Basics

Question 1

Define load, deflection, stiffness, torsional stiffness and compliance

Answer 1

Load - force that acts in a structure
Deflection - response of a structure due to a load (deformation)
Stiffness = load/deflection
Torsional stiffness = torque/angular deflection
Compliance - a structure with low stiffness

Question 2

Explain poisons effect

Answer 2

As you pull a material it will extend in length and decrease in cross-area
When you compact a material it will decrease in length but increase in cross-area
AoLo = A1L1 (as volume is constant)

Question 3

How is poisons effect removed?

Answer 3

Use stress and strain instead

Question 4

Describe shear, torsion and torque

Answer 4

Stress acting in opposite directions can cause shear
Shear modulus = shear stress/shear strain
Torque - force acting away from centre, T = Fr
Torsion - twisting force on material (usually tube)

Question 5

Draw cantilever bending point loaded and cantilever bending uniformally loaded

Answer 5

Point Loaded - One end is fixed, other has load applied

Uniformly - fixed at both ends, load through out = bending in middle

Question 6

Define the area moment of inertia

Answer 6

Resistance to deflection from a shape
Rod - Ix = Iy = 0.25πr^4
Tube - Ix = Iy = 0.25π(r2^4-r1^4)
Column - Ix = Iy = (b.h^3)/12

Question 7

How do asymmetric shapes affects inertia?

Answer 7

Stiffer in wider axis = higher inertia

Question 8

Describe a tensile test (inc measurements needed)

Answer 8

Measure gauge length, and area of specimen, put on 50mm gauge markers, clamp one end into grip - move machine to grip other end in
Constant load increase - measure load vs extension then plot a stress vs strain graph - will be for engineering stress and strain (assumes area and length constant) as necking occurs

Question 9

How do you reliably find Ym of a material?

Answer 9

Strain gauge - when conductors are strained their resistance increases (can be used to give strain), only five measurements for localised strain (2mm)
Extensometers - measure accurate extension of the material and general strain

Question 10

Why is a tensile test not an accurate way of measuring the Young’s modulus?

Answer 10

Machine elastically deforms as well so not reliable values, must use extensometer or strain gauge

Question 11

What values can be calculated from a tensile test?

Answer 11

UTS - highest stress material can withstand
Yield point - when plastic deformation begins
Young’s modulus - gradient of elastic deformation (only estimate)
Elongation to failure = Δx/x - how long material extends before failure

Question 12

What’s the difference between 0.2% proof stress and yield stress?

Answer 12

0.2% proof stress is used when yield point is not obvious, and is found using a tangent at 0.2% strain on stress vs strain graph

Question 13

Define torsional stiffness

Answer 13

Torque per angular displacement

Question 14

Define poisons ratio and it’s significance

Answer 14

Poisons ratio = -Δεx/Δεy
Where x is strain normal to stress axis, y is strain in stress axis
A poison ratio of 0.5 will conserve volume under load (metals have a ratio of roughly 0.3)

Question 15

Define the bulk modulus

Answer 15

The compressibility of a material

K = Young’s modulus/3 - 6.poisons ratio

Question 16

What happens to the modulus of metallic alloys in impacts?

Answer 16

Modulus increases with strain rates - quick strain rates = higher modulus

Question 17

What are the error sources for a tensile test?

Answer 17

Misshapen specimens 
Incorrect alignment (shear occurs)
Poor surface finish (early onset necking)
Poor gripping 
Internal defects (cause weaknesses)

Question 18

How do you measure the Young’s modulus of brittle materials?

Answer 18

Can’t use tensile tests as grips damage material
Would use 3/4 point bending tests (more accurate)
Test gives load vs extension which can be converted to stress vs strain graph

Question 19

Why is hardness testing done?

Answer 19

Most testing is destructive (can’t use material again) but some components need testing before they go into service - hardness testing is used

Question 20

Describe the 3 types of hardness testing

Answer 20

Brunel test - small spherical indenter is pressed into material, indent size = Brunel hardness
Rockwell scale - pyramid diamond used instead as less damage to material
Vickers teat - same process as Rockwell but Hv = 1.854load/mean length^2

Question 21

What are common errors and things to avoid in hardness testing?

Answer 21

operator judgment of indent can be wrong, surface should be strain free, at least 5 indents to measure hardness, causing work gardening around indent = indents should be 5.diameters apart, indents should be 3d in from edge (strain field only in material), leaves small work hardened areas in material

Question 22

Describe the difference between macro and micro hardness tests

Answer 22

Micro hardness tests test coatings/surface layers = reduced force needed = elastic recovery = micro stress appearing > macro stress
Indent load = resistance to penetration . d^metals resistance to strain

Question 23

What factors need to be considered for material selection?

Answer 23

Material properties (Tm, Ym, strength) 
Fixed parameters (length/size)
Variable parameters (mass, area etc)
Then corrosion resistance & processing requirements

Question 24

Describe the process of developing a performance index

Answer 24

Draw loading situation and material shape
Work out lowest strength deformation mode
Write out fixed & variable parameters and what needs to be max/minimised
Eliminate variable parameters from equations
Simplify and separate material properties - select material

Question 25

How does the shape of a component affect material selection?

Answer 25

Different shapes have different failure modes and therefore different material properties are needed

Question 26

Describe the structure and workings of a blast furnace

Answer 26

- iron ore, coke and limestone loaded into top of furnace, CO reacts with FeO to leave Fe - limestone reacts with impurities and excess oxygen to form slag layer but Fe has highest density so pig iron is tapped off at bottom periodically (bottom of furnace is hottest)

Question 27

How is pig iron turned into steel?

Answer 27

Pig iron still contains many impurities - desulphurisation (Mn added to replace S in iron) - O slowly added to remove C from iron - P is removed - iron is degassed - Alloying elements added to give specific composition

Question 28

Compare basic oxygen furnace to electric arc furnace for steel making

Answer 28

BOF - 800tonnes/hour, 50% scrap steel & 50% pig iron | EAF - can use 100% scrap but higher energy requirements

Question 29

Describe aluminium making process

Answer 29

Bayer process: - Al found in bauxite (also contains iron) - heated in furnace, iron rich material removed as ‘red mud’ - Refined using electrolysis (@1000°)- carbon anode, graphite cathode, Al and slag layer in between (Al tapped off)

Question 30

How can metal production process reduce energy requirements

Answer 30

- scrap reduces energy requirements and requires less contaminants to be removed - insulating furnace - using more efficient heating methods

Question 31

Describe the alloying process and the benefits associated

Answer 31

Carried out during liquid phase of metals to get complete diffusion, but solidification can cause segregation Alloying increase Tm, Ym and UTS due to disrupting regular layers of metal (also decreases ductility)

Question 32

What are the two different casting routes and detail some methods

Answer 32

``` Shapes casting (final shape): sand, die, centrifugal, investment casting Semi-finished casting: continuous and ingot (then need processing into final shape) ```

Question 33

Describe alloy solidification

Answer 33

When cooling rate < diffusion rate = homogeneous structure When cooling rate > diffusion = heterogenous structure = worse properties Higher cooling rate = cheaper production = standard production method

Question 34

Explain segregation profiles of an alloy that’s been moulded

Answer 34

Solidification is most likely at an interface (as reduced energy requirements) means that will start at mould wall, fast cooling rate = dendritic growth and solute is pushed out from dendrite as solidifies Means that centre of mould has diff comp to outside = segregation

Question 35

Describe and explain the microstructure of ingot casted material

Answer 35

Chill zone (nucleation points on mould wall), columnar zone region (dendrites have grown), uneaxial zone (secondary dendrite arms swept into melt pool and act as nucleation points = solidification in centre = stops dendrite growth)

Question 36

What grain shape is more beneficial and how is this achieved?

Answer 36

Short and random grains = homogenous = good mechanical properties but not ductile (long grains opposite) Increased equiaxed zone by slower cooling, inoculation (adding ceramic nucleation points), ultrasonic vibration (breaks off dendritic arms) and lower super heat (material less heat to lose before solidifying)

Question 37

Explain gravity segregation

Answer 37

Denser liquid sinks through casting = higher solute at bottom, or denser dendrites move through liquid = increased grain size Stopped by insulating to stop reface solidification

Question 38

Explain ‘A’ and ‘V’ segregation

Answer 38

Dendrites follow flow patterns through material and are re-melted/broken by rich liquid, Liquid then gets trapped and sets up melt pool = solute rich areas when solidify Stopped by preventing flow patterns from establishing

Question 39

Explain shrinkage

Answer 39

Upon solidification elements typically shrink by 3-5% (water/ice exception), exaggerated by solid emitting gas that was held in liquid (gas evolution)

Question 40

Explain the problems with gas evolution in solidifying material

Answer 40

Gas bubbles form at nucleation points, then rise due to low density, fast solidification can form solid around bubbles causing porosity in final material Stopped by: slow solidification or degassing liquid, pores can be removed from hot working processes

Question 41

Explain rate of solidification wanted to avoid/limit gas evolution problems

Answer 41

Very slow - so gas has time to rise to surface and escape Or extremely fast so that larger bubbles don’t have time to form = less porosity Medium cooling is worst for porosity and thus properties

Question 42

Describe plastic deformation

Answer 42

Atoms significantly displaced = no recovery as bond broken and new ones made = high energy needed Defects lower energy needed by decreasing bonds Plastic deformation occurs when dislocations move = energy above CRSS, lower CRSS = lower properties

Question 43

Define a slip system and it’s relevance for plastic deformation

Answer 43

Any combination of plane and direction Slip system with lowest activation energy taken as active slip system, closer packed the system = lower activation energy (CRSS needed in that direction) 5 systems must be active in order to have dislocation movement

Question 44

Draw and describe an edge dislocation

Answer 44

Extra half plane of atoms which alters local spacing around it, causing a compressive and tension strain field around it = reduced CRSS as less bonds need to be broken

Question 45

Explain work hardening

Answer 45

Dislocations move = strain fields interact with eachother forming locks and Jogs = more energy required to overcome, significant in FCC as high dislocation density Only occurs < 0.6Tm, above 0.6Tm annealing occurs which cancels out

Question 46

Explain solid solution strengthening

Answer 46

To form solid solution atoms must have similar size, structure and valence e-, but minute size difference causes strain fields - interact with dislocation strain fields = increased strength

Question 47

Explain interstitial atom strengthening

Answer 47

Atoms much smaller that original atoms so can fit in interstitial sites but causes strain field (interacts with dislocation) However, interstitial can move toward dislocation and cancel our strain field - energy needed to reestablish field for further dislocation movement

Question 48

Explain precipitate strengthening

Answer 48

More solute added than solubility = excess = precipitates forming during solidification, slow cooling = large precipitate = reduced strength Precipitate impeded dislocation movement as strain field around it If coherent = cut by dislocation, if in coherent = dislocation lengthens to move around (both = inc energy)

Question 49

What is meant by coherent particles?

Answer 49

Slip plane is continuous/same alignment

Question 50

How does grain size affect material strength?

Answer 50

Slip planes are incoherent across grain boundaries = dislocations stopped and pile up until strain field affects dislocation in next grain = more energy needed to move dislocation through grains = higher strength if fine grains as more grain boundaries present

Question 51

Describe phase strengthening

Answer 51

Different phases act like large precipitates = incoherent so dislocation lengthens = increased energy needed = increase strength Going through eutectic point = maximum phase strengthening as max α and β present