Distribution of shear stresses in circular Shafts subjected to torsion

Distribution of shear stresses in circular Shafts subjected to torsion : The simple torsion equation is written as This states that the shearing stress varies directly as the distance ‘r' from the axis of the shaft and the following is the stress distribution in the plane of cross section and also the complementary shearing stresses in an axial plane. Hence the maximum strear stress occurs on the outer surface of the shaft where r = R The value of maximum shearing stress in the solid circular shaft can be determined as From the above relation, following conclusion can be drawn (i) t maxm µ T (ii) t maxm µ 1/d 3 Power Transmitted by a shaft: In practical application, the diameter of the shaft must sometimes be calculated from the power which it is required to transmit. Given the power required to be transmitted, speed in rpm ‘N' Torque T, the formula connecting These quantities can be derived as follows     Torsional stiffness: The torsional stiffness k is defined as the torque per radian twist . For a ductile material, the plastic flow begins first in the outer surface. For a material which is weaker in shear longitudinally than transversely – for instance a wooden shaft, with the fibres parallel to axis the first cracks will be produced by the shearing stresses acting in the axial section and they will upper on the surface of the shaft in the longitudinal direction. In the case of a material which is weaker in tension than in shear. For instance a, circular shaft of cast iron or a cylindrical piece of chalk a crack along a helix inclined at 450 to the axis of shaft often occurs. Explanation: This is because of the fact that the state of pure shear is equivalent to a state of stress tension in one direction and equal compression in perpendicular direction. A rectangular element cut from the outer layer of a twisted shaft with sides at 450 to the axis will be subjected to such stresses, the tensile stresses shown will produce a helical crack mentioned. TORSION OF HOLLOW SHAFTS: From the torsion of solid shafts of circular x – section , it is seen that only the material at the outer surface of the shaft can be stressed to the limit assigned as an allowable working stresses. All of the material within the shaft will work at a lower stress and is not being used to full capacity. Thus, in these cases where the weight reduction is important, it is advantageous to use hollow shafts. In discussing the torsion of hollow shafts the same assumptions will be made as in the case of a solid shaft. The general torsion equation as we have applied in the case of torsion of solid shaft will hold good Hence by examining the equation (1) and (2) it may be seen that the t maxm in the case of hollow shaft is 6.6% larger then in the case of a solid shaft having the same outside diameter. Reduction in weight: Considering a solid and hollow shafts of the same length 'l' and density 'r' with di = 1/2 Do Hence the reduction in weight would be just 25%. Illustrative Examples : Problem 1 A stepped solid circular shaft is built in at its ends and subjected to an externally applied torque. T0 at the shoulder as shown in the figure. Determine the angle of rotation q0 of the shoulder section where T0 is applied ? Solution: This is a statically indeterminate system because the shaft is built in at both ends. All that we can find from the statics is that the sum of two reactive torque TA and TB at the built – in ends of the shafts must be equal to the applied torque T0 Thus     TA+ TB = T0                  ------   (1) [from static principles] Where TA ,TB are the reactive torque at the built in ends A and B. wheeras T0 is the applied torque From consideration of consistent deformation, we see that the angle of twist in each portion of the shaft must be same. i.e    qa = q b = q 0 using the relation for angle of twist  N.B: Assuming modulus of rigidity G to be same for the two portions So the defines the ratio of TA and TB So by solving (1) & (2) we get Non Uniform Torsion: The pure torsion refers to a torsion of a prismatic bar subjected to torques acting only at the ends. While the non uniform torsion differs from pure torsion in a sense that the bar / shaft need not to be prismatic and the applied torques may vary along the length. Here the shaft is made up of two different segments of different diameters and having torques applied at several cross sections. Each region of the bar between the applied loads between changes in cross section is in pure torsion, hence the formula's derived earlier may be applied. Then form the internal torque, maximum shear stress and angle of rotation for each region can be calculated from the relation The total angle to twist of one end of the bar with respect to the other is obtained by summation using the formula If either the torque or the cross section changes continuously along the axis of the bar, then the å (summation can be replaced by an integral sign ( ∫ ). i.e We will have to consider a differential element. After considering the differential element, we can write Substituting the expressions for Tx and Jx at a distance x from the end of the bar, and then integrating between the limits 0 to L, find the value of angle of twist may be determined.