Shear Stress in Torsion of Circular Shafts--Not a True Stress, but an Equivalent Stress

Abstract

The present research argues that the shear stress of shaft in torsion is not a true stress but an equivalent stress, based on the phenomena that the stress disproves the plane assumption, that the body part in torsion is unable to maintain equilibrium, and that shear stress on the longitudinal section drawn from reciprocal theorem is not only self-contradictory, but also in conflict with Newton's third law. Namely, the angle strain generated by torque equals that produced by shear stress. The above phenomena are caused by equivalent stresses. Introduction Shear stress is indispensable in designing the torsion of circular shafts in engineering. However, despite the use of the finite element method, which is precise, the shafts often break or lack in rigidity, which has frequently led to accidents and resulted in great loss. The reason lies in the false belief that there is true shear stress in the body of torsion. This fundamental error has mistaken the true shear stress for an equivalent stress in calculation; hence, it constitutes the principle reason for the breaking of shafts and the inadequacy of rigidity. The present paper will illustrate that the shear stress in torsion of shafts is not a true stress but an equivalent stress from the following perspectives: 1. the unequilibrium in torsion of circular shafts; 2. the self-contradiction of torsional shear stresses and 3. the contradiction between torsional shear stresses and the Newton's third law. The Torsional Shear Stress Disproving the Plane Assumption In the deduction of the torsional shear stress, a basic assumption is that during the deformation, cross sections still remain planes without the change in form and size[5, 10, 11]. If this assumption is disproved, it follows that the formula deducted from it is false. Fig. 1(a) shows the distribution of the shear stress on the cross section and the vertical section of the round shaft under the internal torque n M . A wedge is formed by the two intersecting sections through the axis oo' as shown in Fig. 1(b). The stress distribution of the wedge, as shown in Fig. 1(b), is drawn from the torsion deformation formula and the theorem of conjugate shear stresses. To make it clearer, abcd of the square is marked in Fig. 1(c) so as to study the deformation of the square[2, 3]. From the principle stress formula of the pure shear[1, 4, 7, 8], we get     3 1   , . When the principle stress acts on the two diagonals ac and bd, ac is tensioned so that a moves to a' and c to c' along the diagonal. In the same way, bd is compressed so that b moves to b' and d to d'. It could be seen that under the shear stress, the square has been turned into a lozenge. Obviously, a' is not on the original cross section o'ab while c' is not on ocd, as shown in Fig. 1(c). The cross section is slanted and it does not remain the plane it was. The elastic theory is contradictory with itself. Therefore, the shear stress formula and the deformation formula which are deducted from it are false. When the thin-wall circular tube is twisted as shown by Fig. 2(a), mm and nn, the two parallel circumference lines which are perpendicular to the axis x, still remain perpendicular to the axis x. In this International Conference on Mechanics and Civil Engineering (ICMCE 2014) © 2014. The authors Published by Atlantis Press 241 case, the cross section remains the plane with only a change of angle, and the plane assumption is valid. Only torque can account for it because if it is due to the internal shear stress, the effect will be different as shown by Fig. 2(b): instead of being perpendicular to the axis x, circumference lines mm and nn turn to be ellipses m'm' and n'n' slanting to the axis x. Thus we can see that, both theories and experiments proves that the theory of shear stress in torsion cannot guarantee the validity of the plane assumption. Fig. 1 The Shear Stress in Torsion of Circular Shaft Denies the Plane Assumption (a) (b) Fig. 2 The Torsional Shear Stress Violates the Plane Assumption in Torsion of the Thin-wall Circular Tube The Unequilibrium in the Torsion of the Cylinder Example 1 As shown in Fig.3(a), the cylinder with the radius of R, which three force couples act on, is in equilibrium. Let M M M C A   , with the equilibrium condition M M M M C A B 2     , its torque should be what is shown in Fig. 3(b).