Jnp

with <p = angle through which a tangent to the boundary rotates in travelling around the re-entrant position (radians) and r being taken as negative.

For standard thick-walled open sections such as T, 1, Z, angle and channel sections Roark also introduces formulae for angles of twist based upon the same inscribed circle procedure parameters.

5.6. Thin-walled closed tubes of non-circular section (Bredt-Batho theory)

Consider the thin-walled closed tube shown in Fig. 5.6 subjected to a torque T about the Z axis, i.e. in a transverse plane. Both the cross-section and the wall thickness around the periphery may be irregular as shown, but for the purposes of this simplified treatment it must be assumed that the thickness does not vary along the length of the tube. Then, if r is the shear stress at B and r' is the shear stress at C (where the thickness has increased to t') then, from the equilibrium of the complementary shears on the sides AB and CD of the element shown, it follows that xt dz = r Y dz t t = x't'

i.e. the product of the shear stress and the thickness is constant at all points on the periphery of the tube. This constant is termed the shear flow and denoted by the symbol q (shear force per unit length).

The quantity q is termed the shear flow because if one imagines the inner and outer boundaries of the tube section to be those of a channel carrying a flow of water, then, provided that the total quantity of water in the system remains constant, the quantity flowing past any given point is also constant.

Fig. 5.6. Thin-walled closed section subjected to axial torque.

At any point, then, the shear force Q on an element of length ds is Q — xtds = qds and the shear stress is q/t.

Consider now, therefore, the element BC subjected to the shear force Q = qds — xtds. The moment of this force about O

= dT = Qp where p is the perpendicular distance from O to the force Q.

Therefore the moment, or torque, for the whole section

But the area COB = \ base x height = \pds i.e. dA = ^pds or 2dA = pds torque T = 2q j dA T = 2qA

where A is the area enclosed within the median line of the wall thickness. Now, since

where t is the thickness at the point in question.

It is evident, therefore, that the maximum shear stress in such cases occurs at the point of minimum thickness.

Consider now an axial strip of the tube, of length L, along which the thickness and hence the shear stress is constant. The shear strain energy per unit volume is given by

Thus, with thickness t, width ds and hence V = tLds f r2 U = / — tLds j 2 G

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