Piotr Szablewski
Technical University of Łódź
Department of Technical Mechanics and Informatics
ul. Żeromskiego 116, 90-924 Łódź, Poland
Phone: (48) (42) 636 14 29
E-mail: piotr.szablewski@p.lodz.pl
Numerical Modelling of Geometrical
Parameters of Textile Composites
Abstract
This paper describes a simple sinusoidal geometrical model of textile composites. On the
basis of this model, how to obtain certain geometrical parameters that fully characterize
the geometry of the composite structure will be presented. Using geometrical considerations, it is possible to obtain from this model basic mechanical parameters that are very
useful for further strength analysis. Such mechanical considerations and a method for calculating mechanical parameters will be presented in future articles. On the basis of the
above-mentioned theoretical considerations, a special computer program was developed.
The method of calculation presented in this paper can be applied to more complicated
models of textile composites.
Key words: textile composites, textile mechanics, numerical methods, woven fabric, unit
cell model.
textile composites. A symmetric plain
weave unit cell configuration was studied
to evaluate the performance of the procedure. The stress results were compared
with those from other analyses.
Introduction
Textile composites represent a class of
advanced materials that are reinforced
with textile preforms for structural or
load-bearing applications. In general,
composites can be defined as a select
combination of dissimilar materials with
a specific internal structure and external
shape. The unique combination of two
material components leads to particular mechanical properties and superior
performance characteristics not possible
with any of the components alone [1-4].
Many authors in their works have discussed the classical textile studies performed by Peirce, Kawabata and Leaf.
Scida and others [5] presented a micromechanical model called MESOTEX for
the prediction of the elastic behaviour of
composites reinforced with non-hybrid
weave (plain weave, satin weave and
twill weave) and hybrid weave (hybrid
plain weave and hybrid twill weave) fabrics. By using the classical thin laminate
theory applied to each woven structure,
this analytical model takes into account
the strand undulations in the two directions and also integrates the geometrical
and mechanical parameters of each constituent (resin, fill and warp strands). Woo
and Whitcomb in [6] analysed macroelements for detailed stresses of woven
Lomov, Gusakov and others [7] used a
mathematical model of the internal geometry of 2D and 3D woven fabrics as
a unit cell geometry preprocessor for
meso-mechanical models of composite
materials. The model computes a spatial
placement of all the yarns in a fabric repeat for a given weave structure (a special coding algorithm is employed) and
the given warp and weft yarns’ geometrical and mechanical parameters.
In [8], Jiang, Tabiei and Simitses presented a stress and strain averaging procedure for local and global analysis of plain
weave fabric composites. Osada, Nakai
and Hamada [9] considered the final
fractures of composites to be caused by
cumulation of the microfractures, so that
the initiation of a microfracture, namely,
the initial fracture, is an important factor
of which to know the mechanical properties. Microfracture behaviours in textile
composites were decided by the geometry of the textile fabric quantitatively.
First, in order to investigate the geometry
of the textile fabric, the crimp ratio and
aspect ratio were measured. Tensile testing was performed and the knee point on
the stress-strain curve was identified. The
arrangement, properties and structure of
the fibres within the yarn and the yarns
within the fabric generate a complex
mechanism of deformation. Therefore,
Turfaoui and Akesbi [10] developed a
theoretical model of the mechanical behaviour of the plain weave. The modelling of textile structures by the finite element method was a new approach based
Szablewski P.; Numerical Modelling of Geometrical Parameters of Textile Composites.
FIBRES & TEXTILES in Eastern Europe January / December / B 2008, Vol. 16, No. 6 (71) pp. 49-52.
on the combination of geometric and
mechanical models. The finite element
method permits a construction and representation of fabrics by taking into consideration the yarn undulation, the existence
or not of symmetries in the basic cell and
the type of contact between the warp and
weft yarns.
The modelling of textile structures was
also considered by Milašius in [12] and
[13], where considerations are presented
concerning the principles of woven fabric
3D structure formation and cross-section
shapes. Kobza [14] presented the bending theory of multi-layered composites
filled with soft materials.
The range of applications for composite
materials appears to be limitless. Textile
composites can be defined as the combination of a resin system with a textile fibre, yarn or fabric system. They may be
either flexible or quite rigid. This paper
will present how to build a simple geometrical model of textile composites and
how to obtain from this model certain geometrical and mechanical parameters that
are very useful for further strength analysis. As a novelty, in this study, the validity
and limitations of the presented geometric
model will be studied carefully. Besides,
the usability of new iterative procedures
in the Mathematica environment will be
tested to extend the range of applications.
The geometrical model
of textile composites
Let us consider a balanced plain weave
textile composite in which the warp
and fill yarns contain the same number
of fibres n, with all the filaments having the same diameter df , and with the
warp and fill yarns having the same yarn
49
t
(1)
z w ( x) = sin(π x / Lu ) , − Lu / 2 ≤ x ≤ Lu / 2 .
2
t
(1)
z w ( x) = sin(π x / Lu ) , − Lu / 2 t≤ x ≤ Lu / 2 .
2
) = sin(
π x / Lu ) , z− (x
Lu)/ are
2 ≤ x ≤ Lu / 2 .
of
function
The trigonometric functions of the angle β wz win( xterms
w
2
t
z w ( x) =of the
sin(angle
π x / Lβuw) ,in− terms
Lu / 2 ≤ofxfunction
≤ Lu / 2 . z w (x) are
The trigonometric functions
2 −12/ 2
z w (x) are
The
trigonometric
functions
of
the
angle β w in terms of function
′
′
′
, π xs/inβ
,
=
=
tan
β
dz
dx
z
(2)
=
+
cos β w = [1 +z( z(′wx))2 =]−1t/ 2sin(
z
[
1
(
z
)
]
w
w
w.
(1)
Lu )w, − Lwu / 2 ≤ xw≤ Lu / 2 .
w
2 −1 / 2
2 −1 / 2
2
tan β w = [π t (2 Lu )]cos(
tanwβ=c cos(
π+ x( z/ ′wL)u )], of the
tan β w β= dz
dx = of
z ′w function
.
, β trigonometric
, angle
t
cosπ
β wx =L[1u )+, ( z ′w ) but
] tan
sinβ
z ′w [1functions
w =
w
in
terms
z
(x
) are (2)
The
w
1/ 2
( x) = sin(π x / Lu ) , − Lu / 2 ≤ x ≤ Lu / 2 .
′
′w [1 +w ( z ′w ) 2 ]−1 / 2 , tan β w = dz
,β = tan
= z(1)
cos
β wL= )[1, + ( z ′wbut
) 2 ]−tan
sinββw cos(
w dx = z w .
2
(
)
[
]
π
π
π
tan
β
=
t
2
L
cos(
x
x
/
L
)
,
w
u of the angle
u β in terms of w
function cz w (x) are u
The trigonometric
functions
w
tan β w 1).
= [π t (2 L )]cos(π x Lu ) ,
but tan β = tan β cos(π x / L ) ,
where β c is the so-called crimp angle (see cos
Figure
β w = [1 + ( z ′w ) 2u]−1 / 2 ,
sinβ w = z ′w [1 + ( z ′w )w2 ]−1 / 2 , c tan β w = dz uw dx = z ′w .
β
in
terms
of
function
z
(x
)
are
e trigonometric functions of the angle
w
w
where β is the so-called
crimp
(see
2 )−]1cos(
/Figure
2
π xthe
tan
β =
t( z(2′w )L
x 1).
Lβuw) =
, dz wbut
β c cos(of
/ Lpath.
) , These lower and upper c
′ [πangle
portion
= zβ′w .w = tan
dx tan
, π tan
cos β = [1 + ( z ′ )c2 ]−1 / 2 ,
sinβ
u]
u (2)
w =wz wβ[1 +
where
(3)
tan β c = tan β w ( x w= 0 ) = π t w(2 Lu ) .
c is the so-called crimp angle (see Figure 1).
shown
in
Figure
2. The parametric equations o
−1 / 2 π x L ) ,
tan β w= =z ′[π[1t+ ((2z ′L)u 2)]]cos(
tan= βzw′ .= tan β c cos(π x / Lu ) (2)
,
s β w = [1 + ( z ′w ) 2 ]−1 / 2 ,
sinβ
w
wL dx
w
w
wtan β =
w tan β (, x =tan
(2but
)
.
(3)
π =t dz
0 )u=βwhere
c
w
u
β c is
so-called crimp angle (see Figure 1).
t the
)
. x ≤ofL the
β w (πx x=/ 0L) =) ,π−t L(2portion
L2u ≤
path. These
lower
and
upper
curves
n β w = [π t (2 Lu )]cos(π x Lu ) ,
but tan β w = tan β c cos(πtan
.
zxw (/βxLc)u==) ,tansin(
/
/
2
x
(
x
)
=
x
+
(
t
/
2
)
sin
β
x
z1 ( x) = z win( xt
(
) , are depicted
u
u
1
w
where β c is the so-called crimp angle (see2 Figure 1).u
shown in Figure 2. The parametric equations of the lower and upper c
x 2 ( x) = x − (t / 2) sin β w ( x) ,
z 2 ( x) = z w ( x
tan β c = tan β w ( x = 0 ) = π t (2 Lu ) .
ere β c is the so-called crimp angle (see Figure 1).
portion of the
path. These
lower
upper
arez1depicted
in
xof
)=
x + (tand
sinterms
β w (curves
x)of, function
( x)z w(3)
=(xz w) (are
x) the
− (t detail
/ 2) cosof
β wthe
( x) undu
,
β/ 2w )in
functions
the
angle
. trigonometric
tan β c = tan β w ( x = 0 ) = π t (2 Lu )The
1 (x
Using
eqs.
(1)
to
(3)
and
setting
in
the
first
of
shown in Figure 2. The
parametric
equations
of
the
lower
and
upper
curves
are:
portion
of
the
path.
These
lower
and
upper
x 2 ( x) = x(3)
− (t / 2) sin β w ( x) ,
z 2 ( x) = z w ( x) + (t / 2) cos β w ( x) . cur
(x = 0) =1.πCross
n β c = tan β w Figure
t (2 Lsection
u ).
2the
−1 /warp
2
2 shown
−12.
/ 2 Geometry
Figure
of
an
undulation
region
of the representative unit cos
cellβ(RUC)
along
yarn.
in
Figure
2.
The
parametric
equations
of
the
root
of
the
equation
= z ′win
tan
dz depicted
dx
. the detail
, path.
, curves
= z ′wlower
( z ′ ) ] of the
sinβThese
[1 + ( z ′wand
) ] portion
portion
upper
are
of the
of βthe
path.
These
lower
and upper
cu
w =
w = [1 +
x1 ( x) = xw + (t / 2) sin β w ( x) ,w
z1 ( xalong
) = z wthe
( x) warp
− (t / 2yarn.
) cos
β w (wx) ,
Figure
2.
The
parametric
equations
of
lower
and
upper
curves
are:
shown
in2the
Figure
2.
The
parametric
equations
ot
x
(
x
)
=
,
it
ca
eqs.
(1)
to
(3)
and
setting
in
the
first
of
eq.
(7)
0
ofπthe
These
lower
upper
curves
are
depicted
in
the
detail
of
]
π
tan β w = [π shown
t (2portion
Lu )in
cos(
x Using
Lpath.
)
,
but
tan
β
=
tan
β
cos(
x
/
L
)
,
−1 / 2
u β ( x) , 2
(xxa))+=c(tan
tc /cos(
2path.
) sin
z11and
(2xa()φ=aupper
portion
These
lower
cu−0
)]zare:
1. +
tanupper
β c cos
=
x 2 ( x) = shown
x − (t / 2in) sin
β w (u x2.) ,The parametric
z 2 ( x) =wz wxequations
txof
/ +2of
)β(the
cos
1(φ
w ( x)
wφ(ax))[w
Figure
theβ lower
and
curves
the
root
of
the
equation
Figure
2.
The
path.
are
packing density pd. The yarn packing where βc is thex1so-called
angle
( xparametric
) , , b =in−zthe
(1 x(upper
x)φx−acurves
βudepicted
x(. xdetail
) )==zzww(of
x( x)of)t+
(portion
x) = x +ofcrimp
(t the
/ 2) sin
β w These
x) , lower
z1 ( xand
)xshown
=where
zx)w)=(=
xin
((see
+(=t(t/π2/ x2) acos
)sin
sin
βwwand
/L
x2za(1equations
2x
w ( x) x
where1).
β c is theshown
so-called
crimp2.angle
(see Figure
1).
in
Figure
The
parametric
equations
of
the
lower
and
upper
curves
are:
(7)
density is defined as pd = n · Af /A, where
Figure
lower
uppv
,βit
Usingxeqs.
/ path.
2β
x ((1)
x) =tox(3)
+ (tand
2)setting
sin
xcos(
(x=
x2)eq.
(x+xx−)(12t−(((of
2−sin
)10cos
βcan
().x,)be, shown
) ,theφfirst
)/sin
β w β(2xwβ)(c,in
za )2 [(1zx1+)of
cos
(xwx,)These
(z=xw2portion
)β(z(7)
=x)cos
t/x(2/ata2))/the
)=
z ( xthat
)and
= zthe
2 ( x1) = x − (t / 2φ
ww(0
w ( x)
a + tan
xtan
=wc the
x + (path.
t φ/ Figure
2))]sin
β=
xlower
z12(upper
x) =equatio
zcurve
) , parametric
Af = � df2/4 is the cross-sectional area of
1 ( x )shown
w ( The
w ( x)
in
2.
portion
of
These
and
portion
of
the
path.
These
lower
and
upper
curves
are
depicted
Using
eqs.
(1)
to
(3)
and
setting
in
the
first
of
eqi
the root of
the
equation
(xxπ)),,x / L zzand
(xx)0)=)==xxπ+−t((t(twhere
2))sin
sinββ (3)
= zz=w(−(xxx))−+. ((tt//22))cos
cosββ w((xx)), .
w(=
2 (xx))=
tan β c = tan β w x(xx12(=
/2/2L
the filament. The cross-sectional area of
aw in
a
u2.1 (
u ). φ
shown
The
w x −a(t 2.
x 2 ( xxin
)b =Figure
/ 2equations
) sin βparametric
, the equations
z 2 ( x)and
= zof
( xth)
shown
Figure
The
parametric
lower
w (wx )of
wuppe
the
root
of
the
equation
x ((3)
xa )and
= 0setting
, it can in
bethe
shown
eqs.(RUC)
(1) to (3)
andthe
setting
the Using
first ofeqs.
eq. (1)
(7) to
first that
of e
Figure
section
of the representativeUsing
unit
cell
along
warpinyarn.
2
the yarns is given
by1.ACross
2x ( x ) = x − (t / 2) sin
2 β ( x )2 ,
−1 / 2
2) sin
) ,= xz1w( (xx) )=+x(t+/ 1(2t)/cos
β w β( xw)( .x) , (8) z1 ( x) = z
f = (� df n)/(4pd).
wcos (φ )] z 2 (=x0
φa + tan
β2 c cos(
φthe
[equation
1fibres
+to
tan(3)
βand
a )(1)
c the
a
x
(
x
)
=
,
it
can
be
shown
t
Using
eqs.
setting
in
the
first
of
eq.
(7)
0
the
root
of
the
root
of
the
equation
The
volume
fraction
of
the
in
Figure
1.
Cross
section
of
the
representative
unit
cell
(RUC)
along
the
warp
yarn.
xβ) w=( xxx)eqs.
+2, x(t) /(1)
β 1( x)zsin
(22x))the
= zzfirst
)aw,(setting
The representative unit cell (RUC) is a
x1 ( xof2) =thex +representative
(t / 2)xsin
x2β)βw−
/,in
β−w21w/((2(xxx=))of
)−=0, e(,z
1 (Using
to
( x(zt)1warp
=2)xzsin
2 (β
(φcos
φwarp
cos(
φ1−(a(x(3)
)t)[1/w=2+and
tan
Figure
1.
Cross
section
unit
(RUC)
along
the
yarn.
where
and
.
φ
π
=
x
/
L
x
=
−
x
a + tan
ccell
c cos
a )]
The
volume
fraction
of
the
fibres
in
the
RUC
is
given
by
(
4
)
/(
4
)
V
=
p
A
l
a
H
.
For
the
yarns,
RUC
is
by
.
a
a
u
b
a
the
root
of
the
equation
f
d
rectangle consisting of two warp yarns
the
(the
xeqs.
x(1)
teq.
/ 2(3)
)(7)
sin
βx =
xza)w),(=x0)2 +, in
( x)be
= zshown
(
((xsetting
itzt2can
Using
and
2 eqs. (1) to x(3)
/=
2( x
2(to
root
of
the
)−0,of
xtan
) =2xsetting
−cos
(tUsing
/ 22in
)xsin
β−)1wfirst
(and
βw−w1(/(x2x)=)+.0th
2 the
2=
2/φ
φH
x azequation
Lxau))1wwarp
xyarns,
=β c−(cos
x/a 2.2)(cos
,cell
φa the
+ tan
βarc
φa ) [12by
+(of
βrepresentative
Forthe
thefibres
warp
yarns,
the
is
givLu 2
tan
β. π
cos(
[and
1along
+ tan
φyarn.
volume
fraction
of
in
RUC
is length
given
))]φ/(
V
p d (Aφof
a(7)
the
a )=
b the
c cos(
c ( 4first
alwhere
a 4+
cFor
a )]
interlaced with two fillThe
yarns
with resin
f =
Figure
1.
Cross
section
the
unit
(RUC)
warp
2
eq.
x
(x
)
=
0,
it
can
be
shown
thefibres
equation
eqs.
firs
fraction
ofof
is 2given
by
/(4and
) . For in
=a (4(1)
p Atol )(3)
a Hsetting
thethe
warp
2 the
2
−Using
1 / 2 V1a
the arc
length
given by portion
l = 2 ∫of 1 + (The
z ′w ) 2 volume
dx + 2(Lawhere
−the
L )root
. aUsing
this
integrating,
we
,π
(where
)]have
0f is
φ and
)in
[1 +the
tanRUC
β xc acos
φavalue
=
. the
= πβ cxcos(
φ2(1)
=
x a / droot
Lu and
x2b =equa− x a2 .
matrix
filling
the isremaining
u 2 φ au +φtan
−1 / 2
a / Lu aand xb = −that
the
of
the
aof
Using
eqs.
to
(3)
and
setting
in
the
first
of
eq.
cos
(
)]
φ
+
tan
β
cos(
φ
)
[
1
+
tan
β
φ
=
Lu 2 eqs.
2unit cell
Figure
1.
Cross
section
of
the
representative
(RUC)
along
the
warp
yarn.
x
(
x
)
=
, 0it(
Using
(1)
to
(3)
and
setting
in
the
first
of
eq.
(7)
0
the
root
of
equation
a
c
a
c
a
L
2
−
2
a
en
l = 2 fraction
1 + ( zof′wφ2)the
dxπfibres
+x2(/aLin
−L
this
and
we
have
a. For1 the
the arc
is given
byvolume
the volume. Its dimensions
arelength
2a × u2a
2Using
−by
1 / 2 integrating,
u ) . RUC
2x a2given
where
and
.
=
x
=
−
The
the
is
(
4
)
/(
4
)
V
=
p
A
l
a
H
warp
tion
∫
a
a
u
b
f
d
,
cos
(
)]
0
φ
+
tan
β
cos(
φ
)
[
1
+
tan
β
φ
=
′
( zcw )equation
dx where
+root
−φthe
La u=) equation
.π Using
we hy
the arc length
is given by l =the
the
a2 root
a2( aof
and and
x a / Lu this
xb =integrating,
− xa .
for is
the
volume
fraction:
− Lu 2 a the warpc yarn.
∫ 2 1of+ the
Figure
1.itsCross
section
ofdenoted
the representative
andrelationship
thickness
by
H. Theunit cell (RUC) along
Lu 2
2
2
2
2
−1 /
L
−
u
The volume
fraction for
of the
in fraction:
the RUC
is given
4a− xHa .) . Forφthe
=/ L(4 pand
A lx) /(=
warpβyarns,
+ tan
where
φ a by
= πVxf ahave
c cos(φ a ) [·1 + tan β c cos (φ a )]
relationship
the fibres
volume
Using
and integrating,
other parameters are denoted
and shown
l = 2 u∫ adre12+ (b z ′w ) 2 dx
+ 2(a 2−a 2Lu ) . Using
this
the
arcthis
length
given
bywe
2 and integrating,
2
−1 / 2 we h
2
−
1
/
2
relationship
foristhe
volume
fraction:
Lu 2
β(4)
cos(
φ=a )]− x . = 0 ,
2
, xb((9)
φa− L+u tan
β cos(φφaa·)+
[1tan
+ tan
cβ
c0cos
where
φφa a=)(φ[π1a+)]
x atan
/ Lu=βand
[1 −the
] , Vlationship
V f =of[ A
/( a t hshows
( Luidealization
/ ais) ggiven
c cos
in Figure
1,pwhich
e volume fraction
the
in
RUC
4a volume
= (4 p Afor
l ) /(the
H ) . Forfraction:
the warp
2yarns,c
a
dfibres
r )]the
1 (β c )by
((zβ′w ))2] dx
+ 2(a − Lu ) . Using
this andwhere
integrating,
we
have
a x =(4)
the arc length
is pgiven
by)]l[1=−f 2( L∫ / a1d) +
and
.
=
x
/
L
−
x
φ
π
V
=
[
A
/(
a
t
h
g
,
where φ a = π x a / Lu anda xb = −ax a . u
of the real model. LThe
idealizations
arer
b
a
relationship
2u
u 2
f2
d
1 for
c 2 the volume fraction:
g 1 (β c ) = 1 + 2 1 + tan β c E (2m) / π , m = tan
mu ≤/ a1) g1for
1 p+dtan
, ([10−≤( L
(5)
V − Lβ
/( a tβh )]
(β c )]0, ≤ β c ≤ π 2 ),
u=2
c[ A
1 +most
( z ′w ) warp
dx + and
2(a − Lu 2) . f Using
this andcr integrating,
we2 have
a
arc lengthrelatively
is given
by l = 2because
simple
2
∫
relationship
for
the
volume
fraction:
0
≤
m
≤
1
for
0 ≤ β ≤ π 2 ), (5)
g
(
β
)
=
1
+
2
1
+
tan
β
E
(
m
)
/
π
m
=
tan
β
1
+
tan
β
,
(
,
2
2
2
1
c
c
Using
(4)
− Lu 2
weft
yarnsh are
nominally
straight.
The
≤ m ≤ 1 for consid0 ≤ β c ≤ π 2 ),
) =p1d +/(layer)
tan
βLcuc E
πβ c,)]cm
β c 1complicated
+ tan β cof, ( 0cmathematical
[1 − (E(m)
Vg (=β [ A
a2 t h1r+)]and
/ a(is
)mg)a1/(complete
, = tan
where
elliptic
integral
r = 1 + t r /( 2 t ) ( t r is the thickness off1thec resin
erations,
we
can
obtain the relationship
ationship forthickness
the volume
fraction:
of
the
yarn
along
the
centreline
Figure
1.
Cross
section
of
the
representative
unit
cell
(RUC)
along
the
warp yarn.
Figure
2.
Geometry
of an u
where
E(m)
is
a
complete
elliptic
integral
of
(
is
the
thickness
of
the
resin
layer)
and
h
=
1
+
t
/(
2
t
)
t
2
2
2
the second kind.
V f = [ A p d /( a t hrr )] [1 − (rLu / a ) gwhere
(4)≤ 1isofafor
marea
0 ≤the
β celliptic
≤ π 2 integ
βcc ))] ,=hr1=+12+ t1r +
E (the
m) /thickness
π , m = tan
),
β cthe
1 +cross-sectional
tanlayer)
β c , and
( 0 ≤E(m)
acomplete
tip of
11r((β
offor
the
resin
/( 2tan
t ) (βt rc is
of the yarn path is
denoted by t. If the fithe second kind.
fill
yarns
A
in
the
undulating
region,
2
2
2
2
tipπ 2 ),
kind.
volume
[1 − ( Lufraction
= [ A p d /( abre
t hr )]
/ a )gg1 (1β(βcspecified
(4)
≤ in
m(5)
≤ 1 RUC
0given
≤Using
βlayer)
)c )=] ,1 + 2 for
1 + the
tan unit
β c E (mwhere
)the
/ π second
mh ==tan
, ( 0thickness
, volume
complicated
consideratio
The
fibres
the
isresin
by
(4(5)
4a H ) . For
Vand
p d Aismathematical
l )a/(complete
the warp
c ≤
f = E(m)
(oft r the
isβ cthe
offor
theFigure
elliptic
integ
1 +fraction
t rβ/(c 21t +
) tan
r
determined
by
integration
using
the func2.
Geometry
of
an
undulation
region
along
cell2.1.
is too
small,
it
is
necessary
to
add
an
A
in
the
undula
area
of
a
tip
of
the
fill
yarns
L
2
2
2
2
theβundulation
region
tip
u
π
0
≤
m
≤
1
for
0
≤
β
≤
2
(β c ) = 1 + 2 1 + tan The
β c Egeometry
(where
m) / π , hm =of
),
(5)
=1tan
1
+
tan
β
,
(
the
second
kind.
tions
in eqs.integral
(7) to (9)
c
c thickness of the resin layer)
ofthis and integrating, we
+The
t r /( 2cgeometry
t r is the
ttr)to( the
additional resin layer ofr2.1.
thickness
′w ) 2 dxgiven
= 2E(m)
1is+a( zcomplete
+given
2elliptic
(a −inLeqs.
arcundulation
length is given
by land
ofthe
the
region
(7) to (9):
u ) . Using
∫
complicated
mathematical
considerations,
we can obtain the r
2.1. The geometry of theUsing
undulation
region
unit
cell.
The
geometry
of
the
path
of
the
the
second
kind.
− Lu 2
a constant
normal
the
we
can
determine
the equationsL of
ere hr = 1 + t r /(On
is
a complete
integral
( t r basis
is theofthickness
of thickness
the resin layer)
andtoE(m)
2 t )the
where
hr =centreline
1+
tr /(2t)path,
(tr elliptic
isFigure
the
thickness
Lu 2along the
Lu warp
2
u 2 in region
2. aGeometry
undulation
yarn.
Atip
the undulating
region,
determined
area
of
tip of
of theoffillanyarns
warp
or
fill
yarn
is
modelled
using
two
relationship
fornormal
the
volume
fraction:
curves
of the
yarn
in
the
cross
section
of
the
RUC along
the can
undulating
the basis
of warp
a constant
thickness
tothe
the
path,
we
the
equations
ofz ( x)(dx / dx) dx
E(m)
is
acentreline
complete
of
thethe
resin
layer)
and
2.1.
The
geometry
of
undulation
region
second kind. the lower and upperOn
Atipdetermine
= ∫ z1 dx
+ Figure
z
dx
=
On
basis
of
a
constant
thickness
normal
to
the
centreline
path,
we
can
determine
the
1
2
2
1
1 equatio
2. Geometry
of
an und
∫
in eqs.
(7) toof
(9):
assumptions: 1 – the centreline
of the
the lower and
upper elliptic
curves integral
of the warp
in given
the
cross
section
the RUC0 along the∫0 undulating
of
theyarn
second
kind.
xa along for
Using
complicated
mathematical
considerations,
we can
obtain
the
relationship
the
cr
the
lower
and
upper
curves
of
the
warp
yarn
in
the
cross
section
of
the
RUC
the
undu
2.1.
The
geometry
of
the
undulation
region
yarn path consists of undulation portions
Figure
2.
Geometry
of
an
undulation
region
along
the
warp
yarn.
Figure
2.
Geometry
of
an un
L
2
L
2
L
2
L
2
V f the
= [area
A p d of
/(ofaatatip
hconstant
− ( Lfill
/ ayarns
) g1 (βuAcnormal
)] ,
u
u
On
basis
tou the
centreline
path,
we
can determine
the equatio
r )]
uthickness
region,
determined
by
integration
using
of[1the
tip in the undulating
Using
complicated
mathematical
considerations
Now,
the
area
and straight
portions,
with the
centreline
Atip the
=Figure
z1 dx12.
+Geometry
zin2 dx
z1 (undulation
xcross-sectional
)(dx1 / of
dx)the
dx +RUC
z1 (of
x)(the
dx1fill
/ dxyarn,
) dx
.
of∫an
region
the
warp
yar
the lower
and of
upper
curves
of
warp
yarn
the
section
along
the
undu
1. The geometry
of the
undulation
region
2 =cross
∫20considerations,
∫Aalong
2
The
the
undulation
the
ofthe
a2 tip
of( 0the
fill
yarns
On the
basis described
of a constant
centreline
can
of
Figure
2.
Geometry
of
in
(7)
to
π an
≤m
≤
1 for
0in≤consideration
β c ≤undulatin
2for
g 1 (geometry
β cnormal
)given
= 1 +Using
2toeqs.
1the
+complicated
tan
β c(9):
E (mpath,
)region
/ π ∫0,we
),unt
m=
tandetermine
β carea
1Finally,
+ tan
βequations
mathematical
can
obtain
the
Using
mathematical
xacomplicated
xa tiprelationship
c3,we
of the undulating
portions
by thickness
the 2lower
and
upper curves ofOnthe
in
cross
of the
RUC
along
the
undulating
thewarp
basisyarn
ofLu aUsing
nor2. Geometry
ofof
undulation
region
the
yarfu
given
in
(7)
to can
(9):
the
undulating
region,
determined
integration
area
of the
a tip
thesection
fillFigure
Athe
in
thewarp
undulat
area
a tip
ofwe
the
fill
yarnsalong
2 constant
Luof
2thickness
Lyarns
Luan
2eqs.
sine function,
– the
cross-sectional
u 2
tipby
complicated
mathematical
considerations,
obtain
relationship
3is
the basis ofthea constant
thickness
normal
to the centrelinewhere
path, we
equations
ofAtipofinthe
Now,
the
cross-sectional
area
of( the
fill
yarn,
which
the
same asinte
th
isdx
the
thickness
resin
layer)
and
E(m)
is
a. )]complete
elliptic
hr centreline
=can
1 + tdetermine
2 tpath,
)+ ( t r zthe
Using
complicated
mathematical
consideration
mal to the
we
can
deterr /(
A
=
z
dx
=
z
(
x
)(
dx
/
dx
)
dx
+
z
x
)(
dx
/
dx
)
dx
Figure
2.
Geometry
A
=
a
t
[
1
−
(
L
a
)
g
(
β
g
(β
,
where
area
and
the
thickness
of
the
yarn
normal
L
2
L
2
L
2
A
in
the
undulating
region,
determined
by
integratio
area
of
a
tip
of
the
fill
yarns
1 in1 eqs.
2the
1
1tip
lower and upper curves of the warp yarn in the cross section oftip the
RUC
along
undulating
given
to2 (9):
given ∫uin 1eqs. (7)
u 1u to (9):
2
c u
2 of
∫0Using
∫0 (7)
∫x upA
in
the
undulati
area
of
a
tip
of
the
fill
yarns
complicated
mathematical
considerations,
we
can
obtain
the
relationship
fo
the
second
kind.
mine
the
equations
of
the
lower
and
tip
Finally,
xa z dx + Figure
a
to its centreline are uniform along the arc
A
z dx 2.= Geometry
z ( x)(dxof
/ an
dx)undu
dxalo
+
Figure
1Geometry
1
1region
tip = ∫L2.
Lu 2in eqs. (7)
Lu 2 to (9): Lu 2
given
Lu 2 of 2an undulation
Lu 12
∫
∫
u 2 Lu 2 region,
the
undulating
determined
by
integratio
area
of ayarn
tip ofin
the
fill
yarns Atip in given
per
curves
of
the
warp
the
cross
in
eqs.
(7)
to
(9):
Using
complicated
mathematical
consider
0
0 (10) for Axa in eq. (4), we can r
length of the centreline. The centreline of
Substituting
eq.
dx
=− (the
(Now,
dx(1βthe
/ dx
)=dx∫+is
z u+
(same
/ the
dx
)=dx
A
z1where
dx
zarea
(fill
x)(
dx
Now, Athe
cross-sectional
which
the
warp
yarn,
is/ dx
2. A
2 1 + A
Larea
2t [
2x )(
2u a
21of
1dx
tip =
2as
2 )=
1G
1β
tip
=zto
a2x))(gyarn,
1∫the
−.zconsiderations,
(φtipdx
)A/ π)=und
,cross-sectional
∫0 inzL1udx
∫0 Using
∫LzuL1ufill
∫gdx
of
the(7)
undulating
2 mathematical
c )]complicated
23(β cfill
2A
a ,in
c the
ofx∫a1L1tip
of
the
Using
givenalong
eqs.
(9):
Lmathematical
Lyarns
complicated
considerations,
obtain
thw
the warp yarn path in an undulating re- section of the RUC
Lu / area
aL:u)02dx
u 02
ux 2 we can
xa
a
aof
Now,
the
cross-sectional
area
the
fill
yarn,
wh
A
=
z
dx
+
z
dx
=
z
(
x
)(
dx
dx
+
z
(
x
)(
dx
/
dx
)
dx
.
2.1.
The
geometry
of
the
undulation
region
Finally,
yarn,
which
is
the
same
the
yarn,
1 lower
1
2 up2
1 area
1 a tip
1 as
1warp
tip These
portion
of
the
path.
and
A
in
the
undulating
of
of
the
fill
yarns
∫
∫
∫
∫
tip
A
in
the
undulating
region,
determin
area
of
a
tip
of
the
fill
yarns
Atip =given
z1 dx
z 2 dx
= ∫ z1 ( x)(dx1 / dx) dx +
Lu 2
Lu 2
in1Leqs.
to2 (9):
gion is specified by
3Luxa2
tip
u+2 ∫(7)
∫which
0
0
xa
Agiven
=Finally,
2 yarn,
AA
+the
t(a
– Lto
Substituting
(10)
for
in
(4),
we
can
rearrange
thefill
result
as
Now,
the
cross-sectional
area
ofis
the
fill
isuz).
the
same
asx)athe
yarn,
is waA
Now,
cross-sectional
area
ofwarp
the
yarn,
per curves are
depicted
indxthe
detail
of=eq.
tipdx
0eq.
0(Finally,
in
eqs.
(7)
(9):
A
=
z
+
z
dx
z
(
x
)(
dx
/
)
dx
+
x
)(
dx
/
dx
given
in
eqs.
(7)
to
(9):
L
2
L
2
L.u 2
1
1
2
2
1
1
1
1
tip
∫
∫
∫
∫
u
u
φ
π
A
=
a
t
[
1
−
(
L
a
)
g
(
β
)]
g
(
β
)
=
1
−
G
(
,
β
)
/
whereto the
. dx
On
the basis region
of a constant
thickness
determine
the equat
ushown
2 in
2
ccentreline2 path,
a
cwe can
t
LucFigure
a, : normal
the
undulation
2.
Now,
the
cross-sectional
area
of
the
fill
yarn,
which
is
the
same
as
the
warp
yarn,
i
0
0
x
x
Finally,
Finally,
a
a
2
(1)
z w ( x) = sin(π x / Lu ) , − Lu / 2 ≤ x ≤ Lu the
2the
2RUC
A
+(β ∫the
zLu2area
dx
=
( x)(
dx(1βund
/ dx
Lh
L in
2Now,
2( L
lower and upper
theLu cross
section
along
the
cross-sectional
the
w
2 Lof
tipLu =
g/ 12g.parametric
g1 + g 2curves
)(Lu ofa )of
+ 1the
−lower
V fwarp
u 2 / p yarn
u 2where
gyarn,
)Lu
∫0uLzu 1a2dx) g12of
∫ z1fill
2 (Lu a ) − ( equations
r
d = 0 u. A = a t [1 −
c )] ,
2
c (1
The
the
2
t
(1)
0
x
Finally,
(10)
Acan
=dx
=same
z+1 as
( x)(
/ equation
dx
dx
+) di∫
a( x
t
z(4),
+ the
z1+
(L−
x∫)(Gzresult
dx
/2dx
) )]
dx
zdx
)(
dx)1yarn,
1which
2 dx
1warp
cross-sectional
of
fill
yarn,
is
the
∫ =z1dx)∫(1)
Substituting
eq.
rearrange
1 dx1we
2A
1φ
1where
tipin=)]
∫0tipzFinally,
u / 2 .upper
curves
A Now,
= are:
a t [eq.
1the
− (10)
(L
afor
) gAA
)∫a/,πquadratic
,∫ area
where
. ∫the
−=
(1the
(, βas
g/ dx
(1)
( x) = sin(π x / Lu )z,w−( xL)u=/ 22 ≤sin(
x ≤π Lxu/ /L2u .) , − Lu / 2 ≤ x ≤ Land
2 (β cprocedure
4 g=220a(βt c[1parameters
u0 a2)(g
2 (β cx
L2a u . c xa
Equation
(11)
is used in theu iterative
to determine
0
xa
xa
2
L
a
:
Now,
area
thecan
fill rea
ya
Substituting
(10)
eq.
(4),ofwe
Finally,
sin(π x / Lu ) , − Lu / 2The
≤ xtrigonometric
≤ Lu / 2 .
u
of
functions of the angle β w in terms
Afunction
= a t [1 − (zLwu(x(1)
a) )are
g (β c )] ,
g 2 (β cthe
)eq.
= 1cross-sectional
− G2for
(φ a A, βin
where
c ) /π .
The function of the crimp angle
defined
byNow,
G2 2is the
Au =athe
a: area
tcross-sectional
[1 −of
( Lthe
) g 2 yarn,
(βarea
g 2whic
(βas
where
of
the
fill
yarn,
The trigonometric
functions of the angle
L
u afill
c )] ,which
c)
Now,
cross-sectional
is
the
same
t
Finally,
Substituting
eq.
(10)
for
A
in
eq.
(4),
we
can
rearrange
the
result
as
a
quadratic
equa
Substituting
eq.
(10)
for
A
in
eq.
(4),
we
can
re
L
/2.
in(x
terms
ofAfunction
z( L
(x)a )2are
−1 / 2
2
−1
w4
(1)
(terms
xThe
) = trigonometric
sin(
πthe
x / angle
Lu )z,2functions
−
Lw/ u2are
/ 2terms
≤ xof≤the
Lu function
/angle
2 . Gβ w((1)
β
in
of
z
)
are
e utrigonometric
of
φ
π
=
a
t
[
1
−
g
(
β
)]
g
(
β
)
=
1
−
G
(
,
β
)
/
,
where
.
φ
,
β
)
=
I
−
(
I
+
I
)(
1
+
tan
β
)
+
I
(
1
+
tan
β
)
,
where
βw zinwfunctions
of
function
(x)
−
1
2
−
1
/
2
w
2c
c
c
2
a
c
w ]
ca :2
2+ (az ′ )c ] L1, a :tan
2 β 3 = dzu Finally,
4
Finally,
L
′
′
′
2
,
=
dx
z
.
(2)
=
+
=
cos
β
[
1
(
z
)
s
inβ
z
[
1
eq.
u Substituting
w
w terms of function
w z (x
w ) are w
w (10) for
w A in eq. (4), uwe can rearrange the result as a quadratic eq
β w in
ometric
(1) w
− Lu / 2 functions
≤ x ≤ Lu / 2of. the angle
w
Substituting
can re
A = a t [eq.
1 − (10)
( Lu afor
) g 2A(βinc )]eq.
g
, (4), we
where
2 )−]1cos(
/2
2φ
−1 tan
/ ,2 β
Lcutan
a β: πI 2=x=/dz−
φ
I
=
2
cos
2
F
(
,
m
)
(
[
β
=
t
2
L
tan
β
=
cos(
L
)
,
π
π
2 −1 / cos
2tanβ
2x −s1L
/ u2 ) , = z ′but
′
′
′
,
,
=
dx
z
., A in eq. A(4),
(2)
=
+
+
[
1
(
z
)
]
inβ
[
1
(
z
)
]
a
a
1
w
u
w
u
L
a
wzare
w tanwβ = dz
w
w
w (10)
w
:
Substituting
eq.
for
we
can
rearrange
the
result
as
a
quadratic
eq
′
′
′
function
z ww(x=
)
ngle
=
a
t
[
1
−
(
L
a
)
g
(
β
)]
g
(
β
)
=
,
where
(12)
, w sinβ
,
=
dx
z
.
(2)
+
s β w β=w[1in+ terms
( z ′w The
) ]oftrigonometric
[
1
(
z
)
]
u
2
c g 2 (β c ) = 1 − G 2 (φ a2 , β cc ) / π
w
w the angle β
w
w of function
w
z w (x) are A = a t [1 − ( Lu a ) g 2 (β c )] , u where
functions
of
w in terms
1/ 2
′wbut
′w) ,(.=φ atan
,βπ x=tan
β = dzbut
+β( z ′w=)[2π]−t1 / (22, L )]cos(
sinβ
) 2 ]−tan
Idx
=/=[L2βz4E
, m)βL+cu 2cos(
m) − 2 m sin 2φ a (2 − m +Substituting
m cos 2φ a ) −1 / 2eq.
] (m
− 1)for
, A in eq. (4), we(1c
a(m: π− 1x)/FL(uφ)a ,(2)
tanπwβxw= =Lz ′w[π)[1,t+ ((2z L
Ltan
tan
wπ
3x
(10)
u )]cos(
u ) w,β cos(
w
ntions
βwhere
in
terms
of
function
z
(x
)
are
w of the angle
u
u
w
c
u
w
w crimp angle (see Figure 1).
theβso-called
4
/2
2
c is 2tan
Substituting
eq.
(10)
for
A
in
eq.
(4),
rearra
x, βLutan
)=, [β1 w+β=(but
=
tan
β
cos(
x
/
L
)
,
πzt′w [(12+L(uz)′w]cos(
π
L
a
:
−dx
1 / 2w
2
−
1
/
2
′
(2)
=
dz
z
.
(2)
) 2 ]−π1cos
Substituting
(10) m
for((2)
thecan
result
[(1β− m
)(1 − m + m cos
atan[
1A−u in
m)eq.
cos(4),
φa ] we
[(1 -can
m) rearrange
1− m
] . we
c
u I′4 = m cos
′
′ φa )] + meq.
, φ a tan
z ′w )w ] , w sinβ
w
w = z w [1 + ( z w ) ]
w = dz w dx = z w .
4
L
a
:
4
where
β
is
the
so-called
crimp
angle
(see
Figure
1).
L
a
2
−
1
/
2
:
u
here
isβ the
(see
1). = but
= so-called
tan
cos(
) ),]tan
π xc(/2Langle
u
′w [1ββ+cwtan
w
, but βtan
βπw )Figure
scinβ
(=z ′w[crimp
cos(
x==dz
Lπutw) ,(dx
x / Lu ) ,
π(2)
w = ztan
(3)
βπ)c t]=(see
tanL,u uβFigure
2 Lu )z.′w . tan β w = tan β c cos(
w ( x = 01).
s the so-called crimp
angle
of the first and second kind.
F (φ a , m) and E4(φ a , m) are the incomplete elliptic integrals
4
s(π x Lu ) ,
but tan β w = tan β c cos(π x / Lu ) ,
Set of Equations 12.
2
tan
β c = Ltan) .β w ( x = 0 ) = π t (2 Lu ) .
(3)
β c = Figure
tan β w (1).
xwhere
= 0) =
en (see
to the (3)
volume fraction equation (4) and the ya
Assuming
u so-called crimp angle (see Figure
βπct is(2the
1). that 40 ≤ β c << 1 , we obtain approximations
(3)
n β w ( x = 0 ) = π t (2 Lu ) .
2
and4January
higher
in the/ Bseries
shape equation (10). We neglect
terms
of order
β c Europe
FIBRES
& TEXTILES
in Eastern
/ December
2008, Vol.expansions
16, No. 6 (71) to obtain
50 (see Figure 1).
ed crimp angle
4
final(3)
results:
(3)
tan β c = tan β w ( x = 0 ) = π t (2 Lu ) .
4
π t (2 Lu ) .
t = A p /(V h a)(3)
,
L = a (1 − V h / p ) / (1 − 2 / π ) .
(1
a. If 0 < βc ≤ π / 4, then STOP. Convergence of the geometric parameters
1
1 z/t
t, L u and tr has been achieved and
the geometry of the RUC is estab0,5
0,5
lished.
2x / Lu
2x / Lu
b. If βc > π / 4, then set βc = π / 4 and
-1
-0,5
0,5
1
-1
-0,5
0,5
1
assume that the specified value of the
-0,5
-0,5
spacing
is incorrect.
Solve εfor
2g. Calculate the yarn
difference
in the a
iterates
using the measure
= (β c ) i − (β c ) i −1
t,
L
,
t
and
a
in
steps
from
3c
to
3f.
2h.
If
,
then
increase
the
index
.
ε
>
tol
i
→
i
+
1
-1
u r
-1
to step 2a, else, if i > imax , then STOP and print a
2i. If i ≤ imax
c., then
Set tgo
r = 0 or hr = 1 and solve the quadmessage.
ratic equation (11) for the two roots
2j. If ε < tol , then the fixed point iteration is judged to have converged to
Figure 3. Geometry of an undulation region along the warp yarn calculated numerically:
β c = (β c ) i . Lu /a.
a) βc ≤ π/4 – model is valid, b) βc > π/4 – model is not valid (a cusp forms).
Take the positive real root for the ratio
3. Check the constraintofonLthe
crimp angle.
u /a.
, then
Convergence of the geometric parameters t, L
3a. If 0 < β c ≤
/40
Table 1. The output data.
d.π If
< LSTOP.
u /a ≤ 1, then go to step 3e, else,
achieved and the geometry of the RUC is established.
if L /a > 1, then set Lu /a = 1 and solve
3b. If β c > π / 4 , thenu set β c = π / 4 and assume
that the specified value of the
n
a, mm
t, mm
Lu, mm
βc, deg
No. of iterations
eq.
(11) for hr. a in steps from 3c to 3f.
incorrect. Solve for t, Lu , t r and
2000
1.4110
0.0857
0.5869
12.90
4
e. Because βc =the1,quadratic
we determine
from
equation (11) for the two roots
3c. Set t r = 0 or hr = 1 and solve
t
=
(2/π)
(L
/a)a.
Substieq.
(3)
that
10000
1.4110
0.4727
0.7812
43.54
6
u
Take the positive real root for the ratio of Lu a .
this expression for t into the vol14000
1.6260
0.5804
0.9117
45.00
23 3d. If 0 < L / a ≤tute
1 , then go to step 3e, else, if Lu a > 1 , then set Lu a = 1 and
u
ume
fraction equation (4) and solve
hr .
for
the
new
yarn spacing
to obnew that
determine
from eq.a(3)
3e. Because tan β c = 1 , we
t = (2 π ) ( Lu a ) a
tain
From the small crimp angle equations
Substituting eq. (10) for A in eq. (4), we
expression for t into the volume fraction equation (4) and solve for the new
(13), compute the initial values oftothe
can rearrange the result as a quadratic
obtain a new = π d p A[1 − g1 (β c ) ( Lu a )] [2 ( Lu a ) V f hr ] , where β c = π
thickness t and length Lu. The initial
equation in the
ratio
of
L
/
a:
u
2
3f. For anew from step 3e, compute Lu = ( Lu a ) ⋅ a new , t = (2 π ) Lu and t r = 2 t
g1 g 2 (Lu a ) − ( g1 + g22 )(Lu a ) + 1 − V f hr / p d crimp
= 0 . angle (β ) is computed from
where βc(11)
= π/4.
g1 g 2 (Lu a ) − ( g1 + g 2 )(Lu a ) + 1 − V f hr / p d = 0 . c 0
(11)
f.
For
a
eq.
(3).
new from step 3e, compute
(11)
2 Equation (11) is used in the iterative
L
.
procedure
to
determine
parameters
u
− ( g1 + g 2 )(Lu a ) + 1 2− V f hr / p d = 0 .
(11)
Numerical
Lu = (Lu /a)·anew, t = (2/π)Lu and
1 g 2 (Lu a ) +
Lexample
.
(11)
used inL thea )iterative
procedure
parameters
g1 Equation
g 2 (of
a )crimp
− ( gis21 +
V
h1
p d h= /0p.to determine
, Vf = –
Inputtodata:
duf = 0.007 tmm
= 1.411 mm
0.64
r−/ V
) +f Begin
gL1ugthe
− (gg2 1)(+
L+u1 −a2.
(11), pd = 0.75 , tol = π 4 ⋅10 −6 , im
The function
by
Gu2g 2is)(defined
1)., a(11)
iteration
loop: for
imax
2 (L
u a ) angle
f the
r
d = 0.
r = 2t(h
r
The
function
of
the
crimp
angle
is
defined
by
G
L
.
quation
(11)
is
used
in
the
iterative
procedure
to
determine
parameters
Equation
11
is
used
in
the
iterative
proce2
2
−
1
/
2
2
−
1
− ( g1 + g 2 )(LuGa()φ+ 1, β− V) f=hIr / −p d( I= +0 .I )(1 + tan β )
(11)
u
in steps
ofc )determine
one.
+2Iprocedure
tan βto
, where
Lu .
(11)
used in the in
iterative
parameters
a Equation
c
2
1
2 is
4 (1−+
2
LuInput
.
Equation
iterative
parameters
determine
parameters
Gangle
=(11)
I31 − (is.I 2used
+byI 3 )(c1 the
+ tan
) 1 / 2procedure
β1c and
) −1 , where
he function dure
of thetocrimp
defined
Gc2) is
No. of
dataa,dmm
V = 0.64, a = 1.411
2 (φ a , β
tr+=I 40(1or+ tan
hrto=determine
forOutput
the crimp
a.β cSet
f = 0.007,
data:
n
t, mm f Lu , mm β c , deg
The
function
ofdetermine
crimp
angle
by
G2 −1isLdefined
. is defined
is used in the
iterative
procedure
parameters
The
function
of
the
crimp
angle
is
defined
2 to
−I1the
/ 2=of
2) angle
uangle
= 0.75, tol = π/4 · 10–6, imax = 30. iterations
I
=
2
cos
,
−
2
F
(
,
m
,
φ
φ
The
function
the
crimp
by
G
mm,
p
)
solve
the
quadratic
equa(β
2 + I (1 + atan β ) , where
2
G2 (φ a , β c ) = I1 −1 ( I 2 + I 3 )(Ia1 += tan
β
)
d
c
i–1
4
−22cFβ(φ)a−,1m
φcIa ,−50).
/ 2) ,
2
2000
1.4110 0.0857
0.5869
12.90
4
Equation
12
2 + I(11)
of the crimp by
angle
by
φ1 aG,(see
β2 cos
) =page
)(I1+2 +=I tan
tan
β ctwo
) −1 2, roots
G2 Gis2 (defined
where
−1 / 2) −1 ,of
The
output
data is
presented
in Table
1.
the
the
ratio
ctan tion
β1c−) 1=()IF
I21(+φ−aI(,3Im
1m+sin
βa c()2−4−1 /(21m+
+for
Im
+
tan
β
where
10000
1.4110
0.4727
0.7812
43.54
6
I 3 = [22 E
(φ−a1,/ 2m
)2+c(φ2a(,m
2
2
φ
+
cos
2
φ
)
−
1
)
,
(12)
c ] (m
2 ) − 3 )(
4 (1
1
/
2
−
a
,h1 +/ ptan =β0I .2) I=3 −=2+[F2I (Eφ(1(aφ,+m
)m,)2+β2)(−m1 ,−where
1) F (φ a , m) −Lu 2/ a.
m sin 2φ a(11)
(2 − m + m cos 2φ a ) ] (m − 1)14000
,
(12) 0.9117
1 =
1.6260
0.5804
45.00
23
a ,tan
−)2(+Icos
Ia3f)(
−φV
c
c
1a
21 +
4
r
d
I1 φ=E(φ
2 cos
,)(
= −2 φ
2 F)(]φ+a2, m
, , m)−,1m
φm)
F(φ
, m)
and
are
I 4 a=
m cos
[(a1, −
1 −athe
m +incomplete
mI 2cos
m2)(If
) cos φ a ] [(1
1 − m ] . For n = 14000, a maximum crimp angle
/ 2 (1 − mreal
2a)m
cos
Fcos
φatan[
/- m)
aφ exists,
)+
2 (mintegrals
− 1)IFa4 (=Iφ1am=
,m
−φ aφfirst
2[(m,1sin
(2mI−2a+m=m+−cos
mb.
(m − 1)m, (root
cos
−and
m2)(φ1second
φaa2)φno
]a+) positive
m]atan[
1 − mL) ucos
] [(1(12)
- m) 1 − m ] .
a−
3 = [ 2 E (φ a , m
elliptic
of
the
−1 / 2a
I
=
−
2
F
(
,
m
)
,
φ
= [2I E=(φ[a2,parameters
m
)
+
2
(
m
−
1
)
F
(
φ
,
m
)
−
2
m
sin
2
φ
(
2
−
m
+
m
cos
2
φ
)
]
(
m−1−
(12)spacing
/a21) , condition, the yarn
a I 3determine
2
L
.
iterative procedure
to
a is recomthen
STOP
the
program
and
write
a
a
a
2E (φ , m) + 2 (um − 1) F (φ , m) − 2 m sin 2φ ( 2 − m + m cos 2φ )
m − 1) , crimp angle condition,
(12)
a m) cos φ ] [(1 - m)
a 1 − m ] . For n = a14000,]a (maximum
3m cos φ )a] + m atan[ m (1 −
kind.
the yarn spacing a is recomputed.
1(−φ m, )(
1) −and
m +E
−1 / 22elliptic
4 = m cos φ a [(
a are the incomplete
a message.
puted.
(
,
m
)
integrals
of
the
first
and
second
kind.
F
m
φ
warning
m
)
+
2
(
m
−
1
)
F
(
φ
,
m
)
−
2
m
sin
2
φ
(
2
−
m
+
m
cos
2
φ
)
]
(
m
−
1
)
,
(12)
ngle G2 is defined
a a [(1 − m )(1 − m + m
φa )] +2 φ m)elliptic
m (1 − m)of
φmaFigure
] [(1and
m ]kind.
. calculated shape of the undulation region along the warp
a a byI 4 =
a cos
In
numerically
integrals
F (mφI acos
,=mφm
) aand
cos φE (φ[(a1, m
− )mare
)(1 −the
m +incomplete
m
]atan[
+ smallest
m atan[
m (cos
1 −the
)first
cosroot,
φ-3,aa)m)
]a second
[(11-−m)
1− m] .
c. cos
Takea the
positive
real
2
−1 Assuming
/ 2Assuming
2 4 0 −≤
1 β 2 <<a1 , we
two
cases:
model
is valid
(the
π / 4 ), b) model is not valid because a cusp forms i
β c ≤yarn
2 that
we
obtain
approximations
to
the
volume
fraction
equation
(4)
and
that
2 obtain
+[(
tan
β
)
I
(
1
+
tan
β
)
,
where
c
1
−
m
)(
1
−
m
+
m
cos
φ
)
]
+
m
atan[
m
(
1
−
m
)
cos
φ
]
[(1
m)
1
−
m
]
.
c incomplete
approximations
to
the
volume
fraction
equation
(4)3,anda the
yarn
Assuming
that 0 ≤elliptic
β c << integrals
1 , wea obtain
a the
ofand
the
first
and
second
kind.
(φ a , m) cand E (φ a4, m) are
In
Figure
numerically
multiply
the
root
by
the
yarn
spac).
region
close
to
the
end
of
the
undulating
region ( β c > π / 4calculated
2
approximations
theWe
fraction
and
Evolume
(φ aneglect
, m) are
the incomplete
elliptic
integrals
ofthe
theseries
first and
second kind.
F (φ a , mto
)(10).
and
higher
in
expansions
to
obtain
the
shape 2equation
terms
of
order
β
2
c
shape
of
the
undulation
region along
and
E
(
,
m
)
are
the
incomplete
elliptic
integrals
of
the
first
and
second
kind.
F
(
,
m
)
φ
φ
ing aoftoorder
obtain
aequation
newhigher
value
of
the
unand
in
the
series
expansions
to
obtain
the
equation
(10).
terms
β
, wethe
obtain
to
the
volume
fraction
(4)
and
the
yarn
1and
−ssuming
2 F (φ a , mthat
) equation
, 0 ≤ β c <<
a yarnapproximations
aWe neglect
c
(4)shape
shape
equation
E (φ a , m) arefinal
the incomplete
elliptic
first
andapproximations
second kind.
we
obtain
the volume fraction equation
(4) yarn
and the
yarn for two cases: a)
Assuming
that 0integrals
≤ β c2 <<of
12,the
results:
the equation
warp
is and
shown
dulation
Ltoexpansions
2
u.
/02 ≤ β β<<
−1order
final
results:
obtain
to the to
volume
(4)
the yarn
Assuming
that
1, , we
neglect
terms
of
order
and
higher
in approximations
thelength
series
obtainfraction
the
hape
neglect
terms
of
(φ a , mequation
)2− 2(10).
m(10).
sinWe
2φWe
(
2
−
m
+
m
cos
2
φ
] (mc −and
c 1)
2(12)
a
a ) to the
model
is
valid
(β
π/4), b) model is not
L
≤
a,
then
go
to
step
2e,
else,
d.
If
0
<
,t we
obtain
approximations
volume
fraction
equation
(4)
and
the
yarn
t 0 ≤ β c << 1higher
and
obtainc ≤the
shape
equation
(10). LWe
neglect
terms
of) /order
u 2β
2higher in the series expansions to (13)
in
the
series
expansions
to
obtain
c
=
A
p
/(
V
h
a
)
,
=
a
(
1
−
V
h
/
p
(
1
−
/
)
.
π
d
u
f
r
d
nal results:
the series
obtain
the in the filaequation
We neglect
terms
of/ porder
t f=shape
Ar 1p−
h2 r φa) ],(10).
−β c2 L
/and
(13)forms
π )=. higher
d /(
d ) / (1set
validexpansions
because atocusp
then
a and in
solve
cos 2We
φa )the
] + final
m atan[
mV) βfcos
[(1 - m) L1u−=ma](.1if− LVuf >hr a,
u
a
final m
results:
results
n+ m
(10).
neglect
terms
of (order
c and higher in the series expansions to obtain the
final
results:
ment in aofregion
to the end of the
eq.
(11)
hiterative
= A p d /(V f hSmall
Lu = a approximations
(1 − V f hr /approximations
p d ) / (1(13)
− 2 /are
(13)beginning
π ) .used
crimp
angle
in for
the
at the
the close
r. in the procedure
r a) ,
(13)h are
used
procedure
at the beginning
of the
t =Small
A p d /(crimp
V f hr aangle
),
Lu = a (e.
1 −Determine
V
− 2 /thickness
(13)
π ) .iterative
undulating
region
(β
>
π/4).
f
r / p d ) /a(1
new
t
from
the
calculations.
c
6
the incomplete elliptic integrals
kind.
t = A of
p d the
/(V ffirst
hr aand
) , second
/ (1 − 2 / π ) .
(13)
calculations.
(13) L
u = a (1 − V
f hr / p d ) at
crimp angle
procedure
beginning
hmall
Lu =approximations
a (1 − V f hr / p d )(13)
/ (1 −are
2 / πused
) . in the iterative
(13) (4). of the
fibre volume
fractionthe
equation
r a) ,
Small
crimp
angle
approximations
(13)
are
used
in
the
iterative
procedure
at
the
beginning
of
the
we
obtain
approximations
to
the
volume
fraction
equation
(4)
and
the
yarn
alculations. 2.2. Algorithm
to crimp
determine
the
architecture
parameters
f. Calculate
a new
crimp
angle
(βc)i from
Small
angle
approximations
(13)
are
used
in the
iterative
procedure at the beginning of the
2.2.
Algorithm
to determine
the
architecture
parameters
calculations.
Conclusions
2 angle
crimp
approximations
(13)
angle approximations
arehigher
used
in
the
iterative
procedure
atobtain
the beginning
of
the
and
in
the
series
expansions
to
the
eglect
termsSmall
of order
β(13)
eq.
(3)
using
new
L
and
t detercalculations.
c
u
.2. Algorithm
to
determine
the
architecture
parameters
are
used
in
the
iterative
procedure
at
the
Input parameters:
n, d f to
, pdetermine
, a,
Output
parameters:
t, Lu , β ct,, tLr , , aβnew
–presented
new yarnmethod of analysis was
The
in steps
2c to 2e.
dn,
parameters:
d f V, f p, dtol,
,the
a, imax.
Vmined
Output
parameters:
Algorithm
architecture
f , tol, imax. parameters
u
c , t r , a new – new yarn
beginning2.2.
ofInput
the
calculations.
2.2.
Algorithm
to
determine
the
architecture
parameters
o
very
useful
for the modelling of textile
g.
Calculate
the
difference
in
the
iterates
,parameters
Vβ –imax.
vol.o ofV¼– of
the
V yarn
–L(13)
vol.
of the
innew
¼
ofyarn
the
– vol.
=V45, iftol,
Vr RUC,
nput
n,the
donly
,spacing
pif ,βa,
parameters:
t,RUC,
βyarn
ayarns
hm
architecture
c only
vol.
ofRUC,
¼ of the
yarns
in RUC,
¼ of the
Vr – vol.
Lu to
=parameters:
adetermine
(1 − Vspacing
u, V
c , t–
r , vol.
newof –the
c = 45 ,Output
f hr / p d )f / (1 −d 2 / π ) . f
composites.
It can
of course be applied
using
the
measure
Input
parameters:
n, architecd f , pd , a, V f , tol, imax. Output parameters: t, Lu , β c , t r , anew – new
yarn
o toindetermine
Algorithm
the
of
the
resin
¼
of
the
RUC.
of
the
resin
in
¼
of
the
RUC.
,
V
–
vol.
of
¼
of
the
RUC,
V
–
vol.
of
the
yarns
in
¼
of
the
RUC,
pacing
only
if
–
vol.
β
=
45
V
,
Input
parameters:
n,
d
,
p
,
a,
V
,
tol,
imax.
Output
parameters:
t,
L
,
β
,
a
– newmodels
yarn (not only
t
for
more
complicated
.
.mations
Calculate
the
the
measure
ε
=
(
β
)
−
(
β
)
yarn d
cdifference in the iterates using
r
f
f
u
c
r
new
o
c
i
c
i
−
1
in theimax.
iterative
at of
the
the ––new
ers: n, d(13)
pare
a,used
V
t,¼beginning
Lof
β c ,RUC,
t r ,ofaVnew
, V – ovol.
vol. yarn
of the yarns in ¼ with
of thea RUC,
spacing
if Output
–
vol.
β c =procedure
45parameters:
V
f ,ture
f , tol, only
d , parameters
u , the
yarn
r
centreline
h. Ifofε¼>oftol,
then
increase
the ofindex
f. the
of the
RUC.spacing
If resin
then
increase
thecross-sectional
index
ε > tolin, ¼
i→
V – vol.
the
RUC,
the yarns
¼ sinusoidal
of the RUC,
only
if iβ+c 1area
=. 45 A, of
Vr – vol.of the yarn).
2 . Set
.Onin
yarn
1.
Compute
the
A
(π d 2f V
)=r /(–(4π–vol.
pdvol.
.
of
yarn
Compute
cross-sectional
area
AnVyarn
) /(4 phrd )=.1Set
hrthe
= 1basis
– vol.
ofof¼1.
of
then,RUC,
Va,yarn
vol.
of thethe
yarns
infrom
¼+ of
the=from
RUC,
f β c = 45o , V
the
resin
the
RUC.
Input
parameters:
dinf , p¼the
V, –tol,
imax.
df )n
of this model, certain geoiASTOP
→
ithe
1.
d,of
2a,
else,
if
,
then
and
print
a
non-convergence
If i ≤ imax , then go toofstep
>
i
imax
thet,crimp
resin
ofequations
thenew
RUC.(13), compute
Output
parameters:
LA
βcin
,angle
tr¼
, crimp
ayarn
From
theparameters
small
initial
values
ofvalues
the thickness
t and length
metrical
parameters
imax,
then
go to
step
2a,
else,
ifof the
u, small
new –from
From
the
angleAequations
compute
the
initial
thickness
t and lengthwere obtained.
¼Compute
of the
the RUC.
of
the
the
cross-sectional
area
=i. (πIfdi2f ≤n)(13),
/(4 pthe
mine
architecture
message.
d ) . Set hr = 1 .2
.
.
Set
A
of
the
yarn
from
1.
Compute
the
cross-sectional
area
yarn
spacing
only
if
β
=
45°,
V
–
vol.
of
A
=
(
π
d
n
)
/(
4
p
)
h
=
1
.
The
initial
crimp
angle
is
computed
from
eq.
(3).
L
(
β
)
These
parameters
fully characterize the
i
>
imax,
then
STOP
and
print
a
nonc
.
The
initial
crimp
angle
is
computed
from
eq.
(3).
L
(
β
)
f
d
r
u the angle
c 0judged
If ε <the
tolsmall
, thencrimp
fixed
point iteration
is
have
converged
tothickness
theA crimp
cto0initial
2angle
From
equations
(13),
compute
the
of the
length ) . Set h = 1 .
A values
of
the yarn
from
1.Vuyarn
Compute
the
cross-sectional
area
= (πt dand
2in
f n) /( 4 p d geometry
r of the considered composite
–
vol.
of
the
yarns
¼
of
the
RUC,
convergence
message.
From
the
small
crimp
angle
equations
(13),
compute
the
initial
values
of
the
thickness
t
and
length
.
.
Set
A
of
the
yarn
from
he
cross-sectional
area
A
=
(
π
d
n
)
/(
4
p
)
h
=
1
, danew – new
pd L,βu a,
, )tol,
imax.
t, Lufrom
, fβ c eq.
, t r (3).
yarn
=V(βf cinitial
r
crimp Output
angle (βparameters:
) 0 is computed
c. The
i.
¼
the RUC,
–cvol.
of
the
resin
the
small
crimp
equations
thepoint
initial
values structure.
of the thickness
t and length
The method
has shown that
j. 1in
If
εimax
> of
tol,
thencompute
the
fixed
iterarFrom
.VThe
initial
crimp
computed
from
eq.
(3).
(βangle
iangle
=initial
1into¼
2.ofBegin
the
loop:
forthe
imax
steps
one.
2.Luiteration
Begin
the
iteration
loop:
for
to
in(13),
steps
of
one.
c )i0=is
mall
crimp
angle
equations
(13),
compute
values
of
the
thickness
t and
length
– vol. of ¼ of
of the
the RUC.
RUC, V yarn L– vol.
ofinitial
the yarns
inangle
¼ of the
RUC,
Vr – vol.
.
The
crimp
is
computed
from
eq.
(3).
(
β
)
the
presented
model
is
valid
only for a
tion
is
judged
to
have
converged
to
the
u
c
0
or
and
for
the
crimp
angle
(
β
)
solve
the
quadratic
equation
(11)
for
the
2a.
Set
t
=
h
1
0
or
and
for
the
crimp
angle
(
β
)
solve
the
quadratic
equation
(11)
for
the
2a.
Set
t
=
h
=
1
0
tialthe
crimp
angle (on
computed
from
eq.in(3).
β )the
r
c i −1
rrimax
r
c i −1
ck
constraint
iangle.
= 1 to
for
steps
of one.
0 iscrimp
C. Begin the iterationc loop:
crimp
angle
of
less
than
45°
(β
crimp
angle
β
=
(β
)
.
cof
c ≤ π/4).
i
=
1
c
i
2.
Begin
the
iteration
loop:
for
to
imax
in
steps
one.
roots of
theroots
ratioofLthe
. geometric
two
L a.
β c t≤r π
, then
STOP.
Convergence
ofaratio
the
parameters t, L and t r has
. If2a.0 <Set
(11)been
for the
= 0/ 4or
htwo
r = 1 and for the crimpu angle (β cu ) i −1 solve the quadratic uequation
1. Compute
the
cross-sectional
area
A
of for
Ifequation
the crimp
angle
βc > π/4, the model is
i
=
1
2.
Begin
the
iteration
loop:
to
imax
in
steps
of
one.
or
and
for
the
crimp
angle
(
β
)
solve
the
quadratic
(11)
for
the
2a.
Set
t
=
h
=
1
0
c i −the
1 and
i =the
1 If
teration
loop:
for of
to
imax
inis
ofroot
one.r Lreala root
2 positive
2b.
Ifsteps
no
aCheck
exists,
then
STOP
program
and
writenot
a message.
warning
message.
ed
the
of
the
RUC
no
positive
exists,
STOP
the
program
write
a warning
two
roots
ratio
.restablished.
. for
Ageometry
of
the
yarn
from
naland
area
ALu=
dreal
40p doru) . Set
h1r Land
=u 1then
the2b.
yarn
from
valid because
a cusp forms in the fil3.
constraint
the
crimp
thethe
crimp
angle
(for
βon
2a.a(πroots
Set
=the
hrL= the
f nt)r /(
c )the
i −1 solve the quadratic equation (11) for the
two
of
ratio
a
.
and
for
the
crimp
angle
(
β
)
solve
quadratic
equation
(11)
h
=
1
u
4
/
>
,
then
set
β
=
/
4
and
assume
that
the
specified
value
of
the
yarn
spacing
a
is
π
π
r. =If0 βor
r
c
i
−
1
a
to
obtain
Take
the
smallest
positive
real
root,
and
multiply
the
root
by
the
yarn
spacing
2c.
Take
the
smallest
positive
real
root,
and
multiply
the
root
by
the
yarn
spacing
a
to
obtain
a
c If no positive
c 2c.
h
=
1.
2b.
real
root
L
a
exists,
then
STOP
the
program
and
write
a
warning
message.
ament
in
a
region
closeato the end of the
angle.
r
gle equations (13),
compute theu initial
values
the
thickness
length
two
roots
ofof
the
ratio
a . t and
2b.value
Ifand
no
positive
real
root
Lto
aLLuexists,
then
STOP the program and write a warning message.
oots
of the ratio
Lu for
a .new
L
.
new
value
of
the
undulation
length
.
of
the
undulation
length
u 3f.
t
,
L
,
a
in
steps
from
3c
incorrect.
Solve
t
u
u
u
r
a
to
obtain
a
2c.
Take
the
smallest
positive
real
root,
and
multiply
the
root
by
the
yarn
spacing
le (β c ) 0 is computed from eq. 2b.
(3). If no positive real root L a exists, then STOP the program and write a warning message.
u root,
to obtain
the
positive
real
and
the
by
the
spacing
positive
root
L=u1the
aIf
exists,
then
STOP
the
and
aelse,
warning
TEXTILES
Eastern
Europe
January
/≤December
/ B 2008,
Vol.
16,write
No.
6L(71)
2d.
Ifa 0,length
< Lsmallest
then
go
step
Lu message.
>set
a u,L
then
Lu solve
= a yarn
and
eq.ah(11)
for hr .a
51
LTake
0in2c.
<
then
step
2e,to
else,
if2e,
a ,ifmultiply
then
asetand
eq. solve
(11) for
Lauto
.,program
value
undulation
or
and
solve
quadratic
equation
(11)
for
the
roots
Set t rnew
L
au =
. root
=real
h&2d.
0FIBRES
u go
u ≤the
u >two
r.
r of
2c.
Take
the
smallest
positive
real
root,
and
multiply
the
root
by
the
yarn
spacing
a to obtain a
L
.
new
value
of
the
undulation
length
uspacing
efor
the
smallest
positive
real
root,
and
multiply
the
root
by
the
yarn
a
to
obtain
a
t
from
the
fibre
volume
fraction
equation
(4).
2e.
Determine
a
new
thickness
t from the fibre volume fraction equation (4).
2e.asteps
Determine
a new
i = 1If
to0 positive
imax
in
ofgoone.
2d.
< Lu ≤
, then
to step
2e,thickness
else,
Take
the
real
root
for
the
ratio
of LifuofLathe
u. > a , then set Lu = a and solve eq. (11) for hr .
new
undulation
length
2d.(β If)Lu0. <
Lu ≤value
a , then
go to
step 2e, else,
LLu uthe
>. a , then set Lu = a and solve eq. (11) for hr .
length
and of
forthe
theundulation
crimp angle
solve
the
quadratic
equation
(11)iffor
1value
a)
b)
z/t
undulating region. Using geometrical
considerations, it is possible to obtain
from this model basic mechanical parameters that are very useful for further
strength analysis. Such mechanical considerations and a method for calculating
mechanical parameters will be presented
in a future article. Using the Mathematica environment [11], a special computer
program was developed to calculate the
geometrical parameters.
References
1. Cox B. N., Flanagan G., Handbook of
Analytical Methods for Textile Composites, Nasa Contractor Report 4750,
Hampton, Virginia, March 1997.
2. Hull D., Clyne T. W., An Introduction to
Composite Materials, Cambridge University Press, 1997.
3. Jones R. M., Mechanics of Composite
Materials, McGraw-Hill Book Company,
Washington, DC, 1975.
4. Tsai S. W., Hahn H. T., Introduction to
Composite Materials, Technomic Publishing Co., Westport, CT, 1980.
5. Scida D., Aboura Z., Benzeggagh, M.L.,
Bocherens, E., Composites Science and
Technology, vol. 57, 1997, pp. 17271740.
6. Woo K., Whitcomb J. D., Composites
Science and Technology, vol. 60, 2000,
pp. 693-704.
7. Lomov S. V., Gusakov A. V., Huysmans
G., Prodromou A., Verpoest I., Composites Science and Technology, vol. 60,
2000, pp. 2083-2095.
8. Jiang Y., Tabiei A., Simitses G. J., Composites Science and Technology, vol. 60,
2000, pp. 1825-1833.
9. Osada T., Nakai A., Hamada H., Composite Structures, vol. 61, 2003, pp. 333339.
10. Turfaoui M., Akesbi S., Colloids and
Surfaces A: Physicochemical and Engineering Aspects, vols. 187-188, 2001,
pp. 439-448.
11. Wolfram Research Inc., Mathematica,
Version 3.0, Champaign, Illinois, 1996.
12. Milašius V., Some Problems in Technical Design of Woven Textiles, Fibres &
Textiles in Eastern Europe, vol. 5, 2(17),
1997, pp. 40-43.
13. Milašius V., Woven Fabrics CrossSection: Problems, Theory, and Experimental Data, Fibres & Textiles in
Eastern Europe, vol. 6, 4(23), 1998,
pp. 48-50.
14. Kobza W., Equilibrium of the Bent Soft
Filled Laminate, Fibres & Textiles in
Eastern Europe, vol. 9, 2(33), 2001, pp.
42-45.
Received 2.04.2007
52
Reviewed 2.04.2008
FIBRES & TEXTILES in Eastern Europe January / December / B 2008, Vol. 16, No. 6 (71)