BEHAVIOUR OF BUILDINGS FEATURING TRANSFER BEAMS IN THE REGIONS OF LOW TO MODERATE SEISMICITY Alireza Mehdipanah PhD Candidate at The University of Melbourne SUPERVISORS: A/PROF. NELSON LAM DR. ELISA LUMANTARNA
Seismic Standards: Seismic design provisions classify vertical irregularities based on the ratio of mass, stiffness or strength of adjacent storeys. No. Irregularity ASCE 7-10 Eurocode NZS 1170.5 k i k i+1 < 0.7, k i k i+1 < 0.7, 1 Stiffness Irregularity (Soft Storey) k i (k i+1 +k i+2 +k i+3 ) 3 < 0.8 Gradual reduction in the lateral stiffness k i (k i+1 +k i+2 +k i+3 ) 3 < 0.8 2 Strength Discontinuity (Weak Storey) F i F i+1 < 0.8 The ratio of the actual storey resistance to the resistance required by the analysis should not vary disproportionately. F i F i+1 < 0.9 3 Mass Irregularity m i m i+1 > 1.5 Gradual reduction in the mass m i m i+1 > 1.5 Or m i+1 m i > 1.5
In the seismic codes, Linear Static Analysis is limited to the Regular Structures. Dynamic Analysis is suggested for irregular buildings.
According to Eurocode 8 for irregular in elevation buildings Lateral Force Method cannot be used. Chinese Standard: For the buildings with transfer elements, nonlinear methods shall be used.
NZS 1170.5 o H<10 m or o Largest translational period T< 0.4 sec. or o Not irregular and T < 2.5 sec. Use of Linear Static Analysis is permitted. For the irregular buildings, performing Response Spectrum Method is obligatory. Design base shear from dynamic method should be scaled up to the base shear from static method.
ASCE 7-10: for structures o H<50 m o Having Strength Irregularity or Discontinuity in Vertical Lateral Force-Resisting system Use of Linear Static Analysis is permitted.
Literature: There are controversial opinions on the use of linear static analysis in the literature o o for setbacks for stiffness and mass irregularities
Inherent deficiencies of Response Spectrum Analysis: 1. Forces and Moments have no sign. 2. Values of responses do not satisfy equilibrium. Each response should be calculated directly. 3. Challenges exist on use of response spectrum and selection of appropriate real or synthetic scenarios in real engineering practice. o o o Number of required scenarios, Readiness of engineers in industry for the proper selection of records, Difficulty in the regions of low to moderate seismic zone.
Assumption of the Static Analysis Method: Effects of higher modes are negligible. o Force distribution is based on the shape of the fundamental mode o The total mass of building is participating in the first mode Is this class of irregularity violates the assumptions of Equivalent Static Analysis?
Design of shear wall dominant structures featuring transfer beams is not addressed adequately in the regions of low to moderate seismicity.
2 parameters are influencing the responses of buildings; o Fundamental period of structure o Degree of frame action in the response Code provisions are merely based on the estimation of the fundamental mode. A. Chopra and E. Cruz, Evaluation of Building Code Formulas for Earthquake Forces, Journal of Structural Engineering, vol. 112, pp. 1881-1899, 1986.
Case studies: 75 case study models are prepared based on 5 plan types. Each type has variety of heights and different locations of interruptions in the load resisting systems. Standards: AS/NZS 1170.0:2002, AS/NZS 1170.2:2011, AS/NZS 1170.4:2007 and AS3600-2009 Lateral resisting systems: Ordinary moment-resisting systems + Limited ductile shear walls µ= 2.0, Sp= 0.77 Buildings are located in Melbourne TC3 Terrain Category and Region A5
Percentage of Storey Shear Resisted by Moment-Resisting Frames
A B C D E F G H I J K L M 3 55.2 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 Type 1: 2 1 11.2 5.6 5.6 Locations of discontinuous columns 8 A B C D E F G H 41.8 8.4 5.0 5.0 5.0 5.0 5.0 8.4 5.0 7 5.0 6 Type 2: 5 36.5 5.0 6.5 4 5.0 3 5.0 2 5.0 1
4 A B C D E F G H 41.8 8.4 5.0 5.0 5.0 5.0 5.0 8.4 Type 3: 3 2 1 5.0 16.5 6.5 5.0 Locations of discontinuous columns A B C D E F G H I J K 6 42.0 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 4.2 6.0 Type 4: 5 4 3 2 1 6.0 30.5 6.5 6.0 6.0
Type 5: 6 5 4 3 2 1 6 5 4 3 2 1 4.5 4.5 22.5 4.5 4.5 4.5 4.5 4.5 22.5 4.5 4.5 4.5 A B C D E F G H I J 31.5 1.5 3.0 4.5 4.5 4.5 4.5 4.5 3.0 1.5 A B C D E F G H I J 31.5 1.5 3.0 4.5 4.5 4.5 4.5 4.5 3.0 1.5
Periods for the irregular buildings are almost identical to the regular buildings.
Modal Mass Participation Ratio for the irregular buildings are almost identical to the regular buildings.
Mode Shapes:
Inter-storey Drift Ratio Based on the Code Response Spectrum Analysis and The solid lines are the results of dynamic analysis. The dotted lines are static analysis results.
Generalised Lateral Force Method (GLFM) of Analysis Idealising building into a single-degree-of-freedom (SDOF) by calculating the effective dynamic parameters of the buildings. k eff = V δ m eff = σ m 2 eff jδ j σ 2 m j δ j T eff = 2π m eff δ eff = σ m jδ j 2 σ m j δ j a eff = V m eff k eff
Generalised Lateral Force Method (GLFM) of Analysis
Inter-storey Drift Ratio Based on the Response Spectrum Analysis and Generalised Lateral Force Method (GLFM)
Storey Shear Based on the Response Spectrum Analysis and Generalised Lateral Force Method (GLFM)
For the shear wall dominant irregular buildings featuring transfer beams o Soft storey due to the change in the stiffness will not form. o Structural response of irregular buildings with transfer feature is almost identical to the regular building. o Generalised Lateral Force Method (GLFM) of Analysis can have an acceptable estimation of drifts in all cases and a good estimation of shear forces along the height of structures which are shorter than 30 m.
Thank you
Mu(t) ሷ + Cu(t) ሶ + Ku(t) = p(t) Where, M, C and K denote the mass, damping and stiffness matrices, respectively. u(t) and p(t) are the displacement and force vectors. u(t) = y n φ n where φ n is the deflected shape of n th mode of vibration and y n is the time dependent factor describing the displacement of n th mode at the reference node Γ = φt Mr φ T Mφ = σ m jφ j σ m j φ j 2
Generalised Mode Shapes: Γ. φ = p 1 h 3 + p 2 h 2 + p 3 h + p 4 p 1 = 0.9849, p 2 = 1.795, p 3 = 0.6316, p 4 = 0.007134 Goodness of fit: R 2 : 0.9998 Γ. φ = a 1 sin൫b 1. h + c 1 ) + a 2 sin൫b 2. h + c 2 ൯ a 1 = 0.5793, b 1 = 4.283, c 1 = 2.53, a 2 = 0.357, b 2 = 2.462, c 2 = 1.142 Goodness of fit: R 2 : 0.9988