Chapter 3 The applicability of LEFM - Division of Solid Mechanics

Chapter 3
The applicability of LEFM
3.1 Crack tip plasticity
3.1.1 Behaviour at the crack tip;
plane stress, plane strain or mixed conditions
According to linear elasticity (c.f. previous section), we have in front of the
crack tip (θ = 0) the following generalized bi-axial stress state
σ x = σ y =
K I
√
2πr
(3.1)
σ z = 0 (plane stress) or σ z = ν(σ x + σ y )(plane strain) (3.2)
τ xy = τ yz = τ zx = 0 (3.3)
Clearly, due to the high tensile in-plane normal stresses, the material will
contract in the thickness direction. However, the extent to which it will do
so depends on the constraint due to the surrounding material. Considering a
through thickness crack in a plane body, it is clear that a large thickness will
prohibit contraction (plane strain condition), except at the surface, where a
plane stress state will exist. For a thin specimen on the other hand, plane
stress conditions will prevail throughout the thickness, for an illustration, see
Fig.3.1 below. For an inner crack on the other hand, a plane strain situation
will generally prevail at the crack tip.
Now, why is this interesting? Well, the answer is that the condition in the
thickness direction will affect the size of the plastic zone that inevitably will
be found at crack tips in metallic materials.
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Figure 3.1: Plane strain and plane stress conditions
3.1.2 General plastic behaviour at bi-axial loading
for plane stress and plane strain conditions
Let us consider a flat specimen which is subjected to the in-plane bi-axial
applied stress σ 0 , and which obeys the Tresca yield condition (used for simplicity).
With the yield criterion
where
in combination with the prevailing loading
σ eT = σ Y (3.4)
σ eT = σ 1 − σ 3 (σ 1 ≥ σ 2 ≥ σ 3 ) (3.5)
σ 1 = σ 2 = σ x = σ y = σ 0 (3.6)
σ 3 = σ z (3.7)
we get plastic yielding when
⎧
⎨σ Y
plane stress
σ 0 = 1
⎩
1 − 2ν σ Y ≈ 3σ Y plane strain
(3.8)
As can be seen, plane strain conditions will restrict the plastic flow under
bi-axial loading conditions.
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3.1.3 Crack tip behavior and applicability of LEFM;
a first discussion
Since a generalized bi-axial stress state is prevailing at the crack tip, and since
the associated plastic flow behaviour strongly depends on the constraint in
the thickness direction, it follows from the above discussion, that the size
of the plastic zone will be smaller under plane strain conditions than under
plane stress conditions. Furthermore, for the case of a through-thickness
crack in a flat specimen, the plastic zone size will be larger for a thin specimen
than for a thicker one.
Now, as long as the plastic zone is small compared to the crack length, it
may be argued (and motivated by the success of LEFM) that K governs the
occurrence of fracture (and as we shall see later on, the fatigue crack growth
behaviour). Thus, we we would like to find an estimation of the plastic zone
size and, based on that, set up a criterion for the applicability/validity of
LEFM.
3.1.4 The Irwin approach
The first simple approach for finding the plastic zone size, with focus on its
length r p in the direction of the crack (θ = 0), is to start with the linearelastic
solution for the crack tip stresses, and to state that r p is given by
σ el.sol
y (r p ) = ασ Y (3.9)
where the factor α is equal to 1 for plane stress conditions and equal to 3 (or
less, see subsequent discussion) for plane strain conditions, resp. However,
such an approach will be erroneous, since we will then “miss” part of the
area below the stress component curve. Irwin solved this problem, by simply
translating the elastic stress-curve to the right, and calculated the corrected
plastic zone size by the relation
∫ r1
0
σ y (r)dr =
∫ rp
0
ασ Y dr (3.10)
where r 1 is the uncorrected plastic zone length, c.f. Fig.3.2 below. To see
this, we note that
A 1 = A ′ 1
A 3 = A ′ 3
⇒ A 2 = A ′ 2 ⇒ A 1 + A 2 = A ′ 1 + A ′ 2
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Figure 3.2: Approximation of the plastic zone size according to Irwin
By the above relation we get
K I
√
2π
∫ r1
0
K I
1
√ r
(r)dr = ασ Y r p ⇒
√
2π
2 √ r 1 = ασ Y r p ⇒
r p =
√
2KI
√
r1
√ πασY
(3.11)
Furthermore,
σ y (r 1 ) = ασ Y ⇒
K I
√ 2πr1
= ασ Y ⇒
√
r1 =
K I
√
2πασY
⇒
implying
r 1 = 1 ( ) 2 KI
(3.12)
2π ασ Y
r p =
√
2KI
√ πασY
K I
√
2πασY
i.e.
r p = 1 π
(
KI
ασ Y
) 2
(3.13)
As can be seen, the corrected plastic zone size (r p ) is, according to Irwin’s
analysis, twice the size of the uncorrected one (r 1 ).
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3.2 Applicability of LEFM
It was found above, that
r p = 1 π
(
KI
ασ Y
) 2
(3.14)
In order to ensure the applicability of LEFM, we know that the plastic zone
size needs to be small compared to the crack length. More quantitatively,
for LEFM-testing of K Ic -values, the American Society for Testing and Materials
(ASTM) state the following requirements for critical dimensions of the
test specimens (typically, Single Edge Notched Bend/SENB specimens, or
Compact Tension/CT specimens)
the ASTM criterion for K Ic testing
⎧ ⎫
⎨ a ⎬ ( ) 2
t
⎩ ⎭ ≥ 2.5 KIc
(3.15)
σ
W − a
Y
where the second requirement assures plane strain conditions and the third
requirement assures that no excessive plastic flow will take place in the uncracked
ligament.
It may be noted that with the original proposal α = √ 3 by Irwin for plane
strain conditions, the ASTM condition requires that the crack length is to
be approximately 25 times larger than the plastic zone at the time of fracture.
In some literature (some course books and some handbooks) it is argued
that the ASTM condition is to be used as a general criterion for the applicability
of LEFM, where the thickness criterion makes sure that K Ic may be
used as a relevant value for the fracture toughness, and where the size of the
unfractured ligament makes sure that no excessive plastic deformation has
taken place. However, LEFM can also be used for plane stress situations, as
long as the plastic zone is small compared to the crack length, and as long
as a relevant value for the fracture toughness at plane stress K c /K 1c (> K Ic )
is used.
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3.3 Some further comments
Another approach for finding the size of the plastic zone is the one by Dugdale
(the so called strip yield model), where the plastic zone size is found by
superimposing elemental cases; similar results as for the Irwin approach are
found for cases of small scale yielding. By the Dugdale-approach, fracture
formulas based on the Crack (Tip) Opening Displacements/C(T)OD can also
be formulated, which find their major use in Elasto-Plastic Fracture Mechanics
(EPFM).
It should also be mentioned that it has been proposed to use Irwin’s results
to correct the crack length in cases where the crack-length condition
for LEFM is not fulfilled. This may be seen as an attempt to avoid EPFM.
However, we will not enter into that details here.
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3.4 Questions
3.1) Check if LEFM , according to the ASTM criterion, is applicable in the
introductory example in Ch.1, Introduction?
Answer: All critical dimensions are larger than 1.4 cm, thus LEFM is OK!
3.2) Check if LEFM , according to the ASTM criterion, is applicable in
Question 2.4) in Ch.2 ? Assume that t = 1.5cm and that σ Y = 1200MP a.
Answer: All critical dimensions are larger than 1.1 cm, thus LEFM is OK!
3.3) Do problem 2.12 in the book by Dahlberg & Ekberg!
3.4) Calculate the maximum load that the linearly elastic detail shown below
can withstand withoput cracking! The validity of LEFM is to be checked
w.r.t. the ASTM criterion!
Answer: P c = 2.8MN, All critical dimensions are larger than 1.8 cm, thus
LEFM is OK!
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