A numerical model for laminated composite thin-walled members with openings considering shear lag effect

Engineering Structures 185 (2019) 392–399

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A numerical model for laminated composite thin-walled members with openings considering shear lag effect

T

M. Vojnić-Purčar , A. Prokić, M. Bešević ⁎

University of Novi Sad, Faculty of Civil Engineering Subotica, Kozaracka 2A, 24000 Subotica, Serbia

ARTICLE INFO

ABSTRACT

Keywords: Thin-walled laminated beams Composites Openings Shear lag Finite element approximation

A novel model describing the shear lag of laminated composite thin-walled beams with openings has been developed. Warping of open-closed cross- sections is defined by displacements parameters at selected nodes. Vlasov’s assumption of neglecting shear strains in the middle surface is not necessary and stresses can be calculated directly from the strains. The general approach to the solution of the problem is based on the finite element method and linear stiffness matrix has been derived using the principle of virtual displacements. Computing program has been developed and numerical results for thin-walled laminated beams with openings have been presented.

1. Introduction The composite thin-walled beams of closed cross-sections are extensively used in aerospace, marine and civil engineering. It is not uncommon that openings of different sizes can exist in these elements to allow access for services, inspection and repair facility, or to provide openings for windows and passage into the interior of the structure. In addition to that, these opening serve also to reduce the overall weight of the structure. The aim of the present study is to investigate the influence of such openings on shear lag effects in composite thin-walled elements. As is well known, due to shear lag effects, the structural behavior can be different from that predicted by the elementary beam theory, which assumes that cross section remains plane after deformation. The phenomenon of shear lag has been extensively studied in order to develop a reliable model for its analysis. The classical Vlasov theory [23] of thin-walled beams is based on the assumption that the shear strains in the middle surface can be neglected. While this result offers a simple analytical solution, it is unable to reflect a phenomenon such as shear lag. Reissner [16] proposed the procedure, based on the energy variation method, where longitudinal displacements of flanges were assumed to be in a form of parabolic curve. Many researches used this method after combining different types of curves for describing longitudinal displacements. Kuzmanovic and Graham [8] applied Ressiner method in a study of shear lag in the single cell box girder. Dezi and Mentrasti [3] solved this problem by assuming as unknown the longitudinal displacements and three functions describing the warping of

⁎

the horizontal flanges. Song and Scordelis [21,20] assumed that the flanges and web were flexible out of their plane and the stress in the webs could be calculated by the elementary beam theory. They presented formulas for determining the shear lag effects in beams of simple cross section. Empirical method for the determination of shear lag in box girders was presented by Evans and Taherian [5]. They proposed explicit formulas for the determination of shear lag for different types of loads and supports of beams. The influence of the finite element mesh on stress concentration and shear lag on simply supported box beams and continuous beams was presented by Lertisima et al. [11] and Sanguanmanasak et al. [18], respectively. The shear lag analysis by the adaptive finite element method was presented by Lee and Wu [10,9]. In their first paper they dealt with simple plated structures and in the second one the procedure was extended to the shear lag analysis of complex plated structures. The papers dealing with the investigation of shear lag in composite materials are much less represented. Kirstek et al. [7] and Evans and Ahmad [4] proposed a method for shear lag analysis using harmonic analysis for composite (steel-concrete) box girders of various crosssections. In the second paper, harmonic analysis is extended for more complex cross sections. Some solutions to this issue are presented in papers of Takayanagi et al. [22] and Lopez-Anido and GangaRao [12]. They examined the influence of shear lag on the I beams, and the thinwalled prismatic orthotropic composite beams, respectively. Recent papers [25,26] propose solution of single-cell thin-walled compositelaminated box beams under bending loads with consideration of both shear lag and shear deformation. In their work shape of the stress is

Corresponding author. E-mail addresses: [email protected] (M. Vojnić-Purčar), [email protected] (A. Prokić).

https://doi.org/10.1016/j.engstruct.2019.01.142 Received 23 April 2018; Received in revised form 8 January 2019; Accepted 31 January 2019 Available online 07 February 2019 0141-0296/ © 2019 Elsevier Ltd. All rights reserved.

Engineering Structures 185 (2019) 392–399

M. Vojnić-Purčar, et al.

predetermined and it could be used only for single-cell cross-section. Salim and Davalos [17] used the harmonic technique for shear lag analysis of thin-walled open and closed composite beam, simply supported, loaded by shear flow and axial stress resultant. Shear lag model was incorporated into beam model and explicit formulas for various cross sections were developed. New finite elements for thin-walled laminated beams are presented by more authors [24], Barradas Cardoso et al. [2], but their application is based on the adoption of the warping function, which predetermines the distribution of normal stresses that disables the shear lag calculation. Shin et al. [19] derived the sectional shape functions as displacement fields only for rectangular cross sections. In this paper, the novel finite element defined on the basis of the warping function proposed by Prokić [14,15] is applied in the analysis of the shear lag effect in composite thin-walled beams with opening. Beginning of this research is presented in author’s previos paper Prokic et al. [13]. Warping function is defined by displacement parameters of nodal points as unknowns, therefore the assumption of neglecting the shear strain in the middle plane is not necessary, and the stresses can be directly determined from the relevant strains. Computer program has been derived that allows the calculation of longitudinal stresses without any limitations, taking into account the phenomenon of shear lag for an arbitrary cross-section and for any boundary conditions. According to the author’s knowledge, there is a lack of investigation in this area.

Fig. 2. Transversal displacements of the cross-section.

According to the first assumption the cross-sectional behavior can be described by only three displacement components, two translations u and v of point C and an angle of rotation about z-axis (Fig. 2).

u =u y v =v+ x

(1)

Normal and tangential displacements of an arbitrary point with coordinates x and y on the contour, where the angle of rotation is sufficiently small, are

2. Basic theory and assumptions A straight, thin-walled beam with an open or closed cross section is considered (Fig. 1). The midline of cross-section is idealized by a number of straight lines connected by discrete points. Two coordinate systems are adopted in the analysis. The first one is Descartes' coordinate system xyz, of the right orientation, where the z axis is parallel to the axis of the rod, and x and y axes lie in the cross section plane, while the second one is the coordinate system esz, also of the right orientation. The present theory is based on the following assumptions:

= v sin + u cos + hn = v cos u sin + h

(2)

where is the angle between the (e, s, z) and (x, y, z) coordinate systems, hn represents the perpendicular distance from normal at point S to point C given by:

hn = x sin

y cos

(3)

and h represents the perpendicular distance from tangent at an arbitrary point to point C given by:

1. the cross-section is rigid in its own plane, 2. the longitudinal displacements caused by warping vary linearly between any two adjacent nodal points, 3. the part of the shear strains in the middle surface of the wall, due to the bending moment, is negligible. 4. the relative warping in relation to the midline is qualitatively defined with the solution of Saint-Venant’s torque.

h = x cos

+ y sin

(4)

The displacement at z direction of an arbitrary point on contour can be found by using the hypothesis concerning the absence of shearing strain in the middle surface due to bending moment:

w =w

yv

xu + wwarp

Fig. 1. Thin-walled beam of arbitrary cross section.

393

(5)

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6+n degrees of freedom at each ui , vi, wi, vi , ui , i , w1i, w2i, ... , wni Eqs. (1) and (9) can be converted to matrix form = w

cos sin 0

0 sin 0 cos 0 x

0 0

0 0

hn h y 0

0 0 1

end

0 ... 0 ... 0 ... 0 ... i . . . 1 . . .

node

0 0 n

u u v v w w1

Fig. 3. Displacement between nodes.

wi The last term of (5) defines warping of the cross-section suggested by Prokić [14,15]. s e wwarp = wwarp + wwarp

Let us denote the vector of generalized nodal displacement (Fig. 4) in the following way

(6)

where s wwarp

q = [qu , qv , q , qw , qw1, qwi , ... , qwn ]T

=

wi (z )

i (x ,

y)

qu = [u1, u1 , u2, u2 ]T

represents warping along the midline of cross-section. Unknown parameters wi are displacements of arbitrary points on the midline. Those points are nodes of the section and their number determines the number of unknown parameters of displacements. Function i depends on the mode of displacement change between the nodes of polygonal cross-section. If this change is linear, according to the second assumption, which is in conformity with the classical theory of thin-walled beams, then the function i has a simple geometrical meaning, as shown in Fig. 3. The function i exists only along parts between point i, where it takes the value 1, and adjacent nodes, where it takes the value 0. The second term on the right side of (6), determines the relative warping in relation to the midline of the cross-section, and, according to the fourth assumption, is equal to

(x , y )

q = [ 1,

T 2]

iw i i

i = 1, 2, .... ,n

(13)

Hermite polynomials are adopted as interpolation functions for displacements u and v, and a linear displacements function is adopted for , w, w1, ... , wn

Nu = [1 Nv = [1

3

2

3

2

N = [1

+2

3

+2

3

L( L(

2

2

+2

+ 2

3)

3 3)

2

3

2 2

3

2

2

L( 3

L(

+

2

3)] 3)]

(14)

]

Thus, it follows

u = Nu qu v = Nv qv = Nq

Thus, the total longitudinal displacement is

xu +

v2 ]T

qwi = [wi1, wi2 ]T

(9)

(x , y ) = h n e

yv

v1 , v2,

qw = [w1, w2 ]T

where

w =w

qv = [v1,

(8)

(z )

(12)

where

(7)

i

wde =

(11)

wn

w = Nqw wi = Nqwi i = 1, 2, .... ,n

(10)

(15)

Substituting (15) into (11), the displacement of an arbitrary point of cross-section could be obtained in terms of nodal parameters

When observing the previous equation, it can be concluded that there are n + 1 parameters of longitudinal displacements, regarding n parameters of longitudinal displacements of nodal points and one longitudinal displacement of center of cross section. It should be noted that only the n parameters of displacements are independent, but in further derivation n + 1 parameter will be used, provided that: The normal force from the stresses induced by the warping of cross section or bimoments caused by axial deformation is equal to zero.

= w

cos Nu sin Nv sin Nu cos Nv xN u yN v qu qv q qw qw1

hn N hN N

0 0 N

0 ... 0 ... 1N . . .

0 … 0 … iN …

0 0 nN

= Aq

qwi

3. Finite element

qwn

A typical thin-walled element is shown in Fig. 4. The element has

394

(16)

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M. Vojnić-Purčar, et al.

Fig. 4. Finite element.

where

N

u

=

N

v

=

N =

1 [ L 1 [ L 1 [ L

6 +6

2

L (1

6 +6

2

L( 1 + 4

4 +3

2)

6

3 2) 6

6

2

6

L( 2 + 3 2

L (2

3 2 )]

1 1]

0 0

B=

2)]

0

z

xN

A=

s

yN

0

(17)

z

0

v

N ¯ (h + 2e ) N

N

1N

0

1 ,s N

...

iN

…

nN

...

i ,s N

…

n ,s N

(21)

and 3.1. Stiffness matrix The strain components different from zero are:

=[

zs

= =

w z z

=

N

v

=

=

+

w s

(19)

6 + 12

L( 4 + 6 ) 6

6 + 12

L (4

6 ) 6

12 12

L ( 2 + 6 )] L (2

6 )] (22)

z s

=D

(23)

in which with D we denote the matrix of reduced stiffnesses [6]:

Substituting (16) into (18), the following is obtained

= B ·q

1 [ L2 1 [ L2

The constitutive eqations of a lamina in s-z coordinate system are given by:

where z

u

N = [0 0]

(18)

T zs ]

z

N

D=

(20)

where

Q¯11 Q¯16 Q¯ 16 Q¯66

(24)

Linear stiffness matrix is obtained by following equation:

(25)

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M. Vojnić-Purčar, et al.

Table 1 Normal stresses at nodal points. Lay-ups

[15/−15]2S [30/−30]2S [45/−45]2S

Fig. 5. Position of openings.

[60/−60]2S [75/−75]2S

where the values that determine the geometrical properties of the crosssection and submatrices K1 – K11 are given in Appendix A.

[0/90]2S

References

Present Ansys Present Ansys Present Ansys Present Ansys Present Ansys Present Ansys

Points 1

2

3

4

5

376.89 376.16 378.23 366.48 369.46 357.62 366.59 353.76 373.33 352.98 188.11 178.11

357.23 358.66 364.08 350.45 358.80 364.27 356.66 344.71 361.7 346.49 179.13 173.34

343.2 342.28 353.17 335.86 350.24 333.55 348.61 335.04 353.12 337.68 172.59 168.58

325.6 328.16 338.22 327.86 337.96 331.77 336.94 334.41 340.85 335.27 164.18 164.45

316.78 308.22 331.74 315.61 332.43 331.02 330.59 332.57 333.37 332.26 159.01 157.5

4. Numerical examples The adopted warping function enables the analysis of shear lag effect of thin-walled beams combined of open and closed cross sections (thin-walled girders of closed cross section with openings). To test the accuracy of the proposed method, a computing program was developed and numerical examples were analyzed. 4.1. Example 1 A simply supported girder of total length L = 50 cm with two openings in the upper flange, as shown in Fig. 5. was subjected to a vertical concentrated load P = 100 kN acting on the mid-span. The geometry of the characteristic cross-section is given in Fig. 6. Each segment of cross-section is made up of eight layers placed symmetrically with respect to the center line. Material properties of the beam : E1 = 53.78 GPa, E2 = 17.93 GPa, G12 = G13 = 8.96 GPa, G23

Fig. 7. Normal stresses along lower flange due to size of openings.

= 3.45 GPa, = 13 = 23 = 0.25. The beam is modeled with 18 elements. To increase the accuracy of the distribution of normal stresses along the midline of the profile 22 cross-sectional nodal points for closed and 17 for open cross-section are adopted. The boundary conditions are set to prevent the rotation of the cross-section around the z axis and to prevent displacements in directions of x, y and z axes at one end, i.e in the x and y directions at the other end: 12

z=0 u = v = w = =0 z = L u = v = =0

The cross sectional deplalanation is allowed in all nodes of the end cross-section, wi 0. For the beam with openings 2l/ L = 0.6, the normal stresses along the lower flange of the mid-span cross-section are calculated. The results for different orientation of the layers are presented in Table 1, along with the results obtained by the ANSYS [1] finite element model using the shell (4-node SHELL181) elements. Also, to examine the influence of the size of the openings on the shear lag effect, the normal

(26)

Fig. 6. Cross sections.

396

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M. Vojnić-Purčar, et al.

Fig. 8. Position of openings.

Fig. 10. Normal stresses along lower flange due to size of openings.

Table 2 Normal stresses at nodal points. Lay-ups

[15/−15]2S [30/−30]2S [45/−45]2S [60/−60]2S [75/−75]2S [0/90]2S

References

Present Ansys Present Ansys Present Ansys Present Ansys Present Ansys Present Ansys

Points 1

2

3

4

5

6

182.91 184.10 191.22 184.85 187.83 186.79 190.02 187.55 188.80 184.11 91.60 86.70

168.97 170.60 178.33 174.20 176.72 171.86 179.15 174.54 177.63 171.62 84.79 80.41

173.87 179.60 179.47 175.14 175.98 170.75 178.10 172.06 176.93 175.23 86.67 87.09

187.59 186.18 188.29 191.76 181.51 186.41 183.13 184.93 182.52 187.40 92.78 95.13

195.47 197.00 193.59 197.49 185.31 191.09 186.72 189.61 186.35 192.91 96.32 99.97

205.47 206.20 200.25 201.40 190.19 193.98 191.36 191.72 191.23 193.88 100.70 99.99

profile, 23 cross-sectional nodal points are adopted, Fig. 9. Normal stresses along the lower flange of the middle cross-section, in the function of the layers orientation (for2l/ L = 0.6) and opening’s size, are presented in Fig. 10 and Table 2.

Fig. 9. Cross sections.

5. Conclusion A model based on warping function that enables analysis of shear lag effect in the composite thin-walled members with the openings is presented. Through the numerical examples it can be concluded that shear lag and ply angle of the layers significantly affect the normal stress distribution on the flanges of thin-walled box beams. With respect to refined finite element models using shell elements, the present method has the advantage of reducing the amount of degrees of freedom without loss of precision in evaluating stress states.

stresses in mid-span cross-section are calculated and results are shown in Fig. 7. For the size of openings 2l/ L = 0.75, normal stresses in the upper flange are reduced due to large openings. As a result, normal stresses in the lower flange are increased. 4.2. Example 2 In this numerical example we analyze the simply supported two-cell box beam with openings on the both sides, Fig. 8, with the same span, loading, boundary conditions and the material characteristics as in the previous example. The beam is idealized using 18 finite elements. To improve the distribution of normal stresses along the midline of the

Acknowledgments The present work has been supported by The Ministry of Education and Science of the Republic of Serbia (Project No. ON174027).

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M. Vojnić-Purčar, et al.

Appendix A In stiffness matrix (25) the values that determine the geometrical properties of the cross-section are given by:

F= Sx =

Ix

i

Iy

=

A11 A11 A11

i

A11

i

i

j

=

I

j

j

=

Se =

x i2 + x k2 + xi x k 3 yi2 + yk2 + yi yk 3

)l )l

+ B11 (x i + xk ) cos + B11 (yi + yk ) sin

6 2xi + x k 6

)+B )+B )+B )+B

11

xi + 2xk 6

2yi + yk 6 yi + 2yk 6

+

l

Ixe =

x +x A16 i 2 k h¯

ns

Iye =

A16

ns

=

yi + yk 2

h¯ A16 2

i

=

e

I1 i =

11

11

+ B11

cos 2

(

+

xi + xk 2

D11sin2

sin

+

l yi + yk 2

I1 cos

)+D

11 sin

l

sin 11 2

l

h¯ + B16 (yi + yk + h¯ sin ) + 2D16 sin

l

ns ns ns

I 1i

i

=

i

I1

i

j

=

i

I1 j

j

=

I1

i

I1

j

e

xi + xk 2

x +x A16 i 2 k

B16 cos

+ B16 cos

y +y A16 i 2 k

A16

yi + yk 2

B16 sin

+ B16 sin

1 A 2 16 1 A 2 16

1 A i 2 16

= =

i

e

ns

I 3i

i

=

I 3i

j

=

I3

j

=

j

A16

(A66 h¯ + 2B66) l [ A66 h¯ 2B66 ] i

Se1 =

+ B16 (x i + xk + h¯ cos ) + 2D16 cos

ns

=

y j

l

sin 2

l I y1 i = I1

l

cos 2

cos

h¯

A16 2 + B16 l

i

=

x j

(A16 h¯ + 2B16) l

ns

j

x

2xi yi + 2xk yk + xi yk + xk yi

( ( ( (

I D11cos2

l A i 11 3 l A i 11 6 l A i 11 3

=

i i

e

i

=

j

i

A11

i

I

I

ns

=

j

Iy

I

=

+ B11 sin

A11

ns

Ixy =

+ B11 cos

yi + yk 11 2

A11

ns

Iyy =

xi + xk 2

11

ns

Ixx =

i

(A (A

ns

Sy =

Ix

A11 l

ns

[A66 h¯ + 2B66 ] A66 l A66 i l A66 i l

i

l

+ B16 l

where:

Bij =

nl Q¯ (hk k = 1 ij, k 1 nl Q¯ (hk2 k = 1 ij, k 2

Dij =

1 3

Aij =

nl Q¯ (hk3 k = 1 ij, k

hk 1) = hk2 1) =

nl Q¯ t k = 1 ij, k k nl Q¯ e t k = 1 ij, k T , k k

hk3 1) =

In previous expressions with

nl Q¯ k = 1 ij, k

i

eT2, k tk +

tk3 12

is denoted the sum upon the polygon sides connected by point i, and with

polygon sides of the cross-section. Expressions for reduced stiffness are given by following equations: 2

Q¯ 12 Q¯ 22 Q¯ 12 Q¯ 26 Q¯ 22

Q¯11 = Q¯ 11 Q¯ 16 = Q¯ 16

2 Q¯ 26 Q¯ 22

Q¯66 = Q¯ 66

Q¯ 11 = Q11 m4 + Q22 n4 + 2m2n2 (Q12 + 2Q66 ) Q¯ 12 = m2n2 (Q11 + Q22 4Q66) + (m4 + n4 ) Q12 Q¯ 16 = [Q11 m2 Q22 n2 (Q12 + 2Q66 )(m2 n2)] mn Q¯ 22 = Q11 n4 + Q22 m4 + 2m2n2 (Q12 + 2Q66 ) Q¯ 26 = [Q11 n2 Q22 m2 + (Q12 + 2Q66 )(m2 n2)] mn Q¯ 66 = m2n2 (Q11 + Q22 2Q12) + Q66 (m2 n2 )2

Q11 = Q12 = Q22 =

E1

2 E2 12 E 1

1

1

12 E2 2 E2 12 E 1

E2 1

2 E2 12 E 1

Q66 = G12

398

ns

we denote the sum upon each of the

Engineering Structures 185 (2019) 392–399

M. Vojnić-Purčar, et al.

In stiffness matrix (25) submatrices K1, K2 , ... , K11 are given by:

K1 =

K3 =

12 L3 6 L2 12 L3

6 L2 4 L 6 L2

6 L2

2 L

K5 =

4 L 6 L2 2 L

0

1 L

K9 =

4 L

12 L3

12 L3 6 L2

6 L2 2 L

6 L2

6 L2 2 L

12 L3 6 L2

6 L2 4 L

0

0

6 L2

6 L2 12 L3 6 L2

2 L

K4 =

6 L2 4 L

1 L

1 L

0

0 1 L

1 L

1 L

1 L 1 2

1 2

1 2

L 6 L 3

1 L

1 2 1 2

K8 =

K10 =

1 L

1 L

K6 =

0

1 2

L 3 L 6

K2 =

6 L2 4 L

12 L3 6 L2 12 L3

0 1 L

1 L

K7 =

6 L2 2 L 6 L2

12 L3 6 L2

6 L2

12 L3 6 L2 12 L3 6 L2

0

12 L3 6 L2

1 L

1 L

1

0

1 L 1 L

1 L

0

1 2 1 2

1

K11 =

1

1 L

0

1 L

1 L

0

1

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