DIAPHRAGMATIC BEHAVIOUR OF ONESTOREY SPACE FRAME – GENERAL CASE
Figure C.11: Simple onestorey structure comprising four columns whose tops are connected by a rigid slabdiaphragm. Figure C.11: Simple onestorey structure comprising four columns whose tops are connected by a rigid slabdiaphragm.
These rules induce the diaphragm to develop a parallel (translational) displacement by δ_{xo}, δ_{yo} and a rotationθ_{z} about the a point C_{T}(x_{CT}, y_{CT}), the centre of stiffness, in the xC_{T}y coordinate system, which has as origin the point C_{T} and is rotated with respect to the initial coordinate system X0Y by an angle a. To better understand this chapter it is useful to study first §5.4.
Figure C.12 Figure C.12
Χ 0Υ initial coordinate system x’0y’ auxiliary coordinate system xC_{T}y principal coordinate system
Figure C.13: Parallel translation in both directions and rotation of the diaphragm, due to force H Figure C.13: Parallel translation in both directions and rotation of the diaphragm, due to force H
It is useful to recall the relationship between the two sets of components for the vector v(x,y) transferred from one coordinate system x0y to another ζ0η rotated by an angle φ with respect to the original system. This vector can be either a displacement vector, a force vector, or defining the location of a point.
Figure C.2: Relationships of axis transformation due to rotation Figure C.2: Relationships of axis transformation due to rotation
 ζ =x · cos φ & η =x · sin φ ζ =y · sin φ & η =y · cos φ  Application of the above relationships on the stiffnesses of a column with crosssection is inclined by an angle φ_{i} with respect to the initial system x0y: The displacement of a column top i by δ_{xi} is equivalent to the displacements δ_{ζ} _{i}, δ_{η}_{i} in the principal system ζ, η of column i: δ_{ζ} _{i} = δ _{xi} · cos φ _{i}and δ_{η}_{i}=δ_{xi}·sinφ _{i} Displacements δ_{ζ}_{i} and δ_{η}_{i} will result in the corresponding shear forces: V _{ζ} _{i} =K _{ζ} _{i} · δ _{xi} · cos φ _{i}and V_{η}_{i}=K_{η}_{i}·δ _{xi}·sinφ_{i}. V_{xxi}=V _{ζ} _{i} · cos φ _{i} V _{η} _{i} · sin φ _{i} =(K _{ζ} _{i} · δ _{xi} · cos φ _{i} ) · cos φ _{i} (K _{η} _{i} · δ _{xi} · sin φ _{i} ) · sin φ _{i} =K _{ζ} _{i} · δ _{xi} · cos^{2} φ _{i} +K _{η} _{i} · δ _{xi} · sin^{2} φ _{i} → V_{xxi} = δ _{xi} · ( K _{ζ} _{i} · cos^{2} φ _{i} +K _{η} _{i} · sin^{2} φ _{i} ) V_{xyi}=V _{ζ} _{i} · sin φ _{i} +V _{η} _{i} · cos φ _{i} =(K _{ζ} _{i} · δ _{xi} · cos φ _{i} ) · sin φ _{i} +(K _{η} _{i} · δ _{xi} · sin φ _{i} ) · cos φ _{i} =K _{ζ} _{i} · δ _{xi} · (1/2)sin2 φ _{i} K _{η} _{i} · δ _{xi} · (1/2) · sin2 φ _{i} → V_{xyi}= δ _{xi} · (1/2) · ( K _{ζ} _{i} K _{η} _{i} ) · sin2 φ _{i} In a simpler form, these two expressions, as a function of the normalised stiffness coefficients, of column i in direction x, K_{xxi} and K_{xyi}, are: V_{xxoi}= δ _{xi} × K_{xxi}where K_{xxi}= K_{ζ}_{i}×cos^{2}φ_{i} +K_{η}_{i}×sin^{2}φ_{i} (a) V_{xyoi}= δ _{xi} × K_{xyi}where K_{xyi}=(1/2)×(K_{ζ}_{i}K_{η} _{i})×sin2φ_{i} (b) Quantities K_{xxi} and K_{xyi} are called normalised stiffnesses of column i in direction x. In a similar way, the expressions as a function of the normalised stiffness coefficients K_{yxi} and K_{yyi} of column in direction y are: V_{yxoi}= δ _{yi} × K_{yxi}where K_{yxi}=(1/2)×(K_{ζ}_{i}K_{η} _{i})×sin2φ_{i}= K_{xyi} (c) V_{yyoi}= δ _{yi} × K_{yyi}where K_{yyi}=K_{ζ}_{i}×sin^{2}φ_{i} +K_{η}_{i}×cos^{2}φ_{i}_{ }(d)  In every coordinate system, for each column i always K_{xyi}=K_{yxi}.
 Stiffness coefficients K_{xxi}, K_{xyi} and K_{yyi} are identical in all systems parallel to x0y, therefore to x’0y’ as well.
 In rectangular columns the angle φ is irrelevant since K_{ζ}=K_{η} therefore (a) and (d) always give K_{xx}=K_{yy} =K_{ζ}=K_{η} and (b) always gives K_{xy}=0.
The columns of the example have the same E and h, therefore their stiffness can be assumed as Κ _{ji} =C × I_{ji} where C=12E/h^{3} As elasticity modulus is assumed Ε =32.80 GPa corresponding to concrete C30/37 (§6.1) therefore C=12E/h^{3}=12 × 32.80 × GPa/(3.0^{3}m^{3})=14.58 × 10^{9}N/m^{5} . Column stiffnesses with respect to their local coordinate systems ζ , η : C _{1} , C _{2} 400/400 K _{ζ1} = K _{1η} = K _{2ζ} = K _{2η} =14.58·10^{9} N / m ^{5} × 0.400 × 0.400^{3}/12 × m ^{4} =31.1 × 10^{6} N / m , C _{3} 800/300 K _{3ζ} =14.58 × 10^{9} × 0.300 × 0.800^{3}/12=186.6 × 10^{6} N / m , K _{3η} =14.58 × 10^{9}·0.800 × 0.300^{3}/12=26.2 × 10^{6} N / m C _{4} 300/600 K _{4ζ} =14.58 × 10^{9} × 0.600 × 0.300^{3}/12=19.7 × 10^{6} N / m , K _{4η} =14.58 × 10^{9} × 0.300 × 0.600^{3}/12=78.7 × 10^{6} N / m The diaphragmatic behaviour may be considered as superposition of three cases: (a) parallel translation of the diaphragm in X direction due to horizontal force component H_{X}, (b) parallel translation of the diaphragm in Y direction due to horizontal force component H_{Y,} (c) rotation of the diaphragm due to moment M_{CT} applied at the centre of stiffness C_{T} The horizontal seismic forces are applied at each point with mass, while the resultant force is applied at the centre of mass C_{M}. In case the direction of the force H passes through the point C_{T}_{ }as well as C_{M} the moment has zero value and therefore the diaphragm develops zero rotation. C.3 Case 1: Translation of centre of stiffness C_{T} along x direction by δ_{xο}
Figure C.3 Figure C.3
When a horizontal force H_{x} acts on the C_{T} along the x direction, the 3 following equilibrium equations should apply:  The sum of forces acting in x direction is be equal to H_{x}, i.e. H_{x}=Σ (V_{xxoi}) (i)
 The sum of forces acting in y direction is equal to zero, i.e. Σ(V_{xyoi})=0.0 (ii)
 The sum of moments V_{xxoi} and V_{xyoi} about point C_{T} is equal to zero,
i.e. Σ(V_{xxoi}·y_{i}V_{xyoi}· x_{i})=0 (iii) where V_{xyoi}=δ_{xo}×K_{xyi} with K_{xyi}=1/2×(K_{ζ}_{i}K_{η}_{i})×sin2φ’_{i }and φ’_{i}=φ_{i} a Expression (i) yields H_{x}=Σ(δ_{xo}× K_{xxi})=δ_{xo}×Σ(K_{xxi}) →
Because of the identity sin ( w _{1}  w _{2} )=sin w _{1} · cos w _{2} cos w _{1} · sin w _{2}, K_{xyi}=1/2·(K_{ζ}_{i}K_{η}_{i})·(sin2φ_{i}·cos2acos2φ_{i}·sin2a) equation (ii) becomes Σ(K_{xyi})=0 → Σ[(K_{ζ}_{i}K_{η}_{i})·sin2φi·cos2a]=Σ[(K_{ζ}_{i}K_{η}_{i})cos2φ_{i}·sin2a] → cos2a·Σ[(K_{ζ}_{i}K_{η}_{i})·sin2φ_{i}]=sin2a·Σ[(K_{ζ}_{i}K_{η}_{i})·cos2φ _{i}] → Calculation of angle a of the principal system : Using the same values for K _{ζ} , K _{η} previously calculated and since 2 φ _{1} =2 φ _{2} =0˚, 2 φ _{3} =60˚ and 2 φ _{4} =90˚: (3) → tan2a=Σ[(K_{ζ}_{i}K_{η}_{i})×sin2φ_{i}]/Σ[(K_{ζ}_{i}K_{η}_{i})×cos2φ_{i} ] → tan2a=[(31.131.1) × 10^{6} × sin0˚ +(31.131.1) × 10^{6} × sin0˚+(186.626.2) × 10^{6} × sin60˚+ (19.778.7) × 10^{6} × sin90˚]/ /[( 31.131.1) × 10^{6} × cos0˚+(31.131.1) × 10^{6} × cos0˚ + (186.626.2) × 10^{6} × cos60˚+ (19.778.7) × 10^{6}·cos90˚]=[0+0+138.959.0]/[0+0+80.2+0.0]=79.9/80.2=0.996 → 2a=44.88˚ → a=22.44˚ Therefore the angle of the principal system is a=22.44˚ Coordinate system transformation : From the initial system X0Y we are transferred to the auxiliary x’0y’, which is parallel to the principal system xC_{T}y, therefore angle of the initial system is a=22.4˚: C_{1}: from C_{1}(0,0) and φ _{1} =0˚ becomes C_{1}(0,0) φ ’_{1}=22.44˚ C_{2}: from C_{2}(6.0,0) and φ _{2} =0˚ becomes C_{2}(6.0 ×0.924=5.544, 6.0×0.3817=2.290), φ’_{2}=22.44˚ C_{3}: from C_{3}(0.0,5.0) and φ _{3} =30˚ becomes C3(5.0×0.3817=1.9085, 5.0×0.924=4.62), φ’_{3}=30˚22.44˚ =7.56˚ → C_{3}(5.544, 2.290), φ’_{3}=7.56˚ C_{4}: from C_{4}(6.0, 5.0) and φ _{4} =45˚ becomes C_{4}(6.0×0.924+5.0×0.3817=7,4542, 6.0×0.3817+5.0×0.9243=2.331), φ’_{4}=45˚22.44˚ =22.56˚ → C4(7.454, 2.331), φ’_{4}=22.56˚ C_{M}: from C_{M}(3.0, 2.5) becomes C_{M}(3.0 ×0.924+2.5×0.3817=3.727, 3.0×0.3817+2.5×0.9243=1.166) Working on the auxiliary coordinate system x’0y’ x_{i}=x’_{i}x’_{CT} and y_{i}=y’_{i}y’_{CT} therefore equation (iii) yields: δ _{xo} × Σ ((y’_{i}y_{o}) × K_{xxi}(x’_{i} x’_{CT}) × K_{xyi})=0 → Σ (y’_{i} × K_{xxi})y_{o} × Σ (K_{xxi}) Σ (x’_{i} × K_{xyi})+ x’_{CT} · Σ (K_{xyi})=0 → Σ ((x’_{i} x’_{CT}) × K_{yyi}(y’_{i} y’_{CT}) × K_{xyi})=0 →  Σ (y’_{i} × K_{xyi})+ y’_{CT} × Σ (K_{xyi})+ Σ (x’_{i} × K_{yyi}) x’_{CT} × Σ (K_{yyi})=0. Since Σ(K_{xyi})=0, Σ(y’_{i}×K_{xxi}) y’_{CT}×Σ(K_{xxi})Σ(x’_{i}×K_{xyi})=0 →  Σ (y’_{i} × K_{xyi})+ Σ (x’_{i} × K_{yyi}) x’_{CT} × Σ (K_{yyi})=0 →
C.4 Case 2: Translation of centre of stiffness C_{T} along direction y by δ_{yο}
Figure C.4 Figure C.4
Accordingly, taking into account that K_{yxi}=K_{xyi}, the corresponding expressions are: Since C_{1}, C_{2} have a square cross section → K_{xx1}=K_{yy1}=K_{xx2}=K_{yy2}=31.1·10^{6} N/m, K_{xy1}=K_{xy2}=0.0 C_{3}: φ ’_{3}=7.56˚ (a) → K_{xx3}= K_{ζ}_{3}·cos^{2}φ_{3}+K_{η}_{3}·sin^{2}φ_{3} =[186.6·0.9913^{2} +26.2·0.1316^{2}]·10^{6}=183.82· 10^{6} N/m (b)→ K_{xy3}=(1/2)·(K_{ζ}_{3}K_{η}_{3})·sin2 φ_{3}=0.5·(186.626.2)· 10^{6} ·0.261=20.93·10^{6} N/m (d) → K_{yy3}= K_{ζ}_{3}·sin^{2}φ_{i}+K_{η}_{3}·cos^{2}φ_{3} =[186.6·0.1316^{2} +26.2·0.9913^{2}] ·10^{6} N/m =28.98·10^{6} N/m C4: φ ’_{4}=22.56˚ (a) → K_{xx4}=K_{ζ}_{4}·cos^{2}φ_{4}+K_{η}_{4}·sin^{2}φ_{4}=[19.7·0.9235^{2}+78.7·0,3837^{2}] ·10^{6}=28.39·10^{6} N/m (b)→ K_{xy4}=(1/2)·(K_{ζ}_{4}K_{η}_{4})·sin2φ_{4} =0.5·(19.778.7)·10^{6}·0.709=20.92·10^{6} N/m (d) → K_{yy4}= K_{ζ}_{4}·sin^{2}φ_{i}+K_{η}_{4}·cos^{2}φ_{3} =[19.7·0,3837^{2}+78.7·0.9235^{2}] ·10^{6}=70.02·10^{6} N/m Total stiffnesses of the diaphragm and structural moments of the stiffnesses with respect to C_{T} : K_{xx}= Σ (K_{xxi})=( 31.1+31.1+183.82+28.39) ·10^{6}=274.41·10^{6} N/m K_{yy}= Σ (K_{yyi})= (31.1+31.1+28.98+70.02) ·10^{6}=161.2·10^{6} N/m Σ (x’_{i}·K_{yyi})=[0.0·31.1+5.544·31.1+1.9085·28.98+7.4542 ·70.02]·10^{6} =749.55·10^{6} N Σ (x’_{i}·K_{xyi})=[0.0·0+5.544·0+1.9085·20.93+7.4542 ·(20.93)]·10^{6}=116.07·10^{6} N Σ (y’_{i}·K_{xxi})=[0.0·31.12.29·31.1 +4.62·183.82+2.331 ·28.39]·10^{6}=844.21·10^{6} N Σ (y’_{i}·K_{xyi})= [0.0·0.0 2.29·0.0 +4.62·20.93+2.331 · (20.92)]·10^{6}=47.93·10^{6 }N x’_{CK}=[ Σ (x’_{i}·K_{yyi}) Σ (y’_{i}·K_{xyi})]/ Σ (K_{yyi})=( 749.5547.93)/161.2=4.353 m y’_{CK}=[ Σ (y’_{i}·K_{xxi}) Σ (x’_{i}·K_{xyi})]/ Σ (K_{xxi})=( 844.21+116.07)/274.41=3.500 m C.5 Case 3: Rotation of the diaphragm by an angle θ_{z} about C_{T}
Figure C.51: Displacements of a random point i of the diaphragm due to rotation by θz Figure C.51: Displacements of a random point i of the diaphragm due to rotation by θz

Figure C.52: Displacements due to rotation from a moment M applied in CT Figure C.52: Displacements due to rotation from a moment M applied in CT
δ _{xi} = δ _{i} ·sin ω _{i} δ _{xi} =r_{i}· θ _{z} ·y_{i}/r_{i}=y_{i}· θ _{z}  Consider the principal coordinate system xC_{T}y The deformation induced by the external moment M applied in C_{T}_{ }, is examined. In order to examine this deformation a transfer is needed from the auxiliary system x’0y’ to the principal system xC_{T}y by a simple parallel translation. By alsotransferring the centre of mass in the principal system as well, the structural eccentricities e_{ox}, e_{oy} of C_{M} with respect to C_{T}, derive from the expressions:
The deformation of the diaphragm is a rotation θ_{z} about C_{T}. Rotation θ_{z}of diaphragm induces a displacement δ_{i} at the top of each column i with coordinates x_{i},y_{i} with respect to the coordinate system having as origin C_{T}. If the distance of point ifrom C_{T} is r_{i}, the two components of the (infinitesimal) deformation δ_{i} are: δ_{xi}=θ_{z}×y_{i} and δ_{yi} = θ_{z}×x_{i} Coordinate transfer in the principal system and calculation of the structural eccentricities : C_{1}(0.04.353=4.35, 0.03.500=3.50) φ ’_{1}=22.44˚ K_{xx}=31.1 ×10^{6}, K_{yy}=31.1×10^{6}, K_{xy}=0.0 C_{2}(5.5444.353=1.19, 2.2903.500=5.79), φ ’_{2}=22.44˚ K_{xx}=31.1 ×10^{6}, K_{yy}=31.1×10^{6}, K_{xy}=0.0 C_{3}(1.90854.353=2.44, 4.623.500=1.12), φ ’_{3}=7.56˚ K_{xx}=183.82, K_{yy}=28.98, K_{xy}=20.93 C_{4}(7,45424.353=3.10, 2.33123.500=1.17), φ ’_{4}=22.56˚ K_{xx}=28.39, K_{yy}=70.02, K_{xy}=20.92 C_{M}(3.7274.353=0.626, 1.1663.500=2.334) and obviously C_{T}(0,0) e_{ox}=0.626 m, e_{oy}=2.334 m Displacements δ_{xi}, δ_{yi} incuce shear forces V_{xi} and V_{yi} in each column: V_{xi}=K_{xxi} × δ _{xi} = K_{xxi} × ( θ _{z} × y_{i} ) →V_{xi}= θ_{z}×K_{xxi}×y_{i} and V_{yi}=K_{yyi}×δ_{yi}=k_{yyi}×θ_{z}×x_{i} → V_{yi}=θ_{z}× k_{yyi}×x_{i} The resultant of the moments of all shear forces V_{xi}, V_{yi} with respect to the C_{T} should be equal to the external moment M_{CT} i.e.: M_{CT}= Σ [V_{xi} × y_{i} + V_{yi} × x_{i}+K_{zi}] → M_{CT}= θ _{z} × Σ [K_{xxi} × y_{i}^{2} + K_{yyi} × x_{i}^{2}+K_{zi}] →
Calculation of diaphragm torsional stiffness : Κ_{θ} = Σ (K_{xxi} × y_{i}^{2}+K_{yyi} × x_{i}^{2}+Kzi)= Σ [31.1×3.5^{2}+31.1×4.35^{2}+31.1×5.79^{2}+31.1×1.19^{2}+183.82×1.12^{2}+ 28.98×2.44^{2}+28.39×1.17^{2}+70.02×3.10^{2}]×10^{6} N×m= =[381.0+588.5+1042.6+44.0+230.6+172.5+38.9+672.9] ×10^{6} N×m → Κ_{θ}=3174×10^{6} N×m The quantity Κ_{θ}=Σ(K_{xxi}×y_{i}^{2} + K_{yyi} ×x_{i}^{2}+K_{zi}) is called torsional stiffness of the diaphragm, measured in units ofmoment e.g. N×m, by analogy with the quantities K_{x}=Σ(K_{xi}), K_{y}=Σ(K_{yi}) which are called lateral stiffnesses of the diaphragm in direction x and y respectively and measured in N/m. The torsional stiffness of the column itself is usually insignificant and can be ignored (§5.4.3.4). Lateral stiffness Κ _{jk} of diaphragm denotes the force in j direction required to cause displacement of the diaphragm by one unit in the k direction. Torsional stiffness Κ_{θ} of diaphragm denotes the moment required to cause rotation of the diaphragm by one unit. C.6 Torsional stiffness ellipse, torsional radii and equivalent system Question: Create a simple idealised equivalent structural system with the same seismic behaviour as the actual structural system. Solution: Four idealised columns E_{1}, E_{2} and E_{3}, E_{4 }are placed_{ }symmetrically with respect to the centre C _{T} and to the axes x and y, i.e. all four idealised columns have x and y coordinates with the same absolute values. Each idealised column isassumed to have a stiffness equal to K_{x}=1/4×Σ(K_{xi}) and K_{y}=1/4 ×Σ(K_{yi}).
Figure C.6 Figure C.6
This system satisfies the two conditions of the actual system, concerning the translations of all diaphragm columns since: Stiffness by x: 4×1/4×Σ(K_{xi)}=Σ(K_{xi)} and stiffness by y: 4×(1/4) ×Σ(K_{yi)}=Σ(K_{yi)}. To satisfy the third condition, the torsional stiffness of the idealised system should be K _{θ} _{,eq} =[K_{x1} × y^{2}+K_{x2} × y^{2}+K_{y1} × x^{2}+K_{y2} × x^{2}]= Σ (K_{xxi}) × y^{2}+ Σ (K_{yyi}) × x^{2} equal to the torsional stiffness of the actual system K _{θ} _{,re} =K _{θ} = Σ (K_{xxi}·y_{i}^{2} + K_{yyi}·x_{i}^{2} 2K_{xyi}·x_{i}·y_{i}+K_{zi}) , i.e. should K_{θ}_{,eq}= K_{θ}_{,re} → Σ(K_{xxi})·y^{2}+Σ(K_{yyi})·x^{2}=K_{θ} →
The curve equation (8) is an ellipse with centre C_{T} and semiaxes r_{x}, r_{y} oriented along the principal axes Calculation of the torsional stiffness ellipse r_{x}=√(K _{θ} /K_{yy})=√(3134·10^{6}N·m / 161.2 ·10^{6}N/m)=4.41 m r_{y}=√(K _{θ} /K_{xx})=√(3134·10^{6} N·m / 274.41 ·10^{6}N/m)=3.38 m The torsional behaviour of a floor can be described by the torsional stiffness ellipse (C_{T}, r_{x}, r_{y}) which represents the equivalent distribution of the diaphragm stiffness. The radii r_{x}, r_{y}, of the ellipse are called torsional radii. There are infinite solutions of idealised dual system sets, whose most characteristic is the one with the four idealised columns placed in the four ellipse ends. Generally, there are infinite solutions with ntuple diametrically opposed systems, where each idealised column stiffness is equal to 1/n of the total system stiffness.  The torsional stiffness ellipse is better to enclose the mass inertial ring
 The position of ‘walls’ is not significant, provided they are located as close as possible to the perimeter. It is significant, however, to have high stiffness in both directions.
 Significant deviation between C_{M} and C_{T} can be handled by placing walls of high stiffness in the perimeter of the building.
C.7 Superposition of the three cases So far, all the calculations depended on the structure geometry and were not affected by the magnitude of the external loading. For instance the centre of stiffness, the structural eccentricities, and the torsional radii, are independent of the seismic force magnitude. Next, the displacements and stress resultants of the structure, due to external seismic loading H, will be calculated. The relevant seismic force H is applied at the centre of mass C_{M} of the diaphragm. This force is resolved in two forcesH_{x} and H_{y} parallel to the two axes of the principal system. In order to apply the previous analysis, the forcesH_{x}, H_{y} are transferred to the centre of stiffness C_{T} together with the moment M_{CT} according to the expression:
(9) Calculation of moment M_{CT} : The structural eccentricities are e_{ox} =0.626 and e_{oy}= y_{CM}=2.334m For a given force Η =H_{X}=90.6 kN in the initial system X0Y, the equivalent loading in the principal system (a=22.44 ^{ο} ) is: H_{x}=H_{X} × cosa=90.6 ×0.924=83.72 kN and H_{y}=H_{X}×sina=90.6×0.382=34.64 kN and Μ _{CT} =H_{x} × e_{oy}+H_{y} × e_{ox}=83.72kN × (2.334m)+[34.64kN × (0.626m)]=195.4 kNm+21.7kNm=217.1 kNm The following quantities are calculated using the H_{x}, H_{y}, M_{CT}:  the displacements δ_{xo}, δ_{yo} and θ _{z} of C_{T} using the expressions:
Quantities K_{xx}= Σ (K_{xxi})= 161.2 × 10^{6}N/m, K_{yy}= Σ (K_{yyi})= 161.2 × 10^{6} N/m and Κ_{θ} =3134 × 10^{6} N × m are already calculated therefore: δ _{xo} =H_{x}/ Κ _{xx } =83.72 × 10^{3}N/(274.41 × 10^{6}N/m)=0.305 mm, δ _{yo} =H_{y}/ Κ _{yy } =34.46 × 10^{3}N/(161.2 × 10^{6} N/m)= 0.214 mm and θ _{z} =M_{CT}/K _{θ} =217.1kNm/(3134·10^{3}kNm)=0.692·10^{4}  the displacements δ_{xi}, δ_{yi} of the top of each column using the expressions:
 and the displacements δ_{ζ}_{i}, δ_{η}_{i} by transferring δ_{xi}, δ_{yi} to the local system of each column using the expressions:
C_{1}: δ _{x1} = δ _{xo}  θ _{z} × y_{1}=0.305mm0.692 × 10^{4}·(3.50 × 10^{3}mm)=(0.305+0.242)mm=0.547 mm δ _{y1} = δ _{yo} + θ _{z} × x_{1}=0.214mm+0.692 × 10^{4}·(4.35 × 10^{3}mm)=(0.2140.301)mm=0.515 mm By transferring to the local system where φ ’_{1}=0.022.44 ° =22.44 °. δ_{ζ} _{1} = δ _{x1} × cos φ ’_{1}+ δ _{y1} × sin φ ’_{1}=0.547 ×0.9240.515×(0.382)=0.505+0.197=0.702 mm δ_{η} _{1} = δ _{x1} × sin φ ’_{1}+ δ _{y1} × cos φ ’_{1}=0.547 ×(0.382)+(0.515)×0.924=0.2090.476=0.267 mm C_{2}: δ _{x2} = δ _{xo}  θ _{z} × y_{2}=0.305mm0.692 × 10^{4}·(5.79 × 10^{3}mm)=(0.305+0.400)mm=0.705 mm δ _{y2} = δ _{yo} + θ _{z} × x_{2}=0.214mm+0.692 × 10^{4} × 1.19 × 10^{3}mm=(0.214+0.082)mm=0.132 mm By transferring to the local system where φ ’_{2}=0.022.44 ° =22.4 °. δ_{ζ} _{2} = δ _{x2} × cos φ ’_{2}+ δ _{y2} × sin φ ’_{2}=0.705 ×0.924+(0.291)×(0.132)=0.652+0.038=0.701 mm δ_{η} _{2} = δ _{x2} × sin φ ’_{2}+ δ _{y2} × cos φ ’_{2}=0.705 ×(0.382)+(0.132)×0.924=0.2690.122=0.147 mm C_{3}: δ _{x3} = δ _{xo}  θ _{z} × y_{3}=0.305mm0.692 × 10^{4} × 1.12 × 10^{3}mm=(0.3050.078)mm=0.227 mm δ _{y3} = δ _{yo} + θ _{z} × x_{3}=0.214mm+0.692 × 10^{4} × (2.44 × 10^{3}mm)=(0.2140.169)mm=0.383 mm By transferring to the local system where φ ’_{3}=30.022.44 ° =7.56 °. δ_{ζ} _{3} = δ _{x3} × cos φ ’_{3}+ δ _{y3} × sin φ ’_{3}=0.227 ×0.991+(0.383)×0.132=0.2250.050=0.175 mm δ_{η} _{3} = δ _{x3} × sin φ ’_{3}+ δ _{y3} × cos φ ’_{3}=0.227 ×0.132+(0.383)×0.991=0.0300.380=0.410 mm C_{4}: δ _{x4} = δ _{xo}  θ _{z} × y_{4}=0.305mm0.692 × 10^{4} × (1.17·10^{3}mm)=(0.305+0.081)mm=0.386 mm δ _{y4} = δ _{yo} + θ _{z} × x_{4}=0.214mm+0.692 × 10^{4} × 3.10 × 10^{3}mm=(0.214+0.215)mm=0.001 mm By transferring to the local system where φ ’_{4}=45.022.44 ° =22.56 °. δ_{ζ} _{4} = δ _{x4} × cos φ ’_{4}+ δ _{y4} × sin φ ’_{4}=0.386 ×0.923+0.001×0.384=0.355+0.000=0.355 mm δ_{η} _{4} = δ _{x4} × sin φ ’_{4}+ δ _{y4} · cos φ ’_{4}=0.386 ·0.384+0.001·0.923=0.148+0.001=0.147 mm In the example considered, for seismic action in X direction, the deformation due to rotation at column C_{2} gives δ _{x2,} _{θ} =0.400 mm, higher than the deformation due to translation δ _{xo} =0.305 mm. The total displacement is therefore δ _{x2} =0.305+0.400=0.705 mm. 
Figure C.7: Moment distribution (Mji,1 Mji,2=Vji×h) Figure C.7: Moment distribution (Mji,1 Mji,2=Vji×h)
shear forces and bending moments of each column in its local system based on the relationships:
Calculation of shear forces and bending moments : C_{1}: V _{ζ} _{1} = δ_{ζ} _{1} · K _{ζ} _{1} =0.702mm · 31.1 · 10^{6} N/m =21.8 kN V _{η} _{1} = δ_{η} _{1} · K _{η} _{1} =0.267mm · 31.1 · 10^{6} N/m =8.3 kN M _{ζ} _{1,1} =V _{ζ} _{1} · h · 0.50=21.8·3.0·0.50=32.7 kNm M_{ζ}_{1,2}= M_{ζ}_{1,1} =32.7 kNm M _{η} _{1,1} =V _{η} _{1} · h · 0.50=8.3·3.0·0.50=12.5 kNm M_{ηι}_{,2}= M_{η}_{1,1} =12.5 kNm C_{2}: V _{ζ} _{2} = δ_{ζ} _{2} · K _{ζ} _{2} =0.701mm · 31.1 · 10^{6} N/m =21.8 kN V _{η} _{2} = δ_{η} _{2} · K _{η} _{2} =0.147mm · 31.1 · 10^{6} N/m =4.6 kN M _{ζ} _{2,1} =V _{ζ} _{2} · h · 0.50=21.8·3.0·0.50=32.7 kNm M_{ζ}_{2,2}= M_{ζ}_{2,1} =32.7 kNm M _{η} _{2,1} =V _{η} _{2} · h · 0.50=4.6·3.0·0.50=6.9 kNm M_{η}_{2,2}= M_{η}_{2,1} =6.9 kNm C_{3}: V _{ζ} _{3} = δ_{ζ} _{3} · K _{ζ} _{3} =0.175mm · 186.6 · 10^{6} N/m =32.7 kN V _{η} _{3} = δ_{η} _{3} · K _{η} _{3} =0.410mm · 26.2 · 10^{6} N/m =10.8 kN M _{ζ} _{3,1} =V _{ζ} _{3} · h · 0.50=32.7·3.0·0.50=49.1 kNm M_{ζ}_{3,2}= M_{ζ}_{3,1} =49.1 kNm M _{η} _{3,1} =V _{η} _{3} · h · 0.50=10.8·3.0·0.50=16.2 kNm M_{η}_{3,2}= M_{η}_{3,1} =16.2 kNm C_{4}: V _{ζ} _{4} = δ_{ζ} _{4} · K _{ζ} _{4} =0.355mm · 19.7 · 10^{6} N/m =7.0 kN V _{η} _{4} = δ_{η} _{4} · K _{η} _{2} =0.147mm · 78.7 · 10^{6} N/m =11.6 kN M _{ζ} _{4} =V _{ζ} _{4} · h · 0.50=7.0·3.0·0.50=10.5 kNm M_{ζ}_{4,2}= M_{ζ}_{4,1} =10.5 kNm M _{η} _{4,1} =V _{η} _{4} · h · 0.50=11.6·0·0.50=17.4 kNm M_{η}_{4,2}= M_{η}_{4,1} =17.4 kNm GEOMETRIC DATA OF THE DIAPHRAGM (independent of the loading) 2. Calculation of angle a of the main system using expression (3) 3. Calculation of column angles φ ’_{i} in the principal system x,y using the expression φ ’_{i}= φ _{i} a 4. Calculation of column stiffnesses K_{xxi}, K_{xyi}, K_{yyi} in the principal system x,y using expressions (a), (b), (c) 5. Calculation of each column centre of gravity and centre of mass (x’_{CM}, y’_{CM}) using the relationships for the transformation from the principal system X0Y to the auxiliary system x’0y’. 6. Calculation of diaphragm total stiffnesses using the expressions K_{x}= Σ (K_{xxi}), K_{y}= Σ (K_{yyi}). Calculation of the quantities Σ (x’_{i} · K_{yyi}), Σ (y’_{i} · K_{xyi}), Σ (y’_{i} · K_{xxi}), Σ (x’_{i} · K_{xyi}) and the coordinates of the centre of stiffness C_{T}(x’_{CT}, y’_{CT}) using the expressions (4) and (5) 7. Transfer of the coordinates x’, y’ of the diaphragm C_{M} and each column centre of gravity in the principal system. Calculation of the structural eccentricities e_{ox}, e_{oy }using the expressions (6’). 8. Calculation of torsional stiffness K _{θ} using the expression (7) 9. Calculation of torsional radii r_{x}, r_{y} using the expressions (8) Up to this point the calculations are independent of the external loading. Next, the stress distribution and the elastic deformations of the diaphragm elements, due to the external loading, are calculated. FLOOR DISPLACEMENTS AND STRESS RESULTANTS (dependent on the loading) 10. Transfer of horizontal forces H_{x}, H_{y} and moment M_{CT} to the C_{T} using the expression (9). 11. Calculation of displacements δ _{xo} , δ _{yo} and θ _{z} of the C_{T} using the expressions (10). 12. Calculation of the displacements δ _{xi} , δ _{yi} of each column top using the expressions (11). 13. Calculation of the corresponding displacements δ_{ζ} _{i} , δ_{η} _{i} in the local system of each column using the expressions (12) 14. Calculation of principal shear forces V _{ζ} _{i} , V _{η} _{i} of each column using the expressions (13) 15. Calculation of principal bending moments M _{ζ} _{i,1} , M _{ζ} _{i,2} & M _{η} _{i,1} , M _{η} _{i,2} using the expressions (14) C.9 Expressions relating the initial system X0Y and the principal system xC_{T}y Seismic force Η _{X} =H acting in X direction (H_{Υ}=0) in the initial system X0Y: The equivalent forces H_{x}, H_{y} in the principal system xC_{T}y are: H_{x}=H·cosa and H_{y}=H·sina and the equivalent deformations of C_{T}, δ_{x}, δ_{y,} in the principal system are: δ_{x} = δ_{XX}_{ο}·cosa + δ_{ΧΥο}·sina and δ_{y}= δ_{XX} _{ο}·sina +δ_{ΧΥο}·cosa. H_{x}=K_{xx} ·δ _{x} and H_{y}=K_{yy}·δ_{y} therefore: H · cosa = K_{xx} · ( δ _{XX} _{ο} · cosa + δ_{ΧΥο} · sina) → H=K_{xx}·( δ_{XX}_{ο}+tana· δ_{ΧΥο}) (i) H · sina = K_{yy} · ( δ _{XX} _{ο} · sina + δ_{ΧΥο} · cosa) →tana·H=K_{yy}·(tana·δ_{XX} _{ο}+δ_{ΧΥο}) (ii) Seismic force Η _{Y} =H acting in Y direction (H_{X}=0) in the initial system X0Y: The equivalent forces H_{x}, H_{y} in the principal system xC_{T}y are: H_{x}=H·sina and H_{y}=H·cosa and the equivalent deformations of C_{T}, δ_{x}, δ_{y},in the principal system are: δ _{x} = δ_{ΧΥο} · cosa+ δ _{YYo} · sina and δ_{y} =δ_{YXo}·sina+ δ _{YYo}·cosa. H_{x}=K_{xx} · δ _{x} and H_{y}=K_{yy}·δ_{y} due to δ _{YXo} = δ_{ΧΥο} yields: H · sina=K_{xx} · ( δ_{ΧΥο} · cosa+ δ _{YYo} · sina) → tana·H=K_{xx}·( δ_{ΧΥο}+tana ·δ_{YYo}) (iii) H · cosa=K_{yy} · ( δ_{ΧΥο} · sina+ δ _{YYo} · cosa) → H=K_{yy}·(tana·δ_{ΧΥο}+δ _{YYo}) (iv) (iv) → K_{yy}=H/(δ_{YYo}  tana·δ_{ΧΥο}) (i) → K_{xx}=H/(δ_{XX}_{ο} + tana·δ_{ΧΥο}) (iii) → tana·H= H/( δ_{XX}_{ο}+tana·δ_{ΧΥο})·( δ_{ΧΥο}+tana·δ _{YYo}) → tana · ( δ _{XX} _{ο} +tana · δ_{ΧΥο} )= δ_{ΧΥο} +tana · δ _{YYo} ) → tana·δ_{XX}_{ο} +tan^{2}a· δ_{ΧΥο} = δ_{ΧΥο} +tana·δ_{YYo} → δ_{ΧΥο} · (1 tan^{2}a)=tana · ( δ _{XX} _{ο}  δ_{ΧΥο} )→ δ_{ΧΥο}=[tana/(1 tan^{2}a)]·( δ_{XX} _{ο}  δ_{YYo}) → tan2a = 2 δ_{ΧΥο} / ( δ _{XX} _{ο}  δ _{YYo} ) (C.9.1) K_{xx}=H/( δ _{XX} _{ο} +tana · δ_{ΧΥο} ) (C.9.2) K_{yy}=H/( δ _{YYo} tana · δ_{ΧΥο} ) (C.9.3) Κ_{θ} =H · (Y_{CT}Y_{CM})/ θ _{xz} _{ } (C.9.4) r_{x}= √(K _{θ} /K_{yy}) (C.9.5) r_{y}= √(K _{θ} /K_{xx}) (C.9.6) C.10 Onestorey space frame with rectangular columns in random arrangement The method which describes the diaphragmatic behaviour of each floor is presented in Appendix D. Also relevant examples are presented verifying the validity of the algorithms of the general method. 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) assuming using software, assuming actual beam and column torsional stiffnesses.
Figure C.10: Frame and model of a singlestorey space frame Figure C.10: Frame and model of a singlestorey space frame
The names, loadings, coordinates and dimensions of slabs and columns are shown in figure of §C.1. The crosssection of beams is 250/500. The concrete class is C30/37 (Ε=32.80 GPa). C.10.1 Analysis using manual calculations, assuming fixed columns (k=12) This method was described via the example presented to all previous paragraphs of this appendix. C.10.2 Analysis using the Excel file, assuming fixedended columns (k=12) The related <diaphragm_general.xls> spreadsheet is used assuming fixedended columns.
Figure C.10.2: The results are identical to those of the practical calculations.The torsional stiffness ellipse and the equivalent columns of the structure are drafted at the end of the spreadsheet. Figure C.10.2: The results are identical to those of the practical calculations.The torsional stiffness ellipse and the equivalent columns of the structure are drafted at the end of the spreadsheet.
The torsional stiffness ellipse and the equivalent columns of the structure are drafted at the end of the spreadsheet. C.10.3 Analysis using the Excel file, assuming columns with k=6 The centre of stiffness, the torsional radii and the stress resultants derived from this method are exactly the same as those of the previous methods, with the exception of deformations, whose their relative values, however, remain constant 12/6=2.00. C.10.4 Analysis using software, assuming actual beam and column torsional stiffnesses In order to define the diaphragmatic function of a building’s storey, with actual stiffness and not with assumption of fixed function of its columns, it is necessary to use the appropriate software. Structural analysis software is necessary for one storey, as well as multistorey buildings, even in case of rectangular columns in parallel arrangement. The method which describes the diaphragmatic behaviour of each floor is presented in Appendix D and this example is analysed as example 1.
