Equivalent static method
The equivalent static method replaces the dynamic earthquake load on a regular building with a set of horizontal static forces. It works well when the building responds mainly in its first mode: regular in plan and elevation, and not too tall. IS 1893 (Part 1) sets the conditions under which it may be used and requires dynamic analysis beyond them.
The steps are fixed. Estimate the fundamental period from an empirical formula based on height and structural system. Read the spectral acceleration coefficient from the design spectrum for that period and the soil type. Combine it with the zone factor, importance factor and response reduction factor to get the design horizontal coefficient. Multiply by the seismic weight, which is the full dead load plus a specified fraction of the imposed load, to get the design base shear. A minimum base shear also applies.
The base shear is then distributed up the height in proportion to each floor's weight times the square of its height above the base. This parabolic distribution reflects the first-mode shape, in which upper floors move more and therefore carry larger inertia forces. Storey shears, overturning moments and drifts follow from these floor forces.
The method is simple and transparent, which makes it valuable for checking software output even when a full dynamic analysis is used. Torsion from accidental eccentricity must be added, and both principal directions must be considered.
Check that the floor forces add back to the base shear before moving on.
Approximate period, RC moment frame without infill (IS 1893): T_a = 0.075 h^0.75A_h = (Z / 2)(I / R)(S_a / g); V_B = A_h WFloor force: Q_i = V_B W_i h_i² / Σ W_j h_j²Medium soil spectrum (IS 1893:2016): S_a/g = 2.5 for 0.10 ≤ T ≤ 0.55 s; 1.36 / T for 0.55 to 4.0 s
Using the full imposed load in the seismic weight instead of the specified fraction.
Problem. A 4-storey RC moment frame without infill, 12 m tall on medium soil in zone IV, has floor seismic weights of 2 000 kN at 3, 6 and 9 m and 1 500 kN at the roof (12 m). Take I = 1.0 and R = 5. Find the base shear and floor forces.
- T_a = 0.075 × 12^0.75 = 0.075 × 6.447 = 0.484 s, within the plateau, so S_a / g = 2.5.
- A_h = (0.24 / 2)(1.0 / 5)(2.5) = 0.06; W = 3 × 2 000 + 1 500 = 7 500 kN; V_B = 0.06 × 7 500 = 450 kN.
- W h²: 18 000, 72 000, 162 000 and 216 000; sum 468 000.
- Floor forces: Q₁ = 450 × 18/468 = 17.3 kN, Q₂ = 69.2 kN, Q₃ = 155.8 kN, Q₄ = 207.7 kN (sum 450 kN).
Answer. V_B = 450 kN; floor forces 17.3, 69.2, 155.8 and 207.7 kN from first floor to roof
Quick quiz
1Why does the floor force grow with height squared?
It follows the first-mode shape, where upper floors move most.
2What is the seismic weight?
Dead load plus a specified fraction of imposed load.
3When is the equivalent static method not enough?
For tall or irregular buildings, where higher modes matter.
- IS 1893 (Part 1) (India): Criteria for earthquake-resistant design: seismic zones and how to work out earthquake forces.
- EN 1998 (Eurocode 8) (Europe): Design of structures for earthquake resistance.
- ASCE/SEI 7 (USA): Minimum design loads (dead, live, wind, snow, seismic) for buildings and other structures.
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.
- A. K. Chopra, Dynamics of Structures, Pearson, latest edition.
- T. Paulay, M. J. N. Priestley, Seismic Design of Reinforced Concrete and Masonry Buildings, Wiley, 1992.
- P. Agarwal, M. Shrikhande, Earthquake Resistant Design of Structures, PHI Learning, latest edition.
Free and official sources
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