RECIPROCAL FIGURES, FRAMES, AND DIAGRAMS OF FORCES. 35 



This equation is identical with that of the first case, with the exception of the 

 coefficient of the part due to H, which depends on the density of the body, and 

 the value of a-, the ratio of lateral expansion to longitudinal compression. 



Hence, if the external forces are given in the two cases of no stress and no 

 strain in the direction of z, and if the density of the body or the intensity of the 

 force acting on its substance is in the ratio of a to (1 — o-) 2 in the two cases, the 

 internal forces will be the same in every part, and will be independent of the 

 actual values of the coefficients of elasticity, provided the strains are small. 

 The solutions of the cases treated by Mr Airy, as given in his paper, do not 

 exactly fulfil the conditions deduced from the theory of elasticity. In fact, the 

 consideration of elastic strain is not explicitly introduced into the investigation. 

 Nevertheless, his results are statically possible, and exceedingly near to the 

 truth in the cases of ordinary beams. 



As an illustration of the theory of Airy's Function, let us take the case of 



j- _ 1 r 2 P cos 2 p0 (22). 



In this case we have for the co-ordinates of the point in the diagram corres- 

 ponding to {xy) 



£ = ^ = r^cos^-l)^ v = -r' i P- l $m(2p-l)d . . (23), 



and for the components of stress 



*■ = % = ~ (2i>-l)r 2 *- 2 cos (2p-2)0 = - g = - Pvv ] 



*F y (24) - 



If we make 



G = - 1* cosp0 and H = - r* sinpd . . . (25), 

 p p 



then 



"|2 



/0 ,.{dGf dGc\ 2 \ 



(26). 



Hence the curves for which G- and H respectively are constant will be lines of 

 principal stress, and the stress at any point will be inversely as the square of 

 the distance between the consecutive curves G or H. 



If we make 



£ = p cos (f> and v = p sin <£ ~] 



then we must have y ■ (27). 



p = r 2p-l an( j 0=s(2p-l)0 J 



If we put 



?for 2^T then p + l = 2 and (2P-1)(2 2 -1)=-1, 



