LEFM—Linear Elastic Fracture Mechanics
Crack-tip fields for plane stress and plane strain: Mode I and Mode II
Linear, isotropic elastic solids with Young's modulus, E , and Poisson's ratio, ν .
The in-plane stresses in the crack tip fields are the same in plane stress and plane strain,
however, σ 33 = 0 in plane stress and σ 33 = ν ( σ 11 + σ 22 ) in plane strain
Mode I - -Symmetric stresses & strains fields at tip.
Universal behavior (all problems !) as r approaches crack tip:
KI
I
σαβ
(θ )
2π r
σ αβ =
K
uα = uα + I
E
0
r I
uα (θ )
2π
E = E in plane stress
=
E
in plane strain
1 −ν 2
I
The distributions, σαβ
, are given in most texts on LEFM
KI
, σ 12 = 0
2π r
where K I is called the Mode I stress intensity factor
Mode I-- On the plane ahead of the tip: σ 22 =
(standard definition)
Mode II - -Anti - symmetric stresses & strains fields at tip.
II
Universal behavior (all problems !) as r approaches crack tip (σαβ
in the texts):
σ αβ =
K II
II
σαβ
(θ )
2π r
uα = uα0 +
K II
E
r II
uα (θ )
2π
K II
, σ 22 = 0 (standard definition)
2π r
where K II is called the Mode II stress intensity factor
On the plane ahead of the tip: σ 12 =
Mode III - -Out - of - plane shearing (also called "anti - plane shear").
Universal behavior (all problems !) as r approaches crack tip:
(σ 13 , σ 23 ) =
u3 =
2 K III
G
K III
2π r
⎛ ⎛θ ⎞
⎛θ ⎞⎞
⎜ sin ⎜ 2 ⎟ , cos ⎜ 2 ⎟ ⎟
⎝ ⎠⎠
⎝ ⎝ ⎠
r
⎛θ ⎞
sin ⎜ ⎟
2π
⎝2⎠
(σ 12 = K III / 2π r on plane ahead of tip)
(G ≡ E /(2(1 +ν )) is shear modulus)
Derivation of the Mode III fields given in class
By superposition (linear elasticity), conditions at any crack tip can always be
represented as a sum of the three modes.
Some Basic Solutions
x2
3
Tensile crack
x1
σ /σ ∞
or
τ /τ
Exact solution:
σ 22 =
σ ∞ x1
x12 − a 2
∞
2
Exact solution
KI = σ ∞ π a
( x1 > a, x2 = 0)
Asymptotic crack-tip
field
1
K
2π r
Shear crack
0
1
Exact solution:
σ 12 =
τ ∞ x1
x −a
2
1
2
K II = τ ∞ π a
( x1 > a, x2 = 0)
2
3
x1 / a
4
5
For these two problems the K-field gives and accurate
estimate of the stress ahead of the crack for r/a<1/4
There are several excellent compilations of solutions. We will make use of Tada, Paris & Irwin
Some Basic Solutions, continued
ASTM compact tensile specimen
h = 0.6b
h1 = 0.275b
c = D = 0.25b
tensile edge - crack : K I = 1.122σ ∞ π a
thickness ≡ t = b / 2
KI =
periodic tensile cracks :
⎡ 2b
⎛ π a ⎞⎤
K I = σ ∞ π a ⎢ tan ⎜
⎟⎥
⎝ 2b ⎠ ⎦
⎣π a
P
⎛a⎞
a F1 ⎜ ⎟
bt
⎝b⎠
F1
1/ 2
Mode I at all points
of crack edge
a/b
penny shaped crack : K I =
2
π
σ∞ πa
(see Tada, et al page???
An example of results in the
Stress Analysis of Cracks Handbook
by H. Tada, P.C. Paris and G.R. Irwin
ASME Press, New York, NY, 2000
An edge-notched infinite beam under pure
bending. Pages 55-57 of the Handbook
Results are presented on the this page and the
next for the stress intensity factor (a mode I problem),
the crack opening displacement at the surface,
and the additional rotation due to the presence
of the crack.
Energy Release Rate, Prescribed Load vs. Prescribed Displacement, and
Relation to Stress Intensity Factors
Prescribed load/thickness, P. Load-point displacement = ∆
G = energy release rate (J / m 2 )
SE = strain energy of the system/thickness
P∆ = potential energy/thickness of load
PE = energy of the system/thickness
Note that for any linear elastic system, SE = P∆ / 2, and thus
Generic cracked body
for prescribed load:
PE = SE − P∆ = − P∆ / 2
The energy release rate for prescribed load is defined as
⎛ ∂PE ⎞
G = −⎜
⎟ ⇒
⎝ ∂a ⎠ P
G=
1 ∂ ( P∆ )
P ⎛ ∂∆ ⎞
= ⎜
⎟
2 ∂a P 2 ⎝ ∂a ⎠ P
∆
P
which depends only on geometry, including, a, E and ν . Thus,
Define compliance,
C=
dC
1 2 dC
⎛ ∂∆ ⎞
P
G
P
=
⇒
=
⎜
⎟
da
2
da
⎝ ∂a ⎠ P
Graphical interpretation
of energy release rate
due to crack advance
Energy Release Rate, continued: Role of compliance and loading conditions
As indicated in the figure, introduce a linear spring with compliance
CM in series with the cracked body. Let ∆T be the total displacement
through which P works. Now, let ∆T be prescribed.
∆T = ∆ + CM P = ∆ + (CM / C )∆
1
1
1
2
PE = SE + CM P 2 = C −1∆ 2 + CM −1 ( ∆T − ∆ )
2
2
2
and a straight-forward calculation (see pg.8of notes) again gives
G=
1 2 dC
P
2
da
a
Note! The energy release rate does not depend on CM .
In particular, G is the same for both prescribed load and prescribed displacement.
This is not intutive.
Most results for G are obtained either from analytical or numerical calculations (see later).
However, the above formula permits experimental evaluation of G by experimentally
measuring the compliance at two nearly equal crack lengths, a and a+da.
Energy Release Rate, continued: Relation between G and K’s
Mode I situation:
symmetric body and
symmetric loading.
Since G is independent of loading, consider a body under prescribed ∆.
Consider two configurations of the cracked body one with a and the other with a+∆a.
Because the system is elastic, the energy released in advancing the crack ∆a is the same
as the work done in closing the crack from a+∆a to a (see figure). Because ∆ is prescribed,
no work is done by applied loads in closing crack. The work to close the crack is
1 ∆a
σ 22 ( x, 0) = K I (a) / 2π x
∆W = G∆a = ∫ σ 22 ( x, 0) ⎡⎣u2 ( x, 0+ ) − u2 ( x, 0− ) ⎤⎦dx
2 0
8 ∆a − x
u 2 ( x , 0 + ) − u 2 ( x , 0 − ) = K I ( a + ∆a )
∆a
2
∆a − x
E
2π
K I ( a ) K I ( a + ∆a ) ∫
dx
=
E = E in plane stress, E = E /(1 −ν 2 ) in plane strain
0
x
πE
1
= K I ( a ) K I ( a + ∆a ) ∆ a
Under all three modes, one finds
E
1
1
2
2
1 2
G
=
K
+
K
+
K III 2
(
I
II )
⇒ G = K I (Irwin's universal relation!)
E
2G
E
Discuss units
Energy Methods for Determining Energy Release Rate
Double cantilever beam specimen
Compute compliance of specimen treating each arm as a cantilever beam,
and consider the specimen to have unit thickness and P is force/thickness
∆ Pa 3 4 Pa 3
∆ 8a 3
=
=
⇒ C≡ = 3
Eb3
P Eb
2 3EI
1 2 dC 12 P 2 a 2
=
G= P
da
Eb3
2
Since this is a mode I problem (by symmetry), Irwin's relation gives
KI = 2 3
Pa
b3/ 2
See notes pg. 10 for the problem of a thin strip (plane stress)
with a semi-Infinite crack subject to rigid grips.
This is an exact solution. Note it is independent of crack length.
E∆2
1
E∆
G=
,
K
=
I
4(1 −ν 2 )b
2 1 −ν 2 b
Energy Methods for Determining Energy Release Rate, continued
Delamination of stressed thin film on elastic substrate
1D analysis (uniaxial stress in film)
h
σ
1 σ 2h
strain energy/area in film far ahead of tip =
2 E
strain energy/area in film far behind tip = 0
σ
delamination crack
energy released/area due to crack advance:
1 σ 2h
G=
2 E
which is independent of crack length.
This problem does not have symmetry and it is an example where the crack is a combination
of mode I and II. Later in the course we will determine the two stress intensity factors.
The simple result for G is valid when the crack is long enough such that steady-state conditions
apply. In practice this means the crack length has to be long compared to the film thickness.