Part A Flashcards
(256 cards)
1
Q
What is meant by stress equilibrium in statics?
A
The condition where there are no net forces or moments acting on a body.
2
Q
What is strain continuity?
A
The assumption that no damage is introduced and deformation is continuous.
3
Q
What assumptions are made about metals in statics?
A
That they are isotropic and homogeneous.
4
Q
What is tensile stress and how is it defined?
A
Stress = Force / Area.
5
Q
What is tensile strain and how is it defined?
A
Strain = Change in length / Original length.
6
Q
What is the principle of superposition in elasticity?
A
Stresses applied in any order yield the same result — valid only in elastic regimes.
7
Q
Why is the superposition principle not valid in plasticity?
A
Because plastic deformation involves permanent changes and is not linear.
8
Q
What microscopic mechanism underlies plasticity in metals?
A
Dislocations.
9
Q
Why is stress analysis important for engineering applications?
A
To ensure structural integrity and prevent failure.
10
Q
What is typically used as the ‘currency’ in stress-based design?
A
The applied stress.
11
Q
How do design codes help in material selection?
A
By specifying allowable stress levels for safe component performance.
12
Q
What are the key mechanical properties from tensile tests?
A
Yield stress (σ_y) and tensile strength (σ_TS).
13
Q
What are nominal stress and strain?
A
Values based on original dimensions, not accounting for true deformation.
14
Q
How is true stress related to nominal stress?
A
True stress accounts for changing cross-sectional area during deformation.
15
Q
What is the work hardening rate?
A
dσ/dε, described using empirical relations of true stress and strain.
16
Q
What is the most attractive parameter for design from tensile tests?
A
The yield stress.
17
Q
What is a tensor in mechanics?
A
An object describing multilinear relationships, generalizing scalars, vectors, and matrices.
18
Q
What is the rank of stress and strain tensors?
A
They are second-rank tensors.
19
Q
What is the stress tensor in 3D?
A
A 3x3 matrix with 9 components, reducible to 3 principal stresses by transformation.
20
Q
What is the purpose of Mohr’s circle?
A
To graphically represent 2D stress states and principal stress directions.
21
Q
What do eigenvalues represent in the stress tensor?
A
Principal stresses (σ₁, σ₂, σ₃).
22
Q
How can principal stresses be determined mathematically?
A
By solving the characteristic equation (quadratic or cubic) from the stress tensor.
23
Q
How do principal stresses relate to strains in isotropic materials?
A
Principal stresses produce principal strains in the same directions.
24
Q
What elastic constants define the stress-strain relationship in isotropic materials?
A
Young’s modulus (E) and Poisson’s ratio (ν).
25
What components define the stress tensor σ_ij?
Nine components: σ_11 to σ_33, transformable to principal stresses.
26
How is maximum shear stress calculated from principal stresses?
(σ₁ − σ₃) / 2.
27
What controls brittle fracture?
The maximum tensile stress, usually σ₁.
28
What defines brittle fracture in engineering?
Failure occurs at a stress below the design stress, often due to crack-like defects.
29
Why is conventional design based on preventing yielding not sufficient for brittle materials?
Because brittle materials can fracture without significant plastic deformation, especially in presence of defects.
30
What design approach was developed to account for brittle fracture?
Fracture mechanics.
31
At what stress can brittle fracture occur relative to yield stress?
Before reaching 2/3 of the yield stress (σ_y).
32
What mechanical properties are extracted from a uniaxial tensile test?
Yield stress (σ_y) and tensile strength (σ_TS).
33
What are design codes typically formulated around?
Stresses, particularly principal stresses.
34
Which principal stress controls brittle fracture?
Maximum tensile stress, typically σ₁.
35
What failure modes are possible for a beam in bending?
Overload failure and brittle fracture.
36
What defines a desirable failure mode?
General yielding, rather than brittle fracture.
37
What is Kirch’s solution used for?
Describing stress fields around circular holes in an infinite plate under tension.
38
What is the stress concentration factor (Kt) at a circular hole?
Kt = 3 under uniaxial tension.
39
What causes high stresses near circular holes?
Force flow divergence around holes leads to stress concentration.
40
Where is the maximum hoop stress located at a circular hole?
At θ = ±90°, with σ_θθ = 3σ_app.
41
What does the principle of superposition allow in elasticity?
Combining different loading conditions (uniaxial, shear, equibiaxial) to analyze complex stress states.
42
What happens to stress concentration factor under equibiaxial tension?
Kt = 2.
43
What is Kt under pure shear?
Kt = 4.
44
How is Kt affected by plate width?
Finite plate width reduces symmetry and changes stress field — Kt redefined using nominal stress.
45
Why is high mesh resolution needed near holes in FEA?
Due to steep stress gradients requiring fine elements for accuracy.
46
What is Inglis' contribution to fracture mechanics?
Provided the stress field solution for elliptical holes in infinite plates.
47
How does σ_max change as ellipse becomes more crack-like?
It increases sharply; for sharp cracks (ρ → 0), σ_max → ∞.
48
What is the formula for max stress near an elliptical hole?
σ_max = σ_app (1 + 2a/b), where a = half crack length, b = tip radius.
49
How is curvature radius (ρ) used in stress analysis of cracks?
σ_max = σ_app (1 + 2√(a/ρ)).
50
Why is plasticity important in real materials?
It limits stress elevation near crack tips, contrary to infinite elastic prediction.
51
What did McMeeking's FEM study show?
Peak local tensile stress can reach ~3.8–4.5×σ_y depending on work hardening exponent.
52
What is photoelasticity used for?
To visualize stress fields using optical stress patterns.
53
Why are fatigue cracks so dangerous?
They can become very sharp (tip radii < 0.1 mm), concentrating stress.
54
What does the Charpy impact test measure?
The energy absorbed during high strain rate fracture of a notched specimen.
55
What are the three conditions the Charpy test promotes to trigger brittle fracture?
Low temperature, high strain rate, and a triaxial stress state from a notch.
56
How is the Charpy test energy calculated?
E = m g (h1 - h2), the difference in potential energy before and after fracture.
57
What specimen geometry is used in Charpy testing?
55 mm × 10 mm × 10 mm bar with a 2 mm deep, 45° V-notch with 0.25 mm root radius.
58
What is the typical impact velocity in Charpy testing?
About 5 m/s, corresponding to a strain rate of ~10³ s⁻¹.
59
What is the ductile-to-brittle transition curve?
A sigmoidal curve showing how impact energy varies with temperature.
60
What are the upper and lower shelves in the DBTT curve?
Upper shelf = ductile behavior; lower shelf = brittle behavior.
61
What is Fracture Appearance Transition Temperature (FATT)?
The temperature at which fracture surface is 50% fibrous and 50% cleavage.
62
What is the Nil Ductility Temperature (NDT)?
Temperature at which fracture is 100% cleavage.
63
What is the Fracture Transition Plastic temperature (FTP)?
Temperature at which fracture is 100% fibrous.
64
What causes brittle fractures in Liberty ships during WWII?
DBTT exceeded at cold temperatures; cracks initiated at stress concentrators and propagated rapidly.
65
What factors influence the DBTT of ferritic steels?
Grain size, carbon content, and microstructure.
66
What materials do not experience a DBTT?
Low-strength FCC metals (e.g., Al, Cu alloys), most HCP metals.
67
What happens to impact energy in high-strength alloys as T decreases?
Relatively low and constant, reflecting inherent brittleness.
68
How can fracture initiation sites be identified?
Via SEM and energy dispersive spectroscopy (EDS), often on non-metallic inclusions.
69
What features characterize cleavage fracture?
Cleavage steps, river lines, and facets aligned with {001} planes.
70
What is the role of grain size in cleavage?
Facet size is linked to ferrite grain size.
71
What is the classical ductile fracture mechanism?
Void initiation, growth, and coalescence around inclusions.
72
What increases toughness in the microvoid coalescence mechanism?
Energy absorbed during void growth, linked to work hardening and second-phase particle size.
73
What differences are observed in steel tested at room vs. cryogenic temperature?
More cleavage fracture at low temperature; more ductile fracture at room temperature.
74
What surface features are measured on fractured Charpy specimens?
Cleavage area %, lateral expansion, shear lip size, and thumbnail (fibrous) area.
75
Why are Charpy impact results considered qualitative?
They show relative toughness but cannot predict design stresses or crack size.
76
What are the limitations of correlating Charpy energy to fracture toughness?
Scatter in data and test differences limit reliable quantitative correlation.
77
What is the impact of irradiation on Charpy curves?
Reduces upper shelf toughness and raises DBTT.
78
Why is CVN still widely used in industry?
It's simple, cost-effective, and historically linked to many design codes.
79
What does 41 J at −40°C represent in design?
A common design criterion in oil and gas industry.
80
What challenges arise when CVN design codes are violated?
Further analysis or testing may be required to justify continued service.
81
When should the fracture mechanics framework be used?
When a sharp crack may be present or is assumed to be present in a component.
82
What property of fatigue cracks justifies using fracture mechanics?
Fatigue cracks are very sharp, with notch tip radii as small as 0.075 mm.
83
In comparing cracks under stress, which factor most strongly influences proximity to failure?
Both crack size and applied stress determine proximity to failure.
84
What does the linear elastic stress intensity factor (K) represent?
A measure of the intensity of the stress field near the crack tip.
85
What are the standard K equations for common geometries?
Through-thickness edge crack: K = 1.12 σ √(πa); Semi-circular surface crack: K = 0.6 σ √(πa).
86
What is fracture toughness?
The critical value of K at which fracture occurs, determined experimentally.
87
What is the role of NDE in fracture mechanics?
To evaluate potential or actual defect sizes in a component.
88
Which crack loading mode is most damaging?
Mode I (opening mode) is most critical and gives the lowest K at failure.
89
What parameter controls fatigue crack growth rate?
The range of stress intensity factor, ΔK.
90
What is ΔK?
ΔK = K_max − K_min, or ΔK = factor × Δσ × √(πa) for a specific crack geometry.
91
What is the Paris relationship for crack growth?
da/dN = A (ΔK)^m, where A and m are material/environment dependent constants.
92
What is the log form of the Paris law?
log(da/dN) = log(A) + m log(ΔK).
93
Why is the log form of the Paris law useful?
It gives a linear relationship for easier experimental fitting.
94
How can the Paris law be used to predict fatigue life?
By integrating da/dN from initial crack size a₁ to critical size a₂.
95
Why must ΔK be substituted before integration?
Because it depends on both a and Δσ.
96
What is the result of integrating the Paris law over a crack range?
An equation predicting number of cycles N required for crack growth.
97
What is A₁ in the integrated Paris law?
A₁ = A (1.12)^m Δσ^m π^(m/2), assuming constant Δσ.
98
How does fatigue crack growth vary with crack length?
It accelerates with increasing crack length due to rising ΔK.
99
Why is the Paris relationship widely used?
It enables predictive fatigue life assessment and defect tolerance design.
100
What industries rely on the Paris law for fracture mechanics?
Aerospace (airframe, aero-engines) and others requiring damage tolerance certification.
101
What are the primary material types used in fibre reinforced composites?
Polymer matrix composites (PMCs), metal matrix composites (MMCs), and ceramic matrix composites (CMCs).
102
Why are fibre composites considered anisotropic?
Their properties vary significantly with direction due to fibre alignment.
103
What motivates the use of fibre composites in engineering?
Desire for increased stiffness, strength, toughness, and reduced weight for substitution of metals.
104
What are common applications of fibre reinforced composites?
Military, aerospace, sport, transportation, construction, and energy sectors.
105
What is the main challenge of using composites in structural applications?
Anisotropy, inhomogeneity, and variability in mechanical properties.
106
What kinds of fibres are typically used in composites?
Carbon, boron, SiC, Si3N4, Al2O3, Kevlar, and drawn polypropylene.
107
What defines a good fibre for reinforcement?
High modulus, strength, and E/ρ ratio; often high covalent/ionic bonding.
108
What is the function of the matrix in composites?
Bind fibres, transfer load, isolate and protect fibres.
109
What determines the maximum fibre volume fraction (Vf)?
Packing geometry and infiltration limits; typically capped around 0.5.
110
Why might a weak interfacial bond between fibre and matrix be desirable?
It promotes energy dissipation via debonding and fibre pull-out, improving toughness.
111
What is the rule of mixtures for Young’s modulus in the fibre direction?
E_c = E_f * V_f + E_m * V_m
112
What assumption is made for parallel fibre loading?
Equal strain in fibre and matrix.
113
What is the expression for modulus transverse to fibres?
1/E_c = V_f/E_f + V_m/E_m
114
Why is the transverse modulus much lower?
Due to poor load transfer across fibres and weak matrix properties.
115
What controls strength parallel to fibres?
Fibre strength and interfacial bond quality.
116
What is the usable strength parallel to fibres?
0.7 σf*Vf < σc* < ~1.2 σf*Vf
117
What limits transverse tensile strength?
Matrix properties, defects, residual stress, and interfacial bonding.
118
What is the tensile strength anisotropy in carbon-epoxy systems?
Parallel ~1000 MPa, perpendicular ~40 MPa.
119
How can composite performance be improved for multidirectional loading?
Use laminate layups like [0/90], balancing directionality.
120
What is the Weibull distribution used for?
Statistical modeling of fibre strength variability.
121
What influences the Weibull modulus?
Material variability — low for E-glass, higher for ceramics.
122
What mechanisms contribute to toughness in brittle matrix composites?
Fibre debonding, frictional sliding, and fibre pull-out.
123
What is the ideal sequence of events for toughening?
1) Fibre debonding, 2) Fibre fracture near debonded region, 3) Fibre pull-out.
124
Why does weak fibre-matrix bonding increase toughness?
Allows energy-absorbing mechanisms like debonding and pull-out to activate.
125
What design features increase composite toughness?
High Vf, strong fibres, and longer fibre pull-out lengths.
126
What approach is used to formulate fatigue design codes?
Traditional approach based on total life using S-N curves.
127
What does an S-N curve represent?
Stress range (Δσ) vs. number of cycles to failure.
128
Why are S-N curves limited?
They don't differentiate between crack initiation and propagation.
129
How many tests are typically used to define an S-N curve?
Twelve tests under uniform cyclic loading to failure.
130
What is the stress range (Δσ) in a fatigue cycle?
Δσ = σ_max − σ_min.
131
What is the stress amplitude (σ_a)?
σ_a = Δσ / 2.
132
What is the mean stress (σ_m)?
σ_m = (σ_max + σ_min) / 2.
133
What is the stress ratio (R)?
R = σ_min / σ_max.
134
What is the amplitude ratio (A)?
A = σ_a / σ_m = (1 − R) / (1 + R).
135
What is the stress ratio for fully reversed loading?
R = −1.
136
What does fully reversed loading imply for σ_m?
Mean stress is zero.
137
What stress ratio is commonly used in aerospace fatigue testing?
R = 0.1.
138
What does R = 0.1 represent?
A tension-tension cycle where σ_min = 0.1σ_max.
139
How does mean stress affect fatigue life?
Increasing mean stress decreases fatigue life for the same stress range.
140
What diagrams help visualize mean stress effects on fatigue?
S-N diagrams and σ_a vs. σ_m interaction curves.
141
What are common fatigue failure criteria?
Modified Goodman, Gerber, Soderberg, and ASME elliptical.
142
What is Miner’s Rule used for?
Predicting fatigue life under variable amplitude loading.
143
What is the Miner’s Rule equation?
Σ (n_i / N_i) = 1, where n_i is applied cycles and N_i is cycles to failure at stress amplitude i.
144
What is the Elastic Stress Concentration Factor (Kt)?
Kt = σ_max / σ_ref, for non-crack stress raisers like holes or notches.
145
How do notches affect S-N curves?
They shift the curve downward in stress range.
146
What reduces the effective Kt under fatigue loading?
Plasticity and reduced sampled volume near the notch.
147
How are design curves adapted for notched components?
By scaling S-N curves using Kt and drawing curves for defined failure probabilities.
148
What is the metallurgical focus for fatigue-resistant design?
Preventing crack initiation by increasing yield strength and surface strengthening.
149
What techniques are used to enhance fatigue resistance?
Surface hardening, shot peening, using hard outer layers with tough cores.
150
What factors complicate fatigue design?
Surface roughness, corrosion, notch effects, stress relief, mean stress, and variability.
151
Why is fatigue a critical issue in engineering?
Because ~90% of structural failures are due to fatigue and often uninsured.
152
When does creep become significant in materials?
At temperatures above ~0.4 Tm (melting temperature).
153
What is the modern approach to creep design?
A strain-based approach: ε = f(σ, t, T).
154
Where is creep particularly critical?
Chemical plants, nuclear reactors, steam turbines, gas turbines.
155
What is a stress rupture test?
A constant load test at fixed temperature until failure, used to assess creep life.
156
How are stress rupture curves similar to fatigue S-N curves?
They relate applied stress to time-to-failure.
157
Why are stress rupture tests limited?
Require extensive time-temperature-stress data; empirical and component-specific.
158
What is the Larson-Miller parameter used for?
Correlating time-temperature data for creep assessment.
159
What are the three main stages of the creep curve?
(i) Instantaneous strain, (ii) Primary creep (decreasing strain rate), (iii) Secondary creep (steady-state).
160
What causes primary creep?
Work hardening due to dislocation interactions.
161
What characterizes secondary creep?
Constant strain rate as work hardening is balanced by recovery.
162
What is tertiary creep?
Accelerated strain due to internal damage such as voids and grain boundary cracks.
163
Why is minimum creep rate important?
It is used in design to define steady-state creep and can be correlated with microstructural mechanisms.
164
What microstructural feature helps explain secondary creep?
Dislocation-dislocation interactions studied using TEM.
165
What are the main mechanisms of creep?
Dislocation creep (power-law), diffusion creep, and grain boundary sliding/cavitation.
166
What is power-law creep modeled by?
dε/dt = Bσⁿ exp(−Q/kT), where n ~ 3–8, and Q is the activation energy.
167
What is diffusion creep?
Creep controlled by stress-driven atomic diffusion, active at low stress and high temperature.
168
What is Herring-Nabarro creep?
Lattice diffusion creep: dε/dt = σDv / (Td²).
169
What is Coble creep?
Grain boundary diffusion creep: dε/dt = σD_gb / (Td³).
170
How does grain size affect diffusion creep?
Larger grains reduce creep rate due to inverse grain size dependence.
171
How are alloys designed to resist creep?
Raise Tm, add solid solution elements, use stable fine precipitates, and limit grain boundary effects.
172
Why are single crystals used in Ni-based superalloys?
To eliminate grain boundaries and improve creep resistance.
173
Why is extrapolating creep data risky?
Mechanism changes, microstructural evolution, and limited test data at long times.
174
What graphical methods are used for extrapolation?
Log-log plots of minimum creep rate or rupture time; Larson-Miller-type parameters.
175
What is a conservative design limit in creep?
Defining acceptable life as reaching 1% creep strain.
176
Why doesn't the superposition principle apply in plasticity?
Because plastic deformation is path-dependent and non-linear.
177
How many independent stress components are there in 3D stress?
Six independent components from the symmetric 3×3 stress tensor.
178
What does the first subscript in σ_ij refer to?
The plane normal to which the stress is acting.
179
What do the principal stresses represent?
The maximum and minimum normal stresses with no shear components on their planes.
180
How are principal stresses labeled?
σ1 is the maximum tensile stress; σ3 is the most compressive.
181
What are the elastic constants used in 3D stress-strain relationships?
Young’s modulus (E) and Poisson’s ratio (ν).
182
What strains occur under uniaxial tensile loading?
One tensile strain and two lateral compressive strains.
183
What is ν in elastic and plastic regions for metals?
ν ≈ 0.3 in elastic region, ν ≈ 0.5 in plastic region.
184
What simplifications occur in plane stress?
σ3 = 0, stress in the thickness direction is negligible.
185
What simplifications occur in plane strain?
ε2 = 0, strain in the out-of-plane direction is negligible.
186
How can σ1 and σ2 be derived from strain gauge readings?
By using superposition: E(ε1 + νε2) = σ1(1 − ν²), etc.
187
What effect does a notch introduce under uniaxial loading?
A triaxial stress state due to required strain continuity.
188
Why does plane strain promote brittle fracture?
Due to increased constraint and triaxiality inhibiting plastic deformation.
189
What stress governs brittle fracture?
The maximum principal tensile stress (σ1).
190
What stress governs yielding?
Maximum shear stress, derived from principal stress differences.
191
What are the key stresses in thin-walled pressure vessels?
Hoop stress σ_hoop = Pr/t and longitudinal stress σ_longitudinal = Pr/2t.
192
Which stress dominates in thin-walled pressure vessels?
Hoop stress (σ1) is the largest.
193
What does 'leak-before-break' imply?
That the critical crack size exceeds the wall thickness (a_critical > t).
194
What does von Mises' yield criterion state?
(σ1−σ2)² + (σ2−σ3)² + (σ3−σ1)² = 2σ_y².
195
What does the Tresca yield criterion state?
σ1 − σ3 = σ_y.
196
What role do shear stresses play in yield criteria?
Yielding is governed by critical shear stresses, not tensile stress alone.
197
Which criterion includes all three principal stresses?
Von Mises.
198
Which yield criterion ignores σ2?
Tresca.
199
Why is a statistical distribution needed for brittle materials?
Because fracture strength depends on flaw size, shape, orientation, and location, which are randomly distributed.
200
What distribution is commonly used for fracture strength of brittle materials?
The Weibull distribution.
201
What is the key assumption in using statistical methods for brittle fracture?
That the survival of a component depends on the survival of all its elements.
202
Why does the probability of failure increase with volume?
Because a larger volume increases the chance of encountering a critical flaw.
203
How is survival probability PS related to risk of rupture R?
PS = exp(−R), where R is the cumulative risk of failure over the material volume.
204
What is the Weibull risk function g(σ) used for?
To describe how fracture risk increases with applied stress.
205
What form did Weibull propose for g(σ)?
g(σ) = ((σ − σ_L)/σ_o)^m, for σ ≥ σ_L.
206
What are the parameters in the Weibull function?
σ_L (threshold stress), σ_o (scale parameter), and m (Weibull modulus).
207
What does the Weibull modulus m represent?
It characterizes the variability of fracture strength — higher m means lower variability.
208
What is the role of σ_L in Weibull analysis?
It is the limiting stress below which fracture probability is zero.
209
What is the 2-parameter Weibull function?
PS = exp[−(V/Vo)((σ/σ_o)^m)], assuming σ_L = 0.
210
When is the 3-parameter Weibull function used?
When σ_L ≠ 0 and the stress distribution is non-uniform.
211
What happens when σ_L = 0 is assumed?
The distribution becomes more conservative for design purposes.
212
Why is the Weibull function applied to area as well as volume?
In thin materials or surface-loading cases, flaws scale with area rather than volume.
213
What increases reliability according to Weibull theory?
A higher Weibull modulus (m) and lower stressed volume.
214
Why is proof testing important in Weibull design?
It can eliminate weak samples and reduce apparent flaw population.
215
What is Vo in Weibull analysis?
An arbitrary reference volume used for normalization, such as 1 mm³, 1 cm³, or 1 m³.
216
How does changing Vo affect σo?
It changes the numerical value of σo, so Vo must remain consistent throughout an analysis.
217
How is σo interpreted physically?
As the characteristic strength at which 63.2% of unit-volume specimens would fail.
218
What happens to PS and PF when σ = σo and V = Vo?
PS = 1/e ≈ 0.37 and PF = 1 − PS ≈ 0.63.
219
Why is the Weibull function linearized?
To allow estimation of the Weibull modulus m via linear regression.
220
What is the linearized form of the Weibull equation?
lnln(1/PS) = m ln(σ) + constant.
221
What does a plot of lnln(1/PS) vs ln(σ) yield?
A straight line with slope m and y-intercept related to σo and Vo.
222
What is commonly plotted instead of PS?
−lnln(PS) to match the negative slope and data curvature visually.
223
How is PS estimated from ranked data?
Using statistical estimators based only on rank j and total n, not on σ values.
224
Why can't PS be estimated directly from stress values?
Because stresses vary across labs and only rank order is statistically meaningful.
225
What are common estimators for PS?
PS = 1 − j/(n+1), PS = 1 − (j−0.3)/(n+0.4), PS = 1 − (j−0.5)/n, and PS = 1 − (j−3/8)/(n+1/4).
226
What is the most common method for estimating the Weibull modulus m?
Least-squares linear regression on the linearized Weibull equation.
227
Why is a sample size of n > 20 recommended?
To capture enough variability in fracture stresses for a reliable estimate of m.
228
What does the 'weakest link' theory state?
Larger volumes increase the chance of a critical flaw and thus reduce fracture strength.
229
What is the equation for the mean fracture stress σM?
σM = σL + σo * (V⁻¹/m) * Γ(1 + 1/m) * f(m), where Γ is the gamma function.
230
How does the mean fracture strength scale between two volumes?
σ1/σ2 = (V2/V1)^(1/m).
231
How does m affect sensitivity to volume change?
As m increases, the strength becomes less sensitive to volume changes.
232
What is the effect of increasing V1 = 100V2 for m = 10?
σ1 = 0.63, indicating a 37% decrease in strength compared to σ2 = 1.
233
Why does Weibull theory predict decreasing strength with increasing volume?
Because it accounts for the statistical likelihood of encountering larger flaws.
234
What is sub-critical crack growth?
Slow crack growth under constant or variable loading at stresses below critical fracture stress.
235
What causes sub-critical crack growth in ceramics?
Stress-enhanced thermally activated or chemically driven mechanisms at crack tips.
236
What environments promote sub-critical crack growth?
Presence of hydroxyl ions (OH⁻), water for oxide ceramics, elevated temperatures for non-oxides.
237
What is the relationship between crack velocity and stress intensity factor?
v = AK₁ⁿ, where A and n are material/environment dependent constants.
238
What are the three regions of crack growth behavior?
Region 1: v ~ AK₁ⁿ; Region 2: v constant (diffusion-limited); Region 3: v increases rapidly as K₁ → K₁c.
239
What is K₁c?
Critical stress intensity factor for fast fracture (fracture toughness).
240
Why is the double-torsion test useful for crack velocity studies?
It yields v-K₁ data independent of crack length or geometry.
241
What controls Region 1 crack growth?
Stress-enhanced adsorption or reaction rates of corrosive species.
242
What controls Region 2?
Diffusion rate of species to the crack tip.
243
What characterizes Region 3?
v and K strongly interdependent, high rate of crack propagation.
244
What equation models lifetime based on sub-critical growth?
t_f ≈ (2K₁j^(2−n)) / (AD²π), for Region 1 dominant behavior.
245
What affects time to failure t_f the most?
Sub-critical crack growth exponent n and applied stress σ.
246
How does increasing n affect time to failure?
Higher n → longer lifetime.
247
What is an SPT diagram?
A plot relating survival probability to stress and projected lifetime.
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How is σ1 (1-second equivalent strength) calculated?
σ₁ = σ_F (t_c)^1/n.
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What axis scaling is used in SPT diagrams?
log(σ₁) vs. −logln(1/PS) to create parallel lines per decade of life.
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What data are required to construct an SPT diagram?
n-value and failure times under constant or converted stress.
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How can n be determined using strain-rate testing?
By testing at different strain rates and comparing fracture strengths.
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Why might strain-rate testing be preferred over v-K₁ methods?
It avoids assumptions about similarity between artificial and natural flaw propagation.
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What is the equation relating σA, σB, and n+1?
(σA/σB)^(n+1) = ε̇A/ε̇B.
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What is the crack growth rate in terms of K₁ and time?
dK₁/dt = (A/2)(D²π)K₁ⁿ⁻¹.
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What constant relates stress to time-to-failure?
t_f = C / σⁿ, where C = Bσ₀ⁿ⁻².
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How does lifetime change with a change in σ by factor x?
σ₁/σ_F = (t_c)^1/n; or log shift in σ₁ is x/n for x decades in time.
