CHAPITRE 18. Exemple de calcul d ossature en portique en béton armé.

Transcription

1 18-1 CHAPITRE 18. Exemple e calcul ossature en portique en béton armé. Example prepare by PHUNG NGOC DUNG, PhD researcher at ULg. Note 1!!! In this example, notation «ot.» stans for coma,. Example: 1. means 1, Note!!! The example is use with γ c = 1,5 an γ s =1,15. The values are ifferent in each country: France: γ c = 1,3 an γ s =1,0; Belgium: γ c = 1,5 an γ s =1,00. I. INTRODUCTION The esign example of a reinforce concrete builing which is presente hereafter aims at two main goals: - To present the partially esigning proceures of a reinforce concrete frame uner a given seismic excitation accoring to Eurocoe 8 an Eurocoe - To check the behaviours of the reinforce concrete frame which is corresponingly esigne an etaile to Eurocoe 8 uner un-given seismic excitations by using Pushover analysis. In orer to get a fully esigne an etaile reinforce concrete frame, there are several preliminary esigning steps. The final rawings which are use on the sites are the results of series of calculations. To choose the best results, that is, the sectional imensions, material properties, reinforcement areas, etc, the esigning iterations must be carrie out. The following presentation of the total esign proceures is just some parts of the completely iterative processing. The issues which are presente in the esign example are: - To escribe the builing architecture an properties such as materials, loas, - To check the chosen cross sectional imensions in pre-esign. - To analyze the structure uner a given seismic excitation. - To verify the structural elements. - To check the builing uner other seismic excitations. II. GENERAL DESCRIPTION The builing which is chosen to esign is an office an flat builing. The builing has 6 stories. It is a six-story reinforce concrete two-way frame. The floor plan is presente in Figure 1.

2 18- II.1. ain geometry escriptions: There are 3 bays of 5m an 4 bays of 5m. The area of current floor is about 300m ( 0 15 = 300m ). The structure has in-plane an elevation regularity. The story height is 3m, except the groun story height is 3.5m. The cross sectional imensions for all columns are 400mmx500mm. The slab thickness is 150mm; the imensions of all beams are 50x500mm (slab inclue). II.. Exterior an partitioning walls: The perimeter walls are glass an masonry ones. They o not affect the free isplacement of the frame uring earthquakes. II.3. Loas: The characteristic values for the loas are: II.3.1 For the intermeiate floor: - Slab weight 3.75kN/m - Flooring 1.9 kn/m - Live loa 3 kn/m II.3. For the roof floor: - Slab weight 3.75kN/m - Flooring 1.9 kn/m - Live loa 0.75 kn/m - snow loa 0.4 kn/m II.4. Preliminary Consierations: Subsoil Class: C Ductility Level: DC eium level. Important category of the builing is II orinary builing an γ I = 1. The non-structural elements of the builing are fixe in a manner as not to interfere with structural eformations. The structure is rigi fixe in non-eformable founations. The relative esign groun acceleration for the reference return perio is a = 0.15g. II.5. aterials Concrete class: C5/30 Ecm = 31GPa = 31 KN / mm Longituinal ribbe reinforcing steel bars S500 was chosen. Transverse ribbe reinforcing steel bars S500. II.6. Design Proceures: For the R/C multi-story flexible frame builings, the inter-story rift control governs the esign. So, the pre-esign proceures of the cross sectional imensions of the frame members are the checks of horizontal isplacements inuce by the earthquakes. gr III. PREDESIGN III.1. Weights of asses: EC8 In accorance to EC8 [3], the inertial effects of the esign seismic actions shall be evaluate by taking into account the presence of the masses associate to all gravity loas appearing in the following combination of actions: Gkj " + " ψ E, iqk, i Where: ψ E,i : combination coefficients. ψ E,I is etermine as following 4..4 EC8 ψ Ei, = ϕψ.

3 18-3 The value of ϕ is to be from the table EC8 ϕ = 1.0 for the top story. ϕ = 0.8 for the correlate occupancies. ϕ = 0.5 for the roof story. (EC8 table 4. for categories A-C* - omestic an resiential an for stories inepenently occupie) ψ,i : combination coefficients. Determining from the Annex A1:1990:00, table A.1.1 ψ, i = 0.3 for the occupancy (category A). ψ, i = 0 for the snow an win loas. So, the results are: For the intermeiate stories: Wfloori = 3.75* * *0.3*3*300 = 1917KN For the roof story: Wroof = 3.75* * *0.3*0.75* *0*0.4*300= KN All weights of masses are calculate from 4.3.1(10P) EC8 [3] Weight of the beams on floors: Wb = 0.5*( )* 5*31( beams)*5m= 339 KN*6 = 034.4KN for all floors. Weight of the columns on floors: Wc = (0.4*0.5*(3 0.5)*5*0( columns))*5 floors + for all floors. (0.4*0.5*( )*5*0( columns)) = 1550KN Total weights of the builing: W = W + W + W + W = 1917* = 14938KN floori roof floor b c III.. Base shear force: ( EC8 [3]) Accoring to EC8 [3], Base shear force inuce by an earthquake is etermine as the following expression: Fb = S( T1 )* m* λ (4.5 EC8 [3]) Where: - S (T 1 ) the orinate of the esign spectrum at perio T 1 - T 1 : The funamental perio of vibration of the builing for lateral motion in irection consiere. W - m: is the total mass of the builing. m = g - λ: is correction factor. λ = 0.85 if T 1 Tc an the stories of the builings stories λ = 1 if otherwise Accoring to (3) EC8 [3], for the builings with heights up to 40m, the value of T 1 (s) may be approximate by the following expression: 3 T 4 1 = C t * H (4.6 EC8 [3]) Where: - C t =0.75 for the concrete frames. - H is the height of the builing (m); H = 3.5* + 5*3 = 18.5m So, we have: 3 4 T1 = 0.75*18.5 = 0.67( s) S (T 1 ) the orinate of the esign spectrum at perio T 1 is etermine from EC8. With subsoil class C, we have, for the Type 1 spectrum:

4 18-4 S = 1.15 TB ( s) = 0.( s) Table 3. Type EC8 [3] TC ( s) = 0.6( s) TD () s = () s S = 1.15 TB ( s) = 0.( s) For an T 1 = 0.67(s) so we have TC < T1 < TD an S (T 1 ) is calculate from the TC ( s) = 0.6( s) TD () s = () s expression 3.14 (3...5 EC8 [3])..5 TC ag * S* * S ( T1 ) = q T β * ag Where: - a g =γ I *a gr =1*0.15g 3..1(3) - β - is the lower boun factor for the horizontal esign spectrum, β=0.. - S= q is the behaviour factor calculate from 5... EC8. In accorance to 5... EC8 [3], q is to be calculate from the following expression: q = q * k w. q 0 is the basic value of the behaviour factor etermine from table EC8. For the concrete frames with the DC uctility class, q 0 is calculate by the following α α expression: q0 = 3.0* u, in which the ratio, u 1, 3 α1 α =, is calculate from 5... (5a) 1 EC8. So the value of q 0 is 3.0*1.3=3.9 k w is the factor reflecting the prevailing failure moe in structural systems with walls. k w is calculate from 5... (11). So the value of k w is 1.00 So, the value of q is etermine as below: q=3.9*1= g*1.15* * = 0.099g S ( T1 ) = = 0.099g 0.*0.g = 0.04g W Fb = S( T1 )* m* λ = g* *0.85 = W g F = 157KN b III.3. Torsion Effects The torsion effect is taken into account for the transverse current frame in a simplifie manner. ( EC8 [3]). Accoring to EC8 [3], if the lateral stiffness an mass are symmetrically istribute in plan an unless the acciental eccentricity of 4.3.(1)P is taken into account by a more exact metho, the acciental torsion effects may be accounte for by multiplying the action effects in the iniviual loa resisting elements resulting from the application of (4) by a factor δ given by: x δ = *. If the builing is istribute symmetrically in plan an elevation, it can be ivie Le x into plane moels an the factor δ is etermine by δ = 1+ 1.*. L e

5 18-5 Accoring to the builing plan, we can etermine the values of x an L e as follow: x=5m for frame at line, x=10 for frame at line 5; L e =0m. x 5 So, the value of δ is calculate by the following expression: δ = * = * = 1.15 an Le 0 δ=1.3 for line 5. However, we have seen in (Plumier, Construction en zone sismique,[9]) that in fact a realistic δ for such builing is rather δ=1.15. In this pre-esign step the value of δ = 1.15 will be use. It shoul be checke at the final esign state. III.4. Seismic force istribution: Seismic forces istribute to all frames of the builing epen on both their stiffness an their positions in plan, ue to torsion. Force istribution along the height of the builing using the simplifie formula 4.11 or 4.10 EC8 [3]. In accorance to (1) EC8 [3], the funamental moe shapes in the horizontal irections of analysis of the builing may be calculate using methos of ynamics or may be approximate by horizontal isplacements increasing linearly along the height of the builing. Accoring to () EC8 [3], the seismic action effects shall be etermine by applying, to the two planar moels, horizontal forces, F i, to all stories. si* mi Fi = Fb* (4.10) sj* mj Where: - F i is the horizontal force acting on the story i. - F b is the seismic base shear. - s i, s j isplacements of masses m i, m j in the funamental moe shape. - m i, m j - are the story masses. Accoring to (3) EC8 [3], when the funamental moe shape is approximate by horizontal isplacements increasing linearly along the height, the horizontal forces F i are given by: zi* mi Fi = Fb* (4.11 EC8 [3]) zj* mj Where: - F i is the horizontal force acting on the story i. - F b is the seismic base shear. - z i, z j heights of the masses m i, m j above the level of application of the seismic action.. - m i, m j - are the story masses. So, the values of F i can be calculate as the following table: Table III.1 Horizontal seismic force Distribution Weight w i zi* wi Story Height (z i ) m z i *w i zj* wj F i (KN) Fb = 157 [KN]. The above seismic forces, F i, are total seismic forces acting at each story for the whome builing an all frames. Accoring to the stiffness of each frame, we will istribute the seismic forces to each frame linearly incluing torsion effects.

6 18-6 In the action irection of the earthquake (irection Y or transverse irection), there are 5 portal frames. We will istribute the seismic forces to the transverse frame at line as following: F 6= F6* * δ = 31.34* *1.15 = 73.9KN. F 5 5* * 87.0* * = F δ KN 5 = 5 = F 4= F4* * δ = 31.5* *1.15 = 53.3KN. F 3 3* * 175.9* * = F δ KN 5 = 5 = F = F* * δ = 10.4* *1.15 = 7.7KN. F 1 1* * 65.8* * = F δ KN 5 = 5 =. Figure Lateral seismic forces III.5. The limitation of the inter-story rifts: A plane frame is analyse an the isplacements of the frame subjecte to the applie forces F -1 to F - 6, which are compute above, will be etermine by using SAP 000. Version Accoring to 4.3.1(7) EC8[3], the elastic moulus of Concrete E=E/= 15,5KN/mm. Accoring to EC8 [3], the following limits shall be observe: For the builings having non-structural elements of uctile materials attache to the structure: r * v * h Where: - h: is the story height h=3m an h=3.5m - r : is esign inter-story rift as efine in () EC8 [3], evaluate as the ifference of the average lateral isplacements s at the top an bottom of the story uner consieration an calculate accoring to r = s i s. i 1

7 Accoring to EC8 [3], for isplacement analysis, if linear analysis is performe the isplacements inuce by the esign seismic action shall be calculate on the elastic eformations of the structural system by means of the following simplifie expression: s = q * e Where: + s : is the isplacement of a point of the structural system inuce by the seismic action. + e : is the isplacement of the same point of the structural system, as etermine by a linear analysis base on the esign response spectrum accoring to EC8. + q : is the isplacement behaviour factor, assume equal to q unless otherwise specifie. So q = q = ν: is the reuction factor to take into account the lower return perio of the seismic action associate with the amage limitation requirement. The value of ν also epens on the important class of the builing. The important class of the builing is II so the value of ν is 0.5 (accoring to () EC8 [3]). The value of s must be smaller than the value erive from the elastic spectrum. When etermining e, the torsion effects of the seismic actions shall be accounte for. Table of rifts: Table III. Story Drifts Story Story elastic Behaviour Drift from EC8 isplacements. factor of s i r =s i -s i-1 r *ν h mm) ( e - mm) Displacement So, the conition r * v * h is met. The section imensions an material properties which are chosen are satisfie with pre-esigne steps accoring to EC8. IV. PRELIINARY STEPS: Accoring to EC8 4. IV.1. Structural regularity. IV.1.1 Regularity in plan: Accoring to EC8 [3], criteria for the regularity in plan are: The builing structure is symmetrical in plan with respect to two orthogonal irections. The plan configuration is compact. The in-plane stiffness of the floors is sufficiently large to istribute seismic forces among the vertical structural elements. Lmax Lmax 0 The slenerness λ = of the builing is λ = = < 4. Lmin Lmin 15 In accorance to (5) EC8 [3], at each level for each irection of analysis x or y, the structural eccentricity e 0 an the torsional raius r verify the two conitions below, which are expresse for the irection of analysis y: e0 X 0.3* rx (4.1-EC8 [3]) rx > ls

8 18-8 Where: - e 0X : is the istance between the centre of stiffness an the centre of mass, measure along x irection, which is normal to the irection of analysis consiere. - r X : is square root of the ratio between torsional effects an lateral stiffness in y irection (torsional raius) - l s : is the raius of gyration of the floor mass in plan (square root of the ratio of (a) the polar moment of inertia of the floor mass in plan with respect to the centre of mass of the floor to (b) the floor mass). There are not setbacks in the plan. IV.1. Regularity in elevation: Accoring to EC8 [3], criteria for regularity in elevation are: All the lateral resisting systems run without interruption from their founations to the top of the builing. Both lateral stiffness an the mass remain constant or reuce graually without abrupt changes, from the base to top. There are not setbacks. IV.. Structural Analysis: (4.3 EC8 [3]) IV..1 oelling: Because of the in-plan an in-elevation regularity, in accorance to 4.3.1(5) an to the table 4.1(4..3.1) the allowe simplifications are: - The analytical moel: Planar. - The metho of analysis: Using simplifie metho Lateral force metho of analysis can be use because all the conitions of are met: The builing has funamental perio of vibration T 1 in the two main irections smaller than the following values: 4* TC 4*0.6 =.4s T1 T1 = 0.67s.0s.0s All the criteria for regularity in elevation given in EC8 are met. - Behaviour factor: is the reference value. The builing will be analyse with two planar frames using the lateral force metho an comparison with two planar frames using the response spectrum analysis. IV.. Natural Perios: Accoring to EC8 [], the natural perio can be etermine by Reileight etho or approximate metho. The first perio T 1 efine by approximate formula is equal to the value of IV..3 Local effects of infill There is no infill for the current transverse an longituinal frames. IV.3. Verification of structural type: IV.3.1 Torsional Rigiity: In accorance to (4)P EC8 [3], the first four types of systems (i.e. frame, ual an wall systems of both types) shall possess a minimum torsional rigiity that satisfies expression (4.1b) in both horizontal irections. But in accorance to 5...1(5), for frame or wall systems with vertical elements that are well istribute in plan, the requirement specifie in (4)P above may be consiere as being satisfie without analytical verification.

9 IV.4. Selection of uctility class 18-9 The chosen uctility class for esign is DC. So, esigning, imensioning an etailing must ensure a uctile behaviour of the elements meaning that uctile moes of failure shoul precee failure moes with sufficient reliability. The plastic hinges which are evelope in response to the seismic excitation must be able to issipate a meium amount of energy in a stable manner. IV.5. aterial checks IV.5.1 Concrete In accorance to EC8 [3], for uctility class DC the use of concrete class which is lower than C16/0 is not allowe in primary seismic elements. So, we choose the concrete class C5/30. IV.5. Flexural reinforcement steel In accorance to EC8 [3], only ribbe bars are allowe as reinforcing steel in critical sections of primary seismic elements. The reinforcing steel class S500, the high uctility steel that satisfies the aitional requirements in critical regions concerne in table C.1, annex C EC [], is chosen. IV.5.3 Shear reinforcement steel In accorance to (P) EC8 [3], except for the close stirrups or cross-ties, only ribbe bars are allowe as reinforcing steel in critical of primary seismic elements. The reinforcing steel class S500 for flexural reinforcement steel was chosen. So, we also choose S500 for shear reinforcement steel. IV.6. Secon orer Effects(P- ) Accoring to 4... () - EC8 [3], the secon-orer effects (P- effects) nee not to be taken into account if the following conition is fulfille in all stories: θ = Ptot * r 0.1 Vtot * h (4.8 EC8 [3]) Where: θ is the inter-story rift sensitivity coefficient. P tot is the total gravity loa at an above the story consiere in the seismic esign situation. r = s i s - is esign inter-story rift, evaluate as the ifference of the average lateral i 1 isplacements s at the top an the bottom of the story uner consieration. V tot is the total seismic shear at the consiere level. Ptot * r Ptot r At the groun story: θ = = * = * = > 0.1 Vtot * h Vtot h * At the intermeiate story: θ = Ptot r * Vtot * h = = > So, the secon-orer effects cannot be neglecte.

10 18-10 V. BEA DESIGN AND VERIFICATION: Accoring to EC8 [3] 4 an EC[]. V.1. TRANSVERSE CURRENT FRAE OR DIRECTION Y CURRENT FRAE: V.1.1 Action effects: (Accoring to 5.4. EC8 [3]) In accorance to 5.4. EC8 [3], for the beams with uctility DC, the esign values for the builing moments shall be obtaine from the analysis of the structures for the seismic situation accoring to EN Combination of actions for seismic esign situations is calculate as following expression: Gk, j" + " P" + " AE" + " ψ, i* Qk, i j 1 i 1 Where: G k,j is the permanent or persistent action j. P is the pre-stressing action. A E is the esign value of seismic actions for the reference return perio (esign spectrum) The loas were uniformly istribute along the length of the beam. No reistribution of the bening moments was mae. The esign value of the shear forces shall be etermine in accorance with the capacity esign rules, EC (P), consiering the equilibrium of the beam uner: a) the transverse loa acting on it with the seismic esign situation an b) en moment i, (with i=1, enoting the en sections of the beam), corresponing, for each sense of seismic action, to plastic hinge formation at the en of either of the beam or of the vertical elements, which ever takes place first which are connecte to the joint where the beam en an i frame into. The analysis is performe using two planar moels, one for each main irection. The torsion effects were etermine separately by these two imensions accoring to () EC8: If the analysis is performe using two planar moels, one for each main horizontal irection, torsion effects may be etermine by oubling the acciental eccentricity e ai of the expression x eai =± 0.05Li an applying the rules of (1) EC8 with the factor δ, δ = *, Le x replace by the factor δ, δ = 1+ 1.*. Le Because of the symmetry, the actual eccentricity between stiffness centre S an the nominal mass centre, e 0 = 0, is equal to 0, an the aitional eccentricity, e, taking into account of the ynamic effect of simultaneous transitional an torsional vibration, can not be compute. So, the only eccentricity taken into account is acciental torsional effect. e1 = eai =± 0.05Li (When using the Response Spectrum Analysis) Where: e ai is the acciental eccentricity of the storey mass from its nominal location, applie in the same irection at all floors. L i is floor imension perpenicular to the irection of the seismic action. V.1. Action Summary: V.1..1 Gravity actions: DEAD Loa: The self-weight loa. DL slab: The ea loas inuce by the floor an coating weight. LL slab: The live loas inuce by the variable actions LL roof slab: The live loas inuce by the variable roof actions Snow loa: The loas inuce by the snow. Joint loa: The loas acting to the joints of the transverse frame inuce by perpenicular frames. WIND loa.

11 18-11 Figure 3 Gravity Loas V.1.. Seismic actions Seismic actions use to analyse in the frames will be etermine by two methos of analysis: Lateral Force Analysis an Response Spectrum Analysis. The analysis is performe using two planar moels, one for each main irection. Lateral force Analysis: The torsion effects were etermine separately by these two imensions accoring to () EC8: If the analysis is performe using two planar moels, one for each main horizontal irection, torsion effects may be etermine by oubling the acciental eccentricity e ai of the expression eai =± 0.05Li an applying the rules of (1) EC8 with the factor x x δ, δ = *, replace by the factor δ, δ = 1+ 1.*. So we have δ = 1.3 Le Le The above seismic forces are total seismic acting in all of the builing or all frames. Accoring to the stiffness of each frame, we will istribute the seismic forces to each frame linearly incluing torsion effects. In the action irection of the earthquake, there are 5 portal frames. When istributing seismic forces to all floors of the frame, the torsion effects will be taken into account by the factor δ which is calculate from (1) EC8. Factor δ, here, accounts for the analysis moel

12 18-1 with two planar irections. We will istribute the seismic forces to the current transverse frame as following: o 1 1 F 6= F6* * δ = 31.34* *1.3 = 83.54KN. 5 5 o 1 1 F 5 F5* * δ 87.0* * KN 5 5 o 1 1 F 4= F4* * δ = 31.5* *1.3 = 60.KN. 5 5 o 1 1 F 3 F3* * δ 175.9* * KN 5 5 o 1 1 F = F* * δ = 10.4* *1.3 = 31.3KN. 5 5 o 1 1 F 1 F1* * δ 65.8* * KN 5 5 Response Spectrum Analysis: Accoring to EC8 [], the response of all moes of vibration contributing significantly to the global response shall be taken into account. The requirements may be eeme to be satisfie if either of the following can be emonstrate: - the sum of the effective moal masses for the moes taken into account amounts to at least 90% of the total mass of the structure; - All moes with effective moal masses greater than 5% of the total mass are taken into account. When using a spatial moel, the above conitions shoul be verifie for each relevant irection. Accoring to EC8 [], whenever all relevant moal responses may be regare as inepenent of each other, the maximum value EE of a seismic action effect may be taken as: E = E. E Ei Accoring to EC8 [], whenever a spatial moel is use for the analysis, the acciental torsional effects may be etermine as the envelope of the effects resulting from the application of static loaings, consisting of sets of torsional moments - ai about the vertical axis of each storey i: ai = eai Fi Where: ai is the torsional moment applie at storey i about its vertical axis. eai is the acciental eccentricity of storey mass i Fi is the horizontal force acting on storey i, as erive in for all relevant irections. Accoring to () - EC8 [], the effects of the torsional loaings shoul be taken into account with positive an negative signs (the same sign for all storeys). Whenever two separate planar moels are use for the analysis, the torsional effects may be accounte for by applying the rules of () to the action effects compute in accorance with From all things mentione above, SAP000 will be use to analyse the structure with oal Response Spectrum. The analysis will run accoringly to EC8. V.1.3 COBINATIONS: Accoring to EC8 [3], there are 3 combinations etermine from all above actions: Combination 1: DEAD Loa + DL Slab + Joint Loa + Seismic Loa + ψ, * Q, i i k

13 18-13 Combination (in opposite irection): DEAD Loa + DL Slab + Joint Loa - Seismic Loa + ψ, i* Qk, i Combination3: Envelope of Combination 1 an Combination. V.1.4 INTERNAL FORCES: The internal forces will be etermine by SAP000, version 9.0. The internal forces, which are use to esign the section reinforcement of the frames, are etermine by Lateral Force Analysis metho. Figure 4 - Transverse or Direction Y Frame Line

14 18-14 TABLE: Element Forces Transverse Y current Frames First Story Beams Frame Section OutputCase CaseType StepType P= E Axial V=V E Shear 3= E oment Text m Text Text Text KN KN KN-m 43 0 envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in V.1.5 DESIGN OF BEAS OF THE FIRST STORY IN FRAE LINE (members 43-45, an 79-81) V Geometrical Restraints Effective flange with: Accoring to (3) EC8 [3], the effective flange with b eff may be assume as follows: - For primary seismic beams framing into exterior columns, the effective flange with, b eff, is taken to the with b c of the column in the absence of the transverse beam, or equal to this with increase by h f on each sie of the beam if there is transverse beam of similar epth. - For the primary seismic beams framing into interior columns the above length may be increase by h f on each sie of the beam. So, the effective flange with of beams framing to the exterior columns is: bc = 400mm ; hf = 150mm beff = bc + * * hf = 1000mm The effective flange with of beams framing to the interior columns is: bc = 400mm ; hf = 150mm beff = bc + * 4* hf = 1600mm Beam Column centroial axis istance : The beam framing symmetrically into the exterior columns has the eccentricity as following expression: e = 0.00mm inimum with of the beams : - In accorance to EC8 [3], the effective transfer of cyclic moments from a primary seismic beam to a column shall be achieve, by limiting the eccentricity of the beam axis relative to that of the column into which it frames. - A eeme to satisfy rule for EC8 is to limit the istance between the centroial bw min { bc + hw;bc} axes of the two members to less than b c /4. bw 00mm

15 18-15 With to height ratio of the web of the beam: In accorance to EC [], lateral buckling of the slener beams: - (P1), where the safety of beams against lateral buckling is in oubt, it shall be checke by an appropriate metho. - (P), the with to height ratio of the beam s web must be ensure to the following hw = 500mm conition: hw <.5* bw, so 350 <.5*50 = 65 bw = 50 Limitation of the beam with: Accoring to EC8 [3], the beam with has to be checke as following conition: = 750 bw min( bc + hw;* bc) b w = 50. This conition is met. * 50 = 500 V.1.5. Flexural reinforcement Ultimate limit States Bening moment envelope iagram is presente as following: The reinforcement of the sections will be calculate by using EC. The actual strength of the materials: - Concrete: As chosen above, the concrete class is C5/30, accoring to EC [], concrete material properties are: fck = 5Pa = 5 N / mm ; fcm = 33Pa = 33 N / mm ; fctm =.6Pa =.6 N / mm αc* fck fc = = 16.67Pa= N / mm ; αc = 1; γc = 1.5 γ c (Note: In Belgium γ C = 1.5; In France γ C = 1.3) - Reinforcing steel: As chosen above, the steel class is S500, accoring to EC an EC3, steel material properties are: fyk = 500Pa= 500 N / mm * f yk 1*500 f y = α = = 434.8Pa = N / mm γ s 1.15 (Note: In Belgium γ S = 1.0; In France γ S = 1.0) - In accorance to ultimate limit states of bening plus axial force esign proceure of EC an EC [], we will use a rectangular iagram for compresse concrete block. In this case, the value of η is 1, so η* f c = 1* f c = 16.7 Pa. Flexural Reinforcing steel of Left sie of first span of the beam (Axis A) + = 9534Nm = Nm - For + : Beam s imensions: b=50mm; h=500mm. Cover:

16 18-16 c nom = c min + c ev ( Expression 4.1 EC []) Where: c min = max[c min,b ; c min,ur ]. (Accoring to Expression 4. EC[]) c min,b = iameter of bar. Assume 0mm bars an 6 mm hoops - Table 4. EC[]. c min, ur = minimum cover ue to environmental conitions. Assuming that Exposure class is XC1 an Structural Class is S4 c min,ur = 15mm c ev = 10mm c nom = =30 mm Concrete cover thickness + ½ reinforcement iameter = = 43 mm The effective height : = h concrete cover thickness stirrup iameter ½ reinforcement iameter = = 457 mm The effective with of the beam : b eff = 1450mm Accoring to Appenix A1 Concise EC [5], an to How to esign concrete using EC [6]: 3 E 9534*10 K = = = 0.01 b eff * * f ck 1600*457 *5 To restrict the ratio x/ < 0.45 δ = 0.85 an K = z = * K * * = + = 0.95*= x =.5*(-z) = 57.15mm <1.5*h f = 1.5*150=187.5mm On the other han we have the relationship: R = As * fy * z, So the area of reinforcement steel can be etermine as follows: 3 E 9534*10 As = = 490.mm = 4.9cm f * z 434.8* y We choose 4 φ 14 (A s = 6. cm ) for flexural reinforcements The resistance of the + section is R, b, A =+ 119kNm an the over-strength factor is =. The normalise flexural reinforcements are : ρ = As b * 1.4 w = = 5*( ) Checking for spacing of the bars Clear spacing of the bars: (50-*30-*6-4*14)/3 = 40.7mm Accoring to 8.() EC [], minimum clear istance between bars: = max [bar iameter, aggregate +5mm] = max [14, 0+5] = 5mm < For - : Beam s imensions: b=50mm; h=500mm Cover: c nom = c min + c ev ( Expression 4.1 EC []) Where: c min = max[c min,b ; c min,ur ]. (Accoring to Expression 4. EC[]) c min,b = iameter of bar. Assume 0mm bars an 6 mm hoops - Table 4. EC[]. c min, ur = minimum cover ue to environmental conitions. Assuming that Exposure class is XC1 an Structural Class is S4 c min,ur = 15mm c ev = 10mm c nom = =30 mm Concrete cover thickness + ½ reinforcement iameter + stirrup iameter = 60mm

17 18-17 Working height : = h concrete cover thickness ½ reinforcement iameter = =440mm R E ; R = µ * b* * η* fc ; E = Nm 3 E *10 The ULS conition: µ = = 0.31 b* * η * fc 50*440 *1*16.7 x z = 0.33 < 0.45; = 0.85 z = 0.86* = mm; On the other han we have the relationship: R = As * fy * z, So the area of reinforcement steel can be etermine as follows: 3 E *10 As = = 1134mm = 11.3cm f * z 434.8*378.4 y We choose 3 φ 0+*φ10 or 4 φ0 (A s = 1.6 cm ) for flexural reinforcements The resistance of the section is R, b, A = 18kNm an the over-strength factor is =. The normalise flexural reinforcements are : ρ = As b *.0 w = = 5*( ) Checking for spacing of the bars Clear spacing of the bars: (50-*30-*6-4*0)/3 = 3.7mm Accoring to 8.() EC [], minimum clear istance between bars: = max [bar iameter, aggregate +5mm] = max [14, 0+5] = 5mm < 3.7 so this conition is met - Check for the ratio between negative reinforcement an positive reinforcement: Accoring to (4a) EC8 [3], at the compression zone, reinforcement is not less than half of the reinforcement provie at the tension zone. The compression reinforcement area is 60mm an the tension reinforcement area is 160 mm so this conition is met Flexural Reinforcing steel of right sie of first span of beam (Axis B): + = 8191Nm = Nm - For + : Beam s imensions: b eff =1600 mm; h w =500 mm Concrete cover thickness + ½ reinforcement iameter =50mm Working height : = h concrete cover thickness ½ reinforcement iameter = =450mm The ULS conition: ; = µ * b * * η* f ; = 8191Nm R E R eff c E 3 E 8191*10 µ * * η * c 1600*450 *1*16.7 x = = 0.0 = < 0.45 b f z x= 9 < hf = 150; = z = 0.977* = 440mm Area of reinforcement steel can be etermine as follows: 3 E 8191*10 As = = 49mm = 4.3cm f * z 434.8* 440 y

18 18-18 We choose 4 φ 14 (A s = 6. cm ) for flexural reinforcements. The resistance of the + section is R, b, B = 119kNm an the over-strength factor is =. The normalise flexural reinforcements are : ρ = As b * 1.4 w = = 5*( ) Checking for spacing of the bars Clear spacing of the bars: (50-*30-*6-4*14)/3 = 40.7mm. Accoring to 8.() EC [], minimum clear istance between bars:= max [bar iameter, aggregate +5mm] = max [14, 0+5] = 5mm < 40.7 so this conition is met - For - : Beam s imensions: b=50mm; h=500mm Concrete cover thickness + ½ reinforcement iameter =60mm Working height : = h concrete cover thickness ½ reinforcement iameter = =440mm The ULS conition: R s ; R = µ * b* * η* fc 3 s *10 s = Nm µ = = 0.18 b* * η * fc 50* 440 *1*16.7 x z = 0.31 < 0.45 = 0.87 z = 0.87* = 38.8mm Area of reinforcement steel can be etermine as follows: 3 s *10 As = = 106.7mm = 11cm f * z 434.8*38.8 y We choose φ φ 18 + φ10(a s = 11.9 cm ) or φ0+φ18 for flexural reinforcements. The resistance of the section is,, = 08kNm an the overstrength factor is =. The normalise flexural reinforcements are : ρ = As b *.0 w = = 5*( ) Checking for spacing of the bars Clear spacing of the bars: (50-*30-*6-*18-*0)/3 = 34mm Accoring to 8.() EC [], minimum clear istance between bars: = max [bar iameter, aggregate +5mm] = max [14, 0+5] = 5mm < 34mm so this conition is met - Check for the ratio between negative reinforcement an positive reinforcement: Accoring to (4a) EC8 [3], at the compression zone, reinforcement is not less than half of the reinforcement provie at the tension zone. The compression reinforcement area is 60mm an the tension reinforcement area is 1190 mm so this conition is met Check for the eflection: Accoring to 15.7 Concise EC[5], the SLS state of eflection may be checke by using the span to effective epth approach. To use the span to effective epth approach, verify that: Allowable l/ = N*K*F1*F*F3 actual l/ R b B

19 18-19 Where: N = Basic l/: check whether ρ > ρ 0 : ρ = ρ = 0.005; ρ 0 = f ck 0.5 /1000=0.005 use the Exp (7.16a) Concise EC [5] : N = *f ck 0.5 ρ/ρ *f ck 0.5 *(ρ/ρ 0 1) 1.5 = 18.5 K=1.3 (en span) table Concise EC F1 = 1; F=1 F3=310/σ s Where: σ s = (A s,pro /A s,req ) = 18/500<=1.5 l/ = Actual l/ = 5000/457=10.9. So this conition is met. V Specific measures for the flexural reinforcement. in/max reinforcing steel - In accorance to EC8 [3], minimum tension reinforcement ratio shall not fctm.6 excee the value: ρ min = 0.5* = 0.5* = (5.1 - EC8 [3]). f yk 500 The reinforcement content is satisfactory. - Accoring to EC8 [3], within the critical regions, the tension reinforcement ' f ratio shall not excee the value below: max * c ρ = ρ + µ φ * εsy, f y Where: µ φ - Curvature uctility, T 1 = 0.67s > T C = 0.6s µ φ = *q 0-1 = 6.8. f y ε sy, = Es So, ρ = max * * = > The reinforcement content is satisfactory. Longituinal bar iameters: Accoring to EC8 [3], to prevent the bon failure, the iameter of longituinal bars of the beams is limite as the following conitions: - For interior beam column joints: bl 7.5* fctm ν bl fctm * 4.0* *( * ν ) ( ENV8.) h * ' C γ R f ρ y * kd * hc fy ρmax Where: h c is the with of the column parallel to the bars, so h c = 500mm. f ctm : is the mean value of the tensile strength of concrete f ctm =.6N/mm. F y = Pa. ν is the normalise esign axial force in column, taken with its minimum value NE for seismic esign situation. ν = fc * AC N E = N; f c = 16.67Pa; A c = 400x500=00000mm ν = NE fc * A = C 16.7* = k D is the factor reflecting the uctility class equal to 1 for DCH, to /3 for DC. ρ compression steel ratio ρ = ρ max = 0.05.

20 18-0 γ R = 1. bl 7.5* *0.387 So: * = bl = 500*0.056=8mm h 1* c * * The chosen reinforcement is satisfactory. - For exterior beam column joints: bl 7.5* fctm bl fctm *1+ 0.8ν 4.0* *( * ν ) ( ENV8) hc γ R * fy hc fy Where: h c is the with of the column parallel to the bars, so h c = 500mm. f ctm : is the mean value of the tensile strength of concrete f ctm =.6N/mm. F y = Pa. ν is the normalise esign axial force in column, taken with its minimum value NE for seismic esign situation. ν = fc * AC N E = N; f c = 16.67Pa; A c = 400x500=00000mm ν = NE 0.8 fc * A = C 16.7* = k D is the factor reflecting the uctility class equal to 1 for DCH, to /3 for DC. ρ compression steel ratio ρ = ρ max = γ R = 1. bl 7.5*.6 So: *(1 0.8*0.8) h c 1* = bl = 500*0.055=7.5mm The chosen reinforcement is satisfactory. Top reinforcement of the beam. - In accorance to part 1-3 ENV8, one fourth of the maximum top reinforcement shall run along the entire beam length. - Two φ0 bars will run along the entire span. V Shear resistance Design shear forces compute in accorance to the capacity esign criterion: - Accoring to EC8 [3], in the primary seismic beams shear forces shall be calculate in accorance with the capacity esign rule, consiering the equilibrium of the beam uner: a) the transverse loa acting on it in seismic esign situation an b) en moments i, (with i=1, enoting the en sections of the beam), corresponing for each sense of the seismic action, to plastic hinge formation at the ens either of the beams or of the vertical elements which are connecte to the joint where beam en i frames into. - The calculation of shear forces as following the sketch below:

21 18-1 Determining + AR1, - BR1, - AR, + BR, V A0, V B AR1 an + BR The bottom reinforcement area of longituinal bars is 6.cm (4 φ 14), we etermine the value of + AR1 = + BR as following: fy * As 434.8*6.*100 x= = = 64.6mm η * fc * b 1*16.7*50 14 = = 455mm x z = = = z* f * A = 0.95*465*434.8*60 = Nmm = 119KNm AR1 y s AR1 + BR = 119kNm. - - AR : The top reinforcement area of longituinal bars is 4 φ 0 (A s = 1.6 cm ), we etermine the value of - AR as following: Fas = fy * As = 434.8*160 = Nmm Fc = η * fc * b*( h z) = 16.7*50*(500 z) = 4175*(500 z) Fas = Fc = 4175*(500 z) z = 368.8mm AR = Fas * bas + Fc * bc 1 bas = z coating φstirrup φbars bas = = 33.8mm h z bc = = = 65.6mm = * *( )*65.6 = Nmm = 18KNm AR AR - - BR1 : The top reinforcement area of longituinal bars is φ0 + φ18 (A s = 11.9 cm ), we etermine the value of - BR1 as follows:

22 BR1 as as c c 18- F = f * A = 434.8*1190 = 51741Nmm as y s F = η * f * b*( h z) = 16.7*50*(500 z) c c = 4175*(500 z) F = F = 4175*(500 z) as c z = 376.1mm = F * b + F * b 1 bas = z coating φstirrup φbars b = = 340.1mm as h z bc = = = 6 mm = 51741* *( )*6 = Nmm = 08KNm BR1 BR1 - Determining V B0 an V A0 : 3750*5 3850*5 VB0 = VA0 = + = N = 50.4KN * - So we have: + ( AR1+ BR1) V1 = γ R * = 1* = 65 l 5 + ( + ) AR BR V = γ R * = 1* = 67; VB0 = VA0 = 50.4KN l 5 An so: - At support A: V min = V 1 + V A0 = =-14.6 ; V man = V + V A0 = = At support B: V min = -V + V B0 = =-16.6 ; V man = V 1 + V B0 = =115.4 V c an V R Computations : - In the critical sections: V c = 0 - Outsie the critical sections: V c = V R1. - In accorance with EC() an neglecting the axial force influence, the value of V R,c

23 18-3 EN /3 VR, c = * k *(100* ρl * fck ) 0.15* σcp * bw * γ c f ck is compresse strength of the concrete at the age of 8 ays. γ c = k = 1+.0 mm; 0.0 ρ = Asl l bw * where: A sl is the area of tension reinforcements. b w is the minimum with. Ns σ cp = ; N s is the longituinal force. Pa Ac Replacing with the value of f ck is 5Pa, reinforcing steel percentage is 60 50*( ) = , so we have: /3 V R, ct = *1*(100*0.0053*5) *50*467 = 394 N = 3.9 KN 1.5 Computations - The computations shall run in accorance to 6..1() EN199 an the specific rules shall get along with truss moel (EN1998) - Accoring to 6..1() EC [], the shear resistance of a member with shear reinforcement is equal to V R = V R,s + V cc + V t. V t is the esign value of the shear component of the force in the tensile reinforcement, in the case of an incline tensile chor, so V t = 0 an V cc = 0. Asw So: V R = V R,s + V cc VRs = * z* fyw *cosθ s - In critical regions (*height of the beams) shear force will be carrie out only by the stirrups. We choose stirrups with legs, of 6mm in iameter an 80mm spacing. Shear 57 force capacity is: VR = VRs = *467*434.8 = N = 145 KN>> V max Outsie critical regions, shear forces are carrie out by stirrups with legs, 6mm in iameter an 10mm spacing. Shear force capacity is: 57 VR = VR, c + VRs = * 467* = 19400N = 19.4KN 10 The ifference between the shear force values at support A an B is small enough to neglect the possibility of moifying the shear reinforcement. Along the whole beam length, shear force must be less than the value of V R,max which is the esign value of the maximum shear force which can be sustaine by the member, limite by crushing of the compression struts. V R,max = α cw * b w *z *ν 1 *f c / (cotθ + tanθ ) Where: ν 1 is a strength reuction factor for concrete cracke in shear α cw is a coefficient taking account of the state of the stress in the compression fck chor. So: υl = 0.6* 1 = 0.6*0.9 = an α cw = 1. V R,max = 1*50*467*16.7/(1+1)=56000N=56KN>V max - The computations shall run in accorance to 6..1() EN199 an the specific rules shall get along with truss moel (EN1998)

24 V Specific measures 18-4 Detailing: - In accorance to (6P) - EC8 [3], the stirrup minimum iameter within the critical regions is 6mm this requirement is met. - The first hoop is place not more than 50mm from the en cross section of the beam this requirement is met. - Within the critical regions, the spacing of the hoops is not greater than: hw/ 4 = 15 mm; 4* bw = 4*6 = 144mm So, the conition is satisfactory. 5 mm;8* bl = 8*14 = 11mm Casting an Placing for beam: All requirements are met.

25 18-5 Floor Level V.1.6 REINFORCEENT OF OTHER BEAS The reinforcement of other beams of the transverse frame are etermine by using the similar ways as the beams on the first floor. They are summarize in the following tables. Position of column Table V.1- Properties of the section an seismic actions in transverse frame Sections b w b c h w b eff nom + E + R of the (mm) (mm) (mm) (mm) (mm) (knm) (knm) beams - E (knm) - R (knm) 1- External En Internal En Internal ile External En Internal En Internal ile External En Internal En Internal ile Beams of Floor Table V. - Designe Longituinal Reinforcement an specific measures in transverse frame Position Sections Top Bottom ρ ρ ρ max ρ min max of of the Reinforc Reinforc (%) (%) (%) (%) (mm) column beams (mm ) (mm ) R (KNm) 1- External En (4φ14) Internal En (4φ14) Internal ile 68 60(4φ14) External En (4φ14) Internal En (4φ14) Internal ile (4φ14) External En (3φ14) Internal En (3φ14) Internal ile 40 46(3φ14) External En 46 46(3φ14) Internal En 46 46(3φ14) Internal ile 46 46(3φ14) Table V.3 - Designe Stirrup Reinforcement an specific measures of transverse frame Beams of Floor Sections of the beams V max (KN) φ - stirrup (mm) Number of legs Spacing V R (KN) V Rmax (KN) 1- Critical region 1- Outsie critical region 3-4 Critical region 3-4 Outsie critical region 5-6 Critical region 5-6 Outsie critical region

26 18-6 V.. LONGITUDINAL OR DIRECTION X CURRENT FRAE: V..1 Action effects: (Accoring to 5.4. EC8 [3]) All the steps are the same as transverse frames. The loas were uniformly istribute along the length of the beam. No istribution of the bening moments was mae. The esign value of the shear forces shall be etermine in accorance with the capacity esign rules, EC (P). The torsion effects were etermine separately by these two imensions accoring to () EC8: If the analysis is performe using two planar moels, one for each main horizontal irection, torsion effects may be etermine by oubling the acciental eccentricity e ai of the expression eai =± 0.05Li an applying the rules of (1) EC8 with the x x factor δ, δ = *, replace by the factor δ, δ = 1+ 1.*. L L V.. Action Summary: V...1 Gravity actions: e DEAD Loa: The self-weight loa. DL slab: The ea loas inuce by the floor an coating weight. LL slab: The live loas inuce by the variable actions LL roof slab: The live loas inuce by the variable roof actions Snow loa: The loas inuce by the snow. Joint loa: The loas acting to the joints of the longituinal frame inuce by perpenicular frames. WIND loa. V... Seismic actions: The analysis is performe using two planar moels, one for each main irection. The torsion effects were etermine separately by these two imensions accoring to x () EC8: δ = 1+ 1.* = 1+ 1.* = 1.. So we have δ = 1. Le 15 The above seismic forces are total seismic acting in all of the builing or all frames. Accoring to the stiffness of each frame, we will istribute the seismic forces to each frame linearly incluing torsion effects. In the action irection of the earthquake, there are 4 portal frames. When istributing seismic forces to all floors of the frame, the torsion effects will be taken into account by the factor δ which is calculate from (1) EC8. Factor δ, here, accounts for the analysis moel with two planar irections. We will istribute the seismic forces to the current transverse frame as following: FB 6 = F6* * δ = 31.34* *1. = 96.4KN. F 5 5* * 87.0* * = B F δ KN 4 = 4 = FB 4 = F4* * δ = 31.5* *1. = 69.5KN. F 3 3* * 175.9* * = B F δ KN 4 = 4 = FB = F* * δ = 10.4* *1. = 36.1KN. F 1 1* * 65.8* * = B F δ KN 4 = 4 =. e

27 18-7 Response Spectrum Analysis: Response Spectrum Analysis will be carrie out the same as the transverse frame. V..3 COBINATIONS: Accoring to EC8 [3], there are 3 combinations etermine from all actions above: Combination 1: DEAD Loa + DL Slab + Joint Loa + Seismic Loa + ψ, i* Qk, i Combination (in opposite irection): DEAD Loa + DL Slab + Joint Loa - Seismic Loa + ψ, i* Qk, i Combination3: Envelope of Combination 1 an Combination. V..4 INTERNAL FORCES: The internal forces will be etermine by SAP000, version 9.0. The internal forces, which are use to esign the section reinforcements of the frames, will be etermine by Lateral Force Analysis metho. Figure 5 - Longituinal Frame or Direction X Frame in line B

28 18-8 TABLE V.4 : Element Forces Longituinal or Direction X Frames Frame Section OutputCase CaseType StepType P N E Axial V V E Shear 3 E oment Text m Text Text Text KN KN KN-m 31 0 envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in envelopeofseism Combination ax envelopeofseism Combination in V..5 DESIGN OF BEAS OF THE FIRST STORY (members 31-33, 49-51, an ) V..5.1 Geometrical Restraints Effective flange with: Accoring to (3) EC8 [3], the effective flange with b eff may be assume as follows: - For primary seismic beams framing into exterior columns, the effective flange with, b eff, is taken to the with b c of the column in the absence of the transverse beam, or equal to this with increase by h f on each sie of the beam if there is transverse beam of similar epth. - For the primary seismic beams framing into interior columns the above length may be increase by h f on each sie of the beam. So, the effective flange with of beams framing to the exterior columns is: bc = 500mm ; hf = 150mm beff = bc + * * hf = 1100mm The effective flange with of beams framing to the interior columns is: bc = 500mm ; hf = 150mm beff = bc + * 4* hf = 1700mm Beam Column centroial axis istance : The beam framing symmetrically into the exterior columns has the eccentricity as following expression: e = 0.00mm inimum with of the beams :