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Patran / Nastran
Lecture /
Julien Le Fanic
Tp
´nie Me
´canique
Ge
´nierie des Syste
`mes
Inge
February th 
Lectures Scope
Lectures Scope
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1. Lecture 1 deals with basics Finite Elements Method and introduces Nastran and Patran softwares.
A cantilever beam is studied in linear elasticity and then with geometrical non linearity. If time left
students can realize another exercise defined in appendixes §D.
2. Lecture 2 deals with plates and shells. A 2D plate with a hole is studied to assess a KT . Then
buckling modes are computed for the same plate under compressive load. Finally a Guyan static
reduction is performed.
3. Lecture 3 will let students finish Lecture 2 case studies before an assessment of a time dependent
response for a beam and a contact 3D modelization.
4. Lecture 4 deals with Fsm idealization.
Lecture 2/4 - Table of Contents
Table of contents
Lectures Scope
Lecture 2/4 - Table of Contents
The theory of Finite Elements Method
Weak Form
Simple axial element
Simple bending element
Extra case study
Definition
Case study # 3 - Plate with a hole
Definition
Meshing Technique
KT assessment
Closed form solution
Nastran linear static run
Results from Nastran linear static run from .op2
Conclusion & Outlook
Case study # 4 - Buckling Analysis
Theoretical Apart´
e
Definition
Mesh
Nastran EIGRL Card
Nastran buckling run
Results from Nastran buckling run from .f06
Results from Nastran buckling run from .op2
Conclusion & Outlook
Case study # 5 - Guyan Static Reduction
Theoretical Apart´
e
Definition
Nastran ASET Card
Nastran EXTOUT Parameter
Nastran linear static run for reduction
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Lecture 2/4 - Table of Contents
Table of contents
Results from Nastran linear static run from .pch
Reading external element
Nastran linear static run for residual structure
Results from Nastran linear static run from .op2
Conclusion & Outlook
References
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The theory of Finite Elements Method
Weak Form
The theory of Finite Elements Method
Weak form
One solves
−∆u = f
u=0
in
on
Ω
∂Ω
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(17)
It is considered f ∈ L2 (Ω). It is assumed u ∈ H 2 (Ω) . Thus by Green formula
Z
Z
∇u · ∇vdΩ =
Ω
∂u
Z
f vdΩ +
Ω
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(18)
, ∀v ∈ H0 (Ω)
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(19)
∂Ω
∂n
vdΓ
If it is chosen v ∈ H01 (Ω) one has
Z
Z
∇u · ∇vdΩ =
Ω
f vdΩ
1
Ω
The equation (19) is called the variational formulation or the weak form of the differential equation (17).
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The theory of Finite Elements Method
Simple axial element
The theory of Finite Elements Method
Simple axial element
N1 (x)
N2 (x)
y
F1 ,u1
F2 ,u2
x
l
Figure 1:
Simple axial element assumption sketch
Linear form function

 N1 (x) = 1 l − x
l 2
 N2 (x) = 1 x + l
l
2
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u1
u2
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(20)
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Strain worth
ε=B
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u1
u2
=
h
∂N1
∂x
,
∂N2
∂x
i
u1
u2
=
h
− 1l ,
1
l
i
The theory of Finite Elements Method
Simple axial element
The theory of Finite Elements Method
Simple axial element
Stiffness matrix is derived as
l
Z
Z2
T
K =
E
B EBdΩ = S
Ω
l2
−l
2
−1
1
1
−1
dx .
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The simple bar element works merely as∗
F1
F2
=
ES
l
−1
1
1
−1
u1
u2
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with stiffness matrix
K =
ES
l
−1
1
1
−1
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Nota : Mass matrix is derived as
M =
ρSl
6
2
1
1
2
.
.
Nastran’s user can output K, M or any matrix in an .op2 or an .op4 file.
∗
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. . . a F = k u spring
©
The theory of Finite Elements Method
Simple bending element
The theory of Finite Elements Method
Simple bending element
M1 ,θ1
M2 ,θ2
l
F1 ,v1
Figure 2:
F2 ,v2
Simple bending element assumption sketch
For a simple bending element one has the relationship
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



F1
M1
F


 M2
2





6
2EI  −3l

=

−6
3

l


−3l
−3l
2l2
3l
l2
−6
3l
6
3l

−3l
v


 1
l2 
θ1

v
3l  
2


θ2
2l2







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Extra case study
Definition
Extra case study - Clamped curved beam linear analysis
Definition
The framework is linear elasticity.
F
R
2R
y
x
B
$
A%
A load F is applied in C. The beam is clamped in A and supported in B. The beam has a rectangular
3
. The beam is made of
section S = b × h with b = h = 10 mm and R = 1000 mm. Inertia worth Iz = bh
12
steel E = 200 GPa and ν = 0.30. A 100 kg mass has been hung at C.
Aim of Extra Case study : Students have to run a Nastran SOL 101 analysis and assess the deformation of
the structure.
Closed form solution : At C displacement worth
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vC = −
2π F R3
5 EIz
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Case study # 3 - Plate with a hole
Definition
Case study # 3 - Plate with a hole
Definition
The framework is linear elasticity.
The plate is t = 5 mm thick, w = 50 mm wide and L = 100 mm long. Structure is clamped Lhs and
a σ0 = 100 MPa tensile load is applied Rhs. Hole has a diameter D = 10 mm. The plate is made of
aluminium E = 70 GPa and ν = 0.33.
Aim of Case study # 3 : Students have to find the stress
concentration factor KT associated to the hole.
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Meshing Technique
A square patch is to be designed under Patran around the hole.
This will basically allow to have a better assessment of stress around the hole.
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Meshing Technique
Background : In order to mesh a surface the latter may be defined. But it
does exist a more efficient solution.
Technique : A surface may be meshed
only drawing in Patran boundary
curves and using the Mesh > 2
Curves menu form.
Example : Consider the two arcs in
next Patran snapshots. One realizes
the mesh of the surface between the 2
curves without necessity of its definition.
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Meshing Technique
The patch is to be split under Patran in 8 subdomains. The mesh is to be built with Nastran CQUAD4
elements.
Nota : 2 circles are not shown in the above figure but do exist to build the mesh.
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Meshing Technique
With Patran sketch the patches first. Assumptions are :
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1. choose the number of square elements around the hole N with N ≡ 0 mod [6] and N ≡ 0 mod [4]
e.g. N = 72
2. circular patch around the hole for a side length N square elements
3. circular patch around the hole for 3a × 2a rectangular elements to be broken for the transition mesh
4. 2 × R radius circular patch around the hole
5. square patch has a 6 × R = 3 × D side length
Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Mesh
The plate contains coarse CQUAD4 5 × 5 mm2 elements.
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Mesh
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Mesh
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Mesh
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Mesh
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Mesh
The plate contains a disk of transition elements from 1 to 3 CQUAD4.
The transition is done in Patran splitting 24 CQUAD4 with the
Utilities menu.
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Mesh
Background : In order to refine
a mesh one can shape every element. But it does exist a more
efficient solution to derive the
fine mesh from a coarse mesh.
Technique : Enter the Utilities> Fem-Elements > Break
Elements menu form.
Example : Consider the mesh
done between the two arcs. One
splits each CQUAD4 in four smaller
ones with the selection of the desired element and an edge.
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Mesh
The circular boundary of the structure contains 72 CQUAD4.
Nota : Remove duplicate nodes with the Equivalence Patran
function of the Mesh Menu form.
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
Mesh
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Case study # 3 - Plate with a hole
Meshing Technique
Case study # 3 - Plate with a hole
BCs & Loading
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Case study # 3 - Plate with a hole
KT assessment
Case study # 3 - Plate with a hole
KT assessment
KT is defined as
KT =
σmax
σ0
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Nota : when the problem is easy KT definition is straightforward ; but for complex structure σ0 is usually
particularly touchy to recover a universal definition. A reference textbook for KT definition is the Peterson
[1].
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Case study # 3 - Plate with a hole
Closed form solution
Case study # 3 - Plate with a hole
Closed form solution
For a circular hole the mathematical theory of elasticity (refer e.g. to [2]) leads to
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σxx (x =
l
2
(
, y) =
1+
1 R2
2
y2
+
3 R4
2
y4
)
σ0
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(29)
Case study # 3 - Plate with a hole
Nastran linear static run
Case study # 3 - Plate with a hole
Nastran linear static run
Nastran .dat is generated from Analysis menu
Then run the analysis with Nastran
$ nastran case study 3.dat news=n old=n scr=y
User obtains as output to Nastran run
◦ case study 3.log : Control File
◦ case study 3.f04 : Execution Summary Table
◦ case study 3.f06 : ASCII Results file
◦ case study 3.op2 : Binary Results file
27/71
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Students have to plot the reacted force and the reacted moment in
Patran after the import of the .op2.
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - u [mm]
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - σI [MPa]
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - σI [MPa]
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - σI [MPa]
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - freebody diagram form
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - elements and nodes for freebody
diagram
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - freebody diagram
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - freebody diagram
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - σyy [MPa] vs. y [mm]
The error to closed form solution of Nastran assessed from freebody results can be assessed through color
spectrum : ◦ < 4.00 %≤◦ < 5.00 %≤◦ < 7.00 %≤◦< 8.00 %.
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Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2
Case study # 3 - Plate with a hole
Results from Nastran linear static run from .op2 - σyy [MPa] vs. y [mm]
From freebody it is assessed
KT
=
=
=
2F
.
a × t × σ0
.
N ×F
π × R × t × σ0
2.99− .
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From GPSTRESS (grid point stress) in .f06 one derives
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KT
=
=
σ37
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2.86 ± 3.45%−
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σ0
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Case study # 3 - Plate with a hole
Conclusion & Outlook
Case study # 3 - Plate with a hole
Conclusion & Outlook
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◦ One does not match the closed form solution because the studied plate is not an infinite continuum
◦ Error has been assessed after computation of a load F derived from a freebody and a surface computed from plate thickness × sum of mid adjacent quads heigth for comparison with a closed from
solution
◦ Error in regard of closed form solution from theory of elasticty is under 3.00% but is highly mesh
dependent ; stress results have been assessed on a single path at x = 0 in the range y ∈ [ D
, w
]
2
2
◦ A more generalized methodology to assess the convergence of a mesh under Nastran is to use the
GPSTRESS (grid point stress : the extra/interpolation of σ recovery at Gauss points to nodes) output
and the associated discontinuities the GPSDCON
Case study # 3 - Plate with a hole
Conclusion & Outlook
Case study # 3 - Plate with a hole
Conclusion & Outlook
, 0) one
For the mesh used for Case study # 3 illustration purpose at Node 37 whose coordinates are (0, D
2
has
S T R E S S E S
A T
G R I D
P O I N T S
- S U R F A C E
10
SURFACE X-AXIS X NORMAL(Z-AXIS) Z
REFERENCE COORDINATE SYSTEM FOR SURFACE DEFINITION CI
GRID ELEMENT
STRESSES IN SURFACE SYSTEM
PRINCIPAL STRESSES
MAX
ID
ID
FIBER
NORMAL-X
NORMAL-Y
SHEAR-XY
ANGLE
MAJOR
MINOR
SHEAR
VON MISES
37
0
MID
2.860E+02 1.036E+01 -4.517E-03
-0.0009 2.860E+02 1.036E+01 1.378E+02
2.809E+02
and
G R I D
P O I N T
S T R E S S
D I S C O N T I N U I T I E S - S U R F A C E
10
SURFACE X-AXIS X NORMAL(Z-AXIS) Z
REFERENCE COORDINATE SYSTEM FOR SURFACE DEFINITION CI
GRID NORMAL STRESS DISCONTINUITY IN SURFACE SYSTEM
PRINCIPAL STRESS DISCONTINUITY
ERROR
ID
FIBER
NORMAL-X
NORMAL-Y
SHEAR-XY
MAJOR
MINOR
MAX SHEAR
VON MISES
EST.
37
MID
1.413E-03 7.667E-04 5.977E+00 1.831E-01 1.831E-01 1.831E-01 2.696E-01
3.451E+00
Error formula are quoted in Nastran Linear Static Guide [3] ; e.g. for a tensor component
δg =

1
Ne
2
1 X
2
(σi − σg )

Ne 
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i=1
then one derives a mean value depending upon tensor components number.
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Case study # 4 - Buckling Analysis
Theoretical Apart´
e
Case study # 4 - Buckling Analysis
Theoretical Apart´e
The buckling analysis is a bifurcation analysis.
Start point is the functional W that describes the elastic energy of the solid with a parameter λ
associated to the load.
The eigenvalues stem from setting equal to 0 the second derivative of the functional W .
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Case study # 4 - Buckling Analysis
Definition
Case study # 4 - Buckling Analysis
Definition
The framework is linear elasticity.
The plate is t = 3 mm thick. A compressive load σ0 = −100 MPa is applied at one side. The beam is
clamped at one side. The plate is made of aluminium E = 70 GPa and ν = 0.33.
Aim of Case study # 4 : Students have to find the buckling modes associated to the compressive load and
assess the critical load below which the structure remains stable.
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Case study # 4 - Buckling Analysis
Mesh
Case study # 4 - Buckling Analysis
Mesh
The mesh of case study # 3 is suitable for the buckling analysis.
The mesh is to be built with Nastran CQUAD4 elements.
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Case study # 4 - Buckling Analysis
Nastran EIGRL Card
Case study # 4 - Buckling Analysis
Nastran EIGRL Card
Nastran SOL 105 is to be used.The Nastran EIGRL Card has to be called in Case Control Section after the
linear SUBCASE associated to the compressive load by a Nastran METHOD Card.
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1
2
3
4
5
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11
12
13
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15
16
$ Case Control Section
SUBCASE 1001
SUBTITLE = COMPRESSION
LABEL = COMPRESSION
LOAD = 2
$
SUBCASE 1002
SUBTITLE = COMPRESSION
LABEL = COMPRESSION
METHOD = 1
STATSUB ( BUCKLING )=1001
$
BEGIN BULK
$ Bulk Section
EIGRL
1
...
10
Case study # 4 - Buckling Analysis
Nastran buckling run
Case study # 4 - Buckling Analysis
Nastran buckling run
Nastran .dat is generated from Analysis menu
Then run the analysis with Nastran
$ nastran case study 4.dat news=n old=n scr=y
User obtains as output to Nastran run
◦ case study 4.log : Control File
◦ case study 4.f04 : Execution Summary Table
◦ case study 4.f06 : ASCII Results file
◦ case study 4.op2 : Binary Results file
45/71
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .f06
Case study # 3 - Plate with a hole
Results from Nastran buckling run from .f06
Students have to find the eigenvalues associated to the computed modes in the .f06.
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .f06
Case study # 3 - Plate with a hole
Results from Nastran buckling run from .f06
Students have to find the eigenvalues associated to the computed modes in the .f06.
MODE
NO.
1
2
3
4
5
6
7
8
9
10
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EXTRACTION
ORDER
1
2
3
4
5
6
7
8
9
10
EIGENVALUE
2.922786E+00
8.688133E+00
1.231678E+01
1.690665E+01
1.714466E+01
2.491786E+01
2.727503E+01
3.472403E+01
3.905111E+01
4.561754E+01
R E A L
E I G E N V A L U E S
RADIANS
CYCLES
1.709616E+00
2.947564E+00
3.509528E+00
4.111770E+00
4.140611E+00
4.991779E+00
5.222550E+00
5.892710E+00
6.249089E+00
6.754076E+00
2.720938E-01
4.691193E-01
5.585586E-01
6.544085E-01
6.589987E-01
7.944663E-01
8.311947E-01
9.378539E-01
9.945734E-01
1.074944E+00
GENERALIZED
MASS
1.147029E+03
2.579305E+03
4.383950E+02
3.473366E+03
8.870814E+02
1.714101E+03
6.138878E+03
2.523782E+03
8.644048E+03
3.599225E+03
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Students have to plot the buckling mode in Patran after the import of the .op2.
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
Case study # 4 - Buckling Analysis
Results from Nastran buckling run from .op2
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Case study # 4 - Buckling Analysis
Conclusion & Outlook
Case study # 4 - Buckling
Conclusion & Outlook
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◦ Eigenmodes are particularly dependent upon boundary conditions applied
at the loaded plate end : eigenvalues are different if rotations are clamped
or not
Case study # 5 - Guyan Static Reduction
Theoretical Apart´
e
Case study # 5 - Guyan Static Reduction
Theoretical Apart´e
The framework is linear elasticity. One solves
K f f uf = Ff
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(36)
with o-set interior degrees of freedom and a-set exterior degrees of freedom one can split the equation (36)
Koo
K T oa
K oa
K aa
uo
ua
=
Fo
Fa
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(37)
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(38)
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(39)
whose first line of (37) means
Koo uo + K oa ua = Fo
−1
−1
⇔ Koo
(Koo uo + K oa ua ) = Koo
Fo
−1
−1
⇔ uo = Koo
Fo − Koo
K oa ua
−1
−1
⇔ uo = Koo Fo − Koo K oa ua
{z
}
| {z } |
u0
o
−Goa
or
0
uo = uo + Goa ua .
.
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with Goa boundary transformation and u0
o fixed boundary displacement.
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Case study # 5 - Guyan Static Reduction
Theoretical Apart´
e
Case study # 5 - Guyan Static Reduction
Theoretical Apart´e
Thus the second line of (37) means with the definition (39) of uo
K T oa uo +
a
K aa ua = F
+ K aa ua = Fa
⇔ K T oa u0
o + Goa ua


.
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

 T

T
0
⇔ K oa Goa +K aa  ua = Fa − K oa uo
|

{z
}
| {z }
Kaa
Fa
with Kaa boundary stiffness and Fa boundary load. Typically with a classical Nastran run one has
Kaa ≡ KAAX and Fa ≡ PAX.
Note that (40) defines the boundary displacement field associated to the ASET.
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(40)
Case study # 5 - Guyan Static Reduction
Definition
Case study # 5 - Guyan Static Reduction
Definition
The framework is linear elasticity.
The residual structure is selected as the square structure 30 × 30 mm2 centered on the circular hole.
Aim of Case study # 5 : Students have to realize a Guyan Static Reduction of structure associated to
Case study # 2 and run SOL 101 the residual structure. The fill-in of KAAX is to be assessed by the students
and compare to a classical finite elements matrix.
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Case study # 5 - Guyan Static Reduction
Nastran ASET Card
Case study # 5 - Guyan Static Reduction
Nastran ASET Card
Nastran offers a card called ASET to define the degrees of freedom to be omitted in the reduction process.
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Case study # 5 - Guyan Static Reduction
Nastran EXTOUT Parameter
Case study # 5 - Guyan Static Reduction
Nastran EXTOUT Parameter
In order to realize a Guyan Static Reduction one has to use EXTOUT parameter.
1
2
$ Case Control Parameters
PARAM
EXTOUT DMIGPCH
Set equal to DMIGPCH the EXTOUT parameter means a .pch file creation for output of KAAX and PAX.
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Case study # 5 - Guyan Static Reduction
Nastran linear static run for reduction
Case study # 5 - Guyan Static Reduction
Nastran linear static run for reduction
Nastran .dat is generated from Analysis menu
Then run the analysis with Nastran
$ nastran case study 5.dat news=n old=n scr=y
User obtains as output to Nastran run
◦ case study 5.log : Control File
◦ case study 5.f04 : Execution Summary Table
◦ case study 5.f06 : ASCII Results file
◦ case study 5.op2 : Binary Results file
◦ case study 5.pch : PUNCH file
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Case study # 5 - Guyan Static Reduction
Results from Nastran linear static run from .pch
Case study # 5 - Guyan Static Reduction
Results from Nastran linear static run from .pch
Students have to check .pch file contains a stiffness matrix and an
interface load vector. The former is called KAAX and the latter PAX.
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Case study # 5 - Guyan Static Reduction
Reading external element
Case study # 5 - Guyan Static Reduction
Reading external element
In order to read the result of a Guyan Static Reduction one has to use
K2GG and P2G commands.
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1
2
3
4
5
$ Case Control Section
$ Command for reading boundary stiffness
K2GG = ...
$ Command for reading interface load
P2G = ...
Case study # 5 - Guyan Static Reduction
Nastran linear static run for residual structure
Case study # 5 - Guyan Static Reduction
Nastran linear static run for residual structure
Nastran .dat is generated from Analysis menu for the residual
structure.
Note the Selected Group Menu Form of the Analysis menu is
particularly suitable to export a part of a previous model (here the
central square of the plate).
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Case study # 5 - Guyan Static Reduction
Nastran linear static run for residual structure
Case study # 5 - Guyan Static Reduction
Nastran linear static run for residual structure
The Nastran run case study 5 residual.dat has to call the result
of the SOL 101 reduction run case study 5.dat.
1
2
3
4
5
6
$ Case Control Section
INCLUDE ’ case_study_5 . pch ’
$ Command for reading boundary stiffness
K2GG = ...
$ Command for reading interface load
P2G = ...
Then run the analysis with Nastran
$ nastran case study 5 residual.dat news=n old=n scr=y
User obtains as output to Nastran run
◦ case study 5 residual.log : Control File
◦ case study 5 residual.f04 : Execution Summary Table
◦ case study 5 residual.f06 : ASCII Results file
◦ case study 5 residual.op2 : Binary Results file
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Case study # 5 - Guyan Static Reduction
Results from Nastran linear static run from .op2
Case study # 5 - Guyan Static Reduction
Results from Nastran linear static run from .op2
Students have to check with Patran that for the residual structure the
results from the run of case study #3 and case study #5 are in
accordance.
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Case study # 5 - Guyan Static Reduction
Conclusion & Outlook
Case study # 5 - Guyan Static Reduction
Conclusion & Outlook
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◦ The analysis performed is particularly used in dynamic analyis to reduce the degrees of freedom of
a model [4].
◦ The Guyan Static Reduction was particularly used by the past to allow concurrent work between
engineering teams.
◦ One can check that on the opposite of an assembly global finite elements stiffness matrix KAAX is not
sparse.
◦ One can use a DMIGOP2 output for the matrix KAAX and the load vector PAX instead of the Ascii DMIGPCH
output ; reading the binary format in Nastran to run the reduced structure is a little bit more touchy.
Main outcome is the binary file size is significantly smaller than the Ascii one.
References
References
[1] R.E. Peterson.
Stress concentration factors: charts and relations useful in making strength calculations for machine parts
and structural elements.
Wiley-Interscience publication. Wiley, 1974.
[2] B.F. De Veubeke.
A Course in Elasticity.
Number vol. 29 in Applied Mathematical Sciences. Springer-Verlag, 1979.
[3] MSC Software.
Linear Static Analysis User’s Guide: MSC Nastran 2012.
MacNeal-Schwendler Corporation, 2011.
[4] J.F. Imbert.
Analyse des structures par ´
el´
ements finis.
C´
epadu`
es, 1991.
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