lecture 10 Flashcards
slip can not occur in
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
critical shearing stress
stress required to start slip
shear stress =
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
what does the frenkel model predict critical shear stress to be and why is this not very accurate
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
What is the issue with the frenkel model
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
plastic deformation occurs at
forces way below forces holding atoms together - yield shouldnt occur at such low forces
why was plastic deformation occuring at forces way below forces holding atoms together
defects in materials
what is the carpet analogy equivalent to
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
edge dislocation diagram
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
screw dislocation diagram
twisted the sample
types of dislocation
screw dislocation
edge dislocation
combination of both
how can edge dislocation lead to elongation
pushing the plane of atoms out to the end of material
screw dislocation
can be untwisted material gets a little bit longer
when screw dislocation and edge dislocation lead to elongation what forms
slip planes (close packed planes) crystal structure very similar but moved slightly
slip planes are
lowest energy path bonds are relatively weak
slip tends to occur on what plane for FCC
along close packed planes - the (111) plane - 4 of these in a unit cell {111}
how many slip directions are there and therefore how many slipping systems are there in FCC
3 directions therefore 12 slipping systems 4 x 3 (12 directions the planes could slip along)
close packed planes of atoms moving over due
to defects helping them
not all dislocations are perfect may get
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)
what are the three slip directions diagram for FCC
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]
Slip in HCP and BCC
HCP close backed basal plane very complicated to calculate
BCC no defined slip planes
dont need to calculate
Slip bands produced when
object exposed to normal force but at certain angles end up with shearing stress
what is the benefit of resolving forces
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
greek letter phi symbol
angle from normal to slip plane
greek letter Lambda
angle from force to slip direction
shearing stress equal to
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
when resoled shear stress is greater than
critical resolved shear stress then slip will occur
what equation can be used to calculate cos phi =
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
what equation can be used to calculate cos lambda =
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
when finding whether direction is activated
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)
difference between calculating cos phi and cos lambda
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)
what will the slip plane be
slip plane will always be (1 1 1) as we only look at the FCC term
What will the slip directions be
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)
what will the applied force be
will be the vector direction of the force eg [2 1 0]
if resolved stress is greater than critical shear stress but negative
slip will occur but in opposite direction to one specified
ie [1 -1 0] to [-1 1 0]
EBSD image shows
crystal orienation
texturing materials
make crystal structure all similar means slipping that occurs is very controlled
