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]
