Civil Engineering Reference
In-Depth Information
reinforcement required by flexural shear. The calculated torsional longitudinal steel should be
added to the longitudinal steel required by bending moments.
7.2.6 Minimum Longitudinal Torsional Steel
In sections 2.3.1.2 and 2.3.1.3, a pair of equations (2.104) and (2.105), are given for the design
of longitudinal torsional steel. Equation (2.104) was derived to ensure the torsional strength,
while Equation (2.105) was given to calculate the minimum longitudinal torsional steel to
ensure ductility. In this section, we will derive Equation (2.105).
In order to avoid a brittle torsional failure, a minimum amount of torsional reinforcement
(including both transverse and longitudinal steel) is required in a member subjected to torsion.
The basic criterion for determining this minimum torsional reinforcement is to equate the
post-cracking strength
T
n
to the cracking strength
T
cr
:
T
n
=
T
cr
(7.87)
45
◦
, the post-cracking torsional strength
T
n
of both solid and hollow
sections can be predicted from Equation (2.99):
Taking the angle
α
r
=
2
A
o
A
t
f
yt
s
T
n
=
(7.88)
The cracking torque of solid sections subjected to combined torsion, shear and bending
T
cr
can be predicted by the compatibility torsion given in Section 7.2.5:
33
f
c
(MPa)
A
cp
or 4
f
c
(psi)
A
cp
T
cr
=
0
.
(7.89)
p
cp
p
cp
where the ACI notations
A
cp
and
p
cp
are
the same as
A
c
and
p
c
, respectively, in Section
7.1.3.4. The concrete strength
f
c
and
f
c
are in MPa or psi. In the case of hollow sections,
Mattock (1995) suggested a simple relationship between the cracking torque of a hollow
section
T
cr
,
hollo
w
and that of a solid section with the same outer dimensions
T
cr
,
solid
:
T
cr
,
hollo
w
T
cr
,
solid
=
A
g
A
cp
(7.90)
where
A
g
is the cross-sectional area of the concrete only and not including the hole(s), while
A
cp
is the area of the same hollow section including the hole(s). For solid sections,
A
g
=
A
cp
.
The cracking torque of solid and hollow sections can then be expressed by one equation:
33
f
c
(MPa)
A
g
A
cp
p
cp
or 4
f
c
(
psi
)
A
g
A
cp
p
cp
T
cr
=
0
.
(7.91)
Inserting Equations (7.91) and (7.88) into Equation (7.87) gives:
167
f
c
(MPa)
A
g
A
cp
s
A
o
p
cp
0
.
A
t
f
yt
=
(7.92)
