In this article we will discuss about the atomic approach to elastic deformation of metals.
At absolute zero, the two neighbouring atoms in a metal are in stable equilibrium state when the distance between them is r0. Fig. 6.11 illustrates this schematically. At large distance of separation between two atoms, the main force is of attractive nature, FA (negative).
When the distance between them decreases the force of repulsion, FR too become effective and increases rapidly with decreasing distance of separation. At r0, the repulsive and attractive forces exactly balance each other i.e. the net force is zero. At this distance, the potential energy too has a minimum. This magnitude, Eb0 of potential energy is called the bond energy as illustrated in Fig. 6.11 (b).
The generalized equation relating potential energy Eb and the distance of separation, r is,
For example, a tensile force F2 has to be applied to increase the distance of separation from to r0 to r2 and a compressive force F1 has to be applied to decrease the distance from to r0 to r1.
Elastic modulus is a measure of stiffness. Since the force can be related to the stress, and changes in the distance of separation can be related to strain, the elastic modulus can be obtained from the slope of the net force-distance curve in Fig. 6.11 (c). A tangent to this curve drawn at r0 almost coincides with the curve if there are small displacements on either side of r0.
Also is true that elastic strains for both ductile and brittle materials are very small, such as 0.001 to 0.005 to be within the small range of displacements. Thus, Young’s modulus, E of the material for both tensile and compressive stresses is proportional to the negative of this slope at r0 in force-distance curve,
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E ∝ – dF / dr = d2 Eb / dr2 … (6.21)
The values of Young’s modulus are directly related to the interatomic bonding energies. Materials with deep and narrow energy wells have a high elastic modulus as dF/dr is equal to the second derivative of the energy with respect to the distance. Thus, strong bonding results in high values of elastic modulus and vice versa.
The transition metals have high elastic moduli as compared to the alkali metals as the former have partial covalent bonding of atoms, which makes them strongly bonded. Shallow potential well of weakly bonded material, as illustrated in Fig. 6.11 (d), result in small values of elastic modulus.
Stiffness of a material is defined as its ability to resist the elastic deformation. Stiffness is dependent on the shape of the structural part. For similar shapes, it is proportional to the elastic modulus. Under a stress of 200 MNm-2, a steel shaft deforms elastically by 0.001 mm/mm but aluminium shaft deforms by 0.003 mm/mm. Table 6.1 shows some elastic properties of some materials. For applications such as in aircraft or spacecraft, in which lightness of weight is important, the appropriate design factor is
With respect to this factor, boron is superior to steel by ratio 190/25 = 8, according to table 6.1. This means that a structural component can be given the desired resistance to elastic deformation (stiffness) by only one-eighth as much as boron compared to a similar design in steel. But boron is brittle.
Brittle materials cannot withstand accidental overloading during service, and may fail in a catastrophic manner. It can be used as a reinforcing fibre for a ductile matrix such as of aluminium. Fibres have greater stiffening effect, when the stress is along their length. The brittleness of boron is countered by the cushioning effect of ductile aluminum.
The ductile matrix stops a propagating crack, if a fibre embedded in it breaks accidentally. A 40 vol. % of boron increases Young’s modulus from 71 GNm-2 for pure aluminium to 219 GNm-2 for the composite, i.e., this composite shall be as stiff as steel but one third its density. Composites that employ graphite or sapphire as the stiffening fibres offer an even greater advantage.
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Problem 1:
Determine the length of a 1.25 in bar of aluminium to which a stress of 210 MNm-2 is applied. Given modulus of elasticity = 66.7 GNm-2.
Solution:
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According to Hooke’s Law,
e = σ / E
= 210 / 66700 = 0.00315 mm/mm
Thus, the length becomes
= 1.25 + 1.25 x 0.00315
= 1.2539 m