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
