Civil Engineering Reference
In-Depth Information
7.6.3
Case 3—Interior support
Reinforcement required to resist bending moments:
The same approach
discussed for Case 1 is followed. The effective reinforcement depth is
−−
φ
f_bar
d: hc
=⋅
26 in.f3
=
f3
c
2
The longitudinal reinforcement ratio required for bending is
M
1
bd
(
)
u3
ρ
:
⋅
=
0.01132
M
=
349 kipft
⋅
⋅
=
f_req_bend3
u3
−
β
⋅
1
w
f3
0.65 f
⋅⋅
d
⋅
0.15d
fu
f3
f3
2
The minimum reinforcement requirement has to be verified. Equation
(8-8) of ACI 440.1R-06 is used. If the failure is not governed by FRP rup-
ture, this requirement is automatically achieved:
4.9fpsi
f
⋅
′
⋅
bd,
300psi
f
(
)
c
2
A:
f_min3(
⋅
bd
⋅
=
1.706 in
⋅
=
f_min3
w
f3
w
f3
fu
fu
A
bd
f_min3
ρ
:
=
0.004687
=
f_min3
⋅
w
f3
The design reinforcement ratio required for bending is taken as
(
)
ρ
:max
ρ
,
ρ
=
0.0113
=
f_bend3
f_req_bend3
f_min3
Reinforcement ratio required for creep-rupture stress check:
The bend-
ing moment to consider for the creep-rupture stress check is
w .20w
w
+
D
LNeg
M:
M
⋅
=
197.5ftkip
⋅
⋅
=
3_creep0
s3
SNeg
The limit stress is
f
f
_
creep
= 12.8·ksi
The ratio of depth of neutral axis to reinforcement depth, kf,
f
, can be writ-
ten as a function of the reinforcement ratio, ρ
f
_
creep
:
(
)
(
)
2
k
ρ
:
2
⋅ρ
⋅
n
+ρ⋅
n
−ρ
⋅
n
=
f_creep3f_creep
f_creep
f
f_creep
f
f_creep
f
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