ML022030417
| ML022030417 | |
| Person / Time | |
|---|---|
| Site: | Cook  |
| Issue date: | 07/16/2002 |
| From: | Greenlee S Indiana Michigan Power Co |
| To: | Document Control Desk, Office of Nuclear Reactor Regulation |
| References | |
| AEP:NRC:2520, TAC MB3603, TAC MB3604 | |
| Download: ML022030417 (135) | |
Text
SOLVIA Verification Manual EXAMPLE A81 HARMONIC RESPONSE OF A SIMPLY SUPPORTED BEAM Objective To verify the harmonic response method when different boundary conditions of a model are used.
Physical Problem A simply supported beam is exposed to a sinusoidal force as shown in the figure below.
F = F0 sin cot E=2.0.101" N/m 2
- a
=
a0.1lm a
x=7.5m L
p = 7800 kg/m3 "lF,,
= 100 N Finite Element Model The model consists of eight BEAM elements. Due to symmetry of the structure it is enough to model one half of the beam. The sinusoidal load is non-symmetric and, therefore, two runs with different boundary conditions need to be done. This analysis procedure may be of interest if the structure is very large. The procedure is used here for demonstration and verification reasons only.
Solution Results The input data used in the two analyses are shown on pages A81.3 and A81.4.
The nodal results from symmetric and anti-symmetric boundary conditions are saved in the SOLVIA POST database. Mode number 1 and 2 are calculated with symmetric boundary and mode 3 is calcu lated with anti-symmetric boundary. The mode shapes with their boundary conditions are shown in the top figures on page A8 1.2.
The harmonic response for the Z-displacement, node 2 and the stress component stress-rr in element 5, point 1 have been calculated and are shown in the bottom figures on page A81.2.
Numerical results for 9 Hz including corresponding theoretical values based on three mode shapes of an Bernoulli-Euler continuous beam:
Version 99.0 SOLVIA-POST Theory Z-disp, node 2 [m]
1.375.10-3 1.355.10-'
Stress-rr, element 5 5.616.106 5.473.106 point 1 [N/m2]
Linear Examples A81.1
SOLVIA Verification Manual Linear Examples i
I S
gJI So V:
05 H
I0 IS FREOUTENCY SOLVIA ENGINEERING AB Version 99.0 A81 HARMONIC RESPONSE OF A SIMPLY SUPPORTED BEAM EXCITATION IN NCOý 2 DIR 3 I
S I0 IS 2C AS FRECUENCY oL P.S.
I S
E EI SOLVIA-POST 99.0 SOLVIA ENGINEERING AS A81.2
SOLVIA Verification Manual SOLVIA-PRE input HEADING
'A81 HARMONIC RESPONSE OF A SIMPLY SUPPORTED BEAM' DATABASE CREATE MASTER IDOF=010101 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT FREQUENCIES SUBSPACE-ITERATION NEIG:2 SSTOL=1.E-10, MODALSTRESS=YES COORDINATES 1
TO 3
- 5.
MATERIAL 1
ELASTIC E=2.E11 NU=0.
DENSITY-7800.
EGROUP 1
SECTION 1
BEAMVECTOR GLINE Ni=I BEAM RESULTS=STRESSES RECTANGULAR WTOP=0.1 D=0.1
/
1
- 0. 0. 1.
N2=3 AUX=-1 EL=8 First run, symmetric FIXBOUNDARIES FIXBOUNDARIES SOLVIA END SOLVIA-POST input 13 15
/1
/ 3 A81 HARMONIC RESPONSE OF A SIMPLY SUPPORTED BEAM First run, symmetric DATABASE CREATE END Version 99.0 Linear Examples A81.3
SOLVIA Verification Manual Linear Examples SOLVIA-PRE input DATABASE OPEN FREQUENCIES SUBSPACE-ITERATION NEIG=1 SSTOL=i.E-10, MODALSTRESS=YES Second run, anti-symmetric DELETE BOUNDARIES FIXBOUNDARIES 13
/
1 FIXBOUNDARIES 3
/
3 SOLVIA END SOLVIA-POST input A81 HARMONIC RESPONSE OF A SIMPLY SUPPORTED BEAM Second run, anti-symmetric DATABASE ADD WRITE FILENAME='a8ib.lis' FREQUENCIES MASS-PROPERTIES SET PLOTORIENTATION=PORTRAIT SET RESPONSETYPE=VIBRATIONMODE VIEW=Y NSYMBOLS=MYNODES, ORIGINAL=YES MESH DEFORMED=NO EAXES=STRESS-RST SUBFRAME=i2 MESH TIME=i BCODE=ALL NNUMBERS=MYNODES MESH TIME=2 BCODE=ALL ENUMBER=YES SUBFRAME 12 MESH TIME=3 BCODE=ALL DAMPING 1
/
1 2 TO 3 2 EHARMONIC-RESPONSE ELEMENT=5 POINT=i KIND=SRR FSTART=0.5 FSTEP=0.25, FEND=25 AXES=1 OUTPUT=ALL MEND=3 2 3 50.
- 0.
END DATA NHARMONIC-RESPONSE NODE=2 DIRECTION=3 KIND=DISPLACEMENT FSTART=0.5, FSTEP=0.25 FEND=25 AXES=i OUTPUT=ALL MEND=3 2 3 50.
- 0.
END DATA ZONE NAME=EL45 INPUT=ELEMENTS
/
4 5 ELIST ZONENAME=EL45 TSTART=I TEND=3 NLIST TSTART=1 TEND=3 END Version 99.0 A81.4
SOLVIA Verification Manual EXAMPLE A82 STIFFENED PLATE CANTILEVER UNDER TIP MOMENT Objective To verify the use of the linear BEAM element offset capability.
Physical Problem A cantilever of channel cross-section is loaded by concentrated end moments as shown in the figure on page A23.1.
Finite Element Model The model is shown in the top figure on page A82.2. The BEAM element nodes lie in the same plane (Z=O) as the PLATE element nodes. By using the offset option in the local t-direction (TOFFSET) a translation of the BEAM elements from the plane of the nodes is made.
Solution Results The theoretical displacement is ML 2
2EI The input data on page A82.3 gives the following result:
End displacement 6 (inch):
User Hints
"* Note that if the plate cantilever is subjected to concentrated force end loads as described in Example A23 a finer mesh is required due to the constant membrane action of the PLATE element.
"° Note that the specification of the BEAM section offset data refer to the local beam coordinate system.
Version 99.0 Linear Examples A82.1
SOLVIA Verification Manual Linear Examples A82 STIFFENED PLATE CANTILEVER UNDER TIP MOMENT ORIGINAL
- 10.
TIME I
2 Z1 Y MOMENT 10000 t
EAXES=RST SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 Version 99.0 A82 STIFFENED PLATE CANTILEVER UNDER TIP MOMENT ORIGINAL
- 10.
Z MAX DISPL.
0.031497 TIME I
X Y
REACTION 3.9788E-12 DISPLACEMENT MAX 0.031497 0 029528 "O
025s91 0 021654 0 0
,177 17 So013780 9.8427E-3 S.90S6E-3 1 9685E-3 MIN 0 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A82.2
SOLVIA Verification Manual SOLVIA-PRE input HEADING
'A82 STIFFENED PLATE CANTILEVER UNDER TIP MOMENT' DATABASE CREATE COORDINATES 1
/
2
- 40.
/
3
- 40.
120.
/
4
- 0.
120.
MATERIAL 1
ELASTIC E=3.E7 NU
- 0.
EGROUP 1
PLATE GSURFACE 1 2 3 4
EL1=1 EL2=1 EDATA
/
1
- 1.
EGROUP 2
BEAM SECTION 1 RECTANGULAR WTOP=I.
D=6.
TOFFSET=3.5 BEAMVECTOR
/
1
- 1.
ENODES 1
-1 2 3 2
-1 1 4 FIXBOUNDARIES
/
1 2 LOADS CONCENTRATED 3 4 -10000.
4 4 -10000.
MESH NNUMBER=MYNODES EAXES=RST VECTOR=MOMENT SOLVIA END SOLVIA-POST input A82 STIFFENED PLATE CANTILEVER UNDER TIP MOMENT DATABASE CREATE WRITE FILENAME='a82.1is' MESH ORIGINAL=DASHED OUTLINE=YES CONTOUR=DISPLACEMENTS VECTOR=REACTION NLIST NLIST KIND=REACTION SYSTEM 1 CARTESIAN ELIST SYSTEM-i END Version 99.0 Linear Examples A82.3
SOLVIA Verification Manual EXAMPLE A83 FREQUENCY ANALYSIS OF A TORSION SPRING SYSTEM Objective To verify the use of constraint equations in frequency analysis using consistent mass matrix.
Physical Problem A torsion spring system as shown in the figure below is considered.
k2 Gears D2 J2 J3 D3 D1 n1 =-
= 1.5 D
1 k, J1 TD1 D4T J4 34n4-
-=1.5 D3 Torsion spring constant k
100 Nm/rad k2 =150 Nm/rad k 3 = 200 Nm/rad Mass moment of inertia J1 =10 kgm 2 J2 = 5 kgm2 J3 = 7.5 kgm2 J4 = 15 kgm 2 Finite Element Model The spring system is modeled using three SPRING AXIALROTATION elements. The massive gears are modeled with concentrated rotational masses and the speed ratios between different shafts are simulated using constraint equations.
Solution Results The theoretical solution can be found in [1] p. 68. The input data shown on page A83.2 agrees with the theoretical solution:
f- 0.3775 Hz f2= 0.8979 Hz User Hints Note that a consistent mass matrix must, in general, be used when using RIGIDLINKS or CONSTRAINTS in a dynamic analysis or when massproportional or centrifugal loading is employed.
Reference
[1]
Blevins, R.D., Formulas for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company, 1979.
Version 99.0 Linear Examples A83.1
SOLVIA Verification Manual SOLVIA-PRE input HEADING
'A83 FREQUENCY ANALYSIS OF A TORSION SPRING SYSTEM' DATABASE CREATE MASTER IDOF=111011 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT FREQUENCY DETERMINANT-SEARCH NEIG=2 MODALSTRESSES=YES COORDINATES 1
/
2
- 5.
/
3
- 5.
- 5.
/
4
- 10.
- 5.
5
- 10.
/
6
- 15.
MASSES 2
0 0 0 10.
/
3 0 0 0 5 4
0 0 0 7.5
/
5 0
0 0 15 EGROUP 1
SPRING DIRECTION=AXIALROTATION PROPERTYSET 1 K=O00 PROPERTYSET 2 K=150 PROPERTYSET 3 K=200 ENODES 1
12
/
2 34
/
3 56 EDATA
/
1 1
/
2 2
/
3 3 FIXBOUNDARIES
/
1 6 CONSTRAINTS 3 4 2 4
-1.5 4 4 5 4
-1.5 SOLVIA END SOLVIA-POST input A83 FREQUENCY ANALYSIS OF A TORSION SPRING SYSTEM DATABASE CREATE WRITE FILENAME='a83.1is' FREQUENCIES SET RESPONSETYPE=VIBRATIONMODE NLIST TSTART-1 TEND=2 ELIST TSTART=1 TEND=2 END Version 99.0 Linear Examples A83.2
SOLVIA Verification Manual EXAMPLE A84 TWO-DIMENSIONAL HEAT TRANSFER WITH CONVECTION Objective To verify the PLANE conduction element with convective boundary conditions in a steady-state heat transfer analysis.
Physical Problem A rectangular area, as shown in the figure below, is subjected to a fixed temperature of 100°C at the bottom edge. The left edge is insulated. The top and right edges have a convective boundary condition with an environmental temperature of 00C specified. This problem is described in [1].
Convection ID
- _C
-Uniform thickness
" a=0.6m
~b=0.2m c =1.0 m "Convection "Material properties Insulator Conductivity = 52.0 W/m 'C "Convection coefficient = 750 W/m 2 oC E
(convection to ambient temperature "Teprtr-l at 00 C along B-C-D)
.,Temperature 1000C
- I--,'
A La B
Finite Element Model Fifteen 8-node PLANE conduction elements are used to model the rectangular area as shown in the top figure on page A84.3. The convective boundary condition is specified by convection boundary segments along the top and the right edge of the plate. The mesh spacing is performed as prescribed in [1].
Solution Results The input data shown on page A84.4 is used in the SOLVIA-TEMP analysis. The analytical solution to this problem is derived in [2] and is presented in [1 ]. The solution to this problem is a highly variable temperature field as shown in the figure on page A84.3. The analytical value for the temperature at point E is 18.3°C. The temperature calculated by SOLVIA-TEMP at point E (Node 9) is 18.3°C.
A contour plot of temperature distribution and a vector plot of element fluxes and the temperature distribution along the right edge can be seen in the figure on page A84.3.
Version 99.0 Linear Examples A84.1
SOLVIA Verification Manual Linear Examples User Hints
"* This problem is modelled by a relative coarse mesh. A much finer mesh should be used in practice near the large temperature gradient at the right lower corner of the model.
- Note that a contour plot of the flux deviation gives valuable information regarding the quality of the mesh in a SOLVIA-TEMP analysis, see bottom figure of page A84.3.
References
[1]
NAFEMS, Background to Benchmarks, 1993
[2]
Carslaw, H.S. and Jaeger, J.C., Conduction of Heat in Solids, Second Ed., Oxford University Press, 1959.
Version 99.0 A84 TWO DIMENSIONAL HEAT TRANSFER WITH CONVECTION ORIGINAL 0
0.1 Z
ORIGINAL I 0.1 Z
L7 LIy ill6 13
[2
.£.0 S9 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB
-y A84.2
SOLVIA Verification Manual Linear Examples ORIGINAL 0.1 TIME I
A84 TWO DIMENSIONAL HEAT TRANSFER WITH CONVECTION TIME I
z Ly
-J F-0.2 0.4 0.6 0.8 EDGE NODE 7 -
17 SOLVIA POST 99.0 SOLVIA ENGINEERING AD HEAT FLUX 318S9 TEMPERATURE MAX 100.00 7 93. 784 "8 13S2 68.919 56.487
- 44. 055 31.623
- 19.
90 6.7582 MIN 0.5420S
.0 A84 TWO DIMENSIONAL HEAT TRANSFER WITH CONVECTION ORIGINAL C--
0.1 TIME I Z
ORIGINAL 0.1 L
yTIME I
HEAT FLUX MAX 48717 4S710 39695 33680 27665 216S0 i5635 9619.8 3604.8 MIN 597.26 SOLVIA-POST 99.0
~DEVIATION OF HEAT FLUX MAX 3085.7 2507.1 2121.,1 1735.7 S.
... 3 S O. O 964.28 578.57 192.86 MIN 0 SOLVIA ENGINEERING AB Version 99.0
' ' 1 ' 1.
I I *
' 1 '
4-y LY o
o_
SOLVIA-POST 99.0 SOLVIA ENGINEERING AB A84.3
SOLVIA Verification Manual SOLVIA-PRE input HEAD
'A84 TWO DIMENSIONAL HEAT TRANSFER WITH CONVECTION' DATABASE CREATE COORDINATES ENTRIES NODE Y
Z 1
- 0.
- 0.
TO 7
0.6 TO 17 0.6 1.
18
- 0.
1.
T-MATERIAL 1 CONDUCTION K=52.0 T-MATERIAL 2 CONVECTION H=750.
EGROUP 1
PLANE STRAIN MATERIAL=1 GSURFACE 1 7 17 18 EL1=3 EL2=5 NODES=8 T-BOUNDARIES SEGMENTS CON-MATERIAL=2 INPUT=LIN:
7 17 17 18 T-LOADS ENVIRONMENT CONVECTION=YES INPUT=LINES 7 17
- 0.
- 0.
17 18
- 0.
- 0.
T-LOADS TEMPERATURE 1 100.
TO 7 100.
SET NSYMBOLS=MYNODES MESH NNUMBERS-MYNODES SUBFRAME=21 MESH ENUMBER=YES SOLVIA-TEMP END ES SOLVIA-POST input A84 TWO DIMENSIONAL HEAT TRANSFER WITH CONVECTION T-DATABASE CREATE WRITE FILENAME='a84.1is' MESH CONTOUR=TEMPERATURE VECTOR=TFLUX SUBFRAME=21 NPLINE NAME=EDGE 7
TO 17 NLINE LINENAME=EDGE KIND=TEMPERATURE OUTPUT=ALL MESH CONTOUR=TFLUX SUBFRAME=21 MESH CONTOUR=FDEVIATION EMAX NUMBER=3 END Version 99.0 A84.4 Linear Examples
SOLVIA Verification Manual EXAMPLE A85 HARMONIC VIBRATION OF A DAMPED PLATE Objective To demonstrate a COMPLEX-HARMONIC analysis with SHELL elements.
Physical Problem A simply supported rectangular plate is subjected to a harmonic transverse point load (at point A in the figure below). At point B the plate has a concentrated mass and spring/damper elements con nected to the ground.
ka
= 1.5 m b=1.0m M
d = 0.4 m B
d E - 210 CPa rA v v=0.3 b>
b = 7800 kg/rn3 F = F0 sin cot Fo =IN thickness = 3.0 mm Rayleigh-damping Spring to ground Damper to ground Point mass c = 0.3 1/s
[=0 k = 10 kN/m c = 15 Ns/m m=2.5 kg Finite Element Model The top figure on page A85.2 shows the finite element model. It consists of 150 4-node SHELL elements and two SPRING elements.
Solution Results The input data on pages A85.5 and A85.6 has been used. The time average of the power flow in the plate is shown on pages A85.2 and A85.3 at frequencies 9, 11 and 13 Hz. The vector plots of the active power flow show a transport from the load to the damper. A detailed discussion of this example is found in [1] and further examples are found in [2].
The time averages of the input active and reactive power are shown on page A85.4 as well as the displacement amplitude at point A and the element force amplitude in the damper.
Reference
[1]
Alfredsson, K.S., "Influence of Local Damping on Active and Reactive Mechanical Power Flow",
4-th International Congress on Intensity Techniques, Senlis, France, 31 Aug - 2 Sept, 1993.
[2]
Alfredsson, K.S., Josefson, B.L., Wilson, M.A,, "Use of the Energy Flow Concept in Vibration Design", 36th AIAA/ASME/ASCE/AHS/ASC Structures, Structural Dynamics and Materials Conference in New Orleans, LA, April 1995.
Version 99.0 Linear Examples A85.1
SOLVIA Verification Manual Linear Examples Version 99.0 ASS HARMONIC VIBRATION OF A DAMPED PLATE ORIGINAL ;
0.2 Z
TIME I
Y X
D FORCE MAS FP R 000060 B 110l1l C 111101 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB A85.2
SOLVIA Verification Manual Linear Examples A8S HARMONIC VIBRATION OF A DAMPED PLATE ORIGINAL 0.S FREO II Y
L x P-ACTIVE MAX 9.7501E-3
- 9.
407E-3 7.9220E-3
-*6.7032E-3 S. 4844E-3 4.2657E-3 3.0469E-3 1..82851E -3 6.0938E-4 MIN 0
ORIGINAL 0- 5 S
FREO It STRAINENERGY DENSITY L x SHELL TOP MAX
- 1. 3213 1!.
0768 0.9117.
- 0. 74670 0.58166 0.41662 O.2S1S9 S. 0.086549 MIN 4.0303E-3 ORIGINAL 0.S FRED II Y
ORIGINAL F
O.S L x FREO I1 P-ACTIVE 9.9491 E-3 SOLVIA-POST 99.0 P-REACTIVE 0.030991 SOLVIA ENGINEERING AB A8S HARMONIC VIBRATION OF A DAMPED PLATE ORIGINAL i
0.5 FREO 13 Y L x P-ACTIVE MAX 3.8572E-4 3.616tE-4 S3. [340E-4 2.6S18E-4 2.1697E-4 1.687SE-4 t.20S4E-4 7.2322E-S 2.4107E-5 MIN 0 ORIGINAL i H0.5 FREO 13 STRAINENERGY Y
DENSITY L
x SHELL TOP MAX 0.017072 0.04131 r) n,ýo,ýz 0.032368 0 026486 0.02060S 0 014723
- 8. 8415E-3 2 9600E-3 MIN 1.9211E-5 ORIGINAL i-40.5 FREO 13 Y
ORIGINAL 0-i 0.S L
XFREO 13 P-ACTIVE 3.9146E-4 SOLVIA POST 99.0 SOLVIA ENGINEERING AB Version 99.0 Y
Lx Y
Lx P-REACTIVE 2.t467E-3 A85.3
SOLVIA Verification Manual Linear Examples A8S HARMONIC VIBRATION OF A DAMPED PLATE S..
I I
8 10 12 14 FREQUENCY I
/
/
6 8
10 FREQUENCY 12 1 4 C)
p0 H
z U
01 N
C_
C) z N
C)
X J)
I V)
"C)N CL 0 =
C)-1<
C) 8 0
12 14 FREQUENCY S.
T.
{
I 8
10 FREQUENCY 12 1
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7 Ul SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 I0 14
/!
./
Co z
H C)
C.
C)
N z
M CD nJ 0'3 CD =
0 C)
C)Ci C) H C) <
/
N C) cr)
C-1 cC A85.4 I
SOLVIA Verification Manual SOLVIA-PRE input HEADING
'A85 HARMONIC VIBRATION OF A DAMPED PLATE' DATABASE CREATE ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT COMPLEX-HARMONIC INCREMENTATION=LINEAR 72 8 20 12 2 14 RAYLEIGH-DAMPING ALPHA=0.3 BETA=0.
TIMEFUNCTION 1
- 0. 1.
/
100.
- 1.
COORDINATES 1
/
2 1.5
/
3 1.5 1.
/
4
- 0.
- 1.
5 1.1 0.5
/
6 0.4 0.5
/
7 1.1 0.5 0.5 MASSES 5
2.5 2.5 2.5 0.
- 0.
- 0.
MATERIAL 1
ELASTIC E=2.1Ell NU=0.3 DENSITY=7800.
K=0.
EGROUP 1
SHELL THICKNESS 1
3.E-3 GSURFACE 1 2 3 4 EL1=15 EL2=10 NODES=4 EGROUP 2
SPRING DIRECTION=AXIALTRANSLATION PROPERTYSET 1
K=10E3 PROPERTYSET 2
C=15.
ENODES 1
57
/
2 57 EDATA 11 /
22 FIXBOUNDARIES 12356 INPUT:LINES 1 2 3 4 FIXBOUNDARIES 12346 INPUT=LINES
/
2 3/
4 1 FIXBOUNDARIES INPUT=NODES
/
7 LOADS CONCENTRATED 6 3 1.
VIEW ID=i XVIEW=--.
YVIEW--1.2 ZVIEW=0.7 SET NNUMBERS=MYNODES NSYMBOLS=MYNODES VIEW=1 MESH BCODE=ALL VECTOR-LOAD SOLVIA END Version 99.0 Linear Examples A85.5
SOLVIA Verification Manual SOLVIA-POST input A85 HARMONIC VIBRATION OF A DAMPED PLATE DATABASE CREATE WRITE FILENAME='a85.1is' CONTOUR AVERAGING=YES SET RESPONSETYPE=REAL SET VIEW=Z ORIGINAL=YES DEFORMED-NO OUTLINE=YES SET TIME=9 MESH CONTOUR-P-ACTIVE SUBFRAME-22 MESH CONTOUR=ENERGY MESH VECTOR=P-ACTIVE MESH VECTOR=P-REACTIVE SET TIME-1l MESH CONTOUR=P-ACTIVE SUBFRAME=22 MESH CONTOUR=ENERGY MESH VECTOR=P-ACTIVE MESH VECTOR=P-REACTIVE SET TIME=f3 MESH CONTOUR=P-ACTIVE SUBFRAME-22 MESH CONTOUR=ENERGY MESH VECTOR=P-ACTIVE MESH VECTOR-P-REACTIVE SET PLOTORIENTATION=PORTRAIT HEIGHT=0.20 NCOMPLEX-HARMONIC NODE=6 DIRECTION-3 KIND-P-ACTIVE OUTPUT=ALL SUBFRAME=22 NCOMPLEX-HARMONIC NODE-6 DIRECTION-3 KIND=P-REACTIVE NCOMPLEX-HARMONIC NODE=6 DIRECTION=3 KIND=DISPLACEMENT ECOMPLEX-HARMONIC ELEMENT-2 POINT=1 KIND=SXX END Version 99.0 Linear Examples A85.6
SOLVIA Verification Manual EXAMPLE A86 HARMONIC RESPONSE OF A DAMPED TWO-DEGREE-OF-FREEDOM SYSTEM Objective To verify the COMPLEX-HARMONIC method with SPRING elements.
Physical Problem A sinusoidal force is applied to one mass of the damped two-degree-of-freedom system shown in the figure below. This case has been analyzed without damping in Example A77.
F F =F 0 sincot F,
I1 N Fo =INkZ m 1 = 1 kg m 2 =l1kg ki = 1.0.10 4 N/m k2 = 1.5.104 N/m cl = 70.0 Ns/m c2 = 2.0 Ns/m Finite Element Model The model consists of two SPRING elements (DIRECTION = AXIALTRANSLATION) and two concentrated masses. The model is shown on page A86.2.
Solution Results The input data shown on pages A86.5 and A86.6 is used. The displacement amplitude of node 3 has a peak value at frequency 9.4 Hz, see page A86.2. The natural frequencies of an undamped system are 10.3 Hz and 30.1 Hz. For this heavy damped system the second natural frequency has a very slight influence.
The input power and element forces as a function of frequency are shown on page A86.4.
A comparison with the analytical solution for characteristic responses at the frequency 10 Hz is made in the table below.
Analytical SOLVIA Displacement, amplitude, node 3
[ml 3.04.10-4 3.04.10-4 Velocity, amplitude, node 3
[m/s]
1.9.10-1.91.10-2 Acceleration, amplitude, node 3
[m/s 2]
1.20.10-'
1.20.10-'
Element force, amplitude, el. no. 4
[N]
0.99.10-2 1.00.10-2 Active powerflow, element no. 4 **)
[Nm/s]
-8.32. 10-
-8.32.10-'
- )
Input active powerflow, node 2 **)
[Nmrs]
6.95.10-3 6.95.10-3 Strain energy (whole model) **)
[Nm]
1.49.10-4 1.49-10-4
- ) Power flow in opposite element axial direction. **) Time average values.
Version 99.0 Linear Examples A86.1
SOLVIA Verification Manual A86 HARMONIC RESPONSE OF A DAMPED TWO-DEGREE-OF-FREEDOM SYSTEM I
/
/
t0 20 FREQUENCY S...................
30 10 t0 20 FREQUENCY 2 N
, F Version 99.0 A86 HARMONIC RESPONSE OF A DAMPED TWO-DEGREE-OFIFREEDOM SYSTEM MAX D7SPL. F-2.2898E-5 Z
REAL PART FREO 40 L
x 2
3 LOAD I
MAX DISPL.
- 2.2898E-5 Z
REAL PART FREQ 40 L
X
%1J 0 2 03 r
S EAXES=RST MASTER cill[
B RINGAS SOLVTA-POST 99.0 SOLVIA ENGINEERING AB S.............
.. y
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Z.
30 4
SOLV7A-POST 99 0
SOLVIA ENGINEERING AB Linear Examples IiF-A86.2
SOLVIA Verification Manual A86 HARMONIC RESPONSE OF A DAMPED TWO-DEGREE-OF-FREEDOM SYSTEM a
/
i 00\\ 0
/
a 10 20 FRECUENCY SOLYIA-POST 99.0 V
/
/-;
FREQUENCY SOLVIA ENGINEERING AB Version 99.0 Linear Examples
/
/
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,7' 2O 30 4
/
/
- I 71
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to A86.3
SOLVIA Verification Manual Linear Examples A86 HARMONIC RESPONSE OF A DAMPED TWO-DEGREE-OF-FREEDOM SYSTEM N-U Lh hi C')
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Li
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20 FREQUENCY Ld CD 0~
hL
- a.
uj h
CD z wi <
Li A ~/
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S /
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0 20 40 FREQUENCY
-------- J 20 FREQUENCY
() 0 4
- 1-j N
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SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0
- /
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hi hi r hi F tD CD C hi<
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I A86.4
SOLVIA Verification Manual SOLVIA-PRE input DATABASE CREATE HEADING
'A86 HARMONIC RESPONSE OF A DAMPED TWO-DEGREE-OF-FREEDOM SYSTEM' MASTER IDOF=011111 NSTEP=1 COMPLEX-HARMONIC INTERVAL=LIN 1
4 5
50 15 25 40 ANALYSIS MASSMATRIX=CONSISTENT COORDINATES ENTRIES NODE X 1 1 TD 3 3 MASS 2111000 3111000 EGROUP 1 SPRING DIRECTION=AXIALTRANSLATION PROPERTYSET 1 K=1.0E4 PROPERTYSET 2 K=1.5E4 PROPERTYSET 3 C-7.OE1 PROPERTYSET 4 C=2.0E0 ENODES 112/223/312/423 EDATA 11/22/33/44 FIXBOUNDARIES /
1 LOAD CONC 2
1 1.
TIMEFUNCTION 1 0 1 /
1E3 1 SOLVIA MODELSUM=DET END Version 99.0 Linear Examples A86.5
SOLVIA Verification Manual Linear Examples SOLVIA-POST input A86 HARMONIC RESPONSE OF A DAMPED TWO-DEGREE-OF-FREEDOM SYSTEM DATABASE CREATE WRITE FILENAME='a86.1is' SET DIAGRAM=GRID SET RESPONSE=REAL MESH NNUMBERS=YES NSYMBOLýYES VECTOR=LOAD VIEW=-Y SUBFRAME=12 MESH NNUMBERS=YES NSYMBOL=YES BCODE=ALL EAXES=RST SET HEIGHT=0.20 SUBFRAME 21 NCOMPLEX-HARMONIC NODE=3 DIRECTION=1 KIND=DISPLACEMENT OUTPUT=ALL NCOMPLEX-HARMONIC NODE=3 DIRECTION=1 KIND=ACCELERATION OUTPUT=ALL SUBFRAME 21 NCOMPLEX-HARMONIC NODE=2 DIRECTION=1 KIND=P-ACTIVE OUTPUT=ALL NCOMPLEX-HARMONIC NODE=2 DIRECTION=1 KIND=P-REACTIVE OUTPUT=PLOT NCOMPLEX-HARMONIC NODE=3 DIRECTIONýl KIND=VELOCITY OUTPUT=LIST, FSTART=9 FEND=10 SET PLOTORIENTATION=PORTRAIT ECOMPLEX ELEMENT=1 POINT=1 KIND=FR SUBFRAME=22 ECOMPLEX ELEMENT=2 POINT=1 KIND=FR ECOMPLEX ELEMENT=3 POINT=1 KIND=FR ECOMPLEX ELEMENTý4 POINT=1 KIND=FR ELIST SELECTýPOWERFLOW TSTART=10 SUMMATION KIND=ENERGY TSTART=10 CASECOMBINATION 1 STANDARD FUNCTION=SRSS
- 10. 1. REAL
- 10. 1. IMAGINARY SET RESPONSETYPE=CASECOMBINATION ELIST NLIST NLIST KIND=REAC END Version 99.0 A86.6
SOLVIA Verification Manual EXAMPLE A87 MODAL COMBINATION METHODS IN RESPONSE SPECTRUM ANALYSIS Objective To verify the methods of combining modal responses in the response spectrum method.
Physical Problem A beam structure shown in the figure below is subjected to an earthquake base acceleration in the X-direction. The 3-D structure is asymmetrical. This example has been analyzed in reference [1],
ex. E26-8.
z X
(Lx, -Ly, -Lz)
(0, 0, 0) m=l.10-2 kg L =L y =LZ =lm E = 1 N/m 2
v=0 Bending moment of inertia I, = It = 1 m 4 Torsional moment of inertia I1 = 4/3 m 4
The numerical values are only selected so that the relation between stiffness and mass is EI ml- =100 1/sec 2 which is in accordance with reference [1].
The acceleration response spectrum values for the three natural frequencies are specified as
[2.27001 a = 2.4500r m/sec2 16.9800 i with 5% of critical damping Solution Results Using the input data on pages A87.4 and A87.5 the following natural frequencies were obtained:
0.730681 0 0.76897jHz 2.3172J
[ 45910]
or 4.83161rad/sec 14559 The two lowest frequencies are closely spaced with significant cross-correlation factors in the CQC method. The eigenmodes are shown on page A87.3 and the modal displacements in the X-, Y-and Z-directions at node 4 are:
5.79770 p = -1.84954 7.93510 2.60511 (P2 = 9.64852 0.345505
= 7.7 2 0 1 1 1 p3 =-1.86687 1-6.07575]
Version 99.0 Linear Examples Y
L, A87.1
SOLVIA Verification Manual Linear Examples The calculated modal participation factors for base excitation in the X-direction are:
F1 = 57.9770 Fi2 = 26.0511 10-'
F3 = 77.2011.10-3 The maximum relative displacement for each mode i isu (i)
= 1-' S p. / Cw2 giving max i
a i I
i b
=
u 36.202] m
- 7. 1 2 3
=
19.625 ]
u () =/11.549[mm (2) =26.380 mm U(3)
=
-4.746/m m 49.548 max 0.9451j
[-15.445]
The relative displacement response at node 4 shown in the table below was calculated by SOLVIA POST and the values agree with the results of a separate analysis using the formulas given in [2],
command MODALCOMBINATION.
Modal Comb. Method Rel. displ. at node 4 in mm X
Y Z
SRSS 41.79 29.19 51.91 CQC 46.53 19.18 52.53 Ten Percent 47.56 38.22 52.80 Double Sum 46.75 36.80 52.79 (using 100 sec duration)
Algebraic 62.95 10.09 35.05 Absolute Sum 62.95 42.67 65.94 The cross-correlation factors in the CQC method are P12 = P 21 0.79280, P23 P 3 2 0.006383, P 13 = P 31 = 0.005705.
The response in the X-direction is under-estimated using the SRSS method compared to the CQC, Ten Percent and Double Sum methods.
The response in the Y-direction is calculated to be much lower using the CQC method compared to the SRSS, Ten Percent and Double Sum methods.
The response in the Z-direction is about the same for the SRSS, CQC, Ten Percent and Double Sum methods.
The results of the Absolute Sum method can be considered as an upper bound of the response.
A further discussion on the SRSS and CQC methods can be found in [1].
References
[1]
Clough, R. W., and Penzien, J., Dynamics of Structures, Second Edition McGraw-Hill, Inc., 1993.
[2]
SOLVIA-POST 99.0 Users Manual, SOLVIA Engineering AB, Report SE 99-2, 1999.
Version 99.0 A87.2
SOLVIA Verification Manual Linear Examples A87 MODAL COMBINATION METHODS IN RESPONSE SPECTRUM ANALYSIS REFERENCE 0.2 Z
REFERENCE ý-
-- 0.2 j
MAX DISPL. i 10 X
Y MODE 1 FRED 0.73068 2%
3 MASTER 000000 B 111111 REFERENCE v-
-- 0.2 MAX DISPL.I 10 MODE 2 FRED 0.76897 REFERENCE 1
-1 0.2 MAX DISPL.
10 MODE 3 FRED 2.3172 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 A87.3
SOLVIA Verification Manual Linear Examples SOLVIA-PRE input HEADING
'A87 MODAL COMBINATION METHODS IN RESPONSE SPECTRUM ANALYSIS' DATABASE CREATE FREQUENCIES SUBSPACE-ITERATION NEIG=3 ANALYSIS DYNAMIC IMODS=2 NMODS=3 COORDINATES 1
0 0
0 2
0
-1 0
3 0 1 4
1 1 MASS 4 0.01 0.01 0.01 0 0 0 MATERIAL 1 ELASTIC E=I.
NU-0.
EGROUP 1 BEAM SECTION 1 GENERAL RINERTIA=1.333333333 SINERTIA=1 TINERTIA=1 AREA=1E7 ENODES 1312 2123 3234 FIXBOUNDARIES
/
1 SOLVIA END Version 99.0 A87.4
SOLVIA Verification Manual SOLVIA-POST input A87 MODAL COMBINATION METHODS IN RESPONSE SPECTRUM ANALYSIS DATABASE CREATE WRITE FILENAME='a87.1is' FREQUENCIES SET RESPONSETYPE=VIBRATION NLIST N4 DIR=i23 TSTART=i TEND=3 SET PLOTORIENTATION=PORTRAIT HEIGHT=0.20 SET NSYMBOL=YES DMAX=I.
MESH ORIGINAL-YES DEFORMED=NO BCODE=ALL NSYMBOL=YES NNUMBER=YES SUBFRAME=22 SET ORIGINAL=DASHED SET AXES=NO MESH TIME=I MESH TIME=2 MESH TIME=3 RSPECTRUM 1 FACTOR=i 0.7 0.75 0.76 1.0 2.0 3.0
- 0.
2.27 2.27 2.45 2.45 6.98 6.98
- 5.
2.27 2.27 2.45 2.45 6.98 6.98
- 10.
2.27 2.27 2.45 2.45 6.98 6.98 DAMPING 1 1 5.
TO 3 5.
MODAL-COMBINATION MODAL-COMBINATION MODAL-COMBINATION MODAL-COMBINATION MODAL-COMBINATION 1
2 3
4 5
MODAL-COMBINATION 6
RSPECTRUM=i DIRECTION=X RSPECTRUM=-
DIRECTION=X RSPECTRUM=i DIRECTION=X RSPECTRUM-i DIRECTION=X RSPECTRUM=i DIRECTION=X DURATION=-00.
RSPECTRUM=1 DIRECTION=X METHOD=SRSS METHOD=CQC METHOD=ABS METHOD=TENPERCENT METHOD-DOUBLESUM, METHOD=ALGEBRAIC SET RESPONSETYPE=MODAL-COMBINATION NLIST N4 DIR=123 TSTART=i TEND=6 END Version 99.0 Linear Examples A87.5
SOLVIA Verification Manual EXAMPLE A88 SIMPLY SUPPORTED BEAM CONDITIONS USING THE END RELEASE OPTION Objective To verify the BEAM element when using the end release option.
Physical Problem Three beam spans as shown in the figure below are under central loads. The beam spans are simply supported without bending stiffness connection to the neighbouring span. The vertical displacements under the loads are to be calculated.
F IF B
IF C
II L=l m E= 2.0.l10" N/m 2 v=0.3 F=1N Rectangular cross-section 10 x 10 mm Finite Element Model The model is shown on page A88.2 and consists of six BEAM elements where three of the beam elements use the end release option.
The bending stiffness about the local s-axis (global x-axis) is released at node 3 for both connecting elements and at node 5 for the left connecting element.
The selected end release options exemplify two possibilities to achieve simply supported end conditions for the beams connecting to nodes 3 and 5. Since the end release option is used twice at node 3 it is then necessary to also fix the x-rotation at node 3 since it is left without stiffness and would otherwise cause a zero diagonal element in the stiffness matrix.
Solution Results The theoretical displacement at mid-span for a simply supported beam is
-FL 3 48 El The input data shown on page A88.3 gives the following vertical displacements in mm at the mid-span locations:
Version 99.0
,,~~~
F
¢ 1acl Linear Examples I
I A88.1
SOLVIA Verification Manual A88 SIMPLY SUPPORTED BEAM CONDITIONS USING THE END RELEASE OPTION ORIGINAL 1 0.2 MAX DISPL. i 1.2504E-4 TIME I
z Ly REACTION LI SOLVIA ENGINEERING AB SOLVIA-POST 99.0 Version 99.0 A88 SIMPLY SUPPORTED BEAM CONDITIONS USING THE END RELEASE OPTION ORIGINAL 0.2 Z L y t
12 C53 B-6 D7 r
S I
IEAXES=RST MASTER 000000 B 101000 C 101100 O 111010 ORIGINAL 0.2 Z
TIME I 1y 12 3
4 5
FORCE SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB Linear Examples A88.2
SOLVIA Verification Manual SOLVIA-PRE input DATABASE CREATE HEADING
'A88 SIMPLY SUPPORTED BEAM CONDITIONS USING THE END RELEASE OPTION' COORDINATES ENTRIES NODE Y 1 0.
TO 7 3.
MATERIAL 1 ELASTIC E=2.0ET1 NU=0.3 EGROUP 1 BEAM SECTION 1 RECTANGULAR WTOP=0.01 D=0.01 BEAMVECTORS 1 1. 0.
- 0.
ENODES 1 -1 1 2 TO 6 -1 6 7 ENDRELEASE 1 RELI11 ENDRELEASE 2 REL1=5 EDATA ENTRIES EL SECTION ENDRELEASE 1 1 0 211 312 411 510 610 2*
FIXBOUNDARIES 13 /
FIXBOUNDARIES 52 /
FIXBOUNDARIES 4 /
LOAD CONCENTRATED 2
3
-1.
4 3 -1.
6 3
-1.
1357 7
3 SET NSYMBOLS=YES VIEW=X MESH BCODE=ALL NNUMBERS=YES EAXES=RST MESH ENUMBERS=YES VECTOR=LOAD SOLVIA END SOLVIA-POST input SUBFRAME=12 A88 SIMPLY SUPPORTED BEAM CONDITIONS USING THE END RELEASE OPTION DATABASE CREATE WRITE
'a88.lis' NLIST MESH ORIGINAL=YES VECTOR=REACTION VIEW=X END Version 99.0 Linear Examples A88.3
SOLVIA Verification Manual EXAMPLE A89 LAMINATED STRIP Objective To verify the laminated 16-node SHELL element.
Physical Problem A simply supported laminated strip is subjected to a central line load. The laminate consists of seven layers as shown in the figure below. This problem is described in [1].
fiber direction, layer 1 T
10 15 15 10 0.1 0.1 0.1 0.4 0.1 0.1 0.1 0c 900 00 900
- 0.
900 00 t,c poi lay lay poi I lay*
All dimensions in mm.
The line load, P = 10 N/mm at the center of the strip (X=25 mm)
Material properties:
Ea =1.105 MPa Vba = 0. 4 Gab = 3.10' N Eb = 5.103 MPa G,, =2.10' tv nt D er 1, top surface er 2, bottom surface nt E er 1, bottom surface IPa Pa Gbc = 2.10' MPa Version 99.0 Linear Examples A89.1
SOLVIA Verification Manual Finite Element Model One quarter of the strip is modeled using five 16-node SHELL elements. The finite element model with the symmetrical boundary condition is shown on page A89.2.
Solution Results The input data shown on pages A89.3 and A89.4 is used in the SOLVIA analysis. Contour plots of
- a. (bottom of layer 1) and a, (top of layer 1) are shown on page A89.3.
Results compared with NAFEMS target values.
SOLVIA NAFEMS [1]
Central displacement [mm]
-1.06
-1.06 (node 6, Z-direction)
Bending stress cy, point E [MPa]
682.6 683.9 (element 5, layer 1, bottom, node 6)
Interlaminar shear stress u., at point D [MPa]
-4.1
-4.1 (element 5, layer 1, top, node 6)
(element 5, layer 2, bottom, node 6)
-4.1
-4.1 Reference
[1]
Hardy, S.J, "Composite Benchmarks", NAFEMS Report R0031, 1994.
A89 LAMINATED STRIP ORIGINAL
- 2.
z TIME I
EFORCE S
MASTER c
000000 B ClOOI0 C 100011 D 101000 E ilOlii F 1110CC S AillN i
SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB Version 99.0 Linear Examples A89.2
SOLVIA Verification Manual A89 LAMINATED STRIP MAX DISPL.
" 1.0665 TIME I
STRESS-XZ Y
LAYER 2 L
x SHELL BOT MAX 0.014830 1-
~0.22891
-0.77638
-1 3239
-2.4188 i
-2.
9683
- 3. S138 1 :
_1-4.0612 MIN-4.3350 SOLVIA-POST 99.0 SOLVIA ENGINEERING AB Version 99.0 Linear Examples A89.3
SOLVIA Verification Manual SOLVIA-PRE input DATABASE CREATE HEADING
'A89 LAMINATED STRIP' COORDINATES ENTRIES NODE X
Y 1
0 0
2 10 0
3 25 0
4 0
5 5
10 5
6 25 5
MATERIAL 1 MATERIAL 2 1 1 10 0
2 1 10 90 3 1 10 0
4 1 40 90 5 1 10 0
6 1 10 90 7 1 10 0
ORTHOTROPIC EA=lE5 EB=5E3 NUBA=0.4, GAB=3E3 GAC=2E3 GBC=2E3 LAMINATE EGROUP 1 SHELL MATERIAL=2 RESULT=NSTRESS SET NODES=16 GSURFACE 1 2 5 4 EL1=2 GSURFACE 2 3 6 5 EL1-3 THICKNESS 1 1.
LOAD ELEMENT TYPE=FORCE INPUT=LINES 3
6 THICKNESS
-5
-5 FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES 13 INPUT-LINES /
2 5 2/2 156 INPUT=LINES /
3 6 246 INPUT=LINES /
6 4 MESH NNUMBER=MYNODES NSYMBOL=YES VECTOR=LOAD, ENUMBER=YES BCODE=ALL SOLVIA END Version 99.0 Linear Examples A89.4
SOLVIA Verification Manual SOLVIA-POST input A89 LAMINATED STRIP DATABASE CREATE WRITE FILENAME='a89.1is' NLIST N6 DIR=3 SHELLSURFACE LAYER=I PLOT=BOTTOM ELIST EL5 MESH CONTOUR=SXX VIEW=Z SUBFRAME=l2 SHELLSURFACE PLOT=TOP MESH CONTOUR=SXZ SHELLSURFACE LAYER=2 PLOT=BOTTOM MESH CONTOUR=SXZ SUBFRAME=12 ELIST EL5 END Version 99.0 Linear Examples A89.5
SOLVIA Verification Manual EXAMPLE A90 SANDWICH SHELL Objective To verify the laminated 9-node SHELL element in a sandwich plate case.
Physical Problem A simply-supported square sandwich plate is subjected to a uniform normal pressure. The laminate consists of three layers with a thick central core as shown in the figure below. This problem is described in [ 1].
a b - Material directions Y
Ab a
°E 00 C\\
C co
,:5 Cý In N-L = 10 inch Pressure 100 psi T
Z,t,c Face sheet, layer 3 2
x Face sheet, layer 1 All dimensions in inches Material properties Face sheets:
Ea =10.106 psi Vba =0.3 Eb =4.106 psi Gab =1.875.106 psi G3, = t. 105 psi Gk=1.105 psi Core:
Ea =Eb =lpsi (=0)
G ac = 3.0- 104 psi Gkb =l.2.10 4psi Note that assumed values of Gac and Gbc for the sheets are used.
Finite Element Model Due to symmetrical conditions only one quarter of the plate is considered. The model with sixteen 9-node SHELL elements and boundary conditions are shown on page A90.2.
Solution Results The input data on pages A90.4 and A90.5 is used in the SOLVIA analysis.
Version 99.0 LI L
Linear Examples A90.1
SOLVIA Verification Manual Contour plots of stresses in the principal material direction for layer number 1, midsurface location, are shown on page A90.3.
Results compared with NAFEMS' target values.
SOLVIA NAFEMS [1]
Central displacement [inch] (node 3, Z-direction)
-0.1224
-0.123 Stress in the material a-direction at central point [psi]
34609 34449 (element 16, layer 1, midsurface, node 3)
Stress in the material b-direction at central point [psi]
13479 13350 (element 16, layer 1, midsurface, node 3)
Quarter point, (E) in plane shear stress, (jab [psi]
-5067.5 (element 11, layer 1, midsurface, node 142)
-5106 (element 10, layer 1, midsurface, node 142)
-5112 (element 7, layer 1, midsurface, node 142)
-5117 (element 6, layer 1, midsurface, node 142)
-5124 Reference
[1]
Hardy, S.J, "Composite Benchmarks", NAFEMS Report R0031, 1994.
AGO SANDWICH SHELL ORIGINAL
- 1.
z ORIGINAL Y
TIME I
Lx x
y C;
C C<
C<
G4 MASTER B 010101 C OiLOI D 1000. 1 PRESSURE E 1010ý1
<100 F 110111 OG 11:I1I SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB Version 99.0 Linear Examples A90.2
SOLVIA Verification Manual Linear Examples A90 SANDWICH SHELL Y
MAX DISPL.
F-- 0 12241 L
xTIME 1 K-i-k I
1 I
-
N H
V A
MIN-2.S3S8E-12 SOLVIA ENGINEERING AB SOLVIA-POST 99.0 STRESS-AA LAYER I
SHELL MID MAX 34610 32447 28120 23794 19468 15142 10816 6489.3 2163.1 MIN 7.9697E-12 EAXES=
S' ORTHOTROPIC IN\\
TSTRESS -B I ) "*
LAYER I MAX 12479 L t~i:
"9266.7 7S81.9 5897.0 4212.2 2527.3 S8 4 2.A3 Version 99.0 MAX DISPH TIME Ii Ii F--H 0.12241
'I
-H Y
I
- X A90.3
SOLVIA Verification Manual SOLVIA-PRE input DATABASE CREATE HEADING
'A90 SANDWICH SHELL' COORDINATES ENTRIES NODE X
Y 1
0 0
2 5
0 3
5 5
4 0
5 MATERIAL 1 ORTHOTROPIC EA=10.E6 EB=4.E6 NUBA=0.3 GAB=1.875E6 GAC=1.E5 GBC=I.E5 MATERIAL 2 ORTHOTROPIC EA=1.
EB-1. NUBA=O.
GAB=I.
GAC=3.0E4 GBC-1.2E4 MATERIAL 3 LAMINATE 1 1 3.4739 0
2 2 93.0522 0
3 1 3.4739 0
EGROUP 1 SHELL MATERIAL=3 RESULT=NSTRESS STRESSREFERENCE=ORTHOTROPIC SET NODES=9 GSURFACE 1 2 3 4 EL1=4 EL2-4 THICKNESS 1 0.806 LOAD ELEMENT TYPE=PRESSURE INPUT=SURFACE 1 2 3 4 t 100.
100.
FIXBOUNDARIES 1356 INPUT=LINES 1 2 FIXBOUNDARIES 2346 INPUT=LINES /
1 4 FIXBOUNDARIES 246 INPUT=LINES /
3 4 FIXBOUNDARIES 156 INPUT=LINES /2 3
MESH VECTOR=LOAD SUBFRAME=21 MESH NNUMBER=MYNODES NSYMBOL=YES ENUMBER=YES BCODE=ALL VIEW=Z SOLVIA END Version 99.0 Linear Examples A90.4
SOLVIA Verification Manual SOLVIA-POST input A90 SANDWICH SHELL DATABASE CREATE WRITE FILENAME='a90.1is' NMAX DIR=3 SHELLSURFACE LAYER=1 PLOT=MID ELIST ELi6 EZONE E /
6 7 10 11 SHELLSURFACE LIST=MID ELIST E MESH CONTOUR=SAA VIEW=Z SUBFRAME=21 MESH CONTOUR=SBB EAXES=ORTHO MESH CONTOUR=SAB EAXES=ORTHO SUBFRAME=21 MESH CONTOUR=DISP END Version 99.0 Linear Examples A90.5
SOLVIA Verification Manual EXAMPLE A91 LAMINATED SQUARE PLATE UNDER NORMAL PRESSURE Objective To verify the laminated 4-node SHELL element.
Physical Problem A simply supported square laminated plate is subjected to a uniform normal pressure. The laminate consists of 9 layers with lay-up 0°/90°/0O/90°/00/90°/00/90°/0° as shown in the figure below. This example is described in [1].
Y aa, b - Material direction layer 1 b
a L
L = 10 inch Thickness t = 0.1 inch Pressure = 10 psi 1t7,C layer 8, 0.125 t layer 1, 0.1 t Material properties Ea =4.0.107 psi Vba = 0.25 Eb =1.0.10 6 psi Gab = 0.6.106 psi Gac = 0.5.106 psi Gbc = 0.5.106 psi Finite Element Model Due to symmetry only one quarter of the plate is considered. The model with sixtyfour 4-node SHELL elements and the material directions for layers 1 and 2 are shown on pages A91.2 and A91.3.
Solution Results The input data on pages A91.6 and A91.7 is used in the SOLVIA analysis.
Contour plots of the section moments for the laminate and principal stresses for layers 1 and 2 are shown on pages A91.3 and A91.4.
Stresses cF,,,oyz, y and a,. along a line in the thickness direction formed by the integration points nearest the centre are shown on pages A91.4 and A91.5.
Version 99.0 L
lay-up
- 0.
900 0o.
900 0o 0 90°
- 0.
90° 0.
Linear Examples A91.1
SOLVIA Verification Manual Results compared with NAFEMS' target values:
SOLVIA NAFEMS [I]
Central displacement [inch] (node 3, Z-direction)
-0.4479
-0.4486 Section moment, Mrr at centre, node 3
-88.54
-88.79 (SOLVIA value from contour plot)
Reference
[1]
Taig, I.C., "Finite Element Analysis of Composite Materials", NAFEMS Report R0003, 1994.
Version 99.0 A91 LAMINATED SQUARE PLATE UNDER NORMAL PRESSURE ORTGINAL O.S Z
TIME I
~
'~<
~PRESSURE 10 MAST ER 000000 B
101001 C 011101 O 100011 E 101011 F 110111 G 1111ll SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB Linear Examples A91.2
SOLVIA Verification Manual Version 99.0 Linear Examples A91.3
SOLVIA Verification Manual Linear Examples A91 LAMINATED SQUARE PLATE UNDER NORMAL PRESSURE MAX DISPL.
v--
0.44793 TIME I Y
MAX DISPL.
v-0.14793 LX TIME 1
SPRINCIPAL LAYER I
SHELL TOP 67786
-67786 SOLVIA-POST 99.0 SPRINCIPAL LAYER 2
SHELL TOP 45510
-45S10 SOLVIA ENGINEERING AB C
[
0 N
0.
0.00 A91 LAMINATED SQUARE PLATE UNDEF t
I TIME I
1 4
1 N
N I!)
-r U')
Li In I--
0.02 0.04 0.06 0.08 0.10 N
Z NORMAL PRESSURE
.TIM&
1
/
.00 0.02 0.04
.06 0.08
- 0.
0 fH-I K SOLVTA ENGINEERING AS SOLVIA-POST 99.0 Version 99.0 Y
LX X
X In 1/)
wn
'HICK I f
! J i /
VJ O
I A91.4
A91 LAMINATED SQUARE I
TIME 1
Linear Examples PLATE UNDER NORMAL PRESSURE Li Li Li Qf 0.00 0.02 0.04 0.06 0.08 L0.0 THICK SOLVIA-POST 99.0 I
TIME 1
0.00 C. 02 0.0
.0 0.08 0.
0 TH ICK SOLVIA ENGINEERING AB Version 99.0 SOLVIA Verification Manual
)q Li Li Li C -
A91.5
SOLVIA Verification Manual SOLVIA-PRE input DATABASE CREATE HEADING
'A91 LAMINATED SQUARE PLATE UNDER NORMAL PRESSURE' COORDINATES ENTRIES NODE X
Y 1
0 0
2 5
0 3
5 5
4 0
5 MATERIAL 1 ORTHOTROPIC EA=4E7 EB=1E6 NUBA=0.25 GAB=0.6E6 GAC=0.SE6 GBC=0.5E6 MATERIAL 2 LAMINATE 1 1 10 0
2 1 12.5 90 3 1 10 0
4 1 12.5 90 5 1 10 0
6 1 12.5 90 7 1 10 0
8 1 12.5 90 9 1 10 0
EGROUP 1 SHELL MATERIAL=2 TINT=-3 SET NODES=4 GSURFACE 1 2 3 4 EL1=8 EL2=8 THICKNESS 1 1.OE-1 LOAD ELEMENT TYPE=PRESSURE INPUT-SURFACE 1 2 3 4 t 10 10 FIXBOUNDARIES 1356 INPUT=LINES 1 2 FIXBOUNDARIES 2346 INPUT=LINES /
1 4 FIXBOUNDARIES 246 INPUT=LINES /
3 4 FIXBOUNDARIES 156 INPUT=LINES /2 3
MESH NNUMBER=MYNODES NSYMBOL=YES VECTOR=LOAD, ENUMBER=YES BCODE=ALL SHELLSURFACE LAYER=1 SET VIEW=Z NNUMBER=MYNODES NSYMBOL=MYNODES MESH EAXES=ORTHOTROPIC SUBFRAME=21 SHELLSURFACE LAYER=2 MESH EAXES=ORTHOTROPIC SOLVIA END Version 99.0 Linear Examples A91.6
SOLVIA Verification Manual SOLVIA-POST input A91 LAMINATED SQUARE PLATE UNDER NORMAL PRESSURE DATABASE CREATE WRITE FILENAME='a91.lis' NMAX DIR=3 MESH CONTOUR=MRR-SECTION EAXES=RST VIEW=Z SUBFRAME=21 MESH CONTOUR=MSS SHELLSURFACE LAYER=
MESH VECTOR=SPRINCIP SHELLSURFACE LAYER=2 MESH VECTOR=SPRINCIP EPLINE THICK 64 1 2 3 4 5 6 7 8 9 21 22 23 24 25 26 ELINE THICK KIND=SXX ELINE THICK KIND=SYZ ELINE THICK KIND=SYY ELINE THICK KIND=SXY END SUBFRAME=21 10 11 12 13 27 SUBFRAME=21 SUBFRAME=21 14 15 16 17 18 19 20,
Version 99.0 Linear Examples A91.7
SOLVIA Verification Manual EXAMPLE A92 DYNAMIC ANALYSIS OF A BEAM Objective To verify the implicit direct time integration methods available in SOLVIA when used in a simply supported beam analysis.
Physical Problem A simply supported uniform beam subjected to a central step-function loading as shown in the figure below. This problem is also analyzed in Example A93 using the mode superposition procedure.
t P (t)
El, p, L, A L=2m E=2.10t" N/m2 I = 1.01159.10-6 m4 p = 7800 kg/m 3 A=0.01m 2
P0 = 100N t
Finite Element Model The finite element model consists of 20 BEAM elements, see top figure on page A92.3. A consistent mass matrix is used.
The time integration methods used in the analyses are the Newmark method (trapezoidal rule), the Wilson theta-method and the Hilber-Hughes method. Four different time step sizes are used for each method, namely DT = T/50, DT = T/25, DT = T/10 and DT = T/5, where T is the time period for the lowest frequency mode of the beam structure.
Solution Results The theoretical solution for this problem is presented in [ 1 ], pp. 401-403.
Displacement:
u(x,t)= 2P1L3 i
£fl<1cosc t)sin n7tx itEI
= n L
where an =
1 n= 1,5,9,...
n= 3,7,11,....
n =
even numbers n 4Tc4 EI and con -
L pA Velocity:
t Du(x,t)_
2PoL sX.
nmrx at
,2 ElpA n=1 L
a2U~x~tn IP n
~
Acceleration:
i(x,t)
LuA 2xcosP *t
-sin at, LpA *=I L
Version 99.0 L/2 P(t)
' -i P
Linear Examples V
A92.1
SOLVIA Verification Manual Bending moment:
M(x't) EI=
E u(xt)-
2P°L >2
=1n(llcosont)sinntxL Zýa C2it n 2 (ICSOtsn L
Shearforce:
V(x,t)=EI.
(
3u(x, t-2eO
(
tC ax 7C
=
n L
Note that the acceleration series does not converge with increasing mode number n, which is due to the singular nature of the load. Therefore, the acceleration response is not treated further in this example. In practice, of course, the load is associated with a finite area and has a finite rise time.
The theoretical solution (using n = 12000) for the displacement, velocity and bending moment at the midspan of the beam and the shear force at the left support can be seen in the figures on page A92.3.
Note the high frequency content in the shear force response compared to the bending moment and the velocity response, which in turn contain higher frequencies than the displacement response. The theo retical solution curves are also shown for comparison in the subsequent result figures.
The results from the Newmark time integration analysis can be seen on pages A92.4 and A92.5. The time integration parameters used are DELTA = 0.5 and ALPHA = 0.25. This method is also called the trapezoidal rule and the constant-average-acceleration method. Note that the trapezoidal rule has no algorithmic damping properties (non-dissipative) but it causes period elongation.
The results from the Wilson time integration analysis can be seen on pages A92.6 and A92.7 using time integration parameter THETA = 1.4. Note that the Wilson method is highly dissipative for larger time steps [2]. Note also that the Wilson method has larger period elongation as compared with the trapezoidal rule.
The results from the Hilber-Hughes time integration analysis can be seen on pages A92.8 and A92.9 using time integration parameter GAMMA = -0.1. Note that the Hilber-Hughes method equals the trapezoidal rule when GAMMA = 0 and that the algorithmic damping properties of the Hilber-Hughes method are a function of the GAMMA value [2]. The highest damping ratio is found when GAMMA = -1/3. The practical range of GAMMA is therefore -1/3*< GAMMA *0. The solution curves for the Hilber-Hughes method and the trapezoidal rule are in close agreement, but we may note that the Hilber-Hughes method gives smoother curves for the velocity, the bending moment and the shear force when using smaller time steps, DT = T/50 and DT = T/25, due to the algorithmic damping of higher modes.
User Hints
" A commonly used rule-of-thumb for non-dissipative algorithms requires at least 10 time steps per period to be taken for accuracy [2]. Hence, when DT = T/50 the response using non-dissipative algorithms contains frequencies up to 5 times the fundamental frequency with reasonable accuracy.
" Note also that an accurate solution of bending moments and shear forces requires small time steps as compared with the displacement solution in a dynamic analysis.
" The USERCURVE command can be used in SOLVIA-POST to easily include user-supplied curves.
References
[1]
Clough, R.W., and Penzien, J., Dynamics of Structures, Second Edition, McGraw-Hill, 1993.
[2]
Hilber, H.M., Hughes, T.J.R., and Taylor, R.L., "Improved Numerical Dissipation for Time Integration Algorithms in Structural Dynamics", Earthquake Engineering and Structural Dynamics, Vol. 5, pp. 283-292, 1977.
Version 99.0 Linear Examples A92.2
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'A92A DYNAMIC ANALYSIS OF A BEAM,
- NEWMARK, DT=T/50' DATABASE CREATE MASTER IDOF-I00011 NSTEP=80 DT=0.001 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT DELTA=0.5 ALPHA=0.25 TIMEFUNCTION 1
- 0. 1.
/
- 1.
1.
COORDINATES 1
/
2 C. 1.
METHOD=NEWMARK,
/
3
- 0. 2.
MATERIAL 1
ELASTIC E=2.E11 NU=-.
DENSITY=7800.
EGROUP 1
BEAM RESULTS=FORCES BEAMVECTOR 1
- 1. 0.
- 0.
SECTION 1 GENERAL RINERTIA=1.E-6 SINERTIA=1.01159E-6, TINERTIA=1.E-6 AREA=0.01 GLINE 1 3 -1 EL=20 FIXBOUNDARIES 23
/
FIXBOUNDARIES 3
/
LOADS CONCENTRATED 2 3 100.
1 3
SET VIEW=X NSYMBOLS=YES MESH VECTOR=LOAD NNUMBERS=MYNODES BCODE=ALL SUBFRAME=12 MESH EAXES=RST SOLVIA END Version 99.0 Linear Examples A92. 10
SOLVIA Verification Manual SOLVIA-POST input A92A DYNAMIC ANALYSIS OF A BEAM,
- NEWMARK, DT=T/50 DATABASE CREATE AXIS 1 VMIN=0.
VMAX=0.08 LABELSTRING='TIME' AXIS 2 VMIN=0.
VMAX-2.E-4 LABELSTRING='DISPLACEMENT' AXIS 3 VMIN=-0.015 VMAX=0.015 LABELSTRING=IVELOCITY' AXIS 4 VMIN=-100.
VMAX=0.
LABELSTRING='BENDING MOMENT' AXIS 5 VMIN=-200.
VMAX=100.
LABELSTRING='SHEAR FORCE' USERCURVE 1
READ A92DISP.DAT USERCURVE 2
READ A92VEL.DAT USERCURVE 3
READ A92MOM.DAT USERCURVE 4
READ A92FORCE.DAT SET DIAGRAM=GRID NHISTORY NODE=2 DIRECTION=3 XAXIS=i YAXIS=2 SYMBOL-i PLOT USERCURVE 1 XAXIS=--
YAXIS=-2 SUBFRAME=OLD NHISTORY NODE=2 DIRECTION=3 KIND-VELOCITY XAXIS=1 YAXIS=3 SYMBOL=I PLOT USERCURVE 2 XAXIS=-I YAXIS=-3 SUBFRAME=OLD EHISTORY ELEMENT=-0 POINT-2 KIND=MS XAXIS=i YAXIS=4 SYMBOL=i PLOT USERCURVE 3 XAXIS=-1 YAXIS=-4 SUBFRAME=OLD EHISTORY ELEMENT=I POINT-i KIND=FT XAXIS=1 YAXIS=5 SYMBOL=i PLOT USERCURVE 4 XAXIS=-I YAXIS=-5 SUBFRAME=OLD END Version 99.0 Linear Examples A92.11
SOLVIA Verification Manual EXAMPLE A93 MODE SUPERPOSITION ANALYSIS OF A BEAM Objective To verify the mode superposition procedure in SOLVIA in a dynamic beam structure analysis.
Physical Problem A simply supported uniform beam subjected to a central step-function loading as shown in the figure below will be evaluated. This problem has been analyzed using direct time integration methods as described in Example A92.
SP (t)
L/2 El, p, L,A L=2m E=2.10" N/im 2 I = 1.01159.10-6 m4 p = 7800 kg/m 3 A =0.01 M2 P0 =100 N Finite Element Model Twenty BEAM elements are used to model the simply supported beam structure as shown in the top figure on page A93.3. A consistent mass discretization is applied. The modal time integration is per formed with the Newmark method (trapezoidal rule) and the user has no control of it. Two different time step sizes are used, DT = T/25 and DT = T/10 where T is the time period for the lowest fre quency mode of the beam structure. Ten modes are used in the mode superposition procedure. For comparison, a single mode analysis is also performed. The subspace iteration method is used in the evaluation of frequencies and mode shapes.
Solution Results The theoretical solution for this problem is presented in [1 ], pp. 401-403.
S(xt)=2PL23
_n
(
sCo t)sinn(
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4- (,-co s
i L~
{1 n n= 1,5,9,....
where a 1n 1
n=
3,7,11,...
10
, n =
even numbers n 474 El and On-L4 pA DXt u(x, t) -2P 0ýL Velocity:
u~xt 2atP 2-sinwntsm!nii at 7C
.=,n2 Version 99.0 P(t)
PO-I.
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-Zýý t
A93.1
SOLVIA Verification Manual Acceleration:
(
2(x,t)
)
2 at2 LpA.=I L
Bending moment:
M(x,t)= El-a 2 u(Xt)-_
2P 0L j Ln-(1-cos>Ot)sinnTcx Z
X U2 n=1 ng Shearforce:
V(xt)=EI. a'3u(xt) - 2P°
- -(i-coscont)cos nfl X ani n
L Note that the acceleration series does not converge with increasing mode number n, which is due to the singular nature of the load. Therefore, the acceleration response is not treated further in this example. In practice, of course, the load is associated with a finite area and has a finite rise time.
The theoretical solution (using n = 12000) for the displacement, velocity and bending moment at the midspan of the beam and the shear force at the left support can be seen in the bottom figures on page A93.3. Note the high frequency content in the shear force response compared to the bending moment and the velocity response, which in turn contain higher frequencies than the displacement response. The theoretical solution curves are also shown for comparison in the subsequent figures.
The frequency range for the ten modes is between 20.0 and 1608 Hz. The time steps DT = T/25 and DT = T/10 are only selected to allow a comparison with the response obtained by the corresponding direct time integration method shown in Example A92. The modes with higher frequency are in this case excited initially to some degree although the time step is large compared to their periods. Nor mally, about ten time steps per period are used for the highest frequency mode to participate in the modal superposition.
The input for the analysis using 10 modes and DT=T/25 in the mode superposition procedure is shown on pages A93.6 and A93.7. The results can be seen on page A93.4 together with the results for DT = T/I 0. We may note the period elongation of approximately 10 percent for the displacement response when DT = T/10. Using 10 modes we get almost identical results as for the corresponding analysis in Example A92 using the trapezoidal rule.
The solution results from the analysis using a single mode in the mode superposition procedure can be seen on page A93.5. The displacement response is almost the same as the 10 mode case but the velocity, the bending moment history and shear force history show somewhat different results since the higher frequency modes are not participating.
User Hints Note that the higher modes contribute significantly to the bending moment response and even more significantly to the shear force response as compared with the displacement response.
Therefore, when limiting the number of modes used in a mode superposition analysis one should consider the response quantity to be evaluated [1].
"* A small enough time step is also important. In general, at least 10 time steps per period of the highest frequency mode to participate in the modal superposition are required for accuracy.
Reference
[1]
Clough, R.W., and Penzien, J., Dynamics of Structures, Second Edition, McGraw-Hill, 1993.
Version 99.0 Linear Examples A93.2
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SOLVIA Verification Manual SOLVIA-PRE input HEAD
'A93A MODE SUPERPOSITION ANALYSIS OF A BEAM, NMODES=10, DT=T/25' DATABASE CREATE MASTER IDOF=100011 NSTEP=40 DT=0.002 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT IMODS=1 NMODES=10 FREQUENCIES SUBSPACE-ITERATION NEIG=10 SSTOL=1.E-10 TIMEFUNCTION 1
- 0. 1.
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- 1.
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COORDINATES 1
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- 1.
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3
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- 2.
MATERIAL 1 ELASTIC E=2.E11 NU=0.
DENSITY=7800.
EGROUP 1
BEAM RESULTS=FORCES BEAMVECTOR 1
- 1. 0.
- 0.
SECTION 1 GENERAL RINERTIA=1.E-6 SINERTIA=1.01159E-6, TINERTIA=1.E-6 AREA:0.01 GLINE 1 3
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FIXBOUNDARIES 3
/
3 LOADS CONCENTRATED 2 3 100.
SET VIEW=X NSYMBOLS=YES MESH VECTOR=LOAD NNUMBERS=MYNODES BCODE=ALL SUBFRAME:12 MESH EAXES:RST SOLVIA END Version 99.0 Linear Examples A93.6
SOLVIA Verification Manual SOLVIA-POST input A93A MODE SUPERPOSITION ANALYSIS OF A BEAM, NMODES:10, DT=T/25 DATABASE CREATE AXIS 1 VMIN=O.
VMAX=0.08 LABELSTRING='TIME' AXIS 2 VMIN=0.
VMAX=2.E-4 LABELSTRING='DISPLACEMENT' AXIS 3 VMIN=-0.015 VMAX=0.015 LABELSTRING='VELOCITY' AXIS 4 VMIN=-100.
VMAX:0.
LABELSTRING='BENDING MOMENT' AXIS 5 VMIN=-200.
VMAX=100.
LABELSTRING='SHEAR FORCE' USERCURVE 1
READ A92DISP.DAT USERCURVE 2
READ A92VEL.DAT USERCURVE 3
READ A92MOM.DAT USERCURVE 4
READ A92FORCE.DAT SET DIAGRAM=GRID NHISTORY NODE-2 DIRECTION=3 XAXIS=1 YAXIS=2 SYMBOL=1 PLOT USERCURVE 1 XAXIS=-1 YAXIS=-2 SUBFRAME=OLD NHISTORY NODE=2 DIRECTION=3 KIND=VELOCITY XAXIS=1 YAXIS=3 SYMBOL=1 PLOT USERCURVE 2 XAXIS=--
YAXIS=-3 SUBFRAME=OLD EHISTORY ELEMENT=f0 POINT=2 KIND:MS XAXIS=1 YAXIS=4 SYMBOL=1 PLOT USERCURVE 3 XAXIS=-1 YAXIS--4 SUBFRAME=OLD EHISTORY ELEMENT=i POINT=1 KIND=FT XAXIS=1 YAXIS=5 SYMBOL=i PLOT USERCURVE 4 XAXIS=--
YAXIS:-5 SUBFRAME=OLD END Version 99.0 Linear Examples A93.7
SOLVIA Verification Manual EXAMPLE A94 ORTHOTROPIC CYLINDER UNDER INTERNAL PRESSURE Objective To verify the PLANE AXISYMMETRIC element with pressure interpolation in a linear orthotropic analysis.
Physical Problem The figure below shows the cylinder to be analyzed. The cylinder is assumed to be guided so that no axial displacement can occur and it is acted upon by internal pressure. The material constant g is used to control the degree of orthotropy.
Z Eý = 106 gN/mm 2 Vab = 0.25 /
Eb =106 N/mm 2 v 0 =0.25/g
[E l
=106 N/mm 2 Vbc= 0.25 G, = 4. 105 N/mm 2 g = 0.16667 r2 r,= 100 mm
-Y r2 =200 mm p =10 N/mm 2 The orthotropic axes a, b and c correspond to the global axes Y (radial), Z (axial) and X (circumferen tial), respectively.
Finite Element Model The cylinder is modeled using five 9-node PLANE AXISYMMETRIC elements with pressure interpolation included. The internal cylinder pressure is applied along the line between the nodes 2 and 3, see the figure on page A94.3.
Solution Results The material constants are chosen to produce a semi-deformable material when g = 1/6.
When g = 1/6 the bulk modulus of the material becomes infinite and the requirement that the element pressure is finite means that the following sum of strains is zero:
f eaa + ebb + ecc = 0 In our analysis we have used g = 0.16667 which gives E. = 166670 N/mm 2, Vab = Vac = 1.49997 and the bulk modulus K = 2.31486. 1010 N/mm2.
Version 99.0 Linear" Examples A94.1I
SOLVIA Verification Manual The analytical solution to this problem is [1 ]:
1 --C
(\\k I.
1--c k
I.
Pck+ kr f-pck+l (r
)-k-1 (bb Vab 'Jaa + Vcb Cc.o c = r, / r,, r is the radial coordinate and I(1-u Vab z k =
The column "Theory" in the tables below gives the analytical values for g = 1/6.
The input data used in the SOLVIA analysis can be seen on pages A94.5 and A94.6.
Stresses in N/mm2 at nodes 2 and 1:
Y-coord.
0 aa (bb Cy0c (mm)
SOLVIA Theory SOLVIA Theory SOLVIA Theory 100
-9.7519
-10
-9.4300
-9.3333 22.5676 22.6667 200 0.014
- 0.
1.3283 1.3333 5.3281 5.3333 Using ten 4-node elements the following results are obtained:
Y-coord.
(3aa G bb (Ycc (mm)
SOLVIA Theory SOLVIA Theory SOLVIA Theory 100
-8.8169
-10
-9.7678
-9.3333 22.1573 22.6667 200
-0.1171
- 0.
1.3088 1.3333 5.290 5.3333 The element hydrostatic pressure values in N/mm 2 (negative mean stress) at nodes 2 and 1:
Version 99.0 Y-coord.
9-node element 4-node element Theory (mm)
SOLVIA SOLVIA 100
-1.1286
-1.1909
-1.1111 200
-2.2235
-2.1613
-2.2222 Linear Examples A94.2
SOLVIA Verification Manual The distribution of the orthotropic stresses as calculated by SOLVIA for the 9-node element analysis can be seen in the top figures on page A94.4 with symbols included. For comparison the theoretical solution curves are also included.
The solution results for the 4-node element analysis calculated by SOLVIA can be seen in the bottom figures on page A94.4.
User Hints
- Note that the 9-node PLANE element has a linear pressure interpolation assumption. The 4-node PLANE element has a constant pressure assumption.
Reference
[11 Lekhnitski, S.G., "Theory of Elasticity of an Anisotropic Body", Mir, Moscow, 1977.
Version 99.0 Linear-Examples A94.3
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SOLVIA Verification Manual SOLVIA-PRE input HEADING
'A94 ORTHOTROPIC CYLINDER UNDER INTERNAL PRESSURE' DATABASE CREATE MASTER IDOF=100111 COORDINATES
/
ENTRIES NODE Y
Z 1
200.
- 20.
/
2 100.
- 20.
/
3 100.
/
4 200.
MATERIAL 1
ORTHOTROPIC EA=166670.
EB=1.E6 EC=I.E6 NUAB=l.49997, NUAC=1.49997 NUBC=0.25 GAB=4.E5 EGROUP 1 PLANE AXISYMMETRIC RESULTS=NSTRESSES, PRESSURE-INTERPOLATION:YES GSURFACE 3
4 1 2 EL1=5 EL2=1 NODES=9 LOADS ELEMENT INPUT=LINE 2 3
- 10. 10.
FIXBOUNDARIES 3 INPUT=LINE
/
1 2
/
3 4
SET NSYMBOLS=MYNODES NNUMBERS=MYNODES PLOTORIENTATION=PORTRAIT MESH VECTOR=LOAD SUBFRAME=12 MESH EAXES=ORTHOTROPIC SOLVIA END Version 99.0 Linear Examples A94.5
SOLVIA Verification Manual SOLVIA-POST input A94 ORTHOTROPIC CYLINDER UNDER INTERNAL PRESSURE DATABASE CREATE STRESSREFERENCE=ORTHOTROPIC WRITE FILENAME='a94.1is' VMIN=0.
VMIN=-10.
VMIN=-10.
VMIN=5.
VMAX=100.
VMAX=0.
VMAX=5.
VMAX=25.
LABEL='RADIAL' LABEL='STRESS-AA' LABEL='STRESS-BB' LABEL='STRESS-CC' EPLINE NAME=RADIAL
/
1 4 7 3
TO 5 4 7
3 USERCURVE 1
READ A94SAA.DAT USERCURVE 2
READ A94SBB.DAT USERCURVE 3
READ A94SCC.DAT SET PLOTORIENTATION=PORTRAIT DIAGRAM=GRID SUBFRAME 12 ELINE LINENAME=RADIAL KIND=SAA XAXIS=1 YAXIS=2 SYMBOL=1 PLOT USERCURVE 1 XAXIS=--
YAXIS=-2 SUBFRAME=OLD ELINE LINENAME=RADIAL KIND=SBB XAXIS=1 YAXIS=3 SYMBOL=1 PLOT USERCURVE 2 XAXIS=-1 YAXIS=-3 SUBFRAME=OLD SUBFRAME 12 ELINE LINENAME=RADIAL KIND=SCC XAXIS=1 YAXIS=4 SYMBOL=1 PLOT USERCURVE 3 XAXIS=-1 YAXIS=-4 SUBFRAME=OLD CONTOUR AVERAGE=NO MESH CONTOUR=SMEAN ELIST SELECT=S-EFFECTIVE END OUTPUT=ALL OUTPUT=ALL OUTPUT=ALL Version 99.0 AXIS AXIS AXIS AXIS 1
2 3
4 Linear Examples A94.6
SOLVIA Verification Manual EXAMPLE A95 MATERIAL DAMPING IN COMPLEX-HARMONIC ANALYSIS Objective To verify the PLANE, SOLID and SHELL elements with viscous and hysteretic material damping in a COMPLEX-HARMONIC analysis.
Physical Problem A sinusoidal force is applied to a damped one degree-of-freedom system.
I F F/2 m/2 h, k E, a, F/2 Sm/2 B
L Thickness t = 1 L=I m B=1 m m
I 1lkg F=IN Elastic material Lk E=-=
100 N/M 2 Bt V= 0.
Viscous damping c = 5 Ns/m Hysteretic damping h = 30 N/m Viscous damping coefficient L
cr=
.c=0.05 s EBt Hysteretic damping coefficient L
71=--.h=0.3 EBt Finite Element Model One element of each type is used and shown on page A95.3.
Version 99.0 Stiffness k = 100 N/m Linear Examples i
_,_1 A95.1
SOLVIA Verification Manual Solution Results The theoretical displacement amplitude in complex form at the mass point of the spring/damper system is R
HO2+joC
+k) where R
load amplitude in complex form co excitation circular frequency h
C=c++/-
CO The velocity amplitude in complex form at the mass point is CO uz The displacement and velocity at the midpoint of the spring/damper system are Umid =u/2 "miýd =U/2 The forces in the spring and the damper are F~
=ku F
= Cu The time average of the power flow in the spring/damper system at mid point is Sr1 j(Fspring +-Famp)udmid where F
and F5*
are complex conjugate of the corresponding element forces.
The time average of the power flow into the spring/damper system is Sin
-1 F*u 2 -_
where F = F* = F since a sinusoidal force is applied with zero phase angle.
The time average of the strain energy in the spring/damper system is U =l1*k u 4
The time average of the dissipated energy in the spring/damper system is 2--
Version 99.0 Linear Examples A95.2
SOLVIA Verification Manual The numerical solution of time averages obtained at the excitation frequency 2 Hz using the input data on pages A95.4 and A95.5 gives the following results:
Theory PLANE SOLID SHELL element element element Active powerfiow into the structure, 4.872.10-2 4.872.10-2 4.872.10-2 4.872.10-2 Re(S), in Nm/s Contribution from all loaded nodes of the element Active power flow in the structure, 2
2 2
ReSrinNI~nsmi oit 2.436.10-2.436.10-2.436.10-2.436. 10-2 Re (5r) I, in Nmi/(m2,s), mid point Total dissipated power, 1l, in Nm/s 4.872.10-2 4.872.10-2 4.872.10-2 4.872.10-2 Total strain energy, U, in Nm 2.088.10-'
2.088.10 2.088.10-'
2.088.10-3 Note that the input active power into the structure is equal to the dissipated power in the material damping. The active power flow at 2 Hz is shown as vector and contour plots on page A95.4. The dissipated power density and strain energy density are constant over the elements, see contour plots on page A95.5.
A9S MATERIAL DAMPING IN COMPLEX-HARMONIC ANALYSIS ORIGINAL H-H O.S TIME I
z X -
2 S34
'5 2-t S 22 25 21 3-1
[03 SL2FORCE 23 B102---
240.656667 MASTER B 00On-O L OlIJO0 0
C200 E '1, 111 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB Version 99.0 Linear Examples A95.3
SOLVIA Verification Maual Linear Examples ASS MATERIAL DAMPING IN 2
REQUENCY SOLVIA-POST 99.0 0
I, 4
FREOUENCY SOLVIA ENGINEERING AB Version 99.0
'N i
COMPLEX-HARMONIC ANALYSIS J
CL R [
.C !
- 0
/2
/2
-'Jr
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>22-p ASS MATERIAL DAMPING IN COMPLEX-HARMONIC ANALYSIS ORIGINAL )---- 0.2 Z
FREO 2 L
~*
g0
ýlfi 3
P -ACT IVE 0.04323 ORIGINAL I.2 Z
FREO 2 L_
1,z~:~~
P-ACTIVE MAX 0.018721 0.045676
- i*!i*E*
- i~i E*;
W~i~E£;ii*{£:
- 0.039S B6 0.033496 0.027106 0.021316 I
0.015229 9.13S3E-3 a -
ono~~
2 ~
')~3.045!E
'3 MiN 0 SOLVIA-POST 99S0 SOLVIA ENGINEERING AS I
J 0
L{
A95.4
SOLVIA Verification Manual Linear Examples SOLVIA-PRE input DATABASE CREATE HEAD
'A95 MATERIAL DAMPING IN COMPLEX-HARMONIC ANALYSIS' COMPLEX-HARMONIC 0.1 59 6.0 TIMEFUNCTION 1 0 1 /
500 1 SYSTEM 1 Y=1.5 SYSTEM 2 Y=3.0 COORDINTES SYSTEM=0 ENTRIES NODE Y Z 1 1.
- 1. / 2
.0 1_. / 3 COORDINTES SYSTEM=1 11
.0
- 1. 1. /12
.0
.0 15 -1. 1. 1. /16
-1.
0 COORDINTES SYSTEM=2 ENTRIES NODE Y Z 21
- 1. 1. /22 0
1 /
2 MASSES 10.50/
- 20.
11 0.25 /
12 0.
21 0.166666666 /
.0.0 /
4
- 1..0
- 1. /13
.0.0.0 /14
.0 1.
.0
- 1. /17 -1.
.0.0 /18
-1.
- 1.
.0 3
.0
.0 /24
- 1. 0. /
25
.5 1.
50 25 /
15 0.25 /
16 0.25 22 0.166666666 /
25 0.666666666 MATERIAL 1 ELASTIC E=100 DENSITY=0 VISCOUS-DAMPING=0.05 HYSTERETIC-DAMPING=0.3 EGROUP 1 PLANE STRESS2 INT=3 Version 99.0 A9S MATERIAL DAMPING IN COMPLEX-HARMONIC ANALYSIS MAX DISPL.
1.B37SE-3 2
FREI 2 DTSSIP-POWER DENSITY SHElLL TOP MAX 0.048721361 0.087227D3 S)i~i'*
0.018722166 0.018721629
- 0.048721092 o0.08720SSS 0.018723018 S)
' i i: iii~i:i~i:i:i:)¢ 0.018719181 MIN 0.048721361 MAX DISPL.
-i 4.8375E-3 Z
FREG 2 STRAINENERGY DENSITY SHELL TOP MAX 2.08B2SDIE-3
- :::::: ::::::::::::::::::::::::::::::::::::::::::: 2. 0 8 8 3 3 0 7 E - 3 2 08830760 3 2.0882846E-3 2 088216i6E3 2.0882386E-3 2 8882156E-3 2.888192SE3 L _
2.088 !69SE-3 MIN 2 0882SOIE-3 SOLVIA ADST 99.0 SOLVIA ENGINEERING AB A95.5
SOLVIA Verification Manual SOLVIA-PRE input (cont.)
GSURFACE 2
3 4 1 EL1=1 EL2=1 NODES=4 EDATA /
1 1.
EGROUP 2 SOLID TINT=3 RSINT=3 GVOLUME 11 12 13 14 15 16 17 18 EL1=1 EL2:1 NODES=8 EGROUP 3 SHELL RINT-3 SINT=3 GSURFACE 22 23 24 21 EL1=1 EL2:1 NODES=8 THICK 1 1.
FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES FIXBOUNDARIES LOAD 13 11 3 21 3 3 INPUT=LINE/
3 4 /
23 24 3 INPUT=SURFACE/
13 14 18 17
/
4 14 24 1 /
18 2 /
13 CONC 0.50 /
2 3 0.50 0.25 / 12 3 0.25 /
15 3 0.25 /
16 3 0.25 0.166666666 /
22 3 0.166666666 /
25 3 0.666666666 SOLVIA COLOR LINE VIEW ID=C XVIEW=5 YVIEW=2 ZVIEW=l SET HIDDEN=NO NSYMBOL=YES NNUMBER=YES MESH VIEW=C VECTOR-LOAD BCODE=ALL END ENUMBER=GROUP SOLVIA-POST input
- A95 MATERIAL DAMPING IN COMPLEX-HARMONIC ANALYSIS DATABASE CREATE WRITE
'a95.lis' SET RESPONSETYPE=REAL HEIGHT:0.25 VIEW=X NCOMPLEX-HARMONIC NODE-i DIRECTION=3 KIND:DISPLACEMENT SUBFRAME=21 NCOMPLEX-HARMONIC NODE=i DIRECTION=3 KIND=P-ACTIVE OUTPUT=ALL SET TIME=2 MESH VECTOR=P-ACTIVE SUBFRAME=12 MESH CONTOUR:P-ACTIVE MESH CONTOUR=DISSIPATED-POWER SUBFRAME=12 MESH CONTOUR=ENERGY ELIST SELECT=POWER TSTART=2 SUMMATION KIND-ENERGY DETAILS=YES TSTART=2 NCOMPLEX-HARMONIC NCOMPLEX-HARMONIC NCOMPLEX-HARMONIC NCOMPLEX-HARMONIC NCOMPLEX-HARMONIC NCOMPLEX-HARMONIC NCOMPLEX-HARMONIC NCOMPLEX-HARMONIC END NODE=2 NODE=11 NODE=12 NODE=-5 NODE=16 NODE=21 NODE=22 NODE=25 DIRE=3 DIRE=3 DIRE=3 DIRE-3 DIRE=3 DIRE=3 DIRE=3 DIRE=3 KIND=P-ACTIVE KIND=P-ACTIVE KIND=P-ACTIVE KIND=P-ACTIVE KIND=P-ACTIVE KIND=P-ACTIVE KIND=P-ACTIVE KIND=P-ACTIVE OUTPUT-LIST OUTPUT=LIST OUTPUT-LIST OUTPUT=LIST OUTPUT=LIST OUTPUT=LIST OUTPUT=LIST OUTPUT=LIST FEND=2 FEND=2 FEND=2 FEND=2 FEND=2 FEND=2 FEND=2 FEND=2 Version 99.0 Linear Examples A95.6
SOLVIA Verification Manual Linear Examples EXAMPLE A96 FREQUENCIES OF A BEAM WITH U CROSS-SECTION Objective To verify the dynamic behaviour of the BEAM element when the shear center and the center of gravity are not coinciding.
Physical Problem The natural frequencies and mode shapes of a free-free beam with a U cross-section as shown in the figure below are to be calculated.
L L=5 in WTOP = 0.1 m D=0.1 m T1 = 0.01 in T2 =0.01 mm E =2.1.10" N/m2 v=0.3 Di T1 T2
-E t,Z Center of gravity Shear center WTOP 1
p = 7800 kg/m3 Finite Element Model The finite element model consists of twenty BEAM elements, see figure on page A96.3. A consistent mass matrix is used. Rigid links are attached to the beam in the transverse direction to make it easier to show the torsion modes. A SHELL element model of the beam is used as a reference solution.
Solution Results The input data for the BEAM element model is shown on page A96.6 and the input data for the SHELL element model is shown on page A96.7.
Version 99.0 Z L Y
S A96.1
Linear Examples SOLVIA Verification Manual Calculated natural frequencies, Hz:
The first six modes are rigid body modes with zero frequency since the beam has no support.
Mode 7 and 11 are pure bending modes in the plane X = 0. The pure bending modes are possible since both the shear center and the center of gravity have zero X-coordinates. The BEAM and the SHELL element models give almost the same frequencies from the pure bending modes.
The remaining of the listed and displayed modes are coupled modes with both bending and torsion. A pure torsion mode is not possible since the shear center and the center of gravity do not coincide.
The frequencies of the BEAM and the SHELL element models agree reasonably well up to mode 13.
The torsional half wave length in mode 14 is, however, only about 1 meter. The warping of the cross section becomes then significantly constrained which increases the stiffness. This effect is modelled by the SHELL element but not by the BEAM element.
The torsional stiffness of a BEAM element with U cross-section is calculated based on Saint Venant's theory of torsion for thin rectangular areas. Any warping of the cross-section is unrestrained.
User Hints In the command FREQUENCIES SUBSPACE-ITERATION the parameter RIGIDMODES = YES is necessary when rigid body modes are present and parameter FSHIFT = -f, improves the con vergence rate where f, is an approximation of the first non.zero natural frequency in Hz.
Version 99.0 Mode BEAM model SHELL model 7
23.5 23.4 8
25.2 25.4 9
35.5 32.7 10 53.2 55.8 11 64.4 64.2 12 71.2 75.9 13 90.6 92.9 14 97.8 123.4 A96.2
Linear Examples SOLVIA Verification Manual BEAM element model A96 FREQUENCIES OF A BEAM WITH U CROSS-SECTION ORIGINAL 0.2 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB SHELL element model Version 99.0 z
X-
-y A96A FREQUENCIES OF A BEAM WITH U CROSS-SECTION, SHELL ELEMENTS ORIGINAL 0.2 Z
X 1'--y SOLVIA ENGINEERING AB A96.3 J
SOLVIA-PRE 99.0
0 0
6
'0 E41 P
0 0
1 0
0 0
0 0
0 t1 0
- 0 0
(D 0"
w P
SOLVIA Verification Manual SHELL element model, eigenmodes AX6A FROSENCIES OF A BEAM WITH U CR0SS-BECTION.
SHELL ELEONUTS NOLAI -POST AX P SOLVIA ENGINEERING A
ANOA FREP EN TES Or A BEAM UITHU CRSS-EACTAI SHELL ELEPESTS "A*A CR A-A A XAAoo z
HOR E
I C
PSOG PA X7S
[
SOLAIS A5 SO A9 50B1 SORSOA A&GIN AO AS Version 99.0 Linear Examples A96A FREQUENCIES Or A BEAM WITH U CRASS-SECTICN.
SHELL ELEMENTS MAX OPEC 12 TA, SEGVIA-PRST BS C 51RA ENGINEE A S, AS A96A FRECUENCIES OF A 3EAHi WITH U CROSS-SECTICA, SHELL ELEMENTS MAX O!SPLFR-0 2366 z
MOBE t4 REC t23 37 N"ý X 'J So'VIA÷POST,9g Q
SCtLVIA EN*ER*GA.
A96.5 A9RA R AOOASORES OF A NEAM ACITH A EARAS SOOPRAS SEELL ELEPENTS NOAXIA-WE -0R A
SALA A ENSRREEA Z
xN AN--
AR6A FRECUENCIES OF A BEAM WITH u CROSS-SECTIRN.
SHELL 0LEMENTS MAX DI RL. -AZ AM NO DE 9 FREG 32 74 AL A96A FRECUENCAES XA BEAM RITH U CROSS-SECTION, SHELL ELEMENTS MAX DISPL., 0 26664 M.OE 10 'REn SS.!02 x
SOPOAP EOSSGSOSAAS AN I
SOLVIA-POST 99.D 504VIA ENGINEERING AD
SOLVIA Verification Manual SOLVIA-PRE input, BEAM element model HEADING
'A96 FREQUENCIES OF A BEAM WITH U CROSS-SECTION' DATABASE CREATE FREQUENCIES SUBSPACE-ITERATION NEIG=15 SSTOL=-E-10 FSHIFT=-20 NQ=40 RIGIDMODES=YES COORDINATES 1
2
- 0.
- 5.
3
.05
- 0.
4 05
- 0.
MATERIAL 1
ELASTIC E=2.1E11 NU=0.3 DENSITY=7800.
EGROUP 1
BEAM SECTION 1
U WTOP=0.1 D=0.1 T1=0.01 T2=0.01 BEAMVECTOR 1
-1. 0.
- 0.
GLINE 1 2 AUX=-l EL=20 RIGIDLINK ADDZONE=AUX 31/41 TRANSLATE ZONE=AUX COPIES=20 YSTEP=0.25 MESH NNUMBER=MYNODES NSYMBOL=MYNODES SOLVIA END SOLVIA-POST input, BEAM element model A96 FREQUENCIES OF A BEAM WITH U CROSS-SECTION DATABASE CREATE WRITE
'a96.lis' FREQUENCIES MASS-PROPERTIES SET RESPONSE-TYPE=VIBRATIONMODE ORIGINAL=YES MESH TIME=7 MESH TIME=8 MESH TIME=9 MESH TIME=10 MESH TIME=1i MESH TIME12 MESH TIME=13 MESH TIME=14 END Version 99.0 Linear Examples A96.6
SOLVIA Verification Manual SOLVIA-PRE input, SHELL element model HEADING
'A96A FREQUENCIES OF A BEAM WITH U CROSS-SECTION, SHELL ELEMENTS' DATABASE CREATE ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT FREQUENCIES SUBSPACE-ITERATION NEIG=15 RIGIDMODES=YES SSTOL:lE-10 FSHIFT=-20 MEMORY SOLVIA=24 COORDINATES 1
-0.045
- 0.
0.1
/
2
-0.045 5.
0.1 3
-0.045
- 0.
0.005
/
4
-0.045
- 5.
0.005 5
0.045 0. 0.1
/
6 0.045 5.
0.1 9
0 0
0.005
/
10 0
- 5.
0.005 11 0.045 0.
0.005
/ 12 0.045 5.
0.005 MATERIAL 1
ELASTIC E=2.1El1 NU=0.3 DENSITY=7800.
EGROUP 1
SHELL THICKNESS 1
0.01 GSURFACE 6
5 11 12 EL1=10 EL2=4 NODES=16 GSURFACE 12 11 3
4 EL1=10 EL2=6 NODESf16 GSURFACE 4
3 1
2 EL1=10 EL2=4 NODES=16 MESH NNUMBER=MYNODES NSYMBOL=MYNODES SOLVIA MODEL-
SUMMARY
=DETAIL END SOLVIA-POST input, SHELL element model A96A FREQUENCIES OF A BEAM WITH U CROSS-SECTION, SHELL ELEMENTS DATABASE CREATE WRITE
'a96a.lis' FREQUENCIES MASS-PROPERTIES SET RESPONSE-TYPE=VIBRATIONMODE COLOR LINE MESH TIME=7 MESH TIME=8 MESH TIME=9 MESH TIME=10 MESH TIME=11 MESH TIMEf12 MESH TIME=13 MESH TIME-14 END Version 99.0 Linear Examples A96.7
SOLVIA Verification Manual EXAMPLE A97 SEGMENT OF A PIPE PARTLY FILLED WITH WATER Objective To verify the BEAM element load with interpolation of intensity along a global coordinate axis.
Physical Problem The figure below shows a segment of a pipe partly filled with water. The segment length is 1 m and plane stress conditions are assumed.
R=l m Thickness= 0.001 m E=2.10i" N/m 2 v'= 0.3 Ppipe = 0 kg/m3 Pwtr= 1000 kg/m 3 g=9.81 m/s 2 1 =120o Finite Element Model The finite element model is shown on page A97.2 and consists of 144 BEAM elements.
Solution Results The theoretical solution for this example can be found in reference [1]. Input data used in the SOLVIA analysis is found on pages A97.2 and A97.3.
Element results at the fixed point:
SOLVIA Theory element 1, node 1 Axial force, N (r-direction) 7.57.10' 7.30.10' Shear force, N 1.24.104 1.22.104 Bending moment, Nm
-5.54.10'
-5.54.10' Reference
[1]
Roark, R. J., Formulas for Stress and Strain, Fourth Ed., McGraw-Hill Book Company, 1965.
Version 99.0 Linear Examples R
Fixed F!
A97.1
SOLVIA Verification Manual Linear Examples A97 SGEGENr OF A PIPE PARTLY F!LLED WITH WATER 2
z RE AON SOLVIA ENGINEER
!NG A*I DATABASE CREATE HEADING
'A97 SEGMENT MASTER IDOF=100011 SYSTEM 1 CYLINDRIC COORDINATES ENTRIES NODE R THETA 1
- 1. -90.
OF A PIPE PARTLY FILLED WITH WATER' MATERIAL 1 ELASTIC E=2.0E11 NU=0.3 EGROUP 1 BEAM RESULT=FORCE BEAMVECTOR /
1
- 1. 0.
- 0.
GLINE 1 1 AUX=-1 EL=144 SYSTEM=i SECTION 1 RECTANGULAR D=1.E-3 WTOP=1.
LOADS ELEMENT INPUT=LINES INTERPOLATION=Z COORD1=0.5 V1=0.
COORD2=-1.
1 1 t FIXBOUNDARIES /
1 SET VECTOR=LOAD BCODE=ALL NNUMBER=MYNODES NSYMBOL=MYNODES MESH VIEW=X SOLVIA END Version 99.0 ORIGINAL MAX DISPL. -
90 rTNE I SqOLVI.A-POS7 99.0]
A97 SEGMENT NF A PIP2 PARTLY FILLED WITH SATER ORIGNiAL G0.2 TIME t S0LVIA PNIE 99.G OLI A97 SEGMENT OF A%
PIPE PARTLY FILLED WI*TH WATER T ANX D0[SPL.0 092 52 Z
TAX ONISL. -
92.52 Z
TIN HE Ly TIME[
I Ly IT AY S
- ING NO AVENAGING 4A4 1223
\\
jsAN 2230 S
!i129 4 1ý,tt B,93
- -Is
..I
- -3ý1.4
!*:-6*
8!i:-418I 1
MIN-12235 M1IN-5538.1 SOLVIA-PO5T 99.0
$OLVIA E.NGINEERING A3 AN97 SEGMNT OF A PIPE PAART OILL WASITH AATER 505517 POST 99A NOýI 5
ENGINEERINGAT AT SOLVIA-PRE input V2=14715.
VMIN=0.
EFORCE L4715 1-7ER A97.2
SOLVIA Verification Manual SOLVIA-POST input A97 SEGMENT OF A PIPE PARTLY FILLED WITH WATER DATABASE CREATE WRITE
'a97.1is' CONTOUR AVERAGING=NO MESH ORIGINAL=DASHED VECTOR=REACTION VIEW=X MESH CONTOUR=FT SUBFRAME=21 MESH CONTOUR=MS EPLINE RING 1 1 2 TO 144 1 2 ELINE RING KIND=FT SUBFRAME=21 ELINE RING KIND=MS ELIST ELI END Version 99,0 Linear Examples A97.3
SOLVIA Verification Manual EXAMPLE A98 PIPE JOINT Objective To demonstrate the model generation of a pipe joint using SHELL elements.
Physical Problem A pipe joint under internal pressure as shown in the figure below is considered.
t '2, I F,
ýi = 250 mm 11 = 1000 mm t- = 5 mm F_ PRA'l _25.103 N/m 4
42 =500 mm 12 1600 mm t2 -10 mm F2 Pý2 _250.103 N/m 4
E= 2.1.10" N/m 2 v =0.3 p = 2.10 6 N/m 2 Finite Element Model One quarter of the pipe joint is considered due to symmetrical conditions. The SHELL element mesh is generated in two local systems. The location of the local systems, keynodes and the parameters used to define the mesh density are shown in the figure on the next page.
XI Two mathematical surfaces of type cylinder are used to find the inter Y0R2 2 section line between node 2 and S$ELB node 4. Using the command
$ELC GSURFACE, 16-node SHELL ele 10 ments are generated and the mesh is EL.
LB shown on top of page A98.3.
$ELB 2 L
- $ELCYL $ELB Two symmetrical planes and a fixed X-displacement at node 11 are used SELCYL 7
$ELD as boundary conditions.
$ELB X"
Version 99.0 I
I F2 Linear Examples
"-I A98.1
SOLVIA Verification Manual Solution Results A deformation plot of the model is shown in the left bottom figure on page A98.3. Since the cylindri cal shells are rather thin there are considerable ovalizations in the cylinders, which have not damp ened out at the free ends. The influence on the stresses at the intersection from conditions at the free ends is, however, small.
The von Mises effective stress at the intersection is shown in a contour plot in the right bottom figure on page A98.3. Note that averaging of nodal stresses should not be used since there is a natural stress discontinuity due to the different thicknesses of the cylinders.
The fineness of the mesh can be checked by stress deviation plots, see the top figures on page A98.4.
It can be seen that the stress deviation (from the nodal mean values) is of the order one percent of the von Mises effective stress.
The degree of accuracy is also illustrated by the smoothness in stress variations along the intersecting line, see the bottom figures on page A98.4. The stress values in the large cylinder are marked with triangle symbols.
The radial displacement, hoop stress and meridional stress along the line node 2 - 1 (cylinder 1) and along the line node 2 - 6 (cylinder 2) are shown on page A98.5. The ovalizations of the cylinders can be seen to cause a significant portion of the radial displacements. The stresses at the intersection dominate, however, the stress response.
The membrane stresses Ym (meridional) and ah (hoop) for a thin vessel without any disturbance are:
C p
R p.R 2t t
The radial displacement due to this membrane stress state is:
R Version 99.0 Cylinder 1, p = 250 mm Cylinder 2, 4 = 500 mm Linear Examples A98.2
SOLVIA Verification Manual ORIGINAL -*
02 SOLVIA-PRE 99.0 8 PIPE JOINT
-I SLIELL TOP STELL BOTTOM NON-SHELL SOLVIA ENGINEERING AS Linear Examples A98 PIPE JOINT ORIGINAL I--71 MAX DISPL.
4.01t3E-4 TIME I
SOLVIA-POST 99.0 so0LVIA ENGINEERING AS Version 99.0 A98 PIPE JOINT MAX DISPL. -
3.7788E-4 Z
TIME I
ZONE INTER X
Y MISES NO AVERAGING SHELL TOP MAX I 7922E8 4.4950E8
- 3. 900SE8 3.3061EB 2.7117E8
- 2. 1 t72E8 1.S228E9
- 9. 2833E7
- 3. 3388E7 M!N 3 6663E6 SOLVIA-POST 99.0 SOLVIA ENGINEERIN ASB Z
xJ-A98.3
a 0
".0 O0 (0
00 00 (0(0 0:0
- 0 (0
O
>0
- 0
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SOLVIA Verification Manual SOLVIA-PRE input DATABASE CREATE HEAD
'A98 PIPE JOINT' MEMORY SOLVIA=20 PARAMETER
$ELA=16
$ELB=5
$ELCYL=10
$ELC=8
$ELD=16 SET NODES=16 MIDNODES=2 SET NNUMBERS=MYNODES NSYMBOL:MYNODES PLOTORIENTATION=PORTRAIT COLOR EFILL=SHELL SYSTEM 1 SYSTEM 2 MATERIAL CYLINDRICAL X=1.0 CYLINDRICAL Z=0.8 AX=0 AZ=1 1 ELASTIC E=2.1E11 NU=0.3 COORDINATES SYSTEM=1 ENTRIES NODE R
1 0.125 2
0.125 3
0.125 4
0.125 COORDINATES SYSTEM=2 ENTRIES NODE R
5 0.250 6
0.250 7
0.250 8
0.250 9
0.250 10 0.250 11 0.250 SET SYSTEM=1 THETA
- 90.
- 90.
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- 0.
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1 2 4 3 t 2.E6 LOAD ELEMENTS TYPE-FORCE INPUT=LINE /
1 3 out 125.E3 SET SYSTEM-2 LINE CYLINDRIC 5 8 EL=$ELB LINE CYLINDRIC 8 7 EL=$ELB LINE COMBINED 5 7 8
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7 4 ADDZONE=CYL2 GSURFACE 5 6 9 8 LINE COMBINES 7 9
8 Version 99.0 Linear Examples A98.6
SOLVIA Verification Manual SOLVIA-PRE input (cont.)
GSURFACE 7
9 11 10 THICKNESS 1 10E-3 LOAD ELEMENTS INPUT-SURFACE 2
5 7 4 t 2.E6 /
5 6 9 8 t 2.E6 /
7 9
11 10 t 2.E6 LOAD ELEMENTS TYPE=FORCE INPUT=LINE 6
9 out 250.E3 /
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11 MESH AXES=LOCAL MESH BCODE=ALL VECTOR=EFORCE SOLVIA END SOLVIA-POST input A98 PIPE JOINT DATABASE CREATE WRITE
'a98.lis' ZONE NAME=CYL1 INPUT=EGROUP /
1 ZONE NAME=CLOSE GLOBAL-LIMITS XMAX=0.57 ZONE NAME=CYL1 INPUT=ZONE OPERATION=INTERSECT ZONE1=CLOSE ZONE NAME=INTER INPUT=ZONE OPERATION=ADD ZONEl=CYLl CYL2 EGROUP 1
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2 1644 TO 1657 5 1672 TO 1694 6 SET PLOTORIENTATION=PORTRAIT MESH ORGINAL=YES CONTOUR AVERAGING-NO MESH INTER CONTOUR=MISES CONTOUR AVERAGING=ZONE MESH CYL2 CONTOUR=MISES SUBFRAME=12 MESH CYLI CONTOUR=MISES MESH CYL2 CONTOUR=SDEVIATION SUBFRAME-12 MESH CYLl CONTOUR=SDEVIATION Version 99.0 Linear-Examples A98.7
SOLVIA Verification Manual Linear Examples SOLVIA-POST input (cont.)
MESH CYL2 PLINES=INTER2 SUBFRAME=12 MESH CYL1 PLINES=INTER1 AXIS 1 VMIN=O. VMAX=0.20 LABEL='DISTANCE' AXIS 2 VMIN=O. VMAX=500.E6 LABEL='MISES' ELINE INTER2 KIND=MISES SYMBOL=2 XAXIS=1 ELINE INTER1 KIND=MISES SYMBOL=1 XAXIS=-l YAXIS=2 YAXIS=-2 SUBFRAME=OLD MESH EG2 PLINES=N2-N6 SUBFRAME=12 MESH EGI PLINES=N2-Nl AXIS 3 VMIN=-100E-6 VMAX=400E-6 LA8EL='RADIAL DISPLACEMENT, NLINE N2-N6 DIRECTION=1 SYMBOL=1 YAXIS=3 SUBFRAME=12 NLINE N2-N1 DIRECTION=3 SYMBOL=1 YAXIS=3 AXIS 4 VMIN=O VMAX=25OE6 LABEL='HOOP STRESS' ELINE N2-N6 KIND=SYY YAXIS=4 SUBFRAME=12 ELINE N2-N1 KIND=SYY YAXIS=4 AXIS 5 VMIN=
ELINE N2-N6 ELINE N2-N1 w-400E6 VMAX=20OE6 LABEL='MERIDIONAL STRESS' KIND=SZZ YAXIS=5 SUBFRAME=12 KIND=SXX YAXIS=5 SUMMATION KIND=LOAD END Version 99.0 A98.8
SOLVIA Verification Manual EXAMPLE A99 CANTILEVER BEAM, STANDARD T CROSS-SECTION Objective To verify the BEAM element stress calculations when using a standard T cross-section.
Physical Problem A cantilever beam as shown in the figure. One end is fixed and the other end is subjected to a transverse force in the Y-direction acting at the center of gravity.
z X
E 2.1.101 N/m 2 v=0.3 P=2-10 3 N t1 t2 L=I m wtop= 0.1 m d=0.1 m tl = 0.01 m t2 = 0.01 m c = center of gravity sc = shear center Finite Element Model The cantilever beam is modeled using one BEAM element with standard T cross-section. A reference model using 256 cubic SHELL elements, example A99A, has also been analyzed. The reference model can be seen in the figure on page A99.5.
Solution Results If no BEAM section offset is defined the BEAM element nodes are located at the center of gravity.
The concentrated force is defined to act at the center of gravity and can then be directly applied to the BEAM element end node. For a standard T cross-section the shear center does not coincide with the center of gravity and the transverse load will produce a twisting of the beam element. With the orien tation of the local element r-s-t system as shown in the figure above the twisting moment is Mr = P.A, where A is the distance between the shear center and the center of gravity.
Version 99.0 Linear Examples A99.1
SOLVIA Verification Manual a
The torsional rigidity of the T cross-section is rs assumed to be equal to the sum of two narrow rectangular sections. The shear stress distribu tion in the T cross-section due to a twisting ars moment as calculated in SOLVIA can be seen in the figure to the left. The shear stress ars is assumed constant along the flange in the s-direc tion and the shear stress a, is assumed constant along the web in the t-direction.
Jrs 1II I,t I ¶ it a
a 6aa The detailed distribution of the shear stress due to Saint-Venant torsion is, therefore, neglected at the ends and at the re-entrant corners. The figure to the left shows schematically the detailed stress distribution. Example A65 indicates one method to calculate detailed torsional stresses.
The shear stress distribution in the T cross section due to a shear force P in the s-direction can be seen in the figure to the left. A parabolic variation in the s-direction of the flange is assumed with a maximum value of 0r3 (max)-
1.5. P wtop. ti and zero at the comers s = +/-wtop / 2.
The shear stresses at stress points 6 and 7 due to a shear force P in s-direction are. calculated as C r s = y, (m ax) j l
l t2 J2 Version 99.0 Linear Examples A99.2
SOLVIA Verification Manual Transverse shear effects are included in the standard BEAM sections in SOLVIA. The shear areas in the s-and t-direction for the T cross-section are A, = wtop. tl and A, = d.t2, respectively.
The axial stress distribution, arr, in a standard BEAM section is calculated as Fr M,.t_ M,*sj Area I
1, where si, ti are the coordinates for the considered stress point.
Using the input data shown on page A99.10 in the SOLVIA analysis the following displacements at the tip node are obtained:
Uy = 4.02-10. 3 m (p, = 9.29-10-3 The stresses from the beam model at the section X = L/2 are shown in the table below at selected stress points. This section is away from local disturbances at the concentrated load and the boundary conditions.
Stresses [MPa]
Stress point at local node N1 t
2 1
8 7 F6 3
4 4
Note that for the standard BEAM section the von Mises effective stress is calculated as ae =
+ 3 max, where Tmx= Max(aj a)
The nominal values are used without any stress concentration.
Note that the shear stress due to the twisting moment is +/- 7.50 MPa in the web and in the flange. The shear stress due to the shear force is 2.97 MPa in the flange at stress points 6 and 7.
A 16-node SHELL model with 16 x 8 elements for both the flange and the web was analyzed using the input data on pages A99.1 1 and A99.12. The following results were obtained:
The displacements at center of gravity (node 9) are Uy = 4.03-10-' m (px = 9.66"10-3 The distribution of the Y-and Z-displacements in the beam can be seen as contours on page A99.6.
Element point lines called FLANGE and WEB are defined at the middle section of the beam, see the figure of page A99.6.
Version 99.0 Point 1
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-59.46
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60.87 4
-5.95 0
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5.95 10.47
-7.50 19.09 Linear Examples A99.3
SOLVIA Verification Manual 1
-- ine FLANGE BOT 7
4 line WEB TOP line WEB MID mine WEB BOT The axial stress a, and the two shear stresses o., and ua, are shown along these lines for the location TOP, MID and BOTTOM of the shell on pages A99.7 to A99.9.
The axial stress can be seen to vary practi cally linearly over the cross-section. This is in accordance with the beam theory where the normals to the beam axis are assumed to remain straight during deformation. The calculated axial stresses in MPa at the stress point locations 1, 4 and 7 are -59.44,
-5.94 and 5.94, respectively, which are in excellent agreement with the results for the BEAM element.
The variation of the shear stress cy, along the line FLANGE at the MID surface is only due to the transverse shear force and has no torsional component. The maximum value is 2.97 MPa, which agrees with the BEAM element results. The values along the TOP and BOTTOM surface includes both the parabolic distribution due to the shear force and the portion due to Saint-Venant torsion. The value at the location for stress point 7 is 10.54 MPa which agrees very well with the BEAM results of 10.47 MPa.
Neglecting the variations near the end of the flange the value of y due to Saint-Venant torsion is
-7.60 MPa on the TOP surface, which compares well with the value of -7.50 MPa for the BEAM element.
The variation of the shear stress a.,, along the line WEB is practically only due to Saint-Venant torsion and can be seen to be +/- 7.73 MPa with some small variation along the web. This value also compares well with the BEAM result of +/- 7.50 MIPa.
The shear stress a,, in the thickness direction of the flange is generally zero except at the two ends where Saint-Venant shear stress develop also in the Z-direction. A physical picture is provided by the membrane (soap film) analogy, [11 page 268. A uniformly tensioned membrane shaped as the cross section and supported at the boundaries is loaded by pressure. The slope of the membrane is then proportional to the Saint-Venant torsional shear stress.
Another analogy is provided in example A65 where uniform heat generation over the cross-section is used. The heat flux is then proportional to the shear stress.
Note that the shear stress in the thickness direction of a SHELL element is always constant. The reason is that the normal to the midsurface of the SHELL is remaining straight during deformation.
All points along the rotated but straight normal have, therefore, the same shear strain. The actual shear stress distribution in the thickness direction would be parabolic with zero values at the surfaces and 1.5 times the average shear stress value at the midsurface.
Version 99.0 Linear Examples A99.4
SOLVIA Verification Manual User Hints
"* Note that a fine SHELL element model must be used to calculate the shear stress distribution correctly (within the shell theory).
"* Note that the reaction forces and moments are evaluated at the nodal points. If no BEAM section offset is defined the reactions are defined at the center of gravity for the BEAM standard section.
Reference
[1]
Timoshenko, S., and Goodier, J.N., Theory of Elasticity, 2nd edition, McGraw Hill, 195 1.
Version 99.0 STANDARD T CROSS-SECTION, SHELL A99A CANTILEVER
- BEAM, ORIGINAL F
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'A99 CANTILEVER BEAM, STANDARD T CROSS-SECTION' DATABASE CREATE COORDINATES 1 /
2 0.5 / 3 1.0 MATERIAL 1
ELASTIC E=2.1E11 NU=0.3 EGROUP 1
BEAM RESULTS=STRESSES SECTION 1
T WTOP=0.1 D=0.1 T1=0.01 T2=0.01 LIST SECTION BEAMVECTOR 1
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-1 2 3 FIXBOUNDARIES
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i LOADS CONCENTRATED 3 2 2.E3 SOLVIA END SOLVIA-POST input A99 CANTILVER BEAM, STANDARD T CROSS-SECTION DATABASE CREATE WRITE FILENAME='a99.1is' NLIST DIRECTION-246 NLIST DIRECTION=246 KIND-REACTIONS ELIST SELECT=BASICS ELIST SELECT=S-EFFECTIVE END Version 99.0 Linear Examples A99. 10
SOLVIA Verification Manual SOLVIA-PRE input HEADING
'A99A CANTILEVER BEAM, STANDARD T CROSS-SECTION, SHELL' MEMORY SOLVIA=21 DATABASE CREATE COORDINATES 1
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11 7 3 TO 32 4
11 7 3 89 4
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TO 96 4
11 7 3 EPLINE NAME=WEB 177 4
11 7 3 TO 192 4 11 7 3 NLIST N9 VIEW ABC 2 3 1 SET DIAGRAM=GRID OUTLINE=SHELL MESH CONTOUR=DY ORGINAL=YES VIEW=ABC SUBFRAME=12 MESH CONTOUR=DZ ORGINAL=YES COLOR EFILL=SHELL MESH PLINES=FLANGE SUBFRAME=12 OUTLINE=SHELL COLOR LINE MESH PLINES=WEB OUTLINE=SHELL HIDDEN=NO SHELLSURFACE PLOTRESULTS=TOP SET PLOTORIENTATION=PORTRAIT ELINE LINENAME=FLANGE KIND=SXX OUTPUT=ALL SHELLSURFACE PLOTRESULTS=MID ELINE LINENAME=FLANGE KIND=SXX SHELLSURFACE PLOTRESULTS=BOTTOM ELINE LINENAME=FLANGE KIND=SXX OUTPUT=ALL SUBFRAME=12 SUBFRAME=I2 Version 99.0 Linear Examples A99.11
SOLVIA Verification Manual Linear Examples SOLVIA-POST input (cont.)
SHELLSURFACE PLOTRESULTS=TOP ELINE LINENAME=WEB KIND=SXX OUTPUT=ALL SHELLSURFACE PLOTRESULTS=MID ELINE LINENAME=WEB KIND=SXX SUBFRAME=12 SHELLSURFACE PLOTRESULTS=BOTTOM ELINE LINENAME=WEB KIND=SXX OUTPUT=ALL SHELLSURFACE PLOTRESULTS=TOP SET PLOTORIENTATION=PORTRAIT ELINE LINENAME=FLANGE KIND=SXY OUTPUT=ALL SHELLSURFACE PLOTRESULTS=MID ELINE LINENAME=FLANGE KIND=SXY SHELLSURFACE PLOTRESULTS=BOTTOM ELINE LINENAME=FLANGE KIND=SXY OUTPUT=ALL SUBFRAME=12 SUBFRAME=12 SHELLSURFACE PLOTRESULTS=TOP ELINE LINENAME=WEB KIND=SXY OUTPUT=ALL SHELLSURFACE PLOTRESULTS=MID ELINE LINENAME=WEB KIND=SXY OUTPUT=ALL SUBFRAME=12 SHELLSURFACE PLOTRESULTS=BOTTOM ELINE LINENAME=WEB KIND=SXY OUTPUT=ALL SHELLSURFACE PLOTRESULTS=TOP ELINE LINENAME=FLANGE KIND=SXZ OUTPUT=ALL SUBFRAME=12 SHELLSURFACE PLOTRESULTS=MID ELINE LINENAME=FLANGE KIND=SXZ SHELLSURFACE PLOTRESULTS=BOTTOM ELINE LINENAME=FLANGE KIND=SXZ OUTPUT=ALL SUBFRAME=12 SHELLSURFACE PLOTRESULTS=TOP ELINE LINENAME=WEB KIND=SXZ OUTPUT=ALL SHELLSURFACE PLOTRESULTS=MID ELINE LINENAME=WEB KIND=SXZ SUBFRAME=12 SHELLSURFACE PLOTRESULTS=BOTTOM ELINE LINENAME=WEB KIND=SXZ OUTPUT=ALL SHELLSURFACE PLOTRESULTS=TOP ELINE LINENAME=FLANGE KIND=MISES SHELLSURFACE PLOTRESULTS=MID ELINE LINENAME=FLANGE KIND=MISES SHELLSURFACE PLOTRESULTS=BOTTOM ELINE LINENAME=FLANGE KIND=MISES OUTPUT=ALL SUBFRAME=12 OUTPUT=ALL SUBFRAME=12 SHELLSURFACE PLOTRESULTS=TOP ELINE LINENAME=WEB KIND=MISES OUTPUT=ALL SHELLSURFACE PLOTRESULTS=MID ELINE LINENAME=WEB KIND=MISES SUBFRAME=12 SHELLSURFACE PLOTRESULTS=BOTTOM ELINE LINENAME=WEB KIND=MISES OUTPUT=ALL END Version 99.0 A99.12
SOLVIA Verification Manual EXAMPLE A100 FREQUENCY ANALYSIS OF A SPHERICAL DOME, PLANE STRESS3 Objective To verify the PLANE STRESS3 (membrane) element to model a spherical surface in a frequency analysis.
Physical Problem A hemisphere is considered as shown in the figure below. The boundary conditions at Z = 0 are symmetrical.
R= 4.5 m h = 0.08 in E= 2.0.10'0 N/m 2 v= 0.3 p = 3000 kg/m3 Finite Element Model A sector of one radian of the spherical surface is modeled using 160 nine node PLANE STRESS3 elements, see left top figure on page A 100.3. Skew system directions are used to define the boundary conditions in the non-global circumferential directions. A consistent mass matrix assumption is used in the analysis and the subspace iteration method is used to calculate the lowest membrane frequency of the hemisphere.
Solution Results The analytical solution to this problem of natural frequencies of a closed spherical membrane shell can be found in [1], page 338. The lower branch solution with n = 2 gives 2(l
)
=7+3v-V(7+3v) 2 -16(1-v2) where E2-p'03 2.R2 E
Insertion of numerical values gives f =
67.1 Hz 27c The input data of the SOLVIA analysis can be found on page A100.4.
SOLVIA calculates the fundamental frequency f= 67.1 Hz for the PLANE STRESS3 element. The mode shape is shown in the right top figure on page A100.3.
Version 99.0 Linear Examples A100.1
SOLVIA Verification Manual Using a PLANE AXISYMMETRIC finite element model the two lowest frequencies are calculated as ft = 67.1 Hz f2 = 84.7 Hz The second frequency corresponds to the lower branch solution with n = 4 which gives the analytical solution f = 84.3 Hz. The mode shapes are shown in the bottom figures on page A100.3.
User Hints
" The upper branch solution for n = 0 gives the natural frequency solution Q2=
2 This mode is normally referred to as the fundamental mode for a closed spherical membrane shell. To calculate this mode the boundary conditions along the line Tp = 90 need to be antisymmetric. Further discussion can be found in [1].
" A sector of one radian is considered in the PLANE STRESS3 model and in the AXISYMMETRIC model. The translational mass is 4.860.103 kg in the models and can be listed with the command MASSPROPERTIES. The AXISYMMETRIC model has only translational mass in the radial (Y) and the axial (Z) directions.
"* The bending stiffness is not important in this example for the considered modes. Hence, the model with PLAINE STRESS3 elements and the model with PLANE AXISYMMETRIC elements give practically the same results.
Reference
[1]
Kraus, H., Thin Elastic Shells, John Wiley & Sons, 1967, pp. 336-339.
Version 99.0 Linear Examples A 100.2
'0 0'
C 41 2;
Lb LbLb000000 Lb Lb Lb Lb0Lb0Lb000 Lb Lb'
Lb03V LbLt Lb -
Lb.-.LbW.LbLbLb4 03 Lb
Lb'd
'JLbLbLbfl LbLb-.0NLbLbLb LbZ Lb Lb Lb
- - Lb
- LL 0Lb Lb LLb Lb
LbI Lb Lb Lb
*Lb
Lb 0/
0 z
(fj
-3 0
0
Lb Lb Lb Lb
- bLLbLbLb'>
Lb
'
Lb
Lb<'
o
LLbLLb
j
- >
Lb
- DLb'.
.'.'
LLb
Lb r
Lb Lb Lb Lb Lb 0
jUL Lb Lb Lb LbSLbLbLbb Lb
- ILL, Lb Lb Lb---7
>LL
Lb...Lb Lb
.4 CD C11
SOLVIA Verification Manual SOLVIA-PRE input, PLANE STRESS3 model HEADING
'A100 FREQUENCY ANALYSIS OF A SPHERICAL DOME, PLANE STRESS3' DATABASE CREATE MASTER IDOF=000111 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT FREQUENCIES SUBSPACE-ITERATION NEIG=1 SSTOL=I.E-10 IACCN=1 SYSTEM 1 SPHERICAL COORDINATES ENTRIES NODE R
THETA PHI 1
4.5
- 0.
- 0.
2 4.5
- 0.
- 90.
3 4.5 57.2957795 90.
SKEWSYSTEMS VECTORS 1
0.5403023 0.841471 0.
-0.841471 0.5403023
- 0.
MATERIAL 1
ELASTIC E=2.E10 NU=0.3 DENSITY=3000.
EGROUP 1
PLANE STRESS3 EDATA
/
1 0.08 GSURFACE 1 2 3 1 ELI=20 EL2=8 NODES:9 SYSTEM=i BLENDING=ANGLES NSKEWS INPUT=LINE
/
1 3 1
FIXBOUNDARIES 2
INPUT=LINES
/
1 2
/
1 3 FIXBOUNDARIES 3
INPUT=LINES
/
2 3
FIXBOUNDARIES 1
INPUT=NODES
/
SET PLOTORIENTATION=PORTRAIT MESH BCODE=ALL NNUMBERS=MYNODES NAXES=SKEW NSYMBOLS=MYNODES SOLVIA END SOLVIA-POST input A100 FREQUENCY ANALYSIS OF A SPHERICAL DOME, PLANE STRESS3 DATABASE CREATE WRITE
'alOO.lis' MASS-PROPERITES FREQUENCIES SET PLOTORIENTATION:PORTRAIT SET RESPONSETYPE=VIBRATIONMODE ORIGINAL=YES OUTLINE=YES MESH CONTOUR=DISPLACEMENTS END Version 99.0 Linear Examples A100.4
SOLVIA Verification Manual Linear Examples SOLVIA-PRE input, AXISYMMETRIC model HEADING
'A100A FREQUENCY ANALYSIS OF A SPHERICAL
- DOME, AXISYMMETRIC' DATABASE CREATE MASTER IDOF=100111 ANALYSIS TYPE=DYNAMIC MASSMATRIX=CONSISTENT FREQUENCIES SUBSPACE-ITERATIONS NEIG=2 SSTOL:I.E-10 SYSTEM 1 CYLINDRICAL COORDINATES ENTRIES NODE R
THETA XL 1
4.46
- 0.
TO 3
4.54
- 0.
4 4.46
- 90.
TO 6
4.54
- 90.
MATERIAL 1
ELASTIC E=2.E10 NU=0.3 DENSITY=3000.
EGROUP 1
PLANE AXISYMMETRIC GSURFACE 3
6 4 1 EL1=50 EL2=1 SYSTEM=i NODES=8 FIXBOUNDARIES 2
/
4 5 6 FIXBOUNDARIES 3
/
1 2 3
SET HEIGHT=0.25 MESH SMOOTHNESS=YES NSYMBOLS=MYNODES NNUMBER=MYNODES MESH SMOOTHNESS=YES NSYMBOLS=MYNODES BCODE=ALL SOLVIA END SOLVIA-POST input A100A FREQUENCY ANALYSIS OF A SPHERICAL DOME, AXISYMMETRIC DATABASE CREATE WRITE
'al00a.lis' MASS-PROPERTIES FREQUENCIES SET RESPONSETYPE=VIBRATIONMODE SET SMOOTHNESS=YES ORIGINAL=YES MESH TIME=1 MESH TIME=2 END Version 99.0 A100.5
SOLVIA Verification Manual EXAMPLE A101 DYNAMIC EXCITATION OF A BEAM STRUCTURE Objective To verify the use of NHIST-PORTHOLE = COMBINATION in the SOLVIA-POST command DATABASE when different boundary conditions of a model are used in a dynamic analysis.
Physical Problem A beam structure is subjected to a dynamic excitation by two concentrated loads acting at a corner node of the structure in the horizontal directions X and Y, see figure. The time dependence of the loads is described by the time functions shown in figures on page A101.4. The time functions are the same as used in Example A79.
The geometry is also the same as used in Example A79 except that an extra point mass has been added at node 5 to create a convenient geometric symmetry plane at X=0.75.
The response at the structure nodes 5 and 7 are to be calculated.
Geometry L
H =3.0Om L =1.5mi 1 =2.0m B = 1.0m Circular section BEAM elements 7
Diameter D = 0.020 m 8
6 m
Material H
M 5 E=2-10 N/m2 H,
v =0.3 p = 7850 kg/m3 Rayleigh damping cY = 0, 5.10-3 Point mass in = 10 kg iZ X
Loads Fx = 10-(time function 1)
Fy = 10 (time function 2)
Finite Element Model Due to the geometric symmetry a half model of the structure can be used. The load is nonsymmetric so symmetric and antisymmetric boundary conditions at nodes in the plane X=0.75 are used in the runs A101A and A 101 B, respectively. The corresponding models with boundary conditions are shown on page A 101.4 and the input data is shown on pages A 101.8 - A 101.10. Direct integration is used with a time step of 0.005 sec.
Version 99.0 Linear Examples A101.1
SOLVIA Verification Manual Solution results Displacements and accelerations in the horizontal directions X and Y are saved on the nhist-porthole file for the nodes 5 and 8. For all other nodes results are only saved at every 550:th step, thus at 4 solution steps since the total number of steps is 2200. The use of the nhist-porthole file for selective saving of nodal results reduces dramatically the amount of result data.
The nhist-porthole standard file SOLVIA63.DAT is renamed to A10lA.DAT and A101B.DAT, respectively, for the runs A101A and AlOiB using the DISKFILE command. The nhist-porthole files are added in the SOLVIA-POST run A101C using the factor 0.5 for both files, see figure below. Thus total results for node 5 (and 8) are obtained and the displacements, accelerations and response spectra in the X and Y directions at node 5 are shown on pages A101.5 - A101.6.
In run A101D the model response is the mirror image of the unmodelled portion of the structure since the factor 0.5 is used for the nhist-porthole file A101A.DAT and the factor -0.5 for the file AlO1B.DAT, see figure below. The results at the model node 8 are then the mirror image of the results at the structure node 7. Hence, the Y-direction results of the model node 8 are the same as for the structure node 7, but the X-direction results of the model node 8 are the negative of those for structure node 7. The model node 8 results are shown on pages A101.6.
For comparison, the complete structure has been modeled in run A101E, see figure on page A101.4, and results at the structure nodes 5 and 7 are shown on pages A101.7 - A101.8 as a reference solution.
A summary of the analysis runs and the maximum and minimum displacements/accelerations are given in the tables on the next page. The combined results using the symmetric model are identical with the reference solution of the complete structure model when the proper signs of the response are considered.
8 Ms5 0.5
- A101A 6 A0"5 F,
- ,O.5 FY 0.5
- A101A 0.5 Fy 6
+t 0.5 F,
+
S7,
-is5 F, i0.5 F,
- 0i.5F, A101C
-7' F,
- 6
- F, A101D 6
(-0.5
- Ai01B)
Version 99.0 Linear Examples A101.2
SOLVIA Verification Manual Input data Run Name Purpose SOLVIA-PRE SOLVIA-POST input input on pages on pages 1
A101A Symmetric boundaries A101.8 - A101.9 A101.9 2
A101B Anti-symmetric boundaries A101.9 A101.10 3
A101C NHIST COMBINATION A101.10 results at node 5 4
A101D NHIST COMBINATION A101.10 - A101.11 results at node 7 5
A101E Reference solution A101.l1 -A101.12 A101.12 full model Extreme values of the response Reference solution Combined results using half model Node 5 Node 7 A101C:node5 A101D:node8 Min. displacement
-5.13882.10-4
-5.36335.10-4
-5.13882.10-4
-4.96034.10-4 X-direction (note 1)
Max. displacement
+ 5.88536.10-4 4.96034.10-4
+5.88536.10-4 5.36335-10-4 X-direction (note 1)
Min. acceleration
-4.85002-10-4 1.3 7 9 7 4. 10 -'
- 4.85002.10-1
- 1.37974.10-1 Y-direction Max. acceleration
+ 5.10907. 10-'
+ 1.68328.10-'
+ 5.10907.10-1 1.68328.10-'
Y-direction Note 1: Reverse the sign to obtain the extreme X-direction results for node 7 since the results for node 8 of the half model in run A101D are the mirror images of the results for structure node 7.
User Hints The combination of nhist-porthole file results may be of interest in the dynamic analysis of a very large model which has symmetry properties. Only a portion of the full model need then be considered but using analysis cases with different boundary conditions. The results from these cases may then be superimposed to produce the full solution. This procedure may speed up the solution considerably when the symmetry portion can be solved in-core but the full model yields an out-of-core solution.
Version 99.0 Linear Examples A101.3
SOLVIA Verification Manual Linear Examples AI01A DYNAMIC EXCITATICN OF A BEAM STRUCTURE, SYMMETRICAL AIBiA DYNAMIC EXCITATION OF A 3EAM STRUCTURE.
SYMMETRICAL lORIGINAL 0.2 I 2----
TIME I.1 Y\\M B
44 L
MA.STER.
.÷.
r 04 24
-3 i
O{
000 C
2 1
12 1
'9+/-09 IfN C 111l:1 SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 SOLVBA ENGINEERING AB A1010 DYNAMIC EXCITATION OF A BEAM STRUCTURE. ANTI-SYMMETRICAL AIOlE DYNAMIC EXC ITATION OF A BEAM STRUCTURE, REF.
SOLUTION ORIGINAL
-+/-
02 Z
ORIGINAL 0.2 TIME I.
YI4x TIME 0.1 Y\\L-x" B
43 4
B 12
'4 FORCE
- 0. 467484
- 0. 467484 MAS FER4 O[MASTER C 01100 000101 O'till, I B' ifiti SOLVIA-PRE 99 0 SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 SOLVIA ENGINEERING AB Version 99.0 A101.4
SOLVIA Verification Manual Version 99.0 Linear Examples A101.5 AIDIA DYNAMIC EXCITATION OF A BEAM STRUCTURE, SYMMETRICAL i..........
l o
S A
IS IS TI E
...ir *.............. F i " T i ;...............
i............
5 2
'C A
A I
12 TI DE SOLVATAPOST 99.0 SOLVER ENGIES RESSN AR A1E1l SYNAMIC EXCITATION OF A BEAM STRUCTURE. ANTE-SYTMETRTCAL 117..
I--,
-,-..I.,-I-.*,-
- .r* *.---......
S...
- i.-i~ti.! --.*.......
i
' ' l l i.......,.........,..........
v iv TJME SOLVIA-POST 99.0 SOLVIA ENGINEERING AB
SOLVIA Verification Manual AESIC COMPARE RESULTS WITH NODE S IN REF.
SOLUTION 4E44-'44'.,PECRU DA<
44
. 2 4-.0 S.
.................. i........................i i ~ ' :...........
S..
.L " i......... !.......
4, 4
- o. L I'
loo FREQUENCY SOVI-PS T 99.05SOLTIA ENGINEERIN2 AB5 4,
U*
4 R;
UNC SOLVAPOT4.0-LVAEGIER4 GA Version 99.0 Linear Examples A[01D COMPARE RESULTS WITH NODE 7 IN REF.
SOLUTION 1-4, V*
EE' I: *T j 4::
'A t I H '
T TOME 4,
1
!'4j o"Z
- :I 1 *"
.......... i.............................................
'4i li
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[
- A,
,4 S-i J.: L,'..e,... "
. t L
°,
tj 4 S
2 O
E R
SOLVER P05; 99.0 SOLVER ENGINEERING AR so AIDED COMPARE RESULTS WITH NODE 7 IN REF.
SOLUTION R44'AOSE s?4?:44 UA44P X 2 U
SO I
i
ýES.-ONSE~
SRLTU CXý XE 2I C
o "iij i
i I I,4 6 1
. :S : '
.i
.....i 4......! * : !.............i FRECUENCY SOLVIA ENGINEERING AB LVIA-POST 99.0 A101.6 ALOID COMPARE RESULTS WITH NODE 7 IN REF.
SOLUTION SI I~iL......
i................
'* f
......!. i _ _ _ _...*....................... i..............
6 e
40 12 TI"E i ' t....... * " i.....................
- *, *l~
l!.I
- r *l~
~..*.,,,, -*,.v.*.,,..............:,.............
o
- 2.
4 6*
6 1
12 T I.".:E SOLVIA-POST 99.0 SOLVIA ENGINEERING AB
SOLVIA Verification Manual Linear Examples AIOlE DYNAMIC EXCITATION OF A BEAM STRUCTURE. REF. SOLUTION TI. E U
2 A
O 1
12 SOLVIA-POST 99.0 SOLVEA ENGINEERING AB AIJIE DYNAMIC EXCITATION OF A BEAM STRUCTURE, REF.
SOLUTION
- ~
~ ~ ~~....
!li*,
.,l!v'
- n~ ~ ~ ~ ~.
Li AlII IB9S 11f 15 i r i P,
! F..........
i.....:
2 E
iA SOLVIA-POST 99.0 SOLVIA ENGINEERING AS AIOIE DYNAMIC EXCITATION OF A BEAM STRUCTURE, REF.
SOLUTION IESPCNS\\ SPECI:UM ORAM' X 2.0 S.0 Li
.1i iyv O
- 0 100 FREOUENCY
.....i,
- A Y
l.................
!- : 2...........
I rI TA 0.1 t
i
~oo FRERUENCY SOLVIA-POST 99.0 SOLVIA ENGINEERING AB AIBIE DYNAMIC EXCITATION OF A BEAM STRUCTURE, REF.
SOLUTION vi?
- ^ : ~
i i
S4 6
3 to 12 IME
_T M i
99.0-IA,
A
,A'jJi.L5,II
- ~
j
~
,I,,7a S' ! !ii i
i !i ' i A1
[
1 5 LB SOLVIA-POST 99 0 SOLVIA ENGINEERINS AR Version 99.0 A101.7 T
SOLVIA Verification Manual AIOIE DYNAMIC EXCITATION OF A BEAM STRUCTURE, REF SOLUTIJN n~j F
} -..............
T I.E rOTE 5NLVEA-POST 99.0 SOLVIA ENGINEERING AB SOLVIA-PRE input DATABASE CREATE HEAD 'Al01A DYNAMIC EXCITATION OF A BEAM STRUCTURE, SYM7METRICAL' MASTER NSTEP=2200 DT=0.005 ANALYSIS DYNAMIC SPARSE=YES RAYLEIGH-DAMPING ALPHA=0 BETA=5.E-3 PORTHOLE SAVEDEFAULT=NO NHIST-PORTHOLE 5 1 DISP /
8 1 DISP 5 2 DISP /
8 2 DISP 5 1 ACC /
8 1 ACC 5 2 ACC /
8 2 ACC NSAVESTEPS FIRST1=550 LAST1=2200 INCRi=550 ESAVESTEPS FIRST1=550 LAST1=2200 INCRI=550 MATERIAL 1 ELASTIC E=2.0Eii NU=0.3 DENSITY=7850 COORDINATES 10o0
/
40 80 12/
90 41 0.75 0 2 /
42 10/5 0 3 /12 0.75 1 2 O02 013
/ 43 0.750 3
/
44 0.75 1 3 1
BEAM MATERIAL=I Ni=1 N2=5 AUX=41 EL=I6 Ni=5 N2=9 AUX=41 EL=8 Ni=5 N2=41 AUX=9 EL=6 Ni=8 N2=5 AUX=12 EL=8 Nl=44 N2=12 AUX=42 EL=6 Version 99.0 AIlE DYNAMIC EXCITATION OF A BEAM STRUCTURE, REF.
SOLUTION RESINSE ý%,PECA'N SLIP 2.C U.
Li m ~ ~
~ ~ ~.....
i *
° i FREQýUENCY
.,Es;:o-ýE SPýEc_.u R,
cli I
- 2.
/
-I o
FRECUENCY SOLVIA-POST 99.0 SOLVFA ENGINEERING AT EGROUP GLINE GLINE GLINE GLINE GLINE
/ /
/
/
/
GLINE GLINE GLINE GLINE GLINE Ni=4 N1=8 N1=42 N1=9 N1=12 N2=8 N2=12 N2=8 N2=43 N2=9 AUX=42 AUX=42 AUX=44 AUX=5 AUX=41 EL=-6 EL=8 EL=6 EL=6 EL=8 Linear Examples
- J 0
A101.8
SOLVIA Verification Manual SOLVIA-PRE input (cont.)
SECTION 1 CIRCULARSOLID D-20E-3 BOUNDARIES 111111 NODES 1
2 3 4
BOUNDARIES 100011 NODES /
41 42 43 44 MASS /
5 10.
TIMEFUNCTION 1
READ
'A79XDIR.DAT' TIMEFUNCTION 2 /
READ
'A79YDIR.DAT' LOADS CONCENTRATED
- node dir fac timef 5
1
- 10.
1 5
2
- 10.
2 DISKFILE SOLVIA63.DAT
'A101A.DAT' VIEW VI 2 1.5 SET PLOTORIENTATION=PORTRAIT VIEW=V1 NSYMBOL=Y NNNUMBERS=MYNODES MESH BCODE=ALL VECTOR=LOAD TIME=0.100 AXIS 1 VMIN=-1.
VMAX=I.
PLOT TIMEFUNCTION 1 YAXIS=1 SUBFRAME:12 PLOT TIMEFUNCTION 2 YAXIS=1 SOLVIA END SOLVIA-POST input A101A DYNAMIC EXCITATION OF A BEAM STRUCTURE, SYMMETRICAL DISKFILE SOLVIA63.DAT
'A101A.DAT' DATABASE CREATE NHIST-PORTHOLE=YES SET PLOTORIENTATION=PORTRAIT SET DIAGRAM:GRID NHISTORY NODE=5 DIRECTION=1 KIND=DISP SUBFRAME-12 NHISTORY NODE=5 DIRECTION=2 KIND=DISP END SOLVIA-PRE input DATABASE OPEN HEAD 'AI01B DYNAMIC EXCITATION OF A BEAM STRUCTURE, ANTI SYMMETRICAL' BOUNDARIES 011100 NODES
/
41 42 43 44 DISKFILE SOLVIA63.DAT
'A101B.DAT' SET PLOTORIENTATION=PORTRAIT VIEW:V1 NSYMBOL=Y NNNUMBERS=MYNODES MESH BCODE=ALL VECTOR=LOAD TIME=0.100 SOLVIA END Version 99.0 Linear Examples A101.9
SOLVIA Verification Manual SOLVIA-POST input Ai01B DYNAMIC EXCITATION OF A BEAM STUCTURE DISKFILE SOLVIA63.DAT
'Ai01B.DAT' DATABASE CREATE NHIST-PORTHOLE=YES SET DIAGRAM=GRID NHISTORY NODE=5 NHISTORY NODE=5 END SOLVIA-POST input PLOTORIENTATION=PORTRAIT DIRECTION=1 KIND=DISP SUBFRAME=12 DIRECTION=2 KIND=DISP Ai01C DYNAMIC EXCITATION OF A BEAM STRUCTURE DATABASE CREATE 0.5 'Al01A.DAT' 0.5 'A101B.DAT' NHIST-PORTHOLE=COMBINATION HEAD 'Ai01C COMPARE RESULTS WITH NODE 5 IN REF.
SOLUTION' WRITE FILENAME='aI01c.lis' SET PLOTORIENTATION=PORTRAIT NEWPAGE=NO DIAGRAM=GRID NHISTORY NODE=5 DIRECTION=* KIND=DISP SUBFRAME:I2 NHISTORY NODE=5 DIRECTION=2 KIND=DISP NHISTORY NODE=5 DIRECTION=i KIND=ACC NHISTORY NODE=5 DIRECTION=2 KIND=ACC RESPONSE-SPECTRUM KIND=ACCELERATION FSTART=0.5 FEND=100 FINC=1
- 2.
- 5.
RESPONSE-SPECTRUM KIND=ACCELERATION FSTART=0.5 FEND=-00 FINC=1
- 2.
- 5.
END SUBFRAME=12 NODE=5 DIRECTION=i, AXES=4 SUBFRAME=12 NODE=5 DIRECTION=2, AXES=4 SOLVIA-POST input A101D DYNAMIC EXCITATION OF A BEAM STRUCTURE DATABASE CREATE NHIST-PORTHOLE=COMBINATION 0.5
'Al01A.DAT'
-0.5
'A101B.DAT' HEAD 'A01D COMPARE RESULTS WITH NODE 7 IN REF.
SOLUTION' WRITE FILENAME='alOld.lis' SET PLOTORIENTATION=PORTRAIT NEWPAGE=NO SET RESPONSETYPE=TIMERESULT DIAGRAM=GRID Version 99.0 Linear Examples A101.10
SOLVIA Verification Manual SOLVIA-POST input (cont.)
NHISTORY NODE=8 DIRECTION=i KIND=DISP SUBFRAME=12 NHISTORY NODE=8 DIRECTION:2 KIND=DISP NHISTORY NODE=8 DIRECTION=* KIND=ACC SUBFRAME=I2 NHISTORY NODE=8 DIRECTION=2 KIND=ACC RESPONSE-SPECTRUM KIND=ACCELERATION NODE=8 DIRECTION=i, FSTART=0.5 FEND=100 FINC=1 AXES=4 SUBFRAME=12
- 2.
- 5.
RESPONSE-SPECTRUM KIND=ACCELERATION NODE=8 DIRECTION=2, FSTART=0.5 FEND=100 FINC=1 AXES=4
- 2.
- 5.
END SOLVIA-PRE input DATABASE CREATE HEAD 'Al01E DYNAMIC EXCITATION OF A BEAM STRUCTURE, REF.
SOLUTION' MASTER NSTEP=2200 DT=0.005 ANALYSIS DYNAMIC SPARSE=YES RAYLEIGH-DAMPING ALPHA:0 BETA=5.E-3 PORTHOLE SAVEDEFAULT=NO NHIST-PORTHOLE DISPSCAN=ABSMAX VELSCAN=ABSMAX ACCSCAN=ABSMAX 5 1 DISP TO 8 1 DISP 5
2 DISP TO 8 2 DISP 5 1 ACC TO 8 1 ACC 5 2 ACC TO 8 2 ACC NSAVESTEPS FIRST1=550 LAST1=2200 INCR1=550 ESAVESTEPS FIRST1=550 LAST1=2200 INCRI=550 MATERIAL 1 ELASTIC E=2.0E11 NU=0.3 DENSITY=7850 COORDINATES 1 0 00 /
2 1.5 0 0
/3 1.5 1 0 /4 0
1 0 5 0 0 2
/
6 1.5 0 2
/7 1.5 1 2
/8 0
1 2 9 0 0 3 /10 1.5 0 3 /11 1.5 1 3 /12 0 1 3 EGROUP 1
BEAM MATERIAL=1 GLINE Ni=I N2=5 AUX=2 EL=-6 / GLINE N1=2 N2=6 AUX=1 EL=16 GLINE N1=3 N2=7 AUX=4 EL=-6 / GLINE N1=4 N2=8 AUX=3 EL=16 GLINE N1=5 N2=9 AUX=2 EL=8
/ GLINE NI=6 N2=10 AUX=i EL=8 GLINE Ni17 N2=11 AUX=4 EL=8
/ GLINE NI=8 N2=12 AUX=3 EL=8 GLINE N1=5 N2=6 AUX=9 EL=12 / GLINE Ni=6 N2=7 AUX=i0 EL:8 GLINE N1=7 N2=8 AUX=11 EL=12 / GLINE N1=8 N2=5 AUX=12 EL=8 GLINE Ni=9 N2=10 AUX=5 EL=f2 / GLINE Ni=I0 N2=i1 AUX=6 EL=8 GLINE N1=11 N2=12 AUX=7 EL=42 / GLINE Ni=12 N2=9 AUX=8 EL=8 SECTION 1 CIRCULARSOLID D:20E-3 BOUNDARIES 111111 NODES /
1 2 3 4 MASSES 5 10.
6 10.
Version 99.0 Linear Examples A101. Il
Linear Examples SOLVIA Verification Manual SOLVIA-PRE input (cont.)
TIMEFUNCTION TIMEFUNCTION 1 /
READ 'A79XDIR.DAT' 2 /
READ 'A79YDIR.DAT' LOADS CONCENTRATED
- node dir fac timef 5
1
- 10.
1 5
2
- 10.
2 VIEW V1 2 1.5 SET PLOTORIENTATION=PORTRAIT VIEW:Vi NSYMBOL=Y NNNUMBERS=MYNODES MESH BCODE=ALL VECTOR=LOAD TIME=0.100 AXIS 1 VMIN=-1.
VMAX=l.
PLOT TIMEFUNCTION I YAXIS=1 SUBFRAME=12 PLOT TIMEFUNCTION 2 YAXIS=1 SOLVIA END SOLVIA-POST input A101E DYNAMIC EXCITATION OF A BEAM STRUCTURE, REF SOLUTION DATABASE CREATE NHIST-PORTHOLE=YES WRITE FILENAME='al0le.lis' SET DIAGRAM=GRID SET PLOTORIENTATION:PORTRAIT NEWPAGE=NO NHISTORY NODE=5 DIRECTION=1 KIND=DISP SUBFRAME:12 NHISTORY NODE=5 DIRECTION=2 KIND=DISP NHISTORY NODE=5 DIRECTION=1 KIND=ACC SUBFRAME=12 NEISTORY NODE=5 DIRECTION=2 KIND=ACC RESPONSE-SPECTRUM KIND=ACCELERATION NODE=5 DIRECTION=1, FSTART:0.5 FEND=I00 FINC-1 AXES=4 SUBFRAME=12
- 2.
- 5.
RESPONSE-SPECTRUM KIND=ACCELERATION NODE-5 DIRECTION=2, FSTART=0.5 FEND=C00 FINC=1 AXES=4
- 2.
5.
NHISTORY NHISTORY NHISTORY NHISTORY NODE=7 DIRECTION=1 KIND=DISP SUBFRAME=12 NODE=7 DIRECTION=2 KIND=DISP NODE=7 DIRECTION=1 KIND=ACC SUBFRAME:12 NODE=7 DIRECTION=2 KIND=ACC RESPONSE-SPECTRUM KIND=ACCELERATION FSTART=0.5 FEND=I00 FINC=d
- 2.
- 5.
RESPONSE-SPECTRUM KIND=ACCELERATION FSTART=0.5 FEND=100 FINC=1
- 2.
- 5.
E END NODE=7 DIRECTION=1, AXES=4 SUBFRAME=12 NODE=7 DIRECTION=2, AXES=4 Version 99.0 A101.12
SOLVIA Verification Manual EXAMPLE A102 CONDITIONAL CASE COMBINATIONS FOR A MULTISPAN BEAM Objective To demonstrate the use of conditional case combinations in SOLVIA-POST.
Physical Problem A multispan beam structure is subjected to five independent load cases, see figures below. The objective is to determine the combinations of the five load cases that produce the maximum and the minimum moments at point A and to show the distributions of moments, shear forces, displacements and loads that correspond to these combinations.
z A
A oQ-2.
I._
LI5 L/5 L/10 L/10 L/5 L
L = 20 m E = 30.10 9 N/m 2 WTOP = 0.1 m v = 0.2 D =0.2 m P = 10000 N/m P
p P
AP p(
Q Cross section Ds t
WTOP Version 99.0 Load Case 1 Load Case 2 Load Case 3 Load Case 4 Load Case 5 Linear Examples A102.1
SOLVIA Verification Manual Finite Element Model The input data on pages A 102.5 and A102.6 is used in the analysis. The model consists of 20 BEAM elements per span. Node 7 is positioned at the point where the maximum and minimum moments are to be calculated. The model including the boundary conditions and the five load cases is shown on page A 102.3.
Solution results Element 50, point 2 is located at node 7, see figure on page A102.3. Load cases with positive values of the moment MS at element 50, point 2, are added to case combination 1 using the SOLVIA-POST command CASE-COMBINATION N EPK-CONDITIONAL FUNCTION=ADDIFPOSITIVE, see the SOLVIA-POST input data. Load cases with negative moment MS are similarly added to case combination 2. The following moment values for the load cases and the case combinations were obtained:
The distribution of moments and shear forces for the load cases and the case combinations are shown on pages A102.3 and A102.4. The applied load, the reactions and the deformations for the two case combinations are shown on page A 102.4.
User Hints
" For good accuracy in the distribution of moments, shear forces and reactions due to distributed loading it is important to use a sufficiently large number of elements in each span. The distributed loading of BEAM elements are applied using reversed fixed end forces/moments, which are statically equivalent to a linearly varying distributed loading under zero end displacements/
rotations of each element. If such a fixed end force/moment acts in a displacement/rotation degree of-freedom that is fixed to zero, it is ignored when the load vector is assembled. Hence, such a fixed end force/moment is also not included in the reactions. In the present example with 20 elements per span the accuracy of the calculated reaction forces is about 5 percent. The accuracy of the calculated positive moment at node 7 is much better, namely about 0.3 percent.
" The current example consists of a small structure for ease of demonstration. However, when applied to a larger structure with hundreds of load cases, which shall be combined in some "worst" manner, the SOLVIA-POST command CASE-COMBINATION can be very powerful.
Version 99.0 Load Case Moment MS (Nm) at node 7 1
1045.7 2
-3159.7 3
11622.6 4
-3159.7 5
1045.7 Case Moment MS (Nm) at Combination node 7 1
13714.0 2
-6319.4 Linear Examples A 102.2
SOLVIA Verification Manual Linear Examples AG02 ZONDITIONAL CASE ORIGINAL 2.
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SOLVIA Verification Manual Linear Examples COMBINATIONS FOR A MULTISPAN BEAM
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MAX VALUE H 0.01 S7 CASE COSB I ORIGINAL
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,[SPAN BEAM z L LU LOAD RECTO REACTICON 21626 L X LOAD 2000 I
SOLVIA Verification Manual SOLVIA-PRE input HEADING
'A102 CONDITIONAL CASE COMBINATIONS FOR A MULTISPAN BEAM' DATABASE CREATE MASTER IDOF=110101 PARAMETER
$E=20 $E1=10 COORDINATES ENTRIES NODE X
1
- 0.
TO 6
- 20.
7
- 10.
MATERIAL 1 ELASTIC E=30E9 NU=0.2 EGROUP 1 BEAM RESULT=FORCE SECTION 1 RECTANGULAR WTOP=0.1 D=0.2 BEAMVECTOR
/
1
- 0.
-1. 0.
LINE STRAIGHT 3 7 EL=$E*
LINE STRAIGHT 7 4 EL=$E1 LINE COMBINE 3
4 7 GLINE 1 2
-1 EL=$E GLINE 2 3
-1 EL=$E GLINE 3 4
-1 EL=$E GLINE 4 5
-1 EL=$E GLINE 5 6
-1 EL=$E FIXBOUNDARIES 3 /
1 23456 LCASE i LOADS ELEMENT INPUT=LINES /
LCASE 2 LOADS ELEMENT INPUT=LINES /
LCASE 3 LOADS ELEMENT INPUT=LINES /
LCASE 4
LOADS ELEMENT INPUT=LINES /
LCASE 5 LOADS ELEMENT INPUT=LINES /
SET MESH MESH MESH MESH MESH MESH MESH NSYMBOLS=MY NNUMBERS=MY BCODE=ALL SUBFRAME=13 N7+1 ENUM=YES NNUM=YES VECTOR=LOAD TIME=1 VECTOR=LOAD VECTOR=LOAD VECTOR=LOAD VECTOR=LOAD 1 2 T 10.E3 2
3 T 10.E3 3
4 T 10.E3 4 5 T 10.E3 5 6 T 10.E3 VIEW=-Y PLOTORIENTATION=PORTRAIT NSYM=YES BCODE=ALL EAXES=STRESS TIME=2 SUBF=14 TIME=3 TIME=4 TIME=5 SOLVIA END Version 99.0 Linear Examples A 102.5
SOLVIA Verification Manual SOLVIA-POST input:
A102 CONDITIONAL CASECOMBINATIONS FOR A MULTISPAN BEAM DATABASE CREATE WRITE FILENAME='a102.1is' SET PLOTORIENTATION=PORTRAIT CASE-COMBINATION 1 EPK-CONDITIONAL ELEMENT=50 POINT=2 KIND=MS 1 1. TO 5 1.
END DATA CASE-COMBINATION 2 EPK-CONDITIONAL ELEMENT=50 POINT=2 KIND=MS 1 1. TO 5 1.
END DATA LIST CASE-COMBINATION EPLINE NAME=EBEAM SUBFRAME 15 ELINE EBEAM ELINE EBEAM ELINE EBEAM ELINE EBEAM ELINE EBEAM SUBFRAME 15 ELINE EBEAM ELINE EBEAM ELINE EBEAM ELINE EBEAM ELINE EBEAM KIND=MS KIND=MS KIND=MS KIND=MS KIND=MS KIND=FT KIND=FT KIND=FT KIND=FT KIND=FT FUNCTION=ADDIFPOSITIVE FUNCTION=ADDIFNEGATIVE
/
1 1 2 TO 100 1 2 TIME=1 TIME=2 TIME=3 TIME=4 TIME=5 TIME=1 TIME=2 TIME=3 TIME=4 TIME=5 SUBFRAME 14 SET RESPONSETYPE=CASECOMBINATION ELINE EBEAM KIND=MS TIME=1 ELINE EBEAM KIND=MS TIME=2 ELINE EBEAM KIND=FT TIME=1 ELINE EBEAM KIND=FT TIME=2 SUBFRAME 14 SET ORIGINAL=DASHED VIEW=-Y MESH VECTOR=LOAD TIME=1 MESH VECTOR=REAC TIME=1 MESH VECTOR:LOAD TIME=2 MESH VECTOR=REAC TIME=2 NLIST N7 NLIST N7 ELIST E50 ELIST E50 SUMMATION END TSTART=1 TEND=5 RESPONSETYPE=TIMERESULTS TSTART=1 TEND=2 RESPONSETYPE=CASECOMBINATION TSTART=1 TEND=5 RESPONSETYPE=TIMERESULTS TSTART-1 TEND=2 RESPONSETYPE=CASECOMBINATION KIND=REACTIONS TSTART=1 TEND=2 RESPONSETYPE=CASECOMBINATION Version 99.0 Linear Examples AI102.6
SOLVIA Verification Manual INDEX Acoustic cavity A56 Axisymmetric shell A2 Base motion A54 Beam curved All,A12,A57 on elastic foundation A41 Cantilever end loads A7, AS ground motion A54 stiffened A23, A82 tapered A22 truss A13 T-section A99 Z-section A35 Cantilever under distributed load BEAM element A17 ISOBEAM element A18 PLANE STRAIN element A15 PLANE STRESS element A14 PLATE element A19 SHELL element A20 SOLID element A16 Cantilever frequencies BEAM elements A47 ISOBEAM elements A48 off-center masses A37 PLANE STRAIN elements A45 PLANE STRESS elements A44 PLATE elements A49 SHELL 4 node A66 SHELL elements A50 SOLID elements A46 Case combinations conditional A102 NHIST-PORTHOLE A101 dynamic A101 Centrifugal loading A69 Combining results from different boundary conditions static load cases A80 dynamic A101 harmonic response AS1 Complex harmonic response A85, A86, A95 Cylinder ring A5, A6 edge bending A2 fluid-filled A34 hemispherical end A36 intersections A98 orthotropic A94 pinched A32, A33 Scordelis-Lo roof A30, A31, A67 thick Al thin A4 Damping weighted A26 viscous A95 hysteretic A95 Dome, spherical A24, A100 Dynamic analysis direct integration A52, A54, A55, A92, A101 modal superposition A79, A93 Earthquake analysis response spectrum method A76, A87 time history by modal superposition A79 Electric potential A63 End release of BEAM elements A88 Floor response spectra A79 FLUID elements A34, A53 Fracture mechanics A43 Ground motion A54 Harmonic response A77, AS1 Heat convection A84 Heat conduction, transient PLANE element A71, A72 TRUSS element A60, A73 Heat conduction, steady state AXISYMMETRIC element A62 PLANE element A61 Heat flux loading A73 Heat generation A74, A75 Hemisphere gravity load A24 point load A40 Intersection of cylindrical surfaces A98 Laminated shell A89, A90, A91 Membrane shell A24, A100 Offset of BEAM element axis A82 Pipe elbow A68 T connection A98 Version 99.0 Linear Examples I.1I
SOLVIA Verification Manual Plate circular A3 clamped A25 frequencies A51 laminated A91 orthotropic A39, A89, A90 sandwich A90 simply supported A27, A51 skew A38 stiffened A23 triangular A29 twisting A27, A28, A29 Plate frequencies simply supported A51 clamped rhombic A59 Power flow A85 Response spectrum load pulses A78 modal combination methods A87 Rigid body modes A96 Rigid links A23, A37 Ring, pinched BEAM elements AlO ISOBEAM elements A9 PLATE elements A6 SHELL elements A5 Ring under water pressure A97 Roof spherical A24 Scordelis-Lo A30, A31, A67 Sandwich plate A90 Shear center A99 Spherical dome frequencies AlO0 Stiffened shell A23, A82 Strain energy density A85 Stress concentration around circular hole A42 Stress deviation quality measure A27, A42 Thermal eigenvalues A64 Thickness variation using SHELL elements A22 Toroidal shell A68 Torsional stresses A65, A96, A99 Torsional frequencies off-center masses A37 spring system A83 free-free U beam A96 Transition SHELL element A22 Travelling load A58 Truss structure A13, A21 Twisting of plate A27, A28, A29 Wave propagation using FLUID3 elements A53 using GENERAL elements A70 using TRUSS elements A52 Water pressure load A97 Version 99.0 Linear Examples 1.2