260 



CALCULATION OF THE TRANSVERSE STRENGTH 



Further, we find from the first of equations (4) : 



Ma=-— {ip- P/ + zaxA + 2 by a) 



2 



Add to this the identity : 



(II) 



(a= + b-'- d'-V)=o 



-P 



. •. Ma = — (zp" + a' + b^ - Pa" - a" - b' + zuXa + 2by^ 



2 

 But i^ + a? + V^ is the square of the radius of gyration ?'„, of the ring relative to a 

 point O which has the coordinates a and h in the coordinate system, GX, GY . 

 Hence: 



io = i^ + a" + b" (12) 



and 



Ma = ^ {io - Pa'- cC- b'' + zuxa + 2byA) (n') 



2 



Draw a circle with radius ig and with O as center as shown in Fig. 2. This circle is 



FIG. 2. 



Weutce/ A«i& 



the circle of polar inertia. Join OA and draw AT normal to OA to cut the circle at T. 



Then OT=i„ and AT = ?/ - (9^ 

 and 



OA={xA-aY ^ {yA-bY = pA -^ a^ ^ b"" -laXA-^hyA (13) 



If we call the radius vector from O to any point on the neutral axis r we have : 



~0A = r/ 



Hence : 



AT =C- 



r/ 



Substituting in (11'), we find that: 



-P 



Ma = ^AT= J^{ig^-r/) 

 2 2 



