Civil Engineering Reference
In-Depth Information
Minimum reinforcement for crack control (ACI 11.8.4)
According to ACI 318 11.8.4, closed stirrups or ties of area
A
h
parallel to
A
s
shall be provided.
To simplify, assume
N
uc
= 0,
A
n
= 0 in
2
A
st
= max (
A
st2
,
A
sts
,
A
stmin
,
A
st
) = 35.28 in
2
(22,761 mm
2
)
A
h
= 0.5(
A
st
− A
n
) = 17.64 in
2
(11,381 mm
2
)
8 # 5 in 10 layers @ 305 mm (12″) c/c
Determination of the required depth to satisfy the stress limits at nodes 1
or 4 and to check the anchorage:
For nodal zone anchoring one tie b
n
= 0.8
(ACI A.5.2)
f
cu
= 0.85b
n
f
c
′
f
f
cu
= 1.785 ksi
(12.31 MPa)
Required depth
d
reqd
=
P
2/(f
f
cu
b
) = 35 mm (1.375″)
13.2.1.4
Design of the strut
Members 1, 3, 6, 8 (ACI A.2.6 and A.3.2); b
s_
bottle
= 0.75
By providing four two-legged no. 5 rebars as stirrups at 305 mm (12″)
c/c, which is also required for crack control and calculated earlier in accor-
dance to ACI 11.8.4,
A
s
2
= 1548 mm
2
(2.4 in
2
)
s
2
= 305 mm (12″)
g
2
= 86.98°
(
A
s
2
/
bs
2
)sin(g
2
) = 0.003
The stress in these bottle-shaped members will be limited to f
f
cu
= f0.85b
s_
bottle
f
c
′
f
f
cu
= 1673.437 psi (11.538 MPa)
Required depth for members 1 and 6:
d
1_strut
=
P
1/(f
f
cu
b
) = 79 mm (3.129″)
Required width for members 3 and 8:
d
3_strut
=
P
3/(f
f
cu
b
) = 73 mm (2.892″)
Member 4 (ACI A.2.6 and A.3.2)
This member is considered prism strut: b
s_
prism
= 1
The stress in prism-shaped members will be limited to
f
f
cu_
prism
=f0.85b
s_
prism
f
c
′ = 2231 psi (15.383 MPa)
Required width for member 4:
D
4_strut
=
P
1/(f
f
cu_
prism
b
) = 64 mm (2.503″)
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