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110 Copyright c 2006 Tech Science Press CMES, vol.12, no.2, pp.109-119, 2006

nular cases. However, it may encounter difficulty for the

eccentric case. Two requirements are needed: degenerate

kernel expansion must be available and distinction of in-

terior and exterior expression must be separated. There-

fore, the collocation angle of is not in the range 0 to2 in our adaptive observer system. This is the reason

why we can not formulate in terms of Galerkin formula-

tion using orthogonal properties twice. Free of worrying

how to choose the collocation points, uniform collocation

along the circular boundary yields a well-posed matrix.

On the other hand, Bird and Steele [Bird M. D.; Steele

C. R. (1991)] have also used separated solution proce-

dure for bending of circular plates with circular holes in

a similar way of the Trefftz method and addition theorem.

In this paper, the null-field integral equation is utilized

to solve the Saint-Venant torsion problem of a circularshaft weakened by circular holes. The mathematical for-

mulation is derived by using degenerate kernels for fun-

damental solution and Fourier series for boundary den-

sity in the null-field integral equation. Then, it reduces

to a linear algebraic equation. After determining the un-

known Fourier coefficients, series solutions for the warp-

ing function and torsional rigidity are obtained. Numeri-

cal examples are given to show the validity and efficiency

of our formulation.

2 Formulation of the problem

What is given in Figure 1 is a circular bar weakened by

Ncircular holes placed on a concentric ring of radius b.

Figure 1 : Cross section of bar weakened by

N (N= 3) equal circular holes

The radii of the outer circle and the inner holes are R

and a, respectively. The circular bar twisted by couples

applied at the ends is taken into consideration. Follow-

ing the theory of Saint-Venant torsion [Timoshenko S. P.;

Goodier J. N. (1970)], we assume the displacement fieldto be

u=yz, v= xz, w= (x,y), (1)

where is the angle of twist per unit length along the z

direction and is the warping function. According to the

displacement field in Eq. (1), the strain components are

x= y = z = xy = 0, (2)

xz = wx+ uz = (x

y), (3a)

yz =w

y+v

z= (

y+x), (3b)

and their corresponding components of stress are

x= y = z = xy = 0, (4)

xz = (

xy), yz = (

y+x), (5)

where is the shear modulus. There is no distortion in

the planes of cross sections since x= y = z = xy = 0.We have the state of pure shear at each point defined by

the stress components xz and yz. The warping function

must satisfy the equilibrium equation

2

x2+2

y2= 0 in D , (6)

where the body force is neglected and D is the domain.

Since there are no external forces on the cylindrical sur-

face, we have tx= ty= tz = 0. By substituting the normalvector, the only zero tz becomes

tz = xznx+yzny = 0 on B. (7)

By substituting (5) into (7) and rearranging, the boundary

condition is

xnx+

yny = ynxxny = n =

non B, (8)