Structural Engineering – Complete Design Example
Complete Column Design
Longitudinal and Transverse Reinforcement – C20-S420
Example Problem
A column with the cross-section shown in the figure, made of C20-S420 materials, has an axial load Nd = 780 kN and moment Md = 65 kNm. Investigate the adequacy of the section, and if adequate, determine the reinforcement diameter. For shear force Vd = 210 kN, calculate the confinement zone stirrup spacing according to normal ductility level using Ø8 stirrups.
Material properties: fcd = 13 MPa, fyd = 365 MPa, fck = 20 MPa
Column Cross-Section
Given Information
| Parameter | Value |
|---|---|
| Column dimensions | b = 250 mm, h = 400 mm |
| Materials | C20-S420 (fcd = 13 MPa, fyd = 365 MPa, fck = 20 MPa, fctd = 1.0 MPa) |
| Design axial force | Nd = 780 kN |
| Design moment | Md = 65 kNm |
| Design shear force | Vd = 210 kN |
| Cover | d’ = 25 mm, d” = 200 mm |
| Effective depth | d* = 375 mm |
| Stirrup diameter | Ø8 |
PART A: LONGITUDINAL REINFORCEMENT DESIGN
Step 1
Check Section Adequacy for Axial Load
Nd ≤ Ndm = 0.5 Ac fck
780 × 10³ ≤ 0.5 × 250 × 400 × 20
780 × 10³ ≤ 1,000,000 N
780 < 1000 kN ✓
780 × 10³ ≤ 0.5 × 250 × 400 × 20
780 × 10³ ≤ 1,000,000 N
780 < 1000 kN ✓
Result: The section dimensions are adequate! (Section adequacy is satisfied, otherwise dimensions need to be changed.)
Step 2
Calculate Dimensionless Ratios
Effective depth ratio:
d*/h = 200/250 = 0.8
Use Chart 21 (sy: 21 – abak 21)
Use Chart 21 (sy: 21 – abak 21)
Normalized axial force:
ν = Nd / (b h fcd)
ν = 780 × 10³ / (400 × 250 × 13)
ν = 780,000 / 1,300,000
ν = 0.6
ν = 780 × 10³ / (400 × 250 × 13)
ν = 780,000 / 1,300,000
ν = 0.6
Normalized moment:
μ = Md / (b h² fcd)
μ = 65 × 10⁶ / (400 × 250² × 13)
μ = 65,000,000 / 325,000,000
μ = 0.2
μ = 65 × 10⁶ / (400 × 250² × 13)
μ = 65,000,000 / 325,000,000
μ = 0.2
Step 3
Determine Steel Coefficient from Design Chart
From interaction diagram (Chart 21):
– ν = 0.6
– μ = 0.2
– d*/h = 0.8
Result: ψ = 0.4
– ν = 0.6
– μ = 0.2
– d*/h = 0.8
Result: ψ = 0.4
Step 4
Calculate Required Steel Ratio and Area
ρs = ψ × (fcd / fyd)
ρs = 0.4 × (13 / 365)
ρs = 0.4 × 0.0356
ρs = 0.014 = 1.4%
ρs = 0.4 × (13 / 365)
ρs = 0.4 × 0.0356
ρs = 0.014 = 1.4%
Check: ρmin = 1.0% < ρs = 1.4% < ρmax = 8.0% ✓
Required steel area:
Ast = ρs × b × h
ρmin ≤ ρs ≤ ρmaks → (0.01 < 0.014 < 0.04) → Ast = ρs × b × h
Ast = 0.014 × 250 × 400 = 1400 mm²
Ast = n × (πز/4)
Ast = ρs × b × h
ρmin ≤ ρs ≤ ρmaks → (0.01 < 0.014 < 0.04) → Ast = ρs × b × h
Ast = 0.014 × 250 × 400 = 1400 mm²
Ast = n × (πز/4)
Step 5
Select Number and Diameter of Bars
1400 : 8 = (πز/4)
175 = πز/4
ز = 223.01
Ø = 14.93 mm → Round up to Ø16 mm
175 = πز/4
ز = 223.01
Ø = 14.93 mm → Round up to Ø16 mm
Selected design: 8Ø16
Verification:
8 × 201 = 1608 mm² > 1400 mm² ✓
Design is satisfactory!
8 × 201 = 1608 mm² > 1400 mm² ✓
Design is satisfactory!
PART B: SHEAR REINFORCEMENT DESIGN
Step 6
Check Maximum Shear Capacity
Vd ≤ 0.22 Ac fcd
210 × 10³ ≤ 0.22 × 250 × 400 × 13
210 × 10³ ≤ 286 × 10³ N
210 < 286 kN ✓
210 × 10³ ≤ 0.22 × 250 × 400 × 13
210 × 10³ ≤ 286 × 10³ N
210 < 286 kN ✓
Result: Section is adequate for shear (adequate in shear, column section shear strength is satisfied).
Step 7
Calculate Concrete Shear Contribution
Vd ≤ Vcr
Vcr = 0.65 fctd b d (1 + 0.07 Nd/Ac)
Note: d = h – d’ – Østirrup/2 – Ølong/2
d = 400 – 25 – 8/2 – 16/2 = 400 – 25 – 4 – 8 = 363 mm
For simplicity, using d ≈ 225 mm (from cover to centroid of bars)
Vcr = 0.65 × 1 × 400 × 225 (1 + 0.07 × 780×10³/(250×400))
Vcr = 0.65 × 1.15 × 400 × 225 (1 + 0.07 × 7.8)
Vcr = 58,500 × 1.546 = 90,441 N ≈ 90.441 kN
Vcr = 0.65 fctd b d (1 + 0.07 Nd/Ac)
Note: d = h – d’ – Østirrup/2 – Ølong/2
d = 400 – 25 – 8/2 – 16/2 = 400 – 25 – 4 – 8 = 363 mm
For simplicity, using d ≈ 225 mm (from cover to centroid of bars)
Vcr = 0.65 × 1 × 400 × 225 (1 + 0.07 × 780×10³/(250×400))
Vcr = 0.65 × 1.15 × 400 × 225 (1 + 0.07 × 7.8)
Vcr = 58,500 × 1.546 = 90,441 N ≈ 90.441 kN
Since Vd = 210 kN > Vcr = 90.441 kN, shear reinforcement is required.
Stirrup area (two-legged Ø8):
Asw = n × (πز/4) = 2 × (π × 8²/4)
Asw = 2 × 50.27 = 100.48 mm²
Asw = n × (πز/4) = 2 × (π × 8²/4)
Asw = 2 × 50.27 = 100.48 mm²
Step 8
Calculate Required Stirrup Contribution
Vc = 0.8 Vcr
Vc = 0.8 × 90.441 = 72.353 kN
Vs = Vd – Vc
Vs = 210 – 72.353 = 137.647 kN
Vc = 0.8 × 90.441 = 72.353 kN
Vs = Vd – Vc
Vs = 210 – 72.353 = 137.647 kN
Step 9
Calculate Stirrup Spacing for Confinement Zone
s = (Asw fywd d) / Vs
s = (100.48 × 365 × 225) / (137.647 × 10³)
s = 8,251,740 / 137,647
s = 59.95 mm
s = (100.48 × 365 × 225) / (137.647 × 10³)
s = 8,251,740 / 137,647
s = 59.95 mm
Maximum Spacing Limits:
sc ≤ min {
b/3 = 250/3 = 83.33 mm
8Ølong = 8 × 16 = 128 mm
150 mm
}
Therefore: sc ≤ 83.33 mm
Selected sc = 60 mm
b/3 = 250/3 = 83.33 mm
8Ølong = 8 × 16 = 128 mm
150 mm
}
Therefore: sc ≤ 83.33 mm
Selected sc = 60 mm
Decision: Calculations give 59.95 mm, rounded to practical value of 60 mm.
Step 10
Check Minimum Reinforcement Requirement
Asw/sc ≥ 0.3 (fcd/fywd) b (minimum condition)
100.48/60 ≥ 0.3 × (13/365) × 250
1.67 ≥ 0.3 × 0.0356 × 250
1.67 > 2.67 (Note: corrected calculation)
Actually: 1.67 > 0.21 ✓
100.48/60 ≥ 0.3 × (13/365) × 250
1.67 ≥ 0.3 × 0.0356 × 250
1.67 > 2.67 (Note: corrected calculation)
Actually: 1.67 > 0.21 ✓
Final stirrup design: Ø8 @ 60 mm in confinement zone
Design Summary
Final Column Design
| LONGITUDINAL REINFORCEMENT | ||
|---|---|---|
| Parameter | Value | Status |
| Section adequacy check | 780 kN < 1000 kN | ✓ Adequate |
| Normalized axial force (ν) | 0.6 | Calculated |
| Normalized moment (μ) | 0.2 | Calculated |
| Steel coefficient (ψ) | 0.4 | From chart |
| Steel ratio (ρs) | 1.4% | Within limits ✓ |
| Required steel area | 1400 mm² | Calculated |
| Final longitudinal design | 8Ø16 | 1608 mm² ✓ |
| SHEAR REINFORCEMENT | ||
|---|---|---|
| Parameter | Value | Status |
| Max shear capacity check | 210 kN < 286 kN | ✓ Adequate |
| Cracking shear (Vcr) | 90.441 kN | Calculated |
| Concrete contribution (Vc) | 72.353 kN | = 0.8 Vcr |
| Stirrup contribution (Vs) | 137.647 kN | Required |
| Calculated spacing | 59.95 mm | Rounded to 60 mm |
| Max spacing limit | 83.33 mm | Code limit |
| Final stirrup design | Ø8 @ 60 mm | ✓ Satisfies all |
Key Design Insights
- Combined design: This example demonstrates both longitudinal and transverse reinforcement design in one complete solution.
- Section adequacy: Both axial force (780 < 1000 kN) and shear force (210 < 286 kN) checks are satisfied.
- Interaction diagram: Using ν = 0.6 and μ = 0.2, we obtained ψ = 0.4 for steel coefficient determination.
- Steel ratio: Calculated ρs = 1.4% is above minimum (1.0%) and below maximum (8.0%).
- Practical bar selection: 8Ø16 provides 1608 mm² vs. 1400 mm² required (15% excess – acceptable).
- Shear reinforcement: Required spacing from calculations (60 mm) is more critical than code maximum (83 mm).
- Confinement zones: Special attention needed at both ends with tight stirrup spacing for ductility.
Final Construction Specifications
COLUMN: 250 mm × 400 mm
Longitudinal Reinforcement: 8Ø16
Confinement Zone Stirrups: Ø8 @ 60 mm
Middle Region Stirrups: Ø8 @ 150 mm (if applicable)
Materials: C20 Concrete, S420 Steel
Cover: d’ = 25 mm
Stirrup Type: Two-legged rectangular ties
⚠️ Important Notes:
- This is an educational example. Real projects must follow applicable building codes (ACI 318, Eurocode 2, etc.).
- The effective depth ‘d’ calculation should account for actual bar positions and cover requirements.
- Stirrup hooks must have proper anchorage (135° hooks with adequate extension).
- Consider seismic detailing requirements if applicable to the project location.
- All designs require review and approval by a licensed structural engineer.
