Civil Engineering Reference
In-Depth Information
Check ACI maximum stirrup spacing:
s
<
p
h
/
8
p
h
=
2540
−
8(38
+
6)
=
2188 mm
,
p
h
/
8
=
2188
/
8
=
273 mm
>
252 mm OK
Use No. 4 stirrups at 250 mm (10.0 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
.
511)(2188)
1118 mm
2
(1
2
) governs
=
.
73 in
.
42
f
c
(MPa)
A
cp
f
y
A
t
s
p
h
0
.
f
yt
f
y
A
,
min
=
−
42
√
27
0
.
.
6(387
.
1
×
10
3
)
=
−
(0
.
511)(2188)(1)
413
950 mm
2
1118 mm
2
does not govern
=
2068
−
1118
=
<
Use 10 No. 4 longitudinal bars so that spacing will be less than 305 mm (12 in.) specified
by the ACI Code.
1290 mm
2
(2
2
)
1118 mm
2
(1
2
)OK
A
=
10(129)
=
.
0in
.
>
.
73 in
.
This example shows that the minimum torsional longitudinal steel
A
,
min
does not gov-
ern, w
hen the
appl
ied tor
sional stress (
τ
n
)
is
greater
than the cr
acking t
orsional stress
33
f
c
(MPa),
A
,
min
could
become negative, but still have the same meaning: 'does not govern'.
33
f
c
(MPa) (4
f
c
(psi)). When
0
.
τ
n
is much greater than 0
.
Solution (2): T
n
=
56.5 kN m (500 in. kips)
Sectional properties
: same as (1)
Check threshold torque
083
f
c
(MPa)
A
c
10
6
)
T
n
,
threshold
=
0
.
p
c
=
(0
.
436)(59
.
0
×
=
25
.
7kNm
T
n
=
56
.
5kNm
>
25
.
7kNm
Torsion needs to be considered.
33
f
c
(MPa)
The nominal
torsiona
l stress, 0.183
f
c
(MPa), is much less than the compatibility torsional
083
f
c
(MPa)
183
f
c
(MPa)
τ
n
=
.
/
.
.
=
.
<
.
(56
5
25
7)(0
0
0
stress of 0.33
f
c
(MPa).
Calculate
t
d
,
A
o
and
p
o
10
6
)
4
T
n
A
c
f
c
=
4(56
.
5
×
t
d
=
6)
=
21 mm (0
.
83 in
.
)
(387
.
1
×
10
3
)(27
.
1
2
p
c
t
d
=
360 000 mm
2
2
)
A
o
=
A
c
−
387 100
−
2540(21)
/
2
=
(558 in
.
p
o
=
p
c
−
4
t
d
=
2540
−
4(21)
=
2456 mm (96
.
7in
.
)
