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) $\delta _{ii}=3$

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

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 $i\neq j$ and $p\neq q$.

• 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) $e_{qrs}d_{q}d_{s}=0$

${\begin{aligned}e_{qrs}d_{q}d_{s}=&\sum _{q=1}^{3}\sum _{s=1}^{3}e_{qrs}d_{q}d_{s}\\=&({e_{1r1}d_{1}d_{1}})~~~+e_{1r2}d_{1}d_{2}+e_{1r3}d_{1}d_{3}+\\&e_{2r1}d_{2}d_{1}+({e_{2r2}d_{2}d_{2}})~~~+e_{2r3}d_{2}d_{3}+\\&e_{3r1}d_{3}d_{1}+e_{3r2}d_{3}d_{2}+({e_{3r3}d_{3}d_{3}})\end{aligned}}$

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

${\boldsymbol {\mathsf {C}}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}$

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

${\begin{aligned}{\boldsymbol {\mathit {1}}}&=\delta _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\{\boldsymbol {\mathsf {I}}}&={\frac {1}{2}}[\delta _{ik}\delta _{jl}+\delta _{il}\delta _{jk}]~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}\end{aligned}}$

The stress-strain relation is

${\boldsymbol {\sigma }}={\boldsymbol {\mathsf {C}}}:{\boldsymbol {\varepsilon }}$

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

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

Write the stress-strain relations in expanded form.

${\begin{aligned}{\boldsymbol {\sigma }}&={\boldsymbol {\mathsf {C}}}:{\boldsymbol {\varepsilon }}\\&=\left(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}\right):{\boldsymbol {\varepsilon }}\\&=\left(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}\right):{\boldsymbol {\varepsilon }}+2\mu ~{\boldsymbol {\mathsf {I}}}:{\boldsymbol {\varepsilon }}\end{aligned}}$

Now, in dyadic notation

${\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}=(\delta _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j})\otimes (\delta _{kl}~\mathbf {e} _{k}\otimes \mathbf {e} _{l})=\delta _{ij}\delta _{kl}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\otimes \mathbf {e} _{k}\otimes \mathbf {e} _{l}~{\text{and}}~{\boldsymbol {\varepsilon }}=\varepsilon _{kl}~\mathbf {e} _{k}\otimes \mathbf {e} _{l}$

Therefore

${\begin{aligned}({\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}):{\boldsymbol {\varepsilon }}&=\delta _{ij}\delta _{kl}\varepsilon _{kl}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=\delta _{ij}\varepsilon _{kk}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\end{aligned}}$

Similarly,

${\begin{aligned}{\boldsymbol {\mathsf {I}}}:{\boldsymbol {\varepsilon }}&=\left({\frac {1}{2}}[\delta _{ik}\delta _{jl}+\delta _{il}\delta _{jk}]\varepsilon _{kl}\right)~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&={\frac {1}{2}}\left(\delta _{ik}\varepsilon _{kj}+\delta _{il}\varepsilon _{jl}\right)~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&={\frac {1}{2}}\left(\varepsilon _{ij}+\varepsilon _{ji}\right)~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=\varepsilon _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\qquad {\text{(symmetry)}}\end{aligned}}$

The stress-strain law becomes

${\begin{aligned}{\boldsymbol {\sigma }}&=\lambda ~\delta _{ij}\varepsilon _{kk}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}+2\mu ~\varepsilon _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}\\&=\left(\lambda \delta _{ij}\varepsilon _{kk}+2\mu \varepsilon _{ij}\right)\mathbf {e} _{i}\otimes \mathbf {e} _{j}\end{aligned}}$

Expanding the left hand side, we get

$\sigma _{ij}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}=\left(\lambda \delta _{ij}\varepsilon _{kk}+2\mu \varepsilon _{ij}\right)\mathbf {e} _{i}\otimes \mathbf {e} _{j}$

Therefore,

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

Problem 2: Rotating Vectors and Tensors

Let ($\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}$) be an orthonormal basis. Let ${\boldsymbol {A}}$ be a second order tensor and $\mathbf {u}$ be a vector with components

${\begin{aligned}{\boldsymbol {A}}&=5~\mathbf {e} _{1}\otimes \mathbf {e} _{1}-4~\mathbf {e} _{2}\otimes \mathbf {e} _{1}+2~\mathbf {e} _{3}\otimes \mathbf {e} _{3}\\\mathbf {u} &=-2~\mathbf {e} _{1}+3~\mathbf {e} _{3}\end{aligned}}$

Solution

Part 1

Write out ${\boldsymbol {A}}$ and $\mathbf {u}$ in matrix notation.

${\mathbf {A} ={\begin{bmatrix}5&0&0\\-4&0&0\\0&0&2\end{bmatrix}};\qquad \mathbf {u} ={\begin{bmatrix}-2\\0\\3\end{bmatrix}}}$

Part 2

Find the components of the vector $\mathbf {v} ={\boldsymbol {A}}\bullet \mathbf {u}$ in the basis ($\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}$).

The components of $\mathbf {v}$ are

$v_{i}=A_{ij}u_{j}$

Therefore

$v_{1}=(5)(-2)=-10;~~v_{2}=(-4)(-2)=8;~~v_{3}=(2)(3)=6$

${\mathbf {v} =-10~\mathbf {e} _{1}+8\mathbf {e} _{2}+6\mathbf {e} _{3}}$

Part 3

Find the components of the vector $\mathbf {w} =\mathbf {v} \times \mathbf {u}$ in the basis ($\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}$).

The cross product is given by

$\mathbf {w} =\mathbf {v} \times \mathbf {u} ={\begin{vmatrix}\mathbf {e} _{1}&\mathbf {e} _{2}&\mathbf {e} _{3}\\v_{1}&v_{2}&v_{3}\\u_{1}&u_{2}&u_{3}\end{vmatrix}}={\begin{vmatrix}\mathbf {e} _{1}&\mathbf {e} _{2}&\mathbf {e} _{3}\\-10&8&6\\-2&0&3\end{vmatrix}}=(8)(3)\mathbf {e} _{1}-[(-10)(3)-(6)(-2)]\mathbf {e} _{2}-(8)(-2)\mathbf {e} _{3}$

Therefore,

${\mathbf {w} =24~\mathbf {e} _{1}+18\mathbf {e} _{2}+16\mathbf {e} _{3}}$

Part 4

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

The tensor product is given by

$\mathbf {v} \otimes \mathbf {w} =v_{i}w_{j}~\mathbf {e} _{i}\otimes \mathbf {e} _{j}$

Hence, in matrix notation

$\mathbf {C} ={\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}{\begin{bmatrix}w_{1}&w_{2}&w_{3}\end{bmatrix}}={\begin{bmatrix}-10\\8\\6\end{bmatrix}}{\begin{bmatrix}24&18&16\end{bmatrix}}$

${\mathbf {C} ={\begin{bmatrix}-240&-180&-160\\192&144&128\\144&108&96\end{bmatrix}}}$

Part 5

Rotate the basis clockwise by 30 degrees around the $\mathbf {e} _{3}$ direction. Find the components of $\mathbf {u}$, $\mathbf {v}$, $\mathbf {w}$, and ${\boldsymbol {C}}$ in the rotated basis.

The vector transformation rule is

$v_{i}^{'}=l_{ij}v_{i}$

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

In this case, the direction cosines are

${\begin{aligned}l_{11}&=\mathbf {e} _{1}^{'}\bullet \mathbf {e} _{1}=\cos(30)=0.866&l_{12}&=\mathbf {e} _{1}^{'}\bullet \mathbf {e} _{2}=\cos(60)=0.5&l_{13}&=\mathbf {e} _{1}^{'}\bullet \mathbf {e} _{3}=\cos(90)=0\\l_{21}&=\mathbf {e} _{2}^{'}\bullet \mathbf {e} _{1}=\cos(120)=-0.5&l_{22}&=\mathbf {e} _{2}^{'}\bullet \mathbf {e} _{2}=\cos(30)=0.866&l_{23}&=\mathbf {e} _{2}^{'}\bullet \mathbf {e} _{3}=\cos(90)=0\\l_{31}&=\mathbf {e} _{3}^{'}\bullet \mathbf {e} _{1}=\cos(90)=0&l_{32}&=\mathbf {e} _{3}^{'}\bullet \mathbf {e} _{2}=\cos(90)=0&l_{33}&=\mathbf {e} _{3}^{'}\bullet \mathbf {e} _{3}=\cos(0)=1\end{aligned}}$

Therefore,

$\mathbf {u} ^{'}=\mathbf {L} \mathbf {u} ={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}-2\\0\\3\end{bmatrix}}={\begin{bmatrix}-1.732\\1\\3\end{bmatrix}}$

${\mathbf {u} ^{'}=-1.732\mathbf {e} _{1}+1\mathbf {e} _{2}+3\mathbf {e} _{3}}$

Similarly,

$\mathbf {v} ^{'}=\mathbf {L} \mathbf {v} ={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}-10\\8\\6\end{bmatrix}}={\begin{bmatrix}-4.66\\11.93\\6\end{bmatrix}}$

${\mathbf {v} ^{'}=-4.66\mathbf {e} _{1}+11.93\mathbf {e} _{2}+6\mathbf {e} _{3}}$

Also,

$\mathbf {w} ^{'}=\mathbf {L} \mathbf {w} ={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}24\\18\\16\end{bmatrix}}={\begin{bmatrix}29.78\\3.59\\16\end{bmatrix}}$

${\mathbf {v} ^{'}=29.78\mathbf {e} _{1}+3.59\mathbf {e} _{2}+16\mathbf {e} _{3}}$

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

$T_{ij}^{'}=l_{ip}l_{jq}T_{pq}$

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

In matrix form,

$\mathbf {T} ^{'}=\mathbf {L} \mathbf {T} \mathbf {L} ^{T}$

Therefore the components of ${\boldsymbol {C}}$ in the rotated basis are give by

$\mathbf {C} ^{'}={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}-240&-180&-160\\192&144&128\\144&108&96\end{bmatrix}}{\begin{bmatrix}0.866&-0.5&0\\0.5&0.866&0\\0&0&1\end{bmatrix}}={\begin{bmatrix}-138.8&-16.7&-74.6\\355.3&42.8&190.9\\178.7&21.5&96\end{bmatrix}}$

${\mathbf {C} ^{'}={\begin{bmatrix}-138.8&-16.7&-74.6\\355.3&42.8&190.9\\178.7&21.5&96\end{bmatrix}}}$

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

$w_{0}(0)=u_{0}(L/2)=\varphi _{x}(L/2)=0~.$

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

$u_{0}(0)=w_{0}(0)=\varphi _{x}(0)=u_{0}(L/2)=\varphi _{x=L/2}=0.$

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.

Figure 4. Deformed shape under $M=10\pi$ for Problem 3B.1.

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

${\begin{aligned}I&={\cfrac {1}{12}}h^{4}~;&A&=h^{2}\\{\cfrac {EA}{EI}}&={\cfrac {A}{I}}~;&{\cfrac {12h^{2}}{h^{4}}}={\cfrac {12}{h^{2}}}=100\end{aligned}}$

Therefore, $h=0.34641$.

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

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

Nonlinear finite elements/Homework 7