text:
Micromechanics of composites/Proof 3

Surface-volume integral relation 1
Let $$\Omega$$ be a body and let $$\partial{\Omega}$$ be its surface. Let $$\mathbf{n}$$ be the normal to the surface. Let $$\mathbf{v}$$ be a vector field on $$\Omega$$ and let $$\boldsymbol{S}$$ be a second-order tensor field on $$\Omega$$. Show that

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\cdot\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} ~. $$

Proof:

Recall the relation

\boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})] = \mathbf{a}\cdot[\{\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)\}\cdot\mathbf{b}] ~. $$ Integrating over the volume, we have

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \int_{\Omega} \mathbf{a}\cdot[\{\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)\}\cdot\mathbf{b}~\text{dV}~. $$ Since $$\mathbf{a}$$ and $$\mathbf{b}$$ are constant, we have

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \mathbf{a}\cdot\left[\left\{\int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV}\right\}\cdot\mathbf{b}\right]~. $$ From the divergence theorem,

\int_{\Omega} \boldsymbol{\nabla} \bullet \mathbf{u}~\text{dV} = \int_{\partial{\Omega}} \mathbf{u}\cdot\mathbf{n}~\text{dA} $$ we get

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \int_{\partial{\Omega}} [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]\cdot\mathbf{n}~\text{dA} ~. $$ Using the relation

[(\mathbf{v}\bullet\mathbf{a})(\boldsymbol{S}\bullet\mathbf{b})]\cdot\mathbf{n} = \mathbf{a}\cdot[\{\mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})\}\cdot\mathbf{b}] $$ we get

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \int_{\partial{\Omega}} \mathbf{a}\cdot[\{\mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})\}\cdot\mathbf{b}]~\text{dA} ~. $$ Since $$\mathbf{a}$$ and $$\mathbf{b}$$ are constant, we have

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \mathbf{a}\cdot\left[\left\{\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA}\right\}\cdot\mathbf{b}\right] ~. $$ Therefore,

\mathbf{a}\cdot\left[\left\{\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA}\right\}\cdot\mathbf{b}\right]= \mathbf{a}\cdot\left[\left\{\int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV}\right\}\cdot\mathbf{b}\right]~. $$ Since $$\mathbf{a}$$ and $$\mathbf{b}$$ are arbitrary, we have

{ \int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} } \qquad\qquad\qquad\square $$

title: Micromechanics of composites/Proof 3


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Micromechanics of composites/Proof 4

Question
Let $$\partial{\Omega}$$ be a surface. Let $$\mathbf{x}$$ be the position vector of points on the surface and let $$\mathbf{t}$$ be a vector field that are defined on $$\partial{\Omega}$$. If

\int_{\partial{\Omega}} \mathbf{x}\times\mathbf{t}~\text{dA} = \mathbf{0} $$ show that

\int_{\partial{\Omega}} \mathbf{x}\otimes\mathbf{t}~\text{dA} = \int_{\partial{\Omega}} \mathbf{t}\otimes\mathbf{x}~\text{dA} ~. $$

Proof
If we assume a Cartesian basis, we can write the given relation in index notation as

\int_{\partial{\Omega}} e_{ijk}~x_j~t_k~\text{dA} = 0 $$ where $$e_{ijk}$$ is the Levi-Civita (permutation) symbol. Since $$e_{ijk}$$ is does not depend upon the position we can write

e_{ijk} \int_{\partial{\Omega}} ~x_j~t_k~\text{dA} = 0 ~. $$ Define

A_{jk} := \int_{\partial{\Omega}} x_j~t_k~\text{dA} ~. $$ Then,

e_{ijk} A_{jk} = 0 ~. $$ Expanding, we get

e_{i11} A_{11} + e_{i12} A_{12} + e_{i13} A_{13} + e_{i21} A_{21} + e_{i22} A_{22} + e_{i23} A_{23} + e_{i31} A_{31} + e_{i32} A_{32} + e_{i33} A_{33} = 0 \qquad i=1,2,3 ~. $$ Expanding further, we get three equations

\begin{align} e_{112} A_{12} + e_{113} A_{13} + e_{121} A_{21} + e_{123} A_{23} + e_{131} A_{31} + e_{132} A_{32} & = 0 \\ e_{212} A_{12} + e_{213} A_{13} + e_{221} A_{21} + e_{223} A_{23} + e_{231} A_{31} + e_{232} A_{32} & = 0 \\ e_{312} A_{12} + e_{313} A_{13} + e_{321} A_{21} + e_{323} A_{23} + e_{331} A_{31} + e_{332} A_{32} & = 0 \end{align} $$ or,

\begin{align} e_{123} A_{23} + e_{132} A_{32} & = 0 \\ e_{213} A_{13} + e_{231} A_{31} & = 0 \\ e_{312} A_{12} + e_{321} A_{21} & = 0 \end{align} $$ or,

A_{23} = A_{32} ~; A_{13} = A_{31} ~; A_{12} = A_{21} ~. $$ Therefore,

\int_{\partial{\Omega}} x_2~t_3~\text{dA} = \int_{\partial{\Omega}} x_3~t_2~\text{dA} = \int_{\partial{\Omega}} t_2~x_3~\text{dA} ~; \int_{\partial{\Omega}} x_1~t_3~\text{dA} = \int_{\partial{\Omega}} x_3~t_1~\text{dA} = \int_{\partial{\Omega}} t_1~x_3~\text{dA}~; \int_{\partial{\Omega}} x_1~t_2~\text{dA} = \int_{\partial{\Omega}} x_2~t_1~\text{dA} = \int_{\partial{\Omega}} t_1~x_2~\text{dA} ~. $$ Also, by symmetry,

\int_{\partial{\Omega}} x_1~t_1~\text{dA} = \int_{\partial{\Omega}} t_1~x_1~\text{dA} ~; \int_{\partial{\Omega}} x_2~t_2~\text{dA} = \int_{\partial{\Omega}} t_2~x_2~\text{dA} ~; \int_{\partial{\Omega}} x_3~t_3~\text{dA} = \int_{\partial{\Omega}} t_3~x_3~\text{dA} ~. $$ Therefore we may write,

\int_{\partial{\Omega}} x_j~t_k~\text{dA} = \int_{\partial{\Omega}} t_j~x_k~\text{dA} ~. $$ Reverting back to direct tensor notation, we get

{ \int_{\partial{\Omega}} \mathbf{x}\otimes\mathbf{t}~\text{dA} = \int_{\partial{\Omega}} \mathbf{t}\otimes\mathbf{x}~\text{dA} ~. } $$

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Micromechanics of composites/Proof 4


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Micromechanics of composites/Proof 5

Surface-volume integral relation 2
Let $$\Omega$$ be a body and let $$\partial{\Omega}$$ be its surface. Let $$\mathbf{n}$$ be the normal to the surface. Let $$\mathbf{v}$$ be a vector field on $$\Omega$$. Show that

\int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA}~. $$

 Proof:

Recall that

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\cdot\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} $$ where $$\boldsymbol{S}$$ is any second-order tensor field on $$\Omega$$. Let us assume that $$\boldsymbol{S} = \boldsymbol{\mathit{1}}$$. Then we have

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{\mathit{1}}\cdot\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{\mathit{1}} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{\mathit{1}})]~\text{dV} $$ Now,

\boldsymbol{\mathit{1}}\cdot\mathbf{n} = \mathbf{n} ~; \boldsymbol{\nabla} \bullet \boldsymbol{\mathit{1}} = \mathbf{0} ~; \boldsymbol{A}\cdot\boldsymbol{\mathit{1}} = \boldsymbol{A} $$ where $$\boldsymbol{A}$$ is any second-order tensor. Therefore,

\int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA} = \int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} ~. $$ Rearranging,

{ \int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA} \qquad \square } $$

title: Micromechanics of composites/Proof 5


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Micromechanics of composites/Proof 6

Curl of the gradient of a vector - 1
Let $$\mathbf{v}$$ be a vector field. Show that

\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}) = 0 ~. $$

Proof:

For a second order tensor field $$\boldsymbol{S}$$, we can define the curl as

(\boldsymbol{\nabla} \times \boldsymbol{S})\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{S}^T\cdot\mathbf{a}) $$ where $$\mathbf{a}$$ is an arbitrary constant vector. Substituting $$\boldsymbol{\nabla}\mathbf{v}$$ into the definition, we have

[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v})]\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}^T\cdot\mathbf{a}) ~. $$ Since $$\mathbf{a}$$ is constant, we may write

\boldsymbol{\nabla}\mathbf{v}^T\cdot\mathbf{a} = \boldsymbol{\nabla} (\mathbf{v}\cdot\mathbf{a}) = \boldsymbol{\nabla} \varphi $$ where $$\varphi = \mathbf{v} \cdot \mathbf{a}$$ is a scalar. Hence,

[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v})]\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{\nabla \varphi)} ~. $$ Since the curl of the gradient of a scalar field is zero (recall potential theory), we have

\boldsymbol{\nabla} \times (\boldsymbol{\nabla \varphi)} = \mathbf{0} ~. $$ Hence,

[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v})]\cdot\mathbf{a} = \mathbf{0} \qquad\qquad \forall\mathbf{a} ~. $$ The arbitrary nature of $$\mathbf{a}$$ gives us

{ \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}) = \mathbf{0} \qquad \square } $$

title: Micromechanics of composites/Proof 6


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Micromechanics of composites/Proof 7

Curl of the transpose of the gradient of a vector
Let $$\mathbf{v}$$ be a vector field. Show that

\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}^T) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{v}) ~. $$

 Proof:

The curl of a second order tensor field $$\boldsymbol{S}$$ is defined as

(\boldsymbol{\nabla} \times \boldsymbol{S})\cdot\mathbf{a} = \boldsymbol{\nabla} \times (\boldsymbol{S}^T\cdot\mathbf{a}) $$ where $$\mathbf{a}$$ is an arbitrary constant vector. If we write the right hand side in index notation with respect to a Cartesian basis, we have

[\boldsymbol{S}^T\cdot\mathbf{a}]_k = [\mathbf{b}]_k = b_k = S_{pk}~a_p $$ and

[\boldsymbol{\nabla} \times \mathbf{b}]_i = e_{ijk}\frac{\partial b_k}{\partial x_j} = e_{ijk}\frac{\partial (S_{pk}~a_p)}{\partial x_j} = e_{ijk}\frac{\partial S_{pk}}{\partial x_j}~a_p = [(\boldsymbol{\nabla} \times \boldsymbol{S})]_{ip}~a_p ~. $$ In the above a quantity $$[~]_i$$ represents the $$i$$-th component of a vector, and the quantity $$[~]_{ip}$$ represents the $$ip$$-th components of a second-order tensor.

Therefore, in index notation, the curl of a second-order tensor $$\boldsymbol{S}$$ can be expressed as

[\boldsymbol{\nabla} \times \boldsymbol{S}]_{ip} = e_{ijk}\frac{\partial S_{pk}}{\partial x_j} ~. $$ Using the above definition, we get

[\boldsymbol{\nabla} \times \boldsymbol{S}^T]_{ip} = e_{ijk}\frac{\partial S_{kp}}{\partial x_j} ~. $$ If $$\boldsymbol{S} = \boldsymbol{\nabla}\mathbf{v}$$, we have

[\boldsymbol{\nabla} \times \boldsymbol{\nabla}\mathbf{v}^T]_{ip} = e_{ijk}\frac{\partial }{\partial x_j} \left(\frac{\partial v_k}{\partial x_p}\right) = \frac{\partial }{\partial x_p}\left(e_{ijk}\frac{\partial v_k}{\partial x_j}\right) = \frac{\partial }{\partial x_p}\left([\boldsymbol{\nabla} \times \mathbf{v}]_i\right) = [\boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{v})]_{ip} ~. $$ Therefore,

{ \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{v}^T) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{v}) \qquad \square } $$

title: Micromechanics of composites/Proof 7


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Micromechanics of composites/Proof 8

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Relation between axial vector and displacement
Let $$\mathbf{u}$$ be a displacement field. The displacement gradient tensor is given by $$\boldsymbol{\nabla}\mathbf{u}$$. Let the skew symmetric part of the displacement gradient tensor (infinitesimal rotation tensor) be

\boldsymbol{\omega} = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} - \boldsymbol{\nabla}\mathbf{u}^T) ~. $$ Let $$\boldsymbol{\theta}$$ be the axial vector associated with the skew symmetric tensor $$\boldsymbol{\omega}$$. Show that

\boldsymbol{\theta} = \frac{1}{2}~\boldsymbol{\nabla} \times \mathbf{u} ~. $$
 * }

Proof:

The axial vector $$\mathbf{w}$$ of a skew-symmetric tensor $$\boldsymbol{W}$$ satisfies the condition

\boldsymbol{W}\cdot\mathbf{a} = \mathbf{w}\times\mathbf{a} $$ for all vectors $$\mathbf{a}$$. In index notation (with respect to a Cartesian basis), we have

W_{ip}~a_p = e_{ijk}~w_j~a_k $$ Since $$e_{ijk} = - e_{ikj}$$, we can write

W_{ip}~a_p = -e_{ikj}~w_j~a_k \equiv -e_{ipq}~w_q~a_p $$ or,

W_{ip} = -e_{ipq}~w_q ~. $$ Therefore, the relation between the components of $$\boldsymbol{\omega}$$ and $$\boldsymbol{\theta}$$ is

\omega_{ij} = -e_{ijk}~\theta_k ~. $$ Multiplying both sides by $$e_{pij}$$, we get

e_{pij}~\omega_{ij} = -e_{pij}~e_{ijk}~\theta_k = -e_{pij}~e_{kij}~\theta_k~. $$ Recall the identity

e_{ijk}~e_{pqk} = \delta_{ip}~\delta_{jq} - \delta_{iq}~\delta_{jp}~. $$ Therefore,

e_{ijk}~e_{pjk} = \delta_{ip}~\delta_{jj} - \delta_{ij}~\delta_{jp} = 3\delta_{ip} - \delta_{ip} = 2\delta_{ip} $$ Using the above identity, we get

e_{pij}~\omega_{ij} = -2\delta_{pk}~\theta_k = -2\theta_p~. $$ Rearranging,

\theta_p = -\frac{1}{2}~e_{pij}~\omega_{ij} $$ Now, the components of the tensor $$\boldsymbol{\omega}$$ with respect to a Cartesian basis are given by

\omega_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial x_j} - \frac{\partial u_j}{\partial x_i}\right) $$ Therefore, we may write

\theta_p = -\cfrac{1}{4}~e_{pij} \left(\frac{\partial u_i}{\partial x_j} - \frac{\partial u_j}{\partial x_i}\right) $$ Since the curl of a vector $$\mathbf{v}$$ can be written in index notation as

\boldsymbol{\nabla} \times \mathbf{v} = e_{ijk}~\frac{\partial u_k}{\partial x_j}~\mathbf{e}_i $$ we have

e_{pij}~\frac{\partial u_j}{\partial x_i}~ = [\boldsymbol{\nabla} \times \mathbf{u}]_p \qquad \text{and} \qquad e_{pij}~\frac{\partial u_i}{\partial x_j}~ = - e_{pji}\frac{\partial u_i}{\partial x_j} = - [\boldsymbol{\nabla} \times \mathbf{u}]_p $$ where $$[~]_p$$ indicates the $$p$$-th component of the vector inside the square brackets.

Hence,

\theta_p = -\cfrac{1}{4}~\left(-[\boldsymbol{\nabla} \times \mathbf{u}]_p - [\boldsymbol{\nabla} \times \mathbf{u}]_p\right) = \frac{1}{2}~[\boldsymbol{\nabla} \times \mathbf{u}]_p ~. $$ Therefore,

{ \boldsymbol{\theta} = \frac{1}{2} \boldsymbol{\nabla} \times \mathbf{u} \qquad \square } $$

title: Micromechanics of composites/Proof 8


text:
Micromechanics of composites/Proof 9

{| cellspacing="0" cellpadding="0" style="margin:0em 0em 1em 0em; width:80%"
 * colspan="2" style="width:80%; horizontal-align:right; vertical-align:top; border:1px solid Sienna; background-color:White;" |

Relation between axial vector and strain
Let $$\mathbf{u}$$ be a displacement field. Let $$\boldsymbol{\varepsilon}$$ be the strain field (infinitesimal) corresponding to the displacement field and let $$\boldsymbol{\theta}$$ be the corresponding infinitesimal rotation vector. Show that

\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \boldsymbol{\nabla} \boldsymbol{\theta} ~. $$
 * }

Proof:

The infinitesimal strain tensor is given by

\boldsymbol{\varepsilon} = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T) ~. $$ Therefore,

\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \frac{1}{2}[\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}) + \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}^T)] ~. $$ Recall that

\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}) = \mathbf{0}\qquad\text{and}\qquad \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\mathbf{u}^T) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{u}) ~. $$ Hence,

\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \frac{1}{2}[\boldsymbol{\nabla} (\boldsymbol{\nabla} \times \mathbf{u})] ~. $$ Also recall that

\boldsymbol{\theta} = \frac{1}{2} \boldsymbol{\nabla} \times \mathbf{u}~. $$ Therefore,

{ \boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \boldsymbol{\nabla} \boldsymbol{\theta} \qquad \square } $$

title: Micromechanics of composites/Proof 9


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Micromechanics of composites/Proof 10

Rigid body displacement field
Show that, for a rigid body motion with infinitesimal rotations, the displacement field $$\mathbf{u}(\mathbf{x})$$ for can be expressed as

\mathbf{u}(\mathbf{x}) = \mathbf{c} + \boldsymbol{\omega}\cdot\mathbf{x} $$ where $$\mathbf{c}$$ is a constant vector and $$\boldsymbol{\omega}$$ is the infinitesimal rotation tensor.

Proof:

Note that for a rigid body motion, the strain $$\boldsymbol{\varepsilon}$$ is zero. Since

\boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = \boldsymbol{\nabla} \boldsymbol{\theta} $$ we have a $$\boldsymbol{\theta} = $$ constant when $$\boldsymbol{\varepsilon} = 0$$, i.e., the rotation is homogeneous.

For a homogeneous deformation, the displacement gradient is independent of $$\mathbf{x}$$, i.e.,

\boldsymbol{\nabla}\mathbf{u} = \frac{\partial \mathbf{u}}{\partial \mathbf{x}} = \boldsymbol{G}\qquad\leftarrow\qquad\text{constant} ~. $$ Integrating, we get

\mathbf{u}(\mathbf{x}) = \boldsymbol{G}\cdot\mathbf{x} + \mathbf{c} ~. $$ Now the strain and rotation tensors are given by

\boldsymbol{\varepsilon} = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T) = \frac{1}{2}(\boldsymbol{G} + \boldsymbol{G}^T) ~; \boldsymbol{\omega} = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} - \boldsymbol{\nabla}\mathbf{u}^T) = \frac{1}{2}(\boldsymbol{G} - \boldsymbol{G}^T) ~. $$ For a rigid body motion, the strain $$\boldsymbol{\varepsilon} = 0$$. Therefore,

\boldsymbol{G} = -\boldsymbol{G}^T\qquad \implies \qquad \boldsymbol{\omega} = \boldsymbol{G} ~. $$ Plugging into the expression for $$\mathbf{u}$$ for a homogeneous deformation, we have

{ \mathbf{u}(\mathbf{x}) = \boldsymbol{\omega}\cdot\mathbf{x} + \mathbf{c} \qquad \square } $$

title: Micromechanics of composites/Proof 10


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Micromechanics of composites/Proof 11

Identity 1
Let $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$ be two second order tensors. Show that

\boldsymbol{A}:\boldsymbol{B} = (\boldsymbol{A}^T\cdot\boldsymbol{B}):\boldsymbol{\mathit{1}}~. $$

Proof:

Using index notation,

\boldsymbol{A}:\boldsymbol{B} = A_{ij}~B_{ij} = A^T_{ji}~B_{ij} = A^T_{ji}~B_{ik}~\delta_{jk} = [\boldsymbol{A}^T\cdot\boldsymbol{B}]_{jk}~\delta_{jk} = (\boldsymbol{A}^T\cdot\boldsymbol{B}):\boldsymbol{\mathit{1}} ~. $$ Hence,

{ \boldsymbol{A}:\boldsymbol{B} = (\boldsymbol{A}^T\cdot\boldsymbol{B}):\boldsymbol{\mathit{1}} \qquad \square } $$

Identity 2
Let $$\boldsymbol{A}$$ be a second order tensor and let $$\mathbf{a}$$ and $$\mathbf{b}$$ be two vectors. Show that

\boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} ~. $$

 Proof:

It is convenient to use index notation for this. We have

\boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = A_{ij}~a_i~b_j = (A_{ij}~b_j)~a_i = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} ~. $$ Hence,

{ \boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} \qquad \square } $$

Identity 3
Let $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$ be two second order tensors and let $$\mathbf{a}$$ and $$\mathbf{b}$$ be two vectors. Show that

(\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) ~. $$

Proof:

Using index notation,

(\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (A_{ij}~a_j)(B_{ik}~b_k) = (A_{ij}~B_{ik})(a_j~b_k) = (A^T_{ji}~B_{ik})(a_j~b_k) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) ~. $$ Hence,

{ (\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) \qquad \square } $$

Identity 4
Let $$\boldsymbol{A}$$ be a second order tensors and let $$\mathbf{a}$$ and $$\mathbf{b}$$ be two vectors. Show that

(\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) \qquad \text{and} \qquad \mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = [\boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{a})]^T = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T ~. $$

Proof:

For the first identity, using index notation, we have

[(\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b}]_{ik} = (A_{ij}~a_j)~b_k = A_{ij}~(a_j~b_k) = A_{ij}~[\mathbf{a}\otimes\mathbf{b}]_{jk} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) ~. $$ Hence,

{ (\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) \qquad \square } $$

For the second identity, we have

[\mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b})]_{ij} = a_i~(A_{jk}~b_k) = (a_i~b_k)~A_{jk} = (a_i~b_k)~A^T_{kj} = [(\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T]_{ij} ~. $$ Therefore,

\mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T ~. $$ Now, $$\mathbf{a}\otimes\mathbf{b} = [\mathbf{b}\otimes\mathbf{a}]^T$$ and $$(\boldsymbol{A}\cdot\boldsymbol{B})^T = \boldsymbol{B}^T\cdot\boldsymbol{A}^T$$. Hence,

(\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T = (\mathbf{b}\otimes\mathbf{a})^T\cdot\boldsymbol{A}^T = [\boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{a})]^T ~. $$ Therefore,

{ \mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = [\boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{a})]^T = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T \qquad \square } $$

title: Micromechanics of composites/Proof 11


text:
Micromechanics of composites/Proof 12

Question
Let $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$ be two second-order tensor fields. Let the average of any second-order tensor field ($$\boldsymbol{S}$$) over the region $$\Omega$$ (of volume $$V$$) be defined as

\langle \boldsymbol{S} \rangle := \cfrac{1}{V}\int_{\Omega} \boldsymbol{S}~\text{dV} ~. $$ Show that

\langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle = \langle [\boldsymbol{A} - \langle\boldsymbol{A} \rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle] \rangle~. $$

Proof
Expanding out the right hand side, we have

\begin{align} \langle [\boldsymbol{A} - \langle\boldsymbol{A}\rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle] \rangle& = \cfrac{1}{V}\int_{\Omega}[\boldsymbol{A} - \langle \boldsymbol{A} \rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle]~\text{dV} \\ & = \cfrac{1}{V}\int_{\Omega}[ \boldsymbol{A}\cdot\boldsymbol{B} - \langle \boldsymbol{A} \rangle\cdot\boldsymbol{B} - \boldsymbol{A}\cdot\langle \boldsymbol{B} \rangle + \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle]~\text{dV} \\ & = \cfrac{1}{V}\int_{\Omega}\boldsymbol{A}\cdot\boldsymbol{B}~\text{dV} - \cfrac{1}{V}\int_{\Omega}\langle \boldsymbol{A} \rangle\cdot\boldsymbol{B}~\text{dV} - \cfrac{1}{V}\int_{\Omega}\boldsymbol{A}\cdot\langle \boldsymbol{B} \rangle~\text{dV} + \cfrac{1}{V}\int_{\Omega}\langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle~\text{dV} ~. \end{align} $$ Now $$\langle \boldsymbol{A} \rangle$$ and $$\langle \boldsymbol{B} \rangle$$ are constants with respect to the integration. Hence,

\begin{align} \langle [\boldsymbol{A} - \langle\boldsymbol{A}\rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle] \rangle & = \cfrac{1}{V}\int_{\Omega}\boldsymbol{A}\cdot\boldsymbol{B}~\text{dV} - \langle \boldsymbol{A} \rangle\cdot\left(\cfrac{1}{V}\int_{\Omega}\boldsymbol{B}~\text{dV}\right) - \left(\cfrac{1}{V}\int_{\Omega}\boldsymbol{A}~\text{dV}\right)\cdot\langle \boldsymbol{B} \rangle + \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle\left(\cfrac{1}{V}\int_{\Omega}~\text{dV}\right)\\ & = \langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle + \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle \\ & = \langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle ~. \end{align} $$ Therefore,

{ \langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle = \langle [\boldsymbol{A} - \langle\boldsymbol{A}\rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle]\rangle \qquad \square } $$

title: Micromechanics of composites/Proof 12


text:
Micromechanics of composites/Proof 13

Question
Let $$\boldsymbol{\sigma}$$ be the Cauchy stress and let $$\boldsymbol{\nabla}\mathbf{v}$$ be the velocity gradient in a body $$\Omega$$ with boundary $$\partial{\Omega}$$. Let $$\mathbf{n}$$ be the normal to the boundary. Let $$V$$ be the volume of the body. If the skew-symmetric part of the velocity gradient is zero, i.e., $$\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{\nabla}\mathbf{v}^T$$, or if the stress field is self equilibrated, i.e., $$\langle \boldsymbol{\sigma} \rangle = \langle \boldsymbol{\sigma} \rangle^T$$, show that

\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle - \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V} \int_{\partial{\Omega}} [\mathbf{v} - \langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}]\cdot [(\boldsymbol{\sigma} - \langle \boldsymbol{\sigma} \rangle)\cdot\mathbf{n}]~\text{dA} ~. $$

Proof
Taking the trace of each term in the identity

\langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle = \langle [\boldsymbol{A} - \langle\boldsymbol{A} \rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle]\rangle $$ the difference between the average stress power and the product of the average stress and the average velocity gradient can be written as (using either the symmetry of the stress or of the velocity gradient)

\begin{align} \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle - \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle & = \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle - \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle - \langle\boldsymbol{\nabla}\mathbf{v} \rangle:\langle \boldsymbol{\sigma} \rangle + \langle\boldsymbol{\nabla}\mathbf{v} \rangle:\langle \boldsymbol{\sigma} \rangle \\ & = \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle - \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle - \langle\boldsymbol{\nabla}\mathbf{v} \rangle:\langle \boldsymbol{\sigma} \rangle + [\langle\boldsymbol{\nabla}\mathbf{v} \rangle^T\cdot\langle \boldsymbol{\sigma} \rangle]:\boldsymbol{\mathit{1}} \end{align} $$ Recall that

\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}} (\boldsymbol{\sigma}\cdot\mathbf{n})\cdot\mathbf{v}~\text{dV} ~; \langle\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\Omega} \boldsymbol{\nabla}\mathbf{v}~\text{dV} ~; \langle \boldsymbol{\sigma} \rangle = \cfrac{1}{V}~\int_{\partial{\Omega}}\mathbf{x}\otimes\bar{\mathbf{t}}~\text{dA} ~; \cfrac{1}{V}\int_{\Omega}\boldsymbol{\nabla} \mathbf{x}~\text{dV} = \boldsymbol{\mathit{1}} ~. $$ Also, from the divergence theorem

\int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA}~; \int_{\Omega} \boldsymbol{\nabla} \mathbf{x}~\text{dV} = \int_{\partial{\Omega}} \mathbf{x}\otimes\mathbf{n}~\text{dA} ~. $$ Therefore,

\begin{align} \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle - \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle & = \cfrac{1}{V}\int_{\partial{\Omega}} (\boldsymbol{\sigma}\cdot\mathbf{n})\cdot\mathbf{v}~\text{dV} - \langle \boldsymbol{\sigma} \rangle:\left[\cfrac{1}{V}\int_{\partial{\Omega}}\mathbf{v}\otimes\mathbf{n}~\text{dA}\right] - \langle\boldsymbol{\nabla}\mathbf{v} \rangle: \left[\cfrac{1}{V}~\int_{\partial{\Omega}}\mathbf{x}\otimes\bar{\mathbf{t}}~\text{dA}\right] \\ & \qquad\qquad + [\langle\boldsymbol{\nabla}\mathbf{v} \rangle^T\cdot\langle \boldsymbol{\sigma} \rangle]: \left[\cfrac{1}{V}\int_{\partial{\Omega}}\mathbf{x}\otimes\mathbf{n}~\text{dA}\right]~. \end{align} $$ Since $$\langle \boldsymbol{\sigma} \rangle$$ and $$\langle\boldsymbol{\nabla}\mathbf{v} \rangle$$ are independent of $$\mathbf{x}$$, we can take these inside the integrals to get

\begin{align} \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle - \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle & = \cfrac{1}{V}\int_{\partial{\Omega}} \left[(\boldsymbol{\sigma}\cdot\mathbf{n})\cdot\mathbf{v} - \langle \boldsymbol{\sigma} \rangle:(\mathbf{v}\otimes\mathbf{n}) - \langle\boldsymbol{\nabla}\mathbf{v} \rangle:(\mathbf{x}\otimes\bar{\mathbf{t}}) + [\langle\boldsymbol{\nabla}\mathbf{v} \rangle^T\cdot\langle \boldsymbol{\sigma} \rangle]:(\mathbf{x}\otimes\mathbf{n})\right]~\text{dA} \end{align} $$ Using the identity

(\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) $$ we get

[\langle\boldsymbol{\nabla}\mathbf{v} \rangle^T\cdot\langle \boldsymbol{\sigma} \rangle]:(\mathbf{x}\otimes\mathbf{n}) = [\langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}]\cdot[\langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n}] ~. $$ Also, using the identity

\boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} $$ we get

\langle \boldsymbol{\sigma} \rangle:(\mathbf{v}\otimes\mathbf{n}) = [\langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n}]\cdot\mathbf{v} ~; \langle\boldsymbol{\nabla}\mathbf{v} \rangle:(\mathbf{x}\otimes\bar{\mathbf{t}}) = [\langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\bar{\mathbf{t}}]\cdot\mathbf{x} = [\langle\boldsymbol{\nabla}\mathbf{v} \rangle^T\cdot\mathbf{x}]\cdot\bar{\mathbf{t}} = [\langle\boldsymbol{\nabla}\mathbf{v} \rangle^T\cdot\mathbf{x}]\cdot(\boldsymbol{\sigma}\cdot\mathbf{n})~. $$ Since $$\boldsymbol{\nabla}\mathbf{v}^T = \boldsymbol{\nabla}\mathbf{v}$$, we have $$\langle\boldsymbol{\nabla}\mathbf{v} \rangle^T = \langle\boldsymbol{\nabla}\mathbf{v} \rangle$$ (we could alternatively use the symmetry of $$\langle \boldsymbol{\sigma} \rangle$$ to arrive at the following equation). Hence,

\langle\boldsymbol{\nabla}\mathbf{v} \rangle:(\mathbf{x}\otimes\bar{\mathbf{t}}) = [\langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}]\cdot(\boldsymbol{\sigma}\cdot\mathbf{n})~. $$ Plugging these back into the original equation, we have

\begin{align} \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle - \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle & = \cfrac{1}{V}\int_{\partial{\Omega}} \left\{(\boldsymbol{\sigma}\cdot\mathbf{n})\cdot\mathbf{v} - [\langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n}]\cdot\mathbf{v} - [\langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}]\cdot(\boldsymbol{\sigma}\cdot\mathbf{n}) + [\langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}]\cdot[\langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n}]\right\}~\text{dA} \\ & = \cfrac{1}{V}\int_{\partial{\Omega}} \left\{ [(\boldsymbol{\sigma}-\langle \boldsymbol{\sigma} \rangle)\cdot\mathbf{n}]\cdot\mathbf{v} - [(\boldsymbol{\sigma}-\langle \boldsymbol{\sigma} \rangle)\cdot\mathbf{n}]\cdot(\langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}) \right\}~\text{dA} \\ & = \cfrac{1}{V}\int_{\partial{\Omega}} \left\{ [(\boldsymbol{\sigma}-\langle \boldsymbol{\sigma} \rangle)\cdot\mathbf{n}]\cdot[\mathbf{v} - \langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}] \right\}~\text{dA}~. \end{align} $$ Hence

{ \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle - \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}} \left[ [\mathbf{v} - \langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}]\cdot[(\boldsymbol{\sigma}-\langle \boldsymbol{\sigma} \rangle)\cdot\mathbf{n}] \right]~\text{dA}~. } \qquad\qquad\qquad\qquad\square $$

title: Micromechanics of composites/Proof 13


text:
Micromechanics of composites/Proof 14

Question
Let $$\boldsymbol{P}$$ be the first Piola-Kirchhoff stress and let $$\dot{\boldsymbol{F}}$$ be the time rate of the deformation gradient in a body whose reference configuration is $$\Omega_0$$ with boundary $$\partial{\Omega}_0$$. Let $$\mathbf{N}$$ be the normal to the boundary. Let $$V_0$$ be the volume of the body. Let $$\mathbf{X}$$ represent the position of points in the reference configuration. Let $$\dot{\mathbf{x}}$$ be the material time derivative of $$\mathbf{x}$$. Let $$\langle \boldsymbol{A} \rangle$$ represent the unweighted volume average of a quantity $$\boldsymbol{A}$$. Show that

\begin{align} \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle \cdot\langle \boldsymbol{P} \rangle & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}} - \langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N} \right\}~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}} - \langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes (\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \dot{\mathbf{x}}\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N}\right\}~\text{dA} ~. \end{align} $$

Proof
Recall the identity

\langle \boldsymbol{A}\cdot\boldsymbol{B} \rangle - \langle \boldsymbol{A} \rangle\cdot\langle \boldsymbol{B} \rangle = \langle [\boldsymbol{A} - \langle\boldsymbol{A} \rangle]\cdot[\boldsymbol{B} - \langle \boldsymbol{B} \rangle]\rangle ~. $$ Therefore,

\begin{align} \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle \cdot\langle \boldsymbol{P} \rangle & = \langle [\dot{\boldsymbol{F}} - \langle \dot{\boldsymbol{F}}\rangle ]\cdot[\boldsymbol{P} - \langle \boldsymbol{P} \rangle] \rangle \\ & = \cfrac{1}{V_0}\int_{\Omega_0}[\dot{\boldsymbol{F}} - \langle \dot{\boldsymbol{F}}\rangle ]\cdot[\boldsymbol{P} - \langle \boldsymbol{P} \rangle]~\text{dV}\\ & = \cfrac{1}{V_0}\int_{\Omega_0}\dot{\boldsymbol{F}}\cdot\boldsymbol{P}~\text{dV} -\cfrac{1}{V_0}\int_{\Omega_0}\dot{\boldsymbol{F}}\cdot\langle \boldsymbol{P} \rangle~\text{dV} -\cfrac{1}{V_0}\int_{\Omega_0}\langle \dot{\boldsymbol{F}}\rangle \cdot\boldsymbol{P}~\text{dV} +\cfrac{1}{V_0}\int_{\Omega_0}\langle \dot{\boldsymbol{F}}\rangle \cdot\langle \boldsymbol{P} \rangle~\text{dV} \\ & = \cfrac{1}{V_0}\int_{\Omega_0}\dot{\boldsymbol{F}}\cdot\boldsymbol{P}~\text{dV} -\left(\cfrac{1}{V_0}\int_{\Omega_0}\dot{\boldsymbol{F}}~\text{dV}\right)\cdot\langle \boldsymbol{P} \rangle -\langle \dot{\boldsymbol{F}}\rangle \cdot\left(\cfrac{1}{V_0}\int_{\Omega_0}\boldsymbol{P}~\text{dV}\right) +\langle \dot{\boldsymbol{F}}\rangle \cdot\left(\cfrac{1}{V_0}\int_{\Omega_0}\boldsymbol{\mathit{1}}~\text{dV}\right) \cdot\langle \boldsymbol{P} \rangle~. \end{align} $$ We want express the volume integrals above in terms of surface integrals. To do that, recall that

\begin{align} \int_{\Omega_0}\dot{\boldsymbol{F}}\cdot\boldsymbol{P}~\text{dV} & = \int_{\partial{\Omega}_0}\dot{\mathbf{x}}\otimes(\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \\ \int_{\Omega_0}\dot{\boldsymbol{F}}~\text{dV} & = \int_{\Omega_0}\boldsymbol{\nabla}_0~ \dot{\mathbf{x}}~\text{dV} = \int_{\partial{\Omega}_0}\dot{\mathbf{x}}\otimes\mathbf{N}~\text{dA} \\ \int_{\Omega_0}\boldsymbol{P}~\text{dV} & = \int_{\partial{\Omega}_0}\mathbf{X}\otimes(\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \\ \int_{\Omega_0}\boldsymbol{\mathit{1}}~\text{dV} &=\int_{\Omega_0}\boldsymbol{\nabla}_0~ \mathbf{X}~\text{dV} = \int_{\partial{\Omega}_0}\mathbf{X}\otimes\mathbf{N}~\text{dA}~. \end{align} $$ Therefore,

\begin{align} \cfrac{1}{V_0}\int_{\Omega_0}\dot{\boldsymbol{F}}\cdot\boldsymbol{P}~\text{dV} & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0}\dot{\mathbf{x}}\otimes(\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \\ \left(\cfrac{1}{V_0}\int_{\Omega_0}\dot{\boldsymbol{F}}~\text{dV}\right)\cdot\langle \boldsymbol{P} \rangle & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0}(\dot{\mathbf{x}}\otimes\mathbf{N})\cdot\langle \boldsymbol{P} \rangle~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0}\dot{\mathbf{x}}\otimes[\langle \boldsymbol{P \rangle^T\cdot\mathbf{N}]}~\text{dA} \\ \langle \dot{\boldsymbol{F}}\rangle \cdot\left(\cfrac{1}{V_0}\int_{\Omega_0}\boldsymbol{P}~\text{dV}\right) & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \langle \dot{\boldsymbol{F}}\rangle \cdot[\mathbf{X}\otimes(\boldsymbol{P}^T\cdot\mathbf{N})]~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes(\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \\ \langle \dot{\boldsymbol{F}}\rangle \cdot\left(\cfrac{1}{V_0}\int_{\Omega_0} \boldsymbol{\mathit{1}}~\text{dV}\right) \cdot\langle \boldsymbol{P} \rangle & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \langle \dot{\boldsymbol{F}}\rangle \cdot(\mathbf{X}\otimes\mathbf{N})\cdot\langle \boldsymbol{P} \rangle~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes[\langle \boldsymbol{P} \rangle^T\cdot\mathbf{N}]~\text{dA} ~. \end{align} $$ Collecting the terms, we have

\begin{align} \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle \cdot\langle \boldsymbol{P} \rangle&= \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \left\{\dot{\mathbf{x}}\otimes(\boldsymbol{P}^T\cdot\mathbf{N}) -\dot{\mathbf{x}}\otimes[\langle \boldsymbol{P \rangle^T\cdot\mathbf{N}]} -[\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes(\boldsymbol{P}^T\cdot\mathbf{N}) +[\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes[\langle \boldsymbol{P} \rangle^T\cdot\mathbf{N}]\right\}~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \left\{\dot{\mathbf{x}}\otimes[\boldsymbol{P}^T\cdot\mathbf{N}-\langle \boldsymbol{P \rangle^T\cdot\mathbf{N}]} -[\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes[\boldsymbol{P}^T\cdot\mathbf{N}-\langle \boldsymbol{P} \rangle^T\cdot\mathbf{N}]\right\}~\text{dA}\\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}}-\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes[\boldsymbol{P}^T\cdot\mathbf{N}-\langle \boldsymbol{P \rangle^T\cdot\mathbf{N}]}~\text{dA}~. \end{align} $$ Therefore,

{ \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle \cdot\langle \boldsymbol{P} \rangle = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}} - \langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N} \right\}~\text{dA} ~. } $$ From the above, clearly

\left(\cfrac{1}{V_0}\int_{\Omega_0}\dot{\boldsymbol{F}}~\text{dV}\right)\cdot\langle \boldsymbol{P} \rangle = \langle \dot{\boldsymbol{F}}\rangle \cdot\left(\cfrac{1}{V_0}\int_{\Omega_0}\boldsymbol{P}~\text{dV}\right) = \langle \dot{\boldsymbol{F}}\rangle \cdot\left(\cfrac{1}{V_0}\int_{\Omega_0} \boldsymbol{\mathit{1}}~\text{dV}\right) \cdot\langle \boldsymbol{P} \rangle~. $$ Therefore,

\cfrac{1}{V_0}\int_{\partial{\Omega}_0}\dot{\mathbf{x}}\otimes[\langle \boldsymbol{P \rangle^T\cdot\mathbf{N}]}~\text{dA} = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes(\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes[\langle \boldsymbol{P} \rangle^T\cdot\mathbf{N}]~\text{dA} ~. $$ Thus we can alternatively write the expression for the difference as

\begin{align} \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle \cdot\langle \boldsymbol{P} \rangle&= \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \left\{\dot{\mathbf{x}}\otimes(\boldsymbol{P}^T\cdot\mathbf{N}) -[\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes(\boldsymbol{P}^T\cdot\mathbf{N}) -\left[\dot{\mathbf{x}}\otimes[\langle \boldsymbol{P}^T \rangle\cdot\mathbf{N}] -[\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes[\langle \boldsymbol{P}^T \rangle \cdot\mathbf{N}]\right]\right\}~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}}-\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes(\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \end{align} $$ or,

\begin{align} \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle \cdot\langle \boldsymbol{P} \rangle&= \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \left\{\dot{\mathbf{x}}\otimes(\boldsymbol{P}^T\cdot\mathbf{N}) -\dot{\mathbf{x}}\otimes[\langle \boldsymbol{P}^T \rangle \cdot\mathbf{N}] -\left[\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes(\boldsymbol{P}^T\cdot\mathbf{N}) -[\langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes[\langle \boldsymbol{P}^T \rangle\cdot\mathbf{N}]\right]\right\}~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \dot{\mathbf{x}}\otimes[\boldsymbol{P}^T\cdot\mathbf{N} - \langle \boldsymbol{P}^T \rangle\cdot\mathbf{N}]~\text{dA} ~. \end{align} $$ Hence,

{ \begin{align} \langle \dot{\boldsymbol{F}}\cdot\boldsymbol{P} \rangle - \langle \dot{\boldsymbol{F}}\rangle \cdot\langle \boldsymbol{P} \rangle & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}} - \langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N} \right\}~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} [\dot{\mathbf{x}} - \langle \dot{\boldsymbol{F}}\rangle \cdot\mathbf{X}]\otimes (\boldsymbol{P}^T\cdot\mathbf{N})~\text{dA} \\ & = \cfrac{1}{V_0}\int_{\partial{\Omega}_0} \dot{\mathbf{x}}\otimes \left\{ [\boldsymbol{P} - \langle \boldsymbol{P} \rangle]^T\cdot\mathbf{N}\right\}~\text{dA} ~. \end{align} } $$

title: Micromechanics of composites/Proof 14


text:
User:Shukraraj~enwikiversity

I am an electrical engineering students in Nepal I.O.E., TU.I like mainly in reserch and scocial works.I like to meet many technical fields student in the worlds and share to experinces.Please mail me at shukraraj@gmail.com for disscussion.

title: User:Shukraraj~enwikiversity


text:
MediaWiki/Engine/index.php

index.php is the main web entry point for the MediaWiki engine. The source code is at trunk/phase3/index.php/. The documentation is at. A brief description is at mw:Manual:Index.php.

variables
(documentation: mw:Manual:Global variables,mw:category:MediaWiki configuration settings,mw:category:variables, )

In rough order of appeareance:

$mediaWiki $maxLag $action $wgRequest $title $wgTitle $wgOut $wgContLang $wgUseAjax - mw:Manual:$wgUseAjax $IP - mw:manual:$IP ("include path") $dispatcher $wgLoadBalancer $wgServer $wgDisableInternalSearch $wgSquidMaxage $wgEnableDublinCoreRdf $wgEnableCreativeCommonsRdf $wgCommandLineMode $wgUseExternalEditor $wgDisabledActions $wgArticle $wgUser $wgRequest $wgPostCommitUpdateList $wgLoadBalancer

requires

 * webstart.php
 * wiki.php (mw:manual:wiki.php)
 * AjaxDispatcher.php


text:
User:Nishikant cp

Assurance means pre-acting to any situation to ensure the availability.

Software Quality Asurance Test comprises of Controlling & Assuring Activities.

title: User:Nishikant cp


text:
BCP/Sagittaria latifolia

Sagittaria latifolia

title: BCP/Sagittaria latifolia


text:
BCP/Sagittaria latifolia/Logs

This page contains recent bloom logs for the plant Sagittaria latifolia (Broadleaf Arrowhead, duck potato, Indian potato, wapato).

Recent Logs

 * Jomegat 01:25, 13 August 2007 (UTC)

=====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).
 * (no archived logs yet)

title: BCP/Sagittaria latifolia/Logs


text:
User:Gurvat ram

Software Quality Assurance

title: User:Gurvat ram


text:
User:KTucker/My Sandbox

This page is for experiments with wiki markup. Some of these explorations may become useful components (or templates) for use on Wikiversity and other Mediawiki based projects such as WikiEducator.

= Page and Section Layout Templates =

 * Section boxes and page templates.
 * Testing area for navigation templates.
 * Use Case Template
 * Chat - insert a chat area on a wiki page (not implemented)

=Portal Header=

- this will list several portals

=Portal Templates=

Tabbed
I needed a portal framework similar to the one on Wikipedia for Free Software. So, I implemented an almost identical equivalent on Wikiversity:

 * Portal Instructions: Wikipedia and Wikiversity.
 * Example Portal:Social Entrepreneurship

A variation has been set up which I use in this project:

 * Portal:FLOSS4Science

=Login Link=

After several failed experiments ... success (thanks to SB Johnny):

Code:

Login

(note the extra 'E' in 'FULLPAGENAMEE').

Result: Login

=Sub-pages=

=Multilingual Pages=

 * /Multilingual/

=Tests=

Temp

 * /Temp/

Add a section

 * Add an Activity

Template:Test Preload

Wikipedia Search
Wikipedia search on a term such as "Social sciences".

Alija Hamiti Born in Gjilan Kosovo it is taken with Karate is a champion of many fold since the child but later began to Kikbox But She has achieved many successes in karate after and kikbox. It now begins with Fubtoll this year which is an achievement so much for them and now plays for Albania is also in the women's national team football in Albania and Albania team plays in the defensive position it plays swing arm and right hand

Properties
A page/template as an object (holding the data about something, e.g. a learning activity), with sub-page "method" templates to display it:

 * Some Thing
 * Some Thing/Properties
 * Some Thing/Viewer

Table View
Enough to show that there is potential:

Incorporating Text from another page

 * Content of Template:Social Entrepreneurship Tasks is incorporated below:

Conditions
Check whether a parameter has been specified:

Template "/Animals/" with a name and attribute:

mouse size: elephant _:

frog _

frog size:

frog weight:

_ size:

_ _:

mouse weight:

Mind Maps - Freemind

- apparently, the required extension is not installed.

Images

 * Libre Learning Cycle

Aim: mouseover text (alt'?) to be different from caption:

=Bookmarks=

 * Help:Template - Template on WikiNews w:Help:Template
 * Help:Substitution - w:Help:Substitution
 * Colloquium
 * Help:Icons
 * Icons by subject
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title: User:KTucker/My Sandbox


text: File:Kt.png

title: File:Kt.png


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User talk:Swaroopa

Hello Swaroopa, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --HappyCamper 16:45, 22 August 2007 (UTC)

French
I thought you might find these links interesting:

 * http://fr.wikiversity.org
 * French
 * French stream

Cheers! --HappyCamper 16:45, 22 August 2007 (UTC)

title: User talk:Swaroopa


text:
User talk:Gurvat ram

Hello Gurvat ram, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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title: User talk:Gurvat ram


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User talk:Ailurophile~enwikiversity

Hello Ailurophile, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 17:05, 22 August 2007 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Ailurophile. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Ailurophile~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:26, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:02, 19 April 2015 (UTC)

title: User talk:Ailurophile~enwikiversity


text:
User talk:Leftmostcat

Hello Leftmostcat, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 17:39, 22 August 2007 (UTC)

title: User talk:Leftmostcat


text:
Talk:Teletraffic engineering/How is telephony traffic simulated

I think this is an excellent starting point for building a sound simulation course. I hope to add to it whin I have time and after organizing my thoughts as where can I add.

But why is the misleading title (telephony...). This is just an example of a much broader subject that is reviewed on the page.

Bani 18:52, 22 August 2007 (UTC)


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Technical writing/Sasha Sladeckova

Back to Technical Writing

title: Technical writing/Sasha Sladeckova


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Talk:Technical writing/Sasha Sladeckova

What do you want the name of this page to be? --JWSchmidt 19:28, 22 August 2007 (UTC)

Sasha: Hi! So far,I want to leave it as it is. Do you know how to change the page name?
 * Yes, at the top of this page is a tab labeled "move" which you can click on - but I think it will only show up after 4 days of your first day of editing. --HappyCamper 16:01, 23 August 2007 (UTC)

title: Talk:Technical writing/Sasha Sladeckova


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MediaWiki talk:Anonnotice

If MediaWiki:Anonnotice is blank or contains a single - then anons will see MediaWiki:Sitenotice.

Edit page to only contain something like     to prevent anons from seeing sitenotice.

See mw:Manual:Interface/Sitenotice for more info.

title: MediaWiki talk:Anonnotice


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File:KinnetiK Seduced by the Dark Side Screenplay.pdf

Used with the WikiU Film School Lesson on how to format a movie script


text:
File:KinnetiK Seduced by the Dark Side Screenplay(Animated version).pdf

Used with the WikiU Film School Lesson on how to format a movie script


text:
Nonlinear finite elements/Homework 4/Hints

Important
If you have a body containing regions made up of different materials it is not sufficient to mesh each region separately. The regions also have to share boundary nodes. The figure shows you an example.


text: File:MeshContinuity.png

title: File:MeshContinuity.png


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Nonlinear finite elements/Euler Bernoulli beams

Displacements

\begin{align} u_1 & = u_0(x) - z \cfrac{dw_0}{dx} \\ u_2 & = 0 \\ u_3 & = w_0(x) \end{align} $$

Strains

\varepsilon_{11} = \varepsilon_{xx}= \varepsilon_{xx}^0 + z \varepsilon_{xx}^1 $$

\begin{align} \varepsilon_{xx}^0 & = \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \\ \varepsilon_{xx}^1 & = -\cfrac{d^2w_0}{dx^2} \end{align} $$

Strain-Displacement Relations

\varepsilon_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right) + \frac{1}{2}\left(\frac{\partial u_m}{\partial x_i}\frac{\partial u_m}{\partial x_j}\right) $$

\begin{align} \varepsilon_{11} & = \frac{1}{2}\left(\frac{\partial u_1}{\partial x_1} + \frac{\partial u_1}{\partial x_1}\right) + \frac{1}{2}\left(\frac{\partial u_1}{\partial x_1}\frac{\partial u_1}{\partial x_1}+ \frac{\partial u_2}{\partial x_1}\frac{\partial u_2}{\partial x_1}+ \frac{\partial u_3}{\partial x_1}\frac{\partial u_3}{\partial x_1} \right) \\ \varepsilon_{22} & = \frac{1}{2}\left(\frac{\partial u_2}{\partial x_2} + \frac{\partial u_2}{\partial x_2}\right) + \frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}\frac{\partial u_1}{\partial x_2}+ \frac{\partial u_2}{\partial x_2}\frac{\partial u_2}{\partial x_2}+ \frac{\partial u_3}{\partial x_2}\frac{\partial u_3}{\partial x_2} \right) \\ \varepsilon_{33} & = \frac{1}{2}\left(\frac{\partial u_3}{\partial x_3} + \frac{\partial u_3}{\partial x_3}\right) + \frac{1}{2}\left(\frac{\partial u_1}{\partial x_3}\frac{\partial u_1}{\partial x_3}+ \frac{\partial u_2}{\partial x_3}\frac{\partial u_2}{\partial x_3}+ \frac{\partial u_3}{\partial x_3}\frac{\partial u_3}{\partial x_3} \right) \\ \varepsilon_{23} & = \frac{1}{2}\left(\frac{\partial u_2}{\partial x_3} + \frac{\partial u_3}{\partial x_2}\right) + \frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}\frac{\partial u_1}{\partial x_3}+ \frac{\partial u_2}{\partial x_2}\frac{\partial u_2}{\partial x_3}+ \frac{\partial u_3}{\partial x_2}\frac{\partial u_3}{\partial x_3} \right) \\ \varepsilon_{31} & = \frac{1}{2}\left(\frac{\partial u_3}{\partial x_1} + \frac{\partial u_1}{\partial x_3}\right) + \frac{1}{2}\left(\frac{\partial u_1}{\partial x_3}\frac{\partial u_1}{\partial x_1}+ \frac{\partial u_2}{\partial x_3}\frac{\partial u_2}{\partial x_1}+ \frac{\partial u_3}{\partial x_3}\frac{\partial u_3}{\partial x_1} \right) \\ \varepsilon_{12} & = \frac{1}{2}\left(\frac{\partial u_1}{\partial x_2} + \frac{\partial u_2}{\partial x_1}\right) + \frac{1}{2}\left(\frac{\partial u_1}{\partial x_1}\frac{\partial u_1}{\partial x_2}+ \frac{\partial u_2}{\partial x_1}\frac{\partial u_2}{\partial x_2}+ \frac{\partial u_3}{\partial x_1}\frac{\partial u_3}{\partial x_2} \right) \end{align} $$

The displacements

u_1 = u_0(x_1) - x_3 \cfrac{dw_0}{dx_1}~; u_2 = 0~; u_3 = w_0(x_1) $$ The derivatives
 * $$\begin{align}

\frac{\partial u_1}{\partial x_1} & = \cfrac{du_0}{dx_1} - x_3\cfrac{d^2w_0}{dx_1^2} ~; & \frac{\partial u_1}{\partial x_2} & = 0 ~; &\frac{\partial u_1}{\partial x_3} & = - \cfrac{dw_0}{dx_1} \\ \frac{\partial u_2}{\partial x_1} & = 0 ~; & \frac{\partial u_2}{\partial x_2} & = 0 ~; &\frac{\partial u_2}{\partial x_3} & = 0\\ \frac{\partial u_3}{\partial x_1} & = \cfrac{dw_0}{dx_1}~; & \frac{\partial u_3}{\partial x_2} & = 0 ~; &\frac{\partial u_3}{\partial x_3} & = 0 \end{align}$$

von Karman strains
The von Karman strains

\begin{align} \varepsilon_{11} & = \cfrac{du_0}{dx_1} - x_3\cfrac{d^2w_0}{dx_1^2} + \frac{1}{2}\left[ \left(\cfrac{du_0}{dx_1}-x_3\cfrac{d^2w_0}{dx_1^2}\right)^2 + \left(\cfrac{dw_0}{dx_1}\right)^2\right] \\ \varepsilon_{22} & = 0 \\ \varepsilon_{33} & = \frac{1}{2}\left(\cfrac{dw_0}{dx_1}\right)^2 \\ \varepsilon_{23} & = 0 \\ \varepsilon_{31} & = \frac{1}{2}\left(\cfrac{dw_0}{dx_1}-\cfrac{dw_0}{dx_1}\right) - \frac{1}{2}\left[\left(\cfrac{du_0}{dx_1}-x_3\cfrac{d^2w_0}{dx_1^2}\right) \left(\cfrac{dw_0}{dx_1}\right)\right] \\ \varepsilon_{12} & = 0 \end{align} $$

Balance of forces

\begin{align} \cfrac{dN_{xx}}{dx} + f(x) & = 0 \\ \cfrac{d^2M_{xx}}{dx^2} + q(x) + \cfrac{d}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right) & = 0 \end{align} $$

Stress Resultants

 * $$\begin{align}

N_{xx} & = \int_A \sigma_{xx}~ dA \\ M_{xx} & = \int_A z\sigma_{xx}~ dA \end{align}$$

Stress-Strain equation

\sigma_{xx} = E \varepsilon_{xx} $$

Stress Resultant - Displacement relations

\begin{align} N_{xx} & = A_{xx} \varepsilon_{xx}^0 + B_{xx} \varepsilon_{xx}^1 = A_{xx}\left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \right] - B_{xx}\cfrac{d^2w_0}{dx^2} \\ M_{xx} & = B_{xx} \varepsilon_{xx}^0 + D_{xx} \varepsilon_{xx}^1 = B_{xx}\left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \right] - D_{xx}\cfrac{d^2w_0}{dx^2} \end{align} $$

Extensional/Bending Stiffness

\begin{align} A_{xx} & = \int_A E~dA\qquad \leftarrow \qquad \text{extensional stiffness}\\ B_{xx} & = \int_A zE~dA\qquad \leftarrow \qquad \text{extensional-bending stiffness}\\ D_{xx} & = \int_A z^2E~dA\qquad \leftarrow \qquad \text{bending stiffness} \end{align} $$

If $$E$$ is constant, and $$x$$-axis passes through centroid

\begin{align} A_{xx} & = E \int_A ~dA = EA \\ B_{xx} & = E \int_A z~dA = 0 \\ D_{xx} & = E \int_A z^2~dA= EI \end{align} $$

Axial Equation

\begin{align} \int_{x_a}^{x_b} \cfrac{d(\delta u_0)}{dx} \left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2\right] A_{xx}~dx & = \int_{x_a}^{x_b} (\delta u_0) f~dx +\\ & \delta u_0(x_a) Q_1 + \delta u_0(x_b) Q_4 \end{align} $$ where

\begin{align} \delta u_0 & := v_1\\ Q_1 & := -N_{xx}(x_a)\\ Q_4 & := N_{xx}(x_b) \end{align} $$

Bending Equation

\begin{align} \int_{x_a}^{x_b} \left\{\cfrac{d(\delta w_0)}{dx} \right. & \left[\cfrac{du_0}{dx} + \cfrac{1}{2}~\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{dw_0}{dx} A_{xx} + \left.\cfrac{d^2(\delta w_0)}{dx^2} \left(\cfrac{d^2w_0}{dx^2}\right) D_{xx} \right\}~dx= \\ & \int_{x_a}^{x_b} (\delta w_0) q~dx + \delta w_0(x_a) Q_2 + \delta w_0(x_b) Q_5 + \delta \theta(x_a) Q_3 + \delta \theta(x_b) Q_6 ~. \end{align} $$ where
 * $$\begin{align}

\delta w_0 & := v_2 & \delta \theta & := \cfrac{dv_2}{dx} \\ Q_2 & := -\left[\cfrac{dM_{xx}}{dx} + N_{xx}\cfrac{dw_0}{dx}\right]_{x_a} & Q_5 & := \left[\cfrac{dM_{xx}}{dx} + N_{xx}\cfrac{dw_0}{dx}\right]_{x_b} \\ Q_3 & := -M_{xx} (x_a) & Q_6 & := M_{xx} (x_b) \end{align}$$

Finite Element Model

 * $$\begin{align}

u_0(x) & =u_1 \psi_1(x) + u_2 \psi_2(x) \\ w_0(x) & =w_1 \phi_1(x) + \theta_1 \phi_2(x) + w_2 \phi_3(x) + \theta_2 \phi_4(x) \end{align}$$ where $$\theta = -(dw_0/dx)$$.

Finite Element Equations

\begin{bmatrix} \mathbf{K}^{11} & \vdots & \mathbf{K}^{12} \\ & \vdots &\\ \mathbf{K}^{21} & \vdots & \mathbf{K}^{22} \end{bmatrix} \begin{bmatrix} \mathbf{u} \\\\ \mathbf{d} \end{bmatrix} = \begin{bmatrix} \mathbf{F}^1 \\ \\ \mathbf{F}^2 \end{bmatrix} $$ where
 * $$\begin{align}

\mathbf{u} & = [u_1 \quad u_2]^T \\ \mathbf{d} & = [w_1 \quad \theta_1 \quad w_2 \quad \theta_2]^T \end{align}$$

 * $$\begin{align}

\mathbf{K}^{11} & = 2 \times 2; \qquad & \mathbf{K}^{12} & = 2 \times 4\\ \mathbf{K}^{21} & = 4 \times 2; \qquad & \mathbf{K}^{22} & = 4 \times 4 \end{align}$$

Symmetric Stiffness Matrix

\begin{align} K_{ij}^{11} & = \int_{x_a}^{x_b} A_{xx}\cfrac{d\psi_i}{dx}\cfrac{d\psi_j}{dx}~dx \\ K_{ij}^{12} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx}\cfrac{dw_0}{dx}\right) \cfrac{d\psi_i}{dx} \cfrac{d\phi_j}{dx}~dx\\ K_{ij}^{21} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx} \cfrac{dw_0}{dx}\right) \cfrac{d\phi_i}{dx}\cfrac{d\psi_j}{dx}~dx\\ K_{ij}^{22} & = \int_{x_a}^{x_b}\left\{\frac{1}{2} A_{xx} \left[\cfrac{du_0}{dx}+\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_i}{dx}\cfrac{d\phi_j}{dx} + D_{xx}\cfrac{d^2\phi_i}{dx^2}\cfrac{d^2\phi_j}{dx^2}\right\}~dx \end{align} $$

Load Vector

\begin{align} F_i^1 & = \int_{x_a}^{x_b} \psi_i f~dx + \psi_i(x_a) Q_1 + \psi_i(x_b) Q_4 \\ F_i^2 & = \int_{x_a}^{x_b} \phi_i q~dx + \phi_i(x_a) Q_2 + \phi_i(x_b) Q_5 + \cfrac{d\phi_i}{dx}(x_a) Q_3 + \cfrac{d\phi_i}{dx}(x_b) Q_6 \end{align} $$

Newton-Raphson Solution

\mathbf{K}(\mathbf{U}) \mathbf{U} = \mathbf{F} $$ where
 * $$\begin{align}

U_1 & = u_1, ~ U_2= u_2, ~ U_3= d_1, ~ U_4= d_2, ~ U_5= d_3, ~ U_6= d_4 \\ F_1 & = F^1_1, ~ F_2= F^1_2, ~ F_3= F^2_1, ~ F_4= F^2_2, ~ F_5= F^2_3, ~ F_6= F^2_4 \end{align}$$ The residual is

\mathbf{R} = \mathbf{K} \mathbf{U} - \mathbf{F} ~. $$ For Newton iterations, we use the algorithm

\mathbf{U}^{r+1} = \mathbf{U}^r - (\mathbf{T}^r)^{-1} \mathbf{R}^r $$ where the tangent stiffness matrix is given by

\mathbf{T}^r = \frac{\partial \mathbf{R}^r}{\partial \mathbf{U}}; \quad\text{or}\quad T_{ij} = \frac{\partial R_i}{\partial U_j}, \qquad i=1 \dots 6, j=1 \dots 6~. $$

Tangent Stiffness Matrix

\begin{align} i=1\dots2;~j=1\dots2 &: \\ & { T^{11}_{ij} = K^{11}_{ij}} \\ \\ i=1\dots2;~j=1\dots4 &: \\ & { T^{12}_{ij} = 2 K^{12}_{ij}} \\ \\ i=1\dots4;~j=1\dots2 &: \\ & { T^{21}_{ij} = 2 K^{21}_{ij}} \\ \\ i=1\dots4;~j=1\dots4 &: \\ & { T^{22}_{ij} = K^{22}_{ij} + \frac{1}{2} \int_{x_a}^{x_b} A_{xx}\left[\cfrac{du_0}{dx} + 2\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_i}{dx}\cfrac{d\phi_j}{dx}~dx } \end{align} $$

Load Steps
Recall

N_{xx} = A_{xx}\left[\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \right] - \cfrac{d^2w_0}{dx^2} $$
 * Divide load into small increments.
 * $$F = \sum_{i=1}^N \Delta F_i $$

 * Compute $$\mathbf{u}$$ and $$\mathbf{d}$$ for first load step,
 * $$\mathbf{K}(\mathbf{U}_0) \mathbf{U}_1 = \Delta F_1 $$

 * Compute $$\mathbf{u}$$ and $$\mathbf{d}$$ for second load step,
 * $$\mathbf{K}(\mathbf{U}_1) \mathbf{U}_2 = \Delta F_1 + \Delta F_2 $$

 * Continue until F is reached.

Membrane Locking
Recall

\begin{bmatrix} \mathbf{K}^{11} & \vdots & \mathbf{K}^{12} \\ & \vdots &\\ \mathbf{K}^{21} & \vdots & \mathbf{K}^{22} \end{bmatrix} \begin{bmatrix} \mathbf{u} \\\\ \mathbf{d} \end{bmatrix} = \begin{bmatrix} \mathbf{F}^1 \\ \\ \mathbf{F}^2 \end{bmatrix} $$ where

\begin{align} K_{ij}^{11} & = \int_{x_a}^{x_b} A_{xx}\cfrac{d\psi_i}{dx}\cfrac{d\psi_j}{dx}~dx \\ K_{ij}^{12} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx}\cfrac{dw_0}{dx}\right) \cfrac{d\psi_i}{dx} \cfrac{d\phi_j}{dx}~dx\\ K_{ij}^{21} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx} \cfrac{dw_0}{dx}\right) \cfrac{d\phi_i}{dx}\cfrac{d\psi_j}{dx}~dx\\ K_{ij}^{22} & = \int_{x_a}^{x_b}\left\{\frac{1}{2} A_{xx} \left[\cfrac{du_0}{dx}+\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_i}{dx}\cfrac{d\phi_j}{dx} + D_{xx}\cfrac{d^2\phi_i}{dx^2}\cfrac{d^2\phi_j}{dx^2}\right\}~dx \end{align} $$

For Hinged-Hinged
Membrane strain:

\varepsilon_{xx}^0 = 0 $$ or

\cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 = 0 $$ Hence, shape functions should be such that

\cfrac{du_0}{dx} \approx \left(\cfrac{dw_0}{dx}\right)^2 $$ $$u_0$$ linear, $$w_0$$ cubic $$\implies$$ Element Locks! Too stiff.

Selective Reduced Integration

 * Assume $$u_0$$ is linear ; $$w_0$$ is cubic.
 * Then $$\cfrac{du_0}{dx} \equiv \cfrac{d\psi_i}{dx}$$ is constant, and $$\cfrac{dw_0}{dx} \equiv \cfrac{d\phi_i}{dx}$$ is quadratic.
 * Try to keep $$\varepsilon_{xx}^0 = $$ constant.

\begin{align} K_{ij}^{11} & = \int_{x_a}^{x_b} A_{xx}\cfrac{d\psi_i}{dx}\cfrac{d\psi_j}{dx}~dx\\ K_{ij}^{12} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx}\cfrac{dw_0}{dx}\right) \cfrac{d\psi_i}{dx} \cfrac{d\phi_j}{dx}~dx\\ K_{ij}^{22} & = \int_{x_a}^{x_b}\left\{\frac{1}{2} A_{xx} \left[\cfrac{du_0}{dx}+\left(\cfrac{dw_0}{dx}\right)^2\right] \cfrac{d\phi_i}{dx}\cfrac{d\phi_j}{dx} + D_{xx}\cfrac{d^2\phi_i}{dx^2}\cfrac{d^2\phi_j}{dx^2}\right\}~dx \end{align} $$
 * $$\mathbf{K}^{11}$$ integrand is constant, $$\mathbf{K}^{12}$$ integrand is fourth-order, $$\mathbf{K}^{22}$$ integrand is eighth-order

Full integration

n_{\text{gauss pt}} = \text{int}[(p+1)/2] + 1 $$ Assume $$A_{xx}$$ = constant.

\begin{align} K_{ij}^{11} & = A_{xx} \int_{x_a}^{x_b} \cfrac{d\psi_i}{dx}\cfrac{d\psi_j}{dx}~dx \\ & = A_{xx} \int_{-1}^{1} \left(J^{-1}\cfrac{d\psi_i(\xi)}{d\xi}\right) \left(J^{-1}\cfrac{d\psi_j(\xi)}{d\xi}\right)J~d\xi = A_{xx} \int_{-1}^{1} F(\xi)~d\xi \\ & \approx A_{xx} W_1 F(\xi_1) \leftarrow { \text{one-point integration}} \end{align} $$

\begin{align} K_{ij}^{12} & = \frac{1}{2} \int_{x_a}^{x_b} \left(A_{xx}\cfrac{dw_0}{dx}\right) \cfrac{d\psi_i}{dx} \cfrac{d\phi_j}{dx}~dx\\ & = \cfrac{A_{xx}}{2} \int_{-1}^{1} \left(\sum_{i=1}^4 w_i J^{-1}\cfrac{d\phi_i(\xi)}{d\xi}\right) \left(J^{-1}\cfrac{d\psi_i(\xi)}{d\xi}\right) \left(J^{-1}\cfrac{d\phi_j(\xi)}{d\xi}\right)~dx \\ & \approx A_{xx}\left[ W_1 F(\xi_1) + W_2 F(\xi_2) + W_3 F(\xi_3)\right] \leftarrow { \text{full integration}} \\ & \approx A_{xx}\left[ W_1 F(\xi_1) + W_2 F(\xi_2)\right] \leftarrow { \text{reduced integration}} \end{align} $$


text: File:EulerBeamFE.png

title: File:EulerBeamFE.png


text: File:HermiteShape.png

title: File:HermiteShape.png


text: File:EulerBeamStiff.png

title: File:EulerBeamStiff.png


text: File:EulerBeamMembraneLock.png

title: File:EulerBeamMembraneLock.png


text:
Nonlinear finite elements/Timoshenko beams

Displacements

\begin{align} u_1 & = u_0(x) + z \varphi_x \\ u_2 & = 0 \\ u_3 & = w_0(x) \end{align} $$

Strains

\begin{align} \varepsilon_{xx}& = \varepsilon_{xx}^0 + z \varepsilon_{xx}^1 \\ \gamma_{xz}& = \gamma_{xz}^0 \end{align} $$

\begin{align} \varepsilon_{xx}^0 & = \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \\ \varepsilon_{xx}^1 & = \cfrac{d\varphi_x}{dx} \\ \gamma_{xz}^0 & = \varphi_x + \cfrac{dw_0}{dx} \end{align} $$

Principle of Virtual Work

\delta W_{\text{int}} = \delta W_{\text{ext}} $$ where
 * $$\begin{align}

\delta W_{\text{int}} & = \int_{x_a}^{x_b}\int_A (\sigma_{xx}\delta \varepsilon_{xx} + \sigma_{xz}\delta \gamma_{xz}) dA~dx\\ & = \int_{x_a}^{x_b} (N_{xx}\delta \varepsilon_{xx}^0 + M_{xx}\delta \varepsilon_{xx}^1 + Q_{x}\delta \gamma_{xz}^0) ~dx \\ \delta W_{\text{ext}} & = \int_{x_a}^{x_b} q\delta w_0~dx + \int_{x_a}^{x_b} f\delta u_0~dx \end{align}$$

Q_x = { K_s} \int_A \sigma_{xz}~dA $$ $$K_s$$ = shear correction factor

Taking Variations

\varepsilon_{xx}^0= \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 $$ Take variation

\delta \varepsilon_{xx}^0= \cfrac{d(\delta u_0)}{dx} + \frac{1}{2}\left(2\cfrac{dw_0}{dx}\right) \left(\cfrac{d(\delta w_0)}{dx}\right) = \cfrac{d(\delta u_0)}{dx} + \cfrac{dw_0}{dx}\cfrac{d(\delta w_0)}{dx} ~. $$

\varepsilon_{xx}^1= \cfrac{d\varphi_x}{dx} $$ Take variation

\delta \varepsilon_{xx}^1= \cfrac{d\delta \varphi_x}{dx} $$

\gamma_{xz}^0= \varphi_x + \cfrac{dw_0}{dx} $$ Take variation

\delta \gamma_{xz}^0= \delta \varphi_x + \cfrac{d(\delta w_0)}{dx} $$

Internal Virtual Work

 * $$\begin{align}

\int_{x_a}^{x_b} N_{xx}\delta \varepsilon_{xx}^0~dx & = \int_{x_a}^{x_b} N_{xx}\left[ \cfrac{d(\delta u_0)}{dx}+\cfrac{dw_0}{dx}\cfrac{d(\delta w_0)}{dx}\right]~dx \\ \int_{x_a}^{x_b} M_{xx}\delta \varepsilon_{xx}^1~dx & = \int_{x_a}^{x_b} M_{xx}\left[\cfrac{d\delta \varphi_x}{dx}\right]~dx \\ \int_{x_a}^{x_b} Q_{x}\delta \gamma_{xz}^0~dx & = \int_{x_a}^{x_b} Q_{x} \left[\delta \varphi_x + \cfrac{d(\delta w_0)}{dx}\right]~dx \end{align}$$

\begin{align} \delta W_{\text{int}} = \int_{x_a}^{x_b} & \left\{ N_{xx}\left[ \cfrac{d(\delta u_0)}{dx}+\cfrac{dw_0}{dx}\cfrac{d(\delta w_0)}{dx}\right]\right. + \\ &M_{xx}\left[\cfrac{d\delta \varphi_x}{dx}\right] + \left. Q_{x} \left[\delta \varphi_x + \cfrac{d(\delta w_0)}{dx}\right]\right\}~dx \end{align} $$

Integrate by Parts
Get rid of derivatives of the variations.
 * $$\begin{align}

\int_{x_a}^{x_b} N_{xx}&\left[ \cfrac{d(\delta u_0)}{dx}+\cfrac{dw_0}{dx}\cfrac{d(\delta w_0)}{dx}\right]~dx = \left[N_{xx}\delta u_0\right]_{x_a}^{x_b} - \int_{x_a}^{x_b} \cfrac{dN_{xx}}{dx}\delta u_0~dx + \\ & \qquad \qquad \qquad \left[N_{xx}\cfrac{dw_0}{dx}\delta w_0\right]_{x_a}^{x_b} - \int_{x_a}^{x_b} \cfrac{d}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right)\delta w_0~dx \\ \\ \int_{x_a}^{x_b} M_{xx}&\left[\cfrac{d(\delta \varphi_x)}{dx}\right]~dx = \left[M_{xx}\delta \varphi_x\right]_{x_a}^{x_b} - \int_{x_a}^{x_b} \cfrac{dM_{xx}}{dx}\delta \varphi_x~dx \\ \\ \int_{x_a}^{x_b} Q_{x} &\left[\delta \varphi_x + \cfrac{d(\delta w_0)}{dx}\right]~dx = \int_{x_a}^{x_b} Q_{x} \delta \varphi_x~dx + \left[Q_{x}\delta w_0\right]_{x_a}^{x_b} - \int_{x_a}^{x_b} \cfrac{dQ_{x}}{dx}\delta w_0~dx \end{align}$$

Collect terms

 * $$\begin{align}

&\left[N_{xx}{\delta u_0}\right]_{x_a}^{x_b} - \int_{x_a}^{x_b} \cfrac{dN_{xx}}{dx}{\delta u_0}~dx + \\ &\left[N_{xx}\cfrac{dw_0}{dx}\delta w_0 + Q_{x}\delta w_0\right]_{x_a}^{x_b} - \int_{x_a}^{x_b} \left[\cfrac{d}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right) + \cfrac{dQ_{x}}{dx}\right]\delta w_0~dx + \\ &\left[M_{xx}\delta \varphi_x\right]_{x_a}^{x_b} - \int_{x_a}^{x_b} \left(\cfrac{dM_{xx}}{dx} - Q_x \right)\delta \varphi_x~dx \\ &= \int_{x_a}^{x_b} q~\delta w_0~dx + \int_{x_a}^{x_b} f{\delta u_0}~dx \end{align}$$

Euler-Lagrange Equations

 * $$\begin{align}

\int_{x_a}^{x_b} \left(\cfrac{dN_{xx}}{dx} + f\right){\delta u_0}~dx & = \left[N_{xx}{\delta u_0}\right]_{x_a}^{x_b} \\ \int_{x_a}^{x_b} \left[\cfrac{d}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right) + \cfrac{dQ_{x}}{dx} + q \right]~\delta w_0~dx& = \left[N_{xx}\cfrac{dw_0}{dx}~\delta w_0 + Q_{x}~\delta w_0\right]_{x_a}^{x_b} \\ \int_{x_a}^{x_b} \left(\cfrac{dM_{xx}}{dx} - Q_x \right)~\delta \varphi_x~dx &= \left[M_{xx}~\delta \varphi_x\right]_{x_a}^{x_b} \end{align}$$

\begin{align} \cfrac{dN_{xx}}{dx} + f & = 0 \\ \cfrac{d}{dx}\left(N_{xx}\cfrac{dw_0}{dx}\right) + \cfrac{dQ_{x}}{dx} + q& = 0 \\ \cfrac{dM_{xx}}{dx} - Q_x& = 0 \end{align} $$

Constitutive Relations

\sigma_{xx} = E\varepsilon_{xx} ~; \qquad \sigma_{xz} = G\gamma_{xz} $$ Then,

\begin{align} N_{xx} & = A_{xx}~\varepsilon_{xx}^0 + B_{xx}~\varepsilon_{xx}^1 \\ \\ M_{xx} & = B_{xx}~\varepsilon_{xx}^0 + D_{xx}~\varepsilon_{xx}^1 \\ \\ Q_{x} & = S_{xx}~\gamma_{xz}^0 \end{align} $$ where

S_{xx} = K_s\int_A G~dA \qquad \leftarrow \qquad \text{shear stiffness} $$

Equilibrium Equations

 * $$\begin{align}

\cfrac{d}{dx}\left\{ A_{xx} \left[ \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \right] \right\} + f & = 0 \\ \cfrac{d}{dx}\left\{ S_{xx} \left( \cfrac{dw_0}{dx} + \varphi_x \right) + A_{xx} \cfrac{dw_0}{dx} \left[ \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \right] \right\} + q & = 0 \\ \cfrac{d}{dx}\left(D_{xx}\cfrac{d\varphi_x}{dx}\right) + S_{xx}\left( \cfrac{dw_0}{dx} + \varphi_x \right)& = 0 \end{align}$$

Weak Form

 * $$\begin{align}

\int_{x_a}^{x_b} A_{xx} \cfrac{d(\delta u_0)}{dx} \left[ \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \right]~dx &= \int_{x_a}^{x_b} f\delta u_0~dx + \left[N_{xx}\delta u_0\right]_{x_a}^{x_b} \\ \\ \int_{x_a}^{x_b} A_{xx}\cfrac{d(\delta w_0)}{dx}\cfrac{dw_0}{dx} \left[ \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \right]~dx &+ \int_{x_a}^{x_b} S_{xx}\cfrac{d(\delta w_0)}{dx}\left(\cfrac{dw_0}{dx}+\varphi_x\right)~dx = \\ &\int_{x_a}^{x_b} q\delta u_0~dx + \left[\left(N_{xx}\cfrac{dw_0}{dx}+Q_x \right)\delta w_0\right]_{x_a}^{x_b} \\ \\ -\int_{x_a}^{x_b} S_{xx}\delta \varphi_x\left(\cfrac{dw_0}{dx}+\varphi_x\right)~dx & + \int_{x_a}^{x_b} D_{xx}\cfrac{d(\delta \varphi_x)}{dx}\cfrac{d\varphi_x}{dx}~dx= \left[M_{xx}\delta \varphi_x\right]_{x_a}^{x_b} \end{align}$$

Trial Solution

 * $$\begin{align}

u_0(x) & = \sum_{j=1}^m u_j~\psi^{(1)}_j~; \qquad \qquad \delta u_0 = \psi^{(1)}_i \\ \\ w_0(x) & = \sum_{j=1}^n w_j~\psi^{(2)}_j~; \qquad \qquad \delta w_0 = \psi^{(2)}_i \\ \\ \varphi_x(x) & = \sum_{j=1}^p s_j~\psi^{(3)}_j~; \qquad \qquad \delta \varphi_x = \psi^{(3)}_i \end{align}$$

Element Stiffness Matrix

\begin{bmatrix} \mathbf{K}^{11} & \vdots & \mathbf{K}^{12} & \vdots & \mathbf{K}^{13} \\ & \vdots & & \vdots & \\ \mathbf{K}^{21} & \vdots & \mathbf{K}^{22} & \vdots & \mathbf{K}^{23} \\ & \vdots & & \vdots & \\ \mathbf{K}^{31} & \vdots & \mathbf{K}^{32} & \vdots & \mathbf{K}^{33} \\ \end{bmatrix} \begin{bmatrix} \mathbf{u} \\\\ \mathbf{w} \\\\ \mathbf{s} \end{bmatrix} = \begin{bmatrix} \mathbf{F}^1 \\ \\ \mathbf{F}^2 \\\\ \mathbf{F}^3 \end{bmatrix} $$

Choice 1

 * $$\psi^{(1)}$$ = linear ($$m=2$$)
 * $$\psi^{(2)}$$ = linear ($$n=2$$)
 * $$\psi^{(3)}$$ = linear ($$p=2$$).

Nearly singular stiffness matrix ($$6 \times 6$$).

Choice 2

 * $$\psi^{(1)}$$ = linear ($$m=2$$)
 * $$\psi^{(2)}$$ = quadratic ($$n=3$$)
 * $$\psi^{(3)}$$ = linear ($$p=2$$).

The stiffness matrix is ($$7 \times 7$$). We can statically condense out the interior degree of freedom and get a ($$6 \times 6$$) matrix. The element behaves well.

Choice 3

 * $$\psi^{(1)}$$ = linear ($$m=2$$)
 * $$\psi^{(2)}$$ = cubic ($$n=4$$)
 * $$\psi^{(3)}$$ = quadratic ($$p=3$$)

The stiffness matrix is ($$9 \times 9$$). We can statically condense out the interior degrees of freedom and get a ($$6 \times 6$$) matrix. If the shear and bending stiffnesses are element-wise constant, this element gives exact results.

Example Case
Linear $$u_0$$, Linear $$w_0$$, Linear $$\varphi_x$$.

\cfrac{dw_0}{dx} = \text{constant}. $$ But, for thin beams,

\cfrac{dw_0}{dx} = \text{slope} = -\varphi_x \leftarrow (\text{linear!}) $$ If constant $$\varphi_x$$

\cfrac{d\varphi_x}{dx} = 0 $$

Also
 * 1) $$Q_x = S_{xx}\varphi_x \ne 0 \implies$$ Non-zero transverse shear.
 * 2) $$M_{xx} = D_{xx} \cfrac{d\varphi_x}{dx} = 0 \implies $$ Zero bending energy.

Result: Zero displacements and rotations $$\implies$$ Shear Locking!

Recall

\cfrac{d}{dx}\left\{ A_{xx} \left[ \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \right] \right\} + f= 0 $$ or,

\cfrac{d}{dx}\left\{ A_{xx} \varepsilon_{xx}^0 \right\} + f= 0 $$ If $$f = 0$$ and $$A_{xx} = $$ constant

A_{xx}\cfrac{d}{dx} (\varepsilon_{xx}^0)= 0 \qquad \implies \qquad \varepsilon_{xx}^0 = ~\text{constant}. $$ If there is only bending but no stretching,

\varepsilon_{xx}^0 = 0 = \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 $$ Hence,

\cfrac{du_0}{dx} \approx \left(\cfrac{dw_0}{dx}\right)^2 $$

Also recall:

\cfrac{d}{dx}\left\{ S_{xx} \left( \cfrac{dw_0}{dx} + \varphi_x \right) + A_{xx} \cfrac{dw_0}{dx} \left[ \cfrac{du_0}{dx} + \frac{1}{2}\left(\cfrac{dw_0}{dx}\right)^2 \right] \right\} + q = 0 $$ or,

\cfrac{d}{dx}\left\{ S_{xx} \gamma_{xz}^0 + A_{xx} \cfrac{dw_0}{dx} \varepsilon_{xx}^0 \right\} + q = 0 $$ If $$q = 0$$ and $$S_{xx} = $$ constant, and no membrane strains

S_{xx}\cfrac{d}{dx} (\gamma_{xz}^0)= 0 \qquad \implies \qquad {\gamma_{xz} = ~\text{constant}}~ = \cfrac{dw_0}{dx} + \varphi_x $$ Hence,

\varphi_x \approx \cfrac{dw_0}{dx} $$

Shape functions need to satisfy: $$ \cfrac{du_0}{dx} \approx \left(\cfrac{dw_0}{dx}\right)^2 ~; \qquad\text{and}\qquad \varphi_x\approx \cfrac{dw_0}{dx} $$

Example Case 1
Linear $$u_0$$, Linear $$w_0$$, Linear $$\varphi_x$$.

 * First condition $$\implies$$ constant $$=$$ constant. Passes! No Membrane Locking.
 * Second condition $$\implies$$ linear $$=$$ constant. Fails! Shear Locking.

Example Case 2
Linear $$u_0$$, Quadratic $$w_0$$, Linear $$\varphi_x$$.

 * First condition $$\implies$$ constant $$=$$ quadratic. Fails!  Membrane Locking.
 * Second condition $$\implies$$ linear $$=$$ linear. Passes! No Shear Locking.

Example Case 3
Quadratic $$u_0$$, Quadratic $$w_0$$, Linear $$\varphi_x$$.

 * First condition $$\implies$$ linear $$=$$ quadratic. Fails! Membrane Locking.
 * Second condition $$\implies$$ linear $$=$$ linear. Passes! No Shear Locking.

Example Case 4
Cubic $$u_0$$, Quadratic $$w_0$$, Linear $$\varphi_x$$.

 * First condition $$\implies$$ quadratic $$=$$ quadratic. Passes!  No Membrane Locking.
 * Second condition $$\implies$$ linear $$=$$ linear. Passes!  No Shear Locking.

Option 1

 * Linear $$u_0$$, linear $$w_0$$, linear $$\varphi_x$$.
 * Equal interpolation for both $$w_0$$ and $$\phi_x$$.
 * Reduced integration for terms containing $$\phi_x$$ - treat as constant.

Option 2

 * Cubic $$u_0$$, quadratic $$w_0$$, linear $$\varphi_x$$.
 * Stiffness matrix is $$9\times9$$.
 * Hard to implement.


text: File:TimoshenkoBeam.png

title: File:TimoshenkoBeam.png


text:
Nonlinear finite elements/Homework 7/Hints

Hints 1: Index notation
Index notation:

\sigma_{ij} = 2\mu~\varepsilon_{ij} + \lambda~\varepsilon_{kk}~\delta_{ij}~. $$

If $$j=i$$
 * $$\begin{align}

\sigma_{ii} & = 2\mu~\varepsilon_{ii} + \lambda~\varepsilon_{kk}~\delta_{ii} \\ & = 2\mu~\varepsilon_{kk} + 3\lambda~\varepsilon_{kk} \\ & = (2\mu + 3\lambda)~\varepsilon_{kk} \end{align}$$

\sigma_{kk} = (2\mu + 3\lambda)~\varepsilon_{kk} $$

Dummy indices are replaceable.

Hint 2: Index notation
Index notation:

\sigma_{ij} = 2\mu~\varepsilon_{ij} + \lambda~\varepsilon_{kk}~\delta_{ij}~. $$

Multiply by $$\delta_{ij}$$:
 * $$\begin{align}

\sigma_{ij}~\delta_{ij} & = 2\mu~\varepsilon_{ij}~\delta_{ij} + \lambda~\varepsilon_{kk}~\delta_{ij}~\delta_{ij}\\ \implies \sigma_{jj}& = 2\mu~\varepsilon_{ii} + \lambda~\varepsilon_{kk}~\delta_{ii}\\ \implies \sigma_{kk}& = 2\mu~\varepsilon_{kk} + 3\lambda~\varepsilon_{kk} \\ \implies \sigma_{kk}& = (2\mu + 3\lambda)~\varepsilon_{kk} \end{align}$$

Multiplication by $$\delta_{ij}$$ leads to replacement of one index.

A_{ij}~\delta_{kl} = ? \qquad A_{ij}~\delta{jl} = ? $$

Hint 3: Index notation
Index notation:
 * $$\begin{align}

\boldsymbol{\sigma} & =\sigma_{ij}~\mathbf{e}_i\otimes\mathbf{e}_j \\ \boldsymbol{\varepsilon} & =\varepsilon_{ij}~\mathbf{e}_i\otimes\mathbf{e}_j \end{align}$$ From the definition of dyadic product, we can show
 * $$\begin{align}

(\mathbf{a}\otimes\mathbf{b}):(\mathbf{u}\otimes\mathbf{v}) & = (\mathbf{a}\bullet\mathbf{u})(\mathbf{b}\bullet\mathbf{v}) \end{align}$$ Contraction gives:
 * $$\begin{align}

\boldsymbol{\sigma}:\boldsymbol{\varepsilon} & = (\sigma_{ij}~\mathbf{e}_i\otimes\mathbf{e}_j): (\varepsilon_{kl}~\mathbf{e}_k\otimes\mathbf{e}_l) \\ & = \sigma_{ij}~\varepsilon_{kl}~(\mathbf{e}_i\otimes\mathbf{e}_j): (\mathbf{e}_k\otimes\mathbf{e}_l) \\ & = \sigma_{ij}~\varepsilon_{kl}~ (\mathbf{e}_i\bullet\mathbf{e}_k)(\mathbf{e}_j\bullet\mathbf{e}_l) \\ & = \sigma_{ij}~\varepsilon_{kl}~\delta_{ik}~\delta_{jl} \\ & = \sigma_{ij}~\varepsilon_{ij} \end{align}$$

Hint 4: Tensor product
Index notation:
 * $$\begin{align}

\boldsymbol{\varepsilon} & =\varepsilon_{ij}~\mathbf{e}_i\otimes\mathbf{e}_j \\ \boldsymbol{\mathsf{C}} & = C_{ijkl}~\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l \end{align}$$ Definition of dyadics products:
 * $$\begin{align}

(\mathbf{a}\bullet\mathbf{b})\otimes\mathbf{x} & = (\mathbf{b}\bullet\mathbf{x})\mathbf{a} \\ (\mathbf{a}\bullet\mathbf{b}\otimes\mathbf{c})\otimes\mathbf{x} & = (\mathbf{c}\bullet\mathbf{x}) (\mathbf{a}\otimes\mathbf{b}) \\ (\mathbf{a}\bullet\mathbf{b}\otimes\mathbf{c}\otimes\mathbf{d})\otimes\mathbf{x} & = (\mathbf{d}\bullet\mathbf{x}) (\mathbf{a}\otimes\mathbf{b}\otimes\mathbf{c}) \end{align}$$ We can show that
 * $$\begin{align}

(\mathbf{a}\otimes\mathbf{b}\otimes\mathbf{c}\otimes\mathbf{d}):(\mathbf{u}\otimes\mathbf{v}) & = ((\mathbf{a}\bullet\mathbf{b}\otimes\mathbf{c}\otimes\mathbf{d})\otimes\mathbf{v})\bullet\mathbf{u} \end{align}$$ Contraction gives:
 * $$\begin{align}

\boldsymbol{\mathsf{C}}:\boldsymbol{\varepsilon} & = (C_{ijkl}~\mathbf{e}_i\otimes\mathbf{e}_j \otimes \mathbf{e}_k\otimes\mathbf{e}_l): (\varepsilon_{mn}~\mathbf{e}_m\otimes\mathbf{e}_n) \\ & = C_{ijkl}~\varepsilon_{mn}~(\mathbf{e}_i\otimes\mathbf{e}_j \otimes \mathbf{e}_k\otimes\mathbf{e}_l):(\mathbf{e}_m\otimes\mathbf{e}_n) \\ & = C_{ijkl}~\varepsilon_{mn}~ ((\mathbf{e}_i\bullet\mathbf{e}_i\otimes\mathbf{e}_k\otimes\mathbf{e}_l)\otimes\mathbf{e}_n) \bullet\mathbf{e}_m \\ & = C_{ijkl}~\varepsilon_{mn}~ (\mathbf{e}_l\bullet\mathbf{e}_n)(\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k) \bullet\mathbf{e}_m \\ & = C_{ijkl}~\varepsilon_{mn}~\delta_{ln} (\mathbf{e}_k\bullet\mathbf{e}_m)(\mathbf{e}_i\otimes\mathbf{e}_j) = C_{ijkl}~\varepsilon_{mn}~\delta_{ln}~\delta_{km}~\mathbf{e}_i\otimes\mathbf{e}_j \\ & = C_{ijkl}~\varepsilon_{kl} \end{align}$$

Hint 5 : Tensor product
Tensor Product of two tensors:
 * $$\begin{align}

\boldsymbol{A} & =A_{ij}~\mathbf{e}_i\otimes\mathbf{e}_j \\ \boldsymbol{B} & =B_{kl}~\mathbf{e}_k\otimes\mathbf{e}_l \end{align}$$

Tensor product:
 * $$\begin{align}

\boldsymbol{A}\otimes\boldsymbol{B} & = (A_{ij}~\mathbf{e}_i\otimes\mathbf{e}_j) \otimes (B_{kl}~\mathbf{e}_k\otimes\mathbf{e}_l) \\ & = A_{ij} B_{kl} \mathbf{e}_i\otimes\mathbf{e}_j \otimes \mathbf{e}_k\otimes\mathbf{e}_l \end{align}$$

Hint 6: Vector transformations
Change of basis: Vector transformation rule

v^{'}_i = L_{ij} v_j $$

$$L_{ij}$$ are the direction cosines.
 * $$\begin{align}

L_{11} & = \mathbf{e}_1^{'}\bullet\mathbf{e}_1; & L_{12} & = \mathbf{e}_1^{'}\bullet\mathbf{e}_2; & L_{13} & = \mathbf{e}_1^{'}\bullet\mathbf{e}_3 \\ L_{21} & = \mathbf{e}_2^{'}\bullet\mathbf{e}_1; & L_{22} & = \mathbf{e}_2^{'}\bullet\mathbf{e}_2; & L_{23} & = \mathbf{e}_2^{'}\bullet\mathbf{e}_3 \\ L_{31} & = \mathbf{e}_3^{'}\bullet\mathbf{e}_1; & L_{32} & = \mathbf{e}_3^{'}\bullet\mathbf{e}_2; & L_{33} & = \mathbf{e}_3^{'}\bullet\mathbf{e}_3 \end{align}$$

In matrix form

\mathbf{v}^{'} = \mathbf{L}~\mathbf{v}; \mathbf{v} = \mathbf{L}^T~\mathbf{v}^{'}; \mathbf{L}\mathbf{L}^T = \mathbf{I} \implies \mathbf{L}^T = \mathbf{L}^{-1} $$

Other common form: Vector transformation rule

v^{'}_i = Q_{ji} v_j $$
 * $$\begin{align}

Q_{11} & = \mathbf{e}_1\bullet\mathbf{e}_1^{'}; & Q_{12} & = \mathbf{e}_1\bullet\mathbf{e}_2^{'}; & Q_{13} & = \mathbf{e}_1\bullet\mathbf{e}_3^{'} \\ Q_{21} & = \mathbf{e}_2\bullet\mathbf{e}_1^{'}; & Q_{22} & = \mathbf{e}_2\bullet\mathbf{e}_2^{'}; & Q_{23} & = \mathbf{e}_2\bullet\mathbf{e}_3^{'} \\ Q_{31} & = \mathbf{e}_3\bullet\mathbf{e}_1^{'}; & Q_{32} & = \mathbf{e}_3\bullet\mathbf{e}_2^{'}; & Q_{33} & = \mathbf{e}_3\bullet\mathbf{e}_3^{'} \end{align}$$

In matrix form

\mathbf{v}^{'} = \mathbf{Q}^T~\mathbf{v}; \mathbf{v} = \mathbf{Q}~\mathbf{v}^{'}; \mathbf{Q}\mathbf{Q}^T = \mathbf{I} \implies \mathbf{Q}^T = \mathbf{Q}^{-1} $$

Hint 7: Tensor transformations
Change of basis: Tensor transformation rule

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} $$

Other common form

T^{'}_{ij} = Q_{pi} Q_{qj} T_{pq} $$

In matrix form,

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


text:
Nonlinear finite elements/Buckling of beams

Newton-Raphson
Standard Newton-Raphon methods perform poorly for bucking problems.

Arc length method

 * Also called Modified Riks Method.
 * Control the size of the load step using a parameter $$\lambda$$.
 * Solve for both $$\lambda$$ and $$\Delta u$$ in each Newton iteration.

Assume $$F$$ = independent of geometry. Then

F = \lambda~\bar{F} $$

$$\lambda$$ can be thought of as a normalized load parameter.

\text{Residual}~= r(u, \lambda) = K(u)~u - \lambda~\bar{F} $$

The load increment is computed using

\lambda = \pm\sqrt{\Delta s^2 - \Delta u_n^2} $$

The reference arc length

\Delta s_0 = \cfrac{F}{n_{\text{loadstep}}} $$


text: File:Buckling.png

title: File:Buckling.png


text: File:BucklingNR.png

title: File:BucklingNR.png


text: File:BucklingArc.png

title: File:BucklingArc.png


text: File:Isoparametric.png

title: File:Isoparametric.png


text:
MediaWiki/Engine/wiki.php

This file / (reference:) defines the class MediaWiki(reference:) - to be the base class

Functions

 * MediaWiki
 * setVal( $key, &$value )
 * getVal
 * initialize ( &$title, &$output, &$user, $request)
 * checkInitialQueries
 * preliminaryChecks
 * initializeSpecialCases
 * articleFromTitle
 * initializeArticle
 * finalCleanup ( &$deferredUpdates, &$loadBalancer, &$output )
 * doUpdates ( &$updates )
 * doJobs
 * restInPeace
 * performAction( &$output, &$article, &$title, &$user, &$request )


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Technical Writing/Romeo

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User talk:Ivanicina

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Interests
ICT in education, knowledge management, systems thinking and soft operations research. On WikiVersity I'm contributing to an initiative on Social Entrepreneurship.

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User:Robert hawkins

World Bank Institute
Telephone: +27-12-431-3130 email: rhawkins@worldbank.org

Mailing Address: PO Box 12629, Hatfield, 0028 Pretoria, South Africa

Physical Address: 1st Floor, Pro Equity Court Building 1250 Pretorius St. Hatfield Pretoria

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User:Grahamlash

=Welcome=

My interest is entrepreneurial education
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User:Mmarais/My Sandbox

Playtime!!!

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User:Moolla

=Welcome=

Who am I
I work at the Community University Partnerships office at the University of the Witwatersrand, Johannesburg.

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User:Ujorwic/My Sandbox

Let's get started ...

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User:Bobday

=Rubbish= [] My sandbox

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User:Rodneylsmith9/My Sandbox

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User:Tnswettenham

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My thoughts

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User:Robert hawkins/My Sandbox

[|Social Entrepreneurship Netowrk]

How do we build a community of curriculum developers and curriculum sharers?

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title: User:Robert hawkins/My Sandbox


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Technical writing/Overview/Technical writing enrollment template here

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User:Desbo~enwikiversity/My Sandbox

This is where my interests lie

1. Health

I am currently conducting sleep research towards an MSc degree at University of the Witwatersrand, and working in the private sector, attempting to treat sleep disorders. To make an appointment for a sleep study, please call 072 907 9004, or visit http://www.pretoriasleepcentre.com

2. Education

Volunteering with the Student Sponsorship Programme to grant academically talented, but financially disadvantaged, High School kids scholarships to study a some of Johannesburg's finest schools. To find out more about how you can contribute, or for applications, please visit: http://www.sspsa.org

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User:Grahamlash/My Sandbox

My Practice Sessions

Here is The Wheel:

The Wheel shows how The Karatara Project works

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Talk:Social Entrepreneurship Curriculum

Are these the main headings that should be included on this curricular page? What is the ideal structure? Robert hawkins 10:09, 23 August 2007 (UTC)

Brainstorming on Headings
For Professors/Educators --> access to learning materials and resources for your curriculum. Contest announcement --> add to these resources

For Students --> I want to be a social entrepreneur -- links to franchise ideas? Links to communities? Links to sources of support?

For the community --> Im a community member and I want to start an initiative. Where do I begin?

Partners page --> South African partners --> tertiary institutions; NGOs CBOs; Government; International partners?

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Technical writing/Bartek

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Technical writing/Zdenka

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Portal:Social Entrepreneurship Network/Social Entrepreneurship Network news/Activities

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Collaborative Project for Social Entrepreneurship

Objectives

 * Common framework for students/universities to collaborate on projects
 * Opportunity for peer review of projects
 * Electronic format to capture data on communities and information on projects
 * Means of identifying most promising ideas and potentially funding these ideas.

Brainstorming of Ideas
Entrepreneurs' basic functions include interactions with the environment, Opportunity Scanning and Identification. Social entrepreneurs are no diifrent. They also interact with the environment, and find opportunities to act, they are usualy innovative in their solutions.Innovation are not very difficult to make. Contrary to a gen. perception, innovation is not invention. It is (commercial) application of the idea. To initiate participants into active interactions with the environment and identification of opportunities, simple projects can be designed. For example, if some visitors to this page suggest a solution to any environmental problem in their locality, and a disscussion on the topics will certainly help everybody. In one of my visits in rural India, a small village hospital had a strange problem. People would consume chewing tobaco and spit in the corners. Many solutions were tried, including positioning spit toons at different places, making appeals, even administrative orders, but things did not change. A social entrepreneur thought of an innovative idea. He palced religious masseges, symbols etc. in all corners of the hospital. People in India have a great respect for all religions. The problem got solved by a simple actions. Do you have a similar story to tell? --Prof. Vivek Sharma 06:57, 10 January 2008 (UTC)


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Technical writing/Jana Sindelarova

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User:Alexandra~enwikiversity

During the weekend, I'll write here why I had decided to join the wikiversity community.

Take part in ourTechnical Writing course!

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Art History/Reflections/Connor

Connor's Reflections So far, I have a rough draft of what I am going to turn in. I have a page of information about Rodin, his sculpture of Eve, and why the Musee Rodin needs it. In addition, I also have a lot of information about the Gates of Hell, which is what this statue of Eve is part of. I also have a picture of the sculpture that I got off of Artstor. I think that I could turn it in in a few days. 23 August 2007

Sep. 9, 2007 I presented my project and did my self- evaluation last week. I also just got back from discovery, so I'm still catching up. At the moment I'm looking for those 3 museums. I'm also getting some background information on Serra.

Sep. 12, 2007 Last class we had a long argument about whether Serra's work is really art. I was on the side of that it is absolutely not art. However, after seeing a video of his work, and a video of walking through it, I have changed my mind. The optical illusions of some of the pieces are very cool, and how he changes the sense of reality in his statues is definitely worthy of being considered art. However, there are a still a few of his sculptures that I do not like, and a few that I still don't consider art, even if I have seen them in video.

Sep. 14, 2007 I just finished my review of Serra's work and I have a new appreciation for some of it. However, that is only some of it. I still do not think a row of belts is art, but to each his own.

I still need to add my bibliography and edit my review and then I'll be done.

Sep.17 2007 I liked the new yorker review better than the other one. Plus, that guy had an awesome last name. He also actually had an opinion, while the other review just seemed like random background information.

Sep. 21 2007 We had some trouble thinking of in depth questions, but we eventually got some.

Skidway Joe Why is it blurry? Who is it? What does the title mean? What is he doing? Why is it the only unfocused picture in your collection?

Gonna Walk All Over What was the meaning of this picture? What were you trying to portray? Why is it titled Gonna Walk All Over? Do you know this person?

Cake Walk What is a cake walk? Who are the people? Why did you include it in your collection? Where does this take place? Have you ever participated in a cake walk?

American Sharecroppers Why do they look so happy, despite their poverty? Why is the kid not happy? Why did you feel like this was picture-worthy? During what time period was this picture taken?

Ghetto Baby What’s the point of this picture? Where is this? Why did you choose to take a picture of a baby in a go-kart? What is the setting of this picture? Where are his parents and why is he in a go-kart? Do you know this baby or did you just take a picture of him?

September 28 Mr. Anderson's second presentation was very helpful. I am getting a deeper understanding of the greater meaning in his photos. However, I do not understand what we are trying to do with this though

October 5th I am working with Rawson for the museum project. The theme we chose was Power and Authority. There should be a lot of works about this.

October 9th As I predicted, there are a lot of works relating to power. For the five works we chose Korean gold that only royalty could wear, a ceromonial Mesopotamian sword, a bust of Caesar, a painting of Napolean, and a picture of Mao Zedong. We tried to concentrate on different figures, and tried to also use dictators, like Mao as part of the collection.

October 12 Questions about Power and Authority What are some typical examples of power and authority? Do you think that overexaggerating things to make leaders look more powerful is right? Do you think that it is impossible to show a work involving power and authority without it going into the realm of propaganda? Is it possible to have authority without power? Or vice versa?

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