finite element simulations with ansys workbench 13.pdf

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Chapter 1 Introduction 1

Chapter 1Introduction

1.1 Case Study: Pneumatically Actuated PDMS Fingers

1.2 Structural Mechanics: A Quick Review

1.3 Finite Element Methods: A Conceptual Introduction

1.4 Failure Criteria of Materials

1.5 Review



Chapter 1 Introduction Section 1.1 Case Study: Pneumatically Actuated PDMS Fingers 2

Section 1.1Case Study: Pneumatically Actuated

PDMS Fingers

[1] The pneumatic

fingers are part of a

surgical parallel robot

system remotely

controlled by a surgeon

through the Internet.

[2] A single

finger is studied in

this case.

Problem Description




[6] Undeformedshape.

[5] As air pressureapplies, the finger bends

downward.

0

1

2

3

4

5

0 0.2 0.4 0.6 0.8 1.0

S t r e s s ( M P a )

Strain (Dimensionless)

[4] The strain-stresscurve of the PDMSelastomer used in

this case.

[3]Geometric

model.




Static Structural Simulations

[1] Preparematerial

properties.

[2] Creategeometric model.

[3] Generate finiteelement mesh.

[4] Set up loadsand supports.

[5] Solve themodel.

[6] View theresults.




[7] Displacements.

[8] Strains.




Buckling and Stress-Stiffening

[2] The upper surface wouldundergo compressive stress.

It in turn reduces the bendingstiffness.

[1] If we applyan upward

force here...

• Stress-stiffening : bending stiffness increases with increasing axial tensile stress, e.g., guitar string.

• The opposite also holds: bending stiffness decreases with increasing axial compressive stress.

• Buckling : phenomenon when bending stiffness reduces to zero, i.e., the structure is unstable.

Usually occurs in slender columns, thin walls, etc.

• Purpose of a buckling analysis is to find buckling loads and buckling modes.




Dynamic Simulations

• When the bodies move and

deform very fast, inertia effect

and damping effect must beconsidered.

• When including these

dynamic effects, it is called a

dynamic simulation.




Modal Analysis

• A special case of dynamic

simulations is the simulation of free

vibrations, the vibrations of astructure without any loading.

• It is called a modal analysis.

• Purpose of a modal analysis is to

find natural frequencies and mode

shapes.




Structural Nonlinearities

-30

-25

-20

-15

-10

-5

0

0 40 80 120 160 200

D e f l e c t i o n

( m m )

Pressure (kPa)

[1] Solution of thenonlinear simulationof the PDMS finger.

[2] Solution of thelinear simulation pfthe PDMS finger.

• Linear simulations assume that

the response is linearly

proportional to the loading.

• When the solution deviates from

the reality, a nonlinear simulation

is needed.

• Structural nonlinearities come

from large deformation, topology

changes, nonlinear stress-strain

relationship, etc.



Chapter 1 Introduction Section 1.2 Structural Mechanics: A Quick Review 10

Section 1.2Structural Mechanics: A Quick Review

• Engineering simulation: finding the responses of a problem domain subject to

environmental conditions.

• Structural simulation: finding the responses of bodies subject to

environmental conditions.

• The bodies are described by geometries and materials.

• Environment conditions include support and loading conditions.

• Responses can be described by displacements, strains, and stresses.




Displacements

X

Y

[1] The body beforedeformation.

[2] The body afterdeformation.

[4] After thedeformation, the

particle moves to anew position.

[5] The displacementvector {u} of the particle isformed by connecting thepositions before and after

the deformation.

[3] An arbitrary particleof position ( X, Y, Z ), before

the deformation.

u{ } = u X

uY

u Z { }




Stresses

X Y

Z

!

Y

! YX

! YZ

! X

! XY

! XZ

!

Z

! ZX

! ZY

[1] The referenceframe XYZ .

[2] This face iscalled X-face, since the X -direction is normal

to this face.

[4] The X -componentof the stress on X -face.

[3] This face is callednegative X -face.

[5] The Y -

component of thestress on X -face.

[6] The Z -componentof the stress on X -face.

! { } =! X

" XY

" XZ

" YX

! Y

" YZ

" ZX

" ZY

! Z

#

$%%

&%%

'

(%%

)%%

! XY

= ! YX

, ! YZ

= ! ZY

, ! XZ

= ! ZX

! { } = ! X

! Y

! Z

" XY

" YZ

" ZX { }




Strains

X

Y

! A !B

!C

!!B

!!C

A B

C

!!!C

!!!B

D

[2] Original

configuration ABC .

[3] Afterdeformation,

ABC moves to! A !B !C .

[4] To compare withoriginal configuration,rotate ! A !B !C to a new

configuration ! A !!B !!C .

[5] Translate ! A !!B !!C so

that ! A coincides with A.

The new configuration is A !!!B !!!C . Now C !!!C is the

amount of stretch of ABC inY -face.

[6] The vector BD describes the stretch of

ABC in X -face.

[7] And the vectorD !!!B describes the twist

of ABC in X-face.

[1] The referenceframe.

Strain on X -face =B !!!B

AB

! X

=

BD

AB, "

XY =

D ###B

AB




! { } =! X

" XY

" XZ

" YX

! Y

" YZ

" ZX

" ZY

! Z

#

$%%

&%%

'

(%%

)%%

! XY

= ! YX

, ! YZ

= ! ZY

, ! XZ

= ! ZX

! { } = ! X

! Y

! Z

" XY

" YZ

" ZX { }

• Physical meaning of strains:

• The normal strain ! X

is the

percentage of stretch of a fiber which

lies along X -direction.

• The shear strain ! XY

is the angle

change (in radian) of two fibers lying

on XY -plane and originally forming a

right angle.

• We can define other strain components

in a similar way.




Governing Equations

u{ } = u X

uY

u Z { }

! { } = ! X

! Y

! Z

" XY

" YZ

" ZX { }

! { }

= ! X ! Y ! Z " XY " YZ " ZX

{ }Totally 15 quantities

• Equilibrium Equations (3 Equations)

• Strain-Displacement Relations (6 Equations)

• Stress-Strain Relations (6 Equations)




Stress-Strain Relations: Hooke's Law

! X

=

" X

E# $

" Y

E# $

" Z

E

! Y =

" Y

E# $

" Z

E# $

" X

E

! Z

=

" Z

E

# $ " X

E

# $ "

Y

E

% XY

=

& XY

G , %

YZ =

& YZ

G , %

ZX =

& ZX

G

G =E

2(1+! )

• For isotropic , linearly elastic materials,

Young's modulus (E) and Poisson's ratio (! )

can be used to fully describe the stress-

strain relations.

• The Hooke's law is called a material

model .

• The Young's modulus and the Poisson's

ratio are called the material parameters

of the material model.




! X =

" X

E#$

" Y

E#$

" Z

E+% &T

! Y =

" Y

E#$

" Z

E#$

" X

E+% &T

! Z =

" Z

E

#$ " X

E

#$ "

Y

E

+% &T

' XY

=

( XY

G , '

YZ =

( YZ

G , '

ZX =

( ZX

G

• If temperature changes (thermal loads)

are involved, the coefficient of thermal

expansion, (CTE, ! ) must be included.

• If inertia forces (e.g., dynamicsimulations) are involved, the mass

density must be included.



Chapter 1 Introduction Section 1.3 Finite Element Methods: A Conceptual Introduction 18

Section 1.3Finite Element Methods: A Conceptual

Introduction

• A basic idea of finite element methods is to divide the structural body into small and

geometrically simple bodies, called elements, so that equilibrium equations of each

element can be written, and all the equilibrium equations are solved simultaneously

• The elements are assumed to be connected by nodes located on the elements' edges

and vertices.

Basic Ideas




In case of the pneumatic finger, thestructural body is divided into 1842

elements. The elements are connected by10856 nodes. There are 3x10856 unknown

displacement values to be solved.

•Another idea is to solve unknown

discrete values (displacements at the

nodes) rather than to solve unknown

functions (displacement fields).

• Since the displacement on each node

is a vector and has three components

(in 3D cases), the number of total

unknown quantities to be solved is

three times the number of nodes.

• The nodal displacement components

are called the degrees of freedom (DOF's) of the structure.




• In static cases, the system of equilibrium equations has following form:

K !" #$ D{ } = F { }

• The displacement vector {D} contains displacements of all degrees of

freedom.

• The force vector {F } contains forces acting on all degrees of freedom.

• The matrix [K ] is called the stiffness matrix of the structure. In a special

case when the structure is a spring, {F } as external force, and {D} as the

deformation of the spring, then [K ] is the spring constant.




Basic Procedure of Finite Element Method

1. Given the bodies' geometries, material properties, support conditions, and loading

conditions.

2. Divide the bodies into elements.

3. Establish the equilibrium equation: [K ] {D} = {F }

3.1 Construct the [K ] matrix, according to the elements' geometries and the material

properties.

3.2 Most of components in {F } can be calculated, according to the loading conditions.

3.3 Most of components in {D} are unknown. Some component, however, are known,

according to the support conditions.

3.4 The total number of unknowns in {D} and {F } should be equal to the total number

of degrees of freedom of the structure.




4. Solve the equilibrium equation. Now, the nodal displacements {d} of each element are

known.

5. For each element:

5.1 Calculate displacement fields {u}, using an interpolating method, {u} = [N] {d }. The

interpolating functions in [N] are called the shape functions.

5.2 Calculate strain fields according to the strain-displacement relations.

5.3 Calculate stress fields according to the stress-strain relations (Hooke's law).




Shape Functions

d 1

d

2

d

3

d

4

d

5

d

6

d

7

d

8

X

Y

[1] A 2D 4-nodequadrilateral element

[2] This element'snodes locate atvertices.

• Shape functions serve as interpolating

functions, allowing the calculation of

displacement fields (functions of X , Y ,

Z ) from nodal displacements (discrete

values).

u{ } = N!" #$ d { }

• For elements with nodes at vertices,

the interpolation must be linear and

thus the shape functions are linear (of

X, Y, Z ).




• For elements with nodes at vertices as well as at middles of edges, the interpolation

must be quadratic and thus the shape functions are quadratic (of X, Y, Z ).

• Elements with linear shape functions are called linear elements, first-order elements, or

lower-order elements.

• Elements with quadratic shape functions are called quadratic elements, second-order

elements, or higher-order elements.

• ANSYS Workbench supports only first-order and second-order elements.




Workbench Elements

[1] 3D 20-nodestructural solid.Each node has 3

translationaldegrees of

freedom: D X , DY ,and D Z .

[2] Triangle-based

prism.

[3] Quadrilateral-based pyramid.

[4] Tetrahedron.3D Solid Bodies




2D Solid Bodies

[5] 2D 8-nodestructural solid.Each node has 2

translationaldegrees of

freedom: D X and DY .

[6] DegeneratedTriangle.




3D Surface Bodies

[7] 3D 4-nodestructural shell.Each node has 3

translational and 3rotational degreesof freedom: D X , DY ,

D Z , R X , RY , and R Z .

[8] DegeneratedTriangle

3D Line Bodies

[9] 3D 2-Node

beam. Each node has3 translational and 3rotational degrees offreedom: D X , DY , D Z ,

R X , RY , R Z .



Chapter 1 Introduction Section 1.4 Failure Criteria of Materials 28

Section 1.4Failure Criteria of Materials

Ductile versus Brittle Materials

• A Ductile material exhibits a large amount of strain before it

fractures.

• The fracture strain of a brittle material is relatively small.

• Fracture strain is a measure of ductility.




S t r e s s

Strain

!

y

[3] Yieldpoint.

[2] Fracturepoint.

[1] Stress-straincurve for a ductile

material.

• Mild steel is a typical ductile material.

• For ductile materials, there often exists an

obvious yield point, beyond which the

deformation would be too large so that the

material is no longer reliable or functional;

the failure is accompanied by excess

deformation.

• Therefore, for these materials, we are most

concerned about whether the material

reaches the yield point ! y

.

Failure Points for Ductile Materials




[2] Fracturepoint.

[1] Stress-straincurve for a

brittle material.

Failure Points for Brittle Materials

• Cast iron and ceramics are two examples

of brittle materials.

• For brittle materials, there usually doesn'texist obvious yield point, and we are

concerned about their fracture point ! f .

S t r e s s

Strain

!

f




Failure Modes

• The fracture of brittle materials is mostly due to

tensile failure.

•The yielding of ductile materials is mostly due to shear

failure




Principal Stresses

!

X !

X

!

Y

!

Y

! XY

! XY

! XY

! XY

X

Y

[1] Stressstate.

!

(!

X ,"

XY )

(!

Y ,"

XY )

!

[2] Stress inthe basedirection.

[3] Stress in

the direction thatforms 90

o withthe basedirection.

[4] Other stresspairs could bedrawn.

[6] Point ofmaximumnormal

stress.

[7] Point ofminimum

normalstress.

[8] Point ofmaximum

shear stress.

[9] AnotherPoint of

maximum shearstress.

[5]Mohr'scircle.

• A direction in which the shear

stress vanishes is called a

principal direction.

• The corresponding normal stress

is called a principle stress.




• At any point of a 3D solid, there are three principal directions and

three principal stresses.

• The maximum normal stress is called the maximum principal stress

and denoted by ! 1.

• The minimum normal stress is called the minimum principal stress and

denoted by ! 3 .

• The medium principal stress is denoted by ! 2.

• The maximum principal stress is usually a positive value, a tension;

the minimum principal stress is often a negative value, a

compression.




Failure Criterion for Brittle Materials

• The failure of brittle materials is a tensile failure. In other words, a

brittle material fractures because its tensile stress reaches the

fracture strength ! f .

•We may state a failure criterion for brittle materials as follows: At a

certain point of a body, if the maximum principal stress reaches the

fracture strength of the material, it will fail.

• In short, a point of material fails if

!

1"!

f




Tresca Criterion for Ductile Materials

• The failure of ductile materials is a shear

failure. In other words, a ductile material yields

because its shear stress reaches the shear

strength ! y

of the material.

• We may state a failure criterion for ductile

materials as follows: At a certain point of a

body, if the maximum shear stress reaches the

shear strength of the material, it will fail.

• In short, a point of material fails if

! max

" ! y

• It is easy to show (using

Mohr's circle) that

! max

=

" 1#"

3

2

! y =

" y

2

• Thus, the material yields if

!

1"!

3 #!

y

• (! 1"! 3) is called the stress

intensity.




Von Mises Criterion for Ductile Materials

• In 1913, Richard von Mises proposed a theory for predicting the yielding of ductile

materials. The theory states that the yielding occurs when the deviatoric strain energy

density reaches a critical value, i.e.,

w d !w yd

• It can be shown that the yielding deviatoric energy in uniaxial test is

w yd =

(1+! )" y 2

3E

• And the deviatoric energy in general 3D cases is

w

d =

1+!

6E " 1#" 2( )

2

+ " 2 #" 3( )2

+ " 3 #" 1( )2$

%&'

()




•After substitution and simplification, the criterion reduces to that the yielding

occurs when

1

2 !

1"!

2( )2

+ ! 2 "!

3( )2

+ ! 3 "!

1( )2#

$%

&

'( ) !

y

• The quantity on the left-hand-side is termed von Mises stress or effective stress, and

denoted by ! e ; in ANSYS, it is also referred to as equivalent stress,

! e =

1

2 !

1"!

2( )2

+ ! 2 "!

3( )2

+ ! 3 "!

1( )2#

$%

&

'(

• The equivalent strain, or effective strain ! eis defined by

! e =

" e

E



Chapter 2 Sketching

Chapter 2

Sketching

2.1

Step-by-Step: W16x50 Beam2.2 Step-by-Step: Triangular Plate

2.3

More Details

2.4 More Exercise: M20x2.5 Threaded Bolt



Chapter 2 Sketching

Secti

Section 2.1

W16x50 Beam 7.07

Problem Description



Chapter 2 Sketching

Secti

• Start up <D

• Sketching/M

• Draw>Rec

• Draw>Poly

• Dimension

• Dimension

• Dimension

• Dimension

• Modify>Co

• Modify>Tr

• Modify>Fil

Techniques/Concepts



Chapter 2 Sketching

Secti

[1] Click: singleselection

[2] Control-click: add/remove selection

[5] Rightbox

[6] Scroll-whee

in/out.

[7] Middle-click-drag:

rotation.

Basic Mouse Operations



Chapter 2 Sketching

Sectio

Section 2.2

Triangular Plate

Problem Description



Chapter 2 Sketching

Sectio

Techniques/Concepts

• Draw>A

• Dimensi

• Modify>

• Modify>

• Constra

• Weak/St

• Weak/St

• Selection

• Single/Bo



Chapter 2 Sketching

Sectio

2D Graphics Controls

[8] <Undo>.

[2] <Zoom to Fit>.

[4] <Box Zoom>.[5] <Zoom>.[6] <Previous View>.

[7] <Next View>.[3] <Pan>.



Chapter 2 Sketching

Sec

• Pull-down Menus

and Toolbars

• Mode Tabs

•Tree Outline

• Sketching

Toolboxes

• Graphics area

Section 2.3 More De



Chapter 2 Sketching

Sec

Model Tree

• A Model tree is a tree representation of a

geometric model.

• A model tree consists of features and a part

branch.

• The parts are the only objects that will be

exported to <Mechanical>.

• The order of the objects is relevant.

<DesignModeler> renders the geometryA



Chapter 2 Sketching

Sec

[1] Currentlyactive plane.

Sketching Planes

• A sketch must be created on a

sketching plane; each plane may

contain multiple sketches.

• In the very beginning of a

<DesignModeler> session, three

planes are automatically created:

<XYPlane>, <YZPlane>, and

<ZXPlane>.



Chapter 2 Sketching

Sec

Sketches

•A sketch consists of

points and

edges; edges may be straight lines or

• Multiple sketches may be created on a plane.

[1] To create a new sthe active sketchin

click <New Sket

[2] Currentlyactive sketch.



Chapter 2 Sketching

Sec

Auto Constraints

• C

- The cursor is coincident with a line.

• P - The cursor is coincident with another point.

• T - The cursor is a tangent point.

•

- The cursor is a perpendicular foot.

• H

- The line is horizontal.

• V - The line is vertical.

• //

- The line is parallel to another line.



Chapter 2 Sketching

Sec

Sketching Toolboxes



Chapter 2 Sketching

Section 2.4 M

Section 2.4

M20x2.5 Threaded Bolt

Problem Description

[1] Metricsystem.

[2] Nominaldiameter

d = 20 mm.

[3] Pitchp = 2.5 mm.



Chapter 2 Sketching

Section 2.4 M

32 1

1

p

=

2 7 . 5

d 1

d

Extern

thread(bolt

H

4

p



Chapter 2 Sketching

Section 2.4 M

• Dimensions>Angle

• Modify>Replicate

•Revolve

Techniques/Concepts



Chapter 2 Sketching

Se

Section 2.5

Spur Gears

Problem Description

• To satisfy the fundamental law



Chapter 2 Sketching

Se

[6] Contactpoint (pitch

point).

[8no

T

[3] Pitch circler p = 2.5 in.

[9] Addendumr a = 2.75 in.

[1] The drivinggear rotatesclockwise.

[2] The driven

gear rotatescounter-

clockwise.

[5] Line ofcenters.



Chapter 2 Sketching

Se

Techniques/Concepts

• Draw>Construction Point



Chapter 2 Sketching

Sec

Section 2.6

Microgripper

Problem Description

[1] Grippingdirection.

[3] SMAactuator.

[4] Gbe



Chapter 2 Sketching

Sec

144

176 4 0 0

1 4 0

2 1 2

7 7

4 7

8 7

20

R25R45

32

92

D30

Unit: μ m

Thickness: 300 μ m



Chapter 2 Sketching

Sec

Techniques/Concepts

• Constraints>Equal Radius

• Copy bodies (Mirror)

• Create new sketch

• Constraints>Tangent

• Multiple parts



Chapter 3 2D Simulations

Chapter 3

2D Simulations

3.1

Step-by-Step: Triangular Plate

3.2 Step-by-Step: Threaded Bolt-and-Nut

3.3

More Details

3.4 More Exercise: Spur Gears




Sectio

Section 3.1

Triangular Plate

Problem Description

• The plate is made of steel and designed to

withstand a tensile force of 20,000 N on each




Sectio

Techniques/Concepts • Project Schematic

• Concepts>Surface

• Analysis Type (2D)

• Plane Stress Proble

• Generate 2D Mesh

• 2D Solid Elements

• <Relevance Cente

<Relevance>

• Loads>Pressure

• Weak Springs

• Solution>Total De




Section 3.2 T

Section 3.2

Threaded Bolt-and-Nut

Problem Description

[1] Bolt.[2] Nut.




Section 3.2 T

The plane of symmetry

T h e a xi s of s y mm e t r y

17 mm

[1] Thsimul

mod




Section 3.2 T

Techniques/Concepts

• Hide/Show Sketches

• Display Model/Plane

• Add Material/Frozen

• Axisymmetric Problems

• Contact/Target

• Frictional Contacts

• Edge Sizing

• Loads>Force

• Supports>Frictionless Support




Sec

Section 3.3

More Details

Plane-Stress Problems

• Plane stress

Z = 0,

• The Hook's

X

=

Y =

Z

=

XY

=

XY

G ,

• A problem m

plane-stress

X

X

Y

XY

XY




Sec

Plane-Strain Problems

X

Y

Z

Y

X

XY

X

Y

XY

• Plane strain condition

Z

= 0, ZX

=

• The Hook's law beco

X =

E

(1+ )(1 2 )

Y =

E

(1+ )(1 2 )

Z

=E

(1+ )(1 2

XY

= G XY

, YZ

• A problem may assum

condition if its Z -direc




Sec

R

R

Z

Z

RZ

RZ

R

Z

RZ

RZ

[1] Strain

state at apoint of aaxisymmetric

structure.

[2] Stress

Axisymmetric Problems

• If the geomet

loading of a s

axisymmetric

then all respo

independent

• In such a cas

R =

R =

• both

and




Sec

Mechanical GUI

• Pull-down Menus

and Toolbars

• Outline of Project

Tree

•Details View

• Geometry

• Graph

• Tabular Data




Sec

Project Tree

• A project tree may contain one or more

simulation models.

• A simulation model may contain one or more

<Environment> branches, along with other

objects. Default name for the <Environment>

branch is the name of the analysis system.

• An <Environment> branch contains <Analysis

Settings>, environment conditions, and a




Sec

Unit Systems [1] Built-in unitsystems.

[2] Unfor c

pro

• Consistent versus Inconsistent

Unit Systems.

• Built-in versus User-Defined Unit

Systems.

•Project Unit System.

• Length Unit in <DesignModeler>.

• Unit System in <Mechanical>.

• Internal Consistent Unit System.




Sec

Environment Conditions




Sec

Results Objects

View Results

[5] You can controlhow the contour

displays.

[6] Some resultscan display with

vectors.

[3] Label.




Se

Section 3.4

Spur Gears

Problem Description




Se

Techniques/Concepts

• Copy bodies (Translate)

• Contacts

• Frictionless

•Symmetric (Contact/Target)

• Adjust to Touch

• Loads>Moment

• True Scale




Se

100 100

R15

Section 3.5

Filleted Bar

Problem Description

[1] The bar ismade of steel.




Se

Part A. Stress Discontinuity

Displacement field iscontinuous over the

entire body.




Se

[2] Originalcalculated stresses

(unaveraged) are notcontinuous across

element boundaries,i.e., stress at boundary

has multiple values.

[4] By default, stresses areaveraged on the nodes, and thestress field is recalculated. That

way, the stress field iscontinuous over the body.




Se

Part B. Structural Error

• For an element, strain energies calculated using averaged stresses an

stresses respectively are different. The difference between these twocalled <Structural Error> of the element.

• The finer the mesh, the smaller the structural error. Thus, the struct

used as an indicator of mesh adequacy.




Se

Part C. Finite Element Convergence

0.0781

0.0782

0.0783

0.0784

0.0785

0.0786

0.0787

D i s p l a c e m e n t ( m m )

[1] Quadrilateralelement.

[2] Triangularelement.




Se

Part D. Stress Concentration

[1] To accuratelyevaluate the

concentrated stress,finer mesh is needed,

particularly around thecorner.




Se

Part E. Stress Sigularity

The stress in thiszero-radius filletis theoretically

infinite.

• Stress singu

to sharp co

• Any locatio

of infinity a

points.

• Besides a c

radius, a po

forces is al

Chapter 4 3D Solid Modeling 1



Chapter 4 3D Solid Modeling 1

Chapter 43D Solid Modeling

4.1

Step-by-Step: Beam Bracket

4.2 Step-by-Step: Cover of Pressure Cylinder

4.3 Step-by-Step: Lifting Fork

4.4 More Details

4.5 More Exercise: LCD Display Support

4.6

Review

Chapter 4 3D Solid Modeling Section 4.1 Beam Bracket 2




Section 4.1Beam Bracket

Problem Description

The beam bracket ismade of WT8x25 steel.

X

Y

Z





• Local coordinate systems

• Sketching with plane view

versus in 3D view

• Use of Triad

• Add Material

• Rounds/Fillets

• Turn on/off edges display

Techniques/Concepts

Chapter 4 3D Solid Modeling Section 4.2 Cover of Pressure Cylinder 4



p g y

Section 4.2Cover of Pressure Cylinder

Problem Description

[1] Pressurecylinder.

[2] Cylinder

Cover.

[3] Back view ofthe cover.




p g y

30.3

25.3

21.0 1.3

3 1 . 0

3.010.0

R8.5

R7.5

R19.0

Unit: mm.62.0

2.3 1.6 7.4

R4.9

R3.2

R9.0 R14.5

R18.1

R25.4

R27.8

7.4

6 2 . 0

R3.4




p g y

Techniques/Concepts

• Create new planes

• Set up local coordinate systems

• Plane with boundary

• Modify>Duplicate

• Cut Material

Chapter 4 3D Solid Modeling Section 4.3 Lifting Fork 7



p g g

Section 4.3Lifting Fork

Problem Description

[1] Fork (steel).

[2] Glass panel (1.0mm).




p g g

Unit: mm.

2 2 0 0

2500

2400

200

200

1 6 0

0

[1] The crosssection here is

160x40 mm.


130x20 mm.


100x10 mm.




Techniques/Concepts

• Skin/Loft

• Lofting guide line

• Add Frozen

• Copy bodies (Pattern)

• Boolean

• Create 3D surface bodies

Chapter 4 3D Solid Modeling Section 4.4 More Details 10



• Triad

• Isometric View

• Rotation

• Selection Filters

• Extend Selection

• Selection Panes

• Edge Display

• Tools for 3D

features

Section 4.4 More Details




Triad

[1] Click anarrow willorient the

view normalto that arrow.

[2] A black

arrow representsa negativedirection.

[4] Click thecyan sphere toreturn to the

isometric view.

[3] If the cyan spherecoincides with the origin,that means the view is an

isometric view.




Rotations

[1] Hold the middle mouse buttondown while moving around the graphic

area, you can rotate the model.

[2] Freerotation.

[3] Roll,rotation aboutscreen Z-axis.

[4] Yaw,rotation about

screenY-axis.

[5] Pitch,rotation aboutscreen X-axis.

[6] The type of rotation depends onthe location of the cursor.




Selection Aides

• Selection Filters

• Extend Selectin

• Selection Panes




Bodies and Parts

• A body is entirely made of one kind of

material and is the basic building

blocks of a model.

• A 3D body is either a solid body, a

surface body, or a line body.

• A part is a collection of same type of

bodies. All bodies in a part areassumed to be bonded together with

one another.

• In <Mechanical>, parts are meshed

independently

• A model may consist of one or more

parts.

•In <Mechanical>, connections

(contacts, joints) among parts must be

established to complete a model.

This is the only

geometric

entities that will

be attached to

<Mechanical> for

simulations.




FeaturesFeatures

•Based Features

• Extrude

• Revolve

• Sweep

• Skin/Loft

• Surface

• Lines

• Point

• etc.• Placed Features

• Thin/Surface

• Blend

• Chamfer

• etc.

• Planes

•Operations

• etc.

Chapter 4 3D Solid Modeling Section 4.5 LCD Display Support 16



Section 4.5LCD Display Support

Problem Description




200

90

60

44

1 0

5 0

4 2

2 0

1 7




• Revolve

• Skin/loft

• Thin/Surface

Techniques/Concepts




Chapter 5

3D Simulations5.1


5.2 Step-by-Step: Cover of Pressure Cylinder

5.3

More Details

5.4 More Exercise: LCD Display Support

Ch 5 3D S l S




Sect

Section 5.1

Beam Bracket

Problem Description[2] The

bracket is designedwithstand a load of 2uniformly distributed

the seat plate.

Ch 5 3D S l S




Sect

Techniques/Concepts

• Engineering Data

• Material Assignment

• Stress Tool>Safety Factor

• Structural Error

• Mesh Control>MultiZone

• 3D Solid Elements

Ch 5 3D Si l i S i 5 2 C




Section 5.2 Cove

Section 5.2

Cover of Pressure Cylinder

Problem Description

[2] The cover isdesigned to hold

up an internalpressure of 0.5

MPa.

Ch 5 3D Si l i S i 5 2 C




Section 5.2 Cove

Techniques/Concepts

• Add a new material in<Engineering Data>

• Isotropic Elasticity

• Material Assignment

• Loads>Pressure

• Create a new coordinate

system

Ch t 5 3D Si l ti S




Sec

Global Mesh Controls

• Relevance Center

• Relevance

Section 5.3

More Details

Ch t 5 3D Si l ti S




Sec

Mesh with MultiZone Method

• Generally, hexahedral elements are more

desirable than tetrahedral.

• A simple idea of creating hexahedra is to

mesh faces (source) of a body with

quadrilaterals and then "sweep" along a path

up to other end faces (target) of the body.

• Not all bodies are sweepable.

• The idea of <MultiZone> method is to

Cha ter 5 3D Sim lati ns Sec




Sec

Coordinate Systems

• When defining an environment

condition or a solution object by

<Components>, you need to refer

to a coordinate system. By default,

<Global Coordinate System> is

used, which is a Cartesian

coordinate system.

• To define a new coordinate

[1] Type ofthe coordinate

system.

[2] Origin.

[3] Axes.

Chapter 5 3D Simulations Sec




Sec

[1] Increase/

decrease contourbands.

[4] Number ofsignificant digits.

[3] Double-clickto edit value.

[5] T

the

[6ind

[2] The divider

can be dragged.

Legend Controls

Chapter 5 3D Simulations Sec




Sec

Adaptive Meshing

• Workbench provides a tool to

automate the mesh refinement

until a user-specified level of

accuracy is reached.

• This idea is termed adaptive

meshing .

• Internally, Workbench exploits

the structural errors to help

Chapter 5 3D Simulations Section 5 4




Section 5.4

Section 5.4

LCD Display Support

Problem Description[2] The design load(40 N) applies on

the trough.

Chapter 5 3D Simulations Section 5 4




Section 5.4

Techniques/Concepts

• Loads>Bearing Load

Chapter 6 Surface Models




Chapter 6

Surface Models6.1

Step-by-Step: Bellows Joints

6.2


6.3

More Exercise: Gearbox

6.4 Review

Chapter 6 Surface Models Sect




Sect

Section 6.1

Bellows Joints

Problem Description

• With the internal pressure, the

engineers are concerned about the

radial deformation (due to an





Sect

R315 R315 28

2 0





Sect

Techniques/Concepts

• Create surface bodies

using <Revolve>.

• Top/Bottom of a surface

body

• Shell Elements





Sect

Section 6.2

Beam Bracket

Techniques/Concepts

• Create surface bodies

using <Mid-Surface>





Section 6.3

Gearbox

Problem Description

[1] The flangedbearing is madeof gray cast iron.





Unit: mm.

3 5 5

1 5

3 0

1 7 0 R30

R50

R20

R40

170 200 70

(R170)

(R70)





Techniques/Concepts

• Create surface bodies by

<Thin/Surface>

• Loads>Bearing Loads

• Set up <Bonded>

connections.

Chapter 7 Line Models 1



Chapter 7Line Models

7.1

Step-by-Step: Flexible Gripper7.2 Step-by-Step: 3D Truss

7.3 More Exercise: Two-Story Building

7.4 Review

Chapter 7 Line Models Section 7.1 Flexible Gripper 2



Section 7.1Flexible Gripper

Problem Description

[3] Actuationdirection (input).

[2] The ends areconnected to a rigidground (preventing

translations androtations).

[4] Grippingdirection(output).

[1] The gripper is

made of POM.

P 1(!70,0)

P 2(!90,40)

P 3(!69,120)

P 4(

!35,160)

P 5(!34,100)

P 6(!24,60)

P 7(0,50)

X

Y




Techniques/Concepts

• Line bodies

• Cross Sections

• Cross Section Alignments

• Cross Section Solids

• Beam Elements

• Symmetry Conditions

• Geometric Advantage

0

10

20

30

40

50

60

0 5 10 15 20 25 30 35 40 45 50 H o r i z o n t a l D i s p l a c e m e n t ( m m )

Input Displacement (mm)




Convergence Study of Beam Elements

52.32

52.34

52.36

52.38

52.40

0 125 250 375 500 625

O u t p u t D i s p l a c e m e n t ( m m )

Number of Elements

[1] In this exercise, wemeshed with 34

elements, resulting 52.335mm of displacement.

[2] The displacementconverges to 52.381 mm.

Chapter 7 Line Models Section 7.2 3D Truss 5



200"

200"

75"

P 1 P 2

P 3

P 4

P 5 P 6

P 7

P 8

P 9 P 10

1

2

3 4

5

6 7 8 9

10

11

12

13

14

15

16

17

18 19

20

21

22

23

24

25

1 0 0

"

1 0 0

"

X Y

Z

75"

Section 7.23D Truss

Problem Description

Design Loads for the Transmission Tower

Joint F X (lb) F Y (lb) F Z (lb)

P 1 1,000 -10,000 -10,000

P 2 0 -10,000 -10,000

P 3 500 0 0

P 6 600 0 0

Chapter 7 Line Models Section 7.2 3D Truss 6



Techniques/Concepts

• Create points

• Concepts>Lines From Points

• Convergence of straight beam

elements

Chapter 7 Line Models Section 7.3 Two-Story Building 7



Section 7.3Two-Story Building

Problem Description

[1] All beams andcolumns are madeof structural steel,

with a crosssection of W16x50. [2] The floor slabs

are made of reinforcedconcrete, with athickness of 5".

[3] Eachfloor-to-floorheight is 10'.

2 0

'

20' 20'20'

Chapter 7 Line Models Section 7.3 Two-Story Building 8



Techniques/Concepts

• Adjust Cross Section

Alignments

• Concepts>Surface From

Edges

• Use of Selection Panes• Flip Surface Normal

• Form New Part

• Import Engineering Data

• Inertial>Standard Earth

Gravity

• Inertial>Acceleration

Chapter 8 Optimization



Chapter 8

Optimization8.1

Step-by-Step: Flexible Gripper

8.2

More Exercise: Triangular Plate

8.3

Review


Section



Section 8.1Flexible Gripper

Problem Description

P2( 90 40)

P

3(

69,120)

• Positions of the P 2, P 3, and P 6 are free to be changed.

• The idea is to fix the X-coordinates of these points and

adjust their Y-coordinates to achieve a better GA value.


Section



Techniques/Concepts

• Filtering Prefixes and Suf fixes

• Input Parameters

• Output Parameters

• Design Points

• Goal Driven Optimization

• Design of Experiments

• DOE Tables

• Response Surfaces

• Optimization


Sectio



Section 8.2Triangular Plate

Problem Description

W

[2] The initial valueof the width of the

bridge is 30 mm andits allowable range is

20-30 mm.

[ch

red


Sectio



Techniques/Concepts

• No additional techniques/concepts are introduced.

Chapter 9 Meshing 1



Chapter 9Meshing

9.1 Step-by-Step: Pneumatic Fingers

9.2 More Exercise: Cover of Pressure Cylinder

9.3 More Exercise: Convergence Study of 3D Solid Elements

9.4 Review

Chapter 9 Meshing Section 9.1 Pneumatic Fingers 2



Section 9.1Pneumatic Fingers

Problem Description

Unit: mm.

80

5

1

2

5.1

4

3 3.2 1

(19.2)

Plane ofsymmetry.

Chapter 9 Meshing Section 9.1 Pneumatic Fingers 3



Techniques/Concepts

• Mesh Metric: Skewness

• Hex Dominant Method

• Sweep Method

• MultiZone Method

• Section View

• Tools>Freeze

• Create>Slice

• Nonlinear Simulations

• Line Search

•Displacement Convergence

Chapter 9 Meshing Section 9.2 Cover of Pressure Cylinder 4



Section 9.2Cover of Pressure Cylinder

Techniques/Concepts

• Patch Conforming Method

• Patch Independent Method

Chapter 9 Meshing Section 9.3 Convergence Study of 3D Solid Elements 5



Section 9.3Convergence Study of 3D Solid Elements

Problem Description

100 mm

10 mm

[1] The beam ismade of steel.

[2] The width of the beamis 10 mm. A uniform loadof 1 MPa applies on theupper face of the beam.

[3] We willrecord thevertical tipdeflection.




Lower-Order Elements

0.60

0.64

0.68

0.72

0.76

0 3000 6000 9000 12000 15000

T i p D e f l e c t

i o n ( m m )

Number of Nodes

[1] Lower-ordertetrahedron.

[2] Lower-orderperpendicular

prism.

[3] Lower-orderparallel prism.

[4] Lower-orderhexahedron.




Higher-Order Elements

0.746

0.747

0.748

0.749

0.750

0.751

0.752

0 2000 4000 6000 8000 10000

T i p D e f l e c t i o n ( m m )

Number of Nodes

[1] Higher-ordertetrahedron.

[2] Higher-orderperpendicular prism.

[3] Higher-orderparallel prism.

[4] Higher-orderhexahedron.




Hexahedra

0.746

0.747

0.748

0.749

0.750

0.751

0.752

0 2000 4000 6000 8000 10000

T i p D e f l e c t i o

n ( m m )

Number of Nodes

[2] Higher-orderhexahedron.

[1] Lower-orderhexahedron.




Tetrahedra

0.600

0.640

0.680

0.720

0.760

0 2000 4000 6000 8000 10000

T i p D e f l e c

t i o n ( m m )

Number of Nodes

[1] Lower-ordertetrahedron.

[2] Higher-ordertetrahedron.




Parallel Prisms

0.66

0.68

0.70

0.72

0.74

0.76

0 2000 4000 6000 8000 10000


Number of Nodes

[2] Higher-orderparallel prism.

[1] Lower-orderparallel prism.




Perpendicular Prisms

0.66

0.68

0.70

0.72

0.74

0.76

0 2000 4000 6000 8000 10000


Number of Nodes

[2] Higher-orderperpendicular prism.

[1] Lower-orderperpendicular prism.




Guidelines

• Never use lower-order tetrahedra or triangles.

• Higher-order tetrahedra or triangles can be as good as other elements as long as

the mesh is fine enough. In cases of coarse mesh, however, they perform poorly

and are not recommended.

• Lower-order prisms are not recommended.

• Lower-order hexahedra and quadrilaterals can be used, but they are not as

efficient as their higher-order counterparts.

• Higher-order hexahedra, prisms, and quadrilaterals are among the most efficient

elements so far we have discussed. Mesh your models with these elements

whenever possible. If that is not possible, then at least try to achieve a higher-

order hexahedra-dominant or quadrilateral-dominant mesh.

Chapter 10 Buckling and Stress Stiffening



Chapter 10

Buckling and Stress Stiffen10.1

Step-by-Step: Stress Stiffening

10.2

Step-by-Step: 3D Truss

10.3

More Exercise: Beam Bracket

10.4 Review


Section



Section 10.1Stress Stiffening

Problem Description

[1] The beam is made ofsteel and has a uniform

cross section of 10x10 mm.

[2] A uniformly distributedload of 0.1 N/mm appliesdownward on the beam.

[3] Anon thefree to


Section



Stress Stiffening Effects

5

6

7

8

9

10

M a x i m u m D

e f l e c

t i o n ( m m )

This is the point with zero axialforce. On the right, the beam issubject to tensile force. On the

left, the beam is subject tocompressive force.


Section



Linear Buckling Analysis


Section



Section 10.23D Truss

Problem Description

2EI

2(29,000,000)(0.13852)




Results

Buckling will occur

when 23% of designloads apply on the

structure. Themultiplier can be viewed

as safety factor. Thestructure is not safe.


Sectio



Section 10.3Beam Bracket

Problem Description

• It is a good practice that an

engineer always checks the

t t l t bilit h


Sectio



Results

The <LoadMultiplier> can beviewed as a safetyfactor. It predictsthat 203 times ofdesign load will

initiate a buckling.The structure is

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! 9#$ 5$*' -0$12$%37 /, $12*; ') '#$ -0$12$%37 ./--$0$%3$ )- '#$ '() (*+$,8

! "#$% ($ :;*7 . *%. / ')<$'#$06 '#$ 5$*' -0$12$%37 /, =>8?= @A :;<=5>> ? ;>)5>=@6 (#/3# /,

* #*0,# 52AA *%. 2%:;$*,*%' -)0 )20 $*0,8

! "#$% ($ :;*7 . *%. 0 ')<$'#$06 '#$ 5$*' -0$12$%37 /, B=?8=C @A :=<;5AA ? ;>)5>=@6 (#/3# /,

3;),$ ') * #*04)%/3 -0$12$%37 )- '#$ 4/..;$ . D>EB8E= @AF8

Chapter 12 Structural Dynamics 1



Chapter 12Structural Dynamics

12.1 Basics of Structural Dynamics

12.2 Step-by-Step: Lifting Fork

12.3 Step-by-Step: Two-Story Building

12.4 More Exercise: Ball and Rod

12.5 More Exercise: Guitar String

12.6 Review

Chapter 12 Structural Dynamics Section 12.1 Basics of Structural Dynamics 2

Section 12 1



Section 12.1Basics of Structural Dynamics

Key Concepts

• Lumped Mass Model

• Single Degree of Freedom Model

• Undamped Free Vibration

• Damped Free Vibration

•Damping Coefficient

• Damping Mechanisms

• Viscous Damping

• Material Damping

• Coulomb Friction• Modal Analysis

• Harmonic Response Analysis

• Transient Structural Analysis

• Explicit Dynamics

• Response Spectrum Analysis• Random Vibration Analysis


Lumped Mass Model: The Two-Story Building



p y g

k1

c 1

k2

c 2

m1

m

2

[1] A two-degrees-of-freedom model for findingthe lateral displacementsof the two-story building.

[2] Totalmass lumped atthe first floor.

[3] Total masslumped at the roof

floor.

[4] Total bendingstiffness of the

first-floor's beamsand columns.

[5] Totalbending stiffnessof the second-floor's beamsand columns.

[6] Energy dissipatingmechanism of the first

floor.

[7] Energy dissipatingmechanism of the

second floor.


Single Degree of Freedom Model



g g

F ! = ma

p! kx ! c ! x = m!! x

m!! x + c ! x + kx = p

k

c

m

p

x

• We will use this single-degree-of-freedom lumped mass model to

explain some basic behavior of dynamic response.

• The results can be conceptually extended to general multiple-

degrees-of-freedom cases.


Undamped Free Vibration



If no external forces exist, the equation for the

one-degree-of-freedom system becomes

m!! x + c ! x + kx = 0

If the damping is negligible, then the equation

becomes

m!! x + kx = 0

The

x = Asin ! t +B( )

Natural frequency: ! =k

m(rad/s) or f =

!

2" (Hz)

Natural period: T = 1f

p

D i s p l a c e

m e n t ( x )

time (t)

T =2!

"


Damped Free Vibration



p

D i s p l a c e m e n

t ( x )

time (t)

T d =

2!

" d

T d

m!! x + c ! x + kx = 0

If the damping c is small (smaller than c c

),

then the general solution is

x = Ae

!"# t sin #

d t + B( )

Where

!

d =! 1"# 2 , ! =

c

c c

, c c = 2m!

The quantity c c is called the critical damping

coefficient and the quantity ! is called the

damping ratio.


Damping Mechanisms



p g

• Damping is the collection of all energy dissipating mechanisms.

• In a structural system, all energy dissipating mechanisms come down

to one word: friction. Three categories of frictions can be identified:

• friction between the structure and its surrounding fluid, called

viscous damping ;

• internal friction in the material, called material damping , solid

damping, or elastic hysteresis;

• friction in the connection between structural members, called dry

friction or Coulomb friction.


Analysis System



y y

The foregoing concepts may be generalized to multiple-degrees-of-freedom cases,

M!" #$ !!D{ }+ C !

" #$

!D{ }+ K !" #$ D{ } = F { }

Where {D} is the nodal displacements vector, {F } is the

nodal external forces vector, [ M] is called the mass

matrix , [C ] is called the damping matrix , and [K ] is the

stiffness matrix.

Note that when the dynamic effects (inertia effect

and damping effect) are neglected, it reduces to a static

structural analysis system,

K !

" #$ D{ } = F { }


Modal Analysis



M!" #$ !!D{ }+ C !" #$ !D{ }+ K !" #$ D{ } = 0

For a problem of n degrees of freedom, it has at most n solutions, denoted by

{Di }, i =1,2,...,n . These solutions are called mode shapes of the structure. Each mode

shape {Di } can be excited by an external excitation of frequency

!

i , called the natural

frequency of the mode.

In a modal analysis, since we are usually interested only in the natural frequencies

and the shapes of the vibration modes, the damping effect is usually neglected to

simplify the calculation,

M!" #$ !!D{ }+ K !

" #$ D{ } = 0


Harmonic Response Analysis



M!" #$ !!D{ }+ C !

" #$ !D{ }+ K !

" #$ D{ } = F { }

<Harmonic Response> analysis solves a special form of the equation, in which the

external force on i th degree of freedom is of the form

F i = A

i sin(!t +"

i )

where Ai is the amplitude of the force, !

i is the phase angle of the force, and ! is

the angular frequency of the external force. The steady-state solution of the

equation will be of the form

D

i = B

i sin(!t +"

i )

The goal of the harmonic response analysis to find the magnitude Bi and the

phase angle ! i , under a range of frequencies of the external force.


Transient Structural Analysis



M!" #$ !!D{ }+ C !

" #$

!D{ }+ K !" #$ D{ } = F { }

<Transient Structural> analysis solves the general form of the equation. External

force {F } can be time-dependent forces. All nonlinearities can be included. It uses

a direct integration method to calculate the dynamic response.

The direct integration method used in <Transient Structural> analysis is

called an implicit integration method .


Explicit Dynamics



M!" #$ !!D{ }+ C !

" #$

!D{ }+ K !" #$ D{ } = F { }

Similar to <Transient Structural>, <Explicit Dynamics> also solves the general

form of equation. External force {F } can be time-dependent forces. All

nonlinearities can be included. It also uses a direct integration method to

calculate the dynamic response.

The direct integration method used in <Explicit Dynamic> analysis is called

an explicit integration method .

Chapter 12 Structural Dynamics Section 12.2 Lifting Fork 13

Section 12 2



Section 12.2Lifting Fork

Problem Description

During thehandling, the fork

accelerates upwardto a velocity of 6 m/s

in 0.3 second, andthen decelerates to

a full stop in another0.3 second, causingthe glass panel to

vibrate.


Static Structural Simulation



The maximumstatic deflection

is 15 mm.


Transient Structural Simulation



[2] Themaximumdeflection.

[3] The vibration damps

out fast and reduces toless than 7 mm in about

0.6 second.

[1] History oftip-deflection.

Chapter 12 Structural Dynamics Section 12.3 Two-Story Building 16

Section 12.3



Section 12.3Two-Story Building

Problem DescriptionHarmonic loadswill apply on this

floor deck.

Two scenarios are investigated:

• Harmonic load of magnitude of 10

psf due to the dancing on the floor.

• Harmonic load of magnitude of 0.1

psf due to rotations of a machine.


Modal Analysis



[1] The firstmode (1.55 Hz).

[2] The sixth

mode (9.59 Hz).

[3] The eighthmode (10.33

Hz)




• The dancing frequency is close to the fundamental mode

(1.55 Hz), that's why we pay attention to this mode,

which is a side sway mode (in X-direction).

• For the rotatory machine, we are concerned about the

floor vibrations in vertical direction. That's why we pay

attention on the sixth and eighth modes.


Side Sway Due to Dancing



At dancing frequency of

1.55 Hz, the structure is

excited such that the

maximum X-displacement

is 0.0174 in (0.44 mm).

This value is too small to

be worried about.

Amplitude of side swaydue to harmonic loadof magnitude of 1 psf.


Vertical Deflection of the Floor Due to Rotatory



Machine

Amplitude of verticaldeflection of the floordue to harmonic loadof magnitude of 1 psf.

Although high frequencies do excite the floor, but the values are

very small. At frequency of 10.3 Hz, the excitation reaches a

maximum of 0.0033 in (0.1 times of 0.033 in), or 0.084 mm. The

value is too small to cause an issue.

Chapter 12 Structural Dynamics Section 12.4 Disk and Block 21

Section 12.4



Section 12.4Disk and Block

Problem Description

[2] Right before the

impact, the disk movestoward the block with avelocity of 0.5 m/s.

[1] Before theimpact, the block

rests on thesurface.

[3] Both the disk and theblock are made of a verysoft polymer of Young's

modulus of 10 kPa,Poisson's ratio of 0.4, and

mass density of 1000 kg/m3.


Results




Solution Behavior



Chapter 12 Structural Dynamics Section 12.5 Guitar String 24

Section 12.5



Section 12.5Guitar String

The main purpose of this exercise is to demonstrate how to use theresults of a static simulation as the initial condition of a transient

dynamic simulation

Chapter 13 Nonlinear Simulations 1

Chapter 13



Chapter 13

Nonlinear Simulations

13.1 Basics of Nonlinear Simulations

13.2

Step-by-Step: Translational Joint

13.3 Step-by-Step: Microgripper

13.4 More Exercise: Snap Lock

13.5 Review

Chapter 13 Nonlinear Simulations Section 13.1 Basics of Nonlinear Simulations 2

Section 13.1



Section 13.1Basics of Nonlinear Simulations

Key Concepts

• Nonlinearities

• Causes of Structural Nonlinearities

• Steps, Substeps, and Iterations

• Newton-Raphson Method

• Force/Displacement Convergence

• Solution Information

• Line Search

• Contact Types

• Contact versus Target• Contact Formulations

• Additional Contact Settings

• Pinball Region

• Interface Treatment

• Time Step Controls• Update Stiffness


Nonlinearities



F o r c e { F }

Displacement {D}

F o r c e { F }

Displacement {D}

[1] In a linearsimulation, [K ]

(slope of the line)is constant.

[2] In a nonlinearsimulation, [K ] (slope

of the curve) ischanging with {D}.

• In a nonlinear simulation, the

relation between nodal force {F } and

nodal displacement {D} is nonlinear.

• we may write

K (D)!"

#$ D{ } = F { }

• Challenges of nonlinear simulations

come from the difficulties of solving

the above equation.




Steps, Substeps, and Iterations[1] Number ofsteps can be



• Steps (Load Steps)

• Each step can have its own analysis settings.

• Substeps (Time Steps)

• In dynamic simulations, time step is used

for integration over time domain.

• In static simulation, dividing into substeps is

to achieve or enhance convergence.

• Iterations (Equilibrium Iterations)

• Each iteration involves solving a linearized

equilibrium equation.

steps can bespecified here.

[3] Each step

can have itsown analysis

settings.

[2] To switchbetween steps,

type a step number

here.


Newton-Raphson Method



Displacement D{ }

F o r c e

F {

}

D

0 D

1 D

2 D

3 D

4

F 0

F 1

F 2

F 3

F 0 + !F

P 0

P 1

P 2

P 3

P 4

P 1

! P 2

! P 3

! P 4

!

[1] Actual responsecurve, governed by

K (D)!" #$ D{ } = F { }

[2] Displacements atcurrent time step

(known).

[5] Displacements at nexttime step (unknown).

[3] External

force atcurrent timestep (known).

[4] Externalforce at next

time step(known).


Suppose we are now at P 0

and the time is increased one substep further so that

th t l f i i d t F + !F and t t fi d th di l t



the external force is increased to F 0 + !F , and we want to find the displacement

at next time step D4.

Starting from point P 0

, <Workbench> calculates a tangent stiffness [K ], the

linearized stiffness, and solves the following equation

K !" #$ %D{ } = %F { }

The displacement D0 is increased by !D to become D

1. Now, in the D-F space,

we are at (D1,F

0 + !F ) , the point P

1

! , far from our goal P 4. To proceed, we need to

"drive" the point P 1

! back to the actual response curve.

Substituting the displacement D1

into the governing equation, we can

calculate the internal force F 1 ,

K (D

1)!" #$ D

1{ } = F 1{ }

Now we can locate the point (D1,F

1) , which is on the actual response curve. The

difference between the external force (here, F 0 + !F ) and the internal force (here,

F 1) is called the residual force of that equilibrium iteration,

F 1

R= (F

0 + !F )" F

1

If the residual force is smaller than a criterion, then the substep is said to be

converged, otherwise, another equilibrium iteration is initiated. The iterations

repeat until the convergence criterion satisfies.


Force/Displacement

Convergence



[1] You can turnon <Force

Convergence> andset the criterion.

[2] You can turn

on <DisplacementConvergence> andset the criterion.

[3] When shellelements or beamelements are used,

<MomentConvergence> can be

activated.

[4] When shellelements or beamelements are used,

<RotationConvergence> can be

activated.

Convergence


Solution Information




Line Search

o r c e



D0 D1

F 0

F 0 + !F

Calculated !D

Goal

F o

Displacement

[1] In some cases, when the F-D curve is highly nonlinear or

concave up, the calculated !D

in a single iteration mayovershoot the goal.

[2] Line search can beturned on to scale

down the incrementaldisplacement. By

default, it is <ProgramControlled>.





Contact Types



• Bonded

• No Separation

• Frictionless

• Rough

• Frictional

• Linear versus Nonlinear Contacts


Contact versus Target [1] To specify a contactregion, you have to select a setof <Contact> faces (or edges),



( g ),and select a set of <Target>

faces (or edges).

[2] If <Behavior> is set to<Symmetric>, the roles of

<Contact> and <Target> willbe symmetric.

• During the solution, <Workbench> will

check the contact status for each point

(typically a node or an integration

point) on the <Contact> faces against

the <Target> faces.

• If <Behavior> is set to <Symmetric>,

the roles of <Contact> and <Target>

will be symmetric.

• If <Behavior> is set to <Asymmetric>,

the checking is only one-sided.


Contact Formulations



[1] Workbenchoffers several

formulations toenforce contactcompatibility.

[2] <Normal Stiffness> is input here.

The input value (default to 1.0) isregarded as a scaling factor to multiply astiffness value calculated by the program.

• MPC (multi-point constraint)

• Pure Penalty

• Normal Lagrange

• Augmented Lagrange


Additional Contact

Settings



Settings

• Pinball Region

• Interface Treatment

• Time Step Controls

• Update Stiffness

Chapter 13 Nonlinear Simulations Section 12.2 Translational Joint 16

Section 13.2



60

20

20

40

Translational Joint

Problem Description

[3] All connectorshave a cross sectionof 10x10 mm.

[1] Thetranslational jointis used to connect

two machinecomponents, sothat the relativemotion of thecomponents is

restricted in thisdirection.

[2] All leaf springshave a cross section

of 1x10 mm.

Chapter 13 Nonlinear Simulations Section 12.2 Translational Joint 17

Results



0

30

60

90

120

0 10 20 30 40

F o r c e ( N )

Displacement (mm)

[1] NonlinearSolution.

[2] Linear Solution.

101.73

74.67



Chapter 13 Nonlinear Simulations Section 13.3 Microgripper 19

Results



[1] contactstatus. [2] contact

pressure.

Chapter 13 Nonlinear Simulations Section 13.4 Snap Lock 20

Section 13.4



Snap Lock

Problem Description

7

20

20

7

10

30

17

7

5

10

5

8

The purpose of this

simulation is to find out

the force required to push

the insert into the

position and the force

required to pull it out.


[2] I i

Results (Without Friction)



[2] It requires236 N to pull

out.

[1] It requires 189 N

to snap in.

[3] The curve isessentially symmetric.

Remember that wedidn't take the

friction into account.


Results (With Friction)



[1] It requires 328 Nto snap in.

[2] It requires 305 N to

pull out.

[3] Because offriction, the curve is

not symmetric.



Chapter 14 Nonlinear Materials Section 14.1 Basics of Nonlinear Materials 2

Section 14.1



Basics of Nonlinear Materials

Key Concepts

• Linear versus Nonlinear Materials

• Elasticity

• Linear Elasticity

• Hyperelasticity

• Plasticity

• Plasticity

• Yield Criteria

• Hardening Rules• Plasticity Models

• Hyperelasticity

• Required Test Data

• Strain Energy Functions

• Hyperelasticity Models


Linear/Nonlinear Materials

/ A r e a )



• When the stress-stain relation of a

material is linear, it is called a linear

material , otherwise the material is

called a nonlinear material .

• For a linear material, the stress-strain

relation is expressed by Hooke's law,

in which two independent material

parameters are needed to

completely define the material.

• Orthotropic elasticity is also available

in <Workbench>.

S t r e s s ( F o

r c e /



Elastic/Plastic Materials

/ A r e a )

[1] Elasticmaterial.



• If the strain is totally recovered after

release of the stress, the behavior is

called elasticity.

• On the other hand, if the strain is not

totally recoverable (i.e., there is no

residual strain after release of the

stress), the behavior is called plasticity

and the residual strain is called the

plastic strain.

S t r e s s ( F o

r c e /


S t r e s s ( F o r c e / A r e a )


[2] Plastic

material.

[3] Plastic strain.


Hysteresis



S t r e s s

Strain

• The term hysteresis is used for the energy

loss in a material during stressing and

unstressing.

• Most of materials have more-or-less hysteresis

behavior. However, as long as it is small

enough, we may neglect the hysteresis

behavior.

S t r e s s

Strain


Hyperelastic

Hyperelasticity





Hyperelasticmaterial.

•Nonlinear non-hysteresis elasticity are characterized

by that the stressing curve and the unstressing curve

are coincident: the energy is conserved in the cycles.

• Challenge of implementing nonlinear elastic material

models comes from that the strain may be as large

as 100% or even 200%, such as rubber under

stretching or compression.

• Additional consideration is that, under such large

strains, the stretching and compression behaviors

may not be described by the same parameters.

•This kind of super-large deformation elasticity is

given a special name: hyperelasticity .


PLASTICITY

[1] Idealized[2] Initial yield



Idealized Stress-Strain Curve



stress-strain

curve.

point (or

elastic limit).

[3] The stress-strain relation isassumed linear

before Yield

point, and theinitial slope is theYoung's modulus.

[4] When thestress is released,

the strain

decreases with aslope equal to theYoung's modulus.

• Plasticity behavior typically occurs in ductile

metals subject to large deformation. Plastic strain

results from slips between planes of atoms due to

shear stresses. This dislocation deformation is a

rearrangement of atoms in the crystal structure.

• A stress-strain curve is not sufficient to fully

define a plasticity behavior. There are two

additional characteristics that must be described: a

yield criterion and a hardening rule.


Yield Criteria



• <Workbench> uses von Mises criterion as the yield criterion, that is, a stressstate reaches yield state when the von Mises stress !

e is equal to the current

uniaxial yield strength ! y " , or

1

2 !

1"!

2( )2

+ ! 2 "!

3( )2

+ ! 3 "!

1( )2#

$%

&

'( =!

y )

• The yielding initially occurs when ! y " = !

y , and the "current" uniaxial yield

strength ! y " may change subsequently.

• If the stress state is inside the cylinder, no yielding occurs. If the stress state is on

the surface, yielding occurs. No stress state can exist outside the yield surface.


!

3 This is a von Mises yield surface, which

is a cylindrical surface aligned with the



!

1

!

2

!

1 = !

2 = !

3

axis ! 1 = !

2 = !

3 and with a radius of

2! y

" , where ! y

" is the current yield

strength.


Hardening Rules



• If the stress state is on the yield surface and the stress state continues to "push" the

yield surface outward, the size (radius) or the location of the yield surface will

change. The rule that describes how the yield surface changes its size or location is

called a hardening rule.

• Kinematic hardening assumes that, when a stress state continues to "push" a yield

surface outward, the yield surface will change its location, according to the "push

direction," but preserve the size of the yield surface.

• Isotropic hardening assumes that, when a stress state continues to "push" a yield

surface, the yield surface will expand its size, but preserve the axis of the yield

surface.


[1] Kinematic hardeningassumes that the difference

[2] Isotropic hardeningth t th t il



!

y 2!

y

S t r e s s

Strain

between tensile yield

strength and thecompressive yield strength

remains a constant of 2! y

.

!

y "

!

y

" S t r e s s

Strain

assumes that the tensile

yield strength and thecompressive yield strengthremain equal in

magnitude.


Plasticity Models in Workbench



[2] To complete adescription of plasticity

model, you must include itslinear elastic properties.

[1] Currently,<Workbench>

provides sixplasticity models.




• It is possible that a set of test data is obtained by superposing two sets of other test

data. For example, the set of uniaxial compressive test data can be obtained by adding a



set of hydrostatic compressive test data to a set of equibiaxial tensile test data.

[1] Uniaxialcompressive test.

[2] Equibiaxialtensile test.

[3] Hydrostaticcompressive test.

= +


300



0

60

120

180

240

0 0.2 0.5 0.7

S t r e s s ( p s i )


[1] Uniaxialtest data.

[2]Equibiaxial test

data.

[3] Shear testdata.


Hyperelasticity

Models in



Workbench



Chapter 14 Nonlinear Materials Section 14.2 Belleville Washer 18

40 mm

22 mm• We will compress the Belleville



The Belleville washer is

made of steel, withthickness of 1.0 mm.

spring by 1.0 mm and then

release it completely.

• A force-displacement curve will

also be plotted.

• We will examine the residual

stress after the spring is

completely released.


80

Force-versus-Displacement Curve



-80

-60

-40

-20

0

20

40

60

80

0 0.2 0.4 0.6 0.8 1.0

C o m

p r e s s i v e F o r c e ( N )

Displacement (mm)

[1] The curve isquite different

between loadingand unloading.[3] Let's explore the

residual stress at thispoint when the external

force is completelyreleased.

[2] There is no practice use of thissection. It is the force required to pullthe spring back to its original position.


Residual Stress

[1] Residuali l



equivalent stress.

[2] Residual hoop stress. Notethat the top surface is

dominated by tension, whilethe bottom surface is

dominated by compression.

Chapter 14 Nonlinear Materials Section 14.3 Planar Seal 21

Section 14.3



Planar Seal

Problem Description

0

40

80

120

160

200

0 0.1 0.2 0.3

S t r e s s ( p s i )

Engineering Strain (Dimensionless)

[1] Uniaxialtest.

[2] Biaxialtest.

[3] Shear test.• The seal is used in the door of a

refrigerator. The seal is a long

strip, and we will model it as a

plane strain problem.




.800

1.100

.333 .500

.133

.133

.867

R.150

R.150

R.200

R.200

R.050

R.050

Unit: in.

[2] Steelplate.

[1] Rubberseal.

[3] Steelplate.

[4] The upperplate is displaced0.85" downward.


Results



A force-versus-displacement curve. Notethat the force unit should

be read lbf/in instead of lbf.

Chapter 15 Explicit Dynamics 1

Chapter 15



p

Explicit Dynamics

15.1 Basics of Explicit Dynamics

15.2

Step-by-Step: High-Speed Impact

15.3

Step-by-Step: Drop Test

15.4 Review

Chapter 15 Explicit Dynamics Section 15.1 Basics of Explicit Dynamics 2

Section 15.1



Basics of Explicit Dynamics

Key Concepts

• Implicit Integration Methods

• Explicit Integration Methods

• Solution Accuracy

• Integration Time Steps

• Automatic Mass Scaling

• Static Damping


Implicit Integration Methods



M!" #$ !!D{ }+ C !" #$ !D{ }+ K !" #$ D{ } = F { }

• <Transient Structural> solves the above equation using the following algorithm:

!Dn+1 =

!Dn + !t " !!D

n+1+(1# " )!!D

n$%

&'

Dn+1 =Dn + !t !

Dn +

1

2 !t 2

2" !!

Dn+1+ (1# 2" )!!

Dn$%

&'

• The parameters ! and ! are chosen to control characteristics of the algorithm such as

accuracy, numerical stability, etc.

• It is called an implicit method because the response at the current time step depends on

not only the historical information but also the current information; iterations are

needed in a single time step.


Explicit Integration Methods



M!" #$ !!D{ }+ C !

" #$

!D{ }+ K !" #$ D{ } = F { }

• <Explicit Dynamics> solves the above equation using the following algorithm:

!Dn+1

2

= !Dn!1

2

+ !!Dn"t

Dn+1

= Dn + !D

n+1

2

!t

• It is called explicit methods because the response at the current time can be calculated

explicitly; no iterations within a time step is needed.


Solution Accuracy



• <Explicit Dynamics> uses the principle of conservation of energy to monitor the solution

accuracy.

(Reference Energy) + (Work Done)Reference!Current = (Current Energy)

• It calculates overall energy at each cycle. If the energy error reaches a threshold, the

solution is regarded as unstable and stops. The default threshold is 10%.

• The Energy Error is defined by

Energy Error =

(Current Energy)-(Reference Energy) - (Work Done)Reference!Current

max Current Energy , Reference Energy , Kinetic Energy( )




The red curve isthe energy error.

In this case, thesolution is quite

stable.


Integration Time Steps



• With explicit methods, the integration time step needs to be small enough to ensure

stability and accuracy of the solution. The German mathematicians, Courant, Friedrichs,

and Lewy, suggested that, in a single time step, a wave should not travel further than the

smallest element size, i.e.

!t "h

c

where h is the smallest element size, c is the wave speed in the element.

• Because of the CFL condition, when generating meshes for <Explicit Dynamics>, make

sure that one or two very small elements do not control the time step. In general, a

uniform mesh size is desirable for <Explicit Dynamics> simulations.


Automatic Mass Scaling



• The wave speed in an element is c = E ! , where E is the Young's modulus

and ! is the mass density of the element. Further, ! = m V , where m is the

mass and V is the volume of the element. Therefore the CFL condition yields

!t " fhm

VE

• The idea of mass scaling is to artificially increase the mass of small elements,

so that the stability time step can be increased.


Static Damping



• <Explicit Dynamics> is primarily

designed for solving transient dynamic

problems.

• Using <Static Damping> option, a steady-

state solution can also be obtained.

• The idea is to introduce a damping

force, to critically damp the lowest mode

of oscillation.

Chapter 15 Explicit Dynamics Section 15.2 High-Speed Impact 10

Section 15.2



High-Speed Impact

Chapter 15 Explicit Dynamics Section 15.3 Drop Test 11

Section 15.3



Drop Test

10

R3

20!

5 m/s

120

R20

[1] The phonebody is made of an

aluminum alloy.

finite element simulations with ansys workbench 13.pdf

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