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Torsion of an elastic layer by two circular dies
Printed inbat Btiti ,ti1.&nrsci,~7. Vd.15, pp. 171475. Pm#mm Prets.
TORSION OF AN ELASTIC LAYER BY TWO CIRCULAR DIES BRIJ MOHAN SINGHt and RANJIT S. DHALIWAL Departmentof Mathematicsand Statistics,TbeUniversityof Calgary,Calgary,Canada (Communicatedby I. N. SNEDDDN) Abstract-Two circulardiscs of rigidmaterialand of differentradiiare bondedto the opposite faces of an Infiniteelastic layer.The torques T, and Tzhave been calculatedwhen the discs are rotated&roughdiffered angles.
1. INTRODUCTION
THEPROBLEMof the torsional oscillationsof the elastic layer by one disc attached over a face while the other face is either stress free or rigidly clamped has been considered by Bycroft[l] and Gladwell[Z].Collins[3f has considered the torsional os~~ations of an elastic layer by two equal discs attached symmetricallyto opposite faces of the layer and also the angles through which the discs are rotated are equal but opposite in sign. In this paper we consider the generalization of the static part of Gladwell[2] and Collins[3] problems. Two discs are taken of different sizes and the angles of rotation of the discs are also taken as different. The method used here is different from Collins[3], who has solved the symme~c~ problem. The solution of the problem is reduced to a pair of simultaneous Fredholm integral equations which are then solved by the method of iteration as well as numerically. Analytic expressions are obtained for the torques and their numerical values are displayed graphically. 2. FORMULATION OF THE PROBLEM AND ITS SOLUTION
We consider an infinite, isotropic, homogeneous elastic layer bounded by the planes z = 0, z = h of a cylindricalcoordinate system (r, 0, z). Let the rigid discs of radii 6 and a be attached to the faces z = 0, h respectively such that the line joining the centres of the discs be the z-axis, When the discs are rotated through different angles @ and o respectively, and the rest of the faces z = 0, h are stress free, the displacement component u”assumes the form (0, u, 0), and the equation of equ~i~um is given by a%
1 av
v
ar’+;;--#+~=o.
a% a2
The only non-zero components of stress are given by uw = CLavla2,o;b
=
p(avlar - v/r),
(2)
where CCis the modulus of rigidity of the material. The displacement v as well as the stress components are functions of r, 2 only. The funds conditions may be written as v (r, 6) = cur,
O a,
aiYc(r,h)=O, v(r, 0) = #Jr,
cr, (I; 0)= 0,
0<
r-C 6, r>b,
where Q and /3 are arbitrary angles of rotation of the discs. tpermaaeataddress:30/3 Gandhi Colony, Mu&far-Nagar,U.P., India. 171
(31
172
B.M. SINGH andR.S.DHALIWAL
The solution of eqn (1) may be written in the form
I
u(r, z) = 0m1(co,:, (uh) [A sinh {u(h - z)} + B sinh (uz)]J,(ur) du,
(4)
where A(u) and B(u) are arbitrary functions of u to be determined from the boundary conditions and J,( ) is a Bessel function of the first kind and of order one. Making use of (2) and (4), the boundary conditions (3) lead to the following pair of dual integral equations
(5)
I
w
0
C(u)J,(ur) du = 0,
r > a,
~m[~[l+si~~~~)]-us~~~uuh)]I’(u’)du
I
O
=-Br,
(7)
m
0
where
(6)
D(u)J,(ur) du = 0, r > b,
(8)
C(u) = B(u)-A(u)/cosh(uh), D(u) = -A(u)+ B(u)/cosh (uh).
(9)
If we introduce the representations C(u)=~u~~O(r)sin(ut)dt, 0
(10)
D(u) = 3 u .* $(t) sin (ut) dt, I
(11)
the eqns (6) and (8) are identically satisfied and the eqns (5) and (7) are also satisfied if 4(t) and $(t) satisfy the following simultaneous Fredholm integral equations =2cwt, O
#(t)-la~(u)K(u,t)du-lbJI(u)G(u,~)dll 0 0
O
(12) (13)
where
e-’ -sin($sin(F)dr, sinh (r) G(u, t) = f
_/Om&
sin (F) sin (z) dr.
(14)
(15)
3.ITERATIVESOLUTIONOFTHESIMULTANEOUSFREDHOLM INTEGRAL EQNS (12) AND (13)
To obtain the iterative solution of the integral equations for values of h > max (a, b), we replace the sine terms in the integrands of eqns (14) and (15) by their power series expansion and find that K(u, C)= -
il: K”(U,
t)z”/h2”+‘,
n-1
G(u, t) =
2 n-1
K.(u, t)H,/hZ”+‘,
(16)
Torsion of an elastic layer by two circular dies
173
where K”(U, t) = (-l)“[(u
mee-rr2”dr sinh(r)=
-,)*” -(u + 1)‘“]. [(2n + 1)
n_ , 2
’
822”
-
’
(17) ,.a.,
in which l(n) denotes the Reimann Zeta-function. We now assume the solution of eqns (12) and (13) in the following form m
W) = “p”(rYh”,
$0) =
“go Ib”W/h”.
(19)
Now substituting for K(u, t), G(u, t), 4(t) and S(t) from (16) and (19) in (12) and (13) and equating the coefficients of like powers of h, we obtain f$&) = 2at,
$00) = -2/3t
(20) du.
n 2
1,
O< t< a, K,,,(u, t)~“-2,-,(u)-H,~K,(u,
tbL&)]
du,
(21)
n 3 1, O
(22)
where [n] denotes the integral part of n. From (19)-(22), the iterative solution can be obtained upto any order of h. Making analytical calculations upto n = 5, we find that 4(t) =
40)
=
1 2t 2at - --hS 3r [(3)(aa3+7/3b3)
-w
+
L2’((3)(7~a3+@b3) h’ 3r
-~~~(5)[31aa3(3a2+5~*)+/3b3(36’+5t2)]+0(~~),
(23)
where f(3) = 1.202, 3(S) =c1.037. 4. EXPRESSIONS
(24)
FOR THE TORQUES
The expressions for the torque about the centre of the disc at the faces z = 0 and I = h are respectively given by T, = -2s
I
b
0
t*r,.z,(r,0) dr,
T2 = -2~
I
a
0
r*u,(r, h) dr.
m
Using (2), (4), (9)-(ll), we find that eqns (25) may be written as T, = -8~
I
b
0
t$(t) dt,
T2 = -8~
a@(t)dr. I0
(26)
For h > max (a, b), the expressions (19) for 4(t) and t,%(t)may be substituted in (26) and from IES Vd. 15. No. Z-B
B. M. SINGH and R. S. DHALIWAL
174
(23) and (26), we find that 8 T1 =j’“b’
2 28 -,,1,~(3)t7ua3+Bb3)+5h,a -L~(5~2~bJ+31aa3(a2+b2)}
1
+O(h-6), (27)
When the face z = h is stress free and the face z = 0 has a disc of radius b attached to it, we let a tend to zero and find that T2 tends to zero and the expression (27) for Tl agrees with the expression for the torque obtained by Gladwell ([2], p. 1021).while noting the change in notation. If we take a =b and a =-/I, we find that T, = T2 and u(h/2, r) = 0. In this case the expression for T, from (27) agrees with the one obtained by Collins ([31, p. 226). 5. NUMERICAL
RESULTS
Numerical values of the torques T, and T2 have been calculated by numerical integration of the expressions (26) for 6 = 1, j3 = 10m2,a = 1,2,3,4,5, h = 1,2,3,4,5 and a = - 10m2,-2 x 10b2, -3 x 10m2.Numerical values of S(t) and 4(t) required in (26) have been evaluated from the
Fig. 1. Variationof T, and T2againsth for b = 1, @= lo-‘.
.-
38-------
74
3.n2.6-
3
4
-h Fis. 2. Variation
of T, and T, againsth for b = 1, fi = lo-‘.
5
Torsion of an elastic layer by hvo circulardies
175
3.8 .
3.4 -
------______ __________----------
____________-----------02c
2
3
4
5
-h
Fig. 3. Variationof T, and T2againsth for b = I, fi = IO-‘.
eqns (12) and (13) by reducing these integral equations to a set of algebraic equations. The values of T, and T, are displayed against h, the layer thickness, for various values of a and a in Figs. l-3. REFERENCES [l] G. N. BYCROFT,PM. Tmns. R. Sm. Ser. A 48, 327 (1955). [2] G. M. L. GLADWELL, ht. L Engng Sci. 7, 1011(1%9). 131W. D. COLLINS, Pm. Land. M& Sm. ThirdSer. 12, 226 (1962). (Received 9 July 1976)
TORSION OF AN ELASTIC LAYER BY TWO CIRCULAR DIES BRIJ MOHAN SINGHt and RANJIT S. DHALIWAL Departmentof Mathematicsand Statistics,TbeUniversityof Calgary,Calgary,Canada (Communicatedby I. N. SNEDDDN) Abstract-Two circulardiscs of rigidmaterialand of differentradiiare bondedto the opposite faces of an Infiniteelastic layer.The torques T, and Tzhave been calculatedwhen the discs are rotated&roughdiffered angles.
1. INTRODUCTION
THEPROBLEMof the torsional oscillationsof the elastic layer by one disc attached over a face while the other face is either stress free or rigidly clamped has been considered by Bycroft[l] and Gladwell[Z].Collins[3f has considered the torsional os~~ations of an elastic layer by two equal discs attached symmetricallyto opposite faces of the layer and also the angles through which the discs are rotated are equal but opposite in sign. In this paper we consider the generalization of the static part of Gladwell[2] and Collins[3] problems. Two discs are taken of different sizes and the angles of rotation of the discs are also taken as different. The method used here is different from Collins[3], who has solved the symme~c~ problem. The solution of the problem is reduced to a pair of simultaneous Fredholm integral equations which are then solved by the method of iteration as well as numerically. Analytic expressions are obtained for the torques and their numerical values are displayed graphically. 2. FORMULATION OF THE PROBLEM AND ITS SOLUTION
We consider an infinite, isotropic, homogeneous elastic layer bounded by the planes z = 0, z = h of a cylindricalcoordinate system (r, 0, z). Let the rigid discs of radii 6 and a be attached to the faces z = 0, h respectively such that the line joining the centres of the discs be the z-axis, When the discs are rotated through different angles @ and o respectively, and the rest of the faces z = 0, h are stress free, the displacement component u”assumes the form (0, u, 0), and the equation of equ~i~um is given by a%
1 av
v
ar’+;;--#+~=o.
a% a2
The only non-zero components of stress are given by uw = CLavla2,o;b
=
p(avlar - v/r),
(2)
where CCis the modulus of rigidity of the material. The displacement v as well as the stress components are functions of r, 2 only. The funds conditions may be written as v (r, 6) = cur,
O
aiYc(r,h)=O, v(r, 0) = #Jr,
cr, (I; 0)= 0,
0<
r-C 6, r>b,
where Q and /3 are arbitrary angles of rotation of the discs. tpermaaeataddress:30/3 Gandhi Colony, Mu&far-Nagar,U.P., India. 171
(31
172
B.M. SINGH andR.S.DHALIWAL
The solution of eqn (1) may be written in the form
I
u(r, z) = 0m1(co,:, (uh) [A sinh {u(h - z)} + B sinh (uz)]J,(ur) du,
(4)
where A(u) and B(u) are arbitrary functions of u to be determined from the boundary conditions and J,( ) is a Bessel function of the first kind and of order one. Making use of (2) and (4), the boundary conditions (3) lead to the following pair of dual integral equations
(5)
I
w
0
C(u)J,(ur) du = 0,
r > a,
~m[~[l+si~~~~)]-us~~~uuh)]I’(u’)du
I
O
=-Br,
(7)
m
0
where
(6)
D(u)J,(ur) du = 0, r > b,
(8)
C(u) = B(u)-A(u)/cosh(uh), D(u) = -A(u)+ B(u)/cosh (uh).
(9)
If we introduce the representations C(u)=~u~~O(r)sin(ut)dt, 0
(10)
D(u) = 3 u .* $(t) sin (ut) dt, I
(11)
the eqns (6) and (8) are identically satisfied and the eqns (5) and (7) are also satisfied if 4(t) and $(t) satisfy the following simultaneous Fredholm integral equations =2cwt, O
#(t)-la~(u)K(u,t)du-lbJI(u)G(u,~)dll 0 0
O
(12) (13)
where
e-’ -sin($sin(F)dr, sinh (r) G(u, t) = f
_/Om&
sin (F) sin (z) dr.
(14)
(15)
3.ITERATIVESOLUTIONOFTHESIMULTANEOUSFREDHOLM INTEGRAL EQNS (12) AND (13)
To obtain the iterative solution of the integral equations for values of h > max (a, b), we replace the sine terms in the integrands of eqns (14) and (15) by their power series expansion and find that K(u, C)= -
il: K”(U,
t)z”/h2”+‘,
n-1
G(u, t) =
2 n-1
K.(u, t)H,/hZ”+‘,
(16)
Torsion of an elastic layer by two circular dies
173
where K”(U, t) = (-l)“[(u
mee-rr2”dr sinh(r)=
-,)*” -(u + 1)‘“]. [(2n + 1)
n_ , 2
’
822”
-
’
(17) ,.a.,
in which l(n) denotes the Reimann Zeta-function. We now assume the solution of eqns (12) and (13) in the following form m
W) = “p”(rYh”,
$0) =
“go Ib”W/h”.
(19)
Now substituting for K(u, t), G(u, t), 4(t) and S(t) from (16) and (19) in (12) and (13) and equating the coefficients of like powers of h, we obtain f$&) = 2at,
$00) = -2/3t
(20) du.
n 2
1,
O< t< a, K,,,(u, t)~“-2,-,(u)-H,~K,(u,
tbL&)]
du,
(21)
n 3 1, O
(22)
where [n] denotes the integral part of n. From (19)-(22), the iterative solution can be obtained upto any order of h. Making analytical calculations upto n = 5, we find that 4(t) =
40)
=
1 2t 2at - --hS 3r [(3)(aa3+7/3b3)
-w
+
L2’((3)(7~a3+@b3) h’ 3r
-~~~(5)[31aa3(3a2+5~*)+/3b3(36’+5t2)]+0(~~),
(23)
where f(3) = 1.202, 3(S) =c1.037. 4. EXPRESSIONS
(24)
FOR THE TORQUES
The expressions for the torque about the centre of the disc at the faces z = 0 and I = h are respectively given by T, = -2s
I
b
0
t*r,.z,(r,0) dr,
T2 = -2~
I
a
0
r*u,(r, h) dr.
m
Using (2), (4), (9)-(ll), we find that eqns (25) may be written as T, = -8~
I
b
0
t$(t) dt,
T2 = -8~
a@(t)dr. I0
(26)
For h > max (a, b), the expressions (19) for 4(t) and t,%(t)may be substituted in (26) and from IES Vd. 15. No. Z-B
B. M. SINGH and R. S. DHALIWAL
174
(23) and (26), we find that 8 T1 =j’“b’
2 28 -,,1,~(3)t7ua3+Bb3)+5h,a -L~(5~2~bJ+31aa3(a2+b2)}
1
+O(h-6), (27)
When the face z = h is stress free and the face z = 0 has a disc of radius b attached to it, we let a tend to zero and find that T2 tends to zero and the expression (27) for Tl agrees with the expression for the torque obtained by Gladwell ([2], p. 1021).while noting the change in notation. If we take a =b and a =-/I, we find that T, = T2 and u(h/2, r) = 0. In this case the expression for T, from (27) agrees with the one obtained by Collins ([31, p. 226). 5. NUMERICAL
RESULTS
Numerical values of the torques T, and T2 have been calculated by numerical integration of the expressions (26) for 6 = 1, j3 = 10m2,a = 1,2,3,4,5, h = 1,2,3,4,5 and a = - 10m2,-2 x 10b2, -3 x 10m2.Numerical values of S(t) and 4(t) required in (26) have been evaluated from the
Fig. 1. Variationof T, and T2againsth for b = 1, @= lo-‘.
.-
38-------
74
3.n2.6-
3
4
-h Fis. 2. Variation
of T, and T, againsth for b = 1, fi = lo-‘.
5
Torsion of an elastic layer by hvo circulardies
175
3.8 .
3.4 -
------______ __________----------
____________-----------02c
2
3
4
5
-h
Fig. 3. Variationof T, and T2againsth for b = I, fi = IO-‘.
eqns (12) and (13) by reducing these integral equations to a set of algebraic equations. The values of T, and T, are displayed against h, the layer thickness, for various values of a and a in Figs. l-3. REFERENCES [l] G. N. BYCROFT,PM. Tmns. R. Sm. Ser. A 48, 327 (1955). [2] G. M. L. GLADWELL, ht. L Engng Sci. 7, 1011(1%9). 131W. D. COLLINS, Pm. Land. M& Sm. ThirdSer. 12, 226 (1962). (Received 9 July 1976)