SIMPLE STRAINS AND STRESSES 81 
____ In stating that the shear strain on the small square element or thin 
layer MN is equal to 2’//’, it is assumed that only the edges of that 
- layer are subjected to shear stress, and that there is no stress on the 
square faces, It is obvious that there cannot be any stress on the outer 
square face, and since the angular movement of each particle of the shaft 
about the axis of the shaft is proportional to its distance from that axis, 
particles which are on the same radial line when the shaft is unstrained 
will remain on that line as the latter revolves ; there can therefore be no 
relative movement between the layer MN and the layer next it within 
the shaft, and therefore there cannot be any stress on the inner face of 
the layer MN. | 
96. Moment of Resistance of a Shaft to Torsion—It has been 
shown that the angle of twist of a circular shaft is 6 =2/1/Cd or fl/Cr. 
Hence f= 0Cr//, and for given values of 0, C, and /, f is proportional to r. 
Now for all parts of a given shaft subjected to a given torque, 9, C, and Z 
are the same, therefore if a circular shaft be conceived to be made up of a 
number of thin tubes, the Shear stress on any one of them will be pro- 
portional to its radius. Let, be the mean radius of one of these tubes, a, 
_ the area of its cross section, and f, the shear stress on it. Then 7,/f=7,/7, 
 orf,=r,f/r. The total shear stress on the cross section of this tube is 
Ji%, and the moment of this about the axis is far, Hence the 
moment of resistance of this tube to torsion is Layri =t, where I, is 
the polar moment of inertia of the cross section of the tube about its axis. 
In like manner the moment of resistance to torsion of each of the other 
_ tubes is the factor f/r multiplied by the polar moment of inertia of its 
ross section about the axis. Hence M, the moment of resistance of the 
___ whole of the tubes, or of the solid shaft, is f/r multiplied by the sum of 
_ the polar moments of inertia of the separate annular parts of the cross 
section, which is equal to f/7 multiplied by the polar moment. of inertia 
_ of the whole cross section. But the polar moment of a circle of radius r 
' A 
about an axis through its centre and perpendicular to its plane is as 
_ therefore M=5r5f= ig? 
The following is another way of determining the moment of resistance 
of a circular shaft to torsion. Consider a small sector OAB of the cross 
- ---- d-----» 
Fig. 101. 
section of the shaft (Fig. 101). The full section of the shaft is shown 
at (a). The part in the neighbourhood of the sector is shown enlarged 
at (0), and an oblique view of this is shown at (c). Let f denote the 
¥F ¢ 
