Advances in Physics Theories and Applications ISSN 2224-719X (Paper) ISSN 2225-0638 (Online) Vol 1, 2011
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u, v, w are displacements along radial, circumferential and axial directions respectively.
Where
rr , , zz
are the normal stress components and r , z , zr are the shear stress components ,
err , e , ezz are normal strain components and er , e z , ezr are shear strain components. Substituting the equations(2) ,(3)and (4) in equation(1),gives the following three displacement equations of
motion :
2 u,rr r 1u,r r 2u r 2u, u, zz r 1 v,r r 2 3 v, w,rz 2u u,tt v,rr r 1v,r r 2v r 2 2 v, v, zz r 2 3 u, r 1 u,r r 1 w, z v,tt
2 w, zz w,rr r 1w,r r 2 w, u,rz r 1 v r 1 u, z w,tt
, z
(4)
To solve equation (4),we take
1 u , ,r r Using
Eqs
(5)
1 v , , r in
Eqs
(1),
we
w ,z find
, ,T
that
satisfies
2 2 2 (( 2 ) 2 2 2 ) ( ) 2 0 z t z
the
equations.
2
2 1
( 1 ( 2 ) 2
(1 2
2 2 2 ) ( )1 0 2 2 z t
2 2 2 ) 0 2 2 z t
Equation (5c) in
(5a)
(5b)
(5c)
gives a purely transverse wave. This wave is polarized in planes perpendicular to the
z-axis. We assume that the disturbance is time harmonic through the factor e i t . 3. Solution to the problem
The equation (5) is coupled partial differential equations of the three displacement components. To uncouple equation(6),we can write three displacement functions which satisfies the simply supported boundary conditions followed by Sharma [8]
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