Doubly Reinforced Concrete Beam — Worked Example with Non-Yielding Compression Steel
Architecture fixes a 12 × 20 in beam that must carry Mu = 260 kip-ft. As a singly reinforced section it falls just short of the ACI 318-19 tension-controlled limit, so adding tension steel only makes it worse. Two #7 bars in the top fix it, but they never reach yield, so the textbook "assume f′s = fy" shortcut overstates their contribution.
The beam
| Section | b = 12 in, h = 20 in (depth fixed by the ceiling), rectangular |
| Materials | f′c = 4,000 psi (β1 = 0.85), Grade 60, 1.5 in clear cover, #3 stirrups, ¾-in aggregate |
| Demand | Mu = 260 kip-ft (positive moment) |
RC beam tool: bw 12, h 20, aggregate 0.75, 4-#9 bottom, then add 2-#7 compression bars, Mu 260.
Open the RC beam calculator →What the section can carry while tension-controlled
At εt = 0.00507: c = 0.003 × 17.56/0.00807 = 6.53 in → a = 0.85c = 5.55 in
As,max = 0.85 × 4 × 12 × 5.55/60 = 3.77 in²
φMn,max = 0.9 × 3.77 × 60 × (17.56 − 5.55/2)/12 = 251 kip-ft < 260
| Singly reinforced | εt | φ | φMn (kip-ft) | ACI 318-19 §9.3.3.1 |
|---|---|---|---|---|
| 3-#9 (3.00 in²) | 0.0072 | 0.900 | 207.3 | OK, but too weak |
| 4-#9 (4.00 in²) | 0.0046 | 0.862 | 252.0 | Not tension-controlled — not permitted |
Try 4-#9 bottom and 2-#7 top
Clear spacing of 4-#9 in 12 in: (12 − 2 × 1.875 − 4 × 1.128)/3 = 1.25 in ≥ max(1 in, db = 1.128, 4/3 × 0.75 = 1.0) ✓
The textbook first trial — and why it fails
ε′s = 0.003(4.96 − 2.31)/4.96 = 0.00160 < εy = 0.00207 → assumption wrong
Solve for c with f′s = Esε′s
Solving: c = 5.33 in, a = 4.53 in
f′s = 87(5.33 − 2.31)/5.33 = 49.3 ksi (elastic) · Cc = 40.8 × 4.53 = 184.8 kip · Cs = 1.20(49.3 − 3.4) = 55.1 kip
εt = 0.003(17.56 − 5.33)/5.33 = 0.0069 ≥ 0.00507 → tension-controlled, φ = 0.90
Mn = 184.8(17.56 − 4.53/2) + 55.1(17.56 − 2.31) = 2,827 + 840 = 3,667 kip-in = 305.6 kip-ft
φMn = 275.1 kip-ft ≥ 260 ✓ (0.95)
The compression bars carry 55 kip instead of the 68 kip the yield assumption would give them. Here the yield assumption would overstate Mn by only 0.8% (307.9 vs 305.6 kip-ft), but it also puts c and εt in the wrong place, and in shallower beams or with larger d′ the error grows quickly.
More top steel, diminishing returns
| Top bars | c (in) | f′s (ksi) | εt | φMn (kip-ft) |
|---|---|---|---|---|
| none | 6.92 | — | 0.0046 | 252.0 (φ = 0.862) |
| 2-#6 | 5.67 | 52.5 | 0.0063 | 273.2 |
| 2-#7 | 5.33 | 49.3 | 0.0069 | 275.1 |
| 2-#8 | 5.00 | 45.6 | 0.0075 | 276.6 |
Even 2-#6 works. Compression steel mostly buys ductility (it moves the beam into the tension-controlled zone); the moment gain beyond that is small because the lever arm hardly changes. 2-#7 is chosen to give a little margin and stiffer stirrup corners.
Tools used in this example
RC beam design (ACI 318-19). Guide: How to design a reinforced concrete beam. Related: RC T-beam floor beam.
Frequently asked questions
When do you need compression reinforcement in a beam?
When the section is too shallow to develop Mu while staying tension-controlled. Under ACI 318-19 the tension steel must reach εt ≥ εty + 0.003 (0.00507 for Grade 60). Compression bars carry part of the compression force, which raises the neutral axis, increases εt and lets the section reach φ = 0.90 again.
How do you know if compression steel yields?
Compute its strain from the neutral axis: ε′s = 0.003(c − d′)/c. It yields if ε′s ≥ fy/Es = 0.00207. Shallow beams, small c and a large d′ all keep the compression bars elastic. In this example ε′s = 0.0017, so f′s = 49.3 ksi.
Why does adding tension steel not help here?
More tension steel deepens the compression block and lowers εt, which drops φ below 0.90. For this beam the largest tension-controlled singly reinforced capacity is 251 kip-ft, which is less than 260, no matter how the steel is chosen.
Do compression bars need ties?
Yes. ACI 318-19 §9.7.6.4 requires compression reinforcement in beams to be enclosed by ties or stirrups, sized and spaced like column ties (§25.7.2), so the bars cannot buckle outward.