EML 4230 Introduction to Composite Materials

Chapter 4 Macromechanical Analysis of a Laminate

Hygrothermal Loads

Dr. Autar KawDepartment of Mechanical Engineering

University of South Florida, Tampa, FL 33620

Courtesy of the TextbookMechanics of Composite Materials by Kaw

Mechanical Strains

T

T

T

=

kxy

y

x

kxy

y

x

k

Txy

Ty

Tx

kxy

y

x

k

Mxy

My

Mx

Hygrothermal Stresses

k

Mxy

My

Mx

k662616

2621

161211

k

Txy

Ty

Tx

QQQ

QQQ

QQQ

=

22

Zero resultant stresses

02

2

dz =

τ

σ

σ

-h/

h/

Txy

Ty

Tx

011

dzτ

σ

σ

k =

n

Txy

Ty

Tx

k

h

h

k

k -

Zero resultant stresses

0 =dz

QQQ

QQQ

QQQ

h

h

1 = k

n

k

Mxy

My

Mx

k662616

262212

161211

1 - k

k

Deriving final formula

0 =dz

QQQ

QQQ

QQQ

h

h

1 = k

n

k

Txy

Ty

Tx

kxy

y

x

k662616

262212

161211

1 - k

k

0 =dz z +

QQQ

QQQ

QQQ

h

h

1 = k

n

k

Txy

Ty

Tx

xy

y

x

0xy

0y

0x

k662616

262212

161211

1 - k

k

Deriving final formula

dz

QQQ

QQQ

QQQ

h

h

1 = k

n

=dz z +

QQQ

QQQ

QQQ

h

h

1 = k

n

Txy

Ty

Tx

k662616

262212

161211

1 - k

k

xy

y

x

0xy

0y

0x

k662616

262212

161211

1 - k

k

Deriving final formula

dz T

QQQ

QQQ

QQQ

h

h

1 = k

n

=dz z +

QQQ

QQQ

QQQ

h

h

1 = k

n

kxy

y

x

k662616

262212

161211

1 - k

k

xy

y

x

0xy

0y

0x

k662616

262212

161211

1 - k

k

Deriving final formula

N

N

N

=

BBB

BBB

BBB

+

AAA

AAA

AAA

Txy

Ty

Tx

xy

y

x

662616

262212

161211

oxy

oy

ox

662616

262212

161211

)(1

1

662616

262212

161211

hh

α

α

α

QQQ

QQQ

QQQ

k =

nT =

N

N

N

] = N[ k-k

xy

y

x

k k

Txy

Ty

Tx

T

Other three equations – from zero resultant moments

M

M

M

=

DDD

DDD

DDD

+

BBB

BBB

BBB

Txy

Ty

Tx

xy

y

x

662616

262212

161211

oxy

oy

ox

662616

262212

161211

)(1

21 2

12

662616

262212

161211

hh

α

α

α

QQQ

QQQ

QQQ

k =

nT =

M

M

M

] = M[ k - k

xy

y

x

k k

Txy

Ty

Tx

T

Final formula

0

T

T

D| BB|A =

MN

)(1

1

662616

262212

161211

hh

α

α

α

QQQ

QQQ

QQQ

k =

nT =

N

N

N

] = N[ k-k

xy

y

x

k k

Txy

Ty

Tx

T

)(1

21 2

12

662616

262212

161211

hh

α

α

α

QQQ

QQQ

QQQ

k =

nT =

M

M

M

] = M[ k - k

xy

y

x

k k

Txy

Ty

Tx

T

Final formula

T

T

T

=

kxy

y

x

kxy

y

x

k

Txy

Ty

Tx

kxy

y

x

k

Mxy

My

Mx

Example 4.5Calculate the residual stresses at the bottom surface of the 90o ply

in a two ply [0/90] Graphite/Epoxy laminate subjected to a

temperature change of -75oC. Use the unidirectional properties of

Graphite/Epoxy lamina from Table 2.1. Each lamina is 5 mm thick.

Example 4.5

C/ m/m.

.

=

α

α

αo-

-

0

102250

1020004

7

12

2

1

C/ m/m .

.

=

α

α

αo-

-

xy

y

x

o

0

102250

1020004

7

0

C/ m/m.

.

=

α

α

αo-

-

xy

y

x

o

0

102000

1022507

4

90

Example 4.5

GPa

.

..

..

= Q

17700

035108972

089728181

][ 0

GPa

.

..

..

= Q

17700

081818972

089723510

][ 90

Example 4.5

)]0050(0000[

0

)10(2250

)10(2000

)10(

17700

035108972

089728181

)75( 4

7

9 .- - . .

.

.

..

..

= +

N

N

N-

-

Txy

Ty

Tx

]00000050[

0

)10(2000

)10(2250

)10(

17700

081818972

089723510

)75( 7

4

9 .. .

.

.

..

..

+

Pa - m. .

.

=

0

101311

1013115

5

Example 4.5

])0050()0000[(

0

102250

102000

)10(

17700

035108972

089728181

)75(21 224

7

9 . . .

.

.

..

..

=

M

M

M-

-

Txy

Ty

Tx

])000.0()0050[(

0

102000

102250

)10(

17700

035108972

089723510

)75(21 227

4

9

. .

.

.

..

..

+ -

-

m Pa .

.

=

0

105381

1053812

2

Example 4.5

Pa -m

.

..

..

[A] =

10170700

0106089108972

0108972106089

7

87

78

m Pa - .

.-

[B] = 26

6

000

01014320

00101432

m Pa -

.

..

..

[D] = 3

2

32

23

10975500

0100078104142

0104142100078

Example 4.5

0

T

T

DB

BA =

MN

κ

κ

κ

γ

ε

ε

.

...

...-

.

...

.-..

=

.

.-

.-

.-

xy

y

x

xy

y

x

0

0

0

2

326

236

7

687

678

2

2

5

5

10975500000

010007810414201014320

010414210007800101432

00010170700

010143200106089108972

001014320108972106089

0

105381

105381

0

101311

101311

/m

m/m

.

.-

.-

.-

=

κ

κ

κ

γ

ε

ε

-

-

-

-

xy

y

x

xy

y

x

1

0

102761

102761

0

109073

109073

1

1

4

4

0

0

0

Example 4.5

0

102761

102761

)0050(

0

109073

1090731

1

4

4

90

-

-

-

xy

y

x

.

.

. + .

.

=

γ

ε

ε

o

m/m .

.

= -

-

0

104752

1002914

3

Example 4.5

)75(

0

102000

1022507

4

.

.

=

γ

ε

ε-

-

Txy

Ty

Tx

m/m.

.

= -

-

0

10150000

101687505

2

Example 4.5

0

1015000

10168750

0

104752

1002915

2

4

3

-

-

-

-

Mxy

My

Mx

.

.

.

.

=

γ

ε

ε

0

1024900

10658503

3

-

-

.

.

=

0

1024900

1065850

)10(

17700

081818972

0897235103

3

9

900

-

-

xy

y

x

.

.

.

..

..

=

τ

σ

σ

Pa..

.

=

0

107184

1053577

6

Example 4.5

Ply #

Position

x

y

xy

1 (00)

Top

Middle

Bottom

2.47510-4

-7.16010-5

-3.90710-4

-1.02910-3

-7.09810-4

-3.90710-4

0.0

0.0

0.0 2 (900)

Top

Middle

Bottom

-3.90710-4

-7.09810-4

-1.02910-3

-3.90710-4

-7.16010-5

2.47510-4

0.0

0.0

0.0

Global Strains for Example 4.3

Example 4.5

Ply #

Position

y

y

xy

1(00)

Top

Middle

Bottom

4.718 107

-9.912 106

-6.701 107

7.535106

9.912106

1.229107

0.0

0.0

0.0

2(900)

Top

Middle

Bottom

1.229107

9.912106

7.535106

-6.701107

-9.912106

4.718107

0.0

0.0

0.0

Global Stresses for Example 4.3

Coefficients of Thermal and Moisture ExpansionTo find coefficients of thermal and moisture expansion of laminates• Symmetric laminates [B] = 0• No bending occurs under thermal hygrothermal loads• Assuming ∆T = 1 and ∆C = 0

N

N

N

+

N

N

N

=

BBB

BBB

BBB

+

AAA

AAA

AAA

Cxy

Cy

Cx

Txy

Ty

Tx

xy

y

x

662616

262212

161211

oxy

oy

ox

662616

262212

161211

N

N

N

AAA

AAA

AAA

=

Txy

Ty

Tx

*66

*26

*16

*26

*22

*12

*16

*12

*11

0xy

0y

0x

1 = T

0 = Cxy

y

x

(4.62)

(4.72)

Coefficients of Thermal and Moisture ExpansionTo find coefficients of thermal and moisture expansion of laminates• Symmetric laminates [B] = 0• No bending occurs under thermal hygrothermal loads• Assuming ∆T = 0 and ∆C = 1

N

N

N

+

N

N

N

=

BBB

BBB

BBB

+

AAA

AAA

AAA

Cxy

Cy

Cx

Txy

Ty

Tx

xy

y

x

662616

262212

161211

oxy

oy

ox

662616

262212

161211

N

N

N

AAA

AAA

AAA

=

Cxy

Cy

Cx

*66

*26

*16

*26

*22

*12

*16

*12

*11

xy

0y

0x

1 = C

0 = T xy

y

x

0

Example 4.6Find the coefficients of thermal and moisture expansion of a Graphite/Epoxy laminate. Use the properties of unidirectional Graphite/Epoxy lamina from table 2.1.

Example 4.6

Pa-m

.

..-

.-.

] = A[-

--

--

* 1

10298900

0108869102972

0102972103535

9

1011

1110

Example 4.6

)( 1

662616

262212

1612113

1h - h

α

α

α

QQQ

QQQ

QQQ

T =

N

N

N

k - k

kxy

y

x

k

k = Txy

Ty

Tx

)]. - (-. [-.

.

) (

.

..

..

) = ( -

-

0075000250

0

102250

102000

1017700

035108972

089728181

1 4

7

9

)]00250(00250[

0

102000

102250

)10(

17700

0351818972

089723510

1 7

4

9 .- - . .

.

.

..

..

) + ( -

-

]0025000750[

0

102250

102000

)10(

17700

035108972

089728181

)1( 4

7

9 . - . .

.

.

..

..

+ -

-

Pa - m.

.

=

0

106732

1085213

3

∆T = 1°C

Example 4.6

0

106732

108521

10298900

0108869102972

01029721035353

3

9

1011

1110

.

.

.

..-

.-.

= -

--

--

0xy

0y

0x

m/m..

.

= -

-

0

106002

1030396

7

C m/m.

.

= -

-

0xy

0y

0x

1 = T

0 = Cxy

y

x

/

0

106002

1030396

7

Example 4.6

)h - h(

QQQ

QQQ

QQQ

C =

N

N

N

1 - kk

xy

y

x

662616

262212

161211

k

3

1 = kCxy

Cy

Cx

)]. - (-. [-.) (

.

..

..

) = ( 0075000250

0

60

0

1017700

035108972

089728181

1 9

)]. - (-. [

.

) (

.

..

..

) + ( 0025000250

0

0

60

1017700

0351818972

089723510

1 9

]. - . [.) (

.

..

..

) + ( 0025000750

0

60

0

1017700

035108972

089728181

1 9

Pa - m.

.

=

0

100777

1084247

7

∆C = 1 kg/kg

Example 4.6

0

100777

108424

10298900

0108869102972

01029721035357

7

9

1011

1110

.

.

.

..-

.-.

= -

--

--

0xy

0y

0x

m/m.

.

= -

-

0

108856

1043022

2

kgkg m/m.

.

-

-

0xy

0y

0x

= T

= Cxy

y

x

//

0

108856

1043022

2

0

1

END