G2 G. F. BECKER FINITE STRAIN IN ROCKS. 



Thus one may express e,/and g in terms of b — 



e = c = - — • f= Ah 



•10i/3' * ""IO1/3' 



The line of unaltered direction is then given hy — 



tan y. =«t- e = — -=- = ton 10° 4'. 

 6 2 ,/ 8 



In the case of finite strain it has been shown that this angle preserves 

 its initial value unchanged : so that if b = -- 1, / — e = — 0.2887. Since 

 also e = — //4, the following is a consistent set of displacements for 



f = 30° : 



b = — 1 ; c = 0.0577 ; / = — 0.2308 ; # ■ = e. 



Knowing the displacements, the corresponding values of v and a* may be 

 determined as has been shown in the earlier part of this paper. If these 

 angles only are required they may be found from the following formulas. 

 For any value of b when Poisson's ratio obtains — 



tan u> = 



l/(2— 3 ef + b l — y 25 e + 1/ . 



2v/(l + e ) 2 (l-4e) 



^ 2y = - ( i-4i)*-(it°«r-y 



For the present displacements the formulas give — 



m = 28° 43'; > = — 22° 37'. 



For the spacing of the two possible sets of fissures, therefore— 

 w = I cos (w ± y) = (1 - 4 e) cos (« ± v), 



which for this set of displacements gives 0.765 and 0.481. 



Supposing that both systems of fissures form, the following diagram 

 (figure 10) shows their disposition at the moment of rupture* 



* Example for <j> = 00°.— It may be of interest to some readers to give a second example of a similar 

 strain. In the diagram, figure 11, the force is supposed to art at t>0° to the plane of support. If. 



also, 6 %, the following values result: e = 0.0866 ;/= — 0.3464 ; gr = e; xo = 36°35'; w=32°4'; 



v = — 16° 32'; ju. = — :J2° 34'; h = 0.9172; A = 1.230; 5 = 0.575; G = 1.087 ; D = orf = 0.869 : » 

 0.630 or 0.432. The range for one set of planes of maximum tangential strain is o° 21', and for the 

 other set 16° 26'. 



