RECIPROCAL FIGURES, FRAMES, AND DIAGRAMS OF FORCES. 37 



where a is a constant introduced in integration, and depends on the manner in 

 which the beam is supported. From this we obtain the values of the vertical, 

 horizontal, and shearing stresses, 



<J = g + ir=!»-^<W-V)- • ■ • Pi). 



d*¥ _„h + k 



dxcly ~~ b 3 



u = - izi:. = 6 -w x v ( h ~y) ( 33 )- 



The values of Q and of U, the vertical and the shearing stresses, as given by 

 these equations, are perfectly definite in terms of h and k, the load and the 

 weight of the beam per unit of length. The value of P, the horizontal stress, 

 however, contains an arbitrary function Y, which we propose to find from the 

 condition that the beam was originally unstrained. We therefore determine 

 a and /3, the horizontal and vertical displacement of any point (x, y), by the 

 method indicated by equations (13), (14), (15) 



2n{* + l)a = * + * { (3a 2 * - «•) (b - 2y)-*x(3by* - 2tf) } - a \vy + »^+ T (34), 

 2n(a + l)/3=- ] ^{^f-ly^ + 3a(a^-^Xby-f)}+llf-a d ^ + X' (35), 



where X' is a function of x only, and Y' of y only. Deducing from these dis- 

 placements the shearing strain, and comparing it with the value of the shearing 

 stress, U, we find the equation 



* + *{6aV»-2 B P+l&(By r .^)}+«rJ« = flj p+g+^ . (36). 

 Hence 



**-12 ^(fiy-f) (37), 



f-!»Ji(IA-^ + ^ 1 f = . . (33). 



If the total longitudinal stress across any vertical section of the beam is zero, 



d¥ 

 the value of -j- must be the same when y = and when y = b. From this con- 

 dition we find the value of P by equation (32) 



-p _ h + Jc 



| 3(« 2 - x 2 ) + 2y 2 - 2by - b 2 } (b - 2y) . . . (39). 



The moment of bending at any vertical section of the beam is 



f\ydy={K + k)Q 2 {x' i -a 2 ) + \vy . . . (40). 



VOL. XXVI. PART I. K 



