Mechanical loads on a MASTER’S THESIS turbofan …1030140/FULLTEXT01.pdfengine failure, due to the...

MASTER’S THESIS

2009:101 CIV

MASTER’S THESIS

Universitetstryckeriet, Luleå

2009:101 CIV

Dženan Hozić

Mechanical loads on a turbofan engine structure

at blade-off

MASTER OF SCIENCE PROGRAMME Space Engineering

Luleå University of Technology Department of Applied Physics and Mechanical Engineering

Division of Solid Mechanics

MASTER’S THESIS


2009:101 CIV

Dženan Hozić


at blade-off




MASTER’S THESIS


2009:101 CIV

Dženan Hozić


at blade-off




Foreword/Acknowledgments

This thesis represents the final step in acquiring a Master of Sciencedegree in Space Engineering at Luleå University of Technology. Thethesis was conducted at Volvo Aero Corporation in Trollhättan, Swedenbetween July and December of 2008, at the Performance and ControlSystems department.

I would like to start by thanking my thesis supervisor at Volvo Aero,Anders Lundbladh for his enormous wealth of knowledge and guidanceduring the thesis. Further I would like to express my gratitude to RobertReimers, Johan Forsman, engineers at Volvo Aero and professorJan –Olov Aidenpää at Luleå University of Technology for helping me inunderstanding rotor dynamics and the problem in general.

A special thanks to the examiner of this thesis professor Hans-ÅkeHäggblad at Luleå University of Technology how has taken upon himselfto be my examiner.

Last and most importantly a huge acknowledgement and gratitude Iwould like to send to my family,Vahid, Nazima, Belma how havesupported and encourage me through out my studies and this thesis.With out your support it would have been hard to accomplish all this.

Thank you!

___________________

Dženan Hozic

i

Abstract

Aircraft engine development is an expensive and time consumingbusiness. The engines have to be designed to withstand loads wellabove those encountered during normal operating conditions. To meetthe certification requirements manufacturers conduct tests where theengines are subjected to extreme conditions and loads. One such ex-treme condition is fan blade-off (FBO).

Using finite element methods (FEM) computational software onecan simulate FBO conditions and predict what loads it will inflict onthe engine structure. These simulations involve large models that con-tain many degrees of freedom. As such they are time consuming tocreate and evaluate. Also they demand a detailed knowledge of theengine geometry to create.In preliminary design studies of engines there exists a need for know-ing the loads an engine is subjected to. This provides the design teamwith valuable information that is used in decisions regarding enginearchitecture.

The aim of this thesis is to create a scaled-down, simpler model of aturbofan engine, that can be used to predict bearing and engine mountloads, for a fan blade-off event. In addition the model can be used toestimate or check other parameters like weight and to compare differ-ent engine architectures.

The turbofan model created was based on rotor dynamics, this modelwas created in MATLAB and DyRoBeS rotor dynamic simulation soft-ware. The simulations were done on three turbofan engines; a refer-ence engine to validate the turbofan model and later applied on twoproposal turbofan engines for the coming Next generation Single Aisle(NSA) aircraft.

The results for the simulations show that the turbofan model canbe used in the conceptual design stage. It can provide the design teamwith an initial estimate for bearing and engine mount loads. Takinginto account the simplicity of the turbofan model, a conclusion on theaccuracy of the results is that they are within acceptable limits.

ii

NomenclatureFEM Finite Element Method

GC Geometric Centre

GE General Electric

ISO International Organization for Standardization

LP-shaft Low Pressure shaft

NSA Next generation Single Aisle

P & W Pratt & Whitney

RR Rolls-Royce

VAC Volvo Aero Corporation

C Symmetric Damping matrix

G Skew-Symmetric Gyroscopic matrix

H Skew-Symmetric Circulatory matrix

K Symmetric Stiffness matrix

M Symmetric Mass matrix

q(t) Generalized coordinates vector

BPR Bypass Ratio

CM Centre of Mass

CR Centre of Rotation

DoF Degrees of Freedom

FBO Fan Blade-Off

LPT Low Pressure Turbine

iii

CONTENTS CONTENTS

Contents1 Introduction to the thesis 1

1.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Purpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Model of the turbofan 32.1 Background to Rotor dynamics . . . . . . . . . . . . . . . . . 42.2 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2.1 Translational equation of motion . . . . . . . . . . . . 82.2.2 Rotational equation of motion . . . . . . . . . . . . . 92.2.3 Summary of the equations of motion . . . . . . . . . . 13

2.3 Stiffness matrix of the model . . . . . . . . . . . . . . . . . . 142.3.1 Shaft flexibility . . . . . . . . . . . . . . . . . . . . . . 142.3.2 Bearing flexibility . . . . . . . . . . . . . . . . . . . . . 212.3.3 Stiffness matrix of the turbofan model . . . . . . . . 27

2.4 Summary of the turbofan model . . . . . . . . . . . . . . . . . 29

3 Using the turbofan model 303.1 Parameters of the Engines . . . . . . . . . . . . . . . . . . . . 30

3.1.1 Reference Engine . . . . . . . . . . . . . . . . . . . . . 303.1.2 NSA-1HPT and NSA-2HPT . . . . . . . . . . . . . . . 37

3.2 Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.2.1 Whirl speed . . . . . . . . . . . . . . . . . . . . . . . . 403.2.2 Campbell diagram . . . . . . . . . . . . . . . . . . . . 403.2.3 Critical Speed . . . . . . . . . . . . . . . . . . . . . . . 43

3.3 Unbalance response . . . . . . . . . . . . . . . . . . . . . . . . 443.4 Engine structure loads . . . . . . . . . . . . . . . . . . . . . . 45

3.4.1 Bearing loads . . . . . . . . . . . . . . . . . . . . . . . 473.4.2 Engine mount loads . . . . . . . . . . . . . . . . . . . . 48

4 DyRoBeS 504.1 Creating a model . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.1.1 Material . . . . . . . . . . . . . . . . . . . . . . . . . . 514.1.2 Shaft Elements . . . . . . . . . . . . . . . . . . . . . . 524.1.3 Disks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.1.4 Unbalance . . . . . . . . . . . . . . . . . . . . . . . . . 554.1.5 Bearings . . . . . . . . . . . . . . . . . . . . . . . . . . 564.1.6 Turbofan model in DyRoBeS . . . . . . . . . . . . . . . 57

4.2 Simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.2.1 Critical Speed Analysis . . . . . . . . . . . . . . . . . . 59

iv

CONTENTS CONTENTS

5 Results 605.1 Eigenvalue analysis . . . . . . . . . . . . . . . . . . . . . . . 60

5.1.1 Reference engine . . . . . . . . . . . . . . . . . . . . . 605.1.2 NSA 1-HPT engine . . . . . . . . . . . . . . . . . . . . 615.1.3 NSA 2-HPT engine . . . . . . . . . . . . . . . . . . . . 62

5.2 Bearing loads . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.2.1 Reference engine . . . . . . . . . . . . . . . . . . . . . 635.2.2 NSA 1-HPT engine . . . . . . . . . . . . . . . . . . . . 635.2.3 NSA 2-HPT engine . . . . . . . . . . . . . . . . . . . . 64

5.3 Engine mount loads . . . . . . . . . . . . . . . . . . . . . . . 665.3.1 Reference engine . . . . . . . . . . . . . . . . . . . . . 665.3.2 NSA 1-HPT . . . . . . . . . . . . . . . . . . . . . . . . 665.3.3 NSA 2-HPT . . . . . . . . . . . . . . . . . . . . . . . . 67

6 Discussion 68

7 Conclusions 69

8 Future work 70

References 71

A MATLAB code 72A.1 Reference engine . . . . . . . . . . . . . . . . . . . . . . . . . 72A.2 NSA 1-HPT engine . . . . . . . . . . . . . . . . . . . . . . . . 77A.3 NSA 2-HPT engine . . . . . . . . . . . . . . . . . . . . . . . . 82

v

1 INTRODUCTION TO THE THESIS

1 Introduction to the thesisWhen the jet-age reached commercial aviation in the beginning of the1950s with aircraft like the de Havilland Comet, the worlds first jet pow-ered commercial aircraft and Boeing 707 which was introduced a few yearslater, started a new era in human aviation. These early aircrafts were pow-ered by turbojet engines that propelled the aircraft to speeds close to thespeed of sound. But they were fuel inefficient consuming a lot of fuel forevery pound of thrust produced. To improve the fuel efficiency engineersintroduced the turbofan engine to commercial aviation in the beginning ofthe 1960s.

A turbofan engine can be described as a turbojet engine with a largeducted fan mounted in front of the compressor stage. This configurationmakes it possible to divide the airflow into the engine where only aportion of the air mass goes into the engine core, and the rest of the airmass bypasses the engine core. To generate the same thrust byaccelerating a larger mass of air to a lower velocity requires less energy,and makes the engine more fule efficient.[1]This is one of the main reasons why turbofan engines are still some 50years later the most widely used aircraft engines, not just in commercialaviation but also in military aircraft across the world.

In early turbofan engines the fan configuration that was used created alow bypass ratio, BPR. These fan configurations were of smallerdiameters and consisting of larger number of fan blades. As commercialaviation increased and more aircraft were beginning to be used, issues offuel consumption and noise suppression from the aircraft engines beganto be even more important. As a responds to this the high BPR turbofanengines were developed. High BPR turbofans delivered even more thrustand at the same time decreased the engines noise output, they are themost common turbofan engines mounted on today’s commercial aircraftand are characterized with having larger fan diameters and a smallernumber of fan blades.

Due to the very conservative certification requirements for commercialaircraft and their engines, a fan of a turbofan engine is of vitalimportance. One of the most important requirements during thecertification of an engine is the engines ability to withstand catastrophicengine failure, due to the loss of one of the fan blades. This sort of failureis known as fan blade-off, FBO. As turbofan engine development isconstantly moving towards larger fan diameters with fewer heavier fanblades, a loss of a fan blade generates larger loads on engines and theirstructures.To ensure that there is no further damage to the aircraft, the engine

1

1.1 Background 1 INTRODUCTION TO THE THESIS

components must be able to handle the loads generated by the loose bladeand the imbalance following a fan blade off.

1.1 BackgroundVolvo Aero Corporation,VAC is a partner in almost every large transportaircraft engine program today, the components that VAC manufacturesare for the most part made for the large aircraft engine manufacturers,GE, P&W and RR. These manufacturers are currently in the process ofdeveloping new types and concepts of aircraft engines that are going toreplace the existing turbofan engines. In order for VAC to continue beinga strong partner in future engine programs, the need to independently beable to estimate loads in future engines in order to increase the knowledgeof how they would affect VAC components, is high.

Today the loads that a component from VAC has to be able to withstand aregiven by the respective engine manufacturer. These loads can accuratelybe estimated using FEM computational software. But this requires largeamounts of man hours and a detailed knowledge of the engine geometry toperform.As the specifications of future engines is still not fully defined, VAC is in-terested in developing a simpler model that can estimate loads in a aircraftengine, when the engine is subjected to FBO.

1.2 PurposeThe purpose of this thesis is to derive a model that can be used to estimatethe initial unbalance load of the fan, when FBO occurs, the model shouldalso be able to estimate what kind of bearing and engine mount loads theunbalance of the fan would give rise to.

The purpose is to use this model on the coming NSA aircraft engine, togive VAC an initial understanding of the load requirements for the newengine designs(architectures).

2

2 MODEL OF THE TURBOFAN

2 Model of the turbofanModern turbofan engines are designed so that the fan is attached to the lowpressure shaft (LP-shaft) which is in turn suspended on bearings. To con-trol deflections at normal operating conditions, the LP-shaft is suspendedon three bearings.

When FBO occurs a fan blade is ejected in a tangential trajectory fromthe fan. This blade is collected by the fan casing surrounding the fan. Therest of the fan receives an impulse towards the opposite direction of theloose fan blade. This will give rise to large mechanical loads within theengine structure. To lower the amplitude of these mechanical loadsengineers have added a fuse to the front bearing. This means that thebearing will breake free from the structure when the system is operatingabove normal operating conditions and thus make the whole system moreflexible. All this will happen within a few milliseconds from the initial fanblade separation.

The model in this thesis will describe the system after the initial fanblade separation and the collapse of the fuse holding the front bearing.

To create a simple model that could adequately represent a FBO case, amodel was derived that included the fan disk and LP-shaft of the turbofanengine. These two parts are connected together to form an overhungrotor, to represent the situation when the front bearing has broken free. Agraphical representation of this model is shown in figure 1 below.

Figure 1: A graphic representation of the model

3

2.1 Background to Rotor dynamics 2 MODEL OF THE TURBOFAN

As can be seen in figure 1 the overhung rotor model is suspended on twobearings. The fan disk is represented by a hollow cylinder for the hub andthin rectangular sheets representing the fan blades. This simplificationwas made due to the complex shape of the fan blades, also the fan disk isassumed to be rigid. Both the LP-shaft and the bearings of the rotorsystem are flexible.The model of the turbofan engine represented in figure 1 can be seen as aclassical rotor dynamic model. Using the governing equations of rotordynamics a mathematical representation of the model can be derived.

2.1 Background to Rotor dynamicsRotor dynamics is a branch of general dynamics which describes the dy-namic behaviour of mechanical system, were at least one part is rotatingwith significant angular momentum. These parts are referred to as rotors.The definition of a rotor by ISO is

”a rotor is a body suspended through a set of cylindrical hinges or bear-ings that allows it to rotate freely about an axis fixed in space” [3]

Despite the definition, rotors can be divided into two subgroups, namelyfixed rotors and free rotors. Free rotors are objects and mechanicalsystems were the axis of rotation is not fixed in space. Examples of suchrotors are celestial bodies, space vehicles and spinning projectiles. Theother subgroup of rotors, the fixed rotors, is generally best described bythe ISO definition for rotors. Turbo machinery is one of the most commonexamples of this subgroup. One distinct difference between fixed and freerotors is the spin speed at which they rotate. A spin speed of a fixed rotoris generally considered to be constant or imposed by a driving device. Inthe case of free rotors the spin speed is governed by the conservation ofthe angular momentum of the rotor. [3]

Technological studies related to rotor dynamics goes back to thenineteenth century, this was when the rotational speed of machinesincreased and it became important to analyze there dynamic behaviour.The first study into rotor dynamics was made by Scottish engineerWilliam Rankine in 1869. His model of a flexible rotor system correctlystated that the system had a speed at which large amplitudes due tovibration were observed. He named this speed ”critical speed”. Rankineincorrectly stated that running rotors stably above the critical speed wasimpossible, but this was nevertheless demonstrated to be possible by theSwedish engineer Gustaf de Laval. He realized that the behaviour ofrotors running above the critical speeds i.e., super-critical conditiondiffered from those running below the critical speed i.e., sub-criticalcondition.

4

2 MODEL OF THE TURBOFAN 2.1 Background to Rotor dynamics

Figure 2: Rankine rotor Figure 3: Jeffcott rotor

It was not until 1919 when Henry H. Jeffcott derived his model of a rotorsystem, that a theoretical explanation to super-critical running rotorscould be found. This model, that became known as the Jeffcott rotor gaveexplanations to many important features of rotor dynamics. Among otherthe self-centring that occurs in a rotor that is running at super-criticalconditions. [3, 10]

5

2.2 Equations of motion 2 MODEL OF THE TURBOFAN

2.2 Equations of motionThe rotor dynamical model in figure 1 is restricted by the bearings torotate around one axis, and as the fan disk is driven by the low pressureturbine (LPT) through the LP-shaft, then the spin speed of the fan disk isconstant. The bearings can be modelled as four springs acting in pair inthe two directions orthogonal to the principle axis of rotation. This isdepicted in figure 4 below.

Figure 4: The model with 4 springs representing bearings

When FBO accrues the centre of mass (CM) for the fan disk will movefrom the centre of rotation (CR), this will induce an unbalance in the fandisk.

The model in figure 4 can be viewed as an axisymmetric rotor systemwith no damping acting on it. This configuration of the rotor system infigure 4 will give a rotor dynamical model with four degrees of freedom(DoF), two translational and two rotational DoF corresponding to theplane of motion perpendicular to the axis of rotation in figure4.

The mathematical equations describing a rotor dynamical system arethree dimensional and fairly complex. In particular when dealing withrotational DoF, they do not allow the use of any direct linear model. Onecan assume a number of simplifying conditions that would allow alinearized mathematical model to be used. This linearized model wouldretain the basic features of rotor dynamics and allow the rotor dynamicmodel to be described in both a qualitative and quantitative manner.[2, 3]

6

2 MODEL OF THE TURBOFAN 2.2 Equations of motion

By assuming simplifying conditions, that all displacements and velocities,both angular and linear to be small, except the spin velocity around theaxis of rotation. This will give a linearized model that describes the rotordynamical behaviour of a system, using the general equation of motion forrotor dynamical systems below.

Mq(t) + (C + G) q(t) + (K + H) q(t) = f(t) (1)

Where M is the symmetric mass matrix, C is the symmetric dampingmatrix, G is the skew-symmetric gyroscopic matrix, K is the symmetricstiffness matrix, H is the skew-symmetric circulatory matrix, q(t) are thegeneralized coordinates of the system, and f(t) are the external forcesacting on the system.[3]

As damping was not taken into consideration in the model of this thesis,then the damping matrix C and circulatory matrix H would become nullmatrices i.e., C = H = 0. And as there is no external force acting on themodel then the time-dependent forcing function vector is also equal tozero, f(t) = 0. Inserting this into equation 1, the general equation ofmotion for a rotor dynamical system, would give the governing equationof motion for the rotor dynamical model describing the turbofan modelused in this thesis.

Mq(t) + Gq(t) + Kq(t) = 0 (2)

The four DoF that describe the motion of the turbofan model can beinserted into the generalized coordinates vector q(t) of equation 2, thenthe generalized coordinates becomes

q(t) =

xyϕxϕy

(3)

7


2.2.1 Translational equation of motion

The translational equation of motion corresponds to the fixed coordinateframe (XYZ), also known as the inertial frame of reference. Origin of thisframe of reference is set to the centre of the rear bearing and the Z-axispasses through the LP-shaft and along the geometric centre (GC) of thefan disk. This axis corresponds to the rotational axis of the rotor.

Figure 5: Deflected turbofan Figure 6: XY-plane

Figure 5 is representing the deflected turbofan model when it is subjectedto an unbalance in the fan. From the XY-plane view of figure 6 a positionvector ~r is obtained that represents the translational motion from theorigin to the CM of the unbalanced fan in the inertial frame. In figure 6 εrepresents the eccentricity, which is the length of how far the CM isdisplaced from the GC when the fan disk is unbalanced.

~r = (x+ ε cosωt) X + (y + ε sinωt) Y (4)

differentiation of equation 4 with respect to time will give the final twoposition vectors, assuming ω is constant:

~r = (x− εω sinωt) X + (y + εω cosωt) Y (5)

~r =(x− εω2 cosωt

)X +

(y − εω2 sinωt

)Y (6)

The translational motion of the rotor does not induce any gyroscopiceffects on the system and thus this part of the gyroscopic matrix G inequation 2 is a null matrix. The stiffness matrix K is acting on the fanfrom the GC and the mass matrix M is acting on the fan at its CM. Using

8


this and inserting equation 6 into equation 2 will give the translationalequations of motion for the turbofan model.

Mtrans

(xy

)+ K

(xy

)= Mtrans

(εω2 cosωtεω2 sinωt

)(7)

The elements of the translational mass matrix Mtrans of equation 7 arecomposed of the fan disk mass mfan along the diagonal of the matrix, i.e.

Mtrans =[mfan 0

0 mfan

](8)

The elements of the stiffness matrix K acting on the fan disk are governedby the stiffness of the LP-shaft and the bearings, these elements are goingto be derived in the subsequent sections of this thesis (see sec. 2.3).

2.2.2 Rotational equation of motion

The rotational equation of motion corresponds to the rotational coordinateframe(xyz). The origin of this rotational frame of reference is set at thecentre of the front bearing, the x and y axes of this frame of reference areparallel to the X and Y axis of the inertial frame of reference (XYZ). Thez axis of the rotational frame of reference (xyz) is defined as the spin axisof the fan disk, and is always perpendicular to the fan disk. The Z axes ofinertial frame of reference (XYZ) is defined as the principle rotational axisof the system.When the fan disk is unbalanced the spinning motion of itwill cause the LP shaft to bend. This bending of the shaft will displace thespin axis from the principle axis of rotation of the system.

Figure 7: Turbofan model with the rotational frame of reference (xyz)

The spin of the fan disk about the new plane as seen in figure 7, will giverise to a gyroscopic moment on the fan disk. This gyroscopic phenomenonthat accrues can not be neglected in the model as it gives rise to someartificial stiffening within the system. Including this gyroscopic

9


phenomenon in the model is done by describing the precession motion ofthe fan disk in figure 7. Using the angular moment and figure 8 below[6]the gyroscopic moment of the fan disk can be described as follows.

Figure 8: Precession motion of the fan disk

The angular moment of the fan disk about its spin axis is:

Hz = Jpω (9)

From figure 8 the angular moment about the spin axis is geometricallydescribed as:

H2xz +H2

yz +H2Zz = H2

z (10)

Where Hxz and Hyz are given trigonometric as:

Hxz = HZz tanϕx (11)

Hyz = HZz tan−ϕy (12)

Inserting equations 11,12 into equation 10 and implementing thepreviously stated assumption of small displacements gives:

H2Zz ≈ H2

z (13)

10


The physical meaning of equation 13 ] is that when displacements aresmall the angular moment for the fan disk about the rotational axis canbe assumed equal to the angular moment about the spin axis. Insertingequation 9 into equation 13 gives the angular moment about therotational axis:

HZz ≈ Jpω (14)

The angular moment about the remaining two axes are given by:

Hx = Jtωx +Hzϕy (15)

Hy = Jtωy −Hzϕx (16)

Inserting equation 9 into equations 15,16 with ωx = ϕx and ωy = ϕy give:

Hx = Jtϕx + Jpωϕy (17)

Hy = Jtϕy − Jpωϕx (18)

In order to get the angular moments of the fan disk that causes thegyroscopic phenomena a time differentiation of angularmoment

{ddtH = H

}is needed[6].

Taking the time derivative of equations 14,17,18 yields[2, 3]:

Hx = Jtϕx + Jpωϕy (19)

Hy = Jtϕy − Jpωϕx (20)

Hz = 0 (21)

11


Using the derived angular moment equations 20,19 above and insertingthem into the governing equation of motion for the rotor dynamical model2, will give the corresponding rotational equation of motion for theturbofan model,

Mrot

(ϕxϕy

)+ G

(ϕxϕy

)+ K

(ϕxϕy

)= 0 (22)

where the rotational mass matrix Mrot and the gyroscopic matrix G are:

Mrot =[Jt 00 Jt

](23)

G = ω

[−Jp 0

0 Jp

](24)

The stiffness matrix K of equation 22 will, as in the case of thetranslational equation of motion 7, be derived in subsequent sections (seesec. 2.3)

12


2.2.3 Summary of the equations of motion

Summarizing the equations of motion derived in subsections 2.2.1 and2.2.2 is done by first substituting the generalized coordinate vector 4 intothe governing equation of motion 2 for the rotor dynamical model.

M

xyϕxϕy

+ G

xyϕxϕy

+ K

xyϕxϕy

= 0 (25)

Then insert equations 7-8 and 22-24 into equation 25:

mfan 0 0 0

0 mfan 0 0

0 0 Jt 0

0 0 0 Jt

x

y

ϕx

ϕy

+ ω

0 0 0 0

0 0 0 0

0 0 −Jp 0

0 0 0 Jp

x

y

ϕx

ϕy

+

+ K

x

y

ϕx

ϕy

= mrotεω

2

cosωt

sinωt

0

0

(26)

13

2.3 Stiffness matrix of the model 2 MODEL OF THE TURBOFAN

2.3 Stiffness matrix of the modelIn section 2 it was stated that both the bearings and LP-shaft used inthe turbofan model were flexible. To incorporate both flexibilities into thestiffness matrix of the turbofan model the flexibility will be divided intotwo steps, the first corresponding to the flexibility of the LP-shaft and thesecond step will correspond to the flexibility of the bearings. These twoflexibilities will then be superimposed to create a general flexibility matrixfor the turbofan model. In the final stage by using linear algebra the flexi-bility matrix will be transformed into a stiffness matrix that is used in theturbofan model.

2.3.1 Shaft flexibility

To derive the shaft flexibility of the LP-shaft as viewed in figure 4, thebearings in this figure will be assumed infinitely rigid while the shaft isseen as a beam with varying cross-section. As the LP-shaft has varyingcross-section the best way to derive its flexibility is to divide the shaft intotwo parts were each part will be given a constant cross-section, this canbe viewed in figure 9 below. Also the two parts of the shaft have differentmaterial properties.

Figure 9: A cross-section view of the shaft with rigid bearings

As the fan disk is rigidly connected to the shaft, the shaft will apply aforce and a moment on to the fan disk.In figure 9 the shaft is supportedby the rigid bearings, this will give the boundary conditions used to getthe flexibility for the part of the shaft that is located between thebearings. The boundary condition for the remaining overhung part of theshaft is that one end of it is fixed to the portion of the shaft that is locatedbetween the bearings. Using elementary cases for Euler - Bernoulli beamtheory, expressions for the deflection and deflection angle of the shaft canbe derived when subjected to a reaction force and moment.

14

2 MODEL OF THE TURBOFAN 2.3 Stiffness matrix of the model

Starting with the case when the shaft is subjected to an unbalance forceF , then the total deflection ∆F and the deflection angle ϕ

Fof the shaft

are given by summarizing the deflections for each part of the shaft. Agraphic representation of the deflection for the shaft parts are given infigure 10 below.

Part A

Part B

Figure 10: Shaft parts describing the deflections when subjected to Funb

As we are assuming small deflections, the total deflection of the shaftcomposed of the shaft parts in figure 10 is:

∆F = ∆A + ∆B (27)

In equation 27 the deflection ∆A due to the flexibility of shaft part A isgiven by:

∆A = ϕAβL (28)

15


Were the deflection angle ϕA in equation 28 is given together withM1 = FβL as:

ϕA =[

αL

3EAIA

]M1 =

[αβL2

3EAIA

]F (29)

The deflection ∆B of shaft part B in figure 10 is:

∆B =[β3L3

3EBIB

]F (30)

Inserting equations 28-30 into equation 27 will give the total deflection ofthe shaft when subjected to a reaction force F :

∆F =[β2L3

3

(α

EAIA+

β

EBIB

)]F (31)

The total deflection angle of the shaft when subjected to a reaction force Fis given by:

ϕF

= ϕA + ϕB (32)

In equation 32 the deflection angle ϕA of the part A in figure 10 is givenby equation 29 above:

ϕA =[αβL2

3EAIA

]F

The deflection angle of part B ϕB , in figure 10 is:

ϕB =[β2L2

2EBIB

]F (33)

Substituting equations 29 and 33 into equation 32 will give an expressionfor the deflection angle ϕ

Fof the shaft when subjected to a reaction force

F :

ϕF

=[βL2

6

(2α

EAIA+

3βEBIB

)]F (34)

16


Looking at the case when the fan disk is subjected to a moment M due tothe unbalance of the fan disk, expressions for the total deflection 4M anddeflection angle ϕ

Mof the shaft are derived using the same approach as in

the previous case with the reaction force F .

Part A

Part B

Figure 11: Shaft parts describing the deflections when subjected to M

Using the elementary cases of Euler - Bernoulli beam theory and figure11 above, expression for the total deflection 4M is:

4M = ∆A + ∆B (35)

where ∆Ais given by :∆A = ϕAβL (36)

In equation 36 the deflection angle ϕA of part A in figure 11 is :

ϕA =[

αL

3EAIA

]M (37)

17


The expression for the deflection ∆B of shaft part B in equation 35 is :

∆B =[β2L2

2EBIB

]M (38)

Substituting equations 36 - 38 into equation 35 yields the expression forthe deflection of the shaft when subjected to a reaction moment M :

∆M =[βL2

6

(2α

EAIA+

3βEBIB

)]M (39)

The expression for the deflection angle ϕM

of the shaft due to reactionmoment M is

ϕM

= ϕA + ϕB (40)

as in the previous case the deflection angle ϕA of shaft part A is given byequation 37 :

ϕA =[

αL

3EAIA

]M

The expression for the deflection angle ϕB of part B in figure 11 is :

ϕB =[βL

EBIB

]M (41)

Substituting equations 37 and 41 into equation 40 will give the totaldeflection angle of the shaft when subjected to a reaction moment M :

ϕM

=[L

3

(α

EAIA+

3βEBIB

)]M (42)

Summarizing the expressions for the deflection and deflection anglesderived in equations 31, 34, 39 and 42 above for the shaft when subjectedto unbalance, will show that the expressions within the brackets ofequations 34 and 39 are equal. With this equality the expressions can besummarized as follows:

∆F =[β2L3

3

(α

EAIA+ β

EBIB

)]F ⇐⇒ asf1F

ϕF

=[βL2

6

(2α

EAIA+ 3β

EBIB

)]F ⇐⇒ asf2F

∆M =[βL2

6

(2α

EAIA+ 3β

EBIB

)]M ⇐⇒ asf2M

ϕM

=[L3

(α

EAIA+ 3β

EBIB

)]M ⇐⇒ asf3M

(43)

18


The deflections and deflection angles summarized in equation 43 are allone dimensional. As the shaft used in this thesis is of a three dimensionalnature, it will have to be represented in a way that allows the use of theexpressions in equation 43 to derive the shafts flexibility, thatcorresponds to its three dimensional nature. Using the shafts axialsymmetry, gives the ability to represent the shaft along its symmetryplanes, that are along the rotational axis. With these symmetry planesexpressions in equation 43 can be used to derive the deflection anddeflection angles of the three dimensional shaft.

Figure 12: Two symmetric planes describing the 3-D displacement of thefan disk[8]

Figure 12 along with the expressions summarized in equation 43 will givethe following deflections and deflection angles in the symmetry planes :

x = asf1Fx + asf2My

y = asf1Fy − asf2Mx

ϕx = − asf2Fy + asf3My

ϕy = asf2Fx + asf3Mx

(44)

19


Expressing equation 44 in matrix form :

x

y

ϕx

ϕy

=

asf1 0 0 asf2

0 asf1 −asf2 0

0 −asf2 asf3 0

asf2 0 0 asf3

Fx

Fy

Mx

My

(45)

were the shaft flexibility is given by :

Asf =

asf1 0 0 asf2

0 asf1 −asf2 0

0 −asf2 asf3 0

asf2 0 0 asf3

(46)

20


2.3.2 Bearing flexibility

The second step in defining the flexibility matrix for the turbofan model itto derive expressions for the flexibility of the bearings. To be able to do thisthe shaft of the turbofan model in figure 4 will be assumed infinitely rigid.Utilizing the symmetry in the model, a similar approach can be used as inthe derivation of the shaft flexibility in subsection 2.3.1.As with the shaft flexibility we will start with the case of a reaction forceF acting on the rigid shaft. From figure 4 we get a free-body diagram of asymmetry plane of the model, that is shown in figure 13 below.

Figure 13: Free-body diagram when subjected to Funb

The free-body diagram of figure 13 gives the following relations:

α+ β = 1 (47)

↑ F + FRB − FFB = 0 (48)

γL is the distance from the rear bearing to the point of rotation of theshaft when subjected to unbalance.

When subjected to a unbalance the shaft as showed in figure 13 will startto rotate around a point of rotation as a rigid body. This rotation gives thedeflections and deflection angles of the shaft that correspond to theflexibility of the bearings.

21


Figure 14: Rigid body rotation when subjected to F

Figure 14 together with the relation in equation 47 gives:

y FRBγL+ FFB (α− γ)L− P (1− γ)L = 0 (49)

were the spring forces of the bearings are :

FRB = kRB4

RB(50)

FFB = kF B4

F B(51)

using the fact that the angle of rotation is constant along the wholelength of the shaft and that all displacements are small, the followingrelations for deflection and deflection angles are derived

ϕF

=∆F

(1− γ)L(52)

∆RB

γL=

∆F

(1− γ)L(53)

∆F B

(α− γ)L=

∆F

(1− γ)L(54)

Inserting equations 50, 51 into equation 48 and 49 gives :

F = kF B4

F B− k

RB4

RB(55)

F (1− γ) = kRB

∆RBγ + k

F B∆

F B(α− γ) (56)

22


Relations in equations 53- 54 in equation 55 and 56 :

F =[k

F B(α− γ)− k

RBγ

1− γ

]∆F (57)

F (1− γ) =

(k

RBγ2 + k

F B(α− γ)2

1− γ

)∆F (58)

Inserting equation 57 into equation 58 will give an expression for γ :

γ =k

F Bα (1− α)

kRB

+ kF B

(1− α)(59)

As the expression for γ in equation 59 is only dependent on the springstiffnesses representing the bearings, inserting the expression intoequation 57 gives an expression for the deflection of the shaft as afunction of the spring stiffnesses when subjected to a reaction force F .

∆F =

[k

RB+ k

F B(α− 1)2

kRBk

F Bα2

]F (60)

Using the relations in equations 52, 59 in equation 60 gives also theexpression for the deflection angle as a function of the spring stiffnesses.

ϕF

=[k

RB+ k

F B(1− α)

kRBk

F Bα2L

]F (61)

To derive the expressions for the deflection and deflection angle of theshaft when subjected to a reaction moment M , the same method ofapproach was used as when the shaft was subjected to a reaction force F .The free-body diagram of the shaft when subjected to a reaction momentM can be seen below.

23


Figure 15: Free-body diagram and rigid body rotation when subjected toM

From figure 15 above, the following equations are given :

↑ FRB − FFB = 0 (62)

y FRBγL+ FFB (α− γ)L−M = 0 (63)

Inserting the relations of equation 50, 51 and 53, 54 into equation 62gives : [

kRBγ

1− γ

]∆M =

[k

F B(α− γ)

1− γ

]∆M (64)

The relation in equation 64 will give an expression for γ :

γ =k

F Bα

kRB

+ kF B

(65)

Inserting the relations in equation 50, 51 and 53, 54 into equation 63gives the reaction moment M :

M =

[k

RBγ2 + k

F B(α− γ)2

1− γ

]L∆M (66)

24


Inserting the expression for γ in equation 65 into equation 66 gives thedeflection of the shaft as a function of the bearing stiffnesses, when theshaft is subjected to an reaction moment M :

∆M =

[(k

RB+ k

F B) (k

RB+ k

F B(1− α))(

kRBk2

F B+ k

F Bk2

RB

)α2L

]M (67)

To derive an expression for the deflection angle of the shaft whensubjected to a reaction moment M we will insert the relations ofequations 52 and 65 into equation 67 above :

ϕM

=

[(k

RB+ k

F B)2(

kRBk2

F B+ k

F Bk2

RB

)α2L2

]M (68)

The expressions within the brackets of equations 61 and 67 are for anyarbitrary numerical value equal. With this equality the deflection anddeflection angles in equations 60 - 61 and 67 - 68 above, can besummarized as follows:

∆F =[k

RB+k

F B(α−1)2

kRB

kF B

α2

]F ⇐⇒ abs1F

ϕF

=[k

RB+k

F B(1−α)

kRB

kF B

α2L

]F ⇐⇒ abs2F

∆M =[k

RB+k

F B(1−α)

kRB

kF B

α2L

]M ⇐⇒ abs2M

ϕM

=[

(kRB+k

F B )2

(kRBk2

F B+k

F Bk2

RB)α2L2

]M ⇐⇒ abs3M

(69)

Using the same principle as in subsection 2.3.1 for deriving theexpressions for the shaft flexibility in three dimensions, will give thefollowing expressions for the bearing flexibility using figure 12 andexpressions in equation 69 :

x = abs1Fx + abs2My

y = abs1Fy − abs2Mx

ϕx = − abs2Fy + abs3My

ϕy = abs2Fx + abs3Mx

(70)

25


The set of expressions in equation 70 expressed in matrix form :

x

y

ϕx

ϕy

=

abs1 0 0 abs2

0 abs1 −abs2 0

0 −abs2 abs3 0

abs2 0 0 abs3

Fx

Fy

Mx

My

(71)

were the bearing flexibility matrix Abs is :

Abs =

abs1 0 0 abs2

0 abs1 −abs2 0

0 −abs2 abs3 0

abs2 0 0 abs3

(72)

26


2.3.3 Stiffness matrix of the turbofan model

To derive a stiffness matrix for the turbofan model, one first has to createa general flexibility matrix for the turbofan. As we have defined that boththe shaft and bearings of the turbofan model are flexible, the best way tocreate the general flexibility matrix in to use the derived shaft and bearingflexibilities in subsections 2.3.1 and 2.3.2 and superimpose them to createthe general flexibility matrix for the turbofan model.

The general flexibility of the turbofan model is given by :

Atf = Asf + Abs (73)

Equations 46 and 72 inserted into equation 73 give [9]:

Atf =

asf1 + abs1 0 0 asf2 + abs2

0 asf1 + abs1 − (asf2 + abs2) 0

0 − (asf2 + abs2) asf3 + abs3 0

asf2 + abs2 0 0 asf3 + abs3

⇒

⇒ Atf =

atf1 0 0 atf2

0 atf1 −atf2 0

0 −atf2 atf3 0

atf2 0 0 atf3

(74)

The stiffness matrix of the turbofan model is taken as the inverse of thegeneral flexibility matrix in equation 74, using linear algebra this isdescribed as [4] :

K = A−1tf (75)

were A−1tf is given by :

A−1tf =

1atf1atf3 − a2

tf2

atf3 0 0 −atf2

0 atf3 atf2 0

0 atf2 atf1 0

−atf2 0 0 atf1

⇒ (76)

27


Inserting equation 76 into equation 75 will give the stiffness matrix of theturbofan model :

K =

k1 0 0 −k2

0 k1 k2 0

0 k2 k3 0

−k2 0 0 k3

(77)

the elements of the stiffness matrix in equation 77 are given by equation76 as:

k1 = atf3atf1atf3−a

2tf2


2tf2


2tf2

(78)

28

2 MODEL OF THE TURBOFAN 2.4 Summary of the turbofan model

2.4 Summary of the turbofan modelSubstituting the stiffness matrix of equation 77 into the equation of mo-tion for the turbofan model from equation 26 in subsection 2.2.3, will com-pletely define the turbofan model used in this thesis :

mfan 0 0 0

0 mfan 0 0

0 0 Jt 0

0 0 0 Jt

x

y

ϕx

ϕy

ω

0 0 0 0

0 0 0 0

0 0 −Jp 0

0 0 0 Jp

x

y

ϕx

ϕy

+

+

k1 0 0 −k2

0 k1 k2 0

0 k2 k3 0

−k2 0 0 k3

x

y

ϕx

ϕy

= mfanεω

2

cosωt

sinωt

0

0

(79)

As the equation 79 is rather large and bulky it is possible to define a set ofcomplex coordinates that allow equation 79 to be written in a morecompact form [2]. The defined set of complex coordinates are:{

r = x+ iyϕ = ϕy − iϕx

(80)

For equation 79 to be written on the complex form defined by the set ofcomplex coordinates in equation 80, then the second row of equation 79 ismultiplied by the imaginary unit i and then added to the first row. In thesame way the third row is multiplied by the imaginary unit −i and thenadded to the fourth row. This will give the following compact formequation that describes the turbofan model: mfan 0

0 Jt

r

ϕ

+ iω

0 0

0 −Jp

r

ϕ

+

+

k1 −k2

−k2 k3

r

ϕ

= mfanεω2

exp (iωt)

0

(81)

29

3 USING THE TURBOFAN MODEL

3 Using the turbofan modelThe model of the turbofan derived in section 2 will be used on three turbo-fan engines, one is a reference engine to verify the derived turbofan model.For the reference engine some of the parameters used were given by across section drawing of the engine and some were calculated using thedimensions from the cross section drawing. The reference engine parame-ters were then inserted into the derived turbofan model to calculate someresponse values of the reference engine. These calculated response valueswere then verified against known values of the reference engine, and sim-ulations done in DyRoBeS see section 4.The two remaining turbofan engines used in the derived turbofan model ofthis thesis are NSA-1HPT and NSA-2HPT. These two engines are propos-als from VAC on how turbofan engines might be designed for the comingNSA class of aircraft. The two NSA turbofan engines are very similar. Theonly difference between the engines is that NSA-1HPT has one stage inthe high pressure turbine, while NSA-2HPT has two stages in the highpressure turbine which makes this engine slightly longer. The geometryof these engines was derived through a case study of existing turbofan en-gines by engineers at VAC. For the NSA engines some of there parameterswere taken from the cross section drawings of there geometry. But someimportant parameters for the fan disks were scaled from the parametersof the reference engine.

3.1 Parameters of the EnginesIn this subsection the parameters of the turbofan engines used in the tur-bofan model will be defined.

3.1.1 Reference Engine

The calculated parameters for the reference engine are the mass of the fandisk mfan, the mass moments of inertia Jp, Jt of the fan disk, the areamoments of inertia IA, IB of the LP-shaft and the eccentricity ε of CM forthe fan disk.

As stated in section 2 the fan disk consists of a hub and a number ofairfoils. The airfoil mass mafone

of one airfoil for the fan disk are definedas 90% of the fan blade mass, the remaining 10% is the fan blade rootmass. For the reference engine used in this thesis the mass of a fan bladeis defined and thus the mass of a airfoil mafone

is also defined. Thenumber of fan blades nfb on the reference engine is eighteen, and whenthe engine is subjected to a FBO a total mass loss corresponding to twoairfoils is expected. As the fan blade root is made from the same materialas the hub, and due to the way the fan blades are connected to the hub

30

3 USING THE TURBOFAN MODEL 3.1 Parameters of the Engines

the fan blade roots can be seen as part of the hub. The mass of the hubmhub was not known for the reference engine. But as the material of thehub was known using the cross section drawing of the reference engine amass for the hub can be estimated.

Figure 16: Reference engine cross section view of the fan disk hub

As the hub can be seen as a hollow cylinder the mass of the hub can beestimated from figure 16 :

mhubD1 = ρtitanπ(R2

1 −R23

)D1


2 −R23

)D2


2 −R23

)D4

(82)

mhub = mhubD1 − 2mhubD2 − 2mhubD4 (83)

Total mass of the airfoils after a FBO is :

maf = (nfb − 2)mafone(84)

The total mass of the fan disk mfan after FBO is then given by equations83, 84 :

mfan = mhub +maf (85)

31

3.1 Parameters of the Engines 3 USING THE TURBOFAN MODEL

The mass moments of inertia for the fan disk consist of the massmoments of inertia for the hub and airfoils respectively. Using figure 16with equation 82 the following mass moments of inertia for the hub aregiven[6]:

Polar mass moment of inertia of the hubJpD1 = 1

2mhubD1

(R2

1 +R23

)JpD2 = 1

2mhubD2

(R2

2 +R23

)JpD4 = 1

2mhubD4

(R2

2 +R23

)Jphub

= JpD1 − 2JpD2 − 2JpD4

(86)

Transversal mass moment of inertia of the hubJtD1 = 1

12mhubD1

[3(R2

1 +R23

)+D2

1

]JtD2 = 1

12mhubD2

[3(R2

2 +R23

)+D2

2

]+mhubD2

(D2+3D3+2D4

2

)2JtD4 = 1

12mhubD4

[3(R2

2 +R23

)+D2

4

]+mhubD2

(D2+D3

2

)2Jthub

= JtD1 − 2JtD2 − 2JtD4

(87)

The mass moments of inertia for the airfoils for the fan disk are derivedby using the elementary formula of mass moment of inertia with figures17, 18 below representing the airfoil.

Figure 17: Cross section view Figure 18: Rot. plane view

32


The centre of mass CM of the airfoil in figures 17, 18 is set at 42% of theairfoil length and the thickness t of the airfoil is approximated to be 5% ofthe width B. The reference engine has as previously stated eighteenairfoils and when FBO occures a mass loss corresponding to two airfoils isexpected, and for simplicity the remaining airfoils are assumed be evenlyspaced around the hub of the fan disk.

The polar mass moment of inertia is given by figure 17 and the followingmass moment equations [6] :

Jpaf= (naf − 2)

[JpCM

+mafone(R1 + 0.42Laf )2

](88)

where JpCMis :

JpCM=ˆr2dm = mafone

(B2

12+ 0.0897L2

af

)(89)

inserting equation 89 into equation 88 give the expression for the polarmass moment of inertia of the airfoils :

Jpaf= (naf − 2)mafone

[(B2

12+ 0.0897L2

)+ (R1 + 0.42Laf )2

](90)

The transversal mass moment of inertia of the airfoils is given by usingfigure 18. As any of the airfoils could potentially come loose asimplification is made for the approximation of the transversal massmoment of inertia. This simplification implies that after FBO theremaining airfoils will be evenly spread out around the hub, this willcreate a symmetry condition that will simlify the calculation of Jtaf

. Thefollowing expressions of the transversal mass moment of inertia are given:

Jtafone=[JCM +mafone

((R1 + 0.42Laf ) sin θ)2]

(91)

where JCM and θ is given by :

JCM = mafone

(t2

12+ 0.0897L2

af

)sin θ (92)

θ for 22.5o, 45o, 67.5o, 90o (93)

33


Inserting equations 92 and 93 into equation 91 gives :

Jtaf=

∑[mafone

(t2

12 + 0.0897L2af

)sin θ+

+mafone ((R1 + 0.42Laf ) sin θ)2]

for 22.5o, 45o, 67.5o, 90o (94)

The total mass moments of inertia for the fan disk is given by adding themass moments of inertia for the hub and airfoils, the total polar andtransversal mass moment of inertia for the the fan disk is given by :

Jp = Jphub+ Jpaf

(95)

Jt = Jthub+ Jtaf

(96)

A rotational plane view for the shaft parts A and B from figure 9 can beviewed in figure 19 below.

Part A and B

Figure 19: Rotational plane view of the shaft

Using the area moment of inertia for a cylindrical shell gives[5] :

Ix =ˆ

A

y2dA (97)

to introduce cylindrical coordinates the following substitutions must bemade :

dA = rdrdφy = r sinφ (98)

34


substituting the expressions of equation 98 in equation 97 gives :

Ix =

2π

0

R2ˆ

R1

r3 sin2 φdrdφ (99)

as the shaft is symmetrical then Ix = Iy, this together with theintegration of equation 99 gives the area moment of inertia for the shaftparts viewed in figure 19 as :

IA =π

4[R4A2−R4

A1

](100)

IB =π

4[R4B2−R4

B1

](101)

When the fan disk in intact the CM of the fan disk is located along therotational axis, but when airfoils are lost the balance of the fan disk isdisturbed as the mass of the airfoil opposite the lost one will cause theCM to move from the rotational axis and thus create an eccentricity ε,this can be viewed in figure 20 below.

Figure 20: The eccentricity of CM of the fan disk. Only the two lost and twoopposing blades are depicted. The remaining fan blades are not shown.

An expression for the eccentricity can be derived with the followingformula [11] :

xcm =

∑i

mixi∑i

mi(102)

35


using equation 102 in conjunction with figure 20 will give an expressionfor the eccentricity ε:

ε =

[(Dfan

2

)+ (0.42− 1)Laf

]2maf

(mhub + nafmaf )(103)

36


3.1.2 NSA-1HPT and NSA-2HPT

The fan disk for the two NSA engines have the same dimensions and thusis assumed to be identical. As there are no large changes in fan disk designfor turbofan engines some important parameters of the NSA engines fandisk is going to be scaled from the reference engine in subsection 3.1.1. Theparameters that are going to be scaled from the reference engine is turbo-fan mass mfan, the mass moments of inertia Jp, Jt and the shaft radii RB1

and RB2 of shaft part B. Together with the mentioned parameters above,also the bearings stiffnesses kRB and kFB have to be scaled. The remain-ing parameters will be given from the respective cross section drawings ofthe engines.

As with the reference engine the mass of the fan disk mfan is composed ofthe fan disk hub mass mhub and the fan disk airfoil mass maf . For theNSA engines, the mass of one airfoil mafone

is known and the number ofairfoils naf is projected to be eighteen. For an intact fan disk of a turbofanengine, the ratio between the fan disk hub mass mhub and the fan diskairfoil mass maf is assumed constant. With this assumption the fan diskhub mass mhub can the NSA engines can be scaled as follows :[

mhub

nafmafone

]NSA

=[

mhub

nafmafone

]RE

⇒

⇒ [mhub]NSA =[

mhub

nafmafone

]RE

[nafmafone]NSA (104)

As it is not clear what kind of material the fan disk hub of the NSAengines is going to be made from, nor are the dimensions clearly defined.In that case the best way to estimate the mass moments of the inertia forthe fan disk is to scale them from the reference engine. As in the casewith the mass mfan of the fan disk the ratio between transversal andpolar mass moments of inertia are assumed to be constant. To estimatethe relation between the polar mass moment of inertia Jp for the NSA andreference engines an assumption, that the ratio between polar massmoment of inertia Jp for a fan disk and mass moment of inertia for asphere with a diameter and mass equal to that of a fan disk is constant,was made.

The mass moment of inertia for a sphere is [7] :

J = mD2

4(105)

37


the relation between polar mass moment of inertia Jp of the NSA andreference engine using equation 105 is : Jp

mfanD2

fan

4

NSA

=

Jp

mfanD2

fan

4

RE

⇒

⇒ [Jp]NSA =

Jp

mfanD2

fan

4

RE

[mfan

D2fan

4

]NSA

(106)

the relation between the polar and transversal moments of inertia isgiven by : [

JtJp

]NSA

=[JtJp

]RE

⇒

⇒ [Jt]NSA =[JtJp

]RE

[Jp]NSA (107)

As there was no estimation on how thick part B of the LP-shaft was forthe NSA engines but there was a estimate on the thickness of part A.Using the thickness of the LP-shaft for the reference engine, gives thefollowing relation that can approximate the thickness of part B of theLP-shaft : [

RBRA

]NSA

=[RBRA

]RE

⇒

⇒

[RB1 ]NSA =

[RB1RA1

]RE

[RA1 ]NSA

[RB2 ]NSA =[RB2RA2

]RE

[RA2 ]NSA

(108)

The bearing stiffness for the NSA engines are scaled against the referenceengine using the static unbalance force Funbstatic produced by the fan disk.The static unbalance force is given by :

Funbstatic= mfanεω

2 (109)

38


As the stiffness for both the front and rear bearing of the reference engineare equal, then the same is assumed for the NSA engines and togetherwith equation 109: [

kRBFunbstatic

]NSA

=[

kRBFunbstatic

]RE

⇒

[kRB = kFB ]NSA =[

kRBmfanεω2

]RE

[mfanεω

2]NSA

(110)

The NSA engine parameters that are not derived using equations 84, 85,100, 101 and 103 or scaled using equations 104, 106 - 108 and 110 are :

Parameter NSA-1HPT NSA-2HPT Unit CommentL 2.2806 2.2810 m Length of

LP-shaftα 5/7 73/92 Fraction of the

length for part Aβ 2/7 19/72 Fraction of the

length for part BDfan 1.77 1.77 m Fan diameterLaf 0.6160 0.6160 m Lenght of airfoilRA1 0.045 0.045 m Inner radius of

shaft part ARA2 0.055 0.055 m Outer radius of

shaft part Amafone

5.7889 5.7889 kg Mass of one airfoilEA 185 185 GPa E-modulus of

shaft part AEB 120 120 GPa E-modulus of

shaft part BkRB = kFB 0.110 0.110 GN/m Bearing stiffnessω 448.9336 448.9336 rad/s Rotational speed

of fan disknaf 18 18 st Number of airfoils

Table 1: Some additional parameters for the NSA engines

39

3.2 Eigenvalues 3 USING THE TURBOFAN MODEL

3.2 EigenvaluesThe eigenvalues for the turbofan model derived in section 2 represent thenatural frequency of the turbofan model. When the rotational speed ofthe turbofan model equals the natural frequency of the turbofan model itwill induce large amplitudes of oscillation of the unbalanced fan disk. Thefrequency at which the rotational speed of the turbofan model correspondto the natural frequency is known as the critical speed. As the dynamicbehavior for this kind of model depends on the spin speed of the fan disk,then the dynamic behavior is usually summarized by a plot of the naturalfrequencies as a function of the rotational speed, this plot is known asCampbell diagram.

3.2.1 Whirl speed

Whirl speed is defined as the rotational vibration speed of the rotor centre,in this case the fan disk centre. Related to the spin the whirl speed caneither be forward or backward. If the whirl and spin speeds of the fan diskare in the same direction, this is called a forward whirl [2].

3.2.2 Campbell diagram

The rotational vibration speed of the undamped turbofan model can bestudied by solving the homogeneous part of equation 81. The homogeneouspart of equation 81 is : mfan 0

0 Jt

r

ϕ

+ iω

0 0

0 −Jp

r

ϕ

+

+

k1 −k2

−k2 k3

r

ϕ

= 0 (111)

To solve the homogeneous equation the following type of solution isintroduced : r = r0 expiλt

ϕ = ϕ0 expiλt(112)

where λ in equation 112 corresponds to the eigenvalue of the turbofanmodel. Differentiating equation 112 two times and inserting it intoequation 111 gives the following characteristic equation : −mfanλ

2 + k1 −k2

−k2 −Jtλ2 + Jpλω + k3

r0

ϕ0

= 0 (113)

40

3 USING THE TURBOFAN MODEL 3.2 Eigenvalues

The determinant of the matrix for the characteristic equation 113 is :

det

−mfanλ2 + k1 −k2

−k2 −Jtλ2 + Jpλω + k3

(114)

solving the determinant using linear algebra [4], yields the followingequation :

λ4−(ωJpJt

)λ3−

(k1

mfan+k3

Jt

)λ2+

(ω

k1JpmfanJt

)λ+(k1k3 − k2

2

mfanJt

)= 0 (115)

For each spin speed ω, equation 115 will give four real rootscorresponding to four eigenvalue modes that occurs at differentfrequencies. Plotting the eigenvalue modes given by equation 115 as afunction of spin speed ω produces a Campbell diagram that contains threehorizontal asymptotes and one asymptote that diverges to λ = ω

Jp

Jt.

41

3.2 Eigenvalues 3 USING THE TURBOFAN MODEL

Figure 21: Example of a Campell diagram[2]

The above figure 21 gives an example of a Campell diagram, that wascreated for an Jeffcott roter viewed on the right side in figure 21. Thethree curves with horizontal asymptotes correspond to two backwardwhirling motions (below zero in figure 21) and one to forward whirling(above zero in figure 21). As defined in subsection 3.2.1 the turbofanmodel is in synchronous forward whirling motion as is the Jeffcott rotor infigure 21 above. Thus, for the turbofan model, similar to the Jeffcott rotorabove, the two backward whirling motions corresponding to the lower twocurves will not be excited.

42

3 USING THE TURBOFAN MODEL 3.2 Eigenvalues

3.2.3 Critical Speed

The critical speed of the turbofan model is as stated previously a point werethe whirl speed of the turbofan correspond to a eigenvalue mode derivedfrom equation 115.

Critical speeds have the following defining features [2]:

• They correspond at well-defined values of the spin speed.

• At the critical speed the amplitude grows linearly in time if no damp-ing is present. It amplitude can be maintained within reasonablelimits and, as a consequence, a critical speed can be passed.

• The value of the critical speed is fixed, but the value of the maximumamplitude depends on the amplitude of the perturbation that causesit. In particular, the main flexural critical speeds do not depend onthe amount of unbalance, but the amplitude increases with increas-ing unbalance.

To get an equation for the critical speed of the turbofan model, one makesthe substitution λ = ω into equation 115 that gives a quadratic equationfor ω2 which yields :

ωcr = ±

√√√√k1 (Jp − Jt)−mfank3 ±√

[k1 (Jp − Jt) +mfank3]2 − 4mfan (Jp − Jt) k22

2mfan (Jp − Jt)(116)

where we will only consider positive rotational speed of equation 116.

Inserting the critical speed line λ = ω into the Campbell diagram aspresented in figure 21 in subsection 3.2.2 will give a graphicrepresentation of where a critical speed is.

As can be seen in figure 21 the turbofan model has only one critical speedwhich is when the critical speed line intersects the eigenvalue curvecorresponding to the forward whirling motion of the turbofan model.

43

3.3 Unbalance response 3 USING THE TURBOFAN MODEL

3.3 Unbalance responseThe unbalance response of the turbofan model will be given by the am-plitudes of the displacement and displacement angle. They are derivedusing equation 81 and a type of solution that is very similar to the oneused in subsection 3.2.2 under equation 112, the only difference is that theeigenvalue term λ is replaced with the whirl speed ω of the turbofan modeldescribed in subsection 3.2.1.[2, 3]

The type of solution for the unbalance response is : r = r0 expiωt

ϕ = ϕ0 expiωt(117)

using the type of solution given by 117 in equation 81 will yield the thefollowing :

−ω2mfanr0 + k1r0 − k2ϕ0 = mfanεω2

−ω2Jtϕ0 + ω2Jpϕo + k3ϕ0 − k2r0 = 0(118)

Rearranging the terms in equation 118 will give two relations thatdescribe the amplitudes of the displacement and displacement angle ofthe fan disk, in response to the unbalance in the turbofan model.

The amplitude of the displacement is given as :

r0 = mfanεω2

((Jp − Jt)ω2 + k3

χ

)(119)

and the amplitude of the displacement angle is given by :

ϕ0 = mfanεω2 k2

χ(120)

In equations 119 and 120 the variable χ is given as :

χ = − (Jp − Jt)mfanω4 + [(Jp − Jt) k1 −mfank3]ω2 + k1k3 − k2

2 (121)

44

3 USING THE TURBOFAN MODEL 3.4 Engine structure loads

3.4 Engine structure loadsTo estimate the mechanical loads on the engine structure when the en-gines fan loses a fan blade, it is necessary to know what kind of unbalanceloads the dynamic behavior of the fan disk would give rise to. One un-balance load created by the fan disk is the static unbalance force which isrepresented by equation 109 in subsection 3.1.2 as :

Funbstatic= mfanεω

2

this unbalance load is a result of the fan disk CM being displaced from theGC of the turbofan model. The static unbalance force is a rapidly grow-ing force with increasing spin speed, but due to the dynamic behavior ofthe fan disk the static unbalance force as displayed above is not valid forthe entire FBO envelope. When the fan disk is rotating with sub-criticalspin speed the direction of the displacement of the fan disk from the GCof the turbofan model will be in phase with the eccentricity ε of the fandisk, i.e. they will be acting in the same direction and the unbalance loadincreases. But when the fan disk is rotating with super-critical spin speedthe fan disk will flip and continue to rotate around the CM created by theunbalance. This means that with increasing spin speed the displacementwill decrease, eventually approching −ε as spin speed tends to infinity. Asa result of this the unbalance load will start decrease when super-criticalconditions are fulfiled. This rotor dynamic phenomena that accrues whenthe fan disk passes the critical speed is known as self-centering.[10, 2]

Sub-critical conditions Super-critical conditions

Figure 22: A illustration of self-centering[2, 3]

To take into account the dynamic effects of self-centering, of the fan

45

3.4 Engine structure loads 3 USING THE TURBOFAN MODEL

disk at super-critical conditions, on the unbalance load the amplitude ofthe displacement given by equation 119 is subsection 3.3 is added to theeccentricity ε of the static unbalance force given by equation 109 in sub-section 3.1.2, also displayed above. This will yield the dynamic unbalanceforce of the turbofan model represented as :

Funbdynamic= mfan (ε+ ro)ω2 (122)

Together with the unbalance forces there is also a gyroscopic momentacting on the fan disk. The gyroscopic moment of the fan disk is opposingthe displacement of the fan disk from the GC of the turbofan model, thusalways trying to force the fan disk to its neutral position. This gyroscopicmoment is derived from equations 19 and 20 in subsection 2.2.2. Bytransforming equations 19 and 20 to complex coordinates as defined byequation 80 gives :

Mϕ = Jtϕ− iJpωϕ (123)

by implementing the type of solution given by equation 117 fromsubsection 3.3 onto equation 123 above, will give the gyroscopic momentequation :

Mgyro = (Jp − Jt)ω2ϕ0 (124)

Apart from the loads mentioned above in this subsection there is oneadditional load that has to be considered, this is the force exerted on thebearings by the bending of the LP-shaft. As the span of the LP-shaftbetween the bearings is large and the mass of the shaft is not negligible abending of the shaft will exert some loads on the bearings as well. Notethat this shaft bending force was not used for the critical speed estimates,it was only used for the calculation of the bearing and engine mount loads.

Figure 23: Bending of the LP-shaft

The loads that are exerted on the bearings due to the shaft bending asseen in figure 23 are given by :

FSB =mshaftpartA

4SBω2

2⇒

46


⇒ FSB =Mgyroρsteelπ

(R2A2−R2

A1

)L3α3ω2

32EAIA(125)

The forces from the fan dominate with respect to the LP-shaft part A, andthe centrifugal force of this part is of a smaller magnitude. Therefore theshape of the deformed shaft was taken from elementary beam theory for abeam supported at two points (the bearings) with the bending momentfrom the fan applied at the front bearing. As the exact shaft mean line(including its own centrifugal force) was not determined, the maximumdeflection was taken as a conservative (high) estimate of the deformedshafts center of mass.

The above derived unbalance and shaft loads can now be implemented toestimate the bearing and engine mount loads, which are the two enginestructure loads that were considered in this thesis.

3.4.1 Bearing loads

To estimate the bearing loads a free-body diagram of the turbofan modelwas used. With the free-body diagram an estimate could be made of howthe unbalance loads of the fan disk would be distributed between thefront and rear bearings. The model in the free-body diagram is seen as aquasi-static system.

Figure 24: Free-body diagram, bearing

Using the free-body diagram displayed in figure 24 above, the followingset of equations is given :

↑ Funb + FFB + FRB − 2FSB = 0 (126)

yFB FunbβL+ FSBαL− FRBαL−Mgyro = 0 (127)

47

3.4 Engine structure loads 3 USING THE TURBOFAN MODEL

From the set of equations displayed by equations 126 and 127, thefollowing relations are derived :

FFB =Mgyro

αL+ FSB −

Funbα

(128)

FRB = Funbβ

α+ FSB −

Mgyro

αL(129)

these relations displayed in equations 128 and 129 above are used toestimate the distribution of the unbalance loads between the front andrear bearings.

3.4.2 Engine mount loads

To estimate the engine mount loads, a model of a engine was used thattogether with the unbalancing loads also included forces on the mountscreated by the engine thrust, engine weight and shaft bending of the LP-shaft. As with the bearing loads described in subsection 3.4.1 the model isseen as a quasi-static system and a free-body diagram of this engine modelis displayed in figure 25 below.

Figure 25: Free-body diagram, engine mounds

Using the free-body diagram in figure 25 the following set of equations isgiven :

↑ Funb + FFM + FT sin θ + FRM − gmengine − 2FSB = 0 (130)

48


→ FT cos θ − FN = 0 (131){yT FunbL1 + FFML3 + FSBL5 + FNL7+

+gmengineL4 − FSBL2 − FRML6 −Mgyro = 0(132)

Equation 131 gives :

FT =FN

cos θ(133)

inserting equation 133 into equation 130 and utilizing equation 132 willgive relations for the front and rear engine mount loads as follows :

FFM =Mgyro + FSB (2L6 + L2 − L5) + gmengine (L6 − L4)

L3 + L6+

+−Funb (L1 + L6)− FN (L6 tan θ + L7)

L3 + L6(134)

FRM =Funb (L1 − L3) + FSB (2L3 + L5 − L2) + gmengine (L3 + L4)

L3 + L6+

++FN (L7 − L3 tan θ)−Mgyro

L3 + L6(135)

49

4 DYROBES

4 DyRoBeSDyRoBeS stands for Dynamics of Rotor Bearing Systems, and is a commer-cial simulation software specifically developed for advanced rotor dynamicanalysis. The structure of the software is such that it consists of a majorcode enabling the user to carry out eigenfrequency analysis and state re-sponse analysis. There are also minor packages that contain a number ofauxiliary codes for calculating bearing properties and other relevant rotorsystem data. The DyRoBeS code is essentially based on well establishedrotor dynamic and bearing theory available in the literature.

As with any simulation software the first step is to create a model. Ageometric model in DyRoBeS is created in much the same way as with aordinary FEM software, the input of material properties like stiffness,damping and mass is also very similar to a FEM software. HoweverDyRoBeS has one important difference when compared to ordinary FEMsoftware, the difference lies in that DyRoBeS has a function for inputinggyroscopic rotor data which plays a very important role in rotor dynamicanalysis. Together with the mentioned functions DyRoBeS also facilitatespowerful pre-and post processors.

The solver used in DyRoBeS utilizes a large system of differentialequations which are transformed into a matrix representation. Thesubsequent iteration process in order to find a solution is similar to theprocedures used in FEM codes.

With the DyRoBeS software one can study in detail lateral and torsionalrotor motion as well as axial vibration systems. The steady state responseanalysis is not only restricted to systems subjected to rotor speedsynchronous periodic loads, it is also possible to use the software toanalyze models subjected to non synchronous loads like sub synchronousor super synchronous periodic disturbances. Besides the steady stateresponse analysis, DyRoBeS provides a special feature for time transientresponse analysis.

In order to give some insight of how a model is created using DyRoBeS, ashort description will follow showing how a model is created. Also adescription of the different analysis options will be given.

50

4 DYROBES 4.1 Creating a model

4.1 Creating a modelTo create a model in DyRoBeS a set of steps are required to be done. Figure26 displays the Data Editor which shows a number of folders with differenttabs, in these tabs and folders data can be entered, which will define themodel.

Figure 26: Modeling/Data Editor

To create a model in DyRoBeS that describes the turbofan model used inthis thesis, it was only necessary to utilize some of the folders in the DataEditorn shown in figure 26. The folders in the Data Editor that were usedto create the model are as follows :

1. Material

2. Shaft Elements

3. Disks

4. Unbalance

5. Bearings

Each of these will be summarized in the subsequent subsections.

4.1.1 Material

In the material folder the material properties of the materials used weredefined. In the model used in this thesis the only material property thatwas of interest was the Young’s modulus.

51

4.1 Creating a model 4 DYROBES

4.1.2 Shaft Elements

The Shaft Elements folder is where the shaft(s) of the model are defined.A shaft in the Data Editor is made up of Elements with the numberingfor each shaft starts at the left end. At each elemets end there is a Sta-tion located. Note that in rotordynamics the term Station is commonlyused instead of the term Node which is commonly used in FEM litera-ture. In DyRoBeS one can choose between three types of Elements whichare: Cylindrical, Conical and User’s Supplied elements. Each element canhave several subelements that allow for resonable flexibility when model-ing systems with geometric discontinuities. In turn each subelement cancontain a number of layers. The use of subelements when modeling com-plex systems is prefered as it saves computational time, but at a cost oflittle accuracy in the result.

Figure 27: Shaft Elements folder

52


The different fields in figure 27 are used to input data to create a shaft.The most important fields are commented shortly below.

• Starting Station #: The starting station ,also element number ofthe current shaft. This number is updated automatically and is usedto remind the user of the starting element number of the particularshaft .

• Ele: Element number. The Element number in the first row must beequal to the Starting Station #.

• Sub: Subelement number. Each element can contain a max of 20subelements.

• Lev: Level for the subelement. Each subelement can have a max of5 levels.

• Mat: Material number. Defines the material properties the elementinherits, the material properties are defined in the Material folder.

• Length: Defines the length of a subelement.

• Mass ID: Defines The inner mass diameter of a element.

• Mass OD: The outer mass diameter of a element.

• Stiff ID: The stiffness inner diameter of a element.

• Stiff OD: The stiffness outer diameter of a element.

For a standard cylindrical element which was used in this thesis, the MassID & OD are used for calculating the kinetic energy of the system whichcorrespond to the mass and gyroscopic matrices. The Stiffness ID & ODare used for calculating the systems potential energy that is the stiffnessmatrix. In the model plot created, upper half represents the Mass modeland the lower half corresponds accordingly to the Stiffness model.

4.1.3 Disks

Like the Shaft Elements folder in figure 27 was used to define the LP-shaftof the turbofan model in DyRoBeS, likewise the Disks folder will be used todefine the fan disk of the turbofan model in DyRoBeS. In DyRoBeS a diskcan be located at any element station, but the disk can not be placed at asubstation within a element. If needed more than one disk can be placedat the same station.

53


Figure 28: Disks folder

The input data fields in figure 28 used to create the fan disk of theturbofan model are explained shortly below:

• Type: Defines whether the disk is to be seen as a Rigid or Flexibledisk.

• Stn: The station number at which the disk is located.

• Mass: Here the mass of the disk is defined.

• Dia. Inertia: Gives the disk a diametral moment of inerta, which isalso known as transveral moment of inertia.

• Ploar Inertia: Gives the disk a polar moment of inertia.

• Lenght, ID, OD: These values are used to define the geometry thedisk.

• Density: Defines the density of the material used in the disk.

54


If the Density of the disk in defined as non-zero then additional mass prop-erties will be calculated based on the disk geometry, and will be added tothe specified mass properties. Due to the complex shape of the fan diskthis function was not utilized in the setup of the fan disk in DyRoBeS.

For a rigid disk as was the case with the fan disk, will in the model ofDyRoBeS have four degrees-of-freedom of which are two rotational andtwo translational degrees-of-freedom. These degrees-of-freedom willcorrespond to the element stations to which the disk is attached to.

4.1.4 Unbalance

Using the Unbalance folder one can define the amount and placement ofthe unbalance loads. These mass unbalances are discrete and located atdifferent planes with a magnitude equal to the mass multiplied the eccen-tricity (m ∗ ε) or a constant synchronous excitation by a force independentof the rotational speed with amplitude equal to F0.

Figure 29: Unbalance folder

The input data fields in the unbalance folder are brifely explained be-low:

• Ele: Element number at which the unbalance is placed.

• Sub: Subelement number at which the unbalance is placed, if anysubelements exist.

• Type: What type of unbalance the system is subjected to. 0 = Typicalmass unbalance force, 1 = Constant synchronous excitation.

• Left Unb Amp: Left end unbalance amplitude, either (m ∗ ε) or F0.

55


• Left Ang: Left end unbalance phase angle measured from x-axis indegrees.

• Right Unb Amp: The amplitude of the unbalance at the right end,either (m ∗ ε) or F0.

• Right Ang: Phase angle of the right end.

For an unbalanced disk at station i, the unbalance can be placed at the leftend of element i. However if the unbalanced disk is located at the rightend (last station) of a shaft, there is no element number corresponding tothe last station number of the shaft. This unbalance can then be placed atthe right end of the last subelement instead. The unbalance is displayedin the model as a red arrow.

4.1.5 Bearings

All bearing/damper/seals/support forces acting on, or interacting between,the rotating assemblies and non-rotating structures fall into the Bearingsfolder. There exist eleven types of bearings to choose from, amoung othersone can choose linear/non-linear, real/pseudo, fluid film, rolling elements,seals and active magnetic bearings. Non-linear bearings can only be usedin time transient analysis with the exception for squeeze film dampersand generalized non-linear isotropic bearings, these can be also used insteady state synchronous response analysis as well. Also different types ofbearings can be attached to the same station at the same time. In figure30 the following input data fields were used to create the bearings :

• Add Brg/Del Brg: Adds or deletes a bearing from the system.

• Previous/Next: Scrolls between the existing bearings in the system.

• Station I and Station J: The two stations that the bearings areconnected to.

• Foundation: If this box is checked then the bearing is connected tothe foundation. The input data for the foundation is provided underthe Foundation folder as seen in figure 26.

• Angle: Measures the degrees beween the global fixed coordinate sys-tem and the local coordinate system were the bearing coefficients aredefined. Measured in the counterclockwise direction.

• Type: Defines what kind of bearing is used. All the available bear-ings types supported by DyRoBeS can be viewed in figure 30 above.

• Coefficients: Kxx,Cxx,Kyy,Cyy,Kxy,Cxy, ect. are damping and stiff-ness coefficients for the bearings.

56

4 DYROBES 4.2 Simulations

Figure 30: Bearings folder

The bearings used to create the turbofan model was a Linear ConstantBearing type. A foundation was not used instead the linear bearings werefixed to the rigid ground, this is indicated by putting Station J = 0.

4.1.6 Turbofan model in DyRoBeS

Using the Data Editor of DyRoBeS as described under subsections 4.1.1through 4.1.5 the following generic model was created: The model shownabove in figure 31 can now be used to preform simulations both for thereference engine and the NSA engines, just by changeing some numericalvalues.

4.2 SimulationsAs mentioned in the introduction of DyRoBeS, the software can preformsimulations of lateral and torsional rotor motion as well as axial vibration

57

4.2 Simulations 4 DYROBES

Figure 31: Turbofan model created in DyRoBeS

systems. The simulations performed in this thesis were of lateral nature.With the lateral rotor motion simulations one can preforme a couple ofdifferent analysis as shown in the Lateral Analysis Option tab displayedin figure 32below.

Figure 32: Lateral Analysis Option

The analysis that are to be preformed were the Critical Speed Anal-ysis and Steady State Synchronous Repsonse Analysis. But due to someunforeseen licence problems with the DyRoBeS software, only the Critical

58

4 DYROBES 4.2 Simulations

Speed Analysis was possible to be preformed.

4.2.1 Critical Speed Analysis

Initiating a Critical Speed Analysis is fairly straight forward as can beseen i figure 32. There are only three options to be choosen:

• Spin/Whirl Ratio: Describes the type of critical speed under calcu-lation. The typical values for the Spin/Whirl Ratio are:

– 1 = Forward synchronous critical speeds– -1 = Backward synchronous critical speeds– 0 = Planar critical speeds for non-rotating systems– 2 = Half frequency whirl (Subsynchronous criticals), etc.

• No. of Modes: This parameter is used to calculate the selected modeshapes. In practice, only the lowest 3 critical speed modes are ofimportance.

• Stiffness: An undamped critical speed calculation assumes a isotropicsystem, but the bearing stiffness (Kxx and Kyy) in the bearing datamay not be the same. With this option one can select the stiffness tobe used in the calculation.

59

5 RESULTS

5 Results

5.1 Eigenvalue analysis

The results of the eigenvalue analysis with the turbofan model describedin section 2 will, together with the critical speed analysis done by the Dy-RoBeS software described in section 4, be presented for each engine sepa-retly.

5.1.1 Reference engine

For the reference engine in addition to the results from eigenvalue analy-sis with the turbofan model and critical speed analysis done by DyRoBeS,there is a reference value of the critical speed for the fan provided by aFEM model of the reference engine.

Using equation 115 given in sub-subsection 3.2.2 and doing the criticalspeed analysis with DyRoBeS as described in subsection 4.2, one will getthe Campbell diagram and a figure of the critical speed analysis showingthe first eigenmode of the reference engine as follows:

Figure 33: Campell diagram Figure 34: First eigenvalue mode

With equation 116 from sub-subsection 3.2.3 an exact value for thecritical speed can be extrapolated from the Campbell diagram in figure33. We relate the calculated critical speeds of the turbofan model andDyRoBeS to the reference value provided by the FEM model of thereference engine. Note that the reference value in table 2 below isnormalized and that no actual values are provided.

60

5 RESULTS 5.1 Eigenvalue analysis

ValueReference value FEM critical speed ωcrF EM/ωcrF EM

1Turbofan model critical speed ωcr/ωcrF EM

0.9690DyRoBeS model critical speed ωcr/ωcrF EM

0.9685

Table 2: Comparison of critical speeds for the reference engine

5.1.2 NSA 1-HPT engine

Applying equations 115 and 116 from sub-subsections 3.2.2 and 3.2.3, to-gether with the DyRoBeS simulation for the NSA 1-HPT engine gives thefollowing results.

The Campbell diagram and first eigenvalue mode for the NSA 1-HPT are:

Figure 35: Campbell diagram Figure 36: First eigenvalue mode

the critical speeds are summarized as follows:

Value UnitSpin/Rotational speed ω 448.93 rad/sTurbofan model critical speed ωcr 135.98 rad/sDyRoBeS critical speed ωcr 136.03 rad/sDifference Turbofan model/DyRoBeS < 0.1 %

Table 3: Critical speeds for the NSA 1-HPT

61

5.1 Eigenvalue analysis 5 RESULTS


Using the same approach as for the NSA 1-HPT engine is sub-subsection5.1.2, the following Campbell diagram and first eigenvalue mode is pro-duced:

Figure 37: Campbell diagram Figure 38: First eigenvalue mode

and the critical speeds corresponding to figures 37 and 38 are :

Value UnitSpin/Rotational speed ω 448.93 rad/sTurbofan model critical speed ωcr 191.28 rad/sDyRoBeS critical speed ωcr 191.32 rad/sDifference Turbofan model/DyRoBeS < 0.1 %

Table 4: Critical speeds for the NSA 2-HPT

62

5 RESULTS 5.2 Bearing loads

5.2 Bearing loadsThe results of the bearing load analysis done by the turbofan model, will bepresented for each engine separatly. For the reference engine the resultsproduced by the turbofan model are going to be compared with referencevalues given by the manufacturer of the engine.


At the rotational speed of the fan were the fan is subjected to a FBO, equa-tions 109, 122, 124 and 125 give the initial forces on the reference engine.Applying them to equations 128 and 129 in sub-subsection 3.4.1 will givethe bearing loads when subjected to static unbalance loads Funbstatic

anddynamic unbalance loads Funbdynamic

. These are going to be compared tothe maximum transiant bearing loads from the FEM model provided bythe manufacturer. Table 5 provides the difference between the results ofthe turbofan model and the FEM model from the manufacturer, in percent.

Static unbalance load Value UnitFront Bearing load difference −7.52 %Rear Bearing load difference 8.98 %Dynamic unbalance load Value UnitFront Bearing load difference −73.94 %Rear Bearing load difference −31.35 %

Table 5: Comparison of bearing loads for the reference engine


Using equations 109, 122, 124 and 125 at the rotational speed for the NSA1-HPT engine were FBO accures, gives the FBO forces displayed in table6 below:

Value UnitSpin/Rotational speed ω 448.93 rad/sStatic unbalance force Funbstatic

1161.1 kNDynamic unbalance force Funbdynamic

333.95 kNGyroscopic moment Mgyro 99.14 kN/mShaft bending force FSB 92.62 kN

Table 6: FBO initiated forces and moment of the NSA 1-HPT engine

63

5.2 Bearing loads 5 RESULTS

Bearing loads for the NSA 1-HPT engine, for both the static unbalanceload Funbstatic

and dynamic unbalance load Funbdynamicfrom table 6 give

the bearing loads in table 7 below. From the results for the referenceengine in table 5 one could conclude that the bearing loads correspondingto the static unbalance are probably closer to the real value of the loads.

Static unbalance load Value UnitTurbofan model Front Bearing load FFB 1471.1 kNTurbofan model Rear Bearing load FRB 495.24 kNDynamic unbalance load Value UnitTurbofan model Front Bearing load FFB 313.81 kNTurbofan model Rear Bearing load FRB 165.09 kN

Table 7: Bearing loads for the NSA 1-HPT engine


Using the same approach as for the NSA 1-HPT engine in subsection 5.2.2gives the FBO forces as :

Value UnitSpin/Rotational speed ω 448.93 rad/sStatic unbalance force Funbstatic

1161.1 kNDynamic unbalance force Funbdynamic

731.69 kNGyroscopic moment Mgyro 120.47 kN/mShaft bending force FSB 154.11 kN

Table 8: FBO initiated forces and moment of the NSA 2-HPT engine

The static and dynamic unbalance forces Funbstaticand Funbdynamicfrom

table 8 respectivly, gives the following bearing loads in table 9. From theresults for the reference engine in table 5 one could conclude that thebearing loads corresponding to the static unbalance are probably closer tothe real value of the loads.

64

5 RESULTS 5.2 Bearing loads

Static unbalance load Value UnitTurbofan model Front Bearing load FFB 1242.6 kNTurbofan model Rear Bearing load FRB 389.69 kNDynamic unbalance load Value UnitTurbofan model Front Bearing load FFB 701.42 kNTurbofan model Rear Bearing load FRB 277.95 kN

Table 9: Bearing loads for the NSA 2-HPT engine

65

5.3 Engine mount loads 5 RESULTS

5.3 Engine mount loads

As for the results of the bearing loads in subsection 5.2 the engine mountloads are going to be presented separetly for each engine. Also as in thecase with bearing loads, the reference engine mount loads are going to becompared against reference values provided for from the manufacturer.


Using the initial forces like in subsection 5.2.1 and applying them to equa-tions 135 and 134 will give the engine mount loads when subjected to staticunbalance loads Funbstatic and dynamic unbalance loads Funbdynamic

. Theseare going to be compared to the maximum transient mount loads from theFEM model provided by the manufacturer. Table 10 provides the differ-ence between the results of the turbofan model and the FEM model fromthe manufacturer, in per cent.

Static unbalance load Value UnitFront Mount load difference 233.27 %Rear Mount load difference 23.03 %Dynamic unbalance load Value UnitFront Mount load difference −7.15 %Rear Mount load difference −23.68 %

Table 10: Comparison engine mount loads for the reference engine

5.3.2 NSA 1-HPT

With the FBO forces of Table 6 inserted into equations 135 and 134 willgive the engine mount loads for the NSA 1-HPT engine as:

Static unbalance load Value UnitTurbofan model Front Mount load FFM 1641.2 kNTurbofan model Rear Mount load FRM 679.96 kNDynamic unbalance load Value UnitTurbofan model Front Mount load FFM 361.99 kNTurbofan model Rear Mount load FRM 227.87 kN

Table 11: Engine mount loads for NSA 1-HPT engine

Looking at the results for the reference engine in table 10 the enginemount loads corresponding to the dynamic unbalance in table 11 are prob-ably closer to the real values for the engine mount loads.

66

5 RESULTS 5.3 Engine mount loads

5.3.3 NSA 2-HPT

As with the NSA 1-HPT engine in subsection 5.3.2 above, using the FBOforces from table 8 in subsection 5.2.3 with equations 135 and 134 givesthe following engine mount load for NSA 2-HPT engine:

Static unbalance load Value UnitTurbofan model Front Mount load FFM 1334.3 kNTurbofan model Rear Mount load FRM 493.56 kNDynamic unbalance load Value UnitTurbofan model Front Mount load FFM 755.98 kNTurbofan model Rear Mount load FRM 346.62 kN

Table 12: Engine mount loads of NSA 2-HPT engine

Looking at the results for the reference engine in table 10 the enginemount loads corresponding to the dynamic unbalance in table 12 are prob-ably closer to the real values for the engine mount loads.

67

6 DISCUSSION

6 DiscussionThe turbofan model that was created contained the main features of a rotordynamic model. As such it was well suited to be implemented for use inproblems like the one stated in this thesis. The results of the eigenvaluesimulations show that the rotor dynamical features of the turbofan modelwere in agreement with the commercial rotor dynamic software DyRoBeS,and the agreement with reference data was also high.

The bearing and engine mount loads used as reference data was the sumof vertical loads on the bearings and engine mounts of the referenceengine. This proved to be a good comparison to the quasi-static bearingand engine mount systems given by figures 24 and 25. From the resultsin subsections 5.2.1 and 5.3.1, which are the result of the bearing andengine mount simulations done on the reference engine by the turbofanmodel, it can be seen that there is an inconsistency in the results. Thisinconsistency is related to the dynamic and static unbalance loads used inthe simulations.

Table 5 in subsection 5.2.1 show that for the bearing loads a goodapproximation to the reference values is given when applying the staticunbalance force, while implementing the dynamic unbalance force in thebearing loads simulations resulted in poor comparison with the referencevalues.In the case of engine mount loads the inconsistency is vice versa as can beseen in table 10 in subsection 5.3.1. When applying the dynamicunbalance force to the system it provides a good approximation to thereference values of the engine mount loads, but when applying the staticunbalance force the approximation is very poor.

This inconsistency could be explained by viewing the static unbalanceforce as an transient overshoot force which occurs at the instant of FBO,this force is then transmitted and absorbed in the bearings. Within a veryshort time frame the dynamic effects of self-centring comes into effect,and the transient overshoot force fades out and is transformed into asteady-state force that correspond to the dynamic unbalance force. Thisdynamic unbalance force continues to act on the engine and induces theloads on the engine mounts.

The accuracy of the results in all of the simulations is very much affectedby input values used in the governing equations. Most of the input valuesfor the different turbofan engines like masses, moments of inertia andlengths used in this thesis are approximations or scaled values. Thiscould most certainly contribute to the accuracy of the results and shouldbe improved upon in order to improve accuracy of the results.

68

7 CONCLUSIONS

7 ConclusionsThis thesis has shown that it is possible to use a scaled down model of aturbofan engine to predict FBO loads for a preliminary design study. It hasbeen shown through simulations and comparison with existing data thatthe overall accuracy of the turbofan model is reasonably good.

As the turbofan model produced satisfactory results in the simulations ofeigenvalue problems, it can be concluded that the model can be used withhigh accuracy for solving these kinds of problems.

The turbofan model gives an easy method to predict bearing and enginemount loads at a preliminary design stage. That can be used inpreliminary dimensioning of engine components.

Bearing loads given by the turbofan model are lower than the maximumvalue of the real loads in the front bearing, and higher than the FEMmodel value in the rear bearings.

Engine mount loads computed by the turbofan model are underestimatedboth for front and rear engine mounts when compared to the FEM modelvalues of the real loads.

Peak values for bearing and engine mount loads do not occur within thesame time frame.

69

8 FUTURE WORK

8 Future workTo improve on the turbofan model one should try to enhance the accuracyof the input data, by finding new methods that could make better predic-tions of masses, moments of inertia, lengths and other important inputparameters.One should try adding some additional components, like the low pressurecompressor or high pressure turbine, to investigate how this would affectthe accuracy of the turbofan model. As this turbofan model did not incor-porate the fan casing which is present on all civil turbofan engines, oneshould try to add this component to the turbofan model to examine whateffect it would have on the results.

To better understand how the loads in the engine mounts are distributeda more realistic mechanical system should be derived, to take intoaccount the three dimensional nature of the engine mounts.

A further investigation on how the dynamic effects of self-centring affectthe system is also required. A better understanding on how the actualprocess of self-centring occurs would greatly contribute to theunderstanding of how the unbalance forces at FBO should beimplemented.

It would be of great interest to expand the existing turbofan model to beable to handle time-transient simulations. With this expansion the modelcould be used to better understand the time development after fanblade-off.

70

REFERENCES REFERENCES

References[1] Anderson, John D. 2005. Introduction to Flight. 5th ed. New York:

MacGraw-Hill/Irwin. ISBN 007-123919-2.

[2] Genta, Giancarlo. 1999. Vibration of Structures and Mechaines: Prac-tical Aspects. 3rd ed. New York: Springer-Verlag. ISBN 0-387-98506-9.

[3] Genta, Giancarlo. 2005. Dynamics of rotating systems. 1 st ed. NewYork: Springer-Verlag. ISBN 0-387-20936-0.

[4] Lay, David C. 2003. Linear Algebra and Its Applications. 3rd ed.Boston: Adisson Westey Publishing Company. ISBN 0-201-70970-8.

[5] Lundh, Hans. 2005. Grundläggande hållfasthetslära. 1st ed.Södertälje: Fingraf Tryckeri AB. ISBN 91-972860-2-8.

[6] Meriam, James L. and Kraige Glenn L. 2003. Engineering Mechaines:Dynamics. 5th ed. New York: John Wiley & Sons. ISBN: 0-471-26606-x.

[7] Nordling, Carl and Jonny Österman. 2004. Physics Handbook for Sci-ence and Engineering. 7th ed. Lund: Studentlitteratur. ISBN 91-44-03152-1.

[8] Olsson, Fredrik. 2006. Rotordynamic Model of a Fiber Refinery inBEAST. Master thesis Luleå University of Technology Luleå: . ISSN:1402-1617.

[9] Råde, Lennart and Bertil Westergren. 2004. Mathematics Handbookfor Science and Engineering. 5th ed. Lund: Studentlitteratur. ISBN91-44-03109-2.

[10] Vance, John M. 1988. Rotordynamics of Turbomachinery. 1st ed. NewYork: John Wiley & Sons. ISBN 0-471-80258-1.

[11] Young, Hugh D. and Roger A. Freedman. 2004. University Physicswith Modern Physics. 11th ed. San Francisco: Addison Wesley. ISBN:0-8053-8684-X.

71

A MATLAB CODE

A MATLAB codeA.1 Reference engine%% Reference engine

clcclear allclose allwarning off

%% Initialvärden

Lbf = Classified %Avstånd från bakre lager till främre lager (m)Lff = Classified %Avstånd från främre lager till rotorns CM (m)L = Lbf + Lff; %Längden på rotoraxeln (m)alfa = Lbf/L; %Andel av rotoraxel längd mellan bakre och

främre lagerbeta = Lff/L; %Andel av rotoraxel längd mellan främre lager

och fläkt CMr1 = Classified %Inre radien av rotoraxeln 1 (m)r2 = Classified %Yttre radien av rotoraxeln 2(m)r3 = Classified %Inre radien av rotoraxeln 3 (m)r4 = Classified %Yttre radien av rotoraxeln 4(m)Df = Classified %Diametern på fläkten (m)Lbl = Classified %Längden på ett fläktblad (m)Maf = Classified %Massan av fläktbalden (kg)Mnv = Classified %Massan av navet (kg)rho = 8000; %Densiteten på stål (kg/m^3)E1 = 185E+9; %E-modul för axelmaterial(Stål vid T=200 C)

(Pa)E2 = 120E+9; %E-modul för axelmaterial Titan (Pa)Omega = Classified %Redline varvtalet för FBO (rad/s)obFaktor = 1; %Andel vikt av en airfoil av ett fläktbladafLoss = 2; %Antal airfoils förloradeNaf = Classified; %Antal airfoils på fläktenpro = 0.42; %Procentsats av bladlängden där bladets CM

finnskb = 1.7E+8; %Styvheten för bakre lagerkf = 1.71E+8; %Styvheten för främre lagerite=500; %Antalet iterationer

72

A MATLAB CODE A.1 Reference engine

%% Areatröghetsmoment för rotoraxeln

I1 = (pi/4)*((r2^4-r1^4)); %Konstant tvärsnitt av ettcylindriskt skal (m^4)


%% Massan på Rotorsystemet

Mob = afLoss * obFaktor*(Maf/Naf); %Massan av obalanskraften (kg)Msys = Maf + Mnv; %Massa på Rotorsystemet (kg)

%% Excentriciteten av fläktens masscentrum efter FBO

ex = (((Df/2)+Lbl*(pro-1))*Mob) / (Msys + Mob); %Excentriciteten avCM (m)

%% Tröghetsmoment för Fläkten

Jp = Classified %Polärt tröghetsmoment verkar längstrotationsaxeln (kg m^2)

Jt = Classified %Transversellt tröghetsmoment verkar längstsymmetriaxlarna (kg m^2)

%% Flexibilitetsmatrisen för axeln

aa1 = ((L^3*beta^2)/3) * ( (alfa/(E1*I1)) + (beta/(E2*I2)) );aa2 = ((L^2*beta)/6) * ( ((2*alfa)/(E1*I1)) + ((3*beta)/(E2*I2)) );aa3 = (L/3) * ( (alfa/(E1*I1)) + ((3*beta)/(E2*I2)) );

Aa =

[aa1 0 0 aa2;0 aa1 -aa2 0;0 -aa2 aa3 0;aa2 0 0 aa3];

Flexibillitetsma-trisen

%% Flexibilitetsmatrisen för lagren

ab1 = (kb+kf*(alfa-1)^2) / (kb*kf*alfa^2);ab2 = (kb+kf*(1-alfa)) / (kb*kf*alfa^2*L);ab3 = (kb+kf)^2 / ((kb*kf^2+kf*kb^2)*alfa^2*L^2);

Aa =

[ab1 0 0 ab2;0 ab1 -ab2 0;0 -ab2 ab3 0;ab2 0 0 ab3];


73

A.1 Reference engine A MATLAB CODE

%% Styvhetsmatrisen för hela systemet

Asys = Aa + Ab; % Superponering av flexibilitetsmatrisen föraxeln och lagren resp.

K = inv(Asys); % Styvhetsmatrisen för systemet

%% Iterering av Egenfrikvens kurvor

B = 1;C = -(Jp/Jt);D = -(((K(1,1))/Msys) + ((K(4,4))/Jt));E = (((K(1,1))*Jp) / (Msys*Jt)) ;F = (((K(1,1))*(K(4,4)) - ((K(1,4))^2)) / (Msys*Jt));for x =0:ite

P = [B (x*C) D (x*E) F];x = x + 1;rotter = roots(P);for i=1:4lamda(x,i) = rotter(i);endend

%% Iterering av Amplitud kurvor

for x = 1:iteY(:,x) = Msys*ex*x^2 * ( ((Jp-Jt)*x^2 + (K(4,4)))/(-(Jp-Jt)*Msys*x^4 + ((Jp-Jt)*(K(1,1)) - Msys*(K(4,4)) )*x^2 + ((K(1,1))*(K(4,4))) - (K(1,4)^2)));Z(:,x) = Msys*(ex+Y(x))*x^2 ;x = x + 1;end

74

A MATLAB CODE A.1 Reference engine

%% Plottning av Egenfrikvens kruvor

% Campbell diagram av egenfrikvenserna

y1 = lamda(:,1);y2 = lamda(:,2);y3 = lamda(:,3); %Forward whirl modey4 = lamda(:,4);t=0:ite; % Definierar y-axelnz=0:ite;y=0:ite;y=z;

figure (1)grid onhold onplot(t,y1,’k’,t,y3,’r’,t,y2,’g’,t,y4,’g’)plot(z,y,’–’)

title(’Reference engine’,’FontSize’,15)xlabel(’Spin/Rotational speed rad/s’,’FontSize’,15)ylabel(’Eigenvalue \lambda rad/s’ ,’FontSize’,15)legend(’\Omega(Jp/Jt)’,’Forward whirl mode’,’Backward whirlmode’,’Backward whirl mode’,’\lambda=\Omega Crit.speed’)

%% Kritiskt varvtal lamda = Omega

Wcr=(sqrt(((K(1,1))*(Jp-Jt)-Msys*(K(4,4))+sqrt(((K(1,1))*(Jp-Jt)+Msys*(K(4,4)))^2-4*Msys*(Jp-Jt)*(K(1,4))^2))/(2*Msys*(Jp-Jt))))Delta = (-(Jp-Jt)*Msys*Omega^4 + ( (Jp-Jt)*(K(1,1)) - Msys*(K(4,4)))*Omega^2 + ((K(1,1))*(K(4,4))) - (K(1,4)^2));

%% Obalanskrafter och moment

Fstatisk = Msys*ex*Omega^2 % Statiskobalanskraft

Fob = abs( Msys*(ex+Y(317))*Omega^2 ) % Dynamiskobalanskraft

Mgyro = abs((Jp-Jt)*Omega^2*(Msys*ex*Omega^2 * ( K(4,4) /Delta )) )

% Gyroskopisktmoment

Fsb = abs( (Mgyro*rho*pi*(r2^2-r1^2)*L^3*alfa^3*Omega^2) / (32*E1*I1))

% Axelutböjn-ingskraft

75

A.1 Reference engine A MATLAB CODE

%% Lager krafter

Ffb = (Mgyro/(alfa*L))+ Fsb - (Fstatisk/alfa) % Främre lagerlasten

Frb = (Fstatisk*(beta/alfa)) + Fsb - (Mgyro/(alfa*L)) % Bakre lagerlasten

%% Motorfästeskrafter

L1 = 1.235;L2 = 0.4006;L3 = 0.2737;L4 = 0.8026;L5 = 2.1216;L6 = 2.4575;L7 = 0.3661;angle = 0.1396;Fn = 295.77E+3;Mn = 5443;g = 9.82;Ffm = ( Mgyro + Fsb*(2*L6+L2-L5) + g*Mn*(L6-L4) - Fob*(L1+L6) -Fn*(L6*tan(angle)+L7) ) / (L3+L6)Frm = ( Fob*(L1-L3) + Fsb*(2*L3+L5-L2) + g*Mn*(L3+L4) +Fn*(L7-L3*tan(angle)) - Mgyro) / (L3+L6)

76

A MATLAB CODE A.2 NSA 1-HPT engine

A.2 NSA 1-HPT engine%% Reference engine


%% Initialvärden

Lbf = 1.63; %Avstånd från bakre lager till främre lager (m)Lff = 0.6506; %Avstånd från främre lager till rotorns CM (m)L = Lbf + Lff; %Längden på rotoraxeln (m)alfa = Lbf/L; %Andel av rotoraxel längd mellan bakre och


och fläkt CMr1 = 0.09/2; %Inre radien av rotoraxeln 1 (m)r2 = 0.11/2; %Yttre radien av rotoraxeln 2(m)r3 = 0.1713; %Inre radien av rotoraxeln 3 (m)r4 = 0.1917; %Yttre radien av rotoraxeln 4(m)Df = 1.77; %Diametern på fläkten (m)Lbl = 0.6160; %Längden på ett fläktblad (m)Maf = 92.6224; %Massan av fläktbalden (kg)Mnv = 98.6040; %Massan av navet (kg)rho = 8000; %Densiteten på stål (kg/m^3)E1 = 185E+9; %E-modul för axelmaterial(Stål vid T=200 C)

(Pa)E2 = 120E+9; %E-modul för axelmaterial Titan (Pa)Omega = 448.9336; %Redline varvtalet för FBO (rad/s)obFaktor = 1; %Andel vikt av en airfoil av ett fläktbladafLoss = 2; %Antal airfoils förloradeNaf = 16; %Antal airfoils på fläktenpro = 0.42; %Procentsats av bladlängden där bladets CM


77

A.2 NSA 1-HPT engine A MATLAB CODE









Jp = 31.8992; %Polärt tröghetsmoment verkar längstrotationsaxeln (kg m^2)

Jt = 18.3463; %Transversellt tröghetsmoment verkar längstsymmetriaxlarna (kg m^2)



Aa =





Aa =



78










79


%% Plottning Egenfrikvens kruvor




title(’NSA 1-HPT’,’FontSize’,15)xlabel(’Spin/Rotational speed rad/s’,’FontSize’,15)ylabel(’Eigenvalue \lambda rad/s’ ,’FontSize’,15)legend(’\Omega(Jp/Jt)’,’Forward whirl mode’,’Backward whirlmode’,’Backward whirl mode’,’\lambda=\Omega Crit.speed’)










80


%% Lager krafter




L1 = 0.9415;L2 = 0.2919;L3 = 0.1130;L4 = 0.1000;L5 = 1.3275;L6 = 1.4028;L7 = 0.3954;angle = 0.1396;Fn = 117E+3;Mn = 3161;g = 9.82;Ffm = ( Mgyro + Fsb*(2*L6+L2-L5) + g*Mn*(L6-L4) - Fob*(L1+L6) -Fn*(L6*tan(angle)+L7) ) / (L3+L6)Frm = ( Fob*(L1-L3) + Fsb*(2*L3+L5-L2) + g*Mn*(L3+L4) +Fn*(L7-L3*tan(angle)) - Mgyro) / (L3+L6)

81


A.3 NSA 2-HPT engine%% Reference engine


%% Initialvärden

Lbf = 1.81; %Avstånd från bakre lager till främre lager (m)Lff = 0.4710; %Avstånd från främre lager till rotorns CM (m)L = Lbf + Lff; %Längden på rotoraxeln (m)alfa = Lbf/L; %Andel av rotoraxel längd mellan bakre och


och fläkt CMr1 = 0.09/2; %Inre radien av rotoraxeln 1 (m)r2 = 0.11/2; %Yttre radien av rotoraxeln 2(m)r3 = 0.1713; %Inre radien av rotoraxeln 3 (m)r4 = 0.1917; %Yttre radien av rotoraxeln 4(m)Df = 1.77; %Diametern på fläkten (m)Lbl = 0.6160; %Längden på ett fläktblad (m)Maf = 92.6224; %Massan av fläktbalden (kg)Mnv = 98.6040; %Massan av navet (kg)rho = 8000; %Densiteten på stål (kg/m^3)E1 = 185E+9; %E-modul för axelmaterial(Stål vid T=200 C)

(Pa)E2 = 120E+9; %E-modul för axelmaterial Titan (Pa)Omega = 448.9336; %Redline varvtalet för FBO (rad/s)obFaktor = 1; %Andel vikt av en airfoil av ett fläktbladafLoss = 2; %Antal airfoils förloradeNaf = 16; %Antal airfoils på fläktenpro = 0.42; %Procentsats av bladlängden där bladets CM


82










Jp = 31.8992; %Polärt tröghetsmoment verkar längstrotationsaxeln (kg m^2)

Jt = 18.3463; %Transversellt tröghetsmoment verkar längstsymmetriaxlarna (kg m^2)



Aa =





Aa =



83










84


%% Plottning Egenfrikvens kruvor




title(’NSA 1-HPT’,’FontSize’,15)xlabel(’Spin/Rotational speed rad/s’,’FontSize’,15)ylabel(’Eigenvalue \lambda rad/s’ ,’FontSize’,15)legend(’\Omega(Jp/Jt)’,’Forward whirl mode’,’Backward whirlmode’,’Backward whirl mode’,’\lambda=\Omega Crit.speed’)










85


%% Lager krafter




L1 = 0.7061;L2 = 0.2354;L3 = 0.1036;L4 = 0.1000;L5 = 1.5440;L6 = 1.6570;L7 = 0.4425;angle = 0.1396;Fn = 117E+3;Mn = 3111;g = 9.82;Ffm = ( Mgyro + Fsb*(2*L6+L2-L5) + g*Mn*(L6-L4) - Fob*(L1+L6) -Fn*(L6*tan(angle)+L7) ) / (L3+L6)Frm = ( Fob*(L1-L3) + Fsb*(2*L3+L5-L2) + g*Mn*(L3+L4) +Fn*(L7-L3*tan(angle)) - Mgyro) / (L3+L6)

86

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