Handout 9: Failures

Question 1

Explain the Griffith criterion.

Answer 1

• The failure of engineering components and structures often involves fracture. Recall that for a component containing a pre-existing crack of length a (edge crack) or 2a (internal crack), fast fracture will occur at a stress σ given by
• This is the Griffith criterion, where K_IC is the fracture toughness of the material (for mode I crack opening) and Y is a geometrical factor (≈ 1).
• The component will fail by fracture if the criterion is satisfied – i.e if - the applied stress rises to the critical level;
• or - the crack grows and reaches the critical length; or
• the fracture toughness falls.

Question 2

What is the equation that gives the local stress at a distance r from the end of a sharp edge crack of length a?

Answer 2

Question 3

Explain what residual stresses are.

Answer 3

• Residual stresses are common in manufactured components – often caused by plastic deformation, thermal expansion or contraction, or phase changes with associated volume or shape change.
• This can be expressed as the spring model, where there are internal compressive forces which need to be balanced out.

Question 4

Explain how the process shot-peening exploits the effect of residual stresses formed from deformation.

Answer 4

• A shot deforms metal surface, the metal yields in tension, material further below remains elastic.
• The shot is removed, the elastically deformed material relaxes, causing ‘springback’, it cant return to original shape.
• The result is a permanent indentation in the surface, with a compressive residual stress close to the surface, balanced by tensile residual stress deeper into the material.
• Multiple impacts of shot all over the surface produce a near-surface region containing compressive residual stress. This enhances the fatigue life of the component by reducing the tensile stress acting on short surface cracks, so making it more difficult for them to propagate.

Question 5

Explain how thermal stresses arise.

Answer 5

• Occurs when a material cools unevenly .e.g Steel I beams (central web cools down faster).
• Example: The top surface of a metal slab is heated. The hot region expands (a). Because its yield stress is reduced when it is hot, it can deform plastically to accommodate the expansion, so the stress in the hot region falls. (b).
• (c) As the hot region cools, it contracts, but its yield stress has now gone back to its original higher value so it cannot deform plastically. It thus exerts a force on the neighbouring cold region. The final tensile residual stress in the region that had been heated can be up to the material’s yield stress, and is balanced by compressive stress elsewhere.

Question 6

Explain how residual stresses arise in polymer injection moulding.

Also explain how this can cause problems.

Answer 6

• In polymer injection moulding, the material close to the mould wall solidifies and contracts first, with material in the centre still molten (a). This material then cools and shrinks (b), putting the surface regions into compression and leaving the core under tension (c).
• If cooling is asymmetric this can lead to warping and distortion of the part.
• The same process is done with glass, the outside is cooled fast leaving a low density glass structure, while the inside shrinks and puts the outside into compression.

Question 7

Explain the different types of fatigues.

Answer 7

Question 8

Where do fatigue cracks typically nucleate at?

Answer 8

• Fatigue cracks in an uncracked component typically nucleate at
  
  • Stress concentrations (e.g. holes; welds; section changes; inclusions etc.)
  • Surface features (especially machining marks; corrosion sites etc.)

Question 9

Explain the term ‘endurance limit’ for steels.

Answer 9

• Steels typically show an endurance limit (S_o) of about half their tensile stress. Below this stress, the fatigue life of an uncracked specimen is infinite. Some alloys (e.g. Al alloys) do not show an endurance limit.

Question 10

Why is the fatigue life a welded components dictated by crack growth rather than by crack initiation?

Answer 10

There are frequently cracks in and around welds (and perhaps slag inclusions which also act like cracks). The fatigue life of welded components is therefore controlled by crack growth rather than by crack initiation.

Question 11

What factors initiate cracks in welds?

Answer 11

• Cracks in welds generally result from a combination of:
• temperature gradients causing thermal stresses
• variations in composition in the weld metal/HAZ giving differences in contraction
• segregation during solidification
• hydrogen embrittlement
• inability of the weld metal to contract during cooling (similar to hot tearing of castings)

Question 12

What are measures that can be taken to reduce cracks and residual stresses due to welding?

Answer 12

• modify design of joint to minimize thermal stresses from shrinkage during cooling
• change welding process parameters, procedures and/or sequence
• preheat components being welded
• avoid rapid cooling after welding
• Induce residual compressive stress in weld metal by shot peening

Question 13

What issues does corrosion cause?

Answer 13

Corrosion in metals occurs when metal reacts to form a corrosion product (often an oxide).

The product has inferior properties to the metal, so causes problems. e.g.

• Lower mechanical strength;
• Electrical insulator rather than conductor

Question 14

For corrosion to occur, we need two reactions, anodic and cathodic, explain these two reactions.

Answer 14

• Anodic reactions liberate electrons, e- ; a metal atom M is oxidized to form an ion Mn+ by the removal of electrons.
• M = Mⁿ⁺ + ne^-
• The metal ions will often be soluble in water or react with oxygen ions or hydroxyl ions to form compounds. Removal of the metal in these ways constitutes corrosion.
• Cathodic reactions consume electrons (called reduction), with various possible processes e.g.
• 2 H⁺ + 2e^- = H₂

Question 15

What is bimetallic corrosion?

Answer 15

• If two dissimilar pieces of metal are put in contact into an aqueous medium (e.g. water, or dilute acid) then one of the metals becomes the anode while the other becomes the cathode. We need both types of reactions before corrosion can happen.
• Oxidation (the anodic process) will occur for the reaction with the lowest SEP, while reduction (the cathodic process) will occur for the reaction with the highest SEP.

Question 16

Explain why iron is galvanised zinc, giving the relevant equations.

Answer 16

Consider zinc and iron in contact with each other in clean water (pH neutral) in the presence of air. The standard electrode potentials and processes are as follows:

Zn = Zn²⁺ + 2e^- E_o = - 0.76V (Anodic process)

[Fe = Fe²⁺ + 2e^- , E_o = -0.76 V Alternative Anodic process - doesn’t take place]

O₂ + 2H₂O + 4e^- = 4(OH^-) E_o = +0.40V (Cathodic process).

• The zinc is oxidized to form Zn²⁺ ions, releasing electrons which flow into the iron, which becomes the cathode. On the iron surface, oxygen (dissolved in the water) reacts with water to form hydroxyl ions (OH- ). This uses up electrons, so the corrosion of the zinc continues. If the zinc (with the lower SEP) were not present, the anodic reaction would be oxidation of the iron, which would therefore corrode. The presence of the zinc protects the iron.

Question 17

What is differential aeration in steels and how does it lead to crevice corrosion?

Answer 17

Fe = Fe²⁺ + 2e^- (Anode)

O₂ + 2H₂O + 4e^- = 4(OH- ) (Cathode)

• The two reactions occur at different regions, electrons are transported between the two through the metal.
• The lowest oxygen levels are deep inside cracks and crevices. This means that steels are liable to form deep cracks as a result of the waters presence, because once a crack has formed corrosion (the anodic process) will be concentrated at the growing tip of the crack where the oxygen concentration is lowest.
• Rust has lower density than the metal from which it forms, and is associated with expansion which can wedge cracks apart.

Question 18

Why are marine conditions far more corrosion prone?

Answer 18

• Chloride ions (present in sea water) are particularly effective in assisting this mechanism, so that pitting corrosion is particularly prevalent in marine applications.
• The use of salt (sodium chloride) on roads in winter to avoid ice formation can also lead to severe corrosion of vehicle components. Local concentration corrosion can also result from broken or scratched paint coatings.

Question 19

Describe how differential energy corrosion leads to sensitization(and hence weld in chromium steels.

Answer 19

• Features causing a local increase in energy in a metal (e.g. grain boundaries, dislocations, precipitate interfaces) act as anodic regions and can dissolve rapidly; other regions form cathodes.
• If the steel also contains carbon, heating during welding can cause the carbon to react with chromium to form precipitates of chromium carbide on grain boundaries in the HAZ, depleting the neighbouring metal of chromium. The steel is then said to be ‘sensitized’. The regions close to the grain boundaries do not contain enough chromium to be able to repair the protective surface film of chromium oxide when it is damaged in service.

Question 20

Explain how Cathodic inhibitors work

Answer 20

• Cathodic inhibitors work by forming surface layers which inhibit the cathodic reaction, and are intrinsically safe – a reduction in concentration will lead to an increased corrosion, but it will still be less than the rate in the absence of the inhibitor.

Question 21

Explain how sodium nitrite works as a anodic inhibitor.

Answer 21

• Acts by forms a continuous protective film of iron oxide on the steel surface which acts as a barrier to further corrosion.
• Process called passivation.
• Inhibitor encourages oxidation of the steel, and unless the film is sufficiently thick the corrosion rate of the steel is considerably greater than the corrosion rate with no inhibitor present at all.
• The concentration of the anodic inhibitor must be above the critical level – if it falls below this level then rapid corrosion will result.
• If the anodic film is incomplete, then this corrosion will be localised in the unprotected regions, and lead to pitting.

Question 22

Explain what pourbaix diagrams are.

Answer 22

• It can be useful to represent the behaviour in aqueous solutions on a ‘map’. Pourbaix diagrams are used to plot the electrochemical potential against the pH of the solution.
• The Pourbaix diagrams can be altered dramatically by the presence of certain ions. e.g. stainless steel in aerated water shows a very large passive region because of the formation of a stable protective Cr2O3 layer. In the presence of chloride ions (e.g. as present in sea water) the film breaks down as a soluble complex chromium chloride forms, and no passive region is found. The diagrams below show the anodic reactions in water with and without oxygen, and the cathodic reactions for iron, titanium and aluminium.

Question 23

Explain what ‘Immunity’ and ‘Passivation’ means in terms of Pourbaix diagrams.

Answer 23

• ‘Immunity’ is a range of pH and potential where corrosion of the metal is thermodynamically impossible.
• ‘Corrosion’ implies that there is a thermodynamic driving force tending to dissolve the metal as ions
• ‘Passivation’ shows that there is a driving force to form a stable film (e.g. oxide or hydroxide) on the metal surface, but this may or may not form an effective barrier to further corrosion.
  *

Question 24

Explain what anodizing is in in aluminium and titanium.

Answer 24

• Aluminium and titanium can be ‘anodised’ to build up a thick protective layer of oxide.
• The hollow cells that are part of the structure of the thicker layers can be used to incorporate dyes or other surface modifiers such as PTFE (e.g. non-stick coating on saucepans).
• The tops of the cells are sealed after filling.
  *

Question 25

Explain stress corrosion cracking.

Answer 25

• SCC is characterised by cracks propagating at a stress well below the normal failure stress with little macroscopic plastic deformation in certain types of environment.
• Cracking can be transgranular or intergranular and is associated with particular metal/environment combinations.
• SCC is a common cause of industrial failures (particularly in the chemical industry).
• Essential ingredients: A susceptible material, sufficient tensile stress (which can be residual stress),a specific environment.

Question 26

Name the three different types of SCC mechanisms.

Answer 26

SCC can be caused by three different mechanisms:

• active path dissolution
• film-induced cleavage
• hydrogen embrittlement.

Question 27

Explain how active path dissolution occurs in SCC.

Answer 27

• Active path dissolution involves rapid corrosion along a narrow path (such as a grain boundary) with the rest of the material being passive.
• An example is the process of weld decay in unstabilized stainless steel discussed earlier.
• Cracks will be intergranular.

Question 28

Explain how film induced cleavage occurs in SCC.

Answer 28

• In film induced cleavage, a brittle surface film (e.g. an oxide) on a ductile metal cracks, and the crack then propagates a short distance into the metal (~ 1 µm) before it is blunted.
• The brittle film then reforms by corrosion at the crack tip, and the process repeats.

Question 29

Why does SCC tend to be more severe in high strength materials?

Answer 29

High strength tends to be associated with limited ductility)

Question 30

For the following alloys give the environments that cause SCC:

• Al alloys
• Cu alloys (e.g. brass, bronzes)
• Carbon steels
• Stainless steels
• Titanium alloys

Answer 30

• Al alloys: Chloride solutions, seawater
• Cu alloys: Ammonia vapour and solutions, other nitrogen sources
• Carbon steels: Sodium hydroxide solution, nitrates, mixed acids (sulphuric/nitric), hydrogen sulphide, seawater, chlorides.
• Stainless steels: Chloride solutions, seawater, hydrogen sulphide.
• Titanium alloys: Concentrated nitric acid, seawater

Question 31

Explain how hydrogen embrittlement causes failure in alloys.

Answer 31

• Hydrogen embrittlement occurs when hydrogen atoms diffuse towards regions of high hydrostatic tension (e.g. just ahead of a crack tip).
• Hydrogen atoms are very small and therefore can diffuse rapidly in some metals. The hydrogen lowers the fracture toughness.
• Hydrogen diffuses much more rapidly in ferritic iron than in austenite; austenitic steels are therefore almost immune to hydrogen embrittlement.

Question 32

How is the hydrogen absorbed by the alloys in the first place to cause hydrogen embrittlement?

Answer 32

• Atomic hydrogen is necessary, usually generated chemically or electrochemically (‘nascent hydrogen’).
• Typically, damp/wet conditions in conjunction with electric currents or even small amounts of corrosion; e.g. pickling; electroplating; MMA welding with damp electrodes.

Question 33

What sort of failure does hydrogen embrittlement cause?

How do you avoid hydrogen embrittlement?

Answer 33

• Under tensile stress (applied stress or residual stress) failure occurs by brittle fracture. The fracture is often initially intergranular.
• Preventing hydrogen cracking:
  
  • Use a lower-strength alloy, which is less susceptible. Steels with yield stress lower than about 700 MPa are generally resistant to hydrogen cracking.
  • Avoid treatments (e.g. electroplating) and service conditions which promote hydrogen absorption
  • Heat the component after the treatments to remove any dissolved hydrogen (e.g. 150-200 ^oC for 1 - 2 hours).
  • Reduce residual stresses. - Reduce stress concentrators (e.g. reduce notch severity).

Question 34

Describe liquid metal embrittlement.

Answer 34

• Liquid metal embrittlement is another type of SCC. As with aqueous SCC it is favoured by a tensile stress, and exposure to a specific environment, in this case of a liquid (molten) metal.
• The symptoms are that the metal component which is in contact with the liquid metal suddenly fails by intergranular fracture.
• In some metal-liquid metal combinations, the molten metal is able to penetrate along grain boundaries forming cracks. It then diffuses a limited distance into the host metal. It reduces the bond strength at the crack tip, and so the fracture toughness falls dramatically.
• Because it is particularly easy for metal to diffuse along grain boundaries, this type of LME is nearly always intergranular.

Question 35

Describe common scenarious leading to liquid metal embrittlement.

Answer 35

• Exposure of welded carbon steel to molten metal (e.g. hot dip galvanising, tinning).
• Residual stresses in the weld can then cause initiation of cracks or propagation of existing cracks.
• Brazing (Cu-Zn filler) and soldering (Sn-Pb filler) of structures which contain residual stresses •
• Process vessels containing liquid metal (e.g. baths for molten zinc made of carbon steel; baths for molten aluminium made of almost any metal; containment of mercury by carbon steels)
• Accidental contact between liquid metal and susceptible alloy (e.g. mercury in contact with aluminium alloys in aircraft structure) •
• Overheating of components coated with a metal, above the melting point of the coating or plating (e.g. Cd - plated bolts made from high-chromium steel; galvanised steel bolts).

Question 36

Briefly explain microbial corrosion.

Answer 36

• Micro-organisms (generally bacteria) can be responsible for corrosion of metals by various different mechanisms.
• For example Desulfovibrio desulfuricans thrives under damp or wet anaerobic (i.e. no oxygen) conditions. It produces sulphide ions which accelerate anodic dissolution and stress corrosion cracking.
• Common to occur in container and tanker ships.

Question 37

Why do ferritic steels undergo a transition from ductile behaviour at high temperatures to brittle at low temperatures?

Answer 37

• Such behaviour is common in metals with bcc crystal structures: dislocation mobility falls as the temperature drops, and the yield stress rises rapidly; the stress at the crack tip then reaches a level at which cleavage between grains occurs in preference to ductile void formation and coalescence, with a lower dissipation of energy, and hence lower toughness. The fracture surface shows characteristic crystalline facets.

Question 38

Draw a graph of impact energy vs temperature for different materials.

Why is impact energy used as a measure for fracture toughness?

Answer 38

• Because of the difficulties of performing fracture toughness measurements over a wide range of temperatures, the effect is often described in terms of energy absorbed in impact tests.
• Note that polymers also show ductile-brittle transitions (as shown here for nylon, a thermoplastic): here the transition is associated with the reducing mobility of the polymer chains below the glass transition temperature (Tg for nylon is ~50 °C)

Question 39

Explain how corrosion of glasses occurs due to moisture.

Answer 39

• Many glasses contain sodium as a network modifier (breaks up silicate –O-Si-O-Si-O- network, so reduces softening temperatures and promotes a wide temperature range in which glass can be worked).
• If the glass is in contact with moisture (particularly in an acid environment), the sodium atoms which terminate the silicate chains can be replaced by hydrogen atoms.
• This is associated with shrinkage of the surface layers of the glass. The glass is put into tension and surface cracks form.

Question 40

List the 5 different causes of degradation of polymers and explain how the cause failure.

Answer 40

• Photo-degradation - UV photons (e.g. in sunlight) which cause breakage of covalent bonds. May lead to three distinct effects: reduction in polymer chain length; depolymerisation (i.e. chain breakage leading to monomer formation); or increase in cross-linking. Often some discolouration.
• Oxidising atmospheres - (e.g. air, various fluids, ozone (= O3)). Similar effects to photodegradation.
• High temperatures - Reversible effects: softening of thermoplastics. Irreversible effects: increase in crystallinity; covalent bond damage as above; discolouration; charring (due to oxidation or chemical decomposition = ‘pyrolysis’).
• Solvent Damage - Some small solvent molecules can penetrate between the polymer chains, reducing the elastic modulus (plasticisation), and swelling the material. For example nylon can absorb >5% water. This effect is reversible. Irreversibility includes leaching of additives.
• Environmental Stress Cracking - Premature failure under stresses below the conventional design stress in certain environments. Amorphous polymers are particularly susceptible. Early stages of ESC can lead to the formation of multiple very fine cracks: ‘crazing’.
• The stress is often provided by residual stresses arising from the manufacturing process,
  
  • Uneven shrinkage in an injection moulding leading to internal stresses and often distortion
  • Parts which have been subjected to elastic stresses during joining, or assembly.

Question 41

List the different types of degredation in composites and explain how they cause failure.

Answer 41

• Matrix degredation - All the different things that can happen to polymers, e.g. water/solvent swelling, ESC.
• Fibre degredation - Polymers: Kevlar may be affected by UV and oxidation, leading to loss of strength and toughness
  
  • Glass suffers leaching (surface dissolution), so degrades in water.
  • Carbon fibres do not degrade
• Interface degradation:
  
  • Swelling stresses can result in cracking of interfaces.
  • Capillary flow (‘wicking’) of solvent along cracked interfaces can dramatically increase degradation rates, because it allows a ‘short-cut’ route for solvent to reach the interior of structures without having to diffuse through the matrix.
  • ESC resulting in rapid and specific attack of fibre or matrix at the interface. Plasticisation of matrix at interface, leading to creep and distortion

Question 42

Explain how the fatigue failure of composites is due to external stress cycling.

Answer 42

• Composites are full of stress concentrators and internal stresses.
• When subjected to alternating load, damage builds up in the form of fractured fibres and failure at the fibre-matrix interface.
• As a result, they show progressive degradation of mechanical properties. The elastic modulus falls, and the strength drops.
• Failure tends to be by the gradual linkage of many sub-critical cracks, rather than by the catastrophic fast fracture seen in metals.

Question 43

Explain how thermal cycling leads to failure of composites.

Answer 43

• If composites are heated, the fibres and matrix expand by different amounts, often resulting in elastic or plastic deformation, sometimes accompanied by fracture.
• Exactly which combination of effects is seen depends on the fibre-matrix mix, the lay-up sequence, and whether the temperatures involved are above the softening temperature of the matrix:
• (a) Build-up of internal stresses
• (b) Distortion of composite (particularly for multi-ply composites)
• (c) Fracture of fibres (reduction in effective fibre length; not for Kevlar which is tough)
• (d) Fracture of matrix between fibres (e) Fracture of fibre-matrix interface