LECTURE 19
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 (ii) t
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
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 45
A rectangular element cut from the outer layer of a twisted shaft with sides at 45
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
Considering a solid and hollow shafts of the same length 'l' and density 'r' with d
Hence the reduction in weight would be just 25%.
A stepped solid circular shaft is built in at its ends and subjected to an externally applied torque. T
Thus T [from static principles] Where T From consideration of consistent deformation, we see that the angle of twist in each portion of the shaft must be same. i.e q using the relation for angle of twist
So the defines the ratio of T So by solving (1) & (2) we get
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 T |