fem problems with solutions

5/28/2018 FEM Problems with solutions

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FEM Assignment 2014

NIT-TRICHY

FEM Assignment

5th SemesterSubmitted by:

NITINVARMAN

111111060

4/4/2014

Software used: Ansys APDL

This document contains an abstract about the work on Ansys to analyse different problems.


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Problem 1An Aluminum cantilever beam shown in Figure 1 is subjected to a point load at the end with the

other end fixed. Using one dimensional elements, discretize the model and perform the

numerical analysis for the prediction of displacement von-mises stresses in the beam using

ANSYS software. With beam theory validate the numerical results obtained in ANSYS.

Figure 1

Nomenclature:

L =110m Length of beam

b =10m Cross Section Base

h =1 m Cross Section Height

P=1000N Point Load

E=70GPa Youngs Modulus of Aluminum at Room Temperature

=0.33 Poissons Ratio of Aluminum

Solution:

Using Numerical Analysis in Ansys:

The results obtained are

Maximum Von-mises stress: 62.7KPa

Maximum Deflection = 7.59mm

Von mises Result:


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Deflection result:

Ansys Macro/NOPR

KEYW,PR_SET,1

KEYW,PR_STRUC,1

KEYW,PR_THERM,0

KEYW,PR_FLUID,0

KEYW,PR_ELMAG,0

KEYW,MAGNOD,0

KEYW,MAGEDG,0

KEYW,MAGHFE,0

KEYW,MAGELC,0

KEYW,PR_MULTI,0

KEYW,PR_CFD,0

/GO

!*

!*

/PREP7

K,1,0,0,0,

K,2,110,0,0,LSTR, 1, 2

!*


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ET,1,BEAM188

!*

!*

MPTEMP,,,,,,,,

MPTEMP,1,0

MPDATA,EX,1,,7e10

MPDATA,PRXY,1,,0.33SECTYPE, 1, BEAM, RECT, , 0

SECOFFSET, CENT

SECDATA,10,1,0,0,0,0,0,0,0,0,0,0

TYPE, 1

MAT, 1

REAL,

ESYS, 0

SECNUM, 1

!*

LMESH, 1

LMESH, 1

/UI,MESH,OFF

ESIZE,10,0,

LMESH, 1

LCLEAR, 1

LMESH, 1

/UI,MESH,OFF

FLST,2,1,3,ORDE,1

FITEM,2,1

!*

/GO

DK,P51X, ,0, ,0,ALL, , , , , ,

FLST,2,1,3,ORDE,1

FITEM,2,2

!*

/GO

FK,P51X,FY,-1000

FINISH

/SOL

/STATUS,SOLU

SOLVE

FINISH

/POST1

!*

/EFACET,1

PLNSOL, U,Y, 0,1.0

!*

PRNSOL,S,PRIN!*

PRESOL,S,PRIN

)/GOP


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Theoretical validation:

Von Mises Stress

For Plane stress, Von Mises equivalent stress can be expressed as:

= (x2

- xy + y2

+ 3xy2)

In the top most plane => xy=0, y=0.

Therefore = x.

Bending stress M(x) = Where I =b*h

3/12 and c = h/2.

M(x) = P ( x- L) x = 6P(x-L)/bh2 Maximum stress = 6P(L)/bh2=66 KPa

Maximum Von Mises stress = 66 KPaDeflection:From Euler Bernoulli Equation

() From previous derivations

( )

Integrating once we get

At the fixed end => x=0, y=0, Thus C1=0

Integrating again we get

EIy=P(x3/6Lx

2/2)+C2

Again at the fied end, y(0) = 0 Thus C2=0.

Therefore deflection = y

== (P/EI) * (x3/6Lx

2/2) = Px

2/6EI * (x-3L)


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Maximum deflection is at x=L

max= PL3/3EI = 7.61 mm.

Thus the theoretical values and the Numerical analysis values obtained from Ansys are

almost same.

Problem 2:

Determine the force in each member of the following truss. Indicate if the member is in tension

or compression. The cross-sectional area of each member is 0.01 m2 and the Youngs modulus is

200 X 109 N/m2.Figure 2 Plane Truss

Solution:

From the numerical analysis using Ansys, The forces on each nodes are given

below


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Figure 1 : Representation of the forces and constraints on links

Figure 2 : Forces at different nodes


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Element 2 (or member AB) applies a force of 1500 N in the x-direction and 800

N in the negative y -direction on node 1 (or pin A). This means that the total

force in AB is (15002+800

2) = 1700 N . The resultant acts from A to B i.e. the

member is pulling on pin A. So it must be in tension. Similarly, the force in

Element 1 (AC) is 2000 N (tension) and in Element 3 (BC) is 2500 N(compression).

The result is shown in the table below

Link Identification Force(N) Tension / Compression

AC 2000 Tension

AB 1700 Tension

BC 2500 Compression

Ansys Macro:

/NOPR

KEYW,PR_SET,1

KEYW,PR_STRUC,1

KEYW,PR_THERM,0

KEYW,PR_FLUID,0

KEYW,PR_ELMAG,0

KEYW,MAGNOD,0

KEYW,MAGEDG,0

KEYW,MAGHFE,0

KEYW,MAGELC,0KEYW,PR_MULTI,0

KEYW,PR_CFD,0

/GO

!*

!*

/PREP7

!*

ET,1,LINK180

!*

R,1,.01, ,0

!*

!*

MPTEMP,,,,,,,,MPTEMP,1,0

MPDATA,EX,1,,200e9

MPDATA,PRXY,1,,

N,1,0,2.8,,,,,

N,2,1.50,2,,,,,

N,3,0,0,,,,,

FLST,2,2,1

FITEM,2,1

FITEM,2,3

E,P51X

FLST,2,2,1

FITEM,2,2

FITEM,2,3

E,P51X


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FLST,2,2,1

FITEM,2,2

FITEM,2,1

E,P51X

FLST,2,1,1,ORDE,1

FITEM,2,1

!*/GO

D,P51X, , , , , ,UX,UY, , , ,

FLST,2,1,1,ORDE,1

FITEM,2,3

!*

/GO

D,P51X, , , , , ,UX, , , , ,

FLST,2,1,1,ORDE,1

FITEM,2,2

!*

/GO

F,P51X,FY,-2800

FINISH

/SOL

/STATUS,SOLU

SOLVE

FINISH

/POST1

!*

PRESOL,FORC

)/GOP

Problem 3 :

Determine the nodal deflections, reaction forces, and stress for the truss

system shown in Figure 3 below (E = 200GPa, A = 3250mm2).

Solution:

The numerical solution using ansys for the given problem is given below.

The nodes, links and constraints given in Ansys are shown pictorially

below:


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The contour plot of the deflection is shown below:


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The contour plot of stresses is shown below:

The list of reaction forces are shown below:


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Ansys Macro:

/NOPR

KEYW,PR_SET,1

KEYW,PR_STRUC,1

KEYW,PR_THERM,0

KEYW,PR_FLUID,0

KEYW,PR_ELMAG,0

KEYW,MAGNOD,0

KEYW,MAGEDG,0

KEYW,MAGHFE,0

KEYW,MAGELC,0

KEYW,PR_MULTI,0

KEYW,PR_CFD,0

/GO

!*

!*

/PREP7

!*

ET,1,LINK180

!*

R,1,.00325, ,0

!*

!*

MPTEMP,,,,,,,,

MPTEMP,1,0

MPDATA,EX,1,,200e9MPDATA,PRXY,1,,

N,1,,,,,,,

N,2,1.8,3.118,,,,,

N,3,3.6,0,,,,,

N,4,5.4,3.118,,,,,

N,5,7.2,0,,,,,

N,6,9,3.118,,,,,

N,7,10.8,0,,,,,

FLST,2,2,1

FITEM,2,1

FITEM,2,2

E,P51X

FLST,2,2,1FITEM,2,2


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FITEM,2,3

E,P51X

FLST,2,2,1

FITEM,2,3

FITEM,2,4

E,P51X

FLST,2,2,1FITEM,2,4

FITEM,2,5

E,P51X

FLST,2,2,1

FITEM,2,5

FITEM,2,6

E,P51X

FLST,2,2,1

FITEM,2,6

FITEM,2,7

E,P51X

FLST,2,2,1

FITEM,2,1

FITEM,2,3

E,P51X

FLST,2,2,1

FITEM,2,3

FITEM,2,5

E,P51X

FLST,2,2,1

FITEM,2,5

FITEM,2,7

E,P51X

FLST,2,1,1,ORDE,1

FITEM,2,1

!*

/GO

D,P51X, , , , , ,UX,UY, , , ,

FLST,2,1,1,ORDE,1

FITEM,2,7

!*

/GO

D,P51X, , , , , ,UY, , , , ,

FLST,2,2,1

FITEM,2,2

FITEM,2,4

E,P51X

FLST,2,2,1FITEM,2,4

FITEM,2,6

E,P51X

FLST,2,1,1,ORDE,1

FITEM,2,1

!*

/GO

F,P51X,FY,-280000

FLST,2,1,1,ORDE,1

FITEM,2,3

!*

/GO

F,P51X,FY,-210000FLST,2,1,1,ORDE,1


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FITEM,2,5

!*

/GO

F,P51X,FY,-280000

FLST,2,1,1,ORDE,1

FITEM,2,7

!*/GO

F,P51X,FY,-360000

FINISH

/SOL

/STATUS,SOLU

SOLVE

FINISH

/POST1

!*

PRESOL,FORC

PLDISP,0

PLDISP,0

!*

!*

/EFACET,1

PLNSOL, U,Y, 0,1.0

!*

/EFACET,1

PLNSOL, S,EQV, 0,1.0

!*

PLESOL, S,EQV, 0,1.0

etable,saxl,ls

pretab,saxl

AVPRIN,0,1,

ETABLE,1,S,EQV

!*

AVPRIN,0,1,

ETABLE, ,S,X

!*

!*

/EFACET,1

PLNSOL, S,Y, 0,1.0

!*

PLESOL, S,INT, 0,1.0

!*

PLESOL, S,Y, 0,1.0

!*

PLESOL, S,X, 0,1.0!*

PLESOL, S,EQV, 0,1.0

GPLOT

FINISH

/SOL

FINISH

/PREP7

)/GOP


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Problem 4:

The problem to be modeled in this example is a simple bicycle frame shown in

the following figure. The frame is to be built of hollow aluminum tubing having

an outside diameter of 25mm and a wall thickness of 2mm for the main part of

the frame.

Solution:

The numerical analysis of the given problem is done using Ansys. The pictorial

representation of the approach is shown below


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The pink colour arrows in the picture indicates the constraints on the nodes

The red colour arrows indicate the direction of forces on the nodes.

The contour plot for the deformation is shown below:


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From the above result we could infer that the maximum deformation of the

nodes in the bicycle frame is around 0.13 mm

The contour plot for the von-mises stress is shown below:


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The forces on each element is shown below:


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PRINT FORC ELEMENT SOLUTION PER ELEMENT

***** POST1 ELEMENT NODE TOTAL FORCE LISTING *****

LOAD STEP= 1 SUBSTEP= 1

TIME= 1.0000 LOAD CASE= 0

THE FOLLOWING X,Y,Z FORCES ARE IN GLOBAL COORDINATES

ELEM= 1 FX FY FZ MX MY MZ

1 -402.73 280.34 0.14818E-11-0.25143E-09-0.24405E-09 5738.6

3 402.73 -280.34 -0.14818E-11 0.25699E-09 0.24405E-09 -4228.4


3 -402.73 280.34 0.14818E-11-0.25661E-09-0.24405E-09 4228.4

4 402.73 -280.34 -0.14818E-11 0.26216E-09 0.24405E-09 -2718.1


4 -402.73 280.34 0.14818E-11-0.26229E-09-0.24405E-09 2718.1

5 402.73 -280.34 -0.14818E-11 0.26785E-09 0.24405E-09 -1207.9


5 -402.73 280.34 0.14818E-11-0.26771E-09-0.24405E-09 1207.9

6 402.73 -280.34 -0.14818E-11 0.27327E-09 0.24405E-09 302.36


6 -402.73 280.34 0.14818E-11-0.27346E-09-0.24405E-09 -302.36

7 402.73 -280.34 -0.14818E-11 0.27901E-09 0.24405E-09 1812.6

ELEM= 6 FX FY FZ MX MY MZ7 -402.73 280.34 0.14818E-11-0.27893E-09-0.24405E-09 -1812.6

8 402.73 -280.34 -0.14818E-11 0.28449E-09 0.24405E-09 3322.9


8 -402.73 280.34 0.14818E-11-0.28453E-09-0.24405E-09 -3322.9

9 402.73 -280.34 -0.14818E-11 0.29008E-09 0.24405E-09 4833.1


9 -402.73 280.34 0.14818E-11-0.29024E-09-0.24405E-09 -4833.1

10 402.73 -280.34 -0.14818E-11 0.29580E-09 0.24405E-09 6343.3


10 -402.73 280.34 0.14818E-11-0.29578E-09-0.24405E-09 -6343.311 402.73 -280.34 -0.14818E-11 0.30134E-09 0.24405E-09 7853.6

Ansys Macro:

/NOPR

KEYW,PR_SET,1

KEYW,PR_STRUC,1

KEYW,PR_THERM,0KEYW,PR_FLUID,0


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KEYW,PR_ELMAG,0

KEYW,MAGNOD,0

KEYW,MAGEDG,0

KEYW,MAGHFE,0

KEYW,MAGELC,0

KEYW,PR_MULTI,0

KEYW,PR_CFD,0/GO

!*

!*

/PREP7

!*

ET,1,PIPE288

!*

!*

MPTEMP,,,,,,,,

MPTEMP,1,0

MPDATA,EX,1,,70000

MPDATA,PRXY,1,,0.33

!*

!*

SECTYPE,1,PIPE, ,1

SECDATA,25,2,0,0,1,0,0,0,

SECOFFSET,0,0,

SECCONTROL,0,

SOCEAN,0

!*

/REPLOT,RESIZE

/REPLOT,RESIZE

/REPLOT,RESIZE

K,1,0,325,0,

K,2,0,400,0,

K,3,500,400,0,

K,4,500,0,0,

K,5,825,0,50,

K,6,825,0,-50,

LSTR, 1, 2

LSTR, 2, 3

LSTR, 3, 4

LSTR, 4, 1

/ANG,1,30,YS,1

/REP,FAST

LSTR, 6, 3

LSTR, 5, 3

LSTR, 6, 4LSTR, 4, 5

/ANG,1,30,XS,1

/REP,FAST

GPLOT

/ANG,1,-30,YS,1

/REP,FAST

/ANG,1,-30,YS,1

/REP,FAST

/ANG,1,-30,YS,1

/REP,FAST

/ANG,1,-30,YS,1

/REP,FAST

/ANG,1,-30,YS,1/REP,FAST


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/ANG,1,-30,YS,1

/REP,FAST

/ANG,1,-30,YS,1

/REP,FAST

TYPE, 1

MAT, 1

REAL,ESYS, 0

SECNUM, 1

!*

ESIZE,0,20,

FLST,2,8,4,ORDE,2

FITEM,2,1

FITEM,2,-8

LMESH,P51X

/UI,MESH,OFF

FLST,2,1,3,ORDE,1

FITEM,2,2

!*

/GO

DK,P51X, , , ,0,ALL, , , , , ,

FLST,2,2,3,ORDE,2

FITEM,2,5

FITEM,2,-6

!*

/GO

DK,P51X, , , ,0,UY,UZ, , , , ,

FLST,2,1,3,ORDE,1

FITEM,2,3

!*

/GO

FK,P51X,FY,-600

FLST,2,1,3,ORDE,1

FITEM,2,4

!*

/GO

FK,P51X,FY,-900

/REPLOT,RESIZE

FINISH

/SOL

/STATUS,SOLU

SOLVE

FINISH

/POST1

!*/EFACET,1

PLNSOL, U,SUM, 0,1.0


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Problem 5

The Simple Conduction Example is constrained as shown in the following figure 5.

Thermal conductivity (k) of the material is 10 W/m*C and the block is assumed to

be infinitely long.

Solution:

The numerical analysis of the given problem is done using Ansys as

shown below.

The model with the thermal constraints on its faces is modelled as

shown below:


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The contour plot for the thermal deformation is shown below:


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Ansys Macro:

KEYW,PR_SET,1

KEYW,PR_STRUC,0

KEYW,PR_THERM,1

KEYW,PR_FLUID,0

KEYW,PR_ELMAG,0

KEYW,MAGNOD,0

KEYW,MAGEDG,0

KEYW,MAGHFE,0

KEYW,MAGELC,0

KEYW,PR_MULTI,0

KEYW,PR_CFD,0

/GO

/PREP7

*SET,length,1.0

*SET,height,1.0

blc4,0,0,length, height

ET,1, PLANE55

MP,KXX,1,10 ESIZE,length/20 AMESH,ALL

FINISH

/SOLU


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ANTYPE,0

NSEL,S,LOC,Y,height D,ALL,TEMP,500

NSEL,ALL

NSEL,S,LOC,X,0

NSEL,A,LOC,X,length

NSEL,A,LOC,Y,0

D,ALL,TEMP,100

NSEL,ALL

SOLVE

FINISH

/POST1

PLNSOL,TEMP,,0,

Problem 6:

A distributed load of 1000 N/m (1 N/mm) will be applied to a solid steel beam

with a rectangular cross section as shown in the figure below. The cross-

section of the beam is 10mm x 10mm while the modulus of elasticity of the

steel is 200GPa.


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Solution:

The numerical solution for the above problem is done using Ansys.

The Finite element model with the forces and constraints is shown below:

The contour plot

for the

deformation isshown below:


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The contour plot for the von mises stress is shown below:

Ansys Macro:

KEYW,PR_SET,1

KEYW,PR_STRUC,1

KEYW,PR_THERM,0

KEYW,PR_FLUID,0

KEYW,PR_ELMAG,0

KEYW,MAGNOD,0

KEYW,MAGEDG,0

KEYW,MAGHFE,0

KEYW,MAGELC,0


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KEYW,PR_MULTI,0

KEYW,PR_CFD,0

/GO

K,1,0,0

K,2,1000,0

L,1,2

ET,1,BEAM3

R,1,100,833.333,10

MP,EX,1,200000

MP,PRXY,1,0.33

ESIZE,100

LMESH,ALL

FINISH

/SOLU

ANTYPE,0

DK,1,UX,0,,,UY

DK,2,UY,0

SFBEAM,ALL,1,PRES,1

SOLVE

FINISH

)/GOP

fem problems with solutions

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