The story modeling "--" corresponds to the known modeling from RFEM. No floor set or load transfer surface is generated.
Nevertheless, the building model results in a wide range of improved and simplified information:
- Centers of mass and rigidity
- Mass per story
- Mass center
- Cumulative mass / center
- Story actions
- Story forces
- Delta story forces
- Location of resultant story forces
- Interstory drifts
- Displacement
- Delta interstory drift
- Gravity center of story
Center of Mass and Rigidity
The center of mass, just like the mass in the table Static Analysis – Results by Story – Centers of Mass and Rigidity, refers to the mass including the applied vertical loads. For the example here, a surface load of 2.7 kN/m² is applied in Load Combination 2. With a surface of 6 m x 6 m, this results in a mass of 9.72 metric tons. The center of mass is located in the center of the building. This output is helpful when there are many different loads on a story.
The center of mass and the mass are calculated for each defined story.
Mass of the first floor for the example used here:
- 2 concrete walls, 0.9 metric t each
- 2 CLT walls, 0.151 metric t each
- Concrete ceiling, 14.4 metric t
- 2*0.9+2*0.151+14.4=16.5 metric t
The cumulative mass and its center of gravity refer to the story above.
Story Center of Gravity
The story center of gravity is the center of gravity in each story. The point is displayed in the Building Stories dialog box and in the Structure – Building Model – Building Story table.
For the first floor of this example in the Y-direction:
The following image shows the result from the building model as a comparison:
Story Actions
The Story Actions result table includes the story forces, the difference of the story forces by story, and the position of the resulting story forces.
The story forces are always displayed for the top and bottom points of a story. The example in this chapter using Load Case 2 results in a horizontal force in the Y-direction of 2 kN/m x 6 m = 12 kN, which is introduced on each floor. With a correspondingly refined FE mesh setting, this result is easily achieved, as shown in the following image.
Minor deviations compared to the manual calculation result from the warping or rotation of the walls at the line hinges and line supports.
Interstory Drifts
Similarly to the Story Action table, the Interstory Drift table lists the displacements of each story and the difference in the displacement in each story.
In the attached example, the maximum displacement on the top floor LC2 is 8.25 mm in the global Y-direction. Since the displacement in the lower floor is 3.4 mm, a value of 4.85 mm is displayed as the difference.
Vertical Result Lines
Once a Building Story is defined, Vertical Result Lines are available for this story in the Navigator – Results. This allows all deformations and forces in the respective floor to be displayed graphically.
Walls
The definition of walls and deep beams is described in the previous chapters of this manual. In this example, a shear wall is defined on the cross-laminated timber wall of Surfaces 11 and 12.
As soon as the shear walls are defined, two additional result tables are displayed.
The -Forces in Walls- table shows the total forces and the forces per unit length.
These forces are converted into member forces in the Member Forces in Walls table by using a result member. These forces are also used for the design of the shear walls in the Concrete Design or Timber Design add-ons. Furthermore, these forces are available in the Navigator – Results like normal member internal forces.
Moreover, the shear walls automatically generate result sections at the top and bottom points of the story. Among other things, the resultant force of a wall is also displayed graphically here.
The Member Forces in Walls table shows the percentage critical compression load as well as the compression. In this example, it is calculated in Load Case 1 as follows:
ηNcr = N / Ncr = 22.3 kN / 1,737 kN = 1.28%
The ηNc column is calculated from the allowable compression force divided by the existing compression force.
ηNc = N / Nc = 22.3 kN / 2,520 kN = 0.89%
Nc = fck = 2.1 kN/cm² * 1,200 cm² = 2,520 kN
