FIRST AND SECOND MOMENTS 31 



total area is A = BH+ bh, we have, by the principle of moments, 

 x^BH-\- bTi) = BH x +- bh x - , whence 



BH* + bh 2 



Having found X Q , the /for the centroidal axis GG is determined by 

 the relation 



T T 



J-Q - J. Q 



APPLICATIONS 



51. A uniform rod 18 in. long weighs 8 Ib. and has weights of 2 lb., 3 lb., 4 lb., 

 and 5 lb. strung on it at distances of 6 in. apart. Find the point at which the rod 

 will balance. 



Solution. Since the rod is uniform, its weight may be assumed to be concen- 

 trated at its center. If, then, x denotes the distance of the center of gravity from 

 the end at which the 2-lb. weight is hung, by taking moments about this end 



2x0+3x6+8x9+4x12+5x18 

 2+3+8+4+5 



= 10 T 4 T in. 



52. A section like that shown in Fig. 29 has the dimensions b = 3 in., d = 5 in., 

 t = \ in. Locate its center of gravity, or centroid. 



Solution. To locate the gravity axis 

 1 1, take moments about any parallel 

 line as a base, say AB. Then, dividing 

 B the figure into two rectangles, since the 

 center of gravity of each rectangle is at 



-6 



A 1 r--*-H- 



Tl 



its center, we have 



_ 



3 



Similarly, to determine the gravity axis 

 2 2, by taking moments about CD we 

 have 



\D 



41 x 



8 1 

 FIG. 



53. The section shown in Fig. 30 is 

 made up of two 10-in. channels 30 Ib./ft. 

 and a top plate 9 in. x | in. Locate its 

 gravity axes and determine its moment of inertia with respect to the axis 11. 



Solution. From Table IV the area of each channel is 8.82 in. 2 To determine 

 the gravity axis 1 1, take' moments about the lower edge of the section. Then 



_ 2 x 8.82 x 5 + 9 x x 10 _ 

 2/0 ~ 2 X 8.82 + 9 x \ 



