Final Flashcards
Compare tresca and von Moses criterion under equi-biracial loading
Both predict the same yielding
Compare tresca and von Moses under pure shear (s1 = -s2, s3 =0)
Yield stress in von misss yield surface is higher th an yield stress in tresca yield surfsce
The maximum tensile stress in a pressure vessel is its…
Hoop stress
The fracture toughness of a material is found using
Kic = sigma_fYsqrt(pi*a) where a = crack radius = crack length /2
The material that can withstand thelargest internal crack…
Is the one with the highest fracture toughness
In the plot of ln(ln(1/(1-F))) vs ln(sigma)…
Weibull modulus m = slope
Characteristic strength sigma_0 = intercept
From the strength data, estimate the stress for a failure probability F
Use eqn 31, solve for stress sigma and plug in values of m and sigma_0 (from plot) and F (given)
Weibull-What would you expect the stress to be for the same probability of failure F(V) = (2V) where the specimen has twice the volume and were broken in 3 pt bending rather than uniax?
Even though sample is Two times as large, and larger volume = more flaws, consideration of how much of the volume is subjected to high stress and eqn 32 show that3pt bending has higher yield strength
The lifetime of a component can be reliably increased by…
Improving the flaw detection to detect smaller cracks
If you apply a cyclic tensile/compressive stress with an amplitude if σ_max = |σ_min|, where σ_a
Yes
Solid solution strengthening can be either ____ or ____. ______ generally offers higher stiffness.
Substitutions or interstitial. Interstitial offers higher stiffness because less distortion to lattice?
Shear stress to overcome slip-plane obstacles ______ with increasing φ_c
Decreases
Pierles force
Resistance to dislocation movement resulting from the ‘drag’ of compressive and tensile distortion in the lattice around the dislocation
Materials with wide dislocations have ____ Peierl’s forces.
Low (since distortion is spread out over a large volume and is much less intense at its core)
Temperature dependence on peierl’s force:
Wide dislocation = temperature dependence is Low = small temperature dependence of the yield stress e.g. fcc metals
Materials with narrow and intense dislocation fields have _____ high Peierel’s forces and ______ temperature sensitivity of the yield stress.
High peierl’s forces, large temperature sensitivity (because higher T facilitates dislocation mobility, thus reducing the yield stress. Can lead to brittle fracture.
In a screw dislocation…
The dislocation line is || to Burgers vector
Single xtals tend to slip in their…
…most closely-packed planes because:
- distances between planes are maximum so most loosely bound
- slip in cp directions minimizes distance over which the stresses need to displace the slipping atoms
Slip occurs when…
…the stress on the slip plane and in the slip direction reaches the critical resolves shear stress
τ_crss = σcosθcosφ = =σ/m
Structure factor dependent on the orientation of a slip system relative to the applied tensile stress
m - Schmidt factor
Yield stress will generally be ____ in polyxtalline a materials since many of the grains will be oriented unfavorably (have high m)
Higher
Yield
Onset of plastic deformation
The most favorable slip plane/direction has…
Slip plane with highest m has…
High m (lowest shear stress τ_c)
Highest resolved shear stress
If you know τ_c = τ_crss, solve for all m values
Want to find min σ_y because one with highest m (denominator) is activated first
σ_y = τ_c/m
Most materials have…
Low τ_c —> high m —-> yielding
Below this threshold, there will be no yielding. At is the onset of yielding.
Yield stress:
σ_y = τ_c/(cosφcosλ)
Where τ_c = τ_crss
Φ is angle between stress and planenormal directions
λ is angle between stress and specific slip plane
The strain energy surrounding a dislocation in an elastic stress field is equal to…
The area under the area under the elastic region of the stress/strain curve
Energy of dislocations (because perfect lattice is perturbed)
U = Gb^2/2 is the energy used to keep the dislocation straight
Peach Koehler
Force/stress required to bow a dislocation
Shear… to bow a dislocation out to a certain R
τ = Gb/2R
… increases with G, increases with b, decreases with R
Brittle fracture is dominated by
Flaws in the material
Types of time-dependent failure
Creep failure: time dependent plastic deformation leads to failure
Fatigue: cyclic loading accumulates damage
Area under force-atomic plane separation curve =
Work needed to separate the planes of atoms = energy needed to create 2 new surfaces
If equilibrium spacing ~= plane spacing, σ_max (max tensile strength) =
~ 1/3 of E needed to separate the two planes (E/3)
Does material absorb impact energy? If so then ____ leads to failure.
Do cracks propagate instead? Then ____ leads to failure.
Plastic deformation.
Brittle fracture.
Plastic deformation is always correlated with ____ energy
High
Below ductile to brittle transition temp, _____ dominates. Above, _____ dominates.
Below- bond breaking takes less energy than moving dislocations —> brittle failure dominates
Above- moving dislocations takes less energy than bond breaking —> ductile failure dominates
Charpy impact testing measures…
Primary function is to determine…
…Energy absorbed by sample that leads to failure. (Notch defines failure path)
…determine whether a material experiences a ductile-brittle transition with decreasing temperature
Shear stresses must act on and activate _____ for plastic deformation to occur
Specific slip systems
Slip system consists of
(Slip plane)[direction]
{family of planes}
Single xtal yield strength is ___ than polycrystalline yield strength
Lower, since in polyxtal, have to activate more slip systems
Single xtal = 1 slip system
Polyxtal >= 5 slip systems
According to yield surfaces, plastic deformation occurs ____
Outside the yield surfaces, as soon as one is encountered.
[Elastic region enclosed by the surfaces]
Brittle fracture is
Sudden, rapid, unstable
Thin samples
Plane stress; ε_thickness direction is not zero, unrestricted and allowed to deform. σ_thickness direction(s) = 0
I.e. STRESSES CONFINED to crack face PLANE
Thick samples
Plane strain. STRAINS CONFINED TO CRACK FACE PLANE since deformation is restricted in the thickness direction.
σ_t = 0, ε not 0
In the linear elastic region true and Engineering stress
Are equal
In the plastic regime, true stress is…
Greater than engr. Stress since dividing by a smaller area.
OPPOSITE for COMPRESSION
In the plastic regime, true strain is
Less than Engineering strain since dividing by a greater length
OPPOSITE for COMPRESSION
True strain =
ln(1+engr strain)
The 2 linear elastic constants absolutely necessary to describe behavior under uniaxial tension are
E and v (others can be derived from these two; E and v can be obtained from uniax. Tension tests)
0th tank tensor
Scalar i.e. distance, temperature
1st rank tensor
Vector i.e. force, velocity
2nd rank tensor
Relates two vectors i.e. stress (relates force and area normal), strain (relates displacement and position vectors)
Normal Deformation =
e_ij = e_ii = Change in displacement / change in position = δu_i/δx_i
Shear deformation =
e_ij = Change in displacement / change in position = δu_i/δx_j
Deformation tensor composed of …
Derivatives (coefficients) of displacement vectors
Strain tensor
ε_ij = (1/2)[δu_i/δx_j + δu_j/δx_i]
Strain tensor is…
Symmetric
Rotation tensor
ω_ij = (1/2)[δu_i/δx_j - δu_j/δx_i]
Rotation tensor is
Antisymmetric
In mixed strain/rotation deformations, the deformation tensor =
e_ij = ε_ij + ω_ij
For normal deformations, ω_ii =
0
For normal deformations ε_ii =
e_ii
Convention for positive stresses
Both face normal and stress directions are the SAME SIGN
Stress invariants
Any rotation or coordinate system will yield the same stress values
Principal stresses are the…
Three roots of the characteristic equation
Plane stress condition
there is one direction In which all stresses are zero
Principal direction
Shear stresses are zero
In a cubic system, there is no
Coupling between normal and shear stresses. So if no shear stress applied, will not get shear strains out
Voight approximation of polyxtalline material stiffness
UPPER BOUND; parallel springs each feel same ε; yields higher stiffness
E = ΣV_i*E_i
V_i is VOLUME FRACTION of the ith element in a material
Reuss approximation
LOWER BOUND; springs in series feel same σ; lower stiffness
1/E = ΣV_i/E_i
V_i is VOLUME FRACTION of the ith element in a material
Curvature if potential well tells about
Elastic modules. Higher curvature = higher stiffness
Depth of potential well tells about
Melting point. Deeper well=stronger bonds= higher MP= more energy needed to completely separate atoms
Symmetry of well tells about
Coeff. of thermal expansion. Less symmetric = higher α = addition of energy has larger effect on equilibrium spacing
Stiffness ____ with temperature
Decreases since average volume per atom as it oscillates increases
In a unidirectional fiber composite, will be stiffest in the direction ___ to fibers
Parallel.
Matrix and fibers like springs in parallel.
Rule of Mixtures:
Stiffness E = E_mV_m + E_fV_f
Stress σ = Εε
When stress is applied perpendicular to the unidirectional fibers in a composite, the ____ will deform more Han the _____.
Matrix more than fibers; FIBERS ADD NOTHING TO STIFFNESS AGAINST STRESS APPLIED PERPENDICULAR
When the FILM is in TENSION
curves UP
When FILM is under COMPRESSION
Curves DOWN
Biaxial modulus M_s
Analogous to E in uniaxial tension.
Radius of film experiencing residual stresses ______ with stiffer substrate, _____ with thicket substrate, ______ with stress in film, ______ with film thickness
Increases, increases, decreases, decreases
_____:strengthening::______:toughening
_____: both
Metals: already have mechs. If dislocation movement present to dissipate stresses so use strengthening mechs.
Ceramics: don’t have the mechs. for stress dissipation so stress at the crack tip will lead to cat. Failure, hence, toughening mechanisms are used
Polymers: addition if secondary phase both toughens and strengthens
Strengthening mechanisms in metals serve to…
Increase the yield strength by reducing the mobility of dislocations
In SUBSTITUTIONAL solid solution strengthening, ______ impurities introduce tensile lattice strains. ______ impurities introduce compressive lattice strains.
Small impurities —>tensile
Large impurities —> compressive
Adding more solute atoms tends to ______ τ_y
Increase, increasing concentration of solute atoms increases Δτ
Adding solute atoms of a very different size will ______ τ_y
Increase, large difference in size increases the strain introduced by addition of solute atom (A) so Δτ increases
High dislocation density tends to _____ τ_y
Decrease, more dislocations = more defects?
Will see a greater increase in strength by_______ substitution
Interstitial because LARGER STRESS FIELDS AROUND THEM.
Subs. atoms are not actually obstacles, so they don’t impede dislocation movement like interstitials do
Strong obstacle
Φ_c = 0
Δτ_max = Gb/x
Dislocation has to completely wrap itself around obstacle to rejoin itself
Precipitation strengthening involves
Precipitation of stable/meta stable particles out of the matrix
Dispersion strengthens involves
The introduction of (harder) secondary phase particles into the (soft) matrix
Non-deforming particles
Like strong obstacles. Dislocation has to completely bend around particle
Deforming particles
Cases where it is less energetically costly to break up s particle than it is to bow a dislocation around it.
Applying shear stress to the area between the particle’s atoms (2r*x, x = inter atomic spacing)
Energy to cut particle in half relation shows that…
Higher atomic packing density = smaller x = larger Δτ = stronger material
MORE PARTICLES INCREASES STRENGTH
The necessary shear stress to bow a dislocation _____ with particle size
DECREASES
The shear stress necessary to break up a deforming obstacle ____ with particle size
INCREASES
Complete coherency is when
Matrix and particle have aligned lattice & spacing
Incoherency is when
There is a mismatch between matrix and particle lattices/atomic spacing
When complete coherency between particles and matrix, dislocations tend to
CUT THROUGH the particle
When incoherency between particles and matrix, dislocations tend to
BOW AROUND the obstacle
Grain boundaries _____ dislocation movement
Impede
A material can be strengthened by _____ the grain size because ______.
Decreasing, because smaller grains = more grain boundaries= greater impedance to dislocation movement = greater resistance to plastic deformation = higher strength
Cold working _______ yield strength, _______ tensile strength, _______ ductility
Increase, increase, decrease
Increased defect interaction impedes dislocation movement
Toughening mechanisms involve…
Reduction of localized stresses at crack tip since toughness in brittle materials is limited by flaws
A material with larger flaw size will have ____ fracture strength
Lower
A material with more chaotic grain boundaries will have _____ toughness
Higher, requires more stress to fail
Bridging fibers increase toughness in brittle materials by ______ & ______
Sharing the load and reducing local stresses at the crack tip.
Absorbing energy through frictional sliding and debonding of fibers
Decreasing flaw size will _______ toughness
Increase
Bumper and indices of slip system in FCC
12: {111}<1-10>
Number and indices of slip planes in BCC
12: {110} (and others depending on symmetry?)
In uniaxial tension, the plane in which τ is maximized is _____ degrees from acialbdirextion, and the shear stress there = _____
45 deg., (σ_11)/2
Minimum stress to begin yielding occurs when φ = _____ = ______ degrees.
λ, 45
Region II of the stress/strain plot for ductile single crystals corresponds to_____.
Work hardening: multiple slip systems involved, dislocations interact on intersecting slip planes
Region III of the stress/strain plot for ductile single crystals corresponds to_____.
Cross slip: alleviates the hardening process since plastic deformation now in competition with fracture
Region I of the stress/strain plot for ductile single crystals corresponds to_____.
Single slip system, easy glide
Burgers vector is
The displacement produced by a line defect.
The vector to displace to the nearest atom. (Is why expect slip systems to have smallest shear strains; since smallest displacement needed -close packed planes, though not necessarily true in ionic solids pike rock salt)
In an edge dislocation, b is _____ to the defect line
Perpendicular, otherwise there wouldn’t be dislocation movement
The edge line (extra/missing plane of atoms) moves _____ of the resolved shear stress
In the direction of
Screw dislocation, b is ______ to defect line, and the dislocation line moves ______ to the resolved shear stress
…Parallel…
…Normal to…
Transformation toughening
At a crack, stresses can induce a phase transformation to a LARGER COLUME PHASE
Results in COMPRESSIVE STRESSES locally at the crack tip which reduces overall tensile stress - This is good since ceramics are stronger under compression
Also, some of the input energy went into the transformation rather than into crack propagation
Microcracking
Stress-induced microcracks at the secondary phase help to relieve stresses locally and impede crack advance (long crack=low strength)
Want the micro cracks to form at the secondary phase DURING loading in order for the E tp be absorbed
Compressive surface strengthening
Aluminosilicate glass with Na ions
Ion exchange process - larger K ions diffuse in thus introducing compressive stresses at the surface
Same overall fracture strength, but the amount of stress it can take is higher since compressive (-) σ_surf
Which test yields lower tensile strength? Uni- or multi-axial tension?
Multi will be less since higher probability of probing a flaw in a critical orientation
What is the pressure given a principal stress state?
P = -I/3 = -(σ1+σ2+σ3)/3
You apply a normal strain in the [001] direction of a Pt single xtal (cubic). No other applied stress. What is the strain component in both Vought and Einstein notation?
ε3 and ε33
You apply a normal stress on the (001) face of the Pt single xtal in the [010] direction. No other applied stress. What is the strain component in both Vought and Einstein notation?
σ32 = σ4
For a rheological model subject to constant stress σ0 from t=0 to t =t1 where stress is removed, what is this time-dependent response called?
Applying a stress and observing strain response is CREEP
Like stress on airplane wing and observing diffusional atom motion
For a Kelvin-voight model subject to constant stress σ0 from t=0 to t =t1 where stress is removed, what is the initial strain at t = 0+?
0 since dashpot cannot instantaneously respond
For a Kelvin-voight model subject to constant stress σ0 from t=0 to t =t1 where stress is removed, what is the strain in the system just before t is removed?
Looks like spring in parallel with broken dashpot so ε = σ/Ε
For a Kelvin-voight model subject to constant stress σ0 from t=0 to t =t1 where stress is removed, does the strain change immediately after t1?
No, because dashpot cannot instantaneously respond
For a Kelvin-voight model subject to constant stress σ0 from t=0 to t =t1 where stress is removed, what does ε vs t plot look like?
ε(0+) = 0, concave down positive slope curve up to ε(t1) = ε_max= σ0/Ε, then concave up negative slope curving back down to 0 where t2 approaches infinity
What are the equilibrium equations needed to derive a constitutive equation?
Equilibrium - deals with stresses only
Compatibility - deals with strains only
Slow loading is equivalent to _____ stiffness
HIGH
Time-temperature equivalence:
Compliance_____, and stiffness ____ with time
Increases/decreases
Time-temperature equivalence:
Compliance_____, and stiffness ____ with temperature
Increases/decreases
Standard linear solid rheological model
Spring in parallel with a spring+dashpot in series
σ = σΑ + σΒ = σΑ+σC ; σC = σB ηε_dotC = Eb*εb
ε = εΑ=εΒ + εC
For a Kelvin-voight model subject to constant strain (uniaxial displacement) ε0 from where stress is removed at t=0+, what is this time-dependent response called?
Stress relaxation: applying a strain and observing how material relaxes
The more realistic model for thermal stresses in a material predicts a ______ final temperature allowed before failure
Colder
For failure in thermal shock, set the max surface stress equal to the _______ stress
Failure stress, because highest tensile stresses occur at surface so will fail there
In _____ strength steels (BCC), the transition temperature is sensitive to _____ and ______
Composition and microstructure
____ strength FCC (and some Au and Cu alloys) ______ experience a DBTT
Low strength, Do NOT
Brittle high strength materials are also ______ _______ to changes in temperature
Relatively INSENSITIVE
There is no ______ associated with fatigue failure
Gross plastic deformation
Fatigue failure is ______ in nature
Brittle, even for ductile materials
Fatigue involves the _____ and _____ of cracks under an applied ______ stress
Nucleation, growth, tensile
Three stages of fatigue:
- Crack initiation
- Crack propagation
- Final failure
