536 Dr. J. Prescott on the Torsion of 



We have now shown that, wherever the shear lines are 

 nearly parallel either in a thin straight strip, or in a thin 

 curved strip where the width is small compared with the 

 radius of curvature, the shear stress is approximately pro- 

 portional to the distance from the middle line of the strip 

 and is in contrary directions on the two sides of the middle 

 line. 



Torque in a Long Thin Section. 



Any particular shear line in a long thin section such as a 

 lono- thin rectangle, or an unclosed tube, or a long narrow 

 ellipse, is a closed curve which crosses the width of the strip 

 at any point at distance x from the centre on one side and 

 — x on the other side. Then the area enclosed by this shear 

 line is 



A=f2a?<fa, (54) 



ds being an element of the middle line. 



This integral extends- between the two points where the 

 line bends round and turns back. In the integral for the 

 area x must be regarded as a function of s, since the length 

 x increases as the section broadens out and decreases as it 

 narrows again. 



Now, by equation (30), 



Q = 2njAdf 



= 2n^2xdsd% 



= ±^xds.Sdx 



= 4 \\ xds . 2nrxdx 



= §nr^x 2 dxds (55) 



The limits for x in this integral are to %t, not — -§-£ to -j£, 

 since the shear lines enclosing the area A expand from the 

 centre outwards. Therefore 



Q=Snr^(itfds 

 = inr^fds (56) 



We have thus expressed the torque by means of an integral 

 of & (which must be regarded as a function of s) with 

 respect to s, the length measured along the middle Curve of 

 the section. The result in (56) is accurate for a strip in 

 which the shear lines are parallel curves running in a strip 

 whose width is everywhere infinitesimal in comparison with 

 the radius of curvature of the central line. For the appli- 

 cation of this equation it is important that the two parts of 



