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IBRACON Structures and Materials Journal • 2012 • vol. 5 • nº 6
Design of compression reinforcement in reinforced concrete membrane
5.2 Design reinforcement in case III
Assuming a membrane that is subjected to forces which respects the con-
ditions imposed by equations 48 and 50, the intention is to calculate the
amount of reinforcement necessary to be positioned in the y direction in
order to compressive stress in concrete is equal to the maximum strength
f
c2max
. This calculation method was based that presented by Jazra [12].
Therefore, with equations 43, 44, 45, 46 and 47 the intention is to find the
value of θ which is the solution to the problem. First, if nxy = 0, then θ = 0°.
If θ
1
exists and 0 < θ ≤ |θ
1
|, so:
(51)
θ= arcsen
(
2.n
xy
f
cd1
.h
)
2
If θ
1
exists and |θ
1
| < θ < |θ
2
| or if θ
1
does not exist and
0º <
θ < |θ
2
|, then:
(52)
θ= arcsen
(
2.n
xy
f
cd1
.h
0,8+170.
[
[
2.ε
yd
-ε'
c
.(1-cos2θ)
(1+cos2θ)
)
2
Finaly, if |θ
2
| ≤ θ < |θ*|, then:
(53)
θ= arcsen
(
2.n
xy
f
cd2
.h
)
2
Therefore, to find θ, it should follow the steps.
1. If n
xy
= 0, θ = 0.
2. If n
xy
≠ 0, use the iterative method to find θ through equation 52.
3. If converges, for CA-25, two solutions can be found, but it is
only valid that one which it respects θ <θmax.
4. If converges, for CA-25, check if θ ≤ θ1. Because equation 52 is
not valid for this domain, θ must be found using equation 51.
5. If converges, for CA-50 and CA-60, θ found is solution.
6. If does not converge, find the solution using equation 53.
With values
of ε
x
= ε
yd
, ε
2
= ε’
c
and θ, it is possible to obtain the
value of ε
1
and ε
y
. Taking back equation 31 and 39, then:
(54)
ε
y
= 2.ε
yd
-ε'
c
.(1-cos2θ)
(1+cos2θ) +ε'
c
-ε
yd
Table 2 – Maximums values for
θ
in case III
max
(‰)
yd
o
θ
( )
max
CA-25
1,04
CA-50
2,07
CA-60
2,48
33,76
35,03
33,74
It is possible to calculate the forces in the reinforcement using
equations 1 and 2. The reinforcements are given by:
(55)
a
sx
= n
sx
σ
x
= n
sx
E
cs
.ε
x
= n
sx
E.ε
yd
= n
sx
f
yd
(56)
a
sy
= n
sy
σ
y
= n
sy
E
cs
.ε
y
5.3 Design reinforcement in case II
In this item, the formulation for designing compression reinforcement in
case II will be presented. As already exposed, the demonstration of equa-
tions is the same for case III and only the final equations will be presented
here. Thus, for case II, the design limits presented in item 5.1 also must
be considered, but the domain of functions of θ is |45°| ≤ θ ≤ |90°|.
In this case, ε
y
= ε
yd
. By adapting expression 36, f
c2max
is repre-
sented by equation 57.
(57)
f
c2max
=
f
cd1
0,8+170.
[
2.ε
yd
-ε
' c
.(1+cos2θ)
(1-cos2θ)
]
Thus, for case II, the limits of θ assume the values
shown in Table 3.
For CA-25, it follows that:
(58)
.
(
(
sen(2.|θ
max
|)
0,8+170.
[
[
2.ε
yd
-ε
' c
.(1+cos (2.|θ
max
|))
(1-cos(2.|θ
max
|))
|n
xy
|≤ f
cd1
.h 2
For CA-50 e CA-60, it follows that:
(59)
|n
xy
|≤ f
cd2
.h.sen(2|θ
max
|)
2
If the shear force to which the membrane is subjected respects
condition 58 or 59, then the value of θ must be found by using the
Table 3 – Values of
θ
,
θ
e
θ
* and
θ
1 2
max
for steels prescribed by NBR 6118 in case II
(‰)
yd
o
|
θ
| ( )
1
o
|
θ
| ( )
2
o
|
θ
*| ( )
o
θ
( )
max
CA-25
CA-50
CA-60
1,04
77,83
47,26
50,93
56.24
2,07
DOES NOT EXIST
58,26
54,97
54,97
2,48
DOES NOT EXIST
63,21
56,26
56,26