CHE 333 Class 18

Fracture of Materials

Ductile or Brittle FailureFollowing 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 FailureSvS stops here

Fracture ProcessesFor ductile materials which have plasticityonce the neck starts, a sequence of structural changes occurs as the metal proceeds to failure. Internally voids areinitiated . Both the true stress and strainare both still increasing. With furtherincrease in strain the voids becomelarger, as they become circular orellipsoidal in shape. The small ligamentsof metal between the voids eventuallytears and an internal crack is initiated.With further true stress and strain increasethe last areas to fail are those connecting the internal crack to the surface. A “shear lip”at 45o to the surface characterizes this regionleading to “cup and cone” fracture in rod.

Ligament of material Stretches between voids

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

s

45o

Second Phase

In two phase material, several differentmicro processes can lead to void initiationgrowth and failure. If the second phase is either large or continuous,then a dislocation pile up in one phase will leadto a void formation. This will then grow in phase. The second phase will have strengthenedthe material but will also have eventuallyinitiated the failure process.Examples include pearlitic steels and other platelet structures such as those in titanium alloys Void from dislocation

pile up on slip plane

Micro Failure Example

Example of toughened material?

Micro Failure ProcessesWhen 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 singlecrystal 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

I II IIIMode

Plane Strain Fracture Toughness

Ductile to Brittle TransitionIn 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.

Carbon content effect on DBTT for SteelsShear ratio to impact energy

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 25 60 35 55

b 55 50 60 60

c 55 60 35 45

d 45 55 45 35

Data units MPa.