Question 15.1: Bond Stiffness and the Elastic Modulus Consider a uniaxial l......
Bond Stiffness and the Elastic Modulus
Consider a uniaxial load applied to a simple cubic unit cell and demonstrate that the elastic modulus E of a material is directly proportional to the bond stiffness.
Assume that the equilibrium distance between atoms is the lattice parameter a. The engineering stress S that develops when a load P is applied to a cross-sectional area a² is
S =\frac{ P}{a^2}
For elastic loading, Hooke’s law applies such that
S =\frac{ P}{a^2}=Ee
where e is the engineering strain. The engineering strain can also be expressed as
e =\frac{Δa}{a}
where Δa is the extension, and the load P can be written as
P = kΔa
where k is the stiffness of the bonds, which are modeled as linear springs (see Figure 15-1). Solving for E,
E =\frac{ P}{a^2 e}=\frac{kΔa}{a^2} \ \frac{a}{Δa} =\frac{k}{a}
This equation is a rough approximation and should not be considered to be strictly correct, but it does show that the elastic modulus is proportional to the stiffness between atoms or ions in a solid. As the bond stiffness increases, the elastic modulus increases.
