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GP3 > Structural Failure of Materials - Failure > Flashcards

Structural Failure of Materials - Failure Flashcards

1
Q

Derive an expression for the theoretical tensile strength of a material using YM.

A

Check slide 2

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2
Q

Derive an expression for the theoretical tensile strength of a retrial in terms of surface energy when fracture occurs.

A

Check slide 3

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3
Q

Why do the experimental tensile strengths of materials not follow the theoretical.

A

Fracture does not occur by uniform separation of atoms in a plane.
It happens by:
Shear: plastic deformation - ductile failure
Or
Fracture initiates locally at a defect and spreads across the cross section - brittle failure

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4
Q

Which two methods can be used for understanding crack tips?

A

Griffith energy balance criterion
Linear elastic fracture mechanics (based on stress)
Both are equivalent

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5
Q

Derive the Griffith criterion for fixed grip load.

A

Check slide 6

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6
Q

Define toughness.

A

The quality of a material to resist crack propagation

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7
Q

Why is σ(f) less than theoretical strength?

A

Cracks concentrate stress

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8
Q

Why does σ(f) increase as c decreases?

A

Less stress concentrated at crack tip

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9
Q

Why does the strength of a material vary from specimens to specimens?

A

Different c in each material and occurs and fracture occurs at the single largest flaw.

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10
Q

Why doers σ(f) increase as E and γ increase?

A

Stronger bonding.

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11
Q

Show that Griffith equation applies to fracture stresses under all loading conditions.

A

Check slide 8

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12
Q

What is the mechanical energy releases rate, G?

A

Energy released per unit crack area

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13
Q

Derive the mechanical energy release rate, G.

A

Check slide 8

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14
Q

Why are the surface energy values, γ, deduced from fracture always lower than the true value od surface energy?

A

Crack surface not flat and crack surrounded by micro cracks that increase the surface area without knowing.
Other mechanisms to absorb energy, eg plastic deformation around the crack tip.
Polymer exhibit crazing which involves drawing molecules from the surface during fracture.
Grain boundary segregation can lower surface energy deduced
Corrosive environments can also lower surface energy deduced.

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15
Q

What is G(c)?

A

The total energy (new surface + plastic work + work of pullout etc) required to extend a crack through a unit area.

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16
Q

What assumptions are made in defining G(c).

A

The extra energy of dissipating mechanisms act like a surface energy ie a single value material property not affected by geometry of crack or loading system.
The dissipation at the crack tip does not affect the energy release when the crack propagates eg, the plastic zone must be small compared to the size of the crack.

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17
Q

How is G(c) related to true surface energy?

A

for ideal brittle fracture:

G(c) = 2γ(s)

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18
Q

What names does G(c) go by?

A

Critical strain release rate (don’t use)
Critical mechanical energy release rate
Toughness
Crack resistance energy, R

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19
Q

What condition must be satisfied for fracture involving G(c)?

A

G ≥ G(c)

Mechanical energy driving the crack forward is greater than the critical mechanical energy for fracture.

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20
Q

Are G and G(c) material properties?

A

G(c) is, G is just the mechanical energy driving a crack forward under certain conditions.

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21
Q

When is LEFM valid?

A

For near field solutions (close to the crack tip) but not very close to the crack tip as stress would tend ∞

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22
Q

What general assumption is made about the type of crack propagation for LEFM?

A

Mode ! fracture is most important and dominates because cracks tend to propagate on planes with the maximum principle stress on it.

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23
Q

State the general solution for crack propagation in LEFM.

A

Check slide 11

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24
Q

What is K in LEFM?

A

The stress intensity factor

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25
Q

What are the units of K?

A

MNm^-3/2
or
MPam^1/2

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26
Q

What are the units of G?

A

Jm^-2
or
Nm^-1

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27
Q

What is K a measure of?

A

Stress at the crack tip.

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28
Q

What is K(Ic)?

A

The critical; stress intensity factor for mode I fracture. It is also known as the toughness.

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29
Q

What is the difference between K and K(c)?

A

For brittle materials K(c) is a material property and has nothing to do with crack geometry or loading conditions whereas K is a function of crack geometry and loading conditions.

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30
Q

Relate G and K(I)

A

Check slide 14

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31
Q

How does superposition work for K and G.

A

For K, different contributions towards the same fracture mode should have their K contributions added together.
For G, contributions in different modes should have G added together.

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32
Q

Show how the Griffith equation changes in plane strain (thick plate)

A

Check slide 15

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33
Q

How does specimen thickness influence fracture?

A

For thin specimens use plane stress, whereas in thick plane strains used.

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34
Q

State K(I) for uniform loading for a straight through crack in ∞ specimens and an edge crack in a semi-infinite specimens.

A

Check slide 15

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35
Q

Sketch plots of K against c showing what happens under stable and unstable growth.

A

Check slide 16

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36
Q

Why is fracture stable when a line force acts at the mouth of a crack or a posit force acts at the centre of a semicircular edge crack?

A

K will decrease as c increases as the crack tip gets further from the force.

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37
Q

State the Weibull equation for a statistical approach to dealing with the unpredictability of brittle fracture.

A

Check slide 17

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38
Q

What are plot on the axis for Weibull plot to find the Weibull modulus?

A

Plot ln(ln(1/P(s))) = lnV - ln(-V(0)σ(0)^m) + mlnσ

m is the Weibull modulus and can be found using the gradient when lnln(1/P(s)) is plotted against lnσ

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39
Q

How can we increase the mean strength of brittle materials?

A 
Increase toughness
Decrease c (by improved processing and or/ avoiding damage in service)
Proof testing (stress to design level)
Build in compressive stresses into the surface as bending usually max at surface.
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40
Q

State an expression from simple beam theory that shows how stress Varys with load in a 3 point bend test.

A

Check slide 19

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41
Q

Sketch the set of a SENB under a 3 point bend test.

A

Check slide 19

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42
Q

What are the methods of producing a sharp crack for a SENB?

A

Saw the notch with a razor blade and 0.25µm diamond paste.
Bridge anvil compression (compressive loading produces a small crack by stable growth)
Single Vickers indentation then polish plastic zone away
Chevron notch
Grow a fatigue crack (usually from a notch) to avoid sudden brittle fracture)

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43
Q

Sketch a schematic of apparent K(Ic) against notch tip radius to demonstrate why crack tips in testing must be as sharp as possible.

A

Check slide 19

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44
Q

State expressions for G at the point of crack growth for a double cantilever beam under constant displacement, constant load and constant moment conditions

A

Check slide 21

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45
Q

For a double cantilever beam test, what kinds of crack propagation do the conditions of constant displacement, constant load and constant moment cause?

A

Stable crack propagation
Unstable crack propagation
Neutral, independent of c

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46
Q

What is the main advantage of compact tension specimens for testing toughness and fatigue?

A

It is a standardised shape

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47
Q

What kinds of cracks do vickers indentations cause?

A

Semicircular, radial-median cracks

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48
Q

Sketch a cross section of a Vickers indentation.

A

Check slide 22

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49
Q

State an expression for finding K(c) of a material using a Vickers indentation.

A

Check slide 24

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50
Q

What are the advantages and disadvantages in using a Vickers indentation to measure hardness?

A 
Ads:
Cheap and easy
Small flaw like real flaws
Very small amount of material required
Disads:
Very difficult to measure c accurately
Approximate analysis as c not >> plastic zone.
Lateral or Palmqvist cracking, chipping etc not accounted for
Not valid for porous materials
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51
Q

What are the differences between calculating the K(Ic) or G(c) for a sample of simple and more complicated geometry?

A

Simple:
Analytical calculation
Complex:
More sophisticated numerical calculations.
G for other test pieces can be found experimentally with compliance calibration

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52
Q

Derive an expression for G involving dλ/dc that can be used for compliance calibration.

A

Check slide 24

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53
Q

How is compliance calibration used to find G(c)?

A

If P is plotted against c for various loads and the gradient of each slope is measured to be 1/λ, λ can then be plotted against c with dλ/dc being found at the crack length where fracture is occurring.

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54
Q

Give an expression for a first approximation for the size of the plastic zone ahead of a crack tip.

A

Check slide 25

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55
Q

What assumptions are made in the first approximation of size of plastic zone at a crack tip?

A

Plane stress
Cylindrical plastic zone
No work hardening
Tresca criterion.

56
Q

For a second estimate of plastic zone ahead of a crack tip, what is usually considered?

A

The plastic relaxation expanded the platic zone and simply moved the elastic stress field forward ahead of it by a distance r(y) such that the load supported per unit thickness would be the same as that for a fully elastic crack.

57
Q

Derive an expression for the size of a plastic zone ahead of a crack tip for a second estimate.

A

Check slide 25

58
Q

Derive an expression for an estimate of crack tip opening displacement.

A

Check slide 26

59
Q

What is the dominant determining factor in the toughness of most metals?

A

The work done in plastic zone deformation per unit area of crack advance is very high in most metals so cracks tend not to propagate.

60
Q

Sketch the shape of the plastic zone ahead of a crack tip in plane stress and plane strain conditions.

A

Check slide 27

61
Q

Why is the plastic zone smaller ahead of the crack tip in the plane strain case compared to plane stress?

A

The plastic zone is smaller in plane strain compared to plane stress (for the same K) as there is the requirement for there to be a tensile stress in the z direction. This reduces the shear stresses which drive plastic deformation.

62
Q

Sketch a plot of apparent K(c) and % flat fracture against specimen thickness.

A

Check slide 27

63
Q

Why is the percentage of flat fracture high for thick specimens?

A

Thick so plane strain meaning less plastic deformation and less energy needed to propagate crack.

64
Q

How do specimens usually fail in plane strain?

A

By shear rather than mode I, giving 45° lips on fracture surface. This is due to large shear stresses on planes inclined to the crack plane and crack front by 45° (Mohr’s Circle)

65
Q

Why is there a higher toughness in plane stress than plane strain?

A

It results from the larger plastic zone.

66
Q

Why is it preferable to measure the plane strain fracture toughness as opposed to the pale stress?

A

Firstly it is a material property, secondly it provides a pessimistic answer so we design on the side of safety.

67
Q

How can we achieve plane strain for fracture toughness testing?

A

Plastic zone must be less than the thickness of specimen.
Form LEFM:
r(y) &laquo_space;distance over which the near field solution is valid (c/2)
The near field (and plastic zone) are well away from any boundaries.
Basically if the normally the plastic zone must be much smaller than any of the specimen’s dimensions

68
Q

State the criterion for valid plane strain toughness testing.

A

Check slide 28

69
Q

What methods can we use when LEFM is not practically possible to determine toughness by plane strain (eg specimen required for validity is too big)?

A

Crack Tip Opening Displacement

The J-Integral

70
Q

How can crack tip opening displacement be used to find the toughness of a material?

A

With the relation on slide 30.
δ(c) is a material parameter that characterises failure so even when LEFM is not valid, δ(c) can be measured to find the fracture toughness.

71
Q

What is J in the J-integral?

A

The rate of change of mechanical potential energy with crack width.
It is considered the non-linear version of G.

72
Q

Define J-integrak.

A

It is an integral about a path, s, starting on the lower surface of the crack and ending on the upper.

73
Q

Write the summation convention of J integral.

A

Check slide 31

74
Q

How does transformation toughening work and apply it to Yttria Stabilised Zirconia.

A

Change in microstructure can compress crack tips to toughen the material.
In Yttria Stabilised Zirconia, Tetragonal phase is metastable formed when cooling to RT due to Yttria stabaliser keeping small grain size.
Monoclinic is the stable RT phase and the transformation has a 4% increase in volume.
Tensile stress ahead of crack tip causes local transformation.

75
Q

Sketch a plot of load against extension for a polymer at varying temperatures between well below T(g) and greater than T(g)

A

Check slide 3

76
Q

How does the experimental value of G(c) compare to the theoretical for polymers and explain.

A

The experimental G(c) is 2-3 times greater than estimates from breaking of bonds.
This high value results from shear yielding and crazing which absorb a lot of energy.

77
Q

Explain what happens to polymers when crazing and eventually cracking.

A

Randomly aligned molecules far from craze.
Molecules align during plastic deformation at craze tip due to the weak intermolecular forces being overcome by stress.
Molecules are drawn into fibrils 1-100nm in diameter.
Voids then form between fibrils.
Eventually craze leads to fracture.
Crazes tend to occur in regions of high tensile stress so tend to nucleate at surface scratches.

78
Q

How can crazes be healed?

A

By high temperature > T(g) and/or compressive stress.

79
Q

Sketch a failure map in principle stress space for the effect of stress state on crazing.

A

Check slide 5

80
Q

How does temperature affect crazing?

A

Increasing temperature will often decreases the craze growth rate as it reduces the adoption of penetrant at the tip.

81
Q

How can rubber toughen polymers?

A

The inclusion of rubber particles to polymers concentrates stress and induces subsidiary crazing and shear deformation ahead of the crack tip.
Rubber particles restrain further opening of grazers/shearing preventing interlinkage and development into cracks.
Dissipates much more energy.

82
Q

What conditions must be met for rubbers to toughen polymers?

A

The rubber must have a T(g) below the operating T.
Rubber is a discrete 2nd phase, insoluble in the polymer matrix
There must be good adhesion between rubber particles and the matrix which is usually achieved by block co-polymers or grafting during polymerisation.

83
Q

Example of a rubbery material in polymer to strengthen.

A

HIPS

84
Q

Describe how crack bridging mechanisms slow crack propagation.

A

By forcing the crack to propagate a long and torturous path through a material with many direction changes it will slow down. Sintered materials often good at this where phases have fused together.

85
Q

What happens in fibre pullout to crack propagation if the interface between fibre and matrix is strong?

A

The crack will propagate passing through fibre and matrix.

86
Q

What happens in fibre pullout to crack propagation if the interface between fibre and matrix is weak?

A

The fracture process is thought and energy is dissipated away from the crack tip. This can be by debonding with the crack propagating on the other side of the fibre and leaving the fibre intact then the fibre cracking away from the crack and pulling out, all absorbing energy.

87
Q

Derive an expression for the work in fibre pullout.

A

Check slide 10

88
Q

State the expression for critical transfer length of a fibre.

A

Check slide 10

89
Q

Define critical transfer length of a fibre.

A

The half length of a fibre for which sufficient load is transferred from the matrix just to fracture the fibre.

90
Q

Derive an expression for the total contribution of fibre pull-out to the fracture energy.

A

Check slide 11

91
Q

What type of fracture usually happens in ceramics and metals?

A

Ceramics have a mix of intergranualar and trans granular cracking whereas metals are usually trans granular along specific planes due to grain boundaries being tougher than grains.

92
Q

When will metals fracture in an intergranualar manner?

A

If the toughness of the gbs is lowered by segregation of solute atoms to the gbs or if a 2nd phase forms at the gbs.

93
Q

Discuss temper embrittlement in steels.

A

It happens in many steels after quenching and during tempering.
It is the intergranualar fracture with raised ductile-brittle transition temperature and lowered toughness.
Due to segregation of trace impurities of groups IVB-VIB, increased in IVB and VB by co-segregation with Ni or Mn.

94
Q

How can we control temper embrittlement in steels?

A

Temper above critical temperature range.
Cool at maximum rate after tempering.
Careful scrap selection
Austenitiese at the minimum temperature to maintain fine grain size (more gbs so lower conc of segregants and higher toughness).
Add Mo or Ti to scavenge P in Ni-Cr steels

95
Q

Describe how overheating of low alloy steels leads to failure.

A

When low alloy steels are preheated at high Ts before hot rolling, quenching and tempering, inter granular ductile facets are seen in the subsequent RT fracture surface of impact specimens (in austenite not ferrite). The facets cause a reduction in impact energy and reduction in ductility.
The inter granular failure wrt the high T austenite gbs is known as overheating. At very high T, large amounts of MnS are formed which is known as burning. The result of dissolution of sulphides at high T and subsequent re-pption at gbs during cooling.

96
Q

How to avoid overheating low alloy steels.

A

Use low S and Mn contents
Avoid intermediate cooling rates where worst effected.
Low T ageing treatment before final quench and temper to redistribute alloy elements.

97
Q

List pre-existing defects that can initiate fracture in a material.

A

Microcontacts (ceramics - metals too tough)
Macrocontacts from improper handing.
Machining damage
Processing defects:
-Pores (powder processed material)
-Inclusions of 2nd phase particles and impurities

98
Q

State how micro contacts can initiate cracks in ceramics.

A

Considering the elastic contact of a sphere in planar surface (blunt indenter)
Max tensile stress and crack initiation stress on slide 18
Sharp indenters more damaging due to plastic deformation

99
Q

Describe how machining damage can initiate cracks.

A

Abrasive particles are like small indenters

Flaw size tends to increase with severity of machining, although microstructure often determines crack arrest.

100
Q

Where can cracks arrest in a material’s microstructure?

A 
Gbs
Triple lines
2nd phases
this implies that c α L(grain size) for mild machining
and σ(f) α L^-1/2
101
Q

Describe how pores can arise in powder processing.

A

Can arise from incomplete sintering
Imperfect powder packing in green body
Burnout of binder or melting sintering aids, esp if not well dispersed
(Even small, sub-critical pores can degrade strength bu reducing E and G(c))

102
Q

State an equation that describes fracture strength due to pores in a material.

A

Check slide 20

103
Q

Describe how inclusions of 2nd phases and impurities affect fracture initiation.

A

Act as/initiate critical flaws.
In brittle materials can cause micro cracks or concentrate stress by having:
-low toughness or weak interface with matrix.
-big thermal expansion coefficient mismatch with the matrix leading to residual stress.
-a big difference in stiffness.

104
Q

How can residue stresses arise in single phase ceramics that could cause crack initiation?

A

If the ceramic has anisotropic α then residual stress.

105
Q

How can brittle inclusions initiate fracture in metals?

A

Plasticity relaxes residual stresses too effectively for fracture, but brittle inclusions can initiate cracks by fracturing under the stress of dislocation pile ups. Eg, steels with alumina particles, silicates and Fe3C

106
Q

State the process by which fracture occurs at dislocation pile-ups.

A

Plastic deformation to cause pileups where slip is blocked.
Nucleation of crack
Growth of crack.

107
Q

Give an approximate derivation for the stress to nucleate a crack at the tip of a pileup.

A

Check slide 23

Follows Hall-Petch type relationship

108
Q

Give an approximate expression for the fracture stress when controlled by crack growth.

A

Check slide 24

109
Q

What evidence is there that brittle fracture in metals can be controlled by crack growth?

A

Microcracks can often be seen prior to failure which are not present before plastic deformation.
Brittle fracture is sensitive to hydrostatic stress state (tensile stress favours brittle fracture). Only the externally applied shear stress is involved in nucleation so final failure must be controlled by crack growth.

110
Q

Describe how fracture in metals can be controlled by yield.

A

Nucleation cannot occur below the yield stress as there would be a lack of pileups. If the yield stress is larger than the stress for cracks to nucleate and grow, fracture takes place as soon as yield occurs.

111
Q

Give an expression that shows ductile-brittle transition (DBT).

A

check slide 25

112
Q

Sketch a plot of brittle fracture stress against yield stress showing T(DBTT)

A

Check slide 26

113
Q

What is the only way to simultaneously increase yield strength and lower DBTT?

A

Reduce grain size.

114
Q

What test can be used to observe DBT?

A

-Charpy or Izod.
Pendulum swung against notched specimen.
Energy absorbed so doesn’t swing as high.
Large energy absorbed if ductile, small if brittle fracture.
Ductile will fracture by dimpled surface while brittle will lead to faceted surface.
-Tensile testing.
Lower strain rates.
Ductile fracture is initiated by cup and cone surfaces with 30-60% reduction in area.
Vary T, when fracture is brittle, the reduction in area rapidly drops to almost zero.

115
Q

How do coarse and fine slip differ?

A

When shearing a grain by displacement u, coarse slip is a single slip with large displacement =u, whereas fine slip is smaller displacements of u/n on n slip planes.
The number of dislocations in an individual pileup and therefore stress conc, is smaller in fine slip, therefore a larger total strain is required for a crack to form.

116
Q

Which is more likely to cause crack initiation, coarse or fine slip?

A

Coarse as larger pileups for the same stress, which lead to nucleation and growth of cracks.

117
Q

How does the size and distribution of 2nd phase particles influence coarse/fine slip and ultimately ductile and brittle failure.

A

Small coherent particles that are cut by dislocations lead to strain softening, so existing slip bands continue to operate and slip is coarse so the alloy is more likely to be brittle.
Larger more widely spaced coherent particles, or strong and well bonded incoherent particles are circumnavigated by bowing. This leads to work hardening and the operation of sources on new slip planes that have not been hardened, thus fine slip and more ductility.

118
Q

What is the process of ductile fracture of alloys and particulate composites when they neck down and fracture?

A

Holes nucleate at inclusions
Growth
Coalescence
Premature fracture

119
Q

What does a rough fracture surface imply when an ally necks down and fractures?

A

Implies cavities.

120
Q

Describe the nucleation of cavities around hard particles and the critical stress criterion.

A

When the matrix is sheared compressive stresses develop in two quadrants adjacent other particle and tensile stress in the other two, with corresponding arrays of interstitial and vacancy prismatic dislocation loops to retain the shape of the undeformed particle.
Stress increases linearly with strain and particle diameter.
See sketch on slide 1
Assuming k does not vary rapidly, large particles dominate the nucleation and the strain to nucleation will be greater in alloys with smaller particles.
This is due to larger particles causing more plastic flow for mismatch in matrix to be accommodated.

121
Q

Derive an expression for the number of loops to accommodate a rigid particle in a matrix.

A

Check slide 2

122
Q

Describe the growth and coalescence of voids around a rigid particle in a matrix under stress.

A

The spherical void concentrates stress and elongates, initially at ≈ twice the overall strain rate so becomes ellipsoidal and eventually elongates at the overall strain rate.
At some critical strain rate, plasticity becomes localised (voids coalesce) fracture with no further elongation.

123
Q

What simple criterion can be used for void coalescence and failure around an inclusion in a metal?

A

Brown & Emburey: Void coalescence and failure when 45° shear bands form between voids.
Sketch on slide 2

124
Q

Derive an expression for the true strain in the specimens at coalescence.

A

Check slide 3

125
Q

The model derived for true strain in specimens at coalescence is based on the assumption that deformation is uniform in the specimens on a larger scale, what actually happens and what is the effect?

A

Necking actually creates enhanced local shear which can influence the nucleation and growth of holes, and holes coalesce earlier than the model would predict by the mechanism of void sheeting.
This mechanism is for transgranular ductile fracture which is the usual path.

126
Q

Give expressions for fibre failure first strength, at high and low fibre concentrations along with an expression for volume fraction of fibre at the lowest strength.

A

Check slide 6

127
Q

Give expressions for matrix failure first strength, at high and low fibre concentrations along with an expression for volume fraction of fibre at the lowest strength.

A

Check slide 7

128
Q

Why is failure transverse to the fibres in a composite less than that of the matrix strength for weak fibre-matrix interfaces?

A

Failure can occur through the matrix and matrix-fibre interface without need for any fibre fracture.
If the fibre-matrix interface is weak, then fibres readily debone and the transverse strength is controlled by the width and strength of matrix ligaments.

129
Q

Derive an expression for the transverse strength of a fibre composite with weak fibre-matrix interface.

A

Check slide 7

130
Q

How do fibre composites behave under shear stress?

A

Dependant on fibre-matrix interface properties and matrix properties.
Simple estimate is that shear strength is the same as the matrix shear strength.
Fibres will provide stress and strain concs which reduce the composite shear strength, particularly at high fibre volume fractions.

131
Q

State the Tsai-Hill criterion (anisotropic version of Von Mises).

A

Check slide 9

132
Q

What is the simplest failure criterion that can be used for fibres under combined loading?

A

Maximum stress criterion. This compares the applied stress in each mode of loading to the failure stress. Whichever reaches 1 first causes failure (slide 9)

133
Q

State the stress components for off axis loading of a fibre composite.

A

Check slide 9

134
Q

Give an expression for the critical transfer length of a short fibre composite under longitudinal loading.

A

Check slide 10

135
Q

State the failure strength of a fibre and composite for L>L(c) for longitudinal loading.

A

Check slide 10

136
Q

State the failure strength of a fibre and composite for L

A

Check slide 10

GP3 (5 decks)

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