Verification Example - Seismic Forces Vertical Distribution
Description
This verification example represents the vertical distribution of seismic forces of a special steel moment frame structure utilizing Tekla Tedds. This example is based on Design Example 32 of the 2018 IBC SEAOC Structural / Seismic Design Manual Volume 1 (Pages 87 through 90). Comparisons and contrasts are tabularized and discussed regarding the results from Tedds and the SEAOC Design Example.
Problem statement
Determine the following:
- Seismic base shear, V
- Vertical distribution exponent, k
- Vertical distribution factor Cvx and lateral seismic force Fx at each level.
Tedds calculation
Seismic forces (ASCE7) - Compared using version 3.1.05
Running the example in Tedds
The Tedds verification examples referenced in this document can be run in Tekla Tedds from the Engineering library index, in the Verification Examples\Seismic forces (ASCE7) folder.
References
2018 IBC SEAOC Structural / Seismic Design Manual Volume 1
ASCE 7-16 Minimum Design Loads and Associated Criteria for Buildings and Other Structures
Example information
Lateral force resisting system: Special steel moment-resisting frame (the design example does not specifically state the steel moment frame is a special steel moment frame, but based off of R=8, it can be concluded that the lateral force resisting system is a special steel moment frame).
W = 3,762 kips
Cs = 0.062
R = 8.0
Ω0 = 3.0
Ie = 1.0
T = 1.06 sec.
See Figure 32-1 of the SEAOC design example for story height and story weight.
Notes and assumptions
- Site Class D
- TL = 8 sec. (assumed)
- SDS = 0.496g (based on equation 12.8-2 with provide R, Cs, and Ie values)
- SS = 0.546g (based on calculated SDS, and Equations 11.4-1 and 11.4-3 and Table 11.4-1)
- SD1min = 0.526g (minimum value of SD1 to satisfy Equation 12.8-3)
- S1min = 0.42g (based on calculated SD1min and Equations 11.4-2 and 11.4-4 and Table 11.4-2)
- Since S1 > 0.2, a ground motion hazard analysis is required per
Section 11.4.8(3). However, exception (2) can be taken because:
- Cs is determined by Equation 12.8-2
- T ≤ 1.5Ts
- Ts = SD1 / SDS = 0.526 / 0.496 = 1.06 sec.
- 1.06 ≤ 1.5*1.06 = 1.59 sec.
Comparison of Results between Tedds and SEAOC Design Example 32 | |||
---|---|---|---|
Component | Tedds Result | SEAOC Design Example 32 | % Difference |
Base Shear, V | 233.3 kips | 233.2 kips | 0.0% |
Vertical Distribution Exponent, k | 1.28 | 1.28 | 0.0% |
Table 1: Vertical force distribution table (Tedds results)
Table 2: Vertical force distribution table (SEAOC Design Example 32)
Conclusion
Upon reviewing the results above, the determination of vertical distribution of seismic forces within Tedds match the SEAOC design example (apart from minor differences due to rounding and precision).