# Seismic Base Shear Calculator

Estimate the period, get the coefficient, multiply by seismic weight.

- Page: https://jianzhujisuan.com/en/seismic-base-shear
- Category: Structural Design (https://jianzhujisuan.com/en/c/structural-design)
- Chinese version: https://jianzhujisuan.com/zh/seismic-base-shear.md
- Data checked: 2026-10-25

## Quick result

With the default inputs (Structure height hn 20 m, Structural system Reinforced concrete frame, Design spectral SDS 1.0 g, Design spectral SD1 0.6 g, Long-period TL 8 s, Response modification R 5, Importance factor I 1.0, Seismic weight W 10000 kN): **Base shear V 1737 kN**

## Inputs

| Parameter | Type | Default | Unit | Note |
|---|---|---|---|---|
| Structure height hn | number | 20 | m |  |
| Structural system | select | Reinforced concrete frame | — |  |
| Design spectral SDS | number | 1.0 | g |  |
| Design spectral SD1 | number | 0.6 | g |  |
| Long-period TL | number | 8 | s |  |
| Response modification R | number | 5 | — |  |
| Importance factor I | number | 1.0 | — |  |
| Seismic weight W | number | 10000 | kN |  |

## Formula

- **Fundamental period**: T = Ct·hn^x
- **RC frame**: Ct = 0.0466, x = 0.9
- **Steel frame**: Ct = 0.0724, x = 0.8
- **Shear wall**: Ct = 0.0488, x = 0.75
- **Response coefficient (T ≤ TS)**: Cs = SDS/(R/I)
- **(TS < T ≤ TL)**: Cs = SD1/(T·R/I)
- **(T > TL)**: Cs = SD1·TL/(T²·R/I)
- **Plateau corner**: TS = SD1/SDS
- **Minimum**: Cs,min = 0.044·SDS·I ≥ 0.01
- **Base shear**: V = Cs·W

## Worked example

20 m reinforced concrete frame, SDS 1.0, SD1 0.6, R 5, I 1.0, seismic weight 10000 kN: T 0.691 s exceeds TS 0.60 s so the descending branch applies, Cs 0.174 and 1737 kN base shear, or 0.17W.

## Questions

### How do I calculate seismic base shear?

V = Cs W, and Cs has three branches. Below the plateau corner TS = SD1/SDS it is SDS/(R/I); between TS and TL it is SD1/(T R/I); beyond TL it is SD1 TL/(T² R/I). With SDS 1.0, SD1 0.6, R 5 and I 1.0, TS is 0.6 s, so a 20 m frame at T 0.69 s has already passed the corner and gets 0.6/(0.69 x 5) = 0.174, which on 10000 kN is 1737 kN. At T 0.4 s it would still be on the plateau at 0.2 and 2000 kN. The total is then distributed as Cvx = wx hx^k / sum(wx hx^k), with k = 1 for T up to 0.5 s and k = 2 beyond 2.5 s. The top floor has the largest share but there is no fixed ratio: on a ten-storey uniform building it is only about 0.18V.

### What value of R should I use?

The one your code assigns to the system and detailing level — it is not something to be raised for economy. In ASCE 7, special moment steel frames take R 8, intermediate 4.5, special moment concrete frames 8, ordinary concrete frames 3 and special shear walls 6. The crucial point is that a high R is permission to compute a smaller force in exchange for the ductile detailing that lets the structure survive: closed hoops at close spacing, strong column weak beam, strong shear weak flexure. Claiming the high R without building in that detailing simply spends the safety margin.

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