Seismic base shear calculator
The equivalent lateral force procedure turns a design spectrum, a ductility factor and an importance factor into one coefficient, then applies it to the building weight. This engine computes the calculated, capped and minimum coefficients, shows which one governs, and reports the base shear.
| Value | Result | Status |
|---|---|---|
| Calculated SDS/(R/Ie) | 0.167 | overridden |
| Maximum SD1/(T·(R/Ie)) | 0.125 | used |
| Minimum 0.044·SDS·Ie | 0.044 | below calculated |
| Governing Cs | 0.125 | maximum (period) |
| Base shear V | 1250 kN | — |
Vertical distribution, torsional amplification, drift and diaphragm forces are separate checks.
Cs = SDS / (R/Ie) · Elastic demand divided by the response modification factor.
Cs,max = SD1 / (T·(R/Ie)) · Controls for long-period structures.
Cs,min = max(0.044·SDS·Ie, 0.01) · Prevents unrealistically low design forces.
V = Cs·W · W is the seismic weight, typically dead plus a share of live.
A clear path from inputs to results.
The equivalent lateral force procedure turns a design spectrum, a ductility factor and an importance factor into one seismic coefficient, then applies it to the building weight. This calculator computes the calculated, capped and minimum coefficients, shows which one governs, and reports the base shear.
How to use
- Enter SDS and SD1 from the site hazard for the selected risk category.
- Set the response modification factor R for the structural system.
- Set the importance factor Ie.
- Enter the fundamental period and the seismic weight.
The formula
Reference: ASCE 7 equivalent lateral force procedure (§12.8); site coefficients from the hazard maps.
Worked example
SDS 1.0, SD1 0.6, R 6, Ie 1.0, T 0.8 s, W 10,000 kN
- SDS 1.0.
- SD1 0.6.
- R = 6, Ie = 1.0.
- T = 0.8 s.
- Seismic weight 10,000 kN.
- 1
Calculated coefficient: 1.0 / 6 = 0.167.
- 2
Period cap: 0.6 / (0.8 × 6) = 0.125.
- 3
Minimum: max(0.044 × 1.0, 0.01) = 0.044.
- 4
The cap governs, so Cs = 0.125.
- 5
Base shear: 0.125 × 10,000 = 1,250 kN, or 12.5% of the building weight.
Cs 0.125 (period-capped) · base shear 1,250 kN.
Interpreting the result
The cap exists because long-period structures attract less force than the short-period spectrum implies. Notice that R does most of the work: moving from R = 3 to R = 6 halves the design force, which is why the choice of structural system — and the detailing that justifies it — is the single biggest seismic decision on a project.
Assumptions & limitations
- Regular structures that qualify for the equivalent lateral force procedure.
- The period is either the code empirical value or a properly computed fundamental period.
- Seismic weight includes dead load plus the code-specified share of other loads.
- Vertical distribution, torsional amplification, drift limits and diaphragm forces are separate checks.
Frequently asked questions
Which coefficient governs?
The smaller of the calculated and period-capped values, but never below the minimum. The engine labels the governing one.
What R value should I use?
It depends on the seismic force-resisting system and its detailing, from the ASCE 7 table. Using a high R without the corresponding detailing is unsafe.
Why does a longer period reduce the force?
The design spectrum falls with period, so a flexible structure attracts less acceleration — at the cost of larger displacements.
Does this cover drift?
No. Drift and P-delta checks are separate, and often govern tall or soft-storey buildings.
Methodology last reviewed 2026-10-21. Results are informational and not financial or engineering advice — see our disclaimer.