Reinforced Concrete T-Beam Floor Beam — ACI 318-19 Worked Example
A simply supported 24-ft concrete beam cast with a 5-in slab, designed from loads to bars. The flange makes the compression block so shallow that the beam behaves like a very wide rectangle, the shear is low enough that only minimum stirrups are needed, and the cracked-section properties feed a deflection check that passes easily.
The beam
| Geometry | Simple span L = 24 ft (clear span 24 ft), beams at 10 ft on center, web bw = 14 in, total depth h = 24 in, 5-in slab |
| Materials | f′c = 4,000 psi normal weight, Grade 60 bars, 1.5 in clear cover, #3 stirrups, 1-in aggregate |
| Loads | Self-weight of slab and stem, superimposed dead 20 psf, live 100 psf × 0.8 = 80 psf (office, reduced) |
RC beam tool: T-beam, bw 14, h 24, bf 86, hf 5, 3-#8, #3 stirrups at 10 in, Mu 187.4, Vu 26.5, Ma 137.0.
Open the RC beam calculator →Factored load and demands
Live 80 × 10 = wL = 0.800 k/ft
wu = 1.2(1.102) + 1.6(0.800) = 2.60 k/ft → Mu = 2.60 × 24²/8 = 187.4 kip-ft
Service Ma = (1.102 + 0.800) × 24²/8 = 137.0 kip-ft
ACI 318-19 Table 6.3.2.1
bf = 14 + 2(36) = 86 in
Assume the block stays in the flange
Rn = Mu/(φbfd²) = 187.4 × 12,000 / (0.9 × 86 × 21.63²) = 62.1 psi
ρ = (0.85f′c/fy)[1 − √(1 − 2Rn/0.85f′c)] = 0.001045 → As = 0.001045 × 86 × 21.63 = 1.94 in²
As,min = 200bwd/fy = 200 × 14 × 21.63/60,000 = 1.01 in² (governs over 3√f′c, 0.96 in²)
| Bars | As (in²) | φMn (kip-ft) | Mu/φMn | Clear spacing |
|---|---|---|---|---|
| 2-#9 | 2.00 | 192.2 | 0.97 | 7.99 in (little reserve) |
| 3-#8 | 2.37 | 228.0 | 0.82 | 3.63 in |
| 4-#7 | 2.40 | 231.6 | 0.81 | 2.25 in |
| 2-#10 | 2.54 | 242.7 | 0.77 | 7.71 in |
Choose 3-#8: three bars give a symmetric layout with a corner bar in each stirrup bend and comfortable spacing. With them, a = 2.37 × 60/(0.85 × 4 × 86) = 0.49 in < hf = 5 in, confirming rectangular behavior; c = 0.57 in and εt = 0.110, far beyond the 0.00507 tension-controlled limit, so φ = 0.90.
Minimum stirrups, and where they can stop
φVc = 0.75 × 2√4,000 × 14 × 21.63 = 28.7 kip > Vu → no strength stirrups needed
But Vu > φλ√f′cbwd = 14.4 kip → Av,min required (§9.6.3.1)
#3 two-leg at s = 10 in: Av = 0.22 in² ≥ Av,min = 50 × 14 × 10/60,000 = 0.117 in²; s ≤ d/2 = 10.8 in ✓
φVn = 0.75(38.3 + 28.5) = 50.1 kip
Stirrups may stop where Vu = 14.4 kip: x = 12 − 14.4/2.60 = 6.5 ft from each support
Icr, Ie and the live-load deflection
Icr = 86(2.88)³/3 + 19.1(21.63 − 2.88)² = 7,384 in⁴ · Ig = 32,563 in⁴ (ȳ = 7.09 in from top)
Mcr = 7.5√4,000 × 32,563 / 16.91 / 12,000 = 76.1 kip-ft < Ma → cracked
Ie = Icr / [1 − ((2/3)Mcr/Ma)²(1 − Icr/Ig)]: at D (79.3 kip-ft) 10,797 in⁴; at D + L (137.0 kip-ft) 8,261 in⁴
ΔD = 0.21 in, ΔD+L = 0.48 in → ΔL = 0.27 in ≤ L/360 = 0.80 in ✓
Result
3-#8 bottom bars, #3 two-leg stirrups at 10 in for at least 6.5 ft from each support (continued full length in practice). Flexure 0.82, shear 0.53, live-load deflection L/1,085. Remaining checks not shown: development length of the #8 bars, top bars to hold stirrups, and long-term deflection with λΔ.
Tools used in this example
RC beam design (ACI 318-19), rebar sizes and development length. Guide: How to design a reinforced concrete beam.
Frequently asked questions
When does a T-beam act like a rectangular beam?
When the depth of the compression block a is no more than the flange thickness. Then all the compression is in the flange and the beam is analyzed as a rectangle of width bf. In this example a = 0.49 in against a 5-in slab, which is typical of floor beams in positive moment.
How wide is the effective flange of a T-beam?
ACI 318-19 Table 6.3.2.1: for a flange on both sides, the overhang on each side is the least of 8hf, half the clear distance to the next web, and ln/8. Here that is min(40, 53, 36) = 36 in each side, so bf = 14 + 72 = 86 in.
Where can stirrups be stopped?
Minimum shear reinforcement is required wherever Vu exceeds φλ√f′c·bw·d (ACI 318-19 §9.6.3.1). For this beam that is 14.4 kip, which the shear drops below 6.5 ft from each support. In practice many designers continue minimum stirrups full length to hold the top bars and simplify placing.
Why is a floor beam deflection check based on Ie and not Ig?
Service moments exceed the cracking moment, so the beam is cracked over much of its length and stiffness falls toward Icr. ACI 318-19 Table 24.2.3.5 gives the effective moment of inertia Ie between Icr and Ig. Using Ig would underestimate this beam's deflection by about a factor of three.