# Numerical Modeling of the Motion of the Flat Textile Structures

Authors: Piotr Szablewski
Journal: ISRN Mechanical Engineering
 (2011)

---

## Abstract

In many problems from the field of textile engineering (e.g., fabric folding, motion of the sewing thread) it is necessary to investigate the motion of  the objects in dynamic conditions, taking into consideration the influence of the forces of inertia and changing in the time boundary conditions. This paper deals with the model analysis of the motion of the flat textile structure using Lagrange's equations in two variants: without constraints and with constraints. The motion of the objects is under the influence of the gravity force. Lagrange's equations have been used for discrete model of the structure.


---

## Body

## 1. Introduction

In most cases of dynamic analysis of textile structures, it is necessary to use the model of heave elastica, as a one-dimensional body of linear weightq and bending rigidity C in the state of large deflections. In this paper, it is assumed that during the run of bending effect the flat textile structure (e.g., fabric) will be represented as its longitudinal section. The mathematical model will be described as a flat deflection curve; this will be treated as a heavy elastica, as shown in Figure 1 and in [1].Figure 1
The model of fabric approximate to elastica.The word “heavy” underlines the decisive influence of the gravity forces during the motion. It is assumed that the particular longitudinal sections do not act on each other by internal forces. Furthermore, the constancy of properties along the whole width of the bending strip is assumed. It will be assumed also that the elastica is inextensible. The analysis was made using Lagrange's equations, describing the motion of the system ofn particles in conservative force field. Two variants of Lagrange's equations were considered: equations without constraints and equations with constraints.


## 2. Discrete Model of the Object and  the  Lagrange's Equations

The Lagrange's equations, describing the motion of the system ofn particles in conservative force field, studyed in detail among other things in [2–5], are presented below:(1)ddt(∂T∂φ̇k)-∂T∂φk+∂U∂φk=0,k=1,2,…,n,
where T-kinetic energy, U-potential energy, φk,φ̇kkth generalized coordinate and velocity. If we have n generalized coordinates, then we can write n Lagrange's equations (1). The problem reduces to appropriate choosing of generalized coordinates φk, deriving the formula for kinetic and potential energy, and next deriving and solving the Lagrange's equations.During the analysis, we replace often continuous systems by discrete systems using partition inton elements. In considered problem partition was made as follows. The elastica of length L, mass M and bending rigidity C fixed at point A was replaced by system of n masses of identical value m=M/n, connected by n straight, inextensible, and weightless segments of length l=L/n, as shown in Figure 2 and work [6]. The other end B of elastica is free. If ρ is linear mass of elastica, then M=ρL. The position of ith mass is described by generalized coordinate as an inclination angle φi(i=1,2,…,n) of ith segment to the vertical direction. Coordinates of ith mass and its derivatives are as below:(2)xi=l∑j=1isinφj,ẋi=l∑j=1icos⁡φj⋅φ̇j,yi=l∑j=1icos⁡φj,ẏi=-l∑j=1isinφj⋅φ̇j.
The kinetic energy of ith mass is Ti=1/2m(ẋi2+ẏi2).Figure 2
Discrete model in coordinate system.Therefore, the total kinetic energy of the system ofn mass is(3)T=12ml2∑i=1n[(∑j=1icos⁡φj⋅φj)2+(∑j=1isinφj⋅φj)2].
The total potential energy of the system U is the sum of two components(4)U=V+Vb,
where V is the potential energy of gravity forces and Vb is the potential energy of the bending.The potential energy of gravity forces isV=∑i=1nmgyi=mgl∑i=1n∑j=1icosφj and the potential energy of the bending Vb=1/2C∫0L(dφ/ds)2ds. After substitution the derivative by difference quotient, we have(5)Vb=12C∑i=1n-1(φi+1-φil)2l=C2l∑i=1n-1(φi+1-φi)2.
Thus, the total potential energy of the system U is(6)U=V+Vb=mgl∑i=1n∑j=1icos⁡φj+C2l∑i=1n-1(φi+1-φi)2.
The formula for derivative of kinetic energy T in respect of φk after transformations is as follows:(7)∂T∂φk={ml2∑i=1n∑j=1isin(φj-φk)⋅φ̇jφ̇k,fori≥k,0,fori<k.
After changing the limits of summation, we can write(8)∂T∂φk=ml2∑i=kn∑j=1isin(φj-φk)⋅φ̇jφ̇k.
Similarly, the other derivatives can be written as follows:(9)ddt(∂T∂φ̇k)=ml2∑i=kn∑j=1i[cos⁡(φj-φk)⋅φ̈j-sin(φj-φk)⋅(φ̇j-φ̇k)φ̇j],(10)∂V∂φk=-mgl∑i=knsinφk=-(n-k+1)mglsinφk,(11)∂Vb∂φk=Cl∑i=1n-1[(φi+1-φi)⋅∂∂φk(φi+1-φi)].
The derivative ∂(φi+1-φj)/∂φk depending on the value of index i is given by(12)∂∂φk(φi+1-φi)={1,fori=k-1,-1,fori=k,0,fori<k-1,i>k.
Using summation from i=1 to i=n-1 the formula (11) can be written as follows:(13)∂Vb∂φk={Cl(φ1-φ2),fork=1,Cl(-φk-1+2φk-φk+1),fork=2,3,…,n-1,Cl(φn-φn-1),fork=n.
Therefore, the derivative of total potential energy in respect of φk is(14)∂U∂φk=∂V∂φk+∂Vzg∂φk=-mgl∑i=knsinφk+Cl∑i=1n-1[(φi+1-φi)⋅∂∂φk(φi+1-φi)].
Using (8), (9), and (14) in Lagrange's equations (1) and dividing by ml2, we have(15)∑i=kn∑j=1i[cos⁡(φj-φk)φ̈j-sin(φj-φk)φ̇j2]-(n-k+1)glsinφk+Cml3×∑i=1n-1[(φi+1-φj)⋅∂∂φk(φi+1-φj)]=0fork=1,2,…,n.
Finally, it was obtained system of n 2nd-order ordinary differential equations with unknown generalized coordinates φi(i=1,2,…,n). Formula (15) represents kth Lagrange's equation, which describes the motion of elastica as in Figure 2.Interpretation of boundary conditions results from formula for total potential energy of the system (6). In the presented problem, the end B of elastica is free; therefore, it is not loaded by any forces. The end A is fixed by joint; therefore, the bending moment M=0, and displacements in point A are equal zero (it has not influence on potential energy).


## 3. The Lagrange's Equations in the Matrix Form

For further analysis, the Lagrange’s equations (15) were written in the matrix form. The following denotations were introduced:(16)Φ=[φ1φ2⋮φn],dΦdt=Φ̇=[φ̇1φ̇2⋮φ̇n],d2Φdt2=Φ̈=[φ̈1φ̈2⋮φ̈n].
In the matrix form the Lagrange's equations (15) are(17)AΦ̈+B+C+D=0,
where A is an appropriate matrix and B,C,D are vectors as follows:(18)A=A(Φ)=[A11A12⋯A1nA21A22⋯A2n⋯⋯⋯⋯An1An2⋯Ann],B=B(Φ,Φ̇)=[B1B2⋮Bn],C=C(Φ)=[C1C2⋮Cn],D=D(Φ)=[D1D2⋮Dn].
On the basis of (15) we can write the matrix A and vectors B,C and D as follows:(19)Akj=Skjcos⁡(φj-φk),Bk=-∑j=1n[Skjsin(φj-φk)φ̇j2],Ck=-(n-k+1)glsinφk,Dk={Cml3(φ1-φ2),fork=1,Cml3(-φk-1+2φk-φk+1),fork=2,3,…,n-1,Cml3(φn-φn-1),fork=n,
where the matrix S of dimension n×n is defined by(20)S=[nn-1n-2⋯321n-1n-1n-2⋯321n-2n-2n-2⋯321⋯⋯⋯⋯⋯⋯⋯333⋯321222⋯221111⋯111].
The element of kth row and jth column of matrix S can be written as(21)Skj=n-k+1forj≤k,Skj=n-j+1forj>k.
Its task is to realize the double summation from i=k to i=n and from j=1 to j=i.Finally, it was obtained system ofn 2nd-order ordinary differential equations with unknown generalized coordinates Φ in matrix form(22)Φ̈=-A-1(B+C+D).


## 4. Numerical Solution of the Problem

We can write the system of (22) in the form of system 2n 1st-order ordinary differential equations, introducing additional vector of generalized velocities Φ̇=Ω=[ω1ω2⋯ωn]T. In this way, we obtain(23)Φ̇=Ω,Ω̇=-A-1(B+C+D).
The system of (23) has been solved using a standard Runge-Kutta-Fehlberg 4th-order integration method. To solve this problem numerically, the following denotations were introduced.The vector of unknowns(24)Y=[Y1Y2⋮Y2n],
where(25)Yi=φifori=1,2,…,n,Yi=ωi-nfori=n+1,n+2,…,2n.
The right side of (22) has been denoted by vector F; that is,(26)F=-A-1(B+C+D),F=[F1F2⋮Fn].
Finally, the following system of differential equations was solving:(27)Ẏ1=Yn+1,⋮Ẏn=Y2n,Ẏn+1=F1,⋮Ẏ2n=Fn,
with initial conditions for t=t0(28)Φ=Φ0=[φ10⋮φn0],Ω=Ω0=[ω10⋮ωn0].


## 5. The Lagrange's Equations with Constraints

In a previous  section the elastica fixed only at point A was considered. The other end B of elastica was free (Figure2). In this case, generalized coordinates φk were completely independent in themselves. In many problems connected with the motion of elastic, there are some constraints concerning angles φk. Then, the Lagrange's equations undergo modification.In this  section the motion of elastica with two ends fixed at two points A and B using joints as in Figure3 will be considered. This case may be used for example to simulate fabric folding and so on. Then, the boundary conditions for the end B (which previously was free) are as follows:(29)xn=0,yn-H=0.Figure 3
The elastica with two ends fixed at points A and B.In accordance with (2) for xn and yn we can rewrite the following two conditions of constraints (29) obtaining(30)f1≡l∑k=1nsinφk=0,f2≡l∑k=1ncos⁡φk-H=0.
Thus, we can write the Lagrange's equations in the following form:(31)ddt(∂T∂φ̇k)-∂T∂φk+∂U∂φk+λ1∂f1∂φk+λ2∂f2∂φk=0,k=1,2,…,n,
where λ1,λ2 are the unknown Lagrange's multipliers (their interpretations are the appropriate reaction forces in point B-horizontal and vertical).Derivatives∂f1/∂φk and ∂f2/∂φk are as below:
(32)∂f1∂φk=lcos⁡φk,∂f2∂φk=-lsinφk.
Let right side of the kth (15) be denoted by Ψk(φ,φ̇). Therefore, using (31) and (32), we have the following system of equations:
(33)Ψk(φ,φ̇)+λ1mlcos⁡φk-λ2mlsinφk=0,k=1,2,…,n.
Using matrix notation, we can write new matrix form of the system (33)
(34)AΦ̈+B+C+D+λ1⋅E+λ2⋅G=0,
where(35)E=E(Φ)=[E1E2⋮En],Ek=cos⁡φkml,k=1,2,…,n,G=G(Φ)=[G1G2⋮Gn],Gk=-sinφkml,k=1,2,…,n.
The system of (34) with the (30) presents the system of n+2 differential-algebraic equations with n+2 unknowns, that is, n generalized coordinates φi(i=1,2,…,n) and two Lagrange's multipliers λ1,λ2. To solve such problem, (30) have been differentiated two times in respect of time(36)∑k=1n(cos⁡φk⋅φ̈k-sinφk⋅φ̇k2)=0,∑k=1n(-sinφk⋅φ̈k-cos⁡φk⋅φ̇k2)=0.
Next, from (34), the multipliers λ1 and λ2 have been eliminated successively, using twice Gaussian elimination. After this, (34) together with (36) present the system of n 2nd-order ordinary differential equations in respect of φi(i=1,2,...,n). Using matrix notation, we have(37)ÂΦ̈+B̂+Ĉ+D̂=0,
similarly as (17) in  Section 3. The components of new matrix Â and vectors B̂,ĈiD̂ are as below:Fork=1,2,…,n-2(38)Âkj=(Akjcos⁡φk+1-Ak+1jcos⁡φk)sin(φk+1-φk+2)-(Ak+1jcos⁡φk+2-Ak+2jcos⁡φk+1)×sin(φk-φk+1),forj=1,2,…,n,B̂k=(Bkcos⁡φk+1-Bk+1cos⁡φk)sin(φk+1-φk+2)-(Bk+1cos⁡φk+2-Bk+2cos⁡φk+1)sin(φk-φk+1),Ĉk=(Ckcos⁡φk+1-Ck+1cos⁡φk)sin(φk+1-φk+2)-(Ck+1cos⁡φk+2-Ck+2cos⁡φk+1)sin(φk-φk+1),D̂k=(Dkcos⁡φk+1-Dk+1cos⁡φk)sin(φk+1-φk+2)-(Dk+1cos⁡φk+2-Dk+2cos⁡φk+1)sin(φk-φk+1),Ân-1j=cos⁡φj,Ânj=-sinφj,forj=1,2,…,n,B̂n-1=-∑j=1n(sinφj⋅φ̇j2),B̂n=-∑j=1n(cos⁡φj⋅φ̇j2),Ĉn-1=Ĉn=0,D̂n-1=D̂n=0.
The system of (37) has been transformed to the form similarly as in Section 5; that is,(39)Φ̈=-Â-1(B̂+Ĉ+D̂).
Next, we can write(40)Φ̇=Ω,Ω̇=-Â-1(B̂+Ĉ+D̂).
Finally, the system of (40) has been solved using a standard Runge-Kutta-Fehlberg 4th-order integration method.The considerations presented above concerned with the elastica for which the end B was fixed by joint. Of course, the other conditions of constraints may be introduced for this one.If we want to allow vertical motion of the end B, then we have only one constraint for coordinatex of the end B:x must be always equal to value a= const., and we have only one algebraic equation (conditions of constraints) in the form of(41)f1≡xn-a=0.
The Lagrange's equations are simple in this case, because we have only one Lagrange's multipliers λ1.If we want to have the end B fixed at point B at an angle ofφ=α, where x=a,y=b, then there are three conditions of constraints(42)f1≡xn-a=0,f2≡yn-b=0,f3≡φn-α=0,
and three Lagrange's multipliers λ1, λ2, and λ3.


## 6. The Results of Calculations

To present the results of calculations and accuracy of numerical method two numerical cases were considered.The first one concerned the motion of heavy elastica under the influence of the gravity force in conservative force field. The elastica was fixed byjoint at one end only as in Figure2.In the second one, the elastica was fixed by joint at two ends as in Figure3.


### 6.1. The Motion of the Elastica Fixed at One Point

In this case, the elastica fixed at one point by joint as in Figure2 will be considered. The second end is free. The motion is under the influence of the gravity and elastic forces.The properties of the elastica are as follows:(i)
bending rigidity,C=0.0025 Nm2,(ii)
linear mass density,ρ=0.01 kg/m,(iii)
length,L= 1 m.Too many pointsn of division can result in long time of calculations. In the example below, it has been assumed that n=8.The initial conditions (28) are as follows:(43)φk0=π2rad,ωk0=0rad/s,fork=1,2,…,n.
In the initial instant, the elastica has rectilinear, horizontal shape. The velocity of all points is zero. After solving the (23) with the time step Δt=0.005 s, we have obtained the shape of elastica in the following time instant t0,t1,t2,…. The shapes are presented in Figure 4. Figure 5 shows variations of generalized coordinates φ1,φ2,…,φ8 in the respect of time t. Figure 6 shows variations of generalized coordinates φ1,φ2,…,φ8 in the respect of time t for elastica of bending rigidity 10 times less than previously (C=0.00025 Nm2).Figure 4
The shape of elastica in the following time instant fromt0 to t20.Figure 5
The variation of generalized coordinatesφ1,φ2,…,φ8 in the respect of time t.Figure 6
The variation of generalized coordinatesφ1,φ2,…,φ8 in the respect of time t for bending rigidity of C=0.00025 Nm2.In Figures5 and 6, the heavy dashed line represents graph of angular displacement of rigid rod (physical pendulum) fixed in the same point as elastica (Figure 7).Figure 7
The rigid rod as physical pendulum.From the graphs, it may be inferred that in the initial state of the motion the points of elastica oscillate around the dynamic equilibrium (as for physical pendulum). The rigid rod (Figure7) can be treated as the elastica of bending rigidity C=∞ for which all points have permanently the same angular displacements φk=φ dla k=1,2,…,n. As the time goes, deviations from dynamic equilibrium become bigger and bigger.In order to check accuracy of calculations, we follow as below.For chosen coordinateφ8, its values marked by φ8Δt1 with given time step Δt1=0.005 s have been calculated. Next, the calculations have been repeated for the less time step Δt2=1/2Δt1. The difference Δφ8=φ8Δt1-φ8Δt2 has been determined for following tj=t0+i·Δt1.The calculations have been made forC=0.0025 Nm2. The graph of differences Δφ8 for following time instant is presented in Figure 8.Figure 8
Test of accuracy of generalized coordinateφ8 in the following time instant.The differencesΔφ8 are not big at all (between −0.00019 and 0.00022 rad), even after large number of iterations (i=400). The accuracy turned out to be satisfactory.


### 6.2. The Motion of the Elastica Fixed at Two Points

In this case, the elastica fixed at two points by joints as in Figure3 will be considered. The motion is under the influence of the gravity and elastic forces.The properties of the elastica are as follows:(i)
bending rigidity,C=0.0015 Nm2,(ii)
linear mass density,ρ=0.01 kg/m,(iii)
length,L=1.2 m.Number of points of division isn=16. The system of (40) has been solved using Runge- Kutta-Fehlberg 4th-order integration method with time step Δt=0,005 s. The initial conditions (28) are as follows:(44)φk0=π2[1-2(k-1)n-1]rad,ωk0=0rad/s,fork=1,2,…,n.
Finally, the shape of the elastica in the following time instant tj has been obtained. It is present in Figure 9.Figure 9
The shape of the elastica in the following time instant.In order to check accuracy of calculations, we tested how the boundary conditions (29) change as time goes by. In this way, we can check how the algorithm secures fulfillment of the boundary conditions(45)f1≡xn=0,f2≡yn-H=0,
As a measure of deviation of boundary conditions, we can take Σ=|f1|+|f2|, which, if the boundary conditions are fulfilled, should be equal zero. The calculations have been made for two time steps: Δt1=0.005 s i Δt2=1/2Δt1. The variation of Σ in respect of number of iterations are presented in Figure 10 for two time steps.Figure 10
The analysis of fulfillment of the boundary conditions∑=|f1|+|f2|.From Figure10, we can conclude that for two times less time step deviation of boundary conditions was significant less. For Δt1 after 300 iterations, deviation was about 2·10-4 m, and for Δt2=1/2Δt1, deviation was about 2·10-5 m (10 times less).


## 6.1. The Motion of the Elastica Fixed at One Point

In this case, the elastica fixed at one point by joint as in Figure2 will be considered. The second end is free. The motion is under the influence of the gravity and elastic forces.The properties of the elastica are as follows:(i)
bending rigidity,C=0.0025 Nm2,(ii)
linear mass density,ρ=0.01 kg/m,(iii)
length,L= 1 m.Too many pointsn of division can result in long time of calculations. In the example below, it has been assumed that n=8.The initial conditions (28) are as follows:(43)φk0=π2rad,ωk0=0rad/s,fork=1,2,…,n.
In the initial instant, the elastica has rectilinear, horizontal shape. The velocity of all points is zero. After solving the (23) with the time step Δt=0.005 s, we have obtained the shape of elastica in the following time instant t0,t1,t2,…. The shapes are presented in Figure 4. Figure 5 shows variations of generalized coordinates φ1,φ2,…,φ8 in the respect of time t. Figure 6 shows variations of generalized coordinates φ1,φ2,…,φ8 in the respect of time t for elastica of bending rigidity 10 times less than previously (C=0.00025 Nm2).Figure 4
The shape of elastica in the following time instant fromt0 to t20.Figure 5
The variation of generalized coordinatesφ1,φ2,…,φ8 in the respect of time t.Figure 6
The variation of generalized coordinatesφ1,φ2,…,φ8 in the respect of time t for bending rigidity of C=0.00025 Nm2.In Figures5 and 6, the heavy dashed line represents graph of angular displacement of rigid rod (physical pendulum) fixed in the same point as elastica (Figure 7).Figure 7
The rigid rod as physical pendulum.From the graphs, it may be inferred that in the initial state of the motion the points of elastica oscillate around the dynamic equilibrium (as for physical pendulum). The rigid rod (Figure7) can be treated as the elastica of bending rigidity C=∞ for which all points have permanently the same angular displacements φk=φ dla k=1,2,…,n. As the time goes, deviations from dynamic equilibrium become bigger and bigger.In order to check accuracy of calculations, we follow as below.For chosen coordinateφ8, its values marked by φ8Δt1 with given time step Δt1=0.005 s have been calculated. Next, the calculations have been repeated for the less time step Δt2=1/2Δt1. The difference Δφ8=φ8Δt1-φ8Δt2 has been determined for following tj=t0+i·Δt1.The calculations have been made forC=0.0025 Nm2. The graph of differences Δφ8 for following time instant is presented in Figure 8.Figure 8
Test of accuracy of generalized coordinateφ8 in the following time instant.The differencesΔφ8 are not big at all (between −0.00019 and 0.00022 rad), even after large number of iterations (i=400). The accuracy turned out to be satisfactory.


## 6.2. The Motion of the Elastica Fixed at Two Points

In this case, the elastica fixed at two points by joints as in Figure3 will be considered. The motion is under the influence of the gravity and elastic forces.The properties of the elastica are as follows:(i)
bending rigidity,C=0.0015 Nm2,(ii)
linear mass density,ρ=0.01 kg/m,(iii)
length,L=1.2 m.Number of points of division isn=16. The system of (40) has been solved using Runge- Kutta-Fehlberg 4th-order integration method with time step Δt=0,005 s. The initial conditions (28) are as follows:(44)φk0=π2[1-2(k-1)n-1]rad,ωk0=0rad/s,fork=1,2,…,n.
Finally, the shape of the elastica in the following time instant tj has been obtained. It is present in Figure 9.Figure 9
The shape of the elastica in the following time instant.In order to check accuracy of calculations, we tested how the boundary conditions (29) change as time goes by. In this way, we can check how the algorithm secures fulfillment of the boundary conditions(45)f1≡xn=0,f2≡yn-H=0,
As a measure of deviation of boundary conditions, we can take Σ=|f1|+|f2|, which, if the boundary conditions are fulfilled, should be equal zero. The calculations have been made for two time steps: Δt1=0.005 s i Δt2=1/2Δt1. The variation of Σ in respect of number of iterations are presented in Figure 10 for two time steps.Figure 10
The analysis of fulfillment of the boundary conditions∑=|f1|+|f2|.From Figure10, we can conclude that for two times less time step deviation of boundary conditions was significant less. For Δt1 after 300 iterations, deviation was about 2·10-4 m, and for Δt2=1/2Δt1, deviation was about 2·10-5 m (10 times less).


## 7. Conclusion

After carrying out series of numerical tests, it can be concluded that Lagrange's equations are practically useful for the analysis of the motion of heavy elastica. It can be used for simulation of many problems from the field of textile mechanics, for example, fabric folding (Figure11), motion of the sewing thread and so on. Similar investigations, but not using Lagrange's equations, have been described by Lloyd in [7].Figure 11
The simulation of fabric folding using Lagrange's equations.The results of calculations show clearly wave nature of phenomenon. The bending rigidity has not influence on the convergence of the results. The analyzed motion of elastica is stable owing to initial conditions.



---

*Source: 457462-2011-06-01.xml*