*GThe structural frame with 6 levels, project <B_d1>*

In appendix Β’, the crossbar was used to assess the behaviour of the multistorey plane frame.

In this chapter, the composition of space frames, through beams and slabs, is considered. The diaphragm is the structural unit for the assessment of coupled space frames.

A diaphragm is the horizontal part of the floor consisting of slabs, connecting beams and columns. A floor may comprise more than one diaphragm, as shown in the example building of the present paragraph.

The most characteristic point of the diaphragm is the centre of stiffness.

The centre of stiffness C_{T} of a specific floor diaphragm is defined as the point about which the diaphragm rotates, under horizontal seismic force *H*.

The C_{T} point depends only on the geometry of the floor and is independent of the loadings, meaning that regardless the magnitude of the force, which may become equal to 2H, or 3H, or any other value, the centre of stiffness will remain the same. Of course, it depends on the geometry of the overlying and the underlying floors.

Principal axis system of diaphragm

is defined as the orthogonal axis system xC_{T}y, in which when a horizontal seismic force H is applied on C_{T}, along axis x (or y), it results in translating C_{T} only along axis x (or y respectively). The angle *a *of the principal system, as to the initial system X0Y, is called principal angle of the diaphragm.

(The cross-sections of columns, walls and beams are 500/500, 2000/300 and 300/500 respectively)

The above diaphragm data as well as the torsional stiffness *K** _{θ}* and torsional ellipse stiffness ellipse (

The behaviour of diaphragm on the left at the 5^{th} level, analysed next is directly affected by the L-type frame C_{21}-C_{26}-C_{27}-C_{28} connected to it and indirectly by column C_{6} and the diaphragm on the right, through the frames of other floors.

A general method for the calculation of the diaphragm data is presented next.

**This method creates a condition allowing the diaphragm only to rotate about the centre of stiffness C _{T}, which remains stationary with respect to the ground.**

The method comprises four steps, the first three of which include a space frame analysis.

*Loading: Force H _{X}=100 and moment M_{CM,X} on centre of mass C_{M} of diaphragm i*

Results: total displacements of structure and displacements of diaphragm i

The two displacements δ_{XX1}, δ_{XY1} of point 1 of diaphragm i and its angle θ_{XZ} are required.

Point 1 corresponds to column Κ2 joint, but could be any point of the diaphragm.

The only results needed from this analysis are the translations of point 1 *δ*_{XX1}*=2.681 mm*, *δ*_{XY1}*=-0.231 mm* and the angle *θ*_{XZ}*=4.5613**×10 ^{-5}* of the diaphragm. The displacements of point 2 will be used only to verify the generality of the method.

All displacements are absolute with respect to the ground.

*Loading: H _{X}=100 on diaphragm i with rotational restraint*

*Results: the total displacements of structure and the two parallel translations of diaphragm*

The two translations of point 1 δ_{XXo}, δ_{XYo}, are only required, being identical for any point of the diaphragm i, thus also for C_{T}, since the angle of rotation of the diaphragm is zero.

All diaphragm points, thus also CT, have identical displacements

*Diaphragm is restrained against rotation, therefore **θ*_{XZ}*=0*

Due to zero rotation of diaphragm i, point 1 has only two parallel translations, δ_{XXo}=2.103 mm and δ_{XYo}=-0.047mm, being identical for all diaphragm points, thus also for C_{T}. In other diaphragms located in the same, upper or lower level, small, yet measurable rotations exist, thus every point on them has different displacements.

*Loading: H _{Y}=100 on diaphragm i with rotational restraint
Results: the total displacements of structure and the two parallel translations of diaphragm*

The two translations of point 1 δ_{XXo}, δ_{XYo}, are only required, being identical for any point of the diaphragm i, thus also for C_{T}, provided that the angle of rotation of the diaphragm is zero*.*

All diaphragm points, thus also C_{T}, have identical displacements

*Diaphragm is restrained against rotation, therefore θ _{XZ}=0*

Due to zero rotation of diaphragm i, point 1 has only two parallel translations, δ_{YXo}=-0.047 and δ_{YYo}=1.668 mm, being identical for all diaphragm points, thus also for C_{T}. As in step 2, in other diaphragms located in the same, upper or lower level small, yet measurable rotations exist, thus every point on them has different displacements.

The secondary displacements δ_{XYo} of loading 1 and δ_{YXo} of loading 3 are always equal, i.e. δ_{XYo}=δ_{YXo} should always derive from the space frame analysis.

Analysis 3 is performed only to obtain translation δ_{YYo} needed for the calculation of angle a of the principal system.

The angle a of the principal system of diaphragm i is calculated by means of the equation C.9.1 of §C.9 using the 2^{nd} and 3^{rd} analysis results:

*The only load acting on diaphragm i is moment on C _{T}*

*Diaphragm displacements are induced only due to rotation*

*The centre of stiffness C _{T} remains stationary with respect to the ground*

Using this trick, namely by subtracting analysis 2 results from analysis 1 results, the diaphragm i develops only rotation, while the remaining diaphragms develop both translations and rotation. However only diaphragm i is examined here. The most important result of this trick is that the diaphragm i is rotated about C_{T}, which remains stationary with respect to the ground, allowing the calculation of its precise location. The rotation angle of diaphragm i is the rotation angle θ_{XZ} calculated in step 1 under the corresponding loading.

The displacements of an arbitrary point j of the diaphragm, due to rotation *θ** _{XZ}* derive form the expressions:

*δ _{Χ}*

The C_{T} coordinates corresponding to these displacements are:

*X _{CT}=X_{j}-*

*Y _{CT}=Y_{j}+*

In the example considered, for point 1:

*δ _{Χ}*

*δ*_{Yt1}*=**δ*_{XY1}*-**δ*_{XYo}*=-0.211-(-0.047)=-0.164 mm*

The diaphragm rotation is identical to that of analysis 1, i.e. *θ*_{XZ}*=**4.5613·10 ^{-5}*.

The coordinates of point 1 are (6.0, 0.0), thus:

*X _{CT}=X_{1}-*

*Y _{CT}=Y_{1}+δ_{Xt1}/_{θXZ}=0.0+0.578·10^{-3}/(4.5613·10^{-5})=12.7 m*

Each diaphragm has two lateral stiffnesses in the two principal directions.

Lateral stiffness of diaphragm *K _{xx}* (or

*K _{xx}=H_{xx}/*

Using analysis 2, the external force *H _{X}=100* of the initial system X0Y is resolved into two equivalent principal forces

Respectively, the two translations of the centre of stiffness in X0Y, derived from analysis 2, are *δ*_{XXo}*=2.103*, *δ _{ΧΥο}*

*δ*_{xxo}*=**δ*_{XXo}**·***cosa-**δ*_{XYo}**·***sina=2.103*·*0.994-(-0.047)*·*(-0.107)=2.090-0.005=2.085 mm *

*δ*_{xyo}*=-**δ*_{XXo}**·***sina+**δ*_{XYo}**·***cosa=-2.103*·*(-0.107)+(-0.047)*·*0.994=0.225-0.047=0.178 mm*→

*K _{xx}=H_{xx}/*

Torsional stiffness of diaphragm *K** _{θ}* is defined as the ratio of moment

Moment acting on C_{T} is equal to

*M _{XCT}=100.0*

Torsional stiffness distribution of diaphragm is the curve on which, if the idealised columns with the same lateral stiffnesses as those of the diaphragm, are placed symmetrically with respect of the centre of stiffness, the torsional stiffness derived is the same as that of the diaphragm.

Torsional stiffness ellipse of diaphragm is defined as the ellipse having C_{T} as centre and *r _{x}=√(K*

*r _{x}=√K_{θ}/K_{yy}=√[5178×10^{6}Νm/60.1×10^{6}N/m]=9.28 m*

r_{y}=√K_{θ}/K_{xx}=√[5178×10^{6}Nm/47.7×10^{6}N/m]=10.42 m

The equivalent building should comprise same number of floors and diaphragms. The columns of each floor should have specific dimensions, thus for each floor relative lateral and torsional stiffnesses are used.

Each diaphragm is replaced by an equivalent one consisting of *4* fixed-ended columns placed symmetrically with respect to its centre of stiffness.

The equivalent diaphragm *i* is assumed to behave as an one-storey structure of fixed-ended columns. The four relative translations of the centre of stiffness along *X, Y* axes *δ*_{XXoΖ}_{,i}*, δ*_{XYoΖ}_{,i}* , δ*_{YXoΖ}_{,i}*, δ*_{YYoΖ}* _{,i}* as well as the rotation angle

*δ*_{XXo}_{Ζ}_{,i}*= **δ*_{XXo,i}*-**δ*_{XXo,i-1}* , **δ*_{XYo}_{Ζ}_{,i}*=**δ*_{XYo,i}*-**δ*_{XYo,i-1}* ,
*

Therefore, the data of the equivalent diaphragm are: rotation angle a_{z,i}=2δ_{XYo}_{Ζ}_{,i}/( δ_{XXo}_{Ζ}_{,i}- δ_{YYo}_{Ζ}_{,i}),

torsional stiffness K_{θ}_{Z,i}=H**·**c_{Y}/ θ_{XZ}_{Μ}_{Z,i} , lateral stiffnesses K_{xxZ,i}= H/(δ_{XXoZ,i}+δ_{XYoZ,i}**·**tana_{Z,i}) και K_{yy}_{Ζ}_{,}_{ι}=H/(δ_{YYoZ,i}-δ_{XYoZ,i}**·**tana_{Z,i}) (equations C.9.2 and C.9.3 of §C.9).

Also r_{xZ,i}=√K_{θ}_{Z,i}/K_{yyZ,i}, r_{yZ,i}=√K_{θ}_{Z,i}/K_{xxZ,i}

The location of the equivalent diaphragms is determined by translating the centre of stiffness to point 0.0, 0.0. In this way the equivalent building is created and for its diaphragms i the following relations apply:

*δ*_{XXo_equal,i}*=**Σ**(**δ*_{XXo}_{Ζ}_{,k}*)** *όπου* **k=i **έως** 1** *à* **δ*_{XXo_equal,i}*=**Σ**(**δ*_{XXo,i}*-**δ*_{XXo,i-1}*)= (**δ*_{XXo,i}*-**δ*_{XXo,i-1}*)+ (**δ*_{XXo,i-1}*-**δ*_{XXo,i-2}*)+…+ (**δ*_{XXo,1}*-0.0)=**δ*_{XXo,i}* *à* **δ*_{XXo_equal,i}*=**δ*_{XXo,i}* *

Using the same logic* **δ*_{XYo_equal,i}*=**δ*_{XYo,i }*, **δ*_{YYo_equal,i}*=**δ*_{YYo,i}* **και** **θ*_{XZ_equal,i}*= **θ*_{XZ}_{Μ}_{,i}

Therefore all diaphragm quantities *K _{xx_equal,i}* ,

*The seismic assessment of the actual building can be performed in an easy, direct and descriptive way by means of the equivalent building.*