The elastic properties of a homogeneous crystal are usually anisotropic. Even in the cubic crystal, the relationship between the stress and strain based on the orientation of the crystal axes relative to the stress. In common, the number of elastic constants characterizing a body is big. Though, this number is considerably decreased due to the symmetric nature of both strain and stress tensors.
Definition: Elasticity is basically the study of the capability of crystals to incorporate changes or adapt to new circumstances without difficulty.
Experimental determination of elastic constants:
The classic techniques for the measurement of the elastic constants of crystals are illustrated in the review by Hearmon in the year 1964). In this process, quartz transducer is transmitted via the test crystal and reflected from the rear surface of the crystal back to the transducer. The elapsed time between the initiation and receipt of the pulse is evaluated by standard electronic techniques. The velocity is obtained by dividing the roundtrip distance via the elapsed time. In a representative arrangement the experimental frequency might be 15 s-1, and the pulse length 1 µsec. The wavelength is of the order of 3 x 10-4 cm. The crystal specimen might be of the order of 1 cm in length. The elastic stiffness constants C11, C12, C44 of a cubic crystal might be determined from the velocities of three waves. A longitudinal wave propagates all along a cube axis having velocity (C11/ρ)1/2, here 'ρ' is the density. A shear wave propagates all along a cube axis having Velocity (C44/ρ)1/2, whereas a shear wave having particle motion all along a 11‾0 direction propagates all along a 110 direction with velocity [(C11 - C12)/2ρ]1/2
Elastic waves in cubic crystals:
By considering the forces acting on an element of volume in the crystal we determine for the equation of motion in the x-direction:
ρü = ∂Xx/∂x + ∂Xy/∂y + ∂Xz/∂z
With identical equations for the y and z directions; 'ρ' is the density and 'u' is the displacement and ü = d2u/dt2
From the equation cij = cji it follows, taking the cube edges as the x, y, z directions, that
ρü = C11 (∂exx/∂x) + C12 (∂eyy/∂x) + C44 (∂exy/∂y + ∂ezx/∂z)
This transforms to:
ρü = C11(∂2u/∂x2) + C44[(∂2u/∂y2) + (∂2u/∂z2)] + (C12 + C44) [(∂2v/∂x∂y) + (∂2w/∂x∂y)]
Here u, v and w are the components of displacement. One solution is represented via a longitudinal wave:
u = uoei(ωt - kx)
Moving all along the x cube edge:
- ω2ρ = - k2C11
Here k = 2π/λ where 'λ' is wave vector and ω = 2πv is the angular frequency in such a way that the velocity is:
v = ω/k = (C11/ρ)1/2
The other solution is represented by a transverse or shear wave moving all along the y cube edge having the particle motion in the x-direction:
v = voe[i(ωt - ky)]
- ω2ρ = - k2C44
So that: v = (C44/ρ) 1/2
There is as well a solution represented by a shear wave moving in the z direction having particle motion in the x direction. In common there are three kinds of wave motion for a particular direction of propagation in the crystal; however only for a few special directions can the waves be categorized as pure longitudinal or pure transverse.
By minor manipulations we might represent the equation as:
ρü = C11(∂2u/∂x2) + C44[(∂2u/∂y2) + (∂2u/∂z2)] + (C12 + C44) [(∂2v/∂x∂y) + (∂2w/∂x∂y)] as
ρoü = (C11 - C12 - 2C44)(∂2u/∂x2) + C44 ∇2u + (C12 + C44) ∂/∂x div ρ
Here the displacement ρ = ui + vj + wk is not to be confused by means of density now written as ρo. If:
C11 - C12 = 2C44
The first term on the right in the above equation drops out, and we can write on summing with the equations for the y and z motions:
ρ.. = C44 ∇2 ρ + (C12 + C44) grad div ρ
This equation consists of the significant property that it is invariant under rotations of the reference axes, as each and every term in the equation is an invariant. Therefore the relation is the condition that the crystal must be elastically isotropic; that is, that waves must propagate in all directions having equal velocities. Though, the longitudinal wave velocity is not essentially equivalent to the transverse wave velocity. The anisotropy factor A in the cubic crystal is stated as:
A = 2C44/(C11 - C12)
And is unity for the elastic isotropy.
There are among the elastic stiffness constants certain relations first achieved by Cauchy. The relations reduce to:
C12 = C44
in a crystal of the cubic symmetry. If this is satisfied, the isotropy condition becomes C11 = 3C44. If then a cubic crystal were elastically isotropic and the Cauchy relation is satisfied, the velocity of the transverse waves would be equivalent to the velocity of the longitudinal waves.
The conditions for the validity of the Cauchy relations are as follows:
1) All forces should be central, that is, act all the along lines joining the centers of the atoms. This is not usually true of covalent binding forces, nor of the metallic binding forces.
2) Each and every atom should be at a center of symmetry; that is, replacing each inter atomic vector must not change the structure.
3) The crystal must be primarily under no stress. In metallic lattices the nature of the binding is not such that we would anticipate the Cauchy relation to work out well. In ionic crystals the electrostatic interaction of the ions is the principal interaction and is central in nature. This is not surprising that the Cauchy relation is fairly well satisfied in the alkali halides.
Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)
Expand your confidence, grow study skills and improve your grades.
Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.
Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.
Free to know our price and packages for online physics tutoring. Chat with us or submit request at firstname.lastname@example.org
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!