Structural Concrete - Hassoun

144 Chapter 3 Flexural Analysis of Reinforced Concrete Beams
( f
′
)( )
ρ b = 0.85β c 600
1
f y 600 + f y
( 0.005
ρ max =
)
ρ b =
= 0.0325
( 5
8)
(0.0325) =0.0203 ρ min = 1.4
400 = 0.0035
0.008
Since ρρ min , it is a tension-controlled section and φ = 0.9. Let a = A s f y /
(0.85f c ′ b) = 1470 × 400/(0.85 × 30 × 300) = 77 mm, c = 90 mm. Also φM n = φA s f y (d − a/2) =
0.9 × 1470 × 400(490 −77/2) × 10 −6 = 238.9 KN⋅m. Also ε t = 0.003(d t − c)/c = 0.003
(490 −90)/90 = 0.01333 > 0.005, φ = 0.9 as assumed.
3. The internal design moment strength is greater than the external factored moment. Therefore, the
section is adequate.
Example 3.19
Calculate the design moment strength of a rectangular section with the following details: b = 250 mm,
d = 440 mm, d ′ = 60 mm, tension steel is six bars 25 mm in diameter (in two rows), compression steel
is three bars 20 mm in diameter, f c ′ = 20MPa, and f y = 350 MPa.
Solution
1. Check if compression steel yields:
A s = 490 × 6 = 2940 mm 2 A ′ s = 314 × 3 = 942 mm2 A s − A ′ s = 1998 mm2
ρ =
2940
250 × 440 = 0.0267 942
ρ′ =
250 × 440 = 0.00856
ρ − ρ ′ = 0.01814
For compression steel to yield:
ρ − ρ ′ ≥ 0.85 × 0.85 ×
ρ − ρ ′ = 0.01814 > 0.01351.
Therefore, compression steel yields.
2. Calculate M n :
( )( )( )
20 60 600
350 440 600 − 350 = 0.01351
a = A s − A s ′
0.85f c ′ b = 1998
= 164 mm
0.85 × 20 × 250
[ (
M n = 1998 × 350 440 − 164 )
]
+ 942 × 350(440 − 60) × 10 −6 = 417.3KN ⋅ m
2
3. Check φ based on ε t ≥ 0.005.
ε t = 0.003(d t − c)
c
d t = h − 65 mm = d + 25 mm
d t = 440 + 25 = 465 mm
a = 164 mm c = 164 = 193 mm
0.85
for two rows of tension bars
Let ε t = 0.003(465 −193)/193 = 0.04228, which is less than 0.005, but greater than the 0.004
limit. Also φ = 0.48 +83 × ε t , = 0.831, and φM n = 0.831 (417.3) = 346.8 KN⋅m.