Civil Engineering Reference
In-Depth Information
Solution (1): T
n
=
136 kN m (1200 in. kips)
Sectional properties
b
=
0
.
508 m(20 in
.
)
h
=
0
.
762 m(30 in
.
)
387 100 mm
2
(600 in
2
)
A
c
=
A
cp
=
bh
=
(508)(762)
=
.
p
c
=
p
cp
=
2(
b
+
h
)
=
2(508
+
762)
=
2540 mm (100 in
.
)
A
c
(387 100)
2
2540
10
6
mm
3
p
c
=
=
59
.
0
×
Check threshold torque
083
√
27
083
f
c
(MPa)
τ
n
,
threshold
=
0
.
=
0
.
.
6
=
0
.
436 MPa
083
f
c
(MPa)
A
c
10
6
)
T
n
,
threshold
=
.
p
c
=
.
×
=
.
0
(0
436)(59
25
7kNm
T
n
=
136 kN m
>
25
.
7kN m
.
Torsion needs to be considered.
33
f
c
(MPa)
The
nominal
torsional stress 0.439
f
c
(MPa) is greater than the cracking torsional stress
083
f
c
(MPa))
439
f
c
(MPa)
τ
n
=
/
.
.
=
.
>
.
(136
25
7)(0
0
0
0.33
f
c
(MPa).
Calculate
t
d
,
A
o
and
p
o
10
6
)
4
T
n
A
c
f
c
=
4(136
×
=
6)
=
.
.
t
d
51 mm (2
00 in
)
10
3
)(27
(387
.
1
×
.
1
2
p
c
t
d
=
322 000 mm
2
(499 in
2
)
A
o
=
A
c
−
387 100
−
2 540(51)
/
2
=
.
p
o
=
p
c
=
4
t
d
=
2540
−
4(51)
=
2336 mm (92
.
0in
.
)
Design of stirrups
10
6
2(322 000)(413)
(cot
×
A
t
s
=
T
n
2
A
o
f
ty
136
cot
α
r
=
α
r
)
α
r
)mm
2
/mm (0
2
/in
=
0
.
511(cot
.
0201(cot
α
r
)in
.
.
)
The angle of cracking direction
α
r
can be chosen to suit the designer's purpose. The ACI
Code requires that 30
◦
≤
α
r
60
◦
to prevent excessive cracking. For best crack control,
≤
α
r
should be taken as 45
◦
.
Select
45
◦
, cot
α
r
=
α
r
=
1:
A
t
s
=
511 mm
2
2
0
.
/
mm (0
.
0201 in
.
/
in
.
)
129
0
Select No. 4 bars
s
=
511
=
252 mm (10.0 in.)
.
