Two examples are analysed applying the general method on the same simple structure of Appendix C.1. The choice of a simple structure is useful for easy monitoring of the results and the comprehension of the diaphragmatic floor behaviour of floors. In the onestorey structure of project <B_d91> the two translations of node 5 (column C1) and the diaphragm rotation θ _{XZ} should be calculated, using either the related software or any other relevant software. Optionally, the translations of the remaining points of the diaphragm may be calculated. All diaphragm data can derive based on these displacements.
Figure D.7.11: The simple structure of 4 columnsC1:400/400, C2:400/400, C3:800/300 φ=30º, C4:300/600 φ=45º, h=3.0 m, beams 250/500 Figure D.7.11: The simple structure of 4 columnsC1:400/400, C2:400/400, C3:800/300 φ=30º, C4:300/600 φ=45º, h=3.0 m, beams 250/500
Figure D.7.12: Output image of the software Figure D.7.12: Output image of the software
The mass inertial ring is the same as in the previous case, since it depends only on the loads. However all other results are different, as expected. The values of the coordinates of the centre of stiffness C_{T} are (2.806, 4.193), and the torsional radii arer_{x}=4.390 m, r_{y}=3.842 m versus the values 2.688, 4.897, r_{x}=4.411 and r_{y}=3.381 obtained from the assumption of fixedended columns in Appendix C. All results are displayed in detail by selecting from the menu “View” and then “Diaphragm results”, “report”. θ_{XZ} =11.952·10^{5}. The remaining results are better displayed in 3D, by selecting from the menu “View”, “Diaphragm results”, “3D floor” combined with “free rotation analysis” or “fixed rotation analysis” or “rotation only”, as presented in the two following pages. In the onestorey structure considered the “only rotation” condition can derive directly from the analysis by applying only moment as external loading, the reason being that the centre of stiffness C_{T} in onestorey diaphragms remains, by definition, stationary with respect to the ground. Calculation [*]Note The software performs the calculations of the diaphragm automatically. The verification of the algorithms using the software tools is presented here. of the diaphragmatic behaviour
Figure D.7.13 Figure D.7.13

Figure D.7.14 Figure D.7.14

Figure D.7.15 Figure D.7.15
  Loading 2 : H_{X}=90.6 kN Diaphragm restrained against rotation  Loading 3 : H_{Y}=90.6 kN Diaphragm restrained against rotation 
Figure D.7.16 Figure D.7.16

Figure D.7.17 Figure D.7.17

Figure D.7.18 Figure D.7.18
 Analysis results: The translations of point 1 are δ _{XX1} =1.178, δ _{XY1} =0.395mm and the diaphragm rotation is θ _{XZ} =11.952 ·10^{5}  Analysis results: The diaphragm develops only parallel translations in X,Y directions, being restrained against rotation. Thus all diaphragm points (including C_{T}) develop the same displacements: δ _{XXo} =0.677 mm , δ _{XYo} =0.060.  Analysis results: The diaphragm develops only parallel translations in X,Y directions, being restrained against rotation, which are: δ _{YXo} =0.0600, δ _{YYo} =0.839 mm The angle of the principal system derives from the expression: tan(2a)=2 δ _{XYo} /( δ_{ΧΧο}  δ _{YYo} )= =2 ·(0.060)/(0.6770.839)= =0.741 → 2a=36.5° → a=18.2°  Calculation of the diaphragmatic behaviour (continued)
Figure D.7.19 Figure D.7.19

Figure D.7.110 Figure D.7.110
 Loading 1 minus loading 2: H_{X}=0, M_{XCT}=90.6 · (Y_{CT} Y_{CM} ) +90.6 
Figure D.7.111 Figure D.7.111
 Subtraction results: The diaphragm only rotates by θ _{XZ} about the centre of stiffness C_{T}. The translations of the first point due to rotation: δ _{Xt,1} = δ _{X,1}  δ _{XXo} =1.1780.677 =0.501 mm, δ _{Yt,1} = δ _{Y,1}  δ _{XYo } =0.395+0.060 =0.335mm and X_{CT}=X_{1} δ _{Yt,1} / θ _{XZ} = =0.0+0.335 ·10^{3} /11.952 ·10^{5} =2.803 m Y_{CT }=Y_{1}+ δ _{Xt,1} / θ _{XZ} = 0.0+0.501 ·10^{3} /11.952 ·10^{5} =4.192 m  Determination of stiffnesses and torsional radii : The lateral stiffnesses K_{xx}, K_{yy} are calculated using the expressions C.9.2 and C.9.3 of §C.9, with a=18.186° and tana=0.329: K_{xx}=H/(δ_{XXo}+δ_{XYo}·tana)=[90.6/(0.6770.060·0.329)]· 10^{6}N/m= = 137.8·10^{6} N/m K_{yy}=H/(δ_{YYo}_{XYo}·tana)=[90.6/(0.839+0.060·0.329)]·10 ^{6}N/m= =105.5·10^{6} N/m M_{XCT} [*]NoteMoment, rotation and torsional stiffness Kθ_{ }have the same values in both the initial system X0Y and the principal system xCTy. It is preferable to work in the initial system, as the calculations are simpler. =90.6 · ( Υ _{CT} Y_{CM})+90.6 · c_{Y}=90.6 ·(4.1932.509)+90.6·1.0= =243.2 kNm, K _{θ} = M_{XCT}/ θ _{XZ} =243.2/11.952 ·10^{5}=20.3·10^{5 }kNm r_{x}=√K _{θ} /K_{yy}=√[20.3·10^{8}Nm/105.5·10^{6}N/m]=4.39 m r_{y}=√K _{θ} /K_{xx}=√[20.3·10^{8}Nm/137.8·10^{6}N/m]=3.84 m  The expressions determining the C_{T} coordinates are general and they apply to any point of the diaphragm. For instance, from column 4: X_{CT}=X_{4}δ_{Yt,4}/θ_{XZ}=6.00.382·10^{3}m/11.952·10^{5}=6.03.20=2.80 m Y_{CT}=Y_{4}+δ_{Xt,4}/θ_{XZ}=5.00.096·10^{3}m/11.952·10^{5}=5.00.80=4.20 m
Figure D.7.112: Equivalent system of 4 columns of 424/371 crosssection (at the “ Equivalent system” field enter k=1 n=4k=4) Figure D.7.112: Equivalent system of 4 columns of 424/371 crosssection (at the “ Equivalent system” field enter k=1 n=4k=4)
 In this example considered, since the structure is onestorey (the same applies to the ground floor of any multistorey building), the torsional stiffness ellipse of the equivalent system is identical to that corresponding to the actual floor. Selecting “Equivalent system/Draw”=ON, the torsional stiffness ellipse, verified previously, with the 4 equivalent columns of 424/373 crosssection located on its vertices, are displayed by the software (see figure D.712). The equivalence of these 4 fixedended columns is checked next:  The calculations are effected in the principal coordinates system, where each column stiffness in its local system, is the same with that of theprincipal. K_{x}=12Ε·Ι_{x}/h^{3} and K_{x}=12Ε·Ι_{y}/h^{3} (see §5.1.1), since in the example considered the shear effect, in any case insignificant had been ignored, and therefore k_{va}=1. I_{x}= 0.373·0.424^{3}/12= 23.693 ·10^{4} m^{4}, I_{y}= 0.424 ·0.373^{3}/12=18.336·10^{4} m^{4} Given E=32.8 GP and h=3.0 m → K_{x}=12·32.8·10^{9}Pa·23.693· 10^{4}m^{4}/3.0^{3}m^{3}=34.54·10^{6} N/m K_{y}=12 · 32.8 · 10^{9} Pa · 18.336 · 10^{4}m^{4}/3.0^{3}m^{3}=26.73·10^{6} N/m For 4 equivalent columns→ K_{xx}=Σ(K_{x})=4·34.54=138.2·10^{6} N/m, K_{yy}= Σ (K_{y})=4·26.73=106.9·10^{6} N/m , Therefore equal to the actual stiffness (slight differences are justified by the need to use integer mm). The torsional stiffness of the equivalent diaphragm is K_{θ}=Σ(K_{xi}·y_{i}^{2}+_{Kyi}_{·}x_{i}^{2}+0.0) (expression 7, §C.5) → K_{θ}=2·Κ_{x}·3.842^{2}+2·K_{y}·4.390^{2}=2·34.54·10^{6}N/m·14.76m^{2} + 2·26.73· 10^{6}N/m·19.271m^{2}= (10.2+10.3)·10^{5} kNm =20.5·10^{5} kNm , therefore equal to the actual torsional stiffness. It is obvious that r_{x}, r_{y} values are also identical, being equal to thesquare root of the ratio of equal quantitiesr_{x}_{Ζ}=√(K_{θ} /K_{yy})=4.39 m, r_{y}_{Ζ}=√(K_{θ}/K_{xx})=3.84 m. The equivalent building may comprise only 4 columns, or any number of columns n=4k, where κ is nonzero integer, e.g. 4, 8, 12, 16, 20, … These columns by groups of four, are placed symmetrically with respect to the centre of stiffness. For instance, the case of 8 columns of 356/312 crosssection, yields: I_{x}= 0.312·0.356^{3}/12= 11.731 ·10^{4} m^{4}, I_{y}= 0.356 ·0.312^{3}/12=9.010·10^{4} m^{4} Given E=32.8 GP and h=3.0 m → K_{x}=12·32.8·10^{9}Pa·11.731·10^{4}m^{4}/3.0^{3}m^{3}=17.01·10^{6} N/m, K_{y}=12·32.8·10^{9}Pa·9.010· 10^{4}m^{4}/3.0^{3}m^{3}=13.13·10^{6} N/m. For 8 equivalent columns → K_{xx}=Σ(K_{x})=8·17.01=136.8·10^{6} N/m and K_{yy}= Σ(K_{y})=8·13.13=105.1·10^{6} N/m. To calculate the torsional stiffness, the coordinates (±3.105 m, ±2.717 m) of the 4 intermediate points are used, which are displayed both at the screen and in the “report” (if “coords” selected). Alternative equivalent structures
Figure D.7.113: Case of 8 columns (k=2, n=4k=8)of 356/312 equivalent crosssection Figure D.7.113: Case of 8 columns (k=2, n=4k=8)of 356/312 equivalent crosssection

Figure D.7.114: Case of 16 columns (k=4, n=4k=16)of 299/262 equivalent crosssection Figure D.7.114: Case of 16 columns (k=4, n=4k=16)of 299/262 equivalent crosssection

Figure D.7.115: Case of 24 columns (k=6, n=4k=24)of 271/237 equivalent crosssection Figure D.7.115: Case of 24 columns (k=6, n=4k=24)of 271/237 equivalent crosssection

Figure D.7.116: Case of 48 columns (k=12, n=4k=48)of 227/199 equivalent crosssection Figure D.7.116: Case of 48 columns (k=12, n=4k=48)of 227/199 equivalent crosssection
 K _{θ} = Σ (K_{xi} · y_{i}^{2}+_{Kyi} · x_{i}^{2})= 2 · Κ _{x} · 3.842^{2}+4 · Κ _{x} · 2.717 ^{2} +2 · K_{y} · 4.390^{2}+4 · K_{y} · 3.105^{2}= =[17.01 ·(29.52+29.53)+13.13·(38.54+38.56)]·10^{3} kNM=[10.1+10.1]·10^{5} kNM =20.2·10^{5} kNM Thus, the system of 8 fixedended columns is also equivalent to the actual structure. In this way all systems with n=4k are verified to be equivalent to the initial actual structure. In the threestorey structure of project <B_d92> using the related software or any other relevant software, for each of the three loadings in this particular example the two translations of node 9 (column C1) and the rotation of the diaphragm of level 2 are calculated. Optionally, the translations of the other diaphragm points may be calculated. All diaphragm data are computed based only on the displacements of node 9.
Figure D.7.21: The simple structure of 3 storeys and 4 columns.The floor is typical, identical to the floor of first example Figure D.7.21: The simple structure of 3 storeys and 4 columns.The floor is typical, identical to the floor of first example
After entering into “Element Input”, select “Tools” from the menu and then “Diaphragm calculation”. In the dialog opened, enter H=90.6 kN, c _{Y}=1.0 m, select “Use fixed columns=OFF” and press “ΟΚ” and the diaphragm data of the current floor are displayed. Floor 1, that corresponds to level 2 is then selected..
Figure D.7.22: Output image of the software Figure D.7.22: Output image of the software
The coordinates of the centre of stiffness C _{T} are ( 2.847, 3.785), and the torsional radii are r_{x}=4.305 m, r_{y}=3.979 m (versus the coordinates 2.688, 4.897, r_{x}=4.411 and r_{y}=3.381 derived by the assumption of fixedended columns). The crosssection of the equivalent columns is 335/321 (versus 521/399 derived by the assumption of fixedended columns). All results are displayed analytically by selecting from the menu “View”, “Diaphragm results”, “report”. In level 2 θ_{XZ} =30.2962·10^{5}. The rest of results are better displayed in 3D, by selecting from the menu “View”, “Diaphragm results”, “3D floor” combined with “free rotation analysis” or “ restrained rotation analysis” or “rotation only”, as presented in the two following pages. In the structure considered, being multistorey, the “only rotation” condition of a diaphragm, i.e. with C_{T} remaining stationary with respect to the ground, may only derive using the trick of the following general method,. Calculation of the diaphragmatic behaviour of level 2
Figure D.7.23 Figure D.7.23

Figure D.7.24 Figure D.7.24

Figure D.7.25 Figure D.7.25
 Loading 1 : H_{X}=90.6 kN with loading eccentricity c_{Y}=1.0 m resulting in moment M_{XCM}=90.6 kNm  Loading 2 : H_{X}=90.6 kN Diaphragm restrained against rotation  Loading 3 : H_{Y}=90.6 kN Diaphragm restrained against rotation 
Figure D.7.26 Figure D.7.26

Figure D.7.27 Figure D.7.27

Figure D.7.28 Figure D.7.28
 Analysis results: The translations of point 1 are δ _{XX1} =3.318, δ _{XY1} =0.986mm and the diaphragm rotation is θ _{XZ} =30.2962 ·10^{5}  Analysis results: The diaphragm develops only parallel translations in X,Y directions, being restrained against rotation. Thus all diaphragm points (including C_{T}) develop the same displacements: δ _{XXo} =2.1712 mm, δ _{XYo} =0.1235  Analysis results: The diaphragm, being restrained against rotation, develops only parallel translations in X,Y directions, which are: δ _{YXo} =0.1235, δ _{YYo} =2.4362 mm The angle of the principal system derives from the expression: tan(2a)=2 δ _{XYo} /( δ_{ΧΧο}  δ _{YYo} )= 2 ·(0.1235)/(2.17122.4362)= 0.932 → 2a=43.0° → a=21.5°  Calculation of the diaphragmatic behaviour of level 2 (continued)
Figure D.7.29 Figure D.7.29

Figure D.7.210 Figure D.7.210
 Loading 1 minus loading 2: H_{X}=0, M_{XCT}=90.6 · (Y_{CT} Y_{CM} ) +90.6 
Figure D.7.211 Figure D.7.211
 Subtraction results: The diaphragm only rotates by θ _{XZ} about the centre of stiffness C_{T}. The translations of the first point due to rotation: δ _{Xt,1} = δ _{X,1}  δ _{XXo} = =3.31802.1712 =1.1468, δ _{Yt,1} = δ _{Y,1}  δ _{XYo } = =0.9859+0.1235 =0.8624 and X_{CT}=X_{1} δ _{Yt,1} / θ _{XZ} = 0.0+0.8624 ·10^{3} /30.2962 ·10^{5} =2.847 m Y_{CT }=Y_{1}+ δ _{Xt,1} / θ _{XZ} = 0.0+1.1468 ·10^{3} /30.2962 ·10^{5} =3.785 m  Determination of stiffnesses, torsional radii and equivalent system : The lateral stiffnesses K_{xx}, K_{yy} are calculated using the expressions C.9.2 and C.9.3 of §C.9 with a=21.49° → tana=0.393: K_{xx}=H/(δ_{XXo}+δ_{XYo}·tana)= =[90.6/(2.17120.1235·0.393)]·10^{6}N/m= 42.7·10^{6} N/m K_{yy}=H/(δ_{YYo}_{XYo}·tana)= =[90.6/(2.4362+0.1235·0.393)]·10^{6}N/m= 36.5·10^{6} N/m M_{XCT}=90.6 · ( Υ _{CT} Y_{CM})+90.6 · c_{Y}=90.6 ·(3.7852.525)+90.6·1.0= 204.8 kNm, K _{θ} = M_{CT},_{X}/ θ _{XZ} =204.8/30.2962 ·10^{5}=6.759·10^{5 }kNm r_{x}=√K _{θ} /K_{yy}=√[6.759·10^{8}Nm/36.5·10^{6}N/m]=4.30 m r_{y}=√K _{θ} /K_{xx}=√[6.759·10^{8}Nm/42.7·10^{6}N/m]=3.98 m  The expressions determining the C_{T} coordinates are general and they apply to any point of the diaphragm. For instance, from column 4: X_{CT}=X_{4}δ_{Yt,4}/θ_{XZ}=6.00.9553·10^{3}m/30.2962·10^{5}=6.03.153=2.847 m Y_{CT}=Y_{4}+δ_{Xt,4}/θ_{XZ}=5.00.3681·10^{3}m/30.2962·10^{5}=5.01.215=3.785 m
Figure D.7.212: Equivalent diaphragm of level 2 with 4 columns of 335/321 crosssection (at the “ Equivalent system” field enter k=1 n=4k=4) Figure D.7.212: Equivalent diaphragm of level 2 with 4 columns of 335/321 crosssection (at the “ Equivalent system” field enter k=1 n=4k=4)
 The equivalent system of the threestorey building comprises three diaphragms, equivalent to the actual ones. The results for both the actual and the equivalent diaphragms are given in the “report” of the related software. The two following tables of the equivalent diaphragms are taken from the report.  Equivalent system displacements Elevation  dXXo,i mm  dXYo,i mm  qXZM,i 1.0e5  dYYo,i mm  dXXoZ,i mm  dXYoZ,i mm  dYYoZ,i mm  qXZMZ,i 1.0e5  1  0.612  0.079  4.2009  0.773  0.612  0.079  0.773  4.2009  2  2.171  0.123  13.4045  2.436  1.559  0.044  1.663  9.2036  3  4.198  0.072  24.6910  4.437  2.026  0.052  2.001  11.2865  Equivalent system crosssections Elevation  hi m  αZ,i deg  MXM,i KNm  KqZ,i MNm  KxxZ,i MN/m  KyyZ,i MN/m  rxZ,i m  ryZ,i m  AxZ,i mm  AyZ,i mm  1  3.00  22.3069  90.60  2156.7  156.30  112.45  4.379  3.715  441  374  2  3.00  20.1371  90.60  984.4  58.72  53.95  4.271  4.094  335  321  3  3.00  38.0593  90.60  802.7  43.83  46.21  4.168  4.279  306  314  Verification of the tabulated results using the theory presented in §D.6: δ _{XXo} _{Ζ} _{,1} = δ _{XXo,1} 0.0=0.612 , δ _{XYo} _{Ζ} _{,1} = δ _{XYo,1} 0.0=0.079, δ _{YYo} _{Ζ} _{,1} = δ _{YYo,1} 0.0=0.773, θ _{XZ} _{Μ} _{Z,1} = θ _{X} _{ΖΜ} _{,1} –0.0=4.2009 δ _{XXo} _{Ζ} _{,2} = δ _{XXo,2}  δ _{XXo,1} =2,1710.612=1.559, δ _{XYo} _{Ζ} _{,2} = δ _{XYo,2}  δ _{XYo,1} =0.123(0.079)=0.044 δ _{YYo} _{Ζ} _{,2} = δ _{YYo,2}  δ _{YYo,1} =2.4360.773=1.663, θ _{XZ} _{Μ} _{Z,2} = θ _{X} _{ΖΜ} _{,2} – θ _{XZ} _{Μ} _{,1} =13.40454.2009=9.2036 δ _{XXo} _{Ζ} _{,3} = δ _{XXo,3}  δ _{XXo,2} =4.1982.171= 2.026, δ _{XYo} _{Ζ} _{,3} = δ _{XYo,3}  δ _{XYo,2} =0.072(0.123)=0.051 δ _{YYo} _{Ζ} _{,3} = δ _{YYo,3}  δ _{YYo,2} =4.4372.436=2.001, θ _{XZ} _{Μ} _{Z,3} = θ _{X} _{ΖΜ} _{,3} – θ _{XZ} _{Μ} _{,2} =24.691013.4045=11.2865 Diaphragm data for all levels derive from these quantities, therefore specifically for level 2: tan(2a_{z,2})=2 · δ _{XYo} _{Ζ} _{,2} /( δ _{XXo} _{Ζ} _{,2}  δ _{YYo} _{Ζ} _{,2} )=2 · (0.044)/(1.5591.663) → tan(2a_{z,2})=0.8462 → a_{z,2}=20.12º and tana_{z,2}=0.3667 K _{θ} _{Z,2} = Μ_{ΧΜ} _{,2} / θ _{XZ} _{Μ} _{Z,2} =90.6 ·1.0/ 9.2036=984.4, K_{xxZ,2}= H/( δ _{XXoZ,2} + δ _{XYoZ,2} · tana_{Z,2})=90.6/(1.559+(0.044) · 0.3667)=58.7 and K_{yy} _{Ζ} _{,2} =H/( δ _{YYoZ,2}  δ _{XYoZ,2} · tana_{Z,2})=90.6/(1.663(0.044) ·0.3667)=53.95 r_{xZ,2}=√K _{θ} _{Z,2} /K_{yyZ,2}=√(984.4/53.95)=4.27, r_{yZ,2}=√K _{θ} _{Z,2} /K_{xxZ,2 }=√(984.4/58.72)=4.094 quod erat demonstrandum . [1] If “fixed columns=ON” then the results are based on the assumption of fixedended columns and the results are identical with those of the first two cases. The slight differences in the results versus the ones obtained by the manual calculations, as well as the ones resulting from the excel file are due to the small differences of the centre of mass, due to the uneven loads from the columns selfweight. [2] The software performs the calculations of the diaphragm automatically. The verification of the algorithms using the software tools is presented here. [3] The horizontal seismic force acts on the centre of mass C_{M}. The loading eccentricity, c_{y}=1.0, may also be given as equivalent moment M_{CM,X}=H_{X}·c_{Y}. In this case M_{CM,X}=90.6·1.0=90.6 kNm. This additional eccentricity increases the effect of rotation producing larger translations due to rotation, thus leading to a more accurate calculation of diaphragm torsional data. Besides that, the moment induced by the eccentricity moment provides solutions even in cases that the centres of mass and stiffness are close or coincide. [4] Moment, rotation and torsional stiffness Kθ_{ }have the same values in both the initial system X0Y and the principal system xCTy. It is preferable to work in the initial system, as the calculations are simpler.
