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.'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,IlI the principal directions of the stress tensor E for a given point.The corresponding matrix is 0 0 0 Ou 0 0 0 GIL 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 dWou=(+uen+mem)dv TSee failure of composite materials in Section 5.3.2. 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. sI 0 0 0 sII 0 0 0 sIII dWtotal 1 2 -- sij eij dV j Âi = Â dWtotal 1 2 -- sIeI + + sIIeII sIIIeIII = ( ) dV TX846_Frame_C14 Page 273 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
Figure 14.1 Principal Directions for the Stress Tensor EEm are the principal strains that one can express as functions of stresses using the constitutive Equation 10.1 as: e=l'z-nae(②l E This leads to: )=兰所+6+-a+a+a (Note that (dwldv)represents energy per unit volume). The total elastic deformation above is due to the dilatation and distortion of the material.The Von Mises criterion postulates that the material resists to a state of isotropic (or spherical)stress,but will plastify when the distortion energy per unit volume reaches a critical value.One notes that: $)-(供（傺） spherical stress The isotropic portion of the stress state here is written as:It creates a state of isotropic dilatation (Equation 10.1): e=++四） 总c+om+cmd then: d aw 2003 by CRC Press LLC
eI eII eIII are the principal strains that one can express as functions of stresses using the constitutive Equation 10.1 as: This leads to: (Note that (dW/dV) represents energy per unit volume). The total elastic deformation above is due to the dilatation and distortion of the material. The Von Mises criterion postulates that the material resists to a state of isotropic (or spherical) stress, but will plastify when the distortion energy per unit volume reaches a critical value. One notes that: The isotropic portion of the stress state here is written as: . It creates a state of isotropic dilatation (Equation 10.1): then: Figure 14.1 Principal Directions for the Stress Tensor e 1 + n E ------------S n E = – --trace( ) S I dW dV-------- Ë ¯ Ê ˆ total 1 2 -- 1 + n E ------------ sI 2 sII 2 sIII 2 ( ) + + n E -- sI + + sII sIII ( )2 – Ó ˛ Ì ˝ Ï ¸ = dW dV-------- Ë ¯ Ê ˆ distorsion dW dV-------- Ë ¯ Ê ˆ total dW dV-------- Ë ¯ Ê ˆ spherical stress = – sI + + sII sIII 3 -------------------------------- e 1 + n E ------------ sI + + sII sIII 3 ------------------------------- Ë ¯ Ê ˆ = n E -- sI + + sII sIII – ( ) dW dV-------- Ë ¯ Ê ˆ spherical stress 1 2 -- 3 sI + + sII sIII 3 ------------------------------- Ë ¯ Ê ˆ ¥ ¥ e Ó ˛ Ì ˝ Ï ¸ = dW dV-------- Ë ¯ Ê ˆ spherical stress 1 2 -- 1 + n E ------------ sI + + sII sIII ( )2 3 -------------------------------------- n E -- sI + + sII sIII ( )2 – Ó ˛ Ì ˝ Ï ¸ = TX846_Frame_C14 Page 274 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
One obtains then by replacing: 7 W d元 listorsion -去栏+听+-o生+ .-1+y(g+c+C E 3 then: aw 1 (14.1) d元 distorsion 3 One can rewrite as following the quantity in brackets: 影d++品-oa-ou-no} 3(c+i+m-3c0i+m+omo} d 6G(+Ou+om)2... distorsion (14.2) …-3(C10+00m+0o)} Remarks:If one denotes as n the direction making the same angle with each of the principal directions (following Figure 14.1),one observes on the face with the normal n a stress o such that:=(n) that is: Gl3 {o}= o/3 cm/3 which can be decomposed as: ■A normal stress:on=言.7 then: O= 1+Ou+Ou 3 The above consists of the average or isotropic part of the stress tensor.2 ■A shear stress: t=-on Recall the expressionthat constitutes the first scalar invariant of the stress tensor. 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: A normal stress: sn = ◊ The above consists of the average or isotropic part of the stress tensor.2 A shear stress: 2 Recall the expression sI + sII + sIII that constitutes the first scalar invariant of the stress tensor. dW dV-------- Ë ¯ Ê ˆ distorsion 1 2 -- 1 + n E ------------ sI 2 sII 2 sIII 2 ( ) + + n E -- sI + + sII sIII ( ) – 2º Ó Ì Ï = º 1 + n E ------------ sI + + sII sIII ( )2 3 -------------------------------------- n E -- sI + + sII sIII ( )2 – + ˛ ˝ ¸ then: dW dV-------- Ë ¯ Ê ˆ distorsion 1 4G ------- sI 2 sII 2 sIII 2 ( ) + + sI + + sII sIII ( )2 3 – -------------------------------------- Ó ˛ Ì ˝ Ï ¸ = 2 3 -- sI 2 sII 2 sIII 2 { } + + – sIsII – sIIsIII – sIIIsI 2 3 -- sI + + sII sIII ( )2 3 sIsII + + sIIsIII sIIIsI { } – ( ) dW dV-------- Ë ¯ Ê ˆ distorsion 1 6G ------- sI + + sII sIII ( ) = { 2º º 3 sIsII + sIIsIII + sIIIsI – ( ) } n n s s n { } s sI/ 3 sII/ 3 sIII Ó / 3˛ Ô Ô Ì ˝ Ô Ô Ï ¸ = s n then: sn sI + + sII sIII 3 = ------------------------------- t s2 sn 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, aw distorsion The shear stress t also appears as the shear characteristic of the distortion energy. 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: (o1102-t2)+(o203-t23)+(03011-ti) One then has for any coordinate system: distorsion …-3(o1102-t2)+(o203-t)+(03011-ti)} then: aw d元 =元o-a+(aa- …+(o8-0)+6(t2+3+1)} The elastic domain (where the distortion energy is below a certain critical value) can then be characterized by the condition: a{(c11-022)2+(022-03)+(03-011)2. …+6(2+t3+亏i)}<1(14.3) To determine the constant,a uniaxial test is sufficient;in effect if one denotes by o.the elastic limit obtained from a tension-compression test,one has: a2o6=1 then: a=112o2 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. 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 se the elastic limit obtained from a tension–compression test, one has: then: t 2 1 3 -- sI 2 sII 2 sIII 2 sI + + sII sIII 3 ------------------------------- Ë ¯ Ê ˆ 2 + + – Ó ˛ Ì ˝ Ï ¸ = dW dV-------- Ë ¯ Ê ˆ distorsion 1 2G ------- 3 2 --t 2 Ë ¯ Ê ˆ = s11s22 t 12 2 ( ) – s22s33 t 23 2 ( ) – s33s11 t 31 2 + + ( ) – dW dV-------- Ë ¯ Ê ˆ distorsion 1 6G ------- s11 + + s22 s33 ( ) = { 2º º 3 s11s22 t 12 2 ( ) – s22s33 t 23 2 ( ) – s33s11 t 31 2 – ( ) + + ( ) – } dW dV-------- Ë ¯ Ê ˆ distorsion 1 12G ---------- s11 – s22 ( )2 s22 – s33 ( )2 = { + º º s33 – s11 ( )2 6 t 12 2 t 23 2 t 31 2 + } + ( ) + + a s11 – s22 ( )2 s22 – s33 ( )2 s33 – s11 ( ) { + + 2º º 6 t 12 2 t 23 2 t 31 2 + } ( ) + + < 1 a 2se 2 ¥ = 1 a 1/2se 2 = TX846_Frame_C14 Page 276 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
Figure 14.2 Principal Axes for an Orthotropic Ply 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: For an orthotropic material,the principal directions for the stresses do not coincide with the orthotropic directions,unlike the isotropic case. 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. For the fiber/resin composites,the elastic limit corresponds with the rupture limit. The rupture strengths are very different when loading is applied along the I direction or along the t direction. The rupture strengths are different in tension as compared with in compression. One can then write in the orthotropic coordinates ,t,z-shown in Figure 14.2- an expression similar to Equation 14.3,as: a(G(-0)2+b(0-0:)2+c(o:-0)2+dr+eri+fs 1 (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.'The constants a,b,c,d,e,f in Equation 14.4 above will be determined using the results of the following tests: Test along the longitudinal direction ( 1 a+c=) rupture For an orthotropic material,the procedure is identical. 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: For an orthotropic material, the principal directions for the stresses do not coincide with the orthotropic directions, unlike the isotropic case. A uniaxial test is not enough to determine all the terms of the equation for the criterion because the mechanical behavior changes with the direction of loading. For the fiber/resin composites, the elastic limit corresponds with the rupture limit. The rupture strengths are very different when loading is applied along the l direction or along the t direction. The rupture strengths are different in tension as compared with in compression. One can then write in the orthotropic coordinates ,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: Test along the longitudinal direction : Figure 14.2 Principal Axes for an Orthotropic Ply 3 For an orthotropic material, the procedure is identical. a s – st ( )2 b st – sz ( )2 c sz – s ( )2 dt z 2 et tz 2 ft t 2 + + + ++ £ 1 a c + 1 s 2 ------ rupture = TX846_Frame_C14 Page 277 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
Test along the transverse direction t: 1 a+b=- Or rupture Test along the transverse direction z: due to transverse isotropy: b+c= 1 Oi rupture then: 1 a=C=- 267 rupture 1 1 b=3 O1 rupture 267 rupture ■Shear tests: →f=月 Tit rupture 1 T红→e=乏ooc 、1 ■ta→d= Tirupture due to transverse isotropy. Replacing in Equation 14.4: {o-+(o-1 …- -1o,-+1(++s1 2 rupture Or rupture rupture Tiz rupture 2003 by CRC Press LLC
Test along the transverse direction t: Test along the transverse direction z: due to transverse isotropy: then: Shear tests: due to transverse isotropy. Replacing in Equation 14.4: a b + 1 st 2 ----- rupture = b c + 1 st 2 ----- rupture = a = c 1 2s 2 ---------- rupture = b 1 st 2 ----- rupture 1 2s 2 --------- rupture = – t t Æ f 1 t t 2 ----- rupture = ttz Æ e 1 ttz 2 ----- rupture = t z Æ d 1 t t 2 ----- rupture = 1 2s 2 -------- rupture s – st ( )2 s – sz ( )2 { } + º º 1 2s rupture 2 ----------------------- 1 st rupture 2 – ------------------ Ë ¯ Ê ˆ st – sz ( )2 – 1 t t rupture 2 -------------------- t t 2 t z 2 ( ) + ttz 2 ttz rupture 2 + + -------------------- £ 1 TX846_Frame_C14 Page 278 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
and in developing': +0_ 60 1 2 rupture npure 2 (0,+0)+20 (rupture rupture ++ (14.5) 2 t21 Tetrupture izrupture 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: O/rupt- 12 -≤1 iz rupt. Lz(rupt. 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 t,t (see Figure 14.2),one has 0z=Tt红=Tz=0 Equation 14.5 is simplified,and one obtains what is called "the Hill-Tsai criterion" for a ply subject to stresses within its plane: oi o OO: <1 (14.6 2 Oi rupture Remarks: The rupture strengths of the "fiber/matrix"plies are different in tension and in compression.'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 Attention,this is not valid for a fabric that is not transversely isotropic!(see Application 18.2.10). s See values in Section 3.3.3. 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 ,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: 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 2 s rupture 2 ------------------- s t 2 s z 2 + s t rupture 2 ------------------- s s rupture 2 + – ------------------- st + sz ( ) szst 1 s rupture 2 ------------------- 2 s t rupture 2 – ------------------ Ë ¯ Ê ˆ + º º t t 2 t z 2 + t t rupture 2 -------------------- t tz 2 t tz rupture 2 + + -------------------- £ 1 s 2 s rupt. 2 ------------- st 2 strupt. 2 ------------ sz 2 sz rupt. 2 ------------- 1 s rupt. 2 ------------- 1 strupt. 2 ------------ 1 sz rupt. 2 + – ------------- Ë ¯ Ê ˆ + + – sst º º 1 strupt. 2 ------------ 1 sz rupt. 2 ------------- 1 s rupt. 2 + – ------------- Ë ¯ Ê ˆ stsz 1 sz rupt. 2 ------------- 1 s rupt. 2 ------------- 1 strupt. 2 + – ------------ Ë ¯ Ê ˆ – – szs º º t t 2 t t rupt. 2 -------------- ttz 2 ttz rupt. 2 -------------- t z 2 t z rupt. 2 +++ -------------- £ 1 sz = == t z ttz 0 s 2 s rupture 2 ------------------ st 2 st rupture 2 ------------------ sst s rupture 2 ------------------ t t 2 t 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). Safety factor:Let a<1 the Hill-Tsai expression found for a state of stress o,o,Te.One can then increase the load by means of a multiplication coefficient k to reach the limit as: (ka (ke)(o(ko)()=d=1 o 01 o rupture rupture rupture rupture The margin of safety can then be defined as the expression: (ko)-g!=k-1 O which can also be written as: safety factor 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 ox that one can apply on a ply in the direction x in Figure 14.3.The stresses 0 o,ta in the orthotropic axes are given by Equation 11.4 as: 「c2 s2 -2cs 6￥ c2 2cs 0 -cs (c2-3) 0 Figure 14.3 Direction of Loading Distinct from Orthotropic Axes 2003 by CRC Press LLC
corresponding to the type of loadings in the numerators (tension or compression). Safety factor: Let a2 < 1 the Hill–Tsai expression found for a state of stress s, st , tt. 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 sx that one can apply on a ply in the direction x in Figure 14.3. The stresses s, st , tt in the orthotropic axes are given by Equation 11.4 as: Figure 14.3 Direction of Loading Distinct from Orthotropic Axes ks ( )2 s 2 --------------- kst ( )2 s t 2 -------------- ks ( ) kst ( ) s 2 -------------------------- kt t ( )2 t t 2 + – + --------------- k2 a2 = = 1 rupture rupture rupture rupture ks ( ) – s s ------------------------- = k – 1 safety factor 1 a = --- – 1 s st Ót t˛ Ô Ô Ì ˝ Ô Ô Ï ¸ c 2 s2 2– cs s 2 c2 2cs cs cs – c 2 s 2 ( ) – sx 0 0 Ó ˛ Ô Ô Ì ˝ Ô Ô Ï ¸ = TX846_Frame_C14 Page 280 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
where one recalls that c=cose and s sine.Thus, Ot=COx 0,=s0 Tu=CSOx Replacing in the expression of the Hill-Tsai criterion of Equation 14.6,we have c22 6 rupt.rupt.rupt.rupt. then: xupure c 1 1 N't rupt. O:rupt. Remarks: If x is in tension,then oupure and orupu are the limit stresses in tension (tensile strengths).In effect,when 0=0: Ox rupture=/rupture and whenθ=90°： Ox rupture=Or rupture The evolution of the upre,when varies,was discussed in Section 3.3.2. y Figure 14.4 Pure Shear in x,y Axes 2003 by CRC Press LLC
where one recalls that c = cosq and s = sinq. Thus, Replacing in the expression of the Hill–Tsai criterion of Equation 14.6, we have then: Remarks: If sx is in tension, then s rupture and st rupture are the limit stresses in tension (tensile strengths). In effect, when q = 0: sx rupture = s rupture and when q = 90∞: sx rupture = st rupture The evolution of the sx rupture, when q varies, was discussed in Section 3.3.2. Figure 14.4 Pure Shear in x,y Axes s c 2 = sx st s 2 = sx t t = cssx sx 2 c 4 s 2 ------ s 4 s t 2 ------ c 2 s 2 s 2 --------- c 2 s 2 t t 2 + – + --------- Ó ˛ Ì ˝ Ï ¸ £ 1 rupt. rupt. rupt. rupt. sxrupture 1 c 4 s rupt. 2 ---------------- s 4 st rupt. 2 -------------- c 2 s 2 1 tt rupt. 2 --------------- 1 s rupt. 2 – --------------- Ë ¯ Ê ˆ + + = ----------------------------------------------------------------------------------------------------------- TX846_Frame_C14 Page 281 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
14.3.2 Shear Strength For a state of pure shear represented in Figure 14.4,one will have in an analogous manner: 「c2 s2 -2cs 0 c2 2cs 0 CS -cS (c2-s Ot=-2cstxy 0:2cstxy (14.7) ta =(c2-s)tsy Using this in the Hill-Tsai criterion in Equation 14.6: 4c22 4c2 4c2s2 c- ≤1 Of rupture then: 1 rupture 4c22 2 (c2-s37 Of rupture Or rupture Trupture Remarks:Here,taking into account Figure 14.4(Tx>0)and Equations 14.7, upure will be the limit stress in compression,and o,rupure the limit stress in tension for0°≤0≤90°. 2003 by CRC Press LLC
14.3.2 Shear Strength For a state of pure shear represented in Figure 14.4, one will have in an analogous manner: (14.7) Using this in the Hill–Tsai criterion in Equation 14.6: then: 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∞ £ q £ 90∞. s st Ót t˛ Ô Ô Ì ˝ Ô Ô Ï ¸ c 2 s2 –2cs s 2 c2 2cs cs cs – c 2 s 2 ( ) – 0 0 Ótxy ˛ Ô Ô Ì ˝ Ô Ô Ï ¸ = s = –2cstxy st = 2cstxy t t c 2 s 2 = ( ) – txy txy 2 4c 2 s 2 s rupture 2 ------------------- 4c 2 s 2 st rupture 2 ------------------ 4c 2 s 2 s rupture 2 ------------------- c 2 s 2 ( ) – 2 t t rupture 2 ++ + --------------------- Ó ˛ Ì ˝ Ï ¸ £ 1 txy rupture 1 4c 2 s 2 2 s rupture 2 ------------------- 1 st rupture 2 + ------------------ Ë ¯ Ê ˆ c 2 s 2 ( ) – 2 t t rupture 2 + --------------------- = ---------------------------------------------------------------------------------------------------- TX846_Frame_C14 Page 282 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
