506 



Mr. C. A. Carus-Wilson on the Distribution 



each element the points c l5 c 2 , c 3 , &c. will continue to lie on 

 a line parallel to ox, since the displacement of each of these 

 points, due to the distortion of the corresponding element, is 

 parallel to o x. But if as we descend from the first element 

 we find $i becoming greater than s 2 , the points c 2 , c B , &c. will 

 have a displacement to the left which increases as we descend, 

 so the line passing through c l3 c 2 , c 3 , &c. will be bent ; and if 

 the shearing-strains become equal again on the lowest element, 

 this line will be concave to the right. 



It thus appears that any * 1 ^-- 



horizontal line will be curved 

 concave downwards if the sum 

 of s 1 + s 2 is greater at the end 

 of the line than in the centre, 

 and vice versa, but that the line 

 will remain straight if % + s 2 is 

 constant along its length, and 



Also that a vertical line will 

 remain so if s Y is everywhere 

 equal to s 2 ; but if s 1 is > s 2y 

 the line will be curved and 

 concave towards the right, and 

 vice versa. 



We can now proceed to de- 

 termine the curvature of the 

 lines in the case before us. 



Commencing at the upper 

 extremity of the bar, fig. 2, a 

 line such as a b will remain 

 straight, because p is every- 

 where constant, and l x = l 2 . 

 The line cdef will be curved 

 concave downwards, since from 

 c to d and from e to/ p is less 

 than it is from d to e (where it 

 is constant and l x = l 2 ) ; d e will 

 therefore remain straight, and 

 cd and ef curved as shown. 

 The line g h i will be concave 

 downwards and curved along 

 its entire length, since p gets 

 less as we move away from h 

 on both sides. 



The line j k I m n will 

 remain straight from j to k, 

 since p is there constant and l } = l 2 ; from k to I p becomes 



