538 Dr. J. Prescott on the Torsion of 



Suppose b is small compared with a. Then we get 



= 4nTl,(l-£, + ...), .... (60) 



\ y being the moment of: inertia about the major axis. The 

 proportional error in equation (57), when applied to this 



b 2 

 section, is therefore approximately -,. Thus there is an 



error of about 10 per cent, when a = 'db, and about 1 per 

 cent, when a = 10b. For this section then our formula gives 

 more accurate results than for the rectangular section. We 

 may expect in general that (57) will give better results for 

 a strip with tapered and rounded ends than for one with 

 rectanguhir ends. 



Uniform Circular Tube split Longitudinally. 



If r is the mean radius of the tube and t the uniform 

 thickness, equation (56) gives 



Q='JnTl this 

 Jo 

 = lnn d x2irr (Ql) 



The ratio of the torque in the split tube to that in the 



.It 2 

 complete tube is — -^ for the same angle of twist. But this 



is not a fair comparison because the maximum shear stresses 

 are not the same in she two cases. The maximum shear 

 stresses are, for the split tube vrt, and for the complete tube 

 nr(r-\-\t). If we make these shear stresses equal by giving 

 different values to t in the two cases, then the ratio of the 

 torque in the split tube to that in the complete tube is 



lt(r + it) It , , co , 



o - — o^-^ = - nearly. .... (62 

 3 r 3 r J * 



The split tube is therefore much weaker than the complete 

 tube under torsion, and very much less rigid. 



ft is interesting to find the relative axial displacement of 

 the open ends of the split tube. For this purpose we may 

 use equation (33) and regard S in that equation (which is 

 the mean shear stress across the section) as zero for the split 

 tube. Then »=-T$pds 



= -2tA, (63> 



