# Page:EB1911 - Volume 25.djvu/1048

1022
STRENGTH OF MATERIALS

elasticity, a radius CD turns round to CD', and a line AD drawn at any distance r froin the axis, and originally straight, changes into the helix AD'. Let be the angle which this helix makes with lines parallel to the axis, or in other words the angle of shear at the distance

Fig. 39. r from the axis, and let i be the angle of twist DCD'. Taking two sections at a distance dx from one another, we have^ the arc Odx = rdi. Hence q, the intensity of shearing stress in a plane of cross-section, varies as r, since q = C0 = Cr di/dx. The resultant moment of the whole shearing stress on each plane of cross-section is equal to the twisting moment M. Thus

J2xr 2 gdr = M. Calling ri the outside radius (where the shearing stress is greatest) and 91 its intensity there, we have g = rgi/n, and hence, for a solid shaft, Si = 2M/nTi s . For a hollow shaft with a central hole of radius U the same reasoning applies : the limits of integration are now n and ri, and

2Mn 91 -*<nÂ« -ftf)- The lines of principal stress are obviously helices inclined at 45 to the axis.

If the shaft has any other form of section than a solid or sym- metrical hollow circle, an originally straight radial line becomes warped when the shaft is twisted, and the shearing stress is no longer proportional to the distance from the axis. The twisting of shafts of square,- triangular and other sections has been investigated by Saint- Venant. In a square shaft (side = Js) the stress is greatest at the middle of each side, and its intensity there is g, = M/o-28ifc 3 .

For round sections the angle of twist per unit of length is

in hollow shafts.

Oi 2M . ... . 2M

• = cV, = 7U;* m solld and Â«C<r,Â«-fiÂ«)

In what has been said above it is assumed that the stress is within the limit of elasticity. When the twisting couple is increased so that this limit is passed, plastic yielding begins in the outermost layer, and a larger proportion of the whole stress falls to be borne by layers nearer the centre. The case is similar to that of a beam bent beyond the elastic limit, described above. If we sup- pose the process of twisting to be continued, and that after passing the limit of elasticity the material is capable of much distortion without further increase of shearing stress, the distribution of stress on any cross section will finally have an approximately uniform value q', and the moment of torsion will be

f r i l 2*r>q'dr = WW-ri i ).

In the case of a solid shaft this gives for M a value greater than it has when the stress in the outermost layer only reaches the intensity g', in the ratio of 4 to 3. It is obvious from this consideration that the ultimate strength of a shaft to resist torsion is no more deducible from a knowledge of the ultimate shearing strength of the material than the ultimate strength of a beam to resist bending is deducible from a knowledge of the tensile and crushing strength. It should be noticed also that as regards ultimate strength a solid shaft has an important advantage over a hollow shaft of the same elastic strength, or a hollow shaft so proportioned that the greatest working intensity of stress is the same as in the solid shaft.

Twisting Combined with Bending. â€” This important practical case is realized in a crank-shaft (fig. 40). Let a force P be applied at the

crank-pin A at right angles to the plane of the crank. At any section of the shaft C (between the crank and the bearing) there is a twisting moment Mi = P. AB and a bending mo- ment M 2 = P.BC. There is also a direct shearing force P, but this does not _ require to be taken into account in calculating the stress at points at the top or bottom of the circumference (where the intensity is greatest), since the direct shearing stress is distributed

so that its intensity is zero at these points. The stress there is consequently made up of longitudinal normal stress (due to bending), pt =4M 2 /irri 3 , and shearing stress (due to torsion), gi = 2Mi/xn 3 . Combining these, as in Â§ 64, we find for the prin- cipal stresses r = 2JM 8 Â±V(Mi 2 -r-M s 2 )}/irri 3 ,orr = 2P(BCÂ±AC)/irn 3 . The greatest shearing stress is2P. ACjirri 3 , and the axes of principal stress are inclined so that tan 20 =Mi/M 2 = AB/BC. The axis of greater principal stress bisects the angle ACB. 