Seismic Base Shear (ASCE/SEI 7-22 Equivalent Lateral Force)
Finds the total horizontal earthquake force at the base of a building, the design base shear, by the equivalent lateral force procedure of ASCE/SEI 7-22 §12.8 [1]. Structural engineers use it for the lateral design of regular low- and mid-rise buildings.
How it is calculated
V = Cs·W (Eq. 12.8-1)
Cs = SDS / (R/Ie) (Eq. 12.8-3)
Cs ≤ SD1 / [T·(R/Ie)] for T ≤ TL
where W is the effective seismic weight, SDS and SD1 the design spectral accelerations at short periods and at 1 s (in g), R the response modification coefficient, Ie the importance factor and T the fundamental period. Above the long-period transition TL the cap is SD1·TL/[T2·(R/Ie)]. That is Method 2. Method 1, new in 7-22, takes Cs = Sa/(R/Ie), with Sa read from the multi-period design spectrum at T (or its peak, if T is below the peak’s period). It may not be used where Exception 2 of §11.4.5 applies. Either way, Cs is held at or above 0.044·SDS·Ie and 0.01, and where the mapped 1 s acceleration S1 ≥ 0.6 also at or above 0.5·S1/(R/Ie).
R, Ω0 and Cd come from Table 12.2-1 for the chosen system, and Ie from Table 1.5-2. The approximate period is Ta = Ct·hnx (Eq. 12.8-8, Table 12.8-2), with hn the structural height in meters. SDS, SD1, Sa, S1 and TL are site values from Chapter 11 or the ASCE Hazard Tool.
Limits and common mistakes
- A period from analysis may not exceed Cu·Ta (Table 12.8-1). Using a long computed period without that cap gives too low a force.
- Check Table 12.2-1 for the system’s height limits and the seismic design categories where it is permitted. The page does not.
- §12.6 limits where this procedure may be used; tall or irregular buildings may need modal analysis.
- W is the dead load plus the parts of other loads §12.7.2 names, not the full live load.
- V is a strength-level force. The ASCE 7 load combinations apply factors to it.
Worked example
A four-story office of Risk Category II is braced by steel special concentrically braced frames. Its structural height is 16 m and its effective seismic weight 20,000 kN. The site gives SDS = 0.8, SD1 = 0.4, S1 = 0.35 and TL = 8 s. Find the design base shear.
- Keep Seismic Response Coefficient Cs Method 2 and Risk Category II.
- Enter Design Spectral Acceleration, Short 0.8, Design Spectral Acceleration, 1 s 0.4 and Mapped MCE Spectral Acceleration, 1 s 0.35.
- Choose Seismic Force-Resisting System Steel special concentrically braced frames (R 6).
- Choose Fundamental Period Approximate period, Structure Type All other structural systems and enter Structural Height 16 m.
- Keep Long-Period Transition 8, enter Effective Seismic Weight 20000 and press Solve. The Solution shows:
Approximate fundamental period Ta = Ct·hn^x: 0.3904 s
Cs = SDS/(R/Ie): 0.133333
Cs,max: 0.170765
Cs,min: 0.0352
Controlling limit: Cs (SDS)
Seismic base shear V = Cs·W: 2,666.67 kN
Overstrength factor Ω0 (Table 12.2-1): 2
Deflection amplification factor Cd (Table 12.2-1): 5
Check by hand. 160.75 = 8, so Ta = 0.0488 × 8 = 0.390 s. Cs = 0.8 / 6 = 0.1333, below the cap 0.4 / (0.390 × 6) = 0.171. V = 0.1333 × 20,000 = 2,667 kN, about 13 % of the building’s weight.