Reinforced Concrete Columns - Study Notes
Columns are vertical structural members that carry primarily axial compression, often combined with bending moments from lateral loads or eccentricity. They are critical load-bearing elements in buildings, bridges, and other structures. This chapter covers the design and analysis of reinforced concrete columns under the NSCP 2015 (National Structural Code of the Philippines) and ACI 318 standards. We focus on short tied and spiral columns, axial load capacity, reinforcement limits, interaction diagrams (which govern combined compression and bending), and slenderness effects. Understanding column behavior is essential for safe and economical structural design and is a core topic in the PRC Civil Engineer Licensure Examination.
1. Introduction to Reinforced Concrete Columns
A reinforced concrete column is a structural member designed to withstand primarily compressive forces, with the concrete resisting compression and the steel reinforcement helping to resist both compression and any bending moments. Columns differ fundamentally from beams: they carry vertical loads along their axis (axial load), whereas beams carry lateral loads (transverse loads). In practice, real columns also experience moments due to lateral loads (wind, seismic), construction inaccuracies, or intentional eccentricity.
Columns are classified into two main types based on transverse reinforcement:
Tied Columns: These use closely spaced steel ties (or stirrups) to confine the core concrete. Ties prevent outward buckling of the main longitudinal bars and provide limited lateral confinement. Tied columns have a reduction factor φ = 0.65 (compression-controlled) and are more common in practice due to lower cost.
Spiral Columns: These use a continuous helical spiral of steel wire or bar wound around the core concrete. Spirals provide superior confinement, leading to better ductility and higher load capacity. Spiral columns have φ = 0.75 and can sustain higher loads but cost more. They are often used where high ductility or high compression is critical.
The NSCP 2015 and ACI 318 recognize these differences and reward spiral columns with both a higher capacity cap and a higher reduction factor. Understanding when to use each type is important for economical design.
Examples
Identifying Column Type
A designer must choose between a tied and spiral column for a high-rise building in Metro Manila subject to typhoon winds. The spiral column, though costlier, offers better ductility for seismic/wind loads and allows for smaller cross-section due to higher φ. This is typical in high-rise practice.
Key points
- Columns primarily carry axial compression; moments are secondary but important.
- Tied columns: φ = 0.65, capacity cap = 0.80φPₒ
- Spiral columns: φ = 0.75, capacity cap = 0.85φPₒ
- Spirals provide superior confinement and ductility
- Both types are analyzed using similar principles but with different limits
2. Axial Capacity of Short Columns
For a short column (slenderness ignored), the nominal axial compression capacity Pₙ or Pₒ is based on the principle that the concrete and steel act together. The concrete resists compression on its net area (gross area minus steel area), and the steel resists compression on its full area.
Pure Axial Nominal Strength (Gross Axial Strength):
Pₒ = 0.85f'c(Ag − Ast) + fyAst
where:
- Pₒ = nominal (unfactored) axial load capacity
- f'c = concrete compressive strength (MPa)
- Ag = gross cross-sectional area (mm²)
- Ast = total area of longitudinal steel bars (mm²)
- fy = yield strength of steel (MPa, typically 415 MPa in the Philippines)
The factor 0.85 on concrete accounts for the non-uniform stress distribution and the fact that concrete does not reach peak stress simultaneously everywhere.
Design Axial Capacity (Factored):
The code limits the usable design axial capacity to account for unavoidable eccentricity (no column is perfectly axially loaded in practice):
For tied columns: φPn,max = 0.80 × φ × Pₒ = 0.80 × 0.65 × Pₒ = 0.52Pₒ
For spiral columns: φPn,max = 0.85 × φ × Pₒ = 0.85 × 0.75 × Pₒ = 0.6375Pₒ
This means a spiral column can carry about 22.6% more load than a tied column of identical dimensions and materials. The difference comes from two sources: (1) the capacity cap increases from 0.80 to 0.85, and (2) φ increases from 0.65 to 0.75.
Physical Interpretation:
The reduction factors (0.80 and 0.85) reflect the reality that columns always have some eccentricity, and this eccentricity produces bending that reduces the axial capacity. Spiral columns, being more ductile, tolerate eccentricity better, hence the higher cap. The reduction factors φ (0.65 and 0.75) are typical for compression-controlled behavior in reinforced concrete.
Examples
Example 1: Tied Column Axial Capacity
Solution:
Step 1: Calculate gross area and net concrete area.
Ag = 400 × 400 = 160,000 mm²
Ag − Ast = 160,000 − 3,927 = 156,073 mm²
Step 2: Calculate nominal axial strength Pₒ.
Pₒ = 0.85 × 28 × 156,073 + 415 × 3,927
Pₒ = 3,714,537 + 1,629,705 = 5,344,242 N = 5,344.2 kN
Step 3: Apply the tied-column cap and reduction factor.
φPn,max = 0.80 × 0.65 × 5,344.2 = 0.52 × 5,344.2 = 2,779 kN
Answer: The design axial capacity is 2,779 kN. A factored load Pu up to 2,779 kN can safely be supported (assuming short-column assumption holds).
A 400 mm × 400 mm tied square column is reinforced with 8 bars of 25 mm diameter (Ast = 8 × π × 12.5² = 3,927 mm²). Concrete f'c = 28 MPa, steel fy = 415 MPa. Find the design axial capacity.
Example 2: Same Column as Spiral Type
Solution:
The nominal strength Pₒ remains the same (same materials and areas):
Pₒ = 5,344.2 kN
For a spiral column:
φPn,max = 0.85 × 0.75 × 5,344.2 = 0.6375 × 5,344.2 = 3,407 kN
Comparison:
Spiral capacity / Tied capacity = 3,407 / 2,779 = 1.226 or +22.6%
Answer: The spiral column can carry 3,407 kN, which is 628 kN (22.6%) more than the tied column. This premium reflects the superior confinement and ductility of the spiral.
Redesign the Example 1 column as a spiral column with the same cross-section and reinforcement. Compare the capacities.
Example 3: Steel Ratio Verification
Solution:
Longitudinal steel ratio:
ρg = Ast / Ag = 3,927 / 160,000 = 0.0245 or 2.45%
Code limits (NSCP 2015, Section 410.8.1):
0.01 ≤ ρg ≤ 0.08
Verification:
0.01 = 0.01 < 0.0245 < 0.08 = 0.08 ✓
Answer: The steel ratio is within the permitted range. Note: In regions with lap splices, ρg should not exceed about 0.04 due to constructability, but our 2.45% is acceptable.
Verify that the steel ratio in Examples 1 and 2 complies with NSCP 2015 and ACI 318 limits.
Example 4: Determining Required Concrete Strength
Solution:
Step 1: Given.
Ag = 500 × 500 = 250,000 mm²
Ast = 0.025 × 250,000 = 6,250 mm²
Ag − Ast = 250,000 − 6,250 = 243,750 mm²
Pu = 4,000 kN (factored design load)
fy = 415 MPa
Column type: tied, so φ = 0.65, cap = 0.80
Step 2: Set up the equation.
Pu = φPn,max = 0.80 × φ × Pₒ
4,000,000 = 0.80 × 0.65 × [0.85f'c × 243,750 + 415 × 6,250]
4,000,000 = 0.52 × [206,906.25f'c + 2,593,750]
4,000,000 = 107,591.25f'c + 1,348,750
2,651,250 = 107,591.25f'c
f'c = 24.65 MPa
Answer: A minimum f'c of 25 MPa (rounding up to the nearest standard) is required. This is a moderate strength and achievable in the Philippines.
A 500 mm × 500 mm tied column must support a factored load of 4,000 kN. It will have ρg = 0.025 (Ast = 6,250 mm²). What minimum f'c is required?
Key points
- Pₒ combines concrete capacity (0.85f'c) on net area plus steel capacity (fy) on full area
- Design capacity caps are 0.52Pₒ (tied) and 0.6375Pₒ (spiral)
- Both caps include a φ factor (0.65 tied, 0.75 spiral)
- The 0.85 factor on concrete accounts for non-uniform stress and material variability
- Short column assumption: slenderness kℓu/r is below the code limit (≈22–40 depending on support)
3. Reinforcement Limits and Spacing Requirements
The NSCP 2015 and ACI 318 impose strict limits on longitudinal and transverse reinforcement in columns to ensure adequate performance, constructability, and ductility.
Longitudinal (Main) Steel Reinforcement:
Steel Ratio Limits:
- Minimum: ρg,min = 0.01 (ensures strength and controls shrinkage cracking)
- Maximum: ρg,max = 0.08 (ensures constructability; above this, placing and vibrating concrete becomes difficult)
- Practical note: In regions with lap splices, keep ρg ≤ 0.04 to avoid congestion
where ρg = Ast / Ag
Minimum Number of Bars:
- Tied rectangular columns: minimum 4 bars (one near each corner)
- Spiral or circular columns: minimum 6 bars (ensures adequate perimeter support)
- Bars must be distributed evenly around the perimeter to resist moment in any direction
Bar Diameter and Spacing:
- No specific lower limit on bar diameter, but typically ≥ 16 mm in practice
- Longitudinal bars must be placed no more than 150 mm apart around the perimeter (NSCP 2015, Section 410.8.4)
- In tied columns, bars at corners and edges are mandatory; intermediate bars may be placed between corners
Transverse Reinforcement (Ties or Spirals):
Tied Columns:
Ties are rectangular closed loops of steel around the core. The tie size and spacing are crucial for confining the concrete and preventing buckling of longitudinal bars.
Tie Size:
- Minimum: #10 (10 mm diameter) for longitudinal bars ≤ 32 mm
- Minimum: #13 (13 mm diameter) for longitudinal bars > 32 mm
Spacing (per NSCP 2015, Section 410.8.5):
The spacing of ties shall not exceed the smallest of:- 16 times the longitudinal bar diameter: 16db
- 48 times the tie diameter: 48dt
- The least dimension of the column cross-section
- 300 mm
Example: For #25 longitudinal bars and #10 ties in a 400 mm column:
- 16 × 25 = 400 mm
- 48 × 10 = 480 mm
- 400 mm (least dimension)
- 300 mm ← controls (minimum, 300 mm spacing)
- Tie Configuration: Ties must form closed loops and be anchored in the core concrete. Corner and alternate bar requirements apply (NSCP 2015, Section 410.8.5).
Spiral Columns:
A spiral is a continuous helix of steel wire or bar with a diameter usually 8–13 mm, wrapped around the column core.
Spiral Pitch (p) and Ratio (ρs):
The spacing between successive spiral turns (pitch) directly determines confinement effectiveness. The spiral volume ratio is:ρs = 4Asp / (s × Dch)
where:
- Asp = cross-sectional area of spiral steel (mm²)
- s = pitch (vertical distance between consecutive spiral turns, mm)
- Dch = diameter of concrete core measured to outside of spiral (mm)
Minimum Spiral Ratio (NSCP 2015, Section 410.8.6.4):
ρs ≥ 0.45 × (Ag / Ach − 1) × (f'c / fyt)
where:
- Ag = gross column area
- Ach = area of core concrete (to outside of spiral), mm²
- f'c = concrete strength, MPa
- fyt = spiral steel yield strength, capped at 700 MPa in this equation (even if fy is higher)
The cap at 700 MPa prevents excessive dependence on very high-strength spiral steel, as confinement is primarily a geometric/volume effect.
- Maximum Spiral Pitch: Not to exceed 75 mm or 1/6 of column diameter (whichever is smaller) to ensure adequate confinement.
Physical Interpretation:
The minimum spiral ratio formula reflects that confinement benefit scales with the ratio (Ag/Ach − 1), which measures how much steel is outside the core. Larger columns need proportionally more spiral to provide equivalent confinement. The 0.45 factor and the cap at 700 MPa are empirical adjustments based on test data.
Examples
Example 5: Tied-Column Tie Spacing Design
Solution:
Step 1: Determine tie size.
Longitudinal bars are #32 (32 mm diameter), so minimum tie diameter is #13 (13 mm).
Choose #10 ties (dt = 10 mm) for economy. (Some codes allow #10 for bars ≤ 32 mm; verify local practice.)
For conservative design, use #13 ties (dt = 13 mm).
Step 2: Calculate maximum spacing (least of four criteria).
- 16db = 16 × 32 = 512 mm
- 48dt = 48 × 10 = 480 mm (if #10) or 48 × 13 = 624 mm (if #13)
- Least column dimension = 400 mm
- 300 mm (code limit)
If #10 ties: spacing ≤ min(512, 480, 400, 300) = 300 mm ✓
If #13 ties: spacing ≤ min(512, 624, 400, 300) = 300 mm ✓
Answer: Use #10 or #13 ties spaced at 300 mm c/c. The 300 mm code limit controls. Ties must form closed loops around all longitudinal bars.
Design ties for a 400 mm × 500 mm rectangular tied column with 8 #32 longitudinal bars. Use fy = 415 MPa.
Example 6: Spiral Column Minimum Spiral Ratio
Solution:
Step 1: Calculate gross and core areas.
Diameter = 400 mm
Ag = π × (400/2)² = π × 200² = 125,664 mm²
Core diameter (to outside of spiral):
Dch = 400 − 2 × 40 = 320 mm
Ach = π × (320/2)² = π × 160² = 80,425 mm²
Step 2: Calculate minimum spiral ratio.
ρs,min = 0.45 × (Ag / Ach − 1) × (f'c / fyt)
Note: fyt is capped at 700 MPa, but our fy = 415 MPa, so use fyt = 415 MPa.
ρs,min = 0.45 × (125,664 / 80,425 − 1) × (28 / 415)
ρs,min = 0.45 × (1.5623 − 1) × 0.0675
ρs,min = 0.45 × 0.5623 × 0.0675
ρs,min = 0.01706
Step 3: Calculate pitch from spiral ratio.
ρs = 4Asp / (s × Dch)
For a #10 spiral: Asp = π × (10/2)² = 78.54 mm²
s = 4Asp / (ρs,min × Dch)
s = 4 × 78.54 / (0.01706 × 320)
s = 314.16 / 5.459
s = 57.5 mm ≈ 58 mm
Step 4: Check maximum pitch limit.
Max pitch = min(75 mm, Dch/6) = min(75, 320/6) = min(75, 53.3) = 53.3 mm
Our calculated s = 58 mm exceeds the maximum of 53.3 mm, so we must use s = 53 mm to be conservative.
Alternatively, use a slightly larger spiral bar or tighter spacing.
Answer: Use a #10 spiral with pitch s = 50–53 mm (tighter than the minimum ratio requires, to meet the maximum pitch limit).
A 400 mm diameter spiral column has a 40 mm concrete cover and uses #10 spiral (10 mm diameter) with fy = 415 MPa. The concrete f'c = 28 MPa. Calculate the minimum spiral pitch needed.
Example 7: Verifying Longitudinal Steel Ratio
Solution:
Step 1: Calculate the steel ratio.
Ag = 350 × 350 = 122,500 mm²
ρg = 2,513 / 122,500 = 0.0205 or 2.05%
Step 2: Verify against code limits.
Code: 0.01 ≤ ρg ≤ 0.08
Our ρg = 0.0205
0.01 < 0.0205 < 0.08 ✓
Step 3: Check constructability.
8 bars in a 350 mm square column: arrange in a square pattern (e.g., corners and midpoints of sides).
Space between bars ≈ 350 / 2.83 ≈ 124 mm (rough estimate for 8 bars)
This is adequate for concrete placement and vibration. ✓
Note: If the column had been 300 mm × 300 mm, 8 #20 bars might be tight; the designer would consider 6 bars or a larger column.
Answer: The column is acceptable in terms of steel ratio and constructability.
A 350 mm × 350 mm column has 8 #20 bars (Ast = 2,513 mm²). Check if it meets ρg limits and can be constructed.
Key points
- Longitudinal steel: 0.01 ≤ ρg ≤ 0.08 (practically ≤0.04 with laps)
- Minimum 4 bars in tied columns, 6 in spiral columns
- Tied-column spacing: least of 16db, 48dt, least column dimension, or 300 mm
- Spiral minimum ratio: ρs ≥ 0.45(Ag/Ach − 1)(f'c/fyt), with fyt ≤ 700 MPa
- Spiral pitch: ≤ 75 mm or 1/6 of diameter (whichever smaller)
- These limits ensure ductility, prevent bar buckling, and allow proper concrete consolidation
4. Axial–Moment Interaction Diagram
Real columns are rarely subjected to pure axial compression. They always experience moments due to:
- Lateral Loads: Wind, seismic, or live loads create moments.
- Eccentricity: The load may not be perfectly centered, creating a moment M = P × e.
- Slenderness: In slender columns, P-delta effects amplify moments.
- Continuity: In frame structures, negative moments from beam-column joints create biaxial bending.
The interaction diagram is a graphical representation of all combinations of axial load (Pn) and bending moment (Mn) that a column section can support. It is fundamental to column design.
Building the Interaction Diagram:
The diagram is constructed by varying the strain distribution across the column section from:
- Compression-Controlled Failure (top): pure axial compression, no moment
- Balanced Point (middle): concrete reaches maximum compression strain (0.003) exactly when steel reaches yield strain (fy/Es)
- Tension-Controlled Failure (bottom): steel yields well before concrete crushes, acting like a beam
Key Points on the Diagram:
- Pure Axial Load (ε = 0 throughout):
At the top of the diagram:
Pn = Pₒ = 0.85f'c(Ag − Ast) + fyAst (for short columns; applied with cap 0.80 or 0.85)
Mn = 0 (no moment) Balanced Point (εc = 0.003, εst = fy/Es):
This is the critical transition point:- Concrete has reached its maximum usable compression strain (0.003 or "crushing")
- Steel has reached yield strain (usually εy = 415/200,000 ≈ 0.002075 for fy = 415 MPa)
- Both materials are at their limits simultaneously
- Any further increase in eccentricity will fail the steel in tension; any decrease will fail the concrete in compression
- The balanced point gives the maximum moment the section can sustain
The neutral axis location at balance (cb) is found from strain compatibility:
εc / cb = εy / (d − cb)
0.003 / cb = 0.002075 / (d − cb)
cb = 0.003 × d / (0.003 + 0.002075) ≈ 0.591d (for fy = 415 MPa)The balanced axial load Pb and moment Mb are computed using moment equilibrium about the centroid.
- Pure Bending (P = 0, M = Mn):
At the bottom of the diagram, the column acts like a beam. The neutral axis is located by strain compatibility, and the section is analyzed for maximum moment with zero axial load. The capacity is the same as a doubly-reinforced beam.
Failure Regions:
- Compression-Controlled Region (above the balanced point):
The concrete crushes (ε = 0.003) before the steel yields. Failure is sudden and brittle. The reduction factor φ = 0.65 (tied) or 0.75 (spiral) applies. - Tension-Controlled Region (below the balanced point, typically εst > 0.005):
The steel yields well before the concrete crushes. Failure is ductile with visible cracks and deformation. The reduction factor φ = 0.9 applies (same as flexure). - Transition Region (between 0.002 ≤ εst ≤ 0.005):
Intermediate behavior. The code allows linear interpolation of φ between the compression and tension extremes, but columns rarely operate here.
Design Interaction Equation:
For any combined load (Pu, Mu), the interaction relationship is:
f = (Pu / (φPn)) + (Mu / (φMn)) ≤ 1.0
Or rearranged:
Pu / (φPn) + Mu / (φMn) ≤ 1.0
where Pn and Mn are read from the column section (unfactored), and φ is the appropriate reduction factor. Every point (Pu, Mu) must satisfy this inequality.
Simplified Biaxial Bending (Bresler Equation):
When a column is bent about both axes (biaxial bending), an approximate criterion is:
1/Pu = 1/(φPnx) + 1/(φPny) − 1/(φPo)
where Pnx and Pny are the axial capacities when the section is bent about the x and y axes respectively, and Pₒ is the pure axial capacity. This is a conservative approximation useful for quick design checks.
Code Interaction Diagram Formula (ACI 318 Appendix D):
For a more refined analysis, the ACI provides the following bilinear approximation (often used in practice):
For Pu > 0.1f'c Ag (compression-controlled):
Pu / (0.1f'c Ag) + Mu / (φMn,max) ≤ 1.0
For Pu ≤ 0.1f'c Ag (tension-controlled or near balance):
Pu / (φPn) + Mu / (φMn) ≤ 1.0 (linear interpolation)
These formulas simplify design without loss of safety.
Examples
Example 8: Locating the Balanced Point
Solution:
Step 1: Calculate neutral axis at balance.
For fy = 415 MPa:
Es = 200,000 MPa (modulus of elasticity of steel)
εy = fy / Es = 415 / 200,000 = 0.002075
Strain compatibility at balance:
0.003 / cb = 0.002075 / (d − cb)
0.003(d − cb) = 0.002075 × cb
0.003d − 0.003cb = 0.002075cb
0.003d = 0.005075cb
cb = 0.003d / 0.005075 = 0.591d
Assuming the centroid of steel is at d ≈ 400 − 40 (cover) − 12.5 (bar radius) = 347.5 mm ≈ 350 mm
cb = 0.591 × 350 = 206.9 mm ≈ 207 mm
Step 2: Analyze forces at balance.
Compression in concrete:
cc = 0.85f'c × b × a
where a = β₁ × cb = 0.85 × 207 = 175.95 mm (β₁ = 0.85 for f'c ≤ 30 MPa)
cc = 0.85 × 28 × 400 × 175.95 = 1,674,504 N
Tension in steel:
Ts = fy × Ast = 415 × 3,927 = 1,629,705 N
(Note: cc ≈ Ts, as expected at balance; small difference due to rounding.)
Step 3: Calculate balanced moment about the gross centroid.
Assuming bars are at depth d = 350 mm from top (distance from neutral axis c = d − cb = 350 − 207 = 143 mm)
Mb = Cc × (distance from neutral axis to centroid of compression) + Ts × (distance to steel)
A rough estimate (for demonstration):
Mb ≈ 1,630 kN × (0.35 − 0.088) m ≈ 1,630 × 0.262 ≈ 427 kN·m
(Exact calculation requires proper moment arm calculation, which is involved. The balanced moment is typically 1.5 to 2.5 times the pure-bending moment for a column.)
Answer: The balanced neutral axis is at cb ≈ 207 mm, and the balanced moment is approximately 400–450 kN·m (order of magnitude). This point sits on the interaction diagram and represents the maximum moment the section can sustain.
For the tied column in Example 1 (400 × 400 mm, 8 #25 bars, f'c = 28 MPa, fy = 415 MPa), find the neutral axis depth (cb) and balanced moment (Mb) at the balanced point.
Example 9: Checking Combined Loading Against Interaction Diagram
Solution:
Step 1: Identify the column and its capacity.
From Example 1: φPn,max = 2,779 kN (tied column)
We need the moment capacity Mn at Pu = 2,000 kN.
Step 2: Read (or calculate) the moment capacity from the interaction diagram.
At pure axial (Pu = 2,779 kN): Mn ≈ 0 (at the top of the diagram)
At balanced point (Pu ≈ Pb ≈ 1,200–1,400 kN): Mn ≈ 400–450 kN·m (maximum moment)
At pure moment (Pu = 0): Mn ≈ 150–200 kN·m (as a flexural member)
At Pu = 2,000 kN, by linear interpolation (compression-controlled region):
Mn ≈ (2,000 − 2,779) / (1,200 − 2,779) × 450 ≈ rough estimate of 100–150 kN·m
For a more accurate answer, the interaction diagram for this specific section would be plotted, or a computer program used. Let's assume (from the diagram or calculation) that Mn ≈ 120 kN·m at Pu = 2,000 kN.
Step 3: Apply the design criterion.
(Pu / (φPn)) + (Mu / (φMn)) ≤ 1.0
(2,000 / 2,779) + (200 / (0.65 × 120)) ≤ 1.0
0.719 + (200 / 78) ≤ 1.0
0.719 + 2.564 ≤ 1.0
3.283 ≤ 1.0 ✗ FAILS
The column is overstressed. The moment of 200 kN·m is too large for the available moment capacity at this load level. The designer must either:
- Increase the column size
- Increase the steel reinforcement
- Reduce the applied moment (by bracing, or reducing eccentricity)
Answer: As proportioned, the column is inadequate. A larger section or more steel is required.
The column from Example 1 must support a factored load Pu = 2,000 kN and factored moment Mu = 200 kN·m (about one axis). Check if it is safe.
Example 10: Biaxial Bending Check (Bresler Equation)
Solution:
The Bresler reciprocal formula for biaxial bending is:
1/Pu = 1/(φPnx − δPnx × Mux/(φMnx)) + 1/(φPny − δPny × Muy/(φMny)) − 1/(φPo)
A simplified version often used is:
1/Pu ≤ 1/(φPnx) + 1/(φPny) − 1/(φPo)
Let's use this simplified form (conservative):
Step 1: Calculate the reciprocal terms.
1/(φPnx) = 1/4,500 = 0.0002222
1/(φPny) = 1/4,500 = 0.0002222
1/(φPo) = 1/5,000 = 0.0002000
Step 2: Apply Bresler's equation.
1/Pu,available = 0.0002222 + 0.0002222 − 0.0002000
1/Pu,available = 0.0002444
Pu,available = 1/0.0002444 ≈ 4,090 kN
Step 3: Compare with applied load.
Pu,applied = 3,500 kN < 4,090 kN ✓ SAFE (in terms of axial capacity with biaxial bending)
Note: This is a simplified check. A full interaction surface or detailed computation of (Mnx, Mny) at Pu = 3,500 kN would be more accurate. The Bresler equation is conservative but useful for quick design.
Answer: The column is adequate under biaxial loading according to the Bresler approximation.
A 500 mm × 500 mm column carries Pu = 3,500 kN, Mux = 250 kN·m (about x-axis), Muy = 200 kN·m (about y-axis). Use the Bresler reciprocal equation to check adequacy. Assume φPnx = φPny = 4,500 kN, φPo = 5,000 kN.
Key points
- The interaction diagram plots all (Pn, Mn) combinations a section can support.
- Compression-controlled (top): concrete crushes, φ = 0.65 or 0.75, brittle.
- Balanced point: concrete strain = 0.003, steel strain = fy/Es, maximum moment.
- Tension-controlled (bottom): steel yields, φ = 0.90, ductile (rare for columns).
- Design check: (Pu / φPn) + (Mu / φMn) ≤ 1.0 for any factored load combination.
- Biaxial bending: use Bresler equation or 3D interaction surface (advanced).
- The interaction diagram is the key tool for column design; all load cases must plot inside the φ-reduced diagram.
5. Slenderness and Moment Magnification
Short columns (low slenderness) can be designed using the interaction diagram directly. However, slender columns experience additional moments due to lateral deflection under load, reducing their capacity. These are called P-δ (P-delta) and P-Δ (P-Delta) effects.
What Is Slenderness?
The slenderness ratio is defined as:
Slenderness Ratio = kℓu / r
where:
- k = effective length factor (depends on support conditions and frame stability)
- ℓu = unsupported length of the column
- r = radius of gyration = √(I/Ag)
For a rectangular section: r = depth / √12
For a circular section: r = diameter / 4
Limits for Short Columns (Slenderness Ignored):
The NSCP 2015 (and ACI 318) define a column as "short" if:
For Non-Sway Frames (Braced, columns cannot sway horizontally):
kℓu/r ≤ 34 − 12(M₁/M₂)
but not more than 40, where M₁/M₂ is the ratio of the smaller to larger end moment (positive if moments bend the column in single curvature, negative if double curvature). If M₂ = 0 (cantilever), the limit becomes 22.
For Sway Frames (Unbraced, columns can sway):
kℓu/r ≤ 22
If the slenderness ratio exceeds these limits, moment magnification must be applied.
Why Moment Magnification?
Consider a column with axial load Pu and initial moment M₀. As the column deflects laterally by amount δ, an additional moment is created:
M_secondary = Pu × δ
This secondary moment causes more deflection, which causes more moment, and so on. The total moment becomes:
M_total = M₀ + Pu × δ
For slender columns, δ can be significant, and the total moment can far exceed M₀. The moment magnifier method approximates this iterative process.
Moment Magnification Formula:
The design moment is amplified as:
M_amplified = δ × M₀
where the magnifier δ (delta) depends on the loading and frame type:
For Non-Sway (Braced) Frames:
δ = Cm / (1 − (Pu / (0.75 × φ × Pc)))
where:
- Cm = moment coefficient (typically 0.6 − 0.4(M₁/M₂) for non-uniform moment, or 1.0 for uniform moment)
- Pc = Euler buckling load = π²EI / (kℓu)²
- φ = reduction factor (0.65 or 0.75 depending on column type)
For Sway (Unbraced) Frames:
δ_s = 1 / (1 − ΣPu / (0.75 × φ × ΣPc))
where the sums are over all columns in the story, accounting for the frame's overall stability.
Or, more simply, if the frame is braced laterally (shear walls, diaphragms):
δ_s = 1.0 (sway is prevented)
Design Procedure for Slender Columns:
- Calculate slenderness ratio kℓu/r
- Check if it exceeds the limit for short columns
- If yes, calculate Pc = π²EI / (kℓu)² and Cm
- Calculate the moment magnifier δ
- Amplify the moment: M_design = δ × M₀
- Check the interaction diagram with (Pu, M_design)
Simplified Approach (Conservative):
For hand calculations, a conservative approximation is:
δ_approx = 1 + (Pu / (Ag × f'c)) × (kℓu/r)² / (40,000)
If δ > 1.5, the column is quite slender and may be uneconomical.
Effective Length Factor k:
The factor k accounts for end support conditions:
- Pinned-Pinned (both ends free to rotate, no sway): k = 1.0
- Fixed-Fixed (both ends rigidly fixed, no sway): k = 0.5
- Fixed-Pinned (one end fixed, one pinned, no sway): k = 0.7
- Cantilever (fixed at base, free at top, sway): k = 2.0
- Unbraced Frame (sway possible): k = 1.2 − 2.0 depending on frame geometry and stiffness
For typical building frames, if columns are connected to rigid foundations and beams, k ≈ 0.6 − 0.9 for the lower stories and increases toward the top.
Physical Interpretation:
Slenderness matters when Pu / Pc (or Pu / (0.75φPc)) approaches 1. If the load is much less than the buckling load, deflections remain small and moment magnification is negligible (δ ≈ 1.0). For heavily loaded columns, δ can be 1.5 to 3 or more, significantly reducing capacity.
Examples
Example 11: Checking Short-Column Criterion
Solution:
Step 1: Calculate the radius of gyration.
For a square section: r = b / √12 = 400 / √12 = 400 / 3.464 = 115.5 mm
Step 2: Calculate the slenderness ratio.
kℓu / r = 1.0 × 4,500 / 115.5 = 38.9
Step 3: Apply the short-column criterion for non-sway frames.
M₁/M₂ = 50/100 = 0.5
Limit = 34 − 12(M₁/M₂) = 34 − 12(0.5) = 34 − 6 = 28
But the limit is capped at 40, so the limit is min(28, 40) = 28.
Step 4: Compare.
kℓu/r = 38.9 > 28 ✗
Answer: The column exceeds the short-column limit. Moment magnification must be applied; it cannot be treated as short. The designer should either:
- Increase the column size (larger r)
- Reduce the unsupported length (additional lateral supports)
- Add more lateral bracing to reduce k
A 400 mm × 400 mm tied column is part of a building with lateral bracing (non-sway frame). Its unsupported length is 4,500 mm, and it is pinned at both ends (k = 1.0). The column carries moments M₁ = 50 kN·m and M₂ = 100 kN·m (bending in single curvature). Check if the column can be treated as short.
Example 13: Effect of Lateral Bracing on Slenderness
Solution:
With lateral bracing at midheight, the unsupported length is now:
ℓu = 2,250 mm (half the original length)
Slenderness ratio:
kℓu / r = 1.0 × 2,250 / 115.5 = 19.5
Short-column criterion (same as before):
Limit = 34 − 12(0.5) = 28 ≤ 40
Comparison:
kℓu/r = 19.5 < 28 ✓
Answer: With lateral bracing at midheight, the column meets the short-column criterion (kℓu/r = 19.5 < 28). It can now be designed as a short column without moment magnification. This demonstrates how lateral bracing (shear walls, moment-resisting frames, or diagonal braces) dramatically improves column capacity by reducing effective length.
Instead of a 4,500 mm unsupported length, add lateral bracing at midheight (h = 2,250 mm) to the column in Example 11. Recalculate the slenderness ratio and check the short-column criterion.
Key points
- Short column: kℓu/r ≤ 22 (sway frames) or ≤ 40 (non-sway frames); slenderness ignored.
- Slender column: requires moment magnification to account for P-δ and P-Δ effects.
- Moment magnifier δ = Cm / (1 − Pu/(0.75φPc)) for non-sway; larger for sway frames.
- Pc = π²EI/(kℓu)² is the Euler buckling load; high Pu/(0.75φPc) increases δ.
- Design moment = δ × M₀; use amplified moment in the interaction diagram.
- Effective length factor k depends on support conditions and frame type.
- Slender columns are uneconomical; prefer lateral bracing (k ≤ 0.9) or larger sections.
6. Design Examples and Board-Exam Problem Types
This section provides realistic design examples that combine the concepts from previous sections and are representative of PRC Civil Engineer Licensure Examination problems.
Design Philosophy for Columns:
- Select a trial section based on the estimated axial load and allowable stress (typically assuming ρg ≈ 0.015–0.03).
- Calculate the required steel area to satisfy the axial-load requirement.
- Verify the steel ratio is within 0.01–0.08 (or ≤0.04 with laps).
- Check tie or spiral reinforcement and spacing requirements.
- Assess slenderness: if kℓu/r exceeds limits, either enlarge the section, add lateral bracing, or accept moment magnification.
- Prepare the interaction diagram (or use code tables) and verify all factored load combinations (Pu, Mu) plot inside the φ-reduced diagram.
- Check constructability and cost; iterate if needed.
Common Board-Exam Problem Types:
- Pure Axial Capacity: Given section dimensions, materials, and reinforcement, find design capacity φPn,max.
- Required Section Size: Given axial load and moments, find the minimum column size.
- Steel Ratio Verification: Confirm ρg is within code limits.
- Tie or Spiral Design: Size and space transverse reinforcement.
- Interaction Diagram: Plot or use to check combined load adequacy.
- Slenderness Effect: Determine if moment magnification applies; calculate amplified moment.
- Biaxial Bending Check: Use Bresler equation or 3D interaction surface.
Typical Given Data (From PRC Exam):
- Column dimensions (mm × mm or diameter in mm)
- Concrete strength f'c (MPa): 20, 25, 28, 35 MPa (common in Philippines)
- Steel yield fy (MPa): 415 (common), 500, 700 (special cases)
- Unsupported length ℓu (mm)
- End conditions (pinned, fixed, cantilever) → determine k
- Axial load Pu (kN) and moment Mu (kN·m) (factored)
- Request: Design the column (select size, reinforcement) OR verify adequacy
Examples
Example 14: Design of a Tied Column for Pure Axial Load
Solution:
Step 1: Estimate the required Pₒ.
For a tied column:
φPn,max = 0.52 × Pₒ = Pu = 4,000 kN
Pₒ = 4,000 / 0.52 = 7,692 kN
Step 2: Determine required gross area.
Assuming ρg = 0.02 (middle of practical range):
Ag = Pₒ / (0.85f'c(1 − ρg) + fy × ρg)
Ag = 7,692,000 / (0.85 × 28 × 0.98 + 415 × 0.02)
Ag = 7,692,000 / (23.268 + 8.3)
Ag = 7,692,000 / 31.568
Ag = 243,607 mm²
For a square section: b = √Ag = √243,607 ≈ 494 mm
Round to standard size: b = 500 mm (square column)
Revised Ag = 500 × 500 = 250,000 mm²
Step 3: Calculate required steel area.
Ast = ρg × Ag = 0.02 × 250,000 = 5,000 mm²
Select bars: 8 #25 bars (8 × π × 12.5² = 3,927 mm²) — too small
Select bars: 10 #25 bars (10 × π × 12.5² = 4,909 mm²) — close
Select bars: 8 #28 bars (8 × π × 14² = 4,923 mm²) — acceptable
Actual Ast = 4,923 mm² (use 8 #28 bars)
Revised ρg = 4,923 / 250,000 = 0.0197 ≈ 0.02 ✓
Step 4: Verify the capacity.
Pₒ = 0.85 × 28 × (250,000 − 4,923) + 415 × 4,923
Pₒ = 0.85 × 28 × 245,077 + 2,042,845
Pₒ = 5,843,570 + 2,042,845 = 7,886,415 N ≈ 7,886 kN
φPn,max = 0.52 × 7,886 = 4,101 kN > 4,000 kN ✓
The column is slightly overdesigned, but acceptable for economy.
Step 5: Design ties.
Longitudinal bar diameter: 28 mm
Tie diameter: #13 (13 mm minimum for bars > 32 mm, but use #10 for bars ≤ 28 mm)
Tie size: #10 (10 mm diameter)
Spacing limits:
- 16db = 16 × 28 = 448 mm
- 48dt = 48 × 10 = 480 mm
- Least column dimension = 500 mm
- Code maximum = 300 mm ← controls
Tie spacing: 300 mm c/c
Step 6: Bar arrangement.
For an 8-bar column in a 500 mm square:
- 4 corner bars (one at each corner)
- 4 intermediate bars (one at midpoint of each side)
Clear cover: 40 mm
Bar spacing around perimeter: approximately 500 mm / 4 sides ≈ 125 mm (adequate; limit is 150 mm)
Step 7: Verify steel ratio and constructability.
ρg = 0.0197 ✓ (within 0.01–0.08)
Bars: 8 ≥ 4 (minimum) ✓
Ties: #10 @ 300 mm c/c ✓
Final Design Summary:
- Column dimensions: 500 mm × 500 mm (square)
- Longitudinal steel: 8 #28 bars (4,923 mm²)
- Ties: #10 @ 300 mm c/c (closed loops, all around)
- Clear cover: 40 mm
- Design capacity: φPn = 4,101 kN (exceeds 4,000 kN requirement)
- Short-column assumption: valid (assume kℓu/r < 22 for braced frame)
Design a square tied column to carry a factored load Pu = 4,000 kN. Use f'c = 28 MPa, fy = 415 MPa, and assume ρg = 0.02. The column is part of a braced frame, so treat it as short. Provide final dimensions and reinforcement sketch.
Example 15: Verification of a Slender Column Under Combined Loading
Solution:
Step 1: Check slenderness.
kℓu / r = 1.2 × 6,000 / 115.5 = 7,200 / 115.5 = 62.4
Limit for unbraced frame: 22
62.4 > 22 → Column is slender; moment magnification required.
Step 2: Calculate moment magnification.
E = 4,700√28 ≈ 24,860 MPa
I = 400⁴ / 12 = 2.133 × 10⁹ mm⁴
Euler buckling load:
Pc = π² × 24,860 × 2.133 × 10⁹ / (1.2 × 6,000)²
Pc = 133,154 × 10⁹ / 51,840,000
Pc ≈ 2,570 kN
(For an unbraced frame, use sway moment magnifier; simplified, multiply moment by ~1.5–2.0 for high slenderness.)
Or, use the formula:
δ_s = 1 / (1 − Pu / (0.75 × φ × Pc))
δ_s = 1 / (1 − 2,500 / (0.75 × 0.65 × 2,570))
δ_s = 1 / (1 − 2,500 / 1,254)
δ_s = 1 / (1 − 1.993) = 1 / (−0.993)
This gives a negative or very large magnifier, indicating the column is near instability (Pu ≈ Pc). The column is likely overstressed or the section is inadequate.
Step 3: Verdict.
The column cannot support the combined loading (2,500 kN + 150 kN·m) given its slenderness. The designer must:
- Increase the column size (larger I, lower slenderness)
- Add lateral bracing (reduce k, lower kℓu/r)
- Reduce the applied moment (brace the structure laterally)
This is a typical board-exam scenario: identify the slenderness problem and recommend design modifications.
A 400 mm × 400 mm tied column is 6 m tall (unbraced frame, k = 1.2) and carries Pu = 2,500 kN and Mu = 150 kN·m (from lateral load). Verify adequacy using NSCP 2015. Assume f'c = 28 MPa, fy = 415 MPa, 8 #25 bars.
Example 16: Spiral Column Design for High Capacity
Solution:
Step 1: Verify axial capacity of spiral column.
Longitudinal steel: 8 #28 bars (same as Example 14), Ast = 4,923 mm²
Ag = 250,000 mm²
Pₒ = 0.85 × 28 × (250,000 − 4,923) + 415 × 4,923 = 7,886 kN (same as tied)
For spiral column:
φPn,max = 0.85 × 0.75 × 7,886 = 5,051 kN
Compare with tied column:
Tied: φPn,max = 0.52 × 7,886 = 4,101 kN
Spiral: φPn,max = 0.6375 × 7,886 = 5,051 kN
Increase: (5,051 − 4,101) / 4,101 = 23.2% ✓
The spiral column carries 950 kN more, validating the premium for spirals.
Step 2: Design the spiral.
Gross area: Ag = 250,000 mm²
Core diameter (concrete cover = 40 mm): Dch = 500 − 2 × 40 = 420 mm
Core area: Ach = π × (420/2)² = π × 210² = 138,544 mm²
Minimum spiral ratio:
ρs ≥ 0.45 × (Ag / Ach − 1) × (f'c / fyt)
Note: fyt is capped at 700 MPa; our fy = 415 MPa, so fyt = 415 MPa.
ρs ≥ 0.45 × (250,000 / 138,544 − 1) × (28 / 415)
ρs ≥ 0.45 × (1.8044 − 1) × 0.0675
ρs ≥ 0.45 × 0.8044 × 0.0675
ρs ≥ 0.02439
Step 3: Calculate pitch and spiral bar size.
Spiral ratio formula:
ρs = 4Asp / (s × Dch)
Choose a #10 spiral (Asp = 78.54 mm²):
s = 4 × 78.54 / (0.02439 × 420)
s = 314.16 / 10.24
s = 30.7 mm
Round to a standard pitch: s = 30 mm (slightly tighter for safety)
Step 4: Verify pitch limits.
Maximum pitch:
- 75 mm, OR
- Dch / 6 = 420 / 6 = 70 mm ← controls
Our s = 30 mm < 70 mm ✓
Step 5: Check the spiral ratio with the chosen pitch.
Actual ρs = 4 × 78.54 / (30 × 420) = 314.16 / 12,600 = 0.02494 > 0.02439 ✓
Final Design Summary (Spiral):
- Column dimensions: 500 mm × 500 mm (square)
- Longitudinal steel: 8 #28 bars (4,923 mm²)
- Spiral: #10 (10 mm diameter) with 30 mm pitch
- Clear cover: 40 mm
- Design capacity: φPn = 5,051 kN (23% more than tied)
- Advantage: Better ductility for seismic loading; smaller design possible for same load.
Redesign the column from Example 14 as a spiral column with the same cross-section (500 mm × 500 mm). Calculate the minimum spiral ratio and pitch using f'c = 28 MPa, fy = 415 MPa. Compare capacity with the tied column.
Example 17: Board-Exam Style: Design Selection
Solution:
Step 1: Check slenderness.
r = 450 / √12 = 129.9 mm
kℓu / r = 0.8 × 5,000 / 129.9 = 30.8
Limit for braced frame: 34 − 12(M₁/M₂) ≈ 28–40 (assume 34 for no moment ratio given)
30.8 < 34 ✓ → Column is short; no moment magnification needed.
Step 2: Estimate and verify required Ag and steel.
Assuming 8 #25 bars (Ast = 3,927 mm²):
Ag = 450 × 450 = 202,500 mm²
ρg = 3,927 / 202,500 = 0.0194 ✓ (within 0.01–0.08)
Pₒ = 0.85 × 28 × (202,500 − 3,927) + 415 × 3,927
Pₒ = 0.85 × 28 × 198,573 + 1,629,705
Pₒ = 4,716,414 + 1,629,705 = 6,346,119 N ≈ 6,346 kN
For a tied column:
φPn,max = 0.52 × 6,346 = 3,300 kN
Step 3: Check pure axial capacity.
φPn,max = 3,300 kN > Pu = 3,500 kN?
NO — the column lacks 200 kN of pure axial capacity. But this is before considering the moment. The interaction diagram check (next step) will be more revealing.
Step 4: Check combined loading (using interaction diagram).
At Pu = 3,500 kN (above the pure-axial cap of 3,300 kN), the point lies outside the interaction diagram on the compression side. This indicates the column cannot sustain any moment at this load level.
Estimate moment capacity at Pu = 3,500 kN:
Since Pu > φPn,max, the column is in the compression-failure region with minimal moment capacity. Assuming a typical interaction diagram, Mn ≈ 0 or very small at high compression.
Interaction check:
(Pu / φPn,max) + (Mu / φMn) ≤ 1.0
(3,500 / 3,300) + (180 / φMn) ≤ 1.0
1.061 + (180 / φMn) ≤ 1.0
This immediately fails (left side > 1).
Step 5: Verdict and recommendation.
The proposed 450 mm × 450 mm section is INADEQUATE. The axial capacity is insufficient, and the section cannot sustain the combined load (3,500 kN + 180 kN·m).
Recommended Modifications:
- Increase section size: Try 500 mm × 500 mm (from Example 14), which has φPn,max ≈ 4,100 kN.
- Increase steel ratio: Use 10 #25 bars instead of 8 (ρg ≈ 0.024). This increases Pₒ and both axial and moment capacities.
- Use spiral reinforcement: Same 450 mm section as a spiral column would yield φPn,max ≈ 3,900 kN (still insufficient for 3,500 kN but better than 3,300 kN).
- Reduce applied loads: If loads are preliminary, confirm final design loads (may be conservative).
Preferred solution: Enlarge to 500 mm × 500 mm with 8 #28 bars (spiral), achieving φPn,max ≈ 5,050 kN and good moment capacity.
This is typical of board exams: identify the deficiency, explain the physics, and recommend engineering solutions.
A building column in Metro Manila must support Pu = 3,500 kN and Mu = 180 kN·m. The column is 5 m tall in a braced frame (k = 0.8). The designer proposes a 450 mm × 450 mm section with f'c = 28 MPa, fy = 415 MPa. Perform a complete check and recommend if this design is adequate. If not, suggest modifications.
Key points
- Design is iterative: select trial section, calculate steel, verify limits, check interaction diagram.
- Always check both axial and combined loading (with moments).
- For slender columns, calculate moment magnification; don't ignore P-δ effects.
- Verify constructability (clearances, bar spacing, concrete placement).
- Tied columns are cheaper but less ductile; spirals are costlier but preferred in seismic zones.
- Board exams often combine multiple concepts: check slenderness, verify steel ratio, and validate interaction diagram.
7. Code Requirements and Design Checklist
The NSCP 2015 (National Structural Code of the Philippines) and ACI 318-14/19 establish minimum requirements for reinforced concrete columns to ensure safety, constructability, and adequate performance. This section summarizes key code articles and provides a design checklist for exams and practice.
Key NSCP 2015 and ACI 318 Code Sections:
NSCP 2015, Section 410 — Columns:
- 410.1 — Scope: Applies to columns that carry primarily axial compression with or without bending.
- 410.2 — Design Assumptions: Columns are assumed to act compositely (concrete + steel); strain compatibility applies.
- 410.3 — Strength Requirements: Design strength must equal or exceed factored loads:
φPn ≥ Pu and φMn ≥ Mu (or combined interaction check) - 410.4 — Minimum Eccentricity: Even if no moment is applied, assume a minimum eccentricity emin to account for construction inaccuracy. This is implicitly handled by the 0.80 and 0.85 capacity caps.
410.7 — Longitudinal Reinforcement:
- Minimum: ρg,min = 0.01 (1% of gross area)
- Maximum: ρg,max = 0.08 (8% of gross area)
- In lap-splice regions: ρg ≤ 0.04 (to avoid congestion)
- Minimum 4 bars in tied columns, 6 in spiral columns
- All bars must be continuous (no splicing in critical zones unless detailed)
410.8 — Transverse Reinforcement (Ties):
- Size: #10 (10 mm) for longitudinal bars ≤ 32 mm; #13 (13 mm) for bars > 32 mm
- Spacing (least of): 16db, 48dt, least column dimension, 300 mm
- Closed loops, anchored in core
- Corner and alternate bars must be enclosed by ties
410.9 — Spiral Reinforcement:
- Minimum ratio: ρs ≥ 0.45(Ag/Ach − 1)(f'c/fyt), with fyt ≤ 700 MPa
- Maximum pitch: 75 mm or Dch/6 (whichever smaller)
- Minimum pitch: typically ≥ 20 mm (constructability)
410.10 — Slenderness Effects:
- Short columns: kℓu/r ≤ 22 (unbraced) or ≤ 34 − 12(M₁/M₂) ≤ 40 (braced)
- Slender columns: apply moment-magnifier method or second-order analysis
- Buckling loads: Pc = π²EI/(kℓu)² (Euler formula)
410.11 — Design with Combined Loads:
- All combinations of (Pu, Mu) must satisfy the interaction equation
- If biaxial bending: use 3D interaction surface or Bresler reciprocal equation
- Design moment = δ × M₀ (moment-magnified if slender)
Reduction Factors φ:
- Compression-controlled (columns): φ = 0.65 (tied), 0.75 (spiral)
- Tension-controlled (beams, tension-flexure): φ = 0.90
- Shear/Torsion: φ = 0.75
Capacity Caps for Axial Load:
- Tied: φPn,max = 0.80 × 0.65 × Pₒ = 0.52Pₒ
- Spiral: φPn,max = 0.85 × 0.75 × Pₒ = 0.6375Pₒ
Design Checklist for Board Exams:
✓ Material Selection:
- [ ] f'c specified (typical: 20, 25, 28, 35 MPa)
- [ ] fy specified (typical: 415 MPa)
- [ ] E = 200,000 MPa (steel)
- [ ] E = 4,700√f'c (concrete, MPa)
✓ Section Dimensions:
- [ ] Gross area Ag calculated
- [ ] Radius of gyration r = √(I/Ag) computed
- [ ] Slenderness ratio kℓu/r checked against code limits
- [ ] If slender, moment magnification δ applied
✓ Longitudinal Reinforcement:
- [ ] Steel area Ast selected (or required area calculated)
- [ ] Steel ratio ρg = Ast/Ag verified: 0.01 ≤ ρg ≤ 0.08 (or ≤0.04 with laps)
- [ ] Number of bars ≥ 4 (tied) or ≥ 6 (spiral)
- [ ] Bar diameters and spacing around perimeter checked (max 150 mm)
- [ ] Clear cover ≥ 40 mm (typical)
✓ Transverse Reinforcement (Ties or Spiral):
- [ ] Tied: Size (#10 or #13) and spacing (least of 16db, 48dt, least dimension, 300 mm) specified
- [ ] Spiral: Diameter and pitch selected; ρs ≥ minimum ratio verified; pitch ≤ 75 mm or Dch/6
✓ Axial Capacity:
- [ ] Pₒ = 0.85f'c(Ag − Ast) + fyAst calculated
- [ ] φPn,max = 0.52Pₒ (tied) or 0.6375Pₒ (spiral) determined
- [ ] φPn,max ≥ Pu verified
✓ Combined Loading (Interaction Diagram):
- [ ] Interaction diagram prepared or consulted (hand-sketched or computed)
- [ ] All factored load combinations (Pu, Mu) plot inside the φ-reduced diagram
- [ ] Or: (Pu/(φPn)) + (Mu/(φMn)) ≤ 1.0 for each load case
✓ Special Cases:
- [ ] Biaxial bending: Bresler equation or 3D surface used if Mux and Muy both significant
- [ ] Slender column: Moment magnifier δ computed and applied
- [ ] Lap splices: If present, ρg ≤ 0.04 and proper splice length provided
- [ ] Earthquake/seismic: If applicable, BNBC or NSCP seismic requirements (ductility, confinement) met
Common Code Violations (and How to Avoid Them):
- ρg out of bounds: Check 0.01 ≤ ρg ≤ 0.08 every time. Underreinforced (ρg < 0.01) risks sudden tensile failure; overreinforced (ρg > 0.08) causes construction delays.
- Tie spacing too large: Remember the four criteria (16db, 48dt, least dimension, 300 mm max). The smallest one controls.
- Ignoring slenderness: High kℓu/r requires moment magnification. Omitting it can overestimate capacity.
- Wrong φ: Columns are compression-controlled (φ = 0.65 or 0.75), NOT 0.90 (that's for flexure).
- Forgetting the cap (0.80 or 0.85): The capacity is NOT φPₒ, but φ × (0.80 or 0.85) × Pₒ.
- Using gross area for steel stress: Steel is on full area Ast, not net area. Concrete is on net area (Ag − Ast).
- No interaction diagram for combined loading: A column with only axial load is rare. Always prepare the diagram or use an interaction table.
- Ignoring constructability: A 300 mm × 300 mm column with 8 #32 bars is theoretically possible but practically impossible to vibrate properly. Verify spacing and constructability.
Examples
Code Checklist Application: Example 18
Design Checklist for Example 18:
Materials:
✓ f'c = 28 MPa (typical Philippine concrete)
✓ fy = 415 MPa (standard Philippines steel)
✓ E (steel) = 200,000 MPa
✓ E (concrete) = 4,700√28 ≈ 24,860 MPa
Section Dimensions:
✓ Ag = 400 × 400 = 160,000 mm²
✓ I = 400⁴/12 = 2.133 × 10⁹ mm⁴
✓ r = √(I/Ag) = √(2.133 × 10⁹ / 160,000) = 115.5 mm
✓ Slenderness ratio (assume kℓu = 4,500 mm): kℓu/r = 38.9
Compare to limit (non-sway, M₁/M₂ = 0.5): 28
VIOLATION: 38.9 > 28 → Slender column; moment magnification required
Mitigation: Add lateral bracing or increase section size.
Longitudinal Reinforcement:
Ast = 8 × π × 12.5² = 3,927 mm²
ρg = 3,927 / 160,000 = 0.0245 or 2.45%
✓ 0.01 ≤ ρg ≤ 0.08? Yes, 2.45% is within range.
✓ Minimum 4 bars (tied)? Yes, 8 bars provided.
✓ Bar spacing around perimeter: ~(4 × 400) / 8 ≈ 200 mm > 150 mm limit?
*Check: 8 bars in square arrangement (4 corners + 4 midpoints of sides) ≈ 200 mm.
VIOLATION (marginal): Spacing exceeds 150 mm slightly.
Mitigation: Reduce spacing or increase section.
Transverse Reinforcement (Ties):
Longitudinal bar diameter: 25 mm → use #10 ties (10 mm) or #13 (13 mm)
Tie spacing limits:
- 16 × 25 = 400 mm
- 48 × 10 = 480 mm (if #10)
- 400 mm (least dimension)
- 300 mm ← controls
✓ Tie spacing: 300 mm c/c (meets code)
Axial Capacity:
Pₒ = 0.85 × 28 × 156,073 + 415 × 3,927
Pₒ = 3,714,537 + 1,629,705 = 5,344,242 N = 5,344.2 kN
φPn,max = 0.52 × 5,344.2 = 2,779 kN
✓ Capacity computed (adequate for example load, but verify against actual Pu)
Combined Loading:
✗ Interaction diagram NOT prepared in this checklist step. Would require moment capacity calculation and combined load check.
SUMMARY OF VIOLATIONS:
- Slenderness: kℓu/r = 38.9 > limit of 28 (non-sway) or 22 (sway). Moment magnification needed. CRITICAL
- Bar spacing (marginal): ~200 mm exceeds 150 mm slightly. Acceptable with minor adjustment.
RECOMMENDATIONS:
- Add lateral bracing at midheight (reduce kℓu to 2,250 mm, then kℓu/r ≈ 19.5 < 28 ✓)
- OR increase section to 450 mm × 450 mm (larger r, lower slenderness)
- OR apply moment magnification method (if loads allow)
- Verify bar spacing is within 150 mm (may require 9 or 10 bars)
This checklist example shows how systematic verification catches code violations early, allowing timely corrections before design approval.
A designer proposes a 400 mm × 400 mm column with 8 #25 bars (tied). Before finalizing, apply the design checklist to identify any code violations.
Key points
- NSCP 2015, Section 410 governs column design; ACI 318 provides detailed guidance.
- Minimum ρg = 0.01, maximum = 0.08 (≤0.04 with splices); minimum 4 or 6 bars depending on type.
- Tied column tie spacing: least of 16db, 48dt, least dimension, 300 mm.
- Spiral minimum ratio: ρs ≥ 0.45(Ag/Ach − 1)(f'c/fyt); pitch ≤ 75 mm or Dch/6.
- φ = 0.65 (tied), 0.75 (spiral); capacity caps are 0.80 and 0.85 respectively.
- Slenderness limit: kℓu/r ≤ 22 (sway) or ≤ 34−12(M₁/M₂) ≤ 40 (non-sway).
- All factored load combinations must satisfy interaction: (Pu/φPn) + (Mu/φMn) ≤ 1.0.
- Check material properties (f'c, fy, E), section geometry, and all code requirements systematically.
Diagrams
Reinforced Concrete Columns - Concept Overview
This mind map provides a hierarchical overview of the main topics in reinforced concrete column design, showing how column types, capacity calculations, reinforcement requirements, interaction diagrams, and slenderness effects are interconnected. It serves as a visual study guide for understanding the overall structure of column design principles.
Reinforced Concrete Column Design Process
This flowchart illustrates the step-by-step design process for reinforced concrete columns, from initial material selection through final code verification. It shows decision points (short vs. slender, safe vs. unsafe) and feedback loops where adjustments are needed, mirroring real design practice and examination scenarios.
Column Failure Modes and Interaction Diagram Regions
This state diagram illustrates the progression from compression-controlled to tension-controlled failure modes as a column transitions from pure axial compression to pure bending. It shows how strain conditions (εc and εst) determine the failure mode and the applicable reduction factor φ, which is central to understanding the interaction diagram.
Typical Steel Ratio Distribution in Column Design
This pie chart shows the typical distribution of longitudinal steel ratios in practical column design. Most designs fall in the economic range of 1.5–2.5% to balance material cost and structural capacity. Minimum ratios are rare (over-constrained), and high ratios are avoided due to constructability and cost. This chart helps reviewees understand where their designs should typically lie on the spectrum of allowable ratios.
In summary
Reinforced concrete columns are vertical structural members that carry primarily axial compression, often combined with bending moments. This chapter covered the fundamental design principles and code requirements for short and slender columns according to NSCP 2015 and ACI 318.
Key Concepts:
- Axial Capacity: The nominal strength Pₒ = 0.85f'c(Ag − Ast) + fyAst combines concrete and steel contributions. Design capacity is capped at 0.52Pₒ (tied) or 0.6375Pₒ (spiral) to account for unavoidable eccentricity. Spiral columns carry ~23% more due to superior confinement.
- Reinforcement Limits: Longitudinal steel ratio must satisfy 0.01 ≤ ρg ≤ 0.08 (≤0.04 with splices). Minimum 4 bars in tied columns, 6 in spiral. Tied-column spacing is limited to the least of 16db, 48dt, least dimension, or 300 mm. Spiral ratio ρs ≥ 0.45(Ag/Ach − 1)(f'c/fyt) with pitch ≤ 75 mm.
- Interaction Diagram: Real columns carry both P and M. The interaction diagram plots all (Pn, Mn) combinations. Compression-controlled (above balanced point): brittle, φ = 0.65 or 0.75. Tension-controlled (below): ductile, φ = 0.90. Design check: (Pu/φPn) + (Mu/φMn) ≤ 1.0.
- Slenderness: If kℓu/r exceeds ~22 (sway) or ~28–40 (non-sway), the column is slender and requires moment magnification (δ method). Slenderness accounts for P-δ and P-Δ effects that amplify moments and reduce capacity.
- Design Practice: Select a trial section, calculate steel area, verify all limits (ρg, bar spacing, ties/spirals), check slenderness, prepare the interaction diagram, and validate that all factored loads (Pu, Mu) plot inside the φ-reduced diagram. Use checklists to avoid code violations.
Board Exam Tips:
- Always verify slenderness; don't assume short-column behavior.
- Double-check the 0.80 and 0.85 capacity caps (common errors).
- Remember φ = 0.65 or 0.75 for columns (not 0.90).
- Plot or consult the interaction diagram for combined loading; pure-axial capacity rarely governs.
- Use worked examples and systematic checklists to ensure code compliance.
- In high-rise or seismic zones, spiral columns are preferred for ductility; tied columns are economical for low-rise, non-seismic buildings.