Calculation Procedure:
- Compute the values of q^ qmax, and pmax for a singly
reinforced beam
As the following calculations will show, it is necessary to reinforce the beam both in ten-
sion and in compression. In Fig. 6, let A 5 = area of tension reinforcement, in^2 (cm^2 ); A's =
area of compression reinforcement, in^2 (cm^2 ); d' = distance from compression face of
concrete to centroid of compression reinforcement, in (mm); fs = stress in tension steel,
lb/in^2 (kPa);/' = stress in compression steel, lb/in^2 (kPa); es= strain in compression steel;
p â As/(bd);p' = A's/(bd); q = pfylfc\ M 14 =? ultimate moment to be resisted by member,
in-lb (N-m); Mul = ultimate-moment capacity of member if reinforced solely in tension;
Mu2 = increase in ultimate-moment capacity resulting from use of compression reinforce-
ment; Cui = resultant force in concrete, Ib (N); Q 2 = resultant force in compression steel,
Ib (N).
If/' âfy9 the tension reinforcement may be resolved into two parts having areas of A 8
-Ay and Ay. The first part, acting in combination with the concrete, develops the moment
A/Ml. The second part, acting in combination with the compression reinforcement, devel-
ops the moment M 52.
To ensure that failure will result from yielding of the tension steel rather than crushing
of the concrete, the ACI Code limits/? -p' to a maximum value of 0.75/? 6 , where pb has
the same significance as for a singly reinforced beam. Thus the Code, in effect, permits
setting/' =fy if inception of yielding in the compression steel will precede or coincide
with failure of the concrete at balanced-design ultimate moment. This, however, intro-
duces an inconsistency, for the limit imposed onp-p
f
precludes balanced design.
By Eq. 9, qb = 0.85(0.80)(87/137) = 0.432; <?max = 0.75(0.432) = 0.324; pmax =
0.324(5/50) - 0.0324.
- Compute Mu1, Mu2r and Cu2
Thus, Mu = 690,000(12) = 8,280,000 in-lb (935,474.4 N-m). Since two rows of tension
bars are probably required, d = 24 - 3.5 = 20.5 in (520.7 mm). By Eq. 6, Mul =
0.90(14)(20.5)
2
(5000) x (O'.324)(0.809) = 6,940,000 in-lb (784,081.2 N-m); Mu2 =
8,280,000 - 6,940,000 - 1,340,000 in-lb (151,393.2 N-m); Cu2 = MJ(d - d') =
1,340,0007(20.5 - 2.5) = 74,400 Ib (330,931.2 N).
(a) Section (b) Strains (c) Resultant
forces
FIGURE 6. Doubly reinforced rectangular beam.
