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7 CONTENTS 1. GENERAL Eurocodes Units of measurement Symbols LIMIT STATES DESIGN 2.1 General Actions on structures Classification of actions Nominal/ Characteristic values of loads Materials self-weights Floor coverings self-weights Brick masonry loads Imposed loads Snow loads and wind loads Design values of actions Non-seismic actions Seismic actions Combinations of actions Wind influence Effects of indirect actions Effects of actions Deformation-stress definition of structures for combinations without seismic actions (STR/GEO) [EC0, ] Deformation-stress definition of structures for combinations with seismic (or accidental) actions Deformation-stress envelope definition Strength of structural elements Exercise STRUCTURAL MODEL AND ANALYSIS Structural model Effective span of beams and slabs Effective width of flanges Rigit bodies... 38

8 3.1.4 Diaphragmatic behaviour Seismic loading model Deformations - Stresses Frame model Slab analysis Frame analysis The effect of torsional stiffness on indirect beam-beam supports Modelling floor diaphragms using two-dimensional finite elements Modelling slabs using members and finite elements The frame behaviour in regions of columns Deflection of beams Torsional stiffness of beams Final conclusion SLABS General Two-dimensional finite elements Assumptions Stress resultants and deflections Unfavourable loadings and envelopes of stresses deflections Clarifications Analyses using tables Assumptions Application width of concentrated load on slab Distribution width of concentrated load Support moments of continuous slabs Contribution of pinned slab supports Cantilevers Static analysis Deflection One-way slabs Static analysis Deflection Effective of live load on the static analysis of one-way slabs Accurate analysis method

9 Simplified method for envelope estimation Two-way slabs Shear forces and support reaction forces Simplified method Czerny s elasticity method Fundamental support and span moments Deflections MARCUS method CZERNY s elasticity method Bending moments in continuous two-way slabs Continuous strip method Accurate method (manual) Practically accurate method Influence of live load Slabs supported on three edges Slabs supported on two adjacent edges Exercises SEISMIC BEHAVIOUR OF FRAMES One-storey plane frames (or coupled columns) Bending and shearing effect on deformations and stresses The degree of fixity effect of columns The effect of columns moment of inertia The effect of columns differential height Frame column stiffness One-storey frame structural systems One-storey dual structural systems Coupled one-storey plane frames Multistorey plane frames Multistorey plane frame systems Multistorey plane dual systems Comparison of multistorey frame and dual systems Space frames Diaphragmatic behaviour Centre of mass and radius of gyration Centre of stiffness and elastic displacements of the diaphragm Subject description Translation of centre of stiffness CT along x direction

10 Translation of centre of stiffness CT along y direction Rotation of the diaphragm by an angle θz about CT Torsional stiffness ellipse, torsional radii and equivalent system Work methodology Assessment of building torsional behaviour One-storey space frame with rectangular columns in parallel arrangement Analysis using manual calculations, assuming fixed-ended columns (k=12) Analysis using the Excel file, assuming fixed-ended columns (k=12) Analysis using the Excel file, assuming columns with k= Analysis using software, assuming actual beam and column torsional stiffnesses Multistorey space frame of rectangular columns in parallel arrangement Analysis using manual calculations, assuming fixed-ended columns (k=12) Analysis using the Excel file, assuming fixed-ended columns (k=12) Analysis using the Excel file, assuming columns with k= Analysis using software, assuming actual beam and column torsional stiffnesses Exercises SEISMIC ACCELERATIONS AND LOADINGS OF BUILDINGS Seismic response of buildings Seismic zones [EC8, 3.2.1] Importance factors [EC8, 4.2.5] Ground type [EC8, 3.1.2] Viscous damping Behaviour factor q Structural types [EC ] Regularity in plan [EC ] Regularity in elevation [EC ] Ductility classes [EC ] Basic value of the behaviour factor qo [EC ] Failure mode factor kw [EC ] Conclusion Design spectrum of horizontal seismic actions [EC8, & 3.2.1(3)] Design spectrum of vertical seismic actions [EC8, (5)] Dynamic analysis και natural periods of structure

11 6.3 Seismic stresses Seismic accelerations Seismic forces, shear forces, bending moments Approximate method for the calculation of seismic accelerations [EC8, ] Cracking and plasticity effects Applications Frame type structure Wall-equivalent dual type structure Approximate analysis of the frame and wall type structures Loading envelopes Special cases Columns not belonging in a diaphragm Vertical component of seismic action [EC8, ] APPENDICES APPENDIX A: MODELLING SLABS WITH FINITE ELEMENTS Α.1 Modelling slabs with finite elements in a structural frame Α.1.1 The frame behaviour in the regions close to columns Α Model using members Α Model with finite elements Α First Conclusion Α.1.2 Deflection of beams Α Model using members Α Model using finite elements Α Second Conclusion Α.1.3 Torsional stiffness of beams Α Model using members Α Model using finite elements Α Third Conclusion Α FINAL CONCLUSION APPENDIX B: STIFFNESS OF MULTISTOREY PLANE FRAMES Β.1 Stiffness of frame crossbar Β.2 Equivalent multistorey frame crossbar relative stiffness Β.3 Relative stiffness of crossbar column Β.4 Effect of slabs on stiffnesses using finite elements Β.5 Effect of walls on stiffnesses using finite elements

12 APPENDIX C: DIAPHRAGMATIC BEHAVIOUR OF ONE-STOREY SPACE FRAME GENERAL CASE C.1 Subject description C.2 Axis transformation C.3. Case 1: Translation of centre of stiffness C T along x direction by δ xο C.4 Case 2: Translation of centre of stiffness C T along y direction by δ yο C.5 Case 3: Rotation of the diaphragm by an angle θ z about C T C.6 Torsional stiffness ellipse, torsional radii and equivalent system C.7 Superposition of the three cases C.8 Work Methodology C.9 Expressions relating the initial system X0Y and the principal system xc T y C.10 One-storey space frame with rectangular columns in random arrangement C.10.1 Analysis using manual calculations, assuming fixed columns (k=12) C.10.2 Analysis using the Excel file, assuming fixed-ended columns (k=12) C.10.3 Analysis using the Excel file, assuming columns with k= C.10.4 Analysis using software, assuming actual beam and column torsional stiffnesses APPENDIX D: DIAPHRAGMATIC BEHAVIOUR OF MULTISTOREY SPACE FRAME GENERAL CASE D.1 Subject description D.2 General method for the calculation of the diaphragm i data D.3 Diaphragm lateral stiffnesses D.4 Diaphragm torsional stiffness D.5 Torsional stiffness distribution D.6 Equivalent system Relative lateral stiffness Relative torsional stiffness D.7 Examples

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14 Static and Dynamic Analysis 2.5 Exercise The residential building of the sketch includes a basement of area m 2 and height 3 m, a ground floor and four storeys of identical dimensions and a top floor of area 4 6 m 2 and height 2.5 m. The masses at levels 0, 1, 2, 3, 4 are equal to M G =220 t and M Q =44 t, at level 5 to M G =180 t and M Q =44 t, while at the top level to M G =20 t and M Q =4 t. The building is situated in the seismic area Z 1 and the distribution of seismic accelerations is triangular. The design seismic acceleration of magnitude 0.12g is applied at the center of mass of the building. The calculation of the seismic and wind forces as well as a comparison between them is asked. Figure 2.5-1: The geometrical and loading model of the building's wind and earthquake actions M i [t]: masses, w [kn/m 2 ]: wind loads, a i [m/sec 2 ]: seismic accelerations EARTHQUAKE RESISTANT BUILDINGS Ι 29

15 Volume B Since the building is residential ψ 2 =0.30 and consequently during an earthquake the dynamic masses are evaluated as M=M G M Q. Thus, the dynamic masses at levels 0, 1, 2, 3 and 4 are equal to M G+0.30Q,i=0-4 = =233 t, at level 5 is equal to M G+0.30Q,5 = =193 t, while at the top level is equal M G+0.30Q,6 = =21 t. Assessment of seismic forces Figure 2.5-2: Wind forces Fw are less significant comparing to earthquake forces Fs. W [kn]: gravity loads F w [kn]: wind forces F s [kn]: seismic forces The total mass of the building during earthquake is M= =1146 t, while the CM (mass center) is located at distance z ο from the ground floor basis: tm z 0 = = = 9.0 m t 30 Apostolos Konstantinidis

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17 Volume B Diaphragmatic behaviour In general, when there is eccentric loading at a floor, e.g. imposed by the horizontal seismic action, the in-plane rigidity of the slab forces all the in-plane points (therefore all column heads 1 on the slab) to move the same way. The 3 displacements δ z, φ x, φ y of each node belonging to the diaphragm are independent of each other, while the rest δ x, δ y, φ z are depended on the 3 displacements of point C T called Center of Elastic Torsion of the diaphragm. Displacements δ xi, δ yi, φ zi at the point i of the horizontal diaphragm are expressed as: φ zi =φ z δ xi =δ xct -y i φ z δ yi =δ yct +x i φ z Figure : Diaphragmatic behaviour of floor Figure : The displacements of a random point i of the diaphragm due to φz In a floor diaphragm of 20 main and 14 slave nodes, the number of the unknown displacements (degrees of freedom) is equal to =63. 1 The term column refers to columns as well as to walls. 40 ΙApostolos Konstantinidis

18 Static and Dynamic Analysis is indicated by the red dashed line at the Interface of the related software so that the engineer is able to check the order of magnitude of the distribution of the seismic accelerations. Furthermore, using this method there is no need for additional dynamic response spectrum analysis. Notes The main comparison is performed on seismic accelerations. This is based on the fact that seismic forces provide the same fast visual outcome with respect to their vertical distribution, provided that all floor masses are identical. When the principle system is inclined, for earthquake in x direction only, the seismic accelerations are developed in both x, y directions. Τhe same stands for earthquake in y direction only. The earthquake in x direction generates components in x direction only when the structure is symmetrical, otherwise it generates components in both directions. Earthquake in Y Direction Masses a/g H[kN] V[kN] Figure 3.2-4: Ten-storey building of a non-symmetrical dual system (project <B_547-3b>) Figure 3.2-5: Distribution of Seismic Accelerations-Forces-Shears EARTHQUAKE RESISTANT BUILDINGS Ι 43

19 Static and Dynamic Analysis Figure : The structural model of the frame considering loadings: g and q and the seismic W The general analysis of the frame subjected to vertical uniform load w, from table 1 is: I k = I 1 h = l = l ( 5.0m ) H = H A = H B = w = w = 0.60m w, 4h ( k + 2 ) 4 3.0m ( ) V A M M = V A CA B = M = M l 5.0m = w = w = 2.50m w 2 2 B CD h 3.0m = H = 0.60m w = 0.60m 3 3 = M DC = M DB 2 w 2 2 = h H = 3.0m 0.60m w = 1.20m 3 3 The general analysis of the frame subjected to horizontal load W, from table 39 is: H A = H B V = V A B W = = 0.50W 2 3 h k 3 3.0m 1.47 = W = W = 0.27 W l (6k + 1) 5.0m ( ) 2 w EARTHQUAKE RESISTANT BUILDINGS Ι 47

20 Static and Dynamic Analysis 1 st combination: w=1.35g+1.50q= =59.55 kn/m M A =-M B =1.35M A,g +1.50M A,q = =35.7 knm M CA =M CD =M DC =-M DB =1.35M CA,g +1.50M CA,q = =-71.5 knm, H A =-H B =1.35H A,g +1.50H A,q = =35.7 kn, V A =V B =1.35H A,g +1.50H A,q = =148.9 kn V CD =w l/2+(m DC -M CD )/l= /2+( )/5.0=148.9 kn, V DC =V CD -w l= = kn 7 N Α =-V CD (self-weight of column)= = kn N B =V DC ( self-weight of column)= = kn x=v CD /w=148.9/59.55=2.50 m 8, M max = M CD + (V CD x)/2=-71.5+( )/2=114.6 knm 9, w l 2 /8= /8=186.1 knm Figure Figure Figure Figure Τhis stress resultant, as well as most of the following, could be calculated by the simple observation of the symmetry both of structure and loading. However this general process is handling asymmetries as well, e.g. as in the 2 nd loading. 8 x is the point of zero shear force corresponding to position of maximum bending moment. 9 From structural analysis it is known that the maximum bending moment Mmax in a span i,j is located at the point m at distance x from the end i, where shear forces become zero. The moment at that point is given by the expression Mmax=Mi,j + AV where AV is the area under the shear forces diagram from the point i to the point m. In this example, since there is only uniform load, the area is AV=(Vi,j x)/2. EARTHQUAKE RESISTANT BUILDINGS Ι 49

21 Volume B 2 nd combination: w=g q + Ε x = E x =36.0 kn/m + E x M A =M A,g +0.30M A,q + M A,W = =-79.2 knm, M B = = knm M CA = M CD = =39.0 knm, M DC = = knm, M DB = =125.4 knm, H A = H A,g +0.30H A,q + H A,W = =-39.4 kn, H B = =-82.6 kn, V A = V A,g +0.30V A,q + V A,W = =57.1 kn, V B = =122.9 kn V CD =w l/2+(m DC -M CD )/l= /2+( )/5.0= =57.0 kn, V DC =V CD -w l= = kn N Α =-V CD - self-weight of column = =-69.0 kn N B =V DC - self-weight of column = = kn x=v CD /w=57.0/36.0=1.58 m, M max = M CD + (V CD x)/2=39.0+( )/2=84.0 knm, w l 2 /8= /8=112.5 knm Figure Figure Figure Figure ΙApostolos Konstantinidis

22 Volume B Combinations and Envelopes of stress resultants Figure Figure Figure ΙApostolos Konstantinidis

23 Static and Dynamic Analysis 3 rd combination: g+0.30q-e X (additional combination U2) Figure : Bending moment diagram and elastic line Combinations and Envelopes of Bending Moments Figure : Shear force diagram Figure : The three combinations of bending moments Figure : Envelope of bending moments. The values given per 0.20 m are needed for the reinforcement design and the reinforcement detailing EARTHQUAKE RESISTANT BUILDINGS Ι 55

24 Volume B Modelling slabs using members and finite elements Finite element method assists in defining slabs behaviour with significant accuracy. However, in order for the results to represent the reality, suitable assumptions should be adopted. The effects of the following factors are examined thoroughly in Appendix A: 1) The frame behaviour at the regions close to columns 2) The deflection of beams 3) The torsional stiffness of beams The above factors affect the behaviour of slabs. In order to investigate this effect, a simple structure is being modelled in two ways: (i) Using members, according to which the slab is modelled as a grid of main and secondary joists, without the assumption of rigid bodies. (ii) Using triangular finite elements. The summary of Appendix A is presented below: Figure : The structure of project <B_331> of the related software (column sections 400/400, beam sections 300/500, slab thickness 170 mm) 64 ΙApostolos Konstantinidis

25 Static and Dynamic Analysis Figure : Slab modelled with members and the displaced structure (project <B_336>) Figure : Slab modelled with triangular finite elements (project <B_331>, pi-fes) (a=beam-slab common deflection curve) EARTHQUAKE RESISTANT BUILDINGS Ι 65

26 Volume B The frame behaviour in regions of columns The model using both members and more accurate two-dimensional finite elements, takes into account the slab frame behaviour in regions close to columns, in contrast to the inexpensive approach of simply supported slab throughout its length. However, in order for the slab to behave as a common frame with the columns in the actual structure, the slab-columns connections (where strong negative bending moments are developed) should be reinforced with strong, correctly placed and well anchored negative top reinforcement at slabs. For this reason, the slab analysis using finite elements in common worksheets should consider pinned supports on columns. Figure : Bending moment diagrams of joists of structure modelled with members (project <B_336>) In case of members, the two main side joists (of slab) forming a common frame with the columns, have greater torsional rigidity than the intermediate nearly simply supported main joists and bear heavier loads, thereby to develop strong negative bending moments at their supports and relatively low positive bending moments at their spans. The interim main joists develop strong positive bending moments at their spans, while being supported on the end joists through the secondary joists stressed by significant positive bending moments at their spans. 66 ΙApostolos Konstantinidis

27 Static and Dynamic Analysis In the more accurate model using two-dimensional finite elements, the main side strips behave intensively as frames. Τhe results are similar to those of using members with the following differences: (a) In the main side strips (corresponding to the main side joists) the frame behaviour is more intensive, since the moments at the supports are greater and moments at the spans are smaller, (b) In the interim main strips (corresponding to the interim main joists) the span moments are smaller, (c) The span moments of the secondary strips (corresponding to secondary joists) end up to be greater. This is due to the fact that the internal torsional stiffness of the slab elements (torsion) is stronger than the respective of members. Figure : Bending moment diagrams of slab strips modelled with triangular finite elements (project <B_331>, pi-fes) EARTHQUAKE RESISTANT BUILDINGS Ι 67

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29 Volume B Example (project <Β_422-1>) The project deals with a simple one-storey building consisting of 9 columns, 12 beams and 4 slabs, as illustrated in the figure. Figure The four slabs have identical dimensions of 4.0 mx6.0 m, thickness of 150 mm, covering load g e =1.0 kn/m 2 and live load q=5.0 kn/m 2. Concrete class: C30/ Ι Apostolos Konstantinidis

30 Volume B Figure : Button Show All 1 re-diplays the model. Button Contours shows, using color gradation,the displacements contours. Figure Each color (on the 3D color scale bar) corresponds to a range of displacements (mm). For many years, the method of color gradation display of displacement contours was a 2D method of representing 3D information. Today, provided the 3D abilities, the direct 3D or 4D display is preferable, especially for stereoscopic display. 82 Ι Apostolos Konstantinidis

31 Static and Dynamic Analysis Figure : The following button sequence displays the 3D deformation of the structure: Details 1 in FEM results, Diagrams at Dx=Dy=0.1m, OK, Selection 2, Displacements Z & Diagrams 3. For better viewing Light 2 7 is switched on. Figure : In the previous screen button sequence Menu 4, Full Screen Mode 5 and 4D 6, provides stereoscopic display using blue-red glasses. Displacements which induse stresses help the engineers understand better the structural behaviour (engineering perception). When the deflected structure is concave upwards, the bending moments are positive, considering that the fibers under tension are located at the bottom face of the slabs. EARTHQUAKE RESISTANT BUILDINGS Ι 83

32 Static and Dynamic Analysis Figure : In Slab Results section: Button Displacements 1 for 2nd time hides deformations. In FEM Results section: Button Details 2 defines distribution per 0.1 m. Button sequence Selection 3, Bending, V11, Diagrams 4 displays the 3D view of the distribution of shear forces [Vx]. Figure : Button sequence M, View, Options, Front, OK, Quad displays the front view of 3D distribution of shear forces [Vx]. Figure : Button sequence View, Options, Rear displays the rear view of the 3D distribution of shear forces [Vx]. EARTHQUAKE RESISTANT BUILDINGS Ι 87

33 Static and Dynamic Analysis Figure : Button Selection 1, Bending, M11 Diagrams displays the 3D view of distribution of bending moments [Mx]. Figure : Button sequence Selection 1, Bending, M11, Diagrams displays the 3D view of distribution of bending moments [Mx]. Figure : Button sequence View, Options, Rear displays the rear view of the 3D distribution of bending moments [Mx]. EARTHQUAKE RESISTANT BUILDINGS Ι 89

34 Volume B Figure : Button sequence Selection, Bending, M22 Diagrams displays the 3D view of distribution of bending moments [My]. Figure : Button sequence View, Options, Front, Quad displays the front view of the 3D distribution of bending moments [My]. Figure : Button sequence View, Options, Rear displays the rear view of the 3D distribution of bending moments [My]. 90 Ι Apostolos Konstantinidis

35 Volume B Figure : The actual stucture in 3D. Figure : The deformation of slabs in 3D. 102 Ι Apostolos Konstantinidis

36 Volume B Figure : Distribution of shear forces [Vx] ([V11] in FEM Results, detailed per 0.10 m) in 3D. Figure : Front view of the 3D distribution of shear forces [Vx] ([V11] in FEM Results ). Figure : Side view of the 3D distribution of shear forces [Vx] ([V11] in FEM Results ). 104 Ι Apostolos Konstantinidis

37 Static and Dynamic Analysis Figure : Distribution of bending moments [Mx] ([M11] in FEM Results, details per 0.10 m) in 3D. Figure : Front view of the 3D distribution of bending moments [Mx] ([M11] in FEM Results ). Figure : Side view of the 3D distribution of bending moments [Mx] ([M11] in FEM Results ). EARTHQUAKE RESISTANT BUILDINGSΙ 107

38 Static and Dynamic Analysis Figure : Shear forces [Vy] ([V22] in FEM Results, detailed per 0.1 m) extend only in the regions of middle supports. Figure : Front view of the 3D distribution of shear forces [Vy] ([V22] in FEM Results ). Figure : Side view of the 3D distribution of shear forces [Vy] ([V22] in FEM Results ). At the points of the slab (in this case of the cantilever) where fixed transverse support exist, numerous peaks are created, mainly for shears (in this case and kn) and secondarily for moments. These regions are forced to carry large part of the load of the adjacent slabs mainly near the supports. This is the reason why the occurrence of high shears shortly before and shortly after the support. However, these shears are taken into account in detail by considering their average values in a width, e.g. 1.0 m, which equals to 47.0 kn ( Slab Results ). EARTHQUAKE RESISTANT BUILDINGSΙ 111

39 Static and Dynamic Analysis Figure : Distribution of bending moments [Mx] ([M11] in FEM Results, detailed per 0.10 m) in 3D. Figure : View of the 3D distribution of bending moments [Mx] ([M11] in FEM Results ). Figure : Side view of the 3D distribution of bending moments [Mx] ([M11] in FEM Results ). The positive peak moments forming a sharp hole, at the region of the support exactly behind the cantilever, results from the negative load created at this region by the cantilever s moment. Notice that the influence of the strong cantilever, both in negative and positive moments, decreases in a small distance from the cantilever support. The insignificant differences in values of moments M x between FEM results and Slab results are due to the relatively small curvature of the 3D moment diagram M x. EARTHQUAKE RESISTANT BUILDINGSΙ 113

40 Volume B Unfavourable loadings and envelopes of stresses deflections The minimum load applied on a slab is equal to g whereas the maximum load is equal to p=(γ g - 1) g i + γ q q i. Τhe general problem concerns the way that the slabs should be loaded for developing maximum stresses. This is a complicated issue since, even for a simple case of six slabs arranged in a grid seven unfavourable loadings are required, as illustrated in the following figure. The analysis of such a simple example via tables can be performed only for equally grid axes max M1, max M4, max M5, min M2, min M3, min M6 max M2, max M3, max M6, min M1, min M4, min M5 max M1-2 max M5-6 min M3-4 max M1-3, min M2-4 max M2-4, min M1-3 max M3-5, min M4-6 max M4-6, min M3-5 The problem becomes even more complicated when slabs are not arranged in a grid. General and accurate analysis can only be achieved by means of the the finite element method. However, the resources for such an analysis require sophisticated algorithms and modern computers. The related software performs parallel processing of such algorithms on all available cores of a personal computer providing solutions to such complicated problems within seconds. The unfavourable loadings on the three previously resolved examples will be examined next. Calculations, taking into account the unfavourable loadings, are performed by selecting Adverse\ Slabs 1 in the tab Loads of pi- FES interface. Results are displayed by pressing, Adverse 2 and the stress resultant required e.g. Shears 3. Figure Figure Ι Apostolos Konstantinidis

41 Volume B Figure : Bending moment envelopes The most unfavourable bending moments are equal to: x direction: M x =12.3 (10.8), M x,erm =-22.8 (-22.0) [knm] y direction: M y =4.9 (4.1), M y,erm =-18.8 (-16.3) [knm] Figure : Front view of 3D bending moment diagrams, corresponding to the envelope of [Mx] Figure : Side view of 3D bending moment diagrams, corresponding to the envelope of [My] Note that although the live load has a relatively high value, the differences between the bending moments are small and less than15%. 118 Ι Apostolos Konstantinidis

42 Static and Dynamic Analysis Figure : Deflection envelopes The largest slab deflection is equal to y=-2.00 (-1.53) mm and y=+0.24 mm. It is concluded that the slab is lifted, which is in contrast with the responseof the global loading case, due to symmetry of course. Figure : Front view of 3D deflection diagrams Figure : Side view of 3D deflection diagrams Notice that the deflections are as high as 30% while opposite sign deformations also arise (blue lines). EARTHQUAKE RESISTANT BUILDINGS Ι 119

43 Volume B The BUILDING module Figure : In module BUILDING, the interaction of the slabs with the structure is taken into account resulting from point to point more or less favourable stresses compared to the module SLABS.. As explained in detail in chapter 3 and Appendix Α, the influence of beams and columns on the behaviour of a building is very important. In this example, the bending moment of the slabs S 1 - S 2 is significantly smaller compared to the SLABS module. The support moment, in this module, is provided directly on the side faces of the beams, wherein the detailing takes place. However, the peak moment in the middle of the support (in the middle of the beam) is much higher. Project <B_411-1> (Adverse\Slabs Active module\building), dense meshing gives: triangular elements, 16,417 nodes, 3856 linear members, system of 98,502 equations, memory 680 MB, time 12 sec, FPS= Ι Apostolos Konstantinidis

44 Static and Dynamic Analysis Multistorey buildings and envelopes Figure : The actual structure in 3D Figure : The structure s model in 3D For the slab analysis, the unfavourable loadings are created per floor, i.e. loading of any floor slab has no effect on slabs of other floors. In SLABS module this effect is obvious while in BUILDING module is rather insignificant. Project <Japan5>, running only in the professional version, consists of 10 floors with an area of 200 m 2. The following measurements are obtained using the following meshing parameters: Overall size = 0.20 m, Perimeter size = 0.10 m. Adverse\Slabs Active module\ SLABS : 134,292 triangular elements, 77,131 nodes, 6,507 linear members, 462,786 equations, memory 3.8 GB, time 125 sec, FPS=37 Adverse\Slabs Active module\ BUILDING : 183,217 triangular elements, 120,239 nodes, 34,727 linear members, 721,434 equations, memory 5.3 GB, time 209 sec, FPS=28 EARTHQUAKE RESISTANT BUILDINGS Ι 133

45 Volume B 4.5 One-way slabs One-way slabs are those supported on two opposite edges, such as slab s1 in figure of 4.1. If a one-way slab is supported on more than two edges and its aspect ratio, i.e. the ratio of the larger to the smaller theoretical span, is greater than 2.0 (such as slab s3 in the same figure), it is considered as one-way slab in the principal direction while taking into account the secondary stresses in rest edges Static analysis Continuous one-way slabs are analysed considering a frame of continuous member s of rectangular shape cross-section, having width equal to 1.00 m and height equal to the thickness of the slab. The strip loads comprise self-weight, dead and live loads. Analysis is performed: α) approximately, by applying all design loads p=1.35g+1.50q (when live load is relatively small) β) accurately, by taking into account unfavourable loadings. Figure : Three-span continuous slab 146 Ι Apostolos Konstantinidis

46 Static and Dynamic Analysis Example: The three slabs (previous figure) have L 1 =4.50 m, h 1 =180 mm, g 1 =10.0 kn/m 2, q 1 =2.0 kn/m 2, L 2 =4.00 m, h 2 =140 mm, g 2 =5.0 kn/m 2, q 2 =2.0 kn/m 2, L 3 =4.00 m, h 3 =140 mm, g 3 =5.0 kn/m 2, q 3 =2.0 kn/m 2, where loads g include self-weight. Perform static analysis considering global loading in ultimate limit state. Design load in each slab is equal to p i =γ g g i +γ q q i =1.35 g i q i, thus on 1.00 m wide strip, it is: p 1 = =16.5 kn/m p 2 =p 3 = =9.75 kn/m The three-span continuous slab will be solved through Cross method. Fundamental design span moments (table b3) M 10 =-p 1 L 1 2 /8= /8=-41.8 knm M 12 =M 21 =-p 2 L 2 2 /12= /12=-13.0 knm M 23 =-p 3 L 3 2 /8= /8=-19.5 knm Moments of inertia I I 01 =I c = /12= m 4 I 12 =I 23 = /12= m 4 =0.47I c Stiffness factors k, distribution indices υ k 10 = 3I 10 4I c L 01 = = υ 01= k 12 = 4I 12 = I c = υ 4I c L 12 4I c = k 21 =k 12 = υ 21 = k 23 = 3I 23 4I c L 23 = I c 4I c 4.0 = υ 01= [ ] [+3.6] [+0.3] M 1 =-22.6 knm M 2 =-13.9 knm [-( )] [-(-0.8)] EARTHQUAKE RESISTANT BUILDINGS Ι 147

47 Volume B V 01 = /2-22.6/4.50=32.1 kn V 10 = /2-22.6/4.50=-42.1 kn V 12 = /2+( )/4.00=21.7 kn V 21 = /2+( )/4.00=-17.3 kn V 23 = /2+13.9/4.00=23.0 kn V 32 = /2+13.9/4.00=-16.0 kn maxm 01 = /(2 16.5)=31.2 knm maxm 12 = /(2 9.75)-22.6=1.5 knm maxm 23 = /(2 9.75)=13.1 knm Figure : Shear force diagram Figure : Bending moment diagram 148 Ι Apostolos Konstantinidis

48 Volume B Deflection Slab member AB of length L, moment of inertia I, elasticity modulus E, is subjected to uniform load p. Given shear force V A,R (at left support) and bending moment M A, calculate equation of elastic line due to bending and maximum deflection. Figure : General case of bending of member (slab or beam) Considering coordinate z origin at the left end: V( z ) = V A, R M ( z ) = M A p z + V A,R z p z d y( z ) The basic equation of elastic line E I = M ( z ) is solved in two steps: 2 dz Step 1 dy( z ) 1 1 φ( z ) = = = M ( z )dz ( M dz E I E I VA,R z p z φ( z ) = ( M A z + + C1 ) E I 2 6 Hence, the equation of the elastic line tangents is: 1 p 3 VA,R 2 φ( z ) = ( z z M A z + C1 ) E I 6 2 ( 1 ) A V A,R p z z ) dz 150 Ι Apostolos Konstantinidis

49 Static and Dynamic Analysis Step 2 1 ( z ) = φ( z )dz = ( E I 1 p 4 VA,R ( z ) = ( z E I 24 6 p 3 VA,R 2 z z M A z + C M A 2 z z + C1 z + C ) 2 y 1 y 2 y(0)=0 C 2 =0 Hence, the equation of the elastic line is: 1 p 4 VA,R 3 M A 2 y( z ) = ( z z z + C1 z ) ( 2 ) E I ) dz y(l)= p L VA,R L M L p L V L A A,R M A L 0 = ( + C1 L ) C1 = + + (3) E I Thus, the equations of the elastic line tangents (1) and deflections (2) are determined. The maximum deflection is at the location where the first derivative of the elastic line equation is zero, i.e. at the point z where φ(z) = p z VA,R z (1) M A z + C1 = 0 (4) 6 2 The real positive root of the cubic equation (3) gives the desired point z max, which replaced in equation (2) yields the maximum deflection y max. Example: Deflection of first slab (example of 4.3.1): For L=4.5 m, p=16.5 kn/m, V A,R =32.1 kn and M A =0.0, expression (3) yields: C 1 = kn m 2 =45.7 kn m 2 6 (4) (16.5/6) z 3 -(32.1/2) z =0 2.75z z =0 z max =2.112 m (2) y(z)= 1 E I ( z z z) (1.2) y(2.112)= 1 E I ( ) 10 3 N m 3 = 59.8 E I 10 3 N m 3 For slab thickness h=180 mm and modulus of elasticity for concrete E=32.80 GPa: I=(b h 3 )/12=( )/12= m 4 E I= N/m m 4 = N m 2, therefore, y 1,max =y(2.112)= N m = m=3.75 mm N m2 EARTHQUAKE RESISTANT BUILDINGS Ι 151

50 Static and Dynamic Analysis The elastic line of the continuous slab given by expressions (1.2), (2.2), (3.2) is illustrated in the following figure: Figure : The elastic line of the three slabs (from the equations) Project <B_451> (pi-fes) produces identical deflections: Figure : Front view of the elastic line (from pi-fes with active module\slabs) EARTHQUAKE RESISTANT BUILDINGS Ι 153

51 Static and Dynamic Analysis Maximum support moments (void loading on adjacent spans and alternating with the rest) Example: Figure The continuous slab shown in the figure, of span length L=5. 00 m and of thickness h=160 mm, is subjected to covering load g e =1.0 kn/m 2 and live load q=5.0 kn/m 2. Concrete class C50/60. Calculate the shear forces and bending moments envelopes for the three slabs, in ultimate limit state. Solution: Self-weight: g o = 0.16m 25.0kN/m 3 = 4.00 kn/m 2 Covering load: g e = 1.00 kn/m 2 Total dead loads: g = 5.00 kn/m 2 Total live loads: q = 5.00 kn/m 2 The design dead load for each slab is g d = =5.0 kn/m and the total design load is p d =γ g g+γ q q= =14.25 kn/m. Manual calculations: I=(b h 3 )/12=( )/12= m 4 Modulus of elasticity for concrete C50/60 is equal to E=37.3 GPa. E I= N/m m 4 = N m 2 For I 10 =I 12 =I 23 =I c, stiffness factors k distribution indices υ are: k k 3I10 = 4Ic L 4I12 = 4I L 01 3 = = = = Figure υ 01 = = υ 12 = = c Due to the symmetry of the structure : υ 21 = και υ 23 = EARTHQUAKE RESISTANT BUILDINGS Ι 155

52 Volume B Loading 1: w 1 =w 3 =p d =14.25 kn/m, w 2 =g d =5.0 kn/m (V 01,max, M 01,max, M 12,min, V 32,max, M 23,max ) Principal support moments from table b3 Μ 10 =Μ 23 =-w 1 L 2 /8= /8=-44.5 knm, Μ 12 =Μ 21 =-w 2 L 2 /12= /12=-10.4 knm [ ] [ ] [ ] M 1 =-24.1 knm M 2 =-24.1 knm [-( )] [-(-3.6)] [-(-0.3)] V 01 = /2-24.1/5.0= =30.8 kn V 10 = =-40.5 kn V 12 = /2=12.5 kn M 01,max =V 01 2 /(2 w 1 )= /( )=33.3 knm w 1 L 2 /8= /8=44.5 knm M 12,min =V 12 2 /(2 w 2 )+M 1 = /(2 5.0)-24.1= = =-8.5 kn 12 w 2 L 2 /8= /8=15.6 knm 01: (3) C 1 =( / /6)=54.1 kn m 2 (4) (14.25/6)z 3 -(30.8/2)z = z z =0 gives z max =2.347 m (2) y(z)=1/ [(14.25/24) (30.8/6) )] y(2.335)=6.18 mm 12: Due to symmetry of both structure and loading z max =2.5 0m C 1 =( / / /2)kN m 2 = =-34.2 kn m 2 (2) y(z)=1/ [(5.00/24) (12.5/6) / ) y(2.50)=-2.72 mm Figure The alternative calculation is: Μ12=w1 L 2 /8+M1= /8-24.1= =-8.5 knm 156 Ι Apostolos Konstantinidis

53 Static and Dynamic Analysis Loading 2: w 1 =w 3 =g d =5.0 kn/m, w 2 =p d =14.25 kn/m (V 01,min, M 01,min, M 23,max, V 32,min, M 23,min ) Principal support moments from table b3 Μ 10 =Μ 23 =-w 1 L 2 /8= /8=-15.6 knm, Μ 12 =Μ 21 =-w 2 L 2 /12= /12=-29.7 knm [ ] [-5.2] [+0.5] M 1 =-24.1 knm M 2 =-24.1 knm [-( )] [-(+1.5)] [-(+0.2)] Figure V 01 = /2-24.1/5.0= =7.7 kn V 10 = =-17.3 kn V 12 = /2=35.6 kn M 01,max =V 01 2 /(2 w 1 )=7.7 2 /(2 5.0)=5.9 knm w 1 L 2 /8= /8=15.6 knm M 12,max =V 12 2 /(2 w 2 )+M 1 = /( )-24.1= =20.4 knm w 2 L 2 /8= /8=44.5 knm 01: (3) C 1 =( / /6)kN m 2 =6.0 kn m 2 (4) (5.00/6)z 3 -(7.7/2)z = z z =0 gives z max,1 =1.53 m και z max,2 =4.21 m (2) y(z1)=1/ [(5.00/24) (7.7/6) ) y(1.53)=0.45 mm (2) y(z2)=1/ [(5.00/24) (7.7/6) ) y(4.21)=-0.39 mm 12: Due to symmetry of both structure and loading z max =2.50 m C 1 =( / / /2)kN m 2 = =13.9 kn m 2 (2) y(z)=1/ [(14.25/24) (35.6/6) / ) y(2.50)=3.18 mm EARTHQUAKE RESISTANT BUILDINGS Ι 157

54 Static and Dynamic Analysis Loading 4: w 1 =g d =5.0 kn/m w 2 =w 3 =p d =14.25 kn/m This loading is an antisymmetric case of loading 3 with respect to the middle. Envelopes of the results of all loadings: Figure : Envelopes of shear forces bending moments - deflections Analysis using table b4 g d /p d =5.0/14.25=0.35 m 1 =10.695, m B =-9.025, m 2 =17.425, p 1A =2.315, p 1B =-1.635, p 2B =1.805 V 01,max =p d L/p 1A = /2.315=30.8 kn V 10,min =p d L/p 1B = /1.635=-43.6 kn V 12,max =p d L/p 2B = /1.805=39.5 kn M 01,max =p d L 2 /m 1 = /10.695=33.3 knm M 1,min =p d L 2 /m B = /9.025=-39.5 knm M 12,max =p d L 2 /m 2 = /17.425=20.4 knm Analysis using the table is easy, however it fails to provide the negative value of bending moment at middle span. EARTHQUAKE RESISTANT BUILDINGS Ι 159

55

56 Volume B Figure : Four-storey frame type structure comprising three columns In case of a basement, seismic forces developed at its roof level have a zero value. However, fixed end condition applies only to the base of the columns along the basement perimeter walls. If the frame also comprises walls, as presented in the following paragraph, the stiffnesses and the moment distribution of the walls differ between them. This difference becomes more distinctive as the number of stories increase. The total joint displacements and column stress resultants (shear forces and bending moments) are obtained from the frame analysis. Quantities K, k and a derive from the previous results. The apparent stiffness K i of storey i derives from the expression K i =V i /δ i, while the apparent stiffness of column j of storey i from the expression K i,j =V i,j /δ i. 238 Ι Apostolos Konstantinidis

57 Static and Dynamic Analysis Seismic forces Shear force diagram Bending moment diagram Elastic line Figure : Column frame type structure with triangular distribution of seismic forces (practical graphical representation) Example (3 rd storey): Column stiffnesses: K 3,1 =9.6/0.823=12, K 3,2 =15.8/0.823=19, K 3,3 =9.6/0.823=12. Storey stiffness: Κ 3 = (20+15)/0.823=43. The same value is obtained if calculated as the summation of the column stiffnesses of the storey, i.e. K 3 = K 3,1 +K 3,2 + K 3,3 =43. If shear effect is taken into account (Shear effect=on), the displacements are δ 1 =0.85, δ 2 =1.05, δ 3 =0.84, δ 4 =0.50 mm, i.e. the difference is insignificant. If rigid bodies effect is taken into account (Rigid body=on), the displacements are δ 1 =0.80, δ 2 =0.97, δ 3 =0.77, δ 4 =0.45 mm, i.e. both decrease of displacements and increase of stiffnesses are small but measurable. EARTHQUAKE RESISTANT BUILDINGS Ι 239

58 Volume B Eight-storey frame type structure under triangular seismic loading Figure : FRAME type structure comprising three column with cross-section 400/400 Figure : Equivalent structure of one fixedended column per storey Notes: In all types of structures, frame or dual, the sum of column shear forces of a storey is equal to the sum of the seismic forces of all the above storeys. Indicatively, for the first storey the sum is =180, while for the last =40. The middle column of the first storey carries the 71.4/180=40% of the total shear force, while each of the end columns carries 30%. In the last storey the middle column carries the 18.6/40=46%, while each of the end columns carries 27%. In both frame and dual systems, for each column M o -M u =V h, where M o is the moment at the top, M u is the moment at the base, V is the shear force and h is the height of the column. For instance, for the middle column of the previous structure 98.7-(-115.6)= ( ). 246 Ι Apostolos Konstantinidis

59 Static and Dynamic Analysis Eight-storey dual type structure under triangular seismic loading Figure : DUAL type structure comprising two columns and one wall with cross-sections 400/400 and 2000/300 respectively. Figure : Equivalent structure of one fixed-ended column per storey Notes: In the first storey, the sum is =180. The wall carries 158.9/180=88% of the total shear force, while each column carries 11%. In the last storey, the sum is =40. The wall carries 12.5/40=32% of the total shear force, while each column 34%. It is concluded that the wall has a favourable effect on the first storey columns, in contrast to that corresponding to the last storey. The expression M o -M u =V h, applies for both columns and wall. Indicatively, for the first storey wall (-786.4)= ( ), while for that of the last storey = (37.5=37.5). The maximum displacement of the dual type structure is equal to 8.08 mm, i.e. almost three times smaller than the one corresponding to the frame type structure (22.51 mm). EARTHQUAKE RESISTANT BUILDINGS Ι 247

60 Volume B Fifteen-storey frame type structure under triangular seismic loading Figure : FRAME type structure comprising three columns with cross-section 400/400 Figure : Equivalent structure of one fixed-ended column per storey Note: It should be emphasised that in all previous examples, the comparison of the two structural systems is important rather than the absolute quantities, which after all derive from specific values of the seismic forces. These values have been selected arbitrarily, yet satisfying the triangular distribution rule. 248 Ι Apostolos Konstantinidis

61 Static and Dynamic Analysis Fifteen-storey dual type structure under triangular seismic loading Figure : DUAL type structure comprising one wall with a cross section of 2000/300 and two columns with a cross section of 400/400 with cross-sections 400/400 and 2000/300 respectively Note: Figure : Equivalent structure of one fixed-ended column per storey The maximum displacement of the frame type structure is equal to 160 mm, i.e. almost twice of that corresponding to the dual type structure (75 mm). EARTHQUAKE RESISTANT BUILDINGS Ι 249

62 Volume B Centre of stiffness and elastic displacements of the diaphragm The current paragraph examines the special case of orthogonal columns in parallel arrangement. The general case is examined in Appendix C Subject description Centre of stiffness Centre of mass Figure : Simple one-storey structure comprising four columns, whose tops are connected by a rigid slab-diaphragm. Figure : Parallel translation of the diaphragm in both directions and rotation, due to a force Η applied to the centre of mass CM (Χ0Υ initial coordinate system, xcty main coordinate system) 254 Ι Apostolos Konstantinidis

63 Static and Dynamic Analysis When a horizontal force H acts on a storey level, all the points of the slab including the column 9 tops move in accordance with the same rules due to the in-plane rigidity of the slab. These rules induce the diaphragm to develop a parallel (translational) displacement by δ xo, δ yo and a rotation θ z about the centre of stiffness C T (x CT, y CT ) in xc T y coordinate system, which is parallel 10 to the initial coordinate system X0Y and has as origin the point C T. The diaphragmatic behaviour may be considered as a superposition of three cases: (a) parallel translation of the diaphragm along the X direction due to horizontal force component H X, (b) parallel translation of the diaphragm along the 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 mass point, 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 Translation of centre of stiffness C T along x direction Figure : Parallel translation along the x direction due to force Hx applied at CT 9 Henceforth the term column accounts for terms column and wall. 10 In the general case, i.e. in the case of columns with inclined local principal axes with respect to the initial system X0Y, the inclination angle of the principal system with respect to the initial system is a 0 (see Appendix C). Therefore, when the system of orthogonal columns is parallelly arranged then KX=Kx, VX=Vx, KXY=Kxy=0, meaning that a horizontal force applied at the centre of stiffness in x direction results in a displacement only along x (the same applies for y direction). EARTHQUAKE RESISTANT BUILDINGS Ι 255

64 Volume B In case a horizontal force H x is applied at C T in x direction, the following 2 equilibrium equations apply: The sum of forces in x direction is equal to H x, i.e. H x =Σ(V xoi ) (i). The sum of moments V xoi about the point C T is equal to zero, i.e. Σ(V xoi y i )=0 (ii). Each column i carries a shear force V xoi =δ xo K xi. Σ(V xoi )=Σ(δ xo K xi )=δ xo Σ(K xi ), expression (i) gives H x =δ xo Σ(K xi ) H x =K x δ xo where K x =Σ(K xi ). Expression (ii) gives Σ(V xoi [Y i -Y CT ])=0 Σ(V xoi Y i )-Σ(V xoi Y CT )=0 Y CT Σ(V xoi )= Σ(V xoi Y CT ) Y CT =Σ(δ xo K xi Y i )/Σ(δ xo K xi ) Y CT =Σ(K xi Y i )/Σ(K xi ) Translation of centre of stiffness C T along y direction Figure : Parallel translation in y direction due to force Hy applied at CT Accordingly, the corresponding expressions are derived for direction y. H y =K y δ yo where K y =Σ(K yi ) and X CT =Σ(K yi X i )/Σ(K yi ) Summarising, the centre of stiffness and the lateral stiffnesses are defined by the following expressions: Centre of stiffness and lateral stiffnesses: ( X i K yi ) X CT =, Hx = Kx δxo where Kx = ( Kxi ) (4 ) ( K ) Y yi (Yi K xi ) = ( K ) H CT, y y yo xi = K δ where Ky = ( Kyi ) (5 ) 256 Ι Apostolos Konstantinidis

65 Static and Dynamic Analysis Rotation of the diaphragm by an angle θ z about C T Figure : Displacements due to rotation developed from moment M applied at CT To determine the deformation developed by external moment M, applied at the centre of stiffness C T, the initial system X0Y is transferred (by parallel translation) to the principal system xc T y. The centre of mass is transferred to the principal system along the structural eccentricities 11 e ox, e oy in accordance with the following expressions: Principal coordinate system x i= X i X CT, yi Yi YCT =, e ox= xcm, e oy= ycm (6 ) The displacement of the diaphragm consists essentially of a rotation θ z about the C T, inducing a displacement δ i at each column top i with coordinates x i,y i in respect to the coordinate system with origin the C T. If the distance between the point i and the C T is r i, the two components of the (infinitesimal) deformation δ i are equal to δ xi =-θ z y i and δ yi =θ z x i. The shear forces V xi and V yi in each column developed from the displacements δ xi, δ yi are: V xi =K xi δ xi =K xi (-θ z y i ) V xi =-θ z K xi y i and V yi =K yi δ yi =K yi (θ z x i ) V yi =θ z k yi x i The resultant moment of all shear forces V xi, V yi about the centre of stiffness is 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 xi y i 2 +K yi x i 2 +K zi ) Torsional stiffness K zi of column i Columns resist the rotation of the diaphragm by their flexural stiffness expressed in terms K xi y i 2, K yi x i 2 (in N m), and their torsional stiffness K zi, which is measured in units of moment e.g. N m. 11 The eccentricities eox, eoy are called structural because they depend only on the geometry of the structure and not on the external loading. As presented in chapter 6, besides structural eccentricities, accidental eccentricities also exist. EARTHQUAKE RESISTANT BUILDINGS Ι 257

66 Volume B Assessment of building torsional behaviour The degree of the torsional stiffness of a diaphragmatic floor is determined by the relation between the equivalent mass inertial ring (C M, l s ) and the torsional stiffness ellipse (C T, r x, r y ). The optimal location of the two curves is where the torsional stiffness ellipse encloses the mass inertial ring. Figure 5.4.4: Equivalent mass inertial ring (CM, ls) and torsional stiffness ellipse (CT, rx, ry) A building is classified as torsionally flexible [EC ] if either r x <l s or r y <l s is satisfied in at least one diaphragm storey level. In this example both conditions are satisfied. For a building to be categorized as being regular in plan, the two structural eccentricities e ox, e oy at each level shall satisfy both conditions e ox 0.30r x & e oy 0.30r y [EC ]. In this particular example the first condition is satisfied e ox =0.94 m 0.30r x (= =1.173 m), whereas the second one is not e ox =1.34 m 0.30r x (= =0.924 m). Therefore the building that comprises that specific floor diaphragm is not regular in plan. Simplified seismic analysis may be performed, provided that the following conditions are met for each x, y direction: r x 2 > l s 2 + e ox 2 r y 2 > l s 2 + e oy 2 [EC (8) d)]. In this example the first condition is satisfied (=15.3)> (= =8.8), whereas the second one is not (=9.5)< (= =9.7). We therefore conclude that the simplified seismic analysis may not be performed at the building including this particular floor diaphragm. 268 Ι Apostolos Konstantinidis

67 Volume B Calculation 22 of the diaphragmatic behaviour 1 st (and unique) floor level Figure Figure Figure st Loading: H X =90.6 kn 2 nd Loading: H X =90.6 kn (1 st Loading) minus (2 nd Loading): H X =0 eccentricity 23 c Y =1.0 m Diaphragm restrained M CM,X =90.6 knm against rotation M CT, X =90.6 y CM c Y Figure Figure Figure The displacements of each point i δ X,i, δ Y,i and the rotation angle of the diaphragm θ XZ = The diaphragm develops zero rotation and moves parallelly 24 to the axes X, Y. 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. The C T derives from the expressions: X CT 25 =X 1 -δ Yt,1 /θ XZ =3.646 m Y CT =Y 1 +δ Xt,1 /θ XZ =3.316 m 22 The analysis of the diaphragmatic floor is performed automatically by the software. Algorithms are verified using the tools provided by the software. In this example with zero angle a of the principal system, all the diaphragm data may be calculated by two simple analyses and by the equations of the special case a=0, already presented in the previous paragraphs. Here, the general case of columns arranged randomly is been used, which applies even in the special case of the rectangular columns in parallel arrangement. The method is explained in detail in Appendix D. 23 The horizontal seismic load is applied at the CM. The eccentricity of the loading can be given also as equivalent torsional moment MCM,X=HX cy, which in this case is equal to MCM,X= =90.6 knm. This additional eccentricity aims to increase the effect of the rotation, i.e. to give larger displacements due to rotation, in order to calculate the torsion related data of the diaphragm more accurately. 24 In the special case of an one-storey building comprising only rectangular columns arranged parallelly to the axes X,Y, the horizontal force acting in X or Υ displaces the diaphragm only in X or Υ. 25 The equations determining the CT coordinates are general and may be applied for each point. Indicatively, for column 4: 272 Ι Apostolos Konstantinidis

68 Static and Dynamic Analysis Calculation of the diaphragmatic behaviour (continued) 1 st (and only) floor level Figure rd Loading: H Y =90.6 kn Diaphragm restrained against rotaion Figure Figure Analysis results: 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. Definition of the principal system 26, of the torsional stiffness radii and of the equivalent system (see C.6): tan(2a)=2δ XYo /(δ XXo -δ YYo )=0.0 2a=0 a=0 δ xxo =δ XXo =0.684 mm, δ yyo =δ YYo =0.824 mm K xx =H x /δ xxo = m/ m= N/m K yy =H y /δ yyo = m/ m= N/m M CT,X =90.6 y CM c Y =90.6 ( ) =164.5 knm K θ =M CT, X /θ XZ =164.5/ = knm r x = K θ /K yy = N/m/ N/m=3.931m r y = K θ /K xx = N/m/ N/m=3.582 m XCT=X4-δYt,4 /θxz= m/( )=6.0-2,355=3.645 m YCT=Y4+δXt,4/θXZ= m/( )= =3.316 m. 26 In this example, it is already determined that the angle of the principal system is zero, if the type of the structure is considered and the 2 nd analysis (according to which δxyo=0). The calculation has been performed for the sake of generality. To this end, other quantities have also been calculated, such as the centre of stiffness, which in this case is obtained from the simple application of moment at the point CM. EARTHQUAKE RESISTANT BUILDINGS Ι 273

69 Volume B The area of the shear forces diagram and heights represents the total moment of the floors and is larger in the triangular distribution than in the rectangular one. The maximum displacement, developed at the 10 th level in column c13, is 70/10=7.0 times greater than the one of the ground floor, i.e. δ xx,10,13 = =39.2 mm and δ xy,10,13 = =6.8 mm. The behaviour of the actual structure under orthogonal and triangular seismic force distribution is subsequently considered. In the related software, in <project B_547-1>, the seismic forces are input in the dialog Seismic Forces located at Parameters, Horizontal Forces. For orthogonal distribution input H x =200 at all levels, while for triangular distribution input values from to Always check Apply seismic forces =ON in order to use in the analysis the given seismic forces, instead of the default derived from the modal response spectrum analysis,. To perform the analysis press Solve Building and finally to review the results press Analysis Results. Figure : The actual structure of the building with the wireframe model Figure : Seismic action with base shear 2000 kn and orthogonal distribution of seismic actions Due to the bisymmetric geometry, in each diaphragm, the center of stiffness C T is located almost at the centre of the floor and therefore its displacement is almost equal to the average of the displacements of columns c4, c13. To compare all cases, the displacements are divided by a= mm. The displacements of the center of stiffness C T and of columns c4 and c13 are listed in the following table: 280 Ι Apostolos Konstantinidis

70 Static and Dynamic Analysis 6.3 Seismic stresses Seismic accelerations The seismic accelerations a XX, a XY, a XZ 10 of each diaphragm are obtained under the X horizontal component of the seismic action using appropriate methods (e.g. the CQC method applied by the related software) from the combination of the modal analysis results. Their application point of the seismic accelerations is the diaphragm s centre of mass. For nodes not belonging to a diaphragm, such as the nodes of the corner column of the building illustrated in the figure, the application point of the seismic accelerations is the node position. Seismic action in X direction Mass a/g H[kN] V[kN] t t t t t t t t t t Figure : Ten-storey building of mixed system. Diaphragmatic and non-diaphragmatic nodes. Project <B_547-2> Figure : Page from the software report Distribution of seismic accelerations-forces-shear forces The nodal seismic forces H XX, H XY και H XZ are obtained from the nodal seismic accelerations multiplied by the nodal mass. The seismic accelerations and forces due to a horizontal component (X and Y) of the seismic action are not only developed in the direction considered but also in the other horizontal direction and the vertical. 10 Seismic acceleration axx means acceleration in X due to seismic action in X, axy means acceleration in Y due to seismic action in X and axζ means acceleration in Ζ due to seismic action in X. EARTHQUAKE RESISTANT BUILDINGS 311 Ι

71 Static and Dynamic Analysis Modal analysis results of the frame type structure Fixed condition at the ground level (project <Β_641-1>) Figure : 1 st mode in X: T=0.975 sec, participation 84% Figure : 2 nd mode in X: T=0.321 sec, participation 10% Figure : 3 rd mode in X: T=0.189 sec, participation 3.5% Figure : 4 th mode in X: T=0.134 sec, participation 1.6% The sum of the effective modal masses of the first four modes amounts to 99% of the total mass. The first mode is the fundamental one as its effective mass is the 84% of the total mass. All four modes are translational and not torsional, as expected, due to the double symmetry of the structure. EARTHQUAKE RESISTANT BUILDINGS 317 Ι

72 Static and Dynamic Analysis Case 2: Foundation with footing beams (project <Β_641-2>) Figure : Structure and model Frame system with q=3.60 Figure : Seismic acceleration-forces-shear forces 1 st fundamental period:t1=1.012 sec, participation 85% Bending moments of ground floor columns Figure : Displacements under seismic action in x δmax=25.7 mm Figure : Ground floor column 0c2 (400/400) Figure : Ground floor column 0c6 (500/500) Due to stronger cross-sections of footing beams compared to those of columns, the displacement of the structure is slightly larger than that of assumed fixity at the base (25.7 against 24.5). The structural system remains intact while the behaviour factor q is taken equal to The bending moments of columns at the footing neck are roughly the same as in the fixed condition. EARTHQUAKE RESISTANT BUILDINGS 319 Ι

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74 Volume B Wall-equivalent dual type structure Figure : Structure comprising only columns and perimeter walls, project <B_642-1> The structure derives from the frame type structure of by replacing columns c2, c5, c8 and c11 of cross-section 400/400 with four perimeter walls of cross-section 2000/300. The main mode shapes of the modal analysis of the first case of the wall system are illustrated in the following page (The modes of the other three cases are similar). In the next four pages, the characteristic quantities for all cases are presented. It is extremely useful to compare between the variants of the wall system, but also between the frame and wall systems. General conclusions of the wall system: 1) In all cases, the natural period of the first mode shape is of the order of 0.70 sec. If however the stiffnesses of the elements are taken as being 100% of the elastic, the value of periods is of order of 0.50 sec (see 6.3.3). 2) The wall system behaviour is clearly better than the behaviour of the frame system, particularly when in the presence of a basement with perimeter walls. 322 ΙApostolos Konstantinidis

75 Static and Dynamic Analysis Modal analysis results of the wall type structure Fixed condition at the ground level (project <Β_642-1>) Figure : 1 st mode: Tx=0.644 sec, Cx=6.6%, Cy=70.7% Figure : 2 nd mode: Tx=0.639 sec, Cx=70.5%, Cy=6.6% Figure : 3 rd mode: Tx=0.456 sec, Cx=0.5%, Cy=0.0% Figure : 4 th mode: Tx=0.187 sec, Cx=1.8%, Cy=10.7% In the absence of symmetry in y direction, the fundamental mode shape in x (the 2 nd ) has also a component in y, meaning that both translational and torsional responses are developed. The sum of the effective modal masses of the 12 first modes amounts to 99.5% and 98.1% of the total mass of the structure in x and in y directions respectively. EARTHAQUAKE RESISTANT BUILDINGS 323 Ι

76 Static and Dynamic Analysis Case 4: Basement with perimeter walls (project <Β_642-4>) Figure : Structure and model Ductile wall system in X and Y with q=3.60 Figure : Seismic acceleration-forces-shear forces 1 st fundamental period:t1x=0.643 sec, participation 61% Bending moments of ground floor columns Figure : Displacements under seismic action in x δmax=16.4 mm Figure : Ground floor wall 0c2 (2000/300) Figure : Ground floor column 0c6 (500/500) The overall behaviour of the structure is much better compared to that of the strong foundation at the ground floor approaching that of the assumed fixed conditions at the base. EARTHAQUAKE RESISTANT BUILDINGS 327 Ι

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78 Static and Dynamic Analysis D.1 Subject description APPENDIX D DIAPHRAGMATIC BEHAVIOUR OF MULTISTOREY SPACE FRAMES GENERAL CASE The assessment of the behaviour of one-storey plane frames under horizontal seismic forces was presented in paragraph 5.1, where the structural unit examined was the column (or wall). In appendix Β, the crossbar was used to assess the behaviour of the multistorey plane frame. In this chapter, the composition of space frames, through beams and slabs, is considered. The diaphragm is the structural unit for the assessment of coupled space frames. Studying paragraph 5.4 and Appendices B and C is recommended in order to better understand this chapter. Figure D.1-1: The structural frame with 6 levels, project <B_d1> EARTHQUAKE RESISTANT BUILDINGS Ι 395

79 Volume B A diaphragm is the horizontal part of the floor consisting of slabs, connecting beams and columns. A floor may comprise more than one diaphragm, as shown in the example building of the present paragraph. Figure D.1-2: The space frame structural model with bars of columns and beams and diaphragms The most characteristic point of the diaphragm is the centre of stiffness, described in paragraphs 3.1.4, and C.5. The centre of stiffness C T of a specific floor diaphragm is defined as the point about which the diaphragm rotates, under horizontal seismic force H. The C T point depends only on the geometry of the floor and is independent of the loadings, meaning that regardless the magnitude of the force, which may become equal to 2H, or 3H, or any other value, the centre of stiffness will remain the same. Of course, it depends on the geometry of the overlying and the underlying floors. Principal axis system of diaphragm is defined as the orthogonal axis system xc T y, in which when a horizontal seismic force H is applied on C T, along axis x (or y), it results in translating C T only along axis x (or y respectively). The angle a of the principal system, as to the initial system X0Y, is called principal angle of the diaphragm. The above diaphragm data as well as the torsional stiffness K θ and torsional ellipse stiffness ellipse (C T, r x, r y ), should be calculated in a general way. The method to be used should deal with more than one diaphragm per floor. Moreover, the general solution should account for the influ- 396 Ι Apostolos Konstantinidis

80 Volume B D.2 General method for the calculation of the diaphragm i data Step 1: Analysis under force H Χ and concentrated moment M CM,X applied at the centre of mass of diaphragm i. Figure D.2-1: Loading: Force HX=100 and moment MCM,X on centre of mass CM of diaphragm i Results: total displacements of structure and displacements of diaphragm i The two displacements δ XX1, δ XY1 of point 1 of diaphragm i and its angle θ XZ are required. Any value may be given to force H Χ, as long as it is the same for the analyses of the three first steps. In this example, H Χ is given equal to 100 kn. The application point of force H is irrelevant, but for higher accuracy in calculations, the force is assumed applied at an arbitrary eccentricity c y, e.g. 2.0 m, in relation to the centre of mass, in order to produce a significant rotation of the diaphragm and thus achieve a higher accuracy in the calculations. That eccentricity is essential, especially in cases where the centre of stiffness is close or coincides with the centre of mass. Therefore, in addition to force H Χ, moment M CM,x =100kN 2.00m= 200 knm is also applied at C M. 398 Ι Apostolos Konstantinidis

81 Static and Dynamic Analysis The magnitude of force H needs to be important enough to result in significant diaphragm translations that can be distinguished from the translations due to moment. The application of the sole moment in the absence of force H results in rotation, but also in translations, which cannot be calculated since the centre of stiffness is still unknown. Point 1 corresponds to column Κ2 joint, but could be any point of the diaphragm. Figure D.2-2: The diaphragm displacements in plan, due to 2 parallel translations δxxo, δxyo and one rotation θxz The only results needed from this analysis are the translations of point 1 δ XX1 =2.681 mm, δ XY1 = mm and the angle θ XZ = of the diaphragm. The displacements of point 2 will be used only to verify the generality of the method. All displacements are absolute with respect to the ground. EARTHQUAKE RESISTANT BUILDINGS Ι 399

82 Volume B Step 4: Analysis results from loading 1 minus analysis results from loading 2, meaning that only moment and only rotation exists about the centre of stiffness C T, which remains stationary with respect to the ground. Figure D.2-7: The only load acting on diaphragm i is moment on CT Diaphragm displacements are induced only due to rotation The centre of stiffness CT remains stationary with respect to the ground Using this trick, namely by subtracting analysis 2 results from analysis 1 results, the diaphragm i develops only rotation, while the remaining diaphragms develop both translations and rotation. However only diaphragm i is examined here. The most important result of this trick is that the diaphragm i is rotated about C T, which remains stationary with respect to the ground, allowing the calculation of its precise location. The rotation angle of diaphragm i is the rotation angle θ XZ calculated in step 1 under the corresponding loading. 404 Ι Apostolos Konstantinidis

83 Static and Dynamic Analysis Also K θ =M XH /θ XZH =M XM /θ XZM = Nm. The constant moment value in all diaphragms and the respective calculation of torsional stiffness is utilized in the determination of the equivalent system of D.6. D.5 Torsional stiffness distribution Torsional stiffness distribution of diaphragm is the curve on which, if the idealised columns with the same lateral stiffnesses as those of the diaphragm, are placed symmetrically with respect of the centre of stiffness, the torsional stiffness derived is the same as that of the diaphragm. Figure D.5 Torsional stiffness ellipse of diaphragm is defined as the ellipse having C T as centre and r x = (K θ /K yy ), r y = (K θ /K xx ) as radii. r x = K θ /K yy = [ Νm/ N/m]=9.28 m r y = K θ /K xx = [ Nm/ N/m]=10.42 m EARTHQUAKE RESISTANT BUILDINGS Ι 407

84 Volume B Example D.7.2: In the three-storey structure of project <B_d9-2> 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.7.2-1: The simple structure of 3 storeys and 4 columns. The floor is typical, identical to the floor of first example 416 Ι Apostolos Konstantinidis

85 Volume B Calculation of the diaphragmatic behaviour of level 2 Figure D Figure D Figure D 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 Figure D Figure D Analysis results: The translations of point 1 are δ XX1 =3.318, δ XY1 =-0.986mm and the diaphragm rotation is θ XZ = 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 = mm, δ XYo = Analysis results: The diaphragm, being restrained against rotation, develops only parallel translations in X,Y directions, which are: δ YXo = , δ YYo = mm The angle of the principal system derives from the expression: tan(2a)=2δ XYo /(δ ΧΧο -δ YYo )= 2 ( )/( )= a=43.0 a= Ι Apostolos Konstantinidis

86 Static and Dynamic Analysis Calculation of the diaphragmatic behaviour of level 2 (continued) Figure D Loading 1 minus loading 2: H X =0, M XCT =90.6 (Y CT - Y CM )+90.6 Figure D 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 = = =1.1468, δ Yt,1 = δ Y,1 -δ XYo = = = and X CT =X 1 -δ Yt,1 /θ XZ = / =2.847 m Y CT =Y 1 +δ Xt,1 /θ XZ = / =3.785 m Figure D 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/( )] 10 6 N/m= N/m K yy =H/(δ YYo - XYo tana)= =[90.6/( )] 10 6 N/m= N/m M XCT =90.6 (Υ CT -Y CM )+90.6 c Y =90.6 ( ) = knm, K θ =M CT, X /θ XZ =204.8/ = knm r x = K θ /K yy = [ Nm/ N/m]=4.30 m r y = K θ /K xx = [ Nm/ N/m]=3.98 m Note: 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 = m/ = =2.847 m EARTHQUAKE RESISTANT BUILDINGS Ι 419