FTA of Main Landing Gear Systems

FTA of Main Landing Gear SystemsFTA OF MAIN LANDING GEAR SYSTEMSMain Landing Gear SystemFor any aircraft landing paraphernalia is the undercarriage which corroboration the craft when its not flying, until it to take off and during landing. Landing gear hold net weight of whole aircraft during taxing without any damage.Fig A380 master(prenominal) landing gear configurationComponents of landing gearThe materials used to construct gear components are of great importance and are selected as per their propertiesThe of import components of landing gear areDown-lock and drag braceRetraction actuators, Rotational actuatorsTrunnionForward trunnion bracesMetering pin extensionRotational LockpinsAft bracesOleo piston chamberOleo pistonAxle beam fold and compensation actuatorBrake assemblyTires and wheelsSensing wheelAxle beam assembly blow of lading gearIn this part we discussed about the Ductile and Brittle Failure, Stress Corrosion Cracking, Stress Rapture, Fatigue Cracking Failure Dynam ic Failure, Landing gear Spring Failure and Wheel Failure. We analyzed these possible problem and construct suitable error tree diagram diagram analysis in order to discern ill condition in brief.The objective of constructing the fault tree is to investigate and analyze the possible failures and different components and systems of the landing gear with their consequences and solutions.In the case of mechanical failuresThere are 12 types of failuresExcessive, deflection , thermal shock, impact creep, relaxation, brittle fracture, ductile fracture, wear, outflow, failure, corrosion, stress corrosion cracking, and various type of fatigue. On the basis of this problem that place occurs in briny landing gear we construct the fault tree to identify the failure condition and met the requirements.Materials for landing gearThe materials used for the landing gear are richly strength steelTitaniumAluminumFailure mechanism of landing gearFatigue cracking failureMostly aircrafts and militar y experience grave damage and the fatigue.Stress corrosion crackingThis SCC is caused by synergy between a corrosive environment and a mechanical tensile streeDynamic failureWhen aircraft land on the ground tricycle landing gear and load affected by the ground/pavement response are distributed on the and can cause problem.Landing gear spring failureHere micro cracks acted as stress concentration as well initiation crack order leading the spring to fracture due to fatigueWheel failureDuring landing wheel experienced a lot of pressure. Due to this over pressure that could damage4.2 The fault tree of typical main landing gear systemThe main purpose of constructing fault tree is, in order to identify the possible failure of any system which can occurs in any earthner.In case of main landing gear, fault tree can be construct on the basis of two main conditionsFault tree analysis for failure condition 511 one or both MLG fail to acquit and pour down lock with dark down and locked i ndication.Fault tree analysis for failure condition 511 one or both MLG fail to extend and down lock.Between these two failure conditions of main landing gear, I choose to analyzed fault tree analysis for failure comdititon511 one or both MLG fail to extend and down lock with false down and locked indication.The analysis based on fault treeMy main object of constructing this fault tree is to identify the typical failure condition of main landing gear in order to save aircraft from its hazardous failureAs we inquire to take safety procedure for every parts of aircraft, landing gear is also aspect which cause aircraft in a dangerous modeSo in order to get the better of this failure of landing gear and for the safety of aircraft, I finally decide and came to know the purpose of failing landing gear and what can be the cause for this.With the help of fault tree analysis, we analyzed the each and every problem inside the landing gear and try to overcome by constructing fault treeIn cas e of LH or RH or both MLG fail to extend and down lock with false down and locked indication, pilot receive alert. So while construction fault tree we take as a main typeface by utilize event symbolisation as shown in fault tree.After we decided main event than we analyzed for the cause of it by using the Boolean symbol called AND gate. We use AND gate so that we could analyze all next truthful put down level condition which can possibly occur. Possible cause of man landing gear are false LG down position indication and LH or RH MLG fail to extend and down lock. Still in that location are some cause to fail these two components, which we ordain doctor by Boolean logic symbol call OR gate. The purpose of OR gate is to identify among of various possible problem if any of one or more of the next lower level event are true for failure of above condition.In fault tree analysis we solved the problem and come to the conclusion with what if, and with the help of various Boolean logi c symbols. The main event that I mention in the top is not only the problem that cause landing gear to fail but also the lower level parameters which fails landing gear fails. When we look from external and if landing gear is not extend during landing than we conclude that the landing gear doesnt work. unless besides engineer who work in the field of safety assessment, other observer will just guess just the landing gear fail. We never thought of what cause the landing gear fails. As per landing gear consists various components so the probability of failure also high. If small components for physical exertion spring fails, than the landing gear fail.so in order to overcome these all possible failure fault tree is constructed.Among of various symbols and representation, AND and OR gate plays vital portion to make all possible decision for failure.Determination of stripped rebuff set for fault tree analysisMinimal repel set is define as a combination of primary events sufficient for the top event, on other words intersection of primary event. The main objective of representing a fault tree in terms of various Boolean equations is that these equations can then be used to determine the fault trees minimal cut sets and minimal path sets. While we obtain the minimal cut set, the quantification of the fault tree is more or less straightforward.For every fault tree will consists of finite number of minimal cut sets that are unique for that event. There are two kind of minimal cut sets which can occurs the top event to occur. One components minimal cut sets, if there are any, represent those single failures that will cause the top event to occur. Whereas two-components minimal cut sets represents the double failures that in concert will cause the top event to occur. Similarly for an n- components minimal cut set, all n components in the cut set must fail in order for the top event to occur.The calculation of minimal cut-setsThe minimal cut sets expression for the top event can be written as in the general formT = M1 + M2 ++Mk,Here the terms T is the top event and Mi are the minimal cut sets.Whereas for the each minimal cut sets for n-components minimal cut set can be represents asMi = X1X2., XnHere the X1 and X2 , etc represents the basic components failure in the fault tree analysisI represents my above failure events as ABC, T is top event caused by all lower level events. Here I want to mention the example of top event expression isT = A + B.CHere A, B and C are components failures. In this example A represents the one-components minimal cut set whereas B.C represents the tow-components minimal cut set.In order to determine the minimal cut sets of a fault tree, the tree should be translated first to its equivalent Boolean equations.Here I want to mention the example of calculation the minimal cut set to my fault treeThe procedure to calculate the minimal cut sets of fault treeT = E1.E2 being AND gateE1 = A+ E3 being OR gateE3 = B+C bei ng OR gateE2 = C+E4 being OR gateE4 = A.B being OR gateSubstituting the top down first the expression of minimal cut can be expressed as belowT = (A+E3) . (C+E4)= (A.C) + (E4.C) + (E4.A) + (E3.E4)Substituting for E3,T = (A.C) + (B+C).C + E4.A + (B+C). E4= A.C + B+C + C.C + E4.A + E4.B + E4.CHere according to idempotent law, C.C = C, substituting this value to above equationsT = A.C + B.C + C + E4.A + E4.B + E4.C. over again according to the law of absorption twice, A.C + B.C + C + E4.C = CThen the above results become,T = C + E4.A + E4.BHence, substituting for E4, applying the law of absorption twice,T = C + (A.B).A + (A.B).B= C + A.BThe minimal cut sets are thus C and A.B where, C is one- components and A.B is two-components minimal cut sets.