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LecturesModule 21

RC — Introduction & Material Properties

Concrete f'c, rebar fy, stress–strain behavior, and ACI 318 strength-design framework.

AISC Reference Box
  • ACI 318-19 Ch. 19 — Concrete: Design and Durability Requirements
  • ACI 318-19 Ch. 20 — Steel Reinforcement Properties and Durability
  • ACI 318-19 §21.2 — Strength Reduction Factors φ
  • ACI 318-19 §22.2 — Design Assumptions for Strength
  • ASTM A615 / A706 — Deformed Reinforcing Bars

Lecture Notes

Reinforced concrete (RC) combines plain concrete — strong in compression, weak in tension — with deformed steel reinforcing bars that carry tensile stresses. ACI 318-19 governs the design of structural concrete in the United States, and all formulas in this chapter cite ACI 318-19 section numbers.

Concrete compressive strength f'c is measured on 6×12 in. (or 4×8 in.) standard cylinders at 28 days per ASTM C39. Common building values are f'c = 3,000 to 6,000 psi; high-strength concretes reach 8,000–15,000 psi. The modulus of elasticity is Ec = 57,000·√f'c (psi) for normal-weight concrete (ACI §19.2.2.1). The modulus of rupture (cracking stress) is fr = 7.5·λ·√f'c (psi) (ACI §19.2.3.1).

Stress–strain behavior of concrete is non-linear: stress rises to a peak at strain ≈ 0.002, then descends to a useful ultimate strain εcu = 0.003 (ACI §22.2.2.1 — the assumed crushing strain used in flexural design). The descending branch is steeper for higher-strength concretes, which is why φ-factors and detailing rules tighten as f'c increases.

Steel reinforcement (ASTM A615 / A706) is specified by GRADE, which is the yield strength fy in ksi. Grade 60 (fy = 60,000 psi) is standard for beams, columns, and slabs; Grade 80 and Grade 100 are increasingly used for columns and walls (ACI §20.2.2.4). The steel modulus is Es = 29,000,000 psi (ACI §20.2.2.2). Bars are sized by NUMBER — a #N bar has a diameter of N/8 inch (so a #8 bar is 1.000 in.); areas come from ACI Appendix A.

ACI Strength Design (also called LRFD for concrete): φ·Sn ≥ U, where U is the factored load effect using ASCE 7 combinations (1.2D + 1.6L is typical), Sn is the nominal strength computed from material properties and assumed strain compatibility, and φ is the strength-reduction factor. φ depends on the limit state (ACI Table 21.2.2): flexure tension-controlled = 0.90; compression-controlled tied column = 0.65; spiral column = 0.75; shear and torsion = 0.75; bearing on concrete = 0.65.

Cover requirements (ACI §20.5.1) protect rebar from corrosion and fire: 3 in. clear when cast against earth, 1.5 in. for beams/columns exposed to weather, 0.75 in. for slabs not exposed to weather. Cover is measured to the OUTERMOST steel (typically the stirrup or tie). Bar spacing minimums (ACI §25.2): clear spacing ≥ db, 1 in., or 4/3 × max aggregate size — whichever is greatest.

Project case study — Cardinal Square RC — 4-story cast-in-place concrete office

Every chapter's worked example is one step in the design of the same building: Plan: 4 bays N–S × 3 bays E–W @ 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Floor: 7 in one-way slab on RC beams (b = 14 in, h = 24 in) framing to RC columns (16 in × 16 in tied) on grid. Roof: same beam grid + 6 in slab. Materials: Normal-weight concrete f'c = 4,000 psi (λ = 1.0). Reinforcement Grade 60 (fy = 60,000 psi); ties/stirrups Grade 60 #3 / #4. Clear cover: 1.5 in interior beams/slabs, 2 in columns, 3 in footings (ACI §20.5.1).

Chapter 21 — Materials & ACI 318 framework
Project-wide concrete and reinforcement
Demand carried forward
All factored demands derived below use ASCE 7 combos with material constants from this chapter: Ec = 57,000·√f'c = 57,000·√4,000 ≈ 3,605 ksi fr = 7.5·√4,000 ≈ 474 psi εy = fy/Es = 60/29,000 = 0.00207 εcu = 0.003.
This chapter contributes
Commits the project to f'c = 4,000 psi NWC and Grade 60 rebar, fixes φ-factors (0.90 tension-controlled, 0.65 tied column, 0.75 shear/spiral), and sets the Whitney β1 = 0.85 used in every later chapter.
Feeds into next chapter
These constants propagate to Chapters 22–26 (flexure, shear, ld, columns, slabs).
Strain ε (in/in)Stress σ (ksi)Fy = 50 ksiεy ≈ 0.0017Fu = 65 ksiElastic (E = 29,000 ksi)Yield plateauStrain hardeningNeckingRupture
Concrete reaches εcu = 0.003 at crush; rebar yields at εy = fy/Es (~0.00207 for Gr 60).

Formula Sheet

NameEquationAISC Ref
Modulus of elasticity (NWC)Ec = 57000·√f'c (psi)ACI §19.2.2.1
Modulus of rupturefr = 7.5·λ·√f'c (psi)ACI §19.2.3.1
Ultimate concrete strainεcu = 0.003ACI §22.2.2.1
Steel modulusEs = 29,000,000 psiACI §20.2.2.2
Design inequalityφ Sn ≥ UACI §5.3.1

Worked Example

Compute Ec, fr, and yield strain for an RC beam

Given
f'c = 4,000 psi normal-weight concrete (λ = 1.0); rebar Grade 60 (fy = 60,000 psi).
Load combination
Material check only — no load combo.
Required strength
Establish material constants used in flexural design.
Limit states
  • Modulus of elasticity Ec
  • Modulus of rupture fr
  • Yield strain εy
AISC reference
ACI 318-19 Chapters 19 and 20
Solution steps
  1. 1. Given
    f'c = 4000 psi. fy = 60000 psi. λ = 1.0 (normal-weight).
  2. 2. Modulus of elasticity
    Ec = 57000·√f'c. Ec = 57000·√4000. Ec = 57000·63.25. Ec = 3,604,997 psi. Ec ≈ 3,605 ksi.
  3. 3. Modulus of rupture
    fr = 7.5·λ·√f'c. fr = 7.5·(1.0)·√4000. fr = 7.5·63.25. fr = 474 psi.
  4. 4. Yield strain
    εy = fy / Es. εy = 60000 / 29,000,000. εy = 0.00207.
  5. 5. Compare with εcu
    εcu = 0.003 > εy = 0.00207. Therefore steel yields before concrete crushes — tension-controlled if εt ≥ 0.005.
Final design decision
Use Ec = 3,605 ksi and fr = 474 psi in deflection/cracking checks. Tension-controlled section requires εt ≥ 0.005 at nominal strength.
Common mistakes in this example
  • Using f'c in ksi inside the Ec = 57000√f'c formula (the constant 57000 is for psi).
  • Forgetting λ for lightweight concrete (λ = 0.75 to 0.85).
  • Using Es = 30,000 ksi instead of the ACI value 29,000 ksi.

FE-Style Worked Examples (5)

Each example mirrors the NCEES FE Civil Reference Handbook style: brief givens, a labeled figure, AISC section reference, step-by-step numeric solution, and a single boxed answer.

Given
Normal-weight concrete, f'c = 4,000 psi, wc ≈ 145 pcf.
AISC Reference
ACI 318-19 §19.2.2.1
Step-by-step solution
  1. 1. Problem Statement
    Compute Ec for normal-weight concrete with f'c = 4,000 psi.
  2. 2. Variables & Values
    f'c = 4,000 psi λ = 1.0 (NWC) use Ec = 57,000·√f'c (psi).
  3. 3. Approach
    Apply the standard NWC modulus expression — wc-dependent form is optional.
  4. 4. Formula Setup
    Ec = 57,000·√f'c (psi)
  5. 5. Substitution
    Ec = 57,000·√4,000 = 57,000·63.246 = 3,605,000 psi
  6. 6. Solution Steps
    Ec ≈ 3,605 ksi (3.6 × 10⁶ psi).
Answer Ec ≈ 3,605 ksi.
PuLcAxially loaded column

Practice Problems

  1. [E] State typical building f'c values (3,000–6,000 psi).
  2. [E] State Ec = 57,000√f'c (psi) for NWC.
  3. [E] State fr = 7.5λ√f'c (psi).
  4. [E] State εcu = 0.003 (ACI §22.2.2.1).
  5. [E] State Es = 29,000 ksi (ACI §20.2.2.2).
  6. [M] Compute Ec for f'c = 5,000 psi NWC.
  7. [M] Compute fr for f'c = 4,000 psi NWC.
  8. [M] Compute εy for Grade 60 and Grade 80; discuss tension-controlled implications.
  9. [M] List clear-cover for beams cast against earth vs exposed to weather vs interior (§20.5).
  10. [M] State min clear bar spacing for an interior beam with 1.5 in. cover.
  11. [H] Why is Ec = 57000√f'c empirical? Impact on cracked vs uncracked section analysis.
  12. [H] Lightweight concrete 110 pcf: recompute Ec using §19.2.2.1b.
  13. [H] Modular ratio n = Es/Ec for f'c = 4,000 psi; discuss transformed-section analysis.
  14. [H] Write the φ vs εt transition equation for the intermediate region (§21.2.2).
  15. [H] Research why ACI uses 0.85·f'c (not f'c) in the Whitney block — rectangular-block calibration.
Structured Clues
  • Use psi units when applying 57000·√f'c and 7.5λ√f'c.
  • ψg for Grade 60 = 1.0; for Grade 80 = 1.15; for Grade 100 = 1.30.
  • Cover is measured to the OUTERMOST steel (usually the stirrup).
Code References
  • ACI 318-19 Ch. 19, 20
  • ACI §21.2 (φ)
  • ASTM A615/A706

Quiz

1. ACI 318-19 takes the ultimate compressive strain of concrete equal to:
2. The modulus of elasticity of normal-weight concrete in psi is:
3. A #8 reinforcing bar has a nominal diameter of:
4. The strength-reduction factor φ for a tension-controlled flexural section is:
5. Compute Ec (ksi) for NWC with f'c = 4000 psi.
6. Compute the modulus of rupture fr (psi) for f'c = 5000 psi, λ = 1.0.
7. Yield strain εy of Grade 60 reinforcement is:
8. A #6 bar has a nominal diameter of:
9. Required clear cover for a beam cast against and permanently exposed to earth is:
10. φ for a compression-controlled tied column is:
11. Modular ratio n = Es/Ec for f'c = 4000 psi (Es = 29,000 ksi) is approximately:
12. For f'c = 6000 psi, β1 equals:
13. Minimum clear spacing between parallel #8 bars in a single layer is the LARGEST of db, 1 in., and:
14. If LL = 60 psf and DL (including self-weight) = 90 psf, factored load wu (psf) under 1.2D + 1.6L is:

Common Student Mistakes

  • Mixing ASD and LRFD load combinations in the same problem.
  • Using nominal strength Rn instead of design strength φRn.
  • Forgetting to check every limit state listed in the AISC chapter.

"Professor Explains" Script

Today we're talking about rc — introduction & material properties. Think of this topic as one step in the LRFD workflow: identify the demand, identify the limit states from the relevant AISC chapter, then check that φ·Rn is at least equal to Ru. We'll walk through the failure modes, the equations, and a worked example. Pay close attention to where the resistance factor changes — that's where students lose points on exams.