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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
ELEM= 2 FX FY FZ MX MY MZ
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
ELEM= 3 FX FY FZ MX MY MZ
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
ELEM= 4 FX FY FZ MX MY MZ
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
ELEM= 5 FX FY FZ MX MY MZ
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
ELEM= 7 FX FY FZ MX MY MZ
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
ELEM= 8 FX FY FZ MX MY MZ
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
ELEM= 9 FX FY FZ MX MY MZ
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