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MEC132 JD > Lecture 3 and 4 > Flashcards

Lecture 3 and 4 Flashcards

1
Q

Density is linked to

A

atomic weight - but is not always the case

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2
Q

As increase temperature density does what and why

A

the density decreases as atoms;
amplitude of vibration increases
atoms gets further apart as it would require more energy to stay at the same distance and vibrate with greater amplitude see lennord jones
same mass for larger volume therefore lower density
thermal expansion

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3
Q

At certain temperature some materials undergo

A

a phase change ie FCC to BCC different structure

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4
Q

what can happen in a phase change

A

change in density
magnetic properties alter
a lot of mechanical properties
at certain temperature property you are relying on may disappear due to phase change

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5
Q

BCC

A

body centered cubic atoms touch across internal diagonal not a close packed system atoms all corners one in the middle not a stable structure

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6
Q

FCC

A

face centered cubic close packed structure based on layers of atoms, atoms touch across diagonal of face packing sequence ACB ACB ACB

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7
Q

Most materials are

A

crystalline structure (not are but focus on crystalline atm)

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8
Q

crystalline structures can be descirbed as

A

regular array of atoms can be described as unit cells which is a basic building block to make the whole structure

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9
Q

lattice spacing

A

a - length width and depth of unit cell (volume of unit cell = a^3)

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10
Q

sharing of atoms between unit cells meaning

A

if you had a unit cell of simple cubic structure at each corner was an atom then each atom would be shared with 8 other unit cells therefore an eighth of that atom would be inside the unit cell
total atoms in unit cell = 1/8 * 8 = 1

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11
Q

What can we tell from unit cell of an atom

A

everything dont care about the number of unit cells as it just makes the crystal larger repeat of the same unit cell can characterize the crystal by characterizing the unit cell

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12
Q

atomic packing fraction

A

number of particles x volume of particle (4/3pi r^3) / volume of unit cell

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13
Q

method for finding APF

A

find lattice structure in terms of radius of atoms (see what plane atoms touching on)
find volume of unit cell in terms of radius of atom
find number of atoms in unit cell (remember some shared)
insert into APF equation, radius cancels giving %

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14
Q

simple cubic structure

A

one atom at each corner of cube

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15
Q

What does the APF tell you

A

% of unit cell filled with atomic volume

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16
Q

HCP

A

hexagonal close packed close packed structure based on layers of atoms packing structure AB AB AB

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17
Q

number of atoms in FCC unit cell

A

1/2 * 6 + 1/8 * 8 = 4

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18
Q

types of materials FCC

A

aluminium copper

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19
Q

FCC close packed plane

A

(1 1 1)

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20
Q

FCC a in terms of r equation

A

a^2 + a^2 = (4R)^2
a^2 = 8R^2
a= 2root2 * R

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21
Q

perfect dislocation is where

A

atom crystal moved but maintained the structure of the crystal - hoped from one to the other

22
Q

HCP materials

A

cobalt alpha titanium

23
Q

HCP structure unit cell

A

hexagonal prism each face is a place A and A with the B layer in the middle unit cell is a third of this structure

24
Q

HCP unit cell

A

2 atoms per unit cell V = (root3 * a^2 *c)/2

where c is the distance between two layers of the same structure ie A and A

25
Q

BCC

A

not a close packed system atoms all corners one in the middle not a stable structure

26
Q

BCC materials

A

alpha iron

27
Q

Some materials will have different properties in

A

different directions - easy to split in one direction hard in another direction

28
Q

miller indices are

A

a way of describing a crystals orientation

29
Q

miller indices used to describe

A

planes

30
Q

brackets for planes groups of planes

directions and groups of directions

A

(planes) {collection of planes} [directions]

goes x y z

31
Q

what is a plance

A

a surface of a cube, due to symmetry will be the same as another surface this is how we group together (1 0 0) same as (0 1 0) (0 0 1) therefore group {1 0 0} or {0 1 0}
where it cuts the axis that we put the point same crystal structure no difference in information from plane mechanically respond the same

32
Q

diagram if (1 1 1) plane and (2 1 0)

A

see powerpoint

33
Q

miller indices equation

A 
lattice constant (a) / distance at which it cuts as fraction of a
ie a/a = 1
a/infinity = 0 ie parallel to this axis
a/ 0.5 a = 2
34
Q

how to do negative in miller indices

A

put bar over top of number

35
Q

miller indices directions vs planes

A

the plane and direction with same miller indices will perpendicular to each other take start point of direction as origin and indices with respect to that

36
Q

direction in miller indices is

A

how far it travels in a certain direction

37
Q

(1 1 1) plane for FCC

A

is close packed plane as its plane in which atoms are in contact with each other slip plane

38
Q

orientation of stresses defined in

A

similar way to miller indicies x axis is 1 y axis is 2 z axis is 3

39
Q

normal stress

A

acting out of surface

40
Q

shear stress

A

acting parallel to surface

41
Q

stress notation

A

sigma ij where i is the plane and j is the direction the force acts

42
Q

how many components of stress on a cube

A

9 but gets cut down to 6 unique ones as you cannot generate stress on one surface of cube without having equal and opposite on other side

43
Q

each surface has what kind of stresses on it

A

2 shear and one normal (first number will all be the same if on the same plane, the normal the number will be the same, the shear stresses will be the other two directions)

44
Q

1 0 0 plane also known as

A

x plane or yz plane

45
Q

how are shear stresses written

A

with a tau instead of a sigma

46
Q

what the three normal stresses and 3 shearing stresses

A

sigma 11 sigma 22 sigma 33

tau 12 tau 13 and tau 23 these cover all forces due to equal and opposite reactions

47
Q

show that there are only 3 unique shear stresses and state why

A

see powerpoint due to equal and opposite and static equilibrium - material will push back in opposite direction therefore tau 13 = tau 31
tau 23 = tau 32
tau 12 = tau 21

48
Q

why can we cancel the z stresses

A

as we often work in 2 dimensions as forces in third are negligible (walls and thin plates) interested in stresses in surface not thickness

49
Q

shear is rotating clockwise this is

A

positive if anticlockwise motion it is negative

50
Q

normal is tension this is

A

positive if compression it is negative

51
Q

in static equilibrium stresses are balanced shear stresses are

A

complimentary (see powerpoint)

MEC132 JD (9 decks)

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