7 Failure Theories

Under the heading of “Failure Theories” we discuss a number of topics relevant to

characterizing the strength of materials. Due to their high strength-to-weight ratio
and low cost, aluminum alloys have been extensively tested in the past. Steel and
titanium alloys are used for high strength components at the expense of weight in
the case of steel and cost in the case of titanium. All of these metals are ductile
in nature and their failure is characterized by yielding and plastic deformation.
The early structural material, namely, wood, and the new epoxy/fiber composite
materials are characterized as brittle. In these, there is hardly any perceptible
plastic deformation and they fail abruptly.

7.1 Brittle Failure

From the available experimental data, brittle failure is governed by one of the
three accepted theories that follow.

7.1.1 Maximum Principal Stress Theory

On the basis of uni-axial tension and compression tests, we assign ultimate
strengths σut in tension and σuc in compression. Numerically σuc is positive. We
extend these findings to two- and three-dimensional situations where three princi-
pal stresses σ1 , σ2 , and σ3 are present. According to the maximum principal stress
theory, when any one of the three principal stresses reaches ultimate stress, the
material would fail. Fig. 7.1 shows the failure envelope in a two-dimensional case.
σ2

σut

σ1
σuc σut

σuc

Figure 7.1 The failure envelope according to the maximum principal stress theory when σ3 = 0.

159

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 160 Mechanics of Aero-structures

Usually, the ultimate stress in compression is numerically higher than that in

tension. Fig. 7.1 represents this unsymmetry.

7.1.2 Mohr-Coulomb Theory

The maximum principal stress theory only involves the highest absolute value of
one stress. For some materials, the presence of a second stress alters the ultimate
stress. Mohr-Coulomb theory is an attempt to take into account this coupling.
In the case of the bi-axial stress, this theory states that failure occurs if in the

σ1 σ2
First quadrant: or = 1, (7.1)
σut σut
σ1 σ2
Third quadrant: or = −1, (7.2)
σuc σuc
σ1 σ2
Second quadrant: + = 1, (7.3)
σuc σut
σ1 σ2
Fourth quadrant: + = 1. (7.4)
σut σuc

As shown in Fig. 7.2, the effect of stress coupling is present in the second and
fourth quadrants.

σ2

σut

σ1
σuc σut

σuc

Figure 7.2 The failure envelope according to the Mohr-Coulomb theory when σ3 = 0.

7.1.3 Maximum Principal Strain Theory

According to this, failure occurs when the principal strain reaches a critical value.
In the two-dimensional case,

1
1 = [σ1 − νσ2 ],
E
1
2 = [σ2 − νσ1 ]. (7.5)
E

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 161 Failure Theories

Then the failure envelope in the stress plane is governed by σ1 − νσ2 = a constant.
This can be sketched as shown in Fig. 7.3.

σ2

σut

σ1
σuc σut

σuc

Figure 7.3 The failure envelope according to the maximum principal strain theory when σ3 = 0.

7.2 Failure Theories for Composites

Our discussion here concerns epoxy matrix composites with glass, boron, or
graphite fibers and with well-defined symmetry axes. Failure in this group of
composites can be characterized as brittle. In intermetallic composites in which
two or more metals form the constituents and, in general, in composites with
metal components, ductile failure through yielding has to be taken into account.
As shown in Fig. 7.4, we denote the fiber direction by x1 and the mutually
orthogonal directions perpendicular to the fiber directions by x2 and x3 .

y x2
x

x1

Figure 7.4 Coordinates for a fiber composite.

Uniaxial tests with a specimen oriented in the fiber direction x1 and in the
perpendicular directions x2 and x3 determine the failure strengths X 1 , X 2 , and X 3 ,
respectively. Additional shearing tests are conducted along the planes tangential
to the fibers to obtain shear strengths S12 and S13 . With these data, the objective
of a failure theory is to predict failure in a multicomponent stress situation.
Table 7.1 shows the elastic and failure properties of glass/epoxy, boron/epoxy, and
graphite/epoxy composites. The subscripts in X 1t and X 1c indicate the values in
tension and compression, respectively. We note the failure stresses in compression
can be quite different from those in tension for some of these materials.

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 162 Mechanics of Aero-structures

Table 7.1. Elastic and failure properties of selected composites from Jones (1975)
Property Unit Glass/epoxy Boron/epoxy Graphite/epoxy
E1 (GPa) 53.78 206.84 206.84
E2 (GPa) 17.93 20.68 5.17
ν12 0.25 0.3 0.25
G (GPa) 8.96 6.89 52.59
X1t (MPa) 1034.22 1379.00 1034.21
X1c (MPa) 1034.22 2757.90 689.48
X2t (MPa) 27.58 82.74 41.37
Y2c (MPa) 137.90 275.79 117.21
S12 (MPa) 41.37 124.11 68.95

7.2.1 Maximum Stress Theory for Composites

If the stress components are given in a rotated coordinate system, (x, y, z), where
the x-axis is at an angle θ from the x1 axis, by using the rotation matrix Q from
Chapter 2, we can find the components in the (x1 , x2 , x3 ) system. The maximum
stress theory predicts that to avoid failure,
σ11 < X 1 , σ22 < X 2 , σ33 < X 3 , τ12 < S12 , τ13 < S13 . (7.6)

In the simple case of a single stress component σx x acting in the x direction, we

have the conditions
X1 X2 S12
σx x < , σx x < , σx x < . (7.7)
cos2 θ sin2 θ cos θ sin θ
Experiments have shown that the maximum stress theory could result in large
errors.

7.2.2 Maximum Strain Theory for Composites

Using the anisotropic constitutive relations, the stresses at failure can be converted
to strain values. The failure criterian can be expressed as conditions on maximum
strains. Again, experiments do no agree well with this approach.

7.2.3 Tsai–Hill Theory

A modified version of Hill’s criterion for anisotropic plastic yield was suggested
by Tsai as a failure criterion for composites. This theory has the form

(G + H )σ11
2
+ (H + F)σ22
2
+ (F + G)σ33
2
− 2(H σ11 σ22 + Gσ22 σ33 + Fσ33 σ11 ),
 2 
+ 2 Lσ23 + Mσ13
2
+ N σ12
2
= 1. (7.8)

The three separate conditions we have seen before are now reduced to a single
condition. This adds one more constant. By applying the stress components one

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 163 Failure Theories

at a time, we have, when σ12 and σ13 act alone, failure results if
1 1
2N = 2
, 2M = 2
. (7.9)
S12 S13
Using σ11 , σ22 , and σ33 , one at a time, we find
1 1 1
G+H = , H+F= , F +G = . (7.10)
X 12 X 22 X 32
We need one more shear test in the 23-direction to find
1
L= 2 . (7.11)
S23
If we impose the additional condition that the directions 2 and 3 are indistin-
guishable, we have
X 2 = X 3, S12 = S13 . (7.12)
This reduces the six constants to four. For the case of plane stress, this failure
criterion reduces to
σ11
2
σ11 σ22 σ22
2
σ12
2
− + + = 1. (7.13)
X 12 X 12 X 12 2
S12
With a single component of stress σx x , we get
 
cos4 θ 1 1 sin4 θ 1
+ − cos2 θ sin2 θ + = 2 . (7.14)
2
X1 2
S12 X12
X 22 σx x
This one condition replaces the three conditions of the maximum stress or strain
theories. When all the normal stresses are tensile, the Tsai–Hill theory gives
results close to the experimental values. In many cases, the failure strength in
compression is different from that in tension. To account for this, linear terms in
normal stresses are added to the Tsai–HIll criterion to obtain the Tsai-Wu tensor
criterion:
F11 σ11
2
+ F22 σ22
2
+ F33 σ33
2
+ 2(F12 σ11 σ22 + F13 σ11 σ33 + F31 σ33 σ11 )
+ F1 σ11 + F2 σ22 + F3 σ22 = 1. (7.15)
This calls for additional compressive failure tests to determine the added three
constants.

7.3 Ductile Failure

In aerospace applications, ductile materials such as metals are deemed failed

if they reach plastic yield. In a uni-axial experiment some materials show a
clear value of yield stress. In some, such as steels, the yield point is not clearly
observable and the elastic stress corresponding to a 0.2% strain is considered the
yield stress. There are two commonly used theories to extend uni-axial behavior
to the three-dimensional stress state: the Tresca theory and the von Mises theory.
They are both based on the assumption that hydrostatic stress has no effect on
yielding.

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 164 Mechanics of Aero-structures

7.3.1 Tresca Theory

The Tresca theory states that yielding begins when the maximum shear stress
reaches a critical value. If the three principal stresses are σ1 , σ2 , and σ3 , the
maximum shear is (see Chapter 2),
1
τmax = Max [|σ1 − σ2 |, |σ2 − σ3 |, |σ3 − σ1 |]. (7.16)
2
Then the shear stress at yield for the one-dimensional case is
σy
τmax = τ y = . (7.17)
2
When σ3 = 0, we find
Max.[|σ1 |, |σ2 |, |σ1 − σ2 |] = σ y . (7.18)
A plot of these edges in the σ1 , σ1 -plane, forming the yield envelope, is shown in
Fig. 7.5.

σ2
σy σ2

σy
σ1
σy σ3 σ1

σy

(a) (b)

Figure 7.5 The yield envelope according to the Tresca theory when σ3 = 0.

As the hydrostatic pressure is proportional to σ1 + σ2 + σ3 , the pressure inde-

pendence of the yield condition implies that on any plane σ1 + σ2 + σ3 = con-
stant, the yield envelope would appear the same. These planes are called the
π-planes. If we view these planes from a perpendicular line of sight, the Tresca
envelope would appear as a perfect hexagon. This is shown in Fig. 7.5(b). The
corners in the hexagon may create certain problems if we need to draw normal
directions to the yield surface.

7.3.2 Von Mises Theory

The yield theory of von Mises is based on the portion of the elastic energy stored
in the material in the form of distortion energy. The total strain energy density is
given by
1
[σ1 1 + σ2 2 + σ3 3 ],
U= (7.19)
2
where we use the principal axes to avoid shear stresses. Using Hooke’s law,
1
1 = [σ1 − ν(σ2 + σ3 )], . . . . (7.20)
E

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 165 Failure Theories

Eliminating the strain components in terms of stress components, the total energy
density becomes

1 ! 2 "
U= σ1 + σ22 + σ32 − 2ν(σ1 σ2 + σ2 σ3 + σ3 σ1 ) . (7.21)
2E

The volume change or dilatation of a unit volume is given by

dv = 1 + 2 + 3. (7.22)

The work done by the hydrostatic stress due to the volume change is a part of the
total energy, called the dilatational energy. This is obtained as

1 σ1 + σ2 + σ3
Udil = ( 1 + 2 + 3 ),
2 3
σ1 + σ2 + σ3 1 − 2ν
= (σ1 + σ2 + σ3 ),
6 E
1 − 2ν
= (σ1 + σ2 + σ3 )2 . (7.23)
6E

If we subtract Udil from U , what remains is the distortion energy Udis . We may
imagine a unit volume undergoing a volume change with stored energy Udil and
then a distortion, which stores an additional amount of energy Udis .

Udis = U − Udil ,
1   2 
= 3 σ1 + σ22 + σ32 − 6ν(σ1 σ2 + σ2 σ3 + σ3 σ1 )
6E
1 − 2ν  2
− σ1 + σ22 + σ32 + 2(σ1 σ2 + σ2 σ3 + σ3 σ1 ) ,
6E
1+ν 2
= σ1 + σ22 + σ32 − (σ1 σ2 + σ2 σ3 + σ3 σ1 ) . (7.24)
3E

According to the von Mises theory, a ductile material yields when the distortion
energy Udis reaches a critical value. From a uni-axial test, the critical value for
Udis is

1+ν 2
Udis = σ . (7.25)
3E y

For a bi-axial stress state with σ3 = 0, we have

σ12 + σ22 − σ1 σ2 = σ y2 . (7.26)

This is an equation for an ellipse, with its major axis making 45◦ with the σ1 -
direction. Fig. 7.6 shows the von Mises ellipse superposed on the Tresca hexagon.
The ellipse becomes a circle when projected on to the π-plane. The yield behavior
of steel is often modeled using either Tresca or von Mises theory.

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 166 Mechanics of Aero-structures

σ2

σ2
σy
σ3 σ1

σy
σ1
σy

σy

Figure 7.6 The yield envelopes according to the Tresca theory and the von Mises theory when σ3 = 0.

7.4 Fatigue

Experiments have shown that the strength of metals under incremental loading
in tension (or compression) is substantially different compared to the case of
cyclic loading. Anyone who has broken a paper clip by bending it back and

σm
Δσ

Figure 7.7 An oscillating stress with a mean value of σm .

forth knows this. Fig. 7.7 shows an oscillating stress amplitude around a steady
mean stress of σm . The difference between the maximum and minimum stress of
the oscillation is usually denoted by σ . First, let us consider the case of zero
mean stress and the applied stress varying between (+σ/2) and (−σ/2). A
plot, (Fig. 7.8), of the stress difference σ versus log N , called an S − N –plot,
shows the number of load cycles N before a specimen fails for a given σ .

Δσ

Δσe

0 log N

Figure 7.8 A plot of the stress amplitude versus number of cycles to failure for a typical metal.

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 167 Failure Theories

Most metals (notably, steel) show a threshold for σ called the endurance
limit, σe , below which the specimen can withstand an unlimited number
of cycles without failure. Above this threshold value, we see a linear range,
which can be expressed as

σ = σ1 − k log N , (7.27)

where (−k) is the slope of the S − N curve and σ1 is twice the failure stress in
noncyclic loading.
Typical aluminum alloys do not show an endurance limit.

7.4.1 Palmgren-Minor Rule of Cumulative Damage

Assuming that a specimen under a stress difference of σi has a life of Ni cycles,
if it is subjected to n i cycles, we conclude that n i /Ni fraction of its life has been
used up. This is the basis of the Palmgren-Minor rule of cumulative damage. If
we subject a structure to M different stresses σi , i = 1, 2, . . . , M, each lasting
for n i cycles, then the Palmgren-Minor rule states,
n1 n2 nM
+ + ··· + = 1. (7.28)
N1 N2 NM
This approximate rule can be of use in predicting the life of a structure.

7.4.2 Goodman Diagram

So far we have assumed a purely oscillating stress with no mean value. When
there is a positive mean value σm , the maximum tensile stress in the specimen
is σm + 0.5σ and this high tensile stress has a damaging effect and decreases

Δσ
Δσ0
p
>
1
p
=
1
p
<
1

σm /σu

Figure 7.9 The effect of mean stress on the effective stress amplitude according to the Goodman formula.

the fatigue life of a specimen. If σu is the ultimate stress for failure, the Goodman
diagram, shown in Fig. 7.9, shows
σ σm
=1− , (7.29)
σ0 σu

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 168 Mechanics of Aero-structures

where σ is the effective stress difference to be used in the S − N curve. This

equation was modified to include the variability in the experimental data, to read
 p
σ σm
=1− , (7.30)
σ0 σu

where the index p is a constant that is material dependent.

7.5 Creep

Time-dependent deformation of structures under stress is called creep. Creep is

particularly significant at elevated temperatures in metals. Although Hooke’s law
gives an instantaneous response to the applied stress, when creep is present the
structure continues to deform as a function of time. Immediately after applying
the load the strain increases in the form
d
= Ct β σ m , (7.31)
dt
which is called primary creep. Here, C, β, and m are material constants and t is
time. In the second stage, we see the strain increase in the form

d
= Cσ n , (7.32)
dt
which is called the secondary creep. In the third stage, the strain, again, rapidly
increases resulting in material failure. This stage is called the tertiary creep.
All of the constants listed in the creep formulas are temperature dependent.
Accurate calculation of creep deformation is crucial in applications involving
high temperatures, such as turbine blades in a jet engine.

7.6 Stress Concentration Factor

This topic is adequately covered in a first course in strength of materials. When

the geometry of a stressed body has discontinuities such as abrupt changes in
cross-sectional areas or internal holes, we multiply the nominal stress by a factor
called stress concentration factor, C, to account for the actual stress at a point.
As shown in Fig. 7.10, in a large plate under tension σ0 , the presence of a small
elliptical hole increases the stress at points A and B by
σ a
=C =1+2 , (7.33)
σ0 b
where a and b are semi-axes of the ellipse.
When b → 0, the stress concentration factor goes to infinity while the ellipse
becomes a sharp-edged crack. Cracks are considered in the next section.
It is a recommended design practice to provide generous radii at corners to
reduce stress concentration when the geometry demands such corners.

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 169 Failure Theories

σ0

A B 2b
2a

σ0

Figure 7.10 An elliptic hole in an infinite plate.

7.7 Fracture Mechanics

Materials tend to develop sharp-edged discontinuities called cracks for various

reasons. Local material failure, corrosion, and dislocation accumulation in poly-
crystalline metals are some of the reasons for the origin of cracks. When a cracked
material is stressed, one or more stress components at the tip of a crack can reach
very high values – in fact, they reach infinity for a sharp crack tip according to
the linear elasticity theory. The loading at a crack tip is grouped into three modes:
Mode I, Mode II, and Mode III. Fig. 7.11 shows the three modes of loading in a
split beam.

Mode I Mode II Mode III

Figure 7.11 The three modes of loading affecting the crack tip stresses.

Of course, every crack has two surfaces, like the upper and lower surfaces
shown in the figure. In Mode I, the displacements on these surfaces are per-
pendicular to the surfaces and in Mode II and Mode III they are parallel to the
surfaces. As stated earlier, according to the linear elasticity solutions, some stress
components at the tip of a crack may go to infinity. Using a polar coordinate
system with its origin at the crack tip (see Fig. 7.12), the stress components and
displacements very close to the crack tip (when r is small compared to the crack
length) have been found using the theory of elasticity. These are listed in the
following sections for the three modes of loading. The proportionality constants
in these expressions, K I , K I I , and K I I I , are called the stress intensity factors,
which depend on the shape of the body, length of the crack, and the nature of
loading. When there is a single load P as shown in Fig. 7.11, the stress intensity
factors are proportional to P.

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 170 Mechanics of Aero-structures

y
r
θ
2a x

Figure 7.12 Crack tip coordinates.

7.7.1 Mode I
 
KI θ θ 3θ
σx x = √ cos 1 − sin sin ,
2πr 2 2 2
 
KI θ θ 3θ
σ yy = √ cos 1 + sin sin ,
2πr 2 2 2
KI θ θ 3θ
σx y = √ sin cos cos ,
2πr 2 2 2
*  
KI r θ θ
u= cos 1 − 2ν + sin2 ,
G 2π 2 2
*  
KI r θ θ
v= sin 2 − 2ν − cos2 ,
G 2π 2 2
w = 0, σzz = ν(σx x + σ yy ), σx z = σ yz = 0. (7.34)

On the crack plane y = 0, we have

KI
σ yy = √ , x > a, (7.35)
2πr
*
2(1 − ν)K I r
v= , x < a. (7.36)
G 2π

An infinite panel with uniform tensile stress σ0 at infinity with a central crack of
length 2a is shown in Fig. 7.13. For this geometry and loading, we have
√
K I = σ0 πa. (7.37)

For a panel with finite width W , with the crack symmetrically located, the approx-
imation
√ +
K I = σ0 πa sec(πa/W ) (7.38)

is used.

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 171 Failure Theories

σ0

2a

σ0

Figure 7.13 Infinite panel with a central crack under tension.

7.7.2 Mode II

 
KI I θ θ 3θ
σx x = − √ sin 2 + cos cos ,
2πr 2 2 2

KI I θ θ 3θ
σ yy = √ sin cos cos ,
2πr 2 2 2
 
KI I θ θ 3θ
σx y = √ cos 1 − sin sin ,
2πr 2 2 2
*  
KI I r θ 2 θ
u= sin 2 − 2ν + cos ,
G 2π 2 2
*  
KI I r θ θ
v= cos −1 + 2ν + sin2 ,
G 2π 2 2

w = 0, σzz = ν(σx x + σ yy ), σx z = σ yz = 0. (7.39)

On the crack plane y = 0, we have

KI I
σx y = √ , x > a, (7.40)
2πr
*
2(1 − ν)K I I r
u= , x < a. (7.41)
G 2π

For an infinite plate with a central crack of length 2a (shown in Fig. 7.14)
under a shear stress τ0 , we have
√
K I I = τ0 πa. (7.42)

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 172 Mechanics of Aero-structures

τ0

τ0
2a

Figure 7.14 Infinite panel with a central crack under shear.

7.7.3 Mode III

KI I I θ
σx z = − √ sin ,
2πr 2
KI I I θ
σ yz = √ cos ,
2πr 2
*
KI I I r θ
w= sin ,
G 2π 2
σx y = 0, σx x = σ yy = σzz = 0,

u=0 v = 0, σx y = 0. (7.43)

On the crack plane y = 0, we have

KI I I
σ yz = √ , σx z = 0, x > a, (7.44)
2πr
*
KI I I r
w= , x < a. (7.45)
G 2π
All of the preceding expressions for Mode I and Mode II are applicable to
the case of plane strain. By changing Poisson’s ratio ν to ν/(1 + ν), they can be
converted for use in plane stress cases.

7.7.4 Crack Propagation

When a crack propagates in a brittle material, new surfaces are created. A certain
amount of energy is needed to create new surfaces. We use γ to denote the surface
energy density per unit area of the surface, which is dependent on the material.
For simplicity, let us consider a material with a crack of length a with an applied
load P and the displacement under the load ū. Next, let us imagine the crack
extending by a. The change in potential energy is given by

U − Pū − T S + 2γ ta = 0, (7.46)

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 173 Failure Theories

where U is the strain energy, T temperature, S entropy, and t crack depth. If we

neglect the temperature effect, we can drop the entropy term. The strain energy
U depends on the displacement ū and on the crack length a. Thus,

∂U ∂U
ū + a − Pū + 2γ a = 0. (7.47)
∂ ū ∂a

Setting the coefficients of ū and a to zero, we have

∂U ∂U
= P, = −2γ t, (7.48)
∂ ū ∂a

where the first relation is Castigliano’s second theorem and the second is a
condition to propagate the crack.

a Δa
(a)

σyy

(b)
a + Δa x

Figure 7.15 Stress in front of a crack of length a (a) and the displacement inside a crack of length a + a (b).

Now, to find the rate of change in strain energy with respect to the crack length,
we consider as an example Mode I loading and the state of stress in front of the
original crack of length a and the displacement between x = a and x = a + a.
These are sketched in Fig. 7.15. If we isolate the upper half above the crack and
use a coordinate x starting from the old crack tip, before we had the tensile stress
distribution

KI
σ yy = √ , x > 0, (7.49)
2π x

between x = 0 and x = a. After the crack has extended, this stress has reduced
to zero and a displacement distribution,

*
2(1 − ν)K I a − x
v= , (7.50)
G 2π

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 174 Mechanics of Aero-structures

represents the opened crack. The transition from “before” to “after” can be viewed
as decreasing the tensile stress to zero while this stress moves by the displacement
v. The work done on the upper and lower half of the material is
 a
1
U = −2 σ yy vtd x,
0 2
 √
(1 − ν)K I2 a a − x
=− √ td x,
2π G 0 x


∂U (1 − ν)K I2 1 1−x
=− t d x, (7.51)
∂a 2π G 0 x
where we have replaced x/a by x in the last line. Using x = sin2 θ,

 1  π/2
1−x
dx = 2 cos2 θdθ = π. (7.52)
0 x 0

Finally,
(1 − ν)K I2
γ = . (7.53)
4G
This value of K I , which is responsible for extending the crack, is called the
√
critical stress intensity factor K I c , which has the dimension of Pa m as can be
verified from the stress expressions. As the applied load P is increased from 0,
there comes a point when K I reaches the critical value K I c and the crack begins
to propagate.
2Gγ
K I2c = . (7.54)
1−ν
Using the energy method we also get
4Gγ
K I2I c = , K I2I I c = 4Gγ . (7.55)
1−ν
When the loading is such that all the three modes are present, by computing
 a
(σ yy v + σx y u + σ yz w)td x, (7.56)
0

we find
 
(1 − ν) K I2 + K I2I + K I2I I = 4Gγ . (7.57)
The total energy dissipated by creating new surfaces is also known as the energy
release rate. They are denoted by G I , G I I , and G I I I for the three modes. By
multiplying 2γ t by the number of active (where the crack tip is moving) crack tips,
we can find the energy release rate. The energy used for creating the new surfaces
can be regained, in principle, by exerting pressure and by waiting for diffusive
healing of the crack in time. For all practical purposes, the term “dissipated
energy” aptly describes the energy spent in creating the surfaces during the crack
tip motion.

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 175 Failure Theories

7.7.5 Example: Energy Method for a Split Cantilever

P

a
P

Figure 7.16 Split cantilever.

Fig. 7.16 shows a split cantilever beam with Mode I loading. For the loading
shown, the deflection ū of the tip is given by
Pa 3 3E I ū
, P=
ū = , (7.58)
3E I a3
where I is the second moment of area for one of the beams. The energy stored in
the upper and lower beams totals to
3E I ū 2
U = P ū = . (7.59)
a3
Using the energy release rate
2(1 − ν)t K I2 ∂U 9E I ū 2 P 2a2
2γ t = =− = = . (7.60)
G ∂a a4 EI

Pa
KI = + . (7.61)
4(1 − ν 2 )t I

7.7.6 Ductility at the Crack Tip

In ductile materials, the magnitude of stress is limited by the yield stress. Then
the infinite stress predicted by the theory of elasticity does not occur. If the crack
is away from any material boundaries, there is a small region near the crack tip
where yielding takes place. As the crack advances, energy is dissipated in the
yield zone. It is known that this plastic energy dissipation is much higher than
the surface energy spent. However, experiments show that a critical value K I c
can be used to characterize crack growth even when crack tip yielding is present.
Dugdale has postulated that an effective crack length can be defined by adding an
approximate width of the plastic zone to the physical crack length. Later, Irwin
estimated the added crack length as the radius of an approximate plastic zone,
(1 − 2ν)2 K I2
rp = , (7.62)
2πσY2
for plane strain cracks in Mode I.

7.8 Fatigue Crack Growth

Poly-crystalline solids contain a large number of microscopic flaws called dislo-

cations where the regular crystal arrangement is broken. They also contain grain

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 176 Mechanics of Aero-structures

boundaries where crystals with different orientations meet. When these materi-
als are subjected to oscillating loads, the defects could move and coalesce into
microscopic cracks. It takes a number cycles of loading for this microcracks to
grow into visible range. If we denote by a0 the minimum size detectable, we can
predict their further growth using a relation advanced by Paris:
da
= C(K )m , (7.63)
dN
where C and m are material constants and
K = K max − K min . (7.64)
Experimental data are often presented in a log–log plot to obtain the constants.
√
As an example, with K in the units of MPa m and da/d N in m/cycle, austenitic
steel has C = 5.6 × 10−12 and m = 3.25.

FURTHER READING
Anderson, T. L., Fracture Mechanics: Fundamentals and Applications, CRC
Press (1995).
Bannantine, J., Comer, J., and Handrock, J., Fundamentals of Metal Fatigue
Analysis, Prentice Hall (1990).
Cherepanov, G. P., Mechanics of Brittle Fracture, McGraw-Hill (1979).
Jones, R. M., Mechanics of Composite Materials, McGraw-Hill (1975).
Lubliner, J., Plasticity Theory, Macmillan (1990).
Suresh, S., Fatigue of Materials, Cambridge University Press (1998).

EXERCISES
7.1 A brittle material fails at 80 MPa in a uni-axial test. A plate made of
this material is subjected to a bi-axial tension in the x, y-plane with the
proportional loading σx x = 2α and σ yy = α with α being a load control
parameter. Obtain the value of α at failure, (a) if the material fails according
to the maximum stress criterion, and (b) if it fails under the maximum strain
criterion. Assume the Poisson’s ratio is 1/3.
7.2 The yield stress for a steel alloy is 1.5 GPa. Obtain the internal pressure in
a thin-walled cylindrical pressure vessel with radius 1 m and thickness 0.2
cm for yielding to occur if (a) Tresca yield criterion is used and (b) von
Mises criterion is used.
7.3 A shaft of radius 5 cm carries a torque of 100 Nm and a bending moment
of 50 Nm. Obtain the maximum shear stress and the maximum Mises stress
in the shaft.
7.4 A circular cross-section fuselage is idealized as a thin-walled shell of radius
4 m with 48 booms of radius 2 cm each. The booms are distributed uniformly
around with the first and the 25th booms being located the farthest from the
neutral axis. If the maximum allowable compressive stress in the booms is
1 GPa, what is the allowable bending moment for this structure?

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

 177 Failure Theories

7.5 A boron/epoxy composite lamina is subjected to a uni-axial stress in a

direction 30◦ to the fiber direction. Using the strength values given Table 7.1,
obtain the maximum tensile and compressive stress it can carry.
7.6 A thin sheet is in plane stress in the x, y-plane with a central crack of length
5 mm, aligned with the x-axis. It is subjected to a constant shear stress,
σx y = 80 MPa and zero normal stress σx x . It was observed that the crack
begins to propagate when σ yy = 100 MPa. Obtain the surface energy density
γ for this material if it has a shear modulus of 80 GPa and a Poisson’s ratio
of 0.3. We may use the infinite plate approximation for this case.
7.7 A structure in plane strain has a crack of length 1 cm perpendicular to the
direction of an applied stress of 100 MPa. If the yield stress for this material
is 200 MPa and the Poisson’s ratio is 0.3, compute the effective crack length
including the plastic zone length (according to Irwin’s approximation) with
K I based on the original crack length. Repeat this calculation using the
effective crack length.
7.8 In the Paris law for crack growth under cycling loading,
da
= C(K )n ,
dN
the material constants C and n are given by
C = 5.6 × 10−12 , n = 3.25,
for austenitic steel. These constants correspond to stress intensity factors
√
in the units of MPa m and da/d N in m/cycle. Compute the number of
cycles for a crack of initial length 2 mm to grow to 10 mm under an applied
cyclic stress from 0 to 100 MPa. Assume the crack is located in an infinite
plate perpendicular to the direction of stress.
7.9 In the Paris law for crack growth under cycling loading,
da
= C(K )n ,
dN
the material constants C and n are given by
C = 5.6 × 10−12 , n = 3.25,
for austenitic steel. An infinite plate subjected to an applied stress cycle
from 0 to 100 MPa for 100,000 cycles showed, during inspection, a crack
of length 1 cm. Compute the initial crack length before the loading.
7.10 A plot of the stress amplitude S in cyclic loading for a material follows the
law
S = 20 − k log N ,
where S is in kPa, log has the base 10, k is a constant, and N is the number
of cycles to failure. If a specimen made of this material fails in 100,000
cycles at a stress amplitude of 100 kPa, find the value of k. A structure made
of this material is first subjected to 100 kPa for 40,000 cycles and then to
50 kPa load cycles. How many cycles would it last at this second load?

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press

https://doi.org/10.1017/CBO9781139871983.008 Published online by Cambridge University Press