Design of a Singly Reinforced Concrete Beam (NSCP 2015 / ACI 318): Complete Worked Example

A singly reinforced concrete beam — relying on tension steel alone, with no compression reinforcement — is the default starting point for most rectangular beams. A doubly reinforced or larger section is only needed once the factored moment exceeds what a tension-controlled singly reinforced section can carry. This article walks through the complete flexural design procedure under NSCP 2015 (which mirrors ACI 318 strength design), then works a full numerical example down to the selected bars and stirrup spacing, with every arithmetic step shown.

Design Basis: NSCP 2015 and ACI 318

NSCP 2015 Chapter 4 (Structural Concrete) is based on ACI 318-14 and adopts the ACI 318 strength design method, so the load combinations, strength reduction factors, and design equations below apply under either reference. Design starts with the governing factored moment, usually the larger of several combinations, most commonly U = 1.2D + 1.6L for gravity beams. Nominal strengths are reduced by φ and compared to demand: φM_n ≥ M_u, φV_n ≥ V_u. Per the ACI 318-19 provisions on strength reduction factors, φ = 0.90 for tension-controlled flexure, φ = 0.75 for shear.

A section stays singly reinforced when its steel ratio ρ falls between a minimum, set to prevent brittle cracking failure, and a maximum tied to the tension-controlled strain limit, which guarantees ductile behavior before the concrete crushes.

Step-by-Step Flexural Design Procedure

1. Determine M_u at the critical section from analysis (e.g., 1.2D + 1.6L).
2. Assume tension-controlled: set φ = 0.90, to be verified later.
3. Resistance factor:
   R_n = M_u / (φ b d²)
4. Required steel ratio from the quadratic flexure equation:
   ρ = (0.85 f'_c / f_y) × [1 − √(1 − 2R_n / (0.85 f'_c))]
5. Minimum steel ratio (against sudden cracking failure); use this value if ρ from step 4 is lower:
   ρ_min = max [ 0.25√f'_c / f_y , 1.4 / f_y ]
6. Tension-controlled limit. ε_t ≥ 0.005 is required for φ = 0.90, which — with a 0.003 crushing strain and d_t = d — gives c/d ≤ 0.375 and:
   ρ_max = 0.85 β₁ (f'_c / f_y) × 0.375
   β₁ = 0.85 for f'_c ≤ 28 MPa, decreasing 0.05 per 7 MPa above that (not below 0.65). (The 0.005 limit is the NSCP 2015 / ACI 318-14 value; ACI 318-19 writes it as ε_t ≥ ε_ty + 0.003, about 0.0051 for Grade 415/420 steel — a negligible difference for this example.)
7. Required area: A_s = ρ b d; select bars meeting or slightly exceeding it.
8. Check bar spacing and cover against code minimums for the exposure condition.
9. Verify with the provided A_s: a = A_s f_y / (0.85 f'_c b), recheck tension-controlled, confirm φM_n ≥ M_u via M_n = A_s f_y (d − a/2).

Our Concrete Beam Design (ACI 318-19M) calculator automates this sequence, and the RC Beam Section Design tool is handy for cross-checking a trial section.

Worked Example

Design the flexural reinforcement for a rectangular beam with the following inputs, then check shear.

Effective Depth

Assuming a single layer of 25 mm main bars behind 10 mm stirrups and 40 mm clear cover:

Resistance Factor and Steel Ratio

With φ = 0.90 assumed:

Minimum and Maximum Steel Ratio Checks

ρ = 0.01024 > ρ_min = 0.00337 — OK. With β₁ = 0.85 (f'_c = 28 MPa):

ρ = 0.01024 < ρ_max = 0.01828 — tension-controlled, φ = 0.90 confirmed.

Required Steel Area and Bar Selection

A single 25 mm bar: π/4 × 25² = 490.9 mm². Bars needed = 1,344 / 490.9 = 2.74, round up to 3–25 mm bars: A_s,provided = 3 × 490.9 = 1,472.6 mm².

Bar Spacing and Cover Check

Clear width inside the stirrups: 300 − 2(40) − 2(10) = 200 mm. With 3–25 mm bars occupying 75 mm, the two gaps total 125 mm, or 62.5 mm each — well above the minimum clear spacing per NSCP 2015 / ACI 318-19 (the greatest of 25 mm, d_b = 25 mm, and 4/3 the maximum aggregate size). A single layer of 3–25 mm bars fits comfortably in the 300 mm web.

Final Capacity Verification

Depth to neutral axis c = a/β₁ = 85.6/0.85 = 100.7 mm, giving a net tensile strain ε_t = 0.003(d−c)/c = 0.003(437.5−100.7)/100.7 ≈ 0.0100, which is well above the 0.005 tension-controlled limit — φ = 0.90 stands confirmed.

Shear Design

For the same beam, take V_u = 150 kN at the critical section (distance d from the support face). With λ = 1.0, using the NSCP 2015 concrete shear strength expression (ACI 318-19 permits the same value when A_v ≥ A_v,min, which is confirmed below):

Since V_u (150 kN) > φV_c (88.6 kN), stirrups are required:

This is well below V_s,max ≈ 458 kN (section adequate without enlarging) and below (1/3)√f'_cb_wd ≈ 231.5 kN, so maximum stirrup spacing is governed by d/2 = 218.8 mm (not the tighter d/4 limit), or 600 mm, whichever is smaller.

For 10 mm diameter, 2-legged stirrups (f_yt = 275 MPa), A_v = 2 × (π/4 × 10²) = 157.1 mm²:

This exceeds the 218.8 mm ceiling, so the spacing limit governs. Use 10 mm stirrups at 200 mm on center near the support. Checking minimum shear reinforcement at s = 200 mm:

A_v,provided (157.1 mm²) exceeds A_v,min, so the spacing is adequate. Widen spacing toward midspan as V_u drops, re-checking each section against the actual shear diagram.

For a fuller check, including service-load deflection (not covered here), see the companion RC Beam Design calculator on rhces.com.

Assumptions & Limitations

• Normal-weight concrete assumed (λ = 1.0); lightweight concrete needs its own λ reduction.
• Main bars Grade 415 MPa; stirrups Grade 275 MPa — substitute your project's actual grades.
• Single layer of tension steel assumed (d_t = d); multiple layers change d and the strain check.
• M_u = 200 kN·m and V_u = 150 kN are assumed for illustration; use your own structural analysis in practice.
• Excludes deflection, torsion, development length/anchorage, and splice detailing — each needs separate verification.
• 40 mm cover is typical for interior beams; confirm the exposure-specific requirement for your project.
• Educational reference only — not a substitute for a licensed engineer's sealed calculations.

Frequently Asked Questions

1. When does a beam need compression reinforcement instead of staying singly reinforced?

When the required ρ would exceed ρ_max for a singly reinforced section — the concrete would crush before the tension steel yields enough, producing a brittle failure. The usual first fix is to increase depth or width, since that is more economical; compression steel is reserved for sections with fixed dimensions.

2. What happens if ρ_min and ρ_max checks are skipped?

Skipping ρ_min risks a beam whose cracking moment exceeds its reinforced capacity — it can fail suddenly the first time it cracks. Skipping ρ_max risks a compression-controlled or transition-zone failure: sudden and brittle, with little warning before collapse — the opposite of the ductile behavior strength-design provisions are built around.

3. Why does the tension-controlled strain limit (ε_t ≥ 0.005) matter for the φ factor?

ACI 318 and NSCP 2015 scale φ with ductility: clearly tension-controlled sections (steel yields well before the concrete crushes) get the full φ = 0.90, while sections nearer compression-controlled behavior get a lower φ, down to 0.65. Checking ε_t confirms which φ actually applies.

For more free structural and civil engineering calculators like the ones used in this example, browse all free web tools, and check the downloads page for spreadsheet templates you can adapt for routine beam design checks.