# 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) ${\displaystyle A_{ms}=b_{m}(c_{r}-d_{r})}$

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

2) ${\displaystyle A_{ms}=b_{m}(c_{s}-d_{s})}$

Valid. Both ${\displaystyle m}$  and ${\displaystyle s}$  are free indices.

3) ${\displaystyle t_{i}=\sigma _{ji}n_{j}}$

Valid. The free index is ${\displaystyle i}$  and the dummy index is ${\displaystyle j}$ .

4) ${\displaystyle t_{i}=\sigma _{ji}n_{i}}$

Invalid. The free index is ${\displaystyle i}$  on the LHS and ${\displaystyle j}$  on the RHS.

5) ${\displaystyle x_{i}x_{i}=r^{3}}$

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

6) ${\displaystyle B_{ij}c_{j}=3}$

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

#### Part 2

Show the following:

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

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

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

The LHS is

${\displaystyle e_{ijk}e_{pqk}=e_{ij1}e_{pq1}+e_{ij2}e_{pq2}+e_{ij3}e_{pq3}}$

If ${\displaystyle i=j}$ , then we have

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

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

• Case 1: ${\displaystyle i=1}$ , ${\displaystyle j=2}$ , ${\displaystyle p=1}$ , ${\displaystyle q=2}$ .
${\displaystyle LHS=({e_{121}e_{121}})~~+({e_{122}e_{122}})~~+({e_{123}e_{123}})~~=1~;~~RHS=({\delta _{11}\delta _{11}})~~-({\delta _{12}\delta _{12}})~~=1}$
• Case 2: ${\displaystyle i=1}$ , ${\displaystyle j=2}$ , ${\displaystyle p=2}$ , ${\displaystyle q=1}$  or

${\displaystyle i=2}$ , ${\displaystyle j=1}$ , ${\displaystyle p=1}$ , ${\displaystyle q=2}$ .

${\displaystyle LHS=({e_{121}e_{211}})~~+({e_{122}e_{212}})~~+({e_{123}e_{213}})~~=-1~;~~RHS=({\delta _{12}\delta _{21}})~~-({\delta _{11}\delta _{22}})~~=-1}$
• Case 3: ${\displaystyle i=2}$ , ${\displaystyle j=1}$ , ${\displaystyle p=2}$ , ${\displaystyle q=1}$ .
${\displaystyle LHS=({e_{211}e_{211}})~~+({e_{212}e_{212}})~~+({e_{213}e_{213}})~~=1~;~~RHS=({\delta _{22}\delta _{11}})~~-({\delta _{21}\delta _{12}})~~=1}$
• Case 4: ${\displaystyle i=2}$ , ${\displaystyle j=3}$ , ${\displaystyle p=2}$ , ${\displaystyle q=3}$ .
${\displaystyle LHS=({e_{231}e_{231}})~~+({e_{232}e_{232}})~~+({e_{233}e_{233}})~~=1~;~~RHS=({\delta _{22}\delta _{22}})~~-({\delta _{23}\delta _{23}})~~=1}$
• Case 5: ${\displaystyle i=2}$ , ${\displaystyle j=3}$ , ${\displaystyle p=3}$ , ${\displaystyle q=2}$  or

${\displaystyle i=3}$ , ${\displaystyle j=2}$ , ${\displaystyle p=2}$ , ${\displaystyle q=3}$ .

${\displaystyle LHS=({e_{231}e_{321}})~~+({e_{232}e_{322}})~~+({e_{233}e_{323}})~~=-1~;~~RHS=({\delta _{23}\delta _{32}})~~-({\delta _{22}\delta _{33}})~~=-1}$
• Case 6: ${\displaystyle i=3}$ , ${\displaystyle j=2}$ , ${\displaystyle p=3}$ , ${\displaystyle q=2}$ .
${\displaystyle LHS=({e_{321}e_{321}})~~+({e_{322}e_{322}})~~+({e_{323}e_{323}})~~=1~;~~RHS=({\delta _{33}\delta _{22}})~~-({\delta _{32}\delta _{23}})~~=1}$
• Case 7: ${\displaystyle i=3}$ , ${\displaystyle j=1}$ , ${\displaystyle p=3}$ , ${\displaystyle q=1}$ .
${\displaystyle LHS=({e_{311}e_{311}})~~+({e_{312}e_{312}})~~+({e_{313}e_{313}})~~=1~;~~RHS=({\delta _{33}\delta _{11}})~~-({\delta _{31}\delta _{13}})~~=1}$
• Case 8: ${\displaystyle i=3}$ , ${\displaystyle j=1}$ , ${\displaystyle p=1}$ , ${\displaystyle q=3}$  or

${\displaystyle i=1}$ , ${\displaystyle j=3}$ , ${\displaystyle p=3}$ , ${\displaystyle q=1}$ .

${\displaystyle LHS=({e_{311}e_{131}})~~+({e_{312}e_{132}})~~+({e_{313}e_{133}})~~=-1~;~~RHS=({\delta _{31}\delta _{13}})~~-({\delta _{33}\delta _{11}})~~=-1}$
• Case 9: ${\displaystyle i=1}$ , ${\displaystyle j=3}$ , ${\displaystyle p=1}$ , ${\displaystyle q=3}$ .
${\displaystyle LHS=({e_{131}e_{131}})~~+({e_{132}e_{132}})~~+({e_{133}e_{133}})~~=1~;~~RHS=({\delta _{11}\delta _{33}})~~-({\delta _{13}\delta _{31}})~~=1}$

Hence the ${\displaystyle ~e-\delta }$  relation is satisfied for all cases.

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

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

4) ${\displaystyle e_{qrs}d_{q}d_{s}=0}$

{\displaystyle {\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 ${\displaystyle r=1}$ ,

${\displaystyle e_{qrs}d_{q}d_{s}=({e_{112}})~~~d_{1}d_{2}+({e_{113}})~~~d_{1}d_{3}+({e_{211}})~~~d_{2}d_{1}+({e_{213}})~~~d_{2}d_{3}+({e_{311}})~~~d_{3}d_{1}+({e_{312}})~~~d_{3}d_{2}=0}$

For ${\displaystyle r=2}$ ,

${\displaystyle e_{qrs}d_{q}d_{s}=({e_{122}})~~~d_{1}d_{2}+({e_{123}})~~~d_{1}d_{3}+({e_{221}})~~~d_{2}d_{1}+({e_{223}})~~~d_{2}d_{3}+({e_{321}})~~~d_{3}d_{1}+({e_{322}})~~~d_{3}d_{2}=0}$

For ${\displaystyle r=3}$ ,

${\displaystyle e_{qrs}d_{q}d_{s}=({e_{132}})~~~d_{1}d_{2}+({e_{133}})~~~d_{1}d_{3}+({e_{231}})~~~d_{2}d_{1}+({e_{233}})~~~d_{2}d_{3}+({e_{331}})~~~d_{3}d_{1}+({e_{332}})~~~d_{3}d_{2}=0}$

Hence shown.

#### Part 3

The elasticity tensor is given by

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

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

{\displaystyle {\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

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

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

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

Write the stress-strain relations in expanded form.

{\displaystyle {\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}}}

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

Therefore

{\displaystyle {\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,

{\displaystyle {\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

{\displaystyle {\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

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

Therefore,

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

## Problem 2: Rotating Vectors and Tensors

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

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

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

#### Part 2

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

The components of ${\displaystyle \mathbf {v} }$  are

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

Therefore

${\displaystyle v_{1}=(5)(-2)=-10;~~v_{2}=(-4)(-2)=8;~~v_{3}=(2)(3)=6}$
${\displaystyle {\mathbf {v} =-10~\mathbf {e} _{1}+8\mathbf {e} _{2}+6\mathbf {e} _{3}}}$

#### Part 3

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

The cross product is given by

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

Therefore,

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

#### Part 4

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

The tensor product is given by

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

Hence, in matrix notation

${\displaystyle \mathbf {C} ={\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}{\begin{bmatrix}w_{1}&w_{2}&w_{3}\end{bmatrix}}={\begin{bmatrix}-10\\8\\6\end{bmatrix}}{\begin{bmatrix}24&18&16\end{bmatrix}}}$
${\displaystyle {\mathbf {C} ={\begin{bmatrix}-240&-180&-160\\192&144&128\\144&108&96\end{bmatrix}}}}$

#### Part 5

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

The vector transformation rule is

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

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

In this case, the direction cosines are

{\displaystyle {\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,

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

Similarly,

${\displaystyle \mathbf {v} ^{'}=\mathbf {L} \mathbf {v} ={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}-10\\8\\6\end{bmatrix}}={\begin{bmatrix}-4.66\\11.93\\6\end{bmatrix}}}$
${\displaystyle {\mathbf {v} ^{'}=-4.66\mathbf {e} _{1}+11.93\mathbf {e} _{2}+6\mathbf {e} _{3}}}$

Also,

${\displaystyle \mathbf {w} ^{'}=\mathbf {L} \mathbf {w} ={\begin{bmatrix}0.866&0.5&0\\-0.5&0.866&0\\0&0&1\end{bmatrix}}{\begin{bmatrix}24\\18\\16\end{bmatrix}}={\begin{bmatrix}29.78\\3.59\\16\end{bmatrix}}}$
${\displaystyle {\mathbf {v} ^{'}=29.78\mathbf {e} _{1}+3.59\mathbf {e} _{2}+16\mathbf {e} _{3}}}$

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

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

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

In matrix form,

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

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

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

## Problem 3: More Beams

### Part A

Consider a beam of length ${\displaystyle L}$  = 100 in., cross-section 1 in. ${\displaystyle \times }$  1 in., and subjected to a uniformly distributed transverse load ${\displaystyle q_{0}}$  lbf/in. Model one half of the beam using symmetry considerations.

#### Part 1

Hinged-Hinged Beam

The boundary conditions are

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

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 ${\displaystyle U_{y}}$  at ${\displaystyle x=L/2}$
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

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

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 ${\displaystyle U_{y}}$  at ${\displaystyle x=L/2}$
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 ${\displaystyle U_{x}}$  ${\displaystyle U_{y}}$  Rotation
${\displaystyle M=10\pi }$  -0.040666 6.3205 -3.1287
${\displaystyle M=20\pi }$  9.9578 0.12729 -6.2577

The deformation plots are shown in Figure 4 and 5.

 Figure 4. Deformed shape under ${\displaystyle M=10\pi }$  for Problem 3B.1.
 Figure 5. Deformed shape under ${\displaystyle M=20\pi }$  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:${\displaystyle R=100}$ , ${\displaystyle \varphi =145^{o}}$ , ${\displaystyle EI=10^{6}}$ , and ${\displaystyle EA=100EI}$ . We assume a square cross section. Then

{\displaystyle {\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, ${\displaystyle h=0.34641}$ .

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