1DMeshing Hypermesh

- 1D Meshing - 1D Meshing

1D Meshing1D Meshing1D Meshing

This chapter includes material from the book Practical FiniteThis chapter includes material from the book Practical Finiteadditional material added by Matthias Goelke.additional material added by Matthias Goelke.

When To Use 1D ElementsWhen To Use 1D ElementsWhen To Use 1D Elements

ne of the dimensions is very large in comparison to the other ne of the dimensions is very large in comparison to the other

510

5

YY

XX

ZZ

Element shape LineElement shape Line

Additional data from user - The remaining two dimensions, the cross Additional data from user - The remaining two dimensions, the cross

Element type Rod, bar, beam, pipe, axi-symmetric shell etc.Element type Rod, bar, beam, pipe, axi-symmetric shell etc.

Practical applications - Long shafts, beams, pin joint, connection elementsPractical applications - Long shafts, beams, pin joint, connection elements

Types of 1D Types of 1D

Rod Bar BeamRod Bar Beam

Tension/ All 6 dofsU , U , U , R , R

Same as bar compression

(and Torque for

All 6 dofsU , U , U , R , R

x y z x y, R (123456)

but also support unsymmetric (and Torque for

some softwares)

x y z x y, R (123456)

zApplicable for

unsymmetric sections i.e.some softwares)

U , R (1,4)

zApplicable forsymmetric c/s

sections i.e.shear center and U , R (1,4)

x xsymmetric c/s shear center and

warpagewarpage

Tension Shaft Same as barTension compression

Shaft subjected to

Same as bar+compression

members subjected to multiaxial

+for members

(truss), Shafts multiaxial loading,

for unsymmetric (truss), Shafts

subjected to loading,bolted,

unsymmetric c/s

Torque, Connection

welded joints, connection Connection

elementsconnection elementselements elements

11

1D Meshing -1D Meshing -

Finite Element Analysis. It also has been reviewed and has Finite Element Analysis. It also has been reviewed and has

ther two:ther two:

10001000x>>> y, z x>>> y, z

1000>>> 5, 101000>>> 5, 10

oss sectional area. oss sectional area.

elementselements

of 1D elements:of 1D elements:

Beam Pipe Axisymmetric shellBeam Pipe Axisymmetric shell

Same as U , U , R (1,3,5)x z y

t beam. Except it has internal

x z yZ- axis of symmetry, X as

radial axis.has internal non zero

radial axis.For objects symmetric

er and non zero diameter

For objects symmetric about the axis ofer and diameter about the axis of

rotation and subjected rotation and subjected to the axisymmetric to the axisymmetric boundary conditions.boundary conditions.

Piping Thin shell pressurePiping systems,

Thin shell pressurevessels, cylindrical,systems,

Structural vessels, cylindrical,conical objectsStructural

analysisconical objectsetc.analysis etc.


Stiffness Matrix DerivationStiffness Matrix DerivationStiffness Matrix Derivation

What Is Stiffness And Why Do We Need It In FEA?What Is Stiffness And Why Do We Need It In FEA?

Stiffness K is defined as Force/length (units N/mm). Physical inStiffness K is defined as Force/length (units N/mm). Physical ina unit displacement. The stiffness depends on the geometry as wa unit displacement. The stiffness depends on the geometry as w

Cast Iron Mild SCast Iron Mild S

FF

Consider 3 rods of exactly the same geometrical dimensions Consider 3 rods of exactly the same geometrical dimensions required to produce a 1 mm displacement then the Cast Iron would required to produce a 1 mm displacement then the Cast Iron would respectively, indicating K > K > K

CI MS Alrespectively, indicating K > K > K

CI MS Al

Mild Steel Mild SteelMild Steel Mild Steel

Now consider 3 different cross sectional rods of the same materialNow consider 3 different cross sectional rods of the same materialdifferent. Therefore, stiffness depends on the geometry as well asdifferent. Therefore, stiffness depends on the geometry as well as

Importance of the stiffness matrix - For structural analysis, stiffnessImportance of the stiffness matrix - For structural analysis, stiffnessanalysis is [F] = [K] [D]. The force is usually known, the displacementanalysis is [F] = [K] [D]. The force is usually known, the displacementthe element. This means if we formulate the stiffness matrix forthe element. This means if we formulate the stiffness matrix foranalysis of any geometry could be performed by meshing it and thenanalysis of any geometry could be performed by meshing it and then

Methods for formulating the stiffness matrix Methods for formulating the stiffness matrix

1) Direct Method1)2)

Direct MethodVariational Method2)

3)Variational MethodWeighted Residual Method3) Weighted Residual Method

The direct method is easy to understand but difficult to formulate usingResidual Methods are difficult to understand, but easy from a programmingResidual Methods are difficult to understand, but easy from a programmingeither use the Variational or Weighted Residual Method formulationeither use the Variational or Weighted Residual Method formulation

Rod Element Stiffness Matrix Derivation By The Direct MethodRod Element Stiffness Matrix Derivation By The Direct Method

Methodology for derivation of stiffness matrix by the direct methodMethodology for derivation of stiffness matrix by the direct method

Assume there are n dofs for a given element (for example, a quad4 elements Assume there are n dofs for a given element (for example, a quad4 elements

Step 1) Assume the 1st dof 0, and all the other dof = 0. This will Step 1) Assume the 1st dof 0, and all the other dof = 0. This will

Step 2) Assume the 2nd dof 0, and all the other dof = 0. This willStep 2) Assume the 2 dof 0, and all the other dof = 0. This will::::

22


interpretation Stiffness is equal to the force required to produceinterpretation Stiffness is equal to the force required to producewell as the material properties.well as the material properties.

Mild Steel AluminiumMild Steel Aluminium

F FF F

Cast Iron, Mild Steel, and Aluminium. If we measure the force Cast Iron, Mild Steel, and Aluminium. If we measure the force ould require the maximum force, followed by Steel and Aluminiumould require the maximum force, followed by Steel and Aluminium

Mild SteelMild Steel

erial. Again, the force required to produce a unit deformation will beerial. Again, the force required to produce a unit deformation will beas the material.as the material.

fness is a very important property. The equation for linear staticfness is a very important property. The equation for linear staticdisplacement is unknown, and the stiffness is a characteristic property ofdisplacement is unknown, and the stiffness is a characteristic property of

or a given shape, like line, quadrilateral, or tetrahedron, then theor a given shape, like line, quadrilateral, or tetrahedron, then thethen solving the equation F = K D.then solving the equation F = K D.

using computer programming. While the Variational and Weightedogramming point of view. Thats the reason why all software codesogramming point of view. Thats the reason why all software codes

ormulation.ormulation.

MethodMethod

thod:thod:

uad4 elements total dofs = 4*6 = 24). uad4 elements total dofs = 4*6 = 24).

will lead to equation 1.will lead to equation 1.

will lead to equation 2.will lead to equation 2.


Step n) Assume the nth dof 0, and all the other dof = 0. This will Step n) Assume the nth dof 0, and all the other dof = 0. This will

Step n+1) Sum all the equations, 1 + 2 + 3 + 4 ..+ n.Step n+1) Sum all the equations, 1 + 2 + 3 + 4 ..+ n.

Step n+1 will give us the most generalized formulation of the stiffness Step n+1 will give us the most generalized formulation of the stiffness

Rod ElementRod Element

ii

No. of nodes :- 2 1 dof at each nodeNo. of nodes :- 2 1 dof at each node

Stiffness matrix formulation (Direct Method)Stiffness matrix formulation (Direct Method)

UUi

FFi

ii

Case 1:Case 1:

u >0,i

u =0j

u >0,i

u =0j

F = 0 F + F =0 F = -FF = 0x

F + F =0i j

F = -Fi j

F = 0x i j i j

= F/A = u/LF = 0

x = F/Ax

= u/L x

F = 0F/A= Eu/L = E

F = 0yF/A= Eu/L = E

xy

F = 0F = 0z

F = (AE/L)*u F = -F = - (AE/L)*uF = (AE/L)*ui i

F = -F = - (AE/L)*uj i i

.....(A)i i j i i

.....(A)

Case 2: u =0 u >0Case 2: u =0 u >0i J

F = (AE/L)*u F = -F = - (AE/L)*u ...... F = (AE/L)*uj j

F = -F = - (AE/L)*ui j j

......

ii

F = -Fi ji j

Case 3: General case - u , u > 0i j

Case 3: General case - u , u > 0i j

From (A) and (B)From (A) and (B)

F = (AE/L)*u (AE/L)*uF = (AE/L)*u (AE/L)*ui. i j

F = -(AE/L)*u + (AE/L)*uF = -(AE/L)*u + (AE/L)*uj. i j

33


will lead to equation n. will lead to equation n.

fness matrix.fness matrix.

ElementElement

jj

L

A = Area of c/sE = Modulus of ElasticityE = Modulus of Elasticity

jj

STATICSSTATICS

M = 0M = 0xx

M = 0M = 0yy

M = 0M = 0z

...... (B)...... (B)

UUj

j Fj Fj


1Fi

1Fi

= (AE/L)= (AE/L)

F -1Fj

-1(2x1)

Force Matrix Stiffness

(2x1)

Force Matrix Stiffness

Properties of the stiffness matrixProperties of the stiffness matrix

The order of the stiffness matrix corresponds to the to The order of the stiffness matrix corresponds to the to

A singular stiffness matrix means the structure is unconstrained A singular stiffness matrix means the structure is unconstrained

Each column of the stiffness matrix is an equilibrium s Each column of the stiffness matrix is an equilibrium s

A symmetric stiffness matrix shows the force is directly A symmetric stiffness matrix shows the force is directly

Diagonal terms of the matrix are always positive meaning Diagonal terms of the matrix are always positive meaningdisplacement in the right direction. Diagonal terms willdisplacement in the right direction. Diagonal terms will

Rod elements support only tension or compression and no shearRod elements support only tension or compression and no shearorder of the stiffness matrix is 2x2, where the number of unknowns order of the stiffness matrix is 2x2, where the number of unknowns

Number of unknowns = no. of dofs - no. of dofs constraint by a SingleNumber of unknowns = no. of dofs - no. of dofs constraint by a Singleuser as 0)user as 0)

Usually in comparison to the total dofs for the model, the constraintsUsually in comparison to the total dofs for the model, the constraintsapproximately the total dofs.approximately the total dofs.

The order of the stiffness matrix = total dof x total dofThe order of the stiffness matrix = total dof x total dof

Quad4 element:Quad4 element:

dof/node = 6

total dofs = 6X4total dofs = 6X4

order of the stiorder of the sti

|F| = |K| |||F| = |K| ||24x1 24x24 24x1

This means to solve a single quad4 element problem, the software This means to solve a single quad4 element problem, the software

Tetra10 element:Tetra10 element:

|F| = |K| |||F|30x1

= |K| ||30x30 30x1

44


-1 ui

-1 ui

1 u1 uj(2x2) (2x1)

fness Matrix Displacement Matrix

(2x2) (2x1)

fness Matrix Displacement Matrix

otal dofs.otal dofs.

structure is unconstrained and has rigid body motion.structure is unconstrained and has rigid body motion.

set of nodal forces required to produce the unit respective dof.set of nodal forces required to produce the unit respective dof.

directly proportional to displacement.directly proportional to displacement.

meaning a force directed in say the left direction cannot produce ameaning a force directed in say the left direction cannot produce awill be zero or negative only if the structure is unstable.will be zero or negative only if the structure is unstable.

shear force, bending moment and torque. In the above equation, the shear force, bending moment and torque. In the above equation, the wns is 2.wns is 2.

Single Point Constraints (at fixed nodes, dofs are specified by theSingle Point Constraints (at fixed nodes, dofs are specified by the

nstraints are negligible and therefore the total no. of unknowns is nstraints are negligible and therefore the total no. of unknowns is

6X4 = 24,6X4 = 24,

the stiffness matrix = 24 x 24the stiffness matrix = 24 x 24

are will have to solve 24 equations internally. are will have to solve 24 equations internally.

dof /node = 3dof /node = 3

total dofs = 3*10 = 30

order of stiffness matrix = 30 x30order of stiffness matrix = 30 x30


For A Given FE Model, How Many Equations Does A Software

Assume there are 20,000 nodes for a mesh model consisting of only Assume there are 20,000 nodes for a mesh model consisting of only

Total dofs = 20000*6 = 120,000.Total dofs = 20000*6 = 120,000.

Stiffness matrix order = 120,000x120,000Stiffness matrix order = 120,000x120,000

Number of equations the FEA software will solve internally = 120,000Number of equations the FEA software will solve internally = 120,000

Table On The Equivalence Of Variational FEM And WeightedTable On The Equivalence Of Variational FEM And Weighted

The problem: A one dimensional rod element is subjected to a concentraThe problem: A one dimensional rod element is subjected to a concentrais fixed as discussed above.is fixed as discussed above.

The governing differential equation and boundary conditions:The governing differential equation and boundary conditions:

AE ( d2u/dx2 ) = 0 u | =0x =0

AE ( d u/dx ) = 0 u | =0x =0

AE ( du/dx) | = PAE ( du/dx) | = Px=Lx=L

Rayleigh-Ritz FEM

1. Starting point Strong 1. Starting point

X

Strong

AE(d = [ AE/2 (dy/dx)2 dx] P u

Xj AE(d

Weak = [ AE/2 (dy/dx)2 dx] P uj

Xi

Weak X

jXi j

XXi

w =w =iIntegrating Integrating

x

AE

x

AE x

2. FE discretization step Where2. FE discretization step Where

u = u N = u N + u Ni j i i j j

L = (x u = u N = u N + u Ni j i i j j

N =(x -x)/L , N =(x-x )/Li j j i

L = (x du/dx =

i j j i

3. Substituting the derivative in the functional 3. Gale3. Substituting the derivative in the functional

x

3. Gale

= AE/2 (u -u /L)2 dx-P ux

j

= AE/2 (u -u /L)2 dx-P uj i j

xi

AE xi

4. Minimizing the functional 4. E4. Minimizing the functional 4. E

=AE/2L[u 2-2u u +u 2]-Pui i j j j

=AE/2L[u -2u u +u ]-Pui i j j j

/u = AE/2L[2u -2u ] = 0i i j

/u = AE/2L[-2u +2u ] -P = 0K =

ei i j

/u = AE/2L[-2u +2u ] -P = 0j i j

e

= AE= AE

= AE / = AE /

(as (as

55


are Solve?

only thin shell elements (6 dof/node).only thin shell elements (6 dof/node).

120,000120,000

eighted Residual FEM In Solid Mechanicseighted Residual FEM In Solid Mechanics

concentrated tensile force at one end (free end) and the other endconcentrated tensile force at one end (free end) and the other end

Galerkin FEM

ong formong form

AE(d u/dx ) = 0 2 2AE(d u/dx ) = 0 eak from

2 2

eak from

w [AE ( d2u/dx2 )]dx = 0ii

Weighting functionWeighting functionegrating by partsegrating by partsx

j LE dw/dx du/dx dx = [w AE du/dx]

i i

xj L

E dw/dx du/dx dx = [w AE du/dx]i i

xi

oi

WhereWhere

(x - x )j i(x - x )j idu/dx = (u -u )/L

j ij i

Galerkin stepGalerkin step

xx

jU

i

xj

E dN /dx dN /dx dxi j

xj

x

Ui

U= N AE du/dxi

xi jx

i jU

i

xi

Evaluate the element stiffness matrixEvaluate the element stiffness matrix

xjj

K = AE dN /dx dN /dx dxi j

xi j

xi

xj j

x

dN /dx dN /dx dx dN /dx dN /dx dxi i i j

j j

x x dN /dx dN /dx dx dN /dx dN /dx dx

i i i jx

ix

i

AEx x

AE

dN /dx dN /dx dx dN /dx dN /dx dx

xj

xj

dN /dx dN /dx dx dN /dx dN /dx dxj i j j

x xxi

1 -1

xi

1 -1AE / LAE / L

-1 1-1 1

(as dN /dx = -1/L , dN /dx = 1/L )(as dN /dx = -1/L , dN /dx = 1/L )i j


5. Write down the final algebraic form of 5. W5. Write down the final algebraic form ofequations

5. Wequations

2 -2 U 0 = AE/L= AE/2L

2 -2 U 0i

=

= AE/L= AE/2L

-2 2 Uj

P=-2 2 Uj

P

1 -1 U 0= AE/L

1 -1 Ui

0= AE/L

-1 1 U P=-1 1 Uj

P

Observation: Both of the formulations lead to the same Observation: Both of the formulations lead to the same

Exercise 1: Rod element subjected to a tensile force.Exercise 1: Rod element subjected to a tensile force.

50

500500

E= 2.1 * 105 N/mm2, = 0.30 (Steel)E= 2.1 * 10 N/mm , = 0.30 (Steel)

Analytical solution:Analytical solution:

Stress = F/A = 100,000/(3.14*25*25) = 50.95 N/mm2Stress = F/A = 100,000/(3.14*25*25) = 50.95 N/mm

Displacement = FL / AE = 100,000*500 / (3.14*25*25*2.1*10Displacement = FL / AE = 100,000*500 / (3.14*25*25*2.1*10

How to solve using software: The following basic steps are requiredHow to solve using software: The following basic steps are requireddata format may vary from software to softwaredata format may vary from software to software

Step 1 Beam SectionStep 1 Beam Section Define the circular cross section (diameter 50 mm) Define the circular cross section (diameter 50 mm)

Step 2 MeshingStep 2 Meshing Create nodes Create nodes Manually enter in the coordinates of the nodes (0,0,0), Manually enter in the coordinates of the nodes (0,0,0), Create rod element by picking the above 2 nodes (assign Create rod element by picking the above 2 nodes (assign Create material and assign it to the element Create material and assign it to the element

Step 3 Boundary ConditionsStep 3 Boundary Conditions Specify the type of analysis (Linear Static analysis) Specify the type of analysis (Linear Static analysis) Apply force in x-direction = 100,000 N Apply force in x-direction = 100,000 N Apply constraint = x translations Fixed Apply constraint = x translations Fixed

Step 4 SolutionStep 4 Solution Create solution set / load step Create solution set / load step Solve Solve

PostprocessingStep 5 PostprocessingStep 5 Display displacement and stress result Display displacement and stress result Animate for deflection Animate for deflection Compare analytical and software results Compare analytical and software results

66


Write down the final algebraic formWrite down the final algebraic form

-AE du/dx |1 -1 U 0-AE du/dx | x=xiAE/L

1 -1 Ui

0

AE du/dx |x=xi

AE/L-1 1 U

j P=

x=xi-1 1 Uj P

same final algebraic form of the stiffness matrixsame final algebraic form of the stiffness matrix

1,00,000 N1,00,000 N

500500

0 ) = 0.12 mm05) = 0.12 mm

uired in any commercial software. The sequence of steps and inputuired in any commercial software. The sequence of steps and input

the nodes (0,0,0), (500,0,0)the nodes (0,0,0), (500,0,0)nodes (assign circular cross section)nodes (assign circular cross section)


How Does The Software Solve This Problem Internally?

FFi

= (AE/L)= (AE/L)

Fjj

-100000-100000

= (3.14*502 *2.068E05)/(4*500)= (3.14*50 *2.068E05)/(4*500)

100000100000

-100000-100000

= 8.1169 E05= 8.1169 E05

100000

u = 0.12 mm = u / L = 0.12/500 = 0.00024638u = 0.12 mm = u / L = 0.12/500 = 0.00024638j j

= E * = 2.068 *105 * 0.00024638 = 50.9 N/mm2 = E * = 2.068 *105 * 0.00024638 = 50.9 N/mm2

Exercise 2: Rod element cannot resist shear forceExercise 2: Rod element cannot resist shear force

30,000

10001000

Everything is the same as Exercise - 1, except the force is appliedEverything is the same as Exercise - 1, except the force is appliedcalculations. The FEA will show a very large displacement and zecalculations. The FEA will show a very large displacement and zerequired dof to resist the vertical force (i.e. to keep the system in erequired dof to resist the vertical force (i.e. to keep the system in e

y ?y ?xx

zz

77


1 -1 u1 -1 ui

-1 1 ujj

1 -1 01 -1 0

*2.068E05)/(4*500)*2.068E05)/(4*500)

-1 1 u-1 1 uj

-uj

-uj

uj

= My/I = My/I

I = bd3/12I = bd3/12

30,000 N = PL3/3EI

100

100

1

applied along the y-axis. Compare the FEA results with the handapplied along the y-axis. Compare the FEA results with the handzero stress. This is because the rod element does not have thezero stress. This is because the rod element does not have theequilibrium).equilibrium).

FF


What do you feel, which dofs are required to resist the vertical fo

RRz

Uyy

To resist the vertical force (F = 0 and M = 0) we need an elementy z

To resist the vertical force (F = 0 and M = 0) we need an elementy z

about Z.about Z.

These dofs are supported by Beam and Bar elements and if theThese dofs are supported by Beam and Bar elements and if thewould get correct results.

In real life, very rarely is a single element used for analysis. It isIn real life, very rarely is a single element used for analysis. It isinvolved, the software combines the stiffness matrices of the individualinvolved, the software combines the stiffness matrices of the individualequal to the total dofs of the model.equal to the total dofs of the model.

Stiffness Matrix - Assembly Of 2 Rod ElementsStiffness Matrix - Assembly Of 2 Rod Elements

A E L A

1 2

A E L1 1 1

A

1 2

Because of a force at point 3, what would the force be at Points 1Because of a force at point 3, what would the force be at Points 1try to visualize the forces at the various points.try to visualize the forces at the various points.

For equilibrium F = 0. The reaction force at point 1 is F and at For equilibrium F = 0. The reaction force at point 1 is F and at x

The free body diagram is:The free body diagram is:

-F F -FA E L-F F -F1 2a

A E L1 1 11 2a1

Total force at point 2 = - F + F = 0Total force at point 2 = - F + F = 0

In any finite element model, the summation of forces and momentsIn any finite element model, the summation of forces and momentsrestrained and at which the external force and moment is applied)restrained and at which the external force and moment is applied)complete model is 0. (The external forces and moments = the reactioncomplete model is 0. (The external forces and moments = the reactionfor ensuring correct results.for ensuring correct results.

Assembly

A E A E L1 1 1

A E L1 1

K =1

A E11

1 2a

A E L1 1 1

A E L1 1 1

K =1 11

L1 1 2a1

L1

F1

k1

-k1

0 u11 1 1 1

F = -k k 0 uF2a

= -k1

k1

0 u2a

0 0 0 0 00 0 0 0 0

88


orce?

FF

lement which supports translation along the Y axis and rotationlement which supports translation along the Y axis and rotation

same exercise is repeated with a beam or bar element then onesame exercise is repeated with a beam or bar element then one

is a combination of many elements. When multiple elements areis a combination of many elements. When multiple elements areindividual elements. The combined stiffness matrix has an orderindividual elements. The combined stiffness matrix has an order

ElementsElements

FA E L

3

FA E L

2 2 2

3

1 and 2? Before reading the answer, please shut your eyes and1 and 2? Before reading the answer, please shut your eyes and

at point 2 = 0.at point 2 = 0.

F FA E LF F2 2 2

A E L

2

32b 32b

moments is zero at the internal nodes (except the nodes which aremoments is zero at the internal nodes (except the nodes which areapplied). The overall summation of the forces and the moments for theapplied). The overall summation of the forces and the moments for the

reaction forces and moments). This is one of the important checksreaction forces and moments). This is one of the important checks

Assembly of elements

A EA E LA E LA E

K = 22A E L2 2 2A E L2 2 2

K = 222 L2

2

2b 3L

2

0 0 0 0 0

F = 0 k -k uF2b

= 0 k2

-k2

u2b

F3

0 -k2

k2

u3

F3

0 -k2

k2

u3


F1

k1

-k1

F1

k1

-k1

F + F = -k k +kF + F2a 2b

= -k1

k +k1 22a 2b 1 1 2

F 0 -kF3

0 -k2

F k -kF1

k1

-k1

0 = -k1

k +k1 2

0 = -k1

k +k1 2

F3

0 -k23 2

F k kF1

k11

k12

F = k kF2

= k21

k22

F3

k31

k323x1

F3

k31

k323x1

Exercise 3: Assembly of 2 rod elementsExercise 3: Assembly of 2 rod elements

A E L A A E L1 1 1

A 2 1 1 1 2

1 21 2R = 2 mmR = 2 mm

L = 259.74 L =L = 259.741

L =

K = A E / L = 104 N/mmK = A E / L = 104 N/mm1 1 1 1

F 1 -1F1

1 -1

0 = 104 -1 30 = 10 -1 3

1000 0 -21000 0 -2

0 3 -20 3 -2

= 104= 104

1000 -2 21000 -2 2

99


0 u1

0 u1

-k u + u-k2

u + u2a 2b2 2a 2b

k uk2

u3

0 u 0 u

u

1

u-k2 2

-k2 2

k2

u32 3

k uk13

u1

k u k23

u 2

u33 33x3 3x1

k

u33 33x3 3x1

k

A E L F = 1000 NA E L2 2 2

F = 1000 N2 2 2

33R = 4 mm

L = 519.48R = 4 mm

L = 519.48

K = A E / L = 2 * 104 N/mmK = A E / L = 2 * 104 N/mm2 2 2 2

0 00 0

-2 u2

-2 u2

2 u32 u3

2 u2 u2

u = 0.1 mmu = 0.1 mm22

u = 0.15 mm2 u

3

u = 0.15 mm32 u

3


Beam ElementBeam ElementBeam Element

The beam element is the most general type of element (6 dofs perThe beam element is the most general type of element (6 dofs perfor various applications.for various applications.

123456123456

ii

2 Nodes2 Nodes

[F]12 X1

= [K]12X12

[F]12 X1

= [K]12X12

FFix

K11

K12

K13

K14

K15

K16

K17

K18

K19

K110

K111

Fiy

K22

K23

K24

K25

K26

K27

K28

K29

K210

K2iy

K22

K23

K24

K25

K26

K27

K28

K29

K210

K2

Fiz

K33

K34

K35

K36

K37

K38

K39

K310

K3

Fiz

K33

K34

K35

K36

K37

K38

K39

K310

K3

M K K K K K K K KMix

K44

K45

K46

K47

K48

K49

K410

K4

M K K K K K K KMiy

K55

K56

K57

K58

K59

K510

K5

Miz

K66

K67

K68

K69

K610

K6iz 66 67 68 69 610 6

Fjx

= K77

K78

K79

K710

K7

Fjx

= K77

K78

K79

K710

K7

F K K K KF

F

jyK

88K

89K

810K

8

K K KFjz

K99

K910

K9

Mjx

K1010

K10

Mjy

K1111jy 1111

Mjz

Mjz

12x1

* Beam element can resist force and moment about x, y, and z axis.* Beam element can resist force and moment about x, y, and z axis.

Exercise 4: Importance of boundary condition: Simply supported Exercise 4: Importance of boundary condition: Simply supported

30,000 N

Create 3 nodes and 2 beam elements. Apply a force on the centralCreate 3 nodes and 2 beam elements. Apply a force on the centralnodes. Solve the problem and compare the results. If the appliednodes. Solve the problem and compare the results. If the appliedwith following values:with following values:

= (1/48)*{PL3/EI} and = M*y/I (M = PL/4) = (1/48)*{PL3/EI} and = M*y/I (M = PL/4)

If your results do not match with the above values, then try followingIf your results do not match with the above values, then try following

= (2/384)*{PL3/EI} and = M*y/I (M = PL/8) = (2/384)*{PL /EI} and = M*y/I (M = PL/8)

If your answer matches with the 2 set of formulas, then it meansIf your answer matches with the 2nd set of formulas, then it meanssimply supported.simply supported.

1010


per node, 3 translational and 3 rotational) and is used frequentlyper node, 3 translational and 3 rotational) and is used frequently

123456123456

jj

6 dof per node (3T + 3R)6 dof per node (3T + 3R)

12X1[]

12X1[]

111K

112u

ix

211K

212v

iy211K

212v

iy

311K

312w

iz311K

312w

iz

K 411

K412

ix

K 511

K512

iy

611K

612

iz611 612 iz

711K

712u

jx711K

712u

jx

K v811

K812

vjy

K w911

K912

wjz

1011K

1012

jx

1111K

1112

jy1111 1112 jy

K1212

jz

K1212

jz

12x12 12x1

axis.axis.

ed beamed beam

100

central node and the appropriate boundary conditions on the endcentral node and the appropriate boundary conditions on the endapplied boundary conditions are correct, then the results would matchapplied boundary conditions are correct, then the results would match

wing formulas:wing formulas:

ans the fix-fix boundary conditions have been applied instead of theans the fix-fix boundary conditions have been applied instead of the

1010


Simply supportedSimply supported

Y12345 12345

Y

X

12345 12345X

ZZ

Fix-FixFix-Fix

Y123456 12345

X

123456 12345X

ZZ

The difference between the above two boundary conditions is simThe difference between the above two boundary conditions is simof simply supported and fixed for fix-fix). There is a drastic changeof simply supported and fixed for fix-fix). There is a drastic changefactor of 2 and displacements by a factor of 4. Now imagine a realfactor of 2 and displacements by a factor of 4. Now imagine a realby a CAE engineer as fix-fix and the analysis shows that stress =by a CAE engineer as fix-fix and the analysis shows that stress =of the stress is less than the yield stress (250 N/mm2), a colorfulof the stress is less than the yield stress (250 N/mm2), a colorfulsupported by animation files, would be submitted to a design engineersupported by animation files, would be submitted to a design engineersince it is simply supported, the magnitude of stress will be doublesince it is simply supported, the magnitude of stress will be doublestructure will fail soon..structure will fail soon..

1111


Please observe rotation of scale at the Please observe rotation of scale at the end (R = free) and deformation profile

z

Fix- Fix is all the dofs fix (clamped, just likeFix- Fix is all the dofs fix (clamped, just likecemented in the wall)cemented in the wall)

Stress DisplacementStress DisplacementN/mm2 mm

76.43 0.6412345

76.43 0.6412345

38.22 0.1656

38.22 0.1656

mply because of one dof, the rotation about the z axis (free in casemply because of one dof, the rotation about the z axis (free in casechange in the stress and displacement results. The stress differs by achange in the stress and displacement results. The stress differs by a

real life steel structure which is simply supported, being modelledreal life steel structure which is simply supported, being modelled200 N/mm2 and displacement = 2.5 mm. Since the magnitude200 N/mm2 and displacement = 2.5 mm. Since the magnitude

ful impressive report with a wonderful power point presentationful impressive report with a wonderful power point presentationengineer with the conclusion that the design is safe. But in real life,engineer with the conclusion that the design is safe. But in real life,

double (400 N/mm2 ) and the displacement 4 times (10 mm) and thedouble (400 N/mm2 ) and the displacement 4 times (10 mm) and the

1111


Boundary conditions take least amount of time to define, but it is4 cylinder Engine block requires 300 hours and defining the bounda4 cylinder Engine block requires 300 hours and defining the boundain vain if the boundary conditions are not applied properly.in vain if the boundary conditions are not applied properly.

How To Apply The Correct Boundary Conditions? Who Can THow To Apply The Correct Boundary Conditions? Who Can T

The application of boundary conditions is a tricky subject and is basedThe application of boundary conditions is a tricky subject and is based

No university, consultant, or book can teach these things. ThisNo university, consultant, or book can teach these things. This100% involvement, thinking about the behavior of the component100% involvement, thinking about the behavior of the componentthe FE model is behaving in the same way as in reality (displacementthe FE model is behaving in the same way as in reality (displacementfrequencies and response comparison for dynamic analysis, etc.)frequencies and response comparison for dynamic analysis, etc.)boundary conditions are needed. Discussions with experienced CAEboundary conditions are needed. Discussions with experienced CAEare also needed. In short, making mistakes and reworking and modifyingare also needed. In short, making mistakes and reworking and modifyingis achieved is the true teacher.is achieved is the true teacher.

Special Features Of Beam/Bar ElementsSpecial Features Of Beam/Bar Elements

a. Orientation of cross sectiona. Orientation of cross section

Changing the orientation angle of the cross section is supported bChanging the orientation angle of the cross section is supported bfurther below.further below.

YY

X Z

b. Taper elementsb. Taper elements

Regular bar elements cannot take into account a variation in theRegular bar elements cannot take into account a variation in theInput data for a taper beam is two cross sections (front and rear).Input data for a taper beam is two cross sections (front and rear).

1212


is the most important step in FEA. Typically, the brick meshing of aboundary condition requires 8 hours. 300 hours hard work would goboundary condition requires 8 hours. 300 hours hard work would go

Teach Us Quickly?Teach Us Quickly?

based on pure engineering judgment, skill, and common sense.based on pure engineering judgment, skill, and common sense.

is something to be learned from practical and field experience.is something to be learned from practical and field experience.ponent due to various forces, and checking after every iteration whetherponent due to various forces, and checking after every iteration whether

(displacement plots and strain gauge results for static analysis or natural(displacement plots and strain gauge results for static analysis or natural). If the model is not behaving correctly, then modifications in the). If the model is not behaving correctly, then modifications in the

CAE engineers, field engineers, and testing department engineersCAE engineers, field engineers, and testing department engineersmodifying the FE model again and again until satisfactory correlationmodifying the FE model again and again until satisfactory correlation

ElementsElements

by most commercial software and will be discussed in some detailby most commercial software and will be discussed in some detail

YY

X Z

the cross section. This could be achieved by using a taper beam.the cross section. This could be achieved by using a taper beam.rear).rear).

1212


c. Curved beams: Employed to model curvaturesc. Curved beams: Employed to model curvatures

d. Beam end shortening: To represent the geometry accurately at d. Beam end shortening: To represent the geometry accurately at

Without end shortening (interference at joint)Without end shortening (interference at joint)

1313


ely at jointsely at joints

End shorteningEnd shortening

1313


e. Beam cross section offset: Useful in representing the geom

OriginalOriginal

Beam Beam

f. Beam end releasef. Beam end release

A beam element has 6 dofs. There are special situations, like a roA beam element has 6 dofs. There are special situations, like a rospecific direction is free. The use of a regular beam element andspecific direction is free. The use of a regular beam element andrepresent the exact situation. The end release is a special commandrepresent the exact situation. The end release is a special command

For example, consider the door above. Its rotation about the verFor example, consider the door above. Its rotation about the verbeam element. The end release (rotation along vertical axis) wouldbeam element. The end release (rotation along vertical axis) would

g. Shear centerg. Shear center

It is applicable only for non symmetric cross sections (like angle orIt is applicable only for non symmetric cross sections (like angle orto a symmetric cross sectional beam, it will bend. But for an unsymmto a symmetric cross sectional beam, it will bend. But for an unsymmas well twisting. Twisting is due to unbalanced shear forces in theas well twisting. Twisting is due to unbalanced shear forces in theapplied, will nullify the effect of torsion.applied, will nullify the effect of torsion.

1414


geometry as per a real life situation

OriginalOriginal

Beam offsetBeam offset

rotating shaft or hinge joint, where the rotation or translation in arotating shaft or hinge joint, where the rotation or translation in aand its spider connection, using Rigid or RBE2 elements, might notand its spider connection, using Rigid or RBE2 elements, might not

command to release one or more dofs of a beam.command to release one or more dofs of a beam.

rtical axis is free. Suppose the pin of the hinge is simulated by artical axis is free. Suppose the pin of the hinge is simulated by aould be helpful in simulating the door in real life.ould be helpful in simulating the door in real life.

or channel cross sections). When a vertical shear force is appliedor channel cross sections). When a vertical shear force is appliedunsymmetric cross section, the same force would result in bendingunsymmetric cross section, the same force would result in bending

the flange. The shear center is the point at which when a force isthe flange. The shear center is the point at which when a force is

1414


1 1P P1 1P P

FF

CC

P1

P1

P

For symmetric cross sections, the centroid and shear center coincide,For symmetric cross sections, the centroid and shear center coincide,

Shear flow for an I - section shows that the forces in the flangeShear flow for an I - section shows that the forces in the flangechannel, there is an unbalanced couple which gives rise to torsionchannel, there is an unbalanced couple which gives rise to torsion

The shear center is a property of the given cross section and it doesThe shear center is a property of the given cross section and it does

Beam elements are recommended for symmetric as well unsymmBeam elements are recommended for symmetric as well unsymmBar elements have a limitation on the shear center and therefore are nBar elements have a limitation on the shear center and therefore are n

Rigid ElementsRigid ElementsRigid Elements

Rigid elements were devised to simplify the input of constraint eqRigid elements were devised to simplify the input of constraint eqelements, but are equations that define dependencies between degreeselements, but are equations that define dependencies between degreeselements which lead to a large percentage of input errors thereelements which lead to a large percentage of input errors thereperfectly rigid materials in nature, a rigid element is usually a poorperfectly rigid materials in nature, a rigid element is usually a pooris simply a constraint equation or a set of constraint equations ofis simply a constraint equation or a set of constraint equations of

A1 u1 + A2 u2 + A3 u3 ...=A1 u1 + A2 u2 + A3 u3 ...=

0;0;

where the A = constant coefficients and the u = degrees of freedomwhere the A = constant coefficients and the u = degrees of freedomothers.others.

RBAR ElementRBAR Element

The RBAR element rigidly connects from one to six dependent degreesThe RBAR element rigidly connects from one to six dependent degreesof freedom. The six independent degrees of freedom must be capable of freedom. The six independent degrees of freedom must be capable

1515


P1

P1

PP

hF = P

hF = P

P1

eP

1

e

coincide, therefore there is no twisting.coincide, therefore there is no twisting.

are in equilibrium and thus there is no twisting. While for a Care in equilibrium and thus there is no twisting. While for a Crsion.rsion.

does not depend on the magnitude of the force.does not depend on the magnitude of the force.

unsymmetric cross sections, as it takes into account the shear center. unsymmetric cross sections, as it takes into account the shear center. ore are not recommended for unsymmetric cross sections.ore are not recommended for unsymmetric cross sections.

quations for commonly used configurations. They are not structuralquations for commonly used configurations. They are not structuraldegrees of freedom. There are many illegal configurations of rigiddegrees of freedom. There are many illegal configurations of rigidfore they should be used with care. Because there really are nofore they should be used with care. Because there really are no

poor representation of any structural component. In general, a rigidpoor representation of any structural component. In general, a rigidthe form:the form:

freedom. Usually, one degree of freedom is dependent on all of thefreedom. Usually, one degree of freedom is dependent on all of the

degrees of freedom (the m -set) to exactly six independent degrees degrees of freedom (the m -set) to exactly six independent degrees be capable of describing the rigid body properties of the element.be capable of describing the rigid body properties of the element.

1515


If a rigid element is long or of similar length as the structure and is ststructure. In this case a simplified representation using structuralstructure. In this case a simplified representation using structuralsuch as bolts typically results in overly stiff joints. The use of a beamsuch as bolts typically results in overly stiff joints. The use of a beamallows easy output of force quantities.allows easy output of force quantities.

Proper Application Of Rigid ElementsProper Application Of Rigid ElementsProper Application Of Rigid Elements

When part of a structure is very stiff relative to and lightly coupledWhen part of a structure is very stiff relative to and lightly coupledapplication of rigid elements. An example might be an engine connecapplication of rigid elements. An example might be an engine connecconsidered rigid if its deflections have little effect on the frame.considered rigid if its deflections have little effect on the frame.

Rigid elements are also appropriate when modeling a mechanism,Rigid elements are also appropriate when modeling a mechanism,are placed at the same location and only coupled degrees of freedomare placed at the same location and only coupled degrees of freedomrigid element is a very important application of rigid element. Seerigid element is a very important application of rigid element. See

Hinge using zero length rigidHinge using zero length rigid

Webinar On RBE And MPC ElementsWebinar On RBE And MPC Elements

We recorded a webinar about how to use RBE and MPC elements. LiWe recorded a webinar about how to use RBE and MPC elements. LiIn addition, tips are given on how to avoid such errors.In addition, tips are given on how to avoid such errors.

(http://www.altairhyperworks.de/ResLibDownload.aspx?file_id=854&(http://www.altairhyperworks.de/ResLibDownload.aspx?file_id=854&service=All&category=All&order_by=date_creaservice=All&category=All&order_by=date_crea

1616


strongly coupled to the structure, it is probably overly stiffening thestructural elements would be preferred. Using rigid elements for fastenersstructural elements would be preferred. Using rigid elements for fasteners

beam for modeling a bolt not only gives a better representation butbeam for modeling a bolt not only gives a better representation but

coupled to, the structure under investigation, it is a candidate forcoupled to, the structure under investigation, it is a candidate forconnected through soft mounts to a frame. The engine might beconnected through soft mounts to a frame. The engine might be

mechanism, such as a hinge as part of a structure. In this case, two nodesmechanism, such as a hinge as part of a structure. In this case, two nodesfreedom are constrained together. In fact, this type of zero lengthfreedom are constrained together. In fact, this type of zero length

See the following figure depicting a hinge.See the following figure depicting a hinge.

length rigid elementslength rigid elements

Likely errors and their impact on modeling results are discussed. Likely errors and their impact on modeling results are discussed.

wnload.aspx?file_id=854&keywords=rbe&industry=All&product_wnload.aspx?file_id=854&keywords=rbe&industry=All&product_e_created&order_by_da=desc&resource_id=1916)e_created&order_by_da=desc&resource_id=1916)

1616


FastenersFastenersFasteners

Modeling of fasteners and welds using the finite element methodModeling of fasteners and welds using the finite element methodcomplex and often exhibit nonlinear behavior. The best approach tcomplex and often exhibit nonlinear behavior. The best approach tversus poor attachment. An absolute evaluation of fastener peversus poor attachment. An absolute evaluation of fastener pemethods. The following section will discuss some of the methodsmethods. The following section will discuss some of the methods

Spot WeldsSpot Welds

Spot weld connections are normally used between relatively thinSpot weld connections are normally used between relatively thinRigid elements or common nodes (for very thin members) are usedRigid elements or common nodes (for very thin members) are usedrigid element is normal to the surfaces it attaches. It should be norigid element is normal to the surfaces it attaches. It should be nowill allow the welds to virtually spin in flat areas of the structurewill allow the welds to virtually spin in flat areas of the structurecodes will usually ignore nodes attached to rigid elements, so constraintscodes will usually ignore nodes attached to rigid elements, so constraintswill be added in the normal direction of the element. This approachwill be added in the normal direction of the element. This approach

If beam elements are used instead of rigid elements, the singularitiesIf beam elements are used instead of rigid elements, the singularitiesbenefit of using beams for spot welds involves the ease of recobenefit of using beams for spot welds involves the ease of recodiameter of the spot weld and parent material properties woulddiameter of the spot weld and parent material properties wouldillustrates a typical modeling scenario for spot welds.illustrates a typical modeling scenario for spot welds.

Spot Weld Modeling

There should be at least one node on the parts between spot wThere should be at least one node on the parts between spot wjoined. If there are no degrees of freedom between neighboringjoined. If there are no degrees of freedom between neighboringsimulated. The forces recovered in this type of model can be comsimulated. The forces recovered in this type of model can be comand placement of welds will be adequate.and placement of welds will be adequate.

Fillet Weld

Fillet welds normally occur between heavy gage plates orFillet welds normally occur between heavy gage plates orconnecting shell elements if the parts are also modeled withconnecting shell elements if the parts are also modeled withaspect ratios. The stresses will not be accurate but will giveaspect ratios. The stresses will not be accurate but will giveweld is too complex to capture in a finite element model. Theweld is too complex to capture in a finite element model. Theup using notch sensitivity tables for the material in questionup using notch sensitivity tables for the material in question

1717


is an arguable procedure. The details around a fastened joint areis an arguable procedure. The details around a fastened joint areto adopt is one in which the techniques give an indication of goodto adopt is one in which the techniques give an indication of good

performance would be extremely difficult using current standardperformance would be extremely difficult using current standardused in modeling for a variety of connections.used in modeling for a variety of connections.

sheet metal components such as those used in car body design.sheet metal components such as those used in car body design.used to model spot welds. Care must be taken to ensure that eachused to model spot welds. Care must be taken to ensure that each

oted that the lack of drilling degrees of freedom in some solversoted that the lack of drilling degrees of freedom in some solversture causing matrix singularities. Automatic singularity eliminationture causing matrix singularities. Automatic singularity elimination

constraints will need to be added manually or a soft rotational springconstraints will need to be added manually or a soft rotational springoach is sometimes referred to as fictitious drilling stiffness option.oach is sometimes referred to as fictitious drilling stiffness option.

singularities will be found and automatically eliminated. A secondarysingularities will be found and automatically eliminated. A secondaryovering forces in the welds. A round beam with the approximateovering forces in the welds. A round beam with the approximate

uld work well as long as it had a non-zero length. The figure belowuld work well as long as it had a non-zero length. The figure below

eld Modeling

weld nodes to allow relative motion between the two parts to beweld nodes to allow relative motion between the two parts to beneighboring spot welds, then there is, in effect, a continuous seam weldneighboring spot welds, then there is, in effect, a continuous seam weld

mpared to test data on similar welds to determine if the numbermpared to test data on similar welds to determine if the number

or stampings. These welds can be effectively modeled byor stampings. These welds can be effectively modeled bywith shell elements. Care must be taken to avoid excessivewith shell elements. Care must be taken to avoid excessive

relative nominal values. The irregular geometry of the actualrelative nominal values. The irregular geometry of the actualThe nominal values can be correlated to test results or scaledThe nominal values can be correlated to test results or scaled

uestion. The following figure depicts this method of modeling.uestion. The following figure depicts this method of modeling.

1717


Fillet Weld ModelingFillet Weld Modeling

It is also common to use rigid elements similar to spot welds for fillIt is also common to use rigid elements similar to spot welds for fill

Rivets And BoltsRivets And Bolts

To model riveted or bolted joints with a linear code, we must assumeTo model riveted or bolted joints with a linear code, we must assumeother or that they will always be able to slip. In most cases, it isother or that they will always be able to slip. In most cases, it isthe joined surfaces is allowed to slip. This assumption requiresthe joined surfaces is allowed to slip. This assumption requires(conservative assumption), but that there are no prying forces(conservative assumption), but that there are no prying forcesdesign joints so that there is no slipping under the peak loads. Thisdesign joints so that there is no slipping under the peak loads. Thiscomparing it to the intended installation tension multiplied by acomparing it to the intended installation tension multiplied by athe application of basic engineering principals and the answers arethe application of basic engineering principals and the answers are

Pre-Loaded JointsPre-Loaded Joints

To properly account for the pre-load in a joint, the stiffness of theTo properly account for the pre-load in a joint, the stiffness of thewill normally require the construction of a detailed solid model ofwill normally require the construction of a detailed solid model ofbeams or also be represented with a full solid model. The pre-loadbeams or also be represented with a full solid model. The pre-loadapplication of equal and opposite forces on a segment of the fasapplication of equal and opposite forces on a segment of the fasachieve the desired pre-load after an initial run determines the jointachieve the desired pre-load after an initial run determines the jointmerited because slipping during bolt up and under load, and pryingmerited because slipping during bolt up and under load, and pryingstudied a nonlinear code would probably be required. Today therestudied a nonlinear code would probably be required. Today thereas a force or a prescribed displacement.as a force or a prescribed displacement.

8. 1D Element Mesh Creation In HyperMesh8. 1D Element Mesh Creation In HyperMesh

1D elements perform a critical function in Finite Element Analysis1D elements perform a critical function in Finite Element Analysismeshes, distribute loads and in general provide a quick and easymeshes, distribute loads and in general provide a quick and easy

There are numerous types of 1D elements ranging from infinitelyThere are numerous types of 1D elements ranging from infinitelythat can be stressed.that can be stressed.

HyperMesh currently supports bar2, bar3, rigid links, rbe3, plot, rigid,HyperMesh currently supports bar2, bar3, rigid links, rbe3, plot, rigid,The following is a list of the different element types and their purposeThe following is a list of the different element types and their purpose

bar2: supports complex beams bar2: supports complex beams

bar3: supports complex beams (bar3 elements contain bar3: supports complex beams (bar3 elements contain

gap: supports gap elements gap: supports gap elements

joint: supports kinematic joint definitions supplied with joint: supports kinematic joint definitions supplied with

1818


eld Modelingeld Modeling

fillet welds.fillet welds.

assume that the joined surfaces either will never slip relative to eachassume that the joined surfaces either will never slip relative to eachis assumed that the heads do not slip but the interface betweenis assumed that the heads do not slip but the interface betweenthat the fastener carry the full shear load between the surfacesthat the fastener carry the full shear load between the surfacesin the joint (usually non-conservative assumption). It is best toin the joint (usually non-conservative assumption). It is best toThis can be checked by looking at the shear force resultant andThis can be checked by looking at the shear force resultant andconservative (lowest) friction coefficient. These methods requireconservative (lowest) friction coefficient. These methods require

are not printed in the finite element output.are not printed in the finite element output.

the joint in the direction of the fastener axis must be included. Thisthe joint in the direction of the fastener axis must be included. Thisof at least the joint area. The fastener can still be modeled withof at least the joint area. The fastener can still be modeled with

load can then be applied by a thermal shrink of the fastener or byload can then be applied by a thermal shrink of the fastener or byastener. The thermal or opposite forces will need to be adjusted toastener. The thermal or opposite forces will need to be adjusted tojoint to fastener stiffness ratio. Usually this amount of detail is notjoint to fastener stiffness ratio. Usually this amount of detail is notying forces are usually ignored. If the full details of a joint are to beying forces are usually ignored. If the full details of a joint are to be

there are solvers that permit automatically apply the pre-tension loadthere are solvers that permit automatically apply the pre-tension load

HyperMeshHyperMesh

Analysis as they can be used to connect nodes together, attach dissimilarAnalysis as they can be used to connect nodes together, attach dissimilarway to attach things together.way to attach things together.

ely rigid simple connections to complex cross sectioned elementsely rigid simple connections to complex cross sectioned elements

rigid, rod, spring, weld, gap, and joint one dimensional elements.rigid, rod, spring, weld, gap, and joint one dimensional elements.pose.pose.

contain a third node designed to support second order beams)contain a third node designed to support second order beams)

with safety analysis codeswith safety analysis codes

1818


plot: supports display type elements

rbe3: supports NASTRAN RBE3 elements rbe3: supports NASTRAN RBE3 elements

rigid link: supports rigid elements with multiple nodes rigid link: supports rigid elements with multiple nodes

rigid: supports rigid elements rigid: supports rigid elements

rod: supports simple beams rod: supports simple beams

spring: supports springs or damper spring: supports springs or damper

weld: supports weld elements

HyperMesh also uses an entity called a connector which definesHyperMesh also uses an entity called a connector which definesfastened to each other. These connectors are used to create or realizefastened to each other. These connectors are used to create or realize

Some Modeling Hints Regarding 1D Elements WithSome Modeling Hints Regarding 1D Elements With

While working with 1D elements, sooner or later you will come acWhile working with 1D elements, sooner or later you will come ac

CROD CROD

CBAR CBAR

CBEAM CBEAM

In brief, CROD elements support (allow) tension and compressionIn brief, CROD elements support (allow) tension and compressionCBAR element is a kind of simplified CBEAM element and shouldCBAR element is a kind of simplified CBEAM element and shouldproperties is constant and symmetrical. The CBAR element requiresproperties is constant and symmetrical. The CBAR element requiresrequirement, CBAR elements are not useful for modeling structuresrequirement, CBAR elements are not useful for modeling structuresis not present in CBEAM elements.is not present in CBEAM elements.

Thus, CBEAM elements are used to model more complicated geomThus, CBEAM elements are used to model more complicated geomelements the neutral axis and shear center are noncoincident. Oelements the neutral axis and shear center are noncoincident. Obeam theory.beam theory.

Furthermore, we recommend the beginner to use the predefinedFurthermore, we recommend the beginner to use the predefined(provided you are using OptiStruct as your finite element solver)(provided you are using OptiStruct as your finite element solver)the OptiStruct library is that the stress recovery points which arethe OptiStruct library is that the stress recovery points which areOf course, in addition to the cross-sections depicted further beloOf course, in addition to the cross-sections depicted further belo1D cross-sections with HyperMesh.1D cross-sections with HyperMesh.

CROD ElementsCROD Elements

In the following the general modeling building process based onIn the following the general modeling building process based onalso be applied to CBAR and CBEAM elements, too.also be applied to CBAR and CBEAM elements, too.

This element understands tension and compression loads (axialThis element understands tension and compression loads (axialforces) only. In other words, the nodes of a CROD element onlyforces) only. In other words, the nodes of a CROD element onlytranslation degrees of freedom (still this element has a torsionaltranslation degrees of freedom (still this element has a torsionalstiffness).stiffness).

1919


nodesnodes

defines how different components or assemblies in the model aredefines how different components or assemblies in the model arerealize one-dimensional elements in the model.realize one-dimensional elements in the model.

With HyperMeshWith HyperMesh

across the following element types:across the following element types:

pression only, whereas CBARS and CBEAMS allow bending as well. Apression only, whereas CBARS and CBEAMS allow bending as well. Ashould be used whenever the cross-section of the structure and itsshould be used whenever the cross-section of the structure and its

uires that its shear center and the neutral axis coincide. Due to thisuires that its shear center and the neutral axis coincide. Due to thisstructures that may warp, such as open channel-sections. This limitationstructures that may warp, such as open channel-sections. This limitation

geometries with varying cross sections. Important to note, in CBEAMgeometries with varying cross sections. Important to note, in CBEAMOverall, this type of element demands a deeper understanding ofOverall, this type of element demands a deeper understanding of

predefined 1D element cross-sections available with the OptiStruct librarypredefined 1D element cross-sections available with the OptiStruct libraryer). Another benefit of employing the cross-sections available wither). Another benefit of employing the cross-sections available with

needed for postprocessing stress results are predefined already.needed for postprocessing stress results are predefined already.ow (OptiStruct library) you can create your very own and generalow (OptiStruct library) you can create your very own and general

on CROD elements is documented. Most of working the steps canon CROD elements is documented. Most of working the steps can

(axial(axialhavehave

sionalsional

1919


How To Model CRODs

As we are going to use the FEM program OptiStruct, make sure that As we are going to use the FEM program OptiStruct, make sure that

You may start by building CROD elements manually (with/withoutYou may start by building CROD elements manually (with/withoutpanel is found in the menu bar by selecting Mesh > Create > 1D Elements >panel is found in the menu bar by selecting Mesh > Create > 1D Elements >

Make sure that the element type to be created is set to CROD. InMake sure that the element type to be created is set to CROD. Inbounding nodes.bounding nodes.

Other more general options to create many CRODs in a single step Other more general options to create many CRODs in a single step panel. This panel can be found through the menu bar by selecting Mesh >panel. This panel can be found through the menu bar by selecting Mesh >

Again make sure and specify the element configuration to be rod.Again make sure and specify the element configuration to be rod.

It may happen that the newly created elements are displayed as CWELD It may happen that the newly created elements are displayed as CWELD

If this is the case, then the element type must be updated. ThisIf this is the case, then the element type must be updated. Thisthe menu bar by selecting Mesh > Assign > Element Type.the menu bar by selecting Mesh > Assign > Element Type.

The element types are updated by setting rod= to CROD, then selecting the corresponding The element types are updated by setting rod= to CROD, then selecting the corresponding

Up until now the element specific information, such as cross-sectionUp until now the element specific information, such as cross-sectionas a line:as a line:

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that the corresponding User Profile is activated.that the corresponding User Profile is activated.

(with/without any property definition) by making use of the Rods panel. This(with/without any property definition) by making use of the Rods panel. ThisElements > Rods.Elements > Rods.

In the simplest case, a CROD element is created by selecting its In the simplest case, a CROD element is created by selecting its

ep exist of course. For instance, by making use of the Line Mesh ep exist of course. For instance, by making use of the Line Mesh selecting Mesh > Create > Line Mesh.selecting Mesh > Create > Line Mesh.

CWELD elements (not what we want).CWELD elements (not what we want).

This is done using the Element Type panel, which is found throughThis is done using the Element Type panel, which is found through

then selecting the corresponding elements and the selecting update.then selecting the corresponding elements and the selecting update.

ction data, is missing. The CROD elements are thus simply displayedction data, is missing. The CROD elements are thus simply displayed

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In order to define the elements cross-sectional properties, we recommendIn order to define the elements cross-sectional properties, we recommendselecting Properties > HyperBeam (also accessible from the 1D page >selecting Properties > HyperBeam (also accessible from the 1D page >

In the HyperBeam panel, select the standard section subpanel:In the HyperBeam panel, select the standard section subpanel:

As stated before, we will make use of the readily available crossAs stated before, we will make use of the readily available crosssection library, OPTISTRUCT is activated. From the list of availablesection library, OPTISTRUCT is activated. From the list of availableAbove it is set to Rod (Box, H at etc. are also valid of course). NAbove it is set to Rod (Box, H at etc. are also valid of course). Nabout the 1D element type!

Clicking the create button then starts up HyperBeam which allowsClicking the create button then starts up HyperBeam which allows

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recommend using HyperBeam which is found in the menu bar byrecommend using HyperBeam which is found in the menu bar bypage > HyperBeam).page > HyperBeam).

oss sections in OptiStruct. Therefore, make sure that for standardoss sections in OptiStruct. Therefore, make sure that for standardcross-sections (standard section type:), chose the one of interest.cross-sections (standard section type:), chose the one of interest.

Note, the cross section is named Rod it doesnt say anythingNote, the cross section is named Rod it doesnt say anything

ws to interactively changing the radius of the rod cross-section.ws to interactively changing the radius of the rod cross-section.

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Note, that all the relevant cross-sectional data such as area, momentsbar of HyperBeam. This information is stored as a beamsection collecbar of HyperBeam. This information is stored as a beamsection collecthe beamsection will be referenced inside the mandatory propertythe beamsection will be referenced inside the mandatory property

In order to switch back to the HyperMesh GUI, just activate the iconIn order to switch back to the HyperMesh GUI, just activate the iconBrowser).Browser).

Before we proceed creating a property collector, we need to defineBefore we proceed creating a property collector, we need to defineMaterial). In many cases, especially while getting started with FEM,Material). In many cases, especially while getting started with FEM,elastic isotropic temperature independent material behavior). Theelastic isotropic temperature independent material behavior). TheModulus, Poissons ratio and density.Modulus, Poissons ratio and density.

Then, in the next step the property collector for the CROD elementsThen, in the next step the property collector for the CROD elements> Property). In the pop-up window of the property collector, the> Property). In the pop-up window of the property collector, thepreviously defined cross-section, namely the beamsection rod andpreviously defined cross-section, namely the beamsection rod and

Note, if you are using CBEAM or CBAR elements, the property CarNote, if you are using CBEAM or CBAR elements, the property Carthe Library).Activate the Material tab next to Property, and assignthe Library).Activate the Material tab next to Property, and assign

Eventually, the property collector is created. Its card image (solver Eventually, the property collector is created. Its card image (solver

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moments of inertia are automatically calculated and listed in the sidecollector (here named rod; listed in the side bar on the left). Latercollector (here named rod; listed in the side bar on the left). Laterty collector. Complicated? Not really, as you will see ty collector. Complicated? Not really, as you will see

icon in the upper left menu bar (beneath Model in the Modelicon in the upper left menu bar (beneath Model in the Model

define a material (right mouse click in the Model Browser > Create >define a material (right mouse click in the Model Browser > Create >FEM, the material card image will be MAT1 (MAT1 describes linearFEM, the material card image will be MAT1 (MAT1 describes linear

The material is named steel using standard values for YoungsThe material is named steel using standard values for Youngs

elements is created (right mouse click in the Model Browser > Createelements is created (right mouse click in the Model Browser > CreateCard Image is chosen as PROD. In here we also reference theCard Image is chosen as PROD. In here we also reference the

and the material steel.and the material steel.

rd Image would either be PBEAML or PBARL (L = as we are using rd Image would either be PBEAML or PBARL (L = as we are using the previously created material steel to this property collector.the previously created material steel to this property collector.

er syntax) then reads aser syntax) then reads as

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Where PID = property collector ID, MID = material collector, A is the area and J Where PID = property collector ID, MID = material collector, A is the area and J

In the general analysis work flow (not restricted to 1D elements),In the general analysis work flow (not restricted to 1D elements),interest.interest.

To visually check the attributes of the 1D elements, make sure thatTo visually check the attributes of the 1D elements, make sure thatis activated. Then the CROD element will be displayed as:is activated. Then the CROD element will be displayed as:

CBAR ElementsCBAR Elements

Simple beams with constant properties (symmetrical andSimple beams with constant properties (symmetrical andCBEAM elements. However, what you need to recall is that iCBEAM elements. However, what you need to recall is that iis centered at the shear center of the cross-section (shear centeris centered at the shear center of the cross-section (shear centerand the shear center is not accounted for. As a consequence, CBARand the shear center is not accounted for. As a consequence, CBARbe the case with open channel sections. This is because the cross-sebe the case with open channel sections. This is because the cross-sewhenever you are modeling open or nonsymmetrical sectiowhenever you are modeling open or nonsymmetrical sectiogoing to use.going to use.

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the area and J is the torsional constant.the area and J is the torsional constant.

the property collector must now be assigned to the elements ofthe property collector must now be assigned to the elements of

that Shaded Elements and 3D Element Representationthat Shaded Elements and 3D Element Representation

d constant cross-section), may be modeled with CBAR ord constant cross-section), may be modeled with CBAR orin CBAR elements, the origin of the element coordinate systemin CBAR elements, the origin of the element coordinate system

er and neutral axis coincide). Any offsets between the neutral axiser and neutral axis coincide). Any offsets between the neutral axisCBAR elements are not useful for modeling beams that warp, as mayCBAR elements are not useful for modeling beams that warp, as may

sections of CBAR elements remain plane. In other words,sections of CBAR elements remain plane. In other words,sections, be especially cautious with the element type you aresections, be especially cautious with the element type you are

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The followin,lcross-&eetlons areavailable:The followin,lcross-&eetlons areavailable:

r er e

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p e C H A Np e C H A N

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or ...... or ...... or ......I. . CHAH2. . CHAH2

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The cross-sectional properties, shear flexibility factors, and stressThe cross-sectional properties, shear flexibility factors, and stressTYPE and DIMi as shown below.TYPE and DIMi as shown below.

As the CBAR and CBEAM elements also account for bending, the orientationAs the CBAR and CBEAM elements also account for bending, the orientationimportant (as it automatically defines the location of the pre-defined stress important (as it automatically defines the location of the pre-defined stress

CBARCBAR

Meshing (building) of CBAR or CBEAM elements is just as simple as Meshing (building) of CBAR or CBEAM elements is just as simple as

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stress recovery points (C, D, E, and F) are then computed using thestress recovery points (C, D, E, and F) are then computed using the

orientation of the element and thus of its cross-section becomesorientation of the element and thus of its cross-section becomesdefined stress recovery points; see discussion below).defined stress recovery points; see discussion below).

CBEAMCBEAM

as with the creation of CROD elements (e.g. Line Mesh panel)as with the creation of CROD elements (e.g. Line Mesh panel)

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This time the element config is set to bar2 (if you miss this step, you This time the element config is set to bar2 (if you miss this step, you

In order to define the cross-section, we again make use of HyperBeam In order to define the cross-section, we again make use of HyperBeam

1. Building (manually) CBAR / CBEAM elements:1. Building (manually) CBAR / CBEAM elements:

Start the meshing process with the Bars panel (Mesh > Create > 1DStart the meshing process with the Bars panel (Mesh > Create > 1D

The element type is set to CBAR.The element type is set to CBAR.

2. Specify cross sectional properties (with HyperBeam)2. Specify cross sectional properties (with HyperBeam)

The standard section library OPTISTRUCT is used.The standard section library OPTISTRUCT is used.

Again, the cross section properties can be interactively changedAgain, the cross section properties can be interactively changedbeamsection collector is named rodbeamsection collector is named rod

3. In the property collector, the Card Image PBARL is chosen i.e. L3. In the property collector, the Card Image PBARL is chosen i.e. L

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you still can update the element configuration later). you still can update the element configuration later).

HyperBeam and the OptiStruct library.HyperBeam and the OptiStruct library.

1D Elements > Bars)1D Elements > Bars)

in HyperBeam. The beamsection is named rod_library and the in HyperBeam. The beamsection is named rod_library and the

L indicates that we reference the cross section from the libraryL indicates that we reference the cross section from the library

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Recall that the material named steel was created before ...

The Card Image of the property collector now reads as:The Card Image of the property collector now reads as:

Please compare this card image with the one belonging to PRODPlease compare this card image with the one belonging to PROD(CStype) is ROD as defined in HyperBeam, DIM1 corresponds to(CStype) is ROD as defined in HyperBeam, DIM1 corresponds tois the ID of the material collector, and beamsec references the beis the ID of the material collector, and beamsec references the be

4. As mentioned earlier, care must now be taken regarding the orientation4. As mentioned earlier, care must now be taken regarding the orientationlocation of the stress recovery points).location of the stress recovery points).

The local element x-direction is given by its first (A) and secondThe local element x-direction is given by its first (A) and secondof the element, is then defined by a vector v and the local x-axisof the element, is then defined by a vector v and the local x-axissystem, however. Note, more information is also available in the helpsystem, however. Note, more information is also available in the help> CBEAM or CBAR). To review and update the elements orientation> CBEAM or CBAR). To review and update the elements orientationBars).Bars).

In this example, the v vector parallels the global y-direction i.e. yIn this example, the v vector parallels the global y-direction i.e. ythe upper side of the rod.the upper side of the rod.

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OD (from the example above). Here, the type of the cross-sectionOD (from the example above). Here, the type of the cross-sectionthe radius of the rod, PID is the ID of the property collector, MIDthe radius of the rod, PID is the ID of the property collector, MID

before created cross-section.before created cross-section.

orientation of the cross-section (orientation of the element and thus orientation of the cross-section (orientation of the element and thus

node (B). Its overall orientation, i.e. the local xy plane (Plane 1)node (B). Its overall orientation, i.e. the local xy plane (Plane 1)axis. The v vector is defined with respect to the global coordinateaxis. The v vector is defined with respect to the global coordinatehelp documentation (\help\hwsolvers\hwsolvers.htm, then Indexhelp documentation (\help\hwsolvers\hwsolvers.htm, then Index

orientation (if needed) use the Bars panel (Mesh > Edit >1D Elements >orientation (if needed) use the Bars panel (Mesh > Edit >1D Elements >

ycomp=1. Consequently, the stress recovery point C is located atycomp=1. Consequently, the stress recovery point C is located at

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A note regarding the stress recovery points: Its location essentially

M = Moment, c distance from neutral fiber (i.e. location of the stressM = Moment, c distance from neutral fiber (i.e. location of the stresscalculates the stresses in the rod or on its surface. Actually, thecalculates the stresses in the rod or on its surface. Actually, thelocations to give the maximum stress for the cross-section.locations to give the maximum stress for the cross-section.

Visualization Of Element StressesVisualization Of Element Stresses

Evaluating element stresses opens up a very different aspect. InEvaluating element stresses opens up a very different aspect. Inwith respect to the stress recovery points C, D, E, F (see explanationswith respect to the stress recovery points C, D, E, F (see explanationsCBAR/CBEAM element.CBAR/CBEAM element.

In HyperView, node A or node B and the stress recovery point of inIn HyperView, node A or node B and the stress recovery point of inrecovery point C.recovery point C.

What Is The Meaning Of Pins a, Pins b, Offset a And OffsetWhat Is The Meaning Of Pins a, Pins b, Offset a And Offset

So far the Bars panel was used to manually create CBAR (or CBEAM)So far the Bars panel was used to manually create CBAR (or CBEAM)

pins a and pins bpins a and pins b

You may pin certain translations or rotations (degrees of freedom)You may pin certain translations or rotations (degrees of freedom)supposed to transmit all translations and rotations, then pinssupposed to transmit all translations and rotations, then pinsof freedom must be eliminated, then the corresponding pin at nodeof freedom must be eliminated, then the corresponding pin at node

As an example, rotation with respect to the x-axis (dof 4) should nAs an example, rotation with respect to the x-axis (dof 4) should nb=4. This still enables the 1D element to rotate around its x-axis,b=4. This still enables the 1D element to rotate around its x-axis,element attached to node B.element attached to node B.

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entially defines the c in the equation = Mc/I :

stress recovery point), I = Moment of Inertia. Based on c, Optistructstress recovery point), I = Moment of Inertia. Based on c, Optistructthe stress recovery points are automatically calculated at specificthe stress recovery points are automatically calculated at specific

In the case of CBAR (CBEAM) elements, stresses are determinedIn the case of CBAR (CBEAM) elements, stresses are determinedxplanations above). This stress value is then used to contour the entirexplanations above). This stress value is then used to contour the entire

interest must be specified, e.g. longitudinal stress at node A andinterest must be specified, e.g. longitudinal stress at node A and

set b?set b?

CBEAM) elements.CBEAM) elements.

freedom) of the 1- D element. In other words, if the 1D element isfreedom) of the 1- D element. In other words, if the 1D element isa and pins b must be set to 0 (zero). In case a certain degreea and pins b must be set to 0 (zero). In case a certain degree

node a or node b (eventually even both) needs to be pinned.node a or node b (eventually even both) needs to be pinned.

not be transferred from node b to the attached element, then pinsnot be transferred from node b to the attached element, then pinsBUT as requested, this rotation is not transmitted to the neighborBUT as requested, this rotation is not transmitted to the neighbor

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offset a, offsetbb

Because bar elements are one-dimensional representations ofBecause bar elements are one-dimensional representations ofbetween the real-world edge of a bar and its center axis. For exambetween the real-world edge of a bar and its center axis. For examdetermine where the edge of the bar is attached to another elementdetermine where the edge of the bar is attached to another elementrelation to the connection points.relation to the connection points.

Note, in this model, the 1D elements Note, in this model, the 1D elements

In the model shown next, the 1D elements form a straight line (elementsIn the model shown next, the 1D elements form a straight line (elements

Element Representation ).Element Representation ).

However, internally the elements nodes a and b are offset in orHowever, internally the elements nodes a and b are offset in or

visualized by activating the 3D Element Representation .visualized by activating the 3D Element Representation .

Reviewing the elements provides insight on how the offset is defined:Reviewing the elements provides insight on how the offset is defined:

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three-dimensional objects, these values represent the distancethree-dimensional objects, these values represent the distancemple, if the bar was cylindrical in shape, node a and node b wouldmple, if the bar was cylindrical in shape, node a and node b would

element. The offsets would identify where the bars central axis lay inelement. The offsets would identify where the bars central axis lay in

elements physically follow the given geometry.elements physically follow the given geometry.

(elements are displayed with Shrink Mode and Traditional(elements are displayed with Shrink Mode and Traditional

rder to account for the true geometry. This offset can only berder to account for the true geometry. This offset can only be

defined:defined:

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The offset of node B of the selected element is 4 mm (hence by= 4).The offset of node B of the selected element is 4 mm (hence by= 4).can be modeled without physically translating the elements.

In the image above the displacement contour plot is shown. In thisIn the image above the displacement contour plot is shown. In thisfor the true geometry.for the true geometry.

How To Apply Pressure On 1D Elements?How To Apply Pressure On 1D Elements?

Pressure on CBAR or CBEAM elements may be applied as load typeP

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