How to Design a Reinforced Concrete Beam — ACI 318-19 Step by Step
A practical order of operations for sizing a reinforced concrete beam: pick the section, find the steel, choose bars that fit, confirm the strain limit, design the stirrups, then check service behavior. Each step uses the 12 × 24 in beam from the free RC beam calculator (Mu = 250 kip-ft, Vu = 45 kip, f′c = 4,000 psi, Grade 60).
Get factored demands
Combine loads per ASCE 7 (usually 1.2D + 1.6L for floors) and find Mu at midspan and Vu at a distance d from the face of the support. For continuous beams, use an envelope with pattern live load — the continuous beam tool gives both. Keep the unfactored service moment Ma for Step 8.
Choose a trial section
Start with the ACI 318-19 Table 9.3.1.1 minimum depth (ℓ/16 simply supported, ℓ/18.5 one end continuous, ℓ/21 both ends continuous, ℓ/8 cantilever). A width of about half the depth, rounded to a form size (10, 12, 14, 16 in), keeps bars and stirrups buildable. Beams cast with a slab are T-beams in positive moment; get bf from Table 6.3.2.1.
Estimate the effective depth
Compute the required steel
ρ = (0.85f′c/fy)[1 − √(1 − 2Rn/0.85f′c)] = 0.05667 × (1 − 0.8066) = 0.01096
As,req = 0.01096 × 12 × 21.63 = 2.84 in² · As,min = 200bwd/fy = 0.865 in²
Choose bars and check they fit
4-#8 gives 3.16 in². Clear spacing in one layer = (12 − 2 × 1.875 − 4 × 1.0)/3 = 1.42 in, which must be at least the largest of 1 in, db, and 4/3 of the maximum aggregate (1.33 in for 1-in stone) — it fits. If it doesn't, try fewer larger bars, a wider beam, or two layers (then recompute d to the bar-group centroid).
Confirm strength and the strain limit
εt = 0.003(21.63 − 5.47)/5.47 = 0.0089 ≥ εty + 0.003 = 0.00507 → φ = 0.90
φMn = 0.9 × 3.16 × 60 × (21.63 − 4.65/2)/12 = 274.5 kip-ft ≥ 250 ✓
Design the stirrups
Vs,req = 45/0.75 − 32.8 = 27.2 kip → s = Avfytd/Vs = 0.22 × 60 × 21.63/27.2 = 10.5 in
smax = d/2 = 10.8 in (Vs ≤ 4√f′cbwd) → #3 two-leg stirrups at 9 in: φVn = 48.4 kip ≥ 45 ✓
Away from the supports, increase spacing where the shear drops; minimum stirrups are required wherever Vu > φλ√f′cbwd (§9.6.3.1).
Check service: cracking, deflection, crack control
Mcr = 7.5√f′cIg/yt = 45.5 kip-ft · at Ma = 160 kip-ft: fs = 31.9 ksi
Crack control smax = 15(40/fs) − 2.5cc = 14.1 in ≥ 2.4 in bar spacing ✓
If the section is shallower than Table 9.3.1.1, compute deflection with Ie from Table 24.2.3.5 and compare with the limits of Table 24.2.2.
Finish the detailing
- Development length of the bottom bars past the critical sections (development length card).
- At least one-third of positive steel continuous into simple supports, 6 in minimum embedment (§9.7.3.8).
- Top bars to hold the stirrups; closed stirrups if torsion or compression steel is present.
- Skin reinforcement on the side faces when h > 36 in (§9.7.2.3).
Pick bars and stirrups; get φMn, required As, Vc, stirrup spacing, kd, Icr and crack control in one place.
Open the RC beam calculator →Worked examples
RC T-beam floor beam · Doubly reinforced beam with non-yielding compression steel
Frequently asked questions
How deep should a concrete beam be?
ACI 318-19 Table 9.3.1.1 sets the minimum depth to skip a deflection calculation: ℓ/16 for simply supported beams, ℓ/18.5 with one end continuous, ℓ/21 with both ends continuous and ℓ/8 for cantilevers (Grade 60, normal-weight concrete). Strength usually wants a little more; a span/12 to span/16 first guess is common.
What is the formula for required steel area?
With Rn = Mu/(φbd²), ρ = (0.85f′c/fy)[1 − √(1 − 2Rn/(0.85f′c))] and As = ρbd. It assumes the steel yields and φ = 0.90, both of which must be confirmed after the bars are chosen.
What if the beam is not tension-controlled?
Under ACI 318-19 a beam with small axial load must reach εt ≥ εty + 0.003. If it does not, deepen or widen the section, raise f′c, or add compression steel. Adding more tension steel makes it worse.
How do I pick stirrup spacing?
Find Vs = Vu/φ − Vc, then s = Av·fyt·d/Vs. Take the smallest of that, the spacing that still provides Av,min, and the maximum spacing (d/2 or 24 in, halved when Vs exceeds 4√f′c·bw·d). Round down to a practical increment.