lecture 10

Question 1

slip can not occur in

Answer 1

a pure normal stress requires shear stress

can occur in a tensile test as force applied by normal stress causes shear stress at certain directions within material

Question 2

critical shearing stress

Answer 2

stress required to start slip

Question 3

shear stress =

Answer 3

tau c* sin (2 pi x /b)

where tau c is critical shear stress and b is the diameter of atoms or nearest neighbour distance

Question 4

what does the frenkel model predict critical shear stress to be and why is this not very accurate

Answer 4

critical shear modulus tau c = shear modulus/2pi ie shear stress to cause slip is way above what it should be
but experimentally tc <= G/10^3 ie way lower

Question 5

What is the issue with the frenkel model

Answer 5

way too simplistic something missing - overestimates the critical shear modulus by huge amount
sees it as breaking all bonds at once which requires alot of energy but can just move a defect along by only breaking a few bonds
linear defect seperating slipped and unslipped regions can move at much lower forces than bonding energy

Question 6

plastic deformation occurs at

Answer 6

forces way below forces holding atoms together - yield shouldnt occur at such low forces

Question 7

why was plastic deformation occuring at forces way below forces holding atoms together

Answer 7

defects in materials

Question 8

what is the carpet analogy equivalent to

Answer 8

fenkel model sees it as breaking all bonds at once (moving the whole carpet) which requires alot of energy but can just move a defect along by only breaking a few bonds
linear defect seperating slipped and unslipped regions can move at much lower forces than bonding energy

Question 9

edge dislocation diagram

Answer 9

extra plane of atoms somewhere in the middle
forces atoms around it out causing associated elastic energy
size of dislocation = nearest neighbour distance - one in powerpoint perfect unit disloaction

Question 10

screw dislocation diagram

Answer 10

twisted the sample

Question 11

types of dislocation

Answer 11

screw dislocation
edge dislocation
combination of both

Question 12

how can edge dislocation lead to elongation

Answer 12

pushing the plane of atoms out to the end of material

Question 13

screw dislocation

Answer 13

can be untwisted material gets a little bit longer

Question 14

when screw dislocation and edge dislocation lead to elongation what forms

Answer 14

slip planes (close packed planes) crystal structure very similar but moved slightly

Question 15

slip planes are

Answer 15

lowest energy path bonds are relatively weak

Question 16

slip tends to occur on what plane for FCC

Answer 16

along close packed planes - the (111) plane - 4 of these in a unit cell {111}

Question 17

how many slip directions are there and therefore how many slipping systems are there in FCC

Answer 17

3 directions therefore 12 slipping systems 4 x 3 (12 directions the planes could slip along)

Question 18

close packed planes of atoms moving over due

Answer 18

to defects helping them

Question 19

not all dislocations are perfect may get

Answer 19

partial or imperfect dislocations where the passage of disloaction does not leave the structure unchange
locally change structure may start with FCC end with HCP (dont consider impartial dislocations)

Question 20

what are the three slip directions diagram for FCC

Answer 20

putting negatives put should be bar
[1 0 -1] same as [-1 0 1] (reverse direction)
[1 -1 0] same as [-1 1 0]
[0 1 -1] same as [0 -1 1]

Question 21

Slip in HCP and BCC

Answer 21

HCP close backed basal plane very complicated to calculate
BCC no defined slip planes
dont need to calculate

Question 22

Slip bands produced when

Answer 22

object exposed to normal force but at certain angles end up with shearing stress

Question 23

what is the benefit of resolving forces

Answer 23

design orientation of cut of material and orientation of force applied such that material fails at much higher load as slip planes are never activated

Question 24

greek letter phi symbol

Answer 24

angle from normal to slip plane

Question 25

greek letter Lambda

Answer 25

angle from force to slip direction

Question 26

shearing stress equal to

Answer 26

stress applied * cos phi * cos * Lambda
critical shear stress = 0 if lambda or phi equal 90 degrees (normal stress no shearing stress)
critical shear stress = maximum if lambda = phi = 45 degrees

Question 27

when resoled shear stress is greater than

Answer 27

critical resolved shear stress then slip will occur

Question 28

what equation can be used to calculate cos phi =

Answer 28

direction of force (dot) slip plane= magnitude of direction of forces * magnitude of slip plane * cos phi
if look at formula sheet cos phi is slip plane and force

Question 29

what equation can be used to calculate cos lambda =

Answer 29

direction of force (dot) slip direction = magnitude of direction of forces * magnitude of slip direction* cos phi
if look at formula sheet cos lambda is slip direction against force

Question 30

when finding whether direction is activated

Answer 30

calculate cos phi and cos lambda (cos lambda will be specific to this direction)
shear stress = applied stress * cos phi * cos lambda
if critical shear stress is lower than calculated value then plane is active (if its negative its in opposite direction)

Question 31

difference between calculating cos phi and cos lambda

Answer 31

cos phi only need to calculate once (direction of force in comparison to slip plane is constant need to calculate cos lambda for all 3 directions)

Question 32

what will the slip plane be

Answer 32

slip plane will always be (1 1 1) as we only look at the FCC term

Question 33

What will the slip directions be

Answer 33

they will be the three slip direction for FCC [1 -1 0] [0 1 -1] and [1 0 -1] (one zero one and negative one in three combinations)

Question 34

what will the applied force be

Answer 34

will be the vector direction of the force eg [2 1 0]

Question 35

if resolved stress is greater than critical shear stress but negative

Answer 35

slip will occur but in opposite direction to one specified

ie [1 -1 0] to [-1 1 0]

Question 36

EBSD image shows

Answer 36

crystal orienation

Question 37

texturing materials

Answer 37

make crystal structure all similar means slipping that occurs is very controlled