27. RC — One-Way Slabs

01

Engineering story

A century of steel — from concept to skyline

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

Concrete crushes; steel yields. Together they carry a column.

φPn,max = 0.80·φ·[0.85·f'c·(Ag − Ast) + fy·Ast] for tied; ×0.85 for spiral.

Chapter 26 covers short RC columns under axial and P–M interaction: tied vs spiral detailing, φPn,max, and the P–M interaction diagram.

Iconic steel structures built on engineering excellence

Whitney short-column formula
1937
ACI unified strength design
1971
ACI 318-19
2019

Load pathService D+LPu (Combo 2)φPn (tied 0.65 / spiral 0.75)P–M interactionSlenderness check

02

Learning objectives

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

Objective 01

Compute φPn,max (short column)

Tied: 0.80·φ·[0.85 f'c(Ag−Ast) + fy·Ast], φ=0.65. Spiral: 0.85·φ·[…], φ=0.75.

Why it matters: Baseline pure-axial capacity.
Where it is used: Every column check.
Connects to: ACI 22.4.2.

Objective 02

Tied vs Spiral detailing

Ties resist rebar buckling laterally; spirals also confine the core and boost ductility.

Why it matters: Spiral columns get φ=0.75 and 0.85·Pmax factor.
Where it is used: Circular columns.
Connects to: ACI 10.7.6, 22.4.2.

Objective 03

Enforce reinforcement ratio ρg

1% ≤ ρg = Ast/Ag ≤ 8% (typically ≤ 4% for constructability).

Why it matters: Keeps concrete workable and steel effective.
Where it is used: Every column design.
Connects to: ACI 10.6.1.

Objective 04

P–M interaction diagram

Points: pure axial (top), balanced (P = Pb, M = Mb), pure flexure (bottom).

Why it matters: Combined loading check.
Where it is used: Every real column.
Connects to: ACI 22.4.

Objective 05

Slenderness (short vs slender)

Short if klu/r ≤ 22 (unbraced) or ≤ 34 − 12·M1/M2 (braced).

Why it matters: Above the limit, use moment magnification.
Where it is used: Long or lightly restrained columns.
Connects to: ACI 6.2.5.

Objective 06

Design a column (Pu, Mu → section)

Pick c, compute As, plot on P–M interaction, iterate.

Why it matters: Rare that Pu alone controls — Mu usually joins.
Where it is used: Every real design.
Connects to: ACI 22.4.

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

Concrete crushing / rebar buckling

Cover spalls; longitudinal bars buckle between ties.

Root cause

Ties too widely spaced.

Lesson learned

Enforce tie spacing s ≤ 16·db, 48·dt, least column dim.

§ACI 25.7.2

Failure mechanism

Slender column buckling

Long column bows out under axial + lateral load.

Root cause

Slenderness > 22 (unbraced) not treated.

Lesson learned

Use moment magnification per ACI 6.6.4.

§ACI 6.6.4

How failure propagates

The five-stage failure progression

1Compressive load

2Cover spalls first

3Rebar buckles

4Confined core takes over

5Ductile crushing

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 one-way slab is really a wide, shallow beam. Why do we still add temperature and shrinkage steel perpendicular to the main reinforcement, and what cracking pattern would you see if we left it out?

One-Way Slab — Design Strip Concept

Design a 1-ft (12-in) wide strip as a rectangular beam with b = 12 in and d = h − cover − d_b/2. Every equation from Chapter 22 (flexure) still applies; only the units change (moment per foot, steel area per foot).

Design strip — 1 ft wide, thickness h, primary steel at d

Minimum Thickness (ACI Table 7.22.2.1) — Skips Deflection Check

End condition	Minimum h
Simply supported	ℓ / 20
One end continuous	ℓ / 24
Both ends continuous	ℓ / 28
Cantilever	ℓ / 10

Values above assume normal-weight concrete and Grade 60 steel. Multiply h by 1.65 − 0.005·w_c (pcf) for lightweight concrete; by (0.4 + f_y/100,000) for higher-grade steel.

Temperature & Shrinkage (T&S) Steel — ACI 24.4.22.3

(1)

A_{s,T&S} = 0.0018 \cdot b \cdot h (Grade 60)

Provided perpendicular to primary steel to control cracking from temperature and shrinkage strains. Spacing s ≤ min(5h, 18 in).

⚠ Common Mistakes in Slab Design

Designing for shear stirrups. One-way slabs typically don't have stirrups — instead the depth is chosen so V_u ≤ φV_c.
Forgetting T&S steel in the perpendicular direction. Cracking will show.
Using primary-bar spacing rule (3h) for T&S bars (which allow 5h).
Ignoring live-load pattern loading for continuous slabs — use ACI moment coefficients (§6.5) or a stiffness analysis.

06

Professional practice, safety & ethics

One-way slab practice

Professional practice

Provide temperature and shrinkage steel in the perpendicular direction; it is a code minimum, not optional.
Show bar chairs, cover, and top-steel support explicitly — top bars walked down during the pour lose their lever arm.
Check minimum thickness (ACI Table 7.3.1.1) before running a deflection calculation.

Safety in design & construction

Slabs are thin, so a small placement error is a large percentage of d.
Construction loads (stacked material, pumping equipment) on a young slab are a frequent collapse trigger.
Cantilever slab top steel misplacement is a recurring cause of balcony collapses.

Engineering ethics

Do not accept 'the mesh floated up' — require re-pour or an engineered repair.
Insist on pre-pour reinforcement inspection and document it.
Report any balcony or cantilever deficiency to the owner immediately, in writing.

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

Slab cost per square foot

Approach

Slab thickness is the dominant cost variable: every extra inch adds concrete, weight, and foundation load.
Temperature and shrinkage steel is a fixed cost per area — budget it, don’t discover it.
Formwork reuse across identical bays is the biggest single saving in slab construction.

Worked cost example — One-way slab installed cost

Basis: 7″ slab, per 1,000 ft²

Line item	Qty	Rate	Cost
Concrete (7″)	21.6 yd³	$165	$3,564
Formwork + shoring	1000 ft²	$8	$7,500
Reinforcement (main + T&S)	1.1 ton	$2,200	$2,420
Place, finish, cure	1000 ft²	$3	$2,600
Estimated total			$16,084

Takeaway. ≈$15.9/ft² installed — shaving 1″ of thickness saves about $0.55/ft² plus foundation savings.

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

Tied vs spiral column confinement

Under axial load, tied cores spall and lose capacity abruptly (φ=0.65). Spiral cores stay confined and remain ductile (φ=0.75).

09

Engineering figures

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

§26.6.1

Tied column

Fig. 26.1Rectangular tied cage
Tied columns: 0.80·Pmax factor, φ=0.65.

§26.6.2

Spiral column

Fig. 26.2Continuous spiral cage
Spirals confine and boost ductility.

§26.6.3

Crushing failure

Fig. 26.3Cover spalled, bars buckled
Tie spacing critical.

§26.6.4

Slender bowing

Fig. 26.4Buckled slender column
P-δ magnifies moments.

10

Worked examples

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

Example 26.1

φPn,max for a 16×16 tied column (4 #8)

Compute the accidental-eccentricity-capped design axial capacity.

Problem statement

A short 16×16 in tied column has 4 #8 bars (Ast = 3.14 in²), f'c = 4 ksi, fy = 60 ksi. Compute φPn,max.

Tied RC column — FIG. 26.1 — a tied rectangular column.

DIM16×16 in tied column with 4 #8 longitudinal bars.

Given

b = h = 16 in → Ag = 256 in²
Ast = 3.14 in²
f'c = 4 ksi, fy = 60 ksi
Tied → φ = 0.65, cap = 0.80·Pn

Find

φPn,max

Assumptions

Short column (no slenderness effect)
Pure axial (no Mu)

Code references

ACI 318-19 22.4.2
ACI 318-19 21.2.2

Theory & approach

Pure Pn is capped by 0.80 (tied) or 0.85 (spiral) to account for accidental eccentricity; then multiplied by φ.

Step-by-step solution

1
Pure Pn
FormulaACI 22.4.2.2
P_n = 0.85·f'_c·(A_g − A_st) + f_y·A_st
P_n = 0.85·4·(256 − 3.14) + 60·3.14
P_n = 0.85·4·252.86 + 188.4 = 859.7 + 188.4 = 1,048 k
2
Accidental-e cap
Formula
P_n,max = 0.80·P_n (tied)
P_n,max = 0.80·1,048 = 838 k
3
φ
FormulaACI 21.2.2
φ = 0.65 (tied, compression-controlled)
4
Design capacity
φP_n,max = 0.65·838 = 545 k
5
ρg check
ρ_g = 3.14/256 = 0.0123 = 1.23%
1% ≤ 1.23% ≤ 8% ✓

Verification

ρg in allowable band and Pn cap correctly applied.

Final answer

φPn,max = 545 k (short tied column).

Common mistakes

Using φ = 0.90 (that's flexure TC).
Skipping the 0.80 accidental-e cap.
Forgetting to subtract Ast from Ag.

References

· ACI 318-19

11

Guided practice

Compute the governing variables — hints unlock as you need them

A 14-in-diameter circular SPIRAL column has 6 #6 longitudinal bars (A_st = 2.64 in²), f'c = 4 ksi, f_y = 60 ksi. Compute A_g, ρ_g, P_o, and φP_n,max per ACI §22.4.2, then compare with an equivalent tied section.

Your turn

A_g =in^2rho_g =P_o =kipphi*P_n,max =kip

Hints

1.A_g = π·D²/4 = π·14²/4 = 153.94 in².

12

Independent practice

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

Design task

A 16 in × 16 in tied RC column is reinforced with eight #8 longitudinal bars (Ast = 6.32 in²). Compute the nominal axial capacity at zero eccentricity P0, the code cap Pn,max = 0.80·P0 (tied), and the design axial strength φPn,max.

Given

Ag = 16·16 = 256 in²
Ast = 6.32 in²
f'c = 4000 psi, fy = 60,000 psi
Tied column: αc = 0.80, φ = 0.65

Approach

P0 = 0.85·f'c·(Ag − Ast) + fy·Ast (ACI 318 Eq. 22.4.2.2).
For a tied column, Pn,max = 0.80·P0 (ACI 318 §22.4.2.1).
φPn,max = 0.65·Pn,max for a tied compression-controlled section.

Submit your answer

P₀ =kipphi*P_n,max =kip

13

Mini design challenge

Pick the column configuration that satisfies strength, ρg, and detailing

Brief

Design a short RC column for P_u = 480 k, f'c = 4 ksi, f_y = 60 ksi. Compare a 16×16 tied square against a 14-in spiral. Check φP_n,max, ρ_g limits, minimum bar count, tie/spiral detailing, AND ductility posture.

Requirements

φP_n,max ≥ P_u = 480 kip
1 % ≤ ρ_g ≤ 8 % (ACI §10.6.1.1)
Tied: cap = 0.80·P_o, φ = 0.65; min 4 bars (ACI §10.7.3.1)
Spiral: cap = 0.85·P_o, φ = 0.75; min 6 bars; ρ_s ≥ 0.45·(A_g/A_ch − 1)·f'c/f_yt (ACI §25.7.3)
Tie spacing s ≤ min(16·d_b, 48·d_t, least column dim) — ACI §25.7.2

14

Chapter summary

A mind map of how every concept connects

Graded Chapter Quiz(8 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 / 12

For a 16×16 tied column, Ast=3.14, f'c=4, fy=60 ksi. φPn,max is closest to:

◆◆ MediumACI 318-19 §22.4.2.1 · Table 22.4.2.1 · Table 21.2.2

A. 400 k
B. 500 k
C. 545 k
D. 700 k