chapter 2research.iaun.ac.ir/pd/atrian/pdfs/uploadfile_6849.pdf · question: for a graphite/epoxy...

58

Chapter Chapter 22 MacromechanicalMacromechanical Analysis of a LaminaAnalysis of a Lamina

•Stress, Strain•Stress–strain relationships for different types of materials•Stress–strain relationships for a unidirectional/bidirectional lamina

Islamic Azad University, Najafabad Branch, A. Atrian

Islamic Azad University, Najafabad Branch 59

MacromechanicalMacromechanical Analysis of a LaminaAnalysis of a [email protected]@gmail.com

Stress on an arbitrary surface

1.Stress and Strain:

Applied loads on an infinitesimal surface on y-z plane


Stress state at any point can berepresented by 9 components, writingequations of equilibrium:

MacromechanicalMacromechanical Analysis of a LaminaAnalysis of a Lamina

It means that just with 6 independentcomponents, one can show stress state atany point:


MacromechanicalMacromechanical Analysis of a LaminaAnalysis of a LaminaStrain:


And, so on:


1 ( )2

ji

ijj i

uux x

1 12 2

1 12 21 12 2

x xy xz

yx y yz

zx zy z


2.Stress & Strain Transformation:



3.Stress-Strain Relations (Constitutive Equations or Hook’s Law):


Stiffness Matrix Compliance Matrix

There are 36 constants relating stress to strain components whichreduce to 21 constants due to the symmetry of the stiffnessmatrix [C].


Question: Prove that stiffness and compliance matrix are symmetric?

Stress–strain relationships for different types of materials:A. Anisotropic MaterialsB. Monoclinic MaterialsC. Orthotropic MaterialsD. Transversely Isotropic MaterialsE. Isotropic Materials



A. Hook’s law for anisotropic (triclinic) materials: 21 independent constants

There are not any planes of symmetry for the material properties in this kind of materials.

It’sIt’s clearclear thatthat subjectingsubjecting anisotropicanisotropic materialsmaterials toto uniaxialuniaxial stressstress makesmakes thethebodybody underunder bothboth axialaxial andand shearshear strainsstrains (Coupling)(Coupling)



B. Hook’s law for monoclinic materials: 13 independent constants

If there is one plane of material property symmetry (for example plane of symmetry is z=0 or 1-2 plane). E.g Feldspar

MaterialMaterial symmetrysymmetry impliesimplies thatthat thethe materialmaterial andand itsits mirrormirror imageimage aboutabout thetheplaneplane ofof symmetrysymmetry areare identicalidentical




Deformation of a cubic element made of monoclinic material under uniaxial stress in 3-direction(plane 1-2 is plane of symmetry)

• Apply tensile stress in 3-direction:Using the Hooke’s law Equation and thecompliance matrix for the monoclinicmaterial, one gets:


C. Hook’s law for orthotropic materials: 9 independent constants

If there are two orthogonal planes of material property symmetry for amaterial, symmetry will exist relative to a third mutually orthogonal plane


AA unidirectionalunidirectional laminalamina asas aa orthotropicorthotropic materialmaterialwithwith fibers,fibers, arrangedarranged inin aa rectangularrectangular arrayarray.. AAwoodenwooden bar,bar, andand rolledrolled steelsteel areare otherother examplesexamples..

Islamic Azad University, Najafabad Branch, A. Atrian 70

Note that, for orthotropic, there is nono interactioninteraction between normal stressesand shearing strains such as occurs in anisotropic materials (no(no coupling)coupling)


Deformation of a cubic element made of orthotropic material

• Thus, the cube will not deform inshape under any normal loadapplied in the principaldirections. This is unlike themonoclinic material, in whichtwo out of the six faces of thecube changed shape.

• A cube made of isotropicmaterial would not change itsshape either; however, thenormal strains, ε1 and ε2, willbe different in an orthotropicmaterial and identical in anisotropic material.


D. Hook’s law for transversely isotropic materials: 5 independent constants

If at every point of a material there is one plane in which the mechanical properties are equal in all directions (or if one of 3 symmetry plane is isotropic)


AA unidirectionalunidirectional laminalamina asas aa transverselytransversely isotropicisotropicmaterialmaterial withwith fibersfibers arrangedarranged inin aa squaresquare arrayarray (plane(plane 22––33isis thethe planeplane ofof isotropy)isotropy)

72

E. Hook’s law for isotropic materials: 2 independent constants (E, v)

If there is an infinite number of planes of material property symmetry

Question: Why for isotropic material:




Example: Show that there are 13 constants for monoclinic materials in stress–strain relations?

Solution:

If direction 33 is perpendicular to the plane of symmetry. Now, in thecoordinate system 1–2–3, strain-stress relation with Cij = Cji is:


1


Also, in the coordinate system 1′–2′–3′:

Because there is a plane of symmetry normal to direction 3, the stresses andstrains in the 1–2–3 and 1′–2′–3′ coordinate systems are related by


2

3


Substituting eq.3 in eq.5:

Subtracting eq.6 from eq.4:

Because and are arbitrary:

Similarly, one can show that:


1 4

6

52

Thus, only 13 independentelastic constants arepresent in a monoclinicmaterial.


Example: Show that there are 9 constants for orthotropic materials in stress–strain relations?

?



4.Stiffnesses, compliances, & engineering constants fororthotropic materials:

9 constants in these materials are:33 elasticity (Young) moduli E1, E2 & E3 in 1, 2 & 3 directions (material axis)33 shear moduli G12, G23 & G31 in the 1-2, 2-3 & 3-1 planes33 Poisson’s ratio v12, v23 & v31


4-2

4-1


Prove:

Apply uniaxial tensile stress

So:

The Young’s modulus in direction 1, E1, is defined as:

The Poisson’s ratio, ν12 & v13 are defined as:

Note: νij is defined as the ratio of the negative of the normal strain indirection j to the normal strain in direction i, when the load is applied in thenormal direction i.



Similarly for uniaxial tensile stresses in 2 & 3 directions…

Then apply

So:

The shear modulus in plane 2–3 is defined as:

And, similarly:



Note: When an orthotropic material is stressed in principal materialcoordinates (the 1, 2, and 3 coordinates), it doesdoes notnot exhibitexhibit either shear-extension or shear-shear coupling.

Due to symmetry of stiffness & compliance matices (Sij=Sji), last 1212 obtainedconstants will be reduced to 99:

((Betti’sBetti’s law)law)


4-3

82

Obtaining stiffness matrix from compliance matrix for orthotropic materials: Because these two matrices are mutually inverse, using matrix algebra

where,


4-4

4-5

4-6

4-7



5.Restrictions on engineering constants:

A. Isotropic Materials:

When an isotropic body is subjected to hydrostatic pressure( ):


5-1

Volumetric strain 5-2

5-4

Bulk modulus

K is positive. if it wasnegative, a hydrostaticpressure would causeexpansion of a cube ofisotropic material.

5-3


B. Orthotropic Materials:Based on the first law of thermodynamicsfirst law of thermodynamics, the stiffness and the compliance matrix must be positive definite (i.e, have positive principal values):

the positive nature of the stiffnessess leads to:


5-5

5-7

5-6

5-8

5-9 5-10


In order to obtain a constraint on one Poisson's ratio, Poisson's ratio, vv2121, , in terms of twoothers, v32 and v13, from Eq.(5-12):


5-7

5-11

5-12

5-13


Applications & importance of these restrictions on engineeringconstants for orthotropic materials

1. To optimizeoptimize propertiesproperties of a composite because they show that the nineindependent properties cannot be varied without influencing the limits of theothers.

2. To examineexamine experimentalexperimental datadata to see if they are physically consistentwithin the framework of the mathematical elasticity model.



Example 5-1: Dickerson and DiMartino test: They measured Poisson's ratiosfor boron-epoxy composite materials as high as 1.97 for the negative of thestrain in the 2-direction over the strain in the 1-direction due to loading in the1-direction (v12). The reported values of the Young's moduli for the twodirections are E1=11.86 X 106 psi (81.77 GPa) and E2=1.33 x 106 psi (9.17GPa).


(For isotropic materials this ratio is nearly 1)

1.97 < 2.99: v12 is a reasonable numbereven though our intuition based onisotropic materials (v < 1/2) rejects sucha large number.

Also, the 'converse‘ (or minor) Poisson's ratio, v21' wasreported as 0.22. This value satisfies the reciprocal relations ineq. (5-9).


Example 5-2: Find the compliance and stiffness matrix for a graphite/epoxylamina. The material properties are given as

Solution:



Solution (continue):


(or by using eq. 4-6)


6. STRESS-STRAIN RELATIONS FOR PLANE STRESS IN ANORTHOTROPIC MATERIAL

Note: A plane stress state on a lamina isis notnot merelymerely anan idealizationidealization ofreality, but instead is a practicalpractical andand achievableachievable objectiveobjective of how we must usea lamina with fibers in its plane.


Plane stress conditions for a thin plate Unidirectionally Reinforced Lamina

6-1


A unidirectionally reinforced lamina mainly can withstand in plane loads especiallyin its fibers direction. So, it would need 'help''help' carrying in-plane stressperpendicular to its fibers, but that help can be provided by other (parallel) layersthat have their fibers in the direction of the stress using laminate

For orthotropic materials, imposing a statestate ofof planeplane stressstress results in impliedoutout--ofof--planeplane strainsstrains of:

The normal strain, ε3, is notnot anan independentindependent strainstrain because it is a function ofthe other two normal strains, ε1 and ε2. Therefore, the normal strain, ε3, canbe omitted from the stress–strain relationship. Also, the shearing strains, γ23

and γ31, can be omitted because they are zero:


6-2

6-3

92


6-6 6-7

6-4

Reduced stiffnesses for a plane stress state in the 1-2plane which are determined by:1. as the components of the inverted compliance matrix ineq. (6.3) or2. from the Cij directly by applying the condition to thestrain-stress relations to get an expression for andsimplifying the results to get:

6-5



Relationship of Compliance and Stiffness Matrix to EngineeringElastic Constants of a Lamina:

For a unidirectional laminaunidirectional lamina, these engineering elastics constants are:E1= longitudinal Young’s modulus (in direction 1)E2= transverse Young’s modulus (in direction 2)V12= major Poisson’s ratioG12= in-plane shear modulus (in plane 1–2)



The unidirectionalunidirectional oror bidirectionalbidirectional laminalamina is a specially orthotropic laminabecause normal stresses applied in the 1–2 direction do not result in anyshearing strains in the 1–2 plane because Q16 = Q26 = 0 = S16 = S26. Also,the shearing stresses applied in the 1–2 plane do not result in any normalstrains in the 1 and 2 directions because Q16 = Q26 = 0 = S16 = S26.


(Minor Poisson’s Ratio)

This equation is verified…


For planeplane stressstress on isotropicisotropic materialsmaterials:

Question: Obtain stiffness & compliance matrices for orthotropic & isotropic materials under plane stress condition?

Question: Show that for an orthotropic material Q11 ≠ C11. Explain why. Also, show Q66 = C66. Explain why


6-8 6-9


Question: For a graphite/epoxy unidirectional lamina, find the following1. Compliance matrix2. Minor Poisson’s ratio3. Reduced stiffness matrix4. Strains in the 1–2 coordinate system

Use the properties of unidirectional graphite/epoxy lamina from next table.



7. STRESS-STRAIN RELATIONS FOR A LAMINA OF ARBITRARY ORIENTATION


1-2 coordinate system: local axes: material axes1-direction: Longitudinal direction L2-direction: Transverse direction T

x–y coordinate system: global axes: off-axes



7-1

7-3

The transformation matrix7-2

7-4

&

Reuter Matrix

7-5


The elements of the transformed reduced stiffness matrix are given by:


7-6

7-3

7-5

7-7

Note: The transformed reduced stiffness matrix has terms in all ninepositions in contrast to the presence of zeros in the reduced stiffnessmatrix. However, there are still only fourfour independentindependent material constantsbecause the lamina is orthotropic



Note:- From Eqs. (6-3) & (6-4), for a unidirectional lamina loaded in the material axesdirections, no coupling occurs between the normal and shearing terms of strainsand stresses (specially orthotropic).- However, for an angle lamina, from Eqs. (7-6) & (7-8), coupling takes placebetween the normal and shearing terms of strains and stresses.- Therefore, Equation (7-6) & (7-8) are stress–strain equations for what is calleda generally orthotropic lamina.

7-8

7-9

Transformed compliance matrix:

102

MacromechanicalMacromechanical Analysis of a LaminaAnalysis of a LaminaNote:- The only advantage associated with generally orthotropic as opposed toanisotropic laminae is that generally orthotropic laminae areare easiereasier totocharacterizecharacterize experimentallyexperimentally.- In fact, there is no difference between solutions for generally orthotropiclaminae and those for anisotropic laminae whose stress-strain relations,under conditions of plane stress, can be written as:

where the anisotropic compliances in terms of the engineering constantsare:

7-11

7-12

7-10



MacromechanicalMacromechanical Analysis of a LaminaAnalysis of a LaminaAnisotropic elasticity relations:1.LekhnitskiiLekhnitskii coefficients or mutual influence coefficients or shearshear--extensionextensioncoupling coefficients:

2.ChentsovChentsov coefficients or shearshear--shearshear coefficients (not for in-planeloading):

7-15

7-14

7-13


Reciprocal relation for Chentsov coeff. (like Betti’s law):

3. The out-of-plane shearing strains of an anisotropic lamina due to in-planeshearing stress and normal stresses are:

Note: Neither of these shear strains arise in an orthotropic material unless itis stressed in coordinates other than the principal material coordinates.


7-17

7-16


The apparent engineering constants apparent engineering constants for an orthotropic lamina that is stressed in non-principal x-y coordinates (from Eqs. 7-9 & 7-11 & 7-12):

Question: Obtain Eqs. (7-18)?


7-18


From Eqs. (7-18):


Normalized Moduli for Boron-Epoxy

Normalized Moduli for Glass-Epoxy

These results were summarized by JonesJones as a simple theorem: theextremum (largest and smallest) material properties do not necessarilyoccur in principal material (1-2) coordinates: Vxy or Gxy


Visual interpretation:Why Gxy>G12 & E1>E45 for a composite material with fiber modulus muchgreater than matrix modulus?



Know more…


Auxetic (Negative Poisson’s Ratio) Structures

Re-entrant

Chiral

Higher Poisson’s Ratio

Dickerson & Di Martinohave published data forcross-plied boron-epoxycomposites in which thePoisson's ratios rangefrom 00..024024 toto 00..878878 inthe orthotropic case andfrom --00..414414 toto 11..9797 for a+/-25 ° laminate

Variation of Poisson's ratio with orientation for the woven roving laminate panel P. D. Craig & J. Summerscalest Experiments


Example 7-1: Find the following for a 60° angle lamina of graphite/epoxy.Use the properties of unidirectional graphite/epoxy lamina from the lasttable.1. Transformed compliance matrix2. Transformed reduced stiffness matrixIf the applied stress is σx=2 MPa, σy=–3 MPa, and τxy=4 MPa, also find3. Global strains4. Local strains5. Local stresses6. Principal stresses7. Maximum shear stress8. Principal strains9. Maximum shear strain



Solution:1: From example 5-2, then using Eq.(7-9):



2: by inverting Transformed compliance matrix or using Eq.(7-7):

3: from Eq.(7-8):


112

4: using Eq.(7-4):

5: using Eq.(7-1):




6:

7:



8:

9:


Note: The axes of principal normal stressesand principal normal strains do not match,unlike in isotropic materials.


Example 7-2: A uniaxial load is applied to a 10° ply. The linear stress–straincurve along the line of load is related as σx = 123εx, where the stress ismeasured in GPa and strain in m/m. Given E1 = 180 GPa, E2 = 10 GPa andν12 = 0.25, find the value of(1)shear modulus, G12(2) modulus Ex for a 60° ply.

Solution:



8. Strength and Failure Criteria for An Orthotropic Lamina

• A successful design of a structure requires efficient and safe use ofmaterials.• Theories need to be developed to compare the state of stress in amaterial to failure criteria.• It should be noted that failure theories are only stated and theirapplication is validated by experiments.• The first step in such a definition is the establishment of allowablestresses or strengths in the principal material directions.

•• ForFor aa laminalamina stressedstressed inin itsits ownown plane,plane, therethere areare fivefive fundamentalfundamentalstrengthsstrengths::

XtXt = axial or longitudinal strength in tension= axial or longitudinal strength in tensionXcXc= axial or longitudinal strength in compression= axial or longitudinal strength in compressionYtYt = transverse strength in tension= transverse strength in tensionYcYc = transverse strength in compression= transverse strength in compressionS S = shear strength= shear strength



Experimental Determination of Strength and Stiffness:


Or

Stiffness

Strength


Measuring E1 & V12 & X:Uniaxial tension loading in the 1-direction


ASTM D 638 tension test specimen

119

Measuring E2 & V21 & Y:Uniaxial tension loading in the 2-direction




Measuring G12:Uniaxial tension loading at 45° to the 1-direction


G12


Measuring G12 & S:Torsion tube test


Sandwich crossSandwich cross--beam beam test (By Shockey and described by Waddoups)

Rail shearRail shear test (By Whitney, Stansbarger, & Howell)“It’s widely used in aerospaceaerospaceindustry”


Maximum Stress Failure Theory:


Maximum normal stress theory by RankineRankine

Maximum shearing stress theory by TrescaTresca

Like theories for isotropic materialsLike theories for isotropic materials

oo TheThe stressesstresses actingacting onon aa laminalamina areare resolvedresolved intointo thethe normalnormalandand shearshear stressesstresses inin thethe locallocal axesaxes..oo FailureFailure isis predictedpredicted inin aa lamina,lamina, ifif anyany ofof thethe normalnormal oror shearshearstressesstresses inin thethe locallocal axesaxes ofof aa laminalamina isis equalequal toto oror exceedsexceeds thethecorrespondingcorresponding ultimateultimate strengthsstrengths ofof thethe unidirectionalunidirectional laminalamina..oo GivenGiven thethe stressesstresses oror strainsstrains inin thethe globalglobal axesaxes ofof aa lamina,lamina,oneone cancan findfind thethe stressesstresses inin thethe materialmaterial axesaxes byby usingusingEquationEquation ((77--11))oo ThesesTheses 55 criteriacriteria actact independentlyindependently andand don’tdon’t interactinteract withwiththethe othersothers..

8-1


Example 8-1: Find the max. value of S>0 if a stress of σx=2S, σy=–3S, andτxy =4S is applied to the 60° lamina of graphite/epoxy. Use maximum stressfailure theory and the properties of a unidirectional graphite/epoxy laminagiven in Table of page 97.

Solution:


(7-1)

Table p.93

(8-1)

• The preceding inequalities also show that the angle lamina will fail in shear.• The maximum stress that can be applied before failure is:


Strength Ratio:• In a failure theory such as the maximum stress failure theory of Section, it can bedetermined whether a lamina has failed if any of the inequalities of Equation (8-1) areviolated.• However, this does not give the information about how much the load can be increasedif the lamina is safe or how much the load should be decreased if the lamina has failed.



Examples 8-2: Assume that one is applying a load of σx=2MPa, σy=−3MPa,τxy=4MPa to a 60° angle lamina of graphite/epoxy. Find the strength ratiousing the maximum stress failure theory.

Solution:If the strength ratio is R, then the maximum stress that can be applied is

σx=2R, σy=−3R, τxy=4RFollowing Example 8-1 for finding the local stresses gives

σ1=0.1714×10 R, σ2=−0.2714×10 R, τ12=−0.4165×10 RUsing the maximum stress failure theory as given by Equation (8-1) yields

R=16.33Thus, the load that can be applied just before failure isσx=16.33×2=32.66 MPa, σy=16.33×(−3)=-48.99 MPa, τxy=16.33×4=65.32 MPa

NoteNote thatthat allall thethe componentscomponents ofof thethe stressstress vectorvector mustmust bebe multipliedmultiplied byby thethe strengthstrengthratioratioIf SRSR >> 11: then the lamina is safe and the applied stress can be increased by a factor ofSR.If SRSR << 11: the lamina is unsafe and the applied stress needs to be reduced by a factor ofSR.SRSR == 11: implies the failure load.



Failure Envelopes:A failure envelope is a threethree--dimensionaldimensional plotplot of the combinations of the normal andshear stresses that can be applied to an angle lamina justjust beforebefore failurefailure.

DrawingDrawing threethree dimensionaldimensional graphsgraphs cancan bebe timetime consumingconsuming:: developingdeveloping failurefailure envelopesenvelopesforfor constantconstant shearshear stressstress ττxyxy ((22 normalnormal stressesstresses σσxx andand σσyy asas thethe twotwo axes)axes):: Then,Then, ifif thetheappliedapplied stressstress isis withinwithin thethe failurefailure envelope,envelope, thethe laminalamina isis safesafe;; otherwise,otherwise, itit hashas failedfailed..


Failure envelope for the 60° lamina ofgraphite/epoxy for a constant shear stress of τxy= 24 Mpa (Example p.145, Kaw)

Failure Criteria for Metals


Maximum Strain Failure Theory:


oo ThisThis theorytheory isis basedbased onon thethe maximummaximum normalnormal strainstraintheorytheory byby StSt.. VenantVenant andand thethe maximummaximum shearshear stressstresstheorytheory byby TrescaTresca asas appliedapplied toto isotropicisotropic materialsmaterials..oo TheThe strainsstrains appliedapplied toto aa laminalamina areare resolvedresolved totostrainsstrains inin thethe locallocal axesaxesoo FailureFailure isis predictedpredicted inin aa lamina,lamina, ifif anyany ofof thethe normalnormaloror shearingshearing strainsstrains inin thethe locallocal axesaxes ofof aa laminalamina equalequaloror exceedexceed thethe correspondingcorresponding ultimateultimate strainsstrains ofof thetheunidirectionalunidirectional laminalamina

8-2

NoteNote:: In fact, if the Poisson’s ratio is zero in the unidirectional lamina, thetwo failure theories will give identical results.


Example 8-3: Find the maximum value of S>0 if a stress, σx=2S, σy=–3S,and τxy=4S, is applied to a 60° graphite/epoxy lamina. Use maximum strainfailure theory. Use the properties of the graphite/epoxy unidirectional laminagiven in Table p.97

Solution:Considering “[S]” and “local stresses” from example 2.6 (Kaw) and example8-1 (here):



Assume a linear relationship between all the stresses and strains until failure;then the ultimate failure strains are

Using the inequalities (8-2) and recognizing that S>0:



oo TheThe maximummaximum valuevalue ofof SS beforebefore failurefailure isis 1616..3333 MPaMPa.. TheThe samesame maximummaximumvaluevalue ofof SS == 1616..3333 MPaMPa isis alsoalso foundfound usingusing maximummaximum stressstress failurefailure theorytheory..

oo ThereThere isis nono differencedifference betweenbetween thethe twotwo valuesvalues becausebecause thethe modemode ofof failurefailureisis shearshear.. However,However, ifif thethe modemode ofof failurefailure werewere otherother thanthan shear,shear, aa differencedifference ininthethe predictionprediction ofof failurefailure loadsloads wouldwould havehave beenbeen presentpresent duedue toto thethe Poisson’sPoisson’sratioratio effect,effect, whichwhich couplescouples thethe normalnormal strainsstrains andand stressesstresses inin thethe locallocal axesaxes..

oo NeitherNeither thethe maximummaximum stressstress failurefailure theorytheory nornor thethe maximummaximum strainstrain failurefailuretheorytheory hashas anyany couplingcoupling amongamong thethe fivefive possiblepossible modesmodes ofof failurefailure..

ooTheThe followingfollowing theoriestheories areare basedbased onon thethe interactioninteraction failurefailure theorytheory::Tsai-HillTsai-Wu



Tsai–Hill Failure Theory

-This theory is based on the distortiondistortion energyenergy failurefailure theorytheory ofof VonVon--MisesMises’’ distortionaldistortionalenergyenergy yieldyield criterioncriterion forfor isotropicisotropic materialsmaterials as applied to anisotropic materials.

-The strain energy in a body consists of two parts:1. Due to a change in volume and is called the dilationdilation energyenergy2. Due to a change in shape and is called the distortiondistortion energyenergy

-It is assumed that failure in the material takes place only whenwhen thethe distortiondistortion energyenergy isisgreatergreater thanthan thethe failurefailure distortiondistortion energyenergy of the material.

“Hill”“Hill” adopted the Von-Mises’ distortional energy yield criterion to anisotropic materials.“Tsai”“Tsai” adopted it to a unidirectional lamina. Based on the distortion energy theory, heproposed that a lamina has failed if below inequality is violated:


8-3



The components G1, G2, G3, G4, G5, and G6 of the strength criterion depend on thefailure strengths and are found as follows:

1. Apply to a unidirectional lamina; then, the lamina will fail. Thus, Equation(8-3) reduces to


3. Apply to a unidirectional lamina and, assuming that the normal tensilefailure strength is same in directions (2) and (3), the lamina will fail. Thus, Equation (8-3) reduces to


8-7

8-6

8-5

8-4



From Equation (8-4) to Equation (8-7):

Because the unidirectional lamina is assumed to be under plane stress (σ3=τ31=τ23=0),then Equation (8-3) reduces through Equation (8-8) to:

8-8

8-9


Example 8-4: Find the maximum value of S > 0 if a stress of σx = 2S, σy =–3S, and τxy =4S is applied to a 60° graphite/epoxy lamina. Use Tsai–Hillfailure theory. Use the unidirectional graphite/epoxy lamina properties givenin Table slide 97.

Solution:From example 8-1:

Using the Tsai–Hill failure theory from Equation (8-9):



Remarks:1. Unlike the maximum strain and maximum stress failure theories, the Tsai–Hill failure theory considers the interaction among the three unidirectional lamina strength parameters.

2. The Tsai–Hill failure theory does not distinguish between the compressiveand tensile strengths in its equations. This can result in underestimation ofthe maximum loads that can be applied when compared to other failuretheories. For the load of σx = 2 MPa, σy =–3 MPa, and τxy = 4 MPa, as foundin Example 8-2, Example 8-3, and Example 8-4, the strength ratios are givenbySR = SR = 1616..33 33 (maximum stress failure theory)(maximum stress failure theory)SR = SR = 1616..33 33 (maximum strain failure theory)(maximum strain failure theory)SR = SR = 1010..94 94 (Tsai(Tsai––Hill failure theory)Hill failure theory)

Tsai–Hill failure theory underestimates the failure stress because the transversetransverse tensiletensilestrengthstrength of a unidirectional lamina is generally much less than its transversetransversecompressivecompressive strengthstrength and the compressive strengths are not used in the Tsai–Hill failuretheory, but it can be modified to use corresponding tensile or compressive strengths inthe failure theory as follows


8-10

8-9

136

Where:

For Example 8-4, the modified Tsai–Hill failure theory given by Equation (2.10) nowgives:

the strength ratio is SRSR == 1616..0606 (modified(modified TsaiTsai––HillHill failurefailure theory)theory)




3. The Tsai–Hill failure theory is a unified theory and thus doesdoes notnot givegive thethemodemode ofof failurefailure like the maximum stress and maximum strain failure theoriesdo. However, one can make a reasonable guess of the failure mode bycalculating and . The maximum of these threevalues gives the associated mode of failure.In the modifiedmodified TsaiTsai––HillHill failurefailure theorytheory, calculate the maximum of |σ1/X1|,|σ2/Y|, and |τ12/S| for the associated mode of failure.



Tsai–Wu Failure Theory- This failure theory is based on the totaltotal strainstrain energyenergy failurefailure theorytheory ofof BeltramiBeltrami.- Tsai-Wu applied the failure theory to a lamina in planeplane stressstress.- A lamina is considered to be failed if below inequality is violated:

- This failure theory is moremore generalgeneral than the Tsai–Hill failure theory because itdistinguishes between the compressivecompressive and tensiletensile strengthsstrengths of a lamina.

The components H1, H2, H6, H11, H22, and H66 of the failure theory are found usingthe five strength parameters of a unidirectional lamina as follows:

1- Apply to a unidirectional lamina; the lamina will fail.Equation (8-11) reduces to:


8-11

8-12

139

2- Apply to a unidirectional lamina; the lamina will fail.Equation (8-11) reduces to

From Equation (8-12) and (8-13):

3- Apply to a unidirectional lamina; the lamina will fail. Equation (8-11) reduces to



8-13

8-15

8-14

8-16

140



5- Apply to a unidirectional lamina; it will fail. Equation (8-11) reduces to



8-17

8-18

8-19

8-20

141



And, from experiments:



8-21

8-23

8-22

8-24

142

Example 8-5: Find the maximum value of S > 0 if a stress σx=2S, σy=–3S,and τxy = 4S are applied to a 60° lamina of graphite/epoxy. Use Tsai–Wufailure theory.

Solution:From example 8-1:

From above equations:



143

Using the Mises–Hencky criterion for evaluation of H12:

Substituting these values in Equation (8-11):

Summarizing the four failure theories:




Problems Chapter 2:

1. Derive Equation (7-6).

2. Prove

3. Derive Equation (7-8).

4. Prove

5. Plot the apparent engineering constants Ex, Ey, Gxy, Vxy, & asfunctions of θ from θ=0 to θ=90 for high-modulus graphite-epoxy, anorthotropic material with E1=30 X 10^6 psi (207 GPa), E2=0.75 x 10^6psi (5.2 GPa), G12 =0.375 x 10^6 psi (2.59 GPa), and V12 =0.25.

6. Find the engineering constants of a 60° graphite/epoxy lamina. Use theproperties of a unidirectional graphite/epoxy lamina from Table of slide 95.



7. A 60° angle graphite/epoxy lamina is subjected only to a shear stress τxy= 2 MPa in the global axes. What would be the value of the strains measuredby the strain gage rosette — that is, what would be the normal strainsmeasured by strain gages A, B, and C? Use the properties of unidirectionalgraphite/epoxy lamina from table of slide 97.

8. What are the values of stiffness matrix elements C11 and C12 in terms of the Young’s modulus and Poisson’s ratio for an isotropic material?



9. Consider an orthotropic material with the stiffness matrix given by

Find:a. The stresses in the principal directions of symmetry if the strains in theprincipal directions of symmetry at a point in the material are ε1 = 1 μm/m,ε2 = 3 μm/m, ε3 = 2 μm/m; γ23 = 0, γ31 = 5 μm/m, γ12 = 6 μm/mb. The compliance matrix [S]c. The engineering constants E1, E2, E3, ν12, ν23, ν31, G12, G23, G31

10. Find the strains in the 1–2 coordinate system (local axes) in aunidirectional boron/epoxy lamina, if the stresses in the 1–2 coordinatesystem applied to are σ1 = 4 MPa, σ2 = 2 MPa, and τ12 = –3 MPa. Use theproperties of a unidirectional boron/epoxy lamina from Table of slide 97.



10.Find the reduced stiffness [Q] and the compliance [S] matrices for aunidirectional lamina of boron/epoxy. Use the properties of a unidirectionalboron/epoxy lamina from Table page 97

11. The reduced stiffness matrix [Q] is given for a unidirectional lamina isgiven as follows:

What are the four engineering constants, E1, E2, ν12, and G12, of the lamina?

12. Find the off-axis shear strength and mode of failure of a 60° boron/epoxylamina. Use the properties of a unidirectional boron/epoxy lamina from Tablepage 97. Apply the maximum stress failure and maximum strain failuretheories.



13. Find the maximum biaxial stress, σx = –σ, σy = –σ, σ > 0, that one canapply to a 60° lamina of graphite/epoxy. Use the properties of aunidirectional graphite/epoxy lamina from Table of page 97. Use maximumstrain failure theories.


149

• Concepts of volume and weight fraction• Elastic moduli for the composite

تحليل ميکرومکانيکي يک تک اليهتحليل ميکرومکانيکي يک تک اليه: : فصل سومفصل سومMicromechanical Analysis of a LaminaMicromechanical Analysis of a Lamina



1.Introduction, Definition:

• Macromechanics of a lamina discussed about the “apparent” properties of alamina.• In previous chapter, a large enough piece of the lamina was considered sothat the fact that the lamina is made of two or more constituent materialscannot be detected.• In previous chapter, we were able to say that a unidirectional lamina (boronfibers in epoxy) has certain stiffnesses and strengths that we measured invarious directions. but, we didn’t find out how can the stiffnesses & strengthsof a boron-epoxy composite material be varied by changing the proportion ofboron fibers to epoxy matrix?

Micromechanical Analysis of a LaminaMicromechanical Analysis of a Lamina

Basic Question of Micromechanics


Unlike in isotropic materials, experimental evaluation of parameters forunidirectional lamina is quite costly and time consuming because they arefunctions of several variables like:• The individual constituents of the composite material• Fiber volume fraction• Packing geometry• Processing• Etc

Thus

“The need and motivation for developing analytical models to find these “The need and motivation for developing analytical models to find these parameters are very important”parameters are very important”



Some important definitions:

•• VolumeVolume fractionfraction::

Consider a composite consisting of fiber and matrix:

•• MassMass fractionfraction (weight(weight fraction)fraction)::


fiber volume fraction

matrix volume fraction

1-3

1-2

1-1

1-4



mass fraction of the fibers

mass fraction of the matrix 1-6

1-7

1-8

1-5

1-9

1-11

1-10

1-121-4



Table 3-1



Table 3-2


Example 1: A glass/epoxy lamina consists of a 70% fiber volume fraction.Use properties of glass and epoxy from Table 3.1 & Table 3.2, respectively, todetermine the:1. Density of lamina2. Mass fractions of the glass and epoxy3. Volume of composite lamina if the mass of the lamina is 4 kg4. Volume and mass of glass and epoxy in part (3)

Solution:

1.


Table 3-1

Table 3-2

1-11


2.

3.

4.


0.7=0.8294


2.Evaluation of the Four Elastic Moduli:

there are four elastic moduli of a unidirectional lamina:

• Longitudinal Young’s modulus, E1• Transverse Young’s modulus, E2• Major Poisson’s ratio, ν12• In-plane shear modulus, G12


159

The objective of all micromechanics approaches is to determine the elasticmoduli or stiffnesses or compliances of a composite material in terms ofthe elastic moduli of the constituent materials:

Three approaches to determine the four elastic moduli:

A- Strength of materials approach

B-Semi-empirical models

C-Elasticity approach




3. Strength of materials approach:


-From a unidirectional lamina, take arepresentative volume element thatconsists of the fiber surrounded by thematrix.-The fiber, matrix, and the compositeare assumed to be of the same width,h, but of thicknesses tf , tm, and tc,respectively.-The area of the fiber, the matrix, & thecomposite:

3-1


The following assumptions are made in the strength of materials approachmodel:

• The bond between fibers and matrix is perfect.• The elastic moduli, diameters, and space between fibers are uniform.• The fibers are continuous and parallel.• The fibers and matrix follow Hook’s law (linearly elastic).• The fibers possess uniform strength.• The composite is free of voids.



3.1. Longitudinal Young’s Modulus


Assuming that the fibers, matrix, and composite follow Hooke’s law andthat the fibers and the matrix are isotropic, the stress–strain relationshipfor each component and the composite is:

3-2

3-3

3-4

(3-3) & (3-4) in (3-2)

3-63-5



The rule of mixturesThe rule of mixtures3-7

(3-3) & (3-4)3-8

Fraction of load of composite carried byfibers as a function of fiber volume fractionfor constant fiber to matrix moduli ratio.


Example 2: Find the longitudinal elastic modulus of a unidirectionalglass/epoxy lamina with a 70% fiber volume fraction. Use the properties ofglass and epoxy from Table 3.1 and Table 3.2, respectively. Also, find theratio of the load taken by the fibers to that of the composite.

Solution:

From Table 3.1 & 3.2:

From Eq. (3-7):

From Eq. (3-8):




Longitudinal Young’s modulus as function of fiber volume fraction and comparison withexperimental data points for a typical glass/polyester lamina. (Experimental data pointsreproduced with permission of ASM International.)


3.2. Transverse Young’s Modulus


3-9

3-10

3-11

by using Hook’s law3-12

(3-11) & (3-12) in (3-10) & using (3-9)3-13


Example 3: Find the transverse Young’s modulus of a glass/epoxy lamina with a fiber volume fraction of 70%. Use the properties of glass and epoxy from Table 3.1 and Table 3.2, respectively.

Solution:


Transverse Young’s modulus as a function of fiber volume fraction for constant fiber to matrix moduli ratio.



Theoretical values of transverse Young’s modulus as a function of fibervolume fraction for a Boron/Epoxy unidirectional lamina (Ef = 414 GPa,νf = 0.2, Em = 4.14 GPa, νm = 0.35) and comparison withexperimental values.

169

3.3. Major Poisson’s Ratio


The major Poisson’s ratio is definedas the negative of the ratio of thenormal strain in the transversedirection to the normal strain in thelongitudinal direction, when anormal load is applied in thelongitudinal direction…

Assume a composite is loaded in thedirection parallel to the fibers:

The deformations in the transverse direction of the composite

3-14




3-15

(3-15) in (3-14)3-16

3-17 3-18

3-19


3.4. In-Plane Shear Modulus


3-21

3-22

3-20

The resulting shear deformations

3-23

(3-20)(3-21)(3-22)

assumption


Example 4: Find the major and minor Poisson’s ratio and the in-plane shearmodulus of a glass/epoxy lamina with a 70% fiber volume fraction. Use theproperties of glass and epoxy from Table 3.1 and Table 3.2, respectively.

Solution:

From Table 3.1 & 3.2:

From Eq. 3-19:


(3-23)



Theoretical values of in-plane shear modulus as a function of fiber volume fraction andcomparison with experimental values for a unidirectional glass/epoxy lamina (Gf = 30.19GPa, Gm= 1.83 GPa).


4. Semi-empirical models


Poor agreement of transverse Young’s modulus (E2) and in-plane shear modulus (G12) with experimental results

Better modeling techniques: finite element, finite differenceand boundary element methods, elasticity solution, and variationalprincipal models

semi-empiricalmodels: Halphinand Tsai Model(simple equations by curve fittingto results that are based on elasticity)

Unfortunately, these models are available only as complicatedequations or in graphical form


4.1. Longitudinal Young’s Modulussame as that obtained through the strength of materials approach

4.2. Transverse Young’s Modulus


4-1 4-2

reinforcing factor•Fiber geometry•Packing geometry•Loading conditionsDepends on

square packing geometry

S For circular fiber

S

hexagonal packing geometry

For rectangular fiber

(a: length/ b: width)(b is in the direction of loading)



Concept of direction of loading for calculation oftransverse Young’s modulus by Halphin–Tsaiequations


Example 5: Find the transverse Young’s modulus for a glass/epoxy laminawith a 70% fiber volume fraction. Use the properties for glass and epoxyfrom Table 3.1 and Table 3.2 (slide 147-148), respectively. Use Halphin–Tsaiequations for a circular fiber in a square array packing geometry.

SolutionBecause the fibers are circular and packed in a square array, the reinforcingfactor ξ = 2.From Table 3.1: Ef = 85 GPa.From Table 3.2: Em = 3.4 GPa.From Equation (4-2)

From Equation (4-1), the transverse Young’s modulus of the unidirectionallamina is:

(This modulus using the mechanics of materials approach for the same problem, from Example 3, was found to be 10.37 Gpa)




Theoretical values of transverse Young’s modulus as afunction of fiber volume fraction and comparison withexperimental values for boron/epoxy unidirectional lamina(Ef = 414 GPa, νf= 0.2, Em = 4.14 GPa, νm = 0.35)

179

4.3. Major Poisson’s Ratiosame as that obtained through the strength of materials approach

4.4. In-Plane Shear Modulus


4-3 4-4

reinforcing factor•Fiber geometry•Packing geometry•Loading conditionsDepends on

square packing geometry

SFor circular fiber

S

hexagonal packing geometry

For rectangular fiber

(a: length/ b: width)(b is in the direction of loading)




Concept of direction of loading tocalculate in-plane shear modulus byHalphin–Tsai equations

Note: The value of ξ = 1 for circular fibers in a square array givesreasonable results only for fiber volume fractions of up to 0.5

Hewitt and Malherbe function 4-5


Example 6: Using Halphin–Tsai equations, find the shear modulus of aglass/epoxy composite with a 70% fiber volume fraction. Use the propertiesof glass and epoxy from Table 3.1 and Table 3.2, respectively. Assume thatthe fibers are circular and are packed in a square array. Also, get the value ofthe shear modulus by using Hewitt and Malherbe’s formula for the reinforcingfactor.

SolutionFor Halphin–Tsai’s equations with circular fibers in a square array ξ=1.From Example 4, Gf=35.42 GPa, Gm=1.308 GPa.

From Eq. (4-4):

From Eq. (4-3):

(This modulus using the mechanics of materials approach for the same problem, from Example 4, was found to be 4.013 Gpa)

Because the volume fraction is greater than 50%, Hewitt and Malherbesuggested a reinforcing factor




Theoretical values of in-plane shear modulus as a function of fibervolume fraction compared with experimental values forunidirectional glass/epoxy lamina (Gf = 30.19 GPa, Gm = 1.83GPa).


5. Elasticity approach

• Elasticity accounts for equilibrium of forces, compatibility, and Hooke’s lawrelationships in three dimensions• The strength of materials approach may not satisfy compatibility and/oraccount for Hooke’s law in three dimensions,• Semi-empirical approaches are just as the name implies — partly empirical

The elasticity models described here are called composite cylinderassemblage (CCA) models:

In a CCA model, one assumes the fibers are circular in cross-section, spread in a periodic arrangement, and continuous


Periodic arrangement of fibers in a cross-section of a lamina (CCA model)


The composite can be considered to be made of repeating elements calledthe representative volume elements (RVEs):

• The RVE is considered to represent the composite and respond the same as the wholecomposite does• Appropriate boundary conditions are applied to this composite cylinder based on theelastic moduli being evaluated


Composite cylinder assemblage (CCA) model usedfor predicting elastic moduli of unidirectionalcomposites

The representative volume elements (RVE)

5-1


5.1 Longitudinal Young’s Modulus

Apply an axial load, P, in direction 1

To find E1 in terms of elasticelastic modulimoduli ofof thethe fiberfiber andand thethe matrixmatrix, and thethegeometricalgeometrical parametersparameters suchsuch asas fiberfiber volumevolume fractionfraction, we need to relate theaxial load, P, and the axial strain, ∈1, in these terms:

Assuming the response of a cylinder is axisymmetric, the equilibriumequation in the radial direction is given by


5-2

5-3

radial stresshoop stress


The normal stress–normal strain relationships in polar coordinates, r–θ–z, foran isotropic material with Young’s modulus, E, and Poisson’s ratio, ν, aregiven by:

Note: The shear stresses and shear strains are zero in the r–θ–z coordinatesystem for axisymmetric response.


5-4


The strain-displacement equations for axisymmetric response are


5-6

(5-5) in (5-4)

& ∈z = ∈1

5-5

IAUN, Dr. A. Atrian 188

Substituting Equation (5-6) in the equilibrium equation (5-3) gives:

The solution to the linearlinear ordinaryordinary differentialdifferential equationequation is found by assumingthat


5-7

5-8

(5-7) in (5-8)

5-9

189

The preceding expression (5-9) requires that

Therefore, the form of the radial displacement is

The preceding equations are valid for a cylinder with an axisymmetricresponse. Thus, the radial displacement, uf and um, in the fiber and matrixcylinders, respectively, can be assumed of the form

However, because the fiber is a solid cylinder and the radial displacement uf isfinite, Bf = 0; otherwise, the radial displacement of the fiber uf would beinfinite. Thus


5-10

5-11

5-12 5-13

5-14



Differentiating Equation (5-13) and Equation (5-14) gives:

Using Equation (5-15) in Equation (5-6), the stress–strain relationships forthe fiberfiber and matrixmatrix are:


5-15

5-195-18

5-175-16


How do we now solve for the unknown constants How do we now solve for the unknown constants AAff, A, Amm, , BBmm, and ε, and ε11??The following four boundary and interface conditions will allow us to do that:

1-The radial displacement is continuous at the interface, r = a,

2- The radial stress is continuous at r = a:


From (5-13) & (5-14)5-20

From (5-16) & (5-18)

5-21


3- Because the surface at r = b is traction free, the radial stress on theoutside of matrix, r = b, is zero:

4- The overall axial load over the fiber-matrix cross-sectional area indirection 1 is the applied load, P, then:


Then, (5-16) gives5-22

Because the axial normal stress, σz, is independent of θ,

5-23

5-24


Then, from Equation (5-16) and (5-18):

Solving Equation (5-20), Equation (5-21), Equation (5-22), and Equation (5-25), we get the solution to Af, Am, Bm, and ε1.Using the resulting solution for ∈1, and using Equation (5-2):


5-25

5-26


Example 7: Find the longitudinal Young’s modulus for a glass/epoxy laminawith a 70% fiber volume fraction. Use the properties for glass and epoxyfrom Table 3.1 and Table 3.2, respectively. Use equations obtained using theelasticity model.

Solution:From Table 3.1 & 3.2

Using Equation (5-26), the longitudinal Young’s modulus:

For the same problem, the longitudinal Young’s modulus was found to be60.52 GPa from the mechanics of materials approach as well as the Halphin–Tsai equations



5.2 Major Poisson’s Ratio

5.3 Transverse Young’s Modulus

5.4 Axial Shear Modulus

Study from page Study from page 239239, , Mechanics of Composite Materials, Mechanics of Composite Materials, AutarAutar K. KawK. Kaw



Problems Chapter 3:

1. A hybrid lamina uses glass and graphite fibers in a matrix of epoxy for itsconstruction. The fiber volume fractions of glass and graphite are 40 and20%, respectively. The specific gravity of glass, graphite, and epoxy is 2.6,1.8, and 1.2, respectively. Finda. Mass fractionsb. Density of the composite

2.Determine the expression for the modulus of a composite material thatconsists of matrix material reinforced by a slab of constant thickness in thedirection in which the modulus is desired as in below Figure


3. A resin hybrid lamina is made by reinforcing graphitefibers in two matrices: resin A and resin B. The fiberweight fraction is 40%; for resin A and resin B, theweight fraction is 30% each. If the specific gravity ofgraphite, resin A, and resin B is 1.2, 2.6, and 1.7,respectively, finda. Fiber volume fraction b. Density of composite


4. Show that ,if the fibers are much stiffer than the matrix — thatis, Gf >> Gm

5. A unidirectional glass/epoxy lamina with a fiber volume fraction of 70% isreplaced by a graphite/epoxy lamina with the same longitudinal Young’smodulus. Find the fiber volume fraction required in the graphite/epoxylamina. Use properties of glass, graphite, and epoxy from Table 3.1 and Table3.2.

6. Sometimes, the properties of a fiber are determined from the measuredproperties of a unidirectional lamina. As an example, find the experimentallydetermined value of the Poisson’s ratio of an isotropic fiber from the followingmeasured properties of a unidirectional lamina:a. Major Poisson’s ratio of composite = 0.27b. Poisson’s ratio of the matrix = 0.35c. Fiber volume fraction = 0.65



7. Using elasticity model equations, find the elastic moduli of a glass/epoxyunidirectional lamina with 40% fiber volume fraction. Use the properties ofglass and epoxy from Table 3.1 and Table 3.2, respectively (slide 144).Compare your results with those obtained by using the strength of materialsapproach and the Halphin–Tsai approach. Assume that the fibers arecircularly shaped and are in a square array for the Halphin–Tsai approach.


199

• Code for laminate stacking sequence• Relationships of mechanical loads applied to a laminate to strains andstresses in each lamina.

تحليل ماکرومکانيکي يک چند اليهتحليل ماکرومکانيکي يک چند اليه: : فصل چهارمفصل چهارمMacromechanicalMacromechanical Analysis of a LaminateAnalysis of a Laminate



1. Introduction:

The two basic questions of laminate analysis are:1. what are the conditions that the laminae must meet to be a laminate?2. how Will a laminate respond to loading, i.e., imposed forces and

moments?

The main reason for lamination is to combine The main reason for lamination is to combine laminaelaminae to create a laminate to create a laminate for achieving the largest possible bending stiffness for the materials used.for achieving the largest possible bending stiffness for the materials used.

MacromechanicalMacromechanical Analysis of a LaminateAnalysis of a Laminate


2. Laminate Code:


[0/–45/90/60/30]: It consists of five plies, each ofwhich has a different angle to the reference x-axis. Aslash separates each lamina. The code also implies thateach ply is made of the same material and is of thesame thickness. Sometimes, mayalso denote this laminate, where the subscript T standsfor a total laminate.


MacromechanicalMacromechanical Analysis of a LaminateAnalysis of a [email protected]@gmail.com

IAUN, Dr. A. Atrian 203

3. Classical Lamination Theory (CLT):

3.1. 1D Linearly Elastic & Isotropic Beam Stress–Strain Relation


4-1

Distance z, from the centroidal line

the radius of curvature of the beam

4-2

4-3



This shows that, under a combined uniaxial and bending load, thestrain varies linearly through the thickness of the beam

curvature of the beam

4-4


3.2. Strain-Displacement Equations

Consider a plate under in-plane loads such as shear and axial forces, andbending and twisting moments:


u0,v0,w0: displacements in the x, y & zdirections, respectively, at the mid-plane.

u, v, w: the displacements at any point in theX, y, & z directions, respectively.


AssumptionsAssumptions::

1. Each lamina is orthotropic.2. Each lamina is homogeneous.3. A line straight and perpendicular to the middle surface remains straightand perpendicular to the middle surface during deformation

4. The laminate is thin and is loaded only in its plane (plane stress)

5. Displacements are continuous and small throughout the laminate, where h is the laminate thickness

6. Each lamina is elastic

7. No slip occurs between the lamina interfaces


207

At any point other than the mid-plane, the two displacements in the x–yplane will depend on the axial location of the point and the slope of thelaminate mid-plane with the x and y directions:

the displacement u in the x-direction is:

Similarly, taking a cross-section in the y–z plane would give the displacementin the y-direction as:


4-5

4-6

4-8

4-7



All these Eqs. Can be written in matrix form as:


the definitions of the mid-plane strains

the mid-plane curvatures

4-9

This eq. shows:• the linear relationship of the strains in alaminate to the curvatures of thelaminate.• the strains are independent of the xand y coordinates.• Also, note the similarity between Eq.(4.10) & Eq.(4.4), which was developedfor the one dimensional beam.

4-10


Example 4.1: A 0.010 in. thick laminate is subjected to in-plane loads. If themidplane strains and curvatures are given as follows, find the global strainsat the top surface of the laminate?

Solution:


4-10


3.3. Strain and Stress in a Laminate:

If the strains are known at any point along the thickness of the laminate, thestress–strain eq. calculates the global stresses in each lamina

The reduced transformed stiffness matrix, corresponds to that of the plylocated at the point along the thickness of the laminate…


4-11

(4-10) in (4-11)

4-12


It’sIt’s clearclear fromfrom eqeq.. ((44--1212))::• The stresses vary linearly only through the thickness of each lamina.• The stresses, however, may jump from lamina to lamina because thetransformed reduced-stiffness matrix changes from ply to ply because thismatrix depends on the material and orientation of the ply.



3.4. Force and Moment Resultants Related to Midplane Strains &Curvatures (how to find the midplane strains and curvatures if the loads applied tothe laminate are known)

Consider a laminate made of n plies:


213

The z-coordinate of each ply k surface (top and bottom) is given by

Ply 1:

Ply k: (k = 2, 3,…n – 2, n – 1):

Ply n:


4-13



Integrating the global stresses in each lamina gives the resultant forces perunit length in the x–y plane through the laminate thickness as

Similarly, integrating the global stresses in each lamina gives the resultingmoments per unit length in the x–y plane through the laminate thickness as


4-14

4-15


Which gives


4-16

(4-12) in (4-16)

4-18

4-17


Due to independency of midplane strains and plate curvatures to z coordinate & to be constant of the transformed reduced stiffness matrix for each ply:

Knowing that:


4-19

4-20


It gives:


4-22

4-21

4-23

Extensional MatrixExtensional Matrixrelating the resultant in-planeforces to the in-plane strains

Coupling MatrixCoupling Matrixcoupling the force and momentterms to the midplane strains and midplane curvatures.

Bending Stiffness MatrixBending Stiffness Matrixrelating the resultant bending moments to the plate curvatures


Combining these eqs. gives 6 simultaneous linear eqs. & six unknown as:


4-24


Steps for analyzing a laminated composite subjected to the appliedforces and moments:

1.Find the value of the reduced stiffness matrix [Q] for each ply using its fourelastic moduli, E1, E2, v12 & G12 (Eq.6-7 chap.2)

2. Find the value of the transformed reduced stiffness matrix for each plyusing the [Q] matrix calculated in step 1 and the angle of the ply (Eq.7-7Chap.2).

3. Knowing the thickness, tk, of each ply, find the coordinate of the top andbottom surface, hi, i = 1…, n, of each ply, using Eqs. (4-13).

4. Use the matrices from step 2 and the location of each ply from step 3 tofind the three stiffness matrices [A], [B], and [D] from Eqs. (4-23).

5. Substitute the stiffness matrix values found in step 4 and the appliedforces and moments in Eq. (4-24).



6. Solve the six simultaneous Eqs. (4-24) to find the midplane strains andcurvatures.

7. Now that the location of each ply is known, find the global strains in eachply using Eq. (4-10).

8. For finding the global stresses, use the stress–strain (Eq. 7-6 chap.2).

9. For finding the local strains, use the transformation (Eq. 7-4 chap.2).

10. For finding the local stresses, use the transformation (Eq. 7-1 chap.2).


221

Example 4.2: Find the three stiffness matrices [A], [B], and [D] for a three-ply [0/30/–45] graphite/epoxy laminate. Use the unidirectional propertiesfrom Table of slide 95 of graphite/epoxy. Assume that each lamina has athickness of 5 mm.

Solution:

the reduced stiffness matrix for the 0° graphite/epoxy ply is:




the transformed reduced stiffness matrix for each of the three plies is (Eq.7-7 chap.2):



the locations of the ply surfaces are (Eq. 4-13):



Example 4.3: For previous example if the laminate is subjected to a load of Nx = Ny = 1000 N/m, find:(a) Mid-plane strains and curvatures(b) Global and local stresses on top surface of 30° ply(c) Percentage of load, Nx, taken by each ply

Solution:


Solving this system of eqs. gives


The strains at the top surface of the 30° ply (z=h1=-0.0025 m) (Eq. 4-10):

The global stresses from Eq. 7-6 chap.2:

The local strains (Eq. 7-4 chap.2) and local stresses (Eq. 7-1 chap.2) in the30° ply at the top surface:




Global Strains (m/m)


MacromechanicalMacromechanical Analysis of a LaminateAnalysis of a LaminateGlobal Stresses (Pa)

Local Strains (m/m)


MacromechanicalMacromechanical Analysis of a LaminateAnalysis of a LaminateLocal Stresses (Pa)

The portion of the load, Nx, taken by each ply can be calculated byintegrating the stress, σxx, through the thickness of each ply. However,because the stress varies linearly through each ply, the portion of the load,Nx, taken is simply the product of the stress, σxx, at the middle of each plyand the thickness of the ply:


The sum total of the loads shared by each ply is 1000 N/m, (223.2+ 531.5 +245.2), which is the applied load in the x-direction, Nx.


231

Problems of Chap.4:

1.Condense (summarize) the following expanded laminate codes:a. [0/45/–45/90]b. [0/45/–45/–45/45/0]c. [0/90/60/60/90/0]d. [0/45/60/45/0]e. [45/–45/45/–45/–45/45/–45/45]

2.Expand the following laminate codes:




3. A laminate of 0.015 in. thickness under a complex load gives the followingmidplane strains and curvatures:

Find the global strains at the top, middle and bottom surface of the laminate.

4. Show that, for a symmetric laminate, the coupling stiffness matrix is equalto zero.



5. The global stresses in a three-ply laminate are given at the top and bottomsurface of each ply. Each ply is 0.005 in. thick. Find the resultant forces andmoments on the laminate if it has a top cross section of 4 in. × 4 in.

6. Demonstrate that the force per unit width on a two-layered laminate withorthotropic laminae of equal thickness oriented at +alpha and -alpha to theapplied force is

What are A11, A12 & B16 in terms of the transformed reduced stiffnessesof a lamina and the lamina thickness, t?


chapter 2research.iaun.ac.ir/pd/atrian/pdfs/uploadfile_6849.pdf · question: for a graphite/epoxy...

Documents