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Natural Sciences Tripos Part II

II

C16: Composite Materials

Prof. T. W. Clyne

Name............................. College..........................

-

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Part II Materials Science. Course C16: Composite Materials - Student Handout C16H1

TWC - Michaelmas 2015

Course C16: Composite Materials

Summary

This course covers various aspects of the performance and usage of composite materials. It isprimarily oriented towards “conventional” composites, which comprise long fibres (usually of

glass or carbon) in a polymeric matrix. However, there is some coverage of other types ofreinforcement and also of composites based on metals or ceramics. The treatment includes bothelastic and plastic deformation, plus fracture characteristics, and indications are given as to howcomposites can often offer highly attractive combinations of lightness, stiffness, strength andtoughness, and hence why their usage continues to expand. Certain thermal characteristics arealso covered. Towards the end of the course, information is provided about the mechanics of(thick and thin) surface coatings, which can be treated as a special type of composite system.

Synopsis (12 lectures)

Lecture 1 - Overview of Types of Composite SystemOverview of Composites Usage. Types of Reinforcement and Matrix. Carbon and Glass Fibres. PMCs,MMCs and CMCs. Aligned Fibre Composites, Woven Rovings, Chopped Strand Mat, Laminae andLaminates.

Lecture 2 - Elastic Properties of Long Fibre CompositesUse of the Slab Model. Halpin-Tsai Expressions. Poisson Ratios. Elastic Loading of a Lamina. Matrix

Notation. Kirchoff Assumptions. Axial and Transverse Loading. Effect of Material Symmetry on the Number of Independent Elastic Constants.

Lecture 3 - Off-axis Elastic Properties of Laminae & LaminatesLoading at an Arbitrary Angle to the Fibre Axis. Derivation of Transformed Stress-Strain Relationship.Effect of Loading Angle on Stiffness and Poisson Ratio. Tensile-Shear Interaction Behaviour. Obtainingthe Elastic Constants of a Laminate.

Lecture 4 – Classification of LaminatesStiffness of Laminates. Tensile-Shear Interactions and Balanced Laminates. In-plane Stresses within aLoaded Laminate. Coupling Stresses and Symmetric Laminates.

Lecture 5 - Short Fibre & Particulate Composites – Stress DistributionsThe Shear Lag Model for Stress Transfer. Interfacial Shear Stresses. The Stress Transfer Aspect Ratio.Stress Distributions with Low Reinforcement Aspect Ratios. Numerical Model Predictions. HydrostaticStresses and Cavitation.

Lecture 6 - Short Fibre & Particulate Composites – Stiffness & Inelastic BehaviourLoad Partitioning and Stiffness Prediction for the Shear Lag Model. Fibre Aspect Ratios needed toapproach the Long Fibre (Equal Strain) Stiffness. Inelastic Interfacial Phenomena. Interfacial Sliding and

Matrix Yielding. Critical Aspect Ratio for Fibre Fracture.

Lecture 7 - The Fibre-Matrix InterfaceInterfacial Bonding Mechanisms. Measurement of Bond Strength. Pull-out & Push-out Testing. Controlof Bond Strength. Silane Coupling Agents. Interfacial Reactions and their Control during Processing.

Lecture 8 - Fracture Strength of CompositesAxial Tensile Strength of Long Fibre Composites. Transverse and Shear Strength. Mixed Mode Failureand the Tsai-Hill Criterion. Failure of Laminates. Internal Stresses in Laminates. Failure Sequences.Testing of Tubes in combined Tension and Torsion.

Lecture 9 - Fracture Toughness of CompositesEnergies absorbed by Crack Deflection and by Fibre Pull-out. Crack Deflection . Toughness of Different

Types of Composite. Constraints on Matrix Plasticity in MMCs. Metal Fibre Reinforced Ceramics.

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Lecture 10 - Compressive Loading of Fibre CompositesModes of Failure in Compression. Kink Band Formation. The Argon Equation. Prediction ofCompressive Strength and the Effect of Fibre Waviness. Failure in Highly Aligned Systems. Possibility ofFibre Crushing Failure.

Lecture 11 - Thermal Expansion of Composites and Thermal Residual StressesThermal Expansivity of Long Fibre Composites. Transverse Expansivities. Short Fibre and Particulate

Systems. Differential Thermal Contraction Stresses. Thermal Cycling. Thermal Residual Stresses.

Lecture 12 - Surface Coatings as Composite SystemsMisfit Strains in Substrate-Coating Systems. Force and Moment Balances. Relationship between ResidualStress Distribution and System Curvature. Curvature Measurement to obtain Stresses in Coatings.Limitations of Stoney Equation. Sources of Misfit Strain. Driving Forces for Interfacial Debonding.

Booklist

D.Hull & T.W.Clyne, "An Introduction to Composite Materials", Cambridge University Press,(1996) [AN10a.86]

Web-based Resources

Most of the material associated with the course (handouts, question sheets, examples classesetc) can be viewed on the web and also downloaded. This includes model answers, which arereleased after the work concerned should have been completed. In addition to this text-basedmaterial, resources produced within the DoITPoMS project are also available. These includelibraries of Micrographs and of Teaching and Learning Packages (TLPs). The following TLPs aredirectly relevant to this course:

• Mechanics of Fibre Composites • Bending and Torsion of Beams

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Lecture 1: Overview of Composites & Types of Composite System

Stiff, Light, Corrosion-Resistant Structures – The Attractions of Composites

Fig.1.1 Data for some engineering materials, in the form of a map of Young’s modulus againstdensity

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Fibres used in Composite Materials

Carbon Fibres

Fig.1.2 Effect of heat treatment temperature on the strength and Young’s modulus of carbon fibres produced from a PAN precursor

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Glass Fibres

Polymeric Fibres

Fig.1.3 Structures of (a) cellulose & (b) Kevlar (poly paraphenylene terephthalamide) molecules

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Other Reinforcements

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Fibre Distributions and Orientations

Fig.1.4 A fibre laminate (stack of plies), illustrating the nomenclature system

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Lecture 2: Elastic Properties of Long Fibre Composites

Use of the Slab Model

Fig.2.1 Schematic illustration of loading geometry and distributions of stress and strain, andeffects on the Young’s moduli and shear moduli, for a uniaxial fibre composite and forthe slab model representation

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Halpin-Tsai Expressions

! H-T gives good approximation

! H-T gives slight overestimate

Fig.2.2 Predicted dependence on fibre volume fraction, for the epoxy-glass fibre system, of(a) transverse Young’s modulus and (b) shear moduli of long fibre composites

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Poisson Ratios

Fig.2.3 Schematic representation of the three Poisson ratios of an aligned composite

Fig.2.4 Predicted dependence on fibre volume fraction, for the epoxy-glass fibre system, of thethree Poisson ratios

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Elastic Loading of a Lamina

Matrix Notation

Kirchoff Assumptions

Axial and Transverse Loading

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Effect of Material Symmetry on the Number of Independent Elastic Constants

Fig.2.5 Indication of the form of the S pq and C pq matrices (matrix notation for S ijkl and C ijkl tensors), for materials exhibiting different types of symmetry. All of the matrices aresymmetrical about the leading diagonal.

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Lecture 3: Off-axis Elastic Properties of Laminae and Laminates

Loading at an Arbitrary Angle to the Fibre Axis

Fig.3.1 (a) Relationship between the fibre-related axes in a lamina (1, 2 & 3) and the co-ordinate system (x, y & z) for an arbitrary in-plane set of applied stresses.(b) Illustration of how such an applied stress state ! ’ij (! x, ! y & " xy) generates stresses inthe fibre-related framework of ! ij (! 1, ! 2 & " 12)

Derivation of Transformed Stress-Strain Relationship

For a thin lamina, stresses and strains in the through-thickness (3) direction are neglected, sothat the 3, 4, and 5 components in matrix notation are of no concern. Therefore, when a lamina isloaded parallel or normal to the fibre axis, the strains that interest us are given by:

! 1

! 2

" 12

#

$

%%%

&

'

(((= S [ ]

) 1

) 2

* 12

#

$

%%%

&

'

(((=

S 11

S 12

0

S 12

S 22

0

0 0 S 66

#

$

%%%

&

'

(((

) 1

) 2

* 12

#

$

%%%

&

'

((( (3.1)

in which, by inspection of the individual equations, it can be seen that:

S 11

=

1

E 1

S 12

= !

" 12

E 1

= !

" 21

E 2

S 22

=

1

E 2

S 66

=

1

G12

The first step in establishing the lamina strains for off-axis loading is to find the stresses,

referred to the fibre axis (! 1, ! 2 and " 12), in terms of the applied stress system (! x, ! y and " xy).This is done using the standard transform equation:

! ij = aik a jl "! kl

in which aik is the direction cosine of the (new) i direction referred to the (old) k direction.Obviously, the conversion will work in either direction provided the direction cosines are definedcorrectly. For example, the normal stress parallel to the fibre direction ! 11, sometimes written as! 1, can be expressed in terms of the applied stresses ! '11 (= ! x), ! '22 (= ! y) and ! '12 (= " xy)

! 11

= a11a11

"! 11

+ a11a12

"! 12

+ a12a11

"! 21

+ a12a12

"! 22

The angle # is that between the fibre axis (1) and the stress axis ( x). Referring to the figure:

a11 = cos! (= c) a12 = cos 90 "! ( ) = sin! (= s)

a21 = cos 90 +! ( ) = " sin! (= "s) a22 = cos! (= c)

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Carrying out this operation for all three stresses:

! 1

! 2

" 12

#

$

%%%

&

'

(((= T [ ]

! x

! y

" xy

#

$

%%%%

&

'

((((

in which T [ ] =

c2

s2

2cs

s2

c2 )2cs

)cs cs c2 ) s2( )

#

$

%%%%

&

'

((((

(3.2)

The same matrix can be used to transform tensorial strains, such as obtaining $ 1, $ 2 and $ 12 from $ x,$ y and $ xy. However, to use engineering strains (% xy = 2$ xy etc), T [ ] must be modified (by halvingthe elements t 13 and t 23 and doubling elements t 31 and t 32), to give:

! 1

! 2

" 12

#

$

%%%

&

'

(((= T '[ ]

! x

! y

" xy

#

$

%%%%

&

'

((((

in which T '[ ] =

c2

s2

cs

s2

c2 )cs

)2cs 2cs c2 ) s 2( )

#

$

%%%%

&

'

((((

(3.3)

The procedure is now a progression from the on-axis stress-strain relationship to a general one

involving a transformed compliance matrix, S !" #$ , which will depend on . Firstly, the inverse of

Eqn.(3.3) is required, giving strains relative to the loading direction (the information required), in

terms of strains relative to the fibre direction, involving the inverse of T '[ ] , written as T '[ ]!1

! x

! y

" xy

#

$

%%%%

&

'

((((

= T '[ ])1

! 1

! 2

" 12

#

$

%%%

&

'

((( in which T '[ ]

)1=

c2

s2 )cs

s2

c2

cs

2cs )2cs c2 ) s2( )

#

$

%%%%

&

'

((((

Strains relative to the fibre axis can be expressed in terms of corresponding stresses (Eqn.(3.1)):

! x

! y

" xy

#

$

%%%%

&

'

((((

= T '[ ])1

S [ ] * 1

* 2

+ 12

#

$

%%%

&

'

(((

Finally, the original transform matrix of Eqn.(3.2) can be used to express these stresses in termsof those being externally applied, to give the result:

! x

! y

" xy

#

$

%%%%

&

'

((((

= T '[ ])1

S [ ] T [ ]

* x

* y

+ xy

#

$

%%%%

&

'

((((

= S #$ &'

* x

* y

+ xy

#

$

%%%%

&

'

((((

(3.4)

The elements of S !" #$ are therefore obtained by concatanation (the equivalent of multiplication)

of the matrices T '[ ]!1

, S [ ] and T [ ] . The following expressions are obtained:

S 11

= S 11c4+ S

22s4+ 2S

12 + S

66( )c2s2

S 12

= S 12

c4+ s

4( )+ S 11 + S 22 ! S 66( )c2s2

S 22

= S 11s4+ S

22c4+ 2S

12 + S

66( )c2s2

S 16

= 2S 11 ! 2S

12 ! S

66( )c3s ! 2S

22 ! 2S

12 ! S

66( )cs3

S 26

=

2S 11 ! 2S 12

! S 66( )cs3

! 2S 22

! 2S 12

! S 66( )c3s

S 66

= 4S 11 + 4S

22 ! 8S

12 ! 2S

66( )c2s2 + S

66 c

4+ s

4( )

(3.5)

It can be seen that S !" #$ ! S [ ] as # ! 0, as required.

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Effect of Loading Angle on Stiffness and Poisson Ratio

Fig.3.2 Variation with loading angle # of (a) Young’s modulus E x and shear modulus G xy and(b) Poisson ratio & xy (using equal stress model), for a lamina of epoxy-50% glass fibre

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Tensile-Shear Interaction Behaviour

Fig.3.3 Variation with loading angle # of the tensile-shear interaction compliance S 16 , for alamina of rubber-5% Al fibre, and photos of 4 specimens (between crossed polars) underaxial tension, lined up at the appropriate values of # , showing tensile-shear distortions

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Obtaining the Elastic Constants of a Laminate

Fig.3.4 Schematic depiction of the loading angle ' between the x-direction (stress axis) and thereference direction (# =0˚), for a laminate of n plies. Also shown is the angle # k betweenthe reference direction and the fibre axis of the k th ply (1k direction)

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Lecture 4: Classification of Laminates

Stiffness of Laminates

Fig.4.1 Variation with loading angle ' (between the stress axis and the reference (# =0˚)direction) of the Young’s modulus of a single lamina and of two simple laminates, madeof epoxy-50% glass fibre. (The equal stress model was used to obtain the transverseYoung’s modulus of the lamina, E 2.)

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Tensile-Shear Interactions and Balanced Laminates

Fig.4.2 Variation with loading angle ' (between the stress axis and the reference (# =0˚)

direction) of the interaction ratio, ( xyx (ratio of the shear strain % xy to the normal strain$ x) of a single lamina and of three simple laminates, made of epoxy-50% glass fibre.(The equal stress model was used to obtain the transverse Young’s modulus of thelamina, E 2.)

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In-plane Stresses within a Loaded Laminate

Fig.4.3 (a) Predicted stresses within one ply of a loaded crossply laminate (epoxy-50%glass)and (b) a schematic of these stresses for loading parallel to one of the fibre axes

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Coupling Stresses and Symmetric Laminates

Fig.4.4 Elastic distortions of a crossply laminate as a result of (a) uniaxial loading and(b) heating

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Lecture 5: Short Fibre & Particulate Composites - Stress Distributions

The Shear Lag Model for Short Fibre Composites

Displacements of Fibre and Matrix

Fig.5.1 Schematic illustration of the basis of the shear lag model, showing (a) unstressed system,(b) axial displacements, u, introduced on applying tension parallel to the fibre and(c) variation with radial location of the shear stress and strain in the matrix

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Derivation of Equations

The model is based on assuming that the build-up of tensile stress along the length of the fibreoccurs entirely via the shear forces acting on the cylindrical interface. This leads immediately tothe basic shear lag equation:

d! f

d x =

"2# i

r 0

(5.1)

The interfacial shear stress, "i, is obtained by considering how the shear stress in this directionvaries within the matrix as a function of radial position. This variation is obtained by equating theshear forces on any two neighbouring annuli in the matrix:

2! r 1

" 1 d x = 2! r

2

" 2 d x ie

" 1

" 2

=

r 2

r 1

#" = " i

r 0

r

$ % & ' ( )

The displacement of the matrix in the loading direction, u, is now considered. The shear strainat any point in the matrix can be written both as a variation in this displacement with radialposition and in terms of the local shear stress and the shear modulus of the matrix, Gm

! ="

Gm

=

" i

r0

r

# $ %

& ' (

Gm

and ! = du

dr

It follows that an expression can be found for the interfacial shear stress by considering thechange in matrix displacement between the interface and some far-field radius, R, where thematrix strain has become effectively uniform (du/dr ! 0).

duur0

u R! = " i

r0

Gm

dr

rr0

R

!

!

" i

=

u R # u

r0( )Gm

r0

ln R

r0

$ % &

' ( )

(5.2)

The appropriate value of R is affected by the proximity of neighbouring fibres, and hence by thefibre volume fraction, f . The exact relation depends on the precise distribution of the fibres, butthis needn't concern us too much, particularly since R appears in a log term. If an hexagonal arrayof fibres is assumed, with the distance between the centres of the fibres at their closest approachbeing 2 R, then simple geometry leads to

R

r0

!

" #$

% &

2

=

'

2

f

3

(

1

f

Substituting for "i in the basic shear lag equation now gives

d! f

d x =

"2 u R " ur0

( )Gm

r0

2

1

2ln

1

f

# $ %

& ' (

The displacements u R and ur0 are not known, but their differentials are related to identifiablestrains. The differential of ur0 is simply the axial strain in the fibre (assuming perfect interfacialadhesion and neglecting any shear strain in the fibre - which is taken as being much stiffer thanthe matrix)

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dur0

d x = !

f =

" f

E f

while the differential of u R, ie the far-field axial strain of the matrix, can be taken as themacroscopic strain of the composite

du R

d x

!

" 1

Differentiating the expression for the gradient of stress in the fibre and substituting these tworelations into the resulting equation, with the shear modulus expressed in terms of Young'smodulus and Poisson's ratio [ E m=2 Gm (1+ #m)], leads to

d2!

f

d x2 =

n2

r0

2 !

f " E

f #

1( ) (5.3)

in which n is a dimensionless constant (for a specified composite), given by

n =2 E

m

E f 1+ !

m( ) ln 1

f

" # $

% & ' (5.4)

This is a second order linear differential equation of a standard form, which has the solution

! f = E

f "

1 + Bsinh

n x

r0

#

$%

&

'( + Dcosh

n x

r0

#

$%

&

'(

and, by applying the boundary condition of $f = 0 at x = ± L (the fibre half-length), the constants B and D can be solved to give the final expression for the variation in tensile stress along the lengthof the fibre

! f

=

E f

" 1

1# cosh n x

r0

$

%&

'

()sech n s( )

*+,

-,

./,

0, (5.5)

in which s is the aspect ratio of the fibre (= L/r0). From this expression, the variation in interfacialshear stress along the fibre length can also be found, using the basic shear lag equation, bydifferentiating and multiplying by (-r0/2),

! i =

E f n "

1

2sinh

n x

r0

#

$%

&

'( sech n s

( ) (5.6)

An estimate can now be made of the axial modulus of the composite. This is done by usingthe Rule of Averages ($1 = f $f

_ + (1- f ) $m

_ ), with the average matrix stress taken as its Young's

modulus times the composite strain and the average fibre stress obtained by integrating the aboveexpression for $f over the length of the fibre. This leads to

E 1 =

! 1

" 1

= f E f 1#

tanh n s( )

n s

$

%&

'

() + 1# f ( ) E m (5.7)

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The Stress Transfer Length (Aspect Ratio)

Fig.5.2 Predicted (shear lag) variations in (a) fibre tensile stress and (b) interfacial shear stressalong the axis of a glass fibre in a polyester-30% glass composite subject to an axialtensile strain of 10-3, for two fibre aspect ratios

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Fibre End Regions - Hydrostatic Stresses and Cavitation

(a) (b)

Fig.5.3 Photoelastic (“frozen stress”) models under applied axial load, showing the stress fieldin the matrix around two stiff reinforcements having the same aspect ratio, with(a) cylindrical and (b) ellipsoidal shapes

Fig.5.4 Predicted (finite element) hydrostatic stress fields around sphere and cylinder (s=5) ofSiC in an Al matrix, with an applied axial tensile stress of 100 MPa (and differentialthermal contractions stresses corresponding to a temperature drop of 50 K)

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Lecture 6:Short Fibre & Particulate Composites - Stiffness & Inelastic Behaviour

Shear Lag Model Predictions for Stiffness

Fig.6.1 Predicted composite/matrix Young’s modulus ratio, as a function of fibre/matrix Young’smodulus ratio, for aligned short fibre composites with 30% fibre content and fibre aspect

ratio (s) values of (a) 30 and (b) 3. Shear lag model predictions are reliable when s isrelatively large. For very short fibres, the predictions become inaccurate, due to neglectof the stress transfer across the fibre ends, which is more significant for shorter fibres

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Approach to Rule of Mixtures (Long Fibre) Stiffness

Fig.6.2 A set of four (rubber – 5% Al fibre) photoelastic models under axial load, showing howthe stress field and the axial extension change as the aspect ratio and degree ofalignment of the fibres are changed

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Interfacial Sliding and Matrix Yielding

Fig.6.3 Plots of the dependence of peak fibre stress, ! f0, (at the onset of interfacial sliding ormatrix yielding) on the critical shear stress for these phenomena, " i*. Plots are shown

for different fibre aspect ratios, with n values typical of polymer- and metal-basedcomposites. Also indicated are typical value ranges for fracture of fibres and for matrix

yielding and interfacial debonding

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Critical Fibre Aspect Ratio

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Lecture 7: The Fibre-Matrix Interface

Bonding Mechanisms and Residual Stresses

Bonding Mechanisms

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Residual Stress Distributions

Fig.7.1 Predicted stress distribution around and within a single fibre, in a polyester-35% glasslong fibre composite, as a result of differential thermal contraction (T drop of 100 K).

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Silane Coupling Agents for Glass Fibres

Fig.7.4 Depiction of the action of silane coupling agents, which are used to generate improved fibre-matrix bonding for glass fibres in polymeric matrices. The silane reacts withadsorbed water to create a strong bond to the glass surface. The R group is one whichcan bond strongly to the matrix.

Objectives for MMCs and CMCs

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Bond Strength Measurement

Single Fibre Pull-out Testing

Fig.7.2 Schematic stress distributions and load-displacement plot during single fibre pull-out

testing. The interfacial shear strength, " *, is obtained from the pull-out stress, ! 0,*

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Single Fibre Push-out Testing

Fig.7.3 Schematic stress distributions and load-displacement plot during the single fibre push-

out test. One difference from the pull-out test is that the Poisson effect causes the fibreto expand (rather than contract), which augments (rather than offsets) the radialcompressive stress across the interface due to differential thermal contraction

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Lecture 8: Fracture Strength of Composites

Axial, Transverse and Shear Strengths of Long Fibre Composites

Fig.8.1 Schematic depiction of the fracture of a unidirectional long fibre composite at criticalvalues of (a) axial, (b) transverse and (c) shear stresses

Axial Strength

Transverse and Shear Strengths

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Failure Criteria for Laminae subject to In-plane Stresses

Maximum Stress Criterion

Mixed Mode Failure and the Tsai-Hill Criterion

Fig.8.2 Single ply failure stresses, as a function of loading angle: (a) maximum stress criterion, for polyester-50%glass (! 1*=700 MPa, ! 2*=20 MPa, " 12*=50 MPa) and (b) maximumstress and Tsai-Hill criteria, plus experimental data, for epoxy-50%carbon(! 1*=570 MPa, ! 2*=32 MPa, " 12*=56 MPa)

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Experimental Data for Single Laminae

Fig.8.3 Schematic illustration of how a hoop-wound tube is subjected to simultaneous tensionand torsion in order to investigate failure mechanisms and criteria

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Failure of Laminates

Failure Sequences in Laminates

Fig.8.4 Loading of the crossply laminate of Fig.4.4 parallel to one of the fibre directions:(a) cracking of transverse plies as ! 2 reaches ! 2*, (b) onset of cracking parallel to fibresin axial plies as ! 2 (from inhibition of Poisson contraction) reaches ! 2* and (c) final

failure as ! 1 in axial plies reaches ! 1*

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Failure of Laminates under Uniaxial and Biaxial Loading

Fig.8.5 Stresses within an angle-ply laminate of polyester-50%glass fibre, as a function of the ply angle: (a) stresses within one of the plies, as ratios to the applied stress. and(b) applied stress at failure (maximum stress criterion, with ! 1*=700 MPa, ! 2*=20 MPaand " 12*=50 MPa)

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Fig.8.6 Stresses within an angle-ply laminate of polyester-50%glass fibre, as a function of the ply angle, when subjected to biaxial loading, with !

x

=2! y

: (a) stresses within one of the plies, as ratios to the applied ! x. and (b) applied stress, ! x, at failure (maximum stresscriterion, with ! 1*=700 MPa, ! 2*=20 MPa and " 12*=50 MPa)

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Lecture 9: Fracture Toughness of Composites

Fracture Energies of Reinforcements and Matrices

Crack Deflection at Interfaces – Planar Systems

Fig.9.1 Schematic load-displacement plots for 3-point bend testing of monolithic SiC and a SiClaminate with (weak) graphitic interlayers

Fig.9.2 SEM micrographs showing the layered structures of (a) a mollusc and (b) a SiC laminate

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Energy of Interfacial Debonding in Fibre Composites

Fig.9.3 Schematic representation of the advance of a crack in a direction normal to the fibreaxis, showing interfacial debonding and fibre pull-out processes

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Energy of Fibre Pull-out

Gcpo = N d x0 L

0

L

! " rx02# i* = f

" r2$ % &

' ( )

" r# i* L

$ % &

' ( )

L3

3

$

% &

'

( ) =

fs2r# i*3

(9.1)

Effects of Fibre Flaws and Weibull Modulus

Fig.9.4 Schematic depiction of stress distribution, and associated probability of fracture, along a fibre bridging a matrix crack, for (a) fixed fibre strength ! * (m=!) and (b) strengthwhich varies along the length of the fibre, due to the presence of flaws (finite m).

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Fracture Energy of a Metal Fibre Reinforced Ceramic Composite

There have been many attempts to produce ceramic-matrix composites with high toughness,mostly with limited success. A promising approach, however, is based on a network of metallic

fibres, as in a commercial product called “Fiberstone” (Fig.9.5). The fibres are usually about0.5 mm diameter, although finer fibres can be used.

Fig.9.5 Optical micrograph of a polished section of a composite (“Fiberstone”) comprisingcoarse stainless steel fibres in an alumina matrix.

During fracture, fibres bridge the crack and energy is absorbed by both frictional pull-out andplastic deformation (Fig.9.6). These mechanisms dominate the work of fracture.

Fig.9.6 Schematic of Fiberstone fracture: (a) overall geometry, (b) debonding, possibly fracture,and then frictional pull-out and (c) debonding, plastic deformation and then fracture

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The composite work of fracture, Gcnet, is thus the sum of contributions from pull-out, Gcpo, andplastic deformation, Gcfd, of fibres bridging the crack plane, assuming a fraction g of them undergothe former (and (1-g) the latter). The expression for fibre pull-out, Eqn.(9.1), can be used, but therelationship between N and f depends on fibre orientation distribution, and that referred to alignedfibres. For an isotropic (random) distribution, N is reduced by a factor of two†, leading to

Gcpo

=

gfs2r! i

*

6 (9.2)

where s is the ratio of ) (average fibre length extending beyond crack plane) to fibre radius, r.

Gcfd is obtained assuming interfacial debonding to a distance x0 from the crack plane(Fig.9.6(c)), treating the work done as if it were a tensile test with an original fibre length 2 x0

Gcfd = (1! g)2 x0 NU fd = (1! g)2 x0

f

2" r2

# $ %

& ' ( W

fd" r2

= (1! g) x0 fW fd (9.3)

where U fd and W fd are the work of deformation of the fibre, expressed respectively per unit length(J m-1) and per unit volume (J m-3). The latter is the area under the stress-strain curve of the fibre.

The value of ) is in this case given by the product of x0 and $ *, the fibre strain to failure, leading to

Gcfd

= (1! g) "

# *

$

% &'

( ) fW

fd =

(1! g)srfW fd#

*

(9.4)

Fig.9.7 Experimental fracture energy values of “Fibrestone”, as a function of fibre volume fraction, and predictions obtained using Eqns.(9.3) amd (9.5), for fine and coarse fibres.

Note that, for a given fibre protrusion aspect ratio, s (= ) /r), both pull-out and plasticdeformation contributions increase linearly with the absolute scale (fibre diameter). Compositesreinforced with coarser fibres therefore tend to be tougher, particularly for this type of composite.It’s clear that refining the scale of the microstructure does NOT always give benefits!

† see EE Underwood, Quantitative Stereology. 1970, Addison-Wesley

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Lecture 10: Compressive Loading of Fibre Composites

Euler Buckling

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Kink Band Formation

This is the most common form of failure in (axial) compression. A kink band is shown inFig.10.1. Prediction of the applied stress at formation is based on the schematic in Fig.10.2.

Fig.10.1 Optical micrograph of an axial section of a carbon fibre composite after failure underuniaxial compression, showing a kink band.

Fig.10.2 Schematic of stresses and fibre alignments at failure, with a corresponding Mohr circle.

! k * =

" 12*

# =

" 12*

# 0 +

" 12*

G12

$ % &

' ( )

=1

G12

+# 0

" 12*

$

% &'

( )

*1

(10.1)

Fig.10.3 Predicted kinking stress (Eqn.(10.1), as a function of misalignment angle, for epoxy-60%carbon composites, with two different interfacial shear strengths

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Failure by Fibre Crushing in Highly Aligned Systems

(a) (b)

Fig.10.4 (a) Fragment of SiC monofilament extracted from a Ti-35%SiC composite after loadingunder axial compression and (b) schematic of the crushing process

Fig.10.5 Stresses in Ti-35%SiC monofilament composite (average axial values for fibre, matrixand composite) as axial strain increases from external loading. At zero strain, stressesare only from differential thermal contraction. The matrix yields at ! mY . Matrix workhardening is neglected. Failure occurs when the fibre stress reaches a critical value ! f*.

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Neglecting matrix work hardening, the composite failure stress can be expressed (Fig.10.5) as

! c* = E

1c" cmY

+ E 1c

'" c* # "

cmY( ) (10.2)

in which the composite moduli before and after matrix yielding are given by

E 1c = fE

f + 1! f ( ) E

m E

1c

'= fE

f

The strains at matrix yield, and at final failure, can be written (Fig.10.5) as

! cmY

="

mY +"

m#T

E m

! c* =

" f* + "

f #T

E f

Substituting into Eqn.(10.2), and applying the residual stress force balance ( f ! f %T + (1- f )! m%T = 0):

! c* = f !

f* + 1" f ( )!

mY

A (small) correction for the effect of misalignment leads to:

! c* =

f ! f* + 1" f ( )!

mY

cos2# 0

(10.3)

This predicted strength is independent of the thermal residual stresses (whereas the strain at which

failure occurs will depend on them).

Fig.10.6 Experimental strength data, as a function of the initial angle between fibre and loadingaxes, during compression of Ti-35%SiC specimens. Also shown are predictions for

failure by kink band formation (Eqn.(10.1)) and by fibre crushing (Eqn.(10.3)).

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Lecture 11:Thermal Expansion of Composites & Thermal Residual Stresses

Thermal Expansivity Data for Reinforcements and Matrices

Fig.11.1 Thermal expansion coefficients for various materials over a range of temperature

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Derivation of Expression for Composite Axial Expansivity

Fig.11.2 Schematic showing thermal expansion in the fibre direction of a long fibre composite,using the slab model

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Transverse Thermal Expansivities

Fig.11.3 Predicted thermal expansivities of Al-SiC uniaxial fibre composites, as a function of

fibre content

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Thermal Stresses in Composite Systems

Magnitudes of Thermal Residual Stresses

Stresses in Composites during Thermal Cycling

Fig.11.4 Neutron diffraction data for an Al-5vol%SiC whisker (short fibre) composite, showinglattice strains (& hence stresses) within matrix & reinforcement during unloadedthermal cycling. (111) reflections were used for both constituents. The gradients shownare calculated values for elastic behaviour, assuming a fibre aspect ratio of 10.

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Lecture 12: Surface Coatings as Composite Systems

A Substrate-Deposit System with a Uniform Misfit Strain

Fig.12.1 Schematic depiction of the generation of curvature in a flat bi-material plate, as a resultof the imposition of a uniform misfit strain, *$ . The strain and stress distributions shownare for the case indicated, obtained using Eqns.(I.8) & (I.9) from Appendix I.

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Relation between Curvature and Misfit Strain

The misfit strain, %$ , generates a bending moment, creating a curvature. A fairlystraightforward mathematical treatment of this (see Appendix I) leads to the following relationship

! =

6 E d E

s h + H ( )h H "#

E d

2h

4+ 4 E

d E

sh

3 H + 6 E

d E

sh

2 H

2+ 4 E

d E

sh

H 3+ E

s

2 H

4 (12.1)

Note that, for a given deposit/substrate thickness ratio, h/ H , the curvature is inversely proportional

to the substrate thickness, H . This scale effect is important in practice, since it means that

relatively thin substrates are needed if curvatures are to be sufficiently large for accurate

measurement. Predicted curvatures, obtained using this equation, are shown in Fig.12.2.

Curvatures below about 0.1 m-1 (radius of curvature, R > 10 m) are difficult to measure accurately.

Fig.12.2 Predicted curvature, as a function of the fall in temperature, for four differentsubstrate/deposit combinations.

Biaxial Stresses

In practice, there are often stresses in all in-plane directions. For an isotropic in-plane stressstate, there is a stress (= ! x) in the z-direction, which induces a Poisson strain in the x-direction.For isotropic stiffness and no through-thickness stress (! y = 0), the strain in the x-direction is:

! x E = "

x # $ "

y + "

z( ) = " x 1# $ ( )

so the relation between stress and strain in the x-direction can be expressed

! x

" x

=

E

1# $ ( )

= E ' (12.2)

and this modified form of the Young’s modulus, E’, is usually applicable in expressions referring

to substrate/coating systems having an equal biaxial stress state.

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Stoney’s Equation – the Thin Coating Limit

A simplified form of Eqn.(12.1) applies for coatings much thinner than the substrate (h

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Accuracy of the Stoney Equation

The Stoney equation is easy to use and, indeed, is widely used. However, it does have thelimitation of being accurate only in a regime in which the curvatures are relatively small. In someapplications – such as with semiconductor wafers – surfaces are very smooth, so that highlyaccurate optical methods of curvature measurement are feasible and this is not such a problem.However, when curvatures are high (or need to be high for reliable measurement), the Stoneyequation should not be used.

Fig.12.3 Predicted dependence on thickness ratio of (a) curvature and (b) stress in deposit(coating), obtained using Eqns.(I.7), (I.8) and (I.9), and the Stoney equation(Eqn.(12.3).) The Poisson ratios of substrate and deposit were both taken as 0.2

Driving Force for Interfacial Debonding

The presence of residual stresses in a substrate/coating system constitutes a driving force fordebonding (spallation), since they will almost certainly be at least partially relaxed when thisoccurs, releasing stored elastic strain energy.

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Fig.12.4 Predicted effect of substrate thickness, H (for a fixed coating thickness h), on the averagestress levels in coating and substrate, created by a given misfit strain, and also on thestrain energy release rate for debonding (showing the contributions from stresses incoating and substrate). The effect of curvature adoption on the stresses is neglected.

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Appendix I – Relation between Curvature and Misfit Strain

Referring to Fig.12.1, the forces P and –P generate an unbalanced moment, given by

M = P

h + H

2

! " #

$ % & (&.1)

where h and H are deposit and substrate thicknesses respectively. Since the curvature, + ,(through-thickness strain gradient) is given by the ratio of moment, M , to beam stiffness, '

! =

M

" (&.2)

P can be expressed as

P =2 ! "

h + H (&.3)

The beam stiffness is given by

! = b E yc( )

" H "#

h"#

$ yc2 d yc = b E d h h

2

3" h# + # 2

% & '

( ) * + b E

s H

H 2

3+ H # + #

2% & '

( ) * (&.4)

where ) (distance between neutral axis ( yc = 0) and interface ( y = 0)) is given (see Appendix II) by

! =h

2 E

d " H

2 E

s

2 hE d + HE

s( ) (&.5)

The magnitude of P is found by expressing the misfit strain as the difference between the

strains resulting from application of the P forces (ie by writing a strain balance):

!" = " s # "

d =

P

HbE s

+P

hbE d

, $P

b= !"

hE d HE

s

hE d + HE

s

%

& '(

) * (&.6)

Combination of this with Eqns.(I.3)-(I.5) gives a general expression for the curvature, ( , arising

from imposition of a uniform misfit strain, %$

! =6 E

d E

s h + H ( )h H "#

E d

2h

4+ 4 E

d E

sh

3 H + 6 E

d E

sh

2 H

2+ 4 E

d E

sh

H 3+ E

s

2 H

4 (&.7)

The stress at the interface ( y = 0), and at the free surfaces ( y = h or - H ), can be written in termsof the base level in each constituent (arising from the force balance) and the change due to thestress gradient (= curvature ) stiffness):

! d

y=h="P

b h+ E

d # h " $ ( ) & !

d y=0

="P

b h " E

d # $ (I.8)

! s y=" H

=P

b H " E

s # H + $ ( ) & !

s y=0

=P

b H " E

s # $ (I.9)

Of course, since the gradient is linear in each constituent, this gives the complete stress profile.

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Appendix II – Location of the Neutral Axis

Fig.12.4 Location of the Neutral Axis of a Bi-Material Beam

The force balance

b ! y( )" H

h

# d y = 0 (&&.1)

can be divided into contributions from the two constituents and expressed in terms of the strain

b E d ! ( y)0

h

" d y + b E s! ( y)# H

0

" d y = 0 (&&.2)

which can then be written in terms of the curvature (through-thickness strain gradient) and the

distance from the neutral axis

b E d ! y " # ( )0

h

$ d y + b E s! y " # ( )" H

0

$ d y = 0 (&&.3)

Removing the width, b, and curvature, ( , which are constant, this gives

E d y

2

2! " y

#

$%

&

'(0

h

+ E s y

2

2! " y

#

$%

&

'(! H

0

= 0

) E d h

2

2! " h

* + ,

- . / + E s

! H 2

2! " H

* + ,

- . / = 0

)" E d h + E s H ( ) =1

2 E d h

2 ! E s H 2( )

)" =h

2 E

d ! H 2 E

s

2 hE d + HE

s( )

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Property Data (Room Temperature)

Fibres

FibreDensity

(Mg m-3)

Axial

Modulus

E 1 (GPa)

Transverse

Modulus

E 2 (GPa)

Shear

Modulus

G12 (GPa)

Poisson

Ratio

#12

Axial

Strength

$* (GPa)

Axial

CTE

*1 (µ+ K-1)

Transverse

CTE

*2 (µ+ K-1)

Glass 2.45 76 76 31 0.22 3.5 5 5

Kevlar 1.47 154 4.2 2.9 0.35 2.8 -4 54

Carbon (HS) 1.75 224 14 14 0.20 2.1 -1 10

Carbon (HM) 1.94 385 6.3 7.7 0.20 1.7 -1 10

Diamond 3.52 1000 1000 415 0.20 4 3 3

Boron 2.64 420 420 170 0.20 4.2 5 5

SiC(monofilament)

3.2 400 400 170 0.20 3.0 5 5

SiC(whisker)

3.2 550 350 170 0.17 8 4 4

Al2O3

(* continuous)

3.9 385 385 150 0.26 1.4 8 8

Al2O3

(, staple)

3.4 300 300 120 0.26 2.0 8 8

W 19.3 413 413 155 0.33 3.3 5 5

Matrices

MatrixDensity

(Mg m-3)

Young's

Modulus

E (GPa)

Shear

Modulus

G (GPa)

Poisson

Ratio

#

Tensile

Strength

$* (GPa)

Thermal

Expansivity

* (µ+ K-1)

Epoxy 1.25 3.5 1.27 0.38 0.04 58

Polyester 1.38 3.0 1.1 0.37 0.04 150

PEEK 1.30 4 1.4 0.37 0.07 45

Polycarborate 1.15 2.4 0.9 0.33 0.06 70

PolyurethaneRubber 1.2 0.01 0.003 0.46 0.02 200

Aluminium 2.71 70 26 0.33 0.07 24

Magnesium 1.74 45 7.5 0.33 0.19 26

Titanium 4.51 115 44 0.33 0.24 10

Borosilicateglass

2.23 64 28 0.21 0.09 3.2

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Question Sheet 1

[Can be attempted after lecture 8: property data on C16H62 may be used if necessary.]

1. Show that the Young's modulus of a composite lamina (having the elastic constants, referredto the fibre axis, given below) falls by about 50% if it is loaded at 7˚ to the fibre axis,

compared with the on-axis value. What is the minimum Young's modulus that the lamina canexhibit and at what loading angle does this occur ?[ E 1 = 200 GPa, E 2 = 7 GPa, G12 = 3 GPa, v12 = 0.3]

2. Explain what is meant by tensile-shear interactions in composite laminae. Using information

in the Data Book, derive an expression for the tensile-shear interaction compliance S 16

of alamina. State how this is used in describing the elastic deformation of the lamina under anapplied uniaxial tensile load. For a lamina of an epoxy-glass composite, with the elastic

constants given below, calculate the loading angles for which the lamina will show no shearstrains under such a load.[ E 1 = 40 GPa, E 2 = 8 GPa, #12 = 0.3, G12 = 3 GPa]

3. Fig.1 shows the stresses (parallel and transverse to the fibre direction) within one ply of anangle-ply laminate subjected to unequal biaxial tension ($ x = 2 $ y). The stresses (ratios to $ x)are shown as a function of the ply angle, # (measured relative to the direction of $ x). Thecritical stresses for failure of this composite axially, transversely and in shear, ie $1*, $2* and"12*, are respectively 700 MPa, 50 MPa and 30 MPa. Using Fig.1, and the maximum stresscriterion for failure, find the pressure at which an internally pressurized tube (radius = 50 mm,

wall thickness = 2 mm) of this composite, wound with a ply angle, # of ±40˚ (to the hoopdirection) is predicted to fail.

Fig.1 Stresses (ratios to ! x) within one ply of an angle-ply laminate, as a function of the plyangle, when subjected to biaxial loading, with ! x=2! y.

Find the failure pressure using the Tsai-Hill failure criterion, which can be expressed as

! 1

! 1*

"

# $%

& '

2

+!

2

! 2*

"

# $%

& '

2

( !

1!

2

! 1*

2 +

) 12

) 12*

"

# $%

& '

2

*1

Explain any difference between this value and that obtained previously. Using the criterion

you consider most reliable in this case, obtain an approximate estimate of the ply angle thatwould give the largest failure pressure.

{from 2014 Tripos}

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4. An angle-ply (±50˚) laminate of a polyester-50%glass composite is subjected to an increasingtensile stress in the ! x (' =0˚) direction. Use the facility at the end of the section entitled“Failure of Laminates and the Tsai-Hill Criterion”, within the “Mechanics of FibreComposites” TLP (www.doitpoms.ac.uk/tlplib/fibre_composites/index.php ), to establish theapplied stress at which the laminate will fail (according to the maximum stress criterion),

given that ! 1* = 700 MPa, ! 2* = 20 MPa and " 12* = 50 MPa. Carry out the same calculation,

using simple analytical equations, for one of the two plies (ie ignore the presence of the other)and compare this value with the first result. Account for any difference between the two.

5. Candidate materials for a gas pipeline are steel and a glass fibre reinforced polymercomposite. The diameter of the pipeline will be 1 m and the maximum gas pressure will be100 bar (10 MPa). The composite would be filament-wound, at ±45˚ to the hoop direction.

There are no concerns about stiffness, so the key design criterion is to avoid phenomena whichcould lead to failure (which would be likely to be plasticity in the case of the steel and sometype of microstructural damage in the case of the composite). The main design variable will

be the wall thickness. Using the von Mises yield criterion (steel) and the Tsai-Hill failure

criterion (composite), and ignoring the issue of safety factors, estimate the minimum wallthickness in each case and hence deduce which material would allow the lighter pipeline.

Comment on the assumptions and sources of error in your calculation and on whether theremight be a danger of any other types of failure. Without carrying out any further calculations,indicate whether and how you would recommend changing the fibre winding angle of the

composite in order to make it more effective for this application.

[The von Mises yield criterion can be written

! 1 " !

2( )2

+ ! 2 "!

3( )2

+ ! 3 "!

1( )2

2# !

Y

where ! 1, ! 2 and ! 3 are the principal stresses and ! Y is the uniaxial yield stress. The latterhas a value of 150 MPa for the steel. The density of the steel is 7.8 Mg m-3.

The Tsai-Hill criterion for failure of a composite ply under plane stress conditions can be

expressed as:

! 1

! 1*

" # $

% & '

2

+!

2

! 2*

" # $

% & '

2

( !

1!

2

! 1*

2 +

) 12

) 12*

" # $

% & '

2

* 1

where ! 1, ! 2 and " 12 are the stresses parallel, transverse and in shear relative to the fibre

axis and ! 1*, ! 2* and " 12* are corresponding critical values (measured respectively to be900 MPa, 30 MPa and 40 MPa). The composite density is 1.8 Mg m-3.

The stresses within a lamina, subject to ! x, ! y and " xy, are given by

! 1

! 2

! 12

"

#

$$$

%

&

'''=

c2

s2

2cs

s2

c2 (2cs

(cs cs c2 ( s 2

"

#

$$$

%

&

'''

! x

! y

) xy

"

#

$$$$

%

&

''''

where c = cos# and s = sin# , and # is the angle between x and 1 (fibre) directions.]

{from 2012 Tripos}

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Question Sheet 2

[Can be attempted after lecture 12; property data on C16H62 may be used if necessary.]

1. (a) A 1 mm thick unidirectional ply of epoxy-25vol% glass fibre composite is bonded at120˚C to a steel plate with the same dimensions, and curing goes to completion at this

temperature. The bonded pair is then cooled (elastically) to room temperature (20˚C).Describe the out-of-plane distortion that arises and calculate the associated curvature(s).

(b) When the bonded pair is loaded in compression parallel to the fibre axis of the ply, it isobserved that the curvature(s) it exhibits starts to reduce. Account for this effect. Calculate

the applied stress at which the specimen would become flat and comment on whether this islikely to be achievable.[For glass fibres: E = 76 GPa, * = 5 ) 10-6 K-1, # = 0.22

for epoxy resin: E = 3.5 GPa, * = 58 ) 10-6 K-1, # = 0.40for steel: E = 210 GPa, * = 11.4 ) 10-6 K-1, # = 0.26

For an aligned long fibre composite. axial and transverse thermal expansivities, , c, tr and , c, tr,

are given by the following (force balance and Schapery) expressions

! c, ax

=

! m

1" f ( ) E m +! f fE f 1" f ( ) E m + fE f

, ! c, tr

=

! m

1" f ( ) 1+# m( )+! f f 1+ # f ( )"! c, ax# 12c

The curvature, ( , exhibited by a pair of bonded plates, each of thickness h, when there is amisfit strain %+ between their natural (stress-free) lengths, is given by

! =12 "#

h E

1

E 2

+14 + E

2

E 1

$

% & &

'

( ) )

]

{from 2008 Tripos}

2. (a) A “vibration-damped” sheet material is made by bonding a 1 mm thick rubber layerbetween two steel plates of thickness 1 mm. The sheet is pushed against the surface of a largecylindrical former, which has a radius of 0.5 m. Sketch the through-thickness distributions of

strain and stress in the sheet, assuming that both the steel and the rubber remain elastic.

(b) This forming operation is actually designed to generate plastic deformation, creating ashaped component with a uniform curvature in one plane. Taking the steel to have a yield

stress of 300 MPa (in compression or tension), and assuming that the rubber remains elastic,show that the above operation would in fact induce plastic deformation in outer layers of bothmetal sheets and calculate the thickness of the layers that would yield in this way and the

plastic strain at the free surfaces.

(c) Show that, if the width of the sheet (length along the axis of the cylinder) is 0.5 m, then thebeam stiffness (' = EI ) of the sheet is 216.7 N m2 and the bending moment that would beneeded in order to bring the sheet into contact with the cylindrical former would be 433 N m,assuming that the steel remained elastic. Calculate the required bending moment for the actualcase, with the steel undergoing plastic deformation at a yield stress of 300 MPa (but

neglecting any work hardening).

[Steel: Young’s modulus, E = 200 GPa;

Rubber: Young’s modulus value more than 4 orders of magnitude lower]

{from 2011 Tripos}

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3. (a) The curvature, ( , arising from a misfit strain, %+, between a coating (deposit) of thicknessh and a substrate of thickness H is given by

! =

6 E d E

s h + H ( )h H "#

E d

2h

4+ 4 E

d E

sh

3 H + 6 E

d E

sh

2 H

2+ 4 E

d E

sh

H 3+ E

s

2 H

4

where E d and E s are the Young’s moduli of deposit and substrate. Show that, in the limit ofh

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5. (a) The residual stresses created during rolling of steel sheet down to a thickness of 1 mm,which are exclusively in the rolling direction, approximate to a compressive stress of 40 MPa

in the 0.1 mm layers at the free surfaces and a uniform tensile stress in the central 0.8 mm. Alarge square plate is cut from the sheet, with sides parallel to the rolling and transversedirections. It is initially flat. It is then exposed to a chemical on one side, which uniformly

dissolves the steel on that side to a depth of 0.1 mm. Sketch a sectional drawing in the planecontaining the rolling and through-thickness directions, showing the internal stresses andstrains before and after the dissolution. Neglecting edge effects, describe the shape that the

plate will adopt, calculating the magnitude of any curvature and commenting on whether itwould be noticeable.

(b) It’s suggested that it might be possible to return the plate to being completely flat by shot

peening the surface that had been chemically attacked, so as to put it into compression.Without carrying out any calculations, predict the qualitative effect of such a treatment.

[The curvature, ( , arising from imposition of a uniform misfit strain, %$ , between the

two layers in a bi-layer system (with thicknesses h and H , and having the same stiffness)

is given by

! =

6 h + H ( )hH "#

h4+ 4h

3 H + 6h

2 H

2+ 4hH

3+ H

4

Young’s modulus of the steel = 200 GPa ]

{from 2015 Tripos}

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Examples Class I

[Property data on C16H62 may be used if necessary.]

1. (a) The components of the compliance tensor of an epoxy-glass fibre composite lamina,referred to the fibre axis direction and the transverse direction, can be written

S =

S 11 S 12 0

S 12

S 22

0

0 0 S 66

=

1 / E 1 !

" 12 / E 1 0

!" 21

/ E 2

1 / E 2

0

0 0 1 / G12

Using information in the Data Book, show that the interaction compliance giving the shearstrain arising from a normal stress, when the lamina is loaded at an angle - to the fibre axis, is

S 16

= 2S 11 ! 2S

12 ! S

66( )c3s ! 2S

22 ! 2S

12 ! S

66( )cs3

in which c = cos- and s = sin-.

(b) Using the following measured values of elastic constants of the composite E

1 = 50 GPa, E

2 = 5 GPa, !

12 = 0.3, G

12 = 10 GPa

calculate the shear strain induced in the lamina when a normal tensile stress of 100 MPa isapplied at an angle of 30˚ to the fibre axis.

(c) The dependence of this interaction compliance on - is shown below for a differentcomposite. Sketch the corresponding plot for a 0/90 crossply laminate of the same material,obtained by assuming that the laminate compliance, at any given -, can be taken as theaverage of those for the constituent plies at their corresponding - values.

{from 2009 Tripos}

[The questions below involve use of the DoITPoMS TLP “Mechanics of Fibre Composites”]

2. On the page “Stiffness of Laminates”, use the facility at the end to create an epoxy-50% glasscomposite (dragging the materials icons concerned to the matrix and reinforcement boxes) and

to estimate the ratio of maximum to minimum Young’s modulus it exhibits when loaded atdifferent angles to the fibre axis. Now create a 0/90 (cross-ply) laminate of the samecomposite and repeat the operation. Find a sequence giving complete in-plane isotropy and

confirm that the Young’s modulus in this case is about 22 GPa for all in-plane directions.

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3. On the page “Failure of Laminates and the Tsai-Hill criterion”, use the facility at the end tocreate a polyester-50%glass angle-ply laminate (±40˚). Taking this to be a filament-wound

tube, with the plies at ±40˚ to the hoop direction, and a radius/wall thickness ratio of 20,subjected to internal pressure, P, estimate the value of P at which failure will occur, accordingto the Tsai-Hill criterion, given that $1* = 700 MPa, $2* = 20 MPa and "12* = 50 MPa. Usinganalytical equations, carry out the same calculation for one of the two plies (ignoring the

presence of the other). Account for the difference between this value and the one youobtained treating the laminate as a whole (using the numerical procedure in the TLP).

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Examples Class II

[Property data on C16H62 may be used if necessary.]

1. (a) For a small aircraft, a choice must be made between an Al alloy and a composite for thefuselage material. The fuselage will approximate to a cylinder of diameter of 2 m and will

experience internal pressures up to 0.6 atm (0.06 MPa) above that of the surroundingatmosphere, axial bending moments of up to 500 kN m and torques of up to 600 kN m. Thecomposite fuselage would be produced by filament-winding at ±45˚ to the hoop direction. Itmay be assumed that this is a strength-critical application, with the airframe stiffness expected

to be adequate in any event. Using the Tresca yield criterion (Al) and the Tsai-Hill failurecriterion (composite), and ignoring the issue of safety factors, estimate the minimum wallthickness in each case and hence deduce which material would allow the lighter fuselage.

(b) Comment on the main sources of error in your calculation and also on whether there mightbe a danger of any other type of failure.

[For the Al alloy, the yield stress in uniaxial tension = 250 MPa

For the composite, failure stresses for loading transverse and in shear relative to the fibreaxis are both 50 MPa: the possibility of failure by fracture of the fibres can be neglected.

Densities: Al = 2.70 Mg m-3, composite = 1.50 Mg m-3 The peak axial stress in a thin-walled cylinder subjected to a bending moment M is R M / I ,where R is the radius and I is the moment of inertia, which is given by " R3 t , where t is the

wall thickness.]{from 2006 Tripos}

2. (a) A thick metal sheet was held at 1000˚C in air for several hours, after which time an oxide

film had formed (on both sides), with a thickness of 100 µm. No significant stresses werecreated in metal or oxide during this process. During subsequent cooling, much of this oxide

spalled off from the substrate when the temperature reached 300˚C. Estimate the fractureenergy of the interface between the metal and the oxide, stating your assumptions.

(b) The above thermal treatment was repeated on a different sheet of the same metal, in theform of a relatively narrow strip of a thinner sheet and in a configuration such that only one

side of the strip was exposed to air. In this case, it was observed that spallation did not occur,even after cooling to ambient temperature (20˚C), and that the strip exhibited noticeablecurvature at this stage. Give a qualitative explanation of the fact that spallation occurred in the

first experiment (part (a)), but not in the second.

(c) In the curved strip obtained after the above experiment (part (b)), would the oxidized sidebe expected to be convex or concave? The residual thickness of the metal was found to be

1 mm. What magnitude of curvature would be expected? Is this significantly different fromthe value that would be obtained if the Stoney approximation were used?

[The curvature, ( , arising from imposition of a uniform misfit strain, %$ , between the twolayers in a bi-layer system (with thicknesses h and H ) is given by

! =

6 E d E

s h + H ( )h H "#

E d

2h

4+ 4 E

d E

sh

3 H + 6 E

d E

sh

2 H

2+ 4 E

d E

sh H

3+ E

s

2 H

4

where E d and E s are the corresponding Young’s moduli (and biaxial versions of these apply ifthe same misfit strain is also being created in the other in-plane direction).

Thermal expansivities: metal = 15 ) 10-6 K-1 oxide = 7 ) 10-6 K-1 Young’s moduli: metal = 100 GPa oxide = 200 GPaPoisson ratios: metal = 0.3 oxide = 0.2]

{from 2014 Tripos}

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3. (a) Steel sheet of thickness 1 mm is given a thin protective layer of vitreous enamel. Thiscoating is created by adding glassy powder to the surface and holding at around 700-800˚C,

causing the powder to fuse and form a layer of uniform thickness. The sheet is then furnacecooled, taking several hours to reach room temperature, such that the thermal misfit strain iscompletely relaxed by creep down to about 220˚C, after which cooling is elastic. Assuming

that the coating / substrate thickness ratio, h/ H , is sufficiently small for the Stoney equation to

be valid, estimate the elastic strain in the coating, stating your assumptions.

(b) The adhesion of the enamel to the steel is excellent, so the system is highly resistant to

debonding, but it’s found that, if the coated sheet is progressively bent in one plane (with thesteel undergoing plastic deformation), then through-thickness cracks appear in the enamellayer (on the convex side) when the local radius of curvature reaches 60 mm. Assuming that

such cracking starts when the tensile strain in the enamel reaches a certain level, use thisinformation to estimate this critical strain.

(c) A fabrication procedure requires bending of the coated sheet to a radius of curvature of

50 mm. The suggestion is made that, instead of furnace cooling the sheet after formation ofthe coating, it should be removed from the furnace and cooled more quickly, such that elastic

cooling occurs below about 420˚C (and stress relaxation is complete until this point). Wouldyou expect this measure to result in the elimination of through-thickness cracking duringbending of the sheet to this curvature?

(d) For the latter case (ie the rapidly cooled sheet), what are the principal stresses within the

coating, before and after the bending operation? (The deformation of the steel sheet can betaken as entirely plastic.)

[Property data:

Steel: , = 14 10-6 K-1 Enamel: , = 5 10-6 K-1; E = 70 GPa; & = 0.2

where , is the thermal expansivity, E is the Young’s modulus, and & is the Poisson ratio]{from 2010 Tripos}

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