CELE Reinforced & Prestressed Concrete — Reinforced Concrete ColumnsRevision Notes
Final-week revision notes for Reinforced Concrete Columns. If you have already studied the full chapter, this page is your go-to refresher before sitting the CELE. Compact, high-yield, and aligned with what Professional Regulation Commission (PRC) — Board of Civil Engineering tests in the Reinforced & Prestressed Concrete subtest.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Reinforced & Prestressed Concrete subtest is marked as "Core" in the official pattern, and Reinforced Concrete Columns appears in position 4th of 7 in the CELE Reinforced & Prestressed Concrete review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Reinforced Concrete Columns - Revision Notes
Columns are vertical compression members that transmit loads from beams and slabs down to the foundation. In the PRC Civil Engineer Licensure Examination, RC columns consistently appear in both the morning (theory) and afternoon (design) sets. This chapter covers the two fundamental column types — tied and spiral — their axial-load capacity limits under NSCP 2015 (aligned with ACI 318), reinforcement ratio requirements, the axial–moment interaction diagram, and slenderness effects. Mastery of the capacity equations, the correct φ factors, and the cap multipliers (0.80 vs 0.85) is non-negotiable for exam success.
Sections
Exam Tips
- Memorize the two cap-and-phi combos: Tied → 0.80 × 0.65 = 0.52 Po; Spiral → 0.85 × 0.75 = 0.6375 Po.
- Quick check: spiral column capacity ≈ 23% higher than tied for identical dimensions and materials.
- In multiple-choice problems, if the question states 'spiral column,' immediately switch to φ = 0.75 and cap = 0.85.
Key Points
- A column is classified as a compression member whose primary load is axial compression, often combined with bending moment.
- Tied columns use individual rectangular or circular ties at regular intervals to confine the longitudinal bars and resist bar buckling.
- Spiral columns use a continuous helical spiral wire that provides superior confinement, making them more ductile (they give warning before collapse).
- Short columns: slenderness effects are negligible — failure is by material strength alone.
- Long (slender) columns: additional moments arise from lateral deflection (P-δ and P-Δ effects); moment magnification is required.
- Classification by control zone on the interaction diagram: compression-controlled (above balanced point) vs. tension-controlled (below balanced point).
- NSCP 2015 Section 410 governs the design of compression members.
Definitions
Term
Tied Column
Definition
A column confined laterally by individual rectangular, square, or circular ties placed at intervals along the column height.
Importance
Most common in Philippine construction; uses φ = 0.65 and a 0.80 axial-load cap — lower than spiral columns.
Term
Spiral Column
Definition
A column confined laterally by a continuous helical spiral of wire or bar wound around the longitudinal reinforcement.
Importance
More ductile due to continuous confinement; uses φ = 0.75 and a 0.85 axial-load cap. Preferred in seismic zones.
Term
Short Column
Definition
A column where slenderness ratio kℓu/r is low enough that secondary (P-delta) moments are negligible — typically kℓu/r ≤ 22 for unbraced frames or ≤ 34 − 12(M1/M2) ≤ 40 for braced frames.
Importance
Short columns can be designed using the direct capacity equations without moment magnification.
Term
Gross Area, Ag
Definition
Total cross-sectional area of the column including both concrete and steel (bh for rectangular, πD²/4 for circular).
Importance
Used in both the capacity formula and in computing the longitudinal steel ratio ρg.
Term
Core Area, Ach
Definition
The area of the concrete core of a spiral column measured to the outside edge of the spiral wire.
Importance
Used in computing the minimum spiral reinforcement ratio ρs.
Section Title
Column Classification and Behavior
Common Mistakes
- Confusing Ag (gross area) with Ach (core area) when computing spiral ratio.
- Using Ag instead of (Ag − Ast) for the concrete contribution in the capacity formula — the steel displaces concrete.
- Applying the wrong φ factor: tied columns use φ = 0.65, spiral columns use φ = 0.75.
- Forgetting that the 0.80/0.85 multiplier AND the φ factor both apply: ϕPn,max = 0.80(0.65)Po for tied.
Formulas
Example
400×400 mm tied column, 8-25mm bars (Ast=3927 mm²), f'c=28 MPa, fy=415 MPa: Ag=160,000 mm²; Po = 0.85(28)(160,000−3,927) + 415(3,927) = 0.85(28)(156,073) + 1,629,705 = 3,714,537 + 1,629,705 = 5,344,242 N ≈ 5,344 kN
Formula
Po = 0.85 f'c (Ag − Ast) + fy Ast
Variables
Po = nominal pure-axial capacity (N); f'c = concrete compressive strength (MPa); Ag = gross cross-sectional area (mm²); Ast = total area of longitudinal steel (mm²); fy = steel yield strength (MPa)
Application
Compute the theoretical upper-bound axial capacity of any short RC column before applying code reduction factors.
Example
Continuing the example: φPn,max = 0.80 × 0.65 × 5,344,242 = 0.52 × 5,344,242 = 2,779,006 N ≈ 2,779 kN
Formula
φPn,max = 0.80 φ Po (tied, φ = 0.65)
Variables
φPn,max = design axial capacity (N); 0.80 = eccentricity cap for tied columns; φ = 0.65 = strength reduction factor for tied columns
Application
Final design axial load limit for tied columns. Any factored axial load Pu must satisfy Pu ≤ φPn,max.
Example
Same section as spiral: φPn,max = 0.85 × 0.75 × 5,344,242 = 0.6375 × 5,344,242 = 3,406,954 N ≈ 3,407 kN. The spiral column carries ~23% more load.
Formula
φPn,max = 0.85 φ Po (spiral, φ = 0.75)
Variables
φPn,max = design axial capacity (N); 0.85 = eccentricity cap for spiral columns; φ = 0.75 = strength reduction factor for spiral columns
Application
Final design axial load limit for spiral columns.
Exam Tips
- Board shortcut: For tied → multiply Po by 0.52; for spiral → multiply by 0.6375.
- When asked 'how much more capacity does a spiral column have compared to a tied column?' → ratio is 0.6375/0.52 ≈ 1.226, so about 22–23% more.
- Always check whether the problem says 'tied' or 'spiral' before selecting φ.
Key Points
- The nominal pure-axial strength Po represents the theoretical maximum if the load were applied with zero eccentricity.
- Code reduces Po by a cap factor (0.80 or 0.85) to account for unavoidable minimum eccentricity in real columns.
- The design (factored) axial capacity is φPn,max = cap × φ × Po.
- Spiral columns get a higher cap (0.85 vs 0.80) because their confinement delays core crushing.
- Spiral columns also get a higher φ (0.75 vs 0.65) because their failure mode is more ductile.
- Both advantages compound: the spiral column is significantly stronger and safer under the same geometry.
Definitions
Term
Nominal Axial Capacity (Po)
Definition
The theoretical maximum axial load a column cross-section can carry assuming perfectly concentric loading and no eccentricity.
Importance
Base value from which all code-reduced design capacities are computed.
Term
Strength Reduction Factor (φ)
Definition
A factor less than 1.0 applied to nominal strength to account for material variability, construction tolerances, and mode of failure. For columns: φ = 0.65 (tied), φ = 0.75 (spiral) per NSCP 2015.
Importance
Critical exam distinction: columns are NOT beam-flexure (φ = 0.90); they are compression-controlled.
Section Title
Axial Capacity of Short Columns
Common Mistakes
- Using φ = 0.90 (beam value) for columns — this is the most common and costliest error.
- Using Ag instead of (Ag − Ast) in the concrete term; the steel bars physically displace concrete.
- Applying only the φ factor without the cap multiplier (0.80 or 0.85), or vice versa.
- Computing Po but forgetting to multiply by both 0.80 and 0.65 for tied columns (the product is 0.52).
Formulas
Example
400×400 mm column, 8-25mm bars (Ast = 3,927 mm²): ρg = 3,927/160,000 = 0.0245. Since 0.01 ≤ 0.0245 ≤ 0.08 → OK.
Formula
ρg = Ast / Ag
Variables
ρg = longitudinal steel ratio (dimensionless); Ast = total area of longitudinal steel (mm²); Ag = gross cross-sectional area (mm²). Code limits: 0.01 ≤ ρg ≤ 0.08
Application
Verify that the selected steel area satisfies code minimum and maximum limits.
Example
400 mm diameter column, 40 mm cover: core diameter = 400 − 2(40) = 320 mm; Ach = π(320)²/4 = 80,425 mm²; Ag = π(400)²/4 = 125,664 mm². With f'c = 28 MPa, fyt = 415 MPa: ρs,min = 0.45(125,664/80,425 − 1)(28/415) = 0.45(0.5628)(0.0675) = 0.01708 ≈ 1.71%
Formula
ρs,min = 0.45 (Ag/Ach − 1)(f'c / fyt)
Variables
ρs = volumetric spiral ratio = (volume of spiral per turn)/(volume of core per turn); Ag = gross column area (mm²); Ach = core area measured to outside of spiral (mm²); f'c = concrete strength (MPa); fyt = spiral yield strength (MPa), capped at 700 MPa in this formula
Application
Find the minimum spiral wire size/pitch for a spiral column. The spiral must be strong enough to compensate for shell spalling.
Example
Column with 25 mm longitudinal bars and 10 mm ties: s ≤ min(16×25, 48×10, 400) = min(400, 480, 400) = 400 mm
Formula
Tie spacing ≤ min(16db,long ; 48db,tie ; least column dimension)
Variables
db,long = diameter of longitudinal bar (mm); db,tie = diameter of tie bar (mm); least column dimension = smaller side of rectangular column (mm)
Application
Determine maximum allowable spacing of ties along column height.
Exam Tips
- In spiral ratio problems, always compute Ach using the core diameter = D − 2(cover), measured to the OUTSIDE of the spiral.
- ρg = 0.01 to 0.08 is a hard code limit — even if a column has enough strength with ρg = 0.005, it still fails the code check.
- Common board problem: compute Ast,min and Ast,max for a given column size, then check if a specific bar arrangement is acceptable.
Key Points
- Longitudinal steel ratio: 0.01 ≤ ρg = Ast/Ag ≤ 0.08. This range is mandated by NSCP 2015 Section 410.
- Minimum ρg = 0.01 (1%) ensures the column can resist unintended bending and prevents creep redistribution problems.
- Maximum ρg = 0.08 (8%) prevents congestion of bars; in practice, ρg ≤ 0.04 is preferred at lap-splice locations.
- Minimum number of longitudinal bars: 4 bars for tied rectangular/square columns, 6 bars for spiral/circular columns.
- Ties: minimum diameter is No. 10 (10 mm) for longitudinal bars ≤ 32 mm; spacing ≤ 16 × longitudinal bar diameter, ≤ 48 × tie diameter, and ≤ least column dimension.
- Spiral reinforcement: must meet a minimum volumetric ratio ρs to ensure the spiral replaces the load-carrying capacity lost when the concrete shell spalls.
Definitions
Term
Longitudinal Steel Ratio (ρg)
Definition
Ratio of total longitudinal steel area to gross cross-sectional area of the column.
Importance
Must fall between 1% and 8% per NSCP 2015. Values outside this range are code violations regardless of stress calculations.
Term
Spiral Ratio (ρs)
Definition
Volumetric ratio of spiral reinforcement to the volume of the concrete core enclosed by the spiral.
Importance
Ensures the spiral provides enough confinement to maintain post-spalling ductility. The minimum is explicitly given by the NSCP spiral ratio formula.
Section Title
Reinforcement Limits
Common Mistakes
- Using fyt > 700 MPa in the spiral ratio formula — the code caps fyt at 700 MPa for this equation even if actual yield is higher.
- Measuring Ach to the center of the spiral wire instead of to the outside of the spiral.
- Forgetting the minimum bar count: 6 bars for spiral columns (not 4).
- Applying the 4-bar minimum to circular tied columns — circular columns need at minimum 6 bars when spirally reinforced.
Formulas
Example
Column with d = 350 mm, fy = 415 MPa: cb = (600 × 350)/(600 + 415) = 210,000/1,015 = 206.9 mm
Formula
At balanced point: c_b = (600 × d) / (600 + fy)
Variables
cb = balanced neutral axis depth (mm); d = effective depth to tension steel (mm); fy = yield strength of steel (MPa); 600 is derived from εcu = 0.003 and Es = 200,000 MPa
Application
Locate the balanced neutral axis depth to find the balanced point on the interaction diagram.
Example
Conceptual: at the balanced point, the compression block depth a = β1 cb; compute each steel layer's stress from strain compatibility, then sum forces.
Formula
Pb = 0.85 f'c β1 cb b + (compression steel forces) − (tension steel forces)
Variables
Pb = nominal axial load at balanced condition (N); β1 = stress block factor (0.85 for f'c ≤ 28 MPa, reduced by 0.05 per 7 MPa above 28 MPa, min 0.65); cb = balanced neutral axis depth (mm); b = column width (mm)
Application
Compute the axial load at the balanced point of the interaction diagram.
Exam Tips
- Board shortcut: if asked whether a (Pu, Mu) point is 'safe,' check if it falls inside the φPn − φMn envelope.
- The maximum moment on the interaction diagram occurs at the balanced point, NOT at pure flexure.
- For exam problems giving a pre-drawn interaction diagram, simply plot the point and state inside (safe) or outside (unsafe).
Key Points
- Real columns almost always carry both axial load P and bending moment M simultaneously — pure axial load is a theoretical idealization.
- The interaction diagram is the complete strength envelope: each point (Pn, Mn) on the curve is a combination of axial load and moment that the section can just sustain at nominal strength.
- Any factored (Pu, Mu) combination that plots INSIDE the φ-reduced interaction diagram is acceptable (adequate capacity).
- Any point OUTSIDE the φ-reduced diagram represents inadequate capacity — the section must be redesigned.
- Key points on the diagram: (1) Pure axial: (Po, 0) at the top; (2) Balanced point: (Pb, Mb), largest moment capacity; (3) Pure flexure: (0, Mn) at the bottom.
- Above the balanced point: compression-controlled failure (concrete crushes before steel yields) — more brittle.
- Below the balanced point: tension-controlled failure (steel yields before concrete crushes) — more ductile.
- The balanced point is defined by simultaneous concrete crushing (εc = 0.003) and tension steel yielding (εs = εy = fy/Es).
- φ transitions from 0.65 (compression-controlled) to 0.90 (tension-controlled) as the section moves from above to below the balanced point.
Definitions
Term
Interaction Diagram
Definition
A plot of all (Pn, Mn) combinations at nominal strength that a given column section can resist. It is the complete strength envelope for combined axial load and bending.
Importance
The primary design and check tool for columns under combined loading. Every board problem on column moment capacity uses this concept.
Term
Balanced Point (Pb, Mb)
Definition
The specific (Pn, Mn) point where concrete extreme fiber simultaneously reaches εcu = 0.003 AND the tension steel just reaches yield strain εy = fy/Es.
Importance
The balanced point marks the boundary between compression-controlled and tension-controlled behavior; it is also the point of maximum moment on the interaction diagram.
Term
Compression-Controlled Zone
Definition
Region of the interaction diagram above the balanced point where failure initiates by concrete crushing. εs < εy at failure. φ = 0.65 for tied columns.
Importance
Most column designs fall in this zone for high axial loads.
Term
Tension-Controlled Zone
Definition
Region of the interaction diagram below the balanced point where failure initiates by steel yielding. εs > 0.005 at failure. φ = 0.90.
Importance
Applies to columns with low axial load and high moment — essentially behaving like beams.
Section Title
Axial–Moment Interaction Diagram
Common Mistakes
- Treating columns as pure axial members and ignoring moment — even small eccentricities (code minimum = 15 mm or 0.1h) can control design.
- Plotting (Pu, Mu) on the nominal diagram instead of the φ-reduced diagram.
- Confusing Pb (balanced axial) with the maximum usable axial load — they are different quantities.
- Incorrectly applying a single φ across the entire interaction diagram — φ varies from 0.65 to 0.90 depending on the zone.
Formulas
Example
Rectangular 400×400 mm column, ℓu = 4,000 mm, both ends fixed to rigid beams (assume k = 0.70): r = 0.30(400) = 120 mm; kℓu/r = 0.70(4,000)/120 = 23.3. For braced frame, limit = 34 − 12(M1/M2). If M1/M2 = 0.5, limit = 34 − 6 = 28. Since 23.3 < 28 → short column.
Formula
Slenderness ratio = kℓu / r
Variables
k = effective length factor (depends on end conditions); ℓu = unsupported (clear) column height (mm); r = radius of gyration (mm): 0.30h for rectangular, 0.25D for circular
Application
Determine if a column is short (slenderness effects negligible) or slender (moment magnification required).
Example
EI = 0.4 Ec Ig for columns (simplified). Used as denominator of the magnifier formula.
Formula
Pc = π² EI / (kℓu)²
Variables
Pc = Euler critical buckling load (N); E = modulus of elasticity of concrete (MPa); I = effective moment of inertia (mm⁴); kℓu = effective length (mm)
Application
Compute the critical buckling load needed for the moment magnifier δns.
Example
If Pu = 1,500 kN, Pc = 8,000 kN, Cm = 0.80: δns = 0.80/(1 − 1500/(0.75×8000)) = 0.80/(1 − 0.25) = 0.80/0.75 = 1.067. Mc = 1.067 × M2.
Formula
δns = Cm / (1 − Pu/(0.75 Pc)) ≥ 1.0
Variables
δns = moment magnification factor for braced frames; Cm = equivalent moment correction factor = 0.6 + 0.4(M1/M2) for members with no transverse loads (≥ 0.40); Pu = factored axial load (N); Pc = critical load (N)
Application
Amplify the design moment to account for P-δ effects in slender braced columns.
Exam Tips
- Quick check for short column (braced, equal end moments): kℓu/r ≤ 34 − 12(1.0) = 22.
- r ≈ 0.30h for rectangular section is a very common board exam approximation — memorize it.
- If the problem does not specify end conditions, assume k = 1.0 for braced and k = 1.2 for unbraced as conservative estimates.
Key Points
- Slender columns deflect laterally under load, creating additional eccentricity and amplifying the moment — this is the P-δ (member curvature) and P-Δ (story drift) effect.
- Braced (non-sway) frames: column ends are prevented from significant relative lateral displacement. Slenderness limit: kℓu/r ≤ 34 − 12(M1/M2) but ≤ 40.
- Unbraced (sway) frames: column ends can move laterally relative to each other. Slenderness limit: kℓu/r ≤ 22.
- r = radius of gyration: r = 0.30h for rectangular sections, r = 0.25D for circular sections (approximate).
- If the slenderness ratio exceeds the short-column limit, the design moment must be amplified using the moment magnifier δ.
- Moment magnifier for braced frames: δns = Cm / (1 − Pu/(0.75 Pc)) ≥ 1.0, where Pc = π²EI/(kℓu)².
- Effective length factor k: k = 1.0 for both ends pinned (braced), k < 1.0 for fixed-fixed (braced), k = 1.0–2.0 for sway frames.
- Magnified design moment: Mc = δ × M2, where M2 is the larger factored end moment.
Definitions
Term
Effective Length Factor (k)
Definition
A factor that converts the actual unsupported column length into an equivalent pin-ended length for buckling analysis. k depends on the rotational and translational restraints at column ends.
Importance
k = 1.0 for pin-pin braced; k = 0.5 for fixed-fixed braced; k = 2.0 for fixed-free (cantilever). Wrong k selection leads to wrong slenderness classification.
Term
Moment Magnifier (δ)
Definition
A multiplier greater than or equal to 1.0 applied to the first-order design moment to account for the additional moment caused by lateral deflection of the column.
Importance
Required for all slender columns. Neglecting it underestimates the design moment and leads to unconservative (unsafe) design.
Term
Unsupported Length (ℓu)
Definition
The clear distance between lateral supports (floors, beams, or other restraints) along the column height.
Importance
Used directly in the slenderness ratio calculation. Using the wrong length (e.g., center-to-center instead of clear) gives incorrect results.
Section Title
Slenderness Effects and Moment Magnification
Common Mistakes
- Using r = 0.30h for circular columns — should be r = 0.25D.
- Using center-to-center story height instead of the clear unsupported length ℓu.
- Treating a sway frame as braced — sway frames have a more stringent slenderness limit (22 instead of 34−12M1/M2).
- Setting δ < 1.0 when the formula gives a value less than 1.0 — the code requires δ ≥ 1.0.
- Forgetting to check both P-δ (member) and P-Δ (story) effects for sway frames.
Connections
- RC Beams (Flexure): Columns under low axial load and high moment behave like beams; the bottom portion of the interaction diagram merges with beam flexural theory. The same Whitney stress block and strain compatibility apply.
- Foundation Design: Column axial loads and moments are the primary design inputs for isolated footings, combined footings, and pile caps. Column φPn,max directly sets the maximum load that the foundation must support.
- Structural Analysis / Load Combinations: Pu and Mu for column design come from NSCP 2015 load combinations (1.2D + 1.6L, 1.2D + 1.0E + 1.0L, etc.). The interaction diagram check requires the most critical (Pu, Mu) pair from all governing combinations.
- Seismic Design (NSCP 2015 Section 418): In high seismic zones, columns in special moment frames must meet additional confinement requirements (closely spaced hoops/spirals) over potential plastic hinge zones — directly related to spiral ratio and tie spacing concepts.
- Steel Columns (AISC 360 / NSCP for Steel): Both steel and RC columns use effective length kℓu and slenderness ratio for stability checks. The interaction equation format (P/Pc + M/Mc ≤ 1) parallels the RC interaction diagram concept.
- Prestressed Concrete: Prestressed columns (e.g., precast piles) also use interaction diagrams, but the prestress force shifts the axial axis and changes the balanced point location.
- Construction Materials: f'c and fy selection directly affects Po and all downstream capacities. Philippine standard grades (f'c = 21, 28, 35 MPa; fy = 275, 415 MPa) are the most common exam values.
- RA 544 (Civil Engineering Law): Licensed Civil Engineers in the Philippines are legally responsible for structural design adequacy. Column design errors are a direct professional liability under RA 544.
Exam Strategy
For PRC board exam problems on RC columns: (1) FIRST identify the column type — 'tied' or 'spiral' — to immediately lock in the correct φ and cap factor. (2) ALWAYS use the net concrete area (Ag − Ast) in the Po formula — never Ag alone. (3) Apply the combined multiplier: 0.52 for tied, 0.6375 for spiral. (4) Check ρg bounds (1%–8%) as a routine step even when not explicitly asked. (5) For spiral columns, use Ach measured to the OUTSIDE of the spiral wire. (6) For interaction diagram problems, plot (Pu, Mu) on the φ-reduced diagram — inside = safe, outside = unsafe. (7) For slenderness, compute kℓu/r and compare to the appropriate limit (22 for unbraced, 34 − 12M1/M2 ≤ 40 for braced); only if slender do you need moment magnification. (8) In time-limited exams, recall the shortcut: spiral column capacity ≈ 23% higher than tied for identical sections. (9) Watch for trap questions that give a large ρg (e.g., 10%) — immediately flag as code violation before computing capacity. (10) Cite NSCP 2015 Section 410 (compression members) in any written justification on the board exam.
Quick Review Questions
A 400 × 500 mm tied column has Ast = 4,800 mm², f'c = 30 MPa, fy = 415 MPa. What is the design axial capacity φPn,max?
Ag = 400×500 = 200,000 mm². Po = 0.85(30)(200,000 − 4,800) + 415(4,800) = 0.85(30)(195,200) + 1,992,000 = 4,977,600 + 1,992,000 = 6,969,600 N. φPn,max = 0.80 × 0.65 × 6,969,600 = 0.52 × 6,969,600 ≈ 3,624,192 N ≈ 3,624 kN. (Note: recompute carefully — 0.85×30×195,200 = 4,977,600; +1,992,000 = 6,969,600 N; ×0.52 = 3,624 kN. Corrected answer: ≈ 3,624 kN.)
What are the two φ values for columns, and which column type uses each?
Per NSCP 2015 (ACI 318), tied columns are compression-controlled with φ = 0.65, reflecting their less ductile behavior. Spiral columns, being more ductile due to confinement, receive a higher φ = 0.75. This is a critical exam distinction.
A 500 mm diameter spiral column has the same Po as a 500 mm diameter tied column. By approximately what percentage does the spiral column's φPn,max exceed the tied column's?
Tied: φPn,max = 0.80 × 0.65 × Po = 0.52 Po. Spiral: φPn,max = 0.85 × 0.75 × Po = 0.6375 Po. Ratio = 0.6375/0.52 = 1.226. The spiral column carries about 22.6% more factored axial load for identical geometry and materials.
For a tied column, what is the valid range of the longitudinal steel ratio ρg?
NSCP 2015 mandates a minimum of 1% to ensure adequate resistance to unintended bending and prevent creep redistribution, and a maximum of 8% to prevent steel congestion. In practice, ρg ≤ 4% is preferred at lap splice locations.
What is the minimum number of longitudinal bars for (a) a tied rectangular column and (b) a spiral column?
NSCP 2015 requires at least 4 longitudinal bars for tied columns (one at each corner of a rectangular arrangement) and at least 6 bars for spiral columns to ensure uniform confinement around the circumference.
At the balanced point of a column interaction diagram, what two strains occur simultaneously?
The balanced condition is the unique loading combination where concrete crushes (εcu = 0.003) simultaneously as the tension steel reaches its yield strain. This point marks the boundary between compression-controlled (above) and tension-controlled (below) behavior and also corresponds to the maximum moment on the interaction diagram.
A 350 mm square column (k=1.0, braced frame) has ℓu = 4,500 mm and equal end moments (M1/M2 = 1.0). Is it a short column?
r = 0.30 × 350 = 105 mm. kℓu/r = 1.0 × 4,500/105 = 42.86. Short column limit for braced frame = 34 − 12(M1/M2) = 34 − 12(1.0) = 22. Since 42.86 > 22, this is actually a SLENDER column — moment magnification is required. (This is a trick question to test careful application of the limit.)
In the spiral ratio formula ρs,min = 0.45(Ag/Ach − 1)(f'c/fyt), what is the maximum value of fyt permitted in this calculation?
NSCP 2015 caps the yield strength of spiral reinforcement at 700 MPa in the minimum spiral ratio formula, regardless of the actual yield strength of the spiral wire. This prevents over-reliance on very high-strength spiral steel that may not fully develop its strength within the code model.
What does it mean when a factored load combination (Pu, Mu) plots OUTSIDE the φ-reduced interaction diagram?
The φ-reduced interaction diagram is the design strength envelope. Any (Pu, Mu) point must plot inside (or on) this envelope to be acceptable. A point outside means the section is overstressed and must be redesigned with larger dimensions, higher f'c, or more steel.
What is the formula for the balanced neutral axis depth cb, and what does it represent?
This formula comes from similar triangles in the strain diagram: εcu/(εcu + εy) = cb/d. With εcu = 0.003 and Es = 200,000 MPa (so εy = fy/200,000), substituting: cb/d = 0.003/(0.003 + fy/200,000) = 600/(600 + fy). It locates the neutral axis at the balanced condition.
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