« Centre of stiffness and elastic displacements of the diaphragm Assessment of building torsional behaviour » Work methodology Summarising the theory for the calculation of displacements and stress resultants of a floor diaphragm the following computational procedurecan be formulated. This procedure has also been applied in the related file [*] Note There is also the related file , supporting the general case, presented in the Appendix C. This Excel file may be used to verify the calculations of this and the following examples, but also of any other floor diaphragm. GEOMETRIC DATA OF THE DIAPHRAGM (independent of the loading) 1. Calculation of stiffnesses Kxi, Kyi and torsional stiffness Kzi of each column. 2. Calculation coordinates (XCT,YCT) of the centre of stiffness CT using the expressions (4') and (5'). 3. Transfer of the coordinates X, Y of the centre of mass of the diaphragm CM and each column centre of gravity to the principal axis system. Calculation of the structural eccentricities eox, eoy 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 rx, ry 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 Hx, Hy and the moment MCT to the centre of stiffness CT using the expression (9'). 7. Calculation of the displacements δ xo , δ yo and θ z of the centre of stiffness CT 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 Vxi, Vyi of each column using the expressions (13'). 10. Calculation of the principal bending moments Mxi,1, Mxi,2 & Myi,1, Myi,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 Kij of each column i in direction j is: Κ ji =kji · ( EI ji / h 3 ) · kji , v α (§4.1.5) where kji=12 (columns are assumed to be fixed-ended) Therefore, Κ ji =12 · EI ji / h 3 . Moreover, we assume that all columns have the same torsional stiffness Kzi=0. The columns of the example have the same magnitudes E and h, therefore their stiffness can be assumed to be Κ ji =C · Iji , where C=12E/h3. Modulus of Elasticity is taken as Ε =32.80 GPa, corresponding to concrete class C30/37 (volume C §1.1). Therefore, C=12E/h3=12 · 32.80 · GPa/(3.03m3)=14.58 ·109N/m5. C1, C2 400/400: K1x=K1y=K2x=K2y=14.58 · 109N/m5 · 0.400 · 0.4003/12 · m4=31.1 ·106 N/m C3 800/300: K3x=14.58 ·109·0.300·0.8003/12=186.6·106 N/m K3y=14.58 ·109·0.800·0.3003/12=26.2·106 N/m C4 300/600: K4x=14.58 ·109·0.600·0.3003/12=19.7·106 N/m K4y=14.58 ·109·0.300·0.6003/12=78.7·106 N/m Therefore based on the expressions (3): Calculation of the Centre of Stiffness and the Lateral Stiffnesses : expressions (4') and (5') Kx= Σ (Kxi)=(31.1+31.1+186.6+19.7) ·106 N/m=268.5·106 N/m Ky= Σ (Kyi)=(31.1+31.1+26.2+78.7) ·106 N/m=167.1·106 N/m Χ CT =Σ( Xi · Kyi )/ Ky =[(0.0·31.1+6.0·31.1+0.0·26.2+6.0·78.7)·106]/(167.1· 106)=658.8/167.1=3.94 m Υ CT = Σ (Yi · Kxi)/Kx=[(0.0·31.1+0.0·31.1+5.0·186.6+5.0·19.7)·106]/(268.5· 106)=1031.5/268.5=3.84 m Coordinate transfer to the principal system xCTy : expressions (6') C1(-3.94,-3.84), C2(2.06,-3.84), C3(-3.94,1.16), C4(2.06,1.16), CM(-0.94,-1.34), C T(0.0,0.0) eox=-0.94 m, eoy=-1.34 m Calculation of the diaphragm's torsional stiffness : expression (7') K θ = Σ [Kxi · yi2+Kyi · xi2+Kzi]= (31.1·3.842+31.1·3.942+31.1·3.842+31.1·2.062+186.6·1.162+26.2·3.942+19.7·1.162+78.7·2.062)· ·106N/m·m2=(1194.8+1355.4)·106Nm=2550·106 Nm [*] Note The torsional stiffnesses Kz of columns are equal to: Kz1=Kz2=0.5E × Id/h=0.5×32.80×109Pa×0.1404×0.44m4/(3.0m)=20× 106 N × m Kz3= 0.5E × n × b3 × d/h=0.5×32.80×109Pa×0.251×0.33m3×0.8m/(3.0m)=30× 106 Nm Kz4= 0.5E × n × b3 × d/h=0.5×32.80×109Pa×0.229×0.33m3×0.6m/(3.0m)=20× 106 Nm, therefore Kθ=(2550+90)×106 Nm=2640×106 Nm, i.e. an increase of the floor torsional stiffness by 3.5%. Diaphragm's torsional radii : expressions (8') rx=√(K θ /Ky)=√[2550 · 106Nm/(167.1 · 106N/m)]=3.91 m ry=√(K θ /Kx)=√[2550 · 106Nm/(268.5 · 106N/m)]=3.08 m (b) DISPLACEMENTS AND STRESS RESULTANTS (independent of the loading) Given the seismic force H=90.6 kN [*] Note This force corresponds to seismic acceleration (applied at mass position) a=0.20g,therefore Η=Σ(mi)·0.20g=45.3·103kgr×0.20×10m/sec2=90.6 kN. and the interstorey height h=3.0 m, calculate the displacements of the centre of stiffness CT and those of each column, the shear forces carried by each column and the corresponding bending moments for the following three loading cases: (A) Hx=90.6 kN, Hy=0, (B) Hx=0 kN, Hy=90.6 kN and ( Γ ) Hx=90.6 kN, Hy=-27.2 kN Case Α : Hx=90.6 kN, Hy=0.0 Seismic moment at the centre of stiffness : expression (9') MCT=-Hx · eoy+Hy · eox=-90.6kN · (-1.34m)=121.4 kNm Displacements of the centre of stiffness : expressions (10') δ xo =Hx/Kx=90.6 · 103N/(268.5 · 106N/m)=0.337 · 10-3m=0.337 m θ z =MCT/K θ =121.4 · 103Nm/( 2550 · 106Nm)=47.6 ·10-6 Displacements of columns : (expressions 11') δ xi = δ xo + δ x θ i = δ xo - θ z · yi δ yi =0.0+ θ z · xi= θ z ·xi To compare the quantities, the displacements due to rotation are calculated separately. Column C1: δ x θ 1 =- θ z · y1=-47.6 ·10-6·(-3.84·103mm)=0.183 mm δ y θ 1 = θ z · x1=-0.188 mm Summarising, C1: 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: C2: 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 C3: δ x θ 3 =-0.055 mm, δ y θ 3 =-0.188 mm, δ x3 =0.337-0.055=0.282 mm, δ y3 =-0.188 mm C4: δ x θ 4 =-0.055 mm, δ y θ 4 = 0.098 mm, δ x4 =0.337-0.055=0.282 mm, δ y4 = 0.098 mm Seismic shear forces : expressions (13') C1: Vx1= δ x1 · Kx1=0.520·10-3m·31.1·106N/m=16.17 kN, Vy1=δy1 ·Ky1=-0.188·31.1=-5.85 kN C2: Vx2= δ x2 · Kx2=0.520·31.1= 16.17 kN, Vy2=δy2·Ky2= 0.098· 31.1= 3.05 kN C3: Vx3= δ x3 · Kx3=0.282·186.6=52.62 kN, Vy3=δy3·Ky3=-0.188· 26.2=-4.93 kN C4: Vx4= δ x4 · Kx4=0.282·19.7= 5.56 kN, Vy4=δy4·Ky4= 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) axi=ayi=0.50, and therefore (1-axi)=(1-ayi)=0.50. C1: Mx1,1=0.5 ·16.17·3.0=24.3 kNm, Mx1,2=-24.3 kNm, My1,1=0.5 · (-5.85) ·3.0=-8.8 kNm, My1,2= 8.8 kNm C2: Mx2,1=0.5 ·16.17·3.0=24.3 kNm, Mx2,2=-24.3 kNm, My2,1=0.5 ·3.05·3.0= 4.6 kNm, My2,2= -4.6 kNm C3: Mx3,1=0.5 ·52.62·3.0=78.9 kNm, Mx3,2=-78.9 kNm, My3,1=0.5 ·(-4.93)·3.0=-7.4 kNm, My3,2= 7.4 kNm C4: Mx4,1=0.5 ·5.56·3.0= 8.3 kNm, Mx4,2= -8.3 kNm, My4,1=0.5 ·7.71·3.0= 11.6 kNm, My4,2=-11.6 kNm Case B: Hx=0.0 kN, Hy=90.6 Seismic moment at the centre of stiffness : expression (9') MCT=-Hx·eoy+Hy·eox=90.6kN ·(-0.94m)=-85.2 kNm Displacements of the centre of stiffness : expressions (10') δ xo =0 δ yo =Hy/Ky=90.6 ·103N/(167.1·106N/m)=0.542 mm θ z =MCT/K θ =-85.2 ·103Nm/(2550·106Nm)=-33.4·10-6 Only the quantities for column C3 are calculated here. The results for the remaining columns can be derived anyway from the related file . Displacements of columns (C3) : expressions (11') δ x θ 3 =- θ z · y3=-(-33.4) ·10-6·1.16·103mm=0.039 mm δ y θ 3 = θ z · x3=(-33.4) ·(-3.94·103mm)=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 (C3) : expressions (13') Vx3= δ x3 · Kx3=0.039 ·186.6=7.28 kN Vy3= δ y3 · Ky3=0.674 ·26.2=17.66 kN Seismic bending moments (C3) : expressions (14') Mx3,1=0.5 ·7.28·3.0=10.9 kNm Mx3,2=-10.9 kNm My3,1=0.5 ·17.66·3.0=26.5 kNm My3,2=-26.5 kNm Seismic moment in the center of stiffness : expression (9') MCT=-Hx · eoy+Hy · eox=-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 =Hx/Kx=90.6/268.5mm=0.337 mm δ yo =Hy/Ky=-27.2/167.1=-0.163 mm θ z =MCT/K θ =147 ·103Nm/(2550·106Nm)=57.65·10-6 Next, only the data of column C3 will be calculated. The other results can be derived, from the related file . Displacements of columns (C3) : expressions (11') δ x θ 3 =- θ z · y3=-57.65 ·10-6·1.16·103mm=-0.067 mm δ y θ 3 = θ z · x3=57.65 ·(-3.94·103mm)=-0.227 mm δ x θ 3 =-0.067 mm, δ y θ 3 =-0.227 mm, δ x3 =0.337-0.067=0.270 mm, δ y3 =-0.163-0.227=-0.390 mm Seismic shear forces (C3) : expressions (13') Vx3= δ x3 ·Kx3=0.270 ·186.6=50.38 kN Vy3= δ y3 ·Ky3=-0.390 ·26.2=-10.22 kN Seismic bending moments (C3) : expressions (14') Mx3,1=0.5 ·50.38·3.0=75.6 kNm Mx3,2=-75.6 kNm My3,1=0.5 ·(-10.22)·3.0=-15.3 kNm My 3,2 =15.3 kNm Notes 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 cross-sectional dimensions, etc. For seismic action in the x direction, the displacement of the centre of stiffness CT 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 CT 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 fixed-ended 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 one-storey 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 cross-sectional dimensions e.g. 400/400 instead of 200/200, the displacement of the diaphragm would be (400/200)4=16 times greater i.e. δyo=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. « Centre of stiffness and elastic displacements of the diaphragm Assessment of building torsional behaviour » 