Simplified Design for Beams and One-Way Slabs - Simplified Design of Reinforced Concrete Buildings - page 95

(5) Determine flexural reinforcement for the beams at the 1st floor level

(a) Top bars at exterior columns

Check governing load combination:

￿ gravity loads
M u = 326.7 ft-kips ACI Eq. (9-2)

￿ gravity + wind load
M u = 1.2(3.9)(28.58) 2 /16 + 0.8(99.56)

= 317.9 ft-kips ACI Eq. (9-3)

or

M u = 1.2(3.9)(28.58) 2 /16 + 0.5(1.09)(28.58) 2 /16
+ 1.6(99.56) = 426.1 ft-kips ACI Eq. (9-4)

￿ also check for possible moment reversal due to wind moment:
M u = 0.9(3.9)(28.58) 2 /16 ± 1.6(99.56) = 338.5 klf, 19.9 ft-kips

ACI Eq. (9-6)

M u 426.1

4(17) =

A s =

4d =

6.27 in. 2

From Table 3-5: Use 8-No. 8 bars (A s = 6.32 in. 2 )

minimum n = 36[1.5 + 0.5 + (1.0/2)] 2 /57.4 = 4 bars < 8 O.K.

Check ρ = A s /bd = 6.32/(36 17) = 0.0103 > ρ min = 0.0033 O.K.

(b) Bottom bars in end spans:

A s = 373.4/4(17) = 5.49 in. 2

Use 8-No. 8 bars (A s = 6.32 in. 2 )

(c) Top bars at interior columns:

Check governing load combination:

￿ gravity load only
M u = 522.8 ft-kips ACI Eq. (9-2)

￿ gravity + wind loads:
M u = 1.2(3.9)(28.54) 2 /10 + 0.8(99.56)

= 460.8 ft-kips ACI Eq. (9-3)

or

M u = 1.2(3.9)(28.54) 2 /10 + 0.5(1.09)(28.54) 2 /10
+ 1.6(99.56) = 584.9 ft-kips ACI Eq. (9-4)