analysis of laminates -...
TRANSCRIPT
Analysis of laminates
Since the unidirectional lamina has poor transverse properties, it is hardly used
in form of a single ply in the applications. Normally, several laminae are stacked
together to form laminate. The fiber angle of each lamina can be designed such that the
desired properties of the laminate are achieved.
1
2
34
x y
z
If the fiber angles of plies 1, 2, 3 and 4 are 1, 2, 3, and 4, respectively, the
stacking sequence of this laminate can be written as [1/2/3/4]. For example,
[0/30/60/90] is a four-ply laminate with the fiber angle of 0, 30, 60, and 90,
respectively. Additional rules are used to describe stacking sequence of laminate, as
the following.
If two consecutive plies have the same fiber angle, subscripted number may be used.
[0/30/30/90] [0/302/90]
or can be used for consecutive plies
[45/-45/30/90] [45/30/90]
Subscribed “s” is used in case of symmetric laminate
[-30/60/60/-30] [-30/60]s
[0/90/0/90/45/45/90/0/90/0] [(0/90)2/45]s
Overbar “” indicates center ply for laminate with odd number of plies
2 s0/90 /45 [0/90/0/90/45/90/0/90/0]
69
Theory of laminated beam in pure flexure
MM
No shear force
The following analysis is based on a paper by Pagano (1967)
Assumption:
1. Plane section remain plane
2. Symmetric stacking sequences (N.A. on midplane)
3. 0or 90 fiber angle only (no shear couple)
4. Perfectly bonded plies continuous strain
Strain at distance z from N.A.,
x
z z
where is a radius of curvature
It is noticed that the strain x is similar to isotropic beam. It is independent of material
properties. However, stress x x xE depends on material properties
At the jth ply, x x x xj j j j
zE E
(1)
Considering moment,
N.A.
zx
dz
Force from stress x can be determined from xbdz , where b is the depth of beam.
Thus, moment can be determined from zxbdz
The total moment 2
0
2
h
xM bzdz
70
2
2
0
2h
x j
bE z d
z
3 2
0
2
3
h
x j
b zE
2
3 31
1
2
3
N
x j jjj
bE z z
(2)
For isotropic beam: or Mz I
MI z
But f f
zE E
Thus, 3
12f fE I E bh
M
(3)
Compare (2) and (3) 2
3 313
1
8N
f x jjj
E E zh
jz
Theory of laminated plates with coupling
Fundamental of strain-displacement relationship
A
D C
B
A
DC
B
x u
dx dx v
vdx
x
udx
x
71
Displacement of A is u and v. So B displaces by and u v
u dx v dx x
x
in
the x and y direction, respectively. u and v are in terms of x and y or and
.
,u u x y
,v v x y
Thus, x
A B AB
AB
or 1xA B AB
2 2
221x
u vA B dx dx dx dx
x x
2 2
2 2 22 1 1 2x x
u u vdx dx
x x x
Assume that product of displacement derivative 0
So, x
u
x
Similarly, y
v
y
and z
w
z
From figure above, tan
vdx
xu
dx dxx
Since is very small, that is tan or
vdx
xu
dx dxx
u
x
is very small compared t unity, that is u
dx dxx
That is v
x
Similarly, u
y
So, share strain xy
u v
y x
72
Similarly, xz
u w
z x
and yz
v w
z y
Classical Lamination Theory (CLT)
Objective: To develop the stress-strain analysis for a multi-ply laminate. CLT is a
combination of orthotropic elasticity, and classical thin plate theory.
Thin Plate Theory.
a b
t
x y
z
x-y plane is on the mid-plane of the plate. Plate is assumed as “thin” if
and 10 10
a bt
Allowable plate loading
Nx
Ny
Nxy
Nx, Ny, Nxy are stress resultants with a unit of force/length. The stress resultant can be determined from ,x x avgN t
Allowable moment loading
Mxx
Myy
Mxy
Myx
73
Mxx, Myy, Mxy, Myx are moment resultants with a unit of force×length
plate length such as
N-m
m. Mxx, Myy are bending moment and Mxy, Myx are twisting moment.
Relating stress/moment resultants to plate stresses
z
Nx
Mxx Mxx
Nx t
Mxx
Nx
x
From equilibrium of force, 0xF
We obtain 2
2
t
x xt
N d
z
From equilibrium of moment, 0xM
We obtain 2
2
t
xx xt
M zdz
Similarly; 2
2
t
y yt
N d
z , 2
2
t
xy xyt
N dz
yxN
2
2
t
yy yt
M zdz
, 2
2
t
xy xyt
yxM zdz M
Deformation of a thin plate (Kirchhoff Hypothesis)
74
Basic assumptions:
1. Plate is thin, i.e. t << a, b
2. Displacement u, v and w are small compared to thickness t
3. x, y, and xy are small compared to unity
4. xz and yz are negligible.
5. u and v are linear functions of z
6. z is negligible.
7. xz and yz vanish on the plate surfaces.
Kirchhoff Hypothesis: “A straight line initially perpendicular to mid plane of plate
remains straight and perpendicular after deformation”
t
+z
zc
B
c D
O
uo
x
c
O zc
From figure above, uo is displacement in the x-direction of point O or displacement of
the mid plane.
Displacement of point c, sincoc zuu
For small angle , sin = tan =
But plane mid of slope tan dx
dw
Thus, dx
dwzuu coc or
dx
dwzuu o
Similary, dy
dwzvv o
75
From stress-strain relationship,
xox
ox z
dx
wdz
x
u
x
u
2
2
where 2
2
dx
wdx is the curvature of the middle surface.
Similarly, yoyy z
y
v
where 2
2
dy
wdy
and x
v
y
uxy
yx
wz
x
v
yx
wz
y
u oo
22
yx
wz
x
v
y
u oo
2
2
where xyoxy z
yx
wxy
2
2
Thus, the strain at any positions can be written in a matrix form as
xy
y
x
oxy
oy
ox
xy
y
x
z
This equation is important for determining strains at any points on the plate. If
the mid-plane strains and curvatures are known, strains at any positions can be
determined.
Example: A small segment of a four-layer laminate deformed such that a point Po on
the mid-plane has an extension strain in the x-direction of 1000 strain. Radius of
curvature Ro is 0.2 m. Determine x through the thickness of the laminate if the
laminate thickness is 0.6 mm.
76
Ro=0.2 m
z
x
Thus, the curvature 152.0
11 mRo
x
Strain zz xoxx 51000
At upper surface, 50003.01010000003.0 6 zx
strain500 μ
At lower surface, 50003.01010000003.0 6 zx
strain2500 μ
2500
-500
Notice: Strain in laminates linearly varies with z and independent of material properties
Laminate stress
From plate theory:
xy
y
x
oxy
oy
ox
xy
y
x
z
Recall the ply stress-strain relationship
77
xy
y
x
xy
y
x
Q
xy is function of z and ][Q depends on material properties and fiber angle
Thus, more appropriate stress-strain relationship is
kxy
y
x
k
kxy
y
x
Q
where k refers to ply number
Thus, we obtain
xyoxy
yoy
xox
kkxy
y
x
z
z
z
QQQ
QQQ
QQQ
662616
262212
161211
……………………………………**
This equation is not practical because curvature x, y, and xy are not
generally known. So the mid-plane strains and curvatures should be related to the
applied forces and bending moments using the equilibrium equation.
For example:
2
2
161211161211
2
2
t
txyyx
oyx
oy
ox
t
txx dzQzQzQzQQQdzN
2
2
16
2
2
16
2
2
12
2
2
11 ....
t
txy
t
t
oxy
t
t
oy
t
t
ox dzQzdzQdzQdzQ
nz
nz
nz
z
z
z
z
z
ox dzQdzQdzQdzQ
111
3
2
311
2
1
211
1
0
111 ..........
nz
nz
nz
z
z
z
z
z
oy dzQdzQdzQdzQ
112
3
2
312
2
1
212
1
0
112 ..........
nz
nz
nz
z
z
z
z
z
oxy dzQdzQdzQdzQ
116
3
2
316
2
1
216
1
0
116 ..........
78
nz
nz
nz
z
z
z
z
zx zdzQzdzQzdzQzdzQ
111
3
2
311
2
1
211
1
0
111 ..........
nz
nz
nz
z
z
z
z
zy zdzQzdzQzdzQzdzQ
112
3
2
312
2
1
212
1
0
112 ..........
nz
nz
nz
z
z
z
z
zxy zdzQzdzQzdzQzdzQ
116
3
2
316
2
1
216
1
0
116 ..........
Evaluate: ………………..** xyyxoxy
oy
oxx BBBAAAN 161211161211
where 11
mm
n
m
mijij zzQA and 2
12
12
1
mm
n
m
mijij zzQB
n is number of ply of the laminate.
The position of each ply zi are graphically shown below
zn zn-1
zn-2 zn-3
z0 z1 z2 z3
z
-t/2
t/2
Consider the stress resultant in the y-direction and the shear stress resultant.
From 2
2
t
y yt
N d
z , and 2
2
t
xy xyt
N dz
yxN
We obtain ……………..** xyyxoxy
oy
oxy BBBAAAN 262212262212
……………..** xyyxoxy
oy
oxxy BBBAAAN 662616662616
Similarly;
2
2
t
txx zdzM
79
2
2
162
2
2
16
2
2
12
2
2
11 ....
t
txy
t
t
oxy
t
t
oy
t
t
ox dzQzzdzQzdzQzdzQ
……………...** xyyxoxy
oy
ox DDDBBB 161211161211
where 31
3
13
1
mm
n
m
mijij zzQD
Write in matrix form
xy
y
x
oxy
oy
ox
xy
y
x
xy
y
x
BBBBBB
BBBBBB
BBBBBB
BBBAAA
BBBAAA
BBBAAA
M
M
M
N
N
N
662616662616
262212262212
161211161211
662616662616
262212262212
161211161211
or
o
DB
BA
N
M
where A Extensional stiffness, relate o to N
B Coupling stiffness, relate to N, or o to M
D Bending stiffness, relate to M
Analytical Approach:
1. Given material properties: E1, E2, G12, v12, and ply thickness
2. Stacking sequences.
3. Measured , , , , , and o o ox y xy x y xy
4. Analyze
a. Calcualte [Q] for each material
b. Calcualte [Q ] for each ply
c. Calculate [ABD]
80
5.
o
DB
BA
N
M
Coupling effects: Two types
1. Lamina shear coupling: Layer level
Term 16 and 26 in 11 12 16
12 22 26
16 26 66
x x
y y
xy x
Q Q Q
Q Q Q
Q Q Q y
These couplings are couplings of normal stresses to shear strain and shear stress
to normal strain. These couplings lead to terms A16, A26, D16, and D26 in [ABD]
2. Laminate-level coupling: All Bij
These couplings relate force resultants to curvatures and moment resultants to
strains. For isotropic materials, A16, A26, D16, and D26 equal zero, but Bij may not be
zero if their material properties are not symmetric with respect to the middle plane.
Stiffness characteristics of laminate
1. Symmetric laminate: Both geometric and material properties are symmetric
about the middle surface
From 21
2
12
1
mm
n
m
mijij zzQB
For symmetric laminates, all Bij = 0
So, In-plane loads will not cause bending or twisting and bending or twisting
moments will not cause mid-plane extension or contraction.
2. Antisymmetric laminate: Fiber angles of plies at the distance +z and –z are +
and -, respectively
- For example: [-45/45/-45/45] or [45]2
81
- can not be 0 or 90
- A16 = A26 = D16 = D26 = 0
- B11 = B12 = B22 = B66 = 0
3. Quasi-isotropic laminate: Laminates which exhibit some isotropic behavior.
- The sub-matrix [A] in [ABD] is in the form of
11 12
12 11
11 12
0
0
0 02
A A
A A
A A
- To obtain above property, it is required that
1. Laminate is required to have three or more identical laminae
2. Each lamina is /N apart where N is number of plies
4. Cross-ply laminate: Laminates which plies orient at either = 0 or 90
5. A balance cross-ply laminate: Cross-ply laminate which has number of 90
plies equal to number of 0 plies.
6. Angle-ply laminate: Laminate which has ply angle as + or - only.
Derivation and use of laminate compliances
The where[ABD] may be called the laminate stiffness
matrix is not practical since the applies loads or moments are usually known. It is more
appropriate to write the equation in the form of
o
DB
BA
N
M
-1
where =o a b M a b A B
b d N b d B D
82
The [abd] matrix can be approximate from the sub-matrix [A], [B], and [D]. However,
with the advance in computation tool, one can perform the matrix inversion directly.
For a given applied loads and moments, mid-plane strains and curvature can be
determined. Then, stresses in the laminate are determined from
o
k kQ z
Example Determine the stiffness matrix [ABD] for [45] laminate consisting of 0.25-
mm thick unidirectional A/S 3501 graphite/epoxy plies.
Solution:
It can be shown that 45
45.22 31.42 32.44
31.42 45.22 32.44 GPa
32.44 32.44 35.6
Q
and 45
45.22 31.42 32.44
31.42 45.22 32.44 GPa
32.44 32.44 35.6
Q
+45 z0 = -t
z
z2 = t
z1 = 0 -45
Thus, 1 45 451
where 0.25 mmn
mij ij m m ij ij
m
A Q z z Q t Q t t
11 2245.22 0.25 45.22 0.25 22.61 GPa mm = A A
12 31.42 0.25 31.42 0.25 15.71 GPa mmA
16 26 0A A
66 35.6 0.25 35.6 0.25 17.8 GPa mmA
83
So, ……………………………………….** 22.61 15.71 0
15.71 22.61 0 GPa mm
0 0 17.8
A
Similarly, 21
2
12
1
mm
n
m
mijij zzQB
21
224511
20
21451111 2
1zzQzzQB
222222 0025.022.4525.0022.45
2
1B
02
1 21
224512
20
21451212 zzQzzQB
2222216 mm-GPa 027.2025.044.3225.0044.32
2
1B
21626 mm-GPa 027.2 BB
So, …………………………….** 2mm-GPa
0027.2027.2
027.200
027.200
B
Similarly, 31
3
13
1
mm
n
m
mijij zzQD
3333311 mm-GPa 471.0025.022.4525.0022.45
2
1D
with similar calculation
…………………………………….** 2mm-GPa
371.000
0471.0327.0
0327.0471.0
D
Example The [45] laminate described in the previous example is subjected to a
uniaxial force per unit length Nx = 30 MPa-mm. Find the resulting stresses and strains
in each ply along the x and y directions.
84
Nx = 30 MPa-mm
x
y
Solution:
From the previous example
371.0
0471.0
0327.0471.0
0027.2027.28.17
027.200061.22
027.200071.1561.22
Sym
ABD
387.6
0959.4
0986.1959.4
03386.03386.0333.1
3379.00001034.0
3379.00000415.01034.0
1
Sym
ABDabd
The mid-plane strains can be calculated as
1-mm 01017.0
0
0
0
mm/mm 001183.0
mm/mm 003043.0
0
0
0
0
0
30
abd
xy
y
x
oz
oy
ox
Therefore, stresses are calculated as follow
85
Bottom
z
Top
BottomTop
12
Strain distribution on the laminate (both plies)
zz
z
z
xyoxy
yoy
xox
xy
y
x
01017.0
001183.0
003043.0
Next, determine stress distribution on each ply
Ply number 1(+45 ply )
zSym
Q
xy
y
x
01017.0
001183.0
003043.0
60.35
44.3222.45
44.3241.3122.45
45
45
………………………………….…..**
z
z
z
3621.00603.0
3297.00421.0
3299.01004.0
Thus, stresses at the bottom of ply number 1 (z = -0.25 mm) are
GPa
03018.0
00403.0
01795.0
and stresses at the top of ply number 1 (z = 0) are
GPa
0603.0
0421.0
1004.0
Ply number 2 (-45 ply )
z
z
z
Q
xy
y
x
3621.00603.0
3297.004212.0
3299.010043.0
45
45
………………………………..**
86
Thus, stresses at the bottom of ply number 2 (z = 0) are
GPa
0603.0
00421
1004.0
and stresses at the top of ply number 1 (z = 0.25 mm) are
GPa
03017.0
04031.0
01795.0
In conclusion, stress distribution through the thickness of the laminate can be shown as
Ply No. z-coordinate x (MPa) y (MPa) xy (MPa)
1 -0.25 17.95 -40.3 -30.18
(+45) 0 100.4 42.1 60.3
2 0 100.4 42.1 -60.3
(-45) 0.25 17.95 -40.3 30.17
Laminate Engineering Constant
For a symmetric laminate subjected to in-plane loads, the stress resultants and
strains are related according to
oyx
oy
ox
xy
y
x
AAA
AAA
AAA
N
N
N
662616
262212
161211
or
where
xy
y
x
oyx
oy
ox
N
N
N
AAA
AAA
AAA
662616
262212
161211
ijA are members of the [abd]
Therefore, effective longitudinal Young modulus, only apply xN
ox
xxE
87
11
/
AN
tN
x
x
tA11
1
where t is the thickness of the laminate
Similarly, tA
E
yN
oy
yy
11only apply
1
Effective laminate shear stress, tA
G
xyN
oxy
xyxy
66only apply
1
Effective laminate Poisson’s ratio, 11
12
only apply A
Av
xN
ox
oy
xy
Similarly: 11
16, A
Axyx
(xy due to x)
66
26, A
Ayxy
(y due to xy)
Hygrothermal Effects in laminate
Assumptions: 1. No interaction between mechanical and hygrothermal effects
2. Uniform temperature and moisture distribution
Thin plate theory:
where t refers to “total”
xy
y
x
oxy
oy
ox
txy
ty
tx
z
Strains with Hygrothermal effects may be written as
t
k k kkkS T C
or t
k kk kQ T k
C
88
tx x x
ty y yk
txy xy xy xyk
T C
Q T
T C
x
yC
…………………………………………**
With Hygrothermal effects; Resultant force 2
2
t
x xt
N d
z
2
11 12 16 11 12 16
2
2 2
11 12 16 11 12 16
2 2
t
o o ox x y yx x y xy
t
t t
x y xy x y xyt t
N Q Q Q zQ zQ zQ dz
T Q Q Q dz C Q Q Q dz
Considering each integral
1st Integral 11 12 16 11 12 16o o ox y xy x yA A A B B B xy (the same as the previous
case)
2nd Integral 1 11 12 161
nk k k k k k T
k k x y xyk
T z z Q Q Q
xN . This integral is
called the thermal stress resultant
3rd Integral 1 11 12 161
nk k k k k k M
k k x y xy xk
C z z Q Q Q
N
M
. This integral is
called the hygroscopic stress resultant
Therefore,
11 12 16 11 12 16o o o T
x x y xy x y xy x xN A A A B B B N N
Calculate other stress resultants, we obtain
89
Mechanical To
o T Mx x x x
o T My y y y
o T Mxy xy xy xy
T Mx x x x
T My y y y
T Mxy xy xy xy
N N N
N N N
N N NA B
M M MB D
M M M
M M M
tal Temp. Moisture
Resultant Strain Resultant Resultant
Thermal stress and moment resultants (NT, MT ) are numerically equal to the
mechanical stress moment resultant which would be required to induce strains exactly
equal to the thermal strain.
Laminate Failure Analysis
Concept: 1. Failure of each ply occurs at a different load level.
2. After a ply fails, stiffness matrices [ABD] have to be modified.
3. Laminate can carry additional load even after many failures have occurred
Design Philosophies:
1. “First ply failure” design philosophy: Service loading does not cause any ply failure.
2. “Last ply failure” design philosophy: Service loading does not cause final ply
failure.
Consider a stress resultant-strain curve of a laminate under uniaxial loading.
A 1st ply failureB 2nd ply failureC Ultimate laminate failureA
B
C
Strain, x
Load, Nx
x(total)
x(2)x
(1) x(4)x
(3)
Nx(2)
Nx(1)
Nx(4)
Nx(3)
Nx(total)
90
The total forces and moment is the summation of forces and moment from each
section
k
nn
n
Total M
N
M
N
1
Similarly, strains and curvature may be determined from
k
nn
no
Total
o
1
The load deformation relationship for each section is
n
no
nn
nn
n
n
DB
BA
M
N
where is the modifies stiffness matrices after the (n-1) ][ nnn DBA th ply failure
Modified stiffness matrices: 2 concepts
1) Some stiffnesses of the failed ply are remained
Failure
In this case of failure, E2, G12, and v12 = 0but E1 0
2) All stiffnesses are zero.
Example
A [0/90/0]s laminate consisting of AS/3501 graphite/epoxy laminae is
subjected to uniaxial loading along the x direction. Use the maximum strain criterion to
find the loads corresponding to first ply failure and ultimate failure, then plot the load-
strain curve.
Solution:
Material properties of AS/3501 are: E1 = 138 GPa, E2 = 9.0 GPa, G12 = 6.9 GPa,
v12 = 0.3, thickness = 0.25 mm
MPa 1448LS , MPa 3.48
TS
91
Thus,
9.600
005.972.2
072.28.138
0Q and
9.600
08.13872.2
072.205.9
90Q
Aij of [0/90/0]s is 900][25.02][25.04 QQAij
900][5.0][ QQ
Without ply failures;
GPa-mm
35.1000
045.7808.4
008.43.1431A
Failure strains are:
0.01051
E
Se L
L and
0.00542
E
Se T
T
Therefore, first ply failure occurs at 90 ply when 0.00541 Tx ee
The laminate force-deformation is
1
1
1 0054.0
35.1000
045.7808.4
008.43.143
0
0
xy
y
xN
Solve,
0
000281.0
mmGPa 773.0
1
1
1
xy
y
xN
Modify [A] to account for first ply failure; 2 approaches
1. All stiffnesses of 90 ply are zero; 0][90
Q
So,
9.600
005.972.2
072.28.138
][][0
2QA GPa-mm
The second strain increment is
0051.00.0054-0.010512 xLx e
92
The incremental force-deformation
2
2
2 0051.0
9.600
005.972.2
072.24.138
0
0
xy
y
xN
Solve,
0
00153.0
mmGPa 704.0
2
2
2
xy
y
xN
The ultimate failure load,
mm-GPa 477.1704.0773.021 xxTotal
x NNN ……………Ans
2. Assume: E2 = G12 = v12 = 0 but E1 = 138 GPa for the failed 90 ply
So,
000
01380
000
90Q
and GPa-mm
9.600
005.7872.2
072.28.138
][ 2A
So,
2
22
2 0051.0
0
0
xy
y
x
A
N
Solve;
0
000178.0
mmGPa 707.0
2
2
2
xy
y
xN
The ultimate failure load,
mm-GPa 48.1707.0773.021 xxTotal
x NNN ……………Ans
The load-displacement curve is
93
Strain, x
Load, Nx
0.01050.0054
1.477
0.773
1.4801
2
Delamination due to interlaminar stress
The in-plane failure of laminated plate is introduced in the previous section.
Another type of failure is failure in form of separation between plies which is called
“delamination.” This mode of failure is caused by the so-called interlaminar stress. To
derive the interlaminar stresses, consider the stress equilibrium.
x
y
z
dx dy
dz x
xx dx
x
xy
xyxy dy
y
xz
xzxz dz
y
Nx
Nx
Equilibrium equation: 0xF
0
xyxx x xy xy
xzxz xz
dydz dx dydz dxdz dy dxdzx y
dxdy dz dxdyz
Rearrange:
0xyx xz dxdydzx y z
or
94
0xyx x
x y zz
………………………………………………………**
By considering the equilibrium in the other direction, we obtain
0xy y yz
x y z
………………………………………………………**
and 0yzxz z
x y z
………………………………………………………**
These are three equation called “stress equilibrium equation.” Consider a
laminate plate under uniaxial loading,
At any point: 0xyx x
x y zz
x is assumed to be constant along the x-direction. That is 0x
x
.
So, to satisfy the equilibrium equation, 0xy xz
y z
Or xyxz z d
yz
From CLT, xy is constant in the interior region. This leads to the conclusion that xz
has to be zero according to the previous integral. However xy is vanished on the free
edges. Thus there is a region which xy is decreased or 0xy
y
. Consequently, xz 0
on the boundary region. This stress may cause delamination of the plates