Shear walls
A shear wall is a reinforced concrete wall that resists lateral load by acting as a deep vertical cantilever fixed at the foundation. Because its stiffness in its own plane is far greater than that of a column, a few well-placed walls can control the drift of a building in wind and earthquakes and keep partitions, services and the occupants safe.
Under lateral load, a slender wall bends: one edge goes into tension and the other into compression, and shear runs through the web. Vertical reinforcement resists the bending tension, horizontal reinforcement resists shear, and both are distributed in two curtains in walls of normal thickness. At the edges, where compression is highest, boundary elements with closely spaced confining hoops prevent crushing and bar buckling. Openings for doors and windows turn a solid wall into coupled walls linked by short deep beams, which need diagonal reinforcement to work as fuses.
Placement matters as much as strength. Walls arranged symmetrically in plan, preferably near the perimeter, resist torsion; a single wall core on one side causes the building to twist. Walls should run continuously from the foundation to the top; stopping a wall at the first floor to open up a lobby creates a dangerous discontinuity.
Foundations of shear walls see large overturning moments and need a raft, piles or a heavy combined footing. In analysis, cracked-section stiffness is used for drift checks. Indian rules for ductile walls are in IS 13920, with overall seismic design in IS 1893.
Start with walls near the perimeter on both axes; then check torsional irregularity.
Cantilever wall, top load P, height H: base moment M = P H; flexural tip deflection δ = P H³ / (3 E I)Wall in-plane I = t l_w³ / 12Short-term modulus (IS 456): E_c = 5 000 √f_ck MPa
Stopping walls above the ground floor, which creates a soft and weak storey.
Problem. A wall 4 m long and 200 mm thick is 12 m tall and carries a lateral load of 300 kN at the top. Concrete M25. Using the gross section, find the base moment and the flexural tip deflection.
- Base moment M = 300 × 12 = 3 600 kNm.
- E_c = 5 000 √25 = 25 000 MPa = 25 × 10⁶ kPa. I = 0.2 × 4³ / 12 = 1.067 m⁴, so E I = 2.667 × 10⁷ kNm².
- δ = 300 × 12³ / (3 × 2.667 × 10⁷) = 518 400 / 8.0 × 10⁷ = 0.00648 m.
- δ ≈ 6.5 mm. With cracked stiffness and shear deformation the real value is larger.
Answer. Base moment 3 600 kNm; gross-section flexural deflection about 6.5 mm
Quick quiz
1What resists shear in the web of a wall?
Horizontal reinforcement (with the concrete).
2Why provide boundary elements?
To confine highly compressed wall edges against crushing and bar buckling.
3Why should walls be placed symmetrically?
To avoid torsion of the building in plan.
- IS 13920 (India): Ductile design and detailing of reinforced concrete structures for seismic forces.
- IS 1893 (Part 1) (India): Criteria for earthquake-resistant design: seismic zones and how to work out earthquake forces.
- IS 456 (India): Plain and reinforced concrete: general rules for materials, durability, design and construction.
- ACI 318 (USA): Building code requirements for structural concrete.
- EN 1998 (Eurocode 8) (Europe): Design of structures for earthquake resistance.
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- S. Unnikrishna Pillai, Devdas Menon, Reinforced Concrete Design, McGraw Hill Education (India), latest edition.
- J. K. Wight, J. G. MacGregor, Reinforced Concrete: Mechanics and Design, Pearson, latest edition.
- A. H. Nilson, D. Darwin, C. W. Dolan, Design of Concrete Structures, McGraw Hill, latest edition.
- R. Park, T. Paulay, Reinforced Concrete Structures, Wiley, 1975.
Free and official sources
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