STRENGTH AND ELASTICITY OF REINFORCED CONCRETE. 163 



iv = the uniform load per unit of length. 

 I = the span. 



If a = the area of the horizontal reinforcement in square 

 inches required at the centre of the beam : — 



Then a is proportional to the central bending moment 



7 7 2 



when oc =— i.e., -^ — and we may write the equation in 



2 o 



terms of a thus : — 



y =^B (lx-DC 3 ) 

 l 



For any length of beam denoted by x a - x i , the area required 



to resist the shearing stresses arranged in a vertical plane 



is: — 4a J , v /■»»-.»! 



y a -!/i=-y- j p-^) - f — j— 



and the area of rods arranged to slope at an angle of 45° 

 away from the beam, is clearly the value of (y 2 -3/1) Sec. 

 45° for the length of beam included between (xz-Xi). 



If we make the distance between the ordinates of the 

 parabola = 1, the equation may be written : — 



4a f i /X2+X! 



The shearing stress between one foot from the ends and 

 the ends of the span is : — 



4a A 1\ 4a /7 ^ 



2/2 -j /l = T (i_ T j=_ (I -i) 



and the area of rods inclined at 45° is = ~ (I - 1) Sec. 45° 



1 



Between 2 feet and 1 foot : — 



4a /., 3 \ 4a /7 \ 



the area of rods inclined at 45° is : — 



Area = ^ (1—3) Sec. 45° 



It should be clearly understood that the actual stress 

 upon the inclined rod is the difference between the total 

 stress given by the above equations and the shearing resist- 



