26. RC — Compression Members (Columns)

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

Tied and spirally reinforced columns have similar axial capacity but very different failure modes. Which one would you specify in a seismic zone, and why does the confinement geometry change the ductility so dramatically?

9.1 Introduction — Compression Members

This chapter opens the study of reinforced concrete columns, with particular emphasis on short, stocky columns subjected to small bending moments. Such columns are often said to be "axially loaded." Short, stocky columns with large bending moments are covered later, while long or slender columns are treated in the slenderness chapter. Concrete columns can be classified into three broad compression categories:

Short compression blocks / pedestals. If the height of an upright compression member is less than three times its least lateral dimension, it is a pedestal. ACI (§2.2, §14.1.3) permits a pedestal to be designed as unreinforced or plain concrete, with a maximum design compressive stress of 0.85·φ·f'_c (φ = 0.65). If the required load exceeds 0.85·φ·f'_c·A_g, either enlarge the cross-section or design the member as a reinforced column.
Short reinforced concrete columns. A column that fails through the strength of its materials, not through instability, is classed as a short column. Load capacity is controlled by cross-section dimensions and material strengths.
Long / slender reinforced concrete columns. As columns become more slender, bending deformations grow, so secondary moments P·Δ from the lateral deflection Δ add to the primary moments. When these are large enough to reduce axial capacity, the column is called long or slender. ACI classifies short vs slender by a slenderness ratio; the effects are ignorable in roughly 40% of unbraced and 90% of braced columns. Secondary moments must not exceed 5% of the primary moments if the member is to be designed as short.

The lateral deflection Δ under primary moment M produces a secondary P·Δ moment that adds to M.

9.2 Types of Columns

A plain concrete column carries very little load, but capacity is greatly increased when longitudinal bars are added. Further gains follow from providing lateral restraint for those bars: under compression the concrete tends to shorten lengthwise and expand laterally (Poisson effect), and closely spaced closed ties or helical spirals wrapping the longitudinal steel supply that restraint.

Tied columns — longitudinal bars restrained by a series of closed ties (Fig. 9.2a). Ties are inexpensive and effective in increasing column strength: they prevent displacement during construction and resist outward buckling of the bars once the shell begins to spall. Tied columns are usually square or rectangular but can also be octagonal, round, or L-shaped.
Spiral columns — longitudinal bars wrapped by a continuous helical spiral of small-diameter bar (Fig. 9.2b). Closely spaced spirals do a better job of confining the longitudinal steel, sharply increase resistance to axial compression, and — because they pass into hoop tension after the shell spalls — provide a warning failure rather than a sudden one. Spirals are commonly round but may also be square or rectangular.
Composite columns — a structural steel shape (pipe or W-section) embedded in reinforced concrete (Fig. 9.2c, d).

Tied, spiral, and composite column cross-sections (schematic).

9.3 Axial Load Capacity of Columns

Perfectly axially loaded columns do not exist in practice, but the theoretical pure-axial capacity is an excellent starting point. Decades of testing show that the ultimate stress the concrete can reach in a column is well approximated by 0.85·f'_c. It does not matter much whether concrete or steel approaches its ultimate strength first: if one is stressed close to its limit, its large deformation transfers stress to the other. Therefore the pure axial nominal capacity of a short column is

Nominal axial capacity (pure axial)

(1)

P_{n} = 0.85 \cdot f_{c}^{'} \cdot (A_{g} - A_{s t}) + f_{y} \cdot A_{s t}

where A_g is the gross cross-section area (concrete + steel) and A_st is the total area of longitudinal reinforcement.

9.4 Failure of Tied and Spiral Columns

Tied columns. As a short tied column is loaded to failure, the outer concrete shell spalls off. Unless the ties are closely spaced, the longitudinal bars then buckle outward once their lateral support (the covering concrete) is lost — the failure can be sudden, and such failures have been reported frequently in structures subject to earthquake loading.

Spiral columns. When a spiral column is loaded to failure, the shell spalls, but the confined core continues to stand. If the spiral is closely spaced, it develops hoop tension and confines the core, allowing the column to resist additional load beyond what caused spalling. The closely spaced spiral and longitudinal bars form a cage that keeps the core intact, so the spalling of the shell provides a warning that failure is going to occur if the load is further increased. American practice is to neglect any excess capacity after the shell spalls off.

Shell strength contribution

(2)

S h e l l s t r e n g t h = 0.85 \cdot f_{c}^{'} \cdot (A_{g} - A_{c})

Effective spiral strength (2× longitudinal steel)

(3)

S p i r a l s t r e n g t h = 2 \cdot ρ_{s} \cdot f_{y t} \cdot A_{c}

Here A_c is the core area (out-to-out of the spiral). Tests show the lateral hoop pressure produced by the spiral increases the confined core capacity by roughly twice that of the same weight of longitudinal steel — the origin of the 2·ρ_s·f_yt term used in the ACI minimum-spiral-ratio expression.

How this chapter is organized. Real columns rarely carry pure axial load — they are always bent as well. We build up from that fact: (1) the six eccentricity cases that trace the boundary of a column's strength, (2) the plastic centroid that the resultant load must pass through at failure, (3) strain compatibility → P_n, M_n for any strain distribution, (4) the P–M interaction diagram, (5) the ACI code modifications (φ, 0.80/0.85 caps), (6) how to read the standardized ACI interaction charts, and (7) tied vs spiral confinement.

1. Axial Load + Bending — the Six Eccentricity Cases

All columns are subject to some bending. The code approximates this by requiring minimum eccentricities (0.10h tied, 0.05h spiral) even for "axial" columns. As eccentricity e = M/P grows from zero to infinity, the failure mode migrates through six regimes:

Case	Load / eccentricity	Failure mode at ε_cu = 0.003
(a)	Large P, negligible M	Concrete crushes uniformly; all bars at f_y in compression
(b)	Large P, small M — full cross-section still in compression	Concrete crushes on the more compressed face at 0.85f'_c
(c)	Larger e — tension develops on the far face but steel there is below f_y	Concrete still crushes first (compression-controlled)
(d)	Balanced condition — tension steel just reaches f_y as concrete crushes	Simultaneous ε_s = ε_y and ε_c = 0.003
(e)	Large M, small P	Tensile bars yield first, then concrete crushes (tension-controlled)
(f)	Large M, negligible P	Behaves as a beam in flexure

Left → right: eccentricity grows, failure migrates from concrete crushing to steel yielding.

2. The Plastic Centroid — Where P_n Must Act

The plastic centroid is the point through which the resultant column force must pass to produce uniform strain at failure — all concrete at 0.85f'_c and all steel at f_y. Eccentricity e is measured from the plastic centroid, not the geometric center. For a symmetrical section the two coincide; for a T- or unsymmetrical section they don't.

Plastic-centroid axial resultant

(4)

P_{o} = 0.85 \cdot f_{c}^{'} \cdot (A_{g} - A_{s t}) + f_{y} \cdot A_{s t}

Location x̄ from a chosen edge

(5)

\overset{x}{ˉ} = Σ C_{i} \cdot x_{i} / P_{o}

Worked Example 26.1 — T-section plastic centroid. Composite unsymmetrical section: left rectangle 6″ wide × 16″ tall (b₁ = 6, h₁ = 16), right rectangle 8″ wide × 8″ tall (b₂ = 8, h₂ = 8) attached to the right face and bottom-aligned. 4 #9 bars centered near the interface (x_s = 7 in), A_st = 4.00 in², f'_c = 4 ksi, f_y = 60 ksi. Total width = 14 in.

Two-rectangle composite section. C₁ and C₂ are the concrete compressive resultants of each rectangle (uniform 0.85 f'_c at plastic-centroid conditions); C'ₛ is the steel resultant. The plastic centroid x̄ is the line of action of the resultant P_o = C₁ + C₂ + C'ₛ.

Step 1 — Concrete resultants (formula first). At the plastic-centroid state every fibre of concrete is at 0.85 f'_c. For each rectangle the resultant equals stress × area and acts through its geometric centroid.

Rectangle-i concrete force

(6)

C_{i} = 0.85 \cdot f_{c}^{'} \cdot A_{c, i} = 0.85 \cdot f_{c}^{'} \cdot (b_{i} \cdot h_{i})

(7)

C_{1} = 0.85 \cdot (4) \cdot (6 \cdot 16) = 0.85 \cdot (4) \cdot (96) = 326.4 k a t x_{1} = b_{1} /2 = 3.00 in

(8)

C_{2} = 0.85 \cdot (4) \cdot (8 \cdot 8) = 0.85 \cdot (4) \cdot (64) = 217.6 k a t x_{2} = b_{1} + b_{2} /2 = 6 + 4 = 10.00 in

Step 2 — Steel resultant. Deduct the concrete stress that the bars displace (bars occupy concrete area that would otherwise carry 0.85 f'_c):

Steel force (formula)

(9)

C_{s}^{'} = A_{s t} \cdot (f_{y} - 0.85 \cdot f_{c}^{'})

(10)

C_{s}^{'} = (4.00) \cdot (60 - 0.85 \cdot 4) = (4.00) \cdot (60 - 3.4) = (4.00) \cdot (56.6) = 226.4 k a t x_{s} = 7.00 in

Step 3 — Nominal axial capacity P_o.

Formula

(11)

P_{o} = Σ C_{i} + C_{s}^{'} = 0.85 \cdot f_{c}^{'} \cdot (A_{g} - A_{s t}) + f_{y} \cdot A_{s t}

(12)

P_{o} = 326.4 + 217.6 + 226.4 = 770.4 k

Step 4 — Plastic centroid x̄. Sum moments of the three resultants about the left edge and divide by P_o:

Formula (moment of resultants about x = 0)

(13)

\overset{x}{ˉ} = (C_{1} \cdot x_{1} + C_{2} \cdot x_{2} + C_{s}^{'} \cdot x_{s}) / P_{o}

(14)

\overset{x}{ˉ} = [(326.4) (3) + (217.6) (10) + (226.4) (7)] /770.4

(15)

\overset{x}{ˉ} = [979.2 + 2176.0 + 1584.8] /770.4 = 4740.0/770.4 = 6.15 in f r o m t h e l e f t e d g e

P_o = 770.4 k; plastic centroid at x̄ = 6.15 in from the left face (compare with geometric centroid x_g ≈ 5.29 in — the two differ because the section is unsymmetrical). All column eccentricities e in Sections 3–5 below are measured from this line.

3. Strain Compatibility → P_n and M_n

Fix ε_c = −0.003 on the compression edge, pick any strain on the far edge, and the strain field is linear (plane sections). From there you get c, ε_s, ε'_s, then the resultants C_c, C'_s, T_s. Statics gives one (P_n, M_n) point.

Neutral-axis depth from linear strain

(16)

c = [0.003/ (0.003 + ε_{t})] \cdot d

Whitney block

(17)

a = β_{1} \cdot c

Force resultants

(18)

C_{c} = 0.85 \cdot f_{c}^{'} \cdot a \cdot b C_{s}^{'} = A_{s}^{'} \cdot (f_{s}^{'} - 0.85 \cdot f_{c}^{'}) T_{s} = A_{s} \cdot f_{s}

Statics (about the plastic centroid)

(19)

P_{n} = C_{c} + C_{s}^{'} - T_{s} M_{n} = Σ F_{i} \cdot (a r m t o pl a s t i cce n t r o i d)

Worked Example 26.2 — one interaction-diagram point. 14×24 in tied column, 6 #9 bars (3 top, 3 bottom, A_s = A'_s = 3.00 in²), 2.5-in cover to centroids, f'_c = 4 ksi, f_y = 60 ksi. Assumed strains: ε_c = −0.003 on compression edge, +0.002 on far edge.

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

Step 1 — Neutral-axis depth from linear strain (formula first).

Formula

(20)

c = [0.003/ (0.003 + ε_{t})] \cdot d w i t h d = 21.5 in, ε_{t} = 0.002

(21)

c = [0.003/ (0.003 + 0.002)] \cdot (21.5) = (0.6) \cdot (21.5) = 12.90 in

Whitney block depth

(22)

a = β_{1} \cdot c = 0.85 \cdot (12.90) = 10.97 in

Step 2 — Steel strains & stresses (formula first).

Compression steel strain

(23)

ε_{s}^{'} = 0.003 \cdot (c - d^{'}) / c

(24)

ε_{s}^{'} = 0.003 \cdot (12.90 - 2.5) /12.90 = 0.00242 > ε_{y} = 0.00207

A_{s}^{'} y i e l d s

f_{s}^{'} = 60 k s i

Tension steel strain

(25)

ε_{s} = 0.003 \cdot (d - c) / c

(26)

ε_{s} = 0.003 \cdot (21.5 - 12.90) /12.90 = 0.00200 < ε_{y}

d oes n o t y i e l d

f_{s} = E_{s} \cdot ε_{s} = 29, 000 \cdot 0.00200 = 58.0 k s i (c a pp e d a t f_{y} i f e x cee d e d)

(27)

S in ce ε_{s} \approx ε_{y} t h esec t i o ni s a t ba l an ce d / co m p r ess i o n - co n t r o l l e d b o u n d a r y

ϕ = 0.65 (t i e d) .

Compression edge at ε_c = 0.003; far-edge strain = 0.002. NA at c = 12.90". Whitney block: 0.85 f'_c across a = 10.97", giving C_c. Compression steel adds C'_s (deducting concrete it displaces); tension steel gives T_s.

Step 3 — Force resultants (formula first).

Concrete compressive resultant

(28)

C_{c} = 0.85 \cdot f_{c}^{'} \cdot a \cdot b

(29)

C_{c} = 0.85 \cdot (4.0) \cdot (10.97) \cdot (14) = 522.2 k

Compression-steel resultant (deduct displaced concrete)

(30)

C_{s}^{'} = A_{s}^{'} \cdot (f_{s}^{'} - 0.85 \cdot f_{c}^{'})

(31)

C_{s}^{'} = (3.00) \cdot (60 - 0.85 \cdot 4) = (3.00) \cdot (56.6) = 169.8 k

Tension-steel resultant

(32)

T_{s} = A_{s} \cdot f_{s}

(33)

T_{s} = (3.00) \cdot (58.0) = 174.0 k

Step 4 — Axial equilibrium (ΣV = 0).

Formula

(34)

P_{n} = C_{c} + C_{s}^{'} - T_{s}

(35)

P_{n} = 522.2 + 169.8 - 174.0 = 518.0 k

Design axial

(36)

ϕ P_{n} = ϕ \cdot P_{n} = 0.65 \cdot 518.0 = 336.7 k

Step 5 — Moment equilibrium about the plastic centroid (ΣM = 0).

Formula (symmetric section, x̄ = h/2 = 12″)

(37)

M_{n} = C_{c} \cdot (h /2 - a /2) + C_{s}^{'} \cdot (h /2 - d^{'}) + T_{s} \cdot (d - h /2)

(38)

M_{n} = 522.2 \cdot (12 - 5.485) + 169.8 \cdot (12 - 2.5) + 174.0 \cdot (21.5 - 12)

(39)

M_{n} = 522.2 \cdot (6.515) + 169.8 \cdot (9.5) + 174.0 \cdot (9.5) = 3402.1 + 1613.1 + 1653.0 = 6668.2 in - k = 555.7 f t - k

Design moment

(40)

ϕ M_{n} = 0.65 \cdot 555.7 = 361.2 f t - k

One point on the P–M diagram: (φP_n, φM_n) = (336.7 k, 361.2 ft-k). Sweep ε_t from 0 → ∞ (compression-controlled → tension-controlled) to trace the whole envelope.

Repeat with different far-edge strains to sweep the whole diagram: ε_t → ∞ gives point (P_o, 0); ε_t = ε_y gives the balanced point; P_n = 0 gives the pure-flexure point.

4. P–M Interaction Diagram — Reading the Column's Full Story

Nominal curve → apply 0.80 (tied) or 0.85 (spiral) cap → apply φ variation → design envelope. Every (P_u, M_u) must fall inside.

5. Three ACI 318-19 Modifications to the Nominal Curve

φ variation. Compression-controlled: φ = 0.65 tied / 0.75 spiral. Tension-controlled (ε_t ≥ 0.005): φ = 0.90. Linear interpolation between.
Multiply P_o by 0.80 (tied) or 0.85 (spiral) to cover accidental eccentricity. This produces the horizontal cap on the top of the design curve.
Combined design cap — you never use a P_n larger than φP_n,max:

Tied (ACI 22.4.2.2)

(41)

ϕ P_{n, ma x} = 0.65 \cdot 0.80 \cdot [0.85 \cdot f_{c}^{'} \cdot (A_{g} - A_{s t}) + f_{y} \cdot A_{s t}]

Spiral

(42)

ϕ P_{n, ma x} = 0.75 \cdot 0.85 \cdot [0.85 \cdot f_{c}^{'} \cdot (A_{g} - A_{s t}) + f_{y} \cdot A_{s t}]

Example 26.3 — Design a square tied column (McCormac Ex 9.1)

Design a square tied column to support an axial dead load D = 130 k and an axial live load L = 180 k. Try ρ_g ≈ 2% longitudinal steel; f'_c = 4,000 psi, f_y = 60,000 psi.

Factored load (ACI 5.22.2)

(43)

P_{u} = 1.2 \cdot D + 1.6 \cdot L = 1.2 (130) + 1.6 (180) = 444 k

Size Ag with ρg = 0.02 (tied)

(44)

ϕ P_{n} = ϕ \cdot 0.80 \cdot [0.85 \cdot f_{c}^{'} \cdot (A_{g} - A_{s t}) + f_{y} \cdot A_{s t}] (A C I E q .10 - 2) 444 = (0.65) (0.80) \cdot [(0.85) (4) (A_{g} - 0.02 \cdot A_{g}) + (60) (0.02 \cdot A_{g})]

A_{g} = 188.4 i n^{2} U se 14 \cdot 14 in (A_{g} = 196 i n^{2})

Re-solve Ast for the chosen Ag

(45)

444 = (0.52) \cdot [(0.85) (4) (196 - A_{s t}) + (60) \cdot A_{s t}]

A_{st} = 3.31 in^{2} Use 6 #7 bars (A_{st} = 3.61 in^{2})

Tie spacing (#3, ACI 25.7.2)

(46)

s_{max} = min(48 \cdot d_{t}, 16 \cdot d_{b}, least dim) = min(48 \cdot ⅜, 16 \cdot ⅞, 14) = min(18, 14, 14) = 14 in Use #3 ties @ 14 in.

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

Example 26.4 — Design a round spiral column: size and bars (McCormac Ex 9.2)

Design a round spiral column for P_D = 240 k, P_L = 300 k. Try ρ_g ≈ 2%; f'_c = 4,000 psi, f_y = 60,000 psi.

Factored load

(47)

P_{u} = 1.2 (240) + 1.6 (300) = 768 k

Size Ag with ρg = 0.02 (spiral)

(48)

ϕ P_{n} = ϕ \cdot 0.85 \cdot [0.85 \cdot f_{c}^{'} \cdot (A_{g} - A_{s t}) + f_{y} \cdot A_{s t}] (A C I E q .10 - 1) 768 = (0.75) (0.85) \cdot [(0.85) (4) (A_{g} - 0.02 \cdot A_{g}) + (60) (0.02 \cdot A_{g})]

A_{g} = 266 i n^{2} T r y Ø = 18 in

A_{g} = 254.5 i n^{2} (< 266 i n^{2})

ρ_{g} m u s t b e abi t hi g h er .

Re-solve Ast for Ø = 18 in

(49)

768 = (0.6375) \cdot [(0.85) (4) (255 - A_{s t}) + (60) \cdot A_{s t}]

A_{st} = 5.97 in^{2} Use 6 #9 bars (A_{st} = 6.00 in^{2})

Example 26.5 — Spiral reinforcement design for Ex 26.4

Continuation of Ex 26.4. Ø 18 in column with 1½ in cover → core diameter D_c = 15 in; core area A_c = π(15)²/4 = 177 in². Try #3 spiral (d_b = 0.375 in, a_s = 0.11 in²).

Min spiral ratio (ACI 25.7.22.4)

(50)

ρ_{s, min} = 0.45 \cdot (A_{g} / A_{c} - 1) \cdot (f_{c}^{'} / f_{y t}) = 0.45 \cdot (255/177 - 1) \cdot (4/60) = 0.0132

Pitch from ρs definition

(51)

ρ_{s} = 4 \cdot a_{s} \cdot (D_{c} - d_{b}) / (s \cdot D_{c}^{2}) 0.0132 = 4 \cdot (0.11) \cdot (15 - 0.375) / (s \cdot 1 5^{2})

s = 6.435/(0.0132 \cdot 225) = 2.17 in Use #3 spiral @ 2 in pitch (checked with Appendix A, Table A.14).

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

6. Using the ACI Standardized Charts

Doing strain-compatibility for every column is tedious, so ACI SP-17 publishes normalized interaction diagrams keyed on:

(52)

K_{n} = P_{n} / (f_{c}^{'} \cdot A_{g}) R_{n} = P_{n} \cdot e / (f_{c}^{'} \cdot A_{g} \cdot h) = M_{n} / (f_{c}^{'} \cdot A_{g} \cdot h)

Figure 10.16 — ACI rectangular column interaction diagram R4-60.7. Read (Rn, Kn) → interpolate between adjacent ρg curves → obtain ρg → As = ρg·Ag.

How to read this chart (the exam procedure).

Confirm the chart matches your column. The title block gives f'_c, f_y, γ = (h − 2·cover)/h, and the bar arrangement ("bars on 2 faces" = L-chart, "on 4 faces" = R-chart). If your γ falls between two charts (e.g. γ = 0.75 with charts at 0.70 and 0.80), read both and linearly interpolate ρ_g.
Compute the two dimensionless coordinates K_n and R_n from your factored (P_u, M_u). Divide by an assumed φ (0.65 tied / 0.75 spiral) to convert to (P_n, M_n) first.
Plot the point. Move right by R_n from the y-axis, then up by K_n. The point must fall inside the envelope of the family — anywhere outside the top curve (ρ_g = 0.08) means the section is too small.
Interpolate ρ_g between the two curves that bracket your point. Any point that lands on a ρ_g curve gives that exact steel percentage.
Check where you are on the diagram. Points above the "balance" radial line are compression-controlled (φ = 0.65); below → tension-controlled (φ = 0.90); the diagonal e/h radial lines let you read eccentricity by inspection.
Compute A_st = ρ_g·A_g and pick a bar arrangement so 1% ≤ ρ_g ≤ 8% (ACI 26.9.1.1; keep ≤ 4–5% for buildable splices).

Reverse use — given a section, find P_n at an eccentricity. Draw the radial line for the known e/h from the origin. Where it crosses the ρ_g curve of the given reinforcement, read K_n (and R_n) → back out P_n = K_n·f'_c·A_g. This is exactly the workflow of Example 26.10 below.

Pick the chart that matches your bar layout — "on two faces" (L-charts) vs "on all four faces" (R-charts). Wrong chart → wrong ρ_g.
Select the chart for your γ = (h − 2·cover)/h. Interpolate between γ values if needed.
Compute K_n and R_n from your factored P_u, M_u (using φ from the strain check).
Plot the point and read ρ_g off the family of curves. A_st = ρ_g·A_g.
Keep 1% ≤ ρ_g ≤ 8% (ACI 26.9.1.1). Values above 4–5% cause splice congestion.

7. Tied vs Spiral — Confinement & the Different φ

Why the different φ. A spiral column confines its core: when the concrete cover spalls at ultimate, the spiral engages the interior concrete and the column keeps carrying load in a ductile manner. Tied columns simply crush once the cover pops. Same mechanism, better outcome for spirals → 15% higher φ and 20% larger P_n,max coefficient.

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

P_n vs φP_n — do not confuse these.

P_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.
φP_n (design) = the strength the code allows you to count on. Always < P_n. This is what you compare with the factored demand P_u.
Design check: φP_n ≥ P_u. If you write P_n ≥ P_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.

8. Worked Example 26.6 — Rectangular Tied Column, Bars on Two Faces

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

14 × 20 in tied column, cover to bar centroid = 2.5 in, f'_c = 4 ksi, f_y = 60 ksi. Loads: P_D = 125 k, P_L = 140 k, M_D = 75 ft-k, M_L = 90 ft-k. Select A_st using ACI SP-17 charts (bars on two end faces).

Factored demand (ACI 5.22.2, load combo U = 1.2D + 1.6L)

(53)

P_{u} = 1.2 (125) + 1.6 (140) = 374 k M_{u} = 1.2 (75) + 1.6 (90) = 234 f t - k

Assume φ = 0.65 (compression-controlled)

(54)

P_{n} = P_{u} / ϕ = 374/0.65 = 575.4 k M_{n} = 234/0.65 = 360 f t - k e = (12 \cdot M_{n}) / P_{n} = (12) (360) /575.4 = 7.51 inγ = (h - 2 \cdot co v er) / h = (20 - 2 \cdot 2.5) /20 = 0.75

Chart coordinates (ACI SP-17)

(55)

K_{n} = P_{n} / (f_{c}^{'} \cdot A_{g}) = 575.4/ (4 \cdot 14 \cdot 20) = 0.513 R_{n} = P_{n} \cdot e / (f_{c}^{'} \cdot A_{g} \cdot h) = (575.4) (7.51) / (4 \cdot 280 \cdot 20) = 0.193

Interpolate between γ = 0.70 chart (ρ_g ≈ 0.0220) and γ = 0.80 chart (ρ_g ≈ 0.0185) → ρ_g ≈ 0.0202.

(56)

A_{st} = \rho _{g} \cdot b \cdot h = 0.0202 \cdot (14)(20) = 5.66 in^{2} Use 6 #9 bars (A_{st} = 6.00 in^{2}) — 3 in each end face.

Verify: because the SP-17 curves used correspond to f_s/f_y < 1.0 (ε_t < ε_y), the section is indeed compression-controlled → the assumed φ = 0.65 is correct. Then check ACI 25.7 tie detailing.

Example 26.6 — plot (Rn = 0.193, Kn = 0.513) on the R4-60.7 chart. The point falls between the ρg = 0.02 and 0.03 curves → ρg ≈ 0.0202.

9. Worked Example 26.7 — Square Column, Bars on Four Faces

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

Design a short square tied column. P_u = 600 k, M_u = 80 ft-k, f'_c = 4 ksi, f_y = 60 ksi. Bars uniformly around all four faces.

Trial size — average stress ≈ 0.6·f'c

(57)

A_{g, r e q} \approx P_{u} / (0.6 \cdot f_{c}^{'}) = 600/2.4 = 250 i n^{2} T r y 16 \cdot 16 in (A_{g} = 256 i n^{2})

Demand/eccentricity

(58)

e = (12) (80) /600 = 1.60 in P_{n} = 600/0.65 = 923.1 k K_{n} = 923.1/ (4 \cdot 256) = 0.901 R_{n} = (923.1) (1.6) / (4 \cdot 256 \cdot 16) = 0.0901 γ = 11/16 = 0.6875

Interpolating "bars on four faces" ACI charts (γ = 0.60 → ρ_g ≈ 0.025; γ = 0.70 → 0.022) gives ρ_g ≈ 0.023.

(59)

A_{s t} = (0.023) (256) = 5.89 i n^{2}

Use 8 #8 bars (A_{st} = 6.28 in^{2})

Example 26.7 — plot (Rn = 0.090, Kn = 0.901) for the 16×16 trial. The point sits between ρg = 0.02 and 0.03 → ρg ≈ 0.023 (interpolated between the γ = 0.60 and 0.70 charts for bars on four faces).

10. Worked Example 26.8 — Spiral Round Column

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

Short round spiral column, 20-in diameter, f'_c = 4 ksi, f_y = 60 ksi, P_u = 500 k, M_u = 225 ft-k. Cover to bar centroid = 2.5 in.

(60)

A_{g} = π \cdot (20)^{2} /4 = 314 i n^{2} e = (12) (225) /500 = 5.40 in P_{n} = P_{u} / ϕ = 500/0.75 = 666.7 k (ϕ = 0.75 s p i r a l, co m p r ess i o n - co n t r o l l e d) K_{n} = 666.7/ (4 \cdot 314) = 0.531 R_{n} = (666.7) (5.40) / (4 \cdot 314 \cdot 20) = 0.143 γ = 15/20 = 0.75

Interpolating between ACI round-column charts → ρ_g ≈ 0.0235.

(61)

A_{s t} = (0.0235) (314) = 7.38 i n^{2}

Use 8 #9 bars (A_{st} = 8.00 in^{2})

Example 26.8 — round spiral column: plot (Rn = 0.143, Kn = 0.531). For a round section use the round-column charts (Graphs 11 & 12 of Appendix A); the reading procedure is identical.

11. Worked Example 26.9 — Iterating Column Depth for a Target ρ_g

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

14-in-wide short tied rectangular column, bars on two end faces, P_u = 500 k, M_u = 250 ft-k, f'_c = 4 ksi, f_y = 60 ksi, target ρ_g ≈ 2%.

(62)

e = (12) (250) /500 = 6.00 in P_{n} = 500/0.65 = 769.2 k

Trial size	14 × 20	14 × 22	14 × 24
K_n = P_n/(f'_c·A_g)	0.687	0.624	0.572
R_n = P_n·e/(f'_c·A_g·h)	0.206	0.170	0.143
γ = (h − 5)/h	0.750	0.773	0.792
ρ_g (from chart)	0.0315	0.020	0.011

The 14 × 22 in section hits the target ρ_g ≈ 2%.

(63)

A_{s t} = (0.020) (14 \cdot 22) = 6.16 i n^{2}

Use 8 #8 bars (A_{st} = 6.28 in^{2}), 4 on each end face.

Design lesson. Doubling the depth cuts ρ_g nearly threefold — depth is the strongest lever for a moment-heavy column. Try to keep 1% ≤ ρ_g ≤ 4% (ACI 26.9.1.1) for buildable splices.

Example 26.9 — three trial depths plotted on one chart. As h grows, the point marches down-and-left, and ρg drops. The 14 × 22 point sits right on ρg ≈ 0.02 → target hit.

12. Worked Example 26.10 — Reverse Use: P_n at a Given Eccentricity

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

The 14 × 20 in tied column of Example 26.9 (Figure 10.20 of McCormac) is reinforced with 6 #10 bars on the two end faces. Compute the nominal load P_n that the column can support at an eccentricity of 10 in with respect to the x-axis. f'_c = 4 ksi, f_y = 60 ksi.

Geometry & reinforcement

(64)

A_{g} = 14 \cdot 20 = 280 in^{2} A_{st} = 6 \cdot 1.27 = 7.62 in^{2} (6 #10) \rho _{g} = A_{st}/A_{g} = 7.62/280 = 0.0272 \gamma = (20 − 2 \cdot 2.5)/20 = 0.75 e/h = 10/20 = 0.50

The trick. On the interaction diagram, R_n/K_n = (P_ne / f'_cA_gh) / (P_n/f'_cA_g) = e/h. So any (P_n, M_n) with the given eccentricity lies on the straight radial line from the origin with slope K_n/R_n = h/e = 2.0. Just walk that radial line until it crosses the ρ_g = 0.0272 curve.

Read from the chart

(65)

O n t h e " ba r so n 2 f a ces " c ha r t f or γ = 0.75, t h ee / h = 0.50 r a d ia l hi t s ρ_{g} \approx 0.027 a t a pp r o x ima t e l y K_{n} \approx 0.80, R_{n} \approx 0.40

K_{n} / R_{n} = 2.0 ✓

Back-substitute for Pn, Mn

(66)

P_{n} = K_{n} \cdot f_{c}^{'} \cdot A_{g} = 0.80 \cdot (4) (280) = 896 k M_{n} = R_{n} \cdot f_{c}^{'} \cdot A_{g} \cdot h = 0.40 \cdot (4) (280) (20) /12 = 747 f t - k C r oss - c h ec k : e = 12 \cdot M_{n} / P_{n} = 12 \cdot 747/896 = 10.0 in ✓

Design capacity (φ from strain check)

(67)

B ec a u se t h e p o in t f a l l s in t h eco m p r ess i o n - co n t r o l l e d r e g i o n o f t h ec ha r t (ab o v e ba l an ce), ϕ = 0.65 : ϕ P_{n} = 0.65 \cdot 896 = 582 k \cdot ϕ M_{n} = 0.65 \cdot 747 = 486 f t - k

Example 26.10 — draw the radial e/h = 0.50 line from the origin. Where it crosses the ρg = 0.027 curve → read Kn and Rn, then back out Pn and Mn.

Take-away. Interaction charts work both ways: forward (given loads → find reinforcement) and reverse (given reinforcement → find capacity at any eccentricity). The reverse workflow is the fastest way to check an existing column against a new load combination.

🧭 Master Recipe — Short-Column P + M Design

Compute factored demand P_u, M_u; find eccentricity e = M_u/P_u. Enforce e ≥ 0.10h (tied) or 0.05h (spiral).
Estimate a trial section from A_g ≈ P_u / (0.45·f'_c) for gravity-heavy or ≈ P_u / (0.55·f'_c) for moment-heavy columns.
Compute K_n, R_n. Pick the correct ACI SP-17 chart (bars on 2 vs 4 faces, correct γ).
Read ρ_g and pick bar sizes so 1% ≤ ρ_g ≤ 4–5%.
Verify with strain compatibility at the specific (P_u, M_u) point; confirm φ used matches the strain state.
Check the 0.80/0.85 cap — P_u ≤ φP_n,max.
Detail ties or spirals (ACI 25.7): tie size ≥ #3 for longitudinal ≤ #10, spacing ≤ min(16·d_b,long, 48·d_b,tie, least dim).
Slenderness check. Non-sway: kℓ_u/r ≤ 34 − 12·M₁/M₂. Sway: kℓ_u/r ≤ 22. Otherwise apply moment magnifiers δ_ns, δ_s.

⚠ Common Mistakes in RC Columns

Designing for P_u alone and ignoring M_u. Almost no column is purely axial.
Using P_o instead of 0.80·P_o (tied) / 0.85·P_o (spiral). The reduction covers accidental eccentricity even in "pure" axial.
Measuring eccentricity from the geometric center of an unsymmetrical section instead of the plastic centroid.
Picking the wrong SP-17 chart — bars on 2 faces vs 4 faces gives materially different ρ_g.
Assuming φ = 0.90 for a compression-controlled section. Read ε_t — most gravity columns are at φ = 0.65.
Adding compression steel where f'_s already carries — forgetting to subtract 0.85·f'_c·A'_s for concrete displaced by the compression bars.
Skipping the slenderness check. Long columns need moment magnifiers δ_ns, δ_s — a "short-column" design can grossly under-estimate M_u.

06

Professional practice, safety & ethics

RC column practice

Professional practice

Provide the full P–M interaction basis for each column; a single axial check is not a design.
Show ties/spirals, splice locations, and joint detailing — column capacity depends on confinement.
Coordinate column sizes with architecture early; late size reductions invalidate the interaction check.

Safety in design & construction

Spiral columns fail in a ductile, warned manner; tied columns can fail abruptly — confinement is a life-safety detail.
Columns are non-redundant gravity elements: one loss can trigger progressive collapse.
Slenderness effects (moment magnification) must not be dropped for tall unbraced columns.

Engineering ethics

Never reduce tie spacing requirements to ease placement.
Report deteriorated or spalled columns immediately; concealed corrosion has caused catastrophic collapses.
Do not certify a column retrofit without verifying the load path during the temporary condition.

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

Column reinforcement economics

Approach

Concrete carries axial load far more cheaply than steel: a larger section with ρ near 1–2% usually beats a small section at ρ = 4%.
Spiral columns cost more to fabricate than tied columns but earn a higher φ and much better ductility.
Repeating one column size over several floors saves formwork more than the steel it wastes.

Worked cost example — ρ = 1.5% vs ρ = 3.5% for Pu = 900 kip

Basis: f′c = 5 ksi

Line item	Qty	Rate	Cost
Option A — 20″×20″, Ast = 6.0 in²	0.082 ton rebar	$2,200	$180
Option A concrete + forms (12 ft)	1.23 yd³ installed	$460	$566
Option B — 16″×16″, Ast = 9.0 in²	0.123 ton rebar	$2,200	$271
Option B concrete + forms (12 ft)	0.79 yd³ installed	$460	$363
Estimated total			$1,380

Takeaway. Nearly a wash (≈$746 vs ≈$634) — choose on floor area, formwork repetition, and constructability, not the spreadsheet alone.

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(10 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 / 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

RC — Compression Members (Columns)

Compute φPn,max (short column)

Tied vs Spiral detailing

Enforce reinforcement ratio ρg

P–M interaction diagram

Slenderness (short vs slender)

Design a column (Pu, Mu → section)

Concrete crushing / rebar buckling

Slender column buckling

9.1 Introduction — Compression Members

9.2 Types of Columns

9.3 Axial Load Capacity of Columns

9.4 Failure of Tied and Spiral Columns

1. Axial Load + Bending — the Six Eccentricity Cases

2. The Plastic Centroid — Where Pn Must Act

3. Strain Compatibility → Pn and Mn

4. P–M Interaction Diagram — Reading the Column's Full Story

5. Three ACI 318-19 Modifications to the Nominal Curve

Example 26.3 — Design a square tied column (McCormac Ex 9.1)

Example 26.4 — Design a round spiral column: size and bars (McCormac Ex 9.2)

Example 26.5 — Spiral reinforcement design for Ex 26.4

6. Using the ACI Standardized Charts

7. Tied vs Spiral — Confinement & the Different φ

8. Worked Example 26.6 — Rectangular Tied Column, Bars on Two Faces

9. Worked Example 26.7 — Square Column, Bars on Four Faces

10. Worked Example 26.8 — Spiral Round Column

11. Worked Example 26.9 — Iterating Column Depth for a Target ρg

12. Worked Example 26.10 — Reverse Use: Pn at a Given Eccentricity

🧭 Master Recipe — Short-Column P + M Design

⚠ Common Mistakes in RC Columns

Tied column

Spiral column

Crushing failure

Slender bowing

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

Pure Pn

Accidental-e cap

φ

Design capacity

ρg check

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

Upload your worked solution (PDF)

2. The Plastic Centroid — Where P_n Must Act

3. Strain Compatibility → P_n and M_n

11. Worked Example 26.9 — Iterating Column Depth for a Target ρ_g

12. Worked Example 26.10 — Reverse Use: P_n at a Given Eccentricity