Summary of Classical Lamination Theory (CLT) Calculationscourses.washington.edu/mengr450/CLT_Summary.pdf · Summary of Classical Lamination Theory (CLT) Calculations Numerical examples

Summary of Classical Lamination Theory (CLT) Calculations


Numerical examples illustrating discussion in:



Section 6.8.1: A CLT Analysis When Loads Section 6.8.1: A CLT Analysis When Loads Are Known




Section 6.8.2: A CLT Analysis When MidplaneStrains and Curvatures are Known





(Sections 6.8.1 and 6.8.2 are nearly identical…)






Section 6.8.1: A CLT Analysis When Loads Are Known

1. Define the problem:a) Specify number of different materials usedb) Specify properties for each materialb) Specify properties for each material


1. Define the problem:a) Specify number of different materials usedb) Specify properties for each materialb) Specify properties for each material

Example: Suppose two materials are used…graphite-epoxy and glass epoxy. From Table 3.1, typical properties are

Mat’l Name

Mat’l Number

E11 (psi) E22 (psi) ν12 G12 (psi) α11 (in/in-ºF)

α22 (in/in-ºF)

Ply thick (in)

Gr/Ep 1 25 x 106 1.5 x 106 0.30 1.9 x 106 -0.5x10-6 15 x10-6 0.005 Gl/Ep 2 8.0 x106 2.3 x 106 0.28 1.1 x106 3.7x10-6 14 x10-6 0.005


1. Define the problem:a) Specify number of different materials usedb) Specify properties for each materialb) Specify properties for each materialc) Specify laminate description



T]30/90/30/0[ − A 4-ply laminate. This description is adequate if same material is used for all plies.

Example:



T]30/90/30/0[ − A 4-ply laminate. This description is adequate if same material is used for all plies. For illustrative purposes assume Gr/Ep used for 0º and 90º plies and Gl/Ep used for ±30º plies

Graphite/Epoxy

Glass/Epoxy

Example:



Example:

inintzz

inintzz

inintzz

inmtzz

inintz

inin

010.0005.0005.0

005.0005.0000.0

000.0005.0005.0

005.0005.001.0

010.02/)020(.2/

020.0)4(0.005 thicknesslaminate Total

434

323

212

101

0

=+=+==+−=+==+−=+=−=+−=+=

−=−=−===


1. Define the problem:a) Specify number of different materials usedb) Specify properties for each materialb) Specify properties for each materialc) Specify laminate descriptiond) Specify mechanical and thermal loads


1. Define the problem:a) Specify number of different materials usedb) Specify properties for each materialb) Specify properties for each materialc) Specify laminate descriptiond) Specify mechanical and thermal loads

Example:

FTFTFT

MN

MN

MN

servicecure

xyxy

yyyy

xxxx

°−=∆⇒°=°=

−−==

−==−−==

275 75 350

in/inlbf 0854.0 lbf/in 4.64

in/inlbf 22.0 lbf/in 377

in/inlbf 0.4 lbf/in 520


2. Calculate the [ABD] matrix:


2. Calculate the [ABD] matrix:a) Calculate the [Q] matrix for each material



( )

( )

( ) ( ) ( )

−

−

−

−

=

=

12

2221211

2211

2221211

221112

2221211

221112

2221211

211

66

2212

1211

00

0

0

00

0

0

][

G

EE

EE

EE

EE

EE

EE

EE

E

Q

QQ

QQ

Qνν

νν

νν



Example: xx 66 010452.01014.25Example:

psi

x

xx

xx

Q

psi

x

xx

xx

Q

EpGl

EpGr

=

=

6

66

66

/

6

66

66

/

1010.100

010353.210659.0

010659.010184.8

][

1090.100

010508.110452.0

010452.01014.25

][


2. Calculate the [ABD] matrix:a) Calculate the [Q] matrix for each materialb) Calculate the matrix for each ply][Qb) Calculate the matrix for each ply][Q



[ ]

=

662616

262212

161211

QQQ

QQQ

QQQ

Q



)sin(cossincos)22(

sincos)2(sincos)2(

cossincos)2(2sin

sincos)2(sincos)2(

sincos)4()sin(cos

sinsincos)2(2cos

4466

226612221166

3661222

36612116226

422

226612

41122

3661222

36612116116

22662211

44122112

422

226612

41111

θθθθ

θθθθ

θθθθ

θθθθ

θθθθ

θθθθ

++−−+=

−−−−−==

+++=

−−−−−==

−+++==

+++=

QQQQQQ

QQQQQQQQ

QQQQQ

QQQQQQQQ

QQQQQQ

QQQQQ


2. Calculate the [ABD] matrix:a) Calculate the [Q] matrix for each materialb) Calculate the matrix for each ply][Qb) Calculate the matrix for each ply

Example: For ply 1,

][Q

psi

x

xx

xx

Q plyEpGr

=°

6

66

66

0/

1090.100

010508.110452.0

010452.01014.25

][



Example: For ply 2,

][Q

psi

xxx

xxx

xxx

Q plyEpGl

=°

666

666

666

30/

1000.210741.010784.1

10741.010907.210563.1

10784.110563.110823.5

][



Example: For ply 3,

][Q

psi

x

xx

xx

Q plyEpGr

=°

6

66

66

90/

1090.100

01014.2510452.0

010452.010508.1

][



Example: For ply 4,

][Q

psi

xxx

xxx

xxx

Q plyEpGl

−−−−

=°−

666

666

666

30/

1000.210741.010784.1

10741.010907.210563.1

10784.110563.110823.5

][


2. Calculate the [ABD] matrix:a) Calculate the [Q] matrix for each materialb) Calculate the matrix for each ply][Qb) Calculate the matrix for each plyc) Calculate the [Aij], [Bij], and [Dij] matrices

][Q



][Q

Example:

{ }in

lbf

x

xx

xx

zzQAn

kkkkijij

=−=∑=

−3

33

33

11

1004.3900

0103.1621015.20

01015.20104.191

)(



][Q

Example:

{ } lbfzzQBn

kkkkijij

−−−−−

=−= ∑=

−59.20.372.89

0.373308.27

2.898.27778

)(2

1

1

21

2



][Q

Example:

{ } inlbfzzQDn

kkkkijij −

−−−−

=−= ∑=

−30.1185.0446.0

185.046.2672.0

446.0672.034.9

)(3

1

1

31

3


2. Calculate the [ABD] matrix:a) Calculate the [Q] matrix for each materialb) Calculate the matrix for each ply][Qb) Calculate the matrix for each plyc) Calculate the [Aij], [Bij], and [Dij] matricesd) Assemble the [ABD] matrix

][Q



][Q

[ ]

=

=

662616662616

262212262212

161211161211

662616662616

262212262212

161211161211

DDDBBB

DDDBBB

DDDBBB

BBBAAA

BBBAAA

BBBAAA

DB

BAABD

ijij

ijij



][Q

Example:

−−−−−−−−−−−

−−−−−

=

30.1185.0446.059.20.372.89

185.046.2672.00.373308.27

446.0672.034.92.898.27778

59.20.372.89100.3900

0.373308.27010162101.20

2.898.277780101.2010191

][3

33

33

x

xx

xx

ABD


3. Calculate the [abd] = [ABD]-1 matrix:



161211161211 bbbaaa

[ ] [ ]

== −

662616662616

262212622212

161211612111

666261662616

262221262212

161211161211

1

dddbbb

dddbbb

dddbbb

bbbaaa

bbbaaa

bbbaaa

ABDabd



Example: Example:

−−−−−

−−−−−

=

−−−

−−−

−−−

−−−−

−−−−−

−−−−−

870.0025.0114.010226.010203.010862.0

025.0586.0044.010454.01016.110129.0

114.0044.0188.010381.010196.010827.0

10226.010454.010381.0109.2600

10203.01016.110196.001064.810811.0

10862.010129.010827.0010811.01009.9

][

343

333

343

3336

43466

33366

xxx

xxx

xxx

xxxx

xxxxx

xxxxx

abd


4. Calculate thermal stress and moment resultants:


4. Calculate thermal stress and moment resultants:a) Calculate effective thermal expansion

coefficients for each plycoefficients for each ply



coefficients for each plycoefficients for each ply

))(sin()cos(2

)(cos)(sin

)(sin)(cos

2211

222

211

222

211

ααθθαθαθαα

θαθαα

−=

+=

+=

xy

yy

xx



coefficients for each plycoefficients for each plyExample:

Ply Number

Mat’l Number

Fiber angle (deg)

αxx (in/in- ºF)

αyy (in/in- ºF)

αxy (in/in- ºF)

1 1 0 -0.5x10-6 15 x10-6 0 2 2 30 6.28 x10-6 11.4 x 10-6 -8.92 x 10-6 3 1 90 15 x10-6 -0.5x10-6 0 4 2 -30 6.28 x10-6 11.4 x 10-6 8.92 x 10-6



coefficients for each plycoefficients for each plyb) Calculate thermal stress & moment resultants

[ ] [ ]{ }

[ ] [ ]{ }

[ ] [ ]{ }∑

∑

∑

=−

=−

=−

−++∆≡

−++∆≡

−++∆≡

n

kkkkxyyyxx

Txy

n

kkkkxyyyxx

Tyy

n

kkkkxyyyxx

Txx

zzQQQTN

zzQQQTN

zzQQQTN

11662616

11262212

11161211

ααα

ααα

ααα



coefficients for each plycoefficients for each plyb) Calculate thermalstress & moment resultants

[ ] [ ]{ }

[ ] [ ]{ }

[ ] [ ]{ }∑

∑

∑

=−

=−

=−

−++∆≡

−++∆≡

−++∆≡

n

kkkkxyyyxx

Txy

n

kkkkxyyyxx

Tyy

n

kkkkxyyyxx

Txx

zzQQQT

M

zzQQQT

M

zzQQQT

M

1

21

2662616

1

21

2262212

1

21

2161211

2

2

2

ααα

ααα

ααα



coefficients for each plycoefficients for each plyb) Calculate thermal stress & moment resultants

Example:

ininlbfMN

ininlbfMinlbfN

ininlbfMinlbfN

Txy

Txy

Tyy

Tyy

Txx

Txx

/ 0246.0 0

/ 0005.0 /123

/ 401.0 /129

−−==

−=−=

−−=−=


5. Calculate midplane strains and curvatures



+ T

xxxxoxx NNbbbaaa 161211161211ε

+

++

+

++

=

Txyxy

Tyyyy

Txxxx

Txyxy

Tyyyy

xxxx

xy

yy

xx

oxy

oyy

oxx

MM

MM

MM

NN

NN

NN

dddbbb

dddbbb

dddbbb

bbbaaa

bbbaaa

bbbaaa

662616662616

262212622212

161211612111

666261662616

262221262212

161211161211

κκκγεε



Example:

0o

xxε

−

−

−

=

−

−

−

−

−

1

1

1

6

6

20.0

40.0

50.0

rad 10900

/ 101300

0

in

in

in

x

ininx

xy

yy

xx

oxy

oyy

xx

κκκγεε


6. For each ply: a) Calculate ply strains in x-y coordinate system

(usually at ply interfaces)

+

=

xy

yy

xx

oxy

oyy

oxx

xy

yy

xx

z

κκκ

γεε

γεε


6. For each ply: a) Calculate ply strains in x-y coordinate system

(usually at ply interfaces) LAMINATE PLY STRAINS, x-y COORDINATE SYSTEM:

PLY NO Z-COORD EPSxx EPSyy GAMxy

------- -0.100E-01 0.005000 -0.005300 0.0029001

------- -0.500E-02 0.002500 -0.003300 0.0019002

------- 0.000E+00 0.000000 -0.001300 0.0009003

------- 0.500E-02 -0.002500 0.000700 -0.0001004

------- 0.100E-01 -0.005000 0.002700 -0.001100

Example:


6. For each ply: a) …usually also transform strains to 1-2

coordinate system

[ ]z

xy

yy

xx

zT

=

2/2/12

22

11

γεε

γεε


6. For each ply: a) …usually also transform strains to 1-2

coordinate system LAMINATE PLY STRAINS, 1-2 COORDINATE SYSTEM:

PLY NO Z-COORD EPS11 EPS22 GAM12

Example:

PLY NO Z-COORD EPS11 EPS22 GAM12

--------------------------------------------------------------0.10000E-01 0.005000 -0.005300 0.002900

1-0.50000E-02 0.002500 -0.003300 0.001900

--------------------------------------------------------------0.50000E-02 0.001873 -0.002673 -0.004073

20.00000E+00 0.000065 -0.001365 -0.000676

-------------------------------------------------------------0.00000E+00 -0.001300 0.000000 -0.000900

30.50000E-02 0.000700 -0.002500 0.000100

-------------------------------------------------------------0.50000E-02 -0.001657 -0.000143 -0.002821

40.10000E-01 -0.002599 0.000299 -0.007218

--------------------------------------------------------------


6. For each ply: b) Calculate ply stresses in x-y coordinate system



ply)each (for

662616

262212

161211

∆−∆−∆−

=

xyxy

yyyy

xxxx

xy

yy

xx

T

T

T

QQQ

QQQ

QQQ

αγαεαε

τσσ



LAMINATE PLY STRESSES, x-y COORDINATE SYSTEM:

PLY NO Z-COORD SIGxx SIGyy TAUxy

Example:

PLY NO Z-COORD SIGxx SIGyy TAUxy

---------------------------------------------------------------0.10000E-01 0.12169E+06 0.42794E+03 0.55100E+04

1-0.50000E-02 0.59756E+05 0.23131E+04 0.36100E+04

---------------------------------------------------------------0.50000E-02 0.23372E+05 0.57335E+04 0.63146E+04

20.00000E+00 0.10155E+05 0.69006E+04 0.13317E+04

--------------------------------------------------------------0.00000E+00 0.55707E+04 -0.34266E+05 0.17100E+04

30.50000E-02 0.27052E+04 0.14874E+05 -0.19000E+03

--------------------------------------------------------------0.50000E-02 -0.27044E+04 0.82159E+04 0.32505E+04

40.10000E-01 -0.12352E+05 0.10865E+05 0.42260E+04

--------------------------------------------------------------


6. For each ply: b) usually also transform stresses to 1-2 coord

system

[ ]

=

xy

yy

xx

T

τσσ

τσσ

12

22

11


6. For each ply: b) usually also transform stresses to 1-2 coord

system LAMINATE PLY STRESSES, 1-2 COORDINATE SYSTEM:

PLY NO Z-COORD SIG11 SIG22 TAU12

Example:

PLY NO Z-COORD SIG11 SIG22 TAU12

---------------------------------------------------------------0.10000E-01 0.12169E+06 0.42794E+03 0.55100E+04

1-0.50000E-02 0.59756E+05 0.23131E+04 0.36100E+04

---------------------------------------------------------------0.50000E-02 0.24431E+05 0.46744E+04 -0.44802E+04

20.00000E+00 0.10495E+05 0.65610E+04 -0.74342E+03

--------------------------------------------------------------0.00000E+00 -0.34266E+05 0.55707E+04 -0.17100E+04

30.50000E-02 0.14874E+05 0.27052E+04 0.19000E+03

--------------------------------------------------------------0.50000E-02 -0.27893E+04 0.83009E+04 -0.31034E+04

40.10000E-01 -0.10208E+05 0.87202E+04 -0.79402E+04

--------------------------------------------------------------




Section 6.8.2: A CLT Analysis When Midplane Strains and Curvatures are Known


Section 6.8.2: A CLT Analysis When Midplane Strains and Curvatures Are Known

1. Define the problem:a) Specify number of different materials usedb) Specify properties for each materialb) Specify properties for each materialc) Specify laminate descriptiond) Specify midplane strains/curv & thermloads


1. Define the problem:a) Specify number of different materials usedb) Specify properties for each materialb) Specify properties for each materialc) Specify laminate descriptiond) Specify midplane strains/curv & thermloads

Example:

FTFTFT

in

ininin

in

servicecure

xyoxy

yyoyy

xxoxx

°−=∆⇒°=°=

−==

=−=

−==

−

−

−

275 75 350

20.0 rad 900

40.0 /1300

50.0 0

1

1

1

κµγ

κµε

κε


2. Calculate the [ABD] matrixa) Calculate the [Q] matrix for each materialb) Calculate the matrix for each ply][Qb) Calculate the matrix for each plyc) Calculate the [Aij], [Bij], and [Dij] matricesd) Assemble the [ABD] matrix

(in this case do not need the [abd] matrix)

][Q



coefficients for each plycoefficients for each plyb) Complete calculation of thermal stress and

moment resultants



coefficients for each plycoefficients for each plyb) Complete calculation of thermal stress and

moment resultants

ininlbfMN

ininlbfMinlbfN

ininlbfMinlbfN

Txy

Txy

Tyy

Tyy

Txx

Txx

/ 0246.0 0

/ 0005.0 /123

/ 401.0 /129

−−==

−=−=

−−=−=


5. Calculate stress and moment resultants

T

Txx

o

oxxxx

N

NBBBAAAN ε161211161211

−

=

Txy

Tyy

Txx

Txy

Tyy

xy

yy

xx

oxy

oyy

xy

yy

xx

xy

yy

M

M

M

N

N

DDDBBB

DDDBBB

DDDBBB

BBBAAA

BBBAAA

M

M

M

N

N

κκκγε

662616662616

262212262212

161211161211

662616662616

262212262212



T

Txx

o

oxxxx

N

NBBBAAAN ε161211161211

−

=

Txy

Tyy

Txx

Txy

Tyy

xy

yy

xx

oxy

oyy

xy

yy

xx

xy

yy

M

M

M

N

N

DDDBBB

DDDBBB

DDDBBB

BBBAAA

BBBAAA

M

M

M

N

N

κκκγε

662616662616

262212262212

161211161211

662616662616

262212262212

(this is the only difference in calculation….)



−

−−

12902.898.27778031.203191 EENxx

−

−

−

−

−

−

−

−−−−−−−−−−−

−−−

=

0246.0

0005.0

401.0

0

123

20.0

40.0

50.0

900

1300

30.1185.0446.059.20.372.89

185.046.2672.00.373308.27

446.0672.034.92.898.27778

59.20.372.8930.3900

0.373308.270316231.20

µµ

E

EE

M

M

M

N

N

xy

yy

xx

xy

yy



lbf/in 520xxN

−

−=

in/in-lbf 085.0

in/in-lbf 22.0

in/in-lbf 0.4

lbf/in 4.64

lbf/in 377

xy

yy

xx

xy

yy

M

M

M

N

N


6. For each ply: a) Calculate ply strains in x-y and 1-2 coordinate systems (at ply interfaces)b) Calculate ply stresses in x-y and 1-2 coordinate systems (at ply interfaces)

Analysis complete!

Summary of Classical Lamination Theory (CLT) Calculationscourses.washington.edu/mengr450/CLT_Summary.pdf · Summary of Classical Lamination Theory (CLT) Calculations Numerical examples

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