Civil Engineering Reference
In-Depth Information
Design of stirrups
Select
45
◦
, cot
α
r
=
α
r
=
1:
10
6
2(360 000)(413)
A
t
s
=
T
n
2
A
o
f
ty
=
56
.
5
×
190 mm
2
2
=
0
.
/
mm (0
.
0075 in
.
/
in
.
)
Select No. 3 bars
71
s
=
190
=
373 mm (14
.
7in
.
)
0
.
Check ACI maximum stirrup spacing:
s
<
p
h
/
8
p
h
=
2540
−
8(38
+
6)
=
2188 mm
,
p
h
/
8
=
2188
/
8
=
273 mm governs
s
max
=
305 mm (12 in.)
Use No. 3 stirrups at 270 mm (10.6 in.) spacing.
Design of longitudinal steel
A
t
s
p
h
A
t
s
p
h
(1)(1)
f
yt
f
y
tan
2
A
=
α
r
=
=
(0
.
190)(2188)
416 mm
2
(0
2
)
=
.
64 in
.
42
f
c
(MPa)
A
cp
f
y
A
t
s
p
h
.
0
f
yt
f
y
A
,
min
=
−
42
√
27
10
3
)
0
.
.
6(387
.
1
×
=
−
(0
.
190)(2188)(1)
413
1652 mm
2
(2
2
) governs
=
2068
−
416
=
.
56 in
.
Use 10 No. 5 longitudinal bars so that spacing will be less than 305 mm (12 in.) specified
by the ACI Code.
2000 mm
2
(3
2
)
1652 mm
2
(2
2
)OK
A
,
min
=
10(200)
=
.
1in
.
>
.
56 in
.
This example shows that the minimum torsional longitudinal steel
A
,
min
will gove
rn, when
the
applied
torsional stress (
33
f
c
(MPa)
τ
n
)
is
less
than the cracking torsional stress of 0
.
(4
f
c
(psi)). In this low range of torsional stress
τ
n
, the ACI Code allows the volume of stirrups
0.190 mm
2
/mm. However, a compensating volume
of longitudinal steel must be increased to
A
,
min
to be reduced with decreasing
τ
n
to
A
t
/
s
=
1652 mm
2
in order to ensure torsional
ductility. This way of treating limit design is economical, because the cost of stirrups is high
while the cost of longitudinal steel is low.
=
