1. Structural Linear 1

7/25/2019 1. Structural Linear 1

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BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

COMPUTER AIDED DESIGN[ME F 342]

Training on ANSYS

[P-1 Structural Linear Analysis (1-D, 2-D, 3-D)]

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Model Example-1

A rectangular plate of size: Length 4, Width 2 and Thickness 2,is subjected to a load of 900 unit along the length direction. E

= 2X105, Poissions Ratio = 0.33.

Find out the deflection and tensile Stress in X-

direction. Compare your results with a theoretical solution.

Theoretical Solution:

2/22522

900)( mmN

A

Px =

==

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mmAE

PLx 0045.0

102)22(

4900)(

5=

==


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/TITLE,2-D PLATE STRESS ANALYSIS!PRE-PROCESSING BEGINS

/PREP7

!DEFINE ELEMENT

ET,1,82,,,3 !ACTIVATE THICKNESS OPTION BY GIVING KOPT(3)=3R,1,2 !DEFINE THICKNESS AS 2 (YOU CAN CHANGE AND CHECK RESULTS!

!(HOWEVER, YOU NEED TO CHANGE THE LOAD PER NODE)

!DEFINE MATERIAL PROPS

MP,EX,1,2E5 !DEFINE YOUNG'S MODULUS

MP,NUXY,1,0.33 !DEFINE POISSON RATIO!DEFINE SOLID MODELLING

RECTG,,4,,2 !MODELLING A RECTANGLE

APLOT !PLOT AREAS (HERE, PLOT OR SHOW THE RECTANGLE)

Define Model & Material Properties | Pre-Processing

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Define Mesh | Pre-Processing

!DEFINE MESH SIZEKESIZE,ALL,0.5

!MESH THE AREA

AMESH,1 !MESH AREA '1'

EPLOT !PLOT ELEMENTSFINI

SAVE

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PLOT CONTROLS: NUMBERING

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PLOT CONTROLS: NUMBERING

KPLOT & KPT NUMBERING

LPLOT & LINE NUMBERING

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7/42BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

How to find Element Numbers?

EPLOT !With Element numbers activated

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How to find out Node-Numbers?

NPLOT !With Node numbers activated

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ELIST !LIST ALL ELEMENT DETAILS

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NLIST !LIST ALL NODE DETAILS

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/SOLU !ACTIVATE SOLN PHASE!DEFINE SUPPORT

NSEL,S,LOC,X,0 !SELECT ALL NODES WITH X=0 LOCATION

D,ALL,UX,0 !CONSTRAIINT ALL NODES IN X DIRN

D,1,UY,0 !CONSTRAIINT NODE#1 IN Y DIRN

NSEL,ALL !SELECT ALL NODES

NSEL,S,LOC,X,4 !SELECT ALL NODES WITH X=4 LOCATION

F,ALL,FX,100 !APPLY FORCE ON ALL NODES 100 UNIT PER NODE (TOT=

9NODES X 100N/NODE = 900 N)

SOLVE

Solut ion Phase

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Deformed Shape with Un-deformed

Click: General Postprocessor Plot Results Deformed Shape Def + Undeformed

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Tensile Stress in X-Direction

Average Stress = 900 / 2X2 = 225

Local Stresses because of local nodal loads. Can be reduced by

fine mesh at corners

Click: General Postprocessor Plot Results Contour Plot Nodal Solu Stress X-Comp

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Comparative Results based on Various Mesh Size

Mesh = 0.125

33 Nodes

27.27 Unit load per node

You can observe the refinement in results, local stresses (red, orange, yellow

and green zones) are reduced because of fine mesh.

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Mesh = 0.25

17 Nodes

53 Unit load per node

Mesh = 0.5

9 Nodes

100 Unit load per node



Comparative Results based on Various Mesh Size

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mmAE

PLlSolutionTheoritica x 0045.0

102)22(

4900)(:

5=

==

FEA

Solution



Model Example-2

A cantilever beam Length 80, Width 20 and Thickness 1, issubjected to a load of 100 unit at the end in ve y-direction. E

= 2X105, Poissions Ratio = 0.33.

Find out the deflection of end point and tensile Stress

in X-direction. Compare your results with a theoretical

solution.

Theoretical Solution: 80 Length

2

0Width

P = 100

Thickness 1(you can change and check the results)

2

3 /120

2

20

12

201

80100)( mmNy

I

Mbxb =

==mm

EI

PL128.0

12

2011023

80100

3 35

33

=

==

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Define Model and Material Props

/TITLE,2-D BEAM STRESS ANALYSIS!PRE-PROCESSING BEGINS

/PREP7

!DEFINE ELEMENT!

ET,1,82,,,3 !ACTIVATE THICKNESS OPTION BY GIVING KOPT(3)=3

R,1,1 !DEFINE THICKNESS AS 1 (YOU CAN CHANGE AND CHECK RESULTS)

!DEFINE MATERIAL PROPS!

MP,EX,1,2E5 !DEFINE YOUNG'S MODULUS

MP,NUXY,1,0.33 !DEFINE POISSON RATIO

!DEFINE SOLID MODELLINGRECTG,,80,,20 !MODELLING A RECTANGLE

APLOT

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Define Mesh

!DEFINE MESH SIZE

KESIZE,1,2 !SPECIFY ELEMENT SIZE NEAR KEYPOINT '1'





EPLOT !PLOT ELEMENTS

FINISAVE

!PRE-PROCESSING ENDS

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Solut ion Phase

!ENTER SOLUTION PHASE/SOLU

!DEFINE SUPPORT

NSEL,S,LOC,X,0 !NODES IN LINE X=0 IS SELECTED

D,ALL,ALL,0 !ALL DOF CONSTRAINEDNSEL,ALL

!DEFINE LOAD

FK,3,FY,-100 !FORCE DEFNED ON KEYPOINT

SOLVE

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Deformation Plot


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Check the deformation at free end (UY)


mmEI

PL128.0

12

2011023

80100

3 35

33

=

==

FEM Solution: 0.13 mm

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Bending Stress Plot

Click: General Postprocessor Plot Results Contour Plot Nodal Solu Stress X-Comp

2

3 /120

2

20

12

201

80100)( mmNy

I

Mbxb =

==

FEM Solution: 0.128 N/mm2

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Model Problem-3

A 2-D simply supported beam of 80X20 with thickness 20 issubjected to a central load of 10,000 unit load. E= 200X105.

Find out the central deflection of end point and tensile

Stress in X-direction. Compare your results with a theoretical

solution.

2

3 /150

2

20

12

2020

405000)( mmNy

I

Mbxb =

==

Theoretical Solution:

mmEI

PL04.0

12

202010248

8010000

48 35

33

=

==

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Model Problem-3

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Define Mesh

/TITLE,2-D BEAM STRESS ANALYSIS!PRE-PROCESSING BEGINS

/PREP7

!STEP1! DEFINE ELEMENT!

ET,1,82,,,3 !ACTIVATE THICKNESS OPTION BY GIVING KOPT(3)=3

R,1,20 !DEFINE THICKNESS AS 1 (YOU CAN CHANGE AND CHECK RESULTS)!DEFINE MATERIAL PROPS!

MP,EX,1,2E5

MP,NUXY,1,0.33

!DEFINE SOLID MODELLING

RECTNG,,80,,20

APLOT

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Define Mesh

HPTCREATE,LINE,3,5,COORD,40,20!DEFINE MESH SIZE

KESIZE,ALL,2

!MESH THE AREA

AMESH,1 !MESH AREA '1'EPLOT !PLOT ELEMENTS

FINI

SAVE

!PRE-PROCESSING ENDS

!ENTER SOLUTION PHASE

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Solut ion Phase

/SOLU

!DEFINE SUPPORT

DK,1,UY,0

DK,1,UX,0

DK,2,UY,0

FK,5,FY,-10000

SOLVE

!SOLUTION PHASE ENDS

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Deformed with Un-Deformed Plot

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Deflection Plot (UY)

FEM Solution: 0.055 mm

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mmEI

PL

04.0

12

202010248

8010000

48 35

33

=

==


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Stress Plot (SigmaX)

2

3 /150

2

20

12

2020

405000)( mmNy

I

Mbxb =

==

FEM Solution: 143 N/mm2

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Model Problem-4

2-D Hollow Cylinder of internal radius 5 and external radius 7,

is subjected to an internal pressure of 10.

(a) Determine using 1/4th solid domain:

1] Radial deflection

2] Hoop (Tangential) Stress

E= 200X105.

(b) Solve the same problem with solid domain

(c) Solve the same problem with 3/4th solid domain

Hoop stress = pidi/2t= 25 N/mm2

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Model Problem-5

!INTERNALLY PRESSURIZED CYLINDER

/PREP7

PCIRC,5,7,0,90 !CREATING A ANNULAR QUARTER CIRCLE (Chk with 0,180)

ET,1,PLANE82

ESIZE,0.25

AMESH, ALL

MP,EX,1,200E5

MP,PRXY,1,0.3

FINISH

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Model Problem-5

/SOLU

ANTYPE,0 ! STATIC ANALYSIS

DL,2,1,UX,0

DL,4,1,UY,0

SFL,3,PRES,10

SOLVE

FINI

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Model Problem-5

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Stress Plot: Sigma-X & Sigma-Y

Hoop stress = pdi/2t= 25 N/mm2

| Radial stress = pressure = 10 N/mm2

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Stress Plot: Principal Stress

Hoop stress = pdi/2t= 25 N/mm2

| Radial stress = pressure = 10 N/mm2

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Model Problem-5

A compression disc of diameter 5, is compressed by a

pressure of 500 on the upper half of its circumference.

Determine contact stress. E= 200X105.

Solution is given for the above problem.

Solve for load to be applied at the topmost point of disc &

supported at the bottom most point.

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Define a Solid Model

/TITLE,2-D DISC COMPRESSION STRESS ANALYSIS

/PREP7

ET,1,82,,,3

R,1,3

MP,EX,1,2E5

MP,NUXY,0.33

PCIRC,0,5

APLOT

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Define Mesh

KESIZE,ALL,0.5


EPLOT !PLOT ELEMENTS

FINI

SAVE

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Solution | Boundary Conditions

/SOLU

!DEFINE SUPPORT

SFL,1,PRES,100

SFL,2,PRES,100

DL,3,1,SYMM

DL,4,1,SYMM

SOLVE

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Deformed Plot


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Stress Plot: VonMises

1. Structural Linear 1

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