Page 86 - Electrical Properties of Materials
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68                            Bonds

                    T              branches of engineering this phenomenon is known as elasticity. So if we man-
                                   age to obtain the E(r) curve, we can calculate all the elastic properties of the
                                   solid. Let us work out as an example the bulk elastic constant. We shall take a
                                   cubical piece of material of side a (Fig. 5.2) and calculate the energy changes
                                   under isotropic compression.
                               T
                                     If we regard E(r) as the energy per atom, the total energy of the material is
                                      3
                                   N a a E(r 0 ) in equilibrium, where N a is the number of atoms per unit volume.
                                   If the cube is uniformly compressed, the interatomic distance will decrease by
      T
                                                                      3
                                    r, and the total energy will increase to N a a E(r 0 – r). Expanding E(r 0 – r)
              a                                                          = 0, we get
                                   into a Taylor series and noting that (∂E/∂r) r = r 0
                                                                    2
     Fig. 5.2                                  E(r 0 –  r)= E(r 0 )+  1    ∂ E     ( r) + ···  (5.1)
                                                                               2
     A cube of material, side a,is                              2  ∂r 2
                                                                        r = r 0
     isotropically compressed.
                                   Hence, the net increase in energy is equal to
                                                                2
                                                       1    3    ∂ E       2
                                                         N a a         ( r) .                (5.2)
                                                       2       ∂r 2
                                                                   r = r 0
                                   This increase in energy is due to the work done by moving the six faces of the
                                   cube. The total change in linear dimensions is (a/r 0 ) r; thus we may say
                                   that each face has moved by a distance (a/2r 0 ) r. Hence, while the stress
                                   is increasing from 0 to T, the total work done on the piece of material is
                                                               1  2  a r
                                                            6 × Ta     .                     (5.3)
                                                               2    2r 0
                                     From the equality of eqns (5.2) and (5.3) we get

                                                                    2
                                                   3 a 3     1 a 3    ∂ E      2
                                                     T   r =               ( r) ,            (5.4)
                                                   2 r 0     2 r 3  ∂r 2
                                                                0      r = r 0
                                   whence
                                                                 2
                                                            1    ∂ E      r
                                                       T =        2        .                 (5.5)
                                                           3r 0  ∂r      r 0
                                                                     r = r 0
                                     Defining the bulk elastic modulus by the relationship of stress to the volume
                                   change caused, that is

                                                            a 3   3 a    3 r
                                                      T = c    ~      = c    ,               (5.6)
                                                               = c
                                                            a 3    a      r 0
                                   we can obtain c with the aid of eqn (5.5) in the form
                                                                  2
                                                             1    ∂ E
     It is worth noting that most ma-                    c =       2      .                  (5.7)
                                                            9r 0  ∂r
     terials do obey this Hooke’s law                                 r = r 0
     for small deformations, but not for  So we have managed to obtain both Hooke’s law and an expression for the
     large ones. This is in line with the  bulk elastic modulus by considering the interaction of atoms. If terms higher
     assumptions we have made in the  than second order are not negligible, we have a material that does not obey
     derivation.                   Hooke’s law.
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