2008 - Belt Angle Stress Paper - Mc Ginty et al

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. D. McGinty,1 T. B. Rhyne,2,3 and S. M. Cron2,4

nalytical Solution for the Stresses Arisingn �/� Angle Ply Belts of Radial Tires

REFERENCE: McGinty, R. D., Rhyne, T. B., and Cron, S. M., ‘‘Analytical Solution for theStresses Arising in �/� Angle Ply Belts of Radial Tires,’’ Tire Science and Technology,TSTCA, Vol. 36, No. 4, October – December 2008, pp. 244-274.

ABSTRACT: Stresses arising in the belts of radial ply tires, particularly those at the belt edge,are known to be critical to tire durability. Belt edge stresses are commonly calculated usingfinite element �FE� methods that provide estimates of the levels but do not necessarily givesignificant insight into the underlying mechanics. In contrast, analytical models can providephysical insight into the mechanisms affecting tire durability but are currently incomplete dueto the challenges faced in obtaining closed-form mathematical solutions. Nevertheless, ana-lytical solutions remain important to tire design and development because they can expose theentire design space, show the mathematical relationships between the variables, and allowrapid parameter studies.

This work develops an analytical description of the belt deformations and stresses, par-ticularly at the belt edge. The formulation captures all the first-order mechanics pertinent tofinite width, antisymmetric �/� angle belt packages present in radial tires. It incorporatesinterply shear stresses already recognized in the literature and adds to that a new mechanismcontrolling the interaction of the plies via a Poisson effect. The analytical model is validatedby comparison to FE simulations and is also contrasted with a classical analytical model in theliterature. The design space for the belt composite is then explored by parameter variation.Finally, since all these solutions depend on homogenization of the belt layers, the analyticalsolution is compared to a FE model of discrete cables embedded in rubber to explore theaccuracy of the homogenization step.

KEY WORDS: belt stresses, interply shear, belt composite, belt parameters, tire durability,finite element, discrete model

Steel belted radial tires introduced by Michelin in 1949 cured a number ofhe ills afflicting the bias ply tires that came before them. Contact patch defor-ation �squirm� was greatly reduced by stabilization from the steel belts, al-

owing much improved wear. The main durability issue with bias ply tires,nterply shear in the carcass, was eliminated because the belts ended beforeignificant shear was generated and no interply shear arises in the radial carcass.oth of these improvements led to much lower rolling resistance and longer

ervice life of tires. Along with these and other positive improvements came aew durability issue arising from the existence of the free edge of the �/�

Mercer Engineering Research Center, Warner Robins, Georgia 31088, USA. Electronic mail:[email protected] Americas Research and Development Corporation, Greenville, South Carolina 29605,USA.Presenter and corresponding author. Electronic mail: [email protected]
Electronic mail: [email protected]
244

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MCGINTY ET AL. ON SOLUTION FOR THE STRESSES 245

ngle ply belts. In classical composite mechanics, assuming perfect couplingetween plys, the interply shear stress is singular at the edge. Fortunately, inadial tire belts, the coupling is not perfect but takes place through a rubberoupling layer. The imperfect coupling limits the magnitude of the edge stresso a finite, but typically large, value. The management of this stress is critical tohe design of radial tires. We emphasize that this stress is an edge effect that isarasitic in nature; it does not contribute to the function of the tire.

Today, finite element �FE� methods are typically used to calculate belttresses in tire design. Unfortunately these methods just give point solutions ando not expose the design space, do not indicate what parameters are importanto the magnitude of the stresses, and give little insight into the underlyingechanics. Thus, analytical solutions are welcome, even given the simplifying

ssumptions necessary to obtain them, because they can show the relationshipsetween the parameters and allow mapping of the entire design space easily.nalytical solutions also give insight into the fundamental behavior of the belt

omposite.The first published analytical solution for the belt edge shear problem ap-

licable to tires was by Puppo and Evensen in 1970 �1�. Their solution uses anlegant superposition technique to reach the final result of a flat �/� angle plyomposite under generalized plane stress. The Puppo and Evensen model isarticularly applicable to the tire belt composite because it is formulated with anite thickness isotropic shear layer between the anisotropic belt layers. We willhow that due to an unnecessary assumption this solution leaves out somemportant physics that significantly alter the results. Another analytical modelnown as the Kelsey model is found in a paper by Turner and Ford �2�. Thisodel is specifically formulated for tire belt composites but makes more restric-

ive assumptions than Puppo and Evensen.In addition to the two models mentioned above, there is rich literature

urrounding the calculation of the interply stress at the free edge of angle plyomposites since 1970. Most of this literature is more applicable to fiber-einforced plastic composites than to cable-reinforced rubber as used in tires.imple closed form solutions usable in tire design are rare. Two recent surveyapers give a good bibliography of the literature on the general subject �3,4�.he authors believe that the Puppo and Evensen solution is still the best closed-

orm, simple model applicable to tire belt composites.An analytical solution containing all the first-order physics relative to the

elt stresses arising in radial tires is derived below. The results will be comparedith an FE solution and the Puppo-Evensen solution. The belt design space willext be explored by parameter variation. All of these solutions depend on someorm of homogenization of the belt layers. A detailed FE analysis of discreteables embedded in rubber is carried out to study the implications of this ho-
ogenization assumption.

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246 TIRE SCIENCE AND TECHNOLOGY

implifications and Assumptions

As might be expected, rather severe assumptions will be made to the prob-em to make the mathematics tractable. It is believed that these restrictions willot affect the fundamental character of the results or their usefulness in helpingnderstand tire behavior.

The radial tire belt geometry is complicated by curvature and complexeformation in the contact patch. The belts do not act in isolation but areoupled to the radial carcass and often to circumferential nylon plys. The beltsre loaded in tension by the inflation pressure and by transverse and circumfer-ntial bending in the contact patch. All these complications will be simplified tohe case of two flat angled ply laminas, separated by a rubber coupling layer,oaded in tension. The lamina and the resulting laminate are assumed thin suchhat the problem can be treated as generalized plane stress. The tension twistingoupling that arises in the case of planar extension of the belts will be neglected.his is equivalent to solving the axisymmetric cylindrical belt case that occurs

n tires.The belts will be modeled as steel cables embedded in incompressible

ubber. The cables will be considered incompressible as well, because thetresses involved will be too small to generate significant strains in the steel.ecause the cable modulus is much greater than the rubber modulus, the strain

n the cable direction is considered to be a function of the cable properties onlynd is assumed to be approximately zero. The cable is also assumed not to shearr deform due to lateral or shear stresses. Finally, the rubber is considered to beinear elastic and small strain assumptions will be made.

erivation of the Lamina Equations

Figure 1 gives a schematic representation of the lamina. The material co-rdinates are labeled 1, 2, 3 where the cable follows the “1” direction. The bodyoordinates are labeled X, Y, Z, where the “X” direction represents the circum-erential direction in the tire and primary loading direction, “Y” represents theidth direction of the belts, and “Z” represents the thickness direction of theelts, the radial direction in the tire. The lamina can be described by six param-ters: Ec, the cable modulus; P, the cable pace; �, the angle of the cable relativeo the X body coordinate; D, the cable diameter and assumed lamina thickness;, the modulus of the rubber; and �, Poisson’s ratio of the rubber.

The derivation of the lamina equations is well-known and will not be car-ied out in detail here. The equations can be found in composite materials textsuch as Jones �5�. The strain transformation equations between the local lamina
nd global coordinates can be written as

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Fmbf


�x = �1 cos2 � + �P − D

P��2 sin2 � − �12

2 cos � sin �� ,

�y = �1 sin2 � + �P − D

P��2 cos2 � + �12

2 cos � sin �� , �1�

�xy = �1 sin 2� + �P − D

P��− �2 sin 2� + �12 cos 2�� .

ote that these equations are the same as the two-dimensional orthotropic strainransformation equations, Jones Eq �2.68�, for example, except for the additionf the ��P−D� / P� factor. The strains �2 and �12 apply to the rubber only be-ause we assumed that the cable does not strain in these directions. Thesetrains are thus scaled by the ��P−D� / P� factor to give the correct averageesponse of the composite lamina.

The strain stress equations for the lamina can be written as

�x =cos2 �

VcEc��x cos2 � + �y sin2 � + 2�xy sin � cos �� +

1 +

E�P − D

P�

��x sin2 ��2 − �1 + �sin2 �� − �y�1 + �cos2 � sin2 �

2

IG. 1 — Lamina description. Note that the lamina is described by five parameters, the rubberodulus, the cable modulus, the cable pace, the cable diameter, and the cable angle relative to theody coordinate X. The lamina thickness is equal to the cable diameter. The material coordinate (1)ollows the cable direction.

− �xy sin 2��1 − �1 + �sin �� ,

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�y =sin2 �

VcEc��x sin2 � + �y cos2 � − 2�xy sin � cos �� +

1 +

E�P − D

P�

��y cos2 ��2 − �1 + �cos2 �� − �x�1 + �cos2 � sin2 �

− �xy sin 2��1 − �1 + �cos2 �� , �2�

�xy =sin 2�

VcEc��y − �x�sin � cos � + �xy�cos2 � − sin2 �� +

1 +

E�P − D

P�

�− �y sin 2��1 − �1 + �cos2 �� − �x sin 2��1 − �1 + �sin2 ��

+ �xy�2 − �1 + �sin2 2�� .

hese equations can be obtained from Jones Eq �2.82�, for example. The quan-ity Vc is the volume fraction of the cable. Note that the first term in each ofhese equations represents the cable contribution to the strain and the seconderm represents the rubber contribution.

At this point it would be interesting to check these results against a FEnalysis of a lamina. Using the first of Eqs �2� we can compute the longitudinalodulus of a lamina, ELAMINA, by dividing the longitudinal stress by the lon-

itudinal strain while letting the lateral stress, �y, and shear stress, �xy, be zero.he result is

�x

�x= ELAMINA =

1

cos4 �

VcEc+

1 +

EP − D

P�2 − �1 + �sin2 ��sin2 �

. �3�

The FE simulation is made with a two-dimensional axisymmetric cylinderf a large radius made up of a single element. The loading is an imposedncrease in the cylinder radius, which results in an imposed elongation in theircumferential direction. The rubber part is defined as a hyperelastic incom-ressible material described by Treloar’s law. The steel is given a linear elastic,omogeneous, orthotropic material definition. The cable angle is varied fromero to 90 degrees. The FE model compared with the analytical solution, Eq �3�,s given in Fig. 2 along with the details of the model. A base set of beltarameters commonly found in radial tires is used and will be used for furtheralculations. The agreement is very good, which gives confidence to continue to
model of the laminate belts.

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iased Belt Composite

Now we will construct a composite made up of two laminas as describedbove having �/� cable angles and separated by a homogeneous isotropicubber layer. This laminate will represent the belt package found in modernadial tires. The laminate is shown in Fig. 3. One additional parameter is intro-uced, the rubber layer thickness t.

For a given elongation, �x, the resulting stress, �x, in the first belt is muchreater if the second belt is present than if the first belt is alone. Yet Eq �2� musttill apply to each belt, whether the other is present or not. Each belt transmitsts presence to the other through stresses generated in the rubber layer. Typically

IG. 2 — Lamina modulus versus the belt cable angle. The analytical solution, Eq (3), is comparedith a finite element solution. The belt parameters are typical of those used in modern radial tireelts. The agreement is very good over the full range of possible belt cable angles.

he assumption is made that only the shear stresses in the rubber layer are

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ignificant. The shear stresses are certainly important, but the lateral stress in theubber layer is also of first order. We begin with the shear stress.

n-Plane Shear Stress, �xy

The stress equilibrium equation in the x direction is

��x

�x+

��xy

�y+

��xz

�z= 0. �4�

or uniaxial elongation the derivative ��x /�x=0. Then Eq �4� becomes

��xy

�y= −

��xz

�z. �5�

or the second belt, the derivative on the right side of Eq �5� can be written as

��xz

�z=

�xz,belt 2 top − �xz,belt 2 bottom

D, �6�

ut the shear stress on top of belt 2 is zero because it is a free surface. Therefore

IG. 3 — Laminate description. Note that the laminate description adds only one additional pa-ameter to the problem, the rubber layer thickness. The rubber in the lamina and laminate aressumed to have the same properties. The lamina, however, are laid at plus and minus angleselative to the X axis. In the tire, the X axis represents the circumferential direction, the Y axisepresents the belt width direction, and the Z axis represents the radial direction.

e can write, using Eqs �5� and �6�,

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��xy

�y=

�xz,belt 2 bottom

D. �7�

he shear stress at the bottom of belt 2 equals the shear stress in the rubberayer, which is

�xz,rubber layer = G�xz,rubber layer. �8�

nserting Eq �8� into Eq �7� gives

��xy

�y= G

�xz,rubber layer

D, �9�

ut the shear strain in the rubber layer is related to the x-direction displacementsf belt 1 and belt 2 such that

�xz,rubber layer =ubelt 2 − ubelt 1

t. �10�

or the plus and minus angled composite

ubelt 2 = − ubelt 1. �11�

sing Eq �11�, Eq �10� becomes

�xz,rubber layer = 2ubelt 2

t. �12�

nsert Eq �12� into Eq �9�, giving

��xy

�y=

2G

tDubelt 2. �13�

ecognize that

ubelt 2 = �0

y �u

�ydy . �14�

nsert Eq �14� into Eq �13� and differentiate to obtain

�2�xy

�y2 =2G

tD

�u

�y. �15�

his equation relates the in-plane shear stress in the composite to its x-directionisplacement. Recall the small strain definitions and the fact that in our problem
�u /�x�=0, then

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�xy =�u

�y.

hus Eq �15� becomes

�2�xy

�y2 =2G

tD�xy . �16�

Note that Eqs �12� and �13� could be combined to give

�xz,rubber layer =D

G

��xy

�y. �17�

his turns out to be a convenient form to use later in the calculation of thenterply shear strain.

ateral Stress, �y

To this point the derivation is straightforward and does not add significantlyo the existing literature. It has incorporated the effects of intraply stresses asell as interply shear stresses. The classical solution of Puppo and Evansen also

ncludes all these aspects but neglects the effects of lateral stresses in the rubberayer. It will now be shown that this assumption is unnecessary and reduces theccuracy of the results.

The lateral stresses in the belt composite come from two sources. The firsts simply an imposed stress from an external source. We will name this stress

y,BE, for “y-stress imposed at the belt edge.”The second source of lateral stress is generated internally from the fact that

he rubber layer is sandwiched between the two orthotropic belt laminas. Con-ider that a belt composite is stretched to �x=1%. Since the rubber in the rubberayer is incompressible, ��=0.5�, it will, without interference from the beltayers, become narrower such that �y =−0.5%. The belt lamina acting alonender the same extension will become much more narrow than that at theelatively small cable angles used in tires, ��y��0.5%. The rubber layer, con-trained above and below by the belt layers, will therefore be put in compres-ion. This compression will be reacted by tension in the belt layers such that theet internal force is zero.

Recall Hooke’s law for the plane stress lateral strain in isotropic materials,

�y =1

E��y − ��x� .

nvert this equation to express stress in terms of strain,

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�y =E

1 − 2 ��y + ��x� . �18�

ext write the equilibrium of the internal lateral forces in the belts,

�2 Belts��y,belt��D� + ��y,rubber layer��t� = 0.

olve this expression for the belt lateral stress,

�y,belt = −1

2� t

D��y,rubber layer. �19�

Note that the strains, �x and �y, of the belts and rubber layer must all be theame because they are one laminate. Then �y,rubber layer is given by Eq �18� ands substituted into Eq �19� to give the lateral stress in each belt,

�y = �y,BE −E

2�1 − 2�� t

D��y + �x� . �20�

pply Inextensibility to the Cables �1=0, Ec→�While the lamina equations derived above contain a finite cable modulus,

e will now enforce the assumption that the cable modulus, Ec, is infinite andhus, the strain in the cable direction is zero in order to make the laminatequations tractable. Thus, the first terms of Eqs �1� and �2� are eliminated. Thetrain transformation equations, Eq �1�, become

�x = �P − D

P��2 sin2 � − �12

2 cos � sin �� ,

�y = �P − D

P��2 cos2 � + �12

2 cos � sin �� ,

�xy = �P − D

P��12 cos 2� − �2 sin 2�� .

ombine these equations to eliminate �2 and �12, resulting in

�x cos2 � + �y sin2 � + �xy sin � cos � = 0. �21�

ow let the cable modulus, Ec, go to infinity in Eq �2�, leaving only the rubberomponent of the strain.

�x =1 + �P − D� �x sin2 ��2 − �1 + �sin2 �� − �y�1 + �cos2 � sin2 �−

2 � ,
E P �xy sin 2��1 − �1 + �sin ��

E

N

S

d

w

W

N

�s


�y =1 +

E�P − D

P� �y cos2 ��2 − �1 + �cos2 �� − �x�1 + �cos2 � sin2 �−

�xy sin 2��1 − �1 + �cos2 �� ,

�22�

�xy =1 +

E�P − D

P�

− �y sin 2��1 − �1 + �cos2 �� − �x sin 2��1 − �1 + �sin2 ��+�xy�2 − �1 + �sin2 2�� .

quations �22� can be expressed in matrix form as

� �x

�y

�xy� = �A11 A12 A13

A21 A22 A23

A31 A32 A33��x

�y

�xy� . �23�

ote that the Aij matrix is symmetric.

imultaneous EquationsUsing the results to this point we can write a system of equations that

escribe the deformations of �/� angle ply composite belts,

� �x

�y

�xy� = �A11 A12 A13

A21 A22 A23

A31 A32 A33��x

�y

�xy� , �23��

here the coefficients are given by Eq �22�, and

�x cos2 � + �y sin2 � + �xy sin � cos � = 0, �21��

�2�xy

�y2 =2G

tD�xy , �16��

�y = �y,BE −E

2�1 − 2�� t

D��y + �x� . �20��

e thus have six equations for the six dependent variables,

�x,�y,�xy,�x,�y,�xy .

ote that the only differential equation is Eq �16�.The equations will be solved assuming a known constant extensional strain,

x, and a known imposed lateral stress at the belt edge, �y,BE, if any. Using this
et of equations eliminate �xy from Eq �16�, giving

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�2�xy

�y2 � tD

2G�A11 + �A12A31 − A11A32�� t

D� E

2�1 − 2�1

tan2 �

− �A11A33 − A13A31��xy

= �A11A32 − A12A31��y,BE

+ �xA31 − �A12A31 − A11A32�E

2�1 − 2�� t

D�� 1

tan2 �− � . �24�

quation �24� is a second-order differential equation for �xy. The solution hashe form

�xy = Aesy + Be−sy + C .

he boundary conditions are1. ��xy /�y�=0 at y=0, the belt composite centerline.2. �xy =0 at y=w, the edge of the belts. The solution of Eq �24� can then

e written as

�xy = �y,BE

tan �+ �x

E

�1 − 2� P

P − D+

1

2� t

D� 1

tan �� 1

tan2 �− �� 1 −

cosh sy

cosh sw� ,

�25�

here

s =�2�1 − �tD

�P − D

P� sin2 �

2 − �1 + �sin2 � + � t

d��P − D

P�cos2 �

. �26�

elt Edge Shear Strain, �xz

We are now in a position to calculate the interply shear strain in the belts,ne of the main objectives of this work. Using Eq �25� in Eq �17� we see thathe interply shear strain can be written as

�xz = − 4 cos �� 1

2�1 − ��D

t��P − D

P� 1

2 − �1 + �sin2 � + � t

d��P − D

P�cos2 �

�1 − 2�E

�y,BE + �x� P

P − D� +

1

2� t

D� 1

tan2 �− � sinh sy

cosh sw� , �27�

here s is given again by Eq �26�. This result will be used extensively later, but
rst the derivation is completed by calculating the longitudinal stress.

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ongitudinal Stress, �x

Given the solution for �xy, the equation set can be solved for the longitu-inal stress,

�x =�y,BE

tan2 �+ �x

E

2�1 − �2�� P

P − D�2 − �1 + ��sin2 2�

sin4 �

+ � t

D� 1

tan2 �� 1

tan2 �− ��

+ �xyE

2�1 − �2�� P

P − D� sin 2��1 − �1 + ��sin2 ��

sin4 �+ � t

D� 1

tan3 � .

�28�

n expression for �xy to use in Eq �28� can be obtained from Eqs �16� and �25�s

�xy = � P

P − D� �1 − 2�

E�y,BE + �x1 +

1

2� t

D�� P

P − D� 1

tan2 �− �

� sin 2�

2 − �1 + �sin2 � + � t

D�� P

P − D�cos2 �� 1 −

cosh sy

cosh sw� . �29�

hen the modulus at the centerline, y=0, of the belt laminate can be written as

�x,CL

�x=

E

2�1 − �2�� P

P − D�1 − �1 + ��sin2 2�

sin4 �+ � t

D� 1

tan2 �� 1

tan2 �− �� ,

�30�

ssuming that the imposed belt edge lateral stress is zero. Equation �30� can besed to compare with the centerline belt modulus using FE analysis. Figure 4ives the result for the base case of laminate parameters and allowing the beltngle to cover its full range. Note that above a belt angle of about 15 degreeshe agreement with the FE analysis is very good. Below 15 degrees the analyti-al and FE results separate due to the infinite cable modulus assumption madeor the analytical case. The analytical lamina modulus is shown in this figure foromparison. At angles greater than 54.7 degrees �the minimum of the curve�,he belts are effectively uncoupled and act essentially as independent lamina.he minimum of the curve represents the belt angle at which there is no Pois-on’s mismatch between the rubber layer and the lamina.

Now we will compare the expression derived for the interply shear strain,q �27�, with the FE analysis. Figure 5 looks along the Y axis at the edge of the
elts and defines the interply shear strain. The base case of the parameters is

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sed for the FE comparison. Figure 6 gives the result. The agreement is veryood. The FE analysis gives a lower peak value at the belt edge by a fewercent, as might be expected. Note that the interply shear strain is an edgeffect and essentially disappears 10–15 mm inboard of the belt edge. The in-erply shear strain is effectively independent of the width of the belts; thus,elatively narrow belt models can be used to study this edge effect. This factecomes important later in the discrete model discussion. The FE analysis is an

IG. 4 — Laminate centerline modulus versus the belt cable angle. The analytical solution, Eq (30),s compared with a finite element solution. The belt parameters are typical of those used in modernadial tire belts. The agreement is very good for belt cable angles greater than about 15 degrees.elow that angle the infinite cable modulus assumption becomes increasingly invalid. Most radial

ire belts use angles from about 15 to 30 degrees. For belt cable angles greater than about 55egrees the belt plys become effectively uncoupled and act like individual lamina. The laminaodulus, Eq (3), is also plotted for comparison purposes.

baqus model using quadratic elements and embedded reinforcements.

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IG. 5 — Interply shear strain representation. This diagram is looking along the Y axis at the edgef the belts. The angle describing the interply shear strain in the rubber layer is shown.

IG. 6 — Interply shear strain versus belt width. The analytical solution, Eq (27), is compared withfinite element solution. The belt parameters are typical of those used in modern radial tire belts.

he agreement is very good over the full width of the belts. The finite element solution underpredictshe peak strain at the belt edges by a few percent relative to the analytical solution. Note that thenterply strain is an edge effect that effectively disappears 10 mm from the edge. Thus, to study beltdge shear strains the belt models do not need to be very wide. The decay of shear strain at the belt
dges is effectively independent of the belt width.

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omparison with Published SolutionThe classical solution by Puppo and Evensen can be formulated in terms of

he nomenclature of this study so that a direct comparison can be made. Thelassical solution does not include the stiffening mechanism due to the inter-ally generated lateral stress as explained above. Therefore it underpredicts thetress and strain values generated by FE analysis and the current analyticalolution. The key equations, Eqs �25�, �26�, �28�, �27�, and �29�, are shownelow. The terms missing from the classical solution are circled:

( ) ⎭⎬⎫

⎩⎨⎧ −⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ −⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

−−+=

swsy

Dt

DPPE

xy

xy coshcosh1

tan1

tan1

21

1tan 22BE, υ

ααυε

ασ

τ

( )( ) ααυ

αυ22

2

cossin12

sin12

⎟⎠⎞

⎜⎝⎛ −⎟⎠⎞

⎜⎝⎛++−

⎟⎠⎞

⎜⎝⎛ −−

=

PDP

dtP

DPtD

s

( )( )

( )( )[ ]

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

+−⎟⎠⎞

⎜⎝⎛

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umerical examples of the difference between the present solution and Puppond Evensen are given in Fig. 7 for the centerline modulus and in Fig. 8 for
nterply shear strain using the base case parameters.

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esign Space ExposureAt this point we have some confidence that the analytical model reasonably

epresents the behavior of the laminate. Now we can expose the design spaceelative to the magnitude of the interply shear strain. Parameter variation will besed, varying each parameter in the analytical solution about a base case. Howhould the belts be loaded for the parameter study? The functioning mode of theelts in a tire is complicated. Upon inflation the belts probably work mostly inn imposed stress mode. In the contact patch an imposed strain state might beore appropriate. In some regions an imposed strain energy state may be re-

uired. The equations we have derived are naturally set up to look at imposedtrain, but it is easy to set them up for imposed stress or imposed energy. Weill look at both the imposed strain and imposed stress conditions; imposed

nergy will fall between these two extremes.The parameters for the base case are given in Table 1. The design space is

hown in Fig. 9 with the darker curves representing the imposed strain case.

IG. 7 — Comparison of laminate centerline modulus versus the belt cable angle for the publishedolution of Puppo and Evensen, the present theory, and finite element model. The published solutionacks the stiffening mechanics due to lateral stress in the laminate.

ariation of the cable pace, cable diameter, rubber layer thickness, and belt

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ABLE 1 — Base case parameter values for belt design space exposition. The values for the basease are typical of those used in modern tire belt composites. The extensional strain value used isrbitrary and simply applies a linear scaling to the shear strain.

P= 1.6 mm �Cable Pace�D= 0.56 mm �Cable Diameter�E= 7.5 MPa �Rubber Modulus��= 25 �/� deg �Belt Angle�t= 0.9 mm �Rubber Layer Thickness�

x= 0.01 �Extensional Strain�w= 30 mm �Belt 1 /2 Width��= 0.5 �Poisson’s Ratio Rubber�

�y,BE= 0 MPa �Imposed Lateral Stress�s= 0.2982

IG. 8 — Comparison of laminate interply shear strain versus belt width for the published solutionf Puppo and Evensen, the present theory, and FE analysis. Typical base case laminate parametersre used. The published solution underpredicts the peak value at the belt edge by about 20% in thisase. Smaller belt angles produce larger differences.

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able angle show complicated nonlinear behavior. Depending on the parameterr the location in the parameter space, the imposed strain and imposed stressesults can trend in the same or opposite directions. Variation of Poisson’s ratios included for completeness. The interply shear strain is not very sensitive tooisson’s ratio, is linear, and does not depend on the deformation mode. Finally,e note that the interply shear strain is sensitive to the value of the imposed belt

dge lateral stress, is linear in this parameter, and is independent of the func-ioning mode.

Note that parameter variation was not applied to the rubber modulus. Ex-mination of Eq �27� indicates that the interply shear strain is not sensitive toubber modulus in the imposed strain case, except that it interacts with themposed belt edge lateral stress. The imposed lateral stress required to obtain a

IG. 9 — Parameter variation example, interply shear strain at the belt edge versus laminate pa-ameters. Each laminate parameter is varied about a base case. The imposed strain case is repre-ented by the darker curve. The imposed stress case is represented by the lighter curve. Theariation of Poisson’s ratio and imposed lateral stress gives the same result regardless of theeformation mode of the laminate. The other parameters show complicated nonlinear behaviorependent on the deformation mode. The case of imposed strain energy density is not shown butalls between the limiting curves in this figure.

iven interply shear strain is scaled by the inverse of the rubber modulus.

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quation �30� indicates that the modulus of the laminate is linear in the rubberodulus. Thus, in the imposed stress condition the interply shear strain is an

nverse function of the rubber modulus.As an example of how these results could be used, consider that belts are

ypically designed to a specified strength. The strength is specified in N/m,hich represents the breaking strength of a lamina loaded in the cable directioner unit width. For a given steel material the strength can be calculated based onhe cable diameter and cable pace. Then these two parameters become coupled.igure 10 plots lines of constant belt strength versus interply shear strain andable diameter at imposed strain. Clearly there is an optimum cable diameter for

IG. 10 — Interply shear strain versus cable diameter, lines of constant belt strength. Strain ismposed at 1%. Note that for each belt strength there is an optimum cable diameter that willinimize the interply shear strain. The imposed stress case does not show defined minimums but

ndicates that small cables and high belt strength is the direction to lower interply shear strain.

particular belt strength that minimizes the interply shear strain.

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iscrete Finite Element Simulation

ackground

The analytical solution and the FE analysis to this point depend on homog-nization of the lamina. An important question is whether the analytical solutionr the homogenized FE analysis for that matter is indicative of the real stresseshat occur in the discrete, cable in rubber, physical belts. To try and answer thatuestion, an FE model will be developed that treats the cables and the rubberatrix as distinct continua.

odel Description

The entire model is shown in Fig. 11, while Fig. 12 shows a magnified viewf the internal construction of the model. The model is nominally 60 mm widend 180 mm long.

The model is constructed from three uniquely meshed layers. The uppernd lower reinforced layers are meshed with quadratic elements aligned diago-ally along the cable directions as shown. The rubber layer separating the rein-orced layers is also meshed with quadratic elements but they are aligned withhe transverse and longitudinal axes of the complete ply. The geometry of these

IG. 11 — Complete view of the half million degree of freedom discrete finite element model of awo-ply cable reinforced laminate.

hree layers is arranged such that all of the nodes lying in a plane that adjoins

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djacent layers are coincident. Figure 13 shows a small fragment of the modelhich highlights this arrangement.

Obviously, such an arrangement introduces a degree of discontinuity intohe computed displacement field, because elements faces of adjoining layers doot share common nodes. However, the overall stiffness of the individual layerss correct and thus the global behavior of the ply is captured. Further, at loca-ions where the nodes coincide, the displacements and resulting strain fields aremooth. Away from these locations the displacements across the layer bound-ries are discontinuous. The authors recognize that this mesh is a less-than-erfect representation of the structure. However, the results appear to show thator small deformations the arrangement is useful for insight into the mechanics.

aterialsDefinition of material constitutive laws are actually simplified in the dis-

rete model because there is no need to apply volume fraction techniques to theaterial moduli which describe the reinforced layers. For the rubber, a neo-ookean law was used where the initial tensile modulus was taken to be 6 MPa.ulk modulus properties were set such that Poisson’s ratio was equal to 0.495.

IG. 12 — Zoomed view of internal construction of discrete finite element model of a two-ply cableeinforced laminate.

As shown previously in Fig. 13, the cable bundles have been homogenized

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nto solid elements oriented along the cable direction. A variety of techniquesould be applied in the generation of material properties for the cables. In theresent case, we have chosen to define the cables as orthotropic with the mate-ial 11-direction aligned with the cable axis. This provides a simple frameworkor adjusting the material properties independently in the various material di-ections. For the cable orthotropic elasticity matrix, we have set E11,22,33

180000 MPa, E12,13,23=0 MPa, G11,22,33=1 MPa.With these properties the bending stiffness of the cable is very low. In fact

he cables behave essentially as Timoshenko beams. If Gij is reduced belowMPa, the model becomes numerically unstable. On the other hand, if Gij is

ncreased to 100 MPa, the cables behave essentially as classical beams �planeections remain planar and normal to the cable neutral fiber�.

oundary ConditionsReferring to Fig. 14, the axial strain is introduced into the ply by applying

isplacements onto the end faces of the rubber layer separating the reinforcedayers. The tendency of the ply to twist when subjected to the axial load isesisted by requiring the outer face of one of the reinforced layers to remain flat.he length of the model has been established such that the end effects of

oading are not seen at the center of the model. The main disadvantage of thiset of boundary conditions is that the longitudinal strain at the edges of the plys less than the longitudinal strains at the centerline. For the models used in this

IG. 13 — Zoomed view of internal element arrangement. The rubber layer separating the rein-orced layers is meshed with quadratic elements that are aligned with the transverse and longitu-inal axes of the complete ply. The geometry of these three layers are arranged such that all of theodes lying in a plane that adjoins adjacent layers are coincident.

tudy, the difference is nominally 20%.

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ominal Case Comparison with Homogeneous TheoryOur comparison with the homogeneous theory will involve looking at the

nterply shear strain in the rubber layer connecting the reinforced layers. Figure5 shows the two locations, A and B, at the midplane of the rubber layer wherehe interply shear strain will be evaluated as a function of lateral position.

Figure 16 shows the interply shear strain values for the nominal case wherehe cable diameter is 0.56 mm, cable pace is 1.6 mm, rubber layer thickness is.9 mm, and belt angle is 25°. The centerline axial strain is prescribed at 0.1%edge strain is therefore 0.08%�.

Recalling Eq �27�, the homogeneous theory indicates that the interply sheartrain is a linear function of �x. With this in mind, it seems reasonable to scalehe discrete model FE results to account for the reduced values of �x at the plydges. In Fig. 17 we see the comparison between the theoretical response andhe corrected discrete model response.

At this point we can make the following observations regarding the re-ponse predicted by the discrete FE model.

BC to maintain flatness

Axial Deformation

IG. 14 — Zoomed view of the discrete model boundary conditions. Axial strain is introduced intohe ply by applying displacements onto the end faces of the rubber layer separating the reinforcedayers. The tendency of the ply to twist when subjected to the axial load is resisted by requiring theuter face of one of the reinforced layers to remain flat.

IG. 15 — View of the locations, A and B, at the midplane of the rubber layer where the interply
hear strain will be evaluated as a function of lateral position.

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1. The mean discrete FE response �strain corrected� appears to agree withthe theoretical response at the ply edge.

2. The shear gradient with respect to width is much less in the discrete FEcase than the response predicted by theory.

3. The shearing response mentioned in point 2 allows the shear to persistfurther inboard in the ply than does the theoretical response.

Beyond the response described above, the discrete model also allows us tobserve the variation in interply shear strain between cables. Figure 18 showshe variation in strain �uncorrected� at the ply edge along the two paths indi-ated for the nominal case.

Clearly the discrete model, in addition to doing a reasonable job of captur-ng the homogenized prediction, is capable of capturing additional strain con-entrations of potential importance to tire design.

arameter Sensitivity ResultsGiven the variation in the interply shear strain indicated above, it should be

seful to observe the sensitivity of the belt edge interply shear strain to varia-ions in the key parameters of the model, namely, rubber layer thickness, cable

IG. 16 — Comparison of discrete model and homogeneous theory interply shear strain values forhe nominal case where the cable diameter is 0.56 mm, cable pace is 1.6 mm, rubber layer thicknesss 0.9 mm, and belt angle is 25°. The centerline axial strain is prescribed at 0.1% (edge strain isherefore 0.08%).

ace, and cable diameter.

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Figure 19 first shows the sensitivity of interply shear strain to variation inable diameter. All other parameters are held equal to the nominal case de-cribed above. Points labeled “max” correspond to position “B” in Fig. 15bove. Points labeled “min” correspond to position “A.” Also note that in allases the results have been corrected for 0.1% axial edge strain as outlinedbove.

For small cable diameters, the mean discrete model response agrees wellith the homogenized theory. However, as the cable diameter becomes rela-

ively large, the discrete model predicts more interply shear strain than theomogenized theory. It appears that the nonlinearity seen in the homogenizedesponse is exaggerated by the presence of the discrete cable.

The relative variation of the maximum and minimum strains can be clearlyeen in a contour plot �Fig. 20� of the strains at one of the ply edges. It is easyo see that an increase in the cable diameter causes an increase in the sheartrain at both points A and B.

Figure 21 shows the sensitivity of the interply shear strain to variation inable pace. Similar to the case of cable diameter shown above, agreementetween the mean discrete model response and the theoretical response is bettern the linear range and becomes poorer as the pace decreases and the homoge-

IG. 17 — Comparison of corrected discrete model and homogeneous theory interply shear strainalues for the nominal case where the cable diameter is 0.56 mm, cable pace is 1.6 mm, rubberayer thickness is 0.9 mm, and belt angle is 25°. The centerline axial strain is prescribed at 0.1%.

eous behavior becomes more nonlinear. The variation of the max and min,

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ShearStrain

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IG. 18 — Interply shear strain variation at the ply edge between cables and along the midplane.

IG. 19 — Sensitivity of interply shear strain to variation in cable diameter. All other parametersre held equal to the nominal case. Points labeled “max” correspond to position “B.” Pointsabeled “min” correspond to position “A.” Also note that all of the discrete model interply strains
ave been corrected as previously described.

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IG. 20 — Contour plots of the interply shear strain at one of the ply edges for three values of cableiameter. The centerline axial strain is prescribed at 0.1%. Deflections in the load direction haveeen amplified by 100. The minimum and maximum points are labeled “A” and “B,” respectively.

IG. 21 — Sensitivity of interply shear strain to variation in cable pace. All other parameters areeld equal to the nominal case. Points labeled “max” correspond to position “B.” Points labeledmin” correspond to position “A.” Also note that all of the discrete model interply strains have
een corrected as previously described.

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owever, exhibit quite a different response. In this case, the minimum strainalue �point A� decreases much more rapidly than the maximum strain valuepoint B�. Again, by looking at the contour plots of shear strain at the ply edgehown in Fig. 22, we can see that the strain maximum between the cablesemains relatively constant �point B� as the cable pace is reduced, but theinimum value �point A� increases more rapidly.

Finally, the plot in Fig. 23 shows the sensitivity of the interply shear straino variation in the rubber layer thickness. We now see reasonably good match ofhe mean to the theoretical homogeneous response, but the maximum and theinimum values diverge as the rubber layer thickness is decreased. Looking at

he contour plots in Fig. 24, we can see that the shear strain tends to concentrateetween the cables for small rubber layer thickness, while for larger rubberayer thickness the overall variation in the strain is greatly reduced relative tohe mean.

onclusions

The analytical solution developed here matches very well homogenized FEodels of angled ply belts. Experience shows that those models are relevant to

eal world tire durability behavior. The conclusion, then, is that the analytical

IG. 22 — Contour plots of the interply shear strain at one of the ply edges for three values of cableace. The centerline axial strain is prescribed at 0.1%. Deflections in the load direction have beenmplified by 100. The minimum and maximum points are labeled “A” and “B,” respectively.

olution should be valuable in tire design due to its ease of use and ability to

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IG. 23 — Sensitivity of interply shear strain to variation in rubber layer thickness. All otherarameters are held equal to the nominal case. Points labeled “max” correspond to position “B.”oints labeled “min” correspond to position “A.” Also note that all of the discrete model interply
trains have been corrected as previously described.
IG. 24 — Contour plots of the interply shear strain at one of the ply edges for three values ofubber layer thickness. The centerline axial strain is prescribed at 0.1%. Deflections in the loadirection have been amplified by 100. The minimum and maximum points are labeled “A” and “B,”
espectively.

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xpose the entire design space. The optimum belt parameters can easily beetermined for any proposed functioning mode.

The discrete FE models give trends and mean values that generally supporthe homogenized analytical solution. The stress field in the discrete model isuch more complicated, as might be expected. Local stress concentrations are

een to significantly exceed the stresses predicted by the analytical model, es-ecially in the nonlinear part of the design space. Further work on discreteodels is warranted.

eferences

1� Puppo, A. H., and Evensen, H. E., “Interlaminar Shear in Laminated Composites Under Gen-eralized Plane Stress,” Journal of Composite Materials, Vol. 4, 1970, pp. 204–220.

2� Turner, J. L., and Ford, J. L., “Interply Behavior Exhibited in Compliant Filamentary Compos-ite Laminates,” Rubber Chemistry and Technology, Vol. 55, 1982, pp. 1078–1094.

3� Kant, T., and Swaminathan, K., “Estimation of Transverse/Interlaminar Stresses in LaminatedComposites—A Selective Review and Survey of Current Developments,” Composite Struc-tures, Vol. 49, 2000, pp. 65–75.

4� Mittelstedt, C., and Becker, W., “Free-Edge Effects in Composite Laminates,” Applied Me-chanics Reviews, Vol. 60, Issue 5, 2007, pp. 217–245.

5� Jones, R. M., Mechanics of Composite Materials, Hemisphere Publishing Corp., 1975.

http://dx.doi.org/10.1177/002199837000400206

http://dx.doi.org/10.1115/1.2777169

http://dx.doi.org/10.1115/1.2777169

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2008 - Belt Angle Stress Paper - Mc Ginty et al

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