Composite Materials Design and Applications Part 10 ppsx

(13.8) 13.3 CASE OF A PLY One can observe from Equation 13.7 that the stress–strain relations in the plane x,y appear decoupled in the case when s zz = 0. We suppose that this applies for the plies making a thin laminate. Each ply will be characterized in its plane by the following relations which are extracted from relations 13.5 10 and 13.7: Ⅲ In the orthotropic axes ᐉ,t: (13.9) Ⅲ In the x,y axes, making an angle q with the orthotropic axes: ➡ ➡ ; ; 10 The orthotropic axes of Equation 13.5 are denoted as l,t,z for a ply (see Section 3.3.1). 1 G xz s 2 21 n +() E t c 2 G ᐉt += G xz q () 1 s 2 21 n +() E t c 2 G ᐉt + = 1 G xy 4c 2 s 2 1 E ᐉ 1 E t 2 n tᐉ E t ++ ˯ ʈ c 2 s 2 –() 2 G ᐉt += G xy q () 1 4c 2 s 2 1 E ᐉ 1 E t 2 n tᐉ E t ++ ˯ ʈ c 2 s 2 –() 2 G ᐉt + = h xy G xy 2cs c 2 E ᐉ s 2 E t º – Ó Ì Ï –= º c 2 s 2 –() n tᐉ E t 1 2G ᐉt – ˯ ʈ ˛ ˝ ¸ + m xy G xy 2cs s 2 E ᐉ c 2 E t º – Ó Ì Ï –= º c 2 s 2 –() n tᐉ E t 1 2G ᐉt – ˯ ʈ ˛ ˝ ¸ – z xy G xy 2cs nn tᐉ –() E t –= z xz G xz cs 1 G ᐉt 21 n +() E t – ˯ ʈ –= e ᐉ e t g ᐉt Ó ˛ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ Ï¸ 1 E ᐉ n – tᐉ E t 0 n – ᐉt E ᐉ 1 E t 0 00 1 G ᐉt s ᐉ s t t ᐉt Ó ˛ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ Ï¸ = TX846_Frame_C13 Page 270 Monday, November 18, 2002 12:29 PM © 2003 by CRC Press LLC (13.10) For the constants, the values were shown in detail in 13.8. e xx e yy g xy Ó ˛ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ Ï¸ 1 E x n yx E y – h xy G xy n – xy E x 1 E y m xy G xy h x E x m y E y 1 G xy s xx s yy t xy Ó ˛ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ Ï¸ = TX846_Frame_C13 Page 271 Monday, November 18, 2002 12:29 PM © 2003 by CRC Press LLC 14 THE HILL–TSAI FAILURE CRITERION There are many failure criteria for orthotropic materials. The most commonly used for design calculations is the so-called “Hill-Tsai” criterion. 1 This criterion can be interpreted as analogous to the Von Mises criterion which is applicable to isotropic material in elastic deformation. We will review at the beginning the principal aspects of the Von Mises criterion. 14.1 ISOTROPIC MATERIAL: VON MISES CRITERION The material is elastic and isotropic. In Figure 14.1, one denotes by I,II,III the principal directions of the stress tensor S for a given point. The corresponding matrix is The general form of the deformation energy dW for an elementary volume dV surrounding the point considered can be written as: which can be reduced to 1 See failure of composite materials in Section 5.3.2. s I 00 0 s II 0 00 s III dW total 1 2 s ij e ij dV j  i  = dW total 1 2 s I e I s II e II s III e III ++() dV= TX846_Frame_C14 Page 273 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC One obtains then by replacing: (14.1) One can rewrite as following the quantity in brackets: (14.2) Remarks: If one denotes as the direction making the same angle with each of the principal directions (following Figure 14.1), one observes on the face with the normal , a stress such that: = S() that is: which can be decomposed as: Ⅲ A normal stress: s n = ◊ The above consists of the average or isotropic part of the stress tensor. 2 Ⅲ A shear stress: 2 Recall the expression s I + s II + s III that constitutes the first scalar invariant of the stress tensor. dW dV ˯ ʈ distorsion 1 2 1 n + E s I 2 s II 2 s III 2 ++() n E s I s II s III ++() 2 º – Ó Ì Ï = º 1 n + E s I s II s III ++() 2 3 n E s I s II s III ++() 2 +– ˛ ˝ ¸ then: dW dV ˯ ʈ distorsion 1 4G s I 2 s II 2 s III 2 ++() s I s II s III ++() 2 3 – Ó ˛ Ì ˝ ϸ = 2 3 s I 2 s II 2 s III 2 s I s II s II – s III s III – s I –++{} 2 3 s I s II s III ++() 2 3 s I s II s II s III s III s I ++()–{} dW dV ˯ ʈ distorsion 1 6G s I s II s III ++() 2 º {= º 3 s I s II + s II s III + s III s I ()– } n n s s n s {} s I / 3 s II / 3 s III / 3 Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ = s n then: s n s I s II s III ++ 3 = ts 2 s n 2 –= TX846_Frame_C14 Page 275 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC then: which can be compared with Equation 14.1. Thus, The shear stress t also appears as the shear characteristic of the distortion energy. Ⅲ One recognizes in Equation 14.2 the presence of the first and second scalar invariants of the stress tensor independent of the coordinate system. In coordinate axes other than the principal directions, the second invariant can be written as: One then has for any coordinate system: then: The elastic domain (where the distortion energy is below a certain critical value) can then be characterized by the condition: (14.3) To determine the constant, a uniaxial test is sufficient; in effect if one denotes by s e the elastic limit obtained from a tension–compression test, one has: then: t 2 1 3 s I 2 s II 2 s III 2 s I s II s III ++ 3 ˯ ʈ 2 –++ Ó ˛ Ì ˝ ϸ = dW dV ˯ ʈ distorsion 1 2G 3 2 t 2 ˯ ʈ = s 11 s 22 t 12 2 –() s 22 s 33 t 23 2 –() s 33 s 11 t 31 2 –()++ dW dV ˯ ʈ distorsion 1 6G s 11 s 22 s 33 ++() 2 º {= º 3 s 11 s 22 t 12 2 –() s 22 s 33 t 23 2 –() s 33 s 11 t 31 2 –()++()– } dW dV ˯ ʈ distorsion 1 12G s 11 s 22 –() 2 s 22 s 33 –() 2 + º {= º s 33 s 11 –() 2 6 t 12 2 t 23 2 t 31 2 ++()++} a s 11 s 22 –() 2 s 22 s 33 –() 2 s 33 s 11 –() 2 º ++{ º 6 t 12 2 t 23 2 t 31 2 ++()+}1< a 2 s e 2 ¥ 1= a 1/2 s e 2 = TX846_Frame_C14 Page 276 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC 14.2 ORTHOTROPIC MATERIAL: HILL–TSAI CRITERION 14.2.1 Preliminary Remarks A parallel with the Von Mises criterion can be seen with the following remarks: Ⅲ For an orthotropic material, the principal directions for the stresses do not coincide with the orthotropic directions, unlike the isotropic case. Ⅲ A uniaxial test is not enough to determine all the terms of the equation for the criterion because the mechanical behavior changes with the direc- tion of loading. Ⅲ For the fiber/resin composites, the elastic limit corresponds with the rupture limit. Ⅲ The rupture strengths are very different when loading is applied along the l direction or along the t direction. Ⅲ The rupture strengths are different in tension as compared with in compression. One can then write in the orthotropic coordinates ᐉ,t,z — shown in Figure 14.2 — an expression similar to Equation 14.3, as: (14.4) 14.2.2 Case of a Transversely Isotropic Material In the following, we will limit ourselves, for the purpose of simplification, to the case of a transversely isotropic material. 3 The constants a, b, c, d, e, f in Equation 14.4 above will be determined using the results of the following tests: Ⅲ Test along the longitudinal direction ᐉ: Figure 14.2 Principal Axes for an Orthotropic Ply 3 For an orthotropic material, the procedure is identical. a s ᐉ s t –() 2 b s t s z –() 2 c s z s ᐉ –() 2 d t ᐉ z 2 e t tz 2 f t ᐉ t 2 1£+++++ ac+ 1 s ᐉ 2 rupture = TX846_Frame_C14 Page 277 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC Ⅲ Test along the transverse direction t: Ⅲ Test along the transverse direction z: due to transverse isotropy: then: Ⅲ Shear tests: due to transverse isotropy. Replacing in Equation 14.4: ab+ 1 s t 2 rupture = bc+ 1 s t 2 rupture = a = c 1 2 s ᐉ 2 rupture = b 1 s t 2 rupture 1 2 s ᐉ 2 rupture –= ᭿ t ᐉ t Æ f 1 t ᐉ t 2 rupture = ᭿ t tz Æ e 1 t tz 2 rupture = ᭿ t ᐉ z Æ d 1 t ᐉ t 2 rupture = 1 2 s ᐉ 2 rupture s ᐉ s t –() 2 s ᐉ s z –() 2 +{} º º 1 2 s ᐉ rupture 2 1 s t rupture 2 – ˯ ʈ s t s z –() 2 – 1 t ᐉ t rupture 2 t ᐉ t 2 t ᐉ z 2 +() t tz 2 t tz rupture 2 1£++ TX846_Frame_C14 Page 278 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC and in developing 4 : (14.5) Remark: For the case of a “three-dimensional” orthotropic material, an analogous reasoning to the previous presentation leads to a more general criterion, which can be written as: 14.2.3 Case of a Unidirectional Ply Under In-Plane Loading When the stress state is plane-stress, in the plane defined by the axes ᐉ ,t (see Figure 14.2), one has Equation 14.5 is simplified, and one obtains what is called “the Hill–Tsai criterion” for a ply subject to stresses within its plane: (14.6) Remarks: Ⅲ The rupture strengths of the “fiber/matrix” plies are different in tension and in compression. 5 Do not forget to place in the denominator of each of the first three terms of Equation 14.6 the values of the rupture strengths 4 Attention, this is not valid for a fabric that is not transversely isotropic! (see Application 18.2.10). 5 See values in Section 3.3.3. s ᐉ 2 s ᐉ rupture 2 s t 2 s z 2 + s t rupture 2 s ᐉ s ᐉ rupture 2 –+ s t s z +() s z s t 1 s ᐉ rupture 2 2 s t rupture 2 – ˯ ʈ +º º t ᐉ t 2 t ᐉ z 2 + t ᐉ t rupture 2 t tz 2 t tz rupture 2 ++ 1£ s ᐉ 2 s ᐉ rupt. 2 s t 2 s t rupt. 2 s z 2 s z rupt. 2 1 s ᐉ rupt. 2 1 s t rupt. 2 1 s z rupt. 2 –+ ˯ ʈ s ᐉ s t º –++ º 1 s t rupt. 2 1 s z rupt. 2 1 s ᐉ rupt. 2 –+ ˯ ʈ s t s z 1 s z rupt. 2 1 s ᐉ rupt. 2 1 s t rupt. 2 –+ ˯ ʈ s z s ᐉ º –– º t ᐉ t 2 t ᐉ t rupt. 2 t tz 2 t tz rupt. 2 t z ᐉ 2 t z ᐉ rupt. 2 1£+++ s z t ᐉ z t tz 0=== s ᐉ 2 s ᐉ rupture 2 s t 2 s t rupture 2 s ᐉ s t s ᐉ rupture 2 t ᐉ t 2 t ᐉ t rupture 2 +–+ 1< TX846_Frame_C14 Page 279 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC corresponding to the type of loadings in the numerators (tension or compression). Ⅲ Safety factor: Let a 2 < 1 the Hill–Tsai expression found for a state of stress s ᐉ , s t , t ᐉ t. One can then increase the load by means of a multiplication coefficient k to reach the limit as: The margin of safety can then be defined as the expression: which can also be written as: 14.3 VARIATION OF RESISTANCE OF A UNIDIRECTIONAL PLY WITH RESPECT TO THE DIRECTION OF LOADING 14.3.1 Tension and Compression Resistance We propose to evaluate the maximum stress s x that one can apply on a ply in the direction x in Figure 14.3. The stresses s ᐉ , s t , t ᐉ t in the orthotropic axes are given by Equation 11.4 as: Figure 14.3 Direction of Loading Distinct from Orthotropic Axes k s ᐉ () 2 s ᐉ 2 k s t () 2 s t 2 k s ᐉ ()k s t () s ᐉ 2 k t ᐉ t () 2 t ᐉ t 2 +–+ k 2 a 2 1== rupture rupture rupture rupture k s ᐉ () s ᐉ – s ᐉ k 1–= safety factor 1 a 1–= s ᐉ s t t ᐉ t Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ c 2 s 2 2cs– s 2 c 2 2cs cs cs– c 2 s 2 –() s x 0 0 Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ = TX846_Frame_C14 Page 280 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC where one recalls that c = cos q and s = sin q . Thus, Replacing in the expression of the Hill–Tsai criterion of Equation 14.6, we have then: Remarks: Ⅲ If s x is in tension, then s ᐉ rupture and s t rupture are the limit stresses in tension (tensile strengths). In effect, when q = 0: s x rupture = s ᐉ rupture and when q = 90∞: s x rupture = s t rupture Ⅲ The evolution of the s x rupture , when q varies, was discussed in Section 3.3.2. Figure 14.4 Pure Shear in x,y Axes s ᐉ c 2 s x = s t s 2 s x = t ᐉ t cs s x = s x 2 c 4 s ᐉ 2 s 4 s t 2 c 2 s 2 s ᐉ 2 c 2 s 2 t ᐉ t 2 +–+ Ó ˛ Ì ˝ ϸ 1£ rupt. rupt. rupt. rupt. s x rupture 1 c 4 s ᐉ rupt. 2 s 4 s t rupt. 2 c 2 s 2 1 t ᐉ t rupt. 2 1 s ᐉ rupt. 2 – ˯ ʈ ++ = TX846_Frame_C14 Page 281 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC [...]... rupture Remarks: Here, taking into account Figure 14.4 (txy > 0) and Equations 14.7, s rupture will be the limit stress in compression, and st rupture the limit stress in tension for 0 Ê q Ê 90 â 2003 by CRC Press LLC TX846_Frame_C15 Page 283 Monday, November 18, 2002 12:30 PM 15 COMPOSITE BEAMS IN FLEXURE Due to their slenderness, a number of composite elements (mechanical components or structural pieces)... means that the equilibrium and behavior relations are formally identical to those that characterize the behavior of classical homogeneous beams Utilization of these relations for the calculation of stresses and displacements then leads to expressions that are analogous to the common beams We will limit ourselves to the composite beams with constant characteristics (geometry, materials) in any cross section,... (geometry, materials) in any cross section, made of different materialswhich we call phases that are assumed to be perfectly bonded to each other To clarify the procedure and for better simplicity in the calculations, we will limit ourselves in this chapter to the case of composite beams with isotropic phases The extension to the transversely isotropic materials is immediate When the phases are orthotropic,... (evaluation of stresses and displacements) becomes a very complex problem when one gets into three-dimensional aspects In this chapter, we propose a monodimensional approach to the problem in an original method It consists of the denition of displacements corresponding to the traditional stress and moment resultants for the applied loads This leads to a homogenized formulation for the exureand for torsion... the relation: dv T y = ã GSề ấ q z + ậ dx â 2003 by CRC Press LLC h x Gi dS y D (15 .10) TX846_Frame_C15 Page 292 Monday, November 18, 2002 12:30 PM 15.1.5 Technical Formulation 15.1.5.1 Simplications We extend to the composite beams the simplications made for the homogeneous beams as: 1 syy and szz . and 15.7, and the simplification 2, at the beginning of Section 15.1.5.1, one can write and with the displacement field in Equation 15.3: Putting h x in the form: (15.12) Figure 15.4 Normal and. traditional stress and moment resultants for the applied loads. This leads to a homogenized formulation for the flexure and for torsion. This means that the equilibrium and behavior relations are. stresses and displacements then leads to expressions that are analogous to the common beams. We will limit ourselves to the composite beams with constant characteristics (geometry, materials)