Chapter 2: Elastic Constitutive Equations of a Laminate

Home , Constitutive equation, Orthotropic material, Transverse isotropy

Chapter 2: Elastic Constitutive Equations of a Laminate 2.0 Introduction • Equations of Motion • Symmetric of Stresses • Tensorial and Engineering Strains • Symmetry of Constitutive Equations

2.1 Three-Dimensional Constitutive Equations • General Anisotropic Materials • Orthotropic Materials • Transversely Isotropic Materials • Isotropic Materials

2.2 Relation Between Mathematical & Engineering Constants • Isotropic Materials • Orthotropic Materials

2.3 Constitutive Equations for an Orthotropic Lamina • Plane Strain Condition • Plane Stress Condition

2.4 Constitutive Equations for an Arbitrarily Oriented Lamina • Coordinate Transformation • Stress Transformation • Strain Transformation • Stiffness and Compliance Matrix Transformation

2.5 Engineering Constants of a Laminate • Lamina • Laminate

2.6 Hygrothermal Coefficients of a Lamina

2.7 Summary 2.0 INTRODUCTION

x2 u 2 2.0.1 Equations of Motion of Elastic Solids

xP(,xxx123 , ) • Equations of Equilibrium (Kinetics) 2 x1 u1 u σρ += f i ij , = 123 , , ij, j i ∂ t 2 x σ 3 u3 22 ∂ x2 u2 σ σ 12 23 σ • Equations of Kinematics σ 12 32 σ σ 11 31 (strain-displacement)

σ x u ε =+12(uu) 33 1 1 ij i,, j j i ε 22 x3 u3 x u ε 2 2 21 • Constitutive Equations (stress-strain) ε 23 ε 12 σε= ε ε ij Cijklijkl kl , , , = 123 , , 31 ε 11 ε 13 ε 32 x1 u1 33

x3 u3 2.0.2 Symmetry of Stresses

Consider a plane 1-2. Equilibrium x σ 2 22 σσσσ∗∗− ∗∗+ ∗∗− ∗∗= σ in x1 1111110tttt 11 21 21 21 σ σσ−∗∗−−∗∗= σσ 12 in x2 ()()22 22110tt 12 12 1 σ σ σσ∗∗− ∗∗= 11 11 Moment about A: 12110tt 21 1 ∴=σσ σ 12 21 12 A σσ= . Similarly we can show, from 2-3 plane 23 32 σ σσ= 21 σ 1-3 plane 13 31 22 x σσ= 1 Therefore, ij ji ij , = 123 , , Stress tensor is Symmetric. Tensorial and Contracted Notation Tensorial Contracted σ σ σ11 σ1 σ22 σ2 στ33 = στ3 23 23 = 4 or 4 στ = = στ or στ31= 31 στ5 5 12 12 = 6 or 6 2.0.3 Tensorial and Engineering Strains

x2 Tensorial Strains: ε ε =+12(uu) 21 ij i,, j j i ε ε = 12 iiuij i, i = normal strains. 1 ε =+12(uu) ij≠ tensorial shear strain. ij i,, j j i 1 ε Engineering shear strain 12 .A ε γεε=+=(uu +)= Total shear strain 21 ij ij ji i,, j j i x1 ε Engineering Strains 22 εγ==u2 +u3 u 44∂ εε==1 x ∂x x u ε 111∂ ∂ 3 2 2 2 21 x1 ∂ ε ∂ u u 23 ε u2 εγ==3 +1 12 εε==∂ 55∂ ε 222 x ∂x ε 11 x2 ∂ 1 3 31 ε ∂ ∂ ε 13 u u u 32 εε==∂ 3 εγ==∂ 1 +2 333 66 ∂ x u x3 x2 x1 ε 1 1 ∂ ∂ 33 ∂

x3 u3 Generalized Hooke’s Law (3-D Constitutive Equation) σε Stress-Strain Equation iijj = C i, j = 1,2,3,4,5,6

σ ε 1 CCCCCC11 12 13 14 15 16 1 σ ε 2 CCCCCC21 22 23 24 25 26 2 σ CCCCCC ε 3 = 31 32 33 34 35 36 3 4 τ CCCCCC41 42 43 44 45 46 4γ

5 τ CCCCCC51 52 53 54 55 56 5 γ

6 τ CCCCCC61 62 63 64 65 66 6 γ C is called the stiffness matrix. εσ Strain-Stress Equation iijj = S i, j = 1,2,3,4,5,6 ε σ 1 SSSSSS11 12 13 14 15 16 1 ε σ 2 SSSSSS21 22 23 24 25 26 2 ε SSSSSS σ 3 = 31 32 33 34 35 36 3 4γ SSSSSS41 42 43 44 45 46 4 τ

5 γ SSSSSS51 52 53 54 55 56 5 τ

6 γ SSSSSS61 62 63 64 65 66 6 τ

S is called the compliance matrix. 2.0.4 Symmetry of Constitutive Matrix 1 Strain energy density, U = σε - - - -(1) 0 2 ii 1 UC= εε 0 2 ij j i σ ==∂ U0 i Cij j ∂εi ε ∂ 2U 0 = C ∂ε ∂ε ij ji 1 Eqn.(1) can be written as U = σε 0 2 jj 1 UC= εε 0 2 ji i j σ ==∂ U0 j Cji i ∂εj ε ∂ 2U 0 = C ∂ε ∂ε ji ij

Since the order of differentiating a scalar quantity U0 shouldnot change the result. Therefore, Cij = Cji .Stiffness matrix is symmetric.

Similarly, Sij = Sji 2.1 3-D CONSTITUTIVE EQUATIONS (a) General Anisotropic Material (no plane of material symmetry). σ ε 1 1 CCCCCC11 12 13 14 15 16 σ2 ε2 CCCCCC21 22 23 24 25 26 σ3 3ε = CCCCCC31 32 33 34 35 36 4 τ 4γ CCCCCC41 42 43 44 45 46 5 τ 5 γ CCCCCC51 52 53 54 55 56 6 τ 6 γ CCCCCC61 62 63 64 65 66 • Number of unknowns = 6x 6 = 36 • Because symmetry of Cij, number of unknowns = 6x 7/ 2 = 21 (b) Specially Orthotropic Materials (3 mutually perpendicular planes of material symmetry). Reference coordinate system is parallel to the material coordinate system. σ ε 1 1 C11 σ2 ε2 C21 C 22 Sym σ3 3ε = CCC31 32 33 4 τ 4γ 000C44 5 τ 5 γ 0000C55 6 τ 6 γ 0000 0C66 Number of unknowns = 9 Features

• No interaction between normal stresses (σ1, σ2, σ3) and shear strains (γ4, γ5, γ6 ). Normal stresses acting along principal material directions produce only normal strains.

• No interaction between shear stresses (τ4, τ5, τ6) and normal strains (ε1, ε2, ε3). Shear stresses acting on principal material planes produce only shear strains.

• No interaction between shear stresses and shear strains on different planes. That is shear stress acting on a principal plane produces a shear strain only on that plane. (c) Transversely Isotropic Material An orthotropic material is called transversely isotropic when one of its principal plane is a plane of isotropy. At every point on this plane, the mechanical properties are the same in all directions.

(2-3): Plane of Isotropy

σ 1 C ε 11 1 σ 2 CC21 22 ε2 σ3 C12 C 23 C 22 Sym ε = − 3 CC22 23 4τ 000 γ 2 4 5 τ 000 0 C55 5 γ 6 τ 000 0 0C55 6 γ Number of unknowns = 5 (d) Isotropic Material A material having infinite number of planes of material symmetry through a point. σ  1  C11  1   σ      2  C12 C 11 Sym  2   σ3  CCC12 12 11  3    =  ε   4τ 000C44  4    ε      5 τ 0000C44 5     ε  6 τ  0000 0C44  6  CC− γ where C = 11 12 44 2 γ Number of unknowns = 2 γ Summary Material Independent Elastic constants 1. Anisotropic material 36 2. Anisotropic elastic materials 21 3. Orthotropic material 9 4. Orthotropic material with 5 transverse isotropy 5. Isotropic material 2 2.2 Relations Between Mathematical and Engineering Constants

(a) Isotropic Materials (E & ν) x2 ενσ=− 31/ E ενσ=− σ 21/ E 1 εσ= 11/ E x1 x3 Definition: σ /ε Elastic Modulus (E) = Stress/Strain = 1 1 ν ε /ε Poisson’s Ratio ( ) = - Transverse strain/Applied strain = - 2 1 x

3 ενσ 31 x =− 2 Shear Strains / Normal Strains

E εσ 11 =

nx in x in γ γ γ x in 6 5 4 − − − /

3 1 2 E ενσ 21

=−

− − σ

νσ νσ 1

Planes

/ σ E 1 1 1 / / E E

/ τ x

E 23 τ 2 x Applied Stresses 23 -x 1 / Shear stresses G

− σ −

νσ

νσ 2

σ τ / x E 31 2 2 2

3 τ / / - x / 31 E E G

σ − −

νσ 3 νσ

x τ

/ σ 1 12

- x 3 3 τ 3 12 / / / E E G 2 x2

Constitutive Equation σ 1

x1 εσ= iijjS x3

ε 1 −−νν  1   EEE000 1   νν   ε  −−1   2 EE E000 2    νν   −− 1  3ε   000 3    =  EEE  σ  εγ= 1 στ= 423 000G 00 423    σ   =εγ  1  στ=  5 31 0000G 0 5 31     σ  = 1 =  612εγ  00000G 612στ

−1 {}σε= []{} {}σε= []S {} or C Restrictions of Elastic Constants

E Shear modulus G = 21( + ν)

for Shear modulus to be positive, ν > - 1

E Bulk modulus K = 31( − 2ν)

for Bulk modulus to be positive, ν < 1/2

−1 < ν < 1/2 (b) Orthotropic Materials

x 2 E2

ενσ=− 31311/ E σ 1 ενσ=− 21211/ E x E εσ= 1 1 111/ E

x3 E3

Definition: σ /ε Elastic Modulus (E1) = Stress/Strain = 1 1 ν ε /ε Poisson’s Ratio ( 12) = - Transverse strain/Applied strain = - 2 1 x

3 ενσ Normal Strains 31311 =− Shear Strains x nx in x in nx in 2

− − −

γ γ γ 1 εσ 3 2 6 5 4 111 / = E

/ − − σ

νσ νσ 11 Planes

311 1 1 13 1 12

/ τ

σ 323 23 ενσ 1 21211 x / / /

=− G τ E E 2 23 -x Shear stresses 3,

Applied Stresses τ

131 31 − σ −

x E

νσ

3 νσ 22 /

τ - x σ G 322 2 23 2 2 21 / 31 2 E x 1 1 / /

E E τ x 212 12 1

- x τ / 12

G σ − −

νσ 33 νσ 233 3 3 32 3 31 /

E σ 3 / / E E from Symmetry of S- matrix: Constitutive Equation Therefore

{}

σε {}

        

εσ

εγ εγ εγ SS 423 612 5 221 12 =

= ε = = ε ε [] 2 1 3 [] = S

S νν EE 31 ij − i {} 1          = {} This is the well known Betti’s reciprocal law of orthotropic material properties. orthotropic material law of known Betti’s reciprocal This is the well =           ji

− −−

ν νν 1 12 13 SS or EEE 0000 000 00000 331 13 11

EEE EE EE or 11 122 11

ν ν ij ji −

= ν 1 23 21

E {} σε i j 22 233 22 SS 332 23 = That is = [] − −

SS ν ν 1 32 jji ij 31 {} =

33 33 νν EE 12 1 1 G 000 000 000 = 23 when i 21 2 1 , G 00 ≠ 31

j νν EE 13 1 1 = G 0 12 31           3         

, and

στ στ στ

423 612 5

σ σ σ = = = 2 1 3

31 νν EE 23          2 = 32 3 Stress-Strain Equation Coefficients ofCoefficients C are given by: Where

{} Where σε CG C C C 11 33 22 423 44 = ∆= = = = = [] C 1 1 1 − [] − EE − EE EE

νν {}

νν νν 23 12 EE EE 13 332 23 221 12 331 13 123 , 1 = ∆ ∆ ∆ [] −− −− −

CG νν νν 1 1 232 12 323 13 55 =

−− νν 1 13 131 21 , C C C 13 12 23 and ==

== ==

ν νννννν ν νννννν ν νννννν 31 23 12 13 11 23 13 21 21 31 12 32 EEE E EE EE EE EEE EE ++ ++ ++ 12 23 13 CG 612 66 ∆∆ ∆∆ ∆∆ = 21 32 13 12 12 32 21 31 32 31 21 23 23 13 12 − Transversely Isotropic Material (:Plane x23 x ) = EE23 = GG12 13 νν= 12 13 E G = 2 23 + 21( 23) ν

Restrictions on Elastic Constants of Orthotropic Materials From Energy Principles, Lempriere showed that the Strain Energy is Positive if the Stiffness and Compliance Matrices are Positive Definite.

Mathematical Argument

(a) If only one stress is applied at a time, then the work done is positive if and only when the corresponding direct strain is positive. That is when > Sii 0 > Therefore: E1231223,,, E E G , G , and G 13 0 (b) Under suitable constraints, it is possible to deform a body in one-direction. > Then the work done will be positive if only when Cii 0 −νν =>1 23 32 C11 ∆ 0 EE23 −>νν νν < 1023 32 or 23 32 1

or ν < E2 23 E3

ν < Ei In general ij Ej

Note all through ∆ was assumed to be greater than 0. This condition would give additional equations. (refer to R. M. Jones.) 2.3 Constitutive Equations of a Thin Orthotropic Lamina

Two-Dimensional Bodies: Variation in stress and strain can be defined by two-coordinates. There are two types of problems. (a) Plane strain - Thick bodies εγ=== γ 0 zxzyz xyv2,, ∴==ττ xzw3,, xz yz 0

(b) Plane Stress - Thin bodies xxu1,, στ=== τ zxzyz0 ∴=εγ = γ = zxzyz0 Strain-Stress Equation:  ε  SS 0   xyv,,  1  11 12  1  2 ε =   xzw,,  2  SS21 22 0  2  3   σ   γ   00S   12 66 σ12

Or σ xxu1,, −ν  12   ε  1 0   1  EE1ν 1  1   −   =  12 1   ε2  0  2   EE12  σ   1    12γ   00  12   G12  σ σ Stress -Strain Equation: σ ′′ σ  QQ 0    1   EE1120  1   1  11 12  1       σ =   σ = ′′  2  QQ21 22 0  2  Or  2  EE12 2 0  2     ε     ε   σ   00Q    σ   00G   12 66 12ε 12 12 12ε

γE E ν γ E′ = 1 ′ = 2 21EE 1 12 2 Where: 1 1−νν E2 1−νν E′ ==−−ν 12 21 12 21 12 11νν12 21 12 21 νν 2.4 Stress-Strain Relations for Arbitrary Orientation of a Lamina y (a) Transformation of coordinates P(x,y) − xx12- Material coordinate system y xy− - Reference coordinate system x x1 x2 Consider a point P(x,y), its coordinates in x P(x,y) − xx12system is x1 x=+ xCosθθ ySin 1 y =− + θ θ x2 xSinθ yCos x2 x or αα x   Cosθθ Sin x x1  11xyx  1  =     =    −  or x  ααy x2   Sinθθ Cos y 2  22xy α Direction cosine matrix ij where I = 1,2 and j = 1, 2 (b) Stress Transformation

We use tensors transfer stresses between the two coordinate systems σαασ= ij ik jl kl ij,,= 12 and kl,,= xy

Example: i=j=1 σ αασ=+++ αασ αασ αασ 11 11 11xx 11 12 xy 12 11 yx 12 12 yy

If m = Cos θ and n = Sin σσ=+22θ σσ + 11 mmnnxxThen2 xy yy

Similarly we can establish the other two stress components. Finally we can write

σ   mn222 mn   11    xx  =−22  σ22   nm2 mn yy    −−22σ   12σ   mn mn m n  xy  σ

{}σσ−−= []T {} xx12 xy σ or −1 {}σσ−−= []Tσ {} []TTσ = [] xy x12 x where []Tσ - is the stress transformation matrix. mn22−2 mn −1  22  []TTσ = []=−[] T()θ = nm2 mn  22 mn−− mn m n  (c) Strain Transformation

{}εε x−− yTenTσ x x Ten = 12 []{}

 100  11      {}ε − = []Tσ 010  x yTen   ε 22 1    00 2   ε 12 Eng

{}εε−−= []THσ []{} γ x yTen x12 x Eng  100 − {}εε= []1[][]{} −1   x− yEngHTHσ x− x Eng []H = 010 12    002

{}εε−−= []Tε {} x yEng x12 x Eng

Where strain transformation matrix is:

 mn22− mn   = 22 []Tε  nm mn  22mn−− mn m22 n  Let us start with stress equation Let {} [] {} {}

QTQT

{} σσ σε σε

(d) Stiffness Transformation σε xy xy yxy xy xy xy ======[] [] [] [] [] QT TQ QT TQ TQ T

[]

σσ

σε σσ σ σ Q [] [] [] xy xy xy [] {} 12 {} 12 12 − − − 12 12 [] [] [] xy − −

{} T ε − T 1 {} in x - y coordinate system 12 {} − xy

{} {}

σε εε 12 12 − − = = [] []

Q T ε 12 − − 1 {} {} 12 xy −   QQQxx xy xs QQ11 21 0      = [] []T QQQyx yy yx  TσσQQ12 22 0  T     QQQsx sy ss   00Q66  where = E11 Q11 − νν 1 12 21

= E22 Q22 − νν 1 12 21

ν = 12EE 22 = 21 11 Q12 − −ν 1112νν 21 12 21 νν

QG66 = 12 Elements pf [Q]xy matrix =++4 4 22 +22 QmQnQmnQmnQxx 11 22 2412 66 =+4 4 +22 +22 QnQmQmnQmnQyy 11 22 2412 66 =+++22 22 44 − 22 Qxy mnQ11 mnQ22 m( n Q)12 mnQ4 66

=−+−3 3 33 +−33 Qxs m nQ11 mn Q22 mn( m n Q)12 mn2( m n Q 66) =−+−3 3 33 +−33 Qys mnQ11 mnQ22 mn( mnQ12) mn2( mnQ66) =+−22 22 22 +−222 Qss mnQ11 mnQ22 mnQ2 12 m( n Q)66 [] Notice in the Q xy matrix • It is fully populated - means normal-shear coupling. • Although 4 independent constants were used; we have ‘6’ unknowns. (e) Compliance Matrix {}      {}

{}

εε

εσ

ε ε xy x y yxx x xy yxxxy x x x xy yx xy xy xy      = = = = = [] [] [] [] T      ST TS TS

SSS ε SSS SSS

εσ ε xy ys xs yy xy yx xx xs ss sy sx {} [] [] −− {} xxx xx 12 12 212 12 − − −−

[] {} σ      −     

σ σ

σ {} where xx yy xy      − [] STST xy −− =

[] εε [] 12 [] T =++4 4 22 +22 SmSnSmnSmnSxx 11 22 2 12 66

=+4 4 +22 +22 yySnSmSmnSmnS11 22 2 12 66

=+++22 22 44 −22 Sxy mnS11 mnS22 m( n S12) mnS66

=−+−3 3 33 +−33 Sxs 22 m nS11 mn S22 2 mn( m n S12) mn( m n S 66 )

=−+−3 3 33 +−33 Sys 222 mnS11 mnS22 mn( mnS12) mn( mnS66 )

=+−+−22 22 22 222 Sss 448 mnS11 mnS22 mnS12 m( n S66) 2.5 Engineering Constants of an Arbitrarily Oriented Laminate

Arbitrarily Oriented Lamina σ Let us examine what happens when you apply x in x direction. We get ... y ε x - in x ε ε y y - in y

ε γ x and shear strain, xy γ xy σ θ = x ε = σ x x Ex or x εx Ex

ν =−ε y ενε=− =− xy ∴ = 1 = 1 xy yxyxν x Sxx Ex x Ex E S ε σ x xx γ xy ν Shear coupling coefficient η = =− xy ν =− xs Syx xySE yx x εx Ex γηε==xs or xy xs x η x η Ex = xs η = σ Ssx xsSE sx x Ex Shear Coupling Coefficients:

1 1 η ->Ratio of shear strain γ S = E = xs xy xx x ε Ex Sxx to normal strain x due to applied σ . ν x =− xy ν =−SE Syx xy yx x η ε Ex sx ->Ratio of normal strain x γ to shear strain xy due to η applied τ . S = xs η = SE xy sx E xs sx x x η η Similarly we have: ys, sy

  1 − yx sx  ν   ε   EEGx y η xy    x    xx  = − xy 1 sy  εy   νη yy     EEGxyxy σ   γxy   xy   ys 1  xsη σ  η   EEGx yxy σ Engineering Constants of an Arbitrarily Oriented Lamina

2 2 22 1 =−m 22νν+−n 22 mn+ (mn12 ) nm( 21 ) Ex E 1 E 2 G 12 n 2 m 2 mn22 1 =−nm( 22νν)+−mn(22 )+ E E 12 E 21 G y 1 2 12 222 mn22 mn22 mn( − ) 14=+( νν)++4 ( )+ 1 12 1 21 Gxy E 1 E 2 G 12 νν 2 2 22 xy ==yx m 2 −2 +−n 2 2 mn+ mn( 12 ) nm( 21 ) EEx y E 1 E 2 G 12 νν 33 ηη mn( m− n ) sx xs 22mn 22 mn 22 ==mn( −12 )−−nm( 21 )+ GE12x E 1 E2 G12 νν ηη mn33 mn− sy ys 22mn 22 mn 22 ( ) ==nm( −12 )−−mn( 21 )+ GE12y E 1 E2 G12 νν Variation of Ex and Ey with Fiber Angle

Material: E1 = 10E2 & G12=0.45 E2 12 ν 12=0.35 and E2 = 2 Msi

8 Ex Ey E2 E 6 2

0 0 102030405060708090 Angle θ Variation of Gxy with Fiber Angle

0.8

Gxy 0.6 E2

0.4

0.2

0 0 102030405060708090 Angle θ Variation of ν xy with Fiber Angle

0.6

0.5

0.4 ν xy 0.3

0.2

0.1

0 0 153045607590 Angle θ η η Variation of xs and ys with Fiber Angle

0.5

0 η ys -0.5

-1 η xs -1.5

-2

-2.5 0 153045607590 Angle θ Engineering Constants of a Laminate

• N-Layers • Each Layer can have different Thickness, Orientation, and Material

N = Tt∑ i i=1

{}σε= []{} {}σε= 1 []{} Stress-Strain in ith Layer i C i i av T ∫ Cdz Assumption: Strain is constant through out the laminate {}σε= []dz{} Average Stress in the laminate is: av ∫ C t  N  = 1 [] {}ε T ∑ Cti i  i=1  {}σε= []{} {}εσ= []S {} av C av or av xy− av For 3-D model stress-strain are six For 2-D model stress-strain are three Engineering Constants are: S = 1 = 1 ν =− yx Ex Gyz v xy S Sxx S44 xx = 1 = 1 =−Szx Ey Gxz v νxz Syy S55 Sxx = 1 = 1 S Ez Gxy v =− zy Szz S66 yzν Syy

MmLamCode: micromechanics and laminate analysis unidirectional code mmTEXlam: micro and laminate analysis of textile fabric composite code 2.6 Hygrothermal Coefficients of a Lamina

2.6.1 Coefficients of Thermal Expansion

(a) Isotropic Materials

y Original

b’ b x l

l’ Expanded due to ∆T

T TTll' − Coefficient of thermal expansion, ααα=== x y lT∆ Units: in/in/oF or m/m /oC (B) Orthotropic Materials

x2 Deformed

b’ b

l x1 l’

Original

Coefficient of thermal expansion α T = ll' − In x1-direction 1 lT∆ − In x -direction α T = bb' 2 2 b∆T  T  α 1  {}ε = T ∆ Thermal strains: α2  T    0  2.6.2 Coefficients of Moisture Expansion

All organic composites absorbs moisture. The absorption depends on the relative humidity to which it is exposed and its moisture content. For a given RH, temperature, and atmospheric pressure composite will have a saturation value. This is moisture content that the material will reach, if it is exposed for a very long time. This is a fixed value for a material. The moisture content is expressed as percent change in weight of the material. Like thermal expansion,increase in moisture would also expands the material. The orthotropic materials have two coefficients of moisture expansion, one along the fiber and the other across the fiber. x 2 Deformed Change in moisture ∆Μ

b’ b

l x1 l’

Original Coefficient of moisture expansion −  M  β T = ll' 1 In x1-direction β  1 lM∆ ε M = M ∆ Moisture strains: {} 2  M In x -direction β T = bb' − β  2 2 bM∆  0  2.6.3 Coefficients of Thermal & Moisture Expansion for Lamina in Arbitrary Orientation y x 2 x1 Recall the strain transformation:

T T {}εε= []Tε {} xy 12−

x Where  mn22− mn  22  []Tε =  nm mn  22 22mn−− mn m n  m=cos θ and n=sin Thermal strains in x-y due to ∆T are: θ  ε   mn22− mn T  α T   x    1   x  = 22 T ∆ = T  εy   nm mn 2  T  αy     −−22α   T   xyγ  22mn mn m n  0  αxy  α Coefficients of thermal expansion in x-y: αααTTT=+2 2 ααTT=+2 2 α T αααTTT=− x mn1 2 y nm1 2 xy 2mn()12 Coefficients of thermal expansion in x-y: αααTTT=+2 2 x mn1 2 ααTT=+2 2 α T y nm1 2 αααTTT=− xy 2mn()12

Coefficients of moisture expansion in x-y: βββMMM=+2 2 x mn1 2 MM=+2 2 M ββy nm1 β2 MMM=− xyβββ2(mn 12 ) Summary

2.0 Introduction • Equations of Motion • Symmetric of Stresses • Tensorial and Engineering Strains • Symmetry of Constitutive Equations