268 STRENGTH OF MATERIALS 



hu = distance from compression face to center of gravity of 



metal reenforcement, 



r = ratio of area of metal to total area of cross section, 

 JS lt E c , E t = moduli of elasticity, of steel, of concrete in compression, 

 and of concrete in tension respectively, 



* = ?' *= 



A t S t 



p t = unit stress in metal reenforcement, 



p c = unit compressive stress in outer fiber of concrete, 



p t = unit tensile stress in outer fiber of concrete) 



The reenforcement is supposed to be in the tension flange, as shown 

 in Fig. 160, and // , I', .ire measured at stresses p c , p t respectively. 



Equating the sum of the horizontal forces to zero, and remembering 

 that the strain diagrams are arcs of parabolas, there results tin- 

 equation 



(111) \vp e = f(l-/);>, + *7v 



If the cross section remains plane during tlcxuiv. we also have 



/,,\ '"'/'' 



(1U ) A-Ii' P.- l-j-f 1 - 



Substituting these values in (111), and solving the resulting quadratic 

 for v, we have 



(U3) g = - ( + r + 7i + r + r + u(n - 1)| 



- ( " 1) 



which determines the position of the neutral axis. Taking moments 

 about this neutral axis, the moment of resistance of the cross section 

 is found to be 



Equations (113) and (114) suffice for the solution of problems in 

 which the beam has not been bent sufficiently to produce cracks on 

 the tension side.* When such cracks occur the validity of the para- 

 bolic law for the variation of stress on the tension side of the neutral 



* These formulas have been recommended by a committee of the American Railway 

 Engineering and Maintenance of Way Association. 



