An orthotropic laminate composite containing a completely broken layer is considered. The problem is formulated in terms of integral transforms and then reduced to a singular integral equation which is solved numerically. The strength of stress singularity at the crack tip is determined from a characteristic equation which is obtained by studying the dominant part of the singular integral equation near the end points. The stress intensity factors are given for various material properties

Arin, K.

English

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https://ntrs.nasa.gov/search.jsp?R=19750016783 2020-03-22T21:56:32+00:00Z
IFSM-74-59
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LEHIGH 0 :
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A NOTE ON A BROKEN LAYER IN AN
ORTHOTROFIC LAMINATE COMPOSITE
(tiASA — C'- 1 += )U 3)	 A E, .l
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BY
K. ARIN
N75-
U11C IdS3/24	 221f,6
TECHNICAL REPORT NAS' TR-7/1-4
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MARCH 1974
NASA NGR-39-097-911
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7
A NOTE ON A BROKEN LAYER
IN AN ORTHOTROPIC LAMINATE COMPOSITE
K. ARIN
Lehigh *University, Bethlehem, Pa., 18015, U.S.A.
ABSTRACT
An orthotropic laminate composite containing a completely
broken layer is considered. The problem is formulated in
terms of integral transforms and then reduced to a singular
integral equation which is solved numerically. The strength
of stress singularity at the crack tip is determined from a
characteristic equation which is obtained by studying the dom-
inant part of the singular integral equation near the end points.
The stress intensity factors are given for various material
properties.
1. INTRODUCTION
In a previous work by Arin [1), the problem of an ortho-
tropic laminate composite containing a layer with a crack and
bonded to two half-planes of dissimilar materials has been
solved and the effect of the material properties as well as
various crack sizes on the stress intensity factor at the tip
of the crack illustrated. It has been shown that there are
basically two types of•orthotropic materials. The results were
given for two different cases (Materials Type I and II). It
was concluded that the decrease in the stress intensity factors
1
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{ y
i7a
I
I^
4J
a^
t^
ill
rtp
.,kk
N
.y
o.
I	 a^
was apparent in the case of a matrix of stiffer material.
This is meaningful from the point of view of crack arrest.
In this paper, the limiring case for which the crack
reaches the interface will be investigated. It can be shown
that the Fredholm kernels in [1] become unbounded as the crack
touches the interface, hence requiring a separate treatment
of the problem. Extracting the additional singularities, the
singular integral equation of [1] can be solved in a similar
manner. Here, only the material Type I will be considered.
The solution for material Type II can be obtained similarly.
The notation and certain results of [1] wiwl be used wherever
necessary. Both the plane strain and the generalized plane
stress conditions are studied simultaneously.
The isotropic counterpart of this problem has been solved
by Gupta [2] and Ashbaugh [3].
2. FORMULATION OF THE PROBLEM AND THE SOLUTION
The singular integral equation of the problem of an or-
thotropic laminate composite containing a layer with a crack
of length ?.a is given as (1]:
1 a f(t)dta	 (1-VxyVyx)
F fa	 t-x + fa k(x,t)f(t)dt = - 2Y2 E	 p(x), JxJ<a0 y
subject to
fa ^(t)dt = 0
	
(2.1)
2
aav1(t10)
^(t) =
	 a t
	,	 ltl<a	 y = 0 (2.2)
r	 where
Y'
ll
t9
t^
w
"	 P
I	 .^
i
u
I yY
rr
ofrI	 ^
and v1 (x,y) is the displacement component in y direction corre-
sponding to the layer. Here, Ex, Ey are the Young's Moduli;
vxyP Vyx are Poisson's ratios and Gxy is the shear modulus. The
bielastic constants Y i
 are given in Appendix A and [1) and the
kernel k(x,t) is given in Appendix B. The input function p(x)
is the crack surface pressure and corresponds to the perturbation
problem. (2.1) is given for the generalized plane stress. The
plane strain case is obtained by simply interchanging certain
constants (see [11).
However, the Fredholm kernel k(x,t) becomes unbounded for
a = h which represents a broken laminate. It can easily be
shown that in thie case, the part of k(x,t) which contributes
to further singularities can be expressed as (see Appendix B)
k (x,t) = 1 f[k*(x,n)e-(h—t)n Wl wlI
s	
xY20 o 1
+ k*(x,n)e-(h-t)n s5/jw3lldn	 (2.3)
2
where k*(x,n) and k,*^(x,n) are the asymptotic expressions of
kl(x,n) and k 2 (x,n) as n + ^.
Hence, equations (2.1) can now be written as
1 f  ^Wdt + f'`' k `.,t)^(t)dt + f  k (x,t)^Mdt
r -h	 t-x	 , L	 h f
(1•VxyVyx)
P(x)	 lxl<h2v20Ey 
Ih ^(t)dt = 0	 (2.4)
3
_V
i
5.	 X
_	 1
kf (x,t) = k( x ,t) - ks (x,t) = 1ry
	
t ( [kl ( x ,n) - k*(x,n)l
20 0	 1
e-
 
(h-t)n 05/ jw1I + [k2 (x,n) - k2(x , n)le-(h-t ) n^/Iwgl
t
where
9
w
'H
A^,
S
} do	 :)
It is quite easy to show that k f (x,t) is a bounded kernel and
thus can be evaluated numerically from (2.5).
On the other hand, k s (x,t) can be obtained in closed form.
From the Appendix S, the asymptotic expressions of k l (x,n) and
k2 (x,n) can be given as follows:
k*(x,n) = Y
51e
-Iwllnhcosh ( wlnx ) + Y52e-Iw3lnhcosh(w3nx)
-Iw l lnh	 -Iw31nh
k*(x,n) = Y 53 e	 coch(wlnx) + Y 54 e	 cosh(w3nx) (2.6)
2
Hence, after intermediate manipulations we arrive at
ks (x,t) = 2rrY20 3=1 
e j [ t- ( aj h+b j x) + t- aj -bjx)l
	
j= 1,..4
	
(2.7)
where bielastic constants a j ,b j and cj are given in Appendix A.
In dimensionless variables ( 2.4) becomes
3
7T I1 {TlX + 2y 2o. =1cj[ T- (aj+b j X) + T-(a.-bjX)l}
00 (T)dT + hfi k f (hX, hr) ^0 ( T ) dT = 9(X)	 ,	 IXI^1
f l ^0(T)dT = 0
	 (2.8)
4
iwhere
	
'I
t^
i
L_	 11, . A
T = t/h , X = x/h
(1-vxyvyx)
(2.9)(hX)^o (T) = a(t)	 g(X) _ -	 2y F	 P	 r20 y
The solution of (2.8) can be expressed as follows [5]:
	
F(T)	 h2YF(t/h)e:riy
^o ( T ) =	 2	 _ 	 r	 'TI<1	 (2.10)
	
(].-T )
	
.(t- h)Y(t+h)Y
where the unknown function F(T) is Holder continuous in the in-
terval -1<T<l and o<Re(y)<l.
The strength of singularity Y can be determined by investi-
gating the dominant part of (2.8) or (2.4) near the end b ints
(see [51). Defining the sectionally holomorphic function
(D(z) = 1 fh I(t)dt	 (2.11):r -h	 t-z
following results can be obtained:
Y
_ (2) cot^ry[ F(-l) _ F(1) Y] + (1(zl)	 h<zl=x<h^(zl)
(h+x)
	
(h-x)
	
Y	 h < z	 ..^(z2j ) -
 (-25j ) F(1)	
+ fi 2 ( z2 )	 2j =ah+bx <7	 7
 sinny(h+x)
(1 + 2bj)h
h(D (z	 Y	 F (1)	 + ID (z )
	
h < z 3j=a jh-b j x <37 ) -	 (2bj) sinRy(h-x)Y
	
3 3j
(1 + 2b j )h	 (2.12)
A = 1,2,3
where (P i , i = 1,2,3 are bounded functions everywhere except the
d of is where the behave as follows:en p n	 Y	 ) .,
^^ (z
 D1)
^1 (z l ) < z+Lh a ' 0 2 (z 2j ) < z -h a
	
1-	 1 2i	 2
5
i+
i^
11
i
^i((
I'1
t.
d
i
f
I
`d
0 3 (z 3j )	 < z D2
—
h a	 (2.13)
37	
3
where aj and Dj are real constants and a j <Re(y)• Also, ,D*(zl)
satisfies the Holder condition near and at the end points.
Substitution from (2.12) into the dominant part of either
(2.4) or (2.8) and considering that F(T) is an odd function,
the following characteristic equation is obtained to determine y:
cosnY + 1 E	 fL = 0
	 (2.14)
2Y 20 j=1 bjY
which can be shown to have at least one real root.
After determining y, (2.8) can be solved using the numeri-
cal method given in (4), i.e.
N
7 E	
F(Tj)Wj(T 
j 
1X 
i 
+ 7rhk s (hX i ,hr j ) + frhkf(hXi,hTj)l
j=1
9(Xi)	 i = 1,..,N-1
N
E	 W.F(Tj) = 0
j=l	 (2.15)
where
PN(-Y,-Y)(Tj) = 0	 j = 1,..,N
PN-1
(1-Y,l-Y) (X i ) = 0	 i = 1,..,N-1	 (2.16)
and the Wj are the corresponding weights. From (2.15) N unknowns
F(T j ), j = 1,..,N can be determined.
3. THE STRESS INTENSITY FACTOR
The stress intensity factor at the crack tip will be de-
FI s
fined as follows	 [txy2 (x,0)	 = 0]:
J	 K = lim	 [2(x-h)] ya	 (x,0)	 (3.1)
x+h
x>h a,
y
l^	 where ay2 (x,y) corresponds to the matrix.	 aY2 (x,0)	 can be ob-
tained in terms of ^(t) starting from the expressions given in
][1].	 Hence, after intermediate steps we obtain,
'Y 19 (1-vxYVyx )	 h	 do
-	
aY2(x,0) = fh f(t)dt o fT]
I
E*
Y i
-(h-t)nS5/Iw 1 I	 -Iw3I(x-h)n
	 Y15 	-Iw11(x-h)n a
{e	 [Y16fl3(n)e	 + Y21 f17(n)e	 ]
- (h-t)n 05/I w 3 I	 -Iw3I(x-h)n 	 Y15
+ e
	 [Y* fl4(n)e	 + . — f1.(n)
Y2116
-Iw*I(x-h)n	 (3.2)
@	
e	 1	 ] }.8 
where
^^ 5y fll(n) = Y21 +Y22 tanh(winh)
5
f12(n) = — + Y22 tanh(w3nh)
21
f 13 (11)	 = Y 53 f(n)	 + f 7 (n)f11 (n)	 + f8(n)f12(n)
f 14 (n?	 = Y 54 f(n)	 + f9 (TO f ll( n )	 + f 10 (n)f12 (n)	 (3.3)
f15(n) = R7 + S8 sign(w3)'tanh(winh)
f16 (TI) = Sa + p**	 sign(w3)	 tanh(w3nh)
t	 f17 (n)	 = Y 51 f(TO	 -	 f 7 (n)f15 (n)	 -	 f$(n)f16(n)
a
f 16 (n)	 = 'Y	 f(n)	 -	 f 9 (n)f 15 (n)	 -	 f 10 (n)f 16 (n)	 (3.4)52
I
7
l
1i
^f
!I
However, the right hand side of (3.2) becomes unbounded
near the tip of the crack. Hence, the dominant part which con-
tributes to the singular behavior can be extracted in a manner
similar to the procedure used to obtain (2.4). Thus, ay2(x,0)
can be expressed as:
(1-
vx^x) a (x,o) = E  	 !h — (t)dt - + a° (x,o)
Gy	 Y2	 j=1 r _h t-(ash+box)
	
Y2
(3.5)
where aY 2 (x,0) is a bounded function.
To determine the stress intensity factor K, the behavior
of the Cauchy integral in (3.5) near the end points has to be
investigated. Following the method given by Muskhelishvili (5),
the sectionally holomorphic function defined in (2.11) can be
expressed near x = h as follows (h<x<-):
4)(z	 ) _ _ ( h )Y	 F(1)	 + (D (z	 )	 r
4	 2b3*.	 sinTry (x-h)Y	 o 4i
h < z 4j =ash+box < m	 (3.6)
Here, 4)o (z4. j ) is bounded everywhere except the end points, and
near x = h
$° (z4 ) I <	 C a 	 ao <Re (Y)Iz4.1-h^ o
where C and a  are real constants. Hence from (3.1), (3.5) and
(3.6) we obtain
i
8
-	 OF(1)E*	 4	 c*
K = * --Y
-- £ ---- , generalized (3.7)
(1-vxyvYx ) siniry	 j=1 (bj ) y 	 plane stress
1-vXYVYXFor plane strain	 *	 = Ex 0 should be replaced by 1/A22
M	 (see (11).•'	 EY^ 
4. NUMERICAL RESULTS AND CONCLUSION
The numerical results will be given for p(x) = po = constant
which corresponds to uniform crack surface pressure. K/pohy
values as well as y will be given for different material combi-
nations.
The following materials will be selected:
(A) Boron-Epoxy:
Ex = 3.5 x 10 6psi	 Ey = 3.24 x 107psi
vyx = 0.23
	
G,Y = 1.23 x 106psi
and for the plane strain
E z = 3.5 x 106psi	
vzx = v
zy = 0.23
(B) Boron-Epoxy:
Ex = 2.72 x 10 7psi, EY = 5.5 x 106psi
vyx = 0.22
	 GxY.= 7.0 x 10 5 psi
and for the plane strain
E Z = 2.72 x 10 7psi v zX = vzy = 0.22
(C) Boron-Epoxy:
Ex = 5.5 x 10 6psi , By = 2.72 x 107psi
vyx = 0.1084	 Gxy = 7.0 x 105psi
and for the plane strain
E z = 5.5 x 10 6psi , vzx = vzY = 0.1084
(D) Glass-Fiber (208 volume fraction):
Ex = 6.6 x 10 5psi	 Ey = 2.52 x 106psi
vyx = 0.32	 Gxy = 2.9 x 105psi
and for the plane strain
Ez = 6.6 x 10 5 psi, vzX = vzy = 0.32
9
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Both y and K/pohy
 are given in Table 1 for various material
combinations, As seen from the equation (3.7), the stress in-
tensity factor is determined in terms of F(1). Since the un-
knowns obtained from (2.15) are F(T j )(j = 1,..,N), F(1) is found
by an extrapolation. Hence, three different values for K/pohy
are given in Table 1 to illustrate the effect of extrapolation.
Columns 1, 2 and 3 are obtained by extrapolating F(T j ) ( j = 1,..,N),
(j = 1,2,3) and (j = 2,3,4) respectively. From these results, it
appears that F(T 1) (as well as F(TN )) has a significant effect on
the stress intensity factor. It is also observed that the y
and the K/pohy
 values are quite sensitive to the material properties.
ACKNOWLEDGEMENT
The author gratefully acknowledges the support of this work
by the National Aeronautics and Space Administration under the
Grant NGR-39-007-011.
REFERENCES
1. K. Arin, "An Orthotropic Laminate Composite Containing
a Layer with a Crack", NASA Technical Report TR-74-1,
Lehigh University, IFSM-74-57, March 1974.
2. G.D. Gupta, "A Layered Composite with a Broken Laminate",
Int. J. Solids Structures, Vol. 9, pp. 1141-1154, 1973.
3. N.E. Ashbaugh, "Stresses in Laminated Composites Containing
a Broken Layer", ASME Paper No. 72-WA/APM-14.
4. F. Erdogan, G.D. Gupta, T.S. Cook, "The Numerical Solutions
of Singular Integral Equations", Methods of Analysis and
Solutions to Crack Problems, ed., G.C. Sih, Wolters-Noordhoff
Publishing, 1972.
i
5. N.I. Muskhelishvili, Singular Integral Equations, Wolters-
Noordhoff Publishing, Groningen, 1958.
11
L
a5
o :9
APPENDIX P,
For Material Type I:
Dimensionless bielastic constants y. (i = 46,70):
(see reference [1] for the constants 
Y 1 - Y45)'
Y46 = IY 27
 + Y26 sign( w l )][Y25 + Y24 sign(w3)]
- [Y 29 + y28 sign(w 3 )][Y 23 + Y 24 sign(wl))
Y47 = Y 38 + Y 39 pign(w3)
Y48 = Y40 + Y41 sign(wl)
Y49 = Y42 + Y43 sign(w3)
Y50 = Y44 + Y45 sign(wl)
Y 51 = Y 7
 Y47/(Y19•Y46)
Y52	 '8 Y48/(y19y46)
Y53 = Y7 Y49/(Y19y46)
Y54 = Y8 Y50/(Y19y46)
Y 55 = R9 sign(w l )	 -	 R$ sign(w3) a
k
Y56 YllY12
IR8 sign(w3)	
- R10 sign(w3)]
d
S	 Y57
_
- Y22 - Y1 R9 sign (w l)21
Y 58 = y1112
I- Y22 +1
	
R10	 sign(w 3 )I21
ryry
S
7
_s
Y 59 y21
+ Y22 sign(wl)
r12
"]
i^
0
{
t
i
8
j
i
--- -,--
Y60 = t3^Y21 + Y22 sign(w 3)
e
Y61 = 07 + 0 * slgn(w*3 ) sign(w1)
Y62 = $8 + 0** sign(w3) sign(w3)
!	 Y63 = Y57 Y46 + Y47 Y59 + Y48Y60
Y64 = Y58 Y46 + Y49 Y59 + Y50Y60
i Y65 = Y55 Y46 - Y61 r 47 - Y62Y48
Y66 = Y 56 Y46 - Y61Y49 - Y62Y50
X67 = Y16Y63/(Y46Y19)
Y68 - Y65•Yi5/(Y46Y19Y21)
Y69 - Y16 Y64 / (Y46Y19)
Y70 - Y66Y15/(Y46Y19Y21)
Bielastic constants c i , b i , and a  (i = 1,3):
w1
	 12
c 1 = - S Y51 , b^
S	 5
^ w1 Y 52 + 1 31 Y 53	 _ wl fD3c2
 =	 5	
^ b 2
	
5
X31
	
w32
c3 - S5 Y54 ' b3 =
5
ai=1+-Di
13 6
I ;a
i9
MK-
R
f
1
i
1
Sielastic constants c, bi, ai (i = 1,4):
w	 ^
C*
	
^
= 	
Y67 , b1 =	
1	
x
3
41-5 	 5
02 = - B Y68 b2 = ^^ U5
°3 = -	 Y69	 b3 = *-31-1-3-1
5	 5
C* L= -	 l Y70 , b4 = 
L 3  Ij w 1
5	 5
^1 = 1-bi
14
AP?DNDIX B
Fredholm kernel (Bielastic constants Y  are defined in [13):
k(x,t) = ny	 l [kl(x,n)e
20 0
-(h-t)n s5/Iw3I
+ k 2 (x,n)e	 ] do
k 1 (x,n) = 2y19 n [Y 7f5 (x ,n) f 7 (n) + Y8f6(x,n)f8(n)I
k2 (x,n) = 2y19 (n [Y 7 f5 (x,n) f9 (n) + Y8f6(x,n)f10(n)1
f(n) = f 3 (n)f1 (n) - f2(n)f4(n)
f1 (n) = Y25 + Y24tanh(w3nh)
f 2 (n) = Y29 + Y28tanh(w3nh)
f3 (n) = Y 27 + Y26tanh(wlnh)
a
f4(n) = Y 23 + Y24tanh(wlnh)i,
f5(n,x) = cosh(wlnx)/cosh(wlnh)
f6 (n,x) = cosh(w3nx)/cosh(w3nh)
f 7 (n) = Y 38 + Y39tanh(w3nh)
f 8 (n) = Y40 + Y4ltanh(wlnh)
f9 (n) = Y42 + Y43tanh(w3nh)
i
f10(n) = Y 94 + Y45tanh(wlnh)
15
.'30,
x
F..
i
Figure 1. Geometry of the problem.
^,
ii


A note on a broken layer in an orthotropic laminate composite

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