Chapter 5 - Mechanical Properties Flashcards

Question

What is the TE

Answer 1

These are stresses that remain in a solid material after the original cause of the stresses has been removed. Sources include inelastic (plastic) deformation, temperature gradients (during thermal cycle) or structural changes (phase transformation).

Answer 2

Strength is the ability of a material to withstand an applied load while toughness is the ability to absorb energy without fracture

Answer 3

1. Elastic 2. Stable Plastic (work hardening) 3. Unstable Plastic (necking)

Answer 4

The true stress-strain curve because the cross sectional area as the experiment proceeds is taken into account meaning A goes down and sigma=P/A=goes up

Answer 5

applied loads to non permanent displacements and strains.

Answer 6

six; three normal and three shear

Answer 7

The fractional displacements

Answer 8

stress vs strain

Answer 9

- equilibrium conditions ensuring force balance on a volume element - displacement and strain relations that ensure the solid remains continuous as it defors - constitutive equations that related stresses and straines

Answer 10

less than the yield stress and strain less than a few percent (removable, non-permanent deformation).

Answer 11

mechanical property that measures the tensile stiffness of a solid material. It quantifies the relationship between tensile stress and axial strain in the linear elastic region, E = sigma/e where sigma is stress and e is strain. In other words it is the slope of the linear part of the stress strain curve.

Answer 12

measures the phenomenon in whic ha material tends to expand in directions perpedincular to the direction of compression.

Answer 13

0.5 and 0 meaning cork has very little compression.

Answer 14

it is the ratio of shear tress to sheear strain.

Answer 15

The atomic binding

Answer 16

the extent at which an object resists deformation in response to an applied force.

Answer 17

Isotropic mechanical properties meaning same properties in al direction.

Answer 18

anisotropic mechanical properties meaning different properties depending on direction

Answer 19

it states that the force (F) needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance—that is, F = kx where k is a constant factor characteristic of the spring (i.e., its stiffness), and x is small compared to the total possible deformation of the spring.

Answer 20

Exponentially go down.

Answer 21

In length axis due to the azimuthal strain

Answer 22

``` Type I: Material with no YS Type II: Material with YS Type III: Upper and lower YS Type IV: Ideal plastic behavior Type V: Small Plastic Behavior Type VI: portevin le chatellier effect. ```

Answer 23

Is a measure of the resistance of a materials against permanent shape change.

Answer 24

It is the reversible displacement of crystal planes relative to each other. Original structure is recovered when the stress is removed.

Answer 25

is the permanent slippage of crystal planes, which occurs above a critical stress. When many parallel atomic planes slip relative to each other, macroscopic deformation results

Answer 26

Because of dislocations?

Answer 27

The single crystal will plastically deform by shear on | the plane with the highest resolved shear stress.

Answer 28

The Schmid factor describes the relationship between external normal stresses and the shear stresses caused in a slip system!

Answer 29

Melting temperature is also an indicator of atomic bonding strength and there is relationship with Young's modulus. The general trend is that a higher melting temperature indicates a higher modulus and vice versa.

Answer 30

Young’s modulus, aka elastic modulus is more or less the stiffness of a material. In rough terms, it is the change in stress caused or divided by a change in strain. The bigger the modulus, the stiffer the material. Coefficient of thermal expansion is the change in strain caused by a change in temperature. The two concepts tend to be inverse.

Answer 31

A point at which Maximum load or stress required to initiate the plastic deformation of material such point is called as Upper yield point. And a point at which minimum load or stress required to maintain the plastic behavior of material such a point is called as Lower yield point.

Answer 32

Since there is no define slip system due to the random orientation of grains in the pollycristalline system then it isusually stronger than signle crystals. The crystal with the largest resolved shear stress (Tr) yields first and others later.

Answer 33

The Hall–Petch relationship tells us that we could achieve strength in materials that is as high as their own theoretical strength by reducing grain size.

Answer 34

Impurities like carbon and nitrogen can pint dislocations leading to an upper yield point. Once dislocations break loose from pinning points they start moving but still encountering pinning pints.

Answer 35

Ductile, Brittle, Fatigue, Creep

Answer 36

Necking, formation of pores at impurities/inclusions, joining of small pores to a crack, and final rupture at the highest shear stress in an angle (max shear stress). smooth

Answer 37

Initial dislocation motion, moving dislocations fruther costs more energy than nucleating a crack, brittle fracture-cracking. Dislocations start to pile up. sudden break withouth the plastic section.

Answer 38

Compression tests.

Answer 39

weakening of am aterial caused by cyclic loading that results in progressive and localized structural damage and the growth of cracks. Once a fatigue crack has initiated, it will grow a small amount with each loading cycle, typically producing striations on some parts of the fracture surface.

Answer 40

When steel has been strained (deformed plastically) and then allowed to age, it has been subjected to what is known as strain ageing

Answer 41

Aging is an essential step that ensures that the materials in the alloy do not revert to their original configuration after a time period. Aging is performed under controlled conditions so that the resultant grain structure will create a greater tensile strength in the metal than in its former state.

Answer 42

In materials, the motion of dislocations is a discontinuous process. When dislocations meet obstacles during plastic deformation (such as particles or forest dislocations), they are temporarily arrested for a certain time. During this time, solutes (such as interstitial particles or substitutional impurities) diffuse around the pinned dislocations, further strengthening the obstacles' hold on the dislocations. Eventually these dislocations will overcome the obstacles with sufficient stress and will quickly move to the next obstacle where they are stopped and the process can repeat

Answer 43

Dynamic strain aging (DSA) is a process where aging is sufficiently rapid to occur during straining and it produces a variety of inhomogeneous deformations which are characterized by terms such as Portevin-le Chatelier effect, serrated yielding, jerky or serrated flow, blue brittleness, etc.

Answer 44

wohler diagram plots the stress and the number of cycles to failure

Answer 45

Creep is a time (also stress and temperature) dependent mechanical deformation of solids which occurs slowly at stresses below the ultimate tensile stress. The time dependence differentiates creep from plastic deformation.

Answer 46

Yes, they can climb due to the vacancy movement and this allows dislocations to climb over obstacles.

Answer 47

A form of diffusion creep in which atoms migrate along grain boundaries

Answer 48

refers to the deformation of crystalline solids by the diffusion of vacancies through their crystal lattice. Diffusion creep results in plastic deformation rather than brittle failure of the material.

Answer 49

Diffusion creep is caused by the migration of crystalline defects through the lattice of a crystal such that when a crystal is subjected to a greater degree of compression in one direction relative to another, defects migrate to the crystal faces along the direction of compression, causing a net mass transfer that shortens the crystal in the direction of maximum compression. The migration of defects is in part due to vacancies, whose migration is equal to a net mass transport in the opposite direction.

Answer 50

is a mode of deformation of crystalline materials (and amorphous materials) that occurs at low stresses and held at elevated temperatures in fine-grained materials. In Nabarro–Herring creep (NH creep), atoms diffuse through the crystals, and the creep rate varies inversely with the square of the grain size so fine-grained materials creep faster than coarser-grained ones.

Answer 51

Time, temperature, stress

Answer 52

is a means of predicting the lifetime of material vs. time and temperature using a correlative approach based on the Arrhenius rate equation

Answer 53

grain size since grain boundary sliding and grain rotation is resposible for this phehonema due to diffusion at grain boundaries.

Answer 54

is a standardized high strain-rate test which determines the amount of energy absorbed by a material during fracture. Absorbed energy is a measure of the material's notch toughness. The qualitative results of the impact test can be used to determine the ductility of a material. If the material breaks on a flat plane, the fracture was brittle, and if the material breaks with jagged edges or shear lips, then the fracture was ductile. Usually, a material does not break in just one way or the other and thus comparing the jagged to flat surface areas of the fracture will give an estimate of the percentage of ductile and brittle fracture

Answer 55

No, they are uniaxial.

Answer 56

The von Mises stress is used to predict yielding of materials under complex loading from the results of uniaxial tensile tests.

Answer 57

No, there is a gradient from center line to surface.

Answer 58

Macrooscopic plasticity or cracking

Answer 59

Radiallly oriented (perpendicular to the angle direction, perpendicular to the cladding surface point) since the sstress going outwards is higher than the stress going on height.

Answer 60

is defined as the change in dimensions of a material (based on the original dimensions) as a result of an applied stress.

Answer 61

Strains induced by pellet thermal expansion.

Answer 62

Hour-glassed

Answer 63

ratio of thermal stress to failure stress where failure is either fracture for ceramics or yielding for ductile metals.

Answer 64

a parameter that mesaures the resisntace to thermal stress failiure which is a function of the failiure stress (yield or fracture), the thermal conductivity and the thermal expansion coeffcieitn. The larger the more thermal stress resistant

Answer 65

Large failure stress, high thermal conductivity and l ow thermal expansion coefficient

Answer 66

force applied per unit area, typically measured in 𝑀𝑃𝑎=106 𝑁/𝑚2:

Answer 67

stress (sigma) = F/A where: F = the force applied normal to the sample’s cross-sectional area A = the cross-sectional area of the sample

Answer 68

sample displacement per unit length, unitless (often quoted as %)

Answer 69

e = deltaL/L where L is the length and deltaL is the change in length.

Answer 70

positive strain

Answer 71

negative strain

Answer 72

completely recoverable deformation where the body returns to its original shape once the external forces are removed. Atomic bonds are stretched, but not broken.

Answer 73

linear, hookes law stress (sigma) = E*strain where E is the youngs modulus/elastic modulus

Answer 74

by the interatomic forces in the material, and is dependent on crystal structure and orientation in the material, however it is not very affected by alloying, cold working, or heat treatment.

Answer 75

The Young’s modulus does decrease with increasing temperature as the interatomic forces become weaker.

Answer 76

When a material is stretched in one direction, it tends to contract in the other two directions (or vice versa for compression). This relationship is captured with Poisson’s ratio, 𝜈.

Answer 77

v = - strain on x / strain on z if pulled on the z direction

Answer 78

permanent deformation, which does not recover once the forces are removed.

Answer 79

through dislocation motion which facilitates the slipping of lattice planes against one another

Answer 80

popular method of studying short-term mechanical behavior (strength and ductility) of a material. The test is performed with a quasi-static uniaxial tensile stress state (the sample is being stretched along one direction at a sufficiently slow and usually constant rate – aka, no sudden changes)

Answer 81

does not take into account the fact that the cross sectional area and length of the specimen are not constant throughout the test.

Answer 82

1. elastic regime 2. yield point 3. plastic regime (uniform strain) 4. plastic regime (nonuniform strain)

Answer 83

the initial linear regime of the curve where the material experiences elastic deformation only.

Answer 84

Youngs modulus

Answer 85

The point where the deformation is nonlinear. In other words is marks the transition from elastic to plastic deformation. The yield points value is the yield stress

Answer 86

where plastic deformation occurs also reffered to as strain hardening regime. Here, the cross sectional area decreases in a uniform manner and ocntinues until it reaches a maximum aat the Ultimate Tensile Strength (UTS).

Answer 87

The transition between uniform and non uniform plastic deformation. It also marks the point where necking occurs.

Answer 88

This occurs when dislocation movement gets impeded by interstitial atoms such as carbon, nitrogen, and so on forming solute atmospheres around the dislocations. However, at very high stress, the dislocations break away from the solutes and thus require less stress to move at this point.

Answer 89

The elongation that occurs at constant load or stress at the lower yield point

Answer 90

During yield point elongation, a type of deformation bands known as “Lüders bands” are formed across the yield elongation regime

Answer 91

After the Lüders bands cover the whole gauge length of the specimen, the usual strain hardening regime sets in.

Answer 92

true stress strain takes into account the change in cross sectional area and length as the experiment carries on.

Answer 93

the true stress to continue increasing past necking as the test is continued. meaning the true curve is higher and the fracture usually happens with more strain.

Answer 94

due to the decreased cross sectional area that is now being put in tension.

Answer 95

The brittle has a steeper elastic regime and a higher fracture point but all happening at much less strain

Answer 96

1. Square like sudden break 2. elongated diagonal break 3. pen-like and middle flaw

Answer 97

is a measure of the degree of plastic deformation that can be sustained at fracture.

Answer 98

has very low ductility, meaning it tends to fracture at low strains after little plastic deformation. Brittle materials are considered to be those with a fracture strain of

Answer 99

1. equilibrium conditions 2. relations between displacements and strains 3. constitutive equations

Answer 100

method to determine which combination of stress components is equivalent to a uniaxial stress

Answer 101

It is the PR/tt pressure * tube radius/tube thickness

Answer 102

the azimuthal force is two times larger than the axial force

Answer 103

Because the azimuthal force is 2 times larger than the axial force.

Answer 104

The azimuthal force the normal force, the reference temperature and the thermal expansion coefficient

Answer 105

he distribution of crystallographic orientations in a polycrystalline material. no texture = the orientations are fully random, there is no preferred orientation strong texture = material has a preferred orientation

Answer 106

with X-ray diffraction, EBSD (electron backscatter diffraction) in an SEM (scanning electron microscope), and other diffraction techniques

Answer 107

Textrue introduction and elongated grains along the rolling direction resulting in direciton dependent properties

Answer 108

resistance to indentation

Answer 109

characterize mechanical properties in materials, particularly the resistance to deformation. It does not involve total destruction of the sample as needed in tensile testing, and generally requires only a small volume of material.

Answer 110

Hardness is not the same as strength, however hardness values are generally proportional to the strength values obtained in tension or compression tests.

Answer 111

If dislocations are able to move in a crystal with relative ease, it means the material does not intrinsically offer any resistance to the dislocation movement and thus would be less strong. But if the microstructure is laden with various obstacles, dislocation movement will be effectively impeded and this movement obstruction will be translated into an increase in strength.

Answer 112

Obstacles to dislocations can be dislocations themselves (strain hardening), grain boundaries (Hall-Petch and grain size strengthening), solute atoms (solid solution strengthening), precipitates (precipitation strengthening), and dispersions of fine stable particles (dispersion strengthening).

Answer 113

The hardneing of a material by the introduction of dislocations. When a metal is cold-worked, its strength increases. As dislocation density is increased, the movement of dislocations becomes increasingly difficult due to the interfering effect of the stress fields of other dislocations. In the early stages of plastic deformation, slip is generally limited on primary glide planes and the dislocations tend to form coplanar arrays. However, as the deformation proceeds, cross-slip takes place and dislocation multiplication mechanisms start to operate. The cold worked structure then forms high dislocation density regions or tangles that soon develop tangled networks

Answer 114

annealed crystal contains a dislocation density of about 10E8 m−2. Heavily cold worked materials may contain 10E14 to 10E16 m−2.

Answer 115

Grain boundaries can act as obstacles to dislocation motion. As grain orientation changes at the grain boundary, slip planes in a grain get disrupted at the grain boundary. Thus the dislocations gliding on a slip plane cannot burst through the grain boundary. Instead, they get piled up against it therefore limiting dislocation movement and causing hardening.

Answer 116

A greater stress concentration is created at a larger grain, which has enough energy to push through to the next grain. For smaller grains, the stress concentration is not great enough to produce slip in the next grain readily, thus fine grained materials exhibit higher yield strength.

Answer 117

mathematical relationship between yield stress and grain size whic his a fucntio of the friction stress or yield strengh at infinite grain size and the unlocking parameter which measures relative hardenign contributio of grain boudaries and the grain diameter.

Answer 118

Dislocations encounter barriers on their path from the solutes present in the host lattice.

Answer 119

Dispersed particles impart significant obstacles on dislocation motion. When the particles are small (<5 nm), and/or soft and coherent, dislocations can cut and deform the particles (“particle shearing”). For larger precipitates, the phenomenon of Orowan bypassing occurs, in which the dislocations form loops around the particles.

Answer 120

in which the dislocations form loops around the particles.

Answer 121

less than 5 nm

Answer 122

When the particles are small (<5 nm), and/or soft and coherent, dislocations can cut and deform the particles (“particle shearing”).

Answer 123

are used to measure the resistance to failure of a material under a sudden applied force. They are intended to study the most severe conditions relative to the potential of fracture

Answer 124

Impact tests measure the impact energy, or the energy absorbed prior to failure.

Answer 125

The elastic stress field around a crack tip it depends on the geometry of the crack-containing solids, the size and location of the crack, and the magnitude and distribution of the loads applied.

Answer 126

Rapid unstable failure

Answer 127

1. Opening mode displacement perpendicular to crack faces 2. displacement parallel to crack faces but perpendicular to the edge 3. tearing mode parallel to the crack faces and to the leading edge

Answer 128

The critical stress intensity factor Kic Kic = Y*sigma*sqrt(pi*ac) It depends on the specimen and crack geometry, the applied stress (sigma), and the critical crack length (ac).

Answer 129

the maximum crack length

Answer 130

In load bearing structure exposed to elevated temperatures for a long duration of time.

Answer 131

Time dpeendent plastic strain at constant temperature and stress

Answer 132

above half the melting temperature.

Answer 133

A creep test is generally done under uniaxial tensile stress. A creep curve is plotted as creep strain vs. time at a constant load and temperature

Answer 134

1. Primary stage (transient creep) 2. secondary or second stage creep (minimum creep rate) 3. tertiary stage

Answer 135

1. the strain hardening tthat occurs when something is plastically deformed due to dislocations getting tangled in each other. 2. softening that occurs as a result of the creep which causes easy plastic deformation to occur.

Answer 136

In the first stage work hardening during plastic deformation rules over recovery/softening mechanisms therefore exhibiting decreasing strain rate with time.

Answer 137

This strengthening occurs because of dislocation movements and dislocation generation within the crystal structure of the material.

Answer 138

1. dislocation creep 2. nabarro=herring creep 3. coble creep

Answer 139

At high stresses, creep is controlled by dislocation motion. This is accomplished by both dislocation glide and climb, however climb is the rate-controlling element in creep. The glide motion of dislocations is impeded by long range stresses due to dislocation interactions. These stresses can be relieved by dislocation climb and subsequent annihilation.

Answer 140

Creep due to the diffusion of vacancies through the lattice (through grains)

Answer 141

Creep due to the diffusion of vacancies along grain boundaries.

Answer 142

Because creep mechanisms like nabarro herring and coble creep mechanisms depend on the diffusion of vacancies. Vacancy concentration increases with irradiation.

Answer 143

With high temperatures sinc eit requires energy to transprot vacancies through grains than along grain boundaries. Irradiation can help.

Answer 144

weakening of a material caused by repeated applied loads (cyclic loading)

Answer 145

If the stress amplitude (S) is plotted against the number of cycles to failure (N). It represents the number of cycles needed for failiure given a stress.

Answer 146

For some ferrous materials and titanium alloys, the S-N curve becomes horizontal at higher N values, or there is a limiting stress level called the “fatigue limit” or “endurance limit”, which is the largest value of fluctuating stress that will not cause failure for essentially an infinite number of cycles.

Answer 147

As the stress increases the number of cycles decreases. As the stress decreases the number of cicles increases.

Chapter 5 - Mechanical Properties Flashcards

(173 cards)