Table of Contents
ISRN Mechanical Engineering
Volume 2011 (2011), Article ID 457462, 8 pages
Research Article

Numerical Modeling of the Motion of the Flat Textile Structures

Department of Technical Mechanics and Informatics, Technical University of Lodz, Ulica Zeromskiego 116, 90 924 Lodz, Poland

Received 22 January 2011; Accepted 24 March 2011

Academic Editor: W.-H. Hsieh

Copyright © 2011 Piotr Szablewski. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


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.

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 weight 𝑞 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 of 𝑛 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 of 𝑛 particles in conservative force field, studyed in detail among other things in [25], are presented below:𝑑𝑑𝑡𝜕𝑇𝜕̇𝜑𝑘𝜕𝑇𝜕𝜑𝑘+𝜕𝑈𝜕𝜑𝑘=0,𝑘=1,2,,𝑛,(1) where 𝑇-kinetic energy, 𝑈-potential energy, 𝜑𝑘,̇𝜑𝑘𝑘th 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 into 𝑛 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:𝑥𝑖=𝑙𝑖𝑗=1sin𝜑𝑗,̇𝑥𝑖=𝑙𝑖𝑗=1cos𝜑𝑗̇𝜑𝑗,𝑦𝑖=𝑙𝑖𝑗=1cos𝜑𝑗,̇𝑦𝑖=𝑙𝑖𝑗=1sin𝜑𝑗̇𝜑𝑗.(2) The kinetic energy of ith mass is 𝑇𝑖=1/2𝑚(̇𝑥2𝑖+̇𝑦2𝑖).

Figure 2: Discrete model in coordinate system.

Therefore, the total kinetic energy of the system of 𝑛 mass is1𝑇=2𝑚𝑙2𝑛𝑖=1𝑖𝑗=1cos𝜑𝑗𝜑𝑗2+𝑖𝑗=1sin𝜑𝑗𝜑𝑗2.(3) The total potential energy of the system 𝑈 is the sum of two components𝑈=𝑉+𝑉𝑏,(4) where 𝑉 is the potential energy of gravity forces and 𝑉𝑏 is the potential energy of the bending.

The potential energy of gravity forces is 𝑉=𝑛𝑖=1𝑚𝑔𝑦𝑖=𝑚𝑔𝑙𝑛𝑖=1𝑖𝑗=1cos𝜑𝑗 and the potential energy of the bending 𝑉𝑏=1/2𝐶𝐿0(𝑑𝜑/𝑑𝑠)2𝑑𝑠. After substitution the derivative by difference quotient, we have𝑉𝑏=12𝐶𝑛1𝑖=1𝜑𝑖+1𝜑𝑖𝑙2𝐶𝑙=2𝑙𝑛1𝑖=1𝜑𝑖+1𝜑𝑖2.(5) Thus, the total potential energy of the system 𝑈 is𝑈=𝑉+𝑉𝑏=𝑚𝑔𝑙𝑛𝑖𝑖=1𝑗=1cos𝜑𝑗+𝐶2𝑙𝑛1𝑖=1𝜑𝑖+1𝜑𝑖2.(6) The formula for derivative of kinetic energy 𝑇 in respect of 𝜑𝑘 after transformations is as follows:𝜕𝑇𝜕𝜑𝑘=𝑚𝑙2𝑛𝑖𝑖=1𝑗=1𝜑sin𝑗𝜑𝑘̇𝜑𝑗̇𝜑𝑘,for𝑖𝑘,0,for𝑖<𝑘.(7) After changing the limits of summation, we can write𝜕𝑇𝜕𝜑𝑘=𝑚𝑙2𝑛𝑖𝑖=𝑘𝑗=1𝜑sin𝑗𝜑𝑘̇𝜑𝑗̇𝜑𝑘.(8) Similarly, the other derivatives can be written as follows:𝑑𝑑𝑡𝜕𝑇𝜕̇𝜑𝑘=𝑚𝑙2𝑛𝑖𝑖=𝑘𝑗=1𝜑cos𝑗𝜑𝑘̈𝜑𝑗𝜑sin𝑗𝜑𝑘̇𝜑𝑗̇𝜑𝑘̇𝜑𝑗,(9)𝜕𝑉𝜕𝜑𝑘=𝑚𝑔𝑙𝑛𝑖=𝑘sin𝜑𝑘=(𝑛𝑘+1)𝑚𝑔𝑙sin𝜑𝑘,(10)𝜕𝑉b𝜕𝜑𝑘=𝐶𝑙𝑛1𝑖=1𝜑𝑖+1𝜑𝑖𝜕𝜕𝜑𝑘𝜑𝑖+1𝜑𝑖.(11) The derivative 𝜕(𝜑𝑖+1𝜑𝑗)/𝜕𝜑𝑘 depending on the value of index 𝑖 is given by𝜕𝜕𝜑𝑘𝜑𝑖+1𝜑𝑖=1,for𝑖=𝑘1,1,for𝑖=𝑘,0,for𝑖<𝑘1,𝑖>𝑘.(12) Using summation from 𝑖=1 to 𝑖=𝑛1 the formula (11) can be written as follows:𝜕𝑉𝑏𝜕𝜑𝑘=𝐶𝑙𝜑1𝜑2𝐶,for𝑘=1,𝑙𝜑𝑘1+2𝜑𝑘𝜑𝑘+1𝐶,for𝑘=2,3,,𝑛1,𝑙𝜑𝑛𝜑𝑛1,for𝑘=𝑛.(13) Therefore, the derivative of total potential energy in respect of 𝜑𝑘 is𝜕𝑈𝜕𝜑𝑘=𝜕𝑉𝜕𝜑𝑘+𝜕𝑉𝑧𝑔𝜕𝜑𝑘=𝑚𝑔𝑙𝑛𝑖=𝑘sin𝜑𝑘+𝐶𝑙𝑛1𝑖=1𝜑𝑖+1𝜑𝑖𝜕𝜕𝜑𝑘𝜑𝑖+1𝜑𝑖.(14) Using (8), (9), and (14) in Lagrange's equations (1) and dividing by 𝑚𝑙2, we have𝑛𝑖𝑖=𝑘𝑗=1𝜑cos𝑗𝜑𝑘̈𝜑𝑗𝜑sin𝑗𝜑𝑘̇𝜑2𝑗𝑔(𝑛𝑘+1)𝑙sin𝜑𝑘+𝐶𝑚𝑙3×𝑛1𝑖=1𝜑𝑖+1𝜑𝑗𝜕𝜕𝜑𝑘𝜑𝑖+1𝜑𝑗=0for𝑘=1,2,,𝑛.(15) 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:𝜑𝚽=1𝜑2𝜑𝑛,𝑑𝚽=̇𝑑𝑡𝚽=̇𝜑1̇𝜑2̇𝜑𝑛,𝑑2𝚽𝑑𝑡2=̈𝚽=̈𝜑1̈𝜑2̈𝜑𝑛.(16) In the matrix form the Lagrange's equations (15) are𝐀̈𝚽+𝐁+𝐂+𝐃=0,(17) where A is an appropriate matrix and 𝐁,𝐂,𝐃 are vectors as follows:𝐴𝐀=𝐀(𝚽)=11𝐴12𝐴1𝑛𝐴21𝐴22𝐴2𝑛𝐴𝑛1𝐴𝑛2𝐴𝑛𝑛,̇𝚽=𝐵𝐁=𝐁𝚽,1𝐵2𝐵𝑛,𝐶𝐂=𝐂(𝚽)=1𝐶2𝐶𝑛,𝐷𝐃=𝐃(𝚽)=1𝐷2𝐷𝑛.(18) On the basis of (15) we can write the matrix A and vectors 𝐁,𝐂 and 𝐃 as follows:𝐴𝑘𝑗=𝑆𝑘𝑗𝜑cos𝑗𝜑𝑘,𝐵𝑘=𝑛𝑗=1𝑆𝑘𝑗𝜑sin𝑗𝜑𝑘̇𝜑2𝑗,𝐶𝑘𝑔=(𝑛𝑘+1)𝑙sin𝜑𝑘,𝐷𝑘=𝐶𝑚𝑙3𝜑1𝜑2𝐶,for𝑘=1,𝑚𝑙3𝜑𝑘1+2𝜑𝑘𝜑𝑘+1𝐶,for𝑘=2,3,,𝑛1,𝑚𝑙3𝜑𝑛𝜑𝑛1,for𝑘=𝑛,(19) where the matrix 𝐒 of dimension 𝑛×𝑛 is defined by𝐒=𝑛𝑛1𝑛2321𝑛1𝑛1𝑛2321𝑛2𝑛2𝑛2321333321222221111111.(20) The element of 𝑘th row and 𝑗th column of matrix 𝐒 can be written as𝑆𝑘𝑗𝑆=𝑛𝑘+1for𝑗𝑘,𝑘𝑗=𝑛𝑗+1for𝑗>𝑘.(21) 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 of 𝑛 2nd-order ordinary differential equations with unknown generalized coordinates Φ in matrix form̈𝚽=𝐀1(𝐁+𝐂+𝐃).(22)

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 ̇[Φ=Ω=𝜔1𝜔2𝜔𝑛]𝑇. In this way, we obtaiṅ̇𝚽=𝛀,𝛀=𝐀1(𝐁+𝐂+𝐃).(23) 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 unknowns𝑌𝐘=1𝑌2𝑌2𝑛,(24) where𝑌𝑖=𝜑𝑖𝑌for𝑖=1,2,,𝑛,𝑖=𝜔𝑖𝑛for𝑖=𝑛+1,𝑛+2,,2𝑛.(25) The right side of (22) has been denoted by vector 𝐅; that is,𝐅=𝐀1𝐹(𝐁+𝐂+𝐃),𝐅=1𝐹2𝐹𝑛.(26) Finally, the following system of differential equations was solving:̇𝑌1=𝑌𝑛+1,̇𝑌𝑛=𝑌2𝑛,̇𝑌𝑛+1=𝐹1,̇𝑌2𝑛=𝐹𝑛,(27) with initial conditions for 𝑡=𝑡0𝚽=𝚽0=𝜑01𝜑0𝑛,𝛀=𝛀0=𝜔01𝜔0𝑛.(28)

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 (Figure 2). 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 Figure 3 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:𝑥𝑛𝑦=0,𝑛𝐻=0.(29)

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) obtaining𝑓1𝑙𝑛𝑘=1sin𝜑𝑘𝑓=0,2𝑙𝑛𝑘=1cos𝜑𝑘𝐻=0.(30) Thus, we can write the Lagrange's equations in the following form:𝑑𝑑𝑡𝜕𝑇𝜕̇𝜑𝑘𝜕𝑇𝜕𝜑𝑘+𝜕𝑈𝜕𝜑𝑘+𝜆1𝜕𝑓1𝜕𝜑𝑘+𝜆2𝜕𝑓2𝜕𝜑𝑘=0,𝑘=1,2,,𝑛,(31) 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: 𝜕𝑓1𝜕𝜑𝑘=𝑙cos𝜑𝑘,𝜕𝑓2𝜕𝜑𝑘=𝑙sin𝜑𝑘.(32) Let right side of the 𝑘th (15) be denoted by Ψ𝑘(𝜑,̇𝜑). Therefore, using (31) and (32), we have the following system of equations: Ψ𝑘𝜆(𝜑,̇𝜑)+1𝑚𝑙cos𝜑𝑘𝜆2𝑚𝑙sin𝜑𝑘=0,𝑘=1,2,,𝑛.(33) Using matrix notation, we can write new matrix form of the system (33) 𝐀̈𝚽+𝐁+𝐂+𝐃+𝜆1𝐄+𝜆2𝐆=0,(34) where𝐸𝐄=𝐄(𝚽)=1𝐸2𝐸𝑛,𝐸𝑘=cos𝜑𝑘𝐺𝑚𝑙,𝑘=1,2,,𝑛,𝐆=𝐆(𝚽)=1𝐺2𝐺𝑛,𝐺𝑘=sin𝜑𝑘𝑚𝑙,𝑘=1,2,,𝑛.(35) 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 time𝑛𝑘=1cos𝜑𝑘̈𝜑𝑘sin𝜑𝑘̇𝜑2𝑘=0,𝑛𝑘=1sin𝜑𝑘̈𝜑𝑘cos𝜑𝑘̇𝜑2𝑘=0.(36) 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 have𝐀̈𝚽+𝐁+𝐂+𝐃=0,(37) similarly as (17) in Section 3. The components of new matrix 𝐀 and vectors 𝐃𝐁,𝐂i are as below:

For 𝑘=1,2,,𝑛2𝐴𝑘𝑗=𝐴𝑘𝑗cos𝜑𝑘+1𝐴𝑘+1𝑗cos𝜑𝑘𝜑sin𝑘+1𝜑𝑘+2𝐴𝑘+1𝑗cos𝜑𝑘+2𝐴𝑘+2𝑗cos𝜑𝑘+1𝜑×sin𝑘𝜑𝑘+1𝐵,for𝑗=1,2,,𝑛,𝑘=𝐵𝑘cos𝜑𝑘+1𝐵𝑘+1cos𝜑𝑘𝜑sin𝑘+1𝜑𝑘+2𝐵𝑘+1cos𝜑𝑘+2𝐵𝑘+2cos𝜑𝑘+1𝜑sin𝑘𝜑𝑘+1,𝐶𝑘=𝐶𝑘cos𝜑𝑘+1𝐶𝑘+1cos𝜑𝑘𝜑sin𝑘+1𝜑𝑘+2𝐶𝑘+1cos𝜑𝑘+2𝐶𝑘+2cos𝜑𝑘+1𝜑sin𝑘𝜑𝑘+1,𝐷𝑘=𝐷𝑘cos𝜑𝑘+1𝐷𝑘+1cos𝜑𝑘𝜑sin𝑘+1𝜑𝑘+2𝐷𝑘+1cos𝜑𝑘+2𝐷𝑘+2cos𝜑𝑘+1𝜑sin𝑘𝜑𝑘+1,𝐴𝑛1𝑗=cos𝜑𝑗,𝐴𝑛𝑗=sin𝜑𝑗𝐵,for𝑗=1,2,,𝑛,𝑛1=𝑛𝑗=1sin𝜑𝑗̇𝜑2𝑗,𝐵𝑛=𝑛𝑗=1cos𝜑𝑗̇𝜑2𝑗,𝐶𝑛1=𝐶𝑛𝐷=0,𝑛1=𝐷𝑛=0.(38) The system of (37) has been transformed to the form similarly as in Section 5; that is,̈𝐀𝚽=1𝐃𝐁+𝐂+.(39) Next, we can writė̇𝐀𝚽=𝛀,𝛀=1𝐃.𝐁+𝐂+(40) 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 coordinate 𝑥 of the end B𝑥 must be always equal to value 𝑎= const., and we have only one algebraic equation (conditions of constraints) in the form of𝑓1𝑥𝑛𝑎=0.(41) 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 of 𝜑=𝛼, where 𝑥=𝑎,𝑦=𝑏, then there are three conditions of constraints𝑓1𝑥𝑛𝑓𝑎=0,2𝑦𝑛𝑓𝑏=0,3𝜑𝑛𝛼=0,(42) 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 Figure 2.

In the second one, the elastica was fixed by joint at two ends as in Figure 3.

6.1. The Motion of the Elastica Fixed at One Point

In this case, the elastica fixed at one point by joint as in Figure 2 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 points 𝑛 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:𝜑0𝑘=𝜋2𝜔rad,0𝑘=0rad/s,for𝑘=1,2,,𝑛.(43) 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 from 𝑡0 to 𝑡20.
Figure 5: The variation of generalized coordinates 𝜑1,𝜑2,,𝜑8 in the respect of time 𝑡.
Figure 6: The variation of generalized coordinates 𝜑1,𝜑2,,𝜑8 in the respect of time 𝑡 for bending rigidity of 𝐶=0.00025 Nm2.

In Figures 5 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 (Figure 7) 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 coordinate 𝜑8, its values marked by 𝜑Δ𝑡18 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=𝜑Δ𝑡18𝜑Δ𝑡28 has been determined for following 𝑡𝑗=𝑡0+𝑖Δ𝑡1.

The calculations have been made for 𝐶=0.0025 Nm2. The graph of differences Δ𝜑8 for following time instant is presented in Figure 8.

Figure 8: Test of accuracy of generalized coordinate 𝜑8 in the following time instant.

The differences Δ𝜑8 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 Figure 3 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 is 𝑛=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:𝜑0𝑘=𝜋212(𝑘1)𝜔𝑛1rad,0𝑘=0rad/s,for𝑘=1,2,,𝑛.(44) 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 conditions𝑓1𝑥𝑛𝑓=0,2𝑦𝑛𝐻=0,(45) 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 Figure 10, 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 2104 m, and for Δ𝑡2=1/2Δ𝑡1, deviation was about 2105 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 (Figure 11), 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.


  1. P. Szablewski and W. Kobza, “Numerical analysis of Peirce’s Cantilever test for the bending rigidity of textiles,” Fibres and Textiles in Eastern Europe, vol. 11, no. 4, pp. 54–57, 2003. View at Google Scholar · View at Scopus
  2. J. Leyko, Dynamika Układów Materialnych, vol. II, PWN, Warszawa, Poland, 1961.
  3. W. Greiner, Classical mechanics: Systems of Particles and Hamiltonian Dynamics, Springer, New York, NY, USA, 2003.
  4. O.M. O'Reilly, Intermediate Dynamics for Engineers: A Unified Treatment of Newton Euler and Lagrangian Mechanics, Cambridge University Press, New York, NY, USA, 2008.
  5. J. R. Taylor, Classical Mechanics, University Science Books, Mill Valley, Calif, USA, 2005.
  6. P. Szablewski, “Modelling of dynamical behaviour of flat textile structure using lagrange’s equations,” in Proceedings of the 10th International Conference (IMTEX '09), pp. 35–38, Łódź, Poland, 2009.
  7. D. W. Lloyd, W. J. Shanahan, and M. Konopasek, “The folding of heavy fabric sheets,” International Journal of Mechanical Sciences, vol. 20, no. 8, pp. 521–527, 1978. View at Google Scholar · View at Scopus