F. K. Kong MA, MSc, PhD, CEng, FICE, FIStructE, R. H. Evans CBE, DSc, D ès Sc, DTech, PhD, CEng, FICE, FIMechE, FIStructE (auth.)-Reinforced and Prestressed Concrete-Springer US (1987)

360 Prestressed concrete simple beams
Tendon strain at collapse is (see eqn 9.5-2)
i.e.
Comments
Cpb = cpa+ Cpc (= 0) = Cpa
= 3·045 - 0.0035
X
X = _...::3.:..:.0~45""==
Cpb + 0.0035
[pb = 2.945x as before
Eliminating x from eqns (9.5-11) and (9.5-12),
8.969
/pb = cpb + 0.0035
(9.5-11)
(9.5-12)
which is solved with the stress/strain curve in Fig. 9.5-3, giving
[pb = 1046 N/mm 2 ; cpb = 0.0051
Using eqn (9.5-12), x = 355.2 mm,
Mu = (1046)(3300)[870 - (0.45)(355.2)] Nmm
= 2451 kNm
(a) The site error in part (b) only leads to a small reduction in the Mu
value, from 2922 kNm to 2451 kNm. However, the reduction in the
cracking moment may be serious, and so may the increase in the
working load deflection.
(b) The general principles as illustrated in this example can of course be
applied to flanged beams. With reference to Fig. 9.5-4, if at collapse
the neutral axis is within the flange thickness, then the method of
solution is as for a rectangular beam; in particular, eqns (9.5-5) to
(9.5-7) can be applied without modification. If the neutral axis depth
x exceeds the flange thickness he, the compatibility condition is still
represented by eqn (9.5-5), but the equilibrium condition has to be
worked out from Fig. 9.5-4. Using the rectangular stress block of
Fig. 4.4-5, the compression in the shaded area (1) in Fig. 9.5-4 is
0.405fcubwx and that in the areas (2) is 0.45fcu(b - bw)hc.
Therefore, the equilibrium condition is
[pbAps = 0.405fcubwX + 0.45fcu(b - bw)hc (9.5-13)
Eliminating x from eqns (9.5-13) and (9.5-5) gives the following
equation which relates the unknown tendon stress [pb to the unknown
tendon strain cpb:
~ b = 0.45fcu . [ 0.9/hccu bw
p (! Cpb - Cpe - /J1cc + /Jzccu . b
+ (1- bbw)~] (9.5-14)