Structural Concrete - Hassoun

178 Chapter 4 Flexural Design of Reinforced Concrete Beams
Assume a = 2.0 in. and calculate As:
(
M u = φA s f y d − a )
2
587.8 × 12 = 0.9 × 60A s (17.5 − 1.0) and A s = 7.92 in. 2
Check assumed a = A s f y ∕(0.85f ′ cb)=7.92 × 60∕(0.85 × 4 × 60) =2.33in. Revised As =
587.8 × 12/(0.9 × 60 × 16.33) = 7.99 in. 2 Check revised a: a = 7.99 × 2.33/7.92 = 2.35 in.,
which is very close to 2.33 in.
Alternatively, Eq. 4.2 can be used to get ρ and As. Choose eight no. 9 bars in two rows
(area = 8.0 in. 2 ), (b min = 11.8 in.). Extend four no. 9 bars on both sides to the columns. The
other four bars can terminate where they are not needed, beyond section F; see the longitudinal
section in Fig. 4.13.
2. Design of section at B: M u =−452.2 K⋅ft. The section acts as a rectangular section, b = 16 in.
and d = 17.5 in. The main tension reinforcement lies in the flange:
ρ max = 0.01806 and R u,max = 820psi (Table 4.1)
Check the maximum moment capacity of the section as singly reinforced:
φM n,max = R u,max bd 2 = 0.82(16)(17.5)2 = 334.8K⋅ ft
12
which is less than the applied moment. Compression steel is needed:
A s1 = 0.01806(16)(17.5) =5.06 in. 2
M u2 = 452.2 − 334.8 = 117.4K⋅ ft
M u2 = φA s2 f y (d − d ′ )
Assumed ′ = 2.5in.
117.4 × 12 = 0.9A s2 (60)(17.5 − 2.5) and A s2 = 1.74 in. 2
Total tension steel = 5.06 + 1.74 = 6.8 in. 2 Use seven no. 9 bars in two rows (area used 7.0 in. 2 ,whichis
adequate). For compression steel, use two no. 9 bars (area 2.0 in. 2 ), extended from the positive-moment
reinforcement to the column. Actually, four no. 9 bars are available; see the longitudinal section in
Fig. 4.13.
The seven no. 9 bars must extend in the beam BC beyond section F into the compression zone and
also must extend into the column BA to resist the column moment of 452.2 K ⋅ ft without any splices at
joints B or C.
Check if compression steel yields by using Eq. 3 or Table 3. Assume K = 0.01552 (d ′ /d) =
0.1552(2.5)/(17.5) = 0.02217 >ρ 1 = 0.01806. Therefore, compression steel yields, and f s ′ = 60 ksi, as
assumed.
Stirrups are shown in the beam to resist shear (refer to Chapter 8), and two no. 5 bars were placed
at the top of the beam to hold the stirrups in position. Ties are used in the column to hold the vertical
bars (refer to Chapter 10). To determine the extension of the development length of bars in beams or
columns, refer to Chapter 7.
4.7 EXAMPLES USING SI UNITS
Example 4.13
Design a singly reinforced rectangular section to resist a factored moment of 280 kN ⋅ musingthe
maximum steel percentage for tension-controlled sections. Given: f ′ c = 20N∕mm2 , f y = 400 N/mm 2 ,and
b = 250 mm.