robin/Robin_PHD
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

.

Browse Source
This commit is contained in:
Robin Clark 2010-06-07 16:49:54 +01:00
parent a404a9210b
commit a9a4c254e1
1 changed files with 13 additions and 10 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

23
component_failure_modes_definition/component_failure_modes_definition.tex
View File

@ -249,9 +249,9 @@ take a given component $C$ and return its set of failure modes $F$.
$$ FM : C \mapsto F $$ $$ FM : C \mapsto F $$
\begin{definition} \begin{definition}
We can define a set $U$ which is a set of sets of failure modes, where We can define a set $\mathcal{U}$ which is a set of sets of failure modes, where
the component failure modes in each of its members are unitary~state. the component failure modes in each of its members are unitary~state.
Thus if the failure modes of $F$ are unitary~state, we can say $F \in U$. Thus if the failure modes of $F$ are unitary~state, we can say $F \in \mathcal{U}$.
\end{definition} \end{definition}
\section{Component failure modes:\\ Unitary State example} \section{Component failure modes:\\ Unitary State example}
@ -271,7 +271,7 @@ Thus because both fault modes cannot be active at the same time, the intersectio
$$ R_{SHORTED} \cap R_{OPEN} = \emptyset $$ $$ R_{SHORTED} \cap R_{OPEN} = \emptyset $$
therefore therefore
$$ FM(R) \in U $$ $$ FM(R) \in \mathcal{U} $$
We can make this a general case by taking a set $F$ (where $f_1, f_2 \in F$) representing a collection We can make this a general case by taking a set $F$ (where $f_1, f_2 \in F$) representing a collection
@ -283,7 +283,7 @@ We can say that if any pair of fault modes is active at the same time, then the
unitary state: unitary state:
we state this formally we state this formally
\begin{equation} \begin{equation}
\forall f_1,f_2 \in F \dot ( f_1 \neq f_2 \wedge \mathcal{ACTIVE}({f_1}) \wedge \mathcal{ACTIVE}({f_2}) ) \implies F \not\in U \forall f_1,f_2 \in F \dot ( f_1 \neq f_2 \wedge \mathcal{ACTIVE}({f_1}) \wedge \mathcal{ACTIVE}({f_2}) ) \implies F \not\in \mathcal{U}
\end{equation} \end{equation}
@ -293,7 +293,7 @@ we state this formally
% \end{equation} % \end{equation}
That is to say that it is impossible that any pair of failure modes can be active at the same time That is to say that it is impossible that any pair of failure modes can be active at the same time
for the failure mode set $F$ to exist in the family of sets $U$. for the failure mode set $F$ to exist in the family of sets $\mathcal{U}$.
Note where that are more than two failure~modes, Note where that are more than two failure~modes,
by banning any pairs from being active at the same time by banning any pairs from being active at the same time
we have banned larger combinations as well. we have banned larger combinations as well.
@ -382,8 +382,8 @@ for each component in the functional group under analysis.
For example: were we to have a simple functional group with two components R and T, of which For example: were we to have a simple functional group with two components R and T, of which
$$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$. $$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$.
This means that a functional~group $FG=\{R,T\}$ will have a component failure modes set % $FM_{cfg} $ This means that a functional~group $FG=\{R,T\}$ will have a component failure modes set
of $FM_{cfg} = \{R_o, R_s, T_o, T_s, T_h\}$ of $FG_{cfg} = \{R_o, R_s, T_o, T_s, T_h\}$
For a cardinality constrained powerset of 2, because there are 5 error modes ( $|{FG_{cfg}}|=5$), For a cardinality constrained powerset of 2, because there are 5 error modes ( $|{FG_{cfg}}|=5$),
applying equation \ref{eqn:ccps} gives :- applying equation \ref{eqn:ccps} gives :-
@ -432,9 +432,12 @@ where :
that are members of the functional group $FG$ that are members of the functional group $FG$
i.e. $ \forall j \in J | C_j \in FG $ i.e. $ \forall j \in J | C_j \in FG $
\item Let $|{C}_{j}|$ \item Let $|{C}_{j}|$
indicate the number of mutually exclusive fault modes each component has indicate the number of mutually exclusive fault modes of each component
\item Let $SU$ be a set of unitary state failure modes from the functional group \item Let $FG_{cfg}$ be the collection of all failure modes
nder analysis $SU = FM(FG)$ from all the components in the functional group.
\item Let $SU$ be a set of failure modes from the functional group,
where all contributing components $C_j$
are guaranteed to be `unitary state' i.e. $(SU = FG_{cfg}) \wedge (\forall j \in J | C_j \in \mathcal{U}) $
\end{itemize} \end{itemize}
%} %}