GIRDERS AND TRUSSES 105 



the diagram for bending moment due to a uniformly distributed 

 load is a parabola; therefore the reinforcing planks need not ex- 

 tend the full length of the beam. They would have to extend the 

 full length if there were danger of a longitudinal shearing failure, 

 and the thickness of the reinforcement would also be governed 

 by the requirement for shear. 



Dividing; 182,933 -r 300,000 = 0.61, which shows that the resist- 

 ing moment of the beam is 61 per cent of the bending moment 

 created by the load. The ends of the reinforcing planks must extend 

 each side of the middle of the span to the point where the bend- 

 ing moment is 61 per cent of the bending moment at the middle 

 of the span. This may be obtained graphically by constructing 

 a parabola with a base equal to 20 and a height about equal to 

 this, the height divided decimally to any scale. At a height equal 

 to 61 on the middle ordinate draw a horizontal line to intersect 

 the parabola. From the point of intersection drop a perpendicular 

 to the base. This defines the point where, theoretically, the rein- 

 forcement may end. Practically it should extend a little further. 



The lengths of the reinforcing planks may be calculated by men 

 who can solve a quadratic equation. The bending moment on a 

 uniformly loaded beam at any point distant x from one end is as 



follows: . wLx wx* 



Substitute the values for M x , w, and L and solve for x. 

 W - ^r~ = 523 Ibs. per lineal foot (in even numbers). 



182 933 

 MX = ^ = 15,244 ft. Ibs. (in even numbers). This moment 



the resisting moment of the beam without reinforcement. 



i;au 523 x 2Qj 523s a 

 Then 15,244 = -- ^ --- 55 



Clearing of fractions, 



2 x 15,244 - 30,488 = 523 X 20x - 523x>. 

 Dividing by the coefficient of x 1 , 



58.3 - 20z - X s . 



Transposing, x 2 - 20x - -58.3. 

 Extracting the square root, _ 



x - + ?? ^(~J- 58.3 - 3.55 ft. 



