RESOLUTION OF STRAINS FROM DISPLACEMENTS. 29 



The foregoing synthesis shows how a plane undilatio'nal strain maybe 

 resolved when the displacements are given. Cases also arise in which it 

 is desirable to find the displacements a, 6, e and / from a known shear 

 and values of v and //.. If the ratio of the shear is «, the values of «r and s 

 can he derived from it, and these two values, together with the values of 

 v -f ;>■ and •> — ,"■ constitute four equations from which a, b, e and / can 

 be deduced. They give — 



a = s sin (y + //) + n sin (y — //) ; 1 -p e = cos Q> — /j.) + s cos (* 4- ;i) • 



b = s sin (v + //.) — (7 sin (y — /1) ; 1 +/= cos (y — //.) — s cos (v -j- //.). (S) 



These values, substituted in the formulas of preceding paragraphs, show 

 to what simplest strain system a given rotation and shear are referable. 



Strain due /<> Pressure. — For the sake of keeping the discussion of strains 

 together, it may be assumed here by anticipation that a pressure produces 

 a cubical compression of ratio h * and two ecpial shears of ratio a at right 

 angles to one another. For brevity, let — 



2 t = a — or 1 ; 2 r = a + aT\ 



Then the displacement formulas for a strain due to a pressure in the 

 direction -V are — 



x' = j (r — t cos 2 <*/) — ■'! t sin 2 # ;y' = \(T + t cos 2 >'/) — j t sin 2 < f / ; 



Z 

 h 



^==?(r + 0. 



It will be observed that these formulas are analogous to those for simple 

 shear. 



When the pressure is vertical, so that # — 90°, 



, %a . y za 



x z '-j ] y ^ ; z = r 



If a vertical strain of this kind is combined with a scission or shearing 

 motion in a horizontal direction, the values of x only will be modified by 

 the second strain. If x" is the final value of x and 2 s x is the amount of 

 the shear produced by the scission — 



x ' - j ~ ^h ; y" = v' ; z " = z ' ; and ,i,n < v - *> = * ^ T - *>• 



*- Here h is taken greater than unity, and is the reciprocal of the value which in a given case 

 would satisfy (5). 



