OpenLLM-France/wikimedia huggingface dataset:
# Electric Circuit Analysis/Nodal Analysis/Answers

Exercise 7: Answers
Model Answer
KCL @ Node b:
$i_{2}=i_{1}+i_{6}$
Thus by applying Ohms law to above equation we get.
${\begin{matrix}\ {\frac {V_{c}-V_{b}}{R_{2}}}-{\frac {V_{b}}{R_{1}}}-{\frac {V_{b}-V_{d}}{R_{6}}}=0\end{matrix}}$
Therefore
${\begin{matrix}\ \ -V_{b}({\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+{\frac {1}{R_{6}}})+V_{c}({\frac {1}{R_{2}}})+V_{d}({\frac {1}{R_{6}}})=0\end{matrix}}$ ............... (1)
KCL @ Node c:
$i_{3}=i_{2}+i_{4}$
Thus by applying Ohms law to above equation we get.
${\begin{matrix}\ {\frac {V_{s}-V_{c}}{R_{3}}}-{\frac {V_{c}-V_{b}}{R_{2}}}-{\frac {V_{c}-V_{d}}{R_{4}}}=0\end{matrix}}$
Therefore
${\begin{matrix}\ \ -V_{b}({\frac {1}{R_{2}}})-V_{c}(({\frac {1}{R_{2}}}+{\frac {1}{R_{3}}}+{\frac {1}{R_{4}}})+V_{d}({\frac {1}{R_{4}}})={\frac {V_{s}}{R_{3}}}\end{matrix}}$ ............... (2)
KCL @ Node d:
$i_{5}=i_{4}+i_{6}$
Thus by applying Ohms law to above equation we get.
${\begin{matrix}\ {\frac {V_{s}-V_{c}}{R_{3}}}-{\frac {V_{c}-V_{b}}{R_{2}}}-{\frac {V_{c}-V_{d}}{R_{4}}}\end{matrix}}$
Therefore
${\begin{matrix}\ \ -V_{b}({\frac {1}{R_{3}}})-V_{c}({\frac {1}{R_{4}}})+V_{d}({\frac {1}{R_{3}}}+{\frac {1}{R_{4}}}-{\frac {1}{R_{5}}})=0\end{matrix}}$ ............... (3)
$G_{1}={\frac {1}{R_{1}}}$ etc thus equations 1; 2 & 3 will be re-written as follows:
${\begin{matrix}\ V_{b}(-G_{1}-G_{2}-G_{6})+V_{c}(G_{2})+V_{d}(G_{6})&=&0....(1)\\\ \\\ V_{b}(-G_{2})+V_{c}(G_{2}+G_{3}+G_{4})+V_{d}(-G_{4})&=&(V_{s}\times G_{3})....(2)\\\ \\\ V_{b}(-G_{3})+V_{c}(-G_{4})+V_{d}(G_{3}+G_{4}+G_{5})&=&0....(3)\end{matrix}}$
Now we can create a matrix with the above equations as follows:
${\begin{bmatrix}(-G_{1}-G_{2}-G_{6})&(G_{2})&(G_{6})\\(-G_{2})&(G_{2}+G_{3}+G_{4})&(-G_{4})\\(-G_{4})&(G_{4})&(G_{3}+G_{4}+G_{5})\end{bmatrix}}.{\begin{bmatrix}V_{b}\\V_{c}\\V_{d}\end{bmatrix}}={\begin{bmatrix}0\\V_{s}\times G_{3}\\0\end{bmatrix}}$
The following matrix is the above with values substituted:
$A.{\vec {X}}={\vec {Y}}$ → ${\begin{bmatrix}-0.056&0.05&0.001\\-0.05&0.0503&-0.0002\\-0.0001&-0.0002&0.000367\end{bmatrix}}.{\begin{bmatrix}V_{b}\\V_{c}\\V_{d}\end{bmatrix}}={\begin{bmatrix}0\\0.0009\\0\end{bmatrix}}$
Now that we have arranged equations 1; 2 & 3 into a matrix we need to get Determinants of the General matrix, and Determinants of alterations of the general matrix as follows:
Solving determinants of:
Matrix A  : General matrix A from KCL equations
Matrix A1 : Genral Matrix A with Column 1 substituted by ${\vec {Y}}$.
Matrix A2 : Genral Matrix A with Column 2 substituted by ${\vec {Y}}$.
Matrix A3 : Genral Matrix A with Column 3 substituted by ${\vec {Y}}$.
As follows:
$detA=-9.8\times 10^{-8}$
$detA1=-1.7\times 10^{-8}$
$detA2=-1.8\times 10^{-8}$
$detA3=-1.5\times 10^{-8}$
Now we can use the solved determinants to arrive at solutions for Node voltages $V_{b};V_{c}andV_{d}$ as follows:
1. $V_{b}={\frac {detA1}{detA}}=0.17V$
2. $V_{c}={\frac {detA2}{detA}}=0.19V$
3. $V_{d}={\frac {detA3}{detA}}=0.15V$
Now we can apply Ohm's law to solve for the current through $R_{3}$ as follows:
$I_{R_{3}}={\frac {V_{s}-V_{c}}{R_{3}}}={\frac {9V-0.19V}{10k\Omega }}=0.881mA$ :
# Electric Circuit Analysis/Passive Sign Convention/Answers

* Exercise 1: Answers
* It is a resistor because it dissipates +16 Watts.
* It is a source because it is dissipating -2.25 milli-Watts (or supplying 2.25 milli-Watts).
* It is a source because it is supplying 225 Watts (or dissipating -225 Watts).
* It is a resistor because it is dissipating +30 milli-Watts.
# Electric Circuit Analysis/Resistors in Parallel/Answers

* Exercise 4: Answers
* 1. $R_{eq}=2.5\Omega$.
* 2.1 $R_{eq}=1.024\Omega$.
* 2.2 $I_{1}=7.68A$.
* 2.3 $I_{2}=5.12A$.
* 2.4 $I_{3}=2.19A$.
* 2.5 $V_{s}=15.36Volts$.
* 3. $R_{eq}=3.377\Omega$.
* 4 Voltage Sources can be connected in parallel if they have the same voltage output.
# Electric Circuit Analysis/Resistors in Series/Answers

* Exercise 3: Answers
* V1 = V2 = 10V since R1 = R2 and I = 2A
* V1 = 2.5V ; V2 = 3.75V ; V3 = 8.75V and I = 1.25A
* V$_{Stotal}$ = 9.5V ( Summation of Volatge Sources in Series ) then: V1 = 1.5833V ; V2 = 2.3750V and V3 = 5.5417V .
# Electric Circuit Analysis/Simple Resistive Circuits/Answers

* Exercise 2: Answers
* 600 V, 180,000W.
* 0.04A, more commonly written 40mA.
* 134.164V, 0.447A.
* No. There is not enough information. The only thing that is given is the voltage. With two equations there can only be two unknowns, one unknown is great, but we have three.
* How did you do?
# Elasticity/Torsion of thin walled open sections
## Torsion of thin-walled open sections

Examples are I-beams, channel sections and turbine blades.
We assume that the length $b\,$ is much larger than the thickness $t\,$, and that $t\,$ does not vary rapidly with change along the length axis $\xi \,$.
Using the membrane analogy, we can neglect the curvature of the membrane in the $\xi \,$ direction, and the Poisson equation reduces to
    ${\frac {d^{\phi }}{d\eta ^{2}}}=-2\mu \alpha$
which has the solution
    $\phi =\mu \alpha \left({\frac {t^{2}}{4}}-\eta ^{2}\right)$
where $\eta$ is the coordinate along the thickness direction.
The stress field is
    $\sigma _{3\xi }={\frac {\partial }{\partial }}{\phi }{\eta }=-e\mu \beta \eta ~~;~~~\sigma _{3\eta }=0$
Thus, the maximum shear stress is
    $\tau _{\text{max}}=\mu \beta t_{\text{max}}\,$
# Elasticity/Torsion of triangular cylinder
## Example: Equilateral triangle

* Torsion of a cylinder with a triangular cross section
The equations of the three sides are
    ${\begin{aligned}{\text{side}}~\partial S^{(1)}~:~~&f_{1}(x_{1},x_{2})=x_{1}-{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(2)}~:~~&f_{2}(x_{1},x_{2})=x_{1}+{\sqrt {3}}x_{2}+2a=0\\{\text{side}}~\partial S^{(3)}~:~~&f_{3}(x_{1},x_{2})=x_{1}-a=0\end{aligned}}$
Let the Prandtl stress function be
    $\phi =Cf_{1}f_{2}f_{3}\,$
Clearly, $\phi =0\,$ at the boundary of the cross-section (which is what we need for solid cross sections).
Since, the traction-free boundary conditions are satisfied by $\phi \,$, all we have to do is satisfy the compatibility condition to get the value of $C\,$. If we can get a closed for solution for $C\,$, then the stresses derived from $\phi \,$ will satisfy equilibrium.
Expanding $\phi \,$ out,
    $\phi =C(x_{1}-{\sqrt {3}}x_{2}+2a)(x_{1}+{\sqrt {3}}x_{2}+2a)(x_{1}-a)$
Plugging into the compatibility condition
    $\nabla ^{2}{\phi }=12Ca=-2\mu \alpha$
Therefore,
    $C=-{\frac {\mu \alpha }{6a}}$
and the Prandtl stress function can be written as
    $\phi =-{\frac {\mu \alpha }{6a}}(x_{1}^{3}+3ax_{1}^{2}+3ax_{2}^{2}-3x_{1}x_{2}^{2}-4a^{3})$
The torque is given by
    $T=2\int _{S}\phi dA=2\int _{-2a}^{a}\int _{-(x_{1}+2a)/{\sqrt {3}}}^{(x_{1}+2a)/{\sqrt {3}}}\phi dx_{2}dx_{1}={\frac {27}{5{\sqrt {3}}}}\mu \alpha a^{4}$
Therefore, the torsion constant is
    ${\tilde {J}}={\frac {27a^{4}}{5{\sqrt {3}}}}$
The non-zero components of stress are
    ${\begin{aligned}\sigma _{13}=\phi _{,2}&={\frac {\mu \alpha }{a}}(x_{1}-a)x_{2}\\\sigma _{23}=-\phi _{,1}&={\frac {\mu \alpha }{2a}}(x_{1}^{2}+2ax_{1}-x_{2}^{2})\end{aligned}}$
The projected shear stress
    $\tau ={\sqrt {\sigma _{13}^{2}+\sigma _{23}^{2}}}$
is plotted below
* Stresses in a cylinder with a triangular cross section under torsion
The maximum value occurs at the middle of the sides. For example, at $(a,0)$,
    $\tau _{\text{max}}={\frac {3\mu \alpha a}{2}}$
The out-of-plane displacements can be obtained by solving for the warping function $\psi$. For the equilateral triangle, after some algebra, we get
    $u_{3}={\frac {\alpha x_{2}}{6a}}(3x_{1}^{2}-x_{2}^{2})$
The displacement field is plotted below
* Displacements $u_{3}\,$ in a cylinder with a triangular cross section.
# Elasticity/Warping of circular cylinder
## Example 1: Circular Cylinder

Choose warping function
    $\psi (x_{1},x_{2})=0\,$
Equilibrium ($\nabla ^{2}{\psi }=0$) is trivially satisfied.
The traction free BC is
    $(0-x_{2}){\frac {dx_{2}}{ds}}-(0+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
Integrating,
    $x_{2}^{2}+x_{1}^{2}=c^{2}~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
where $c$ is a constant.
Hence, a circle satisfies traction-free BCs.
* There is no warping of cross sections for circular cylinders
The torsion constant is
    ${\tilde {J}}=\int _{S}(x_{1}^{2}+x_{2}^{2})dA=\int _{S}r^{2}dA=J$
The twist per unit length is
    $\alpha ={\frac {T}{\mu J}}$
The non-zero stresses are
    $\sigma _{13}=-\mu \alpha x_{2}~;~~\sigma _{23}=\mu \alpha x_{1}$
The projected shear traction is
    $\tau =\mu \alpha {\sqrt {(x_{1}^{2}+x_{2}^{2})}}=\mu \alpha r$
Compare results from Mechanics of Materials solution
    $\phi ={\frac {TL}{GJ}}~~\Rightarrow ~~\alpha ={\frac {T}{GJ}}$
and
    $\tau ={\frac {Tr}{J}}~~\Rightarrow ~~\tau =G\alpha r$
# Elasticity/Transformation example 1
## Example 1

Derive the transformation rule for second order tensors ($T_{ij}^{'}=l_{ip}l_{jq}T_{pq}$). Express this relation in matrix notation.

### Solution

A second-order tensor $\mathbf {T}$ transforms a vector $\mathbf {u}$ into another vector $\mathbf {v}$. Thus,
    $\mathbf {v} =\mathbf {T} \mathbf {u} =\mathbf {T} \bullet \mathbf {u}$
In index and matrix notation,
    ${\text{(1)}}\qquad v_{i}=T_{ij}u_{i}\leftrightarrow v_{p}=T_{pq}u_{q}~{\text{or,}}~\left[v\right]=\left[T\right]\left[u\right]$
Let us determine the change in the components of $\mathbf {T}$ with change the basis from ($\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3}$) to ($\mathbf {e} _{1}^{'},\mathbf {e} _{2}^{'},\mathbf {e} _{3}^{'}$). The vectors $\mathbf {u}$ and $\mathbf {v}$, and the tensor $\mathbf {T}$ remain the same. What changes are the components with respect to a given basis. Therefore, we can write
    ${\text{(2)}}\qquad v_{i}^{'}=T_{ij}^{'}u_{i}^{'}~{\text{or,}}~\left[v\right]^{'}=\left[T\right]^{'}\left[u\right]^{'}$
Now, using the vector transformation rule,
    ${\begin{aligned}{\text{(3)}}\qquad v_{i}^{'}&=l_{ip}v_{p}~;~u_{i}^{'}=l_{ip}u_{p}~{\text{or,}}~\left[v\right]^{'}=\left[L\right]\left[v\right]~;\left[u\right]^{'}=\left[L\right]\left[u\right]\\v_{q}&=l_{iq}v_{i}^{'}~;~u_{q}=l_{iq}u_{i}^{'}~{\text{or,}}~\left[v\right]=\left[L\right]^{T}\left[v\right]^{'}~;\left[u\right]=\left[L\right]^{T}\left[u\right]^{'}\end{aligned}}$
Plugging the first of equation (3) into equation (2) we get,
    ${\text{(4)}}\qquad l_{ip}v_{p}=T_{ij}^{'}u_{i}^{'}~{\text{or,}}~\left[L\right]\left[v\right]=\left[T\right]^{'}\left[u\right]^{'}$
Substituting for $v_{p}$ in equation~(4) using equation~(1),
    ${\text{(5)}}\qquad l_{ip}T_{pq}u_{q}=T_{ij}^{'}u_{i}^{'}~{\text{or,}}~\left[L\right]\left[T\right]\left[u\right]=\left[T\right]^{'}\left[u\right]^{'}$
Substituting for $u_{q}$ in equation (5) using equation (3),
    ${\text{(6)}}\qquad l_{ip}T_{pq}l_{iq}u_{i}^{'}=T_{ij}^{'}u_{i}^{'}~{\text{or,}}~\left[L\right]\left[T\right]\left[L\right]^{T}\left[u\right]^{'}=\left[T\right]^{'}\left[u\right]^{'}$
Therefore, if $\mathbf {u} \equiv \left[u\right]$ is an arbitrary vector,
    $l_{ip}T_{pq}l_{iq}=T_{ij}^{'}\Rightarrow T_{ij}^{'}=l_{ip}l_{jq}T_{pq}~{\text{or,}}~\left[T\right]^{'}=\left[L\right]\left[T\right]\left[L\right]^{T}$
which is the transformation rule for second order tensors.
Therefore, in matrix notation, the transformation rule can be written as
    $\left[T\right]^{'}=\left[L\right]\left[T\right]\left[L\right]^{T}$
# Elasticity/Warping functions
## Warping Function and Torsion of Non-Circular Cylinders

Warping functions are quite useful in the solution of problems involving the torsion of cylinders with non-circular cross sections.
For such problems, the displacements are given by
    $u_{1}=-\alpha x_{2}x_{3}~;~~u_{2}=\alpha x_{1}x_{3}~;~~u_{3}=\alpha \psi (x_{1},x_{2})$
where $\alpha \,$ is the twist per unit length, and $\psi \,$ is the warping function.
The stresses are given by
    $\sigma _{13}=\mu \alpha (\psi _{,1}-x_{2})~;~~\sigma _{23}=\mu \alpha (\psi _{,2}+x_{1})$
where $\mu \,$ is the shear modulus.
The projected shear traction is
    $\tau ={\sqrt {(\sigma _{13}^{2}+\sigma _{23}^{2})}}$
Equilibrium is satisfied if
    $\nabla ^{2}{\psi }=0~~~~\forall (x_{1},x_{2})\in {\text{S}}$
Traction-free lateral BCs are satisfied if
    $(\psi _{,1}-x_{2}){\frac {dx_{2}}{ds}}-(\psi _{,2}+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
or,
    $(\psi _{,1}-x_{2}){\hat {n}}_{1}+(\psi _{,2}+x_{1}){\hat {n}}_{2}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
The twist per unit length is given by
    $\alpha ={\frac {T}{\mu {\tilde {J}}}}$
where the torsion constant
    ${\tilde {J}}=\int _{S}(x_{1}^{2}+x_{2}^{2}+x_{1}\psi _{,2}-x_{2}\psi _{,1})dA$
# Elasticity/Torsion of thin walled closed sections
## Torsion of thin-walled closed sections

The Prandtl stress function $\phi \,$ can be approximated as a linear function between $\phi _{1}\,$ and $0\,$ on the two adjacent boundaries.
The local shear stress is, therefore,
    $\sigma _{3s}={\frac {\phi _{1}}{t}}$
where $s\,$ is the parameterizing coordinate of the boundary curve of the cross-section and $t\,$ is the local wall thickness.
The value of $\phi _{1}\,$ can determined using
    $\phi _{1}={\frac {2\mu \alpha A}{\oint _{S}{\frac {dS}{t}}}}$
where $A\,$ is the area enclosed by the mean line between the inner and outer boundary.
The torque is approximately
    $T=2A\phi _{1}\,$
# Elasticity/Transversely loaded wedge
## Sample homework problems
### Wedge loaded transversely by a concentrated load

Given:
A wedge of infinite length with a concentrated load $\mathbf {P} =P~{\widehat {\mathbf {e} }}_{2}$ per unit wedge thickness at the vertex. Plane stress/strain.
* Wedge loaded transversely by a concentrated load
Find:
The stress field in the wedge.

### Solution

From the Flamant solution, we know that the stress field in the wedge is
    ${\begin{aligned}\sigma _{rr}&={\frac {2}{r}}\left(C_{1}\cos \theta +C_{2}\sin \theta \right)\\\sigma _{r\theta }&=0\\\sigma _{\theta \theta }&=0\end{aligned}}$
The constants $C_{1}\,$ and $C_{2}\,$ can be found by using the equilibrium conditions
    ${\begin{aligned}2\int _{-\beta }^{\beta }\left(C_{1}\cos \theta -C_{2}\sin \theta \right)\cos \theta ~d\theta &=0\\P+2\int _{-\beta }^{\beta }\left(C_{1}\cos \theta -C_{2}\sin \theta \right)\sin \theta ~d\theta &=0\end{aligned}}$
or,
    ${\begin{aligned}C_{1}\left[2\beta +\sin(2\beta )\right]&=0\\P+C_{2}\left[\sin(2\beta )-2\beta \right]&=0\end{aligned}}$
Therefore,
    $C_{1}=0~;~~C_{2}={\frac {P}{2\beta -\sin(2\beta )}}$
Hence, the stresses are
    ${\begin{aligned}\sigma _{rr}&={\frac {2P\sin \theta }{r\left[2\beta -\sin(2\beta )\right]}}\\\sigma _{r\theta }&=0\\\sigma _{\theta \theta }&=0\end{aligned}}$
# Elasticity/VonMises stress
## Von Mises Stress or Equivalent Stress

Von Mises is a criteria used in predicting the onset of yield in ductile materials. The von Mises stress/equivalent stress/distortional strain energy is given by the relation
    $\sigma _{\text{eq}}={\sqrt {I_{1}^{2}-3I_{2}}}={\sqrt {\sigma _{11}^{2}+\sigma _{22}^{2}+\sigma _{33}^{2}-\sigma _{11}\sigma _{22}-\sigma _{22}\sigma _{33}-\sigma _{33}\sigma _{11}+3\sigma _{12}^{2}+3\sigma _{23}^{2}+3\sigma _{31}^{2}}}$

## Related Content

Introduction to Elasticity
# Elasticity/Warping of elliptical cylinder
## Example 2 : Elliptical cylinder

Choose warping function
    $\psi (x_{1},x_{2})=kx_{1}x_{2}\,$
where $k\,$ is a constant.
Equilibrium ($\nabla ^{2}{\psi }=0$) is satisfied.
The traction free BC is
    $(kx_{2}-x_{2}){\frac {dx_{2}}{ds}}-(kx_{1}+x_{1}){\frac {dx_{1}}{ds}}=0~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
Integrating,
    $x_{1}^{2}+{\frac {1-k}{1+k}}x_{2}^{2}=a^{2}~~~~\forall (x_{1},x_{2})\in \partial {\text{S}}$
where $a\,$ is a constant.
This is the equation for an ellipse with major and minor axes $a\,$ and $b\,$, where
    $b^{2}=\left({\frac {1+k}{1-k}}\right)a^{2}$
The warping function is
    $\psi =-\left({\frac {a^{2}-b^{2}}{a^{2}+b^{2}}}\right)x_{1}x_{2}$
The torsion constant is
    ${\tilde {J}}={\frac {2b^{2}}{a^{2}+b^{2}}}I_{2}+{\frac {2a^{2}}{a^{2}+b^{2}}}I_{1}={\frac {\pi a^{3}b^{3}}{a^{2}+b^{2}}}$
where
    $I_{1}=\int _{S}x_{1}^{2}dA={\frac {\pi ab^{3}}{4}}~~;~~I_{2}=\int _{S}x_{2}^{2}dA={\frac {\pi a^{3}b}{4}}$
If you compare ${\tilde {J}}$ and $J\,$ for the ellipse, you will find that ${\tilde {J}}<J$. This implies that the torsional rigidity is less than that predicted with the assumption that plane sections remain plane.
The twist per unit length is
    $\alpha ={\frac {(a^{2}+b^{2})T}{\mu \pi a^{3}b^{3}}}$
The non-zero stresses are
    $\sigma _{13}=-{\frac {2\mu \alpha a^{2}x_{2}}{a^{2}+b^{2}}}~~;~~\sigma _{23}=-{\frac {2\mu \alpha b^{2}x_{1}}{a^{2}+b^{2}}}$
The projected shear traction is
    $\tau ={\frac {2\mu \alpha }{a^{2}+b^{2}}}{\sqrt {b^{4}x_{1}^{2}+a^{4}x_{2}^{2}}}~~\Rightarrow ~~\tau _{\text{max}}={\frac {2\mu \alpha a^{2}b}{a^{2}+b^{2}}}~~(b<a)$
* Shear stresses in the cross section of an elliptical cylinder under torsion
For any torsion problem where $\partial S\,$ is convex, the maximum projected shear traction occurs at the point on $\partial S\,$ that is nearest the centroid of $S\,$.
The displacement $u_{3}\,$ is
    $u_{3}=-{\frac {(a^{2}-b^{2})Tx_{1}x_{2}}{\mu \pi a^{3}b^{3}}}$
* Displacements ($u_{3}\,$) in the cross section of an elliptical cylinder under torsion
# ASE Certification

To be respected and recognized as a mechanic you will want certification from the National Institute for Automotive Service Testing and certification is organized as follows

#### Automoblie and Light Truck

* A1 Engine Repair
* A2 Automatic Transmission & Transaxle
* A3 Manual Drive Train & Axles
* A4 Suspension & Steering
* A5 Brakes
* A6 Electrical/Electronic Systems
* A7 Heating & Air Conditioning
* A8 Engine Performance
* A9 Light Vehicle Diesel Engines

#### Collision

* B2 Painting & Refinishing
* B3 Non-Structural Analysis & Damage Repair
* B4 Structural Analysis & Damage Repair
* B5 Mechanical & Electrical Components
* B6 Damage Analysis & Estimating
* C1 Automobile Service Consultant

#### Truck Equipment

* E1 Truck Equipment Installation & Repair
* E2 Electrical/Electronic Systems Installation & Repair
* E3 Auxiliary Power Systems Installation & Repair
* F1 Alternate Fuels

#### Transit Bus Systems

* H1 Compressed Natural Gas(CNG) Engines
* H2 Diesel Engines
* H3 Drive Train
* H4 Brakes
* H5 Suspension & Steering
* H6 Electrical/Electronic Systems
* H7 Heating Ventilation & Air Conditioning (HVAC)
* H8 Preventive Maintenance & Inspection (PMI)
* L1 Advanced Engine Performance Specialist
* L2 Electronic Diesel Engine Diagnosis Specialist
* P1 Medium-Heavy Truck Dealership Parts Specialist
* P2 Automobile Parts Specialist
* P4 General Motors Parts Consultant

#### School Bus Certification

* S1 Body Systems & Special Equipment
* S2 Diesel Engines
* S3 Drive Train
* S4 Brakes
* S5 Suspension & Steering
* S6 Electrical/Electronic Systems
* S7 Air Conditioning Systems & Controls

#### Medium-Heavy Truck Certification

* T1 Gasoline Engines
* T2 Diesel Engines
* T3 Drive Train
* T4 Brakes
* T5 Suspension & Steering
* T6 Electrical/ Electronic Systems
* T7 Heating, Ventilation & Air Conditioning (HVAC)
* T8 Preventive Maintenance Inspection
* X1 Undercar Specialist Exhaust Systems Test
# ATP mentor training/Assessment of academic proficiency, motivation and goals

Print version as picture:
Please specify goals in vocational education and/or university education you might want to pursue.
* 1.
* 2.
* 3.
* 4.
* 5.
* 6.
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| Subject | Grade | How difficult is the subject? | How is your motivation? | How is your interest? | Do you want to improve? | Is the subject relevant for your academic goals? |
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| Wikiversity assistant teacher program: mentor training | http://en.wikiversity.org/wiki/ATP_mentor_training |
# ASCII art/Examples
## Original source

Art deleted from w:Wikipedia:Bad Jokes and Other Deleted Nonsense
# Complex socio-ecological systems/List of Participants

General introduction to the group
We have a very heterogeneuos and interdisciplinary group in this class, representing a diversity of students, research associates and faculty, including ecologists, anthropologists, sociologists, with people working in different parts of Latin America, Africa and the United States. We have also people accompanying and contributing from a far, including students in Guatemala, Costa Rica and Brazil and a non UF contributor from Germany interested in systems theory (Erkan). Some expectations expressed by participants relate to interest in pursuing a better understanding of how resilience might be applied to social contexts and, more broadly, to research in the social sciences. people also mentioned interest in exploring the applications of resilience thinking to climate change and natural disasters. Claudia mentioned her interest in understanding better the concept of robustness as applied to socio-ecological systems.
PRELIMINARY LIST OF PARTICIPANTS
1. Adam Watts – UF/SNRE
2. André Baby – UF – Visiting Scholar (ACLI)
3. Arika Virapongse – UF/SNRE/TCD
4. Caitlin Baird – UF/Anthropology
5. Claudia Monzon – UF/SNRE/TCD
6. Claudia Romero – UF/Biology
7. Danielle Watts – UF/SNRE
8. David Hanson – UF
9. Erkan Yilmaz – Germany
10. Flavia Leite – UF/Sociology
11. Franklin Paniagua – UF/SNRE/TCD
12. Grenville Barnes - UF/SFRC/TCD/African Studies
13. Jennifer Arnold – UF/SNRE
14. Jeremy Radachowsky – Guatemala
15. Julia Quaedvlieg – UF/Sociology
16. Laura Avila – UF/SNRE/TCD
17. Lisandro Juno – UF/ Visiting Scholar (ACLI)
18. Marliz – UF/Sociology
19. Raissa Guerra – UF/SNRE/TCD
20. Renato Farias – UF – Visiting Scholar (ACLI)
21. Ricardo Mello – UF/SNRE
22. Sam Schramski – UF/SNRE
23. Shelly Johnson – UF/SNRE/SFRC
24. Thaissa Sobreiro – UF/SNRE
25. Wendy-Lin Bartels - UF/Florida Climate Institute/ACLI
26. Xanic Rondon – UF/Sociology
# Complex socio-ecological systems/Systems modeling

Discussants: Andre and Ricardo
Janssen, Marco A. and Elinor Ostrom. 2006. Empirically Based, Agent-Based Models. Ecology and Society 11(2): 37.
Expore modeling tools such as NetLogo.
# Complex socio-ecological systems/Resilience-based management

Discussants: Flavia and Xanic
Scheffer, M, Willliam Brock and Frances Westley. 2000 Socioeconomic Mechanisms Preventing Optimum Use of Ecosystem services. Ecosystems 3: 451-471