“awn... a .nu . .< ”.4 - ”33". 1 . _‘.,.‘ <. ,. 'f"§5!w?‘i' 2-; ,4... ,, ".y‘uxt'v v m >—d>.<"> . (n) (n) (n) (n) (n) (n) 1:61;!) = C44 Yer ’ 1x9!) = C55 Yxr ’ 1x91.) = C66 7x0 where the superscript (n) ranges from 1 to N, and N is the number of constituents in the composite system. In Eq. (2.2.1), trim = [:‘otimdr, (i = x,0,r) (2.2.2) is the thermal strain, T0 and T1 are the reference (stress free) and final (current) tempera- tures respectively, and or, is the coefficient of thermal expansion in the r"h -direction. Sim- ilarly, WM) = [SBAMMM (i = x.e.r) (2.2.3) is the strain due to moisture, Mo and M 1 are the reference (stress free) and final(current) moisture levels respectively, and [3‘- (M) is the coefficient of expansion in the r"h ~direction due to moisture permeation. For transversely isotropic materials, the material stiffness coefficients Ci]. (T, M) have the following relationships: 18 The assumptions (3 - c, 1‘) ensure that the strains are in a state of generalized plane strain, therefore the displacements take the form: a?” (x, 0, r) = u‘") (9, r) +xéx ué") (x, 0, r) = v(") (6, r) (2.2.5) WW (9. r) u,"" (x, 9, r) where Ex is the uniform strain in the lamina in the fiber direction (along X, axis). The strain - displacement relations are as follows: (n) (n) .. 3v v‘") law I ) (n) 2(a) _ 13" n Wm (n) -.- a“ 9 r55 r x' r ( ) ( ) 8.0:) 3‘” n 7(a) _ la“ n 8’ 8r ’9 r 0 The equilibrium equations in cylindrical coordinates for generalized plane stress are given as: (n) ( ) ( ) latxa a192+102 = 0 (227) PT +5—0r r . . 13600!) 313:) 210!) r39 +5 + r =0 1+31.x) 80’ (n) GEM—0'3") 0 :56 “a: ““7“- The governing equations in terms of displacements are obtained by substituting Eq. (2.2.1) and Eqs. (2.2.5) - (2.2.6) into Eq. (2.2.7): 2 (n) (n) (n) C‘§’[ rza—l; “$5 ]+ c‘” ’[3-2—2- “ ]= 0 3r r 302 l9 n 32 (n) Och (n) a“, (n) r __§ _ x _ x + C( ) r W _ __r_ _ r 39 as 39 23 3739'" 39 a? r 2 ,, ,, ( ) (n) +C(n) a v( )+8w( )_ra¢on _ra‘¥o ’0] 22 (392 BE '55 6 2 (n) (n) (n) (n) 3 av (n)+ raw +aw _ * C44 \ a? ’37 " ”W a" J ' 0 (n) ~ on (n) (n) av”) (n) (n) (n) C12 r(—ex+rl>x +‘I’x )‘I’sz -a—9 —w +rd>x +r‘l’1c (2.2.8) aé‘ ad") arm (n) (n) (n) x x x -t-C13 {£— -<1>x -‘I’x +rE—r—ar —r—ar ] (n) (n) (n) av (n) (n) (n) (n)_ 3% We +C23r r—z—[am ae — ure +, +\r ra—r 4.5—; 2 (n) (n) ad) (I!) a? (n) .cgg).[.:_r_‘; .gt; —,‘"’—\P,‘"’—r$’ 457' b.) av av‘") 32w”) _ ”’44 (+8861? +52— ’0 In the Eqs. (2.2.8), the first one governs the deformation due to longitudinal shear, and the latter two govern the behavior due to normal, transverse shear, and hygrotherrnal loading. The first one is decoupled from the other ones. Moreover, considering the number of planes of geometric symmetry in the fiber geometry, the solution to the last two of Eq. (2.2.8) can be divided into a symmetric part (due to normal and hygrotherrnal loads) and an antisymmetric part (due to transverse shear loads). The solution to the first equation may be separated into a response due to shear in the X1 - X2 plane and a response due to shear in the X1 - X3 plane. 20 2.3 Solution for Symmetric Loading The solution to the last two of Eq. (2.2.8) for deformations due to applied normal strains (t:1 = €1,82 = €2,83 = é3)and hygrotherrnal loads (AM, AT) are: v‘"’(e, r) = 2 vi(")(r)sin(2i9) ‘j‘ (2.3.1) w‘”’(e, r) = Z w,‘"’(r)eos(2te) i=1 and: ej"’(e, r) = 2¢:‘")(r)cos(2i9), ‘11:")(9, r) = 2?:‘n)(r)cos(2i9) (2.3.2) 1: 0 I: 0 ¢§"’(e, r) = 2 egf’(r)eos (2:9), v§"’(e, r) = 2 ‘11:" (r)cos (2:9) i= 0 i8 0 (Doom _ .. (n) . (n) _ .. (n) . r , r) — 2 d)" (r)cos (219), ‘1’, (0, r) .. Z ‘1’” (r)cos (210) i= 0 r= 0 eye, r) = 2 exp) cos (2:9) t'= 0 The following equations are obtained by substituting Eq. (2.3.1) and Eq. (2.3.2) into the last two of Eq. (2.2.8). .. ( ) 2 ( ) ( ) ( ) dwm . . n . n . n u . i .Eosrn (210) C22 (- 41 vi — 2rwi )+ C23 -2rr$ ”‘ (2.3.3) 2 (n) (n) (n) d v. dv. dw. } (n) 2 t r (n) . t . (n) _ '1' C44 7' 'd? +r'd—r- —V'- -21r$ -21Wi — 0 21 .. . (n) . (n) Won (:0 .- - .20 cos (216) {C22 (— 21vi )+ C33 r -—2- + r-d—r (2.3.4) 3: (n) (n) dv. dv. + Cg) [2ir—dr' ]+ C3) [Zir—dr' - 2ivim — 41' 2mm] (n) (n) .. (n) (n) (n) (n) +r(C12 -Cl3 (—£xr+d)xr +‘I’x‘ )+r(C22 —C33 ) . ((1);?) + if" ) + r( cg" — egg) )(eff’ + \Pff’) } = 0 When i = 0, Eq. (2.3.3) vanishes, and the solution to Eq. (2.3.4) for orthotropic materials can be obtained as follows: wé") = 33:5“; 33;" r” +H‘"’r (2.3.5) where Bx) and 35;) areconstants, kg) = JCSVCS), kg) = -,/C$)/C3(") ,and (n) ___ 1 (n) (n) . (n) (n) H {W(Cl3 -C12 )(-Exo+¢xo +1130 ) (2..36) +(Cz(3")- C‘"’)(d>2+§"’+\ré"’) (C‘"’- *C("))(¢r(n)+‘*’r(n))} 0 0 0 0 For transversely isotropic or isotropic materials: H (n) = 0, 3.3:) = 1, kg) = - When 1' >0, the solution to Eqs. (2.3.3) and (2.3.4) is written in the form: 4 v“) = 2A‘”’r "' (2.3.7) 22 where A5") and B 5.") are constants and 1.3.") are the eigenvalues of the solution. By sub- stitution of Eq. (2.3.7) into Eqs. (2.3.3) and (2.3.4), the solution can be written as: v,""(e r) = 2 2:9“)3‘5" msin(2i9) ”‘1“ (2.3.8) 1(a) w,""(e r) = B‘"’r° (”Nagy 1” +H‘”’r + 2 23.9%” "' eos(2ie) i=lj=l where: (u) (n) (n) (n) " —i,22..‘.'"(c‘"’+c‘"’)— 2iIc:”+C‘”’) 2.“) __ + _ brm 1 (btn))2 —4a‘"’e‘"’ 2'1 - 20.90.1200!) 5 i (n) b 2 If,” _ I m 1(a) (bite) Jar-MCI") +aitIz 2a. (2.3.10) (n) b 2 (n) ___1___ (n) (n) (n) 793 ="’ (n) (n) (bi I’4ai Ci 2a 2a a, (n) (n) _ I b.- l ( (n))2 (n) (n) 2a,. 20‘. (n) (n) (n) “1’ = C33 C44 (n) .2 W (n) (n) (n) (n) (n) .2 (n) 2 (2.3.11) + 8i 2C” )c‘” (n) (n) (n) .2 2 C, = C22 C44 (41 —1) 23 B 1.5.") are the unknown coefficients to be determined as follows. To eliminate displacement and stress singularity in fiber, some coefficient should be set to zero. That is: (l) (1) (1) 302 =Bi3 =Bi4 =0 At each constituent interface r" , displacements and normal stress components must be continuous: v”) (0, r”) = v(M 1) (9. 7,.) WW (0, r") = w(M I) (0, r") ( ) ( 1) (2.3.12) II n + or (6’ r") = a (e, r") 1' t3? (9, r") = 1332+” (e, r") In the diamond array, the outer boundary conditions of a cell are determined, when a cell is deformed, by assuming that [36, 37, 38]: 1. a cell must retain the symmetry with respect to both axes (two fold symmetry), 2. the deformed shape must remain typical of all the other cells, 3. the repeating adjacent cells are congruent. In addition, the displacements and stresses must be continuous across the repeating cell boundaries, so the following conditions are obtained: (142),,4- (112),” = é2d2 é3‘13 (0,), = (0,9,. (1...), = (1,9,. (u3),+ ((43),. (2 313) 24 where (on) P and (12",) P are the normal and shearing stresses at the boundary AB, and the £2 and (3.3 are the applied strains in the X2 and X3 directions, respectively. The points P and P” are located on boundary AB as shown in Figure 3, and the locations are defined as follows: If the point P has the polar coordinates (r,, 0,.) , then due to the two fold symmetry and congruency of the adjacent repeating cells, the coordinate of point P” will be: r,. = J(d2—r,,cosep)2+ ((13-1',,sin01,)2 . 2. .14 6 _ amn[d3—rpsmep] ( 3 ) P" - d -r cost) 2 P P 2.4 Solution for Transverse Shear Loading Since the last two of Eqs. (2.2.8) are also the governing equations for the deformations caused by applied transverse shearing strain (723 ), and the solution procedure is the same as the aforementioned symmetric loading, only the final form of solution will be given here. The solution can be written as follows: "' 4 (I) (n) __ (n) ('01 (n) (n) 2., . v (0, r) — D01 r+D02 ;+ 2 2;]. Di]. r cos (219) ”12"” (2.4.1) (It) 4 (n) (n) 1"" w (9,r) = 2 2g]. Di]. r" sin(2i0) i=1j=l where 2.5.") are defined in Eq. (2.3.10) and Eq. (2.3.11). and 25 x3 I T B d3 ‘1’ i > ' A x2 |< .. >| P, Figure 3. The determination of boundary condition from congruency and symmetry of repeating cells 26 (,0 = 2 W’Icz‘g’ +6”) 2 (C?) +C‘”’) ‘7 ( ) ( ) 3(3 ) ( ) (2'42) '1 n n n D 1.3.") are constants, which are the unknown coefficients to be determined as follows, as in the symmetric loading case. In order for the solution to be bounded at r = 0 , some of D II") are set to zero: DI? _ D_(31) _ Duo) = 0 The rest of coefficients are determined by the continuity condition Eq. (2.3.12) at each interface and the boundary conditions at the boundary of the repeating cell. In this case, the boundary conditions at the edge AB are: (“2)P'I’ (”2) pa = %?23d3 1 (2(3) ,+ (143) P. = 512342 (2.4.3) (0,), = (0,9,. (2,9, = (0,9,. where points P and P” are defined in Eq. (2.3.14), and 723 is the applied shearing stain in the XZ‘Xg plane. 2.5 Solutions for Longitudinal Shear Loading The deformation under longitudinal shear loading is governed by the first of Eqs. (2.2.8). In this case, the solution is divided into one for applied loading in the X, - X3 plane and one for applied loading in the X1 - X2 plane. 27 The solution for longitudinal shear loading in the X, - X3 plane (713 = 7:3, 712 = 0) is: u x0) 10') a”) (e, r) = 2 (F35 " +F,f;"r ‘2 )sinUcG) (2.5.1) Ic=l (n) (n) (n) (n) (n) (n) (n) (n) where 7L“ = kJC66 /C55 ,1” = ’kn/Cso /C55 ,and F,l ,sz areconstants. To eliminate the singular contribution, the following coefficient are set to zero: Féé’ = 0 The rest of coefficient F X ) and F3) are determined by the similar procedure like previ- ous cases. The continuity conditions are at each interface are: not) (6, r") = u“H 1) (9, r") (n) (n + 1) (2.5.2) In (9, r") = I” (9, r") The boundary condition at the boundary AB are: (“2) + (“2) = ?13d P P" 3 (2.5.3) (11;)P = (111),,» where 713 is the applied shearing strain in the X,-X3 plane, and 1 indicates X, direction. The solution for longitudinal shear loading in the X, - X2 plane (712 = 7:2, 713 = 0) is: ” 10') (l) a“) (e, r) = Z (63% *' +03)?“ )cos(k9) (2.5.4) k = l (n) (n) . (n) (n) where A.“ , A.” are the same ones as grven above, and (3,1 , Gm are constants. To eliminate the singular contribution, the following coefficient are set to zero: 28 0;,” = 0 The rest of coefficient 0,: f) and 03) are determined by the similar procedure like previ- ous cases. The continuity conditions are at each interface are: a“) (e,r,,) = u‘"*” «mg (n) (n + 1) (2.5.5) I” (9, r") = 1x, (6, r”) The boundary condition at the boundary AB are: (u ) + (u ) = 712d 2 p 2 P 3 (2.5.6) (111)!) = (11‘) Pp where 712 is the applied shearing strain in the X,-X3 plane, and 1 indicates X, direction. 2.6 Computation of Effective Composite Properties For the composite material with a hexagonal array of the fibers, the composite consti- tutive relations in material coordinates are: 61 C11 C12 C12 51" ¢r ‘ “’1 0'2 = C12 C22 C23 82 ‘ 4’2 ’ V2 03 _C12 C23 C22 33 “ 4’2 ' ‘V2 C22 ‘ C23 123 = T723, t13= 55713, ”‘12: 55712 (2-6-1) The effective composite properties are calculated by applying a single unit strain in one direction, setting all other strains and hygrothermal loads to zero, and calculating the resulting average stresses. For example, if t»:l = l and all other strains are zero, then: 29 Cu=ovcu=62 as» Likewise the other effective composite properties can be determined in this way. The aver- age stresses are calculated as follows: an» a, = diz O’o,(0,r)dr 1:13 — i 021:“ (O, r) dr where .. M _ He) .. 42$in (“H e)’ a) _. atan(./§) (2.6.4) If the C i J. are obtained at a certain temperature and moisture level, the coefficients of ther- mal expansion and moisture expansion can be calculated by considering only constant thermal expansion or constant moisture expansion over their ranges (with all other strains zero). The procedures are very similar to that of getting Ci]. . Then the coefficients of ther- mal expansion are obtained by setting AT = l with all other strains set to zero: {a} = -[C1“{o) (2.6.5) The coefficient of moisture expansion are obtained by setting AM = l with all other strains set to zero: 30 {M} = —[C] ‘1 {a} (2.6.6) where AM is a small change in moisture content. CHAPTER III MICROMECHANICAL FAILURE THEORY 3.1 Introduction The failure theory developed here utilizes the micromechanical elasticity solution obtained in the previous chapter, and the well-known failure criteria for isotropic as well as orthotropic materials to predict the failure in a given constituent. In this chapter, a brief illustration will be given of a simple example of micromechan- ics based failure predictions. Detailed explanations will be given concerning the concept of the micromechanical failure theory and how the micromechanical elasticity solution and failure criteria are utilized here. Finally, the procedure for curve fitting of failure enve- lopes obtained by the micromechanical failure theory to polynomial type failure envelopes will be discussed. 3.2 An Example of Micromechanical Failure Consider the following example where a compressive force is acting on composite material through a rigid body as shown in Figure 4. The composite material is composed of material A with cross sectional area S A and material B with cross sectional area 8;. The total cross sectional area and length are S, and L, respectively. The Young’s modulus of composite, material A and material B are E,, EA and EB respectively. From the equilibrium 31 32 Material A h Load P = CtA, Rigid Body T Material B Base PA # Figure 4. The illustration of a simple composite body subjected to compressive force andl the free body diagram 33 of horizontal forces, we can get: 2PA+PB = P (3.2.1) The strain and displacement of each constituent are: for constituent A, 8A = — = _ = _ 5, = (3.2.2) for constituent B, EB = — = — = —— a, = — (3.2.3) Since material A and B are compressed through a rigid body, it can be assumed that the displacements are same. Therefore, by equating Eq. (3.2.2) and Eq. (3.2.3): EASA P = —P (3.2.4) A EBSB B The substitution of Eq. (3.2.4) into Eq. (3.2.1) leads to: EBSB P8 = ZEASA+EBSB P (3.2.5) 34 Substituting of Eq. (3.2.5) into Eq. , PA is obtained: P = 5‘5“ P A ZEASA+EBSB (3.2.6) Then, the stress state of each constituent and the combined material, respectively, when only the compressive load P is acting is obtained as follows: a, = g (3.2.7) P E P A A 0' = — = (3.2.8) A A ZEASA-t-EBSB P E P a, = —B = 3 (3.2.9) If the applied load is increased, the stress in constituent A and/or constituent B will even- tually reach its ultimate value. Failure of the composite may be defined as failure of either constituent A or B. The stress in the composite at that time is: of = P: (3.2.10) and we have either P“ E P“ u A A o - — (3.2.11) A SA ZEASA-t-EBSB Ol' 35 2 "U a: PP“ OB=— B = .212 SB 2EASA+EBSB (3 ) where of, P", 0;, Pg, 6;, and P; are the stresses and loads in the composite material and its constituents, respectively, when the failure occurs. In this case, the stress in each constituent can be calculated, and a proper failure criterion for each constituent has to be chosen to predict failure. Here, it can be observed that the (average) stress state of the composite material is different from that of each constituent from Eq. (3.2.10), Eq. (3.2.10) and Eq. (3.2.12), when the failure occurs in some constituent. In order adopt this concept to failure of fiber reinforced composite materials, we need to 1) set up an appropriate RVE (Representative Volume Element), 2) perform stress anal- ysis in RVE for a given set of external loads, and 3) choose the proper failure criterion for each constituent to predict the onset of failure. Steps 1 and 2 were discussed in the previ- ous chapter. Below, some common failure criteria are discussed. 3.3 Failure Criteria Failure criteria can be roughly divided into those for isotropic materials and those for orthotropic materials. Failure criteria play a very important role in the micromechanical failure theory, because failure criteria predict whether or not failure occurs at the selected points in the R.V.E. Thus, the several failure criteria useful for prediction of the failure in fiber, matrix and fiber/matrix interface will be described briefly. 3.3.1 Failure Criteria for Isotropic Material 36 Maximum Principal Stress Theory, or Rankine Theory The maximum principal stress criterion assumes that failure occurs when one of the prin- cipal stresses is equal to or greater than the value of the uniaxial yield stress in tension, 6:, or is equal to or less than the value of the uniaxial yield stress in compression, of. This criterion asserts that yielding will occur when any one of the following conditions is reached: I . 2 T > T 2 T ntensron, 61 a“, or 62.6", or 0'3 on I . 2 C > C 2 C ncompressron, (31 a“, or 02.0“, or 03 a, This theory assumes that only the maximum principal stress causes failure, and the contri- bution of stress interaction is disregarded [39]. Maximum Principal Strain Theory, or Saint- Venant Theory The maximum principal strain criterion assumes that failure occurs when one of the prin- cipal strains is equal to or greater than the value of the uniaxial yield strain in tension, 8:, or is equal to or less than the value of the uniaxial yield strain in compression, sf . The 1' C . a“ and an are related to the ultrrnate stresses by: where E is the elastic modulus of the material. This criterion asserts that yielding will occur when any one of the following conditions is reached: T . T T In tensron, £1 2 a“, or 82 2 e“, or £3 2 e“ . C C C In compressron, e, 2 a“, or £2 2 an, or 83 2 e, 37 In this theory, the role of the interaction of the principal stress is accounted for failure [39]. Maximum Shear Stress Criterion, or Tresca Criterion The maximum shear stress criterion (sometimes called the Coulomb theory) assumes that failure occurs when the maximum shear stress is equal to or greater than the value of the maximum shear stress, 1:“ = - whrch rs occumng under unraxral tensron. The man- 2°u’ mum shear stress is equal to half of the difference between the maximum and minimum principal stresses. This criterion asserts that failure will occur when any one of the follow- ing conditions is reached: '01 " c’2' = c’u . lo2 — 63] o “9 [03—01]: a In this theory, some interaction of the principal stresses is included, but this theory is not applicable for the material whose tensile and compressive strengths are difi‘erent [39]. Distortion Energy Criteria, or the Von Mises Criterion The distortion energy theory assumes that failure begins when the distortion energy is equal to or greater than the distortion energy at failure in uniaxial tension. This criterion asserts that yielding will occur when the following condition is reached: 1 2 2 2 §[(°1 -02) + (oz-03) +(03’01)2] = on The distortional energy criterion accounts for interaction of the principal stresses [39]. 38 3.3.2 Failure Criteria for Orthotropie Material Independent Maximum Stress Criterion The independent maximum stress criterion assumes that failure will occur when any of the stress components referred to the principal material axes is equal to or greater than its cor- responding allowable strength in that direction. Thus, this criterion asserts that failure would occur when any one of the following conditions is reached: c"11 = X,T. “611 = Xi: 022 = X; ”322 = X: (’33 = X; ‘033 = X: c’23 = X23, c’31 = X311 (’12 = X12 where 0,]. are the stress components referred to the material axes, and X? , X? and Xijare the tensile, compressive and shear strengths, respectively, of the material in its i"I princi- pal direction. This theory does not account for stress interaction [40]. Independent Maximum Strain Criterion The independent maximum strain criterion assumes that failure will occur when any of the strain components referred to the principal material axes is equal to or greater than its cor- responding allowable strain in that direction. Thus, this criterion asserts that failure would occur when any one of the following condition is reached: _ T _ C 811' e,. “311 " er _ T __ C 322 " ‘32, ’322 ' 32 823 = £23, 531 = 331’ 512 = 912 . . T C where 8,]. are the strarn components referred to the materral axes, and e, , e, and e ,1. are . . . . . . .th . . the tensrle, compressrve and shear strengths, respectrvely, of the materral 1n rts l prrncr- pal direction. This theory accounts for the stress interaction but disregards any strain inter- action [40]. 3.4 Micromechanical Failure Theory Failure criteria for composite materials have been deve10ped on the basis of the afore- mentioned micromechanics elasticity solution and failure criteria for isotropic and ortho- tropic materials. They can be cast in several forms in either stress- or strain- space, and require only stiffness properties and uniaxial strength data of each constituent The failure of composite materials is predicted when stress or strain states of a point in any of the constituents (e.g., fiber, matrix and interface) exceed a critical level as pre— dicted by a properly chosen failure criterion for each constituent. When a lamina (compos- ite) load gives rise to a micromechanical stress state that causes failure in some constituent, then that load state is assumed to lie on the failure envelope for the composite material. When a failure criterion for each constituent is chosen, the mechanical properties of each constituent and the failure characteristics of each constituent in the composite material should be taken into account. Since the primary emphasis here was on polymer matrix composite materials, the details to be discussed are most suitable for these material systems. 40 Fiber Fibers in composite materials are used as a reinforcing agent. In polymer matrix com- posite materials, the longitudinal Young’s modulus of the fibers is often 10 times greater than that of the matrix [4]. If a unidirectional lamina is subjected to a tensile or a compres- sive longitudinal load, the failure of the lamina is usually a result of the fracture of fiber in tension or fiber buckling/kinking in compression, and the failure modes and the failure stresses for each case are quite different. When a tensile longitudinal load is acting on a lamina, the axial strain of each constituent in the lamina is the same, since it was assumed that fibers are perfectly bonded to the matrix through fiber/matrix interface: 5’ = e’" = a’ (3.4.1) where 8;, a: and 8:! are the longitudinal strains of fibers, matrix and lamina, respec- tively. However, generally, the fibers have a lower ultimate strain than matrix, and lamina fracture occurs at the fiber ultimate strain [41]. When a lamina is subjected to a compres- sive longitudinal load, the failure of a lamina is often caused by buckling or kinking of fibers in out-of phase mode or in-phase [42]. But, according to Hashin [22], the depen- dence of both fiber failure modes on the axial shear stresses is not identified clearly. Hence, it is concluded that axial tensile and compressive strains of fiber dominate the composite longitudinal strength. Thus, the maximum strain criterion was chosen as the fiber failure criterion in this research. Failure modes such as fiber buckling/kinking are assumed to initiate at some critical level of axial strain, and so are accounted for in indi- rectly. In addition, it is assumed that the fiber never fails due to transverse and shear strains (82, 83, 723, 731 and 712 ), therefore, the much bigger values than that of 41 longitudinal strength are assigned to transverse and shear strength components. F iber/matrix interface Generally, the fiber/matrix interface is a vanishingly thin region in the composite between the fiber and matrix, and is formed during the composite manufacturing process. It transfers the load between the fiber and matrix in composites, so that it affects the mechanical response of composite materials. If the interface region has a small but finite thickness, it is often referred to as an inter- phase. The interphase may have distinct elastic properties, and it was shown in references [43, 44] that the elastic moduli and the thickness of the interphase have an influence on the effective elastic properties. But the elastic moduli and the thickness are often not available for many composite systems. Thus, in the current study, an interphase region was not con- sidered, though the model is sufficiently general to have considered one. Furthermore, its normal and shearing strengths are difficult to measure experimentally. So, in the current research, the normal and shearing strengths were determined using the following procedure [45]. With all other applied loads being zero, the lamina ultimate transverse strength was applied to the micromechanical model. Then, normal and shearing stresses were computed in the matrix at the fiber/matrix interface region, and principal stresses were computed elsewhere in the matrix. If the maximum principal stress in the matrix due to the load was less than the tensile strength of the matrix material in the bulk state, then it was assumed that failure occurred at the fiber/matrix interface and the maxi- mum normal stress at the fiber/matrix interface region represents the normal strength of interface. Otherwise, it was assumed that failure occurred in the matrix, and the normal strength of interface is equal to or greater than the computed maximum normal stress at 42 the interface. The shear strength of interface was calculated in a similar way by applying the lamina ultimate inplane shearing stress. It was assumed that interface was perfectly bonded to both fiber and matrix, and failed due to tensile transverse normal stress or shear stresses (are, on ). Hence independent maximum stress criterion was applied to predict the failure in the interface. [28, 32] Matrix: When a lamina is subjected to a longitudinal load, the matrix plays the role of both protecting the fibers and distributing load between and among the fibers. However, when a transverse tensile load is applied to the lamina, the fibers do not serve as a load-carrying constituent in the lamina, but act as solid inclusions in matrix. Due to the existence of solid inclusions in the matrix, the local stresses and strains in the matrix are higher than the applied stress [4]. In addition, even though the region with maximum stress in the composites is often located in some other phase, failure generally occurs in the matrix first. Furthermore, a uniaxial load acting on a composite material in any direction can cause a three dimensional stress state in the matrix, which causes the stress components to interact. So, it is assumed that the transverse and shear strength of composite material is dominated by the strength of the matrix. The suitability of a matrix failure criterion depends on the properties of the matrix - ductile, brittle, etc. In this work, the following criteria are considered: maximum principal stress, maximum principal strain, Von-Mises and Tresca criterion, wherein the Tresca failure criterion is applicable to matrices of which the tensile strength and the compressive strength are the same. The stress evaluation points are carefully selected as shown Figure 5 in order not to miss the locations of high stress. The stress evaluation points at the interface region and 43 near the interface region in the matrix should be densely distributed, because not only do the highest stresses often exist in that region, but also the stress gradient in that region is very large. In the current research, 249 points in the RVE were chosen at which to evaluate the stresses as shown in Figure 5. The interface is treated as a finite region, but is so thin that the stress evaluation points in that region appear to overlap with those in the other regions. Lamina Strength Components The uniaxial normal strengths of a lamina are determined if a uniaxial lamina load act- ing along a principal material of a lamina creates a micromechanical stress state that causes some constituent to fail, and the lamina load is assumed to be the unidirectional strength of the lamina in that direction. Constituent failure due to the micromechanical stress state caused by a lamina load is predicted by the failure criterion chosen for the con- stituent. The shear strengths of a lamina are obtained in the same way as the longitudinal normal strength are found. But more attention and insight should be paid to determine the transverse normal strength and the longitudinal shear strength. In reality, the arrangement of fibers in contin- uous fiber-reinforced composite materials is random, therefore these materials can be regarded to be transversely isotropic in both stiffness and strength. The assumption of hex- agonal packing of fibers, as used in the current elasticity solution, gives a rise to transverse isotropy of elastic moduli, but not strength. Thus, the transverse normal load and the lon- gitudinal shear load causing failure of some constituent should be applied and averaged over a range of loading directions from 0 degrees loading direction to 360 degrees (or 0 degrees to 30 degrees due to symmetry) to obtain the transverse normal and the r 9 "m f + : interface 0 : matrix Figure 5. The stress evaluation points in RVE 45 longitudinal shear strength of the lamina. To perform these processes, a uniaxial stress load acting on the model at a certain angle should be transformed to the reference axes by using the stress transformation law. When a stress load is acting on the lamina at a certain angle as shown in Figure 6, by stress transformation laws, the stress tensor 6 in the Cartesian coordinate PX,XZX3 , of which components are 6,]. i, j = 1, 2, ..., 6 , can be expressed in terms of the stress tensor 0' in the Cartesian coordinate PX1X2X3 , which has 6,}. i, j = l, 2, ..., 6 as components, as follows [47]: o = ATOA (3.4.2) where A is a proper transformation matrix, and AT is the transpose matrix of A. The A T . . and A matrices are wrrtten as: l 0 0 A = [0,1] = 0 cos (--6) sin (—9) (3'43) 0 -sin (-6) cos (-6) T 1 o o T [4.1] = 0 cos (-6) -sin (-6) (3.4.4) 0 sin (-9) cos (-6) >- II The tensors 0', 6 are: 011 0’12 031 311 312 831 O = [0:1] = c’12 c’22 “23 ' 6 = [31,] = 312 322 323 (3°45) c’31 023 033 631 623 t’33 46 X3 x, A X2 P , x2 x, x, (b) Figure 6. (a) The RVE subjected to an off-axis load (b) The reference axes and the rotated axes 47 Substituting Eq. (3.4.3), Eq. (3.4.4) and Eq. (3.4.5) into Eq. (3.4.2), stress tensor 6 in the Cartesian coordinate PX1X2X3 is transformed into the stress tensor 0' in the Cartesian coordinate PX 1X2X3 as follows: [ail = 21.-Hal [a] 1 O 0 611 512 (’31 1 0 0 = 0 cos (—6) —sin (-9) 3,2 022 523 0 cos (-6) sin (—6) _0 sin (-6) cos (-6) 331 523 533 0 -sin (—0) cos (-6) 011 = 1"11 0‘12 = 612c036+63lsrn8 622 = {5‘22cosfi2 + 2623 sinOcosB + 633 sin 62 (3.4.6) 623 = —622cosesin9 + 623( cos62 — sinez) + 633cos9sin0 0’33 = 622 sine2 — 2623 cos6sin6 + 633 cost)2 Eq. (3.4.6) gives us the relationships between the stresses in reference axes and those in certain rotated axes. Therefore, an off-axis stress load can be transformed into the refer- ence axes (the Cartesian coordinate PX1X2X3 shown in Figure 6) by using Eq. (3.4.6). The transformed stresses are taken as a stress load acting on the model, and cause a micro- mechanical stress state in each constituent. If the failure of some constituent is predicted by the failure criterion chosen for that constituent, then the stress load is taken as the strength of lamina at that angle, and all the lamina loads at each angle from 0 degrees to 360 degrees are obtained. The lamina transverse normal and longitudinal shear strengths are obtained by averaging the obtained lamina loads from 0 degrees to 360 degrees. Failure Envelopes 48 The failure envelopes in X, - X2 plane, i.e., when the biaxial loads in X, and X2 direc- tions are acting together, are obtained in the following way. First, the longitudinal lamina load is fixed at some value which is less than or equal to the longitudinal normal strength of the lamina, then the transverse lamina load giving rise to failure in some constituent is sought and averaged from 0 degrees to 360 degrees. Accordingly, the lamina fails at the biaxial load, and the longitudinal and transverse normal loads are assumed to lie on the failure envelope. Then, the longitudinal lamina load is incremented, and the transverse normal lamina load that causes failure is again obtained. After the above processes are repeated over the range from the tensile longitudinal strength to compressive longitudinal strength of the lamina, the failure envelope for biaxial loading in stress space is obtained. The failure envelope in strain space can be obtained by converting the stress states in fail- ure envelope to strain states by means of the constitutive law for an orthotropic body [48]: l v12 V13 811 - forr‘fazz'faaa 1 1 1 l v12 v23 822 - Eazz-EGn-E—zoaa (3.4.7) 1 V13 v23 8 = —o -—o -—o 33 33 11 2 E3 51 52 28 --—-—lo 28 -—lo 28 -——lo 23 " 23» 13 ‘ 13' 12 ' 12 023 613 012 where 0,]. is Cauchy stress tensor, 2,} is infinitesimal strain tensor, and E, and 6,1. are Young’s moduli and shear moduli, respectively. When converting a failure envel’ope from stress space to strain space, it should be noted that a biaxial stress state creates a tri - axial strain state due to Poisson’s effect. This phenomenon can be easily verified from Eq. (3.4.7). 49 If we expand the concept for obtaining the failure envelope for biaxial loading, we can get the failure envelope when the longitudinal normal, transverse normal and inplane shear loads are acting on the lamina at the same time. In this case, the maximum loads in longitudinal and transverse directions should be obtained first. Then, similar to the biaxial case, the longitudinal shear load is applied, with the longitudinal normal load fixed at some value equal to or less than the maximum load in the longitudinal direction. Then the transverse load causing failure in some constituent is averaged from 0 degrees to 360 degrees. Thus the transverse load is determined by averaging the obtained transverse loads from 0 degrees to 360 degrees as before. The longitudinal normal and shear loads, and the averaged transverse load are assumed to be on the 3 dimensional failure surface. 3.5 Curve Fitting of Failure Envelopes The failure envelope obtained by the micromechanical failure theory for biaxial lam- ina loading can be fit to a polynomial type failure theory by using the predicted unidirec- tional strengths and nonlinear regression. Among the polynomial type failure criteria for orthotropic materials, Tsai - Wu’s failure criterion is widely used, and includes interaction among the stress components analogous to the Von - Mises criterion for isotropic materi- als [21]. The general form of Tsai - Wu’s failure criterion is [12]: Fifi-”yap,- = 1 131' = 1,2.....6 (3.5.1) where summation on the repeated subscripts is implied, a, are the components of the Cauchy stress tensor and F i , F i]. are strength parameters. If we expand the Eq. (3.5.1), then we have: 50 F161 + F262 + F303 + F464 + F565 + F666 + Pnof + F226: (3.5.2) + F336: + F440: + F550: + [765635 4» 217120162 + 2F130'10'3 + 2F1 40104 + 217150105 + 2F1651°62F2392°3 + 217240204 + 2F250'20'5 + 217266266 + 217346304 + 217350365 + 2F360’3O'6 + 217456465 + 217460406 + 217560566 = 1 In the above equation, the linear stress terms provide for strength difference and sign reversal of normal suesses in tension and compression, respectively, which is extremely important for a lamina. But the sign reversal is unsubstantial for the (shear stresses, because the strength of a lamina should not be affected by the direction or the sign of the shear stress components. So the terms pertaining to the first order shear stresses must van- ish. The relevant terms are: F464, F505, F606, 217146164, 2F150'los, 217160106, 217240204, 217256265 2F260206, 2F34o3o4, 217350365, 217360306, 217450405, 217460406, 2F560506 Since the stress components are in general not zero, the only way to make the relevant terms vanish is to set the strength parameters zero: F4=F5=F6=Fr4=Frs=F16=F24=F25 (3.5.3) =F26=F34=F35=F36= 45=F46=F56=0 Now the Eq. (3.5.2) can be rewritten as: 2 2 2 2 Flo1 + F202 + F363 + F1101 + F2262 + F3303 + F4464 (3.5.4) 2 2 + F5565 + F6606 + 217120162 + 217130103 + 2F230'20’3 = 1 1 l l 1 1 l F=—--—— F=—-— F=-———— 1 , 2 r 3 ' XIT Xf x: Xf x; x: 1 l 1 F =—— F = F = 11 C. 22 T c, 33 c, (3.5.5) Xfxr X2X2 3X3 1 l l F44=—, F55=—, F66:— The other strength parameters will be discussed later. For the case of biaxial stress state, (i, j = 1, 2), the only non - zero stress components are 61 and 62. Consequently, the Eq. (3.5.4) is simplified to: 2 2 F191 + F262 + F1 lo1 + F2202 + 217126162 = 1 (3.5.6) The non - zero strength parameters F 1 and F :1 can be determined in terms of strength measured by simple tension or compression or shear tests. The non - zero strength param- eters of Fij (i ¢j) i.e., F12, F23 and F31 give an account of the interaction between the two normal stress components. For example, among the interaction terms, F12 can be obtained from Eq. (3.5.6): F - 1 1F 1 61F 021? 357 12' ‘ _1+—F2+—11+;l' 22 (..) l 2 0'2 e1 02 The Eq. (3.5.7) shows us that F1] are functions of the ratio of a, to of . The only way that the interaction parameter can be evaluated experimentally is to perform the biaxial tests with every possible ratio of o". to O’j . However, this experimental task is unhappily not as easy and simple as the uniaxial test or shear test. In addition, even if F1} are very small, they play an important role in the failure criterion in that small changes in F”. can signifi- 52 cantly affect the predicted strength and its failure envelope. In Figure 7, the influence of F12 among the interaction terms is shown. To ensure that the failure envelope is closed and finite, Tsai and Wu suggest [12]: F1,“ = _L (3.5.8) F ,,F j, where —1 $17,)?“ $1. Tsai and Hahn suggest [19]: F..* = -O.5 (3.5.9) '1 to make the failure enve10pe similar to that of Von Mises failure criterion. In this study, the problematical interaction strength parameters, F ,j , are obtained by making the failure envelope obtained by micromechanical failure theory fit to the polyno- mial type failure theory by means of the nonlinear regression method[15, 49]. For exam- ple, the interaction strength parameter F12 is obtained as follows. Let the left hand side of Eq. (3.5.6) be Y, and the right hand side of Eq. (3.5.6) be y, : 2 2 Y, = F,c,, + F262,- + F11°11+ F2202: + ”1261,62.- (3.5.10) yi=1 F12 which is the regression coefficient in this nonlinear regression procedure can be cal- culated by minimizing a deviation function S defined as follows: N s = X (Y,-—y,)2 (3.5.11) i=1 SIGMA2 53 0.5 ' l Figure 7. The variation of failure Envelopes due to change in the interaction term 1",, in a, — 02 domain X10 54 2 = 2(F1611+F2621+F11°ii+F22°§1+2171291552," 1) (3-5-12) i=1 where N is the number of data points. Using the method of calculus, we can minimize the function given by Eq. (3.5.11) as shown below: as _ (3.5.13) 'apj - 0 Substituting Eq. (3.5.11) into Eq. (3.5.13) gives: N as a 2 2 2 a!"12 = 3F,2(F161i+F2°2i+F1161:+F22°21+2F12°11°21' 1) = 0 (3-5-14) 1 If we take the partial derivative of Eq. (3.5.14), we have: N 8S 2 2 5F” = 2 {46“02‘(F1°li+F2°2i+F11011+F22°2FI) (35.15) i=1 2 +8F12(oli62i) } = 0 If we solve for the regression coefficient, F12, from Eq. (3.5.15), then we obtain the coeffi- cient as follows: (3.5.16) :1 N u «I >< where N N N N N 2 2 3 3 X = 2 olio2i- 2 FloliOZi— 2 FZGIIOZi" 2 Fllclioli- 2 F220,,0'2, i=1 i=1 i=1 i=1 i=1 N 2 2 Y: 2201302: i=1 55 Hence, the problematical interaction terms are determined without the need for any biaxial experiments. With the determined strength parameters and the uniaxial strengths of lamina, it is pos- sible to plot the polynomial type failure envelope in two ways, when a biaxial stress state is applied to the lamina. One of the approaches is to find a set of a, and 02 to satisfy Eq. (3.5.6) by trial and error method. That is, a, (say) is fixed at some value first which is equal to or less than the ultimate strength in X, or X2 direction, and then 0'2 is sought to satisfy the Eq. (3.5.6) by varying it from its uniaxial tensile strength to its compressive strength. By repeating the above process, the set of points on the failure envelope is calcu- lated. The other method is to obtain an equation for a, or 0'2 from Eq. (3.5.6) by means of the discriminant formula for a quadratic equation. Rearranging Eq. (3.5.6) with regard to 0'2 in the descending order gives: 2 2 F2262 + (F2 + 2F12crl)o'2 + Flo, + Flo, — 1 = 0 (3.5.17) Solving Eq. (3.5.17) for 0'2 gives: 4L 2 I 2 a, = 21, (3.5.18) 22 A 2 I 2 — (F2 + 2171261) - Jar, + 217126,) — 4172411716l + F161 -1) 62 = 2F (3.5.19) 22 At a value of a, , Eq. (3.5.18) represents points in one half of the failure envelope, and Eq. (3.5.19) represents points in the other half of the failure envelope. Therefore, a set of 62 56 and a, on the failure envelope, which satisfy Eq. (3.5.6), is obtained by varying a, from the tensile strength to the compressive strength of the lamina. CHAPTER IV INDEPENDENT MODE FAILURE CRITERIA 4.1 Introduction A lot of researches have been carried out to predict the failure envelope more realisti- cally, on the assumption that the failure envelope is a piecewise smooth curve, since the mechanical behavior of composite material is anisotropic. Some researchers established separate failure criteria in each quadrant of stress space by defining a difl‘erent functions, in order to obtain piecewise smooth curve [23, 16]. In this chapter, it will be discussed how to obtain a more realistic and applicable failure criterion by considering the failure mode of each constituent [22]. 4.2 The Concept of Independent Mode Failure Criteria The stress distribution in each constituent is affected by the magnitude of applied load and/or the ratio of the applied loads in each direction. The change in the magnitude of applied loads causing failure in some constituent leads to changing only the magnitude of all the existing stress components at the same rate, but does not afl‘ect the relative magni- tude of all the existing stress components. Thus it makes a contribution to the first occur- rence of failure in the same constituent. But the change in the ratio of applied loads causing failure changes the distribution of stress components in the constituents, and may 57 the 8111 Eli): SUE Em} 58 induce a change in the failure mode. This fact implies that the failure mode of all the points on the failure envelopes are not the same. In other words, the failure envelope of a composite material is represented by theintersected area of the failure envelopes of each constituent. Moreover, it would be unreasonable that the entire failure envelope can be represented by a single equation, which gives a mathematically smooth curve, since each failure mode is entirely different. Thus, it is very useful to establish a separate failure cri- terion for each constituent in terms of lamina loads by supposing that failure takes place in only the constituent under consideration, rather than using a single equation for entire fail- ure envelope. 4.3 Independent Mode Failure Criteria The failure envelope of each constituent in a given composite material system can be written as function of macro (lamina) load. This function is obtained via the microme- chanical model by allowing failure to occur in only the constituent under consideration. All other constituents are assigned strength values that are orders of magnitude larger than the actual strength to ensure that failure will not be predicted to occur in these constitu- ents. Then the failure envelopes are predicted in terms of lamina loads, using the micro- mechanical failure theory. By repeating this process for each constituent, the failure envelope of each constituent can be calculated as a function of the macro (composite) stress state. The inner intersected area of the failure envelopes is regarded as the failure envelope for the lamina. Since both the failure load and the mode of failure are predicted in the current scheme, it is possible to cast the failure model in a simple mathematical form with separate criteria 59 for each mode of failure. For example, if a polynomial form is desired, the criteria can be written as follows: For fiber: 1 t l Ffo, + Fgopj = 1 (4.3.1) If the lamina loading is biaxial, Eq. (4.3.1) is expanded to: Pfia’1 +P26’2+P{ (a ,2) +P§2(a‘2) +2P§2a" 02 = 1 (4.3.2) For matrix: 1","6, +F'."o.o (4.3.3) rji If the lamina load is biaxial, Eq. (4.3.3) is expanded to: 2 2 Fret, + F2012 + Ff,(o',) + F’2"2(ol2) + 2F'l"2o',ol2 = 1 (4.3.4) For fiber/matrix interface: '11 n P“, 6’.+ P’Uo’a , (4.3.5) If the lamina load is biaxial, Eq. (4.3.5) is expanded to: F',"o 1+ F'2no 2+ 173(01): + F’2’2(o2)2 + 2F‘1'2011012- = 1 (4.3.6) where the superscripts m, f, in and I refer to matrix, fiber, interface and lamina, respec- tively, Ff and Ff are the strength parameters in terms of the lamina load as shown rn Eq. 60 (3.5.5), and 0': represents the components of stress in the lamina. The strength parameters of each constituent in Eqs. (4.3.1 — 4.3.6) should be computed as functions of the lamina loads using the micromechanical failure theory. Then, the strength parameters (i.e., F, and F ,,) can be calculated as in Eq. (3.5.5). Alternatively, other mathematical forms of the failure criterion may be used. For example, an appropriate form for fiber failure may be: in tension: a? _I_ = 1 (4.3.7) slur in compression: ; = 1 (4.3.8) slur Once the shape of the failure envelope for each constituent is known, an appropriate mathematical expression to describe each surface can be chosen. CHAPTER V THE INDEPEDENT MODE FAILURE CRITERION BASED ON “IN-SITU” CONSTITUENT STRENGTH 5.1 Introduction The accuracy of the micromechanical failure theory depends largely on the degree of error in the material prOperties of each constituent in the composite system. The geometry, stiffness and strength of each constituent in a composite system are inherently randomly varied during the manufacturing processes of a composite system, and are usually differ- ent from those of each constituent in bulk state [31]. For this reason, the use of bulk mechanical properties in the micromechanical failure theory may lead to poor failure pre- dictions in composite systems. Especially, the mechanical properties of polymers which are commonly used as matrix are time - dependent, so that the mechanical properties of polymers in a composite system are possibly different from its bulk mechanical properties. For example, the creep function of an epoxy matrix is given by the four-parameter model [52]: 1 1( 4:) t J t = —+— l—e +— 5.1.1 () E0 Er “o ( ) where E0, E,, h and “o are constants, and t is time. A plot of the creep function for 934 resin and the epoxy matrix used in glass/epoxy composites is shown in Figure 8. If the 61 applied loading, a , is constant with time, the strain will be [53]: 62 3 I I I I I 934 2.5 - _____________ . ............................. . ............ , ./ . I 2 - . a? E 5 EPOXY 1.5 " d :; 1 I 0.5 - . o I I I I I o 20 40 so so 100 120 TIM E (min.) Figure 8. The creep behavior of some materials used as matrix 63 £(t) = O’J(t) (5.1.2) This visco-elastic behavior of matrix leads to poor prediction of the transverse strength of a lamina, since matrix failure dominates the transverse strength of a lamina. Another obstacle to using micromechanical models is that some mechanical properties of constituents are not commonly available. For example, the compression strength of a matrix material, which is sometimes different from the tension strength, is often not readily available. Hence, the current model includes the option to extract average “in-situ” constituent strength properties using unidirectional composite strength data, so that it indi- rectly accounts for the above variations as well as thermal residual stresses, voids, and other processing effects. 5.2 The In-situ Constituent Strengths To obtain the in-situ strength of each constituent, a failure mode must be assumed for a given applied uniaxial loading state. For example, transverse lamina strength is often dominated by matrix strength, while failures due to longitudinal loads usually occur in the fiber. Thus, the processes to obtain the in-situ strength of each constituent are different. Fiber Strength Since the longitudinal tension and compression strengths of a composite material are dominated by the tension and compression strengths of fiber, the longitudinal strength of a composite material is applied as a load. The resulting maximum strain in the fiber at that 64 load is taken as the ultimate strain of the fiber. Matrix Strength The transverse tensile and compressive strength of a composite are dominated by the tensile and compressive strengths of the matrix or by the strengths of the interface. According to Kominar and Wagner [50], the failure of unidirectional composites with very stiff resins can take place without significant preliminary interfacial debonding but as the result of matrix crack nucleation and growth. Moreover, Folias [51] found that if the deb- onding at the interface does not occur, there exists a stress magnification factor in the matrix which attains a maximum between the fibers. If a crack does initiate at the inter- face, then it commonly propagates through the matrix to connect with cracks at other interfaces, eventually leading to lamina failure. When a transverse load is applied to a uni- directional lamina, it is difficult to distinguish between the load at initial failure and the ultimate load, because the cracks often propagate so rapidly. Consequently, in the current model, the transverse strength of the composite is applied as a load in the model at each angle from 0 degrees to 360 degrees, and the maximum principal stress in the matrix is determined for each angle, and averaged from 0 degrees to 360 degrees. The averaged maximum principal stress is taken as the ultimate strength of the matrix. Note that this value of matrix strength may also be related to the interface strength, depending upon the mode of failure in the experimental tests employed for mechanical characterization of the lamina. 65 5.3 The Independent Mode Failure Model Using “In-Situ” Properties The obtained in-situ strengths are used as constituent strengths in the failure criterion for each constituent. In this independent mode failure criterion, the interface mode of fail- ure is excluded, because the transverse strength of the composite was assumed to be dom- inated by matrix strengths. In polynomial form, the failure criteria can be written as: Fiber Pfof +P’o’o’. (5.3.1) rjio where Ff and Ff are obtained in terms of macro by substituting‘ ‘in- -situ” strengths of the matrix into Eq. (3.5.5) and the interaction terms F; are obtained by nonlinear regression like Eq. (3.5.16). Matrix F76, +F’."oo (5.3.2) '1‘]: where 15’"I and F" are obtained by substituting‘ ‘in-situ” strengths of the matrix into Eq. (3.5.5) and the interaction terms P1; are obtained by nonlinear regression like Eq. (3.5.16). This technique enables us to match the predicted and experimental unidirectional strengths as well as to obtain more realistic failure envelope under multi-axial loading conditions, because much of the inherent randomness and processing effects are efl‘ec- tively accounted for. ' CHAPTER v1 NUMERICAL IMPLEMENTATION AND DISCUSSION OF RESULTS 6.1 Numerical Algorithms The numerical procedure used in this research consists of conversion of macro stress loading to macro strain loading, micromechanical stress analysis, micromechanical failure prediction, prediction of uniaxial strengths, and the failure envelope for multiaxial load- ing, and curve fitting of the failure envelopes. The used algorithm is depicted in Figure 9 in detail and is described as follows: 1. Using the micromechanical elasticity solutions, compute the effective lamina material properties and store the micromechanical stresses due to each of the six single components of unit macro strain (lamina strain). 2. Convert 1 MPa of unidirectional lamina load to the corresponding lamina strain state and apply the strain state to the micromechanical model. Compute stresses at the desired points within each constituent (i.e., fiber, matrix, or interface) of the material system due to the uniaxial lamina load. 3. Evaluate the chosen failure criterion at each point within each constituent of the mate- rial, store the failure information, and determine the maximum failure index (mum) for all the points. 4. Check for failure (tymx ). If the failure criterion is not Von - Mises criterion among 66 67 START l. 0 Compute the lamina effective moduli. - Calculate and store micromechanical stresses and strains due to unit macro strains. Yes 2. 0 Convert lamina load(s) to lamina strains. 0 Apply the strains to the model. V 0 Compute stresses at the desired points within each constituent. l in-situ properties 9 7. - Apply the longitudinal or transverse strength as a load. . OCheckthemaximumstressorstrain — 3. 2:33:16 31:1: $033?“ {allure within the related constituent. - Determine the maximum failure 0‘ index (trim) for all points. t . Increase or decrease the lamina load. 4. 0 Check for failure. 0.99 zwms l Applied lamina load Yes 5 Failure Index - Uniaxial Strength I 5. 0 Fix longitudinal stress at some value less than the tensile strength. -Choose 1MPaasatransversestless. olncreaseordecrease transverse stress. 'The longitudinal and the transverse stress are assumed to be on the failure envelope. ' F“ the @1013 envelope (0 polynomial - Repeat this process until the longitudinal "7 type failure envelope or Other form. stress reaches the longitudinal strength. 5‘ Figure 9. The flow chart 68 the aforementioned failure criteria, the uniaxial strength is determined as: Applied lamina load (6.1.1) U'Nax‘al Strength = Failure Index If Von — Mises criterion is chosen as the failure criterion, the acting lamina load is regarded as the uniaxial strength, when 1 - e < wmax < 1 + e, where e is a failure toler- ance. If Wm“ < 1 —e or wmax> 1 + a, increase or decrease the lamina load, and return to step 2. Thus, the uniaxial lamina strength is determined in an iteration fashion. 5. On the basis of the uniaxial strengths of the lamina, calculate the biaxial strength as follows. First, vary the longitudinal stress of the lamina from the value of tensile strength to the compressive strength. Then find the transverse stress for the fixed longitudinal stress thatsatisfies 1-e<\vmax< 1 +e,using steps 2, 3 and 4. If 1—a<\ymax< l +e,thelon- gitudinal stress and the transverse stress are considered to be on the failure envelope. 6. Using a nonlinear regression procedure and the uniaxial strengths of the lamina, the failure envelope is fit to a polynomial type failure envelope or some other mathematical form. 7. To get the in-situ strengths of each constituent, perform step 1 first. Then apply the longitudinal or transverse strengths as a load to the model, and check the maximum princi- pal stress or maximum strain for the load among each related constituent. The failure envelope and uniaxial strength of the lamina based on the in-situ strengths of each constituent are obtained by inputting the in-situ strengths as the strength of each constituent and performing steps 1 - 6 again. 69 6.2 The Numerical Results and Discussion The Verification of the Micromechanics Solution In the simulation, the composite material T300/934 system was used, and the material prOperties are listed in the Tables 1 - 3. The solution for each loading case consists of a series summation, so that it is very important to see how many terms are needed for the solution to converge. From Figure 10, it can be observed that 9 terms are not sufficient for 03 to converge. But, using 19 terms, all stress components converge with acceptable errors. Here, the convergence of the solu- tion is checked along line 2 in Figure 11 in terms of stresses, not displacements, because the stress is of greatest interest. The stresses at or near the interface also converge within about 19 terms. The point matching technique or collocation was applied to obtain some constant terms in the solution for each loading case, at some selected points along the diagonal boundary AB of the RVE. Therefore, it needs to be assessed whether or not the behavior of stress components along the diagonal boundary AB changes abruptly or becomes sin- gular. Figure 12 shows that the predicted principal stresses on the diagonal boundary and lines parallel to the diagonal boundary as shown Figure 11 are changing very smoothly as expected. The effective material properties are determined by Eqs. (2.6.1) - (2.6.6), based on the average stresses. The obtained effective material properties are compared with those pro- posed by Chamis [2]. The formulae used by Chamis are: En = EflVf-t- Emu --V,) (6.2.1) TABLE 1. Material properties of fiber used in calculation 70 T300 AS4 E, (GPa) 0 233.000 235.0000 E2 (GPa) 23.1000 14.0000 F4 (GPa) 23.1000 14.0000 nun 0.4000 0.250 nu,3 0.2000 0.2000 nun 0.2000 0.2000 023 (GPa) 8.2700 5.5000 6,3 (GPa) 8.9600 28.0000 on (GPa) ’ 8.9600 28.0000 alphal (lldeg. °C) -5.4000E-07 -3.6000E-07 alpha2 (lldeg. °C) 1.0080E-05 1.80005-05 alpha3 (lldeg. °C) 1.0080E-05 1.8000E-05 allow. axial strain (tension) 0.0118 0.0153 allow. axial strain (comp.) 0.0094 0.0145 T300, AS4: as used in [46] TABLE 2. Material properties of matrix used in calculation 71 934 3501-6 E, (GPa) 4.6500 4.3000 E, (GPa) 4.6500 4.3000 E, (GPa) 4.6500 4.3000 nu23 0.3630 0.3400 nu,3 0.3630 0.3400 nun 0.3630 0.3400 623 (GPa) 1.7000 1.6000 (3,, (GPa) I 1.7000 1.6000 6,2 (GPa) 1.7000 1.6000 alphal (lldeg. °C) 0.414E-04 040005-04 alpha2 (lldeg. °C) 0.414E—04 0.4000E-04 alpha3 (lldeg. °C) 0.414E-04 0.4000E-4 ult. tensile stress (MPa) 58.8000 83.0000 ult. comp. stress (MPa) 58.8000 207.0000 934, 3501-6: as used in [46] 72 TABLE 3. Material properties of composites measured in experiment AS4/3501-6 E, (GPa) 148.0000 145.0000 E, (GPa) 9.6500 10.6000 nun 0.3000 0.2700 0,; (GPa) 4.5500 7.6000 tens. long. strength (MPa) 1314.0000 2090.0000 comp. long. strength (MPa) 1220.0000 1440.0000 tens. trans. strength (MPa) 43.0000 64.0000 comp. trans. strength (MPa) 168.0000 228.0000 long. shear strength (Mpa) 48.0000 71.0000 fiber vol. fraction (%) 60.0 65.0 T300/934 [54] AS4/3501-6 [56] SIGMA (PA) 73 ............ : TermS‘: 9 -------- :Terms= 19 ----- : Terms: 29 : Terms: 39 4.0 5D A N G L E (DEG.) 20 30 60 7O 80 .. Figure 10. The convergence of stresses according to the number of terms 74 1.732 >X2 0'8” |L>1.0 :Linel 0.98:Line2 ~—> 0.96:1.ine3 ——->0.94:Line4 Figure 11. The stress evaluation position SIGMA (PA) 75 20 30 40 50 60 7O 80 90 A N G L E (056.) Figure 12. The Stress distribution near boundary 76 E E E'" 22' 33-1—JVIU-Em/Ef2) G :0 = "' ‘2 ‘3 l—JVfU—Gm/Gfl) Gm 1— Eu — rim/0,23) v12 = v13 = vfl2Vf+vm (l - Vf) G23 V23 = 522-: -- 1 where E, G and v are Young’s modulus, shear modulus and Poisson’s ratio, the arabic subscripts stand for the material directions, and the subscript f and m mean fiber and matrix, respectively. It is shown in Figure 13 that the prediction of longitudinal and transverse Young’s modulus, E1 and E2 , respectively, by the current model has a better agreement with experiments than that by Chamis. It is observed in Figure 14 and Figure 15 that the longi- tudinal shear modulus and Poisson’s ratio, 612 and v12 , respectively, predicted by the current model have some deviation with the experimental results. But the transverse shear modulus and Poisson’s ratio, G23 and v23 are difficult to measure reliably, so the source of the difference is difficult to identify. Results using the Micromechanical Failure Theory In Figure 16, the lamina loads leading to failure in some constituent applied in the direction varying from 0 degrees to 360 degrees are plotted by adopting the maximum principal stress criterion for the matrix. It can be observed that there exists an axes of sym- metry every 30 degrees, and the axes of symmetry for transverse normal strengths coin- cides with that for longitudinal shear strength. The absolute maximum value of transverse 77 x 1011 (a) 2 I I I I I I I I I I —— : micromechanics 15’ ---- :Chamis E1 (Pa) 0 I I I I I 4 I I I 10 15 20 25 30 35 4o 45 50 55 Fiber Volume Fraction (%) 8% o : Tsai [54] x : Chang-Kutlu [55] Figure 13. The Variations of El and E2 with fiber volume fraction 78 1(109 . (a) 8 I I I I I I I j I 1 : micromechanics 5' ---- :Chamis : micromechanics , - -— " ' fl ' ' v ’ ‘ mi‘ 1’ - - - - :Chamis " - fl — ' ” - — 623 (Pa) I I I I I I I I I O I 10 15 20 25 30 35 40 45 50 55 60 Fiber Volume Fraction (%) o : Tsai [54] x : Chang-Kutlu [55] Figure 14. The Variations of 612 and 623 with Fiber Volume Fraction 79 0.5 . a . . . . . . . . a 1 0.4 - - g 0.3 - _____________________ I_ _ _ - 0'2 ' : micromechanics ' 0.1 - - — - - :Chamis - 010 1'5 2'6 55 3.0 35 4'6 45 53 55 66 (b) 0.5 .. . a . . . . . . r 1 . 0.4 :_ ____________________________________ 1 g 0.3 - 0 - 0'2 ' :micromechanics ' 0.1' ---- :Chamis ' 0 . l L . . . . . 10 15 20 25 30 35 40 45 50 55 60 Fiber Volume Fraction (%) o : Tsai [54] x : Chang-Kutlu [55] Figure 15. The Variations of v12 and v23 with Fiber Volume Fraction SIGMA (Pa) 80 x 107 T300/934 2" _"‘3 X2T II ---- : X2C ------- : X12&X13 0" d -2. I .4 \ ’/\\ ’~\ I"\ l—.‘ ”~\ [a \\’l \\”l \\v’/ \\v’/ \\\’/ \\ll -6' n n 1 n n n J 0 50 100 150 200 250 300 350 A N G L E (DEG) Figure 16. The Variation of maximum loads causing failure in transverse direction as the applied angle changes 81 tensile and compressive lamina loads causing failure are found out at 30, 90, 150... degrees, but the maximum value of longitudinal shearing lamina load bringing about fail- ure are found at 0, 60, 120... degrees. The axes of symmetry can be expected from the assumed arrangement of fibers in the composite. The predicted unidirectional strengths are compared with the experimental results in Figures 17 - 22. To determine the unidirectional strengths of a lamina, the maximum strain criterion was adopted for fiber, the maximum stress criterion was used for interface, and one of the following failure criteria were used for matrix as discussed in Chapter 3: maxi- mum principal strain criterion, maximum principal stress criterion, Tresca criterion, and Von - Mises criterion. Hereafter, the only failure criterion for matrix will be mentioned, since the failure criteria for the other constituents are fixed. In each case, bulk matrix prop- erties used. As shown in Figures 17 and 20, the predicted transverse tensile strengths for both T300/934 and AS4/3501-6 by maximum principal stress criterion, Tresca criterion and Von - Mises criterion were in good agreements with experimental results, but by maxi- mum principal strain criterion were relatively poor. The maximum principal stress crite- rion rendered the strength closest to the experimental results for both composite systems. For AS4I3501-6, Tresca criterion is not applicable, because the tensile strength of matrix is different from the compressive strength. It was observed that all the predicted transverse tensile strengths were lower than the those of matrices. This phenomenon comes from the fact that the fibers are acting as solid inclusions and give rise to higher stress than the applied stress load, when loading is applied in the transverse direction. No matter what criterion was used, the prediction of transverse compressive strength as shown in Figures 18 and 21 was very poor, when compared with the experimental 82 “or x 2 T (T300/934) 1. Maximum principal strain criterion 2. Maximum principal stress criterion 3. Tresca criterion 4. Von - Mises criterion 5. Experimental result [54] Figure 17 . The predicted tensile transverse strengths by micromechanical failure theory ('I‘300l934) Pa 83 “or x 2 c (1'300/934) -14 '- -16 - 1. Maximum principal strain criterion 2. Maximum principal stress criterion 3. Tresca criterion 4. Von - Mises criterion 5. Experimental result [54] Figure 18. The predicted compressive transverse strengths by micromechanics failure theory ('I‘300/934) 84 “or x 1 2 (ram/934) 4.5 I 3.5 a 2.5 1.5- 1. Maximum principal strain criterion 2. Maximum principal stress‘criterion 3. Tresca criterion 4. Von - Mises criterion 5. Experimental result [54] Figure 19. The predicted longitudinal shear strengths by micromechanics failure theory ('1‘300/934) 85 “or x 2 T (AS4I3506) 0 0.5 1 1.5 2.5 3 3.5 4 4.5 1. Maximum principal strain criterion 2. Maximum principal stress criterion 3. Von - Mises criterion 4. Experimental result [56] Figure 20. The predicted tensile transverse strengths by micromechanics failure theory (AS4I350 1 -6) Pa 86 0,5108 x 2 C (AS4I3506) -0.5 - -1 .- -1.5 - -2 . '2 5 0:5 I 1.5 2 2:5 3 3.5 4 415 1. Maximum principal strain criterion 2. Maximum principal stress criterion 3. Von - Mises criterion 4. Experimental result [56] Figure 21. The predicted compressive transverse strengths by micromechanics failure theory (AS4I3501-6) 87 “07 X1 2 (AS4I3506) I I I j I I I I 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 1. Maximum principal strain criterion 2. Maximum principal stress criterion 3. Von - Mises criterion 4. Experimental result [56] Figure 22. The predicted longitudinal shear strengths by micromechanics failure theory (AS4I3501-6) 88 results. This may originate from the fact that the compressive strength of the matrix in a composite is different from that in bulk state. For T300l934, Von - Mises criterion pre- dicted better transverse compressive strength than the other criteria. For AS4l3501-6, the prediction by maximum principal suess criterion was much better than any other criteria, because the absolute compressive strength of matrix is greater than the absolute tensile strength. In Figures 19 and 22, the predicted longitudinal shear strength is depicted. The obtained longitudinal shear strengths by maximum principal strain criterion, maximum principal stress criterion and Tresca criterion are exactly same, but the shear strength by Von - Mises criterion is slightly bigger than those by the other method. When the predicted shear strength was compared with the experimental results for both cases, the error was less than about 15 percent. Since it was supposed that the longitudinal strength of a lamina is dominated by the fibers, and fibers are brittle, the ultimate longitudinal strain of the fiber was taken to be the same as that of the lamina. Therefore, the longitudinal strength predicted from the micro- mechanics failure theory is exactly the same as the experimental result. In the process of obtaining unidirectional strengths, it was found that the matrix strength can influence the axial strength of composite. When the axial load is applied, ten- sile or compressive stress is developed in the matrix due to the Poisson’s effect, and a lam- ina may fail by failure of matrix, not in the fiber. This implies that, to obtain higher strength of composite, the strengths of all constituents should be considered. Thus, this micromechanical failure criterion can be used in designing composite systems. The failure envelopes for T300l934 in stress space predicted by the micromechanical failure theory are shown in Figure 23, and the failure envelope by maximum principal N SIGMA2 (Pa) 0 x10 89 T300l934 -2 . . -4 - - —6 I I —2 -1 .5 -1 —O.5 0 0.5 1 1 .5 2 S'GMAI (Pa) X109 — : Maximum principal strain criterion - - - :Maximum principal stress criterion _._ :Trescacriterion ----- :Von-MisesCriterion Figure 23. Failure envelope in stress space (T300l934) 90 strain criterion is most conservative among the predicted failure envelopes. Considering the different scales of 0', axis and 62 axis in the plot, it is noticed that the failure enve- lopes are long and thin along the 6, -axis. The conversion of the failure envelopes in biax- ial stress state to strain space gives rise to three-dimensional strain state due to Poisson’s effect by stresses in X, and X2 directions. The failure envelopes in 3 dimensional strain space are shown in Figure 24, and are inclined to 83 axis. If they are replotted in the 8, and 82 space, the enve10pes will be the same as shown in Figure 25. For AS4/3501-6, the failure envelopes in stress space are shown in Figure 26, and the failure envelopes in the two-dimensional strain space are shown in Figure 27. Since the compressive strength of the matrix in this composite system is much larger than its tensile strength, the maximum principal stress criterion gives better results than any other criteria in this research, while the maximum principal strain failure criterion provided poor results. In Figure 28, the failure envelope for T300l934 was plotted by choosing the Von- Mises failure criterion for the matrix, then it was fit to the polynomial type failure enve- lope by using nonlinear regression and unidirectional strengths of the lamina as discussed in Chapter 3, Section 5. In this process, the interaction strength parameter, F12, was obtained. The failure envelopes obtained by the micromechanical failure theory and the nonlinear regression are compared with the Tsai-Wu failure envelope which is obtained using experimental results and the interaction strength parameter suggested by Tsai-Hahn as in Eq. (3.5.9). The interaction strength parameters predicted by the two methods are shown in Table 4 below: EPSILON3 91 T300l934 3‘ 2~ 0.01 0.01 '°'°°5 —o.005 -°-°1 —0.01 EPS'LONi EPSILON2 Figure 24. Failure envelopes in 3 dimensional strain space EPSILON2 92 T300/934 0.01 ' 0-005l‘ ~—~ ....... o b -0.005 I' -0.01 '- -0.01 —0.005 0 0.005 0.01 E P S I L O N 1 — :Maximumprincipalstraincriterion --- :Maximumprincipalsuesscriterion _.- :Treseacriterion ----- :Von-MisesCriterion Figure 25. Failure envelopes in strain space (T‘300/934) 93 x m" AS4/3501—6 2 I I I I I I I I 1.5 - . 1 " II A 0.5 I f... I ----------------------- I m r—— 0- l V I N O ' i ' < i 2 I- . , . .- (D 0.5 l ........... 1 I- ............................. (D —1 I I II I I -1 .5 I I ____________________________ u —2 I 1 -2.5 I I I I I I J I -2 -1.5 -‘I -0.5 0 0.5 1 . 1.5 2 2.5 SIGMA1 (Pa) ms — : Maximum principal strain criterion - - - : Maximum principal stress criterion --- :Von-MisesCriterion Figure 26. Failure envelopes in stress space (AS4I3501-6) 94 AS4/3501 -6 0.01 r EPSILON2 —0.02 ' -0.01 0.01 0.02 0 EPSILON1 -— :Maximumprincipalstraincriterion --- :Maximum principal stresscriterion --- :Von-MisesCriterion Figure 27. Failure envelopes in strain space (AS4I3501-6) SIGMA2 (Pa) 95 “on T300l934 2 I I I I I I I I I 1.5 - '- 1 ' I 0.5 '- - 0 . . . _O.5 I I". I I" ,l' / .1 _1 I '/' .l ' a, o’.’ -1.5 - ! ,-" - \ ,1 .\. .I". -2P \'\ .................. d -225 I I -3 I I J J I I I I I -3 —2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 SIGMA1 (Pa) X109 — :Von-MisesCriterion --- :Cuve fitting ofmicromodel prediction -'- :Tsai-Wu Figure 28. Fitting failure envelope to polynomial type failure envelope and the failure envelopes using Tsai-Wu theory and experimental results (T300l934) 96 TABLE 4. Comparison of interaction strength parameter Nonlinear regression Strength Parameter -0.235339655944E-l7 -O.46462603093E- 17 In tension - tension space the failure envelope obtained by the micromechanical failure theory is very close to the Tsai-Wu’s failure envelope obtained using experimental values. The deviation between the computed Tsai-Wu type and the experimental Tsai-Wu enve- lopes mainly comes from the differences in interaction term F,2 for each envelope and the transverse compressive strength. Figure 29 shows the failure envelopes for AS413501-6 where the maximum principal stress failure criterion was used for the matrix to get the failure envelope from the micro- mechanical failure theory. In tension - tension space, the failure envelopes from the micro- mechanical failure theory are in good agreement with the Tsai-wu’s failure envelope using experimental results. Thus, unidirectional strengths as well as strengths under multiaxial loading of the lam- ina are determined. The predicted strengths of the lamina by the model are in reasonably good agreement with experimental results. The fiber volume fraction in composites has an influence on the strengths of compos- ites, so the failure envelope varies as the fiber volume fraction changes. The variations in failure envelope due to the changes in fiber volume fraction are shown in Figures 30 and 31 for T300l934. From Figure 30, it can be observed that the longitudinal strength of a lamina is proportional to the fiber volume fraction, that is, the longitudinal strength increases as fiber volume fraction increases. But the transverse strength is not influenced as much as the longitudinal strength as shown in Figure 30. The transverse strength SIGMA2 (Pa) O I ..h 97 I I I I I I ..3 —2 —1 O 1 2 SIGMA1 (P8) x10 ”0.1 — :Maximum principal stress criterion --- :Curve fitting ofmicromodel predictions "" :Tsai-Wu Figure 29. Fitting failure envelope to polynomial type failure envelope and the failure envelopes using Tsai-Wu theory and experimental results (AS4I3501-6) 98 T300/934 x107 1.5 0.5 0 SIGMA1 (Pa) -0.5 -1.5 -6- 4 2 0 2 .4 _ 32229.6. x10“ due to the change in fiber volume fraction Figure 30. The changes in failure envelope in stress space EPSILON2 99 T300l934 00015 i I I U i 0.01 '- 0.005 '- —0.01 - .0015 - - -0.015 -0.01 -0.005 0 0.005 0.01 EPSILON1 Figure 31. The changes in failure envelope in strain space due to the change in fiber volume fraction 0.015 100 increases a little bit, and then decreases slightly as fiber volume fraction increases. The failure strain in the transverse direction decreases as fiber volume fraction increases. When a lamina is subjected to a plane stress state, the failure envelope is plotted in three-dimensional space which takes longitudinal, transverse, and longitudinal shear stresses as axes. In this case, since the strength of a lamina is finite, the failure surface should be closed. It is shown in Figure 32 that the failure surface is closed and symmetric about the plane in which the longitudinal shear stress is zero. The failure surface is obtained by using the maximum principal stress criterion for the matrix. When the shear- ing load is applied on the lamina, the influence of shearing stress on lamina strength is shown in Figure 33. The longitudinal strength is not affected up to about 20 MPa of longi- tudinal shear stress, but when the longitudinal shearing suesses exceed about 20 MPa, the longitudinal strength decreases very fast. On the other hand, the transverse strength decreases steadily as the shearing load increases. The Results of Independent Mode Failure Criteria The failure envelope of each constituent in a lamina is predicted in terms of lamina loads by using the micromechanical failure theory and allowing failure to occur in only the constituent under consideration. A typical result is shown in Figure 34, which was obtained by choosing the Von - Mises criterion for the matrix. The intersected part of each constituent is regarded as the failure envelope for the composite (T300l934). To see more clearly, the failure envelope is replotted in Figure 35, and it can be ascertained that fiber dominates the longitudinal strength, while interface and/or matrix govern the transverse strength. If Tresca failure criterion is used for matrix, only the matrix dominates the trans- verse strength as shown in Figure 36. For AS4I3SOl-6, the independent failure envelopes 101 T300l934 ‘ \ ,a. < ,,____ c3; 2‘ I /\ g" f: 0. ;::::::;, 4‘,¢——-—".____.——r <“IIIII’ 5 ‘ / - _4 M x107 SIGMA2 (Pa) '5 ‘15 SIGMA1 (Pa) Figure 32. The failure envelopes in three dimensions when shear loading is applied. x 107 T300l934 102 SIGMA2 (Pa) O N l N -1 .5 -1 -0.5 O 0.5 1 1.5 Shear load: L1: 0 MPa L2: 10 MPa L3: 20 MPa L4: 30 MPa L5: 35 MPa L6: 38.5 MPa Figure 33. The failure envelopes in two dimensions when shear loading is applied. SIGMA2 (Pa) -12 .3 103 7 T300l934 Figure 34. Independent mode failure envelope in stress space(T300/934) j- ’I I / u I I ,z’ :fiber II '* I,’ ---- :interface -—- :matrix 1 -2 -1 0 1 2 3 4 5 6 7 SIGMA1 (P8) x1010 104 x107 T300l934 8- H 6- d [ .. - .. ..-: '.".'.'_'.:.:.:..-.'.-..' 2‘: ""~ _________ 4.. ’./"’ \\ . w" \o 75 I ’ \- 9: 2” I' \ + cl. ‘ N i < O- I d 2 ,' (D .I' —-2_ 'I -. CD \ I I .\. I'l- -4r- \-\.‘ O’ I. d -6, j _ ..3. q ‘2 -1-5 -1 -0.5 0 0.5 1 1.5 2 SIGMA1 (Pa) x10. — :Ftber --- :Interfaoe --— :Matrix Figure 35. The zoom of the independent mode failure enve10pes x107 T300l934 8- . 6~ [ q ——————————— .-;,‘-_.'.".": :.T..’_T_".:.:.._-_._.__\ 4r l/ -\. - .’ is 75 -’ '\ Q; 2” .1" '\ " .’ \ N l-’ .\ < O- I' .' _, 2 .\ _/" \ Ll c_D_2- \ I-’ .. (D \ ‘l' \ .1 0“ .’ .4 - -\ -------------------------- I.’ - -6 - I . -aL . -2 -1.5 -1 -O.5 O 0.5 1 1.5 2 SIGMA1 (Pa) X10“ -— :Fibel’ "" :Interfaoe - - — :Matrix 105 Figure 36. The independent failure envelopes (Tresca criterion) 106 are shown in Figures 37 and 38 by adopting maximum principal stresses failure criterion for matrix, and it is also confirmed that the matrix dominates the transverse strength of the composite. Making use of the uniaxial strengths of each constituent in terms of lamina load, and fitting the failure envelope of each constituent to a polynomial type failure criterion, the strength parameters are obtained for the independent mode failure criteria as functions of macro loads. The obtained suength parameters of each constituent for a plane stress state are listed in Table 5. But the strength parameters for matrix can be changed by the choice of failure criterion for the matrix. If the strength parameters are substituted into Eqs. (4.3.2), (4.3.4) and (4.3.6), the failure envelope of each constituent are obtained as shown in Figures 39 and 40. In the figures, as expected, the fiber dominates the longitudinal strength of lamina, and interface and/or matrix dominates the transverse strength. The pre- diction in tension and tension space has good agreement with the Tsai - Wu failure enve- lope based on experimental results. The independent mode failure criterion based on “in-situ ” constituent strengths Thus far, the AS4I3501-6 and T300l934 composite systems were simulated to assess theory, but the prediction of transverse strengths did not have good agreement with the experimental results. This may be due to the reasons discussed in Section 5.1. Thus, to make the failure envelopes more realistic, the in-situ matrix strength can be used instead of the bulk strength. Table 6 shows the comparison between the in-situ strength and the bulk strength of 3501-6 and 934 resins. In Figure 41, the failure envelope based on in-situ strength of matrix are shown and compared with Tsai-Wu failure enve10pe using experi- mental results. In the tension - tension region, the current failure envelope based on in-situ 107 1.5 r j 1 t I l r A 0.5- - 3. (U . e. l N ... < O- -=======-_====-1=.- :============== -l 2 (D (”—05% —— fiber - —-- :lnterface ------ :matrix -1- -+ _1.5 r r r r r r r -4 -3 -2 -1 O 1 2 3 4 SIGMA1 (Pa) mm Figure 37. The independent failure envelopes in stress space (AS4I3501-6) SIGMA2 (Pa) 108 3 l I I l I 2- - 1- . L-.-..-. =. -: .-_.-..- ...-. =.1. .-.-_.-. :. : .: .—_—_". :. : ’— ."..T_-. :. 2 ZT‘T‘ “““““ O'- "-\'-t i l i -1" g. ! .............. - — —.— —.-.—.—.- -.—.—.-.--—.—.—.—.—o-o—o—-n' -2» _ -3 . . . . l - -2 -1 O 1 2 SIGMA1 (Pa) “0. — :Frber --- :Interface Figure 38. The zoom of the independent mode failure envelopes 109 TABLE 5. The strength parameters for independent failure criterion AS4/350l-6 F: 0.586385108445310 0.215978453212309 pg 1 0.623829307604318 0332273910265318 I“; 0.230417004521308 0.529573725026310 ng 0.248058914102315 0.199772075123519 qu 0.242261961673318 0.818557039821319 F26 0.910411205632315 0.568826873132323 Fin 0.201801474273310 0.0000000000005400 FY; 0.512993473221321 0.865685248847321 F2" 0.104603918702307 0.835030761877308 F; 0.201807946708315 0.845712457806316 Flt; 0.225731392658318 0.623613958904519 I72; 0.604565155100315 0.258876386798316 Fin 0.000000000me 0.210148107289309 [fl 0.318131154099318 0.493464634521519 F: 0.00000000000E+00 0.935305955771308 F’znz 0.353761427543315 0.977490324149316 Finz 0.360073205736317 0.455226648330318 ’7'; 0.9104079092589355 0.227530749253360 110 no“ T300l934 1 I U ' i U ,W ,”' ‘\ ,” \ ,, : f 0.5- ,’ - ‘j- / I / OI- I, I I A I’ E; c. —0.5- ,r - i" N ,’ < I’ 2 ' x’ (D -1- \\ ”” I (.5 _____ -‘LS- II -2. u _2.5 l l j l l l l —6 -4 -2 0 2 4 6 8 10 SIGMAt (Pa) “010 — :F‘rber --- :Interface _ . ._ :Matrix ----- :Tsai-Wu failure envelope Figure 39. The curve fitting of independent mode failure envelopes . (T300l934) 111 x 100 T300l934 2 . : fiber ' - - - — : interface 1'5 ' ------- : matrix I 1 .- .......... : Tsai-Wu (experiment) . ) .0 tn SIGMA2 (Pa .L . .5 01 d 0| 0 I N I .. .0 . .U a. o. ''''' ''''''''' in Figure 40. The zoom of curve fitting of independent mode failure envelopes (T300l934) O SIGMA1 (Pa) ‘0) x10 112 strength is very close to the Tsai - Wu failure envelope. In Figure 42, the independent mode failure envelopes are shown. In the tension - tension region, the matrix fails at a lower transverse stress as the longitudinal load increases. TABLE 6. The comparison between bulk- and in-situ strength Tension (MPa) Compression (MPa) Bulk strength In-situ strength Bulk strength In-situ strength Verification of the independent mode failure theory and in-situ strengths In order to examine the proposed micromechanical approach for the prediction of the strength of composites under combined loading, the unidirectional lamina (AS/3501) under off—axis loading as shown in Figure 43 was simulated. The unavailable properties were determined by backing out “in-situ” properties. As shown in Figure 44, the predicted strengths are in good agreement with the experimental results in both tension and com- pression loading cases. The composite strength was predicted as follows. First, the failure envelopes and unidirectional strengths were predicted by the micromechanical theory, and the interaction parameters were obtained by fitting to the Tsai-Wu type polynomial failure. Then using the tensor transformation law, the strength parameters were transformed into the off-axis coordinate. The independent mode failure curves were obtained by applying Eqs. 4.3.7 and 4.3.8 to fiber and Eq. (5.3.2) to matrix. As the loading angle varies, it is shown that the failure mode changes from the fiber to the matrix. The used material prop- erties are listed in TABLE 1. 113 x10“ T300/934 2 I f I I I 1 I I I 1.5 - —- : maximum principal stress r ._ - - - - :Tsai-Wu based on experiment 1 - ‘ 0.5 ' r , .— ‘‘‘‘‘ m -\ I g 0 p I , ’ - ’ ’ ..\ I N ./". < .415 :- /°’ I u 2 .I. I l I (2 -1 " I I." ' CD I .I. -1.5 " l I I —J 'I \ I .’ .\. ’ a " ' -2 - "~ .................. 4 _2.5 I It .3 J j l l l I J l A -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 SIGMA1 (Pa) X109 Figure 41. The failure enve10pe based on in-situ strength SIGMA2 (Pa) 1 P or 1.5 .0 or O I d -1.5 114 T300l934 x10 ' -— :flber - - - :matrix I ‘ 5 I l I ' I l I _ I I I . ------------------------------- 'l -1.5 -1 -0.5 0 0.5 1 1.5 SIGMA1 (Pa) Figure 42. The independent mode failure envelope based on in-situ strength ////AL. I ooooooooooooo 116 AS/ 3501 I I I I — :eomposite - - - :fiber ~--- : matrix 0 : experimental + c 4.0 5.0 6.0 7.0 A N G L E (DEG.) Figure 44. The predicted strengths of AS/3501 under off-axis loading on the basis of in-situ strength 80 117 TABLE 7. Material properties used in calculation AS 3501 E1 (GPa) 213.000 3.45000 32 (GPa) 13.8000 3.45000 E3 (GPa) 13.8000 3.45000 nu23 0.2500 0.3500 nun 0.2000 0.3500 nun 0.2000 0.3500 623 (GPa)*** 5.5200 1.3000 GB (GPa) 13.800 1.3000 6,; (GPa) 13.800 1.3000 Tensile Strength (Pa)* 0.01017321'" 0.7009612E+08 Compression Strength (Pa)* 0.01017321‘” -0.2792998E+09 I ASI3501 XlT (MPa) 1447.000 x1c (MPa) 1447.000 X2T (MPa) , 51.7000 xzc (MPa) 206.000 *: The in-situ strengths, “z The maximum allowable strain E ***Z The assumed property from 623 = W 23 AS, 3501: [31], ASI3501: [20] CHAPTER VII CONCLUSIONS A model for predicting failure in fiber-reinforced composites has been developed and assessed. The following conclusions have been reached: 0 On the basis of micromechanics elasticity solutions, the material properties of the lamina can be obtained with good agreement to the experimental results. 0 The predicted unidirectional strengths of the lamina based on the micromechanical failure theory are in good agreement with experimental results except the u'ansverse com- pressive strength. - To estimate the strength of a lamina for the purpose of designing composite systems, the micromechanical failure theory needing only the fiber volume fraction and the mate- rial properties in bulk state as input can be employed. 0 The interaction terms in the Tsai-Wu type failure theory can be determined by non- linear regression and fitting the failure envelope from the current (or other) micromechan- ics failure theory to a Tsai-Wu type failure criterion. 0 When the compressive strength of matrix is different from the tensile strength, the maximum principal stress criterion for matrix gives the best prediction of transverse ten- sile strength and compressive strength of composites. 118 119 ° To reflect the difference in tensile and compressive strengths of constituent materials in the micromechanical failure theory, it is better to adopt a failure criterion that can accommodate such a difference in strength. 0 For a plane stress state, it was shown that the failure surface is closed and symmetric about the plane where shear load is zero. The longitudinal strength was not affected by a small shear load. 0 It is possible to cast the failure model in a simple mathematical form with separate criteria for each mode of failure as a function of lamina load. The strength parameters can be obtained by the micromechanical failure theory. 0 The independent mode failure criterion based on “in-situ” constituent strength tech- nique shows excellent promise for inexpensive and efficient characterization of compos— ites under multi-axial loading conditions. 0 In the current work, primary attention has been given to the case of plane stress load- ing. 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