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EML 4230 Introduction to Composite Materials. Chapter 4 Macromechanical Analysis of a Laminate Hygrothermal Loads Dr . Autar Kaw Department of Mechanical Engineering University of South Florida, Tampa, FL 33620 Courtesy of the Textbook Mechanics of Composite Materials by Kaw. - PowerPoint PPT Presentation
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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
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