Structural Concrete - Hassoun

6.3 Long-Time Deflection 233
Then calculate I g :
[
bt
3
I g =
(y
12 + bt − t ) ] 2
+
2
[b w
(y − t) 3
3
]
+
(h − y)
[b 3 ]
w
2. Cracked moment of inertia, I cr :Letx be the distance of the neutral axis from the extreme
compression fibers (x = kd).
a. Rectangular section with tension steel, A s , only:
i. Calculate x from the equation
ii. Calculate
bx 2
2 − nA s(d − x) =0 (6.11)
l cr = bx3
3 + nA s(d − x) 2 (6.11a)
b. Rectangular section with tension steel A s and compression steel A ′ s:
i. Calculate x:
x = bx2
2 +(n − 1)A′ s(x − d ′ )−nA s (d − x) =0 (6.12)
ii. Calculate
l cr = bx3
3 +(n − 1)A′ s(x − d ′ ) 2 + nA s (d − x) 2 (6.12a)
c. T-sections with tension steel A s :
i. Calculate x:
(
x = bt x − t )
(x − t) 2
+ b
2 w − nA
2
s (d − x) =0 (6.13)
ii. Calculate I cr :
[
bt
3
I cr =
(x
12 + bt − t ) 2
]
+
2
[b w
(x − t) 3
3
3
]
+ nA s (d − x) 2 (6.13a)
6.3 LONG-TIME DEFLECTION
Deflection of reinforced concrete members continues to increase under sustained load, although
more slowly with time. Shrinkage and creep are the cause of this additional deflection, which
is called long-time deflection [1]. It is influenced mainly by temperature, humidity, age at time
of loading, curing, quantity of compression reinforcement, and magnitude of the sustained load.
The ACI Code, Section 24.2.4.1, suggests that unless values are obtained by a more comprehensive
analysis, the additional long-term deflection for both normal and lightweight concrete flexural
members shall be obtained by multiplying the immediate deflection caused by sustained load by
the factor
ζ
λ Δ =
(6.14)
1 + 50ρ ′
where
λ Δ = multiplier for additional deflection due to long-term effect.
ρ ′ = A ′ s∕bd for section at midspan of simply supported or continuous beam or at support of
cantilever beam
ζ = time-dependent factor for sustained loads that may be taken as shown in Table 6.2.