A novel approach for the three-point flexure test of multidirectional laminates

Abstract

The complementary energy of a multidirectional laminate is obtained as a function of mechanical and hygrothermal force and moment resultants. Three-point flexure is then analysed assuming that the specimen is in contact with the support rollers at four points. As this problem is statically indeterminate, the redundant force is obtained by applying Engesser's second theorem. Lift-off at supports is assumed to occur when the redundant unknown is zero. The displacement field of a multidirectional laminate in three-point flexure is then obtained by applying Engesser's first theorem, including transverse shear and hygrothermal effects. The displacements related to bending and twisting moments, hygrothermal effects and transverse shear forces are analysed separately.

Keywords

Engesser's theorems multidirectional laminates three-point flexure test

Introduction

The displacement field of a composite plate constituted by two plies oriented at -θ and θ angles subjected to a uniform bending moment was calculated by Reissner and Stavsky.¹ This solution was used by Whitney and Dauksys² to obtain the displacement field in a unidirectional off-axis composite beam subjected to a uniform bending moment. These authors discussed the inherent problems related to bending–twisting coupling in an off-axis flexure test, due to the increasing interest at that time in bending experiments as a means of characterizing off-axis unidirectional composites. This interest was due to the end-constraint problem introduced by off-axis tensile specimens, discussed by Pagano and Halpin.³ In their conclusions, Whitney and Dauksys pointed out that the effect of bending–twisting coupling on middle-point displacement may dissipate for large span-to-width ratios for the case of uniform bending moment applied along the beam.² Furthermore, Halpin⁴ suggested that beams loaded by concentrated forces can be expected to display, at least qualitatively, similar response characteristics to those obtained for uniform bending. Following these conclusions, classical laminated beams theory⁵ provides a solution for the displacements considering only bending moments along the length of the beam, confirming its validity for high span-to-width ratios.

Grédiac analysed four-point bending tests on off-axis composites taking into account the lift-off effect, and proposed rotating dowel pins in order to avoid this effect and the consequent twisting moments.⁶

Pagano analysed the problem of pinned edge laminates, starting from the exact elastic solution for cylindrical bending of composite laminates, when the laminate is constituted of orthotropic layers.⁷ The next step was to consider the finite boundaries of a rectangular plate for the same kind of laminates, considering the plate pinned on all four edges.⁸ The third step included the effect of shear coupling due to off-axis layers in the laminate for the case of cylindrical bending.⁹ In allthree cases, comparisons between the exact solutions and those obtained by classical laminated plates theory were carried out in order to establish the range of validity of classical laminated plates theory.

Some works have dealt with the three-point bending test of unidirectional off-axis composites. A method was proposed for determining in-plane shear modulus by three-point flexure of unidirectional composites oriented at 45°.¹⁰ The problem was that the conditions for lift-off of the specimen were not determined. In order to solve this problem, the displacement field of an off-axis composite in three-point flexure was determined, including in-plane and out-of-plane stresses.¹¹ In this work both lift-off and no lift-off cases were analysed. When it was assumed that lift-off did not occur, the specimen was supported at four points. The analytical approach was compared with experimental results concerning the displacement of the middle point in 50 different experimental conditions.¹² Five specimens with different fibre orientations were tested at 10 different spans, and load-displacement curves corresponding to the middle point of the specimen were obtained. In this way, it was possible to generalize the method proposed by Mujika et al.,¹⁰ obtaining G_LT for different fibre orientation angles. Recently, the displacement and stress field of unidirectional off-axis composites in three-point flexure has been analysed by the finite element method.¹³ The agreement between analytical, experimental and numerical results is adequate. After having proposed a method for obtaining G_LT, the idea was extended to use the same test for obtaining in-plane shear strength.¹⁴ The analysis was carried out for the case when lift-off occurs. In addition to the stress analysis for determining the critical point of failure, an error analysis was carried out in order to measure the influence of elastic parameters and fibre orientation angle on the determination of G_LT. Recently, off-axis flexure and tensile tests have been carried out in order to compare results of in-plane shear strengths obtained by both methods.¹⁵

With respect to multidirectional laminates in flexure, a new test method for determining in-plane shear properties by [±45] antisymmetric laminates has been proposed.¹⁶ In this case, there is an initial lift-off due to hygrothermal effects, which is compensated by opposite twisting moments when the load is increasing during the test.

The aim of this work is to extend the analysis carried out for unidirectional laminates to the case of multidirectional laminates in three-point flexure. The analysis includes the effect of in-plane stresses due to plate moments, out-of-plane shear stresses due to shear forces, and hygrothermal effects. In this case, as well as the bending–twisting coupling that appears in the unidirectional off-axis case, membrane-plate coupling effects may appear. Due to membrane-plate coupling, bending and twisting moments originate normal and shear strains in the laminate. Nevertheless, supports do not restrict normal and shear deformation modes if friction effects are considered as negligible. As a consequence, it can be assumed that membrane forces are null. This means that the support and load configuration is the same as in the unidirectional case. Figure 1 shows three-point flexure of a multidirectional laminate with lift-off at the supports. Therefore, the moments and the derivatives of moments and shear forces can be obtained in the same manner as in the case of unidirectional off-axis composites. In the present study, after having obtained the complementary strain energy of a multidirectional laminate, Engesser's theorems,¹⁷ also called generalized Castigliano's theorems, are applied in order to determine lift-off conditions and the displacement field.

Figure 1.

Lift-off at supports in a laminate subjected to three-point bending.

Moments, shear forces and equilibrium equations

Figure 2 shows the sign convention adopted for force and moment resultants, where the notation for twisting moments and shear forces corresponds to Daniel and Ishai,¹⁸ i.e. q subscript corresponds to yz and r corresponds to zx.

Figure 2.

Sign convention adopted for force and moment resultants.

In the absence of distributed loads, plate equilibrium equations are given by:¹⁰

(1)

As explained above, the specimen is assumed to be supported at four corner points. In this case, the problem is statically indeterminate to the first degree. By assuming the applicability of the superposition principle, the load configuration can be decomposed into a symmetric and a skew-symmetric configuration, as shown in Figure 3.

Figure 3.

Decomposition of loads and reactions: (a) overall configuration; (b) symmetric configuration; (c) skew-symmetric configuration.

In the symmetric configuration, bending moments perunit length M_x and M_y, twisting moments per unit lengthM_s, and shear forces per unit length V_r, V_q are assumed to be uniformly distributed along cross-sections that are normal with respect to x and y axes, respectively. Because of the equilibrium of moments with respect to x and y axes, bending moments per unit length are:

(2)

Where $L_{1} = \frac{L}{2} L_{1}^{'} = \frac{L^{'}}{2} b_{1} = \frac{b}{2}$ ; being L, L′ and b thedimensions shown in Figure 1.

Because of the equilibrium of forces in z axis, shear forces per unit length are:

(3)

Replacing bending moments and shear forces of Equations (2) and (3) in Equations (1), it is seen that the distribution of M_s must be uniform. As the resultant twisting moment in any cross-section is 0, M_s twisting moment is 0 in the symmetric configuration.

The skew-symmetric configuration of Figure 3(c) is equivalent to a system of twisting moments per unit length uniformly distributed along the edge surfaces of the plate, in the volume among the four reaction forces.^10,11The value of twisting moment is one half of the reaction forces, being:

(4)

Therefore, the symmetric configuration is related to bending moments and shear forces, and the skew-symmetric configuration is related to twisting moments. Differential equilibrium conditions in Equations (1) are satisfied by assuming uniform distributions of moments and forces. Nevertheless, in the proximity of reactions and applied load, these distributions will change. On the other hand, point forces at corners are related to twisting moments and shear forces;¹⁹ in this analysis, it is supposed that in the symmetric configuration M_s, V_r and V_q are 0 on the edge surfaces, even in the proximity of concentrated reactions. Otherwise, edge effects that depend on laminate configuration generating interlaminar shear and normal stresses are not considered.¹⁸

Laminated plates theory with shear effects

In-plane strain and stresses

Laminated plates theory, taking into account transverse shear effects, is considered. The reference plane is the middle plane of the laminate, and the system ofreference is depicted in Figure 2. In matrix form, in-plane components of the strain field expressed in expanded and abbreviated form are:

(5)

Where ${ε^{0}}$ are strains of the reference plane; {κ} are plate and twisting curvatures. The stress–strain relation is given by:

(6)

Where {σ}_k are in-plane stresses at [lamina kQ]_k are reduced stiffness coefficients; {e}_k are hygrothermal strains given by:

(7)

Where α_i is the coefficient of thermal expansion ; β_i is the coefficient of moisture expansion; ΔT is the variation of temperature; ΔC is the moisture concentration.

Force and moment resultants concerning in-plane stress components in expanded and abbreviated form are given by:

(8)

Where {N} are membrane forces and {M} are plate moments.

The laminate stiffness matrices are:

(9)

Hygrothermal force and moment resultants are:

(10)

The inverse relation of Equation (8) is:

(11)

Where [a], [b], [c] and [d], are the laminate compliance matrices. As the inverse of a symmetric matrix is also symmetric, this means that [c] = [b]^t.

Out-of-plane stresses

Out-of-plane stresses can be obtained from equilibrium equations,²⁰ being:

(12)

In spite of the in-plane stress–strain relation being affected by hygrothermal effects, these effects will be assumed uniform in the whole specimen, their derivatives being zero. Thus, hygrothermal effects will not beconsidered in this section. The stress–strain relation for the lamina k according to Equation (6) can be written as:

(13)

Where keys are used to indicate a 3 × 1 column matrix, its transposed matrix being a 1 × 3 row matrix. Brackets are used to indicate a square matrix. From Equations (12) and (13) and taking into account Equation (5) it results that:

(14)

Otherwise, according to Equation (11), the strainsand the curvatures of the reference plane are given by:

(15)

Complementary strain energy, strain energy and Engesser's theorems

The complementary strain energy per unit volume is given by:

(16)

In the case that the stress–strain relation is linear, strains are:

(17)

where S_ijkl are the components of the compliance tensor and e_ij are initial strains. Then:

(18)

Due to the symmetry of the compliance tensor and as i and j are dummy indexes,²¹ it results that:

(19)

On the other hand, e_ij are independent with respect to stresses. Taking into account Equations (18) and (19):

(20)

Taking into account Equation (17):

(21)

The result given in Equation (21) coincides with the grey area in Figure 4. It is worth noting that when initial strains are taken into account, strain energy and complementary strain energy are not the same, in spite of the linear elastic behaviour of the material. This assertion will be shown in the following paragraphs.

Figure 4.

Strain energy and complementary strain energy in the case of initial strains e_ij and initial stresses t_ij.

The stress–strain relation in terms of stiffness coefficients is given by:

(22)

Where t_ij = C_ijkle_kl are initial stresses; C_ijkl are stiffness coefficients. The definition of strain energy per unit volume is:

(23)

After following similar steps as in Equations (18) to (21), the density of strain energy is given by:

(24)

It can be shown that Equation (24) is equivalent to the pointed area in Figure 4 given by:

(25)

Otherwise, Engesser's first theorem or the generalized form of Castigliano's second theorem states that:¹⁷

(26)

Where P_k is a generalized load applied at point k and δ_k is the generalized displacement at point k in the direction of P_k. The generalized load concept includes load and moments, and generalized displacement includes displacements and rotations.²²

Otherwise, if X is a redundant force its associated displacement is null. Then, Engesser's second theorem or the generalized form of Castigliano's theorem of least work states that:

(27)

Derivative of the complementary strain energy in multidirectional laminates

In-plane stresses

According to Equation (21) the complementary strain energy per unit area due to in-plane stresses is:

(28)

Where the superscript (NM;HT) indicates the complementary energy related to membrane (N), plate (M) and hygrothermal (HT) effects, respectively; and L_z indicates the length in thickness direction. Replacing strains from Equation (5) gives:

(29)

Equation (29) can be written as:

(30)

Equation (30) in matrix form is:

(31)

Concerning Equation (31), on the one hand, ${ε^{0}}^{t}$ and {κ}^t do not depend on z. On the other hand, hygrothermal strains vary from ply to ply. According tothe definition of force and moment resultants, Equation (31) can be written as:

(32)

Taking into account the definition of hygrothermal force and moment resultants of Equation (10), Equation (32) can be written as:

(33)

Where

(34)

Equation (33) can be written as:

(35)

Replacing Equation (11) in Equation (35) and taking into account the symmetry of the compliance matrix, gives:

(36)

Developing matrix products of Equation (36) gives:

(37)

Due to the properties of the transposed matrix, the following identity is satisfied:

(38)

The first equality in Equation (38) is satisfied because that matrix product is a scalar. The second equality is satisfied because [c] = [b]^t. Then, Equation (37) can be written as:

(39)

Thus, the complementary strain energy in the laminate due to in-plane stresses is given by:

(40)

Where L_x and L_y indicate length and width directions, respectively. Derivating Equation (40) with respect to P_k gives:

(41)

Taking into account that hygrothermal forces and moment resultants do not depend on applied forces, Equation (41) can be written as:

(42)

Since [a] and [d] are symmetric and matrix products in Equation (42) are scalars, this gives:

(43)

According to Equation (43), if forces and moments per unit length and their derivatives were known as a function of applied loads, displacements or redundant unknowns could be obtained.

Out-of-plane stresses

Out-of-plane stress–strain relations are not affected by hygrothermal effects. Thus, assuming linear elastic behaviour, strain energy and complementary strain energy per-unit volume are the same, being:

(44)

Stress–strain relation is given by:¹⁸

(45)

Where < > indicate a column matrix of two components. According to Equations (44) and (45), the complementary strain energy per unit area in extended and abbreviated forms is:

(46)

Then, the complementary energy due to shear is:

(47)

Derivating Equation (47) with respect to P_k and taking into account Equation (46), gives:

(48)

In this section, complementary strain energy and its derivative in Equation (48) have not been expressed as a function of force and moment resultants from Equations (14) and (15). This is because it is considered more comprehensible to use particular relations that correspond to three-point flexure, which will be discussed in the next section.

Three-point flexure

In-plane stresses

In the case of three-point flexure, if friction effects are considered negligible, membrane forces are null, even in the most general case where all kind of coupling effects appear. For instance, normal and shear strains in the laminate induced by plate-membrane coupling are not restricted by support conditions. The specimen will deform freely in membrane mode, as the unique constraints are downwards displacements at supports along the width of the specimen. Thus:

(49)

Replacing Equation (49) in Equation (43) gives:

(50)

Taking into account Equation (11), Equation (50) can be written as:

(51)

Where {κ^HT} are the curvatures when only hygrothermal effects are considered. Developing the term due to applied moments in Equation (51) gives:

(52)

The term due to hygrothermal effects is given by:

(53)

Out-of-plane stresses

In three-point flexure, as {N} = {0} for the whole specimen, replacing Equation (15) in Equation (14) gives:

(54)

According to moments and shear forces in Equations (2), (3) and (4), the derivatives of moments are:

(55)

Otherwise, compliance matrixes can be decomposed in column matrixes as:

(56)

Then, Equations (54) can be written as:

(57)

Integrating Equation (57) with respect to z, gives:

(58)

Integration constants are obtained taking into account boundary conditions. Analysing the case of τ_q, as they are zero on the external faces of the specimen:

(59)

Imposing the continuity in the k-th interlamina for laminas k and k+1:

(60)

Applying Equation (59) for lamina 1, according to Equation (58) gives:

(61)

Applying Equation (60) in a recursive manner, gives:

(62)

In an analogous manner, imposing the boundary conditions for τ_r:

(63)

After having obtained integration constants, Equations (58) can be expressed as:

(64)

being

(65)

According to Equation (65), the coefficients h_ij where i, j are q or r can be written as:

(66)

being

(67)

Taking into account Equations (46) and (64), the complementary strain energy per unit area is:

(68)

(69)

where

(70)

Then, the shear compliance coefficients of the laminate K_ij are:

(71)

where the terms I_ij/kl are given by:

(72)

where i, j, k, lcan be q or r; and $ζ_{p} = \frac{z_{k}^{p} - z_{k - 1}^{p}}{p}$ being p = 1,…,5

The derivative of the complementary energy per unit surface, from Equation (69) is:

(73)

Thus the derivative of the complementary strain energy is:

(74)

Redundant forces and lift-off condition

In order to obtain the redundant force, Engesser's second theorem given in Equation (27) is applied. The derivative of the complementary energy must be calculated with respect to the redundant unknown X. Taking into account Equations (2), (3) and (4) gives:

(75)

Replacing Equation (75) in Equations (52) and (53), the derivative of the complementary energy given in Equation (51) is:

(76)

The specimen is divided into four parts, A, B, C and D, according to Figure 5. It was shown by Mujika and Mondragon¹¹ that the strain energy is the same in the four parts in the case of a unidirectional off-axis laminate. Concerning complementary strain energy U^* of a multidirectional laminate, the part that does not include hygrothermal effects is equal to the strain energy. Since a multidirectional laminate is a set of unidirectional lamina, this part of U^* is the same in the four parts of the specimen. Concerning the part of U^* related to hygrothermal effects, as these effects are assumed to be uniformly distributed, it is also the same in the four parts.

Figure 5.

Upper view of the specimen, showing parts A, B, C and D, applied load and reactions.

In the same reference it was also shown that the vertical displacement field of a unidirectional composite satisfies central symmetry conditions. It can be shown for the present case by similar arguments as in Mujika and Mondragon.¹¹ Thus, parts A and D have the same displacement field, and the same occurs in parts B and C. Therefore, Equation (75) can be written as:

(77)

Replacing moments from Equation (2) and the derivative of Equation (75) the redundant unknown is obtained as:

(78)

where $c = \frac{L}{b}$ is the span-to-width ratio. Only positive values of X have physical meaning, as negative values would mean that the support is adhered to the specimen. Moreover the term related to hygrothermal effects must be negative because otherwise there would be a support reaction without applied load. Its meaning is related to the initial lift-off of the specimen due to hygrothermal effects. Then, the last term is the attraction force that the support should exert for lift-off not to occur.

The lift-off condition can be established imposing X = 0. Thus, the critical span-to-width ratio corresponding to lift-off is given by:

(79)

According to Equations (4) and (78), twisting moment is given by:

(80)

According to Equation (4), when lift-off occurs M₈ = P/8. The twisting ratio k is defined as the ratio between the twisting moment when lift-off does not occur and when it does.¹¹ According to Equation (80) it is:

(81)

In Equation (81) only k ≥ 1 values have physical sense, as k =1 corresponds to the limit case of lift-off. If k > 1, it means that lift-off occurs. In this case it is necessary to impose k = 1 to obtaining M_s = P/8.

Derivatives of moments and forces: the unit load method

The derivatives of moments and forces can be obtained by applying the unit load method in the statically determinate system in the same manner as Mujika and Mondragon.¹¹ In order to get moment equilibrium with respect to the support diagonal, two unitarial loads, one at each side of the diagonal and at the same distance, are applied. Symmetric points with respect to the centre of the specimen satisfy this condition and, moreover, they have equal displacements as explained above. In the following development, points that satisfy central symmetry will be named symmetric points. The displacement at any point of the specimen calculated by applying two unitarial loads at symmetric points will be twice the actual displacement of any of the two points, as derivatives will be twice in all cases.

Figure 6(a) shows an upper view of the system of two symmetric unitarial loads applied in parts A and D and its decomposition in symmetric (Figure 6(b)) and skew-symmetric (Figure 6(c)) configurations. According to the assumptions pointed out for obtaining bending moments in Equations (2), the derivatives of bending moments are:

Figure 6.

Decomposition of unitarial loads applied on parts A and D: (a) Overall configuration; (b) symmetric configuration; (c) skew-symmetric configuration.

(82)

where x₁, y₁ are the distances from support points to unitarial load application points; and x, y are the coordinates of any point.

The skew-symmetric configuration in Figure 6(c) can be decomposed into two configurations of uniform twisting moments. The first is generated by the loads at the corners in the volume among the loads, value 1/4. The second corresponds to the twisting moments generated by the loads inside the specimen in the volume among these loads, which value is -1/4. Adding both configurations, the derivatives of twisting moments for parts A and D are:

(83)

When unitarial loads are applied in parts B and C, as shown in Figure 7, the symmetric load configuration in Figure 7(b) is the same as in Figure 6(b); only loads inside the specimen in the skew-symmetric configuration change in Figure 7(c) with respect to Figure 6(c). These loads inside the specimen generate a state of uniform twisting moments in the volume among the loads, whose value is 1/4. Therefore, the derivatives of bending moments do not change in parts B, C with respect to parts A, D. Thus, only twisting moment derivatives included among the internal four loads change, being:

Figure 7.

Decomposition of unitarial loads applied on parts B and C: (a) Overall configuration; (b) symmetric configuration; (c) skew-symmetric configuration.

(84)

Taking into account the same considerations as for shear forces in Equations (3), from Figures 6 and 7 it can be seen that shear forces derivatives are the same for parts A, B, C and D except for the changes of sign, being:

(85)

Displacements caused by bending and twisting

The integrals of Equation (52) can be calculated replacing moments from Equation (2), and derivatives from Equations (82) and (83) for parts A and D and from Equations (82) and (84) for parts B and C. Twisting moments will be determined according to Equation (81). Then, if lift-off occurs being c ≥ c_LO, k =1 and M_s = P/8. If c ≥ c_LO then k < 1 and $M_{s} = k \frac{P}{8}$ . Using normalized coordinates $x_{0} = \frac{x_{1}}{L}$ and $y_{0} = \frac{y_{1}}{b}$ the common integral terms for the four parts are:

(86)

where I_ij' integral is related to M_i moment and M'_j derivative, with i, j = x, y, s. These integrals are related to the displacement due to j curvature caused by the i moment, as shown in Figure 8.

Figure 8.

Relations between I_ij' integrals and bending and twisting curvatures.

According to Equations (52) and (86), and taking into account that two unitarial loads are applied, the displacement at any point of parts A, D due to bending and twisting w_M1 is:

(87)

Replacing results from Equations (86) in Equation (87) gives:

(88)

for $0 < x_{0} < \frac{1}{2} 0 < y_{0} < \frac{1}{2}$

When x₀= y₀ = 0, displacement is null. This point represents the contact points in parts A and D. As only M'_s derivatives change for displacement field in parts B and C, only I_xs', I_ys' and I_ss' integrals change for these zones. The additional term of displacement for parts B and C is related to the non-zero value of twisting moment derivatives in the central part of the specimen, as seen in Equations (84). The new integral terms that must be added in order to obtain the displacement field for parts B, C are:

(89)

The additional displacement Δw_M at any point of parts B and C is:

(90)

Replacing Equations (89) in Equation (90):

(91)

Therefore, w₂ displacement at parts B, C is given by:

(92)

Replacing Equations (88) and (91) in Equation (92), the displacement field in parts B, C is:

(93)

for $0 < x_{0} < \frac{1}{2} 0 < y_{0} < \frac{1}{2}$

In this case, when x₀ = y₀ = 0 and k = 1, displacement is not null. When k < 1, replacing Equation (81) without taking into account the term due to hygrothermal effects, displacements are null. Otherwise, it can be verified that the increment of displacement in Equation (91) for points belonging to the middle lines (x₀, 1/2) and (1/2, y₀) are null.

Displacements due to hygrothermal effects

The displacement field due to hygrothermal effects is given in Equation (53). Replacing moment derivatives from Equations (82) and (83), the integral terms in parts A and D are:

(94)

The displacement field in parts A and D due to hygrothermal effects is given by:

(95)

Replacing Equation (94) in Equation (95) gives:

(96)

The additional term that corresponds to parts B andC, replacing the moment derivative from Equation (84), is:

(97)

The increment of displacement that corresponds to parts B and C is:

(98)

Adding Equations (96) and (98), the displacementfield in parts B and C due to hygrothermal effectsis:

(99)

Displacements caused by shear

Developing the matrix product of Equation (74), the derivative of complementary strain energy due to transverse shear is:

(100)

Replacing force derivatives from Equation (85), the integral terms are:

(101)

The displacement field in every part due to transverse shear is given by:

(102)

Replacing Equation (101) in Equation (102) gives:

(103)

Displacement of the load application point

The displacement of the load application point can be obtained replacing x₀ = y₀ = 1/2 in Equations (88), (96) and (103). As δ_M, δ_HT and δ_V are the displacements due to moments, hygrothermal effects and transverse shear forces, respectively, they are:

(104)

In the case of a unidirectional laminate, $d_{i j} = \frac{12}{h^{3}} S_{i j}$ being i, j = x, y, x. Otherwise, hygrothermal effects are null and $K_{m n} = \frac{6}{5 h} S_{m n}$ being m, n = q, r. It can be seen that the first and third of Equations (104) coincide with Equations (7) and (8) of Mujika et al.¹²

Stress field

If lift-off occurs, the problem is statically determinate. In other cases, redundant unknown X has to be determined from Equation (78) after checking the suitability of the hypotheses concerning concentrated reactions and load application. This aspect can be carried out by determining level curves of displacements at supports and at the load application line.¹²

After determining moments from Equations (2) and (4) and hygrothermal force and moment resultants from Equation (10), middle plane strains ${ε^{0}}$ and curvatures {κ} can be obtained from Equation (11). Then, using Equations (5) and (6) in-plane stresses can be calculated.

Interlaminar stresses can be determined from Equations (64) and (65) after having obtained shear forces from Equation (3) and integration constants from Equations (62) and (63).

Conclusions

The displacement field of a multidirectional laminate in three-point flexure has been determined using Engesser's first and second theorems, after having determined the complementary strain energy of a multidirectional laminate. Bending, twisting, transverse shear and hygrothermal effects have been included in the analysis.

The approach can be used for determining level curves at support lines and at the load application line in order to check the suitability of assumptions concerning concentrated reactions and load application. Moreover, for a given lay-up, the lift-off condition can be determined for obtaining a statically determinate case in a particular laminate configuration.

After having determined the displacement of the middle point including all effects and the stress field, the approach can be used for analysing experimentally multidirectional composites subjected to three-point flexure.

Footnotes

Funding

The financial support of the Basque Country Government on the research project TILAM of the program SAIOTEK 2009 is gratefully acknowledged.

Conflict of interest statement

None declared.

Nomenclature

References

Reissner

Stavsky

. Bending and stretching of certain types of heterogeneous aeolotropic elastic plates. J Appl Mechan 1961; 28: 402–408.

Whitney

Dauksys

. Flexure experiments on off-axis composites. J Compos Mater 1970; 4: 135–137.

Pagano

Halpin

. Influence of end constraint in the testing of anisotropic bodies. J Compos Mater 1968; 2: 19–31.

Halpin

. Primer on composite materials analysis. Lancaster, PA: Technomic Publishing Co Inc, 1992.

Whitney

. Structural analysis of laminated anisotropic plates. Lancaster, PA: Technomic Publishing Co Inc, 1987.

Grédiac

. Four-point bending test on off-axis composites. Compos Struct 1993; 24: 89–98.

Pagano

. Exact solutions for composite laminates in cylindrical bending. J Compos Mater 1969; 3: 398–411.

Pagano

. Exact solutions for rectangular bidirectional composites and sandwich plates. J Compos Mater 1970; 4: 21–35.

Pagano

. Influence of shear coupling in cylindrical bending of anisotropic laminates. J Compos Mater 1970; 4: 330–343.

10.

Mujika

Berglund

Varna

Mondragon

. 45° flexure test for measurement of in-plane shear modulus. JCompos Mater 2002; 36: 2313–2337.

11.

Mujika

Mondragon

. On the displacement field for unidirectional off-axis composites in 3-point flexure – Part 1: Analytical approach. J Compos Mater 2003; 37: 1041–1066.

12.

Mujika

de Benito

Mondragon

. On the displacement field for unidirectional off-axis composites in 3-point flexure – Part 2: Numerical and experimental results. J Compos Mater 2003; 37: 1191–1217.

13.

Romera

Adarraga

Cantera

Mujika

. Numerical analysis of displacement and stress fields of off-axis composites in three-point flexure. J Compos Mater 2011; 45: 671–682.

14.

Mujika

Valea

Ganan

Mondragon

. Off-axis flexure test: A new method for obtaining in-plane shear properties. J Compos Mater 2005; 39: 953–980.

15.

Vargas

Mujika

. Determination of in-plane shear strength of unidirectional composite materials using the off-axis three-point flexure and off-axis tensile tests. JCompos Mater 2010; 44: 2487–2507.

16.

Vargas

Mujika

. Determination of in-plane shear properties by three-point flexure test of ±45° anti-symmetric laminates. Polymer Testing 2011; 30: 204–215.

17.

Oden

Ripperger

. Mechanics of elastic structures. New York: McGraw-Hill, 1981.

18.

Daniel

Ishai

. Engineering mechanics of composite materials. New York: Oxford University Press, 2006.

19.

Timoshenko

Woinowsky-Krieger

. Theory of plates and shells. New York: McGraw-Hill Inc, 1959.

20.

Berthelot

J-M

. Composite materials. Mechanical behavior and structural analysis. New York: Springer, 1998.

21.

Karasudhi

. Foundations of solid mechanics. Dordrecht. The Netherlands: Kluwer Academic Publishers, 1991.

22.

Shames

Dym

. Energy and finite element methods in structural mechanics. Bristol, PA: Taylor & Francis, 1985.