22. RC — Flexural Design of Beams

01

Engineering story

A century of steel — from concept to skyline

Engineer reviewing blueprints against a steel-frame construction site at sunrise

For a tension-controlled beam, φMn = φ·As·fy·(d − a/2).

The Whitney block turns non-linear concrete into a rectangle 0.85·f'c wide and a = β1·c deep.

Chapter 22 covers singly reinforced, doubly reinforced, and T-beams per ACI 318-19 Ch. 9 & 22, using the Whitney stress block and εcu = 0.003.

Iconic steel structures built on engineering excellence

Whitney stress block
1937
Unified design ACI 318-71
1971
Strain-based φ (ACI 318-02)
2002
ACI 318-19
2019

Load pathService D+LMu (Combo 2)Whitney blockAs, a, c, εtφMn ≥ Mu

02

Learning objectives

What you will be able to do after finishing Chapter 1 — and why each objective matters in practice

Objective 01

Compute φMn for a singly reinforced beam

a = As·fy/(0.85·f'c·b); φMn = φ·As·fy·(d − a/2).

Why it matters: Governs every rectangular beam.
Where it is used: Ch 22 flexure.
Connects to: ACI 318-19 §22.2.

Objective 02

Check strain εt and φ

εt = 0.003·(dt − c)/c. Tension-controlled if εt ≥ 0.005.

Why it matters: Sets φ = 0.90 vs 0.65.
Where it is used: Every RC flex check.
Connects to: ACI 21.2.

Objective 03

Enforce As,min

As,min = max(3√f'c/fy·b·d, 200·b·d/fy) [psi units].

Why it matters: Prevents brittle first-crack failure.
Where it is used: Detailing.
Connects to: ACI 9.6.1.2.

Objective 04

Design T-beams (effective flange be)

be = min(L/4, bw + 16·hf, ½·clear span to next beam).

Why it matters: Real slab-beam systems act compositely.
Where it is used: Every floor-beam design.
Connects to: ACI 6.3.2.

Objective 05

Doubly reinforced beams

Add compression steel A's when b·d is constrained; ensure compression steel yields (c/d ratio check).

Why it matters: Recovers φ = 0.90 in tight sections.
Where it is used: Renovations, deep beams.
Connects to: ACI 22.2.

Objective 06

Detailing (spacing, cover, hooks)

Clear spacing ≥ max(db, 1 in, 4/3·dagg); hook development per Ch. 25.

Why it matters: Bond and concrete placement.
Where it is used: Working drawings.
Connects to: ACI 25.

03

Engineering motivation

What each part of a steel-frame building actually does — and why it exists

Before you design any single member, you have to see the whole system. A steel-frame building is not a collection of independent shapes bolted together — it is a deliberate load path, engineered so that every kilonewton of gravity, wind, or seismic demand has a continuous route from where it starts to the ground where the earth can resist it.

The photograph below shows a typical steel framing detail. Drag each labelled chip onto the structural element it names — the drop is only accepted when it lands inside the correct element's outlined region. A correct answer locks in with a green outline and a short explanation; a wrong answer flashes the region red, tells you what you actually hit, and returns the chip so you can try again. Press Reveal expected placements to see the reference solution (that attempt is then marked as assisted).

Structural steel framing — identify each element by dragging the labels — Fig. 1.3 · Drop a chip inside the outlined element it names. Green = correct and locked; red flash = wrong element, chip returns.

04

Failure mechanisms

Why we design the way we do — six ways steel structures have failed, and what each disaster taught the profession

Every provision in AISC 360 is a scar. Behind each equation, load factor, and detailing rule is a bridge, a walkway, or a tower whose failure cost lives and rewrote the profession. The six case studies below trace the mechanisms that motivate the code you are about to learn.

Read each one as an engineer, not a spectator: identify the load, the limit state, the missing check, and the specific clause that exists today because that check was missed. When you meet those clauses again in Chapters 5–17, they will read as answers, not rules.

Failure mechanism

Compression crushing (brittle)

Over-reinforced beam crushes concrete before rebar yields.

Root cause

εt < 0.005 (transition or CC).

Lesson learned

Keep εt ≥ 0.005 for φ = 0.90.

§ACI 21.2

Failure mechanism

Under-reinforced ductile flexure

Wide midspan cracks precede plastic-hinge formation — the safe mode.

Root cause

εt >> 0.005; steel yields long before concrete crushes.

Lesson learned

This is the intended mode — always target it.

§ACI 22

How failure propagates

The five-stage failure progression

1Elastic cracking

2Rebar yield at εs=εty

3Neutral axis rises

4εcu = 0.003

5φMn reached

Design codes intervene at the transition from yield to instability. Everything before yield is elastic and reversible; everything after instability is a race to collapse. LRFD keeps the demand well below the first transition.

05

Lecture notes

The full textbook chapter — figures, equations, and engineering narrative

Reflection · Think before you read

For a singly-reinforced beam, adding more tension steel increases capacity — until it doesn't. What physically limits how much steel is useful, and why does ACI cap ρ near the tension-controlled boundary?

How this chapter is organized. First we build intuition — the three stages an RC beam passes through as load grows (Ch. 2). Then we quantify each stage: the cracking moment (Stage 1→2), the ultimate strength model with the Whitney stress block (Stage 3), the ε_t-based classification and φ factor, and finally the ρ_min / ρ_max limits that keep every design ductile. Six worked examples take you all the way from analyzing a given section to sizing a brand-new beam.

1 · Three Stages of Beam Behavior (McCormac §22.1)

A reinforced-concrete beam does not behave like a homogeneous elastic beam. As the load grows from zero to failure, the cross section passes through three distinct stages. Every design equation you will use — cracking moment, service-load stresses, ultimate strength — belongs to one of these stages. Get the stages straight and the rest of the chapter is bookkeeping.

Figures 22.1 – 2.3 (McCormac): the three stages of loading of an RC beam.

Stage 1 — Uncracked Concrete

Tensile bending stress is less than the modulus of rupture f_r. The full gross section resists bending — compression on one face, tension on the other — and steel is almost inactive because its strain (and therefore stress f_s = n·f_c,tension) is small. Analyze it as a plain elastic beam using I_g:

Elastic flexure formula, gross section

(1)

f = M \cdot y / I_{g}

Stage 2 — Cracked, Elastic Stresses (Service Loads)

Once M exceeds the cracking moment M_cr, tension cracks propagate up toward the neutral axis. The concrete below the neutral axis is done — steel now carries all the tension. Because service stresses remain below about 0.5·f'_c, the compression stress block on top is still a triangle (straight-line) and the transformed-area method applies. This is the range that governs deflections and crack-width checks (Chapter 6 of McCormac).

Stage 3 — Ultimate Strength

Load is pushed further until steel yields (f_s = f_y) and the compressive fiber strain reaches ε_cu ≈ 0.003. The concrete compression stress is no longer linear — it follows the parabolic stress-strain curve of concrete. Whitney's equivalent rectangular block (0.85·f'_c over depth a = β₁c) replaces the parabola while preserving total force and centroid. This is the stage every strength equation in Chapters 3–5 of McCormac operates in.

2 · Moment–Curvature Diagram (Figure 2.4)

Moment–curvature for a tensile-reinforced RC beam. Three slopes = three stages. Curvature θ = ε / y at any fiber a distance y from the neutral axis.

Curvature

(2)

θ = ε / y

Reading the diagram. Between 0 and M_cr the curve is nearly vertical (I_g is large). At M_cr tension cracks open and the slope drops (I_cr < I_g) but stays nearly linear up to yielding. After the steel yields, the curve flattens almost horizontally — huge extra curvature (rotation, deflection) for very little extra moment — until concrete crushes and the beam fails. Ductility is that long, flat tail. Under-reinforced beams give lots of warning; over-reinforced beams do not.

3 · Cracking Moment M_cr (§2.2)

Because the reinforcement is a small fraction of the section (usually ≤ 2%) and the concrete is still uncracked, treat the beam as homogeneous. Bending stress at any fiber a distance y from the centroid of the gross section is

Gross-section bending stress

(3)

f = M \cdot y / I_{g}

The section cracks when the extreme-fiber tensile stress reaches the modulus of rupture f_r. ACI 318-19 §19.2.3 (equation 19.2.22.2) gives

Modulus of rupture — US customary

(4)

f_{r} = 7.5 \cdot λ \cdot f_{c}^{'} (p s i, w i t h f_{c}^{'} in p s i)

Modulus of rupture — SI

(5)

f_{r} = 0.7 \cdot λ \cdot f_{c}^{'} (M P a, w i t h f_{c}^{'} in M P a)

Setting the extreme-fiber stress = f_r and solving for M gives the cracking moment (ACI Eq. 9-9):

Cracking moment (ACI 24.2.3.5 / Eq. 9-9)

(6)

M_{cr} = f_{r} \cdot I_{g} / y_{t}

y_t is the distance from the centroid of the gross section to the extreme fiber in tension. λ = 1.0 for normal-weight concrete; λ < 1.0 for lightweight (see §1.12).

Example 22.1 — Uncracked stress & cracking moment

Concrete cross-section — reinforcement, cover, stirrups/ties and b/h/d dimensions.

Given. Rectangular beam, b = 12 in, h = 18 in, 3 #9 bars near the bottom (A_s = 3.00 in²), f'_c = 4,000 psi (normal-weight). Applied moment M = 25 ft·k.

(a) Extreme-fiber stress on the uncracked section.

(7)

I_{g} = (1/12) \cdot b \cdot h^{3} = (1/12) \cdot (12) \cdot (18)^{3} = 5, 832 i n^{4}

(8)

y_{t} = h /2 = 9.00 in

(9)

f = M \cdot y / I_{g} = (12 in / f t \cdot 25, 000 f t \cdot l b) (9.00) /5, 832 = 463 p s i

(b) Modulus of rupture.

(10)

f_{r} = 7.5 \cdot (1.0) \cdot 4, 000 = 474 p s i

Since 463 psi < 474 psi, the section has not cracked → Stage 1 analysis was valid.

(c) Cracking moment.

(11)

M_{cr} = f_{r} \cdot I_{g} / y_{t} = (474) (5, 832) /9.00 = 307, 152 in \cdot l b

M_cr ≈ 25.6 ft·k

Any applied moment above 25.6 ft·k puts the section into Stage 2 and the transformed-area (cracked) analysis must be used.

4 · Effective Depth and Cross-Section Drawing

Every worked example starts with a labelled sketch of the section: overall width b, total height h, effective depth d = h − cover − d_tie − d_b/2, and every bar with its area A_s. The effective depth d — not h — is what appears in every strength equation.

Standard singly-reinforced section — mark b, h, d and A_s before writing any equation.

5 · Strength (Ultimate) Model — Whitney Stress Block (§22.5)

At ultimate the code fixes the extreme-fiber concrete strain and lets strains vary linearly with depth (plane sections remain plane). Whitney's equivalent rectangular block reproduces the total compression force and its centroid, replacing the real parabolic curve.

Ultimate concrete strain (ACI 22.2.22.1)

(12)

ε_{c u} = 0.003

Horizontal equilibrium: C = T ⟹ 0.85·f'c·b·a = As·fy

(13)

a = A_{s} \cdot f_{y} / (0.85 \cdot f_{c}^{'} \cdot b) = β_{1} \cdot c

Neutral-axis depth

(14)

c = a / β_{1}

Whitney factor β1 (ACI 22.2.2.22.6)

(15)

β_{1} = 0.85 f or f_{c}^{'} \leq 4, 000 p s i

d r o p s 0.05 p er + 1, 000 p s i

\geq 0.65

Nominal moment — take moments about the tension steel

(16)

M_{n} = A_{s} \cdot f_{y} \cdot (d - a /2)

6 · Strains in Flexural Members (§3.5)

Because strains vary linearly across the depth and ε_c = 0.003 at the top fiber, similar triangles give the net tensile strain ε_t at the extreme layer of tension steel:

Net tensile strain — by similar triangles

(17)

ε_{t} = (d - c) / c \cdot 0.003

Yield strain for Grade 60 is ε_y = 60,000 / 29,000,000 ≈ 0.00207 (≈ 0.002). Once ε_t is known, φ comes straight from ACI Table 21.2.2 — see §7 below.

Example 22.2 — Compute a, c and ε_t

Concrete cross-section — reinforcement, cover, stirrups/ties and b/h/d dimensions.

Given. b = 14 in, d = 21 in, A_s = 3.00 in², f_y = 60 ksi, f'_c = 3 ksi.

(18)

a = A_{s} \cdot f_{y} / (0.85 \cdot f_{c}^{'} \cdot b) = (3.00) (60) / [(0.85) (3) (14)] = 5.04 in

(19)

β_{1} = 0.85 (s in ce f_{c}^{'} = 3, 000 p s i \leq 4, 000 p s i)

(20)

c = a / β_{1} = 5.04/0.85 = 5.93 in

(21)

ε_{t} = (d - c) / c \cdot 0.003 = (21 - 5.93) /5.93 \cdot 0.003 = 0.00762

ε_t = 0.00762 ≫ 0.005 → tension-controlled, ductile, φ = 0.90

7 · Strength Reduction Factor φ vs ε_t (§3.7, Figure 3.5)

φ is not a single number — it varies with how ductile the section is at ultimate. ACI 318-19 §21.2.2 defines three regions bounded by ε_ty = f_y/E_s and ε_ty + 0.003. For Grade 60 those bounds simplify to ε_t = 0.002 and ε_t = 0.005.

Figure 3.5 — φ varies linearly across the transition zone for Grade 60 reinforcement.

Section classification (from ε_t at nominal strength)	Strain range	φ (tied / spiral)
Compression-controlled	ε_t ≤ ε_ty ≈ 0.002 (Gr 60)	0.65 / 0.75
Transition	ε_ty < ε_t < 0.005	linear interpolation, e.g. φ = 0.65 + (ε_t−0.002)·(250/3)
Tension-controlled (ductile — preferred)	ε_t ≥ 0.005	0.90 / 0.90

Design implication. Sections designed with ρ near ρ_min land far into tension-controlled territory (φ = 0.90) automatically. Sections with ρ close to ρ_max fall into the transition zone — you get a smaller φ, so the section becomes uneconomical. McCormac's advice: increase depth and reduce ρ until ε_t ≥ 0.005 — you get φ = 0.90 back, less steel, less crowding, and better bond.

8 · Minimum & Maximum Steel — Preventing Brittle Failure (§3.8)

If a beam is too lightly reinforced, its ultimate moment is less than its cracking moment: the first crack pops and the beam breaks in half with no warning. If it is too heavily reinforced, concrete crushes before steel yields — also no warning. ACI puts a floor and a ceiling on ρ.

Minimum steel — US customary (ACI Eq. 9.6.1.2 / old Eq. 10-3)

(22)

A_{s, min} = (3 \cdot f_{c}^{'} / f_{y}) \cdot b_{w} \cdot d \geq (200 \cdot b_{w} \cdot d) / f_{y}

Minimum steel — SI

(23)

A_{s, min} = (f_{c}^{'} / (4 \cdot f_{y})) \cdot b_{w} \cdot d \geq (1.4 \cdot b_{w} \cdot d) / f_{y}

As a percentage

(24)

ρ_{min} = 3 \cdot f_{c}^{'} / f_{y} \geq 200/ f_{y}

The 200/f_y floor governs up to about f'_c = 4,440 psi; the √f'_c form governs above. Values of ρ_min are tabulated (Appendix A, Table A.7) — read the row for your f_y and f'_c.

Maximum steel — set by εt ≥ 0.004 for gravity beams (ACI 9.3.22.2)

(25)

ρ_{ma x} \approx 0.0181 f or f_{c}^{'} = 4 k s i, f_{y} = 60 k s i (T ab l e A .7)

9 · Master Recipe — Compute φM_n of a Given Section

Step 1 — Draw the section. Mark b, h, d, and A_s.

Step 2 — Check ρ. ρ = A_s/(b·d). Confirm ρ_min ≤ ρ ≤ ρ_max using Table A.7.

Step 3 — Compute a. a = A_s·f_y/(0.85·f'_c·b).

Step 4 — Compute c. β₁ from ACI Table 22.2.2.22.6, then c = a/β₁.

Step 5 — Compute ε_t. ε_t = (d − c)/c · 0.003. Classify:

ε_t ≥ 0.005 → tension-controlled, φ = 0.90 ✓
0.002 < ε_t < 0.005 → transition, interpolate
ε_t ≤ 0.002 → compression-controlled, section is not permitted as a flexural member (ACI 9.3.22.2)

Step 6 — M_n. M_n = A_s·f_y·(d − a/2).

Step 7 — φM_n. Multiply Step 6 by φ from Step 5. This is the value you compare with M_u: φM_n ≥ M_u.

M_n vs φM_n — do not confuse these.

M_n (nominal) = the strength the section can theoretically carry using specified material properties (f'_c, f_y). Computed from strain compatibility. It is not the design capacity.
φM_n (design) = the strength the code allows you to count on. Always < M_n. This is what you compare with the factored demand M_u.
Design check: φM_n ≥ M_u. If you write M_n ≥ M_u, you have skipped safety.

Same rule applies to shear (φV_n ≥ V_u), axial (φP_n ≥ P_u), and torsion — always insert φ before comparing.

10 · Worked Examples — Analysis of Given Sections

Example 22.3 — Tension-controlled section, φ = 0.90

Concrete cross-section — reinforcement, cover, stirrups/ties and b/h/d dimensions.

Given. b = 15 in, d = 24 in (total h = 27 in, 3-in cover), 4 #9 bars (A_s = 4.00 in²), f'_c = 4,000 psi, f_y = 60,000 psi.

(26)

ρ = A_{s} / (b \cdot d) = 4.00/ [(15) (24)] = 0.0111

(27)

ρ_{min} = 0.0033, ρ_{ma x} = 0.0181 (A pp e n d i x A, T ab l e A .7)

ρ_{min} < ρ < ρ_{ma x} ✓

(28)

a = A_{s} \cdot f_{y} / (0.85 \cdot f_{c}^{'} \cdot b) = (4.00) (60, 000) / [(0.85) (4, 000) (15)] = 4.71 in

(29)

β_{1} = 0.85 (f_{c}^{'} = 4, 000 p s i)

(30)

c = a / β_{1} = 4.71/0.85 = 5.54 in

(31)

ε_{t} = (d - c) / c \cdot 0.003 = (18.46/5.54) \cdot 0.003 = 0.0100 > 0.005

t e n s i o n - co n t r o l l e d, ϕ = 0.90

(32)

M_{n} = A_{s} \cdot f_{y} \cdot (d - a /2) = (4.00) (60) (24 - 4.71/2) = 5, 194.8 in \cdot k = 432.9 f t \cdot k

(33)

ϕ M_{n} = (0.90) (432.9) = 389.6 f t \cdot k

φM_n = 389.6 ft·k — this is the ACI design moment capacity.

Example 22.4 — Section that fails the ductility limit

Concrete cross-section — reinforcement, cover, stirrups/ties and b/h/d dimensions.

Given. b = 12 in, d = 15 in, 3 #11 bars (A_s = 4.68 in²), f'_c = 4,000 psi, f_y = 60,000 psi.

(34)

ρ = 4.68/ [(12) (15)] = 0.026 > ρ_{ma x} = 0.0181

(35)

a = (4.68) (60, 000) / [(0.85) (4, 000) (12)] = 6.88 in

(36)

c = 6.88/0.85 = 8.09 in

(37)

ε_{t} = (15 - 8.09) /8.09 \cdot 0.003 = 0.00256 < 0.004

Section is not ductile — ε_t falls below the ACI 9.3.22.2 limit of 0.004 required for flexural members. This section may not be used as designed. Increase d (deepen the beam) or reduce A_s.

Example 22.5 — Transition zone (0.004 < ε_t < 0.005)

Concrete cross-section — reinforcement, cover, stirrups/ties and b/h/d dimensions.

Given. b = 10 in, d = 15 in, 3 #9 bars (A_s = 3.00 in²), f'_c = 4,000 psi, f_y = 60,000 psi.

Steel ratio (formula)

(38)

ρ = A_{s} / (b \cdot d)

(39)

ρ = 3.00/ [(10) (15)] = 0.020 (ρ_{min} < ρ

ρ > ρ_{ma x, t c} = 0.0181 f or ε_{t} = 0.005)

Whitney block depth (formula)

(40)

a = A_{s} \cdot f_{y} / (0.85 \cdot f_{c}^{'} \cdot b)

(41)

a = (3.00) (60, 000) / [(0.85) (4, 000) (10)] = 180, 000/34, 000 = 5.29 in

Neutral-axis depth (formula)

(42)

c = a / β_{1} (β_{1} = 0.85 f or f_{c}^{'} \leq 4 k s i)

(43)

c = 5.29/0.85 = 6.22 in

Tensile strain (formula, plane sections)

(44)

ε_{t} = 0.003 \cdot (d - c) / c

(45)

ε_{t} = 0.003 \cdot (15 - 6.22) /6.22 = 0.00423

0.004 < ε_{t} < 0.005

t r an s i t i o n z o n e

Plane-section strain (0.003 crush, 0.00423 at steel) and Whitney block (0.85 f'_c over a = 5.29").

φ in transition zone (formula, ACI 21.2.2)

(46)

ϕ = 0.65 + (ε_{t} - 0.002) \cdot (250/3)

(47)

ϕ = 0.65 + (0.00423 - 0.002) \cdot (250/3) = 0.65 + 0.186 = 0.836

Nominal moment (formula)

(48)

M_{n} = A_{s} \cdot f_{y} \cdot (d - a /2)

(49)

M_{n} = (3.00) (60) \cdot (15 - 5.29/2) = 180 \cdot (12.355) = 2, 223.9 in - k = 185.3 f t - k

Design moment (formula)

(50)

ϕ M_{n} = ϕ \cdot M_{n}

(51)

ϕ M_{n} = (0.836) (185.3) = 154.9 f t - k

φM_n = 154.9 ft·k. Note φ < 0.90 — this section is uneconomical; deepening the beam would recover full φ.

11 · Worked Examples — Sizing a New Beam (McCormac Ch. 4)

Example 22.6 — Choose b, d and A_s for a target M_u

Concrete cross-section — reinforcement, cover, stirrups/ties and b/h/d dimensions.

Given. M_u = 600 ft·k, target ρ ≈ 0.0120, f_y = 60 ksi, f'_c = 4 ksi. Assume φ = 0.90.

Design equation solved for b·d²

(52)

M_{u} / (ϕ \cdot b \cdot d^{2}) = ρ \cdot f_{y} \cdot (1 - ρ \cdot f_{y} / (1.7 \cdot f_{c}^{'}))

(53)

(12 in / f t \cdot 600, 000 f t \cdot l b) / (0.90 \cdot b \cdot d^{2}) = (0.0120) (60, 000) \cdot [1 - (0.0120) (60, 000) / (1.7 \cdot 4, 000)]

(54)

b \cdot d^{2} = 12, 427 i n^{3}

t r y 14 in \cdot 33 in (d \approx 30.00 in)

(55)

A_{s} = ρ \cdot b \cdot d = (0.0120) (14) (30) = 5.04 i n^{2}

use 4 #10 (A_{s} = 5.06 in^{2})

Check. ρ_actual = 5.06/[(14)(30)] = 0.01205 → ρ_min < ρ < ρ_max ✓. Recompute φM_n (via Table A.13 or the master recipe) → φM_n = 610.3 ft·k > 600 ft·k ✓.

Final section: 14 in × 33 in with 4 #10 bars, d = 30 in, 2½-in cover.

Example 22.7 — Include beam self-weight iteratively

Concrete cross-section — reinforcement, cover, stirrups/ties and b/h/d dimensions.

Given. Simple span L = 25 ft, D = 2 k/ft (excluding self-weight), L = 3 k/ft, f_y = 60 ksi, f'_c = 3 ksi, target ρ ≈ 0.18·f'_c/f_y = 0.009.

Iteration 1 — assume beam weight 400 lb/ft

(56)

w_{u} = 1.2 \cdot (2 + 0.4) + 1.6 \cdot (3) = 7.68 k / f t

(57)

M_{u} = w_{u} \cdot L^{2} /8 = (7.68) (25)^{2} /8 = 600 f t \cdot k

(58)

b \cdot d^{2} = M_{u} / (ϕ \cdot R_{n}) w i t h R_{n} f r o m T ab l e A .12

b \cdot d^{2} = 16, 577 i n^{3}

t r y 18 \cdot 33 in (d = 30.5 in)

(59)

B e am w t = (18) (33) /144 \cdot 150 = 619 l b / f t — g r e a t er t han 400 l b / f t a ss u m e d

i t er a t e .

Iteration 2 — assume 650 lb/ft

(60)

w_{u} = 1.2 \cdot (2 + 0.650) + 1.6 \cdot (3) = 7.98 k / f t

M_{u} = 623.4 f t \cdot k

b \cdot d^{2} = 17, 223 i n^{3}

t r y 18 \cdot 34 in (d = 31.0 in)

(61)

B e am w t = (18) (34) /144 \cdot 150 = 637.5 l b / f t < 650 l b / f t ✓

(62)

A_{s} = ρ \cdot b \cdot d = (0.009) (18) (31) = 5.02 i n^{2}

try 4 #9 (A_{s} \approx 4.00 in^{2}) — slightly less, so verify capacity.

(63)

a = (5.00 \cdot 60) / (0.85 \cdot 3 \cdot 18) = 6.54 in

ϕ M_{n} = 0.9 \cdot 5.0 \cdot 60 \cdot (31 - 6.54/2) = 7, 487.6 in \cdot l b

623.9 f t \cdot k > M_{u} ✓

Final section: 18 in × 34 in with the chosen bar layout — self-weight is now consistent with the loading used.

Why the iteration matters. Beam self-weight is a dead load that depends on the beam size, which you don't know until you've solved the design equation. Assume a size, size the beam, recompute self-weight, and repeat once or twice until the assumed and computed weights match to within a few percent.

12 · Structural Safety Framework (§22.4) & Advantages of Strength Design (§22.3)

Side of the inequality	Factor	Uncertainty it covers
Demand (loads)	Load factors γ > 1 (ASCE 7-22 §2.3)	Variability in D, L, W, E, S values
Capacity (strength)	Strength-reduction factor φ < 1 (ACI 21.2)	Variability in f'_c, f_y, dimensions, workmanship

Master strength design inequality

(64)

ϕ \cdot (n o mina l s t r e n g t h) \geq Σ γ_{i} \cdot (n o mina l l o a d)_{i}

Written for flexure: φM_n ≥ M_u. For shear: φV_n ≥ V_u. For axial: φP_n ≥ P_u. Never compare a nominal strength to a factored demand — always insert φ first.

Why strength design beat working-stress design. (1) It captures the nonlinear concrete stress-strain curve → better strength prediction. (2) It uses one consistent theory for beams, slabs, and columns. (3) Load-specific γ factors recognise that dead load is more predictable than live/wind/seismic — a more rational safety margin. (4) It fully utilizes high-strength steels. (5) It allows a much wider range of steel percentages, giving the designer more flexibility.

⚠ Common Mistakes in RC Flexure

Comparing M_n to M_u instead of φM_n to M_u.
Using ε_t from cracked-section elastic analysis instead of ε_t at ultimate (ε_c = 0.003).
Assuming φ = 0.90 without checking ε_t ≥ 0.005 — a heavily reinforced section may be in the transition zone.
Using β₁ = 0.85 for high-strength concrete (f'_c > 4,000 psi). β₁ decreases with strength.
Forgetting ρ_min = max(3√f'_c/f_y, 200/f_y) → under-reinforced brittle failure.
Using d = h (ignoring cover and bar radius). Effective depth is always less than total depth.

13 · Using ACI 318-19 Tables & Design Aids

Table / Chart	Gives you	How to read
ACI Table 21.2.2	φ vs. ε_t	Enter with computed ε_t → read φ.
ACI Table 22.2.2.22.6	β₁ vs. f'_c	Enter with concrete strength → read β₁ (0.85 for ≤ 4 ksi, then decreasing).
ACI SP-17 Flexure Charts	Trial section ρ for given M_u	Enter with M_u/(φ·b·d²) → read ρ → compute A_s = ρ·b·d.
Reinforcing bar tables (Appendix)	Bar area, weight, spacing per foot width	Convert required A_s into bar-count × bar-area combination that fits within b − 2·cover.

06

Professional practice, safety & ethics

RC flexure practice

Professional practice

Show bar size, count, cover, and cutoff points on the drawings; the Whitney block assumes the steel is where you drew it.
Coordinate top-bar congestion with the beam-column joint; bars that cannot be placed are not reinforcement.
Specify cover by exposure condition — cover is a durability and fire requirement, not a detailing preference.

Safety in design & construction

Design tension-controlled (εt ≥ 0.005) so the section warns before it fails; over-reinforced sections crush without warning.
Bars placed in the wrong face (a classic field error) can halve capacity — require inspection before pour.
Never allow bar cutoffs to be shortened in the field.

Engineering ethics

Do not raise φ or lower cover to make a section work.
If a pour was made with misplaced steel, investigate and document — do not rely on 'reserve capacity'.
Report any deviation between placed and detailed reinforcement.

Inspector verifying reinforcing bar size, spacing, and cover before a concrete pour — Field inspection: rebar size, spacing, and cover verified before placement.

Engineers reviewing sealed structural drawings across a conference table — Design review: documenting assumptions before the drawings are sealed.

ABET / licensure link. These points map to ABET Student Outcomes 2 and 4 — engineering design within realistic constraints, and recognition of ethical and professional responsibilities. Expect NCEES FE and PE exam questions on the NSPE Code of Ethics, OSHA construction requirements, and the engineer's standard of care.

07

Cost analysis

Flexural design economics

Approach

Deeper beams need less steel: doubling d roughly halves As for the same Mu, and steel costs far more per pound than concrete.
Doubly reinforced sections are a last resort — compression steel is expensive and congests the section.
Round bar selections to common sizes (#5–#9) and repeatable layouts.

Worked cost example — Singly vs doubly reinforced for Mu = 420 k-ft

Basis: b = 14″

Line item	Qty	Rate	Cost
Option A — h = 30″, As = 3.16 in² (4-#8)	0.107 ton rebar	$2,200	$235
Option A concrete + forms	1.1 yd³-equivalent	$320	$352
Option B — h = 22″, As = 4.74 + A′s = 1.58 in²	0.214 ton rebar	$2,200	$471
Option B congestion/placement premium	1 ls	$250	$250
Estimated total			$1,308

Takeaway. The deeper singly reinforced beam costs ~$590 less — use depth, not compression steel, whenever headroom allows.

Unit rates are representative US averages for teaching purposes. On a real project, price with current local rates (RSMeans, fabricator quotes, or contractor pricing) and state the estimate date.

08

Animated concepts

Key mechanics visualised — watch the strain profile, stress block, or buckled shape evolve

Strain diagram — εcu = 0.003 and εt vs c

Linear strain profile pivots about the neutral axis. As c shrinks, εt at the tension steel grows past 0.005 — the section becomes tension-controlled and φ = 0.90.

Whitney stress block — 0.85 f'c × a

The parabolic concrete stress is replaced by a rectangle of intensity 0.85·f'c and depth a = β1·c. Compression resultant C = 0.85·f'c·b·a balances tension T = As·fy.

09

Engineering figures

Full-page reference diagrams — the visual vocabulary you will use for the rest of the course

§22.6.1

Flexural test

Fig. 22.1RC beam under 4-pt load
Flexural cracks at midspan.

§22.6.2

T-beam formwork

Fig. 22.2Rebar cage before pour
Slab casts monolithically with the web.

§22.6.3

Ductile flexure

Fig. 22.3Wide midspan cracks
Under-reinforced → warning cracks.

§22.6.4

Compression crush

Fig. 22.4Top crushed
Over-reinforced → brittle failure.

10

Worked examples

Full textbook solutions — problem, theory, step-by-step, verification, interpretation

Example 22.1

φMn for singly reinforced beam (b=12, d=20, As=3.0 in²)

Compute a, c, εt, φ, and φMn line-by-line.

Problem statement

A rectangular beam has b = 12 in, d = 20 in, As = 3.0 in² (three #9), f'c = 4,000 psi, fy = 60 ksi. Compute φMn.

RC beam under load — FIG. 22.1 — flexural cracks form at Mu ≈ φMn.

DIMWhitney block: 0.85·f'c over depth a = β1·c.

Given

b=12 in
d=20 in
As=3.0 in²
f'c=4,000 psi
fy=60 ksi

Find

φMn

Assumptions

Rectangular section
Grade 60 rebar

Code references

ACI 318-19 §22.2
ACI 318-19 §21.2

Theory & approach

Force balance C = T gives a directly. Then classify εt and pick φ.

Step-by-step solution

1
Whitney depth a
FormulaACI 22.2
a = A_s·f_y / (0.85·f'_c·b)
a = 3.0·60 / (0.85·4·12) = 180 / 40.8 = 4.41 in
2
Neutral axis c
Formula
c = a / β_1, β₁ = 0.85
c = 4.41 / 0.85 = 5.19 in
3
Strain εt
FormulaACI 21.2.2
ε_t = 0.003·(d − c)/c
ε_t = 0.003·(20 − 5.19)/5.19 = 0.003·(14.81/5.19) = 0.00857
0.00857 ≥ 0.005 → tension-controlled, φ = 0.90
4
Nominal Mn
Formula
M_n = A_s·f_y·(d − a/2)
M_n = 3.0·60·(20 − 2.205) = 180·17.795 = 3,203 k-in = 267 k-ft
5
Design φMn
φM_n = 0.90·267 = 240 k-ft
6
Check As,min
Formula
A_s,min = max(3·√f'_c/f_y·b·d, 200·b·d/f_y)
= max(3·√4000/60,000·12·20, 200·12·20/60,000)
= max(0.76, 0.80) = 0.80 in² ≪ As = 3.0 in² ✓

Verification

Under-reinforced (εt ≫ 0.005): ductile mode as intended.

Final answer

φMn = 240 k-ft (tension-controlled, φ = 0.90).

Common mistakes

Confusing a with c (a = β1·c).
Using φ = 0.90 without checking εt.
Forgetting As,min.

References

· ACI 318-19

11

Guided practice

Compute the governing variables — hints unlock as you need them

A singly-reinforced beam has b = 10 in, d = 18 in, A_s = 2.0 in², f'c = 4,000 psi, f_y = 60 ksi. Compute a, c, ε_t, classify the section (tension-controlled / transition / compression-controlled), and report φM_n.

Your turn

a =inc =ineps_t =phi*M_n =kip*ft

Hints

1.Whitney block: a = A_s·f_y / (0.85·f'c·b) = 2.0·60 / (0.85·4·10) = 3.53 in.

12

Independent practice

Solve the chapter's design task — compute each governing variable

Design task

A singly-reinforced rectangular beam has b = 12 in, d = 20 in, and As = 3.00 in² (three #9 bars). Compute the Whitney stress-block depth a, the net tensile strain εt, and the design flexural strength φMn. Verify the section is tension-controlled.

Given

f'c = 4000 psi (NWC), β1 = 0.85
fy = 60,000 psi
b = 12 in, d = 20 in, As = 3.00 in²

Approach

Force equilibrium: a = As·fy / (0.85·f'c·b).
c = a/β1; εt = 0.003·(d − c)/c. Tension-controlled if εt ≥ εty + 0.003.
Mn = As·fy·(d − a/2). φ = 0.90 when tension-controlled.

Submit your answer

a =inepsilon_t =phi*M_n =k-ft

13

Mini design challenge

Select the option that satisfies every code and serviceability requirement in the brief

Brief

Design a singly-reinforced rectangular beam for M_u = 200 k·ft with b = 12 in, d = 20 in, f'c = 4,000 psi, f_y = 60 ksi. Solve for A_s, verify the section is tension-controlled, check A_s,min and A_s,max, and pick a practical bar arrangement that fits in one layer.

Requirements

Solve φM_n = M_u for A_s (iterate on a = A_s·f_y/(0.85·f'c·b))
Confirm ε_t ≥ 0.005 → φ = 0.90 (ACI §21.2)
A_s ≥ A_s,min = max[3√f'c/f_y, 200/f_y]·b·d ≈ 0.79 in² (ACI §9.6.1.2)
Verify bar clear spacing (ACI §25.2) fits in b = 12 in
Report ρ = A_s/(b·d); compare to ρ_max at ε_t = 0.005

Section

Wt (lb/ft)

Δ (in)

Ru/Rn

Cost

Pick

14

Chapter summary

A mind map of how every concept connects

Graded Chapter Quiz(10 FE-style questions · ACI 318-19 required)

These questions reference ACI 318-19 — sections, equations, and tables are cited explicitly. Use a calculator. Each question offers a clue you may reveal before answering. Submissions are recorded to your account once signed in.

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16

FE exam preparation

NCEES-style practice with timer, equation sheet, and mastery tracking

Exam mode

30:00 Calculator

Question 1 / 13

b=12, d=20, As=3.0 in², f'c=4000, fy=60 ksi. a is closest to:

◆◆ MediumACI 318-19 §22.2.2.4.1

A. 2.0
B. 3.0
C. 4.4
D. 6.0

RC — Flexural Design of Beams

Compute φMn for a singly reinforced beam

Check strain εt and φ

Enforce As,min

Design T-beams (effective flange be)

Doubly reinforced beams

Detailing (spacing, cover, hooks)

Compression crushing (brittle)

Under-reinforced ductile flexure

1 · Three Stages of Beam Behavior (McCormac §22.1)

Stage 1 — Uncracked Concrete

Stage 2 — Cracked, Elastic Stresses (Service Loads)

Stage 3 — Ultimate Strength

2 · Moment–Curvature Diagram (Figure 2.4)

3 · Cracking Moment Mcr (§2.2)

Example 22.1 — Uncracked stress & cracking moment

4 · Effective Depth and Cross-Section Drawing

5 · Strength (Ultimate) Model — Whitney Stress Block (§22.5)

6 · Strains in Flexural Members (§3.5)

Example 22.2 — Compute a, c and εt

7 · Strength Reduction Factor φ vs εt (§3.7, Figure 3.5)

8 · Minimum & Maximum Steel — Preventing Brittle Failure (§3.8)

9 · Master Recipe — Compute φMn of a Given Section

10 · Worked Examples — Analysis of Given Sections

Example 22.3 — Tension-controlled section, φ = 0.90

Example 22.4 — Section that fails the ductility limit

Example 22.5 — Transition zone (0.004 < εt < 0.005)

11 · Worked Examples — Sizing a New Beam (McCormac Ch. 4)

Example 22.6 — Choose b, d and As for a target Mu

Example 22.7 — Include beam self-weight iteratively

12 · Structural Safety Framework (§22.4) & Advantages of Strength Design (§22.3)

⚠ Common Mistakes in RC Flexure

13 · Using ACI 318-19 Tables & Design Aids

Flexural test

T-beam formwork

Ductile flexure

Compression crush

φMn for singly reinforced beam (b=12, d=20, As=3.0 in²)

Whitney depth a

Neutral axis c

Strain εt

Nominal Mn

Design φMn

Check As,min

Graded Chapter Quiz(10 FE-style questions · ACI 318-19 required)

Upload your worked solution (PDF)

3 · Cracking Moment M_cr (§2.2)

Example 22.2 — Compute a, c and ε_t

7 · Strength Reduction Factor φ vs ε_t (§3.7, Figure 3.5)

9 · Master Recipe — Compute φM_n of a Given Section

Example 22.5 — Transition zone (0.004 < ε_t < 0.005)

Example 22.6 — Choose b, d and A_s for a target M_u