Flap Strut Fairing System Expluatation and Critical Path Method Use

Network planning methods can be used in logistics as well as engineering sciences to exhaust and optimise work planning, order and failure control mechanisms and cost control, each as a separate project with the failure characteristics of the composite material. Since the failure problem is related to the operation of the logistics department and the information of the engineering department in this network, we will look at all the related processes both from the logistics’ point of view and also from the technical aspect of the engineers.

In the network process, both the starting and ending positions can be represented, which gives a more or less satisfactory result.

There are several ways in which, in addition to the use of the CPM, the result is influenced by widespread methods, such as the Monte Carlo method, Markov theory and others. However, there are some critiques for CPM and Program Evaluation and Review Technology Method (PERT).

In PERT, there are some assumptions for simplifying the model, for example use of the distribution. Also, to provide an appropriate distribution for activity times, we need historical data. PERT method is less used in the aviation as CPM.

To use the PERT model, there are some assumptions to simplify the model, one of them is to use distribution. To provide an appropriate distribution, we need to have historical data, but not always they are available and, in these cases, we are using human judgement, not the stochastic assumptions. To solve problems better, sometimes, we can use the fuzzy set theory instead of the probability theory. In this case, we can use the fuzzy numbers. One of the most important tools available to the scheduler during a maintenance outage is the CPM of scheduling.

CPM is the quickest and the most accurate method available for scheduling and managing large, complex work packages, optimising and modifying schedules of disruptions of work packages. Critical path scheduling is a technique used for illustrating activity sequences when the actions are made step by step. Many activities, such as receiving a request from the technical department, placing and controlling logistics orders and concluding contracts with the supplier, can take place simultaneously, while there is a set of activities that cannot be started until the initial activities are completed. The relationship between parallel and priority activities is represented by a network graph. Initially, for each of the events in the schedule, the occurrence time of the earlier events is determined, as well as the occurrence time of the later events is determined and the spare time is calculated for each stage. The total delivery time is calculated and the critical path connects these stages [1].

The task of creating the given grid schedule is as follows: to develop a grid schedule for the failure of the flange rail cladding, so that the component procurement process ensures everything in the most efficient and fastest way by reducing the influencing factors, which would ensure the requirements of a safe flight.

In order to be able to determine the critical path, their designations are given below: T₀ – the path of the first activity or the optimistic time, T_M – the possible time, Tp – the pessimistic time and Te – the possible operating time of each activity is shown in Table 1.

Calculation of the critical path:

0–1–2–3–4–5–6–7–8 = 6 + 3.8 + 4.3 + 6 + 7.8 + 5.8 + 5.1 + 7 + 6.8 = 52.6 (Critical path network)

Fig. 1.

The network graph shows the activities with arrows if they consume time and resources, and shows which activity must be completed to move on to the next activity. In events, the result of the previous action and the beginning of the following action are recorded in a unit of time, while events do not consume time or resources, they only record a certain state. Since a network graph represents activities in the order in which they are executed, then steps 2–3 cannot be started until steps 0–2 are completed.

The single biggest factor in determining a flange rail-facing problem is the path, or the sequence of activities that moves through the schedule from start to finish.

Evaluating the graph, it can be concluded that several paths, 0–1–2–3–4–5–6–7–8 and 0–1–2–7–8, are possible, which differ in length expressed in time units. This length is obtained by adding up the time grain of the actions along the entire length of the path.

The length of the longest path indicates the total time required to execute the network schedule for the supply of the flange cladding components if the process is bottlenecked. If the specified time consumption limits are not respected on the longest route, or deliveries and inspections are delayed, all other activities are delayed in terms of time unit or resource consumption.

It is for these reasons that the longest path of the schedule is called the critical path, and the activities on this path are called the critical activities. The other paths, which are allowed with time delays for the shortest paths, are called reserve time.

As a typical error in the schedule, there may be duplication of one or more activities. In this case, I looked at the actions so that they do not duplicate. If, however, one of the actions is duplicated, then I use fictitious actions to determine the interrelationship of the events and their duration in the time unit is equal to zero. Next, we will consider how to calculate the probability of lining the flanges, how long it will take and with what probability we will be able to successfully complete the delivery and replacement project.

The flap is closed by the system process action plan and probabilistic determination.

To calculate the probability, we must first calculate the standard deviation according to the formula. 1 Pt=Tp−T06, $${P_t} = {{{T_p} - {T_0}} \over 6},$$ where P_t – time to performing job, T_p – pessimistic time, T₀ – possible time.

To calculate the σ – probability, we must first calculate the P_n – standard deviation, according to the formula below: 2 σ=Pn. $$\sigma = \sqrt {{P_n}} .$$

Looking at this graph and the Table 2, we can calculate the probability of the longest path and it will be 5.64 days, while the standard deviation σ will be √ (5.64 = ) 2.37 days. Table 2 shows the probability of a path segment, which ultimately shows the total path deviation if all critical path conditions are met [2].

Therefore, we can clearly see if we have to do additional inspections. If we do not detect the damage to the flange rail coating in time, then additional time and inspection steps come. In my case, it can take up to 8 days to wait for a response from the manufacturer on the next steps, resulting in delays in ordering parts and increased aircraft downtime costs.

In order to not create such a situation, when the damage of the flange rail material shell is the reason for the increase of the total path, then let’s take a deeper look at how the structure of the component itself and the mathematical justification of the technical condition of the structure can reduce the length of the critical path.

As we know, from the basics of the theory of probability, we can use several theories in connection with composite materials, for example, at the beginning, if we do not know the state of crack development, we classify it as a structurally significant damage and a vectoral value is equated to this damage, which is a randomness vector (T_d,T_c), where T_c is taken as the critical failure index for the component, Td is the service time when a failure, for example a crack is contacted to the material by looking at the development of all cracks to determine the t_SL of the composite material, or the specific time when it cannot be used due to system failure [1].

In the case mentioned above, we use p-set or p-series function calculation. We can also use the Monte Carlo method in the calculation of the flange rail structure, which defines the situation when we cannot determine the damage size of the flange rail structure, because no structural inspections have been performed in the time period, T_d,T_c.

These two methods are mentioned above in the descriptive part of the theory, because apart from these two application methods, we can also use the third method.

The third method, Markov is increasingly used not only in aviation, but also in other industries to perform in-depth calculations on the dependence of one process on another, which will affect the failure of the system over a period of time and, if not paying attention to the defect, can lead to a disaster.

This method offers us opportunities to determine the nature of failure and the frequency of component replacement if we use routine maintenance as a basic monitoring model. The graph and Table 3 show the most frequent types of failure, but they only create a condition for our system when a part is replaced and do not cause downtime, do not pose a threat of the need for a longer period of time for the work process [3].

At first look, the n + 1 state represents the maintenance of the aircraft in the corresponding interval between two consecutive inspections, while the three additional states represent the termination of the aircraft operation due to the end of successful operation, when the specific maintenance time has expired in the event of defect elimination. Therefore, we can divide:

E_i aircraft service time, t, with i-th, inspection interval, t∈(t_(i – 1)), t_i, i = 1,2,..(n + 1);

All these processes can be viewed as one whole matrix. Denote the failure to detect i during the test as v_i; failure probability in the service time interval t∈(t(i – 1), where q_i and probability of successful service continuation as u_i. Since all these things are completed, then u_i + v_i + q_i = 1.

We assume that the model as the defect is resolved and no additional operating measures are required t_SL, unless no defects are found at the moment of time, t_SL. Inspection interval, (n + 1) th, with condition, E(n + 1), does not change the reliability, but we do it to know the state of the aircraft, t_(i – 1). There was no failure, and no failure was detected during the inspection period, t(i – 1). But it is enough that it is equal to P(T_d > t(i – 1)).

The transition probability matrix for this process can be constructed as mentioned above. E_i can be reached only when the condition that u_i + v_i + q_i =1 is fulfilled and the cycle has ended without stalling.

This means when, as an example, we can accept works performed in a unit of time and as a result of which we have as per Table 3.

Below is information in a graphic form, about the most frequently replaced parts for the mechanical part of the structure, which are related to the operation of the flange rail linings and do not hinder the nature of the operation.

Therefore, the condition that u_i + v_i + q_i =1 is fulfilled and the cycle is completed without a hitch.

Fig. 2.

According to the given information, we can say that more often we change PN E0091-10-250NN braid, then PN D5754052520200A cover pitot bolt, PN D5754057420200A cover and PN D5754057720000 seal, which means that these parts of the mechanism are quickly damaged and in case of failure, we can predict that.

Since the cladding of the flange rails consists of a shell of carbon fibre-reinforced polymer material, we also need to look at the evolution of the defect in this material. After the calculation of the critical path, the untimely failure of the flange rail cladding complicates the work execution.

Let us consider the case of flanged rail cladding, n_c = 1, where the carbon fibre-reinforced polymer can be considered as a series system with some internal links damaged and two links with and without defects. Let us denote the random number of damage links, (K)l,0 ≤ K_l ≤ n_l, by strength, Fy(x); we say that K_l is of Y-type and let us denote it by Fz(x); the strength, (n_L – K_l), is denoted without defects say Type Z. We assume that there are two ways to calculate this process. In the first part of the process, we use K_l connected to Y-type. They can appear before maintenance according to some prior distribution, πL = (πL1, πL2,…., π(L(nL + 1)), where πLk = P(K_L = k – 1), or in the maintenance process, where the stress level, L_I, exceeds the fault excitation voltage, F_K(x). Then, the second stage and the process of accumulation of elementary damage in the transverse direction until the failure of the specimen occurs. Two levels of differences between L_I with and without defects, and three differences in strength describe accuracy groups in these groups from the six types of corresponding probability structures. We further assume that in Table 4.

In the case of A-type, it is assumed that the difference between the bond strength of Y- and Z-type relativity is small, and failure of the specimens can be caused by any type of bond failure. In B-type, it is assumed that the difference between the bond strength of Y- and Z-types is very large and we need to consider the bond strength of type Z only when there is no Y-type bond. In some ways, the description of the B-type group is the limit of the A-type group, when the difference between the strength of Y- and Z-type bonds increases. Indeed, any description different from A₁ is a kind of approximation to the description of the group, we think that the usefulness of considering this different set is determined by the difference in materials, different requirements for calculation accuracy and different test sample sizes.

It also determines the crack or component failure in different materials and their stress concentration patterns [4].

Sometimes, it is acceptable to assume that sample failure can only occur in items with defects. This leads to the structures shown below. It is very important that in this case, we know the L_I defects. But, some limitations appear. For K_L, we should only use a probability mass function such that P(K_L = 0) = 0, as in the example below, where the function AC calculates X = min, which depends on the strength of the Y links. And, if BC is calculated, then Y links should be greater than 0 value.

It is easy to connect the strength of the samples and the strength of the individual, L_I . As an example, we can use A₁. and B₃. We can use the formula below:

where {p_k, k = 0,1,….,n_L} is the probability distribution, K_L. This probability distribution can be considered as a function of the applied nominal voltage, i.e. it is a binomial distribution b(k,pL,nL)=pLk(1−pL)nL−knl!/k!(nL−k)!$b(k,{p_L},{n_L}) = p_L^k{(1 - {p_L})^{nL - k}}nl!/k!({n_L} - k)!$, then it can be defined as p_L = F_k(x), kur F_k(x), stress point of the start of the defect. In this case, we can add F. And, we get p0=(1−Fk(x))nL${p_0} = {(1 - {F_k}(x))^{{n_L}}}$. This means that if carbon fibre-reinforced polymer undergoes routine maintenance without noticing cracks in the shell, an initial stress point develops, which then turns into a problem starting point. The process of gradual (underload) accumulation of defects and serial system failure can be characterised by a Markov chain. Then, we use the additional letter M for the designation: M_A1, M_A2 and so on. If the monotonic tensile loading process is described by an increasing (to infinity) sequence {x₁, x₂, X_n…}, then the number of Y links and sample strength are random functions of time, K_L(t) and X(t). Let’s use M_A1 as an example. We have (t) = min(Y₁,Y₂,…,Y_KL(t), Z₁, Z₂,…,Z_nL – K_L(t)). We assume that the Markov chain (n_L + 2) defines Markov chain i denotes (i = 1) of Y-type, where i = 1,…, n_L + 1, to determine the i(n_L + 2) absorption state corresponding to the fracture of the sample. The state change process and the corresponding process, K_L(t), are shown by the matrix of transition probabilities: 8 P=[ p11p12p13p23……P1(nc+1)0p22p23p24……P2(nc+1)00p33p34……P3(nc+1) ] $$P = [\matrix{ {{p_{11}}} \hfill {{p_{12}}} \hfill {{p_{13}}} \hfill {{p_{23}} \ldots \ldots {P_{1({n_c} + 1)}}} \hfill \cr 0 \hfill {{p_{22}}} \hfill {{p_{23}}} \hfill {{p_{24}} \ldots \ldots {P_{2({n_c} + 1)}}} \hfill \cr 0 \hfill 0 \hfill {{p_{33}}} \hfill {{p_{34}} \ldots \ldots {P_{3({n_c} + 1)}}} \hfill \cr } ]$$ 9 P=[ 0.120.130.140.5900.230.280.79000.341 ] $$P = [\matrix{ {0.12} \hfill {0.13} \hfill {0.14} \hfill {0.59} \hfill \cr 0 \hfill {0.23} \hfill {0.28} \hfill {0.79} \hfill \cr 0 \hfill 0 \hfill {0.34} \hfill 1 \hfill \cr } ]$$

In the Markov chain matrix shown above, P is a function t = 1, 2, whereas the primary use K_L and it is taken as a vectoral quantity.

Defects can appear in the first stage before the manufacturing process (technological defect) and during operation. We can naturally assume a binomial of the number of defects for a certain number of elements, n. K_L, b(k,pL,nL)=pLk(1−pL)nL−knl!/k!(nL−k)!$b(k,{p_L},{n_L}) = p_L^k{(1 - {p_L})^{nL - k}}nl!/k!({n_L} - k)!$, where p_L is the probability that a defect appears in one link. Let’s consider that in this case, B3 p0=(1−pL)nL${p_0} = {(1 - {p_L})^{{n_L}}}$.

If n is large enough, the Poisson’s law is usually used as an approximation of the binomial distribution: 10 pk=p(k,λ)=exp(−λ)λkk $${p_k} = p(k,\lambda ) = {{\exp ( - \lambda ){\lambda ^k}} \over k}$$ 11 pk=p(k,λ)=exp(−2)211=0.27 $${p_k} = p(k,\lambda ) = {{\exp ( - 2){2^1}} \over 1} = 0.27$$

The parameters, p_L = 2 and λ = 2, can be taken regardless of the technological defect or p_l = 2, λ = 2, where F_k(x) is the fault initiation voltage at the beginning, but it has its own limitations [4].

We make an additional assumption that the distribution of the number of cross-sections with defects is a conditional Poisson distribution with a zero-probability condition that this number is zero. We refer to this distribution as the multiple system failure section. We choose an assumed size, K_L, for a random number of broken cross-links. 12 P(KL=1)=pk=e−λ1−e−λ∗λkk! $$P({K_L} = 1) = {p_k} = {{{e^{ - \lambda }}} \over {1 - {e^{ - \lambda }}}} * {{{\lambda ^k}} \over {k!}}$$ 13 P(KL=1)=pk=5−21−5−2∗111!=0.66 $$P({K_L} = 1) = {p_k} = {{{5^{ - 2}}} \over {1 - {5^{ - 2}}}} * {{{1^1}} \over {1!}} = 0.66$$ k = 1,2,…. We again denote the strength of this damaged multi-link section by F_y(x).

Then, after that sample strength: F(x)=1−∑k=1∞pk(1−Fy(x))k=1−e−λFy(x)1−e−λ$F(x) = 1 - \mathop \sum \limits_{k = 1}^\infty pk{(1 - Fy(x))^k} = {{1 - {e^{ - \lambda {F_y}(x)}}} \over {1 - {e^{ - \lambda }}}}$, let us denote a specific assumed feature of this distribution by K_L. In this case, it is easy to specify the values of K_L below E(K_L) = λ/(1 – e^(–λ)) we assume that E(K_L) = 2/(1 – 5^(–2)) while the value of a random variable with a normal Poisson distribution is equal to λ, or in our case 2, remember, that the natural assumption: λ = λ₁ × L, where λ₁ is the defect intensity (frequency) or λ = 5 × 10, L is the length of the sample. This means that E(K_L), is not proportional to the length of the sample as opposed to models where p_k is defined by the usual binominal or position distribution. For models based on the Markov chain theory for priority distribution (technological) defect numbers, K_L can again use Binomial or Poisson elation p(k, λ), should be distributed at point n; this means that pk = p(k, λ), where k ≤ n; p(n + 1) = 1 – ∑1^n = p_k and p_k = 0, where k ≥ n + 2. In my case, you can use p_k = p(0.5; 2), where k ≤ n or 0.5 ≤ 1; p1+1=1−∑11□∗pk${p_{1 + 1}} = 1 - \mathop \sum \limits_1^1 * {p_k}$, and p_k = 0, where k ≥ 1 + 2. This procedure is necessary because it is used in the Markov chain theory. As a primary distribution, π = (1,0,..0) means that the failure rate is zero at the very beginning, but a positive feature develops as the crack or fracture slowly develops. This means that if we do not follow the development of the crack over time, we can reach a breaking point, which is the basis for replacing the entire shell, but to predict the stages of development, as mentioned above, we use the Markov model [4].

Therefore, looking at the above example, it can be concluded that it is very important to follow the condition of the material directly and to perform all inspections and maintenance as part of the technical operation or as recommended by the manufacturer. If we do not comply with these conditions, we will have to lose time and money due to the late identification of the defect. If we are using flap track fairing assembly and component maintenance tasks according to Structural Repair Manual (SRM), we can step by step replace the Moveable-Fairing Cover Retainers as per the steps below.

Put the WARNING NOTICE(S) in position in the cockpit to tell persons not to operate the flaps or slats.

Fig. 3.