Onestorey space frame with rectangular columns in parallel arrangement
The analysis of the onestorey space frame illustrated in the figure, under horizontal seismic force H=90.6 kN is performed by means of four methods: (i) analysis using manual calculations, assuming fixedended columns, (ii) analysis using the Excel file, assuming fixedended columns, (iii) analysis using the Excel file, assuming columns with k=6, (iv) analysis using software, assuming actual beam and column torsional stiffnesses.
Figure 5.4.5: The structure and the model of the onestorey space frame Figure 5.4.5: The structure and the model of the onestorey space frame
The names, the loadings, the coordinates and the dimensions of slabs and columns are shown in figure of §5.4.1. The crosssection of beams is 250/500. The concrete class is C30/37 (Ε=32.80 GPa). Analysis using manual calculations, assuming fixedended columns (k=12) This method was described via the example presented to all previous paragraphs of this chapter. Analysis using the Excel file, assuming fixedended columns (k=12) The related <diaphragm_ortho.xls> spreadsheet is used assuming fixedended columns.
Figure 5.4.5.2: The results are identical to those of the practical calculations.The torsional stiffness ellipse and the equivalent columns of the structure are illustrated at the end of the spreadsheet. Figure 5.4.5.2: The results are identical to those of the practical calculations.The torsional stiffness ellipse and the equivalent columns of the structure are illustrated at the end of the spreadsheet.
Analysis using the Excel file, assuming columns with k=6 The centre of stiffness, the torsional radii and the stress resultants are exactly the same as those of the previous cases, with the exception of deformations, whose relative values, however, remain constant 12/6=2.00. Analysis using software, assuming actual beam and column torsional stiffnesses The analysis may only be performed using the software. The relevant project is <B_545>. The theory for the analytical determination of the diaphragmatic behaviour at a storey level is described in detail in Appendix D.
Figure 5.4.5.41 Figure 5.4.5.41
The equivalent mass inertial ring is the same as in the two previous methods, since it depends only on the loadings. On the contrary all other quantities are different, as expected. The centre of stiffness C_{T}coordinates are (3.646, 3.314), whereas the torsional stiffness radii are equal to r_{x}=3.920 m και r_{y}=3.572 m (instead of 3.910 and 3.080 respectively, assuming fixedended columns). The crosssection of the equivalent columns is 415/378 (instead of 524/406). All results are listed in details by pressing "Display", "Diaphragm results", "report". These results are presented below, along with the corresponding results (in parentheses) of the 1^{st} and 2^{nd} method considering columns fixed at both ends. Seismic action in X: θ_{XZ}= 4.3186·10^{5} (4.765·10^{5} ) mm, δ_{XXo}=0.6840 (0.3375) mm, δ_{XYo}=0.0 (0.0) mm Seismic action in Y: θ_{YZ}=3.4940·10^{5 }(3.345·10^{5}) mm, δ_{YXo}=0.0 (0.0) mm, δ_{YYo} =0.8239 (0.5420) mm
Figure 5.4.5.42 Figure 5.4.5.42

Figure 5.4.5.43 Figure 5.4.5.43

Figure 5.4.5.44 Figure 5.4.5.44
  Diaphragm restrained against rotation  (1^{st} Loading) minus (2^{nd} Loading): M_{CT},_{X}=90.6 · y_{CM}+90.6 · c_{Y} 
Figure 5.4.5.45 Figure 5.4.5.45

Figure 5.4.5.46 Figure 5.4.5.46

Figure 5.4.5.47 Figure 5.4.5.47
 The displacements of each point i δ _{X,i} , δ _{Y,i} and the rotation angle of the diaphragm  Each point of the diaphragm (therefore the C_{T }as well_{ }) has the same principal displacements δ _{XXo} =0.684 mm , δ _{XYo} =0.  The diaphragm develops only a rotation θ _{XZ} about C_{T}. The displacements of each point i due to rotation are equal to: δ _{Xt,i} = δ _{X,i}  δ _{XXo} , δ _{Yt,i} = δ _{Y,i}  δ _{XYo } .  Calculation of the diaphragmatic behaviour (continued) 1^{st} (and only) floor level
Figure 5.4.5.48 Figure 5.4.5.48

Figure 5.4.5.410 Figure 5.4.5.410
 Diaphragm restrained against rotaion 
Figure 5.4.5.49 Figure 5.4.5.49
 The diaphragm is not rotated, but only translated in parallel to the axes X, Y. Each point of the diaphragm (therefore and the C_{T}) has the same principal displacement: δ _{YXo} =0, δ _{YYo} =0.824 mm. The 3 ^{rd }analysis completes the necessary series of analyses for the determination of all diaphragm data.  tan(2a)=2 δ _{XYo} /( δ _{XXo}  δ _{YYo} )=0.0 à 2a=0° à a=0° δ _{xxo} = δ _{XXo} =0.684 mm, δ _{yyo} = δ _{YYo} =0.824 mm K_{xx}=H_{x}/ δ _{xxo} =90.6·10^{3}m/0.684·10^{3}m=132.5·10^{6} N/m K_{yy}=H_{y}/ δ _{yyo} =90.6·10^{3}m/0.824·10^{3}m=110.0·10^{6} N/m M_{CT,X}=90.6 · y_{CM}+90.6 · c_{Y}=90.6 ·(3.3162.500)+90.6·1.0=164.5 kNm K _{θ} = M_{CT},_{X}/ θ _{XZ} =164.5/9.681 ·10^{5}=17.0·10^{5}kNm r_{x}=√K _{θ} /K_{yy}=√17.0·10^{8}N/m/110.0·10^{6}N/m=3.931m r_{y}=√K _{θ} /K_{xx}=√17.0·10^{8}N/m/132.5·10^{6}N/m=3.582 m 
