Jtam-A4.dvi


JOURNAL OF THEORETICAL

AND APPLIED MECHANICS

52, 3, pp. 617-628, Warsaw 2014

THERMAL STRESSES AROUND TWO UPPER CRACKS PLACED

SYMMETRICALLY ABOUT A LOWER CRACK IN AN INFINITE

ORTHOTROPIC PLANE UNDER UNIFORM HEAT FLUX

Shouetsu Itou

Kanagawa University Rokkakubashi, Department of Mechanical Engineering, Kanagawa-ku, Yokohama, Japan

e-mail: itous001@kanagawa-u.ac.jp

Two upper collinear cracks are placed parallel to a lower crack in an infinite orthotropic
plane under uniform heat flux perpendicular to the cracks. The surfaces of the cracks are
assumed to be thermally insulated. The mixed boundary value conditions with respect to
the temperature field and those with respect to the stress field are reduced to dual integral
equations using the Fourier transform technique. In order to satisfy the boundary conditions
outside the cracks, the differences in temperature and displacement at each crack surface are
expanded in a series of functions that are zero outside the cracks. The unknown coefficients
in each series are evaluated using the Schmidtmethod. The stress intensity factors are then
calculated numerically for selected crack configurations.

Keywords: heat flux, three parallel cracks, stress intensity factor, infinite orthotropic plane

1. Introduction

Fiber-reinforced composite materials have been widely used as structural members in airpla-
nes, automobiles and high-speed trains because they are both strong and lightweight. In their
construction, a matrix is reinforced with fibers, which are stiffer and stronger than the matrix.
Therefore, it follows that the composite materials are orthotropic.When thematerials are sub-
jected to repeated cycles of stress, some cracks may develop in the matrix material because the
fibers are stronger than thematrix. If cracked compositematerials are used in high-temperature
or low-temperature environments, heat flows through the materials. In this case, it is useful to
evaluate the stress intensity factors that are caused by the disturbance in heat flux around the
cracks.

The stress intensity factor was first determined for a crack in an infinite plate in which
heat flowed perpendicular to the crack by Sih (1962). Later, Sekine (1977) evaluated the stress
intensity factor for a crack in a half-plane under heat flux. Sekine (1979) also determined the
thermal stresses for two cracks in an infinite plate under heat flux.The two cracks were situated
arbitrarily in the infinite plate, and heat flowed perpendicular to one of the two cracks (Sekine,
1979). Itou (1991) evaluated the thermal stresses for a crack in an infinite elastic layer the upper
surface of which was heated to maintain a constant temperature T0, and the lower surface of
which was cooled to maintain a constant temperature −T0. Itou and Rengen (1993) evaluated
the thermal stresses around two parallel cracks in two bonded dissimilar elastic half-planes the
upper crack of which lied in the upper half-plane, while the other crack was in the lower half-
plane. The stress intensity factors were obtained for a crack in an adhesive layer sandwiched
between two dissimilar elastic half-planes under heat flux perpendicular to the cracks by Itou
(1993). Later, a similar problemwas also solved for the case in which two collinear cracks were
situated in the adhesive layer sandwiched between two dissimilar elastic half-planes by Itou and
Rengen (1995).



618 S. Itou

As for orthotropic elastic problems that are related to thermal stresses, Tsai (1994) deter-
mined the stress intensity factors for a crack in an infinite orthotropic plate under uniform heat
flow. Chen and Zhang (1988) evaluated stress intensity factors around two collinear cracks in
an orthotropic plate under heat flux. Later, Chen and Zhang (1994, 1995) evaluated stress in-
tensity factors caused by a disturbance in heat flux from three coplanar cracks in an infinite
orthotropic plate. The stress intensity factors were also evaluated for two parallel cracks in an
infinite orthotropic plate under heat flux by Itou (2001).
The solutions for an infinite plate are ineffective if a crack exists near the plane surface of a

structuralmember.ChenandZhang (1993) determined the thermal stresses around two collinear
cracks in an orthotropic strip. In their solution, two cracks were placed in the middle surface
of the strip. If the cracks were situated near the stress-free surface, the thermal stress intensity
factors would have been affected by the presence of the surface. Itou (2000) estimated the stress
intensity factors around a crack in an orthotropic layer the upper surface of which maintained
a constant temperature T0, while the lower surfacemaintained a constant temperature −T0. In
the paper by Itou (2000) the stress intensity factors were also determined for a crack under heat
flux.
Since stresses around a crack are very high, a parallel crack may also develop above the

original crack in compositematerials. If a tensile stress is applied to thematerial, a stress increase
and stress shielding occur around the crack ends due to the position of the upper cracks (Kamei
andYokobori, 1974). It is necessary to clarify whether or not the same phenomenon also occurs
in a cracked orthotropic material under heat flux. In the present paper, the thermal stresses
around three insulated cracks in an infinite orthotropic plate are evaluated under heat flux.
Two upper cracks are situated symmetrically on either side of the central crack, and heat flows
perpendicular to the cracks. Themixed boundary value conditions concerning the temperature
field are reduced to two pairs of dual integral equations. To solve the equations, the differences
in temperature at each crack surface are expanded in a series of functions that are zero outside
the cracks. The unknown coefficients in the series are determined from the conditions inside
the cracks using the Schmidt method (Yau, 1967). Next, the mixed boundary value conditions
concerning the stress field are reduced to a set of dual integral equations. The differences in
displacement at the upper and lower cracks are also expanded in a series of functions that are
zero outside the cracks. The Schmidtmethod is used to solve for the unknown coefficients so as
to satisfy the conditions inside the cracks, and the stress expressions are represented by infinite
integrals. From the character of the integrands, the stress intensity factors are defined in the
usual manner and are computed for steel and ceramic-fiber-reinforced ceramic (Tyrannohex)
infinite planes.

2. Fundamental equations

With respect to rectangular coordinates (x,y), as shown in Fig. 1, a crack is situated along the
x-axis from −c to c at y=−h, and two collinear cracks are situated along the x-axis from −b
to −a and from a to b at y =0. For convenience, we refer to −h¬ y ¬ 0 as layer (1), 0¬ y
as upper half-plane (2), and y¬−h as lower half-plane (3).
The fundamental equations for an orthotropic material were derived by Nowinski (1978).

For convenience, the basic equations are shown here. If a state of plane stress is assumed, the
stresses can be expressed by

τxx =Q11εxx+Q12εyy−β1T τyy =Q12εxx+Q22εyy−β2T τxy =Q66γxy (2.1)
with

Q11 =
Exx

1−νyxνxy
Q22 =

Eyy
1−νyxνxy

Q12 =
Eyyνxy
1−νyxνxy

=
Exxνyx
1−νyxνxy

Q66 =Gxy β1 =Q12αyy +Q11αxx β2 =Q12αxx+Q22αyy

(2.2)



Thermal stresses around two upper cracks placed symmetrically... 619

Fig. 1. Coordinate system and three parallel cracks

where Exx, Eyy are Young’s moduli, Gxy is the shear modulus, νxy, νyx are Poisson’s ratios,
and αxx,αyy are the coefficients of linear expansion. The relationships between the strains and
displacements are given as follows

εxx =
∂u

∂x
εyy =

∂v

∂y
γxy =

∂u

∂y
+
∂v

∂x
(2.3)

In equation (2.1), the temperature T satisfies

∂2T

∂x2
+k2
∂2T

∂y2
=0 (2.4)

with

k2 =
ky
kx

(2.5)

where ky, kx are the thermal conductivities.

Substituting equation (2.1) into the equations of equilibrium for the forces reduces these
equations to the forms

Q11
∂2u

∂x2
+Q66

∂2u

∂y2
+L
∂2v

∂x∂y
−β1
∂T

∂x
=0

Q66
∂2v

∂x2
+Q22

∂2v

∂y2
+L
∂2u

∂x∂y
−β2
∂T

∂y
=0

(2.6)

with

L=Q12+Q66 (2.7)

3. Boundary conditions

Uniform heat flow (heat flux) q is applied perpendicular to the three cracks as shown in Fig. 1.
Stresses are caused when the heat flow is disturbed by the insulating effect of the cracks. The
temperature field can be provided using the following boundary conditions



620 S. Itou

∂T1
∂y
=











































∂T2
∂y

at y=0, |x| °

−t at y=0, a¬ |x| ¬ b

∂T3
∂y

at y=−h, |x| ¬∞

−t at y=−h, 0¬ |x| ¬ c

(3.1)

T1 =

{

T2 at y=0, 0 |x|  a, b |x| °
T3 at y=−h, c¬ |x| ¬∞

(3.2)

with

t=
q

ky1
(3.3)

The variables with the subscript “1” are for layer (1). The variables for upper half-plane (2) and
lower half-plane (3) are denoted with the subscripts “2” and “3”, respectively.
The stress field can be evaluated using the following boundary conditions

τyy1 =























τyy2 at y=0, |x| ¬∞
0 at y=0, a¬ |x| ¬ b
τyy3 at y=−h, |x| ¬∞
0 at y=−h, 0¬ |x| ¬ c

(3.4)

τxy1 =























τxy2 at y=0, |x| ¬∞
0 at y=0, a¬ |x| ¬ b
τxy3 at y=−h, |x| ¬∞
0 at y=−h, 0¬ |x| ¬ c

(3.5)

u1 =

{

u2 at y=0, 0 |x|  a, b |x| °
u3 at y=−h, c¬ |x| ¬∞

(3.6)

v1 =

{

v2 at y=0, 0 |x|  a, b |x| °
v3 at y=−h, c¬ |x| ¬∞

(3.7)

4. Analysis

4.1. Fundamental equations in Fourier domain

To find the solutions, the Fourier transforms are introduced as

f(ξ)=

∞
∫

−∞

f(x)exp(iξx) dx f(x)=
1

2π

∞
∫

−∞

f(ξ)exp(−iξx) dξ (4.1)

Applying equation (4.1)1 to equation (2.6) results in

Q66
d2u

dy2
− ξ2Q11u− iLξ

dv

dy
+iβ1ξT =0

Q22
d2v

dy2
− ξ2Q66v− iLξ

du

dy
−β2
∂T

∂y
=0

(4.2)



Thermal stresses around two upper cracks placed symmetrically... 621

Eliminating u or v from equation (4.2), the ordinary differential equations are obtained as

ζ1
d4u

dy4
+ ζ2
d2u

dy2
+ ζ3u= iη1

d2T

dy2
+iη2T

ζ1
d4v

dy4
+ ζ2
d2v

dy2
+ ζ3v= η3

d3T

dy3
+η4
dT

dy

(4.3)

with

ζ1 =
Q22Q66
L

ζ2 =−(Q266+Q11Q22−L2)
ξ2

L

ζ3 =Q11Q66
ξ4

L
η1 = ξ

(

β2−
β1Q22
L

)

η2 =β1Q66
ξ3

L
η3 =

Q66β2
L

η4 = ξ
2
(

β1−
β2Q11
L

)

(4.4)

The Fourier-transformed stresses are found to be

τxx =Q11(−iξ)u+Q12
dv

dy
−β1T τyy =Q12(−iξ)u+Q22

dv

dy
−β2T

τxy =Q66
du

dy
− iξQ66v

(4.5)

Equation (2.4) can now be expressed in the Fourier domain as

d2T

dy2
−
(ξ

k

)2

T =0 (4.6)

4.2. Temperature field

The solutions for equation (4.6) have the following forms for layer (1), upper half-plane (2),
and lower half-plane (3), respectively

T1 =A1 sinh
|ξ|y
k
+B1cosh

|ξ|y
k

T2 =A2exp
(

−
|ξ|y
k

)

T3 =A3exp
(|ξ|y
k

)

(4.7)

where A1,B1,A2, and A3 areunknowncoefficients.Boundaryconditions (3.1)1,3,whicharevalid
for −∞<x<+∞, can be easily satisfied. In order to satisfy equations (3.2), the temperatures
at y=0 and y=−h are expanded by the series

π(T01 −T02)=



















∞
∑

n=1

cn
1

2n
sin
[

nsin−1
(a+b−2|x|
b−a

)

− nπ
2

]

for a¬ |x| ¬ b

0 for 0¬ |x| ¬ a,
b |x| °

(4.8)

π(T−h1 −T
−h
3 )=







∞
∑

n=1

c′ncos
[

(2n−1)sin−1 x
c

]

for 0¬ |x| ¬ c
0 for c |x| °

where cn and c
′
n are the unknown coefficients, and the superscripts “0” and “−h” denote the

values at y = 0 and y = −h, respectively. The Fourier transforms of equations (4.8) can be
expressed by



622 S. Itou

T
0

1−T
0

2 =
∞
∑

n=1

cn
1

ξ
sin
[(a+b)ξ

2
− nπ
2

]

Jn
[(b−a)ξ
2

]

T
−h
1 −T

−h
3 =

∞
∑

n=1

c′n
(2n−1)
ξ
J2n−1(cξ)

(4.9)

where Jn(ξ) is the Bessel function. Then, it can be easily shown that remaining boundary
conditions (3.2) and (3.5) are reduced to the forms

∞
∑

n=1

cnEn(x)+
∞
∑

n=1

c′nFn(x)=−t for a¬x¬ b

∞
∑

n=1

cnGn(x)+
∞
∑

n=1

c′nHn(x)=−t for 0¬x¬ c
(4.10)

where the expressions of the known functions En(x), Fn(x), Gn(x), and Hn(x) are omitted.
Now, equation (4.10) can be solved for the unknown coefficients cn and c

′
n using the Schmidt

method (Yau, 1967). Here, the temperature has been determined completely.

4.3. Stress field

Next, the stress field is evaluated. It can be seen that the solutions to equation (4.3) take
the following forms for i (i=1,2,3)

u1 =C1 sinh(α1y)+D1cosh(α1y)+E1 sinh(α2y)+F1 cosh(α2y)

+ iA1
f1
ξ
sinh
(|ξ|y
k

)

+iB1
f1
ξ
cosh
(|ξ|y
k

)

v1 = iγ1D1 sinh(α1y)+iγ1C1cosh(α1y)+ iγ2F1 sinh(α2y)+ iγ2E1cosh(α2y)

+B1
f2
|ξ|
sinh
(|ξ|y
k

)

+A1
f2
|ξ|
cosh
(|ξ|y
k

)

u2 =C2exp(−α1y)+E2exp(−α2y)− iA1
f1
ξ
exp
(

−|ξ|y
k

)

v2 =−iγ1C2exp(−α1y)− iγ2E2exp(−α2y)+A1
f2
|ξ|
exp
(

−|ξ|y
k

)

u3 =C3exp(α1y)+E3exp(α2y)+ iA1
g1f1
ξ
exp
(|ξ|y
k

)

+iB1
g2f1
ξ
exp
(|ξ|y
k

)

v3 =−iγ1C3exp(α1y)+ iγ2E3exp(α2y)

+A1
g1f2
|ξ|
exp
(|ξ|y
k

)

+B1
g2f2
|ξ|
exp
(|ξ|y
k

)

(4.11)

where Ci,Di,Ei, and Fi are unknown coefficients, and α1 and α2 are the roots of the following
equation

ζ1α
4+ ζ2α

2+ ζ3 =0 (4.12)



Thermal stresses around two upper cracks placed symmetrically... 623

In equations (4.11), γ1, γ2, f1, f2, g1, and g2 are expressed by

γ1 =
ξ2Q11−Q66α21
Lξα1

γ2 =
ξ2Q11−Q66α22
Lξα2

f1 =
[β1(Q66k

2−Q22)+β2L]k2
f3

f2 =
Q66β2k+k

3(β1L−Q11β2)
f3

f3 =Q66Q22−k2(Q266+Q11Q22−L)+Q11Q66k4

g1 =
cosh
(

− |ξ|h
k

)

exp
(

− |ξ|h
k

) g2 =
sinh
(

− |ξ|h
k

)

exp
(

− |ξ|h
k

)

(4.13)

Substituting equations (4.7) and (4.11) into equation (4.5), the stress expressions are obtainable
in the Fourier domain. Equations (3.8) and (3.11) are valid in the entire region of x, and these
can be easily satisfied.

To satisfy equations (3.10) and (3.13), the differences in displacement at y = 0 and at
y=−h are expanded in the following series

π(u01−u02)=



















∞
∑

n=1

dn
1

2n
sin
[

nsin−1
(a+ b−2|x|
b−a

)

−
nπ

2

]

sgn(x) for a¬ |x| ¬ b

0 for 0¬ |x| ¬ a,
b |x| °

π(v01 −v02)=



















∞
∑

n=1

en
1

2n
sin
[

nsin−1
(a+ b−2|x|
b−a

)

−
nπ

2

]

for a¬ |x| ¬ b

0 for 0¬ |x| ¬ a,
b |x| °

π(u−h1 −u
−h
3 )=











∞
∑

n=1

d′n sin
[

2nsin−1
x

c

]

for 0¬ |x| ¬ c

0 for c |x| °

π(v−h1 −v
−h
3 )=











∞
∑

n=1

e′ncos
[

(2n−1)sin−1 x
c

]

for 0¬ |x| ¬ c

0 for c |x| °
(4.14)

where dn, en, d
′
n, and e

′
n are the unknown coefficients to be determined, and sgn(x) is the

signum function. The Fourier transformed expressions of expressions (4.14) are

u01−u02 =−i
∞
∑

n=1

dn
1

ξ
cos
[(a+ b)ξ

2
− nπ
2

]

Jn
[(b−a)ξ
2

]

v01−v02 =
∞
∑

n=1

en
1

ξ
sin
[(a+ b)ξ

2
−
nπ

2

]

Jn
[(b−a)ξ
2

]

u−h1 −u
−h
3 = i

∞
∑

n=1

d′n
2n

ξ
J2n(cξ) v

−h
1 −v

−h
3 =

∞
∑

n=1

e′n
2n−1
ξ
J2n−1(cξ)

(4.15)

Then, the stress field can be expressed by the unknown coefficients dn, en, d
′
n, and e

′
n and the

known coefficients cn and c
′
n.

Finally, the remaining boundary conditions inside the cracks are reduced to the following
forms:



624 S. Itou

— for a¬x¬ b

∞
∑

n=1

dnKnu(x)+
∞
∑

n=1

enLnu(x)+
∞
∑

n=1

d′nKnu(x)+
∞
∑

n=1

e′nLnu(x)=−U(x)

∞
∑

n=1

dnMnu(x)+
∞
∑

n=1

enNnu(x)+
∞
∑

n=1

d′nMnu(x)+
∞
∑

n=1

e′nNnu(x)=−V (x)
(4.16)

— for 0¬x¬ c

∞
∑

n=1

dnKnl(x)+
∞
∑

n=1

enLnl(x)+
∞
∑

n=1

d′nKnl(x)+
∞
∑

n=1

e′nLnl(x)=−W(x)

∞
∑

n=1

dnMnl(x)+
∞
∑

n=1

enNnl(x)+
∞
∑

n=1

d′nMnl(x)+
∞
∑

n=1

e′nNnl(x)=−Z(x)
(4.17)

where the expressions of the known functions Knu(x),Lnu(x), ...,W(x), and Z(x) are omitted.
Equation (4.17) can be solved for the coefficients dn, en, d

′
n, and e

′
n using the Schmidtmethod

(Itou and Haliding, 1997).

5. Stress intensity factors

Using the relationship for a¬x

∞
∫

0

Jn(aξ)
[

cos(ξx), sin(ξx)
]

dξ

=
[

− a
n

√
x2−a2

(

x+
1√
x2−a2

)−n
sin
nπ

2
,

an√
x2−a2

(

x+
1√
x2−a2

)−n
cos
nπ

2

]

(5.1)

the stress intensity factors can be determined as follows

K1a = lim
x→a−

√

2π(a−x)τ0yy1 =
∞
∑

n=1

en
−QL4

√

2π(b−a)

K1b = lim
x→b+

√

2π(x− b)τ0yy1 =
∞
∑

n=1

en
(−1)nQL4
√

2π(b−a)

K2a = lim
x→a−

√

2π(a−x)τ0xy1 =
∞
∑

n=1

dn
QL9

√

2π(b−a)

K2b = lim
x→b+

√

2π(x− b)τ0xy1 =
∞
∑

n=1

dn
(−1)n+1QL9
√

2π(b−a)

K1c = lim
x→c+

√

2π(x− c)τ−hyy1 =
∞
∑

n=1

e′n
(1−2n)(−1)nQL18√

πc

K2c = lim
x→c+

√

2π(x− c)τ−hxy1 =
∞
∑

n=1

d′n
2n(−1)nQL23√

πc

(5.2)

where the expressions of the known constants QL4 ,Q
L
9 ,Q

L
18, and Q

L
23 are omitted.



Thermal stresses around two upper cracks placed symmetrically... 625

6. Numerical examples

The effect of orthotropy on the stress intensity factors can be illustrated by examining Tyran-
nohex as the orthotropic material. Tyrannohex is a ceramic-fiber-reinforced ceramic material
developed by Ishikawa, Kajii, Matsunaga, Hogami, Kohtoku and Nagasawa (1998) and its ma-
terial properties are listed in Table 1.

Table 1.Material properties

Constants Steel Tyrannohex

Exx [GPa] 205.9 135.0

Eyy [GPa] 205.9 87.0

µxy [GPa] 79.2 50.0

νxy 0.3 0.15

νyx 0.3 ·1.01 0.09667
αxx [×10−5/◦C] 1.14 0.32
αyy [×10−5/◦C] 1.14 0.32
kx [W/(m

◦C] 48.6 3.08

ky [W/(m
◦C] 48.6 3.08

For an isotropic material, it holds that

Exx =Eyy νxy = νyx αxx =αyy kx = ky (6.1)

This casepresentsnoproblems in solving the temperaturefield.However, equation (4.12) has two
kinds of multiple roots. Therefore, equations (4.11) cannot be used as the solution for equation
(4.3). The analysis presented here holds, even for an isotropic material, if the value of νyx is
replaced by a value slightly larger than νxy. In this example, steel is selected as a representative
isotropic material, and the constants used for the calculation are also given in Table 1.
The known functions Fn(x) and Gn(x) in equation (4.10) and Knu(x), Lnu(x), ..., Nnl(x),

and Z(x) in equation (4.17) contain semi-infinite integralswith respect to the integral variable ξ.
If the integrands of the integrals donotdecrease rapidly, these aremodified so as todecay rapidly
as ξ increases. Then, numerical integration can be performed precisely using Filon’s method.
First, the Schmidt method is applied to solve for the coefficients cn and c

′
n in equation

(4.10) by taking the first 12 terms in an infinite series. Next, the coefficients dn, en, d
′
n, and e

′
n

in equation (4.17) are solved for. It has been verified that the left-hand side of equation (4.10)
coincides with the right-hand side of equation (4.10). The same applies to equation (4.17).
The length of the upper two cracks is fixed to b− a = c. The stress intensity factors for

steel are calculated numerically against a/c. The results for h/c=0.5, 1.0, and 2.0 are plotted
in Figs. 2a, 2b, and 2c, respectively. In these figures, the stress intensity factors are divided by
Exxαxx

√
πc3t/4 to show the values as non-dimensional quantities. The stress intensity factors

for Tyrannohex are also plotted in Figs. 3a through 3c.

7. Discussion

It is clear that, at the least, one of the values of K1a,K1b, and K1c is negative. Therefore, the
crack surfaces come into contact with each other at one or more of the crack ends. In this case,
the boundary conditions with respect to the temperature field fail to be valid. The same applies
to the boundary conditions with respect to the stress field. In the present paper, it is assumed
that the crack surfaces do not come into contact with each other due to the existence of a thin
gap.



626 S. Itou

Fig. 2. Stress intensity factors versus a/c for h/c=0.5 (a), h/c=1.0 (b), h/c=2.0 (c)
and b−a= c (steel)

Fig. 3. Stress intensity factors versus a/c for h/c=0.5 (a), h/c=1.0 (b), h/c=2.0 (c)
and b−a= c (Tyrannohex)



Thermal stresses around two upper cracks placed symmetrically... 627

8. Conclusions

Based on the numerical calculations outlined above, we can draw the following conclusions:

• It is clear that the values of K2c are not affected by the presence of the upper parallel
cracks for a value larger than h/c=2.0. However, if the value of h/a decreases, the stress
intensity factor K2c appears to increase. For h/c = 0.5, the peak value of K2c occurs
near a/c=0.8, and stress shielding occurs for a/c¬ 0.3. In the present paper, the stress
intensity factors for h/c< 0.5 were not successfully calculated. However, it can probably
be estimated that a severe increase in the stresses occurs if h/c decreases.

• For h/c=2.0, the lowest value of K2b appears near a/c=0.8. However, the value of K2b
has a large value near a/c = 1.0 for h/c = 0.5. Therefore, it can be estimated that the
value of K2b can be very large if h/c decreases.

• Thevalues of K2c forTyrannohexare somewhat smaller than those for steel, andthecurves
forTyrannohexare similar to those for steel. Therefore, the safety of the cracked structural
member will be ensured by comparing the fracture toughness value of Tyrannohex with
the values obtained from the curves for steel.

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Manuscript received October 21, 2013; accepted for print January 16, 2014