Landing Gear Analysis

Structures - Materials Project Work
By
NARASIMHA PRASAD Nagesh & PENKULINTI Sai Sreenivas
Submitted to
MAHAJAN Prasad Baliram

ABSTRACT
Landing gear is one of the critical subsystems of an aircraft. The need to design landing gear with
minimum weight, minimum volume, high performance, improved life, and reduced life cycle cost have
posed many challenges to landing gear designers and practitioners. Further, it is essential to reduce the
landing gear design and development cycle time while meeting all the regulatory and safety requirements.
Many technologies have been developed over the years to meet these challenges in design and
development of landing gear. This project presents the study of lower brace of a CANADAIR CL-215
landing gear.
CONTENTS:
List of figures. & tables.
1. Introduction 1
2. Aim and Objective 1
3. Study and Analysis 3
3.1 Kinematic load estimation 3
3.2 Stress estimation by RDM method 6
3.2.1 Cut method, assumptions and equations 7
3.2.2 Global coordinate system to local coordinate system 8
3.2.3 Forces in beam 10
3.2.4 Equivalent Von Mises Stress 10
3.2.5 Code and Results 12
3.3 Stress estimation by ABAQUS (FEM) 13
4. Comparison between RDM & FEM 15
CONCLUSION 16
REFERENCE 17
APPENDIX 18

LIST OF FIGURES & TABLES.
 Fig. 1: Canadair CL-215.
 Fig. 2: Canadair CL-215 Landing gear Linking graph (a) and physical representation (b).
 Fig. 3: Landing Gear lower brace element representation (a) and links configurations (b).
 Table 1: Basic link systems.
 Table 2: Link configuration and tensors.
 Table 3: Element force and moment balance.
 Fig. 4: 2D diagram to obtain elemental length.
 Fig. 5: MATLAB code for solving simplified system of equation.
 Fig. 6: Top and side views of element BF cross section.
 Fig. 7: Front view of element BF.
 Fig. 8: Cut method to obtain moment & loads.
 Fig. 9: Transformation in z axis 𝜃1 =
𝜋
2
.
 Fig. 10: Transformation in y axis 𝜃2 = 37.058°
.
 Fig. 11: Transformation in z axis 𝜃3 = 𝜋.
 Fig. 12: Stress-Strain curve of ductile material.
 Fig. 13: MATLAB code for Von Mises equation solving.
 Fig. 14: Graph plotting x coordinate with stress.
 Fig. 15: Constrained couple applied on surfaces.
 Fig. 16: Load and Boundary condition application.
 Fig. 17: Meshing of part (Within limits of student version).
 Fig. 18: Final Von Mises contour representation.

Structures-Materials Project
Narasimha Prasad Nagesh & Penkulinti Sai Sreenivas Page | 1
I. Analysis of CANADAIR CL-215 retractable landing gear.
1. INTRODUCTION
Of the many components that must be defined in an aircraft, the Landing gear is one of the most crucial.
It must be placed in the right position and proper structural orientation in order to ensure smooth and
effective realization of its purpose. The landing gear supports the craft when it is not flying, allowing it to
take off, land, and taxi without damage. Faster aircraft have retractable undercarriages, which fold away
during flight to reduce drag.
Fig. 1: Canadair CL-215.
The Canadair CL-215 is a twin-engine, high-wing general-purpose amphibious aircraft. It features an
atypically spacious fuselage for an amphibian, which is designed to accommodate for the operational
needs of various roles that the aircraft was developed to perform. The CL-215 can be used as an airborne
firefighting platform, in which capacity it is used as a water bomber; it has been claimed to be the first
aircraft designed to withstand the severe aerodynamic and hydrodynamic loads imposed by such usage.
Beyond the water bomber role, the CL-215 was designed for use in other capacities, such as a search and
rescue platform, passenger transport, and freighter; for this purpose, the cabin can be configured in various
different ways, including a flexible combi configuration. Under typical operations, these applications
would harness the aircraft's ability to land and take-off from the water, the hull having been designed to
enable its use upon the open seas.
2. AIM AND OBJECTIVE
A landing gear assembly consists of various components viz. Lower side stay, Upper side stay, Locking
actuators, Extension actuators, Tyres and Locking pins to name a few. Each unit having a specific
operation to deal with, in this project the main unit being studied is the lower brace.

Fig. 2: Canadair CL-215 Landing gear Linking graph (a) and physical representation (b).
The aircraft’s built up mass is 17100Kgs which should be supported by the landing gear during operation
which may vary according to operational conditions. More specifically the lower brace is reconstructed as
in the Fig. 3(a). Few assumptions are considered to make the problem simple which are,
1. All the joints are perfect with no friction.
2. No distributed loads across individual elements.
3. Joints are permanent.
4. Whole unit follows linear elastic behaviour.
Fig. 3: Landing Gear lower brace element representation (a) and links configurations (b).
The primary objective is to analyse stresses in any one element of the lower brace unit using strength of
materials or RDM method and Finite Element Method (FEM) and compare the both. Using the obtained
data a suitable material is proposed for the component. The approach used here is to study the overall
behaviour of the element by taking up each aspect, finally summing up the total effect of all the aspects in
the functioning of the element.
The step wise approach is as below,
a. Evaluation of loads
b. Evaluation of loads using Strength of materials concept.
c. Evaluation of loads using Finite Element Methods.
d. Evaluation of Material Selection.

3. STUDY AND ANALYSIS
3.1 KINEMATIC LOAD ESTIMATION.
To make a load calculation for any mechanical component one should also obtain the Degree of static
indeterminacy. Number of unknowns are equal to number of equations hence the system is externally
determinate.
By the Fig. 2(a,b) the tensor for each link can be tabulated. The basic three types of links are as follows,
Link Torseur Link diagram
Pivot [T] = [
𝑿 𝟎
𝒀 𝑴
𝒁 𝑵
]
Rotule [T] = [
𝑿 𝟎
𝒀 𝟎
𝒁 𝟎
]
Spherical Cylinder [T] = [
𝟎 𝟎
𝒀 𝟎
𝒁 𝟎
]
Table 1: Basic link systems.
Inspecting the lower brace of the structural system one can make comparison with one of the three basic
links. This is done in order to ease the problem solving technique where all loads and moments should be
defined with respect to their direction.
Sl No. Link. Configuration Tensor.
1 B(4-1) Pivot
[
𝑋41 0
𝑌41 𝑀41
𝑍41 𝑁41
]
2 C(4-2) Spherical Cylinder
[
0 0
𝑌42 0
𝑍42 0
]
3 C(4-3) Rotule
[
𝑋43 0
𝑌43 0
𝑍43 0
]
4 D(0-3) Rotule
[
𝑋03 0
𝑌03 0
𝑍03 0
]
5 E(0-2) Pivot
[
𝑋02 0
𝑌02 𝑀02
𝑍02 𝑁02
]

6 F(0-1) Pivot
[
𝑋01 0
𝑌01 𝑀01
𝑍01 𝑁01
]
Table 2: Link configuration and Tensors.
Being defined all the tensors for each links, considering these a system of equation can be formulated to
solve for the unknown forces and momentum values. In order to do so we need to consider the ground
force tensor as below,
TA = [
4𝑚𝑔 0
0 0
3𝑚𝑔 0
] where, m = 8550Kgs. And g = 10 m/s2
.
With all the data available consider each element to give the Force and Moment balance which can be
achieved by the summation of the Tensor values and equating those to the external force or moment. In
turn forming the system of equations required to obtain unknowns,
Force balance by,
F1 + F2 = 0
Moment balance by,
M + Lx(F) = 0
Cross products of length and forces are solved utilizing an online tool (link provided in reference).
Element Force Balance Moment Balance (Moment acting)
BF (1)
X41 + X01 = 0
Y41 + Y01 = 0
Z41 + Z01 = 0
BFy×Z01 – BFz×Y01 = 0
M41 + M01 + BFz×X01 = 0
N41 + N01 - BFy×X01 = 0
CE (2)
X02 = 0
Y42 + Y02 = 0
Z42 + Z02 = 0
CEy×Z01 – CEz×Y01 = 0
M02 + CEz×X01 = 0
N02 - CEy×X01 = 0
CD (3)
X43 + X03 = 0
Y43 + Y03 = 0
Z43 + Z03 = 0
CDy×Z01 – CDz×Y01 = 0
CEz×X01 = 0
- CDy×X01 = 0
A BC (4)
4mg - X43 – X41 = 0
-Y41 – Y43 - Y42 = 0
3mg - Z41 + Z43 - Z42 = 0
ABy×Z41 -ABz×Y41 +ACy×Z42 -ACz×Y42
+ACy×Z43 - ACz×Y43 = 0
M41 + ABz×X41 + ACz×X43 = 0
N41 - ABy×X41 - ACy×X43 = 0
Table 3: Element force and moment balance.
By solving the set of equations in Table 2 the 23 unknown force and moment values can be obtained. The
respective lengths for the given elemental distance are obtained from the provided drawing named
“Maquette Canadair CL215 – 3” the same is provided in the Fig. 4.

Fig. 4: 2D diagram to obtain elemental length.
The basic equations simplification is done by hand and been attached in appendix. The remaining
unknowns are solved using a MATLAB program, the same has been saved in the name as given below,
PROGRAM NAME: matfin.m (present in ZIP file)
 Initially syms command is used to declare variables
 Gravity constant g and mass m are stated
 Linear equations which are to be solved are declared under eqns
 Syntax [A,b] = equationsToMatrix(eqns) is used to convert the equations to matrix form
 Next, vars=symvars(eqns) syntax is used that automatically detects the variables in the equations
and returns the coefficient matrix which follows the order determined by symvar
 x = Ab is used the find the unknown variables.
Fig. 5: MATLAB code for simplified solving system of equation.

𝐴 =
[
0 1 0 0 1.323 0 0 0 0
0 0 0 0 1 0 1 0 0
0 0 1 0 0 1.732 0 0 0
0 0 0 0 0 1 0 1 0
1 0 0 1 0 0 0 0 0
1.864 0 0 0 0 0 0 0 1
0 −1 −1 −1 0 0 0 0 0
0 0 0 0 0 0 −1 −1 −1
0 −0.816 −1.338 −1.338 0 0 0.504 0.588 0.588]
𝑏 =
(
0
0
0
0
0
0
0
−256500
0 )
, 𝑥 =
(
𝑌03
𝑌41
𝑌42
𝑌43
𝑍01
𝑍02
𝑍41
𝑍42
𝑍43)
Results, in KN 𝑥 =
(
−245
−329
84
245
248.6
−48.4
−248.6
48.47
456 )
The final torseurs are,
[T01] = [
−342 0
329 531
248.6 −506
] [T41] = [
342 0
−329 −279
−248.6 172
] [T02] = [
0 0
−84 0
−48.4 0
]
[T42] = [
0 0
84 0
48.47 0
] [T03] = [
0 0
−245 0
−456 0
] [T43] = [
0 0
245 0
456 0
]
Hence the obtained values of forces and moments which are concerning our scope of work (Element BF
or 1, end torseurs) are as given below,
[T41]B = [
𝟑𝟒𝟐 𝟎
−𝟑𝟐𝟗 −𝟐𝟕𝟗
−𝟐𝟒𝟗 𝟏𝟕𝟐
] [T01]F = [
−𝟑𝟒𝟐 𝟎
𝟑𝟐𝟗 𝟓𝟑𝟏
𝟐𝟒𝟗 −𝟓𝟎𝟔
]
Note: Forces are in KN and Moments are in K-Nm.
Now the coordinates must be rectified and transformed into local system in order to ease the solving
methodology.
3.2 STRESS ESTIMATION BY RDM METHOD.
RDM Method is the analytical method used to determine the stresses and deformations of the structure
and is used to compare it with other solutions and also validate the model. The link or component BF
can be considered as beam and stresses are estimated and some of the assumptions made are i.e., the
material is
i. Elastic.
ii. Linear.
iii. Homogeneous.
iv. Isotropic.

Fig. 6: Top and Side views of element BF cross section.
Fig. 7: Front view of element BF.
3.2.1 Cut Method, Assumptions And Equations.
Résistance des Matériaux (RDM) method or Strength of Materials (SOM) is used to calculate the
stresses and deformations in the given structure. The objective is to design the structure according to
criteria of resistance, admissible deformation and acceptable financial cost.
Fig. 8: Cut section method to obtain moment & loads.
Assumptions,
 Elastic – The material returns to its initial shape after a loading & unloading cycle.
 Linear – The deformations are proportional to the stresses.
 Homogeneous – The material is of the same nature throughout its mass.
B G
L-x x
F
Arbitrary section G
N
L

 Isotropic – The properties of the material are identical in all directions.
Static or Cohesion torseur is used to describe the Mechanical actions in a member,
𝑇⃗ = [
𝑁 𝑀𝑡
𝑇𝑦 𝑀 𝑦
𝑇𝑧 𝑀𝑧
]
Where, N is Normal Force, Ty shear force in y direction and Tz shear force in z direction. Mt is the
moment of torsion and its vector has direction x, My is the bending moment along y and Mz bending
moment along z.
Note:
 The Neutral axis of the beam lies along the x-axis. It is the mean fibre where centre of gravity
lies.
 The principal axes are represented by Gy and Gz and they cross through the CG.
 The y-axis lies along the width of the beam and the z-axis lies along the depth of the beam.
 𝑀𝑧 = 𝑀𝑧 + 𝑇𝑦 × (𝐿 − 𝑥)
Stresses in beam are given by,(in MPa.)
𝝈 𝒙𝒙 =
𝑵
𝒃 × 𝑯(𝒙)
+
[𝑴 𝒛 + 𝑻 𝒚 × (𝑳 − 𝒙)] × 𝒚
𝑰 𝒛
𝝈 𝒙𝒚 =
𝑻 𝒚
𝒃 × 𝑯(𝒙) × (𝟏 + 𝜼)
+
𝑴𝒕 × 𝒛
𝑱
𝝈 𝒙𝒛 =
𝑴𝒕 × 𝒚
𝑱
+
𝑻 𝒛
𝒃 × 𝑯(𝒙) × (𝟏 + 𝜼)
Where, y is
𝐻(𝑥)
2
and z is
𝑧
2
.
3.2.2 Global Coordinate System To Local Coordinate System.
In 3-dimensional analysis it is necessary to convert section properties, forces and deflections between
coordinate systems defined by the individual structural members (local coordinates) and the common
coordinate system defining the entire structure (global coordinates).
 Rz (𝜃1 =
𝜋
2
)
 Ry (𝜃2 = 37.058°)
 Ry (𝜃3 = 𝜋)
Rotational matrix along z axis,
Rz = [
cos 𝜃 sin 𝜃 0
−sin 𝜃 cos 𝜃 0
0 0 1
]

Rotational matrix along y axis,
Ry= [
cos 𝜃 0 −sin 𝜃
0 1 0
sin 𝜃 0 cos 𝜃
]
Fig. 9: Transformation in z axis 𝜽 𝟏 =
𝝅
𝟐
.
Fig. 10: Transformation in y axis 𝜽 𝟐 = 𝟑𝟕. 𝟎𝟓𝟖°
.
Fig. 11: Transformation in z axis 𝜽 𝟑 = 𝝅.

Matrix multiplication,
= [
−1 0 0
0 1 0
0 0 −1
] × [
0.797 0 −0.602
0 1 0
0.602 0 0.797
] × [
0 1 0
−1 0 0
0 0 1
]
∴ Final Rotational Matrix is,
[
𝟎 −𝟎. 𝟕𝟗𝟕 𝟎. 𝟔𝟎𝟐
−𝟏 𝟎 𝟎
𝟎 −𝟎. 𝟔𝟎𝟐 −𝟎. 𝟕𝟗𝟕
]
Transformation of a torseurs is done to ease the visualisation of forces acting on the element in a simple
standard coordinate system. Rather than transforming the magnitude of the force accordingly at each
point, it is prefered to rotate the element to local (standard) axis system.
3.2.3 Forces In Beam.
Torseurs T01 and T41 after the transformation,
𝑻⃗⃗ 𝟎𝟏 = [
−𝟏𝟏𝟑. 𝟏𝟗𝟑 −𝟕𝟐𝟗. 𝟒𝟓𝟗
𝟑𝟒𝟐 𝟎
−𝟑𝟗𝟕. 𝟔𝟗𝟐 𝟖𝟒. 𝟑𝟗𝟑
]
𝑻⃗⃗ 𝟒𝟏 = [
𝟏𝟏𝟑. 𝟏𝟗𝟑 𝟑𝟐𝟔. 𝟑𝟓𝟖
−𝟑𝟒𝟐 𝟎
𝟑𝟗𝟕. 𝟔𝟗𝟐 𝟑𝟎. 𝟗𝟖𝟏
]
Note: Forces are in KN and Moments are in K-Nm.
3.2.4 Equivalent Von Mises Stress.
Von Mises stress is a value used to determine if a given material will yield or fracture. It is mostly used
for ductile materials such as metals. The Von Mises criterion states that if the von Mises stress of material
under load is equal or greater than the yield limit of the same material under simple tension, which is easy
to determine experimentally, then the material will yield.
Note: Complex loads/stress tensors that act in a member : 𝜎̿(𝑛) = (
𝜎𝑥𝑥 𝜎𝑥𝑦 𝜎𝑥𝑧
𝜎 𝑦𝑥 𝜎 𝑦𝑦 𝜎 𝑦𝑧
𝜎𝑧𝑥 𝜎𝑧𝑦 𝜎𝑧𝑧
)
Fig. 12 represents the stress-strain curve for the ductile material, 𝜎 𝑦𝑖𝑒𝑙𝑑 is the yield stress and if the 𝜎 𝑎𝑥𝑖𝑎𝑙
(axial stress) is less than the yield stress then it is in elastic region and if 𝜎 𝑦𝑖𝑒𝑙𝑑 = 𝜎 𝑎𝑥𝑖𝑎𝑙 then it represents
the start of plasticity.
Ry (𝜃3 = 𝜋) Ry (𝜃2 = 37.058°) Rz (𝜃1 =
𝜋
2
)

Fig. 12: Stress-Strain curve of ductile material.
The general form of von mises criterion is:
𝝈 𝒆𝒒 =
𝟏
√𝟐
[(𝝈 𝒙𝒙 − 𝝈 𝒚𝒚)
𝟐
+ (𝝈 𝒚𝒚 − 𝝈 𝒛𝒛)
𝟐
+ (𝝈 𝒛𝒛 − 𝝈 𝒙𝒙) 𝟐
+ 𝟔(𝝈 𝒙𝒚
𝟐
+ 𝝈 𝒙𝒛
𝟐
+ 𝝈 𝒚𝒛
𝟐
)]
𝟏
𝟐
Here, 𝜎𝑥𝑥, 𝜎 𝑦𝑦 and 𝜎𝑧𝑧 are the Normal stresses. 𝜎𝑥𝑦, 𝜎𝑥𝑧 and 𝜎 𝑦𝑧 are the Shear stresses. Few tensors are
to be neglected for which the definition and subsequent explanation in the following paragraph provides
clarification.
It is a tensor that consists of nine components 𝜎𝑖𝑗that completely define the state of stress at a point
inside a material in the deformed state or configuration. It is of Rank 2 or it has 2 basis vectors per
component, for example 𝜎𝑖𝑗 here i represents direction the area is perpendicular to and j represents the
direction of the force.
In our case we consider x plane and direction of the force also in x axis, hence the stress and shear
components that contemplates are𝜎𝑥𝑥, 𝜎𝑥𝑦, 𝜎𝑥𝑧, 𝜎 𝑦𝑥 𝑎𝑛𝑑 𝜎𝑧𝑥. From the Cauchy’s second law of motion
𝜎𝑖𝑗 = 𝜎𝑗𝑖which leads to the conclusion that stress tensor is symmetric.
If equivalent von mises stress 𝜎𝑒𝑞 is equal to the applied stress then plasticity starts.
𝝈̿(𝒏) =
[
𝑵
𝒃 × 𝒉(𝒙)
+
[𝑴 𝒛 + 𝑻 𝒚 × (𝑳 − 𝒙)] × 𝒚
𝑰 𝒛(𝒙)
𝑻 𝒚
𝒃 × 𝒉(𝒙) × (𝟏 + 𝜼)
+
𝑴𝒕 × 𝒛
𝑱(𝒙)
𝑴𝒕 × 𝒚
𝑱(𝒙)
+
𝑻 𝒛
𝒃 × 𝒉(𝒙) × (𝟏 + 𝜼)
𝑻 𝒚
𝒃 × 𝒉(𝒙) × (𝟏 + 𝜼)
+
𝑴𝒕 × 𝒛
𝑱(𝒙)
𝟎 𝟎
𝑴𝒕 × 𝒚
𝑱(𝒙)
+
𝑻 𝒛
𝒃 × 𝒉(𝒙) × (𝟏 + 𝜼)
𝟎 𝟎
]

σxx =
−1048.084
ℎ(𝑥)
+
51.785 ×106
2 × ℎ(𝑥)2
−
38 × 103
2 × ℎ(𝑥)2
σxy =
−112.08 × 103
ℎ(𝑥)
σxz =
−12.370 × 106
ℎ(𝑥)2
−
2.83 ×103
ℎ(𝑥)
𝝈 𝒆𝒒 =
𝟏
√𝟐
× [(𝝈 𝒙𝒙
𝟐) + (−𝝈 𝒙𝒙) 𝟐
+ 𝟔 × (𝝈 𝒙𝒚
𝟐
+ 𝝈 𝒙𝒛
𝟐
)]
𝟎.𝟓
Note: The above Stresses are in MPa.
3.2.5 Code And Results.
Determining point where maximum von mises stress acts. Algorithm and code for the same is attached
below.
PROGRAM NAME: ma2.m (in zip file attached).
 Initially syms syntax is used to declare the variables.
 Area function (H) is declared.
 Normal and shear stresses are declared namely, sxx, sxy and sxz.
 Equivalent Von Mises stress equation is stated i.e. sigmaeq.
 A simple syntax, fplot (f, xinterval) is used wherein it plots a curve for the specified interval
which helps in determining the point where maximum Von Mises stress is acting. Where,
f=sigmaeq and xinterval is [0,1116].
Fig. 13: MATLAB code for Von Mises equation solving.

Fig. 14: Graph plotting x coordinate with stress.
The result or the Von Mises Stress value which we got is ~840 MPa that can be seen from the Fig. 14.
3.3 STRESS ESTIMATION BY ABAQUS (FEM).
Finite Element Method (FEM) is computational approach for solving problems governed by differential
equations. The concept is more useful for complicated geometries, loadings and material properties where
analytical solution is difficult or cannot be obtained. A large object is divided into many elements which
is called discretization. Each element will have a simple equation defining the geometry and other physical
characters, equations of all elements form together a global system to define the entire system.
Various FEM solving tools are used viz. ANSYS, Abaqus, DIANA and Nastran to name some. Here
Abaqus CAE tool is used, to model, define and solve the problem. The three main steps in any FEM
problem are,
i. Idealization.
ii. Discretization
iii. Solution.
Taking up each step, the process of idealization & discretization in this particular case is elaborated below,
1. The geometric modelling of the lower brace is accomplished using Abaqus CAE tool. Properties
of the model is defined as 200GPa and 0.3 which are Young’s modulus and Poisions ratios
respectively.
2. In order to apply the loads and boundary conditions appropriately, a constrained couple is created
on a reference point on the side phase of the model. This is done in order to bind the surface to a
reference point in turn assisting the application of moment load as represented in Fig. 15.

Fig. 15: Constrained Couple applied on surfaces.
3. Now, the loads are applied on their respective application points i.e. as the load on the lower brace
effects by the holes through which the connecting rod passes and links the next element. As the
hole is not defined in this model due to many limitations, the loads are applied on the face edges
where the holes would be present. The moments are applied on the coupled reference points,
graphical representation of the same is given in Fig. 16.
Fig. 16: Load and boundary condition application.
4. The next step is discretization, using Abaqus student version the node numbering limit is 1000.
Hence using seed part instance the global element size is defined as 75, this leads in creating n
number of elements within the node limits.

Fig. 17: Meshing of part (Within limits of Student Version).
After this the model is submitted to analysis in Abaqus solver, the principle aim is to compare the value
obtained in RDM and FEM methods. To do so the Von Mises stress value can be considered. Explaining
the concept,
1. The taper in the section was considered while defining the varying cross section of the element
for which the location on x coordinate for the maximum von mises stress is 0 i.e. which is almost
the starting point of the taper.
2. The value of the Von Mises obtained at x=0 in RDM method is ~840 MPa. (elaboration in previous
section)
3. In the same way the Von Mises value in FEM is around in the range of 715MPa – 915MPa. As
the whole region of taper starting is one element. The contour has varying values.
Fig. 18: Final Von Mises contour representation.
4. COMPARISION BETWEEN RDM & FEM.
To make a conclusion between both the methods used, it’s important to understand the basic difference
on how the stress calculations are obtained.
 As explained in FEM section the object is discretized into many elements which are solved later
by combining into a global equation form. This might lead to high concentration of errors that
might sum up as the analysis is carried out.

 In RDM method the geometry is simplified to ease the calculations, thus reducing the accuracy.
But in FEM the structure is meshed which gives results at particular node and element but as the
mesh finess increases the results obtained may not be as desired since even for a small
perturbation, displacement or stress values are obtained at undesired region.
 As the elements and nodes increases CPU cost and also the processing time increases but when it
comes to RDM method those above stated constraints doesn’t matter.
 RDM method is used to solve simple structures but FEM is used to solve simple and complex
structures. In our case the Von mises stress result percentage error between RDM and
FEM(abaqus) is 4.54.
CONCLUSION.
The whole purpose of this project is to study the structural credibility of selected elements in CANADAIR
CL-215 landing gear system. To do so one should have a knowledge on geometry, mechanical properties,
loading, stress variation and material behaviour. Taking into consideration all the aspects mentioned
before, study of the structure is performed.
In order to analyse the stress variation and distribution, the first step is to define the joints and obtaining
the unknown forces and moments. Using these values and the geometric variation the principal stress
matrix is generated. Considering the Von Mises stress theory, which states that “If Von Mises stress is
equal to or more than yield stress plastic deformation starts making the design fail”. Therefore to provide
a stable and safe structure the Von Mises stress value should be respected. According to the same a suitable
material must be selected. Having a simple geometry in this case RDM method is incorporated but the
same cannot be done if a highly complex and non-homogenous system is considered, then Finite Element
Method is used.
Using FEM the model is created with proper mechanical properties and orientation, the Von Mises value
obtained is almost nearby to one calculated in RDM, showing the credibility of comparing both the
techniques. This also confirms the validity of FEM in structural problems which makes it easy to solve
complex geometries.
Summarizing all the results the main objective is to study and select the most suitable material for this
particular element for which a specific tool is required. Hence with the available possibilities it can be
suggested that the Titanium alloy of grade 6, Ti-6Al-2Sn-4Zr-6Mo can be used. Having yield strength of
1100MPa. It would be a safe select. But it should be noted that this is not the final material as cost
effectiveness is not considered in this process.

REFERENCES
1. http://dsk.ippt.pan.pl/docs/abaqus/v6.13/books/usi/default.htm?startat=pt03ch19s07hlb02.html
2. http://www.dhondt.de/ccx_2.15.pdf
3. https://academy.3ds.com/en/software/abaqus-student-edition
4. https://abaqus-docs.mit.edu/2017/English/SIMACAEOUTRefMap/simaout-c-
output.htm#simaout-c-output-t-TheDataFile-sma-topic1
5. “Introduction to Finite Element Method (FEM) for Structural mechanics” reference book by
Marianne Beringhier.
6. https://en.wikipedia.org/wiki/Canadair_CL-215
7. https://www.symbolab.com/solver/vector-cross-product-calculator
8. https://abaqus-docs.mit.edu/2017/English/SIMACAECSTRefMap/simacst-c-coupling.htm
9. http://dsk.ippt.pan.pl/docs/abaqus/v6.13/books/usi/default.htm?startat=pt03ch16s09hlb02.html
10. https://www.simscale.com/docs/content/simwiki/fea/what-is-von-mises-stress.html
11. ps://bertoldi.seas.harvard.edu/files/bertoldi/files/abaqusinputfilemanualv1.pdf?m=1444417191
12. https://www.simscale.com/docs/content/simwiki/fea/what-is-von-mises-stress.html
13. https://en.wikipedia.org/wiki/Cauchy_stress_tensor

APPENDIX

Landing Gear Analysis

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