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IBRACON Structures and Materials Journal • 2012 • vol. 5 • nº 6
Design of compression reinforcement in reinforced concrete membrane
Thus, the forces in reinforcements are calculated
(89)
n
sx
=n
x
+n
xy
.tgθ
n
sx
=320+200.tg8,27°=329,07 kN/m
(90)
n
sy
=n
y
+n
xy
.cotgθ
n
sy
=-2000+200.cotg8,27°=-623,78 kN/m
Finally, the reinforcement areas are given by:
a
sx
= n
sx
f
yd
= 329,07 43,49 =7,57cm²
a
sy
= n
sy
σ
y
= n
sy
E.ε
y
= -623,78
21000.(-1,91‰) =15,52 cm²
8. Conclusions
Methods are presented herein to determine compression reinforcement
in design case II, III and IV provided by CEB [3]. The limits for this design
were also presented; in other words, cases in which it is possible to adopt
compression reinforcement so that compressive stress in concrete is re-
duced to its strength are delimited. In all the cases, these limits are only
related to the shear force to which the membrane is subjected.
Furthermore, a model for concrete strength was used that interpo-
lates the values
of strength between f
cd1
and f
cd2
according to the
curve obtained by Vecchio and Collins [2], so that there is no discon-
tinuity of strength values between cases II and IV and cases III and
IV. Also, we presented how to evaluate whether the compressive
stress in the concrete is lower than the limit to this strength model.
Due to this model adopted for concrete, for cases II and III, the re-
inforcement design became more complex and iterative methods
were necessary for resolution. However, it leads to fewer amounts
of reinforcement than those found when using only the values
sug-
gested by CEB [3] for strength.
About case IV, it was found that there are infinite solutions for re-
inforcement design, although just one leads to the minimum rein-
Table 4 – Iterative calculation of
θ
i
o
θ
( )
i
(‰)
1
f
(MPa)
c2max
(Mpa)
c
o
θ
( )
f
1
1,000
2,072
11,856
95,512
8,164
2
8,164
2,154
11,714
11,856
8,266
3
8,266
2,156
11,710
11,714
8,269
4
8,269
2,156
11,710
11,710
8,269
5
8,269
2,156
11,710
11,710
8,269
6
8,269
2,156
11,710
11,710
8,269
forcement required. This occurs due to the smaller number of fixed
variables as compared to cases II and III. It is possible to find the
most economic solution design through the trial and error method.
9. References
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