264 MECHANICS 



SOLUTION. Divide the section into the rectangles ABC*! 

 and HEFG and the triangle CDC', and assume the lines X-X 

 and Y Y as reference lines. Then, 



and x= 



01+02+03 



01*1 + 02*2 + 03*3 



01+02+03 

 From the illustration, 01 = 100X20 = 2, 000, yi = 70, *i = 20; 



= 1.587. ys = 43. xs = 45.33; a 3 = 86X20 = 1720,^3= 10, 



and #3 = 43. Substituting these values, 



2,000X70+1,587X43+1,720X10 



2,000+1,587+1,720 

 2,000X20+1,587X45.33 + 1,720X43 



= 42.48 



EXAMPLE 2. Find the center of gravity of the bridge chord 

 section shown in Fig. 6. 



SOLUTION. The center of gravity is on the line YY, which 

 is an axis of symmetry. To find the distance y, divide the 

 section into angles and plates and take moments about XX. 

 The areas and centers of gravity of 

 the angles might be located by the 

 preceding principles or taken from 

 a manufacturer's handbook. They 

 are: for the 4"X4"Xi" angle, 

 area = 3.75 sq. in. and distance from 

 center of gravity to back of angle 

 = 1.18 in.; for the 3J"X3i"Xi" 

 angle, area = 3.25 sq. in. and dis- 

 tance from center of gravity to 



back of angle = 1.06 in.. Hence, the moment of the bottom 

 angles is 2X3.75X1.18 = 8.85 and that of the top angles is 

 2X3.25 (15-1.06) = 90.61. The moment of the two web plates 

 is 2X15XiX 7.5 = 112.5, and that of the cover-plate, 24X*X 

 15.25= 183.00. The sum of the moments is. 8.85+90.61 + 112.5 

 + 183.00 - 394.96. The sum of the areas is 2 X 3.25 + 2 X 3.75 

 +24Xi+2X15Xi = 41 sq. in. Then, y = 394.96-*-41 = 9.63 in. 





