THE UNIVERSITY OF MICHIGAN COLLEGE OF ENGINEERING Department of Engineering Mechanics Department of Mechanical Engineering Tire and Suspension Systems Research Group Technical Report No. 9 INTERNAL-STRESS ANALYSIS TECHNIQUES FOR CORD-RUBBER LAMINATES S. K. Clark Project Directors: S. K. Clark and R. A. Dodge ORA Project 02957 administered through: OFFICE OF RESEARCH ADMINISTRATION ANN ARBOR August 1961

K? IN,. 1 4

The Tire and Suspension Systems Research Group at The University of Michigan is sponsored by: FIRESTONE TIRE AND RUBBER COMPANY GENERAL TIRE AND RUBBER COMPANY B. F. GOODRICH TIRE COMPANY GOODYEAR TIRE AND RUBBER COMPANY UNITED STATES RUBBER COMPANY iii

TABLE OF CONTENTS Page LIST OF TABLES vii LIST OF FIGURES ix NOMENCLATURE xi I. FOREWORD 1 II. SUMMARY 3 III. CALCULATION OF INTERPLY STRESSES 5 A. Membrane Effects 5 B. Bending Effects 8 C. Effects of Twist 14 D. Shear Effects 18 E. Extension to Multi-Ply Structures 20 IV. CALCULATION OF CORD LOADS 41 V. EXAMPLES 49 VI. ACKNOWLEDGMENT 59 VII. REFERENCES 61 VIII. DISTRIBUTION LIST 63 V~~~~~~~6

LIST OF TABLES Table Page I. Values of the Function A(m,k) For k up to 12 32 II. Values of the Function B(m,k) For k up to 12 35 III. Values of the Function C(m,k) For k up to 12 37 IV. Summary of Interply Stress 39 vii

LIST OF FIGURES Figure Page 1. Typical element. 6 2. Bending of an orthotropic element. 9 3. Distribution of secondary shearing stresses due to pure bending. 13 4. Element acted upon by a varying bending moment. 18 5. Multi-ply numbering terminology. 21 6. Assumed and ideal normal stress distributions due to bending. 25 7. Typical laminate under load. 26 8. Resolution of bending moments. 42 9. Distribution of cord loads in bending. 44 10. Assumed and ideal induced secondary-stress distributions. 46 11. Shearing-stress intensity in a six-ply structure due to bending. 47 12. Flat belt on pulley. 50 13. Assumed geometry of deformation of a pneumatic tire. 53 14. Interply shear-stress distribution in two- and four-ply structures. 56 ix

NOMENCLATURE English Letters: aij Constants associated with generalized Hooke's law. A(m,k),B(m,k),C(m,k) Symbols used to represent lengthy algebraic expressions. D A plate stiffness. d A dimension from the mid-plane of the structure to the cord centerline. ES,EJ, F Elastic constants. EF, E Elastic constants. Gel Shear modulus. h Ply thickness. k An integer equal to one-half the number of plies in a laminate. 62w k a'w etc. m Bending moment per unit length. m (As a subscript) indicates location of interply stress. M Bending moment. N A cord-load influence coefficient. n Cord count. P Cord load. PB Cord load induced by bending effects. Q Shearing force per unit length. w Deflection in ~ direction. xyz Coordinates, x and y in the plane of the sheet and z normal to it. xi

NOMENCLATURE (Concluded) Greek Letters: of One-half the included angle between cords in adjoining plies in a two-ply laminate. E Strain. t,~,~ Orthogonal co-ordinates aligned along and normal to the principal axes of elasticity, or orthotropic axes, in an orthotropic laminate. These directions are the bisectors of the cord angles. a Stress.,7' Interply stress based on unit ply thickness. An influence coefficient. C " Interply stress. p Radius of curvature. xii

I. FOREWORD The Tire and Suspensions Research Group at The University of Michigan has spent some time in devising methods for calculating stresses in pneumatic tires under certain simple loading conditions. These stresses are useful only if they can be translated into interply stresses in the carcass of the tire and cord loads in the textile or wire fibers. Expressions are thus necessary to determine these interply stresses and cord loads due to all effects present in a real tire. This report presents techniques developed for this purpose. In some cases these techniques are exact; in others they appear to be acceptably close approximations. This discussion is limited to structures which are plane-orthotropic and of bias-ply construction.

II. SUMMARY There are three basic causes for interply stresses in a pneumatic tire carcass. The interply stress necessary to maintain equilibrium in each ply is due to the presence of membrane forces. This stress, discussed extensively in previous technical reports, is well understood and may be determined easily by reference to graphical data presented in Refs. 1 and 2. The distribution of this stress through the thickness of a ply, discussed in Ref. 3, is linear. A completely different type of interply stress, discussed in this report,may arise due to the bending moments which a laminated structure must sustain. The magnitude of this stress can be readily calculated if the bending moments imposed on the structure are known. In general, this stress has its maximum value arising at the faces or boundaries of the outer plies of a multi-ply structure. For example, in a ten-ply laminate, this effect would be most severe between the first and second and between the ninth and tenth plies. A third type of interply stress is that usually associated with shear in simple beam problems. The basic difference between this stress and the other two is that it is not internally self-equilibrating. This interply stress is, and must be, an active stress necessary for the maintenance of equilibrium due to the gradient, or derivative, of bending moment with respect to position. This interply stress is largest at the mid-plane of the 3

laminate, where its value is exactly that determined for any simple isotropic structure. This stress might be expected to be severe in regions of high local bending stresses, such as occur in a tire while traversing a small bump or such as occur in localized regions near the edges of the contact patch in a normal pneumatic tire. The analysis of interply stresses presented here shows that in general a group of thinner plies bonded into a laminate exhibit lower interply stresses than do a smaller number of thicker plies having the same over-all section thickness. An exception is a two-ply structure, which has no interply stress due to bending. These results have direct application to structures where two, four, or six plies may be used, such as in a pneumatic tire. Expressions previously derived in Ref. 4 are reviewed and clarified so that cord loads due to membrane forces in multi-ply laminates may be readily obtained. A different type of cord load may be generated by a bending moment imposed upon a laminate made up of several plies. An approximate expression is developed for the magnitude of the cord load induced due to this effect. Finally, a secondary cord load can be computed due to the anisotropy of the bent sheet. Having the contribution to cord load from all these phenomena, it is possible to calculate them separately and add them together algebraically to determine the total cord load in a ply. 4~

III. CALCULATION OF INTERPLY STRESSES A. MEMBRANE EFFECTS The problem of determining interply stresses due to membrane forces in a multi-ply laminate was discussed in Refs. 2 and 3, where graphs of c'/a A, Ci/cTx, -al3/a33 and -a23/a33 were presented as functions of the cord half angle a. These values are essentially influence coefficients for interply stresses based on a unit thickness of the ply, as explained in Ref. 3. For completeness, Eqs. (1) are presented here giving these expressions: Cx a12a13-a1 a23 cx V alla22-al2 J ac = f(alal a-a3lla223+ a13 /~ 2 /12 al1 ajja22-a12 all_ a a = -j23;'s L a33 a33 where all aij are evaluated at +o. The aij combinations appearing in Eqs. (1) are presented graphically in Figs. 2-5 of Ref. 3, as well as Fig. 4-7 of Ref. 2. The role of interply stresses may be clarified to some extent by considering a simple two-ply laminate such as shown in Fig. 1. The plane 0 represents the boundary between the two plies and is presumed to be the midplane. Considering an element made up of unit area but of thickness dz lying at the top face of this section, one may form a relation between the interply stresses necessary to hold this thin layer in equilibrium and the stresses 5

h Fig. 1. Typical element. previously denoted as the interply stresses associated with unit ply thickness, in the fonn of Eqs. (2). dad = AC dz da = Ca dz (2) T! T1 dc"r = cax dz Since the form of these three equations is identical, it will be sufficient to deal with the first one. Integration through some length will give the interply stress at any location z from the mid-plane. This equation takes the form 6

h oal(z) = o Idz = o [h-z] (3) where a' is given in Eq. (1). Similar relations hold for &T and anr. This equation shows that a linear distribution of interply stress is present through the thickness of each individual ply. The maximum value of the interply stress occurs at the interply bond when the dimension z is 0. At this location, r(o0) = A' h (4) Similarly, Cr i (0) = h T (0) = & ~ h This means that the distribution of interply stress through the thickness of a ply is that given in Ref. 3 as Fig. 7. This entire argument is based on the orthotropic elastic nature of the individual plies which go to make up a multi-ply laminate and, as such, implicitly assumes that in the structure involved here the orthotropy is uniformly distributed throughout the thickness and width of the ply. Thus, it may be imagined that the structure being analyzed is made up of a rubber sheet into which textile filaments are uniformly dispersed but lie, of course, in the direction that the real cords actually lie. Under these assumptions, no stress concentration factor exists in the interply stresses due the proximity of one cord with another in adjacent plies, so that the results just given appear to be valid for membrane forces up to the inclusion of that 7

effect. Since the forces just discussed are membrane forces and as such are presumed to be uniformly distributed throughout the thickness of the specimen, the interply stresses determined on the basis of these expressions will be valid for any laminate having an even number of plies. The discussion in Ref. 3 concerning the interply stresses in alternate plies is still valid. B. BENDING EFFECTS To find interply stresses due to bending, consider a two-ply laminate in bending under conditions in which both plies are in such a stress state that the total strain in their cords is of the same sign, either tension or compression. This has previously been defined as a Type 1 laminate. Referring to Fig. 2, let the bending be about the orthotropic axis of the two-ply laminate, and assume that twist vanishes. From Ref. 1, Hooke's Law relating stress and strain for any material is C allo + al2Cr + al3as, 6, = a21r + a22Ca + a23a (5),E;r = a3la, + a32_rl + a33::p, These stress-strain relations for each ply are general ones. However, due to the assumption of the absence of twist, the shear strain Aid must vanish. Hence, Eqs. (9) of Ref. 1 become, when at and a are applied, CE = a11ao + al2a1 + a13alo ~n = a21.) + a22ca + a23oDT (6) o = a3lc + a32~, + a33oa 8

AX Ply I _I N " _ h NA Ply 2 -h p p Fig. 2. Bending of an orthotropic element.

Equations (6) are applicable to either ply 1 or ply 2 of Fig. 2, using the proper angles oa. Notice that ax' is a shearing stress generated by the ac and the a stresses. a' acts oppositely on each ply. It does not contribute directly to the bending moment about the neutral axis, since ax' can be written in terms of ca and a, to give = a13 a 23 q - a33 a 33 from which, by substitution into the first two of Eqs. (6), one obtains = (all - -) c + a2 - a13a32)C = + a33 12 a33 ES F~ (7) E -— a21 a23a31- +' + 22- a23 - C a a33 a33 F5 E7 as shown in Eqs. (15) of Ref. 1. Now assume that Eqs. (7) are inverted to give d r= Ese + Fe (8) Ti- = Ca = ECrl + Fee where E F2 (Fn - E5Ei) -E2 = T (9) (F - E Er ) F,= -F - EW2i

For pure bending, with the usual assumption of planes remaining planes and small slopes, a2w ce = _ z-2 = - zkg = - = - zkz = - zkn (10) 62w d = - 2zk = - 2z where the ke, kp, and keg are changes in curvature and twist in the i, q directions, respectively. For displacements that are small compared with the dimensions of the shell, these curvature changes may be approximated by the second derivatives of displacement w with respect to position. Using these, the stresses become _ = - z(Ek5 + )Pk) (11) = - z(kn + k) From this, it may be seen at once that both ae and a are linear functions of the normal distance z from the neutral or mid-plane under conditions of orthotropic bending. This fact is sufficient to determine the distribution of shearing stresses necessary due to bending alone. The response of a two-ply laminate to pure bending deformation may be obtained most easily by imagining that the normal stresses given by Eqs. (11) are first applied to the structure alone. The resulting shearing distortions are, from Eqs. (5): 11

Ply 1, Cord Angle + a: z >0 ent1 = -a13(+)l[E kg + Pk]z - a23(e+s) [EkTI + Fk]z Ply 2, Cord Angle - 1' (12) z<O cSln = -al3(-)I[Egkg + k ]z - a23(-)) [Enkn + Fk]z = +al3(+) [E k5 + Fk ]z + a23(+)) [Enkn + Fk ]z The shear distortion of Ply 1 is identical to that of Ply 2 since Eqs. (12) are identical when it is noticed that the upper one applies only for z > 0, the lower one only for z < O. Since these distortions are the same, no shearing components of interply stress are generated by bending in this two-ply structure. But it is necessary to provide external shearing stresses from the boundaries to eliminate the shearing distortion (called'twist' in a plate or shell-like structure) described by Eqs. (12). The magnitude and distribution of these shearing stresses may be obtained by applying Eqs, (2) directly to a two-ply laminate in bending. First, imagine the existence of linearly varying bending stresses ad and Ad as just discussed. Substituting these into the last of Eqs. (1), it is seen that one may write _ al13(+sa) (Ek + + a2F(+c) (Ekn + Fk z (13a) 33(+ a) a33(+c) This represents the intensity of shearing stress necessary at any position 12

It should be noted that for z < O, the angle of the lower ply changes to -_, the functions al3 and a23 change signs, and the required shearing stress becomes aglt = F-3(++a) (a3ka + +kr) + a23(+) + ort L[33(+O) k + a3( -+) (..Z (13b) z< O which is of exactly the same sign and magnitude as given by Eq. (13a) for z > O. The required external shear-stress intensity is illustrated in Fig. 3, for a two-ply structure, and it is seen that shear-stress intensities of the Fig. 3. Distribution of secondary shearing stresses due to pure bending. same sign are needed on both plies, in a fashion varying linearly from zero at the midline to a maximum at the outer surface. Physically, this type of structure, which is orthotropic with respect to bisecting axes under membrane loads, is not orthotropic with respect to the

same bisecting axes under the action of pure bending moments. Bending without twist can be induced only by the action of bending moments in conjunction with shearing stresses, or their equivalent twisting moments, The magnitude of the shearing stresses necessary for the bending without twist of a two-ply structure, along axes bisecting the cord angles, as given by Eqs. (13) and Fig. 3, will obviously influence the cord loads and will be considered later in this report. C. EFFECTS OF TWIST The other possible mode of deformation is twisting. This type of displacement of a sheet is described in Section 10 of Timoshenko and WoinowskyKrieger,5 and the reader may find it helpful to refer to their book for a somewhat more complete description than can be given here. During twisting, normal strains vanish while shearing strains eG do not. Hence, adapting Eqs. (5) to this case gives 0 = a, alc + a12a& + a13a 0 = a21l' + a22a' + a23o (14) CT e = a31al + a32ca + a33a where, from the nature of the geometry of twist, it is recognized that shearing stresses must be provided to induce twist, while any normal stresses will have their origin in either some interply effect or in an external source. These will be denoted by Ad and an. Equations (14) may be simplified by considering 05T as a known quantity which must be specified, thus allowing the first two of Eqs. (14) to be 14

solved for ac and a' in terms of ca, giving 1-al3 a12 a23 =a22 = _2a23_-a3a22C all a12 a11a22-a12 / a21 a22 (15) all -a13 aa21 -a23f a21a13-a11a23>. T all a12 -a2 Substituting these into the last of Eqs. (14) gives ET = 83a1 2a23-al3a22 + a22 3-alla23 a11a22-a12 alla22-a12 (16) + a33j| CY C From the geometry of deformation of a plate as discussed in Ref. 5, the twisting strain may be related to deflection w normal to the surface of the plate by the relation ti= -2z = a a 2zk2- (17) Hence 0ceT = -(2G knT) z (18) 15

the shearing stresses of Eq. (18), to a two-ply structure, in a fashion analogous to Eqs. (12). This gives normal extensions in the two plies as represented by Eqs. (19). Ply 1, Cord Angle + a: z >0 e = -a13(+C). z. (2Gekk) = -a23(+a)z(2G kq) (19) Ply 2, Cord Angle - a: z<0 CE = -a13(-C). z.(2G kin) = +a13(+C).z.(2GSnkeT) EC = -a23(-a)- z.(2G0 kS) = +a23(+a)- z-(2G0Tk~T) Once again the extensions are identical in both plies. Thus, no normal components of interply stress are generated due to twisting of a two-ply laminate. Since it was previously shown that no shearing components of interply stress were generated due to bending, it may finally be concluded that neither bending nor twisting induces interply stresses in a two-ply laminate where both plies have the same elastic properties. Notice that the distributions through the thickness of the laminate of both the normal stresses in bending, as given by Eqs. (11), and the shearing stresses due to twist, as given by Eq. (18), are identical in form. Hence, the developments leading up to Eqs. (12) and (19) are the same both in logic and in the algebra. This fact will be drawn upon later during the discussion 16

of interply stresses in multi-ply structures, at which time one set of interply components may be found with the knowledge that the others are identical in form. Returning to Eq. (18) the intensities of the normal components of stress, generated by these shearing stresses necessary for twisting, are given by C = (a2a23-al3a22 (2Gkr) o z a11a22-a12 (20) a, - 21al3-ajl1a23 (2G0k)z alla22a 12 z 0 Noting that for the lower ply the cord angle becomes -a, inspection shows that the quantities involving aij in parentheses in Eqs. (20) change sign so that for the lower ply the sign of the required normal stresses is exactly the same as for the upper ply. Hence, the distribution of these normal stresses through the thickness of the laminate is identical in form to that shown in Fig. 3. Thus, they are of exactly the same fundamental nature as those stresses associated with pure bending, and, as such, do not act as interply stresses but do influence cord loads, Their origin must also be at the boundaries of the material. It may finally be concluded from this examination of bending and twist that, for a two-ply structure, the following conditions hold: (a) Neither bending nor twist induce interply stresses, although cord loads are induced, 17

(b) Pure bending deformation cannot be obtained from the application of bending moments alone, since additional external shearing stresses are needed. Similarly pure twist cannot be induced by the application of shearing stresses alone, since additional normal stresses are needed. (c) The origin of these additional stresses lies in the boundary forces. D. SHEAR EFFECTS The second type of interply stress arises not due to the moment alone, but due to the derivative of the moment with respect to position, or shearing force. This is most commonly encountered in the study of beams, plates, or shells subjected to lateral loads. As Fig. 4 illustrates, if the magnitude of the bending moment changes from point to point on the beam, shearing forces are necessary for equilibrium of elements in the beam, such as ABCD. A B D M2 Fig. 4. Element acted upon by a varying bending moment. 18

Since, from force equilibrium conditions, the change in bending moment is equivalent to the shearing force, it is usually most convenient to associate this type of shear stress with a shearing force carried by the structure. From Fig. 4, it may be seen that the existence of the shear stresses necessary for equilibrium is completely independent of the type of structure involved in the beam. As a matter of fact, it is also completely independent of the elastic constants which enter into the problem. For that reason, such a shear stress is a necessity from the point of view of static equilibrium, and therefore may be calculated on just exactly the same basis as in an isotropic beam. Under those conditions, it is shown in such standard works as Timoshenko and Woinowsky-Krieger5 that the shearing stresses generated in this manner take the form a = Q h (1 - ) (21a) where is the shearing force per unit length measured in the Q direction, normal to ~. A similar expression could be written concerning the interply stresses at right angles to this, namely, = Q 1 - I) (21b) 71 1Q 2h where now Qr is the shearing force per unit length measured in the ~ direction. Once again, it is seen that the maximum value of these interply stresses occurs at the mid-line or neutral plane of the laminate. Throughout the rest of the laminate, the distribution is parabolic. 19

The first type of interply stress discussed here, that type due to membrane forces, is self-equilibrating in the sense that equal but opposite interply stresses are set up on each of the two plies and that the role of the interply stress in this case is that of providing forces sufficient to maintain strain compatibility on the structure, from the physical point of view. The second type of interply stress, set up due to shearing forces acting on the structure, is statically necessary, not from the point of view of strain compatibility but from that of the actual equilibrium of forces. Looked at in this way, this latter interply stress is of a somewhat different nature from the one previously determined. However, the two may be combined algebraically in their proper vector directions and the entire sum of them taken by the technique of using Mohr's circle for these interply stresses, as previously suggested in Ref. 1. E. EXTENSION TO MULTI-PLY STRUCTURES So far this discussion has used a two-ply structure to illustrate the existence of interply stresses and the techniques for calculating such stresses. It is of considerable interest to generalize each of the types of interply stress to its role in a general multi-ply structure. Figure 5 illustrates the assumed geometry of an orthotropic laminate made up of 2k identical plies, each laid up with alternate angle +t and -a, the thickness of each ply being the same and equal to h. As discussed in Ref. 3, membrane forces generate interply stresses between pairs of plies. For example, in the construction of Fig. 5, membrane20

+(K-l) K + a + (K-0) a + (K-2) 4 \~ ~ ~~~~~(-)______- h (typical)') +3 -— a + + 2 t a +1 0+ a I,~~~~~~~~~~~~~~~~~~ -(K-2)(-l a + -a IN'TERPLY STRESS PLY DESIGNATION CORD ANGLE LOCATI ON Fig. ~,Multi-ply numbering terminology.

induced interply stresses are present at the following locations, using the notation of Fig. 5: ~(k-l), +(k-3), +(k-), ----- +1 if k is even ~(k-l), ~(k-3), ~+(k-5), ----- 0 if k is odd No membrane-induced interply stresses exist at the following locations: ~(k-2), ~(k-4), ~(k-6), --— 0 O if k is even +(k-2), ++(k-4), +~(k-6), - ~ — 1 if k is odd The value of the interply stress induced by membrane stresses is also given by Eq. (4), and depends on the membrane-stress magnitude and the individual ply thickness. h Thus, the interply stresses in a multi-ply laminate due to membrane forces are exactly the same as in a two-ply laminate provided that the stress level remains the same and that the ply thickness is unchanged. The interply stresses due to shear, given by Eqs. (21) for the case of a two-ply structure, may be adapted directly to multi-ply laminates by noting that these equations use h as a dimension representing half the thickness of the material. For multi-ply structures having 2k plies, the half thickness is kh. This value should be substituted for h in these equations so that they become Q2kh - (kh (22) 1 Q 2.kh h j 22

where the rest of the notation is the same as in Eqs. (21). Thus, Eqs. (22) represent a generalization of Eqs. (21) and the special case of two plies may be obtained by letting k = 1. The maximum value of this quantity occurs at the mid-line of Fig. 5, where z = O, at which time 3Q~ ~max 2kh (23) 3Q 1max 2kh This stress becomes smaller as the number of plies 2k is increased, provided that the shear forces and ply thickness remain the same. Note that these shear stresses depend on the total section thickness 2kh and are independent of the number of plies and their individual thickness h. It can be shown that bending and twist induce interply stresses in multi-ply laminates. Since this is a rather lengthy computation it is desirable to trace through one or two cases carefully before giving the results. Consider first a structure of 2k plies, acted upon by certain bending moments which can be represented by linear distributions of normal stress such as given by Eqs. (11). It is recognized that twist, or shearing distortion, of the entire structure will result. The interply stresses, however, can be obtained only by letting the proper value of normal stress, as given by Eqso (11), act on each ply separately and by requiring the interply stresses to insure that each ply undergoes some common shear distortion, not necessarily zero. If the common shear distortion is zero, then the structure as a whole is orthotropic and no additional or secondary shearing stresses are required 23

from the boundaries. On the other hand, if the common shear distortion is not zero, then the structure as a whole is not orthotropic and pure bending without twist will require additional shearing stresses which must come from boundary forces. These additional shearing stresses act on the structure as a whole and neither contribute to nor modify the interply stresses. An identical argument could be made for twisting of a structure, since the shearing stresses causing this twist are linearly distributed through the thickness of the laminate, as shown in Eq. (18). Since the two arguments are the same, the relationships for pure bending will be worked out, and from this the corresponding relations for twist may be inferred. In discussing the distribution of normal stress which represents a bending moment, as given by Eq. (11), it should be recalled that this exact linear distribution is based on the assumption of a uniform distribution of anisotropy throughout the thickness of each ply. This could be achieved practically only by dismenbering the cords into their individual filaments and dispersing them throughout the ply, still keeping their direction the same. Since the cords actually represent concentrated anisotropy, it is difficult to determine the exact distribution of normal stresses, but it seems certain that an acceptably close approximation can be obtained by assuming that an average, effective stress acts on each ply, this stress producing a total force in the ply equal to that produced by the actual distribution. Figure 6 shows these ideal and approximate distributions of normal stresses due to bending, where a six-ply structure is used as an example. Notice that the approximate distribution may be obtained by evaluating the 24

-_______Assumed distribution for real I ealized distribution',Eq_ l )] L _ h._ __ _ __ _ __ _ __ _ _ L(typical) Fig. 6. Assumed and ideal normal stress distributions due to bending. idealized distribution at the mid-points of each of the plies. Next, we consider the state of strain of an element of unit cross-sectional area but of thickness h, equal to the ply thickness, taken from one ply. This is shown in Fig. 7 as being acted upon by normal stresses A and aT arising from bending moments. The total shearing forces acting on this element, which is of unit area, are (ca I)(h)(l). If conditions are such that this shearing force must come from the face of this element, as in calculating the interply stresses, the intensity of these interply stresses must be such that the shearing forces they produce are equal. Thus, denoting interply stresses by a", Il'lt = Ca.h (24) 25

Fig. 7. Typical laminate under load. Using Eqs. (11) and (5), along with Fig. 6, equations may be written expressing compatibility of distortion for various multi-ply structures, assuming in each case the existence of an interply stress acting equally but oppositely on each neighboring ply. The value of the interply stress is determined from the resulting equations. 1. Two Plies A. Ply + 1, Cord Angle + a: = -a13(*x). (Eh k5 + Pkl) -a23(+x), (Enk + Fk) + a33(+c)aj, 26

B. Ply - 1, Cord Angle - o: Eg = -al3(-O)(-2)(E k~ + ~ h T -a23(C)(h k + k) () (Ek + ) - a33(-cx) r = -al3(+ X) 2( Ek + Fk) -a23( k)[(E~k1 + F[) - 33( )dil From this, axt = 0, hence (25) = 0 2. Four Plies A. Ply + 2, Cord Angle + a: ogl = -a13( +x) -h(Ek + Fk) -a23(+).2h(Ek + Tk) + a33(+A) (-)* B. Ply + 1, Cord Angle - a: cea = -al3(-c).^h(Egk +T ka) -a23(-_a) _(ELk + Fkg) + a33(-o )(ai - ao) C. Ply - 1, Cord Angle + ca: egg = -al3( )(-)() Ek + Ad) -a23(+a) (-) (Ek + kg) + a33(+x)(Co -' j) *For simplicity of notation the subscript En will be dropped temporarily and a numerical subscript given which indicates the location of the interply stress. For example, a[ indicates the interply stress acting at the top of first upper ply, c! the mid-plane interply stress, ac' the interply stress at the bottom of the first lower ply, etc. See Fig. 5. 27

D. Ply - 2, Cord Angle - o: Ec1 = -a13(-c) (-I-h) ( k + Tk) -a23(-Cx) (I2') (k + ykZ) + a33(-a) (muI) Solution of these equations gives a' = 0 Cr{ = = h a13 (Ek + kQa &3(++) (Ek + a() from which = - -1 h [33(+)3333(+o k(26) This type of analysis can also be carried out for twisting of a laminate, in which a shear-stress distribution such as given by Eq. (18) acts on the structure. For example, consider a four-ply laminate under these conditions. In general, normal components of interply stress exist in both the e and n directions because compatibility of strain must be maintained in both of those directions. This case is analyzed below. 3. Four Plies (shear stress) A. Ply + 2, Cord Angle + cf: cE = al((+) [-c ] + a2(-I) [-a;] +a1a(+3) (23) (-2G nk l) E = a2l(C) [-a ] + a22(+CX) [-c1 ] +a23() (32) (-2 G nk8 ) 28

B. Ply + 1, Cord Angle -' =E - all(-aO) [acr - aCo] + al2(-c) [T- ao- ] +al(3(-a) (2) (-2G~ qkg,) cn a21(-O) [19' - ci ] + a22(-Q) [a] - ao ] +a23( -C) (-) ( -2q rlk~ T) C. Ply - 1, Cord Angle +.o: E = all(+cX) [crt - a1 ] + a,2(4+ ) [r -' ] +al3(-P+) (-h) (-2GETlk6,r) C:= a2o1( )O[a - aT11] + a22) [CoT - +a23( +) (-7) ( -2h G- k_ ) D. Ply - 2, Cord Angle - a: = all(-Oa) [a:r ] + al2(-( ) [C-'T +a 3( _:) ( _3h) (-2G -lk:-) = a2l(-O) [a'1, ] + a22(-CO) [c'1j, ] +a23('0) (-,3h) ( -2GE!,ke]) Solution of these equations gives - =alh2G,lk ~l2( (+X) a23( c) -a.3( 4) a22( +a) ~ 1G5k all( 4+) a22( + ) -a~2( 2:) a't =ah t-2 kgTIll(z(x) a-3(+::) -al2(+a) al3 (a) al =1 h1 a, )as(+)-a2 ) 29

Similar equations may be written and solved for structures made from six, eight, and on up to 2k plies, both for bending and for twist. Since the work necessary to do this becomes quite lengthy, only the results will be given here. Before presenting these it might be noted that solution of a large number of sets of equations similar to those just given indicates that the resulting interply-stress components take the following forms. (I) Due to bending deformation kg and ki, a shearing component of interply stress exists: (ag A(mk)h 7a(2 a) (k + a- 23(+Q) (k —' —-7 —, — Laa33(...)) (E kk + r. (27) where m indicates the location as shown by Fig. 5, and k is one-half the total number of plies. (II) Due to shearing deformation (twist) kg, normal components of interply stress exist: (a)m = A(m,k)h2 -2G rlkr [ 12() a2( ) -a13( C) a22( —) (28),"m =A(m,k)h2 -2Gl kgl t(-)-'(Qa(] all(+ ) a2a(+U) -"a'2(+) -] (III) The interply stress components occurring at places equally distant or either side of the mid-line are equal but opposite in sign. Thus, (,' ) m = -('a:,) -m (a:) m -(a)-m 3o

or, in brief, A(m,k) = -A(-m,k) (29) (IV) No interply stress acts at the mid-plane, or zero, location. It is convenient to give values of the parameter A(m,k) as a function of m and k. This is done in Table I which, when used in conjunction with Eqs. (24) and (25) may be used to determine interply stress components in any laminate up to 24 plies when curvature changes kg, k1, and kh are known. From the practical point of view, study of Table I indicates that the severity of bending or twist-induced interply stresses increases as the number of plies increases, for the same curvature changes. This means that when tires are actually forced to conform to a certain geometry, such as in rolling over a sharp bump or stone, the tire with the fewest number of plies has the advantage of generating the lowest interply stress. It is often desirable to determine interply stresses as functions of bending and twisting moments rather than as functions of curvature changes. Reference 6 shows that moments and curvatures are related through the expressions me = -DekS - Dk mr = -D k1 - DkS where D = E2 (kh)3 2 D = F (kh) 3 31

TABLE I VALUES OF THE FUNCTION A(m,k) FOR k UP TO 12 No. of m Plies 0 1 2 3 4 5 6 7 8 9 10 11 (2k) k _ _ _.... 2 1 0 4 2 0 1 6 3 0 0 2 8 4 0 1 0 3 10 5 0 0 2 0 4 12 6 0 1 0 3 0 5 14 7 0 0 2 O 4 0 6 16 8 0 1 0 13 0 5 0 7 18 9 0 0 2 0 4 0 6 0 8 20 10 0 1 0 3 0 5 0 7 0 9 22 11 0 0 2' 0O 4 0 6 0 8 0 10 24 12 0 1 0 3 0 5 0 7 0 9 0 11 32

and m~ n G~ Gakes ~4 (kh) 3 By inverting these, one obtains ( D DD-2) (Dm. - Dm) k (_ 2 (33) and for twist m~ kSn 4 - G[ (kh)3 from Eqs. (30) and (32). From Eqs. (31) and (33), - TI-E-mt (kh) 3 ( E 2)J k,= (h3) (34) (kh)3 kqes d (kh) 3 G Equations (34) may be substituted directly into Eqs.' (27) and (28)~ This gives 33

1(+) 8 a3(+ S) Fm3(E) it |FmE j a33(+Q) a23(+Qe) a2 3 a) = B(m,k) h 12 111 m as33( -+C) E +a( ) (ss+C) =B Bmm) -Emm (a3(m) B h2311(+))- a2 (+)Fm - aE l La3s(+a as3( +U) (Cr B~mk_)1 3 a12 ) a23(-k:) -al3(+(::z) a22(4a) B(m,k) = k m k Table II gives B(m,k) as a function of k and m up to k = 12, or 24 plies. Again, the relation B(m,k) = -B(-mk) holds. Finally, the role of ply thickness in interply stresses may be examined by fixing the over-all secticn thickness 2kh as a constant, but varying the number of plies 2k. If this is done, Eqs. (24) and (25) become fti) = -C(m,k) h) (E ke + Fk 23(+) (E k + Fk 0 0Las3(+U) a 3a( +) a 1 ( +c) a23( +CX) -a.12( + c) (226 () = C(m,k)h2 -2Gk a2(+)a23(+) -a123(+ )a(+) 36) =2, -! 1 + a 23(4+O) -al2(+a) a13(42_ T] al(4 +a) a22( +cU) -a 2(+a) where A(m, k) ks ~l-m~~~~~~~~~~~; ~ ho~~~3

TABLE II VALUES OF THE FUNCTION B(m,k) FOR k UP TO 12 No,.of m Plies 0 1 2 3 4 5 6 7 8 9 10 11 (2k) k 2 1 0 4 2 _ _ 1/8 6 3 o o 2/27 8 4 0 1/64 o 3/64 10 5 0 0 2/125 o0 _4/12 12 6 1 1/216 0 0 13/216 0 5/216 ___ 14 7 0 O 2/343 0 4/343 0 6/343 16 8 7 I1/512 T /512 0 7/512 18 99 0 2/729 0 4/729 0 /729 2 0 0 1 0 8/729 20 o10 o 01/1000 o0 3/oo1000 0 /looo o 7/1000 o 9/1ooo 22 11 0 0 2/1331 0 4/1331 0 6/1331 0 8/1331 0 10/1331 24 12 0 /1728o 0o 3/1728 O I 5/1728 0 {7/1728 0 7 9/1728 0 11/1728

since it is presumed that kh = const. = ho C(m,k) is given in Table III. C(m,k) may be assumed to be an efficiency factor in so far as interply stresses are concerned. With the exception of the two ply structure, interply stresses may generally be lowered by increasing the number of plies while keeping the section thickness the same. If a consistent sign convention is used, such as in Ref. 5, membraneinduced and bending-induced interply stress components may be combined algebraically. In general, bending conditions causing tension in a ply in the g or q directions will produce interply stresses on the interfaces of that ply which are of the same sign as membrane-induced interply stresses caused by tensile loads. This provides a simple, ready check on the proper signs of these two types of interply stresses. The interply stress caused by external shear should always be added to the interply stresses caused by membrane and bending phenomena, without regard to sign. Finally, all the various components of interply stress may be combined by use of the Mohr's circle technique advocated in Refs. 1 and 2. A brief examination of the nature of these interply stresses indicates that they have quite different characteristics in one respect, namely, the position at which they occur with maximum intensity. In this regard interply stresses caused by external shear forces reach a maximum at the mid-plane of the structure, membrane-induced interply stresses are effectively distributed 56

TABLE III VALUES OF THE FUNCTION C(m,k) FOR k UP TO 12 No. of m Plies 0 1 2 3 4 6 7 8 9 10 11 (2k) k 2 1 0 4 2 0 1/4 6 3 0 0 2/9 8 4 0 1/16 O 3/16 - 10 9 5 1 0 2/25 1 4/25 12 6 o 1/36 O 3/36 0 5/36 14 7 0 0 2/49 4/49 0 6/49 16 8 o / 34 /64 0 /64 7/64 18 9 0o O 4/81 o 6/81 0 8/81 20 10 1/ioo o 1/1O0 /oo 0 7/100 0 19/1001 22 1 11 O 0 2/121 O 4/121 0 6/121 0 8/121 O 10/121 24 12 0 1/144 O 3/144 0 5/144 0 7/144 0 9/144 0 11/144 o __ _/__ __ _ o ~i __ _ _ _ _ _ _ _ _/_ _ _ _ _ _ -__ _ _ _ _-_ _ _ __ _-_ _ _ _ _

throughout the structure on alternate plies, and bending-induced interply stresses reach a maximum at the junction of the outermost and innermost plies with the rest of the structure. A study of the frequency of occurrence of interply failures on actual tires, at these locations, should allow evaluation of the severity in service of each type of interply stress. Table IVbriefly reviews all the interply stresses discussed here. 38

TABLE IV SUMMARY OF INTERPLY STRESS Type of Interply Stress Generated Type of Shearing Normal Normal Remarks Loading Component Component Component Membrane Eq. (4) Eq. (4) Eq. (4) 1. Proportional to ply thickness and stress level. 2. Independent of total section thickness. 3. Probably primarily fixed by inflation stresses in a real tire. 4. Maximum value occurs at the interface of pairs of plies with opposite angles, all through the the structure. Shear None Eq. (22) Eq. (22) 1. Independent of ply thickness. 2. Inversely proportional to total section thickness. 3. Probably only occurs near edges of the contact patch region and near the bead in a real tire. 4. Maximum value occurs at midplane of entire structure. Distribution is parabolic through remainder of the laminate thickness. Bending Eqs. (27) Eqs. (28) i Eqs. (28) 1. Proportional to bending stress level. Eqs. (35) Eqs. (35) Eqs. (35) 2. Decreases as the number of plies increases, for the same section thickness. 3. Decreases rapidly as section thickness increases, for the same applied moments. 4. Increases rapidly as section thickness increases, for the same curvature changes. 5. Probably primarily fixed by contact patch condi_..... _ tions in a real tire.. ^.~~~~~~~~~~~

IV. CALCULATION OF CORD LOADS The calculation of cord loads due to the presence of membrane forces acting in a laminate of any even number of plies was discussed in Ref. 41 That discussion applies equally well to this report and should be reviewed by the reader. It was shown that cord loads could be expressed graphically in terms of influence coefficients based on external stresses acting on the structure. These stresses were, in turn, based upon external loads and total cross-sectional area of the laminate. In calculating these influence coefficients, it was assumed that the cords carried the whole load in the cord direction, This assumption is probably only slightly conservative for most technically important structures. We next turn to the problem of calculating cord loads induced by bending moments carried by the laminate. Referring to Fig. 8, it may be seen that resolution of moments per unit length m. and me may be made in other directions, such as normal to the cord directions, by means of the expression mi = mgcos2a + m sin2a (37) where bending moments per unit length are expressed as vectorso In dealing with cord loads that arose due to membrane forces, it was assumed that the entire load, resolved in the direction of the cords, was carried by the cords, and none by the rubber. Here, an analogous assumption will be made that the entire moment mx, causing tension or compression in the cords directly, is carried by the cords without consideration of the 41

x Mf~~~ 77/ ~m7 mx sin a a a (a) (b) X-,,n PaL o==~~~ C~Uopper h/2 Cord mid plane h/2 TVTTTWTTTVT7 7 Lower " p 2 Cord (c) 42

surrounding rubber. Again, this assumption is probably only slightly conservative for most technically important applications. Using this principle, the cord loads due to bending may be immediately calculated for a two-ply structure as mx h - noP - 2 B 2 from which P -i 1 [mecos2c + msin2] (38) To compute cord loads in multi-ply structures, some principle must be found for assigning severity of strain to various cords. Strain compatibility requires that, for bending, strains be proportional to distance from the neutral axis or midplane, in this case. This is simply a restatment of the experimentally observable fact that during bending planes originally normal to the laminate's neutral axis remain normal. Hence, Fig. 9 illustrates the situation which must prevail in a multi-ply orthotropic structure under bending, where 2k is the total number of plies involved. Assuming that each ply is of total thickness h, the dimensions di are h = /2 o dj = ( 2 )h o ~2k-1 dk = (2 )h

dK _-P _ P\!\\\\\\K\\\\\ \X \\\ \\,7Ply+2 t2 _ // // // // // // //'// //'/ f/ // { Ply+ I Neutral Pi 9\\ \\ \\ \\ \\ \\ \\ \\ \\ \in Ply-I P2 -// // // // // // // // // // // // Ply-2 Ply-K PK / / // / Fig. 9. Distribution of cord loads in bending.

The cord loads are proportional to their distances from the neutral axis so that PB1 PB2 PBn d, d2 dn Using PBk, the cord load in the outer ply, as a base, one may write dk-l k2 dd nlBk'ddk+ dk PBk + dk + d PBdk-2 + d PBk d d dk k - from which [me cos2c+m1 sin2W ] (2k-1) ~k nh (2k-1)2+(2k-3)2+.. -+(3)2+() k>O Again, this expression is based on an even number of plies in an orthotropic laminate. For the lower half of the structure, Pk = -PkEquation (39) may be used to predict the cord loads generated by the application of bending moments alone, since a twisting moment does not contribute directly to cord loadso However, for an orthotropic structure such as under consideration here, it is often more convenient to work directly from curvatures, which are geometric quantities not involving the stiffnesseso Equations (30) through (32) allow the moments to be expressed in terms of the curvatures, so that when curvatures are given, moments may be found. This will allow the primary component of cord load due to bending to be found directly from Eq. (39). The secondary component of cord load induced by bending comes from the shear and normal stresses which are necessary for pure bending or pure twist.

Equations (13) and (20) indicate that the distribution of such stresses varies linearly with distance z from the mid-plane of a laminate's thickness. This linear variation comes about from the assumed uniform dispersion of anisotropy, in the form of a uniformly distributed filamentary structure, through each ply. Actually, of course, the anisotropy is concentrated in the twisted cords. For that reason the actual distribution of external stress, such as given by Eqs. (13) and (20), is not linear but similar to Fig. 10 for a two-ply structure. Actual induced stress distri bution Two-ply structure Idealized induced stress distribution Fig. 10. Assumed and ideal induced secondary-stress distributions. The structure itself is required to carry a total shearing force equal to one-half the maximum calculated shearing stress times the area of the M'V% 4- 4- I - - - -z A. -?......,'

cross-section. An alternate way of looking at this is that the average shear-' ing stress is one-half the maximum value for a two-ply structure. Cord loads should be calculated on the basis of this average shearing stress, since that is what the structure must actually resist. For a multi-ply structure having 2k plies, each of thickness h as shown in Fig. 9, it may be seen at once that Eqs. (13) and (20) may still be used but should be evaluated at the mid-point of each ply to determine the average stresses needed to find the cord load in that ply. For example, the cord load in the jth ply from the mid-plane would be evaluated by using z = (j-!)h in Eqs. (13) and (20), followed, of course, by use of Ref. 4, where graphs are presented for determining cord loads when shearing and normal stresses are given. For example, in a six-ply structure subjected to pure bending, the distribution of shearing stresses given by Eq. (13) is shown in Fig. 11. Here it Ply Angle Shearing Stress Intensity from Eq(13) 2 t-a? 4F x5 + /Fg6 11.Sharnsrsneiyn sipytcu d

is seen that widely different external shearing stresses act on various plies and induce secondary cord loads. Cord loads due to membrane forces and those due to bending (both primary and secondary components) may be added algebraically to obtain the total cord load in each ply. As would be expected, the cord loads due to membrane forces, as previously calculated, are uniform in each ply of a multi-ply structure under conditions of symmetry, equality of stiffness of each ply, and the existence of an even number of plies. In contrast, the cord loads generated due to bending vary with distance away from the mid-plane of the laminate, with those plies lying at the inner and outer surfaces having the smallest and largest values of cord load due to bending. Thus, in general, either the inner or outer ply of a multi-ply laminate will show the greatest cord load. In summary, cord loads may be generated by: (a) membrane forces; (b) bending moments; and (c) shearing or normal forces required due to the existence of bending and twisting deformations. The complete cord load is made up of the sum of these three effects. They may be added algebraically.

V. EXAMPLES (I) Consider the problem of determining the interply stresses and cord loads in a flat, two-ply belt made of rubberized fabric with the dimensions shown in Fig. 12, and passing over a 12-ino-diameter pulley while carrying a load of 2000 lb. It is necessary to know certain elastic properties of the belt materials before the stresses can be computed. Values will be assumed for these properties which are close to those found for pneumatic tire materials. Accordingly, let G0y = l_2E = 1.2 x 105 psi k~ = - 6i Ex for +z upward. n = 10 cords/in. F = 1.5 x 104 psi k = 0 since no curvature is induced in the belt at right angles to kS =0 the pulley. Interply Stresses (a) Membrane j'T = - ( a)(A) o( h) = 0.050[0 27x2000] = 27 psi from Fig, 6 of Refo 2. (b) Shear To a first approximation no shearing forces are carried by the belto Thus, interply forces due to this effect are zero.

__,o —---- ~_ jSEC. A-A CORD Figs 12. Flat belt on pulleye ~o

(c) Bending It has been shown in Table 1 that no interply stresses are generated in a two-ply laminate due to bending or twist. The maximum interply stress is amax = 27.0 psi Cord Loads (a) Membrane P = (-)(2000) (1.10) = 11.00 lb from Eqo (8) and Fig. 2 of Refo 4o (b) Bending-Primary Component. PB = (m)(cos2a) n-h from Eq. (39) of this report. Combining Eqs. (30) and (31), (2h)3 1(153 +0 m = -(EikS + Tl)(2 (1-2 x 105)()1 + 0 12 12 = 1o67 Thus., = (.67)(lOx305)(.953) = 351 lb This is positive, or tensile, on the outer ply and compressive on the inner ply. (c) Bending-Secondary Component. Since the twist kB is zero, only the terms of EqO (13) are of 51

interest since Eq. (20) vanishes. Equation (13) becomes, when evaluated at z = +.050, 1mhax = [-0.27 x 1.2x105 x(~ max +.075x1.5x104(- ) ](.050) = 260.6 psi This shearing stress, due to its linear distribution, is equivalent to an average shearing stress of half this value acting over the cross section of the structure, as previously discussed. Hence, V = 130.3 psi ~ 1AV Using Eq. (6) of Ref. 4, the secondary component of cord load becomes = 0505 [3.8x13j0.3] - 2.5 lb BS 10 This is positive, or tensile, in the upper ply. Due to the form of the quantity N of Ref. 4, in which N(a) = -N(-Ua), the lower ply has induced in it a secondary bending load of BS = 10 [-3.8x130.3] - 2.5 lb. Now all cord loads may be summed in each ply to give the cord loads in that region of the belt contacting the pulley: Ply 1 - Upper: total cord load PT = 11.0+3.1+2.5 = 16.6 lb Ply 2 - Lower: total cord load PT = 11.0-3.1-2.5 = 5.4 lb. Away from the pulley, in the straight portion, only the membrane component 52

of cord load is present so that the cord load becomes: Ply 1 - Upper: PT = 11.0 lb Ply 2 - Lower: PT = 11.0 lb. Notice that these calculations indicate tension loads in all cords. Hence the assumption of an orthotropic material is justified in this case. The stretch of this belt may be determined approximately by using the value E - 90x103 from which = E 2x103 ~e~ U90x0b3 =.022 ES F 90x103 = 2.2% Hence, its length under load will be 2.2% greater than its original length. (II) Consider the problem of estimating the effects of a two-ply versus a four-ply structure in a pneumatic tire, in so far as interply stresses are concerned. Assume dimensions such as shown in Fig. 13, and elastic properties as measured for a 20 —end-count-rayon-ply stock. Unde -- - - Undeflected Deflected Fig. 13. Assumed geometry of deformation of a pneumatic tire. 53

Wall thickness t = 0.2 Four ply-each ply has thickness h = 0.05 Two ply —each ply has thickness h = 0.1 Cord angle c' — 35~ Q direction is taken circumferentially ~ direction is taken meridianally. Estimate, based on test data for rayon cords, E E = x = 104 =E~ = 2.5x104 F = 5x104 Inflation pressure = 50 psi. From unpublished work by this group, the membrane stresses of the crown may be estimated as those due to inflation only, giving o~~ 3590 psi 00 ~a 780 psi For the calculation of interply stresses due to bending it is necessary to evaluate the terms in Eq. (27). For this, it is noted that from the assumed geometry of deflected shape, as given in Fig. 9, 1 1 k = k - 0.25 k = kG 0. Thus, using Figs. 6 and 7 of Ref. 2, the quantity in brackets in Eq. (27) becomes [0. 65( lOxO+5x. 25) +0. 13( 2.5xo. 25+5xO) ]x104 = 0. 894x104

For the zTw>o...._l-y Structure: h = 0.1 Menmrane-i.rnduced interply stresses c T [( [(O 65) (780) + (o i )(39) ]0 1 = 56 psi from Eq. (4) at interface. Bending ~induced interply stresses "T = 0 from Eqo (2i7) an-d Table I. 1K'j-' F,>KIi' Structure: h 0.05 e:,lr:iranvre~. nduced interply stresses rT)l = [(o.65)(780) + (.j 3) (390) ](0.05) a" ) 0= at the mid-pilan.e O.....;d. j, 7i1.....:ed interply stresses (r ) - (o,94xlO4)(.i.!,, x... 2 22 psi from Eq. ( 27) and Table I. (a,) = o ( = - 2 psi VThile 7su:rzet! f the various interply strens:>i:,. >:+i:ulated here are compar~i." ili Figo iio nTj the basis of the particulau!.:n::.itlons selected, the maximw:r int e> rpry stI Ilss is about the same in the tw-;, —1L9y- and four-ply tires. Note, VTi e4~r,. that; in the four-ply tire the m> i.9. tierply stress ocnclr2 act.. %oundary ofr t;he owutermost ply to the rest f'"!;T,h,-~ earcass, 55

OUTSIDE 56~U lo 456 MEMBRANE BENDING TOTAL INSIDE Interply stresses in two-ply structure. OUTSIDE { 28,2 MEMBRANE BENDING TOTAL INSIDE above. Fig. 14. Interply shear-stress distribution in two- and four-ply structures.

The two-ply structure might have some inherent advantages here since the bending-induced interply stresses are always zero; in the four-ply tire, situations might arise in which the radius of curvature underwent a severe change and caused large interply stresses due to bending. In addition, the bending-induced interply stress is cyclic in nature since it occurs at each passage through the contact patch. It may thus result in fatigue of interply bonds. On the other hand, the membrane-induced interply stress is approximately constant. 57

VI. ACKNOWLEDGMENT The writer acknowledges the assistance of Mr. Richard N. Dodge, Mro No Lo Field, and Mr. D. H. Robbins in checking the conclusions and equations contained herein.

VII. REFERENCES lo S. K. Clark, The Plane Elastic Characteristics of Cord-Rubber Laminates, Univo of Mich. Res. Insto Report 02957-3-T, Ann Arbor, Michigan. 2. S. Ko Clark, Interply Shear Stresses in Cord-Rubber Laminates, Univo of Mich. Res. Insto Report 02957-4-T, Ann Arbor, Michigan. 3. R. N. Dodge, No L. Field, and So K. Clark, Interply Stress and Load Distribution in Cord-Rubber Laminates, Univo of Micho Res. Insto Report 02957-8-T, Ann Arbor, Michigan. 4. S. K. Clark, Cord Loads in Cord-Rubber Laminates, Univo of Micho Res. Inst. Report 02957-5-T, Ann Arbor, Michigano 5 S. Timoshenko and S. Woinowsky-Krieger, Theory of Plates and Shells, McGraw-Hill Book Co,, New York, 1959, p. 167. 6. S. K. Clark and Do H. Robbins, Bending Characteristics of Orthotropic Laminates, Univ. of Mich. Res. Insto Report 02957-9-T, Ann Arbor, Michigan. 61

VIII. DISTRIBUTION LIST Name No. of Copies General Tire and Rubber Company 6 Akron 9, Ohio UoSo Rubber Company 6600 E. Jefferson Avenue Detroit 32, Michigan 6 BoFo Goodrich Tire Company Akron, Ohio 6 The Goodyear Tire and Rubber Company Akron 16, Ohio 6 The Firestone Tire and Rubber Company Akron 17, Ohio 6 Ro A. Dodge 1 S. S. Attwood 1 S. K. Clark 1 The University of Michigan ORA File 1 Project File 11

UNIVERSITY OF MICHIGAN I9111 11 1 1111i I IIlJlI IIII 3 9015 02828 5396