As we can see that a brittle material almost fails at elastic limit. But brittleness is not an absolute property of a material. For example, tungsten is brittle at room temperature and is ductile at high temperatures. The three factors contributing to brittle fracture are : triaxial state of stress, low temperature and high strain rate. All these factors need not be present for a material to show brittle cleavage. A triaxial stress which is present at notch along with low temperature may lead to brittle fracture. Also high strain rate assists those two factors.

For most of the engineering materials ductility is an important criteria. Thus, little amount of ductility is needed for most of the material. Also a material being ductile has its own advantage. A ductile material before failure undergoes plastic deformation thus indicating that it is going to fail. Whereas a brittle fracture is sudden without any warning. Thus most of the place ductile materials are preferred over brittle materials.

Usually ductile facture is preceded by gross plastic deformation known as necking which begins from the maximum stress. Necking begins when increase in strength due to strain hardening fails to compensate by the decrease in cross-sectional area. A rupture occurs when a ductile material is drawn down to a line or a point and then fails. The necking introduces a triaxial state of stress in that region. Many cavities are also formed in that region which on continued strain grows into a large crack. This crack grows in a direction perpendicular to the axis of the specimen. It then propagates along 45° i.e. the shear plane. This type of fracture is also known as cup and cone fracture. While the crack growth direction id outwards i.e. transverse to the tensile axis of the specimen, in a microscopic scale the crack actually moves in a zig-zag manner. The preferred sites of void formation are usually in inclusion, secondary phase particles etc. , whereas for pure metal they are grain boundary triple point. Ductility decreases as the void fraction increases, as the strain-hardening exponent n.

In metallurgy, the Ellingham diagram is used to find out a suitable reducing agent. In the Ellingham diagram the highly stable oxides are found at the bottom and the less stable oxides are found at the top. An element occupying a lower position in the diagram can reduce the oxides of another element present above it. For example, Mg lies below Si thus Mg can be used as a reducing agent for oxides of Si.

Why Ellingham diagram has all straight lines?

Reason for upward slope of lines

The Ellingham diagram is based on a formula given by:

ΔG = ΔH – TΔS

If we draw a graph with ΔG as y-axis and T as x-axis the slope of the curve represents the entropy(ΔS) and the intercept is enthalpy(ΔH). From the above equation we know that enthalpy is independent of temperature and entropy alone depends on temperature. The condition for a reaction to occur or to be spontaneous is to have negative ΔG. Most of the oxides forming reactions are exothermic i.e. ΔH is negative. And if we consider the reaction:

2Mg_(s) + O_2(g) = 2MgO_(s) + Heat

Here we can see that there is a phase transformation taking place from gaseous phase to solid phase. Thus due to this reason the entropy decreases and becomes negatives(-ΔS).Thus if we consider this in the free energy equation, we get :

ΔG = ΔH – T(-ΔS)

ΔG = ΔH + T(ΔS)

Here, if we consider ΔG as y(since we considered it as y-axis), ΔH as c, ΔS as m, T as x(since we considered it as x-axis) we get the straight line equation:

y = mx + c

Thus, this explain why the Ellingham diagram has all straight lines rather than curves. Also for this condition if we consider in terms of graph we get a straight line with positive slope i.e. sloping upwards.

Another important point to note is that, the Gibbs free energy(ΔG) here is considers as Standard free energy of formation of oxides kJ/mole for O₂ . Thus, that’s the reason why only one mole of O₂ is considered for all reactions.

Exceptions in Ellingham diagram

There are two exceptions in the Ellingham diagram. They are as follows.

Let us consider the following reaction :

C_(s) + O_2(g)  \= CO_2(g)

Here, we can see that one molecule of gas(O_2(g)) produces one molecule of gas(CO_2(g)).As the entropy of solids are the lowest thus it can be considered as negligible. Thus, there is no net entropy and the line is almost horizontal.

Let us consider another reaction :

2C_(s) + O_2(g)  \= 2CO_(g)

Here, we can see that one molecule of gas produces two molecule of gas i.e. the entropy increases and becomes positive(+ΔS). Thus if we consider this in the Gibbs free energy equation, we get :

ΔG = ΔH - T(ΔS)

This above equation is a representation of negative slope i.e. we get a line sloping downwards.

Limitations of Ellingham diagram

1. The Ellingham diagram doesn’t explain the kinetics of the reduction.

2. In Ellingham diagram all the reactants and products are considered to be in equilibrium. As we know that, it’s not always true.