Parasismic verification of a building according to Eurocode 8 Make - PowerPoint PPT Presentation

Parasismic verification of a building according to Eurocode 8 Make the seismic verification of the new building of the sport center located in Liege according to the Eurocode EN 1998‐1 The building characteristics are: • Rectangular building of dimensions 28.2 X 12 m • 1 ground level + 1 level and an underground level  3 levels • Columns and central nucleus in concrete. All the other walls are in masonry except the underground walls which are in concrete • The slabs are in concrete • The building is located in Liege with a soil class B according to EN 1998‐1 • The building is an office building with meeting rooms • Concrete is C 30/37 with an instantaneous non cracked Young’s modulus of 35 000 N/mm² • The dead masses of the floor are 700 kg/m² and 600 kg/m² for the roof • The live loads are 500 kg/m² • The vertical component of the earthquake can be neglected

Plane view ‐ underground

Plane view – level 0

Plane view – level 1

Plane view – roof

Elevation

Elevation

Facades

Facades

1. Simplify the static scheme Ask yourself the following questions: • In which direction do the earthquake loads act? • What are the resisting elements under earthquake ? • Do the columns sustain earthquakes loads ? • What are the flexible parts of the building ? • Which elements must be modelled ? • The masonry has a very small resistance to horizontal loads  no resistance • Make the simplest model to represent the structure behaviour under earthquake, minimise the dofs Model the building with beam elements with only a few nodes, minimise the number of nodes

Static scheme – which element sustains the horizontal loads ? Only the concrete central core !

Static scheme – improve the behaviour under horizontal loads Big torsion in the central core. What can be done to reduce the torsion ?

Static scheme – improve the behaviour under horizontal loads To add a concrete wall at the opposite side to increase the lever arm

Static scheme – underground level not taken into account masonry concrte Very stiff In the soil  not taken into account

Static scheme – plane view 510 x 200 510 x 20 Th = 20 12 m E concrete,cracked = E concrete /2 25 m 3.3 m

Static scheme – elevation view 1200 x 40 4 m 1200 x 40 5 m E concrete,cracked = E concrete /2 25 m 3.3 m

Static scheme – elevation view No effect on the horizontal stiffness  not modelled But masses to take into account  M node = m.l 1200 x 40 4 m 1200 x 40 5 m 25 m 3.3 m

Static scheme – Masses M = 600 kg/m² + 0.24 * 500 kg/m² = 720 kg/m² 4 m M = 700 kg/m² + 0.24 * 500 kg/m² = 820 kg/m² 5 m Masses = dead loads + 0.24 live loads  2 = 0.3 offices  = 0.8 Storeys with correlated occupancies 25 m 3.3 m

2. Frequencies and eigen modes • Establish and Compute the stiffness matrix of the system • Establish and Compute the mass matrix of the system • Compute the eigen modes and the frequencies • Draw the deformed shape of the first 2 modes (by hand) • How many modes do you have ?

Beam Stiffness Matrix Matrix in the local axe of the element L : length A : section Iz : in‐plane inertia Iy : out‐of‐plane inertia J: torsional stiffness

Beam Static Characteristics b A = b.h h I = b.h³ /12 J = h.b³ /3 b A = b.h – (b‐2.t).(h‐2.t) I = b.h³ /12‐(b‐2.t).(h‐2.t)³ /12 h t �.� � �� ��� .������ � 𝑒𝑡 = J = S = surface inside the red line � �� �. ��� � ��. ��� � � �� � � = length of the red line divided by the thickness

Local axes  Global axes Apply a rotation matrix 1 2 3 4 5 6 1 C S 0 0 0 0 2 ‐S C 0 0 0 0 3 0 0 1 0 0 0 r = 4 0 0 0 C S 0 5 0 0 0 ‐S C 0 6 0 0 0 0 0 1 r 0 R = 0 r K glo = R T * K loc * R

Matrix assembly 2 2 4 6x6 6x6 K2= 6x6 6x6 6x6 6x6 6x6 6x6 K1= K3= 1 3 6x6 6x6 6x6 6x6 1 3 NODES 1 2 3 4 1 6x6 6x6 0 0 NODES 2 6x6 6x6 0 6x6 3 0 0 6x6 6x6 4 0 6x6 6x6 6x6

Supports 2 4 1 3 Suppress the dofs related to supported nodes NODES 1 2 3 4 1 6x6 6x6 0 0 NODES 2 6x6 6x6 0 6x6 3 0 0 6x6 6x6 4 0 6x6 6x6 6x6

Mass Matrix 1 2 3 4 5 6 7 8 9 10 11 12 y 1 m.L/2 x m,L 2 m.L/2 3 m.L/2 m = mass by unit length L = length 4 0 5 0 6 0 M loc  M glo ; the same as the stiffness matrix 7 m.L/2 8 m.L/2 Assembly of matrices; the same as the stiffness matrix 9 m.L/2 10 0 11 0 12 0

3. Modal properties • Compute the modal stiffness and mass of each mode Depend of the mode • Compute the effective modal mass in both horizontal direction Depend of the mode and the seism direction • Compute the modal share ratio for both horizontal direction Depend of the mode and the seism direction Where: • M is the mass matrix {u d } i is the i th eigen mode • • {e}k is a vector with 1 for each dof in the considered direction and 0 for the others

4. Parasismic calculation • Establish the design acceleration spectrum according to EN 1998‐1 The building is located in Liege with a soil class B according EN 1998‐1 The building is an office building with meeting rooms The q factor is taken equal to 1.5 • Compute the response (displacements) of each mode in each direction • Compute the maximum response in each direction (SSRS) of the building top • Which modes govern the total seismic response for each direction ? • Compute the support forces • Combine the support forces in the X and Y direction • Find a linear combination of the modes to obtain concomitant value of support forces at the foot of the �� � ∑ 𝑆 � central core 𝛽 � � � ���� 𝑥𝑗𝑢ℎ 𝑆 ���� � 𝑆 ���� � ∑ 𝛽 � . 𝑆 �

Design Spectrum definition According to EN 1998‐1: chapter 3.2.2.5

Design Spectrum definition

Design Spectrum definition a gr =0.1.g a g =  i *a gr

Design Spectrum definition design Spectrum Response 3,000 2,500 Behaviour factor : q = 1.5 2,000 Sd (m/s²) 1,500 1,000 0,500 0,000 0,000 0,200 0,400 0,600 0,800 1,000 1,200 1,400 1,600 1,800 2,000 T (s)

� � Seismic response of one mode 1 dof system 𝜑� � 2𝜊𝜕𝜑� � 𝜕 � 𝜑 � �𝜑 � acc max = Sd(  ,  ) design Spectrum Response 3,000 d max = Sd(  ,  )/  ² 2,500 2,000 Sd (m/s²) 1,500 a n dof system 1,000 0,500 0,000 0,000 0,500 1,000 1,500 2,000 T T (s) acc max,i = RM ik * Sd(  ,  ) � � 𝜕 � 𝜃 � � �𝑆𝑁 �� ∗ 𝜑 � 𝜃 � � � 2𝜊𝜕𝜃 � d max,i = RM ik * Sd(  ,  )/  ²

Seismic response of all modes For each mode: the maximum displacement = d i All the mode are not maximum at the same time �  total displacement ≠ ∑ 𝑒 � ��� � �  total displacement = ∑ 𝑒 � ; Square Root of the Sum of the Squares = SRSS ���

Support Forces For each modes: [K glo ] * {u d } i = {F i } where F i = nodal forces If we have a support, the nodal force F i : support forces R i The maximum support force under earthquake: � � R SRSS = ∑ 𝑆 � ��� To combine, the 2 horizontal directions :