Abstract
The mechanical properties of materials are related to the fundamental bonding strengths of the constituent atoms in the solid and any bonding defects which might form. A molecular model is presented so that primary bond formation mechanisms (ionic, covalent, and metallic) can be better understood. How these bonds form and respond to mechanical stress/loading is very important for engineering applications. A discussion of elasticity, plasticity and bond breakage is presented. The theoretical strengths of most molecular bonds in a crystal are seldom realized because of crystalline defects limiting the ultimate strength of the materials. Important crystalline defects such as vacancies, dislocations, and grain boundaries are discussed. These crystalline defects can play critically important roles as time-to-failure models are developed for: creep, fatigue, crack propagation, thermal expansion mismatch, corrosion and stress-corrosion cracking.
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Notes
- 1.
Classical description is—two bodies cannot occupy the same space.
- 2.
In order that ionic contributions are comprehended from both near and far, the potential for anion-pair is often written as: φ(r) = −αe2/r, where α is the Madelung constant. In cubic crystalline structures, α = 1 to 2.
- 3.
The classical oscillator will oscillate until all its energy is finally dissipated. The quantum oscillator, however, will dissipate its energy in quantum amounts (n + 1/2) ℏω until it finally reaches its ground state. In the ground state (n=0), the quantum oscillator will still have zero-point energy oscillation: (1/2)ℏω.
- 4.
The stress gradient is given by: \( \overrightarrow{\nabla}\sigma =\left[\widehat{x}\frac{\partial }{\partial x}+\widehat{y}\frac{\partial }{\partial y}+\widehat{z}\frac{\partial }{\partial z}\right]\sigma \left(x,y,z\right). \)
- 5.
Atom movement is opposite to vacancy movement. Atoms tend to move from relative compressive regions to relative tensile regions. Such atom movement tends to reduce both the relative compressive stress and the relative tensile stress. Atom (or vacancy) movement due to stress gradients is a stress-relief mechanism.
- 6.
Specific density represents the number of atoms per unit area.
- 7.
Other dislocation types can exist, such as screws dislocations (not discussed here). These can also be important in the mass-flow/creep process.
- 8.
The ramp-to-failure/rupture test is described in detail in Chap. 11.
- 9.
The usefulness of ramp-voltage-to-breakdown test for capacitor dielectrics is highlighted in Chap. 12.
- 10.
Recall from Chap. 9, one expects the relaxation-rate constant to be thermally activated: k = k0 exp [−Q/(KBT)].
- 11.
The load/force is constant. The average stress is only approximately constant during testing due to some expected cross-sectional area changes.
- 12.
Recall, from Chap. 11, that the stress-migration/creep bakes for aluminum were generally done at temperatures above 100 °C.
- 13.
Historically, these localized stresses at crack tips have been referred to as either stress raisers or stress risers. The terms will be used interchangeably.
- 14.
The elastic energy Uelastic of the material reduces with crack propagation, thus we have defined ΔUreleased such that it is always positive, i.e., ΔUreleased = −ΔUelastic.
- 15.
Note that when the crack size a goes to zero, the apparent rupture stress goes to infinity. However, in these situations, where the right-hand side of the equation becomes extremely large, the rupture stress will be limited by the normal crack-free rupture mechanisms and σrupture will assume the crack-free rupture strength.
- 16.
Recall that specific energy density is the energy per unit area.
- 17.
Historically, this has been referred to as Griffith’s equation which was developed for brittle materials. More recently, Irwin is usually credited for developing the failure in terms of a strain energy release rate G [(Eq. (13.56)], which incorporates both elastic and plastic deformations when new surfaces or interfaces are formed.
- 18.
Linear coefficients of thermal expansion are listed for several material types (in units of 10−6/°C): αpolymers ≅ 50, αmetals ≅ 10, aceramics ≅ 2, αglass ≅ 0.5.
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Problems
Problems
-
1.
Two atoms are bonded and the bonding potential can be described by the (9, 1) bonding potential, with an equilibrium bond energy 3.0 eV and equilibrium bonding distance of 2.0\( {\AA} \). Calculate the value of the spring/stiffness constant for small relative displacements of the two atoms.
Answer: 6.75 eV/(\( {\AA} \))2 = 108 N/m
-
2.
The bond energy for two atoms is 2.2 eV and the equilibrium bond distance is 1.9 \( {\AA} \). Assuming that the bond can be described by a (9, 2) bonding potential:
-
(a)
What is the maximum tensile force that the bond can support?
-
(b)
What is the maximum bond extension, from equilibrium bonding distance, before the bond fails?
Answers: a) 1.24 eV/\( {\AA} \) = 1.98 × 10−4 dynes b) 0.36 \( {\AA} \)
-
3.
If the Young’s modulus for a solid material is E = 500 GPa, what is the estimated single-bond energy for two atoms in the solid? Assume that the bonding can be described by the (9, 2) potential with an equilibrium bonding distance for the two atoms of 2 \( {\AA} \).
Answer: 222 GPa(\( {\AA} \))3 = 1.4 eV
-
4.
If the Young’s modulus of a solid material is E = 250 GPa, estimate the elastic energy density in the material when the material is tensile strained by 1 %(ε = ΔL/L0 = 0.01).
Answer: 1.25 × 10−2 GPa = 7.81 × 1019 eV/cm3
-
5.
The stress-strain curve for a material with modulus of E = 400 GPa is very similar to that shown in Fig. 13.14. If the elastic region extends to 1 % strain and the fracture strain is 22 %, then calculate the toughness of this material. Assume that the power-law model, which describes the stress versus strain relation in the plastic region, is given by n = 0.3.
Answer: 1.7 GPa = 1.1 ×1022 eV/cm3
-
6.
Using the vacancy density results from Example Problem 4, show that the flux J of vacancies, described by Eq. (4.10), has an activation energy of Q = (Q)formation + (Q)diffusion.
-
7.
Creep can occur in metals due to dislocation movement along slip planes due to shear stress. If a pure tensile stress σT is applied, as illustrated in Fig. 13.10, then a shearing stress τ is generated: τ = σ sin(θ)cos(θ). Show that the maximum shear stress occurs at θ = 45°.
-
8.
Creep, under constant tensile-stress conditions, can be an important failure mechanism for gas turbines due to high angular speeds and high temperatures during operation. To make sure that the turbine blades can withstand the expected creep, a random selection of turbines was stressed to failure by using an angular speed of 2 × the expected operation conditions and at a temperature of 800 °C versus the expected operating temperature of 600 °C. The turbines started to fail after one week under these accelerated conditions. How long would the turbine blades be expected to last (due to creep) at the expected operational conditions? Assume a creep exponent of at least n = 4, an activation energy of at least 1.2 eV, and all stresses are well above the yield strength of the material.
Answer: 96 years
-
9.
Creep, under constant strain, can be an important failure mechanism for clamps/fasteners. To make sure that a clamp is reliable at 200 °C, accelerated data was taken for clamps tightened to 2X their normal stress level while stored at 300 °C. The clamps lose their effectiveness after one week under these accelerated conditions. Find the time-to-failure for the clamps under the operational conditions at 200 °C. Assume a creep exponent of at least n = 4, an activation energy of at least 1.2 eV and that all stresses are well above the yield point of the material.
Answer: 26 years
-
10.
Time-zero cracks are found on the outside of a stainless steel storage vessel. If the depth of the cracks is 20 mm, determine if fast facture is expected as the vessel is pressurized to a level such that 400 MPa of tensile stress exists in the metal. Assume that the stress concentration factor for the stainless steel is Kcrit = 75(MN/m3/2).
Answer: Yes, since 400 MPa > σrupture = 299 MPa, fast rupture is expected as the vessel is being pressurized.
-
11.
Aluminum-alloy rods were randomly selected and ramped-to-rupture at the intended use temperature with a linear ramp rate of tensile stress of 50 MPa/ day. The breakdown distribution was described by a Weibull distribution with (σrupture)63 = 600 MPa and a Weibull slope of β = 6. Assuming that the tensile stress during normal operation is σop = 100 MPa, a time-to-failure power-law with a stress exponent of n = 6, and that the aluminum-alloy has no well defined yield point:
-
(a)
What fraction of the Al-alloy rods will fail immediately (<0.3 day) when loaded with a tensile stress of 100 MPa?
-
(b)
What fraction of the Al-alloy rods will fail after 10 years with a 100 MPa loading?
Answers: a) 0.0021 % b) 7 %
-
12.
In a certain batch of the aluminum-alloy rods, described in problem 11, some of the rods were found to have small cracks. While the characteristic rupture strength (σrupture)63 = 600 MPa showed little/no change, the Weibull slope β degraded to 4. Assuming that the tensile stress during normal operation is σop = 100 MPa and a time-to-failure power-law with a stress exponent of n = 6 for the aluminum-alloy:
-
(a)
What fraction of the metal rods will fail immediately (<0.3 day) when loaded with a tensile stress of 100 MPa?
-
(b)
What fraction of the metal rods will fail after 10 years with a 100 MPa loading?
Answers: a) 0.08 % b) 16 %
-
13.
Steel rods were selected for a high-temperature and high tensile-stress application. During a ramp-testing determination of the rupture strength of steel, at the intended application temperature of 600 °C and with a ramp rate of 200 MPa/day, the following data was obtained: (σrupture)63 = 1,600 MPa and βrupture = 10. The yield strength of the steel is 600 MPa and the intended application is 700 MPa. Assuming a time-to-failure power-law with a stress exponent of n = 6:
-
(a)
What fraction of the metal rods will fail immediately (<0.07 day) when loaded with a tensile stress of 700 MPa?
-
(b)
What fraction of the metal rods will fail after 10 years with a 700 MPa loading?
Answers: a) 0.03 % b) 0.8 %
-
14.
On a certain batch of steel rods described in Problem 13, small cracks were discovered on some of the rods. While the characteristic Weibull strength (σrupture)63 = 1,600 MPa was virtually unchanged, the Weibull slope degraded to β = 6. The yield strength of the steel is 600 MPa and the intended application is 700 MPa. Assuming a time-to-failure power-law with a stress exponent of n = 6:
-
(a)
What fraction of the metal rods will fail immediately (<0.07 day) when loaded with a tensile stress of 700 MPa?
-
(b)
What fraction of the metal rods will fail after 10 years with a 700 MPa loading?
Answers: a) 0.7 % b) 5.3 %
-
15.
Metal poles, that are intended to support signs, undergo continual cyclical stressing at the base plate due to changing wind conditions. To better understand their reliability, such poles were randomly selected for cyclical stress testing. Under an accelerating cyclical stress of Δσ = 800 MPa, the poles started to crack at the base plate after 5,000 cycles. How many cycles are the poles expected to last at the effective operating condition of Δσ = 200 MPa? Assume m = 4.
Answer: 1.28 × 106 cycles.
-
16.
The poles described in Problem 15 will be used to support an extended sign which puts a mean stress tensile stress of 200 MPa in addition to cyclical stress of Δσ = 200 MPa. Assuming that the tensile strength in these poles is σTS = 800 MPa, calculate the expected number of cycles to failure.
Assume m = 4.
Answer: 4.0 × 105 cycles
-
17.
A metal rod has a thermal expansion coefficient of α = 24 × 10−6/°C and a modulus of E = 70 GPa.
-
(a)
If the metal rod is free to expand from 25 °C to 300 °C, what fractional change in rod length would be expected?
-
(b)
If the rod is fully constrained (cannot move), how much thermal stress would be generated in the rod?
Answers: a) 0.66 % b) 462 MPa
-
18.
A metal component, in a certain application, will be thermal cycled from room temp to an oxidizing ambient of 300 °C. To prevent oxidation of the metal at the high temperatures, a thin ceramic coating is used on the metal component. The concern is that cracks will develop in the ceramic layer during thermal cycling thus exposing the metal to oxidation. To accelerate the cracking, the components were thermal cycled from room temperature to 600 °C. If cracks start to develop in the ceramic layer after 500 thermal cycles, how many crack-free cycles would one expect from room temp to 300 °C? Assume that the ceramic material is hard/brittle with a temperature-cycling power-law exponent of m = 9.
Answer: 382,000 cycles
-
19.
If left unprotected, how much faster will a scratch in the paint of your new car oxidize at 80 vs. 40 % relative humidity. Assume an exponential model with parameter: a = 0.12 %RH.
Answer: 122 times faster
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McPherson, J.W. (2019). Time-to-Failure Models for Selected Failure Mechanisms in Mechanical Engineering. In: Reliability Physics and Engineering. Springer, Cham. https://doi.org/10.1007/978-3-319-93683-3_13
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DOI: https://doi.org/10.1007/978-3-319-93683-3_13
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