Nonlinear finite elements/Homework 7/Solutions

Problem 1: Index NotationEdit

SolutionsEdit

Part 1Edit

  1. Determine whether the following expressions are valid in index notation. If valid, identify the free indices and the dummy indices.

1)  

Invalid. The free indices are   on the LHS and   on the RHS.

2)  

Valid. Both   and   are free indices.

3)  

Valid. The free index is   and the dummy index is  .

4)  

Invalid. The free index is   on the LHS and   on the RHS.

5)  

Valid. The dummy index is  . So the sum is a scalar which is equal to  .

6)  

Invalid. There is one free index on the LHS and no free index on the RHS.

Part 2Edit

Show the following:

1)  

 

2)  

The LHS is

 

If  , then we have

 

We get the same result if  . The only nonzero LHS and RHS occur when   and  .

  • Case 1:  ,  ,  ,  .
 
  • Case 2:  ,  ,  ,   or

 ,  ,  ,  .

 
  • Case 3:  ,  ,  ,  .
 
  • Case 4:  ,  ,  ,  .
 
  • Case 5:  ,  ,  ,   or

 ,  ,  ,  .

 
  • Case 6:  ,  ,  ,  .
 
  • Case 7:  ,  ,  ,  .
 
  • Case 8:  ,  ,  ,   or

 ,  ,  ,  .

 
  • Case 9:  ,  ,  ,  .
 

Hence the   relation is satisfied for all cases.

3)  

 

4)  

 

For  ,

 

For  ,

 

For  ,

 

Hence shown.

Part 3Edit

The elasticity tensor is given by

 

where   are Lame constants,   is the second order identity tensor, and   is the fourth-order symmetric identity tensor. The two identity tensors are defined as

 

The stress-strain relation is

 

Show that the stress-strain relation can be written in index notation as

 

Write the stress-strain relations in expanded form.

 

Now, in dyadic notation

 

Therefore

 

Similarly,

 

The stress-strain law becomes

 

Expanding the left hand side, we get

 

Therefore,

 

Problem 2: Rotating Vectors and TensorsEdit

Let ( ) be an orthonormal basis. Let   be a second order tensor and   be a vector with components

 

SolutionEdit

Part 1Edit

Write out   and   in matrix notation.

 

Part 2Edit

Find the components of the vector   in the basis ( ).

The components of   are

 

Therefore

 
 

Part 3Edit

Find the components of the vector   in the basis ( ).

The cross product is given by

 

Therefore,

 

Part 4Edit

Find the components of the tensor   in the orthonormal basis.

The tensor product is given by

 

Hence, in matrix notation

 
 

Part 5Edit

Rotate the basis clockwise by 30 degrees around the   direction. Find the components of  ,  ,  , and   in the rotated basis.

The vector transformation rule is

 

where   are the direction cosines.

In this case, the direction cosines are

 

Therefore,

 
 

Similarly,

 
 

Also,

 
 

From the handout from Slaughter's book, the tensor transformation rule is

 

where   are the direction cosines.

In matrix form,

 

Therefore the components of   in the rotated basis are give by

 
 

Problem 3: More BeamsEdit

Part AEdit

Consider a beam of length   = 100 in., cross-section 1 in.   1 in., and subjected to a uniformly distributed transverse load   lbf/in. Model one half of the beam using symmetry considerations.

Part 1Edit

Hinged-Hinged Beam

The boundary conditions are

 

Compute a plot similar to that shown in Figure 4.3.4 for this case using Beam188 elements. What do you observe?

The result is shown in Table 1.

Table 1. Deflections of a hinged-hinged beam
Load   at  
1 -0.520746
2 -1.04086
3 -1.55922
4 -2.07510
5 -2.58768
6 -3.09622
7 -3.60000
8 -4.09835
9 -4.59065
10 -5.07636

Part 2Edit

Clamped-Clamped Beam

The boundary conditions are

 

Compute a plot for this case using Beam188 elements. Comment on your plot.

The result is shown in Table 2.

Table 2. Deflections of a clamped-clamped beam
Load   at  
1 -0.103456
2 -0.202476
3 -0.294220
4 -0.377753
5 -0.453387
6 -0.521968
7 -0.584455
8 -0.644185
9 -0.696440
10 -0.745243

Listed below is the ANSYS input code for Problem 3A.1 and 3A.2.

/prep7

b = 1
h = 1

et,1,188
sectype,1,beam,rect
secdata,b,h

MP,EX,1,30e6
MP,PRXY,1,0.3

K,1,0,0,0
K,2,50,0,0
k,3,0,50,0

L,1,2,50

latt,1,,1,,3,3,1

LMESH,ALL

!change this section to d,1,all,0 for Problem 3A.2
d,1,uy,0
d,1,uz,0
d,2,rotz,0
nsel,all

sfbeam,all,,pres,10

fini

/solu
nlgeom,on
autots,on
nsubst,10,100,10
outres,all,all
solve
finish

Part BEdit

Part 1Edit

Simulate the unrolling of a cantilever beam from Section 4.1.1 of Ibrahimbegovic (1995) and compare your results with the results shown in the paper.

The result is shown in Table 3.

Table 3. Cantilever free-end displacement components
Load     Rotation
  -0.040666 6.3205 -3.1287
  9.9578 0.12729 -6.2577

The deformation plots are shown in Figure 4 and 5.

 
Figure 4. Deformed shape under   for Problem 3B.1.
 
Figure 5. Deformed shape under   for Problem 3B.1.

The ANSYS input code for this problem is listed below.

/prep7

et,1,beam188
sectype,1,beam,rect
secdata,1,1

mp,ex,1,1200
mp,prxy,1,0

l = 10
pi = 4*atan(1)
r = L/2/pi

K,1,0,-r
K,2,-r,0
K,3,0,r
K,4,r,0
K,5,0,-r

k,6,0,0,10

k,7,0,0,0

larc,1,2,7,r,5
larc,2,3,7,r,5
larc,3,4,7,r,5
larc,4,5,7,r,5

latt,1,,1,,6,6,1
lmesh,all

dk,5,all,0

fk,1,mz,-20*pi

/solu
nlgeom,on
cnvtol,f,5,0.001
outres,all,all
arclen,on
nsubst,100
solve
fini

Part 2Edit

Simulate the clamped-hinged deep circular arch from Example 7.3 of Simo and Vu Quoc (1986) and compare you results with the results shown in the paper.

The inputs are: ,  ,  , and  . We assume a square cross section. Then

 

Therefore,  .

The deformed shape (unconverged) for a load of 905 is shown in Figure 6.

 
Figure 6. Deformed shape of arch.

The load-displacement curve (up to the last converged solution) is shown in Figure 7.

 
Figure 7. Load-displacement plot for circular arch.

The buckling load is 900.925 compared to 905.28 in Simo and Vu Quoc.

The ANSYS file use for the calculations is shown below.

/prep7  
!*
!* Total load
!*
load = 905.0
!*  
!*  Element type
!*
et,1,beam188
keyopt,1,1,0
keyopt,1,2,0
keyopt,1,3,0
keyopt,1,4,0
keyopt,1,6,0
keyopt,1,7,0
keyopt,1,8,0
keyopt,1,9,0
keyopt,1,10,0   
keyopt,1,11,0   
keyopt,1,12,0   
!*
!*  Beam cross-section type
!*
sectype, 1, beam, rect, , 0   
secoffset, cent 
secdata, 0.34641, 0.34641, 0,0,0,0,0,0,0,0 
!*  
!*  Material properties
!* 
mptemp,,,,,,,,  
mptemp,1,0  
mpdata,ex,1,,8.3e8  
mpdata,prxy,1,,0.33 
!*
!* Keypoints
!*
k, 1,   0.000,   0.000, 0.000
k, 2, -95.372, -30.071, 0.000
k ,3,  95.372, -30.071, 0.000 
k ,4,   0.000, 100.000, 0.000
!*  
!* Arcs
!*
larc,3,4,1,100, 
larc,4,2,1,100, 
!*  
!* Element size = 20 elements per arc
!*
lesize,all, , ,20, ,1, , ,1,
!*
!* Mesh the arcs
!*
lmesh, all
!*
!* Plot the nodes
!*
nplot   
/pnum,node,1
/number,0   
/replot 
!*
!* Apply displacement BCs
!*
!* Hinged end
!*
d, 22, ux, 0
d, 22, uy, 0
d, 22, uz, 0
!*
!* Clamped end
!*
d, 1, all, 0
!*
!* Apply load
!*
f, 2, fy, -load
finish  
!*
!* Solve
!*
/solu
antype, static
nlgeom, on
!autots, on
!solcontrol, on
!*
!* Load step 1
!*
time, 1.0
! f, 2, fy, -load
nsubst,100,0,0  
kbc, 0
neqit, 100
outres, ,1
arclen,on,100.0,0.0
lswrite
solve
finish  
!*
!* See solution
!*
/post26
!*
!* Save solution in variables 2 and 3
!*
nsol, 2, 2, u, x               ! Save ux at node 2
nsol, 3, 2, u, y               ! Save uy at node 2
!*
!* Scale solution
!*
prod, 4, 1, , , Load, , ,load  ! Scale time to get load
prod, 5, 2, , ,     , , ,-1    ! Make disp +ve
prod, 6, 3, , ,     , , ,-1    ! Make disp +ve
prvar, 4, 5, 6                 ! Print load, ux, uy
!*
!* Plot solution
!*
/axlab, x, Deflection
/axlab, y, Load
/grid, 1
xvar, 5                     
plvar, 4                       ! plot ux vs load
/noerase
xvar, 6
plvar, 4                       ! plot uy vs load
/erase

Here is another version of solution to this problem.

 
Figure 8. Force-displacement diagram for Problem 3B.2.
 
Figure 9. Shape deformation at the last load step for Problem 3B.2.

The ANSYS input code for Problem 3B.2 is listed below.

/prep7

A = 1
I = A/100
E = 1e6/I
nu = 0.3

et,1,beam188                ! Element type - BEAM188
sectype,1,beam,asec
secdata,A,I,,I,,2*I

mp,ex,1,E
mp,prxy,1,nu

pi = 4*atan(1)
phi = 35/2/180*pi
x = 100*cos(phi)
y = 100*sin(phi)

k,1,0,0,0
k,2,x,-y,0
k,3,0,100,0
k,4,-x,-y

k,5,0,0,100

larc,2,3,1,100
larc,3,4,1,100
latt,1,,1,,5,5,1

lesize,all,,,40
lmesh,all

dk,2,all,0
dk,4,ux,0
dk,4,uy,0
dk,4,uz,0

fk,3,fy,-900

/solu
nlgeom,on
nsubst,100,0,0
outres,all,all
arclen,on
solve
finish

Part 3Edit

Simulate the buckling of a hinged right-angle frame under both fixed and follower loads from Example 7.4 of Simo and Vu Quoc (1986) and compare your results with those shown in the paper.

The force-displacement diagram is shown in Figure 10. The deformation is illustrated in Figure 11 and 12.

 
Figure 10. Force-displacement diagram for Problem 3B.3.
 
Figure 11. Deformation (fixed load) at the last load step for Problem 3B.3.
 
Figure 12. Force-displacement diagram (follower load) for Problem 3B.3.

The ANSYS input code for Problem 3B.3 is listed below.

/prep7

et,1,188

E = 7.2e6
I = 2
A = 6

sectype,1,beam,asec
secdata,A,I,,I,,2*I

mp,ex,1,7.2e6
mp,prxy,1,0.3

k,1,0,0,0
k,2,0,120,0
k,3,23,120,0
k,4,26,120,0
k,5,120,120,0

k,6,200,0,0
k,7,0,200,0
l,1,2,5
l,2,3,1
l,3,4,2
l,4,5,4

latt,1,,1,,6,6,1
lmesh,1
latt,1,,1,,7,7,1
lmesh,2,4

d,1,ux,0
d,1,uy,0
d,1,uz,0

d,18,ux,0
d,18,uy,0
d,18,uz,0

esel,s,elem,,7,8

!replace the line below with f,15,fy,-40000 for fixed load

sfbeam,all,,pres,40000/2
esel,all

fini

/solu
nlgeom,on
outres,all,all
arclen,on
nsubst,200
solve
fini

Warning: The arc length method no longer converges with Ansys 13. Try using the stabilization option instead of arclen, on:

stabilize, constant, energy, 0.001, anytime, 0