Design horizontal coefficient Ah and total base shear Vb by the IS 1893:2016 equivalent static method.
Reference only. Approximate period Ta per IS 1893 (Part 1):2016 Cl. 7.6.2, equivalent-static spectrum Sa/g per Cl. 6.4.2(a) (5% damping — flat 2.5 down to T = 0; the 1 + 15T ramp is for the response spectrum method only), design coefficient Ah per Cl. 6.4.2, and Vb = AhW (Cl. 7.2.1) not less than the Table 7 minimum ρW (Cl. 7.2.2). The storey table builds W per Cl. 7.3 / 7.4 and Table 10 and distributes Vb per Cl. 7.6.3. A complete design also requires torsion and dynamic-analysis scaling (Cl. 7.7.3). Read the clauses in the Code Clause Compass.
How the Seismic Base Shear Calculator works
This calculator gives the design horizontal acceleration coefficient Ah and the total design base shear Vb by the equivalent static method of IS 1893 Part 1:2016. You choose the seismic zone, importance factor I (1.0, 1.2 or 1.5), response reduction factor R (3 for OMRF or 5 for SMRF), soil type, and the period formula, then enter the building height h and the seismic weight W in kN — or switch W to the storey table and let the page build W and h from the floors.
The approximate fundamental period Ta comes from Cl 7.6.2: 0.075h^0.75 for an RC moment frame, 0.085h^0.75 for a steel moment frame, or 0.09h/√d for a frame with masonry infill, where d is the base dimension in the direction considered. The spectral acceleration Sa/g is read from the 5 % damped spectrum for the equivalent static method, Cl 6.4.2(a): a flat 2.5 up to the corner period (0.40 s for rock, 0.55 s for medium soil, 0.67 s for soft soil), then 1.00/T, 1.36/T or 1.67/T up to 4 s, and held at the printed 0.25, 0.34 or 0.42 beyond 4 s.
Ah is then (Z/2) × (I/R) × (Sa/g), and Vb = Ah × W. The result is not allowed to fall below the minimum design lateral force of Cl 7.2.2 (Table 7): 0.7 %, 1.1 %, 1.6 % and 2.4 % of W for Zones II to V. When that minimum controls, the page says so.
The optional storey table works out the seismic weight floor by floor, listed from the base up, with one row allowed to stand for several identical floors. Each floor takes its full dead load (in kN/m² over the floor area, or a total in kN) plus 25 % of the imposed load when the imposed load is 3.0 kN/m² or less and 50 % when it is more (Cl 7.3.1, Table 10). Imposed load on the roof is left out (Cl 7.3.2). W is the sum of the floor weights (Cl 7.4.2) and h is the height of the top level. The base shear is then shared out over the levels as Qi = Vb × Wi·hi² / Σ Wj·hj² (Cl 7.6.3), and the table lists each Qi with the storey shear Vi below that level.
- Period, RC moment frame
- Ta = 0.075 × h^0.75
- Period, steel moment frame
- Ta = 0.085 × h^0.75
- Period, frame with masonry infill
- Ta = 0.09 × h / √d
- Spectral acceleration (medium soil)
- Sa/g = 2.5 for Ta ≤ 0.55 s; 1.36 / Ta for 0.55 s < Ta ≤ 4 s
- Design horizontal coefficient
- Ah = (Z / 2) × (I / R) × (Sa/g), not less than ρ
- Design base shear
- Vb = Ah × W
- Seismic weight of floor i (storey table)
- Wi = DL + f × IL, f = 0.25 for IL ≤ 3.0 kN/m², 0.50 above, 0 on the roof
- Lateral force at level i
- Qi = Vb × Wi·hi² / Σ Wj·hj²
Follows: IS 1893 Part 1:2016 Table 3 (zone factor Z) · IS 1893 Part 1:2016 Cl 6.4.2 (Sa/g and Ah) · IS 1893 Part 1:2016 Cl 7.2.2 and Table 7 (minimum lateral force) · IS 1893 Part 1:2016 Cl 7.3.1, Cl 7.3.2 and Table 10 (imposed load in the seismic weight) · IS 1893 Part 1:2016 Cl 7.4 (seismic weight) · IS 1893 Part 1:2016 Cl 7.6.1 and Cl 7.6.2 (base shear and period) · IS 1893 Part 1:2016 Cl 7.6.3 (distribution of base shear over the height)
Worked example: base shear for a five-storey RC frame building in Zone III
Given
- Seismic zone III, Z = 0.16
- Importance factor I = 1.0, response reduction R = 5 (SMRF)
- Soil type II (medium), RC moment frame without infill
- Building height h = 15 m
- Seismic weight W = 20,000 kN
Steps
- Period: Ta = 0.075 × 15^0.75 = 0.075 × 7.622 = 0.572 s.
- Ta is beyond the 0.55 s corner for medium soil, so Sa/g = 1.36 / 0.572 = 2.379.
- Ah = (0.16 / 2) × (1.0 / 5) × 2.379 = 0.08 × 0.2 × 2.379 = 0.0381.
- Minimum for Zone III is ρ = 1.1 % = 0.011. 0.0381 is larger, so the spectrum value governs.
- Vb = 0.03806 × 20,000 = 761 kN.
Result: Ta = 0.572 s, Sa/g = 2.379, Ah = 0.0381 and design base shear Vb = 761 kN.
Enter the same values in the calculator above to check each step.
Frequently asked questions
How to calculate design base shear as per IS 1893:2016?
Find the approximate period Ta from Cl 7.6.2, read Sa/g for that period and soil type, then work out Ah = (Z/2)(I/R)(Sa/g) and multiply by the seismic weight: Vb = Ah × W. For example, Zone III, I = 1.0, R = 5, medium soil, a 15 m RC frame and W = 20,000 kN give Ah = 0.0381 and Vb = 761 kN.
What is the formula for the fundamental natural period in IS 1893?
For a bare RC moment frame Ta = 0.075h^0.75, for a bare steel moment frame Ta = 0.085h^0.75, and for RC frames with masonry infill Ta = 0.09h/√d, with h and d in metres. A 15 m building with a 16 m base dimension gives Ta = 0.09 × 15 / 4 = 0.338 s by the infill formula, against 0.572 s as a bare RC frame.
What is the minimum design base shear in IS 1893:2016?
Cl 7.2.2 and Table 7 set a floor on the base shear of ρW, with ρ = 0.7 % in Zone II, 1.1 % in Zone III, 1.6 % in Zone IV and 2.4 % in Zone V. It usually governs tall, flexible buildings in low zones: a 60 m RC frame on rock in Zone II has Ah = 0.0062 from the spectrum, so the 0.007 minimum controls and Vb = 700 kN for W = 100,000 kN.
Why is Sa/g taken as 2.5 for short-period buildings in the static method?
For the equivalent static method, Cl 6.4.2(a) keeps Sa/g flat at 2.5 from T = 0 up to the corner period of the soil type. The 1 + 15T ramp below 0.1 s applies only to the response spectrum method, so this calculator does not use it.
What does the seismic base shear calculator not check?
It does not check torsion or scale results from a response spectrum analysis (Cl 7.7.3), and it covers only the moment-frame R values of 3 and 5. The storey table needs the dead load of each floor from you, including half the walls and columns of the storeys above and below it (Cl 7.4.1); it does not take off quantities from a model.
How is the seismic weight calculated as per IS 1893?
Each floor counts its full dead load plus a share of its imposed load: 25 % when the imposed load is 3.0 kN/m² or less and 50 % when it is more (Cl 7.3.1, Table 10). Imposed load on the roof is not counted (Cl 7.3.2). For a G+4 frame of 300 m² floors with 10 kN/m² dead and 3.0 kN/m² imposed, each floor is (10 + 0.25 × 3.0) × 300 = 3,225 kN, the roof at 8 kN/m² is 2,400 kN, and W = 4 × 3,225 + 2,400 = 15,300 kN.
How is the base shear distributed to each floor in IS 1893?
Cl 7.6.3 gives Qi = Vb × Wi·hi² / Σ Wj·hj², with hi measured from the base. For the G+4 building above with 3.2 m storeys, Zone III, medium soil, I = 1.2 and R = 5, Vb = 665.9 kN and the roof takes 665.9 × 60,000 / 156,750 = 254.9 kN, the fourth floor 219.2 kN and the first floor only 13.7 kN. The storey shear below each level is the sum of the forces at and above it, so the ground storey carries the full 665.9 kN.
Step-by-step guides
- Base Shear Calculation as per IS 1893:2016 — Worked ExampleCalculate seismic base shear by the IS 1893:2016 equivalent static method: period Ta, Sa/g, Ah and Vb, with a worked 7-storey infilled frame in Zone IV.
- Wind Load Calculation as per IS 875 Part 3 — Worked ExampleWork out design wind speed Vz and wind pressure pd to IS 875 Part 3:2015, with the k1–k4 factors, a k2 table and worked examples for Hyderabad and Bhubaneswar.
- SBC from SPT N Value (IS 6403) — Worked ExampleHow to find safe bearing capacity from SPT N value: N corrections, the IS 6403 shear check, Meyerhof, Bowles and Burland & Burbidge settlement methods, water table effects and IS 1893 liquefaction screening, with a full worked example.
- Masonry Wall Design (IS 1905) — Worked ExampleMasonry wall design to IS 1905 step by step: slenderness ratio, basic stress fb, stress reduction factor ks, ka, kp, permissible vs actual stress, eccentricity, and IS 4326 seismic bands, with a fully checked 230 mm G+2 brick wall example.
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