IA: 1P2: Materials Flashcards

Question

What must be true about the internal forces **and** stresses for equilibrium to occur?

Answer 1

The shaded surface must carry components of force and stress both normal and parallel to the surface: * The forces may be related by resolving, however this is **NOT** true for the stresses as stresses on an inclined plane act on a different area to the axial stress ## Footnote Shear stresses arise in almost every loading situation, but may be off-axis from the obvious orientations chosen

Answer 2

Shear strain is defined as the change in angle between two initially perpendicular lines (or planes) in the material due to applied shear stres. This is because shear stress distorts the shape of a volume element, rather than changing its axial dimension. ## Footnote It is represented by γ

Answer 3

## Footnote Technically is is tan(γ), however for small strains tan(γ) = γ

Answer 4

It is the ratio of shear stress to shear strain, it characterises the elastic stiffness in shear:

Answer 5

ν = Poisson's ratio

Answer 6

Pure shear strain is a type of deformation where a material is distorted in shape but not in volume. It is caused by pure shear stress

Answer 7

A material that has the same properties in all directions

Answer 8

Only 2, given any two elastic constants the other 2 can be found

Answer 9

During plastic deformation: Yielding takes place at constant volume

Answer 10

In region 2-4 of the load-extension curve: * Yield load increasing and area decreasing -> actual true stress (F/A) to yield the material is increasing * Increase in resistance to yielding is called work hardening * This stabilises yielding, so the whole cross-section reduces uniformly At maximum load, dF/du = 0 (point 4): * the material's ability to work harden in tension has reduced * At any local narrowing of the section the true stress is higher, but work hardening is now insufficient to compensate * The plastic strain localises in this section: necking * as the load falls (4-5), the sample unloads elastically, but intense strain in the neck leads to internal damage (voids) and ductile failure

Answer 11

Nominal stress = Force / **Initial** Area

Answer 12

The initial gauge length Lₒ is used

Answer 13

Describes how much a material stretches or deforms **plastically** before breaking. It is marked on the stress-strain curve as εf:

Answer 14

A metal which has been softened by heat treatment

Answer 15

A metal which has been previously hardened, by stretching

Answer 16

Using the vickers hardness test: * A diamond pyramid is pressed into the surface under a constant load * Local plastic deformation occurs until the load is supported * Load removed and the resulting indent size, d, is measured * A = (d/√2)² = d²/2 Hardness is defined as: H = load / projected area of indent = F / A

Answer 17

**Hardness is approximately equal to 3 x yield stress, H ≈ 3σᵧ** Under the indenter, a steep strain gradient exists - the deformation and work hardening is greatest near the indenter

Answer 18

How stress relates to strain under different conditions, such as an applied load.

Answer 19

* True stress uses the actual cross-sectional area * Nominal stress uses the intial cross-sectional area

Answer 20

σₜ = True stress σₙ = Nominal stress εₙ = Nominal strain

Answer 21

As far as the onset of necking in tension (beyond this point deformation is no longer uniform)

Answer 22

each incement of extension (dl) is divided by the current length (not the initial value), to give an increment of the true strain (dε). The total true strain is then found by integrating between original and final length giving: **Ln(L / L₀)**

Answer 23

εₜ = True strain εₙ = Nominal strain

Answer 24

For εₙ > 0: * True stress > nominal stress * Nominal strain > true strain

Answer 25

When εₙ <<< 1

Answer 26

True strain is additive (nominal is not). If sequential plastic strains are applied to a material the total true strain is the sum of each separate true strain.

Answer 27

An empirical fit for the stress-strain response of a material without a distinct yield point:

Answer 28

**ε₁ = σ₁(1-v²) / E σ₁/ε₁ = E / (1-v²)** The contraint prevents the cube expanding in the 3-direction (ε₃ = 0) and hence there is a transverse stress induced in the cube preventing this expansion. This causes the apparent material stiffness to increase by a factor of 1/(1-v²).

Answer 29

* Constrained expansion or contraction (inducing thermal stresses) * When temperature gradients exist across a product (again inducing thermal stresses) * When dissimlar materials are joined and then subjected to temperature change, giving **differential thermal expansion/contraction** (and again inducing thermal stresses) - this is exploited directly in a bimettalic strip

Answer 30

α = thermal expansion coefficient

Answer 31

No not directly for unrestrained expansion, however, thermal strain can cause thermal stress **if the material’s expansion is constrained**

Answer 32

**stress = E(α₁ - α₂)ΔT**

Answer 33

* Polycrystalline microstructure: Atoms make up many small crystals, or grains, stuck together at grainboundaries to make polycrystalline microstructures * Amorphous microstructure: non-crystalline

Answer 34

They are all roughly a similar size (0.1-0.2nm)

Answer 35

Mixtures of atoms of different elements can pack together efficiently into crystal lattice structures, forming solid solutions (a stype of compound). There are 2 types of solid solutions: Substitutional and Interstitial

Answer 36

When atoms of similar size replace one another in a lattice (i.e. Cu-Zn in brass)

Answer 37

When small atoms can fit into the gaps between metal atoms (i.e. C in Fe: in carbon steels)

Answer 38

* Primary bonds * Secondary bonds Primary bonds tend to be 100x stronger than secondary bonds

Answer 39

* Metallic bonds * Ionic bonds * Covalent bonds

Answer 40

As stiff springs between atoms (or ions), with a non-linear force-separations characteristic (below). The atoms (or ions) have an equilibrium separation rₒ, governed by the balance between attractice and repulsive forces. At the dissociation separation the atoms (or ions) can be separated completely.

Answer 41

Atoms vibrate about the equilibrium separation, with kinetic energy ≈ KʙT, where T is the absolute temperature (in K) and Kʙ is Boltzmann's constant (1.38x10⁻²³ J per atom K⁻¹). The bonds break down when this kinetic energy exceeds the bond energy.

Answer 42

* Face-centred cubic (FCC) * Close-packed hexagonal (CPH) * Body-centred cubic (BCC)

Answer 43

A type of atomic arrangement in solids where atoms are packed together as tightly as possible, minimizing empty space. Around each atom there are 6 locations in which atoms can sit, but only 3 of these can be occupied at once ("odd" or "even"). This gives 2 possivle stacking methods, ABA or ABC Face-centred cubic (FCC) and close-pack hexagonal (CPH) are 2 types of close-packed structures: * FCC = ABC stacking * CPH = ABA stacking

Answer 44

The smallest unit which can be replication by translation in all directions to build up the 3D crystal structure. Unit cells are drawn with the atoms reduced in size, for clarity.

Answer 45

The straight lines through the centres of touching atoms (there are 3 in a close-packed plane)

Answer 46

the unit cell dimensions

Answer 47

**The FCC structure is described by a cubic unit cell with one atom at each corner and one at the centre of each face.**

Answer 48

Any diagonal of any face is a close-packed direction

Answer 49

* They are very ductile when pure, work hardening rapidly but softening again when annealed, allowing them to be rolled, forged, drawn or otherwise shaped by deformation processing. * They are generally tough, i.e. resistant to crack propagation * They retain their ductility and toughness to absolute zero, something very few other materials allow

Answer 50

**The CPH structure is described by a prismatic hexagonal unit cell with an atom in each corner, one in the centre of each hexagonal face, and three in the middle.**

Answer 51

Perpendicular to the axis of the prism

Answer 52

The length of an edge of the cubic unit cell

Answer 53

There are 2 lattice constants: * The side length of the hexagonal base, a * The height of the prism, c

Answer 54

* They are reasonably ductile (at least when hot), allowing them to be forged, rolled, and draw, but in a more limited wat than FCC metals * Their structure makes them more antisotopic than FCC metals (i.e. crystal properties vary with direction)

Answer 55

**The BCC structure is described by a cubic unit cell with one atom at each corner and one in the middle of the cube.** The structure is not-close packed!!

Answer 56

**Along the diagonals** The structure is not close-packed as it is made from stacking planes of atoms in a square array (not hexagonal) however it still has close-packed directions

Answer 57

* They are ductile, particularly when hot, allowing them to be rolled, forged, drawn or otherwise shaped by deformation processing * They are generally tough, and resistant to crack propagation at and above room temperature * They become brittle at low temperatures. The change happens at the "ductile-brittle transition temperature", limiting their use below this

Answer 58

Metal components are commonly manufactured by casting. The solidification mechanism involves the formation of many solid crystalline nuclei, which grow by attachment of atoms to the crystal at the interface between liquid and solid. Solidification is completed when adjacent crystals impinge on one another. But because the orientation of the packing in each crystal is random, there is a misfit in the atomic packing at the interface between two crystals. **The individual crystals are called grains, and the region of imperfect packing is called a grain boundary**

Answer 59

The fraction of space occupied by atoms (assuming a "hard sphere" model) in a unit cell

Answer 60

It's the fraction of space occupied by atoms in a unit cell. **When calculating this you must take account of atoms on the corners and faces being shared between adjacent unit cells.**

Answer 61

n = number of atoms per unit cell A = Atomic mass Vc = Volume of unit cell Nᴀ = Avogadro's numbe

Answer 62

The space between the atoms or molecules. There are 2 types of interstitial space: Tetrahedral and octahedral. Small foreign atoms can fit into these gaps.

Answer 63

A sphere with radius **0.22** of that of the host

Answer 64

A sphere with radius **0.414** of that of the host

Answer 65

Ceramics are a large family of materials made of metallic (or silicon) and non-metallic elements bonded together, primarily through ionic and covalent bonds

Answer 66

* Diamond Cubic (DC) * Halite * Corundum * Fluorite

Answer 67

The DC unit cell has an FCC structure with 4 additonal atoms located in half of the tetrahedral interstitial spaces (those labelled 1-4 on the top/the carbon atoms on the bottom). As the tetrahedral hole is far too small to accommodate a full-sized atom, the others are pushed further apart, lowering the density. Examples of materials with a DC structure is diamond (the hardest ceramic), silicon and germanium (used in semiconductors), and silicon carbide (another hard ceramic).

Answer 68

Many metal oxides have the formula M₂O₃ and form a Corunsum structure. The Oxygen atoms for a close-packed hexagonal (CPH) structure where the metallic atoms occupy 2/3 of the octahedral holes in the lattice.

Answer 69

When crystalline materials melt, the atoms lose their regular packing but are still loosely held together; on solidification crystals usually form readily. Glasses are based on silica, SiO₂, for which crystallisation is difficult. In the solid state silica usually has an amorphous structure, and only crystallises if cooled very slowly. This amorphous structure gives transparency, with the colour and refractive index of the glass readily being customised by alloying.

Answer 70

Polymers are long-chain molecules of carbon with side bonds to other atoms and groups. * Primary bonding between the Carbon atoms is by strong covalent bonds - both along the chains and at cross-links (where 2 chains are bonded together) * Secondary bonding acts between the chains (via the side-groups) by weak van der Waals bonds.

Answer 71

* Thermoplastics * Thermosets * Elastomers

Answer 72

Thermoplastics contain no cross-links, but are divided into two sub-groups: amorphous and semi-crystalline. In pure form amorphous thermoplastics are transparent and semi-crystalline thermoplastics are translucent.

Answer 73

The long-chain molecules are arranged entirely at random, with occasional entanglement points between chains. At these points there is no additional bonding, but they do restrain the deformation and sliding of the molecules. In pure form amorphous thermoplastics are transparent.

Answer 74

They are partly amorphous, and partly ordered in crystalline regions (known as "spherulites"). In the crystalline regions, the long chains line up and pack to give an ordered, repeating structure, just like any other crystal. The low symmetry of the individual molecules means that the unit cell is usualyl defined by three dimensions. Crystallisation can lead to significant shrinkage as the crystalline regions are far more tightly packed than the amorphous regions. In pure form semi-crystalline thermoplastics are translucent.

Answer 75

Elastomers contain a small number of cross-links, between simple chain molecules.

Answer 76

Thermosets contain a large number of cross-links, between chains

Answer 77

* **Technical ceramics**: alumina, silicon carbide, silicon nitride * **Glasses**: Based on silica * **Porous Ceramics**: Brick, concrete, pottery

Answer 78

The gradient of the F-r response at the equilibrium spacing r₀ is the bond stiffness S

Answer 79

**F = Su** Where: S = Stiffness of bond u = displacement from equilibrium

Answer 80

**E = S/r₀** Where: S = Bond Stifness r₀ = equilibrium spacing

Answer 81

It will form densities between those of the pure elements

Answer 82

Solid solutions contain a mixture of different bond stiffnesses (A-A, A-B, B-B): Therefore the Young's modulus of A-B solution are somewhere between pure A and pure B. This relationship is approximately linear according to the "rule of mixtures".

Answer 83

Compounds have stiffer bonds and hence higher Young's modulus

Answer 84

Metals crystallise very easily, however they can be forced to retain an amorphous structure if cooled very quickly. Alternatively bulk amorphous metals can be produced by solidifying at a conventional rate, but with a very unusual composition where the alloys contain atoms of widely different sized, making regular crystal packing difficult. They have usual properties: * Mechanically hard * Very low damping (little energy lost in elastic collisions)

Answer 85

**Copolymers:** More than one monomer polymerised together **Polymer blends:** Molecular-scale mixture of two polymer chains, without cross linking

Answer 86

**The temperature at which the weak secondary bonds are overcome by thermal energy.** This is far lower than the melting point of a crystalline material or glass.

Answer 87

* **Amorphous thermoplastics:** Melt to a viscous liquid (entangled molecules slide over one another). * **Semi-crystalline thermoplastics:** Amorphous regions melt, crystalline regions survive to a higher melting point, Tₘ (typically Tₘ ≈ 1.5 Tg in K), above which a viscous liquid forms. * **Elastomers and thermosets:** secondary bonds melt at Tg but cross-links do not - on heating the polymer does not melt, but decomposes or burns.

Answer 88

* Thermoplastics: Easy to re-mould, weld, and recycle. Viscocity falls with T and can be remoulded above Tg * Elastomers/Thermosets: Mould only once. Cannot recycle (limited-reuse)

Answer 89

The rate of loading: * Deformation above and around Tg relies on molecular sliding, which is sensitive to the rate of deformation * Hence E (and thus Tg) depend on the loading rate * Tg must be defined at a reference loading rate

Answer 90

The Young's modulus of polymers can be manipulated far more in polymers than in metals. This can be by changing the molecular weight, polymer chemistry, and degree of crystallinirt or cross-linking.

Answer 91

They are porous solids

Answer 92

By their **relative density**

Answer 93

Materials that combine 2 different materials to produce new property profiles, exploiting separate qualities of the individual components

Answer 94

* Particulate * Fibres * Laminates

Answer 95

Micron-scale particles are added to the melt before casting or moulding

Answer 96

* Short chopped fibres miced with resin, shape in a mould * Long continuous fibres: Lay out fibre mats in a mould, and infiltrate with resin. Or combine fibre+resin in "prepreg" plies and then laminate

Answer 97

* Wood etc: Stack and glue thin layers * Long-fibre composites: Stack multiple layers of prepreg with different fibre orientations, hot form to shape, curing the resin

Answer 98

By their volume fraction, Vf, of one of their components. e.g. the volume fraction of the particle or fibre reinforcement added to a matrix material.

Answer 99

They follow the rule of mixtures:

Answer 100

It is the rule of mixtures for E

Answer 101

It is the rule of mixtures for 1/E

Answer 102

The **strain** is the same in both layers

Answer 103

The **stress** is the same in both layers

Answer 104

They are **anisotropic**. This means that the material properties differ depening on the orientation of the composite. For example, the stiffness differs depending on if the load is parallel or perpendicular to the layers.

Answer 105

They define the upper and lower limits for the Young's modulus of a composite: The composite moduli must lie between (or on) these limits

Answer 106

* Fibres and laminates are anisotropic * Particulate are isotropic

Answer 107

The **nominal stress** at the **elastic limit**

Answer 108

Due to fracture

Answer 109

Due to crushing

Answer 110

The compressive strength tends to be far high than the tensile strength

Answer 111

By a fraction of their equilibrium spacing

Answer 112

It involves the relative movement of material over very large multiples of the atomic spacing

Answer 113

By applying a shear force

Answer 114

**σₑₗ = E / (10(1+v))** ≈ E/15

Answer 115

The estimate for the ideal strengths of crystalline materials is based on the estimate for the upper limit of the strength from atomic bonds. However there are yielding mechanisms that allow the material to fail at a far lower stress: * **Ceramics: Fracture**, from inherent flaws and cracks * **Metals: Plastic yielding**, enabling deformation that keeps the material intact, at stresses well below the ideal strength The key atomic origin of this yielding mechanism is the **dislocation**.

Answer 116

Dislocations are line defects in a crystal lattice. They are like extra half planes of atoms inserted into the structure.

Answer 117

As there is an extra half-plane of atoms inserted into the structure, the structure must acomodate this atomic displacement: * Part of the "interface" between the top and bottom half of the crystal has "slipped", and part has not * The top half contains an extra half-plane of atoms

Answer 118

It is found at the boundary between slipped and unslipped regions of the crystal

Answer 119

The slip step produced by the passage of one dislocation. **b** is smaller than r₀.

Answer 120

As dislocations cross the lattice, they enable **incremental slip** by extending (and breaking) a few bonds at a time, this is why the stress required to yield crystalline solids is so much less than the ideal strength

Answer 121

The unit vector that points along the line of the dislocation

Answer 122

It describes the closure failure of a circuit throught the crystal that encloses the dislocation line. It is the same as the Burgers vector **b**

Answer 123

Using the **SF** (Start-Finish) Vector: * Start drawing a circuit through the crystal enclosing the dislocation line * Make an equal number of jumps left, down, right, and up * If the circuit is open, directly join the Start atom to the Finish atom with the **SF** (Burgers) Vector Note: If the circuit does not enclose the dislocation line, then there is no slip in this section so the circuit will be closed without a Burgers Vector

Answer 124

* Screw * Edge * Mixed

Answer 125

Dislocations where: * The shear sterss and burgers vector are both at right angles to the dislocation * The dislocation moved in the direction of the stress

Answer 126

Dislocations where: * The shear stress and Burgers vector are both parallel the dislocation * The dislocation moves at right angles to the stress * **The same slip step produced as for edge dislocation** The slip step is in the direction of the Burgers vector (and the shear stress). However, the dislocation itself glides sideways through the crystal. This is how the material shears: each time the dislocation passes through, it leaves behind a shear displacement (the "slip step") in the direction of b.

Answer 127

More generally dislocations are mixed: * Curved, and varying between pure edge and pure screw * Move in a direction normal to the curve under the actions of a shear stress (curved sections expand) * **Net effect remains a slip step in the direciton of the shear stress**

Answer 128

It is best to forget about the atoms and simply think of dislocations as line defects "gliding" across slip planes under the action of imposed shear forces

Answer 129

A dislocation crossing a lattice leads to an incremental slip step (in shear) of the order of one atomic spacing. This can lead to greater plastic strains as: * Crystals contain very many dislocation, with many different planes on which they can glide * In (virtually) any stress state, shear stresses exist to move dislocations The replicating of this incremental slip over thousands of dislocations on multiple slip planes produces continuum bulk plasticity.

Answer 130

Blocks of material slip past one another but the crystal packing is unaffected

Answer 131

The intrinsic lattice resistance to the dislocation motion that comes from additional bond stretching as the dislocation moves each Burgers vector step.

Answer 132

The type of bonding: * Ionic/covalent bonds: High τ₀: high hardness * Metallic bonds: low τ₀: annealed pure metals are soft, metallic alloys are much stronger than pure metals as they provide obstacles to dislocation motions

Answer 133

The energy stored in the lattice due to the presence of a dislocation - atoms around a dislocation are displaced from their equilibrium spacing and thus have a higher energy.

Answer 134

It is like a tensile force acting along the dislocation as to minimise its length

Answer 135

Line tension x dislocation length

Answer 136

* Dislocations try to be as short as possible - it is like they are under tension * Line tension governs how dislocations interact with obstacles

Answer 137

When a gliding dislocation meets obstacles in its slip plane: * It is pinned by the obstacles, and is forced to bow out between them, increasing the resistance per unit length (and thus the dislocation line) * An additional force is required to overcome this resistance and move the dislocation further

Answer 138

* **"Weak" obstacles: Dislocations cut through or pass the obstacle**. The force > obstacle strength (θ > 0°). Resistance force < 2T. * **"Strong" obstacle: Dislocations escape by leaving a loop of dislocation round the obstacle**. The dislocation forms a semi-circle (θ = 0°). Resistance force = 2T.

Answer 139

τ = Contribution to the yield stress due to dislocation pinning α = Obstacle strength G = Elastic shear modulus b = Burgers vector (atomic spacing) L = Obstacle spacing

Answer 140

* α = Obstacle strength * L = obstacle spacing Both of these are variable values which can be manipulated by composition and processing

Answer 141

* **Work hardening**: This uses other dislocation * **Solid solution hardening**: This uses solute atoms * **Precipitation hardening**: This uses particles of another solid (e.g. a compound) * **Grain boundary hardening**

Answer 142

Shear stress = intrinsic resistance + dislocation pinning

Answer 143

k = shear yield stress σᵧ = longitudinal stress needed to move dislocations parallel to a slip plane

Answer 144

Pinning contributions to τᵧ directly translate into constributions to the yield stress σᵧ **by a factor of 3**

Answer 145

Gliding dislocations on different slip planes interact: pinning occurs due to the additional bond distortion at the intersection. The gliding dislocation bows out until the (weak) pinning point gives way, greating a **jog** in the pinning dislocation. Jogs then reduce the mobility of the other dislocations.

Answer 146

L = dislocation separation ρd = dislocation density

Answer 147

By a deformation process (e.g. rolling, wire drawing), to increase the dislocation density while shaping the product.

Answer 148

Most mixtures of metal + other elements for solid solutions. The solute atoms have different size and local bonding to the host atoms in the lattice - they may be considered as toughening the slip plane: * Interstitial solid solutions can also provide hardening by displacing the host atoms from their equilibrium positions - i.e. a similar effect on the slip plane * Solid solutions provide a weak obstacle to dislocations, which bow out until the line tension pulls the dislocation past the solute atom.

Answer 149

L = solute spacing C = solute concentration

Answer 150

Alloying elements also form compounds. When distributed as small **particles** (not individual atoms but rather clusters of them) within a lattice, they provide pinning points for dislocations. Particles provide strong obstacles: the dislocation cannot pass over them, and (usually) the particle lattice is unreated to the surrounding lattice.

Answer 151

The maximum shear stress required to pass (strong) particles is when the dislocation bows out into a semi-circle, leaving a dislocation loop behind.

Answer 152

By their size and volume fraction

Answer 153

Casting is used to mix elements together in liquid state, enabling solid solutions to be manufactured

Answer 154

It is very difficult to manufacture solid particles this asmall, and to mix them into a melt before casting. Therefore the metal may be heat treated in the solid state, forming fine precipitates

Answer 155

Lattice orientation changes at a grain boundary. As a result: * Dislocations cannot slip directly from grain to grain * Dislocations pile-ups occur at the boudnaries * Additional stress from pile-up nucleates dislocations in the adjoining grain

Answer 156

It is given by the Hall-Petch relationship: ## Footnote d = grain size

Answer 157

* The ability of the chain molecules to unravel and slide * Temperature (relative to the glass transition) and the strain-rate * Crystallinity (in thermoplastics), cross-linking (thermosets, elastomers)

Answer 158

Elastic-brittle: * Chain sliding limited * Brittle fracture from ingerent flaws in material * Little or no ductility

Answer 159

Elastic-plastic: * Chain mobility around the glass transition temperature as van der Waals bonds melt * Yielding takes place by **crazing**, **shear yielding**, or **cold drawing**

Answer 160

Microcracks open in tension, bridged by stiff fibres of material aligned with molecules, preventing immediate fracture

Answer 161

Shear bands form, and are stabilised by alignment of molecules; multiple bands form, giving greater ductility

Answer 162

Polymers which do no craze can often be cold drawn. Necking occurs, but the neck is stable: the molecules align in the neck and strengthen it, so the neck spreads along the specimen.

Answer 163

* Below Tg: elastic-brittle * Above Tg: limited shear yielding

Answer 164

* Below Tg: elastic-brittle * Above Tg: non-linear elastic, very large elastic strains to failure little. Little or no ductility: elastic strain is recovered. Elongation to failure is zero.

Answer 165

The stress concentrated at the edge (tip) of a flaw of defect is larger than the remote applied stress σ₀.

Answer 166

For **blunt features** (e.g. drilled (circular) holes). The stress concentration factor for a circular hole (a=r) is 3. Stress concentrations occur at changes in section or threads.

Answer 167

The release of elastic (strain) energy stored in a stressed material. As a crack gets longer the volume of stress-free material "shielded" by it from the applied stress increases. Therefore as a crack grows, it shields more material from stress, releasing energy.

Answer 168

**It is the "rate" at which stored elastic strain energy is released with respect to the creation of new crack area** Note: It is usually given by the symbol G so do not confuse it with shear modulus

Answer 169

## Footnote Units: J/m²

Answer 170

It is a measure of the crack tip loading. It describes the stress near the edge of an infinitely sharp crack

Answer 171

When K reached a critical value, **Kɪᴄ, the critical stress intensity factor**, which is often termed the fracture toughness

Answer 172

When G reaches a critical value, **Gɪᴄ, the critical strain energy release rate**, also known as the fracture energy

Answer 173

* Mode I: Opening (or tensile) mode * Mode II: Sliding (in plane shear) mode * Mode III: Tearing mode Throughout the IA course the opening mode, mode I, is the crack loading mode referred to

Answer 174

Plastic flow at the crack tip can blunt the crack tip. This: * Increases tip radius * Reduces stress concentration * Limits stress to ~yield stress (σᵧ) * Absorbs energy

Answer 175

rₚ = radius of process zone

Answer 176

If the fracture process zone, rₚ, is small compared with specimen and crack dimensions. When rₚ becomes the order of the crack length or specimen size, then the stresses in the process zone are no longer characterised by K

Answer 177

In general, as yield stress increases, fracture toughness decreases

Answer 178

**The temperature below which a material behaves in a brittle manner and above which it behaves in a ductile manner**. At low temperatures, dislocations are less mobile, making ductile materials behave in a brittle manner. At higher temperatures, the dislocations are more mobile, allowing the material to behave in a ductile manner. As temperature falls, dislocations are less mobile, so fracture toughness falls

Answer 179

Dislocation movement allows a material to plastically deform, which helps relieve stress at a crack tip. This plastic deformation can blunt the crack and slow its growth, making the material tougher and delaying fracture. While dislocations don’t stop fracture entirely, they help prevent brittle, sudden failure by allowing the material to absorb more energy before breaking. As dislocations become more mobile, the fracture toughness Kɪᴄ increases.

Answer 180

Ductile fracture: Plasticity concentrates stress on inclusions, nucleating voids that grow and coalesce, ultimately causing fracture. Extensive plastic flow is observed.

Answer 181

Brittle fracture: There is no significant plastic flow. The energy absorbed during fracture is likely to be relatively low

Answer 182

Fracture of ceramics is dominated by their brittle nature. Failure starts from flaws or pores which are introduced during processing. * Tensile failure occurs from the "worst flaw"

Answer 183

Compressive failure is initiated in mode II. This will occur on the planes of maximum shear stress, which are 45° to the loading axis. Failure under compressive loads require a much higher driving force than in tension. For a brittle material the compressive strength is typically much greater than the tensile strength

Answer 184

The main energy-absorbing mechanism raising the toughness of fibre composites is the pulling of fibres out of their sockets in the matrix during crack advance (frictional sliding). This process can make large contributions to the fracture energy of the composite.

Answer 185

The strength at which brittle materials fail depends on the presence of flaws. Consider a chain of links, the survival of the chain under load requires that ALL of the lnks survive. Conversely, failure occurs in the weakest link. The strength of the chain = strength of its weakest link. Designing with brittle materials is about ensuring an acceptably low **risk** of failure

Answer 186

It is a measure of the variability of failure stress: * Low m = Wide spread

Answer 187

According to the weakest link theory: **Pₛ(V) = {Pₛ(V₀)}^(V/V₀)**

Answer 188

On average, a larger sample is more likely to contain one of the larger flaws and therefore fail at a lower stress.

Answer 189

Given that 2 specimens have the same failure (or survival) probability, the specimen with a smaller volume will have the greater failure stress.

Answer 190

Any part of the specimen in compression should be **omitted** from the integral as its strength will be far greater than the tension part.

Answer 191

**The damage or failure of materials due to cycling loading**. Fluctuating loads are more dangerous than monotonic loads

Answer 192

By Δσ-Nf ('Stress-life') curves where: * Δσ = stress range (peak-to-peak) * Δσ/2 = stress amplitude (peak to mean) * σₘ = σₘₐₓ - σₘᵢₙ * Mean stress σ = (σₘₐₓ + σₘᵢₙ) / 2 * Nf = Number of fatigue cycles to failure (N = number of fatigue cycles)

Answer 193

## Footnote R = 1 for a static load R = 0 means that the stress cycles from 0 to σₘₐₓ (zero-tension fatigue) R = -1 means that mean stress σₘ is zero (fully reversed loading)

Answer 194

* Low cycle fatigue * High cycle fatigue

Answer 195

The applied stress amplitude σₑ, about zero mean stress below which fracture does not occur at all, or occurs only after a very large number of cycles (Nf > 10⁷)

Answer 196

The low cycle part of the S-N curve can be expressed using Coffin-Manson law: **Δεᵖˡ (Nf)^β = C₂** ## Footnote It is an empirical description of fatigue for **uncracked components**

Answer 197

The high part of the S-N curve can be expressed using Basquin's law: **Δσ (Nf)^α = C₁** ## Footnote It is an empirical description of fatigue for **uncracked components**

Answer 198

Using: * The Coffin-Manson law for low cycle fatigue * Basquin's law for high cycle fatigue

Answer 199

The greater the mean stress value, the lower the fatigue life

Answer 200

Goodman's empirical rule allows for the effect of mean stress on S-N data. The rule relates the stress range Δσ(σm) for failure under a mean stress σₘ to that for failure at stress range Δσ₀ and zero mean stress, according to the relationship: **Δσ(σm) = Δσ₀ (1 - σₘ/σₜₛ)** where σₜₛ is the tensile strength, giving a correction to the stress range. It describes how the mean stress and alternating stress combine to affect the fatigue life of a material - an increase in mean stress reduces the allowable alternating stress before fatigue failure occurs. That means a smaller cyclic (alternating) stress can become just as damaging as a larger one if it rides on top of a higher mean stress.

Answer 201

Most components are subjected to varying cyclic stress ranges, a simple criterion for predicting the fatigue life in a loading sequence consisting of various blocks of different stress ranges (all about zero mean) is predicted by Miner's rule. Miner's cumulative damage rule states that the specimen fails when the proportion of the life time used up by each block adds up to: **Σ Ni/Nfi = 1** where Ni is the number of cycles corresponding to the ith block of constant stress range Δσi, and Nfi is the number of cycles to failure at that stress range. ## Footnote Miners rule ignores load sequence events and is applicable primarily for high-cycle fatigue

Answer 202

You must use Goodman's rule to account for mean stress effects! Miner's rule is for various blocks of different stress all about a zero mean. Therefore if a block of stress has a non-zero mean, then you must use Goodman's law to determine the equivalent alternating stress with zero mean which has an equal fatigue life (this will be greater range).

Answer 203

A constant Δσ gives an increasing ΔK as the crack length increases

Answer 204

**You do NOT use goodman's rule** to account for mean stress. This is because Paris's law describes the crack propagation stage, and during the crack propagation stage the dominant factor governing fatigue behavior is not the mean stress, but the range of stress intensity at the crack tip. Therefore, Goodman's rule should be used for the crack initiation stage, but not for the crack propagation stage (when Paris' law is used).

Answer 205

Where: a = crack length N = cycles ΔK = tensile stress intensity range A, n = constants

Answer 206

The empirical relationship that describes how a fatigue crack grows in a material under cyclic loading. It applies after a crack has already initiated and is in the steady state crack propagation stage - before fast fracture.

Answer 207

* **Region I - Crack Initiation**: Crack growth per cycle is zero below a threshold cyclic stress intensity factor ΔKₜₕ. Above it, crack growth per cycle is very small (smaller than one lattice spacing). * **Region II - Steady-State Crack Propagation** described by the Paris law: da/dN = AΔKⁿ where A and n are constants * **Region III - Fast Fracture:** At high ΔK, crack growth rate increases rapidly. As Kₘₐₓ approaches the Kɪᴄ, fast fracture occurs.

Answer 208

The total number of cycles to failure. This can be taken as the sum of the number of cycles for crack initiation Ni, and crack propagation Np: ## Footnote Note: The contribution of the final failure step to the total fatigue life is insignificant since it occurs so rapidly.

Answer 209

* For low cycle fatigue, stress levels are high (σₘₐₓ or σₘᵢₙ > σᵧ) so intitiation is easy. Crack propagation takes up most of the fatigue life (Np > Ni) * For high cycle fatigue, stress levels are low (σₘₐₓ & σₘᵢₙ < σᵧ) so intiation is hard. A large fraction of the fatigue life is utilised in crack intiation (Ni > Np)

Answer 210

* Shot peening: Work hardening + residual compressive stresses * Surface hardening: increase hardeness (i.e. yield stress)

Answer 211

* "Clamshell marks" * Striations Both of these features indicate the postion of teh crack tip at some point time and appear as concentric ridges that expand away from the crack initiation site(s), frequently in circular or semi-circular pattern

Answer 212

Clamshell marks are a characteristic of a fracture surface formed during fatigue crack propagation. They are of macroscopic dimensions and may be observed with the unaided eye. They are produced by a change in crack growth conditions, such as a change in stress level or a pause in stress cycling growth rate.

Answer 213

Striations are a characteristic of a fracture surface formed during fatigue crack propagation. They are microscopic in size and can be observed under high magnification. The striations are produced as the crack advances over one cycle, i.e. each striation corresponds to da. Clamshell marks may contain thousands of striations.

Answer 214

* σₕ = Pr/t * σz = Pr/2t

Answer 215

In pressurised cylinders, the main concern is to avoid catastrophic rupture, particularly when the pressured fluid is a gas - in which case an explosion is likely to result. Such rupture commonly occurs in the form of fast crack propagation. Therefore the "l**eak-before-break**" criterion is often used: * If a𝒸ᵣᵢₜ (critical flaw size) < t (vessel wall thickness) - then fast fracture will occur * But if a𝒸ᵣᵢₜ (critical flaw size) > t (vessel wall thickness) - then gas will leak out through the the crack before the crack is large enough to propagate under fast fracture conditions ## Footnote a𝒸ᵣᵢₜ: critical flaw size for fast fracture (K = Kɪᴄ)

Answer 216

* Yσ √(πt) ≤ Kɪᴄ and when t is substituted for the higher (hoop) stress: * Y √(σπPr) ≤ Kɪᴄ ## Footnote NOT IN DATA BOOK

Answer 217

1. The maximum possible initial crack size a₀ is estimated using fracture mechanics 2. An estimate of the minimum number of cycles Nf that the vessel can last before cracks of length a₀ (previously estimated) would grow to a length af at which fast fracture or leakage occurs 3. This marks the point where the crack becomes unstable and rapid fracture or leakage occurs.

Answer 218

The true contact area Aₜ is considerably lower than the nominal area of contact Aₙ. This is because surfaces are never flat at a microscopic level and so contact between surfaces intially occurs only at a few points called asperities.

Answer 219

The contacting asperities in metals flatten by localised yielding. The area of contact aₖ at each asperity is related to the local load wₖ by wₖ = Haₖ (H = hardness). This leads to a relationship between the macroscopic load W and the true contact area Aₜ: **W = HAₜ** ## Footnote Aₜ ≤ Aₙ

Answer 220

The contacting asperities flatten elastically - they do not yield. At small loads, AₜThe contacting asperities flatten elastically - they do not yield. At small loads, Aₜ increases non-linearly with load. At high loads, Aₜ can approach Aₙ

Answer 221

The contacting asperities fracture due to their brittle nature

Answer 222

When the contact points in metals flatten they form adhesive junctions. The surfaces will start sliding if the shear stress τₖ on each junction reaches the material shear strength τₛ of the material (if the surfaces are made of different materials, then the shear stress should be equal to the shear strength of the softer material). Considering that H ≈ 3σᵧ, τₛ ≈ k = σᵧ/2, the total shear force F can be written as:

Answer 223

* Shear force (frictional force) F is proportional to the normal load W * Frictional force F is independent of the nominal area of contact Aₙ μ = F/W

Answer 224

They form very large frictional forces. This is because after sliding has occured, the true area of contact becomes very large (approaches the nominal area of contact). If the area of contact approaches the nominal area of contact, the frictional shear stress tend to the yield shear stress; this limit is called sticking friction.

Answer 225

Oxide layers: The stable state of all metals (except gold) is in the form of oxides. So it is the oxide layers that are resposnible for the shear resistance:

Answer 226

In friction welding, mechanical frictions is used to increase the temperature at the interface between the two objects to be welded. Combined with the imposition of a large nomal loading at the interface, the two parts fuse together * **Spin welding**. Most commonly used with metals. A piece is rapidly rotated and pressed against a static part. In the first phase, friction causes heat and removes the oxide layer. Sticling friction then takes place and remodels the interface until they fully fuse. This process is favoured when melting would have a damaging effect on the metal microstructure. * **Vibration and ultrasonic welding**. Common techniques for welding thermoplastics. Both methods use vibration to create friction, but the direction, amplitude anf requency of vibration are different.

Answer 227

Q is proportional to W/H (i.e. to the true area of contact Aₜ) ## Footnote K = wear coefficient, which depends on how the material is pulled off at the asperity contacts

Answer 228

In adhesive wear small bits shear off every time yield occurs at the asperity tips. This usually occurs when the materials are the same/similar.

Answer 229

Abrasive wear is when two surfaces of very different hardness are brought into contact with each other. Material comes off as hard asperities plough through the soft material. The rate of abrasive wear is usually much larger than the rate of adhesive wear.

Answer 230

Films of low shear srength deliberately inerposed between moving surfaces in contact to reduce friction and wear. Coefficients of friction tend to be in the range 0.001-0.1 . The lubricant film can be solid (e.g. PTFE, graphite), liquid (e.g. oil) or a combination of both. Besides reducing wear, lubrication also ensures a decrease in energy dissipation (and thus heat generation)

Answer 231

In static or low velocity situations, there is contact between the shaft and the bearing. This is called boundary lubrication.

Answer 232

As velocity increases, the shaft becomes increasingly supported by the oil phase. The thickness of the oil layer increases, as well as the oil pressure. Eventually the pressure is so high that the shaft is fully supported by the oil. This regime is called hydrodynamic lubrication.

Answer 233

Optimal lubrication is reached when the stribeck number reaches an optimal point. Friction decreases with the Stribeck number as the oil layer is getting thicker and thicker, taking progressively a larger proportion of the shaft's load. The friction coefficient will reach its minimum (optimum) and there is a transition to hydrodynamic lubrication. At this point, the load on the interface is entirely supported by the fluid film. It is also the point at which the wear rate goes almost to zero. Note: The coefficient of friction increases in the hydrodynamic region due to fluid drag.

Answer 234

η = oil viscosity V = sliding speed at the interface P = pressure due to shaft loading

IA: 1P2: Materials Flashcards

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