Moments and Products of Inertia in a Plane Section. 455 



and are therefore easily constructed. This completes the 

 parallelogram, giving all data for the gyration-circle belonging 

 to G. 



Fig. 2. 



Second Method of representing the Relations (1), (2), and (3) 



Let UA = the mo- 

 ment of inertia about 

 OA, and UB=AY= 

 the moment about OB, 

 OA and OB being, as 

 before, the principal 

 axes at : so that UY 

 is the sum of the prin- 

 cipal moments, and AB 

 their difference. 



Describe the circle 

 AOB with centre T. 

 Let any axis OK cut 

 the circle in K, and 

 from K draw KM perpendicular to UY. 



Then UM is the moment of inertia about OK, 



VM is the moment about OL (perpendicular to OK), 

 and KM is the product of inertia about OK, OL. 



For, taking the area of the section as unity, which does 

 not affect the relations between the quantities, 



UM = UT + TM=i(a 2 + 5 2 ) + i(a 2 -5 2 )cos 20, 



= a 2 cos 2 e + b 2 s'mW = k 2 ; 

 YM = YT-TM = a 2 sin 2 + 6 2 cos 2 6 = l 2 ; 

 KM = TKsin20 = i(a 2 -Z> 2 )sin20 =L 



These relations hold in whatever position the circle OAB is, 

 and therefore may be in the same straight line with MK, 

 in which case OM = MK. Hence, if the moments and product 

 about OK, OL are given, we have the following construction 

 for the principal axes : — 



On OK take OM = the given product of inertia, measuring 

 OM in the positive or negative direction according as the 

 product is positive or negative. From M draw MU parallel to 

 OL in the positive direction equal to the moment about OK, 

 and in the opposite direction draw MV equal to the moment 

 about OL, so that UY is the sum of the moments. Bisect 

 UY in T, and with centre T, radius TO, describe a circle 

 cutting UY in A and B. Then OA, OB are the principal 

 axes, and UA, UB the moments about them. 



2 12 



