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Physics for Civil Engineering

This is an introduction to Electricity, Strength of Materials and Waves.

Lecture 12 (Brittle and Plastic Failure)

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.

stress vs strain diagram

Brittle Fracture

Brittle solids tend to be crystalline (covalent network) or amorphous. Generally, they have a steep straight-line 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 straight-line, 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 worked-out from the critical distance to break a bond.

The general form of the interatomic potential is:

interatomic potential function

The force function and graph are as shown below.

interatomic force function
interatomic force graph

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.

calculus for turning point

Comparing this with the equilibrium separation (see Lecture 6):

critical distance calculation

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.

critical force equation

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).
Normal strain at the critical distance is:

homogeneous separation graphic

Using the general values shown above:

critical strain for homogeneous separation

This again depends only on the slope parameters "m" and "n".

With representative values:
• m=9 and n=1 gives:

critical strain examples

• m=12 and n=6 gives εn=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%.



σc MPa

εc %









SiO2 (glass) Drawn fibre




SiO2 (glass) Aged rod




Microcrack Model

This model says that microcracks cause bonds to break sequentially rather than simultaneously.

The diagram shows that a microcrack involves pre-existing broken bonds along a partial plane of atoms.

broken bonds along crack

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.

stresses transferred to crack tip

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 cross-section as an ellipse. This gives a calculation for the stress concentration factor.

The diagram shows half an ellipse with a semi-major axis "c" and semi-minor axis "d".

crack as a half ellipse

It can be shown (in another course) from the geometry that the stress at the crack tip is given by:

stress concentration factor

Now the effective radius at the elliptical tip is:

effective radius


stress concentration, crack radius and length

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.
It can be seen as two FCC structures displaced by 1/4 of a body diagonal.

The unit cube has lengths, a, along each side.
From Pythagoras' theorem, the body diagonal is length of body diagonal long.

The interatomic separation, b, is 1/4 of the length of the body diagonal, i.e.

interatomic spacing

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

stress vs strain diagram

Plastic solids have a complex stress/strain relationship. As illustrated in the middle graph there is:
• a linear elastic region.
• a yield point or elastic limit.
• a non-linear, non-elastic region.
• a plastic flow or "easy glide" region.
• a work-hardening region.
• an ultimate tensile strength (top of graph).
• a ductile rupture point.

Hooke's law holds for strains less than 1%.
After the yield point, there is permanent slippage, i.e. deformation along shear planes that are at 450 to the tensile stress.
The slippage becomes extensive in the easy glide region.
Then there is resistance to slip as slip planes and defects tangle. This is called "work hardening" because the material stiffens up.
Multiple slipping planes in 3D around the axis, causes necking, as shown on the right.
Ductile rupture happens soon after necking.

Any model for plastic yield will involve shear slipping.

necking diagram

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 cross-section 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.
If the atoms are pushed past the critical point, shown on the far right, then they will all fall into the next valley and there is irreversible change.

homogeneous slip diagram

In cross-section the unstressed state, shows that the atom planes have their centres a distance, bcos30°, apart.
In the critical shear strain position, the atoms have moved half a diameter, b/2, from the unstressed position to where the atoms are directly on top of each other. In the critical shear strain position the atom planes are now a distance b apart.

critical stress geometry

Shear strain and normal strain can be calculated for this situation.

share and normal strain calculation

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.



σc MPa

εc %





Iron (soft)
















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.

The atoms on the extreme left and right edges are in their unstressed position in a cubic array.

At the dislocation edge there are two triangles. One is formed with the red and blue coloured atoms, the other is the blue and black atoms. This configuration puts the bonds between the red atoms and also the black atoms at critical shear strain.

dislocation slip 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 "easy-glide" 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 "work-hardening" 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, curvey letter l and lattice spacing, b, is found from the definition of shear stress as transverse force over area.

effective force and shear strain

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.


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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