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Deflection Analysis of Woven Composite Planes under In-Plane Loading
Available Available online online at at www.sciencedirect.com www.sciencedirect.com
Procedia Engineering 00 (2011) 000–000 Procedia Engineering 14 (2011) 2839–2847
Procedia Engineering www.elsevier.com/locate/procedia
The Twelfth East Asia-Pacific Conference on Structural Engineering and Construction
Deflection Analysis of Woven Composite Planes under InPlane Loading D. DERAKHSHAN1a, R.T. Faal2b 1,2
Faculty of Engineering, Zanjan University, Zanjan, Iran
Abstract Deflection analysis of yarns of a bi-axial woven composite plate under tensional loading is investigated in this article. A unit part of the warp yarn which is limited by two adjacent weft yarns is modeled by a curved beam. Using the Winkler theory of curved beams, the strain energy under the tensile loading is derived. Initial shape of yarns is chosen to be arc shaped. The deflection analysis is accomplished by making use of energy method. First, the external work of the horizontal (or tensile) force and also the vertical contacting force is evaluated. Second, using the variational method the governing equations and boundary conditions are derived and then solved to find out the displacement fields exactly. Finally, making use of the continuity of the deflection of two contacting yarns at the contacting point the vertical contacting force is obtained. It must be mentioned that shear forces and slipping of contacting yarns are disregarded here. The validation of the work is performed by comparison of results with the semi circle beam (yarn) which is found in some usual text books.
© 2011 Published by Elsevier Ltd. Keywords: (Woven composites, In-plane loading, Curved beam, Winkler theory, Yarn)
1
INTRODUCTION
Among various kinds of materials which have superior properties such as lightness, strength, corrosion resistance, thermal and electrical insulation, composite materials have magnificent situation in mechanical engineering. During few decays ago, woven composite is under consideration because of better strength, rigidity, toughness and creep resistance performance. Literature is replete with studies related to
a b
Corresponding author: Email: [email protected] Presenter: [email protected]
1877–7058 © 2011 Published by Elsevier Ltd. doi:10.1016/j.proeng.2011.07.357
2840
D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
2
applications, fabrications and numerical analysis of woven composites. Buckling, slipping and stretching of fibers of woven composite plates which play an important rule in the behavior of those were the subject of some pertinent studies. A great deal of work (for example) (Ishikawa and Chou 1982; Naik and Shembekar 1992; Duosheng et a.l. 2005) had been done in analytical manner to obtain models for elastic deformation, stiffness, strength behaviors and even stability (buckling) predictor. The majority of these analytical models had been developed by rather simple approximating assumptions so that the reliability of models may not be confirmed completely. Also, a few investigations had been presented based on the finite element models (Haan et al. 2001; Boisse et al. 2005; Iannucci 2006). Some ones introduced an elastic woven unit cell which simulates fabric woven composite including yarns and matrix. Others accomplished works related to forming and impacting of woven composites were also analyzed by finite element models. Recently, anti-plane loading (or bending) of multi-layer woven composite was taken under consideration (Li et al. 2008). Nowadays, the experimental methods are another subject of investigation of woven composites. Picture frame and bias extension tests are well-known experimental methods which have been utilized in many papers (Harrison et al. 2004). In this work a robust analytical method is used to evaluate displacement fields based on Winkler’s theory (Langhaar 1962) which develops a linear elastic curved beam formulation. The theory incorporates the hypothesis which states that area cross section remains plane and perpendicular to centroidal line before and after deformation. Also, Hook’s law associates with variational method to derive the relations of the displacements and internal forces. The work has been compared with analytical solutions which are brought in usual mechanical engineering text books. 2
DERIVATION OF GOVERNING EQUATION OF CURVED BEAM
In the woven composite plates, fabrics play an important rule in the behavior of those. Therefore initially we analyze the fabric yarn of a woven composite plate which is considered a curved beam. Analysis was done for constant curvature based on Winkler theory (Langhaar 1962). Consider an arc shaped yarn, which u, v and ρ denotes to radial and circumferential displacement components and also the radius of curvature of a point on the centroidal axis, respectively. These displacements are functions of s, the arc length measured on the centroidal axis. Suppose an ordinate, z in a cross section of beam with area A and moment of inertia I , measured from centroidal axis (positive outward). According to Winkler theory of curved beams, the dimensionless constant Z is defined by (Langhaar 1962) dA
A
∫ ρ + z = ρ (1 + Z )
(1)
A
∫
∫
The net axial tension of beam N = E ε dA and the bending moment M = E zε dA about the axial strain, are as follows (LanghaarA1962) centroidal axis of cross section which ε designates A N = ( EA / ρ )[u + ρv ′ + Z (u + ρ 2 u ′′)] M = − EAZ (u + ρ 2 u ′′)
(2)
where prim denotes to differentiation with respect to argument s. The strain energy of beam is
∫
U = ( EA / 2 ρ 2 )[(u + ρv ′) 2 + Z (u + ρ 2 u ′′) 2 ]ds
(3)
In view of Eqs. (2), the strain energy of beam may be rewritten in terms of net axial tension of beam and the bending moment of beam as
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
∫
U = (1 / 2 ρ 2 EA)[( M + ρN ) 2 + M / Z ]ds
(4)
We consider a composite plate under in-plane tensional loading P which is only applied on two opposite sides of plate. Therefore this loading is applied to parallel yarns which are only in the direction of loading. The upper yarn (as a curved beam) is subjected to axial load P at the right end and vertical load Q which is applied by other yarn on the mid point of beam length shown in Fig. 1. Consider the axial load P and lateral load Q (due to contacted yarn). By virtue of Eq. (3), the total potential energy of beam is readily written as V = − Pv(α ) cos +
α /2
∫
0
α 2
− Pu (α ) sin
α 2
− Qu (α / 2) +
α /2
∫
0
( EA / 2 ρ 2 )[(u + v ′) 2 + Z (u + u ′′) 2 ]ρdθ
(5)
2
( EA / 2 ρ )[(u + v ′) 2 + Z (u + u ′′) 2 ]ρdφ
Where prim designates to derivation with respect to θ and φ . Because of existence of point load Q , the dummy variables of the first and second integrals are θ and φ respectively.
y s
Q
u
lower yarn upper yarn
v
P
x
Figure 1: Curved beam model for woven yarns under axial and lateral loads
Setting δV = 0 leads to the following governing equations u + v ′ + Z (u ′′′′ + 2u ′′ + u ) = 0 u ′ + v ′′ = 0
(6)
Integration of the second equation with respect to θ gives u + v ′ = − A6 Z where A6 is constant. Substituting this equation into the first equation of (6) results in u ′′′′ + 2u ′′ + u = A6 which the solution for u, v are readily given as
= u A1 sin θ + A2 cos θ + A3θ sin θ + A4θ cos θ + A6 ;
0 <θ <α /2
v = A1 cos θ − A2 sin θ + A3 (sin θ − θ cos θ ) + A4 (cos θ + θ sin θ ) + A5 = u B1 sin φ + B2 cos φ + B3φ sin φ + B4φ cos φ + B6 ;
0 <φ <α /2
v = B1 cos φ − B2 sin φ + B3 (sin φ − φ cos φ ) + B4 (cos φ + φ sin φ ) + B5
(7)
3
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
4
The following forced and natural boundary conditions and the continuity and jump conditions are attained by letting δV = 0
u (0) 0;= v(0) 0; = M (0) 0 = at θ 0 = = (u )θ α = (u= )φ 0= ; (v )θ α= (v= )φ 0 ;= (u ′)θ α /= (u ′)φ = /2 /2 2 (S )
= θ α /2
Q + ( S )φ 0= = ; ( N )θ =
cos(= α / 2)(u )φ sin(α / 2)( S )
α /2
= ( N ) ; (M )
0
= (M )
(8)
α /= 2 φ 0= θ α /= 2 φ 0
= sin(= α / 2)(v)φ
; (M )
α= /2 φ α /2
+ cos(α / 2)( N )
= φ α= /2 φ α /2
P =
The equations (8) are also derived by applying equilibrium conditions for an infinitesimal arc subjects to internal moment M , axial force N and shear force S . Continuity of displacements u and v and rotation u ′ are important conditions which must be impose to this analysis. The shear force S has a relation with the bending moment M as S = dM ρ dθ = − EAZ (u ′ + u ′′′) ρ which helps us to attain S in terms of coefficients Ai and Bi , i = 1,2,...,6 . It is worth to mention it that there are twelve conditions associate with twelve coefficients Ai and Bi which are solved simultaneously as follows A2 = 2 A3 = − A6 = − B6 = −2 K [Q sin(α / 2) + 2 P cos(α / 2)]
A4 = K [Q cos(α / 2) − 2 P sin(α / 2)]
(9)
B3 = −2 B4 = − B5 = 2 KP T
{ A1 A5 B1 B2 } = K{P[2(1 + Z )α cos(α / 2)χ 1 (α ) + χ 2 (α )]
+ 0.5Q[2(1 + Z )α sin(α / 2)χ 1 (α ) + χ 3 (α )]}
where K = ρ /( 4 AEZ ) and χ 1 (α ), χ 2 (α ) and χ 3 (α ) are the four dimensional vectors as χ 1 (α ) = {1, − 1, cos(α / 2), sin(α / 2)}T χ 2 (α ) = {−2 sin(α / 2), 0, α - 3 sin α , 1 + 3 cos α }T
(10) T
χ 3 (α ) = {2[cos(α / 2) − 2], 4, 2 cos(α / 2)[3 cos(α / 2) − 2], 2 sin(α / 2)[3 cos(α / 2) − 2] + α }
Substituting the above coefficients into the displacements (7), the horizontal and vertical displacements of beam ( u and Δ respectively) at the end of beam, φ = α / 2 and the midpoint θ = α / 2 can be evaluated in the following forms {(u )θ =α / 2 , (Δ ) φ =α / 2 }T = K{P[2(1 + Z )α cos(α / 2)χ 4 (α ) + χ 5 (α )] + 0.5Q[2(1 + Z )α sin(α / 2)χ 4 (α ) + χ 6 (α )]}
(11)
where χ 4 (α ), χ 5 (α ) and χ 6 (α ) are two dimensional vectors as χ 4 (α ) = {sin(α / 2), 2 cos(α / 2)}T χ 5 (α ) = 2{3 cos 2 (α / 2) − 2 cos(α / 2) − 1, α - 3 sin α }T
(12)
χ 6 (α ) = {α + 3 sin α − 8 sin(α / 2) , 4[ 3 cos 2 (α / 2) − 2 cos(α / 2) − 1]}T
In the special case, Q = 0 and α = π , the above-mentioned displacements are simplified to (u )θ α= = / 2 =− ( Δ )φ α / 2 π =− ρ P / 2 EAZ . For the small values of ρ / z , the truncated series
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
1 /( ρ + z ) ≈ (1 / ρ )[1 − ( ρ / z ) + ( ρ / z ) 2 ] is used to evaluate the integral (1) and we arrive 3 (u )θ α= at Z ≈ I /( Aρ 2 ). Therefore = / 2 =− ( Δ )φ α / 2 π =− ρ P / 2 EI . The lower yarn (weft) is only under the vertical contacting force −Q and for upper yarn (warp) of course both P and Q. Thus we may rewrite the Eqs. (11) for the lower yarn, Fig. 1, by setting P = 0 and replacing Q by −Q. The vertical displacements of two yarns at the contacting point, θ = α / 2 are identical which is implies that Q = λ (α ) P where
λ (α ) =
(1 + Z ) α sin α + 2 ⎡⎣3cos 2 (α / 2 ) − 2 cos (α / 2 ) − 1⎤⎦ 2 (1 + Z ) α sin 2 (α / 2 ) + α + 3sin α − 8sin (α / 2 )
(13)
Substituting Q = λ (α ) P into Eqs. (11) gives the displacements in terms of external force P as follows
{(u )θ =α / 2 , (Δ) φ =α / 2 }T = KP{ [2(1 + Z )α cos(α / 2)χ 4 (α ) + χ 5 (α )]
(14)
+ 0.5λ (α )[2(1 + Z )α sin(α / 2)χ 4 (α ) + χ 6 (α )]}
Moreover, the internal forces such as axial and shear forces and also the internal moment can be easily derived at any points using Eqs. (7) and (9). After some manipulations, the results are 1 ⎧ 1 1 ⎪ξ = S / P = sin(θ − 2 α ) + 2 λ (α ) cos(θ − 2 α ) ⎪ 1 ⎪ 1 1 for 0 < θ < 12 α ⎨η = N / P = cos(θ − 2 α ) − λ (α ) sin(θ − 2 α ) 2 ⎪ 1 ⎪ 1 1 1 1 ⎪μ = M / ρP = cos 2 α − cos(θ − 2 α ) + 2 λ (α )(sin(θ − 2 α ) + sin 2 α ) ⎩ 1 ⎧ ⎪ξ = S / p = sin φ − 2 λ (α ) cos φ ⎪ 1 ⎪ for 0 < φ < 12 α ⎨η = N / P = cos φ + λ (α ) sin φ 2 ⎪ 1 ⎪ 1 1 ⎪μ = M / ρP = cos 2 α − cos φ + 2 λ (α )(− sin φ + sin 2 α ) ⎩
(15)
where the dimensionless terms ξ , η and μ can be useful to analysis of the magnifications of the internal moments and forces to external applied force. The results of Eqs. (15) are also proved by satisfying the equilibrium conditions for the beam. (see Fig. 2). S P
A
Q/2
M
Q M
N N
S
B
P
Q/2
Figure 2: Internal forces and support reactions coming into view at the equilibrium conditions
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
6
3
NUMERICAL EXAMPLES AND RESULTS
As it has mentioned before, the validity of displacement relations are confirmed by comparing with special case of those available in the text books. In what follows, the displacement of single warp yarn and bi-axial woven yarns subjected to an in-plane simple tension are evaluated and compared with each other. Also, the bending moment, axial and shear forces diagrams along the centroidal axis are depicted. Specifications of steel yarns such as material properties, dimensions, number and amount of loading are given in table 1, which are used in all figures. The plot of dimensionless displacements at end (Δ / PK ) and mid-point (δ / PK ) versus α are shown in Fig. 3. Table 1 Specifications of steel yarns
Yong’s modulus
Yarn’s diameter
length × Width
Number of yarns
Axial load /yarn
E = 207 GPa
d = 0.5[mm]
1000 × 1000[mm 2 ]
400
P = 10[kN ]
3.5
0.2
3
0
Dimensionless Vertical Displacment (δ/PK)
Dimensionless Horizontal Displacment (Δ/PK)
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2.5
2
1.5
1
0.5
0
−0.2
−0.4
−0.6
−0.8
−1
0
50
100 α
150
−1.2
0
50
100 α
150
Figure 3: Left: Dimensionless end horizontal displacement, Right: Dimensionless midpoint vertical displacement
The variation of the ratio of the end horizontal displacement of upper yarn (warp) in presence of lower yarn (weft) to that, in the absence of weft versus arc angle (0 < α < 80) is shown in Fig. 4. This variation is almost slow for arc angles bigger than 70 o . Similar geometry and construction of warp and weft yarns imply that the vertical distance of mid-point from each support, Fig 1. equals to yarn half diameter. Also, the horizontal distance of supports of curved beam or spans of the curved beam (L), Fig. 1 is specified using the ratio of width of composite plane to the number of yarns. So, the radius ρ may be calculated in the terms of L and arc angle α as ρ = L /(2 sin(α / 2)) and also in view of Fig. 1 we may easily write d = 2 ρ[1 − cos(α / 2)]. Therefore, the arc angle α may be evaluated using two previous relations as α = 4 tan −1 (d / L) where d denotes to the
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
cross section diameter of curved beam, respectively. Using the data of table 1, we obtain the span length L = 1000 /(400 − 1) = 2.506 [mm], and the arc angle α = 45.13o and also ρ = 3.265 [mm] . 1 0.95 0.9
Displacment Ratio
0.85 0.8 0.75 0.7 0.65 0.6 0.55 0.5
0
20
40
60
80
α
100
120
140
160
180
Figure 4: Horizontal displacement ratio when weft is present to that when weft is absent 0.3
1.015 1.01
0.2 1.005 1
Dimensionless Axial Force
Dimensionless Shear Force
0.1
0
−0.1
0.995 0.99 0.985 0.98
−0.2
0.975 −0.3 0.97 −0.4
0
10
20
θ
30
40
0.965
0
10
Figure 5: Dimensionless shear and axial force versus θ
20
θ
30
40
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
8 0
−0.005
−0.01
Dimensionless Bending Moment
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−0.015
−0.02
−0.025
−0.03
−0.035
−0.04
0
5
10
15
20
θ
25
30
35
40
45
Figure 6: Dimensionless bending moment versus angle θ
Making use of equations (15), the dimensionless displacements are evaluated which are plotted in Fig. 5. Jump in the left graph and extremums in the right are compatible with physical features of problem. The similar trend of dimensionless axial force can be seen for the dimensionless bending moment along the centroidal axis. The extremums are located at the angles nearly θ = 15 o and φ = 30 o (see Fig. 6). So, it is expected that the maximum normal stress may occur at these points. Accurate solution is obtained by first differentiation of η or μ with respect to θ (or φ ). 4
CONCLUSIONS
The main conclusions of this paper may be listed as follows: The end dimensionless horizontal displacement is increased rapidly with the increasing the arc angle and reversely the midpoint dimensionless vertical displacement is reduced. 2. The variation of the ratio of the end horizontal displacement of warp in the presence of weft to that in the absence of weft is almost slow for arc angles bigger than 70 o and in this bond the weft leads to more stiffening or rigidity. 3. Maximum normal stress of yarn cross section occurs at the angles before and after the contacting point. (Look at Fig. 6). 1.
References [1]
Van der Geer J, Hanraads JAJ, Lupton RA. The art of writing a scientific article. J Sci Commun 2000;163:51–9.
[2]
Strunk Jr W, White EB. The elements of style. 3rd ed. New York: Macmillan; 1979.
[3]
Mettam GR, Adams LB. How to prepare an electronic version of your article. In: Jones BS, Smith RZ, editors. Introduction to the electronic age, New York: E-Publishing Inc; 1999, p. 281–304
D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000 [1]
Boisse P, Zouari B and Gasser A (2005). A mesoscopic approach for the simulation of woven fiber composite forming,
[2]
Duosheng X, Rajamohan G, Suong VH (2005). Buckling analysis of tri-axial woven fabric composite structures, Part I: Non-
[3]
Haan PG, Charalambides PG, Suri M (2001). A specialized finite element for study of woven composites, Computational
[4]
Harrison P, Clifford MJ and Long AC (2004). Shear characterization of viscous woven textile composites: A comparison
[5]
Iannucci L (2006). Progressive failure modelling of woven carbon composite under impact, International Journal of impact
[6]
Ishikawa T, Chou TW ,(1982) Elastic behavior of woven hybrid composite. J. Mater. Sci. 17:3211-3220.
[7] [8]
Langhaar HL (1962). Energy Methods in Applied Mechanics, John Wiley and Sons, Inc., New York, London. 2-3, 3-6. Li L, Kim SM, Song SH, Ku TW, Song WJ, Kim J, Chong MK, Park JW, Kang BS (2008). Finite element modeling and
Composites Science and Technology 65, 429–436. linear finite element formulation, composite structures 67, 37-55. Mechanics 27, 445-462. between picture frame and bias extension experiments, Composites Science and Technology 64, 1453–1465. Engineering 32, 1013–1043.
simulation for bending analysis of multi-layer printed circuit board, using woven fiber composite, Journal of Material Processing Technology 201, 746-750. [9]
Naik NK, Shembekar PS. (1992). Elastic behavior of woven fabric composites: I-lamina analysis. J. composite Mater. 26(15): 2196-2225.
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Procedia Engineering 00 (2011) 000–000 Procedia Engineering 14 (2011) 2839–2847
Procedia Engineering www.elsevier.com/locate/procedia
The Twelfth East Asia-Pacific Conference on Structural Engineering and Construction
Deflection Analysis of Woven Composite Planes under InPlane Loading D. DERAKHSHAN1a, R.T. Faal2b 1,2
Faculty of Engineering, Zanjan University, Zanjan, Iran
Abstract Deflection analysis of yarns of a bi-axial woven composite plate under tensional loading is investigated in this article. A unit part of the warp yarn which is limited by two adjacent weft yarns is modeled by a curved beam. Using the Winkler theory of curved beams, the strain energy under the tensile loading is derived. Initial shape of yarns is chosen to be arc shaped. The deflection analysis is accomplished by making use of energy method. First, the external work of the horizontal (or tensile) force and also the vertical contacting force is evaluated. Second, using the variational method the governing equations and boundary conditions are derived and then solved to find out the displacement fields exactly. Finally, making use of the continuity of the deflection of two contacting yarns at the contacting point the vertical contacting force is obtained. It must be mentioned that shear forces and slipping of contacting yarns are disregarded here. The validation of the work is performed by comparison of results with the semi circle beam (yarn) which is found in some usual text books.
© 2011 Published by Elsevier Ltd. Keywords: (Woven composites, In-plane loading, Curved beam, Winkler theory, Yarn)
1
INTRODUCTION
Among various kinds of materials which have superior properties such as lightness, strength, corrosion resistance, thermal and electrical insulation, composite materials have magnificent situation in mechanical engineering. During few decays ago, woven composite is under consideration because of better strength, rigidity, toughness and creep resistance performance. Literature is replete with studies related to
a b
Corresponding author: Email: [email protected] Presenter: [email protected]
1877–7058 © 2011 Published by Elsevier Ltd. doi:10.1016/j.proeng.2011.07.357
2840
D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
2
applications, fabrications and numerical analysis of woven composites. Buckling, slipping and stretching of fibers of woven composite plates which play an important rule in the behavior of those were the subject of some pertinent studies. A great deal of work (for example) (Ishikawa and Chou 1982; Naik and Shembekar 1992; Duosheng et a.l. 2005) had been done in analytical manner to obtain models for elastic deformation, stiffness, strength behaviors and even stability (buckling) predictor. The majority of these analytical models had been developed by rather simple approximating assumptions so that the reliability of models may not be confirmed completely. Also, a few investigations had been presented based on the finite element models (Haan et al. 2001; Boisse et al. 2005; Iannucci 2006). Some ones introduced an elastic woven unit cell which simulates fabric woven composite including yarns and matrix. Others accomplished works related to forming and impacting of woven composites were also analyzed by finite element models. Recently, anti-plane loading (or bending) of multi-layer woven composite was taken under consideration (Li et al. 2008). Nowadays, the experimental methods are another subject of investigation of woven composites. Picture frame and bias extension tests are well-known experimental methods which have been utilized in many papers (Harrison et al. 2004). In this work a robust analytical method is used to evaluate displacement fields based on Winkler’s theory (Langhaar 1962) which develops a linear elastic curved beam formulation. The theory incorporates the hypothesis which states that area cross section remains plane and perpendicular to centroidal line before and after deformation. Also, Hook’s law associates with variational method to derive the relations of the displacements and internal forces. The work has been compared with analytical solutions which are brought in usual mechanical engineering text books. 2
DERIVATION OF GOVERNING EQUATION OF CURVED BEAM
In the woven composite plates, fabrics play an important rule in the behavior of those. Therefore initially we analyze the fabric yarn of a woven composite plate which is considered a curved beam. Analysis was done for constant curvature based on Winkler theory (Langhaar 1962). Consider an arc shaped yarn, which u, v and ρ denotes to radial and circumferential displacement components and also the radius of curvature of a point on the centroidal axis, respectively. These displacements are functions of s, the arc length measured on the centroidal axis. Suppose an ordinate, z in a cross section of beam with area A and moment of inertia I , measured from centroidal axis (positive outward). According to Winkler theory of curved beams, the dimensionless constant Z is defined by (Langhaar 1962) dA
A
∫ ρ + z = ρ (1 + Z )
(1)
A
∫
∫
The net axial tension of beam N = E ε dA and the bending moment M = E zε dA about the axial strain, are as follows (LanghaarA1962) centroidal axis of cross section which ε designates A N = ( EA / ρ )[u + ρv ′ + Z (u + ρ 2 u ′′)] M = − EAZ (u + ρ 2 u ′′)
(2)
where prim denotes to differentiation with respect to argument s. The strain energy of beam is
∫
U = ( EA / 2 ρ 2 )[(u + ρv ′) 2 + Z (u + ρ 2 u ′′) 2 ]ds
(3)
In view of Eqs. (2), the strain energy of beam may be rewritten in terms of net axial tension of beam and the bending moment of beam as
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
∫
U = (1 / 2 ρ 2 EA)[( M + ρN ) 2 + M / Z ]ds
(4)
We consider a composite plate under in-plane tensional loading P which is only applied on two opposite sides of plate. Therefore this loading is applied to parallel yarns which are only in the direction of loading. The upper yarn (as a curved beam) is subjected to axial load P at the right end and vertical load Q which is applied by other yarn on the mid point of beam length shown in Fig. 1. Consider the axial load P and lateral load Q (due to contacted yarn). By virtue of Eq. (3), the total potential energy of beam is readily written as V = − Pv(α ) cos +
α /2
∫
0
α 2
− Pu (α ) sin
α 2
− Qu (α / 2) +
α /2
∫
0
( EA / 2 ρ 2 )[(u + v ′) 2 + Z (u + u ′′) 2 ]ρdθ
(5)
2
( EA / 2 ρ )[(u + v ′) 2 + Z (u + u ′′) 2 ]ρdφ
Where prim designates to derivation with respect to θ and φ . Because of existence of point load Q , the dummy variables of the first and second integrals are θ and φ respectively.
y s
Q
u
lower yarn upper yarn
v
P
x
Figure 1: Curved beam model for woven yarns under axial and lateral loads
Setting δV = 0 leads to the following governing equations u + v ′ + Z (u ′′′′ + 2u ′′ + u ) = 0 u ′ + v ′′ = 0
(6)
Integration of the second equation with respect to θ gives u + v ′ = − A6 Z where A6 is constant. Substituting this equation into the first equation of (6) results in u ′′′′ + 2u ′′ + u = A6 which the solution for u, v are readily given as
= u A1 sin θ + A2 cos θ + A3θ sin θ + A4θ cos θ + A6 ;
0 <θ <α /2
v = A1 cos θ − A2 sin θ + A3 (sin θ − θ cos θ ) + A4 (cos θ + θ sin θ ) + A5 = u B1 sin φ + B2 cos φ + B3φ sin φ + B4φ cos φ + B6 ;
0 <φ <α /2
v = B1 cos φ − B2 sin φ + B3 (sin φ − φ cos φ ) + B4 (cos φ + φ sin φ ) + B5
(7)
3
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
4
The following forced and natural boundary conditions and the continuity and jump conditions are attained by letting δV = 0
u (0) 0;= v(0) 0; = M (0) 0 = at θ 0 = = (u )θ α = (u= )φ 0= ; (v )θ α= (v= )φ 0 ;= (u ′)θ α /= (u ′)φ = /2 /2 2 (S )
= θ α /2
Q + ( S )φ 0= = ; ( N )θ =
cos(= α / 2)(u )φ sin(α / 2)( S )
α /2
= ( N ) ; (M )
0
= (M )
(8)
α /= 2 φ 0= θ α /= 2 φ 0
= sin(= α / 2)(v)φ
; (M )
α= /2 φ α /2
+ cos(α / 2)( N )
= φ α= /2 φ α /2
P =
The equations (8) are also derived by applying equilibrium conditions for an infinitesimal arc subjects to internal moment M , axial force N and shear force S . Continuity of displacements u and v and rotation u ′ are important conditions which must be impose to this analysis. The shear force S has a relation with the bending moment M as S = dM ρ dθ = − EAZ (u ′ + u ′′′) ρ which helps us to attain S in terms of coefficients Ai and Bi , i = 1,2,...,6 . It is worth to mention it that there are twelve conditions associate with twelve coefficients Ai and Bi which are solved simultaneously as follows A2 = 2 A3 = − A6 = − B6 = −2 K [Q sin(α / 2) + 2 P cos(α / 2)]
A4 = K [Q cos(α / 2) − 2 P sin(α / 2)]
(9)
B3 = −2 B4 = − B5 = 2 KP T
{ A1 A5 B1 B2 } = K{P[2(1 + Z )α cos(α / 2)χ 1 (α ) + χ 2 (α )]
+ 0.5Q[2(1 + Z )α sin(α / 2)χ 1 (α ) + χ 3 (α )]}
where K = ρ /( 4 AEZ ) and χ 1 (α ), χ 2 (α ) and χ 3 (α ) are the four dimensional vectors as χ 1 (α ) = {1, − 1, cos(α / 2), sin(α / 2)}T χ 2 (α ) = {−2 sin(α / 2), 0, α - 3 sin α , 1 + 3 cos α }T
(10) T
χ 3 (α ) = {2[cos(α / 2) − 2], 4, 2 cos(α / 2)[3 cos(α / 2) − 2], 2 sin(α / 2)[3 cos(α / 2) − 2] + α }
Substituting the above coefficients into the displacements (7), the horizontal and vertical displacements of beam ( u and Δ respectively) at the end of beam, φ = α / 2 and the midpoint θ = α / 2 can be evaluated in the following forms {(u )θ =α / 2 , (Δ ) φ =α / 2 }T = K{P[2(1 + Z )α cos(α / 2)χ 4 (α ) + χ 5 (α )] + 0.5Q[2(1 + Z )α sin(α / 2)χ 4 (α ) + χ 6 (α )]}
(11)
where χ 4 (α ), χ 5 (α ) and χ 6 (α ) are two dimensional vectors as χ 4 (α ) = {sin(α / 2), 2 cos(α / 2)}T χ 5 (α ) = 2{3 cos 2 (α / 2) − 2 cos(α / 2) − 1, α - 3 sin α }T
(12)
χ 6 (α ) = {α + 3 sin α − 8 sin(α / 2) , 4[ 3 cos 2 (α / 2) − 2 cos(α / 2) − 1]}T
In the special case, Q = 0 and α = π , the above-mentioned displacements are simplified to (u )θ α= = / 2 =− ( Δ )φ α / 2 π =− ρ P / 2 EAZ . For the small values of ρ / z , the truncated series
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1 /( ρ + z ) ≈ (1 / ρ )[1 − ( ρ / z ) + ( ρ / z ) 2 ] is used to evaluate the integral (1) and we arrive 3 (u )θ α= at Z ≈ I /( Aρ 2 ). Therefore = / 2 =− ( Δ )φ α / 2 π =− ρ P / 2 EI . The lower yarn (weft) is only under the vertical contacting force −Q and for upper yarn (warp) of course both P and Q. Thus we may rewrite the Eqs. (11) for the lower yarn, Fig. 1, by setting P = 0 and replacing Q by −Q. The vertical displacements of two yarns at the contacting point, θ = α / 2 are identical which is implies that Q = λ (α ) P where
λ (α ) =
(1 + Z ) α sin α + 2 ⎡⎣3cos 2 (α / 2 ) − 2 cos (α / 2 ) − 1⎤⎦ 2 (1 + Z ) α sin 2 (α / 2 ) + α + 3sin α − 8sin (α / 2 )
(13)
Substituting Q = λ (α ) P into Eqs. (11) gives the displacements in terms of external force P as follows
{(u )θ =α / 2 , (Δ) φ =α / 2 }T = KP{ [2(1 + Z )α cos(α / 2)χ 4 (α ) + χ 5 (α )]
(14)
+ 0.5λ (α )[2(1 + Z )α sin(α / 2)χ 4 (α ) + χ 6 (α )]}
Moreover, the internal forces such as axial and shear forces and also the internal moment can be easily derived at any points using Eqs. (7) and (9). After some manipulations, the results are 1 ⎧ 1 1 ⎪ξ = S / P = sin(θ − 2 α ) + 2 λ (α ) cos(θ − 2 α ) ⎪ 1 ⎪ 1 1 for 0 < θ < 12 α ⎨η = N / P = cos(θ − 2 α ) − λ (α ) sin(θ − 2 α ) 2 ⎪ 1 ⎪ 1 1 1 1 ⎪μ = M / ρP = cos 2 α − cos(θ − 2 α ) + 2 λ (α )(sin(θ − 2 α ) + sin 2 α ) ⎩ 1 ⎧ ⎪ξ = S / p = sin φ − 2 λ (α ) cos φ ⎪ 1 ⎪ for 0 < φ < 12 α ⎨η = N / P = cos φ + λ (α ) sin φ 2 ⎪ 1 ⎪ 1 1 ⎪μ = M / ρP = cos 2 α − cos φ + 2 λ (α )(− sin φ + sin 2 α ) ⎩
(15)
where the dimensionless terms ξ , η and μ can be useful to analysis of the magnifications of the internal moments and forces to external applied force. The results of Eqs. (15) are also proved by satisfying the equilibrium conditions for the beam. (see Fig. 2). S P
A
Q/2
M
Q M
N N
S
B
P
Q/2
Figure 2: Internal forces and support reactions coming into view at the equilibrium conditions
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
6
3
NUMERICAL EXAMPLES AND RESULTS
As it has mentioned before, the validity of displacement relations are confirmed by comparing with special case of those available in the text books. In what follows, the displacement of single warp yarn and bi-axial woven yarns subjected to an in-plane simple tension are evaluated and compared with each other. Also, the bending moment, axial and shear forces diagrams along the centroidal axis are depicted. Specifications of steel yarns such as material properties, dimensions, number and amount of loading are given in table 1, which are used in all figures. The plot of dimensionless displacements at end (Δ / PK ) and mid-point (δ / PK ) versus α are shown in Fig. 3. Table 1 Specifications of steel yarns
Yong’s modulus
Yarn’s diameter
length × Width
Number of yarns
Axial load /yarn
E = 207 GPa
d = 0.5[mm]
1000 × 1000[mm 2 ]
400
P = 10[kN ]
3.5
0.2
3
0
Dimensionless Vertical Displacment (δ/PK)
Dimensionless Horizontal Displacment (Δ/PK)
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2.5
2
1.5
1
0.5
0
−0.2
−0.4
−0.6
−0.8
−1
0
50
100 α
150
−1.2
0
50
100 α
150
Figure 3: Left: Dimensionless end horizontal displacement, Right: Dimensionless midpoint vertical displacement
The variation of the ratio of the end horizontal displacement of upper yarn (warp) in presence of lower yarn (weft) to that, in the absence of weft versus arc angle (0 < α < 80) is shown in Fig. 4. This variation is almost slow for arc angles bigger than 70 o . Similar geometry and construction of warp and weft yarns imply that the vertical distance of mid-point from each support, Fig 1. equals to yarn half diameter. Also, the horizontal distance of supports of curved beam or spans of the curved beam (L), Fig. 1 is specified using the ratio of width of composite plane to the number of yarns. So, the radius ρ may be calculated in the terms of L and arc angle α as ρ = L /(2 sin(α / 2)) and also in view of Fig. 1 we may easily write d = 2 ρ[1 − cos(α / 2)]. Therefore, the arc angle α may be evaluated using two previous relations as α = 4 tan −1 (d / L) where d denotes to the
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
cross section diameter of curved beam, respectively. Using the data of table 1, we obtain the span length L = 1000 /(400 − 1) = 2.506 [mm], and the arc angle α = 45.13o and also ρ = 3.265 [mm] . 1 0.95 0.9
Displacment Ratio
0.85 0.8 0.75 0.7 0.65 0.6 0.55 0.5
0
20
40
60
80
α
100
120
140
160
180
Figure 4: Horizontal displacement ratio when weft is present to that when weft is absent 0.3
1.015 1.01
0.2 1.005 1
Dimensionless Axial Force
Dimensionless Shear Force
0.1
0
−0.1
0.995 0.99 0.985 0.98
−0.2
0.975 −0.3 0.97 −0.4
0
10
20
θ
30
40
0.965
0
10
Figure 5: Dimensionless shear and axial force versus θ
20
θ
30
40
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D. DERAKHSHAN and R.T. Faal / Procedia Engineering 14 (2011) 2839–2847 Author name / Procedia Engineering 00 (2011) 000–000
8 0
−0.005
−0.01
Dimensionless Bending Moment
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−0.015
−0.02
−0.025
−0.03
−0.035
−0.04
0
5
10
15
20
θ
25
30
35
40
45
Figure 6: Dimensionless bending moment versus angle θ
Making use of equations (15), the dimensionless displacements are evaluated which are plotted in Fig. 5. Jump in the left graph and extremums in the right are compatible with physical features of problem. The similar trend of dimensionless axial force can be seen for the dimensionless bending moment along the centroidal axis. The extremums are located at the angles nearly θ = 15 o and φ = 30 o (see Fig. 6). So, it is expected that the maximum normal stress may occur at these points. Accurate solution is obtained by first differentiation of η or μ with respect to θ (or φ ). 4
CONCLUSIONS
The main conclusions of this paper may be listed as follows: The end dimensionless horizontal displacement is increased rapidly with the increasing the arc angle and reversely the midpoint dimensionless vertical displacement is reduced. 2. The variation of the ratio of the end horizontal displacement of warp in the presence of weft to that in the absence of weft is almost slow for arc angles bigger than 70 o and in this bond the weft leads to more stiffening or rigidity. 3. Maximum normal stress of yarn cross section occurs at the angles before and after the contacting point. (Look at Fig. 6). 1.
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