Design of Long (Slender) Column as per IS 456:2000 - Complete Step-by-Step Procedure

 Design of Long (Slender) Column as per IS 456:2000 - Complete Step-by-Step Procedure

When designing reinforced concrete structures, columns are critical load-bearing members that transfer loads from beams and slabs to the foundation. Understanding how to design long columns (also called slender columns) is essential for structural engineers. This comprehensive guide will walk you through the complete design procedure for long columns according to IS 456:2000.

Problem Statement

Let's design a rectangular RCC column with the following specifications:

·        Column dimensions: 400 mm × 600 mm

·        Floor to floor height: 6.0 m

·        Concrete grade: M25 (fck = 25 N/mm²)

·        Steel grade: Fe 500 (fy = 500 N/mm²)

·        Beam dimensions: 300 mm × 500 mm

·        Effective cover: 40 mm

·        Effective length factor: k = 1.0

Analysis Results:

·        Axial load: Pu = 2000 kN

·        Moment about major axis: Mux = 60 kN-m

·        Moment about minor axis: Muy = 50 kN-m

Basic Calculations:

·        Floor to floor height = 6.0 m

·        Unsupported length of column, L = 6.0 - 0.5 = 5.5 m

·        Effective length factor, k = 1.0 (Given)

·        Effective length about both axes, Lex = Ley = 1.0 × 5.5 = 5.5 m

Step 1: Column Classification (IS 456:2000, Cl. 25.1.2)

To determine if the column is short or long, we need to check the slenderness ratio:

For X-axis (Major axis):
Lex/D = 5500/600 = 9.17 < 12
Hence, it is a short column about x-axis

For Y-axis (Minor axis):
Ley/b = 5500/400 = 13.75 > 12
Hence, it is a long column about y-axis

Note: Since the column is long about one axis, we need to consider slender column effects for the entire design.

Step 2: Additional Moments Due to Long Column Effect (IS 456:2000, Cl. 39.7)

For long columns, additional moments arise due to lateral deflection under axial load. These are calculated using the following formulas:

Additional moment about X-axis:
Max = Pu × D/2000 × (Lex/D)²
Max = 2000 × 600/2000 × (5500/600)²
Max = 2000 × 0.3 × (9.17)²
Max = 50.39 kN-m

Additional moment about Y-axis:
May = Pu × b/2000 × (Ley/b)²
May = 2000 × 400/2000 × (5500/400)²
May = 2000 × 0.2 × (13.75)²
May = 75.63 kN-m

Calculation of Reduction Factors (Kx and Ky)

Assume reinforcement percentage, Pt = 2.5%

Gross area of section:
Ag = 600 × 400 = 240,000 mm²

Area of reinforcement:
Asc = 240,000 × 2.5/100 = 6,000 mm²

Area of concrete:
Ac = Ag - Asc = 240,000 - 6,000 = 234,000 mm²

Ultimate axial load capacity:
Puz = 0.45 × fck × Ac + 0.75 × fy × Asc
Puz = 0.45 × 25 × 234,000 + 0.75 × 500 × 6,000
Puz = 2,632.5 + 2,250 = 4,882.5 kN

Finding Pbx and Pby values:

Given effective cover d' = 40 mm

About X-axis: d'/D = 40/600 = 0.067
About Y-axis: d'/b = 40/400 = 0.1

Using SP-16 Table 60 for finding K1 and K2 values:

For X-axis (d'/D = 0.067, approximately 0.05):

·        K1 = 0.222, K2 = 0.104

·        Pbx/(fck × b × D) = K1 + K2 × (Pt/fck)

·        Pbx/(25 × 400 × 600) = 0.222 + 0.104 × (2.5/25)

·        Pbx = (0.222 + 0.0104) × 25 × 400 × 600 = 1,394.4 kN

For Y-axis (d'/b = 0.1):

·        K1 = 0.207, K2 = 0.082

·        Pby/(fck × b × D) = K1 + K2 × (Pt/fck)

·        Pby/(25 × 400 × 600) = 0.207 + 0.082 × (2.5/25)

·        Pby = (0.207 + 0.0082) × 25 × 400 × 600 = 1,294.92 kN

Reduction factors:
Kx = (Puz - Pu)/(Puz - Pbx) = (4882.5 - 2000)/(4882.5 - 1394.4) = 0.826
Ky = (Puz - Pu)/(Puz - Pby) = (4882.5 - 2000)/(4882.5 - 1294.92) = 0.803

Final additional moments:
Max = 0.826 × 50.39 = 41.62 kN-m
May = 0.803 × 75.63 = 60.73 kN-m

Step 3: Primary Moments (IS 456:2000, Cl. 25.4)

Moments due to minimum eccentricity:

About X-axis:
ex = L/500 + D/30 = 5500/500 + 600/30 = 11 + 20 = 31 mm > 20 mm
Mux,min = Pu × ex = 2000 × 31 = 62,000 kN-mm = 62 kN-m

About Y-axis:
ey = L/500 + b/30 = 5500/500 + 400/30 = 11 + 13.33 = 24.33 mm > 20 mm
Muy,min = Pu × ey = 2000 × 24.33 = 48,660 kN-mm = 48.66 kN-m

Comparison with analysis moments:

·        Analysis moment Mux = 60 kN-m vs Minimum Mux = 62 kN-m → Use 62 kN-m

·        Analysis moment Muy = 50 kN-m vs Minimum Muy = 48.66 kN-m → Use 50 kN-m

Step 4: Final Design Moments

Total design moments:

·        Mux,total = Mux + Max = 62 + 41.62 = 103.62 kN-m

·        Muy,total = Muy + May = 50 + 60.73 = 110.73 kN-m

Step 5: Section Adequacy Check (IS 456:2000, Cl. 39.6)

The biaxial bending check equation:
[Mux/Mux1]^αn + [Muy/Muy1]^αn ≤ 1

Calculate αn:
αn = 1 + (Pu/Puz - 0.2)/(0.8 - 0.2) × 1
αn = 1 + (2000/4882.5 - 0.2)/(0.8 - 0.2) × 1
αn = 1 + (0.41 - 0.2)/0.6 = 1.35

Finding Mux1 and Muy1 from SP-16 Charts:

For X-axis (Chart 32):

·        d'/D = 0.067, Pt/fck = 2.5/25 = 0.1

·        Pu/(fck × b × D) = 2000 × 10³/(25 × 400 × 600) = 0.333

·        From chart: Mux1/(fck × b × D²) = 0.16

·        Mux1 = 0.16 × 25 × 400 × 600² = 576 kN-m

For Y-axis (Chart 34):

·        d'/b = 0.1, Pt/fck = 0.1

·        Pu/(fck × b × D) = 0.333

·        From chart: Muy1/(fck × D × b²) = 0.14

·        Muy1 = 0.14 × 25 × 600 × 400² = 336 kN-m

Biaxial check:
[103.62/576]^1.35 + [110.73/336]^1.35 = [0.18]^1.35 + [0.33]^1.35
= 0.134 + 0.253 = 0.387 < 1 ✓

The section is adequate!

Step 6: Final Reinforcement Details

Area of steel required:
Asc = 2.5% of 240,000 = 6,000 mm²

Using 25 mm diameter bars:
Area of one bar = π × 25²/4 = 490.87 mm²
Number of bars required = 6000/490.87 = 12.22 ≈ 12 bars

Actual steel provided:
Asc,provided = 12 × 490.87 = 5,890 mm² (2.45%)

Step 7: Lateral Ties Design (IS 456:2000, Cl. 26.5.3.2)

Diameter of lateral ties:
φ = 25/4 = 6.25 mm → Use 8 mm ties

Spacing of lateral ties:

·        Least lateral dimension = 400 mm

·        16 × φmain = 16 × 25 = 400 mm

·        Maximum spacing = 300 mm

Provide 8 mm ties at 300 mm c/c

Design Summary

Parameter	Value
Column Size	400 mm × 600 mm
Main Reinforcement	12 - φ25 mm bars
Lateral Reinforcement	φ8 mm @ 300 mm c/c
Steel Percentage	2.45%
Concrete Cover	40 mm

 

Key Design Points

1.      Column Classification: Always check slenderness ratio (Le/h) for both axes

2.     Additional Moments: Long columns require consideration of P-Δ effects

3.     Minimum Eccentricity: Compare with analysis moments and use the greater value

4.     Biaxial Bending: Use interaction equation for combined bending about both axes

5.     Reinforcement Detailing: Follow IS 456 requirements for spacing and cover

Conclusion

This comprehensive example demonstrates the complete procedure for designing long columns according to IS 456:2000. The key difference from short column design is the inclusion of additional moments due to slenderness effects, which significantly influence the final design moments and reinforcement requirements.

Remember to always verify your calculations and consult the latest version of IS 456 for any updates or modifications to the design procedure. Proper understanding of long column behavior is crucial for safe and economical structural design.

This design procedure follows IS 456:2000 guidelines and should be used in conjunction with current building codes and professional engineering judgment.