Nonlinear finite elements/Homework 7/Solutions

Problem 1: Index Notation

Solutions

Part 1

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

1) $A_{ms}=b_m(c_r-d_r)$

Invalid. The free indices are $m,s$ on the LHS and $m,r$ on the RHS.

2) $A_{ms}=b_m(c_s-d_s)$

Valid. Both $m$ and $s$ are free indices.

3) $t_i={\sigma }_{ji}{n}_j$

Valid. The free index is $i$ and the dummy index is $j$.

4) $t_i={\sigma }_{ji}{n}_i$

Invalid. The free index is $i$ on the LHS and $j$ on the RHS.

5) $x_i{x}_i=r^3$

Valid. The dummy index is $i$. So the sum is a scalar which is equal to $r^3$.

6) $B_{ij}{c}_j=3$

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

Part 2

Show the following:

1) ${\delta }_{ii}=3$

${\delta }_{ii}={\delta }_{11}+{\delta }_{22}+{\delta }_{33}=1+1+1=3⟹{\delta }_{ii}=3$

2) $e_{ijk}{e}_{pqk}={\delta }_{ip}{\delta }_{jq}-{\delta }_{iq}{\delta }_{jp}$

The LHS is

$e_{ijk}{e}_{pqk}=e_{ij1}{e}_{pq1}+e_{ij2}{e}_{pq2}+e_{ij3}{e}_{pq3}$

If $i=j$, then we have

$LHS=(e_{iik})(e_{pqk})=0;RHS={\delta }_{ip}{\delta }_{iq}-{\delta }_{iq}{\delta }_{ip}=0$

We get the same result if $p=q$. The only nonzero LHS and RHS occur when $i\ne j$ and $p\ne q$.

• Case 1: $i=1$, $j=2$, $p=1$, $q=2$.

$LHS=(e_{121}{e}_{121})+(e_{122}{e}_{122})+(e_{123}{e}_{123})=1;RHS=({\delta }_{11}{\delta }_{11})-({\delta }_{12}{\delta }_{12})=1$

• Case 2: $i=1$, $j=2$, $p=2$, $q=1$ or

$i=2$, $j=1$, $p=1$, $q=2$.

$LHS=(e_{121}{e}_{211})+(e_{122}{e}_{212})+(e_{123}{e}_{213})=-1;RHS=({\delta }_{12}{\delta }_{21})-({\delta }_{11}{\delta }_{22})=-1$

• Case 3: $i=2$, $j=1$, $p=2$, $q=1$.

$LHS=(e_{211}{e}_{211})+(e_{212}{e}_{212})+(e_{213}{e}_{213})=1;RHS=({\delta }_{22}{\delta }_{11})-({\delta }_{21}{\delta }_{12})=1$

• Case 4: $i=2$, $j=3$, $p=2$, $q=3$.

$LHS=(e_{231}{e}_{231})+(e_{232}{e}_{232})+(e_{233}{e}_{233})=1;RHS=({\delta }_{22}{\delta }_{22})-({\delta }_{23}{\delta }_{23})=1$

• Case 5: $i=2$, $j=3$, $p=3$, $q=2$ or

$i=3$, $j=2$, $p=2$, $q=3$.

$LHS=(e_{231}{e}_{321})+(e_{232}{e}_{322})+(e_{233}{e}_{323})=-1;RHS=({\delta }_{23}{\delta }_{32})-({\delta }_{22}{\delta }_{33})=-1$

• Case 6: $i=3$, $j=2$, $p=3$, $q=2$.

$LHS=(e_{321}{e}_{321})+(e_{322}{e}_{322})+(e_{323}{e}_{323})=1;RHS=({\delta }_{33}{\delta }_{22})-({\delta }_{32}{\delta }_{23})=1$

• Case 7: $i=3$, $j=1$, $p=3$, $q=1$.

$LHS=(e_{311}{e}_{311})+(e_{312}{e}_{312})+(e_{313}{e}_{313})=1;RHS=({\delta }_{33}{\delta }_{11})-({\delta }_{31}{\delta }_{13})=1$

• Case 8: $i=3$, $j=1$, $p=1$, $q=3$ or

$i=1$, $j=3$, $p=3$, $q=1$.

$LHS=(e_{311}{e}_{131})+(e_{312}{e}_{132})+(e_{313}{e}_{133})=-1;RHS=({\delta }_{31}{\delta }_{13})-({\delta }_{33}{\delta }_{11})=-1$

• Case 9: $i=1$, $j=3$, $p=1$, $q=3$.

$LHS=(e_{131}{e}_{131})+(e_{132}{e}_{132})+(e_{133}{e}_{133})=1;RHS=({\delta }_{11}{\delta }_{33})-({\delta }_{13}{\delta }_{31})=1$

Hence the $e-\delta$ relation is satisfied for all cases.

3) ${\delta }_{ij}{e}_{ijk}=0$

${\delta }_{ij}{e}_{ijk}=e_{iik}=0.$

4) $e_{qrs}{d}_q{d}_s=0$

$\begin{array}{cc}{e}_{qrs}{d}_q{d}_s=& \sum _{q=1}^3\sum _{s=1}^3{e}_{qrs}{d}_q{d}_s\\ =& (e_{1r1}{d}_1{d}_1)+e_{1r2}{d}_1{d}_2+e_{1r3}{d}_1{d}_3+\\ & e_{2r1}{d}_2{d}_1+(e_{2r2}{d}_2{d}_2)+e_{2r3}{d}_2{d}_3+\\ & e_{3r1}{d}_3{d}_1+e_{3r2}{d}_3{d}_2+(e_{3r3}{d}_3{d}_3)\end{array}$

For $r=1$,

$e_{qrs}{d}_q{d}_s=(e_{112})d_1{d}_2+(e_{113})d_1{d}_3+(e_{211})d_2{d}_1+(e_{213})d_2{d}_3+(e_{311})d_3{d}_1+(e_{312})d_3{d}_2=0$

For $r=2$,

$e_{qrs}{d}_q{d}_s=(e_{122})d_1{d}_2+(e_{123})d_1{d}_3+(e_{221})d_2{d}_1+(e_{223})d_2{d}_3+(e_{321})d_3{d}_1+(e_{322})d_3{d}_2=0$

For $r=3$,

$e_{qrs}{d}_q{d}_s=(e_{132})d_1{d}_2+(e_{133})d_1{d}_3+(e_{231})d_2{d}_1+(e_{233})d_2{d}_3+(e_{331})d_3{d}_1+(e_{332})d_3{d}_2=0$

Hence shown.

Part 3

The elasticity tensor is given by

$𝖢=\lambda 1\otimes 1+2\mu 𝖨$

where $\lambda ,\mu$ are Lame constants, $1$ is the second order identity tensor, and $𝖨$ is the fourth-order symmetric identity tensor. The two identity tensors are defined as

$\begin{array}{cc}1& ={\delta }_{ij}{𝐞}_i\otimes {𝐞}_j\\ 𝖨& =\frac{1}{2}[{\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk}]{𝐞}_i\otimes {𝐞}_j\otimes {𝐞}_k\otimes {𝐞}_l\end{array}$

The stress-strain relation is

$\mathit{\sigma }=𝖢:\mathit{\epsilon }$

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

${\sigma }_{ij}=2\mu {\epsilon }_{ij}+\lambda {\epsilon }_{kk}{\delta }_{ij}.$

Write the stress-strain relations in expanded form.

$\begin{array}{cc}\mathit{\sigma }& =𝖢:\mathit{\epsilon }\\ & =(\lambda 1\otimes 1+2\mu 𝖨):\mathit{\epsilon }\\ & =(\lambda 1\otimes 1):\mathit{\epsilon }+2\mu 𝖨:\mathit{\epsilon }\end{array}$

Now, in dyadic notation

$1\otimes 1=({\delta }_{ij}{𝐞}_i\otimes {𝐞}_j)\otimes ({\delta }_{kl}{𝐞}_k\otimes {𝐞}_l)={\delta }_{ij}{\delta }_{kl}{𝐞}_i\otimes {𝐞}_j\otimes {𝐞}_k\otimes {𝐞}_l\text{and}\mathit{\epsilon }={\epsilon }_{kl}{𝐞}_k\otimes {𝐞}_l$

Therefore

$\begin{array}{cc}(1\otimes 1):\mathit{\epsilon }& ={\delta }_{ij}{\delta }_{kl}{\epsilon }_{kl}{𝐞}_i\otimes {𝐞}_j\\ & ={\delta }_{ij}{\epsilon }_{kk}{𝐞}_i\otimes {𝐞}_j\end{array}$

Similarly,

$\begin{array}{cc}𝖨:\mathit{\epsilon }& =(\frac{1}{2}[{\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk}]{\epsilon }_{kl}){𝐞}_i\otimes {𝐞}_j\\ & =\frac{1}{2}({\delta }_{ik}{\epsilon }_{kj}+{\delta }_{il}{\epsilon }_{jl}){𝐞}_i\otimes {𝐞}_j\\ & =\frac{1}{2}({\epsilon }_{ij}+{\epsilon }_{ji}){𝐞}_i\otimes {𝐞}_j\\ & ={\epsilon }_{ij}{𝐞}_i\otimes {𝐞}_j\text{(symmetry)}\end{array}$

The stress-strain law becomes

$\begin{array}{cc}\mathit{\sigma }& =\lambda {\delta }_{ij}{\epsilon }_{kk}{𝐞}_i\otimes {𝐞}_j+2\mu {\epsilon }_{ij}{𝐞}_i\otimes {𝐞}_j\\ & =(\lambda {\delta }_{ij}{\epsilon }_{kk}+2\mu {\epsilon }_{ij}){𝐞}_i\otimes {𝐞}_j\end{array}$

Expanding the left hand side, we get

${\sigma }_{ij}{𝐞}_i\otimes {𝐞}_j=(\lambda {\delta }_{ij}{\epsilon }_{kk}+2\mu {\epsilon }_{ij}){𝐞}_i\otimes {𝐞}_j$

Therefore,

${\sigma }_{ij}=\lambda {\delta }_{ij}{\epsilon }_{kk}+2\mu {\epsilon }_{ij}$

Problem 2: Rotating Vectors and Tensors

Let (${𝐞}_1,{𝐞}_2,{𝐞}_3$) be an orthonormal basis. Let $𝑨$ be a second order tensor and $𝐮$ be a vector with components

$\begin{array}{cc}𝑨& =5{𝐞}_1\otimes {𝐞}_1-4{𝐞}_2\otimes {𝐞}_1+2{𝐞}_3\otimes {𝐞}_3\\ 𝐮& =-2{𝐞}_1+3{𝐞}_3\end{array}$

Solution

Part 1

Write out $𝑨$ and $𝐮$ in matrix notation.

$𝐀=\left[\begin{array}{ccc}5& 0& 0\\ -4& 0& 0\\ 0& 0& 2\end{array}\right];𝐮=\left[\begin{array}{c}-2\\ 0\\ 3\end{array}\right]$

Part 2

Find the components of the vector $𝐯=𝑨\bullet 𝐮$ in the basis (${𝐞}_1,{𝐞}_2,{𝐞}_3$).

The components of $𝐯$ are

$v_i=A_{ij}{u}_j$

Therefore

$v_1=(5)(-2)=-10;v_2=(-4)(-2)=8;v_3=(2)(3)=6$

$𝐯=-10{𝐞}_1+8{𝐞}_2+6{𝐞}_3$

Part 3

Find the components of the vector $𝐰=𝐯\times 𝐮$ in the basis (${𝐞}_1,{𝐞}_2,{𝐞}_3$).

The cross product is given by

$𝐰=𝐯\times 𝐮=\left|\begin{array}{ccc}{𝐞}_1& {𝐞}_2& {𝐞}_3\\ v_1& v_2& v_3\\ u_1& u_2& u_3\end{array}\right|=\left|\begin{array}{ccc}{𝐞}_1& {𝐞}_2& {𝐞}_3\\ -10& 8& 6\\ -2& 0& 3\end{array}\right|=(8)(3){𝐞}_1-[(-10)(3)-(6)(-2)]{𝐞}_2-(8)(-2){𝐞}_3$

Therefore,

$𝐰=24{𝐞}_1+18{𝐞}_2+16{𝐞}_3$

Part 4

Find the components of the tensor $𝑪=𝐯\otimes 𝐰$ in the orthonormal basis.

The tensor product is given by

$𝐯\otimes 𝐰=v_i{w}_j{𝐞}_i\otimes {𝐞}_j$

Hence, in matrix notation

$𝐂=\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right]\left[\begin{array}{ccc}{w}_1& w_2& w_3\end{array}\right]=\left[\begin{array}{c}-10\\ 8\\ 6\end{array}\right]\left[\begin{array}{ccc}24& 18& 16\end{array}\right]$

$𝐂=\left[\begin{array}{ccc}-240& -180& -160\\ 192& 144& 128\\ 144& 108& 96\end{array}\right]$

Part 5

Rotate the basis clockwise by 30 degrees around the ${𝐞}_3$ direction. Find the components of $𝐮$, $𝐯$, $𝐰$, and $𝑪$ in the rotated basis.

The vector transformation rule is

$v_i^{\text{'}}=l_{ij}{v}_i$

where $l_{ij}$ are the direction cosines.

In this case, the direction cosines are

$\begin{array}{cccccc}{l}_{11}& ={𝐞}_1^{\text{'}}\bullet {𝐞}_1=\cos (30)=0.866& l_{12}& ={𝐞}_1^{\text{'}}\bullet {𝐞}_2=\cos (60)=0.5& l_{13}& ={𝐞}_1^{\text{'}}\bullet {𝐞}_3=\cos (90)=0\\ l_{21}& ={𝐞}_2^{\text{'}}\bullet {𝐞}_1=\cos (120)=-0.5& l_{22}& ={𝐞}_2^{\text{'}}\bullet {𝐞}_2=\cos (30)=0.866& l_{23}& ={𝐞}_2^{\text{'}}\bullet {𝐞}_3=\cos (90)=0\\ l_{31}& ={𝐞}_3^{\text{'}}\bullet {𝐞}_1=\cos (90)=0& l_{32}& ={𝐞}_3^{\text{'}}\bullet {𝐞}_2=\cos (90)=0& l_{33}& ={𝐞}_3^{\text{'}}\bullet {𝐞}_3=\cos (0)=1\end{array}$

Therefore,

${𝐮}^{\text{'}}=𝐋𝐮=\left[\begin{array}{ccc}0.866& 0.5& 0\\ -0.5& 0.866& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}-2\\ 0\\ 3\end{array}\right]=\left[\begin{array}{c}-1.732\\ 1\\ 3\end{array}\right]$

${𝐮}^{\text{'}}=-1.732{𝐞}_1+1{𝐞}_2+3{𝐞}_3$

Similarly,

${𝐯}^{\text{'}}=𝐋𝐯=\left[\begin{array}{ccc}0.866& 0.5& 0\\ -0.5& 0.866& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}-10\\ 8\\ 6\end{array}\right]=\left[\begin{array}{c}-4.66\\ 11.93\\ 6\end{array}\right]$

${𝐯}^{\text{'}}=-4.66{𝐞}_1+11.93{𝐞}_2+6{𝐞}_3$

Also,

${𝐰}^{\text{'}}=𝐋𝐰=\left[\begin{array}{ccc}0.866& 0.5& 0\\ -0.5& 0.866& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}24\\ 18\\ 16\end{array}\right]=\left[\begin{array}{c}29.78\\ 3.59\\ 16\end{array}\right]$

${𝐯}^{\text{'}}=29.78{𝐞}_1+3.59{𝐞}_2+16{𝐞}_3$

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

$T_{ij}^{\text{'}}=l_{ip}{l}_{jq}{T}_{pq}$

where $l_{ij}$ are the direction cosines.

In matrix form,

${𝐓}^{\text{'}}=𝐋𝐓{𝐋}^T$

Therefore the components of $𝑪$ in the rotated basis are give by

${𝐂}^{\text{'}}=\left[\begin{array}{ccc}0.866& 0.5& 0\\ -0.5& 0.866& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-240& -180& -160\\ 192& 144& 128\\ 144& 108& 96\end{array}\right]\left[\begin{array}{ccc}0.866& -0.5& 0\\ 0.5& 0.866& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-138.8& -16.7& -74.6\\ 355.3& 42.8& 190.9\\ 178.7& 21.5& 96\end{array}\right]$

${𝐂}^{\text{'}}=\left[\begin{array}{ccc}-138.8& -16.7& -74.6\\ 355.3& 42.8& 190.9\\ 178.7& 21.5& 96\end{array}\right]$

Problem 3: More Beams

Part A

Consider a beam of length $L$ = 100 in., cross-section 1 in. $\times$ 1 in., and subjected to a uniformly distributed transverse load $q_0$ lbf/in. Model one half of the beam using symmetry considerations.

Part 1

Hinged-Hinged Beam

The boundary conditions are

$w_0(0)=u_0(L/2)={\phi }_x(L/2)=0.$

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	$U_y$ at $x=L/2$
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

Clamped-Clamped Beam

The boundary conditions are

$u_0(0)=w_0(0)={\phi }_x(0)=u_0(L/2)={\phi }_{x=L/2}=0.$

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	$U_y$ at $x=L/2$
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

Part 1

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	$U_x$	$U_y$	Rotation
$M=10\pi$	-0.040666	6.3205	-3.1287
$M=20\pi$	9.9578	0.12729	-6.2577

The deformation plots are shown in Figure 4 and 5.

File:NFE HW7 3B1a.png Figure 4. Deformed shape under $M=10\pi$ for Problem 3B.1.

File:NFE HW7 3B1b.png Figure 5. Deformed shape under $M=20\pi$ 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

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:$R=100$, $\phi =145^o$, $EI=10^6$, and $EA=100EI$. We assume a square cross section. Then

$\begin{array}{cccc}I& =\frac{1}{12}{h}^4;& A& =h^2\\ \frac{EA}{EI}& =\frac{A}{I};& \frac{12h^2}{h^4}=\frac{12}{h^2}=100\end{array}$

Therefore, $h=0.34641$.

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

File:NFE HW7Prob3 2 def.png Figure 6. Deformed shape of arch.

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

File:NFE HW7Prob3 2 load.png 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.

File:NFE HW7 3B2xy.png Figure 8. Force-displacement diagram for Problem 3B.2.

File:NFE HW7 3B2.png 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

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.

File:NFE HW7 3B3xy.png Figure 10. Force-displacement diagram for Problem 3B.3.

File:NFE HW7 3B3a.png Figure 11. Deformation (fixed load) at the last load step for Problem 3B.3.

File:NFE HW7 3B3b.png 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