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IBRACON Structures and Materials Journal • 2012 • vol. 5 • nº 6
Design of compression reinforcement in reinforced concrete membrane
Then:
(72)
|n
xy
|≤ f
cd1
.h 2
Equation 72 means an absolute limit to n
xy
As the assumptions define only one fixed strain, there are infinite
solutions within a range. The parameter that defines this interval is
θ. Thus, it is interesting to delimit the angles θ which are possible
to be assigned to the problem. Thus, using equation 71, it can be
demonstrated that:
(73)
θ
c1
≤θ≤θ
c2
In which:
(74)
θ
c1
= arcsen
(
2.n
xy
f
cd1
.h
)
2
o
o
to 0 ≤ θ ≤ |45 |
(75)
o
o
to |45 | ≤ θ ≤ |90 |
θ
c2
= arcsen
(
2.n
xy
f
cd1
.h
)
2
Besides this criterion, by hypothesis the membrane is always in
biaxial compression state, the strains in any direction are always
negative. Thus, in order to not obtain reinforcement area with a
negative sign, which is an incongruity, the reinforcement forces
should also be negative. Hence, in order to n
sx
≤ 0, θ must respect
the following premise.
(76)
θ
≤
θ
x
=arctg
(
n
xy
n
x
+n
c
)
Similarly, in order to n
sx
≤ 0, the following criteria must be followed.
(77)
θ≥θ
y
=arctg ( n
c
+n
y
n
xy
)
6.2 Reinforcement design
The method that will be presented was based on Jazra [12].
First, the forces to which the membrane is submitted must re-
spect the equation 72. Once this criterion is verified, it must
arbitrate a value of θ such that it respects the limits imposed
by inequations 73, 76 and 77. There will be infinite values pos-
sible to θ, but just one will lead to a minimal reinforcement area.
Thus, it follows that:
(78)
n
sx
=n
c
+n
x
-n
xy
.cotgθ
(79)
n
sy
=n
c
+n
y
-n
xy
.tgθ
With
n
sx
and n
sy
, it is possible to calculate n’
c
through expression 70.
As the direction of the principal stress is the same of the direction
of the principal strain by hypothesis, force n’
c
is related to strain ε
1
.
These terms are related by the constitutive model for the concrete.
Therefore:
(80)
ε
1
=ε`
c
.
(
1- 1- n
c '
f
cd1
.h
)
Obtaining the value of ε
1
and as the value of ε
2
= ε’
c
and θ is known,
it is possible to calculate ε
x
and ε
y
through expressions 81 and 82,
obtained by the Mohr circle.
(81)
ε
x
= ε
1
+ε
2
2 +
(
ε
1
-ε
2
2
)
.cos2θ
(82)
ε
y
= ε
1
+ε
2
2 -
(
ε
1
-ε
2
2
)
.cos2θ
Thus, reinforcement is designed by:
a
sx
= n
sx
σ
x
= n
sx
E
cs
.ε
x
a
sy
= n
sy
σ
y
= n
sy
E
cs
.ε
y
The reinforcement areas obtained are not necessarily the mini-
mum. Therefore, attempts must be made to find this minimum.
7. Example of design
Assuming a membrane with 12cm thickness, f
ck
equal to 25 MPa,
CA-50 and subjected to forces per unit length as shown in Figure 5.
First, it must be verified which design case this problem belongs to.
In order to do that, it must calculate forces n
sx
and n
sy
from equa-
tions 83 and 84.