Shear Wall Design in Buildings | Basics, Types & IS Code Guidelines

Complete Guide to RCC Shear Wall Design: Step-by-Step Manual Approach as per IS 13920:2016

In modern high-rise and earthquake-resistant buildings, shear walls play a crucial role in maintaining stability and safety. A shear wall is a vertical structural element designed to resist lateral forces such as wind, seismic loads, and horizontal pressure, ensuring that the structure remains strong and durable. These walls not only enhance the strength and stiffness of a building but also reduce sway and vibration during natural forces. In this blog, we will explore the basics of shear wall design, types, IS code provisions, step-by-step design procedure, and reinforcement details, making it a complete guide for civil engineering students, professionals, and construction enthusiasts

Shear walls are critical structural elements in multi-story buildings that resist lateral forces from earthquakes and wind loads. This comprehensive guide demonstrates the detailed design procedure for a reinforced concrete shear wall following Indian Standards IS 13920:2016, with practical calculations and real-world considerations.

RCC Shear Wall Design

Project Overview and Given Data

We'll design a shear wall for a G+3 RCC building with the following specifications:

·        Wall Length (Lw): 3500 mm

·        Wall Thickness (tw): 300 mm

·        Total Height (hw): 12000 mm (12 m)

·        Concrete Grade: M30 (fck = 30 MPa)

·        Steel Grade: Fe 500 (fy = 500 MPa)

Design Loads from Structural Analysis

Load Combination	Axial Load (kN)	Bending Moment (kN-m)	Shear Force (kN)
DL + LL + EQ	3200	6800	850

 

Step 1: Preliminary Calculations and Code Compliance

1.1 Effective Depth Calculation

As per IS 13920:2016 (Clause 10.2.1):

Effective depth of wall, dw = 0.8 × Lw
dw = 0.8 × 3500 = 2800 mm

1.2 Factored Forces

Factored Bending Moment, Mu = 1.2 × 6800 = 8160 kN-m
Factored Shear Force, Vu = 1.2 × 850 = 1020 kN
Factored Axial Load, Pu = 1.2 × 3200 = 3840 kN

1.3 Minimum Thickness Check

According to IS 13920:2016 (Clause 10.1.2):

·        Minimum thickness = 150 mm

·        Provided thickness = 300 mm > 150 mm ✓

1.4 Wall Classification

Calculate the aspect ratio:

hw/Lw = 12000/3500 = 3.43 > 2

Classification: Since hw/Lw > 2, this is a slender wall

Note: Wall classification criteria:

·        hw/Lw < 1 → Squat Wall

·        1 ≤ hw/Lw ≤ 2 → Intermediate Wall

·        hw/Lw > 2 → Slender Wall

Step 2: Shear Reinforcement Design (Horizontal)

2.1 Shear Stress Calculation

Factored shear stress, Tv = Vu/(tw × dw)
Tv = 1020 × 10³/(300 × 2800) = 1.21 N/mm²

2.2 Two-Layer Reinforcement Requirement

As per IS 13920:2016 (Clause 10.1.7):
Since Tv > 0.25√fck OR tw > 200 mm, provide steel in two curtains/layers

0.25√fck = 0.25√30 = 1.37 N/mm²
tw = 300 mm > 200 mm

Both conditions satisfied → Two-layer reinforcement required

2.3 Bar Diameter Selection

As per IS 13920:2016 (Clause 10.1.8):

Minimum bar diameter = tw/10 = 300/10 = 30 mm

However, since 30mm bars are not commonly available, we'll use T25 mm bars.

2.4 Maximum Spacing Limitations

The maximum spacing shall not exceed the smallest of:

1. Lw/5 = 3500/5 = 700 mm
2. 3 × tw = 3 × 300 = 900 mm 
3. 450 mm (code limit)

Governing spacing = 450 mm

2.5 Shear Strength Check

Maximum permissible shear stress:

Tc,max = 0.63√fck = 0.63√30 = 3.45 N/mm²
Tv (1.21) < Tc,max (3.45) ✓ → Section adequate for shear

Design shear strength from IS 456:2000 Table 19:

For pt = 0.25%, Tc = 0.48 N/mm²
Tv (1.21) > Tc (0.48) → Shear reinforcement required

2.6 Horizontal Reinforcement Calculation

Minimum reinforcement ratio:

ρh,min = 0.0025 + 0.5(hw/Lw - 2)(ρv,web - 0.0025)
ρh,min = 0.0025 (for this case)

Required area per meter run:

Ah = 0.0025 × 300 × 1000 = 750 mm²/m (in two layers)

Shear force carried by steel:

Vus = (Tv - Tc) × tw × dw
Vus = (1.21 - 0.48) × 300 × 2800 = 612.6 kN

Required reinforcement:

Ah/Sv = Vus/(0.87 × fy × dw)
Ah/Sv = 612.6 × 10³/(0.87 × 500 × 2800) = 0.503

Using T10 mm bars (Area = 78.54 mm²):

Spacing = (2 × 78.54)/(0.503 × 300) = 1.04 → Provide 200 mm c/c

Final check:

Provided: Ah/Sv = (2 × 78.54)/200 = 0.785 > 0.503 required ✓

Provide T10 @ 200 mm c/c horizontal bars in two layers

Step 3: Vertical Reinforcement Design

3.1 Minimum Vertical Reinforcement

ρv,net,min = 0.0025 + 0.01375 × (tw/Lw)
ρv,net,min = 0.0025 + 0.01375 × (300/3500) = 0.00368 or 0.368%

Required area:

Av = 0.00368 × 300 × 1000 = 1104 mm²/m (in two layers)

Using T12 mm bars (Area = 113.09 mm²):

Spacing = (2 × 113.09 × 1000)/1104 = 205 mm

Provide T12 @ 200 mm c/c vertical bars in two layers

Step 4: Boundary Element Requirement Check

4.1 Stress Analysis

Calculate maximum and minimum stresses:

f = Pu/A ± Mu/Z

Where:
A = tw × Lw = 300 × 3500 = 1,050,000 mm²
Z = (tw × Lw²)/6 = (300 × 3500²)/6 = 612.5 × 10⁶ mm³

f = 3840 × 10³/1,050,000 ± 8160 × 10⁶/612.5 × 10⁶
f = 3.66 ± 13.32

fmax = 16.98 MPa
fmin = -9.66 MPa

4.2 Boundary Element Criterion

flim = 0.2 × fck = 0.2 × 30 = 6.0 MPa
fmax (16.98 MPa) > flim (6.0 MPa) → Boundary elements required

Step 5: Boundary Element Design

5.1 Boundary Element Dimensions

Provide boundary elements of size 300 × 600 mm at both ends of the wall.

5.2 Flexural Strength Check Using Annex-A Method

Steel percentage calculation:

Total vertical steel area = (113.09 × 2) × 3500/200 = 3963.15 mm²
pt = Av/(tw × Lw) = 3963.15/(300 × 3500) = 0.00378

Load parameter:

λ = Pu/(fck × tw × Lw) = 3840 × 10³/(30 × 300 × 3500) = 0.122

Steel parameter:

φ = (0.87 × fy × pt)/fck = (0.87 × 500 × 0.00378)/30 = 0.055

Neutral axis calculation:

Xu/Lw = (φ + λ)/(2φ + 0.36) = (0.055 + 0.122)/(2 × 0.055 + 0.36) = 0.379

Limiting neutral axis:

Xu*/Lw = 0.0035/(0.0035 + 0.87fy/Es) = 0.0035/(0.0035 + 0.87 × 500/200000) = 0.617

Since Xu/Lw < Xu*/Lw, the section is under-reinforced.

Moment capacity calculation:
After solving the complex equation from Annex-A:

Mu,capacity = 6950 kN-m < 8160 kN-m required

5.3 Extra Moment for Boundary Elements

Extra moment = 8160 - 6950 = 1210 kN-m
Lever arm = 3500 - 300 - 300 = 2900 mm = 2.9 m

Forces in boundary elements:

Fraction of B.E area = (300 × 600)/(300 × 3500) = 0.171
Axial load share = 0.171 × 3840 = 657 kN

Compression B.E force = 657 + 1210/2.9 = 1074 kN
Tension B.E force = 657 - 1210/2.9 = 240 kN

Step 6: Boundary Element Reinforcement Design

6.1 Longitudinal Reinforcement

As per IS 13920:2016 (Clause 10.4.2), design as short column with minimum 0.8% steel:

Ast,min = 0.8/100 × 300 × 600 = 1440 mm²

Capacity check:

Pu,capacity = 0.4 × fck × (Ac - Ast) + 0.67 × fy × Ast
Pu,capacity = 0.4 × 30 × (180000 - 1440) + 0.67 × 500 × 1440
= 2150 + 482 = 2632 kN > 1074 kN ✓

Provide 8-T16 mm bars (Ast = 8 × 201 = 1608 mm²)

6.2 Confinement Reinforcement (Stirrups)

Required stirrup area:

Ash = 0.05 × Sv × h × fck/fy

Where h = larger dimension measured from outer faces
h = 600 - 2 × 40 = 520 mm (considering 40mm cover)

Using T8 mm stirrups:

Sv = (Ash × fy)/(0.05 × h × fck)
Sv = (2 × 50.27 × 500)/(0.05 × 520 × 30) = 64.4 mm

Maximum spacing limits:

1. h/3 = 520/3 = 173 mm
2. 6 × φ of main bar = 6 × 16 = 96 mm
3. 100 mm (code limit)

Provide T8 @ 90 mm c/c throughout the boundary element

Step 7: Design Summary and Reinforcement Schedule

Main Wall Reinforcement

Component	Vertical Reinforcement	Horizontal Reinforcement
Diameter & Spacing	T12 @ 200 mm c/c	T10 @ 200 mm c/c
Layout	Two layers (each face)	Two layers (each face)
Steel Ratio	0.378%	0.262%

 

Boundary Element Details

Parameter	Specification
Dimensions	300 × 600 mm
Longitudinal Bars	8-T16 mm
Confinement	T8 @ 90 mm c/c
Steel Ratio	0.895%

 

Step 8: Detailing Requirements

8.1 Development Length and Anchorage

·        All reinforcement must be properly anchored as per IS 456:2000

·        Lap lengths for vertical bars: 50 × bar diameter

·        Hook details for boundary element stirrups: 135° bends with 6db extension

8.2 Construction Joints

·        Horizontal construction joints at every floor level

·        Surface preparation and bonding agent application required

·        Continuous vertical reinforcement through joints

8.3 Quality Control Measures

·        Regular inspection of reinforcement placement

·        Concrete compaction using mechanical vibrators

·        Curing for minimum 28 days for design strength achievement

Conclusion

This comprehensive shear wall design demonstrates the systematic approach required for seismic-resistant construction as per IS 13920:2016. The key aspects covered include:

1.      Proper classification of wall type based on aspect ratio

2.     Two-layer reinforcement due to thickness and shear stress requirements

3.     Boundary element provision due to high compressive stresses

4.     Ductile detailing for earthquake resistance

5.     Quality control measures for construction

The design ensures adequate strength, ductility, and seismic performance while maintaining practical construction feasibility. Regular code updates and site-specific conditions should always be considered in actual projects.

Key Design Ratios Summary

·        Vertical reinforcement ratio: 0.378% (> 0.25% minimum)

·        Horizontal reinforcement ratio: 0.262% (> 0.25% minimum)

·        Boundary element steel ratio: 0.895% (0.8-6% range)

·        Shear stress ratio: Tv/Tc,max = 1.21/3.45 = 0.35 (< 1.0 ✓)

This design provides a robust foundation for safe, earthquake-resistant construction while meeting all codal requirements and practical construction constraints.

❓ Frequently Asked Questions (FAQs)

Q1. What is shear wall design in buildings?
Shear wall design refers to the process of analyzing and detailing reinforced concrete (RC) walls that resist lateral forces like wind and earthquake loads, ensuring stability and safety of tall and slender structures.

Q2. Why are shear walls important in high-rise buildings?
Shear walls are essential in high-rise structures because they reduce lateral sway, prevent collapse during earthquakes, and increase stiffness and strength, making the building safer.

Q3. Which IS code is used for shear wall design in India?
In India, IS 456:2000 (Plain and Reinforced Concrete Code of Practice) and IS 13920:2016 (Ductile Detailing of RC Structures for Seismic Forces) are commonly used for shear wall design.

Q4. What are the types of shear walls?
The main types of shear walls are:

• Simple rectangular shear walls

• Coupled shear walls

• Flanged shear walls (L-shaped, T-shaped, U-shaped)

• Core type shear walls

Q5. How do you calculate shear wall thickness?
The thickness of a shear wall depends on the height of the building, design loads, and IS code provisions. Generally, it ranges from 150 mm to 400 mm in reinforced concrete structures.

Q6. What software is used for shear wall design?
Popular software for shear wall design includes ETABS, STAAD Pro, SAP2000, and SAFE, which provide accurate modeling and analysis for lateral load resistance.

Q7. What are the advantages of using shear walls?
Advantages include:

• Higher lateral load resistance

• Increased stiffness and stability

• Reduced structural vibrations

• Cost-effective for tall buildings

• Enhanced earthquake safety