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IBRACON Structures and Materials Journal • 2012 • vol. 5 • nº 6
T. F. SILVA | J. C. DELLA BELLA
(83)
n
sx
=n
x
+|n
xy
|
(84)
n
sy
=n
y
+|n
xy
|
Then:
n
sx
=n
x
+|n
xy
|=320+|200|=500 kN/m
n
sy
=n
y
+|n
xy
|=-2000+|200|=-1800 kN/m
From inequations 15 and 16, as n
sx
> 0 and n
sy
≤ 0, it is not neces-
sary to use tensile reinforcement in y direction, which characterizes
design case III. Then, first, concrete stress must be verified.
θ=arctg
(
(
-- - n
xy
n
y
)
)
=arctg 200 -2000 =5,71°
n
c
=n
xy
.(tgθ+cotgθ)
n
c
=200.(tg5,71°+cotg5,71°)=2020 kN/m
σ
c
= n
c
h = 2020 0,12 =16833,33 kNm
2
=16,83 MPa
f
cd1
=0,85.
(
(
1- f
ck
250
)
)
. f
cd
=0,85. 1- 25 250 . 25 1,4 =13,66 MPa
f
cd2
=0,60.
(
(
1- f
ck
250
)
)
. f
cd
=0,60. 1- f
ck
250 . 25 1,4 =9,64 MPa
Figure 5 – Example of design
In this case σ
c
> f
cd1
, the compressive stress in the concrete is
higher than the maximum limit of strength. Then, it will check if it
is possible design compression reinforcement in the y direction to
decrease the compressive stress in concrete.
First, because steel CA-50 is being used, it must verify if n
xy
re-
spects inequation 85.
(85)
|n
xy
|≤ f
cd2
.h.sen(2|θ
*
|)
2
Then, θ* will be calculated through equation 86.
(86)
θ
*
= arccos
(
ε
yd
ε
yd
+2.2
)
2
θ
*
= arccos
(
(
2,07 2,07+2.2
)
)
2
= arccos 2,07 6,07 2 =35,03°
Now, it is possible to find the limit to n
xy
:
|n
xy
|≤ 9642,9.0,12.|sen(2.35,03)|
2
=543,7 kN/m
Therefore, because the n
xy
is lower than the limit, it is possible to
calculate the compression reinforcement for this case. Thus, using
equation 87, it follows that:
(87)
θ= arcsen
(
(
2.n
xy
f
cd1
.h
0,8+170.
[
2.ε
yd
+2‰.(1-cos2θ)
(1+cos2θ)
]
2
(
(
θ= arcsen
2.200
13660,7.0,12
0,8+170.
[
2.2,07‰+2‰.(1-cos2θ)
(1+cos2θ)
]
2
Iteratively, it is possible to obtain the value of θ. It can be seen from
Table 4 that, in this case, value θ converges. In this example, the
steel is assumed to be CA-50, it is not necessary to check if θ ≤ θ
1
,
because θ
1
does not exist.
Going forward, it is now possible to calculate ε
y
.
(88)
ε
y
= 2.ε
yd
-ε'
c
.(1-cos2θ)
( 1+cos2θ) +ε'
c
-ε
yd
ε
y
= 2.2,07‰+2‰.(1-cos(2.8,27°))
( 1+cos(2.8,27°) )
-2‰-2,07‰= -1,91‰