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MS2 > Mechanical Properties > Flashcards

Mechanical Properties Flashcards

1
Q

What is strength usually thought of as?

A

The resistance to shape change.

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2
Q

What is usually used to test for plastic behavior?

A

A universal tensile testing machine.

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3
Q

Why are samples for a universal tensile testing machine waisted?

A

To ensure constant CSA and give an accurate sample size.

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4
Q

What equipment is available for measuring strains?

A

Clip gauges
Strain gauges
Full field measurement of displacement (DIC)

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5
Q

What process is necessary before using digital image correlation (DIC)?

A

The sample needs a random high contrast surface so is sprayed with black paint.

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6
Q

How to test the strength of ceramics?

A

Compression or 3 or 4 point bend tests.

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7
Q

Difference between engineering and true stress?

A

Engineering stress is force over initial CSA whereas true stress is force over CSA in that instant.

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8
Q

Difference between engineering and true strain?

A

True is ln(li/l0) whereas engineering is (l1-l0)/l0.

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9
Q

State equation to convert engineering stress to true stress.

A

Check.

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10
Q

State equation to change engineering strain to true strain.

A

Check

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11
Q

Which type of stress strain gives rise to UTS.

A

Engineering.

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12
Q

How is a hardness test conducted?

A

A hard ball or pyramid is pushed into a surface and the area indented is measured.

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13
Q

How is hardness usually related to yield stress?

A

Hardness ≈ yield stress x 3

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14
Q

What does hardness mean?

A

The quality of being solid, stiff, and difficult to break.

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15
Q

State Schmid’s law.

A

Yield will occur when the resolved shear stress on a slip system reaches a critical value.

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16
Q

Slip occurs on planes with what?

A

The greatest spacing.

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17
Q

State OILS rule.

A

Check

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18
Q

How does the width of a dislocation core affect its ease of motion through a crystal?

A

Wider cores have lower lattice resistances due to less distortion of the position of each lattice point.

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19
Q

How does temperature affect CRSS?

A

Increased T reduces CRSS due to increased energy to overcome lattice resistance.

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20
Q

How does interstitial strengthening work?

A

Small atoms in interstices but large enough to induce a strain in the lattice that impedes dislocation motion.

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21
Q

How can interstitial solutes cause an upper yield stress?

A

They can diffuse the core of a dislocation lowering the strain energy and pinning it increasing the stress required to activate the dislocation.

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22
Q

What are Lüder’s bands?

A

When a band of deformed grains crosses the specimen at ≈ 45° then widens as slip is triggered in neighbouring grains.

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23
Q

What happens when all grains have yielded?

A

Work hardening begins.

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24
Q

How does substitutional strengthening work?

A

Impurity atoms cause lattice distortion and cause a strain field impeding dislocation motion.

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25
Q

Where can dislocations move from to under substitutional strengthening?

A

From one energy minima to another.

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26
Q

How does solid solubility affect hardening?

A

Large solid solubility leads to minimal hardening.

Small solid solubility leads to stronger hardening.

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27
Q

When is precipitation hardening relevant?

A

When a system has one element that has limited solubility in the other and a two phase region exists.

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28
Q

Outline how heat treatment for strengthening via precipitation hardening works.

A

Quenched from single solid phase (homogenised) below the solvus reaching a metastable state.
Aged with a moderate temperature to decompose the metastable solid solution giving precipitates.

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29
Q

What is the first stage of aging from heat treatment?

A

Solute atoms begin to cluster and vacancies annihilate (leads to drop in yield stress).

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30
Q

What is the second stage of aging from heat treatment?

A

Clusters of solute increase in size forming GP (Guinier Preston) zones which are fully coherent with the matrix.

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31
Q

What is the third stage of aging from heat treatment?

A

Precipitates form using GP zones as nucleation sites.

Not the equilibrium precipitate a metastable one which may be coherent and become a semi coherent precipitate.

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32
Q

What happens after long aging times from heat treatment?

A

The intermetallic will form which is fully incoherent with the matrix and a drop from peak yield stress.

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33
Q

What happens as precipitates from aging get bigger?

A

The cutting through of precipitates by dislocations becomes more difficult as they are less coherent but as the particles get bigger they become further apart (due to fixed amount of solute) so dislocations can bow around obstacles.

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34
Q

Derive the critical condition for dislocation motion around precipitates.

A

Check

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35
Q

Von Mises criterion for slip in polycrystals.

A

Each grain must have five independent slip systems for slip to occur in a polycrystal.

36
Q

Is it easier for slip to occur in hcp polycrystals with c/a < 1.63 or c/a > 1.63?

A

c/a < 1.63 as slip can occur on basal and prismatic planes so Von Mises criterion is satisfied.

37
Q

State the equation of the Hall-Patch effect.

A

Check

38
Q

What effect does grain size have on yield stress?

A

Smaller grains cause increase in yield stress.

39
Q

When can a dislocation cross a grain boundary and how likely is it to?

A

Very unlikely as slip planes in both grains must have a common line and the Burgers vector must lie in the grain boundary plane.

40
Q

Why do polycrystals with bigger grains have lower yield stresses?

A

Dislocations pile up at grain boundary and stress concentrates. There is critical stress that activates sources in the next grain. A greater number of dislocations can pile up in a larger grain before the source in the grain is blocked so lower external stress will activate the source in the next grain meaning larger grained polycrystals have lower yield stresses than smaller.

41
Q

Rules of thumb for yield stress variation with temperature.

A

Pure metals have weaker variations than solutions.
fcc have weaker variation than bcc.
hcp metals and oxide ceramics have behaviour controlled y the secondary slip systems needed to provide 5 independent slip systems.
Covalently bonded ceramics have weak temperature dependence.

42
Q

What are the stages of work hardening in a single crystal?

A

Stage I, II, and III

43
Q

What happens in stage I work hardening?

A

Primary slip in one system.

44
Q

Use Diehl’s rule to find primary and secondary slip systems for tensile axis [1-25] in a copper system.

A

Check

45
Q

What happens in stage II work hardening?

A

More than one slip system operates as dislocations interact forming jogs and locks.

46
Q

What happens in stage III of work hardening?

A

Cross slip.

47
Q

What dislocations can cross slip?

A

Screw only.

48
Q

What stages of work hardening operate in polycrystals?

A

Stage II and III

49
Q

Stage I work hardening tends to be shorter in which of fcc or bcc?

A

Shorter in bcc as more slip systems so stage II activates at lower stress.

50
Q

How easy is cross slip in hcp metals?

A

Easy in low c/a due to slip on both basal and prismatic planes, high c/a are brittle due to no cross slip.

51
Q

How easy is cross slip in bcc metals?

A

Easy as large number of slip planes.

The dislocations are undissociated.

52
Q

How easy is cross slip in fcc metals?

A

Lower the stacking fault energy the more difficult cross slip is as dislocations dissociate.

53
Q

What effect does bowing have on work hardening?

A

Enhances as it decreases the effective precipitate spacing making the next bowing event harder.

54
Q

What two processes operate during annealing?

A

Recovery and recrystallisation.

55
Q

Describe the recovery process.

A

Dislocations annihilate by climbing and gliding leaving residual dislocation density.
Sub grain boundaries then form to minimise strain energy (polygonistation).
This softens the material.

56
Q

Describe the recrystallisation process.

A

Nucleation of new grains at grain boundaries free form dislocations.
Perfect structure forms with larger grains and no dislocations.

57
Q

When do materials deform by twinning?

A

Homologous temperature is low
Strain rate very high
The material does not obey VM criterion
The material is single crystal and orientated to give no RSS on any slip system (unlikely in metals but may be in ionic)

58
Q

What is yield criterion?

A

A hypothesis concerning the limit of elasticity under any combination of stresses.

59
Q

What direction will a purely hydrostatic stress lie on in principle stress space?

A

The [111] direction

60
Q

VM yield criterion equation.

A

Check

61
Q

Tresca yield criterion equation.

A

Check

62
Q

Apply both yield criterion to a combined torsion tension system.

A

Check

63
Q

Apply both yield criterion to thin walled cylindrical pressure vessels.

A

Check

64
Q

What assumptions are made in VM/Tresca yield criterion?

A

Yield strengths are the same in compression and tension.
Volume is conserved in plastic deformations
The mean hydrostatic stress does not affect yield.

65
Q

Why do polymers not follow VM/Tresca yield criterion?

A

Yield is not independent of shape change.

66
Q

What model is used for yield criterion in polymers?

A

Coulomb yield criterion.

67
Q

Equations for the Coulomb yield criterion.

A

Check

68
Q

Equations for the upper and lower bounds of yield for a composite.

A

Check.

69
Q

Derive the yield of a unidirectional fibre composite under longitudinal loading.

A

Check derivation.

70
Q

Derive the yield of a unidirectional fibre composite under transverse loading.

A

Check derivation

71
Q

Derive an expression for the lower bound of the strength of a composite using RoM approach.

A

Check derivation.

72
Q

Derive the theoretical strength of a material using a stress analysis

A

Check derivation.

73
Q

Derive the theoretical strength of a material using an energy analysis

A

Check derivation

74
Q

Derive fracture stress using Griffith analysis.

A

Check derivation.

75
Q

Derive the fracture stress using stress analysis.

A

Check derivation

76
Q

Derive the rate of fracture using Griffith analysis.

A

Check derivation.

77
Q

State the general equations for fracture toughness (mode I).

A

Check

78
Q

Relate energy and stress analysis of fracture.

A

Check equation.

79
Q

What causes brittle to ductile transitions with temperature?

A

Thermal activation of slip systems in bcc and hcp (not in fcc as small Perierls stress).

80
Q

Methods of toughening materials.

A

Micro-cracking
Fibre pullout
Crack deflection
Transformation toughening.

81
Q

Where does micro-cracking have to occur to toughen a material?

A

Behind the crack to redistribute stress and provide crack tip shielding.

82
Q

What is crack deflection?

A

Crack deflected into forming other cracks redistributing the stress.

83
Q

How does fibre pullout work?

A

If fibre doesn’t crack with matrix but breaks elsewhere it will slide against the interface as it is pulled out absorbing some energy from the crack.

84
Q

How does transformation toughening work?

A

Uses volume change to induce the stress at the crack tip and arrest crack propagation.

85
Q

Method for material selection.

A

Identify function
Identify the objective (what equation captures it).
Identify constraints (dimensions or properties that are fixed)
Identify free variables and eliminate them.
Write down the figure of merit (maximise)
Use the Ashby diagram to suggest materials.

MS2 (6 decks)

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