C16H

Natural Sciences Tripos Part II

MATERIALS SCIENCE

II

MATERIALS SCIENCE

C16: Composite Materials

Prof. T. W. Clyne

Lent Term 2014 15

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

Lent Term 2014-15

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

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Course C16: Composite Materials

Synopsis (12 lectures)

Lecture 1 - Overview of Types of Composite System Overview 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 and Laminates.

Lecture 2 - Elastic Constants of Long Fibre Composites Recap of Axial and Transverse Young’s Moduli for an Aligned Long Fibre Composite, derived using the Slab Model. Errors for Transverse Loading and Use of Halpin-Tsai Equations. Derivation of Shear Moduli and Poisson Ratios. Number of Elastic Constants for Systems with different Degrees of Symmetry.

Lecture 3 - Elastic Loading of a Lamina Plane Stress Loading of a Uniaxial Lamina and the Kirchoff Assumptions. Off-axis loading of a Lamina. Elastic Constants as a Function of Loading Angle. Tensile-shear Interactions and Lamina Distortions.

Lecture 4 - Elastic Loading of a Laminate A Laminate considered as a Stack of Laminae. Elastic Properties of Laminates as a Function of Loading Angle. Elastic Constants of Some Simple Laminates. Balanced Laminates. Coupling Stresses and Symmetric Laminates.

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

Lecture 6 - Short Fibre & Particulate Composites – Stiffness & Inelastic Behaviour Load Partitioning and Stiffness Prediction for the Shear Lag Model. Fibre Aspect Ratios needed to approach 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 Interface Interfacial Bonding Mechanisms. Measurement of Bond Strength. Pull-out & Push-out Testing. Control of Bond Strength. Silane Coupling Agents. Interfacial Reactions and their Control during Processing.

Lecture 8 - Fracture Strength of Composites Axial Tensile Strength of Long Fibre Composites. Transverse and Shear Strength. Mixed Mode Failure and 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 Composites Energies 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.

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

Lecture 11 - Thermal Expansion of Composites and Thermal Residual Stresses Thermal 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 Systems Misfit Strains in Substrate-Coating Systems. Force and Moment Balances. Relationship between Residual Stress Distribution and System Curvature. Curvature Measurement to obtain Stresses in Coatings. Limitations of Stoney Equation. Sources of Misfit Strain. Driving Forces for Interfacial Debonding.


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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 classes

etc) can be viewed on the web and also downloaded. This includes model answers, which are released after the work concerned should have been completed. In addition to this text-based material, resources produced within the DoITPoMS project are also available. These include libraries of Micrographs and of Teaching and Learning Packages (TLPs). The following TLPs are directly 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 against density


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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 Constants of Long Fibre Composites

Use of the Slab Model

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


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Halpin-Tsai Expressions

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 the three Poisson ratios


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Stress, Strain, Stiffness & Compliance Tensors

Fig.2.5 Examples of how 2-D relative displacement components can represent different combinations of shear and rigid body rotation


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Lecture 3: Elastic Loading of a Lamina

Symmetry & Use of Matrix Notation for Matter Tensors

Matrix Notation

Effect of Material Symmetry on the Number of Independent Elastic Constants

Fig.3.1 Indication of the form of the Spq and Cpq matrices (matrix notation for Sijkl and Cijkl tensors), for materials exhibiting different types of symmetry. All of the matrices are symmetrical about the leading diagonal


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Off-axis Elastic Constants of Laminae Loading Parallel and Normal to Fibre Axis

Loading at Arbitrary Angles to Fibre Axis

Fig.3.2 (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 in the fibre-related framework of !ij (!1, !2 & "12)


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Derivation of Transformed Stress-Strain Relationship For a thin lamina, stresses and strains in the through-thickness (3) direction are neglected, so

that the 3, 4, and 5 components in matrix notation are of no concern. Therefore, when a lamina is loaded parallel or normal to the fibre axis, the strains that interest us are given by

!1!2" 12

!=! S !#1

# 2

$12

!=!S11 S12 0S12 S22 00 0 S66

!#1

# 2

$12

(3.1)

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

S11 !=!1E1!!!!!!!!!!S12 !=!!

"12E1!=!!"21

E2

S22 !=!1E2!!!!!!!!!!S66 !=!

1G12

!

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 equation expressing any second rank tensor with respect to a new coordinate frame

! ij !=!aikajl "! 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 defined correctly. 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, these direction cosines take the values

a11 =!cos! !(= c)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!a12 !=!cos 90 "!( )!=!sin! !(= s)a21 =!cos 90 + !( ) = "!sin! !(= "s)!!!!!a22 !=!cos! !(= c)

Carrying out this operation for all three stresses

!1

! 2

"12

!=! T !! x

! y

" xy

! (3.2)

where

T !=!c2 s2 2css2 c2 !2cs

!cs cs c2 ! s2( )

The same matrix can be used to transform tensorial strains, so that

!1!2!12

!=! T !! x! y! xy

!


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However, to use engineering strains ($xy = 2%xy etc), T must be modified (by halving the elements t13 and t23 and doubling elements t31 and t32 of the matrix T ), so as to give

!1!2" 12

!=! #T !! x! y" xy

! (3.3)

in which

!T !=!c2 s2 css2 c2 "cs

"2cs 2cs c2 " s2( )

The procedure is now a progression from the stress-strain relationship when the lamina is loaded along its fibre-related axes to a general one involving a transformed compliance matrix, S , which will depend on #. The first step is to write the inverse of Eqn.(3.3), giving the strains

relative to the loading direction (ie the information required), in terms of the strains relative to the fibre direction. This involves using the inverse of the matrix !T , written as !T "1

! x! y" xy

!=! #T $1 !!1!2" 12

!

in which

!T "1 !=!c2 s2 "css2 c2 cs

2cs "2cs c2 " s2( )

Now, the strains relative to the fibre direction can be expressed in terms of the stresses in those directions via the on-axis stress-strain relationship for the lamina, Eqn.(3.1), giving

! 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 terms of 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


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S11 !=!S11c4 + S22s

4 + 2S12 + S66( )c2s2S12 !=!S12 c

4 + s4( ) + S11 + S22 ! S66( )c2s2S22 !=!S11s

4 + S22c4 + 2S12 + S66( )c2s2

S16 !=! 2S11 ! 2S12 ! S66( )c3s!!! 2S22 ! 2S12 ! S66( )cs3S26 !=! 2S11 ! 2S12 ! S66( )cs3 !!! 2S22 ! 2S12 ! S66( )c3sS66 !=! 4S11 + 4S22 ! 8S12 ! 2S66( )c2s2 !+!S66 c4 + s4( )

(3.5)

It can be seen that S & S as # & 0.


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Effect of Loading Angle on Stiffness and Poisson Ratio

Fig.3.3 Variation with loading angle # of (a) Young’s modulus Ex and shear modulus Gxy 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.3.4 Variation with loading angle # of the tensile-shear interaction compliance S16, for a lamina of rubber-5% Al fibre, and photos of 4 specimens (between crossed polars) under axial tension, lined up at the appropriate values of #, showing tensile-shear distortions


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Lecture 4: Elastic Loading of a Laminate Obtaining the Elastic Constants of a Laminate

Fig.4.1 Schematic depiction of the loading angle % between the x-direction (stress axis) and the reference direction (#=0˚), for a laminate of n plies. Also shown is the angle #k between the reference direction and the fibre axis of the k th ply (1k direction)


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Stiffness of Laminates

Fig.4.2 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, made of epoxy-50% glass fibre. (The equal stress model was used to obtain the transverse Young’s modulus of the lamina, E2.)


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Tensile-Shear Interactions and Balanced Laminates

Fig.4.3 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 the lamina, E2.)


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In-plane Stresses within a Loaded Laminate

Fig.4.4 (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.4.5 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 fibre

occurs entirely via the shear forces acting on the cylindrical interface. This leads immediately to the basic shear lag equation:

d! f

dx!=!"2# i

r0 (5.1)

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

2! !r1 !"1 !dx!=!2! !r2 !" 2 !dx!!!!!!!!!ie!!"1" 2!=! r2

r1

#" = " i !r0r

$%&

'()

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

! != "Gm

!=" i !

r0r

#$%

&'(

Gm

!!!!!!!!and!! !=!dudr

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

duur0uR! !=!" i !r0

Gm

drrr0

R

!

!!" i !=!uR # ur0( )Gm

r0 !lnRr0

$%&

'()

(5.2)

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

Rr0

!"#

$%&

2

!=! '2! f ! 3

!(! 1f

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

d! f

dx!=!

"2 uR " ur0( )Gm

r02 !12ln 1

f#$%

&'(

!

The displacements uR and ur0 are not known, but their differentials are related to identifiable strains. The differential of ur0 is simply the axial strain in the fibre (assuming perfect interfacial adhesion and neglecting any shear strain in the fibre - which is taken as being much stiffer than the matrix)


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dur0dx!=!! f =

" f

Ef!

while the differential of uR, ie the far-field axial strain of the matrix, can be taken as the macroscopic strain of the composite

duRdx!!!"1 !

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

d2! f

dx2!=!n

2

r02 ! ! f !"!Ef !#1( ) (5.3)

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

n = 2Em

Ef 1+ !m( )!ln 1f

"#$

%&'

(5.4)

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

! f !=!Ef !"1 !+!Bsinhn!xr0

#

$%

&

'(!+!Dcosh

n!xr0

#

$%

&

'(

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 length of the fibre

! f !=!Ef !"1 ! 1# coshn!xr0

$

%&

'

()sech n!s( )*

+,

-,

./,

0, (5.5)

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

! i !=!Ef !n!"12

sinh n!xr0

#

$%

&

'(!sech n!s( ) (5.6)

An estimate can now be made of the axial modulus of the composite. This is done by using the 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 above expression for !f over the length of the fibre. This leads to

E1 !=!!1

"1=! f !Ef 1#

tanh n!s( )n!s

$%&

'()+ 1# f( )!Em (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 stress along the axis of a glass fibre in a polyester-30% glass composite subject to an axial tensile 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 field in 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) of SiC in an Al matrix, with an applied axial tensile stress of 100 MPa (and differential thermal 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’s modulus 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 is relatively large. For very short fibres, the predictions become inaccurate, due to neglect of 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 how the stress field and the axial extension change as the aspect ratio and degree of alignment 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 or matrix 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-based composites. 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% glass long 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 with adsorbed water to create a strong bond to the glass surface. The R group is one which can 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 fibre to expand (rather than contract), which augments (rather than offsets) the radial compressive 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 critical values 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

Maximum stress:

!1 =!1* !"!!# =!1*c2

!2 =!2* !"!!# =! 2*

s2

$12 = $12* !"!!# =$12*cs

Tsai-Hill:

!" =c2 c2 # s2( )

!1*2+ s4

! 2*2 + c

2s2

$12*2%

&''

(

)**

#1/2

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) maximum stress 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 tension and 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 fibres in 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 MPa and "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 stress criterion, 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 SiC laminate 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 fibre axis, showing interfacial debonding and fibre pull-out processes


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Energy of Fibre Pull-out

Gcpo !=!Ndx0L0

L

! "rx02# i* =f

"r2$%&

'()"r# i*L

$%&

'()L33

$

%&

'

() =

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) strength which 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

(a) (b)

Fig.9.5 Microstructure of a composite (“Fiberstone”) comprising coarse stainless steel fibres in a matrix which is predominantly alumina, illustrated by (a) an X-ray tomograph, showing the fibres only, and (b) an optical micrograph of a polished section

There have been many attempts to produce ceramic-matrix composites with high toughness, but with limited success. Probably the most promising approach is to introduce a network of metallic fibres, and this is the basis of a commercial product (“Fiberstone” – see Fig.9.5). The fibres are often about 0.5 mm diameter, although finer fibres can be used. During fracture, fibres bridge the crack and energy is absorbed by both frictional pull-out and plastic deformation - see Fig.9.6. These mechanisms are likely to dominate any other contributions to the work of fracture.

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


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The work of fracture can thus be estimated by summing the energy absorbed via both processes, assuming that a fraction g of the fibres bridging the crack plane undergo pull-out and the remainder (1-g) undergo plastic deformation and rupture.

Gcnet = Gcpo +Gcfd (9.2)

An expression for the fibre pull-out work was derived previously (Eqn.(9.1)), but the relationship between N and f depends on fibre orientation distribution, and that treatment referred to aligned fibres. For this type of composite, it can be taken as isotropic (random), in which case N is half that for the aligned case (see EE Underwood, Quantitative Stereology. 1970, Addison-Wesley)

N =f

2!r2 (9.3)

leading to

Gcpo !=!gfs2r! i*6

(9.4)

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

Fig.9.7 Data from tensile testing of single 304 stainless steel fibres, showing (a) a set of 10 stress-strain curves and (b) the distribution of corresponding work of deformation values

The work done during plastic deformation and rupture of fibres can be estimated by assuming that interfacial debonding extends a distance x0 from the crack plane - see Fig.9.6(c). The energy is obtained by summing the work done on each fibre, as if it had an original length 2x0 and were being subjected to a simple tensile test

Gcfd = (1! g)2x0NUfd = (1! g)2x0f

2"r2#$%

&'(Wfd"r

2 = (1! g)x0 fWfd (9.5)

where Ufd and Wfd 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 given by the area under the stress-strain curve of the fibre. Some such curves, for the fibres used in “Fiberstone”, are shown in Fig.9.7, together with corresponding Wfd values. The value of ( is in this case given by the product of x0 and %*, the fibre strain to failure, leading to

Gcfd = (1! g)"#*

$%&

'()fWfd =

(1! g)srfWfd

#* (9.6)


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Use of Eqns.(9.2), (9.4) and (9.6) allows prediction of the composite fracture energy, although it requires measurements or assumptions to be made concerning several parameters. In addition to the single fibre work of deformation, Wfd, and the failure strain, %*, estimates are required for the proportion of fibres undergoing pull-out, g, the interfacial frictional sliding stress, "i*, and the (average) length of fibre extending beyond the crack plane, (, and hence the “protrusion” aspect ratio, s (= (/r) Nevertheless, predictions can be made, based on experimental data or on plausible assumptions, and compared with measured composite fracture energies. An example is shown in Fig.9.8, where it can be seen that, even with the relatively low fibre content (~10-15%) that is normally present, the work of fracture is both predicted and observed to be substantial. The experimental Gc values were obtained by impact (Izod) testing.

Fig.9.8 Comparison between experimental data for the fracture energy of “Fibrestone” composites, as a function of fibre volume fraction, and predictions obtained using Eqns.(9.4) amd (9.6), for fine and coarse fibres

The value of s can be estimated from observation of fracture surfaces. However, it’s difficult to be sure whether particular fibres have predominantly undergone pull-out, rather than plastic deformation and rupture - of course, some fibres could deform plastically and then pull out. In any event, very strong bonding may be undesirable, since this will tend to inhibit both pull-out and plasticity, although very poor bonding may allow fracture to take place without the fibres being significantly involved in the process. An intermediate bond strength is likely to give optimal toughness.

It also worth noting that, for a given fibre protrusion aspect ratio, s (= (/r), both pull-out and plastic deformation contributions increase linearly with the absolute scale (fibre diameter). Composites reinforced 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!


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Lecture 10: Compressive Loading of Fibre Composites

Euler Buckling


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Kink Band Formation

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

Fig.10.2 Predicted kinking stress, 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.10.3 (a) Fragment of SiC monofilament extracted from a Ti-35%SiC composite after loading under axial compression and (b) schematic of the crushing process

Fig.10.4 Stresses in Ti-35%SiC monofilament composite (average axial values for fibre, matrix and composite) as axial strain is increased by external loading. At zero strain, stresses in fibre and matrix are from differential thermal contraction. The matrix yields when the stress in it reaches !mY. It is assumed that no matrix work hardening occurs during plastic straining. Failure occurs when the fibre stress reaches the critical value !f*

Failure is expected when the fibre stress reaches !f*, taken to be a single, fixed value. The composite strength !c* can readily be predicted, provided it can be assumed that the matrix yields before composite failure and that matrix work hardening is negligible, since it is then given by

! c* = E1c"cmY + E1c' "c* # "cmY( ) (10.1)

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


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E1c = fEf + 1! f( )EmE1c' = fEf

Now, the strains at matrix yield and at final failure can be written as

!cmY ="mY +"m#T

Em

!c* =" f* +" f#T

Ef

Substituting into Eqn.(10.1), and applying the residual stress force balance condition

f! f"T + 1# f( )!m"T = 0

then leads to

! c* = f! f* + 1" f( )!mY

A correction should be applied for the effect of misalignment in reducing the stress parallel to the fibre axis, leading to

! c* =f! f* + 1" f( )!mY

cos2#0 (10.2)

This predicted strength is independent of the thermal residual stresses (whereas the strain at which failure occurs will depend on them).

Fig.10.5 Experimental strength data, as a function of the initial angle between fibre and loading axes, during compression of misaligned Ti-35%SiC specimens. Also shown are predicted curves for failure by kink band formation and by fibre crushing, obtained by substitution of the values shown into the kinking equation and Eqn.(10.2) respectively


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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, showing lattice strains (& hence stresses) within matrix & reinforcement during unloaded thermal cycling. (111) reflections were used for both constituents. The gradients shown are calculated values for elastic behaviour, assuming a fibre aspect ratio of 10


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Lecture 12: Surface Coatings as Composite Systems

Force and Moment Balances

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 result of the imposition of a uniform misfit strain, )(. The strain and stress distributions shown are for the case indicated, obtained using Eqns.(12.10) & (12.11)


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Relation between Curvature and Misfit Strain The forces P and –P generate an unbalanced moment, given by

M !=!P! h + H2

!"#

$%& (12.1)

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"

(12.2)

P can be expressed as

P =2!!!"h + H

(12.3)

The beam stiffness is given by

! = b E yc( )"H "#

h"#

$ !yc2 !dyc != b!Ed !h

h2

3" h# + # 2

%&'

()*+ b!Es !H

H 2

3+ H# + # 2

%&'

()*

(12.4)

where +, the distance from the neutral axis (yc = 0) to the interface (y = 0) is given (see Appendix on p.64) by

! =h2Ed " H

2Es2 hEd + HEs( ) (12.5)

The magnitude of P is found by expressing the misfit strain as the difference between the strains resulting from application of the P forces.

!" = "s # "d =P

HbEs+

PhbEd

!Pb= "# hEdHEs

hEd + HEs

$%&

'()

(12.6)

Combination of this with Eqs.(12.3)-(12.5) gives a general expression for the curvature, *, arising from imposition of a uniform misfit strain, +(

! =6EdEs h + H( )h!H !"#

Ed2h4 + 4EdEsh

3H + 6EdEsh2H 2 + 4EdEsh!H

3 + Es2H 4 (12.7)

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.


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Biaxial Stresses In practice, there are often in-plane stresses other than those in the x-direction. For an isotropic

in-plane stress state, there is effectively another stress equal to !x in a direction at right angles to it (z-direction); this induces a Poisson strain in the x-direction. Assuming isotropic stiffness and negligible through-thickness stress (!y = 0), the net strain in the x-direction can be written

! xE = " x #$ " y +" z( ) = " x 1#$( )

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

! x

" x=

E1#$( ) = E ' (12.8)

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.

Stoney’s Equation – the Thin Coating Limit A simplified form of Eq.(12.7) applies for coatings much thinner than the substrate (h << H).

The substrate stress becomes negligible and that in the deposit varies little as a result of curvature adoption, so the misfit strain can be converted to a deposit stress. For an equal biaxial case,

! d =Ed1"#d( )$%

Substituting in Eqn.(12.7) for +% then leads to

! =6Ed

' hEs'H 2

" d 1#$d( )Ed

!" =6h 1#$s( )EsH

2 % d (12.9)

This is Stoney’s equation, which is commonly used to relate (biaxial) stress to (biaxial) curvature for thin coatings. The properties required (Es and 's) are only those of the substrate. This is convenient, since these are usually more readily accessible than those of the coating. Unfortunately, the Stoney equation is only accurate in a regime (h<<H) where curvatures are often very small (and hence difficult to measure) - see below.

Stress Distributions in Thick Coating Systems When the condition h << H does not apply, then stresses and stress gradients are often

significant in both constituents. Stress distributions are readily found for the simple misfit strain case outlined above, from the values of P and *, using the expressions

! d y=h="Pb!h

+ Ed !# h " $( ) (12.10a)

! d y=0="Pb!h

!"!Ed !# !$ (12.10b)

! s y="H=

Pb!H

!"!Es !# ! H + $( ) (12.11a)

! s y=0=

Pb!H

!"!Es !# !$ (12.11b)


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The stress distributions in Fig.12.1, 12.3 and 12.4 were obtained using these equations. The adoption of curvature can effect substantial changes in stress levels and high through-thickness gradients can result. It may be seen from Eqns.(12.10) and (12.11) that (for a given value of h/H), since P is proportional to H and * is inversely proportional to H, the stresses (at y=-H, 0 and h) do not depend on H, ie the stress distribution is independent of scale. However, the curvature is not. Substrates must be fairly thin if measurable curvatures are to be generated, although the maximum thickness could be as small as 50 µm, or as large as 50 mm, depending on various factors.

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

Curvature Measurement Techniques


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Accuracy of the Stoney Equation

Fig.12.3 Predicted dependence on thickness ratio of (a) curvature and (b) stress in deposit (coating), obtained using Eqns.(12.7), (12.10) and (12.11), and the Stoney equation (Eqn.(12.9).) The Poisson ratios of substrate and deposit were both taken as 0.2

Possible Sources of a Misfit Strain


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Driving Force for Interfacial Debonding

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


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Appendix – 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

# !dy!= 0 (12.,.1)

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

b Ed!(y)0

h

" !dy!+!b Es!(y)#H

0

" !dy!= 0 (12.,.2)

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

b Ed! y " #( )0

h

$ !dy!+!b Es! y " #( )"H

0

$ !dy!= 0 (12.,.3)

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

Edy2

2! "y

#

$%

&

'(0

h

+!Esy2

2! "y

#

$%

&

'(!H

0

!= 0

)Edh2

2! "h

*+,

-./+ Es

!H 2

2! "H

*+,

-./= 0

)" Edh + EsH( ) = 12Edh

2 ! EsH2( )

)" =h2Ed ! H

2Es2 hEd + HEs( )


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Property Data (Room Temperature)

Fibres

Fibre Density

(Mg m-3) Axial

Modulus E1 (GPa)

Transverse Modulus E2 (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

Matrix Density

(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

Polyurethane Rubber 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

Borosilicate glass 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 C16H65 can be used if necessary.]

1. Show that the Young's modulus of a composite lamina (having the elastic constants, referred to 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 can exhibit and at what loading angle does this occur ? [E1 = 200 GPa, E2 = 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 S16 of a lamina. State how this is used in describing the elastic deformation of the lamina under an applied 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 shear strains under such a load. [E1 = 40 GPa, E2 = 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 an angle-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). The critical 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 stress criterion 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 hoop direction) is predicted to fail.

Fig.1 Stresses (ratios to !x) within one ply of an angle-ply laminate, as a function of the ply

angle, 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 that would 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 increasing tensile 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 Fibre Composites” TLP (www.doitpoms.ac.uk/tlplib/fibre_composites/index.php), to establish the applied 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 polymer composite. The diameter of the pipeline will be 1 m and the maximum gas pressure will be 100 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 which could lead to failure (which would be likely to be plasticity in the case of the steel and some type 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 wall thickness 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 there might 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( )22

# !Y

where !1, !2 and !3 are the principal stresses and !Y is the uniaxial yield stress. The latter has 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 be 900 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 2css2 c2 "2cs"cs cs c2 " s2

! 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 C16H65 can be used if necessary.]

1. A strut is in the form of a hollow cylinder with an outside diameter of 25 mm and a bore of 20 mm. It is manufactured from MMC material composed of 70 vol.% SiC monofilaments in a titanium alloy matrix, with the SiC fibres aligned approximately parallel to the axis of the strut. However, the limitations of the manufacturing process are such that fibre misalignments of up to 4˚ are present. The crushing strength of the SiC monofilaments is known to be about 8 GPa and the yield stress of the Ti alloy is 600 MPa, while the critical shear stress of the composite, on planes parallel to the fibre axis, is measured to be about 200 MPa.

(i) Estimate the shear modulus of the composite and hence the stress for failure by kink band formation. Would failure of this type occur under an axial compressive load of 25 kN?

(ii) Would any other type of failure or deformation be expected under this applied load? [Shear moduli: Ti alloy; G = 44 GPa: SiC monofilament; G = 170 GPa Young’s moduli: Ti alloy; E = 115 GPa: SiC monofilament; E = 400 GPa] {from 2006 Tripos}

2. (a) A 1 mm thick unidirectional ply of epoxy-25vol% glass fibre composite is bonded at

120˚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 is observed 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 is likely 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, ' = 0.40 for 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( )Em +! f fEf

1" f( )Em + fEf

! 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 a misfit strain +% between their natural (stress-free) lengths, is given by

! =12 "#

hE1

E2

+ 14 +E2

E1

$

% & &

'

( ) )

]

{from 2008 Tripos}


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3. (a) During formation of a coating on a substrate, it’s common for a misfit strain, +%, to be created, representing the difference between the (stress-free) in-plane dimensions of the two constituents. For example, this often arises during deposition and/or subsequent cooling. This misfit creates stresses and stains in the coating (and possibly in the substrate). Show that the relation between the stress and strain in the coating, in any (in-plane) direction, can be expressed !"

=E1#$( )

where E is the Young’s modulus and ' is the Poisson ratio. (b) The curvature, *, arising from a misfit strain, +%, between a coating (deposit) of thickness

h and a substrate of thickness H is given by

! =6EdEs h + H( )h!H !"#

Ed2h4 + 4EdEsh


3 + Es2H 4

where Ed and Es are the Young’s moduli of deposit and substrate. Show that, in the limit of h<<H, this reduces to the Stoney equation, giving the curvature in terms of the stress in the deposit, its Poisson ratio, the Young’s modulus of the substrate and the thicknesses of the two constituents.

(c) A glass sheet of thickness 3 mm has a 10 µm layer of Al evaporated onto one side, to form a mirror. The production process generates negligible stress in the coating. The sheet is subsequently heated from room temperature (20˚C) to 170˚C. Calculate the curvature exhibited by the sheet after heating, assuming that the system remained elastic.

(d) Decide, stating any assumptions, whether yielding is in fact likely to occur in the Al layer during heating, given that it has a uniaxial yield stress at 170˚C of 100 MPa.

(e) Hence give an opinion as to whether any detectable distortion of the reflective characteristics of the mirror is likely to be present after it has cooled to room temperature.

[For the glass: E = 75 GPa, - = 8.5 / 10-6 K-1, for the Al: E = 70 GPa, ' = 0.33, - = 24.0 / 10-6 K-1] {from 2009 Tripos}

4. (a) Show that the curvature, *, of a beam (reciprocal of the radius of curvature, R) is equal to

the through-thickness gradient of the strain, with the strain being zero at the neutral axis. {15%}

(b) A “vibration-damped” sheet material is made by bonding a 1 mm thick rubber layer between two steel plates of thickness 1 mm. The sheet is pushed against the surface of a large cylindrical 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.

{25%}

(c) This forming operation is actually designed to generate plastic deformation, creating a shaped 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 both metal sheets and calculate the thickness of the layers that would yield in this way and the plastic strain at the free surfaces.

{20%}


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(d) Show that, if the width of the sheet (length along the axis of the cylinder) is 0.5 m, then the beam stiffness () = EI) of the sheet is 216.7 N m2 and the bending moment that would be needed 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 actual case, with the steel undergoing plastic deformation at a yield stress of 300 MPa (but neglecting any work hardening).

{40%}

[Steel: Young’s modulus, E = 200 GPa; Rubber: Young’s modulus value more than 4 orders of magnitude lower]

{from 2011 Tripos} 5. John Harrison, the famous clock-maker credited with developing a time-keeping system

sufficiently reliable to establish longitude at sea, was reportedly the first to create a bi-metallic strip (for compensation of the effects of temperature change), which he did by casting a thin brass layer onto a thin steel sheet. Show that, if both layers have a thickness of 0.1 mm, and the strip is 100 mm long, then the temperature change required to generate a lateral deflection of 1 mm at its end is about 4.6 K, assuming that the system remains elastic.

Sketch the (approximate) through-thickness distributions of stress and strain within the above strip, after it had been heated by 100 K. Give your view as to whether such heating would be likely to cause plastic deformation within the strip, given that the yield stresses of both constituents are expected to be of the order of 100 MPa.

[The curvature, *, of a bi-material strip comprising two constituents of equal thickness (h), arising from a misfit strain of +( between them, is given by

! =12 "#

h E1

E2

+14 + E2

E1

$%&

'()

where E1 and E2 are the Young’s moduli of the constituents. The relationship between curvature, *, end deflection, y, and length, x, of a bi-material strip may be expressed as

! =2sin tan"1 y

x( )#$%

&'(

x2 + y2( )

For steel: Young’s modulus, E = 200 GPa; thermal expansivity, , = 13 / 10-6 K-1

For brass: Young’s modulus, E = 100 GPa; thermal expansivity, , = 19 / 10-6 K-1] {from 2012 Tripos}


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Examples Class I [Property data on C16H65 can 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 !!=!S11 S12 0S12 S22 00 0 S66

!=1 / E1 !"12 / E1 0

!"21 / E2 1 / E2 00 0 1 /G12

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

S16 !=! 2S11 ! 2S12 ! S66( )c3s!!! 2S22 ! 2S12 ! S66( )cs3

in which c = cos# and s = sin#. (b) Using the following measured values of elastic constants of the composite

E1 = 50!GPa,!! !E2 = 5!GPa,!!!12 = 0.3,!!G12 = 10!GPa

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

(c) The dependence of this interaction compliance on # is shown below for a different composite. 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 the average 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% glass composite (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 at different angles to the fibre axis. Now create a 0/90 (cross-ply) laminate of the same composite 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.


TWC - Lent 2015

3. On the page “Failure of Laminates and the Tsai-Hill criterion”, use the facility at the end to create 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, according to the Tsai-Hill criterion, given that !1* = 700 MPa, !2* = 20 MPa and "12* = 50 MPa. Using analytical 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 you obtained treating the laminate as a whole (using the numerical procedure in the TLP).


TWC - Lent 2015

Examples Class II [Property data on C16H65 can be used if necessary.]

1. (a) For a small aircraft, a choice must be made between an Al alloy and a composite for the fuselage 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 surrounding atmosphere, axial bending moments of up to 500 kN m and torques of up to 600 kN m. The composite fuselage would be produced by filament-winding at ±45˚ to the hoop direction. It may 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 failure criterion (composite), and ignoring the issue of safety factors, estimate the minimum wall thickness 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 might be 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 fibre axis 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 were created 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 fracture energy 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 the form 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 noticeable curvature 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 side be 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 from the value that would be obtained if the Stoney approximation were used?

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

! =6EdEs h + H( )h!H !"#

Ed2h4 + 4EdEsh


3 + Es2H 4

where Ed and Es are the corresponding Young’s moduli (and biaxial versions of these apply if the 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 GPa Poisson ratios: metal = 0.3 oxide = 0.2]

{from 2014 Tripos}


TWC - Lent 2015

3. (a) Steel sheet of thickness 1 mm is given a thin protective layer of vitreous enamel. This coating 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 furnace cooled, taking several hours to reach room temperature, such that the thermal misfit strain is completely 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 the steel undergoing plastic deformation), then through-thickness cracks appear in the enamel layer (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 this information 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 of the 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). Would you expect this measure to result in the elimination of through-thickness cracking during bending 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 be taken 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}

C16H

Documents

aligned fibre composites

fibre fracture

fibre pullout

possibility of fibre

elastic loading

fibre aspect ratios

aligned long fibre composite

failure of laminates