GEOMETRIC DATA OF THE DIAPHRAGM (independent of the loading) 1. Calculation of stiffnesses K_{xi}, K_{yi} and torsional stiffness K_{zi} of each column. 2. Calculation coordinates (X_{CT},Y_{CT}) of the centre of stiffness C_{T} using the expressions (4') and (5'). 3. Transfer of the coordinates X, Y of the centre of mass of the diaphragm C_{M} and each column centre of gravity to the principal axis system. Calculation of the structural eccentricities e_{ox}, e_{oy }of the structure using the expressions (6'). 4. Calculation of the torsional stiffness K _{θ} of the diaphragm using the expression (7'). 5. Calculation of the torsional radii r_{x}, r_{y} using the expression (8'). So far the calculations are independent of the external loading. Next, are calculated the stress distribution and the elastic deformations of the diaphragm elements, due to external loading. FLOOR DISPLACEMENTS AND STRESS RESULTANTS (dependent on the loading) 6. Transfer of the horizontal forces H_{x}, H_{y} and the moment M_{CT} to the centre of stiffness C_{T} using the expression (9'). 7. Calculation of the displacements δ _{xo} , δ _{yo} and θ _{z} of the centre of stiffness C_{T} using the expressions (10'). 8. Calculation of the displacements δ _{xi} , δ _{yi} of each column top using the expressions (11'). 9. Calculation of the seismic shear forces V_{xi}, V_{yi} of each column using the expressions (13'). 10. Calculation of the principal bending moments M_{xi,1}, M_{xi,2} & M_{yi,1}, M_{yi,2} using the expressions (14'). Example 5.4.3.6 (the simple structure used so far in the theory) (a) GEOMETRIC DATA OF THE DIAPHRAGM (independent of the loading) Torsional stiffness calculation The stiffness K_{ij} of each column i in direction j is: Κ _{ji} =k_{ji} · ( EI _{ji} / h ^{3} ) · k_{ji} _{,} _{v} _{α} (§4.1.5) k_{ji}=12 (columns are assumed to be fixedended) Therefore, Κ _{ji} =12 · EI _{ji} / h ^{3} . Moreover, we assume that all columns have the same torsional stiffness K_{zi}=0. The columns of the example have the same magnitudes E and h, therefore their stiffness can be assumed to be Κ _{ji} =C · I_{ji} , where C=12E/h^{3}. Modulus of Elasticity is taken as Ε =32.80 GPa, corresponding to concrete class C30/37 (volume C §1.1). Therefore, C=12E/h^{3}=12 · 32.80 · GPa/(3.0^{3}m^{3})=14.58 ·10^{9}N/m^{5}. K_{1x}=K_{1y}=K_{2x}=K_{2y}=14.58 · 10^{9}N/m^{5} · 0.400 · 0.400^{3}/12 · m^{4}=31.1 ·10^{6} N/m K_{3x}=14.58 ·10^{9}·0.300·0.800^{3}/12=186.6·10^{6} N/m K_{3y}=14.58 ·10^{9}·0.800·0.300^{3}/12=26.2·10^{6} N/m K_{4x}=14.58 ·10^{9}·0.600·0.300^{3}/12=19.7·10^{6} N/m K_{4y}=14.58 ·10^{9}·0.300·0.600^{3}/12=78.7·10^{6} N/m Therefore based on the expressions (3): Calculation of the Centre of Stiffness and the Lateral Stiffnesses : expressions (4') and (5') K_{x}= Σ (K_{xi})=(31.1+31.1+186.6+19.7) ·10^{6} N/m=268.5·10^{6} N/m K_{y}= Σ (K_{yi})=(31.1+31.1+26.2+78.7) ·10^{6} N/m=167.1·10^{6} N/m Χ _{CT} =Σ( X_{i} · K_{yi} )/ K_{y} =[(0.0·31.1+6.0·31.1+0.0·26.2+6.0·78.7)·10^{6}]/(167.1· 10^{6})=658.8/167.1=3.94 m Υ _{CT} = Σ (Y_{i} · K_{xi})/K_{x}=[(0.0·31.1+0.0·31.1+5.0·186.6+5.0·19.7)·10^{6}]/(268.5· 10^{6})=1031.5/268.5=3.84 m Coordinate transfer to the principal system xC_{T}y : expressions (6') C_{1}(3.94,3.84), C_{2}(2.06,3.84), C_{3}(3.94,1.16), C_{4}(2.06,1.16), C_{M}(0.94,1.34), C _{T}(0.0,0.0) e_{ox}=0.94 m, e_{oy}=1.34 m Calculation of the diaphragm's torsional stiffness : expression (7') K _{θ} = Σ [K_{xi} · y_{i}^{2}+K_{yi} · x_{i}^{2}+K_{zi}]= Diaphragm's torsional radii : expressions (8') r_{x}=√(K _{θ} /K_{y})=√[2550 · 10^{6}Nm/(167.1 · 10^{6}N/m)]=3.91 m r_{y}=√(K _{θ} /K_{x})=√[2550 · 10^{6}Nm/(268.5 · 10^{6}N/m)]=3.08 m (b) DISPLACEMENTS AND STRESS RESULTANTS (independent of the loading) (A) H_{x}=90.6 kN, H_{y}=0, (B) H_{x}=0 kN, H_{y}=90.6 kN and ( Γ ) H_{x}=90.6 kN, H_{y}=27.2 kN Case Α : H_{x}=90.6 kN, H_{y}=0.0 Seismic moment at the centre of stiffness : expression (9') M_{CT}=H_{x} · e_{oy}+H_{y} · e_{ox}=90.6kN · (1.34m)=121.4 kNm Displacements of the centre of stiffness : expressions (10') δ _{xo} =H_{x}/K_{x}=90.6 · 10^{3}N/(268.5 · 10^{6}N/m)=0.337 · 10^{3}m=0.337 m θ _{z} =M_{CT}/K _{θ} =121.4 · 10^{3}Nm/( 2550 · 10^{6}Nm)=47.6 ·10^{6} Displacements of columns : (expressions 11') δ _{xi} = δ _{xo} + δ _{x} _{θ} _{i} = δ _{xo}  θ _{z} · y_{i} δ _{yi} =0.0+ θ _{z} · x_{i}= θ _{z} ·x_{i} To compare the quantities, the displacements due to rotation are calculated separately. δ _{x} _{θ} _{1} = θ _{z} · y_{1}=47.6 ·10^{6}·(3.84·10^{3}mm)=0.183 mm δ _{y} _{θ} _{1} = θ _{z} · x_{1}=0.188 mm C_{1}: δ _{x} _{θ} _{1} = 0.183 mm, δ _{y} _{θ} _{1} =0.188 mm, δ _{x1} =0.337+0.183=0.520 mm, δ _{y1} =0.188 mm Likewise, for the rest of the columns: C_{2}: δ _{x} _{θ} _{2} = 0.183 mm, δ _{y} _{θ} _{2} = 0.098 mm, δ _{x2} =0.337+0.183=0.520 mm, δ _{y2} = 0.098 mm C_{3}: δ _{x} _{θ} _{3} =0.055 mm, δ _{y} _{θ} _{3} =0.188 mm, δ _{x3} =0.3370.055=0.282 mm, δ _{y3} =0.188 mm C_{4}: δ _{x} _{θ} _{4} =0.055 mm, δ _{y} _{θ} _{4} = 0.098 mm, δ _{x4} =0.3370.055=0.282 mm, δ _{y4} = 0.098 mm Seismic shear forces : expressions (13') C_{1}: V_{x1}= δ _{x1} · K_{x1}=0.520·10^{3}m·31.1·10^{6}N/m=16.17 kN, V_{y1}=δ_{y1} ·K_{y1}=0.188·31.1=5.85 kN C_{2}: V_{x2}= δ _{x2} · K_{x2}=0.520·31.1= 16.17 kN, V_{y2}=δ_{y2}·K_{y2}= 0.098· 31.1= 3.05 kN C_{3}: V_{x3}= δ _{x3} · K_{x3}=0.282·186.6=52.62 kN, V_{y3}=δ_{y3}·K_{y3}=0.188· 26.2=4.93 kN C_{4}: V_{x4}= δ _{x4} · K_{x4}=0.282·19.7= 5.56 kN, V_{y4}=δ_{y4}·K_{y4}= 0.098· 78.7= 7.71 kN Verification: Σ (Vxi)=90.52 kN≈90.6 kN and Σ (Vyi)=0.02≈0.0 as expected Seismic bending moments : expressions (14') It is assumed (for all columns) a_{xi}=a_{yi}=0.50, and therefore (1a_{xi})=(1a_{yi})=0.50. C_{1}: M_{x1,1}=0.5 ·16.17·3.0=24.3 kNm, M_{x1,2}=24.3 kNm, M_{y1,1}=0.5 · (5.85) ·3.0=8.8 kNm, M_{y1,2}= 8.8 kNm C_{2}: M_{x2,1}=0.5 ·16.17·3.0=24.3 kNm, M_{x2,2}=24.3 kNm, M_{y2,1}=0.5 ·3.05·3.0= 4.6 kNm, M_{y2,2}= 4.6 kNm C_{3}: M_{x3,1}=0.5 ·52.62·3.0=78.9 kNm, M_{x3,2}=78.9 kNm, M_{y3,1}=0.5 ·(4.93)·3.0=7.4 kNm, M_{y3,2}= 7.4 kNm C_{4}: M_{x4,1}=0.5 ·5.56·3.0= 8.3 kNm, M_{x4,2}= 8.3 kNm, M_{y4,1}=0.5 ·7.71·3.0= 11.6 kNm, M_{y4,2}=11.6 kNm Case B: H_{x}=0.0 kN, H_{y}=90.6 Seismic moment at the centre of stiffness : expression (9') M_{CT}=H_{x}·e_{oy}+H_{y}·e_{ox}=90.6kN ·(0.94m)=85.2 kNm Displacements of the centre of stiffness : expressions (10') δ _{yo} =H_{y}/K_{y}=90.6 ·10^{3}N/(167.1·10^{6}N/m)=0.542 mm θ _{z} =M_{CT}/K _{θ} =85.2 ·10^{3}Nm/(2550·10^{6}Nm)=33.4·10^{6} Only the quantities for column C_{3 }are calculated here. The results for the remaining columns can be derived anyway from the related file <diaphragm ortho.xls>. Displacements of columns (C_{3}) : expressions (11') δ _{x} _{θ} _{3} = θ _{z} · y_{3}=(33.4) ·10^{6}·1.16·10^{3}mm=0.039 mm δ _{y} _{θ} _{3} = θ _{z} · x_{3}=(33.4) ·(3.94·10^{3}mm)=0.132 mm δ _{x} _{θ} _{3} =0.039 mm, δ _{y} _{θ} _{3} =0.132 mm, δ _{x3} =0.039 mm, δ _{y3} =0.542+0.132=0.674 mm Seismic shear forces (C_{3}) : expressions (13') V_{x3}= δ _{x3} · K_{x3}=0.039 ·186.6=7.28 kN V_{y3}= δ _{y3} · K_{y3}=0.674 ·26.2=17.66 kN Seismic bending moments (C_{3}) : expressions (14') M_{x3,1}=0.5 ·7.28·3.0=10.9 kNm M_{y3,1}=0.5 ·17.66·3.0=26.5 kNm Seismic moment in the center of stiffness : expression (9') M_{CT}=H_{x} · e_{oy}+H_{y} · e_{ox}=90.6 ·(1.34)+(27.2)·(0.94m)=121.4+25.6=147.0 kNm Displacements of the centre of stiffness : expressions (10') δ _{xo} =H_{x}/K_{x}=90.6/268.5mm=0.337 mm δ _{yo} =H_{y}/K_{y}=27.2/167.1=0.163 mm θ _{z} =M_{CT}/K _{θ} =147 ·10^{3}Nm/(2550·10^{6}Nm)=57.65·10^{6} Next, only the data of column C_{3 }will be calculated. The other results can be derived, from the related file <diaphragm ortho.xls>. Displacements of columns (C_{3}) : expressions (11') δ _{x} _{θ} _{3} = θ _{z} · y_{3}=57.65 ·10^{6}·1.16·10^{3}mm=0.067 mm δ _{y} _{θ} _{3} = θ _{z} · x_{3}=57.65 ·(3.94·10^{3}mm)=0.227 mm δ _{x} _{θ} _{3} =0.067 mm, δ _{y} _{θ} _{3} =0.227 mm, δ _{x3} =0.3370.067=0.270 mm, δ _{y3} =0.1630.227=0.390 mm Seismic shear forces (C_{3}) : expressions (13') V_{x3}= δ _{x3} ·K_{x3}=0.270 ·186.6=50.38 kN V_{y3}= δ _{y3} ·K_{y3}=0.390 ·26.2=10.22 kN Seismic bending moments (C_{3}) : expressions (14') M_{x3,1}=0.5 ·50.38·3.0=75.6 kNm M_{y3,1}=0.5 ·(10.22)·3.0=15.3 kNm  The analysis of such problems is practically performed by a table format. In this respect, using the related Excel files, one can define his own structures and examine the effect of various factors e.g. the elasticity modulus of the columns, the stiffness factors, the crosssectional dimensions, etc.
 For seismic action in the x direction, the displacement of the centre of stiffness C_{T} is 0.337 mm, while the diaphragm maximum displacement due to rotation is 0.520 mm, i.e. 54% greater than the average value. For seismic action in y direction, the C_{T} displacement is 0.520 mm while the diaphragm maximum displacement due to rotation is 0.674 mm, i.e. 30%, greater than the average value.
 The actual displacement is much greater than the maximum of 0.674 mm which is indicative and idealised. This is due to the fact that the actual stiffness [EC8 §4.3.1(6),(7)] due to cracking should be considered reduced by an order of 50%, therefore resulting to a double displacement 2·0.674=1.35 mm. If all columns are considered fixedended simultaneously, the stiffness factor would be taken equal to k=3 instead of k=12 and due to the previous reduction by 50% k=1.5. Therefore in accordance with this stiffness factor, the maximum elastic displacement should be 12/1.5=8 times greater, i.e. δ=8·0.0.674=5.4 mm (results can be verified introducing the changes in the Excel file).
 For the calculation of the actual displacement, EC8 imposes the multiplication of the elastic displacement with the behaviour factor q. Using an average value of q=3.50 the previous displacement would be 3.5·5.4=19mm. Obviously considering the small loads of an onestorey building of these dimensions, the size of its columns is very large, and for this reason the displacement is relatively small. In case of columns of half crosssectional dimensions e.g. 400/400 instead of 200/200, the displacement of the diaphragm would be (400/200)^{4}=16 times greater i.e. δ_{yo}=16·19=304 mm. This displacement is extremely large and prohibited from other provisions of EC8. In case that this building would be a multistorey one, then, the need for strong columns would be even more obvious.
