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Peter's Index Physics Home Lecture 10 Course Index Lecture 12
An introduction to Electricity and Strength of Materials with Peter Eyland
Lecture 11 (Brittle and Plastic Solids)
In this lecture:
• Brittle solids and the critical distance for breaking a bond are introduced,
• the homogeneous separation and microcrack models of brittle failure are compared,
• Plastic solids are introduced,
• the homogeneous shear and dislocation models of plastic failure are compared.
Stress/Strain relationships
Graphs of stress vs strain for different materials enables us to distinguish a number of different types of behaviour.
Brittle Fracture
Brittle solids tend to be crystalline (covalent network) or amorphous. Generally, they have a steep straightline slope on a stress/strain graph that terminates abruptly. This is because the bonds are highly directional and when they break they cannot reform easily in a different direction.
In the brittle fracture graph,
• the solid is elastic (i.e. it will recover its original shape), and
• linear elastic because the gradient is a straightline, and
• fractures at its elastic limit, which is its ultimate tensile strength.
Critical Distance to break a bond
To find the theoretical strength of a brittle solid we need to know the theoretical critical strain that will break a single bond. This is workedout from the critical distance to break a bond.
The general form of the interatomic potential is: 
The force function and graph are as shown below. 
The graph shows how the interatomic force responds to an applied force. Equilibrium separation is represented as being at 10 units of distance. An applied force that increases the separation from equilibrium, will be matched by a negative (attractive) force by the bond. In the illustration, the bond's force will reach a maximum negative size around 15 units and after this the force decreases in size to zero, i.e. the bond will break. This critical distance for breaking the bond is found at the turning point of the force graph.
To find the turning point, find the point where the slope becomes zero. 
Comparing this with the equilibrium separation
(see Lecture 6): 
This is the result that is needed to work out critical strains.
The critical distance is some factor greater than the equilibrium separation distance.
That factor depends only on the slope parameters "m" and "n".
The force to break the bond is found by substituting the critical distance into the force function.
Homogeneous Separation Model
The result above is now applied to the homogeneous separation model for material failure. This model says that all the bonds across a separation plane are simultaneously stretched to the critical distance.
A simple cubic structure is shown at equilibrium and then stretched to the critical distance for breaking,
(other crystal structures are the same in principle). 

Using the general values shown above: This again depends only on the slope parameters "m" and "n". 
With representative values: • m=12 and n=6 gives ε_{n}=12% 
With various slope parameters, this model predicts strains of 12% to 30% to cause fracture, but in practice <1% causes fracture! The homogeneous separation model doesn't fit the facts very well so we need to find a better one.
The table below shows measured critical stresses and strains. Notice that the critical strains are already given in percentages and only one is above 1%.
Material 
E GPa 
σ_{c} MPa 
ε_{c} % 
Al_{2}O_{3} 
355 
300 
0.08 
SiC 
414 
600 
0.14 
SiO_{2 }(glass) Drawn fibre 
72 
1000 
1.39 
SiO_{2 }(glass) Aged rod 
72 
250 
0.35 
Microcrack Model
This model says that microcracks cause bonds to break sequentially rather than simultaneously.
The diagram shows that a microcrack involves preexisting broken bonds along a partial plane of atoms. An applied stress is not felt by the broken bonds and it is transferred to the first available bond. 
Microcracks thus create a stress concentration because the bond at the crack tip has to take up the stresses (blue lines) from the broken bonds.

As shown, the bond at the tip is stretched by the stress concentration and is at the point of breaking. When it breaks, the stresses are passed to the next bond further increasing the concentration of stress, and so once started, the crack will run through the material causing it to break apart.
Theory for the elliptical crack
A microcrack can be modelled in crosssection as an ellipse. This gives a calculation for the stress concentration factor.
The diagram shows half an ellipse with a semimajor axis "c" and semiminor axis "d". 
It can be shown (in another course) from the geometry that the stress at the crack tip is given by: 
Now the effective radius at the elliptical tip is: 
so 
This shows two important features. The stress concentration depends directly on (the square root of) the length of the crack, and depends inversely on (the square root of) the radius of the crack.
The stress concentration increases with the length of the crack, and with the decrease of the crack tip radius.
For brittle materials the radius at the crack tip is equal to the interatomic separation so microcracks are especially dangerous.
Example S9
GaAs is a brittle material that forms a cubic zinc sulphide crystal structure.
If the lattice constant, a, for GaAs is 0.565nm find the stress concentration for a crack 8.8 nm long
Answer S9
The cubic zinc sulphide structure is the diamond structure with alternating atoms of Ga and As, as shown in the unit cube below. 
The unit cube has lengths, a, along each side. The interatomic separation, b, is 1/4 of the length of the body diagonal, i.e. The stress concentration is given by 
A crack of length of 8.8nm (invisible to the naked eye) could have 8.8/0.24 = 37 broken GaAs bonds. This would multiply the applied stress by 12 times.
Plastic solids
Plastic solids have a complex stress/strain relationship. As illustrated in the middle graph there is: 

Hooke's law holds for strains less than 1%. 
Homogeneous shear model
This model says that all the atoms along a slip plane are pushed simultaneously to the critical shear strain.
Assume for convenience that the atoms are close packed hard spheres each with the same diameter, b.
As seen in crosssection the atoms form equilateral triangles.
Small shear stress will make a plane's atoms ride up on each other and relax back when the stress is released. 

In crosssection the unstressed state, shows that the atom planes have their centres a distance, bcos30°, apart. 

Shear strain and normal strain can be calculated for this situation. 
The model predicts that a shear strain of 60% is necessary to produce yield. The table below shows measured critical stresses and strains. As <1% is needed, the model does not fit the facts.
Notice that the critical strains are already given in percentages.
Material 
E GPa 
σ_{c} MPa 
ε_{c} % 
Aluminium 
70 
26 
0.04 
Iron (soft) 
206 
150 
0.07 
Steel 
206 
650 
0.32 
Brass 
110 
75 
0.07 
Copper 
123 
45 
0.04 
Dislocation slip model
This model says that dislocations cause a line of bonds to break sequentially rather than simultaneously.
An edge dislocation is shown as half plane of blue coloured atoms in the diagram. 
A small shear stress pushing the top atoms to the right and the bottom atoms to the left, will weaken the bond between the two black atoms and strengthen the bond between the two red atoms. As the black atoms are in critical strain, the bond between them can easily break and a new bond form between the lower black atom and the approaching blue atom at the dislocation edge. This has the effect of moving the dislocation edge to the top right black atom with the same geometry occuring. As the geometry has just moved one plane to the right there is nothing to stop the dislocation edge running through the material.
The model predicts usual elastic behaviour at <1% shear strain, which is not enough to move the dislocations.
Once the yield point is passed large strains (10% to 20%) would be produced by small increases in stress as dislocation movement cascades.
This is called the "easyglide" region.
Dislocations require relatively good crystalline structure, so when dislocations start to tangle,
by contact with other dislocations or grain boundaries, then their movement would be hindered.
This means that the material becomes stronger.
This is the "workhardening" region that was well known to ancient metalurgists, who used repeated heating, hammering and quenching to stiffen swords and other implements.
With additional work hardening after dislocation movement is stopped, voids form as bonds start to break under further strain.
As voids multiply and join, tearing or "ductile rupture" divides the material.
The model thus fits the observed results.
Effective force to move a dislocation
The effective force to move a dislocation with length, and lattice spacing, b, is found from the definition of shear stress as transverse force over area. 
Example S10
A solid has rigidity modulus of 90Gpa and lattice spacing 160pm.
Assume it yields plastically at a shear strain of 1% by a dislocation that moves by 144nm.
Find
(a) the force per unit length moving a dislocation,
(b) the work done per unit length by the external shear force.
Answer S10
Shear stress is related to shear strain through the Rigidity Modulus
Viscoelastic materials
Viscoelastic materials behave elastically on short time scales and are viscous for long times.
Summarising:
Theoretical strengths from the homogeneous separation and homogeneous shear models are much greater than those found in practice.
The homogeneous separation model has to be replaced with the microcrack model and its stress multiplication at the crack tip.
The homogeneous shear model has to be replaced by the dislocation slip model where critical strain is built into the structure.
The weakness of brittle solids is due to microcracks.
The weakness of plastic solids is due to dislocations.
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