(e-book composite) airbus composite stress manual us mts006 b

Technical Manual MTS 006 Iss. BOuthouse distribution authorised

Composite stress manual1 5 2 1 2 5 4 4

Structural Design Manual

Purpose

To list and homogenise the calculation methods and the allowable values for the composite materials used at the Aerospatiale Design Office. To be used as reference document for all Aerospatiale and subcontractors' stressmen.

Scope

Data processing tool supporting this Manual

Summary

See detailed summary

Document responsibility

Dept. code : BTE/CC/SC Name : P. CIAVALDINI

Validation

Name : JF. IMBERT Function: Deputy Department Group Manager Dept. code : BTE/CC/A Date : 06/05/99 Signature

This document belongs to AEROSPATIALE and cannot be given to third parties and/or be copied without prior authorisation from AEROSPATIALE and its contents cannot be disclosed. AEROSPATIALE - 1999

Composite stress manual

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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 Z are updated with each issue of the manual. This means that different values may be found in the stress dossiers prior to latest issue. The data processing tools are given for information purposes only.

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SUMMARY OF CHAPTERSIss. DETAILED SUMMARY B INTRODUCTION - COMPOSITE MATERIAL PROPERTIES COMPOSITE PLATE THEORY MONOLITHIC PLATE - MEMBRANE ANALYSIS MONOLITHIC PLATE - BENDING ANALYSIS MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS MONOLITHIC PLATE - TRANSVERSAL SHEAR ANALYSIS B MONOLITHIC PLATE - FAILURE CRITERIA MONOLITHIC PLATE - FATIGUE ANALYSIS MONOLITHIC PLATE - DAMAGE-TOLERANCE MONOLITHIC PLATE - BUCKLING MONOLITHIC PLATE - HOLE WITHOUT FASTENER ANALYSIS B MONOLITHIC PLATE - FASTENER HOLE MONOLITHIC PLATE - SPECIAL ANALYSIS B SANDWICHIC - MEMBRANE / BENDING / SHEAR ANALYSIS SANDWICH - FATIGUE ANALYSIS SANDWICH - DAMAGE-TOLERANCE APPROACH SANDWICH - BUCKLING ANALYSIS SANDWICH - SPECIFIC DESIGNS BONDED JOINTS B BONDED REPAIRS BOLTED REPAIRS B THERMAL CALCULATIONS ENVIRONMENTAL EFFECT NEW TECHNOLOGIES STATISTICS B MATERIAL PROPERTIES G H I J K L M N O P Q R S T U V W X Y Z * * * ** B Apr 99 P. Ciavaldini * * * * A B A B Jan 98 P. Ciavaldini Apr 99 P. Ciavaldini Jan 98 P. Ciavaldini Apr 99 P. Ciavaldini * B Apr 99 P. Ciavaldini * ** * B B Apr 99 P. Ciavaldini Apr 99 P. Ciavaldini B Apr 99 P. Ciavaldini B Apr 99 P. Ciavaldini A B C D E F * A A A B Jan 98 P. Ciavaldini Jan 98 P. Ciavaldini Jan 98 P. Ciavaldini Apr 99 P. Ciavaldini A B Date Editor

Jan 98 P. Ciavaldini Apr 99 P. Ciavaldini

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

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HOW TO USE THE COMPOSITE MANUAL?title(s) of subchapter(s) reference of chapter title of chapter reference(s) of subchapter(s)

SANDWICH Effect of normal load Ny4.2.1 . Effect of normal load Ny

N 4.2.11/2

page number

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:

reference of relation

n3

=

Ny b (EMi ei + EMc ec + Ems es )

This elongation this unduces: - in the lower skin, a stress i = Emi , - in the core, a stress c = Emc , - in the upper skin, a stress s = Ems . The equivalent membrane modulus of the sandwich beam may be determined by the relationship m14.

Remark: In the case of a sandwich beam in which Emc ec 0 Mx > 0 Mxy > 0

z R= 1 2 w 2 x

x

tg() =

w x

w x, y

z

z

u, v uo, vo

w

wo x, y

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MONOLITHIC PLATE - BENDINGDesign method

D

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If the displacements from a point at position Z are defined as u, v and w in the coordinate system (x, y, z), then we may write: u = uo - zw o xw o y

v = vo - z

w = wo where uo, vo et wo represent displacements from the neutral plane in the coordinate system (x, y, z). We deduce (by deriving with respect to coordinates) the corresponding non-zero strains: x = ox - z2 w o x 2

d1

y = oy - z

2 w o y 2

xy = oxy - 2 z

2wo x yz

x neutral plan

z ox o2 w 2 x

tg() =

x

where ox, oy and oxy rerepresent strains at a point located on the neutral plane and x, y and xy represent strains at any point at position z.

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Dh 2 h 2

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From the general expression for the bending moment: M =

z dz , we obtain the

relationship between the bending load tensor (M) and the rotation tensor (): d2 (M) = (C) x ()2wO x 2 2wO y 2 2 w O 2 x y

Mx My M xy

C11

C12 C22 C32

C13 C 23 C 33

= C 21C 31

where

d3

Cij =

n k = 1

3 3 k zk zk 1 ij 3

with d4 11() = c4 l + s4 t + 2 c2 s2 (tl l + 2 Glt) 22() = s4 l + c4 t + 2 c2 s2 (tl l + 2 Glt) 33() = c2 s2 (l + t - 2 tl l) + (c2 - s2)2 Glt 12() = 21() = c2 s2 (l + t - 4 Glt) + (c4 + s4) tl l 13() = 31() = c s {c2 l - s2 t - (c2 - s2) (tl l + 2 Glt)} 23() = 32() = c s {s2 l - c2 t + (c2 - s2) (tl l + 2 Glt)}

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

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D

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with d5 l =El 1 tl lt Et 1 tl lt

t =

If the tensor of angles formed by the strain diagram in each direction is defined by (): (x, y, xy) we may write in a simplified form the relationship: d6 () = tg () By convention, we shall assume that () is negative when the upper fibre is in tension. We have: d7 ()z = - () x zz z

ply No. k zk zk - 1 h neutral plan

ply No. 1

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 (Kirchoffs assumption). In this case, the neutral plane shall also be used as a reference for the overall load pattern.

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MONOLITHIC PLATE - BENDINGEquivalent mechanical properties

D

4

4 . DEFORMATIONS AND EQUIVALENT MECHANICAL PROPERTIES Monolithic plates are microscopically heterogeneous. It is sometimes necessary to find their equivalent bending stiffness properties in order to determine the passing loads and resulting deformations. Equivalent bending elasticity moduli are directly derived from the laminate stiffness matrix (C):1 E xx bending equi.

x 1 E yy bending equi. x

x x 1 Gxy bending equi.

d8

(C)-1 =

12 e3

x x

If reference axes (o, x, y) are coincident with the axes of orthotropy of the laminate, we obtain:C11 C 22 (C12 )2 e3 C 22 C 11 C 22 (C 12 ) 2 e 3 C 11 C 66 e3

Exxbending equi. = 12

Eyybending equi. = 12

Gxybending equi. = 12

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MONOLITHIC PLATE - BENDINGExample

D

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5 . EXAMPLE Let a T300/BSL914 laminate (new) be laid up as follows: 0: 2 plies 45: 2 plies 135: 2 plies 90: 2 plies Stacking from the external surface being as follows: 0/45/135/90/90/135/ 45/0.

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

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

Mechanical properties of the unidirectional ply are the following: El = 13000 hb Et = 465 hb lt = 0.35 tl = 0.0125 Glt = 465 hb ep = 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 the reference coordinate system (x, y):

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MONOLITHIC PLATE - BENDINGExemple

D

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Mx = 10 daN My = 0 daN/mm Mxy = - 5 daN/mmz

y

x

Mx = 10 daN

Mxy = - 5 daN

1st step: calculation of stiffness coefficients for the unidirectional ply: {d5} l =13000 = 13057 daN/mm2 1 0.35 0.0125 465 = 467 daN/mm2 1 0.35 0.0125

t =

2nd step: For each ply, stiffness coefficients ij expressed in daN/mm2 are calculated. {d4} ply at 0 11(0) = 13057 22(0) = 467 33(0) = 465 12(0) = 21(0) = 0.0125 x 13000 = 163 13(0) = 31(0) = 0 23(0) = 32(0) = 0

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ply at 45 11(45) = 0.7074 13057 + 0.7074 467 + 2 x 0.7072 0.7072 (0.0125 x 13057 + 2 x 465) = 3925 22(45) = 0.7074 13057 + 0.7074 467 + 2 x 0.7072 0.7072 (0.0125 x 13057 + 2 x 465) = 3925 33(45) = 0.7072 0.7072 (13057 + 467 - 2 x 0.0125 x 13057) = 3297 12(45) = 21(45) = 0.7072 0.7072 (13057 + 467 - 4 x 465) + (0.7074 + 0.7074) x 0.0125 x 13057 = 2995 13(45) = 31(45) = 0.707 x 0.707 {0.7072 13057 - 0.7072 467} = 3146 23(45) = 32(45) = 0.707 x 0.707 {0.7072 13057 - 0.7072 467} = 3146 ply at 135 11(135) = 3925 22(135) = 3925 33(135) = 3297 12(135) = 21(135) = 2995 13(135) = 31(135) = - 3146 23(135) = 32(135) = - 3146 ply at 90 11(90) = 467 22(90) = 13057 33(90) = 465 12(90) = 21(90) = 163 13(90) = 31(90) = 0 23(90) = 32(90) = 0

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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 be calculated for the laminate upper half, then they shall be multiplied by 2. {d3} 90 . 013 0 C11 = 2 467 3 3 3

135+ 3925 0.263

453

0 0.26 33

013 . 3

+ 3925

0.39

3

+ 13057

0.52 3 0.39 3 3

. . 013 3 0 3 0.26 3 013 3 0.39 3 0.26 3 0.52 3 0.39 3 + 2995 + 2995 + 163 C12 = 2 163 3 3 3 3 0.13 3 0 3 0.26 3 0.13 3 0.39 3 0.26 3 0.52 3 0.39 3 3146 + 3146 + 0 C13 = 2 0 3 3 3 3 . 013 3 0 3 0.26 3 0.13 3 0.39 3 0.26 3 0.52 3 0.39 3 + 2995 + 2995 + 163 C21 = 2 163 3 3 3 3 . . 013 3 03 0.26 3 0133 0.39 3 0.26 3 0.52 3 0.39 3 + 3925 + 3925 + 467 C22 = 2 13057 3 3 3 3 0.13 3 0 3 0.26 3 0.13 3 0.39 3 0.26 3 0.52 3 0.39 3 3146 + 3146 + 0 C23 = 2 0 3 3 3 3 0.13 3 0 3 0.26 3 0.13 3 0.39 3 0.26 3 0.52 3 0.39 3 3146 + 3146 + 0 C31 = 2 0 3 3 3 3 0.13 3 0 3 0.26 3 0.13 3 0.39 3 0.26 3 0.52 3 0.39 3 3146 + 3146 + 0 C32 = 2 0 3 3 3 3 . . 013 3 0 3 0.26 3 013 3 0.39 3 0.26 3 0.52 3 0.39 3 + 3297 + 3297 + 465 C33 = 2 465 3 3 3 3

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

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Thus, the following matrix is obtained:858 123 194 55 55 55 151

(C) = 12355

4th step: Search for the rotation tensor {d2} 2 wo x 2 2 wo 55 = y 2 2 wo 2 151 x y55

Mx My Mxy

858

123 194 55

= 12355

hence

2 wo x 2 2 wo y 2 2 wo 2 x y

1287 E 3 .

7.617 E 46.199 E 3

1913 E 4 . 198 E 3 .7.414 E 3

Mx

=

7.617 E 4 1913 E 4 .

=

My Mxy

198 E 3 .

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D10

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2wo x 2 2wo y 2 2 w o 2 x y

1287 E 3 .

7.617 E 46.199 E 3

1913 E 4 . 198 E 3 .7.414 E 3

=

7.617 E 4 1913 E 4 .

=

0

198 E 3 .

5

Thus, we find:2wo x 2 2wo y 2 2 w o 2 x y13.82 E 3

=

2.283 E 3

38.98 E 3

which is the rotation tensor ().

5th step: We now propose to calculate strains (0) for the ply at 0 (at the external line of the layer). {d7} x(0) = 2 w o h x 2 2 x 2 w o h x 2 2 y 2w o h x 2 x y

y(0) = -

xy(0) = - 2

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hence: x(0) = - 1 x 13.82 E-3 x 0.52 = - 7186 d y(0) = - 1 x 2.283 E-3 x 0.52 = - 1187 d xy(0) = - 1 x - 38.98 E-3 x 0.52 = 20270 d Stresses in the layer may be determined afterwards. To do this, refer to chapter C.

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MONOLITHIC PLATE - BENDINGReferences

D

BARRAU - LAROZE, Design of composite material structures, 1987 GAY, Composite materials, 1991 VALLAT, Strength of materials 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

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EMONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS

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MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS Notations

E

1

1 . NOTATIONS (o, x, y): reference coordinate system (o, l, t): coordinate system specific to the unidirectional fibre x, y, xy: material strains at any point wo: displacement from plate neutral plane (N): normal flux tensor (M): bending moment tensor (): membrane type strain tensor (): curvature tensor (A): laminate stiffness matrix (membrane) (B): laminate stiffness matrix (membrane/bending coupling) (C): laminate stiffness matrix (bending) : fibre orientation k: fibre coordinate system El: longitudinal elasticity modulus of unidirectional fibre Et: transversal elasticity modulus of unidirectional fibre lt: longitudinal/transversal poisson coefficient tl: transversal/longitudinal poisson coefficient Glt: shear modulus of unidirectional fibre ep: ply thickness

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MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS Introduction

E

2

2 . INTRODUCTION We have seen in chapter C that there is a relationship which binds membrane strains and loading of the same type. This relationship may be formulated as follows: (N) = (A) x ().

We also saw in chapter D that there is a relationship which binds the curvature tensor and the moment tensor. This relationship may be formulated as follows: (M) = (C) x ().

If lay-up has the mirror symmetry property, then both phenomena are dissociated and independent. In other words, the overall relationship which binds the set of strains and the set of loadings may be formulated as follows:Nx Ny Nxy A11 A 21 A 31 A12 A 22 A 32 0 0 0 A13 A 23 A 33 0 0 0 0 0 0 C11 C21 C31 0 0 0 C12 C 22 C 32 0 0 0 C13 C23 C33x y xy 2wo x 2 2wo y 2 2 wo 2 x y

=Mx My Mxy 0 0 0

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

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MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS Analysis method

E

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3 . ANALYSIS METHOD If lay-up is non-symmetrical, then all zero terms of the previous matrix become non-zero and there is a membrane/bending coupling. Both phenomena become dependent. The relationship between loadings and strains is thus:Nx Ny Nxy A11 A 21 A 31 A12 A 22 A 32 B12 B22 B32 A13 A 23 A 33 B13 B23 B33 B11 B21 B31 C11 C21 C31 B12 B22 B32 C12 C 22 C 32 B13 B23 B33 C13 C23 C33 x y xy 2wo x 2 2wo y 2 2 wo 2 x y

e1Mx My Mxy

=B11 B21 B31

where

e2

Bij = -

n k = 1

2 2 k zk zk 1 Eij 2

ply No. k zk zk - 1 neutral plane

ply No. 1

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MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS Analysis method

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with e3 11() = c4 l + s4 t + 2 c2 s2 (tl l + 2 Glt) 22() = s4 l + c4 t + 2 c2 s2 (tl l + 2 Glt) 33() = c2 s2 (l + t - 2 tl l) + (c2 - s2)2 Glt 12() = 21() = c2 s2 (l + t - 4 Glt) + (c4 + s4) tl l 13() = 31() = c s {c2 l - s2 t - (c2 - s2) (tl l + 2 Glt)} 23() = 32() = c s {s2 l - c2 t + (c2 - s2) (tl l + 2 Glt)} where c cos() where is the fibre direction in the reference coordinate system (o, x, y). s sin() where is the fibre direction in the reference coordinate system (o, x, y). with e4 l =El 1 tl lt Et 1 tl lt

t =

Remark: The terms Bij and Cij must be determined with relation to the laminate neutral line (Kirchoffs assumption). In this case, the neutral plane shall also be used as a reference for the overall load pattern.

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MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS Example

E

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4 . EXAMPLE Let a T300/BSL914 laminate (new) be laid up as follows: 0: 1 ply 45: 1 ply 135: 1 ply 90: 1 ply Stacking from the external surface being as follows: 0/45/135/90.

k = 4 (0) k = 3 (45) k = 2 (135) k = 1 (90)

z4 = 0.26 z3 = 0.13 z2 = 0 z1 = - 0.13 z0 = - 0.26 neutral plane

Mechanical properties of the unidirectional fibre are the following: El = 13000 hb Et = 465 hb lt = 0.35 tl = 0.0125 Glt = 465 hb ep = 0.13 mm

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The purpose of this example is to search for strains at the laminate internal and external surfaces, knowing that the laminate is globally subject to the following fluxes in the reference coordinate system (x, y): Nx = 5 daN/mm Ny = 0 daN/mm Nxy = 0 daN/mm Mx = 0 daN mm daN My = - 0.15 daN mm Mxy = 0 daN

z

y

My = - 0.15 daN

x

Nx = 5 daN/mm

1st step: calculation of stiffness coefficients for the unidirectional fibre: {e4} l =13000 = 13057 daN/mm2 1 0.35 0.0125 465 = 467 daN/mm2 1 0.35 0.0125

t =

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2nd step: For each fibre direction, stiffness coefficients ij expressed in daN/mm2, are calculated. {e3} fibre at 0 11(0) = 13057 22(0) = 467 33(0) = 465 12(0) = 21(0) = 0.0125 x 13000 = 163 13(0) = 31(0) = 0 23(0) = 32(0) = 0 fibre at 45 11(45) = 0.7074 13057 + 0.7074 467 + 2 x 0.7072 0.7072 (0.0125 x 13057 + 2 x 465) = 3925 22(45) = 0.7074 13057 + 0.7074 467 + 2 x 0.7072 0.7072 (0.0125 x 13057 + 2 x 465) = 3925 33(45) = 0.7072 0.7072 (13057 + 467 - 2 x 0.0125 x 13057) = 3297 12(45) = 21(45) = 0.7072 0.7072 (13057 + 467 - 4 x 465) (0.7074 + 0.7074) x 0.0125 x 13057 = 2995 13(45) = 31(45) = 0.707 x 0.707 {0.7072 13057 - 0.7072 467} = 3146 23(45) = 32(45) = 0.707 x 0.707 {0.7072 13057 - 0.7072 467} = 3146 fibre at 135 11(135) = 3925 22(135) = 3925 33(135) = 3297 12(135) = 21(135) = 2995 13(135) = 31(135) = - 3146 23(135) = 32(135) = - 3146

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fibre at 90 11(90) = 467 22(90) = 13057 33(90) = 465 12(90) = 21(90) = 163 13(90) = 31(90) = 0 23(90) = 32(90) = 0

3rd step: Calculation of laminate (membrane) stiffness matrix (A) coefficients Aij expressed in 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 = 2779 A12 = 821 A13 = 0 A21 = 821 A22 = 2779 A23 = 0 A31 = 0 A32 = 0 A33 = 978

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4th step: Calculation of laminate (bending) inertia matrix (C) coefficients Cij expressed in daN mm. {d3} 90 135 45 0 3 3 3 3 3 3 3 3 0 ( 013) 013 0 0.26 013 ( 013) ( 0.26) . . . . + 3925 + 3925 + 13057 C11 = 467 3 3 3 3 3 3 3 3 3 3 3 3 0 ( 0.13) 013 0 0.26 0.13 ( 013) ( 0.26) . . + 2995 + 2995 + 163 C12 = 163 3 3 3 3 3 3 3 3 3 3 ( 0.13) 3 ( 0.26) 3 0 ( 0.13) 0.13 0 0.26 0.13 3146 + 3146 + 0 C13 = 0 3 3 3 3 3 3 3 3 3 3 3 3 0 ( 013) 013 0 0.26 0.13 ( 013) ( 0.26) . . . + 2995 + 2995 + 163 C21 = 163 3 3 3 3 3 3 3 3 3 3 3 3 0 ( 013) 013 0 0.26 013 ( 013) ( 0.26) . . . . + 3925 + 3925 + 467 C22 = 13057 3 3 3 3 3 3 3 3 3 3 ( 0.13) 3 ( 0.26) 3 0 ( 0.13) 0.13 0 0.26 0.13 3146 + 3146 + 0 C23 = 0 3 3 3 3 3 3 3 3 3 3 ( 0.13) 3 ( 0.26) 3 0 ( 0.13) 0.13 0 0.26 0.13 3146 + 3146 + 0 C31 = 0 3 3 3 3 3 3 3 3 3 3 ( 0.13) 3 ( 0.26) 3 0 ( 0.13) 0.13 0 0.26 0.13 3146 + 3146 + 0 C32 = 0 3 3 3 3 3 3 3 3 3 3 3 3 0 ( 013) 0.13 0 0.26 013 ( 013) ( 0.26) . . . + 3297 + 3297 + 465 C33 = 465 3 3 3 3

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

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5th step: Calculation of membrane - bending coupling coefficients Bij expressed in daN. {e2} 90 B11 = - 467 B12 = - 163 B13 = - 0 ( 0.13 ) 2 ( 0.26 )

1352+ 3925

452+ 3925

02+ 13057

0

2

( 0.13 )

0.13

2 2

0

0.26

2

0.13

2

2 ( 0.13) 2 ( 0.26 )

2 2+ 2995

2 0

0

2

( 0.13)

2+ 2995

0.13

2 2

2+ 163

0.26

2

0.13

2

2 2 ( 0.26 )

2 2 3146

2 2+ 0

( 0.13 )

0

2

( 0.13 )

2+ 3146

0.13

2 2

0

0.26

2

0.13

2

2 ( 0.13) 2 ( 0.26 )

2 2+ 2995

2 0

2

B21 = - 163

0

2

( 0.13)

2+ 2995

0.13

2 2

2+ 163

0.26

2

0.13

2 ( 0.13) 2 ( 0.26)

2 2+ 3925

2 0

2

B22 = - 13057 B23 = - 0 B31 = - 0 B32 = - 0

0

2

( 0.13)

2+ 3925

0.13

2 2

2+ 467

0,26

2

0.13

2 2 ( 0.26 )

2 2 3146

2+ 0

( 0.13 )

0

2

( 0.13 )

2+ 3146

0.13

2 2

0

2

0.26

2

0.13

2

2 ( 0.13 ) 2 ( 0.26 )

2 2 3146

2 0

2

0

2

( 0.13 )

2+ 3146

0.13

2 2

2+ 0

0.26

2

0.13

2

2 ( 0.13 ) 2 ( 0.26 )

2 2 3146

2 0

0

2

( 0.13 )

2+ 3146

0.13

2 2

2+ 0

0.26

2

0.13

2

2 ( 0.13 ) 2 ( 0.26 )

2 2+ 3297

2 0

B33 = - 465

0

2

( 0.13)

2+ 3297

0.13

2 2

2+ 465

0.26

2

0.13

2

2

2

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

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6th step: Expression of stiffness overall matrix {e1}Nx Ny Nxy A 11 A 21 A 31 A 12 A 22 A 32 B 12 B 22 B 32 A 13 A 23 A 33 B 13 B 23 B 33 B 11 B 21 B 31 C 11 C 21 C 31 B 12 B 22 B 32 C 12 C 22 C 32 B 13 B 23 B 33 C 13 C 23 C 33 x y xy 2wo x 2 2wo y 2 2 wo 2 x y

=Mx My Mxy B 11 B 21 B 31

thenNx Ny Nxy 2779 821 0 821 2779 0 0 319 53.2 0 0 978 53.2 53.2 0 319 0 53.2 75.1 6.06 0 0 319 53.2 6.06 75.1 0 53.2 53.2 0 0 0 9.59 x y xy 2wo x 2 2wo y 2 2 wo 2 x y

=Mx My Mxy 319 0 53.2

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E3.61 x E 3 3.61 x E 3 0

48/9

By reversing the relationship, we find:

x y

1.15 x E 3 5.0 x E 4

5.0 x E 4

3.80 x E 4 3.8 x E 4

5.00 x E 3 2.0 x E 3

1.99 x E 3 5.0 x E 3

N N N

x y

1.15 x E 3 3.8 x E 4

xy o=

3.80 x E 4

1.28 x E 3

2.33 x E 3

2.33 x E 3

xy

2w

5.00 x E 3

2.0 x E 3

2.33 x E 3

3.57 x E 2

7.23 x E 3

1.67 x E 2

M

x 2 2w o y 2 2w

x

2.00 x E 3

5.0 x E 3

2.33 x E 3

7.23 x E 3

3.57 x E 2

1.67 x E 2

M

y

2

o

x y

3.61 x E 3

3.61 x E 3

0

1.67 x E 2

1.67 x E 2

1.44 x E 1

M

xy

7th step: Search for the strain tensor By replacing loading by values quoted at the beginning of the example in the previous relationship, we find:x y xy 2wo x 2 2wo y 2 2 wo 2 x y 5.44 E 3 mm / mm 174 E 3 mm / mm . 154 E 3 mm / mm . (5440 d) ( 1740 d) (1540 d)

=2.38 E 2 mm 1 4.57 E 3 mm 1 2.05 E 2 mm 1

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8th step: Search for strains in the upper fibre at (0) To do this, membrane strains (x, y, xy) are added to strains resulting from the bending 2 2wo 2wo h wo h h x , x ,2 x effect 2 2 2 y 2 x y 2 x {d7} x(0) = x 2 w o h x 2 2 x 2 w o h x 2 2 y 2wo h x 2 x y

y(0) = y -

xy(0) = xy - 2

hence: x (0) = 5.44 E-3 + (-1) x 2.38 E-2 x 0.26 = - 748 d y (0) = - 1.74 E-3 + (-1) x 4.57 E-3 x 0.26 = - 2928 d xy (0) = - 1.54 E-3 + (-1) x 2.05 E-2 x 0.26 = - 3790 d For the lower fibre, we would find: x (90) = 5.44 E-3 + 2.38 E-2 x 0.26 = 11628 d y (90) = - 1.74 E-3 + 4.57 E-3 x 0.26 = - 552 d xy (90) = + 1.54 E-3 + 2.05 E-2 x 0.26 = 6870 d

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MONOLITHIC PLATE - MEMBRANE + BENDING ANALYSIS References

E

BARRAU - LAROZE, Design of composite material structures, 1987 GAY, Composite materials, 1991 VALLAT, Strength of materials M. THOMAS, Analysis of a laminate plate subjected to membrane and bending loads, 440.227/79 J.C. SOURISSEAU, 40430.030 J. CHAIX, 436.127/91

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FMONOLITHIC PLATE - TRANSVERSAL SHEAR ANALYSIS

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MONOLITHIC PLATE - TRANSVERSAL SHEARNotations

F

1

1 . NOTATIONS (o, x, y): reference coordinate system (o, l, t): coordinate system specific to the unidirectional fibre El: longitudinal elasticity modulus of unidirectional fibre Et: transversal elasticity modulus of unidirectional fibre lt: longitudinal/transversal Poisson coefficient tl: transversal/longitudinal Poisson coefficient Glt: shear modulus of unidirectional fibre ep: ply thickness Ek: longitudinal elasticity modulus with relation to x-axis of ply No. k n: total number of laminate layers : fibre orientation 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 k to n : shear stress Txy, Tyz, T(): shear load flux

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MONOLITHIC PLATE - TRANSVERSAL SHEARIntroduction

F

2

2 . INTRODUCTION The purpose of this chapter is to determine interlaminar shear stresses in a monolithic plate subject to a shear load flux. For simplification purposes, we shall assume that the laminate is made up of n identical fibres but with different directions.z Tyz > 0 y z

Txz > 0 x

y k=n ep

k=1 x

Layer No. k in direction has the following longitudinal elasticity modulus with relation to the reference coordinate system (o, x, y): f1 Ek =1 1 c s + + c 2 s2 2 tl El Et Et Glt4 4

see chapter C3.

with c cos() s sin()

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F

2 3

1/5

We shall assume that shear load Txz (direction z load shearing a plane perpendicular to xaxis) creates stress xz and, based on the reciprocity principle, stress zx. Similarly, we shall assume that shear load Tyz (direction z load shearing a plane perpendicular to y-axis) creates stress yz and, based on the reciprocity principle, stress zy. These shear stresses are called interlaminar stresses.

z

zy zx xz Tyz yz y

Txz

x

3 . ANALYSIS METHOD To calculate interlaminar stresses xz (zx) generated by shear load Txz (Tyz), use the following methodology. We shall only consider the case of a laminate subject to shear load Txz. The analysis principle is the same for Tyz. In this case, inertias (El) and static moments (E W k) are measured with relation to y-axis. Elasticity moduli (Ek) are measured with relation to x-axis.

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F

32/5

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

f2

zg =

2n

k =1 n

2 2 Ek z k zk 1

(

k =1

Ek z k zk 1

(

)

)z

ply No. n ply No. k zg zk - 1 z1 z0 = 0 ply No. 1

zk

y

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

f3

El =

n

k =1

Ek

(z

k

zk 1 12

)

3

+

n

k =1

Ek z k z k 1

(

)

zk + zk 1 zg 2 z

2

ply No. k

zk

zk - 1

zg

y

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3rd step: Then the (elasticity moduli weighted) static moment E W k (of the material surface located above the line where interlaminar stress is to be calculated), is determined. This static 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 =

n

zi + z i 1 Ei z i z i 1 zg i=k 2

(

)

z

ply No. k

zk

zk - 1

zg

y

If the line is situated at the centre of a fibre at z =

zk + zk 2

1

, the relationship becomes:

f5

E Wk =

n i = k

Ei zi zi

(

1

)

z i + zi 2

1

zg 1

zk + z k Ek 2

1

zk

1

zk + z k 4

+

zk

1

2

zg z

z +z

ply No. k

k

k 1

2

zk

zk - 1

zg

y

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4th step: Shear stress xzk is determined, so that: f6 xzk =Txz . E Wk El

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 for greater accuracy), it is possible to plot the interlaminar shear stress diagram over the entire plate width. The previous relationship shows that the shear stress is maximum when the static moment is maximum as well, i.e. at the neutral axis (z = zg).z z

xzk ply No. k zxk zg

xz

y

Remark: The previous analysis is based on a shear load flux Txz applied to a section perpendicular to x-axis. In the case of any section forming an angle in the coordinate system (o, x, y), the shear load flux in this new section may be expressed as a function of Txz and Tyz.

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y

T( + /2) ds

T() -Txz

x -Tyz

As shown in the drawing above, the z equilibrium of the hatched material element leads to the following relationship: T() ds - Txz ds cos() - Tyz ds sin() = 0 hence: T() = cos() Txz + sin() Tyz Tyz It is easy to show that for = Arctg , a modulus extremum T() (called main shear Txz load flux) is reached that is equal to:

f7

l T() l =

Txz 2 + Tyz 2

Example: if shear load fluxes Txz and Tyz are equal, then the maximum shear load flux is located in the plane with a direction = 45. Its modulus equals2 Txy (or 2 Tyz).

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B

MONOLITHIC PLATE - TRANSVERSAL SHEARExample

F

41/9

4 . EXAMPLE Let a T300/BSL914 laminate (new) be laid up as follows: 0: 1 ply 45: 1 ply 135: 1 ply 90: 1 ply Stacking from the external surface being as follows: 0/45/135/90.0 45 135 90

Mechanical properties of the unidirectional fibre are the following: El = 13000 hb (130000 MPa) Et = 465 hb (4650 MPa) lt = 0.35 tl = 0.0125 Glt = 465 hb (4650 MPa) ep = 0.13 mm e = 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/mmz

y

Txz = 0.7 daN/mm x

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Knowing the mechanical properties of the unidirectional fibre, elasticity moduli of each fibre 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 =1 0.707 0.707 + 13000 4654 4

0.0125 1 + 0.707 2 0.707 2 2 465 13000

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}465 (0.13 2 0 2 ) + 925 (0.26 2 0.13 2 ) + 925 (0.39 2 0.26 2 ) + 13000 (0.522 0.39 2 ) 2 ( 465 (0.13 0) + 925 (0.26 0.13) + 925 (0.39 0.26 ) + 13000 (0.52 0.39 ))

zg =

zg = 0.42 mm

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z

z4 = 0.52 z3 = 0.39 zg = 0.42 z2 = 0.26 z1 = 0.13 z0 = 0

2nd step: Analysis of the laminate bending stiffness El with relation to the neutral axis {f3}(0.26 0.13 ) 3 (0.13 0) 3 + 925 + 12 12 (0.52 0.39 ) 3 (0.39 0.26 ) 3 + 13000 + 12 122

El = 465

925

0.13 + 0 0.42 + 465 (0.13 0) 2 0.26 + 0.13 0.42 + 925 (0.26 0.13) 2 0.39 + 0.26 0.42 + 925 (0.39 0.26) 2 0.52 + 0.39 0.42 13000 (0.52 0.39) 2 2 2 2

El = 0.085134 + 0.169352 + 0.169352 + 2.380083 + 7.618211 + 6.087656 + 1.085256 + 2.07025 El = 19.67 daN.mm

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3rd step: Analysis of static moments E W k (with relation to the neutral line) at the base and center of each ply. At the top of ply at 0 {f4} E W 4 = 0 daNz

0 45 135 90

y

At the center of ply at 0 {f5} 0.52 + 0.39 0.42 E W 4 = 13000 (0.52 0.39 ) 2 0.52 + 0.39 0.39 0.52 + 0.39 0.39 + 0.42 13000 4 2 2

E W 4 = 59.15 - 2.11 E W 4 = 57.04 daNz

0

y

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At the base of ply at 0 {f4} 0.52 + 0.39 0.42 E W 4 = 13000 (0.52 0.39 ) 2

E W 4 = 59.15 daNz

0

y

At the center of ply at 45 {f5} 0.52 + 0.39 0.39 + 0.26 0.42 + 925 (0.39 0.26 ) 0.42 E W 3 = 13000 (0.52 0.39) 2 2 0.39 + 0.26 0.26 0.39 + 0.26 925 0.26 + 0.42 2 2 2

E W 3 = 59.15 - 11.42 + 7.67 E W 3 = 55.4 daNpz

45 y

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At the base of ply at 45 {f4} 0.52 + 0.39 0.39 + 0.26 0.42 + 925 (0.39 0.26 ) 0.42 E W 3 = 13000 (0.52 0.39) 2 2

E W 3 = 59.15 - 11.42 E W 3 = 47.73 daNz

45 y

At the center of ply at 135 {f5} 0.52 + 0.39 0.39 + 0.26 0.42 + 925 (0.39 0.26 ) 0.42 + E W 2 = 13000 (0.52 0.39) 2 2 0.26 + 0.13 0.42 925 (0.26 0.13) 2 0.26 + 0.13 0.13 0.26 + 0.13 0.13 + 0.42 925 4 2 2

E W 2 = 59.15 - 11.42 - 27.06 + 15.48 E W 2 = 35.35 daNz

135 y

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At the base of ply at 135 {f4} 0.52 + 0.39 0.39 + 0.26 0.42 + 925 (0.39 0.26 ) 0.42 + E W 2 = 13000 (0.52 0.39) 2 2 0.26 + 0.13 0.42 925 (0.26 0.13 ) 2

E W 2 = 59.15 - 11.42 - 27.06 E W 2 = 19.87 daNz

135 y

At the center of ply at 90 {f5} 0.52 + 0.39 0.39 + 0.26 0.42 + 925 (0.39 0.26 ) 0.42 + E W 1 = 13000 (0.52 0.39 ) 2 2 0.26 + 0.13 0.13 + 0 0.42 + 465 (0.13 0) 0.42 925 (0.26 0.13) 2 2 0.13 + 0 0 0.13 + 0 0 + 0.42 465 4 2 2

E W 1 = 59.15 - 11.42 - 27.86 - 21.46 + 11.71z

E W 1 = 10.12 daN

90

y

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At the base of ply at 90 {f4} E W 1 = 0 daNz

90

y

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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. at the base of the ply at 0. Its value equals at E W 0 = 59.15 daN, which gives stress xz0: {f6} xz0 =0.7 x 59.15 = 2.1 hb (21 MPa) 19.67

If these interlaminar shear stresses are analysed for each fibre, stresses are distributed along the laminate thickness as follows: xzk =0.7 E Wk 19.670.52

0.455

0.39

0.325

z (mm)

0.26

0.195

0.13

0.065

0 0 0.5 1 1.5 2 2.5

(hb)

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MONOLITHIC PLATE - TRANSVERSAL SHEARReferences

F

BARRAU - LAROZE, Design of composite material structures, 1987 GAY, Composite materials, 1991 VALLAT, Strength of materials

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GMONOLITHIC PLATE - FAILURE CRITERIA

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FAILURE CRITERIANotations

G

1

1 . NOTATIONS l (l): longitudinal stress in unidirectional fibre t (2): transversal stress in unidirectional fibre lt (6): shear stress in unidirectional fibre l (l): longitudinal strain in unidirectional fibre t (2): transversal strain in unidirectional fibre lt (6): shear strain in unidirectional fibre Rl: allowable longitudinal stress Rlt: allowable longitudinal tension stress Rlc: allowable longitudinal compression stress Rt: allowable transversal stress Rtt: allowable transversal tension stress Rtc: allowable transversal compression stress S: allowable shear stress

AEROSPATIALE - 1999

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FAILURE CRITERIAInventory

G

2

2 . INVENTORY OF STATIC FAILURE CRITERIA The purpose of this chapter is to describe various failure criteria of the unidirectional fibre within a laminate. The following criteria shall be presented in chronological order (this is not an exhaustive list): - 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 - Tsa - Wu's criterion For three-dimensional criteria, we shall assume that the composite material is subjected to the following stress tensor and strain tensor: () = (1, 2, 3, 4, 5, 6) () = (1, 2, 3, 4, 5, 6)

For two-dimensional criteria, we shall assume that the unidirectional fibre is subjected to the following stress tensor and strain tensor: (lt) = (l, t, lt) (lt) = (l, t, lt)

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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 i = Xi or i = - X'i for compression stresses for tension stresses

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

t

l

lt

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FAILURE CRITERIAMaximum strain criterion

G

2.2

2.2 . Maximum strain criterion This criterion is applicable for orthotropic materials only. The criterion anticipates failure of the material if: for 1 i 6 g2 i = Yi or i = - Y'i for compression strains for tension strains

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

t

l

lt

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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. The criterion anticipates failure of the material if:

6

1

C i ( i ) 2 = 1

Coefficients Ci depend on the material used. For the two-dimensional case, the expression becomes: g3 C1 (l)2 + C2 (t)2 + C6 (lt)2 = 1 The failure envelope may be represented as follows: t

l

lt

This is the first criterion which calls for stress dependency.

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FAILURE CRITERIAPuck's criterion

G

2.4

2.4 . Puck's criterion This two-dimensional criterion is valid for orthotropic materials only. The criterion anticipates failure of the material if: 1 = X1 or 1 = - X'1 and 1 + 2 + 12 = 1 X1 X2 X6 2 2 2

for tension stresses for compression stresses

g4

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

l

lt

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

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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. The criterion anticipates failure of the material if: F (2 - 3)2 + G (3 - 1)2 + H (1 - 2)2 + L (4)2 + M (5)2 + N (6)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 (t)2 + G (l)2 + H (l - t)2 + N (lt)2 = 1

t

l

lt

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FAILURE CRITERIANorris's criterion

G

2.6

2.6 . Norris's criterion This two-dimensional criterion is valid for orthotropic materials only. The criterion anticipates failure of the material if: F (2 - 3)2 + G (3 - 1)2 + H (1 - 2)2 + L (4)2 + M (5)2 + N (6)2 = 1 and for 1 i 6 i = Xi or i = - X'i for compression stresses for tension stresses

Coefficients F, G, H, L, M and N depend on the material used. For a two-dimensional analysis, the expression becomes: g6 F (t)2 + G (l)2 + H (l - t)2 + N (lt)2 = 1 - X'1 l X1 and - X'2 t X2 and - X'6 lt X6

t

l

lt

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FAILURE CRITERIAFischer's criterion

G

2.7

2.7 . Fischer's criterion This two-dimensional criterion is valid for orthotropic materials only. The criterion anticipates failure of the material if: l + t K l t + lt = 1 X1 X 2 X 6 X1 X2 2 2 2

g7

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

l

lt

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FAILURE CRITERIA Hoffman's criterion FAILURE CRITERIA2.8 . Hoffman's criterion This criterion is valid for orthotropic materials only. The criterion anticipates failure of the material if:

G

2.8

C1 (2 - 3)2 + C2 (3 - 1)2 + C3 (1 - 2)2 + C4 (4)2 + C6 (6)2 + C5 (5)2 + C'1 1 + C'2 2 + C'3 3 = 1 Coefficients C1, C2, C3, C4, C5, C6, C'1, C'2 and C'3 depend on the material used. For a two-dimensional analysis, the expression becomes: g8 C1 (t)2 + C2 (l)2 + C3 (l - t)2 + C6 (lt)2 + C'1 l + C'2 t = 1 t

l

lt

Very good tension accuracy, but very bad compression results.

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FAILURE CRITERIATsa - Wu's criterion

G

2.9

2.9 . Tsa - Wu's criterion This criterion intends to be as general as possible and then, there is, a priori, no particular hypothesis. This criterion anticipates failure of the material if: For 1 i 6 Fi i + Fij i j + Fijk i j k + = 1 For a two-dimensional analysis, there is: g9 F1 l + F2 t + F6 lt + F11 (l)2 + F22 (t)2 + F66 (lt)2 + 2 F12 l t + 2 F26 t lt + 2 F16 l lt = 1 Coefficient F1, F2, F6, F11, F22, F66, F12, F26 and F16 depend on the material used.

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FAILURE CRITERIAAerospatiale's criterion

G

31/2

3 . Failure criterion used at Aerospatiale: Hill's criterion As seen previously, Hill's criterion is, in its general form, formulated as follows: F (t)2 + G (l)2 + H (l - t)2 + N (lt)2 = 1 This non-interactive criterion is applicable at the elementary ply only. There is a laminate failure when the most highly loaded layer is broken. If the expression is developed, we obtain: (G + H) (t)2 + (F + H) (l)2 - 2 H l t + N (lt)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. There is a failure if h = l Rl 2 2

g10

+ t Rt

2

+ lt S

2

l 2 t = 1 Rl

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FAILURE CRITERIAAerospatiale's criterion

G

32/2

Thus, the following Reserve Factor is deduced: g11 RF =1 = h 12 l + t + lt + l 2 t S Rl Rt Rl 2 2

This criterion is the one used by Aerospatiale. In order to avoid having a premature theoretical failure in the resin, the transversal modulus Et was considerably reduced (by a coefficient 2 approximately) with relation to the average values measured. This "design" value is determined so that the transversal strain is greater than the longitudinal one. The allowable plane shear value S of the unidirectional fibre was determined for having, a good test/calculation correlation and significant tension and compression failures of notched or unnotched laminates.

B

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FAILURE CRITERIAExample

G

41/4

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 corresponding chapter (C.6) and quoted in the following pages. Let a T300/BSL914 laminate (new) be laid up as follows: 0: 6 plies 45: 4 plies 135: 4 plies 90: 6 plies Mechanical properties of the unidirectional fibre are the following: El = 13000 hb (130000 MPa) Et = 465 hb (4650 MPa) lt = 0.35 Glt = 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 reference coordinate system (x, y) (see chapter C.6) : Nx = 30.83 daN/mm Ny = - 2.22 daN/mm Nxy = 44.92 daN/mm

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42/4

Reminder of stresses applied to the fibre with a 0 direction l = 29.42 hb t = 0.06 hb lt = 5.03 hb {g10} 29.42 0.06 5.03 29.42 x 0.06 h = + + =1 120 2 120 5 7 .5 2

2

2

2

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

{g11} Reserve Factor: R.F. =1 h2 = 1 0.51 = 14 .

Margin = 100 (R.F. - 1) 40 %

Reminder of stresses applied to the fibre with a 45 direction l = 80.17 hb t = - 1.14 hb lt = - 1.36 hb {f10} 80.17 h2 = 120 2

. 114 + 12

2

. 136 + 7.5

2

. 80.17 x ( 114) 2 120

h2 = 0.45 + 0.009 + 0.033 + 0.006 = 0.498

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{g11}

Reserve Factor: R.F. =

1 0.498

= 142 .

Margin 42 %

Reminder of stresses applied to the fibre with a 135 direction l = - 59.17 hb t = 2.14 hb lt = 1.36 hb {g10} 59.17 h2 = 100 2

2.14 + 5

2

. 136 + 7.5

2

59.17 x 2.14 100 2

h2 = 0.35 + 0.183 + 0.033 + 0.0126 = 0.579

{g11} Reserve Factor: R.F. =1 0.579 = 131 .

Margin 31 %

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Reminder of stresses applied to the fibre with a 90 direction l = - 8.42 hb t = 0.95 hb lt = - 5.03 hb {g10} 8.42 h = 100 2 2

0.95 + 5

2

5.03 + 7.5

2

8.42 x 0.95 100 2

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

{g11} Reserve Factor: R.F. =1 0.494 = 142 .

Margin 42 %

Conclusion: the laminate overall margin is therefore 31 %

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MONOLITHIC PLATE - FAILURE CRITERIAReferences

G

BARRAU - LAROZE, Design of composite material structures, 1987 GAY, Composite materials, 1991 VALLAT, Strength of materials Comparative analysis of composite material damaging criteria BOUNIE, Failure criteria of mechanical bonds in composite materials, 1991, 440.180/91

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HMONOLITHIC PLATE - FATIGUE ANALYSIS

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IMONOLITHIC PLATE - DAMAGE TOLERANCE

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MONOLITHIC PLATE - DAMAGE TOLERANCENotations

I

1

1 . NOTATIONS (o, x, y): panel reference frame Nx: x-direction normal flow Ny: y-direction normal flow Nxy: shear flow i: orientation of fibre i li: longitudinal strain of fibre i ti: transverse strain fibre i lti: angular slip of fibre i adm: permissible longitudinal strain of unidirectional fibre adm: permissible slip of unidirectional fibre li: longitudinal stress of fibre i ti: transverse stress of fibre i lti: shear stress of fibre i Rl: permissible longitudinal stress of unidirectional fibre Rt: permissible transverse stress of unidirectional fibre S: permissible shear stress of unidirectional stress R: reduction coefficient for permissible longitudinal stress S: reduction coefficient for permissible shear stress

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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 5.8 requests resistance to ultimate loads with "realistic" impact damage susceptible to be produced in production and in service. This damage must be at the limit of the detectability threshold defined by the selected inspection procedure. Also, static strength must be demonstrated after application of mechanical fatigue ( 5.2) and test specimens must 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.Detection threshold (impact depth in mm) Damage at detectability threshold limit Damage Tolerance Range

Low thickness

Static strength range

High thickness Impact energy

Static cut-off energy

Damage-tolerance cut-off energy

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MONOLITHIC PLATE - DAMAGE TOLERANCEIntroduction

I

2 3 3.1

Distinction is made between the range above the detectability threshold where all damage will be detectable and the range above the static cut-off energy and below the detectability threshold where the damage will never be detected. In this "Damage tolerance" section, we shall discuss both manufacturing defects and impact 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 which occurs in service.

3.1 . Manufacturing damage of flaws Manufacturing damage or flaws include porosities, microcracks and delaminations resulting from anomalies, during the manufacturing process and also edge cuts, unwanted routing, 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 during the 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 by routine quality inspections. For all composite parts, the acceptance/scrapping criteria must be defined by the Design Office. Acceptable damage/flaws are incorporated into the ultimate load justification by analysis and into the test specimens to demonstrate the tolerance of the structure to this damage throughout the life of the aircraft.

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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 three types 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 mechanical properties are such that the static dimensioning requirements naturally cover the fatigue dimensioning requirements. This is valid for a laminate submitted to plane loads, less than 60 % of ultimate load. However, complex areas or areas with a sudden variation in rigidity may favour the appearance of delaminations under triaxial loads. Today, it is very difficult to (analytically or numerically) model the growth rate of a possible flaw. This is why a "safe-life" justification philosophy has been adopted. It is based on two principles 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" curves associated with them (relatively flat curve with low gradient), the factor 5 normally used on metallic structures for the number of lives to be simulated during a fatigue test, was replaced by a load factor. All these points will be discussed in detail in section O (MONOLITHIC PLATE FATIGUE).

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MONOLITHIC PLATE - DAMAGE TOLERANCECorrosion damage - Environmental effects3.2.2 . Corrosion damage and environmental effects a) Corrosion

I

3.2.2

Composites are insensitive to corrosion. Nevertheless, their association with certain metallic materials can cause galvanic coupling liable to damage certain metal alloys. For information purposes, the table below shows various carbon/metal pairs over a scale ranging from A to E. We consider that type A and B couplings are correct and that those of types C, D and E are not.

Coupling with carbon correct

A A B B B B C D D D D

Anodised titanium, protected titanium fasteners Titanium and gold, platinium and rhodium alloys Chromiums, chrome-plated parts Passivated austenitic stainless steels Monel, inconel Martensitic stainless steels Ordinary steels, low alloys steels, cast irons Anodic or chemically oxidised aluminium and light alloys Cadmium and cadmium-plated parts Aluminium and aluminium-magnesium alloys Aluminium-copper and aluminium-zinc alloys

b) Environmental effects At high temperatures, aggressions by hydraulic fluids may cause damage such as separation, delamination, translaminar cracks, etc. Rain can cause damage by erosion, etc. All these points will be discussed in detail in section W (INFLUENCE OF THE ENVIRONMENT).

Coupling with carbon to be avoided

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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 into question 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 is damage detection. This is true both during manufacture and in service. In service, the detectability threshold depends on the type of scheduled in-service inspection. There are four types of inspections: Inspection - Special detailed (ref: Maintenance Program Development: MPD): An intensive examination of a specific location similar to the detailed inspection except for the following differences. The examination requires some special technique such as non-destructive test techniques, dye penetrant, high-powered magnification, etc., and may required disassembly procedures. This type of inspection is mainly conducted during production but can be used exceptionally in service. Inspection - Visual Detailed (ref: Maintenance Program Development: MPD): An intensive visual examination of a specified detail, assembly, or installation. It searches for evidence of irregularity using adequate lighting and, where necessary, inspection aids such as mirrors, hand lens, etc. Surface cleaning and elaborate access procedures may be required. This type of inspection enables BVID (Barely Visible Impact Damage) to be detected.

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MONOLITHIC PLATE - DAMAGE TOLERANCEInspection of damage

I

4 4.1 4.2

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 accuracy of 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 geometrical detectability criteria are as follows (cf. ref. 22S 002 10504): Depth of flaw "" Mean "A" value Inside box structure (broken fibres) 0.1 mm 0.2 mm

Outside box structure 0.3 mm 0.5 mm

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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 geometrical detectability criteria are as follows (cf. ref. 22S 002 10504):

Depth of flaw "" 2 mm or thickness of structure if < 2 mm

Size of perforation 20 mm

4.4 . Minimum damage detectable by a Walk Around Check This type of damage is called Large VID (Large Visible Impact Damage). The geometrical detectability criteria are not explicitly defined but the damage must be detectable without ambiguities 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 the damage.Special detailed inspection Detailed visual inspection BVID General visual inspection Minor VID

Walk around Large VID Size of damage

Depth of indent diameter

= 0.3 mm

= 2 mm 20 mm 50/60 mm

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

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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:Damage susceptible to be detected DVI BVID Minor VID WA Large VID Inevitably detectable damage

Undetectable damage

For General Visual Inspection:Damage susceptible to be detected GVI BVID Minor VID WA Large VID Inevitably detectable damage

Undetectable damage

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MONOLITHIC PLATE - DAMAGE TOLERANCEInfluence of damage - Porosity

I

4.5 5 5.1 5.1.1

Remark: Note that certain authors define the BVID notion according to the type of inspection selected. In this case, for a general inspection: MINOR VID BVID In our document, we will conserve the initial definition related to the visual detailed 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 lack of inter- or intra-layer cohesion, generally small in size, but distributed uniformly or almost throughout the complete thickness of the laminate. Note that for unidirectional tapes the porosities have a tendency to be located between the layers whereas, for fabrics, they are more generally located where the weft and warp threads cross. The porosity ratio given is a surface porosity ratio measured by the ultrasonic attenuation method. The permissible absorption level is fixed at 12 dB irrespective of the thickness inspected (cf. note 440.241/90 issue 2 - SIAM curve). All absorption areas above this limit will be considered as 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 be porous.

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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 various porosity ratios distributed in all interply areas to determine the influence on the mechanical characteristics for a 3 % ratio considered as the acceptable limit. This ratio combined with the fatigue, ageing and residual test effects at 80 C, led to the following losses in mechanical characteristics: 3 % porosity Loss of characteristics after F + VC1 + 80 C - 15 % - 47 % - 20 % - 20 %

T300/N5208 BENDING INTERLAMINAR SHEAR COMPRESSION TENSILE (high bearing stress) joint not supported

Loss of characteristics after F + VC1 + 80 C - 19 % - 33.5 % - 19 % - 19 %

Example of porosity acceptance criteria The 3 % acceptance criterion appears therefore as being non-conservative for interlaminar shear. However, let us recall: - that the spar boxes of the wings, movable surfaces or fin are subjected to very low interlaminar stresses, - only T300/N5208 had porosities, - that the 3 % porosity criterion distributed at all interply areas is today no longer applied to primary structures. The permissible porosity ratio depends on the thickness of the laminate.

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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 transverse tensile 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. Natural delaminations appear during manufacture (surface contamination). A foreign body left in the 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 of compression strength of around 10 % for the two materials (T300/N5208 and T300/BSL914) tested in new condition at = 80 C. In aged/fatigue condition the drop in strength is 20 % for T300/N5208 and 13 % for T300/BSL914 in relation to the new state/80 C reference. Fatigue leads to no growth of the flaw.

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5.1.2.12/5

Fastener areas Description As for the skin bottom delaminations, the lack of cohesion in these areas occurs between two well-defined plies, sometimes at several levels but generally adjacent. These flaws come through to the bore. They are created during the drilling operations. The ultrasonic inspections conducted after each test case showed no evolution of existing flaws. The parameter representing the size of the damage is the number given by: =damage fastener damage fastener

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

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

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I

5.1.2.13/5

Two cases can occur: - if 1.1: no reduction will be made on the initial reserve factor RF, - if < 1.1: after reduction, the new reserve factor is equal to RF = RF 1.1

The values of are given by the graphs in section Z for the prepreg epoxy carbon fibre T300/914. Generally speaking, the graphs gives the values of for the flaw (delamination) but also for repairs which may be made on it (injection of resin, NAS cup). They enable you to find therefore: - 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 surface area is greater than: S mm2 75 3.2 120 4.1 160 4.8 285 6.35 440 7.92 440 9.52

These permissible delamination values are valid only for isolated delaminations. For delaminated hole concentrations and irrespective of the size of the delaminations, the flaw must be covered by a concession if: - for aligned fasteners, more than 20 % of the holes are delaminated and/or two flaws are less than 5 fastener pitches apart, - for a delamination at a fastener or of another skin bottom area, they are less than 120 mm apart. - in designated area, permissible delamination is defined as follows: S mm2 50 3.2 80 4.1 110 4.8 200 6.35 400 7.92 400 9.52

These permissible delamination values are valid only for isolated delaminations.

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5.1.2.14/5

For delaminated hole concentrations and irrespective of the size of the delaminations, the flaw must be covered by a concession if: - for aligned fastener, more than 10 % of the holes are delaminated and/or two flaws are less than 5 fastener pitches apart, - for a delamination at a fastener or of another skin bottom area, they are less than 120 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 concession if 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 damage ratio.fastener

The choice of the solution is governed by the following rules: - for a pure load, the repair or untreated delamination must resist ultimate loads under the most severe environmental conditions, - for a pure bearing stress test, the calculation value Vc is taken as reference. The Vb repair will not be acceptable if is lower than 1. Vc The validation range of the acceptable solutions given in the table below is damage 6.fastener

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INormal injection -

5.1.2.15/5

Load

Condition New

Untreaded Injection via delamination vent hole Acceptable Acceptable Acceptabledamage fastener

NAS cup Acceptable

Unacceptable Unacceptable

Pure tensile Aged-wet

New Bearing Tensile Aged-wet

< 4

Acceptable Acceptabledamage fastener < 4.5

-

Acceptable

Acceptabledamage fastener < 4.5

-

Acceptable Acceptable

New Pure compression Aged-wet

Unacceptable Unacceptable Unacceptable Acceptable Unacceptable Acceptabledamage fastener < 2.5

damage fastener

< 5

Acceptable Acceptable Acceptabledamage fastener < 5.25 damage fastener < 5.25

Acceptabledamage fastener < 2

Acceptabledamage fastener < 5.25

New Bearing compression Aged-wet Without bending JOINT tensile Bending 1000 d Bending 2500 d Without bending JOINT compression Bending 1000 d Bending 2500 d

damage fastener

< 4.75

Acceptabledamage fastener < 4

Unacceptable Unacceptable

Acceptable

Acceptable Acceptable Acceptable

Unacceptable Unacceptable Unacceptable

-

Acceptable Acceptable in "hollow" Unacceptable Acceptable Acceptable in "hollow" Acceptable

Unacceptable Unacceptable Unacceptable Unacceptable Unacceptable Acceptable

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MONOLITHIC PLATE - DAMAGE TOLERANCEDelaminations in stiffener area

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5.1.2.21/5

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 runouts are made during moulding without later machining operations, these fairly tortured areas may include lacks of cohesion either in the base, or in the stiffener itself.

U-section

Half core Baseplate U-section Wedge

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

L e

l

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Loss of characteristics due to crater Loss of characteristics due to flaw

Size of flaw

Test conducted Tensile Compression (stiffener runouts not protected) Tensile Compression (stiffener runouts protected) Compression (with reinforcement) Compression (without reinforcement)

Conditions

ATR 72 T300/914 L = 10 mm l = 4 mm e = 1 mm

New = 20 C

- 28 %

New = 20 C

0%

-4% Aged = 70 C - 12 %

ATR 72 HTA/EH25

For unprotected stiffener runouts (that is, when it was impossible to thicken the skin to make structure relatively simple to manufacture), this flaw must be covered by a concession. When it is located at protected stiffener runouts (that is with a significant skin overthickness at stiffener runout), this flaw will be covered by a concession only if its size is greater than the following values: L = 10 mm Punching Description This flaw is due to an imperfect Mosite cut leading to flaws at stiffener ends. l = 2 mm e = 0.5 mm

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I

5.1.2.23/5

L e

l

e

Loss of characteristics due to punching Loss of characteristics due to flaw

Size of flaw

Test conducted Tensile Compression (stiffener runouts not protected) Tensile Compression (stiffener runouts protected)

Conditions

ATR 72 T300/914 L = 10 mm e = 1 mm

New = 20 C

- 20 %

New = 20 C

0%

Must be covered by a concession when located at unprotected stiffener runouts. When located at protected stiffener runouts, it will be covered by a concession only if it size is greater than the following values: L = 10 mm l = 2 mm e = 0.5 mm

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Flaws "E", "B", "AB" and "BC" Description These flaws are located at various levels: FLAW E Delamination in radius between U-sections and base FLAW B Delamination under wedge FLAW AB Delamination at skin midthickness

Flaws BC correspond to one or more lacks of cohesion of stiffener wedge as shown on diagram below:Flaw BC A B

C Wedge

U-section Half core

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5.1.2.25/5

Loss of characteristics due to flaw Loss of characteristics due to flaw

Type of flaw V10F T300/N5208 200 mm2 (flaw B) ATR 72 T300/914 (flaw BC) ATR 72 T300/914 (flaw BC)

Test conducted

Conditions

Tensile (between wedge and base skin) Tensile (unprotected stiffener runouts) Compression (unprotected stiffener runouts)

New = 20 C Wet ageing = 50 C Wet ageing = 50 C

- 17 %

- 20 %

0%

Stiffener top Lack of interlayer cohesion at top of stiffener between the U-section and the wedge does not seem to modify the mechanical characteristics.

Stiffener base Lack of interlayer cohesion in stiffener base hardly modifies the mechanical characteristics. Within the scope of the V10F programme, the greatest drop is less than 10 % in standard stiffener compression case with a type BC flaw.

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MONOLITHIC PLATE - DAMAGE TOLERANCEDelamination in spar radii Delamination on edge of spar flanges5.1.3 . Delamination in spar radii

I

5.1.3 5.1.4

This flaw correspond to lack of cohesion between two well-defined plies in the web/flange blend-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 250 mm2, including delaminations and foreign bodies. In designated areas: maximum local surface area between 2 ribs for a radius is

150 mm2, including delaminations and foreign bodies.

5.1.4 . Delamination on spar flange edgesl

L

Delamination

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. Any other flaws shall be covered by a concession.

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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 A340 aileron, the 2000 fin, the A300/A310 (cf. note 494.048/91); however there are none on the flight V10F (cf. note 494.007/91). These are elongated flaws due to the use of a corrosive stripper (MEK, Methyl Ethyl Ketone). Currently, baltane is used. T300/914 and G803/914 have these flaws; the tests conducted on IM7/977-2 and HTA/EH25 showed no translaminar cracks (cf. note 494.056/91). These cracks are detected by ultrasonic inspection in the fastener areas (the back surface echo totally disappears). They concern all ply directions but do not touch between two plies with different orientations. It is in the high crack density area that the ultrasonic signal is totally damped. There a transition zone between this area and the healthy part of the laminate 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 of these 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 be plotted is coherent with the size of acceptable delaminations.

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MONOLITHIC PLATE - DAMAGE TOLERANCEDelaminations consecutive to a shock

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5.1.71/4

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 of the impact.Delaminated area Impactor indent Damaged area

Loss of characteristics due to a delamination Generally speaking, a composite material with a non-through delamination is much more sensitive from a structural strength viewpoint to compression or shear loads (resin) than to tensile 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 graph below (the points of the static and fatigue test specimens are combined on this curve as it has 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 temperature and aged/fatigue states/residual test at ambient temperature is shown on the curve below by comparison at static test specimen and fatigue test specimen points.

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5.1.72/4

Behaviour to impact damage V10F Static test specimen (CES) Fatigue test specimen (CEF)0 - 1000 500 1000 1500 2000 2500

- 1500 - 2000Allongement de compression (d)

- 2500

i32 - (- 2800 d) Arrt CEF i22 - (- 3108 d) Rupture CES

- 3000 - 3500

- 4000

COURBE ACTUELLE VALEURS DE CALCUL Etat neuf/temprature ambiante ou Etat vieilli/fatigue 20 C/temprature ambiante

- 4500

CES CEF

- 5000

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 having a 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 the presence of a delamination (without ply failure) in surface area Sd, what is the strength of the plain composite plate? Today, there are three methods for evaluating the residual strength in the presence of a delamination (established from experimental results) which call on the stresses and/or strains of the unidirectional fibre and not those of the laminate considered as a homogeneous plate. Each fibre direction must therefore be justified.

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MTS 006 Iss. B



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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 of each fibre in relation to their intrinsic frame (o, l, t). By referring to the "plain plate - calculation method" section, it is possible to calculate the strains in the various layers of the composite from the global flows Nx, Ny and Nxy applied to the laminate and from the characteristics of the material used. For layer "i" defined by its orientation i, the strains of the fibre "i" in its own frame are defined by the following strain tensor: (li, ti, lti). We can define the following failure criterion C1 for each layer "i": l i + lt i adm adm 2 2

i1

C1 =

where adm and adm are the permissible strains (longitudinal and shear) of the unidirectional fibre (equivalent). These values (obtained from the test results) depend on the material and the surface area Sd 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 (dossier 22S00210460). 2nd method: This second method consists in calculating a failure criterion C2 (Hill type criterion in which the permissible stresses are reduced by coefficients R and S) calculated from the stresses in each fibre in relation to their intrinsic frames (o, l, t).

AEROSPATIALE - 1999

MTS 006 Iss. B



I

5.1.74/4

By referring to the "plain plate - calculation method" section, it is possible to calculate the strains in the various layers of the composite from the global flows Nx, Ny and Nxy applied to the laminate and from the characteristics of the material used. For layer "i" defined by its orientation 1, the stresses of the fibre "i" in its own frame are defined by the following stress tensor: (li, ti, lti). We can define the following failure criterion C2 for each layer "i": li R R l li t i + t i + lt i S R ( R R l )2 s t 2 2 2

i2

C2 =

where Rl, Rt and S are the permissible longitudinal, transverse and shear stresses of the unidirectional fibre respectively (equivalent) and where R and s are the reduction coefficients for these permissible stresses. These coefficients depend on the material used and the surface area of the delamination considered 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 aileron panels. 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 i, the strains of the fibre "i" in its own frame are defined by the following strain tensor: (li, ti, lti). We can define the following failure criterion C3 for each layer "i" : li a + lt i adm 2

i3

C3 =

+ li t i ( ab )2

2

where: if 2 adm adm

AEROSPATIALE - 1999

MTS 006 Iss. B


MONOLITHIC PLATE - DAMAGE TOLERANCEDelaminations consecutive to a shock Visual flaws - Sharp scratches1 2 ( adm ) 32

I

5.1.7 5.2 5.2.1

i4

a =

1

( adm )26

2

i5

ab =

2 ( adm )

3

2

( adm )2

else a = adm ab = + The particularity of this method is that it takes into

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