Chapter 8: Fracture of Cracked Members
8.1 - Introduction Pre-existing cracks, or flaws that can be treated like cracks, exist in many materials. These flaws could consist of voids in the material, voids in welds, gouges, foreign material (impurity), delamination of layered materials, separation of reinforcing bar from concrete, etc. Thermal stresses may lead to a crack, even with no load on the material.
8.2.1 Cracks as Stress Raisers
Cracks can be considered infinitely sharp elliptical holes that give rise to infinite stress concentration factors, or stress raisers, as they are often called.
(see Section 2.18. Stress Concentrations, p. 107-8, Mechanics of Materials, 4th ed., Beer, Johnston, and DeWolf, McGraw-Hill, 2006, your deforms book)
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⎛
c⎞
⎛
c⎞
σ y = S ⎜1 + 2 ⎟ = S ⎜1 + 2 ⎟ d⎠ ρ⎠ ⎝ ⎝ Kt =
σy S
=1+ 2
(8.1)
c c =1+ 2 d ρ
As d approaches zero or ρ approaches zero the ellipse is slit-like or crack-like and the stress concentration factor Kt approaches infinity.
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8.2.2 Behavior at Crack Tips in Real Materials Define δ = crack tip opening displacement (CTOD)
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8.2.3 Effect of Cracks on Strength To account for the effect of a crack on the strength of a material, the stress intensity factor K is defined. The quantity K characterizes the severity of the crack and is a function of crack size, geometry of the cracked structural member, and loading type (i.e., tension, torsion, …).
Example: Consider a center-cracked plate loaded in tension. The stress intensity at the tip of crack is defined as K = S πa
a is the half crack length S is the gross stress, a force/area calculation If K < Kc, then no failure. Kc = critical stress intensity factor, which is a material property. The values for Kc for various materials can be found in Table 8.1, 8.2 on p. 318-19 in the text. The relation K = S π a is a result of sophisticated analysis from the theory of linear elasticity. The study of fracture based on this approach is called Linear Elastic Fracture Mechanics (LEFM). Rearranging the equation, the critical level of the gross stress, Sc is given by Sc =
Kc πa
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(Repeating formula for convenience) Sc =
Kc πa
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8.2.4 Effects of Cracks on Brittle versus Ductile Behavior Consider the crack length a = at (t = “transition”) where the critical level of gross stress S equals the yield stress of the material, σo .
Kc 1⎛K ⎞ Sc = σ 0 = → at = ⎜ c ⎟ π ⎝ σ0 ⎠ π at
2
(eq. 8.4)
at is material property since Kc and σ0 are material properties
For a > at → Failure by fracture For a < at → Little strength reduction due to the crack since the material yields anyway If low σ0 and high Kc , then at large. Component will fail by yielding. If high σ0 and low Kc , then at small. Component will fail by fracture.
Figure 8.6 Transition crack length at for a low-strength, hightoughness material (a), and for high-strength, low-toughness material (b). If (b) contains internal flaws ai, its strength in tension σut is controlled by brittle fracture.
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8.2.5 Internally Flawed Materials
If ai = represents inherent (i) half flaw size, then the ultimate strength of the material is actually given by
σ ut = K c π ai
(8.5)
even though σ u from a material property table has a different value of σ u determined from testing the material with no flaws, e.g., Table 4.2 on p. 121. Material is insensitive to cracks longer than 2ai since the flaws inherent in the material are worse than the crack itself. Pretty bad quality control of the material! Or, for some materials, it is the character of the material. Other Loading Modes
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8.3.1 Strain Energy Release Rate, G
G=−
1 dU , strain energy release rate t da
Units of G:
(8.6)
N⋅m N⋅m i.e., U per crack length = m⋅m m2
Gc = critical strain energy release rate, a material property If G = Gc, then the crack grows Using a numerical stress analysis, e.g., the finite element method, for a particular structure, crack length a, loading, and material, the crack can be allowed to grow virtually an amount Δa and G computed and compared to Gc.
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8.3.2 Stress Intensity Factor K (More Formal View for Mode I Loading)
σx =
KI θ⎡ θ 3θ ⎤ cos ⎢1 − sin sin ⎥ + ... 2⎣ 2 2⎦ 2π r
σy =
KI θ⎡ θ 3θ ⎤ cos ⎢1 + sin sin ⎥ + ... 2⎣ 2 2⎦ 2π r
τ xy =
KI θ θ 3θ cos sin cos + ... 2 2 2 2π r
(8.7)
τ yz = 0
τ xz = 0
σz = 0 for plane stress, σz = ν(σx + σy) for plain strain (εz = 0) Formally K I ≡ lim
r →0,θ →0
(σ
All stresses are proportional to KI and
y
2π r
)
1 (square-root singularity). r
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Generalization of stress intensity equation For the cracked plate of Fig. 8.5 the stress intensity factor was given by K = S πa
Though not stated explicitly, to derive this formula it was assumed that a << b and that the plate dimension was long in the loading direction, i.e., the plate is long and wide compared to crack length so the crack does not 'feel' the influence of the edges or ends of the plate. What if a is only somewhat less than b, i.e., plate is “narrow”, or the plate is “short”? What if the crack is at the edge of the specimen? If any of these situations is the case, then we use the same form for the stress intensity factor, but we modify the form slightly. The stress intensity factor is then defined as K = FS π a ↑ accounts for geometric effects
a b (see Figs. 8.12 and 8.13 on pp. 327 and 329, for example, replicated the
F is a function of geometry, where the crack is, and α, where α =
on next two pages, and other figures on p. 330, 331, 339, 340, 340, etc.)
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8.4 Application of K (and F) to Design and Analysis Key equation stress to cause fracture
S=
Kc
fracture toughness (a material property
F πa crack length
Example of using key equation Problem 8.6 - A center-cracked plate of AISI 1144 steel has dimensions, as defined in Fig. 8.12(a), of b = 40 mm and t = 15 mm. Assume the plate is long. For a safety factor of three against brittle fracture, what is the maximum permissible load P on the plate if (a) a = 10 mm, and (b) a = 24 mm? b = 40 mm
Use Fig. 8.12a
t = 15 mm a = 10, 24 mm
K = FS g π a
(a) a = 10 mm → α =
a 10 = = 0.25 b 40
From plot 8.12(a) F ≈ 1 Kc (Table 8.1, p. 318) = 66 MPa m Therefore P Kc 66 × 106 Sg = = = = 372.4 × 106 2bt F π a 1 ⋅ π ⋅ 0.010
Solving for P gives P = 446,838N, for X = 3 → P = 148,900 N
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(b) a = 24mm → α =
24 = 0.6 40
From plot 8.12(a) and formula (a) 1 − 0.5α + 0.326α 2 F= = 1.292 1−α
P Kc 66 × 106 Sg = = = = 186 × 106 2bt F π a 1.292 ⋅ π ⋅ 0.025
Solving for P gives P = 223,200 N, for X = 3 → P = 74, 400 N
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8.4.3 Circular/Semi-Circular Cracks
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For a circular crack in an infinite body: Exact solution (from theory of elasticity) K=
2
π
S πa
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Elliptical Cracks
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8.4.4 Cracks Growing From Holes, Notches, Fillets
Short crack at the edge of a hole
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Long crack at the edge of a hole
Care in Translation of Nomenclature
K B = Fd S π (c + A)
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8.5.3 Superposition for Combined Loads
From Fig. 8.12c and Fig. 8.13a (pp. 11 and 12 in these notes) K1 = F1S1 π a
6 M 6 Pe (8.27, 8.28, 8.29) = 2 b 2t bt P⎛ 6F e ⎞ K = K1 + K 2 = ⎜ F1 + 2 ⎟ π a bt ⎝ b ⎠
S1 =
P bt
K 2 = F2 S 2 π a S2 =
F1 and F2 from figures based on the value of
a b
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8.5.5 Leak-Before-Break Design of Pressure Vessels
Note: Stress applied parallel to crack has no influence on crack to open. Two failure scenarios: #1. Crack grows slowly, penetrate wall, causing leak #2
Crack unzips (grows rapidly), explosive release of pressure
Assuming crack is circular, #1 occurs when c = t. We would therefore like the c that causes crack to unzip to be greater than t, i.e., cc ≥ t. Using Fig. 8.12(a) and K = FS π a → K Ic = 1 ⋅ σ t π a cc =
1 ⎛ K Ic ⎞ ⎜
2
⎟ > t for #1 to occur
π ⎝ σt ⎠
(8.32)
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