0

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B32 EXCAVATION OF A TUNNEL' DATABASE CREATE MASTER IDOF=100111 NSTEP=4 TIMEFUNCTION 1

0.

0.

/

1. 80.

/

3.

80.

/

4.

120.

COORDINATES

/

ENTRIES NODE Y

Z 1

0.

56.

/

2

42.

56.

/

3

60.

56.

4

0.

44.

/

5

42.

44.

/

6

60.

44.

7

0.

30.

/

8

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30.

/

9

60.

36.

10

0.

12.

/

11

36.

12.

/

12

60.

12.

13

0.

0.

/

14

36.

0.

/

15

60.

0.

MATERIAL 1

ELASTIC E=4.32E4 NU=0.

MATERIAL 2

ELASTIC E-4.32E6 NU=0.

EGROUP 1

TRUSS MATERIAL=2 ENODES

/

1 9 12 EDATA

/

ENTRIES EL AREA TBIRTH

/

1 1.5 2.9 EGROUP 2

PLANE STRAIN MATERIAL-i ENODES

/

1 2

1 4

5 TO 4

11 10 13 14 5

3 2

5 6

/

6 6

5 8

9 7 12 11 14 15 LOADS ELEMENT 1 S 1. 1. /

5 S 1. 1.

EGROUP 3

PLANE STRAIN MATERIAL=i ENODES

/

1 9 8 11 12 EDATA

/

ENTRIES EL TDEATH

/

1 1.1 FIXBOUNDARIES 3

/

13 TO 15 FIXBOUNDARIES 2

/

1 STEP 3 TO 13

/

3 STEP 3 TO 15 SET VIEW=X NSYMBOLS=YES MESH NNUMBERS=YES VECTOR=LOAD SUBFRAME=21 MESH ENUMBERS=GROUP BCODE=ALL SOLVIA END Version 99.0 Nonlinear Examples B32.6

SOLVIA Verification Manual SOLVIA-POST input B32 EXCAVATION OF A TUNNEL DATABASE CREATE WRITE FILENAME='b32.1is' TOLERANCES SET PLOTORIENTATION=PORTRAIT SET ORIGINAL:DASHED VIEW=X MESH DMAX=0.5 TIME=i SUBFRAME=12 MESH GSCALE=OLD TIME=2 MESH GSCALE=OLD TIME=3 SUBFRAME=12 MESH GSCALE=OLD TIME=4 ZONE NAME=EG2-3 INPUT=EGROUP

/

2 3 ZONE NAME=EGl-2 INPUT-EGROUP

/

1 2 SET ORIGINAL=YES DEFOPMED=NO OUTLINE=YES MESH EG2-3 VECTOR=SPRINCIPAL TIME-i SUBFRAME-12 MESH EG2-3 CONTOUR=SPMIN TIME:1 MESH EG2 VECTOR=SPRINCIPAL TIME=2 SUBFRAME-12 MESH EG2 CONTOUR=SPMAX TIME=2 MESH EGi-2 VECTOR=SPRINCIPAL TIME=3 SUBFRAME=12 MESH EGI-2 CONTOUR=SPMAX TIME=3 MESH EGl-2 VECTOR=SPRINCIPAL TIME-4 SUBFRAME-12 MESH EG1-2 CONTOUR=SPMAX TIME:4 SET ZONENAME=WHOLE NLIST TSTART=i TEND=4 ELIST TSTART=I TEND-4 END Version 99.0 Nonlinear Examples 1332.7

SOLVIA Verification Manual EXAMPLE B33 CABLE/FRAME STRUCTURE WITH CABLE FAILURE Objective To verify the nonlinear elastic TRUSS element when used to simulate cable failure in a cable/frame structure.

Physical Problem The cable/frame structure shown in the figure below is subjected to a concentrated downward load at the midspan. The cable is assumed to break at the joint when a strain of 3% is reached. The behaviour of the structure before and after the failure of the cable is to be analyzed.

Y b

a = 20 ft b = 12 ft Sa i

a

'I Frame Cable E=4.32.10 6 kips/ft2 E=4.32.10 6 kips/ft 2 v= 0.28 I = 2.778-10-2 ft4 A =0.3333 ft2 A = 6.944.10-3 ft 2 E 4 -32 x 106 kips/ft 2

E- 0.

0.03 Finite Element Model The finite element model used for the analysis is shown in the left figure on page B33.2. The frame is modeled with linear elastic BEAM elements and the cable is modeled using a nonlinear 2-node TRUSS element with the stress-strain relationship shown above. Note that the stress-strain curve is slightly modified in the input data so that the transition to no load in the cable is reasonably smooth.

The automatic iteration method is employed in order to be able to trace the solution during the unloading phase following the cable failure.

Solution Results The input data on page B33.3 and B33.4 is used in the finite element analysis. The time history of the vertical displacement at midspan and the structural configuration at the end of the solution are shown in the right figure on page B33.2.

The history of the axial force in the left horizontal beam member near the load application is shown on page B33.3 together with the traced stress-strain curve for the cable.

Version 99.0 E

Nonlinear Examples B3 3.1

SOLVIA Verification Manual Nonlinear Examples User Hints

"* The nonlinear elastic material can be used to model material failures as shown in this example. Of course, general stress/strain reversals are then not possible. The tension failure strain may only be reached from below.

It is important with rather tight tolerances in this kind of problem with a sudden unloading due to material failure.

"* A conventional iteration method (BFGS or full Newton) may be used if the unloading portion of the solution is of no interest. Larger load steps may then be taken so that the solution jumps over the unloading phase.

B33 CABLE/FRAME STRUCTURE WTTH CABLE FAILURE ORIGINAL A

I X

OýRCE Y

L 36C 0~

B33 CABLE/FRAME STRUCTURE WITH CABLE FAILURE ORIGINAL S

MAX DISPL.

. 0316 TIME 40 REACTION 119 39 FORCE R

.40 AVERAGING MAX 12. 0 85,714 N

28.571

-28 571 1-42B N-20R.20 MIN-98ý253 Reo- 

- ---p---.

- ---.-

"1 SOLVIA-PO 211 N00 00 300

!000 2C:

3,OA 'ULT-IL ER GAMBDA ST 99.0 SOLVIA ENGINEERING AB Version 99.0 Y

-t 1333.2 4*

SOLVIA Verification Manual 333 CABLE/FRA3E STRUCTURE WZTc CABLE ýAILURE 10' 0

200 o00 00 00oo o000 1200 LOAD MULTIPLIE' LAMBDA 0

... D' a &0.3 43 ff,'-...

-E-4) -0 C...' _a-&. 0 & 19 0

.C 3.02 00' 0.06 0.08 0.20 ES 2 E 1 P 1 STRA0N-RR SCLVIA-POST 99.0 SOLVIA ENGINEERING AB SOLVIA-PRE input HEADING

'B33 CABLE/FRAME STRUCTURE WITH CABLE FAILURE' DATABASE CREATE MASTER IDOF=001110 NSTEP=100 KINEMATICS DISPLACEMENTS=LARGE AUTOMATIC-ITERATION NODE=3 DIR=2 DISPLACEMENT=-0.01, DISPMAX=1.0 CONT=YES TOLERANCES ETOL=I.E-8 COORDINATE 1 /

2 MATERIAL MATERIAL

-1.

0.0 0.028 0.0285 0.0288 0.0291 0.0292 0.0294 0.0296 0.0298 0.0299 0.03

1.

Version 99.0 S

2(0.

12.

3 20.

4 40.

1 ELASTIC E-4.32E6 NU=0.28 2

NONLINEAR-ELASTIC

-4.32E6 0.0 129556.8 129556.8 129000.

120000.

110000.

90000.

60000.

20000.

2000.

0.

0.

B33.3 Nonlinear Examples

.7 Lo

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

EGROUP 1

BEAM MATERIAL=i RESULTS=FORCES SECTION 1

GENERAL RINERTIA=1.

SINERTIA=1.

TINERTIA=0.02778, AREA=0.3333 GLINE 1 3 AUX=2 EL:3 GLINE 3 4 AUX=2 EL=3 GLINE 2 1 AUX=3 EL=3 GLINE 4 2 AUX=3 EL=3

• ENODES

/

1 2 1 3

/

2 2

3 4 3

3 2 1

/

4 3 4 2 EGROUP 2

TRUSS MATERIAL=2 ENODES

/

1 2 3 EDATA

/

1 6.944E-3 FIXBOUNDARIES 12/

FIXBOUNDARIES 2

/

4 LOADS CONCENTRATED 3 2 -1.

SET VIEW=Z PLOTORIENTATION=PORTRAIT NSYMBOLS=YES MESH NNUMBERS=MYNODES VECTOR=LOAD SUBFRAME-12 MESH ENUMBERS=GROUP BCODE=ALL SOLVIA END SOLVIA-POST input B33 CABLE/FRAME STRUCTURE WITH CABLE FAILURE DATABASE CREATE WRITE FILENAME='b33.1is' TOLERANCES SET PLOTORIENTATION=PORTRAIT VIEW=Z SUBFRAME 12 CONTOUR AVERAGING=NO VMAX=200 VMIN=-200 MESH ORIGINAL=DASHED CONTOUR=FR VECTOR=REACTION NHISTORY NODE=3 DIRECTION=2 SYMBOL-i OUTPUT=ALL SUBFRAME 12 EHISTORY EL=3 POINT-2 KIND=FR SYMBOL=1 OUTPUT=ALL EGROUP 2 EXYPLOT EL=1 POINT=1 XKIND=ERR YKIND-SRR SYMBOL=i OUTPUT=ALL END Version 99.0 Nonlinear Examples 1333.4

SOLVIA Verification Manual EXAMPLE B34 ELASTIC-PLASTIC CANTILEVER PLATE UNDER END MOMENT Objective To verify the PLANE STRESS element when employed in an elastic-plastic analysis.

Physical Problem The cantilever plate shown in the figure below is subjected to a bending moment at its free end. The bending moment is increased until most of the plate cross-section is plastic and then unloaded. The residual stresses in the plate are to be determined.

M

[106 Nmm]

4*1 4 E = 1.7-105 N/mm 2 v = 0. 325 Gy = 465 N/mm 2 ET = 1.7.10' N/mm 2 h= 40mm L=100mm 1

2 3

Time Finite Element Model The finite element model considered is shown in the figure on page B34.2. The cantilever plate is modeled using four 9-node PLANE STRESS elements which are subjected to a moment. Constraint equations are used to transfer the moment into the plane elements. The material of the plate is assumed to obey an elastic-plastic isotropic hardening material law. Twelve solution steps are employed to calculate the finite element response and the variation of the applied loading is shown in the figure above. In the analysis, stiffness reformation and equilibrium iteration using the Full Newton method are employed at each step of the solution. The finite element matrix is calculated using 6 x 6 Gauss integration.

Solution Results The input data on pages B34.4 and B34.5 is used in the finite element analysis.

Version 99.0 Nonlinear Examples B34.1

SOLVIA Verification Manual The top figure on page B34.3 shows the calculated response as symbols and the analytical results using beam theory are shown as a continuous curve.

Results at element 4, integration point nearest node 4:

SOLVIA Beam theory Max. stress Tyy at time 2, N/mm 2 537.9 538.0 Residual stress ayy at time 3, N/mm 2

-217.8

-217.7 Total strain FYY at time 3 0.04120 0.04125 Accumulated plastic strain at time 3 0.04248 0.04253 The left bottom figure on page B34.3 shows the cross-section distribution of the stress ayy at time 2 and the residual stress at time 3. The line ABC starts at the top integration point of element 4 and extends to the bottom integration point of element 1. The right bottom figure on page B34.3 shows the accumulated effective plastic strain in the cantilever at time 3 and the relation between stress and strain at the integration point nearest node 4.

User Hints

• Note that a high enough integration order in the thickness direction of the cantilever plate is required to predict the spread of the plasticity through the thickness.

834 ELASTIC-PLASTIC CANTILEVER PLATE UNDER END MOMENT ORIGINAL t o.

TIME 2 z

I X

¥ 88 MOMENT

• 0 MASTER 1000 31 EB 1 1J0OI

ý2C

'.I lo0t SCL"TA

-NGINEERTNG-AB SOLVIA-PRE9..

C Version 99.0 Nonlinear Examples B34.2

SOLVIA Verification Manual Nonlinear Examples 3 L.ASTIC-PLAS1 CANTILEVER PATE UNDER E'ND MOMENT

'A -r

-~j

'7'

-7 IA PHT SOLVIA 7'CST 99.0 COLV:A ENGTNEEPING AB I

SCVA-LA S

A 99 0

Version 99.0 B34 ELASUTICPLASTIC CANTILEVER PLATE UNDER END MOSENT S1AX DISPL.

-A it.478 TIME 3 L,

r.-r "ew EAC C SAX 0.0432S8 0.124333 0. 08925 0.03S3t8 SN0 d,

f ItC X

A A ST9.'IA PTT 99.0 A-i P

A PA-AL, SO C7A 5.IGNE`SING AX B 34.3 SOLVIA tNG BER~ING 4B i

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B34 ELASTIC-PLASTIC CANTILEVER PLATE UNDER END MOMENT' DATABASE CREATE MASTER IDOF=-00011 NSTEP=12 DT=0.25 TOLERANCES TYPE=F RNORM=I.E4 RMNORM=4.E6 ITERATION METHOD=FULL-NEWTON LINE-SEARCH=YES TIMEFUNCTION 1

0.

0.

/

2.

1.0

/

3.

0.

COORDINATES

/

ENTRIES NODE Y

Z 1

TO 3

0.

40.

/

4 100.

40.

TO 12 100.

MATERIAL 1

PLASTIC E=1.7E5 NU=0.325 YIELD=465 ET=1.7E3 EGROUP 1

PLANE STRESS2 INT=6 GSULFACE 1 12 4 3 EL1=1 EL2-4 NODES=9 EDATA

/

1 20.

FIXBOUNDARIES 2 INPUT=LINES

/

1 3 FIXBOUNDARIES 3 INPUT=NODES

/

2 LOADS CONCENTRATED 18 4 -4.1E6 CONSTRAINTS 4 2 8 4 -20.

TO 7 2 8 4

-5.

9 2 8 4

5.

TO 12 2 8 4 20.

COORDINATES

/

ENTRIES NODE Y

Z 18 100.

20.

EGROUP 2 SPRING DIRECTION=GLOBAL PROPERTYSET 1 K=IE8 ENODES /

1 8 4 18 4 FIXBOUNDARIES 23 /

18 MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES ENUMBERS=YES BCODE=ALL VECTOR=MLOAD TIME=2 SOLVIA END Version 99.0 Nonlinear Examples B34.4

SOLVIA Verification Manual SOLVIA-POST input B34 ELASTIC-PLASTIC CANTILEVER PLATE UNDER END MOMENT DATABASE CREATE WRITE FILENAME='b34.1is' USERCURVE 1 SORT=NO /

READ AXIS 1 VMIN=O.DO VMAX=0.25 AXIS 2 VMIN=0.DO VMAX=5.E6

'B34.DAT' LABEL-'PHI' LABEL='BENDING MOMENT' NVARIABLE XROT DIRECTION=4 KIND=DISPLACEMENT EGROUP 2 EVARIABLE XMOM TYPE=SPRING KIND=MX RESULTANT ROTATION

-XROT' RESULTANT MOMENT

'XMOM, NPOINT N8 NODE=8 EPOINT SPRING ELEMENT=1 POINT=i RXYPLOT XPOINT=N8 XRESULT=ROTATION YPOINT=SPRING YRESULT=MOMENT, SYMBOL=i XAXIS=1 YAXIS=2 OUTPUT=ALL PLOT USERCURVE 1 SUBFRAME=OLD XAXIS=1 YAXIS-2 SET PLOTORIENTATION=PORTE EGROUP 1 EPLINE ABC 4

1 2 3 4 5 6 TO 1 1 2 3 ELINE ABC KIND-SYY TIME=2 ELINE ABC KIND=SYY TIME=3 MESH CONTOUR=EPACC TIME=_i EGROUP 1 EXYPLOT ELEMENT=4 POINT=1 ELIST ELIST END 456 OUTPUT=ALL SUBFRAME=12 OUTPUT-ALL VIEW=X SUBFRAME=12 XKIND=EYY YKIND-SYY SYMBOL=i EL4 SELECT=PLASTIC EL4 STRAIN=YES Version 99.0 Nonlinear Examples B34.5

SOLVIA Verification Manual EXAMPLE B35 CRACKING IN A STEEL-LINED CONCRETE CYLINDER Objective To demonstrate the use of the concrete material model in a PLANE AXISYMMETRIC analysis.

Physical Problem The steel-lined concrete cylinder shown in the figure below is considered. The cylinder is subjected to a uniform internal pressure and the cylinder is assumed to be guided so that no axial displacement can occur. The pressure at which the concrete section is completely cracked is to be determined. The material parameters are given in the table below.

Steel Concrete a = 0.9 in b=0.1 in c=1.0in d = 0.1 in h= 1.0 in MATERIAL PARAMETERS STEEL:

Young's modulus Poisson's ratio Initial yield stress Strain hardening modulus 30000 ksi 0.3 44 ksi 300 ksi CONCRETE:

Initial tangent modulus Poisson's ratio Uniaxial cut-off tensile strength Uniaxial maximum compressive stress (SIGMAC)

Compressive strain at SIGMAC Uniaxial ultimate compressive stress Uniaxial ultimate compressive strain Finite Element Model The finite element model considered is shown in the left bottom figure on page B35.2. The cylinder is modeled using six 9-node PLANE AXISYMMETRIC elements. The solution response is evaluated in eleven load steps. BFGS iterations are applied at each step with a force tolerance.

Solution Results The input data on pages B35.4 and B35.5 is used in the finite element analysis. The right top figure on page B35.3 shows the radial displacement of the inner steel cylinder during the step-by-step solution and also the variation of the hoop stress for the innermost layer of integration points in the concrete.

Version 99.0 6100 ksi 0.2 0.458 ksi

-3.74 ksi

-0.002 in/in

-3.225 ksi

-0.003 iniin Nonlinear Examples B35.1

SOLVIA Verification Manual Nonlinear Examples In the finite element solution, cracks in the hoop direction are first occurring at the integration points close to the inner surface of the concrete section at a pressure of 700 psi. The cracks propagate in the radial direction during the load history and the concrete section is completely cracked at a pressure of 1600 psi (load step 11) as shown in the right bottom figure on this page and the left top figure on page B35.3. The steel lining remains elastic throughout the entire solution.

User Hints

" The solution is much dependent on the value of the tension stiffening parameter KAPPA. In B35A KAPPA is changed from the default value of 10 to the value 2. A softer solution is then obtained, see the bottom diagrams on page B35.3. The concrete is almost completely cracked already at a pressure of 1300 psi (load step 8).

"* Note that the axial stress, az,, is not uniform in the elements of the concrete section when the cracks have been initiated. There is no variation, however, in the axial (Z) direction and the mid side nodes at Z=0.5 in. are not necessary.

335 CRACK:NG IN A STEEL-LINED CONCRETE CYL7NDER ORKGINAL N0.2 TIM E I S 6!

ORIGINAL

" 0,2 SOLVIA-PRE 99R0 z

LY R PRESSURE 350 EAXESýRST R

SOL/IA ENGINEERING AB 335 CRACKING TN A STEEL-LINED CONCRETE CYLINDER MAX DISPL.

.054SE-4 Z

TIME 3 L

LR MAX DISPL -,

I $30-4 TIME 6 7i

,OLV1A-03ST 99.

CRACK NORMAL L R CRACK NORMAL CONCRETE CRUSHED

\\'NORMAL

""CSED I CRACK 2 CRACKS 3 CRACKS SOLVIA ENGINEERING A8 Version 99.0 B35.2

SOLVIA Verification Manual Nonlinear Examples B3S CRALK'NG IN A STEEL-LINED CONCIREE CYLNKDER MAX D:SPL

2. 0945E T1ME 9 lAX DISPL I-- 2 527KE-4 TIME I I SOLVIA-POST 99.0 z

Y

.R CRACK NORMAL z

ZL_

CRACK NORMAL MISES NO AVERAGING AX 8889. 1 8364.S 7315 2 026509 5216 6

• >" 4167.2 iii! 3 1'7

.9 2068 6

!019:3 MIN 494.69 SOLVIA ENGINEERING AB 835A CRACKING IN A STEE--L:NFD CONCRETE CYLINDER, KAPPA=2 MAX DSPL.-

1.0732E-4 Z

,!ME 3 LY R

MAX DISPL

-- 2.3583E-4 TIME 8

OLVIA-POST 99 0 CRACK NORMAL R

CRACK NORAL CONCRETE CRUSHED NORMAL CLOSED 1 CRACK 2 CRACKS 3 CRACKS SOLVIA ENGINEER:NG AB 635A CRACKING IN A STEEL-LINED CONCRETE CYLINDER.

KAPPA=2 "00 0

VALUE 3F TIMEFjNCCIN 'L 200 400 600 8A0 CD 1200

400 i600

/ALJE O"- 'TtI-EUNC-ION I SO/220A-OST 99.C SOLVIA ENGINEE-RNG AB B35.3 Version 99.0 B35 CRACKING AN STEEL-LTNE)

CONCRE7E CYL' NDER 0D 60 0

00 200 12010 1200 14W01. 6c; VALUE OF T7MECiOC"C7 0 1 Sr 0

R00 62 0

m Q

I0 1200

!600 VALUE OF TIME-UNC7TION. 1 SOVI-PS 99. //I ENGINEERING..

u*

o_

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B35 CRACKING IN A STEEL-LINED CONCRETE CYLINDER' DATABASE CREATE MASTER IDOF=i00111 NSTEP-ll ITERATION METHOD=BFGS TOLERANCES TYPE=F RNORM=100.

RTOL=0.01 TIMEFUNCTION 1

0.

0.

/

2.

700.

/

11. 1600.

COORDINATES

/

ENTRIES NODE Y

Z 1

0.9 0.

/

2

1. 0.

/

3

2.

0.

/

4 2.1 0.

5 0.9 1.

/

6

1. 1.

/

7

2.

1.

/

8 2.1 1.

MATERIAL 1

PLASTIC E-3.E7 NU=0.3 YIELD=44000.

ET-3.E5 MATERIAL 2

CONCRETE EO=6.1E6 NU=0.2 SIGMAT=458.

SIGMAC=-3740.,

EPSC=-0.002 SIGMAU=-3225.

EPSU=-0.003 EGROUP 1

PLANE AXISYMMETRIC MATERIAL=1 GSURFACE 6 5 1 2

EL1=1 EL2=1 NODES=9 GSURFACE 8 7 3 4 EL1=1 EL2=1 NODES=9 LOADS ELEMENT INPUT=LINES 1 5

1. 1.

EGROUP 2

PLANE AXISYMMETRIC MATERIAL=2 GSURFACE 7

6 2 3 ELI=4 EL2=1 NODES=9 FIXBOUNDARIES 3 INPUT=LINES 56

/

67

/

78

/

12

/

23

/

34 SET PLOTORIENTATION=PORTRAIT NSYMBOL=MYNODES MESH NNUMBERS=MYNODES EAXES-RST VECTOR=LOAD SUBFRAME=12 MESH ENUMBERS-GROUP BCODE=ALL SOLVIA END Version 99.0 Nonlinear Examples B35.4

SOLVIA Verification Manual SOLVIA-POST input B35 CRACKING IN A STEEL-LINED CONCRETE CYLINDER DATABASE CREATE WRITE FILENAME='b35.1is' SET PLOTORIENTATION=PORTRAIT SET VECTOR=CRACKNORMAL CONTOUR AVERAGE=NO MESH TIME=3 SUBFRAME=I2 MESH CONTOUR=CONCRETE TIME:6 MESH TIME=9 SUBFRAME=i2 MESH CONTOUR-MISES TIME=1i SUBFRAME 12 NHISTORY NODE=1 DIRECTION=2 XVARIABLE=i SYMBOL=I OUTPUT=ALL EGROUP 2

EHISTORY ELEMENT=4 POINT=i KIND=SXX XVARIABLE=I OUTPUT=ALL ELIST SELECT-CONCRETE ELIST ZONENAME=EGi NLIST ZONENAME=MYNODES END Version 99.0 Nonlinear Examples B35.5

SOLVIA Verification Manual EXAMPLE B36 DYNAMIC ANALYSIS OF A PLASTIC SPHERICAL CAP Objective To verify the geometric and materially nonlinear behaviour of the PLANE AXISYMMETRIC element when subjected to uniform pressure in dynamic analysis.

Physical Problem The spherical cap shown in the figure below is at rest and is suddenly subjected to a constant and uniform pressure over its top surface.

P (psi) 600 Time cL=26.67' h = 0. 41 in R = 22.27 in E= 10.5.10 6 psi v=0.3

-y = 24000 psi ET =0.21.10 6 psi p = 2.54"10-4 lbf's 2 /in 4 Finite Element Model The finite element model considered is shown in the left figure on page B36.2. One radian of the cap is modeled using ten 8-node PLANE AXISYMMETRIC elements with 6 x 6 integration points. A skew coordinate system is used to model the boundary conditions at the support. The solution re sponse is calculated using the Hilber-Hughes integration method with a total of 100 time steps and a step size of 10-5 sec. Large displacements are specified for the solution and the material is described as elastic-plastic with isotropic hardening. The consistent mass matrix assumption is used. Stiffness reformation and BFGS equilibrium iterations are applied at each step in the analysis with the tolerance ETOL = 10-6.

Version 99.0 Nonlinear Examples B36.1

SOLVIA Verification Manual Nonlinear Examples Solution Results The input data on pages B36.3 and B36.4 is used in the analysis. The time histories of the central deflection, velocity and acceleration are shown in the figures below and on page B36.3. The time history of the von Mises effective stress in element 4, point 3, is shown in the lower left figure on page B36.3. The right figure shows the time history of the reaction forces at the support. The central displacement solution predicted in the analysis is in reasonable agreement with the result reported in

[1], which is marked by symbols in the figure below.

User Hints In nonlinear dynamic analysis it is generally important to iterate in each time step, although only relatively few iterations may be required for convergence.

Reference

[1]

Nagarajan, S., and Popov, E.P., "Nonlinear Dynamic Analysis of Axisymmetric Shells", Int. J.

Num. Meth. Eng., Vol. 9, 1975, pp. 535-550.

136 DYNAMIC ANALYSIS OF A PLASTIC SPHERICAL CAP ORIGINAL

1.

Z Y

R NAXES SK4Q OAISINAL I

Z TIME IE-L y

R PRESSURE A0l A IIN AS uLV A-OAF A9 0 SOLAIA ENSINo ERI't Al B36 DYNAMIC ANALYSIS OF A PLASTIC SPHERICAL CAP F->

Q I:

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SOLVIA-POST 99.0 SCLVIA ENGINEERING Al Version 99.0

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B36.2

SOLVIA Verification Manual Nonlinear Examples 336

)YNAMIC ANALYSTS OF A PLASTIC SPHERICAL CAP I

Ct S

0.2 0 4 0.6

,D8, 1.0 10 -3 02ME

[0......0 021T

.. = -:-

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T TIE SOLVIA-POST 99.0

'10-1 SOLVIA ENGINEERING AB B36 DYNAMIC ANALYSTS OF A -LASTIC SPHERICAL

-AP 0.0 1

0 t

TIME 0.2 04 A

03 0

• [03 SOLVIA-POST 99.0 SOLVTA ENG0NEER:NG AB SOLVIA-PRE input HEADING

'B36 DYNAMIC ANALYSIS OF A PLASTIC SPHERICAL CAP' DATABASE CREATE MASTER IDOF=100111 NSTEP=100 DT=1.E-5 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT METHOD=HILBER, GAMMA=-0.2 KINEMATICS DISPLACEMENT=LARGE TOLERANCES ETOL=1.E-6 SYSTEM 1

CYLINDRICAL COORDINATES ENTRIES NODE R

1 22.065 2

22.065 4

22.475 5

22.475 THETA 90 63.33 63.33 90 TO SKEWSYSTEM 1 -26.67 NSKEWS 2

1 TO 4

EULERANGLES 1

MATERIAL 1

PLASTIC E=1.05E7 NU=0.3 YIELD=2.4E4, ET=2.1E5 DENSITY=2.45E-4 Version 99.0 IC i

,-4 c*

i 1336.3

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

EGROUP 1

PLANE AXISYMMETRIC INT=6 GSURFACE 1 2 4 5 EL1=10 EL2=1 NODES=8 SYSTEM=i LOADS ELEMENT INPUTTLINES 4 5 600.

600.

FIXBOUNDARIES FIXBOUNDARIES 2

INPUT=LINES 3

INPUT=NODES

/

15

/

24

/

3 SET MESH MESH NSYMBOLS=MYNODES PLOTORIENTATION-PORTRAIT SMOOTHNESS=YES NNUMBERS=MYNODES NAXES=SKEW SUBFRAME=i2 ENUMBERS=YES VECTOR-LOAD BCODE=ALL SOLVIA END SOLVIA-POST input B36 DYNAMIC ANALYSIS OF A PLASTIC SPHERICAL CAP DATABASE CREATE WRITE FILENAME -'b36.1is' MASS-PROPERTIES USERCURVE 1 /

READ B36.DAT AXIS 1 VMIN=0 VMAX=i.E-3 LABEL='TIME' AXIS 2 VMIN=-0.08 VMAX=0 LABEL='NODE 1 Z-DIR DISPL.'

SET PLOTORIENTATION=PORTRAIT SUBFRAME 12 NHISTORY NODE=i DIRECTION=3 OUTPUT=ALL XAXIS=i YAXIS=2 PLOT USERCURVE 1 SYMBOL=-1 SUBFRAME=OLD XAXIS=-i YAXIS=-2 NHISTORY NODE=i DIRECTION=3 KIND=VELOCITY SUBFRAME NHISTORY EHISTORY SUBFRAME NHISTORY NHISTORY END 12 NODE=1 DIRECTION=3 KIND=ACCELERATION ELEMENT=4 POINT=6 KIND=MISES OUTPUT=ALL 12 NODE=3 NODE=2 DIRECTION=3 KIND=REACTION DIRECTION=2 KIND=REACTION Version 99.0 Nonlinear Examples 1336.4

SOLVIA Verification Manual EXAMPLE B37 REMOVAL AND ADDITION OF ELEMENTS, BIRTH/DEATH OPTION Objective To verify the behaviour of all the structural elements in SOLVIA when using the birth/death option.

Physical Problem Two points are connected by successively adding/deleting three bars as follows:

At time t=0.0:

At time t=1.8:

At time t=2.0:

The structure consists of bars 1 and 2, both stress free.

Bar 2 is removed.

Bar 3 is added stress-free.

The bars are equal in cross-section and material properties.

One end point of the bar structure is fixed while the other is loaded by a force in the direction of the bars. The force is linearly increasing to P=300 lb at time t=3.0, see figure below. The displacements at times 1.0, 2.0 and 3.0 are to be determined.

P Ibs) 300 1-200 100 Time (s)

- 2.2 s, birth option, e(. no. 3 ETIME =1.8s, death option, e(. no. 2 The bar structure, when one of the bars is present, is shown in figure below.

F 2

y A

L K= EA= 100 F

Version 99.0 Nonlinear Examples B3 7.1

SOLVIA Verification Manual Finite Element Model The bar structure is modeled in nine different models, where each model employs only one of the structural finite element types available in SOLVIA, namely TRUSS, PLANE, SOLID, BEAM, ISOBEAM, PLATE, SHELL, PIPE and SPRING, and a time step At=1.0 is used.

Bar 1 is modeled by employing the birth option with TBIRTH = 0.0.

Bar 2 is modeled by employing the death option with TDEATH = 1.8.

Bar 3 is modeled by employing the birth option with TBIRTH = 2.2. Note that this way the element is active first time for the solution time 3.0, but its stress free configuration corresponds to solution time 2.0.

Two element groups per element type are used, which results in a total of 18 element groups, see figures on page B37.4.

Solution Results The equilibrium equations in an incremental step-by-step solution are (without using equilibrium iteration)

SK U = týMR - tF (1) t+Atu = tU + U (2) where tK

= tangent stiffness matrix at time t U

= incremental nodal point displacement vector t+AtR

= externally applied point force vector at time t + At TF

= internal nodal point force vector corresponding to the element stresses at time t t+AtU

= nodal point displacement vector at time t + At An analytical solution for the simple bar problem gives the following results:

Time t = 1 Bars 1 and 2 are active. Bar 3 not yet added.

OK = OKI + °K, = 200 OF= OF, + OF, = 0

'R = 100 OU=0 Eq. (1):

0KU= R-°F gives U=100/200=0.5 Eq. (2):

tu =

1 U + U = 0.5 Hence ICI = 162 0.041667 101 =nc 2 =50.

Version 99.0 Nonlinear Examples B37.2

SOLVIA Verification Manual Time t = 2 Bar 1 active, bar 2 removed (bar 3 is added but is stress-free and does not yet enter the solution).

'K = 'Ki = 200 2R = 200 F = 'F = 50

'U = 0.5 Eq. (1):

'KU = 2R-TF gives U = 150/100 = 1.5 Eq. (2):

2U= 'U + U = 2.0 Hence 2Pc = 0.166667 2(

= 200 20 = 0 (stress - free state at time t = 2)

Time t = 3 Bar I active, bar 2 removed and bar 3 active.

2 K 2

2 EA EA 2K = 2K+

K3----

+

2

= 100+85.71 = 185.71 L

2L

'R = 300 2F = 2F, + 2F3 = 200 + 0 = 200 2U=2.0 Eq. (1):

2KU='R-2F gives U=100/185.71=0.538462 Eq. (2):

3U= 2U+U = 2.538462 Hence 3P, = 0.211538 3 a, = 253.846 3 63=0.038462 (0.538462/14) 3 03 = 46.1538 The input data on pages B37.5 to B37.7 gives the same results in the finite element analysis as the analytical calculations.

User Hints

"* For general remarks regarding the use of the birth and death option, see Example B32.

" Note that if a nodal point is only connected to an element in the finite element model that is inactive (death option or not born) during some time steps of the analysis, that nodal point is automatically constrained not to move during the inactive time period of the element. Otherwise such a node would have zero stiffness and a solution would not be possible.

"* Note that a different response is calculated when a different time step is chosen, and hence the time step and element birth time selection are important in establishing an appropriate model for the actual physical situation.

Version 99.0 Nonlinear Examples B37.3

SOLVIA Verification Manual Nonlinear Examples 937 R-MOVAL AND ADDIO:N OF ELEMENTS, BRTH/DEATH OPTION CRIGINAL 2

Z ZONE 9RTHIy 91 X

i 17-1 lI I

s 1012 132

',72 41 731

ý123 SOLV.A-PRE 99.0 SOLVIA 9-I

.52

-2

,2 E2 N12 AENGINEERING AB B37 REMOVAL AND ADC 3RIGINAL -

2.

T:ON OF ELEMENTS. BTRTH/DE 73 15-2 7-2 1-2 62 2

K 2

SOLVIA-?RE 99.0 23 2!

ý2 22 SOLVIA ENGINE B37 REMOVAL 55ND ADDITION OF ELEMENTS. BIRTH/DEATH OPTION ORTINAL

2.

Z ZONE DEATH 91!

851 62 19'-2 492

'82 7

Y i52 74!I S

SOLI70A-RE 99 121

2I L7 22 SOL'IA ENGINEER1NG B37 REMOVAL AND ORIGINAL 2--

TIME I

ADDIT:ON OF ELEMENTS.

Li K I U

BIRTH/DEATH OPTION Z

x.

[1 SOLVIA-PRE 990 PREE SOLVIA ENGINEERING SSURE 100 FORCE A B Version 99.0 ATH OPTION 19-2 12 I7 2 S1

-2 I

Si"

ý4 1337.4

-,-NG AB

SOLVIA Verification Manual SOLVIA-PRE input HEAD

'837 REMOVAL AND ADDITION OF ELEMENTS, BIRTH/DEATH OPTION' DATABASE CREATE MASTER IDOF=110111 NSTEP=3 TIMEFUNCTION 1

0 0 /

10 10 COORDINATES 11 0 18 12

/

12 0 18 21 0 14 12

/

22 0 14 0 31 1

9 12

/

32 1

9 35 1

8 12

/

36 1

8 41 0

4 12

/

42 0

4 51 0 18 30

/

52 0 18 18 61 0 14 30

/

62 0 14 18 71 0

9 30

/

72 0

9 18 81 0

4 30

/

82 0

4 18 91 0

0 30

/

92 0

0 18 MATERIAL 1

ELASTIC E=1.2E3 EGROUP 1

TRUSS ENODES ADDZONE=BIRTHi

/

1 11 12 ENODES ADDZONE=BIRTH2

/

2 11 12 EDATA

/

ENTRIES EL AREA TBIRTH

/

23 0 13 12

/

24 0 13

/

33 0

9 12

/

34 0

9

/

37 0

8 12

/

38 0

8

/

63 0 13 30

/

64 0 13 18

/

73 0

8 30

/

74 0

8 18 1

1.

0.

/

2

1.

2.2 EGROUP 2 TRUSS ENODES ADDZONE=DEATH

/

1 11 12 EDATA

/

ENTRIES EL AREA TDEATH

/

1

1. 1.8 EGROUP 3

PLANE STRESS2 ENODES ADDZONE=BIRTH1

/

1 23 24 22 21 ENODES ADDZONE=BIRTH2

/

2 23 24 22 21 EDATA

/

ENTRIES EL THICK TBIRTH

/

1

1. 0.

/

2

1. 2.2 LOADS ELEMENT

/

1 R -100.

EGROUP 4

PLANE STRESS2 ENODES ADDZONE=DEATH

/

1 23 24 22 21 EDATA

/

ENTRIES EL THICK TDEATH

/

1

1. 1.8 EGROUP 5

SOLID ENODES ADDZONE=BIRTH1

/

1 35 36 32 31 37 38 34 33 ENODES ADDZONE=BIRTH2

/

2 35 36 32 31 37 38 34 33 EDATA

/

ENTRIES EL TBIRTH

/

1

0.

/

2 2.2 LOADS ELEMENT

/

1 R -100.

EGROUP 6

SOLID ENODES ADDZONE=DEATH

/

1 35 36 32 31 37 38 34 33 EDATA

/

ENTRIES EL TDEATH

/

1 1.8 EGROUP 7

BEAM ENODES ADDZONE=BIRTHi

/

1 11 41 42 ENODES ADDZONE=BIRTH2

/

2 11 41 42 SECTION 1

RECTANGULAR WTOP=1.

D=1.

Version 99.0 Nonlinear Examples B37.5

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

EDATA

/

ENTRIES EL TBIRTH

/

1

0.

/

2 2.2 EGROUP 8

BEAM ENODES ADDZONE=DEATH

/

1 11 41 42 SECTION 1

RECTANGULAR WTOP=1.

D=1.

EDATA

/

ENTRIES EL TDEATH

/

1 1.8 EGROUP 9

ISOBEAM RESULTS=TABLES ENODES ADDZONE=BIRTHI

/

1 61 51 52 ENODES ADDZONE:BIRTH2

/

2 61 51 52 SECTION 1

SDIM=1.

TDIM=1.

EDATA

/

ENTRIES EL TBIRTH

/

1 0.

/

2 2.2 STRESSTABLE 1

111 121 122 131 132 EGROUP 10 ISOBEAM RESULTS=TABLES ENODES ADDZONE=DEATH

/

1 61 51 52 SECTION 1

SDIM=1. TDIM-1.

EDATA

/

ENTRIES EL TDEATH

/

1 1.8 STRESSTABLE 1

111 121 122 131 132 EGROUP 11 PLATE ENODES ADDZONE:BIRTHi

/

1 64 62 61

/

2 61 63 64 ENODES ADDZONE=BIRTH2

/

3 64 62 61

/

4 61 63 64 EDATA

/

ENTRIES EL THICK TBIRTH

/

1

1.

0.

/

2

1.

0.

3

1.

2.2 /

4

1.

2.2 EGROUP 12 PLATE ENODES ADDZONE=DEATH

/

1 64 62 61

/

2 61 63 64 EDATA

/

ENTRIES EL THICK TDEATH

/

1

1.

1.8

/

2

1.

1.8 EGROUP 13 SHELL ENODES ADDZONE=BIRTHI

/

1 73 74 72 71 ENODES ADDZONE-BIRTH2

/

2 73 74 72 71 EDATA

/

ENTRIES EL TBIRTH

/

1

0.

/

2 2.2 THICKNESS 1

1.

EGROUP 14 SHELL ENODES ADDZONE=DEATH

/

1 73 74 72 71 EDATA

/

ENTRIES EL TDEATH

/

1 1.8 THICKNESS 1

1.

EGROUP 15 PIPE RESULTS=TABLES SECTION 1

DIAMETER=2.

THICKNESS=0.17435473 ENODES ADDZONE=BIRTHI

/

1 71 81 82 ENODES ADDZONE=BIRTH2

/

2 71 81 82 EDATA

/

ENTRIES EL TBIRTH

/

1

0.

/

2 2.2 STRESSTABLE 1

1101 1102 1103 1104 1201 1202 1203 1204 EGROUP 16 PIPE RESULTS=TABLES SECTION 1

DIAMETER=2.

THICKNESS=0.17435473 ENODES ADDZONE=DEATH

/

1 71 81 82 EDATA

/

ENTRIES EL TDEATH

/

1 1.8 STRESSTABLE 1

1101 1102 1103 1104 1201 1202 1203 1204 Version 99.0 Nonlinear Examples B37.6

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

EGROUP 17 SPRING AXIALTRANSLATION PROPERTYSET 1

K=100.

PROPERTYSET 2

K=85.7142857 ENODES ADDZONE=BIRTHT

/

1 91 92 ENODES ADDZONE=BIRTH2

/

2 91 92 EDATA

/

ENTRIES EL PROPERTYSET TBIRTH

/

1 1 0.

/

2 2

2.2 EGROUP 18 SPRING AXIALTRANSLATION PROPERTYSET 1 K=l00.

ENODES ADZONE=DEATH

/

1 91 92 EDATA

/

ENTRIES EL TDEATH

/

1 1.8 FIXBOUNDARIES

/

12 22 24 32 34 36 38 42 52 62 64 72 74 82 92 LOADS CONCENTRATED ii 3 i00.

41 3 100.

51 3 100.

61 3

50.

63 3

50.

71 3

50.

73 3

50.

81 3 100.

91 3 100.

VIEW ID=1 XVIEW=2 YVIEW=1 ZVIEW=i SET VIEW-i NSYMBOLS=YES NNUMBERS=YES PLOTORIENTATION=PORTRAIT MESH ZONENAME-BIRTHI ENUMBERS=GROUP MESH ZONENAME=BIRTH2 ENUMBERS=GROUP MESH ZONENAME=DEATH ENUMBERS=GROUP MESH NSYMBOLS=NO NNUMBERS=NO VECTOR=LOAD SOLVIA END SOLVIA-POST input B37 REMOVAL AND ADDITION OF ELEMENTS, BIRTH/DEATH OPTION DATABASE CREATE WRITE FILENAME-'b37.1is' NLIST DIRECTION=3 TSTART=i TEND=3 ELIST TSTART=i TEND=3 END Version 99.0 Nonlinear Examples B37.7

SOLVIA Verification Manual EXAMPLE B38 ANALYSIS OF TEMPERATURES AND THERMAL STRESSES Objective To demonstrate the capability of performing a temperature and thermal stress analysis using the SOLVIA program.

Physical Problem A cantilever beam is subjected to a linearly varying temperature distribution in the Z-direction, see figure below. No other loads are applied to the beam. The same physical problem is also analyzed in Example B 17.

a y

x NEUTRAL AXIS 1254 if, b-'t-

- a 4,0 L

Temp OF:

70°F 200°F L = 6 in.

E lb/in 2:

3.0.10' 3.0.107 a= I in.

v:

0.3 0.3 a 1/OF:

6.0.10-6 6.0.10.

Finite Element Model The finite element model considered is shown in the figure on the top of page B38.2. The cantilever beam is modeled using six 20-node SOLID elements. In the temperature analysis the temperature distribution through the thickness is calculated. A thermal stress analysis is then performed using a thermo-elastic material model.

Solution Results The SOLVIA-PRE input data on page B38.3 is used. Practically the same result is obtained for this finite element model as for the model described in Example B 17. The calculated and theoretical displacements at the free end are:

Theory SOLVIA (node 1)

-5.400.10-'

-5. 400_10-3 The bottom figure on page B38.2 shows the deformed element mesh and a contour plot of the temperature distribution as displayed by SOLVIA-POST using the input data on page B38.3. The solution is stress-free, which is verified by the command EMAX.

Version 99.0 Nonlinear Examples B38.1

SOLVIA Verification Manual 338 ANALYSIS OF TEMPERATURES AND THERMAL STRESSE ORIG:NAL v------ -

0.S SOLVIA-PRE 99.0 z

x X

Y Kga~

000 11 i B Cl I D1 C 1 i0i1 1 D 1 ! 1 1 SOLVIA ENGINEERING AB B38 ANALVSIS OF TEMPERATURES AND THERMAL STRESSES ORIGINAL v*

0.5 MAX DISPL.

5. 475E-3 TIME I

z i

TEMPERATURE MAX 1SO. OOC.

1I40.6$25 S128.9 25 21.875 S[: i"5.625 S: :09.375

'03

!2 MIN 100.000 SOLVIA ENGINEERING AB SOLV!A-POST 99.0 Version 99.0 B38.2 Nonlinear Examples

SOLVIA Verification Manual SOLVIA-PRE input HEAD

'B38 ANALYSIS OF TEMPERATURES AND THERMAL STRESSES' DATABASE CREATE MASTER IDOF=000111 TEMPERATURE=UNCOUPLED COORDINATES 1

0.

6.

1.

/

2

0. 0. 1.

/

3

1.

0. 1.

4

1.

6.

1. /

5

0. 6.

/

6 7

1. 0.

/

8

1.

6.

MATERIAL 1

THERMO-ELASTIC TREF=125.

70.

3.E7 0.3 6.E-6 200.

3.E7 0.3 6.E-6 T-MATERIAL 1

CONDUCTION K=i.

EGROUP 1 SOLID GVOLUME 1 2 3 4 5 6 7 8 ELI=3 EL2=1 EL3=2 NODES=20 FIXBOUNDARIES 123 SURFACE /

2 3 7 6 FREEBOUNDARIES 13 LINE

/

2 6 /

3 7

FREEBOUNDARIES 3

LINE

/

2 3 /

6 7

T-LOADS TEMPERATURE INPUT=SURFACE 1 2 3 4

150.

5 6 7 8

100.

MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES BCODE=ALL SOLVIA END SOLVIA-POST input B38 ANALYSIS OF TEMPERATURES AND THERMAL STRESSES DATABASE CREATE WRITE FILENAME='b38.1is' MESH ORIGINAL=DASHED OUTLINE=YES CONTOUR=TEMPERATURE MESH ORIGINAL=DASHED VECTOR=DISPLACEMENT ZONE NAME=TIPNODES INPUT=NODES 1458 NLIST ZONENAME=TIPNODES EMAX END Version 99.0 Nonlinear Examples 1338.3

SOLVIA Verification Manual EXAMPLE B39 CREEP ANALYSIS OF A CANTILEVER Objective To demonstrate the possibility of performing creep analysis with variable time step in SOLVIA by using the restart option.

Physical Problem Same as in Example B5, see figure below.

Z NEUTRAL.ŽAIý--------------yH M

b = 0.3 in E = 3.0.107 psi h = 4.0 in v = 0.3 L = 40.0 in tc = 6.4. 10-"

1.3.1 in/ in / hr M = 6000.0 lb - in Finite Element Model The finite element model is shown in the figure on page B39.2. In Example B5 the time step size is At = 10 hrs. In this example, three different time step sizes are employed:

1.

The first six time steps in the step-by-step analysis use a time step size At = 10 hrs.

2.

The next three time steps use the step size At = 20 hrs.

3.

The last two time steps use the step size At = 40 hrs.

The change of the time step size is performed by using the restart option in SOLVIA.

Solution Results The input data on pages B39.4 to B39.5 corresponding to the three time step sizes 10 hrs, 20 hrs and 40 hrs, respectively, is used in the finite element analysis.

First the SOLVIA-PRE input data for the first run on page B39.4 is used in a normal execution of SOLVIA and the restart information written on file SOLVIA08.DAT is saved.

Then, the SOLVIA-PRE input data for the second run on page B39.4 is employed in a restart analysis and the restart information previously saved is used as the starting conditions. At the end of this execution the restart information corresponding to step 9 is saved.

Version 99.0 Nonlinear Examples B39.1

SOLVIA Verification Manual Finally, the SOLVIA-PRE input data for the third run on page B39.5 is executed in another restart analysis using the restart information last saved on SOLVIA0S.DAT.

After each execution of SOLVIA the results are saved in the SOLVIA-POST database by using the input data on page B39.5. Results from every timestep are then available from SOLVIA-POST.

Practically the same results are obtained in this analysis as in Example B5. The bending stress in element 3, integration point 9 at time t = 200 hrs is (ycy - 5701 psi.

The theoretical steady-state value, given in Example B5, is 49yy = 5688 psi.

The top figure on page B39.3 shows the deformed finite element mesh at time t = 200 hrs. The bottom left figure shows the time history response of a tip node deflection and the time history of a reaction force at the built-in end.

The bottom right figure shows the time history of the bending stress and strain in element 3, point 9.

B39 CREEP ANALYSIS CF A CANTILEVER ORIGINAL H

S z

Ly

3 EA)

B ORIGINAL 5-.

TIME 10 r

XES=RST IOl 11.4 10 Dill II z

!-y PRESSURE 7A00 SOLV:A ENGINEERING AB SOL/IA-RE 99.0 Version 99.0 Nonlinear Examples B39.2

SOLVIA Verification Manual Nonlinear Examples B39 CREEP ANALYSIS OF A CANTILEVER RI I\\'AL -

-H T_ýE 20 DTSPLACEMENT 0.4824-M.A X D7S2 _.ý

0. 482S TIME' 200 2

REACTION i 058.

SOLVIA ENGINER7RNG AB SOLVIA-OCST 99.0 B39 CREEP ANALYSIS OF A CANTILEVER

.0 AOo 2 00 T N 7~1 ZOLVA-APC7 -9 >5 ASO IA ENUINEERING RB B39 CREEP ANALYSIS OF A CANTILEVER 3

500 150 20 4,

4 A

.0-1ME 0 /:IA-057 99,0 Version 99.0 B3.

Y SOCVA ENGINEERI'.C 05

,ýO 200 B39.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B39 CREEP ANALYSIS OF A CANTILEVER' First run DATABASE CREATE MASTER IDOF=100111 MODEX=EXECUTE NSTEP=6 DT=10.

ITERATION METHOD=BFGS TOLERANCES ETOL-1.E-7 COORDINATES

/

ENTRIES NODE Y

Z 1

/

2

40.

/

3

40.

2.

/

4

0.

2.

INITIAL TEMPERATURES TREF=200.

MATERIAL 1 PLASTIC-CREEP ISOTROPIC XKCRP=1 TREF=200.

ALPHA=-,

XISUBM=1 AO=6.4E-18 A1=3.15 A2=1.

0.

3.E7 0.3 3.E7 3.E6

0.

400.

3.E7 0.3 3.E7 3.E6 0.

EGROUP 1 PLANE STRESS2 GSURFACE 3 4 1 2 EL1=4 EL2=2 NODES=8 EDATA

/

1 0.3 LOADS ELEMENT INPUT=LINE 2

3

0.

-7500.

FIXBOUNDARIES 2 INPUT=LINES

/

FIXBOUNDARIES 3 INPUT=NODES

/

LOADS TEMPERATURES TREF=200.

1 2 1

/

1 4 NSYMBOL=MYNODES NNUMBERS=MYNODES BCODE-ALL VECTOR=LOAD EAXES=RST SUBFRAME=12 SOLVIA END SOLVIA-PRE input HEADING

'B39 CREEP ANALYSIS OF A CANTILEVER' Second run DATABASE OPEN MASTER IDOF=100111 SOLVIA END MODEX=RESTART NSTEP=3 DT-20.

TSTART-60.

Version 99.0 SET MESH MESH Nonlinear Examples 1339.4

SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING

'B39 CREEP ANALYSIS OF A CANTILEVER' Third run DATABASE OPEN MASTER IDOF=100111 MODEX=RESTART NSTEP=2 DT=40.

TSTART=120.

SOLVIA END SOLVIA-POST input B39 CREEP ANALYSIS OF A CANTILEVER First run DATABASE CREATE END SOLVIA-POST input B39 CREEP ANALYSIS OF A CANTILEVER Second run DATABASE RESTART END SOLVIA-POST input B39 CREEP ANALYSIS OF A CANTILEVER Third run DATABASE RESTART WRITE FILENAME='b39b.lis' MESH ORIGINAL=DASHED VECTOR=DISPLACEMENT SUBFRAME-12 MESH VECTOR REACTION SET PLOTORIENTATION=PORTRAIT NHISTORY NODE=3 DIRECTION=3 OUTPUT=ALL SUBFRAME=i2 NHISTORY NODE=4 DIRECTION=2 KIND=REACTION EHISTORY ELE*ENT=3 POINT=9 KIND=SYY OUTPUT=ALL SUBFRAME=12 EHISTORY ELEMENT=3 POINT=9 KIND=EYY OUTPUT=ALL END Version 99.0 B39.5

SOLVIA Verification Manual EXAMPLE B40 ANALYSIS OF SNAP-THROUGH OF AN ARCH STRUCTURE Objective To perform a postbuckling analysis of an arch structure using the automatic step incrementation method.

Physical Problem An assemblage of two identical bars subjected to an apex load is considered as shown in the left figure below.

P 2

A L=10 E=2.1. 106 AREA = 1.0 k = 2.1-105 Finite Element Model From symmetry, the arch structure is modeled using one 2-node TRUSS element as shown in the right figure above.

The analytical solution for the load deflection relationship is:

P = 2kL

-21 1-2 A sinl5 o+(A J

L L

sinl5°--

The input data on page B40.3 is used in the finite element analysis. The obtained numerical solution is given in the figures on page B40.2 and an excellent comparison with the analytical solution is observed.

The deformed finite element model at step 10 and 19 are shown in the top figure on page B40.2. The Z-displacement of node 2 and the axial force in the truss as function of the load multiplier are shown together with the analytical solution in the bottom figure on page B40.2.

User Hints LOAD x 103 If a static solution is attempted by using g either the modified Newton iteration, full 18 Newton iteration, or BFGS method, the displacement response can be traced in the 12 stable region from A to B. Beyond this region the program execution is terminated 6

because the arch stiffness matrix is not positive definite (unless a large step from B\\

to D is taken to avoid the difficulties).

Version 99.0 Y

Solution Results Nonlinear Examples B40.1

SOLVIA Verification Manual Nonlinear Examples 540 ANALYSIS OF SNAP THROUGH OF AN ARCH STRUCTURE ORIGINAL

2.

MAX

ýISPL

3. 0987 TIME 0O SOLVIA-POST 99.3 B40 ANALYSIS OF SNAP Z

ORIGINAL

2.

MAX DISPL. 5.7422 STIME 19 42 REACTION LOAD 68724 1O050 SOL\\/A ENGINEERING AB THROUGH OF AN ARCH STRUCTURE

!I r

S.,j F

/

T

-s----



ý OCj 0OO O0 LOL ENC-TNEER\\NG A7 Version 99.0 47 7d C1000

'00 o0003

-1000O J

/

) A "

B40.2

SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEAD

'B40 ANALYSIS OF SNAP THROUGH OF AN ARCH STRUCTURE' DATABASE CREATE MASTER IDOF:100111 NSTEP=20 KINEMATIC DISPLACEMENT=LARGE AUTOMATIC-ITERATION NODE=2 DIR=3 DISPLACEMENT=-.1 DISPMAX=5.5, CONTINUATION=YES TOLERANCES TYPE=EF RNORM=1000 SYSTEM 1

CARTESIAN PHI=iS COORDINATES 1 /

2 0 10 MATERIAL 1

ELASTIC E=2.1E6 EGROUP 1

TRUSS ENODES

/

1

] 2 EDATA

/

ENTRIES EL AREA 1

1.

FIXBOUNDARIES FIXBOUNDARIES 2/

2 LOADS CONCENTRATED 2 3 -1.

SOLVIA END SOLVIA-POST input B40 ANALYSIS OF SNAP THROUGH OF AN ARCH STRUCTURE DATABASE CREATE WRITE FILENAME=-'b40.1is SUBFRAME 21 SET NNUMBERS=YES NSYMBOLS=YES ORIGINAL=YES VIEW=X MESH VECTOR=REACTION TIME=10 MESH VECTOR=LOAD SUBFRAME 21 AXIS 1 VMIN=-l.E4 VMAX=l.2E4 LABEL='LOAD' AXIS 2 VMIN=-6 VMAX=0 LABEL='DISPLACEMENT' AXIS 3 VMIN--8.E4 VMAX=4.E4 LABEL='ELEMENT FORCE' USERCURVE 1 SORT=NO / READ

'B40LOAD.DAT' USERCURVE 2 SORT=NO / READ

'B40EL.DAT' NHISTORY NODE=2 DIRECTION=3 SYMBOL=6 OUTPUT=ALL XAXIS=1 YAXIS=2 PLOT USERCURVE 1 XAXIS=-1 YAXIS=-2 SUBFRAME=OLD EHISTORY ELEMENT=1 POINT=1 KIND=FR SYMBOL=2 OUTPUT=ALL XAXIS=1 YAXIS=3 PLOT USERCURVE 2 XAXIS=-1 YAXIS=-3 SUBFRAME=OLD END Version 99.0 B40.3

SOLVIA Verification Manual EXAMPLE B41 SNAP-BACK OF A BAR STRUCTURE Objective To verify the AUTOMATIC ITERATION method for post-collapse analysis of structures.

Physical Problem The bar structure shown in the figure below is considered. A concentrated force P is applied to the structure at its left end and the large displacement response is to be analyzed. The bar between nodes 1 and 2 is almost horizontal since the slope ca is 0.01. The axial stiffness of the bar is 19999 which is much larger than the stiffnesses of the three springs.

The parameters of the model are such that a snap-through buckling occurs when the load is consid ered a function of the Y-displacement at node 1. This is similar to the snap-through buckling in Ex ample B40.

A snap-back buckling occurs when the load is considered a function of the Y-displacement at node 5.

The term snap-back is used since the displacement decreases when the load capacity is decreasing following the buckling.

This example is also analyzed in reference [1] and the same notation as in [1] is used here.

N3 v

Q1L aL N4 K3 L

Truss stiffness EA K2 = L+

= 19999 Ll l+a Spring stiffnesses K1 =1.5 K3 = 0.25 K4 = 1.0 Finite Element Model The structure is modeled using one 2-node TRUSS element and three spring elements.

Version 99.0 L = 2500 ca= 0.01 A=50 E = 1.0-106 Nonlinear Examples p

N5

^

1341.1

SOLVIA Verification Manual Solution Results The input data is shown on the pages B41.4 and B41.5.

The load-displacement curves for displacements at nodes 1. 2 and 5 are shown on page B41.3 as well as the relation between the load and the reaction force at node 2.

The snap-through buckling at node 1 is displayed clearly. The stable regions with positive tangent are very stiff while the intermediate unstable portion with negative tangent shows much more flexibility.

If the bar would be rigid the negative tangent stiffness can be calculated to K 3 - K1 = -1.25 which is very close to the value in the diagram.

When the Y-displacement at node 1 is 2500, thus when the bar is vertical, the load is only due to the compressive force in the spring between nodes 1 and 4, which is K3 *2500 = 625. The load-displace ment curve is antisymmetric with respect to the load value 625 and the displacement value 2500.

The variation of the tangent in the force-displacement curve for the Z-displacement at node 5 is much more gradual than the previously discussed curve for the Y-displacement at node 1. We may note the symmetry about the load value 625.

The snap-back buckling at node 5 is associated with positive tangents for the load-displacement curve except at the two transition regions at the maximum and minimum points with rapid changes in the tangent stiffness. The middle portion of the curve, the snap-back portion, has a slope approximately equal to (K3 - KI)/(K 3 -K 1 - K4) =5 which is the limiting value when the bar is considered to be rigid. The initial and final slopes are close to K4 = 1, which applies to the case when the bar is rigid. The anti-symmetry with respect to the load value 625 is again noted.

The initial and final slope for the load-reaction curve at node 2 is very close to unity since almost all load increase is taken as reaction at node 2. During the intermediate portion of the curve a part of the change in reaction is also taken at node 4.

A theoretical solution can be obtained from the equilibrium equations K2Q1 (wv

-I++o)

~vK,(Q -ca)= 0 K2(1-Q2)(

I-j-I)

+ K3Q2 + K,(Q 2 -Q 3) = 0

-K 4 (Q2 -Q3)-

P/L =0 where yV= V(1-Q2

+QQ,

= v/L+a, Q2= u/L and Q3 = u 2 /L This is a nonlinear set of equations which has been solved in an iterative manner by a separate small program. The results have been written to four data files and then included as continuous curves without symbols in the load-displacement and load-reaction diagrams of page B41.3. The SOLVIA solution points lie on the continuous curves as can be seen in the diagrams.

Version 99.0 Nonlinear Examples B41.2

SOLVIA Verification Manual Nonlinear Examples User Hints

"* The AUTOMATIC-ITERATION method is required in order to trace the complete response for this tough problem which involves both snap-through and snap-back buckling.

"* RESULTANTS have been used in the post-processing in SOLVIA-POST in order to place the load on the Y-axis. The NHISTORY and EHISTORY commands would have been simpler to use but then the load would have been placed on the X-axis.

"* Both energy and force tolerances were prescribed in this Example. The force tolerance is of interest in the stiff regions while the energy tolerance is useful in the more flexible regions with small changes in the forces.

Reference

[1]

Crisfield, M.A., Duxbury, P.G. and Hunt, G.W., "Benchmark Tests for Solution Procedures for Geometric Non-Linearity", NAFEMS Report SPGLN, October 1987, pp. 81-91 and 151-167.

B11 SNAP-BACK OF A BAR STRUCTURE

°(D 2

0000 2000 3000 4000 s000 o 000 NODE I N-DrS.LA.

- EN

.S 0

I0

SOo 2000 2500 ncC N0DE 2 7-DI P: AFFIF'T LV0A-POST

95.

S OLVIA ENGINEERNG AS B4[

SNAP-BACK OF A BAR STRUCTURE

6 2GG0 3000 4000 SOOO 6000 NODC S Y-DISPLAC*Eý-NT 0 ;

0 Uo O -1c 0

300 0

000 b3O3C 4oA NODE. 2 ý-kFACT:0N SOLVIA-POST 99 9 SOLVIA ENGINEERING A

Version 99.0

-L I

-C LVIA-POST 99 SOLVIA ENGIiNEERING AB B41.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B41 SNAP-BACK OF A BAR STRUCTURE' DATABASE CREATE MASTER IDOF=100111 NSTEP=50 KINEMATICS DISPLACEMENTS=LARGE AUTOMATIC-ITERATION NODE=5 DIRECTION=2 DISPLACEMENTS-300.,

DISPMAX=5000.

CONTINUATION=YES ALFA=0.2 TOLERANCES TYPE=EF ETOL=I.E-6 RNORM=100.

RTOL=0.001 COORDINATES

/

ENTRIES NODE Y

Z 1

/

2 2500.

25.

/

3 2500.

50.

/

4 100.

/

5

-100.

MATERIAL 1

ELASTIC E=1.E6 EGROUP 1

TRUSS ENODES 1

12 EDATA

/

1 50.

EGROUP 2

SPRING DIRECTION=AXIALTRANSLATION PROPERTYSET 1

K=1.5 PROPERTYSET 2

K=0.25 PROPERTYSET 3

K=1.0 ENODES

/

1 2 3

/

2 1 4

/

3 1 5 EDATA

/

1 1

/

2 2

/

3 3 FIXBOUNDARIES 2

/

2 3 4 FIXBOUNDARIES 3

/

1 3 4 5

LOADS CONCENTRATED

/

5 2 1.

SOLVIA END Version 99.0 Nonlinear Examples B41.4

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SOLVIA Verification Manual EXAMPLE B42 NONLINEAR RESPONSE OF A SHALLOW SHELL Objective To analyze the large displacement response of a doubly curved shell using the automatic iteration method.

Physical Problem A shallow shell subjected to a uniform pressure is considered as shown in figure below. In an X - Y plane view, the structure is square (2a x 2a) in shape. All four edges are hinged.

S a=785 Thickness =100 E=69 57

""-v=0.3 The shell mid surface is defined in Cartesian coordinates with the function Z = b[X(2a - X)+ Y(2a - Y)]

where b =2.0285.10-The surface is almost spherical. The problem is described in [2] but no units are indicated. A similar problem was analyzed in [I].

Finite Element Model Assuming a symmetrical response, only one quarter of the shell structure is discretized using sixteen 16-node SHELL elements, see figure on page B42.2. Symmetrical boundary conditions are applied at nodes along lines 13 - 169 and 169 - 157. For nodes along lines I - 157 and 1 - 13, all displacement degrees of freedom are deleted. The Fortran code on page B42.5 is used to generate a file with nodal coordinates which is read by SOLVIA-PRE.

Solution Results The input data on pages B42.4 and B42.5 is used. The SOLVIA numerical results obtained for the applied pressure vs. displacement relationship are given in the bottom figure on page B42.3. The reference solution in [2] is shown with symbols and a reasonable agreement can be noted. The deformed element mesh at the last solution step is shown in the top figure on page B42.3.

User Hints

• A full model needs in general to be used if both symmetrical and antisymmetrical buckling can occur.

Version 99.0 Nonlinear Examples B42.1

SOLVIA Verification Manual

• The load vector in the automatic iteration method is not updated for deformation dependent pressure effects. The first portion of the solution can include deformation dependent pressure effects if the AUTOSTEP method is used.

References

[1]

Leicester, R.H., "Finite Deformations of Shallow Shells", Proceedings of American Society of Civil Engineers, Vol. 94, (EM6), pp. 1409-1423, 1968.

[2]

Prinja, N.K. and Clegg, R.A., "A Review of Benchmark Problems for Geometric Non-linear Behaviour of 3-D Beams and Shells (

SUMMARY

)", NAFEMS Report R0024, 1993.

B42 NONLINEAR RESPONSE OF A SHALLOW SHELL ORIGINAL

!00.

z KY x

>zl 169 MASTE B

1 C I00C i Y ill Ci G 1

'1 CI SOLV A ENGCNEE\\CNG AB CLV,/A-PRE 99 Version 99.0 Nonlinear Examples B42.2

SOLVIA Verification Manual Nonlinear Examples B42 \\CNLL\\iEAR AESPCNSE OF A SHALLO CSHE 2

MlAX DTSPLi 250 23 TT ME 42 SCLV'IA POST 99.0 SCLVIA E50GNEERING AB B42 NONLTNEAR RESPONSE

ýFA SHALLOW SHELL 03 3 )4 AF-K. i SO -V 1A P05 S9.0 LCLVIA ENGINEERING AB Version 99.0 B4.

X LJ 77

'4 0 U 3.02 0 05 I

I

ýSURE B42.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B42 NONLINEAR RESPONSE OF A SHALLOW SHELL' DATABASE CREATE MASTER NSTEP=50 KINEMATIC DISPLACEMENTS=LARGE AUTOMATIC-ITERATION NODE=-69 DIRECTION=3 DISPLACEMENT=-2 ALPHA=i, DISPMAX=280 CONTINUATION=Y TOLERANCES TYPE=F RTOL=1.E-4 RNORM=1.E3 RMNORM=I.E3 PORTHOLE SAVEDEFAULT=NO COORDINATES /

READ

'B42COOR.DAT' MATERIAL 1

ELASTIC E=69.

NU=0.3 EGROUP 1

SHELL ENODES 1 1 4 43 40 2 17 42 27 3 30 41 14 15 16 29 28 EGENERATION TIME=3 NSTEP1=3 /

1 EGENERATION TIME=3 ETSTEP=4 NSTEP1=39 /

1 TO 4 THICKNESS 1

100.

FIXBOUNDARIES 246 INPUT=LINE 13 169 FIXBOUNDARIES 156 INPUT=LINE

/

157 169 FIXBOUNDARIES 123 INPUT=LINE

/

157 1 /

1 13 LOAD ELEMENT INPUT=ELEMENT TYPE=PRESSURE 1 -t

1. TO 16 -t 1

VIEW ID=i X-1 Y=-1 Z=0.5 MESH VIEW=i BCODE=ALL ZONE CORNER INPUT=NODES /

1 13 157 169 MESH CORNER GSCALE=OLD SUBFRAME=OLD NNUMBER=YES TEXT=NO SOLVIA END Version 99.0 Nonlinear Examples B42.4

SOLVIA Verification Manual SOLVIA-POST input B42 NONLINEAR RESPONSE OF A SHALLOW SHELL DATABASE CREATE WRITE FILENAME='b42.1is' VIEW ID=i X=i Y=-i Z=0.5 MESH ORIGINAL=YES VIEW=i Y=-2.5 USERCURVE 1 SORT=NO /

READ 'B42.DAT' AXIS 1 VMIN=0 VMAX=0.07 LABEL='APPLIED PRESSURE' AXIS 2 VMIN=0 VMAX=-300 LABEL='CENTRAL DEFLECTION' NHISTORY NODE=I69 DIRECTION=3 XAXIS=i YAXIS=2 OUTPUT=ALL PLOT USERCURVE 1 XAXIS=-i YAXIS=-2 SYMBOL=-1 SUBFRAME=OLD END Fortran code to generate node coordinates PROGRAM B42 C

IMPLICIT DOUBLE PRECISION(

A-H, O-Z)

OPEN ( UNIT=ii, FILE='B42COOR.DAT' IMAX 13 A = 785.DO B

2.0285D-4 DL = A /

( REAL ( IMAX ) -i.D0 X =

DL NODE = 0 DO i00 I = 1, IMAX X = X + DL Y =

DL DO i00 J = 1, IMAX Y = Y + DL NODE

= NODE +

1 Z = B

(

(

X *

( 2.

• A X

) + Y

2.

• A Y

WRITE ( ii, 900 ) NODE, X,

Y, Z

i00 CONTINUE C

900 FORMAT(

13,3(IPE15.7 STOP END Version 99.0 Nonlinear Examples B42.5

SOLVIA Verification Manual EXAMPLE B43 LINEARIZED BUCKLING ANALYSIS OF A STIFFENED PLATE Objective To evaluate the linearized buckling load and corresponding mode shape for a stiffened plate subjected to longitudinal compression.

Physical Problem A simply supported plate stiffened by a longitudinal rib is considered, see figure below. The rib is located symmetrically with respect to the plate. The plate-stiffener assembly is subjected to a longi tudinal compressive force. This force is increased until buckling of the plate occurs.

CROSS-SECTION in cm 1.073 4.0 0.50 50 50 100 E=2.l.1010 NI/m2 STIFFENED PLATE v=0.30 (EDGES SIMPLY SUPPORTED)

L=1lm TOTAL COMPRESSIVE LOAD, P Finite Element Model The buckling mode is expected to be symmetric [1]. Thus, one quarter of the plate is modeled using nine 9-node SHELL elements and three 3-node ISOBEAM elements, see left top figure on page B43.3. Symmetrical boundary conditions are applied along the lines 1-2 and 1-3. The option of large displacements is specified. Full Newton iteration with line searches is used.

Version 99.0 Nonlinear Examples 1343.1

SOLVIA Verification Manual Solution Results The SOLVIA program performs a linearized buckling analysis using the input data on pages B43.4 and B43.5 in the following steps:

" For the given total load corresponding to a uniaxial compressive stress of 100.106 N/m 2 the displacement response of the stiffened plate is calculated. The structure is in a state of uniaxial compression.

" Using the deformed geometry corresponding to the external load, an eigenvalue problem is solved to obtain the critical buckling load factor kv = 1.5996. Hence, the critical value of the compressive stress is oc7 = 1.5996.100.06.106 = 160.06.106 N/m 2. Here, the calculated 2nd Piola-Kirchhoff stress of 100.06.106 N/m 2 at time I has been used, see the top right figure of page B43.3. The stress is some what larger than the target value of 100.106 N/m 2 due to the effects of displacements and follower loads.

From ref. [1], the analytical solution is cr = 159.5. 106 N/m 2.

The static deformed element mesh is shown in the right top figure on page B43.3. The buckling mode shape is shown in the bottom figure on page B43.3.

User Hints

"* If the linearized buckling solution is performed with the plate subjected to a compressive load reduced to 1/10 of the original load, almost the same critical stresses are predicted, i.e.,

0cr = 160.03.106 N/m 2.

• If there is no a priori knowledge of the buckling mode shape, the full structure must be discretized to include the possibility of a non-symmetric buckling mode shape.

"* Note that it is necessary to specify large displacements in a buckling analysis.

"* Note that SOLVIA-PRE automatically assigns GLOBAL rotation for a SHELL midsurface node when the node has a boundary condition for rotation specified [2].

"* Limitations for application of the linearized buckling method are discussed in Example B88.

References

[1]

Timoshenko, S.P., and Gere, J.M., Theory of Elastic Stability, 2nd edition, Mc-Graw Hill, 1961.

[2]

SOLVIA-PRE 99.0 Users Manual, Stress Analysis, Report SE 99-1, page 7.25.

Version 99.0 Nonlinear Examples B43.2

SOLVIA Verification Manual Nonlinear Examples

\\4

EAR:Z 3UBCKL:'.R ANALYS:S OF A STIFFENED 3R:

INAL 0.5

11E ORIG7NAL 0.

SOLVIA-PFE RR 0 SOLVCA 1RRG

ý'A7E z

X EORCE SOIIOO FORCE 211R0 z

X EAXES=A-AX

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!NE-RINýG AB B-INEARIZED RUCKLNG ANALYSIS OF A STIFFENED PLATE OR IG:NAL I'

MAX DISPL. -

2.'CR61 A TIME I X

REACT CR X

y ORIGINAL 2~

'!AX DIPL.

2.4096E-4 TIMEI AC VIA-POIT 99.0 SPR INC IPAL SHELý TI)P ACLITA ENGINEERING AR B43 LINEARIZED BUCKJ--NG ANALYSIS OF A STIFFENED PLATE REFERENCE -

0 Cs MAX DISL.'

'.7288E-3 MODE I LAMS A I.S996 Z

X Y

Z-D-R DISPL

",A 17288E 7 I) 6 3 4

2 i.0805E MI1N 0

OLV A ENG-INEERING AB 3

3 3

3

,/:A S

3q.'D 0

Version 99.0 B4.

B43.3

SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING

'B43 LINEARIZED BUCKLING ANALYSIS OF A STIFFENED PLATE, DATABASE CREATE KINEMATICS DISPLACEMENTS=LARGE ITERATION METHOD=FULL-NEWTON LINE-SEARCH=YES TOLERANCE TYPE=F RTOL=1.E-5 RNORM=10000 RMNORM=10 BUCKLINGLOAD NEIG=1 COORDINATES 1

/

2

0.

0.25

/

3 0.50

/

4 0.50 0.25 MATERIAL 1

ELASTIC E=2.1E11 NU=0.3 EGROUP 1

SHELL THICKNESS 1

0.005 GSURFACE 4 2 1 3

EL1=3 EL2=3 NODES=9 LOADS ELEMENT TYPE=FORCE INPUT=LINE 3 4 out -500000.

-500000.

EGROUP 2

ISOBEAM SECTION 1

SDIM=0.005365 TDIM=0.040 GLINE Ni=i N2=3 AUX=4 EL=3 NODES=3 FIXBOUNDARIES 156 INPUT=LINE

/

2 1 FIXBOUNDARIES 246 INPUT=LINE

/

1 3 FIXBOUNDARIES 36 INPUT=LINE

/

3 4

/

4 2

LOADS CONCENTRATED 3 1 -21460 SET NSYMBOLS=YES PLOTORIENTATION=PORTRAIT MESH NNUMBERS:MYNODES VECTOR=LOAD SUBFRAME=12 MESH EAXES=RST BCODE=ALL SOLVIA END Version 99.0 1343.4

SOLVIA Verification Manual SOLVIA-POST input B43 LINEARIZED BUCKLING ANALYSIS OF A STIFFENED PLATE DATABASE CREATE WRITE FILENAME-'b43.lis' BUCKLINGLOAD SET PLOTORIENTATION=PORTRAIT ORIGINAL=DASHED SUBFRAME 12 MESH DMAX=0.5 VECTOR=REACTION MESH DMAX=0.5 VECTOR=SPRINCIPAL NLIST ZONENAME=MYNODES DIRECTION=12 SUMMATION KIND=LOAD SUMMATION KIND=REACTION SET RESPONSETYPE=BUCKLINGLOAD PLOTORIENTATION=LANDSCAPE MESH CONTOUR=DZ END Version 99.0 Nonlinear Examples B43.5

SOLVIA Verification Manual EXAMPLE B44 STIFFENED PLATE WITH INITIAL IMPERFECTIONS Objective To perform a large displacement analysis of a plate structure with initial imperfections using the automatic iteration method.

Physical Problem The same simply supported stiffened plate subjected to a longitudinal compressive force as in Example B43 is considered. The initial imperfection in the stiffened plate geometry is chosen to be:

oximp

+X-c where O

=

vector with initial nodal point coordinates Oximp

=

vector with nodal point coordinates including imperfections c

=

a scalar T

=

vector with lowest buckling mode shape of the stiffened plate subjected to longitudinal compression An analysis is performed to calculate the relationship between the applied compressive force and the lateral (out-of-plane) center displacement for the plate.

Finite Element Model Due to the assumed symmetry only one quarter of the plate is modeled using nine 9-node SHELL elements and three 3-node ISOBEAM elements (same as Example B43).

The user input for the structure geometry consists of two parts, namely:

The coordinates for the perfect geometry of the stiffened plate.

The buckling mode shape, (p.

To obtain (p, a linearized buckling analysis was performed (see Example B43) and the calculated buckling mode shape was saved on the file SOLVIA09.DAT. For the present analysis, the mode shape (p which was written onto SOLVIA09.DAT is used as input for the initial imperfection geometry data.

Solution Results The input data is shown on pages B44.3 and B44.4. The Z-displacement component of the initial imperfection at node 1 is prescribed in the input as equal to 0.1. 10-3 m. The numerical solution shows that the lateral displacement at the center of the plate increases rapidly when n approaches j,.

The theoretical value from a linearized buckling analysis is a, = 1595, see Example B43 and ref. [1].

Version 99.0 Nonlinear Examples B44.1

SOLVIA Verification Manual Nonlinear Examples The deformed mesh at the last load step and the membrane part of the stress in the load direction (stress-rr) are shown in the figure below. The top figure on page B44.3 shows the displacement for node 1 as function of load multiplier and stress-rr at element 1, point 1 as function of the Z-displace ment for node 1.

User Hints

• As the specified initial imperfection is small, the bending moments generated in the plate due to the applied compressive force are small and thus the plate is primarily in a state of uniaxial compression as long as u < 0.80 0 cr.

"* If a large initial imperfection is specified, the lateral displacement at the center of the plate increases more gradually with increase in the applied compressive load.

"* With the SOLVIA-POST command GLIST TSTART = 0 the coordinates of the model including initial imperfections are listed.

Reference

[1]

Timoshenko, S.P. and Gere, J.M., Theory of Elastic Stability, 2nd Edition, Mc-Graw Hill, 1961.

B44 ST<

ENED PLATE

'ITH INITIAL IMPERFECTIONS CRIGINAL

-- 0 05 MAX DISPL

-- 2.504E-3 TIME 30 Z

X RE, STRES SHELL MAX-1 AC-7ON 442C8 S-RR

.52402E8 S2994E8 54: 78E8 55362E8 56546E8 5773CL8 58911418 60098E8 61282E8 M:N-61874E8 SCLVIA ENG <EERING AB

OLV-A-]PCT 99.A Version 99.0 B44.2

SOLVIA Verification Manual 4

STI-FFE-NED PA'ATE J4.TH INT IAL Nonlinear Examples TMPERFECTION2 0D N N

N (D

q.,

T; 0L

• 2 C

"-4

-4

-J IC I

2

• 2<,*

[D j ~li 0.0

t.

1.0 1.5 2.0 LAD MU TIPL I LAMIBLA IS.

B 3.0 C 5 1.0 IcS 2.0 2

3.r NODE I

CSP7 SOLVIA-POST 99.0 SOLVIA E-NGINEERING AB SOLVIA-PRE input HEADING

'B44 STIFFENED PLATE WITH INITIAL IMPERFECTIONS' The buckling mode calculated in Example B43 must be available on file SOLVIA09.DAT.

DATABASE CREATE MASTER NSTEP=30 KINEMATICS DISPLACEMENT=LARGE AUTOMATIC-ITERATION NODE=3 DIRECTION=-

DISPLACEMENT:-0.05E-3, DISPMAX=0.005 CONTINUATION=YES TOLERANCES TYPE=F RTOL=-.E-5 RNORM=10000.

RMNORM=-0.

COORDINATES 1

/

2

0.

0.25 3

0.50 4

0.50 0.25 INITIAL IMPERFECTIONS 1 1 3 0.1E-3 MATERIAL 1

ELASTIC E=2.lEll NU=0.3 EGROUP 1

SHELL THICKNESS 1

0.005 GSURFACE 4 2 1 3 ELi=3 EL2=3 NODES=9 LOADS ELEMENT TYPE=FORCE INPUT=LINE 3 4 out -500000.

-500000.

Version 99.0 i

• 2_

4-i Li N

• 2

-C (i;

r*

f-

+ i 0 -3 B44.3

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

EGROUP 2

ISOBEAM RESULTS=FORCES SECTION 1

SDIM=0.005365 TDIM=0.040 GLINE Ni-i N2=3 AUX=4 EL=3 NODES=3 FIXBOUNDARIES 156 FIXBOUNDARIES 246 FIXBOUNDARIES 36 LOADS CONCENTRATED 3 1 -21460.

SOLVIA END SOLVIA-POST input INPUT=LINE INPUT=LINE INPUT=LINE B44 STIFFENED PLATE WITH INITIAL IMPERFECTIONS DATABASE CREATE WRITE FILENAME-'b44.1is' SET NSYMBOLS=MYNODES SHELLSURFACE MID MESH ORIGINAL=DASHED DMAX=1 SUBFRAME 21 NHISTORY NODE=1 DIRECTION=3 CONTOUR=SRR VECTOR=REACTION SYMBOL=1 OUTPUT=ALL NVARIABLE NAME=DISPZ DIRECTION=3 KIND=DISPLACEMENT EVARIABLE NAME=STRESSRR TYPE=SHELL KIND=SRR RESULTANT NAME=DISPZ STRING='DISPZ' RESULTANT NAME=STRESSRR STRING='STRESSRR' EPOINT NAME=ELi ELEMENT=i POINT=i NPOINT NAME=NODE1 NODE=1 RXYPLOT XPOINT=NODE1 XRESULT=DISPZ YPOINT=ELi YRESULT=STRESSRR, SYMBOL=2 OUTPUT=ALL NMAX EMAX GLIST END TSTART=0.

Version 99.0

/ /

/

2 1 1 3 3 4 4 2 Nonlinear Examples B44.4

SOLVIA Verification Manual EXAMPLE B45 LINEARIZED BUCKLING ANALYSIS OF A CIRCULAR ARCH Objective To evaluate first in-plane buckling mode shape and the corresponding linearized buckling loads for a circular arch subjected to a uniform pressure loading.

Physical Problem A circular arch with hinged ends is considered as shown in figure below. The arch is subjected to a distributed load, p = 1.0 N/mm. An analysis is performed to calculate estimates of the applied distributed load intensities corresponding to the first and second in-plane buckling modes of the arch.

p ct=22.5' R = 64.85 mm E=2.1.10 5 N/mm 2 Rectangular cross-section b= I mm h= I mm Finite Element Model The first buckling mode is expected to be skew-symmetric [1]. The full arch is modeled using twenty 2-node ISOBEAM elements, see top figure on page B45.3. For each node, the X-displacement, Y rotation and Z-rotation degrees of freedom are deleted. At each of the support nodes, all degrees of freedom are deleted except the X-rotation.

Solution Results The theoretical solution for this problem is presented in [1]. The critical value of pressure per unit length is P criticali = E Using input data on pages B45.4 and B45.5 the SOLVIA program performs a linearized buckling as follows:

0 For the given distributed load, p = 1.0 N/mm, the displacement response of the circular arch is calculated. The Z-displacement and rotation along the arch as well as the axial force and bending moment are shown in the bottom figures on page B45.3. The magnitudes of the Z-displacement, the rotation and the moment are small, which allows application of the linearized buckling method.

Version 99.0 Nonlinear Examples B45.1

SOLVIA Verification Manual

• Using the deformed geometry corresponding to the prescribed external loading, an eigenvalue problem is solved to obtain the following values of the first linearized buckling load:

Buckling Mode Intensity of Applied Distributed Load (in-plane)

Theory [1]

SOLVIA 1

4.04 4.08 The two lowest buckling modes are shown in the top figure on page B45.4.

User Hints If a linearized buckling analysis is performed for the circular arch with fixed ends the following results are obtained:

Buckling Mode Intensity of Applied Distributed Load (in-plane)

Theory [1]

SOLVIA 8.38 8.54 The magnitude of the moment is increased due to the built-in ends but the application of the linearized buckling method is still reasonable.

" Note that out-of-plane buckling occurs at a lower buckling load than in-plane buckling in this example. Out-of-plane buckling requires a full 3-D analysis.

" Even though the arch structure has an axis of symmetry, the full arch is discretized because a mode shape of interest is skew-symmetric.

" The stiffness matrix of the 2-node ISOBEAM elements is integrated using the default integration order (lx4x4). If a higher order of integration along the beam axis is employed, the element behaviour in bending deformation is too stiff.

Imperfections from the nominal shape may lower the buckling load. The effects of imperfections can be calculated using the procedure shown in Example B44.

" Limitations for application of the linearized buckling method are discussed in Example B88.

Reference

[1]

Timoshenko, S.P. and Gere, J.M., Theory of Elastic Stability, 2nd Edition, Mc-Graw Hill, 1961.

Version 99.0 Nonlinear Examples B45.2

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0 qý)IdwrxE ný)uijuoN IBnuLIN UOTIL'OUPOA VIA-109

SOLVIA Verification Manual 345 L7NEARIZED BUCKLING ANALYSIS OF A CIRCULAR ARCH REFERENCE

-H.

7 M'AX DZSPL 0

99142 4

M1CDE 1

,MBDA

.4.0 52 REFERENCE IS.Z MAX DISPL. F- 0.057879 MODE 2 LAMBDA 8.9876 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB SOLVIA-PRE input HEADING

'B45 LINEARIZED BUCKLING ANALYSIS OF A CIRCULAR ARCH' DATABASE CREATE MASTER IDOF=I00011 KINEMATICS DISPLACEMENT=LARGE BUCKLINGLOAD NEIG=2 SSTOL=I.E-12 TOLERANCES ETOL=i.E-8 SYSTEM I CYLINDRICAL COORDINATES ENTRIES NODE R

1 64.85 2

64.85 3

60.

THETA 67.5 112.5 90.

MATERIAL 1

ELASTIC E=210000 NU=.3 EGROUP 1

ISOBEAM RESULTS=FORCES SECTION 1

SDIM=i.

TDIM=I.

GLINE Ni=i N2=2 AUX=3 EL=20 NODES=2 SYSTEM=i LOADS ELEMENT TYPE=FORCE INPUT=LINE 1 2 S i.

i.

Version 99.0 Nonlinear Examples B45.4

Nonlinear Examples SOLVIA Verification Manual SOLVIA-PRE input (cont.)

FIXBOUNDARIES 23 /

1 2 3 SET MESH MESH NSYMBOLS=MYNODES VIEW=X EAXES=RST NNUMBERS=MYNODES SUBFRAME=12 VECTOR=LOAD BCODE=ALL SOLVIA END SOLVIA-POST input B45 LINEARIZED BUCKLING ANALYSIS OF A CIRCULAR ARCH DATABASE CREATE WRITE FILENAME='b45.lis' BUCKLINGLOAD SUMMATION SUMMATION KIND=LOAD KIND=REACTION NPLINE N-ARC 1 101 TO 119 2 NLINE N-ARC DIRECTION=3 SYMBOL=I NLINE N-ARC DIRECTION=4 SYMBOL=i EPLINE E-ARC 1 1 2 TO 20 1 2 AXIS 1 VMIN=-65 VMAX=0 LABEL='FORCE-R' ELINE E-ARC KIND=FR SYMBOL=I YAXIS=i ELINE E-ARC KIND=MT SYMBOL=i VIEW=X RESPONSETYPE=BUCKLINGLOAD TIME=1 SUBFRAME=12 TIME=2 ORIGINAL=YES NSYMBOLS=YES Version 99.0 SET SET MESH MESH END B45.5

SOLVIA Verification Manual EXAMPLE B46 LINEARIZED BUCKLING ANALYSIS OF A COLUMN Objective To use the option of linearized buckling analysis to obtain the lowest Euler buckling load for a column.

Physical Problem The column A-B shown in the figure below is considered. A concentrated load P is applied at end B.

LOAD, P CROSS-SECTION in m

-0.2 E=30 GPa 2 BEAM ELEMENTS Finite Element Model The column considered is modeled using two 2-node BEAM elements.

Solution Results The input data is shown on pages B46.2 and B46.3. The obtained numerical result is compared with the theoretical Euler buckling load as follows:

Buckling load (MN)

Theory SOLVIA 2

= 2. 4 6 7 2.471 A good comparison between the analytical and the numerical solution is observed. The buckling mode is shown in the figure on page B46.2.

User Hints

"* To predict the higher buckling loads accurately, it is necessary to employ a finer finite element discretization.

"* Limitations for application of the linearized buckling method are discussed in Example B88.

Version 99.0 Nonlinear Examples B46.1

SOLVIA Verification Manual 946 LINEARIZED 9 CKINCJý ANALYSTS OF A COLJMN MAX D'SPL TIMqE I

SOLVCA-3 3333E-3 3

Z REFERENCE C -.

L iMAX D1SPL.

5.7729E-4

,!OD MC L

AMBDA 2.47" 2

12 LOAD MA0STEL tIa Ce B

0 i[ll t

.POST 99.0 I

SCLVIA Z

ENCGNEERING AB SOLVIA-PRE input HEADING

'B46 LINEARIZED BUCKLING ANALYSIS OF A COLUMN' DATABASE CREATE MASTER IDOF=100011 KINEMATICS DISPLACEMENT=LARGE BUCKLINGLOAD NEIG=1 COORDINATES 1

TO 3

0.

0.

4.

MATERIAL 1

ELASTIC E=30.E9 EGROUP 1

SECTION 1

BEAMVECTOR ENODES

/

BEAM RESULT=FORCES RECTANGULAR WTOP=0.2 D=0.2

/

1

0. 1. 0.

1

-1 1 2

/

2

-1 2 3 FIXBOUNDARIES 23 FIXBOUNDARIES 2

LOADS CONCENTRATED 3 3 -1.E6 SOLVIA END Version 99.0 1

3 I

Nonlinear Examples V

1346.2

SOLVIA Verification Manual SOLVIA-POST input B46 LINEARIZED BUCKLING ANALYSIS OF A COLUMN DATABASE CREATE WRITE FILENAME='b46.1is' BUCKLINGLOAD SUBFRAME 21 SET VIEW=X NSYMBOLS=YES MESH NNUMBERS=YES ENUMBERS=YES VECTOR=LOAD BCODE=ALL SET RESPONSETYPE=BUCKLINGLOAD MESH ORIGINAL=YES END Version 99.0 Nonlinear Examples B46.3

SOLVIA Verification Manual EXAMPLE B47 DYNAMIC RESTART FROM STATIC CONFIGURATION Objective To verify the mode superposition procedure in restart from a pre-loaded condition calculated in a static analysis. Two kinds of pre-loads are analyzed, namely, gravity load and axial compressive load.

The dynamic restart is linear in the gravity load case and nonlinear in the axial load case.

Physical Problem A simply supported beam subjected to a central step loading as shown in the figure below. This problem has been analyzed without any pre-load in Examples A92 and A93.

P(t)

El, p, L, A L=2mr F

E=2'10 N/m 2 I = 1.01159."10--' M4 Xp

= 7800 kg/m 3 L2A

= 0.01 m 2 P

P(t)

PO - 100 N PO g = 9.81 m/s 2 (neg. Z-dir.)

F=-350-103 N 1.0 t

Finite Element Model Twenty BEAM elements are used to model the simply supported beam structure as shown in the top figure on page B47.4. The pre-load is applied in the first run. In the restart run the natural frequencies are calculated for the deformed configuration. A lumped mass discretization is used. The modal time integration is performed with the Newmark method. The time step size used is DT = T/10 where T is the time period for the highest considered frequency mode of the unloaded beam structure. Three modes are used in the mode superposition procedure. The subspace iteration method is used in the calculation of frequencies and mode shapes.

Solution Results The theoretical solution for the additional dynamic response due to the central step load is given below for t >_.

Displacement in Z-direction:

u(x, t)P=-

tpL

-( -

sin c

=

41 FL2

(- c t1)))sin

=l 2

l+

2 Version 99.0 Nonlinear Examples B47.1

SOLVIA Verification Manual l n = 1,5,9

....-- n 2 TE rEI where c=l

=

3,7,11....

and co=

L2 1+

0 n= even numbers E~n27 2 F is negative if compressive.

Velocity in Z-direction:

=u(x,t) 2P0L3 a n (on sin(o. (tl1))sin nlx at rT1EI n=1

_n(J I+Eý*2L Bending moment:

M*,t=E 2u (x t)_2PoL

• lcos (u).(t -l)))sin n~rx Dx 2 TC2 L

n" 1+ Fý2 L

Shear force:

V(x, t)= El D'u(x, t)-2P° cc 0 (lCos((,)"(t_l)))Co.snrUx ax2 i* l F:

L

-=I n l+

2*2 In the static case B47 and its dynamic restart case B47A a constant gravity load acts in the negative Z-direction and gives the following theoretical solution to which the step load response is added.

Displacement from gravity load:

(X=pgAt 4 rx

)(X 1 +(X )4

) 24E L

L

÷L Bending moment from gravity load:

M,(W)=P pgA (Lx-x 2) 2 Shear force from gravity load:

V (x)= pgA (L-2x) gI 2

The combined theoretical solution (displacement, velocity, moment and shear force) for case B47A including the constant gravity load and the central step load is given in the figures on page B47.4. The response curves with symbols in the figures are the corresponding SOLVIA results using the input data on pages B47.6 and B47.7. Three modes have been used in the mode superposition analysis. The frequencies are 20.0 Hz, 80.0 Hz and 180.0 Hz as in Example A93.

In the static case B47B and its dynamic restart case B47C a constant axial compressive load is acting on the beam but no gravity load. In addition, the central step load is applied at time t = 1.

Version 99.0 Nonlinear Examples B47.2

SOLVIA Verification Manual The combined theoretical solution is given in the figures on page B47.5 (without symbols) together with the SOLVIA result curves (with symbols). The input data used is given on pages B47.8 and B47.9. The axial compressive force lowers the natural frequencies to 10.9 Hz, 72.6 Hz and 172.8 Hz, thus a significant reduction, see lower right figure on page B47.5. The axial load has also a large in fluence on the magnitude of the dynamic response from the step load as can be seen in the figures.

User Hints "Since a lumped mass matrix is specified, half of the mass of an element is lumped to each of the two nodes of the element. A consequence is that half of the boundary element mass is lumped to the pinned support and is not acted upon by the gravity load. Hence, the error in the gravity load shear force at each support is

-poAL, 7800.9.81. 0.01 -0.1 = 38.259 N where Le is the element length. Since the shear force is about 800 N the error is of the order 5%.

The error can be reduced by increasing the number of elements.

" In a dynamic restart from a static configuration the difference between the restart load and the load corresponding to the static configuration is the dynamic load. If the constant gravity load acts during the entire solution then the gravity load must be defined with a time function that is con stant over the entire solution time. The additional response due to the dynamic load is then added to the static gravity load response during the dynamic portion of the response.

" In a mode superposition analysis it is important that a sufficient number of modes are included so that the effect of the dynamic load is properly modelled. The number of modes to be included depends on the type of response to be calculated. A displacement response requires in general fewer modes than a shear force response, for example. The required number of modes depends also on the frequency content of the dynamic load, in what direction it is applied and its geometric distribution.

The constant static portion of the response is not affected by the number of modes.

• A small enough time step is also important. In general, at least 10 time steps per period of the highest frequency mode to participate in the modal superposition are required for accuracy.

" In a large displacement analysis the geometric stiffness matrix is calculated. In the calculation of the modes and frequencies in case B47C the effects of the axial compressive load on the modes and frequencies are included since the calculation is based on the configuration at the start of the restart analysis. The mode superposition analysis in the dynamic restart is nonlinear and, for every time step, the modes are used to find the deformed configuration that satisfy the specified tolerance.

" The combined effects of the gravity load and the axial load on the dynamic response due to the step load need in general to be analyzed in a separate analysis. Since the effect of the axial load is nonlinear, superposition of the static gravity load response and the step load response under axial load is in general not meaningful.

References

[11 Clough, R.W., and Penzien, J., Dynamics of Structures, Second Edition, McGraw-Hill. 1993.

[2]

Blevins, Robert D., Formulas for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company, 1979.

Version 99.0 Nonlinear Examples B47.3

-4

• J z

CD C)

D CD I~A 0

C) 0 0

NC 6

'I UNI A C -)

(j)

C)

C)

0)

F m

Cz A>

I C) 0)

)L C-) QC)

[C)

-K 0

CD

E cl 0

z i

SAL

- Tlý T -171,1 lllý xv -0

,,x" ý, j,

, ýleýl ýý

SOLVIA Verification Manual SOLVIA-PRE input HEAD

'B47 DYNAMIC RESTART FROM STATIC CONFIGURATION, GRAVITY LOAD' DATABASE CREATE MASTER IDOF=I00011 NSTEP-1 DT=1 ANALYSIS TYPE=STATIC MASSMATRIX-LUMPED TIMEFUNCTION 1

0.

1. /

2.

1.

TIMEFUNCTION 2

0.

0.

/

1. 0.

/ 1.001 1. /

2.

1.

COORDINATES 1 /

2

0. 1. /

3

0. 2.

MATERIAL 1

ELASTIC E=2.E11 NU=O.

DENSITY 7800.

EGROUP 1

BEAM RESULTS=FORCES BEAMVECTOR 1

1. 0.

0.

SECTION 1 GENERAL RINERTIA=1.E-6 SINERTIA=1.01159E-6, TINERTIA=1.E-6 AREA=0.01 GLINE 1

3 -1 EL=20 FIXBOUNDARIES 23 1

FIXBOUNDARIES 3

/

3 LOAD MASSPROPORTIONAL ZFACTOR=-1 ACC=9.81 TIMEFUNCTION=1 LOADS CONCENTRATED 2 3

-100.

2 SET VIEW=X NSYMBOLS=YES MESH VECTOR=LOAD NNUMBERS=MYNODES BCODE=ALL SUBFRAME=12 MESH EAXES-RST SOLVIA END SOLVIA-POST input B47 DYNAMIC RESTART FROM STATIC CONFIGURATION, GRAVITY LOAD DATABASE CREATE END Version 99.0 Nonlinear Examples 1347.6

SOLVIA Verification Manual SOLVIA-PRE input B47A HEAD

'B47A DYNAMIC RESTART FROM STATIC CONFIGURATION, GRAVITY LOAD' DATABASE OPEN MASTER IDOF=i00011 NSTEP=160 TSTART=i DT=5.E-4 MODEX=RESTART ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED IMODS=l NMODES=3 FREQUENCIES SUBSPACE-ITERATION NEIG=3 SSTOL=i.E-10 SOLVIA END SOLVIA-POST input B47A B47A DYNAMIC RESTART FROM STATIC CONFIGURATION, GRAVITY LOAD DATABASE RESTART HEAD

'B47A DYNAMIC RESTART FROM STATIC CONFIGURATION, GRAVITY LOAD' WRITE 'b47a.lis' FREQUENCIES AXIS 1 VMIN 1.

VMAX:1.08 LABELSTRING='TIME' AXIS 2 VMIN=-1000.E-6 VMAX=-750.E-6 LABELSTRING='DISPLACEMENT' AXIS 3 VMIN=-0.015 VMAX=0.015 LABELSTRING='VELOCITY' AXIS 4 VMIN=350.

VMAX=500.

LABELSTRING='BENDING MOMENT' AXIS 5 VMIN=500.

VMAX=i000.

LABELSTRING='SHEAR FORCE' USERCURVE 1

READ B47DISP.DAT USERCURVE 2 /

READ B47VEL.DAT USERCURVE 3 /

READ B47MOM.DAT USERCURVE 4 /

READ B47FORCE.DAT SET DIAGRAM=GRID NHISTORY NODE=2 DIRECTION=3 XAXIS=d YAXIS=2 SYMBOL=i TSTART=

PLOT USERCURVE 1 XAXIS=-i YAXIS=-2 SUBFRAME=OLD NHISTORY NODE=2 DIRECTION=3 KIND=VELOCITY XAXIS=i YAXIS=3

/

SYMBOL=I TSTART=i.0005 PLOT USERCURVE 2 XAXIS=-1 YAXIS=-3 SUBFRAME=OLD EHISTORY ELEMENT=-0 POINT=2 KIND-MS XAXIS=I YAXIS=4 SYMBOL=1 PLOT USERCURVE 3 XAXIS=-i YAXIS=-4 SUBFRAME=OLD EHISTORY ELEMENT=i POINT=i KIND=FT XAXIS-1 YAXIS=5 SYMBOL-i PLOT USERCURVE 4 XAXIS=-1 YAXIS--5 SUBFRAME=OLD END Version 99.0 Nonlinear Examples 1347.7

SOLVIA Verification Manual SOLVIA-PRE input B47B HEAD

'B47B DYNAMIC RESTART FROM STATIC CONFIGURATION, AXIAL LOAD' DATABASE CREATE MASTER IDOF=100011 NSTEP=1 DT=1 ANALYSIS TYPE=STATIC MASSMATRIX=LUMPED KINEMATIC DISPLACEMENT=LARGE TIMEFUNCTION 1

0.

1. /

2.

1.

TIMEFUNCTION 2

0.

0.

/

1. 0.

/ 1.001 1. /

2.

1.

COORDINATES 1

/

2

0.

1. /

3

0.

2.

MATERIAL 1

ELASTIC E=2.E11 NU=O.

DENSITY=7800.

EGROUP 1

BEAM RESULTS=FORCES BEAMVECTOR 1

1. 0.

0.

SECTION 1 GENERAL RINERTIA=1.E-6 TINERTIA=1.E-6 GLINE 1 3

-1 EL=20 FIXBOUNDARIES FIXBOUNDARIES LOADS 23 32 23 3

CONCENTRATED

-100.

2

-350000.

1

/ /

SINERTIA=1.01159E-6, AREA=0.01 1

3 SOLVIA END SOLVIA-POST input B47B B47B DYNAMIC RESTART FROM STATIC CONFIGURATION, AXIAL LOAD DATABASE END CREATE Version 99.0 Nonlinear Examples 1347.8

SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input B47C HEAD

'B47C DYNAMIC RESTART FROM STATIC CONFIGURATION, AXIAL LOAD' DATABASE OPEN MASTER IDOF=100011 NSTEP=i60 TSTART=1 DT=5.E-4 MODEX=RESTART ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED IMODS=I NMODES=10 FREQUENCIES SUBSPACE-ITERATION NEIG=I0 SSTOL=i.E-10 SOLVIA END SOLVIA-POST input B47C B47C DYNAMIC RESTART FROM STATIC CONFIGURATION, AXIAL LOAD DATABASE RESTART HEAD

'B47C DYNAMIC RESTART FROM STATIC CONFIGURATION, AXIAL LOAD' WRITE 'b47c.lis' FREQUENCIES AXIS 1 VMIN=I.

VMAX=1.08 LABELSTRING='TIME' AXIS 2 VMIN=-600.E-6 VMAX=0 LABELSTRING='DISPLACEMENT' AXIS 3 VMIN=-0.02 VMAX=0.02 LABELSTRING-'VELOCITY' AXIS 4 VMIN=0.

VMAX=300.

LABELSTRING-'BENDING MOMENT' AXIS 5 VMIN=-100.

VMAX=500.

LABELSTRING='SHEAR FORCE' USERCURVE 1

READ B47CDISP.DAT USERCURVE 2 /

READ B47CVEL.DAT USERCURVE 3 /

READ B47CMOM.DAT USERCURVE 4 /

READ B47CFORCE.DAT SET DIAGRAM=GRID NHISTORY NODE=2 DIRECTION=3 XAXIS=1 YAXIS-2 SYMBOL=i TSTART=1 PLOT USERCURVE 1 XAXIS=--

YAXIS=-2 SUBFRAME=OLD NHISTORY NODE=2 DIRECTION-3 KIND=VELOCITY XAXIS=1 YAXIS=3 SYMBOL=i TSTART=1.0005 PLOT USERCURVE 2 XAXIS=-1 YAXIS=-3 SUBFRAME=OLD EHISTORY ELEMENT-l0 POINT-2 KIND=MS XAXIS=i YAXIS=4 SYMBOL=i PLOT USERCURVE 3 XAXIS=-I YAXIS=-4 SUBFRAME=OLD EHISTORY ELEMENT=I POINT=1 KIND=FT XAXIS=1 YAXIS=5 SYMBOL=1 PLOT USERCURVE 4 XAXIS=-i YAXIS=-5 SUBFRAME=OLD EHISTORY ELEMENT=10 POINT=1 KIND=FR SYMBOL=I SET RESPONSETYPE=VIBRATION ORIGINAL YES VIEW-X MESH TIME=i SUBFRAME=-3 MESH TIME=2 MESH TIME=3 END Version 99.0 B47.9

SOLVIA Verification Manual EXAMPLE B48 CANTILEVER SUBJECTED TO DEFORMATION DEPENDENT LOADING Objective To demonstrate the ISOBEAM and PLANE STRESS 2 elements subjected to deformation dependent loading.

Physical Problem A cantilever structure subjected to a distributed deformation dependent pressure loading is considered, see figure below. The structure experiences large in-plane bending displacements due to the applied loading.

P lb/in

/

1.-bA L=10in E = 1.2.10 4 lb/in2 h=l in v=0.2 b=1 in Finite Element Model The cantilever considered is modeled using six 2-node ISOBEAM elements, see figure on page B48.2.

The applied pressure loading is in the Y-Z plane and is assumed to act at the axis of the cantilever. The solution response is traced in ten steps using the full Newton iteration without line search.

To provide a reference solution the cantilever is also modeled using 4 x 40 PLANE STRESS2 ele ments, see figure on page B48.3. The loading is divided between the top and the bottom surface of the cantilever in order to simulate the load application at the middle surface used for the ISOBEAM model.

The response is traced in 10 steps in a large displacement/small strain solution (total Lagrangian TL analysis, B48A) and in 200 steps in a large displacement/large strain solution (updated-Lagrangian Jaumann ULJ analysis, B48B).

Solution Results The input data used for the ISOBEAM model is shown on pages B48.6 and B48.7. The deformed configuration at times 5 and 10 are shown on page B48.3. The tip displacements in the Y-and Z directions as function of time are shown in the right top figure on page B48.3. The reference solutions obtained in the ULJ analysis of the PLANE STRESS2 model are shown for comparison as continuous curves without symbols.

The input data for the TL analysis using the PLANE STRESS2 model is shown on pages B148.8 and B48.9. The deformed configuration at time 10 with the applied loads and contours of the von Mises effective stress and of a measure for the effective strain are shown on pages B48.4 and B48.5. The variations of the tip displacement components and the variation of the ayy stress vs. the Green-Version 99.0 Nonlinear Examples B48.1i

SOLVIA Verification Manual Lagrange strain yy ~during the solution are shown in the bottom figure on page B48.5, again with the corresponding ULJ solution shown without symbols for comparison. The ULJ solution was obtained using the input data of page B48.10.

User Hints

" The thickness of the cantilever is such that strains of the order 20% develop during the solution.

The ISOBEAM element is formulated for large displacements but small strains. The constant Young's modulus is used in the relation between 2nd Piola-Kirchhoff stress and Green-Lagrange strain. In spite of these assumptions a reasonable agreement between the ISOBEAM results and the reference solution obtained in the ULJ analysis can be observed.

" The TL analysis of the PLANE STRESS2 model is also based on a constant Young's modulus in the material relation between the 2nd Piola-Kirchhoff stress and the Green-Lagrange strain. The solution is similar to the ISOBEAM solution. The different application of the deformation depend ent load in the two models gives a small difference in the results.

" The 4 - E diagrams for the top and bottom integration points closest to the constrained end of the cantilever show the typical stiffening and softening in tension and compression, respectively, for the TL analysis using a constant Young's modulus. The corresponding ULJ curves show instead softening in tension and stiffening in compression when the Green-Lagrange strain measure is used as in the figure. The ULJ solution is based on a constant Young's modulus in the material relation (Cauchy) stress vs. logarithmic (natural) strain.

B48 CANT7LEVER SJBJECTED TO DFFORNATION DEPENDENT LOADING ORIGINAL Z

TIME I

S 110 102

.034 105 12 3

ORIGINAL I

K EFORCý r"

,T EAXES=RST MAS FF 5

ING AB SOLVA ENGCNEERING AB OLVIA-PRE 99.0 Version 99.0 Nonlinear Examples Pý'

B48.2

SOLVIA Verification Manual Nonlinear Examples E08 CANTILEVE SUBECTED TO DEORIIATX D5EPENDENT LOADING 18A LOAD

.3328 ORIGINAL 2

MAX,ISPL

128ý TIME 10 SOLVIA 3051 99.0 LOA SOLVIA ENGINEERING AB B48 CANTILEVER SUSJ-7ED 70 3EFORMAT:ON DEPENDENT LADING LA o

2

-:ME SX,9/'A 'ST 99.0 SOL'/!A ENGINEE.RING AB B48A CANTILEVER SUBJE C7 TO DEFORMATC7ON DEPENDCENT LOADING CRIG NAL TIME' I£Jl]

JlI.

.1

vy JVV VI\\.VV VV V VV/

tV2 /

y zVV

~

PRESSURF0 C AST 001 0 011 0 1 -A-PR 099.

SOLV7A ENGINEERING AB Version 99.0B4.

z Lyi I

B48.3

SOLVIA Verification Manual Nonlinear Examples ORIGINAL MAX DIS!

TIME 10 8 4 8A CANT!

_ 1.202 SOLVIA-PCDST 99.0 EVER SUBJECTED TO DEF3RMATION DEPENDENT LOADING LOAD 0 9885 SOLVIA ENGINEERING AB 848A CANTILEVER SUBJECTED TO DEFORMATION DEPENDENT ORIGINAL 4 1.

MAX DISPL.

I 202 TIME 10 LOADING z

I L v REACTION 3',8.94 MISES MAX 3178.4 S2979.8 A 2582.5

-;2185.3

788.0 1390.

-,993.47 596.22:

98.95 MIN 8.31496 SOLVIA ENGINEERING A Version 99.0 VIA-POST 99 B48.4

SOLVIA Verification Manual Nonlinear Examples

B48,

-ANTI-VER SiBJ C7-TO DEFORMA--CN DEPENDENT CLN CRI.;TNAL z

MlAX DSIS-'

1 2-C T7E I-EFF. STRAIN MAX 0.17917

0. 16797 ol0. IeS58

0. 12318

0. 10079 O.078397L

- 0 06004 0.0336.11 0.011218 MO1N 2.0997E. 5 S0L0/IA ENGINEERING AS SOLV 1A-PO5 99.0 B -.A CAN1TLOEN1 SLBJ'EC7.D 05CRTO 0Fý-!N DEPENDENT -CADNON 2

2 5

5 B48A CANTILEVER SUS2OCTED TO DEFORMTIOSN DEPENDENT LOADING CLI 25 2 2 Il.0 0 20 0- 3 N

0.N io aA :55/

0.20O

'p COLVIA EN3INC RING AB

-022 0,0

-0..

0. 1 0

25 c

SOLVI/OS T 990 0' ISA ENGONEERON AB Version 99.0 B4.

7

S

0T2' SOLI A L5 5 99 I

z B48.5

SOLVIA Verification Manual SOLVIA-PRE input HEAD 3B48 CANTILEVER SUBJECTED TO DEFORMATION DEPENDENT LOADING' DATABASE CREATE MASTER IDOF=l00011 NSTEP=W0 KINEMATICS DISPLACEMENT=LARGE ITERATION FULL-NEWTON TOLERANCES ETOL=I.E-6 TIMEFUNCTION 1

0.

0.

/

10. 10.

COORDINATES 1

/

2

0. 10.

/

3

0. 0. -1.

MATERIAL 1 ELASTIC E=1.2E4 NU=.2 EGROUP 1

ISOBEAM RESULT=FORCES SECTION 1

SDIM=1.

TDIM1.

GLINE 1 2 3 EL=6 NODES=2 LOADS ELEMENT TYPE=FORCEINTENSITY INPUT=LINE 1 2 S 1. 1.

FIXBOUNDARIES

/

1 3 SET VIEW-X NSYMBOLS=YES SUBFRAME 12 MESH NNUMBERS=YES VECTOR=LOAD MESH EAXES=RST BCODE=ALL SOLVIA END Version 99.0 Nonlinear Examples B48.6

SOLVIA Verification Manual SOLVIA-POST input B48 CANTILEVER SUBJECTED TO DEFORMATION DEPENDENT LOADING DATABASE CREATE WRITE FILENAME='b48.1is' SET VIEW=X PLOTORIENTATION=PORTRAIT SET ORIGINAL-YES VECTOR=LOAD SUBFRAME 12 MESH TIME=5 MESH TIME=-0 USERCURVE 1

READ B48DISPZ.DAT USERCURVE 2

READ B48DISPY.DAT AXIS 1 VMIN-C.

VMAX=10.

LABELSTRING='TIME' AXIS 2 VMIN=--0.

VMAX=O.

LABELSTRING='Z-DIR TIP DISPLACEMENT' AXIS 3 VMIN=-8.

VMAX=0.

LABELSTRING='Y-DIR TIP DISPLACEMENT, SET DIAGRAM=GRID SUBFRAME 12 NHISTORY NODE=2 DIRECTION=3 XVARIABLE-1 SYMBOL-i, XAXIS=1 YAXIS=2 OUTPUT=ALL PLOT USERCURVE 1 XAXIS=-1 YAXIS=-2 SUBFRAME=OLD NHISTORY NODE=2 DIRECTION=2 XVARIABLE=I SYMBOL=i, XAXIS=i YAXIS=3 OUTPUT=ALL SUBFRAME=NEXT PLOT USERCURVE 2 XAXIS=-I YAXIS=-3 SUBFRAME=OLD ELIST SUMMATION KIND=LOAD SUMMATION KIND=REACTION END Version 99.0 Nonlinear Examples 1348.7

SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input B48A HEAD

'B48A CANTILEVER SUBJECTED TO DEFORMATION DEPENDENT LOADING?

DATABASE CREATE MASTER IDOF=-00111 NSTEP=10 DT=I.

KINEMATICS DISPLACEMENT=LARGE ITERATION FULL-NEWTON LINE-SEARCH=YES TOLERANCES ETOL=1.E-6 TIMEFUNCTION 1

0. 0. /

10. 10.

COORDINATES

/

ENTRIES NODE Y

Z 1

TO 3

0.

1. /

4

10.

TO 6

10. 1.

MATERIAL 1

ELASTIC E=1.2E4 NU=0.2 EGROUP 1

PLANE STRESS2 GSURFACE 6 3 1 4 EL1=40 EL2=4 NODES=8 EDATA

/

1

1.

LOADS ELEMENT INPUT=LINE 3 6 0.5 0.5 1 4 -0.5

-0.5 FIXBOUNDARIES 2

INPUT=LINE

/

1 3 FIXBOUNDARIES 3

INPUT=NODE

/

2 SET NSYMBOLS=MYNODES NNUMBERS=MYNODES MESH VECTOR=LOAD BCODE=ALL SOLVIA END Version 99.0 B48.8

SOLVIA Verification Manual SOLVIA-POST input B48A B48A CANTILEVER SUBJECTED TO DEFORMATION DEPENDENT LOADING DATABASE CREATE WRITE FILENAME='b48a.lis' SET MESH MESH MESH ORIGINAL=YES OUTLINE=YES GSCALE=I.3 VECTOR=LOAD CONTOUR=MISES VECTOR=REACTION CONTOUR=EFFSTRAIN USERCURVE 1

READ B48DISPZ.DAT USERCURVE 2

READ B48DISPY.DAT USERCURVE 3

READ B48SYYP.DAT USERCURVE 4

READ B48SYYN.DAT AXIS 1 VMIN=0.

VMAX=i0.

LABELSTRING='TIME' AXIS 2 VMIN=--0.

VMAX=0.

LABELSTRING='Z-DIR TIP DISPLACEMENT' AXIS 3 VMIN=-8.

VMAX=0.

LABELSTRING='Y-DIR TIP DISPLACEMENT' AXIS 4 VMIN=0.

VMAX=0.25, LABELSTRING='GREEN-LAGRANGE STRAIN-YY AT TOP POINT' AXIS 5 VMIN=0.

VMAX=3000, LABELSTRING='CAUCHY STRESS-YY AT TOP POINT' AXIS 6 VMIN=-0.25 VMAX=0.

LABELSTRING='GREEN-LAGRANGE STRAIN-YY AT BOTTOM POINT' AXIS 7 VMIN=-3000 VMAX=C.

LABELSTRING='CAUCHY STRESS-YY AT BOTTOM POINT' SET DIAGRAM=GRID PLOTORIENTATION=PORTRAIT SUBFRAME 12 NHISTORY NODE=5 DIRECTION=3 XVARIABLE=1 SYMBOL=i, XAXIS=1 YAXIS=2 OUTPUT=ALL PLOT USERCURVE 1 XAXIS=-1 YAXIS=-2 SUBFRAME=OLD NHISTORY NODE=5 DIRECTION=2 XVARIABLE=1 SYMBOL=i, XAXIS=1 YAXIS=3 OUTPUT=ALL SUBFRAME=NEXT PLOT USERCURVE 2 XAXIS=-I YAXIS=-3 SUBFRAME=OLD SUBFRAME 12 EXYPLOT ELEMENT=40 POINT=3 XKIND=EYY YKIND=SYY SYMBOL=i, XAXIS=4 YAXIS=5 OUTPUT-ALL PLOT USERCURVE 3 XAXIS=-4 YAXIS=-5 SUBFRAME=OLD EXYPLOT ELEMENT=G60 POINT=i XKIND=EYY YKIND=SYY SYMBOL=i, XAXIS=6 YAXIS=7 OUTPUT=ALL SUBFRAME:NEXT PLOT USERCURVE 4 XAXIS=-6 YAXIS=-7 SUBFRAME=OLD END Version 99.0 Nonlinear Examples B48.9

SOLVIA Verification Manual SOLVIA-PRE input B48B HEAD

'B48B CANTILEVER SUBJECTED TO DEFORMATION DEPENDENT LOADING' DATABASE CREATE MASTER IDOF=100111 NSTEP=200 DT=0.05 KINEMATICS DISPLACEMENT=LARGE STRAIN=LARGE ITERATION FULL-NEWTON LINE-SEARCH=YES TOLERANCES ETOL=1.E-6 TIMEFUNCTION 1

0.

0.

/

100.

100.

COORDINATES

/

ENTRIES NODE Y

Z 1

TO 3

0.

1. /

4

10.

TO 6

10. 1.

MATERIAL 1

PLASTIC E=1.2E4 NU=0.2 YIELD=8.E3 ET=1.E4 EGROUP 1

PLANE STRESS2 GSURFACE 6 3 1 4 EL1=40 EL2=4 NODES=8 EDATA

/

1

1.

LOADS ELEMENT INPUT=LINE 3 6 0.5 0.5 1 4 -0.5

-0.5 FIXBOUNDARIES 2

INPUT=LINE

/

1 3 FIXBOUNDARIES 3

INPUT=NODE

/

2 SET NSYMBOLS=MYNODES NNUMBERS=MYNODES MESH VECTOR=LOAD BCODE=ALL SOLVIA END SOLVIA-POST input B48B B48B CANTILEVER SUBJECTED TO DEFORMATION DEPENDENT LOADING DATABASE CREATE WRITE FILENAME='b48b.lis' SET ORIGINAL=YES OUTLINE=YES GSCALE=-1.3 MESH VECTOR=LOAD /

MESH CONTOUR=MISES VECTOR=REACTION MESH CONTOUR=EFFSTRAIN SET DIAGRAM=GRID PLOTORIENTATION=PORTRAIT SET NEWPAGE=NO LINESPERPAGE=9000 SUBFRAME 12 /

NHISTORY NODE=5 DIRECTION=3 XVARIABLE=1 OUT=ALL NHISTORY NODE=5 DIRECTION=2 XVARIABLE=1 OUTPUT=ALL SUBFRAME 12 EXYPLOT ELEMENT=40 POINT=3 XKIND=EYY YKIND=SYY OUTPUT=ALL EXYPLOT ELEMENT=160 POINT=i XKIND=EYY YKIND=SYY OUTPUT=ALL END Version 99.0 Nonlinear Examples 1348.10

SOLVIA Verification Manual EXAMPLE B49 PLASTIC ANALYSIS OF A THICK WALLED CYLINDER Objective To verify the AUTOMATIC-ITERATION method for solution of a materially nonlinear problem.

Physical Problem A thick-walled long cylinder subjected to internal pressure is considered, see figure below. The cylinder material is elastic-perfectly plastic.

Elastic-perfectly plastic material f

Plane strain conditions von Mises yield condition E =8.667-10 4 lb/in 2 v=0.3 y =15.26 lb/in 2 CL ET-=0.0 lb/in2 Geometry a= 1 in b=2 in Finite Element Model The model is shown in the figure on page B49.2. Ten 8-node PLANE AXISYMMETRIC elements are employed to discretize a unit length of the cylinder. The axial strain is zero throughout the cylinder.

The elasto-plastic response is traced using the AUTOMATIC-ITERATION method.

Solution Results The input data is shown on page B49.4. The stress distribution through the wall and radial displacement as function of the pressure are shown in figures on page B49.3. A good agreement between the numerical results and the solution given in ref. (I], pages 100-102, is observed.

User Hints

" When using an equilibrium iteration method with prescribed load levels, very small load steps need to be used near the collapse load. However, the automatic load stepping algorithm computes the complete load displacement response automatically without the necessity of specifying the time functions for the variation of the applied external load.

"* Examples B 13 and B 14 describe a similar physical problem but solved using time functions and a constant time step.

Version 99.0 Nonlinear Examples B49.1

SOLVIA Verification Manual Reference

[1]

Prager, W. and Hodge, P.G., Theory of Perfectly Plastic Solids, Dover Publications, New York 1968.

B49 PLASTIC ANALYSES OF A -THCK 0ALLED CYLINDER OR:GINAL 0.2 Z

ORIGINAL

.2 I

y T I-E R

r EAXES=RST 9AS ER B toi t

SCL\\/A-PRE 99.0 SOLVTA ENG:NEER:NG AB Version 99.0 Nonlinear Examples B49.2

SOLVIA Verification Manual 34.9 PLASTIC ANALYSIS OF A. 7-<CK jAI 7:"E~ 26 Nonlinear Examples

, CYLIDER~

T I M 26 "x(,I;

-4

'4 _

ý0

'p 24 _

-P

/

0.0-024.1 PADTAIý 9O..V-A-P0S7 99.1, ep 7,y p

(7 2-'

0.2 0*

0 6

0.8 1.0 RAD: AL SOLVIA ENGINEER~ING AB B49 PLASTIC ANALYS7S OF TC-F2 A THICK WALLED4 CYLINDER

-C

-4 I r' C.,

'4 C

'4

'.7 T-0 0

~

C -

R 7 ;4D I 'L t C

)A LLPIE L-?A>'D[A

CLVIA -DCST q9.0 SO7L/IA ENGINEER 0

NGA AB Version 99.0 B4.

"X

'4 N

N Cf 0

F-I I

I B49.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B49 PLASTIC ANALYSIS OF A THICK WALLED CYLINDER' DATABASE CREATE MASTER IDOF=100111 NSTEP=26 AUTOMATIC-ITERATION NODE=4 DIRECTION=2 DISPLACEMENT-1.E-5, DISPMAX=.5E-3 ALPHA=.5 TOLERANCES TYPE=F RTOL-1.E-3 RNORM=1.

COORDINATES

/

ENTRIES NODE Y

Z 1

2.1.

/

2 1.1. /

3

1. /

4

2.

MATERIAL 1

PLASTIC ISOTROPIC E=8.667E4 NU-0.3 YIELD-15.26 ET=0.

EGROUP 1

PLANE AXISYMMETRIC GSURFACE 1

2 3 4 EL1=10 EL2=1 LOADS ELEMENT INPUT=LINE 2 3

1. 1.

FIXBOUNDARIES 3

INPUT=LINE

/

1 2

/

3 4 SET NSYMBOLS=MYNODES MESH EAXES=RST BCODE=ALL SUBFRAME=21 MESH NNUMBERS=MYNODES ENUMBERS=YES VECTOR=LOAD GSCALE=OLD SOLVIA END SOLVIA-POST input B49 PLASTIC ANALYSIS OF A THICK WALLED CYLINDER DATABASE CREATE WRITE FILENAME='b49.1is' EPLINE NAME=RADIAL

/

10 2

5 8 TO 1

2 5 8 ELINE RADIAL KIND=SXX SYMBOL=i SUBFRAME=21 ELINE RADIAL KIND=SYY SYMBOL=1 ELINE RADIAL KIND=SZZ SUBFRAME=21 NHISTORY NODE=4 DIRE=2 SYMBOL=2 OUTPUT=ALL ELIST EL1 TSTART=25 TEND=26 SELECT-PLASTIC ELIST EL1 TSTART=25 TEND=26 END Version 99.0 Nonlinear Examples B49.4

SOLVIA Verification Manual EXAMPLE BS0 THERMAL BUCKLING OF A RECTANGULAR PLATE Objective To verify the applicability of the linearized buckling analysis when temperature effects are included in the calculation.

Physical Problem A rectangular flat plate with a hinged boundary is considered as seen in the figure below. The initial plate temperature is zero degrees and the temperature is uniform through the plate thickness. The in plane thermal stresses are generated within the plate by increasing the plate temperature. An estimate of the lowest temperature at which the generated thermal stresses will cause plate buckling is to be calculated assuming a linearly elastic material.

Translational displacements are zero along the boundary Initial temperature, TEF = 0° C Plate thickness, h = 2.5 mm a

a = 100 mm b = 50 mm E = 210000 N/mm 2 v=0.3 a* = 12.0 6 1 /!°C b

Finite Element Model Since the buckling mode is expected to be symmetric, only a quarter of the plate is modeled using four 9-node SHELL elements as seen in the figure on page B150.3. At all nodes along lines 2 - 3 and 3 - 4. the displacement degrees of freedom are deleted. Along line 1 - 2, the X-displacement and Y rotation degrees of freedom are deleted and along line 1 - 4, the Y-displacement and X-rotation degrees of freedom are deleted. The material constitutive law is thermoelastic and the material elastic constants do not vary with temperature.

Solution Results The analytical solution for this problem is given in ref. [1] page 445 as Sa-b2 )2 a?2D where Eh3 D

12(1-v 2)

TREF = reference temperature at which the plate is stress-free.

Version 99.0 Nonlinear Examples 1350.1

SOLVIA Verification Manual The obtained SOLVIA numerical solution using input data on page B50.5 is compared with the analytical solution as follows (TREF = 0°C):

Critical temperature, Tcr 'C Theory 164.8 SOLVIA 165.1 A good comparison between the solutions is observed. The buckling mode as displayed by SOLVIA POST is shown on page B50.4.

User Hints

"* The eigenvalue solution is performed using the subspace iteration method. A tight convergence tolerance (SSTOL = 1.0 1. 12) is used to obtain the eigenvalue of interest accurately.

" For a buckling analysis by the command BUCKLINGLOADS, large displacements are specified by default, thus KINEMATICS DISPLACEMENTS = LARGE

" The stiffness matrices evaluated at times TSTART and TSTART+DT are used in the calculation of the buckling eigenvalues. The thermo-elastic material model uses the temperature at time TSTART when evaluating the stiffness matrix at time TSTART. Similarly, the temperature at time TSTART+DT is used for the stiffness matrix at time TSTART+DT. the calculated critical temperature is Tcr =4(T1 -To)= 165.1(1-0)= 165.1 where X 165.1 is the calculated eigenvalue.

"* Limitations for application of the linearized buckling method are discussed in Example B88.

Reference

[1]

Boley, B.A., and Weiner, J.H., Theory of Thermal Stresses, John Wiley and Sons, 1960.

Version 99.0 Nonlinear Examples B50.2

SOLVIA Verification Manual Nonlinear Examples B50 THERMAL BUCKLIN, 3F A RECTANGULAR PLATE ORIGINAL Y

3 K

LX r

EAXES=RST SOLVIA-PRE 99.0 ORIGINAL X

D ~~'

1 '

P'"

F-~*-,-

3 010101 F 1 1 011 A1 N01 SOLVIA ENGINEERING AB BSO THERMAL BUCKLING OF A RECTANGULAR PLATE MAX DISPL. 9.9505E-20 TIME 1

X Y

REACTION I-S-0 SOL/ A-PS 9 9.0 SOLV!A ENGINEEPING AB Version 99.0 I

1350.3

SOLVIA Verification Manual Nonlinear Examples BSO T&ERMAL BUCKLING OF A RECTANGULAR PLATE REFERENCE MAX D:SPL.

O-030208 MODE I

LAMBDA 165.06 X

Z-DIR DISPL.

MAX 0.020800

• {

i:,,*0

8.

11 9500

~0

.... C '.4300

.C1 1700 "9.0998E-3

. 1999E-3 3 8999E-3 1.30O E-3 MIN 0 SOLVIA ENGINEERING AB SOLVIA-POST 99.0 Version 99.0 B504

SOLVIA Verification Manual SOLVIA-PRE input HEAD

'B50 THERMAL BUCKLING OF A RECTANGULAR PLATE' DATABASE CREATE KINEMATICS DISPLACEMENTS=LARGE BUCKLINGLOAD NEIG=1 SSTOL=1.E-12 COORDINATES 1 /

2

0.

50.

/

3

25.

50.

/

4

25.

INITIAL TEMPERATURE TREF=0.

MATERIAL 1

THERMO-ELASTIC TREF=0.

-100 2.1E5 0.3 1.2E-5 200 2.1E5 0.3 1.2E-5 EGROUP 1

SHELL THICKNESS 1

2.5 GSURFACE 4

3 2 1 EL1=2 EL2=2 NODES=9 FIXBOUNDARIES 246 LINES 1 4 FIXBOUNDARIES 156 LINES

/

1 2 FIXBOUNDARIES 1236 LINES

/

2 3

/

3 4 LOADS TEMPERATURE INPUT=SURFACE 1 2 3 4

1.

SET VIEW=Z NSYMBOLS=MYNODES MESH NNUMBERS=MYNODES EAXES=RST SUBFRAME=21 MESH ENUMBERS=YES BCODE=ALL SOLVIA END SOLVIA-POST input B50 THERMAL BUCKLING OF A RECTANGULAR PLATE DATABASE CREATE WRITE FILENAME='bS0.1is' EMAX BUCKLINGLOAD MESH TIME=1 NSYMBOLS=YES VECTOR=REACTION SET RESPONSETYPE=BUCKLINGMODE MESH TIME=i ORIGINAL=DASHED CONTOUR=DZ END Version 99.0 Nonlinear Examples 1350.5

SOLVIA Verification Manual EXAMPLE BS1 LARGE DEFLECTION ANALYSIS OF A CABLE STRUCTURE Objective To verify the Full-Newton iteration method and AUTO-STEP for large displacement analysis of cables.

Physical Problem The figure below shows the cable problem considered. The deformed cable configuration is to be calculated when each end is displaced by 2.5 meters along the Y-direction as indicated in the figure.

u/2 - displacement at each cable end U12 U/2 L=30m A=5.10-3 m 2

,_....-* y E=5.107 N/m2 LI Finite Element Model The cable is modeled using 10 TRUSS elements and it is subjected to a uniformly distributed load simulating the cable self weight, see figure on page 851.2. Large displacements are specified in this analysis.

The Y-displacement component at node 1 and node 11 are prescribed using time function 1, see figure on page B51.3. The forces due to the self weight of the cable are applied using time function 2, see figure on page B51.3. Note that the time functions are such that the self weight forces are applied onto the cable after the cable is prestressed in tension.

Solution Results The theoretical Z-coordinate solution for a cable without tension elasticity and bending stiffness is:

z-H

[coh aw -cosh ywh]

w

(

(Ho)

Ho,)/

where H, is determined from the equation L = 2 H0 sinh(3IY and H,

Horizontal reaction force at the cable ends (N) u Total displacement of both cable ends (m) w =

2 Self weight per unit length (N/m) 3 L = 30.0 Cable length (m) a = L - u = 25 at time 25 Distance between cable ends (m)

Version 99.0 Nonlinear Examples B51.1

SOLVIA Verification Manual Nonlinear Examples The SOLVIA numerical solution is obtained by using the input data on pages B51.4 and B5 1.5. The SOLVIA results at time 25 are compared with the theoretical solution in the bottom figure on page B51.3 and good agreement is observed. The deformed mesh and contour plots of axial forces are given in the figures on page B51.4.

User Hints In the unstressed configuration, the straight cable has no stiffness in the direction normal to its longitudinal axis. Thus, to obtain a positive definite stiffness matrix in the first time step, nonzero initial conditions for displacements at the cable nodes are specified such that tensile stresses are introduced in the cable.

The energy convergence criterion is given by AU1(' [t+-t R-t+ t F (-I)

< ETOL AU(0)

[ItRý -

F]

and the force convergence criterion is given by t+-At R -t+At F(-) 2 <_ RTOL RNORM

(!)

(2)

In this problem, after the self weight forces are applied to the cable nodes, the external force vector for the cable structure does not change for the subsequent time steps. This makes the convergence cri terion for energy more difficult to satisfy than the force convergence criterion (as the denominator of Eq. (1) is a small quantity). Therefore, for this case, the convergence of solution was measured using the force convergence criterion.

B51 LARGE DEFzECTTCN ANALYS7S OF A CABLE STRUCTURE OR1 0 SNAL --

2 TIPE 4

K 2

y.

7

,19 ORIGINAL F-.

3 7

Y 7

8 9

C _. j A -PRE

99. 0 COL,/ A ENGINEVRt Version 99.0 Z

Ly toi I

F 2RCE 2

1 A"-,

NG AB i

12 1351.2

Nonlinear Examples SOLVIA Verification Manual u [m]

Time function 1 0

-5 21 Time function 2 z

[in

-5 1

I 10 is A

SOLVIA Theoretical solution

-6 Version 99.0 25 Time F

(Fzw.e[ement length) 20 25 Time y [Iml 5

S...

B51.3

SOLVIA Verification Manual Nonlinear Examples BS!

LARGE 9E-FL C T7 N ANALYSIS OF A CABLE STRUCTURE 3R7GINAL

ý 5

ZI "AX G7SPL.

26602 L

T:ME 1 37

'SI LARGE DE.LECTIAN ANALSIS3 IF A CA3LE STRUCTURE ORIGINAL 5--------S Z

",AX DISPL.--

2 9602 T IM E I.8 7 8--y "CRT:INAL 1A X GISP 7 3476 TIME 25 SOLViA-?OST 99.0 z

,1' SCLVIA ENGINEERING AB ORIG:NAL GAX DSPL 7.3476 TIME 25 SGL,!A-POST 99.0 FORCE -R NO AVERAG[NG MAX 28 771 S28,682 28 G03 28 32S "28 "17 27 969 27 79t

27. 612 27 434 MIN 27,34 FORCE-R NO AVERAGGING MAX 1.911

,. S2 O' 8.47 io t0 l

9 6352 9 1296 8 6240 8 1183 MGN 7 8655 SOLVIA ENGINEERING AB SOLVIA-PRE input HEADING

'B51 LARGE DEFLECTION ANALYSIS OF A CABLE STRUCTURE, DATABASE CREATE MASTER IDOF=I00111 NSTEP=50 DT=0.5 KINEMATICS DISPLACEMENT=LARGE TOLERANCES TYPE=F RTOL=0.01 RNORM-5.

ITERATION METHOD=FULL-NEWTON LINE-SEARCH=YES AUTO-STEP TIMEFUNCTION TIMEFUNCTION 1 /

2 /

0.0. /

0.0. /

3.

2.

/

3.

0. /

4.

2.

/

4.

2.

/

25.

-5.

25.

2.

COORDINATES 1

0.

-15.

TO 11

0.

15.

INITIAL DISPLACEMENTS 1

0.

0.

TO 11

0.

0.333 MATERIAL 1

ELASTIC E=5.E7 EGROUP ENODES EDATA 1

/

/

TRUSS 1

1 2 TO 1

0.005 10 10 11 Version 99.0

-y B51.4

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

FIXBOUNDARIES 3

/

1 11 LOADS CONCENTRATED 2 3 -1. 2 TO 10 3

-1. 2 LOADS DISPLACEMENT 1 2 -0.5 1

/

11 2 0.5 1 SET MESH MESH NSYMBOLS=YES VIEW=X NNUMBERS=YES VECTOR=LOAD TIME=4 SUBFRAME=12 ENUMBERS=YES BCODE=ALL SOLVIA END SOLVIA-POST input B51 LARGE DEFLECTION ANALYSIS OF A CABLE STRUCTURE DATABASE CREATE WRITE FILENAME='b51.lis' SET PLOTORIENTATION=PORTRAIT SUBFRAME 12 SET VIEW-X ORIGINAL=YES NSYMBOL=YES MESH TIME=12 MESH TIME=25 GSCALE=OLD SUBFRAME 12 CONTOUR AVERAGE=NO MESH CONTOUR=FR TIME=12 MESH CONTOUR=FR TIME=25 NLIST NLIST KIND=REACTION ELIST END GSCALE=OLD Version 99.0 Nonlinear Examples B51.5

SOLVIA Verification Manual EXAMPLE B52 ANALYSIS OF A CANTILEVER SUBJECTED TO END MOMENT Objective To verify the full Newton iteration method for large displacement analysis using ISOBEAM elements.

Physical Problem A thin cantilever beam subjected to an end moment is considered, see figure below. The beam cross section is 0.1 x 0.1 and the beam length is 10. The beam bending deformations take place in the Y-Z plane.

B 1I i

L

) M L=10 b -0.1 h=0.1 E =3.104 Y, v M = 392.7.10-4 Cross - section NT Finite Element Model The structure is modeled using ten 2-node ISOBEAM elements as shown in the top figure on page B52.3.

Solution Results The theoretical solution for the end rotation and the translational displacements at end B is given by 0B ML 27t 2nrEI vB 1_ EI sin ML L

ML El WB = EI (lcos ML)

L M*L El Version 99.0 B52.1 Nonlinear Examples aA

SOLVIA Verification Manual where 03

= X-rotation at end B VB,wB = Y-and Z-components of the total displacement at end B, respectively.

The numerical solution obtained using the input data on page B52.4 gives the following answers:

External 0B vB wB Moment 27t L

L M

Theory SOLVIA Theory SOLVIA Theory SOLVIA 196.35x10- 4 0.125 0.125

-0.0997

-0.0997 0.373 0.373 392.7 X 10-4 0.250 0.251

-0.363

-0.365 0.637 0.639 A good comparison between the numerical and analytical solutions is observed. The deformed element meshes for moments 196.35 x 10-4 and 392.7x lx0-are shown in the left bottom figure on page B52.3 also including a contour plot of the S-FORCE distribution in the cantilever.

Displacement in X-and Y-direction for node 2 as function of the load are shown in the right bottom figure on page B52.3.

User Hints

"* The solution response is traced in four equal load steps. The full Newton iteration is best used in the solution of this problem.

"* The computed values for the S-FORCE are small but can be decreased further by tightening the tolerances.

"* The accuracy of the computed displacements can be increased by using more elements.

Version 99.0 Nonlinear Examples B52.2

SOLVIA Verification Manual Nonlinear Examples 5 2 ANALYSIS OF A CANT iEVER SUBECTED TO END MOM'ENT Z

L7 2

3 4

5 6

7 8

9

0 ORIG0NAL i

EAXES=RST SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 852 ANALYS3S OF A CANTILEVER SUBJECTED TO END MOMENT ORIGINAL

2.

MAX DESPL. 7.3S76 T7'11 4 Z L~y I

MAX DISPL. 7,3576 TIME 4 FORCE-S NO AVERAGDNG MAX 13662E-6 SI 1937E-6ý 2

• 4889E-7

":"5 0ý01E-7 S:591 E-71 i-8So7E-7 "O-S

'O2E-7 8-7537E-71 SiOA E

.2202E-6

ý'!N-ý.39SZE-6:

SOLVIA ENGINEERING AS B52 ANALYSIS OF A CANTILEVER SUB.ECTED TO END MOMENT

3. 3

.0.

i5,3 20 U3 0 35.3 40 VALUJE ýF

.*'04 VAU.F 07'NC LN '

30 5 0 0

15.0 30.0 2S.

30.0 35

'C VSCLWA 0)[ 9 I CLVI EN-N OCI.NCM SOLV:A-POST 99 0 SOLVIA ENGINEERING A!

Version 99.0 R3INAL MASTER 1000:

L Y I.

2 SOLVIA-POST 99 3 2

¢/i 5

z c*

e N

o*

1352.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B52 ANALYSIS OF A CANTILEVER SUBJECTED TO END MOMENT' DATABASE CREATE MASTER IDOF=100011 NSTEP=4 KINEMATICS DISPLACEMENT=LARGE ITERATION METHOD=FULL-NEWTON TOLERANCES TYPE=F RTOL=0.01 RNORM=0.01 RMNORM=0.01 TIMEFUNCTION 1

0.

0.

/

4.

392.7E-4 COORDINATES 1

/

2

0.

10.

/

3

0.

0.

1.

MATERIAL 1

ELASTIC E=3.E4 EGROUP 1

ISOBEAM RESULTS=FORCES SECTION 1

SDIM=0.1 TDIM=0.1 GLINE Ni=1 N2=2 AUX=3 EL=10 NODES=2 FIXBOUNDARIES

/

1 3 LOADS CONCENTRATED

/

2 4 1.

SET VIEW=X NSYMBOLS=YES MESH ENUMBERS=YES BCODE=ALL SUBFRAME=I2 MESH NNUMBERS=MYNODES EAXES=RST SOLVIA END SOLVIA-POST input B52 ANALYSIS OF A CANTILEVER SUBJECTED TO END MOMENT DATABASE CREATE WRITE FILENAME='b52.1is' SET VIEW=X NSYMBOLS-YES PLOTORIENTATION=PORTRAIT MESH ORIGINAL=YES TIME=4 SUBFRAME=12 MESH GSCALE=OLD TIME=2 TEXT=NO AXES=NO SUBFRAME=OLD CONTOUR AVERAGE=NO MESH CONTOUR=FS GSCALE=OLD NHISTORY NODE=2 DIRECTION=2 XVARIABLE=1 SYMBOL=i SUBFRAME=12 NHISTORY NODE=2 DIRECTION=3 XVARIABLE=1 SYMBOL=2 NLIST ZONENAME=MYNODES TSTART=2 TEND=4 END Version 99.0 Nonlinear Examples B52.4

SOLVIA Verification Manual EXAMPLE B53 RECTANGULAR SHAFT SUBJECTED TO TWISTING MOMENT Objective To verify the elastic-plastic response of the ISOBEAM element when subjected to a uniform twisting moment.

Physical Problem A cantilever beam with a square cross-section as shown in the figure below is considered. The beam has a bilinear constitutive law with isotropic hardening behaviour. A torque is applied to the beam at end B.

] --

Geometry:

BA

TORQUE, T

A L=-10 b=1.

h =1.

y CROSS-SECTION

[ ILoad:

UX :h T

20.

Material data:

Greenberg et al. [1]:

SOLVIA:

)n]

E =18600 E

100

/ET

= 900 E 18600 oy = 93.33 E-=18600 v =0.0 n=9 Finite Element Model The cantilever is modeled using one 2-node ISOBEAM element. The number of Gauss integration points in the cross-section is 4x4.

Solution Results The input data on pages B53.3 and B53.4 has been used in the analysis. The numerical results dis played as T/k as a function of GO/(Lk), where k = 100/13 and 0 is the end rotation in radians, are compared with the solution of ref. [1] and a good agreement is observed, see the left figure on page B53.2. The solution of ref. [1] is drawn without node symbols.

The X-rotation at node 2 is shown in the right figure. The histories of the shear stresses Cyrs at integra tion points in one quadrant of the cross-section are shown in the figure on page B53.3.

Version 99.0 Nonlinear Examples B53.1

SOLVIA Verification Manual Nonlinear Examples User Hints

"* The bilinear material model properties used in the present numerical solution are chosen to be such that they are a close approximation to the Ramberg-Osgood material relationship employed in ref.

[1].

The calculated solution is piece-wise linear between break points where yielding is started at some of the 4x4 Gauss integration points. For the studied portion of the solution two break points occur corresponding to simultaneous yielding at 8 and 4 integration points, respectively. The remaining 4 integration points at the corners of the cross-section yield at a later stage in the solution.

Reference

[1]

Greenberg, H.J., Dorn, W.S. and Wetherell, E.H., "A Comparison of Flow and Deformation Theories in Plastic Torsion of a Square Cylinder", in Plasticity, Proc. 2nd Symp. on Naval Structural Mechanics (E.H. Lee and P.S. Symonds, Eds.), Pergamon Press, New York, 1960.

B53 RECTANGULAR SHAFT SUBJET7.D 70 Tý4 STING MOMENT C T 1

-/

2 0

2 2.-Ti-*E7A/L/K VA14UJE

_-F T7'-IEFUNCTI0\\l J

SCL',/TA,-POST 99.0 SOLV7-A ENGI:NEER-NG AB Version 99.0 B53.2

SOLVIA Verification Manual B53 RECTANGULAR SHAF-SUBJETED I, STiNG MOMENT L

F 20 VALUE OF TIMEFUNCTION I

z VALUE OF TIME.ThUTIONL a'

L LL

-Z1 A1ý

' 'y S

F-IN is 20

'.,ALjF OF T:MEFUNCTI3N 1

OLVIA-POST 99.0

-Os 20 VALUE OF TI>EFUNCTION N

SOLV A ENGINEER.NG AB SOLVIA-PRE input HEADING

'B53 RECTANGULAR SHAFT SUBJECTED TO TWISTING MOMENT, DATABASE CREATE MASTER IDOF111011 NSTEP=60 DT=0.05 ITERATION METHOD=FULL-NEWTON TOLERANCES ETOL=1.E-6 TIMEFUNCTION 1

0. 0.

/

1.

12.

/

3.

20.

COORDINATES 1

/

2

10.

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SOLVIA Verification Manual EXAMPLE B54 ANALYSIS WITH PLASTIC -MULTILINEAR MODEL, ISOTROPIC HARDENING Objective To verify the PLASTIC-MULTILINEAR material model with isotropic hardening for TRUSS, PLANE and SHELL elements.

Physical Problem A slender bar structure subjected to cyclic uniaxial loading is considered, see figure below. The material behaviour is modeled by a multilinear stress-strain relationship with isotropic hardening.

2000 U

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[_tu 1500 tp

• io L

"[

500 ET Z 0.2 x 06 ET~0.5.10 8

0.001 0.002 STRAIN 0.003 0.004 A=l.0xl.0 L=10.0 Finite Element Model Three separate models of the same structure shown in figure above are considered as follows, see figure on page B54.2.

Model 1:

Model 2:

Model 3:

The bar structure is discretized using one 2-node TRUSS element.

The bar structure is modeled using one 4-node PLANE STRESS element. At nodes 3 and 5, the Z-displacement degree of freedom is kept active to permit a nonzero ZZ-component of strain.

The bar structure is modeled using one 4-node SHELL element. At nodes 8 and 10, the X-displacement degree of freedom is kept active to permit a nonzero XX-component of strain.

The stiffness matrix of the three elements is reformed at each time step and the BFGS iteration is used to obtain the solution.

Solution Results The numerical solution obtained using input data on pages B54.4 and B54.5 yields the same load deformation and stress-strain relationships for each of the three models as seen in the figures on page B54.3 and B54.4. The numerical results agree with the analytical solution. For example, at time t = 5.0 the numerical solution for the axial displacement is 5u, = -0.019 and the analytical solution is given by Version 99.0 Nonlinear Examples B54.1

SOLVIA Verification Manual i5 2=pIp 3p_2p 4p_p p_0p01 E

E T, E

ET, E>T, IA where 'P is the applied external load at time t.

User Hints

"* Since v = 0.0 for the PLANE STRESS element, the ZZ-component of strain (' ezz ) is zero in the elastic region. Subsequently the ZZ-component of strain is nonzero because the plastic strains are considered to be incompressible and therefore t

ezz It p ezz~

=

x yy where tey is the YY-component of the plastic strain at time t.

Thus, if the Z-displacement were to be constrained at nodes 3 and 5, the PLANE element will not give the same answer; as the truss element because tezz would be enforced to be zero.

A similar argument holds for not constraining the X-displacement at nodes 8 and 10 of the SHELL element.

"* With the BFGS iteration method, the solution response was traced in nine very large time steps. In practical analyses involving more complex stress states, the time step chosen typically needs to be smaller to permit an accurate calculation of the plastic response.

BSI ANALYSIS 'JTTH PLASTIC-MULTIL2NEAR MATERIAL MODEL ORIGI\\NAL 2

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EXYPLOT EL=1 POINT=i XKIND=EYY YKIND=SYY SYMBOL=5 ELIST END STRAINS=YES Version 99.0 B54.5

SOLVIA Verification Manual EXAMPLE B55 ANALYSIS WITH PLASTIC-MULTILINEAR MODEL, KINEMATIC HARDENING Objective To verify the PLASTIC-MULTILINEAR material model with kinematic hardening for TRUSS and SOLID elements.

Physical Problem Same as Example B54, but the material behaviour is modeled by a multilinear stress-strain relationship with kinematic hardening.

Finite Element Model Two separate models of the structure as shown in the figure on page B55.2 are considered as follows:

Model 1:

The bar structure is discretized using one 2-node TRUSS element.

Model 2:

The bar structure is discretized using one 8-node SOLID element. At nodes 4, 5, 8 and 9, the X-displacement degree of freedom is kept active to permit a nonzero XX component of strain. Also, at nodes 3, 4, 8 and 7, the Z-displacement degree of free dom is kept active to permit a nonzero ZZ-component of strain.

The stiffness matrix of the structure is reformed at each time step and the Full-Newton iteration with line search is used to obtain the solution.

Solution Results The numerical solution obtained using the input data on pages B55.4 and B55.5 yields the same load deformation and stress-strain relationships for both models, see figures on page B55.3. Plastic strain components at time 2 are shown as contour plots in the top figure of page B55.4. The numerical results are in agreement with the analytical solution.

User Hints

"* The response is elastic until t = 1.

" At time t =2, the plastic strains 'ep

'eP and e are equal to XX, Ieyy ezz aeeu t

2 =(2P P

(P -,P)=_0.3 '

eyy E

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2 p 2 p I2ePx=

e z=

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p 2

p

=

2 yz

=zx 0.

Version 99.0 Nonlinear Examples B55.1

SOLVIA Verification Manual The yield surface translation tensor components at time 2 are equal to 2'c 2 C 2 e py = 200 2

2C 2e', = -100 2z 2C 2 e = -100 where 2C is the hardening parameter at time 2:

2 C=22 T =0.6666.106 3

2E-2 ET The remaining yield surface translation tensor components are zero (

, crv z

=

a With the Full-Newton iteration, the solution response was traced in 45 time steps using AUTO STEP. In practical analyses involving more complex stress states, the time step chosen needs to be small to permit an accurate calculation of the plastic response.

BS ANALYS:1 ORIGhNAL I

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6 7 8 TOLERANCES TYPE=F RTOL=0.01 RNORM=1.

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SOLVIA Verification Manual SOLVIA-PRE input (cont.)

EGROUP 1

TRUSS ENODES

/

1 1 2 EDATA

/

1

1.

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SOLID ENODES

/

1 3 4 5 6 7 8 9 10 FIXBOUNDARIES 123

/

2 10 FIXBOUNDARIES 1

/

3 FIXBOUNDARIES 3

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5 FIXBOUNDARIES 13

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6 1 FIXBOUNDARIES 12

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9 LOADS CONCENTRATED 1 2 1.

3 2 0.25 TO 6 2 0.25 MESH NNUMBERS-YES NSYMBOLS=YES ENUMBERS-YES BCODE=ALL SOLVIA END SOLVIA-POST input B55 ANALYSIS WITH PLASTIC-MULTILINEAR MATERIAL MODEL DATABASE CREATE WRITE FILENAME='b55.1is' SUBFRAME 21 NHISTORY NODE=i DIRECTION=2 XVARIABLE=1 SYMBOL=I OUTPUT=ALL NHISTORY NODE-3 DIRECTION=2 XVARIABLE=1 SYMBOL=2 OUTPUT=ALL SUBFRAME 21 EXYPLOT EL=i POINT=1 XKIND=ERR YKIND-SRR SYMBOL=i OUTPUT=ALL EGROUP 2 EXYPLOT EL=1 POINT-i XKIND=EYY YKIND=SYY SYMBOL=i OUTPUT=ALL SUBFRAME 22 SET TIME=2 MESH EG2 CONTOUR=EPXX MESH EG2 CONTOUR-EPYY MESH EG2 CONTOUR=EPZZ MESH EG2 CONTOUR=EPXY ELIST TSTART=2 STRAINS=YES ELIST TSTART=2 SELECT=PLASTIC ELIST STRAINS=YES END Version 99.0 Nonlinear Examples B55.5

SOLVIA Verification Manual EXAMPLE B56 ANALYSIS OF HERTZ AXISYMMETRIC CONTACT PROBLEM Objective To verify the 2-D contact surface, CONTACT2, and the AUTO-STEP method in the analysis of axisymmetric contact conditions.

Physical Problem An elastic sphere of radius R=100 is considered to come into contact with a flat rigid surface as shown in figure below. The radius, a, of the region of contact is such that a << R. The distribution of contact tractions in the region of contact is to be calculated assuming frictionless contact conditions (i.e., j = 0.0).

P R = 100 E = 30000 R

Symmetry v= 0.25 Rigid target

\\

Contactor

- 0.0 surface

,sutrface Finite Element Model The elastic sphere is chosen to be the contactor body. Since the anticipated region of contact is small, a simple model of the contactor body is obtained by only discretizing a small portion of the elastic sphere in the vicinity of the contact region. The PLANE AXISYMMETRIC elements are used for the discretization as shown in the figures on page B56.2.

The flat rigid surface must be chosen to be the target body. It is modeled using 2 nodes, each with all degrees of freedom deleted.

To simulate the application of the external load P and the horizontal plane of symmetry, the Z-dis placement on entire line 5 - 35 is prescribed. The total load P corresponding to the prescribed dis placement is calculated by integrating the contact pressures.

Solution Results The input data shown on pages B56.4 and B56.5 is used in the SOLVIA analysis. The numerical solution obtained for contact tractions is compared with the classical Hertz solution (see ref. [1]) and a good agreement between the solutions is observed. The principal stresses and a contour plot of the effective stress at the last step are shown in the bottom figures on page B56.3.

Version 99.0 Nonlinear Examples 1356.1

SOLVIA Verification Manual Nonlinear Examples User Hints

"° From symmetry conditions, the results in the top figure on page B56.3 also correspond to the contact traction distribution when two identical elastic spheres are compressed onto each other.

"* The CONTACT2 AXISYMMETRIC elements only model a wedge of one radian of the sphere.

Thus, to obtain the total contact force P, the contact force predicted by the finite element model needs to be multiplied by 2m

" The 4-node PLANE AXISYMMETRIC element is compatible with the contact surface since a contact segment uses 2 nodes (no midnode). If 8-node PLANE AXISYMMETRIC elements are used then the command EDGECONTINUITY can be used to remove the midside node of all elements connecting to the contact surface so that a layer of 7-node elements is created. Alter natively, contact segments can in this case connect to all nodes of the surface between nodes 3 and 4 (corner nodes as well as midside nodes) since the target surface is rigid. The number of contact segments is then twice the number of 8-node elements along the line 3-4.

Reference

[1]

Timoshenko, S.P., and Goodier, J.N., Theory of Elasticity, 2nd ed., McGraw-Hill, 1951, pp. 372-377.

356 ANALYSIS OF HERTZ AXISYMrETRIC CONTACT PROBLEM 0RIG IN AL

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ZONE CONTACTS SOLVA-'RE 99 C SOLVIA ENGSNEERING AR Version 99.0 B56.2

SOLVIA Verification Manual 2000 1000 5

10 Contact pressure SOLVIA solution Corresponding SHertz solution Distance from center Solution to the axisymmetric Hertz problem when total load P = 3.198 -105 956 ANALYSIS OF HERTZ AX9SYMMETRBC CONTACT PROBLEM MAX D7SPL.

0. 52898 Z

TIME 5 LY ZONE N3+tO R

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'B56 ANALYSIS OF HERTZ AXISYMMETRIC CONTACT PROBLEM' DATABASE CREATE MASTER IDOF:100111 NSTEP=5 DT=1.0 AUTO-STEP TMAX=5 TIMEFUNCTION 1

0

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100

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100

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PLANE AXISYMMETRIC LINE CYLINDRIC 3 4 EL=30 SYSTEM=1 LINE STRAIGHT 3 35 RATIO=6 EL=30 LINE STRAIGHT 4 6 RATIO=6 EL=30 GSURFACE 3 4 5 35 NODES=4 FIXBOUNDARIES 2 INPUT=LINES

/

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CONTACT2 AXISYMMETRIC CONTACTSURFACE 1

INPUT=LINE 34 CONTACTSURFACE 2

INPUT-NODES 21 CONTACTPAIR 1

TARGET=2 CONTACTOR=1 LOADS DISPLACEMENTS 5 3 -1 TO 35 3

-1 SET NSYMBOL:MYNODES NNUMBERS=MYNODES PLOTORIENTATION=PORTRAIT MESH MESH ZONENAME=CONTACTS NSYMBOLS=YES SOLVIA END Version 99.0 Nonlinear Examples B56.4

SOLVIA Verification Manual SOLVIA-POST input B56 ANALYSIS OF HERTZ AXISYMMETRIC CONTACT PROBLEM DATABASE CREATE TOLERANCES INTERMEDIATE=YES WRITE FILENAME='b56.1is' SET PLOTORIENTATION=PORTRAIT MESH N3+10 OUTLINE-YES VECTOR=SPRINCIPAL MESH OUTLINE=YES CONTOUR=MISES VECTOR-REACTION CLIST ZONENAME CONTACTS SUMMATION ZONENAME-Cl KIND=CONTACTS DETAILS=YES SUMMATION ZONENAME-EGI KIND=REACTION END Version 99.0 Nonlinear Examples B56.5

SOLVIA Verification Manual EXAMPLE B57 ANALYSIS OF HERTZ PLANE STRAIN CONTACT PROBLEM Objective To verify the 2-D contact surfaces, CONTACT2, and the AUTO-STEP method in the analysis of plane strain contact conditions.

Physical Problem An elastic cylinder of radius R=10 is considered to come into contact with a flat rigid surface as seen in the figure below. The contact surfaces are assumed to be frictionless.

P R =10 R

E--30000 Symmetryv 0.25 Rigid target Contactor it = 0.0 surface surface Finite Element Model Due to the horizontal and vertical symmetry planes only one quarter of the cylinder is modelled, see figure on page B157.2. The elastic cylinder is chosen to be the contactor body. The model of the contactor body is using a finer mesh in the vicinity of the contact region. PLANE STRAIN elements are used for the discretization and a CONTACT2 STRAIN contactor surface is defined from node 3 to node 4.

The flat rigid surface must be chosen to be the target body. It is modeled using 2 nodes, each with all degrees of freedom deleted. The target surface is defined from node I to node 2.

The application of the external load P is modelled by prescribing the same Z-displacement of the nodes in the horizontal symmetry plane so that the plane gradually moves downwards. The total load P corresponding to the prescribed displacement is calculated by integrating the contact pressures.

Solution Results Input data on pages B57.4 and B57.5 is used. A contour plot of von Mises effective stress is shown in the top figure on page B57.3. The shear stress distribution, the minimum principal stresses and the pressure distribution are shown as contour plots in the bottom figures on page B57.3. A vector plot of the contact forces is also shown on page B57.3.

The numerical results obtained for contact tractions are compared with the classical Hertz solution, see the figure on page B57.4. The agreement between the solutions is reasonable, considering that the contact area is somewhat large for application of the Hertz theory.

Version 99.0 Nonlinear Examples 1357.1

SOLVIA Verification Manual User Hints

"* From symmetry conditions, the results presented in the top figure on page B57.4 also correspond to the contact traction distribution when two identical cylinders are compressed onto each other.

"* The command EDGE-CONTINUITY in SOLVIA-PRE is used to delete all element midnodes adjoining the contact surfaces. Contact surface edges may not contain midnodes. Alternatively, two segments per element edge could be used in this case since the target surface is a rigid plane.

Reference

[1]

Timoshenko, S.P., and Goodier, J.N., Theory of Elasticity, 2nd ed., McGraw-Hill, 195 1, pp. 381-382.

357 ANALYSIS OF HERTZ PLANE STRAIN CONTACT PROBLE1 z

y 2

SOLV/A E\\GINEER7NG AB SOL/IA-PRE 99.0 Version 99.0 CRIGINAL Nonlinear" Examples B57.2

SOLVIA Verification Manual Nonlinear Examples B57 ANALYIS OF HERTZ PLANE STRAIN CONTACT PROBLEM SOLVIA POST 99.0 SOLVIA ENGIi 2787.6 2363.2 1938.7 1 S 1/.

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B57.3

SOLVIA Verification Manual 5000 SOLVIA solution Corresponding Hertz solution 1000 Distance from 1

243center SOLVIA-PRE input HEADING

'B57 ANALYSIS OF HERTZ PLANE STRAIN CONTACT PROBLEM' DATABASE CREATE MASTER IDOF-I00111 NSTEP-21 ITERATION METHOD=FULL-NEWTON TOLERANCES ETOL=1.E-6 RCTOL=0.05 AUTO-STEP TMAX=21 TIMEFUNCTION 1

0.

0.

/

21.

10.

COORDINATES

/

ENTRIES NODE Y

Z 1

0.

-10.

/

2

9.

-10.

/

3 0

-10 4

3.7 -9.3

/

5

10.

0.

TO 17

5.

TO 29 30

0.

-8.

/

31 2.95 -7.44 LINE NODES 5 17

/

6 TO 16 LINE ARC 3 4 NCENTER=29 EL=12 MIDNODES=1 LINE ARC 4

5 NCENTER=29 EL=12 MIDNODES=1 RATIO=4 MATERIAL 1

ELASTIC E=3.E4 NU=0.25 EGROUP 1

GSURFACE GSURFACE GSURFACE PLANE STRAIN 3

4 31 30 EL2=6 EL3=6 NODES=8 ADDZONE=C-REGION 4

5 17 31 NODES-8 30 31 17 29 NODES-8 Version 99.0 Nonlinear Examples 1357.4

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

CGROUP 1 CONTACT2 STRAIN CONTACTSURFACE 1

INPUT=LINES ADDZONE=C-REGION CONTACTSURFACE 2

INPUT=NODES

/

2 1 CONTACTPAIR 1 TARGET-2 CONTACTOR-1 EDGE-CONTINUITY OPERATION=DELETE FIXBOUNDARIES 2

INPUT=LINES FIXBOUNDARIES 23

/

1 2 LOADS DISPLACEMENT 5

3 -0.1 TO 29 3 -0.1 29 30 3 4 30 3 MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES SMOOTHNESS=YES SOLVIA END SOLVIA-POST input B57 ANALYSIS OF HERTZ PLANE STRAIN CONTACT PROBLEM DATABASE CREATE WRITE FILENAME-'b57.1is' MESH MESH MESH OUTLINE=YES CONTOUR=MISES OUTLINE-YES CONTOUR=SYZ OUTLINE=YES CONTOUR=SPMIN MESH OUTLINE=YES CONTOUR=SMEAN MESH ZONENAME=C-REGION CONTOUR=DISPLACEMENT VECTOR=CONTACT CLIST SUMMATION END KIND=D-REACTIONS Version 99.0 Nonlinear Examples B57.5

SOLVIA Verification Manual EXAMPLE B58 CABLE AROUND A FAST PULLEY Objective To verify the 2-D contact surfaces, CONTACT2, to analyze plane stress contact conditions with friction.

Physical Problem A cable around a fast pulley, see figure, is fixed at B. An initial force F. = 20000 N is acting in the tangent direction at A and is assumed to have prestressed the cable so that no friction forces are present. The initial cable force around the fast pulley and at B is then also 20000 N. The cable force at A is gradually increased to 40000 N. The force distribution in the cable is to be calculated when the friction coefficient is p. = 0.25 and when the fast pulley is fixed.

F = cable force R=l m g = 0.25 E = 2.101 N/m 2 A = 1.10-4 M2 Analytical solution when sliding:

Feable =FA e-'

(p is measured in radians along the contact sector.

A p is zero where the contact sector starts at 0 = 10'.

Finite Element Model A plane model in the Y-Z plane is considered, see figure on page B58.2.

The cable is modeled using 41 2-node TRUSS elements. The first 40 elements form also 40 contactor segments. The left end segment between nodes 41 and 92 is in the tangent direction of the fast pulley at node 41 (0 = 170'). The end node 92 is fixed. The right end node 1 is assigned a skew system in the tangent direction of the fast pulley so that the applied force can act in the tangent direction. The prestress of the cable is modeled as an initial strain of 0.001, which balance the initial cable force (E.A

=

-E=2 1.10'.1

= 20000 N). The fast pulley is modeled as 41 rigid target segments using nodes 51-92. Each target segment is positioned so that the contactor node is initially located at the segment midpoint. Each contactor node is then initially in contact with a target segment and can move in the tangent direction of the fast trolley. This modelling technique simulates a continuously curving contact surface when the displacements are small, although the target surface is piecewise linear and forms a polygon. Each target surface node is fixed.

Version 99.0 Nonlinear Examples 1358.1

SOLVIA Verification Manual Solution Results The input data on pages B58.4 and B58.5 is used. Solution time t corresponds to the initial prestress of 20000 N of the cable. Thereafter follows 80 steps of linear increase of the cable force at A to 40000 N. The cable force distribution around the fast pulley at the solution steps 1, 10, 40 and the last step 81 are shown in the right figure below.

The analytical solution for the last step is drawn in the figure as a curve without symbols and excel lent agreement with the values calculated by SOLVIA can be seen.

As the cable force at A is increased the sliding portion of the contact surface is spreading and covers in the last step the entire contact sector of 160 degrees. The distribution of the cable force is shown as a contour diagram in the top figure of page B58.3. The maximum cable force is less than 40000 N, since it corre sponds to the midpoint of the TRUSS element.

The contact forces at the nodes of the contactor surface (zone Cl) and the target surface (zone C2) are shown for the last step in the bottom figures of page B58.3.

User Hints Contact analysis with friction is path dependent. A number of equilibrium configurations are in general possible for a given load. It is, therefore, important that the loading sequence is defined in detail and that a sufficiently small solution step is used.

The individual parts of the structure separated by contact surfaces must in general be supported so that a solution exists also without any contact forces. In this example support is provided by the initial conditions, which enforce contact due to the specified geometry. Support is also provided by the geometric stiffness of the TRUSS elements, which is included since large displacements are specified. The actual displacements are, however, small.

Version 99.0 Nonlinear Examples B58.2

SOLVIA Verification Manual 358

-ABLE AROUND A FAST PULLEY

]RTGINAL -

2 Z

TME 8ý L-y ZONE -ABLE Nonlinear Examples BS8 CABLE AROUND A FAST PUL: E g I 'r TFE at ALRCE 40000 NAXES=SKEW

'IASTER 100*i[

B 1!11.1 ORIGINAL

.2 ZONE (2 196 Is 8B/

s B

52 K Ly B I II SOLVIA ENGINEERING AB SOLVIA-POST 99.0 IS 2 3 0'

CABLE S9LVIA ENGINEERING Version 99.0 SOLVIA-PRE 99 0 3A ASB

• I L*

x <

1358.3

SOLVIA Verification Manual Nonlinear Examples BOS CABL AROUND A. 7A PULLEY MAX D-Sz TIME 81 SOLVIA-COST 99.0 SCLV/IA ENGYINEERNG AB BS8 CABLE ARCUND A FAST PULLEY MAX DISPL. t.2333E-3 Z

TIME 81 y

ZONE CI ZI C NA CT LIA ENGINEERNG A MAX DISPL.

TIME 81 ZONE C2

-4 858 CABL 0

E AROUND A FAST PULLEY

-y N

V7 C0L/'A-POST 99 0 "CCNTACT 2797 6 COLVIA ENOCNEER:NG AB Version 99.0 333E-3 z L~y REACTION 20000 FORCE-R NO AVERAG-NG MAX 39638 S384'1 S35956 33S0o 3 31046 28592 26137 23682 MLN 20000 Z

B58.4

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B58 CABLE AROUND A FAST PULLEY' DATABASE CREATE MASTER IDOF=100111 NSTEP=81 KINEMATICS DISPLACEMENT=LARGE TOLERANCES ITEMAX=200 TIMEFUNCTION 1

0. 0. /

1.

0.5

/

81.

1.

SYSTEM 1 CYLINDRICAL COORDINATES ENTRIES NODE R

THETA 1

1.

10. TO 41

1.

170.

51 1.00060955

8.

TO 92 1.00060955 172.

SKEWSYSTEM EULERANGLES

/

1

10.

NSKEWS

/

1 1

MATERIAL 1

ELASTIC E=2.E11 EGROUP 1

TRUSS ENODES ADDZONE=CABLE

/

1 1 2 TO 40 40 41 / 41 41 92 EDATA ENTRIES EL AREA INIT-STRAIN 1

i.E-4 1.E-3 TO 41 1.E-4 1.E-3 CGROUP 1

CONTACT2 STRESS CONTACTSURFACE 1 ADDZONE=C-REGION 41 TO 1

CONTACTSURFACE 2

51 TO 92 CONTACTPAIR 1 TARGET-2 CONTACTOR=1 FRICTIONCOEFFICIENT=0.25 FIXBOUNDARIES 23

/

51 TO 92 LOADS CONCENTRATED 1 3 -40000 SET PLOTORIENTATION=PORTRAIT SET VIEW=X NSYMBOLS=YES SUBFRAME 12 MESH CABLE BCODE=ALL NAXES=SKEW NNUMBERS=YES VECTOR=LOAD TIME=81 MESH C2 BCODE=ALL NNUMBERS=YES SOLVIA END Version 99.0 Nonlinear Examples B58.5

SOLVIA Verification Manual SOLVIA-POST input B58 CABLE AROUND A FAST PULLEY DATABASE CREATE WRITE FILENAME='b58.1is' TOLERANCES SET VIEW:X SET PLOTORIENTATION=PORTRAIT EPLINE CABLE 1 1 TO 41 1 AXIS 1 VMIN=0.

VMAX=3.

LABEL='CABLE' AXIS 2 VMIN=0.

VMAX=50000.

LABEL='AXIAL FORCE' ELINE CABLE KIND=FR XAXIS=1 YAXIS=2 SYMBOL=i OUT=ALL ELINE CABLE KIND=FR XAXIS:-1 YAXIS=-2 SYMBOL=2 TIME-i.

SUBF-OLD ELINE CABLE KIND=FR XAXIS=-1 YAXIS=-2 SYMBOL=3 TIME=-0.

SUBF=OLD ELINE CABLE KIND=FR XAXIS=-I YAXIS=-2 SYMBOL=4 TIME=40.

SUBF=OLD USERCURVE 1 READ

'B58FORCE.DAT' PLOT USERCURVE 1 XAXIS=-1 YAXIS=-2 SUBFRAME=OLD SET PLOTORIENTATION=LANDSCAPE BEAMPLOTWIDTH=0.4 CONTOUR AVERAGE=NO MESH CONTOUR=FR VECTOR=REACTION SET PLOTORIENTATION=PORTRAIT MESH C1 VECTOR=CONTACT MESH C2 VECTOR=CONTACT CLIST ELIST STRAINS YES NLIST EGi END Version 99.0 Nonlinear Examples B58.6

SOLVIA Verification Manual EXAMPLE B59 DYNAMIC ANALYSIS OF FRICTIONAL SLIDING OF A POINT MASS Objective To verify the use of 3-D contact surfaces, CONTACT3, to model frictional sliding.

Physical Problem The figure below shows the problem considered. A point mass, m = 0.2, is attached to one spring which is anchored to a flat rigid surface. The coefficient of friction between the mass and the rigid surface is 0.15.

The point mass is given an initial velocity of -1.0 in the X-direction and is then released from the unstressed condition. The movement of the mass takes place along the X-direction and the motion is resisted by the developed frictional force of 0.3 in the opposite direction of the velocity. An analysis is performed to evaluate the effect of friction on the vibrations of the mass-spring system considered.

I ACCELERATION DUE z*

10 TO GRAVITY = -10 m=0.2 NN.N\\NNNN\\N\\\\\\\\\\\\\\ \\\\\\\\\\

SPRING: E = 1000.0 A = 2.0 RIGID SURFACE:

i =0.15 Finite Element Model The flat rigid surface is chosen to be the target surface and is modeled using 4 nodes which are fixed.

The mass-spring system is chosen to be the contactor body. An auxiliary node 8 (fixed in space) is used to define the contactor surface consisting of a single contactor segment. The auxiliary node lies outside the region of the rigid target surface and is not in contact during the analysis. As a result, the mass node 7 is treated as a solitary node in contact. The direction of the normal vector at the solitary node is obtained from the geometry of the contactor segment. The finite element model is shown in the bottom figure on page B59.2.

Solution Results The input data on pages B59.4 and B59.5 is used in the SOLVIA analysis and the obtained numerical results are shown on page B59.3. The obtained solution is such that at a generic time t, the work done by the frictional force equals the difference between the initial energy and the current energy of the system.

For example, let A and C be two consecutive positive displacement peaks in the left top figure on page B59.3 and let B be the intermediate negative displacement peak. Then 4.u -+k

.mg(u, -u 5 )

Iku 2 -

2ku m

ku 2

c = -

ma (u Version 99.0 Nonlinear Examples B59.1

Nonlinear Examples SOLVIA Verification Manual where k

= stiffness in X-direction uA = X-displacement corresponding to point A uB

= X-displacement corresponding to point B uc

= X-displacement corresponding to point C g

= acceleration due to gravity

Then,

,4umg (uA-uc)=

t=constant k

Hence, the maximum amplitude of vibration decreases linearly in the successive oscillations.

User Hints

• The acceleration shows some discontinuity at its maximum and minimum peak values since the friction force suddenly reverses when the velocity of the mass changes its direction.

The computed acceleration is in general not zero when the point mass suddenly stops its motion at one of the discrete solution times. The friction force at rest can vary between the limit values for motion in the two directions and balances a computed residual acceleration and the spring force.

Some oscillations in the computed acceleration and velocity may, therefore, occur although the computed displacement is practically constant.

SB9 DYNANKC ANALYSIS OF FRICTIONAL SLIDING OF A POINT MASS ORIGINAL 1-5 Z

ORIGCNAL S.

X N -. y 2

N 2

3-B OS-3 AST SOLV-7A ENGi7NEERZBNG AB SOL,/ I, -PD*

E Version 99.0 ý B59.2

SOLVIA Verification Manual Nonlinear Examples Bs9 D0NAM7C ANALYS.S OF FR.CT:ONAL SLIDING OF 4 POINT MASS A)

All 0ý0 0.2

)O4 06 0'.3 1 D 2

A SOLVIA-POST 99.0 TIME SOLVIA ENG\\NEER[NG AB 859 DYNAMIC ANALYS.S OF FR0CT5ONAL SL:DANG OF A

.CTNT MASS I

00 00 I

06 CA 1

-IME T I.

SOLVIA-POST 99.0 SOLVIA ENGINEERING AB BS9 DYNAMIC ANALvSIS IF FRICTIONAL SLIDING OF A POINT MASS Il II

.1 I

F F I I

0.2

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008 lIE BS9 DYNAMIC ANALYSIS OF FRICTIONAL SLIDING OF A P07NT MASS L

I:

S.

• -2

' IE SOLVIA ENGINEERING AB SOLV

-POST 99.3 SOLVIA ENGINEERING 0B Version 99.0 SOV'IA-P3ST 99 u u <

t-,

B59.3

SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEAD

'B59 DYNAMIC ANALYSIS OF FRICTIONAL SLIDING OF A POINT MASS' DATABASE CREATE MASTER IDOF=010111 DT=5.E-3 NSTEP=250 ANALYSIS TYPE=DYNAMIC METHOD=NEWMARK DELTA=.5 ALPHA=.5 TOLERANCES ETOL=1.E-7 COORDINATES

-11 ii

/

10 0

/

2 11 7

0 0

/ /

3 11 -11 8

0

-12

/

4 11 11 INITIAL VELOCITY 7

-1 0 0 MASSES 7

.2

.2

.2 MATERIAL 1 ELASTIC E=1000.

EGROUP 1 TRUSS ENODES 1

67 EDATA

/

1

2.

CGROUP 1 CONTACT3 CONTACTSURFACE 1

/

1 7 6 8 8 CONTACTSURFACE 2

/

1 1 2 3 4 CONTACTPAIR 1

TARGET=2 CONTACTOR=1 FRICTION=0.15 FIXBOUNDARIES

/

1 TO 6

/

8 LOADS MASSPROPORTIONAL ZFACTOR=1 ACCGRA=-10.

VIEW SET MESH MESH ID=1 XVIEW=0.8 YVIEW=1.0 ZVIEW=0.7 VIEW=I NSYMBOLS=YES NNUMBERS=YES ENUMBERS=YES SUBFRAME=21 GSCALE=OLD BCODE=ALL SOLVIA END Version 99.0 1

6 B59.4

SOLVIA Verification Manual SOLVIA-POST input B59 DYNAMIC ANALYSIS OF FRICTIONAL SLIDING OF A POINT MASS DATABASE CREATE WRITE FILENAME-'b59.1is' SET PLOTORIENTATION=PORTRAIT NHISTORY NODE=7 DIRECTION=l KIND=DISPLACEMENT NHISTORY NODE=7 DIRECTION=-

KIND=VELOCITY NHISTORY NODE=7 DIRECTION=I KIND=ACCELERATION EHISTORY ELEMENT=1 POINT=1 KIND=FR CLIST NLIST NLIST KIND=VELOCITY END Version 99.0 Nonlinear Examples B59.5

SOLVIA Verification Manual EXAMPLE B60 COLLISION BETWEEN A POINT MASS AND A RIGID SURFACE Objective To verify usage of 2-D contact surfaces, CONTACT2. for analysis of dynamic contact.

Physical Problem A point mass moving with the uniform velocity -3 m/s in the Z-direction is assumed to impact with a rigid surface and rebound with no loss of energy as shown in the figure below. The analysis is per formed to obtain the numerical solution for the rebound velocity of the point mass.

MASS, m = 0.50 kg AUXILIARY NODES 3

2 VELOCITY, V = -3 m/s CONTACTOR SURFACE

\\

i/

Y TARGET SURFACE 5

4 RIGID SURFACE Finite Element Model The finite element model is shown in the top figure on page B60.2. At node 1, a concentrated nodal mass m = 0.50 kg is assigned. The mass has initial velocity of -3 m/s in the Z-direction.

The rigid surface is considered to be the target surface and is modeled using two nodes each having no degrees of freedom.

The contactor surface is defined over nodes 3, 1 and 2 where nodes 2 and 3 are auxiliary nodes with no degrees of freedom.

Solution Results The input data on pages B60.2 and B60.3 is used and the numerical solution for the rebound velocity of the point mass is 3 m/s in the Z-direction. It agrees exactly with the analytical solution. Time his tory plots of displacement. velocity and acceleration at node 1 are shown in the bottom figures on page B60.2.

User Hints The solution is performed using the time step DT = 0.05 seconds. However, for this example, identical results are obtained for the rebound velocity irrespective of the time step employed.

"* The time integration of dynamic response is performed using the Newmark method with parameters 6 = 0.5 and a. = 0.5. The energy at the impact is then preserved.

Version 99.0 Nonlinear Examples 1360.1

SOLVIA Verification Manual Nonlinear Examples 560

ý3LLIS3ON 3ETQE=N PAINT MASS AND RIGI 0

D SURFACE

/

t 0

ai 1

04 s

o06 1

o.A TIME B60 CDLL:SION BETEEN POINT MASS AND RIGID SURFACE IL 0

0, 6 0 2.

0 0o 8

TIME

//

0.,

0.3 SOLVAA ENGINEERING AM ORIGINAL -

0 5 1 AX DISPL.

I TIME 0.A SC V!A-POS 99.0 L_

/

/

S/VA NIENSA SOLVIA-PRE input HEADING

'B60 COLLISION BETWEEN POINT MASS AND RIGID SURFACE, DATABASE CREATE MASTER IDOF:100111 NSTEP-15 DT=0.05 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=NEWMARK, DELTA=0.5 ALPHA=0.5 I ENTRIES NODE Y

Z 2

1.

2.

/

3

-1.

2.

4

1.

/

5

-1.

MASSES 1

0.5 INITIAL VELOCITIES 1

0.

0.

-3.

CGROUP 1

CONTACT2 STRESS CONTACTSURFACE 1

/

3 1 2 CONTACTSURFACE 2

/

4 5 CONTACTPAIR 1

TARGET=2 CONTACTOR=1 FIXBOUNDARIES

/

2 TO 5 SOLVIA FIXBOUNDARY=NO END Version 99.0

0. 2

.4 05 r,

7M E SOLViA-OST 99.0 COORDINATES 1

0. 1. /

c*

c* ?

r,,l z

,3

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1360.2

SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input B60 COLLISION BETWEEN POINT MASS AND RIGID SURFACE DATABASE CREATE WRITE FILENAME='b60.1is' SET PLOTORIENTATION=PORTRAIT NHISTORY NODE=I DIRECTION=3 KIND=DISPLACEMENT, SYMBOL=I OUTPUT=ALL SUBFRAME=i2 NHISTORY NODE=i DIRECTION=3 KIND=VELOCITY, SYMBOL=i OUTPUT=ALL NHISTORY NODE=1 DIRECTION-3 KIND=ACCELERATION, SYMBOL=i OUTPUT=ALL SUBFRAME=i2 MESH ORIGINAL=DASHED TIME=0.4 NSYMBOLS=YES NNUMBERS=YES END Version 99.0 B60.3

SOLVIA Verification Manual EXAMPLE B61 LONGITUDINAL IMPACT OF TWO IDENTICAL BARS Objective To verify usage of the 2-D contact surfaces, CONTACT2, in the analysis of impact problems.

Physical Problem The figure below shows the problem considered. Two identical bars are moving towards each other with a uniform velocity. The analysis is to be performed to evaluate the stress generated in the bars due to the impact.

BAR NO. 1 BAR NO. 2 Bars 1 and 2 are identical "N

E=1000.0 p=0.001 Area = 1.0 Length = 100 Y

VELOCITY V VELOCITY -V Velocity in X -direction V = 1.0 Finite Element Model Due to symmetry, only the impact of a single bar with a rigid surface needs to be modeled, see the left bottom figure on page B61.2. Bar no. 1 is discretized using twenty 2-node TRUSS elements.

The rigid surface is considered to be the target surface and it is modeled using nodes 22 and 23 each having no degrees of freedom.

The impacting bar is chosen to be the contactor body. The contactor surface is defined along nodes 25, 21 and 24 where nodes 25 and 24 are auxiliary nodes with no degrees of freedom.

Solution Results The numerical solution is obtained using the input data on pages B61.4 and B61.5. The impact stress in the bar at time 0.04 and the time history of the axial stress in the element closest to the target sur face are shown in the left top figure on page B61.3. The right top figure shows the displacement and velocity time histories for the left end of the bar model. The time step is 0.01.

A second analysis denoted B6IA with a ten times smaller time step, 0.001, gives the results shown in the bottom figures of page B61.3. The period of the highest frequency mode of the free bar model is 27IL p

0.0157 40 E

The smaller time step gives about 15 solution steps per the highest frequency period, which exhausts the capacity of the model.

Version 99.0 Nonlinear Examples B61.1

SOLVIA Verification Manual Nonlinear Examples A reference solution is provided in the analysis B61B where an explicit time integration is used. The time step DT is selected so that the wave propagates one element during one time step. Since the wave velocity is c =

/p = 1000 we obtain DT = 5/c = 0.005. The explicit time integration calculates then the exact solution in this case with equal length elements. Since contact surfaces cannot be used by the program when the explicit time integration method is used the contact is modeled by fixing node 20. This allows analysis of the first portion of the response where the bar is in contact with the rigid surface. The response is shown on page B61.4 together with the input data.

The analytical solution is given, for example, in [1].

User Hints

"* The calculated solutions cover the impact duration of 2L/c = 0.2. Note that the period of the fundamental mode of the free bar is also 0.2.

• The time integration of dynamic response is performed using the Hilber-Hughes method with GAMMA = -0.3. The resulting numerical damping can be seen in particular in the left bottom figure on page B61.3.

"* The time step DT = 0.01 is chosen so that a reasonable number of solution points are obtained during the impact. The solution with the ten times smaller time step gives a somewhat sharper wave front.

"* A similar wave propagation problem is studied in example A52.

Reference

[1]

Johnson, W., Impact strength of materials, Edward Arnold Publishers, 1972.

B6*2 OaT DIAL !ýRACr OF NO B70ARS AZ~L -

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SOLVIA Verification Manual Nonlinear Examples

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a 20 JO

11) 80

.00 SOLVIA -OST 09.0 5033:3A ENGINEERING AB B61 I-3NGI-UDTAL IMPACT OF TWO I, 'TCAL BARS, 3T0.0 30

03s, 01 0s1 S0 0.'Os 0

0 :s 0,20 TIME SOLV00-POST 9903 561A L3ONITUDINAL 1'.PACT OF TWO 7OEOO.34L BARS, 0 0 003 3

203 40 03 so 300 TRUS5 B61A LONG:TOO:NAL IMPACT

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-IM 3OLV0 ENG 32RING AB A

-.3 3.3J "0 313200T 099 333 0

0.

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SLVIA E

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Version 99.0Bl.

SOLVIA ENGINEERING AS i SLVIA-0ST99.

T4 7

1.

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SOLVIA Verification Manual Nonlinear Examples 061B LONG0TUD7NAL :MPACT OF -SO IDENTICAL OARS, -XPL]C0 T 7:>E, 0o 2,.

40 20 00 T RUSS 0.20 06.5 LONG3TUDINAL IMPAC7 OF T5O :5-DT:CAL OARS, EXPL-SOLVOA ENG5NE5RN:G AS SOLVSA-POST 99 0 SOLVIA ENGINEERING AS SOLVIA-PRE input B61B

HEADING,

'B61B LONGITUDINAL IMPACT OF TWO IDENTICAL BARS, EXPLICIT' DATABASE CREATE MASTER IDOF=101111 NSTEP=40 DT=0.005 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=CENTRAL COORDINATES

/

ENTRIES NODE Y

Z 1

TO 21 100.

INITIAL VELOCITIES 1

0. 1.

TO 20

0.

• ¢ 1.

MATERIAL 1

ELASTIC E=1.E3 DENSITY=1.E-3 EGROUP 1

TRUSS ENODES

/

1 1

EDATA

/

1

1.

FIXBOUNDARIES SOLVIA END Version 99.0 2

TO 20 20 21 21 B61.4 7-4" 0 )

7 1ss 0.00 0

3*

,3 SOLVIA-POST 99.0 Z:n I

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0 0.'*

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0 0

1, 0.2

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B61 LONGITUDINAL IMPACT OF TWO IDENTICAL BARS, DT=0.01' DATABASE CREATE MASTER IDOF=-01111 NSTEP=20 DT=0.01 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=HILBER, GAMMA=-0.3 TOLERANCES RCTOL=0.01 COORDINATES

/

ENTRIES NODE Y

Z 1

TO 21 100.

/

22 100.

20.

/

23 100.

-20.

24 i00.

30.

/

25 100.

-30.

INITIAL VELOCITIES 1

0.

1.

TO 21

0.

1.

MATERIAL 1

ELASTIC E=I.E3 DENSITY=I.E-3 EGROUP ENODES EDATA 1

TRUSS

/

1 1 2 TO 20 2C 21

/

1 1.

CGROUP 1

CONTACT2 STRESS CONTACTSURFACE 1

/

25 21 24 CONTACTSURAFCE 2

/

22 23 CONTACTPAIR 1

CONTACTOR=d TARGET=2 FIXBOUNDARIES

/

22 TO 25 SET MESH MESH VIEW=X NSYMBOLS=YES NNUMBERS=YES BCODE=ALL ENUMBERS=YES GSCALE=OLD SOLVIA END SOLVIA-POST input B61 LONGITUDINAL IMPACT OF TWO IDENTICAL BARS, DT=0.01 DATABASE CREATE WRITE FILENAME='b61.1is' SET PLOTORIENTATION-PORTRAIT EPLINE NAME=TRUSS

/

1 1 TO 20 1 SUBFRAME 12 ELINE LINENAME-TRUSS KIND-SRR TIME=0.04 SYMBOL=1 OUTPUT=ALL EHISTORY ELEMENT=20 POINT=1 KIND=SRR SYMBOL=1 OUTPUT=ALL SUBFRAME 12 NHISTORY NODE=1 DIRECTION=2 KIND=DISPLACEMENT SYMBOL=1 NHISTORY NODE=i DIRECTION=2 KIND=VELOCITY SYMBOL=1 END Version 99.0 Nonlinear Examples B61.5

SOLVIA Verification Manual EXAMPLE B62 ANALYSIS OF HERTZ DYNAMIC CONTACT PROBLEM Objective To demonstrate the use of 3-D contact, CONTACT3, surfaces for analysis of dynamic contact.

Physical Problem The figure below shows the problem considered. A sphere moving with a uniform initial velocity of -3 in the Z-direction is considered to impact a flat rigid surface. The initial development of the contact region is to be calculated.

Radius = 5.0 Material properties E = 1000.0 Ceontactor surface V = 0.30 Density p = 0.01 z t Initial velocity v, = 3.0

¥....

riction ýE=

Rigid target surface Finite Element Model Due to symmetry of the sphere geometry, only a thin wedge from the sphere continuum is discretized (see figures on page B62.3). The wedge is bounded by its two semicircular sides and the enclosed wedge angle is 10 degrees. The wedge is discretized using 8-node SOLID elements. For each node on Face 1 of the wedge, the degree of freedom normal to Face 1 (i.e., along the skew coordinate direction a) is deleted. Similarly, for each node on Face 2 of the wedge, the degree of freedom normal to Face 2 (global X-direction) is deleted.

The contactor surface is defined over the lower portion of the wedge boundary. The target surface is defined to be the flat rigid surface with corner nodes 11, 12, 13 and 14. The target nodes are specified to have no degrees of freedom. The time step is set to 0.01 and Hilber-Hughes time integration method with GAMMA = -0.3 is used in the time integration.

Solution Results The input data on pages B62.8 and B162.9 is used for the 3D analysis. The calculated normal contact pressure at time 0.05 is shown in the left top figure on page B62.4. For comparison, the corresponding analytical Hertz solution assuming static conditions is drawn in the same figure.

The distribution of velocity, the a,, stress component and the contact forces at the same time 0.05 are shown in the bottom figures of page B62.4. All these results are calculated under the assumption of small displacements.

Version 99.0 Nonlinear Examples B62.1

SOLVIA Verification Manual A large displacement analysis B62A of the same model gives the contact pressure distribution at time 0.05 as shown in the right top figure of page B62.4. It can be seen that the effect of the assumption of large displacements is significant for the model. Result figures are also shown on page B62.5.

For comparison, a 2D axisymmetric model using CONTACT2 surfaces and PLANE AXISYMMETRIC elements is developed, see bottom figure on page B62.7.

The input data on pages B62.9 and B62.10 was used for the axisymmetric analysis B62B (small displacement assumption). The corresponding results are shown in the bottom figures on page B62.5.

An analysis B62C was also performed with the same input except that large displacements were specified and the analysis was extended to time 0.5 (50 steps). The results for time steps 0.05 and 0.5 are shown on pages B62.6 and B62.7. The velocity and the acceleration time histories are shown for the nodes 1, 4, 5 and 3, which lie on the surface of the sphere.

Node 1 is at the center of the contact area and shows significant oscillations in both the velocity and the acceleration due to the sudden start and end of the impact period. Node 5, which is located on the sphere surface and at a distance of one radius from the contact plane shows the smoothest curves. It can be seen that the velocity reverses from -3 before the impact to an average of +3 after the impact.

The 2D contact pressures at time 0.05 are also drawn in the top figures of page B62.4 and can be seen to almost coincide with the 3D curves. A reason for the differences is that the outer surface of the 3D model is composed of flat surfaces while the 2D model is truly axisymmetric.

User Hints The 3D continuum sound velocity in the sphere is cp E(l - V)

=366.9 Cp (I + v)(1 - 2v)p -

6.

For comparison, the unconstrained ID sound velocity is cI= j-=316.2 An estimate of the period of the smallest natural frequency of the free sphere is then 4R T

= 0.0545 since the wave corresponding to this fundamental mode shape must travel across the sphere to the opposite free surface and then back again, giving a travel distance of 4R. The actual period computed using the free axisymmetric model is 0.0605.

The impact duration can be seen from the figures on page B62.7 to be of the order 0.2. Hence, the impact duration is almost 3 times the period of the fundamental mode of the free model.

Version 99.0 Nonlinear Examples B62.2

SOLVIA Verification Manual When a point on the sphere surface suddenly comes in contact with the rigid target surface, waves of various types are generated. The amplitude of the pressure wave is decreasing as 1 / x2, where x is the distance from the contact point, since the propagation is approximately spherical. There fore, we have geometrical damping of the pressure wave in this example in contrast to the uniax ial case in Example B61 where one sharp wave front is the analytical solution. The significant geometrical damping contributes to a prolonged contact duration compared to a uniaxial case of the same length.

Most of the kinetic energy of the sphere is transferred to strain energy in the lowest modes during the impact. After the impact the sphere vibrates as a free body but the amplitude is small, see the lower right figure on page B62.6. Almost all the energy is transferred back to kinetic energy.

Reference

[1]

Timoshenko, S.P. and Goodier, J.N., Theory of Elasticity, 2nd ed., McGraw-Hill, 1951, pp. 372-377.

z B62 ANALYS:S OF HERTZ DYNAMIC CONTACT PROBLEM ORIGINAL 1

x Y

Contcctor surface Skew coordinate system no 1 a

Face no 1 b

j ni 0

• '*Angie =100 Face no 2 X

Version 99.0 SOL/ A-RE 99.0 NAXES=SKE' SOLVrA -NG:NEERING AB Nonlinear Examples B62.3

SOLVIA Verification Manual Total contact force: 180.2 (SOLID)

Contact pressure 0.2 0.4 U.>

u.0 I.i Small displacement analysis Distance from center Nonlinear Examples Total contact force: 165.5 (SOLID)

Distance from center Large displacement analysis 062 ANALYSIS OF HERTZ DYNAMIC CONTAGT PROBLEM MAX DISP0L.

C !4E5 TVME 0 05 C

I 1 X

VELOCITY

/.

A.

MX 2.4874 S2.3320 "2.0210

.x"

/

'.7101

"" 3992

• ..~

.//!,0882 "0

77732 0 46639

""\\',*

"/*,*0.!

5546 "MIN 0

S0LV A-POST 99.0 SOLVIA ENOINEERING AB 862 ANALYSIS OF HERTZ DYNAMIC CONTACT PROBLEM MAX D0SPL.

0. 0705 z

,'ME C.05 ZONE C-VOLUME X

Y SOLVA-POST 99.3 STRESS-ZZ TAO.

'243 1-65.925

-22.627

-18.5661

-54 695

-70.729

'-86 763 C-102 80

-118 83 TIN-126 85 COLVIA ENGINEERING AB Version 99.0 100 1362.4

1 0

N) ccZ cmt c.a m

> c c

I c I x

/11 a

cc cc/i. <7

cc-

t1cvcmtu ccc2

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cm cc cc-cccmcmacmccmama

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x 0

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0 0

z 0

0 0

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362C ANALYSIS OF HERTZ DYNAMIC CONTACT PROBLEM MAX 37SL C

449:

T-NE 005 ZONE EGI Nonlinear Examples Z L y R

362C ANALYS0 S OF ERTZ DYNAMIC CONTACT 90OBLE3 MAX

, SPL.

.9 91806 TIME 0,5 ZONE EGI SOLVIA-POST 99.0 362C ANALYSIS OF HERTZ DYNAMIC CONTACT PROBLEM MAX D0SPL. *--3.C971 IMN_ 0 05 ZCNE C /OLUME d

R CONTAC7

~3 3,496 STRESS-ZZ MAX 4,6637

-27476 27 670

-32 393

-47 214

-62.03

-76 B6

-91,683 MIN-113.92 SOLVIA E.NG7NE7RNG AB R

VEL CC-Y MAX 3.0642 3.OS23 3.0284 2.9806 2 29567 2.9328 2 9089 MIN 2 8731 SOLVIA ENGINEERING A8 362C ANALYSIS OF HERTZ DYNAMIC CONTACT PROBLEM MAX DSPL. 3 91802 Z

TZ0ME 0O, L Y ZONE C-VOLUME R

SOLC A-POST 99.0 C3rNTACTI STRES-77 MAX 0.017477 SC.313442 9 312301 0 3t10231 S8.1606E-3

' I901E-3

,,1

.397E-3; MIN 9 '412E-4' SCLV A ENGINEERING AB B62.6 SOLVIA Verification Manual Version 99.0

SOLVIA Verification Manual B62C A1,0 TOS OF HERTZ DYNAýIC COTACT PRCLE" N

z 22 0 0.2316 D

.2.

C.

TIME T:11E 0,0 12 04 0ýD0 0.2 04 2

SOLVIA-PCST 99.0 SOLVA oNGIO'EERING AB Nonlinear Examples B62C ANALYSIS OF HERTZ LONAMIC SOT ACT 2909 1 TIME 7 ME 21 2

I II!ME IT THE SOLTIA-POST 99.0 SOLV:A EaSLINEER9NG 29 362C ANAILaSIS OF HERTZ 0YNAMIC CCTA7 PROBLEM ORIGINAL I.

Z L

SOL9.II ENGINEERING AB Version 99.0B2.

SO020910 99.3 B62.7

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B62 ANALYSIS OF HERTZ DYNAMIC CONTACT PROBLEM' DATABASE CREATE MASTER IDOF=-00111 NSTEP=5 DT-0.01 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=HILBER, GAMMA=-0.3 SET PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES NODES-8 SYSTEM 1

SPHERICAL COORDINATES SYSTEM=i

/

ENTRIES NODE R

THETA PHI 1

5.

90.

180.

/

2

0.

90.

0.

/

3

5.

90.

0.

4

5.

90.

165.

/

5

5.

100.

165.

/

6

5.

90.

90.

7

5.

100.

90.

COORDINATES SYSTEM=0

/

ENTRIES NODE X

Y Z

11

3.

3.

-5.

/

12 -3.

3.

-5.

13 -3.

-0.5

-5.

/

14

3.

-0.5

-5.

LINE STRAIGHT 1 2 ELUi5 RATIO=12.

LINE ARC 4 6 NCENTER=2 EL=25 RATIO=5 MATERIAL 1

ELASTIC E=1.E3 NU=0.3 DENSITY=1.E-2 EGROUP 1

SOLID DEGEN=1 GVOLUME 2 2 2 2 1 4 5 1

EL1=15 EL2=1 EL3=15 SYSTEM=i GVOLUME 2 2 2 2 4 6 7 5

EL1=25 EL2=1 EL3=15 SYSTEM=1 GVOLUME 2 2 2 2 3 7 6 3

EL1=15 EL2=1 EL3-15 SYSTEM=1 SKEWSYSTEM EULERANGLES

/

1 90 10 -90 ZONE NAME=SLICE INPUT=SPHERICAL-LIMITS SYSTEM=1, THETAMIN-91.

THETAMAX=-01.

NSKEWS INPUT=ZONE SLICE 1 ZONE NAME=C-VOLUME INPUT=SPHERICAL-LIMITS SYSTEM=-,

RMIN=4.

PHIMIN=i65.

PHIMAX=i80.

FIXBOUNDARIES 2

INPUThLINES

/

1 2

/

2 3

FIXBOUNDARIES 23 INPUT=NODES

/

11 TO 14 INITIAL VELOCITIES

/

1

0.

0.

-3.

TO 1744

0.

0.

-3.

CGROUP 1

CONTACT3 CONTACTSURFACE 1

INPUT=SURFACE ADDZONE=C-VOLUME

/

1 4 5 1 CONTACTSURFACE 2

INPUT=NODES

/

1 ii 12 13 14 CONTACTPAIR 1

CONTACTOR=1 TARGET=2 MESH NNUMBERS=MYNODES NAXES=SKEW MEMORY SOLVIA=-0 SOLVIA END Version 99.0 Nonlinear Examples 1362.8

SOLVIA Verification Manual SOLVIA-POST input B62 ANALYSIS OF HERTZ DYNAMIC CONTACT PROBLEM DATABASE CREATE WRITE FILENAME='b62.1is' SET PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES MESH CONTOUR=VELOCITY OUTLINE=YES MESH ZONENAME=C-VOLUME CONTOUR=SZZ VECTOR=CONTACT CLIST END SOLVIA-PRE input B62B HEADING

'B62B ANALYSIS OF HERTZ DYNAMIC CONTACT PROBLEM' DATABASE CREATE MASTER IDOF=100111 NSTEP=5 DT=0.01 ANALYSIS TYPE-DYNAMIC MASSMATRIX=LUMPED METHOD=HILBER, GAMMA=-0.3 SET PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES NODES=4 SYSTEM 1

CYLINDRICAL COORDINATES SYSTEM=i

/

ENTRIES NODE R

THETA 1

5.

-90.

/

2

0.

0.

/

3

5.

90.

4

5.

-75.

/

5

5.

0.

COORDINATES SYSTEM=-

/

ENTRIES NODE Y

Z 11

0.

-5.

/

12

3.

-5.

LINE STRAIGHT 1 2 EL=15 RATIO=12.

LINE ARC 4 5 NCENTER=2 EL=25 RATIO=5 MATERIAL 1

ELASTIC E=1.E3 NU=0.3 DENSITY=U.E-2 EGROUP 1

PLANE AXISYMMETRIC DEGEN=1 GSURFACE 2 1 4 2 EL1=15 EL2=15 SYSTEM=i GSURFACE 2 4 5 2 EL1=15 EL2=25 SYSTEM=i GSURFACE 2 5 3 2 EL1=15 EL2=15 SYSTEM=i ZONE NAME=C-VOLUME INPUT-CYLINDRICAL-LIMITS SYSTEM=1, RMIN=4.

THETAMIN=-90.

THETAMAX=-75.

FIXBOUNDARIES 2

INPUT=LINES

/

1 2

/

2 3 FIXBOUNDARIES 23 INPUT=NODES

/

11 12 INITIAL VELOCITIES

/

1

0.

0.

-3.

TO 936

0.

0.

-3.

CGROUP 1

CONTACT2 AXISYMMETRIC CONTACTSURFACE 1

INPUT-LINE ADDZONE=C-VOLUME

/

1 4 CONTACTSURFACE 2

INPUT=NODE

/

12 ii CONTACTPAIR 1

CONTACTOR=1 TARGET=2 MESH NNUMBERS=MYNODES SOLVIA END Version 99.0 Nonlinear Examples 1362.9

SOLVIA Verification Manual SOLVIA-POST input B62C B62C ANALYSIS OF HERTZ DYNAMIC CONTACT PROBLEM DATABASE CREATE WRITE FILENAME='b62c.lis' SET PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES MESH EGI CONTOUR=VELOCITY OUTLINE=YES TIME=0.05 MESH EGI CONTOUR=VELOCITY OUTLINE=YES TIME=0.50 MESH ZONENAME=C-VOLUME CONTOUR=SZZ VECTOR=CONTACT TIME=0.05 MESH ZONENAME=C-VOLUME CONTOUR=SZZ VECTOR=CONTACT TIME=0.50 SUBFRAME 22 NHISTORY 1 DIR=3 KIND=VEL NHISTORY 4 DIR=3 KIND=VEL NHISTORY 5 DIR=3 KIND=VEL NHISTORY 3 DIR=3 KIND=VEL SUBFRAME 22 NHISTORY 1 DIR=3 KIND=ACC NHISTORY 4 DIR=3 KIND=ACC NHISTORY 5 DIR-3 KIND=ACC NHISTORY 3 DIR=3 KIND=ACC CLIST TSTART=0.05 END Version 99.0 Nonlinear Examples B62.10

SOLVIA Verification Manual EXAMPLE B63 ANALYSIS OF ELASTIC CREEP - CREEP LAW NO. 1 Objective To verify the TRUSS and PLANE elements in creep analysis using the Bailey-Norton creep law.

Physical Problem The figure below shows the bar considered. It is subjected to a constant longitudinal force P = 7500 and the creep response of the bar is analyzed.

L A

L A=1.0 L=10.0 E=30.0.106 v=0.0 Bailey-Norton creep law:

ec = ao (a, t12 a.=6.40.10-",a 1 =3.15, a2 =1.0 Finite Element Model Two separate models of the bar as shown in the left bottom figure on page B63.2 are considered. The first model consists of a single TRUSS element and the second model consists of a single PLANE STRESS element. The bar material creep properties are modeled using creep law no. 1 [1].

Solution Results The input data on pages B63.3 and B63.4 is used. The numerical solution obtained for the total strains as a function of time is given in the right bottom figure on page B63.2. The time step is DT = 100. The numerical solution is in excellent agreement with the analytical solution which is given by e

p 315 p

A EA = 1.02949-10-5 t +2.5-.10-The first term represents the creep strain and is seen to increase linearly with time since the stress is constant.

The Bailey-Norton creep law is further discussed in [2], page 20.

User Hints 0 The time integration of creep response is performed using Euler backward method (cc = 1). This method is unconditionally stable and thus a larger value of time step DT can also be employed.

The selection of the time step size needs to consider the accuracy of the solution and the conver gence in the iterations.

Version 99.0 Nonlinear Examples B63.1

SOLVIA Verification Manual Nonlinear Examples

" The time integration of creep response can also be performed using Euler forward method (cc = 0).

This method is conditionally stable and thus a proper selection of a time step, which is less than (or equal to) the critical time step limit, is important.

In general, the time step needs to be selected so that the maximum creep strain increment is smaller than (1/a,)th, or conservatively, one tenth of the cun'ent elastic strain.

" Initial displacements are used to model a step load application at time 0. The initial displacements correspond to a stress field that balances the applied force. Without initial displacements a ramp function from 0 up to the force value at the first solution step would be used. The resulting creep strain would be affected.

References

[1]

"SOLVIA-PRE 99.0, Users Manual, Stress Analysis", Report SE 99-1.

[2]

Kraus, H., Creep Analysis, John Wiley & Sons, Inc., 1980.

B53 ANALYS.S OF ELASTIC CREEP -

CREEP LAW NO OR7GINAL 1

TrNE 100 z

C

_6 C33 I

S

• LV A-RE 99 FRC 7001 ao 1 0

101 O III

.1 SOLVI/A ENGINEERING AB B63 ANALYSIS OF ELASTIC CREEP -

CREEP LAW NO t

./r 200 400 6ý0 800 :000

-TME.

00 0

200

`00 600 000 1000 COLVIA-POST 99.0 SOL VIA ENGINEERING AR Version 99.0

,m

a.

oJ m

o B63.2

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B63 ANALYSIS OF ELASTIC CREEP CREEP LAW NO 1' DATABASE CREATE MASTER IDOFI00111 NSTEP=10 DT-100 ITERATION METHOD=BFGS COORDINATES

/

ENTRIES NODE Y

Z 1

/

2 10

/

3 0 -5

/

4 0 -6 5

10 -6

/

6 10 -5 INITIAL DISPLACEMENT 2

0.

2.5E-3 5

0.

2.5E-3 6

0.

2.5E-3 MATERIAL 1

PLASTIC-CREEP XKCRP=1 ALPHA=1, AO6.4E-18 A1=3.15 A2-1 0

3E7 0

3E4 0

0 100 3E7 0

3E4 0

0 EGROUP 1

TRUSS ENODES

/

1 1 2 EDATA

/

1

1.

EGROUP 2

PLANE STRESS2 ENODES

/

1 3 4 5 6 EDATA

/

1

1.

FIXBOUNDARIES 2

/

1 3 4 FIXBOUNDARIES 3

/

1 2 4 5 LOADS CONCENTRATED 2 2 7500 5 2 3750 6

2 3750 LOADS TEMPERATURE TREFE50.

SET NNUMBERS=YES NSYMBOLS=YES PLOTORIENTATION=PORTRAIT MESH VIEW=X ENUMBERS=GROUP BCODE=ALL VECTOR-LOAD SOLVIA END Version 99.0 Nonlinear Examples B63.3

SOLVIA Verification Manual SOLVIA-POST input B63 ANALYSIS OF ELASTIC CREEP -

CREEP LAW NO 1 DATABASE CREATE WRITE FILENAME='b63.1is' SET PLOTORIENTATION=PORTRAIT SUBFRAME 12 EHISTORY EL=1 POINT=I KIND=ERR SYMBOL=i OUTPUT=ALL EGROUP 2

EHISTORY EL=i POINT=I KIND=EYY SYMBOL=i OUTPUT=ALL ELIST STRAINS-YES END Version 99.0 Nonlinear Examples B63.4

SOLVIA Verification Manual EXAMPLE B64 ANALYSIS OF ELASTIC CREEP - CREEP LAW NO. 2 Objective To verify the TRUSS and PLANE elements in the analysis of primary and secondary creep.

Physical Problem Same as in Example B63 except for the creep data, see figure below.

tu

-,- t p A

L A=1.0 L=10.0 E=30.0-10 6 v=0.0 Creep law no. 2:

e, =F(1-e-e R)+Gt Fza 0eaG' R=a2 j-G=a5ea Sa3 )

a0 =3.476.10-4 a,=2.08.10-4 a2 =3.991.10-s a3 =l.00.103 a4 =2.094 a,=l.02.10-"

a6 =7.430,104 a 7 =0.0 Finite Element Model Same as in Example B63. The bar material creep properties are modeled using creep law no. 2 [1]

with constants as shown in figure above.

Solution Results The input data on pages B64.3 and B64.4 is used. The numerical solution obtained for the total strain as a function of time is given in the figure on page B64.2. The time step is DT = 250. The numerical solution is in excellent agreement with the analytical solution which is given by te = 1.654-10-' (i 271310-t) +2.684-9 t+25010 6 The first term respresents the primary creep and the second term the secondary creep. The last term is the static strain due to the applied force.

User Hints The time integration of creep response is performed using Euler backward method (a = 1). This method is unconditionally stable and thus a larger value of time step DT can also be employed.

The selection of the time step size needs to consider the accuracy of the solution and the conver2ence in the iterations.

Version 99.0 Nonlinear Examples B64.1

SOLVIA Verification Manual Nonlinear Examples

" The time integration of creep response can also be performed using Euler forward method (a = 0).

This method is conditionally stable and thus a proper selection of a time step, which is less than (or equal to) the critical time step limit, is important. In general, the time step is selected so that the maximum creep strain increment is of the order one tenth of the current elastic strain.

" Initial displacements are used to model a step load application at time 0. The initial displacements correspond to a stress field that balances the applied force. Without initial displacements a ramp function from 0 up to the force value at the first solution step would be used. The resulting creep strain would be affected.

Reference

[1]

"SOLVIA-PRE 99.0, Users Manual, Stress Analysis", Report SE 99-1.

B6,1 ANALYSS OF ELASTTC NJ CREEP T

C f CREEP LAW NO 2 c

9 F

(P 00 0

300 0

CSO0 SCL',_A-POS-99.C S

000 2000 0 C, 4000 5000 SOLVIA ENCiNEErPNC AB Version 99.0 C

z

'0 0

i i

C I(p C-- -I

- -4

 K C-

-9 NJ

'9 B64.2

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B64 ANALYSIS OF ELASTIC CREEP CREEP LAW NO 2' DATABASE CREATE MASTER IDOF=I00111 NSTEP=20 DT=250 ITERATION METHOD=BFGS TOLERANCE TYPE=F RNORM=1000 RTOL=0.001 COORDINATES

/

ENTRIES NODE Y

Z 1 /

2 10

/

3 0

-5

/

4 0 -6 5

10 -6

/

6 10 -5 INITIAL DISPLACEMENT 2

0.

2.5E-3 5

0.

2.5E-3 6

0.

2.5E-3 MATERIAL 1

PLASTIC-CREEP XKCRP=2 ALPHA=-,

A0=3.476E-4 A1=2.08E-4 A2-3.991E-5 A3=1.E3, A4=2.094 A5=1.02E-11 A6=7.43E-4 0

3E7 0

3E4 0

0 100 3E7 0

3E4 0

0 EGROUP 1

TRUSS ENODES

/

1 1 2 EDATA

/

1

1.

EGROUP 2

PLANE STRESS2 ENODES

/

1 3 4 5 6 EDATA

/

1

1.

FIXBOUNDARIES 2

/

1 3 4 FIXBOUNDARIES 3

/

1 2 4 5 LOADS CONCENTRATED 2

2 7500 5 2 3750 6 2 3750 LOADS TEMPERATURE TREF=50.

SOLVIA END Version 99.0 Nonlinear Examples 1364.3

SOLVIA Verification Manual SOLVIA-POST input B64 ANALYSIS OF ELASTIC CREEP CREEP LAW NO 2 DATABASE CREATE WRITE FILENAME='b64.1is' SUBFRAME 21 EHISTORY EL=i POINT=i KIND-ERR SYMBOL=i OUTPUT=ALL EGROUP 2

EHISTORY EL=i POINT=i KIND=EYY SYMBOL=i OUTPUT=ALL ELIST STRAIN=YES END Version 99.0 Nonlinear Examples B64.4

SOLVIA Verification Manual EXAMPLE B65 LARGE DISPLACEMENT ANALYSIS OF A SPHERICAL SHELL Objective To verify the PLANE AXISYMMETRIC element in large displacements using the AUTOMATIC ITERATION method.

Physical Problem A spherical shell subjected to a concentrated apex load is considered as shown in the figure below.

This problem is also analyzed in Example B 12.

h = 0.01576 in H

0.0859 in R =4.76 in 0= 10.90 E

1.0-10 7 psi v=0.3 z

y Finite Element Model Ten 8-node PLANE AXISYMMETRIC elements are used as shown in the figure on page B65.2. The nodes at the apex can slide in the vertical (Z) direction while the boundary conditions at the other end are defined using skew system directions. The top and bottom nodes can slide in the thickness direc tion while the midnode is fixed.

In the first step, the Z-displacement at the apex of the shell is prescribed to be -0.01 in. The maximum apex displacement of interest is -0.175 in.

Solution Results The input data shown on pages B65.4 and B65.5 is used in the SOLVIA analysis. The numerical solution obtained for the applied load versus apex displacement agrees with the solutions of refs. [I]

and [2]. The deformed mesh for load step 14 and central deflection as function of the load multiplier are shown in the figures on page B65.3. The radial stress. STRESS-RR, closest to the fixed boundary as a function of the load multiplier is also shown on page B65.3.

Version 99.0 Nonlinear Examples B65.1

SOLVIA Verification Manual Nonlinear Examples User Hints

• The summation of reaction forces is always performed in the global coordinate system. The listing of the reaction forces is made in global or skew system directions depending on the value of the SET SKEW parameter.

"* The AUTOMATIC-ITERATION method always uses Full-Newton iterations without line-search when performing equilibrium iterations.

References

[1]

Stricklin, J.A., "Geometrically Nonlinear Static and Dynamic Analysis of Shells of Revolution", High Speed Computing of Elastic Structures, Proceedings of the Symposium of IUTAM, Univ. of Liege, August, 1970.

[2]

Mescall, J.F., "Large Deflections of Spherical Shells Under Concentrated Loads",

J. App. Mech., Vol. 32, pp. 936-938, 1965.

I6S A.RGE DISPLACE ENT ANALYSIS OF A SPHERICAL SHELL ORIGINAL 0.C TINE Z LY R

FORCE r

EAXES=

STRESS-RS7 ORIGN\\AL 0

O.

ZONE L I NAXES=SKE ý Si*

MASTERý

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'B65 LARGE DISPLACEMENT ANALYSIS OF A SPHERICAL SHELL' DATABASE CREATE MASTER IDOF=d00111 NSTEP=20 KINEMATICS DISPLACEMENT=LARGE AUTOMATIC-ITERATION NODE=i DIRECTION=3 DISPLACEMENT=-0.01, DISPMAX=0.175 CONTINUATION=YES SYSTEM 1

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4.76788

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2 4.75212

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3 4.75212 79.10 4

4.76 79.10 5

4.76788 79.10 SKEWSYSTEM EULERANGLES 1

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ELASTIC E=1.E7 NU=0.3 EGROUP 1

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EL1=10 EL2=1 NODES=8 SYSTEM=1 FIXBOUNDARIES FIXBOUNDARIES 2 INPUT=LINES /

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3 5 INPUT=NODES /

4 LOADS CONCENTRATED 1 3 -1 SET MESH MESH NSYMBOLS=MYNODES SMOOTHNESS=YES EAXES=STRESS-RST VECTOR=LOAD NNUMBERS=MYNODES SUBFRAME=12 ZONENAME=ELl BCODE=ALL ENUMBERS=YES NAXES=SKEW SOLVIA END Version 99.0 Nonlinear Examples B65.4

SOLVIA Verification Manual SOLVIA-POST input B65 LARGE DISPLACEMENT ANALYSIS OF A SPHERICAL SHELL DATABASE CREATE STRESSREFERENCE=ELEMENT WRITE FILENAME='b65.1is' TOLERANCES SET NSYMBOLS=MYNODES MESH ORIGINAL=DASHED SMOOTHNESS=YES VECTOR=LOAD SUBFRAME 21 NHISTORY NODE 1 DIRECTION=3 XVARIABLE=LAMBDA SYMBOL=i OUTPUT=ALL EHISTORY ELEMENT=1 POINT=7 KIND=SRR XVARIABLE=LAMBDA SYMBOL=2 OUTPUT:ALL NLIST KIND=REACTION SUMMATION KIND=LOAD SUMMATION KIND=REACTION DETAILS-YES END Version 99.0 Nonlinear Examples B65.5

SOLVIA Verification Manual EXAMPLE B66 ANALYSIS OF ONE-DIMENSIONAL SEEPAGE FLOW Objective To verify the seepage model for analysis of unidirectional seepage flow using the PLANE and SOLID elements.

Physical Problem The figure below shows the problem considered. Three layers, each of different permeability, are placed on top of each other. At the top face (Face A), the total seepage head is prescribed to equal 20.0 m and at the bottom face (Face B), the total seepage head is prescribed to equal 8.0 m.

FACE A TOTAL HEAD (PRESCRIBED) = 20.0 m LAYER 1 I

k, =2.10-5 m/s DIRECTION OF SEEPAGE FLOW LAYER 72?

k,=1 101' mn/s z

LY LAYER 3 k3=05.10-' m/sI

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T 4.0 m 4.0 m 4.0 m I

\\1 FACE B TOTAL HEAD (PRESCRIBED) = 8.0 m Finite Element Model Two separate models are considered (see figure on page B66.2):

Model 1:

The PLANE conduction elements are used to discretize the domain of interest. Each layer is represented by one 4-node element.

Model 2:

The SOLID conduction elements are used to discretize the domain of interest. Each layer is represented by one 8-node element.

Solution Results The input data on pages B66.3 and B66.4 is used and the numerical solution obtained for the seepage head agrees exactly with the analytical solution.

Version 99.0 Nonlinear Examples 1366.1

SOLVIA Verification Manual Darcy's law where dz k = permeability

ý = total seepage head q = velocity of seepage flow gives the continuity equation

-2.0 20-1.0

. 0 2 =1--

_0.5 O2--8 4

4 4

where 01 and ¢ 2 are the total seepage heads at Z = 8.0 m and 4.0 m, respectively.

The solution is 4~

= 128/7 = 18.2857 and p2 104/7 = 14.8571 and q = -6/7 = -0.857143.

The maximum pressure is at level z = 4.0 m where Pma =10000 17--4 4=108571Pa The minimum pressure occurs at both the faces where Pmin = 80000 Pa A vector plot of the flux is shown below and contour plots of the total head and the pressure are shown in the top figures on page B66.3.

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'B66 ANALYSIS OF ONE-DIMENSIONAL SEEPAGE FLOW' DATABASE CREATE COORDINATES 1

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0

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EGROUP 4

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SOLID MATERIAL=2 ENODES

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YVIEW=.5 ZVIEW=.5 MESH VIEW=i NNUMBERS=YES NSYMBOLS=YES ENUMBERS=YES SOLVIA-TEMP END SOLVIA-POST input B66 ANALYSIS OF ONE-DIMENSIONAL SEEPAGE FLOW T-DATABASE CREATE WRITE FILENAME='b66.1is' SET PLOTORIENTATION=PORTRAIT VIEW ID=1 XVIEW=1.

YVIEW=.5 ZVIEW=.5 SET VIEW=i MESH VECTOR=TFLUX MESH CONTOUR-TEMPERATURE MESH CONTOUR=TPRESSURE ELIST END Version 99.0 Nonlinear Examples B66.4

SOLVIA Verification Manual EXAMPLE B67 UNCONFINED SEEPAGE FLOW THROUGH A RECTANGULAR DAM Objective To perform an unconfined seepage flow analysis using the PLANE conduction elements.

Physical Problem An analysis of steady-state free surface seepage through the rectangular dam shown in the figure below is considered. The upstream and downstream water levels are maintained at 16 and 0 feet, respectively, and isotropic conditions with a constant permeability are assumed.

Water level Permeability k = t ft/hr Weight density y = 0.1 16' Dam Water lever V

ImpermeabLe Finite Element Model The rectangular dam is discretized using a 24 x 24 mesh of 2-D conduction elements, see the figure on page B67.2. The boundary conditions of the total head or potential are also shown.

Solution Results The input data on pages B67.5 and B67.6 has been used in the SOLVIA-TEMP analysis. The right bottom figure on page B67.2 shows the position of the free surface reported in refs. [1,2]. Good agreement with the results of this analysis can be observed. A contour plot of the pressure and a vector plot of the nodal flow and the element flux as displayed by SOLVIA-POST are shown in the right top figure on page B67.2.

The pressure and the flux along vertical lines at the upstream, mid-and downstream side of the dam are shown on pages B67.3 and B67.4. On page B67.4 the results are also shown along a horizontal line at the base of the dam.

References

[1]

Herbert, R., "Time variant ground water flow by resistance network analogies", J. of Hydrol.,

Vol. 6, pp. 237-264, 1968.

[2]

France, P.W., Parekh, C.J., Peters, J.C., and Taylor, C., "Numerical analysis of free surface seepage problems", Proceedings of the American Society of Civil Engineers, J. Irrigation and Drainage Division, Vol. 97. pp. 165-179, 1971.

Version 99.0 Nonlinear Examples B67.1

SOLVIA Verification Manual Nonlinear Examples 367 UNCONF94NE SEEPAGE

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