Appendix B

                     Common practice frame                                                               Equivalent system

In multistorey plane frames, the quantity measuring the behaviour under horizontal seismic forces is the crossbar stiffness.

Crossbar is the horizontal part of a frame connecting columns. The crossbar is either usually a flanged beam or a slab strip.

Stiffness K_i, of crossbar i, of a plane frame is the ratio of the horizontal seismic force H acting on the crossbar, parallel to the plane of the frame, to the corresponding crossbar displacement δ_i, i.e. K_i=H/δ_i.

To compare the interstorey stiffnesses, it is very helpful to use a simpler frame type structure, with the same behaviour as the actual frame under horizontal seismic forces.

Equivalent multistorey plane system is a virtual frame consisting of the same number of storeys, with fixed-ended columns, on which when horizontal seismic force acts on a crossbar same it produces the same displacement as the one of the actual crossbar.

Analysis of crossbar i under horizontal force H Example: for Η=100 kN and i=2, δ2=2.589 mm	Analysis of crossbar i-1 under horizontal force H Example: for Η=100 kN and i-1=1, δ1=0.936mm
The relative stiffness of crossbar i is equal to ΚZ,i=H/( δi-δi-1). Example: ΚΖ,2=100.00×103N/[(2.589-0.936)×10-3m=60.5 N/m

The equivalent frame may comprise one, two, or more vertical columns and should result in the same displacements as the actual frame. It is easier to perceive this equivalence using an one-bay frame composed of equivalent fixed-ended columns and eventual segmented crossbars.

Relative stiffness K_Ri, of crossbar i, of a multistorey plane frame is the ratio of the horizontal seismic force Η acting on crossbar i, to the relative displacement between the crossbar i and the underlying crossbar i-1, when identical force Ηacts on the latter, i.e. K_Ri=H/(δ_i-δ_i-1).

The stiffnesses Κ_i and the relative stiffnesses Κ_Zi of all crossbars are calculated tabulating the above procedure

The shear forces V_i,j carried by the column j of level i and
the corresponding relative stiffnesses Κ_Zi,j

Relative stiffness of crossbar column of multistorey plane frame is the ratio of its shear force to the force H multiplied by the relative crossbar stiffness K_Ri.

Important: The examination of the behaviour of multistorey frames, beyond its educational purpose is also the basis of the assessment of the multistorey space frames, which function along both plane frame directions x and y. Of course, the rotation factor, considered in Appendix D, is also involved in the space frame analysis.

Indicatively a four-storey frame of 3.0 m high columns is analysed. Each floor comprises columns of 400/400 cross-section, beams of 6.0 m span and 300/500 cross-section and slabs of 150 mm thickness and 1.0 m width at each side of beam faces.

The deformation of the structure

First analysis: a plane frame is analysed, where the influence of slabs is accounted for assuming the effective flange width in accordance with EC2. In frame analysis, beams and columns are usually modelled with members. The influence of slabs is taken into account by adding a b_eff wide flange to the rectangular beam, and a flanged beam is formed, usually of T section.

First analysis results: as shown in the previous pages, the displacement of the crossbar at the 3^rd level, due the horizontal force H=100 kN, is δ_3,1=5.735 mm, therefore its stiffness isK_3,1=H/δ_3,1=100.0·10³N/(5.735·10^-3m)=17.4·10⁶ N/m.

The following elastic displacements δ_i,1 and stiffnesses K_i,1 are obtained from similar analyses applying force H=100 kN separately on each floor i_.

4^th level: H=100 kN, δ_4,1=8.051 mm, K_4,1=12.4·10⁶ N/m

3^rd level: H=100 kN, δ_3,1=5.735 mm, K_3,1=17.4·10⁶ N/m

2^nd level: H=100 kN δ_2,1=3.500 mm, K_2,1=28.6·10⁶ N/m

1^st level: H=100 kN, δ_1,1=1.358 mm, K_1,1=73.6·10⁶ N/m

The deformation of the structure

Second analysis: frame type structure, members and finite elements,
under a horizontal force H acting at the 3^rd level.

Second analysis results: as shown in the previous page, the displacement of the crossbar at the 3^rd level, due to horizontal force H=100.0 kN, is δ_3,2=4.266 mm, therefore its stiffness yields K_3,2=H/δ_3,2=100.0·10³N/(4.266·10^-3m)=23.4·10⁶ N/m.

The following elastic displacements δ_i,1 and stiffnesses K_i,1 are obtained from similar analyses due to the force H=100 kN separately on each floor i.

4^th level: H=100 kN, δ_4,2=6.073 mm, K_4,2=16.5·10⁶ N/m

3^rd level: H=100 kN, δ_3,2=4.266 mm, K_3,2=23.4·10⁶ N/m

2^nd level: H=100 kN δ_2,2=2.528 mm, K_2,2=39.6·10⁶ N/m

1^st level: H=100 kN, δ_1,2=0.886 mm, K_1,2=112.9·10⁶ N/m

Conclusion: The frame stiffness derived from the finite element analysis is higher.

As the effect of slabs on stiffness was examined in the previous paragraph, in order to obtain comparable results, a two-bay frame without slabs is considered next.
Indicatively a four-storey frame with 3.0 m high columns is analysed. Each floor comprises two columns of 400/400 cross-section, one wall of 2000/300 cross-section and two beams of 6.0 m span and 300/500 cross-section.
Columns and beams are modelled with members, while walls are modelled with members and rigid bodies at their tops, or with triangular finite elements.

The deformation of the structure

First analysis results: The displacement of the crossbar at the 3^rd level, due to horizontal force H=100 kN, is δ_3,1=1.552 mm, therefore its stiffness is K_3,1=H/δ_3,1=100.0·10³N/(1.552·10^-3m)=64.4·10⁶ N/m.

The following elastic displacements δ_i,1 and stiffnesses K_i,1 are obtained from similar analyses applying force H=100 kN separately on each floor i.

4^η level: H=100 kN, δ_4,1=3.033 mm, K_4,1=33.0·10⁶ N/m

3^rd level: H=100 kN, δ_3,1=1.552 mm, K_3,1=64.4·10⁶ N/m

2^η level: H=100 kN δ_2,1=0.629 mm, K_2,1=159.0·10⁶ N/m

1^η level: H=100 kN, δ_1,1=0.146 mm, K_1,1=685.0·10⁶ N/m

The deformation of the structure

Second analysis results: The displacement of the crossbar at the 3^rd level, due to horizontal force H=100.0 kN, is δ_3,2=1.399 mm, therefore its stiffness yields K_3,1=H/δ_3,1=100.0·10³N/(1.399·10^-3m)=71.5·10⁶ N/m.

The following elastic displacements δ_i,1 and stiffnesses K_i,1 are obtained from similar analyses applying force H=100 kN separately on each floor i.

4^th level: H=100 kN, δ_4,2=2.783 mm, K_4,2=35.9·10⁶ N/m

3^rd level: H=100 kN, δ_3,2=1.399 mm, K_3,2=71.5·10⁶ N/m

2^nd level: H=100 kN δ_2,2=0.543 mm, K_2,2=184.0·10⁶ N/m

1^st level: H=100 kN, δ_1,2=0.117 mm, K_1,2=855.0·10⁶ N/m

In the example considered, at the 3^rd level, the stiffness obtained using finite element modelling is (71.5-64.4)/64.4=11% higher, while at the 4^th, 2^nd and 1^st level 9%, 16% and 25% higher, respectively.

Conclusion: The frame stiffness slightly differs when walls are modelled with two-dimensional finite elements.