3.3 Mechanical Strength
3.3.4 Material Toughness
σ γ
f E s
= c
= × ×
× ×
= 4 −
100 10 2
4 5 10 10
12 9
6 12
( 166 12) =0 1. GPa (3.19) This example demonstrates that the existence of a very small crack can sig- nificantly reduce the fracture strength from the theoretical cohesive strength by a factor of 100. In some cases, it can even be reduced by a factor of 1,000.
The level of preexisting cracks in a part is primarily determined by the man- ufacturing process and the quality control system. From a reverse engineer- ing perspective, the reproduced part should be manufactured under such a quality control system that only introduces the same level of or less preexist- ing cracks than what the original part is allowed.
structure, have excellent toughness down to –273°C, with no steep ductile- to-brittle transition. Therefore, the identification of alloy phase and crystallo- graphic structure might be required in some reverse engineering applications to ensure proper ductile–brittle transition behavior.
The modern fracture mechanics goes beyond simple stress and strain measurements. Fracture toughness has become a primary criterion in failure analysis based on fracture mechanics. Its calculation quantitatively integrates stress with existing crack size. The fracture toughness is a material property obtained by a valid test that first introduces a parameter defined as a stress intensity factor. Equation 3.20a is the mathematical formula of stress inten- sity factor for a thin plate of infinite width with an existing crack of length 2c in the center and subject to a tensile stress, σ, as illustrated in Figure 3.9.
The same equation also applies to a wide thin plate under a tensile stress σ with an edge crack of length c. This formula is virtually identical to the mathematical formula for fracture toughness expressed by Equation 3.20b.
Equation 3.20a and b is based on linear elastic fracture mechanics. It assumes a linear relationship between stress and strain and small elastic deformation.
It works best for brittle fracture. The fracture toughness, KIC, is a validated material property that satisfies a set of defined conditions, such as meeting the test specimen thickness requirement, while the stress intensity factor is just a numerical quantity describing the loading condition. Any set of stress and crack length will generate a numerical value of stress intensity factor calculated by Equation 3.20a, but will not always produce a valid fracture toughness value as expressed by Equation 3.20b. In contrast to the material toughness measured by the Charpy impact test, which has a unit of energy
–200 –150 –100 –50 0
Temperature, °C 0
2 4 6
Relative Ductility
Charpy impact toughness Tensile
elongation
Angular torsion displacement
FIgurE 3.10
Ductile –brittle transition of ferric steels.
per unit volume, the fracture toughness has a unit of MPa m . The fracture toughness symbol itself, KIC, reflects the complexity of this material param- eter. The Roman numeral subscript I refers to the externally applied load categorized as mode I, that is, tension; C refers to a critical value. It implies that the fracture toughness is a critical reference parameter, and when the combined effect of stress and crack length is beyond this value, the material will fail. Different mathematical equations with various stress and geometric parameters are integrated together to quantitatively describe fracture tough- ness for different load modes and crack configurations. When the loading mode is a shear or torsion, the symbols for fracture toughness are KIIC and KIIIC, respectively. It is worth noting that fracture mechanics predicts a part to fail at different stress levels depending on the size, shape, and form of the existing crack.
K= σ πc (3.20a)
KIC= σ πc (3.20b)
The following example demonstrates the combined effects of stress and crack size on fracture criterion from a fracture mechanics perspective.
Figure 3.11 shows a wide thin plate with an edge crack of c = 0.002 m long and extending through the full thickness. The width, w, is 0.2 m, and the thickness, t, is 0.001 m. The plate is made of aluminum alloy with a yield strength of σy = 350 MPa, and a fracture toughness value of KIC = 40 MPa m.
Determine the maximal load this plate can sustain under tension. What will be the maximal tensile load this plate can sustain when the edge crack grows to 0.02 m long?
t P
P w
FIgurE 3.11
Edge crack in a wide thin plate.
The maximal load, Pmax, this plate can sustain without yielding can be approximately calculated by the following equation:
σy P t w c
P P
= − =
− =
max max max
( ) ( .0 001 0 2 0 002)( . . ) 0 0001998. = 350 MPa Pmax = 350 MPa × 0.000198 m2 = 69,300 Newtons
For a wide thin plate under tension with an edge crack, the fracture tough- ness can be mathematically described by Equation 3.20b:
KIC= σ πc
In fracture mechanics, the maximal load this plate can sustain without frac- ture can therefore be calculated as
σ= Kπ = π = c
IC 40
0 002 504 6
( . ) . MPa
Pmax = σ × [t × (w – c)] = 504.6 × [0.001 × (0.2 – 0.002)] = 99,991 Newtons With a 0.002 m edge crack, the plate will yield at 69,300 Newtons first, before it fractures. The determining factor is yield strength, and the maximum load this plate can sustain under tension is 69,300 Newtons.
When the crack grows to 0.02 m long, the plate will yield at Pmax = 350 MPa × [(0.001) m × (0.2 – 0.02) m = 63,000 Newtons
However, the allowed maximal stress from a fracture mechanics perspective before fracturing will be
σ= Kπ = π = c
IC 40
0 02 159 6
. . MPa
The plate will fracture at Pmax = σ[t × (w – c)] = 159.6 MPa × [0.001 m × (0.2 – 0.02) m] = 28,729 Newtons. Therefore, based on the fracture mechanics calcula- tion, the maximal load this plate can sustain is only 28,729 Newtons when the crack grows to 0.02 m. In other words, the load-carrying capability of this plate decreases more rapidly according to fracture mechanics as the crack length increases, and the determining factor shifts from yield strength when the crack length is 0.002 m to fracture toughness when the crack length grows to 0.02 m.
The complexity of the fracture toughness test requires specialized engineer- ing expertise to obtain a valid value, and it can be costly. Nonetheless, the test of fracture toughness and life calculation based on fracture mechanics are
warranted for the reverse engineered part that is a critical structure element serving in an environment across the ductile–brittle transition temperature.
In many reverse engineering projects, the determination of fracture tough- ness is yet to become a mandatory requirement despite its being a critical parameter that determines if a part will fail. However, when more and more reverse engineered parts are life-limited or critical parts, and the original part was designed based on fracture toughness, the fracture toughness test should be conducted to demonstrate the equivalency whenever feasible.