How to Calculate Rebar Size in Beam

Use As = Mu / (φ·fy·(d − a/2)), where φ = 0.90 for tension-controlled, fy = 60 ksi for Grade 60, d = effective depth (overall height minus cover minus stirrup minus half the bar diameter), and a = As·fy/(0.85·fc′·b) is the stress-block depth. Solve iteratively: guess a, compute As, recompute a, converge. For a 12×20 in beam (d ≈ 17.5 in), fc′=4000 psi, fy=60 ksi, Mu=120 ft·kip: As = 120×12 / (0.90×60×(17.5 − a/2)) ≈ 0.85 in² → 2 #6 (0.88 in²). Always round UP to the next whole bar combination.

The difference is which face is in tension. Simply supported: bottom tension at mid-span → heavy bottom bars, light top hangers. Continuous: bottom tension at mid-span AND top tension (negative moment) over supports → heavy top steel over supports (≈1.2× the bottom steel) plus heavy bottom at mid-span. Cantilever: TOP tension at the fixed end → heavy top steel at the fixed end, light bottom. The cantilever is the trap: crews place heavy steel at the bottom by simply-supported habit and the cantilever collapses on first load because the tension face has no steel.

Compare the factored shear Vu to the concrete shear capacity φVc (φVc = 0.75×2×√fc′×bw×d for normal-weight concrete). When Vu ≤ 0.5·φVc, no stirrups are required by analysis (but ACI mandates minimum stirrups in beams). When 0.5·φVc < Vu ≤ φVc, provide minimum #3 stirrups at d/2 spacing. When Vu > φVc, spacing is computed as s = As·fy·d / Vu, capped at d/2 (or d/4 when Vu > 2·φVc). For a 12×20 in beam, fc′=4000 psi, d=17.5 in: φVc ≈ 26.6 kip; at Vu=18 kip (0.68·φVc) use #3 @ d/2 = 8 in; at Vu=40 kip (>φVc) compute s = 0.22×60×17.5/40 = 5.8 in → #3 @ 5 in.

ACI sets minimum web width so that bars fit with 1.5 in cover, a #3 stirrup, and at least 1 in clear spacing between bars. For 2 bars: #4–#6 need 8 in, #8 needs 9 in, #11 needs 12 in. For 3 bars: #6 needs 11 in, #8 needs 12 in, #11 needs 15 in. For 4 bars: #6 needs 13 in, #8 needs 15 in, #11 needs 19 in. If the architect demands a narrower beam, you must either drop a bar size, bundle bars in pairs (which complicates splices), or step up to a two-layer arrangement that reduces effective depth d.

Top compression steel is required in three cases: (1) a continuous beam over a support, where negative moment puts the top in tension (it is then tension steel, not compression steel); (2) a doubly-reinforced beam where the moment capacity of a singly-reinforced section is insufficient and added compression steel raises the moment arm; (3) a cantilever, where the top is the primary tension face. Hanger bars (2 #4) on the compression face are always present to hold the stirrups up, but they are not structural compression steel unless they are sized by the moment calculation.

The inflection point (point of contraflexure) is where the moment crosses zero — the tension face flips from bottom (mid-span) to top (over the support). Bars are theoretically cut at the inflection point because the moment, and thus the steel stress, is zero there. The trap: under pattern loading (alternate spans loaded, one span live and the next dead), the inflection point shifts by up to 0.1L. Bars terminated at the theoretical point now fall short of the shifted zero-moment location, leaving the beam ~25% under-reinforced under the live load pattern. ACI requires bars to extend the development length ld (≈ 40 in for #6) past the inflection point precisely to absorb this shift.