# Verification Example - Rigid Diaphragm Force Distribution

## Description

This verification example represents the
horizontal distribution of forces of a rigid diaphragm utilizing Tekla Tedds. This
example is based on Design Example 33 of the *2018 IBC SEAOC Structural / Seismic
Design Manual Volume 1 *(Pages 91 through 95). Comparisons and contrasts are
tabularized and discussed regarding the results from Tedds and the SEAOC Design
Example.

## Problem statement

Determine the following:

- Eccentricity and rigidity properties
- Direct shear in walls A and B
- Plan irregularity requirements
- Torsional shear in walls A and B
- Total shear in walls A and B

## Tedds calculation

Rigid diaphragm force distribution - Compared using version 1.1.05

## 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\Rigid diaphragm force distribution 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

The single-story building (shown in Figure 33-1 of the SEAOC Design Example) has a rigid roof diaphragm. Lateral forces in both directions are resisted by shear walls on four sides. The base shear and center of mass provided here account for the mass and the location of all contributing mass sources (roof and walls).

` `

Design base shear, V = 100 kips acting in the north/south direction

` `

Wall Rigidities:

R_{A} = 300 kip / in

R_{B} = 100 kip / in

R_{C} = R_{D} = 200 kip / in

` `

Center of mass (CM):

X_{m} = 40 ft

Y_{m} = 20 ft

## Notes and assumptions

- Walls A through D and in the design example correspond to Walls 1 through 4, respectively, in the Tedds calculation.
- Currently, accidental torsion caused by eccentricity needs to be reviewed separately. Tedds does not automatically calculate this. Two Tedds calculations are provided: one without accidental torsion, and one with accidental torsion analyzed for the eccentricity of 5% of the building dimension (4 ft).
- The rigidities determined in Tedds varies in magnitude to the values in the design example, but they are still of the same proportion to each other; i.e. SW1 is three times as stiff as SW2.
- The design example determines if the structure has any torsional irregularities (Type 1a or Type 1b). The example then calculates and applies the torsional factor (since it is assumed that the Seismic Design Category is C or greater) to the shear wall forces. These capabilities are currently outside of the scope of the Tedds Rigid Diaphragm Force Distribution module.

Comparison of Results between
Tedds and SEAOC Design Example 33 |
|||
---|---|---|---|

Component |
Tedds Result |
SEAOC Design Example 33 |
% Difference |

Center of rigidity (CR) | (20 ft, 20 ft) | (20 ft, 20 ft) | 0.0%^{} |

Eccentricity, e | 20 ft (x-direction) from CM | 20 ft (x-direction) from CM | 0.0% |

Direct Shear in Wall A, V_{D,A} |
75 kips | 75 kips | 0.0% |

Direct Shear in Wall B, V_{D,B} |
25 kips | 25 kips | 0.0% |

Shear due to accidental torsion in Wall A, V’_{T,A} |
-22.5 kips | -22.5 kips | 0.0% |

Shear due to accidental torsion in Wall B, V’_{T,B} |
-22.5 kips | -22.5 kips | 0.0% |

Total Shear in Wall A, V’_{A} |
52.5 kips | 52.5 kips | 0.0% |

Total Shear in Wall B, V’_{B} |
47.5 kips | 47.5 kips | 0.0% |

## Conclusion

Upon reviewing the results above, the determination of horizontal distribution of forces in a rigid diaphragm within Tedds matches the SEAOC design example.