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Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 9

Problem 1: Part 9: Elastic-plastic tangent modulus
Assume that the elastic response of the material is linear, i.e.,

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

Derive the expression for the elastic-plastic tangent modulus for a von Mises yield condition with Johnson-Cook flow stress for a linear elastic material using the expressions that you have derived in the previous parts.

The elastic-plastic tangent modulus is given by

\boldsymbol{\mathsf{C}}^{\text{ep}} = \boldsymbol{\mathsf{C}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf

{C}}:f_{\boldsymbol{\sigma})}} {f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}}\right)~. $$ From the previous parts

f_{\boldsymbol{\sigma}} = \sqrt{\cfrac{3}{2}}~\mathbf{n} ~; f_{\alpha} = - n~B~\alpha^{n-1} \left[1 - \left(\cfrac{T - T_0}{T_m -T_0}\right)\right] ~; f_T = \left(\cfrac{1}{T_m - T_0}\right) \left[\sigma_0 + B \alpha^n\right] ~. $$ Therefore,

\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} = \sqrt{\cfrac{3}{2}}~(\lambda~\boldsymbol

{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf{I}}):\mathbf{n} = \sqrt{\cfrac{3}{2}}~(\lambda~\text{tr}(\mathbf{n})~\boldsymbol{\mathit{1}} +

2~\mu~\mathbf{n}) = \sqrt{\cfrac{3}{2}}~2~\mu~\mathbf{n} ~. $$

Some of the results used in the above derivation are shown below.

Recall (from previous homework):

\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}}:\mathbf{n} = \delta_{ij}~\delta_{kl}

~n_{kl} = n_{kk}~\delta_{ij} = \text{tr}(\mathbf{n})~\boldsymbol{\mathit{1}} $$ and

\boldsymbol{\mathsf{I}}:\mathbf{n} = \frac{1}{2}~(\delta_{ik}~\delta_{jl} + \delta_{il}

~\delta_{jk}) n_{kl} = \frac{1}{2}~(\delta_{jl}~n_{il} + \delta_{jk}~n_{ki}) = \frac{1}{2}~(n_{ij} + n_{ji}) = n_{ij} = \mathbf{n} $$ (we have used the symmetry of the stress tensor above.)

Also,

\text{tr}(\mathbf{n}) = \text{tr}\left(\cfrac{\mathbf{s}}{\lVert\mathbf{s}\rVert_{}}\right) = \cfrac{\text{tr}(\mathbf{s})}{\lVert\mathbf{s}\rVert_{}} $$ Now,

\mathbf{s} = s_{ij} = \sigma_{ij} - \frac{1}{3}~\sigma_{kk}~\delta_{ij} \qquad \implies \qquad \text{tr}(\mathbf{s}) = s_{ii} = \sigma_{ii} - \frac{1}{3}~\sigma_{kk}~\delta_{ii} = \sigma_{ii} - \frac{1}{3}~\sigma_{kk}~3 = \sigma_{ii} - \sigma_{kk} = 0 $$ Therefore,

\text{tr}(\mathbf{n}) = 0 ~. $$

Hence,

(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf{C}}:f_

{\boldsymbol{\sigma})} = \cfrac{3}{2}~4~\mu^2~\mathbf{n}\otimes\mathbf{n} = 6~\mu^2~\mathbf{n}\otimes\mathbf{n} $$ and

f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} = \cfrac{3}{2}

~2~\mu~\mathbf{n}:\mathbf{n} = 3~\mu~\left(\cfrac{\mathbf{s}}{\sqrt{\mathbf{s}:\mathbf{s}}}\right): \left(\cfrac{\mathbf{s}}{\sqrt{\mathbf{s}:\mathbf{s}}}\right) = 3~\mu~\cfrac{\mathbf{s}:\mathbf{s}}{\mathbf{s}:\mathbf{s}} = 3~\mu ~. $$ Plugging in expression for $$\boldsymbol{\mathsf{C}}^{\text{ep}}$$ we get

\boldsymbol{\mathsf{C}}^{\text{ep}} = \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\boldsymbol{\mathsf

{I}} - \left(\cfrac{ 6~\mu^2~\mathbf{n}\otimes\mathbf{n} } {3~\mu -n~B~\alpha^{n-1}\left[1 - \left(\cfrac{T-T_0}{T_m-T_0}\right)\right] ~\cfrac{\boldsymbol{\varepsilon}^p:\mathbf{n}}{\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \sqrt{\cfrac{3}{2}}~\cfrac{\chi}{\rho~C_p}~ \left(\cfrac{1}{T_m - T_0}\right)\left[\sigma_0 + B \alpha^n\right] ~\boldsymbol{\sigma}:\mathbf{n}}\right)~. $$ Now,

\boldsymbol{\sigma}:\mathbf{n} = \cfrac{(\mathbf{s} + \frac{1}{3}~\text{tr}(\boldsymbol

{\sigma})~\boldsymbol{\mathit{1}}):\mathbf{s}}{\sqrt{\mathbf{s}:\mathbf{s}}} = \cfrac{\mathbf{s}:\mathbf{s}}{\sqrt{\mathbf{s}:\mathbf{s}}} + \frac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\cfrac{\boldsymbol{\mathit{1}}:\mathbf{s}}{\sqrt

{\mathbf{s}:\mathbf{s}}} = \sqrt{\mathbf{s}:\mathbf{s}} + \frac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\cfrac{\text{tr}(\mathbf{s})}{\sqrt{\mathbf

{s}:\mathbf{s}}} = \sqrt{\mathbf{s}:\mathbf{s}} = \lVert\mathbf{s}\rVert_{} $$ Therefore, the elastic-plastic tangent modulus can be written as

\boldsymbol{\mathsf{C}}^{\text{ep}} = \lambda~\boldsymbol{\mathit{1}}\otimes\boldsymbol{\mathit{1}} + 2~\mu~\left[\boldsymbol

{\mathsf{I}} - \cfrac{ 3~\mu~\mathbf{n}\otimes\mathbf{n} } {3~\mu -n~B~\alpha^{n-1}\left[1 - \left(\cfrac{T-T_0}{T_m-T_0}\right)\right] ~\cfrac{\boldsymbol{\varepsilon}^p:\mathbf{n}}{\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \sqrt{\cfrac{3}{2}}~\cfrac{\chi}{\rho~C_p}~ \left(\cfrac{1}{T_m - T_0}\right)\left[\sigma_0 + B \alpha^n\right] ~\lVert\mathbf{s}\rVert_{}}\right]~. $$


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Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 8

Problem 1: Part 8: Consistency condition - 2
The yield stress $$\sigma_y$$ is given by the Johnson-Cook model

\sigma_y(\alpha,T) = \left[\sigma_0 + B \alpha^n\right] \left[1 - \left(\cfrac{T - T_0}{T_m -T_0}\right)\right] $$ where $$\sigma_0$$ is the initial yield stress, $$B, n$$ are constants, $$T_0$$ is a reference temperature, and $$T_m$$ is the melt temperature. Derive expressions for $$\partial f/\partial \alpha$$, and $$\partial f/ \partial T$$ for the von Mises yield condition with the Johnson-Cook flow stress model.

The yield function is

f = \sqrt{\cfrac{3}{2}}~\sqrt{\mathbf{s}:\mathbf{s}} - \sigma_y= \sqrt{\cfrac{3}{2}}~\sqrt{\mathbf{s}:\mathbf{s}} - \left[\sigma_0 + B \alpha^n\right] \left[1 - \left(\cfrac{T - T_0}{T_m -T_0}\right)\right] $$ Therefore,

{ \frac{\partial f}{\partial \alpha} = f_{\alpha} = - n~B~\alpha^{n-1} \left[1 - \left(\cfrac{T - T_0}{T_m -T_0}\right)\right] } $$ and

{ \frac{\partial f}{\partial T} = f_T = \left(\cfrac{1}{T_m - T_0}\right) \left[\sigma_0 + B \alpha^n\right] ~. } $$

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Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 7

Problem 1: Part 7: Flow rule
The $$J_2$$ theory of plasticity also states that the material satisfies the von Mises yield condition

f(\boldsymbol{\sigma},\alpha,T) := \sqrt{\cfrac{3}{2}}\lVert\mathbf{s}\rVert_{} - \sigma_y(\alpha,T) $$ where $$\mathbf{s}$$ is the deviatoric part of the stress $$\boldsymbol {\sigma}$$. Derive an expression for $$\partial f/\partial\boldsymbol{\sigma} $$ in terms of the normal to the yield surface

\mathbf{n} = \cfrac{\mathbf{s}}{\lVert\mathbf{s}\rVert_{}} ~. $$

The von Mises yield function is

f = \sqrt{\cfrac{3}{2}}~\sqrt{\mathbf{s}:\mathbf{s}} - \sigma_y ~. $$

We can alternatively write the yield function as

f = \sqrt{\mathbf{s}:\mathbf{s}} - \sqrt{\cfrac{2}{3}}~\sigma_y ~. $$ in which case the following equations take a slightly different form.

Therefore,

\begin{align} \frac{\partial f}{\partial \boldsymbol{\sigma}} = f_{\boldsymbol{\sigma}} & = \sqrt{\cfrac{3}{2}}~ \frac{\partial }{\partial \boldsymbol{\sigma}}\left(\sqrt{\mathbf {s}:\mathbf{s}}\right) \\ & = \left(\sqrt{\cfrac{3}{2}}\right)\left(\frac{1}{2}\right) \left(\cfrac{1}{\sqrt{\mathbf{s}:\mathbf{s}}}\right) \frac{\partial }{\partial \boldsymbol{\sigma}}\left(\mathbf{s}:\mathbf{s}\right) \\ & = \left(\sqrt{\cfrac{3}{2}}\right)\left(\frac{1}{2}\right) \left(\cfrac{1}{\lVert\mathbf{s}\rVert_{}}\right) \frac{\partial }{\partial \boldsymbol{\sigma}}\left(\mathbf{s}:\mathbf{s}\right) \end{align} $$ The deviatoric part of $$\boldsymbol{\sigma}$$ is

\mathbf{s} = \boldsymbol{\sigma} - \frac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\boldsymbol {\mathit{1}} \qquad~\text{or}~\qquad s_{ij} = \sigma_{ij} - \frac{1}{3}~\sigma_{kk}~\delta_{ij} $$ Therefore,

\begin{align} \mathbf{s}:\mathbf{s} & = \left(\boldsymbol{\sigma} - \frac{1}{3}~\text{tr}(\boldsymbol {\sigma})~\boldsymbol{\mathit{1}}\right): \left(\boldsymbol{\sigma} - \frac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\boldsymbol{\mathit {1}}\right) \\ & = \boldsymbol{\sigma}:\boldsymbol{\sigma} - \cfrac{2}{3}~\text{tr}(\boldsymbol{\sigma}) ~\boldsymbol{\sigma}:\boldsymbol{\mathit{1}} + \cfrac{1}{9}~\left(\text{tr}(\boldsymbol{\sigma})\right)^2~\boldsymbol{\mathit {1}}:\boldsymbol{\mathit{1}} \end{align} $$ Now,

\boldsymbol{\sigma}:\boldsymbol{\mathit{1}} = \sigma_{ij}\delta_{ij} = \sigma_{ii} = \text {tr}(\boldsymbol{\sigma}) $$ and

\boldsymbol{\mathit{1}}:\boldsymbol{\mathit{1}} = \delta_{ij}\delta_{ij} = \delta_{ii} = 3 ~. $$ Therefore,

\mathbf{s}:\mathbf{s} =\boldsymbol{\sigma}:\boldsymbol{\sigma} - \cfrac{2}{3}~\left(\text {tr}(\boldsymbol{\sigma})\right)^2 + \cfrac{1}{3}~\left(\text{tr}(\boldsymbol{\sigma})\right)^2 = \boldsymbol{\sigma}:\boldsymbol{\sigma} - \cfrac{1}{3}~\left(\text{tr}(\boldsymbol {\sigma})\right)^2~. $$ The derivative with respect to $$\boldsymbol{\sigma}$$ is

\frac{\partial }{\partial \boldsymbol{\sigma}}(\mathbf{s}:\mathbf{s}) = \frac{\partial }{\partial \boldsymbol{\sigma}}(\boldsymbol{\sigma}:\boldsymbol{\sigma}) - \cfrac{1}{3}\frac{\partial }{\partial \boldsymbol{\sigma}}\left[\left(\text{tr}(\boldsymbol {\sigma})\right)^2\right] = \frac{\partial }{\partial \boldsymbol{\sigma}}(\boldsymbol{\sigma}:\boldsymbol{\sigma}) - \cfrac{2}{3}~\text{tr}(\boldsymbol{\sigma})~\frac{\partial }{\partial \boldsymbol{\sigma}} \left[\text{tr}(\boldsymbol{\sigma})\right] $$ Let us use index notation to find the derivatives. In index notation,

\frac{\partial }{\partial \boldsymbol{\sigma}}(\boldsymbol{\sigma}:\boldsymbol{\sigma}) = \frac{\partial }{\partial \sigma_{ij}}(\sigma_{kl}\sigma_{kl}) = \frac{\partial }{\partial \sigma_{ij}}\left(\sigma_{11}^2 + \sigma_{12}^2 + \sigma_{13}^2 + \sigma_{21}^2 + \sigma_{22}^2 + \sigma_{23}^2 + \sigma_{31}^2 + \sigma_{32}^2 + \sigma_{33}^2\right) $$ Hence, the components of the second-order tensor are

\begin{align} \frac{\partial }{\partial \sigma_{11}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{11} & & \qquad \frac{\partial }{\partial \sigma_{12}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{12} & & \qquad \frac{\partial }{\partial \sigma_{13}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{13} \\ \frac{\partial }{\partial \sigma_{21}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{21} & & \qquad \frac{\partial }{\partial \sigma_{22}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{22} & & \qquad \frac{\partial }{\partial \sigma_{23}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{23} \\ \frac{\partial }{\partial \sigma_{31}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{31} & & \qquad \frac{\partial }{\partial \sigma_{32}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{32} & & \qquad \frac{\partial }{\partial \sigma_{33}}(\sigma_{kl}\sigma_{kl}) & = 2~\sigma_{33} \end{align} $$ Therefore,

\frac{\partial }{\partial \sigma_{ij}}(\sigma_{kl}\sigma_{kl}) = 2~\sigma_{ij} \qquad \text{or} \qquad \frac{\partial }{\partial \boldsymbol{\sigma}}(\boldsymbol{\sigma}:\boldsymbol{\sigma}) = 2~\boldsymbol{\sigma} ~. $$ Similarly,

\frac{\partial }{\partial \boldsymbol{\sigma}}\left[\text{tr}(\boldsymbol{\sigma})\right] = \frac{\partial }{\partial \sigma_{ij}}(\sigma_{kk}) = \frac{\partial }{\partial \sigma_{ij}}(\sigma_{11} + \sigma_{22} + \sigma_{33}) $$ Hence, the components of the second-order tensor are

\begin{align} \frac{\partial }{\partial \sigma_{11}}(\sigma_{kk}) & = 1 & & \qquad \frac{\partial }{\partial \sigma_{12}}(\sigma_{kk}) & = 0 & & \qquad \frac{\partial }{\partial \sigma_{13}}(\sigma_{kk}) & = 0\\ \frac{\partial }{\partial \sigma_{21}}(\sigma_{kk}) & = 0 & & \qquad \frac{\partial }{\partial \sigma_{22}}(\sigma_{kk}) & = 1 & & \qquad \frac{\partial }{\partial \sigma_{23}}(\sigma_{kk}) & = 0\\ \frac{\partial }{\partial \sigma_{31}}(\sigma_{kk}) & = 0 & & \qquad \frac{\partial }{\partial \sigma_{32}}(\sigma_{kk}) & = 0 & & \qquad \frac{\partial }{\partial \sigma_{33}}(\sigma_{kk}) & = 1 \end{align} $$ Therefore,

\frac{\partial }{\partial \sigma_{ij}}(\sigma_{kk}) = \delta_{ij} \qquad \text{or} \qquad \frac{\partial }{\partial \boldsymbol{\sigma}}\left[\text{tr}(\boldsymbol{\sigma})\right] = \boldsymbol{\mathit{1}} ~. $$ Plugging the above results into the expression for the derivative of $$\mathbf{s}:\mathbf{s}$$ we get

\frac{\partial }{\partial \boldsymbol{\sigma}}(\mathbf{s}:\mathbf{s}) = 2~\boldsymbol{\sigma} - \cfrac{2}{3}~\text{tr}(\boldsymbol{\sigma})~\boldsymbol{\mathit {1}} = 2~\left(\boldsymbol{\sigma} - \cfrac{1}{3}~\text{tr}(\boldsymbol{\sigma})~\boldsymbol {\mathit{1}}\right) = 2~\mathbf{s} ~. $$ Hence, we get

\frac{\partial f}{\partial \boldsymbol{\sigma}} = f_{\boldsymbol{\sigma}} = \left(\sqrt{\cfrac{3}{2}}\right)\left(\frac{1}{2}\right) \left(\cfrac{1}{\lVert\mathbf{s}\rVert_{}}\right)~2~\mathbf{s} = \sqrt{\cfrac{3}{2}}~\cfrac{\mathbf{s}}{\lVert\mathbf{s}\rVert_{}} = \sqrt{\cfrac{3}{2}}~\mathbf{n} $$ The required expression is

\frac{\partial f}{\partial \boldsymbol{\sigma}} = f_{\boldsymbol{\sigma}} = \sqrt{\cfrac{3} {2}}~\mathbf{n} ~. $$


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Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 6

Problem 1: Part 6: Continuum elastic-plastic tangent modulus
The continuum elastic-plastic tangent modulus is defined by the following relation

\dot{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}}^{\text{ep}}:\dot{\boldsymbol

{\varepsilon}} ~. $$ Derive an expression for the elastic plastic tangent modulus using the results you have derived in the previous parts.

The stress rate is given by

\dot{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}} : \left(\dot{\boldsymbol{\varepsilon}} - \dot{\gamma}f_{\boldsymbol {\sigma}}\right)~. $$ From the previous part

\dot{\gamma} = \cfrac{f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol{\varepsilon}}}{ f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}} ~. $$ Plug in expression for stress rate to get

\begin{align} \dot{\boldsymbol{\sigma}} & = \boldsymbol{\mathsf{C}} : \left(\dot{\boldsymbol{\varepsilon}} - \cfrac{f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol{\varepsilon}}}{ f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}}~ f_{\boldsymbol{\sigma}}\right) \\ & =\boldsymbol{\mathsf{C}} : \dot{\boldsymbol{\varepsilon}} - \cfrac{f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol{\varepsilon}}}{ f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}}~ \boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} \end{align} $$ We have to express the above in the form

\dot{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}}^{\text{ep}}:\dot{\boldsymbol {\varepsilon}} \qquad\implies\qquad \dot{\sigma}_{ij} = C^{\text{ep}}_{ijkl}~\dot{\varepsilon}_{kl} ~. $$ Since the denominator is a scalar, we don't have to worry about it for this calculation. In that case we can write

\dot{\boldsymbol{\sigma}}=\boldsymbol{\mathsf{C}} : \dot{\boldsymbol{\varepsilon}} - \left(\cfrac{f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol {\varepsilon}}}{\text{denom.}}\right) \boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} $$ In index notation, we can write

\dot{\sigma}_{ij} = C_{ijkl}~\dot{\varepsilon}_{kl} - \cfrac{f^{\boldsymbol{\sigma}}_{pq}~C_{pqrs}~\dot{\varepsilon}_{rs}}{\text{denom.}}~ C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl} $$ Let us manipulate the numerator of the second term above so that we get what we need. Thus

\begin{align} f^{\boldsymbol{\sigma}}_{pq}~C_{pqrs}~\dot{\varepsilon}_{rs}~C_{ijkl}~f^{\boldsymbol {\sigma}}_{kl} & = \left(C_{pqrs}~f^{\boldsymbol{\sigma}}_{pq}\right)~ \left(C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl}\right)~\dot{\varepsilon}_{rs} \\ & = \left(C_{rspq}~f^{\boldsymbol{\sigma}}_{pq}\right)~ \left(C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl}\right)~\dot{\varepsilon}_{rs} \qquad \text{Major symmetry of}~ \boldsymbol{\mathsf{C}} \implies C_{pqrs} = C_{rspq} \\ & = \left(C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl}\right)~ \left(C_{rspq}~f^{\boldsymbol{\sigma}}_{pq}\right)~ ~\dot{\varepsilon}_{rs} \\ & \equiv A_{ij}~B_{rs}~\dot{\varepsilon}_{rs} \equiv M_{ijrs}~\dot{\varepsilon}_{rs} \\ & = \boldsymbol{\mathsf{M}}:\dot{\boldsymbol{\varepsilon}} ~. \end{align} $$ In the above

M_{ijrs} = A_{ij}~B_{rs} \qquad\implies\qquad \boldsymbol{\mathsf{M}} = \boldsymbol{A} \otimes\boldsymbol{B} $$ and

\begin{align} A_{ij} & = C_{ijkl}~f^{\boldsymbol{\sigma}}_{kl} \qquad\implies\qquad \boldsymbol{A} = \boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} \\ B_{rs} & = C_{rspq}~f^{\boldsymbol{\sigma}}_{pq} \qquad\implies\qquad \boldsymbol{B} = \boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} \end{align} $$ Therefore,

\boldsymbol{\mathsf{M}} = (\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes (\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma})} ~. $$ This gives us

\dot{\boldsymbol{\sigma}}=\boldsymbol{\mathsf{C}} : \dot{\boldsymbol{\varepsilon}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf {C}}:f_{\boldsymbol{\sigma})}} {\text{denom.}}\right): \dot{\boldsymbol{\varepsilon}} $$ or,

\dot{\boldsymbol{\sigma}} = \left[\boldsymbol{\mathsf{C}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf {C}}:f_{\boldsymbol{\sigma})}} {\text{denom.}}\right)\right]: \dot{\boldsymbol{\varepsilon}} ~. $$ Hence

\boldsymbol{\mathsf{C}}^{\text{ep}} = \boldsymbol{\mathsf{C}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf {C}}:f_{\boldsymbol{\sigma})}} {\text{denom.}}\right) ~. $$ The continuum elastic-plastic tangent modulus is therefore

{ \boldsymbol{\mathsf{C}}^{\text{ep}} = \boldsymbol{\mathsf{C}} - \left(\cfrac{(\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}})\otimes(\boldsymbol{\mathsf {C}}:f_{\boldsymbol{\sigma})}} {f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}}\right)~. } $$


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Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 5

Problem 1: Part 5: Consistency condition - 1
Use the consistency condition and the expressions you have derived in the previous parts to derive an expression for $\dot{\gamma}$ in terms of $\partial f/\partial\boldsymbol{\sigma}$, $\partial f/\partial\alpha$, $\partial f/\partial T$, $\boldsymbol{\mathsf{C}}$, and $\dot {\boldsymbol{\varepsilon}}$.|undefined

From the consistency condition

\dot{f} = f_{\boldsymbol{\sigma}}:\dot{\boldsymbol{\sigma}} + f_{\alpha}~\dot{\alpha} + f_T~\dot{T} = 0 $$ From the previous parts

\dot{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}} : \left(\dot{\boldsymbol{\varepsilon}} - \dot{\gamma}f_{\boldsymbol

{\sigma}}\right)~;\qquad \dot{\alpha} = \sqrt{\cfrac{2}{3}}~\dot{\gamma}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} ~;\qquad \dot{T} = \cfrac{\chi~\dot{\gamma}}{\rho~C_p}~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}~. $$ Plug into the consistency condition to get

f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\left(\dot{\boldsymbol{\varepsilon}} - \dot

{\gamma}f_{\boldsymbol{\sigma}}\right) + \sqrt{\cfrac{2}{3}}~\dot{\gamma}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_

{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}}+ \cfrac{\chi}{\rho~C_p}~\dot{\gamma}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}} = 0 ~. $$ Collect terms containing $$\dot{\gamma}$$:

\dot{\gamma}\left[ -f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} + \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}}+ \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}} \right] + f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol{\varepsilon}} = 0 $$ Therefore,

\dot{\gamma} = \cfrac{f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:\dot{\boldsymbol{\varepsilon}}}{ f_{\boldsymbol{\sigma}}:\boldsymbol{\mathsf{C}}:f_{\boldsymbol{\sigma}} - \sqrt{\cfrac{2}{3}}~f_{\alpha}~\cfrac{\boldsymbol{\varepsilon}^p:f_{\boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} - \cfrac{\chi}{\rho~C_p}~f_T~\boldsymbol{\sigma}:f_{\boldsymbol{\sigma}}} ~. $$


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Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 4

Problem 1: Part 4: Consistency condition
The consistency condition during plastic flow requires that

\dot{f}(\boldsymbol{\sigma}, \alpha, T) = 0 ~. $$ Write down an expression for the time derivative of $$f(\boldsymbol{\sigma},\alpha,T) $$ using the chain rule.

The time derivative is

\dot{f}(\boldsymbol{\sigma},\alpha,T) = \frac{\partial f}{\partial t} = \frac{\partial f}{\partial \boldsymbol{\sigma}}:\frac

{\partial \boldsymbol{\sigma}}{\partial t} + \frac{\partial f}{\partial \alpha}\frac{\partial \alpha}{\partial t} + \frac{\partial f}{\partial T}\frac{\partial T}{\partial t} ~. $$ In shorter form

\dot{f} = f_{\boldsymbol{\sigma}}:\dot{\boldsymbol{\sigma}} + f_{\alpha}~\dot{\alpha} + f_T~\dot{T} ~. $$


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Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 3

Problem 1: Part 3: Rate form constitutive relation
Write down the rate form of the elastic stress-strain law. Assume that deformations are small so that objectivity of the rates is not a concern.

The rate form of the elastic constitutive equation is

\dot{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}} : \dot{\boldsymbol{\varepsilon}}^e = \boldsymbol{\mathsf{C}} : (\dot{\boldsymbol{\varepsilon}} - \dot{\boldsymbol

{\varepsilon}}^p) $$ Plugging in the expression for $$\dot{\boldsymbol{\varepsilon}}^p$$ we get

\dot{\boldsymbol{\sigma}} = \boldsymbol{\mathsf{C}} : \left(\dot{\boldsymbol{\varepsilon}} - \dot{\gamma}\frac{\partial f}{\partial \boldsymbol

{\sigma}}\right) $$


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Nonlinear finite elements/Homework11/Solutions/Problem 1/Part 2

Problem 1: Part 2: Energy equation
For an adiabatic process, the rate of change of temperature can be written as

\dot{T} = \cfrac{\chi}{\rho C_p} \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}^p $$ where $$\chi$$ is the Taylor-Quinney coefficient, $$\rho$$ is the density, and $$C_p$$ is the specific heat. Express $$\dot{T}$$ in terms of $$\dot{\gamma}$$ and $$\partial f/\partial \boldsymbol{\sigma}$$. This is the evolution law for $$T$$.

Plugging in the expression for $$\dot{\boldsymbol{\varepsilon}}^p$$, we get

{ \dot{T} = \cfrac{\chi~\dot{\gamma}}{\rho~C_p} ~ \boldsymbol{\sigma}:\frac{\partial f}{\partial \boldsymbol{\sigma}}~. } $$


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Nonlinear finite elements/Homework 11/Solutions/Problem 1/Part 1

Problem 1: Part 1: Evolution rule for plastic flow
Let $$\alpha$$ be the equivalent plastic strain, defined as

\alpha := \sqrt{\cfrac{2}{3}} \lVert\boldsymbol{\varepsilon}^p\rVert_{} \qquad \text{where} \qquad \lVert\mathbf{a}\rVert = \sqrt{\mathbf{a}:\mathbf{a}} ~. $$ Express the time derivative of $$\alpha$$ in terms of $$\dot{\gamma}$$ and $$\partial f/\partial \boldsymbol{\sigma}$$. This is the evolution law for $$\alpha$$.

The time derivative of $$\alpha$$ is given by

\dot{\alpha} = \sqrt{\cfrac{2}{3}}~\frac{\partial }{\partial t}(\sqrt{\boldsymbol{\varepsilon}

^p:\boldsymbol{\varepsilon}^p}) = \sqrt{\cfrac{2}{3}}~\left(\frac{1}{2}\right)~ \left(\cfrac{1}{\sqrt{\boldsymbol{\varepsilon}^p:\boldsymbol{\varepsilon}^p}}\right) \frac{\partial }{\partial t}(\boldsymbol{\varepsilon}^p:\boldsymbol{\varepsilon}^p) = \sqrt{\cfrac{2}{3}}~\left(\frac{1}{2}\right)~ \left(\cfrac{1}{\lVert\boldsymbol{\varepsilon}^p\rVert_{}}\right) \frac{\partial }{\partial t}(\boldsymbol{\varepsilon}^p:\boldsymbol{\varepsilon}^p) $$ Now,

\begin{align} \frac{\partial }{\partial t}(\boldsymbol{\varepsilon}^p:\boldsymbol{\varepsilon}^p) & =

\frac{\partial }{\partial t}( \varepsilon_{11}^p\varepsilon_{11}^p + \varepsilon_{12}^p\varepsilon_{12}^p + \varepsilon_

{13}^p\varepsilon_{13}^p + \varepsilon_{21}^p\varepsilon_{21}^p + \varepsilon_{22}^p\varepsilon_{22}^p + \varepsilon_

{23}^p\varepsilon_{23}^p + \varepsilon_{31}^p\varepsilon_{31}^p + \varepsilon_{32}^p\varepsilon_{32}^p + \varepsilon_

{33}^p\varepsilon_{33}^p)\\ & = 2\varepsilon_{11}^p\frac{\partial \varepsilon_{11}^p}{\partial t} + 2\varepsilon_{12}^p\frac{\partial \varepsilon_{12}^p}{\partial t} + 2\varepsilon_{13}^p\frac{\partial \varepsilon_{13}^p}{\partial t} + 2\varepsilon_{21}^p\frac{\partial \varepsilon_{21}^p}{\partial t} + 2\varepsilon_{22}^p\frac{\partial \varepsilon_{22}^p}{\partial t} + 2\varepsilon_{23}^p\frac{\partial \varepsilon_{23}^p}{\partial t} + 2\varepsilon_{31}^p\frac{\partial \varepsilon_{31}^p}{\partial t} + 2\varepsilon_{32}^p\frac{\partial \varepsilon_{32}^p}{\partial t} + 2\varepsilon_{33}^p\frac{\partial \varepsilon_{33}^p}{\partial t} \\ & = 2\varepsilon_{11}^p\dot{\varepsilon}_{11}^p + 2\varepsilon_{12}^p\dot{\varepsilon}_

{12}^p + 2\varepsilon_{13}^p\dot{\varepsilon}_{13}^p + 2\varepsilon_{21}^p\dot{\varepsilon}_{21}^p + 2\varepsilon_{22}^p\dot{\varepsilon}_{22}^p +

2\varepsilon_{23}^p\dot{\varepsilon}_{23}^p + 2\varepsilon_{31}^p\dot{\varepsilon}_{31}^p + 2\varepsilon_{32}^p\dot{\varepsilon}_{32}^p +

2\varepsilon_{33}^p\dot{\varepsilon}_{33}^p \\ & = 2 \sum_{i=1}^{3}\sum_{j=1}^{3}\varepsilon_{ij}^p\dot{\varepsilon}_{ij}^p = 2

\boldsymbol{\varepsilon}^p:\dot{\boldsymbol{\varepsilon}}^p \end{align} $$ Therefore,

\dot{\alpha} = \sqrt{\cfrac{2}{3}}~\left(\frac{1}{2}\right)~ \left(\cfrac{1}{\lVert\boldsymbol{\varepsilon}^p\rVert_{}}\right) \left(2~\boldsymbol{\varepsilon}^p:\dot{\boldsymbol{\varepsilon}}^p\right) = \sqrt{\cfrac{2}{3}}~\cfrac{\boldsymbol{\varepsilon}^p:\dot{\boldsymbol{\varepsilon}}^p}

{\lVert\boldsymbol{\varepsilon}^p\rVert_{}} $$ Using

\dot{\boldsymbol{\varepsilon}}^p = \dot{\gamma}\frac{\partial f(\boldsymbol

{\sigma},\alpha,T)}{\partial \boldsymbol{\sigma}} $$ we get

\dot{\alpha} = \sqrt{\cfrac{2}{3}}~\dot{\gamma}~\cfrac{\boldsymbol{\varepsilon}^p:\frac{\partial f}

{\partial \boldsymbol{\sigma}}} {\lVert\boldsymbol{\varepsilon}^p\rVert_{}} \qquad\text{where}\qquad \boldsymbol{\varepsilon}^p = \int_0^t \dot{\boldsymbol{\varepsilon}}^p~dt = \int_0^t \dot{\gamma}\frac{\partial f}{\partial \boldsymbol{\sigma}}~dt ~. $$


text:
Nonlinear finite elements/Homework 11/Solutions

 * Problem 1: Small Strain Elastic-Plastic Behavior
 *  Given:
 * For small strains, the strain tensor is given by

\boldsymbol{\varepsilon} = \frac{1}{2}\left[\boldsymbol{\nabla} \mathbf{u} + (\boldsymbol{\nabla} \mathbf{u})^T\right] \qquad\text{or}\qquad \varepsilon_{ij} = \frac{1}{2}(u_{i,j} + u_{j,i}) ~. $$
 * In classical (small strain) rate-independent plasticity we start off with an additive decomposition :of the strain tensor

\boldsymbol{\varepsilon} = \boldsymbol{\varepsilon}^e + \boldsymbol{\varepsilon}^p \qquad\text{or}\qquad \varepsilon_{ij} = \varepsilon_{ij}^e + \varepsilon_{ij}^p ~. $$
 * Assuming linear elasticity, we have the following elastic stress-strain law

\boldsymbol{\sigma} = \boldsymbol{\mathsf{C}} : \boldsymbol{\varepsilon}^e \qquad\text{or}\qquad \sigma_{ij} = C_{ijkl}\varepsilon_{kl}^e ~. $$
 * Let us assume that the $$J_2$$ theory applies during plastic deformation of the material. :Hence, the material obeys an associated flow rule
 * $$\text{(1)} \qquad

\dot{\boldsymbol{\varepsilon}}^p = \dot{\gamma}\frac{\partial f(\boldsymbol{\sigma},\alpha,T)}{\partial \boldsymbol{\sigma}} $$
 * where $$\dot{\gamma}$$ is the plastic flow rate, $$f$$ is the yield function, $$T$$ is the temperature, and $$\alpha$$ is an internal variable.

 * Problem 1: Part 1: Evolution rule for plastic flow
 * Problem 1: Part 2: Energy equation
 * Problem 1: Part 3: Rate form constitutive relation
 * Problem 1: Part 4: Consistency condition
 * Problem 1: Part 5: Consistency condition - 1
 * Problem 1: Part 6: Continuum elastic-plastic tangent modulus
 * Problem 1: Part 7: Flow rule
 * Problem 1: Part 8: Consistency condition - 2
 * Problem 1: Part 9: Elastic-plastic tangent modulus
 * Problem 1: Part 10: Discrete evolution equations
 * Problem 1: Part 11: Discrete Kuhn-Tucker
 * Problem 1: Part 12: Trial elastic stress - 1
 * Problem 1: Part 13: Trial elastic stress - 2
 * Problem 1: Part 14: Return mapping
 * Problem 1: Part 15: Plastic flow parameter
 * Problem 1: Part 16: Newton iterations
 * Problem 2: Billet Upset Forging
 * Problem 3: Taylor Impact Tests


text:
Talk:Technical writing/Style

Second person
Second person is basic to modern professional technical writing--whoever keeps championing third person is wrong. Third person has an occasional role, but most projects require that you write as if speaking directly to the reader: not "The mechanic uses a socket wrench to remove the part," (3rd person) but, "Use a socket wrench to remove the part" (2nd person).

 * This statement is incorrect. When writing instructions the reader should never be directly addressed. Second person is fine as long as the reader is not directly addressed. Doing so looks unprofessional. For example: "Use a socket wrench to remove the part" is fine, but "You will use the socket wrench to remove your part" is unprofessional.

Third person
Third person is less effective in technical writing than second person.
 *

Examples: The user clicks the Add button. (3rd person) Click the Add button. (2nd person)
 * WRONG:
 * You will click the add button (2nd person, directly addressing the reader)

As a rule, stay away from future tense and passive voice.

The use of negatives
Casual documentation and marketing may allow or approve the use of contractions. Technical Writing, which is a form of scientific writing, does not. The words doesn't, wouldn't, couldn't, don't, an't, etc., are unacceptable.

Second person, contractions
The technical communications program at the University of California Berkeley, for one, advocates second person and teaches that contractions are usually fine. I've never talked to a fellow STC member who had a problem with either. "Wouldn't" and "couldn't" don't come up much because technical writing strives for present tense, but "don't" is fine in most document types.

Technical writing isn't a kind of scientific writing—scientific writing is a kind of technical writing. That's not a trivial distinction. There are many types of technical writing, of which scientific writing is only one.

The Microsoft Manual of Style is a poor authority to cite. It's a corporate style guide that—if you must read a lot of MS docs, as I do—even Microsoft writers don't always follow. The Chicago Manual of Style is far better researched and organized.

Comment
Chicago Manual of Style is what the Microsoft style is based on. IBM style is also similar to Microsoft style. We commonly write technical documents in third person. Scripts (training) for speakers may be written in second person.

-
There are many cases where Technical Writers are required to document application functionality of programs that perform scientific functions. Additionally, many Technical Writers perform daily work tasks in a technical environment. For example, technical support, development, and engineering.
 * The class technical communications course you speak of at Berkely suggests several reference manuals for successful college writing. My suggestion is that you read and use the exercise booklet.
 * I am not sure what you learned in college, but your education is shy of the views of most educators. (casual, informal, speak)
 * Sometimes we do use second person to address the audience directly. However, if we did this on a daily basis, throughout our documents, we would not be considered as having a good command of English.
 * Though the media may utilize sentence structure such as: He states, "My sister and me...;" a learned writer, writes: He states, "My sister and I..."
 * Writing sentences with contractions such as: I'll does not make for a powerful speaker, either. Frankly, I would not vote for a politician who presented his speech in this manner.

Example (Microsoft Style)

 * The WpdHelloWorld sample driver supports four objects: a device object, a storage object, a folder object, and a file object. Each object supports corresponding properties. These properties are defined in the file WpdObjectProperties.h.

yield return portInt.Receive;
 * The Port and PortSet queueing primitives, see CCR Ports and PortSets. The Port class is a first in first out (FIFO) queue of items - an item can be any valid Common Language Runtime (CLR) type (only items of that type, either CLR base types or user-defined types, can be posted to the port). In most cases there is also a queue of receivers (user code "guarded" by an Arbiter, which means that the Arbiter controls the execution of tasks by filtering the incoming messages as appropriate).
 * The return value indicates this is a C# iterator over CCR ITask instances and informs the compiler that the method body might contain yield statements.

 * Traditional failure handling schemes follow these patterns:
 * For synchronous invocations of methods, the caller checks one or more return values from the method. The method uses the callers execution context, most often this is a thread, to run.
 * Using structured exception handling, the caller wraps synchronous method invocations in try/catch/finally statements and either relies exclusively on the catch{} block for error handling. It may also use a combination of the catch{} block plus explicit checks on the results from the method.
 * Transactions which rely on a variety of compensation implementations by the components being called, OS infrastructure and usually expensive mechanisms to flow context across threads.

The Science of Computing

User based documentation may not be very scientific, but writing for an engineering (mechanical, development, etc.) community requires some level of knowledge beyond basic English. Writing a course on Technical Writing should include commonalities based on a technical environment.

 * Technical writing is related to a science. The science of the computer, language and its functions.
 * Technical writers, write in many mediums and the audience, technical aspects, and style vary.
 * Technical writers are expected to have a good grasp of English. Editors / reviewers only handle minor punctuation and grammer issues.
 * Terminology is as important to technical writers as is the scientific study of the application design.

Third Person Usage

"UC Berkeley's (Berkeley has) eight museums and its many collections support research and teaching in disciplines including anthropology, architecture, art history, art practice, botany, engineering, entomology, film history, geology, history, linguistics, paleontology, physics, and zoology."

Example 1) A new paper reviewing the impact of the loss of large predators and herbivores high in the food chain confirms that their decline has had cascading effects in marine, terrestrial and freshwater ecosystems throughout the world. The study, co-authored by UC Berkeley researchers, highlights the impact “apex consumers” have on the dynamics of fire, disease, vegetation growth, and soil and water quality.
 * Sarah Yang and Robert Sanders, Media Relations | Berkely Using Contractions in Third Person

Example 2) On Sunday, July 17, the moon will acquire its second new companion in less than a month. That’s (that is) when the second of two probes built by the University of California, Berkeley, and part of NASA’s (NASA has - possessive) five-satellite THEMIS mission will drop into a permanent lunar orbit after a meandering, two-year journey from its original orbit around Earth. Example 3) The new NRC report, "A Data-Based Assessment of Research-Doctorate Programs in the United States," synthesized data about more than 5,000 programs in 62 fields at 212 universities with research-based programs. Originally anticipated in 2007, the rankings are based largely on data from the 2005-06 academic year and analyze doctoral programs in the physical sciences and mathematics, agricultural and life sciences, health sciences, engineering, social sciences, and arts and humanities. The voluminous report is online.
 * Robert Sanders, Media Relations | An example of the technical in writing

Example 4) The R (or regression-based) rankings are based on an indirect way of determining the importance faculty attach to various characteristics. First, groups of randomly selected faculty were asked to rate the quality of a sample of representative programs in their field. Based on the sample program ratings, weights were assigned to each of the 20 characteristics using statistical techniques; again, these weights varied by field. These weights were applied to the data about each program, resulting in a second range of rankings. Each approach yielded a different. Example 5) This, the sixteenth edition of The Chicago Manual of Style, marks the first edition to be prepared and published simultaneously in print and online. As opportunities for publishing have grown dramatically in an era of electronic publication and distribution, the guiding principles for this edition have been twofold: to recognize the continuing evolution in the way authors, editors, and publishers do their work, on the one hand, and to maintain a focus on those aspects of the process that are independent of the medium of publication, on the other.
 * Preface Page of Chicago Manual of Style| Third Person

 * Casual or creative writing (informal) may allow some (positive) contractions.
 * The contractions: wouldn't, can't, musn't, won't, don't, etc., are unacceptable in technical documentation (formal).

Additional Information on Third Person

 * http://www.mrsec.wisc.edu/Edetc/research/slides/Active-passive_handoutA.pdf
 * http://www.ee.uconn.edu/SeniorDesign/designone/Lecture%2007%20Technical%20Writing.ppt

Wrong
I've know some decent technical writers who successfully learned on the job, but many writers wander into the field from engineering, journalism, or other writing or technical disciplines without learning (preferably from a reputable certificate or graduate program) the "technical writing style." You can spot them by things like their fondness for passive voice, third person, and words like usage and utilize.

To be clear, third person has a place, and the Technical Writing Style article should reflect that. Sometimes a document, or an aspect of a document, describes what others did and what those actions made happen. This is common, of course, in science writing, but much less common in the many other forms of technical writing. Marketing text, like the Berkeley museum text you cite, frequently uses third person, but that's not technical writing.

Most technical writing tends to be procedural, and--keeping with the tech comm axiom of "talk to the user," should be in second person. It delivers information to someone who must take action based on that information. You wouldn't give driving directions to someone by saying, "The car goes down this road and turns left at a gas station." You'd say, "Go down this road and turn left at the gas station."

Second person can be overdone if the writer chants "you" multiple times a paragraph--which I've seen in tech writing competition entries.

Comment
If the goal for the technical writing class is to teach an informal technical writing method, then specify it. Otherwise, the global audience you seek will not honor your work. I have never been in a business environment where informal writing, in second person, is acceptable other than in blogs, marketing / advertising, comments, and email(sparingly). Please enlighten us on your experience!

 * The many writers you speak of are more educated (experienced) than the technical writer who just received a certification in technical writing. Their backgrounds are specific to that of engineering, medical science, journalism, etc., this means they are highly technical.  Insisting that second person is the appropriate grammatical choice for every technical environment or preferred, is still incorrect.  In many cases, terminology also dictates a grammatical person.

 * Writing active voice, third person is much more complex than writing in second person for any writer. Additionally, whether you are writing technical documentation or a research paper, writing consistently in active voice it is not always possible.

 * Second person is commonly written using an informal method. We both agree that second person is overdone. This grammactical person has been seen throughout the Wikiversity.

 * I would never write, "You'd say," in a technical document. This is a lazy and bad writing habit.  I am not sure why you feel that this is a form of clear and concise writing.  The contraction itself has two meanings.

 * Here is a question for you, "As technical writers, why do we define acronyms in our documents?" Answer:  Acronyms have more than one meaning.

Recents Edits

 * The statement that "technical writing leans towards second person," is incorrect.
 * Some of the edits were reordered under Application Consistensy for clarity.
 * Some of the content was removed under the topic, Understand your environment.
 * Understand your environment has nothing to do with the use of the pronoun "you."
 * Some of the recently added paragraphs on Chicago Manual of Style under the topic Specifying Gender have been removed. Primarily because the information is unrelated to the topic.  We believe an anonymous editor has done this to prove a point rather than create value in the page.
 * We do not write business communication in second person, informal. Emails, comments, requirements documents, technical specifications, etc., are written in third person.
 * Second person (conversational, informal) is a common writing style in marketing communication. Though studies may indicate that technical documentation written for a global audience benefits from this technique (shown through the publication and popularity of the Dummies books), the use is not accepted inside the business environment.

Break down?
This page is too long, I am going to break it down into a more manageable form. AlistairReece 06:33, 17 August 2007 (UTC)

 * Most of this is inaccurate. I'd like to see the credentials of the original author.

title: Talk:Technical writing/Style


text:
User:Seeder

Seeder

title: User:Seeder


text:
User talk:Seeder

seeder

title: User talk:Seeder


text: Template:LeftTOC

title: Template:LeftTOC


text:
User:Open Research/Learning Resources

Introduction
I have create this page to try to organise the learning resources offered through Wikiversity, and the related face-to-face workshop events that could be held at Sussex. Make your suggestions below.

Virtual Learning Resources
An Introduction to Science Journalism

Sources for Science Journalists

KnowCycle

Face-to-face Learning Events
Place suggestions for real-life workshops that could be held in Sussex here.

Add suggested title of workshop and if you know of someone who could offer it, that would be good too.

Do you have knowledge to pass on to other postgrads? Might you offer a workshop yourself?

Suggested workshop titles and descriptions here:

 * "Writing Social Science: Developing Your Thesis Writing Style" suggested by Jenny Gristock.
 * "Put your suggested workshop title here" [by clicking on 'edit' above]

.

Knowledge Transfer Fellowships
Fancy working with a firm or voluntary organisation? Let me know here.

title: User:Open Research/Learning Resources


text:
Filmmaking Basics/Matte Painting/SBTDS Intro

Overview

{| cellpadding="20" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"

 * style="width: 70%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

"Seduced by the Dark Side!"

 * The opening or closing shot
 * In two years, we are going to film the motion picture Seduced by the Dark Side!. But we will have very little money to film the movie.

 * Without a lot of money, we cannot film the scene in front of a magnificent theater. The only alternative that we have is to film the movie in front of a blank wall and use a matte painting for the actual theater.

 * To plan for the shooting of our movie, I need you to create a matte painting.

 * Normally, the matte painting is made after the scene has been filmed. But for now, please use the storyboard frame and create a matte painting for the storyboard frame.

 * }

{| cellpadding="20" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"

 * style="width: 40%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Matte painting canvas

 * For Photoshop, GIMP or ArtRage 2 FREE
 * To create the matte painting, we will use Photoshop, GIMP or ArtRage 2 FREE.

 * To create the matte painting, we must take one frame from the movie and place it on a huge canvas. I have done this for you.  You can simply download the starter for this website and  put it inside your drawing program.

 * Download the picture from:

 * http://upload.wikimedia.org/wikiversity/en/2/28/Matte_starter_1664_1328_SBTDS.png

 * Do it yourself
 * However, you will find it easier to create your own matte painting canvas. (See the next page.)

 * }

{| cellpadding="20" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 100%; background-color: LemonChiffon; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |

The next page

 * Create a matte painting for your version of "Seduced by the Dark Side!" and send it to me.

 * If you need help, on the next page, you can see how to create starter pages for matte paintings.

 * Building your own starter pages for Seduced by the Dark Side! [[Image:Crystal Clear action forward.png|24px]]

 * or

 * Otherwise go to the next assignment in matte painting which is at Lesson:Matte Painting:SpaceStation:Intro

 * }


text: File:Matte starter 1664 1328 SBTDS.png

title: File:Matte starter 1664 1328 SBTDS.png


text:
Filmmaking Basics/Matte Painting/Build Starters

{| cellpadding="10" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 50%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

An original movie frame

 * Still frame from a locked camera
 * Initially, we need a matte painting for the storyboards. Therefore, start with a frame from your storyboards.

 * Later, when you want to use a matte painting for a scene, you start by grabbing a test frame. The movie camera is locked down therefore the position of the physical movie set will never change in the frame by even a single pixel.

 * From the test frame, it is very easy to decide what part of the physical movie set you want the audience to see and what part you want to hide behind the matte painting.

 * Therefore, it will be very easy to build a mask which protects the good part of the physical movie set so your matte painting will never cover this up.

 * Creating the mask is the critical part of preparing a starter page for Tux Paint.

A frame from your storyboards     or    A still frame from your storyboards.
 * style="width: 40%; background-color: cornsilk; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 200px;" |

 * }

{| cellpadding="10" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 40%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

A picture on a piece of paper

 * Setting up a matte painting canvas
 * The concept is very simple. You take one frame from the movie or storyboard and put it on a large canvas.  That's it.

 * The size of the canvas
 * The size of the canvas can be anything you want. Since the movie will be in the wide screen format (16 by 9), probably the best thing is a canvas that has a 16 by 9 aspect ratio.

 * Zoom factor
 * The size should be determined by how much of a zoom effect you want. If you want a zoom out effect of 2 to 1, then you should create the matte painting which is 2 times 480 pixels tall by 854 pixels wide. (Always work in square pixels.)

 * Dolly factor
 * If you also want a slight dolly effect (dolly left or dolly right), you will also need to extend the widths of your matte painting as well.

Import

 * Select a paint or drawing program
 * Matte paintings can be done in Photoshop, GIMP, Painter.

 * Import and begin
 * Simply import your picture and begin to paint (begin careful not to change any of the original picture where the actors are.)

Place the picture     The movie frame is on a blank page.
 * style="width: 60%; background-color: cornsilk; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 200px;" |

 * }

More details

{| cellpadding="10" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 50%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Use the frame from your storyboards

 * Start with your storyboards
 * For this assignment, start with a frame from your storyboards.

 * Later, when we do 3D animation, you will start with a frame from your animation.

 * When we do the live action version, we will use a frame from the actual movie.

 * FrameForge 3D Studio
 * If you are using FrameForge 3D Studio, here is what you do:

→ →

 * The three steps
 * To export a frame, you need to select EXPORT and then click on the picture of the frame you want. Then specify the height which is 480 pixels.

 * }

{| cellpadding="10" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 50%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Create the matte painting canvas

 * Using GIMP
 * With the paint program GIMP, you can do this:

→ →  →

 * The three steps
 * Decide on the frame size, open the frame in the same program, drag the frame to your new canvas, save the picture as a png (which might require an EXPORT.)

 * }

{| cellpadding="10" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 40%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="1" |

Draw your matte painting

 * Protect the frame
 * Draw your matte painting and send me a copy.

Send your picture to me     A matte painting.
 * style="width: 60%; background-color: cornsilk; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 200px;" |

 * }

{| cellpadding="20" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 100%; background-color: LemonChiffon; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |

Send me your finished matte painting

 * When you are finished with your matte painting, please email a copy to me.

Contact Your Instructor

 * Your instructor for this class is Robert Elliott. To email me, you simply  Click Here. [[Image:Crystal Clear app xfmail.png|24px]]

 * }


text:
Filmmaking Basics/Matte Painting/Compositing

{| cellpadding="10" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 50%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

More instructions later

 * Under construction
 * This section will be expanded shortly.

Things to remember

 * Lock the camera
 * The concept of matte painting is based on a locked camera.

 * No matte painting where the actors roam.
 * Nothing must cover up where the actors are.

 * Additional layers are needed
 * Anything which is in front of the actors must be on a separate layer.

 * Coming together
 * All this comes together in Adobe AfterEffects.

A sample frame     A still frame from the movie.
 * style="width: 40%; background-color: cornsilk; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 200px;" |

 * }

{| cellpadding="10" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 50%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Other layers

 * More than just a matte painting
 * You can also add many other layers to a composite of a matte painting.

Place the picture     The frame is on the blank page.
 * style="width: 40%; background-color: cornsilk; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 200px;" |

 * }

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Voice leading

A chord progression is a group of vertically-placed, simultaneously-played melodies. A chord's vertical placement results from the smooth direction of each of the lines. Traditional notation often hinders the perception of the linear aspects of a chord progression. Play the example above several times and try to correlate both the horizontal and vertical aspects of the progression with your ears and eyes.

After seeing and hearing the example as it occurs in real time, you will see its traditional notation in Example A below. Although the linear aspects of the progression become less apparent, the concept of watching and listening for the direction of individual voices in a chord progression is important in all musicmaking!

EXAMPLE TRADITIONAL NOTATION

STYLE

Chord voicing is a reliable indicator of style. Left-hand piano voicings, for instance, can provide a reliable harmonic blueprint for the entire jazz tradition. In Example B below, the C7 chord is portrayed in various jazz styles and examples of variations of each style can be heard by pushing the "play" buttons.

EXAMPLE. HISTORICAL LEFT HAND PIANO VOICINGS

1. Ragtime and stride piano left-hand voicings often are defined, as notated above, with octaves on the root on beat one, on the fifth on beat 3, with the same voicing of the chord on beats 2 and 4. In reality, however, the early stride pianists also played left-hand octave melodies, chord melodies (one of the notes of the chord, usually the highest, supplying a countermelody with a line in the right hand, meter changes (usually from 2/4 to 3/4), and complicated rhythm and accents--all which require advanced control and techniques. 2. BoogieWoogie patterns are usually played over the 12-bar blues. In contrast to stride playing, the boogie-woogie pianist maintains the same left-hand shuffle rhythm pattern (played on the I, IV and V chords of the blues progression) while showing the independence between the two hands in managing intricate interlocking accents and rhythms. The upper voice of the left-hand pattern is a good example of step-wise voice leading providing the harmonic momentum within the same chord--as the 5th (G) moves through the 6th (A) up to the 7th(Bb) and back down again of the C7 chord.

3. During the bebop period, beginning in the 1940s, the right hand provided harmonic sophistication by running extensions of the chord members as part of the melodic line. The left hand provided punctuation, as much as a harmonic backdrop. (The harmonies also changed much more frequently during this period.) The roots of the left-hand chords began to be left out on occasion, but the resolution of 3rd and 7th of the dominant 7th chord usually were handled carefully in any voicing toward the tonic. (For example, if an F Major chord were to appear in the next measure, the E and Bb of the C7 chord above, would most likely resolve to an F and A.)

4. Voicings gradually became more harmonically ambiguous. After adding extensions and alterations, and by leaving out the chords' roots, pianists soon found that a certain "sound" could be obtained by stacking specific interval combinations. (In fact, for some players the term, "voicing", implies only a specific relationship of vertically-stacked intervals rather than the "voice leading" of individual chord members with both horizontal and vertical implications.) A dominant 7th chord, for example, can be voiced by stacking a tritone plus a perfect fourth above either the 3rd or the 7th of the harmony. Thus, the first chord in the example is built on the third of the C7 chord," E", and the next chord is built on the 7th, "Bb". The altered extension in the first chord, "Eb", (enharmonically spelled as "D#" in a C7 chord extension) sounds like a sharped 9th and the "A" in the 2nd chord is the 13th of the C7 chord.

5. Quartal harmonies, stacked fourths, often occur in modal compositions. Thus, a C7 chord is entirely appropriate for a composition written in C mixolydian. However, it would be played as a sus. 4, thus leaving out the 3rd and omitting the specific tension of the tritone (E to Bb) for the desired ambiguity of quartal harmony. The example above uses pitches from an extended C7 chord, but in no way emphasizes either the root of the dominant 7th quality. With a bass player playing pitches based around C, the notes in the example above can be heard as being derived from C but (depending on the bass player) could just as easily be interpreted as belonging to A or D dorian. Voice leading in this context relies on the frequent stepwise motion of the stacked chords.

title: Voice leading


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User:Ksofttechnology

Computer Learning Center Convert Technical Data into Simple format. Create Interest to learn Technical aspect into soft way. Make Master

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Nonlinear finite elements/Homework 2/Solutions

Problem 1: Classification of PDEs
 Given:

 * $$\begin{matrix}

\text{(1)} \qquad u_{,xx}&=&\alpha u_{,t} \\ \text{(2)} \qquad u_{,xxxx}&=&\alpha u_{,tt} \end{matrix}$$

Part 1
What physical processes do the two PDEs model?

Equation (1) models the one dimensional heat problem where $$u(x,t)$$ represents temperature at $$x$$ at time $$t$$ and $$\alpha$$ represents the thermal diffusivity.

Equation (2) models the transverse vibrations of a beam. The variable $$u(x,t)$$ represents the displacement of the point on the beam corresponding to position $$x$$ at time $$t$$. $$\alpha$$ is $$-\frac{\rho A}{EI}$$ where $$E$$ is Young's modulus, $$I$$ is area moment of inertia, $$\rho$$ is mass density, and $$A$$ is area of cross section.

Part 2
Write out the PDEs in elementary partial differential notation

{ \begin{align} u_{,xx} &= \alpha u_{,t} \qquad \equiv \qquad \frac{\partial ^2u}{\partial x^2}=\alpha \frac{\partial u}{\partial t} \\ u_{,xxxx} &= \alpha u_{,tt}\qquad \equiv \qquad \frac{\partial ^4u}{\partial x^4}=\alpha \frac{\partial ^2u}{\partial t^2} \end{align} } $$

Part 3
Determine whether the PDEs are elliptic, hyperbolic, or parabolic. Show how you arrived at your classification.

Equation (1) is  parabolic for all values of $$\alpha$$.

We can see why by writing it out in the form given in the notes on Partial differential equations.

(1) \frac{\partial^2 u}{\partial x^2} + (0) \frac{\partial^2 u}{\partial x\partial t} + (0)\frac{\partial^2 u}{\partial t^2}+ (0) \frac{\partial u}{\partial x} + (-\alpha)\frac{\partial u}{\partial t} +(0) u = 0 $$ Hence, we have $$a = 1$$, $$b = 0$$, $$c = 0$$, $$d = 0$$, $$e = -\alpha$$, $$f = 0$$, and $$g = 0$$. Therefore,

{ b^2-4ac = 0-(1)(0)=0 \Rightarrow \text{parabolic} } $$

Equation (2) is  hyperbolic for positive values of $$\alpha$$,  parabolic when $$\alpha$$ is zero, and  elliptic when $$\alpha$$ is less than zero.

Problem 2: Models of Physical Problems
 Given:

\frac{d}{dx}\left(a(x)\frac{du(x)}{dx}\right)+c(x)u(x)=f(x)\quad \mbox{for }\quad x\in (O,L) $$

Solution

 * Heat transfer: :$$kA\frac{d^2T}{dx^2}+ap\beta T=f$$
 * Flow through porous medium: :$$\mu\frac{d^2\phi}{dx^2}=f$$
 * Flow through pipe: :$$\frac{1}{R}\frac{d^2P}{dx^2}=0$$
 * Flow of viscous fluids: :$$\mu\frac{d^2v_z}{dx^2}=-\frac{dP}{dx}$$
 * Elastic cables: :$$T\frac{d^2u}{dx^2}=f$$
 * Elastic bars: :$$EA\frac{d^2u}{dx^2}=f$$
 * Torsion of bars: :$$GJ\frac{d^2\theta}{dx^2}=0$$
 * Electrostatics: :$$\varepsilon\frac{d^2\phi}{dx^2}=\rho$$

Problem 4: Least Squares Method
 Given:

The residual over an element is

R^e(x,c^e_1,c^e_2,\dots,c^e_n) := \cfrac{d}{dx}\left(a~\cfrac{du^e_h}{dx}\right) + c~u^e_h - f(x) $$ where

u^e_h(x) = \sum_{j=1}^n c^e_j \varphi^e_j(x) ~. $$ In the least squares approach, the residual is minimized in the following sense

\frac{\partial }{\partial c_i}\int_{x_a}^{x_b} [R^e(x,c^e_1,c^e_2,\dots,c^e_n)]^2~dx = 0, \qquad i=1,2,\dots,n ~. $$

Part 1
What is the weighting function used in the least squares approach?

We have,

\frac{\partial }{\partial c_i}\int_{x_a}^{x_b} [R^e]^2~dx = \int_{x_a}^{x_b} 2R^e\frac{\partial R^e}{\partial c_i}~dx = 0 ~. $$ Hence the weighting functions are of the form

{ w_i(x)=\frac{\partial R^e}{\partial c_i} } $$

Part 2
Develop a least-squares finite element model for the problem.

To make the typing of the following easier, I have gotten rid of the subscript $$h$$ and the superscript $$e$$. Please note that these are implicit in what follows.

We start with the approximate solution

u(x) = \sum_{j=1}^n c_j \varphi_j(x) ~. $$ The residual is

R = \cfrac{d}{dx}\left[a~\left(\sum_j c_j\cfrac{d\varphi_j}{dx}\right) \right] + c~\left(\sum_j c_j\varphi_j\right) - f(x) ~. $$ The derivative of the residual is

\frac{\partial R}{\partial c_i} = \cfrac{d}{dx}\left[a~\left(\sum_j \frac{\partial c_j}{\partial c_i} \cfrac{d\varphi_j}{dx}\right) \right] + c~\left(\sum_j \frac{\partial c_j}{\partial c_i}\varphi_j\right) - f(x) ~. $$ For simplicity, let us assume that $$a$$ is constant, and $$c = 0$$. Then, we have

R = a~\left(\sum_j c_j\cfrac{d^2\varphi_j}{dx^2}\right) - f~, $$ and

\frac{\partial R}{\partial c_i} = a~\left(\sum_j \frac{\partial c_j}{\partial c_i} \cfrac{d^2\varphi_j}{dx^2}\right) ~. $$ Now, the coefficients $$c_i$$ are independent. Hence, their derivatives are

\frac{\partial c_j}{\partial c_i} = \begin{cases} 0 & \text{for} i \ne j \\ 1 & \text{for} i = j  \end{cases} $$ Hence, the weighting functions are

\frac{\partial R}{\partial c_i} = a~\cfrac{d^2\varphi_i}{dx^2} ~. $$ The product of these two quantities is

R\frac{\partial R}{\partial c_i} = \left(a\sum_j c_j\cfrac{d^2\varphi_j}{dx^2} - f\right) \left(a\cfrac{d^2\varphi_i}{dx^2}\right) = \sum_j \left[ \left(a^2\cfrac{d^2\varphi_i}{dx^2}\cfrac{d^2\varphi_j}{dx^2}\right) c_j \right] - af\cfrac{d^2\varphi_i}{dx^2}~. $$ Therefore,

\int_{x_a}^{x_b} R\frac{\partial R}{\partial c_i}~dx = 0 \implies \sum_j \left[ \left(\int_{x_a}^{x_b} a^2\cfrac{d^2\varphi_i}{dx^2}\cfrac{d^2\varphi_j}{dx^2}~dx\right) c_j \right] - \int_{x_a}^{x_b} af\cfrac{d^2\varphi_i}{dx^2}~dx = 0 $$ These equations can be written as

K_{ij} c_j = f_i $$ where

{ K_{ij} = \int_{x_a}^{x_b} a^2\cfrac{d^2\varphi_i}{dx^2}\cfrac{d^2\varphi_j}{dx^2}~dx } $$ and

{ f_i = \int_{x_a}^{x_b} af\cfrac{d^2\varphi_i}{dx^2}~dx~. } $$

Part 3

 * 1) Discuss some functions $(\varphi_j^e)$ that can be used to approximate $u$

For the least squares method to be a true "finite element" type of method we have to use finite element shape functions. For instance, if each element has two nodes we could choose

\varphi_1 = \cfrac{x_b - x}{h} \qquad \text{and} \qquad \varphi_2 = \cfrac{x - x_a}{h}~. $$ Another possiblity is set set of polynomials such as

\varphi = \{1, x, x^2, x^3\}~. $$

Problem 5: Heat Transefer
 Given:

 * $$-\frac{d}{dx}\left(k\frac{dT}{dx}\right)=q\quad\mbox{ for }\quad

x \in (0,L)$$ where $$T$$ is the temperature, $$k$$ is the thermal conductivity, and $$q$$ is the heat generation. The Boundary conditions are
 * $$\begin{matrix}

T(0)&=&T_0\\ k\frac{dT}{dx}+\beta(T-T_\infty)+\hat{q}\bigg |_{x=L}&=&0 \end{matrix}$$

Part 1
Formulate the finite element model for this problem over one element

First step is to write the weighted-residual statement

0=\int_{x_a}^{x_b} w_i^e\left[-\frac{d}{dx}\left( k\frac{dT_h^e}{dx}\right)-q\right]dx $$ Second step is to trade differentiation from $$T_h^e$$ to $$w_i^e$$, using integration by parts. We obtain

0=\int_{x_a}^{x_b} \left(k\frac{w_i^e}{dx}\frac{dT_h^e}{dx}- w_i^eq\right)dx-\left[w_i^ek\frac{dT_h^e}{dx}\right]_{x_a}^{x_b} \text{(3)} \qquad $$ Rewriting (3) in using the primary variable $$T$$ and secondary variable $$k\frac{dT}{dx}\equiv Q$$ as described in the text book to obtain the weak form

0=\int_{x_a}^{x_b} \left(k\frac{w_i^e}{dx}\frac{dT_h^e}{dx}- w_i^eq\right)dx-w_i^e(x_a)Q_a^e-w_i^e(x_b)Q_b^e \text{(4)} \qquad $$ From (4) we have the following
 * $$\begin{matrix}

K_{ij}^e&=&\int_{x_a}^{x_b} k\frac{N_i^e}{dx}\frac{dN_j^e}{dx}dx\\ f_i^e&=&\int_{x_a}^{x_b} qN_i^edx\\ K_{ij}^eT_j^e&=&f_i^e+N_i^e(x_a)Q_a^e+N_i^e(x_b)Q_b^e \end{matrix}$$

Part 2
Use ANSYS to solve the problem of heat conduction in an insulated rod. Compare the solution with the exact solution. Try at least three refinements of the mesh to confirm that your solution converges.

The following inputs are used to solve for the solution of this problem:

ANSYS gives the following results

Part 3
Write you own code to solve the problem in part 2.

The following Maple code can be used to solve the problem.


text:
User:Sergeidave

Hi,

My name is David. I'm married and have 3 children. I'm a compulsive learner on all kinds of knowledge, but the ones that interest me the most are Science, Technology, Music and Computer Science. I'm a musician, composer and pianist mainly. I'm kind of new on the Wiki world but I totally love the Wikimedia idea and all of its branches! ---Sergeidave 18:09, 17 August 2007 (UTC)

title: User:Sergeidave


text:
User talk:Sergeidave

Welcome
Hello Sergeidave, 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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 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 18:40, 17 August 2007 (UTC) on the user page. -->

Re: Wikiversity the Movie (15th August 2009)
Hello, there is a new time frame set for the Wikiversity the Movie: it is 15th August 2009 - our next Wikiversity Day. Are you still interested in helping with the music/soundtrack ? If so, please leave a note at: Talk:Wikiversity the Movie/Music. See you, Erkan Yilmaz uses the Chat (try) 14:31, 23 August 2008 (UTC)

title: User talk:Sergeidave


text:
Nonlinear finite elements/Homework 3/Solutions

Problem: The Patch Test
 Given:

Taylor, R.L., Simo, J.C., Zienkiewicz, O.C., and Chan, A.C.H, 1986, The patch test - a condition for assessing FEM convergence,  International Journal for Numerical Methods in Engineering, 22}, pp. 39-62.

Part 1
What is the patch test and why is it used?

 * What is the patch test?

The standard patch test involves setting up a mesh of elements as shown in Figure 1(a) and checking the predicted solution against a known solution. This mesh is called a patch.

The elements should be arbitrary quadrilaterals rather than rectangles. No body forces should be applied. Material properties should be uniform and linear elastic.

The displacements of nodes $$1$$, $$2$$, $$3$$, and $$4$$ are prescribed. The prescribed value depends on the order of the polynomial that is to be represented exactly by the elements. A similar process is followed for three-dimensional elements.

Suppose we wish to have both constant and linear displacements be represented exactly by the elements. Then the prescribed displacement field is

\text{(2)} \qquad { \begin{align} u_x(x, y) & = c_0 + c_1 x + c_2 y \\ u_y(x, y) & = d_0 + d_1 x + d_2 y \end{align} } $$ where ($$c_0, c_1, c_2$$) and ($$d_0, d_1, d_2$$) are constants set by the user. These constants should be nonzero.

Therefore, the prescribed nodal displacements are

\text{(3)} \qquad \begin{align} u_x(x_i, y_i) & = c_0 + c_1 x_i + c_2 y_i \\ u_y(x_i, y_i) & = d_0 + d_1 x_i + d_2 y_i \end{align} $$ where $$(x_i, y_i)$$ are the coordinates of node $$i$$. The finite element solution for these prescribed displacements should be given by equation (2) throughout the patch.

Hence the displacements at nodes $$5$$, $$6$$, $$7$$, and $$8$$ should be given by equation (3)}.

The element strains should be constant and be given by

\begin{align} \varepsilon_{xx} & = \frac{\partial u_x}{\partial x} = c_1 \\ \varepsilon_{yy} & = \frac{\partial u_y}{\partial y} = d_2 \\ \varepsilon_{xy} & = \frac{1}{2}\left(\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{\partial x}\right) = c_2 + d_1 \end{align} $$

The element stresses should also be constant.

These results should be satisfied with a high degree of precision.

The paper talks about an additional test that can be used to assess stability (see Figure 1(b)).

 * Why is it used?

For a finite element analysis, we seek elements (and therefore shape functions) that behave such that as the mesh is refined the approximate solution converges to the exact solution.

The patch test is a necessary condition for convergence. If a finite element solution is to converge with refinement, the finite elements must pass the patch test.

In practice, the patch test checks whether an element passes the completeness test.

Suppose the approximate solution over an element is given by

u_h = \sum_{i=1}^n u^e_i N_i ~. $$ Suppose that we would like the element to represent exactly linear and constant functions. Then, in two dimensions, the shape functions ($$N_i$$) are said to be  complete if

u^e_i = c_0 + c_1 x^e + c_2 y^e \qquad \text{implies that} \qquad u_h(x,y) = c_0 + c_1 x + c_2 y $$ where $$c_0$$, $$c_1$$ and $$c_2$$ are arbitrary constants.

This means that a  complete element can represent  exactly all rigid motions (constant displacements) and all constant strain states.

The patch test is also used to check the correct implementation of a finite element program.

Part 2
What is the consistency requirement?What is the stability condition?

 * Consistency:

The term consistency condition comes from the finite difference literature where it refers to the formal consistency of the difference equations with the correct differential equation.

In the paper, the consistency requirement implies that as the size of the elements tends to zero, the approximate equation $$\mathbf{K} \mathbf{a} = \mathbf{f}$$ will represent the governing differential equation exactly.

By stability, the authors mean that the solution of the finite element system of equations is unique. For linear systems, this means that the stiffness matrix has to be nonsingular.
 * Stability:

Part 3
What do the authors mean by convergence?

Let the finite element approximation be

u_h = \mathbf{N}^T\mathbf{a} $$ where $$\mathbf{N}$$ is the vector of shape functions, and $$\mathbf{a}$$ is the vector of nodal displacements.

By convergence, the authors mean that the approximate solution $$u_h \rightarrow u$$ as $$h \rightarrow 0$$, where $$u$$ is the exact solution, and $$h$$ is the element size.

Part 4
List the 2-D structural elements available in ANSYS and describe what shape functions and degrees of freedom each uses.

Listed below are some example of the 2-D structural elements used by ANSYS. The descriptions and the shape functions of each of these elements are from the  ANSYS Element Manual.
 * PLANE42: 2-D 4-Node Structural Solid
 * The PLANE42 element can be used either as a plane element (plane stress or plane strain) or as an axisymmetric element. The element is defined by four nodes (see Figure 4) with two translational degrees of freedom at each node. The shape functions for this element are

\begin{align} N_1 & = \cfrac{1}{4}(1-s)(1-t),& N_2 & = \cfrac{1}{4}(1+s)(1-t) \\ N_3 & = \cfrac{1}{4}(1+s)(1+t),&N_4 & = \cfrac{1}{4}(1-s)(1+t) \end{align} $$
 * where $$s \in [-1,1]$$ and $$t \in [-1,1]$$.

 * However, if incompatible elements are used (KEYOPT(2)=0), then the approximate solution is of the form

\begin{align} u & = u_1 N_1 + u_2 N_2 + u_3 N_3 + u_4 N_4 + u_a (1 - s^2) + u_b (1 - t^2) \\ v & = v_1 N_1 + v_2 N_2 + v_3 N_3 + v_4 N_4 + v_a (1 - s^2) + v_b (1 - t^2) \\ \end{align} $$
 * where nodes $$a$$ and $$b$$ are incompatible nodes. The forces at the incompatible nodes are zero. Hence the element stiffness matrix can be statically condensed and $$a$$ and $$b$$ do not have to correspond to any physical nodes.

 *  PLANE82: 2-D 8-Node Structural Solid

 * The PLANE82 element is defined by eight nodes (see Figure 5) with two translational degrees of freedom at each node. The element may be used as a plane elementor as an axisymmetric element. The shape functions of this element are

\begin{align} N_1 & = -\cfrac{1}{4}(1-s)(1-t)(1+s+t),& N_2 & = -\cfrac{1}{4}(1+s)(1-t)(1-s+t) \\ N_3 & = -\cfrac{1}{4}(1+s)(1+t)(1-s-t),& N_4 & = -\cfrac{1}{4}(1-s)(1+t)(1+s-t) \\ N_5 & = \cfrac{1}{2}(1-s^2)(1-t),& N_6 & = \cfrac{1}{2}(1+s)(1-t^2) \\ N_7 & = \cfrac{1}{2}(1-s^2)(1+t),& N_8 & = \cfrac{1}{2}(1-s)(1-t^2) \end{align} $$
 * where $$s \in [-1,1]$$ and $$t \in [-1,1]$$.

 *  PLANE182: 2-D 4-Node Structural Solid

 * PLANE182 is a four-noded quadrilateral element and has the same displacement shape functions as PLANE42. In addition, the element allows for pressure degrees of freedom if  KEYOPT(6) = 1 is chosen (mixed u-p formulation). The hydrostatic pressure has an interpolation function one order lower than the displacement interpolation. If  KEYOPT(1) = 0 is chosen, a $$\bar{B}$$ method with full integration is used. One pressure degree of freedom isused per element with this option. If  KEYOPT(1) = 1 is chosen, a $$\bar{B}$$ method with selective reduced integration (and hourglass control) is used in the computations. If  KEYOPT(1) = 2 is chosen, an  enhanced strain method isused and the pressure is interpolated with linear shape functions. There are three pressure degrees of freedom in the element.

 *  PLANE183 - 2-D 8-Node Structural Solid

 * PLANE183 is a two-dimensional eight-node element similar to PLANE82. This element can also include a mixed u-p formulationwhen  KEYOPT(6) is set to 1.

Part 5
Check whether each of these elements passes the patch test. Show details of your calculations.

 Patch test for PLANE182 elements

In this test we assume that the displacements on the patch have the form

\begin{align} u &= 0.001 + 0.002 x + 0.003 y \\ v &= -0.001 - 0.002 x - 0.003 y \end{align} $$ Then, assuming plane strain, the strains in the patch are

\begin{align} \varepsilon_{xx} &= \frac{\partial u}{\partial x} = 0.002 \\ \varepsilon_{yy} &= \frac{\partial v}{\partial y} = -0.003 \\ \gamma_{xy} &= \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} = 0.003 - 0.002 = 0.001 \\ \end{align} $$ The stresses are given by

\begin{align} \sigma_{xx} &= \cfrac{E}{(1+\nu)(1-2\nu)} \left[(1-\nu)\varepsilon_{xx} + \nu\varepsilon_{yy}\right] \\ \sigma_{yy} &= \cfrac{E}{(1+\nu)(1-2\nu)} \left[(1-\nu)\varepsilon_{yy} + \nu\varepsilon_{xx}\right] \\ \sigma_{zz} &= \cfrac{E}{(1+\nu)(1-2\nu)} \left[\nu(\varepsilon_{xx}+\varepsilon_{yy})\right] \\ \tau_{xy} &= \cfrac{E}{2(1+\nu)}\gamma_{xy} \end{align} $$ If $$E = 1000$$ and $$\nu = 0.3$$, the stresses are

\begin{align} \sigma_{xx} & = 9.615384615384616e-01, \qquad \sigma_{yy} = -2.884615384615385e+00, \\ \sigma_{zz} & = -5.769230769230769e-01, \qquad \tau_{xy} = 3.846153846153846e-01 ~. \end{align} $$ If $$E = 1000$$ and $$\nu = 0.499999$$, the stresses are

\begin{align} \sigma_{xx} & = -1.666651111145334e+05, \qquad \sigma_{yy} = -1.666684444500888e+05, \\ \sigma_{zz} & = -1.666664444487555e+05, \qquad \tau_{xy} = 3.333335555557037e-01 ~. \end{align} $$ Let us apply the patch test to the patch shown in Figure 6. The exact solution for the displacements at the nodes is

\begin{align} u_1 & = 0.001, \qquad u_2 = 0.005, \qquad u_3 = 0.014, \qquad u_4 = 0.007 \\ u_5 &= 0.003, \qquad u_6 = 0.0056, \qquad u_7 = 0.01, \qquad u_8 = 0.0064 \\ v_1 &= -0.001, \qquad v_2 = -0.005, \qquad v_3 = -0.014, \qquad v_4 = -0.007 \\ v_5 &= -0.003, \qquad v_6 = -0.0056, \qquad v_7 = -0.01, \qquad v_8 = -0.0064 \end{align} $$ We apply the exact displacements at nodes 1, 2, 3, and 4 and compute the remaining displacements and the stresses and strains using ANSYS. The ANSYS script we have used is shown below:

The results from ANSYS are compared with the exact values in the tables below.

These results clearly show that the element is passing the patch test. Similar results are obtained by changing the  keyopt values in the input file.

Patch test for PLANE42 The patch test (Test B) is applied to the model shown in Figure 5a in the paper. The same solution given in the handout is considered for this analysis.

u=0.002x\quad\mbox{and}\quad v=-0.0006y \qquad \text{(10)} $$ Listed below are the inputs for the 4-node element. To analyze the incompressible materials, $$\nu$$ is set to 0.4999 and the element type is changed to PLANE182. The  keyopt command should be changed to reflect the options available of PLANE182.

Patch test for PLANE82

For an 8-node element, the similar model is created. However, the displacements must also be applied to the midnodes. Only the inputs for PLANE183 are shown here. To run the analysis using PLANE82, the  ET line should be changed to  et,1,82, and the  keyopt line should be changed to reflect the options available for PLANE82. The input for the model using PLANE183 is shown below

 Results The displacements obtained from ANSYS are compared to (10). In order pass the patch test, the displacement at each node must match (10), and stresses have to be constant everywhere. Similarly for PLANE182 and PLANE183, the results shown in the table below use $$\nu=0.4999$$ to simulate the incompressible material. The verification of the results is also similar to the previous cases. Hence, we conclude that these four elements used by ANSYS, PLANE42, PLANE82, PLANE182, and PLANE183 passed the patch test.


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User:Aaron Walden

About
Aaron Walden is a native of the Cherokee Nation Jurisdictional Area in what is today known as Oklahoma. He is descended from the Ross, Lowery, and Shorey families of Cherokees with Scottish heritage.

title: User:Aaron Walden


text:
User talk:Aaron Walden

Invitation
Dear user of Wikiversity,

Do you like to join a reading club, where you can read and discuss literature? Reading clubs could have active and inactive participants. Active participants could be required to edit on a regularly basis untill a deadline is reached and another topic can be chosen for reading and discussing. Hopefully, reading clubs can ensure a social network on Wikiversity of people cooperating with eachother and becoming friends while discussing knowledge.--Daanschr 09:23, 1 November 2007 (UTC)

title: User talk:Aaron Walden


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User:Naussbur/Maflo

Maflo es un grupo de antiguos estudiantes del colegio Liceo Moderno del Norte en Barranquilla, Colombia. El Maflo como corrientemente se le denominaba fue fundado en dicho colegio en el año 1975 cuando sus integrantes cursaban el segundo nivel de educación media y tuvo presencia en el liceo durante el periodo comprendido entre los años 1975 y 1978, cuando la mayoría de sus integrantes se trasladaron a la Normal Mixta de la Costa Norte, colegio en el cual finalmente recibieron el grado de Bachiller. El grupo estaba formado por estudiantes sobresalientes en las diferentes áreas del conocimiento. Inicialmente fue formado por Julio Cesar Almanya, Hernando Burgos Soto, Luis De León Padilla, Noel Gonzalez Ferreira y Miguel Zapata Ferreira. El Primero de estos fue reemplazado por Fidel Torregrosa en el año 1978.

title: User:Naussbur/Maflo


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Nonlinear finite elements/Homework 4/Solutions

Problem 1: Axially Loaded Bar with Nonlinear Modulus
 Given:

Bar of uniform cross-section under axial body load and compressive axial end load ($$R$$). Nonlinear material with constitutive relation

\sigma = E_0(1-\varepsilon) \varepsilon, \qquad \varepsilon = \cfrac{du}{dx} ~. $$ Finite element model uses quadratic elements with shape functions

\begin{align} N_1(x) & = \cfrac{2(x_2 - x)(x_3 - x)}{h^2} \\ N_2(x) & = \cfrac{4(x - x_1)(x_3 - x)}{h^2} \\ N_3(x) & = \cfrac{2(x - x_1)(x - x_2)}{h^2} \end{align} $$ where $$h = x_3 - x_1$$.

Part 1
Plot the shape functions for an element. Show that they add up to 1 at all points within the element.

A plot of the shape functions for an element is shown in Figure 1(a). The shape functions for a mesh are shown in Figure 1(b). If we add the shape functions we get
 * $$\begin{align}

\sum_i N_i(x) & = \cfrac{2(x_2 - x)(x_3 - x)}{h^2} + \cfrac{4(x - x_1)(x_3 - x)}{h^2} + \cfrac{2(x - x_1)(x - x_2)}{h^2} \\ & = \cfrac{2}{h^2}\left[(x_2-x)(x_3-x) + (x-x_1)(x_3-x) + (x-x_1)(x_3-x) + (x-x_1)(x-x_2)\right]\\ & = \cfrac{2}{h^2}\left[(x_2-x_1)(x_3-x) + (x-x_1)(x_3-x_2)\right]\\ & = \left(\cfrac{2}{h^2}\right)\left(\cfrac{h}{2}\right)(x_3-x_1) = 1 \qquad \square \end{align}$$

Part 2
Plot the displacement, stress, strain, and the error norm when you use 10 elements to discretize the bar. Discuss and comment on your observations.

If we use an isoparametric form of the equations to simplify the integration of the stiffness matrix and load vector terms, the shape functions become

\begin{align} N_1(\xi)\, & = -\frac{1}{2} \left( 1-\xi \right) \xi \\ N_2(\xi)\, & = \, \left( 1+\xi \right)\left( 1-\xi \right) \\ N_3(\xi)\, & = \frac{1}{2} \left( 1+\xi \right) \xi ~. \end{align} $$ The map between $$x$$ and $$\xi$$ is

x(\xi) = x_1 N_1(\xi) + x_2 N_2(\xi) + x_3 N_3(\xi); ~\qquad x_2 = \cfrac{x_1+x_3}{2} ~. $$ Therefore, the Jacobian of the deformation is

J = \cfrac{dx}{d\xi} = \cfrac{h}{2} ~. $$ The terms of the stiffness matrix are given by
 * $$\begin{align}

K_{ij} & = \int_{x_1}^{x_3} A E_0 \left(1 - \sum_{k=1}^3 u_k \cfrac{dN_k(x)}{dx}\right) \cfrac{dN_i(x)}{dx} \cfrac{dN_j(x)}{dx}~dx \\ & = \int_{-1}^{1} A E_0 \left[1 - \sum_{k=1}^3 u_k \left(J^{-1}\cfrac{dN_k(\xi)}{d\xi}\right) \right] \left(J^{-1}\cfrac{dN_i(\xi)}{d\xi}\right) \left(J^{-1}\cfrac{dN_j(\xi)}{d\xi}\right)~J~d\xi ~. \end{align}$$ The element stiffness matrix is

\mathbf{K} = \cfrac{A E_0}{3h^2} \begin{bmatrix} 7h+u_3+15u_1-16u_2 & & -8(h+2u_1-2u_2)& &h-u_3+u_1\\ -8(h+2u_1-2u_2) & & 16(h-u_3+u_1)& & -8(h-2u_3+2u_2)\\ h-u_3+u_1& & -8(h-2u_3+2u_2)& & 7h-15u_3-u_1+16u_2 \end{bmatrix} $$ The terms of the tangent stiffness matrix are given by

T_{ij} = K_{ij} + \sum_{m=1}^3 \frac{\partial K_{im}}{\partial u_j} u_m ~. $$ Therefore, the element tangent stiffness matrix is

\mathbf{T} = \begin{bmatrix} 7h+2u_3+30u_1-32u_2 & & -8(h+4u_1-4u_2)& &h-2u_3+2u_1\\ -8(h+4u_1-4u_2)& & 16(h-2u_3+2u_1)& & -8(h-4u_3+4u_2)\\ h-2u_3+2u_1 & & -8(h-4u_3+4u_2)& & 7h-30u_3-2u_1+32u_2 \end{bmatrix} $$ The terms of the element load vector are given by
 * $$\begin{align}

f_i & = \int_{x_1}^{x_3} axN_i(x)~dx + \left.N_i\left(AE\cfrac{du}{dx}\right)\right|_{x_1}^{x_3} \\ & = \int_{-1}^{1} a\left(\sum_{k=1}^3 x_k N_k(\xi)\right) N_i(\xi)~J~d\xi + \left.N_i(\xi)R_i\right|_{-1}^{1}~. \end{align}$$ Therefore, the element load vector is

\mathbf{f} =\cfrac{ah}{6} \begin{bmatrix} x_1 \\ 2(x_1 + x_3) \\ x_3 \end{bmatrix} + \begin{bmatrix} -R_1 \\ 0 \\ R_3 \end{bmatrix} $$

The stress-strain curve in compression is shown in Figure 2.

If we use a mesh containing $$2$$ equally sized elements, we get the solutions shown in Figure 3.

If we use a mesh containing $$10$$ equally sized elements, we get the solutions shown in Figure 4.

Observations:

 * 1) The two-node and three-node elements give the same solution.
 * 2) The solution using two elements shows that the stresses and strains are interpolated linearly within each element and are non longer constant.
 * 3) There is no jump in the stress/strain at the nodes.
 * 4) The solution converges quite rapidly - four iterations. This is due to the shape of the stress-strain curve in compression.

Part 2
Do the same for 100 elements. What are your observations on the effect of mesh refinement on the solution?

If we use a mesh containing $$100$$ equally sized elements, we get the solutions shown in Figure 5.

Observations:

 * 1) The solution is approximated quite accurately with 10 elements. Using 100 elements does not lead to any significant improvement in the solution.
 * 2) The solution converges in four iterations unlike the case where a tensile load was applied.

Part 4
Now, increase the applied load. At what load does the solution fail to converge? Explain why the solution fails to converge at this load.

The solution converges at all loads unlike in the tension problem. If the load is increased to $$R = -20.1$$, the elements invert and we get a negative Jacobian. Even though we get a solution, that solution is unphysical.

The Matlab code use to compute the results is shown below.

You could alternatively write your own Maple code or use ANSYS for these calculations depending on your interests.

Problem 2: Nonlinear Thermal Conduction
Consider the model chip system shown in Figure 6. Assume that heat is being generated in the silicon carbide region of the chip at the rate of 100 W/cm$$^2$$. The cooling channels contain a fluid at a constant 250 K and (supposedly) can dissipate heat at the rate of 800 W/cm$$2$$. The surroundings are at room temperature (300 K).

The thermal conductivity of the silicon carbide and the aluminum nitride are also given in Figure 6. The thermal conductivity of the epoxy resin is 1.5 W/m-K.

Part 1
First run a linear simulation of the problem using room temperature thermal conductivities of the materials. This will act as a check of the model and setup and will help you determine any additional constraints that you will need. Do the cooling channels really dissipate heat at 800 W/cm$2$?

The following ANSYS input is used for this simulation. Due to symmetry, only half of the model will be meshed Figure 7. A constant temperature of 300K is applied around the edge Epoxy resin. Heat is being generated from areas made up by the Silicon Carbide at the rate of 100 W/cm$$^2$$. Each line forming the cooling channels has constant temperature of 250K.

Shown in Figure 8 and 9 below are the contour plots of temperature and the vector sum of heat flux. From Figure 9 we see that the cooling channels do not dissipate heat at the rate suggested in the problem statement.

Part 2
Fit curves to the thermal conductivity versus temperature data for Silicon Carbide and Aluminum Nitride.

Below are the curve-fits for each material.
 * Silicon Carbide:$$k = 2(10^{-8})T^4 - 2(10^{-5})T^3 + 0.0057T^2 - 0.7605T +87.873$$
 * Aluminum Nitride:$$k = -1(10^{-9})T^4 + 1(10^{-7})T^3 + 0.0003T^2 - 0.0941T +28.634$$

Part 3
Perform a nonlinear simulation using these temperature dependent properties to find the equilibrium temperature inside the silicon carbide region. State all simplifying assumptions that you have made.

The same code that was used in the previous problem is used here. However, the thermal conductivities are tabulated according to their temperature.

Shown in Figure 10 and 11 below are the contour plots of temperature and the vector sum of heat flux.

There are many approaches to create the mesh for this problem. You will have to find the way the work best for you. These two codes listed are just two ways that the meshes can be manipulated.


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Pipe and tabor

The pipe and tabor are two instruments played by a single player. The first instrument is the tabor, which is a small drum, and this is played with a drumstick in one hand. The other instrument is the pipe, which is a three-holed flute with a whistle mouthpiece of the same sort as on a tinwhistle or a recorder.

Playing the pipe and tabor might be seen as similar to playing a keyboard instrument, in that one hand plays the melody, while the other provides the accompaniment.

In English, the player of pipe and tabor may be called a taborer (tă'-bər-ər). The French equivalent name is tambourinaire, and the Spanish equivalent is tamborilero.

History
Medieval art provides ample illustration of the popularity of one-man pipe and tabor playing during that period. The taborer's pipe was played in the British Isles and many parts of Western Europe. Sometimes another instrument was played in place of the tabor. In France and Spain a special psaltery was tuned as drones and beaten with a stick as a sort of stringed-tabor. Others played bell, triangle, or even a second pipe in place of the tabor.

Pipe and tabor was used to accompany clown acts, dances, puppet plays, religious processions, and any outdoor occasion calling for music by a single musician. Pipe and tabor also was seen in bands, along with other instruments, from the Mediaeval period into the 19th Century.

The pipe and tabor waned in popularity as bands became more accessible, but never entirely died out. In the 20th Century the regional forms in Spain, Britain, and France resurged.

In the late 19th and early 20th Century, British folklorist Cecil Sharp helped to repopularize a national dance called morris dancing. The pipe and tabor, having been a traditional accompaniment to these dances, with the drum providing audible cues to the dancers, was revived by many morris dance sides (troupes). Some of the new taborers looked to the continent and adopted a larger form of pipe, adapting it by adding a fourth hole, so they would not need to partially cover the bottom of the instrument to get an extra low note. Others stayed with the smaller three-hole pipes. Metal versions of both were produced, the latter being made by the makers of Generation tinwhistles, and widely available inexpensively.

Method
This treatise will focus on the playing of the English pipe and tabor, though much of the information may apply to other species of the pipe and tabor, as well. Aside from regional musical traditions, other forms of pipe vary in their scales, and other forms of tabor vary in their dimensions and in how they are held.

Playing the tabor
The tabor is often suspended from the arm of the same hand that holds the pipe, and it is beaten with the other hand. Some players use a deeper drum, and hang it from a strap around the neck or attach it to a belt. If it has a snare, the player beats on the snare side, with the snare giving a distinctive buzzing sound.

The light shell and rope-tensioned natural heads give the traditional tabor a relatively low pitch. The tabor should be affixed to the arm in such a way that it will not give problems with swinging or spinning too much, but it should not be held firmly against the body, because this will mute the sound and render an inferior tone.

It is important to practice the tabor. The job of the drumming-hand is to maintain a steady beat. Many players practice by beating with a drumstick to about any music they hear. Playing with recordings of professional musicians, regardless of the style, will teach the aspiring taborer proper rhythm.

Some styles of tabor playing are little more than beating the drum like one would clap one's hand with music. Others incorporate rolls and enhanced rhythmic patterns. It is important to listen to other drummers in order to understand what can be done with the instrument.

Drum rolls
It is quite possible to do rolls on the tabor when using an ordinary drumstick. This is accomplished by using a fulcrum effect, bouncing the head of the stick on the drumhead and the part near the rim of the tabor off the rim.

Playing the pipe
The pipe is narrow, enabling the taborer to play all of his scales using the upper registers. Most woodwind instruments play a scale beginning with the lowest possible note, and then add more octaves by overblowing. However, as the taborer's pipe has only three holes, it is impossible to obtain the notes between the lowest four notes and the second octave. Therefore, the pipe begins its first octave at the first overblow.

The pipe is gripped, below the lowest finger hole, between the ring finger on front and the little finger on back, which leaves two fingers and a thumb completely free for playing the instrument.

The following chart shows the fingering of the pipe. Filled circles represent covered holes. Open circles represent holes left uncovered. Plus signs (+) represent overblown registers.

Lessons
Some players recommend that a beginner first practice blowing the pipe at different pressures, and then learn the fingerings once comfortable with these.

With the above fingerings one can begin playing tunes. The following will provide the beginning piper with some material to practice.

Lesson One: A Carol
Christmas is a time when pipe and tabor is often used, due to its merry and traditional nature. It is thus fitting to begin by learning to play a familiar carol. It may help to sing the words in your mind as you play.

Lesson Two: A Jig
It is important to play a tune through enough times to memorize it well.

This tune is one that is used by morris dancers in England.

Lesson Three: A Reel
A reel is a traditional type of tune in two parts. The A part is played a few times and then the B part a few times, alternating.

Off-site resources

 * Morris Dance Tunes compiled by Cecil Sharp
 * Address to a Society of Morris Dancers, 1914 by Sir Francis Darwin

title: Pipe and tabor