25. RC — Development Length and Splices

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

A rebar can't develop its full yield strength the instant it enters the concrete — it needs a development length ld. What physically transfers force from the bar to the concrete, and why do hooks let us shorten that length?

Development Length ℓ_d — What Anchors a Bar

Simplified tension development (ACI 25.4.2.3)

(1)

ℓ_{d} / d_{b} = (f_{y} \cdot ψ_{t} \cdot ψ_{e} \cdot ψ_{s} \cdot ψ_{g}) / (25 \cdot \lambda \cdot \sqrt f'_{c}) for #6 & smaller

Same, #7 and larger

(2)

ℓ_{d} / d_{b} = (f_{y} \cdot ψ_{t} \cdot ψ_{e} \cdot ψ_{s} \cdot ψ_{g}) / (20 \cdot λ \cdot f_{c}^{'})

Modification factors multiply up the required length when conditions are less favorable:

Factor	Applies when	Value
ψ_t (bar location)	More than 12" of fresh concrete cast below the bar (i.e. top bar)	1.3
ψ_e (coating)	Epoxy-coated bar	1.5 (or 1.2 with good cover)
ψ_s (bar size)	Bars #6 and smaller	0.8
ψ_g (grade)	Gr 60 → 1.0; Gr 80 → 1.15; Gr 100 → 1.3	≥ 1.0
λ (lightweight)	Lightweight concrete	0.75 (all-LW) or 0.85 (sand-LW)

Rule of thumb sanity check. For #4 Gr 60 in f'_c = 4,000 psi normal-weight, no coating, bottom bar: ℓ_d ≈ 24·d_b = 24·0.5 = 12 in. Top #4 same section: 1.3× → 16 in. Real detailer's shortcut: use ACI Table 25.4.2.5 which pre-computes ℓ_d in table form.

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

Standard Hooks — ℓ_dh

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

(3)

ℓ_{d h} = f_{y} \cdot ψ_{e} \cdot ψ_{r} \cdot ψ_{o} \cdot ψ_{c} \cdot d_{b}^{1.5} / (55 \cdot λ \cdot f_{c}^{'}) \geq ma x (8 \cdot d_{b}, 6 in)

A 90° hook adds only about 8 bar diameters of anchorage — use hooks when you run out of straight length at a beam end or column joint, not as a substitute for full development.

06

Professional practice, safety & ethics

Development & splice practice

Professional practice

Show development lengths, lap classes, and hook geometry on the drawings — not just 'lap per code'.
Coordinate splice locations away from maximum-moment regions where possible.
Verify bar spacing and cover assumptions used in the ℓd equation are the ones actually detailed.

Safety in design & construction

Anchorage failure is brittle: the bar pulls out with no warning and no residual strength.
Congestion is a safety issue — bars that block concrete flow create honeycombing at the most critical location.
Field-bent bars, especially Grade 60+ , can fracture; require a documented procedure.

Engineering ethics

Never shorten a lap to fit the bar stock on site.
Report insufficient embedment discovered after placement; retrofits (couplers, headed bars) exist for a reason.
Do not approve mechanical couplers without qualification test data.

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

Development and splice cost

Approach

Lap splices waste steel — a Class B lap can add 30–50% bar length in the splice region.
Mechanical couplers cost more per connection but save steel and congestion in heavily reinforced members.
Hooks and headed bars buy anchorage in tight joints for a modest fabrication premium.

Worked cost example — Lap splice vs mechanical coupler, 8-#9 bars

Basis: f′c = 4 ksi, Gr. 60

Line item	Qty	Rate	Cost
Class B lap — extra bar length 4.5 ft × 8	0.123 ton rebar	$2,200	$271
Mechanical couplers	8 ea installed	$85	$680
Congestion/placement premium for laps	1 ls	$150	$150
Estimated total			$1,101

Takeaway. Couplers ($680) beat laps (≈$421) on material alone but win outright where congestion slows placement.

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(8 FE-style questions · ACI 318-19 required)

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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 — Development Length and Splices

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

Development Length ℓd — What Anchors a Bar

Standard Hooks — ℓdh

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(8 FE-style questions · ACI 318-19 required)

Upload your worked solution (PDF)

Development Length ℓ_d — What Anchors a Bar

Standard Hooks — ℓ_dh