Bogdan Supeł,
Zbigniew Mikołajczyk
Department for Knitting Technology
and Structure of Knitted Fabrics
Technical University of Łódź
ul. Żeromskiego, 90-543 Łódź, Poland
E-mail: katdziew@p.lodz.pl
Model of the Compressing Process
of a One- and Two-Side Fastened
Connector
of the connector may be related to a loose knit
the knitted fabric’s outside layers.
of a 3D ofDistance
Knitted
the connector may
be related Fabric
to a loose knitted fabric with a small cover fact
the knitted fabric’s outside layers.
Physical model of the compressing process
Abstract
A model of the compressing process of a 3D distance knitted fabric was related to a
Physical
model
of the
process
single connector
fastened
on one
sidecompressing
by an articulated
joint,
as well physical
as fastened
on of compr
Below are
presented
models
both sides by fixed joints, considering the connector
as
a
slender
rod
with
an
assumed
shape. and
The mathematical
general assumption
accepted
the fa
shape. The compressing
an elastic physical
rod in themodels
physical
modelselastic
was rod with
Below areofpresented
of compressing
a slender,
both
sides
fixedfabric
[2, 3,as4,well
5, 6,
7].
Figure 1 p
based on assumptions
considering
the
morphology
of
the
knitted
as
the
shape. The general assumption accepted the fastening of the rods as one side art
mechanical properties of threads – monofilaments
placed in theg0internal
layer
of the
designations:
– initial
fabric
thickness,
i
sides fixed
[2, and
3, 4,algorithms
5, 6, 7].
Figure
1 presents
a physical
model
with gth
knitted fabric. both
Calculation
methods
were
developed
for the
determinadeflection
(deformation)
of
the rod, p1, p2 – pl
designations:
g0 – initial
fabric
thickness,
g – fabric
thickness
during defle
tion of the functional
dependencies
between
the difference
force
compressing
the knitted
fabric
in ithe positions
of the
rod in the lon
and deflection,deflection
as well as(deformation)
the description
curves
the shapethe
of knitted
the
fabric’
ofofthetherod,
p1, prepresenting
2 – planes including
the bentthe
rod,rod,
S1 –which
transversal
force of the reactio
compressed rod.
A computer
simulation
connects
difference
in the
positionsofofcompressing
the rod in the
longitudinal
direction both
(y), P1 – force
outside layers of the knitted fabric, was carried out with the use of the ‘Mathematica’
theinto
bent
rod, S1 – transversal
forceparameters
of the reaction
the base.
program, taking
consideration
the variable
of theofmodel.
The considerations carried out in this article are significantly based on our previous publication,
and therefore the assumptions of the physical model and detailed descriptions of the
analysis are not included.
Key words: knitted 3D fabrics, distance fabrics, compression process, connector, articulated joint, mathematical model, physical model.
Introduction
The work presented herein is a continuation and broadening of the theoretical
considerations of compressing a single
connector joining the two outside planes
of a distance knitted fabric carried out
in our previous work. The models described below selectively concern the
following different fastening variants:
two-side articulated joints and one-side
and two-side fixed joints. Considering
the particular cases, we assumed that
no mutual displacement of the outside
planes take place. However, praxis indicated that, in the case of some stitches
of the knitted distance fabric, the outside
planes displace mutually during compression.
The fastening variants of the connector
we assumed were not randomly chosen.
The model of articulated fastenings is
related to empirically stated shapes of a
compressed monofilament in the 3D system.
The model presented in this paper considering the two-side fastening of the
connector by fixed joints is related to a
knitted fabric with a compact structure
of the outside layers, which are sometimes even laminated. On the other hand,
the system with a one-side articulated
joint of the connector may be related to
a loose knitted fabric with a small cover
factor of one of the knitted fabric’s outside layers.
44
Physical model of the
compressing process
Mathematical models
of the compressing process of
aFigure
single1.connector
Scheme of physical models of the co
Below are presented physical models of
one-side
fixed rod,
B – model
of aalboth-sides fix
compressing a Figure
slender,1.
elastic
rod with
an The
above-presented
physical
models
Scheme
of physical
models
of the compressed
and
bent elastic
rod; A
assumed shape.one-side
The general
assumption
low
us
to
formulate
differential
equations
fixed rod, B – model of a both-sides fixed rod.
accepted the fastening of the rods as one of the fourth order that connect the loadMathematical
of the compressing
pro
side articulated and both sides fixed [2-7]. ing of
the rod with models
its deformations.
We
Figure 1 presents
a physical model
the rod
(connector)
discussed
as
Mathematical
modelswith
of the consider
compressing
process
of a single
connector
above-presented
models allow
the following designations: g0 – initial two The
separate
rods formed physical
by its projecfourth
order that
of the ro
fabric thickness,
gi –above-presented
fabric thickness duron the allow
planes
0xztoconnect
and
0yz.the loading
The
physicaltion
models
us
formulate
differential equa
(connector) discussed as two separate rods form
ing deflection, fourth
∆g = deflection
(deformaorder that connect the loading of the rod with its deformations. We cons
including
tion) of the rod,
p1, p2 – planes
The differential
equation
describes
(connector)
discussed
as two separate
rods formed
by itsthat
projection
on the planes
The differential
equation
that describes
the de
the knitted fabric’s loops, y0 – difference the deformations
of the
rod influenced
by
has
the
following
form:
in the positionsThe
of the
rod in theequation
longitu- thatthe
loads has
following form:
differential
describes
thethedeformations
of the rod influenced
dinal directionhas
(y),the
P1 following
– force compressform:
(1) y ( x) = y0 (
EJy′′( x) = − P1 y′( x) + S1 x where
ing the bent rod, S1 – transversal force of
the reaction of the base.
where: y ( x) = y0 ( x) + y1 ( x)
EJy′′( x) = − P1 y′( x) + S1 x where
where:
y0 (x) and y1 (x) are the initial buckling,
where:
force P1 action, respectively;
y0 (x) and y1 (x) are the initial buckling, and the buckling occurring as a
S1 – is the force of reaction of the transvers
force P1 action, respectively;
E – is the Young modulus, and
S1 – is the force of reaction of the transversal, sliding support,
J – is the modulus of inertia of the bent obje
E – is the Young modulus, and
J – is the modulus of inertia of the bent object’s cross section.
Differentiating twice the equation (1) toward ‘x
Differentiating twice the equation (1) toward ‘x’, we obtain subsequently:
Figure 1. Scheme of physical models of the compressed and bent elastic rod; A – model of
a one-side fixed rod, B – model of a both-sides fixed rod.
Supeł B., Mikołajczyk Z.; Model of the Compressing Process of a One- and Two-Side Fastened Connector of a 3D Distance Knitted Fabric.
FIBRES & TEXTILES in Eastern Europe January / December / B 2008, Vol. 16, No. 6 (71) pp. 44-48.
where:
y0 (x) and y1 (x) are the initial buckling,
and the buckling occurring as a result of
the force P1 action, respectively;
S1 – is the force of reaction of the transversal, sliding support,
E – is the Young modulus, and
J – is the modulus of inertia of the bent
object’s cross section.
For model A, we find the dependency
connecting force P1 with deformation
∆g in the form of an implicit function
f(P1, ∆g) = lp by substituting the integration constants C11, C21, C31, and C41 determined from the equation system (3) into
the condition of the length of the bent
rod, and finally obtain (Equation 5).
Similarly as for model A, for model B
we find the dependency connecting force
P1 with deformation ∆g in the form of an Figure 2. Mechanical characteristics P1
implicit function f(P1, ∆g) = lp by substi- = f(∆g) of a compressed monofilament for
P1Py1 ′′y(′′x( )x)
P1Py1 ′y(′x( )x)−−S1S1P y′( x)IV−
three variants: a) two articulated fastenings,
IV S
tutingythe
C11, C21,(1.a)
where:
where:
(1.a)
and
and
y( x(P)x1)=yintegration
=′′y(0yx
(0 x())x)++y1y(1xconstants
( )x). .
y′′′y′′′( x( x) )==
IV
b)
and one fixed fastening,
where:
(1.a)
y
x
y
x
y (articulated
x) .
(
)
=
(
) +one
=
y
x
(
)
y′′′(EJEJ
x) = 1and yy ( x( )x1)== and
EJ
EJ
0
31
41 determined from the equa′′ , and CEJ
PC
P1 y′( xEJ
) − S1
c) two1 fixed fastenings.
IV
1 y ( x)
P1 y′′( x)
where:
.
(1.a)
and
′′′
y
x
y
x
y
x
(
)
=
(
)
+
(
)
=
y
x
y
x
=
(
)
(
)
0
1
tion
where: y ( x) = y0 ( x) +EJ
y1 ( x) ..
where:
(1.a) (1.a)
=
EJ system (4) into the condition of the
Equation(1.a)
(1.a)finally
finally
takes
takesthe
theform:
form:
length
of
the
bent
rod,
and
finally
obtain
′′
y′( x) −EJ
SEquation
(
)
P
y
x
′
′′
) − S1 y ( x) = yIV ( x) + yP1( yx) (.x)
1
andIVIVy IV (yx′′′) (=2x2) 1= P1 y ( xwhere:
and
where:
y ((1.a)
x) = y0 ( x) + y1 ( x) .
01( x ) = 1
2takes
2 yP1P
the number(1.a)
of connectors, and for the force
(Equation
6).
(1.a)
finally
the
form:
′′
′′
EJ
EJ
.
.
where:
where:
k
k
=
=
yyEquation
( x( x)Equation
)++k1k(1.a)
y
y
(
x
(
x
)
)
=
=
0
0
Equation
finally
takes
the
form:
(1.a)EJ
finally takes
1 1 the form:
1
EJ
m:
EJ
EJ
obtained, the mechanical characteristic
P
P
2
2
1 2
1 of:
P1The
2
. the
where:
kwritten
y IV((xx)of
) of
++equation
kequation
=x
The
general
integral
integral
(1)
(1)
may
bewritten
the
form
2form
presents
. of:the mechanical charac- P1 = f(∆g) should be designed.
where:
kinin
y IV
k1 1y2′′y( x′′)(takes
)0 =may
0 be
1 = Figure
finally
takes
the
form:
. general
1 =
1 =
EJ
Equation
(1.a)
finally
the
form:
EJ
.
.
(2)
y
y
(
x
(
x
)
)
=
=
C
C
+
+
C
C
x
x
+
+
C
C
sin(
sin(
k
k
x
x
)
)
+
+
C
C
cos(
cos(
k
k
x
x
)
)
EJ
teristics
of
the
compressed monofilament (2)
11112
21P
21
3131
11
4141
2
1
The
integral
of equation
(1)121mayP1be written in the form of:
.of:
y′′( xbe
) = written
0 where:
k1IVgeneral
=form
2
)1 may
in
the
The data of comparable results obtained
for
the
variants
A
and
B
[9],
as
well
as
the
′′
where:
.
y y ((xx)EJ
)=+Ck1integral
y ( x) = 0of equation
k1 = (1)
The general
(2)
cos(may
k1 x) . be written in the form of:
11 + C21 x + C31 sin( k1 x ) + C41
EJ
.
(2)
+tegral
C41 cos(
k
x
)
by the mathematical simulation and by
variant
described
in
the
previous
article
We
We
determine
determine
the
the
integration
integration
constants
constants
C
C
,
,
C
C
,
,
C
C
,
,
and
and
C
C
for
for
model
model
A
A
in
in
Figure
Figure
1
1
from
from
1
11
11
21
21
31
31
41
41
of equation (1)
may
beCwritten
inxthe
form
of: k x ) + C cos( k x ) .
(2)
y
(
x
)
=
+
C
+
C
sin(
11
21
31
1
41
1
The
general
integral
of
equation
(1)
may
be
written
in
the
form
of:
following
following
boundary
conditions:
conditions:
The
of
(1) may [8] – the rod connected
by two artificial experimental tests obtained with the use
. integration
(2)
+ C21 x the
+the
C31
sin(kgeneral
x) boundary
+determine
Cintegral
x) Equation
1We
41 cos( k1 the
C11, C∆21
from(2) designed measuring stand
(yx0→
)→=the
C
+C21C21
++CC31A31sin(
sin(
k1∆constants
cos(
xg), ).=C=31
ybe
y(21∆
gC
yand
CC
kFigure
gxg)))+++
y0y,,0and
(∆
) )=
+form
+C
∆21
∆gxg+(Equation
sin(
, C41 for model A in Figure
in
of
of a 1specially
joints.
tants C11, C
,gwritten
for
model
in
1CCC
from
31=,y
4111
11
1∆
41
1g
11
1k2).
41
41cos( k1k1∆
the 0following
boundary
conditions:
are not included in this article as the
==0determine
0→
→CC
+=
+11
sin(
)+for
+CC41model
cos(
cos(
)0C)==
0,C
0, 11k, 1∆Cgfrom
yy( g(We
g0 )0 )constants
CC,2121Cg→
g0the
CC
k1kgC
g41
k1kA
g0in
constants
the integration
,C+C
,sin(
and∆
Figure
1111C
0+
31
1g+
0 )0C
41
1g+
21=, yC
31, and C41 for model A in Figure 1 from
21
31integration
0integration
11 + C21constants
31 sin( k
1 ∆g )curves
41 cos(
1 a ) strongly
0,
(3)
(3)
C41We
k1 y∆(g∆)g=)the
y0 y,the
sin(k1 ∆g ) + We
cos(
determine
stand
The
have
non-linear
determine
integration
constants
C
,
C
,
C
,
and
C
for
model
A
in
Figureis 1actually
from undergoing further in11
21
boundary conditions:
cos(
)0 )==cos(
→)C=C2121
+→
+Cboundary
−CC41sin(
yy′(′g(the
g0 )0 )==following
tgtgβ
C31C31k1k1cos(
k1kg1gg0+)conditions:
k1ksin(
k1k)1g+
g0C
tgtgββ,k,31g ) = 0, 41
β(→
0 )−
41
1ksin(
0
+
y
g
C
C
g
C
,
C
,
C
,
and
C
for
model
A
in
Figvestigation,
but they will be the subject
character
(a
great
increase
in
the
bend11
21 0
31
1 0
41
1 0
11
31 0 ∆g )
41
n(→
k1 Cg110 )++CC2141∆cos(
k211following
gsin(
boundary
conditions:
+C
gthe
0 ) =k0,
(3)
2+2 C41 cos( k
1
1 ∆g ) = 2y20 ,
(3)
=31=
→
−
−
∆
∆
−
−
∆
∆
=
=
(∆g1g) )from
0(0y→
cos(
cos(
)
)
sin(
sin(
)
)
0.
0.
yure
y′′(′′∆
C
C
k
k
k
k
g
g
C
C
k
k
k
k
g
g
y
g
y
C
C
g
C
k
g
C
k
g
y
∆
)
=
→
+
∆
+
sin(
∆
)
+
cos(
∆
)
=
,
the
following
boundary
conof
a
subsequent
publication.
at
the
greatest
deflections
of
ing
force
P
′
31
31
1
1
1
1
41
41
1
1
1
1
cos(k1 g 0 ) −
→ C+21C
+ C∆31gk1+21
β0 C
β,
11
31C41k1 sin(
1 1 k1 g 0 ) = tg41
1
0
(∆g( g)g0β
=) ),=y+0tgC→
) −+CC4121k1gsin(
k131gysin(
11
21) = 0, C31 sin( k1 ∆g ) + C41 cos( k1 ∆g ) = y0 ,
→
C
1 gC
0 11
0 ) =ktg
0 +
1 0we
41 cos( k1 gthe
ditions
(Equation
3).
the
rod).
The
mechanical
characteristics
On
Onthe
the
other
other
hand,
hand,
we
determine
determine
integration
constants
constants
C
C
,
,
C
C
,
,
C
C
,
,
and
and
C
C
for
for
model
model
B
B
20theintegration
2
11
11
21
21
31
31
41
41
P1ksin(
y′( x)k−k S∆
P
IV (3)
1 g ))=
2
→ −C
) =→
cos(
y′′()0∆)=g=
C11
kC1g∆y0′′′gsin(
00 →
+
0,
C
k11y′′g( x0)) =towhere:
31gk+
1g
41
+
+f(∆g)
) =(cos(
yy
C
kC
g 01=)sin(
C1411 gcos(
k10.aC
gysimilar
(1.a)
( x) =rod
y0 (described
x) + y1 ( x) .by model B is the
=conditions:
()x−)C
x0,
) =character
sin(
)11=from
0.
21
0 and
)the
→41kC1in
+Figure
−0following
=0 boundary
Ck311∆k1gcos(
k((1gg
g0 0the
Cfollowing
k1C
tg
β− C
β ,31P+
11 +
21
131
041
least
P
have
the yThe
Figure
from
set
set
ofof
boundary
conditions:
21in
41k1 sin(
0)
′
′′
1S1 EJ
−
(
)
y
x
(
)
P
y
x
EJ
IV
(3)
(3)
1
1
′
′′
−
(
)
P
y
x
S
(
)
P
y
x
On
the
other
hand,
we
determine
the
integration
constants
C
,
C
,
C
,
and
C
for
model
B
IV
and
where:
.
(1.a)
′′′
(
)
=
(
)
+
(
)
y
x
y
x
y
x
y
x
=
=
(
)
y
x
(
)
11
21
31
41
1
1
1
2 yOn
2C
0susceptible
1
the
other
hand,
we
determine
the
into
load,
whereas
the
model
characteristics
obtained
empirically
[8].
y
(0)
(0)
=
=
0
0
→
→
C
C
+
+
C
=
=
0,
0,
and
where:
.
(1.a)
′′′
he
integration
constants
C
,
C
,
C
,
and
C
for
model
B
′
(
)
=
(
)
+
(
)
y
x
y
x
y
x
y
x
=
=
(
)
y
x
(
)
=
→
+
−
=
(
)
cos(
)
sin(
)
,
y
g
tg
C
C
k
k
g
C
k
k
g
tg
β
β
11
21
31
41
11
11
41
41
′
0
1
)k′ 1EJ
cos(
) − C1 410k1 sin(
→ −C31k1 cos(k1 ∆
=C
) −( g
)21following
0.21
gyFigure
C
g→
0 41
31+1 C31
EJ k1 g 0 ) = tg β ,
1∆
) −tgSk1β
P
x=sin(
)0 k141g10conditions:
P1kyEJ
xboundary
1′′1(EJ
from
the
set
of
10y (1
constants
C+C
, 2C
,(Equation
described
Notwithstanding
and
where:
(1.a) P yin
′′′in
of boundaryytegration
( xform:
) + ythe
ytakes
xy),=, the
y0that
x)y.′mathemati( xy)y =→
y IV
xkand
)k =
21C
31
2finally
1 (P
y(conditions:
=
+00+C
sin(
)g
)+41)EJ
+21
cos(
)i )41
=∆
=(gyfor
g( gi )yi )=
C)C
C21C21C11
g−g,i C
g∆
gi,C
C
C
ksin(
′′( xthe
(1.a)
) previous article is the
and, we determine
integration
constants
C
,
C
,
and
IV
0the
11
11
3131sin(
1 111
41
41cos(
1k1gg
0)0=model
i+
iC
iC
′′0(y∆→
1 ( x ) − S1
=
→
−
cos(
0.
y
g
k
k
C
k
k
31
2
2
(0)
=
→
+
C
=
0,
EJ
31 141 the follow1
41 1 analysis
1 carriedyout
and(4)
where:
′′′(∆
y ( x) = for
y0 (model
x) + y1 ( x) .
xB
=) =
)g
= 1
ymost
( x)susceptible.
(4)
for model
B∆ingFigure
111 from
For example,
cal
concerns
a
single
′′
=
→
−
∆
−
(
)
0
cos(
)
sin(
0.
y
C
k
k
g
C
k
k
Equation
(1.a)
finally
takes
the
form:
P
31
41 1
1 2
m the following
conditions:
EJ
2
′(0)
11EJ
) −, Sand
P21
y,.′(Cx31
P1 y′′B( x)
, form:
, 1 + CPy1 yIVthe
yEquation
y′(0)
tgof
C
C=
kC
tg
αof
αy→
α
α
=set
=tg
→
+y−+C
(1.a)
the
′finally
′′( x1x))integration
IV model
)hand,
Pboundary
x21
SC
1
31takes
31
1kwe
1==tg
′′
On
the
other
determine
constants
C
,
C
C
for
where:
+
k
y
(
x
)
=
0
k
=
IV g (Equa(
→
+
sin(
)
+
cos(
)
=
,
C
k
g
C
k
g
y
1gyi )( 21
1
11
41
ing
set
boundary
conditions
B
a
force
P
=
20
cN
causes
deformation
monofilament,
thanks
to
the
assumption
and
where:
′′′
(
)
=
y
x
y
1
1
y
x
=
(
)
=
y
x
(
)
11
41y ( x ) = 1y i( x ) P
i
and
.
(1.a)
+1 y01 ( x)EJ
y′′′( xk)1 =g(1.a)
y 21form:
(IVxi) = 31P2 1 where:
0 ( x ) + y1 ( x ) .
1
+ =CEquation
= finally
y0 , 0 hand,
→in(
C11k1+gC
0,cos(
20
i )41
i ) other
(4)
takes
the
the
we
determine
the
integration
constants
,
C
,
and
C
for
model
Bthe
EJ C11, C∆g
EJ whereas
IV=
2EJ
2−)set
′′
′
′′
21
31
41
1(kx
where:
.
−
(
)
y
(
x
+
k
y
(
x
)
=
0
P
y
x
S
)
k
=
P
y
EJ
in
Figure
1
from
the
following
of
boundary
conditions:
)=
cos(
cos(
)
)
sin(
sin(
)
)
.
.
β
β
β
β
→
→
+
+
−
=
=
ytion
y′41(′g(On
gi )′′′iy4).
tg
tg
C
C
C
C
k
k
k
k
g
g
C
C
k
k
k
g
g
tg
tg
(4)
=
7.2
mm,
for
model
A
that
we
have
made
concerning
the
physical
IV
1
1
1
1
1
i
i
i
i
21
21
31
31
1
1
1
1
41
41
1
1
1
1
′′
x=y) ′+(0)
k1 =ytg( α
x) →
= and
0Cwhere:
k)11== =tg
αP, . integral
where: of
(1.a)
) =Equation
+ y(1.a)
( x) . finally
y ( xequation
y0 ( x)EJ
y C( x)(sin(
()xkThe
takes
form:
21 +yC31
general
(1)
be written
inthe
the
form of:
1may
,→ C11 + C21 giin
+
)
+
cos(
=
,
k
g
C
k
g
y
IV
2
2
1
EJ
y((0)
=k1of
0EJ
→
+0the
Cwhere:
1the
following
set
of
boundary
31
41
0= EJ
ithe
i0,
deformation
for
it31,is,and
possible
toconditions:
obtainfrom
afrom
generalised
The
Thedescriptions
descriptions
integration
constants
constants
CC11.11
, model,
,CC2121
, equation
,CC31
and
CC4141may
obtained
obtained
the
the
equation
equation
′′from
yFigure
xy) ′+(of
( integration
xC
) 11
=→
411=
1 y tg
P1 will be ∆g = 7.8 mm, and
general
(1)
written
the
form the
of: form:
cos(
sin(
β)be
=kof:
gyifinally
CThe
C31kk11 may
CC
gsin(
(2)
(yEJ
x′′k)(be
=g) iwritten
C)11−of
+
+
Ck311form
x(4)
+.x)C
kin
) .0takes
yintegral
IV
2 finally
2
(1.a)
i )Equation
21 + form:
1x
41k
1x
21
1tg
41kcos(
1 )x=
(1.a)
takes
the
The
general
integral
of
equation
in
the
′β()x.=) of
− Sβin
P
P
sin(
)
=
gi )C−21system
Csystem
kEquation
kdescriptions
gnot
tg
′′
where:
.
y
(
+
y
(
x
k
=
The
the
integration
conIV (1)
the
model
with
articulated
joints
at
both
characteristic
for
the
whole
knitted
fabric
are
not
presented
presented
in
this
this
paper,
paper,
taking
taking
into
into
consideration
consideration
their
their
very
very
complex
complex
shape.
shape.
,
αk1→
α
+41C
=
k1are
tg
1
1
1
i
131
1
1
1
→
+
C
and
(1.a)
==y00 →
+
sin(
cos(
=yy1C(0kx,1) x.obtained
gi )integral
CC
Cyintegration
CC0,
g+i )C+31Cin
Csin(
g( ,xi41)and
+cos(
y, (C
x) x=,)kC
y+10 C
y′′′(The
xy)y(=(0)
=
(2)
(g41
xi )+=
=may
+be
Ck21
).2
1111
1xwhere:
P1 mm.
of
the
constants
from the
equation
The
equation
(1)
written
thek121form
2 EJ
1141
31of:
(2)
(IVxC,)11
=,31descriptions
C11and
C31C21, 41and
xof
+obtained
C
sin(
kyfrom
C314111P
kwith
ygeneral
Equation
takes
the21
form:
2+
EJ
EJ
1cos(
,finally
C
C3141
obtained
from
sides,
∆g
= 8.2
(4)
1 x ) 2+the
1 x ) .the additional assumption
equation
′′( xcon.
11,kC
21
y IV ( x41integral
) +that
k1 yall
)equation
= 0 where:
kmay
′′not
where:
.
(yC
x′(1.a)
)1C
+g, 21
y
(
x
)
=
0
k
=
)are
sin(
)
.
βconstants
β
→ C21 +CCstants
=
k(0)
C
k
k
g
tg
1
1−
1
ik=
i
31 1ycos(
41
1
1
The
general
of
(1)
be
written
in
the
form
system
presented
in
this
paper,
taking
into
consideration
their
very
complex
shape.
We
determine
integration
C11
,form.
C31,, of
and
A in Figure
1 of:
from
, Cknectors
C
Ck+
kC
tg
+complex
=
α
. Phave
x)y=we
+tgCα21their
x→
+
C
+connecting
C
cos(
xforce
)the
y (IVA,
21
EJ
P
21
31
1)connecting
(we
)find
→
sin(
+same
cos(∆g
)Cthe
=,form
gC system
Csin(
gshape.
k1 11with
gthe
Cconstants
k∆g
yform
EJ
the
equation
are
not
in
the
In
31
1 x
41
1force
per, taking
into
consideration
For
Formodel
model
dependency
dependency
P
with
deformation
inin
of(2)C41 for model
2 = the
221
11presented
31
41geometrical
1 g
0
i 1+obtained
i )deformation
i the
′′ ythe
where:
. integration
y (A,
x)constants
+ 11ikfind
)0 =11
0very
kand
ns of the integration
,
C
C
from
equation
.
(
x
)
=
C
+
C
x
+
C
sin(
k
x
)
+
C
cos(
k
x
)
y
21, Cdetermine
31, the
41the
1 y ( xC
1 =
following
boundary
conditions:
We
constants
C
,
C
,
C
,
and
C
for
model
A
in
Figure
1
from
The
general
integral
of
(1)
may
be
in
the
form
of:
11force
31 ,Figure
41
11 21
21
31equation
1
41
1 written
(4)
Equation
(1.a)
finally
takes
the
form:
The
integral
of
equation
(1)
may
be
written
in
the
form
of:
We
determine
the
integration
constants
C
,
C
,
C
,
and
C
for
model
A
in
1
from
thisgeneral
paper,
taking
into
consideration
their
The
mechanical
characteristics
deterorder
to
do
this,
only
the
compress′
,
,
∆g)
∆g)
=
=
l
l
by
by
substituting
substituting
the
the
integration
integration
constants
constants
C
C
,
,
C
C
,
C
C
,
,
and
and
an
an
implicit
implicit
function
function
f(P
f(P
EJk1 11
) −21Ccomplex
y ( gi )taking
givery
= tg β1 into
1→ C21 +p C
p 31k1 cos(
1111 2121 3131
i ) = tg β .
41k31
1 sin( kshape.
1 g41
presented in thisFor
paper,
consideration
the
following
boundary
conditions:
A,
we
find
the
dependency
connecting
force
P+of:
with
deformation
∆g
in
the
of41the
(0)
,
y=′model
tg
C
C
k
=
α
→
+
α
(their
∆
)11into
=ktg
→
+and
∆
sin(
)the
+the
cos(
∆
=x)y+
y(3)
g=
yxof
C
Cform
g41
C
kbe
gof
CFigure
k1sin(
gk)1form
1for
.
x
)
=
C
+
C
x
+
C
cos(
k
x
)
y
021
11,the
21
31(
1∆
41
1mined
0 ,C
P
21
31
1
.
(2)
the
following
boundary
conditions:
(
x
)
C
+
C
x
+
C
sin(
k
x
)
+
C
cos(
)
y
We
determine
the
integration
constants
C
,
C
,
C
C
model
A
in
from
very
complex
shape.
are
basis
for
further
more
ing
the
whole
fabric
should
divided
by
ncy connecting
force
P
with
deformation
∆g
in
the
form
11
21
31
1
CC4141The
determined
determined
from
from
the
the
equation
equation
system
system
(3)
into
the
the
condition
condition
of
of
the
the
length
length
of
bent
bent
rod,
rod,
IV
2
2
general
integral
of
equation
(1)
may
be
written
in
31
1
1
41 constants
1
The
integration
C11g,+C
,integration
C31k, and
the31equation
where:
. substituting
y an(descriptions
ximplicit
) +111k1 y′′21(function
x) =of031the
41 obtained
(∆g)
∆k,g1sin(
)=
→
+C
sin(
∆g )k+Cconstants
), =C
Cg21
kcos(
y21constants
,C
by
the
Cg11kfrom
, andC11, C21, C31, and C41 for model A in F
We
0C
21∆
41
1∆
0,,=C0,
the
following
boundary
conditions:
(k,lyp1gcos(
)g=,)kC
011C
→
+kC
+ yCdetermine
) +cos(
g31
gC0the
Cintegration
)′(C
=g
→
∆f(P
∆
+
cos(
∆
)0 =
, 1sin(
y0+
gCk+1yC
and
andfinally
finally
by find
substituting
and
.C
((x∆obtain:
)gobtain:
=integration
C=21Ctg
xpresented
++
CC31
)+
+C
C=21ypaper,
yythe
0 31
11into
21
1 their
41
1 g 0 )(2)
11
11constants
21sin(
31
41
1−
031
11connecting
1Px
41EJ
1 x )k
β
β
)
cos(
)
sin(
)
.
y
C
k
g
C
k
k
g
tg
→
=
system
are
not
in
this
taking
consideration
very
complex
shape.
we
the dependency
force
with
deformation
∆g
in
the
form
of
1
i
i
i
21
31
1
1
41
1
1
C
determined
from
the
equation
system
(3)
into
the
condition
of
the
length
of
the
bent
rod, C , C , C , and C (3)
the
following
boundary
conditions:
41
((1)
)the
= 0′kbent
+
cos(
= 0,
yC
gof
C
g31kC
C∆form
kfor
gintegration
∆
)==integral
→CC
∆g+ +C
∆)11
+C+C
cos(
)sin(
C21length
gCwritten
kgand
gC
yof:
constants
system (3)We
intodetermine
of
The
general
of+the
equation
may
be
in
11
21
31
41 for model A in
0,the
0model
0 )k
integration
constants
C
,)11rod,
, 21C
inkk11Figure
1 from
0the
11
310 sin(
1→
41
01 ,gdetermine
→
cos(
sin(
kWe
k)1+gC0 )41−A
C41the
((and
0ysubstituting
→
cos(
=C
0,=
yythe
gl0gp)condition
21β
41
0 )+=Ctg
1 g 0 ) = tg β ,
11 + C21 g
0 integration
31 sin( ky1 (gg
0constants
41
1 21
, ∆g)
=
by
the
C
,g10C+)31
and
nction f(P
11
21,31C131y,( ∆
finally
obtain:
n n1
)
=
→
+
∆
+
sin(
+ C41 cos(
) = y0 ,
g
y
C
C
g
C
k1 ∆g ) from
k1 ∆equation
g(3)
(3)
The
descriptions
of
the
integration
constants
C
,
C
,
C
,
and
C
obtained
the
g
g
g
g
g
g
g
g
g
g
g
.
.
(5)
(5)
31
the
following
conditions:
23111
11
21
41
iy(
i (x
i+
i C
i itg
i i cos(
i 0
i 22boundary
i i2 )2 = 21
.
(2)
)
=
C
+
C
x
+
C
sin(
k
x
)
+
C
cos(
k
x
)
y
the
following
boundary
conditions:
′
(
)
)
sin(
,
=
→
+
−
y
g
C
C
k
k
g
C
k
k
g
tg
β
β
)
=
0
→
+
sin(
)
+
cos(
)
=
0,
g
C
g
C
k
g
C
k
g
2
We
determine
the
integration
constants
C
,
C
,
C
,
and
C
for
model
A
in
Figure
1
from
C
C
C
C
k
k
g
g
j
j
k
k
g
g
j
j
C
C
k
k
g
g
j
j
k
k
g
g
j
j
l
l
{
{
+
+
[
sin(
[
sin(
(
∆
(
∆
+
+
))
))
−
−
sin(
sin(
(
∆
(
∆
+
+
(
(
−
1)
−
1)
)]
)]
+
+
[
cos(
[
cos(
(
∆
(
∆
+
+
))
))
−
−
cos(
cos(
(
∆
(
∆
+
+
(
(
−
1)
−
1)
))]}
))]}
+
+
(
(
)
)
=
=
21
1dependency
0 gy41
31
10 =
1 k0 ∆gp)p =
∑ equation
031)system
11
31 k
1 −
21 21For
1 1 β
10the
1k condition
41
121
1
′( g31model
d from ∑
the
(3)
cos(
sin(
=11tg
+2131Cthe
k1 141length
k−31
g1of
tgbent
β4111 ,1 P0 1rod,
A,→
we
find
force
in+the
form
ofg ) + C cos(k ∆g ) = (2)
∆0 41C
→
) =connecting
021
) −41C1deformation
0.
g11the
gwith
k sin(
1 n n0′′)(of
1C
0n)kthe
ny
nC
ninto
nC
n1 +gC∆g
31
1 cos( k1y∆
j =1j =1
∆Cg31(3)
sin(
Cvery
k1 +∆C
( gy ()∆=g )0n41=n→y1 0C→
+their
sin(
g 0 ) 1= 0, y0 ,
→→
+21presented
∆31gk1+cos(
∆2gC)41+kpaper,
) =β y,into
yyn(′∆
gn 0))== ytg
C11
C+conditions:
C31 sin(
k)1this
Csin(
kntaking
gtg
1121 021
31 k
41 k1 shape.
are
not
in
2
11 + C
1 g 0 )complex
41 cos(
the system
following
0boundary
41 cos(
1)∆=
0g, 0 2 consideration
ain:
(
−
g
C
C
k
g
k
g
221
′′
g iβfunction
g,y
g
g
g
∆
=
→
−
∆
−
∆
=
(
)
0
cos(
)
sin(
)
0.
g
C
k
k
g
C
k
k
g
.
(5)
∆g)
=
l
by
substituting
the
integration
constants
C
,
C
,
C
,
and
an
implicit
f(P
2
2
i
i
i
i
i
0
21
31
1
1
0
41
1
1
0
p
11
21
31
On
the
other
hand,
we
integration
C
,
C
,
C
,
and
C
for model B
31∆
1+
1determine
41)) −
1the
1 g + ( j − 1) constants
′′∆(∑
11
21
31
41
∆
∆
−
=
cos(
)
sin(
)
0.
y
g
k
k
g
C
k
k
g
k
g
j
k
g
j
C
k
g
j
k
l
{)C=21
+ C31−[C
sin(
(
∆
+
))
−
sin(
(
∆
+
(
−
1)
)]
[
cos(
(
∆
+
cos(
(
∆
))]}
+
(
)
=
gi
gi0 →
g
g
.
(5)
2
2
11
41
i
1constants
41∆
1 , cos(
→
+
sin(
) +41C21
∆g1 )C
g)g1)+==j0y→
Cintegration
C
=1 0βA→
+n+CC21311gk10from
+ C31kp1sin(
)=
g) 0=) tg
gC041)k+1 C
C31
,kg1and
inCCFigure
k+1 (31
j −g
l11
+ ( j − 1) )] + CWe
(∆
−C
cos(
∆gC
+21
(∆
1)+
))]}
+ (k1 ik)g
=)12g
sin(
→
g(model
g 0 )k−1rod,
k1 g 0k)1 =g 0tg
β ,0,
nthe
ncos(
41yfor
(k1gas
+model
sin(
+
0,=the
ydetermine
C
gfor
Cnnthe
kdependency
41[ cos(
p ,CC
11
41 cos(
0))
1 0C
41(3)
0y,n′(y
0of
21
0j =determined
11 11
212for
0On
31 31
1into
0) =
C
from
the
equation
system
the
condition
of
the
length
of,ncos(
the
bent
in
Figure
1 from
the
following
set
boundary
conditions:
nSimilarly
nmodel
n1we
other
hand,
we
determine
integration
constants
C
C
,
C
Similarly
for
for
model
A,
A,
model
B
BC
we
find
find
the
the
dependency
connecting
connecting
force
force
P
P
with
41
′′(as
11
21
31, and C41 for model B
∆
=
→
−
∆
−
∆
=
)
0
cos(
)
sin(
)
0.
y
g
C
k
k
g
k
k
g
1 1with
(3)
(3)
On
the
other
hand,
we
determine
the
integration
constants
C
,
C
,
C
,
and
C
for
model
B
31
1
1
41
1
1
11 . (5)
21
31
41 2
boundary
conditions:
2
g i the following
g
g
g
g
2
)g)∆g
==∆g
+C21
sin(
+41gthe
cos(
)1(set
==
C
C
g−k01)C
C
gC2f(P
=0conditions:
→
+C
= tg β ,of
tg
C
C∆g
g 0 )form
βP
i
ik g
i 0,
′(k(1gg(∆00finally
′′=
cos(
)(0)
sin(
)0+41=
→
+21determine
ythe
Cgk31
k+∆1of
k111ik))]}
tg
βinobtain:
β∆g)
=)by
→
−C
(y∆′l(,pglgpC
)0by
y=
C
kfor
k1k∆
gB) −kC1 41gk0 1) −sin(
kk1∆
g ) =kthe
Figure
of
11
0in
31jan
1from
31
1 cos(
41
1 sin(
10.
and
=
0k+function
Cconstants
+g1 f(P
0,
n( k1 ( ∆g +deformation
+
(0
−→
1)
)]
+C
[form
cos(
+an
))
−1kcos(
(∆
(→
− 1)
)∆g)
=, l pboundary
j )) − sin(
jtg
gC
jfollowing
in
the
the
form
implicit
function
,
,
substituting
substituting
the
the
deformation
For
model
A,
we
find
the
dependency
force
with
deformation
in
1 implicit
0y
141
0connecting
31
121cos(
41
1 (of
1
1
in
Figure
1
from
the
following
set
of
boundary
conditions:
On
other
hand,
we
the
integration
C
,
C
,
and
model
11
21
31
41
(3)
n
nas
n for kmodel
n find
n ) =dependency
+ Cmodel
cos(
y (∆Similarly
g ) = y0 →
C11for
C41we
k1 ∆gthe
y0 ,
A,
connecting
21∆g + C
31 sin(
1 ∆2g ) +B
2 force P 1 with
221
(0)
=C
0boundary
C∆
del B integration
we
find
Pdetermined
integration
constants
C
C
,21
,31,0,
CkC31
,y∆
,and
from
from
(4)
into
into
the
thek)1=∆the
+C
−
= the
((0)
)=)dependency
cos(
)41
, equation
g∆constants
tg→
C
C
gCC
C)C1sin(
k+
g0,
tgthe
βequation
′′sin(
On
the
hand,
we
determine
integration
11
11
21
41
= 0integration
→(4)
−
∆g ) = 0. C11, C21, C31, and C41
(∆the
cos(
sin(,kconstants
gsystem
Ccos(
gy0),−
C41k12C
41
0g
1k31
1−
0y
41
1=
00.
in
Figure
1==
the
following
set
of
(→
=11with
→
+) C
y1kl0sin(
C
C31yother
k)system
C
0from
C→
+∆g
C,kCconnecting
=
−function
(deformation
0β→
cos(
)k1force
yyy′′′the
C
gand
kgdetermined
gk)=
31k1 k
,
∆g)
=
by
an
implicit
11substituting
21 g i + f(P
1 g=
41
1 g iconstants
i) +
11
41
31
1 in
1f(P
41
1ian
1conditions:
p
11 1 C21, C31, and
the
form
of
implicit
function
,
∆g)
l
by
substituting
the
(
)
=
0
→
+
+
sin(
)
+
cos(
)
=
0,
y
C
C
g
C
k
g
C
k
g
1
p
ng
0 the
11
21
0
31
1
0
41
1
0
condition
condition
of
of
the
length
length
of
of
the
the
bent
bent
rod,
rod,
and
and
finally
finally
obtain:
obtain:
g
g
g
g
g
g
2
2
an
implicit
function
f(P
,
∆g)
=
l
by
substituting
the
in
Figure
1
from
the
following
set
of
boundary
conditions:
.
(5)
2
2
1
p
ig )other
i
i
yy′′(0)
0yi→
C
+
=+the
0,j k+i1))dependency
(−gC
)−k,=1CC
+∆i )]→
+gC
cos(
=
,+(3)
gsin(
y′(0)
C
C
kjtg
Cand
k( 1j system
g1)i ) determine
y0(4)
y∆
the
we
integration constants C11(4)
, C21, C31, and C
→
−constants
=41gk21[0.
(integration
0hand,
)isin(
)21CC
y∑
C
g+determined
{gC
+B
[11
sin(
∆gcos(
(y∆
(sin(
1)
+
−+
cos(
∆gC
+41
−for
))]}
(B
) the
= lthe
C
kC
g0k,1+→
k411 (hand,
the
other
we
the
integration
constants
C
,0from
,))i,C
model
or model On
A, C
for
find
force
with
11
121
i cos(
α1n41
α
=
)=)21==
→
+1k(411determine
)=jconnecting
+−tg
cos(
)k31
=(∆ksin(
g∆model
C
C
C
k41
CkC
y1gPOn
11
31,equation
1,C
31we
1(3)
p
31
,the
from
equation
system
into
the
condition
the
ofconditions:
the bent
11
31
41
31
11equation
21 gC
31 21
1 g
1+
idetermined
i nsystem
iand
iC
41
(4) rod,
ntg0β the
n the
n 0, of
ninto length
(4)
1g ) =
j′
=(
and C41 determined
from
(4)
into
the
cos(
)
sin(
)
,
→
+
−
=
y
C
C
k
k
g
C
k
k
g
tg
β
y
(0)
=
0
→
C
+
C
=
(4)
0 1 from the following
21
31 1 set of
1 boundary
0
41 1 conditions:
1 0
11 the41
in
Figure
1
from
following
set
of
boundary
in
Figure
∆g in the
form
of
an
implicit
function
f(P
,
∆g)
=
l
by
substituting
the
′
(
)
=
→
+
+
sin(
)
+
cos(
)
=
,
g
y
C
C
g
C
k
g
C
g
y
y
(0)
,
α
α
=
→
+
=
y
tg
C
C
k
tg
On the
other
we +length
determine
integration
C11k,0 cos(
C21,kCg31,) and
C41 forkmodel
Bβ .
1bent
p41 constants
condition
ofobtain:
the
and
finally
obtain:
0α →
21 i of the
31the
i = tghand,
i
i C
21
31 1 +
n
) = tg
y′(1 grod,
Cg41C
gi0,
(0)
yobtain:
C
od, andn finally
and
finally
2i ) = tg β →gC
1
21 2 C31k1 = tgαg,g
gi′(C
g11gC
g1gi i=yi 20−
gi ki 21 2sin(
j21
2→
ythe
(0)
→
i∆g
i ik =
i
+ C(4)
g1)
→
−+C
∆from
0sin(
cos(
) =gj 0.
yg,′′(0)
g−1))1)
kCequation
kconditions:
=C
C
0,kk1kk(11=(jyset
.1g.=(6)
nstants ∑
C∑
,{C
,1)31=031and
determined
the
system
(4)31kk1into
11
Figure
from
the
of−),i C
boundary
0+
31 sin( k1 g i ) + C41 cos( k1 g i ) = y0 ,
i ) =)]}
i (6)
k→
jtg
C41sin(
k1kg+
k(1 g(jy−
j()−
l=p21
l41ptg
{CC
+31
+C
[=
sin(
[→
−
)+1−sin(
sin(
−
)]
)]
+
+
[ cos(
[ cos(
) −cos(
cos(
cos(
1)
)]}
+( (C
)11
)g++=)C
=C
11in
21
41
31) 41
41
1β
1∆
11
21
21y ′(0)
1k
1j jCfollowing
41
1j jC )k−
′
tg
C
α
α
(
sin(
.
=
→
−
g
tg
C
C
k
k
β
The
of
the
integration
constants
, and PC141with
obtained from the equation
31 1k A,
iknngdescriptions
i
21we
as
for
for
model
find
the
force
) 0= →
cos(
) − C41kB
sin(
+=C
tgand
k1n31
tg β1.dependency
βC→
n′n( g
nC21
n21model
ng 1) =
n41 1
nconnecting
n1C11i , C21, C31
1 theSimilarly
1 integration
e lengthj =1j =of
obtain:
n i=rod,
yybent
+
C
0,31i +1gC
)tg=, αand
+
+B
y0→→
C
C
k1 gi ) + C41 cos(k1 gi ) = y0 ,
y ,( gnC
On the
hand,
we
the
constants
Cy11
C
for
model
, sin(
Cg41
C,C
kg1ig=C
tg
αobtained
=i 31
+
g, yC′(0)
21
((0)
+41Cdetermine
sin(
) +ofnot
cos(
=tg
g( gother
ki1 gare
C
kg1 ig) i=) in
gi 11Cfinally
g1the
11
21
31
21
31
2consideration
2i
i 31system
i41
i C
0 →
11C
21Cg
0., j paper,
i ) )=g
iC
presented
this
taking
their
very
shape.
The
descriptions
integration
constants
C
,
C
,
and
from complex
the equation
.
(6)
gi Figure
g31ijk1 cos(
gconstants
11
21into
31
41
sin(
=Cyj tg
→
+
y2descriptions
k
k
k
β
β
∆g
in
the
form
of
an
implicit
function
f(P
,
∆g)
=
l
by
substituting
the
deformation
C
k
k
j
C
k
j
k
j
l
{
+
[
sin(
−
sin(
−
1)
)]
+
[
cos(
)
−
cos(
(
−
1)
)]}
+
(
)
=
2)characteristics
2 i1)( −
n ′presents
i g
Figure
2
presents
the
the
mechanical
mechanical
characteristics
of
of
the
the
compressed
compressed
monofilament
monofilament
for
for
the
the
1
p
The
of
the
integration
C
,
C
,
C
,
and
C
obtained
from
the
equation
i
i
21
1
41
1
1
(4)
∑
p
21
31
1
41
1
1
11
21
31
. C
(6)
g)i = y n,y′41
gC
g ik sin(
ggi )2 = tg β. . (5)
the
set+ g(sin(
of
conditions:
kj1=1)j1= from
k1following
− 1) )] + Cin
cos(
)g−i→
cos(
( jC
− 1) g )]}
) k =g l)p +
2
i boundary
i ++
41 [Figure
n
n
n
n
n
′
(
+
+
cos(
g
y
C
C
k
g
(0)
,
y
tg
C
k
tg
α
α
=
→
=
(
)
cos(
)
=
→
−
g
tg
C
C
k
k
g
C
k
β
′
system
are
not
presented
in
this
paper,
taking
into
consideration
their
very
complex
shape.
(5)
(0)
,
y
tg
C
C
k
tg
α
α
=
→
+
=
C
C
k
g
j
k
g
j
C
k
g
j
k
g
j
lp
{
+
[
sin(
(
∆
+
))
−
sin(
(
∆
+
(
−
1)
)]
+
[
cos(
(
∆
+
))
−
cos(
(
∆
+
(
−
1)
))]}
+
(
)
=
0
11
21
31
1
41
1
0
i
i
i
i
21
31
1
i article
31
1 1 (4)1 into
41 1
1 i
n variants
nas
n,constants
21
1this
constants
C
Cpaper,
C31
, and
C11
determined
from
the
equation
system
31as
131
1 C
41
1
variants
A
Aand
and
B
[9],
as
well
as,the
variant
described
described
the
previous
previous
article
[8]
[8]n–21–the
the
the
rod
rod
are
presented
in
taking
into
consideration
their
very
complex
shape.
Theintegration
descriptions
of
thewell
integration
,C
, in
Cgthe
,2 and
C
from
equation
11
21variant
21in
31
41 obtained
gthe
(4) i the
gi system
g41
y∑
= 021nnot
→
C
0,
nB2g[9],
n
n in the form of
j n
2
i presents
i
inof
1Figure
j =(0)
11 + C41 =the
.
(6)
mechanical
characteristics
the
compressed
monofilament
for
the
For
model
A,
we
find
the
dependency
connecting
force
P
with
deformation
∆g
[
sin(
)
−
sin(
(
−
1)
)]
+
[
cos(
)
−
cos(
(
−
1)
)]}
+
(
)
=
k
j
k
j
C
k
j
k
j
l
′
The
descriptions
of
the
integration
constants
C
,
C
,
C
,
and
C
′
1
(0)
,
α
α
=
→
+
=
y
tg
C
C
k
tg
(
)
cos(
)
sin(
)
.
=
→
+
−
=
y
g
tg
C
C
k
k
g
C
k
k
g
tg
β
β
21 1 31i
41 obtained from
1 characteristics
1connected
connected
by
two
artificial
artificial
joints.
are
not
presented
into
consideration
complex21shape.
condition
theC214121length
of
the
finally
monofilament
for41kand
the
=two
→
+joints.
−1rod,
y′(of
g1by
tgcompressed
C31in31
k11this
k1bent
gi )taking
C
k1 gi ) n=obtain:
tg β . p their very
βof
i
31 1
1 i
41111
i )the
1 cos(
1 sin(
n system
nA
npaper,
ncos(
(
)
=
→
+
+
sin(
)
+
)
=
,
g
y
C
C
g
C
k
g
C
k
g
y
y
variants
and
B
[9],
as
well
as
the
variant
described
in
the
previous
article
[8]
–
the
rod
For
model
A,
we
find
the
dependency
connecting
force
P
with
deformation
∆g
in
the
form
of
,
∆g)
=
l
by
substituting
the
integration
constants
C
,
C
,
C
,
and
an
implicit
function
f(P
0
11
21
31
1
41
1
0
i
i
i
i
system
are
not
presented
in
this
paper,
taking
into
consideration
their
comple
1
1
p
11
21
31
the variantThe
described
in
previous
[8] connecting
– the Crod
For
model
A,
we→
find
the
dependency
P,1tg
with
∆gofinthe
theequation
form(4)of constants C11, C21, C31, and C41 very
descriptions
integration
obtained
fro
′( gi ) =characteristics
cos(compressed
sin(
.for
ydescriptions
tg βthe
C
C31
karticle
kconstants
=
βThe
of
the
integration
, Ckforce
and
Cdeformation
obtained from
21
41the
i )31
21 +
1 the
1 g i ) − C41k
1 11
1 ,gC
ents the mechanical
of
monofilament
connected
by
two
artificial
joints.
C
determined
from
the
equation
system
(3)
into
the
condition
of
the
length
of
the
bent
rod,
,
∆g)
=
l
by
substituting
the
integration
constants
C
,
C
,
C
,
and
an
implicit
function
f(P
′
41
1
p
11
21
31
(0)
,
α
α
=
→
+
=
y
tg
C
C
k
tg
n
ForSimilarly
model
A,
we
find
the
dependency
connecting
force
P,1we
with
deformation
∆g
in
the
form
of
system
are
not
presented
in
this
paper,
taking
into
consideration
their
very
comp
as
model
A,
for
model
B
find
the
dependency
connecting
force
P
with
,
∆g)
l
by
substituting
the
integration
constants
C
,
C
,
C
,
and
an
implicit
function
f(P
21for
31
1 g =paper,
g
1
p
11
21
31
system
are
not
presented
in
this
taking
into
consideration
their
very
complex
shape.
1
g
g
g
g
The
descriptions
of
the
integration
constants
C
,
C
,
C
and
C
obtained
from
the
equation
2
11
21
31
d B [9], as well as{Cthe variant
described
[8] kFor
– system
the41j rod
i
i in the previous
i article
i
i 2
. (6) of theforce
C
determined
equation
(3)
into
the
of
bentProd,
+→
)=−land
sin(
(−j C
−from
1)kobtain:
)]the
+kC
cos(
)constants
−the
cos(
−
1)
)]},dependency
+31(, rod,
) the
= l plength
C31C[ sin(
j k41 cos(
kfinally
kwe
A,length
find
the
connecting
∆g in
(6)
1 with deformation
∑
1substituting
41 [)
1 ( jC
C41
from
thekin
(3)
into
condition
of
the
bent
,1equation
∆g)
by
the
integration
, condition
C∆g)
an
implicit
function
f(P
′( gi ) =21not
pk1system
11
21
= tg 1β j. model
yjdetermined
tg
gform
system
paper,
taking
into
consideration
their
very
complex
shape.
i)
21 +1C
31this
1n the
41 of
1nsin(
1 g iimplicit
nβpresented
noffunction
nC
nand= lp by substituting the
wo artificial
joints.
∆g
in
an
f(P
,
deformation
=1 are
1
and
finally
obtain:
,
∆g)
=
l
by
substituting
the
integration
constants
C
,C
an
implicit
function
f(P
1
p
11∆g
and
finally
obtain:
C
determined
from
the
equation
system
(3)
into
the
condition
of
the
length
of
the
bent
rod,
For
model
A,
we
find
the
dependency
connecting
force
P
with
deformation
41 model
1
For
A, we of
find
dependency
connecting
P1, and
withCdeformation
∆g inthe
theequation
form of
The
descriptions
the integration
constants
C11,, Cforce
obtained
from
n
21, C31
41 compressed
Figure
2 presents
the
mechanical
of
the
monofilament
for
the
g i,characteristics
g i 41
g i from
g i system
g i the
g iinto
. the
(5)
integration
constants
C
,
C
C
and
C
determined
from
the
equation
system
(4)
2 condition
2
C
determined
the
equation
(3)
into
of
the
length
of
11
21
31
41
and
finally
obtain:
{
+
[
sin(
(
∆
+
))
−
sin(
(
∆
+
(
−
1)
)]
+
[
cos(
(
∆
+
))
−
cos(
(
∆
+
(
−
1)
))]}
+
(
)
=
C
C
k
g
j
k
g
j
C
k
g
j
k
g
j
l
,
∆g)
=
l
by
substituting
the
integration
constants
C
an
implicit
function
f(P
∑
21
31
1
1
41
1
1
p
1form
11
∆g)
=paper,
lpgjby
substituting
the integration
C21
,g C
an
implicit
function
system
are A,
not
presented
this
taking
into gconsideration
their
very
complex
shape.
For
model
we
findf(P
the
dependency
connecting
force
Pn1 with deformation
∆g
the
1,in
11,in
31, and
n
n
n C
nof –p the rod
=1 as the
gthe
g
gn
. n(5)
variant
described
inconstants
previous
article
n variants A and B [9], as well
2
i
i
i
i
finally
g 2, 3, from
glength
g bent
g i kfrom
. [8]
(5)
condition
of5 the
the
and
{C21 of
+C
[ sin(
)) − sin(
+ ( j41
− 1)determined
)]obtain:
+obtain:
( ∆2gthe
+ gj i bent
)) equation
− cos(
( j − 1) i ))]}
+ into
( i ) 2 =the
k ( ∆g + rod,
j the
k (g∆and
gfinally
C41[ cos(
k1 ( ∆g +system
l p condition of the length o
(3)
Equations
4,
and
i equation
1of
C
determined
condition
the
{C21 i + C
+the
))∆g)
−∑
sin(
+ ( j 31
− 1) i (3)
)]1 + Cinto
+ j1 i )) −C
cos(of
∆g + (length
))]}
+ (21the
) 2C=31
j 16.
k1 (l∆pgnsystem
k1 ( ∆gintegration
kconstants
j − 1) C
l p, rod,
41implicit
,
=
by
substituting
,
C
,
and
an
function
∑
n
n
n
n
31[ sin( k1 ( ∆gf(P
41[ cos(
1 (n
11
1
j
=
by two artificial
joints. gni
j n=1 connected
gn obtain:
gn
gni and finally obtain:
gni
gni 2
. (5)
2
and
finally
{C21 i + A,
sin(from
( ∆g +the
− sin( k1 ( ∆Similarly
+ ( j −connecting
1) (3)
)] + Cinto
+P1j with
)) − cos(
∆g + (length
− 1) we
))]}
(the
)bent
= l rod,
C31[ we
k1find
j i ))
gsystem
k1 ( ∆gcondition
k1 (the
jB
For
model
the
dependency
force
deformation
∆g
in+the
form
of
C∑
equation
the
of
offind
41[ cos(
41 determined
gi
gni the pdependencyg i connecting force
g P1 with
g
n
n
n as for model A,
n n for model
n
j =1
{
+
[
sin(
(
∆
+
))
−
sin(
( ∆g + ( j − 1) )] + C41[ cos( k1 ( ∆g + j i )) − cos( k1 ( ∆g45
+ ( j − 1) i ))]}2 +
C
C
k
g
j
k1and
FIBRES
&
TEXTILES
in
Eastern
Europe
January
/
December
/
B
2008,
Vol.
16,
No.
6
(71)
∑
21
31
1
and
finally
obtain:
=
l
by
substituting
the
integration
constants
C
,
C
,
C
,
an implicit
function
f(P
n
1, ∆g)
pdeformation
11
21
31
Similarly
as
for
model
A,
for
model
B
we
find
the
dependency
connecting
force
P
with
∆g
in
the
form
of
an
implicit
function
f(P
,
∆g)
=
l
by
substituting
the
n
n
n
n
n
1
1
j
=
1
p
g
n
g
g
g
g
g
Similarly
as
for
model
A,
for
model
B
we
find
the
dependency
connecting
force
P
with
n
1
j
g
g
g
g
g
2
2
i
i
i
i
i
g
g equation system
g i (3) into the condition
gi
g i j 2bent
i
i
i
i
(5)
2 ∆the
.
i
{(C
[jgsin(
−.sin(
∆=
+ l( jequation
− 1)
)]substituting
++C41
[ cos()k1 ( ∆=
+the
)) −(6)
cos(
kof
kj1 (−
gthe
g(4)
j into
k1 ( ∆g + ( j − 1) i ))]}
C∑41 {determined
the
of
rod,
k
k
j
k
l
−
)]
+
[
cos(
(
1)
)]}
(
∑
( ∆gC+31
)) −deformation
sin(
+ ( )j −−
1)sin(
)] +in
[(the
cos(
( ∆g +
))
−C
cos(
∆the
+ (+length
−11)
))]}
+g−
( +cos(
)f(P
=))1l p,k
C21 i{+C
C3121[ sin(from
k1 +
j [ isin(
k1 (j∆gintegration
Cconstants
k11)
j11
kimplicit
g and
jC
21
31
1 ()
C
,
C
,
C
,
C
determined
from
system
the
∆g
form
of
an
function
∆g)
by
21
31
41
41
1
p
p
1
1
41
1
n
n
n
n
n
Similarly
model
A,
for
model
B
we
find
the
dependency
force
P
∆g
in
the
form
of
an
implicit
function
f(P
,
=
l
by
substituting
the
deformation
j =1∆g) connecting
1 with
n as for
n
n
n
n
n
1
p
j n=1
n
n+ ( j −constants
nB system
ntheinto
g
g
g i the
g n2
g i obtain:
. (5)
and {jfinally
2
=1 g i + Cobtain:
length
bent
and
C11
Cg21n
C
and
the
equation
the
Similarly
for
model
model
we find(4)
dependency
connecting fo
[ sin( k ( ∆g + j i )) −integration
sin( k ( ∆gcondition
1) i )] +of
[ cos(
+ ,jof
))
−, cos(
∆rod,
(determined
1) i finally
))]} + ( from
)A,
= lfor
C
C the
k, (∆
k (C
g41+as
j−
31
Differentiating twice the Equation (1) toward ‘x’, we obtain subsequently:
∑
of compressing force P1. Essential here is
the influence of the difference in the fastening positions of the rod in the longitudinal direction (parameter y0 in Figure 1)
on the value of this force. The parameter
y0 takes two different values: if the system is symmetric, then y0 ≈ 0 mm; if it is
asymmetric, then y0 = ± 2 mm.
Figure 3. Simulated deflection lines of the rod formed as the effect of the force P1 impact.
Figure 4 presents two families of curves
that represent the shape of the rod of
equal deformations ∆g, but differentiated
by the parameter y0. For model, A we
have the following lines:
 lines 1 – deflection∆g=3mm(P1 = 1.73 cN
for y0 = 2 mm, P1 = 1.65 cN for
y0 = 0 mm, P1 = 1.57 cN for y0 = -2 mm)
 lines2–deflection∆g=5mm(P1=3.49cN
for y0 = 2 mm, P1 = 3.35 cN for
y0 = 0 mm, P1 = 3.20 cN for y0 = -2 mm).
For mode B, we have:
 lines 1 – deflection ∆g = 3 mm
(P1 = 3.073 cN for y0 = 2 mm, P1 = 3.08 cN
for y0 = 0 mm, P1 = 3.07 cN for y0 = -2 mm)
 lines 2 – deflection ∆g = 5 mm
(P1 = 6.34 cN for y0 = 2 mm, P1 = 6.35 cN
for y0 = 0 mm, P1 = 6.34 cN for y0 = -2 mm).
Figure 4. Family of curves representing the shape of the monofilament with the same
deformations ∆g and differentiated by parameter y0 .
complex considerations concerned with
the structure and strength properties of
3D distance knitted fabrics. The hitherto
discussed calculation model takes into
consideration only a simple loading system that uniformly loads all the monofilaments of the compressed knitted fabric.
While using more complex loading systems, such as convex plates or ball-like
ended pistons, the compressed outside
layer of the 3D knitted fabric deforms.
However, this does not change the fact
that the sum of action of the individual
connectors contributes to the total effect
of compressing the fabric. The connectors are deformed differentially, and take
different loads according to the mechanical characteristic P1 = f(∆g).
Simulation of the compressing
process on the basis
of mathematical models
The first simulation presented shows the
shapes of the monofilament under the action of force P1. Observation of the con-
46
nectors’ geometry, as well as the mathematical analysis, [9] shows that the rod
representing the monofilament takes the
shape of a fragment of a sinusoid. The
shape of this curve depends on the value
of the force P1. With an increase in the
force value, an increase in the amplitude
of the curve occurs, but the curve tends to
be narrower. The cause of increasing P1 is,
from the physical point of view, a decrease
in the curvature radius of the sinusoid
apex. Figure 3 presents a broad spectrum
of deflections caused by force P1.
The compressing curves of both variants
considered do not differ much apart on
the difference that in the model B a curvature of the upper part of the rod occurred,
caused by the fastening. This curvature,
causing the rod to compress according to
model B, is less susceptible to deformation than the rod of model A.
The second simulation was aimed to
check the influence of the geometry of
the distance knitted fabric on the value
The differences in the bending forces
that cause the same deflections result
from the different geometries of the rods
(parameter y0). The rods are differentiated by shape (various functions describe
the shape of these curves), and at the
same time they have different curvature
radii of these curves; the consequence
of this is that different forces cause the
same deformation. From the mathematical analysis results, it can be seen that the
rod compressed according to model B is
practically insensitive to the changes of
parameter y0. Within the range that we
tested, the parameter y0 does not influence
the value of the compressing force P1.
The third simulation concerns the influence of the monofilament’s shape on the
value of the compressing force. This is a
problem of different buckling forms occurring, resulting in equal deflections of
the same rod (and with the same geometry of fastening its ends), causing different values of force P1, and oppositely
the same value of force P1 may cause different deformations. The first case (equal
values of deflection and different values
of force P1) is shown in Figure 5.
For a rod compressed according to model
A, a deflection of ∆g = 5 mm is caused
by the forces P1 of 3.48 cN, 3.62 cN,
10.39 cN, and 10.62 cN, whereas for a rod
FIBRES & TEXTILES in Eastern Europe January / December / B 2008, Vol. 16, No. 6 (71)
of this phenomenon, which was confirmed empirically, may be explained in
the following way: the greater the curvatures of the rod are, the greater value
of force P1 causes the same deflection
∆g. This is clearly directly visible from
the geometry of the compressed rods, as
well as from the mathematical analysis
carried out.
Summary
Figure 5. Different forms of the rod deflection caused by the action of different forces P1.
Figure 6. Different shapes of rod deflection caused by the impact of the same force.
compressed according to model B, a deflection of ∆g = 3 mm is caused by P1 of
3.07 cN, 4.01 cN, 9.93 cN, and 7.58 cN.
The second case (equal values of force
P1 causing different deflections ∆g of the
rod) is presented in Figure 6.
For the rod bent according to model A,
the force P1 = 4 cN causes deflections
∆g of 1.75 mm, 2.05 mm, 5.25 mm, and
5.32 mm, whereas for that bent according to model B, the same force causes
deflections of 3.80 mm, 2.99 mm,
0.81 mm, and 0.31 mm. The occurrence
FIBRES & TEXTILES in Eastern Europe January / December / B 2008, Vol. 16, No. 6 (71)
1. Physical and mathematical models of
the compressing process of an elastic,
slender rod with an assumed shape,
fastened by one- or two-side fixed
joints, are presented in this paper. The
presented models are considered for
compressing a 3D distance knitted
fabric. The mechanical characteristics
of the compressed rods were analysed
and the rod’s shapes discussed considering the aspect of the connector’s
tensions and the deformations of the
knitted 3D fabric. The analysis of the
phenomena of compressing the fabric
included:
 the influence of the fastening geometry of a single monofilament on the
value of the force compressing this
monofilament.
 the influence of different forms of
buckling of the compressed monofilament, which occur on the value of
deflection ∆g under the impact of the
compressing force P1.
2. A computer simulation was carried
out of the process of compressing a
model concerned with the knitted fabric, which indicated that
 a change in the parameter y0 (mutual
displacement of the connectors’ fastenings), in model B does not influence the value of force P1, whereas in
model A, it causes changes within the
range of 8%-10% of the P1 value;
 in model A, for the second buckling
form of the bent rod, the force P1 is
nearly 200% higher in comparison
with the value for the first buckling
form for the same deflection ∆g. In
model B, for the second form of buckling of the bent rod, the force P1 is
nearly 150% higher;
 in model A, for the second buckling
form of the bent rod, the deflection
∆g caused by the constant value of
force P1 is higher by nearly 175% in
comparison with the value for the first
buckling form. In model B, for the
second buckling form, the bending of
the rod is higher by 360%.
47
Conclusions
In the future parts of this research work,
mathematical models will be presented
of connectors considering mutual displacements of the outside layers of the
knitted fabric. In parallel, we will conduct activity into designing and building an experimental model related to the
three models of the 3D distance knitted
fabric that we elaborated. This model
will also be aimed to verify further
mathematical models describing the displacement of the outer layers of the 3D
distance knitted fabric mentioned in this
paper. The next stage of our work will
be the development of an elaboration
model for knitted fabrics loaded nonuniformly and loaded by way of objects
with a given shape.
The solutions presented herein should be
considered as an introduction to further
research work aimed at finding optimum
calculation models for 3D distance knitted fabrics in order to determine the mechanical properties of these fabrics, but
firstly at estimating their susceptibility
to compression, as well as adaptation of
the calculation mechanism created for
solving real problems occurring while
designing and applying 3D distance knitted fabrics.
Institute of Biopolymers and Chemical Fibres
�������������������������������������������������
Director of the Institute: Danuta Ciechańska Ph.D., Eng.
The research subject of IBWCH is conducting scientific and development research
of techniques and technologies of manufacturing, processing, and application, into
in the following fields:
� biopolymers,
� chemical fibres and other polymer materials and related products,
� pulp and paper industry and related branches
R&D activity includes the following positions, among others:
� biopolymers,
� functional, thermoplastic polymers,
� biodegradable polymers and products from recovered wastes,
� biomaterials for medicine, agriculture, and technique,
� nano-technologies, e.g. nano-fibres, and fibres with nano-additives,
� processing of fibres, films, micro-, and nano- fibrous forms, and nonwovens,
� paper techniques, new raw material sources for manufacturing paper pulps,
� environmental protection,
The Institute is active in inplementig its works in the textile industry, medicine,
agriculture, as well as in the cellulose, and paper industries.
The Institute is equipped with unique technological equipment, as the
technological line for fibre (e.g. cellulose, chitosan, starch, and alginate) spinning
by the wet method.
The Institute organises educational courses and workshops in fields related to its
activity.
References
1. Kopias K., „Technologia dzianin kolumienkowych”, Ed. WNT, Warszawa 1986.
2. Żurek W., Kopias K. „Struktura płaskich
wyrobów włókienniczych”, Ed. WNT,
Warszawa 1983.
3. Niezgodziński M. E., Niezgodziński T.,
„Wytrzymałość materiałów”, Ed. PWN,
Warszawa 2000.
4. Misiak J., „Stateczność konstrukcji
prętowych”, Ed. PWN, Warszawa 1990.
5. Bodnar A., „Wytrzymałość materiałów”,
Politechnika Krakowska, Kraków 2004.
6. Misiak J. „Obliczenia konstrukcji
prętowych”, Ed. PWN, Warszawa 1993.
7. Biegus A., „Nośność graniczna stalowych konstrukcji prętowych”, Ed.
PWN, Warszawa-Wrocław 1997.
8. Supeł B., Mikołajczyk Z., „Model procesu ściskania łącznika zamocowanego
przegubowo dzianiny dystansowej 3D”,
Fibres and Textiles in Eastern Europe,
Vol. 16, No. 6 (71), pp. 40-44, 2008.
9. Mosurski R., „Mathematica”, Uczelniane
Wydawnictwa Naukowo-Dydaktyczne,
Kraków 2001.
Received 22.08.2007
48
The Institute’s offer of specific services is wide and differentiated, and includes:
� physical, chemical and biochemical investigations of biopolymers and
synthetic polymers,
� physical, including mechanical investigation of fibres, threads, textiles, and
medical products,
� tests of antibacterial and antifungal activity of fibres and textiles,
� investigation in biodegradation,
� investigation of morphological structures by SEM and ESEM
� investigation and quality estimation of fibrous pulps, card boards, and paper
products, including paper dedicated to contact with food, UE 94/62/EC tests,
among others.
� Certification of paper products.
The Institute is active in international cooperation with a number of corporation,
associations, universities, research & development institutes and centres, and
companies.
The Institute is publisher of the scientific journal ‘Fibres and Textiles in Eastern
Europe’.
�����������������������������������������������������������
Institute of Biopolymers and Chemical Fibres
������������������������������������������������������
������������������������������������������������
�����������������������������������������������������
Reviewed 30.10.2007
FIBRES & TEXTILES in Eastern Europe January / December / B 2008, Vol. 16, No. 6 (71)