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**Twisting of hollow section**

In this section, as in the previous, we consider a bar subjected to twisting moments at its ends. The bar axis is straight and the shape of the cross section is constant along the axis. The bar is assumed to have a hollow cross section, i.e, a multiply connected domain. That is any closed curves in the cross section cannot be shrunk to a point in the domain itself without leaving the domain. Let us assume that the boundary of the cross section is defined by a function, f(x, y) = 0, the set of points constituting this boundary by âˆ‚A_{o}, and the enclosed area by A_{o}. The boundary of the i^{th} void in the interior of the cross section is defined by the function gi(x, y) = 0, the set of points constituting this boundary of the i th void by âˆ‚A_{i} and the area enclosed by the void, A_{i}. Thus, the area of the cross section, A_{cs} is given by

where we have assumed that there are N voids. Thus, for the cross section shown in figure 9.22, with the boundary of the cross section in the form of an ellipse centered about origin and oriented such that the major and minor axis coincides with the e_{x} and e_{y} directions respectively and having two circular shaped voids of radius, r centered at (Â±xo, 0), the functions f(x, y) and gi(x, y) would be: f(x, y) = x^{2}/a^{2}+y^{2}/b^{2}âˆ’ 1, g_{1}(x, y) = (xâˆ’xo)^{2}+y^{2}âˆ’r^{2 }and g_{2}(x, y) = (x+x_{o}) ^{2}+y^{2}âˆ’r^{2} . Consequently, this body is assumed to occupy a region in Euclidean point space, defined by

where L is a constant. Since, we are interested in the case where in the bar is subjected to pure end torsional moment, the traction boundary conditions are:

where

and T is a constant. Here en is obtained such that it is in the xy plane, perpendicular to grad(f) and is a unit vector. Similarly, e_{mi} is obtained such that it is in the xy plane, perpendicular to grad(g_{i}) and is a unit vector. Recognize that grad(f) gives the tangent vector to the cross section for any point in âˆ‚A_{o }and grad(g_{i}) is the tangent vector to the cross section at any point in âˆ‚A_{i} . Further, since one end of this body is assumed to be fixed against twisting but free to displace axially and the other end free to displace in all directions, the displacement boundary conditions for this problem is

u(x, y, 0) = u_{z}(x, y)e_{z, } (9.114)

where u_{z} is any function of x and y. Here we have assumed that the surface of the body defined by z = 0 is fixed against twisting but free to displace axially and the other surfaces are free to displace in all directions. Thus, we find that the boundary value problem is similar to that in the previous section (section 9.3) except for the additional boundary condition (9.108). Hence, we assume that the stress is as given in equation (9.44), in terms of the Prandtl stress function, Ï† = Ï†Ë†(x, y). Following the standard procedure, as outlined in section 9.3, it can be shown that the stress function should satisfy,

where Âµ is the shear modulus, and â„¦ is the angle of twist per unit length. For the assumed stress state, (9.44), the boundary condition (9.107) requires that grad(Ï†) Â· grad(f) = 0 on âˆ‚A_{o}. Thus,

where C_{0} is a constant. As before, since, the stress and displacement fields depend only in the derivative of the stress potential and not its value at a location, it suffices to find this stress function up to a constant. Therefore, we arbitrarily set C_{0} = 0, understanding that it can take any value and that the stress and displacement field would not change because of this. Hence, equation (9.116) reduces to requiring,

The boundary condition (9.108) requires that grad(Ï†) Â· grad(g_{i}) = 0 on âˆ‚A_{i} . Hence,

where C_{i} â€™s are constants. Though Ï† is a constant over all free surfaces, the value of this constant can differ between distinct free surfaces, i.e., âˆ‚A_{i} . Also, recognize that these constants cannot be set to zero, as we have already set C_{0} = 0. The boundary condition (9.109) and (9.110) for the stress state (9.44) require

Appealing to Greenâ€™s theorem for multiply connected domains, (2.278) and (2.279), the above equation (9.119) reduces to

where we have used (9.116) and (9.118) and the fact that

Finally, the boundary conditions (9.111) and (9.112) for the stress state (9.44) yields,

Noting that

where we have used the Greenâ€™s theorem for multiply connected domain (2.278) and (2.279) and equations (9.117) and (9.118). Recognize that from Greenâ€™s theorem,

where A_{i} is the area enclosed by the void. Substituting the above equations (9.123) and (9.124) in (9.122) we obtain,

Thus, we have to find Ï† such that the governing equation (9.115) has to hold along with the boundary conditions (9.117) and (9.118). Then, we use equation (9.126) to find the torsional moment required to realize a given angle of twist per unit length, â„¦.

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