Reinforced Concrete Design – Practical Example
Column Shear Reinforcement Design Example
Complete Solution for C25-S420 Rectangular Column
Example Problem
Design the transverse reinforcement (stirrups) for a C25-S420 material (300/500) mm rectangular column subjected to an axial load of Nd = 540 kN and shear force of Vd = 350 kN according to normal ductility requirements in the confinement and middle zones. Longitudinal reinforcement diameter is Ø16, use Ø8 stirrups.
Column Cross-Section
Given Information
| Parameter | Value |
|---|---|
| Column dimensions | b = 500 mm, h = 300 mm |
| Materials | C25-S420 (fcd = 16.67 MPa, fyd = 365 MPa, fctd = 1.15 MPa) |
| Design axial force | Nd = 540 kN |
| Design shear force | Vd = 350 kN |
| Longitudinal reinforcement | Ø16 bars (from previous design) |
| Stirrup diameter | Ø8 |
| Effective depth | d = h – cover = 500 – 20 = 480 mm |
| Ductility level | Normal (Standard frame) |
SOLUTION
Step 1
Check Maximum Shear Force Capacity
Vd ≤ 0.22 Ac fcd
350 × 10³ ≤ (0.22 × 300 × 500 × 16.67)
350 × 10³ ≤ 561 × 10³ N
350 < 561 kN ✓
350 × 10³ ≤ (0.22 × 300 × 500 × 16.67)
350 × 10³ ≤ 561 × 10³ N
350 < 561 kN ✓
Result: The column dimensions are adequate (adequate for shear). The section’s shear capacity meets the requirements.
Step 2
Calculate Concrete Contribution to Shear Resistance
Vd ≤ Vcr
Vcr = 0.65 fctd b d [1 + 0.07 (Nd/Ac)]
Vcr = 0.65 × 1.15 × 300 × 480 [1 + 0.07 (540×10³)/(300×500)]
Vcr = 0.65 × 1.15 × 300 × 480 [1 + 0.07 × 3.6]
Vcr = 107,640 × 1.252 = 134,765 N ≈ 134.765 kN
Vcr = 0.65 fctd b d [1 + 0.07 (Nd/Ac)]
Vcr = 0.65 × 1.15 × 300 × 480 [1 + 0.07 (540×10³)/(300×500)]
Vcr = 0.65 × 1.15 × 300 × 480 [1 + 0.07 × 3.6]
Vcr = 107,640 × 1.252 = 134,765 N ≈ 134.765 kN
Note: The term [1 + 0.07(Nd/Ac)] accounts for the beneficial effect of axial compression on shear capacity.
Since Vd = 350 kN > Vcr = 134.765 kN, shear reinforcement is required.
Required stirrup area:
Asw = n × (πز/4) = 2 × (π × 8²/4) = 2 × 50.27 = 100.48 mm²
(for two-legged stirrups)
Asw = n × (πز/4) = 2 × (π × 8²/4) = 2 × 50.27 = 100.48 mm²
(for two-legged stirrups)
Concrete contribution for normal ductility:
Vc = 0.8 Vcr
Vc = 0.8 × 134.765 = 107.812 kN
Vc = 0.8 × 134.765 = 107.812 kN
Required stirrup contribution:
Vs = Vd – Vc
Vs = 350 – 107.812 = 242.2 kN
Vs = 350 – 107.812 = 242.2 kN
Step 3
Calculate Required Stirrup Spacing – CONFINEMENT ZONE
From shear requirements:
s = (Asw × fywd × d) / Vs
s = (100.48 × 365 × 480) / (242.2 × 10³)
s = 17,603,904 / 242,200 = 72.68 mm
s = (Asw × fywd × d) / Vs
s = (100.48 × 365 × 480) / (242.2 × 10³)
s = 17,603,904 / 242,200 = 72.68 mm
Maximum Spacing Limits for Confinement Zone:
sc ≤ min {
b/3 = 300/3 = 100 mm
8Ømin = 8 × 16 = 128 mm
150 mm
}
Therefore: sc ≤ 100 mm
Selected sc = 75 mm
b/3 = 300/3 = 100 mm
8Ømin = 8 × 16 = 128 mm
150 mm
}
Therefore: sc ≤ 100 mm
Selected sc = 75 mm
Decision: Although calculations give 72.68 mm, we round to a practical value of 75 mm (multiple of 5 mm).
Check Minimum Reinforcement Requirement:
Asw/sc ≥ 0.3 (fcd/fywd) b (minimum condition)
100.48/75 ≥ 0.3 × (16.67/365) × 300
1.34 ≥ 0.3 × 0.0457 × 300
1.34 > 0.28 ✓
100.48/75 ≥ 0.3 × (16.67/365) × 300
1.34 ≥ 0.3 × 0.0457 × 300
1.34 > 0.28 ✓
Satisfies minimum requirement! Therefore selected: Ø8 / 75 mm
Step 4
Calculate Required Stirrup Spacing – MIDDLE REGION
Maximum Spacing Limits for Middle Region:
so ≤ min {
b/2 = 300/2 = 150 mm
12Ømin = 12 × 16 = 192 mm
200 mm
}
Therefore: so ≤ 150 mm
Selected so = 150 mm
b/2 = 300/2 = 150 mm
12Ømin = 12 × 16 = 192 mm
200 mm
}
Therefore: so ≤ 150 mm
Selected so = 150 mm
Check Minimum Reinforcement Requirement:
Asw/so ≥ 0.3 (fcd/fywd) b (minimum condition)
100.48/150 ≥ 0.3 × (16.67/365) × 300
0.67 ≥ 0.28
0.67 > 0.28 ✓
100.48/150 ≥ 0.3 × (16.67/365) × 300
0.67 ≥ 0.28
0.67 > 0.28 ✓
Satisfies minimum requirement! Therefore selected: Ø8 / 150 mm
Step 5
Define Confinement Zone Length
Confinement zone length from each end:
Lconf = max(b/3, 8Ølong, 150 mm)
= max(300/3, 8×16, 150)
= max(100, 128, 150)
= 150 mm from each end
Lconf = max(b/3, 8Ølong, 150 mm)
= max(300/3, 8×16, 150)
= max(100, 128, 150)
= 150 mm from each end
Design Summary
Stirrup Arrangement
| Parameter | Value | Status |
|---|---|---|
| Maximum shear capacity check | 350 kN < 561 kN | ✓ Adequate |
| Concrete cracking force (Vcr) | 134.765 kN | Calculated |
| Concrete contribution (Vc) | 107.812 kN | = 0.8 Vcr |
| Required stirrup contribution (Vs) | 242.2 kN | = Vd – Vc |
| Stirrup area (Asw) | 100.48 mm² | Two-legged Ø8 |
| Calculated spacing from shear | 72.68 mm | Rounded to 75 mm |
| Confinement zone length | 150 mm from each end | Code requirement |
| Final design – Confinement zone | Ø8 @ 75 mm | ✓ Satisfies all |
| Final design – Middle region | Ø8 @ 150 mm | ✓ Satisfies all |
Important Design Points
- Confinement zones: 150 mm from both top and bottom of column require tighter stirrup spacing (75 mm) for improved ductility at beam-column joints.
- Middle region: Can use wider spacing (150 mm) as it’s less critical, but still must satisfy minimum reinforcement requirements.
- Minimum reinforcement governs: Both zones satisfy the minimum Asw/s ratio requirement.
- Axial load effect: The axial compression (540 kN) increases concrete’s shear capacity through the [1 + 0.07(Nd/Ac)] term.
- Practical spacing: Final spacings (75 mm, 150 mm) are rounded to practical values divisible by 5 or 10 mm for construction ease.
Construction Specifications
COLUMN: 300 × 500 mm
Longitudinal Reinforcement: Ø16 bars
Confinement Zone (150 mm from ends): Ø8 @ 75 mm
Middle Region: Ø8 @ 150 mm
Materials: C25 Concrete, S420 Steel
Stirrup Type: Two-legged rectangular ties
⚠️ Important Notes:
- This is an educational example following general principles. Always check local building codes.
- Actual design must consider all load combinations and seismic requirements if applicable.
- Stirrup hooks must be 135° with adequate extension length for proper anchorage.
- All designs require review and approval by a licensed structural engineer.
