CHE 333 Class 18
Fracture of Materials
Ductile or Brittle Failure
Following elastic deformation, two different
processes can occur – plastic deformation
leading to ductile failure or movement to
brittle fracture with little or no plasticity.
The stress strain curve is shown for ductile
failure. After the UTS the ductility is shown
by the necking of the material. This would
only be for round bar.
The stress strain curve for brittle materials
is indicated, which would be for glass and
ceramics.
Brittle Failure
SvS stops here
Fracture Processes
Ligament of material
Stretches between voids
For ductile materials which have plasticity
once the neck starts, a sequence of
structural changes occurs as the metal
proceeds to failure. Internally voids are
initiated . Both the true stress and strain
are both still increasing. With further
increase in strain the voids become
larger, as they become circular or
ellipsoidal in shape. The small ligaments
of metal between the voids eventually
tears and an internal crack is initiated.
With further true stress and strain increase
the last areas to fail are those connecting
the internal crack to the surface. A “shear lip”
at 45o to the surface characterizes this region
leading to “cup and cone” fracture in rod.
Micro Failure Processes.
In single phase materials, small voids appear
at the cell walls as each dislocation effectively
carries a vacancy with it. These voids then
grow with further stressing
Micro Failure Processes
In two phase material, several different
micro processes can lead to void initiation
growth and failure.
If the second phase is either large or continuous,
then a dislocation pile up in one phase will lead
to a void formation. This will then grow in
phase. The second phase will have strengthened
the material but will also have eventually
initiated the failure process.
Examples include pearlitic steels and
other platelet structures such as those
in titanium alloys
Second Phase
45o
Void from dislocation
pile up on slip plane
s
Micro Failure Example
Example of toughened material?
Micro Failure Processes
When the second phase is small and
non discontinuous, other processes in
addition to dislocation pile up and a void
on the slip plane are possible. The controlling
factors are the strength of the second phase
particle and interface between the second
phase and the matrix material.
If the second phase is weaker or more
brittle than the matrix, then the second phase
particle may crack.
Age hardened aluminum alloys behave in
this manner
Micro Failure Example
Al 7079 Age Hardened alloy showing particle failure.
Micro Failure Examples
Macro View of Failure
Fracture Surface Features
Ductile failure in a titanium alloy which is two phase showing dimples from void formation
Final Failure in Shear Lip
Shear dimples in area of shear lip.
Brittle Failure Fracture Surface
A faceted type of fracture surface is often seen with a brittle failure as cracks are on a single
crystal plane, no shear lip would be found..
Fracture Mechanics.
Originally to explain why glass has low toughness where toughness is the ability to resist crack
growth in a material. Simple energy balance used – energy applied is the stress and it is used up by
creating fresh surface. Works well for brittle materials where no plastic deformation. For ductile
materials energy absorbed is complicated by plastic deformation. Fracture mechanics relates the size
of a defect in the material to the stress is can withstand before failure.
Kc = Ysc (pa)0.5
Mode I
II
III
Plane Strain Fracture Toughness
Ductile to Brittle Transition
In some materials, mainly steels, ductility can decrease very sharply with temperature, so a ductile
materials becomes brittle – know as the ductile brittle transition. The standard test is to use an impact
tester – a pendulum type hammer and the energy absorbed in failure is measured by how far the
hammer swings through after impact – the further the less energy.
Shear ratio to impact energy
Carbon content effect on DBTT for Steels
Homework
1.
For a material with a single crystal yield strength of 300 MPa,
calculate the yield strength for a grain size of 50 micron and for a
grain size of 2.5 microns. Assume k =1
2.
Decide the crack initiation process in a tensile test for the
following conditions:Matrix shear strength Matrix Normal Strength Interface Strength Particle Strength
a
b
c
d
Data units MPa.
25
55
55
45
60
50
60
55
35
60
35
45
55
60
45
35