262 CALCULATION OF THE TRANSVERSE STRENGTH 



Hence the internal resultant is normal to the vector from O to the point. The distance z of 

 the internal resultant from the point under consideration, being the leverage w^ith which it 

 acts in producing a bending couple, must be such that 



„_M_ 2 ^'' ^ _ z/-r' _AT (17) 



R pr ir 2r 



Thus the resultant acts along the radical axis of the point and the circle. The envelope of 

 the internal resultants forms a funicular curve, which, when the pressure surface is of cir- 

 cular section, forms a circle concentric with that surface. Summing up we arrive at the fol- 

 lowing conclusions: 



1. For a ring of any form and flexibility, when subject to a uniform fluid pressure, there 

 exists an important circle called the "nodal circle." 



2. Its center O has the coordinates ;r = a and y = b referred to a system of coordinates 

 with origin at the center of gravity G of the flexibility mass of the ring and with axes co- 

 inciding with the principal axes of inertia of the ring. The coordinates of O are found from : 



L L 



j fxdm j p'^ydm 



a = 



2 j x^dm 2 j y'' dm 



o o 



and the radius of the circle from : 



to = ip + a" + b' 



where i^ is the radius of gyration of the ring with respect to G as a pole. 



3. For any point in the ring with vector r from the center of the nodal circle the internal 

 resultant is given hy R = pr and R is acting normal to the vector along the radical axis of 

 the point and the circle. 



4. The bending moment at any point of the ring is equal to the product of one-half the 

 fluid pressure and the "power" of the point relative to the nodal circle and is always found 

 from: 



M=^ («7 -^) 



2 



SPECIAL CASES. 



We assume again the pressure to be uniform. 



If a ring has two axes of symmetry both as regards form and flexibility, although the 

 flexibilities may not be uniformly distributed, the center of the nodal circle will coincide with 



L L 



the center of gravity — O coincides with G — since in that case J p' xdm and j f ydm 



o o 



and hence a and h are zero. If a ring has only one axis of symmetry, say the Y axis, as 

 generally is the case in submarine vessels, O will be in this axis, a will be zero and only b 

 needs to be calculated. 



When the neutral axis is a circle, although, maybe, of varying flexibility, it can be 

 shown that in case of relatively slender frames, whatever the position of G, the center of the 

 nodal circle O will always be in the center of the circle formed by the neutral axis. In this 



i 



