Nonlinear finite elements/Homework 7/Solutions

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Problem 1: Index Notation[edit]

Solutions[edit]

Part 1[edit]

  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 2[edit]

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 3[edit]

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 Tensors[edit]

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

Solution[edit]

Part 1[edit]

Write out and in matrix notation.

Part 2[edit]

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

The components of are

Therefore

Part 3[edit]

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

The cross product is given by

Therefore,

Part 4[edit]

Find the components of the tensor in the orthonormal basis.

The tensor product is given by

Hence, in matrix notation

Part 5[edit]

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 Beams[edit]

Part A[edit]

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 1[edit]

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 2[edit]

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 B[edit]

Part 1[edit]

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 2[edit]

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 3[edit]

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