Reinforced concrete basics

Why it matters

Reinforced concrete is the most used construction material on Earth: foundations, slabs, beams, columns, walls, bridges, tanks and tunnels. Concrete is strong in compression and weak in tension (roughly a tenth as strong), so steel bars are placed where tension develops. Sizing that steel for bending is the first calculation every structural engineer learns, and it shows how strength design reasons about failure.

The governing code

US building concrete is designed to ACI 318, Building Code Requirements for Structural Concrete (the current edition is ACI 318-25, published in 2025), from the American Concrete Institute (https://www.concrete.org). Your jurisdiction adopts a specific edition through its building code, often an earlier one, so confirm which applies. Bridges use AASHTO; other countries use their own codes (Eurocode 2, CSA A23.3). This lesson uses the strength-design ideas common to these codes, with symbols as in ACI 318. Checked on 2026-10-23.

How a beam carries moment

Under positive moment, the top of a beam is in compression and the bottom in tension. After the concrete cracks, the bottom steel carries all the tension T = As * fs, and the concrete above the neutral axis carries the compression C. The two forces form a couple separated by a lever arm.

At ultimate strength, the code assumes:

• concrete crushes at a strain of 0.003 at the extreme compression fibre;
• strain varies linearly through the depth (plane sections stay plane);
• the curved concrete stress distribution can be replaced by an equivalent rectangular stress block of intensity 0.85 f'c and depth a = beta1 * c, where c is the neutral-axis depth and beta1 is a code factor (0.85 for normal-strength concrete up to 4,000 psi, lower for stronger concrete);
• tension steel has yielded (fs = fy) in a properly designed beam.

Equilibrium (C = T) gives the block depth, and the couple gives the nominal moment:

a  = As * fy / (0.85 * f'c * b)
Mn = As * fy * (d - a/2)

b is the beam width and d the effective depth from the top to the centroid of the tension steel.

Strength reduction and ductility

The design strength is phi * Mn, and it must be at least Mu. ACI 318 sets phi = 0.90 for tension-controlled sections, where the net tensile strain in the steel is large enough (at least 0.005) that the beam would crack and deflect visibly before it fails. Sections with less steel strain get a lower phi, and the code limits how much reinforcement a beam may have so it does not fail by sudden concrete crushing. The strain check is:

c     = a / beta1
eps_t = 0.003 * (d - c) / c

The code also sets a minimum amount of flexural steel, so a lightly loaded beam does not fail abruptly when it first cracks. Look up the minimum, cover, spacing and bar-development rules in the adopted edition.

Worked example (illustrative)

A rectangular beam has b = 12 in, overall height 24 in, and d = 21.5 in, with f'c = 4,000 psi (4 ksi) and Grade 60 steel (fy = 60 ksi). The required moment is Mu = 250 kip-ft.

1. Required nominal moment. Mu / phi = 250 × 12 / 0.90 = 3,333 kip-in.
2. Solve for As. Substitute a into Mn: 3,333 = 60 As (21.5 - 60 As / (1.7 × 4 × 12)). This is a quadratic, 44.12 As^2 - 1,290 As + 3,333 = 0, whose smaller root is As = 2.86 in^2.
3. Choose bars. Three #9 bars give As = 3 × 1.00 = 3.00 in^2.
4. Check strength. a = 3.00 × 60 / (0.85 × 4 × 12) = 4.41 in; phi Mn = 0.90 × 3.00 × 60 × (21.5 - 2.21) / 12 = 260 kip-ft >= 250. OK.
5. Check strain. c = 4.41 / 0.85 = 5.19 in; eps_t = 0.003 × (21.5 - 5.19) / 5.19 = 0.0094 >= 0.005: tension-controlled, so phi = 0.90 was right.

A full design would continue: minimum steel, bar spacing and cover so the three bars fit with their stirrups, shear reinforcement, deflection, crack control, and development and splice lengths.

Common mistakes

• Using overall height instead of d. The lever arm is measured to the steel centroid.
• Mixing psi and ksi: f'c = 4,000 in one term and fy = 60 in another.
• Taking the larger root of the quadratic, which gives an absurd amount of steel.
• Assuming phi = 0.90 without checking the steel strain.
• Forgetting constructability: bars must fit, with cover and clear spacing for the aggregate to flow around them.

Professional responsibility

This lesson teaches the mechanics of flexural design only. Real concrete design includes shear, torsion, serviceability, durability, detailing and seismic requirements. Real designs need review and sealing by a licensed professional engineer (PE) under the codes your jurisdiction adopts.

Recap

• Stress block: a = As fy / (0.85 f'c b); Mn = As fy (d - a/2); design with phi Mn >= Mu.
• Solve a quadratic for the required steel, then round up to real bars.
• Check that eps_t >= 0.005 (tension-controlled) and follow the code's limits and detailing rules.

Try it

Open the Do tab to size reinforcement for a beam and check its strength and steel strain in Python.