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IBRACON Structures and Materials Journal • 2012 • vol. 5 • nº 6
P. W. G. N. TEIXEIRA | C. E. M. MAFFEI | M. C. GUAZZELLI
n
t
is the total number of stirrups that cross the crack,
D
the diame-
ter of the cross section,
x
the neutral axis position in cross section,
c
the covering of the reinforcement,
s
the spacing of the stirrups, θ
the inclination angle of the crack with respect to the beam axis and
A
sw
the cross-sectional area of one circular stirrup. Applying the
above expression, with n
t
=5, we obtain 7.2f
ywd
, i.e., practically the
same value previously calculated.
Therefore, the circular stirrups are less efficient than the vertical
stirrups usually employed in rectangular beams.
But beyond this resistant portion, Merta[4] suggests that there
is another portion of contribution coming from the reinforcement
formed by circular stirrups. This portion could be explained as
being formed by the vertical components of the normal forces that
arise in the region of adhesion of the circular stirrups adjacent
to the crack-would be vertical components of forces due to the
change in direction of the stirrup, determined using the formula of
Coulomb for friction in pulleys for coefficient of friction with steel
-concrete, μ=1.5. In such a vertical stirrup this component does not
exist, because the adhesion tensions would not have such vertical
components. Merta [4], presents the deduction and proposes the
following expression for determining such a parcel:
(10)
V
sw,2
= A
sw
f
yw
× (λ .n
d
+1)
where
(11)
n
d
= INT[
(
D 2 -c
)
s cotg(θ)]
and λ=0.53, n
d
the number of stirrups crossing the crackup to half the
height of the section and the other terms as the previous expressions.
Applying in this particular case comes to n
d
=3 and V
sw,2
=2.072f
ywd
.
That is, the part resulting from deviation forces would be about
29% of the first parcel previously calculated.
2.4 Calculation of V
Rd3
The shear strength related to the concrete failure by diagonal ten-
sile stresses, V
Rd3
, would be obtained by the sum of V
c
and V
sw
.
Taking four samples of the tests of Jensen et al[2] which have
not exhausted the capacity of the compressed struts, values can
be compared, presented in Table 1. It is observed that the test
results are always superior to those obtained with the expressions
of Merta [4] and NBR 6118, using in the latter b
w
=D and d=0.72D.
3. Results and discussion
The numerical results obtained with the expressions used in
previous sections suggest that the determination of the resistant
capacity of reinforced concrete elements with circular section is a
subject that requires further experimental studies so that they can
propose suitable expressions for use in projects of these elements.
The numerical attempts made in order to determine the part of shear
force resisted by complementary mechanisms to truss used two
approaches: the concept of effective area proposed by Li [3] and Merta
Figure 8 – Example of calculation of the vertical
components of the force of stirrups for a circular
beam with D = 83.33 cm, d = 0.72 D = 60cm,
o
25cm x = c = 3 cm,
θ
=45 ; circular stirrups,
2
ϕ
10.0 c. 10cm (A =0.8cm ;
ρ
=0.19%),
sw
sw
following Merta [4]: (a) first and last
circular stirrups that crosses the inclined crack
o
at
θ
=45 with respect to the beam axis, (b) sum of
the contributions of circular stirrups in extension
(D – x – c) cotg(
θ
)
Table 1 – Comparison of shear resistance value in kN – circular section, D=25cm; f =31.7MPa
c
Exemplar
(*)With the proposed adjustments.
A
sw
ρ
=A /(Ds)
sw sw
f (MPa)
yw
Tests [2]
NBR 6118(*)
Merta [4]
SDU5
Φ
8 c. 10
0.4%
573
239
151.7
181.7
SDU6
Φ
10 c. 10
0.64%
584
299
205.2
246.4
SDU8
2
Φ
8 c. 10
0.8%
573
331
246.9
282.4
SDU7
Φ
12.5 c. 10
1%
587
374
293.2
353.0