Analysis of Discontinuous Fiber Reinforced Lamina

• Fiber length effects not previously considered. 3 types to be considered:

1. Aligned discontinuous fibers2. Off-axis aligned discontinuous fibers3. Randomly oriented discontinuous fibers

Analysis begins with simplest case of aligned discontinuous fibers.

Types of Discontinuous Fiber Reinforcement

Representative Volume Elements for Aligned Discontinuous Fiber Composite

L

d D

Fiber

Fiber

Matrix

Matrix

(a) Matrix material included at ends of fiber

(b) Matrix material not included at ends of fiber

Three Dimensional and Two Dimensional Random Orientations of Fibers

(a) Fiber length is less than thickness of part. So fibers are randomly oriented in three dimensions

(b) Fiber length is greater than thickness of part. So fiber are randomly oriented in only two dimensions.

L << tL >> t

Schematic Representation of Matrix Shear Deformation in a Short Fiber Composite

Fiber

Fiber

Before Deformation

After Deformation

Reinforcement of Matrix by Fiber

Fiber

Fiber

Before Loading

After Loading

Fiber stiffness >> Matrix Stiffness

Matrix

Grid Lines

Stresses acting on a differential element of fiber

L

d D

dxx

ff d f

Consider free body diagram in fig. 6.4:

044

22

dxddddF fffx (6.1)

Simplifying and rearranging:

ddxd f 4

(6.2)

Separating variables and integrating

x

f dxd

df

0

4

0

(6.3)

Assume that stress transferred across the end of fiber (o = f @ x = 0) is negligible:

Thus, in order to determine σf, we need to know τ as a function of x.

x

f dxd 0

4 (6.4)

Gm y

Shear strain, Shear strain,

Shea

r stre

ss,

Shea

r stre

ss,

(a) Kelly – Tyson Model (b) Cox Model

Assumed stress – strain curves for matrix material in Kelly – Tyson model and Cox model

Two approaches to estimating (x):

1. Kelly-Tyson Model – assume matrix is rigid – plastic (Fig. 6.5 (a))

2. Cox Model – assume matrix is linear elastic (Fig. 6.5(b))

For illustrative purposes, consider Kelly-Tyson Model.

Matrix shear stress (x) = y = yield stress constant

dx

dxd

yx

yf

44

0

(6.5)

And f is linear function of x.

But f has to be symmetric about x = L/2 (Fig. 6.6)

dLL

dy

yf

22

4max (6.6)

fmax cannot increase indefinitely as L is increased, however, and one limiting value of fmax

is the stress f in a continuous fiber composite.

mfm

c1

Variation of interfacial shear stress, τ , and fiber normal stress, σf, with distance along the

fiber according to Kelly – Tyson model.

y

y

x

L

2L

f

fmax

x

Predicted shear stress distributions along fiber from finite element analysis and Cox model.

From Hwang, 1985.

For continuous fiber composite11 cf

1

1

1

1

c

c

f

f

EE

11

11 c

c

ff E

E

As L Li (Ineffective length)

11

1max c

c

ff E

E

Effect of fiber length on stress distributing along fiber according to Kelly – Tyson model

Variation of interfacial shear stress, , and fiber normal stress, f, with distance along the

fiber according to Cox model.

L

f

x

x

Substituting L = Li ; 11

1max c

ff E

E

In Eq. (6.6)

1

11

2 EdE

Ly

cfi

(6.7)

Li referred to as “ineffective length” because f < fmax over this portion of fiber. Li also referred to as “load transfer length” because interfacial shear load is transferred over this length.

Another limiting value of fmax is the fiber tensile

strength, sf1(+). In this case, the applied composite

stress is such that

From Eq. (6.6), the fiber length for this maximum stress is

the “critical length”

y

fc

dsL

2

)(1

)(fsmaxfE

cfE 11

11

(6.8)

(6.9)

Critical length has important implications for calculation of longitudinal composite strength

Recall rule of mixtures for longitudinal composite stress:

mvmfvfc 111 (3.19)

For discontinuous fibers of length L, the average fiber stress is

21

1

2

0

/L

dxff

/L

(6.10)

For fibers of length iLL

21 /iL

xmaxff

(6.11)

and the average fiber stress is

iL

Lmaxf/L

dx/iL

xmaxf

f

/L

222

1

2

0

(6.12)

For fibers of length iLL , the average fiber stress is

maxfLiL

/L

dxmaxfdx/L

xmaxf

f

/iL /L

/iL

21

22

1

2

0

2

2 (6.13)

)(fmaxf s 1

For fiber failure, substituting the failure conditions)(

1 Lc s 1mfm s

and Li = Lc along with Eq. (6.12) in Eq. (3.19)yields the composite longitudinal strength

fvmfsfvcL

LfsLs 112

1for (6.14)cLL

Similarly, substituting the failure conditions

)(fmaxf s 1 )(

1 Lc s 1mfm s

and Li = Lc along with Eq. (6.13) in Eq. (3.19)yields the composite longitudinal strength

fvmfsfvfsLcL

Ls 11121 for cLL (6.15)

mv

)(mfSfv

)(fS

LS

11

For L>>Lc, Eq. (6.15) Eq. (4.21) for continuous fibers

(4.21)

Interfacial shear strength corresponding to critical length found by rearranging Eq. (6.9) as:

c

fy L

ds2

)(1

(6.16)

Thus, by measuring Lc, d and sf1(+), we can determine

interfacial shear strength. Drzal, et al (MSU).

Lc

Single fiber specimen

Note: Measured values of Lc statistically distributed

Longitudinal Modulus of Aligned Discontinuous fiber Composite

Cox Model: Recall Eq. (6.2) :

Rate of change of axial load in fiber is linear function of . Cox assumed proportional to and thatdx

dP

ddx

d f 4

vuHdxdP

(6.17)

where u = axial displacement at a point in fiber v = axial displacement the matrix would have at

same point with no fiber present H = proportionality constant

(i.e., interfacial shear stress is proportional to mismatch between stiffness of fiber and matrix) differentiating Eq. (6.17):

e

EAPH

dxdv

dxduH

dxPd

ff 12

2

(6.18)

where1ff EA

Pdxdu

(From Mechanics of Materials)

edxdv

Rearranging in standard form

HePdx

Pd 2

2

2

where1

2

ff EAH

(6.12)

Solution hp PPP

Where Pp = particular solution = AfEf1e Ph = homogeneous solution

xSxR coshsinh

Resulting fiber stress is

2cosh

2cosh

11 L

xL

eEAP

ff

f

(6.20)

(6.21)

(R and S determined from B.C.s P=0 at x=0 & x=L)

Average fiber stress is

2

2tanh

1

2

1

2

0

L

L

eEL

dx

f

L

f

f

(6.22)

Recall “rule of mixtures” for stress under longitudinal loading,

mmffc vv 11 (6.23)

Substituting Eq. (6.22) in (6.23) and assuming ecmf

Longitudinal modulus is then

mmffc vEvL

LEE

22tanh111

(6.24)

Longitudinal Modulus Vs. Fiber Length

mmff vEvE 1 (3.23)

mmvE

mmffc vEv

LLEE

22tanh111

(6.24)

mmffc vEvEELas 11,

mmc vEELas 1,0

L

Ec1

Variation of modulus ratio Ec1/Em with fiber aspect ratio, L/d, for several composites.

From Gibson, et al., 1982

Aligned Discontinuous Fiber Composite Test Specimens

L

L

L

L

L

Unidirectional laminate made from prepreq tape cut with a knife

Comparison of measured and predicted (Cox model) longitudinal moduli of aligned discontinuous fiber graphite/epoxy for various

fiber aspect ratios. (L/d)eff = L/d. From Suarez, et al., 1986

Comparison of measured and predicted (Cox model corrected for fiber aspect ratio) longitudinal moduli of aligned discontinuous fiber graphite/epoxy for various fiber aspect ratios. (L/d)eff =

0.03L/d. From Suarez, et al., 1986

Correction of Cox model to obtain better agreement with experimental data

• “Effective fiber aspect ratio”

• Adding matrix material at ends of fiber (Fig. 6.14) “Modified Cox Model”

dLZ

dL

eff

(6.27)

mCm

m

C

C

MC EeLe

EeLL

Ev

Ev

E

11

1

1

1(6.28)

E’L/E’m Vs. fiber aspect ratio for boron/epoxy without curve fitting [E’f = 58 x 106 psi (399.62 GPa), Z = 1, f = 51.62 Hz].

E’L/E’m Vs. fiber aspect ratio for boron/epoxy with curve fitting [E’f = 55.43 x 106 psi (381.913 GPa), Z = 0.4, f = 51.62 Hz].

Finite Element Analysis

LRVE:

matrixfiber

Quarter domain: FEM mesh

c

Effective Modulus:c

ccE

where = average applied stress = average strain = /L = average displacement

cc

Finite Element Analysis - continued

Modified Cox model which includes matrix material at ends of fiber. From Hwang and Gibson, 1987.

Le e

MatrixMatrixMatrix

Fiber Fiber Fiber d

Modified Cox Model

L

Matrix Composite Matrix2e

2e

Comparison of predictions from modified Cox model and finite element analysis with experimental data for

boron/epoxy aligned discontinuous fiber composite at different fiber aspect ratios. From Hwang and Gibson, 1987.

Halpin – Tsai Equations

f

f

m vv

EE

1

11 (6.29)

where

mf

mf

EEEE

/1/

1

1(6.30)

Curve – fitting parameter

dL2

Assume E2, G12, υ12 not significantly affected by fiber length. So, use continuous fiber models for E2, G12, υ12.

Dependence of longitudinal modulus on fiber aspect ratio for aligned discontinuous fiber nylon/rubber composite.

Predictions from Halpin – Tsai equations are compared with experimental results. From Halpin, 1969.

2-D Strength Analysis Off-axis Aligned Discontinuous Fibers -

y2

x

1

x

Recall stress transformations: 2

1 cosx

22 sinx

sincosx12

(4.3)

Substitution of stresses from Eqs. (4.3) in Tsai-Hill Criterion Eq. (4.14)

12

212

2

22

221

2

21

LTTLL SSSS

(4.14)

leads to the equation for the off-axis strength

21

2

422

21

21

2

4

Ts

sincossin

LsLTsLs

cosx

(6.31)

Evaluating Eq. (6.31) for off-axis strengthof discontinous fiber composites:

Assume that only the longitudinal strength, SL, depends on fiber length, and use either micromechanics equation (6.14) or (6.15) forSL

(+), depending on whether the fiber length isless than or greater than the critical length. For other strengths ST and SLT, use micromechanicsequations for continuous fiber composites fromChapter 4.

2-D Off-axis Modulus Analysis

4

2

22

121

124

1

11211

sE

scGE

cE

Ex

(2.39)

And similar equations for Gxy, vxy, Ey. where E1 = Ec1 = longitudinal modulus corrected for

fiber length But E2, G12, v12 assumed to be independent of fiber length

,,,, 1212211 GEEfE cx ,,,, 1212212 GEEfE cy ,,,, 1212213 GEEfG cxy ,,,, 1212214 GEEfv cxy

(6.32)

Comparison of predicted and measured off-axis modulus ratio, Ex/Em for graphite/epoxy. From Suarez, et al., 1986.

Tridimensional plot of Ex/Em as a function of fiber aspect ratio and fiber orientation for graphite/epoxy. From Suarez, et al., 1986

Randomly Oriented Discontinuous Fibers

2-D Strength Analysis

Example: Use Maximum Stress Criterion to find off-axis strength, x, then average over all angles.

1 2

1 20

2

2

)(

2

)(

sincossincos2~

dSdSdS TLTLx

(6.34)

0~dx

x(6.33)

or

where,

for ,0 1

2

)(

cos

Lx

S

(Longitudinal tensile failure)for ,21

cossinLT

xS

(Interfacial shear failure)

for ,22

(Transverse tensile failure)

2

)(

sin

Tx

S

where, ,cot)(

1LT

L

SS

LT

T

SS )(

2tan

Randomly Oriented Discontinuous Fibers

2-D Modulus Analysis

• Analytical approach: Averaging elastic constants over all possible orientations by integration.Cox (1952) – analyzed a planar mat of randomly oriented continuous fibers without matrix material. Equations (6.37), (6.38) found, but not acceptable for design use.Nielsen and Chen (1968) – use same averaging concept, along with micromechanical equations for E1, E2, G12, υ12 and transformation equations for planar isotropic system.

Nielsen and Chen 2-D analysis of isotropic Young’s modulus

0

0~

d

dEE

x

(6.39)

Where Ex is given by Eq. (2.39)

4

2

22

121

124

1

11211

sE

scGE

cE

Ex

And E1, E2, G12, υ12 are found from micromechanics equations. Note: fibers assumed to be continuous

Dependence of modulus ratio on fiber volume fraction for several values of Ef/Em from Nielsen – Chen model. From

Nielsen and Chen. 1968

1~ EEx

Use of Invariants

Integration of Ex() in Eq. (6.39) is cumbersome, and the use of invariants simplifies the integration process.

Example : averaged value 11~Q

1

0321

0

011

11

4cos2cos~ U

dUUU

d

dQQ

(6.40)

Similarly,

111

0321

22~

4cos2cos~ QU

dUUUQ

42112~~ UQQ

0~~2616 QQ

2~ 41

66UUQ

The stress – strain relationships for the planar isotropic lamina are then,

xy

y

x

xy

y

x

UUUUUU

2)(0000

41

14

41

(6.41)

or

xy

y

x

xy

y

x

EEEEE

~12~000)~1(~)~1(~~0)~1(~~)~1(~

22

22

Where and are the engineering

constants for the planar isotropic lamina ~12

~~

EG,~E ~

Solving simultaneously for and (Tsai and Pagano, 1968)

G~,~E ~

Using the definitions of the invariants U1 and U4 in terms of E1, E2, G12, and υ12, Tsai and Pagano developed a set of approximate expressions:

1

4141~U

UUUUE

2~ 41 UUG

1

4~UU

(6.43)

Tsai – Pagano approximations

21 85

83~ EEE

21 41

81~ EEG

(6.44)

Where E1 and E2 are estimated from micromechanics.One approach; use Halpin – Tsai equations for aligned discontinuous fibers (Eqs. (6.29, 6.30)), then use these values of E1 and E2 in Tsai-Pagano equations.

Comparison with exp. data for boron/epoxy in Fig. (6.25)

Manera, 1977, also showed good agreement with exp. data using Eq. (6.43)

21 85

83~ EEE Comparison of with experimental data

(From Primer on Composite Materials, by J. C. Halpin, 1984)

Comparison of with experimental data

for a glass/polyester chopped fiber composite. From M. Manera, J. Composite Materials, Vol. 11, April 1977, pp. 235 – 247

1

4141~U

UUUUE

14~ UUv Comparison of with experimental data for a glass/polyester

chopped fiber composite. From M. Manera, J. Composite Materials, Vol. 11, April 1977, pp. 235 – 247

2-D and 3-D randomly oriented fiber composites

(a) Fiber length is less than thickness of part. So fibers are randomly oriented in 3-D

(b) Fiber length is greater than thickness of part. So fiber are randomly oriented in only 2-D

L << tL >> t

Analysis for random orientation of fibers in 3-D

Christensen and Waals (1972) approach:Integrate over all possible fiber orientations in 3-D space - similar to Nielsen and Chen2-D analysis

Nanofibers and Nanotubes as Reinforcements in Composites

• Due to nanometer-sized dimensions, nanofibers and nanotubes have random 3-D orientation in composites

• Nanofibers and nanotubes are not straight, and waviness must be accounted for in analytical models

Nanofiber or nanotube length, L, is much less than thickness of part, t, so nanofibers or nanotubes are randomly oriented in 3-D space

L<<t

t

Scanning electron microscope image of vapor grown carbon nanofibers in a polypropylene matrix. From Tibbetts and McHugh, Journal of Materials Research, 1999

300 nm

Transmission electron microscope image of multi-walled carbon nanotubes in a polystyrene matrix. From Qian, et al., Applied Physics Letters, 2000

Carbon structures, including nanotube(From Smalley web site, 2002)

RVEs for micromechanical model for predicting Young’s modulus of nanotube-reinforced composites. From Anumandla and Gibson, Composites Part A, 2006

Waviness NTLAw Assumed shape

NTL

xsinAz 2

Anumandla-Gibson micromechanics model for modulus of nanotube-reinforced composite

• Young’s modulus of RVE1 assumed to be the same as that of an element of a locally orthotropic lamina containing wavy fibers, as described by Chan & Wang (1994)and Hsaio & Daniel (1996)

• Locally orthotropic compliances in RVE1 estimated using Simplified Micromechanics equations of Chamis (1987)

• Effective modulus of RVE1 with 3D randomly oriented nanotubes estimated using Christensen-Waals 3D model

• Effective modulus of series arrangement of RVE1 and matrix material in RVE2 estimated using inverse rule of mixtures

0

2

4

6

8

10

12

0 0.05 0.1 0.15 0.2 0.25

True Volume Fraction of Nanotube in RVE2

E3D

-RVE

2 (i

n G

Pa) w = 0

w = 0.05

w = 0.1

w = 0.25

w = 0.5

Exp

Comparison of experimental modulus data for MWNT/polystyrene composite from Andrews, et al. (2002) with micromechanics predictions.from Anumandla and Gibson (2006)

Nanoparticle reinforcement of the matrix in a conventional continuous fiber composite. From Vlasveld, et al., Polymer, 2005.

SEM micrographs of carbon fibers (a) before and (b) after carbon nanotube growth on the fiber surface. From Thostenson, et al., 2002.