text:
Nonlinear finite elements/Homework 7/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\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 $$

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

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 $$

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

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 $$

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

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{align}

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{align}$$ For $$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{align}

\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{align}$$ 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{align}

\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{align}$$ 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{align}

(\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{align}$$ Similarly,
 * $$\begin{align}

\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{align}$$ The stress-strain law becomes
 * $$\begin{align}

\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{align}$$ 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{align}

\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{align}$$

Part 1
Write out $$\boldsymbol{A}$$ and $$\mathbf{u}$$ in matrix notation.

$$

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 $$

$$

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,

$$

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} $$

$$

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{align}

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{align}$$ 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} $$

$$ 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} $$

$$ 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} $$

$$ 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} $$

$$

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.

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.

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

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.

The deformation plots are shown in Figure 4 and 5.

The ANSYS input code for this problem is listed below.

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 $$EA = 100 EI$$. We assume a square cross section. Then
 * $$\begin{align}

I & = \cfrac{1}{12} h^4~; & A & = h^2 \\ \cfrac{EA}{EI} & = \cfrac{A}{I}~; & \cfrac{12 h^2}{h^4} = \cfrac{12}{h^2} = 100 \end{align}$$ Therefore, $$h = 0.34641$$.

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

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

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.

Here is another version of solution to this problem.

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

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.

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

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


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Film School Computer Disks

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:100%" Disks from the Star Movie Shop, Triplicane, Tamil Nadu
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Required disks for the filmmaking, film editing, and film scoring classes

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{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:100%" An introduction to filming and editing a conversation
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"Follow Dave?" Editing Workshop CD

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{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:100%" A look at the filming of dramatic scenes in Hollywood
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"24 Unedited Scenes"

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Hearing Voices

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{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:100%" An in depth look at "L-Cuts"
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Fireworks Display

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 * 1) Emphasizing the person who has just finished speaking. (Reaction)
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Blindman's Bluff

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{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:100%" For film scoring
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Home Wrecker

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Green Tea Smoothie

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Sweet Retrobution

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title: Film School Computer Disks


text:
Nonlinear finite elements/Homework 8/Solutions

Problem 1: Plate Bending
 Given:

Consider the flat plate shown in Figure 1.

Transverse normals are defined as the straight lines perpendicular to the midsurface.

If we assume that the plate is thin, the Kirchhoff-Love hypothesis holds, i.e.,
 * Transverse normals remain straight.
 * Transverse normals remain normal to the midsurface.
 * Transverse normals are inextensible.

Part 1
The displacement field that satisfies the Kirchhoff-Love hypothesis is given by
 * $$\begin{align}

u_1(x_1,x_2,x_3) & = u^0_1(x_1,x_2) - x_3~u^0_{3,1} \\ u_2(x_1,x_2,x_3) & = u^0_2(x_1,x_2) - x_3~u^0_{3,2} \\ u_3(x_1,x_2,x_3) & = u^0_3(x_1,x_2) \end{align}$$ where $$u^0_i$$ are the displacements of the midsurface.

Write the expanded form of the above equations after making the following substitutions.
 * $$\text{(2)} \qquad

\begin{align} & x := x_1 ~; y := x_2 ~; z := x_3 \\ & u := u_1 ~; v := u_2 ~; w := u_3 \\ & u_0 := u^0_1 ~; v_0 := u^0_2 ~; w_0 := u^0_3 ~. \end{align} $$

After making the substitutions we get
 * $$\begin{align}

u(x,y,z) & = u_0(x,y) - z~w_{0,1} \\ v(x,y,z) & = v_0(x,y) - z~w_{0,2} \\ w(x,y,z) & = w_0(x,y) \end{align}$$ The expanded form is

\begin{align} u(x,y,z) & = u_0(x,y) - z~\frac{\partial w_0}{\partial x} \\ v(x,y,z) & = v_0(x,y) - z~\frac{\partial w_0}{\partial y} \\ w(x,y,z) & = w_0(x,y) \end{align} $$

Part 2
Assume that the plate undergoes infinitesimal strains and rotations. Then the strain-displacement relations are

\boldsymbol{\varepsilon} = \frac{1}{2}\left[\boldsymbol{\nabla} \mathbf{u} + (\boldsymbol{\nabla} \mathbf{u})^T\right]~. $$ Show that, in terms of the midsurface displacements, the strain-displacement relations can be written in index notation as
 * $$\begin{align}

\varepsilon_{\alpha\beta} & = \frac{1}{2} \left(u^0_{\alpha,\beta} + u^0_{\beta,\alpha}\right) - x_3~u^0_{3,\alpha\beta} \\ \varepsilon_{\alpha 3} & = 0 \\ \varepsilon_{33} & = 0 \end{align}$$ where $$\alpha = 1,2$$ and $$\beta = 1,2$$.

The strain-displacement relations in index notation are

\varepsilon_{ij} = \frac{1}{2}\left[u_{i,j} + u_{j,i}\right] $$ where $$i,j = 1,2,3$$. If the indices $$\alpha$$ and $$\beta$$ run from 1 to 2, we can write the above strain-displacement relations as
 * $$\begin{align}

\varepsilon_{\alpha\beta} & = \frac{1}{2}\left[u_{\alpha,\beta} + u_{\beta,\alpha}\right] \\ \varepsilon_{\alpha 3}= \varepsilon_{3\alpha} & = \frac{1}{2}\left[u_{\alpha,3} + u_{3,\alpha}\right] \\ \varepsilon_{33}& = \frac{1}{2}\left[u_{3,3} + u_{3,3}\right] = u_{3,3} \end{align}$$ The displacements are given by
 * $$\begin{align}

u_1 & = u^0_1 - x_3~u^0_{3,1} \\ u_2 & = u^0_2 - x_3~u^0_{3,2} \\ u_3 & = u^0_3 \end{align}$$ If the index $$\alpha$$ runs from 1 to 2, the displacements can be written as
 * $$\begin{align}

u_\alpha & = u^0_{\alpha} - x_3~u^0_{3,\alpha} \\ u_3 & = u^0_3 \end{align}$$ The partial derivatives of the displacements are (since $$u^0_1, u^0_2, u^0_3$$ are functions only of $$x_1$$ and $$x_2$$)
 * $$\begin{align}

u_{\alpha,\beta} & = u^0_{\alpha,\beta} - x_3~u^0_{3,\alpha\beta} \\ u_{\alpha, 3} & = u^0_{\alpha, 3} - x_3~u^0_{3,\alpha 3} = 0 \\ u_{3,\alpha} & = u^0_{3,\alpha} = 0 \\ u_{3,3} & = u^0_{3,3} = 0 \end{align}$$ Therefore,

\begin{align} \varepsilon_{\alpha\beta} & = \frac{1}{2}\left[u^0_{\alpha,\beta} + u^0_{\beta,\alpha}\right] - x_3~u^0_{3,\alpha\beta} \\ \varepsilon_{\alpha 3}= \varepsilon_{3\alpha} & = 0 \\ \varepsilon_{33}& = 0 \end{align} $$

Part 3
The internal virtual work in the plate is given by (using index notation)

\delta W_{\text{int}} = \int_{\Omega_0}\left(\int_{-h}^{h} \sigma_{\alpha\beta}~\delta \varepsilon_{\alpha\beta}~dx_3\right) d\Omega_0 ~. $$ Note that if a component of the strain tensor is zero, then the variation of that strain component is also zero.

Show that the internal virtual work can be written as
 * $$\text{(3)} \qquad

\delta W_{\text{int}} = \int_{\Omega_0}\left( N_{\alpha\beta}\delta \varepsilon_{\alpha\beta}^0 + M_{\alpha\beta}\delta \varepsilon_{\alpha\beta}^1\right)~d\Omega_0 $$ where
 * $$\begin{align}

N_{\alpha\beta} & = \int_{-h}^h \sigma_{\alpha\beta}~dx_3 \\ M_{\alpha\beta} & = \int_{-h}^h x_3~\sigma_{\alpha\beta}~dx_3 \\ \varepsilon_{\alpha\beta}^0 & = \frac{1}{2} \left(u^0_{\alpha,\beta} + u^0_{\beta,\alpha}\right) \\ \varepsilon_{\alpha\beta}^1 & = - u^0_{3,\alpha\beta} ~. \end{align}$$ Write down the expanded form of equation (3).

The strains are given by

\varepsilon_{\alpha\beta} = \frac{1}{2}\left[u^0_{\alpha,\beta} + u^0_{\beta,\alpha}\right] - x_3~u^0_{3,\alpha\beta} = \varepsilon_{\alpha\beta}^0 + x_3~\varepsilon_{\alpha\beta}^1 $$ Therefore, the variations of the strains are

\delta \varepsilon_{\alpha\beta} = \delta \varepsilon_{\alpha\beta}^0 + x_3~\delta \varepsilon_{\alpha\beta}^1 $$ The internal virtual work is
 * $$\begin{align}

\delta W_{\text{int}} & = \int_{\Omega_0}\left[\int_{-h}^{h} \sigma_{\alpha\beta}~\left( \delta \varepsilon_{\alpha\beta}^0 + x_3~\delta \varepsilon_{\alpha\beta}^1\right) ~dx_3\right] d\Omega_0\\ & = \int_{\Omega_0}\left[\delta \varepsilon_{\alpha\beta}^0\int_{-h}^{h} \sigma_{\alpha\beta}~dx_3 + \delta \varepsilon_{\alpha\beta}^1\int_{-h}^{h} x_3~\sigma_{\alpha\beta}~dx_3\right]~d\Omega_0 \\ & = \int_{\Omega_0}\left[\delta \varepsilon_{\alpha\beta}^0~N_{\alpha\beta} + \delta \varepsilon_{\alpha\beta}^1~M_{\alpha\beta} \right]~d\Omega_0 \end{align}$$ Therefore,

\delta W_{\text{int}} = \int_{\Omega_0}\left( N_{\alpha\beta}\delta \varepsilon_{\alpha\beta}^0 + M_{\alpha\beta}\delta \varepsilon_{\alpha\beta}^1\right)~d\Omega_0 $$ The expanded form of the above equation is

\begin{align} \delta W_{\text{int}} = \int_{\Omega_0}( & N_{11}\delta \varepsilon_{11}^0 + N_{12}\delta \varepsilon_{12}^0 +  N_{21}\delta \varepsilon_{21}^0 + N_{22}\delta \varepsilon_{22}^0 + \\ & M_{11}\delta \varepsilon_{11}^1 + M_{12}\delta \varepsilon_{12}^1 +  M_{21}\delta \varepsilon_{21}^1 + M_{22}\delta \varepsilon_{22}^1)~d\Omega_0 \end{align} $$ or,

\begin{align} \delta W_{\text{int}} = \int_{\Omega_0}( & N_{11}\delta \varepsilon_{11}^0 + 2 N_{12}\delta \varepsilon_{12}^0 +  N_{22}\delta \varepsilon_{22}^0 + \\ & M_{11}\delta \varepsilon_{11}^1 + 2 M_{12}\delta \varepsilon_{12}^1 +  M_{22}\delta \varepsilon_{22}^1)~d\Omega_0 \end{align} $$

Part 4
The  equilibrium equations for the plate can be written as
 * $$\begin{align}

\frac{\partial N_{xx}}{\partial x} + \frac{\partial N_{xy}}{\partial y} & = 0 \\ \frac{\partial N_{xy}}{\partial x} + \frac{\partial N_{yy}}{\partial y} & = 0 \\ \frac{\partial^2 M_{xx}}{\partial x^2} + 2\frac{\partial^2 M_{xy}}{\partial x \partial y} + \frac{\partial^2 M_{yy}}{\partial y^2} + \mathcal{N}(u_0,v_0,w_0) + q & = 0 \end{align}$$ where

\mathcal{N}(u_0,v_0,w_0) = \frac{\partial }{\partial x} \left(N_{xx}\frac{\partial w_0}{\partial x} + N_{xy}\frac{\partial w_0}{\partial y}\right) + \frac{\partial }{\partial y}\left(N_{xy}\frac{\partial w_0}{\partial x} + N_{yy}\frac{\partial w_0}{\partial y} \right) ~. $$ Write the above equations in index notation.

The equilibrium equations are (written using indices $$1, 2, 3$$):
 * $$\begin{align}

& N_{11,1} + N_{12,2} = 0 \\ & N_{21,1} + N_{22,2} = 0 \\ & M_{11,11} + 2M_{12,12} + M_{22,22} + \left(N_{11}~u^0_{3,1} + N_{12}~u^0_{3,2}\right)_{,1} + \left(N_{21}~u^0_{3,1} + N_{22}~u^0_{3,2}\right)_{,2} + q = 0 \end{align}$$ Using the summation convention, we get
 * $$\begin{align}

& N_{1\beta, \beta} = 0 \\ & N_{2\beta, \beta} = 0 \\ & M_{\alpha\beta,\alpha\beta} + \left(N_{1\beta}~u^0_{3,\beta}\right)_{,1} + \left(N_{2\beta}~u^0_{3,\beta}\right)_{,2} + q = 0 \end{align}$$ The first two equations can then be written as

N_{\alpha\beta, \beta} = 0 $$ and the third equation can be written as

M_{\alpha\beta,\alpha\beta} + \left(N_{\alpha\beta}~u^0_{3,\beta}\right)_{,\alpha} + q = 0 $$ Therefore the equilibrium equations in index notation are

\begin{align} & N_{\alpha\beta, \beta} = 0 \\ & M_{\alpha\beta,\alpha\beta} + \left(N_{\alpha\beta}~u^0_{3,\beta}\right)_{,\alpha} + q = 0 \end{align} $$

Part 5
Show that in index-free  Gibbs notation the equilibrium equations for the plate can be written as
 * $$\begin{align}

\boldsymbol{\nabla} \bullet \boldsymbol{N} & = \mathbf{0} \text{(4)} \qquad \\ \boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M}) + \boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\boldsymbol{\nabla u^0_3)} + q & = 0 \text{(5)} \qquad \end{align}$$ where

\begin{align} \boldsymbol{N} & = N_{\alpha\beta}~\mathbf{e}_{\alpha}\otimes\mathbf{e}_{\beta} ~;& \boldsymbol{M} & = M_{\alpha\beta}~\mathbf{e}_{\alpha}\otimes\mathbf{e}_{\beta}\\ \boldsymbol{\nabla} \varphi & = \varphi_{,\alpha} ~\mathbf{e}_{\alpha} ~;& \boldsymbol{\nabla} \mathbf{v} & = v_{\alpha,\beta} ~\mathbf{e}_{\alpha}\otimes\mathbf{e}_{\beta} ~;& \boldsymbol{\nabla} \boldsymbol{S} & = S_{\alpha\beta,\gamma} ~\mathbf{e}_{\alpha}\otimes\mathbf{e}_{\beta}\otimes\mathbf{e}_{\gamma} \\ \boldsymbol{\nabla} \bullet \mathbf{v} & = v_{\alpha,\alpha} ~;& \boldsymbol{\nabla} \bullet \boldsymbol{S} & = S_{\alpha\beta,\beta}~\mathbf{e}_{\alpha} ~;& \boldsymbol{\nabla} \bullet \boldsymbol{T} & = T_{\alpha\beta\gamma,\gamma} ~\mathbf{e}_{\alpha}\otimes\mathbf{e}_{\beta} \end{align} $$ and $$\alpha=1,2$$, $$\beta=1,2$$, and $$\gamma=1,2$$.

Equation (4) in index notation is

\boldsymbol{\nabla} \bullet \boldsymbol{N} = N_{\alpha\beta,\beta}~\mathbf{e}_{\alpha} = \mathbf{0}\qquad \implies N_{\alpha\beta,\beta} = 0 $$ The above equations have been shown to be equivalent to the first two equilibrium equations in the previous part.

Let us now look at equation (5). Let

\mathbf{p} := \boldsymbol{\nabla} \bullet \boldsymbol{M} \qquad \implies p_{\alpha}\mathbf{e}_{\alpha} = M_{\alpha\beta,\beta} \mathbf{e}_{\alpha} $$ Then

\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M}) = \boldsymbol{\nabla} \bullet \mathbf{p} = p_{\alpha,\alpha} = M_{\alpha\beta,\alpha\beta} $$ For the next term, let

\mathbf{w} := \boldsymbol{\nabla} u^0_3 \qquad \implies w_{\alpha}\mathbf{e}_\alpha = u^0_{3,\alpha}\mathbf{e}_\alpha \implies w_{\alpha} = u^0_{3,\alpha} $$ Then,

\mathbf{q} := \boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3 = \boldsymbol{N}\bullet\mathbf{w} \qquad \implies q_{\alpha}~\mathbf{e}_\alpha = N_{\alpha\beta}~w_{\beta}~\mathbf{e}_\alpha = N_{\alpha\beta}~u^0_{3,\beta}~\mathbf{e}_\alpha $$ Hence

\boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\boldsymbol{\nabla u^0_3)} = \boldsymbol{\nabla} \bullet \mathbf{q} = q_{\alpha,\alpha} = (N_{\alpha\beta}~u^0_{3,\beta})_{,\alpha} $$ Therefore, equation (5) can be written as

M_{\alpha\beta,\alpha\beta}+(N_{\alpha\beta}~u^0_{3,\beta})_{,\alpha} + q = 0 $$ The above equation has been shown to be equivalent to the third equilibrium equations in the previous part.

Part 6
Derive the symmetric weak forms of the equilibrium equations starting with equations (4) and (5). Use the coordinate-free Gibbs notation in your derivation.

For equation 4, we multiply the equation by a vector weighting function $$\mathbf{w}_1$$ and integrate over the midsurface $$\Omega_0$$. Then,
 * $$\begin{align}

\int_{\Omega_0} \mathbf{w}_1\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{N})~d\Omega_0 & = 0 \end{align}$$ Using the identity

\boldsymbol{\nabla} \bullet (\boldsymbol{S}^T\bullet\mathbf{v}) = \mathbf{v}\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{S}) + \boldsymbol{S}:(\boldsymbol{\nabla} \mathbf{v}) $$ we get

\mathbf{w}_1\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{N}) = \boldsymbol{\nabla} \bullet (\boldsymbol{N}^T\bullet\mathbf{w}_1) - \boldsymbol{N}:(\boldsymbol{\nabla} \mathbf{w}_1) $$ From the symmetry to the stress tensor $$\boldsymbol{N} = \boldsymbol{N}^T$$. Hence,

\mathbf{w}_1\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{N}) = \boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\mathbf{w}_1) - \boldsymbol{N}:(\boldsymbol{\nabla} \mathbf{w}_1) $$ Therefore
 * $$\begin{align}

\int_{\Omega_0} \mathbf{w}_1\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{N})~d\Omega_0 & = \int_{\Omega_0} \boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\mathbf{w}_1)~d\Omega_0 - \int_{\Omega_0} \boldsymbol{N}:(\boldsymbol{\nabla} \mathbf{w}_1)~d\Omega_0 \\ & = \int_{\Gamma_0} \mathbf{n}\bullet(\boldsymbol{N}\bullet\mathbf{w}_1)~d\Gamma_0 - \int_{\Omega_0} \boldsymbol{N}:(\boldsymbol{\nabla} \mathbf{w}_1)~d\Omega_0 \end{align}$$ where $$\mathbf{n}$$ is the normal to the boundary $$\Gamma_0$$ of the midsurface.

Hence the symmetric weak form of the first equation is

\int_{\Omega_0} \boldsymbol{N}:\boldsymbol{\nabla} \mathbf{w}_1~d\Omega_0 = \int_{\Gamma_0} \mathbf{n}\bullet(\boldsymbol{N}\bullet\mathbf{w}_1)~d\Gamma_0 $$

For equation 5, we multiply the equation by a scalar weighting function $$w_2$$ and integrate over the midsurface $$\Omega_0$$. Then,

\int_{\Omega_0} w_2\left[\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M}) + \boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\boldsymbol{\nabla u^0_3)} + q\right] ~d\Omega_0 = 0 $$ Using the identity,

\boldsymbol{\nabla} \bullet (\varphi\mathbf{v}) = \varphi (\boldsymbol{\nabla} \bullet \mathbf{v}) + \mathbf{v}\bullet(\boldsymbol{\nabla} \varphi) $$ we get

w_2[\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M})] = \boldsymbol{\nabla} \bullet [w_2(\boldsymbol{\nabla}\bullet\boldsymbol{M})] - (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2) $$ Therefore,
 * $$\begin{align}

\int_{\Omega_0} w_2[\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M})]~d\Omega_0 & = \int_{\Omega_0} \boldsymbol{\nabla} \bullet [w_2(\boldsymbol{\nabla}\bullet\boldsymbol{M})]~d\Omega_0 - \int_{\Omega_0} (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2)~d\Omega_0 \\ & = \int_{\Gamma_0} \mathbf{n}\bullet[w_2(\boldsymbol{\nabla} \bullet \boldsymbol{M})]~d\Gamma_0 - \int_{\Omega_0} (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2)~d\Omega_0 \\ & = \int_{\Gamma_0} w_2~[\mathbf{n}\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{M})]~d\Gamma_0 - \int_{\Omega_0} (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2)~d\Omega_0 \end{align}$$ In order to have natural BCs directly in terms of moments, we have to perform a similar set of operations the last term above. Thus,
 * $$\begin{align}

\int_{\Omega_0} (\boldsymbol{\nabla} \bullet \boldsymbol{M})\bullet(\boldsymbol{\nabla} w_2)~d\Omega_0 & = \int_{\Omega_0} \boldsymbol{\nabla} \bullet [\boldsymbol{M}\bullet(\boldsymbol{\nabla w_2)]}~d\Omega_0 - \int_{\Omega_0} \boldsymbol{M}:[\boldsymbol{\nabla} (\boldsymbol{\nabla}{w_2)}]~d\Omega_0 \\ & = \int_{\Gamma_0} \mathbf{n}\bullet[\boldsymbol{M}\bullet{(\boldsymbol{\nabla} w_2)}]~d\Gamma_0 - \int_{\Omega_0} \boldsymbol{M}:[\boldsymbol{\nabla} (\boldsymbol{\nabla} w_2)]~d\Omega_0 \end{align}$$ Hence,

\int_{\Omega_0} w_2~\boldsymbol{\nabla} \bullet (\boldsymbol{\nabla}\bullet\boldsymbol{M})~d\Omega_0= \int_{\Gamma_0} \left\{w_2~\mathbf{n}\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{M})- \mathbf{n}\bullet\boldsymbol{M}\bullet(\boldsymbol{\nabla} w_2)\right\} d\Gamma_0 + \int_{\Omega_0} \boldsymbol{M}:\left(\boldsymbol{\nabla} \boldsymbol{\nabla}{w_2}\right)~d\Omega_0 $$ Similarly,
 * $$\begin{align}

\int_{\Omega_0} w_2\left[\boldsymbol{\nabla} \bullet (\boldsymbol{N}\bullet\boldsymbol{\nabla u^0_3)}\right]~d\Omega_0&= \int_{\Omega_0} \boldsymbol{\nabla} \bullet [w_2(\boldsymbol{N}\bullet\boldsymbol{\nabla u^0_3)]}~d\Omega_0 - \int_{\Omega_0} (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet(\boldsymbol{\nabla} w_2) ~d\Omega_0 \\ & = \int_{\Gamma_0} \mathbf{n}\bullet[w_2(\mathbf{d}ot{\boldsymbol{N}{\boldsymbol{\nabla} u^0_3})]}~d\Gamma_0 - \int_{\Omega_0} (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet(\boldsymbol{\nabla} w_2) ~d\Omega_0 \\ & = \int_{\Gamma_0} w_2~[\mathbf{n}\bullet(\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)]~d\Gamma_0 - \int_{\Omega_0} (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet(\boldsymbol{\nabla} w_2) ~d\Omega_0 \end{align}$$ Therefore, the symmetric weak form of the second equation is
 * $$\begin{align}

\int_{\Omega_0} & \left[ (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet\boldsymbol{\nabla} w_2 - \boldsymbol{M}:\left(\boldsymbol{\nabla} \boldsymbol{\nabla} w_2\right) \right]~d\Omega_0 = \\ & \int_{\Omega_0} w_2~q~d\Omega_0 + \int_{\Gamma_0} \left[w_2\left\{ \mathbf{n}\bullet(\boldsymbol{\nabla} \bullet \boldsymbol{M}) + \mathbf{n}\bullet\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3\right\} - \mathbf{n}\bullet(\boldsymbol{M}\bullet\boldsymbol{\nabla} w_2) \right]~d\Gamma_0 \end{align}$$ Using the distributive law for vectors

(\mathbf{a}+\mathbf{b})\bullet\mathbf{c} = \mathbf{a}\bullet\mathbf{c} + \mathbf{b}\bullet\mathbf{c} $$ we can also write the symmetric weak form for the second equation as:

\begin{align} \int_{\Omega_0} & \left[ (\boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3)\bullet{\boldsymbol{\nabla} w_2} - \boldsymbol{M}:\left(\boldsymbol{\nabla} \boldsymbol{\nabla} w_2\right) \right]~d\Omega_0 = \\ & \int_{\Omega_0} w_2~q~d\Omega_0 + \int_{\Gamma_0} \mathbf{n}\bullet\left( \boldsymbol{\nabla} \bullet \boldsymbol{M} + \boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3\right)~w_2~d\Gamma_0 - \int_{\Gamma_0} \mathbf{n}\bullet(\boldsymbol{M}\bullet\boldsymbol{\nabla} w_2)~d\Gamma_0 \end{align} $$

Part 7
Write down the natural boundary conditions in Gibbs notation.

From the expressions for the weak form, the weighted natural boundary conditions are

\begin{align} & \int_{\Gamma_0} \mathbf{n}\bullet(\mathbf{d}ot{\boldsymbol{N}{\mathbf{w}_1})}~d\Gamma_0 \\ & \int_{\Gamma_0} \mathbf{n}\bullet\left( \boldsymbol{\nabla} \bullet \boldsymbol{M} + \boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3\right)~w_2~d\Gamma_0 \\ & - \int_{\Gamma_0} \mathbf{n}\bullet(\mathbf{d}ot{\boldsymbol{M}{\boldsymbol{\nabla} w_2})}~d\Gamma_0 \end{align} $$ Therefore, the natural boundary conditions are:

\begin{align} \text{Boundary Membrane Forces}~ & & \mathbf{n}\bullet\boldsymbol{N} \\ \text{Boundary Transverse Forces}~ & & \mathbf{n}\bullet\left(\boldsymbol{\nabla} \bullet \boldsymbol{M} + \boldsymbol{N}\bullet\boldsymbol{\nabla} u^0_3\right)\\ \text{Boundary Moments}~ & & \mathbf{n}\bullet\boldsymbol{M} \end{align} $$

Part 8
What are the essential boundary conditions that can be applied to the plate?

The essential boundary conditions that can be applied are

\begin{align} \text{Boundary Membrane Displacements}~ & & u^0_1, u^0_2 \\ \text{Boundary Transverse Displacements}~ & & u^0_3 \\ \text{Boundary Out-of-plane Rotations}~ & & \boldsymbol{\nabla} u^0_3 \end{align} $$

Problem 2: Transient Heat Conduction
 Given:

The model problem shown in Figure 6.

A 4340 steel casing is used to contain a solid high energy (HE) material. The HE material reacts on the inner (star-shaped) surface and generates gases. Since the gases are confined, the temperature of the gases increases according to the ideal gas law ($$PV = nRT$$).

The inner radius of the steel casing is 5 cm. The outer radius of the steel casing is 6 cm. The inner and outer radii of the star shaped region are 4 cm and 2 cm, respectively.

The high energy material has properties: Density = 1200 kg/m$$^3$$, Young's modulus = 1 GPa, Poisson's ratio = 0.49, Coefficient of thermal expansion = 1.2$$\times$$10$$^{-4}$$ /K, Thermal Conductivity = 5 W/m-K, Specific Heat Capacity = 1000 J/kg-K.

The burn-surface does not change position. The temperature is the same at all points on the burn surface. The temperature at the burn surface increases exponentially with time according to the relation

T = T_0 + A~\exp{(B~t)} $$ where $$T_0$$ is the initial temperature (300 K), $$T$$ is the temperature at time $$t$$, $$A = 2$$ K, and $$B = 3$$ K/s.

Solution
The properties of normalized 4340 steel from Matweb are

Part 1
Assume that the initial temperature of the steel and the HE material is 300 K. Also assume that there is no heat flux on the external surface of the steel casing. Plot the temperature contours and stresses in the two materials at $$t = 1$$ s and $$t = 2$$ s.

The temperature contours for the two materials are shown in Figure 7. The low thermal conductivity of the HE material leads to a slow rate of diffusion of heat into the material. Therefore, the steel casing sees no change in temperature. However, the casing does deform due to the deformation of the HE material.

The stress contours for the two materials are shown in Figures 8, 9, 10, and 11. The principal stress $$\sigma_1$$ is tensile in the steel casing and reaches a values of around 200 MPa (close to its yield point). In the HE material, $$\sigma_1$$ is compressive in some regions and tensile in others and reaches values well beyond the ulimate tensile strength of the material inside some elements.

The ANSYS input file used for this problem is given below. The preprocessing stage is shown in the PDF file NFE_HW8Prob2Ans.pdf. The solution stage is shown below:

Part 2
Assume that the initial temperature of the steel and the HE material is 300 K. Also assume that the external surface of the steel casing is maintained at 300 K. Plot the temperature contours and stresses in the two materials at $$t = 1$$ s and $$t = 2$$ s.

The temperature distribution from the previous problem suggests that the heat flux at the external surface is zero if the outside is maintained at 300 K (because of the low thermal conductivity of the HE material.) We don't expect to see any significant difference in the temperature and stress distributions.

Problem 3: Explicit Finite Elements
 Given:

Figure 13(a) shows a beam that is built-in at one and loaded at the other. Figure 13(b) shows a plate that is built-in at one end and loaded at the other. The load that is applied is transient and has the form shown in Figure 13(c).

Assume that the beam and the plate are made of 4340 steel. The beam has dimensions 12 in. $$\times$$ 0.25 in. $$\times$$ 0.25 in. The plate has dimensions 12 in. $$\times$$ 0.25 in. $$\times$$ 6 in.

Simulate the problem for 10 secs. using LS-DYNA or some other explicit FE code. Use beam elements for the beam and shell elements for the plate. Use an element size of 2 in. for the beam and 2 in. $$\times$$ 2 in. for the shell.

Part 1
Plot the displacement of the end of the beam as a function of time.

Figure 14 shows the displacement of the end of the beam as a function of time. Some numerical damping appears to occur if Hughes-Liu beam elements are used with 2$$\times$$2 quadrature.

Part 2
Plot the axial stress at two points on the top of the beam as a function of time.

Figure 15 shows the axial stress at a Gauss point near the top of the beam for elements 2 and 5.

The LS-Dyna keyword file for the beam is shown below.

Part 3
Plot the displacement of the end of the plate as a function of time.

Figure 16 shows the displacement of the end of the plate as a function of time. Hughes-Liu shell elements with 3 integration points are used.

Part 4
Plot the axial stress at two points on the top of the plate as a function of time.

Figure 17 shows the axial stress at a Gauss point near the top of the plate for elements 8 and 11.

The LS-Dyna keyword file for the beam is shown below.


text:
Nonlinear finite elements/Homework 9/Solutions

Problem 1: Total Lagrangian
 Given:

Consider the tapered two-node element shown in Figure 1. The displacement field in the element is linear.

The reference (initial) cross-sectional area is

A_0 = (1-\xi)~ A_{01} + \xi~ A_{02} ~. $$ Assume that the nominal (engineering) stress is also linear in the element, i.e.,

P = (1-\xi)~ P_1 + \xi~ P_2 ~. $$

Part 1
Using the total Lagrangian formulation, develop expressions for the internal nodal forces.

The displacement field is given by the linear Lagrange interpolation expressed in terms of the material coordinate.

 * $$ \mathbf{u}(X,t) =\frac{1}{l_0}[X_2-X\quad X-X_1]

\begin{bmatrix} u_1(t)\\ u_2(t) \end{bmatrix} $$

where $$l_0=X_2-X_1$$. The strain measure is evaluated in terms of the nodal displacement,

 * $$\boldsymbol{\varepsilon}(X,t)=u_{,X}=\frac{1}{l_0}[-1\quad 1]

\begin{bmatrix} u_1(t)\\ u_2(t) \end{bmatrix} $$

which defines the $$\mathbf{B}_0$$ matrix to be

 * $$\mathbf{B}_0=\frac{1}{l_0}[-1\quad 1].$$

The internal nodal forces are then given by the usual relations.

\mathbf{f}_e^{\mbox{int}} = \int_{X_1}^{X_2} \mathbf{B}_0^T(A_0P) dX = \frac{1}{l_0}\int_{X_1}^{X_2} \left((1-\xi)A_{01}+\xi A_{02}\right)\left((1-\xi)P_1+\xi P_2\right) \begin{bmatrix} -1\\ 1 \end{bmatrix} dX $$

Integrating the above integral with $$\xi=(X-X_1)/l_0$$ to obtain

\mathbf{f}_e^{\mbox{int}} = \frac{1}{6}\left(2A_{02}P_2+A_{02}P_1+A_{01}P_2+2A_{01}P_1\right) \begin{bmatrix} -1\\ 1 \end{bmatrix}} $$

Part 2
What are the internal nodal forces if the reference area and the nominal stress are constant over the element?

\mathbf{f}_e^{\mbox{int}} = A_0P \begin{bmatrix} -1\\ 1 \end{bmatrix}} $$

Part 3
Assume that the body force is constant. Develop expressions for the external nodal forces for that case.

The external body forces arising from the body force, $$b$$, are obtained by the usual procedure.

\mathbf{f}_e^{\mbox{ext}} = \int_{X_1}^{X_2} \mathbf{N}^T\rho_0A_0b dX = \frac{\rho_0b}{l_0}\int_{X_1}^{X_2} A_0 \begin{bmatrix} X_2-X\\ X-X_1 \end{bmatrix} dX $$

\mathbf{f}_e^{\mbox{ext}} = \frac{\rho_0b}{6} \begin{bmatrix} x_1^2A_{02}+2A_{01}x_1^2-2x_1A_{02}x_2-4A_{01}x_1x_2+2A_{01}x_2^2+A_{02}x_2^2\\ A_{01}x_1^2+2x_1^2A_{02}-4x_1A_{02}x_2-2A_{01}x_1x_2+A_{01}x_2^2+2A_{02}x_2^2 \end{bmatrix}} $$

Part 4
What are the external nodal forces if the reference area and the nominal stress are constant over the element?

\mathbf{f}_e^{\mbox{ext}} = \frac{bA_0\rho_0l_0^2}{2} \begin{bmatrix} 1\\ 1 \end{bmatrix}} $$

Part 5
Develop an expression for the consistent mass matrix for the element.

The element mass matrix is

 * $$\mathbf{M}_e = \int_{X_1}^{X_2}

\rho_0A_0\mathbf{N}^T\mathbf{N} dX$$

{\mathbf{M}_e = \frac{\rho_0l_0}{12} \begin{bmatrix} 3A_{01}+A_{02} & A_{01}+A_{02}\\ A_{01}+A_{02}& A_{01}+3A_{02} \end{bmatrix}} $$

Part 6
Obtain the lumped (diagonal) mass matrix using the row-sum technique.

Lumped mass matrix is given by

 * $$M_{ii} = \int_{\xi_1}^{\xi_2}\rho_0A_0N_i dX$$

\mathbf{M}_e = \frac{\rho_0l_0}{6} \begin{bmatrix} 2A_{01}+A_{02} & 0\\ 0& A_{01}+2A_{02} \end{bmatrix}} $$

Part 7
Find the natural frequencies of a single element with consistent mass by solving the eigenvalue problem

\mathbf{K}~\mathbf{u} = \omega^2~\mathbf{M}~\mathbf{u} $$ with

\mathbf{K} = \cfrac{E(A_{01} + A_{02})}{2 l_0} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$ where $$E$$ is the Young's modulus and $$l_0$$ is the initial length of the element.

The above equation can be rewrite as

 * $$(\mathbf{K}-\omega^2\mathbf{M})\mathbf{u} = 0$$

which only has a solution if

 * $$\mbox{det}(\mathbf{K}-\omega^2\mathbf{M})=0$$

Solving the above determinant for $$\omega$$, we have

 * $${\omega=0,\pm\sqrt{\frac{18E(A_{01}^2+2A_{01}A_{02}+A_{02}^2)}

{\rho_0l_0^2(A_{01}^2+4A_{01}A_{02}+A_{02}^2)}}}$$

Problem 2: Updated Lagrangian
 Given:

Consider the tapered two-node element shown in Figure 1.

The current cross-sectional area is

A = (1-\xi)~ A_1 + \xi~ A_2 ~. $$ Assume that the Cauchy stress is also linear in the element, i.e.,

\sigma = (1-\xi)~ \sigma_1 + \xi~ \sigma_2 ~. $$

Part 1
Using the updated Lagrangian formulation, develop expressions for the internal nodal forces.

The velocity field is
 * $$v(X,t) = \frac{1}{l_0}\left[X_2-X\quad X-X_1\right]

\begin{bmatrix} v_1(t)\\ v_2(t) \end{bmatrix}$$

In term of element coordinates, the velocity field is
 * $$v(\xi,t) = \frac{1}{l_0}\left[1-\xi\quad\xi\right]

\begin{bmatrix} v_1(t)\\ v_2(t) \end{bmatrix}\qquad \xi=\frac{X-X_1}{l_0}$$ The displacement is the time integrals of the velocity, and since $$\xi$$ is independent of time

 * $$u(\xi,t)=\mathbf{N}(\xi)\mathbf{u}_e(t)$$

Therefore, since $$x = X + u$$

 * $$x(\xi,t)=\mathbf{N}(\xi)\mathbf{x}_e(t)=\left[1-\xi\quad\xi\right]

\begin{bmatrix} x_1(t)\\ x_2(t) \end{bmatrix}\qquad \xi_{,\xi}=x_2-x_1=l$$

where $$l$$ is the current length of the element. For this element, we can express $$\xi$$ in terms of the Eulerian coordinates by

 * $$\xi = \frac{x-x_1}{x_2-x_1}=\frac{x-x_1}{l},\quad

l=x_2-x_1, \quad \xi_{,x}=\frac{1}{l}$$

So $$\xi_{,x}$$ can be obtained directly, instead of through the inverse of $$x_{,\xi}$$.

The $$\mathbf{B}$$ matrix is obtained by the chain rule

 * $$\mathbf{B}=\mathbf{N}_{,x}=\mathbf{N}_{,\xi}\xi_{,x}=\frac{1}{l}[-1\quad 1]$$

Using (146) in Handout 13, we have

 * $$\mathbf{f}_e^{\mbox{int}}=\int_{x_1}^{x_2}\mathbf{B}^T\sigma

A dx=\int_{x_1}^{x_2}\frac{\sigma A}{l} \begin{bmatrix} -1\\ 1 \end{bmatrix}dx $$

Integrating the above equation to obtain

 * $${\mathbf{f}_e^{\mbox{int}}=\frac{1}{6}\left(A_1\sigma_2

+2(A_2\sigma_2+A_1\sigma_1) +A_2\sigma_1\right) \begin{bmatrix} -1\\ 1 \end{bmatrix}} $$

Part 2
Assume that the body force is constant. Develop expressions for the external nodal forces for that case.

The external forces are given by

 * $${\mathbf{f}_e^{\mbox{ext}}=\frac{\rho b}{6}

\begin{bmatrix} 2x_2A_1+x_2A_2-2x_1A_1-x_1A_2\\ 2x_2A_2+x_2A_1-2x_1A_2-x_1A_1 \end{bmatrix}} $$

Problem 3: Modal Analysis
 Given: Consider the axially loaded bar in problem VM 59 of the ANSYS Verification manual. Assume that the bar is made of Tungsten carbide.

The input file for ANSYS is as shown in VM 59 except the following material properties are used: $$E=100 $$ Msi and $$\rho=0.567/386$$ lb$$\cdot$$s$$^2/$$in$$^4$$.

Solution
Find the fundamental natural frequency of the bar.

 * $${f_1 = 36.963 \mbox{ Hz}}$$

Find the first three modal frequencies for a load of 40,000 lbf.

 * $${f_1 = 33.174\mbox{ Hz}\quad f_2=144.107\mbox{ Hz} \quad f_3=328.443\mbox{ Hz}}$$

 Given: Consider the stretched circular membrane in problem VM 55 of the ANSYS Verification manual. Assume that the membrane is made of OFHC (Oxygen-free High Conductivity) copper.

The input file for ANSYS is as shown in VM 55 except the following material properties are used: $$E=17$$ Msi and $$\rho=0.322/386$$ lb$$\cdot$$s$$^2/$$in$$^4$$, and the modes are expanded to 5.

Solution
Find the fundamental natural frequency of the bar.

 * $${f_1 = 1.509\mbox{ Hz}}$$

Find the first five modal frequencies for a load of 10,000 lbf.

 * $${f_1 = 88.329\mbox{ Hz}\quad f_2=203.114\mbox{ Hz}

\quad f_3=320.174\mbox{ Hz}\quad f_4=441.655\mbox{ Hz}\quad f_5=571.350\mbox{ Hz}}$$


text:
User:K1000~enwikiversity

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title: User:K1000~enwikiversity


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User talk:K1000~enwikiversity

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Portal:Engineering and Technology/to do

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BCP/Callicarpa

Callicarpa

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BCP/Callicarpa/Logs

2008 Logs

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Recent Logs

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title: BCP/Callicarpa/Logs


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BCP/Torenia fournieri

Torenia fournieri

title: BCP/Torenia fournieri


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BCP/Torenia fournieri/Logs

2007 Logs

 * SB Johnny-LM 10:19, 15 August 2007 (UTC)
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2008 Logs

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Recent Logs

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title: BCP/Torenia fournieri/Logs


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BCP/Tanacetum vulgare

Tanacetum vulgare

title: BCP/Tanacetum vulgare


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BCP/Tanacetum vulgare/Logs

This page contains recent bloom logs for the plant Tanacetum vulgare (Tansy).

Recent Logs

 * SB Johnny-LM 10:42, 14 August 2007 (UTC)
 * SB Johnny-LM 09:46, 8 September 2007 (UTC)
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=====Archived Logs===== (If there are more than 5 signatures under the recent logs heading, move the oldest logs to under this heading (generally leave the 3 most recent).
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title: BCP/Tanacetum vulgare/Logs


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BCP/Helianthus tuberosus

Helianthus tuberosus

title: BCP/Helianthus tuberosus


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BCP/Helianthus tuberosus/Logs

This page contains recent bloom logs for the plant Helianthus tuberosus (Jerusalem Artichoke).

Recent Logs

 * SBJ 11:00, 15 August 2007 (UTC)
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 * --Luai lashire 02:13, 23 August 2007 (UTC)
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 * Jomegat 02:47, 23 October 2009 (UTC)

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title: BCP/Helianthus tuberosus/Logs


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User:Sdlszst

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User:Yourdeadin

About the user

title: User:Yourdeadin


text:
Reservoir engineering

Reservoir engineering is a branch of petroleum engineering, typically concerned with maximizing the economic recovery of hydrocarbons from the subsurface.

Of particular interest to reservoir engineers is generating accurate reserves estimates for use in financial reporting to the SEC and other regulatory bodies. Other job responsibilities include numerical reservoir modeling, production forecasting, well testing, well drilling and workover planning, economic modeling, and PVT analysis of reservoir fluids.

Reservoir engineers also play a central role in field development planning, recommending appropriate and cost effective reservoir depletion schemes such as waterflooding or gas injection to maximize hydrocarbon recovery.

Types
Reservoir engineers often specialize in two areas:
 * Surveillance (or production) engineering, i.e. monitoring of existing fields and optimization of production and injection rates. Surveillance engineers typically use analytical and empirical techniques to perform their work, including decline curve analysis, material balance modeling, and inflow/outflow analysis.
 * Simulation modeling, i.e. the conduct of reservoir simulation studies to determine optimal development plans for oil and gas reservoirs.

Almost any person with a basic or advanced degree in a pure or applied science can undergo on the job training to become a reservoir engineer, which normally takes five to ten years, although some universities in the USA, Europe, and around the world do offer specialized undergraduate and graduate degrees in petroleum engineering.

title: Reservoir engineering


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text:
Introduction to petroleum engineering

Reserve Categories
Reserves are estimated remaining quantities of oil and natural gas and related substances anticipated to be recoverable from known accumulations, from a given date forward, based on:

• Analysis of drilling, geological, geophysical and engineering data,

• The use of established technology, and

• specified economic conditions, which are generally accepted as being reasonable and shall be disclosed.

Reserves are classified according to the degree of certainty associated with the estimates. 1.Proved reserves are those reserves that can be estimated with a high degree of certainty to be recoverable. It is likely that the actual remaining quantities recovered will exceed the estimated proved reserves.

2.Probable reserves are those additional reserves that are less certain to be recovered than proved reserves. It is equally likely that the actual remaining quantities recovered will be greater or less than the sum of the estimated proved plus probable reserves. 3.Possible reserves are those additional reserves that are less certain to be recovered than probable reserves. It is unlikely that the actual remaining quantities recovered will exceed the sum of the estimated proved plus probable plus possible reserves. Other criteria that must also be met for the categorization of reserves are provided in the accompanying

Definitions
Some commonly used definitions

Gross thickness: It is the thickness of the stratigraphically defined interval in which the reservoir beds occur, including such non-productive intervals as may be interbedded between the productive intervals.

Net pay (net productive) thickness: It is the thickness of those intervals in which porosity and permeability are known or supposed to be high enough for the interval to be able to produce oil or gas.

Net oil-bearing thickness: It includes those intervals in which oil is present in such saturation that the interval may be expected to produce oil, if penetrated by a properly completed well.

Development and Production Status of a Reservoir
Each of the reserves classifications (proved, probable and possible) may be divided into developed and undeveloped categories.

1.Developed reserves are those reserves that are expected to be recovered from existing wells and installed facilities or, if facilities have not been installed, that would involve a low expenditure ( e.g. when compared to the cost of drilling a well) to put the reserves on production. The developed category may be subdivided into producing and non producing:

2.Developed producing reserves are those reserves that are expected to be recovered from completion intervals open at the time of the estimate. These reserves may be currently producing or, if shut-in, they must have previously been on production, and the date of resumption of production must be known with reasonable certainty. 3.Developed nonproducing reserves are those reserves that either have not been on  production, or have previously been on production, but are shut-in, and the date of resumption of production is unknown. 4.Undeveloped reserves are those reserves expected to be recovered from known accumulations where a significant expenditure (e.g. when compared to the cost of drilling a well) is required to render them capable of production. They must fully meet the requirements of the reserves classification (proved, probable, possible) to which they are assigned. 5.In multi-well pools it may be appropriate to allocate total pool reserves between the developed and undeveloped categories or to subdivide the developed reserves for the pool between developed producing and developed non-producing. This allocation should be based on the estimator's assessment as to the reserves that will be recovered from specific wells, facilities and completion intervals in the pool and their respective development and production status.

Reserve Estimation
Procedure for estimation and classification of reserves

The process of reserves estimation falls into three broad categories: volumetric, material balance and decline analysis. Selection of the most appropriate reserves estimation procedures depends on the information that is available. Generally, the range of uncertainty associated with an estimate decreases and confidence level increases as more information becomes available and when the estimate is supported by more than one estimation method. Regardless of the estimation method(s) employed, the resulting reserves estimate should meet the certainty criteria in Definitions of Reserves. 1. Volumetric Methods

Volumetric methods involve the calculation of reservoir rock volume, the hydrocarbons in place in that rock volume and the estimation of the portion of the hydrocarbons in place that ultimately will be recovered. For various reservoir types at varied stages of development and depletion, the key unknown in volumetric reserves determinations may be rock volume, porosity, fluid saturation or Petroleum Society of CIM definitions-guidelines January 25 2002.doc 9 recovery factor. Important considerations affecting a volumetric reserves estimate are outlined below:

- Rock volume – may simply be determined as the product of a single well drainage area and wellbore net pay or by more complex geological mapping. Estimates must take into account geological characteristics, reservoir fluid properties and the drainage area that could be expected for the well or wells. Consideration must be given to any limitations indicated by geological, geophysical data or interpretations as well as pressure depletion or boundary conditions exhibited by test data. - Elevation of Fluid Contacts - In the absence of data that clearly defines fluid contacts, the structural interval for volumetric calculations of proved reserves should be restricted by the lowest known structural elevation of occurrence of hydrocarbons (LKH) as defined by well logs, core analyses or formation testing. - Porosity and fluid saturation and other reservoir parameters – determined from logs and core and well test data. - Recovery factor - based on analysis of production behavior from the subject reservoir, by analogy with other producing reservoirs and/or by engineering analysis. In estimating recovery factors, the evaluator must consider factors that influence recoveries such as rock and fluid properties, hydrocarbons-in-place, drilling density, future changes in operating conditions, depletion mechanisms and economic factors. 2. Material Balance Methods

Material balance methods of reserves estimation involve the analysis of pressure behavior as reservoir fluids are withdrawn, and generally result in more reliable reserves estimates than volumetric estimates. Reserves may be based on material balance calculations when sufficient production and pressure data is available. Confident application of material balance methods requires knowledge of rock and fluid properties, aquifer characteristics and accurate average reservoir pressures. In complex situations, such as those involving water influx, multi-phase behavior, multilayered or low permeability reservoirs, material balance estimates alone may provide erroneous results. Computer reservoir modeling can be considered a sophisticated form of material balance analysis. While modeling can be a reliable predictor of reservoir behavior, the input rock properties, reservoir geometry and fluid properties are critical. Evaluators must be aware of the limitations of predictive models when using these results for reserves estimation. The portion of reserves estimated as proved, probable or possible should reflect the quantity and quality of the available data and the confidence in the associated estimate. 3. Production Decline Methods

Production decline analysis methods of reserves estimation involve the analysis of production behavior as reservoir fluids are withdrawn. Confident application of decline analysis methods requires a sufficient period of stable operating conditions after the wells in a reservoir have established drainage areas. In estimating reserves, evaluators must take into consideration factors affecting production decline behavior, such as reservoir rock and fluid properties, transient versus stabilized flow, changes in operating conditions (both past and future) and depletion mechanism.

Volumetric Calculations of a Reservoir
The following equation is used in the volumetric calculation of reserves:

\text{STOIIP} = V_b \times \frac{N}{G} \times \phi \times (1-S_w) \times \frac{1}{B_o} \qquad (1) $$

where:
 * $$V_b\,$$	= bulk volume of the reservoir rock, bbl
 * $$N/G \,$$ 	= net/gross ratio of formation thickness, fraction
 * $$\phi \,$$	= porosity, fraction
 * $$S_w \,$$	= water saturation, fraction
 * $$B_o \,$$	= oil formation volume factor, rbbl/STB
 * STOIIP = stock-tank oil initially in place, STB

Equation 2 is used to calculate bulk volume of the reservoir $$(V_b)$$

V_b = A \times h \qquad (2) $$

where:
 * $$A$$ 	= area of the reservoir, ft$$^2$$
 * $$h$$	 = thickness of the reservoir, ft

title: Introduction to petroleum engineering


text:
Talk:Introduction to petroleum engineering

Added some common definations Added a few equations...The usage of mathmetical formulas is killing me ...some one please help me. --Yourdeadin 18:20, 18 August 2007 (UTC)yourdeadin
 * You can look at Help:Formula for examples. The LaTeX commands are fairly intuitive. Banerjee 19:30, 18 August 2007 (UTC)

Added the following
* 1 Reserve Catogories * 2 Definitions * 3 Devolopment and Production Status of a Reservoir * 4 Volumetric Calculations of a Reservoir The order might be a bit up and down...so please feel free to settle it the way you want it to Yourdeadin 18:44, 18 August 2007 (UTC)Yourdeadin

Addded more
I am done.Re arrange if you wish to do so. Yourdeadin 18:47, 18 August 2007 (UTC)YOurdeadin
 * 1) 4 Reserve Estimation
 * 2) 6 References


text:
Category:Reservoir engineering

This category is for learning resources related to extracting oil and gas from productive reservoirs.

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Category:Petroleum engineering

This category is for resources related to all aspects of petroleum engineering - drilling, production, reservoir engineering, mud engineering and chemistry, petroleum geology and geophysics etc.

title: Category:Petroleum engineering


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User:Desigrad

This website is primarily for networking with students looking to study in USA for MS/PhD.

Programs offered by most schools in the USA are typically judged by the quality of research. In desigrad.com most of the schools are sorted by their research funding capability.

How is Desigrad.com different from other websites ?
Search capability with over 800 graduate schools. Compare GRE/TOEFL with other graduate students scores. Over 6000 prospective graduate students in our database Graduate School Financial Aid database. Search for current graduate based on Schools Locate graduate students by Major Read Comments about graduate schools posted by students. Locate graduate students by GRE/TOEFL scores Locate the Top Research funded graduate schools Identify graduate schools based on job prospects Graduate students accepted in graduate schools Graduate students secured funding in graduate schools Locate graduate students granted Visas to graduate school Locate graduate students whose Visas have been rejected

DESIGRAD.COM

title: User:Desigrad


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Talk:Pipe and tabor

Hello all.

I would love to know if the creator(s) of the page used any kind of software to picture the sheet music along with the "tablatures".

I would find a lot useful to do such a thing with some tune I can play on the penny whistle already, to learning the new fingering/blowing pressure required by the new instrument, while having fun practicing a real tune which I know *already* (to make things simpler).

Thanks SOO much!

I saw three ships
The music for 'I saw three ships', fascinatingly enough, is a quaver out the whole way through. The first quaver of the tune is actually an upbeat to the rest of the song. So instead of Da daa da daa, it should be da Daa da daa da. Hope my notation suffices.. :P Gaidheal1 15:09, 14 April 2011 (UTC)

title: Talk:Pipe and tabor


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Video software

This learning project allows participants to explore computer software that is useful for working with digital video files.

Cross platform Video Application
Cross platform Video Applications are available on Linux, MacOSX and Windows. This makes it easy to swap between operating systems without the need to learn a new Video software.

HandBrake
Handbrake is an open source video transcoder. It is especially useful if you want to convert the video in specific format to upload it to a video platform or downsize it for using the video on mobile devices.
 * https://handbrake.fr/

OBS Open Broadcaster Software
In Wikiversity you might want to show, how a specific feature in software can be used. With OBS authors of a learning resource or tutorial can record their activities on the screen and record their audio comment. The final video (a Screencast) will be recorded by OBS as a video.

 * https://obsproject.com/

Macintosh OSX
This part of the page describes Apple and third-party software that can be used to process digital media files.

Apple iMovie
iMovie 7.0 has been released as part of iLife '08 and represents a major change in the iMovie application. The new version of iMovie is not totally backwards compatible with iMovie 6.

iMovie 6.0
This version of iMovie appears on Macintosh computers as "iMovie HD". Reading: tutorials from Apple.

Apple's video tutorials and built-in help menu support provide the basic information you need to use iMovie. Additional tips and tricks are discussed below.

iMovie projects. When you use iMovie to work with video files, you work within iMovie "projects". These "projects" have hidden files associated with them. Each video clip in an iMovie project is present as a hidden digital video file (file extension .dv). Digital video files use large amounts of disk space. Five minutes of video can use several gigabytes of space on your hard drive. You can quickly end up with iMovie project files that are too large to backup to a DVD. See below for how to segment such large iMovie project files for backup.

Image sizes. The default size for iMovie HD video projects is 720 x 528 pixels. Reading: iMovie Techniques, Tips and Information. See in particular the section, "Tips for Creating Images in Photoshop to Use in iMovie".

Activity
Open one of your iMovie project files that contains some video. Position the play head at your favorite frame and

External resources

 * iMovie HD - at Wikibooks (very limited).

Terminal
Terminal is a Macintosh application that has some useful UNIX commands for manipulating the large files that can arise when working with digital video. For example,  can be used to split a large iMovie project into parts that will fit on DVDs. Reading: Make large files span multiple optical discs by Aaron Adams.

Terminal can also be used to convert QuickTime movie files to ogg format video, the only video format allowed for upload at Wikiversity. See Creating Ogg videos.

GraphicConverter
GraphicConverter is a powerful Macintosh application for working with digital images and video files. For example, you can use GraphicConverter to make animated GIF files or QuickTime movies from a series of images such as those generated by DAZ Studio.

Windows Movie Maker
This is a provided by default at part of the Windows operating system.

OpenShot (OpenSource)
If you want to use a VideoEditor on Linux, that has equivalent features like Windows Movie Maker, Open Shot as a video editor might be an option for you on Linux. The OpenSource editor is written Python. The main application will be to compose different videos, audio samples and images into a single video with drag and drop features. Due to the Programming Language Python it is easy to port on other Operating Systems. Furthermore swapping between operating systems without the need to learn a new video editor might be a benefit for you for collaborative work in a team.

 * http://www.openshot.org/

KDEnlive (OpenSource)
Kdenlive is just as OpenShot an open source video editor. The first part of the name "KDE" refers to the KDE Frameworks libraries and a full desktop environment for Linux. Now the project KDEnlive, that was started around 2003, has versions for the operating systems Linux, Windows and Mac. Most of the video processing is performed and based on the MLT Framework, which in turn is build upon many other Open Source frameworks and repositories like FFmpeg, movit, ladspa, sox,

 * https://kdenlive.org

Adobe Premiere Pro (Commercial)
Adobe Premiere Pro is a non-linear video editor. The current version CS4 offers considerable advantages over the previous one, including digital video editing, full HD support and Blu-ray and Flash video exporting. Features comparison

KDEnlive (Open Source)
The OpenSource Software KDEnlive has got its name by an acronym: KDE Non-Linear Video Editor. In the very beginning is primarily aimed at the GNU/Linux platform on the Desktop Environment KDE. But also works on Windows (in beta version), BSD and MacOSX (see KDE Downloads). The current work on the Windows port is beta, so it might contain some issues.

 * https://kdenlive.org/download/

OpenShot (Open Source)
If you want to use a VideoEditor on Linux, that has equivalent features like Windows Movie Maker, Open Shot as a video editor might be an option for you on Linux. The OpenSource editor is written Python. The main application will be to compose different videos, audio samples and images into a single video with drag and drop features.

 * http://www.openshot.org/download/

title: Video software


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Vocal pedagogy

Pronuniciation and Diction in Singing (Music)
When singing it is important to realize you must have unified vowels and precise diction to maximize comprehension and also create a nice tone. As singers we must think of individual syllables in addition to the entire word. This guide will start off discussing the small syllable and eventually move to the whole.

Pure Vowels

When singing you should strive for a nice open tone. A way to achieve this, is to focus on the pure vowels which are: [a], [ε], [i], [o], and [u]. They are taken from the italian vowels which are a universal pronunciation tool used in singing.

Pronuciation of each vowel: the vowels must be arranged from light to dark i,e,a,o,u

[a] as in the English pronunciation of father

[ε] as in the English pronunciation of met, or bet, or set

[i] as in the English pronuciation of me, we, or see (when making this sound round your lips a little so the sound is not spread)

[o] this is a little harder in English so like to compare it to the Spanish pronuciation of no. Try to make it a short sound without adding any [u] to it.

[u] as in the English pronunciation of you or tool

Worksheets

title: Vocal pedagogy


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User talk:Laurence42

My name is Laurence and I am 65 years old. I have got Parkinson's Disease and want to do something with the rest of my life that is interesting. My education is not the best but I can read and write. I love children and have 8 grandchildren and love to write childrens stories. I found Wikiversity through my wife Lynne and I know that I can learn how to write through them. I do proofreading for Distributed Proofreaders who have thought me alot since I started proofreading with them. I am writing my first children's story now. I really enjoy writing and my reading has improved alot since I started proofreading. I would like to say thank you too Wikiversity for all that they are doing to make education free for everbody. Laurence

title: User talk:Laurence42


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User:Pika64eBay

Welcome to the master page of Pika64eBay! here is a current list of classes I am in:

Filmmaking (Day 2) Game Design (Year 1,Month 1, Day 1) Plant Science (Year 5, Month 11, Day 30)

title: User:Pika64eBay