48 PRACTICAL STRUCTURAL DESIGN 



must equal the area of the triangle FEE and that the points of 

 contraflexure lie vertically beneath E and F. 



Assume a rectangle AGDC placed on the span AC. The area 

 of this rectangle times the distance to the center of gravity from 

 either support must equal the area of the triangle A BC times the 

 distance of its center of gravity from the same support. This is 

 known as equating moment areas, or, 



/ A s~t A SV\ " X ^ 



(AC x AG) x y = 2 x V- 



Let AG = x. The span, AC, is common to both sides of the 

 equation and the length, y, of the moment arm is one-half of AC, 

 so may be eliminated, as it also is common to both sides of the 

 equation, which is treated as follows; 



ACxZ 

 Eliminating y,AC X x = ~ 



% 



Eliminating the common factor AC, x = 



2 



>7 



From the similarity of triangles, since AG = DC = x = -=> 







then AF = FB, and CE = EB, for F lies in the line AB and E 

 lies in the line CB, and the line GD, parallel to AC, intersects the 

 line AB in F and the line BC in E. 



The length EF = GF + ED; therefore the area of the triangle 

 FBE = area of triangle AGF + area of triangle CDE. The length 



AC 

 GF = ED = -r-; therefore the points of contraflexure are L 



from the supports. 



PL Z 



Since Z = r- and x = -~ 



PL PL 



then x = negative moment = f X -j- = 



The positive moment 



Z 1 PL PL 



~ 2 ~ 2 * 4 8 ' 



Let -7- = -5-: then PI = -=- and I = -= (eliminating P). 



4 O & 6 



Since I = -^ and I + 2o = L, 



Zi 



L-l L 



