Concrete Beam Reinforcement

01 Three jobs, three bar types

A reinforced concrete beam has three kinds of steel, and each one does a different structural job. Understanding which bar carries which force is the first step to sizing reinforcement correctly.

In a simply supported beam — one that rests on two supports with no continuity — the bottom steel carries the full moment and the top bars are often just hanger bars for the stirrups (2 #4 is typical). In a continuous beam, the moment reverses over the supports: the top steel carries negative moment there, and the bottom steel may be reduced or cut off between support zones.

Stirrups are the shear reinforcement. A stirrup is a loop of deformed bar — usually #3 or #4 — bent into a U or rectangle that wraps around the longitudinal bars. The legs of the stirrup cross the diagonal tension cracks that form under shear, so the stirrup area is perpendicular to the crack. This is why stirrup spacing is tightest near the supports, where shear is highest, and widest at midspan, where shear is lowest.

02 ACI 318-19 minimums and limits

Before calculating the exact steel area, you need to know the floors and ceilings the code sets. ACI 318-19 has five limits that apply to every beam, regardless of the loads:

The minimum As formula is the one that surprises people. For 4,000 psi concrete, Grade 60 steel, a 12 in wide beam with d = 17 in:

Term 13 × √4,000 × 12 × 17 / 60,000 = 0.644 sq in Term 2200 × 12 × 17 / 60,000 = 0.680 sq in As minmax(0.644, 0.680) = 0.680 sq in Bars to meet0.680 / 0.31 (#5) = 2.2 → 3 #5 (0.93 sq in)

The 200·bw·d/fy term governs below about 4,444 psi concrete (where 3√f'c = 200). Above that, the 3√f'c term takes over. The minimum is rarely the controlling value in a designed beam — the loads usually demand more — but it is the floor for any beam that carries calculated load, and it prevents a beam with one #3 bar from being "reinforced" on paper.

The maximum As limit is written as a strain limit, not a number: the net tensile strain εt ≥ 0.004 at nominal strength (ACI 318-19 §9.3.1.1). For Grade 60 steel in a rectangular beam, this works out to roughly ρ = As/(bw·d) ≤ 0.023 — about 0.025 for practical purposes. A 12 in × 20 in beam with d = 17 in would max out at about 0.025 × 12 × 17 = 5.1 sq in of steel, which is 16 #5 bars — far more than a beam that size can physically fit in one layer. The max As limit rarely governs in a typical beam; it governs in heavily loaded transfer beams and in columns.

ACI 318-19 §9.6.1.2, §9.3.1.1, §9.7.6.2.2, §9.7.6.3, Table 20.5.1.3.1. The 200/fy term was originally written for Grade 40 steel; at 60,000 psi it produces a smaller As min, which is correct — higher-grade steel needs less area to reach yield.

03 Sizing the bottom steel

The bottom bars carry positive bending moment — the sagging moment at midspan of a simply supported beam. Sizing them is a four-step process:

1. Compute the factored moment Mu. For a simply supported beam with uniform load, Mu = wu·Ln²/8, where wu is the factored load per foot and Ln is the clear span.
2. Solve for required As. Use As = Mu / (φ·fy·(d - a/2)), where φ = 0.9 for tension-controlled sections, d is the effective depth, and a = As·fy/(0.85·f'c·bw). This is iterative because a depends on As, but for a first estimate you can use d - a/2 ≈ 0.9·d.
3. Check As ≥ As min. If the required As is below the minimum from §9.6.1.2, use the minimum.
4. Pick a bar size and count. Divide As required by the bar area (see the rebar area table) and round up. Check that the bars fit in one layer: bars per layer = floor((bw - 2·cover + s) / (db + s)) where s is the clear spacing (≥ max(db, 1 in, 4/3·dagg) per §25.2.1).

For a quick estimate, the first-pass expression As ≈ Mu / (0.9·fy·0.9·d) = Mu / (0.81·fy·d) gets you within 5–10% of the exact value, which is close enough to pick a bar size before running the iteration.

The highlighted rows (#4, #5, #6) are the typical beam bars. Read across to find how many bars of a given size provide a target As: if you need 1.2 sq in, 4 #5 (1.24) or 6 #4 (1.20) both work. The choice between them is usually about fit — 4 #5 bars fit in a 10 in wide beam at 1.5 in cover with 1 in spacing, while 6 #4 bars need a 12 in width to fit in one layer.

04 Stirrups — spacing from shear

Stirrups carry shear. The mechanism is not vertical — shear causes diagonal tension cracks at roughly 45°, and the vertical leg of a stirrup crosses that crack. The force a stirrup can resist is Vs = Av·fy·d/s, where Av is the area of the stirrup legs crossing the crack, d is the effective depth, and s is the spacing.

The design chain for stirrups is:

1. Compute concrete shear capacity. Vc = 2·λ·√f'c·bw·d (ACI 318-19 §22.5.5.1, simplified). For 4,000 psi, normal weight, bw = 12 in, d = 17 in: Vc = 2·1.0·63.2·12·17 / 1000 = 25.8 kip. φVc = 0.75·25.8 = 19.4 kip.
2. Compare Vu to φVc. If Vu ≤ φVc/2, no stirrups are required by code (but minimum stirrups may still be needed for constructability). If φVc/2 < Vu ≤ φVc, minimum stirrups at max spacing govern. If Vu > φVc, stirrups must be designed for Vu - φVc.
3. Solve for spacing. s = φ·Av·fy·d / (Vu - φVc) when Vu > φVc. This spacing tightens as Vu approaches the support.
4. Apply maximum spacing. s ≤ lesser of d/2 and 24 in. If Vu > 4·√f'c·bw·d, max spacing drops to d/4.

For a #3 U-stirrup (2 legs, Av = 2·0.11 = 0.22 sq in), Grade 60, d = 17 in:

Av·fy·d0.22 × 60,000 × 17 = 224,400 lb φAv·fy·d0.75 × 224,400 = 168,300 lb = 168.3 kip s at Vu = 25 kips = 168.3 / (25 - 19.4) = 30.1 in → capped at d/2 = 8.5 in s at Vu = 22 kipVu < φVc, so min stirrups at s ≤ d/2 = 8.5 in → 8 in s at Vu = 12 kipVu < φVc/2 = 9.7 kip? No — 12 > 9.7, so min stirrups still needed s at midspan (Vu ≈ 0)No stirrups needed by code, but typically placed at d/2 or 24 in for constructability

The result is a stepped spacing pattern: tightest near the supports, gradually widening toward midspan. On a drawing it looks like 6 in @ support → 8 in → 12 in → 24 in @ midspan, with the transitions placed where the shear envelope allows the next wider spacing.

05 Top bars and continuity steel

In a simply supported beam, the top bars have a simple job: they hold the stirrups up. Two #4 bars running full length are typical, and they are called "hanger bars" because they carry no calculated force. Their only structural purpose is to anchor the top legs of the stirrups.

In a continuous beam — one that runs over multiple supports without joints — the moment reverses over each support. The bottom face goes into compression and the top face goes into tension. The top steel over the support carries that negative moment, and it is sized with the same expression as the bottom steel:

As (top)Mu,negative / (φ·fy·(d - a/2)) whereMu,negative is the factored negative moment at the support face typically 0.65·wLn²/12 for interior supports of multi-span beams

Top bars over a support extend past the point of inflection — the point where the moment crosses zero — by a development length. This extension is typically about Ln/4 or Ln/3 past the inflection point, whichever is greater. The bars are often cut off in a staggered pattern: half the bars extend further than the other half, so the steel area steps down as the moment decreases away from the support.

For a continuous beam with equal spans, the typical steel layout is:

• Bottom steel: runs full length, sized for the maximum positive moment (usually at midspan of the end span, which carries more positive moment than the interior spans).
• Top steel over supports: sized for the negative moment at the support face, extending Ln/4 past each inflection point.
• Top steel at midspan of interior spans: minimum steel (2 #4) for stirrup anchorage, not for moment.
• Bottom steel cut-offs: some bottom bars may be cut off between midspan and the support where the positive moment drops below the capacity of the remaining bars.

06 Worked example: 20 ft beam

Let's run a complete beam reinforcement design. The beam: 20 ft clear span, simply supported, 12 in × 20 in cross section, carrying a service live load of 1,000 lb/ft and a superimposed dead load of 200 lb/ft. Concrete: 4,000 psi, normal weight. Steel: Grade 60, deformed bar. Cover: 1.5 in.

Step 1 — Loads and effective depth.

Dead loadbeam self-weight = 12×20/144 × 150 lb/ft³ = 250 lb/ft superimposed DL = 200 lb/ft live load = 1,000 lb/ft wu1.2×(250+200) + 1.6×1,000 = 540 + 1,600 = 2,140 lb/ft d20 - 1.5 cover - 0.375 stirrup - 0.312 (#5 db/2) = 17.8 in

Step 2 — Factored moment and required As.

Muwu·Ln²/8 = 2,140 × 20² / 8 = 107,000 lb-ft = 107.0 kip-ft As (first est.)Mu / (0.81·fy·d) = 107×12 / (0.81×60×17.8) = 1,284 / 864.3 = 1.49 sq in Check As minmax(3√4000×12×17.8/60000, 200×12×17.8/60000) = max(0.674, 0.712) = 0.712 1.49 > 0.712 → As required governs = 1.49 sq in

Step 3 — Bar selection.

5 #55 × 0.31 = 1.55 sq in > 1.49 → OK Check fitfloor((12 - 2×1.5 + 1) / (0.625 + 1)) = floor(10/1.625) = 6 → 5 fits in one layer d checkwith 5 #5 in one layer: d = 20 - 1.5 - 0.375 - 0.312 = 17.8 in (matches) φMn0.9 × 1.55 × 60 × (17.8 - 1.55×60/(2×0.85×4×12)) / 12 = 0.9 × 1.55 × 60 × (17.8 - 1.14) / 12 = 0.9 × 1.55 × 60 × 16.66 / 12 = 116.2 kip-ft > 107.0 kip-ft → OK (φMn ≥ Mu)

Step 4 — Shear and stirrups.

Vu at supportwu·Ln/2 = 2,140 × 20 / 2 = 21.4 kip Vc2×√4000×12×17.8/1000 = 2×63.2×12×17.8/1000 = 27.0 kip φVc0.75 × 27.0 = 20.3 kip Vu vs φVc21.4 > φVc/2 = 10.1 → stirrups required Vu vs φVc21.4 > φVc = 20.3 → design stirrups for Vu - φVc = 1.1 kip s requiredφ·Av·fy·d / (Vu - φVc) = 0.75×0.22×60×17.8/1.1 = 176.2/1.1 = 160 in s maxlesser of d/2 = 8.9 in and 24 in → 8 in governs

Since Vu barely exceeds φVc (by 1.1 kip), the calculated spacing is huge (160 in), so the maximum spacing governs everywhere. The stirrup layout for this beam is:

Step 5 — Bar list and weight.

Bottom bars5 #5 × 20 ft = 100 ft (plus development past supports, see below) Development ld19 in for #5 bottom bar, 4,000 psi, uncoated (see development length) Bar extensionextend 19 in past each support centerline → 20 ft + 2×19/12 = 23.2 ft Cut length24 ft stock → 5 bars × 24 ft = 120 ft Top bars2 #4 × 24 ft = 48 ft (hangers, full length + lap) Stirrups21 #3 U-stirrups, bent to fit 12×20 inside 1.5 in cover Stirrup cut length2×(12-3) + 2×(20-3) + 6×0.375(hook) = 18+34+2.25 = 54.25 in = 4.52 ft Stirrup total21 × 4.52 = 94.9 ft of #3 Weight #5120 ft × 1.043 lb/ft = 125.2 lb Weight #448 ft × 0.668 lb/ft = 32.1 lb Weight #394.9 ft × 0.376 lb/ft = 35.7 lb Total193.0 lb of reinforcing steel

For the full material takeoff including tie wire and chairs, see those pages. The beam cage is tied at every stirrup intersection with the longitudinal bars — that is 21 stirrups × 7 bars (5 bottom + 2 top) = 147 tie points, about 1.3 lb of 16-gauge wire.

07 Detailing rules that catch people

The math gives you the bar sizes and counts, but the beam also has to be buildable. These are the detailing rules from ACI 318-19 Chapter 25 that most often cause problems when they are missed:

• Bar spacing minimum (§25.2.1). Clear space between parallel bars in a layer must be at least max(db, 1 in, 4/3·dagg), where dagg is the aggregate size. A 1 in aggregate mix costs you 1.33 in of clear space per bar, which can push a 5-bar layout out of a 12 in beam.
• Stirrup hook (§25.3.2). A #3 through #5 stirrup uses a 135° standard hook with a 6·db extension. The hook must engage a longitudinal bar — if the stirrup hook is not tied around a main bar, it can open under load.
• Development length at supports (§9.7.2.3). At least one-third of the positive moment steel must extend past the face of the support by at least 6 in. For a simple support, this is usually met by running the bottom bars full length to the beam end.
• Lap splices (§25.5). Tension laps follow the splice length table: a #5 Class B lap is 25 in at 4,000 psi. Laps should be staggered so no more than half the bars are spliced at the same section.
• Stirrup placement (§9.7.6). The first stirrup is placed no more than 2 in from the face of the support. Stirrups are not optional at any point where Vu > φVc/2, even in the middle third of a lightly loaded beam.
• Cover to stirrup (Table 20.5.1.3.1). The 1.5 in cover for beams is measured to the outside of the stirrup, not to the main bar. The main bar sits at 1.5 + dstirrup + db/2 from the face, which is why a #5 bar inside a #3 stirrup at 1.5 in cover sits at 2.19 in to its centre — see concrete cover.

08 Bar count from As required

Once you have As required, the bar count is a lookup. Divide As required by the single-bar area and round up:

Values in parentheses are the As actually provided. The gap between required and provided is the reason beam designs are not exact — you round up to the next whole bar, and on small beams that can mean 20–30% more steel than the calculation demands. This is not waste; it is the difference between the continuous math of As required and the discrete reality of bar counts.

What this page does not do

• It does not design your beam. The worked example uses assumed loads and dimensions. A real beam design requires the actual loads, span conditions, and support details from the structural drawings.
• It does not cover seismic detailing. Special moment frames in ACI 318-19 Chapter 18 have much tighter stirrup requirements, strict beam-column joint rules, and capacity-design principles that this page does not address.
• It does not cover deep beams. Beams with Ln/d < 4 are deep beams (ACI 318-19 §9.9) and are designed by strut-and-tie, not by the expression on this page.
• It does not cover T-beams. A beam with a slab on top acts as a T-beam, with the slab contributing to the compression flange. The effective flange width and the change to the compression block are not covered here.

Related

• Development length — how far a bar must extend past the point of maximum stress
• Splice length — Class A and Class B laps for #3–#11
• Rebar area — nominal area per bar and bar count from As required
• Concrete cover — ACI 318-19 cover requirements including beams
• Rebar size chart — diameters, areas and weights for #3–#18
• Rebar grades — yield and tensile strengths for A615 Grades 40–100
• Rebar spacing — turning a spacing into a bar count
• Deformed rebar — why beam bars are ribbed, not smooth
• Hook & bend calculator — standard hooks and bend deductions for stirrups
• Rebar tie — tie wire per ton and per intersection
• Footing rebar — reinforcement for continuous and spread footings