Text

Attachment to Response to NRC Request for Additional Information Re Containment Structure Conformance to Design Basis Requirements. Solvia Verification Manual, Pages A30.1 - A56.4
	ML022030370
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Site:	Cook
Issue date:	07/16/2002
From:	Greenlee S; Indiana Michigan Power Co
To:	; Document Control Desk, Office of Nuclear Reactor Regulation
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	AEP:NRC:2520, TAC MB3603, TAC MB3604
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SOLVIA Verification Manual EXAMPLE A30 SCORDELIS-LO CYLINDRICAL ROOF, CUBIC SHELL Objective To verify the curved isoparametric SHELL element when subjected to gravity loading.

Physical Problem A cylindrical shell roof subjected to gravity loading is considered, see figure below. The shell roof is supported on diaphragms at the ends and it is free along the longitudinal sides.

R = 300 in E = 3.0. 106 psi L = 600 in v = 0.0 h = 3.0 in (thickness)

= 40° Shell weight = 90.0 lb/sq-ft diaphragm X

Finite Element Model Due to symmetry, only one quarter of the cylindrical shell roof needs to be considered. The part A-B C-D in the figure above is modeled using two cubic isoparametric SHELL elements, see figures on page A30.3. Symmetrical boundary conditions are applied along the two sides defined by nodal points between nodes 4 and 1 (line 4-1) and between nodes 3 and 4 (line 3-4). The nodes corresponding to the diaphragm side (line 1-2) are fixed for translation in the Y-and Z-directions.

Solution Results This example problem has been used extensively as a benchmark problem for shell elements. The analytical shallow shell solution generally quoted for the vertical deflection at the centre of the free edge (point B in the figure above) is -3.703 inches [1] although some authors use -3.696 inches. A deep shell exact analytical solution quoted is -3.53 inches. Input data is shown on page A30.8, for the two element model. The problem has also been analyzed with 8, 32 and 72 SHELL elements.

Version 99.0 Linear Examples A30.1

SOLVIA Verification Manual A contour plot and the circumferential stress (stress-rr) and axial stress (stress-ss) along the line CB are shown in the figures on pages A30.4 to A30.7 for the four models.

User Hints

Note that only two SHELL elements with 4x4x2 Gauss integration points can be used to model this example problem resulting into good agreement with theoretical results for displacements.

"* A further description of this example problem can be found in [2].

References

[1]

Scordelis, A.C., Lo, K.S., "Computer Analysis of Cylindrical Shells", J. Amer. Concr. Inst.,

Vol. 61, pp. 539-560, 1964.

[2]

Larsson, G. and Olsson, H., "An Engineering Error Measure for Finite Element Analysis",

Finite Element News, April 1988.

Version 99.0 Number of Displacement Stress-rr Stress-rr elements Z-direction Point B Point C Point B 2 (2x1)

-3.558 1305.4 24.47 8 (4x2)

-3.598 1302.1

-26.26 32 (8x4)

-3.611 1273.5 0.88 72 (12x6)

-3.615 1271.5 2.82 Linear Examples A30.2

SOLVIA Verification Manual A30 SCORDELIS-LO CYLINDRICAL ROOF, CUBIC SHELL ORIGINAL H

SO.

Linear Examples z

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SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB A30 SCORDELIS-LO CYLINDRICAL

ROOF, CUBIC SHELL ORIGINAL ý i

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z x

Y r

s EAXES=

STRESS-RST SMASTER 000000 B OG1iOI C O1000 D 01i!01 E

100011 F

1101i1 SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 Version 99.0 I

A30.3

SOLVIA Verification Manual Linear Examples A30 SCORDELIS-LO CYLINDRICAL ROOF, CUBIC SHELL MAX DISPL.

40338 Z

TIME I

X Y

REACTION 96001 MISES SHELL TOP MAX 1726 2 S162S.I1 1220.7 816.22 614.00 4 11 77 209.S5 MIN 108.44 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A30 SCORDELIS-LO CYLINDRICAL

ROOF, CUBIC SHELL TIMi I

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SOLVIA Verification Manual MAX DISPL TTME I

A30A SCORDELIS-LO CYLINDRICAL

ROOF, 8 CUBIC SHELL

-4.0696 Linear Examples z

x 1, y REACTION 8Z374 MISES SHELL TOP MAX 1865.2 S1749.9

[ tS 19.2 1288.6 1058.

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SOLVIA Verification Manual Linear Examples A30B SCORDELIS-LO CYLINDRICAL

ROOF, 32 CUBIC SHELL MAX DTSPL.

/-.

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A30.6

SOLVIA Verification Manual Linear Examples A30C SCORDELIS-LO CYLINDRICAL

ROOF, T T TIME 1 1 0

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4. 0866 2

TIME 1

X Y

REACTION S~36889 MISES SSHELL TOP MAX 1909 2 1790.5 1553. I 131S. 7 1078.2 840.83 603.41 366.00 128.59 MIN 9.8815 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB

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In (2

(2 LLi 0*

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A30 SCORDELIS-LO CYLINDRICAL ROOF, CUBIC SHELL' DATABASE CREATE ANALYSIS TYPE=STATIC MASSMATRIX=LUMPED SYSTEM 1 CYLINDRICAL COORDINATES ENTRIES NODE R

THETA XL 1

300 90 300 2

300 50 300 3

300 50 0

4 300 90 0

MATERIAL 1

ELASTIC E=3.E6 NU=-.

DEN=0.208333 EGROUP 1

SHELL STRESSREFERENCE=ELEMENT RESULTS=NSTRESSES THICKNESS 1

3.0 GSURFACE 1 2 3 4 EL1=2 EL2=1 NODES-16 SYSTEM=1 FIXBOUNDARIES 246 INPUT=LINE 4 1 FIXBOUNDARIES 23 INPUT=LINE

/

412 FIXBOUNDARIES 156 INPUT=LINE

/

1 4 LOADS MASSPROPORTIONAL ZFACTOR=1.

ACC=-1.

MESH NNUMBERS=MYNODES NSYMBOLS=YES ENUMBER=YES MESH BCODE=ALL EAXES=STRESS-RST SOLVIA END SOLVIA-POST input A30 SCORDELIS-LO CYLINDRICAL ROOF, CUBIC SHELL DATABASE CREATE WRITE FILENAME='a30.1is' MESH CONTOUR=MISES VECTOR-REACTION EPLINE NAME=LINE-CB 1

4 11 7 3

TO 2

4 11 7 3 ELINE LINENAME=LINE-CB KIND=SRR OUTPUT=ALL SUBFRAME=21 ELINE LINENAME=LINE-CB KIND=SSS OUTPUT=ALL NLIST ZONENAME=N3 DIRECTION=23 MASS-PROPERTIES END Version 99.0 Linear Examples A30.8

SOLVIA Verification Manual EXAMPLE A31 SCORDELIS-LO CYLINDRICAL ROOF, PLATE Objective To verify the PLATE element when applied to a curved shell structure subjected to gravity loading.

Physical Problem Same as for Example A30.

Finite Element Model As for the previous example only one quarter of the structure needs to be modeled. A 12x12x4 mesh of PLATE elements is used, see figures on page A31.2.

Solution Results Using the input data shown on pages A31.4 and A31.5 the vertical deflection at point B of figure on page A30-1 is predicted to be -3.521 inches.

A contour plot of the vertical displacement and the variation of the bending moment about the X-axis (corresponding to bending stresses in the circumferential direction, the x,-direction in the selected Local Cylindrical System) for two lines are shown in the figures on page A31.3. The line "X-0" is identical to line CD and the line "X-12" is parallel to line CD but goes through the midpoints of the quadrilaterals nearest line CD.

User Hints

" Note that even though a large number of PLATE elements is used, the displacement solution for the considered point B is still not as good as the solution obtained in Example A30, where only two isoparametric SHELL elements are used. One reason is that the PLATE element is flat and the cylindrical shell roof is, therefore, approximated by straight segments. Another reason is that the membrane action of the PLATE element is the same as for a constant strain triangle. The mem brane forces are constant over each element, which limits the capability of the PLATE element to describe structures, where the membrane forces vary significantly.

"* The recommended finite element to model this example structure is, therefore, the SHELL element.

Version 99.0 Linear Examples A3 1.1

SOLVIA Verification Manual Linear Examples A31 SCORDELIS-LO CYLINDRICAL ROOF, PLATE 7

X Y

ORIGINAL

20.

SOLVIA-PRE 99.0 MASTER 000000 C 0! 1O0O D 01110.0 E 100011 F 110111 SOLViA ENGINEERING AB ORIGINAL ý-

20.

ZONE C-B SOLVIA-PRE 99.0 A31 SCORDELIS-LO CYLINDRICAL ROOF.

PLATE Z

x y

G0 AB SOLVIA ENGINEERIN Version 99.0

ý I A31.2

SOLVIA Verification Manual Linear Examples A31 SCORDELIS-LO CYLINDRICAL

ROOF, PLATE ORIGINAL So.

MAX DISPL,.-

3.9791 TIME I SOLVIA-POST 99.0 sal z

\\

Y Z-DIR DISPL.

0.52716 S0.27418

-0. 73775

-1 2437

-1.7497

-2 2SS6

-2 7616 S-3. 2676 MIN-3.

206 LVIA ENGINEERING AB A31 SCORDELIS-LO STIME CYLINDRICAL ROOF, PLATE 0

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A31.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A31 SCORDELIS-LO CYLINDRICAL ROOF, PLATE' DATABASE CREATE ANALYSIS TYPE=STATIC MASSMATRIX=LUMPED SYSTEM 1 CYLINDRICAL COORDINATES ENTRIES NODE R

THETA XL 1

300 2

300 3

300 4

300 90 50 50 90 300 300 0

0 MATERIAL 1

ELASTIC E=3.E6 DENSITY=0.208333 EGROUP 1

PLATE STRESSTABLE 1

1 2 3 4 5 6 7 GSURFACE 1 2 3 4 EL1=12 EL2=12 SYSTEM=I EDATA

/

1 3.0 FIXBOUNDARY FIXBOUNDARY FIXBOUNDARY 246 INPUT=LINE 23 INPUT=LINE 156 INPUT=LINE LOADS MASSPROPORTIONAL ZFACTOR=1.

ACCGRA=-1.

SET HEIGHT=0.25 NSYMBOLS=MYNODES NNUMBERS=MYNODES MESH BCODE=ALL ZONE NAME=C-B INPUT=GLOBAL-LIMITS XMAX=25 MESH ZONENAME=C-B ENUMBERS=YES SOLVIA END Version 99.0

/

4 1

/

1 2

/

3 4 Linear Examples A31.4

SOLVIA Verification Manual Linear Examples SOLVIA-POST input A31 SCORDELIS-LO CYLINDRICAL ROOF, PLATE DATABASE CREATE WRITE FILENAME='a31.lis' MESH OUTLINE=YES ORIGINAL-DASHED CONTOUR=DZ EPLINE NAME=X-0 531 2 4 1 STEP 4 TO 575 2 4 1 EPLINE NAME=X-12 532 4 7 3

/

530 3 7 4

/

536 4 7 3

/

534 3 7 4 540 4 7 3

/

538 3 7 4

/

544 4 7 3

/

542 3 7 4 548 4 7 3

/

546 3 7 4

/

552 4 7

3

/

550 3 7 4 556 4 7 3

/

554 3 7 4

/

560 4 7 3

/

558 3 7 4 564 4 7 3

/

562 3 7 4

/

568 4 7 3

/

566 3 7 4 572 4 7 3

/

570 3 7 4

/

576 4 7 3

/

574 3 7 4 SYSTEM 1 CYLINDRICAL ELINE LINENAME=X-0 KIND=M11 OUTPUT=ALL SYSTEM=i SUBFRAME=21 ELINE LINENAME=X-12 KIND=M11 OUTPUT-ALL SYSTEM 1 NLIST ZONENANE=N3 DIRECTION=234 MASS-PROPERTIES END Version 99.0 A31.5

SOLVIA Verification Manual EXAMPLE A32 PINCHED CYLINDRICAL SHELL, SHELL ELEMENTS Objective To verify the membrane and bending behaviour of the SHELL element when applied to a curved structure.

Physical Problem The thin cylindrical shell structure shown in the figure below is analyzed for its static response. The cylinder is freely supported at its ends and is loaded by two centrally located and diametrically opposed concentrated forces.

RIGID DIAPHRAGM SUPPORT (v = w y0)

E =3.0-10' psi v=0.3 t=1.0 in.

"a R/t=100 A

L/R=2 P = 300000 lbf.

/

RIGID DIAPHRAGM SUPPORT (v = w==0)

Finite Element Model Using the three symmetry planes of the structure and the load, only one eighth of the cylinder is analyzed. The 16-node SHELL element is employed using different fineness of the mesh.

Version 99.0 Linear Examples A32.1

SOLVIA Verification Manual Solution Results The analytical solution for this problem is reported in [1].

The input data used for one of the models (thirty-six 16-node SHELL element, integration 4x4x2) is shown on pages A32.8 and A32.9.

The displacement in Z-direction and the membrane stress-ss (global X-direction) at point C for the different models are shown in the table on the following page.

The graphic results from SOLVIA-POST contain deformed mesh, radial displacement and shell stresses along the line BC. Stress-1I is the stress in the circumferential direction (the x,-direction) and membrane stress-22 is the membrane stress in axial direction (the x2-direction).

User Hints

"* In SOLVIA-POST the line coordinates are calculated as a line polygon, i.e. the program assumes straight lines between the nodal points. The coordinates on the abscissa of the diagrams in the figures from SOLVIA-POST are, therefore, not equal to the arc length along the circle BC.

"* A rather fine mesh around the point of load application is necessary when stress results are desired. If only a displacement solution is of interest a coarser mesh can be used.

"* Note that the lower integration order 3x3x2 results in significantly worse stress prediction along the line BC than the 4x4x2 integration order, see figures on page A32.6.

Z-displacement Membrane stress-ss SOLVIA-POST at point C at point C results on page Analytical solution

-1.642

-4.72.104 16-node SHELL 36 elements

-1.395

-6.28-104 A32.4 Integration 4x4x2 LINE RATIO=I.0 16-node SHELL 36 elements

-1.622

-5.83. 104 A32.5 Integration 4x4x2 LINE RATIO=4.0 16-node SHELL 36 elements

-1.673

-11.45.104 A32.6 Integration 3x3x2 LINE RATIO=4.0 16-node SHELL 144 elements

-1.664

-5.27.104 A32.7 Integration 4x4x2 LINE RATIO=4.0 Version 99.0 Linear Examples A32.2

SOLVIA Verification Manual Linear Examples Reference

[1]

Lindberg, G.M., Olson, M.D. and Cowper, E.R., "New Developments in the Finite Element Analysis of Shells", National Research Council of Canada, Quarterly Bulletin of the Division of Mechanical Engineering and the National Aeronautical Establishment, Vol.4, pp. 1-38, 1969.

A32 PINCHED CYLINDRICAL SHELL, SHELL ELEMENTS ORIGINAL H

20.

Z TIME FORCE 75000 SOLVIA ENGINEERING AB Version 99.0 A32 PINCHED CYLINDRICAL SHELL, SHELL ELEMENTS Z

its r-s EAXES=

STRESS-RST SOLVIA ENGINEERING Ai SOLVIA-PRE 99.0 SOLVIA-PRE 99.0 ORIGINAL 20.

A32.3

SOLVIA Verification Manual Linear Examples A32 PINCHED CYLINDRICAL SHELL, SHELL ELEMENT' z

X Y

ORIGINAL I SO MAX DISPL.

1.3946 TIME I

09 N]

0 LOAD 75000 SOLVIA-POST 99.0 T1ME 0

C -r

'S 0

so 100 IS0 200 LINE-BC NODE 3 SOLVIA ENGINEERING AB A32 PINCHED CYLINDRICAL TIME I SHEL N

y 0

0 C'J 4/)

0 Li N

H 0

0 N

0o so t00 ISO 200 SHELL ELEMENTS TIME I S.

1'"

K, 50 100 i5O 200 BC SHELLSURFACE FOP SOLVTA-POST 99.0 BC SHELLSUPFACE MT!

SOLVIA ENGINEERING AB Version 99.0 M o 00

o:

Li C-'

O0 M 0

A32.4 L,

b-

SOLVIA Verification Manual Linear Examples A32A PINCHED CYLINDRICAL SHELL, SHELL ELEMENTS RATIO z

x'-y

_ 6223 Ný ORIGINAL MAX DISPL TIME so0 0

100 LINE-BC NODE 3 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A32A PINCHED CYLINDRICAL I

I TIME

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so 100 ISO 200 BC SH1ELSJRFACE TOP SOLVIA-POST 99.0

SHELL, SHELL ELEMENTS, RATIO=4 TTM*

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200 Cc' If Cc' If)

-- I 0[

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SOLVIA Verification Manual Linear Examples A32B PINCHED CYLINDRICAL

SHELL, SHELL ELEMENTS, RATIO=4 INT 3x3xý ORIGINAL i so.

MAX DISPL.

1o.67 3 2 TINE I z

X Y

U)

CL CL U)

U)

LOAD 75000 SOLVIA-POST 99.0 7/

-7 I

0 50 1O0 15O 200 LTNE-BC NODE 3 -

4 SOLVIA ENGINEERING AB A32B PINCHED CYL _INDRICAL SHELL, SHELL ELEMENTS, RATIO=4, INT.

T7-IE A -'

I L CL 0

so 100 ISO 200 Ln

(-'A

(-'

0 0

100 BC SHiELLSURFACE FOP SOLVIA-POST 99.0 BC SHELLSURFACE MiD SOLVIA ENGINEERING AB Version 99.0 M

U) 71 (n

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cc 3x3x2 ISO 200 t

0 0

t I

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A32.6

SOLVIA Verification Manual Linear Examples A32C PINCHED CY[

_I 1663SO L.

H-1.6637 INDRICAL SHELL.

z X lj'Y SHELL

ELEMENTS, RATIO=4, 12*12 I

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SOLVIA-POST 99.0 C'

SOLVIA ENGINEERING AB A32C PINCHED CYLINDRICAL SHELL, SHELL ELEMENTS, RATIO=4, 12*12 TIME TIMI

/

ci) cic C'

C")

C/C u-c Li Hc-n c'J 0

0 Li 50 100 ISO 200 BC SH-ELLSURFACE 13P SOLVIA-POST 99.0 0

so

10.

5r, BC SHELLSURFACE MID SOLVCA ENGINEERING AB Version 99.0 0

0

/

1)

p F

>-n Li

-J C

U,

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LOAD 75000

200, 0g Cdw ci-)

c-n c-n cc)

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A32 PINCHED CYLINDRICAL SHELL, SHELL ELEMENTS' DATABASE CREATE SYSTEM 1 CYLINDRICAL COORDINATES ENTRIES NODE R

THETA XL 1

100 90 100 2

100 0

100 3

100 0

0 4

100 90 0

LINE CYLINDRICAL N1=3 N2=4 EL=6 MIDNODES-2 NFIRST=5 MATERIAL 1

ELASTIC E=3.E7 NU=0.3 EGROUP 1

SHELL STRESSREFERENCE=ELEMENT RESULTS=NSTRESSES GSURFACE 1 2 3 4 EL1=6 EL2=6 NODES=16 SYSTEM=i THICKNESS 1

1.0 FIXBOUNDARIES 23 INPUT-LINES

/

1 2 FIXBOUNDARIES 345 INPUT=LINES

/

2 3 FIXBOUNDARIES 156 INPUT=LINES

/

3 4 FIXBOUNDARIES 246 INPUT=LINES

/

4 1 LOADS CONCENTRATED 4 3 -75000.

SET NSYMBOLS=MYNODES PLOTORIENTATION=PORTRAIT MESH ENUMBER=YES NNUMBERS-MYNODES VECTOR=LOAD MESH EAXES=STRESS-RST SOLVIA END Version 99.0 Linear Examples A32.8

SOLVIA Verification Manual SOLVIA-POST input A32 PINCHED CYLINDRICAL SHELL, SHELL ELEMENTS DATABASE CREATE WRITE FILENAME='a32.1is' SUBFRAME 21 MESH ORIGINAL=YES VECTOR=LOAD NPLINE NAME-LINE-BC 3

5 TO 21 4

SYSTEM 1 CYLINDRICAL NLINE LINENAME=LINE-BC DIRECTION=3 OUTPUT=ALL SYSTEM-i EPLINE NAME=BC 36 3 7 11 4 TO 31 3

7 11 4 ELINE LINENAME=BC KIND=S11 OUTPUT=ALL SYSTEM=1 SUBFRAME=21 SHELLSURFACE PLOTRESULTS=MID ELINE LINENAME=BC KIND=S22 OUTPUT=ALL SYSTEM=i END Version 99.0 Linear Examples A32.9

SOLVIA Verification Manual EXAMPLE A33 PINCHED CYLINDRICAL SHELL, PLATE ELEMENTS Objective To verify the membrane and bending behaviour of the PLATE element when modeling a cylindrical shell surface.

Physical Problem The same shell structure as described in Example A32 is considered, see figure on page A32. 1.

Finite Element Model As for Example A32 only one eighth of the cylinder need to be considered. The finite element model consists of 576 PLATE elements as shown in the figures on page A33.2 giving a total number of 1739 equations.

Solution Results The input data used for the 12x12x4 diamond mesh is shown on pages A33.4 and A33.5.

The resulting deformation of the finite element model and the radial deflection along line BC is shown in the top figure on page A33.3. The membrane force in the global X-direction (x2 direction of the Local Cylindrical System) and bending moment M, (about global X-direction corresponding to bending stresses in the circumferential direction, the x, direction) at the stresstable points on the line BC are displayed in the bottom figure on page A33.3.

The theoretical solution is given in [1].

5,

= displacement in Z-direction at point C.

o,ý = membrane stress at point C in global X-direction.

8.

[in]

cf"" [psi]

Theory SOLVIA Theory SOLVIA

-1.642

-1.619

-4.72.104

-3.30.104 User Hints

As discussed already for Examples A30 and A3 1, a larger number of elements is required in general when modeling a curved shell structure by PLATE elements.

Version 99.0 Linear Examples A33.1

SOLVIA Verification Manual Reference

[1]

Lindberg, G.M., Olson, M.D. and Cowper, E.R., "New Developments in the Finite Element Analysis of Shells", National Research Council of Canada, Quarterly Bulletin of the Division of Mechanical Engineering and the National Aeronautical Establishment, Vol. 4, pp. 1-38, 1969.

Version 99.0 Linear Examples A33.2

SOLVIA Verification Manual Linear Examples A33 PINCHED CYLINDRICAL SHE MAX DISPL.

1,19 Z

TIME.

X Y

Cn X

LOAD 75000 PLATE ELEMENTS T I.M7

'N 0--I

/

o-so 100 t So LINE-KC NODE 3 -

4 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A33 PINCHED CYLINDRICAL TIME 1

SHELL, PLATE ELEMENTS C,)

C!)

z O0 I-U Li cr; 0o o0 0

so 100 ISO 200 BC-LINE

'-0 50 100 BC-LINE SOLVWA-POST 99.0 SOLVTA ENGINEERING AB Version 99.0 206 0

aJ C

or 01 Or Cf C

C-)

LU Li (n

L-

.bP 200 oo o

o o

i A33.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A33 PINCHED CYLINDRICAL SHELL, PLATE ELEMENTS' DATABASE CREATE

¢ SYSTEM 1 CYLINDRICAL COORDINATES ENTRIES NODE R

THETA XL 1

100 90 100 2

100 0

100 3

100 0

0 4

100 90 0

LINE CYLINDRICAL N1=3 N2=4 EL=12 NFIRST=5 MATERIAL 1 ELASTIC E=3.E7 NU=0.3 EGROUP 1

PLATE GSURFACE 1 2 3 4 EL1=12 EL2=12 SYSTEM=i EDATA

/

1 1.0 STRESSTABLE 1

1 2 3 4 5 6 7 FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES 23 345 156 246 INPUT-LINES INPUT=LINES INPUT=LINES INPUT=LINES

/ /

/

1 2 2 3 3 4 4 1 LOADS CONCENTRATED 4

3 -75000.

SET MESH ZONE MESH PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES NNUMBERS=MYNODES NAME=EDGE INPUT=GLOBAL-LIMITS XMAX=5 ZONENAME=EDGE ENUMBER=YES NSYMBOLS=MYNODES SOLVIA END Version 99.0 Linear Examples A33.4

SOLVIA Verification Manual SOLVIA-POST input A33 PINCHED CYLINDRICAL SHELL, PLATE ELEMENTS DATABASE CREATE WRITE FILENAME='a33.1is' SUBFRAME 21 MESH VECTOR=LOAD NPLINE NAME=LINE-BC 3

5 TO 15 4

SYSTEM 1 CYLINDRICAL NLINE LINENAME=LINE-BC DIRECTION=3 OUTPUT-ALL SYSTEM=i EPLINE NAME-BC-LINE 575 1 4 2 STEP -4 TO 531 1 4 2

ELINE ELINE END Linear Examples LINENAME=BC-LINE KIND=F22 OUTPUT-ALL SYSTEM=i SUBFRAME=21 LINENAME=BC-LINE KIND=M11 OUTPUT=ALL SYSTEM=i Version 99.0 A33.5

SOLVIA Verification Manual EXAMPLE A34 ANALYSIS OF CONCENTRIC FLUID-FILLED CYLINDERS Objective To verify the behaviour of the FLUID2 element in a fluid-structure static analysis.

Physical Problem Five concentric fluid-filled cylinders are analyzed for an axial load applied to the stiff end cap as shown in the figure below.

Total load 2.04.104 N z

L\\

stift cap 4

.4 -

1-3.213 i

3.373-S4.775*

6.060 6.203-6.6855 0

7.925 8.090 Cylinders (635 l-T6 aluminum)

Young's modulus E = 6.895-1010 N/m 2 Poisson's ratio v = 0.33 Fluid (Light hydraulic oil)

Bulk modulus K = 1.64. 10' N/m 2 tine of symmetry Y (cm)

Finite Element Model Because of symmetry only one half of the cylinders need to be considered in the axisymmetric model shown in the figures on page A34.2. The finite element model consists of 8-node PLANE and FLUID2 AXISYMMETRIC elements. Fluid and structural elements are modeled with separate nodes.

The nodal displacement in the Y-direction for adjacent structural and fluid element nodes are constrained. The stiff cap is simulated using rigid links to the cylinder nodes. The top fluid nodes are constrained to move in the Z-direction with the top cylinder nodes. Irrotational displacement conditions are enforced in the fluid elements.

The total load of 20400 N corresponds to the axisymmetric model load of 3246.76 N since the circumferential extension is I radian.

Version 99.0 25.250 Linear Examples A34.1

SOLVIA Verification Manual Solution Results The input data on page A34.3 - A34.5 was used in the finite element analysis. The calculated fluid pressures are compared with the experimental pressures presented in [1].

Cylinder Pressure [N/m2]

Experimental SOLVIA 1

5.57-104 5.23-104 2

4.52.104 4.28-10' 3

3.40.104 3.32.104 4

2.53.104 2.33.104 5

1.41,104 1.25-104 The deformed configuration as well as the radial displacement along the cylinder at radius Y = 3.213 cm (line INNER) and the pressure variation along the line RADIUS are displayed by SOLVIA-POST in the top figures on page A34.3. Note that meter is used in the input data for this example.

Reference

[1]

Munro, M. and Piekarski, K., "Stress Induced Radial Pressure Gradients in Liquid-Filled Multiple Concentric Cylinders", J. of Appl. Mech., Vol. 44, No. 2, pp. 218-221, 1977.

Lo W1O~

I CI \\ Y P0OCO

'lICe TOO0O0O Lo Version 99.0 Linear Examples A34 ANALYSIS OF CONCENrTUC FLUID-FILLED CYLINDERS L*

R 111G11A L

ZONE rDopsoZio 0RIGINAL -0.

OOS ZONE TDPFLUID ECLVIA ENGINEERING 18 ECLýI-RE -0 A34.2

SOLVIA Verification Manual A34 ANALYSIS CF CONCENTRIC FLUII MAX DISPL.

2.58SSE-5 TIME I

SOLVIA-POST 99.0

-FILLED CYLINDERS Y

R LOAD 3246.8 SOLVOA ENGINEERING AB SOLVIA-PRE input HEADING

'A34 ANALYSIS OF CONCENTRIC FLUID-FILLED CYLINDERS, DATABASE CREATE MASTER IDOF=-00111 SET NODES=8 MYNODES=200 PARAMETER $EL=30 COORDINATES

fluid top 1

5

.04775 9

.06820

structure 12

.03213 16

.06050 20

.07925

fluid and NGENERATION 1

2

/

ENTRIES NODE nodes 1

10:

/

2

.03213

/

6

.06050

/

10

.07925 top nodes 12

/

13

.03373

/

17

.06203

/

21

.08090 Y

/ /

3 7

20:

/ 14

/ 18 structure bottom nodes:

ZSTEP=-.25250 NSTEP=100

/

ELASTIC E=6.895E10 NU=0.33 FLUID K=1.64E9

.03373

/

.06203

/

4 8

.04610

.06655

.04610

/

15

.04775

.06655

/

19

.06820 I TO 21 PLANE AXISYMMETRIC MATERIAL=I 13 12 112 113 EL2-$EL 15 14 114 115 EL2=$EL 17 16 116 117 EL2=$EL 19 18 118 119 EL2=$EL 21 20 120 121 EL2=$EL A34.3 MATERIAL MATERIAL EGROUP 1

GSURFACE GSURFACE GSURFACE GSURFACE GSURFACE Version 99.0 Linear Examples

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

Linear Examples EGROUP 2

FLUID2 AXISYMMETRIC GSURFACE 2

1 101 102 EL2=$EL GSURFACE 4

3 103 104 EL2=$EL GSURFACE 6

5 105 106 EL2=$EL GSURFACE 8

7 107 108 EL2=$EL GSURFACE 10 9 109 110 EL2=$EL CONSTRAINTS INPUT=LINES M-INPUT=

snl sn2 sdir mnl mn2 mdir 2

102 2

12 112 2

3 103 2

13 113 2

4 104 2

14 114 2

5 105 2

15 115 2

6 106 2

16 116 2

7 107 2

17 117 2

8 108 2

18 118 2

9 109 2

19 119 2

10 110 2

20 120 2

CONSTRAINTS INPUT=NODES / DELETE CONSTRAINTS INPUT=LINES

n1 n2 dir master mdir 1

2 3

1 3

3 4

3 5

6 3

7 8

3 9 i0 3

1 1

3 3

1 3

MATERIAL=2 IRROTATIONALITY=YES NCOINCIDE=NO NCOINCIDE=NO NCOINCIDE=NO NCOINCIDE=NO NCOINCIDE=NO 1 TO 10 RIGIDLINK INPUT=LINES

n1 n2 master 12 13 1

14 15 16 17 18 19 20 21 1

1 1

1 FIXBOUNDARIES 2

INPUT=LINE /

1 101 FIXBOUNDARIES 2

INPUT=NODES /

2 TO 10 ZONE SYM GLOBAL-LIMITS ZMAX=-.25 FIXBOUNDARIES 3

INPUT=ZONE ZONEl-SYM LOADS CONCENTRATED

/

1 3 -3246.76084 SET MESH ZONE PLOTORIENTATION=PORTRAIT VECTOR=LOAD TOP GLOBAL-LIMITS ZMIN=-0.01 ZONE TOPFLUID ZONE TOPSOLID ZONE TOPFLUID ZONE TOPSOLID SUBFRAME 14 MESH TOPFLUID MESH TOPSOLID MESH TOPFLUID MESH TOPSOLID OPERATION=ADD ZONEl=TOP OPERATION=ADD ZONET=TOP OPERATION=INTERSECT ZONE1=FLUID2 OPERATION=INTERSECT ZONEl=PLANE NNUM=MY NSYM=YES NNUM=MY NSYM=YES GSCALE-OLD NSYM=YES BCODE=ALL NSYM=YES BCODE=ALL GSCALE-OLD LIST LINE 112 12 SOLVIA END Version 99.0 A34.4

SOLVIA Verification Manual SOLVIA-POST input A34 ANALYSIS OF CONCENTRIC FLUID-FILLED CYLINDERS DATABASE CREATE WRITE FILENAME='a34.1is' SET PLOTORIENTATION=PORTRAIT NPLINE NAME=INNER 12 202 TO 260 112 EGROUP 2

EPLINE NAME RADIUS 30 2 4 STEP 30 TO 150 2 4 MESH PLINES=ALL SUBFRAME=2111 NLINE LINENAME:INNER DIRECTION=2 SUBFRAME=2222 ELINE LINENAME=RADIUS KIND=PRESSURE SYMBOL=i SUBFRAME=2221 MESH DMAX=3 VECTOR=LOAD ZONE NAME=FLUID INPUT=ELEMENT 30 STEP 30 TO 150 ELIST ZONENAME=FLUID END Version 99.0 Linear Examples A34.5

SOLVIA Verification Manual EXAMPLE A35 Z-SECTION CANTILEVER UNDER DISTRIBUTED EDGE LOAD Objective To verify the membrane behaviour of the SHELL element when subjected to distributed edge loading.

Physical Problem A Z-section cantilever is subjected to pure torsional loading at its free end as seen in the figure below.

At the end X=0, the cantilever is rigidly built-in. This problem is further described in [1] page 35.

L a

C Thickness 0.1 m L = 10.0 m a:2.5m F 0.6 MN b

1.0 m c =2.0 m Elastic material E=2.1.10" N/m 2 v

0.3 Finite Element Model A uniform mesh of 24 cubic SHELL elements are used in the model as seen in the figure on page A35.2. The edge load is defined by SHELL line forces acting in the local SHELL edge direction.

Solution Results The theoretical solution for this problem is given in [I], pp. 35-41. The stress distribution over the cross-section is defined at a point located 2.5 m along the cantilever from the built-in end. The axial membrane stress varies linearly over the flanges and is constant over the web. The membrane shear stress varies quadratically over the flanges and varies linearly over the web.

Using the input data shown on page A35.5 the following results are obtained:

Line coord.

Cubic SHELL Analytical [MPa]

Stress-rr S tress-rs Stress-rr Stress-rs

0.

-110.3

-0.07

-107.9

0.

1.

36.5

-5.96 36.0

-5.85

3.

36.5 5.96 36.0 5.85

4.

-110.3 0.07

-107.9 0.

Version 99.0 Linear Examples A35.1

SOLVIA Verification Manual The contour plot of stress-rr and the stress distribution at X = 2.5 m can be seen in figures on page A35.3. The cubic SHELL element performs very well in this example and the quadratic distribution of the shear stresses is in good agreement with the analytical solution.

User Hints

° The results of an analysis using the 9-node SHELL element can be found on page A35.4. The distribution of the direct stress, stress-rr, is in good agreement with the analytical solution, but the shear stress distribution is relatively poor.

Reference

[11 NAFEMS, Background to Benchmarks, 1993.

Version 99.0 A35 Z-SECTION CANTILEVER UNDER DISTRIBUTED EDGE LOAD ORIGINAL 0.5 Z

TI NE 1 xY

.7 EFORCE 600000 MASTER

5 000000 B 9iNiGA SOLVIA-PRE 99 0 SDLVIA ENGINEERING AB Linear Examples A35.2

SOLVIA Verification Manual Linear Examples A3S Z-SECTION CANTILEVER UNDER DISTRIBUTED EDGE LOAD ORIG7NAL i

0.5 Z

MAX DISPL.

I 0.042963 STR~ESS-RR SHELL MID MAX 6.6156E7 S

5I0341E7 Si 8711E7

-1.2919E7 S44548E7

-7.6178-7

-1.078!E8

-1.3944E8

-i.7107E8 MIN-1.8688E8 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A35 Z-SECTION CANTILEVER UNDER DISTRIBUTED EDGE LOAD TIME 1 TI.M..

D

ýo u))

/

i o i

. -/'

o r i

/'

o t

N-

-]

L

"//

o 2

3 0

1 s

SECTION SOLVIA-POST 99.0 SSHELLSURFACE MID SECTION SHELLSURFACE MID SOLVIA ENGINEERING AB Version 99.0 A35.3

SOLVIA Verification Manual A3SA Z-SECTION CANTILEVER UNDER DISTRIBUTED EDGE LOAD, 9-NODE SHELL ORIGINAL 0.5 TIME I

z I --

Y EFORCE 600000 r

EAXES=

STRESS-RST MASTER 0000CO B t11111 SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 A3SA Z-SECTION CANTILEVER UNDER DISTRIBL TIME

[ /

0 2!

SECTION SHEL LSRFACE LVD SOLVIA-POST 99.0 U)

L__

2*

UTED EDGE LOAD, 9-NODE SHELL TI,M E I

N 2

N

/

1

/

0 2

.3 SECTION SHELLSURFACE MID SOLVIA ENGINEERING AB Version 99.0 C-1 Uf)

Li_

U--

Linear Examples 0

I A35.4

SOLVIA Verification Manual SOLVIA-PRE input HEAD

'A35 Z-SECTION CANTILEVER UNDER DISTRIBUTED EDGE LOAD' DATABASE CREATE COORDINATES 1

0.

1.

3

0.

-1.

5

10. -1.

7

10.

1.

0.

-1.

0.

/

2

/4

/

6

/8

0.

1i.

0.

-1.

10.

-1.

0.

10.

1.

MATERIAL 1

ELASTIC E=210.E9 NU=0.3 EGROUP 1

SHELL RESULTS=NSTRESSES STRESSREFERENCE=ELEMENT THICKNESS 1

0.1 GSURFACE 6

3 4 5

ELI=8 EL2-1 NODES=16 GSURFACE 7

2 3 6

ELi=8 EL2=1 NODES=-6 GSURFACE 8

1 2 7

ELi=8 EL2=1 NODES=f6 LOADS ELEMENT TYPE-FORCE INPUT=LINE 5

6 edge 0.6E6 0.6E6 7 8 edge -0.6E6 -0.6E6 FIXBOUNDARIES INPUT=LINE 12

/

23

/

34 VIEW SET MESH ID=1 XVIEW=2.

YVIEW--1.

ZVIEW=0.5 VIEW=i ENUMBERS=YES NNUMBERS=MYNODES NSYMBOLS=MYNODES, VECTOR=LOAD BCODE=ALL SOLVIA END SOLVIA-POST input A35 Z-SECTION CANTILEVER UNDER DISTRIBUTED EDGE LOAD DATABASE CREATE WRITE FILENAME='a35.1is' SHELLSURFACE PLOTRESULTS=MID VIEW ID=i XVIEW=2.

YVIEW=-i.

ZVIEW=0.5 MESH VIEW=1 OUTLINE=SHELL CONTOUR=SRR ORIGINAL=YES EPLINE 6

3 ELINE ELINE NMAX END NAME=SECTION 10 6 2

/

14 3 10 6 2

/

LINENAME=SECTION KIND=SRR LINENAME=SECTION KIND=SRS 22 3 10 6 2 OUTPUT=ALL SUBFRAME=21 OUTPUT=ALL

)IRECTION=i23 NUMBER=2 Version 99.0 Linear Examples

[

A35.5

SOLVIA Verification Manual EXAMPLE A36 CYLINDRICAL PRESSURE VESSEL WITH HEMISPHERICAL ENDS Objective To verify the PLANE AXISYMMETRIC element under distributed loading when used for axisymmetric shell bending problems.

Physical Problem A pressurized cylinder is capped by hemispheres of the same thickness as shown in the figure below.

The radius to thickness ratio is 40 so thin shell theory is applicable. This problem is described in [1].

E=2.1.010 N/r 2

v=0.3 a = 1.0m (mean radius) h = 0.025m (thickness) p=l.10 6 N/rM2 Finite Element Model Due to symmetry only half of the pressure vessel is modeled using 38 PLANE AXISYMMETRIC elements as shown in the bottom figure on page A36.2.

Most of the local bending is concentrated to the joint between the cylinder and the cap and a fine mesh is used for that part. The element meshing of the model is performed as described in [1].

Solution Results The theoretical solution for this problem can be found in [2] p. 484. Combining the maximum bending stress and the membrane stress at the outer surface we get stress-rr=.293ap 25.22 10 N/m2 where pa =

and C= - a 2h c 2 a-h/2 where p is the internal pressure applied at radius a - h/2. The factor C is introduced to calculate equivalent loads acting at the midsurface, see Example A2.

Version 99.0 Linear Examples A36.1

SOLVIA Verification Manual SOLVIA calculates maximum axial stress at nodal point 7 to be stress-rr = 25.3.106 N/m2.

The deformed mesh and the distribution of the axial stress and the hoop stress at the outer surface of the cylinder can be seen in the figures on page A36.3.

Input data used in the SOLVIA analysis is found on pages A36.4 and A36.5.

References

[1]

NAFEMS, Background to Benchmarks, 1993.

[2]

Timoshenko, S.P. and Woinowsky-Krieger, S., Theory of Plates and Shells, Second Edition, McGraw-Hill, 1959.

Version 99.0 Linear Examples A36.2

SOLVIA Verification Manual Linear Examples A36 CYLINDRICAL PRESSURE

'FJ1E 1 0

1 2

3 4

VESSEL WITH HEMISPHERICAL ENDS Li

}I F-CUTER SOLVIA-POST 99.0 OUTER SOLVIA ENGINEERING AB Version 99.0 A36 CYLINDRICAL PRESSURE VESSEL WITH HEMISPHERICAL ORIGINAL 0 2 MAX DISPL.

712E-4 TIME I SOLVIA-POST 99.0 SOLVIA ENGINEERIN ENDS I

LOAD 916"40 qG AB INJ

r CO C

LJ u)

('

SOLVIA-POST 99 0 SOLVIA ENGINEERING AB

[

I A36 CYLINDRICAL PRESSURE VESSEL WITH HEMISPHERICAL ENDS ORIGINAL 02 Z

MAX DISPL.

1 712E-4 IY TIME I

R REACTION 3.2S04E5 SOLVIA-POST 99. ISOLVIA ENGINEERING AR A36.3

SOLVIA Verification Manual SOLVIA-PRE input HEAD

'A36 CYLINDRICAL PRESSURE VESSEL WITH HEMISPHERICAL ENDS' DATABASE CREATE MASTER IDOF=100111 SET NODES=8 SYSTEM 1

CYLINDRICAL COORDINATES ENTRIES NODE R

1 1.0125 2

0.9875 3

1.0125 4

0.9875 5

1.0125 6

0.9875 SYSTEM 0

COORDINATES ENTRIES NODE 7

8 9

10 11 12 Y

1.0125 0.9875 1.0125 0.9875 1.0125 0.9875 Z=1. 5 THETA

90.

11.068 11.068

0.

z 1.4034 1.4034 1.1136 1.1136 MATERIAL 1

ELASTIC E=210.E9 NU=0.3 ELEMENT 1.E6 1.E6 1.E6 1.E6 1.E6 FIXBOUNDARIES FIXBOUNDARIES PLANE AXISYMMETRIC 2

4 3

1 EL1=12 4

6 5

3 ELI18 6

8 7

5 EL1=4 8 10 9

7 ELi 6 10 12 11 9

ELi=8 RESULTS=NSTRESSES EL2=1 SYSTEM=I EL2=1 SYSTEM=I EL2=1 SYSTEM=C EL2=1 SYSTEM-C EL2=1 SYSTEM=0 INPUT=LINE 2

INPUT=LINE 3

INPUT=LINE

/ /

1 2 11 12 MESH BCODE=ALL OUTLINE=YES SUBFRAME=21 MESH VECTOR=LOAD SOLVIA END Version 99.0 EGROUP 1

GSURFACE GSURFACE GSURFACE GSURFACE GSURFACE LOADS 2

4 4

6 6

8 8 10 10 12 Linear Examples A36.4

SOLVIA Verification Manual SOLVIA-POST input A36 CYLINDRICAL PRESSURE VESSEL WITH HEMISPHERICAL ENDS DATABASE CREATE STRESSREFERENCE-ELEMENT WRITE FILENAME-'a36.1is' SET PLOTORIENTATION=PORTRAIT MESH ORIGINAL=YES VECTOR=LOAD OUTLINE:YES NSYMBOLS:MYNODES MESH ORIGINAL YES VECTOR=REACTION SET PLOTORIENTATION=LANDSCAPE EPLINE NAME=OUTER 1

4 7 3 TO 38 4 7 3 ELINE LINENAME=OUTER KIND=SRR OUTPUT=ALL SUBFRAME=21 ELINE LINENAME=OUTER KIND=STT SUMMATION KIND=LOAD SUMMATION KIND-REACTION DETAILS:YES END Version 99.0 Linear Examples A36.5

SOLVIA Verification Manual EXAMPLE A37 FREQUENCY ANALYSIS OF CANTILEVER WITH OFF-CENTER MASSES Objective To verify the dynamic behaviour of the BEAM and GENERAL Mass elements and the use of rigid links in frequency analysis.

Physical Problem A cantilever beam with two off-center point masses as shown in the figure below is considered. The point masses are connected to the tip end of the beam using rigid links. The off-center distance is 2 m.

At the other end the beam is rigidly built-in. The six lowest frequencies and eigenmodes of the beam structure are analyzed. This example is described in [ 1].

Y X

Mi0 L

,N a

a a

M2 0-Beam cross section 0

D L=10m a=2m D-0.5m M1 10000 kg M2 1000 kg Elastic material E =2-10"1 N/m 2 v

0.3 p = 8000 kg/rm 3 Finite Element Model The cantilever beam is modeled using five standard BEAM elements with a circularsolid section. The point masses are modeled using GENERAL Mass element with one node per element. Rigid links are connecting the BEAM elements with the Mass elements. A consistent mass matrix assumption must be used in this example, and the subspace iteration method is employed in the eigenvalue calcula tions. The calculational model is shown in the figure on page A37.2.

Solution Results Using the input data shown on pages A37.4 and A37.5 the following results are obtained:

Version 99.0 Linear Examples A37.1

SOLVIA Verification Manual User Hints

" GENERAL Mass elements and consistent mass matrix assumption must be used when concen trated point masses are coupled to a structure via rigid links. If a lumped mass matrix is used or the point masses are defined using concentrated nodal masses the off-diagonal terms from the rigid link transformation are lost and the solution result will be in error.

". The standard section BEAM element is formulated including shear deformation. The NAFEMS results refer to an exact 3-D beam element excluding the shear effects. This can be verified by using the general BEAM section and setting the shear areas in s-and t-directions to zero.

Reference

[1]

NAFEMS, The Standard NAFEMS Benchmarks, TSNB, Rev. 3, October 5, 1990.

A37 FREQUENCY ANALYSIS OF CANTILEVER WITH OFF-CENTER MASSES ORIGINAL Z

X y

r EAXES=RST N

HA S TER 0000ZR SOLVTA-PRE B

l1t1 C 222222 99.0 SOLVIA ENGINEERING AB Version 99.0 Mode Reference freq. fr [1]

SOLVIA 1

1.723 1.722 2

1.727 1.725 3

7.413 7.410 4

9.972 9.946 5

18.155 18.044 5

26.957 26.694 Linear Examples A37.2

SOLVIA Verification Manual Linear Examples A37 FREOUENCY ANALYSIS OF CANTILEVER WITH OFF-CENTER MASSES REFERENCE

-1 2.

MAX D!SPL.

8.3296E-3 MODE I FREO 1.7219 Y

REFERENCE F H I.

MAX DISPL.

F 8_$216E-3 X MODE 2 FRED 1.7254 SOLVIA-POST 99.0 2

X Y

SOLVIA ENGINEERING AB A37 FRECUENCY ANALYSIS OF CANTILEVER WITH OFF-CENTER MASSES REFERENCE H

I.

MAX DISPL.

0.020801 MODE 3 FREO 7.4099 2

X '"Y REFERENCE

-- 2.

Y MAX DISPL.

7. 8288E-3 MODE 4 FRER 9.9462 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 A37.3

SOLVIA Verification Manual A37 FREGUENCY ANALYSIS OF CANTILEVER WITH OFF-CENTER MASSES E

RENCE C

I.

MAX DISPL.

.014271 MODE 5 FREO :8.044 z

X ',y REFERENCE

-1 2.

MAX DISPL. -

0.010945 MODE 6 FREG 26 694 SOLVTA-POST 99 0 SOLVTA ENGINEERING AB SOLVIA-POST input A37 FREQUENCY ANALYSIS OF CANTILEVER WITH OFF-CENTER MASSES DATABASE CREATE WRITE FILENAME='a37.1is' FREQUENCIES MASS-PROPERTIES SET RESPONSETYPE=VIBRATIONMODE ORIGINAL=DASHED SET NSYMBOLS=MYNODES SUBFRAME 21 MESH VIEW=Z TIME=i MESH VIEW=I TIME=2 SUBFRAME 21 MESH VIEW=I TIME=3 MESH VIEW=Z TIME=4 SUBFRAME 21 MESH VIEW=I TIME=5 MESH VIEW=Z TIME=6 END Version 99.0 Y

Lx Linear Examples A37.4

SOLVIA Verification Manual SOLVIA-PRE input HEAD

'A37 FREQUENCY ANALYSIS OF CANTILEVER WITH OFF-CENTER MASSES' DATABASE CREATE ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT FREQUENCIES SUBSPACE-ITERATION NEIG=6 COORDINATES 1

2 3

4

10.

2.

10.

-2.

MATERIAL 1

ELASTIC E=200.E9 NU=0.3 DENSITY=8000.

EGROUP 1

BEAM SECTION 1

CIRCULARSOLID D=0.5 BEAMVECTOR 1

0.

1.

GLINE Ni=i N2=2 AUX=-i EL=5 EGROUP 2

MATRIXSET 1 1.E4 2

i.E4 3

I.E4 4

0. /

MATRIXSET 1

1.E3 2

1.E3 3

1.E3 4

0.

/

EDATA

/

ENODES

/

GENERAL 1

MASSMATRIX 5

2 5

1 3

0.

/

6

0.

MASSMATRIX

0. /

FIXBOUNDARIES RIGIDLINK 32

/

42

/

6 2 2 2 4 1

MESH NSYMBOLS=YES NNUMBERS=MYNODES BCODE=ALL EAXES=RST SOLVIA END Version 99.0 0.

Linear Examples A37.5

SOLVIA Verification Manual EXAMPLE A38 SIMPLY SUPPORTED SKEW PLATE UNDER PRESSURE LOAD Objective To verify the SHELL element under distributed load when using distorted elements.

Physical Problem A simply supported skew plate under pressure load is considered as shown in the figure below. The pressure load acts in the negative Z-direction. The material in the plate is linear elastic isotropic. The problem is described in [1].

Uniform thickness 0.01 m L-lm 150'

[=300 Elastic material E=2.1-10 1' N/m 2 v=0.3 Pressure p - 700 N/m 2 D

C Finite Element Model The skew plate is generated using 4x4 cubic SHELL elements. The Z-displacement is fixed for all edges. The element stresses are evaluated at the element nodal points. The finite element mesh corresponds to the fine mesh described in [1], see figure on page A38.2. However, a good stress solution would require a finer mesh.

Solution Results The theoretical solution to this problem is discussed in [11 and is based on classical thin plate theory.

The primary results are the maximum and minimum principal stresses on the lower surface at the plate centre.

The input data used in the SOLVIA analysis is shown on page A38.5. A finite element model using 8x8 4-node SHELL elements is also used to compare with the NAFEMS results.

Version 99.0 Y

B Linear Examples A38.1

SOLVIA Verification Manual Stresses on the bottom surface NAFEMS Cubic SHELL 4-node SHELL Max. principal stress [MN/m2]

0.802 0.793 0.757 Mid. principal stress IMN/m2 ]

0.456 0.409 0.421 The distribution of stress-xx along the diagonal BD and the distribution of stress-yy along the diagonal AC can be seen in the top figures on page A38.3. Contour plots of stress-xx and maximum principal stress can also be seen on page A38.3.

The cotresponding SOLVIA-POST results for the 4-node SHELL element analysis can be seen on page A38.4. The mesh consists of 8x8 elements. Note the large stress jumps in the stress distribution.

User Hints

"* Note that SOLVIA-PRE gives a warning message regarding the element distortion in the model (skew distortion of 60 degrees).

"* Note the stress concentration at the corners B and D. A more detailed discussion regarding error sources can be found in [1].

"* Simply supported boundary conditions for a square plate are discussed in Example A27.

Reference

[1]

NAFEMS, Background to Benchmarks, 1993.

A38 SIMPLY SUPPORTED SKEW PLATE UNDER PRESSURE LOAD ORIGINAL

-0.2 C)'kI Y

ORIGINAL 0.i L

X TiNE I A

MASTER 000000 B 00!000 C 01 000 D 101000 SOLVIA-PRE 99.0 PRESSURE 700 SOLVIA ENGINEERING AS Version 99.0 A3 8.2 z

LX Linear Examples Version 99.0 A38.2

SOLVIA Verification Manual Linear Examples UNDER PRESSURE LOAD

//

6.00 01 o02 0,3 04 O's 0.6 DIAG-O8 SHELLSURFACE SOT SOLVIA-POST 99 0 SOLVIA ENGINEERING AB Version 99.0 A38 SIMPLY SUPPORTED SKEW PLATE UN DR PRESSURE LOAD o.0 O's

.o0

.5 2.0 DIAG-AC SHELLSURFACE ROT SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A38.3 A38 SIMPLY SUPPORTED SKEW PLATE A38 SIMPLY SUPPORTED SKEW PLATE UNDER PRESSURE LOAD DAX DISPL.

I.2617E-5 Y

TIME I L 5 STRESS-XX SHELL BOT MAX 7-926E5 S5. I034E5

-54258

-6.1886Es S-

!.835E6

.7481E6 2

.a127E6

.}2. 8773E6

-3.44t9E6 MIN-3. 7242E6 SCLVIA-POST 99.0 SOLVIA ENGINEERING AB A38 SIMPLY SUPPORTED SKEW PLATE UNDER PRESSURE LOAD MAX DISPL.

1.2617E-5 Y

TIME I Lx PRINCIPAL STRESS MAX.

SHELL BOT MAX 7. 926SES I

7.431 1ES A. 4403ES S.4496ES 4.4588E5 3 4680ES

2. 4772E5 S-,

t.486sEs 1

49570 MIN 30.981 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB

,n w

SOLVIA Verification Manual Linear Examples A38A SIMPLY SUPPORTED SKEW PLATE UNDER PRESSURE LOAD SI 0'2 0

40.

O RI 02/

03 04

0.

00AD30 OSHSELLSLRFACE ROT SOLVIA POST 90.0 SOURIA ENGINEERING AR A38A SIMPLY SUPPORTED SKý 4

-.4 4

C--i

1

+//

I4 PLATE UNDER PRESSURE LOAD I TIME I

I

/1

/I i'

'/\\

\\

0.0 005'

.i!

s

+

SOLVIA-POST 99.0 3IAG--AC SHELLSURFACE SOT SOLVIA ENGINEERING AB A3RA SIMPLY SUPPORTED SKEW PLA MAX DOISPL. -

I 2997E-S TIME I A

SOLVIA-POST 99.0 TE UNDER PRESSURE LOAD Y Lx STRESS-XX SHELL ROT MAX 7.46S2ES I

6,2301E S3.760LES

1. 290 1ES

--I. 180OOES

-3. 6SOES

-6R. I200ES 5-9.

901 ES I-t l OA0ER MIN-L.E229EE6 SOLVIA ENGINEERING AS 5

Version 99.0 A38A SIMPLY SUPPORTED SKEE PLATE UNDER PRESSURE LOAD MAX DISPL.

1 I.2997E-S Y

TIME I

Lx PRINCIPAL STRESS MAX.

SHELL ROT MAX 7.4RSSES 6

R.9989ES A6.058ES S.132LES 4.t994ES 3.2662ES 2 333**E 1R3998ES 46G65 MIN 5.6721 SOLVIA-POST 99.0 SOLVIA ENGINEERING AS q) kt x X X A38.4

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A38 SIMPLY SUPPORTED SKEW PLATE UNDER PRESSURE LOAD' DATABASE CREATE COORDINATES 1

0.258819045 0.965925826 3

0.258819045 -0.965925826

/ /

2 4

0.

0.51763809 0.

MATERIAL 1

ELASTIC E=2.1E11 NU=0.3 EGROUP 1

SHELL RESULTS=NSTRESSES THICKNESS 1

0.01 GSURFACE 1 2 3 4

ELi 4 EL2=4 NODES=16 LOADS ELEMENT TYPE=PRESSURE INPUT=SURFACE 1 2 3 4 T

700.

FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES 3 INPUT=LINES 1 INPUT=NODES 2 INPUT=NODES

/

1 2

/

1 3

/

2 4

/

23

/

34 VIEW SET MESH MESH ID=i XVIEW=0.

YVIEW=-1.

ZVIEW=0.1 NSYMBOLS=MYNODES NNUMBERS=MYNODES VIEW=Z ENUMBER-YES BCODE=ALL SUBFRAME=21 VIEW=i VECTOR=LOAD SOLVIA END SOLVIA-POST input A38 SIMPLY SUPPORTED SKEW PLATE UNDER PRESSURE LOAD DATABASE CREATE WRITE FILENAME='a38.1is' SET VIEW=Z PLOTORIENTATION=PORTRAIT SHELLSURFACE PLOTRESULTS=BOTTOM LISTRESULTS=BOTTOM NAME-DIAG-DB 14 16 4 STEP 3

NAME=DIAG-AC 15 13 1 STEP -5 LINENAME=DIAG-DB LINENAME=DIAG-AC TO 13 2 14 16 4 TO 1

3 KIND=SXX KIND=SYY CONTOUR=SXX OUTLINE=YES CONTOUR=SPMAX OUTLINE=YES SELECT=S-EFFECTIVE NUMBER=3 15 13 1 OUTPUT=ALL OUTPUT=ALL 3 TYPE=MAXIMUM Version 99.0 4 1 EPLINE 4

2 EPLINE 16 3

ELINE ELINE MESH MESH EMAX END Linear Examples A38.5

SOLVIA Verification Manual EXAMPLE A39 ORTHOTROPIC PLATE UNDER PRESSURE LOAD Objective To verify the bending and twisting behaviour of the SHELL element when the material is orthotropic and the SHELL element is subjected to a uniform pressure load.

Physical Problem A simply supported square plate under uniform loading, as shown in the figure below, is considered.

y, b A

B C=0.30m h = 0.015 m (thickness)

F p = 1.0.10 3 N/m 2 (pressure)

Ea = 1.35548.10"° N/m2 EX, Eb = 1.12950.10 9 N/m2 ZEC 0

Vab = 3.847826.10-2 Gab = 1.17" 109 N/m2 2C Finite Element Model Because of symmetry only one quarter of the plate need to be considered, A-B-C-D. An orthotropic linear elastic material model is used with the principal material axes a and b coinciding with the global coordinate axes X and Y, respectively. Two finite element models have been used, one model consists of 16 cubic SHELL elements as shown in the figure on page A39.3 and the other model consists of 144 4-node SHELL elements.

Solution Results The theoretical solution for this problem (without transverse shear deformation) is discussed in [1],

chapt. 11. The expression for the deflection w in the Z-direction takes the form of a double trigonometrical series,

a. -sin mitx.

niTy nm=1,3,5..

n=1,3,5..

a b

and the expression for the coefficients a.

is given on page 371, [1].

Version 99.0 Linear Examples A39.1

SOLVIA Verification Manual From this expression for the deflection in the Z-direction, the bending and twisting moments in the plate are easily derived using eq. 212, [1].

The input data for the model with 16-node SHELL elements is shown on pages A39.6 and A39.7.

Vertical deflection, w(m):

Stress (N/m2):

Location Stress Theory*

16-node SHELL 4-node SHELL comp.

16 elements 144 elements C

YX:

-9.997.10'

_1.008.106

-1.004.106

-y -9.827.104

-9.816.104

-9.884.104 E

oY:

-7.584-10'

-7.664.10' (el. 14,

-7.217.10' (el. 138,

-y: -7.147.104

-7.145.104 point 3)

-7.032-104 point 3)

E (Yx:

-7.664-10' (el. 15,

-8.010.10' (el. 139, GY:

-7.145.104 point 4)

-7.337-104 point 4)

The theoretical results include terms up to m = n = 29.

The deformed models are shown on pages A39.4 and A39.5. The distribution of stress-xx along side D-C and stress-yy along side A-C of the plate are also shown on pages A39.4 and A39.5.

User Hints Note that a finer mesh is necessary when modeling this simply supported plate when orthotropic material is used than when isotropic material is used. This is evident from the large gradient of the TY-stress occuring along the line A-C, which in turn is due to the very small E-modulus in the Y direction. Detailed shear stresses and reactions would require a finer mesh than used here, see Example A27.

Reference

[1]

Timoshenko, S.P. and Woinowsky-Krieger, S., Theory of Plates and Shells, Second Edition, McGraw-Hill, 1959.

Version 99.0 Location Theory*

16-node SHELL 4-node SHELL 16 elements 144 elements C

3.543.10-4 3.602.10-4 3.591.10-4 E

2.529.10-4 2.573.10-4 2.564.10-4 F

1.935.10-4 1.975-1.04 1.966.10-4 Linear Examples A39.2

SOLVIA Verification Manual Linear Examples A39 ORTHOTROPiC PLATE UNDER PRESSURE LOAD ORIGINAL 0.05 TIME 1

z X

PRESSURE 1000 SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 Version 99.0 A39 ORTHOTROPIC PLATE UNDER PRESSURE LOAD ORIGINAL 0-H 0.05 Z

MASTER i10001 B 110011 C 110101 D 110111 E 111001 F

11 01 !

G 111101 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB A39.3

SOLVIA Verification Manual Linear Examples A39 ORTHOTROPIC PLATE UNDER PRESSURE LOAD ORIGINAL

0. O.

MAX DISPL.

3.6022E-1 TIME I z

X REACTION it.. 189 MISES SHELL TOP MAX 9.6289ES 9.0493E5 4

6. 7307ES 5 57I4ES I.4[ 2 1ES 3.2528E5 2.0934E5 93414 MIN 35449 SOLVIA ENGINEERING AB SOLVIA-POST 99.0 A39 ORTHOTROPIC

, TIME i PLATE UNDER PRESSURf

(

\\

? -

X I

LA I-U")

co I

C Cd A-C SHELLSURFACE FOP SOLVIA-POST 99.0 LOAD

,TIME

-I 0.1 0.2 D-C SHELLSURFACE FOP SOLVIA ENGINEERING AB Version 99.0 L.i oU) o -\\

CI N

0.0 0,1 i

T I

N- -

A39.4

SOLVIA Verification Manual Linear Examples A39A ORTHOTROPIC PLATE TIME 1

KF 0..

I

-...... -. -j - *~---]-J 0.1 0.2 O

UNDER PRESSURE oD x

X

(/

(/4 N

C iC 0

N

LOAD, 4-NODE SHELL TIME 1 I

.4.

1 0

0.1

0.

1 0.2 0.3 A-C SHELLSURFACE TO, SOLVIA-POST 99.0 D-C SHELLSURFACE TOP SOLVIA ENGINEERING AB Version 99.0 A39A ORTHOTROPIC PLATE UNDER PRESSURE

LOAD, 4-NODE SHELL ORIGINAL 0.05 Z

MAX DISPL.

i 3.912E-4 y

TIME I-X REACTION 12.643 MISES SHELL TOP MAX 9.58S2E5 9.010SES 7.8610ES 6.7116 ES S. 5621E5 4.4127ES 3.263265

2. 1138ES 96431 MIN 389S8 SOLVIA-POST 99.0 SOLVIA ENGINEERTNG AR 0

I of C

o 0.0 A39.5

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A39 ORTHOTROPIC PLATE UNDER PRESSURE LOAD' DATABASE CREATE MASTER IDOF=110001 COORDINATES 1

0.3 0.3

/

2 5

0.15 0.

/

6

0.

0.3

/

0.15 0.15 3

/

4 0.3 MATERIAL 1 ORTHOTROPIC EA=1.35548E10 EB=1.1295E9, NUAB=3.847826E-2 GAB-1.17E9 GAC=1.17E9 EGROUP 1

SHELL RESULTS=NSTRESSES GSURFACE 1 2 3 4 EL1=4 EL2=4 NODES=f6 THICKNESS 1

0.015 EDATA

/

ENTRIES EL BETA 1

0.

TO 16

0.

LOADS ELEMENT INPUT=SURFACE 1 2 3 4

T 1000 FIXBOUNDARY FIXBOUNDARY FIXBOUNDARY 4

5 3

INPUT=LINE INPUT-LINE INPUT=LINE

/ /

/

34 23 12

/

41 VIEW ID=i XVIEW=1.

YVIEW=-0.5 ZVIEW-0.5 SET NSYMBOLS=MYNODES NNUMBERS=MYNODES VIEW=i MESH VECTOR=LOAD MESH ENUMBERS=YES BCODE=ALL SOLVIA END Version 99.0 Linear Examples A39.6

SOLVIA Verification Manual SOLVIA-POST input A.39 ORTHOTROPIC PLATE UNDER PRESSURE LOAD DATABASE CREATE WRITE FILENAME='a39.1is' VIEW ID-i XVIEW=1.

YVIEW=-0.5 ZVIEW=0.5 SET VIEW=i OUTLINE=YES NSYMBOLS=MYNODES MESH ORIGINAL=YES CONTOUR=MISES VECTOR=REACTION EPLINE NAME-A-C 4

2 6 10 3 STEP 4 TO 16 2 6 10 3 EPLINE NAME=D-C 13 4 11 7 3 TO 16 4 11 7 3 ELINE LINENAME=A-C KIND=SYY OUTPUT=ALL SUBFRAME=21 ELINE LINENAME=D-C KIND=SXX OUTPUT=ALL ZONE NAME=RESULT INPUT=ELEMENTS 14 15 SHELLSURFACE LISTRESULTS=TOP ELIST ZONENAME=RESULT ZONE NAME=CEF INPUT=NODES

/

3 5 6 NLIST ZONENAME CEF END Version 99.0 Linear Examples A39.7

SOLVIA Verification Manual EXAMPLE A40 HEMISPHERICAL SHELL UNDER POINT LOADS Objective To verify the 4-node SHELL element bending behaviour when applied to a curved structure.

Physical Problem A hemispherical shell is subjected to concentrated radial loads at its free edge as shown in the figure below. One pair of loads is directed inwards towards the centre of the hemisphere and the other pair of loads is directed outwards away from the centre. This problem is described in [1].

E Thickness = 0.04 m R=10m E = 68.25.109 N/m 2 "I-*-------

vv=0.3 XX,*z"'""P

= 4000 N XP"'---

Y C

Ix P

A P

Finite Element Model Due to symmetry only one quarter of the hemisphere need to be considered in the analysis, region A-C-E in the figure above. The finite element mesh is generated in a spherical coordinate system using 4-node SHELL elements. SKEW degree-of-freedom Systems are defined for the symmetry planes X = 0 and Y = 0. Along edge A-E there is symmetry about the XZ-plane. Along edge C-E there is symmetry about the YZ-plane. Edge A-C is free to move. At point E the Z-displacement is fixed. The finite element model is shown in the bottom figure on page A40.2.

Solution Results The analytical solution for this problem is discussed in [1]. The input data used in the SOLVIA analysis is shown on pages A40.4 and A40.5.

The reported target value from NAFEMS is the outward displacement in the X-direction at point A, u, =0.185 m In the SOLVIA analysis the calculated displacement in the X-direction at point A (node 13, radial direction) is u, =0.1832m Version 99.0 Linear Examples A40.1

SOLVIA Verification Manual The figures on page A40.3 show a contour plot of the von Mises stress distribution and a contour plot of the radial displacements (in the x3 direction of the Local Spherical System). Note that the bending in the structure is concentrated to the corners where the loads are acting.

Reference

[1]

NAFEMS, Background to Benchmarks, 1993.

Version 99.0 A10 HEMISPHERICAL SHELL UNDER POINT LOADS ORIGINAL

2.

Z TIME I x

Y B,',

16 B,

7 B,

18 E

t[9 122 FORCE S12 2000 MAST-,P 000000 B 010101 C 100011 D 111111 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB Linear Examples A40.2

SOLVIA Verification Manual Linear Examples AM0 HEMISPHERICAL SHiELL UNDER PO INT LOADS ORIGINAL

2.

MAX DISPL H-0.20451 TiT1L I

z X

y LOAD 2000 MISES SHELL TOP MAX 1.1139E7 8.413006 7.3647E6 6.2863E6 5.208006 4.1297E6 3.051AE6 MIN 2.512306 SOLVIA ENGINEERING AB SOLVIA-POST 99.0 Version 99.0 A4.

A40.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A40 HEMISPHERICAL SHELL UNDER POINT LOADS' DATABASE CREATE SYSTEM 1

SPHERICAL COORDINATES

/

ENTRIES NODE R

THETA PHI 1

10.

0.

TO 13

10.

0.

90.

14

10. 90.

TO 26

10. 90.

90.

DELETE 14 SKEWSYSTEMS EULERANGLES 1

0.

TO 13

0.

90.

14

0.

TO 26

-90.

NSKEWS INPUT=NODES 1 1 TO 26 26 MATERIAL 1

ELASTIC E=68.25E9 NU=0.3 EGROUP 1

SHELL THICKNESS 1

0.04 GSURFACE 1 13 26 1 EL1=12 EL2=8 NODES=4 SYSTEM=-,

BLENDING=ANGLES FIXBOUNDARIES 156

/

1 15 TO 26 FIXBOUNDARIES 246

/

1 TO 13 FIXBOUNDARIES 3

/

1 LOADS CONCENTRATED 13 3 2000 26 3 -2000 MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES VECTOR=LOAD BCODE=ALL SOLVIA END SOLVIA-POST input A40 HEMISPHERICAL SHELL UNDER POINT LOADS DATABASE CREATE SYSTEM 1 SPHERICAL WRITE FILENAME='a40.1is' SET OUTLINE=YES ORIGINAL=YES MESH CONTOUR=MISES VECTOR=LOAD MESH CONTOUR=D3 VECTOR=REACTION SYSTEM=i NLIST ZONENAME=MYNODES DIRECTION-123 SYSTEM=i SUMMATION KIND=ENERGY END Version 99.0 Linear Examples A40.4

SOLVIA Verification Manual EXAMPLE A41 BEAM ON ELASTIC FOUNDATION Objective To verify the capability to model a beam supported by an elastic foundation.

Physical Problem A beam, shown in the figure below, resting on an elastic foundation and subjected to a concentrated transverse load at mid-span is considered.

y I

L12 L/2 Beam properties:

Foundation stiffness:

Concentrated loads data:

L = 100 in.

b= 1 in.

h= 10 in.

k. =5.0.105 psi P = 1.0.10 4 lbf E= 2.1.106 psi v=0.3 Finite Element Model The finite element model used for the analysis is shown in the figure on page A40.2. Thirteen BEAM elements are used to model one-half of the beam. Axial-translational SPRING elements are used to represent the lumped stiffness of the elastic foundation.

Solution Results The input data on pages A40.3 and A40.4 is used in the finite element analysis. The analytical solution for the beam on elastic foundation is found in [1].

Deflection at mid-span, (in.):

Bending moment at mid-span, (lbf-in.):

The distribution of the transverse displacement and the bending moment along the beam predicted by the finite element analysis are shown in the figures on page A40.3.

Version 99.0 Linear Examples A41.1

SOLVIA Verification Manual User Hints

"* The differential equation for the beam on an elastic foundation is of the same form as the differential equation for an axisymmetric cylinder, see Example A.2. The finite element repre sentation is, however, here accomplished by BEAM and SPRING elements.

"* It is important to have a fine enough mesh close to the applied load in this example since the significant displacements occur close to the load and decrease rapidly with increasing distance from the applied load.

" The stiffness of the SPRING elements is calculated based on lumping of the foundation stiffness, so that half of the foundation stiffness under each BEAM element is represented by the SPRING elements at each of the two BEAM end nodes. If an ISOBEAM element with, say, 3 nodes were used, a consistent lumping of the stiffness would be recommended so that, for each ISOBEAM element, 1/6th of the foundation stiffness is attributed to the end nodes, and 4/6th to the midside node.

" Note that a small positive (upward) displacement occurs for a few nodes, see page A40.3. If no tension can develop in the foundation, a nonlinear analysis could be carried out using the nonlinear DISP-FORCE option for the SPRING elements. The foundation material would then be modeled to be elastic in compression but with zero stiffness in tension.

Reference

[1]

Timoshenko, S.P., Strength of Materials, Part II, Third Edition, D. Van Nostrand Comp., 1958.

MAX DISPL. i 1.6348E-3 TIME I

SOLVIA-POST 99.0 A41 BEAM ON ELASTIC FOUNDATION Y

LX 14 LOAD 5000 SOLVIA ENGINEERING AB Version 99.0 Linear Examples A41.2

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A41 BEAM ON ELASTIC FOUNDATION' DATABASE CREATE MASTER IDOF=001110 COORDINATES 1

0.

TO 7

15

0.

-10.

TO 21

15.

0.

TO

15.

-10.

TO 14 28

50.

-10.

MATERIAL 1 ELASTIC E=2.1E6 NU-0.3 EGROUP I

SECTION 1

BEAM RESULT=FORCES GENERAL RINERTIA=3.1233 SINERTIA=0.8333333, TINERTIA=83.333333 AREA=I0.

ENODES 1

15 1 2 TO 13 15 13 14 EGROUP 2

SPRING DIRECTION-AXIALTRANSLATION PROPERTYSET 1

K=6.25E5 PROPERTYSET 2

K=1.25E6 PROPERTYSET 3

K=1.875E6 PROPERTYSET 4

K=2.50E6 ENODES 1

1 15 TO 14 14 28 EDATA

/

ENTRIES EL PROPERTYSET 1 1

/

2 2 TO 6 2

/

7 3

/

8 4 TO 13 4 Version 99.0 14 2 Linear Examples A41.3

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

FIXBOUNDARIES FIXBOUNDARIES 16 LOADS CONCENTRATED 1 2 -5000.

SOLVIA END SOLVIA-POST input Linear Examples

/

15 TO 28

/

1 A41 BEAM ON ELASTIC FOUNDATION DATABASE CREATE WRITE FILENAME='a41.1is' MESH VIEW=Z ENUMBER=YES NSYMBOL=YES VECTOR-LOAD NPLINE NAME=AXIAL

/

1 TO 14 NLINE LINENAME=AXIAL DIRECTION=2 SUBFRAME=21 EGROUP 1

EPLINE NAME=BEAM 1

1 2 TO 13 1 2 ELINE LINENAME=BEAM KIND=MT OUTPUT=ALL NLIST ELIST END Version 99.0 A41.4

SOLVIA Verification Manual EXAMPLE A42 PERFORATED TENSION STRIP Objective To verify the performance of the PLANE STRESS element when employed to model stress concentrations around holes.

Physical Problem The rectangular perforated strip, shown in the figure below, subjected to a uniform tension at two opposite sides, is considered.

L L =56 mm b =20 mm d

10 mm h

1 mm (thickness)

E =7.0.10 4 N/m 2 v = 0.25 p =25.0 N/m 2

Finite Element Model The finite element model considered for the analysis is shown on page A42.3. Using symmetry considerations, one-quarter of the strip is modeled with thirty 8-node PLANE STRESS elements.

Solution Results The analytical solution for the stress az, at location C and D in the figure above is given in [E].

The input data on pages A42.7 and A42.8 gives the following results:

Stress a,, at location C, (N/mm2):

Stress cF, at location D, (N/mm2):

Theory SOLVIA Theory SOLVIA 107.5 108.6 18.75 16.49 The calculated stress values are output at nodes 5 and 6, which correspond to the locations C and D, respectively.

The contour lines of von Mises effective stress and the mean stress (negative hydrostatic pressure) are shown on page A42.4. The mean stress is defined as aY mean 3 (c a,,

+a

-zz)

Version 99.0 A42.1 T P I

b Linear Examples

SOLVIA Verification Manual The contour lines of maximum deviation from nodal mean values of effective stress and pressure are also shown on page A42.4. The left top figure on page A42.5 shows the sum of the effective stress deviation and the pressure deviation. A contour plot of the displacement is also shown on page A42.5.

The stress cF, along the symmetry line C-D and the radial displacement along a line from A to C are shown in bottom figure on page A42.5.

For comparison, the contour lines of effective stress and deviation of effective stress are shown on page A42.9 for a refined mesh. We may note that the maximum value of effective stress deviation have decreased about a factor of 4 compared to the corresponding value on page A42.4 for the coarser model.

The value of a,, for the finer mesh at locations C and D are 108.8 (N/ mm2) and 17.55 (N/ mm2),

respectively.

The figures on page A42.6 show principal stresses and Tresca effective stress in the first finite ele ment model as displayed by SOLVIA-POST when element stresses are output at integration points by SOLVIA.

User Hints The stress deviation plots may be used as a tool for determining whether a good enough mesh has been used. In this example one can see that elements, which are distorted from the rectangular shape, see figure on page A42.3, give a higher stress and pressure deviation. Note that no deviation can occur, for example, at the locations C and D since these nodes are coupled only to one element so that the mean value at these nodes is based only on one stress value. A deviation with value zero can, therefore, not be used as a proof that the corresponding stress value is associated with zero error. An exact solution would have zero stress deviations but requires also that other criteria are satisfied. One such criterion is that the boundary conditions for stresses must be satisfied [2].

References

[1]

Timoshenko, S.P., and Goodier, J.N., Theory of Elasticity, Third Edition, McGraw-Hill, 1970, pp. 94-95.

[2]

Larsson, G., and Olsson, H., "An Engineering Error Measure for Finite Element Analysis",

Finite Element News, April, 1988.

Version 99.0 Linear Examples A42.2

SOLVIA Verification Manual Version 99.0 Linear Examples A42 PERFORATED TENSION STRI[

ORIGINAL 2

TIME 9

SOLVIA-PRE 99.0 SOLVIA E*

Z PRESSURE NGINEER!NG AB A42 PERFORATED TENSION 5TRIP ORIGINAL

}

2.

Z DISTORT ION MAX 45 000 S42. 188

36. 583 330. 938

25. 313 9.688 14 063 i 8. 4375 2.8125 MIN 0 SOLVIA ENGINEERING A8 SOLVIA-PRE 99.0 A42.3

SOLVIA Verification Manual Version 99.0 Linear Examples A42 PERFORATED TENSION STRIP ORIGINAL 2.-

2 MAX DISPL. -

0.012889 L

TIME 1

LOAD MISES 1MAX 108.25 m

101.76 7S 813 62 838 49.863 366888 23.913 10.938 MIN 4.4508 SOLVIA-POST 99.0 SOLVIA ENGINEERING AS A42 PERFORATED TENSION STRIP MAX DISPL.

0,012889 Z

TIME I Ly MEAN STRESS SAX 36.405

33.272 27.007 20.741 14 475 8.2090 1.9438

_I.3219 10588 MIN-13. 721 SOLVIA POST 99.I SOLVIA ENGINEERING AS A42 PERFORATED TENSION STRIP MAX DISPL.

0.012889 Z

TIME I

L i '

PRESSURE MAX 1.2596 S1 1809 S.0235 0.86601 O 70855 0.20110 0.39364 0.236[8

0. 078728 MIN 0 SOLVIA-POST 99.0 SOLVIA ENGINEERING AS A42.4

SOLVIA Verification Manual A42 PERFORATED TENSION STRIP MAX DISPL 0 012889 TIME I SOLVIA-POST 99.0 y

L7 DEVIATION OF MISES AND PRESSURE MAX 3. 9382 S3,6921

3. 1998

2. 707S

2. 2153 1t.

7230 1, 2307

0. 73842 S0.2461I MIN 0 SOLVIA ENGINEERING AB A42 PERFORATED TIME i

.L

-I r

co!\\

C N

I L

TENSION STRIP Co C3 C5 I

I TI,ME I 00 ND I-*..

C]

Ni

\\xN r

\\

C*

2 4

6 LINE--AC NODE 4

5 SOLVIA ENGINEERING AB Version 99.0 A42 PERFORATED TENSION STRIP MAX DISPL.

0.012889 Z

TIME I

L 7 DISPLACEMENT MAX 0.012889 L

0.012083 S0.10472 8.8610E-3 7.2499E-3 5.83880 3 4.0277E-3 2.4166E-3 8 O8SS5E-4 MIN 0 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB F-4 NI LI]

N-)

0 1

2 3

4 5

LINE-CD SOLVIA-POST 99.0 o

i i

r

[

i L A42.5 Linear Examples

SOLVIA Verification Manual Linear Examples Version 99.0 A42A PERFORATED TENSION STRIP, RESULTS=STRESSES MAX DISPL 0.012889 Z

TIME L

TRE S CA 1MAX 108.56 S89. 194

76. 286 63.379 501.71

37. 563 24.656 11.748 MIN 5.2944 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A42A PERFORATED TENSION STRIP, RESULTS=STRESSES MAX DISPL.

-0.012889 Z

TIME KL SPRINCIPAL I

[102.96I 102 96 SOLVIA-POST 99.0 SOLVIA ENGINEERING A8 A42.6

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A42 PERFORATED TENSION STRIP' DATABASE CREATE MASTER IDOF=100111 COORDINATES ENTRIES NODES 1

2 3

4 5

6 7

8 Y

10.

0.

5.

10.

0.

Linear Examples z

28.

10.

5.

0.

10.

0.

4 3

7 3

5 EL-6 7

EL-3 6

EL=3 6 7 MIDNODES=1 MIDNODES=1 MIDNODES=1 NCENTER=8 NFIRST=9 RATIO-0.1 MATERIAL 1

ELASTIC E=7.0E4 NU=0.25 EGROUP 1

PLA GSURFACE 1 2 GSURFACE 3 4 EDATA

/

1 1.

FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES NE STRESS2 RESULTS=NSTRESSES 3 7 EL1=3 EL2=2 NODES=8 5

6 ELI 4 EL2=6 NODES=8 2 INPUT=LINES 3 INPUT=LINES

/

8 LOADS ELEMENT INPUT=LINE 1 2 -25.

SET MESH MESH

/ /

2 3 5

6

/

3 4 PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES NNUMBERS=MYNODES VECTOR=LOAD CONTOUR=DISTORTION ENUMBERS=YES SOLVIA END Version 99.0 LINE LINE LINE LINE ARC STRAIGHT STRAIGHT COMBINED A42.7

SOLVIA Verification Manual SOLVIA-POST input A42 PERFORATED TENSION STRIP DATABASE CREATE WRITE FILENAME='a42.1is' SET PLOTORIENTATION=PORTRAIT SET OUTLINE-YES MESH ORIGINAL=DASHED CONTOUR=MISES VECTOR=LOAD MESH CONTOUR=SDEVIATION MESH CONTOUR=SMEAN MESH CONTOUR=PDEVIATION MESH CONTOUR-SPDEVIATION MESH CONTOUR-DISPLACEMENTS EPLINE NAME=LINE-CD 30 3 7 4 TO 27 3

7 4 SET PLOTORIENTATION=LANDSCAPE ELINE LINENAME=LINE-CD KIND=SZZ SUBFRAME=21 NPLINE NAME=LINE-AC 4

9 TO 19 5

NVARIABLE NAME=Y DIRECTION=2 KIND=COORDINATE NVARIABLE NAME=Z DIRECTION=3 KIND=COORDINATE NVARIABLE NAME=DY DIRECTION=2 KIND=DISPLACEMENT NVARIABLE NAME=DZ DIRECTION=3 KIND=DISPLACEMENT RESULTANT NAME=RADIAL-D STRING='(DY*Y+DZ*Z)/(SQRT(Y*Y+Z*Z))'

RLINE LINENAME=LINE-AC RESULTANTNAME=RADIAL-D ZONE NAME=EDGE INPUT=ELEMENTS

/

27 TO 30 ELIST ZONENAME=EDGE END Version 99.0 Linear Examples A42.8

SOLVIA Verification Manual Linear Examples A42B PERFORATED TiM E I TEN 1

F-2 3

4 5

LINE ED SOLVIA-POST 99.0 SION STI CD*

O C)

IT CT CO

RIP, FINE MESH T TMP I

0 2

4 6

8 LINE-AC NODE A 5

SOLVIA ENGINEERING AS Version 99.0 A42B PERFORATED TENSION STRIP, FINE MESH ORIGINAL

2.

Z MAX DISPL.

0.012899--l TIME I LOAD 27.778 MISES MAX 108.64 101. 9 88.S69 I

78.189 61.809 3.48 0429 SX21.670 8.2898 MIN 1.999 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A42B PERFORATED TENSION STRIP, FINE MESH MAX D!SPL.

-0 0.12899 Z

TIME I L

DEVIATION OF MISES MAX 0.69181 S0.64857 O...

6209 0 47562 0.38914 I 30267 I.21618 I.12971 0.043238 MIN 0 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB

\\

L.J NO C

C

'0 0

0 1

J.

I t \\

'\\

\\

A42.9

SOLVIA Verification Manual EXAMPLE A43 FRACTURE MECHANICS ANALYSIS OF A TENSILE SPECIMEN Objective To verify the performance of the PLANE STRAIN element when used for linear elastic fracture mechanics analysis.

Physical Problem The tensile specimen with an edge crack, as shown in the figure below, is analyzed for its elastic response. The stress intensity factor K, [1] is to be determined for the given geometry and load conditions.

z P

L=0.32m b=0.16m a = 0.040 m h = 1.0 m (thickness)

E = 2.0.10" N/m 2 v = 0.29 p = 1.0.103 N/m 2 P

Finite Element Model The finite element model considered is shown on page A43.4. Twenty 8-node PLANE STRAIN elements are used to model the top half of the specimen. At the crack tip, node 11 in the model, six quarter-point triangular elements with 1 / 7-stress singularities are used.

Version 99.0 I.2 L/2 Linear Examples y

A43.1

SOLVIA Verification Manual Solution Results The analytical solution for K, is as follows:

KI =FB

G and dfl dA where G

= energy release rate E

= Young's modulus v

= Poisson's ratio dA = change in crack area 1H

= total potential energy The strain energy release rate at the crack tip node is obtained as G

DH ai ax t

where xS = coordinates of the crack tip nodal point (node 11 in this example) a' = components of the unit vector in the direction of crack propagation t

= 1.0 = specimen thickness The following SOLVIA numerical solution is obtained by using the input data on pages A43.6 and A43.7:

DI-I 1 -

0.62767710-6 Dy The above value represents only the contribution of the material in the upper half of the specimen.

Thus, the total energy release rate, G, is obtained as G = 2(0.627677.10-6)1'0 = 0.125535-10-5 S1.0 This yields that K1 = 532.7 The value compares very well with the reference solution K1 = 531.7 given in ref. [I].

The deformed mesh is shown on page A43.5 as displayed by SOLVIA-POST.

Version 99.0 Linear Examples A43.2

SOLVIA Verification Manual User Hints

"* Accurate results are obtained in this solution, although a coarse finite element discretization is used.

" Note that the zone definitions made in SOLVIA-PRE prior to the SOLVIA command can also be used in SOLVIA-POST. The zone definitions are transferred by the SOLVIA program to the porthole file. When loading the SOLVIA-POST database the zone definitions are stored in the database.

Reference

[1]

Tada, H., Paris, P.C. and Irwin, G.R., The Stress Analysis of Cracks Handbook, Del Research Corp., Hellertown, PENN., 1973.

Version 99.0 Linear Examples A43.3

SOLVIA Verification Manual Linear Examples Version 99.0 A43 FRACTURE MECHANICS ANALYSIS OF A TENSILE SPECIMEN ORIGINAL Z

TIME I

PRESSURE 1000 DISTORTION MAX S2.9t1 49.604 42.990 36.377

,*,* 29. 763 16.535 9.9209 3 3070 MIN 0 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB A43.4

SOLVIA Verification Manual Linear Examples Version 99.0 A43 FRACTURE MECHANICS ANALYSIS OF A TENSILE SPECIMEN ORIGINAL 0. 02 Z

MAX DISPL.

1.3366E-9 TIME I

y LOAD S3.333 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A43 FRACTURE MECHANICS ANALYSIS OF A TENSILE SPECIMEN MAX DISPL. H 4.8874E-10 2

TIME i ZONE TRIANG L y MISES MAX 12816 1 2023

! 01037 885 1. 2 7265.4 S.

S6 79.6 4093.9 i2 SO.1 922.39 MIN 129.51 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A43.5

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A43 FRACTURE MECHANICS ANALYSIS OF A TENSILE SPECIMEN' DATABASE CREATE MASTER IDOF=100111 COORDINATES

/

ENTRIES SYSTEM 1

CYLINDRICAL COORDINATES

/

ENTRIES NODE Y

1

.08 2

.08 3

.16 4

.16 5

.16 7

.0 8

.0 9

.0 Z

.04

.16

.04 TO Y=0. 04 NODE R

THETA 10

.02 180.

11

.0

0.

12

.02 0.

SET MYNODES=0 NODES=-i LINE CYLINDRICAL 12 10 EL-6 MIDNODES=1 MATERIAL 1 ELASTIC E=2.07E11 NU=0.29 EGROUP 1

PLANE STRAIN GSURFACE 1 2 3 4 ELP=2 EL2=1 GSURFACE 6 1 4 5 ELI=2 EL2-1 GSURFACE 7 8 1 6 ELI=2 EL2=2 LINE STRAIGHT 8 9 EL=2 MIDNODES=1 LINE COMBINED 9 2 8 1 GSURFACE 9

10 12 2 EL1-1 EL2=6 GSURFACE 11 12 10 11 EL1=1 EL2=6 ZONE NAME=TRIANG INPUT=CYLINDRICAL-LIMITS SYSTEM=1 RMAX=0.02 MESH ZONENAME=TRIANG NSYMBOLS=YES NNUMBERS=YES SUBFRAME=21 COORDINATES

/

ENTRIES NODE R

THETA 65

.005 0

67

.005 30 TO 71

.005 150 66

.005 180 MESH ZONENAME=TRIANG NNUMBERS=YES NSYMBOLS=YES STRAINENERGY 11 LOADS ELEMENT 3 R -1.E3 STEP 2 TO 7 R -1.E3 FIXBOUNDARIES 3 INPUT-LINES FIXBOUNDARIES 2 INPUT=NODES

/

11 12

/

11 12 2

/

2 3 MESH VECTOR=LOAD CONTOUR-DISTORTION SUBFRAME11 SOLVIA END Version 99.0 Linear Examples A43.6

SOLVIA Verification Manual SOLVIA-POST input A43 FRACTURE MECHANICS ANALYSIS OF A TENSILE SPECIMEN DATABASE CREATE WRITE FILENAME='a43.1is' ENERGYRELEASERATE MESH VECTOR-LOAD ORIGINAL=DASHED MESH ZONENAME=TRIANG CONTOUR=MISES END Version 99.0 Linear Examples A43.7

SOLVIA Verification Manual EXAMPLE A44 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLANE STRESS Objective To verify the dynamic behaviour of the PLANE STRESS element in frequency analysis when employing SKEW degree-of-freedom Systems.

Physical Problem A cantilever beam of rectangular cross-section as shown in the figure below is considered. The cantilever is the same as previously analyzed for a static load in Example A14.

lb L= 1.0 m b = 0.05 m h=0.10m 0 = 300 E = 2.0.101 1 N/m 2 v = 0.30 p=7800 kg/m3 Finite Element Model The cantilever beam is modeled using ten 8-node PLANE STRESS elements with a consistent mass matrix, see the top figure on page A44.3. Since the model is inclined in the Global System, a SKEW degree-of-freedom System is used. The subspace iteration method is used for the frequency calculations.

Solution Results The lowest natural frequency of a cantilever beam is given in [1], p. 108 as:

X2 FI where X= 1.87510407 i

= mass/unit length In this formula rotary inertia and shear deformations are not considered.

Version 99.0 Linear Examples A44.1

SOLVIA Verification Manual Insertion of the numerical values gives for the fundamental mode in the Y-Z plane:

f = 81.80 Hz The SOLVIA solution using the input data on pages A44.4 and A44.5 gives the following result:

f = 81.19 Hz The bottom figure on page A44.3 and the top figure on page A44.4 show the mode shape of the two lowest frequencies as displayed by SOLVIA-POST.

User Hints

"* Note that the consistent element mass matrix for the PLANE element is always calculated using 3x3 Gauss integration and that the user has no control over it. This ensures that the entire mass of the element is accounted for in the mass matrix.

" For a pinned-pinned beam [I] p. 181 gives the following expression for the ratio between the frequency including rotary inertia and shear deformations and the frequency including flexural effects only:

(f)rot.+shea E

2L A)

K (f )ex.

A-KG where K = 10(1+v) for a rectangular section 12 +lIv Assuming the effects of rotary inertia and shear deformation to be of the same order for a cantilever, we obtain the following estimate (f)rot.+shear

80.43 Hz The SOLVIA solution for a consistent mass matrix gives a higher value for the lowest natural frequency than this analytical estimate.

"* Of course, the determinant search method can also be used and it gives the same frequency results.

Reference

[1]

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

Version 99.0 Linear Examples A44.2

SOLVIA Verification Manual Linear Examples Version 99.0 A44 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLANE STRESS ORIGINAL Z

LY o>b NAXES=SKEW

~~M 0

AST ER t00111 B i101i C

M ill SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB A44.3

SOLVIA Verification Manual REFERENCE

-H 0. i MAX DISPL.

F 0.31853 MODE 2 FREO 488.83 SOLVIA-POST 99.0 A44 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLANE STRESS z

Ly

-A SOLVIA ENGINEERING AB SOLVIA-PRE input HEADING

'A44 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLANE STRESS' DATABASE CREATE MASTER IDOF=100111 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT FREQUENCIES SUBSPACE-ITERATION NEIG=2 SSTOL=1.E-8 SKEWSYSTEMS EULERANGLES 1

30.

SYSTEM 1 CARTESIAN PHI=30 COORDINATES ENTRIES NODE YL ZL 1

0 0

2 0

0.05 3

0 0.1 4

5 1

1 0

0.1 MATERIAL 1

ELASTIC E=2.E11 NU=0.3 DENSITY=7800 EGROUP 1

GSURFACE EDATA

/

1 0.05 PLANE STRESS2 5

3 1 4

EL1=10 EL2=1 NODES=8 ENTRIES EL THICK Version 99.0 Linear Examples A44-4

SOLVIA Verification Manual SOLVIA-PRE input (cont.)

NSKEWS INPUT=LINE 13 1

FIXBOUNDARIES 2

/

1 3 FIXBOUNDARIES

/

2 SET NSYMBOLS=MYNODES MESH BCODE=ALL NAXES=SKEW ENUMBERS=YES SOLVIA END SOLVIA-POST input A44 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLANE STRESS DATABASE CREATE WRITE FILENAME='a44.1is' SET RESPONSETYPE=VIBRATIONMODE MESH TIME=i ORIGINAL=DASHED MESH TIME=2 ORIGINAL=DASHED OUTLINE=YES FREQUENCIES MASS-PROPERTIES END Version 99.0 Linear Examples A44.5

SOLVIA Verification Manual EXAMPLE A45 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLANE STRAIN Objective To verify the dynamic behaviour of the PLANE STRAIN element in frequency analysis and when employing SKEW degree-of-freedom Systems.

Physical Problem Same as in figure on page A44. 1.

Finite Element Model Same as in figure on page A44.3 except that PLANE STRAIN elements and a lumped mass distribution are used. The determinant search method is used for the frequency solution.

Solution Results To obtain the same theoretical solution as for the plane stress case of Example A44 the following material data is used:

1 + 2v E*= l+2..E=l.89349.10 1"N/m 2

(1 + V)2 V*=

V

= 0.230769 where E and v are the Young's modulus and Poisson's ratio, respectively, as used in Example A44.

The theoretical solution given in Example A44 is f = 81.80 Hz The input data on page A45.3 gives the following result:

f = 80.76 Hz The figures on page A45.2 show the mode shape of the two lowest frequencies as displayed by SOLVIA-POST.

User Hints

An analysis using the consistent mass discretization yields the result f = 81.19 Hz which is exactly equal to the result obtained in Example A44 (which should be the case).

Version 99.0 Linear Examples A45.1

SOLVIA Verification Manual A45 FUNDA REFERENCE H-

-1 0.1 MAX DISPL.

0.071133 MODE I FREO 80.762 SOLVIA-POST 99.0 MENTAL FREQUENCY OF CANTILEVER, PLANE STRAIN z L y K

SOLVIA ENGINEERING AB

)

A4S FUNDAMENTAL FREQUENCY OF CANTILEVER, PLANE STRAIN L

REFERENCE i--

-3 0.1 MAX DiSPL.

0.06895 MODE 2 FRED 478.11 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 A45.2 Linear Examples

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A45 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLANE STRAIN, DATABASE CREATE MASTER IDOF=100111 ANALYSIS TYPE-DYNAMIC MASSMATRIX=LUMPED FREQUENCIES DETERMINANT-SEARCH NEIG=2 SKEWSYSTEMS EULERANGLES 1

30.

SYSTEM 1 CARTESIAN PHI=30 COORDINATES ENTRIES NODE YL ZL 1

0 0

2 0

0.05 3

0 0.1 4

1 0

5 1

0.1 MATERIAL 1

ELASTIC E=1.89349E11 NU=0.230769 DENSITY=7800 EGROUP 1

PLANE STRAIN GSURFACE 5 3 1 4 EL1=10 EL2=1 NODES=8 NSKEWS INPUT=LINE 13 1

FIXBOUNDARIES 2

/

1 3 FIXBOUNDARIES

/

2 SOLVIA END SOLVIA-POST input A45 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLANE STRAIN DATABASE CREATE WRITE FILENAME='a45.1is' SET RESPONSETYPE=VIBRATIONMODE MESH TIME=i ORIGINAL=DASHED MESH TIME=2 ORIGINAL=DASHED OUTLINE=YES FREQUENCIES MASS-PROPERTIES END Version 99.0 Linear Examples A45.3

SOLVIA Verification Manual EXAMPLE A46 FUNDAMENTAL FREQUENCY OF CANTILEVER, SOLID Objective To verify the dynamical behaviour of the SOLID element in frequency analysis when employing SKEW degree-of-freedom Systems.

Physical Problem Same as in the figure on page A44. 1.

Finite Element Model The finite element model considered is shown on page A46.2. Ten 20-node SOLID elements are used in the model. A consistent mass discretization and the determinant search method of frequency analy sis are used.

Solution Results The theoretical solution for this problem is given in Example A44.

The input data on pages A46.4 and A46.5 is used in the finite element analysis and gives the following results:

Fundamental frequency for motion in the Y-Z plane (mode 2):

The mode shapes of the two lowest frequencies as calculated in the finite element analysis are shown on page A46.3.

User Hints

Note that the consistent element mass matrix for the SOLID element is always calculated using 3x3x3 Gauss points and that the user has no control over it. This ensures that the entire mass of the element is accounted for in the mass matrix.

Version 99.0 Linear Examples A46.1

SOLVIA Verification Manual FUNDAMENTAL FREQUENCY OF CANTILEVER, SOLID A16 ORIGINAL O.OS Z

E, MAST ER

0001T, B OlOl1 C II 11 1 SOLVIA ENGINEERING AR SOLVIA-PRE 99.0 Version 99.0 Linear Examples

-y A46.2

SOLVIA Verification Manual Linear Examples A46 FUNDAMENTAL FREQUENCY OF CANTILEVER, SOLID REFERENCE

0.

005 MAX DISPL.

0.32035 MODE I FREL 40.893 SOLVIA-POST 99 0 SOLVIA ENGINEERING AR Version 99.0 A46 FUNDAMENTAL FREQUENCY OF CANTILEVER, SOLID REFERENCE

0.05 Z

MAX DISPL.

0.319S MODE 2 FREO 81.16 X

Y DISPLACEMENT MAX 0.319S0 0.299S3 O.2S9S9 S0.21966

0. 17972

0. 13978 0.099843 0.0S9906 0.019969 MIN 0 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Z

I y A46.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A46 FUNDAMENTAL FREQUENCY OF CANTILEVER, SOLID' DATABASE CREATE MASTER IDOF-000111 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT FREQUENCIES DETERMINANT-SEARCH NEIG-2 SKEWSYSTEMS EULERANGLES

/

1 30.

SYSTEM 1

CARTESIAN COORDINATES ENTRIES NODE XL 1

-0.02 2

-0.02 3

0.02 4

0.02 5

-0.02 6

-0C02 7

0.02 8

0.02 9

0.0 10 0.0 11

-0.02f 12 0.02f MATERIAL EGROUP 1

PHI=30 YL 1.0 0.0 0.0 1.0 1.0 0.0 0.0 1.0 0.0 0.0 0.0 0.0 ZL 0.1 0.1 0.1 0.1 0.0 0.0 0.0 0.0 0.0 0.1 0.05 0.05 1

ELASTIC E=2.E11 NU=0.3 DENSITY=7800 SOLID GVOLUME 1 2 3 4 5 6 7 8 EL1=10 EL2=1 EL3=1 NODES=20 NSKEWS INPUT=SURFACE 2367 1

FIXBOUNDARIES FIXBOUNDARIES 123

/

2

/

ii 12 2 3 6 7 9 10 SET NSYMBOLS=MYNODES MESH NNUMBERS=MYNODES BCODE=ALL MESH NAXES=SKEW ENUMBERS=YES SOLVIA END Version 99.0 Linear Examples 5

5 5

5 A46.4

SOLVIA Verification Manual SOLVIA-POST input A46 FUNDAMENTAL FREQUENCY OF CANTILEVER, SOLID DATABASE CREATE WRITE FILENAME-'a46.1is' FREQUENCIES MASS-PROPERTIES SET RESPONSETYPE=VIBRATIONMODE ORIGINAL=DASHED NSYMBOLS=MYNODES MESH TIME=1 OUTLINE=YES MESH TIME=2 CONTOUR=DISPLACEMENT END Version 99.0 Linear Examples A46.5

SOLVIA Verification Manual EXAMPLE A47 FUNDAMENTAL FREQUENCY OF CANTILEVER, BEAM Objective To verify the dynamical behaviour of the BEAM element in frequency analysis when employing SKEW degree-of-freedom Systems.

Physical Problem Same as in figure on page A44. 1.

Finite Element Model The inclined cantilever is modeled using ten BEAM elements as shown on page A47.2. The element stiffness matrices are calculated in closed form and the mass is represented using a lumped mass matrix. The subspace iteration method of analysis is used.

Solution Results The theoretical solution for this problem is presented in Example A44.

The input data on page A47.4 gives the following result:

Fundamental frequency (Hz) for motion in the Y-Z plane (mode 2):

The mode shapes of the fundamental frequencies calculated in the finite element analysis for the X-Y and Y-Z planes are shown on page A47.3 as displayed by SOLVIA-POST.

User Hints

Employing consistent mass matrix in this finite element analysis gives a fundamental frequency, f = 81.15 Hz. Note that the consistent mass matrix for the BEAM element does not include contri butions due to shear deformation.

Version 99.0 Linear Examples A47.1

SOLVIA Verification Manual Linear Examples A47 FUNDAMENTAL FREQUENCY OF CANTILEVER, BEAM ORIGINAL 0--

.05 I

Y 9

2 MASTER 000000 B iii111 SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 ORIGINAL i

0.1 SOLVIA-PRE 99.0 A47 FUNDAMENTAL FREQUENCY OF CANTILEVER, BEAM 2

LY t

r s

EAXES=RST SOLVIA ENGINEERING AB Version 99.0 A47.2

SOLVIA Verification Manual Linear Examples A47 FUNDAMENTAL FREQUENCY OF CANTILEVER, BEAM REFERENCE H 0.05 MAX DISPL. i 0.31809 MODE 2 FREO 80.942 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 A47 FUNDAMENTAL FREQUENCY OF CANTILEVER, BEAM REFERENCE 0

0.0 Z

MAX DISPL. H 0.31878 MODE I FREO 40.6S2 X

Y DISPLACEMENT 0.31878 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB z

Ly

-x A47.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A47 FUNDAMENTAL FREQUENCY OF CANTILEVER, BEAM' DATABASE CREATE ANALYSIS TYPE=DYNAMIC MASSMATRIX-LUMPED FREQUENCIES SUBSPACE-ITERATION NEIG=2 SKEWSYSTEMS EULERANGLES 1 30 SYSTEM 1 CARTESIAN PHI=30 COORDINATES

/

ENTRIES NODE YL ZL 1

0 0 TO 11 1 0 NSKEWS 1 1 TO 11 1 MATERIAL 1

ELASTIC E=2.E11 NU=0.3 DENSITY=7800 EGROUP 1

BEAM SECTION 1

RECTANGULAR WTOP=0.05 D=0.1 BEAMVECTOR

/

1

1.

ENODES 1

-1 1 2 TO 10

-1 10 11 FIXBOUNDARIES

/

1 SET VIEW=X NSYMBOLS=YES MESH ENUMBER=YES BCODE-ALL MESH NNUMBER=YES EAXES=RST SOLVIA END SOLVIA-POST input A47 FUNDAMENTAL FREQUENCY OF CANTILEVER, BEAM DATABASE CREATE WRITE FILENAME='a47.1is' FREQUENCIES MASS-PROPERTIES SET RESPONSETYPE=VIBRATIONMODE ORIGINAL=DASHED NSYMBOLS-YES MESH VIEW=I TIME=i VECTOR=DISPLACEMENT MESH VIEW-X TIME=2 NMAX DIRECTION=1346 NUMBER=5 END Version 99.0 Linear Examples A47.4

SOLVIA Verification Manual EXAMPLE A48 FUNDAMENTAL FREQUENCY OF CANTILEVER, ISOBEAM Objective To verify the dynamical behaviour of the ISOBEAM element in frequency analysis when employing SKEW degree-of-freedom Systems.

Physical Problem Same as in figure on page A44. 1.

Finite Element Model The inclined cantilever beam is modeled using ten parabolic ISOBEAM elements as shown on page A48.2. The element stiffness matrices are evaluated using 2x4x4 Gauss points and the mass is represented in a diagonal mass matrix. The determinant search method of frequency analysis is used.

Solution Results The theoretical solution is the same as in Example A44.

The input data on page A48.4 gives the following result:

Fundamental frequency (Hz):

The mode shapes of the fundamental frequencies in the finite element analysis for the X-Y and Y-Z planes are shown on page A48.3.

User Hints

"* Employing a consistent mass matrix in the finite element analysis gives a fundamental frequency f = 81.24 Hz

"* Note that the shear factor for the ISOBEAM element is always equal to 1.0 and the user has not control over it.

Version 99.0 Linear Examples A48.1

SOLVIA Verification Manual Linear Examples A48 FUNDAMENTAL FREQUENCY OF CANTILEVER, ISOBEAM ORIGINAL

-0.05 z

Ly MASTER 000000 B 1I1lI1 SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 ORIGINAL 22 A48 FUNDAMENTAL FREQUENCY OF CANTILEVER, ISOBEAM z

L y

EAXES=RST SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 Version 99.0 B

A48.2

SOLVIA Verification Manual Linear Examples A48 FUNDAMENTAL FRECUENCY OF CANTILEVER, ISOBEAM REFERENCE

0. 05 MAX DISPL.

0.31975 MODE I FREO 40.801 REFERENCE

0. 1 MAX DISPL.

0.31915 MODE 2 FREG 81.298 I1l VIA-POST 99 n SOLVIA ENGINEERING AB Version 99.0 FY X

7 Y

A48.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A48 FUNDAMENTAL FREQUENCY OF CANTILEVER, ISOBEAM' DATABASE CREATE ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED FREQUENCIES DETERMINANT-SEARCH NEIG=2 SKEWSYSTEMS EULERANGLES 1 30 SYSTEM 1

CARTESIAN PHI=30 COORDINATES 1

TO 21

0.

1. /

22

0.

0.1 0.1 NSKEWS 1 1 TO 22 1 MATERIAL 1

ELASTIC E=2.E11 NU=0.3 DENSITY=7800 EGROUP 1

ISOBEAM SECTION 1

SDIM=0.100 TDIM=0.050 ENODES ENTRIES EL AUX NI N2 N3 1

22 1

3 2

TO 10 22 19 21 20 FIXBOUNDARIES 1 22 MESH VIEW=X ENUMBERS=YES BCODE=ALL MESH VIEW=X NNUMBERS=YES EAXES=RST SOLVIA END SOLVIA-POST input A48 FUNDAMENTAL FREQUENCY OF CANTILEVER, ISOBEAM DATABASE CREATE WRITE FILENAME-'a48.1is' FREQUENCIES MASS-PROPERTIES VIEW VIEW SET ID=i ZVIEW=-.

ROTATION=-90.

ID=2 XVIEW=1.

ROTATION=-30.

RESPONSETYPE-VIBRATIONMODE ORIGINAL-DASHED NSYMBOLS=YES MESH VIEW=1 TIME-I SUBFRAME=12 MESH VIEW=2 TIME=2 NPLINE NLINE NLINE END NAME=BEAM

/

1 TO 21 LINENAME=BEAM DIRECTION=i TIME-i SUBFRAME=21 LINENAME=BEAM DIRECTION=3 TIME=2 OUTPUT=ALL Version 99.0 Linear Examples A49.4

SOLVIA Verification Manual EXAMIPLE A49 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLATE Objective To verify the dynamical behaviour of the PLATE element in frequency analysis when employing SKEW degree-of-freedom Systems.

Physical Problem Same as in figure on page A44. 1.

Finite Element Model The finite element model consists of twenty PLATE elements as shown in the top figure on page A49.2. A lumped mass matrix is employed in the analysis, i.e. 1/3rd of the element mass is attributed to each element translational degree of freedom. The displacements are constrained to be zero in the X-direction. The subspace iteration method of frequency analysis is used.

Solution Results The theoretical solution for this problem is presented in Example A44.

Using the input data on page A49.3 the following result is obtained in the finite element analysis:

Fundamental frequency (Hz):

The bottom figure on page A49.2 shows the mode shape corresponding to the fundamental frequency calculated in the finite element analysis.

User Hints

"* A more refined finite element model (40 PLATE elements) and a lumped mass matrix yields the result f = 81.61 Hz while using a consistent mass matrix for this refined model gives f = 81.76 Hz.

"* Note that the consistent element mass matrix is not a consistent matrix in the usual sense, because linear variations in the displacements over the element are assumed and no mass is attributed to the rotational degrees of freedom.

Version 99.0 Linear Examples A49.1

SOLVIA Verification Manual Linear Examples A49 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLATE ORIGINAL --

0.05 z

I MAST ER 1 00001 8 1111O0 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB A49 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLATE REFERENCE I

0. 05 Z

MAX DISPL.

0.31797 MODE I

FREO 81.1S8 X

Y SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 A49.2

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A49 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLATE' DATABASE CREATE MASTER IDOF=100001 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED FREQUENCIES SUBSPACE-ITERATION NEIG=1 SKEWSYSTEM EULERANGLES 1 30 SYSTEM 1

CARTESIAN PHI=30 COORDINATES 1

/

2 0.05

/

3 0.05 1. /

4

0.

1.

MATERIAL 1

ELASTIC E=2.E11 NU=0.3 DENSITY=7800 EGROUP 1

PLATE GSURFACE 1 2 3 4 EL1=1 EL2=5 EDATA

/

1 0.1 NSKEWS INPUT=SURFACE 1234 1

FIXBOUNDARIES 234

/

1 2 MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES BCODE=ALL SOLVIA END SOLVIA-POST input A49 FUNDAMENTAL FREQUENCY OF CANTILEVER, PLATE DATABASE CREATE WRITE FILENAME='a49.1is' FREQUENCIES MASS-PROPERTIES SET RESPONSETYPE=VIBRATIONMODE ORIGINAL=DASHED MESH OUTLINE=YES NSYMBOL=YES NMAX DIRECTION-345 NUMBER=3 END Version 99.0 Linear Examples A49.3

SOLVIA Verification Manual EXAMPLE A50 FUNDAMENTAL FREQUENCY OF CANTILEVER, SHELL Objective To verify the dynamical behaviour of the SHELL element in frequency analysis when employing SKEW degree-of-freedom Systems.

Physical Problem Same as in figure on page A44. 1.

Finite Element Model Five 9-node SHELL elements are used to model the cantilever beam as shown in figures on page A50.2. The element stiffness matrix is calculated using a shear factor of 5/6. A consistent mass discretization is used. The determinant search method of analysis is employed in the frequency analysis.

Solution Results The theoretical solution for this problem is presented in Example A44.

The input data on page A50.4 gives the following result:

Fundamental frequency (Hz) for motion in the Y-Z plane (mode 2):

The figures on page A50.3 shows the mode shapes corresponding to the fundamental frequencies calculated in the finite element analysis.

User Hints Employing a lumped mass matrix in the finite element analysis yields the result f = 80.73 Hz.

Version 99.0 Linear Examples A50.1

SOLVIA Verification Manual Version 99.0 ASO FUNDAMENTAL FREQUENCY OF CANTILEVER, SHELL ORIGINAL 0.05 z

x y

t EAXES=RST SOLVIA-PRE 99.0 SOLVTA ENGINEERING AB Linear Examples A50.2

SOLVIA Verification Manual ASO FUNDAMENTAL FREQUENCY OF CANTILEVER, SHELL REFERENCE 0.05 MAX DISPL. i 0.31759 MODE I

FREC 40.912 X

X y

DISPLACEMENT MAX 0.31759 0.29771 025804 0.21834 0.17865 0

13895 0.099247 O 059S48 0.019849 MIN 0 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 ASO FUNDAMENTAL FRECUENCY OF CANTILEVER, SHELL REFERENCE

0. 05 Z

MAX DISPL.

0.31794 MODE 2 FRED 81.003 x

Y SOLVIA-POST 99.0 SOLVIA ENGINEERING AS Linear Examples A50.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A50 FUNDAMENTAL FREQUENCY OF CANTILEVER, SHELL' DATABASE CREATE ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT FREQUENCIES DETERMINANT-SEARCH NEIG=2 SKEWSYSTEM EULERANGLES 1

30.

SYSTEM 1

CARTESIAN PHI=30 COORDINATES 1

TO 3

0.05

/

4 0.05 1.

/

5

0.

1.

MATERIAL 1

ELASTIC E-2.E11 NU=0.3 DENSITY-7800 EGROUP 1

SHELL THICKNESS 1

T1=0.1 GSURFACE 1 3 4 5

EL1=1 EL2=5 NODES=9 NSKEWS INPUT=SURFACE 1345 1

FIXBOUNDARIES 12346

/

1 3 FIXBOUNDARIES

/

2 SET NSYMBOLS=MYNODES MESH NNUMBERS=MYNODES ENUMBERS=YES BCODE=ALL MESH EAXES-RST SOLVIA END SOLVIA-POST input A.50 FUNDAMENTAL FREQUENCY OF CANTILEVER, SHELL DATABASE CREATE WRITE FILENAME='a50.1is' FREQUENCIES MASS-PROPERTIES SET RESPONSETYPE=VIBRATIONMODE ORIGINAL=DASHED NSYMBOL=MYNODES MESH TIME=i CONTOUR=DISPLACEMENTS MESH TIME=2 OUTLINE=YES NMAX NUMBER=3 END Version 99.0 Linear Examples A50.4

SOLVIA Verification Manual EXAMPLE A51 FUNDAMENTAL FREQUENCY OF A SIMPLY SUPPORTED PLATE Objective To verify the dynamical behaviour of the SHELL element in frequency analysis.

Physical Problem The figure below shows the simply supported plate to be analyzed.

a E

2.0.10" N/m 2 v =0.3 a= 2.00 m h=0.01 m p = 7850 kg/m 3 a

Finite Element Model The figure on page A51.2 shows the finite element model. Because of symmetry, only one quarter of the plate need to be modeled. Four 9-node SHELL elements are used. A lumped mass matrix is employed and the subspace iteration method is used for the frequency calculation.

Solution Results The first natural frequency of a simply supported plate is given for example in [I] p. 258 as follows:

SE.h3 a2 12.p.h. -vT )

Insertion of the numerical values gives fl = 12.00 Hz The SOLVIA solution using the input data on page A51.4 gives f1 = 12.17Hz The corresponding mode shape and distributions of von Mises effective stress and strain energy density are given in figures on page A51.3.

Version 99.0 Linear Examples A5 1.1

SOLVIA Verification Manual The calculated sum of the strain energy for the first mode is 2923.41 Nm which is equal to T D (2c 2923.41 7-P1 Kq 91 = 1/2(2irf1 )2 =29.4 as it should.

User Hints

"* Note that full integration (default) is necessary when a consistent mass matrix is used since reduced integration (2x2x2) gives a spurious mode of very small frequency for this example.

"° A consistent mass matrix gives 12.15 Hz for the fundamental mode.

"* Simply supported boundary conditions for a plate are discussed in Example A27.

"* A finer mesh should be used if detailed stresses are to be calculated.

Reference

[1]

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

ASI FUNDAMENTAL FREQUENCY OF A SIMPLY SUPPORTED PLATE ORIGINAL v--

0.1 z I x/'-

y MASTER 110001 B 110011 C I1010L SLI 10111 E 111001 F II1011 G i110 1 SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 Version 99.0 Linear Examples A51.2

SOLVIA Verification Manual ASI FUNDAMENTAL FREQUENCY OF A SIMPLY SUPPORTED PLATE REFERENCE

0.

01 MAX DISPL.

0.22527 MODE I

FREO 12.17 SOLVIA-POST 99.0 Linear Examples z

X Y

REACTION 13911 SM MISES SHE LL TOP MAX 8.2203E8 S7.8416E8

...... 7. 0842E8

6. 3269E8 S. 5695E8 4-8122E8

'4.

0518E8

3. 297SE8 2.5401E8 MIN 2.1614E8 SOLVIA ENGINEERING AB f

Version 99.0 ASI FUNDAMENTAL FREQUENCY OF A SIMPLY SUPPORTED PLATE MAX DISPL.i 0.22527 Z

MODE 1 FRE0 12.17 X Y STRAINENERGY DENSITY SHELL TOP MAX 2.0594E6 S1.9392E6 1.6987E6 1.4582E6 1.2177E6

9. 7723ES

7. 367SE5 4 9627ES 2.5579ES MIN 1.3SSSES SOLVTA-POST 99.0 SOLVIA ENGINEERING AB A51.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A51 FUNDAMENTAL FREQUENCY OF A SIMPLY SUPPORTED PLATE' DATABASE CREATE MASTER IDOF=-I0001 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED FREQUENCIES SUBSPACE-ITERATION NEIG=1 MODALSTRESSES=YES COORDINATES 1 /

2 1.

3

1.

/

4

0.

1.

MATERIAL 1

ELASTIC E=2.E11 NU=0.3 DENSITY=7850 EGROUP 1

SH GSURFACE 1 2 THICKNESS 1

FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES ELL 3 4 EL1=2 EL2=2 NODES=9 0.01 3 INPUT=LINES 4 INPUT-LINES 5 INPUT=LINES

/ /

2 3 1 2

/

3 4

/

4 1 SET NSYMBOLS=MYNODES MIDSURFACE=NO MESH ENUMBERS=YES BCODE=ALL SOLVIA END SOLVIA-POST input A51 FUNDAMENTAL FREQUENCY OF A SIMPLY SUPPORTED PLATE DATABASE CREATE WRITE FILENAME='a51.lis' FREQUENCIES MASS-PROPERTIES SET MESH MESH RESPONSETYPE=VIBRATIONMODE CONTOUR=MISES VECTOR=REACTION ORIGINAL=YES CONTOUR=ENERGY NMAX NUMBER=3 SUMMATION KIND=ENERGY END DETAILS-YES Version 99.0 Linear Examples A51.4

SOLVIA Verification Manual EXAMPLE A52 WAVE PROPAGATION IN A ROD Objective To verify the dynamical behaviour of the TRUSS element in wave propagation analysis.

Physical Problem A uniform bar free at both ends is considered, see figure below. An axial step force is applied at one end of the bar at time 0.

P X

i, L

II a E =2.0-109 N/m2 p= 1000 kg/im3 t [s]

Finite Element Model The finite element model is shown in the top figure on page A52.4. The bar is modeled using ten 2-node TRUSS elements. The explicit central difference method is chosen for the time integration and a lumped mass matrix is employed.

The time step At is selected according to At pL Le c

BE Version 99.0 a=0.1 m L= 1.0m P [N]

103 Linear Examples A52.1

SOLVIA Verification Manual where L, = length of one truss element c

= wave velocity in the material p

= mass density E

= Young's modulus Since the step load is applied at time 0 it is necessary to specify the corresponding initial acceleration according to P

a,=-ml where m1 = mass at node 1 P

applied force a,

= acceleration at node I Solution Results The theoretical solution for this problem is presented, for example, in [I].

The input data on page A52.5 gives the following results considering 4 time steps:

Longitudinal stress in bar, cy (N /m2):

Theory SOLVIA Before wave front 0

0 After wave front

-1.0.10 5

-1.0-101 The wave front can be seen in the SOLVIA solution to advance one element length per specified time step, which is in agreement with the theoretical solution.

The velocity and acceleration time history for the typical node 2 is shown in the bottom figure on page A52.4.

User Hints

"* In this example with equal lengths of the TRUSS elements the exact theoretical solution is obtained by selecting the time step to be equal to the critical time step.

"* Note that the load vector input by time functions at the start of the solution (time=0) is not used in SOLVIA. Instead the load vector 'R at time 0 is assumed to be

°R = M °U + C°U +°F Version 99.0 Linear Examples A52.2

SOLVIA Verification Manual where 'U and °0 are the initial velocity and acceleration vectors and M and C are the mass and damping matrices, respectively. F is a vector with the nodal equivalent forces corresponding to the initial displacements. Therefore, the initial acceleration is specified as described above to simulate the applied force at time 0. To start the central difference solution procedure AtU is calculated from the initial conditions as A tU=U+At-OU+ At,-.U 2

" Note that the finite element model is completely free to move in the X-direction. If this model is used for a static problem, no solution would be possible since the model contains rigid body modes (and the solution is non-unique). In the dynamic case, however, the inertia of the bar results in a unique solution, and no solution difficulties are encountered.

" In explicit time integration the total system stiffness matrix is not calculated and no matrix factorization is carried out because with explicit time integration the mass matrix in SOLVIA is required to be diagonal. The solution for one time step is therefore obtained very effectively but a large number of time steps may need to be used, since the time step At must satisfy Tn At_< At,, = T 7C where T, is the smallest period of the finite element assemblage with n degrees of freedom.

Reference

[1]

Zukas, J.A., Nicholas, T., Swift, H.F., Greszczuk, L.B. and Curran, D.R., Impact Dynamics, John Wiley & Sons, 1982.

Version 99.0 Linear Examples A52.3

SOLVIA Verification Manual Linear Examples AS2 WAVE PROPAGATION IN A ROD ORIGINAL

0.O TIME 7.071E-S 1

2 3

4 S

6 7

8 9

I0 I

z Iy 1

2 3

4 5

6 7

8 9

10 FORCE 1000 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB A52 WAVE PROPAGATION IN A ROD

/L

/i t

i

/

/ /

/

/t

/

cc

-J U]

U U

0 z

0.

0 0.0 0.[

0.2 TIME IIME SOLVIA-POST 99.0 0.3

.10-3 SOLVIA ENGINEERING AB Version 99.0 C

C -

U]

aCD a

C o

/

0.0 A52.4

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A52 WAVE PROPAGATION IN A ROD' DATABASE CREATE MASTER IDOF=-01111 NSTEP-4 DT=7.071E-5 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=CENTRAL COORDINATES 1

TO 11

0.

1.

INITIAL ACCELERATION 1

0.

2.E3 0.

MATERIAL 1

ELASTIC E=2.E9 DENSITY=1000 EGROUP 1 TRUSS ENODES 1

1 2 TO 10 10 11 EDATA

/

1 0.01 LOADS CONCENTRATED 1 2 1000.

SET VIEW-X NSYMBOLS=YES MESH NNUMBERS=YES ENUMBERS=YES VECTOR=LOAD SOLVIA END SOLVIA-POST input A52 WAVE PROPAGATION IN A ROD DATABASE CREATE WRITE FILENAME='a52.1is' MASS-PROPERTIES SUBFRAME 21 NHISTORY NODE-2 DIRECTION=2 KIND=VELOCITY SYMBOL=I NHISTORY NODE=2 DIRECTION=2 KIND=ACCELERATION SYMBOL=1 ELIST TSTART=7.071E-5 TEND=2.8284E-4 END Version 99.0 Linear Examples A52.5

SOLVIA Verification Manual EXAMPLE A53 WAVE PROPAGATION IN A WATER COLUMN Objective To verify the dynamical behaviour of the FLUID3 element in wave propagation analysis.

Physical Problem Same as in the figure on page A52.1 except that the rod is replaced by a water column of the same wave velocity c.

Finite Element Model Ten 8-node FLUID3 elements are modeling the water column as shown in the top figure on page A53.2. A lumped mass matrix is used in the analysis and the explicit central difference method is chosen for the time integration. The element stiffness matrices are calculated using one-point integration. Time step and initial conditions are selected as for Example A52.

Solution Results The theoretical solution is the one presented in Example A52.

The input data on page A53.3 gives the following results:

Pressure in water column, p (N/im 2):

Theory SOLVIA Before wave front 0

0 After wave front

-1.0.101

-1.0.10 5 The bottom figure on page A53.2 shows the pressure distribution in the water column at the fourth time step predicted by the finite element analysis which is in excellent agreement with the analytical solution.

User Hints See Example A52.

Version 99.0 Linear Examples A53.1

SOLVIA Verification Manual Linear Examples A53 WAVE PROPAGATION IN A WATER COLUMN z

x y

ORIGINAL j 0.05 TIME 7.071E-9 SOLVIA-PRE 99.0 PRESSURE 100E000 SOLVIA ENGINEERING AB PROPAGATION IN A WATER COLUMN 5

.,0 0.2 0.

TIME 2.828 E-0.6 0.8 1..

COLUMN SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 A53 WAVE A

4 1

1

U

-j LtJ Li Lt Li 71 (D

0 o4 04 o

I 0.0 0

0 Li C0 a-0.1 TIME 0.2 0.3

I-3 0

A53.2 oo

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A53 WAVE PROPAGATION IN A WATER COLUMN' DATABASE CREATE MASTER IDOF=I01111 NSTEP=4 DT=7.0710E-5 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=CENTRAL COORDINATES 1

.1 o..1

/

2

o. o..1

/

3

/

4

.1 INITIAL ACCELERATION 1 0.

2.E3 TO 4 0.

2.E3 NGENERATION NSTEP=4 YSTEP=1.

/

1 TO 4

MATERIAL 1

FLUID K=2.E9 DENSITY=1000.

EGROUP 1

FLUID3 RSINT=1 TINT=1 GVOLUME 1 2 3 4 5 6 7 8

EL1=1 EL2=1 EL3=10 NODES=8 LOADS ELEMENT INPUT=SURFACE 1 2 3 4 i.E5 MESH NSYMBOLS=MYNODES NNUMBERS-MYNODES VECTOR-LOAD SOLVIA END SOLVIA-POST input A53 WAVE PROPAGATION IN A WATER COLUMN DATABASE CREATE WRITE FILENAME='a53.1is' SUBFRAME 21 NHISTORY NODE=4 DIRECTION=2 KIND VELOCITY SYMBOL-i EPLINE NAME=COLUMN

/

1 1 TO 10 1 ELINE LINENAME=COLUMN KIND=PRESSURE SYMBOL=i OUTPUT=ALL END Version 99.0 Linear Examples A53.3

SOLVIA Verification Manual EXAMPLE A54 CANTILEVER SUBJECTED TO GROUND MOTION Objective To verify the dynamical behaviour of the BEAM element and the solution procedure for mass proportional loading due to a ground acceleration.

Physical Problem A water tower subjected to ground acceleration, as shown in the figure below, is considered. The time history of the ground motion is also shown in the figure.

u (M)

M k =2.7-106 lbf/ft M = 3000 lbf. s2 / ft a0 = 32.2 ft/s 2

k X

a (t) f~ IS]

0.050 a [t)

Finite Element Model One BEAM element is used to model the tower and the water tank is modeled as a concentrated mass.

The transverse stiffness is set equal to the spring constant k for the tower. Thus 3EI k =2.7.106 lbf/ft and we select (for ease of computation)

Version 99.0 Linear Examples A54.1

SOLVIA Verification Manual I = 1 ft 4 L=10ft E= 0.9.109 lbf/ ft 2

The Newmark method of time integration is chosen with a. = 0.25 and 5= 0.50. The time step size is 0.005 sec.

Solution Results The input data on page A54.4 is used in the finite element analysis. The table below shows the calculated results for some time steps in the analysis together with the analytical solution and results reported in [1].

Transverse displacement, u(t) [ft]

Time Analytical Ref. [1]

SOLVIA 0.010 0.000214 0.0002 0.000239 0.020 0.00169 0.0017 0.00173 0.030 0.00551 0.0055 0.00554 0.040 0.0113 0.0114 0.0113 0.050 0.0174 0.0176 0.0174 0.075 0.0255 0.0254 The figures on page A54.3 shows the time history of the transverse displacement, velocity and acceleration at the top of the water tower and the bending moment at the built-in end of the BEAM element, as displayed by SOLVIA-POST.

User Hints "Note that the forces to which the model is subjected are calculated as MU + KU = -MdEiig where d, is a vector with zeroes in all components except +1 (plus one) in each of the components corresponding to the X-displacements. Hence, the same acceleration forces are applied to all X displacements degrees of freedom, which means that arrival time differences in the accelerations at different supports are not modeled.

"* Note that a ground acceleration in the negative X-direction corresponds to an applied mass proportional loading in the positive X-direction.

Note also that the X-Y-Z system is moving with the ground and we obtain a response measured in this moving system. Therefore, to obtain, for example, the absolute acceleration response (which is measured in a fixed system) it is necessary to add or subtract the ground acceleration time history depending on whether the ground acceleration acts in the positive or negative coordinate direction.

Reference

[1]

Clough, R.W. and Penzien, J., Dynamics of Structures, McGraw-Hill, 1975, pp. 102-105.

Version 99.0 Linear Examples A54.2

SOLVIA Verification Manual Linear Examples AS4 CANTILEVER SUBJECTED TO GROUND MOTION

,,,I

.i.

/4.

i

/

\\

o

/

\\

[

I 1

CO 0

z U)

LJ C

LU I/

0

/

0.00 0

.05 0

.10 0.15 0.20 0.25 0

Nx N

0

10 0

10

/*

/

.00 TIME 0.05

0. 10 0 15 0.20 0.25 TIME SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A54.3 A54 CANTILEVER SUBJECTED TO GROUND MOTION o2

]

C, S1~

ae

/

'7U"

/

0I i

/

\\

.J o

/

1u/

10 1

0.00 0.05

0. 0 015 0.20 0.25 0.00 0.05

0. 10

0.

i 0.0 0.2 T I E TIME SOLVIA POST 99.0 SOLVIA ENGINEERING AB UJ

(-I (J

0 z Version 99.0 t

t 0

I

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A54 CANTILEVER SUBJECTED TO GROUND MOTION' DATABASE CREATE MASTER IDOF=001110 NSTEP=50 DT=0.005 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=NEWMARK, DELTA=0.5 ALPHA=0.25 TIMEFUNCTION 1

0.

/

.025

1.

/

.050

0.

/

1.

0.

COORDINATES 1 /

2

0. 10.

/

3

-3.

MASSES 2

3000.

MATERIAL 1

ELASTIC E-0.9E9 EGROUP SECTION ENODES 1 1

/

BEAM RESULT=FORCES GENERAL RINERTIA= 1.

1 3

1 2 FIXBOUNDARIES

/

1 3 LOADS MASSPROPORTIONAL SOLVIA END SOLVIA-POST input SINERTIA=I.

TINERTIA=1.

AREA=I.

XFACTOR=1.

ACCGRA=32.2 A54 CANTILEVER SUBJECTED TO GROUND MOTION DATABASE CREATE WRITE FILENAME='a54.1is' 21 NODE=2 DIRECTION=1 NODE=2 DIRECTION=i KIND=DISPLACEMENT OUTPUT=ALL KIND=VELOCITY 21 NODE=2 DIRECTION-i KIND=ACCELERATION EL=i POINT=i KIND=MT Version 99.0 SUBFRAME NHISTORY NHISTORY SUBFRAME NHISTORY EHISTORY END Linear Examples A54.4

SOLVIA Verification Manual EXAMPLE A55 CYLINDRICAL TUBE UNDER STEP LOADING Objective To verify the dynamical behaviour of the PLANE AXISYMMETRIC element and the use of direct time integration.

Physical Problem A cylindrical tube initially at rest is subjected to a uniform circumferential line load at its midspan at time t=0+, as shown in the figure below.

P L/2 0° L = 18.0 in D= 6.0 in h = 0.3 in E = 3.0.107 psi v = 0.3 p = 3.663-10-2 lbf.sec 2/ in.4 t [s]

Finite Element Model The finite element model used in the analysis is shown in the left figure on page A55.2. Using symmetry considerations only one-half of the tube is modeled with sixteen 8-node PLANE AXISYMMETRIC elements. The trapezoidal rule (the Newmark method with cu=0.25 and 5=0.50) is Version 99.0 P

P [lb/in]

103 0

Linear Examples y

A55.1

SOLVIA Verification Manual Linear Examples employed in the analysis to obtain the step-by-step dynamic response. A step size of 0.00001 sec. is used and the solution response is evaluated for 60 steps in the analysis. A consistent mass matrix is used.

Solution Results The theoretical solution for this problem is presented in [1] pp. 55-56. The input data on page A55.3 is used in the finite element analysis.

The right figure on page A55.2 shows the deformed finite element mesh at time t=0.00060 sec. and a time-history curve of the radial displacements at the symmetry boundary. The calculated results are in good agreement with the theoretical solution.

User Hints The trapezoidal rule is unconditionally stable, and the time step size is therefore selected based on accuracy considerations only. Note that a time step for use of the central difference method would be considerably smaller.

Reference

[11 "SAP IV - A Structural Analysis Program for Static and Dynamic Response of Linear Systems", Report EERC 73-11, Univ. of Calif., Berkeley, Calif., 1974 (revised).

Version 99.0 ASS CYLINDRICAL TUBE UNDER STEP LOADING ORIGINAL 0.5 Z

ORIGINAL I

Z 3C2 LY TIME IE-S LT R

R FORCE 1500 MASTER 100111 S.

A*

B 111111r C 110111 EAXES=RST SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB ASS CYLINDRICAL TUBE UNDER STEP LOADING ORIGINAL I-

1.

MAX DISPL.

- 3.2664E-4 TIME 6E-4 VA R

0i 0.0 0, 02 0l3 04 oU 0.4

.IO-a 1 IME SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A55.2

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'A55 CYLINDRICAL TUBE UNDER STEP LOADING' Linear Examples DATABASE CREATE MASTER IDOF=100111 NSTEP=60 DT=I.E-5 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT METHOD:NEWMARK TIMEFUNCTION 1

0.

3000.

/

1 3000.

COORDINATES 1

3.15

/

/ 2 ENTRIES NODE Y

Z 3.15

9.

/

3 2.85

9.

/

4 2.85 MATERIAL 1 ELASTIC E=3.0E7 NU=0.3 DENSITY=3.663E-2 EGROUP 1

PLANE AXISYMMETRIC GSURFACE 1 2 3 4 EL1=16 EL2=1 NODES=8 FIXBOUNDARIES FIXBOUNDARIES 3

2 INPUT=LINES INPUT=NODES

/

14

/

2 LOADS CONCENTRATED 1 2 0.5 SET MESH MESH PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES NNUMBERS=MYNODES ENUMBERS=YES BCODE=ALL SUBFRAME=21 VECTOR=LOAD EAXES=RST SOLVIA END SOLVIA-POST input A55 CYLINDRICAL TUBE UNDER STEP LOADING DATABASE CREATE WRITE FILENAME='a55.1is' SET VIEW MESH PLOTORIENTATION=PORTRAIT ID=-

XVIEW=1 ROT=-90 VIEW=i ORIGINAL=DASHED SUBFRAME=12 NHISTORY END NODE=i DIRECTION=2 OUTPUT=ALL Version 99.0 A55.3

SOLVIA Verification Manual EXAMPLE A56 FREQUENCIES OF A WATER-FILLED ACOUSTIC CAVITY Objective To verify the PLANE conduction element in SOLVIA-TEMP when used for frequency analysis.

Physical Problem The acoustic cavity considered for the analysis is shown in the figure below. The cavity is bounded by rigid walls and filled with water which is assumed to be inviscid [1]. The natural frequencies of the cavity are to be determined.

Rigid boundary a= 12 in b = 20 in Water P 3.16.106 psi (bulk modulus) p = 9.35 10-5 lbsec 2/in4 (density)

Material specification to SOLVIA-TEMP:

3.16. 05 (conductivity) 9.35.10-5 (specific heat per unit volume) i a

A Finite Element Model The wave equation governing the motion of the fluid inside the cavity is 0 2 c

+ 2

_ 1 I 2

C [

ay 2 az 2 c 2 at 2 p

= 0 at the boundary an where 0 is the velocity potential, c is the velocity of sound in water, [3 is the bulk modulus and p is the density of the water. The natural frequencies of the cavity can be obtained by performing a frequency analysis of a planar heat flow problem in which the conductivity (k) and specific heat (c) correspond to [3 and p, respectively. Using this approach, the finite element model shown in the top figure on page A56.3 is employed in the analysis. Four 8-node PLANE conduction elements are used to model the cavity. The degrees-of-freedom of all the nodal points are left free corresponding to the boundary Version 99.0 Linear Examples A56.1

SOLVIA Verification Manual conditions of the cavity. In the finite element frequency analysis, a consistent heat capacity assumption is used.

Solution Results The input data shown on page A56.4 is used in the finite element frequency analysis. The lowest four frequencies predicted by the finite element model are shown in the table below. Analytical solutions evaluated using formulas given in [I] are also shown for comparison.

SOLVIA-TEMP Analytical *[ 1]

co, (n =0,m =0) 0 0

0o2 (n 0,m= 1) 9166 9132 (03 (n =,m 0) 15277 15220 o4 (n =1,m =1) 17763 17749

)

to=iTC-c.

+ý T

j ;m,n = 0,1,2,3,..

Contour plots of temperature eigenvectors can be seen in the bottom figure on page A56.3.

User Hints SOLVIA-TEMP calculates the eigenvalues to the equation K4=CýA where K

C A

heat conductivity matrix heat capacity matrix matrix containing eigenvectors diagonal matrix with eigenvalues To obtain the circular eigenfrequencies coi of the wave equation, we must, therefore, take the square root of the SOLVIA-TEMP eigenvalues.

Since only the gradient of the potential is zero at the boundaries while the potential is free, there is one zero frequency (rigid body) mode in this problem.

Reference

[1]

Blevins, R.D., Formulas for Natural Frequency and Mode Shape, Van Nostrand Reinhold, 1979, pp. 337-341.

Version 99.0 Linear Examples A56.2

SOLVIA Verification Manual AS6 FREQUENCIES OF A WATER-FILLED ACOUSTIC CAVITY ORIGINAL 2

z L y r

S EAXES=RST SOLVIA-PRE 99.0 ORIGINAL H--

2.

z Ly SOLVIA ENGINEERING AB AS6 FREQUENCIES OF A WATER-FILLED ACOUSTIC CAVITY REFERENCE H

2.

MODE 2 EIGENV 8-4018E7 Z

REFERENCE d-----

2.

L/y MODE 4

E:GENV 3.iSSiE8 TEMPERATURE MAX 9.4980 8.3107 5.9362 3.5617 1.1872

- 1 1872

-3.S617

-5.9362

-8.3107 MIN-9.4980 SOLVIA-POST 99.0 STEMPERATURE 1I1.032 7.8797 4.. 7278 I. 5759

-I.i 7S9

-4.7278

-7.8797

-1t.032 MIN-12.608 SOLVIA ENGINEERING AB Version 99.0 y

LY Linear Examples A56.3

SOLVIA Verification Manual SOLVIA-PRE input HEAD

'A56 FREQUENCIES OF A WATER-FILLED ACOUSTIC CAVITY' DATABASE CREATE T-ANALYSIS HEATMATRIX=CONSISTENT T-FREQUENCIES NEIG=4 IRBM=1 COORDINATES ENTRIES NODE Y

Z 1

0.

2

12.

0.

3

12.

20.

4

0.

20.

T-MATERIAL 1

CONDUCTION K=3.16E5 SPECIFICHEAT=9.35E-5 EGROUP 1

PLANE STRAIN GSURFACE 1 2 3 4 EL1=2 EL2=2 NODES=8 SET NSYMBOLS=MYNODES MESH NNUMBERS=MYNODES EAXES=RST SUBFRAME=21 MESH ENUMBERS=YES GSCALE=OLD SOLVIA-TEMP END SOLVIA-POST input A56 FREQUENCIES OF A WATER-FILLED ACOUSTIC CAVITY T-DATABASE CREATE WRITE FILENAME='a56.1is' FREQUENCIES SET RESPONSETYPE=VIBRATIONMODE OUTLINE=YES NSYMBOLS=YES MESH CONTOUR=TEMPERATURE TIME=2 SUBFRAME-21 MESH CONTDUR-TEMPERATURE TIME-4 END Version 99.0 Linear Examples A56.4

v t e Letter Sequence Response to RAI
TAC:MB3603, (Open) TAC:MB3604, (Open)
Initiation Request, Request, Request, Request, Request, Request, Request Acceptance... Supplement Administration Questions 31 May 2002 Answers 13 March 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 Results Approval... Other: L-PI-03-016, Response to Opportunity for Comment on Task Interface Agreement (TIA) 2001-10, Design Basis Assumptions for Ability of Prairie Island, Unit 2, Emergency Diesel Generators to Meet Single-Failure Criteria for External Events, ML013480323, ML020020074, ML020020108, ML020030002, ML020410002, ML020570514, ML021060641, ML021070317, ML021140405, ML021140426, ML021610096, ML021990405, ML022140006, ML022600292, ML023290377, ML023640310, ML031010026, ML032040412, ML050900176, ML051030051
MONTHYEAR2002-05-28 [Table View]

ML022030370

Text

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