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IBRACON Structures and Materials Journal • 2012 • vol. 5 • nº 6
Shear strength of reinforced concrete circular cross-section beams
Applying to the beams tested that did not have stirrups and had
2.56% of the longitudinal reinforcement, is obtained:
(4)
V
c
= 3.7×0.0256+0.18 ×1.00× 31.7×0.7×0.0491m
2
=0.053MN=53kN
According to NBR 6118, the value of V
c
is function only of the
tensile strength of concrete. The standard recommends that one
adopt the lower value of tensile strength. In the case of the test
beams the value of “f
ctk,inf
” could be estimated based on the aver-
age value of compressive strength presented (f
c
=31.7MPa), result-
ing in f
ctk,inf
=0.21kN/cm². Applying to the data of the problem and
adopting b
w
=D and d=0.72D, as recommended by AASHTO(2007)
will find the next value of V
c
.
(5)
V
c
= 0.6 × 0.21 × 25 × 0.72 × 25 = 57kN
Finally, by applying the concept of effective area with the
expression of NBR 6118, one obtains:
(6)
V
c
= 0.6 × 0.21 × 0.7 × π×25
2
4 = 43.29kN
In tests, the value of the shear force resistance to the beams
without stirrups varied between 70kN and 117kN, all of which had
longitudinal reinforcement rate composed by bars distributed along
the perimeter, uniformly, of 2.56%. Therefore, the values obtained
in tests with some clearance exceed the predictions of the stan-
dard and Merta´s [4] model. The two predictions reach values very
close. However, considering that the value of V
c
is estimated as the
product of a tension-resistant ban area, it is observed that these
magnitudes as sumevery different values, as:
n
The “ultimate stress” is 1.55MPain the expression of Merta [4]
and 1.26M Pain the expression of NBR 6118;
n
The “resisting area” is 344cm² on the expression of Merta [4]
and was used with a value of 450cm² for application on the
expression of NBR 6118;
n
The products are close: 0.155x344 ~ 0.126x450 = 53kN a 57kN.
If the value of V
c
would be estimated as the one corresponding to
the formation of the first shear crack, considering Stage I, it would
result according to[10] and adopting
ν
= 0.2 for the concrete:
n
ζ
máx
=1.42 (V / A) = 0.21kN/cm², soon V = 72.5kN.
In the test, we obtained V between 70kN and 117kN, which shows
some dispersion in the results, with an average of 83.75kN. Dis-
carding the value of V=117kN the average changes to 72.7kN.
2.3 Calculation of V
sw
For calculation of V
sw
, part of the shear force resisted by the trans-
verse reinforcement, the model proposed by Merta [4], is quite
complex, because it considers the geometry of the circular stir-
rups properly. The model consists in determining the vertical com-
ponent of forces resulting from (A
s
f
yw
) in each stirrup that crosses
the cracks hear (Figure 7). Therefore, one should define how
many stirrups there are at a distance(
D
–
x
–
c
)cotg
θ
, where
D
is the diameter of the cross section
x
is the neutral axis position
in cross section,
c
the overlay of reinforcement and
θ
the angle
that the crack makes with the longitudinal axis of the beam. The
mechanical model is to imagine that the stirrups crossing the crack
will be mobilized, after cracking, considering the condition of ulti-
mate resistance, which reaches the yield stress (f
ywd
).Compared to
rectangular section, circular stirrups would be less efficient. Take,
for example, a section with stirrups with a diameter of 10mm
spaced 10cm. If the beam section is rectangular, with d=60cm,the
value of V
sw
provided, for example, by NBR6118,would be:
(7)
V
sw
= 0.9 × 60 × 2x0.8 10 × f
ywd
= 8.64f
ywd
Taking a circular section with a diameter “
D
”=60/0.72=83.33cm, so
that it has the same effective height “d” of the rectangular section
above, and considering covering “
c
”=3 cm and that the neutral axis
position is 0.3D=25cm, it would result(
D
–
x
–
c
) cotg
θ
, with value
of 55.33cm, in case of
θ
=45
o
– as indicated in Model I of NBR 6118.
There would be 5stirrupsacross the crack, as illustrated in Figure 8.
Determining the vertical component of the forces at each stirrup, as
shown in Figure 8, results find the value of 7.16f
ywd
– i.e.,17% less
than that of a rectangular section with the same effective height.
Merta [4] proposes the following expression to calculate this
portion:
(8)
V
sw,1
= 1.8 × n
t
× A
sw
f
yw
where
(9)
n
t
= (D-x-c) s cotg(θ)
Figure 7 – Data for calculation
of V – adapted from Merta [4]
sw