of an Elastic Plate under Normal Pressure. 99 



The extensional strain of ds 1 is 



d$i — ds_ {ds^ 2 — {ds) 2 

 ds ~ ds(ds 1 -{-ds) 



(d Sl ) 2 -(ds) 2 

 ~ 2(ds) 2 



dx du dy dv 1 fdw\ 2 

 = d7ds + Js~ds + 2\ds) * * ' W 



Now let jds make an angle 6 with the #-axis. 



Then -j- = cos d, -f - sin d. 



ds ds 



7 ~du 7 "du _ 



Also di<=^^+-^y, 



whence s = ^ cos<? + §^ sln ^ 



. ., . p dv , dw , , 



There are similar expressions tor -7- and -y-, and wlien 



these are substituted in equation (1), that equation gives the 



complete expression for the strain in any direction in the 



middle surface. 



Putting = 0, we find that the strain in the direction of 



the .I'-axis is 



~du 1 f~dw\ 2 

 a= ^ — I- 



~dx 



m < 2 » 



Putting 0=^, we find that the strain in the y direction is 



T*+W » 



Again, the strains in the directions making angles -15° 

 and —45° with the #-axis are 



lfdu -du\ 1/Bi' dtA lfdw , 'dioX 2 



1/^w Bm\ _l/d^_dv\ , \(~dw ~dw\ 2 

 Pl ~2Xbx -by) 2\-bx V^U"*^' 



Therefore a 1 -^ 1 = — + — + ^— . " . . . (4) 



oy d# d^ 7)y K J 



But it is easy to show that («i— &) is the shear strain of 

 the element which was originally a rectangle with sides dx 

 and dy in the middle surface. Let us call this shear strain 7. 



H2 



