110 APPLIED MECHANICS 
tension, and that the concrete carries the compression. It is also assumed 
that as the beam bends the strains in the various layers parallel to the 
-neutral surface are proportional to their distances from that surface, as 
in the case of homogeneous beams discussed in Art. 109, page 103. 
From the latter assumption it is obvious that the neutral axis of a 
cross section will not pass through the centre of gravity of that section, 
since the moduli of elasticity of the steel and concrete are not equal. 
Referring to Fig. 149, y, is the depth of the neutral surface below the 
top or compression surface of the beam, / is the depth of the axes of the 
steel bars from the top surface, and b is the breadth of the beam. Let 
a= total area of cross section of the steel bars, 7, = maximum compressive 
stress in the concrete, and f,=stress 
in the steel. The depth of the 
section of the steel bars being small 
compared with the depth of the beam, 
it is assumed that f, is uniform over 
the section of these bars. 
The resultant of the compressive 
stress in the concrete is R,=3/,by,, 
and this resultant acts at a distance 7 
%y, from the neutral surface. — Fic. 149. 
The resultant of the tensile stress 
in the steel is R,=a/,, and this resultant,acts at a distance h —y, from 
the neutral surface. 
The assumption of proportionality of strain to distance from the 
: a Oy ee oan 
neutral surface already mentioned leads to the equation Hy, Baby) 
where E, and E, are the Young’s moduli for the concrete and steel 
respectively. 
Since R, and R, are the forces which form the couple whose moment 
is the moment of resistance of the section, it follows that R, =R,, there- 
fore $f, by, =af,. = 
Dividing the equation OAS gt by the equation $/,dy, =a/, 
Ey Gms ee 
2 a 
Eby; Eya(h-) 
99, tg ai ARR 
vy s  B”? b ie a 
which is a quadratic equation for determining 7. 
the result is which may be written 
Putting 2 =n, the solution of the equation gives 
1 
/ (an? + 2abhn) — an 
yy = b » 
Having found y,, the moment of resistance of the section is 
Ry(h = 4 + 321) = BS by(h — 31)» 
or RA 9+ 34) = af(h — 54)- 
