# 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
*C*and lateral seismic force_{vx}*F*at each level._{x}

## 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

C_{s} = 0.062

R = 8.0

Ω_{0} = 3.0

I_{e} = 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
- T
_{L}= 8 sec. (assumed) - S
_{DS }= 0.496g (based on equation 12.8-2 with provide R, C_{s}, and I_{e}values) - S
_{S}= 0.546g (based on calculated S_{DS}, and Equations 11.4-1 and 11.4-3 and Table 11.4-1) - S
_{D1min}= 0.526g (minimum value of S_{D1}to satisfy Equation 12.8-3) - S
_{1min}= 0.42g (based on calculated S_{D1min}and Equations 11.4-2 and 11.4-4 and Table 11.4-2) - Since S
_{1}> 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 ≤ 1.5T
_{s}- T
_{s}= S_{D1}/ S_{DS}= 0.526 / 0.496 = 1.06 sec. - 1.06 ≤ 1.5*1.06 = 1.59 sec.

- T

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).