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FTA of Main Landing Gear Systems

Paper Type: Free Essay Subject: Engineering
Wordcount: 1840 words Published: 28th Sep 2017

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Main Landing Gear System

For any aircraft landing gear is the undercarriage which support the craft when it’s 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 main landing gear configuration

Components of landing gear

The materials used to construct gear components are of great importance and are selected as per their properties

The main components of landing gear are:

Down-lock and drag brace

Retraction actuators, Rotational actuators


Forward trunnion braces

Metering pin extension

Rotational Lockpins

Aft braces

Oleo cylinder

Oleo piston

Axle beam fold and compensation actuator

Brake assembly

Tires and wheels

Sensing wheel

Axle beam assembly

Failure of lading gear

In this part we discussed about the Ductile and Brittle Failure, Stress Corrosion Cracking, Stress Rapture, Fatigue Cracking Failure Dynamic Failure, Landing gear Spring Failure and Wheel Failure. We analyzed these possible problem and construct suitable fault tree analysis in order to identify failure 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 failures:

There are 12 types of failures

Excessive, deflection , thermal shock, impact creep, relaxation, brittle fracture, ductile fracture, wear, spring, failure, corrosion, stress corrosion cracking, and various type of fatigue. On the basis of this problem that can occurs in main landing gear we construct the fault tree to identify the failure condition and met the requirements.

Materials for landing gear

The materials used for the landing gear are

  • High strength steel
  • Titanium
  • Aluminum

Failure mechanism of landing gear

Fatigue cracking failure

Mostly aircrafts and military experience serious damage and the fatigue.

Stress corrosion cracking

This SCC is caused by synergy between a corrosive environment and a mechanical tensile stree

Dynamic failure

When 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 failure

Here micro cracks acted as stress concentration as well initiation crack site leading the spring to fracture due to fatigue

Wheel failure

During landing wheel experienced a lot of pressure. Due to this over pressure that could damage

4.2 The fault tree of typical main landing gear system:

The main purpose of constructing fault tree is, in order to identify the possible failure of any system which can occurs in any manner.

In case of main landing gear, fault tree can be construct on the basis of two main conditions:

  1. Fault tree analysis for failure condition 511: one or both MLG fail to extend and down lock with false down and locked indication.
  2. 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 tree

My main object of constructing this fault tree is to identify the typical failure condition of main landing gear in order to save aircraft from it’s hazardous failure

As we need to take safety procedure for every parts of aircraft, landing gear is also aspect which cause aircraft in a dangerous mode

So in order to overcome 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 tree

In case 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 event by using event symbol 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 true lower 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 there are some cause to fail these two components, which we will define 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.

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In fault tree analysis we solved the problem and come to the conclusion with what if, and with the help of various Boolean logic 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 doesn’t work. But 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 example 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 role to make all possible decision for failure.

Determination of minimal cut set for fault tree analysis

Minimal cut 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 tree’s 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 together 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-sets

The minimal cut sets expression for the top event can be written as in the general form

T = 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 as

Mi = X1X2.,…… Xn

Here the X1 and X2 , etc represents the basic components failure in the fault tree analysis

I 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 is

T = A + B.C

Here 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 tree

The procedure to calculate the minimal cut sets of fault tree

T = E1.E2 being AND gate

E1 = A+ E3 being OR gate

E3 = B+C being OR gate

E2 = C+E4 being OR gate

E4 = A.B being OR gate

Substituting the top down first the expression of minimal cut can be expressed as below

T = (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.C

Here according to idempotent law, C.C = C, substituting this value to above equations

T = A.C + B.C + C + E4.A + E4.B + E4.C.

Again according to the law of absorption twice, A.C + B.C + C + E4.C = C

Then the above results become,

T = C + E4.A + E4.B

Hence, substituting for E4, applying the law of absorption twice,

T = C + (A.B).A + (A.B).B

= C + A.B

The minimal cut sets are thus C and A.B where, [C] is one- components and [A.B] is two-components minimal cut sets.


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