1999 mts [aerospatiale aeronautique - composite stress manual mts00

Technical ManualMTS 006 Iss. B

Outhouse distribution authorised

Documentresponsibility

Dept. code : BTE/CC/SC Validation Name : JF. IMBERT

Name : P. CIAVALDINI Function: Deputy DepartmentGroup Manager

Dept. code : BTE/CC/ADate : 06/05/99Signature

This document belongs to AEROSPATIALE and cannot be given to third parties and/or be copied withoutprior authorisation from AEROSPATIALE and its contents cannot be disclosed.

© AEROSPATIALE - 1999

Composite stress manual

Purpose To list and homogenise the calculation methods and theallowable values for the composite materials used at theAerospatiale Design Office.

Scope To be used as reference document for all Aerospatiale andsubcontractors' stressmen.

Data processing toolsupporting this Manual

Summary See detailed summary

1

4

5

4

5

21

2

StructuralDesignManual


© AEROSPATIALE - 1999 MTS 006 Iss. A

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Foreword

This issue is incomplete and existing chapters are liable to change.All allowable values and coefficients related to the various materials described in chapter Zare updated with each issue of the manual. This means that different values may be found inthe stress dossiers prior to latest issue.The data processing tools are given for information purposes only.


© AEROSPATIALE - 1999 MTS 006 Iss. B

SUMMARY OF CHAPTERS

Iss. Date Editor

DETAILED SUMMARY A Jan 98 P. Ciavaldini

INTRODUCTION - COMPOSITE MATERIAL PROPERTIES A B Apr 99 P. Ciavaldini

COMPOSITE PLATE THEORY B *

MONOLITHIC PLATE - MEMBRANE ANALYSIS C A Jan 98 P. Ciavaldini

MONOLITHIC PLATE - BENDING ANALYSIS D A Jan 98 P. Ciavaldini

MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS E A Jan 98 P. Ciavaldini

MONOLITHIC PLATE - TRANSVERSAL SHEAR ANALYSIS F B Apr 99 P. Ciavaldini

MONOLITHIC PLATE - FAILURE CRITERIA G B Apr 99 P. Ciavaldini

MONOLITHIC PLATE - FATIGUE ANALYSIS H *

MONOLITHIC PLATE - DAMAGE-TOLERANCE I ** B Apr 99 P. Ciavaldini

MONOLITHIC PLATE - BUCKLING J *

MONOLITHIC PLATE - HOLE WITHOUT FASTENER ANALYSIS K B Apr 99 P. Ciavaldini

MONOLITHIC PLATE - FASTENER HOLE L B Apr 99 P. Ciavaldini

MONOLITHIC PLATE - SPECIAL ANALYSIS M *

SANDWICHIC - MEMBRANE / BENDING / SHEAR ANALYSIS N B Apr 99 P. Ciavaldini

SANDWICH - FATIGUE ANALYSIS O *

SANDWICH - DAMAGE-TOLERANCE APPROACH P *

SANDWICH - BUCKLING ANALYSIS Q *

SANDWICH - SPECIFIC DESIGNS R *

BONDED JOINTS S A Jan 98 P. Ciavaldini

BONDED REPAIRS T B Apr 99 P. Ciavaldini

BOLTED REPAIRS U A Jan 98 P. Ciavaldini

THERMAL CALCULATIONS V B Apr 99 P. Ciavaldini

ENVIRONMENTAL EFFECT W *

NEW TECHNOLOGIES X *

STATISTICS Y *

MATERIAL PROPERTIES Z ** B Apr 99 P. Ciavaldini

*: chapter not dealt with.**: chapter partially dealt with.

B

B

B

B

B

B

B



HOW TO USE THE COMPOSITE MANUAL?

Effec

n3

n4

4.2.1 . Effect of n

Assuming that allnormal load Ny apthe whole cross-se

ε =+b EMi e Ei(

This elongation th

- in the lower ski

- in the core, a s

- in the upper sk

The equivalent meby the relationship

Remark: In the casEmc ec <

ε ≈+

N

b EMi ei(

pagenumber

reference(s) ofsubchapter(s)title of chapter

reference of chaptertitle(s) of subchapter(s)

ror

Ems

Emc

Emi

SANDWICHt of normal load Ny

N 4.2.11/2

ormal load Ny

layers are in a pure tension or compression condition, aplied at the neutral line results in a constant elongation overction. This elongation may be formulated as follows:

+Ny

Mc e Ems ec s)
eferencefelation
MTS 006 Iss. A

is unduces:

n, a stress σi = Emi ε,

tress σc = Emc ε,

in, a stress σs = Ems ε.

mbrane modulus of the sandwich beam may be determined m14.

e of a sandwich beam in which< Emi ei and Emc ec << Ems es, the relationship becomes:

y

Ems es )

es

ec

ei

Z

Y

b

X

σs

σc

σi ε

Ny



HOW TO USE THE COMPOSITE MANUAL?

5 . EXAMPLE

Let a 10 mm wisequence:

- an upper skin (elasticity Es = 6

- a core (honeycmodulus Ec = 1

- a lower skin (celasticity modu

We shall assumemoment:

- Ny = 800 daN,

- Mx = 2000 daN

- Tz = 250 daN.

1st step: to determ

{n3}

9.04500(10=ε

pnumber

reference(s) ofsubchapter(s)title of chapter

rt

1,04

10

0,9

SANDWICHExample

N 51/7

de sandwich beam be defined by the following stacking

carbon layers) of thickness es = 1.04 mm and of longitudinal000 daN/mm2,

omb) of thickness ec = 10 mm and of longitudinal elasticity5 daN/mm2,

arbon cloths) of thickness ei = 0.9 mm and of longitudinallus Ei = 4500 daN/mm2.

that the beam is subjected to the following two loads and

mm,

ine elongation ε induced by normal load Ny.

Z

Y

10X

Ny = 800 daN

Tz = 250 daN

Mx = 2000 daN mm

referenceofrelation

MTS 006 Iss. A

d7612)04.160001015

800 µ=++

Z

X

Y

ε = 7612 µd

age

eference of chapter
itle(s) of subchapter(s)



DETAILED SUMMARY

A . INTRODUCTION - COMPOSITE MATERIAL PROPERTIES1 . Introduction - General2 . Composition

2.1 . Fibres2.2 . Matrices

3 . Processing methods4 . Composite structure design5 . Assembly6 . Advantages - Disadvantages (environmental parameters)7 . Similitudes with metals

7.1 . System equilibrium7.2 . Load distribution

7.2.1 . Normal load N7.2.2 . Bending moment M7.2.3 . Shear load T

7.3 . Material strength laws - Behavior laws7.4 . General instability

8 . Differences with metals

B . COMPOSITE PLATE THEORY1 . Ply

1.1 . Tapes - Fabrics1.2 . Ply behavior (unidirectional orthotropic)1.3 . Definitions - Notations

2 . Laminate2.1 . Principle2.2 . Assembly

3 . Sandwich3.1 . Principle3.2 . Assembly

C . MONOLITHIC PLATE - MEMBRANE ANALYSIS1 . Notations2 . General definitions

2.1 . Homogeneity - Isotropy2.2 . Coupling phenomena

2.2.1 . Plane coupling2.2.2 . Mirror symmetry

3 . Analysis method4 . Deformations and equivalent properties5 . Graphs

5.1 . Failure envelopes5.1.1 . Theoretical principle5.1.2 . Margin search - Methodology

5.2 . Mechanical properties6 . Example

D . MONOLITHIC PLATE - BENDING ANALYSIS1 . Notations2 . Introduction3 . Analysis method



4 . Deformations and equivalent properties5 . Example

E . MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS1 . Notations2 . Introduction3 . Analysis method4 . Example

F . MONOLITHIC PLATE - TRANSVERSAL SHEAR ANALYSIS1 . Notations2 . Introduction3 . Design method4 . Example

G . MONOLITHIC PLATE - FAILURE CRITERIA1 . Notations2 . Inventory of static failure criteria

2.1 . Maximum stress criterion2.2 . Maximum strain criterion2.3 . Norris and Mac Kinnon's criterion2.4 . Puck's criterion2.5 . Hill's criterion2.6 . Norris's criterion2.7 . Fischer's criterion2.8 . Hoffman's criterion2.9 . Tsaï - Wu's criterion

3 . "Aerospatiale"'s criterion: Hill's criterion4 . Example

H . MONOLITHIC PLATE - FATIGUE ANALYSIS

I . MONOLITHIC PLATE - DAMAGE TOLERANCE1 . Notations2 . Introduction3 . Damage sources and classification

3.1 . Manufacturing damage or flaws3.2 . In-service damage

3.2.1 . Fatigue damage3.2.2 . Corrosion damage and environmental effects3.2.3 . Accidental damage

4 . Inspection of damage4.1 . Minimum damage detectable by a Special Detailed Inspection4.2 . Minimum damage detectable by a Detailed Visual Inspection4.3 . Minimum damage detectable by a General Visual Inspection4.4 . Minimum damage detectable by a Walk Around Check4.5 . Classification of accidental damage by detectability ranges

5 . Effects of flaws/damage on mechanical characteristics5.1 . Health flaws

5.1.1 . Porosity5.1.2 . Delaminations

5.1.2.1 . Delaminations outside stiffener



5.1.2.2 . Delaminations in stiffener area of an integrally-stiffened panel5.1.3 . Delamination in spar radii5.1.4 . Delamination on spar flange edges5.1.5 . Foreign bodies5.1.6 . Translaminar cracks5.1.7 . Delaminations consecutive to a shock

5.2 . Visual flaws5.2.1 . Sharp scratches5.2.2 . Indents5.2.3 . Scaling5.2.4 . Steps

6 . Justification of permissible manufacturing flaws7 . Justification of in-service damage

7.1 . Justification philosophy7.1.1 . Undetectable damage7.1.2 . Readily and obvious detectable damage7.1.3 . Damage susceptible to be detected during scheduled in-service inspections

7.1.3.1 . Aerospatiale semi-probabilistic method7.1.3.1.1 . Process for determining inspection intervals7.1.3.1.2 . Inspection interval calculation software7.1.3.1.3 . Load level K to be demonstrated in the presence of large VID

7.1.3.2 . CEAT semi-probabilistic method7.2 . Examples

7.2.1 . AS method applied to A340 ailerons7.2.2 . CEAT method applied to A340 nacelles

J . MONOLITHIC PLATE - BUCKLING1 . Local buckling

1.1 . Design conditions1.1.1 . General1.1.2 . Specific to composite materials

1.2 . Design rules2 . General buckling

2.1 . Variable inertia2.2 . Off-centering2.3 . Post local buckling

K . MONOLITHIC PLATE - HOLE WITHOUT - FASTENER ANALYSIS1 . Notations2 . Introduction3 . General theory

3.1 . 1st method (Whitney and Nuismer)3.2 . 2nd method (NASA)3.3 . 3rd method (isotropic plate)3.4 . 4th method (empirical)

4 . Associated failure criteria4.1 . Point stress4.2 . Average stress4.3 . Empirical

5 . Examples



L . MONOLITHIC PLATE - FASTENER HOLE1 . Notations2 . General - Failure modes

2.1 . Bearing failure2.2 . Net cross-section failure2.3 . Plane shear failure2.4 . Cleavage failure2.5 . Cleavage and net cross-section failure2.6 . Fastener shear failure

3 . Single hole with fastener3.1 . Pitch p definition3.2 . Membrane design - Short cut method

3.2.1 . Theory3.2.2 . EDP computing program PSG33

3.3 . Bending design - Short cut method3.4 . Justifications3.5 . Nominal deviations on a single hole

3.5.1 . Changing to a larger diameter3.5.2 . Pitch decrease3.5.3 . Edge distance decrease

3.6 . "Point stress" finite element method3.6.1 . Description of the method3.6.2 . Justifications

4 . Multiple holes4.1 . Independent holes4.2 . Interfering holes4.3 . Very close holes

5 . Examples

M . MONOLITHIC PLATE - SPECIAL ANALYSIS1 . Stiffener run-out2 . Bending on border3 . Effect of "stepping"4 . Edge effects

N . SANDWICH - MEMBRANE/BENDING/SHEAR/ANALYSIS1 . Notations2 . Specificity3 . Construction principle4 . Design principle

4.1 . Sandwich plate4.2 . Sandwich beam

4.2.1 . Effect of a normal load Ny4.2.2 . Effect of a shear load Tx4.2.3 . Effect of a shear load Tz - Honeycomb shear4.2.4 . Effect of a bending moment Mx4.2.5 . Effect of a bending moment Mz4.2.6 . Equivalent properties

5 . Example

O . SANDWICH - FATIGUE ANALYSIS



P . SANDWICH - DAMAGE TOLERANCE APPROACH1 . Impact damages

1.1 . Delamination1.2 . Separation1.3 . Design rules

2 . Manufacturing defects2.1 . Porosity/bubbling2.2 . Fissures/cracks

Q . SANDWICH - BUCKLING ANALYSIS1 . Local buckling

1.1 . Dimpling1.2 . Wrinkling

2 . General buckling2.1 . Bending2.2 . Shear load

R . SANDWICH - SPECIAL DESIGNS1 . Densified zones2 . Slopes/ramps

S . BONDED JOINTS1 . Notations2 . Bonded single lap joint

2.1 . Elastic behavior of materials and adhesive2.1.1 . Highly flexible adhesive2.1.2 . General case (without cleavage effect)2.1.3 . General case (with cleavage effect)2.1.4 . Scarf joint

2.2 . Elastic-plastic behavior of adhesive and elastic behavior of materials3 . Bonded double lap joint4 . Bonded stepped joint5 . Software6 . Examples

T . BONDED REPAIRS1 . Notations2 . Introduction3 . Analysis method

3.1 . Analytical method3.2 . Digital method

4 . Example

U . BOLTED REPAIRS1 . Notations2 . Stiffness of fasteners

2.1 . Fastener in single shear2.2 . Fastener in double shear

3 . Assumptions4 . Geometrical characteristics5 . Mechanical properties



6 . Assessment of mechanical distributed in-plane forces on the doubler6.1 . Distribution of flow Nx6.2 . Distribution of flow Ny6.3 . Distribution of shear flow Nxy

7 . Assessment of thermal in-plane forces on the doubler ?8 . Assessment of flows in the panel9 . Assessment of loads per fastener

9.1 . Repair with 1 row of fasteners9.2 . Repair with 2 rows of fasteners9.3 . Repair with 3 rows of fasteners9.4 . Repair with 4 rows of fasteners9.5 . Repair with a number of rows of fasteners greater than 49.6 . General resolution method for direction x

10 . Assessment of loads per fastener due to the transfer of shear loads Nxy11 . Justifications12 . Summary flowchart13 . Examples

V . THERMAL CALCULATIONS1 . Notations2 . Introduction3 . Hooke - Duhamel law4 . Behavior of unidirectional fibre5 . Behavior of a free monolithic plate

5.1 . Calculation method5.2 . Residual curing stresses5.3 . Equivalent expansion coefficients

6 . Theory of the bimetallic strip6.1 . Determining stresses of thermal origin6.2 . Study of the link between two parts

6.2.1 . Bolted or riveted joints6.2.1.1 . Force F taken by one fastener6.2.1.2 . Force F taken by two fasteners6.2.1.3 . Force F taken by three fasteners6.2.1.4 . Force F taken by four or more fasteners

6.2.2 . Bonded joints7 . Influence of temperature on aircraft structures

7.1 . General7.2 . Temperature of ambient air

7.2.1 . Temperature envelope7.2.2 . Variation of ambient air temperature

7.2.2.1 . Ambient temperature on ground7.2.2.2 . Ambient temperature in flight

7.3 . Wall temperature7.3.1 . Influence of solar radiation

7.3.1.1 . Maximum solar radiation7.3.1.2 . Solar radiation during the day

7.3.2 . Influence of aircraft speed7.3.3 . Temperature of structure

7.3.3.1 . Calculation method7.3.3.2 . Thermal characteristics of the materials7.3.3.3 . Temperatures of structure on ground7.3.3.4 . Temperatures of structure in flight

7.4 . Recapitulative block diagram8 . Computing softwares9 . Examples



W . ENVIRONMENTAL EFFECT1 . Temperature2 . Aging3 . Humidity

X . NEW TECHNOLOGIES1 . R.T.M.2 . Thermoplastic

2.1 . Shoft fibres2.2 . Long fibres

3 . Glare-Arall

Y . STATISTICS

Z . MATERIAL PROPERTIES1 . Prepreg unidirectional tapes

1.1 . First generation epoxy high strength carbon1.2 . Second generation epoxy intermediate modulus carbon1.3 . Epoxy R glass1.4 . Bismaleimide carbon

2 . Fabrics2.1 . Epoxy resin prepreg

2.1.1 . Carbon2.1.2 . Glass2.1.3 . Kevlar2.1.4 . Hybrid2.1.5 . Quartz polyester hybrid

2.2 . Phenolic resin prepreg2.2.1 . Carbon2.2.2 . Glass2.2.3 . Kevlar2.2.4 . Fiberglass carbon hybrid2.2.5 . Quartz polyester hybrid

2.3 . Bismaleimide resin prepreg2.3.1 . Carbon

2.4 . Wet lay--up epoxy (for repair)2.4.1 . Carbon2.4.2 . Glass2.4.3 . Kevlar2.4.4 . Fiberglass carbon hybrid2.4.5 . Quartz polyester hybrid

3 . R.T.M.3.1 . Epoxy resin

3.1.1 . Carbon3.2 . Bismaleimide resin3.3 . Phenolic resin

4 . Injection moulded thermoplastics4.1 . Carbon

4.1.1 . PEEK4.1.2 . PEI4.1.3 . Polyamide4.1.4 . PPS4.1.5 . Polyarylamide

4.2 . Glass



4.2.1 . PEEK4.2.2 . PEI

5 . Long fibre thermoplastics5.1 . Carbon

5.1.1 . PEEK5.1.2 . PEI

5.2 . Glass6 . Arall-Glare7 . Metallic matrix composite materials (CMM)8 . Adhesives

8.1 . Epoxy8.2 . Phenolic8.3 . Bismaleimide8.4 . Thermoplastic

9 . Honeycomb9.1 . Nomex

- Hexagonal cells- OX-Core- Flex-Core

9.2 . Fiberglass honeycomb- Hexagonal cells- OX-Core- Flex-Core

9.3 . Aluminium honeycomb10 . Foams



A

INTRODUCTION - COMPOSITE MATERIAL PROPERTIES



INTRODUCTIONGeneral A 1

1 . INTRODUCTION - GENERAL

The importance of using composite materials in aeronautical construction, and specificallywithin the Aerospatiale group, has initiated the need to prepare a document the interest ofpreparing a document gathering all the design methods and mechanical properties of themain composite materials used and/or developed by the composite material DesignOffice.

Each one of these two subjects shall make up one volume of the composite materialdesign manual.

Composite materials result from the association of at least two chemically andgeometrically different materials.

"Composite material" commonly means arrangements of fibres - continuous or not - of aresistant material (reinforcing material) which are embedded in a material with a muchlower strength (matrix), and stiffness.

The bond between the reinforcing material and the matrix is created during thepreparation phase of the composite material and this bond shall have a fundamentaleffect on the mechanical properties of the final material.

Composite materials include:

- wood,

- reinforced concrete,

- fibre-reinforced organic matrices (polymer resins),

- particle or fibre-reinforced metal matrices,

- ceramic fibre-reinforced ceramic matrices.

In the aeronautical industry, the term "composite" is mainly associated with fibre-reinforced polymer resins.



INTRODUCTIONComposition - Fibres A 2.1

2 . COMPONENTS OF COMPOSITE MATERIALS

2.1 . Fibres

Their purpose is to ensure the mechanical function of the composite material. Fibres canbe of very different chemical and geometrical types, and the following properties shall bespecifically searched for:

- high mechanical properties.

- physico-chemical compatibility with the matrix.

- easy to use.

- good repeatability of the properties.

- low density.

- low cost.

They are made up of several thousand filaments (the number of filaments being indicatedby 3K: 3000 filaments, 6K: 6000 filaments or 12K: 12000 filaments) with a diameterbetween 5 and 15 µm, and they are commercialised in two different forms:

- short fibres (a few centimeters long): they are felt, pylons (fabrics in which fibres arelaid out randomly) and injected short fibres,

- long fibres: they are cut during manufacture of the composite material, used as suchor woven,

• high strength fibres: glass, carbon, boron,

• synthetic fibres: aramid (kevlar), nylon, polyester,

• ceramic fibres: silica, alumina.



INTRODUCTIONComposition - Matrices - Implementation A 2.2

31/3

2.2 . Matrices

Their function is:

- to provide a bond between the reinforcing fibres (cohesion of all fibres) whilemaintaining a regular interval between them,

- to protect fibres against their environment,

- to allow stress transfer from one fibre to another,

There are three categories of matrix:

- resin matrices:

• thermoplastics (polyethylene, polysulfone, polycarbonate and polyamide, ...),

• thermosetting (phenolic, epoxy and polyester, ...),

• elastomers (polychloroprene, ethylene, propylene, silicone, ...),

- mineral matrices (silicon carbides, carbon),

- metal matrices (aluminium, titanium and nickel alloys).

3 . PROCESSING METHOD

The reinforcing fibre/resin mix becomes a genuinely resistant composite material onlyupon completion of the last manufacturing phase, i.e; curing of the matrix.



INTRODUCTIONImplementation A 3

2/3

* Material curing cycle

This cycle is achieved following the chemical reaction between the various components -this is the crosslinking phase.

The chemical reaction is initiated as soon as products are in contact, and it is oftenaccelerated by heat: the higher the temperature, the quicker and more explosive is thereaction:

There are two types of chemical reactions:

- the polyaddition reaction for epoxy resins where the weight of reactants is equal tothe weight of the compound,

- the condensation reaction (polycondensation) for phenolic resins where twocompounds are formed (a solid one and a gaseous one).

The curing cycle consist of a number of temperature levels of variable duration:

- a gel level which allows getting a consistent temperature gradient throughout thematerial before full gelation to limit internal stresses,

- a curing level which allows hardening,

- a post-curing level which allows internal stresses to be relieved, and additional curingfor a better temperature resistance.

Note: the glass transition point is the temperature value at which all material propertieschange. This important property must be measured, before and after wet aging.



INTRODUCTIONImplementation A 3

3/3

There are several types of manufacturing facilities and processes:

- Manufacturing facilities:

• Autoclave: parts are produced under pressure and at high temperature.

• Oven: parts are vacuum produced and at high temperature.

• Hot press: pressure is applied by a mechanical device or by hydraulic jacks.

- Manufacturing processes:

• Multiple shots process: laminate are cured separately, then bonding oflaminates to the substructure (ribs, honeycombs, etc.) is performed as asecond operation.

• Semi-cocuring process: the external skin is cured separately, the substructure(rib, or honeycomb + internal skin and stiffeners) is then cocured on theexternal skin with an adhesive film spread, if necessary.

• Single phase process: or "cocuring", skins are cured and bonded to thesubstructure (ribs or honeycomb or stiffeners) in one single operation.


© AEROSP

INTRODUCTIONDesign A 4

4 . COMPOSITE STRUCTURE DESIGN

The choice of the design principle depends on the following criteria:

- element geometry.

- element type.

- level of loads to be transmitted.

- manufactured parts suitability for inspection.

- industrialization suitability of the part.

Composite structures use the same types of design principles as metal ones:

- Solid part type structure :

• Multiple rib box type structure

• Multiple spar type structureB

ATIALE - 1999 MTS 006 Iss. B

• Stiffened or milled out panel type structure



INTRODUCTIONAssembly A 5

- Sandwich type structure:

• Sandwich face sheet box

• Through - the - thickness sandwiches

5 . ASSEMBLY

After being manufactured, the different composite (and metal) elements must beconnected to one another to allow load transfer.

The two most commonly used techniques are bonding and bolting (or riveting).

Bonding techniques are tricky to implement (preparation of surfaces to be bonded)because they are sensitive to environmental conditions: hygrometry, temperature, curedate of adhesives.

They are also difficult to control because even a sound adhesive film is a barrier toultrasounds.

More repetitive and reliable bolting techniques may generate:

- stress concentration at fastener holes,

- delamination during drilling or assembly operations,

- corrosion of fasteners or of metal parts assembled with composite parts.



INTRODUCTIONAdvantages - Disadvantages A 6

6 . ADVANTAGES - DISADVANTAGES OF COMPOSITE MATERIALS

The use of composite materials has four major advantages:

- a weight gain which is reflected by fuel saving and, therefore, by a payload increase,

- the capacity to control stiffness and strength according to the areas of the structure,thanks to the different types of layered materials. Composite materials naturally offermembrane-bending coupling or plane coupling possibilities, which can have importantapplications in the field of aero-elasticity,

- a good fatigue strength, which increases the life of aircraft parts concerned andlightens the maintenance program considerably,

- absence of corrosion, which also lightens the maintenance program.

However, composite materials remain sensitive to environmental conditions. Theirmechanical properties change, due to:

- humidity,

- temperature,

- the various aeronautical fluids such as Skydrol (hydraulic fluid), oils or solvents (MEK)and fuels,

- radiation (ultraviolet).

On the other hand, the effects of lightning strikes (temperature rise, melting, impacts,electronic damages) and shocks (delamination, separation, punctures) must be taken intoaccount in the design and justification of composite parts.



INTRODUCTIONMetal/composite material similitudes - System equilibrium A 7

7.11/4

7 . COMPARISON BETWEEN COMPOSITE STRUCTURES AND METALSTRUCTURES

Composite material and metal material structures obey the same basic rules of structuralmechanics.

On the other hand, composite material behavior laws are slightly different from those formetals.

The purpose of this sub-chapter is to specify the similitudes between metal materials andcomposite materials for the structural justification of structures.

Composite parts and metal parts have the same behavior with respect to:

- static equilibrium.

- load distribution rules among several elements.

- basic rules of structural mechanics.

- general instability problems (buckling).

7.1 . System equilibrium

Whatever the type of system or element under study (metal, composite or combined), it issubject to a set of external loads which may be of several types:

- Solid loads: distributed in the volume of the solid and of gravity (selfweight), dynamic(inertial forces), electrical or magnetic origin.

- Areal loads: distributed over the external surface of the solid, such as normalpressures due to a fluid or tangential loads due to friction phenomena.



INTRODUCTIONSystem equilibrium A 7.1

2/4

- Line loads: distributed over a line and which are, in fact, an idealized density ofsurface load with a much smaller application width than length.

- Concentrated loads (P): acting in one point and which are, in fact, an idealizeddensity of surface load acting on a surface with smaller dimensions with respect tothe dimensions of the solid under study.

- Concentrated moments (M): acting in one point and which are, in fact, an idealizedconcentrated moment.

To reach the equilibrium of the solid, all these external loads (C) must be equilibrated byreactions at the bearing surfaces (R).

Σ (C) = - Σ (R)

Z

Y

X

dl

dvP

ds

M




3/4

Let the solid be defined by its external loads and bearing surfaces:

The general equilibrium is summed up by a system of six equilibrium equations: threeequilibrated forces (F) and three equilibrated moments (Mt).

Σ (Fx) = Σ (Cx) + Σ (Rx) = 0Σ (Fy) = Σ (Cy) + Σ (Ry) = 0Σ (Fz) = Σ (Cz) + Σ (Rz) = 0

Z

Y

X

A

B

a

external loads+

bearing surfaces

external loads+

reactions at bearing surfaces

RA

RB

Z

Y

X

ra

deformed system




4/4

Σ (Mt/x) = 0Σ (Mt/y) = 0Σ (Mt/z) = 0

If the system is isostatic, the solving alone of these six equations allows all reactions atthe bearing surfaces to be found.

If the system is slightly hyperstatic and consisting of a simple geometry, it is necessary tointroduce new equations (the number depends on the degree of redundancy) of thedeformation compatibility type that take element stiffness into account.

If the system is complex or if the degree of redundancy is high, only a point stress or amatrix analysis makes it possible to find reactions at the bearing surfaces and the internalloads they generate.

Whatever the case and whatever the type of structure (composite or metal), the threefollowing rules must always be applied before any stress and deformation calculation:

1) External loading must be accurately defined.

2) Reactions must be fully determined.

3) The system must always be equilibrated.



INTRODUCTIONLoad distribution - Normal load N A 7.2.1

7.2 . Distribution of loads among several closely bound structural elements

7.2.1 . Normal load N

If a system made up of several parts which are connected together, is subject to a normalload N, then, the load distribution within the different elements (whether metal orcomposite) is as follows:

we have:

ε = NE S

NE S

NE S

N

E SNE S

kk

k kk k kk

1

1 1

2

2 2

3

3 3

1

3

1

3

1

3= = = ==

= =

�

� �

a1 hence Ni = N E SE Si i

k kk =� 1

3

a2 we may deduce Eeq. memb. (1 + 2 + 3) = E S

S

k kk

kk

=

=

�

�

1

3

1

3

where Ni: load transferred by layer (i)Ei: layer (i) elasticity modulusSi: layer (i) section

3

2

1

εA.N.

σ3

σ2

σ1

N



INTRODUCTIONLoad distribution - Bending moment M A 7.2.2

7.2.2 . Bending moment M

A bending moment M applied to the neutral axis of the system is picked up in each layerin proportion to its bending stiffness.

The moment M breaks down, in each layer (i), into a bending moment Mi and a normalload Ni, so that:

a3 N M E S vE l

ii i i

k kk

==� 1

3

a4 M M EE l

ii i

k kk

==�

ι

1

3

a5 we may deduce E eq. flex. (1 + 2 + 3) = E l

l

k kk

kk

=

=

�

�

1

3

1

3

where Ni: normal load applied to layer (i)Mi: moment applied to layer (i)li: layer (i) inertia with relation to the system neutral axisι i: layer inertia of layer (i)Si: layer (i) sectionvi: distance between layer (i) neutral axis and system neutral axisEi: layer (i) elasticity modulus

li: inertia + "Steiner" inertia b h S d3

122+

�

��

�

��

ι i : layer inertia b h3

12�

��

�

��

3

2

1

εiA.N.

σi

σe

M

εe

v1



INTRODUCTIONLoad distribution - Shear load T A 7.2.3

7.2.3 . Shear load T

Assuming that layers 1, 2 and 3 are parallel and of the same height, a shear load T isapplied to each layer in proportion to its shear stiffness.

we have:

γ = TG S

1

1 1 = T

G S2

2 2 = T

G S3

3 3 =

T

G S

kk

k kk

=

=

�

�

1

3

1

3 = TG Sk kk =� 1

3

a6 hence T T G SG S

ii i

k kk

==� 1

3

a7 we may deduce G eq. (1 + 2 + 3) = G S

S

k kk

kk

=

=

�

�

1

3

1

3

where Ti: shear load transferred by layer (i)Gi: layer (i) shear modulusSi: layer (i) section

32

1T

γ

τm1

τm2τm3



INTRODUCTIONMaterial strength laws - Behavior laws A 7.3

7.3 . Material strength laws - Behavior laws

Composite materials obey the general rules of structural mechanics.

Stress - deformation relationship for a two-dimensional analysis: Hooke's law applies(σ) = (Aij) (ε), the matrix (Aij) is more complex for composite materials as described inchapter C.

The equation of the elastic line of a bent metal beam ∂∂

2

2y

xMEI

= becomes

∂∂

2

2

1

yx

ME lk kk

n==�

for a composite structure.

Normal stress - normal load relationship: for a stressed or compressed metal beam, the

expression σ = NS

becomes σi = N EE S

i

k kk

n

=� 1

for each layer of a composite beam.

Normal stress - bending moment relationship: for a bent metal beam,

σ = M vl

becomes σi = M E vE li i

k kk

n

=� 1

for each layer of the composite beam.

Shear stress - shear load relationship: for a sheared metal beam, τ = T Wl b

becomes

τi = T E wE l b

i i

k k kk

n

=� 1

for each layer of the composite beam.



INTRODUCTIONGeneral instability A 7.4

7.4 . General instability

For a beam, Euler's law which associates the general instability critical compression loadwith the geometrical and mechanical properties of the beam remains valid, whatever thematerial used (metal/isotropic or composite/orthotropic).

Indeed, the critical load is formulated as follows:

Fc = π2

2E l

l for metal beams,

Fc = π2

12

E l

lk kk

n

=� for composite beams,

where l is the buckling length.

Regarding plates, the approach is more complex for composite materials, although basesare identical.

The differential equation which governs composite plate instability is formulated in itsmost general form:

C wx

C C wx y

C wy

N wx

N wy

N wx yx y xy11

4

4 12 33

4

2 2 22

4

4

2

2

2

2

2

2 2 2∂∂

∂∂ ∂

∂∂

∂∂

∂∂

∂∂ ∂

+ + + = + +( )

where C11, C12, C33 and C22 are the temps of the matrix (Cij) binding the rotation tensorand the bending load tensor (see chapter D).

For isotropic materials such as metals, the relationship is simplified:

E e wx

wx y

wy

N wx

N wy

N wx yx y xy

3

2

4

4

4

2 2

4

4

2

2

2

2

2

12 12

( )−+ +

�

��

�

�� = + +

ν∂∂

∂∂ ∂

∂∂

∂∂

∂∂

∂∂ ∂



INTRODUCTIONMetal/composite material differences A 8

1/3

8 . DIFFERENCES BETWEEN METAL AND COMPOSITE MATERIALS

These differences are actually covered by the composite material manual. A fewexamples are given below:

- Metal material isotropic/composite material anisotropic duality

If metal and composite materials are both macroscopically homogeneous, compositematerials are generally anisotropic. This means that their properties depend on thedirection (see drawing below) along which they are measured.

This difference may be an advantage. Through an optimization of the orientation of fibres,it allows a greater freedom to choose element rigidity and, therefore, a more accuratecontrol of load routing.

1

3

2

y

Isotropic material

xProperties are independent from the

coordinate system direction

F/SF F

l

1, 2, 3

∆l/l

1

3

2

y

Anisotropic material

xProperties depend on the coordinate

system direction

F/SF F

l

1

∆l/l

2

3

0

0



INTRODUCTIONMetal/composite material differences A 8

2/3

- Failure criteria

Because of their microscopic heterogeneity, composite materials do not obeycovariant failure criteria (independent from the coordinate system direction) like metalmaterials. Generally, they must be applied to each layer and are applicable only in apreferential direction (the direction of the fibre to be justified).

- Effect of holes

Sizing of holes in composite materials not only takes into account the net cross-section coefficient (as for metal materials) due to material removal, but also adecrease of the intrinsic material strength.

- Effect of bearing

The presence of bearing due to load transfer at a fastener in a laminate causesmembrane stresses to be artificially increased by part of the bearing stresses and, asa result, residual strength to be decreased.

- Damage tolerance

The presence of impact or manufacturing damages causes a significant decrease tothe laminate static strength.

- Effect of fatigue/damage tolerance

Corrosion and fatigue are the overriding factors of the limited life of metal structures.Metal fatigue is controlled by the number of cycles required, on the one hand, toinitiate a crack and, on the other hand, bring it to its critical length (growth phase).Influent factors of this phenomena are stress concentrations and tension loads.



INTRODUCTIONDifférences composite/métal A 8

3/3

As a general rule, fatigue is not a design factor for composite elements of civil aircraftwith thin thicknesses and no structural irregularities. More specifically, mechanicalproperties are such that static design requirements naturally "cover" fatigue designrequirements. Wohler curves are relatively flat and damaging loads are of thecompression type (R = - 1).

(Impact or manufacturing) Damage growth under mechanical fatigue is not allowedbecause of the high rate of delamination growth. The current inability to controlthrough analysis the damage growth rate in composite materials does not allow adamage tolerance justification based on slow growth. For this reason, allowabledamage tolerance values are low; this makes it possible to avoid any explosiveevolution during the aircraft life.

- Metal material plasticity/composite material "brittleness" duality

Metal materials have an elastic range and a plastic range, in their behavior, whichlead to breaking, breaking occurs in carbon composite materials without plasticizing.

F/SF F

l

breaking

∆l/l

elastic zone

Plastic material(metal)

0

plastic zone

F/SF F

l

breaking

∆l/l

elastic zone

0 Brittle material(composite)



INTRODUCTIONReferences A

BARRAU - LAROZE, Design of composite material structures, 1987

GAY, Composite materials, 1991

VALLAT, Strength of materials



B

COMPOSITE PLATE THEORY



C

MONOLITHIC PLATE - MEMBRANE ANALYSIS



MONOLITHIC PLATE - MEMBRANENotations C 1

1 . NOTATIONS

(o, x, y): reference coordinate system(o, l, t): coordinate system specific to the unidirectional ply

k: fibre coordinate systemθ: fibre orientation

nθ: number of plies in direction θeθ: overall thickness of plies in direction θ: eθ = nθ x ep

e: overall thickness of laminaten: number of plies in laminate

(N): flux tensor(σ): stress tensor(ε): elongation tensor(Q): stiffness matrix of unidirectional ply(R): stiffness matrix of laminate(A): stiffness matrix of laminate

El: longitudinal young's modulus of unidirectional plyEt: transversal young's modulus of unidirectional plyνit: longitudinal/transversal Poisson coefficient

νtl = νlt EE

t

l: transversal/longitudinal Poisson coefficient

Glt: shear modulus of unidirectional plyep: ply thickness

Rlt: allowable longitudinal tension stressRlc: allowable longitudinal compression stressRtt: allowable transversal tension stressRtc: allowable transversal compression stressS: allowable shear stress



MONOLITHIC PLATE - MEMBRANEDefinitions - Homogeneity - Isotropy - Coupling C 2.1

2.2.12.2.2

2 . GENERAL DEFINITIONS

2.1 . Homogeneity - Isotropy

- A material is so-called homogeneous when its properties are independent from the pointconsidered.

- A material is isotropic if it has the same properties in all directions.

- A material is anisotropic if there is no property symmetry, i.e. properties depend on thedirection and on the point considered.

- A material is orthotropic if its properties are symmetrical with relation to twoperpendicular planes. Axes of symmetry are so-called axes of orthotropy.

2.2 . Coupling phenomenon

2.2.1 . Plane coupling

In the case of an orthotropic material, there is a “plane coupling” if the loading axis is notcoincident with one of its axes of orthotropy. In that case, normal loading (σ) generatesshear (γ) and shear loading (τ) generates elongation (ε).

2.2.2 . Mirror symmetry

The laminate must be such that each layer has an identical symmetrical layer with relationto the neutral plane.

This symmetry allows the membrane-bending coupling to be eliminated, i.e. theoccurrence of plate bending, when a tension load is applied in its plane.

x

y

N1 N1



MONOLITHIC PLATE - MEMBRANEDesign method C 3

1/8

3 . DESIGN METHOD

The design method for a flat plate consists in assessing stresses in each ply and indetermining the corresponding Hill’s criterion (see § G.3).

Let’s assume that all plies are made up of the same material, and that the laminate isprovided with the mirror symmetry property.

That is to say the central plane of the laminate (for example: (0°/45°/135°/90°) s =(0°/45°/135°/90°/90°/135°/45°/0°). This property implies that there is no coupling betweenthe membrane effects and the bending effects.

Which means that the membrane flux tensor (Nx, Ny, Nxy) induces εx, εy, and γxy typeelongations only and that, on the other hand, the moment flux tensor (Mx, My, Mxy) inducesχx, χy and χxy type rotations only.

In other words, in the case of a laminate with the mirror symmetry property, therelationship which binds loading and elongation may be formulated as follows:

Nx

Ny

Nxy

Mx

My

Mxy

=

Aij 0

0 Cij

εx

εy

γxy

χx

χy

χxy




2/8

A laminate (as well as the sign convention for membrane type load fluxes) may berepresented as follows:

With each fibre direction (θ = 1, 2 or 3) is associated the number of corresponding pliesnθ.

1st step: Design of the stiffness matrix for the unidirectional layer in its own coordinatesystem (l, t). This matrix shall be called (Ql, t).

c1 (σl, t) = (Ql, t) x (εl, t)

σlEl

lt tl1 − ν νν

ν νtl l

lt tl

E1 −

0 εl

σt =ν

ν νlt t

lt tl

E1 −

Et

lt tl1 − ν ν0 εt

τlt 0 0 Glt γlt

y

x

3

1

θ2

θ1

θ3

z

y

x Nx > 0Nxy > 0

Ny > 0

2

t

l




3/8

2nd step: Design of the stiffness matrix for the unidirectional layer in direction θ in thereference coordinate system (x, y). This matrix shall be called (Qx, y,θ).

c2 (Qx, y,θ) = (Tθ) x (Ql, t) x (T'θ)-1

with:

(Tθ) =

(cos ) (sin ) sin cos

(sin ) (cos ) sin cos

sin cos sin cos (cos ) (sin )

θ θ θ θ

θ θ θ θ

θ θ θ θ θ θ

2 2

2 2

2 2

2

2

−

− −

x x

x x

x x

(T'θ) =

(cos ) (sin ) sin cos

(sin ) (cos ) sin cos

sin cos sin cos (cos ) (sin )

θ θ θ θ

θ θ θ θ

θ θ θ θ θ θ

2 2

2 2

2 22 2

−

− −

x

x

x x x x

Matrix (Tθ) corresponding to the basic transformation matrix for stress condition.

Matrix (T'θ) corresponding to the basic transformation matrix for elongation condition.

y

x

tl

θ




4/8

Note: Let material be defined by the two following drawings:

By obtaining their equilibrium, we get the three following expressions:

σl ds - σx ds (cosθ)2 - σy ds (sinθ)2 + τxy ds sinθ cosθi - τxy ds sinθ cosθ = 0

τlt ds + σx ds sinθ cosθ - σy ds sinθ cosθ - τxy ds (sinθ)2 - τxy ds (cosθ)2 = 0

σt ds - σx ds (sinθ)2 - σy ds (cosθ)2 + τxy ds cosθ cosθ - τxy ds sinθ sinθ = 0

Expressions from which the matrix (Tθ) terms are easily taken.

σy

τyx

σt

τtl

τxyσx

ds

θ

x

l

yt

i σy

τyx

θ

y

i

σx

τxyτlt

σl

ds τxy

τyx

σx

x

σy

t

σl

σl

τlt

τtl

σt

σt

τtl

τlt

l




5/8

Remark: the stiffness matrix (Qx, y, θ) also allows determination of the mechanicalproperties of the unidirectional layer in direction θ in the reference coordinatesystem (o, x, y). For the unidirectional layer, we have:

(σx, y) = (Qx, y, θ) x (εx, y) hence (εx, y) = (Qx, y, θ)-1 x (σx, y)

εx1

Ex ( )θ−

ν θθ

yx

yE( )( )

η θθ

yx

xyG( )( )

σx

εy = −ν θ

θxy

xE( )( )

1Ey ( )θ

µθ

yx

xyG ( )σy

γxyη θ

θx

xE( )( )

µ θθ

y

yE( )( )

1Gxy ( )θ

τxy

where:

Ex(θ) = 11 2

4 42 2c

EsE

c sG El t lt

tl

t+ + −

�

��

�

��

ν

Ey(θ) = 11 2

4 42 2s

EcE

c sG El t lt

tl

t+ + −

�

��

�

��

ν

Gxy(θ) = ( )

1

4 1 1 22 22 2 2

c sE E E

c sGl t

tl

t lt+ +

�

��

�

�� +

−ν

( )ν θθ

νyx

y

tl

t l t ltE Ec s c s

E E G( )( )

= + − + −�

��

�

��4 4 2 2 1 1 1

νxy(θ) = νyx(θ) EE

x

y

( )( )θθ

with c ≡ cos(θ) and s ≡ sin(θ)




6/8

3rd step: Knowing the stiffness matrix of each layer (Qx, y, θ) with relation to the referencecoordinate system (x, y), the laminate stiffness matrix can be calculated in this samecoordinate system: (Rx, y).

For this, the mixture law shall be applied.

c3 (Rx, y) = (Q ), ,x yk

n

k

nθ=� 1 or (Rx, y) =

ep

ex yk

n

k(Q ), , θ=� 1

4th step: Determination of the laminate elongation tensor in the reference coordinatesystem.

c4 (εx, y) = 1e

x (Rx, y)-1 x (Nx, y)

ε

ε

γ

x

y

xy

= 1e

(Rx, y)-1

N

N

N

x

y

xy

or

N

N

N

x

y

xy

= (A)

ε

ε

γ

x

y

xy

where (A) is the laminate membrane stiffness matrix: (A) = e x (Rx, y).

Matrix (A) is the stiffness matrix which binds the stress flux tensor (N) with the elongationtensor (ε).

c5 (N) = (A) x (ε)

N

N

N

x

y

xy

=

A A A

A A A

A A A

11 12 13

21 22 23

31 32 33

x

ε

ε

γ

x

y

xy




7/8

where

c6 Aij = ( )Εijk

k kk

nz z( )− −=� 11

with

Ε11(θ) = c4 Εl + s4 Εt + 2 c2 s2 (νtl Εl + 2 Glt)Ε22(θ) = s4 Εl + c4 Εt + 2 c2 s2 (νtl Εl + 2 Glt)Ε33(θ) = c2 s2 (Εl + Εt - 2 νtl Εl) + (c2 - s2)2 Glt

Ε12(θ) = Ε21(θ) = c2 s2 (Εl + Εt - 4 Glt) + (c4 + s4) νtl Εl

Ε13(θ) = Ε31(θ) = c s {c2 Εl - s2 Εt - (c2 - s2) (νtl Εl + 2 Glt)}Ε23(θ) = Ε32(θ) = c s {s2 Εl - c2 Εt + (c2 - s2) (νtl Εl + 2 Glt)}

c ≡ cos(θ) where θ is the fibre direction in the reference coordinate system (o, x, y)

s ≡ sin(θ) where θ is the fibre direction in the reference coordinate system (o, x, y)

Εl = El

tl lt1 − ν ν

Εt = Et

tl lt1 − ν ν

z

ply No. 1

ply No. k

thickness




8/8

5th step: Determination of elongations in each fibre direction

c7 (εl, t, θ) = (T' - θ) x (εx, y)

ε

ε

γ

θ

θ

θ

l

t

lt

=

22

22

22

)(sin)(coscosxsinx2cosxsinx2

cosxsin)(cos)(sin

cosxsin)(sin)(cos

θ−θθθθθ−

θθ−θθ

θθθθ

ε

ε

γ

x

y

xy

6th step: Determination of stresses in each fibre direction

c8 (σl, t, θ) = (Ql, t) x (εl, t, θ)

σ

σ

τ

θ

θ

θ

l

t

lt

= (Ql, t)

ε

ε

γ

θ

θ

θ

l

t

lt

7th step: Assessment of Hill’s criterion in each fibre direction. Refer to chapter G (failurecriteria).



MONOLITHIC PLATE - MEMBRANEEquivalent properties C 4

4 . DEFORMATIONS AND EQUIVALENT MECHANICAL PROPERTIES

Monolithic plates are microscopically heterogeneous. It is sometimes necessary to findtheir equivalent membrane stiffness properties in order to determine the passing loadsand resulting deformations.

Equivalent membrane young's moduli are directly derived from the laminate stiffnessmatrix (A):

c9 (A)-1 = 1e

1

1

1

E Ex

E Ex

x xG

xx

yx

yy

xy

xx yy

xy

memb equi

memb equi

memb equi

memb equi

memb equi memb equi

memb equi

. .

. .

. .

. .

. . . .

. .

−

−

ν

ν

If reference axes (o, x, y) are coincident with the axes of orthotropy of the laminate, weobtain:

E A A Ae Axxmemb equi. .

( )=

−11 22 122

22

E A A Ae Ayymemb equi. .

( )=

−11 22 122

11

G Aexymemb equi. .

= 66

νxymemb equi

AA. .

= 12

22

νyxmemb equi

AA. .

= 21

11



MONOLITHIC PLATE - MEMBRANEGraphs - Failure envelopes - Theoretical principle C 5.1.1

1/3

5 . GRAPHS

5.1 . Failure envelopes

5.1.1 . Theoretical principle

Let a laminate be made up of plies in the same material and described as follows:

- overall thickness e,

- percentage of plies at 0°,



- percentage of plies at 90°.

If membrane fluxes Nx, Ny and Nxy, are applied to the laminate, so that Nx2 + Ny

2 + Nxy2 = 1,

the design method outlined above allows loads inside each layer to be determined and theoverall plate margin (m) to be found (see § G "Failure criteria").

Ny

Nx

Nxy

o

Nx, Ny, Nxy (margin m)

Nx', Ny', Nxy' (zero margin)




2/3

Let's assume that the three fluxes are multiplied by the coefficient m100

1+ .

In this case, the laminate subject to this new loading (Nx', Ny', Nxy') shall have a zeromargin.

Therefore, it is possible to associate each triplet (Nx, Ny, Nxy) with a flux triplet (Nx', Ny', Nxy')so that the margin associated with it is zero.

If this operation is repeated for the set of points so that Nx2 + Ny

2 + Nxy2 = 1 (sphere S with

radius 1), then, surface S' is obtained, corresponding to the set of points with a zeromargin. This is the material failure envelope.

This three-dimensional representation of zero margin points is not easy to use.

Ny

Nx

Nxy

o

S'

S




3/3

It can be represented in a two-dimensional space (Nx, Ny) in the form of graphs (eachcurve corresponding to the intersection S' with an equation plane Nxy = Nxyi).

If this set of curves is projected onto the plane (o, Nx, Ny), a network of curves is obtainedwhich constitutes the breaking graph of the laminate.

This graph (corresponding to a given material and a specific lay-up) allows the laminatemargin (Hill's criterion) to be determined graphically.

Nx

Ny

o

Nxyi = 0

Nxyi

Nxyn

Nx

Ny

Nxy

plane Nxy = 0

plane Nxyi

plane Nxyn



MONOLITHIC PLATE - MEMBRANEGraphs - Failure envelopes - Margin C 5.1.2

5.1.2 . Margin search - Methodology

Let a laminate be subject to fluxes Nxo, Nyo and Nxyo and the breaking graph associatedwith it.

- Plot the straight line D crossing point o and point A of coordinates Nxo and Nyo.

- Perpendicular to this straight line, plot the value Nxyi segment corresponding to thegraph curve Nxyi. Repeat this operation for each graph curve.

- Plot curve C.

- From point A, plot point B so that AB = Nxyo and AB ⊥ D.

- Determine point C, intersection of the straight line (o, B) and curve C.

- The composite plate margin is equal to 100 o Co B

−�

��

�

��1 .

Nx

Ny

D

Nxoo

Nyoc

A

B

CN

xyo

Nxyi



MONOLITHIC PLATE - MEMBRANEGraphs - Mechanical properties C 5.2

In practice, curves are represented in stress and not in flux values. This makes it possibleto group together some laminates per lay-up class (for example: 3/2/2/1 ≡ 6/4/4/2 ≡

9/6/6/3).

A number of orthotropic laminate failure envelopes in carbon T300/914 layers shall befound in chapter Z “material properties”.

5.2 . Mechanical properties

For a given material, a set of graphs may be created giving the mechanical properties(strength and elasticity moduli) of an orthotropic laminate described by its percentages ofplies in each direction (see drawing below).

A number of those graphs associated with carbon T300/914 layer shall be found inchapter Z “material properties”.

Gxy

% to 45°

% to 90°

%

Gxy



MONOLITHIC PLATE - MEMBRANEExample C 6

1/8

6 . EXAMPLE

Given a laminate made of T300/BSL914 (new) with the following lay-up:

0°: 6 plies45°: 4 plies135°: 4 plies90°: 6 plies

Mechanical properties of the unidirectional ply are the following:

El = 13000 hb (130000 MPa)Et = 465 hb (4650 MPa)νlt = 0.35νtl = 0.0125Glt = 465 hb (4650 MPa)ep = 0.13 mme = 2.6 mm

Rlt = 120 hb (1200 MPa)Rlc = - 100 hb (1000 MPa)Rtt = 5 hb (50 MPa)Rtc = - 12 hb (120 MPa)S = 7.5 hb (75 MPa)

The purpose of this example is to search for stresses applied to each ply (0°, 45°, 135°,90°) knowing that the laminate is globally subject to the three following load fluxes in thereference coordinate system (x, y):

Nx = 30.83 daN/mmNy = - 2.22 daN/mmNxy = 44.92 daN/mm

These load fluxes being the continuation of the example covered in chapter K (Fastenerhole).




2/8

1st step: Design of stiffness matrix (Ml, t) for the unidirectional ply with relation to its owncoordinate system (l, t).

{c1}

(Ql, t) =

130001 0 35 0 0125

13000 0 01251 0 35 0 0125

0

465 0 351 0 35 0 0125

4651 0 35 0 0125

0

0 0 465

− −

− −

. ..

. .

.. . . .

(Ql, t) =

13057 163 0

163 467 0

0 0 465

All values being expressed in daN/mm2.

2nd step: Assessment of stiffness matrix for each unidirectional ply with relation to thereference coordinate system (x, y).

{c2}

(Qx, y, 0°) =

1 0 0

0 1 0

0 0 1

13057 163 0

163 467 0

0 0 465

1 0 0

0 1 0

0 0 1

1−




3/8

(Qx, y, 45°) =

0 5 0 5 1

0 5 0 5 1

0 5 0 5 0

. .

. .

. .

−

−

13057 163 0

163 467 0

0 0 465

0 5 0 5 0 5

0 5 0 5 0 5

1 1 0

1. . .

. . .

−

−

−

(Qx, y, 135°) =

0 5 0 5 1

0 5 0 5 1

0 5 0 5 0

. .

. .

. .

−

−

13057 163 0

163 467 0

0 0 465

0 5 0 5 0 5

0 5 0 5 0 5

1 1 0

1. . .

. . .−

−

−

(Qx, y, 90°) =

0 1 0

1 0 0

0 0 1−

13057 163 0

163 467 0

0 0 465

0 1 0

1 0 0

0 0 1

1

−

−

Thus, we find:

(Qx, y, 0°) =

13057 163 0

163 467 0

0 0 465

(Qx, y, 45°) =

3928 2998 3148

2998 3928 3148

3148 3148 3299

(Qx, y, 135°) =

3928 2998 3148

2998 3928 3148

3148 3148 3299

−

−

− −




4/8

(Qx, y, 90°) =

467 163 0

163 13057 0

0 0 465

All values being expressed in daN/mm2.

3rd step: By applying the mixture law, the overall laminate stiffness matrix (Rx, y) isformulated as follows.

{c3}

(Rx, y) = 120

3299x8465x123148x43148x43148x43148x4

3148x43148x413057x63928x8467x62998x8163x12

3148x43148x42998x8163x12467x63928x813057x6

+−−

−+++

−+++

(Rx, y) =

5628 1297 0

1297 5628 0

0 0 1598

(Rx, y)-1 =

188 4 4 32 5 611 20

4 32 5 188 4 4 02 20

611 20 4 02 20 6 25 4

. . .

. . .

. . .

x E x E x E

x E x E x E

x E x E x E

− − − − −

− − − −

− − − −

All values being expressed in daN/mm2 and mm²/daN.




5/8

4th step: Determination of the laminate strain tensor in the reference coordinate system (x,y).

{c4}

ε

ε

γ

x

y

xy

= 12 6.

188 4 432 5 611 20

432 5 188 4 402 20

611 20 402 20 625 4

. . .

. . .

. . . .

xE xE xE

xE xE xE

xE xE x E

− − − − −

− − − −

− − − −

3083

222

4492

.

.

.

− =

2262 6

673 6

10807 6

xE

xE

xE

−

− −

−

All values being expressed in mm/mm.

5th step: Determination of the strain tensor in each fibre direction.

{c7}

(εl, t, 0°) =

1 0 0

0 1 0

0 0 1

2262 6

673 6

10807 6

x E

x E

x E

−

− −

−

=

2262 6

673 6

10807 6

x E

x E

x E

−

− −

−

(εl, t, 45°) =

0 5 0 5 0 5

0 5 0 5 0 5

1 1 0

.. . .

. . .−

−

2262 6

673 6

10807 6

x E

x E

x E

−

− −

−

=

6198 6

4609 6

2935 6

x E

x E

x E

−

− −

− −




6/8

(εl, t, 135°) =

0 5 0 5 0 5

0 5 0 5 0 5

1 1 0

. . .

. . .

−

−

2262 6

673 6

10807 6

x E

x E

x E

−

− −

−

=

− −

−

−

4609 6

6198 6

2935 6

x E

x E

x E

(εl, t, 90°) =

0 1 0

1 0 0

0 0 1−

2262 6

673 6

10807 6

x E

x E

x E

−

− −

−

=

− −

−

− −

673 6

2262 6

10807 6

x E

x E

x E

All values being expressed in mm/mm.

6th step: With the previous results, stresses in each ply are determined.

{c8}

(σl, t, 0°) =

13057 163 0

163 467 0

0 0 465

2262 6

673 6

10807 6

x E

x E

x E

−

− −

−

=

29 42

0 06

5 03

.

.

.

(σl, t, 45°) =

13057 163 0

163 467 0

0 0 465

6198 6

4609 6

2935 6

x E

x E

x E

−

− −

− −

=

8017

114

136

.

.

.

−

−




7/8

(εl, t, 135°) =

13057 163 0

163 467 0

0 0 465

− −

−

−

4609 6

6198 6

2935 6

x E

x E

x E

=

− 5917

214

136

.

.

.

(εl, t, 90°) =

13057 163 0

163 467 0

0 0 465

− −

−

− −

673 6

2262 6

10807 6

x E

x E

x E

=

−

−

8 42

0 95

5 03

.

.

.

All values being expressed in hb.

7th step: In each direction, the corresponding Hill’s criterion is calculated (see chapter G),which gives the following margins for each ply:

0° → 40 % 45° → 42 % 135° → 31 % 90° → 42 %

The ply at 135° is, therefore, the most brittle ply in this loading case.




8/8

8th step: The laminate margin may be found with the breaking graph corresponding to thismaterial (see chapter Z).

We have: Nx = 30.83 daN/mmNy = - 2.22 daN/mmNxy = 44.92 daN/mm

Giving, for an overall thickness of 2.6 mm, the following stresses:

σx = 11.86 hbσy = - 0.85 hb ≈ 0 hτxy = 17.28 hb

+ T = 22 HBx T = 21 HBY T = 18 HB+ T = 15 HBx T = 12 HBY T = 9 HB

T = 6 HBT = 3 HBT = 0 HB

Scale: 1 cm ↔ 3.33 hb

Marge = 100 o Co B

−�

��

�

�� = −�

��

��1 100 72

511 ≈ 41 %

There is a 10 % error with respect to the analytical method (31 %).



MONOLITHIC PLATE - MEMBRANEReferences C




M. THOMAS, Analysis of a laminate plate subject to membrane and bending loads,440.227/79

J.C. SOURISSEAU, 40430.030

J. CHAIX, 436.127/91



D

MONOLITHIC PLATE - BENDING ANALYSIS



MONOLITHIC PLATE - BENDINGNotations D 1

1 . NOTATIONS

(o, x, y): reference coordinate system(o, l, t): coordinate system specific to the unidirectional fibre

u, v, w: displacement from any point on the beamuo, vo, wo: displacement from the beam neutral planeβ: beam curvature at a given pointR: beam radius of curvature at a given point

εx, εy, γxy: strains at any pointεox, εoy, γoxy: strains neutral plane

(M): bending moment tensor

(χ): rotation tensor(α): tensor of angles formed by the deformation diagram(C): inertia matrix of laminate

k: fibre coordinate system

θ: fibre orientation

El: longitudinal young's modulus of unidirectional plyEt: transversal young's modulus of unidirectional plyνlt: longitudinal/transversal poisson coefficient

νtl = νlt EE

t

l: transversal/longitudinal poisson coefficient

Glt: shear modulus of unidirectional plyep: ply thickness



MONOLITHIC PLATE - BENDINGIntroduction - Design method D 2

31/4

2 . INTRODUCTION

In chapter C, we examined the case of a laminate provided with mirror symmetry subjectto membrane type loading. In the paragraph below, we shall examine the case of alaminate with the same properties but, this time, subject to pure bending type loads.

By convention, we shall consider that any positive moment compresses the laminateupper fibre.

Let’s assume that bending moment flows Mx, My and Mxy generate εx, εy and γxy typestrains.

Let’s assume also (Kirchoff) that the neutral plane is coincident with the neutral line.

3 . DESIGN METHOD

Let a bent plate be represented as follows:

z

x, yw

tg(β) = ∂

∂wx

R = 12

2∂

∂

w

x

z

y

x

My > 0

Mxy > 0Mx > 0

z

x, y

u, v

uo, vo

zw

wo



MONOLITHIC PLATE - BENDINGDesign method D 3

2/4

If the displacements from a point at position Z are defined as u, v and w in the coordinatesystem (x, y, z), then we may write:

u = uo - z ∂∂wx

o

v = vo - z ∂∂wy

o

w = wo

where uo, vo et wo represent displacements from the neutral plane in the coordinatesystem (x, y, z).

We deduce (by deriving with respect to coordinates) the corresponding non-zero strains:

d1 εx = εox - z ∂∂

2

2wx

o

εy = εoy - z ∂∂

2

2wy

o

γxy = γoxy - 2 z ∂∂ ∂

2wx y

o

where εox, εoy and γoxy rerepresent strains at a point located on the neutral plane and εx, εy

and γxy represent strains at any point at position z.

z

x

neutral plan

zεx

εox

o

tg(α) = ∂

∂

2

2w

x




3/4

From the general expression for the bending moment: M = �− σ2h

2h dzz , we obtain the

relationship between the bending load tensor (M) and the rotation tensor (χ):

d2 (M) = (C) x (χ)

M

M

M

x

y

xy

=

C C C

C C C

C C C

11 12 13

21 22 23

31 32 33

∂∂

∂∂∂∂ ∂

2

2

2

2

2

2

wxwyw

x y

O

O

O

where

d3 Cij = Εijk k k

k

n z z31

3

1 3−�

��

�

��

−=�

with

d4 Ε11(θ) = c4 Εl + s4 Εt + 2 c2 s2 (νtl Εl + 2 Glt)Ε22(θ) = s4 Εl + c4 Εt + 2 c2 s2 (νtl Εl + 2 Glt)Ε33(θ) = c2 s2 (Εl + Εt - 2 νtl Εl) + (c2 - s2)2 Glt



c ≡ cos(θ) where θ is the ply direction in the reference coordinate system (o, x, y)

s ≡ sin(θ) where θ is the ply direction in the reference coordinate system (o, x, y)




4/4

with

d5 Εl = El

tl lt1 − ν ν

Εt = Et

tl lt1 − ν ν

If the tensor of angles formed by the strain diagram in each direction is defined by (α):(αx, αy, αxy) we may write in a simplified form the relationship:

d6 (χ) = tg (α)

By convention, we shall assume that (α) is negative when the upper fibre is in tension. Wehave:

d7 (ε)z = - (χ) x z

This relationship makes it possible to determine each ply strain and, therefore, to find(using chapter C) stresses applied to it.

Remark: The terms Cij must be determined with relation to the laminate neutral line(Kirchoff’s assumption). In this case, the neutral plane shall also be used as areference for the overall load pattern.

hzk zk - 1

ply No. k

neutral plan

ply No. 1

σ

z

ε

z

α



MONOLITHIC PLATE - BENDINGEquivalent mechanical properties D 4

4 . DEFORMATIONS AND EQUIVALENT MECHANICAL PROPERTIES

Monolithic plates are microscopically heterogeneous. It is sometimes necessary to findtheir equivalent bending stiffness properties in order to determine the passing loads andresulting deformations.

Equivalent bending elasticity moduli are directly derived from the laminate stiffness matrix(C):

d8 (C)-1 = 123e

1

1

1

Ex x

xE

x

x xG

xxbending equi

yybending equi

xybending equi

.

.

.

If reference axes (o, x, y) are coincident with the axes of orthotropy of the laminate, weobtain:

Exxbending equi. = 12 C C Ce C

11 22 122

322

− ( )

Eyybending equi. = 12 C C Ce C

11 22 122

311

− ( )

Gxybending equi. = 12 Ce

663



MONOLITHIC PLATE - BENDINGExample D 5

1/7

5 . EXAMPLE

Let a T300/BSL914 laminate (new) be laid up as follows:


Stacking from the external surface being as follows: 0°/45°/135°/90°/90°/135°/ 45°/0°.

Mechanical properties of the unidirectional ply are the following:

El = 13000 hbEt = 465 hbνlt = 0.35νtl = 0.0125Glt = 465 hbep = 0.13 mm

The purpose of this example is to search for elongations at the laminate external surface,knowing that the laminate is globally subject to the three following moment fluxes in thereference coordinate system (x, y):

z8 = 0.52z7 = 0.39z6 = 0.26z5 = 0.13z4 = 0

k = 8 (0°)k = 7 (45°)k = 6 (135°)k = 5 (90°)k = 4 (90°)k = 3 (135°)k = 2 (45°)k = 1 (0°)



MONOLITHIC PLATE - BENDINGExemple D 5

2/7

Mx = 10 daNMy = 0 daN/mmMxy = - 5 daN/mm

1st step: calculation of stiffness coefficients for the unidirectional ply:

{d5}

Εl = 130001 0 35 0 0125− . .

= 13057 daN/mm2

Εt = 4651 0 35 0 0125− . .

= 467 daN/mm2

2nd step: For each ply, stiffness coefficients Εij expressed in daN/mm2 are calculated.

{d4}

ply at 0°

Ε11(0°) = 13057Ε22(0°) = 467Ε33(0°) = 465Ε12(0°) = Ε21(0°) = 0.0125 x 13000 = 163Ε13(0°) = Ε31(0°) = 0Ε23(0°) = Ε32(0°) = 0

z

y

x Mx = 10 daN Mxy = - 5 daN




3/7

ply at 45°

Ε11(45°) = 0.7074 13057 + 0.7074 467 + 2 x 0.7072 0.7072 (0.0125 x 13057 + 2 x 465) = 3925Ε22(45°) = 0.7074 13057 + 0.7074 467 + 2 x 0.7072 0.7072 (0.0125 x 13057 + 2 x 465) = 3925Ε33(45°) = 0.7072 0.7072 (13057 + 467 - 2 x 0.0125 x 13057) = 3297Ε12(45°) = Ε21(45°) = 0.7072 0.7072 (13057 + 467 - 4 x 465) + (0.7074 + 0.7074) x 0.0125 x 13057 = 2995Ε13(45°) = Ε31(45°) = 0.707 x 0.707 {0.7072 13057 - 0.7072 467} = 3146Ε23(45°) = Ε32(45°) = 0.707 x 0.707 {0.7072 13057 - 0.7072 467} = 3146

ply at 135°

Ε11(135°) = 3925Ε22(135°) = 3925Ε33(135°) = 3297Ε12(135°) = Ε21(135°) = 2995Ε13(135°) = Ε31(135°) = - 3146Ε23(135°) = Ε32(135°) = - 3146

ply at 90°

Ε11(90°) = 467Ε22(90°) = 13057Ε33(90°) = 465Ε12(90°) = Ε21(90°) = 163Ε13(90°) = Ε31(90°) = 0Ε23(90°) = Ε32(90°) = 0




4/7

3rd step: Calculation of laminate inertia matrix (C) coefficients Cij expressed in daN mm.

The laminate being provided with the mirror symmetry property, coefficients Cij shall becalculated for the laminate upper half, then they shall be multiplied by 2.

{d3}

90° 135° 45° 0°

C11 = 2 467013 0

33925

0 26 0133

39250 39 0 26

313057

0 52 0 393

3 3 3 3 3 3 3 3. . . . . . .−+

−+

−+

−�

��

�

��

C12 = 2 163013 0

32995

0 26 0133

29950 39 0 26

3163

0 52 0 393

3 3 3 3 3 3 3 3. . . . . . .−+

−+

−+

−�

��

�

��

C13 = 2 0013 0

33146

0 26 0 133

31460 39 0 26

30

0 52 0 393

3 3 3 3 3 3 3 3. . . . . . .−−

−+

−+

−�

��

�

��

C21 = 2 163013 0

32995

0 26 0133

29950 39 0 26

3163

0 52 0 393

3 3 3 3 3 3 3 3. . . . . . .−+

−+

−+

−�

��

�

��

C22 = 2 13057013 0

33925

026 0133

39250 39 0 26

3467

052 0 393

3 3 3 3 3 3 3 3. . . . . . .−+

−+

−+

−�

��

�

��

C23 = 2 0013 0

33146

0 26 0 133

31460 39 0 26

30

0 52 0 393

3 3 3 3 3 3 3 3. . . . . . .−−

−+

−+

−�

��

�

��

C31 = 2 0013 0

33146

0 26 0 133

31460 39 0 26

30

0 52 0 393

3 3 3 3 3 3 3 3. . . . . . .−−

−+

−+

−�

��

�

��

C32 = 2 0013 0

33146

0 26 0 133

31460 39 0 26

30

0 52 0 393

3 3 3 3 3 3 3 3. . . . . . .−−

−+

−+

−�

��

�

��

C33 = 2 465013 0

33297

0 26 0133

32970 39 0 26

3465

0 52 0 393

3 3 3 3 3 3 3 3. . . . . . .−+

−+

−+

−�

��

�

��

C11 = 858C12 = 123C13 = 55C21 = 123C22 = 194C23 = 55C31 = 55C32 = 55C33 = 151




5/7

Thus, the following matrix is obtained:

(C) =

858 123 55

123 194 55

55 55 151

4th step: Search for the rotation tensor

{d2}

M

M

M

x

y

xy

=

858 123 55

123 194 55

55 55 151

=

∂∂

∂∂∂∂ ∂

2

2

2

2

2

2

woxwoywo

x y

hence

∂∂

∂∂∂∂ ∂

2

2

2

2

2

2

woxwoywo

x y

=

1287 3 7 617 4 1913 4

7 617 4 6199 3 198 3

1913 4 198 3 7 414 3

. . .

. . .

. . .

E E E

E E E

E E E

− − − − −

− − − − −

− − − − −

=

M

M

M

x

y

xy




6/7

∂∂

∂∂∂∂ ∂

2

2

2

2

2

2

wxwy

wx y

o

o

o

=

1287 3 7 617 4 1913 4

7 617 4 6199 3 198 3

1913 4 198 3 7 414 3

. . .

. . .

. . .

E E E

E E E

E E E

− − − − −

− − − − −

− − − − −

=

10

0

5−

Thus, we find:

∂∂

∂∂∂∂ ∂

2

2

2

2

2

2

wxwy

wx y

o

o

o

=

13 82 3

2 283 3

38 98 3

.

.

.

E

E

E

−

−

− −

which is the rotation tensor (χ).

5th step: We now propose to calculate strains ε (0°) for the ply at 0° (at the external line ofthe layer).

{d7}

εx(0°) = - ∂∂

2

2wx

o x h2

εy(0°) = - ∂∂

2

2wy

o x h2

γxy(0°) = - 2 ∂∂ ∂

2wx y

o x h2




7/7

hence:

εx(0°) = - 1 x 13.82 E-3 x 0.52 = - 7186 µd

εy(0°) = - 1 x 2.283 E-3 x 0.52 = - 1187 µd

γxy(0°) = - 1 x - 38.98 E-3 x 0.52 = 20270 µd

Stresses in the layer may be determined afterwards. To do this, refer tochapter C.



MONOLITHIC PLATE - BENDINGReferences D




M. THOMAS, Analysis of a laminate plate subject to membrane and bending loads,440.227/79


J. CHAIX, 436.127/91



E

MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS



MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSISNotations E 1

1 . NOTATIONS


εx, εy, γxy: material strains at any pointwo: displacement from plate neutral plane

(N): normal flux tensor(M): bending moment tensor

(ε): membrane type strain tensor(χ): curvature tensor

(A): laminate stiffness matrix (membrane)(B): laminate stiffness matrix (membrane/bending coupling)(C): laminate stiffness matrix (bending)


k: fibre coordinate system

El: longitudinal elasticity modulus of unidirectional fibreEt: transversal elasticity modulus of unidirectional fibreνlt: longitudinal/transversal poisson coefficient

νtl: transversal/longitudinal poisson coefficientGlt: shear modulus of unidirectional fibreep: ply thickness



MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSISIntroduction E 2

2 . INTRODUCTION

We have seen in chapter C that there is a relationship which binds membrane strains andloading of the same type.

This relationship may be formulated as follows: (N) = (A) x (ε).

We also saw in chapter D that there is a relationship which binds the curvature tensor andthe moment tensor.

This relationship may be formulated as follows: (M) = (C) x (χ).

If lay-up has the mirror symmetry property, then both phenomena are dissociated andindependent. In other words, the overall relationship which binds the set of strains and theset of loadings may be formulated as follows:

N

N

N

M

M

M

x

y

xy

x

y

xy

=

A A A

A A A

A A A

C C C

C C C

C C C

11 12 13

21 22 23

31 32 33

11 12 13

21 22 23

31 32 33

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

0 0 0

ε

ε

γ

∂∂

∂∂∂∂ ∂

x

y

xy

o

o

o

wxwyw

x y

2

2

2

2

2

2

where coefficients Aij and Cij are defined in chapters C and D.



MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSISAnalysis method E 3

1/2

3 . ANALYSIS METHOD

If lay-up is non-symmetrical, then all zero terms of the previous matrix become non-zeroand there is a membrane/bending coupling. Both phenomena become dependent. Therelationship between loadings and strains is thus:

e1

N

N

N

M

M

M

x

y

xy

x

y

xy

=

A A A B B B

A A A B B B

A A A B B B

B B B C C C

B B B C C C

B B B C C C

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

ε

ε

γ

∂∂

∂∂∂∂ ∂

x

y

xy

o

o

o

wxwyw

x y

2

2

2

2

2

2

where

e2 Bij = - Ez z

ijk k k

k

n2

12

1 2−�

��

�

��

−=�

zk zk - 1

ply No. k

ply No. 1

neutral plane



MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSISAnalysis method E 3

2/2

with

e3 Ε11(θ) = c4 Εl + s4 Εt + 2 c2 s2 (νtl Εl + 2 Glt)Ε22(θ) = s4 Εl + c4 Εt + 2 c2 s2 (νtl Εl + 2 Glt)Ε33(θ) = c2 s2 (Εl + Εt - 2 νtl Εl) + (c2 - s2)2 Glt



where

c ≡ cos(θ) where θ is the fibre direction in the reference coordinate system (o, x, y).

s ≡ sin(θ) where θ is the fibre direction in the reference coordinate system (o, x, y).

with

e4 Εl = El

tl lt1 − ν ν

Εt = Et

tl lt1 − ν ν

Remark: The terms Bij and Cij must be determined with relation to the laminate neutralline (Kirchoff’s assumption). In this case, the neutral plane shall also be used asa reference for the overall load pattern.



MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSISExample E 4

1/9

4 . EXAMPLE


0°: 1 ply45°: 1 ply135°: 1 ply90°: 1 ply

Stacking from the external surface being as follows: 0°/45°/135°/90°.

Mechanical properties of the unidirectional fibre are the following:

El = 13000 hbEt = 465 hbνlt = 0.35νtl = 0.0125Glt = 465 hbep = 0.13 mm

z4 = 0.26

z3 = 0.13

z2 = 0

z1 = - 0.13

z0 = - 0.26

k = 4 (0°)

k = 3 (45°)

k = 2 (135°)

k = 1 (90°)

neutral plane




2/9

The purpose of this example is to search for strains at the laminate internal and externalsurfaces, knowing that the laminate is globally subject to the following fluxes in thereference coordinate system (x, y):

Nx = 5 daN/mmNy = 0 daN/mmNxy = 0 daN/mm

Mx = 0 daN

My = - 0.15 daN mm daNmm

��

��

Mxy = 0 daN

1st step: calculation of stiffness coefficients for the unidirectional fibre:

{e4}

Εl = 130001 0 35 0 0125− . .

= 13057 daN/mm2

Εt = 4651 0 35 0 0125− . .

= 467 daN/mm2

z

y

x Nx = 5 daN/mm

My = - 0.15 daN




3/9

2nd step: For each fibre direction, stiffness coefficients Εij expressed in daN/mm2, arecalculated.

{e3}

fibre at 0°

Ε11(0°) = 13057Ε22(0°) = 467Ε33(0°) = 465Ε12(0°) = Ε21(0°) = 0.0125 x 13000 = 163Ε13(0°) = Ε31(0°) = 0Ε23(0°) = Ε32(0°) = 0

fibre at 45°

Ε11(45°) = 0.7074 13057 + 0.7074 467 + 2 x 0.7072 0.7072 (0.0125 x 13057 + 2 x 465) = 3925Ε22(45°) = 0.7074 13057 + 0.7074 467 + 2 x 0.7072 0.7072 (0.0125 x 13057 + 2 x 465) = 3925Ε33(45°) = 0.7072 0.7072 (13057 + 467 - 2 x 0.0125 x 13057) = 3297Ε12(45°) = Ε21(45°) = 0.7072 0.7072 (13057 + 467 - 4 x 465) (0.7074 + 0.7074) x 0.0125 x 13057 = 2995Ε13(45°) = Ε31(45°) = 0.707 x 0.707 {0.7072 13057 - 0.7072 467} = 3146Ε23(45°) = Ε32(45°) = 0.707 x 0.707 {0.7072 13057 - 0.7072 467} = 3146

fibre at 135°

Ε11(135°) = 3925Ε22(135°) = 3925Ε33(135°) = 3297Ε12(135°) = Ε21(135°) = 2995Ε13(135°) = Ε31(135°) = - 3146Ε23(135°) = Ε32(135°) = - 3146




4/9

fibre at 90°

Ε11(90°) = 467Ε22(90°) = 13057Ε33(90°) = 465Ε12(90°) = Ε21(90°) = 163Ε13(90°) = Ε31(90°) = 0Ε23(90°) = Ε32(90°) = 0

3rd step: Calculation of laminate (membrane) stiffness matrix (A) coefficients Aij expressedin daN/mm.

{c6}

90° 135° 45° 0°A11 = (467 x 0.13 + 3925 x 0.13 + 3925 x 0.13 + 13057 x 0.13)A12 = (163 x 0.13 + 2995 x 0.13 + 2995 x 0.13 + 163 x 0.13)A13 = (0 x 0.13 - 3146 x 0.13 + 3146 x 0.13 + 0 x 0.13)A21 = (163 x 0.13 + 2995 x 0.13 + 2995 x 0.13 + 163 x 0.13)A22 = (13057 x 0.13 + 3925 x 0.13 + 3925 x 0.13 + 467 x 0.13)A23 = (0 x 0.13 - 3146 x 0.13 + 3146 x 0.13 + 0 x 0.13)A31 = (0 x 0.13 - 3146 x 0.13 + 3146 x 0.13 + 0 x 0.13)A32 = (0 x 0.13 - 3146 x 0.13 + 3146 x 0.13 + 0 x 0.13)A33 = (465 x 0.13 + 3297 x 0.13 + 3297 x 0.13 + 465 x 0.13)

hence

A11 = 2779A12 = 821A13 = 0A21 = 821A22 = 2779A23 = 0A31 = 0A32 = 0A33 = 978




5/9

4th step: Calculation of laminate (bending) inertia matrix (C) coefficients Cij expressed indaN mm.

{d3}

90° 135° 45° 0°

C11 = 467013 0 26

33925

0 013

33925

013 0

313057

026 013

3

3 3 3 3 3 3 3 3( . ) ( . ) ( . ) . . .− − −+

− −+

−+

−�

��

�

��

C12 = 163013 0 26

32995

0 013

32995

013 0

3163

0 26 013

3

3 3 3 3 3 3 3 3( . ) ( . ) ( . ) . . .− − −+

− −+

−+

−�

��

�

��

C13 = 0013 0 26

33146

0 013

33146

013 0

30

0 26 013

3

3 3 3 3 3 3 3 3( . ) ( . ) ( . ) . . .− − −−

− −+

−+

−�

��

�

��

C21 = 163013 0 26

32995

0 013

32995

013 0

3163

0 26 013

3

3 3 3 3 3 3 3 3( . ) ( . ) ( . ) . . .− − −+

− −+

−+

−�

��

�

��

C22 = 13057013 0 26

33925

0 013

33925

013 0

3467

0 26 013

3

3 3 3 3 3 3 3 3( . ) ( . ) ( . ) . . .− − −+

− −+

−+

−�

��

�

��

C23 = 0013 0 26

33146

0 013

33146

013 0

30

0 26 013

3

3 3 3 3 3 3 3 3( . ) ( . ) ( . ) . . .− − −−

− −+

−+

−�

��

�

��

C31 = 0013 0 26

33146

0 013

33146

013 0

30

0 26 013

3

3 3 3 3 3 3 3 3( . ) ( . ) ( . ) . . .− − −−

− −+

−+

−�

��

�

��

C32 = 0013 0 26

33146

0 013

33146

013 0

30

0 26 013

3

3 3 3 3 3 3 3 3( . ) ( . ) ( . ) . . .− − −−

− −+

−+

−�

��

�

��

C33 = 465013 0 26

33297

0 013

33297

013 0

3465

0 26 013

3

3 3 3 3 3 3 3 3( . ) ( . ) ( . ) . . .− − −+

− −+

−+

−�

��

�

��

hence

C11 = 75.1C12 = 6.06C13 = 0C21 = 6.06C22 = 75.1C23 = 0C31 = 0C32 = 0C33 = 9.59




6/9

5th step: Calculation of membrane - bending coupling coefficients Bij expressed in daN.

{e2}

90° 135° 45° 0°

B11 = - 4670 13

20 26

2

23925

02

0 132

23925

0 132

02

213057

0 262

0 132

2

( . ) ( . ) ( . ) . . .− − −+

− −+

−+

−�

��

�

��

B12 = - 1630 13

20 26

2

22995

02

0 132

22995

0 132

02

2163

0 262

0 132

2

( . ) ( . ) ( . ) . . .− − −+

− −+

−+

−�

��

�

��

B13 = - 00 13

20 26

2

23146

02

0 132

23146

0 132

02

20

0 262

0 132

2

( . ) ( . ) ( . ) . . .− − −−

− −+

−+

−�

��

�

��

B21 = - 1630 13

20 26

2

22995

02

0 132

22995

0 132

02

2163

0 262

0 132

2

( . ) ( . ) ( . ) . . .− − −+

− −+

−+

−�

��

�

��

B22 = - 130570 13

20 26

2

23925

02

0 132

23925

0 132

02

2467

0 262

0 132

2

( . ) ( . ) ( . ) . , .− − −+

− −+

−+

−�

��

�

��

B23 = - 00 13

20 26

2

23146

02

0 132

23146

0 132

02

20

0 262

0 132

2

( . ) ( . ) ( . ) . . .− − −−

− −+

−+

−�

��

�

��

B31 = - 00 13

20 26

2

23146

02

0 132

23146

0 132

02

20

0 262

0 132

2

( . ) ( . ) ( . ) . . .− − −−

− −+

−+

−�

��

�

��

B32 = - 00 13

20 26

2

23146

02

0 132

23146

0 132

02

20

0 262

0 132

2

( . ) ( . ) ( . ) . . .− − −−

− −+

−+

−�

��

�

��

B33 = - 4650 13

20 26

2

23297

02

0 132

23297

0 132

02

2465

0 262

0 132

2

( . ) ( . ) ( . ) . . .− − −+

− −+

−+

−�

��

�

��

hence

B11 = - 319B12 = 0B13 = - 53.2B21 = 0B22 = 319B23 = - 53.2B31 = - 53.2B32 = - 53.2B33 = 0




7/9

6th step: Expression of stiffness overall matrix

{e1}

N

N

N

M

M

M

x

y

xy

x

y

xy

=

A A A B B B

A A A B B B

A A A B B B

B B B C C C

B B B C C C

B B B C C C

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

ε

ε

γ

∂∂

∂∂∂∂ ∂

x

y

xy

o

o

o

wxwyw

x y

2

2

2

2

2

2

then

N

N

N

M

M

M

x

y

xy

x

y

xy

=

2779 821 0 319 0 53 2

821 2779 0 0 319 53 2

0 0 978 53 2 53 2 0

319 0 53 2 751 6 06 0

0 319 53 2 6 06 751 0

53 2 53 2 0 0 0 9 59

−

−

.

.

. .

. . .

. . .

. . .

ε

ε

γ

∂∂

∂∂∂∂ ∂

x

y

xy

o

o

o

wxwyw

x y

2

2

2

2

2

2




8/9

By reversing the relationship, we find:

xyM1Ex44.12Ex67.12Ex67.103Ex61.33Ex61.3

yxo

w2

2

yM2Ex67.12Ex57.33Ex23.73Ex33.23Ex0.53Ex00.2

2y

ow2

xM2Ex67.13Ex23.72Ex57.33Ex33.23Ex0.23Ex00.52x

ow2

xyN03Ex33.23Ex33.23Ex28.14Ex8.34Ex80.3xy

yN3Ex61.33Ex0.53Ex0.24Ex8.33Ex15.14Ex0.5y

xN3Ex61.33Ex99.13Ex00.54Ex80.34Ex0.53Ex15.1

x

−−−−−−∂∂

∂

−−−−−−−−∂

∂

−−−−−−−=∂

∂

−−−−−−γ

−−−−−−−−−−ε

−−−−−−−ε

7th step: Search for the strain tensor

By replacing loading by values quoted at the beginning of the example in the previousrelationship, we find:

ε

ε

γ

∂∂

∂∂∂∂ ∂

x

y

xy

o

o

o

wxwyw

x y

2

2

2

2

2

2

=

5 44 3 5440

174 3 1740

154 3 1540

2 38 2

4 57 3

2 05 2

1

1

1

. / ( )

. / ( )

. / ( )

.

.

.

E mm mm d

E mm mm d

E mm mm d

E mm

E mm

E mm

−

− − −

−

−

−

−

−

−

−

µ

µ

µ




9/9

8th step: Search for strains in the upper fibre at ε (0°)

To do this, membrane strains (εx, εy, γxy) are added to strains resulting from the bending

effect ∂∂

∂∂

∂∂ ∂

2

2

2

2

2

2 22

2wx

x h wy

x h wx y

x ho o o, ,�

��

�

��

{d7}

εx(0°) = εx - ∂∂

2

2wx

o x h2

εy(0°) = εy - ∂∂

2

2wy

o x h2

γxy(0°) = γxy - 2 ∂∂ ∂

2wx y

o x h2

hence:

εx (0°) = 5.44 E-3 + (-1) x 2.38 E-2 x 0.26 = - 748 µdεy (0°) = - 1.74 E-3 + (-1) x 4.57 E-3 x 0.26 = - 2928 µdγxy (0°) = - 1.54 E-3 + (-1) x 2.05 E-2 x 0.26 = - 3790 µd

For the lower fibre, we would find:

εx (90°) = 5.44 E-3 + 2.38 E-2 x 0.26 = 11628 µdεy (90°) = - 1.74 E-3 + 4.57 E-3 x 0.26 = - 552 µdγxy (90°) = + 1.54 E-3 + 2.05 E-2 x 0.26 = 6870 µd



MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSISReferences E




M. THOMAS, Analysis of a laminate plate subjected to membrane and bending loads,440.227/79


J. CHAIX, 436.127/91



F

MONOLITHIC PLATE - TRANSVERSAL SHEAR ANALYSIS



MONOLITHIC PLATE - TRANSVERSAL SHEARNotations F 1

1 . NOTATIONS


El: longitudinal elasticity modulus of unidirectional fibreEt: transversal elasticity modulus of unidirectional fibreνlt: longitudinal/transversal Poisson coefficientνtl: transversal/longitudinal Poisson coefficientGlt: shear modulus of unidirectional fibreep: ply thickness

Ek: longitudinal elasticity modulus with relation to x-axis of ply No. k

n: total number of laminate layers


El: laminate overall inertia with relation to the (moduli weighted) neutral axis

E Wk: static moment with relation to the (moduli weighted) neutral axis of the set of plies kto n

τ: shear stress

Txy, Tyz, T(β): shear load flux



MONOLITHIC PLATE - TRANSVERSAL SHEARIntroduction F 2

2 . INTRODUCTION

The purpose of this chapter is to determine interlaminar shear stresses in a monolithicplate subject to a shear load flux.

For simplification purposes, we shall assume that the laminate is made up of n identicalfibres but with different directions.

Layer No. k in direction θ has the following longitudinal elasticity modulus with relation tothe reference coordinate system (o, x, y):

f1 Ek = 11 2

4 42 2c

EsE

c sG El t lt

tl

t+ + −

�

��

�

��

ν see chapter C3.

with

c ≡ cos(θ)

s ≡ sin(θ)

k = n

k = 1ep

y

z

x

θ

Txz > 0

Tyz > 0z

y

x



MONOLITHIC PLATE - TRANSVERSAL SHEARIntroduction - Analysis method F 2

31/5

We shall assume that shear load Txz (direction z load shearing a plane perpendicular to x-axis) creates stress τxz and, based on the reciprocity principle, stress τzx.

Similarly, we shall assume that shear load Tyz (direction z load shearing a planeperpendicular to y-axis) creates stress τyz and, based on the reciprocity principle, stressτzy.

These shear stresses are called interlaminar stresses.

3 . ANALYSIS METHOD

To calculate interlaminar stresses τxz (τzx) generated by shear load Txz (Tyz), use thefollowing methodology.

We shall only consider the case of a laminate subject to shear load Txz. The analysisprinciple is the same for Tyz.

In this case, inertias (El) and static moments (E Wk) are measured with relation to y-axis.Elasticity moduli (Ek) are measured with relation to x-axis.

Txz

Tyz

x

y

z

τxz

τyz

τzy

τzx



MONOLITHIC PLATE - TRANSVERSAL SHEARAnalysis method F 3

2/5

1st step: The position of the laminate neutral axis is determined. If the laminate lower fibreis used as a reference, then the neutral axis is defined by dimension zg, so that:

f2 zg = ( )( )

E z z

E z z

k k kk

n

k k kk

n

21

21

112

−

−

−=

−=

�

�

2nd step: The (moduli weighted) bending stiffness of laminate El is determined with relationto the lay-up neutral axis

f3 El = ( ) ( )Ez z

E z zz z

zkk k

k

nkk

nk k

k kg

−+ −

+−

�

��

�

��

−

= = −−

� �1

3

1 1 11

2

12 2

zkzk - 1

ply No. n

ply No. 1

ply No. k

zg

y

z

z1

z0 = 0

zk zk - 1

ply No. k

zg

y

z




3/5

3rd step: Then the (elasticity moduli weighted) static moment E Wk (of the material surfacelocated above the line where interlaminar stress is to be calculated), is determined. Thisstatic moment shall be calculated with relation to the plate neutral axis.

If the line is a fibre interface surface (z = zk - 1), then we have the following relationship:

f4 E Wk = ( )E z zz z

zii k

ni i

i ig= −

−� −

+−

�

��

�

��1

1

2

If the line is situated at the centre of a fibre at z = z zk k+ − 1

2, the relationship becomes:

f5 E Wk = ( )E z zz z

zii k

ni i

i ig= −

−� −

+−

�

��

�

�� −1

1

2

Ez z

zz z z

zkk k

kk k k

g+

−�

��

�

��

++ −

�

��

�

��

−−

− −11

1 1

2 4 2

zk zk - 1

ply No. k

zg

y

z

zk zk - 1zg

y

z

ply No. kz z

k k+

− 1

2




4/5

4th step: Shear stress τxzk is determined, so that:

f6 τxzk = T E WE

xz k

l

.

where Txz is the shear load applied to the laminate.

By using this analysis method for each ply interface (or at the center of each ply forgreater accuracy), it is possible to plot the interlaminar shear stress diagram over theentire plate width.

The previous relationship shows that the shear stress is maximum when the staticmoment is maximum as well, i.e. at the neutral axis (z = zg).

Remark: The previous analysis is based on a shear load flux Txz applied to a sectionperpendicular to x-axis.

In the case of any section forming an angle β in the coordinate system (o, x, y), the shearload flux in this new section may be expressed as a function of Txz and Tyz.

τxz

zg

y

z

ply No. k

z

τxzk

τzxk




5/5

As shown in the drawing above, the z equilibrium of the hatched material element leads tothe following relationship:

T(β) ds - Txz ds cos(β) - Tyz ds sin(β) = 0

hence:

T(β) = cos(β) Txz + sin(β) Tyz

It is easy to show that for β = Arctg TT

yz

xz

�

��

�

�� , a modulus extremum T(β) (called main shear

load flux) is reached that is equal to:

f7 l T(β) l = Txz Tyz2 2+

Example: if shear load fluxes Txz and Tyz are equal, then the maximum shear load flux islocated in the plane with a direction β = 45°. Its modulus equals 2 Txy (or 2 Tyz).

y

x

ds

-Txz

T(β + π/2)

-Tyz

β

T(β)


© AEROSPATIALE - 1999 MTS

MONOLITHIC PLATE - TRANSVERSAL SHEARExample F 4

1/9

4 . EXAMPLE


0°: 1 ply45°: 1 ply135°: 1 ply90°: 1 ply

Stacking from the external surface being as follows: 0°/45°/135°/90°.


El = 13000 hb (130000 MPa)Et = 465 hb (4650 MPa)νlt = 0.35νtl = 0.0125Glt = 465 hb (4650 MPa)ep = 0.13 mme = 0.52 mm

The purpose of this example is to search for interlaminar shear stresses in the laminate,knowing that it is subject to the following shear load flux:

Txz = 0.7 daN/mm

B

0°

45°

135°

90°

z

xTxz = 0.

y

7 daN/mm

006 Iss. B




2/9

Knowing the mechanical properties of the unidirectional fibre, elasticity moduli of eachfibre should be calculated in direction x.

{f1}

For the fibre at 90°: k = 1.

E1 = Et = 465 hb (4650 MPa)

For the fibre at 135: k = 2

E2 = 10 70713000

0 707465

0 707 0 707 1465

2 0 012513000

4 42 2. . . . .+ + −�

��

��

E2 = 925 hb (9250 MPa)

For the fibre at 45°: k = 3

E3 = 925 hb (9250 MPa)

For the fibre at 0°: k = 4

E4 = El = 13000 hb (130000 MPa)

1st step: Analysis of the position of neutral axis zg

{f2}

zg = ))39.052.0(13000)26.039.0(925)13.026.0(925)013.0(465(2

)39.052.0(13000)26.039.0(925)13.026.0(925)013.0(465 22222222

−+−+−+−−+−+−+−

zg = 0.42 mm

B




3/9

2nd step: Analysis of the laminate bending stiffness El with relation to the neutral axis

{f3}

El = +−+−12

)13.026.0(92512

)013.0(46533

+−+−12

)39.052.0(1300012

)26.039.0(92533

+��

��

� −+−2

42.02

013.0)013.0(465

+��

��

� −+−2

42.02

13.026.0)13.026.0(925

+��

��

� −+−2

42.02

26.039.0)26.039.0(925

2

42.02

39.052.0)39.052.0(13000 ��

��

� −+−

El = 0.085134 + 0.169352 + 0.169352 + 2.380083 + 7.618211 + 6.087656 + 1.085256 + 2.07025

El = 19.67 daN.mm

z4 = 0.52

z3 = 0.39

z2 = 0.26

z1 = 0.13

z0 = 0

zg = 0.42

z

B




4/9

3rd step: Analysis of static moments E Wk (with relation to the neutral line) at the base andcenter of each ply.

At the top of ply at 0°

{f4}

E W4 = 0 daN

At the center of ply at 0°

{f5}

E W4 = −��

��

� −+− 42.02

39.052.0)39.052.0(13000

��

��

� −++��

��

� −+ 42.0239.0

439.052.039.0

239.052.013000

E W4 = 59.15 - 2.11

E W4 = 57.04 daN

0°45°135°90° y

z

0°

y

z

B




5/9

At the base of ply at 0°

{f4}

E W4 = ��

��

� −+− 42.02

39.052.0)39.052.0(13000

E W4 = 59.15 daN


{f5}

E W3 = ��

��

� −+−+��

��

� −+− 42.02

26.039.0)26.039.0(92542.02

39.052.0)39.052.0(13000

��

��

� −++��

��

� −+− 42.0226.0

226.039.026.0

226.039.0925

E W3 = 59.15 - 11.42 + 7.67

E W3 = 55.4 daNp

0°

y

z

45°

y

z

B




6/9


{f4}

E W3 = ��

��

��

��

� −+

−+−+

− 42.02

26.039.0)26.039.0(92542.02

39.052.0)39.052.0(13000

E W3 = 59.15 - 11.42

E W3 = 47.73 daN


{f5}

E W2 = +−+

−+−+

− ��

��

��

��

� 42.02

26.039.0)26.039.0(92542.02

39.052.0)39.052.0(13000

−��

��

� −+− 42.02

13.026.0)13.026.0(925

��

��

� −++��

��

� −+ 42.0213.0

413.026.013.0

213.026.0925

E W2 = 59.15 - 11.42 - 27.06 + 15.48

E W2 = 35.35 daN

45°

y

z

135°y

z

B




7/9


{f4}

E W2 = +−+

−+−+

− ��

��

��

��

� 42.02

26.039.0)26.039.0(92542.02

39.052.0)39.052.0(13000

��

��

� −+

− 42.02

13.026.0)13.026.0(925

E W2 = 59.15 - 11.42 - 27.06

E W2 = 19.87 daN


{f5}

E W1 = +−+

−+−+

− ��

��

��

��

� 42.02

26.039.0)26.039.0(92542.02

39.052.0)39.052.0(13000

−��

��

� −+−+��

��

� −+− 42.02

013.0)013.0(46542.02

13.026.0)13.026.0(925

��

��

� −++��

��

� −+ 42.020

4013.00

2013.0465

E W1 = 59.15 - 11.42 - 27.86 - 21.46 + 11.71

E W1 = 10.12 daN

135°y

z

90° y

z

B




8/9


{f4}

E W1 = 0 daN

90° y

z

B




9/9

4th step: calculation of maximum interlaminar shear stress

In the example given, it is located at the point where the static moment is maximum, i.e. atthe base of the ply at 0°. Its value equals at E W0 = 59.15 daN, which gives stress τxz0:

{f6}

τxz0 = 0 7 591519 67

. ..

x = 2.1 hb (21 MPa)

If these interlaminar shear stresses are analysed for each fibre, stresses are distributedalong the laminate thickness as follows:

τxzk = 0 719 67

..E Wk

B

0.065

0

0.13

0.195

0.26

0.325

0.39

0.455

0.52

z (mm)

0.5 1 1.5 2 2.5

ττττ (hb)

0



MONOLITHIC PLATE - TRANSVERSAL SHEARReferences F






G

MONOLITHIC PLATE - FAILURE CRITERIA



FAILURE CRITERIANotations G 1

1 . NOTATIONS

σl (σl): longitudinal stress in unidirectional fibreσt (σ2): transversal stress in unidirectional fibreτlt (σ6): shear stress in unidirectional fibre

εl (εl): longitudinal strain in unidirectional fibreεt (ε2): transversal strain in unidirectional fibreγlt (ε6): shear strain in unidirectional fibre

Rl: allowable longitudinal stressRlt: allowable longitudinal tension stressRlc: allowable longitudinal compression stress

Rt: allowable transversal stressRtt: allowable transversal tension stressRtc: allowable transversal compression stress

S: allowable shear stress



FAILURE CRITERIAInventory G 2

2 . INVENTORY OF STATIC FAILURE CRITERIA

The purpose of this chapter is to describe various failure criteria of the unidirectional fibrewithin a laminate.

The following criteria shall be presented in chronological order (this is not an exhaustivelist):

- maximum stress criterion

- maximum strain criterion

- Norris and Mac Kinnon's criterion

- Puck's criterion

- Hill's criterion

- Norris's criterion

- Fischer's criterion

- Hoffman's criterion

- Tsaï - Wu's criterion

For three-dimensional criteria, we shall assume that the composite material is subjected tothe following stress tensor and strain tensor:

(σ) = (σ1, σ2, σ3, σ4, σ5, σ6)

(ε) = (ε1, ε2, ε3, ε4, ε5, ε6)

For two-dimensional criteria, we shall assume that the unidirectional fibre is subjected tothe following stress tensor and strain tensor:

(σlt) = (σl, σt, τlt)

(εlt) = (εl, εt, γlt)



FAILURE CRITERIAMaximum stress criterion G 2.1

2.1 . Maximum stress criterion

This criterion is applicable for orthotropic materials only.

The criterion anticipates failure of the material if:

for 1 ≤ i ≤ 6

g1 σi = Xi for tension stressesorσi = - X'i for compression stresses

For the two-dimensional case, the failure envelope may be represented as follows:

σt

σl

τlt



FAILURE CRITERIAMaximum strain criterion G 2.2

2.2 . Maximum strain criterion

This criterion is applicable for orthotropic materials only.


for 1 ≤ i ≤ 6

g2 εi = Yi for tension strainsorεi = - Y'i for compression strains

For the two-dimensional case, the failure envelope may be represented as follows:

εt

εl

γlt



FAILURE CRITERIANorris and Mac Kinnon's criterion G 2.3

2.3 . Norris and Mac Kinnon's criterion

This criterion is valid for any material.


Ci i( )σ 21

61=�

Coefficients Ci depend on the material used.

For the two-dimensional case, the expression becomes:

g3 C1 (σl)2 + C2 (σt)2 + C6 (τlt)2 = 1

The failure envelope may be represented as follows:

This is the first criterion which calls for stress dependency.

σt

σl

τlt



FAILURE CRITERIAPuck's criterion G 2.4

2.4 . Puck's criterion

This two-dimensional criterion is valid for orthotropic materials only.


σ1 = X1 for tension stressesorσ1 = - X'1 for compression stresses

and

g4 σ σ τ1

1

22

2

212

6

2

1X X X

�

��

�

�� +

�

��

�

�� +

�

��

�

�� =

Coefficients X1, X2 and X6 depend on the material used.

Accuracy close to that of Norris and Mac Kinnon's criterion.

σt

σl

τlt



FAILURE CRITERIAHill's criterion G 2.5

2.5 . Hill's criterion

This criterion is valid for orthotropic materials or for slightly anisotropic materials only.


F (σ2 - σ3)2 + G (σ3 - σ1)2 + H (σ1 - σ2)2 + L (σ4)2 + M (σ5)2 + N (σ6)2 = 1

Coefficients F, G, H, L, M and N depend on the material used.

For a two-dimensional analysis, the expression becomes:

g5 F (σt)2 + G (σl)2 + H (σl - σt)2 + N (τlt)2 = 1

σt

σl

τlt



FAILURE CRITERIANorris's criterion G 2.6

2.6 . Norris's criterion



F (σ2 - σ3)2 + G (σ3 - σ1)2 + H (σ1 - σ2)2 + L (σ4)2 + M (σ5)2 + N (σ6)2 = 1

and for 1 ≤ i ≤ 6

σi = Xi for tension stressesorσi = - X'i for compression stresses

Coefficients F, G, H, L, M and N depend on the material used.


g6 F (σt)2 + G (σl)2 + H (σl - σt)2 + N (τlt)2 = 1

- X'1 ≤ σl ≤ X1 and - X'2 ≤ σt ≤ X2 and - X'6 ≤ τlt ≤ X6

σt

σl

τlt



FAILURE CRITERIAFischer's criterion G 2.7

2.7 . Fischer's criterion



g7 σ σ σ σ τl t l t lt

X XK

X X X1

2

2

2

1 2 6

2

1�

��

�

�� +

�

��

�

�� − +

�

��

�

�� =

Coefficients X1, X2 and X6 depend on the material used.

σt

σl

τlt



FAILURE CRITERIAHoffman's criterion FAILURE CRITERIA G 2.8

2.8 . Hoffman's criterion

This criterion is valid for orthotropic materials only.


C1 (σ2 - σ3)2 + C2 (σ3 - σ1)2 + C3 (σ1 - σ2)2 + C4 (σ4)2 + C6 (σ6)2 + C5 (σ5)2 + C'1 σ1 + C'2 σ2 +C'3 σ3 = 1

Coefficients C1, C2, C3, C4, C5, C6, C'1, C'2 and C'3 depend on the material used.


g8 C1 (σt)2 + C2 (σl)2 + C3 (σl - σt)2 + C6 (τlt)2 + C'1 σl + C'2 σt = 1

Very good tension accuracy, but very bad compression results.

σt

σl

τlt



FAILURE CRITERIATsaï - Wu's criterion G 2.9

2.9 . Tsaï - Wu's criterion

This criterion intends to be as general as possible and then, there is, a priori, no particularhypothesis.

This criterion anticipates failure of the material if:

For 1 ≤ i ≤ 6

Σ Fi σi + Σ Fij σi σj + Σ Fijk σi σj σk + … = 1

For a two-dimensional analysis, there is:

g9 F1 σl + F2 σt + F6 τlt + F11 (σl)2 + F22 (σt)2 + F66 (τlt)2 + 2 F12 σl σt + 2 F26 σt τlt + 2 F16 σl τlt = 1

Coefficient F1, F2, F6, F11, F22, F66, F12, F26 and F16 depend on the material used.



FAILURE CRITERIAAerospatiale's criterion G 3

1/2

3 . Failure criterion used at Aerospatiale: Hill's criterion

As seen previously, Hill's criterion is, in its general form, formulated as follows:

F (σt)2 + G (σl)2 + H (σl - σt)2 + N (τlt)2 = 1

This non-interactive criterion is applicable at the elementary ply only. There is a laminatefailure when the most highly loaded layer is broken.

If the expression is developed, we obtain:

(G + H) (σt)2 + (F + H) (σl)2 - 2 H σl σt + N (τlt)2 = 1

By definition, we shall assume that:

(G + H) = (1/Rl)2 where Rl is the longitudinal strength of the unidirectional fibre.

(F + H) = (1/Rt)2 where Rt is the transversal strength of the unidirectional fibre.

2 H = (1/Rl)2

N = (1/S)2 where S is the shear strength of the unidirectional fibre.

g10 There is a failure if h2 = σ σ τ σ σl

l

t

t

lt l t

lR R S R�

��

�

�� +

�

��

�

�� + �

��

�� −

�

��

�

�� =

2 2 2

2 1


© AEROS

FAILURE CRITERIAAerospatiale's criterion G 3

2/2

Thus, the following Reserve Factor is deduced:

g11 RF = 1 12 2 2

2

h

R R S Rl

l

t

t

lt l t

l

=�

��

�

�� +

�

��

�

�� + �

��

�� +

�

��

�

��

σ σ τ σ σ

This criterion is the one used by Aerospatiale. In order to avoid having a prematuretheoretical failure in the resin, the transversal modulus Et was considerably reduced (by acoefficient 2 approximately) with relation to the average values measured.

This "design" value is determined so that the transversal strain is greater than thelongitudinal one.

The allowable plane shear value S of the unidirectional fibre was determined for having, agood test/calculation correlation and significant tension and compression failures ofnotched or unnotched laminates.

B

PATIALE - 1999 MTS 006 Iss. B



FAILURE CRITERIAExample G 4

1/4

4 . EXAMPLE

Hill's criterion shall be applied to the example considered in the chapter "plain plate -membrane". Stresses applied to fibres are calculated and presented in the correspondingchapter (C.6) and quoted in the following pages.




El = 13000 hb (130000 MPa)Et = 465 hb (4650 MPa)νlt = 0.35Glt = 465 hb (4650 MPa)

Rlt = 120 hb (1200 MPa)Rlc = 100 hb (1000 MPa)Rtt = 5 hb (50 MPa)Rtc = 12 hb (120 MPa)S = 7.5 hb (75 MPa)

The laminate is globally subjected to the three following load fluxes in the referencecoordinate system (x, y) (see chapter C.6) :

Nx = 30.83 daN/mm

Ny = - 2.22 daN/mm

Nxy = 44.92 daN/mm




2/4

Reminder of stresses applied to the fibre with a 0° direction

σl = 29.42 hbσt = 0.06 hbτlt = 5.03 hb

{g10}

h2 = 1120

06.0x42.295.7

03.5506.0

12042.29

2

222

=��

��

�−��

��

�+��

��

�+��

��

�

h2 = 0.06 + 1.44 E-4 + 0.45 - 1.23 E-4 = 0.51

{g11}

Reserve Factor: R.F. = 1 10 51

142h

= =.

.

Margin = 100 (R.F. - 1) ≈ 40 %


σl = 80.17 hbσt = - 1.14 hbτlt = - 1.36 hb

{f10}

h2 = 8017120

11412

1367 5

8017 114120

2 2 2

2. . .

.. ( . )�

��

�� +

−��

�� +

−��

�� −

−��

��

x

h2 = 0.45 + 0.009 + 0.033 + 0.006 = 0.498




3/4

{g11}

Reserve Factor: R.F. = 10 498

142.

.=

Margin ≈ 42 %


σl = - 59.17 hbσt = 2.14 hbτlt = 1.36 hb

{g10}

h2 = −��

�� + �

��

�� + �

��

�� −

−��

��

5917100

2145

1367 5

5917 214100

2 2 2

2. . .

.. .x

h2 = 0.35 + 0.183 + 0.033 + 0.0126 = 0.579

{g11}


131.

.=

Margin ≈ 31 %




4/4


σl = - 8.42 hbσt = 0.95 hbτlt = - 5.03 hb

{g10}

h2 = −��

�� + �

��

�� +

−��

�� −

−��

��

8 42100

0 955

5 037 5

8 42 0 95100

2 2 2

2. . .

.. .x

h2 = 0.007 + 0.036 + 0.45 + 8 E-4 = 0.494

{g11}


142.

.=

Margin ≈ 42 %

Conclusion: the laminate overall margin is therefore 31 %



MONOLITHIC PLATE - FAILURE CRITERIAReferences G




Comparative analysis of composite material damaging criteria

BOUNIE, Failure criteria of mechanical bonds in composite materials, 1991, 440.180/91



H

MONOLITHIC PLATE - FATIGUE ANALYSIS



I

MONOLITHIC PLATE - DAMAGE TOLERANCE



MONOLITHIC PLATE - DAMAGE TOLERANCENotations I 1

1 . NOTATIONS

(o, x, y): panel reference frame

Nx: x-direction normal flowNy: y-direction normal flowNxy: shear flow

α i: orientation of fibre “i”

εli: longitudinal strain of fibre “i”εti: transverse strain fibre “i”γlti: angular slip of fibre “i”

εadm: permissible longitudinal strain of unidirectional fibreγadm: permissible slip of unidirectional fibre

σli: longitudinal stress of fibre “i”σti: transverse stress of fibre “i”τlti: shear stress of fibre “i”

Rl: permissible longitudinal stress of unidirectional fibreRt: permissible transverse stress of unidirectional fibreS: permissible shear stress of unidirectional stress

κR: reduction coefficient for permissible longitudinal stressκS: reduction coefficient for permissible shear stress



MONOLITHIC PLATE - DAMAGE TOLERANCEIntroduction I 2

2 . INTRODUCTION

The regulatory requirements in terms of structural justification concern, on the one hand,the static strength JAR § 25.305 and, on the other hand, fatigue + damage tolerance JAR§ 25.571. For the latter, three cases are to be considered:

- § 25.571 (b) Damage tolerance- § 25.571 (c) Safe-life evaluation* § 25.571 (d) Discrete Source

For the static strength evaluation, Acceptable Means of Compliance ACJ 25.603 § 5.8requests resistance to ultimate loads with "realistic" impact damage susceptible to beproduced in production and in service. This damage must be at the limit of thedetectability threshold defined by the selected inspection procedure. Also, static strengthmust be demonstrated after application of mechanical fatigue (§ 5.2) and test specimensmust have minimum quality level, that is, containing the permissible manufacturing flaws(§ 5.5) and "realistic" impact damage.

The static strength range is defined therefore for a detection threshold and by a "realistic"cut-off energy leading to "realistic" impacts.

The damage tolerance range is outside the static range.

Low thickness

High thickness

Detection threshold(impact depth in mm)

Static strengthrange

Damage Tolerance RangeDamage atdetectability

threshold limit

Static cut-offenergy

Damage-tolerancecut-off energy

Impact energy



MONOLITHIC PLATE - DAMAGE TOLERANCEIntroduction I 2

33.1

Distinction is made between the range above the detectability threshold where all damagewill be detectable and the range above the static cut-off energy and below the detectabilitythreshold where the damage will never be detected.

In this "Damage tolerance" section, we shall discuss both manufacturing defects andimpact damage for the static justification and the fatigue-damage tolerance justification.

The basic assumption to be retained is the fatigue damage no-growth concept.

3 . DAMAGE SOURCES AND CLASSIFICATION

Distinction is made between damage which may occur during manufacture and that whichoccurs in service.

3.1 . Manufacturing damage of flaws

Manufacturing damage or flaws include porosities, microcracks and delaminationsresulting from anomalies, during the manufacturing process and also edge cuts, unwantedrouting, surface scratches, surface folds, damage attachment holes and impact damage(see § 3.2.3).

Damage, outside of the curing process, can occur a detail part or component level duringthe assembly phases or during transport or on flight line before delivery to the customer.

If manufacturing damage/flaws are beyond permissible limits, they must be detected byroutine quality inspections.

For all composite parts, the acceptance/scrapping criteria must be defined by the DesignOffice. Acceptable damage/flaws are incorporated into the ultimate load justification byanalysis and into the test specimens to demonstrate the tolerance of the structure to thisdamage throughout the life of the aircraft.



MONOLITHIC PLATE - DAMAGE TOLERANCEFatigue damage I 3.2

3.2.1

3.2 . In-service damage

This damage occurs in service in a random manner. Distinction is made between threetypes of damage:

- fatigue,- corrosion and environmental effects,- accidental.

3.2.1 . Fatigue damage

Composite materials are said to be insensitive to fatigue; more exactly, their mechanicalproperties are such that the static dimensioning requirements naturally cover the fatiguedimensioning requirements. This is valid for a laminate submitted to plane loads, less than60 % of ultimate load. However, complex areas or areas with a sudden variation in rigiditymay favour the appearance of delaminations under triaxial loads.

Today, it is very difficult to (analytically or numerically) model the growth rate of a possibleflaw. This is why a "safe-life" justification philosophy has been adopted. It is based on twoprinciples which must be underpinned by experimental results:

- non-creation of fatigue damage (endurance),- no-growth of damage of tolerable size.

On account of the dispersion proper to composites and the form of the "Wohler" curvesassociated with them (relatively flat curve with low gradient), the factor 5 normally used onmetallic structures for the number of lives to be simulated during a fatigue test, wasreplaced by a load factor.

All these points will be discussed in detail in section O (MONOLITHIC PLATE -FATIGUE).



MONOLITHIC PLATE - DAMAGE TOLERANCECorrosion damage - Environmental effects I 3.2.2

3.2.2 . Corrosion damage and environmental effects

a) Corrosion

Composites are insensitive to corrosion. Nevertheless, their association with certainmetallic materials can cause galvanic coupling liable to damage certain metal alloys.

For information purposes, the table below shows various carbon/metal pairs over a scaleranging from A to E.

We consider that type A and B couplings are correct and that those of types C, D and Eare not.

A Anodised titanium, protected titanium fasteners

A Titanium and gold, platinium and rhodium alloys

B Chromiums, chrome-plated parts

B Passivated austenitic stainless steels

B Monel, inconel

B Martensitic stainless steels

C Ordinary steels, low alloys steels, cast irons

D Anodic or chemically oxidised aluminium and light alloys

D Cadmium and cadmium-plated parts

D Aluminium and aluminium-magnesium alloys

D Aluminium-copper and aluminium-zinc alloys

b) Environmental effects

At high temperatures, aggressions by hydraulic fluids may cause damage such asseparation, delamination, translaminar cracks, etc.

Rain can cause damage by erosion, etc.

All these points will be discussed in detail in section W (INFLUENCE OF THEENVIRONMENT).

Cou

plin

g w

ith c

arbo

nco

rrect

Cou

plin

g w

ith c

arbo

n to

be a

void

ed



MONOLITHIC PLATE - DAMAGE TOLERANCEAccidental damage - Inspection of damage I 3.2.3

4

3.2.3 . Accidental damage

This is the most important type of damage and the damage most liable to call intoquestion the structural strength of the part. It can occur during the manufacture of the item(drilling delamination) or in service (drop of a maintenance tool, hail or bird strikes).

4 . INSPECTION OF DAMAGE

One of the main preoccupations concerning the damage tolerance of composites isdamage detection. This is true both during manufacture and in service. In service, thedetectability threshold depends on the type of scheduled in-service inspection. There arefour types of inspections:

Inspection - Special detailed (ref: Maintenance Program Development: MPD):

An intensive examination of a specific location similar to the detailed inspection exceptfor the following differences. The examination requires some special technique such asnon-destructive test techniques, dye penetrant, high-powered magnification, etc., andmay required disassembly procedures.

This type of inspection is mainly conducted during production but can be usedexceptionally in service.

Inspection - Visual Detailed (ref: Maintenance Program Development: MPD):

An intensive visual examination of a specified detail, assembly, or installation. Itsearches for evidence of irregularity using adequate lighting and, where necessary,inspection aids such as mirrors, hand lens, etc. Surface cleaning and elaborate accessprocedures may be required.

This type of inspection enables BVID (Barely Visible Impact Damage) to be detected.



MONOLITHIC PLATE - DAMAGE TOLERANCEInspection of damage I 4

4.14.2

Inspection - General Visual (ref: Maintenance Program Development: MPD):

A visual examination that will detect obvious unsatisfactory conditions/discrepancies.This type of inspection may require removal of fillets, fairings, access panels/doors, etc.Workstands, ladders, etc. may be required to gain access.

Inspection - Walk Around Check (ref: Maintenance Review Board Document: MRB):

A visual check conducted from ground level to detect obvious discrepancies.

In general, the Walk Around check is considered as a general daily visual inspection.

4.1. Minimum damage detectable by a Special Detailed Inspection

These inspections are conducted with bulky facilities: ultrasonic, thermographic, X-rays,etc. Minimum detectable sizes are related to the size of the U.S. probes and the accuracyof the tools used, etc.

4.2 . Minimum damage detectable by a Detailed Visual Inspection

This type of damage is called BVID (Barely Visible Impact Damage). The geometricaldetectability criteria are as follows (cf. ref. 22S 002 10504):

Depth of flaw "δδδδ" Inside box structure(broken fibres) Outside box structure

Mean 0.1 mm 0.3 mm

"A" value 0.2 mm 0.5 mm



MONOLITHIC PLATE - DAMAGE TOLERANCEInspection of damage I 4.3

4.4

4.3 . Minimum damage detectable by a General Visual Inspection

This type of damage is called Minor VID (Minor Visible Impact Damage). The geometricaldetectability criteria are as follows (cf. ref. 22S 002 10504):

Depth of flaw "δδδδ" Size of perforation

2 mmor thickness of structure

if < 2 mm20 mm ∅

4.4 . Minimum damage detectable by a Walk Around Check

This type of damage is called Large VID (Large Visible Impact Damage). The geometricaldetectability criteria are not explicitly defined but the damage must be detectable withoutambiguities during a Walk Around Check.

We generally use a 50/60 mm ∅ perforation as criterion.

The diagram below summarises these four detectability levels according the size of thedamage.

Depth of indent δ = 0.3 mm δ = 2 mmdiameter 20 mm ∅ 50/60 mm ∅

In the remainder of this document, we will consider only visual inspections.

Specialdetailed

inspection

Detailedvisual

inspection

BVID

Generalvisual

inspection

MinorVID

Walk around

LargeVID

Size ofdamage



MONOLITHIC PLATE - DAMAGE TOLERANCEClassification of damage I 4.5

4.5 . Classification of accidental damage by detectability ranges

Depending on the type of visual inspection considered during the maintenance phases(general or detailed), we will define three clearly separate detectability ranges:

a) Damage undetectable by visual means used in service.

b) Damage susceptible to be detected during in-service inspections.

c) Damage "inevitably" detectable that can be placed into two categories:

- Readily detectable damage.- Obvious detectable damage.

These ranges are positioned as follows on the previously defined detectability scale:

→ For Detailed Visual Inspection:

→ For General Visual Inspection:

BVID

Undetectabledamage

DVI

LargeVID

WA

MinorVID

Damagesusceptible tobe detected

Inevitablydetectabledamage

BVID

Undetectabledamage

GVI

LargeVID

WA

MinorVID

Damagesusceptible tobe detected

Inevitablydetectabledamage



MONOLITHIC PLATE - DAMAGE TOLERANCEInfluence of damage - Porosity I

4.555.15.1.1

Remark: Note that certain authors define the BVID notion according to the type ofinspection selected.In this case, for a general inspection: MINOR VID ≡ BVIDIn our document, we will conserve the initial definition related to the visualdetailed inspection.

5 . EFFECT OF FLAWS/DAMAGE ON MECHANICAL CHARACTERISTICS

5.1 . Health flaws

5.1.1 . Porosity

→ Description

By "porosity", we mean a heterogeneity of the matrix leading, more often than not, to lackof inter- or intra-layer cohesion, generally small in size, but distributed uniformly or almostthroughout the complete thickness of the laminate. Note that for unidirectional tapes theporosities have a tendency to be located between the layers whereas, for fabrics, they aremore generally located where the weft and warp threads cross. The porosity ratio given isa surface porosity ratio measured by the ultrasonic attenuation method. The permissibleabsorption level is fixed at 12 dB irrespective of the thickness inspected (cf. note440.241/90 issue 2 - SIAM curve). All absorption areas above this limit will be consideredas a delamination and meet therefore the same criteria as a delamination.

However, only T300/N5208, more fluid than T300/BSL914 has a higher tendency to beporous.



MONOLITHIC PLATE - DAMAGE TOLERANCEPorosity I 5.1.1

→ Loss of mechanical characteristics due to porosity

The test results were interpolated, for the V10F wing, on T300/N5208 with variousporosity ratios distributed in all interply areas to determine the influence on the mechanicalcharacteristics for a 3 % ratio considered as the acceptable limit. This ratio combined withthe fatigue, ageing and residual test effects at 80° C, led to the following losses inmechanical characteristics:

T300/N52083 % porosity

Loss of characteristicsafter F + VC1 + 80° C

Loss of characteristicsafter F + VC1 + 80° C

BENDING - 15 % - 19 %

INTERLAMINAR SHEAR - 47 % - 33.5 %

COMPRESSION - 20 % - 19 %

TENSILE (high bearingstress) joint not supported - 20 % - 19 %

→ Example of porosity acceptance criteria

The 3 % acceptance criterion appears therefore as being non-conservative forinterlaminar shear. However, let us recall:

- that the spar boxes of the wings, movable surfaces or fin are subjected to very lowinterlaminar stresses,

- only T300/N5208 had porosities,

- that the 3 % porosity criterion distributed at all interply areas is today no longerapplied to primary structures. The permissible porosity ratio depends on the thicknessof the laminate.



MONOLITHIC PLATE - DAMAGE TOLERANCEDelaminations outside stiffener I 5.1.2

5.1.2.11/5

5.1.2 . Delaminations

A delamination is a lack of cohesion between the layers caused by a shear or transversetensile failure of the resin or, more simply, by forgetting a foreign body.

5 1.2.1 . Delaminations outside stiffener

�� Skin bottom areas

→ Description

A skin bottom delamination is a lack of cohesion between two well-defined plies. Naturaldelaminations appear during manufacture (surface contamination). A foreign body left inthe laminate (separator) will be considered as a delamination.

→ Loss of characteristics due to a delamination

For the V10F wing, a lack of interlayer cohesion up to 400 mm2 leads to a loss ofcompression strength of around 10 % for the two materials (T300/N5208 andT300/BSL914) tested in new condition at θ = 80° C. In aged/fatigue condition the drop instrength is 20 % for T300/N5208 and 13 % for T300/BSL914 in relation to the newstate/80° C reference. Fatigue leads to no growth of the flaw.



MONOLITHIC PLATE - DAMAGE TOLERANCEDelaminations outside stiffener I 5.1.2.1

2/5

�� Fastener areas

→ Description

As for the skin bottom delaminations, the lack of cohesion in these areas occurs betweentwo well-defined plies, sometimes at several levels but generally adjacent. These flawscome through to the bore. They are created during the drilling operations. The ultrasonicinspections conducted after each test case showed no evolution of existing flaws.

The parameter representing the size of the damage is the number given by: φ = ∅∅

fastenerdamage

The parameter representing the drop in characteristics is the number given by: ν = VcVb

where Vb represents the "B value" (see section Y) relevant to all tests characterising thematerial and where Vc is the calculation value used. Provided that the calculation value islower than the "B value", the integrity of the item is ensured. For safety reasons, we willimpose a minimum margin of 10 % between the calculated value and the "b value".

fastener Ø

damage Ø




3/5

Two cases can occur:

- if ν ≥ 1.1: no reduction will be made on the initial reserve factor RF,

- if ν < 1.1: after reduction, the new reserve factor is equal to RF’ = RF 1.1ν

The values of ν are given by the graphs in section Z for the prepreg epoxy carbon fibreT300/914.

Generally speaking, the graphs gives the values of ν for the flaw (delamination) but alsofor repairs which may be made on it (injection of resin, NAS cup). They enable you to findtherefore:

- whether the flaw is acceptable as such,- what type of repair is to be chosen.

→ Examples of acceptance and concession criteria

- in standard area, the delamination must be covered by a concession if its surfacearea is greater than:

S mm2 75 120 160 285 440 440Ø 3.2 4.1 4.8 6.35 7.92 9.52

These permissible delamination values are valid only for isolated delaminations.

For delaminated hole concentrations and irrespective of the size of the delaminations, theflaw must be covered by a concession if:

- for aligned fasteners, more than 20 % of the holes are delaminated and/or two flawsare less than 5 fastener pitches apart,

- for a delamination at a fastener or of another skin bottom area, they are less than120 mm apart.

- in designated area, permissible delamination is defined as follows:

S mm2 50 80 110 200 400 400Ø 3.2 4.1 4.8 6.35 7.92 9.52

These permissible delamination values are valid only for isolated delaminations.




4/5

For delaminated hole concentrations and irrespective of the size of the delaminations, theflaw must be covered by a concession if:

- for aligned fastener, more than 10 % of the holes are delaminated and/or two flawsare less than 5 fastener pitches apart,

- for a delamination at a fastener or of another skin bottom area, they are less than120 mm apart.

- for areas with several fastener rows:

• if the fasteners are on same row: same as above,

• if the flaws are located on several rows, they must be covered by a concessionif they are less than 175 mm apart.

→ Examples of repairs to be made

The table below summarises the repair solutions to be applied when delaminations are

detected at fastener holes in materials T300/914, G803/914 and HTA/EH25 depending on

the loads and the ∅

∅

fastener

damage ratio.

The choice of the solution is governed by the following rules:

- for a pure load, the repair or untreated delamination must resist ultimate loads underthe most severe environmental conditions,

- for a pure bearing stress test, the calculation value Vc is taken as reference. The

repair will not be acceptable if VcVb is lower than 1.

The validation range of the acceptable solutions given in the table below is ∅

∅

fastener

damage ≤ 6.




5/5

Load Condition Untreadeddelamination

Injection viavent hole

Normalinjection NAS cup

New Acceptable Unacceptable - -Pure tensile

Aged-wet Acceptable Unacceptable - Acceptable

NewAcceptable

4fastener

damage<

∅

∅ Acceptable - AcceptableBearingTensile

Aged-wetAcceptable

5.4fastener

damage<

∅

∅

Acceptable

5.4fastener

damage<

∅

∅ - Acceptable

New Unacceptable Unacceptable UnacceptableAcceptable

5fastener

damage<

∅

∅

Purecompression

Aged-wet Unacceptable

Acceptable

5.2fastener

damage<

∅

∅ AcceptableAcceptable

25.5fastener

damage<

∅

∅

NewAcceptable

75.4fastener

damage<

∅

∅

Acceptable

2fastener

damage<

∅

∅

Acceptable

25.5fastener

damage<

∅

∅

Acceptable

25.5fastener

damage<

∅

∅

Bearingcompression

Aged-wetAcceptable

4fastener

damage<

∅

∅ Unacceptable Unacceptable Acceptable

Withoutbending Acceptable Unacceptable - Acceptable

Bending1000 µd Acceptable Unacceptable - Acceptable in

"hollow"JOINT tensile

Bending2500 µd Acceptable Unacceptable - Unacceptable

Withoutbending Unacceptable Unacceptable - Acceptable

Bending1000 µd Unacceptable Unacceptable - Acceptable in

"hollow"JOINT

compressionBending2500 µd Unacceptable Acceptable - Acceptable



MONOLITHIC PLATE - DAMAGE TOLERANCEDelaminations in stiffener area I 5.1.2.2

1/5

5.1.2.2 . Delaminations in stiffener area of an integrally-stiffened panel

❏ Stiffener runouts

Stiffener runouts represent a critical point for dimensioning. When these stiffener runoutsare made during moulding without later machining operations, these fairly tortured areasmay include lacks of cohesion either in the base, or in the stiffener itself.

�� Crater

→ Description

This flaw is consecutive to too short a wedge which gives, after machining of the stiffenerrunout, a crater at the end of the stiffener.

Le

l

Baseplate U-section Wedge

Half core

U-section




2/5

→ Loss of characteristics due to crater

Size of flaw Test conducted ConditionsLoss of

characteristicsdue to flaw

TensileCompression

(stiffener runoutsnot protected)

Newθ = 20° C - 28 %ATR 72

T300/914L = 10 mml = 4 mme = 1 mm

TensileCompression

(stiffener runoutsprotected)

Newθ = 20° C 0 %

Compression(with reinforcement) - 4 %

ATR 72HTA/EH25 Compression

(withoutreinforcement)

Agedθ = 70° C

- 12 %

For unprotected stiffener runouts (that is, when it was impossible to thicken the skin tomake structure relatively simple to manufacture), this flaw must be covered by aconcession. When it is located at protected stiffener runouts (that is with a significant skinoverthickness at stiffener runout), this flaw will be covered by a concession only if its sizeis greater than the following values:

L = 10 mm l = 2 mm e = 0.5 mm

�� Punching

→ Description

This flaw is due to an imperfect Mosite cut leading to flaws at stiffener ends.




3/5

→ Loss of characteristics due to punching

Size of flaw Test conducted ConditionsLoss of


TensileCompression

(stiffener runoutsnot protected)

Newθ = 20° C - 20 %

ATR 72T300/914

L = 10 mme = 1 mm

TensileCompression

(stiffener runoutsprotected)

Newθ = 20° C 0 %

Must be covered by a concession when located at unprotected stiffener runouts. Whenlocated at protected stiffener runouts, it will be covered by a concession only if it size isgreater than the following values:

L = 10 mm l = 2 mm e = 0.5 mm

Le

l

e




4/5

�� Flaws "E", "B", "AB" and "BC"

→ Description

These flaws are located at various levels:

FLAW E FLAW B FLAW ABDelamination in radius

between U-sections andbase

Delamination under wedge Delamination at skin mid-thickness

Flaws BC correspond to one or more lacks of cohesion of stiffener wedge as shown ondiagram below:

C

BA

U-section

Wedge

Flaw BC

Half core




5/5

� Loss of characteristics due to flaw

Type of flaw Test conducted ConditionsLoss of


V10FT300/N5208

200 mm2

(flaw B)

Tensile(between wedgeand base skin)

Newθ = 20° C - 17 %

ATR 72T300/914(flaw BC)

Tensile(unprotected

stiffener runouts)

Wet ageingθ = 50° C - 20 %

ATR 72T300/914(flaw BC)

Compression(unprotected

stiffener runouts)

Wet ageingθ = 50° C 0 %

❏ Stiffener top

Lack of interlayer cohesion at top of stiffener between the U-section and the wedge doesnot seem to modify the mechanical characteristics.

❏ Stiffener base

Lack of interlayer cohesion in stiffener base hardly modifies the mechanicalcharacteristics. Within the scope of the V10F programme, the greatest drop is less than10 % in standard stiffener compression case with a type BC flaw.



MONOLITHIC PLATE - DAMAGE TOLERANCEDelamination in spar radii

Delamination on edge of spar flangesI 5.1.3

5.1.4

5.1.3 . Delamination in spar radii

This flaw correspond to lack of cohesion between two well-defined plies in the web/flangeblend-in radius.

The maximum permissible surface area for a flaw is 100 mm2.

� In standard areas: maximum local surface area between 2 ribs for a radius is 250mm2, including delaminations and foreign bodies.

� In designated areas: maximum local surface area between 2 ribs for a radius is150 mm2, including delaminations and foreign bodies.

5.1.4 . Delamination on spar flange edges

Delamination acceptable after repair is defined as follows :

- 1 delaminated interface only,- l ≤ 5 mm,- L ≤ 25 mm.

An acceptable flaw will however require a Hysol 9321 sealing operation on edge. Anyother flaws shall be covered by a concession.

l

L

Delamination



MONOLITHIC PLATE - DAMAGE TOLERANCEForeign bodies - Translaminar cracks I 5.1.5

5.1.6

5.1.5 . Foreign bodies

Same criteria as given for delaminations (cf. § 5.1.2.1).

5.1.6 . Translaminar cracks

Translaminar cracks have been detected on the ATR 72 outer wing spar box, the A340aileron, the 2000 fin, the A300/A310 (cf. note 494.048/91); however there are none on theflight V10F (cf. note 494.007/91).

These are elongated flaws due to the use of a corrosive stripper (MEK, Methyl EthylKetone). Currently, baltane is used. T300/914 and G803/914 have these flaws; the testsconducted on IM7/977-2 and HTA/EH25 showed no translaminar cracks (cf. note494.056/91).

These cracks are detected by ultrasonic inspection in the fastener areas (the back surfaceecho totally disappears). They concern all ply directions but do not touch between twoplies with different orientations. It is in the high crack density area that the ultrasonic signalis totally damped. There a transition zone between this area and the healthy part of thelaminate where crack density decreases and the ultrasonic back surface echo reappears.

These cracks are parallel to the fibres leaving the holes. They first affect the plies at 0°,then the plies at ± 45°. Some cracks are observed in the central plies at 90°. The axes ofthese crack networks correspond approximately to the hole diameters.

They do not lead to a drop in the mechanical characteristics (cf. note 437.115/91).

The existence of flaws at fasteners can be masked by high density translaminar cracks.Therefore, the threshold of the surface areas of the translaminar cracks which must beplotted is coherent with the size of acceptable delaminations.



MONOLITHIC PLATE - DAMAGE TOLERANCEDelaminations consecutive to a shock I 5.1.7

1/4

5.1.7 . Delaminations consecutive to a shock (during production and in service)

→ Description

An impact causes lack of interlayer cohesion at several levels depending on the energy ofthe impact.

→ Loss of characteristics due to a delamination

Generally speaking, a composite material with a non-through delamination is much moresensitive from a structural strength viewpoint to compression or shear loads (resin) than totensile loads (fibre).

The drops in characteristics within the scope of the V10F programme are:

- 18 % in tensile strength for a maximum invisible impact,

- 36 % in compression strength for a maximum invisible impact.

All points of the tests conducted on the V10F test specimens were plotted on the graphbelow (the points of the static and fatigue test specimens are combined on this curve as ithas been demonstrated that the ageing effect is not significant for damage tolerance).

The curve used at Aerospatiale for the new states/residual test at ambient temperatureand aged/fatigue states/residual test at ambient temperature is shown on the curve belowby comparison at static test specimen and fatigue test specimen points.

Damaged area

Impactor indent

Delaminated area




2/4

Behaviour to impact damage V10FStatic test specimen (CES)

Fatigue test specimen (CEF)

Delaminated surface area (mm2)

→ Ultimate strength of a delaminated laminate

The problem is generally posed as follows: we take a laminate consisting of a set of tapes(or fabrics) that we will assume to be made of the same material, each one of them havinga specific orientation in relation to the reference frame (o, x, y).

The laminate is submitted to shear forces (of membrane type) Nx, Ny and Nxy. In thepresence of a delamination (without ply failure) in surface area Sd, what is the strength ofthe plain composite plate?

Today, there are three methods for evaluating the residual strength in the presence of adelamination (established from experimental results) which call on the stresses and/orstrains of the unidirectional fibre and not those of the laminate considered as ahomogeneous plate. Each fibre direction must therefore be justified.

0 500 1000 1500 2000 2500- 1000

- 1500

- 2000

- 2500

- 3000

- 3500

- 4000

- 4500

- 5000

CES

CEF

i22 - (- 3108 µd)Rupture CES

COURBE ACTUELLE VALEURS DE CALCULEtat neuf/température ambiante ou

Etat vieilli/fatigue à 20° C/température ambiante

i32 - (- 2800 µd)Arrêt CEF

Allo

ngem

ent d

e co

mpr

essi

on (µ

d)




3/4

We will describe here these three methods in chronological order.

1st method:

This first method consists in calculating a failure criterion determined from the strains ofeach fibre in relation to their intrinsic frame (o, l, t).

By referring to the "plain plate - calculation method" section, it is possible to calculate thestrains in the various layers of the composite from the global flows Nx, Ny and Nxy appliedto the laminate and from the characteristics of the material used.

For layer "i" defined by its orientation α i, the strains of the fibre "i" in its own frame aredefined by the following strain tensor: (εli, εti, γlti).

We can define the following failure criterion C1 for each layer "i":

i1 C1 = 2

adm

lt2

adm

l ii

��

�

�

��

�

�

γγ

+��

�

��

�

�

εε

where εadm and γadm are the permissible strains (longitudinal and shear) of theunidirectional fibre (equivalent).

These values (obtained from the test results) depend on the material and the surface areaSd of the delamination considered and the types of loads.

They are given in section Z (sheets giving calculation values and coefficients used).

This criterion was used for the dimensioning of the ATR 72 wing panels (dossier22S00210460).

2nd method:

This second method consists in calculating a failure criterion C2 (Hill type criterion in whichthe permissible stresses are reduced by coefficients κR and κS) calculated from thestresses in each fibre in relation to their intrinsic frames (o, l, t).




4/4

By referring to the "plain plate - calculation method" section, it is possible to calculate thestrains in the various layers of the composite from the global flows Nx, Ny and Nxy appliedto the laminate and from the characteristics of the material used.

For layer "i" defined by its orientation α1, the stresses of the fibre "i" in its own frame aredefined by the following stress tensor: (σli, σti, τlti).

We can define the following failure criterion C2 for each layer "i":

i2 C2 = ( )2

lR

tl2

s

lt2

t

t2

lR

l

RSRRiiiii

κ

σσ−�

�

�

��

�

�

κτ

+��

�

��

�

� σ+�

�

�

��

�

�

κσ

where Rl, Rt and S are the permissible longitudinal, transverse and shear stresses of theunidirectional fibre respectively (equivalent) and where κR and κs are the reductioncoefficients for these permissible stresses.

These coefficients depend on the material used and the surface area of the delaminationconsidered and are determined from the test results.

They are given in section V (sheets giving calculation values and coefficients used).

This criterion was used for the sizing of the A330/340 inboard and outboard aileronpanels.

3rd method:

This method consists in calculating a failure criterion C3 (similar to the one of method 1)calculated from the strains of each fibre in relation to their intrinsic farmes (o, l, t).

For layer "i" defined by its orientation αi, the strains of the fibre "i" in its own frame aredefined by the following strain tensor: (εli, εti, γlti).

We can define the following failure criterion C3 for each layer "i" :

i3 C3 = ( )2

ab

tl2

adm

lt2

a

l iiii

ε

εε+�

�

�

��

�

�

γγ

+��

�

��

�

�

εε

where:

if 2 �εadm� ≤ �γadm�



MONOLITHIC PLATE - DAMAGE TOLERANCEDelaminations consecutive to a shock

Visual flaws - Sharp scratchesI 5.1.7

5.25.2.1

i4 εa =

( ) ( )2adm

2adm

22

31

γ−

ε

i5 εab =

( ) ( )2adm

2adm

62

31

γ−

ε

else

εa = εadm

εab = + ∞

The particularity of this method is that it takes into account (in a significant manner) theload transverse to the fibre.

Tests have shown that presence of a tensile force perpendicular to the fibre directioncompression increases the ultimate strength of the laminate.

Criterion C3 takes this phenomenon into account. Indeed, if εti is of tensile type and εli ofcompression type, the third term of the criterion C3 becomes negative and tends toincrease the reserve factor and therefore the margin (RF = 1/C3).

Today, it is recommended to use this third finer method based on a high number ofexperimental results.

5.2 . Visual flaws

5.2.1. Sharp scratches

→ Description

Sharp scratches are made by scalpels or cutting tools. Sharp scratches lead to drops intensile characteristics of around 15 %; for compression, we assume that there is no dropin characteristics.



MONOLITHIC PLATE - DAMAGE TOLERANCESharp scratches I 5.2.1

→ Examples of acceptance criteria for sharp scratches

A long anomaly is acceptable without concession within following limits:

� On the ATR 72 outer wing carbon box structure,

→ In standard areas: permissible scratches are defined as follows:- maximum length: 100 mm,- maximum depth: 1 ply irrespective of the thickness.

→ in designated areas: the acceptance criteria are as follows:- maximum length: 100 mm,- maximum depth: 1 ply irrespective of the thickness,- all scratches though to a hole, an hedge or stopping less than 5 mm away must be

covered by a concession.

Any scratch concentrations must be covered by a concessions if the flaws are less than20 mm apart.

� On A330/A340 inboard and outboard ailerons, if length of scratch is less than 100 mmand if its depth is less than 0.15 mm for tapes and 0.3 mm for fabrics, sealing withHysol 9321 will be performed.

� On A330/A340, A320, A319, A321 nose landing gear doors (carbon fabrics G803/914),

→ at fittings, the permissible scratches are defined as follows:- maximum length: 10 mm,- maximum depth: 1 ply irrespective of the thickness.

→ outside fittings: the acceptance criteria are as follows :- maximum length: 250 mm,- maximum depth: 1 ply irrespective of the thickness.



MONOLITHIC PLATE - DAMAGE TOLERANCEIndents - Scaling I 5.2.1

5.2.25.2.3 1/3

� On A330/A340, A320, A319, A321 main landing gear doors (carbon fabrics G803/914),

→ at fittings, the acceptance criteria as follows:

- maximum length: 10 mm,- maximum depth: 1 ply irrespective of the thickness.

→ outside fittings: the acceptance criteria area as follows :

- maximum length: 100 mm,- maximum depth: 1 ply irrespective of the thickness.

5.2.2 . Indents

"Indent" type flaws due, for instance, to abrasion of skin by a rototest are permissible if:

- surface area of indent is ≤ 20 mm2 (∅ 5),

- only the 1st ply is totally damaged, that is 2nd ply visible.

Any flaw concentrations must be covered by a concession if two indents are less than100 mm apart.

5.2.3 . Scaling

→ Description

By "scaling", we mean separation or removal of several fibres (locally) altering only thefirst surface ply on monolith edge or on outgoing side of drilled holes.

→ Examples of scaling acceptance criteria

� On ATR 72 outer wing carbon box structure,

→ in standard areas: the permissible scaling flaws are defined as follows:

Maximum surface area = 30 mm2

Maximum depth: 1 ply for th < 20 plies2 plies for th ≥ 20 plies



MONOLITHIC PLATE - DAMAGE TOLERANCEScaling I 5.2.3

2/3

For scaled hole concentrations, this flaw must be covered by a concession if, for alignedfasteners, more than 20 % of the holes are scaled and/or two flaws are less than5 fastener pitches apart.

→ in designated area: permissible scaling flaws are defined as follows:

Maximum surface area = 20 mm2,Maximum depth: 1 ply irrespective of the thickness.

For scaled hole concentrations, this flaw must be covered by a concession if:

- for aligned fasteners, more than 10 % of the holes are scaled and/or two flaws areless than 5 fastener pitches apart,

- for areas with several fasteners rows (e.g. piano area)

• for fasteners on same row: same as above,

• for flaws on several rows; must be covered by a concession if they are less than175 mm apart.

Flaw 1 Flaw 2

Flaw 1 Flaw 2

Flaw 1175 mm

Flaw 2

Flaw 3

For flaws 1 and 3:to be covered by a

concession

For flaws 1 and 2:if S1 and S2 ≤ permissiblesurface area permissible



MONOLITHIC PLATE - DAMAGE TOLERANCEScaling I 5.2.3

3/3

All scaled areas will be sealed with Hysol 9321 to restore flat surface and avoid scalingdeveloping during later operations.

� On A330/A340 inboard and outboard ailerons, scaling on 1 ply of skin will be sealedwith Hysol 9321. Permissible scaling flaws are defined as follows:

→ panels (delaminations at fasteners)


For flaw concentrations at fasteners, two flaws on same row must be separated by 9fasteners.

Areas with several fastener rows:

- on same row: see above,- between different fastener rows

minimum distance = 175 mm

→ panels (leading edge joints), ribs, spar


Maximum depth: 0.2 mm

For flaw concentrations, 5 flaws maximum on 10 consecutive fasteners.

→ panels (other areas) (scaling at fasteners)


Maximum depth: 0.2 mm

For flaw concentrations, two flaws on a given row must be separated by 9 fasteners.



MONOLITHIC PLATE - DAMAGE TOLERANCESteps I 5.2.4

1/2

5.2.4 . Steps

→ Description

This is a fold of one or more skin plies which may occur between two (spar support)blocks or on a sandwich skin during co-curing or in spar webs.

→ Examples of acceptance criteria

� On ATR 72 outer wing carbon box structure

→ On bearing surfaces (spar, rib passage)

- Standard areas: steps on spar and rib passage areas are acceptable within a limit of0.3 mm. This type of flaw will be compensated for by Filleralu over a width of 50 mmon either side of the step.

- Designated areas: this flaw must be covered by a concession irrespective of itsgeometry.

→ On stiffeners

- standard areas: steps on stiffener flanges are acceptable within a height limit of0.3 mm provided that:

• there are no flaws in stiffener radius,• two flaws are at least 400 mm apart in Y-direction (wing frame),• two adjacent stiffeners are not affected in the same section,• an ultrasonic inspection demonstrates absence of "delamination" type flaws.

- designated areas: steps on stiffener flanges must be covered by a concession.

Filleralu50 mm



MONOLITHIC PLATE - DAMAGE TOLERANCESteps - Justification of permissible manufacturing flaws I 5.2.4

2/2

� On A330/A340 inboard and outboard ailerons under spar and rib bearing surfaces,steps lower than or equal to 0.2 mm and with a width lower than or equal to 3 mm willbe accepted, but:

- they must never be trimmed,- they will be compensated for by Filleralu,- in other areas, acceptable height is 0.4 mm.

� On A330 Pratt et Whitney thrust reverser sandwich skins mainly in areas with highcurvatures, steps with a height less than 0.5 mm are accepted in production. Stepsgreater than 0.5 mm will be examined case by case.

6 . JUSTIFICATION OF PERMISSIBLE MANUFACTURING FLAWS



MONOLITHIC PLATE - DAMAGE TOLERANCEJustification of in-service damage I 7

7.17.1.1

7 . JUSTIFICATION OF IN-SERVICE DAMAGE

7.1. Justification philosophy

A justification philosophy in agreement with European regulations (JAR) is associated witheach damage detectability range § 4.5 (undetectable damage; damage susceptible to bedetected [during inspection]; readily and obvious detectable damage).

7.1.1. Justification philosophy for undetectable damage

ACJ 25.603 § 5.1 :The static strength of the composite design should be demonstrated through aprogramme of component ultimate load tests in the appropriate environment, unlessexperience with similar design, material systems and loadings is available todemonstrate the adequacy of the analysis supported by subcomponent tests, orcomponent tests to agreed lower levels.

ACJ 25.603 § 5.2 :The effect of repeated loading and environmental exposure which may result in materialproperty degradation should be addressed in the static strength evaluation…

ACJ 25.603 § 5.5 :The static test articles should be fabricated and assembled in accordance withproduction specifications and processes so that the test articles are representative ofproduction structure.

ACJ 25.603 § 5.8 :It should be shown that impact damage that can be realistically expected frommanufacturing and service, but not more than established threshold of detectability forthe selected inspection procedure, will not reduce the structural strength below ultimateload capability. This can be shown by analysis supported by test evidence, or by test atthe coupon, element or subcomponent level.



MONOLITHIC PLATE - DAMAGE TOLERANCEJustification of in-service damage I 7.1.2

7.1.31/3

Undetectable damage, whether due to accidental impacts (in-service damageundetectable by a detailed visual inspection and therefore corresponding to BVID) ormanufacturing flaws must be covered by a static justification at ultimate load under themost severe environmental conditions (humidity and temperature) and at end of aircraftlife. During the certification tests, this damage will be introduced into minimum marginareas

7.1.2 . Justification philosophy for readily and obvious detectable damage

As laid down in the regulations, any damage which cannot withstand the limit loads mustbe readily detectable during any general visual inspection (50 flights) or obvious.

� Damage readily detectable within an interval of 50 flights must withstand 0.85 LL.� Obvious damage (engine burst) which occurs in flight with crew being aware of it must

withstand 0.7 LL (get-home loads capability).

7.1.3 . Justification philosophy for damage susceptible to be detected duringscheduled in-service inspections

� Regulatory aspects

ACJ 25.603 § 6.2.1 :Structural details, elements, and subcomponents of critical structural areas should betested under repeated loads to define the sensitivity of the structure to damage growth.This testing can form the basis for validating a no-growth approach to the damagetolerance requirements…

ACJ 25.603 § 6.2.3 :...The evaluation should demonstrate that the residual strength of the structure is equalto or greater than the strength required for the design loads (considered as ultimate)...

ACJ 25.603 § 6.2.4 :...For the case of no-growth design concept, inspection intervals should be establishedas part of the maintenance programme. In selecting such intervals the residual strengthlevel associated with the assumed damage should be considered.




2/3

� General

For metallic structures, the two fundamental damage tolerance parameters are theinitiation of the damage and its growth before detection. Many tests have been conductedtherefore to evaluate the growth speed of the damage and the time required to reach itscritical size and therefore its residual strength (limit load).

The critical loading mode is mainly tensile loading.

In contrast, impact damage to the composite structure of perforation/delamination typecause, when it occurs, a very substantial drop in the mechanical strength but it does notgrow under the fatigue load levels on civil aircraft.

The critical loading mode is mainly compression (and shear) loading

εresidual

εL.L.

εU.L.

METALLIC

GrowthTime

Inspection intervals

Initiationthreshold

Repair

εresidual

εL.L.

εU.L.

COMPOSITE

Time

At time of impact




3/3

Several methods are used for demonstrating conformity with regulations:

a) Semi-probabilistic methods

If the no-growth concept of the flaw is demonstrated (by fatigue test), the size of thedamage no longer depends on an evolving phenomenon but on a random event(accidental).

For the damage range between BVID and VID, the aim of the (analytical) justification willbe to determine an inspection interval so that the probability (Re) of simultaneously havinga flaw and a load greater than its residual load will be a highly improbable event(probability per flight hour less than 10-9).

This probabilistic damage occurrence versus time aspect therefore replaces thedeterministic concept for metallic materials where the occurrence of a flaw depends eitheron fatigue initiation, or, for certain areas, on an accidental impact; the effect of the latterbeing a modification in the threshold.

The complete philosophy can be summarised by the curve below. It expresses the loadlevel to be demonstrated and the type of justification versus the damage rangeconsidered.The portion of the curve between the BVID and the VID depends on the results of theprobabilistic analysis.

These methods are used by AS and CEAT.

BVID VID OBVIOUSREADILY DETECTABLE

ULTIMATELOADS

0.7 εL.L.

0.85 εL.L.

εL.L.

εU.L.

εVID

εBVID

εresidual

Re = E - 9

Probabilistic analysisTOLDOM

≥ L.L.

≥ U.L.



MONOLITHIC PLATE - DAMAGE TOLERANCEAerospatiale semi-probabilistic method

Determining inspection intervalsI 7.1.3

7.1.3.17.1.3.1.1 1/6

b) deterministic method (Boeing)

This method is based on two analysis and test configurations:

- demonstrating positive margins at ultimate load with BVID,- demonstrating positive margins at limit load with extensive damage.

The non-growth aspect of the fatigue damage must be demonstrated.

7.1.3.1 . AEROSPATIALE semi-probabilistic method (cf. note 432.0162/96)

7.1.3.1.1 . Process for determining inspection intervals

As stated above and in § 4.5, certain damage is susceptible to be detected duringinspections which implies that the aircraft may possibly fly between two inspections withdamage in a structure the residual strength of which may be lower than the ultimate loads.

In order not to design composite structures less reliable than metallic ones, an inspectionprogramme has been defined so that the probability of simultaneously having a flaw and aload greater than its residual strength will be a highly improbable event (probability perflight hour less than 10-9).

In mathematical form, this requirement can be written:

probability of occurrence of an impact with given energy (Pat)x

probability of not detecting the resulting flaw (1- Pdat)x

probability of occurrence of loads greater than the residual strength of the damage(Prat) ≤ 10-9/fh

or again:

i6 Pat x (1 - Pdat) x Prat ≤ 10-9/fh

This condition involves several notions that we will specify in the following sections.



MONOLITHIC PLATE - DAMAGE TOLERANCEDetermining inspection intervals I 7.1.3.1.1

2/6

❑ Pdat: probability of detecting the damage.

We defined, in § 4.2 to 4.4, the visual detection criteria for "A" value and "B" valuedamage and the mean value for various types of in-service inspections. Knowing that the"A" values correspond to a detection probability of 99 %, the "B" values to a probability of90 % and the mean values to a probability of 50 %, we can deduce the curve below whichshows the probability of detection versus the depth of the indent and the type ofinspection.

❑ Pat: probability of occurrence of an impact with given energy.

Several sources of impacts can be considered (this list is not restrictive):

- projection of gravel,- removal of the item,- dropping of tools or removable items,- shock with maintenance vehicle.

Each impact source will be defined by its incident energy.

As for detection, we will define an impact source by a statistical distribution (in this case,the Log-normal distribution).

We will therefore speak of the impact probability (or, more precisely, the impact energyrange) that we will call (Pat) and which will be characterised by mean energy Em and astandard deviation (according to Sikorsky, the standard deviation σ has a constant valueequal to 0.217).

The probability of having an impact energy between E et E+2 Joules is equal to

�+

=2E

E

dEx)E(fPat

0,3

0,52

0 0.5

5

0.99 1

Depth of indent(mm)

General visual inspection: *Detailed visual inspection: **

*

**

PdatDetection probability




3/6

We also obtain 1dEx)E(f0

=�∞

The impact energy will generally be limited to 50 J (cut-off energy), except for THS root:140 J corresponding to the energy of a tool box failing from the top of the fin.

Now that the impact has been defined, we must find the relation between the incidentenergy (E), the size of the damage (Sd) and its indentation (f).

Generally, we have :

Sd = Ksd e

E

f = Kf 3.3

eE��

��

�

Test campaigns are however necessary to determine the coefficients Ksd and Kf whichdepends on the types of materials, their thickness and the item bearing conditions.

❑ Prat: probability of having a loading case greater than the residual strength of theimpacted laminate.

As we saw in paragraph § 5.1.7, the residual strength of a laminate with a delaminationdefined by its surface area Sd can be determined by the numbers C1, C2 and C3 that wewill call more generally C in the remainder of this section.

The need to have three variables to characterise the number C (εl, εt, γlt ou σl, σt, τlt)makes all theoretical exploitations of the item loads (or deformations) difficult.

Em 2J

Pat

E

f(E)




4/6

We will therefore define a number εresidual = C

admissibleε which represents the permissible

strain of damage of size Sd under a single compression load.

This residual deformation depends of course on the size of the damage Sd. The generalform of this relation can be represented by the following curve:

It is therefore possible to determine, for each point on the item studied, the probability ofoccurrence of the load leading to the failure of the laminate with a delamination of size Sd

Knowing that the following gust occurrence probabilities are generally admitted:

- 2 x 10-5 for limit loads,- 1 x 10-9 for ultimate loads,

We can plot the curve below associating a probability of occurrence Prat with all residualstrength levels (εresidual = k x εL.L.) such that:

i7 Prat = 10- 8.6 k + 3.9

Sd

εnominalεresidual

Prat

εresidualεL.L. εU.L.

10-9

2 x 10-5




5/6

This curve will in fact be compared to a log-normal type occurrence law (or a firstapproximation linear law) for a larger deformation range.

PROBABILITY DETERMINATION LOGIC DIAGRAM

- to have an impact in a given energy range,- to detect damage,- to encounter a load greater than the residual strength of the laminate.




6/6

The inspection interval must be such that risk of failure in the interval:

Pat x (1 - Pdat) x Prat ≤ 10-9/flight hours

f(E)

Emmean

Impact energyrange

PatEnergy

Known impactsource

f

Energy

Depth ofindent

Kf3.3

e

E��

��

�

f

PdatPdat 1

General visual inspection: *Detailed visual inspection: **

*

**

Sd

Energy

Sd

εresidual

Delaminatedsurface

Ksde

E

Prat

εresidualεL.L. εU.L.

10-9

2 x 10-5

Prat

Detection probability



MONOLITHIC PLATE - DAMAGE TOLERANCEDetermining and calculating inspection intervals I 7.1.3.1.1

7.1.3.1.21/4

This curve, like all statistical distribution curves, is characterised by a mean value and astandard deviation. A simple calculation enables us to obtain the following expressions:

εmean = 10(Log (εU.L.) - 0.5554)

σ = 0.0928

To sum up, it is clear that by choosing a given impact energy range, the values of Pat, f,Pdat, Sd, εresidual and Prat are implicitly determined.

The drawing above shows the links between these various quantities.

7.1.3.1.2 . Inspection interval calculation software

The calculation tool is based on the fundamental principle described above: all damagesusceptible to be detected during an inspection must have a probability of encountering aload greater than its residual strength lower than 10-9 per flight hour (maximum value atend of aircraft life or before last inspection).

This principle involves three probabilities:

❑ Pat: probability of occurrence of an impact with a given energy.

❑ Pdat: probability of detecting the damage.

❑ Prat: probability of occurrence of a loading case greater than the residual strength ofthe impacted laminate.

10-92 x 10-5

0.5

1

εmean εL.L. εU.L.

Log-normal law

Prat

Linear law

εresidual



MONOLITHIC PLATE - DAMAGE TOLERANCECalculating inspection intervals I 7.1.3.1.2

2/4

This principle can be stated in a more useable form:

The probability of having damage susceptible to encounter a load greater than its residualstrength is equivalent to the sum of the probabilities of having:

- damage relevant to an incident energy between 0 and 2 J susceptible to encounter aload greater than its residual strength

and- damage relevant to an incident energy between 2 and 4 J susceptible to encounter a

load greater than its residual strengthand

- damage relevant to an incident energy between 48 and 50 J susceptible to encountera load greater than its residual strength.

By discretizing the incident energy and therefore the type of the damage, each flaw rangecan be dealt with independently of the others.

We can therefore apply the fundamental principle to each energy interval then add theresults.

First of all we will consider an incident energy range between E and E+2 Joules.

The trickiest bit is to determine the probability of existence of damage of a well-definedsize versus time knowing that its probability of occurrence is equal to Pat (per flight hour)and its probability of non-detection during inspections is equal to (1 - Pdat).

E

f(E)




3/4

If Pat is the probability of occurrence of the flaw per flight hour at time t1 (before firstinspection for instance) the probability of existence of the flaw is equal to: 1 - (1 - Pat)t1

.

After the first inspection, the probability of occurrence of the flaw is therefore reduced to:[1 - (1 - Pat)t1] (1 - Pdat) then increases according to same curve as before but with a timeshift as initial probability is no longer zero. We repeat this operation up until the lastinspection.

The form of the function makes the calculations difficult; it is for this reason that wecompare the curve to its tangent: 1 - (1 - Pat)t ≈ t x Pat. This approximation remains validas long as the term t x Pat is small in comparison with 1.

This therefore gives the following configuration:

tIT1 t1 t2IT2 t3IT3 IT4 t4

1

Probability ofoccurrence

of a flaw

1 - (1 - Pat) ̂t1[1 - (1 - Pat) ̂t1] (1 - Pdat)

The curve [1 - (1 - Pat) ^ t] will be comparedto its limited development: t x Pat

t

1

Probability ofoccurrence

of a flaw

ITITITIT

IT x Pat

IT x Pat x (1 - Pdat) ̂2 IT x Pat x (1 - Pdat) ̂3

IT x Pat IT x PatIT x Pat

ERL = n x IT

IT x Pat x (1 - Pdat)




4/4

For constant inspection intervals, the mean probability of occurrence of the flaw is equalto:

i1n

1i)Pdat1(xPatxITx

nin

2PatxIT −−+ �

−

=

The maximum probability of occurrence of the flaw (Rd) is equal to:

i8 Rd = IT x Pat +�−

=

−1n

1i

i)Pdat1(xPatxIT

The mean probability of failure (Rr) of the flaw is therefore equal to:

��

��

��

��

−−+ �−

=

i1n

1iPdat)1(xPatxITx

nin

2PatxITxratP

The maximum probability of failure (Rr) of the flaw is therefore equal to:

i9��

��

��

��

−+= �−

=

i1n

1iPdat)1(xPatxITPatxITxratPRr

To find the mean overall risk per flight hour, all we need to do is to integrate this result intoall possible incident energy ranges.

��

��

��

��

−−+ ��−

=

=

=

i1n

1i

J50E

J0EPdat)1(xPatxITx

nin

2PatxITxratP

The overall maximum risk per flight hour (Re) is equal to:

i10��

��

��

��

−+= ��−

=

=

=

i1n

1i

J50E

J0EPdat)1(xPatxITPatxITxratPRe

This risk must be lower than 10 E-9.

The table below summarises (by giving the mathematical links between the variousvariables) the method used to determine Re.



MONOLITHIC PLATE - DAMAGE TOLERANCELoad level K I 7.1.3.1.2

7.1.3.1.31/5

E f Sd εεεεresidual Prat Pdat Pat Rd Rr

0 - 2 J2 - 4 J4 - 6 J

.

.

.

.

.

.

.

.

.

.44 - 46 J46 - 48 J48 - 50 J

i8

i9

7.1.3.1.3 . Load level K to be demonstrated in the presence of Large VID

The previous analysis can be substantiated by a static test with VID and a load level k.CL(1 ≤ k ≤ 1.5).

❑ First method:

This method consists in initially evaluating the reduction coefficient α on the permissiblestrengths of the material so that the final calculated risk Re is equal to 10-9 per flight hour(this determination can only be done by successive approximations).

This means that we can suppose that the damage tolerance behaviour of the material isdegraded in relation to that really used, that is a material whose strength (undercompression loading after impact) will be equal to a certain percentage, called α, of that ofthe real material.

IT & ERL

Re

i10




2/5

In this case, the (εresidual; Sd) is submitted to a homothety in relation to the x-axis.

The number α1 can therefore legitimately be compared to a reserve factor.

We will thus define a static test with VID (Visible Impact Damage) such that the margin inrelation to the residual strain ε (VID) of the flaw is the one defined above.

We obtain:

.L.LxK)VID(1

εε=

α

hence:

i11 K = α )VID(kx)VID(.L.L

α=ε

ε

value representing the load level K to be demonstrated with VID.

❑ Second method:

Another method would consist in directly considering the probability and load notions.

Sd

εresidual

Reduced curve→ Re = 10-9/fh

Basic curve→ Re

x α




3/5

It is clear that for a static test, we can consider that the probabilities of occurrence of theflaw (Pat) and the probabilities of detecting (Pat) and not detecting (1 - Pdat) the flaw areequal to 1 as we are sure that it is present in the item.

If we write the equivalence between the test and the maximum risk per flight hour from aprobabilistic viewpoint, we obtain: Re = Prat x PatVID x (1 - PdatVID) = Prat.

The method consists therefore in determining a fictive ultimate load level such that theprobability of the flaw residual load level is equal to Re.

The drawing below shows that we must randomly subject the curve (strain level ε; Prat) to

a homothety with a factor η1

so as to move point A to point B level. In this case, it appears

that the permissible load level of the VID has a probability of occurrence Re.

We see that this transformation also moves point A' to point B' which corresponds to thefictive ultimate load level that must be applied to the structure.

By zooming in onto the part of the graph which concerns us and imposing a logarithmicscale on the y-axis, we obtain the following representation:

Prat

εresidual

Re

10-9

2 x 10-5

εL.L.εU.L. fictive εU.L.

ε VID

Permissibledeformation of

VID

A'

AB

B'

/η

1




4/5

We obtain:

)VID(kx6.89.3)Re(Log

)B(axis)A(axis +−==η

We can deduce the fictive ultimate load level to be demonstrated in the presence of VID

i129.3)Re(Log

)VID(kx9.12K+−

=

Load level K must always be between 1 and 1.5.

The graph below represents the previous relation (the maximum risk Re per flight hour onthe x-axis and the load level K to be demonstrated on the y-axis). Each curve is relevantto a residual load level of the flaw K.

- Log(Prat)

- Log(Re)

9

4.7

1 K(VID) 1.5

9.3)Re(Log

)VID(Kx9.12

+− 6.8

9.3)Re(Log +−

.L.L

residualkε

ε=

B'

B

A'

A/η 8.6 x K - 3.9




5/5

90.9

1

1.1

1.2

1.3

1.4

1.5

1

2

3

4

5

6

7

8

9

LIMIT LOAD

ULTIMATE LOAD

K Safety factor to be applied to limit loads

K = 1.17 K = 1.7

Re = E-15

k = damage residualload level

k = 1.

10

k = 1.

1

k = 1.

k

1

= 1.

k

12

= 1.

13

k = 1.

k
= 1.
MT

14

k = 1.

k

S 00

15

= 1.

6 Iss.

k = 20

B

16 17 18 19 20 21

Risk of failure per flight hour in Log

Log(Re)



J

MONOLITHIC PLATE - BUCKLING



K

MONOLITHIC PLATE - HOLE WITHOUT FASTENERANALYSIS



MONOLITHIC PLATE - HOLE WITHOUT FASTENERNotations K 1

1 . NOTATIONS

(o, x, y): reference coordinate system of panel(o, 1, 2): orthotropic axis of laminate

φ: angle formed by loading with the orthotropic axisα: angular position of point to be calculated with the orthotropic coordinate system

Ex: longitudinal modulus of laminate in the reference coordinate systemEy: transversal modulus of laminate in the reference coordinate systemGxy: shear modulus of laminate in the reference coordinate systemνxy: Poisson coefficient of laminate in the reference coordinate system

E1: longitudinal modulus of laminate in the orthotropic coordinate systemE2: transversal modulus of laminate in the orthotropic coordinate systemG12: shear modulus of laminate in the orthotropic coordinate systemν12: Poisson's ratio of the laminate in the orthotropic coordinate system

σ x∞ : stress to infinity

σx (y): stress along y-axisσt (α): tangential stress around circular hole

K T∞ : hole coefficient for an infinite plate width

K TL : hole coefficient for a finite plate width

β: "finite plate width" coefficient

L: plate width∅ : hole diameterR: hole radius



MONOLITHIC PLATE - HOLE WITHOUT FASTENERIntroduction - Theory - First method K 2

3.11/3

2 . INTRODUCTION

The purpose of this chapter is to assess stresses at the edge of a hole without fastener onan axially loaded composite plate and to anticipate failure of a notched laminate.

3 . GENERAL THEORY

3.1 . First method (Withney and Nuismer)

From a theoretical point of view, the problem is formulated as follows: let an infinite platebe subjected to stress flux σ x

∞ and with the diameter hole: ∅ .

The method is valid only if the x-axis is the laminate orthotropic axis. What is the stress σx

(y) distribution along the y-axis?

∅ = 2R

x

σx (y = R)

σ x∞

y

σx (y)



MONOLITHIC PLATE - HOLE WITHOUT FASTENERTheory - First method K 3.1

2/3

First, the following number needs to be considered:

k1stressiniteinf

stressedgehole)Ry(Kx

xT =

σ=σ

= ∞∞

This coefficient expresses hole edge stress concentration for the case of an infinitely largeplate. This is the hole coefficient.

For a composite plate, this term may be formulated as a function of the mechanicalproperties of the laminate as follows:

k2 K EE

EGT

x

yxy

x

xy

∞ = + −�

��

�

�� +1 2 ν

Stress σx (y) evolution along the y-axis may be expressed as follows:

k3 σx (y) = ( )σ xT

Ry

Ry

K Ry

Ry

∞∞+

�

��

�

�� +

�

��

�

�� − −

�

��

�

�� −

�

��

�

��

�

�

��

�

�

��

�

�

��

�

�

��2

2 3 3 5 72 4 6 8

If y = R, then this function is reduced to expression k1.

If the material is near-isotropic, then:

k4 σx (R + do) ≈ σ x∞ 2 3

2

2 4+ +ξ ξ

k5 with: ξ = RR do+



MONOLITHIC PLATE - HOLE WITHOUT FASTENERTheory - First method K 3.1

3/3

If the plate is not infinitely large and has a length L, then:

k6 σx (y) = β ( )σ xT

Ry

Ry

K Ry

Ry

∞∞+

�

��

�

�� +

�

��

�

�� − −

�

��

�

�� −

�

��

�

��

�

�

��

�

�

��

�

�

��

�

�

��2

2 3 3 5 72 4 6 8

with:

k7 β = 2 1

3 1

3

+ − ∅��

��

− ∅��

��

L

L

as a first approximation

or

( )13 1

2 1

12

3 13

6 2

β=

−∅�

��

��

+ −∅�

��

��

+∅�

��

�� − −

∅��

��

�

��

�

��

∞L

L

ML

K MLT as a second approximation

in which: M2 =

1 83 1

2 11 1

2

3

2

−− ∅�

��

��

+ − ∅��

��

−

�

�

��

�

�

��

−

∅��

��

L

L

L

∅ = 2R

x

σx (y = R)

σ x∞

y

σx (y)

L



MONOLITHIC PLATE - HOLE WITHOUT FASTENERTheory - Second method K 3.2

1/3

3.2 . Second method (NASA)

This method is based on a NASA study.

For an infinite plate, it expresses the ratio K T∞ between the loading stress to infinity σ x

∞

and the tangential normal stress at the edge σt (α) around the hole.

The position of the point is defined by the angle α with relation to the orthotropic

It shall be assumed that loading is uniaxial.

∅ = 2R

x

σ x∞

y

σt (α)

2

1

α

φ




2/3

The first step consists in searching for the orthotropic axes (o, 1, 2) of the material. Anglesφ and α are thus determined (α being the angular coordinate of the point to be consideredwith relation to the orthotropic coordinate system).

The hole coefficient expression is the following:

k8

[ ]{K t EE

k n k n kTx

∞∞= = − + + + + − −σ α

σα φ φ α φ φ α( ) ( cos ( ) sin ) cos ( ) cos sin sin1

2 2 2 2 2 21

}n k n( ) sin cos sin cos1 + + φ φ α α

with

k9 k = EE

1

2

k10 EE E

EEG

α

α α ν α1 4 1

2

4 1

1212

2

114

2 2=

+ + −�

��

�

��sin cos sin

K11 n = 12

112

2

1

GE

EE2 +��

�

��

�ν−

where E1, E2, G12 and ν12 are the mechanical properties of the laminate in the orthotropiccoordinate system (o, 1, 2).




3/3

If the laminate lay-up is equilibrated, the expression is simplified and becomes:

{ }K EE

k nTt

x

∞∞= = − + +

σ ασ

α α α( )

cos ( ) sin1

2 21

If the laminate lay-up is nearly-isotropic, the expression is reduced to:

KTt

x

∞∞= = − +

σ ασ

α α( )

cos sin2 23

For a nearly-isotropic lay-up and uniaxial loading, hole coefficients for 0°, 45°, 135° and90° fibre directions are thus: 3, 1, 1 and - 1.

x

σ x∞

y

31

- 1K T∞

111

1



MONOLITHIC PLATE - HOLE WITHOUT FASTENERTheory - Third method K 3.3

3.3 . Third method (isotropic plate theory)

If the material is isotropic (or nearly-isotropic) and if the plate is infinitely large, then thestress tensor may be formulated for any point P (identified by its coordinates r and α) onthe plate as follows:

k12 σr = σ σαx xR

rRr

Rr

∞ ∞

−�

��

�

�� + + −

�

��

�

��

21

21 3 4 2

2

2

4

4

2

2 cos

σt = σ σαx xR

rRr

∞ ∞

+�

��

�

�� − +

�

��

�

��

21

21 3 2

2

2

4

4 cos

τrt = − − +�

��

�

��

∞σαx R

rRr2

1 3 2 24

4

2

2 sin

∅ = 2R

x

σ x∞

y

r

α

r

t

P



MONOLITHIC PLATE - HOLE WITHOUT FASTENERTheory - Fourth method K 3.4

1/2

3.4 . Fourth method (empirical)

This method is simple, fast but conservative. For more details, refer to chapter L(MONOLITHIC PLATE - FASTENER HOLE) by considering the bearing load as zero.

Let a plate of length L be subjected to stress triplet σ x∞ , σ y

∞ , and τ xy∞ .

The first step consists in calculating the principal stresses σ p∞ and σ p'

∞ and in weighting

them with the net cross-section coefficient LL − ∅

.

Thus, the main net stresses σ pN and σ p

N' are obtained.

Both stresses are then divided by coefficients Kt (K tc or K t

t for direction p) and K't (K' tc or

K' tt for direction p').

∅

x σ x∞

y

σ y∞

τ xy∞

L



MONOLITHIC PLATE - HOLE WITHOUT FASTENERTheory - Fourth method K 3.4

2/2

These coefficients (smaller than 1) are a function of the material, the elasticity moduli inthe direction considered (p or p'), the hole diameter (∅ ) and the type of load (tension "t" orcompression "c"). They are found in the form of graphs (for carbon T300/914 layer inparticular) in chapter Z (sheets 3 and 4 T300/914).

The two following final stresses are obtained :

σ pF =

σpN

tK

σ pF

' = σp

N

tK'

'

Both stresses are expressed in the main coordinate system (o, p, p').

∅

p

p'

σ pN

'

L

y

x

σ pN /Kt

σ pN

' /K't



MONOLITHIC PLATE - HOLE WITHOUT FASTENER"Point stress" - Failure criterion K 4.1

1/2

4 . ASSOCIATED FAILURE CRITERIA

4.1 . Failure criterion associated with the "point stress" method (Whitney andNuismer)

To determine the failure of a notched laminate, it is generally allowed (for compositematerials) to search for edge stresses at a certain distance do from the hole edge. Indeed,edge distance stress release through microdamages causes them to be analyzed at theedge distance do in practice. This distance depends on the type of load of the fibreconsidered (compression or tension), on the hole diameter and on the material (seechapter Z sheets 9 and 10 for T300/914).

At the composite material stress office of the Aerospatiale Design Office, one considers("point stress" method) that there is a failure in the laminate when the longitudinal stressof the most highly loaded fibre (located at the edge distance do) tangent to the hole isgreater than the longitudinal stress allowable for the fibre.

k13 There is a failure if: σl (y = R + do) > Rl

σl: longitudinal stress of the fibre tangent to the hole

Rl: longitudinal stress allowable for the fibre

fibre at 135°

fibre

at 9

0°

fibre

at 45

°

do

σlσl

fibre at 0°



MONOLITHIC PLATE - HOLE WITHOUT FASTENER"Point stress" - Failure criterion K 4.1

2/2

For complex loads, there is a software (PSH2 on mx4) which automatically models a finiteelement mesh and finds loads in fibres that are tangent to the hole.

Longitudinal stress analysis is performed in a circle of elements, its center of gravity beinglocated at the hole edge distance do.

2 do



MONOLITHIC PLATE - HOLE WITHOUT FASTENER"Average stress" - Failure criterion K 4.2

4.2 . Failure criterion associated with the "average stress" method (Whitney andNuismer)

This method consists in determining the average stress average σx average (ao) betweencoordinate points (0, R) and (0, R + ao). It is assumed that the plate is infinitely large andthe loading uniaxial.

Based on the previous theory (see K 3.1), the following may be formulated as:

σx average (ao) = 1a

y dyo

xR

R aoσ

+

� ( )

After development, we obtain:

k14 σx average (ao) ≈ σξ ξ

ξx∞ − −

−2

2 1

2 4

( )

k15 with: ξ = RR ao+

∅ = 2R

x

σ x∞

y

σx average (ao)

(ao)



MONOLITHIC PLATE - HOLE WITHOUT FASTENEREmpirical method - Failure criterion K 4.3

It is possible to choose ao so that:

σx average (ao) ≈ σx (R + do)

This condition allows the "point stress" and the "average stress" method to becomeequivalent.

The "average stress" method is rarely used at Aerospatiale, the same failure criterion asfor the "point stress" method may be applied: one considers that there is a failure in thelaminate when the longitudinal stress of the most highly loaded fibre tangent to the hole isgreater than the longitudinal stress allowable for the fibre.

4.3 . Failure criterion associated with the empirical method

After determining stresses σ pF and σ p

F' , a smooth calculation must be performed (see

chapter C) in order to assess longitudinal stresses in fibres tangent to the hole.

The Hill's failure criterion shall be used to each single ply (see chapter G3).

It may be noted that this method is relatively conservative because both coefficients Kt

and K't are assessed for different points, each one being the most critical with relation todirections p and p'.

On the other hand, coefficient Kt and K't values were determined only for diametersbetween ∅ 3.2 to ∅ 11.1. It is, therefore, necessary to use the theory for large diameters.

B



HOLE WITHOUT FASTENERFirst example K 5.1

1/4

5 . Example

5.1 . First example

Let a T300/BSL914 (new) square laminate plate of width L = 120 mm be laid up asfollows:


In the coordinate system (o, x, y), it is subjected to the following loading:

N x∞ = 10 daN/mm

N y∞ = 0 daN/mm

N xy∞ = 0 daN/mm

The plate has a diameter hole ∅ = 40 mm.

∅ = 40

x Nx = 10

y

L = 120

246

4

L = 120




2/4

Let's analyse, along the y-axis, the evolution of stress flux Nx (y).

The mechanical properties of the laminate in the reference coordinate system are thefollowing:

Ex = E1 = 6256 daN/mm2 (62560 MPa)Ey = E2 = 3410 daN/mm2 (34100 MPa)Gxy= Gv12 = 1882 daN/mm2 (18820 MPa)νxy = ν12 = 0.4191νyx = ν21 = 0.2285

The value of K T∞ is deduced as follows:

{k2}

K T∞ = 28.3

188262564191.0

3410625621 =+

��

�

�

��

�

�−+

This number represents the hole edge coefficient for the case of a plate of infinite width.

Since the plate does not have an infinite width L = 120 mm, we are led to calculate thefollowing number :

{k3}

β = 148.1

1204013

1204012

3

=��

��

� −

��

��

� −+




3/4

We thus obtain the evolution of normal stress fluxes along the y-axis:

Nx (y) = ��

�

�

��

�

�

��

�

�

��

�

�

��

��

�−��

�

��

�−−��

�

��

�+��

�

��

�+

8642

y207

y205)328.3(

y203

y202

210148.1

Nx (y) = ��

�

�

��

�

�

��

�

�

��

�

�

��

��

�−��

�

��

�−��

�

��

�+��

�

��

�+

8642

y207

y20528.0

y203

y20274.5

And we obtain, at the plate edge (y = 60) a flux of 12.32 daN/mm and at the hole edge(y = 20) a flux of 37.65 daN/mm.

0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60

Nx (y)

37.65

12.32

10

y




4/4

If one determines the flux at a hole edge distance do = 1 mm (see do in tension for theT300/914), one gets: Nx (y = 20 + 1) = 32.47 daN/mm.

A smooth plate calculation (chapter C) with this flux makes it possible to determine thelongitudinal stress of the most highly loaded fibre (fibre at 0°): σl = 32.41 hb.

On the other hand, as the allowable longitudinal tension stress of the same fibre is equalto Rl = 120 hb, based on the "point stress" failure criterion, we obtain:

Margin: ��

��

� −141.32

120 100 = 270 %

At a hole edge distance do = 1 mm (see tension do for fibre T300/914 in chapter Z), flux Nx

is now only 32.47 daN/mm.

A smooth plate calculation makes it possible to find that fibres with a 0° direction aresubjected to a 32.41 hb longitudinal stress at this particular hole edge distance.

The longitudinal tensile strength of fibre T300/914 being 120 hb, the targeted margin isthus:

Margin = 100 ��

��

� −141.32

120 = 270 %



HOLE WITHOUT FASTENERSecond example K 5.2

1/5

5.2 . Second example

Let a T300/BSL914 (infinitely large) laminate plate be laid up as follows:


In the coordinate system (o, x, y), it is subject to the following loading:

N x∞ = 2.8 daN/mm

N y∞ = - 7.8 daN/mm

N xy∞ = 5.3 daN/mm

The plate has a diameter hole ∅ = 40 mm.

246

4

∅ = 40

x

Nx = 2.8

y

Ny = - 7.8

Nxy = 5.3




2/5

Let's determine the normal stress fluxes of the hole edge at point P (fibre at 0° tangent tohole). To do this, we shall use the second method

First of all, (in order to eliminate the shear flux), let's be positioned in the main coordinatesystem (o, p, p') which forms a 22.5° angle with the reference coordinate system (o, x, y).Stress fluxes then become N p

∞ = 5 daN/mm, N p'∞ = - 10 daN/mm.

Orthotropic axes (o, 1, 2) are coincident with the reference coordinate system (o, x, y).

The plate and its loading may then be described as follows:

In the coordinate system (o, p, p'), the mechanical properties of the laminate are thefollowing:

Ep = 5800 daN/mm2 (58000 MPa)Ep' = 3749 daN/mm2 (37490 MPa)Gpp' = 1788 daN/mm2 (17880 MPa)νpp' = 0.3481νp'p = 0.225

∅ = 40

p

p'

Np' = - 10

y

x

φ' = 112.5°

4 246

2

Np = 5

1

P

φ= 22.5°

α = 90°




3/5

In the reference coordinate system (o, x, y) and in the orthotropic coordinate system (o, 1,2), the laminate properties are the following:

Ex = E1 = 6256 daN/mm2 (62560 MPa)Ey = E2 = 3410 daN/mm2 (34100 MPa)Gxy= G12 = 1882 daN/mm2 (18820 MPa)νxy = ν12 = 0.4191νyx = ν21 = 0.2285

A first step shall consist in calculating the effect of the main flux N p∞ at point P as follows:

We have:

{k9}

k = 354.134106256 =

{k10}

°��

��

� −+°+°=°

90x2sin4191.0x218826256

4190cos

3410625690sin

1E

E2441

90

EE90

1

11

1° = =

{k11}

n = 48.2188262564191.0

341062562 =+

��

�

�

��

�

�−




4/5

{k1}

K { +°°++°−=σ

°=ασ= ∞

∞ 90cos354.1)45sin)48.2354.1(5.22cos(62566256)90( 222

p

tT

°°++−°°−°+ 5.22cos5.22sin)48.2354.11(48.290sin)5.22sin354.15.22cos)48.21(( 222

}sin cos90 90° °

K 773.2)90(

p

tT =

σ°=ασ

= ∞∞

A second step shall consist in calculating the effect of the main flux N p'∞ at point P.

{k1}

K' { +°°++°−=σ

°=ασ= ∞

∞ 90cos354.1)45sin)48.2354.1(5.112cos(62566256)90( 222

'p

tT

°°++−°°−°+ 5.112cos5.112sin)48.2354.11(48.290sin)5.112sin354.15.112cos)48.21(( 222

}sin cos90 90° °

K' 646.0)90(

'p

tT −=

σ°=ασ

= ∞∞

p

p'

y

x

4 246

2

Np = 5

1

P2.773




5/5

The deduction is that the normal stress flux tangent to the hole crossing point P is equalto:

Nt (P) = 2.773 N p∞ + (- 0.646) N p'

∞ = 2.773 x 5 + (- 0.646) x (- 10) = 20.31 daN/mm

p

p'

Np' = - 10

y

x

4 246

2

1

P- 0.646



MONOLITHIC PLATE - HOLE WITHOUT FASTENERReferences K



VALLAT, Résistance des matériaux

S.C. TAN, Finite width correction factors for anisotropic plate containing a central opening,1988

J. Rocker, Composite material parts: Design methods at fastener holes 3 ≤ φ ≤ 100 mm.Extrapolation to damage tolerance evaluation, 1998, 581.0162/98

W.L. KO, Stress concentration around a small circular hole in a composite plate, 1985,NSA TM 86038

WHITNEY - NUISMER, Uniaxial failure of composite laminates containing stressconcentration, American Society for testing materials STP 593, 1975

ERICKSON - DURELLI, Stress distribution around a circular hole in square plate, loadeduniformly in the plane, on two opposite sides of the square, Journal of applied mechanics,vol. 48, 1981

B



L

MONOLITHIC PLATE - FASTENER HOLE



MONOLITHIC PLATE - FASTENER HOLENotations L 1

1/2

1 . NOTATIONS

(o, x, y): initial coordinate system(o, M, M'): coordinate system specific to the bearing load(o, P, P'): stress main coordinate system

F: bearing load∅ : fastener diameterSf: countersink surface of fastenere: actual thickness of laminatee*: thickness taken into account in bearing calculationsp: fastener pitch

σ tN : net cross-section stress at the hole

σm: bearing stressσR: allowable stress of material (general designation)

σxa: allowable normal stress of material in direction xσya: allowable normal stress of material in direction yτxya: allowable shear stress of materialτvisa: allowable shear stress of screw

N xB

N yB gross fluxes in panel

N xyB

N xN

N yN net cross-section fluxes

N xyN

N MN

N MN

' net cross-section fluxes in the coordinate system specific to the bearing loadN MM

N'

N Mm additional flux due to the bearing load

β: bearing load angle with relation to the initial coordinate system



MONOLITHIC PLATE - FASTENER HOLENotations L 1

2/2

N

N

PN

PN

'

net cross-section global fluxes in the main coordinate system

α: main coordinate system angle with relation to the bearing load

N xF

N yF corrected final fluxes

N xyF

K mc : compression bearing coefficient

K mt : tension bearing coefficient

Km : bearing coefficient in the broad meaning of the term

K tc : compression hole coefficient

K tt : tension hole coefficient

Kt: hole coefficient in the broad meaning of the term

Kf: bending hole coefficient



MONOLITHIC PLATE - FASTENER HOLEGeneral - Failure modes L 2.1

2.2

2 . GENERAL/FAILURE MODES

The purpose of this chapter is to assess the structural strength of a notched and loadedlaminate fitted with fastener.

Depending on the loading level and the type of geometry, such a system may fail as perseveral failure modes.

2.1 . Bearing failure

Fe∅

≥ σRm

2.2 . Net cross-section failure

Fb e( )− ∅

≥ σxa

e

∅ ∅F

∅ Fb



MONOLITHIC PLATE - FASTENER HOLEGeneral - Failure modes L 2.3

2.42.5

2.3 . Plane shear failure

FL e2 0 35( , )− ∅

≥ τxya

2.4 . Cleavage failure

F

L e− ∅��

��

2

≥ σya

2.5 . Cleavage: net cross-section failure

σxa (b - ∅ ) + τxya L ≤ 2Fe

45°

e

F

L

∅

e

F∅b

L

∅

e

F∅

L



MONOLITHIC PLATE - FASTENER HOLEFailure modes - Pitch definition L 2.6

3.1

2.6 . Fastener shear failure

42

Fπ ∅

≥ τvisa

where:

σxa is the allowable normal stress of the notched material in direction xσya is the allowable normal stress of the notched material in direction yτxya is the allowable shear stress of the notched materialσRm is the allowable bearing stress of the materialτvisa is the allowable shear stress of the screw

3 . SINGLE HOLE WITH FASTENER

The purpose of this sub-chapter is to outline the justification method of a hole with afastener to which is applied a bearing load in any direction, the laminate being subjectedto membrane type surrounding load fluxes and/or bending moment fluxes. The failuremode associated with this method is a combined net cross-section failure mode in thepresence of bearing (see 2.1 and 2.2).

3.1 . Pitch p definition

If the main loading is in the F1 direction, the pitch taken into account in the calculations

shall be equal to: p = p p1 22+ .

∅ ∅

e

F



MONOLITHIC PLATE - FASTENER HOLEMembrane analysis - Short cut method - Theory L 3.2.1

1/8

If the main loading is direction F2, the pitch (which is more commonly called edgedistance) taken into account in the designs shall be equal to: p = 2 p3.

For complex loading (or for simplification purposes), the following pitch value may beused: p = mini (p1; p2; 2 p3).

It should be noted that for membrane or membrane and bending loading, pitch p is limitedto k ∅ where k depends on the material used. The value of k is generally between 4.5 and5. For pure bending loading, this limitation does not apply.

3.2 . Membrane analysis - Short cut method

3.2.1 . Theory

Generally speaking, a failure is reached at a fastener hole when:

l1 σ tN + Km σm ≥ Kt σR

In the case of a membrane loaded single hole with fastener, the various justification(broadly summed up by relationship I1) steps must be followed:

p3

p1 p2

F1

F2

p = p p1 22+




2/8

1st step: For load introduction zones (fittings, splices), the membrane gross flux NB to betaken into account at fasteners is deduced from the constant flux to infinity N∞ by thefollowing relationship:

l2 NB = p5 ∅

N∞ if p > 5 ∅ NB = N∞ if p ≤ 5 ∅

If the zone to be justified is a typical zone (ribs, spars), then:

NB = N∞

The drawing above shows the difference between the flux to infinity and the actual flux atfasteners for a load introduction zone and highlights the existence of a working strip ateach fastener of a width equivalent to 5 Ø. This phenomenon is comparable to the onedescribed in chapter M.1.

5 Ø

Flux

NB

N∞




3/8

2nd step: It consists in transforming pitch corrected gross fluxes (see previous step) intonet cross-section flux in the initial coordinate system:

l3 N xN = N x

B p

p Sfe

− ∅ −

N yN = N y

B p

p Sfe

− ∅ −

N xyN = N xy

B p

p Sfe

− ∅ −

Thus, the equivalent diameter may be determined:

∅ ' = ∅ + Sfe

y

∅

x pN xB

F

β < 0

y

∅

x

N xN

F

β < 0




4/8

where the countersink surface is equal to:

Sf = b h = h2 tgθ

3rd step: It consists in transforming the previously designed fluxes in the coordinatesystem specific to the bearing load:

l4

N

N

N

MN

MN

MMN

'

'

=

22

22

22

)(sin)(coscosxsincosxsin

cosxsinx2)(cos)(sin

cosxsinx2)(sin)(cos

β−βββ−ββ

ββββ

ββ−ββ

N

N

N

xN

yN

xyN

Angle β is, in the trigonometric coordinate system, the angle leading from the M-axis(bearing coordinate system) to the x-axis (reference coordinate system).

b

θh

∅

∅ '

e

F

x

y

N MN

M' M

β < 0




5/8

4th step: Once positioned in the bearing coordinate system, the flux due to the bearingload N M

N (reduced by coefficient Km), is added to (or subtracted from) flux N Mm .

l5 N Mm = F e

e∅ *

e* = mini (e; 2.6 ∅ ) for double shear.e* = mini (e; 1.3 ∅ ) for single shear.

The bearing height e* is voluntarily reduced for a large thickness to take into accountstress concentration at the element surface.

The resulting fluxes are thus expressed by:

l6

N N K

N

N

MN

Mm

m

MN

MMN

±

'

'

SINGLE SHEARDOUBLE SHEAR

x

y

N MN

M' M

± Km N Mm

F

β < 0




6/8

The values of Km depend on the type of loading, the bearing stress and the material used(see chapter Z).

Three calculations shall be made with the following values: - Km; + Km; 0.

5th step: It consists in transferring fluxes so determined in their main coordinate system:

l7

N

N

PN

PN

'

0

=

22

22

22

)(sin)(coscosxsincosxsin

cosxsinx2)(cos)(sin

cosxsinx2)(sin)(cos

α−ααααα−

αα−αα

αααα

N N K

N

N

MN

Mm

m

MN

MMN

±

'

'

where:

α = 12

2Arctg NN N K N

MMN

MN

Mm

m MN

'

'± −�

��

�

��

α >

0

F

M' M

N PN

P

P'




7/8

Angle α is, in the trigonometric coordinate system, the angle leading from the M-axis(bearing coordinate system) to the P-axis (main coordinate system).

6th step: Fluxes are maximized by coefficient 1Kt

where Kt is the hole coefficient.

Kt values depend on the type of loading (tension or compression), the fastener diameter,the mechanical properties and the material used (see chapter Z).

It should be noted that to each of both main fluxes is associated a hole coefficient whichmay be different. This is why their notation differs from the sign*.

l8

NKNK

PN

t

PN

t

'*

0

α >

0

F

M' M

NK

PN

t

P

P'




8/8

7th step: Fluxes so maximized are recalculated in the initial coordinate system (o, x, y).

l9

N

N

N

xF

yF

xyF

= 2))(sin(2))(cos()cos(x)sin()cos(x)sin(

)cos(x)sin(x22))(cos(2))(sin(

)cos(x)sin(x22))(sin(2))(cos(

α−β−α−βα−βα−βα−βα−β−

α−βα−β−α−βα−β

α−βα−βα−βα−β

NKNK

PN

t

PN

t

'*

0

Angle (α - β) are, in the trigonometric coordinate system, the angle leading from the x-axis(reference coordinate system) to the p-axis (main coordinate system).

8th step: A smooth plate calculation is made with fluxes NF previously determined (seechapter C) to obtain the margin.

y

xN x

FF

α - β > 0

P'

P



MONOLITHIC PLATE - FASTENER HOLEEDP computing program PSG33 L 3.2.2

3.2.2 . Computing program PSG33

This software, which can be used on mx4 or PC, is simply the digital application of thetheory presented above, the eight steps being integrated into the calculation.

Let input data relating to the example covered further in this chapter be as follows.

CARACMF 1 3 4 12 4.8 21.6 4.91 -30. 77. 40.*3 T300 neuf1 0.78 0.52 0.52 0.782 0. 45. -45. 90.3 3.

MAT03 1 13000. 465. 465. .35 120.0 -100. 5.*2 -12. 7.5 .13

*

4PE 8. -6. 20.

The software gives the design margin for each value of Km, as well as all intermediateresults. To allow a quick check of loading, it represents the bearing load and main netfluxes in the reference coordinate system.

Note the bearing load direction (β = - 30°).

^90I

N2 = -20.19 I N1 = 22.19* I ** I ** I ** I * /* I * /* I * /*I*/ FM = 77.

--------------------------------------->0IIIIIII

B



MONOLITHIC PLATE - FASTENER HOLEBending analysis L 3.3

1/5

3.3 . Bending analysis - Short cut method

If the notched plate is subjected to bending moment fluxes Mx, My and Mxy, follow theadditional steps described hereafter:

1st step: Determine stresses on the external and internal surfaces corresponding tobending loads only.

As a first approximation, these stresses may be assessed by the general relationship

σ ≈ M vl

Me

≈6

2 . In that case, the material shall be considered as homogeneous.

It is nevertheless recommended to determine these stresses with the computing softwarePSD48 (stacking homogenizing and analysis) which takes into account stiffness variationswithin the laminate or to refer to chapter D.

Thus, for each design direction (x, y and xy), the following stresses are obtained:

σe xB , σe y

B , τe xyB : gross stresses on external surface.

σi xB , σi y

B , τi xyB : gross stresses on internal surface.

external surface

internal surface

σ eB

σ lB

external surface

internal surface

σ eB

σ lB




2/5

2nd step: From these stresses, "equivalent" membrane gross fluxes are evaluated.

l10

∆

∆

∆

∆

∆

∆

n

n

n

n

n

n

exB

eyB

exyB

ixB

iyB

ixyB

= e

σ

σ

τ

σ

σ

τ

exB

eyB

exyB

ixB

iyB

ixyB

surfaceernalintfor

surfaceexternalfor

3rd step: On the contrary of membrane analysis, no majoration between fluxes to infinityN∞ and gross fluxes NB will be taken into account at load introduction areas.

NB = N∞

4th step: "Equivalent" membrane net fluxes are evaluated from "equivalent" membranegross fluxes.

l11 ∆ne xN = ∆ne x

B p

p Sfe

− ∅ −

∆ne yN = ∆ne y

B p

p Sfe

− ∅ −for external surface (with countersunk fastener head)

∆ne xyN = ∆ne xy

B p

p Sfe

− ∅ −

∆n eB

∆n iB

e

external surface

internal surface




3/5

∆ni xN = ∆ni x

B pp − ∅

∆ni yN = ∆ni y

B pp − ∅

for internal surface (no countersunk fastener head)

∆ni xyN = ∆ni xy

B pp − ∅

Confer to sub-chapter L.3.1 to determine fastener pitch.

5th step: "Equivalent" membrane net fluxes are divided by the coefficient Kf (bending holecoefficient) which depends on the material (in general Kf = 0.9).

Hence, we get the (majorated) "equivalent" membrane net fluxes:

l12 ∆ne xF =

∆nK

e xN

f

∆ne yF =

∆nK

e yN

ffor external surface

∆ne xyF =

∆nK

e xyN

f

∆ni xF =

∆nK

ixN

f

∆ni yF =

∆nK

iyN

ffor internal surface

∆ni xyF =

∆nK

ixyN

f

∆n eN

∆n iN

external surface

internal surface




4/5

6th step: Final membrane fluxes from relation I9 are, then, added to fluxes calculated fromrelation I12.

l13 N xF + ∆ne x

F

N yF + ∆ne y

F for external surface (without bearing)

N xyF + ∆ne xy

F

N xF + ∆ni x

F

N yF + ∆ni y

F for internal surface (with bearing)

N xyF + ∆ni xy

F



MONOLITHIC PLATE - FASTENER HOLEMembrane + bending analysis - Summary table L 3.3

5/5

The overall method for the membrane and bending analysis is summarized in the figurehere below.

External surface Neutral line Internal surfaceMembrane Bending Membrane Membrane Bending

Data in theinitial

coordinatesystem

N xB

N yB

N yB

∆n xB

∆n yB

∆n yxB

N xB

N yB

N yB

N xB

N yB

N yB

∆n xB

∆n yB

∆n yxB

Net cross-sectionanalysis

N xN

= N xB

p

pSf

e− −∅

N yN

= N yB

p

pSf

e− −∅

N xyN

= N xyB

p

pSf

e− −∅

∆n xN

= ∆n xB

p

pSf

e− −∅

∆n yN

= ∆n yB

p

pSf

e− −∅

∆n xyN

= ∆n xyB

p

pSf

e− −∅

N xN

= N xB

p

pSf

e− −∅

N yN

= N yB

p

pSf

e− −∅

N xyN

= N xyB

p

pSf

e− −∅

N xN

= N xB

p

p − ∅

N yN

= N yB

p

p − ∅

N xyN

= N xyB

p

p − ∅

∆n xN

= ∆n xB

p

p − ∅

∆n yN

= ∆n yB

p

p − ∅

∆n xyN

= ∆n xyB

p

p − ∅

Rotation inthe load

coordinatesystem

↓ ↓

N MN

N MN

'

N MMN

'β

N MN

N MN

'

N MMN

'β

↓

Addition ofbearing ↓ ↓

N MN

± Km N Mm

N MN

'

N MMN

'β

N MN

± Km N Mm

N MN

'

N MMN

'β

↓

Rotation inthe main

coordinatesystem

N PN

N PN

'α

↓N P

N

N PN

'α - β

N PN

N PN

'α - β

↓

Holecoefficientmaximizing

N

K

PN

t

N

K

PN

t

'

α

∆n xF

= ∆n

K

xN

t

∆n yF

= ∆n

K

yN

t

∆n xyF

= ∆n

K

xyN

t

N

K

PN

t

N

K

PN

t

'

α - β

N

K

PN

t

N

K

PN

t

'

α - β

∆n xF

= ∆n

K

xN

t

∆n yF

= ∆n

K

yN

t

∆n xyF

= ∆n

K

xyN

t

Rotation inthe initial

coordinatesystem

N xF

N yF

N xyF

↓

N xF

N yF

N xyF

N xF

N yF

N xyF

↓

Addition offluxes

N xF

+ ∆n xF

N yF

+ ∆n yF

N xyF

+ ∆n xyF

N xF

+ ∆n xF

N yF

+ ∆n yF

N xyF

+ ∆n xyF



MONOLITHIC PLATE - FASTENER HOLEJustifications - Nominal deviations L 3.4

3.5.1

3.4 . Justifications

Whatever the type of load (membrane or membrane + bending), make sure that:

- the plain monolithic plate subject to "equivalent" membrane load fluxes (NF + ∆neF) or

(NF + ∆niF) is acceptable from a structural strength point of view (refer to chapter C),

- the allowable bearing stress of material σm (which depends on the material, thefastener diameter and the thickness to be clamped - see chapter Z) is greater orequal to the bearing stress applied corresponding to a laminate thickness that issmaller or equal to 1.3 ∅ for single shear or 2.6 ∅ for double shear (see sub-chapterL.2.1 - 4th step):

3.5 . Nominal deviations on a single hole

This sub-chapter is directly related to concession processing. Here, simple rules areoutlined, that shall allow the stressman to assess the effect of a geometrical deviation,such as a fastener diameter, its pitch or edge distance, on an initial margin.

The following paragraphs are valid only for a hole with fastener subject to membranefluxes.

However, for greater accuracy, it is recommended to redo the calculation or use thesoftware psg33.

3.5.1 . Changing to a larger diameter

Following a drilling fault, it is sometimes necessary to change to a repair size or tooversize the fastener.

B



MONOLITHIC PLATE - FASTENER HOLENominal deviations - Pitch decrease L 3.5.2

Based on the theory we have just presented, any diameter change (∅ changes to ∅ ')shall have an effect on:

- the net cross-section coefficient: the resulting reduction shall be equal to:

l14 k = p Sf

ep Sf

e

− ∅ −

− ∅ −

' '

- the bearing stress: we shall assume that there is no effect on the bearing stress, evenif it tends to decrease (this assumption is conservative),

- the hole coefficient: if we assume that the hole coefficient value is in the mostunfavorable case Kt = 0.003684 ∅ 2 - 0.08806 ∅ + 0.886 (see corresponding curve inchapter Z - material T300/914), the resulting reduction shall be equal to:

k' = '89.0088.00037.0

89.0'088.0'0037.02

2

∅∅≈

+∅−∅+∅−∅

Thus, the general relationship may be given as follows:

l15 RF' ≈ RF k k' ≈ RF ∅∅

− ∅ −

− ∅ −'

' 'p Sfe

p Sfe

3.5.2 . Pitch decrease

If loads are parallel to the free edge, no reduction is necessary on the reserve factor:

RF' ≈ RFF2

p

p'



MONOLITHIC PLATE - FASTENER HOLENominal deviations - Edge distance decrease L 3.5.3

1/2

If loads are perpendicular to the free edge, the reduction on the reserve factor is equal to:

l16 RF' ≈ RF p Sf

ep Sf

e

' − ∅ −

− ∅ −

3.5.3 . Edge distance decrease

If loads are parallel to the free edge, the reduction on the reserve factor is equal to:

RF' ≈ RFp Sf

ep Sf

e

' − ∅ −

− ∅ −

F1

p

p'

F2p p'



MONOLITHIC PLATE - FASTENER HOLENominal deviations - Edge distance decrease L 3.5.3

2/2

If loads are perpendicular to the free edge, the reduction on the reserve factor is equal to:

l17 RF' ≈ RF 54.0

p'p��

��

�→ (100 % ± 45°)

RF' ≈ RF 73.0

p'p��

��

�→ (50 % 0; 50 % ± 45°)

RF' ≈ RF 65.1

p'p��

��

�→ isotrope

Important remarks:

- These empirical relationships are valid only for low edge distance variations (2 ∅ ≤ p' ≤2.5 ∅ ).

- For low edge distances, the fact that the failure mode described in sub-chapter L.2.3 isnot critical shall have to be demonstrated.

p p'

F1



MONOLITHIC PLATE - FASTENER HOLE"Point stress" finite element method - Description L 3.6.1

3.6 . "Point stress" finite element method (membrane analysis)

3.6.1 . Description of the method

Procedure PSH2 allows the calculation of stresses in fibres around a circular hole withfastener in a multilayer composite plate subjected to membrane type surrounding fluxes. Itis based on a finite element display of a drilled plate. Mapping calls for two separate parts:

- the bolt (rivet/screw/bolt),

- the drilled plate.

The drilled hole is modeled by 8-junction quadrangular elements and 6-junction triangularelements. The area adjacent to the hole is modeled by two rings of elements. The ringnearest to the hole is thin and is not utilized directly on issued sheets. Issues arepresented on the second ring, the center of gravity of elements being at a design distancefrom the hole corresponding to the point stress theory (do).

Contact elements between the plate and the bolt (which also simulate clearance betweenthe fastener and the edge distance) are of the variable stiffness type. Their stiffness isvery low when there is no contact with the plate, their stiffness is very high if there is acontact.

Loading is achieved by (normal and shear) fluxes on plate edges.

2 do



MONOLITHIC PLATE - FASTENER HOLE"Point stress" finite element method - Justifications L 3.6.2

3.6.2 . Justifications

Make sure that:

- longitudinal stresses in fibres tangent to the hole edge distance (and located at adistance do) are smaller than the longitudinal stress allowable for fibre Rl,

Fibre a

t 45°

Fibre at 135°

Fib r

e at

90°

do

σlσl

Fibre at 0°

- The allowable bearing stress of the material σm is greater or equal to the bearingstress applied corresponding to a laminate thickness that is smaller or equal to 1.3 ∅for single shear or 2.6 ∅ for double shear (see sub-chapter L.2.1 - 3rd step).



MONOLITHIC PLATE - FASTENER HOLEMultiple holes - Independent holes - Interfering holes L 4.1

4.2

4 . MULTIPLE HOLES

The previous study allowed us to find the structural effect of a single hole with fastener (ordistant enough from others) on a monolithic plate subject to membrane or bending typeloads.

We shall now study the effect of several lined up holes. We shall assume that the plate issubjected to a membrane type uniaxial load flux that is perpendicular to the row offasteners.

If loading is parallel to the row of fasteners, refer to chapter L.3.4.2 calculation.

4.1 . Independent holes

If each fastener pitch is greater of equal to 5 ∅ , each fastener may be considered as asingle hole. Refer to sub-chapter L.3.

4.2 . Interfering holes (0 < d < 3.5 ∅∅∅∅ )

If the distance between two holes is smaller than 5 ∅ , the net cross-section coefficient tobe used changes to:

l18 5

5

∅ −

∅ − − ∅ −

d

d Sfe

pas = 5 ∅5 ∅ 5 ∅ 5 ∅



MONOLITHIC PLATE - FASTENER HOLEMultiple holes - Very close holes L 4.3

1/2

On the other hand, the hole coefficient in tension must also be modified. It changes to:

l19 η Kt ≈ t

2

k625.2d565.0d5065.0��

�

�

��

�

�+�

�

��

�

∅−∅−�

�

��

�

∅−∅ (see values of η on next page)

The hole coefficient in compression is unchanged (cf. note 440.197/84), but theconnection of the holes is ignored for the net section calculation.

These new values are to be taken into account in relationships l3 and l8.

4.3 . Very close holes (d = 3.5 ∅∅∅∅ soit p = 1.5 ∅∅∅∅ )

When holes are very close to each other, the diameter ∅ ' envelope hole shall beconsidered. The net cross-section coefficient then changes to:

l20

eSf'pitch

pitch

−∅−

The hole coefficient is not modified by the number η but applies to diameter ∅ '.

B

pas = 5 ∅ - d

d

d

pitch = 4.25 ∅

∅ '

1.5 ∅



MONOLITHIC PLATE - FASTENER HOLEMultiple holes - Kt correction coefficient L 4.3

2/2

Kt correction coefficient

d

pitch

∅ '

1.5 ∅

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

1 1.5 2 2.5 3 3.5 4 4.5 5

5 ∅ −∅

d

pitch = 5 ∅ - d

d pitch = 5 ∅5 ∅ 5 ∅

TR

OU

E

NV

EL

OP

PE

η



MONOLITHIC PLATE - FASTENER HOLEFirst example L 5.1

1/7

5.1 . First example

Let a T300/BSL914 (new) laminate be laid up as follows:


Total thickness: e = 20 x 0.13 = 2.6 mm

It is subjected to the three following fluxes in the initial coordinate system (o; x; y):

N xB = 8 daN/mm

N yB = - 6 daN/mm

N xyB = 20 daN/mm

and to the bearing load:

F = 185 daNβ = - 30°

The fastener is a ∅ 4.8 mm countersunk head one (100° countersink angle, whichcorresponds to a 4.91 mm2). The fastener pitch is 21.6 mm.

The purpose of the example is to determine the three final fluxes that shall be used for theequivalent smooth plate design, which shall provide the hole margin looked for (thiscalculation shall be covered in chapter C).

64

6

4

x

F = 185 daN

β = - 30°

y




2/7

Design of net cross-section fluxes in the initial coordinate system:

{l3}

mm/daN59.11

6.291.48.46.21

6.21x8NNx =

−−=

mm/daN69.8

6.291.48.46.21

6.21x)6(NNy −=

−−−=

mm/daN97.28

6.291.48.46.21

6.21x20NNxy =

−−=

Flux transfer in the bearing coordinate system (o, M, M'):

{l4}

N MN = 31.61 daN/mm

N MN

' = - 28.71 daN/mmN MM

N' = 5.7 daN/mm

β = - 30°

Bearing flux addition: 3 cases shall be considered

The bearing stress is equal to: σm = 6.2x8.4

185 = 14.82 hb

{l5}

N Mm = 14.82 x 2.6 = 38.54 daN/mm

The flux in the load direction being a tension flux (+ 31.61 daN/mm), the value of K mt is

thus equal to 0.135 (see chapter Z - material T300/914 - Sheet 2).




3/7

If the bearing flux minimized by coefficient K mt , is added to previously determined fluxes,

the three configurations K mt > 0; K m

t < 0 and K mt = 0 are obtained:

{l6}

K mt = 0.135

N MN = 31.61 + 0.135 x 38.54 = 36.81 daN/mm

N MN

' = - 28.71 daN/mmN MM

N' = 5.7 daN/mm

β = - 30°

{l6}

K mt = - 0.135

N MN = 31.61 - 0.135 x 38.54 = 26.4 daN/mm

N MN

' = - 28.71 daN/mmN MM

N' = 5.7 daN/mm

β = - 30°

{l6}

K mt = 0

N MN = 31.61 + 0 = 31.61 daN/mm

N MN

' = - 28.71 daN/mmN MM

N' = 5.7 daN/mm

β = - 30°




4/7

Rotation in the main coordinate system (o ; P ; P'):

{l7}

N PN = 37.3 daN/mm

N PN

' = - 29.2 daN/mmα = 4.9°

{l7}

N PN = 26.98 daN/mm

N PN

' = - 29.29 daN/mmα = 5.8°

{l7}

N PN = 32.14 daN/mm

N PN

' = - 29.24 daN/mmα = 5.4°

Angle α is the angle formed by the main coordinate system and the bearing coordinatesystem.

M

x

y P

β = - 30°

α = 4.9°

N PN

' = - 29.2 daN/mm

N PN = 37.3 daN/mm




5/7

Application of hole coefficients:

Monolithic lay-up under study gives the following elasticity and shear moduli in the mainaxes:

α + β = - 34.9° E = 4470 G = 2078 EG

= 2.151 K tt ≈ 0.6 K t

c ≈ 0.87

α + β = - 35.8° E = 4455 G = 2092 EG

= 2.13 K tt ≈ 0.6 K t

c ≈ 0.87

α + β = - 35.4° E = 4461 G = 2086 EG

= 2.139 K tt ≈ 0.6 K t

c ≈ 0.87

The values are derived from chapter Z (T300/914 sheets 3 and 4).

Which gives the following new values for corrected main fluxes :

{l8}

N PN =

6.03.37 = 62.17 daN/mm

N PN

' = 87.0

2.29− = - 33.56 daN/mm

α - β = 34.9°

{l8}

N PN =

6.098.26 = 44.97 daN/mm

N PN

' = 87.0

29.29− = - 33.67 daN/mm

α - β = 35.8°

{l8}

N PN =

6.014.32 = 53.57 daN/mm

N PN

' = 87.0

24.29− = - 33.61 daN/mm

α - β = 35.4°




6/7

Rotation in the initial coordinate system (o; x; y):

A rotation of angle (β - α) is achieved:

{l9}

N xF = 30.83 daN/mm

N yF = - 2.22 daN/mm

N xyF = 44.92 daN/mm

{l9}

N xF = 18.06 daN/mm



{l9}

N xF = 24.32 daN/mm



These fluxes are then used in a smooth plate design. Calculation shall be continued inchapter C.6.

A 31 % (RF = 1.31) margin shall be found.

x

y


N xF = 30.83 daN/mm





7/7

New, let's assume that, as a result of a defective drilling operation, the fastener diameterhad to be changed to a ∅ 6.35 mm with a 8.62 mm2 countersunk surface.

What would be the new margin?

{l15}

RF' = 1.31 x 92.0

6.291.48.46.216.2

62.835.66.21x

35.68.4 =

−−

−−

Which corresponds to a - 8 % margin, thus non allowable. However, a full manual analysis(or using software PSG33) would have made it possible to find a 0 % margin.

If the calculation is conservative, it is due to the fact that the decrease of the bearingstress corresponding to fastener oversizing was not taken into account (see chapterL.3.5.1).

The preceding example shall also be fully covered in the composite material manual part"Calculation programs" (PSG33 instructions).

B

B



MONOLITHIC PLATE - FASTENER HOLESecond example L 5.2

1/5

5.2 . Second example

Let's assume that three moment fluxes are superposed on membrane fluxes:

Mt xB = - 4 daN mm/mm

Mt yB = 3 daN mm/mm

Mt xyB = 5 daN mm/mm

If the material is considered (as a first approximation) as homogeneous, a strength

moment per unit of length equal to:66.2

vl 2

= = 1.127 mm2 is found.

Assuming that a positive moment flux creates compression stresses on the externalsurface, we obtain:

for the external surface:

σe xB =

127.14 = 3.55 hb (35.5 MPa)

σe yB =

127.13− = - 2.66 hb (- 26.6 MPa)

τe xyB =

127.15− = - 4.44 hb (- 44.4 MPa)

z

y

x Mt xB = - 4 daN

Mt xyB = 5 daN

Mt yB = 3 daN

F = 185 daNβ = - 30°




2/5

For the internal surface:

σi xB =

127.14− = - 3.55 hb (- 35.5 MPa)

σi yB =

127.13 = 2.66 hb (26.6 MPa)

τi xyB =

127.14 = 4.44 hb (44.4 MPa)

The purpose of this example is to determine which bending type fluxes must be added tomembrane type fluxes for the fastener hole calculation.

The "equivalent" gross bending type fluxes necessary for the calculations thus have thefollowing value:

{l10}

for the external skin:

∆ne xB = 3.55 x 2.6 = 9.23 daN/mm

∆ne yB = - 2.66 x 2.6 = - 6.92 daN/mm

∆ne xyB = - 4.44 x 2.6 = - 11.54 daN/mm

y

x

INTERNAL SURFACE

EXTERNAL SURFACE

- 4.44 hb3.55 hb

- 2.66 hb4.44 hb

- 3.55 hb

2.66 hb




3/5

for the internal skin:

∆ni xB = - 3.55 x 2.6 = - 9.23 daN/mm

∆ni yB = 2.66 x 2.6 = 6.92 daN/mm

∆ni xyB = 4.44 x 2.6 = 11.54 daN/mm

The "equivalent" net bending type fluxes thus have the following value:

{l11}


∆ne xN = 9.23

6.291.48.46.21

6.21

−− = 13.37 daN/mm

∆ne yN = - 6.92

6.291.48.46.21

6.21

−− = - 10.02 daN/mm

∆ne xyN = - 11.54

6.291.48.46.21

6.21

−− = - 16.72 daN/mm


∆ni xN = - 9.23

8.46.216.21

− = - 11.87 daN/mm

∆ni yN = 6.92

8.46.216.21

− = 8.9 daN/mm

∆ni xyN = 11.54

8.46.216.21

− = 14.84 daN/mm




4/5

Hole coefficient weighting

{l12}


∆ne xF =

9.037.13 = 14.86 daN/mm

∆ne yF =

9.002.10− = - 11.13 daN/mm

∆ne xyF =

9.072.16− = - 18.58 daN/mm


∆ni xF =

9.087.11− = - 13.19 daN/mm

∆ni yF =

9.09.8 = 9.89 daN/mm

∆ni xyF =

9.084.14 = 16.49 daN/mm

All prior calculations were made in the initial coordinate system (o; x; y). These"equivalent" bending type fluxes are thus to be added to the membrane type fluxes foundin the first example (see summary table on next page).




5/5

This table summarizes the various steps of "equivalent" membrane flux calculation of theprevious example.

External surface Neutral line Internal surfaceMembrane Bending Membrane Membrane Bending

Data in theinitial

coordinatesystem

8- 620

9.23- 6.92- 11.54

8- 620

8- 620

- 9.236.9211.54

Net cross-sectiondesign

11.59- 8.6928.97

13.37- 10.02- 16.72

11.59- 8.6928.97

10.29- 7.7125.71

- 11.878.9

14.84

Rotation inthe bearing

loadcoordinate

system

↓ ↓31.61

- 28.715.7

- 30°

28.06- 25.48

5.06- 30°

↓

Addition ofbearing flux ↓ ↓

+ Km36.81

- 28.715.7

- 30°

- Km26.4

- 28.715.7

- 30°

Km = 031.61

- 28.715.7

- 30°

+ Km33.26

- 25.485.06- 30°

- Km22.86

- 25.485.06- 30°

Km = 028.06

- 25.485.06- 30°

↓

Rotation inthe main

coordinatesystem

32.14- 29.2435.4°

↓37.3

- 29.234.9°

26.98- 29.2935.8°

32.14-29.2435.4°

33.69- 25.9134.9°

22.59- 26.01

36°

28.53- 25.9535.4°

↓

Holecoefficientmaximizing

53.57- 33.6135.4°

14.86- 11.13- 18.58

62.17- 33.5634.9°

44.97-33.6735.8°

53.57- 33.6135.4°

56.15- 29.7834.9°

37.65- 29.90

36°

47.55-29.8335.4°

- 13.199.8916.49

Rotation inthe initial

coordinatesystem

24.32- 4.3641.17

↓30.83- 2.2244.92

18.06- 6.7637.31

24.32- 4.3641.17

27.7- 0.8738.82

14.35- 5.6830.81

21.43- 3.0335.12

↓

Addition offinal fluxes

39.18- 15.4922.59

161 % marging 31 % marging14.519.0255.31

1.164.2147.3

8.246.8651.61

8 % marging

The minimum margin is the only one considered, i.e.:

31 % for membrane design8 % for membrane + bending design



MONOLITHIC PLATE - FASTENER HOLEReferences L




LAFON, Carbon fibre structures: simplified rules for sizing at fastener holes, 1983, PL No.139/83

BOUNIE, Failure criteria of mechanical bonds in composite materials, 1991, 440.181/91

LAFON, Justification of design methods used for carbon fibre structures - thin sheetsubject area, 1983, 440.156/83

J. ROCKER, Composite material parts: design methods at fastener holes, 3 ≤ ∅ ≤

100 mm. Extrapolation to damage tolerance evaluation, 1998, 581.0162/98

LAFON, TROPIS, Structural strength of outer wing - justification of design values, 1989,440.233/89

LAFON - LACOSTE, Synthesis of drilled carbon specimen tests, 1984, 440.197/84issue 2.

B

B



M

MONOLITHIC PLATE - SPECIAL ANALYSIS



N

SANDWICH - MEMBRANE / BENDING / SHEAR ANALYSIS



SANDWICH - MEMBRANE / BENDING / SHEARNotations N 1

1 . NOTATIONS

Ny: normal load fluxMx: moment fluxMz: moment fluxTx: shear load fluxTz: shear load flux

Emi: membrane elasticity modulus of lower skinEfi: bending elasticity modulus of lower skinGi: shear modulus of lower skinei: thickness of lower skin

Emc: membrane elasticity modulus of core materialEfc: bending elasticity modulus of core materialGc: shear modulus of core materialec: thickness of core material

Ems: membrane elasticity modulus of upper skinEfs: bending elasticity modulus of upper skinGs: shear modulus of upper skines: thickness of upper skin

zg: neutral axis position with respect to the lower skin

Σ El: overall inertia of elasticity moduli weighted plate

EW: elasticity moduli weighted static moment

µd: microstrain (10-6 mm/mm)B



SANDWICH - MEMBRANE / BENDING / SHEARSpecificity - Construction principle - Design principle N 2

34

2 . SPECIFICITY

A sandwich is a three-phase structure consisting of a core generally made out ofhoneycomb or foam with a low elasticity modulus and two thin and stiff face sheets.

Sandwich structures have a very high specific bending stiffness.

3 . CONSTRUCTION PRINCIPLE

The face sheets and core are assembled by bonding with synthetic adhesives. There areseveral alternative manufacturing processes:

- multiple phase process: face sheets are cured separately, then bonding of facesheets to the honeycomb is performed as a second operation,

- semi-cocuring process: the external face sheet is cured separately, the honeycomband the internal face sheet are then cocured on the external face sheet,

- single phase or "cocuring" process: face sheets and the honeycomb are cured in onesingle operation.

4 . DESIGN PRINCIPLE

The design rules that shall be developed are derived from the classical elasticity (refer to"distribution of load among several closely bound structural elements" in chapter A.7).

core (honeycomb)

internal face sheet

external face sheetadhesivebondinginterface



SANDWICH - MEMBRANE / BENDING / SHEARSandwich plates - Sandwich beams N 4.1

4.21/3

First of all, we shall consider that the three materials together are completely ordinary.Then, we shall simplify the relationships obtained by considering that face sheets are thinand stiff and that the sandwich core is thick and flexible.

4.1 . Sandwich plates

Like monolithic metal or composite plates, sandwich plates are under the general plateequation (see § A.7.4).

The determination of matrices (Aij), (Bij) and (Cij) which connect the strain tensor to theload tensor is described in chapters C, D and E.

4.2 . Short cut theory - "Sandwich" beams

Here, we shall outline a short cut method applicable to sandwich beams. This methoddoes not take into account transversal loading, transversal effects so-called "Poisson"effects and membrane-bending coupling. This simplification may lead to an error ofapproximately 10 % on results obtained in cases of complex loading.

From the overall deformation point of view, sandwich plates obey the conventionalequations of classical elasticity theory. Stiffness equivalences (with iso-cross-section) withhomogeneous beams are described by relationships n14 to n18.

Let a sandwich beam be made up of:

- an upper skin of thickness es, of membrane elasticity modulus Ems and of equivalentbending elasticity modulus Efs,

- a core thickness ec, of membrane elasticity modulus Emc and of equivalent bendingelasticity modulus Efc,

- a lower skin of thickness ei, of membrane elasticity modulus Emi and of equivalentbending elasticity modulus Efi.



SANDWICH - MEMBRANE / BENDING / SHEARSandwich beams N 4.2

2/3

The bending modulus concept comes from the fact that lower and upper skins aregenerally (in the case of honeycomb sandwiches) laminates with different membrane andbending moduli (see chapters C and D). Its value depends on ply stacking. This conceptwas extended to all three materials.

First of all, we shall develop the full sandwich beam theory while taking into account facesheet thickness and bending stiffnesses, then we shall outline at the end of each sub-chapter, the simplified relationships in which face sheets shall supposedly be thin andsubject to membrane stress only.

The neutral line of the sandwich beam is defined by dimension zg to that:

n1 zg = Em e Em e e e Em e e e e

Em e Em e Em e

ii

c c ic

s s i cs

i i c c s s

2

2 2 2+ +�

��

�� + + +�

��

��

+ +

Remark: In the case of a beam in which Emc ec << Emi ei and Emc ec << Ems es, therelationship becomes:

n2 zg = Em e Em e e e e

Em e Em e

ii

s s i cs

i i s s

2

2 2+ + +�

��

��

+

Emc Efc Gc

Emi Efi Gi

Ems Efs Gs

es

ec

ei

bzg



SANDWICH - MEMBRANE / BENDING / SHEARSandwich beams N 4.2

3/3

We shall assume that the beam is subjected to the following overall load pattern at theneutral axis:

- Ny: normal load in direction y

- Tx: shear load in direction x

- Tz: shear load in direction z

- Mx: bending moment around x-axis

- Mz: bending moment around z-axis

Torsional moment My shall not be taken into account because it does not correspond toany realistic loading.

The purpose of this chapter is to determine the stress and elongation diagram for eachone of these five loads.

We shall study the effects of Ny, Tx, Tz, Mx and Mz one by one.

x

y

z

Tz

Tx

Ny

Mx

Mz



SANDWICH - MEMBRANE / BENDING / SHEAREffect of normal load Ny N 4.2.1

1/2

4.2.1 . Effect of normal load Ny

Assuming that all layers are in a pure tension or compression condition, a normal load Ny

applied at the neutral line results in a constant elongation over the whole cross-section.This elongation may be formulated as follows:

n3 ε = ( )N

b Em e Em e Em ey

i i c c s s+ +

This elongation thus induces:

- in the lower skin, a stress σi = Emi ε,

- in the core, a stress σc = Emc ε,

- in the upper skin, a stress σs = Ems ε.

The equivalent membrane modulus of the sandwich beam may be determined by therelationship n14.

Remark: In the case of a sandwich beam in which Emc ec << Emi ei and Emc ec << Ems

es, the relationship becomes:

n4 ε = ( )N

b Em e Em ey

i i s s+

Ems es

Emc ec

Emi ei

z

y

bx

σs

σc

σi ε

Ny



SANDWICH - MEMBRANE / BENDING / SHEAREffect of normal load Ny N 4.2.1

2/2

By taking into account the remark assumptions of the previous page (ei << ec, es << ec,Emc << Emi and Emc << Ems), it is possible to oversimplify load distribution in the differentsandwich layers.

We shall assume that load Ny applied at the beam neutral axis is fully picked up by twomembrane type normal loads (Fs and Fi) in both face sheets.

Both loads have the following value:

n5 Fi ≈ Ny Em e

Em e Em ei i

i i s s+

Fs ≈ Ny Em e

Em e Em es s

i i s s+

z

y

x

Ny

Fs

Fi



SANDWICH - MEMBRANE / BENDING / SHEAREffect of shear loads Tx N 4.2.2

4.2.2 . Effect of shear load Tx

Generally speaking, shear load Tx is distributed in each of the three materials in proportionwith their shear stiffness.

The maximum shear stress in each of the three layers may then be formulated as follows:

n6 τs = 32

Tb e

G eG e G e G e

x

s

s s

s s c c i i+ +

τc = 32

Tb e

G eG e G e G e

x

c

c c

s s c c i i+ +

τi = 32

Tb e

G eG e G e G e

x

i

i i

s s c c i i+ +

The equivalent shear modulus with relation to the x-axis may be determined by therelationship n15.

es Gs

ec Gc

ei Gi

z

y

bx

τs

τc

τi

Tx



SANDWICH - MEMBRANE / BENDING / SHEAREffect of shear load Tz N 4.2.3

1/3

4.2.3 . Effect of shear load Tz

Generally speaking, shear stress τ in materials may be formulated by the relationship:

τ = T EWb El

z

� (Bredt generalized formula).

where

n7 Σ El = b Ef e b Em e e e e zs ss s i c

sg

3 2

12 2+ + + −�

��

�� +

b Ef e b Em e e e zc cc c i

cg

3 2

12 2+ + −�

��

�� +

b Ef e b Em e e zi ii i

ig

3 2

12 2+ −�

��

��

If we consider three critical points A, B and zg, moduli weighted static moments at thesepoints are equal to:

n8 EWA = b Ems es e ee

zi cs

g+ + −�

��

�

��2

EWzg = b Ems es e ee

z b Eme e z

i cs

g ci c g+ + −�

��

�

�� + + +

�

��

�

��

2 2 2 2

2

EWB = b Emi ei z eg

i−��

��2

Shear stresses at these points are then equal to:

n9 τA = T EWb Elz A

Στzg =

T EW

b Elz zg

ΣτB =

T EWb Elz B

Σ

Stress τzg corresponds to the maximum stress within the core. In the general case of a

honeycomb sandwich material, this stress is the maximum shear stress of the

honeycomb.

Stresses τA and τB correspond to shear of (adhesive bonding) interface between the coreand the skins (force sheets).




2/3

The equivalent shear modulus with relation to the z-axis may be determined by therelationship n16.

Remark 1: In the case of a sandwich beam in which Emc ec << Emi ei and Emc ec << Emses, τA, τzg and τB take the following simplified from:

n10 τA ≈ τzg ≈ τB ≈ T

b e e ez

ic

s

2 2+ +�

��

��

Remark 2: It should be noted that the equivalent shear modulus of a thin face sheetsandwich beam is on the same order of magnitude as the core for thehoneycomb it consists of, thus very low.

For the assessment of a honeycomb sandwich beam (or plate) deflection, it istherefore important to take into account this significant effect with respect tothe deformation due to the bending moment.

es

ec

ei

z

y

bx

τA Tz

τB

τzg

es

ec

ei

z

y

bx

τA Tz

τB

τzg




3/3

For example, for a sandwich beam simply supposed, loaded in its center, the deflectedshape due to the shear load may represent approximately 60 % of the overall deflection.

Let an aluminium beam and a sandwich beam with equivalent bending stiffness be, giving:

Aluminium beam: f1 = 0.3041 mm (99.6 %); f2 = 0.0011 mm (0.4 %)

Sandwich beam: f1 = 0.3041 mm (38 %); f2 = 0.5 mm (62 %)

f1: deflection due to the bending moment f P lE l

148

3

=

f2: deflection due to the shear load SG4lP2.12f =

300

1 daN

3

10AluminiumE = 7400 hb, G = 2840 hbEl = 1.85E6 daN mm2

ES = 2.22E5 daNGS = 8.52E4 daN

10

Sandwich0.5 E = 82009 G = 20.5 E = 8200

El = 1.85E6 daN mm2

ES = 8.2E4 daNGS = 180 daN

f1

f2



SANDWICH - MEMBRANE / BENDING / SHEAREffect of bending moment Mx N 4.2.4

1/2

4.2.4 . Effect of bending moment Mx

A bending moment Mx applied at the neutral line results in the creation of a lineardistribution of elongations along the cross-section. At the outer surfaces, we have:

n11 εs = M vEl

M e e e zEl

x s x i c s g

Σ Σ=

− + + −( )

εi = M vEl

M zEl

x i x g

Σ Σ=

with:

Σ El = b Ef e b Em e e e e zs ss s i c

sg

3 2

12 2+ + + −�

��

�� +

b Ef e b Em e e e zc cc c i

cg

3 2

12 2+ + −�

��

�� +

b Ef e b Em e e zi ii i

ig

3 2

12 2+ −�

��

��

The equivalent bending modulus of the sandwich beam may be determined by therelationship n17.

Mx

Ems Efs es

Emc Efc ec

Emi Efi ei

z

y

bx

σs

σi

εs

εi



SANDWICH - MEMBRANE / BENDING / SHEAREffect of bending moment Mx N 4.2.4

2/2

Remark: In the case of a sandwich beam in which ei << ec, es << ec, Emc << Emi andEmc << Ems, self inertias of both face sheets and honeycomb stiffness may bedisregarded:

Σ El ≈ b Ems es e e e z b Em e e zi cs

g i ii

g+ + −��

�� + −�

��

��

2 2

2 2

We shall assume that moment Mx is fully picked up by two membrane type normal loads(F's and F'i) in both face sheets.

Both loads have the same modulus but are opposite. Their value is equal to:

n12 F'i ≈ - F's ≈ Me e e

x

ic

s

2 2+ +�

��

��

z

y

x

F's

F'i

Mx



SANDWICH - MEMBRANE / BENDING / SHEAREffect of bending moment Mz N 4.2.5

4.2.5 . Effect of bending moment Mz

Generally speaking, bending moment Mz is distributed in each of the three materials inproportion to their natural bending stiffness (with relation to the z-axis).

The maximum normal stress in each of the three materials may then be simply formulatedas follows:

n13 σs = ±+ +

62M

b eEm e

Em e Em e Em ez

s

s s

s s c c i i

σc = ±+ +

62M

b eEm e

Em e Em e Em ez

c

c c

s s c c i i

σi = ±+ +

62M

b eEm e

Em e Em e Em ez

i

i i

s s c c i i

The equivalent bending modulus with relation to the z-axis is identical to the equivalentmembrane modulus with relation to the y-axis (see relationships n14 and n18).

Ems es

Emc ec

Emi ei

z

y

bx

Mz

- σi - ε

σs

σc

ε



SANDWICH - MEMBRANE / BENDING / SHEAREquivalent mechanical properties N 4.2.6

4.2.6 . Deformations and equivalent mechanical properties

Sandwiches are microscopically heterogeneous. It is sometimes necessary to find theirequivalent stiffness properties in order to determine the passing loads and resultingdeformations.

For a sandwich beam, equivalences (with iso-cross-section) with respect to typical loadsare the following:

n14 (1) E equivalent normal load = E e

e

k kk

kk

=

=

�

�

1

3

1

3

n15 (2) G equivalent shear load = G e

e

k kk

kk

=

=

�

�

1

3

1

3

n16 (3) G

1load shear equivalent

=

eG

e

k

kk

kk

�

��

�

��

=

=

�

�

1

3

1

3

n17 (4) E equivalent bending moment = E l

l

k kk

kk

=

=

�

�

1

3

1

3

n18 (5) E equivalent bending moment = E e

e

k kk

kk

=

=

�

�

1

3

1

3

lk: self inertia + "Steiner" inertia e e d3

2

12+

�

��

�

��

(3)

(2)

(1)

(4)

(5)E1G1

E3G3

E2G2

e1

e3

e2



SANDWICH - MEMBRANE / BENDING / SHEARExample N 5

1/7

5 . EXAMPLE

Let a 10 mm wide sandwich beam be defined by the following stacking sequence:

- an upper skin (carbon layers) of thickness es = 1.04 mm and of longitudinal elasticitymodulus Es = 6000 daN/mm2 (the bending modulus being identical),

- a core (honeycomb) of thickness ec = 10 mm and of longitudinal elasticity modulusEc = 15 daN/mm2,

- a lower skin (carbon cloths) of thickness ei = 0.9 mm and of longitudinal elasticitymodulus Ei = 4500 daN/mm2 (the bending modulus being identical).

We shall assume that the beam is subjected to the following two loads and moment:

- Ny = 800 daN

- Mx = 2000 daN mm

- Tz = 250 daN

Tz = 250 daN

z

y

10

x

0.9

10

1.04

Mx = 2000 daN mm

Ny = 800 daN




2/7

The purpose of the first part of the example is to determine inner and outer surfaceelongations of the beam subject to load Ny and moment Mx.

1st step: the neutral axis position has to be determined, this position being referenced withrelation to the inner surface.

{n1}

Zg = 04.1600010159.04500

204.1109.004.16000

2109.01015

29.04500

2

++

��

��

� +++��

��

� ++

Zg = 7.09 mm

2nd step: To determine elongation ε induced by normal load Ny.

{n3}

ε = )04.1600010159.04500(10

800++

= 7612 µd (microstrain)

z

y

x

zg = 7.09




3/7

Remark: If the simplified relationship is used, we obtain:

{n4}

ε = )04.160009.04500(10

800+

= 7774 µd the error is 2 %

3rd step: To determine maximum elongations εi and εs induced by moment flux Mx.

{n11}

εs = El

)09.704.1109.0(2000Σ

−++−

εi = El

09.72000Σ

{n7}

Σ El = +��

��

� −+++23

09.7204.1109.004.1600010

1204.1600010

+��

��

� −++23

09.72

109.010151012

101510

2309.7

29.09.0450010

129.0450010

��

��

� −+

Σ El = 5624 + 1169931 + 12500 + 2124 + 2733 + 1785629 = 2978541 daN mm2

Ny = 800 daN

z

y

x

ε = 7612 µd




4/7

giving maximum elongations :

εs = - 3256 µd

εi = 4761 µd

In fact, elongations (and stresses) are calculated at the center of each face sheet:

εs = - 3256 85.433.4 = - 2906 µd

εi = 4761 09.764.6 = 4459 µd

Remark: If the simplified relationship is used, we obtain:

{n12}

F'i = - F's ≈

204.110

29.0

2000

++ = 182.3 daN

Which corresponds to average elongations in lower and upper face sheets equal to:

εs ≈ 600004.1103.182− ≈ - 2921 µd the error is 0.5 %

εi ≈ 45009.0103.182 ≈ 4501 µd the error is 0.9 %

Mx = 2000 daN mm

εi = 4761 µd

z

y

x

εs = - 3256 µd

- 2906 µd

4459 µd




5/7

4th step: Globally, we have:

- at the lower fibre, an elongation of 7612 + 4761 = 12373 µd,

- at the upper fibre, an elongation of 7612 - 3256 = 4356 µd.

The second part of the example consists in calculating the evolution of shear stress due toshear load Tz, at the neutral axis in particular, at point A (upper face sheet - honeycombinterface) and at point B (lower face sheet - honeycomb interface).

1st step: To calculate the inertia of the elasticity moduli weighted beam.

{n7}

Σ El = 2978541 daN mm2

Remark: If the simplified relationship of a sandwich beam is used (see § M.3.2.4), weobtain the value:

Σ El ≈ 10 6000 1.04 22

09.729.09.045001009.7

204.1109.0 �

�

��

� −+��

��

� −++

Σ El ≈ 2955560 daN mm2 the error is 1 %

Ny = 800 daN

z

y

x

εi = 12373 µd

εs = 4356 µd Mx = 2000 daN mm




6/7

2nd step:

- To calculate the elasticity moduli weighted static moment EWzg (static moment with

relation to the neutral axis of part of the material located above it).

{n8}

EWzg = 10 6000 1.04 2

204.1

210

29.0151009.7

204.1109.0 �

�

��

� +++��

��

� −++

EWzg = 275538 daN mm

- To calculate the elasticity moduli weighted static moment EWA (static moment withrelation to the neutral axis at the upper face sheet).

EWA = 10 6000 1.04 ��

��

� −++ 09.7204.1109.0

EWA = 270192 daN mm

- To calculate the elasticity moduli weighted static moment EWB (opposite of the staticmoment with relation to the neutral axis at the lower face sheet).

EWB = 10 4500 0.9 ��

��

� +− 09.729.0

EWB = 268920 daN mm




7/7

3rd step: to determine shear stresses at the neutral axis (shear stress in honeycomb), atpoint A and point B.

{n9}

τzg = 250 27553810 2978541

= 2.31 hb (23.1 MPa)

τA = 250 27019210 2978541

= 2.26 hb (22.6 MPa)

τB = 250 26892010 2978541

= 2.25 hb (22.5 MPa)

It should be noted that, between point A and point B, the shear stress is practicallyconstant. It would be totally constant if the honeycomb elasticity modulus were zero(which may be considered as such).

Remark: With the simplified formula, we find:

{n10}

τA ≈ τzg ≈ τB ≈ ��

��

� ++204.110

29.010

250 = 2.28 hb (22.8 MPa)

The error is 2 %.

τA = 2.26 hb

Tz = 250 daN

z

y

x

- B

- A

τzg = 2.31 hb

τB = 2.25 hb



SANDWICH - MEMBRANE / BENDING / SHEARReferences N




M. THOMAS, Analysis of a laminate plate subjected to membrane and bending loads,440.227/79


J. CHAIX, 436.127/91



O

SANDWICH - FATIGUE ANALYSIS



P

SANDWICH - DAMAGE TOLERANCE APPROACH



Q

SANDWICH - BUCKLING ANALYSIS



R

SANDWICH - SPECIAL DESIGNS



S

BONDED JOINTS



BONDED JOINTSNotations S 1

1/2

1 . NOTATIONS

F: normal load transferred in bonded materials

Fr: failure load of adhesively bonded joint

M: cleavage moment ��

��

� =2txFM

E1: longitudinal elasticity modulus of material 1e1: thickness of material 1E2: longitudinal elasticity modulus of material 2e2: thickness of material 2E: longitudinal elasticity modulus of materials 1 and 2, if they are similare: thickness of materials 1 and 2, if they are similar

Gc: shear modulus of adhesiveEc: longitudinal elasticity modulus of adhesiveec: thickness of adhesive

h: width of adhesively bonded joint

l: length of adhesively bonded joint

lm: minimum length of adhesively bonded joint

t: thickness of cleavage t e e ec= + +��

��1 2

2 2

λ: design constant

k: design constant

D: design constant

τm: average shear stress in adhesively bonded joint

τM: maximum shear stress in adhesively bonded joint

τx: shear stress in adhesively bonded joint at dimension x

τam: allowable average shear stress of adhesive

τaM: allowable maximum shear stress of adhesive

σm: average peel stress in adhesively bonded joint

σM: maximum peel stress in adhesively bonded joint

σa: allowable peel stress of adhesive



BONDED JOINTSNotations S 1

2/2

F1i: normal load passing through material 1 (at center of step No. i)F2i: normal load passing through material 2 (at center of step No. i)∆Fi: normal load transferred by the adhesively bonded joint (in step No. i)

E1i: longitudinal elasticity modulus of material 1 (in step No. i)e1i: thickness of material 1 (in step No. i)E2i: longitudinal elasticity modulus of material 2 (in step No. i)e2i: thickness of material 2 (in step No. i)

li: length of adhesively bonded joint (in step No. i)

τmi: average shear stress in adhesively bonded joint (in step No. i)τMi: maximum shear stress in adhesively bonded joint (in step No. i)



BONDED JOINTSBonded single joint - Highly flexible adhesive S 2.1.1

2 . BONDED SINGLE LAP JOINT

This technique consists in assembling two (or several) elements by molecular adhesion.The adhesive must ensure load transmission.

Bonding of two flat surfaces only shall be considered.

Four cases shall be examined:

- Single joints:• highly flexible adhesive with respect to bonded laminates,• general case (without cleavage effect),• general case (with cleavage effect).

- Scarf joint.

2.1 . Elastic behavior of materials and adhesive

2.1.1 . Highly flexible adhesive

h

e1 E1 Gc

ec

l

E2 e2

Fτ

τF

τ

τm

- l/2 l/2x if E1 and E2 >> Gc



BONDED JOINTSGeneral case - Without cleavage S 2.1.2

1/3

In the case of an adhesive with a very low stiffness as opposed to the stiffness of thelaminates to be assembled, shear stress may be considered as uniform and equal to:

s1 τm = lxh

F

If τa is the allowable shear stress of the adhesive, the minimum length of the adhesivelybonded joint shall be equal to:

lm = maxh

Fτ

The failure load is equal to:

Fr = λ x τam x h

In practice, check that the average stress (which, in this case, is equal to the maximumstress) is smaller or equal to τam.

2.1.2 . General case (without cleavage effect)

F

τF

τ τM

x if E1 x e1 ≠ E2 x e2l/2- l/2

τ τM

x if E1 x e1 = E2 x e2l/2- l/2

τ




2/3

In the case of any bonded assembly (E1 x e1 > E2 x e2) (see drawing on previous page)subjected to a normal load F, the shear stress in the adhesively bonded joint may beformulated as follows (VOLKERSEN) :

s2 τx = τm ��

��

�

+−

λλ+

λλλ

2211

2211

exEexEexEexEx

)2/lx(cosh)xx(sinh

)2/lx(sinh)xx(coshx

2lx

with:

s3 λ = 2211

2211

c

c

exExexEexEexEx

eG +

and

τm = lxh

F

Remark: If E1 x e1 = E2 x e2 = E x e the joint is so-called equilibratedIf E1 = E2 = E and e1 = e2 = e the joint is so-called symmetrical

In the case of an equilibrated joint, the maximum shear stress may be formulated asfollows:

s4 τM = τm x ��

��

� λλ2

lxcothx2

lx

with

s5 λ = c

c

exexEGx2

and

τm = lxh

F

if λ x l << 0 then τM ≈ τm

if λ x l >> 0 then τM ≈ τm x 2

lxλ




3/3

In practice, check that τM ≤ τaM and that τm ≤ τam

If τa is the allowable shear stress of the adhesive, the minimum length of the adhesivelybonded joint shall be equal to:

lm = Max ��

�

�

��

�

�

τ��

�

�

��

�

�

τλ

λ hxF;

x2xFArcthx2

mM ahxa

The failure load is equal to:

Fr = Min ��

��

�τ

λτ

��

��

� λ hxxl;hxx2

x2

lxthm

Ma

a

The latter relationship makes it possible to establish, for a bonded assembly, the conceptof optimum bonding length. Indeed, the function "th ( )" is asymptotically directed towards1 when "λ x l/2" increases; now, value 1 is practically reached for a value of "λ x l/2" equalto 2.7 (th (2.7) = 0.99).

Thus, we have:

λ x l = 2 x 2.7

hence:

l = c

c

c

c

GexexEx82.3

Gx2exexEx4.54.5 ==

λ

In practice, the following relationship shall be used:

s6 loptimal = 3.16 x c

c

GexexE

F

Fr≈ 0.99.Fr

loptimal

l



BONDED JOINTSGeneral case - With cleavage S 2.1.3

2.1.3 . General case (with cleavage effect)

In the case of a symmetrical assembly, the misalignment of neutral axes of parts to be

assembled causes secondary moments ��

��

� =2

txFM to appear in elements, which tends to

create peeling stresses in the adhesive.

Maximum shear and peeling stresses in the adhesive may, in that case, be formulated asfollows (Bruyne and Houwnik) :

s7 τM = τm x ��

��

� +λ+λ kx31x2

lxcothxkx31x2

lx

and

s8 σM = σ x c

c

eex

EEx6x

2k with σ =

exhF

with

s9 k =

2

22

Dx24Fxl

Dx2Fxl1

1

++

and

s10 D = ( )2

3

n1x12txE−

F

F

FF

Mt

M



BONDED JOINTSScarf joint S 2.1.4

1/2

Check that τM ≤ τaM and that ττ

σσ

M

a

M

aM

�

��

�

�� +

�

��

�

�� ≤

2 2

1.

2.1.4 . Scarf joint

In the case of an angle α, scarf joint, the average shear stress is equal to:

τm = hxl

cosxF α

The maximum shear stress τM may be assessed using graphs on next page:

as abscissa: λ x l with λ2 = ��

��

�+

2211c

c

exE1

exE1x

eG

as ordinate: ττ

m

M

e2

FE2

ec; Gc

α

l

E1

e1

F



BONDED JOINTSScarf joint S 2.1.4

2/2

Each curve is representative of a value of ratio 22

11

exEexE

The peeling stress in the adhesively bonded joint shall be considered as constant. It shallbe equal to the following value:

σm = hxl

sinxF α

In practice, check that τM ≤ τaM and that τm ≤ τam.

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

ττ

m

M

0.5 1 10 20 30λ l

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1



BONDED JOINTSElastic-plastic behavior S 2.2

1/3

2.2 . Elastic-plastic behavior of adhesive and elastic behavior of laminates

Case of an elastic-plastic behavior of adhesive (see drawing below).

As long as maximum stresses at joint ends (τM) have not reached the critical value τp

(plasticizing stress of adhesive), the bonded joint behaves like a flexible joint and stressevolution follows the rules defined in paragraph 1).

If the load increases, a plasticizing zone (with stress τp) is formed at the most highlyloaded end of the joint.

If loading is yet increased, the shear stress of the adhesive in this plasticizing zonereaches the critical value τr (failure stress of adhesive), which causes the adhesivelybonded joint failure.

τ

τr

τp

γp γt

γ

ruptureelasti

cElastic-plastic




2/3

The drawing below illustrates, from a quality standpoint, the shear stress evolution in theadhesively bonded joint as the bonding force increases.

Remark: There is no simple theory for the elastic-plastic behavior of a bonded joint. Afinite element model only would allow justification of the structural strength ofsuch a system in this case.

x

τ

l

Load

τp

τr: failure

τp: beginning of plasticizing




3/3

However, in the case of an equilibrated joint and assuming that the adhesive has anelastic-plastic behavior such as described in the drawing below, it is possible to determine(M.J. DAVIS, The development of an engineering standard for composite repairs, AGARDSMP 1994) the length of plasticized adhesive and, of course, the length of adhesive inelastic behavior.

In the case of such behavior, the shear stress diagram in the adhesive is the following:

Lp ≈ pxhx4

Fτ

Le ≈ c

c

Gx2exexE66 ≈

λ

If the joint is equilibrated, the plasticized length is given by the following relationship:

τp L L L Fp p− − +�

��

��

�

��

�

��

�

��

�

�� =

12 2Φ

Φtanh with Φ = exExe

Gx2

c

c

τ

x

Lp

L

Le Lp

τp

τ

γ

τp

γp

elastic-plastic

elasti

c



BONDED JOINTSBonded double lap joint S 3

3 . BONDED DOUBLE LAP JOINT

For the case of a bonded double lap joint, shear stress distribution in the adhesive film isgiven by the following formula (in replacement of relationship s2) :

τx = τm x λ x l x ��

��

�λβ−−λ��

�

��

�

λβ+

λβ− )xx(sinh)1()xx(cosh

)lx(sinh)lx(tanh1

where β = 1

11

22

exEexE1

−

��

��

�+

In the general case, the maximum shear stress at the joint ends is formulated as follows:

τM = τm x λ x l x )lx(sinh

)lx(coshx')'1(λ

λβ+β−

where β' = max (β; 1 - β)

h

e1 E1 Gc

ec

l

E2 2 x e2

e1 E1

x



BONDED JOINTSBonded stepped joint S 4

1/3

4 . ADHESIVELY BONDED STEPPED JOINT

When the laminates to be bonded are too thick or when the loads to be transmitted aretoo high, the "stepping" or scarfing bonding technique is imperative.

The drawing below shows the general geometry of such a joint (the drawing shows athree-stepped joint (n = 3), a higher number may be considered).

The design method consists in determining, for each adhesively bonded joint portion, theload fraction crossing it, then, in considering each step "i" as elementary.

This so-called "short cut" method is a strictly manual method which gives the order ofmagnitude of average shear stresses per step. For greater accuracy, it is recommendedto use the computing software PSB2 (see § S4 and program PSB2 instructions).

Assumptions: Let's assume that transversal effects are insignificant (εy = 0 or Fy = 0). Let'salso assume that there is no secondary bending (off-centering from the neutral line shallnot be taken into account): joints below are considered as equivalent.

EQUIVALENCE

h

F

Material (2)

Material (1)

F

l1 li ln




2/3

1st step: Determination of loads (F1i et F2i) passing through both laminates (parent material"1" and repair material "2") at the center of each step.

We shall assume that loads are distributed (at the center of each step) in proportion to therigidity of each material:

s11 F1i = F x iiii

ii

2211

11

exEexEexE

+F2i = F - F1i

We shall assume that the load evolution in material 1 (and consequently in material 2) islinear by portions. Which leads to the following configuration:

F

FF1i

F2i

E2i, e2i

E1i, e1i

F

F

F1i

F2i

F11

F21

F1n

F2n

x

F2i

F2i

F2n

F

F2x

Evolution of the load transferred in the repair material




3/3

2nd step: From the previously determined curve, the load (∆Fi) transferred by each step iscalculated.

s12 ∆F1 = 21

2112

llFxlFxl

++

∆Fi = ii

i1i1ii

1ii

1iii1i

llFxlFxl

llFxlFxl

++

−++

−

−−

+

++ 2 ≤ i ≤ n - 1

∆Fn = Fn n1n

n1n1nn

llFxlFxl

++

−−

−−

We have also ( )∆F Fii

n=

=� 1

The diagram below presents the method used visually:

3rd step: Then, the average stress τmi is assessed.

We have:

s13 τmi = 1.05 x i

i

lxhF∆

∆F1 F

F

x∆F1

F

F2x

∆Fi

∆Fn

0

∆Fi

∆Fn



BONDED JOINTSEDP software S 5

where 1.05 is a fixed plus factor (according to the rule) allowing one to be conservativewith respect to results established by EDP software.

4th step: Check for the following condition:

For any step (i) τmi ≤ τam

5 . EDP SOFTWARE

The EDP software PSB2 digitally processes problems with adhesively bonded steppedjoints and, therefore, with adhesively bonded single joints as well.

This computing program is based on a differential analysis of the adhesively bonded jointand not on the "short cut" method outlined in chapter § S3.

The purpose of this software is to compute:

- stresses in any point of a bonded stepped single or double joint (evolution of shearstress and average stress per step),

- the evolution of normal stress in parent and repair laminates.

For more information, refer to instructions for use or to the example in chapter § S6.2.



BONDED JOINTSFirst example S 6.1

1/2

6 . EXAMPLES

6.1 . First example: single joint

Let the following symmetrical bonded joint be:

The allowable average shear value of the adhesive being: τam = 0.8 hb.

Assuming that the joint is subjected to load F = 1000 daN and that there is no cleavageeffect.

{s1} τm = 50001000

50x1001000 = = 0.2 hb (2 MPa)

{s5} λ = 1.0x2x5000

400x2 = 0.9

{s4} τM = 0.2 x ��

��

�

250x9.0cothx

250x9.0 = 4.5 hb (45 MPa)

Check that the average stress τm is smaller than the allowable stress τam (0.8 hb; 8 MPa)and that the maximum stress τM is smaller than τaM (8 hb; 80 MPa).

The margin thus obtained is equal to 77 % (RF = 1.77 = 8/4.5). Within the framework ofthe previous example, let's calculate the optimum bonding length from which any increasebecomes useless over the decrease of maximum shear stress in the adhesive.

{s6} loptimale = 3.16 x 400

1.0x2x5000 = 5 mm

h = 100 mm

e1 = 2 mm E1 = 5000 daN/mm2

ec = 0.1 mm

l = 50 mm

E2 = 5000 daN/mm2 e2 = 2 mm

Gc = 400 daN/mm2



BONDED JOINTSFirst example S 6.1

2/2

This result proves that, concerning the maximum shear stress, a change in the bondinglength from 50 mm to 5 mm increases (after calculations) this stress by only 1 %. The gainis thus insignificant.

Concerning the average stress, the minimum length is equal to:

lm = 100x8.0

1000 = 12.5 mm

The drawing below shows the evolution of the actual stress (smooth curve) and the valueof the average stress (dotted curve) in the example quoted.

- 15- 20- 25 - 10 - 5 0 5 10 15 20 250

0.5

1

1.5

2

2.5

3

3.5

4

4.5

τ

x



BONDED JOINTSSecond example S 6.2

1/5

6.2 . Second example: three-stepped joint

Let the following three-stepped joint be defined by its geometry and mechanicalproperties:

The allowable average shear value of the adhesive being: τam = 0.8 hb (8 MPa).

The allowable maximum shear value of the adhesive being: τaM = 8 hb (80 MPa).

We shall assume that the joint is subjected to load F = 100 daN and that there is nocleavage effect.The first stage consists in calculating, at the center of each step, loads passing througheach material.Concerning the first step:

{s11} F21 = 100 x 7000x78.05250x26.0

5250x26.0+

= 20 daN

F11 = 100 - 20 = 80 daN

h = 10 mm

F = 100 daN

F = 100 daN

l1 = 15 mm l2 = 10 mm l3 = 15 mm

1

e22 = 0.52 mmE22 = 5000 daN/mm2

e21 = 0.26 mmE21 = 5250 daN/mm2

e23 = 0.78 mmE23 = 7000 daN/mm2

2

e12 = 0.52 mmE12 = 5000 daN/mm2

e11 = 0.78 mmE11 = 7000 daN/mm2

e13 = 0.26 mmE13 = 5250 daN/mm2




2/5

Concerning the second step :

{s11} F22 = 100 x 5000x52.05000x52.0

5000x52.0+

= 50 daN

F12 = 100 - 50 = 50 daN

Concerning the third step:

{s11} F23 = 100 x 7000x78.05250x26.0

7000x78.0+

= 80 daN

F13 = 100 - 80 = 20 daN

The determination of these values allows the load evolution curve passing throughmaterial 2, or repair material, to be plotted:

The second stage consists in calculating, from the previous curve, loads transferred byeach step:

{s12} ∆F1 = 1015

20x1050x15++ - 0 = 38 daN

{s12} ∆F2 = 1015

20x1050x151015

80x1050x15++−

++ = 24 daN

{s12} ∆F3 = 100 - 1015

80x1050x15++ = 38 daN

F

F

F12

F22

F11

F21

F13

F23

xF21 = 20 daN

F22 = 50 daN

F23 = 80 daN

F = 100 daN

F2x

0




3/5

The drawing below represents the different loads ∆Fi transferred by each step.

The third stage consists in determining for each step the average and maximum stressesin the adhesively bonded joint, based on ∆Fi calculated previously.

steps 1 and 3 being equivalent for symmetry reasons, only the first two shall be justified.

{s13}

τm1 = 1.05 x 15x10

38 = 0.266 hb (2.66 MPa)

{s13}

τm2 = 1.05 x 10x10

24 = 0.252 hb (2.52 MPa)

and

τm3 = τm1 = 0.266 hb (2.66 MPa)

∆F1 F

F

x

∆F1 = 38 daN

∆F2

∆F3

∆F2 = 24 daN

∆F3 = 38 daN




4/5

The fourth stage consists in checking that average stresses are smaller than τam.

0.266 < 0.8 hb

Only a digital analysis (program PSB2) or a finite element analysis (program PSH14) shallbe able to determine with accuracy the shear stress evolution along each step.

F

F

x

τ

τ = 0.266 hb τ = 0.252 hb τ = 0.266 hb




5/5

For information purposes, we present below the output file of PSB2 corresponding to theprevious example.

Basic data:

I 5) 3 3 10 1 3F10) 10.0000000 10.0000000 ← loadingA 8)BONDED STEPPED JOINT SAMPLE

MAT1 MF 4 3MAT1MAT1 1001) 7000.00000 5000.00000 5250.00000 ← parent materialMAT1 2001) 3000.00000 3000.00000 3000.00000MAT1 3001) 0. 0. 0.MAT1 4001) .780000000 0.52000000 .260000000

MAT2P MF 4 3MAT2PMAT2P 1001) 5250.00000 5000.00000 7000.00000 ← repair materialMAT2P 2001) 3000.00000 3000.00000 3000.00000MAT2P 3001) 0. 0. 0.MAT2P 4001) .260000000 .520000000 .780000000

COLLE MF 2 3COLLECOLLE 1001) .050000000 .050000000 .050000000 ← adhesiveCOLLE 2001) 300.000000 300.000000 300.000000

VF 3

1) 15.0000000 25.0000000 40.0000000 ← step dimensions

Output (average stresses in each step):

AVERAGE STRESS IN THE ADHESIVE FOR EACH STEPSTEPS UPPER STEPS HB

1 .229 ← step No. 12 .313 ← step No. 23 .229 ← step No. 3

It may thus be observed that the short cut method provides (in this example), with respectto the PSB2 method, a difference of:

+ 16 % for external steps- 20 % for the central step

Consequently, it is recommended to use as often as possible the software PSB2, itsanalytical model being "closer" to physical reality. The "short cut" method being mainly amanual method.



BONDED JOINTSReferences S


K. STELLBRINK, Preliminary Design of Composite Joints, DLR-Mitt.92-05

S. ANDRE, Structural strength of a bonded joint, AS 432.178/95

A. TROPIS, Study of the behavior of bonded junctions, AS 432.445/96

S. ANDRE, Elastic-plastic analysis of the behavior of a bonded junction with a bondingfailure accounted for, AS 432.651/96

M.J. DAVIS, The development of an engineering standard for composite repairs, AGARDSMP 1994

NASA CR 112-235

NSA CR 112-236

D.A. BIGWOOD A.D. CROCOMBE, Elastic analysis and engineering design formulae forbonded joints

L.J. HART SMITH, Adhesively bonded joints for fibrous composite structures, Mc DonnellDouglas Corporation

L.J. HART SMITH, The design of repairable advanced composite structures, Mc DonnellDouglas Corporation 1985

L.J. HART SMITH, Adhesive bonded scraf and stepped lap joints, Mc Donnell DouglasCorporation

J.W. VAN INGEN A. VLOT, Stress analysis of adhesively bonded single lap joint

S. MALL N.K. KOCHHAR, Criterion for mixed mode fracture in composite bonded joints,University of Missourri-Rolla, 1986

S. MALL W.S. JONSHON, A fracture mechanic approach for designing adhesivelybonded joints, University of Maine

M. DE NEEF, Study of composite material bonding with edge effects accounted for,Alcatel Espace ; Août 1992

M. THOMAS, Stress distribution in bonded stepped joints, 440.128/77



T

BONDED REPAIRS



BONDED REPAIRSNotations T 1

1 . GENERAL NOTATIONS

(o, x, y): reference coordinate system of panel(o, p, p'): main coordinate system of stress fluxes

Nx∝ : normal flux in direction x

Ny∝ : normal flux in direction y

Nxy∝ : shear flux

Np∝ : principal flux in direction p

Np'∝ : principal flux in direction p'

β: angle

In principal direction p

τmi: average shear stress in adhesively bonded joint (in step No. i)

τMi: maximum shear stress in adhesively bonded joint (in step No. i)

Nsi: normal flux in parent material (in step No. i)

Nri: normal flux in repair material (in step No. i)

In principal direction p'

τ'mi: average shear stress in adhesively bonded joint (in step No. i)

τ'Mi: maximum shear stress in adhesively bonded joint (in step No. i)

N'si: normal flux in parent material (in step No. i)

N'ri: normal flux in repair material (in step No. i)

τam: allowable average shear stress of adhesive

τaM: allowable maximum shear stress of adhesive



BONDED REPAIRSIntroduction T 2

2 . INTRODUCTION

When a panel undergoes a damage (hole, delamination, etc.), two types of repair may beconsidered: a bolted repair (see chapter U) or a bonded repair.

Let the damaged (assuming that the damage is a hole) panel (monolithic skin) besubjected to stress fluxes Nx∝ , Ny∝ , Nxy∝ .

We shall assume that the repair is circular and of its stiffness close to that of the skin (noincrease of parent skin fluxes due to load transfer in a repair that is too stiff).

Ny∝

Nxy∝

Nx∝

y

x



BONDED REPAIRSAnalytical method T 3.1

1/6

3 . DESIGN METHOD

3.1 . Analytical method

This is an extrapolation of the bonded joint method (see chapter S) and, therefore, it is notsuited for shear flux transfer. Thus, it is necessary to work within the principal coordinatesystem in which stress fluxes are Np∝ and Np'∝ to return to the case of a single joint. Thismethod is conservative.

1st step: Calculation of principal fluxes Np and Np' and of main angle β.

t1 Np∝ = N N

N N Nx yx y xy

∞ ∞

∞ ∞ ∞

++ − +

212

42 2( )

t2 Np'∝ = N N

N N Nx yx y xy

∞ ∞

∞ ∞ ∞

+− − +

212

42 2( )

t3 β = 12

2Arctg

NN N

xy

x y

∞

∞ ∞−

�

��

�

��

Np'∝

Np∝

p'

p

y

x

β




2/6

2nd step: For each calculation direction (p and p'), let's consider the repair as a 1 mm widematerial strip.

The drawing below shows that, based on a two-dimensional repair (R), two one-dimensional bonded stepped joints (Jp) and (Jp') are determined (or isolated). Each one ofthese elementary bonded joints must transfer a normal load Fp = 1 Np∝ and Fp' = 1 Np'∝ .

For the determination of flux transfers from the parent material to the repair material, referto the design method for bonded stepped joints (see chapter S) or to the computingsoftware PSB2.

Np'∝

Np∝

p'

p

y

x

β

Jp'

J p

1 mm




3/6

3rd step:

From this analysis, the following results are extracted for each step and each direction (pand p'):

- for direction p:

• average and maximum shear stresses in each step of the adhesively bonded joint: τmi

(average stress in step i) and τMi (maximum stress in step i),

• normal fluxes in the parent material for each step Nsi (step i),

• normal fluxes in the repair material for each step Nri (step i).

1i = 2

34

pτ

Ns i

N r i

τ M i

τ m i

N pN

J p

p




4/6

- for direction p':

• average and maximum shear stresses in each step of the adhesively bonded joint: τ'mi

(average stress in step i) and τ'Mi (maximum stress in step i),

• normal fluxes in the parent material for each step N'si (step i),

• normal fluxes in the repair material for each step N'ri (step i).

1

i = 2

3

4

τ

J p'

N'

N' s i

N' r i

p'

p'

τ ' M i

τ ' m i

Np'




5/6

4th step: It consists of a combination of previously determined shear stresses and normalfluxes.

- Average shear stresses in the adhesively bonded joint calculated for both directions pand p' are vertorially combined (although points are different) and the resulting stress foreach step is compared with the allowable average shear value of the adhesiveconsidered.

t4 ( ) ( ' )τ τ τm m ai i m

2 2+ ≤

- Maximum shear stresses in the adhesively bonded joint calculated for both directions pand p' are vectorially combined (although points are different) and the value found foreach step is compared with the allowable maximum shear value of the adhesiveconsidered.

t5 ( ) ( ' )τ τ τM M ai i M

2 2+ ≤

τmi (p)τ'mi (p')

adhesively bondedjoint step No. i

τMi (p)τ'Mi (p')

adhesively bondedjoint step No. i




6/6

- In a plain plate calculation (see chapter C), normal stress fluxes Nsi and N'si for theparent material are associated (although points are different).

This calculation shall be performed where fluxes are maximum (at the beginning of eachstep).

- In a plain plate calculation (see chapter C), normal stress fluxes Nri and N'ri for the repairmaterial are associated (although points are different).

This calculation shall be performed where fluxes are maximum (at the end of each step).

N'si

Nsi

es

parent materialstep No. i

N'ri

Nri

er

repair materialstep No. i



BONDED REPAIRSDigital method T 3.2

3.2 . Digital method

In the case of a highly loaded bonded repair or with complex loading, the use of finiteelement modeling is preferable.

The software PSH14 has been developed for this purpose. It allows automatic modeling ofa circular bonded repair (see drawing below). This model is subjected to membrane stressonly and does not take cleavage effects into account.

The adhesively bonded joint is represented by type 29 volume elements (with elastic-plastic behavior), the panel and repair by type 80 and 83 elements.

The plotted results represent:

- plane fluxes in the parent material,

- plane fluxes in the repair material,

- shear stresses in the adhesive.

For more information, refer to program instructions.

same as1st

quadrant

same as1st

quadrant

same as1st

quadrant

Y-axis

X-axisdetails of

stepelements

507

407

307

207

107

7161726

2736



BONDED REPAIRSExample T 4

1/8

4 . EXAMPLE

Let the following repair be:

Let's assume (with a view to simplification) that both materials are nearly-isotropic (theirelasticity modulus being equal to 4417 daN/mm2 in all directions) for each step and thatsteps are 12 and 20 mm long.

The parent panel is only subjected to shear flux Nxy∝ = 4 daN/mm.

Nxy∝ = 4 daN/mm

y

x

Parent material: G803/914 (new)Repair material: G803/914 (new)

12 20

i = 1 i = 2

0°/90°45°/135°

0°/90°45°/135°

0°/90°45°/135°

0°/90°45°/135°0°/90°45°/135°0°/90°45°/135°0°/90°45°/135°




2/8

We may deduce that the principal coordinate system has a 45° direction and that principalfluxes are equal to Np∝ = 4 daN/mm and Np'∝ = - 4 daN/mm.

{t1}

Np∝ = + 12

4 42 = 4 daN/mm

{t2}

Np'∝ = - 12

4 42 = - 4 daN/mm

{t3}

β = °==∞=��

��

� 459021)(Arctg

21

04x2Arctg

21

y

x

Np'∝ = - 4 daN/mm Np∝ = 4 daN/mm

p' p

β = 45°J pJp'




3/8

After running the software PSB2 (computation of a bonded stepped joint), the followingresults are found for direction p (results may be multiplied by - 1 for direction p'):

- Shear stresses in the adhesive

0 5 10 15 20 25 30 350

1

2

3

2.2 hb

0.207 hb

0.076 hb

1.21 hb

1.35 hbτ

p




4/8

- Stress fluxes in the parent material (direction p)

5

0.5

10 15 20 25 30 350p

0

1

1.5

2

2.5

3

3.5

4

Nsi

1.512 daN/mm




5/8

- Stress fluxes in the repair material (direction p)

5

1

10 15 20 25 30 350p

0

2

3

4

5

6

Nri

2.472 daN/mm




6/8

• The first check consists of a vectorial combination of average shear stresses for eachstep. In this case, average shear stresses are the same (to the nearest sign) in bothdirections p and p'.

The maximum value is equal to 0.207 daN/mm2 in each direction. We may deduce thevectorial resultant stress:

{t4}

( ) ( ' )τ τm m1 1

2 2 2+ = 0.207 = 0.293 daN/mm2

This value is to be compared to the allowable average shear value of the adhesivelybonded joint that is generally selected equal to 0.8 daN/mm2 (a 173 % margin isobtained).

• The second check consists of a vectorial combination of maximum shear stresses foreach step. In this case, shear stresses are the same (to the nearest sign) in bothdirections p and p'.

The maximum stress is reached at the beginning of the first step. The value reached isequal to 2.20 daN/mm2.

We may deduce the vectorial resultant stress :

{t5}

( ) ( ' )τ τM M1 1

2 2 2+ = 2.20 = 3.11 daN/mm2

This value is to be compared to the allowable maximum shear value of the adhesivelybonded joint: 8 daN/mm2 (a 157 % margin is obtained).




7/8

• The third check consists of a plain plate calculation of the parent material for each step(where the flux is maximum: at the beginning of the step).

At the beginning of the first step, the flux in direction p is Ns1 = 4 daN/mm and the flux indirection p' is N's1 = - 4 daN/mm, which corresponds to a shear flux Nxys1 equal to4 daN/mm in the reference coordinate system. At this location, the parent material ismade out of six fabrics (3 fabrics at 0°/90° + 3 fabrics at 45°/135°) G803/914 (supposednew).

A running of the program PSB3 (plain plate computation) makes it possible to find a Hill'scriterion margin equal to 966 %.

At the beginning of the second step, the flux in direction p is Ns2 = 1.512 daN/mm andthe flux in direction p' is N's2 = - 1.512 daN/mm, which corresponds to a shear flux Nxys2

equal to 1.512 daN/mm in the reference coordinate system. At this location, the parentmaterial is made out of two fabrics (1 fabric at 0°/90° + 1 fabric at 45°/135°) G803/914(supposed new).

A running of program PSB3 (smooth plate computation) makes it possible to find a Hill'scriterion margin equal to 843 %.

• The fourth check consists of a smooth plate calculation of the repair material for eachstep (where the flux is maximum: at the end of the step).

At the end of the first step, the flux in direction p is Nr1 = 2.472 daN/mm and the flux indirection p' is N'r1 = - 2.472 daN/mm, which corresponds to a shear flux Nxyr1 equal to2.472 daN/mm in the reference coordinate system. At this location, the repair material ismade out of four fabrics (2 fabrics at 0°/90° + 2 fabrics at 45°/135°) G803/914 (supposednew).

A running of the program PSB3 (plain plate computation) makes it possible to find a Hill'scriterion margin above 1000 %.




8/8

At the end of the second step, the flux in direction p is Nr2 = 4 daN/mm and the flux indirection p' is N'r2 = - 4 daN/mm, which corresponds to a shear flux Nxyr2 equal to4 daN/mm in the reference coordinate system. At this location, the repair material is madeout of eight fabrics (4 fabrics at 0°/90° + 4 fabrics at 45°/135°) G803/914 (supposed new).

A running of the program PSB3 (plain plate computation) makes it possible to find a Hill'scriterion margin above 1000 %.

In conclusion, the minimum safety margin is assessed at 157 % under maximum stress inthe adhesively bonded joint.

B



BONDED REPAIRSReferences T




TRIQUENAUX, Investigation on the use of bonding and on dimensioning rules of bondedjoints, 1995, DCR/M-62385/F-95

CUQUEL-LEONDUFOUR, Study of bonded repairs, 1995, 432.095/95

CIAVALDINI, Effect of stepped machining on the structural strength of a skin intended toreceive a bonded repair, 1994, 440.133/94

STELLBRINK, Preliminary design of composite joints, 1992, DLR-Mitt.92-05

M. THOMAS, Stress distribution in bonded stepped joints, 440.128/77

M. MAHE - D. GRIMALD, Implementation of a digital model for the finite element design ofbonded repairs on composite materials, 436.0086/95



U

BOLTED REPAIRS



BOLTED REPAIRSNotations U 1

1/2

1 . NOTATIONS

∅ : diameter of damage

NxEF

.meca : flux Nx derived from E.F.s not influenced by the repairNy

EF.meca : flux Ny derived from E.F.s not influenced by the repair

NxyEF

.meca : flux Nxy derived from E.F.s not influenced by the repair

Nx or Nx∞

.meca : mechanical origin flux Nx upstream of the repairNy or Ny

∞.meca : mechanical origin flux Ny upstream of the repair

Nxy or Nxy∞

.meca : mechanical origin flux Nxy upstream of the repair

Nxr meca.: mechanical origin flux Nx crossing the doubler

Nyr meca.: mechanical origin flux Ny crossing the doubler

Nxyr meca.: mechanical origin flux Nxy crossing the doubler

Nxr meca. + therm.: mechanical and thermal origin flux Nx crossing the doubler

Nyr meca. + therm.: mechanical and thermal origin flux Ny crossing the doubler

Nxr therm.: thermal origin flux Nx crossing the doubler

Nyr therm.: thermal origin flux Ny crossing the doubler

Nxs meca.: mechanical origin flux Nx in the panel below the doubler

Nys meca.: mechanical origin flux Ny in the panel below the doubler

Nxys meca.: mechanical origin flux Nxy in the panel below the doubler

L sx and h s

x : panel dimensions for calculation of direction xL s

y and h sy : panel dimensions for calculation of direction y

L rx and h r

x : doubler dimensions for calculation of direction xL r

y and h ry : doubler dimensions for calculation of direction y

n: total number of load-carrying fasteners in the given directionr: elementary stiffness of fasteners.

Exs: longitudinal elasticity modulus (direction x) of panelEys: transversal elasticity modulus (direction y) of panelGxys: shear modulus of paneles: panel thickness



BOLTED REPAIRSNotations U 1

2/2

αxs: coefficient of expansion of panel in direction xαys: coefficient of expansion of panel in direction y

Exr: longitudinal elasticity modulus (direction x) of doublerEyr: transversal elasticity modulus (direction y) of doublerGxyr: shear modulus of doublerer: doubler thickness.αxr: coefficient of expansion of doubler in direction xαyr: coefficient of expansion of doubler in direction y

R sx : stiffness of panel with respect to flux Nx

R sy : stiffness of panel with respect to flux Ny

R sxy : stiffness of panel with respect to flux Nxy

R rx : stiffness of doubler with respect to flux Nx

R ry : stiffness of doubler with respect to flux Ny

R rxy : stiffness of doubler with respect to flux Nxy

η: correcting factor of panel shear stiffness

a and a*: fastener pitchni: number of rows of fasteners on a "unit strip"ri: stiffness of all fasteners on row "i"A: distance between the last load-carrying row of fasteners and the axis of symmetry ofthe repair

Fxi: overall load transferred by row of fasteners "i" in direction xFyi: overall load transferred by row of fasteners "i" in direction yfx/xij: load on fastener identified by rows "i" and "j" due to flux Nx

fy/yij: load on fastener identified by rows "i" and "j" due to flux Ny

fx/xyij: direction x load on fastener identified by rows "i" and "j" due to flux Nxy

fy/xyij: direction y load on fastener identified by rows "i" and "j" to flux Nxy

lxi: position of row of fasteners "i" with relation to the axis of symmetry of the repairlyj: position of row of fasteners "j" with relation to the axis of symmetry of the repair

Remark: Without an exponent, a notation looses its directional nature and thus becomesgeneral and applicable to x and y-axes.



BOLTED REPAIRSStiffness of a fastener in single shear U 2.1

2 . STIFFNESS OF FASTENERS

One of the most important parameters for the justification of a bolted repair is the stiffnessof fasteners which make it up. Their effect on load transfer is direct. The purpose of thissub-chapter is to make an analytical assessment of the stiffness of a fastener.Two cases are considered: single shear (the most common in our case) and doubleshear.

2.1 . Stiffness of a fastener in single shear

Let a fastener of diameter D and longitudinal elasticity modulus E bind two parts ofthickness er and es and of elasticity moduli Er and Es.

The stiffness of the fastener + parts system to be bound may be assessed by one of thethree relationships quoted below.

��

�

�

��

�

�++=

ssrr eE1

eE18.0

ED5

r1 → Mac Donnel Douglas

��

��

�++��

�

��

�+=

E83

E1

e2

E83

E1

e2

r1

ss

De85.0

rr

De85.0 sr

→ Boeing

12

1 1 12

12

2 3

re e

D E e E e e E e Es r

r r s s r s=

+�

��

�

�� + + +

�

��

�

��ξ

/

→ Huth

withξ = 2.2 → rivet on metal jointξ = 3 → screw on metal jointξ = 4.2 → carbon seal

EEs

es

D

er

Er



BOLTED REPAIRSStiffness of a fastener in double shear - Assumptions U 2.2

31/2

In the software (Bolted Repairs), the stiffness of the fastener + parts system to be boundis calculated by the Huth method:

u1 ��

��

�+++��

�

��

� +=

Ee21

Ee21

eE1

eE1

D2ee2.4

r1

srssrr

3/2rs

2.2 . Stiffness of a fastener in double shear

For the case of a load-carrying fastener in double shear, let's assume:

rdouble shear ≈ 1.5 rsingle shear

3 . ASSUMPTIONS

Let's have the five following assumptions:

- a delamination type damage shall be considered as a hole if the load flux it issubjected to is a compression or shear flux. In tension, we shall consider that thepanel retains its initial stiffness,

- the panel is subjected to membrane stress. The bending effects cannot be taken intoaccount in this method,

- the panel and doubler have a constant thickness and all fasteners are similar,

- the "Poisson" effect is ignored on the doubler,

EEs

es

D

er2

Er



BOLTED REPAIRSAssumptions U 3

2/2

- no overload on panel due to the presence of the doubler: to take it into account,fluxes in the skin derived from finite element (N EF

.meca ) should be increased inproportion to stiffnesses of the non-damaged skin alone and of the damaged skin withreinforcing piece.

Nx∞

.meca = NxEF

.meca R RR

rx

sx

sx+

∅ = 0with R s

x x s rx

sx

E e h

Ls

∅ = =0

Ny∞

.meca = NyEF

.meca R RR

ry

sy

sy+

∅ = 0

with R sy y s r

y

sy

E e h

Ls

∅ = =0

Nxy∞

.meca = NxyEF

.meca R R

Rrxy

sxy

sxy+

∅ = 0

with R sxy

xy sG es∅ = =0

All type R βα stiffnesses are explained further.

α: x; y; xy

β: r; s

Generally speaking, a bolted repair attached to a panel subjected to three load fluxesNx

∞.meca , Ny

∞.meca , Nxy

∞.meca may be represented as follows:

y

x

Nx∞

.meca

Nxy∞

.meca

Ny∞

.meca

Nxy∞

.mecaNy

∞.meca

Nx∞

.meca



BOLTED REPAIRSGeometrical data - Mechanical properties U 4

5

The justification method of such a repair shall first consist in calculating the three loadfluxes crossing the doubler (Nxr meca.

, Nyr meca. and Nxyr meca.

) then in assessing loads applied tothe repair fasteners, based on these results. The set of fluxes at each fastener may bedetermined on a unique basis.

Some geometrical and mechanical parameters of the structure shall be required forconducting this study.

4 . GEOMETRICAL CHARACTERISTICS

Below are represented the general geometrical characteristics describing the repair. Allother geometrical characteristics appearing further in the document may be formulatedaccording to these characteristics.

5 . MECHANICAL PROPERTIES

The mechanical properties required are the following:

For the panel: Longitudinal (direction x) and transversal (direction y) elasticity moduli,shear modulus and thickness: Exs; Eys; Gxys; es.

For the doubler: Longitudinal (direction x) and transversal (direction y) elasticity moduli,shear moduli and thickness: Exr; Eyr; Gxyr; er.

y

x

2 h rx

es

er

∅

a

a*

2 h ry



BOLTED REPAIRSDistribution of flux Nx U 6.1

For fasteners: Number of load-carrying fasteners (n) and elementary stiffness in shear (r)of each fastener.

6 . ASSESSMENT OF MECHANICAL ORIGIN FLUXES IN THE DOUBLER

The assessment method of fluxes crossing the doubler (Nxr meca., Nyr meca.

and Nxyr meca.) is

identical for all three fluxes Nx∞

.meca , Ny∞

.meca and Nxy∞

.meca . It consists in calculating for eachone the equivalent stiffness of the panel (Rs) and the equivalent stiffness of the doubler(Rr) and in distributing the flux in proportion to those.

6.1 . Distribution of flux Nx

If Nx∞

.meca is the panel flux "far from the repair", the flux Nxr meca.crossing the doubler is equal

to:

u2 Nxr meca. = Nx x

sxr

xr

.meca RRR+

∞

with

u3 R rx

x r rx

rx

x r rx

rx

E e h n rL

E e hL

n r

r

r

=

+

4

4

and

u4 R sx =

E e hL

x s sx

sx

s without chamfer

R sx = Exs es

aL a

h aLs

xsx

sx

* *−

+−�

��

�

�� with chamfer

where 2 L rx is the distance between the centers of gravity of fasteners located on either

side of the damage (see shaded fasteners on drawing below). The number of columnstaken into account shall never exceed 3.



BOLTED REPAIRSDistribution of flux Ny U 6.2

h hsx

rx= − ∅

2

6.2 . Distribution of flux Ny

If Ny∞

.meca is the panel flux "far from the repair", the flux Nyr meca. crossing the doubler is equal

to:

u5 Nyr meca. = Ny y

syr

yr

.meca RRR+

∞

with

u6 R ry

y r ry

ry

y r ry

ry

E e h n rL

E e hL

n r

r

r

=

+

4

4

and

u7 R sy =

E e hL

y s sy

sy

s without chamfer

R sy = Eys es

aL a

h aLs

ysy

sy−

+−�

��

�

��

* with chamfer

y

x

∅

center of gravity of fasteners

2 L sx

chamfer

h sx

L rx



BOLTED REPAIRSDistribution of flux Nxy U 6.3

where 2 L ry is the distance between the centers of gravity of fasteners located on either

side of the damage (see shaded fasteners on drawing below). The number of columnstaken into account shall never exceed 3.

h hsy

ry= −

∅2

6.3 . Distribution of flux Nxy

If Nxy∞

.meca is the panel flux "far from the repair", the flux Nxyr meca. crossing the doubler is

equal to:

u8 Nxyr meca. = Nxy xy

sxyr

xyr

.meca RRR+

∞

with

u9 R rxy

xy r

xy r

G e n r

G e n r

r

r

=+

8

8

and

u10 R sxy = η Gxys es

Factor η makes it possible to take into account the effect of damage size ∅ on the panelshear stiffness below the doubler.

y

x

2 L sy

h sy

center of gravity of fasteners

L ry

chamfer



BOLTED REPAIRSThermal origin loads in doubler U 7

u11 η = 5.0

22

22

V11

11W

W11

11V

WV

��

�

�

��

�

�

−+−+

��

�

�

��

�

�

−+−

+

with

V = 2 Lsx

∅

and

W = 2 Lsy

∅

7 . ASSESSMENT OF THERMAL LOADS IN THE DOUBLER

In the case of two geometrically different plates (1) and (2) bound by a system withstiffness ℜ , the thermal loads applied to plate (1) are equal to:

v9 F = ( )∆θ α α2 2 1 1

1

1 1 1

2

2 2 2

2L L

Le b E

Le b E

−

+ +ℜ

(cf. § V6.1)

By generalizing this relationship with the case of a bolted repair, we find thermal loads indirections x and y which apply to the doubler:

u12 Fxr therm. =

( )2

4

∆θ α αx sx

x rx

rx

r rx

x

sx

s sx

x

s r

r s

L L

Le h E

Le h E nr

−

+ +

ℜℜ (1)

b2

(2)

b1L2

L1



BOLTED REPAIRSFlux in panel - Loads per fastener due to Nx and Ny U 8

9

Fyr therm. =

( )2

4

∆θ α αys sy

yr ry

ry

r ry

yr

sy

s sy

ys

L L

Le h E

Le h E nr

−

+ +

We may thus deduce the thermal gross fluxes in the doubler:

Nxr therm. =

Fh

x

rx

r therm.

2

Nyr therm. =

F

hy

ry

r therm.

2

8 . ASSESSMENT OF FLUXES IN THE PANEL

Gross fluxes in the panel are deduced immediately form fluxes crossing the doubler:

u13 Nxs meca. = Nx

∞.meca - Nxr meca.

- Nxr therm.

Nys meca. = Ny

∞.meca - Nyr meca.

- Nyr therm.

Nxys meca. = Nxy

∞.meca - Nxyr meca.

9 . ASSESSMENT OF LOAD S PER FASTENER DUE TO THE TRANSFER OFNORMAL LOADS Nx AND Ny

Loads in fasteners are deduced from the geometry and from mechanical and thermalfluxes crossing the doubler and calculated previously.

A half repair may be represented as follows. The analysis being similar for directions xand y, indexes x and y have been removed to make the diagram as general as possible.

There are two possible cases (see drawings on next page):

- straight edge,- edge with chamfer.



BOLTED REPAIRSRepair with 1 row of fasteners U 9.1

9.1 . Repair with 1 row of fasteners

If the number of rows of fasteners is equal to 1, the load transmitted in the doubler isequal to loads transmitted by all fasteners on the row (single). The load per fastener isthen deduced immediately by the relationship:

f/fix. = fastenersofnumber

hxNx2

fastenersofnumberF rrr .therm.meca +=

F1F2F3F4F5F6Fr

Fs

F∞

A

er Er

es Es

2 hr N∞

+ .therm.meca

a a a a a

r

F1

F

F1

i = 1i = 2i = 3i = 4i = 5i = 6

A

er Er

es Es

2 hr N∞

+ .therm.meca

a a a a a

r



BOLTED REPAIRSRepair with 2 rows of fasteners U 9.2

9.2 . Repair with 2 rows of fasteners

The overall stiffness of row "i" fasteners is defined by (ri).

ri = ni x r

where ni is the number of fasteners on row "i".

If F1 and F2 correspond to loads transmitted by the rows of fasteners, the systemdisplacement resolution leads to the two following equations:

2222'2'21

222'2'22 SxE

AxFSxE

ASxE

AFSxE

Ar1

SxEAF =��

�

��

�++��

�

��

�++

11111'1'11

22 SxE

axFr1

SxEa

SxEaF

r1F =��

�

��

�+++��

�

��

� −

F1 + F2

F

F2

E2' S2'

E2 S2 E1 S1

r2 r1

E1' S1'

F1

A a





+��

��

�++��

�

��

�++

22'2'22

322'2'23 SxE

ASxE

AFr1

SxEA

SxEAF

2222'2'21 SxE

AxFSxE

ASxE

AF =��

��

�+

2222'2'21

222'2'22

33 SxE

axFSxE

aSxE

aFr1

SxEa

SxEaF

r1F =��

�

��

�++��

�

��

�+++��

�

��

� −

11111'1'11

22 SxE

axFr1

SxEa

SxEaF

r1F =��

�

��

�+++��

�

��

� −

F1 + F2 - F3

F

F3

E2' S2'

E2 S2 E2 S2

r3 r2

E2' S2'

F2

A a

E1 S1

r1

E1' S1'

F1

a





+��

��

�++��

�

��

�++

22'2'23

422'2'24 SxE

ASxE

AFr1

SxEA

SxEAF

2222'2'21

22'2'22 SxE

AxFSxE

ASxE

AFSxE

ASxE

AF =��

��

�++��

�

��

�+

+��

��

�+++��

�

��

� −

22'2'233

44 SxE

aSxE

ar1F

r1F

2222'2'21

22'2'22 SxE

axFSxE

aSxE

aFSxE

aSxE

aF =��

��

�++��

�

��

�+

2222'2'21

222'2'22

33 SxE

axFSxE

aSxE

aFr1

SxEa

SxEaF

r1F =��

�

��

�++��

�

��

�+++��

�

��

� −

11111'1'11

22 SxE

axFr1

SxEa

SxEaF

r1F =��

�

��

�+++��

�

��

� −

F1 + F2 + F3 + F4

F

F4

E2' S2'

E2 S2 E2 S2

r4 r3

E2' S2'

F3

A a

E2 S2

r2

E2' S2'

F2

a

E1 S1

r1

E1' S1'

F1

a



BOLTED REPAIRSwith a number of rows of fasteners greater than 4 U 9.5

9.5 . Repair with a number of rows of fasteners greater than 4

We may easily find the previous equation system type for a number of rows of fastenersgreater than 4.

However, we shall consider that, for a number of rows of fasteners greater than 6, therows greater or equal to 7 have an insignificant effect on load Fr transfer distribution (seediagram below).

Σ Fi

F

Fn

E2' S2'

E2 S2 E2 S2

rn r(n-1)

E2' S2'

F(n-1)

A a

E2 S2

r2

E2' S2'

F2

a

E1 S1

r1

E1' S1'

a

F1

Fr

F8 = 0 F7 = 0 F6 F5 F4 F3 F2

F

F1



BOLTED REPAIRSGeneral resolution method for direction x U 9.6

9.6 . General resolution method for direction x (see drawing below)

It consists in following the procedure indicated below:

Matrixed resolution of the equation system by assuming F = 1.

Assessment of unit loads Fi per row "i" of fasteners.

↓

Assessment of ratio k

u14 k = � =

+

n

1i i

xrx

F

hxNx2.therm.mecar

Assessment of effective loads transferred by each row "i" of fasteners

u15 Fxi = k x Fi

↓

Assessment of effective loads transferred by each fastener of row "i"

u16 fxi = i

xi

nFx15.1

where ni is the number of fasteners of row "i"

ni = *ahx2 x

r (no chamfer)

or

ni = 2*ahx2 x

r − (with chamfer for the first row)

Nx∞

+ .therm.meca

y

x

i = 1i = 2i = 3i = 4i = 5i = 6

fx/xiFxi



BOLTED REPAIRSAssessment of loads per fastener due to Nxy U 10

The fastener is identified by the letter "i" . "i" being the number of the row perpendicular tothe load. By definition, only full lines shall be considered and it shall be assumed that rownumber 1 is located next to the free edge of the doubler.

10 . ASSESSMENT OF LOADS PER FASTENER DUE TO THE TRANSFER OF SHEARLOADS Nxy

Loads in the fasteners are deduced from the geometry and from the mechanical originfluxes crossing the doubler.

Loads on fasteners due to the transfer of Nxyr meca. flux are equal to:

u17 fx/xyij = � =

n

1j

xy

jly

jlyxaxN.mecar

fy/xyij = � =

m

1i

xy

ilx

ilxx*axN.mecar

The fastener is identified by letters "i" and "j". "i" being the number of the row parallel tothe y-axis and "j" being the number of the row parallel to the x-axis. By definition, only fulllines shall be considered and it shall be assumed that row number 1 is located next to thefree edge of the doubler.

fx/xyij

fy/xyij

i = 1

lx3

i = 3

aly2

a*j = 2

j = 1

y

x

Nxy∞

.meca1 ≤ j ≤ n = 21 ≤ i ≤ m = 5



BOLTED REPAIRSJustifications U 11

1/2

11 . JUSTIFICATIONS

The repair justification consists of a notched and loaded plain plate calculation (for theparent skin and for the repair) and a check of the behavior of the existing damage.

The most highly loaded fastener holes (of the four angles) must be justified bysuperposing loads due to fluxes Nx meca.

(fx/x), Ny meca. (fy/y) and to flux Nxy meca.

(fx/xy and fy/xy).

This resulting load (f) must then be combined with fluxes N xs meca., Nys meca.

and Nxys meca. (for

the initial skin) or Nxr meca., Nyr meca.

and Nxyr meca. (for the repair) at the fastener considered.

On the other hand, the damage in the parent skin in the presence of fluxes Nxs meca., Nys meca.

and Nxys meca. at the repair center shall be justified.

General remark: It should be noted that there are two types of fastener arrangements: aso-called "square" arrangement and a so-called "staggered"arrangement. The "square" arrangement is preferred because the holecoefficient is limited.

fy/y

fy/xy

fx/xy fx/x

Nyr

REPAIR

Nxr

Nxyr

x

y

- fy/y

- fy/xy

- fx/xy- fx/x

Ny∝

PARENT SKIN

Nx∝

Nxy∝

f f



BOLTED REPAIRSJustifications U 11

2/2

Both diagrams below show which value of "a" should be applied in the calculation of loadsper fastener in each of these cases.

This theory was implemented with the desktop computing program "REPBOUL" (refer toinstructions).

a

"Square" arrangement

a

"Staggered" arrangement



BOLTED REPAIRSSummary flowchart U 12

12 . SUMMARY FLOWCHART

step No. 5ASSESS

FLUXES IN THEPANEL

meca. + therm.Nxs Nys Nxys

INITIALLOADING

Nx∝ Ny∝ Nxy∝

step No. 1

ASSESS THEDOUBLER

STIFFNESSESR r

x R ry R r

xy

ASSESS THEPANEL

STIFFNESSR s

x R sy R s

xy

step No. 3step No. 2

step No. 4

step No. 6

step No. 7

JUSTIFY THEFASTENER

HOLES OF THEDOUBLER

JUSTIFY THEDAMAGE

step No. 10step No. 9

JUSTIFY THEFASTENER

HOLES OF THEPARENT SKIN

step No. 8

DOUBLERGEOMETRY

CALCULATETHE STIFFNESSOF FASTENERS

r

ASSESSFLUXES IN THE

PANELmeca. + therm.

Nxr Nyr Nxyr

ASSESS THE RESULTINGLOAD

meca. + therm.fx/x fy/y fx/xy fy/xy

ASSESS THERESULTING

LOADmeca. + therm.

f



BOLTED REPAIRSExample U 13

1/11

13 . Example

Let the following repair be:

Mechanical characteristics original panelLay-up: 6/6/6/6Exs = 4878 daN/mm2

Eys = 4878 daN/mm2

Gxys = 1882 daN/mm2

es = 3.12 mmDoubler Mechanical characteristicsLay-up: 2/4/4/2Exr = 4008 daN/mm2

Eyr = 4008 daN/mm2

Gxyr = 2355 daN/mm2

er = 1.56 mmFastener characteristicsTotal number n = 32Stiffness r = 2000 daN/mm∅ = 3.2 mmFastener Elasticity modulus = 7400 daN/mm2

A =

18Ny∝ = - 32 daN/mm

Nxy∝ = - 20 daN/mm

a = 18

fA 1x xy/

fA 1y y/

fA 1y xy/

A

∅ = 20 mm

a* =

18

2 L

=

72 m

my r

2 L

=

108

mm

y s

h sy = 35 mm

2 L sx = 72 mm

2 h ry = 90 mm

x

y




2/11

Assessment of fluxes in the doubler and the panel below the doubler

This example does not include any thermal loads, we shall only cover mechanical fluxes.Therefore, notations shall not have any index.The first calculation step consists in determining both fluxes crossing the doubler (Nyr;Nxyr) which entails assessing panel and doubler stiffnesses with respect to both of theseload types, first of all. The fact that flux Nx∝ is zero has the following consequence Nxr =Nxs = 0.

Determination of doubler and panel stiffnesses

{u6}

mm/daN5138

42000x30

3645x56.1x4008

2000x3036x4

45x56.1x4008

Ryr =

+=

(30 is the total number of load-carrying fasteners in direction y).

{u7}

mm/daN986454

35x12.3x4878Rys ==

{u9}

mm/daN2518

82000x3256.1x2355

82000x32x56.1x2355

Rxyr =

+=

(32 is the total number of load-carrying fasteners in direction xy).

We have:

V = 7220

= 3.6

W = 10820

= 5.4




3/11

{u11}

η = 5.0

22

22

6.311

114.5

4.511

116.3

4.56.3

��

�

�

��

�

�

−+−+

��

�

�

��

�

�

−+−

+ = 0.94

{u10}

R sxy = 0.94 x 1882 x 3.12 = 5494 daN/mm

Determination of fluxes crossing the doubler

{u5}

Nyr = 51385138 9864+

Ny∝ = 0.34 Ny∝ = 0.34 (- 32) = - 10.96 daN/mm

{u8}

Nxyr = 25182518 5494+

Ny∝ = 0.31 Nxy∝ = 0.31 (- 20) = - 6.28 daN/mm

Determination of fluxes crossing the panel below the doubler

{u13}

Nys = - 32 - (- 10.96) = - 21.04 daN/mm

Nxys = - 20 - (- 6.28) = - 13.72 daN/mm




4/11

Assessment of loads on fastener A

Load due to Ny

Repair with 3 rows of fasteners: if F1, F2 and F3 corresponds to loads transmitted by therows of fasteners, the system displacement resolution leads to the three followingequations:

+��

��

�++

2000x51

90x12.3x487818

90x56.1x400818F3

+��

��

�+

90x12.3x487818

90x56.1x400818F2

90x12.3x487818xF

90x12.3x487818

90x56.1x400818F1 =��

�

��

�+

+��

��

�+++��

�

��

� −2000x51

90x12.3x487818

90x56.1x400818F

2000x51F 23

90x12.3x487818xF

90x12.3x487818

90x56.1x400818F1 =��

�

��

�+

90x12.3x.487818xF

2000x51

90x12.3x487818

90x56.1x400818F

2000x51F 12 =��

�

��

�+++��

�

��

� −

by assuming F = 1, loads F1, F2 and F3 per row "i" of fasteners are deduced from thematrix resolution.

F1 = 0.1366 daN

F2 = 0.0669 daN

F3 = 0.0273 daN




5/11

{u14}


k = 42750273.00669.01366.0

45x96.10)(x2 −=++

−

{u15]

Assessment of effective load transferred by row "i = 1" of fastener

Fyi = - 4275 x 0.1366 = - 584 daN

{u16}

Assessment of effective loads transferred by each fastener of row "i = 1"

fyi = 5

584x15.1 − = 1.15 x - 116.8 = - 134 daN

Load due to Nxy

{u17}

Loads on fasteners due to transfer of Nxyr flux are equal to:

fx/xyij = 18365454.18x28.6

++− = - 56.56 daN

fy/xyij = 36

36x18x28.6− = - 113.13 daN

The resultant shall be equal to ( )( ) 5.022 13413.11356.56 ++ = 254 daN and the slopeangle of the load shall have the value Arctg (- 247.13/- 56.56) = 77° (- 180°) = - 103°.




6/11

At this fastener A, fluxes are:

- on the panel

Nxs = 0 daN/mm

Nys = - 32 daN/mm

Nxys = - 20 daN/mm

- in the doubler

Nxr = 0 daN/mm

Nyr = 18

8.116− = - 6.49 daN/mm

Nxyr = - 6.28 daN/mm

fx/xy = - 57 daN A

fy/xy = - 113 daN

fy = - 134 daN- 103°

254 daNy

x




7/11

Now, let's assume that the carbon reinforcing plate is replaced by an Aluminium plate ofthickness er = 0.84 mm (the thickness was selected so that the doubler stiffness mayremain constant: 0.84 x 7400 = 1.56 x 4008 = 6252 daN/mm) and of coefficient expansionαr = 2.2 E-5/°C.

The expansion factor of the parent skin (isotropic T300/314 laminate) is equal to: αs = 1.4E-6/°C (refer to chapter § V 4.2 for the calculation of equivalent coefficient of expansion ofa laminate).

We shall look for thermal loads for an absolute temperature of + 74° C, which correspondsto a relative temperature with respect to the ambient temperature of ∆T = + 54° C (with aview to simplicity, mechanical loads shall be considered as zero).

By applying the relationship {u12}, we find the thermal loads in the doubler in directions xand y.

{u12}

In direction x :

Fxr therm. =

2000x264

4878x53x12.336

7400x63x84.069.27

)69.27x5E2.236x6E4.1(54x2

++

−−−

Fxr therm. = - 314 daN

(26 is the total number of load-carrying fasteners in direction x)

In direction y :

Fyr therm. =

2000x304

4878x35x12.354

7400x45x84.036

)36x5E2.254x6E4.1(54x2

++

−−−

Fyr therm. = - 260 daN

(30 is the total number of load-carrying fasteners in direction y)




8/11

Thermal fluxes in the doubler are deduced by dividing the previous results by widths 2 h rx

and 2 h ry :

Nxr therm. =

63x2314− = - 2.49 daN/mm

Nyr therm. =

45x2260− = - 2.89 daN/mm

Initial material characteristicsLay-up: 6/6/6/6Exs = 4878 daN/mm2

Eys = 4878 daN/mm2

Gxys = 1882 daN/mm2

es = 3.12 mmDoubler material characteristicsAluminiumExr = 7400 daN/mm2

Eyr = 7400 daN/mm2

Gxyr = 2846 daN/mm2

er = 0.84 mmh s

y = 35 mm

∅ = 20 mm

2 L

=

72 m

my r

2 L

=

108

mm

y s

2 L sx = 72 mm

2 h ry = 90 mm

x

y

2 h

=

126

mm

x r

h

= 53

mm

x s

2 L rx = 55.38 mm

Nyr therm. = - 2.89 daN/mm

Nxr therm. = - 2.49 daN/mm




9/11

Which allows calculation of thermal loads on the most highly loaded fasteners (those inangles).

In direction x:

Repair with two rows of fasteners: if F1 and F2 correspond to loads transmitted by the rowsof fasteners, the system displacement resolution leads to the two following equations (seechapter § U 9.2) :

+��

��

�++

12.3x126x487818

2000x61

84.0x126x740018F2

12.3x126x487818xF

12.3x126x487818

84.0x126x740018F1 =��

�

��

�+

12.3x126x487818xF

2000x71

12.3x126x487818

84.0x126x740018F

2000x61F 12 =��

�

��

�+++��

�

��

� −

By assuming that F = 1, loads F1 and F2 per row "i" of fasteners are deduced from thematrix resolution:

F2 x 1.157 E-4 + F1 x 3.237 E-5 = 9.387 E-6

F2 x (- 8.333 E-5) + F1 x 1.038 E-4 = 9.387 E-6

We find:

F1 = 0.1271 daN

F2 = 0.0455 daN

{u14}


k = 18181271.00455.0

63x)49.2(x2 −=+

−




10/11

{u15}

Assessment of effective load transferred by row "i = 1" of fasteners

Fxi = - 1818 x 0.1271 = - 231 daN

{u16}

Assessment of effective loads transferred by each fastener of row "i = 1"

fxi = daN37)33(x15.17

231x15.1 −=−=−

In direction y:

For calculation in direction y, just use results for the transfer of mechanical origin fluxes(distribution on rows of fasteners being independent from the load value - see § U 13 p.5).

We know that:

mechanical loads:

Nyr meca. = - 10.96 daN/mm → fyi = - 134 daN

therefore:

thermal loads:

Nyr therm. = - 2.89 daN/mm → fyi = daN35

96.1089.2x134 −=

−−−

The resulting load on angle fasteners is equal to:

Fresultant therm. = 37 35 512 2+ = daN




11/11

At this fastener, the resulting load is equal to 51 daN.

Thermal fluxes in the doubler have the following value:

Nxr therm. = ��

�

��

�

+−=

+−

−=−0455.01271.0

1271.0x4.2FF

Fx49.2mm/daN79.115.1x18

3721

1

Nyr therm. = ��

�

��

�

++−=

++−

−=−0273.00669.01366.0

1366.0x89.2FFFFx89.2mm/daN69.1

15.1x1835

321

1

fx therm. = 37 daN

fy therm. = 35 daNFresultant therm. = 51 daN

y

x



BOLTED REPAIRSReferences U




ESCANE - CIAVALDINI - TROPIS, Design of a bolted repair on a composite structure -determination of mechanical loads in the reinforcing piece, 1992, 440.192/92

TROPIS - SAVOLDELLI, Repair of carbon self-stiffened panels - justification incompression, 1989, 420.535/90

TROPIS - SAVOLDELLI, Repair of carbon self-stiffened panels - justification in tension,1990, 420.541/90

M. JANINI - P. CIAVALDINI, Design of a bolted repair on a composite structure -determination of mechanical loads in the reinforcing piece, 440. 129/92



V

THERMAL CALCULATIONS



THERMAL CALCULATIONSNotations V 1

1 . NOTATIONS

∆T, ∆θ: relative temperature (difference between effective and ambient temperatures)Tather., θather.: athermane temperatureTamb., θamb.: ambient temperatureTstruc., θstruc.: structure temperature

L: plates length

r: attachment or link stiffness

b1: plate width (1)e1: plate thickness (1)E1: plate modulus of elasticity (1)α1: plate thermal expansion coefficient (1)

b2: plate width (2)e2: plate thickness (2)E2: plate modulus of elasticity (2)α2: plate thermal expansion coefficient (2)

∆L: thermal expansion

∆L': mechanical elongation

F: global mechanical force of thermal origin

f: force on fasteners

τ: shear stress in adhesive

Gc: adhesive shear modulusec: adhesive thickness



THERMAL CALCULATIONSIntroduction V 2

2 . INTRODUCTION

A homogeneous anisotropic solid subjected to a uniform variation of temperature expandsdifferently in different directions. To characterize its behavior, several expansioncoefficients must be defined. In addition, strains and (or) stresses will appear dependingon the boundary conditions and the temperature range inside the solid.

A free composite plate subjected to a uniform variation in temperature will be subjected tostrains and stresses due to the different expansions of the fiber and the resin. Ascomposite plates are generally manufactured by curing at a temperature greater thanutilization θ, residual curing stresses appear in these plates. Although they can be high,they are generally not explicitly calculated but included implicitly into the calculation valuesdetermined on crosswise laminated plates (see § V 5.2).

As well, a variation in temperature applied to an assembly (attached or bonded) of plateswith different expansion coefficients leads to stresses and strains in these plates whichare added to those induced by mechanical loading.

The aim of this chapter is to study the stresses and strains of thermal origin forunidirectional fibers, composite plates, bimetallic strips and, lastly, aircraft structures,submitted to regulation environmental conditions.



THERMAL CALCULATIONSHooke - Duhamel law V 3

3 . HOOKE - DUHAMEL LAW

If a material is submitted to a mechanical load, the stress - strain relation can be written inits tensorial form:

v1 εij = ηijkl x σkl

ou cf. § V 5σij = λ ijkl x εkl

If a homogeneous, elastic and anisotropic solid which is free to deform is submitted to achange in temperature, the strain - temperature relation can be written in its tensorialform:

v2 εij = αij (T - To)

where

To: original temperature (uniform)T: modified temperature (uniform)

αij is the thermal expansion tensor which is a characteristic of the material.αij is symmetrical.For an orthotropic material in the orthotropy reference frame.

(α) =

α

α

α

l

ll

lll

o o

o o

o o

�

�

��

�

�

��

If the material is submitted to a mechanical load, the stress - strain - temperature relationcan be written:

εtotal = εmechanical + εfree thermal

v3 εij = ηijkl x σkl + αij (T - To)orσij = λ ijkl x εkl + βij (T - To)



THERMAL CALCULATIONSBehavior of unidirectional fiber V 4

where

(η): flexibility tensor(λ): rigidity tensor(α): thermal expansion tensor(β): thermal modulus tensor

T: absolute temperature appliedTo: reference temperature∆T: relative temperature

4 . BEHAVIOR OF UNIDIRECTIONAL FIBER

Let us take a unidirectional fiber defined by its longitudinal direction (l) and by itstransverse direction (t), it is a transverse isotropic and orthotropic material.

The unidirectional fiber will be characterized by two expansion coefficients in theorthotropic axis:

t

l

t (y)

l (x)

t (3)



THERMAL CALCULATIONSBehavior of a monolith plate V 5.1

1/3

The expansion coefficients are different in the two directions l and t:

αl: longitudinal expansion coefficient of unidirectional fiber in x-axis.αt: transverse expansion coefficient of unidirectional fiber (resin) in y- and z-axes.

Generally αl << αt (α) =

α

α

α

l

t

t

o o

o o

o o

�

�

��

�

�

��

5 . BEHAVIOR OF A FREE MONOLITHIC PLATE

5.1 . Calculation methodAs we saw in chapter E 3, the general relation between the strain sensor and the loadsensor of a monolith plate can be written in its matrix form (relation to be compared withthe relation v1):

N

N

N

M

M

M

x

y

xy

x

y

xy

=

A A A B B B

A A A B B B

A A A B B B

B B B C C C

B B B C C C

B B B C C C

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

ε

ε

γ

∂∂

∂∂∂∂ ∂

x

y

xy

o

o

o

wxwyw

x y

2

2

2

2

2

2

The sign conventions are as follows:z

y

x Mx > 0 Mxy > 0

My > 0

z

y

x Nx > 0 Nxy > 0

Ny > 0

SIGN CONVENTIONS FORBENDING

SIGN CONVENTIONS FORMEMBRANE




2/3

If the plate is submitted to a relative uniform temperature ∆T = (T - To), the previousexpression becomes (relation which is to be compared with relation v3):

v4

N

N

N

M

M

M

x

y

xy

x

y

xy

=

A B

B C

ε

ε

γ

∂∂

∂∂∂∂ ∂

x

y

xy

o

o

o

wxwyw

x y

2

2

2

2

2

2

- ∆t

α

α

α

α

α

α

Eh

Eh

Eh

Eh

Eh

Eh

x

y

xy

x

y

xy

2

2

2

where the thermoelastic behavior of the laminate is described by vector (αEh) which termsare equal to:

v5 αα ν α ν α α

ν νEh z z c E s E

x k kl l tl t t lt l t

lt tlk

n= −

+ + +−

�

��

�

��−=� ( ) ( ) ( )

1

2 2

1 1

αα ν α ν α α

ν νEh z z s E c E

y k kl l tl t t lt l t

lt tlk

n= −

+ + +−

�

��

�

��−=� ( ) ( ) ( )

1

2 2

1 1

αν α α α ν α

ν νEh z z c s E c s E

xy k kt lt l t l l tl t

lt tlk

n= −

+ − +−

�

��

�

��−=� ( ) ( ) ( )

11 1

αα ν α ν α α

ν νEh

z z c E s Ex

k k l l tl t t lt l t

lt tlk

n22

12 2 2

1 2 1= −

− + + +−

�

��

�

��

−

=�( ) ( )




3/3

αα ν α ν α α

ν νEh

z z s E c Ey

k k l l tl t t lt l t

lt tlk

n22

12 2 2

1 2 1= −

− + + +−

�

��

�

��

−

=�( ) ( )

αν α α α ν α

ν νEh

z z c s E c s Exy

k t lt l t l l tl t

lt tlk

n k22 2

11

2 1= −

− + + − +−

�

�

��

�

�

��

−

=�( ) ( )

where:

c ≡ cos(θ) where θ is the orientation of the fiber in the basic reference frame (o, x, y)s ≡ sin(θ) where θ is the orientation of the fiber in the basic reference frame (o, x, y)

El: longitudinal modulus of elasticity of the unidirectional fiber

Et: transverse modulus of elasticity of the unidirectional fiber

νlt: longitudinal/transverse Poisson coefficient of the unidirectional fiber

νtl = νlt EE

t

l: transverse/longitudinal Poisson coefficient of the unidirectional fiber

ply No. k

neutral plane

ply No. 1

zk - 1zk

ek

l

x

y

t

θ



THERMAL CALCULATIONSCuring stresses - expansion coefficients V 5.2

5.3

5.2 . Residual curing stresses

For thermosetting resin composites, the plates are made by juxtaposing different layerswith different characteristics.

These layers are manufactured simultaneously to the plate. We assume that when theresin cures, each layer is frozen in the state it is in at that time. Let Tp be the curingtemperature; the stresses in the plate can be considered as zero at this temperature.

To obtain the stresses at ambient temperature after cooling, apply relation v4.

If the plate does not have mirror symmetry, the coupling terms, (B) and αEh2 of v4 arenon zero; therefore, a uniform reduction in the temperature (from curing temperature toambient temperature) will create a strain in the plane and a curvature of the neutral plane.

In the same way, if there is in-plane coupling (terms A16, A26 non zero), that is if plies arenot equal in + or - α, angular distortion will occur during cooling after curing, the plate willbe "parallelogram" shaped.

5.3 . Equivalent expansion coefficients

The equivalent expansion coefficient vector αequi. (αx equi., αy equi., αxy equi.) of an orthotropiccomposite plate, with mirror symmetry without in-plane coupling, can be determined bythe following relation:

v6 (αequi.) = (A)- 1 x (αEh)

where terms Aij of the laminate rigidity matrix (A) can be determined by relation c6 ofchapter C.3.



THERMAL CALCULATIONSBimetallic strip theory V 6.1

1/4

6 . BEHAVIOR OF A SYSTEM CONSISTING OF TWO BEAMS WITH DIFFERENTEXPANSION COEFFICIENTS - BIMETALLIC STRIP THEORY

6.1 . Determining stresses of thermal origin

The aim of this subchapter is to study the mechanical influence of the temperature on asystem consisting of two beam elements with an infinitely rigid connection at their ends.

The following analysis is unidirectional. Furthermore, we shall neglect the secondarybending effects and the Poisson effects.

Let us take, therefore, two long plates (L >> b) with an infinitely rigid connection at theirends and with the following mechanical characteristics:

Plate (1):

- length: L- width: b- thickness: e1

- modulus of elasticity: E1

- expansion coefficient: α1 (§ V 5.3)

Plate (2)

- length: L- width: b- thickness: e2

- modulus of elasticity: E2

- expansion coefficient: α2 (§ V 5.3)




2/4

Initially, let us assume that the two plates are free at their ends. If we submit each one ofthem to a uniform relative temperature ∆T (in relation to the ambient temperature of thesetup), they will expand by the following lengths:

∆L1 = ∆T α1 L

where α1 ≠ α2

∆L2 = ∆T α2 L

Now, as these two plates are rigidly attached at their ends, they will deform by the samelength. Mechanical interaction forces of thermal origin (F1 and F2) are created:

F1 : force of plate (2) on plate (1)

F2 : force of plate (1) on plate (2)

(1)

(2)

Lb

(1)

(2)

∆L2

∆L1




3/4

As the system is globally in equilibrium, the force applied to plate (1) is the same as theone applied to plate (2) except for the sign: F1 = - F2.

The plate elongations due to these forces alone will be therefore:

∆L'1 = F LE e b

1

1 1 1

∆L'2 = F LE e L

F LE e b

2

2 2

1

2 2 2=

−

The equal length principal implies the following relation:

∆L1 + ∆L'1 = ∆L2 +∆L'2

After development and simplification, we obtain:

v7 F1 = ∆Γ

T ( )α α2 1−

F2 = ∆Γ

T ( )α α1 2−

where

v8 Γ = 1 11 1 1 2 2 2E e b E e b

+

(1)

(2)

∆L'2

∆L'1- F1

F1

- F1

F1F1

F2

∆L1

∆L2 ∆L'2 ∆L'1




4/4

Remark: For two plates with totally different geometries attached at the ends by anelement (one or several fasteners for example) of global rigidity ℜ , the previousrelation is slightly modified.

The equal length principle becomes:

∆ ∆ ∆ ∆L L L L F1 1 2 2 1

2 2 2 22

+ = + +ℜ

' '

Force of thermal origin is then equal to:

v9 F1 = ( )∆θ α α2 2 1 1

1

1 1 1

2

2 2 2

2L L

Le b E

Le b E

−

+ +ℜ

ℜℜ (1)

b2

(2)

b1L2

L1

∆L'12∆L'2

2∆L2

2

∆L1

2



THERMAL CALCULATIONSBolted system V 6.2.1

6.2.1.11/2

6.2 . Study of the link between two parts

In practice, the link between two parts is ensured either by fasteners or by bonding.

6.2.1 . Bolted or riveted joints

We shall assume that the plates are sufficiently long so that the thermal expansion cannotbe absorbed by the play (and the rigidity) of the fasteners (see previous remark).

The thermal force, which will have been calculated previously by relation v4, will beexpressed herein by letter F.

Several hypotheses can be put forward concerning the number of fasteners (of rigidity r)likely to take force F.

6.2.1.1 . Force F taken by one fastener

f = F

For information purpose, the table on the next page shows the various forces applied tothe structural elements for a splice and a doubler, these being submitted to tensile orcompression loads and at a positive or negative relative temperature.

L

r

F

- F

fb



THERMAL CALCULATIONSBolted system V 6.2.1.1

2/2

We will consider that the expansion coefficient of the upper material (aluminium forinstance) is higher than that of the lower material (isotropic carbon laminate for instance).

It is important to point out that the forces represented on the drawings are those applied tothe structure by the fasteners.

SPLICE DOUBLER

TEN

SIO

NC

OM

PRES

SIO

N ∆∆ ∆∆

T >

0

upper α > lower α upper α > lower α

∆∆ ∆∆T

< 0

upper α > lower α upper α > lower α

0° 0°

0° 0°

Table V6.2.1.1: Transfer of mechanical and thermal forces to the splices and doublers




6.2.1.3

6.1.2.3 . Force F taken by two fasteners

v10 f1 = F

A aA a rA a

A a r

Γ

Γ

++

+ +2

f2 = F - f1

6.2.1.3 . Force F taken by three fasteners

v11 f1 = Fr

A Aa r a r

rAa r

a

1 1 13

1 1

2+ + −��

��

+�

��

�

��

+ +��

��

ΓΓ

Γ

v12 f2 = Fa r1

3 + Γ

f3 = F - f1 - f2

r

F

- F

f1b

r

f2

A a

F

- F

f1b

f2

A

r

a

rr

a

f3




6.2.1.4 . Force F taken by four or more fasteners

v13 f1 = F rA A

a r a r

rAa r

aδ

1 1 13

1 1

2+ + −��

��

+�

��

�

��

+ +��

��

ΓΓ

Γ

where

v14 δ = 1 + n − 310

where n equals number of fasteners.

Remark: Relations v10 to v14 have been established analytically. It is howeverrecommended for a large number of fasteners to solve the problem by a matrixcalculation or a finite element model.

F

- F

f1b

A

r

a

f2

r

a

f3

rr

f4



THERMAL CALCULATIONSBonded joints V 6.2.2

1/2

6.2.2 . Bonded joints

For a bonded link, the (antisymmetrical) loading of the two plates is continuous from oneend of the system to the other; the main part of the loading is however at the start of thebonded link.

The maximum shear stress at the interface of the two plates can be written as follows:

v15 τMax. = Fb

L L E e E eE e E e

λ λ λcoth tanh2 2

1 1 2 2

1 1 2 2

�

��

�

�� + �

��

�

��

−+

�

��

�

��

where

v16 λ = Ge

E e E eE e E e

c

c

1 1 2 2

1 1 2 2

+

DoublerL

τMax.

- τMax.

F

- F

L

2

τMax.



THERMAL CALCULATIONSBonded joints V 6.2.2

2/2

where Gc is the shear modulus of the adhesive and ec its thickness.

This relation has been established by analogy with the bonded joint theory (see chapter S)where the distribution of the shear stresses in the adhesive joint is of the symmetrical typeand where the value of this stress, although negligible, is not zero in the center.

Symmetrical distribution of shear stresses: Bonded splice

Antisymmetrical distribution of shear stresses: Bonded doubler

++

τ ≈ C

+- τ = 0



THERMAL CALCULATIONSInfluence of temperature V 7.1

7.27.2.1

7 . INFLUENCE OF TEMPERATURE ON AIRCRAFT STRUCTURES

7.1 . General

The influence of the temperature on the aircraft structures is twofold:

- thermal stresses induced by the different expansion coefficients per unit length of thecomponents of the composite and metallic structures (spars/skins/rib, etc.) and alsobetween the fibers and the resin,

- reduction of the mechanical properties especially the resin and the adhesives (certainfibers are also sensitive to the temperature).

The demonstration of the resistance to the ultimate loads must be made in the mostpenalizing association case of the ultimate temperatures of the structure combined withselected design-critical mechanical loads.

We shall first of all define the various types of temperatures involved in the proceduredescribed in this chapter.

7.2 . Temperature of ambient air

7.2.1 . Temperature envelope

The static air temperature envelope to be considered on the ground and in flight are givenfor each aircraft (DBD: Data Basis Design); they depend on changes in regulatoryrequirements and aircraft operational limits.

For example, the maximum temperature to be considered on the ground was increased by10° C between the A320 (45°) and the A340 (55°). The minimum temperature on theground is - 54° C (see curves below).



THERMAL CALCULATIONSInfluence of temperature V 7.2.2

7.2.2.1

We must therefore determine the ultimate temperatures of the structure for all aircraftflight phases (static and fatigue). This is dealt with in this subchapter.

7.2.2 . Variation of ambient air temperature

7.2.2.1 . Ambient temperature on ground

The ambient temperature on the ground changes during the day. We will assume that itsvariation is homothetic to the quantity of heat Qϕ received by the ground (see chapter §V 7.3.1.2).

It therefore depends on the time of the day and the geographical location on earth(latitude ζ/type of atmosphere).

The table and curves below show change of ambient temperature on ground for a tropicalatmosphere (55° C at 12 h ≡ ISA + 40° C) and for a polar atmosphere (regulatory lowerlimit of - 54° C).

- 80 - 60 - 40 - 20 20 40 60- 54°C

10000

20000

25000

30000

35000

40000

45000

15000

5000

41100 ft

flight

ground

12500 ft

OAT(° C)

z(fts)

0- 5000



THERMAL CALCULATIONSInfluence of temperature V 7.2.2.2

We consider however that heat builds up during the day which explains why the "night"temperature (32° C for the tropical atmosphere) starts at 20 h and not at 18 h.

Time 6 h 7 h 8 h 9 h 10 h 11 h 12 h 13 hTamb. (° C) 32 38 43.5 48.3 51.9 54.2 55 54.5

Time 13 h 14 h 15 h 16 h 17 h 18 h 19 h 20 hTamb. (° C) 54.5 52.9 50.5 47.3 43.6 39.7 35.7 32

7.2.2.2 . Ambient temperature in flight

The ambient temperature in flight depends on the ambient temperature on the ground(see previous chapter) and the altitude z. From 0 to 40000 fts (troposphere), we generallyconsider that the temperature decreases on average 0.5° C for every 328 fts increase inaltitude with a lower limit of - 54° C.

The diagram below gives the ambient temperature at a given altitude for all ambienttemperatures on the ground.

Tamb. z ≈ - 2 E-3 x z + Tamb. gnd where ∀ z Tamb. z ≥ - 54° C

- 54° C

0

32° C

6 h 12 h 18 h 20 h 24 ht

T amb. ground

STANDARDATMOSPHERE

15° C = ISA

55° C = ISA + 40° C

Qϕ

TROPICALATMOSPHERE

POLARATMOSPHERE



THERMAL CALCULATIONSWall temperature V 7.3

7.3 . Wall temperature

The combined effects of the solar radiation in flight (optional effect) and the speed of theaircraft (Mach number M) lead to a significant increase in the wall temperatures whencompared with the temperature of the ambient air in flight.

0- 5 5 10 15 20 25- 60

- 40

- 20

0

20

40

60

Tamb. z(° C)

Altitude (x 1000 fts)

- 54 ° C

ISA

ISA + 40°



THERMAL CALCULATIONSInfluence of solar radiation V 7.3.1.1

7.3.1.21/3

7.3.1 . Influence of solar radiation

7.3.1.1 . Maximum solar radiation

It is maximum outside the atmosphere (z ≥ 36000 fts) and is equal to 1360 w/m2. Thisradiation is lower on earth due to the influence of the ozone layer, humidity and otherfactors.

At sea level (z = 0) and at 12-o-clock, Qs ≈ 1010 w/m2 in tropical areas.

Between these two points, we assume that Qs varies in a linear manner as a function ofthe altitude: Qsz ≈ 9.72 E-3 x z + 1010 (see curve below).

Remark: These values are unchanged between ISA + 35° C and ISA + 40° C.

7.3.1.2 . Solar radiation during the day

The quantity of heat Qϕ received by the ground depends on the quantity of heat Qsemitted by the sun and passing through the atmosphere (Qs = 1010 w/m2) and the angleof incidence ϕ between the light rays and the ground.

01000

Qs z(w/m2)

Altitude (x 1000 fts)5 10 15 20 25 30 35

1100

1200

1300

1400

1010

1360




2/3

This angle of incidence ϕ itself depends on time t (represented by angle ω on the drawing)and on the latitude ζ of the point under study.

ϕ = Arc (cos ζ x cos ω) = Arc (cos ζ x cos (15 t - 180)) for 18 h ≤ t ≤ 6 h

In tropical atmosphere (ζ ≈ 0°), this expression is simplified and becomes:

ϕ = Arc (cos 0° x cos ω) = 15 t - 180 for 18 h ≤ t ≤ 6 h

The diagram below shows, between 6 h and 18 h the "theoretical" change in the quantityof heat Qϕ that the ground receives for different types of atmospheres.

Ground

Qϕ

Qsϕ

Qϕ

Qsϕ

ϕ = 90°

ϕ

ζω

sola

r rad

iatio

n




3/3

Nevertheless, we will assume that during the night (from 18 h to 6 h), a certain quantity ofheat (≈ 280 W/m2 for a tropical atmosphere) is exchanged between the outside mediumand the structure ("night irradiation").

The table and curve below show the "regulatory" change in the quantity of heat Qϕ duringthe day in tropical atmosphere and at sea level.

The curve Qϕ = Qs x cosϕ has therefore been (arbitrarily) offset at 280 w/m2 for 6 h and18 h.

Time 7 h 8 h 9 h 10 h 11 h 12 h

Qϕϕϕϕ (w/m2) 636 838 929 980 1002 1010

Time 12 h 13 h 14 h 15 h 16 h 17 h

Qϕϕϕϕ (w/m2) 1010 1002 980 929 838 636

0 18 h6 h 12 h 24 h

STANDARDATMOSPHERE

LATITUDE ζ ≈ 45°

TROPICALATMOSPHERE

LATITUDE ζ ≈ 0°

714 w/m2

1010 w/m2

POLARATMOSPHERE

LATITUDE ζ ≈ 90°

Qϕ

t

0 18 h6 h 12 h 24 h

1010 w/m2

Q

t

Qs

Qϕ

280 w/m2Qs x cosϕ



THERMAL CALCULATIONSInfluence of speed - Temperature of structure V 7.3.2

7.3.3

7.3.2 . Influence of aircraft speed

The effect of the speed of the aircraft (friction of the air) increases the ambienttemperature in flight to a level called the athermane temperature.

The athermane temperature (or friction temperature) is the temperature at which thethermal flow exchange between the wall of the structure and the outside medium is zero.

To find the athermane temperature at structure stagnation point, the ambient temperatureat an altitude z must be multiplied by a coefficient which depends on the speed of theaircraft:

Tath. z = Tam. z x 1 1 2+−�

��

�

��

γγ

Mach

where γ = CpCv

= 1.4 γ: ratio between molar heat capacities (perfect gas constant).

where Cp and Cv are the heat capacities of the gas (in this case, of the air at the altitudeconcerned) at constant pressure and volume.

7.3.3 . Temperature of the structure

In the previous subchapters, we defined the various temperatures outside the structure(ambient temperature on ground, ambient temperature in flight, wall temperature andathermane temperature).

The aim of the next chapter is to determine the temperature of the various structuralelements.



THERMAL CALCULATIONSCalculation method V 7.3.3.1

1/2

7.3.3.1 . Calculation method

The temperature of each structural element depends on:

- the change of the athermane temperature which itself depends on the time, thealtitude, the speed of the aircraft and the type of atmosphere,

- the solar radiation at the altitude in question (generally not taken into account),

- the geometrical and thermal characteristics of the various elements comprising thestructure,

- the color of the exterior paint.

The calculation method consists in breaking down the structure into elements assumed tobe at a uniform temperature at time t and in writing the thermal equilibrium of each ofthese elements assuming that at time t = 0 all the structure has a uniform temperatureequal to the temperature of the ambient air.

A finite difference calculation enables the problem to be solved including in transientphase.The quantity of heat required to vary the temperature of each element by ∆T in the timeinterval ∆t is:

C x V x ∆T = Qa + Qc + Qi + (Qϕ - Qr)

where:

C = heat capacity of the materialV = volume of the elementQc = quantity of heat exchanged with the boundary layer by convectionQa = quantity of heat exchanged by conduction with adjacent elementsQi = quantity of heat exchanged with the inside medium (kerosene)Qϕ = quantity of heat received by solar radiationQr = quantity of heat lost by radiation



THERMAL CALCULATIONSCalculation method - Thermal characteristics V 7.3.3.1

7.3.3.22/2

For "tank" structures, three kerosene levels can be studied:

- all internal elements are in contact with the kerosene,- only the lower surface elements and a section of the spars are in contact with the

kerosene,- only the upper surface elements are not in contact with the kerosene.

This method enables us to find at time t the temperature of each element and therefore todeduce the forces and thermal stresses required for the fatigue and static justification.

Remark: The effects of the radiation of a section of the structure to another section neednot be taken into account in the calculations as this effect tends to make thetemperatures uniform.

7.3.3.2 . Thermal characteristics of the materials

- Conductivity

Carbon fiber

DrawingNomex Lightalloy

Tita-nium Transv.

25/25/25/25 50/20/20/10 10/20/20/50

4.5 5.2 3.8Thermalconductivity

(w/m/° C)0.1 143 6.7 3

440.092/92 22S.002.10502

- Specific heat

Carbon fiberNomex Light

alloyTita-nium Longi. Transv.

2 E6 2 E6Specific heat(J/m3/° C) 54 E3 2.6 E6 2.4 E6 22S.002.10502

440.092/92

B



THERMAL CALCULATIONSTemperatures of structure on ground V 7.3.3.3

- Paints

To calculate the temperatures on the ground (or optionally in flight), the absorptivity andemissivity coefficients of the paint must be defined.

The emissivity coefficient ε is generally taken as equal to 0.85 or 0.9.

The absoptivity coefficient α is related to the color of the paint.

Color white lightgray

lightyellow

darkgray

navyblue

α 0.2 0.5 0.5 0.65 0.8

7.3.3.3 . Temperatures of structure on ground

The first step consists in determining, with software PST2, change in the temperature ofthe structure on the ground during the day in order to evaluate the most critical initial flightconditions.

The study is generally conducted over a complete day from 0 h to 24 h but can beextended over two or three days in order to minimize the influence of the initil conditions(ambient temperature at 0 h: 32° C at ISA + 40° C).

Several sections of the structure with different thermal and geometrical characteristics willbe modeled.

As the absoptivity coefficient corresponds to the color of the paint used and the groundambient temperature and insolation curves, we calculate the maximum temperature onthe ground for each structural item during a day (see drawing below).



THERMAL CALCULATIONSTemperatures of structure in flight V 7.3.3.4

1/3

7.3.3.4 . Temperatures of structure in flight

As we saw previously, software PST2 enables us to define the changes in thetemperatures of the structural elements on the ground. We shall apply the same methodto find the changes of these elements in flight.

For this, we must define the aircraft operating scenarios. These scenarios depend on:

- typical aircraft mission (change of speed M and altitude z versus time),

- distribution of the missions during the day (generally not taken into account for thestatic justification),

- type of atmosphere,

- initial conditions defined previously (see chapter V 7.3.3.3).

- Typical mission

Several flight configurations or "missions" can be taken into account depending on theway in which the aircraft is used.

t

T struc. ground

0 12 h 24 h

ISA + 40° C

TEMPERATURE OFSTRUCTURAL

ELEMENT

AMB. TEMPERATUREON GROUND




2/3

Generally, a typical "mission" can be described as follows (A300-600):

- Daily use of the aircraft

A mean "typical" use is determined. For instance: for the A340, five 75' flights have beenconsidered distributed over 1 day as shown below:

Remark: For the static justification, we will not take into account the influence of theprevious flight on the initial conditions of the mission under study. Each flight willbe considered as isolated during the day. We shall therefore choose the mostpenalizing time for the start of the mission (generally 12 h for positivetemperatures).

STAR

T U

P AN

D T

AXI-O

UT

T.O

. + C

LIM

B

CLI

MB

250

KT

ACC

ELER

ATIO

N 2

50 K

T to

330

KT

CLI

MB

30 K

T/0.

78 M

CR

UIS

E 0.

78 M

DES

CEN

T 0.

82 M

/335

KT

DEC

ELER

ATIO

N 3

35 K

T to

250

KT

DES

CEN

T 25

0 KT

DEC

ELER

ATIO

N 2

50 K

T to

240

KT

HO

LD 5

mm

- 2

40 k

T

DEC

ELER

ATIO

N 2

40 K

T to

210

KT

DES

CEN

T 21

0 KT

APPR

OAC

H 2

10 K

T to

145

KT

LAN

DIN

G

TAXI

-IN

31000 fr

10000 fr

5000 fr

1500 fr

DIS

T.(n

m)

TIM

E(m

in)

5 1.3

2.4

2.3

10.3

10.8

78.8

87.2

664.

9

5.7

42.8

4.4

21.5

521

.6

3.3

12.5

512

5

0 400 1440

60' 60' 60' 60'

75' 75' 75' 75'75'

time (min)




3/3

- Definitions of atmospheres

Three types of atmospheres are to be considered in the analyses:

- standard,

- tropical,

- polar.

Remark: For the fatigue analysis, it is sometimes necessary (if we do not want to be tooconservative) to use a random distribution of the type of atmosphereencountered during one year in service. The "standard" operating time can forinstance be broken down as follows:

- ¼ of operation in polar atmosphere,

- ¼ of operation in standard atmosphere,

- ½ of operation in tropical atmosphere.

For the static justification, we shall choose the most penalizing atmosphere (tropical orpolar).

Com

posite stress manual

© AER

OSPATIALE

- 1999M

TS 006 Iss. B

THER

MAL C

ALCU

LATION

SBlock diagram

V7.4

7.4 . Recapitulative block diagram

- Standard- Tropical- Polar

Typicalatmosphere

Solar radiationon ground

0 h

Qϕ

12 h 24 h

Radiation inflight

Qϕ z

Altitudez

Ambient temp.on ground

Tamb. ground

12 h 24 h

Altitudez

55° C

- 54° C

x (1 + 0.18 Mach2)

Tathermane

Twall

- Conduction- Convection- Radiation

Altitudez

SpeedM

Temperatureof elements

STATICJUSTIFICATION

Altitudez

t

Ambienttemp. in flight

SpeedM

t

T struc.

t

t

Typicalflight

mission

24 h0 h

Dailyfrequency

FATIGUEJUSTIFICATION

: optional

to be superimposed on the mechanical effects

Tamb. z



THERMAL CALCULATIONSComputing softwares V 8

1/2

8 . COMPUTING SOFTWARES

For complex structures, there are three software programs to determine, on the one hand,the temperature ranges in the various structural elements and, on the other hand, theresulting thermal stresses and strains.

Software PST2:

It is used to determine the map of the temperatures of the structure over time versuschanges in external conditions and the speed of the aircraft (during a mission forexample).It is assumed that the temperature of the walls is equal to the athermane temperature, thatis the ambient temperature multiplied by the following factor: ��

�

��

�

γ−γ− 2Machx11 .

Knowledge of the thermal conductivity characteristics of the various materials (and thefluids contained in the structure: air, kerosene, etc.) is required together with the heattransfer coefficients between the various elements in order to evaluate the temperaturemap of the structure and its changes.

Remark: We can, for simplification reasons, consider that the complete structure has auniform temperature equal to the outside temperature or to the athermanetemperature.

Once the temperature range within the part has been determined, we must evaluate thestresses and strains of thermal origin. For this, in addition to the "manual" methodpreviously described (§ V 6), this can be done by two computing software programs.

Software PST1:

It is used to determine for a long structure of the dissimilar beam type (several differentmetals) submitted to any temperature field (uniform or not) the stresses of thermal originand the resulting longitudinal strains (x-direction).



THERMAL CALCULATIONSComputing softwares V 8

2/2

The part will be described by its current section which will be broken down into elementaryparts defined by their positions (center of gravity) and their geometrical and mechanicalcharacteristics (cross-section, inertia, modulus of elasticity, expansion coefficient, etc.).

Software PST4:

This calculation sequence is used to determine the thermoelastic stresses of a structureschematized by finite elements for any temperature range (a temperature is associatedwith each node of the structure).

θ

FINITEELEMENT

MODEL

x

BEAMTYPE PART

cdgSlEα

CURRENTSECTION



THERMAL CALCULATIONSFirst example V 9.1

1/4

9 . EXAMPLE

9.1 . First example: thermal stresses and forces in a bolted repair

For the following bolted repair:

a =

15

b = 15

a =

15A

= 15

b =

15f Ax

fA f Ay

a = 15A = 15

y

x




2/4

Initial material:Material: T300/BSL914Lay-up: 6/6/6/6Exxs = 4878 daN/mm2

Eyys = 4878 daN/mm2

es = 3.12 mmαs = 3.5 E-6 mm/mm/° C

Doubler material:Material: AluminiumExxr = 7400 daN/mm2

Eyyr = 7400 daN/mm2

er = 2 mmαr = 24 E-6 mm/mm/° C

Fasteners:D = 3.2 mmE = 10000 daN/mm2

Rigidity of fasteners:

{u1}

��

��

�++=

12.3x48781

2x740018.0

2.3100005

r1

r = 3800 daN/mm

If there are no loads of mechanical origin, what are the forces on the fasteners and theflows at the center of the doubler and the panel if the panel and its repair are heated to anabsolute temperature of 70° C (∆T = 50° C) ?The fasteners subjected to the highest loads are the ones located in the corners of therepair. We shall therefore study fastener A.

As the thermal load (relevant to a strip of material of width b = 15 mm that we will consideras a bimetallic strip) is (in first approximation) independent of the length (relation v7), thecomponents in the x- and y-directions are equal.




3/4

{v8}

1daN6E88.85x2x7400

15x12.3x4878

1 −−=+=Γ

{v7}

daN1156E8.8

)6E246E5.3(50F −=−

−−−= → compression force applied to the doubler

F = 115 daN

F = 115 daN

y

x

F =

115

daN

F =

115

daN

F = 115 daN

A




4/4

We can deduce the force on the fastener A in x-direction:

{v10}

daN8271.0x115

3800x15x15151526E88.8

3800x15x1515156E88.8

115f xA ==

++−

++−=

and the force on fastener A in the y-direction:

{v11}

daN696.0x1156E88.8x15

38001

1515

38001

3800x15x153800311

15156E88.8x15

38001

115f2

yA ==

��

��

� −++

��

��

�

+��

��

� −+−+=

The global force on fastener A is therefore:

fA = 69 82 1072 2+ = daN

We can deduce the flows Nxr and Nyr of thermal origin in the center of the doubler:

Nxr = mm/daN67.715115 −=−

Nyr = mm/daN67.715115 −=−

We can deduce the flows Nxs and Nys of thermal origin in the parent skin:

Nxs = mm/daN67.715

115 =

Nys = mm/daN67.715

115 =



THERMAL CALCULATIONSSecond example V 9.2

1/2

9.2 . Second example: thermal stresses in a bonded joint

Let us suppose that the doubler is bonded and not bolted with an adhesive with thefollowing mechanical characteristics:

Gc: 300 daN/mm2

ec: 0.05 mm

{v16}

3E89412.3x4878x2x740012.3x48782x7400x

05.0300 −=+=λ

We can deduce the maximum shear stress at point A in x-direction:

{v15}

��

��

��

��

��

��

�τ+−−

+−−

=12.3x48782x740012.3x48782x7400

x2

75x3E894tanh

275x3E894

coth105

3E894x805xMax

hb75.6xMax =τ

where 805 (daN) is the global thermal load on the plate (115 x 7) and 105 (mm) is the totalheight of the plate.

The maximum shear stress in y-direction is equal to:

{v15}

��

��

��

��

��

��

�τ+−−

+−−

=12.3x48782x740012.3x48782x7400

x2

105x3E894tanh

2105x3E894

coth75

3E894x575yMax

hb75.6yMax =τ

where 575 (daN) is the global thermal load on the plate (115 x 5) and 75 (mm) is the totalwidth of the plate.



THERMAL CALCULATIONSSecond example V 9.2

2/2

We can therefore deduce the global shear stress at point A:

τMax. = hb54.975.675.6 22 =+

This value is to be compared with the permissible value for the adhesive which generallyis equal to 8 hb. If the plastic adaptation of the adhesive is not taken into account, therepair will unstick.



THERMAL CALCULATIONSThird example V 9.3

1/7

9.3 . Third example: internal curing stresses in a laminated plate

Let us consider a laminate consisting of four carbon tapes (T300/BSL914) with followinglay-up: 0°/45°/135°/90°/90°/135°/45°/0°.

For simplification reasons, we will determine the internal stresses in the fiber at 0° due tothe increase in temperature on curing then to cooling:

∆T = Tambient - Tcuring = 20 - 180 = - 160° C

The mechanical characteristics of the unidirectional fiber are:

El = 13000 hb (130000 MPa)Et = 465 hb (4650 MPa)νlt = 0.35νtl = 0.0125Glt = 465 hb (4650 MPa)ply thickness = 0.13 mmtotal thickness = 2.6 mmαl = - 1 E-6 mm/mm/° Cαt = 40 E-6 mm/mm/° C

0°45°

135°90°90°

135°45°0°

k = 8k = 7k = 6k = 5k = 4k = 3k = 2k = 1

z = 0.52z = 0.39z = 0.26z = 0.13z = 0z = - 0.13z = - 0.26

z = - 0.52z = - 0.39

tl

x

yz




2/7

Let us calculate the membrane thermoelastic behavior coefficients of the laminate:

{v5}

( ) ( )��

�

�

��

�

�

−

−+−−+−+−−=α

0125.0x35.01

6E40)6E1(x35.0465x216E40x0125.0)6E1(13000x2013.02xEh

( ) ( ) ( )��

�

�

��

�

�

−

−+−−−+−+−−+

0125.0x35.01

6E40)6E1(x35.0465x2707.06E40x0125.0)6E1(13000x2707.013.02

( ) ( )��

�

�

��

�

�

−

−+−−+−+−−+

0125.0x35.01

6E40)6E1(x35.0465x2707.06E40x0125.0)6E1(13000x2707.013.02

( ) ( )��

�

�

��

�

�

−

−+−−+−+−−+

0125.0x35.01

6E40)6E1(x35.0465x206E40x0125.0)6E1(13000x2113.02

= 2 (0.002407375 + 0.000779095 + 0.000779095 - 0.000848713) = 6.232 E-3 daN mm-1 ° C -1

( ) ( )��

�

�

��

�

�

−

−+−−+−+−−=α

0125.0x35.01

6E40)6E1(x35.0465x206E40x0125.0)6E1(13000x2113.02yEh

( ) ( ) ( )��

�

�

��

�

�

−

−+−−+−+−−−+

0125.0x35.01

6E40)6E1(x35.0465x2707.06E40x0125.0)6E1(13000x2707.013.02

( ) ( )��

�

�

��

�

�

−

−+−−+−+−−+

0125.0x35.01

6E40)6E1(x35.0465x2707.06E40x0125.0)6E1(13000x2707.013.02

( ) ( )��

�

�

��

�

�

−

−+−−+−+−−+

0125.0x35.01

6E40)6E1(x35.0465x216E40x0125.0)6E1(13000x2013.02

= 6.232 E-3 daN mm-1 ° C -1

( ) ( )��

��

�

−

−+−−−−+−−=α

0125.0x35.01

6E40x0125.0)6E1(13000x1x06E40)6E1(35.0465x1x013.02xyEh

( ) ( ) ( ) ( )��

��

�

−

−+−−−−−+−−−+

0125.0x35.01

6E40x0125.0)6E1(13000x707.0x707.06E40)6E1(x35.0465x707.0x707.013.02

( ) ( )��

��

�

−

−+−−−−+−−+

0125.0x35.01

6E40x0125.0)6E1(13000x707.0x707.06E40)6E1(x35.0465x707.0x707.013.02

( ) ( )��

��

�

−

−+−−−−+−−+

0125.0x35.01

6E40x0125.0)6E1(13000x0x16E40)6E1(x35.0465x0x113.02

= 2 (- 0.001627552 + 0.001627552) = 0




3/7

The terms Aij of the rigidity matrix of the laminate (in daN/mm) were calculated in chapterE 4:

A11 = 2779 x 2 = 5558A12 = 821 x 2 = 1642A13 = 0A21 = 821 x 2 = 1642A22 = 2779 x 2 = 5558A23 = A31 = A32 = 0A33 = 978 x 2 = 1956

As external loads are zero, the relation v1 can be written as follows:

{v4}

5558 1642 0

1642 5558 0

0 0 1956

ε

ε

γ

x

y

xy

= -160

0

3E232.6

3E232.6

−

−

We can deduce the thermal expansions (in mm/mm) of the laminate in the referenceframe (x, y):

ε

ε

γ

x

y

xy

=

0

5E85.13

5E85.13

−−

−−

The results above are the apparent thermal expansions of the plate in the reference frame(x, y) and the expansions of the fiber at 0° in its own reference frame (l, t):




4/7

(εl, t, 0°) =

0

5E85.13

5E85.13

−−

−−

To determine the internal stresses applied to the fiber at 0°, we must find what theexpansions of this fiber would be if it was isolated and free from all strains:

The thermoelastic coefficients of the unidirectional fiber at 0° in its reference frame (l, t)are equal to:

{v5}

( ) ( ) ( )0125.0x35.01

6E40)6E1(x35.0465x206E40x0125.0)6E1(13000x2113.026.0Eh

l −

−+−−+−+−−−=α

= - 8.487 E-4 daN mm-1 ° C-1

( ) ( ) ( )0125.0x35.01

6E40)6E1(x35.0465x216E40x0125.0)6E1(13000x2013.026.0Eh

t −

−+−−+−+−−−=α

= 2.407 E-3 daN mm-1 ° C-1

( ) ( ) ( )0125.0x35.01

6E40x0125.0)6E1(13000x0x16E40)6E1(x35.0465x0x113.026.0Eh lt −

−+−−−−+−−−=α

= 0

The coefficients Aij of the rigidity matrix of the unidirectional fiber at 0° are:

A11 = 13057 x 0.13 = 1697A12 = 163 x 0.13 = 21.19A13 = 0A21 = 21.19A22 = 60.71A23 = 0A31 = 0A32 = 0A33 = 60.45




5/7

Under the influence of a ∆T = - 160° C, the thermal expansions (in mm/mm) of an isolatedfiber would satisfy the following relation:

45.6000

071.6019.21

019.211697

ε

ε

γ

l

t

lt

= -160

0

3E407.2

4E487.8

−

−

We therefore obtain:

ε

ε

γ

l

t

lt

=

0

3E4.6

5E16

−−

−

It is important to specify that in this case, the fiber is submitted to no internal thermalstresses as free from all strains.

By simply calculating the difference, we find the "expansions" (in mm/mm) of the fiber at0° if it was submitted to the stresses of thermal origin alone:

ε'fiber thermal stresses 0° = εthermal of plate - εthermal of 0° fiber alone

These three types of expansions must be determined in the same reference frame. Forthe fiber at 0°, no change of reference frame is required. If we had wanted to study forinstance the internal stresses in the fiber at 45°, we would have had to determine theexpansions of the plate in a frame oriented at 45° in relation to the reference frame(relation c7).

We therefore obtain the expansions (in mm/mm) of the fiber at 0° due only to the thermalstresses which are applied to it:




6/7

ε

ε

γ

'

'

'

l

t

lt

=

0

5E85.13

5E85.13

−−

−

-

0

3E4.6

5E16

−−

−

=

0

3E259.6

5E85.29

−

−

We can determine, by relation c8, the stresses (in hb) of thermal origin applied to the fiberat 0° (and therefore to all fibers as all directions are equivalent):

{c8}

σ

σ

τ

'

'

'

l

t

lt

=

13057 163 0

163 467 0

0 0 465

0

3E259.6

5E85.29

−

−−

=

0

88.2

88.2−

In fact, these stresses are not to be taken into account in the justification of the laminateas they are indirectly taken into account when determining the permissible values for theunidirectional fiber of the material.

It is also possible, by relation v6, to determine the equivalent membrane expansioncoefficients of the laminate (in mm/mm/° C):

{v6}

.equixy

.equiy

.equix

α

α

α

=

5558 1642 0

1642 5558 0

0 0 1956

1−

0

3E232.6

3E232.6

−

−




7/7

.equixy

.equiy

.equix

α

α

α

=

3E511.000

04E972.14E583.0

04E583.04E972.1

−

−−

−−−

0

3E116.3

3E116.3

−

−

α

α

α

x équi

y équi

xy équi

.

.

.

=

0

7E658.8

7E658.8

−

−

This result can easily be checked:

εx = ∆T αx equi.

- 1.385 E-4 = - 160 8.658 E-7



THERMAL CALCULATIONSFourth example V 9.4

1/8

9.4 . Fourth example

Calculation of the temperatures associated with a typical A340 mission on a section of anA340 aileron at bearing 1 (see note 440.092/92).

In agreement with ACJ 25.603, we must, for the structural justification, associate the mostpenalizing environmental conditions (temperature and humidity) with the calculationcases. Here, we shall deal only with the temperature case. The atmosphere chosen willbe tropical.

The Loop 1A calculation cases corresponding to the various aircraft configurations isgiven in the table below:

Cas Conf. Speed Z(ft) Mn Pdyn

(daN/m2)nz(/q)

m(kg)

Clmax

CLwf(p)

AOA(°)

01210111213141516171819

CLLCCCCCCCCCC

/VFEVFEVCVCVCVCVCVCVDVDVDVD

3500000000

299002990029900

00

2925029250

0.820.280.28

0.4990.4990.4990.860.860.86

0.5520.5520.930.93

1122.2556.1556.11766176617661565156515652161216118851885

10.133

22.521

2.521

2.52

2.52

206911186000186000250000250000250000250000250000250000250000250000250000250000

/2.72.7

1.2251.2251.225

111

1.181.18

11

0.5121.2831.8460.9420.7580.3921.0670.8590.4440.770.6210.9

0.725

3.123.469.8610.037.9

3.648.336.593.1

8.076.369.047.19

We shall choose to study case Vc/Vd for an altitude z ≈ 29500 fts (cases 13, 14, 15, 18,and 19) the typical mission of which can be represented by the following diagram (time,speed, altitude):




2/8

First step: Consists in determining, from the meteorological data, the change in ambienttemperature on the ground for a tropical atmosphere. This temperature depends on thequantity of heat due to the solar radiation Qϕ.

By considering the change in the angle of incidence of the rays during the day, we find avariation of Qϕ versus the angle of incidence ϕ and therefore versus the time (Qϕ beingtaken as equal to 280 between 18 h and 6 h)

t 6 h 6 h 45 7 h 7 h 20 8 h 9 h 10 h 10 h 20 11 h 11 h 30 12 h

Qϕϕϕϕ 280 545 636 727 838 929 980 989 1002 1008 1010

t 12 h 12 h 30 13 h 13 h 40 14 h 15 h 16 h 16 h 40 17 h 17 h 15 18 h

Qϕϕϕϕ 1010 1008 1002 989 980 929 838 727 636 545 280

250 kts

Speed(Mach)

t(mm)

z(fts)

0 3 603540

10000

20000

30000

0.5

1

Altitude32000 fts

Speed

0.72

300 kts




3/8

Second step: We will deduce the maximum ambient temperatures on the ground (z = 0)throughout the day (ISA + 40° C) which corresponds to a maximum temperature at 55° Cat midday in a tropical atmosphere (the non-symmetry of the curve in relation to 12 h isexplained by taking into account the buildup of heat during the day):

t 6 h 6 h 45 7 h 7 h 20 8 h 9 h 10 h 10 h 20 11 h 11 h 30 12 h

Tamb. 32 36.6 38 39.9 43.5 48.3 51.9 52.8 54.2 54.8 55

t 12 h 30 13 h 13 h 40 14 h 15 h 16 h 16 h 40 17 h 17 h 15 18 h 19 h 20h

Tamb. 54.9 54.5 53.6 52.9 50.5 47.3 44.9 43.6 42.7 39.7 35.7 32

30 6 9 12 15 18 21 24

200

0

400

600

800

1000

1200

Qϕ(W/m2)

t(time)




4/8

Third step: Consists in evaluating the temperature on the ground of the various structuralitems during a day in order to determine the most penalizing departure time. Thiscalculation was done with software PST2 over three days so as to eliminate the effects ofthe initial conditions.

The calculation method consists in dividing the structure into elementary sections (uppersurface panel, lower surface panel, spar, leading edge, fittings) which initially have auniform temperature (temperature of the ambient air) and in determining their changesbefore takeoff according to the three following phenomena:

- conduction with adjacent elements,- convection with surrounding media (turbulences, kerosene),- solar radiation (α = 0.5; ε = 0.85)

30 6 9 12 15 18 21 24

10

0

20

30

40

50

60

T amb.(° C)

t(time)




5/8

By integrating all these data, we obtain the curve below which represents, over a period of24 hours, the change in the temperature of the various structural elements.

1 ambient air (ISA + 40)

39

3244

15

49

56 52

Tit.Alu.

11

100

20

T struc.(° C)

0 12 24

32/39/44

52/56

11 15

49

55° C

1

t




6/8

We find (as was predictable) that the most unfavorable time for high temperatures is 12 h.We will therefore consider that the aircraft's mission starts at this time.

Fourth step: consists, again with software PST2, in determining the change in temperatureof each part during the mission itself by putting forward the (conservative) hypothesis thatthe ambient temperature on the ground is constant and equal to 55° C throughout themission (ISA + 40° C) and by taking as initial values for the structural elements thepreviously defined temperatures at 12 o'clock.

The curves below represent the change in temperature of each element of the part duringthe mission considered.




7/8

We can see that all the temperatures of the structural elements tend asymptotically to theathermane temperature (skin temperature) which depends on:

- the ambient temperature on the ground (considered as being independent of time): 55° C,- the speed of the aircraft expressed in Mach number: M,- the altitude z.

0

100

55

1 2

3540

uppersurfacepanel

spar

lower surfacepanel

athermanetemperature

Taxi

Clim

b

Clim

b25

0 kt

s

TO Clim

b30

0 kt

s

Clim

b30

0 kt

s

Clim

b0.

72M

ACC

ACC

Clim

b0.

72M

Des

cent

T struc.(° C)

t(s)




8/8

Tather. ≈ Tamb. x (- 1.88 E-3 x z + 55) x (1 + 0.18 M2)

Fifth step: Consists in combining throughout the mission, the loads of mechanical origin(aerodynamic) and the loads of thermal origin. This analysis (not covered by this chapter)gave two design-critical cases (see previous curve):

- point �: VCM = 0.86t = 3426 s

- point �: VDM = 0.92t = 3510 s

Remark: We could have also conducted a study on the negative temperature range werelower limit imposed by regulations is - 54° C.

We however observed that the effect of the speed of the aircraft on theathermane temperature (1 + 0.18 M2) implied high temperatures in flight. Wetherefore limited the justification to - 54° C.



THERMAL CALCULATIONSReferences V


J. HOEB, No. 22/S002.10502, Environment: justification of the count and of test

PV No. 46534 - DCR/L, Thermal characterization of composite T300/914

PV No. 31/807/69, Study of thermo-optical factors

PV No. 50879/88, Determination of "absortance" and "emittance" factors of ATR 72 finitionpaint

A. TROPIS, A340 ailerons - Substantiation of tests and calculation environmentalconditions, AS 440.092/92

P. MEUNIER, Aircraft S1 and S2 surface definition report A300-600, 26 X 002 10558/032



W

ENVIRONMENTAL EFFECT



X

NEW TECHNOLOGIES



Y

STATISTICS



Z

MATERIAL PROPERTIES



SUMMARY

Z . MATERIAL PROPERTIES1 . Prepreg unidirectional tapes

1.1 . First generation Epoxy high strength carbon1.2 . Second generation Epoxy intermediate modulus carbon1.3 . Epoxy R glass1.4 . Bismaleimide carbon

2 . Fabrics2.1 . Epoxy resin prepreg

2.1.1 . Carbon2.1.2 . Glass2.1.3 . Kevlar2.1.4 . Hybrid2.1.5 . Quartz polyester hybrid

2.2 . Phenolic resin prepreg2.2.1 . Carbon2.2.2 . Glass2.2.3 . Kevlar2.2.4 . Fiberglass carbon hybrid2.2.5 . Quartz polyester hybrid

2.3 . Bismaleimide resin prepreg2.3.1 . Carbon

2.4 . Wet Lay-Up Epoxy (for repair)2.4.1 . Carbon2.4.2 . Glass2.4.3 . Kevlar2.4.4 . Fiberglass carbon hybrid2.4.5 . Quartz polyester hybrid

3 . R.T.M.3.1 . Epoxy resin

3.1.1 . Carbon3.2 . Bismaléimide resin3.3 . Phenolic resin

4 . Injection moulded thermoplastics4.1 . Carbon

4.1.1 . PEEK4.1.2 . PEI4.1.3 . Polyamide4.1.4 . PPS4.1.5 . Polyarylamide

4.2 . Glass4.2.1 . PEEK4.2.2 . PEI

5 . Long fibre thermoplastics5.1 . Carbon

5.1.1 . PEEK5.1.2 . PEI

5.2 . Glass



6 . Arall-Glare7 . Metallic matrix composite materials (CMM)8 . Adhesives

8.1 . Epoxy8.2 . Phenolic8.3 . Bismaleimide8.4 . Thermoplastic

9 . Honeycomb9.1 . Nomex

- Hexagonal cells- OX-Core- Flex-Core

9.2 . Fiberglass honeycomb- Hexagonal cells- OX-Core- Flex-Core

9.3 . Aluminium honeycomb10 . Foams



CLASSIFICATION OF FIBROUS MATERIALS/RESIN MATRICES IN ALPHABETICAL ORDER

Chapters 1 to 5

TRADENAME OF MATERIAL CHAPTER TRADENAME OF MATERIAL CHAPTER

1581/501181 + EPOXY resin240/38/644250/44/7697781/V9137788/54/M14796/95/M14APC2 (AS4/PEEK)CD282/PEI DE TENCATEE4049/RTM6EHA 250-33-50EHA 250-33-60ES03/3752ES36D/90120ES36D/90285ES36D/91581F155/T120G1151/RTM6G803/145-4G803/40/V200G803/501G803/914G806 + T120/5052-9390-9396-501G806/501G806/5052G806/9396G814/501G814/V913

Z 2.4.2.1Z 2.4.3.1Z 2.2.2.1Z 2.2.2.2Z 2.1.2.1Z 2.1.3.1Z 2.1.3.2Z 5.1.1.1Z 5.1.2.1Z 3.1.3

Z 2.2.3.3Z 2.2.3.3Z 2.1.4.2Z 2.1.3.4Z 2.1.3.3Z 2.1.3.5Z 2.1.2.2Z 3.1.1

Z 2.1.1.3Z 2.2.1.1Z 2.4.1.1Z 2.1.1.1Z 2.2.4.1Z 2.4.1.2Z 2.4.1.3Z 2.4.1.4Z 2.4.1.5Z 2.1.1.5

G815/V913G874/V250G973/913G986/RTM6GB305/DA3200 (*)GB305/XB5142 (*)GF520/LY564-1 + HY2954 (*)GF630/RTM6HF360/LY564-1 + HY2954 (*)HTA7 (6K) EH25 (46280/25/42 %)HTA7 EH25IM7/977-2IXEF 1022IXEF C36KEVLAR economic 285 + GENIN 90285/ES36D EPOXY resinKEVLAR economic 285 + BROCHIER 1454/914 resinM14/1237RYTON R04T300/914T300/N5208T800H DA508TULTEM 2310VICOTEX 108/788VICOTEX 145.2/788VICOTEX 145.4/796VOCPTEX 145.4/914VICOTEX 250/788VICOTEX 250/796

Z 2.1.1.4Z 2.2.4.2Z 2.1.4.1Z 3.1.2Z 3.1.4Z 3.1.6Z 3.1.8Z 3.1.5Z 3.1.7

Z 2.1.1.2Z 1.1.3Z 1.2.2

Z 4.1.5.1Z 4.1.5.2Z 2.1.3.10Z 2.1.3.11Z 2.1.5.1Z 4.1.4.1Z 1.1.1Z 1.1.2Z 1.2.1

Z 4.1.1.1Z 2.1.3.6Z 2.1.3.7Z 2.1.3.8Z 2.1.3.9Z 2.2.3.1Z 2.2.3.2



CLASSIFICATION BY TYPE OF FIBROUS MATERIALS/RESIN MATRICES

RESIN TYPE OF MATERIAL CURINGTEMPERATURE TRADENAME OF MATERIAL PQ CHAPTER PAGES

Prepreg carbonunidirectional tapes

1st generation high strength

2nd generation intermediarymodulus

180° C

180° C

180° C

180° C

190° C

T300/914

T300/N5208

HTA7 EH25

T800 H DA508T

IM7/977-2

10139-261-01

10139-250-01

10139-501-01

Z 1.1.1

Z 1.1.2

Z 1.1.3

Z 1.2.1

Z 1.2.2

1-10

1

1

1

1-3

Prepreg carbon fabrics

1st generation high strength

180° C

180° C

120° C

125° C

125° C

G803/914

HTA7 (6K) EH25 (46280/25/42 %)

G803/145-4

G815/V913

G814/V913

10139-353-01

10139-353-02

10139-302-03

DASA

10139-450-01

Z 2.1.1.1

Z 2.1.1.2

Z 2.1.1.3

Z 2.1.1.4

Z 2.1.1.5

1-5

1-2

1-2

1-2

1-2

EPOXY

Prepreg glass fabrics 120° C

125° C

7781/V913

F155/T120

10516-022-04

DASA-BAE

Z 2.1.2.1

Z 2.1.2.2

1

1



CLASSIFICATION OF BY TYPE OF FIBROUS MATERIALS/RESIN MATRICES


Prepreg Kevlar 125° C

125° C

120° C

120° C

120° C

175° C

120° C

125° C

125° C

120° C

120° C

7788/54/M14

796/95/M14

ES36D/90285

ES36D/90120

ES36D/91581

VICOTEX 108/788

VICOTEX 145.2/788

VICOTEX 145.4/796

VICOTEX 145.4/914

KEVLAR economic 285+ GENIN 90285/ES36D EPOXY resin

KEVLAR economic 285+ BROCHIER 1454/914 resin

10139-142-00

10139-140-00

10139-143-02

10139-142-05

10139-162-01

10139-142-01

10139-140-03

10139-143-03

10139-143-01

10139-143-00

Z 2-1-3-1

Z 2-1-3-2

Z 2-1-3-3

Z 2-1-3-4

Z 2-1-3-5

Z 2-1-3-6

Z 2-1-3-7

Z 2-1-3-8

Z 2-1-3-9

Z 2-1-3-10

Z 2-1-3-11

1-2

1-2

1-2

1-2

1-2

1-2

1-2

1-2

1-2

1

1

Prepreg glass - carbonhybrid fabrics

125° C

125° C

G973/913

ES03/3752

10056-300-01

10056-300-02

Z 2.1.4.1

Z 2.1.4.2

1-7

1-2

EPOXY

Prepreg quartz - polyesterhybrid fabrics

125° C M14/1237 Z 2.1.5.1 1-2





Prepreg carbon fabrics 140° C G803/40/V200 10139-500-01 Z 2.2.1.1 1

Prepreg glass fabrics 180° C

135° C

240/38/644

250/44/769

10056-72-01

10056-074-01

Z 2.2.2.1

Z 2.2.2.2

1

1

Prepreg Kevlar 135° C

135° C

125° C

VICOTEX 250/788

VICOTEX 250/796

EHA 250-33-50 or EHA 250-33-60

10139-182-01

10139-180-01

Z 2.2.3.1

Z 2.2.3.2

Z 2.2.3.3

1-2

1-2

1

Wet lay-up glass - carbonhybrid fabrics

70° C - 90° C G806 + T120/5052-9390-9396-501 10139-024-01

10057-002-00

Z 2.2.4.1 1-2

PHENOLIC

Prepeg glass - carbonhybrid fabrics

135° C - 150° C G874/V250 10056-200-01 Z 2.2.4.2 1-2





Wet lay-up carbon fabrics 80° C

90° C

90° C

90° C

70° C

G803/501

G806/501

G806/5052

G806/9396

G814/501

10053-080-01 (r)

10139-024-01

10139-024-01

10139-024-01

Z 2.4.1.1

Z 2.4.1.2

Z 2.4.1.3

Z 2.4.1.4

Z 2.4.1.5

1-2

1-2

1-2

1-2

1-2

Wet lay-up glass fabrics 70° C 1581/501 10057-004-00/01 Z 2.4.2.1 1

Prepreg Kevlar 120° C 181 + EPOXY resin 10139-142-00 Z 2.4.3.1 1EPOXY

RTM 150° C

150° C

180° C

120° C

150° C

120° C

120° C

120° C

G1151/RTM6

G986/RTM6

E4049/RTM6

GB305/DA3200 (*)

GF630/RTM6

GB305/XB5142 (*)

HF360/LY564-1 + HY2954 (*)

GF520/LY564-1 + HY2954 (*)

10139-701-00

10139-702-00

10139-701-00

10139-701-00

10056-350-00

10039-700-01

Z 3.1.1

Z 3.1.2

Z 3.1.3

Z 3.1.4

Z 3.1.5

Z 3.1.6

Z 3.1.7

Z 3.1.8

1-2

1-2

1-2

1-2

1-2

1-2

1-2

1-2





Injected thermoplastic ULTEM 2310

RYTON R04

IXEF 1022

IXEF C36

10058-514-01 Z 4.1.1.1

Z 4.1.4.1

Z 4.1.5.1

Z 4.1.5.2

1

1

1

1THERMOPLASTIC

Long-fiber thermoplastics 390° C

300° C

APC2 (AS4/PEEK)

CD282/PEI DE TENCATE

10139-951-00

10139-950-00

Z 5.1.1.1

Z 5.1.2.1

1

1-2

(*): do not use(r): resin only



CLASSIFICATION OF CORE MATERIALS (HONEYCOMB AND FORMS) IN ALPHABETICAL ORDER

Chapters 9 and 10

TRADENAME OF MATERIAL CHAPTER PAGE TRADENAME OF MATERIAL CHAPTER PAGE

5052 - F40 - 2.1 (Flex - Core)5052 - F40 - 2.5 (Flex - Core)5052 - F40 - 3.1 (Flex - Core)5052 - F40 - 4.1 (Flex - Core)5052 - F40 - 5.7 (Flex - Core)5052 - F80 - 4.3 (Flex - Core)5052 - F80 - 6.5 (Flex - Core)5052 - F80 - 8.0 (Flex - Core)5056 - F40 - 2.1 (Flex - Core)5056 - F40 - 3.1 (Flex - Core)5056 - F40 - 4.1 (Flex - Core)5056 - F80 - 4.3 (Flex - Core)5056 - F80 - 6.5 (Flex - Core)5056 - F80 - 8.0 (Flex - Core)ACG - 1/4 - 4.8ACG - 3/8 - 3.3ACG - 1/2 - 2.3ACG -3/4 - 1.8ACG - 1 - 1.3CR III 2024 - 3/16 - 3.5CR III 2024 - 1/8 - 5.0CR III 2024 - 1/8 - 6.7CR III 2024 - 1/8 - 8.0CR III 2024 - 1/8 - 9.5CR III 2024 - 1/4 - 2.8CR III 5052 - 1/16 - 6.5CR III 5052 - 1/16 - 9.5CR III 5052 - 1/16 - 12.0

Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3Z 9.3

181818191920202032323233333311122333444555

CR III 5052 - 1/16 - 13.8CR III 5052 - 3/32 - 4.3CR III 5052 - 3/32 - 6.3CR III 5052 - 1/8 - 3.1CR III 5052 - 1/8 - 4.5CR III 5052 - 1/8 - 6.1CR III 5052 - 1/8 - 8.1CR III 5052 - 1/8 - 12.0CR III 5052 - 1/8 - 22.1CR III 5052 - 5/32 - 2.6CR III 5052 - 5/32 - 3.8CR III 5052 - 5/32 - 5.3CR III 5052 - 5/32 - 6.9CR III 5052 - 5/32 - 8.4CR III 5052 - 3/16 - 2.0CR III 5052 - 3/16 - 3.1CR III 5052 - 3/16 - 4.4CR III 5052 - 3/16 - 5.7CR III 5052 - 3/16 - 6.9CR III 5052 - 3/16 - 8.1CR III 5052 - 1/4 - 1.6CR III 5052 - 1/4 - 2.3CR III 5052 - 1/4 - 3.4CR III 5052 - 1/4 - 4.3CR III 5052 - 1/4 - 5.2CR III 5052 - 1/4 - 6.0CR III 5052 - 1/4 - 7.9CR III 5052 - 3/8 - 1.0


66677788899910101111111212121313131414141515




Chapters 9 and 10


CR III 5052 - 3/8 - 1.6CR III 5052 - 3/8 - 2.3CR III 5052 - 3/8 - 3.0CR III 5052 - 3/8 - 3.7CR III 5052 - 3/8 - 4.2CR III 5052 - 3/8 - 5.4CR III 5052 - 3/8 - 6.5CR III 5056 - 1/16 - 6.5CR III 5056 - 1/16 - 9.5CR III 5056 - 3/32 - 4.3CR III 5056 - 3/32 - 6.3CR III 5056 - 1/8 - 3.1CR III 5056 - 1/8 - 4.5CR III 5056 - 1/8 - 6.1CR III 5056 - 1/8 - 8.1CR III 5056 - 5/32 - 2.6CR III 5056 - 5/32 - 3.8CR III 5056 - 5/32 - 5.3CR III 5056 - 5/32 - 6.9CR III 5056 - 3/16 - 2.0CR III 5056 - 3/16 - 3.1CR III 5056 - 3/16 - 4.4CR III 5056 - 3/16 - 5.7CR III 5056 - 3/16 - 8.1CR III 5056 - 1/4 - 1.6CR III 5056 - 1/4 - 2.3CR III 5056 - 1/4 - 3.4CR III 5056 - 1/4 - 4.3


15161616171717212122222323232424242525262626272727282828

CR III 5056 - 1/4 - 5.2CR III 5056 - 1/4 - 6.0CR III 5056 - 1/4 - 7.9CR III 5056 - 3/8 - 1.0CR III 5056 - 3/8 - 1.6CR III 5056 - 3/8 - 2.3CR III 5056 - 3/8 - 3.0CR III 5056 - 3/8 - 5.4HRH10 - 1/8 - 1.8HRH10 - 1/8 - 3.0HRH10 - 1/8 - 4.0HRH10 - 1/8 - 5.0HRH10 - 1/8 - 6.0HRH10 - 1/8 - 8.0HRH10 - 1/8 - 9.0HRH10 - 3/16 - 1.5HRH10 - 3/16 - 1.8HRH10 - 3/16 - 2.0HRH10 - 3/16 - 3.0HRH10 - 3/16 - 4.0HRH10 - 3/16 - 4.5HRH10 - 3/16 - 6.0HRH10 - 1/4 - 1.5HRH10 - 1/4 - 2.0HRH10 - 1/4 - 3.1HRH10 - 1/4 - 4.0HRH10 - 3/8 - 1.5HRH10 - 3/8 - 2.0


292929303030313111122233344455566677




Chapters 9 and 10


HRH10 - 3/8 - 3.0HRH10/OX - 3/16 - 1.8HRH10/OX - 3/16 - 3.0HRH10/OX - 3/16 - 4.0HRH10/OX - 1/4 - 3.0HRH10 - F35 - 2.5 (Flex - Core)HRH10 - F35 - 3.5 (Flex - Core)HRH10 - F35 - 4.5 (Flex - Core)HRH10 - F50 - 3.5 (Flex - Core)HRH10 - F50 - 4.5 (Flex - Core)HRH10 - F50 - 5.0 (Flex - Core)HRH10 - F50 - 5.5 (Flex - Core)HRP - 3/16 - 4.0HRP - 3/16 - 5.5HRP - 3/16 - 7.0HRP - 3/16 - 8.0HRP - 3/16 - 12.0HRP - 1/4 - 3.5HRP - 1/4 - 4.5HRP - 1/4 - 5.0HRP - 1/4 - 6.5HRP - 3/8 - 2.2HRP - 3/8 - 3.2HRP - 3/8 - 4.5HRP - 3/8 - 6.0HRP - 3/8 - 8.0HRP/OX - 1/4 - 4.5HRP/OX - 1/4 - 5.5


78889101010111112121112233345556677

HRP/OX - 1/4 - 7.0HRP/OX - 3/8 - 3.2HRP/OX - 3/8 - 5.5HRP - F35 - 2.5 (Flex - Core)HRP - F35 - 3.5 (Flex - Core)HRP - F35 - 4.5 (Flex - Core)HRP - F50 - 3.5 (Flex - Core)HRP - F50 - 4.5 (Flex - Core)HRP - F50 - 5.5 (Flex - Core)ROHACELL 31 AROHACELL 51 AROHACELL 71 AROHACELL 51 WFROHACELL 71 WFROHACELL 110 WFROHACELL 200 WF

Z 9.2Z 9.2Z 9.2Z 9.2Z 9.2Z 9.2Z 9.2Z 9.2Z 9.2Z 10Z 10Z 10Z 10Z 10Z 10Z 10

7889991010101112233



MATERIAL CHARACTERISTICSGlossary

Definitions of the main characteristics of the unidirectional fiber

El (daN/mm2) Longitudinal modulus of elasticity

Et (daN/mm2) Transverse modulus of elasticity

Glt (daN/mm2) Shear modulus

νlt Poisson coefficient

ep (mm) Ply thickness

Rlt (hb) Allowable longitudinal tensile strength

Rlc (hb) Allowable longitudinal compression strength

Rtt (hb) Allowable transverse tensile strength

Rtc (hb) Allowable transverse compression strength

S (hb) Allowable in-plane shear strength

τinter (hb) Allowable interlaminar shear strength

K mc Compression bearing stress coefficient

K mt Tensile bearing stress coefficient

K tc Hole compression coefficient

K tt Hole tensile coefficient

σm (hb) Allowable bearing strength

K tflexion Pure bending hole coefficient

κc Damage tolerance reduction coefficient for Rlc

κt Damage tolerance reduction coefficient for Rlt

κs Damage tolerance reduction coefficient for S

εadm. comp. (µd) Allowable damage tolerance compression strain

εadm. tract. (µd) Allowable damage tolerance tensile strain

γadm. cisail. (µd) Allowable damage tolerance shear

Tg dry (° C) Glass transition temperature in dry environment

Tg wet (° C) Glass transition temperature in wet environment

Cθ (µd/° C) Thermal expansion coefficient



DEFINITION of Tg

Typical diagram of the determination of Tg-onset and Tg-peak

(cf. AITM 1-0003, index 2)

E' (Gpa) elastic modulus E" (Gpa) viscoelastic modulus

Tan Delta

Temperature (° C)

tangent A

∆ E∆ T β

point C

point Lpoint M

tangent B

Tg-onset Tg-loss Tg-peak



MATERIAL CHARACTERISTICSGlossary

Definitions of the main characteristics of honeycomb

Ec (daN/mm2) Compressive modulus direction T

Rc (hb) Compressive strength direction T

Gl (daN/mm2) Shearing modulus direction L

Gw (daN/mm2) Shearing modulus direction W

sl (hb) Shear strength direction L

sw (hb) Shear strength direction w

* Preliminary values are obtained from testing one or two blocks of honeycomb type and often only oneor two specimens for each point or condition tested.

** Predicted values indicate that no mechanical tests have been performed.

Ldirection

Wdirection

T



MATERIAL CHARACTERISTICSPrepreg carbon tapes Z 1.1.1

1/10

Resin %34 % PREPREG CARBON TAPE

T300/914Curing180° C

Surface density200 gr/m2

New

T = 20° C

Agedwet

T = 70° C

Agedwet

T = 106° CReferences

El (daN/mm2) 13000 13000 13000

Et (daN/mm2) 465 465 465

Glt (daN/mm2) 465 465 465

νlt 0.35 0.35 0.35

ep (mm) 0.13 0.13 0.13

Rlt (hb) 120 120 120

Rlc (hb) - 100 - 84 - 78.9

Rtt (hb) 5 5 5

Rtc (hb) - 12 - 12 - 12

S (hb) 7.5 6.75 6.3

τinter (hb) 4.5 4.05 3.8

440.233/89

K mc T300/914 (1) T300/914 (1) T300/914 (1)

K mt T300/914 (2) T300/914 (2) T300/914 (2)

K tc T300/914 (3) T300/914 (3) T300/914 (3)

K tt T300/914E (4) T300/914E (4) T300/914E (4)

σm (hb) T300/914 (5) T300/914 (5) T300/914 (5)

K tflexion 0.9 0.9 0.9

581.0162/98

κc T300/914 (6) T300/914 (6) T300/914 (6)

κt T300/914 (6) T300/914 (6) T300/914 (6)

κs T300/914 (6) T300/914 (6) T300/914 (6)

εadm. comp. (µd) T300/914 (7)

εadm. tract. (µd) T300/914 (7)

γadm. cisail. (µd) T300/914 (7)

432.0026/96

Tg onset dry (° C) 120° C

Tg onset wet (° C) 95° C 95° C

Cθ (µd/° C) 1 (longi.) 40 (transv.)




2/10

SHEET T300/914 (1)

0 5 10 15 20 25 30 35 40 45σm

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

K mc




3/10

SHEET T300/914 (2)

0 5 10 15 20 25 30 35 40 50σm

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.18

0.2

K mt

0.16

45




4/10

SHEET T300/914 (3)

0 1 2 3 4 5 6E/G

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.9

1

K tc

0.8




5/10

SHEET T300/914 (4)

0 1 2 3 4 5 6E/G

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

K tt

∅ 11.1∅ 9.52

∅ 7.9

∅ 6.35

∅ 4.8

∅ 3.2




6/10

SHEET T300/914 (5)

0.8 1 1.2 1.4 1.6 1.8 2∅ /e

30

32.5

35

37.5

40

42.5

45

47.5

50

σm




7/10

SHEET T300/914 (6)

0 1000 2000 3000 4000 6000 8000

Sdmm2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

1

K

0.8

0.9

70005000

Ks

Kt

Kc




8/10

SHEET T300/914 (7)

Sdmm2

µd

- 7500

- 5000

- 2500

0

2500

5000

7500

10000

15000

12500

0 1000 2000 3000 4000 6000 800070005000

γa (s)

εa (t)

εa (c)




9/10

SHEET T300/914 (8)

∅mm

do (t)

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.5

0.4

0 2 4 6 8 12 161410

0.45




10/10

SHEET T300/914 (9)

∅mm

do (c)

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.5

0.4

0 2 4 6 8 12 161410

0.45




1/1

Resin %35 %

CARBON TAPET300/N5208PL/112/79Curing

180° CSurface density

200 gr/m2

New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 14000 14000

Et (daN/mm2) 500 500

Glt (daN/mm2) 500 500

νlt 0.35 0.35

ep (mm) 0.13 0.13

Rlt (hb) 120 108

Rlc (hb) - 100 - 90

Rtt (hb) 5 4.5

Rtc (hb) - 12 - 10.8

S (hb) 6.5 8.85

τinter (hb)

K mc 0.1 0.1

K mt 0.1 0.1

K tc 0.765 0.765

K tt 0.6 0.6

σm (hb)

K tflexion 0.9 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)

γadm. cisail. (µd)

Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/1

Resin %34 % CARBON TAPE

HTA EH25Curing180° C


New

T = 20° C

Aged

T = 70° CReferences

El (daN/mm2) 13000 13000

Et (daN/mm2) 465 465

Glt (daN/mm2) 465 465

νlt 0.35 0.35

ep (mm) 0.13 0.13

Rlt (hb) 132 132

Rlc (hb) - 100 - 84

Rtt (hb) 5 5.5

Rtc (hb) - 12 - 12

S (hb) 8.33 7.5

τinter (hb)

440.337/91440.346/93

K mc

K mt

K tc

K tt

σm (hb) 40

K tflexion 0.9 0.9

581.0162/98

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


432.0026/96

Tg onset dry (° C) 172° C

Tg onset wet (° C) 109° C

Tg peak dry (° C) 187° C

Tg peak wet (° C) 137° C

Cθ (µd/° C)




1/1

Resin % CARBON TAPET800H - DA508T528-082/90-03Curing

180° CSurface density145 gr/m2 (sec)

New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 16200 16200

Et (daN/mm2) 580 580

Glt (daN/mm2) 580 580

νlt 0.35 0.35

ep (mm) 0.135 0.135

Rlt (hb) 190 190

Rlc (hb) - 100 - 84

Rtt (hb) 11 11

Rtc (hb) - 12 - 12

S (hb) 11 10

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % CARBON TAPEIM7/977-2

480.387/92-02Curing190° C/ 7 bars

Surface density145 gr/m2 (sec)

New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 16200 16200

Et (daN/mm2) 580 580

Glt (daN/mm2) 580 580

νlt 0.35 0.35

ep (mm) 0.135 0.135

Rlt (hb) 190 190

Rlc (hb) - 100 - 84

Rtt (hb) 11 11

Rtc (hb) - 12 - 12

S (hb) 11 10

τinter (hb)

K mc T300/914 (1) T300/914 (1)

K mt T300/914 (2) T300/914 (2)

K tc T300/914 (3) T300/914 (3)

K tt T300/914 (4) T300/914 (4)

σm (hb) 50 50

K tflexion

κc

κt

κs

εadm. comp. (µd) IM7/977-2 (1)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

SHEET IM7/977-2 (1)

Number of plies

εa (c)

- 3000

- 2950

- 2900

- 2850

- 2800

- 2750

- 2700

- 2600

0 20 40 60 80 120100

- 2650

BVID = 1 mm E = 50 J



MATERIAL CHARACTERISTICSPrepreg carbon fabric Z 2.1.1.1

1/4

Resin % PREPREG CARBON FABRIC (equiv. tape)G803/914

440.353/87Curing180° C

Surface densityNew

T = 20° C

Agedwet

T = 70° CEl (daN/mm2) 11500 11500

Et (daN/mm2) 500 500

Glt (daN/mm2) 500 500

νlt 0.35 0.35

ep (mm) 0.15 0.15

Rlt (hb) 96.4 85.8

Rlc (hb) - 89.5 - 76

Rtt (hb) 5 5

Rtc (hb) - 10 - 10

S (hb) 9.4 6.5

τinter (hb) 5.6 3.9

K mc 0.25 0.25

K mt 0.25 0.25

K tc 0.85 0.85

K tt 0.65 0.65

σm (hb) 40 40

K tflexion 0.9 0.9

κcG803/914 (1) G803/914 (1)

*

κtG803/914 (1) G803/914 (1)

*

κsG803/914 (1) G803/914 (1)

*εadm. comp. (µd) G803/914 (2)

εadm. tract. (µd) G803/914 (2)

γadm. cisail. (µd) G803/914 (2)

Tg dry (° C) 120° C

Tg wet (° C) 95° C

Cθ (µd/° C) 1 (longi.) 40 (transv.)




2/4

Resin %42 %

PREPREG CARBON FABRIC (fabric)G803/914

440.353/87Curing180° C


New

T = 20° C

Agedwet

T = 70° CEl (daN/mm2) 6027 6027

Et (daN/mm2) 6027 6027

Glt (daN/mm2) 500 500

νlt 0.0292 0.0292

ep (mm) 0.3 0.3

Rlt (hb) 49 43.7

Rlc (hb) - 46.3 - 39.3

Rtt (hb) 49 43.7

Rtc (hb) - 46.3 - 39.3

S (hb) 9.4 6.5

τinter (hb) 5.6 3.9

K mc 0.25 0.25

K mt 0.25 0.25

K tc 0.85 0.85

K tt 0.65 0.65

σm (hb) 40 40

K tflexion 0.9 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C) 120° C

Tg wet (° C) 95° C

Cθ (µd/° C) 3.1 3.1




3/4

SHEET G803/914 (1)

Sdmm2

κ

0 200 400 600 800 200014000

0.1

0.2

0.3

0.4

0.5

0.7

1

0.9

0.8

0.6

1200 1600 18001000

κs*

κt

κc*κc

κs




4/4

SHEET G803/914 (2)

Sdmm2

µd

0 200 400 600 800 20001400- 8000

- 6000

- 2000

0

2000

4000

8000

14000

12000

10000

6000

1200 1600 18001000

γa (s)

- 4000

εa (t)

εa (c)




1/2

Resin % CARBON FABRIC (equiv. tape)HTA7 (6K) EH25

440.027/94Curing180° C

Surface densityNew

T = 20° C

Aged

T = 70° CEl (daN/mm2) 11500 11500

Et (daN/mm2) 500 500

Glt (daN/mm2) 500 500

νlt 0.35 0.35

ep (mm) 0.15 0.15

Rlt (hb) 96.4 85.8

Rlc (hb) - 89.5 - 76

Rtt (hb) 5 5

Rtc (hb) - 10 - 10

S (hb) 9.4 6.5

τinter (hb)

K mc 0.25 0.25

K mt 0.25 0.25

K tc 0.85 0.85

K tt 0.65 0.65

σm (hb)

K tflexion 0.9 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %42 %

CARBON FABRIC (fabric)HTA7 (6K) EH25

440.027/94Curing180° C


New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 6027 6027

Et (daN/mm2) 6027 6027

Glt (daN/mm2) 500 500

νlt 0.0292 0.0292

ep (mm) 0.3 0.3

Rlt (hb) 49 43.7

Rlc (hb) - 47.7 - 39.7

Rtt (hb) 49 43.7

Rtc (hb) - 47.7 - 39.7

S (hb) 9.4 6.5

τinter (hb)

K mc 0.25 0.25

K mt 0.25 0.25

K tc 0.85 0.85

K tt 0.65 0.65

σm (hb)

K tflexion 0.9 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % CARBON FABRIC (equiv. tape)G803/145-4

Hurel Dubois (25 S 002 10388)Curing120° C

Surface densityNew

T = 20° C

Agedwet

T = 70° CEl (daN/mm2) 11000 11000

Et (daN/mm2) 500 500

Glt (daN/mm2) 500 500

νlt 0.35 0.35

ep (mm) 0.15 0.15

Rlt (hb) 95 95

Rlc (hb) - 67 - 20.1

Rtt (hb) 10 10

Rtc (hb) - 5 - 1.5

S (hb) 6.5 6.5

τinter (hb)

K mc 0.2 0.2

K mt 0.2 0.2

K tc 0.85 0.85

K tt 0.6 0.6

σm (hb)

K tflexion 0.9 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %42 %

CARBON FABRIC (fabric)G803/145-4

Hurel Dubois (25 S 002 10388)Curing120° C


New

T = 20° C

Agedwet

T = 70° CEl (daN/mm2) 5777 5777

Et (daN/mm2) 5777 5777

Glt (daN/mm2) 500 500

νlt 0.0304 0.0304

ep (mm) 0.3 0.3

Rlt (hb) 49.6 49.6

Rlc (hb) - 34.6 - 10.4

Rtt (hb) 49.6 49.6

Rtc (hb) - 34.6 - 10.4

S (hb) 6.5 6.5

τinter (hb)

K mc 0.2 0.2

K mt 0.2 0.2

K tc 0.85 0.85

K tt 0.6 0.6

σm (hb)

K tflexion 0.9 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % CARBON FABRIC (equiv. tape)G815/913

DA - DAN 1208Curing125° C

Surface densityAged

T = 80° CEl (daN/mm2) 9450

Et (daN/mm2) 500

Glt (daN/mm2) 336

νlt 0.35

ep (mm) 0.175

Rlt (hb) 100

Rlc (hb) - 64

Rtt (hb) 4.2

Rtc (hb) - 3.3

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)

γadm. cisail. (µd)Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %35 %

CARBON FABRIC (fabric)G815/913



Aged

T = 80° CEl (daN/mm2) 5001

Et (daN/mm2) 5001

Glt (daN/mm2) 336

νlt 0.0352

ep (mm) 0.35

Rlt (hb) 42.28

Rlc (hb) - 32.28

Rtt (hb) 42.28

Rtc (hb) - 32.28

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2


440.104/92Curing125° C/1h30/3.5 bars

Surface densityNew

T = 20° C

Aged

T = 20° C

Aged

T = 80° C

AgedT = 80° C

Hurel DuboisEl (daN/mm2) 10700 9700 9000 9800

Et (daN/mm2) 400 400 400 400

Glt (daN/mm2) 400 380 320 400

νlt 0.35 0.35 0.35 0.35

ep (mm) 0.115 0.115 0.115 0.115

Rlt (hb) 83 77 77 73

Rlc (hb) - 72 - 55 - 44 - 38.4

Rtt (hb) 3.5 3.5 3.5 4.5

Rtc (hb) - 3.5 - 3.5 - 2.5 - 6.5

S (hb) 6 3.4 3.4 5

τinter (hb)

K mc 0.25 0.25 0.25 0.25

K mt 0.25 0.25 0.25 0.25

K tc 0.85 0.85 0.85 0.85

K tt 0.65 0.65 0.65 0.65

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %50 %


440.104/92Curing125° C/1h30/3.5 bars


New

T = 20° C

Aged

T = 20° C

Aged

T = 80° C

AgedT = 80° C


Et (daN/mm2) 5572 5072 4722 5122

Glt (daN/mm2) 400 380 320 400

νlt 0.0252 0.0277 0.0298 0.0275

ep (mm) 0.23 0.23 0.23 0.23

Rlt (hb) 41.7 38.7 38.7 37.4

Rlc (hb) - 36.5 - 27.8 - 22.6 - 20

Rtt (hb) 41.7 38.7 38.7 37.4

Rtc (hb) - 36.5 - 27.8 - 22.6 - 20

S (hb) 6 3.4 3.4 5

τinter (hb)

K mc 0.25 0.25 0.25 0.25

K mt 0.25 0.25 0.25 0.25

K tc 0.85 0.85 0.85 0.85

K tt 0.65 0.65 0.65 0.65

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSPrepreg glass fabric Z 2.1.2.1

1/2

Resin % GLASS FABRIC (equiv. tape)V913/7781528/084/94Curing

120° C/1h/1.8 barSurface density

New

T = 20° C

Aged

T = 80° CEl (daN/mm2) 3800 3800

Et (daN/mm2) 400 280

Glt (daN/mm2) 400 280

νlt 0.35 0.35

ep (mm) 0.115 0.115

Rlt (hb) 51 35

Rlc (hb) - 44 - 30

Rtt (hb) 7 3.5

Rtc (hb) - 6 - 3

S (hb) 7.4 3.7

τinter (hb)

K mc 0.1 0.1

K mt 0.1 0.1

K tc 0.85 0.85

K tt 0.75 0.75

σm (hb) 30 30

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %37 %

GLASS FABRIC (fabric)V913/7781528/084/94Curing


476 gr/m2

New

T = 20° C

Aged

T = 80° CEl (daN/mm2) 2118 2054

Et (daN/mm2) 2118 2054

Glt (daN/mm2) 400 280

νlt 0.0667 0.0667

ep (mm) 0.23 0.23

Rlt (hb) 27.8 18.7

Rlc (hb) - 23.9 - 16.1

Rtt (hb) 27.8 18.7

Rtc (hb) - 23.9 - 16.1

S (hb) 7.4 3.7

τinter (hb)

K mc 0.1 0.1

K mt 0.1 0.1

K tc 0.85 0.85

K tt 0.75 0.75

σm (hb) 30 30

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/1

Resin %38 % GLASS FABRIC (fabric)

F155/T120Curing125° C


Agedwet

T = 70° CEl (daN/mm2) 2070

Et (daN/mm2) 2070

Glt (daN/mm2) 260

νlt

ep (mm) 0.12

Rlt (hb) 36.3(T = 23° C)

Rlc (hb) - 30.9(T = 71° C)

Rtt (hb) 36.3

Rtc (hb) - 30.9

S (hb) 5

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSPrepreg Kevlar Z 2.1.3.1

1/2

Resin % KEVLAR (equiv. tape)788/54/M14

245/AERO/61-ACuring125° C/3h

Surface densityAgedwet

T = 70° CEl (daN/mm2) 4600

Et (daN/mm2) 300

Glt (daN/mm2) 160

νlt 0.35

ep (mm) 0.125

Rlt (hb) 44

Rlc (hb) - 14.9

Rtt (hb) 4

Rtc (hb) - 4

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %54 %

KEVLAR (fabric)788/54/M14



Agedwet

T = 70° CEl (daN/mm2) 2465

Et (daN/mm2) 2465

Glt (daN/mm2) 160

νlt 0.0429

ep (mm) 0.25

Rlt (hb) 23.1

Rlc (hb) - 8

Rtt (hb) 23.1

Rtc (hb) - 8

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2




T = 70° CEl (daN/mm2) 4600

Et (daN/mm2) 300

Glt (daN/mm2) 160

νlt 0.35

ep (mm) 0.05

Rlt (hb) 44

Rlc (hb) - 14.9

Rtt (hb) 4

Rtc (hb) - 4

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %65 %




Agedwet

T = 70° CEl (daN/mm2) 2465

Et (daN/mm2) 2465

Glt (daN/mm2) 160

νlt 0.0429

ep (mm) 0.1

Rlt (hb) 23.1

Rlc (hb) - 8

Rtt (hb) 23.1

Rtc (hb) - 8

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % KEVLAR (equiv. tape)ES36D/90285

440.146/85Curing120° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %55 %

KEVLAR (fabric)ES36D/90285

440.146/85Curing120° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing120° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.05

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %65 %


440.146/85Curing120° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.1

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing120° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %60 %


440.146/85Curing120° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % KEVLAR (equiv. tape)VICOTEX 108/788

440.146/85Curing175° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %45 %

KEVLAR (fabric)VICOTEX 108/788

440.146/85Curing175° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % KEVLAR (equiv. tape)VICOTEX 145.2/788

440.146/85Curing120° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %54 %

KEVLAR (fabric)VICOTEX 145.2/788

440.146/85Curing120° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing125° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.05

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %65 %


440.146/85Curing125° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.1

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing125° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %54 %


440.146/85Curing125° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/1

Resin % KEVLAR ECONOMIC (fabric)285 + GENIN 90285/ES36D EPOXY RESINCuring


New

T = 20° CEl (daN/mm2) 2580 (warp)

Et (daN/mm2) 2640 (weft)

Glt (daN/mm2)

νlt

ep (mm) 0.28

Rlt (hb) 46.9 (warp)

Rlc (hb) - 12.6 (warp)

Rtt (hb) 39.5 (weft)

Rtc (hb) - 12.5 (weft)

S (hb)

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/1

Resin % KEVLAR ECONOMIC (fabric)285 + BROCHIER 1454/914 EPOXY RESINCuring


New

T = 20° CEl (daN/mm2) 3175 (warp)


Glt (daN/mm2)

νlt

ep (mm) 0.28


Rlc (hb) - 11 (weft)

Rtt (hb) 38.1 (warp)


S (hb)

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSPrepreg glass - carbon hybrid fabrics Z 2.1.4.1

1/7

Resin % GLASS - CARBON HYBRID (equiv. tape)G973/913

440.399-02Curing125° C/1h30/2 bars

Surface densityNew

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 6560 6560 6105

Et (daN/mm2) 400 400 360

Glt (daN/mm2) 350 350 315

νlt 0.35 0.35 0.35

ep (mm) 0.105 0.105 0.105

Rlt (hb) 54.6 54.6 46.4

Rlc (hb) - 42.8 - 36 - 23.5

Rtt (hb) 4 4 4

Rtc (hb) - 8 - 7 - 4.5

S (hb) 5 5 5

τinter (hb) G973/913 (1) G973/913 (1)*

G973/913 (1)**

K mc 0.2 0.2 0.2

K mt 0.2 0.2 0.2

K tc 1 1 1

K tt G973/913 (2) G973/913 (2) G973/913 (2)

σm (hb) G973/913 (3) G973/913 (3) G973/913 (3)*


κc G973/913 (4) G973/913 (4) G973/913 (4)

κt G973/913 (4) G973/913 (4) G973/913 (4)

κs G973/913 (4) G973/913 (4) G973/913 (4)

εadm. comp. (µd) G973/913 (5)



Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/7

Resin %54 %

GLASS - CARBON HYBRID (fabric)G973/913

440.399-02Curing125° C/1h30/2 bars


New

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 3501 3501 3251

Et (daN/mm2) 3501 3501 3251

Glt (daN/mm2) 350 350 315

νlt 0.0402 0.0402 0.039

ep (mm) 0.21 0.21 0.21

Rlt (hb) 28.1 28.1 24.3

Rlc (hb) - 22.9 - 19 - 12.4

Rtt (hb) 28.1 28.1 24.3

Rtc (hb) - 22.9 - 19 - 12.4

S (hb) 5 5 4

τinter (hb) G973/913 (1) G973/913 (1)*

G973/913 (1)**

K mc 0.2 0.2 0.2

K mt 0.2 0.2 0.2

K tc 1 1 1

K tt G973/913 (2) G973/913 (2) G973/913 (2)

σm (hb) G973/913 (3) G973/913 (3) G973/913 (3)*


κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




3/7

SHEET G973/913 (1)

Number of fabrics

τinter.

0 25100

0.5

1.5

3

2.5

2

1

5 15 20

**

*




4/7

SHEET G973/913 (2)

% plies à 45°

K tt

0 10 20 30 40 1007060 80 90500

0.1

0.3

0.4

0.5

0.6

0.8

1

0.9

0.7

0.2




5/7

SHEET G973/913 (3)

% plies à 45°

σm

0 10 20 30 40 1007060 80 90500

20

30

40

60

50

10

*




6/7

SHEET G973/913 (4)

Sdmm2

κ

0 20 40 60 80 200140120 160 1801000

0.1

0.3

0.4

0.5

0.6

0.8

1

0.9

0.7

0.2

κt

κs

κc

κc*




7/7

SHEET G973/913 (5)

Sdmm2

µd

0 20 40 60 80 200140120 160 180100- 6000

- 4000

0

2000

4000

6000

10000

14000

12000

8000

- 2000

γa (c)

εa (c*)

εa (c)

εa (t)




1/2

Resin % GLASS - CARBON HYBRID (equiv. tape)ES03/3752440.399-02Curing

125° C/1h30/2 barsSurface density

New

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 6560 6560 6105

Et (daN/mm2) 400 400 360

Glt (daN/mm2) 350 350 315

νlt 0.35 0.35 0.35

ep (mm) 0.105 0.105 0.105

Rlt (hb) 54.6 54.6 46.4

Rlc (hb) - 42.8 - 36 - 23.5

Rtt (hb) 4 4 4

Rtc (hb) - 8 - 7 - 4.5

S (hb) 5 5 4

τinter (hb)

K mc 0.2 0.2 0.2

K mt 0.2 0.2 0.2

K tc 1 1 1

K tt 0.7 0.7 0.7

σm (hb)


κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %52 %

GLASS - CARBON HYBRID (fabric)ES03/3752440.399-02Curing


372 gr/m2

New

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 3501 3501 3251

Et (daN/mm2) 3501 3501 3251

Glt (daN/mm2) 350 350 315

νlt 0.0402 0.0402 0.039

ep (mm) 0.21 0.21 0.21

Rlt (hb) 19.7 19.7 17

Rlc (hb) - 16 - 13.3 - 8.7

Rtt (hb) 19.7 19.7 17

Rtc (hb) - 16 - 13.3 - 8.7

S (hb) 3.5 3.5 2.8

τinter (hb)

K mc 0.2 0.2 0.2

K mt 0.2 0.2 0.2

K tc 1 1 1

K tt 0.7 0.7 0.7

σm (hb)


κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2



Surface densityAged

T = 80° CEl (daN/mm2) 9450

Et (daN/mm2) 500

Glt (daN/mm2) 336

νlt 0.35

ep (mm) 0.175

Rlt (hb) 100

Rlc (hb) - 64

Rtt (hb) 4.2

Rtc (hb) - 3.3

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %35 %




Aged

T = 80° CEl (daN/mm2) 5001

Et (daN/mm2) 5001

Glt (daN/mm2) 336

νlt 0.0352

ep (mm) 0.35

Rlt (hb) 42.28

Rlc (hb) - 32.28

Rtt (hb) 42.28

Rtc (hb) - 32.28

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2


440.104/92Curing125° C/1h30/3.5 bars

Surface densityNew

T = 20° C

Aged

T = 20° C

Aged

T = 80° C

AgedT = 80° C


Et (daN/mm2) 400 400 400 400

Glt (daN/mm2) 400 380 320 400

νlt 0.35 0.35 0.35 0.35

ep (mm) 0.115 0.115 0.115 0.115

Rlt (hb) 83 77 77 73

Rlc (hb) - 72 - 55 - 44 - 38.4

Rtt (hb) 3.5 3.5 3.5 4.5

Rtc (hb) - 3.5 - 3.5 - 2.5 - 6.5

S (hb) 6 3.4 3.4 5

τinter (hb)

K mc 0.25 0.25 0.25 0.25

K mt 0.25 0.25 0.25 0.25

K tc 0.85 0.85 0.85 0.85

K tt 0.65 0.65 0.65 0.65

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %50 %


440.104/92Curing125° C/1h30/3.5 bars


New

T = 20° C

Aged

T = 20° C

Aged

T = 80° C

AgedT = 80° C


Et (daN/mm2) 5572 5072 4722 5122

Glt (daN/mm2) 400 380 320 400

νlt 0.0252 0.0277 0.0298 0.0275

ep (mm) 0.23 0.23 0.23 0.23

Rlt (hb) 41.7 38.7 38.7 37.4

Rlc (hb) - 36.5 - 27.8 - 22.6 - 20

Rtt (hb) 41.7 38.7 38.7 37.4

Rtc (hb) - 36.5 - 27.8 - 22.6 - 20

S (hb) 6 3.4 3.4 5

τinter (hb)

K mc 0.25 0.25 0.25 0.25

K mt 0.25 0.25 0.25 0.25

K tc 0.85 0.85 0.85 0.85

K tt 0.65 0.65 0.65 0.65

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % GLASS FABRIC (equiv. tape)V913/7781528/084/94Curing


New

T = 20° C

Aged

T = 80° CEl (daN/mm2) 3800 3800

Et (daN/mm2) 400 280

Glt (daN/mm2) 400 280

νlt 0.35 0.35

ep (mm) 0.115 0.115

Rlt (hb) 51 35

Rlc (hb) - 44 - 30

Rtt (hb) 7 3.5

Rtc (hb) - 6 - 3

S (hb) 7.4 3.7

τinter (hb)

K mc 0.1 0.1

K mt 0.1 0.1

K tc 0.85 0.85

K tt 0.75 0.75

σm (hb) 30 30

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %37 %

GLASS FABRIC (fabric)V913/7781528/084/94Curing


476 gr/m2

New

T = 20° C

Aged

T = 80° CEl (daN/mm2) 2118 2054

Et (daN/mm2) 2118 2054

Glt (daN/mm2) 400 280

νlt 0.0667 0.0667

ep (mm) 0.23 0.23

Rlt (hb) 27.8 18.7

Rlc (hb) - 23.9 - 16.1

Rtt (hb) 27.8 18.7

Rtc (hb) - 23.9 - 16.1

S (hb) 7.4 3.7

τinter (hb)

K mc 0.1 0.1

K mt 0.1 0.1

K tc 0.85 0.85

K tt 0.75 0.75

σm (hb) 30 30

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/1


F155/T120Curing125° C


Agedwet

T = 70° CEl (daN/mm2) 2070

Et (daN/mm2) 2070

Glt (daN/mm2) 260

νlt

ep (mm) 0.12

Rlt (hb) 36.3(T = 23° C)

Rlc (hb) - 30.9(T = 71° C)

Rtt (hb) 36.3

Rtc (hb) - 30.9

S (hb) 5

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2




T = 70° CEl (daN/mm2) 4600

Et (daN/mm2) 300

Glt (daN/mm2) 160

νlt 0.35

ep (mm) 0.125

Rlt (hb) 44

Rlc (hb) - 14.9

Rtt (hb) 4

Rtc (hb) - 4

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %54 %




Agedwet

T = 70° CEl (daN/mm2) 2465

Et (daN/mm2) 2465

Glt (daN/mm2) 160

νlt 0.0429

ep (mm) 0.25

Rlt (hb) 23.1

Rlc (hb) - 8

Rtt (hb) 23.1

Rtc (hb) - 8

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2




T = 70° CEl (daN/mm2) 4600

Et (daN/mm2) 300

Glt (daN/mm2) 160

νlt 0.35

ep (mm) 0.05

Rlt (hb) 44

Rlc (hb) - 14.9

Rtt (hb) 4

Rtc (hb) - 4

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %65 %




Agedwet

T = 70° CEl (daN/mm2) 2465

Et (daN/mm2) 2465

Glt (daN/mm2) 160

νlt 0.0429

ep (mm) 0.1

Rlt (hb) 23.1

Rlc (hb) - 8

Rtt (hb) 23.1

Rtc (hb) - 8

S (hb) 4

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing120° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %55 %


440.146/85Curing120° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing120° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.05

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %65 %


440.146/85Curing120° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.1

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing120° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %60 %


440.146/85Curing120° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing175° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %45 %


440.146/85Curing175° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing120° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %54 %


440.146/85Curing120° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing125° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.05

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %65 %


440.146/85Curing125° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.1

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing125° C

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %54 %


440.146/85Curing125° C


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/1

Resin % KEVLAR ECONOMIC (fabric)285 + GENIN 90285/ES36D EPOXY RESINCuring


New

T = 20° CEl (daN/mm2) 2580 (warp)


Glt (daN/mm2)

νlt

ep (mm) 0.28


Rlc (hb) - 12.6 (warp)

Rtt (hb) 39.5 (weft)


S (hb)

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/1

Resin % KEVLAR ECONOMIC (fabric)285 + BROCHIER 1454/914 EPOXY RESINCuring


New

T = 20° CEl (daN/mm2) 3175 (warp)


Glt (daN/mm2)

νlt

ep (mm) 0.28


Rlc (hb) - 11 (weft)

Rtt (hb) 38.1 (warp)


S (hb)

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/7

Resin % GLASS - CARBON HYBRID (equiv. tape)G973/913

440.399-02Curing125° C/1h30/2 bars

Surface densityNew

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 6560 6560 6105

Et (daN/mm2) 400 400 360

Glt (daN/mm2) 350 350 315

νlt 0.35 0.35 0.35

ep (mm) 0.105 0.105 0.105

Rlt (hb) 54.6 54.6 46.4

Rlc (hb) - 42.8 - 36 - 23.5

Rtt (hb) 4 4 4

Rtc (hb) - 8 - 7 - 4.5

S (hb) 5 5 5

τinter (hb) G973/913 (1) G973/913 (1)*

G973/913 (1)**

K mc 0.2 0.2 0.2

K mt 0.2 0.2 0.2

K tc 1 1 1

K tt G973/913 (2) G973/913 (2) G973/913 (2)

σm (hb) G973/913 (3) G973/913 (3) G973/913 (3)*


κc G973/913 (4) G973/913 (4) G973/913 (4)

κt G973/913 (4) G973/913 (4) G973/913 (4)

κs G973/913 (4) G973/913 (4) G973/913 (4)

εadm. comp. (µd) G973/913 (5)



Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/7

Resin %54 %

GLASS - CARBON HYBRID (fabric)G973/913

440.399-02Curing125° C/1h30/2 bars


New

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 3501 3501 3251

Et (daN/mm2) 3501 3501 3251

Glt (daN/mm2) 350 350 315

νlt 0.0402 0.0402 0.039

ep (mm) 0.21 0.21 0.21

Rlt (hb) 28.1 28.1 24.3

Rlc (hb) - 22.9 - 19 - 12.4

Rtt (hb) 28.1 28.1 24.3

Rtc (hb) - 22.9 - 19 - 12.4

S (hb) 5 5 4

τinter (hb) G973/913 (1) G973/913 (1)*

G973/913 (1)**

K mc 0.2 0.2 0.2

K mt 0.2 0.2 0.2

K tc 1 1 1

K tt G973/913 (2) G973/913 (2) G973/913 (2)

σm (hb) G973/913 (3) G973/913 (3) G973/913 (3)*


κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




3/7

SHEET G973/913 (1)

Number of fabrics

τinter.

0 25100

0.5

1.5

3

2.5

2

1

5 15 20

**

*




4/7

SHEET G973/913 (2)

% plies à 45°

K tt

0 10 20 30 40 1007060 80 90500

0.1

0.3

0.4

0.5

0.6

0.8

1

0.9

0.7

0.2




5/7

SHEET G973/913 (3)

% plies à 45°

σm

0 10 20 30 40 1007060 80 90500

20

30

40

60

50

10

*




6/7

SHEET G973/913 (4)

Sdmm2

κ

0 20 40 60 80 200140120 160 1801000

0.1

0.3

0.4

0.5

0.6

0.8

1

0.9

0.7

0.2

κt

κs

κc

κc*




7/7

SHEET G973/913 (5)

Sdmm2

µd

0 20 40 60 80 200140120 160 180100- 6000

- 4000

0

2000

4000

6000

10000

14000

12000

8000

- 2000

γa (c)

εa (c*)

εa (c)

εa (t)




1/2

Resin % GLASS - CARBON HYBRID (equiv. tape)ES03/3752440.399-02Curing


New

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 6560 6560 6105

Et (daN/mm2) 400 400 360

Glt (daN/mm2) 350 350 315

νlt 0.35 0.35 0.35

ep (mm) 0.105 0.105 0.105

Rlt (hb) 54.6 54.6 46.4

Rlc (hb) - 42.8 - 36 - 23.5

Rtt (hb) 4 4 4

Rtc (hb) - 8 - 7 - 4.5

S (hb) 5 5 4

τinter (hb)

K mc 0.2 0.2 0.2

K mt 0.2 0.2 0.2

K tc 1 1 1

K tt 0.7 0.7 0.7

σm (hb)


κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %52 %

GLASS - CARBON HYBRID (fabric)ES03/3752440.399-02Curing


372 gr/m2

New

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 3501 3501 3251

Et (daN/mm2) 3501 3501 3251

Glt (daN/mm2) 350 350 315

νlt 0.0402 0.0402 0.039

ep (mm) 0.21 0.21 0.21

Rlt (hb) 19.7 19.7 17

Rlc (hb) - 16 - 13.3 - 8.7

Rtt (hb) 19.7 19.7 17

Rtc (hb) - 16 - 13.3 - 8.7

S (hb) 3.5 3.5 2.8

τinter (hb)

K mc 0.2 0.2 0.2

K mt 0.2 0.2 0.2

K tc 1 1 1

K tt 0.7 0.7 0.7

σm (hb)


κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSPrepreg quartz - polyester hybrid fabrics Z 2.1.5.1

1/2

Resin % QUARTZ - POLYESTER FABRIC (equiv. tape)M14/1237Curing

125° C/3hSurface density

New

T = 20° C

Aged

T = 70° C

Aged

T = 70° C

Aged

T = - 55° CEl (daN/mm2) 1820 1820

Et (daN/mm2) 455 455

Glt (daN/mm2) 193 193

νlt 0.35 0.35

ep (mm) 0.105 0.105

Rlt (hb) 46.5 10

Rlc (hb) - 26.1 - 10.6

Rtt (hb) 11.63 2.5

Rtc (hb) - 6.53 - 2.65

S (hb) 8.38 2.51

τinter (hb) 3 1.04

K mc 0.2 0.2

K mt 0.2 0.2

K tc 0.61 0.61

K tt 0.61 0.61

σm (hb) 40 12.8 15.2 18.1

K tflexion 0.9 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSPrepreg quartz - polyester hybrid fabrics Z 2.1.5.1

2/2

Resin %48 % HYBRIDE QUARTZ - POLYESTER (fabric)

M14/1237Curing125° C/3h


New

T = 20° C

Aged

T = 70° C

Aged

T = 20° C

Aged

T = - 55° CEl (daN/mm2) 1150 1150

Et (daN/mm2) 1150 1150

Glt (daN/mm2) 193 193

νlt 0.14 0.14

ep (mm) 0.21 0.21

Rlt (hb) 28.9 6.66

Rlc (hb) - 17.7 - 9.89

Rtt (hb) 28.9 6.31

Rtc (hb) - 17.7 - 6.45

S (hb) 8.38 2.51

τinter (hb) 3 1.04

K mc 0.2 0.2

K mt 0.2 0.2

K tc 0.61 0.61

K tt 0.61 0.61

σm (hb) 40 12.8 15.2 18.1

K tflexion 0.9 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSPrepreg carbon fabrics Z 2.2.1.1

1/1

Resin %40 % PHENOLIC PREPREG CARBON FABRIC (fabric)

G803/40/V200Curing140° C/1h30/4.75 bars


New

T = 20° C

Agedwet

T = 110° CEl (daN/mm2) 6490 6730

Et (daN/mm2) 6490 6730

Glt (daN/mm2) 600 406

νlt 0.05 0.05

ep (mm) 0.3 0.3

Rlt (hb) 51 44

Rlc (hb) - 51 - 30

Rtt (hb) 51 44

Rtc (hb) - 51 - 30

S (hb) 5 5.9

τinter (hb) 4 2

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSPrepreg glass fabrics Z 2.2.2.1

1/1

Resin %38 % PHENOLIC GLASS FABRIC (fabric)

240/38/644Curing180° C


Agedwet

T = 20° CEl (daN/mm2) 1300

Et (daN/mm2)

Glt (daN/mm2)

νlt

ep (mm) 0.1 - 0.3

Rlt (hb) 23

Rlc (hb) - 23

Rtt (hb) 23

Rtc (hb) - 23

S (hb)

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSPrepreg glass fabrics Z 2.2.2.2

1/1

Resin %44 % PHENOLIC GLASS FABRIC (fabric)

250/44/759Curing

135° C/1h30 or 150° C/1hSurface density

190 gr/m2

Agedwet

T = 20° CEl (daN/mm2) 1300

Et (daN/mm2)

Glt (daN/mm2)

νlt

ep (mm) 0.1 - 0.3

Rlt (hb) 23

Rlc (hb) - 23

Rtt (hb) 23

Rtc (hb) - 23

S (hb)

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing135° C/1h30 or 150° C/1h

Surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.125

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %56 %


440.146/85Curing135° C/1h30 or 150° C/1h


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.25

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


440.146/85Curing135° C/1h30 or 150° C/1h

surface densityAged

T ≥ - 55° CT ≤ 70° C

El (daN/mm2) 4670

Et (daN/mm2) 300

Glt (daN/mm2) 230

νlt 0.35

ep (mm) 0.05

Rlt (hb) 67.5

Rlc (hb) - 18.7

Rtt (hb) 5

Rtc (hb) - 5

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %60 %


440.146/85Curing135° C/1h30 or 150° C/1h


AgedT ≥ - 55° CT ≤ 70° C

El (daN/mm2) 2500

Et (daN/mm2) 2500

Glt (daN/mm2) 230

νlt 0.0423

ep (mm) 0.1

Rlt (hb) 35

Rlc (hb) - 10

Rtt (hb) 35

Rtc (hb) - 10

S (hb) 5

τinter (hb)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/1

Resin %50 % or 60%

KEVLAR (fabric)EHA 250-33-50 or EHA 250-33-60

TN-DE 423-7/83Curing125° C/3h


New

T = 80° CEl (daN/mm2) 2580

Et (daN/mm2) 2530

Glt (daN/mm2) 207

νlt

ep (mm) 0.25

Rlt (hb) 36.6

Rlc (hb) - 10

Rtt (hb) 36

Rtc (hb) - 8.8

S (hb)

τinter (hb) 2.5

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSWet lay-up glass - carbon hybrid fabrics Z 2.2.4.1

1/2

Resin % GLASS - CARBON HYBRID (equiv. tape)G806 + T120/5052 - 9390 - 9396 - 501

440.218/94Curing90° C (70° C - 501) 2h/1 bar

Surface densityNew

T = 20° C

Aged

T = 70° CEl (daN/mm2) 10000 10000

Et (daN/mm2) 300 300

Glt (daN/mm2) 300 300

νlt 0.35 0.35

ep (mm) 0.156 0.156

Rlt (hb) 50 40

Rlc (hb) - 38 - 21

Rtt (hb) 4 4

Rtc (hb) - 7 - 4

S (hb) 5 3

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSWet lay-up glass - carbon hybrid fabrics Z 2.2.4.1

2/2

Resin %50 %

GLASS - CARBON HYBRID (fabric)G806 + T120/5052 - 9390 - 9396 - 501

440.218/94Curing90° C (70° C - 501) 2h/1 bar


New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 5167 5167

Et (daN/mm2) 5167 5167

Glt (daN/mm2) 300 170

νlt 0.0204 0.0204

ep (mm) 0.312 0.312

Rlt (hb) 25.6 20.5

Rlc (hb) - 19.6 - 10.9

Rtt (hb) 25.6 20.5

Rtc (hb) - 19.6 - 10.9

S (hb) 5 3

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/3

Resin % GLASS - CARBON HYBRID (equiv. tape)G874/V250528/070/90Curing


Aged

T = 20° CEl (daN/mm2) 1650

Et (daN/mm2)

Glt (daN/mm2)

νlt

ep (mm) 0.15

Rlt (hb) 20.5

Rlc (hb) - 18

Rtt (hb)

Rtc (hb)

S (hb)

τinter (hb) G874/V250 (1) G874/V250 (1)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/3

Resin %47 %

GLASS - CARBON HYBRID (fabric)G874/V250528/070/90Curing


325 gr/m2

Aged

T = 20° CEl (daN/mm2) (825)

Et (daN/mm2) (825)

Glt (daN/mm2)

νlt

ep (mm) (0.4)

Rlt (hb) (10.2)

Rlc (hb) (- 9)

Rtt (hb) (10.2)

Rtc (hb) (- 9)

S (hb)

τinter (hb) G874/V250 (1) G874/V250 (1)

K mc 0.2

K mt 0.2

K tc 0.85

K tt 0.6

σm (hb) 35

K tflexion 0.9

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




3/3

SHEET G874/V250 (1)

Number of fabrics

τinter.

0 5 10 15 20 30250

0.4

1.2

1.6

2

2.4

2.8

0.8



MATERIAL CHARACTERISTICSWet lay-up carbon fabrics Z 2.4.1.1

1/2


432.308/96Curing80° C/2h/1 bar

Surface densityNew

T = 23° C

Aged

T = 55° C

New

T = 70° C

Aged

T = 70° CEl (daN/mm2) 13700 10900 13900 10900

Et (daN/mm2) 500 500 500 500

Glt (daN/mm2) 313 240 240 240

νlt 0.35 0.35 0.35 0.35

ep (mm) 0.15 0.15 0.15 0.15

Rlt (hb) 111 90 98 86

Rlc (hb) - 81 - 61 - 64 - 55

Rtt (hb) 5 5 5 5

Rtc (hb) - 10 - 10 - 10 - 10

S (hb) 7.38 2.96 5.35 2.3

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSWet lay-up carbon fabrics Z 2.4.1.1

2/2

Resin %50 %


432.308/96Curing80° C/2h/1 bar


New

T = 23° C

Aged

T = 55° C

New

T = 70° C

Aged

T = 70° CEl (daN/mm2) 7128 5727 7228 5727

Et (daN/mm2) 7128 5727 7228 5727

Glt (daN/mm2) 313 240 250 240

νlt 0.0246 0.0307 0.0243 0.0307

ep (mm) 0.3 0.3 0.3 0.3

Rlt (hb) 56 45.66 49.67 44

Rlc (hb) - 42 - 32 - 33.33 - 28.8

Rtt (hb) 56 45.66 49.67 44

Rtc (hb) - 42 - 32 - 33.33 - 28.8

S (hb) 7.38 2.96 5.35 2.3

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


528/068/94Curing90° C/2h/1 bar

surface densityNew

T = 20° C

Aged

T = 80° CEl (daN/mm2) 11250 11250

Et (daN/mm2) 300 300

Glt (daN/mm2) 300 300

νlt 0.35 0.35

ep (mm) 0.078 0.078

Rlt (hb) 97.6 86.9

Rlc (hb) - 81.6 - 34.4

Rtt (hb) 2.4 2.4

Rtc (hb) - 10 - 10

S (hb) 7.9 2.04

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %50 %


528/068/94Curing90° C/2h/1 bar


New

T = 20° C

Aged

T = 80° CEl (daN/mm2) 5792 5792

Et (daN/mm2) 5792 5792

Glt (daN/mm2) 300 300

νlt 0.0182 0.0182

ep (mm) 0.156 0.156

Rlt (hb) 46.2 42.9

Rlc (hb) - 41.7 - 17.9

Rtt (hb) 46.2 42.9

Rtc (hb) - 41.7 - 17.9

S (hb) 7.9 2.04

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2


432.0533/96 issue 1Curing90° C/2h/1 bar

Surface densityNew

T = 20° C

New

T = 80° C

Agedwet

T = 20° C

Agedwet

T = 80° C

Agedwet

T = 70° CEl (daN/mm2) 10450 10450 10450 10450 10450

Et (daN/mm2) 283 283 283 283 283

Glt (daN/mm2) 283 283 283 283 283

νlt 0.35 0.35 0.35 0.35 0.35

ep (mm) 0.078 0.078 0.078 0.078 0.078

Rlt (hb) 105 105 105 93.4 93.4

Rlc (hb) - 72 - 49 - 71 - 42.3 - 47

Rtt (hb) 6 6 6 6 6

Rtc (hb) - 10 - 10 - 10 - 10 - 10

S (hb) 9.8 5.75 7.3 4.3 4.8

τinter (hb)

K mc

K mt

K tc 0.79

(Ø 3.2; E/G = 2.35)0.79

(Ø 3.2; E/G = 2.35)0.79

(Ø 3.2; E/G = 2.35)0.79

(Ø 3.2; E/G = 2.35)

K tt 0.78

(Ø 3.2; E/G = 2.35)0.78

(Ø 3.2; E/G = 2.35)0.78

(Ø 3.2; E/G = 2.35)0.78

(Ø 3.2; E/G = 2.35)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %50 %




New

T = 20° C

Agedwet

T = 70° CEl (daN/mm2) 5383 5383

Et (daN/mm2) 5383 5383

Glt (daN/mm2) 283 283

νlt 0.02 0.02

ep (mm) 0.156 0.156

Rlt (hb) 53.5 47.7

Rlc (hb) - 37.1 - 24.2

Rtt (hb) 53.5 47.7

Rtc (hb) - 37.1 - 24.2

S (hb) 9.8 4.8

τinter (hb)

K mc

K mt

K tc 0.79

(Ø 3.2; E/G = 2.35)0.79

(Ø 3.2; E/G = 2.35)

K tt 0.78

(Ø 3.2; E/G = 2.35)0.78

(Ø 3.2; E/G = 2.35)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2



Surface densityNew

T = 20° C

Aged

T = 70° CEl (daN/mm2) 10450 10450

Et (daN/mm2) 283 283

Glt (daN/mm2) 283 283

νlt 0.35 0.35

ep (mm) 0.078 0.078

Rlt (hb) 105 93.4

Rlc (hb) - 72 - 54

Rtt (hb) 6 6

Rtc (hb) - 10 - 10

S (hb) 9.8 4.8

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %50 %




New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 5383 5383

Et (daN/mm2) 5383 5383

Glt (daN/mm2) 283 283

νlt 0.02 0.02

ep (mm) 0.156 0.156

Rlt (hb) 53.5 47.7

Rlc (hb) - 37.1 - 27.8

Rtt (hb) 53.5 6

Rtc (hb) - 37.1 - 10

S (hb) 9.8 4.8

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % CARBON FABRIC (equiv. tape)G814/501Curing

90° C/2h/1 barSurface density

Aged

T = 70° CEl (daN/mm2) 11250

Et (daN/mm2) 300

Glt (daN/mm2) 300

νlt 0.35

ep (mm) 0.115

Rlt (hb) 86.9

Rlc (hb) - 34.4

Rtt (hb) 2.4

Rtc (hb) - 10

S (hb) 3.4

τinter (hb)

K mc 0.25

K mt 0.25

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %50 % CARBON FABRIC (fabric)

G814/501Curing70° C/2h


Aged

T = 70° CEl (daN/mm2) 5792

Et (daN/mm2) 5792

Glt (daN/mm2) 300

νlt 0.0182

ep (mm) 0.23

Rlt (hb) 42.9

Rlc (hb) - 17.8

Rtt (hb) 42.9

Rtc (hb) - 17.8

S (hb) 3.4

τinter (hb)

K mc 0.25

K mt 0.25

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSWet Lay-up glass fabrics Z 2.4.2.1

1/1


1581/501Curing70° C/2h/1 bar


Aged

T = 25° CEl (daN/mm2) 2070

Et (daN/mm2) 2070

Glt (daN/mm2) 260

νlt

ep (mm) 0.3

Rlt (hb) 30

Rlc (hb) - 25

Rtt (hb) 30

Rtc (hb) - 25

S (hb) 5

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/1

Resin % KEVLAR (fabric)181 + EPOXY RESINCuring


New

T = 20° CEl (daN/mm2)

Et (daN/mm2)

Glt (daN/mm2)

νlt

ep (mm) 0.28

Rlt (hb) 40

Rlc (hb) - 10

Rtt (hb) 40

Rtc (hb) - 10

S (hb)

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSRTM Z 3.1.1

1/2

Resin % RTM (equiv. tape)G1151/RTM6

432.0123/95 & 440.417/94Curing180° C/3h30/2 at 3 bars

Surface densityNew

T = 20° C

Aged

T = 70° C

NewT = 20° C

(sock)

AgedT = 70° C

(sock)El (daN/mm2) 11200 11200 10200 10200

Et (daN/mm2) 527 527 480 480

Glt (daN/mm2) 527 527 480 480

νlt 0.35 0.35 0.35 0.35

ep (mm) 0.3 0.3 0.3 0.3

Rlt (hb) 108 108 81 72.1

Rlc (hb) - 68 - 56 - 65 - 55.2

Rtt (hb) 5 5 5 5

Rtc (hb) - 10 - 10 - 10 - 10

S (hb) 8.6 6.2 8.8 6.2

τinter (hb)

K mc 0.25 0.25 0.25 0.25

K mt 0.25 0.25 0.25 0.25

K tc 1 (Ø 4.8) 1 (Ø 4.8) 1 (Ø 4.8) 1 (Ø 4.8)

K tt 0.7 (Ø 4.8) 0.7 (Ø 4.8) 0.7 (Ø 4.8) 0.7 (Ø 4.8)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin %56 %

RTM (fabric)G1151/RTM6

432.0123/95 & 440.417/94Curing180° C/3h30/2 at 3 bars


New

T = 20° C

Aged

T = 70° C

NewT = 20° C

(sock)

AgedT = 70° C

(sock)El (daN/mm2) 5892 5892 5366 5366

Et (daN/mm2) 5892 5892 5366 5366

Glt (daN/mm2) 527 527 480 480

νlt 0.0315 0.0315 0.03 0.03

ep (mm) 0.6 0.6 0.6 0.6

Rlt (hb) 54.1 54.1 41.5 37.1

Rlc (hb) - 35.6 - 29.4 - 34.1 - 28.9

Rtt (hb) 54.1 54.1 41.5 37.1

Rtc (hb) - 35.6 - 29.4 - 34.1 - 28.9

S (hb) 8.6 6.2 8.8 6.2

τinter (hb)

K mc 0.25 0.25 0.25 0.25

K mt 0.25 0.25 0.25 0.25

K tc 1 (Ø 4.8) 1 (Ø 4.8) 1 (Ø 4.8) 1 (Ø 4.8)

K tt 0.7 (Ø 4.8) 0.7 (Ø 4.8) 0.7 (Ø 4.8) 0.7 (Ø 4.8)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % RTM (equiv. tape)G986/RTM6Curing

150° C/3h30/2 at 3 barsSurface density

Aged

T = 70° CEl (daN/mm2) 10299

Et (daN/mm2) 448

Glt (daN/mm2) 448

νlt 0.35

ep (mm) 0.1675

Rlt (hb) 76.6

Rlc (hb) - 68.1

Rtt (hb) 4.48

Rtc (hb) - 8.96

S (hb) 5.8

τinter (hb)

K mc 0.25

K mt 0.25

K tc 0.85

K tt 0.65

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin % RTM (fabric)G986/RTM6Curing


305 gr/m2

Aged

T = 70° CEl (daN/mm2) 5398

Et (daN/mm2) 5398

Glt (daN/mm2) 448

νlt 0.0292

ep (mm) 0.335

Rlt (hb) 39.1

Rlc (hb) - 35.5

Rtt (hb) 39.1

Rtc (hb) - 35.5

S (hb) 5.82

τinter (hb)

K mc 0.25

K mt 0.25

K tc 0.85

K tt 0.65

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % RTM (equiv. tape)E4049/RTM6440.417/94Curing

180° CSurface density

New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 10200 10200

Et (daN/mm2) 480 480

Glt (daN/mm2) 480 480

νlt 0.35 0.35

ep (mm) 0.3 0.3

Rlt (hb) 81 72.1

Rlc (hb) - 65 - 55.2

Rtt (hb) 5 5

Rtc (hb) - 10 - 10

S (hb) 8.8 6.2

τinter (hb)

K mc

K mt

K tc 1 (Ø 4.8) 1 (Ø 4.8)

K tt 0.7 (Ø 4.8) 0.7 (Ø 4.8)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin % RTM (fabric)E4049/RTM6440.417/94Curing

180° Csurface density

630 gr/m2

New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 5366 5366

Et (daN/mm2) 5366 5366

Glt (daN/mm2) 480 480

νlt 0.03 0.03

ep (mm) 0.6 0.6

Rlt (hb) 41.5 37.1

Rlc (hb) - 34.1 - 28.9

Rtt (hb) 41.5 37.1

Rtc (hb) - 34.1 - 28.9

S (hb) 8.8 6.2

τinter (hb)

K mc

K mt

K tc 1 (Ø 4.8) 1 (Ø 4.8)

K tt 0.7 (Ø 4.8) 0.7 (Ø 4.8)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % RTM (equiv. tape)GB305/DA3200

440.102/94Curing120° C

Surface densityNew

T = 20° C

Aged

T = 70° CEl (daN/mm2) 9800 9800

Et (daN/mm2) 424 424

Glt (daN/mm2) 424 424

νlt 0.35 0.35

ep (mm) 0.15 0.15

Rlt (hb) 178 178

Rlc (hb) - 40 - 34

Rtt (hb) 8 8

Rtc (hb) - 10 - 10

S (hb) 5.8 4.8

τinter (hb)

K mc

K mt

K tc 1 1

K tt 0.8 (Ø 3.2) 0.65 (Ø 3.2)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin % RTM (fabric)GB305/DA3200

440.102/94Curing120° C

Surface densityNew

T = 20° C

Aged

T = 70° CEl (daN/mm2) 5135 5135

Et (daN/mm2) 5135 5135

Glt (daN/mm2) 424 424

νlt 0.029 0.029

ep (mm) 0.3 0.3

Rlt (hb) 89.3 89.3

Rlc (hb) - 21 - 17.7

Rtt (hb) 89.3 89.3

Rtc (hb) - 21 - 17.7

S (hb) 5.8 4.8

τinter (hb)

K mc

K mt

K tc 1 1

K tt 0.8 (Ø 3.2) 0.65 (Ø 3.2)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % RTM (equiv. tape)GF630/RTM6528/082/94Curing


New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 11000 11000

Et (daN/mm2) 480 480

Glt (daN/mm2) 480 480

νlt 0.35 0.35

ep (mm) 0.3 0.3

Rlt (hb) 95.5 95.5

Rlc (hb) - 50 - 39.5

Rtt (hb) 5 5

Rtc (hb) - 10 - 10

S (hb) 8.8 6.9

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin % RTM (fabric)GF630/RTM6528/082/94Curing


New

T = 20° C

Aged

T = 70° CEl (daN/mm2) 5766 5766

Et (daN/mm2) 5766 5766

Glt (daN/mm2) 480 480

νlt 0.0293 0.0293

ep (mm) 0.6 0.6

Rlt (hb) 29 29

Rlc (hb) - 26.2 - 20.7

Rtt (hb) 29 29

Rtc (hb) - 26.2 - 20.7

S (hb) 8.8 6.9

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % RTM (equiv. tape)GB305/XB5142

440.102/94Curing120° C

Surface densityNew

T = 20° C

Aged

T = 70° CEl (daN/mm2) 9800 9800

Et (daN/mm2) 424 424

Glt (daN/mm2) 424 424

νlt 0.35 0.35

ep (mm) 0.15 0.15

Rlt (hb) 167 167

Rlc (hb) - 37 - 27

Rtt (hb) 7.5 7.7

Rtc (hb) - 10 - 10

S (hb) 5.75 4.8

τinter (hb)

K mc

K mt

K tc 1 1

K tt 0.8 (Ø 3.2) 0.65 (Ø 3.2)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




2/2

Resin % RTM (fabric)GB305/XB5142

440.102/94Curing120° C

Surface densityNew

T = 20° C

Aged

T = 70° CEl (daN/mm2) 5135 5135

Et (daN/mm2) 5135 5135

Glt (daN/mm2) 424 424

νlt 0.029 0.029

ep (mm) 0.3 0.3

Rlt (hb) 84 84

Rlc (hb) - 19.3 - 14.3

Rtt (hb) 84 84

Rtc (hb) - 19.3 - 14.3

S (hb) 5.75 4.8

τinter (hb)

K mc

K mt

K tc 1 1

K tt 0.8 (Ø 3.2) 0.65 (Ø 3.2)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % RTM (equiv. tape)HF360/LY564-1 + HY2954

440.171/94Curing120° C

Surface density

NewT = 23° C

AgedT = 80° C

(MEC)

NewT = 23° C

sockeffect

AgedT = 80° C

sockeffect(MEC)

El (daN/mm2) 6830 6830 6830 6830

Et (daN/mm2) 416 416 416 416

Glt (daN/mm2) 416 416 416 416

νlt 0.35 0.35 0.35 0.35

ep (mm) 0.16 0.16 0.16 0.16

Rlt (hb) 82.7 70.3 62.9 53.4

Rlc (hb) - 56.6 - 30.7 - 43 - 23.3

Rtt (hb) 5 5 3.8 3.8

Rtc (hb) - 10 - 10 - 7.6 - 7.6

S (hb) 7.1 (4.3) 5.4 (3.3)

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin % RTM (fabric)HF360/LY564-1 + HY2954

440.171/94Curing120° C


NewT = 23° C

AgedT = 80° C

(MEC)

NewT = 23° C

sockeffect

AgedT = 80° C

sockeffect(MEC)

El (daN/mm2) 3644 3644 3644 3644

Et (daN/mm2) 3644 3644 3644 3644

Glt (daN/mm2) 416 416 416 416

νlt 0.0402 0.0402 0.0402 0.0402

ep (mm) 0.32 0.32 0.32 0.32

Rlt (hb) 42.2 36.3 32.2 27.5

Rlc (hb) - 30 - 16.3 - 22.8 - 12.5

Rtt (hb) 42.2 36.3 32.2 27.5

Rtc (hb) - 30 - 16.3 - 22.8 - 12.5

S (hb) 7.1 (4.3) 5.4 (3.3)

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % RTM (equiv. tape)GF520/LY564-1 + HY2954

440.171/94Curing120° C

Surface density

NewT = 23° C

AgedT = 80° C

(MEC)

NewT = 23° C

sockeffect

AgedT = 80° C

sockeffect(MEC)

El (daN/mm2) 11400 11400 11400 11400

Et (daN/mm2) 500 500 500 500

Glt (daN/mm2) 500 500 500 500

νlt 0.35 0.35 0.35 0.35

ep (mm) 0.25 0.25 0.25 0.25

Rlt (hb) 95.4 81.1 72.2 61.6

Rlc (hb) - 50.2 - 35.1 - 38.1 - 26.7

Rtt (hb) 5 5 3.8 3.8

Rtc (hb) - 10 - 10 - 7.6 - 7.6

S (hb) 9.4 (5.6) 7.1 (4.3)

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)




2/2

Resin % RTM (fabric)GF520/LY564-1 + HY2954

440.171/94Curing120° C


NewT = 23° C

AgedT = 80° C

(MEC)

NewT = 23° C

sockeffect

AgedT = 80° C

sockeffect(MEC)

El (daN/mm2) 5977 5977 5977 5977

Et (daN/mm2) 5977 5977 5977 5977

Glt (daN/mm2) 500 500 500 500

νlt 0.0294 0.0294 0.0294 0.0294

ep (mm) 0.5 0.5 0.5 0.5

Rlt (hb) 48.4 41.4 36.8 31.6

Rlc (hb) - 26.4 - 18.4 - 20 - 14

Rtt (hb) 48.4 41.4 36.8 31.6

Rtc (hb) - 26.4 - 18.4 - 20 - 14

S (hb) 9.4 (5.6) 7.1 (4.3)

τinter (hb)

K mc

K mt

K tc

K tt

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd)

εadm. tract. (µd)


Tg wet (° C)

Cθ (µd/° C)



MATERIAL CHARACTERISTICSInjected thermoplastics Z 4.1.1.1

1/1

Resin %30 %

THERMOPLASTICULTEM 2310440.106/92Curing

Surface densityNew

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 1150 1150 1150

Et (daN/mm2)

Glt (daN/mm2)

νlt

ep (mm)

Rlt (hb) 14.3 12.5 11.6

Rlc (hb) - 24.2 - 23.9

Rtt (hb)

Rtc (hb)

S (hb)

τinter (hb)

K mc 0.17 0.17 0.17

K mt 0.17 0.17 0.17

K tc

K tt 0.6 0.6 0.6

σm (hb)

K tflexion




1/1

Resin %40 %

THERMOPLASTICRYTON R04440.106/92Curing

Surface densityNew

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 1450 1450 1450

Et (daN/mm2)

Glt (daN/mm2)

νlt

ep (mm)

Rlt (hb) 11.4 10.3 11.2

Rlc (hb) - 19.6 - 18.6

Rtt (hb)

Rtc (hb)

S (hb)

τinter (hb)

K mc 0.6 0.6 0.6

K mt 0.6 0.6 0.6

K tc

K tt 0.59 0.59 0.59

σm (hb)

K tflexion




1/1

Resin %50 %

THERMOPLASTICIXEF 1022440.106/92Curing

Surface densityNew

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 1100 1100 1100

Et (daN/mm2)

Glt (daN/mm2)

νlt

ep (mm)

Rlt (hb) 17.2 8.5 5.2

Rlc (hb) - 26 - 12.9

Rtt (hb)

Rtc (hb)

S (hb)

τinter (hb)

K mc 0.09 to 0.57 0.09 to 0.57 0.09 to 0.57

K mt 0.09 to 0.57 0.09 to 0.57 0.09 to 0.57

K tc

K tt 0.6 0.6 0.6

σm (hb)

K tflexion




1/1

Resin %30 %

THERMOPLASTICIXEF C36

440.106/92Curing

surface densityNew

T = 20° C

Aged

T = 20° C

Aged

T = 70° CEl (daN/mm2) 1500 1500 1500

Et (daN/mm2)

Glt (daN/mm2)

νlt

ep (mm)

Rlt (hb) 19 9.9 5.8

Rlc (hb) - 25.1 - 12.6

Rtt (hb)

Rtc (hb)

S (hb)

τinter (hb)

K mc 0.13 to 0.25 0.13 to 0.25 0.13 to 0.25

K mt 0.13 to 0.25 0.13 to 0.25 0.13 to 0.25

K tc

K tt 0.63 0.63 0.63

σm (hb)

K tflexion



MATERIAL CHARACTERISTICSLong fibre thermoplastics Z 5.1.1.1

1/1

Resin %32 %

THERMOPLASTICAPC2 (AS4/PEEK)

581.0053/98Curing395° C/35mn/2 bars


New

T = 20° C

Aged

T = 70° C

Aged

T = 120° CEl (daN/mm2) 13400 13400 13400

Et (daN/mm2) 900 900 900

Glt (daN/mm2) 500 500 500

νlt 0.40 0.40 0.40

ep (mm) 0.13 0.13 0.13

Rlt (hb) 150 150 150

Rlc (hb) - 115 - 100 - 95

Rtt (hb) 11 11 11

Rtc (hb) - 12 - 12 - 12

S (hb) 10 10 9

τinter (hb)

K mc T300/914 (1) T300/914 (1) T300/914 (1)

K mt T300/914 (2) T300/914 (2) T300/914 (2)

K tc T300/914 (3) T300/914 (3) T300/914 (3)

K tt T300/914E (4) T300/914E (4) T300/914E (4)

σm (hb)

K tflexion

κc

κt

κs

εadm. comp. (µd) 3300 (6000 mm2)

εadm. tract. (µd) 9000 (6000 mm2)

γadm. cisail. (µd) 3500 (6000 mm2)

Tg dry (° C)

Tg wet (° C)

Cθ (µd/° C)




1/2

Resin % THERMOPLASTICCD282/PEI DE TENCATE (equiv. tape)

432.0015/95Curing300° C/7 bars

Surface densityNew

T = 20° C

Aged

T = 70° C

Skydrol

T = 70° CEl (daN/mm2) 11500 11500 11500

Et (daN/mm2) 500 500 500

Glt (daN/mm2) 400 400 300

νlt 0.35 0.35 0.35

ep (mm) 0.15 0.15 0.15

Rlt (hb) 90 85.8 85.8

Rlc (hb) - 85 - 76 - 49.5

Rtt (hb) 5 5 5

Rtc (hb) - 10 - 10 - 10

S (hb) 10 9 3.9

τinter (hb)

K mc 0.25 0.25 0.25

K mt 0.25 0.25 0.25

K tc 0.85 0.85 0.85

K tt 0.65 0.65 0.65

σm (hb) 40 40 40





2/2

Resin %42 %

THERMOPLASTICCD282/PEI DE TENCATE (fabric)

432.0015/95Curing300° C/7 bars


New

T = 20° C

Aged

T = 70° C

Skydrol

T = 70° CEl (daN/mm2) 6027 6207 5530

Et (daN/mm2) 6027 6207 5530

Glt (daN/mm2) 400 400 300

νlt 0.03 0.03 0.03

ep (mm) 0.3 0.3 0.3

Rlt (hb) 45.8 43.8 43.8

Rlc (hb) - 44.3 - 39.7 - 26

Rtt (hb) 45.8 43.9 43.8

Rtc (hb) - 44.3 - 39.7 - 26

S (hb) 10 9 3.9

τinter (hb)

K mc 0.25 0.25 0.25

K mt 0.25 0.25 0.25

K tc 0.85 0.85 0.85

K tt 0.65 0.65 0.65

σm (hb) 40 40 40




MATERIAL CHARACTERISTICSNomex honeycomb Z 9.1

1/12

Hexagonal cell1/8 inches

HEXCEL NOMEX HONEYCOMBHRH10 - 1/8 - 1.8

Hexcel – september 1989Curing

Volume density29 Kg/m3

New

T = 23°CEc (daN/mm2)

Rc (hb) 0.065Gl (daN/mm2) 2.6Gw (daN/mm2) 1.0

Sl (hb) 0.052Sw (hb) 0.028



528 – 099/90Curing


New

T = 23°C

New

T = 80°C

Aged

T = 80°CEc (daN/mm2) 13.8 13.8

Rc (hb) 0.15* 0.15 0.15Gl (daN/mm2) 3.2* 3.2Gw (daN/mm2) 2.4 2.4

Sl (hb) 0.075* 0.075 0.075Sw (hb) 0.046 0.046 0.046


HEXCEL NOMEX HONEYCOMBHRH10 - 1/8 - 4.030147 of 21/08/92

Curing


New

T = 23°CEc (daN/mm2) 19.3


Sl (hb) 0.155Sw (hb) 0.076

* ASTA/SOCATA HUREL-DUBOIS use these values increased by 20 %.




2/12





New



Sl (hb) 0.190Sw (hb) 0.103





New

T = 23°CEc (daN/mm2) 41.4


Sl (hb) 0.228Sw (hb) 0.117





New

T = 23°CEc (daN/mm2) 53.8


Sl (hb) 0.276Sw (hb) 0.145




3/12





New

T = 23°CEc (daN/mm2) 62.1


Sl (hb) 0.293Sw (hb) 0.172





New



Sl (hb) 0.041Sw (hb) 0.022





New



Sl (hb) 0.052Sw (hb) 0.028




4/12





New

T = 23°CEc (daN/mm2) 7.6


Sl (hb) 0.062Sw (hb) 0.031



528 – 099/90Curing


New

T = 23°C

New

T = 80°C

Aged

T = 80°CEc (daN/mm2) 13.8 13.8


Sl (hb) 0.075* 0.075 0.075Sw (hb) 0.046 0.046 0.046



Curing


New

T = 23°CEc (daN/mm2) 19.3


Sl (hb) 0.155Sw (hb) 0.076





5/12


HEXCEL NOMEX HONEYCOMBHRH10 - 3/16 - 4.5Hexcel – may 1986

Curing


New



Sl (hb) 0.155Sw (hb) 0.076





New


Rc (hb) 0.596Gl (daN/mm2) 9Gw (daN/mm2) 4.5

Sl (hb) 0.255Sw (hb) 0.138





New

T = 23°CEc (daN/mm2) 4.1


Sl (hb) 0.038Sw (hb) 0.017




6/12





New

T = 23°CEc (daN/mm2) 7.6


Sl (hb) 0.059Sw (hb) 0.028



528 – 099/90Curing


New

T = 23°C

New

T = 80°C

Aged

T = 80°CEc (daN/mm2) 13.8 13.8


Sl (hb) 0.075* 0.075 0.075Sw (hb) 0.046 0.046 0.046



Curing


New

T = 23°CEc (daN/mm2) 19.3


Sl (hb) 0.155Sw (hb) 0.076





7/12





New

T = 23°CEc (daN/mm2) 4.1


Sl (hb) 0.038Sw (hb) 0.017





New

T = 23°CEc (daN/mm2) 7.6


Sl (hb) 0.050Sw (hb) 0.025





New

T = 23°CEc (daN/mm2) 11.7


Sl (hb) 0.110Sw (hb) 0.055




8/12

OX - Core cell3/16 inches

HEXCEL NOMEX HONEYCOMBHRH10/OX - 3/16 - 1.8



New



Sl (hb) 0.031Sw (hb) 0.034



528 – 099/90Curing


New

T = 23°CEc (daN/mm2) 9.5


Sl (hb) 0.053Sw (hb) 0.064





New



Sl (hb) 0.072Sw (hb) 0.090




9/12



528 – 099/90Curing


New

T = 23°CEc (daN/mm2) 9.5


Sl (hb) 0.053Sw (hb) 0.064




10/12

Flex - Core cell HEXCEL NOMEX HONEYCOMBHRH10 – F35 – 2.5

Hexcel – october 1992Curing


New

T = 23°CEc (daN/mm2) 8.3


Sl (hb) 0.062Sw (hb) 0.034




New

T = 23°CEc (daN/mm2) 16.5


Sl (hb) 0.117Sw (hb) 0.062




New

T = 23°CEc (daN/mm2) 22.8


Sl (hb) 0.159Sw (hb) 0.103




11/12




New

T = 23°CEc (daN/mm2) 16.5


Sl (hb) 0.090Sw (hb) 0.052




New

T = 23°CEc (daN/mm2) 22.8


Sl (hb) 0.172Sw (hb) 0.097




12/12




New

T = 23°CEc (daN/mm2) 25.5


Sl (hb) 0.207Sw (hb) 0.117




New

T = 23°CEc (daN/mm2) 29.0


Sl (hb) 0.221Sw (hb) 0.124



MATERIAL CHARACTERISTICSFiberglass honeycomb Z 9.2

1/10


HEXCEL FIBERGLASS HONEYCOMBHRP - 3/16 - 4.0

Hexcel – july 1989Curing


New

T = 23°CEc (daN/mm2) 39.3


Sl (hb) 0.145Sw (hb) 0.090





New

T = 23°CEc (daN/mm2) 65.5


Sl (hb) 0.241Sw (hb) 0.138





New

T = 23°CEc (daN/mm2) 93.8


Sl (hb) 0.345Sw (hb)




2/10





New

T = 23°CEc (daN/mm2) 113.1


Sl (hb) 0.414Sw (hb) 0.255


HEXCEL FIBERGLASS HONEYCOMBHRP - 3/16 - 12.0Hexcel – july 1989

Curing


New

T = 23°CEc (daN/mm2) 179.3*


Sl (hb) 0.562Sw (hb) 0.362

* Preliminary value




3/10





New

T = 23°CEc (daN/mm2) 31.7


Sl (hb) 0.117Sw (hb) 0.069





New

T = 23°CEc (daN/mm2) 48.3


Sl (hb) 0.172Sw (hb) 0.107





New

T = 23°CEc (daN/mm2) 57.9


Sl (hb)Sw (hb)




4/10





New

T = 23°CEc (daN/mm2) 82.7


Sl (hb)Sw (hb)




5/10





New

T = 23°CEc (daN/mm2) 9.0


Sl (hb) 0.062Sw (hb) 0.031





New

T = 23°CEc (daN/mm2) 26.2


Sl (hb) 0.110Sw (hb) 0.059





New

T = 23°CEc (daN/mm2) 44.8


Sl (hb) 0.179Sw (hb) 0.103




6/10





New

T = 23°CEc (daN/mm2) 68.9


Sl (hb) 0.234Sw (hb) 0.145





New

T = 23°CEc (daN/mm2) 103.4*

Rc (hb)Gl (daN/mm2) 21.4*Gw (daN/mm2) 9.0*

Sl (hb)Sw (hb)

* Preliminary values




7/10


HEXCEL FIBERGLASS HONEYCOMBHRP/OX - 1/4 - 4.5Hexcel – july 1989

Curing


New

T = 23°CEc (daN/mm2) 29.6*


Sl (hb) 0.117Sw (hb) 0.131



Curing


New

T = 23°CEc (daN/mm2) 44.8*

Rc (hb)Gl (daN/mm2) 6.9*Gw (daN/mm2) 12.4

Sl (hb)Sw (hb)



Curing


New

T = 23°CEc (daN/mm2) 57.9*

Rc (hb) 0.683*Gl (daN/mm2) 9.7*Gw (daN/mm2) 13.8*

Sl (hb)Sw (hb)




8/10



Curing


New

T = 23°CEc (daN/mm2) 22.1*


Sl (hb)Sw (hb)



Curing


New

T = 23°CEc (daN/mm2) 41.4*


Sl (hb)Sw (hb)





9/10

Flex - Core cell HEXCEL FIBERGLASS HONEYCOMBHRP - F35 - 2.5



New

T = 23°CEc (daN/mm2) 17.2


Sl (hb) 0.065Sw (hb) 0.038




New

T = 23°CEc (daN/mm2) 25.5


Sl (hb) 0.097Sw (hb) 0.052




New

T = 23°CEc (daN/mm2) 33.8


Sl (hb) 0.152Sw (hb) 0.076




10/10




New

T = 23°CEc (daN/mm2) 25.5


Sl (hb) 0.090Sw (hb) 0.045




New

T = 23°CEc (daN/mm2) 33.8


Sl (hb) 0.138Sw (hb) 0.069




New

T = 23°CEc (daN/mm2) 42.1


Sl (hb) 0.228Sw (hb) 0.124



MATERIAL CHARACTERISTICSAluminium honeycomb Z 9.3

1/33

HEXCEL ALUMINIUM HONEYCOMBACG - 1/4 - 4.8

Hexcel – december 1988Perforated hexagonal

cell1/4 inches


New

T = 23°CEc (daN/mm2) 102.0


Sl (hb) 0.231Sw (hb) 0.148



cell3/8 inches


New

T = 23°CEc (daN/mm2) 63.4


Sl (hb) 0.134Sw (hb) 0.090



cell1/2 inches


New

T = 23°CEc (daN/mm2) 27.6


Sl (hb) 0.086Sw (hb) 0.048




2/33



cell3/4 inches


New

T = 23°CEc (daN/mm2) 16.5


Sl (hb) 0.055Sw (hb) 0.041

HEXCEL ALUMINIUM HONEYCOMBACG - 1 - 1.3


cell1 inches


New

T = 23°CEc (daN/mm2) 11*

Rc (hb) 0.048Gl (daN/mm2) 9.7*Gw (daN/mm2) 4.8*

Sl (hb) 0.038*Sw (hb) 0.028*





3/33


HEXCEL ALUMINIUM HONEYCOMBCR III 2024 - 3/16 - 3.5



New

T = 23°CEc (daN/mm2) 59.3


Sl (hb) 0.159Sw (hb) 0.099





New

T = 23°CEc (daN/mm2) 137.9


Sl (hb) 0.276Sw (hb) 0.172





New

T = 23°CEc (daN/mm2) 206.8


Sl (hb) 0.414Sw (hb) 0.259




4/33





New

T = 23°CEc (daN/mm2) 262


Sl (hb) 0.531Sw (hb) 0.324





New

T = 23°CEc (daN/mm2) 330.9


Sl (hb) 0.655Sw (hb) 0.403





New

T = 23°CEc (daN/mm2) 27.6


Sl (hb) 0.097Sw (hb) 0.061




5/33


HEXCEL ALUMINIUM HONEYCOMBCR III 5052 - 1/16 - 6.5Hexcel – october 1988

Curing


New

T = 23°CEc (daN/mm2) 189.6*


Sl (hb) 0.331*Sw (hb) 0.207*



Curing


New

T = 23°CEc (daN/mm2) 289.6**

Rc (hb)Gl (daN/mm2) 72.4**Gw (daN/mm2) 36.5

Sl (hb) 0.576*Sw (hb) 0.359*



Curing


New

T = 23°CEc (daN/mm2) 448.2*

Rc (hb)Gl (daN/mm2) 82.7**Gw (daN/mm2) 44.8**

Sl (hb)Sw (hb)

* Preliminary values** Predicted values




6/33



Hexcel – march 1988Curing


New

T = 23°CEc (daN/mm2) 448.2*

Rc (hb) 1.586*Gl (daN/mm2) 103.4**Gw (daN/mm2) 51.7**

Sl (hb) 0.896**Sw (hb) 0.517**


HEXCEL ALUMINIUM HONEYCOMBCR III 5052 - 3/32 - 4.3Hexcel – march 1988

Curing


New

T = 23°CEc (daN/mm2) 82.7**

Rc (hb) 0.303**Gl (daN/mm2) 42.7**Gw (daN/mm2) 18.6**

Sl (hb) 0.2**Sw (hb) 0.128**



Curing


New

T = 23°CEc (daN/mm2) 172.4**


Sl (hb) 0.365**Sw (hb) 0.228**





7/33



Curing


New

T = 23°CEc (daN/mm2) 51.7


Sl (hb) 0.107Sw (hb) 0.062



Curing


New

T = 23°CEc (daN/mm2) 103.4


Sl (hb) 0.196Sw (hb) 0.359


HEXCEL ALUMINIUM HONEYCOMBCR III 5052 - 1/8 - 6.1Rohr – RHR 95-057

Curing


New

T = 23°C T = 120°CEc (daN/mm2)

Rc (hb) 0.448 0.448Gl (daN/mm2) 53 53Gw (daN/mm2) 21.2 21.2

Sl (hb) 0.310 0.279Sw (hb) 0.186 0.168





8/33



Curing


New

T = 23°CEc (daN/mm2) 241.3


Sl (hb) 0.462Sw (hb) 0.276



Rohr – RHR 85-054 ADD CCuring


New

T = 23°C T = 77°C T = 113°CEc (daN/mm2)

Rc (hb) 1.317 1.224 1.152Gl (daN/mm2) 93.1 91 89.6Gw (daN/mm2) 41.4 40.7 40.0

Sl (hb) 0.862 0.801 0.754Sw (hb) 0.517 0.481 0.452



Rohr – RHR 95-057Curing


New

T = 23°C T = 120°CEc (daN/mm2)

Rc (hb) 3.103 3.103Gl (daN/mm2) 144.8 144.8Gw (daN/mm2) 68.9 68.9

Sl (hb) 1.724 1.255Sw (hb) 1.034 0.503




9/33



Curing


New

T = 23°CEc (daN/mm2) 37.9


Sl (hb) 0.083Sw (hb) 0.048



Curing


New

T = 23°CEc (daN/mm2) 75.8


Sl (hb) 0.148Sw (hb) 0.086



Curing


New

T = 23°CEc (daN/mm2) 134.4


Sl (hb) 0.255Sw (hb) 0.148




10/33



Curing


New

T = 23°CEc (daN/mm2) 196.5

Rc (hb) 0.552Gl (daN/mm2) 78.6Gw (daN/mm2) 32

Sl (hb) 0.372Sw (hb) 0.226



Curing


New

T = 23°CEc (daN/mm2) 255.1


Sl (hb) 0.476Sw (hb) 0.290




11/33



Curing


New

T = 23°CEc (daN/mm2) 23.4


Sl (hb) 0.055Sw (hb) 0.032



Curing


New

T = 23°CEc (daN/mm2) 51.7


Sl (hb) 0.107Sw (hb) 0.062



Curing


New

T = 23°CEc (daN/mm2) 100


Sl (hb) 0.193Sw (hb) 0.110




12/33



Curing


New

T = 23°CEc (daN/mm2) 151.7


Sl (hb) 0.283Sw (hb) 0.168



Curing


New

T = 23°CEc (daN/mm2) 196.5


Sl (hb) 0.372Sw (hb) 0.226



Rohr - RHR 95-057Curing


New

T = 23°C T = 120°CEc (daN/mm2)


Sl (hb) 0.469 0.422Sw (hb) 0.276 0.248




13/33



Curing


New

T = 23°CEc (daN/mm2) 13.8


Sl (hb) 0.041Sw (hb) 0.022



Curing


New

T = 23°CEc (daN/mm2) 31


Sl (hb) 0.069Sw (hb) 0.039



Curing


New

T = 23°CEc (daN/mm2) 62.1


Sl (hb) 0.124Sw (hb) 0.072




14/33



Curing


New

T = 23°CEc (daN/mm2) 96.5


Sl (hb) 0.183Sw (hb) 0.107



Curing


New

T = 23°CEc (daN/mm2) 131


Sl (hb) 0.248Sw (hb) 0.138



Curing


New

T = 23°C T = 120°CEc (daN/mm2)


Sl (hb) 0.303 0.273Sw (hb) 0.183 0.164




15/33



Curing


New

T = 23°CEc (daN/mm2) 234.4


Sl (hb) 0.448Sw (hb) 0.269



Curing


New

T = 23°CEc (daN/mm2) 6.9


Sl (hb) 0.022Sw (hb) 0.014



Curing


New

T = 23°CEc (daN/mm2) 13.8


Sl (hb) 0.041Sw (hb) 0.022




16/33



Curing


New

T = 23°C T = 120°CEc (daN/mm2)


Sl (hb) 0.065 0.059Sw (hb) 0.038 0.034



Curing


New

T = 23°CEc (daN/mm2) 48.3


Sl (hb) 0.1Sw (hb) 0.059



Curing


New

T = 23°CEc (daN/mm2) 72.4


Sl (hb) 0.138Sw (hb) 0.079




17/33



Rohr – RHR 85-054 ADD CCuring


New

T = 23°C T = 77°CEc (daN/mm2) 93.1 84.8


Sl (hb) 0.162 0.151Sw (hb) 0.1 0.093



Curing


New

T = 23°C T = 120°CEc (daN/mm2)


Sl (hb) 0.262 0.23Sw (hb) 0.157 0.122



Curing


New

T = 23°CEc (daN/mm2) 182.7


Sl (hb) 0.345Sw (hb) 0.207




18/33

Flex - Core cell HEXCEL ALUMINIUM HONEYCOMB5052 - F40 - 2.1



New

T = 23°CEc (daN/mm2) 44.8


Sl (hb) 0.043Sw (hb) 0.026




New

T = 23°CEc (daN/mm2) 62.1**


Sl (hb) 0.065**Sw (hb) 0.038**




New

T = 23°CEc (daN/mm2) 86.2


Sl (hb) 0.087Sw (hb) 0.052

** Predicted values




19/33




New

T = 23°CEc (daN/mm2) 127.6


Sl (hb) 0.125Sw (hb) 0.079




New

T = 23°CEc (daN/mm2) 199.9


Sl (hb) 0.193Sw (hb) 0.117




20/33




New

T = 23°CEc (daN/mm2) 134.4


Sl (hb) 0.135Sw (hb) 0.083




New

T = 23°CEc (daN/mm2) 213.7


Sl (hb) 0.212Sw (hb) 0.124




New

T = 23°CEc (daN/mm2) 275.8


Sl (hb) 0.299Sw (hb) 0.179




21/33



Curing


New

T = 23°CEc (daN/mm2) 227.5**


Sl (hb)Sw (hb)



Curing


New

T = 23°CEc (daN/mm2) 344.7*


Sl (hb)Sw (hb)





22/33



Curing


New

T = 23°CEc (daN/mm2) 103.4**


Sl (hb) 0.248**Sw (hb) 0.152**



Curing


New

T = 23°CEc (daN/mm2) 241.3*

Rc (hb) 0.827*Gl (daN/mm2) 68.9*Gw (daN/mm2) 24.1*

Sl (hb) 0.462*Sw (hb) 0.265*





23/33



Reference Aerospatiale 5056 3.20Curing


New

T = 23°CEc (daN/mm2) 66.9


Sl (hb) 0.138Sw (hb) 0.076





New

T = 23°CEc (daN/mm2) 127.6


Sl (hb) 0.241Sw (hb) 0.141





New

T = 23°C T = 120°CEc (daN/mm2)

Rc (hb) 0.448 0.448Gl (daN/mm2) 53 53Gw (daN/mm2) 21.2 21.2

Sl (hb) 0.359 0.323Sw (hb) 0.210 0.189




24/33



Hexcel – october 1988Reference Aerospatiale 5056 3.58Curing


New

T = 23°CEc (daN/mm2) 299.9


Sl (hb) 0.510Sw (hb) 0.303



Curing


New

T = 23°CEc (daN/mm2) 48.3


Sl (hb) 0.105Sw (hb) 0.055



Curing


New

T = 23°CEc (daN/mm2) 96.5


Sl (hb) 0.188Sw (hb) 0.107




25/33



Curing


New

T = 23°CEc (daN/mm2) 165.5


Sl (hb) 0.300Sw (hb) 0.172



Curing


New

T = 23°CEc (daN/mm2) 241.3


Sl (hb) 0.421Sw (hb) 0.248




26/33





New

T = 23° CEc (daN/mm2) 31


Sl (hb) 0.072Sw (hb) 0.034



Rohr – RHR 95-057Reference Aerospatiale 5056 4.28Curing


New

T = 23° C T = 120° CEc (daN/mm2)

Rc (hb) 0.148 0.148Gl (daN/mm2) 22.1 22.1Gw (daN/mm2) 11 11

Sl (hb) 0.107 0.087Sw (hb) 0.062 0.048





New

T = 23° CEc (daN/mm2) 124.1


Sl (hb) 0.234Sw (hb) 0.137




27/33





New

T = 23°CEc (daN/mm2) 186.2


Sl (hb) 0.331Sw (hb) 0.193



Rohr – RHR 95-057Curing


New

T = 23°C T = 120°CEc (daN/mm2)


Sl (hb) 0.510 0.459Sw (hb) 0.296 0.267





New

T = 23°CEc (daN/mm2) 20.7


Sl (hb) 0.054Sw (hb) 0.026




28/33





New

T = 23°CEc (daN/mm2) 40


Sl (hb) 0.09Sw (hb) 0.043





New

T = 23°CEc (daN/mm2) 79.3


Sl (hb) 0.159Sw (hb) 0.09





New

T = 23°CEc (daN/mm2) 118.6


Sl (hb) 0.224Sw (hb) 0.131




29/33



Curing


New

T = 23°CEc (daN/mm2) 158.6


Sl (hb) 0.293Sw (hb) 0.169



Curing


New

T = 23°C T = 120°CEc (daN/mm2)


Sl (hb) 0.352 0.316Sw (hb) 0.207 0.186



Curing


New

T = 23°C T = 120°CEc (daN/mm2)


Sl (hb) 0.448 0.363Sw (hb) 0.269 0.209




30/33





New

T = 23°CEc (daN/mm2) 10.3


Sl (hb) 0.031Sw (hb) 0.017





New

T = 23°CEc (daN/mm2) 20.7


Sl (hb) 0.054Sw (hb) 0.026





New

T = 23°C T = 120°CEc (daN/mm2)


Sl (hb) 0.091 0.082Sw (hb) 0.049 0.044




31/33





New

T = 23°CEc (daN/mm2) 63.4


Sl (hb) 0.131Sw (hb) 0.069



Curing


New

T = 23°C T = 120°CEc (daN/mm2)


Sl (hb) 0.307 0.276Sw (hb) 0.179 0.161




32/33




New

T = 23°CEc (daN/mm2) 44.8


Sl (hb) 0.051Sw (hb) 0.029




New

T = 23°CEc (daN/mm2) 86.2


Sl (hb) 0.103Sw (hb) 0.062




New

T = 23°CEc (daN/mm2) 127.6


Sl (hb) 0.15Sw (hb) 0.091




33/33




New

T = 23°CEc (daN/mm2) 134.4


Sl (hb) 0.162Sw (hb) 0.095




New

T = 23°CEc (daN/mm2) 213.7


Sl (hb) 0.251Sw (hb) 0.147




New

T = 23°CEc (daN/mm2) 282.7


Sl (hb) 0.357Sw (hb) 0.212



MATERIAL CHARACTERISTICSFoam Z 10

1/3

Volume density32 ± 7 Kg/m3

ROHACELL FOAM31 A

RohacellTest standard

DIN 53420 Teststandard

NewT = 23°C

Modulus of elasticity E (daN/mm2) DIN 53457 2.6Compressive strength Rc (hb) DIN 53421 0.03

Tensile strength Rt (hb) DIN 53455 0.1Bending strength Rf (hb) DIN 53423 0.08

Shearing modulus G (daN/mm2) DIN 53294 0.8Shear strength S (hb) DIN 53294 0.03Elongation at break % DIN 53455 3.5

Dimensional stability at high temperature °C DIN 53424 180


ROHACELL FOAM51 A



NewT = 23°C



Shearing modulus G (daN/mm2) DIN 53294 1.3Shear strength S (hb) DIN 53294 0.06Elongation at break % DIN 53455 4



ROHACELL FOAM71 A



NewT = 23°C



Shearing modulus G (daN/mm2) DIN 53294 2.2Shear strength S (hb) DIN 53294 0.09Elongation at break % DIN 53455 4.5





2/3


ROHACELL FOAM51 WF



NewT = 23°C






ROHACELL FOAM71 WF



NewT = 23°C








3/3


ROHACELL FOAM110 WFRohacell

Test standardDIN 53420 Test

standardNew

T = 23°CModulus of elasticity E (daN/mm2) DIN 53457 13.5

Compressive strength Rc (hb) DIN 53421 0.22Tensile strength Rt (hb) DIN 53455 0.37Bending strength Rf (hb) DIN 53423 0.52

Shearing modulus G (daN/mm2) DIN 53294 4Shear strength S (hb) DIN 53294 0.175Elongation at break % DIN 53455 3



ROHACELL FOAM200 WFRohacell

Test standardDIN 53420 Test

standardNew

T = 23°CModulus of elasticity E (daN/mm2) DIN 53457 27

Compressive strength Rc (hb) DIN 53421 0.64Tensile strength Rt (hb) DIN 53455 0.68Bending strength Rf (hb) DIN 53423 1.2

Shearing modulus G (daN/mm2) DIN 53294 10Shear strength S (hb) DIN 53294 0.36Elongation at break % DIN 53455 3.5


Composite stress manual - Annex

© AEROSPATIALE - 1999 MTS 006 Iss. B 38Ann. page

Reference documents C BE 019 : Elaboration du dossier de justification structurale

Documents to be consulted

Abbreviations See Lexique Aerospatiale Aéronautique (AerospatialeAeronautique Lexicon)

Terminology List of words defined in the Lexique Aerospatiale Aéronautique(Aerospatiale Aeronautique Lexicon):

Highlights

Issue Date Pages modified Justification of the changes madeA 03/99 All New document.B 05/99 All Mises à jour Chapitres A, F, G, K, L, N, T, V.

Complément au chapitre Z.Rajout Chapitre I.

B

Composite stress manual - Management Information

© AEROSPATIALE - 1999 MTS 006 Iss. B MI page 1

NOT FOR DISTRIBUTION

List of approval

Dept. code Function Name / First name SignatureBTE/CC/SC Head of Department Mr THOMAS

Key words Calculation, Stressing

Bibliography -

Distribution list

Dept. code Function Name / First name (if necessary)

BQP/TE Diderot archives SIBADE AlainBQP/TE BQP/TE Library SIBADE Alain

BTE/SM/MG BT Technical Library BOUTET Fernand

Distribution list managed in real time by BIO/D (Didocost application)

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