material modelling for design

Analysis of an Orthotropic PlyAnalysis of an Orthotropic Ply

By : Shambhu KumarReg.:-2014DN06

Introduction

STRESS-STRAIN RELATIONSHIP AND ENGINERRING CONSTANTS

HOOK’S LAW AND STIFFNESS AND COMPLIANCE MATRICEs

Transformation of Engineering Constant

Transformation of stiffness and Compliance Matrices

References

A single layer of laminated composite material generally is referred to as a ply or lamina.

It usually contains a single layer of reinforcement ,unidirectional or multidirectional .

Their properties and behaviour are controlled by their microstructure and properties of their constituents.

From the mechanism standpoint , fiber composites are among the class of materials called orthotropic materials whose behaviour lies between that of isotropic and that of aniisotropic materials.

Consider rectangular specimens made of isotropic ,anisotropic, and orthotropic materials.

Isotropic material is direction -independent and is characterised by ‘normal stresses produce normal strains only but no shear strain” and shear stress produces shear

strains only but no normal strains.”strains only but no normal strains.”

Deformation response of an orthotropic material , in general ,is similar to that of anisotropic material . That is, it is in direction –dependent , and normal strain as well as shear strains.

Consider a two-dimensional orthotropic lamina ,these constants are the elastic moduli in the longitudinal and transverse directions EL and ET respectively ,the shear modulus of rigidity associated with the axes of symmetry GLT,

and major Poisson’s ratio ν T L , which is gives longitudinally stress causes by transverse stress. stress causes by transverse stress.

T

L

specially orthotropic lamina

1. (σT= τLT=0 , σL ≠0 )

εL= σL/EL …….(1)εT = -ν LT . εL= -ν LT . σL/εL …(2)γLT = 0 …..(3)

2. (σL= τLT=0 , σT ≠0)εT= σT/ET ……..(4)εL = - ν T L . εT = - ν T L . σT/ET ..(5)

γLT = 0 ……. (6)γLT = 0 ……. (6)3. (σL= σT=0, τLT ≠0)

εL=0 …..(7)εT=0….(8)γLT=τLT / GLT ….(9)

/

32

12

31

23

33

22

11

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

232123132332231223312323233323222311

332133133332331233313323333333223311

222122132232221222312223223322222211

112111131132111211311123113311221111

33

22

11

21

13

CCCCCCCCCCCCCCCCCCCCCCCCCCC

CCCCCCCCCCCCCCCCCC

212121132132211221312123213321222111

132113131332131213311323133313221311

322132133232321232313223323332223211

122112131232121212311223123312221211

312131133132311231313123313331223111

.............................

21

13

32

12

31

23

y

x

y

x

SSSSSS

SSSSSS

SSSSSS

262524232221

161514131211

Same form as anisotropic, with 36 coefficients, but 9 are independent as with specially orthotropic case

xy

zx

yz

z

xy

zx

yz

z

SSSSSS

SSSSSS

SSSSSS

SSSSSS

666564636261

565554535251

464544434241

363534333231

3121

1 2 3

3212

1 11 2 3

2 213 23

10 0 0

10 0 0

10 0 0

E E E

E E E

2 213 23

3 31 2 3

23 23

2331 31

12 12

31

12

10 0 0

10 0 0 0 0

10 0 0 0 0

10 0 0 0 0

E E E

G

G

G

1 1

2 2

3 3

10 0 0

10 0 0

10 0 0

E E E

E E E

E E E

3 3

4 4

5 5

6 6

10 0 0 0 0

10 0 0 0 0

10 0 0 0 0

E E E

G

G

G

Use 3-D equations with,

023133

Plane stress,Plane stress,

0,,, 1221

Or

0,,, xyyx

12

2

1

66

2221

1211

12

2

1

00

0

0

S

SS

SS

12

2

1

66

2221

1211

12

2

1

00

0

0

Q

QQ

QQ

5 Coefficients - 4 independent

0

0

2

1

2221

1211

2

1

QQ

QQ

2200

0

12

2

66

2221

12

2

/Q

QQ

1 SQwhere

22 111 2

12 2111 22 12 1

S EQ

S S S

12 12 212 2

12 2111 22 12 1

S EQ

S S S

11 2S EQ 11 222 2

12 2111 22 12 1

S EQ

S S S

12

66

66

1G

SQ

coordinate axis x or y related to the four independent

engineering constant(EL , ET , GLT & ν T L ) for lamina.

yT L

x

normal stresses σxx, and σyy can be written as:-

σL = σxx cos2 θ + σyy sin2 θ + 2τxy cos θ sin θ …(10)

σT = σ sin2 θ + σ cos2 θ + 2τ cos θ sin θ ……….(11)

Using similar approach, we can also write the equation for shear stress as:

τLT = ‐σxx cos θ sin θ + σyy cos θ sin θ + τxy cos2 θ sin2 θ….(12)τLT = ‐σxx cos θ sin θ + σyy cos θ sin θ + τxy cos2 θ sin2 θ….(12)

Eqs. 10-12, can also be written in matrix form as

13

Similar equations can also be used to transform strains from one

coordinate system to another one. The strain transformation equations

are:-

14

1515

Stress –transformation

matrix

that we have relations which can be used to transform strains from one system to other, we proceed to develop relations which will help us transform engineering constants. Pre‐multiplying by eqn.. (13) [T]^-1 transform engineering constants. Pre multiplying Eq. (13) by [T] on either sides, we get:

[T]^‐1{σ} = [T]^‐1 [T] {σ} or {σ} = [T]‐1{σ} (16) [T]^‐1{σ}L‐T = [T]^‐1 [T] {σ}x‐y or {σ}x‐y = [T]‐1{σ}L‐T …(16)

where, {σ}L‐T and {σ}x‐y are stresses measured in x‐y, and L‐T reference frames, respectively

{σ}x‐y = [T]^‐1 [Q] [T]{ε}x‐y or, {σ}x‐y = [Q]{ε}x‐y …(17)

Equation 17 helps us compute stresses measured in x‐y coordinate system in terms of strains strains measure measure in the same system. Here, [Q] is the transformed stiffness matrix, and its individual components are:

Q11 = Q11 cos4θ + Q22 sin4θ + 2(Q12+2Q66) sin2θ cos2θ

Q22 = Q11 sin4θ + Q22 cos4θ + 2(Q12+2Q66) sin2θ cos2θ

Q12 = (Q11 + Q22 ‐ 4Q66)sin2θ cos2θ + Q12 (cos4θ + sin4θ)

Q66 = (Q11 + Q22 ‐ 2Q12 ‐ 2Q66)sin2θ cos2θ + Q66 (cos4θ + sin4 Q θ) 66 (Q11 Q22 2Q12 2Q66)sin θ cos θ Q66 (cos θ sin θ)

Q16 = (Q11 ‐ Q22 ‐ 2Q66)sinθ cos3θ ‐ (Q22 ‐ Q12 ‐ 2Q66 )sin3θ cos θ

Q26 = (Q11 ‐ Q22 ‐ 2Q66)sin3θ cos θ ‐ (Q22 ‐ Q12 ‐ 2Q66 )sin θ cos3θ

…………………………( 18)

Using a transformation procedure similar to the one used to transform stiffness matrix [Q], we can also transform the compliance matrix [S] to an arbitrary arbitrary coordinate coordinate system. system. The elements elements of transformed transformed compliance compliance matrix [S] are defined below:

S11 = S11 cos4θ + S22 sin4θ + (2S12 + S66) sin2θ cos2θ

S22 = S11 sin4θ + S22 cos4θ + (2S12 + S66) sin2θ cos2θ

S12 = (S11 + S22 ‐ S66)sin2θ cos2θ + S12 (cos4θ + sin4θ)

S66 = 2(2S11 + 2S22 ‐ 4S12 ‐ S66)sin2θ cos2θ + S66 (cos4θ + sin4 S θ) 66 = 2(2S11 + 2S22 4S12 S66)sin θ cos θ + S66 (cos θ + sin θ

)

S16 = 2(2S11 ‐ 2S22 ‐ S66)sinθ cos3θ ‐ 2(2S22 ‐ 2S12 ‐ S66 )sin3θ cos θ

S26 = 2(2S11 ‐ 2S22 ‐ S66)sin3θ cos θ ‐ 2(2S22 ‐ 2S12 ‐ S66 )sin θ cos3θ ….(19)