F100 Turbine Exhaust Case

6.11.2014, הטכניון, למנועי סילון וטורבינות גז13-יום עיון ה, ענף הנעה–חיל האוויר

F100 Turbine Exhaust Case

Stress Analysis of Strut Cracks

Cap. Shani Eitan, Propulsion Branch, IAF


Background

• Turbine Exhaust Case (TEC) is a part of Low Pressure Turbine (LPT) module of the F100-220/220E-DPI engine.

• TECs consist of 8 Struts that straighten the turbine exit flow.

• Some of the struts also contain oil lines for the bearings.

• IAF has experienced a high number of cracks on the Struts’ airfoils, that were beyond T.O. limit.



Objectives

• To estimate the risk of flying with a cracked strut

• The objectives of this investigation is to calculate:• Critical crack length 𝑎𝑐𝑟• Static Margin of safety (M.S.) as a function of crack length (a)

𝑀. 𝑆. 𝑎 =𝜎𝑢𝜎− 1, 𝜎 = 𝑓 𝐿𝑂𝐴𝐷𝑆, 𝑎

ቚ𝑎𝑐𝑟 = 𝑎𝑀.𝑆.=0

= ?


Challenges

• Analysis:• Stress calculation of a cracked part (not crack propagation,

not fatigue, but margin of safety at next flight).

• Boundary conditions

• Thermal loads

• Exotic material

• Aerodynamic cross section

• Operator stand point of view (not a designer):• Design is unknown

• Loads are unknown (at least initially)

• Material properties

• Dimensions


LoadsFree Body Diagram

L – Lift – aerodynamic forceD – Drag – aerodynamic forceM – Moment/torque –aerodynamic loadT – Temperature of turbine exit flow – thermal condition

V – Velocity of turbine exit flow

- Fixed displacement boundary condition


LoadsStrut Airfoil Mechanical Properties@ ~1300𝑂F

• Strut Airfoil is made of Nickel INCONEL alloy AMS5599

• Young’s Modulus is E=200GPa

• Ultimate strength is 𝜎𝑢 =827Mpa

• Thermal Expansion coefficient

• is 𝛼 = 13.410−6

𝐾


LoadsCalculation of Velocity

V



• Velocity represented by Flow Rate formula:

• Flow Rate at the Exhaust is the sum of core air flow rate and fuel flow rate:

ሶ𝑚𝑒𝑥 = ρ ∙ A𝑒𝑥 ∙ V → V =ሶ𝑚𝑒𝑥

ρ ∙ A𝑒𝑥

ሶ𝑚𝑒𝑥 = ሶ𝑚𝑐 + ሶ𝑚𝑓



• Core air flow rate is calculated using F100-220 core-fan bypass ratio:

• Known F100 flow rates (fuel including A/B, Engine Inlet Air flow):

β =ሶ𝑚𝑐

ሶ𝑚𝐹= 0.61

ሶ𝑚𝑓 = 49,964𝑝𝑝ℎ

ሶ𝑚𝐸 = 225𝑝𝑝𝑠

V =ሶ𝑚𝑒𝑥

ρ ∙ A𝑒𝑥



• Core air flow rate is calculated using F100-220 core-fan bypass ratio:

• Fuel flow in seconds:

ሶ𝑚𝑐

ሶ𝑚𝐸 − ሶ𝑚𝑐= 0.61 → ሶ𝑚𝑐 =

0.61

1.61ሶ𝑚𝐸 =

0.61

1.61∙ 225 = 86𝑝𝑝𝑠

ሶ𝑚𝑓 = 49,964𝑙𝑏

60 ∙ 60 ∙ 𝑠= 14𝑝𝑝𝑠

V =ሶ𝑚𝑒𝑥

ρ ∙ A𝑒𝑥



• Gas flow rate thru the TEC can therefore be calculated:

• Cross Sectional Area of the flow thru the TEC:

ሶ𝑚𝑒𝑥 = ሶ𝑚𝑐 + ሶ𝑚𝑓 = 86 + 14 = 100𝑙𝑏

𝑠𝑒𝑐

V =ሶ𝑚𝑒𝑥

ρ ∙ A𝑒𝑥

Aex = π 𝑅𝑜2 − 𝑅𝑖

2 = π 15.6742 − 6.3852 = 643.7 𝑖𝑛2



• Gas density in the TEC area:

V =ሶ𝑚𝑒𝑥

ρ ∙ A𝑒𝑥

ρ =𝑃

ǁ𝑅 ∙ 𝑇=

44.8 𝑃𝑆𝐼

271𝐽

𝐾𝑔 ∙ 𝐾∙ 1362𝑂𝐹

=308,885.1

𝑁

𝑚2

271𝐽

𝐾𝑔 ∙ 𝐾∙ 1012K

= 1.126𝑘𝑔

𝑚3

ρ = 4.068 ∙ 10−5𝑙𝑏

𝑖𝑛3



• Velocity at TEC inlet can therefore be calculated:

V =ሶ𝑚𝑒𝑥

ρ ∙ A𝑒𝑥

𝑉 =ሶ𝑚𝑒𝑥

ρ ∙ A𝑒𝑥=

100

643.7 ∙ 4.068 ∙ 10−5= 3,818

𝑖𝑛

𝑠𝑒𝑐= 97

𝑚

𝑠𝑒𝑐


LoadsCalculation of Lift L =

1

2𝜌𝑉2𝑆𝐶𝐿

L


LoadsCalculation of Lift

• Surface Area of the airfoil:

L =1


S =𝑏 𝑐1 + 𝑐2

2=225.8 ∙ 221 + 163.3

2= 43,388 𝑚𝑚2 = 67.25 𝑖𝑛2



• Lift coefficient 𝐶𝐿 of a finite airfoil the sum of the 2D cross sectional 𝐶𝑙:

• Whereas, c(y) is the cross section length at each section of the airfoil:

L =1


𝐶𝐿 =1

𝑆න0

𝑏

𝐶𝑙 𝑦 ∙ 𝑐 𝑦 𝑑𝑦

𝑐 𝑦 =𝑐2 − 𝑐1

𝑏𝑦 + 𝑐1 = −0.225𝑦 + 221



• 𝐶𝑙 𝑦 is the 2D cross sectional lift coeff. At each section. A 2 sections interpolation will be used :

𝐶𝑙 is calculated using lift-line theorem:

L =1


𝐶𝑙 𝑦 =𝐶𝑙2 − 𝐶𝑙1

𝑏𝑦 + 𝐶𝑙1

𝐶𝑙1 = 2𝜋 𝛼1 +1

𝜋න

0

𝜋𝑑𝑧

𝑑𝑥1

𝑐𝑜𝑠𝜗 − 1 𝑑𝜃



• 2D cross section airfoil line z(x) was derived from a SolidWorks model.

• SolidWorks model was constructed using a swept protrusion of the 2 cross sections:

L =1


𝐶𝑙1 = 2𝜋 𝛼1 +1

𝜋න

0

𝜋𝑑𝑧

𝑑𝑥1




• 2D cross section airfoil line z(x) was interpolated using O(5) order level:

L =1


𝐶𝑙1 = 2𝜋 𝛼1 +1

𝜋න

0

𝜋𝑑𝑧

𝑑𝑥1




• 2D cross section airfoil line z(x) was interpolated using O(5) order level:

L =1


𝐶𝑙1 = 2𝜋 𝛼1 +1

𝜋න

0

𝜋𝑑𝑧

𝑑𝑥1


=5E-10x5-3E-07x4 +8E-05x3-0.0087x2 +0.6387x + 0.211

R² =0.9957

-10

0

10

20

30

40

50

050100150200250

Y1

Y2

YM

Poly. (YM)



• 2D cross section airfoil line 𝑑𝑧

𝑑𝑥 1was analytically

calculated:

L =1


𝐶𝑙1 = 2𝜋 𝛼1 +1

𝜋න

0

𝜋𝑑𝑧

𝑑𝑥1


𝑧1 𝑥

= 5 ∙ 10−10𝑥5 − 3 ∙ 10−7𝑥4 + 8 ∙ 10−5𝑥3 − 0.0087𝑥2 + 0.6387𝑥 + 0.211

𝑑𝑧

𝑑𝑥1

= 25 ∙ 10−10𝑥4 − 12 ∙ 10−7𝑥3 + 24 ∙ 10−5𝑥2 − 0.0174𝑥 + 0.6387



• Integrals were calculated using transformation x-𝜃, and using MAPLE:

L =1


𝐶𝑙1 = 2𝜋 𝛼1 +1

𝜋න

0

𝜋𝑑𝑧

𝑑𝑥1


ሻ𝑥 = Τ𝑐 2 (1 − 𝑐𝑜𝑠𝜃

න

0

𝜋𝑑𝑧

𝑑𝑥1

𝑐𝑜𝑠𝜗 − 1 𝑑𝜃 = −2.93




L =1


𝐶𝑙1 = 2𝜋 𝛼1 +1

𝜋න

0

𝜋𝑑𝑧

𝑑𝑥1





• Angle between two cross sections was measured:

L =1


𝐶𝑙1 = 2𝜋 𝛼1 +1

𝜋න

0

𝜋𝑑𝑧

𝑑𝑥1


𝐶𝑙1 = 2𝜋 𝛼1 − 2.93

𝐶𝑙2 = 2𝜋 𝛼2 − 2.72

𝛼1 − 𝛼2 = 110 = 11 ∙𝜋

180= 0.192.



• Turbine exit velocity angle of attack at the last rotor stage is relatively small:

• 𝛼 1 ≈ 0𝑜

• 𝛼 2 ≅ 10𝑜 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑

L =1


𝛼1, 𝛼2



• Lift force acting on the strut airfoil can now be calculated:

L =1


𝐶𝐿 =1

𝑆න0

𝑏

𝐶𝑙 𝑦 ∙ 𝑐 𝑦 𝑑𝑦 =1

43,388න

0

225.8

0.0112𝑦 − 18.41 −0.225𝑦 + 221 𝑑𝑦 = 17.508

L =1

2𝜌𝑉2𝑆𝐶𝐿 =

1

2∙ 1.126 ∙ 972 ∙ 0.043388 ∙ 17.508 = 4,023 𝑁


Loads interim summery

L=4000 N

V=97 m/s

D

M

T=1300F


Loads Drag force Calculation

D



• Drag force on a finite wing is expressed using the formula:

D =1

2𝜌𝑉2𝑆𝐶𝐷



• Drag coefficient on a finite wing is expressed using the formula:

𝐶𝐷 =8

𝜋2𝐶𝐿2

𝜋𝐴𝑅; 𝐴𝑅 =

𝑏2

𝑆



• Drag force can be therefore calculated:

D =4𝜌 𝑉𝑆𝐶𝐿

2

𝑏2𝜋3=4 ∙ 1.126 97 ∙ 0.043388 ∙ 17.5 2

0.2252𝜋3= 15,500 𝑁


Loads Aerodynamic Moment Calculation

M



• Aerodynamic moment acting on a finite wing is expressed using lift-line theorem according to the formula:

M =1

2𝜌𝑉2𝐶𝑀න

0

𝑏

𝑐2(𝑦ሻ𝑑𝑦



• Aerodynamic moment coefficient of a finite wing formula:

𝐶𝑀 =1

0𝑏𝑐2(𝑦ሻ𝑑𝑦

න

0

𝑏

ሻ𝑐(𝑦 2𝐶𝑚(𝑦ሻ𝑑𝑦



• Aerodynamic moment coefficient as a function of location, using 2 cross section interpolation method:

𝐶𝑚 𝑦 =𝐶𝑚2 − 𝐶𝑚1

𝑏𝑦 + 𝐶𝑚1



• Aerodynamic moment coefficient of 2D cross sections:

𝐶𝑚1,2 =1

2න

0

𝜋𝑑𝑧

𝑑𝑥1,2

𝑐𝑜𝑠2𝜗 − 𝑐𝑜𝑠𝜗 𝑑𝜃



• Aerodynamic moment integrals were calculated using MAPLE:



• Aerodynamic moment integrals were calculated using MAPLE:

𝐶𝑚 𝑦 = −0.0033𝑦 + 0.6416

𝐶𝑀 =1

0𝑏𝑐2(𝑦ሻ𝑑𝑦

න

0

𝑏

𝑐 𝑦 2𝐶𝑚 𝑦 𝑑𝑦 = 0.999



• Aerodynamic moment can now be calculated:

M =1

2𝜌𝑉2𝐶𝑀න

0

𝑏

𝑐2 𝑦 𝑑𝑦 = 58.4 𝑁 −𝑀


Loads summery

L=4000 N

V=97 m/s

D=15500 N

M=58N-M

T=1300F


Stress Analysis

• Sress analysis (calculation of 𝜎ሻ was conducted using 3 different methods.

• Numerical analysis:• Finite Elements Method (FEM) – Ansys Workbench

• Analytical method:• Nominal stress of degraded moments of inertia due to

crack presence in a critical cross section.

• Stress intensity factor K1C method


Stress AnalysisFEM – Physical Model

FREE AFT ENDFREE OUTER

SURFACE

FREE INNER

SURFACE

FIXED SURFACE END

(ATTACHED TO LPT)

• TEC is attached (fixed) to the aft outer perimeter of the LPT. Rest of the surfaces are free.

• Struts are attached (fixed) to the inner and outer case surfaces.

• Struts “feel” air flow aerodynamic forces and high temprature


Stress AnalysisFEM – Computational Model -Geometry

FREE AFT ENDFREE OUTER

SURFACE

FREE INNER

SURFACE

FIXED SURFACE END

(ATTACHED TO LPT)

• One TEC strut was modeled in Solidworks, attached to inner and outer platforms representing the case.

• One strut model saves computational memory in comparison to a full TEC model.

FIXED SURFACE END

(ATTACHED TO LPT)


Stress AnalysisFEM – Computational Model – Crack Geometry

crack• Crack width 0.1mm was modeled.

• Several different crack lengths: 0”, 0.5”, 1.5”, 3.0”


Stress AnalysisFEM – Computational Model - Mesh

• SOLID187 3D Element was used. Element size 0.1 mm in crack and 1 mm in near crack surface.


Stress AnalysisFEM – Computational Model – Loads and Boundary Condition

FIXED

FIXED

LOADS


Stress AnalysisFEM – Computational Model –Material Properties


Stress AnalysisFEM – Computational Model – Results

𝜎 = 163𝑀𝑝𝑎M.S.= +4.07

No Crack (a=0”)

𝜎 = 415𝑀𝑝𝑎M.S.= +0.99

a=0.50”

𝜎 = 538𝑀𝑝𝑎M.S.= +0.54

a=1.50”


Stress AnalysisFEM – Computational Model – Results

𝜎 = 855𝑀𝑝𝑎M.S.= -0.02

a=3.0”


Stress AnalysisAnalytical Approach

• Stress in the strut consists of different loads and stress conditions (moment, shear, bending, torque). And therefore would be calculated with VON-MISES formula:

𝜎 = 𝜎𝐵−𝐿𝐷 + 𝜎𝑇2 + 3 𝜏𝐿𝐷 + 𝜏𝑀

2

𝜎 = 𝜎2 + 3𝜏2



• Strut was considered as a fixed supported beam under a uniform load consists of lift, drag, moment and thermal stress



• Geometric properties were calculated using a CAD model:

I𝑥𝑥1 = 67497.36𝑚𝑚4

I𝑦𝑦1 = 994338.67𝑚𝑚4

𝐽 = 994338.67𝑚𝑚4

Area = 425.08 millimeters^2

I𝑥𝑥2 = 142470.71𝑚𝑚4

I𝑦𝑦2 = 2259760.67𝑚𝑚4

𝐽 = 2259760.67𝑚𝑚4

Area = 561.99 millimeters^2



• Uniform force is the vector sum of Lift and drag:

• Maximum bending moment and shear is at the smallest section profile (least area and inertia):

𝑤 =𝐿2 + 𝐷2

𝑏=

40002 + 155002

225.8= 71

𝑁

𝑚𝑚

𝑀 =𝑤𝑏2

12=71 ∙ 225.82

12= 301,665 𝑁 − 𝑚𝑚

V =𝑤𝑏

2=71 ∙ 225.8

2= 8016 𝑁



• Maximum stresses due to lift and drag at the beam are:

• Torque due to aerodynamic moment:

𝜎𝐵 =𝑀

𝐼𝑥𝑥𝑧 =

301665 ∙ 21

67500= 93 𝑀𝑃𝑎

𝜏𝐿𝐷 =V

𝐴=8016

425.8= 18.825 𝑀𝑃𝑎

𝜏𝑀 =𝑇𝑟

𝐽=58400 ∙ 80

994338.67= 4.7 MP𝑎



• Thermal stress is due to strut’s thermal expansion:

• Inner and outer diameters of the TEC also expand, and therefore compensate the strut’s expansion:

𝜀 =∆𝑙

𝑙= 𝛼∆𝑇



• Strut potential expansion is:

• Outer case perimeter expansion:

∆𝑙 = 225.8 ∙ 13.4 ∙ 10−6 ∙ 1012 =3mm

𝜀 =∆𝑃

𝑃= 𝛼∆𝑇

𝑃1 = 𝑃0 𝛼∆𝑇 + 1 = 2𝜋𝑅0 𝛼∆𝑇 + 1

𝑃1 = 2𝜋 ∙ 15.674 13.4 ∙ 10−6 ∙ 1012 + 1 = 99.81"



• Outer Case radius expansion is therefore:

• Inner case is calculated the same way = 2.2mm

𝑅1 =𝑃12𝜋

∆𝑅 = 𝑅1 − 𝑅0 = 5.23𝑚𝑚



• Expansion superposition summery:Outer case expands 5.23

mm

Inner case expands 2.20 mm

The gap between inner and outer case expands 3.03 mm

The strut expands 3.00 mm

The strut expands additional 0.03 mm due to cases

tension



• Thermal stress due to 0.03mm expansion is:

• Total stress can now be calculated using VON-MISES criteria:

𝜎𝑇 = 𝐸𝜀 = 200,000 ∙0.03

225.8= 26 MPa

𝜎 = 26 + 93 2 + 19 + 5 2 = 121 MP𝑎


Stress AnalysisAnalytical Approach – classic approach

• Stress rise due to loss of geometrical stiffness:

𝑎 ↑ ⇒ 𝐴 ↓ , 𝐼𝑥𝑥 ↓ ⇒ 𝜎 =𝑀

𝐼𝑥𝑥𝑧 ↑ , 𝜏 =

𝑉

𝐴↑


Stress AnalysisAnalytical Approach – classic approach

• Stress rise due to loss of geometrical stiffness:

CrackLength [mm]

CrackLength [inch]

CrackLength

[cm]

Cross section area A [mm^2]

Ixx [mm^4]Bending

Stress riseShear Stress

rise

00.00056214247100

12.70.501.27545.281360895%3%

25.41.002.54528.8612800211%6%

38.11.503.81512.5611867220%10%

50.82.005.08496.210880131%13%

63.52.506.35479.849893044%17%

76.23.007.62463.488905960%21%

88.93.508.89447.127918880%26%


M.S. VS Crack length (a)

M.S

. (M

argi

n o

f Sa

fety

)

CmInch Crack

length (a)


Stress AnalysisAnalytical Approach – Fracture Mechanics

• In Fracture Mechanics theorem we define a critical stress intensity factor K1C.

• K1C is a material property – for INCONEL 625 it’s 59 Mpa 𝑚.

• The stress intensity is calculated using the formula:

K1 = βσ πa

• The Margin of Safety would therefore be

𝐾1𝑐K1

− 1



• Definition of 𝑎 and 𝛽

βd/ba/dBda[inch]

1104.34.30

1.0210.124.34.30.5

1.0510.234.34.31

1.0810.354.34.31.5

1.1310.464.34.32

1.2810.584.34.32.5

1.510.74.34.33



• M.S. calculations:

M.S.

𝐾1𝑐K1

− 1K1CK1βσ[MPa]a[m]a[inch]

-590.0112100

1.3934855924.71.021210.01270.5

0.6440945935.91.051210.02541

0.3068245945.11.081210.0381.5

0.0802455954.61.131210.05082

-0.147035969.21.281210.06352.5

-0.335555988.81.51210.07623


Summery of Results

-1

0

1

2

3

4

5

0 0.5 1 1.5 2 2.5 3 3.5

M.S

.

a [inch]

Results Summery - 3 methods of crack M.S.

Classic Fracture Mechanics FEA


Conculsion

• Construction of a decision making tool for fleet managers having a logistic maintenance issue.

• Calculation of M.S. using three different methods

• Results show good correlation between methods in range of 1”-2” crack length

• M.S.=0 in a=3” according to FEM method

• M.S.=0 in a=2.2” according to K1C method

• Engine manufacturer approved a temporary limit of 3” as long as a visual inspection is done after/before every flight

• Investigation helped in keeping availability of the fleet until logistics was solved (procurement and repair implemented)

Documents

F100 Turbine Exhaust Case