1
Chapter 9 Exercises of Chapters II.1 (Various Types of Damage)
and II.2 (Fracture Mechanics)
9.1 Linear elastic fracture mechanics
Exercise 9.1 Crack in a plate under tension
In linear elasticity, the strain energy release rate G is related to the differential of the compliance C, that is the displacement produced per unit applied
load. G is proportional to the square of the stress intensity factor K, which
fixes the level of the strains and stresses close to the crack tip. This factor is
independent of the properties of the material; it depends only on the applied
load (or displacement), on the crack size and on the geometry of the cracked
part.
a) An H long wide plate of width W and thickness B under tension has a central
crack of length 2a, small with respect to W, inclined at an angle β with respect to
the loading axis (Fig. 9.1.1). Determine the compliance of this plate.
Solution
The compliance C is related to the strain energy release rate G:
∂C 2G
=
∂A P 2
(9.1.1)
where A is the crack area and P the applied load.
The strain energy release rate is related to the stress intensity factors:
G=
1 −ν 2
( K I 2 + K II 2 )
E
where ν is the Poisson ratio and E the Young modulus.
(9.1.2)
2
Fig. 9.1.1. Wide plate with an inclined central crack
As:
K I = σ πa cos 2 β
(9.1.3)
K II = σ πa sin β cos β
where σ is the tensile stress equal to P/BW, we find
∂C
1 − ν 2 cos 2 β
= 4π
a
∂a
E BW 2
(9.1.4)
Introducing the plate compliance in the absence of crack, the integration of
(9.1.4) yields:
C=
2

H 
2 W  a 
2
+
π
−
1
2
1
ν

(
)
  cos β 
BWE 
H W 

(9.1.5)
Note that when the crack is parallel to the load axis, the compliance is that of
the un-cracked plate.
b) Calculate the compliance in compression, assuming first perfect glide with no
friction of the crack faces.
3
Solution
In this case, the normal stress on the crack faces being compressive, the stress intensity factor KI vanishes. In (9.1.5), a factor sin2β must be included in the second
term in the bracket.
c) Find the orientation θ of the facet at the crack tip on which the normal stress is
maximum.
Solution
In mode I, the normal stress on a facet making an angle θ with the crack plane is:
σ θθ =
KI
cos3 θ
(9.1.6)
2πr
In mode II, it is:
σ θθ = −
K II 3
θ
sin θ cos
2
2
2πr
(9.1.7)
(It is a good exercise to demonstrate these relations: Eqs.2.48 and 2.52 of Volume II)
Therefore, for the inclined crack:
r σ θθ  2
θ 3
θ
=  cos β cos3 − cos β sin β sin θ cos 
2 2
2
a σ

(9.1.8)
The maximum is found by putting the derivative of (9.1.8) with respect to θ
equal to 0. The result is:
1


cos 2 β  
sin 2 β  2 
tan =
1− 1 + 8
2 4 sin β  
cos 4 β  


θ
(9.1.9)
Table 9.1.1 gives the variation of the angle θ as a function of the angle β, as
well as the sum of the two, which is the angle between the facet with the highest
normal stress and the perpendicular to the loading axis of the plate.
4
Table 9.1.1. Variation of the angle θ as a function of the angle β
Angle β (°)
Angle θ (°)
Angle β +θ (°)
0
0
0
10
-19.2
-9.2
20
-34.7
-14.7
30
-46.1
-16.1
40
-54.5
-14.5
50
-60.7
-10.7
60
-65.1
-5.1
70
-68.2
1.8
80
-69.9
10.1
90
-70.5
19.5
d) Study the evolution of the stress intensity factor in compression for various angles β assuming a friction coefficient f between the crack faces.
Solution
The stress intensity factor KI vanishes while a friction shear stress opposes the
glide of the crack faces, so that:
K II
tan β − f
= ( sin β cos β − f cos 2 β ) =
1 + tan 2 β
σ πa
(9.1.10)
The factor KII vanishes if tanβ < f. It reaches a maximum equal to
(
1 2 1+ f 1+ f 2
) for tan β = f +
1 + f 2 . Some results are shown in Fig. 9.1.2.
5
Fig. 9.1.2 Reduced stress intensity factor for a central crack inclined at an angle β in a wide plate
in compression.
For given values of the critical reduced stress intensity factor and of the friction
coefficient, the cracks which propagate are those making an angle β with the perpendicular to the loading axis within a certain interval.
e) Find the orientation θ of the facet at the crack tip on which the normal stress is
maximum.
Solution
From Eq. 9.1.7 it is deduced that the angle θ is equal to arccos(1/3) in all cases.
This shows that the cracks which propagate deviate in a direction parallel to the
loading axis. The stress intensity factor decreases and the propagation stops.
f) Calculate the compliance of the wide plate when the crack is perpendicular to
the loading axis and its size 2a is not negligible with respect to the width W. Several approximations of the stress intensity factor are proposed, for instance:
()
1
(2)

πa 
K I = σ πa  cos 
W

K I = σ πa
12
(9.1.11)
1− a W
12
(1 − 2a W )
6
Solution
The strain energy release rate is related to the stress intensity factor: G = KI2/E,
and to the derivative of the compliance.
Approximation (1) yields in plane stress:
C=
H 
W a
πa 4 W
sin
+
1 + 4
WBE 
HW
W π H

πa  
 cos W − 1 

(9.1.12)
Approximation (2), valid for a/W < 0.1, (Tada, Paris, Irwin 1985) yields in
plane stress:
C=
H 
W
1 + π
BWE 
H
1
2 1
3 

1 −log (1 −λ ) − (1 −λ ) + 2 (1 −λ ) + 3 (1 −λ )  

(9.1.13)
2a
=λ
with
W
Note that these expressions for small values of a/W coincides with Eq. 9.1.5
(derived in plane strain). For a/W approaching 1/2, Eq. 9.1.12 does not diverge.
Approximation (2) appears to be better in that respect, in spite of its limited validity.
g) A spring of compliance CR is inserted at the point where the load P is applied.
Find the general expression of the strain energy release rate as a function of the
total displacement of the spring head.
Solution
The total displacement u of the spring head is equal to (C + CR)P. The solution is
found by using this value of the load in the expression of the strain energy release
rate.
Exercise 9.2 Crack emanating from a hole
The stresses around a hole in an infinite plate can be calculated using analytical functions (see François 2009, Sect. 2.3.3 p.81).
σ xx + σ yy = 2 Φ′ ( z ) + Φ′ ( z ) 
σ xx − σ yy + 2iσ xy = 2  z Φ′′ ( z ) + Ψ ′ ( z ) 
(9.2.1)
7
For an ellipsoidal hole of semi-axes R(1+m) and R(1-m) the perturbation
to the stress field is represented by the analytical functions:
1
1
Φ (ζ ) = − σ R (1 + m )
ζ
2
1
1 1 + mζ 2 
Ψ (ζ ) = − σ R (1 + m )  2
+1
ζ  ζ − m 
2
(
(9.2.2)
)
with z = R ζ + m ζ .
Various solutions for the stress intensity factor can be found in Table 2.5
(Volume II p.96).
a) A large plate loaded by a uniform tensile stress σ has an ellipsoidal hole of
semi-axes (1+m)R and (1−m)R. A crack of length a emanates across the whole
thickness of the plate from the end of one of the semi-axes. Propose an approximate stress intensity factor for this crack.
Solution
Solutions exist for either a very small crack or for a very long one. In the first case
the problem can be assimilated to a surface crack in a semi-infinite plate:
K I = 1.122K Tσ πa = 2σ K T a
(9.2.3)
KT being the stress concentration factor at the end of the semi-axis.
For the ellipsoidal hole, at the end of the long semi-axis:
K T = 1+ 2
1+ m
1− m
(9.2.4)
(For the short semi-axis change m to –m).
When the crack is very long, it can be assimilated to a crack of length 2 (1+m)R
+ a, in which case:
(
)
K I = σ π  1+ m R + a 2 
In the case of a circular hole (m = 0), Eqs. 9.2.3 and 9.2.5 reduce to:
(9.2.5)
8
K I = 6σ a
(
KI = σ π R + a 2
)
(9.2.6)
b) How are the above solutions modified when a uniform stress λσ is applied in a
direction perpendicular to preceding loading axis? Calculate the result for a circular hole.
Solution
A positive stress λσ creates a compressive local stress at the end of the semi-axis.
This can be calculated from Eqs. 9.2.1 and 9.2.2.
For a circular hole:
λσ
2ζ 2
λσ
Φ′′ = −
Rζ 3
Φ′ = Φ′ =
Ψ′ =
(9.2.7)
λσ ζ 2 + 3
2 ζ4
so that, from (9.2.1):
σ xx 1 ζ 1 ζ 2 + 3
=
− +
λσ ζ 2 ζ 3 2 ζ 4
(9.2.8)
At the edge of the hole for z = iR, (9.2.8) yields σxx/λσ = −1.
The first equation (9.2.6) becomes:
K I = ( 3 − λ ) 2σ a
(9.2.9)
This is the solution given by Tada, Paris and Irwin (1985) for very small values
of a.
The effect of the stress parallel to long cracks is negligible as the normal stress
away from the hole is zero. Tada, Paris and Irwin show that this is the case as soon
as a/(a + R) is larger than 1.
Exercise 9.3 Double cantilever beam (DCB) test piece
9
A sketch of the double cantilever beam test piece is given in Fig. 2.14 (Volume
II p.37). The deflection of the beams can be calculated (Volume II Sect. 2.2.7.3
p.37) and so the compliance. This enables to determine the strain energy release rate by derivation:
G=
12 P 2 2 3h3
1
a =
Eλ 2 4
2 3
4
B h E
a
(9.3.1)
where P is the applied load on each beam, B is the width, h the height and λ
the deflection of each beam, E the Young modulus and a the length of the
beams that is the crack length ( a ≫ h ).
a) Determine the shape of a similar test piece such that the strain energy release
rate is independent of the crack length.
Solution
If the strain energy release rate is independent of the crack length, so is the derivative of the compliance with respect to the crack area Ba. For a beam the deflection
w (0) where the load is applied is such that:
d 2 w 12 Pz
=
dz 2 EBh3
(9.3.2)
with the boundary conditions: dw / dz(z = a) = 0 and w(z = a) = 0. If the width B is
constant, assuming that the height h varies as zm, the deflection λ = w (z = 0) is
proportional to a3-3m. The derivative with respect to aB of the compliance, which is
equal to w (z = 0) divided by the applied load, must be independent of Ba, or simply of a if B is constant. This means that 3 – 3m = 1.
So that h must be proportional to a2/3 (or z2/3) for G to be independent of the
crack length.
In the same way, if now h is constant, B should vary as a for G to be independent of the crack length. However, it seems better to keep the crack length constant
in the experiment, so that the solution with a constant width looks preferable.
b) Design a test piece to measure the fracture toughness of a tubular material of
radius r and of wall thickness t.
Solution
The tube could be partly split a length a along two opposite generatrices and
wedge opened. If a is large enough, the displacement w of one part of the tube is
such that:
10
2 Pz
d2 w
=
dz
πtr 3 E
(9.3.3)
The compliance of one part of the tube can be deduced:
C=
2 a3
3π tr 3 E
(9.3.4)
If the wedge imposes an opening displacement 2λ, the strain energy release rate
is deduced:
G=
9π r 3
Eλ 2
4 a4
(9.3.5)
Exercise 9.4 Temperature gradient
A plate of thickness W, large with respect to the crack size, in which there is a linear temperature gradient ∆θ/W, is constrained so that it cannot bend. The thickness being small plane stress applies.
a) Determine the stress field.
Solution
Assuming that the temperature is equal to −θ on one side of the plate (x1 = −W/2)
and +θ on the other (x1 = +W/2), the strain field is such that:
2θ
x
W 1
(9.4.1)
E
2θ
α
x
1− ν W 1
(9.4.2)
ε1 = ε 2 = α
α being the thermal expansion coefficient.
The stress field is deduced:
σ1 = σ 2 =
where E is the Young modulus and ν the Poisson ratio.
b) Determine the stress intensity factor.
11
Solution
The stress intensity factor can be found by integration of the solution for two opposite point forces applied on the crack faces. In the case of a central crack of
length 2a:
 a + x1 
KI = ∫ σ 2 

−a
 a − x1 
+a
12
dx1 =
E
a
αθ
1− ν
W
πa
(9.4.3)
In the case of an edge crack of length a, it opens in mode I on the cold side
(x1 = -W/2), and:
KI =
∫
a−W 2
−W 2
2θ 
2
E
α
x1 
1 − ν W  π a − W 2 − x1

(

 dx1

)
(9.4.4)

2 2 E
4 a
=
αθ  1 −
πa
π 1− ν
3 W 

c) Hot water flows inside a tube. Where is a crack likely to appear?
Solution
The wall near the inner radius is in compression while it is the contrary near the
outer radius. This is where a crack is likely to appear.
However, if the water is under a sufficient pressure, the situation could be reversed. The hoop stresses due to the pressure p at the inner radius ri and at the
outer radius re are:
1+ k2
p
1− k2
2k 2
r = re =
p
1− k2
σ θθ (r = ri )=
σ θθ (
(9.4.5)
)
where k is the ratio ri/re.
The hoop stresses due to a linear temperature gradient through the wall are:
1 E
1 + k − 2k 2
α∆θ
3 1− ν
1− k2
1 E
−2 + k + k 2
r = re =
α∆θ
3 1− ν
1− k2
σ θθ (r = ri )=
σ θθ (
)
(9.4.6)
12
A crack is likely to appear on the inner wall if σθθ(r = ri) > σθθ(r = re), resulting
from the addition of the two contributions. The result is simply:
p>
Eα∆θ
1− ν
(9.4.7)
This corresponds to very high pressures except for materials with very low coefficient of thermal expansion like Invar, a FeNi alloy(~ 42%)
9.2 Small scale yielding
At the tip of cracks in elastoplastic materials a plastic zone develops. Its
shape and the stress and strain fields are different in plane stress (thin plates)
and in plane strain (thick plates). Small scale yielding corresponds to situations such that stress and strain fields given by the elastic singularity inbed
the plastic zone.
The J integral is defined as:

∂u 
K2
J = ∫  π dx2 − t ⋅
ds  = G = I (1 −ν 2 )
Γ
∂x1 
E

π being the strain energy density, t and u the stress and the displacement respectively on the close circuit Γ around the crack tip.
The path independence of the J integral in elasticity can be extended to
plasticity insofar as there is radial loading and no unloading.
Exercise 9.5 Small scale plastic zone in mode I
a) Demonstrate Eq. 2.104 (Volume II p.57) and Eq. 2.110 (Volume II p.59) and
draw the contours of the plastic zones in plane stress and in plane strain as shown
in Fig.2.26 p.54 of Volume II.
Solution
a) Plane stress in mode I
13
The stress field of the elastic singularity is that of Eq.2.46 (Sect. 2.2.4.3. p.24 of
Volume II). The yield strength is reached at a distance RY where the yield condition is satisfied.
With the von Mises criterion, according to Eq. 2.46:
2
1  KI 
2θ 
2 θ
RY =
  cos  4 − 3cos 
2π  Rp 
2
2
(9.5.1)
Rp being the yield strength. This equation is the same as Eq.2.104.
The angular dependency of the derivative of RY with respect to θ is given by:
–sinθ [2 – 3cos2(θ/2)]. The maximum of RY is found for θ = 2 arccos (2/3)1/2.
b) Plane strain in mode I
1  KI
RY =

2π  Rp
2

2θ 
2
2θ 
 cos  4 (1 −ν + ν ) − 3cos 
2
2


(9.5.2)
Figure 9.5.1 shows the two contours assuming ν = 1/3. This equation is the
same as Eq. 2.110.
Fig. 9.5.1. Contours of the plastic zone in mode I in plane stress and plane strain according to the
von Mises criterion (ν = 1/3).
b) Determine the slip line field, the strain and the strain energy in the plastic zone.
Solution
The slip lines are parallel to the maximum shear directions; at −45° from the
principal direction corresponding to the maximum principal stress for the α
14
lines; at + 45° for the β lines. The Hencky relations allow calculating the hydrostatic stress σm as a function of the orientation θ of the line:
σ m − 2kθ = Cst for α lines
σ m + 2kθ = Cst for β lines
(9.5.3)
Near the crack tip the Prandtl field of the slip lines consists of two 90° symmetrical fans with respect to the crack plane and a square with an inclination of 45° in
its front (Fig. 2.31 of Volume II p.60). The β slip lines emanate from the crack tip
in the upper fan.
In the fan:

π
σ θθ = σ rr = σ m =  1 + 3  k − 2kθ
2

(9.5.4)
The strains εr and εθ are equal to 0 because the slip lines are radial. Hence:
∂ur
=0
∂r

1  ∂uθ
+ ur  = 0

r  ∂θ

(9.5.5)
so that:
ur = f ′ (θ )
uθ = g ( r ) − f (θ )
(9.5.6)
Thus the shear strain:
γ rθ =
1 ∂ur ∂uθ uθ f ′′ (θ ) + f (θ )
d g(r)
+
− =
+r 

r ∂r
∂r
r
r
dr  r 
(9.5.7)
Along x1, ahead of the crack tip, the displacement is equal to zero. Hence:
π

f ′ θ =  = 0
4

π


f  θ =  = g = Cst
4

(9.5.8)
15
Choosing g = 0, let RY(θ) be defined as:
RY (θ ) =
f ′′ (θ ) + f (θ )
γ0
(9.5.9)
where γ0 is the shear yield strength k/µ.
Thus, near the crack tip
γ rθ = γ 0
RY (θ )
r
(9.5.10)
The plastic strain energy density is:
π = kγ 0
()
RY θ
r
(9.5.11)
This is the expected variation in 1/r.
c) Calculate the J integral and the CTOD
Solution
The J integral can be calculated along a circle of small radius r.
To calculate J we need:
dx1 = −r sin θ dθ + cos θ dr ; dx2 = r cos θ dθ + sin θ dr ; u = f ′ (θ ) er − f (θ ) eθ .
Note that dr = 0.
∂e
∂e


du =  f ′′ (θ ) e r − f (θ ) θ + f ′ (θ ) r − f ′ (θ ) eθ  dθ = γ 0 RY (θ )
∂
θ
∂
θ


 ∂u

∂u
du = r  −
sin θ +
cos θ  dθ
∂x
∂x


1
2
∂u
∂u
0=
cos θ +
sin θ
∂x1
∂x2
(9.5.12)
(9.5.13)
From Eqs. (9.5.9), (9.5.12) and (9.5.13):
γ R
∂u
= − 0 Y sin θ e r
∂x1
r
(9.5.14)
16
Using Eqs. (9.5.4), (9.5.11) and (9.5.14), and assuming for simplification that
the boundary of the plastic zone is a lemniscate such that:
()
RY θ = − Rmax cos 2θ
J = −2kγ 0 Rmax ∫
3π 4
π4


 3π

cos 2θ cos θ +  1 +
− 2θ  sin θ dθ
2




KI2
4 2 π 
=
(1 −ν 2 )
1 +  kγ 0 Rmax =
3 
2
E
(9.5.15)
(9.5.16)
With this representation of the plastic zone boundary, ν = 1/3 and k = Rp/√3,
RY(π/2) = Rmax= 1.3. This is larger than the corresponding value found by Eq.
9.5.2.
The determination of the CTOD (denoted δ) is deduced from the relation:
du2
= γ 0 RY θ sin θ
dθ
()
(9.5.17)
Hence:
J = −2k ∫
3π 4
π 4
du2  1

3π
+1+
− 2θ  dθ
dθ  tan θ
2

(9.5.18)
Integrating by parts and noting that u2 is symmetrical with respect to π/2, Eq.
9.5.18 yields:

J = 1 +

π
kδ
2 
(9.5.19)
And finally, Eqs. (9.5.16) and (9.5.19) yield:
(1 −ν )
2
δ=
3 K I2
1 + π 2 ERp
Exercise 9.6 Small scale plastic zone in mode II
(9.5.20)
17
Solution
Equation 2.51 (Volume II p.25) allows calculating the principal stresses σI:
2
12
1 K  1
θ
σ I =  II   − sin ± ( 4 − 3sin 2 θ ) 


2π  Rp  2 
2

(9.6.1)
The von Mises criterion yields:
RY =
2π
(K
II
Rp )
2


 35
4θ
2
2θ
9sin 2 −  4 +ν −ν  sin 2 + 3




(9.6.2)
If ν = 1/2, (35/4 + ν + ν2 ) = 9 and Eq. 9.6.2 is simplified. Figure 9.6.1 shows
the shape of the plastic zone in that case. If ν = 1/3 the result is about the same.
Fig. 9.6.1. Plastic zone in mode II (ν = ½)
Exercise 9. 7. Small scale plastic zone in mode III
Solution
In mode III the only non-zero displacement is u3; the stress components reduce to
σ13 and σ23. The equilibrium equations are:
∂σ 13 ∂σ 23
+
∂x1
∂x2
(9.7.1)
In the elastic zone σ i3 = 2µε i3 , µ being the shear modulus, so that:
∇ 2 u3 =
∂ 2u3
2
1
∂x
+
∂ 2 u3
∂x22
=0
(9.7.2)
18
u3* being a real harmonic function, such that, if z = x1 +ix2:
µ w ( z ) = u3∗ ( z ) + iu3 ( z )
(9.7.3)
Hence, the displacement u3 can be represented by the imaginary part of a
holomorphic function w(z):
u3 =
1
µ
Im  w ( z ) 
 ∂u3
∂u 
+ i 3  = w′ ( z )
∂
x
∂x2 
 1
(9.7.4)
σ 23 + iσ 13 = µ 
The non-zero components of the stress tensor are σ23 and σ13.
In the plastic zone the von Mises criterion yields:
2
σ 132 + σ 23
= σ r32 + σ θ323 = k 2
(9.7.5)
k being the yield strength in shear. The material is assumed not to strain-harden.
The principal stresses are +k, 0, −k. Hence σ22 = 0 and σ23 = k, σ13 = 0 (draw
the Mohr circles). Thus:
σ 3θ = k
σ 3r = 0
(9.7.6)
u3 = u3 (θ )
Let:
γ 3θ =
1 ∂u3
r ∂θ
(9.7.7)
Then, at the boundary of the plastic zone (r = RY):
γ 3θ = γ 0 =
∂u3
RY (θ ) ∂θ
1
(9.7.8)
with γ0 = k/µ.
Equation 9.7.8 yields:
θ
u3 = γ 0 ∫ RY ( q ) dq
0
(9.7.9)
19
We need now to determine the function RY(θ). Let
ζ=
σ 23 − iσ 13
(9.7.10)
k
Now, σ23 – iσ13 = σθ3 exp (iθ) and as at the plastic zone boundary σθ3 = k in the
ζ plane the plastic zone boundary is a circle of unit radius:
ζ = exp ( iθ )
z = x1 − ix2 = RY (θ ) exp ( −iθ ) = F (ζ )
(9.7.11)
Hence:
RY (θ ) = exp ( iθ ) F exp ( iθ ) 
(9.7.12)
We seek an analytical function such that:
σ 23 − iσ 13 →
K III
(9.7.13)
2πz
when |z| goes to infinity, or else when |ζ| goes to 0:
σ 23 − iσ 13 →
K III
K III
 θ
exp  i  =
2πr
 2
2π ( x1 − ix2 )
F (ζ ) = x1 − ix2 →
K III 2
2π (σ 23 − iσ 13 )
2
=
K III 2
2πk 2ζ 2
(9.7.14)
As σ23 = 0 along the crack length, it is represented by the imaginary axis in the ζ
plane. As x2 = 0 along the crack, F(ζ) is real along the imaginary axis in the ζ
plane. A suitable function is:
F (ζ ) =
K III 2 
1 
1+ 2 
2 
2πk  ζ 
(9.7.15)
Thus:
()
RY θ =
K III2
2πk 2
cos θ
(9.7.16)
20
The plastic zone boundary is a circle of diameter KIII2/πk2. Within the plastic
zone:
γ 3θ =
K III2 cos θ
πµ k r
(9.7.17)
K2
u3 = III sin θ
πµ k
The crack tip opening displacement (CTOD) is:
(
)
(
)
CTOD = u3 + π 2 − u3 − π 2 =
2K III2
πµ k
(9.7.18)
Within the elastic zone:
 
K2  
σ 23 + iσ 13 = K III  2π  z − III2  
2πk  
 
−1 2
(9.7.19)
The elastic singularity is translated so that the fictitious elastic crack tip is at
the centre of the plastic zone.
9.2 Applications
Brittle fracture occurs when the stress intensity factor K equals the fracture
toughness KIc. For a through crack of length 2a, K I = σ πa ; for a surface
crack of depth a and length 2c, K I = 2σ a Q , the geometrical factor Q being a function of c/a. In fatigue, the crack propagation rate follows the Paris
law: da dN = C ∆K I m .
Exercise 9.8 Optimum material
a) The empirical relation between the fracture toughness GIc and the yield
strength Rp of a maraging steel is GIc = 281 − 0.1525Rp , the units being kJ/m2 and
MPa, to reach a plateau at 8.9 kJ/m2 when Rp is more than 1,785 MPa. A 25 mm
21
thick and 200 mm wide plate must be used in an application in which no crack
propagation is allowed. A safety factor of 0.75 on the yield strength is required.
Non-destructive testing is assumed to be able to detect 100% of the cracks of surface length larger than 2.5 mm. Determine the most appropriate yield strength to
be achieved, assuming semi-circular surface cracks.
Solution
The failure of the plate is either plastic collapse when the applied load is:
(
)
P = 0.75 25 × 200 Rp
(N )
(9.8.1)
or brittle fracture when:
 EG 4.5 
Ic

P = ( 25 × 200 ) 
2
 (1 −ν ) 2c 
12
( N)
(9.8.2)
E being the Young modulus in MPa, ν the Poisson ratio, GIc the fracture toughness
in kJ/m2, 2c the crack length of depth a = c in m.
The maximum admissible load is the one found by equating the two expressions. With E = 200,000 MPa, 2c = 0.0025 m, GIc given by the empirical relation
in function of the yield strength Rp and neglecting the Poisson ratio, the desired
yield strength is found to be 1,720 MPa (<1,785 MPa).
b) How could the admissible load be increased?
Solution
The only way is to improve the efficiency of the non-destructive testing so as to
decrease 2c.
c) Is leak before break possible?
Solution
Leak before break would be possible if the critical through crack size would exceed the plate thickness. Now this critical crack size is (neglecting the Poisson ratio):
2
EGIc
1K 
ac =  Ic  =
= 4 × 10−4 m
π  Rp 
πRp2
(9.8.3)
22
for Rp = 1,720 MPa. So, leak before break is not possible.
Exercise 9.9 Leak before break
It is required to build a cylindrical gas-storage tank, 6 m high and 1.8 m in diameter, holding a pressure of 10 MPa (100 bars). The safety factor must be 0.75
on the strength. It is required to achieve leak before break to avoid catastrophic
failure. This means that the critical crack size must be larger than the wall thickness. Discuss the choice of the material to be used.
Solution
A normal design would correspond to a thin wall pressure vessel; it can be considered that:
σ=p
R 9
= ≤ 0.75Rm
e e
(9.9.1)
p being the pressure, R the radius and e the wall thickness.
If ac is the critical crack size, it would be safe to assume:
2
2
2K 
2  K Ic 
2ac =  Ic  = 
 ≥e
π σ 
π  0.75Rm 
(9.9.2)
Figure 9.9.1 shows the area corresponding to leak before break. Note that, to
draw the limit of brittle fracture, the fracture toughness KIc decreases when Rm increases down to a plateau where (KIc/ Rm)2 is parabolic.
Fig. 9.9.1. Wall thickness e as a function of 1/Rm, showing the line corresponding to plastic limit
and the one corresponding to brittle fracture. This shows the area corresponding to leak before
break condition.
23
When the two lines cross:
Rm = 0.094K Ic2
(9.9.3)
This line is reported in Fig. 9.9.2, the materials properties chart, on which the
corresponding variation of e as a function of Rm is superimposed. Leak before
break can be obtained above the KIc2/Rm line.
Fig. 9.9.2. Fracture toughness KIc versus fracture strength Rm on which the limit KIc2=10.6Rm is
drawn. The corresponding e = 12/Rm is superimposed.
It shows that this can be the case for steels, Ni and Ti alloys with reasonable
thicknesses. For instance for Rm = 940 MPa and KIc = 100 MPa m1/2, e = 13 mm.
For Cu alloys the wall thickness lies already in the range of centimetres; for Al alloys it becomes really too large.
Of course, other parameters can be modified. Decreasing the stress enlarges the
possibilities at the cost of wall thickness.
Exercise 9.10. Mixed mode fracture of a brittle orthotropic material –
Example of strongly textured zinc sheets 1– Part I: Isotropic Approach
Fracture under combined tensile and shear stresses has received considerable attention. Most of these theoretical studies and experimental investigations have attempted to elucidate the behaviour of isotropic materials. A large number of experiments have been performed using centrally cracked panels which contain a
crack inclined to the tensile axis. The loading is usually expressed in terms of a
combination of a symmetric mode I and an anti-symmetric mode II (see Fig. 9.10.1
and Exercise 9.1).
1 This exercise refers also to Chapter II-3 : Brittle Fracture
24
In the case of isotropic materials it is found that the direction, θ , along which
crack extension occurs is different from the original plane of the crack, except
when the specimens are subjected to pure mode I loading ( β = π / 2 ) (see also Exercise 9.1 and Fig.9.9.1). Fracture tests on centred cracked panels have been performed on zinc at 77K, temperature at which this material with an HCP structure
is perfectly brittle (see Volume II, Chapter 3, Sect. 3.7.2, p. 55, and Lemant and
Pineau 1981).
σ
y
,K
KI
II
k ,I
k II
θ
x
Figure 9.10.1. Plate with an inclined central crack
Cleavage in this material which is strongly anisotropic occurs on the basal
plane which is also the preferential glide plane. The results of these tests are reported in Table 9.9.1 where Kapp is defined as K app = σ R πa , σ R is the fracture
stress, the stress intensity factor is decomposed into the mode I (KI) and the mode
II components. See also Exercise 9.1.
1) Show that
K I = σ πa sin 2 β
K II = σ πa sin β cos β
Solution
See Volume II, Chapter 2, Table 2.5, p. 96.
(9.10.1)
25
2) Approximate solution (Erdogan and Sih, 1963). It is assumed that the crack will
propagate radially in a direction where the tangential (or opening) stress is
maximum and this will occur when the calculated stress intensity on the bifurcated
branch at a critical distance rc will reach the fracture toughness assumed to be
isotropic and equal to KIc and KIIc (fracture toughness under pure mode II loading). Compare these values with the experimental results reported in Table 9.1.1.
Solution
In cartesian coordinates, the stresses are given by (see Volume II, Chapter 2, Sect.
2.2.4.3, p. 24).
σ xx =
KI
θ
θ
3θ 
cos 1 − sin sin 
2
2
2 
2π r
σ yy =
θ
θ
3θ 
KI
cos 1 + sin sin 
2
2
2 
2π r
σ xy =
3θ
KI
θ
θ
cos sin cos
2
2
2
2π r
(9.10.2)
Similar equations have been given for pure mode II (see Eq. 2.51, p. 25).
 cos θ − sin θ 
Using the rotation matrix 
 one can express these stresses in cy sin θ cos θ 
lindrical coordinates:
cos (θ / 2 )  
θ
2θ 
 K I  1 + sin  + 2 K II ( 3cos θ − 1) tan 
2
2
2πr  
cos (θ / 2 ) 
θ 3

K I cos 2 − K II sin θ 
σ θθ =
2 2
2πr 

cos (θ / 2 )
 K I sin θ + K II ( 3cos θ − 1) 
σ rθ =
2πr 
σ rr =
(9.10.3)
Thus in pure mode I:
σ θθ =
In pure mode II:
θ
KI
cos3
2
2πr
(9.10.4a)
26
σ rθ =
K II cos
θ
2
( 3cosθ − 1)
(9.10.4b)
2πr
∂σ θθ
and making this equal to 0 to find the direction where σ θθ is
∂θ
maximum leads to:
Calculating
K I sin θ + K II ( 3cos θ + 1) = 0
(9.10.5)
One can notice that in this direction σ rθ = 0
Fracture will occur when both Eq. 9.10.3, with σ θθ = σ c (cleavage stress) and
Eq. 9.10.5 will be satisfied. Eliminating θ between these two equations, one obtains a relation between KI, KII and KIc.
In mode II, Eq. 9.10.5 leads to
1
3
θ = −Arccos = −70.5degrees
(9.10.6)
and using Eq. 9.10.3, one obtains:
K IIC = −
K IC
2
= 0.87 K IC
3 cos θ sin θ
(9.10.7)
The experimental results are far from these predictions: (i) The angle θ for
cleavage is null. This is due to the strong anisotropy of cleavage stress in zinc; (ii)
Even when β is “small” ( β = 30 degrees), i.e. when the mode II becomes predominant, the fracture toughness is far from the prediction given by Eq. 9.10.7.
As in all fracture problems, we must also verify if this direction corresponds to
that predicted from the energetic approach. See the following question.
3) The strain energy release rate is given by
G (θ ) =
1 −ν 2 2
kI + kII2
E
(
)
(See Volume II, Chapter 2, Sect. 2.2.5, Eq. 2.56) where
(9.10.8)
k I and k II are the
stress intensity factor in mode I and mode II along the bifurcated crack. Assuming,
as Nuismer (1975)(for the references see the end of the chapter) that these stress
intensity factors can be calculated as the continuity of G(θ) stresses between the
27
initial state (δa = 0) and the condition with an infinitively small bifurcated crack
( δa → 0 ) , one obtains, using Eq. 2.45 and Eq. 2.52 from Chapter 2, Volume II:
kI = K I cos3
θ
θ
θ
− 3K II sin cos 2
2
2
2
θ
θ
θ
θ

kII = K I sin cos 2 + K II cos 1 − 3sin 2 
2
2
2
2
(9.10.9)
Calculate G (θ ) given by Eq. 9.10.8 and the angle θ at which it reaches its
maximum value.
Solution
As the crack propagates in its own plane (θ = 0 ) , kI = K I and k II = K II and thus
G=
K I2 + K II2
1 −ν 2 .
E
(
)
One can easily calculate K I2 + K II2 =
GE
(see Table 9.10.1, last column). Us1 −ν 2
ing the loads at fracture and calculating the associated values for KI and KII (Table
9.10.1) it is possible to plot the variation of G as a function of β . This curve (not
shown here) exhibits a large variation of Gc with a minimum at β = 45 degrees.
Thus in this material, it is impossible to predict the load at fracture assuming a
constant value for Gc as it is usually observed in isotropic materials. The effect of
anisotropy is analysed in some detail in Exercise 9.11.
Table 9.10.1. Zinc tested at 77K. No crack bifurcation
β (degrees)
σR (MPa)
Kapp
KI
1/2)
(KI2+KII2)x
KII
1/2
1/2
(MPa m
(MPa m )
(MPa m )
(1-ν2)/E
30
13.65
3.35
0.835
1.445
2.78
45
7.67
1.76
0.86
0.86
1.48
60
21.27
4.04
3.54
3.54
16.73
90
22.7
4.90
4.90
0
24.01
Exercise 9.11. Mixed mode fracture of a brittle orthotropic material –
Example of strongly textured zinc sheets2 – Part II – Anisotropic approach
2
This exercise refers also to Chapter II-3 : Brittle Fracture
28
In this exercise, an attempt is made to show to what extent the introduction of anisotropy (see Fig. 9.11.1) can account for the experimental results reported in Table 9.10.1 in the previous exercise.
1) Firstly we compute the stress field for the case of an orthotropic material corresponding to the sheets of zinc in order to obtain the stresses σθθ and σrθ which
allows the use of the Nuismer (1975) approximation, i.e.:
kI = 2πr σθθ
(9.11.1)
kII = 2πr σ rθ
Figure 9.11.1 Preferential orientation in rolled polycrystalline zinc and notations. The basal
plane which is the cleavage plane is parallel to the crack plane. (Lemant and Pineau, 1981)
Afterwards we can establish the G = f(θ) curves for various values. In the general anisotropic case a cross-term kI kII is found. Lastly we compute G in terms of
kI and kII for self-similar extension. See also Lemant and Pineau (1981).
Solution
Referring to Fig. 9.11.1 we can write Hooke’s law for plane strain (Sih and Liebowitz, 1969):
29
ε xx = a11σ xx + a12σ yy + a13σ zz
ε yy = a21σ xx + a22σ yy + a23σ zz
ε zz = a31σ xx + a32σ yy + a33σ zz
(9.11.2)
γ yz = a44σ yz
γ zx = a66σ xy
As the plate is assumed to be under plane strain conditions
ε zz = γ yz = γ zx = 0 ⇒ τ yz = τ zx = 0 and σ zz = −
1
a31 σ xx + a32 σ yy
a33
(
)
One can thus write Eq. 9.11.2 under the following form:
ε xx = b11σ xx + b12σ yy
ε yy = b21σ xx + b22σ yy
(9.11.3a)
γ xy = b66σ xy
with:
b11 = a11 −
a13
a33
b22 = a22 −
a23
a33
b12 = a12 −
a13a32
a33
b66 = a66
(9.11.3b)
The elastic constants of zinc at 77K are known [Huntington H.B., 1958]. It is
thus possible to calculate the coefficient bij. One finds:
(
= 0.198188 (10
b11 = 0.075596 10−4 MPa −1
b22
−4
MPa −1
)
)
(
b12 = b23 = −0.0614 10−4 MPa −1
(
b13 = 0.004168 10−4 MPa −1
(
b66 = 0.2222 10−4 MPa −1
)
)
)
(9.11.4)
30
The characteristic equation for plane strain in orthotropic materials writes:
b11s 4j + ( 2b12 + b66 ) s 4j + b22 = 0
(9.11.5)
The roots of Eq. (9.11.5) are:
( β > 0)
s1 = α + iβ
(9.11.6)
s2 = −α + iβ
with
 α 0 − β0 

 2 
1/2
α =
 α + β0 
β = 0

 2 
= 0.6936
(9.11.7a)
1/2
= 1.0680
where
1/2
 b22 
 = 1.6216
 b11 
2b + b
β 0 = 12 66 = 0.6596
2b11
α0 = 
(9.11.7b)
Superposing the solutions for modes I and II, one can calculate the stress field
at the crack tip:
Mode I:
σ xx =
 ss s
KI
s 
Re  1 2  2 − 1  
2πr
 s1 − s2  φ2 φ1  
 1  s1 s2  
KI
Re 
 − 
2πr
 s1 − s2  φ2 φ1  
 s s  1 1 
KI
=
Re  1 2  −  
2πr
 s1 − s2  φ1 φ2  
σ yy =
σ xy
1/2
(9.11.8)
1/2
with φ1 = ( cos θ + s1 sin θ ) ; φ2 = ( cosθ + s2 sin θ )
Mode II:
31
σ xx =
 1  s22 s12  
K II
Re 
 − 
2πr
 s1 − s2  φ1 φ2  
σ yy =
 1  1 1 
K II
Re 
 − 
2πr
 s1 − s2  φ2 φ1  
σ xy =
 1  s1 s2
K II
Re 
 −
2πr
 s1 − s2  φ1 φ2
(9.11.9)


 
The stresses can also be calculated using cylindrical coordinates:
σ rr = σ xx cos 2 θ + σ yy sin 2 θ + 2σ xy sin θ cosθ
σ θθ = σ xx sin 2 θ + σ yy cos 2 θ − 2σ xy sin θ cosθ
(9.11.10)
σ rθ = (σ yy − σ xx ) sin θ cosθ + σ xy ( cos θ − sin θ )
2
2
The results of those calculations are shown for β = 45 degrees (see Fig. 9.11.2).
It is observed that the anisotropy does not introduce drastic modifications in the
stress profiles. This is not the case for the displacements (not shown here).
2) For collinear extension, δa, in an anisotropic material, we can compute G as
the work for crack closure:
Solution
G=−
∂W
1 a +δa
=
σ yy [ v ] + σ xy [u ] dx
∂a 2δa ∫0
(
)
(9.11.11)
where the discontinuities [u] and [v] refer to the state after extension while the
stresses σyy and σxy refer to the state before extension. See Sih and Liebowitz
(1969) for the detailed calculations. We obtain:
G=
b22 2  s1 + s2  b11 2
K I Re  i
K II Im ( s1 + s2 )
+
2
 s1s2  2

 i
1
+ K I K II  b22 Re 

2
 s1s2



 + b11 Im ( s1s2 ) 


The quantities b11, b22, s1 and s2 have already been introduced.
(9.11.12)
32
For the orthotropic case, when the crack lies in a plane of symmetry, both
Re(i/s1s2) and Im(s1 s2) are zero and G depends only on K I2 and K II2 :
1/2
1/2
2b12 + b66 
 b11b22   b22 


G =

 +

2b11 
 2   b11 

1/2
1/2


b 
 K I2 +  11  K II2 


 b22 
(9.11.13)
Numerical calculations are used to draw the curve shown in Fig. 9.11.2 using
the experimental results reported in Table 9.10.1 of the previous exercise. These
results clearly show that the strain energy rate at failure is not constant, as in the
isotropic analysis.
3) Indicate schematically the shape of the resistance curve to satisfy the experimental observations.
Solution
Figure 9.11.2 Schematic diagram illustrating the differences for the R curve between an isotropic
material and an orthotropic material.
The R curve for this material must be very steep and centred around θ = 0 degree
as indicated schematically in Fig. 9.11.2.
The fact that the released energy is minimum in the vicinity of β = 45 degrees
strongly suggests that fracture of this material is controlled not only by the normal
stress to the cleavage plane but also by the shear stress or shear strain in the basal
plane. This has been shown on single crystals (see Fig. 3.3.5 and Sect. 3.7.2 in
Chapter 3 of Volume II).
33
References
Erdogan F, Sih GC (1963) On the crack extension in plates under plane bending and transverse
shear. J Basic Eng, 85D:519-527
François D (2009) Mécanique de la rupture. In Clavel M, Bompard Ph (eds) Endommagement et
rupture des matériaux 1. Lavoisier, Paris
Huntington HB (1958) The elastic constant of crystals. Solid-State Phys, vol 7:213
Nuismer RJ (1975) An energy release rate criterion for mixed mode fracture. Int J Fracture, vol
2:245-250
Sih GC, Liebowitz H (1969) Fracture. Vol 2, Academic Press, New York and London, chapter 2
“Mathematical theories of brittle fracture”:67-190
Tada H, Paris PC, Irwin GR (1985). The stress analysis of cracks handbook. Del Research Corporation, Hellerton Pa.
34
Chapter 10 Exercises of Chapter II-3: Brittle Fracture
Exercise 10.1 – Strength and cleavage of an irradiated low alloy steel
In nuclear industry irradiation embrittlement of pressurised water reactor (PWR)
vessel steels is of great concern. In western countries these PWR vessels are made
of A508 steel the composition of which is given in Table 10.1.1 [Al Mundheri et
al., 1989]. CT type specimens were used to measure the fracture toughness before
and after irradiation (3 x 1019 n/cm2 and 8 x 1019 n/cm2). The results are reported
in Table 10.1.2. The microstructure of the material is shown in Fig. 10.1.1. The
results of tensile tests on smooth specimens are given in Fig. 10.1.2. The cleavage
stress was measured using notched specimen shown in Fig. 10.1.3. The Weibull
stress, σW, (see Eq. 3.33 in Sect. 3.4.1, Volume II) was calculated. The results are
reported in Fig. 10.1.4 where the cumulative probability of failure is plotted as a
function of the Weibull stress, σW.
Figure 10.1.1 Optical micrograph showing the bainitic microstructure of the investigated steel
(A508)
35
Figure 10.1.2 Variation of yield strength as a function of test temperature – Initial and two irradiation conditions
Figure 10.1.3 Specimens used for cleavage stress measurements and determination of the
Weibull stress
Figure 10.1.4 Experimental results. Probability of a failure as a function of the Weibull stress
36
1) Briefly comment the composition of this steel and its microstructure shown in
Fig. 10.1.1.
2) Briefly comment the results of tensile tests reported in Fig. 10.1.2.
3) In the calculation of σW, the representative volume V0 has been taken as equal
to (50µm)3. Discuss this value.
4) The results of Fig. 10.1.4 can be represented by the following expresm
σ 
sion: PR = 1 − exp−  W  . Show that the values of m = 22 and σu = 2575 MPa
 σu 
yield a satisfactory representation of the results reported in this figure.
5) Check that the results of fracture toughness measurements given in Table
10.1.2 satisfy the small scale yielding conditions (Eq. 2.109) (Sect. 2.3.3.4).
6) Using Eq. 3.44 (Sect. 3.4.4.2) for the probability of failure in small scale yielding, plot log K Ic4 B (B specimen thickness) as a function of log σ0 (yield strength)
for PR = 5%, 50% and 95%. Discuss the validity of the Beremin theory (Beremin
1983, Andrieu et al 2012) to interpret the results reported in Table 10.1.2.
Solution
1) The content in C, Mn and Ni is sufficient for achieving the hardenability of this
material. The microstructure corresponds to a tempered upper bainite (see Sect.
A2.9 in Volume I). The carbon content is sufficiently low to avoid the formation
of brittle martensite. The Mo addition should allow to avoid intergranular fracture
due to impurity segregation (in particular, P) (see Sect. 3.8 in Volume II).
The content in impurities (S, P, Cu, etc.) is relatively large compared to the
values present in more modern steels in which S ~ 20 − 40 ppm.
The micrograph corresponds to a bainite with a grain size of the order of 10
µm.
2) In the initial condition, as in the irradiated condition, the yield strength strongly
increases when the temperature is decreased. This corresponds to what is called
the “low temperature viscoplasticity” (see Volume I, Sect. 4.3.1). Irradiation generates point defects and other defects which contribute to the strength of the material. This strengthening effect is an increasing function of the irradiation dose.
3) This elementary volume contains about 25 grains. This size is much larger than
that of the elementary volumes (see Sect. 3.3.2). This allows us to use macroscopic quantities to calculate the Weibull stress.
4) See Fig 10.1.5 where it is noted that the expression for PR gives a satisfactory
representation of the experimental results. Note that σW = σu for PR = 1−e-1 = .632.
5) It can be checked that the ligament condition b > 2.5 (KI/σ0)2 is not satisfied in
most cases. This is the reason why the results are interpreted in terms of KJc. See
the description of the load-line curves in Al Mundheri et al., 1989.
6) The theory predicts that at a given temperature, the fracture toughness decreases when the yield strength is increased, as observed when the initial and the
37
irradiated materials are compared. The theory predicts also that for given condition the fracture toughness decreases when the specimen thickness increases, as
observed in irradiated material (3 1019 n/cm2) at −50°C and –100°C for B = 12.5
and 50 mm. The theory predicts that the fracture toughness should be in the ratio
(50/12.5)1/4 = 1.41, which is close to the observed values.
The theory also predicts that for a given condition and assuming that σu does
not change with temperature, the fracture toughness increases with temperature
through the decrease of the yield strength. For instance in the initial condition between −196°C and 0°C, one should observe a ratio of 15.5 for the fracture toughness ((920/500)m/4 -1= (920/500)4.5 = 15.5).
This predicted ratio is much larger than the observed one (159/26 = 5.9). This
difference is likely due to the fact that the theory does not strictly apply at very
low temperatures where the critical step in fracture might be the formation of a
plastic zone of a given size, that is, crack initiation instead of crack propagationcontrolled. The calculated plastic zone size ZP = 1/3π (KIc/σ0)2 = 1/3π (25/920)2 ~
78 µm, which is close to the value adopted for V0. In those conditions the theory
does not strictly apply.
The results of log (KIc4)B vs. log σ0 are shown in Fig. 10.1.6. The data points
corresponding to initial condition (B = 25 mm) and irradiated at 3x1019n/cm2 (B =
12.5 mm and 50 mm) are reported in this figure. Those corresponding to
8x1019n/cm2 are not used because they have not been considered to be significant.
In Fig. 10.1.6 the straight lines corresponding to PR= 5%, 50 %, 95 % have been
drawn with Cm.= 1.5 x106. This value is very close to those calculated more recently (Beremin, 1983, Andrieu et al., 2012). It is observed that the theory accounts rather well for the experimental results, except for three points for which
the plastic zone size was very small (see specimen 329 in Table 10.1.1). It has already been indicated previously that the theory does not strictly apply to those
conditions.
38
Fig. 10.1.5 Variation of log(KIc4)B vs log σ0 in three conditions. Results of fracture toughness at
– 50°C. Application of the Beremin theory giving the probabilities of 5%, 50%, and 95%.
Table 10.1.1.Chemical composition (Wt %)
C
Mn
Ni
Mo
Cr
Si
Cu
S
P
0.15
1.36
0.68
0.46
0.24
0.24
0.063
0.009
0.009
39
Table 10.1.2 Fracture toughness measurements ( Al Mundheri et al., 1989)
Conditions
Specimen
number
Temperature
(°C)
354
0
153.2
330
-70
96.5
323
-100
93.3
359
-100
325
-160
55.4
328
-160
44.1
329
-196
26.4
Irradiated
4
-50
92.6
3x1019 n/cm2
1
-50
91.4
8
-100
2
-100
3
-160
63.7
5
-160
49
4
0
0
0
2
-50
1
-100
20
50
31
50
22
100
32
100
Unirradiated
Irradiated
19
2
3x10 n/cm
Specimen
Type
CT 25
KIc
KJc
(MPa m1/2)
(MPa m1/2)
77.5
55.2
CT 12.5
65.6
137.5
112
CT 50
47.4
48.4
62
103
CT12.5
128
138
Exercise 10.2 – Intergranular Fracture
Tension-Torsion mechanical tests have been conducted on a low alloy steel (A
508 or 15MND6 steel). The composition of the material is given in Table 10.2.1.
These tests (Kantidis et al 1994) were performed on thin tubular specimens (radius, R = 10 mm, thickness t = 1.5 mm)
Table 10.2.1 Chemical composition (Wt%)
C
Mn
Ni
Cr
Mo
Cu
Si
S
P
0.20
1.43
0.85
0.12
0.50
0.14
0.26
0.004
0.019
The material was austenitised at 1,150°C, quenched and tempered at 650°C
for 2 hours followed by a very slow cooling rate between 650°C and 450°C (which
lasted about one week) down to room temperature. This heat treatment was made
40
to simulate the cooling conditions of very heavy forged components with a thickness of 200 to 300 mm. In all these tests the fracture mode at −196°C was brittle
and purely intergranular.
1) What is the physical phenomenon which can explain this intergranular fracture? Comment the effect of impurities and alloying elements in Table 10.2.1
Solution
This failure mode corresponds to temper embrittlement (see Sect. 3.8 in Volume
II) which is associated to the segregation of impurities along the grain boundaries
during the slow cooling part of the heat-treatment. This phenomenon arises in Mn
and Ni steels. The presence of Mo tends to decrease the embrittlement effect.
Heavy components may contain segregated zones (“ghost lines”) formed during
the solidification of the ingots. These segregated zones are enriched in C, alloying
elements and impurities.
2) Nine tension-torsion tests have been performed (Table 10.2.2): τ0 and σ0 are
the values of the shear and tensile stress at which plasticity occurs in these tests;
τMax and σMax are the values of the shear or tensile stress at which fracture occurs
(Table 10.2.2). The results have been reported in Fig 10.2.1.
Show that the yield locus (plastic flow) can be described by the von Mises criterion, which is not the case for fracture
Assuming that fracture is controlled by the maximum tensile stress, σ1, show
that in a tension-torsion test, σ1 can be expressed as:
σ1 =
1
σ + σ 2 + 4τ 2 

2 
(10.2.1)
Table 10.2.2. Results of tension-torsion tests
Number
τ0 /σ0
τ0 (MPa)
σ0 (MPa)
τMax (MPa)
σMax(MPa)
1
∞
582
0
882
0
2
∞
576
0
973
0
3
0
0
1,079
0
1,167
4
~0.50
398
881
398
1,042
5
~1
563
547
618
710
6
~2
584
294
718
526
7
~2
619
274
730
518
8
~2
583
301
687
521
9
~6
615
124
815
329
41
Show that the experimental results are better represented by Eq 10.2.1 taking
for instance σ1 = 1,167 MPa, which is the value determined in pure tensile test.
How do you explain the small differences between Eq 10.2.1 and the experimental results? Do you have any suggestion to improve the interpretation of the
experimental results?
One assumes that the scatter observed in calculated values for σ1 can be described by a Weibull law (see Fig. 10.2.2):
σ 
PR = 1 − exp −  1 
 σu 
m
(10.2.2)
Plot on Fig. 10.2.2 the curve corresponding to Eq. 10.2.2 with m = 16 and σu =
988 MPa .Comment the difference between Eq. 10.2.2 and the experimental results.
Fig 10.2.1 Variation of tensile and shear stress applied to tubular specimens. A 508 steel.
Solution
The von Mises ellipse σeq= (σ2 + 3τ2)1/2 with σeq = 1,079 MPa (tensile yield
strength) represents rather well the experimental results, except the two pure torsion tests for which some buckling effect can be suspected.
The fracture results cannot be described by this ellipse since, fitting the results
on tensile tests (σMax = 1,167 MPa), one should obtain in pure torsion, τMax =
1,167/ 3 = 679 MPa which is a value much lower than those measured (882 and
973 MPa). One should also note that the plastic strain before fracture is likely
42
more important in torsion than in tension, since τMax /τ0 = 1.50 compared to
σMax/σ0 = 1.08
Using the Mohr circle yields to (10.2.1).: σ 1 =
1
σ + σ 2 + 4τ 2 

2 
The values of the calculated maximum tensile stress are reported in Table
10.2.3.
Table 10.2.3 Maximum tensile stress
Number
1
2
3
4
5
6
7
8
9
σ1(MPa)
882
973
1,168
1,176
1,068
1,027
1,033
988
996
Fig. 10. 2.2 Probability to failure of tubular specimens tested in tension-torsion. A 508 steel
3) Additional fracture tests have been performed on other tubular specimens in
which a through crack of length a = 1mm was introduced. The plane of this crack
was perpendicular to the tube axis.
The first tests were carried out under pure mode I. The measured fracture
stress was σR = 446 MPa. Assuming that the tube can be approximated as an infinite plate with a central crack, determine the value of the fracture toughness, KIc.
Are the standards, a, B > 2.5 (KIc /σ0)2 (see Eq. 2.109) satisfied?
As the austenitisation treatment was conducted at 1,150°C an austenite grain
size close to 0.2 mm was produced. Do you see some limitations in the above approach with such large grains?
Fracture toughness tests have also been performed applying only a torque to
the tube generating pure mode II. It is assumed that fracture occurs in the direction,θ0 , where the opening stress, σθθ is maximum. Calculate the angle of bifurcation for the crack, θ0 and the shear stress at fracture, τR . Compare this value with
τ0. Calculate the couple to apply to fracture the tube.
43
Solution
KIc = 446 (π x 10-3)1/2 = 25 MPa m1/2
This value is rather low when compared to pure cleavage where the fracture
toughness is of the order of 40 MPa m1/2. This suggests that the fracture toughness
is lowered by the presence of impurities along the grain boundaries, which produces intergranular fracture
The rules indicate that the thickness must be larger than 2.5 (KIc /σ0)2 = 1.325
mm. This condition is satisfied.
1
It has already been shown (Exercise 9.1) that θ 0 = Arccos = 70.5degrees and
3
that KIIc = 0.865 KIc. One deduces that τ R πa = 21.75 MPa m and thus τR = 388
MPa which is much lower than the yield strength. This is one condition for using
linear fracture mechanics. The torque can be easily calculated as Τ = 2πR2tτR =
366 Nm
References
Al Mundheri M, Soulat P, Pineau A (1989) Irradiation embrittlement of a low alloy steel interpreted in terms of a local approach of cleavage fracture, Fatigue Fract Engng Mater Struct,
12:19-30
Andrieu A, Pineau A , Besson J, Ryckelynck D, Bouaziz O, (2012) Beremin model: Methodology and application to the prediction of the Euro Toughness data set. Eng. Fract. Mech.
95:102-117
Beremin FM (1983) A local criterion for cleavage fracture of a nuclear pressure vessel steel.
Metall. Trans. 14A:2277-2287.
Kantidis E, Marini B, Allais L, Pineau (1994) A criterion for intergranular brittle fracture of a
low alloy steel, Int J Fracture, 66:273-294.
44
Chapter 11 Exercises of Chapter II-4 : Ductile Fracture
The methods of linear elastic fracture mechanics cannot be used with success on
low alloy steels such as A 508 steel (used in the fabrication of pressurised water
reactors) at the upper shelf temperature. Non-linear methods such as the J-integral
approach or the CTOD approach have to be applied in order to assess the resistance to fracture of this type of material. In these materials ductile fracture is related to the nucleation, growth and coalescence of cavities from inclusions.
The aim of these exercises is not to discuss the merits of these methods but to
study the variations of mechanical parameters derived from these approaches (JIc,
CTOD, dJ / da ) with the microstructural parameters describing the inclusion distribution. In Exercise 11.1 only the ductility of notched specimens is analysed.
Exercise 11.2 deals with the fracture toughness in relation with inclusions distribution.
Exercise 11.1. Critical Void Growth Rate
1) This exercise is largely based on a publication by Lautridou and Pineau (1981).
Four heats of A 508 steel were studied. See Table 11.1.1 for their composition and
Table 11.1.2 for their conventional mechanical properties determined at room
temperature.
Table 11.1.1 Chemical composition (Wt %) of the four investigated materials
Steels
C
S
P
Si
Mn
Ni
Cr
Mo
Cu
Al
A
0.15
0.010
0.009
0.32
1.47
0.714
0.157
0.516
0.082
0.029
B
0.16
0.020
0.009
0.255
1.27
0.730
0.190
0.54
0.110
0.019
C
0.128
0.013
0.008
0.235
1.20
0.715
0.185
0.505
0.060
0.025
D
0.155
0.005
0.09
0.275
1.395
0.61
0.09
0.500
0.13
0.018
These materials were received in the heat-treated condition: austenitised at
875°C ± 25 °C, quenched in agitated water, tempered at 650°C ± 15°C, air cooled
and stress relieved at 550°C for 35hours and then aged at 650°C for 10-20 hours.
Table 11.1.1 shows that these materials contain various amounts of sulphur
which means various amounts in MnS inclusions.
45
Table 11.1.2. Conventional mechanical properties at room temperature
Steels
Tensile axis
direction
Charpy V at
100°C (J/cm2)
Yield Strength
(MPa)
Tensile Strength
(MPa)
Reduction
of area
(%)
A
L
240
466
590
75
S
130
476
595
65
B
T
145
290
477
72
C
S
110
225
434
40
D
L
250
432
566
74
A detailed analysis of MnS inclusion distribution was made on polished sections perpendicular to the three forging directions, longitudinal (L), transverse
(T), and short-transverse (S). For the notations, see Fig 4.39 in Volume II. The results of inclusion characteristics and critical CTOD, δc, are given in Table 4.3, in
Volume II, p. 241. In this table, f represents the volume fraction of inclusions calculated from the expression proposed by Franklin (1969):
f ( % ) = 5.4S(Wt %) -
0.001
Mn (Wt %)
(11.1.1)
Parameters L, T, S are the mean length of the MnS inclusions in the four materials; Na the number of inclusions per unit area in a given direction; Nv the
number of inclusions per unit volume (see Lautridou and Pineau, 1981, for the determination of Nv) and δc is the critical CTOD at crack initiation (see Fig. 4.35,
Volume II, p. 237 for the determination of the CTOD).
Knowing the values of Na and the dimensions of the inclusions, determine the
surface area per unit area occupied by the inclusions, AA , in a given plane. In
quantitative metallography there is one fundamental relationship which states that
on a random section, AA is equal to the volume fraction of inclusions , f (see e.g.
Underwood 1972).
Solution
Simple calculations assuming that the inclusions have a parallelepipedic shape
give: steel A, f = 0.0707%; steel B, f = 0.1236%; Steel C, f = 0.0488%. These values are close to those determined from the chemical analysis using Eq. 11.1.1. The
volume fraction in steel D which contains not only MnS inclusions but also oxide
inclusions has not been calculated.
The number of inclusions per unit area on three faces examined, Na, was also
determined (see Table 4.3, Volume II, p. 241). These values can be used to calcu-
46
late the number of inclusions per unit volume, Nv. The details of the calculations
can be found in the appendix of the article by Lautridou and Pineau (1981).
2) Tests on notched specimens (see Fig 11.1.1): the stress triaxiality is defined as
the ratio between the hydrostatic stress and the equivalent von Mises stress. This
ratio can be determined using finite element calculations. It can also be assessed
using the Bridgman analysis (see Eq. 1.12, Volume I, p. 59)
Fig. 11.1.1 Axi-symmetrically notched specimens
The results of these tests on notched specimens are shown in Fig 11.1.2
Fig 11.1.2 Results on notched specimens (Lautridou and Pineau, 1981)
47
Assuming that the cohesion between inclusions and the steel matrix is very
low, it can be inferred that this ductility at failure is essentially due to cavity
growth and coalescence initiated from the inclusions. What do you suspect for the
variation of the critical calculated void growth (R /R0)c using the Rice and Tracey
expression (Eq. 4.10, Volume II, p. 213)? (R0 is the initial void radius).
Solution
This expression reads:
 3 σm
dR 1 df
=
= 0.283exp 
 2 σ eq
R 3 f


 dε eq

(11.1.2.)
Assuming that σm/σeq is constant during the loading, integrating Eq 11.1.2
yields:
 3 σm
 R
log   = 0.283exp 
 2 σ eq
R
 0 c


 ε c

(11.1.3)
This equation predicts that the strain to failure, log εc varies linearly with σm/σeq
with a slope of −3/2. This is partly verified in Fig 11.1.2 where it can be observed
that the calculated (R/R0)c is not very sensitive to stress triaxiality, except in steel
A tested in the longitudinal direction. This indicates that, within a first approximation, it can be assumed that in those steels failure occurs when the calculated volume fraction of cavities reaches a critical value.
Exercise 11.2 Fracture Toughness in relation with inclusion distribution.
Fracture toughness tests were performed at 100°C on CT type specimens prepared from the four materials presented in Exercise 11.1. The J−∆a resistance
curves were determined using deeply cracked specimens (a/W = 0.63). The crack
extension, ∆a, was measured from direct examination of the crack surface of the
specimens which were broken at liquid nitrogen temperature after unloading.
The results of the CTOD measurements are reported in Table 11.2.1 where it is
observed that for a given steel (A) the JIc value is much lower in the S−T orientation than in the L−T orientation. Moreover for a given orientation, for example
S−T (steels A and C) or L−T (steels A and D) , the CTOD tends to increase when
the volume fraction of inclusions decreases.
48
Table 11.2.1 Results of CTOD measurements
Steels
Specimen orientation
CTOD (µm)
A
L-T
S-T
210
125
B
T-L
190
C
S-T
115
D
L-T
200
The experimental J−∆a curves are shown in Fig. 11.2.1
49
Fig. 11.2.1 J−∆a resistance curves (a) Steel A; (b) Steels B, C and D
It is well to remember that there is a linear relationship between CTOD and JIc
(see Eq 2.97, Volume II, p. 52). This relation can simply be written as:
J Ic = M σ Y CTOD
(11.2.1)
where M is not actually a constant but is a function of the stress-strain state
(plane stress, plane strain conditions), the specimen geometry, and the workhardening of the material. McMeeking (1977) has proposed the following expression:
M=
1
2 
σ
(1 +ν )(1 + N ) Y 
0.54 (1 + N ) 3 
NE 
−N
(11.2.2)
where E and ν are Young’s modulus and Poisson’s ratio, while N is the workhardening rate if it is accepted that the stress-strain curve is written under the
form σ ~(εp)N. In Table 11.2.2 the values of M obtained from the measurement of
JIc, CTOD, σY and Eq. 11.2.2 are given.
50
1) What do you think about the good agreement between the two sets of values for
M reported in Table 11.2.2?
Table 11.2.2 Comparison of experimental and calculated values of M coefficient (Eq. 11.2.2)
Steels
WorkHardening
σY (MPa)
CTOD(µm)
M=J/σYCTOD
M (Eq
11.2.2)
A(L-T)
0.10
466
A(S-T)
0.10
476
220
210
2.25
2.31
140
125
2.35
B
0.15
2.31
290
165
190
2.99
2.92
C
D
0.16
225
80
115
3.09
3.18
0.12
432
220
200
2.55
2.48
JIc
(KPa.m)
Solution
This good correspondence might be fortuitous since in particular, the conditions of
small-scale yielding and of plane strain state which are the basis of McMeeking
results were not satisfied in these experiments. However the close correspondence
between CTOD and JIc allows seeking for crack initiation a relation between only
one mechanical parameter, either CTOD or JIc, and one parameter describing inclusion distribution.
2) Among the various correlations proposed in the literature, the most popular is
certainly that one which relates the CTOD to a metallurgical length, in particular
the mean distance between inclusions. An attempt is made to use as a characteristic distance, ∆2, the distance to the nearest neighbour in a plane normal to crack
front. This distance can simply be expressed as:
∆2 = 1 / 2 N a
(11.2.3)
where Na is the number of inclusions per unit area on this plane. This correlation
between the CTOD and ∆2 is shown in Fig. 11.2.2 where a good one to one correspondence is observed. Do you see some limitations to this approach?
51
Fig. 11.2.2 Correlation between CTOD at crack initiation and the mean distance between inclusions in a random plane section perpendicular to the crack front. Two methods of analysis have
been used (Lautridou and Pineau, 1981).
Solution
In spite of the excellent correspondence observed in Fig 11.2.2, this correlation
has strong limitations since it predicts the same CTOD for specimens cut in such a
way that the thickness direction is the same , for instance LT and TL or LS and
SL. It is clear that differences between these orientations can be observed in materials which are strongly textured.
3) In the litterature much less attention has been given to the relation between
stable crack growth and inclusion distribution, as compared to crack initiation.
In order to compare materials with different conventional mechanical properties, suggest a non-dimensional parameter characterising crack resistance
Solution
Lautridou and Pineau (1981) have suggested introducing the following nondimensional quantity:
m=
CTOD dJ
J Ic da
(11.2.3)
52
In order to define a non-dimensional parameter characteristic of inclusion distribution, the experimental results suggest to introduce the following parameter, ∆
∆ = RS (Nv)1/3
(11.2.4)
where RS is the mean radius of the inclusions and NV is the number of inclusions
per unit volume. The results obtained with this correlation are shown in Fig.
11.2.3 (see also Fig. 4.54, Volume II, p. 257)
Fig 11.2.3. Relation between a non-dimensional crack resistance parameter and the product
RS(NV)1/3 (Lautridou and Pineau, 1981).
N.B. A modified cavity growth approach taking into account the inclusion distribution ahead of a crack tip has also been proposed to model crack initiation
(Lautridou and Pineau, 1981). This approach suggests that a priori predictions of
macroscopic fracture behaviour can be obtained using measurement of inclusion
distribution and analytical or numerical results describing the stress-strain field at
the crack tip. However this conclusion applies only to materials in which cavity
nucleation represents only a small contribution to overall ductility or fracture
toughness.
References
Franklin AG, 1969, J Iron Steel Inst, 207:181-186
Lautridou JC, 1980, PhD thesis Ecole des Mines Paris, France
Lautridou JC, Pineau A, Eng Fract Mechanics, 1981, 15:55-71
McMeeking RM, 1977, J Mech Phys Solids, 25:357-381
Chapter 12 Exercises of Chapter II-6: Fatigue
Fatigue crack propagation follows the Paris law, above a threshold ∆Kth:
da
= C ∆K Im
dN
(12.1)
a being the crack length (m), N the number of cycles, C a constant
(m1/2/MPa per cycle), ∆KI the range of the stress intensity factor (MPa m1/2).
Exercise 12.1 Proof test
A 1.2 m in diameter flywheel runs periodically at 4,000 RPM. The maximum
stress is given by the equation:
σ max =
3 +ν
ρω 2 R 2
4
(12.1.1)
where ν is the Poisson ratio, ρ the density, ω the rotation speed and R the radius.
The fatigue crack propagation law for the ferritic steel is:
da
= 3 × 10−13 ∆K I3
dN
(12.1.2)
in m/cycles and MPa m1/2. The Poisson ratio of the steel is 0.3; the fracture
toughness is 100 MPa m1/2. Show that a proof test at an over-speed equal to 1.1
times the service one can provide a safe service life to be estimated.
Solution
The maximum stress at the inner radius of the flywheel is [(3+0.3)/4] x 7800 x
(2π 4,000/60)2 x 0.62 = 406x106 Pa.
There exists the possibility that a defect of size a0 be present in the wall. The
proof test will produce failure of the vessel in case this size is larger than the critical size ac0. This is:
54
2
2
 K Ic   K Ic 
ac 0 = 
 = 0.039 m
 <
 1.1ασ   1.1ασ max 
(12.1.3)
α being a geometrical factor, for instance 2/π1/2 for a penny-shaped crack.
Integration of the crack propagation law yields:
2
N>
(
3 × 10-13 2
 1
1 



ac 
π σ max 3  ac0
)
3
(12.1.4)
where ac0 is given by Eq. (12.1.3) and ac, the critical crack size in service condition, by the same equation deleting the over-speed factor 1.1.
The fatigue life is then found to be larger than 3x104 cycles.
Exercise 12.2 Fatigue crack propagation
1) A piece of equipment of width W and of thickness B and length L contains an a0
deep long surface crack across the thickness. The cyclic loading is either a periodic load P or a periodic displacement u. Write the evolution of the crack length
as a function of time and compare the results.
Solution
Before any loading the displacement is such that:
u = XP
(12.2.1)
denoting the compliance by X, since the constant of the Paris law is named C
(Eq.12.1).
2) Calculate the compliance.
Solution
The strain energy release rate G depends on the compliance:
2
G=
1 2 ∂X 1  u  ∂X ∆K I 2
P
=  
=
(1-ν 2 )
2
∂A 2  X  ∂A
E
(12.2.2)
where A is the crack area, that is Ba, E is the Young modulus and ν the Poisson ratio
55
with:
K I = 2σ a = 2
P
BW
a =2
u X
BW
a
(12.2.3)
Integration of (12.2.2) yields:
X=
2
L 
2 a 
1
+
4
1ν
(
)


BWE 
WL 
(12.2.4)
3) Find the evolution of the fatigue crack propagation rate and of the crack length
as a function of the number of cycles under load control
Solution
Applying the Paris law (12.1), together with Eq. (12.2.3) and considering that for a
surface crack
K I = 2σ a
(12.2.5)
it is found that (in the case where ∆KI = KI):
m
da
 2P  m 2
= C
 a
dN
 BW 
(12.2.6)
Integration yields:
1
a
m
-1
2
=
1
a0
m
-1
2
m
 m  2 P 
− C  − 1
 N
2
 BW 
(12.2.7)
in the case of load control. The crack propagation rate is an increasing function of
the crack length, hence of the number of cycles. The crack length increases until
the stress intensity factor reaches the fracture toughness of the material.
4) Find the evolution of the fatigue crack propagation rate and of the crack length
as a function of the number of cycles under displacement control.
Solution
Equations.12.2.3 and 12.2.4 yield:
56
m
da
am 2
 2E 
= C
u
2 m
dN
 L  
2 a 
1 + 4 (1 −ν ) WL 


(12.2.8)
This is a decreasing function of the crack length as soon as d2a/dN2 is negative,
that is:
12
 4 (1 −ν 2 ) 

a≥
 3WL 
(12.2.9)
If, for instance W = L = 100 mm, this critical size is very small.
The crack propagation rate decreases until the threshold is reached and the
crack stops.
Exercise 12.3 Fatigue crack initiation and propagation in a plate with a
central crack
This plate is assumed to be very large such as the edge effects can be neglected
and the plate considered as infinite. This plate is submitted to an alternate tension-compression stress (0, σ). We intend to calculate the number of cycles to initiate and propagate a fatigue crack from the hole in the direction θ = 0
Fig 12.3.1 Sketch of a drilled plate
1) Number of cycles to initiate a crack, Ni, from the hole. In a first step, plasticity
is neglected. R = 1 mm, σ = 200 MPa. It is assumed that crack initiation will take
57
place when the stress amplitude ∆σ θθ calculated elastically at a characteristic
distance, d = 60 µm, obeys the following relation:
∆σ θθ (MPa) = 9,458 (Ni)-0.246
(12.3.1)
These values have been determined from tests on an austenitic stainless steel.
Keep in mind that, close to the hole, the stress σ θθ at a distance r from the free
surface is expressed as:
σ θθ =
σ

R2 
R4 
1 + 2 + 1 + 3 4  cos 2θ 
2 r
r



(12.3.2)
Calculate Ni. How this approach can be improved to take into account plasticity?
Solution
We find σ θθ (d = 60 µm) = 526.59 MPa; Ni = (9458/ ∆σ θθ )= 125,500 cycles
The above criterion (Eq. 12.3.1) has two main drawbacks: (i) it does not include an endurance limit (see e.g. Fig 6.24, in Chapter 6, Volume II); (ii) it does
not include the effect of plasticity. This can be done using a constitutive equation
and numerical calculations. The damage law to be used is that of Manson-Coffin
(see Sect. 6.4.2.2 in Chapter 6, Volume II)
2) Calculation of the stress intensity factor (Fig 12.3.2)
Give two approximate solutions for K, one for a<<R, the other for a>>R.
Compare these solutions with the results from the Tada, Paris, Irwin Handbook
(2000). See also Exercise 9.2.
Solution
- For a << R, the local stress σlocal ≈ 3σ and for a semi-infinite plate
K = 1.12 x 3 σ x
πa
(12.3.3)
- For a >> R, the hole has no effect. Therefore
K=σ
π(a + R)
(12.3.4)
These approximations are drawn in Fig 12.3.3 where it can easily be shown that
a* = 0.0972 R.
58
Fig. 12.3.2 Approximations for the calculation of the stress intensity factor (Tada et al, 2000)
In Fig 12.3.3 it is noted that when a →0, i.e. when s→0, one obtains the first
solution (Eq. 12.3.3) since F0 = 3.365. When a >> R, i.e. when s→ 1, one finds the
second approximation (Eq. 12.3.4), since F0 = 1. See also Exercise 9.2.
Fig 12.3.3 Schematic variation of K with crack length
3) Calculation of the number of cycles for crack propagation
Indicate how the number of cycles for crack propagation, Np, can be calculated
using the Paris law:
da
= C ∆K m
dN
(12.3.5)
This equation must be integrated between the crack length at initiation (d = 60
µm) and the final crack length ac which is reached when K = KIc . It is suggested
59
to use Eq 12.3.4 to calculate K. As this is an upper bound, the calculated value for
Np is a lower bound.
Solution
One obtains:
da
= C σ π ( a + R ) 


dN
(12.3.6)
Integrating the above equation, Np is expressed as:
Np =
(m / 2 − 1)
( R + d )1− m/2 − ( R + ac )1− m/2 
Cσ m πm /2 
(12.3.7)
where (R+ac) = 1/π(KIc/σ)2.Taking m = 2.25, KIc = 100 MPa m1/2, one obtains ac =
78.58 mm and Np = 8.48 106 cycles.
References
François D (2009) Mécanique de la rupture. In Clavel M, Bompard Ph (eds) Endommagement et
rupture des matériaux 1. Lavoisier, Paris
Tada H, Paris PC, Irwin GR (1985) The stress analysis of cracks handbook. Del Research Corporation, Hellerton Pa.
60
Chapter 13 Exercises of Chapters II-7 (Environment Assisted
Cracking) and II-8 (Creep – Fatigue - Oxidation Interactions)
Exercise 13.1 Fracture Mechanics – Oxidation - Creep Crack Growth
This exercise and the following one are largely based on two recent publications
by K. Chan (2014) and one paper by S-Q Shi, Puls and Sagat (1994).
In many materials, their life at elevated temperature is limited by creepfatigue-oxidation interactions . These forms of damage are examined successively.
In this exercise the damage due to oxidation is examined first.
The oxidation damage can be envisioned to be the result of an ingress of oxygen from the air environment into the crack tip region, most often located along
grain boundaries (GB) (Fig 13.1.1) to form an oxide MO where M is a metallic
element in a variety of potential reactants including (i) Ni solid solutions (Niss)
matrix containing Ni, Cr, Co, Ti and other metals typically added to Ni-based alloys, (ii) alloyed γ and γ’ precipitates and (iii) carbides and eventually intermetallic or laves phases at GB. For example the formation of oxides in the Ni solidsolution phase can take place according the following reaction:
M + x O → M Ox
(13.1.1)
M can be Ni, Cr, Co, Ti, Al or Nb; x can be 1, 3/2, 2 or 5/2 depending on M.
Note that NbC and Ni3Nb can react with oxygen to form Nb2O5 according to
the following reactions:
7O 2 + 4NbC → 2Nb 2O5 + 4CO ( gaz )
(13.1.2a)
5O 2 + 4Ni3 Nb → 2Nb 2O5 + 12Ni
(13.1.2b)
Pure niobium can also react with oxygen to form Nb2O5 oxide. The oxide formation involves a volume change (transformation strain) and, among others, the
formation of a gaseous phase (CO) that may exert pressure on the crack surface.
In the following only reactions without gas formation are considered. On can thus
define a transformation strain εtr as:
ε tr =
∆V
−1
V0
(13.1.3)
61
Fig. 13.1.1 Schematic description of the formation of oxides and creep cavities along a grain
boundary located ahead of a crack subjected to a static stress, σ, (a) crack tip oxide formation;
(b) grain boundary cavitation (taken from K. Chan, 2014)
where ∆V = Vox – V0 is the volume change associated with the formation of the
MO oxide and V0 is the volume of the unit cell containing the element that reacts
with oxygen to form the oxide product. Examples of reactions with unit cell volumes are given in Tables 13.1.1 and 13.1.2.
Table 13.1.1 Summary of the compounds, crystal structure, unit cell volume, elastic modulus
(E), and fracture toughness (KIc ox) of individual constituent phases pertinent for consideration of
oxidation-induced transformation toughening in Ni-based alloys.
Compound
Crystal Structure
Lattice parameter (nm)
Unit cell Vol
(10-3nm3)
E (GPa)
NbC
Fcc
0.447O
93.91
340 to 400
Ni3Nb
Bct
0.3655;0.7495
100.13
200
6.0 to 8.0
M23C6
Cubic
1.063
1201.2
325 to 334
7.1
Ni3Al
Cubic
0.357 to O.36
45.5
178
30
Cr2O3
Hcp
0.496; 1.36
289.87
143.6
3.6
Al203
Hcp
0.478; 1.299
257.59
362
3.5
Kox
(MPa m1/2)
3.0 to 4.0
Nb205
Hcp
0.3607; 0.392
44.22
125
1.5
CoO
Cubic
0.4254
75.44
189
1.12
NiO
Cubic
0.4173
72.67
230 to 260
1.5
NiCr2O4
Cubic
0.832
573.9
106.6
0.57
FeO
Cubic
0.4334
81.41
130
1.7
α- Fe203
Rhombohedral
0.99; 1.361
338.6
219
1.7
γ- Fe2O3
Cubic
0.840
591.92
208
1.7
62
Table 13.1.2. Summary of reactant, product, unit cell of reactant, unit cell of product, Young’s
modulus of oxide, Eox, transformation strain, εtr, crack growth threshold, Kth
Reactant
εtr
Oxide
product
Unit cell
vol (10-3
nm3) reactant
Unit cell
vol (10-3
nm3)
product
Eox (GPa)
Cr in Niss
Cr2O3
41.61
289.87
143.6
5.96
25.54
Al in Niss
Al2O3
41.61
257.59
361.9
5.19
36.00
Co in Niss
CoO
41.61
75.44
143.6
0.81
9.21
Ni in Niss
NiO
41.61
72.67
260
0.75
11.97
Cr in Niss
CrO
41.61
68.92
143.6
0.66
8.80
Ni,Cr in
Niss
NiCr2O4
41.61
573.9
106.6
12.79
28.26
Nb in Niss
Nb2O5
41.61
44.22
125
0.06
3.60
Cr in
M26C6
Cr2O3
1201.2
289.87
143.6
- 0.76
-1440
Nb in NbC
Nb2O5
93.81
44.22
125
- 0.52
- 3.86
Nb in
Ni3Nb
Nb2O5
98.84
44.22
125
- 0.552
- 4.02
Kth
(MPa m1/2)
1) Using the Eshelby theory for inclusions (see Volume I, Sect. 2.7, p. 127) show
that the transformation strain produces a transformation stress in the matrix
which can be approximated by the following expression:
σ tr = −
β Eox ε tr tox
(1 −ν 2 ) L
(13.1.4)
where Eox, ν, tox and L are, respectively, the Young modulus, the Poisson ratio, the
oxide thickness and the length of the second phase formed at GB; β is a geometric
factor of the second phase formed, β = π/4 for a thin plate, as assumed here, and
β = 8/15 for a spherical particle.
Solution
Further assumptions must be made:
(i) The oxide is a flat disk-shaped inclusion generating only a purely elastic deformation inside the particle.
(ii) The oxide inclusion only generates a stress-free strain normal to the disk
while all other components are zero. This assumption is made for simplification
purposes.
(iii) The isotropic elastic moduli are the same for both matrix and oxide.
63
(iv) The effect of the free surface of the crack on the stress distribution inside
the oxide is first ignored for simplicity. This effect can be taken into account as
shown by Shi and Puls (1994) in Appendix A of their paper.
The Eshelby theory for a sphere (see p. 132, Volume I, Eq. 2.160) gives:
σ ox = −α
E
8
ε
ε 22F with α = and ε 22F = tr
2
15
3
(1 −ν )
(13.1.5)
Compressive stresses are generated by the oxide when the volume change is an
expansion, as expected and as it is very often the situation (see Table 13.1.2).
For a thin plate particle, after some lengthy calculations, one finds:
σ ox = −
F
tox
π E ε 22
2
4 (1 −ν ) 4
(13.1.6)
Please note that for a rectangular plate which would be more representative of
the shape of carbide particles, the stresses are no longer uniform within the particle. This effect of particle shape has been studied by Shi & Puls (1994).
The effect of the proximity with the free surface has also been studied by Shi
and Puls (1994, see Appendix A) using the image stress. These authors have
shown that, generally speaking, the image effect is small.
2) Plot the tensile stress in the particle as a function of distance from crack tip,
taking β = π/4, tox = 0.5 µm, E = 143.6 GPa, εtr = 0.66, K = 10MPa m and
K = 30MPa m . The “residual stresses” resulting from the transformation (Eq.
13.1.6) is superimposed to the stress calculated from linear elastic fracture mechanics, i.e.:
σY =
K
2π r
(13.1.7)
where K is the stress intensity factor (for this geometry K ≃ 1.12 σ πL when the
specimen is submitted to a tensile stress σ ( see Volume II, Chapter 2, p. 98)).
Solution
See Fig. 13.1.2 taken from Chan (2014).
This figure shows that for K = 10MPa m , there is no tensile peak stress ahead
of the crack tip, contrarily to the case where K = 30MPa m . Thus, the stabilisation effect of the compressive stresses associated with the oxide formation is most
64
effective at low K levels and diminishes with K levels. The observed K dependence of the relaxed stress field due to the shielding effect associated with oxide
formation strongly suggests that a threshold stress intensity factor, Kth, must be
exceeded in order to initiate fracture of an oxide layer present at the crack tip; Of
course this shielding effect is linearly dependent on εtr. In particular it is expected
that the shielding effect will be much lower when Niobium oxide, Nb2O5, will be
formed instead of chromium oxide such as Cr2O3 (see Table 13.1.2). This might
explain, at least partly, why Ni-base alloys strengthened by γ” precipitates (Ni3Nb)
are extremely sensitive to oxidation effects (see Chapter 8, Volume 2, Sect.
8.5.2.3).
Fig. 13.1.2 Relaxed stress field associated with oxidation induced transformation toughening
ahead of an elastic crack loaded under a static stress (a) K = 10 MPa (m)1/2 , and (b) K = 30 MPa
(m)1/2. Note that tensile stresses are induced even in the presence of oxide when K is increased
from 10 to 30 MPa (m)1/2.
3) Calculate Kth for crack-tip oxide fracture.
The local stress intensity factor at the crack tip, Ktip, is comprised of the applied K due to the remote load and the shielding term, KS resulting from the com-
65
pressive (tensile) transformation stresses induced by oxide formation ahead of the
crack tip.
The weight function method can be used to calculate KS (see Chapter 2, Volume II, p. 32 and Tada, Paris Irwin (1985); it is simply given by:
KS =
2 L σ ox dx
π ∫0
x
(13.1.8)
Solution
The integration of (13.1.8) is simple since we have assumed that σox = constant
(see Eq. 13.1.6). The length L over which this integration must be made is more
difficult to define. This distance is ruled out by oxidation kinetics which can be either linear or parabolic. In the absence of further information, an oversimplification proposed by K Chan (2014) is made.
At a very small distance from the initial crack tip the stress given by Eq. 13.1.6
can exceed the compressive yield stress, σy. The oxide can be broken under the influence of these compressive stresses. The damage zone, Lα, can thus be approximated as the region where the compressive yield stress is reached and Ld is simply
obtained as:
Ld =
β Eox ε tr tox π Eox ε tr tox
=
(1 −ν 2 )σ Y 4 (1 −ν 2 )σ Y
(13.1.9)
This expression for L is included in Eq. 13.1.8 and one obtains:
1/2
E σ ε t 
K th = K ox + 2sgn ( ε tr )  ox Y tr2 ox 
 2 (1 −ν ) 
(13.1.10)
where fracture of the oxide layer is initiated when the Ktip exceeds the fracture
toughness, Kox, of the oxide layer. In Eq. 13.1.10, sgn(x) is the sign function with
x as the dummy variable; sgn(x) = 1 when x is positive, and sgn(x) = −1 when x is
negative.
Equation 13.1.10 indicates that Kth depends on materials parameters. It is also
an increasing function of the yield stress of Ni-based alloys, when applied to these
materials. This dependence can, at least conceptually, be utilised to consider the
effects of grain size and cooling rate on the time-dependent crack growth threshold observed in these alloys since these parameters control the yield strength of
these materials.
66
In Table 13.1.1, the values of Kox given by K Chan (2014) and obtained from a
survey of the literature are given. It is noted that the spinel type oxide, NiCr2O4
which is porous, exhibits much lower fracture toughness than the dense and protective oxide, Cr2O3. In Table 13.1.2 (last column) it is observed that in alloys
containing niobium and forming Nb2O5 oxide or chromium leading to the formation of Cr2O3 the threshold for crack growth is extremely low and even negative in
some cases. The first case associated with Nb2O5 oxide is met in alloys such as In
718 while the second case with Cr2O3 oxide is extremely efficient for protecting
materials, such as stainless steels and a number of Ni base alloys, against corrosion and oxidation.
Exercise 13.2 Creep Damage
Creep crack growth (CCG) involves the formation, growth and coalescence of
creep cavities along grain boundaries (see Chapter 8, Volume II, Sect. 8.3), as
shown in Fig. 13.1.1b. These cavities contribute to the stress relaxation ahead of
the crack tip. To derive the Kth for CCG, cavity formation at a grain boundary is
considered as a transformation process involving coalescence of nv vacancies to
produce a cavity in the form of an oblate ellipsoid.
1) Calculate the volume of a void, Vvoid, formed by the coalescence of nv vacancies.
Solution
Vvoid =
4
3
β π ( nv a )
3
(13.2.1)
where nv is the number of vacancies, a the radius of a vacancy and β a shape factor close to 1.
2) Calculate the transformation strain associated with the coalescence of nv vacancies into a single void.
Solution
4
The volume V0 occupied by nv vacancies before coalescence is V0 = nv a 3 and
3
the transformation strain is:
67
ε tr =
Vvoid
− 1 = β nv2 − 1
V0
(13.2.2)
π
which becomes ε tr =   nv2 − 1 when the void is taken to be an oblate ellipsoid
4
and β = π / 4 ;
Equation 13.2.2 indicates that the transformation strain associated with this
cavity formation increases with increasing number of vacancies in the cavity formation process.
3) Determine the transformation stress and Kth.
Solution
The transformation stress takes a form similar to Eq. 13.1.6 and is given by:
σ tr = −
β E ε tr tcd
(13.2.3)
(1 −ν ) L
2
cd
where tcd and Lcd are the thickness and the length of the cavitated GB layer, respectively. Using the same procedure as in the previous part, the Kth is obtained
from Eqs. 13.1.8 and 13.1.9, leading to:
1/2
 Eσ ε t 
Y tr cd

K th = K cd + 2 
2
 2 (1 −ν ) 
(13.2.4)
where Kcd is the fracture toughness of the cavitated grain boundary. Figure 13.2.1
compares the Kth values as function of the transformation strain due to oxidation
and cavity formation in Ni-based alloys. These calculations were made using
tox = tcd = 0.5µm, σ Y = 1100 MPa, E = 100GPa, Eox = E = 143.6GPa,
ν ox =ν = 0.33 and K ox = K cd = 3.6MPa m.
This value for Kox is representative of Cr2O3 oxide (see Table 13.1.1).
4) Perform similar calculations with E = 100GPa, K cd = 10 MPa m keeping
tcd = 0.5µm and compare with the experimental results shown in Fig. 13.2.2.
Solution
68
See Fig. 13.2.1 which shows that this simple theory (which theoretically applies to
tests performed under vacuum) predicts values of εtr of the order between 1 and
20, which means that the ratio V0 /a3 ~ naV is of the same order of magnitude. The
above considerations apply essentially to the early stage of intergranular cracking.
Figure 13.2.2 shows that Kth vacuum is much larger than Kth air. See for instance X -750 and IN 738 and IN 939 alloys. This suggests that another mechanism than the shielding effect due to oxide formation contributes to the degradation of the grain boundary resistance at elevated temperature.
Fig. 13.2.1 Predicted Kth values as a function of transformation strain, εtr, for crack-tip oxidation
formation compared to those for cavity formation along a grain boundary located ahead of the
crack tip.
Fig. 13.2.2 Predicted and measured Kth values for creep crack growth by grain boundary cavitation in Alloy 718, IN 738, IN 939 and X-750. Experimental results are taken from literature (after K. Chan, 2014. See this reference for the references placed in the inset)
69
References
Bain K, Pelloux RM, Metall Mater Trans A, 1984, 15 A:381-388
Budiansky B, Amazigo JC, Evans AG, J Mech Phys Solids, 1988, 36:167-187
Chan K, Metall and Mater Trans A, 2014, 45A: 3454-3466
Chan K, Enricht MP, Moody J, Fitch SHK, Metall and Mater Trans A, 2014, 45A:287-301
Hoffelner W, Metall Trans A, 1982, 13A:1245-1255
Sadananda K, Shahinian P, Mater Sci Eng, 1980, 43:159-168
Shi S-Q, Puls MP, J Nucl Mater 1994, 208:232-242
Shi S-Q, Puls MP, Sagat S, J Nucl Mater, 1994, 208:243-250
Tada H, Paris P, Irwin GR, 1985, The stress analysis of cracks handbook, del Research St Louis,
MO.
70
Chapter 14. Exercises of Chapter II-9 : Contact mechanicsFriction and Wear
Exercise 14.1 Disc brakes or How to convert kinetic energy into the work
against friction
The disc brake is a wheel brake which slows rotation of the wheel by the friction
caused by pushing brake pads against brake disc with a set of callipers, as schematically shown in Fig. 14.1.1. The brake disc (or rotor) is usually made of cast
iron, but may in some cases be made of composites. This disc is connected to the
wheel and/or the axle of the vehicle. To stop the wheel, friction material in the
form of broke pads, mounted on a device called a brake calliper (see Fig. 14.1.1),
is forced mechanically, hydraulically, pneumatically or electromagnetically
against both sides of the disc. Friction causes the disc and attached wheel to slow
or stop. Brakes convert motion to heat, and if the brakes get too hot, they become
ineffective (see Sect. 9.3.4, Volume II, p. 524), a phenomenon known as brake
fade.
Fig. 14.1.1 Disc brake
1) Assuming that the forces acting on the inner (FTRI) and the outer (FTRO) of
the disc can be reduced to point forces determine the values of FTRI and FTRO.
Then calculate the brake torque (TB) and the brake distance (BD). One assumes
that the rotor disc dimension is 240 mm, the rotor disc material is carbon ceramic
71
matrix, the pad brake area is 2,000 mm², and the pad brake material is a carbon
composite for which the coefficient of friction (wet) is 0.07 – 0.13 and this coefficient (dry) is 0.3 – 0.5. The maximum temperature is 350°C and the maximum
pressure is 1 MPa (Fig. 14.1.2).
Fig. 14.1.2 Forces acting on rotor disc due to contact with brake pads
Solution
One has simply:
FTRI = FTRO (if they are the same materials) = µ FRI.
P
Assuming that µ = 0.50, one has: FRI = max × A , where A is the area of brake
2
pad. So FTRI = FTRO = 0.5 × 1×106 N / m 2 × 2,00010−6 m 2 = 500N .
With the assumption of equal coefficients of friction and normal forces FR on
the inner and outer forces, the brake torque is given by:
TB = FTR = ( FTRI + FTRO ) R = 1,000 × 120 × 10−3 = 120 N.m
The brake distance, X, can simply be evaluated by writing that at the point of
contact of the brake the kinetic energy of the vehicle is converted into the work
done against friction.
The work done is given by
FT x X
(14.1.1)
72
where FT = FTRI + FTRO and X is the distance travelled in meter by the vehicle
before it comes to rest.
The kinetic energy of the vehicle is equal to
mv2/2
(14.1.2)
where m is the mass of the vehicle and v is its velocity.
In order to bring the vehicle to rest, one must equate (14.1.1) and (14.1.2), i.e.
mv 2
2
One assumes that v = 100 km / h (27.77 m/s) and m = 120 kg (dry weight of
vehicle). So we get:
FT X =
2
mv 2 120 × ( 27.77 )
X=
=
= 46.27 m
2 FT
2 ×1000
N.B. The actual weight of a vehicle is closer to 1,000 kg, i.e. about 10 times
higher than the value taken. But if it is a vehicle of 1,000 kg with four disc brakes,
the distance to stop from 100 km / h (46.27 x 10/4 = 116 m) will be doubled, i.e.
close to 100 m, which is a “reasonable” value.
2) Assessment of wear of the brake pads
It is well accepted that with the existing cars, the drivers have to change the pads
of their vehicle every 50,000 or 100,000 km depending if the vehicle is mainly
used on highway or in the city.
The wear can be assessed using the Archard law (Eq. 9.101, p. 540). The nondimensional quantities introduced in Eqs. 9.83a, b & c (p. 528) can be determined
assuming that for instance the brakes are used at a speed of 100 km / h.
How many such braking conditions are possible with a car after a given mileage, for instance 100,000 km?
Solution
Let us calculate successively vɶ, Pɶ and then Wɶ . With a rotor of 240 mm and a
wheel with a diameter of 1,000 mm, vɶ = 2.89 104 , Pɶ ≃ 10 .
Figure 9.34, (Volume II, Chapter 9, p. 528) indicates that in these conditions
ɶ
W ≃ 10−4 , that is W = 10−4 × 240 = 2.4 10−2 mm 2 . The Archard law Wɶ = kA Pɶ
(
)
allows us to evaluate the Archard coefficient kA =10-4 /10=10-5 which is in good
agreement with the values given in Volume II, Chapter 9, p. 543 for severe plasticity-dominated wear.
The Archard law can also be written in the form:
73
∆h = K A P vt
(14.1.3)
where ∆h is the wear displacement (m), KA is the specific wear rate coefficient
(m3 / Nm), P is the contact pressure (Pa), v is the slide rate (m/s) and t is the running time (sec.). KA can be estimated as 1.78 10-13 m3 / Nm from Jang et al., 2004.
The slide rate is equal to 6.65 m/s. The braking time is estimated to be 30 s. These
numerical values which are taken just as an order of magnitude leads to ∆h = 3.55
10-5 m. If one assumes that during 100,000 km, the driver breaks heavily 100
times, this leads to an approximation of 3.55 mm which is of the right order of
magnitude observed on the front and rear wheel drive of a Renault Master III , as
shown by Gailis and Berjova, 2012.
References
Abhang SR, Bhaskar DP (2014) Int J Eng Trends and Technology (IJETT), vol. 8:165-167
Gailis M, Berjova D (2012) Engineering for rural development, Jelgava, 24-25-05.2012:349-354
Jang H, Ko K, Kim SJ, Bash RH, Fash JW (2004) Wear 256:406-414.
Majcherczak D (2003) Etude thermique d’un contact glissant : approche numérique et expérimentale. Application au freinage, PhD thesis, University of Lille, France
Majcherczak D, Defrénoy P (2006) Arch Appl Mech, vol. 75 :497-512
Wong J (2007) Analyse de l’endommagement par fatigue thermique et modélisation du comportement thermomécanique des couples disques - garnitures de type TGV, PhD thesis Ecole
Centrale de Lille, France