P8 AM Structuralanalyismethods

Seismic Design of Multi-storey Buildings: IS-1893 vs. Eurocode-8 Structural Analysis Methods Abdelghani Meslem, PhD & Dominik Lang, PhD Department of Earthquakes and the Environment NORSAR, Kjeller, Norway IS-1893-1:2002 - NE 1998-1:2004 IS 1893-1: 2002 IS-1893 Provisions Criteria for earthquake resistant design of structures Part 1: General provisions and buildings EN 1998-1: 2004 Eurocode-8 Provisions Design of structures for earthquake resistance Part 1: General rules, seismic actions and rules for buildings A. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 Table of contents IS 1893-1:2002 vs. EN 1998-1:2004 IS 1893-1: 2002 Dynamic Characteristics Seismic Masses Fundamental Natural Period Methods of analysis Design Lateral Force Method Modal Response Spectrum Method Linear Time History Method EN 1998-1: 2004 A. Meslem & D. Lang Components of seismic action Accidental/Torsional Effects Second-order Effects (P-Δ effects) Select and Scale Earthquake Records Contribution of Joint Regions © NORSAR – Kjeller (Norway) 2014 A.0 25 Above 3.  ∑k (DLk + ILk ) Percentage of imposed load (IL) to be considered in seismic weight calculation (IS-1893-1:2002.0 50 the imposed load shall also be considered for roof.    IS 1893-1: 2002.4 DLk + ILk the seismic weight of the whole building is the sum of the seismic weights of all the floors. 7. Table 8) Imposed Uniformity Distributed Floor Loads (kN/m2)  Percentage of Load Up to and icluding 3. Lang © NORSAR – Kjeller (Norway) 2014 DL + IL . Meslem & D.Dynamic Characteristics: Seismic Masses IS 1893-1:2002  the seismic weight of each floor (k) is its full Dead Load (DL) plus appropriate amount of Imposed Load (IL). combination coefficient for variable action Ψ2i .60 Interdependently occupied storeys 0.load type coefficient  ΨEi = φ  Ψ2i Storey φ Occupancy type Ψ2 Roof 1.50 Roof with snow 0.2.Dynamic Characteristics: Seismic Masses EN 1998-1:2004  composed of permanent and participating live loads Gk + ∑i (ΨEi  QKi)  EN 1998-1:2004.80 Public. commercial (shops). staircases 0. Meslem & D. parking 0. libraries.80 A.20 Archives 1.30 storeys with correlated occupancies 0.occupancy type coefficient φ . Lang © NORSAR – Kjeller (Norway) 2014 .00 Archives.00 Residential. office 0.4 with: ΨEi . 3. 075  H 0. in m. A. where basement walls are connected with the ground floor deck or fitted between the building columns. along the considered direction of the lateral force. Meslem & D. including moment-resisting frame buildings with brick infill panels.75 for RC frame building for steel frame building H: Height of building.  Approximate Ta. 7. But it includes the basement stories. in seconds. in m.75  0 . This excludes the basement stories.09 H d d: Base dimension of the building at the plinth level.085  H 0. when they are not so connected.Dynamic Characteristics: Fundamental Natural Period IS 1893-1:2002  for moment-resisting frame building without brick infill panels: Ta  0 . may be estimated by the empirical expression: Ta  0. of all other buildings.6 . Lang © NORSAR – Kjeller (Norway) 2014  IS 1893-1: 2002. total effective area of the shear walls in the first storey (in [m2]) Ai .2 A.effective cross-sectional area of shear wall i (in [m2]) lwi .length of shear wall i parallel to applied forces  EN 1998-1:2004. Meslem & D.075 0.3.050 0.085 0.structural coefficient H .3.building height (in [m]) from foundation or top of a rigid basement for moment-resistant steel frames for moment-resistant concrete frames and eccentrically braced steel frames for all other structures for building with concrete or masonry shear walls with Ac:   Ac   Ai  (0.Dynamic Characteristics: Fundamental Natural Period EN 1998-1:2004  based on any equation coming from structural mechanics (e.75  Ct = Ct = 0. Rayleigh method)  for building heights H  40 m : T1  C t  H0. Lang © NORSAR – Kjeller (Norway) 2014 .075 / √Ac with: Ct . 4.2  (lwi /H)2 ) with: Ac .g. However.1  EN 1998-1:2004. planar (2D) models may be used for each direction X and Y Regularity Plan Elevation ● ● ● ○ ○ ● ○ ○ Model type planar (2D) spatial (3D) Ideally. Lang © NORSAR – Kjeller (Norway) 2014 . Regular/Irregular Configurations  IS 1893-1:2002.2. In some cases the analyst may wish to use two-dimensional (planar) in order to reduce the calculation effort.3 A.Modeling Specifications: Planar (2D) & Spatial (3D)  if regular in plan. Meslem & D. 7. the building should be modelled as three-dimensional. this later may be acceptable for buildings with regular geometries where the response in each orthogonal direction is independent and torsional response is not significant. 4. masses and moments of inertia may be lumped at the centre of gravity  EN 1998-1:2004. 7.2 m4 m3 m2 m1 A.5 times the average displacement of the entire diaphragm.3.3.  IS 1893-1:2002. Meslem & D. 4.1  A floor diaphragm shall be considered rigid if horizontal displacements at any point do not exceed more than 10 % of the rigid diaphragm assumption.  EN 1998-1:2004.1  A floor diaphragm shall be considered to be flexible.7. 4.Modeling Specifications: Masses Lumped System  if floor diaphrams are rigid in plane. Lang © NORSAR – Kjeller (Norway) 2014 m4  m3 k k m2 m1 k k* . if it deforms such that the maximum lateral displacement measured from the chord of the deformed shape at any point of the diaphragm is more than 1. 4.5 m4 m3 m2 m1 A.Modeling Specifications: Masses Lumped System  Buildings with regular. Lang © NORSAR – Kjeller (Norway) 2014 m4  m3 k k m2 m1 k k* . or nominally irregular plan configurations may be modeled as a system of masses lumped at the floor levels.  IS 1893-1:2002. 7.8. Meslem & D. 3 .3. 7.3. 4.2  EN 1998-1:2004.3. 7.8. 4. 4. Meslem & D.7 (2) Response Spectrum Method  IS 1893-1:2002.3.3.3  EN 1998-1:2004. Lang © NORSAR – Kjeller (Norway) 2014  EN 1998-1:2004.Methods of Analysis Analysis methods specified in IS 1893-1:2002 and EN 1998-1:2004 low complexity of computation (1) Design Lateral Force Method  IS 1893-1:2002. 7.8.4 (3) Linear Time History Method  IS 1893-1:2002.3 high complexity of computation A.3. 7. typically defined by a seismic design response spectrum. Lang © NORSAR – Kjeller (Norway) 2014  EN 1998-1:2004.3.5 This approach defines a series of forces acting on a building to represent the effect of earthquake ground motion. Meslem & D.(1) Design Lateral Force Method  Buildings shall be deisgned and constructed to resist the effects of design lateral force as a MINIMUM  IS 1893-1:2002. Criteria :  shall be applied to buildings whose response is principally dominated by the 1st mode:  4  TC T1   2.3.0 sec  and that are regular in elevation A.2 . 4. Lang Plan Elevation Allowed simplification in modeling ● ● planar ○ ● spatial © NORSAR – Kjeller (Norway) 2014 .(1) Design Lateral Force Method  Buildings shall be deisgned and constructed to resist the effects of design lateral force as a MINIMUM  IS 1893-1:2002.5 This approach defines a series of forces acting on a building to represent the effect of earthquake ground motion. Criteria (cont'd):  and that are regular in elevation Regularity A. Meslem & D. typically defined by a seismic design response spectrum. 7. Lang © NORSAR – Kjeller (Norway) 2014 . shall then be distributed to individual lateral load resisting elements depending on the floor diaphgram action A. Meslem & D.(1) Design Lateral Force Method Steps: Step 1: the design lateral force shall first be computed for the building as a whole Step 2: this deisgn lateral force shall then be distributed to the various floor levels Step 3: the overall design seismic force thus obtained at each floor level. Lang  IS 1893-1: 2002.2 ).2).2).2 Z = seismic zone factor (given in Table 2 Clause 6. Sa/g = average response acceleration coefficient.5 design horizontal seismic coefficient for the structure.4.(1) Design Lateral Force Method IS 1893-1:2002 Design base shear VB :  Total design lateral force shall be determined for each horizontal direction by the following expresssion: VB  Ah  W with: Ah W- VB  IS 1893-1:2002. 6. A.1 © NORSAR – Kjeller (Norway) 2014 . 7. using the fundamental period Ta seismic weight of the building.4. R = response reduction factor depending on the perceived seismic damage performance of the structure (given in Table 7 Clause 6.4. I = importance factor depending upon the functional use of the structure (given in Table 6 Clause 6.4. Ah  Z  I Sa  2 R g  IS 1893-1:2002. 6.4. Meslem & D. 55 0.67 0.(1) Design Lateral Force Method IS 1893-1:2002 Calculation of average acceleration coefficient at T=Ta: Horizontal components of the seismic action:  IS 1893-1: 2002.00 For the purpose of determining seismic forces.1 For rocky.10 0.67 ≤ T ≤ 4.00 T 0.10 ≤ T ≤ 0.4.50 g   1.36 T 0.50 g   1. or hard soil sites 1  15  T Sa    2 .10 ≤ T ≤ 0. Lang 0. Meslem & D.40 0.50 g   1. the country is classified into four seismic zones The design acceleration spectrum for vertical motions. 6.10 0.00 For medium soil sites 1  15  T Sa    2 .00 ≤ T ≤ 0. when required.10 ≤ T ≤ 0.00 ≤ T ≤ 0.00 ≤ T ≤ 0.10 0. © NORSAR – Kjeller (Norway) 2014 .67 T A.55 ≤ T ≤ 4.40 ≤ T ≤ 4.00 For soft soil sites 1  15  T Sa    2 . may be taken as twothirds of the design horizontal acceleration spectrum. 5 → 0.00 For medium soil sites 1  15  Ta 0. or hard soil sites 1  15  Ta 0.10 ≤ Ta ≤ 0.36 Ta 0.50 g   1.55   2.00 A.55 .00 ≤ Ta ≤ 0.(1) Design Lateral Force Method IS 1893-1:2002 Calculation of average acceleration coefficient at T=Ta:  IS 1893-1: 2002.40 ≤ Ta ≤ 4.10 Sa  0.10 ≤ Ta ≤ 0.40   2.10 ≤ Ta ≤ 0.67 ≤ Ta ≤ 4.67 Ta 0.10 Sa  0.1 For rocky. 6.4 sec → Sa/g = 2.00 Ta 0.50 g   1. Lang Ta © NORSAR – Kjeller (Norway) 2014 For medium soil site (Soil Type II) Ta = 0.10 Sa  0.67   2.00 ≤ Ta ≤ 0. Meslem & D.50 g   1.4.00 ≤ Ta ≤ 0.00 For soft soil sites Example: 4-story building 1  15  Ta 0.55 ≤ Ta ≤ 4.10 ≤ Ta ≤ 0. Meslem & D. 7. Lang © NORSAR – Kjeller (Norway) 2014 .(1) Design Lateral Force Method IS 1893-1:2002 Vertical distribution of base shear to different floor levels:  calculation of horizontal design forces Qi to all storey levels can be done as per the following expression h3 Qi  VB  Wi  hi2 n W j  h j 2 h2  IS 1893-1:2002. hi = height of floor i measured from base. A. Wi = seismic weight of floor i.7 Q3 W3 Q2 W2 h1 Q1 W1 j 1 Qi = design lateral force at floor i. and n = number of storeys in the building (the number of levels ayt which the masses are located). 3.2 Fb ordinate of the design spectrum at T1 total mass of the building correction factor = 0.(1) Design Lateral Force Method EN 1998-1:2004 Base shear force Fb :  shall be calculated for each horizontal direction Fb  Sd (T1 )  m   with: Sd (T1) = M= =  EN 1998-1:2004. 4.0 A. Meslem & D.3.85 if T1  2TC and the building has more than 2 storeys. Otherwise  = 1. Lang © NORSAR – Kjeller (Norway) 2014 . 5  TC    q  T1  2. Lang 2.5 q 2. 4.3.d ( T1 )  a g  S  A.d ( T1 )  a g  S  for TC ≤ T1 ≤ TD : S a .(1) Design Lateral Force Method EN 1998-1:2004 Calculation of design spectral acceleration Sa.2 q=1 q=2 q=4 TB TC T1 Period T [sec] TD .5 2  S a .d  EN 1998-1:2004.d :  dependent on soil class and behavior factor q for 0 ≤ T1 ≤ TB :  2 T 2.3. Meslem & D.d ( T1 )  a g  S  for TD ≤ T1 ≤ 4 s : S a .d ( T1 )  a g  S    1  (  ) q 3   3 TB for TB ≤ T1 ≤ TC : S a .5  TC  TD   2  q  T1  © NORSAR – Kjeller (Norway) 2014 Spectral acceleration Sa. Lang m3 F3 © NORSAR – Kjeller (Norway) 2014 F2 F1 m2 m1 s3 s2 s1 .2 F3 z3 z2 m3 F2 m2 z1 F1 m1 with: zi  height of the respective mass i Type B (dependent on absolute horizontal displacement of masses): s m Fi  Fb  i i  s j  mj  EN 1998-1:2004.3.(1) Design Lateral Force Method EN 1998-1:2004 Distribution of horizontal seismic forces:  calculation of horizontal forces Fi to all storey levels can be done by two ways Type A (dependent on height of masses): z m Fi  Fb  i i  z j  mj  EN 1998-1:2004. 4.3.2 with: si  lateral displacement of mass i in the 1st mode A. Meslem & D.3. 4.3. Meslem & D. Seismic masses: Gk + ∑i (ΨEi  QKi) residential use → Ground type C ΨEi = φ  Ψ2i = 0.79 1kN  1000N 152.24 Level G [kN] Q [kN] G+ Ψ  Q [kN] Mass mi [tons] 3 260 120 289 29.8  0.3 = 0. Lang © NORSAR – Kjeller (Norway) 2014 Units: kg  m s2 .44 2 350 140 384 39.(1) Design Lateral force Nethod: EN 1998-1:2004 Tutorial 1 m3 3-story RC frame building (residential use) m2 3 x 3.33 1N  1 Total seismic mass m A.10 1 750 300 822 83.5 m m1 1. design spectral acceleration: residential use → γI = 1.85 = 274 kN © NORSAR – Kjeller (Norway) 2014 .075  10.base shear force Fb: (since T1 < 2  TC →  = 0. Lang TB T1 TC TD Period T [sec] → Fb = Sa.85) A.33  0.943 m/s2 → Sa.fundamental period: (with Ct = 0.12 m/s2 Spectral acceleration Sa .75 = 0.075 for RC frames) → ag = agR  I = 2.3 g behavior factor q = 4.12  152.0 → T1 = Ct  H 0.d = ag  S  2. Base shear force Fb : .44 s .75 = 0.0 ground motion agR = 0.(1) Design Lateral force Method: EN 1998-1:2004 Tutorial 1 2.5 0.d (T1)  m   = 2. Meslem & D.5/q = 2. 12 96.68 1015.72 321.33 876.0 152.5 83.3   7.1 2 7.44 309.2 1 3. Lang M res 1935.79 293.06 m Fres 274.0 © NORSAR – Kjeller (Norway) 2014 m2 heff Mres m1 .3 Totals m3 4. Effective height of the resultant lateral force: Fres heff  A.60 599.09 274.(1) Design Lateral force Method: EN 1998-1:2004 3.10 273.5 29.70 85.27 91.0 39.0 1935. Load distribution and moment calculation: F3 m3 z m Fi  Fb  i i  z j  mj F2 m2 F1 m1 Tutorial 1 Level Height z [m] Mass mk [tons] zk  mk [mtons] Force Fk [kN] Moment = Fk  zk [kNm] 3 10. Meslem & D. Meslem & D. or by a sitespecific design spectrum mentioned. Lang © NORSAR – Kjeller (Norway) 2014 .  the Response spectrum method shall be performed using the design spectum.(2) Modal Response spectrum method This approach permits the multiple modes of response of a building to be taken into account. A. Lang © NORSAR – Kjeller (Norway) 2014 .g. all the response quantities (e.4 shall be performed for the following buildings: • Regular buildings – those greater than 40 m in height in Zone IV and V. • Irregular buildings – all framed buildings heigher than 12 m in Zones IV and V. Meslem & D. A. story shears and base reactions) shall be multiplied by 𝑉𝐵 /VB. story forces. displacements. • Where VB is less than 𝑉𝐵 . Member forces. 7. the resulted design base shear (VB) shall be compared with a base shear (𝑉𝐵) calculated using a fundamental period TB.8. and those greater than 90 m in height in Zones II and III. and those greater than 40 m in height in Zones II and III.(2) Modal Response Spectrum method Criteria:    IS 1893-1:2002. Lang ≥ 90 % of the total seismic mass and missing mass correction beyond 33 % © NORSAR – Kjeller (Norway) 2014 .8. Meslem & D.(2) Modal Response Spectrum method Criteria (cont'd):   IS 1893-1:2002. 7.4 the number of modes to be used in the analysis should be such that: The sum total of modal masses of all modes considered A. 4.3. Lang © NORSAR – Kjeller (Norway) 2014 Fb 1st mode .(2) Modal Response Spectrum method Criteria:  EN 1998-1:2004.3 Regularity  Allowed simplification Plan Elevation Model ● ○ planar ○ ○ spatial shall be applied if the criteria for analysis method (1) are not fulfilled.3. Meslem & D.0 sec A. this means if:  4  TC T1   2. 4.e. Meslem & D.(2) Modal Response Spectrum method Criteria (cont'd):  EN 1998-1:2004.3  response of all modes shall be considered that contribute significantly to the global building response (i.3.05  mtot ..3. important for buildings of a certain height)  those modes shall be considered for which: (1) the sum of the modal masses is at least 90% of the total building mass or (2) the modal mass is larger than 5% of the total building mass A.9  mtot mi ≥ 0. Lang © NORSAR – Kjeller (Norway) 2014 mi ≥ 0. Meslem & D.number of modes taken into account n . Lang 0.02 s . those modes shall be considered for which: k ≥ 3  √n and Tk ≤ 0.20 s with: n=4 k .story number (from above foundation to top) Tk .20 s → six modes shall be considered !!  Period Tk : A. 4.3 if the '90%' and the '5%' criteria is not fulfilled (e.27 s 0.(2) Modal Response Spectrum method Criteria (cont'd):   EN 1998-1:2004.3.002 s ≤ 0.23 s 0.16 s © NORSAR – Kjeller (Norway) 2014 0.period of vibration of mode k Mode shape: 1 2 3 4 Example: 4-story building k ≥ 3  √4 = 6 and T12 = 0. for buildings prone to torsional effects).g.3. . c22 . k 1n   k 11 . k nn   n1 =>   km ...  .    k  ..  c  n1 ...   cnn  M u  C u  K u  0  modal segmentation:  derive circular frequencies i / periods Ti and mode shapes i A. Assumption: [C] = zero matrix ! – Undamped system K   2  M  0 c1n   .   ... k 33 . mode shapes & modal participating factors: undamped free vibration analysis of the entire building to obtain natural periods and mode shapes for the modes of vibration that need to be considered.(2) Modal Response Spectrum method Obtain natural periods. Lang © NORSAR – Kjeller (Norway) 2014 . .   Differential equation: with: M u  C u  K u  0 0 0  m1 0   0  0 m2 0 M   0 0 m3 0     0  0 0 m i   c11  .. .. ... c 33 .. .. K   .. C   . .   .. Meslem & D..... k 22 . i m j   j2.i W j   j .1  j+1.3 . mode shapes & modal participating factors: Given: .mode shapes i n.i  n 2 j 1  IS 1893-1:2002.i j 1 n j 1  EN 1998-1:2004.1    1    j1.(2) Modal Response Spectrum method Obtain natural periods.1 T1 Modal participation factor of mode k: n Pi  W j 1 j n   j . Lang © NORSAR – Kjeller (Norway) 2014 i  m j   j .1     n.4 A.8. Meslem & D.3.circular frequencies i / periods Ti .1 j. 4.3.1   j. 7. 1 .5 0 0 0 0 0 0 0 0 0 14.113 0.1 0.081 0.1 Example 1.129 0. Lang Mode.137 0.7 28.103 0.098 0.104 0.398 0.6 0 0 1 3.092 0. Meslem & D.6 0 3.1 5.096 0.316 0.8 0 0 0 0 1.7 0 2.6 0 0 0 2.2 1. [-] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 T [s] 0.5 2.084 0.092 0.088 0.136 0.116 0.09 0.118 0.11 0.6 6 0 7 0 4.08 mx' [%] 0 52 8 0 0.5 0 0.(2) Response Spectrum analysis Tutorial 2.264 0.9 0 0 0 0 0 0 0 0 0 0 0 0 0 © NORSAR – Kjeller (Norway) 2014 mz' [%] 0 0 0 39.086 0.5 3.095 0 0.17 0.5 8.134 0.4 2.5 0 0 0 0 0 4.19 0.124 0.105 0.Modal analysis results: Effective masses  building response ’purely ’ translational  first eigenmode is translational A.096 0.7 0.4 0 1 .8 0 0 0 1.6 0 0 0 0 0 0 my' [%] 52.7 0 0 0 0 0 0. 1 .7 0 0 0 0 0 0.118 0.1 5.11 0.Modal analysis results: Effective masses  response of all modes shall be considered that contribute significantly to the global building response A.398 0.4 2.096 0.092 0.6 0 0 0 2.6 6 0 7 0 4.105 0.136 0.264 0.(2) Response Spectrum analysis: EN 1998-1:2004 Tutorial 2.8 0 0 0 0 1.08 mx' [%] 0 52 8 0 0.5 0 0 0 0 0 4.088 0.1 0.7 0 2.092 0.113 0.124 0.096 0.098 0.134 0.6 0 0 1 3.095 0 0.081 0.129 0.316 0.7 0.8 0 0 0 1.084 0. [-] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 T [s] 0.5 0 0.5 0 0 0 0 0 0 0 0 0 14.9 0 0 0 0 0 0 0 0 0 0 0 0 0 © NORSAR – Kjeller (Norway) 2014 mz' [%] 0 0 0 39. Lang Mode.6 0 3.17 0.2 1.6 0 0 0 0 0 0 my' [%] 52.116 0.103 0.137 0.5 2.5 8.7 28.5 3.19 0.4 0 1 .086 0.09 0.1 Example 1.104 0. Meslem & D. 6 0 3.6 my' [%] 52. Meslem & D.5 0 0.098 19 0.5 0 0 0 0 0 0 0 0 0 14.264 4 0.7 0.8 0 0 0 1.398 2 0.4 0 1 .7 28.19 5 0.081 30 0.136 8 0.5 0 0 0 0 0 4.1 5.096 20 0.09 26 0.11 15 0.2 1.118 12 0.084 29 0.095 22 0 23 0.0 © NORSAR – Kjeller (Norway) 2014 mz' [%] 0 0 0 39.08 Sum  mx' [%] 0 52 8 0 0.092 25 0.9  mtot or mi ≥ 0.5 2.6 0 0 0 0 0 0 91.05  mtot A.096 21 0.134 9 0.5 3.116 13 0.105 16 0.9 0 0 0 0 0 0 0 0 0 0 0 0 0 90.137 7 0. T [-] [s] 1 0.6 0 0 1 3.129 10 0. Lang Mode.113 14 0.124 11 0.5 8.103 18 0.Modal analysis results: Effective masses  those modes shall be considered for which: mi ≥ 0.6 6 0 7 0 4.1 .104 17 0.4 2.1 Example 1.316 3 0.6 0 0 0 2.092 24 0.8 0 0 0 0 1.17 6 0.7 0 2.7 0 0 0 0 0 0.1 0.088 27 0.086 28 0.(2) Response Spectrum analysis: EN 1998-1:2004 Tutorial 2. Lang Mode [-] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 … T [s] 0.117 0.095 0.9 1.093 0.111 0.096 0.7 1.077 … mx' [%] 1.Modal analysis results: Effective masses  building strongly prone to torsional response  first eigenmode is torsional: A.7 0 0 0 0 0 0 0 0 0 0 0 0 11 4.087 0.144 0.8 7 0 0 0 0 0 0 0 0 0 0 0 0 … my' [%] 2 0.078 0.9 0 3.15 0.131 0.14 0.1 0.109 0.8 0 0 0 0.094 0.11 0.8 … .114 0.077 0.2 0 13.123 0. Meslem & D.(2) Response Spectrum analysis: EN 1998-1:2004 Tutorial 2.5 1 0 0 0 0 0 0 0 2.6 11.302 0.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … © NORSAR – Kjeller (Norway) 2014 mz' [%] 0 12.138 0.2 .3 53.3 2.106 0.084 0.118 0.183 0.6 0 5 0 0 0 0 0 0.6 0.7 0 0.2 0.5 55.082 0.083 0.122 0.2 10.1 0 0 0 0 1.2 Example 1.142 0.092 0. 092 0.096 0.7 1.11 0.094 0.3 53.122 0.15 0.118 0.144 0.2 .302 0. Lang Mode [-] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 … T [s] 0.078 0.084 0.2 10.9 0 3.7 0 0.3 2.5 1 0 0 0 0 0 0 0 2.142 0.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … © NORSAR – Kjeller (Norway) 2014 mz' [%] 0 12.183 0.2 Example 1.8 0 0 0 0.111 0.1 0 0 0 0 1.106 0. Meslem & D.082 0.6 0 5 0 0 0 0 0 0.2 0.117 0.138 0.077 … mx' [%] 1.8 … .2 0 13.123 0.(2) Response Spectrum analysis: EN 1998-1:2004 Tutorial 2.083 0.7 0 0 0 0 0 0 0 0 0 0 0 0 11 4.Modal analysis results: Effective masses  response of all modes shall be considered that contribute significantly to the global building response A.1 0.109 0.093 0.131 0.114 0.6 11.5 55.14 0.087 0.6 0.9 1.077 0.095 0.8 7 0 0 0 0 0 0 0 0 0 0 0 0 … my' [%] 2 0. Lang Mode T [-] [s] 1 0.144 5 0.106 18 0.5 1 0 0 0 0 0 0 0 2.5 55.095 21 0.302 2 0.Modal analysis results: Effective masses  those modes shall be considered for which: mi ≥ 0.077 … … Sum  mx' [%] 1.117 13 0.8 … .131 9 0.9 1.122 11 0.3 53.078 29 0.093 23 0.6 0 5 0 0 0 0 0 0.1 0 0 0 0 1.7 0 0.8 7 0 0 0 0 0 0 0 0 0 0 0 0 … 77.183 3 0.7 0 0 0 0 0 0 0 0 0 0 0 0 11 4.142 6 0.118 12 0.092 24 0.084 26 0.083 27 0.2 10. Meslem & D.1 19 0.2 0.6 11.8 0 0 0 0.123 10 0.9 0 3.(2) Response Spectrum analysis: EN 1998-1:2004 Tutorial 2.138 8 0.6 0.2 .8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … 76.11 16 0.9  mtot or mi ≥ 0.05  mtot A.7 1.3 2.111 15 0.114 14 0.094 22 0.14 7 0.096 20 0.5 © NORSAR – Kjeller (Norway) 2014 mz' [%] 0 12.2 Example 1.077 30 0.2 my' [%] 2 0.109 17 0.15 4 0.082 28 0.2 0 13.087 25 0. (2) Response Spectrum analysis: EN 1998-1:2004 Tutorial 2.2 Example 1.2 - Modal analysis results: Effective masses  since both criteria are not fulfilled, those modes shall be considered for which: k ≥ 3  √n and Tk ≤ 0.20 sec  k ≥ 3  √4 = 6 modes A. Meslem & D. Lang Mode [-] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 … T [s] 0.302 0.183 0.15 0.144 0.142 0.14 0.138 0.131 0.123 0.122 0.118 0.117 0.114 0.111 0.11 0.109 0.106 0.1 0.096 0.095 0.094 0.093 0.092 0.087 0.084 0.083 0.082 0.078 0.077 0.077 … mx' [%] 1.7 0 0 0 0 0 0 0 0 0 0 0 0 11 4.3 53.2 0.8 7 0 0 0 0 0 0 0 0 0 0 0 0 … my' [%] 2 0.5 55.3 2.6 11.6 0 5 0 0 0 0 0 0.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … © NORSAR – Kjeller (Norway) 2014 mz' [%] 0 12.6 0.9 0 3.2 0 13.1 0 0 0 0 1.2 10.7 1.7 0 0.8 0 0 0 0.5 1 0 0 0 0 0 0 0 2.9 1.8 … (2) Response Spectrum analysis: EN 1998-1:2004 Example 2.1 – 3-Story RC Frame System  Define Masses Seismic...G + 0.3 ∙ Q A. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 SAP2000 (2) Response Spectrum analysis: EN 1998-1:2004 Example 2.1 – 3-Story RC Frame System  Define Number of Modes select the number of modes to be considered A. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 SAP2000 1 – 3-Story RC Frame System  Run Model Analysis A.(2) Response Spectrum analysis: EN 1998-1:2004 Example 2. Lang © NORSAR – Kjeller (Norway) 2014 SAP2000 . Meslem & D. 1 – 3-Story RC Frame System SAP2000  Deformed shape. Mode 1 A..(2) Response Spectrum analysis: EN 1998-1:2004 Example 2.. Lang Mode 2 © NORSAR – Kjeller (Norway) 2014 Mode 3 . Meslem & D.. (2) Response Spectrum analysis: EN 1998-1:2004 SAP2000 Example 2.90 A...1 – 3-Story RC Frame System  Modal Information.98 . Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 0.. first torsional mode is 3rd Σ = 0. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 SAP2000 .2 – 3-Story RC Dual System first eigenmode is torsional A.(2) Response Spectrum analysis: EN 1998-1:2004 Example 2. Lang © NORSAR – Kjeller (Norway) 2014 0.83 A.2 – 3-Story RC Dual System first eigenmode is torsional Σ = 0.74 .(2) Response Spectrum analysis: EN 1998-1:2004 SAP2000 Example 2. Meslem & D. a response is read from the design spectrum.(2) Modal Response spectrum method Steps: Step 1: for each mode of vibration. story shears and base reactions) to obtain the total response of the structure (total response at each floor). Step 3: modal combination of the resulted peak response quantities (e. Lang © NORSAR – Kjeller (Norway) 2014 . based on the modal frequency and the modal mass. for each floor.g. A. Meslem & D. displacements. story forces. Lang Z  I S a Ti   2 R g © NORSAR – Kjeller (Norway) 2014 n.d (T3) T1 T2 T3 Period T [sec] .2 j.2 Sa.1  n.d (T2) Sa.d (T1) Sa.3 j.1 3 j+1. 7.4 Procedure: Mode shape i: 1 n. Meslem & D.(2) Modal Response spectrum method IS 1893-1:2002  IS 1893-1:2002.8.3 j+1.2 Spectral acceleration Sa  Design spectral accelerations Sa(Ti )/g for each mode i : 2 j+1.1 Design seismic coefficient for each mode i: Ai  A.3 j. i) at floor j in mode i is given by: Q j .8.1 Q j.3 Q j+1.4 Qn.(2) Modal Response spectrum method IS 1893-1:2002 Procedure:  Design lateral force at each floor in each mode – the peak lateral force (Q j.2 Q j+1. 7.2 Q j.1 Q j+1.1  IS 1893-1:2002.3 n where A. Lang © NORSAR – Kjeller (Norway) 2014 Pk  W  i 1 n i ik Wi  ik  i 1 2 .2 Qn.i  Pi  W j Qn.i  Ai   j .3 Q j. Meslem & D. i  IS 1893-1:2002.8. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 .4 Story shear forces due to all modes considered– the peak story shear force (Vj) in story j due to all modes considered is obtained by combining those due to each mode A.(2) Modal Response spectrum method IS 1893-1:2002 Procedure:  Story shear forces in each mode – the peak shear force (Vj.i   n Q j i 1 j .i) acting in story j in mode i is given by: V j . 7. Member forces.(2) Modal Response spectrum method IS 1893-1:2002 Procedure: Modal Combination  the peak response quantities (e. A. story forces. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014  IS 1893-1:2002. Response quantity in mode i (including sign). displacements. 7.7 . Response quantity in mode j (including sing). Cross-modal coefficient. story shears and base reactions) shall be combined as per Complete Quadratic Combination (CQC) method (here the modes are assumed to be closely-spaced):  r r     i 1 j 1 r  ij i j i ij j Number of modes being considered.g. 4 . Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014  IS 1893-1:2002. Froof and Fj.8. 7.7 Lateral Forces at each Story due to all modes considered  Lateral forces at each story due to all modes considered – the design lateral forces. then the peak response quantities due to all modes considered shall be combined using the following expresssion:  r    k 1 2 k  IS 1893-1:2002. 7.(2) Modal Response spectrum method IS 1893-1:2002 Procedure: Modal Combination  the If the buildings does not have closely-spaced modes. at roof and at floor j: Froof  Vroof F j  V j  V j 1 A. Meslem & D. 7.i) acting in story j in mode i is given by: V j . 7.4  Story shear forces due to all modes considered– the peak story shear force (Vj) in story j due to all modes considered is obtained by combining those due to each mode  Lateral forces at each story due to all modes considered – the design lateral forces. Froof and Fj.(2) Modal Response spectrum method IS 1893-1:2002 Procedure:  Story shear forces in each mode – the peak shear force (Vj.4 .8. Lang © NORSAR – Kjeller (Norway) 2014  IS 1893-1:2002.i  n Q j i 1 j .8. at roof and at floor j: Froof  Vroof F j  V j  V j 1 A.i  IS 1893-1:2002. 3 j.d (T1) Sa.(2) Modal Response spectrum method EN 1998-1:2004  EN 1998-1:2004.2 Spectral acceleration Sa Design spectral accelerations Sa(Ti )/g for each mode i : 2 j+1. Meslem & D.3 Procedure: Mode shape i: 1 n. Lang n.2 T2 T3 Period T [sec] .1 © NORSAR – Kjeller (Norway) 2014 j.1  n.3 j+1.3.2 j.d (T3) T1 A. 4.1 3 j+1.3 Sa.d (T2) Sa.3. d (Ti ) Mode shape i: n.(2) Modal Response spectrum method EN 1998-1:2004 Procedure: Fj.2 Fj+1. Lang j.1 3 j+1.3 j.1  2 n.2 Fn.2 Fn.2 Fn. Meslem & D.i  m j   j.3 Fj+1.3 .3 n  i1 Fb2.3 j+1.3 Fj.1 resulting shear forces Fb.2 Fj.m  n.i © NORSAR – Kjeller (Norway) 2014 Fj+1.1 j.i   i  S a.1 1 A.2 Fj.3.m .m : Fb .3 j+1.3. 4.1  EN 1998-1:2004. 1  108 kN/m2 I = 2.0 m k = 12  EI/h3 m = 50 tons = 50 kNs2/m m3 = m h m2 = 1.3 g residential use: γI = 1. Meslem & D.679  10-5 m4 h = 3.0 structural parameters: E = 2. Setting up the differential equation of motion: M u  C u  K u  0 0   m1 0 2 0 0     M   0 m2 0   m   0 1.5m h m1 = 2m h k3 = k k2 = 2k k1 = 3k 1. Lang if [C] = 0 : M u  K u  0  k2  k1  k2  K    k 2 k 2  k 3  0  k3  © NORSAR – Kjeller (Norway) 2014 0   5 2 0      k3   k    2 3  1  0 1 1  k 3    .5 0   0 0 0 1 0 m3     A.3 3-story RC frame building (residential use) behavior factor q = 4 ground motion: agR = 0.(2) Response spectrum analysis: EN 1998-1:2004 Tutorial 2. (2) Response spectrum analysis: EN 1998-1:2004 Tutorial 2.30    1    0.676    2     0.00    0  2.5m 2 k  0 k k  m 2 3. Meslem & D.601   1.97 s-1 3 = 13. Lang   0. Eigenmodes:  0. Modal segmentation: K   2  M  0  5k  2m 2  2k 0  2k 3k  1.644   1.70 sec T3 = 0. Modal circular frequencies ωi and periods Ti : 1 = 4.47    3     2.57   1.50 sec T2 = 0.00    A.3 s-1 → → → T1 = 1.47 sec 4.00    © NORSAR – Kjeller (Norway) 2014 .3 2.19 s-1 2 = 8. 0 = 0.511 → 3 = 104.601 + 50  1.3 + 75  0.3 5.0 kNs2/m M2* = 100  0.3 / 1155.7 / 122.426 → 2 = -62.0 kNs2/m → 1 = 128.6442 + 50  1.472 + 75  2.02 = 122.0 = -62. Meslem & D.02 = 1155.(2) Response spectrum analysis: EN 1998-1:2004 Tutorial 2.57 + 50  1. Modal participation factors i : n i  m  j .6762 + 75  0.0 = 128.47 – 75  2.8 = –0.32 + 75  0.i m  2 j .0 = 104.02 = 90. Lang © NORSAR – Kjeller (Norway) 2014 .6012 + 50  1.8 kNs2/m M3* = 100  2.090 A.3 kNs2/m 2 = –100  0.i j1 n j1 j j  i Mi* 1 = 100  0.0 = 1.3 / 90.572 + 50  1.676 – 75  0.7 kNs2/m 3 = 100  2.644 + 50  1.3 kNs2/m M1* = 100  0. 5  0.0  1.5  (2.5 2.47 sec : A.943  1. Lang Sa.d (T) = 1.5  TC  2.0)  1.5  TC  2.20 ∙ 2.20 ∙ 2. Design spectral accelerations Sa(Ti ) for each mode i : T1 = 1.943  1.0)  1.15   2.115 m / s2 q 4.15    0 .943 = 0.0)  1.d (T) = 0.15      1.d (T)  ag  S  2.0  0.d (T)  ag  S  2.5886 m/s2 T2 = 0.846 m/s2 ≥ β ∙ ag = 0.70 sec : Sa. 846 m / s q  T1  4.3 6.943 = 0.7  Check: Sa.943  1.6  2     (2.813 m/s2 ≥ β ∙ ag = 0.813 m / s2 q  T1  4.(2) Response spectrum analysis: EN 1998-1:2004 Tutorial 2.50  Check: Sa.5  0.5886 m/s2 T3 = 0.6      (2.50 sec : Sa.d (T)  ag  S  2.0 © NORSAR – Kjeller (Norway) 2014 . Meslem & D. 115 = 47.8 F1.1 = 100  0.47  0.8 kN F3. Meslem & D.3 kN F2.115 = 9.1 = 58.00  0.846 = 60.115 = –36.7 kN F3.3 = 9.2 kN F2.3 F1.2 = 50  1.2 F3.3 kN F3.3 5.3 kN F2.5 F1.6 kN F2.i  m j   j.5 kN A.511)  1.511)  1.2 = 75  (–0.1 = 36.3 = 50  1.2 = 41.676)  (–0.57)  0.2 = 100  (–0. Lang © NORSAR – Kjeller (Norway) 2014 F2.3 = 100  2.3 F1.3 F1.3 = 47.2 = 62.3 = 75  (–2.846 = 58.7 F1.426  0.0 .(2) Response spectrum analysis: EN 1998-1:2004 Tutorial 2.i   i  S a.426  0.1 = 50  1.090  2.d (Ti ) F3.1 = 75  0.813 = 62.3 = –36.601)  (–0.00  1.846 = 36.i : Fj.0 kN F2.813 = –46.644  1. Lateral story loads Fj.30  1.1= 60.813 = 41.6 F3.2 = –46.00  (–0.090  2.511)  1.426  0.090  2. 8  -27.1 76.(2) Response spectrum analysis: EN 1998-1:2004 Tutorial 2.6 58.5 Fb .i .2 -4.5 118.m .3 60.3 154.m  A. Meslem & D. Maximum shear forces Fb : 9.3 5. Lang © NORSAR – Kjeller (Norway) 2014 n F i1 2 b .7 19.8 166.6 121.5 -46. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 SAP2000 .(2) Modal Response spectrum analysis SAP2000 – 3-Story RC Frame System  Define Response Spectrum to be used A. Lang © NORSAR – Kjeller (Norway) 2014 . Meslem & D.(2) Modal Response spectrum analysis SAP2000 – 3-Story RC Frame System SAP2000  Define Response Spectrum to be used Acceleration is in g unit we can move the cursor on the grave to obtain the coordinartes at any point A. (2) Modal Response spectrum analysis SAP2000 – 3-Story RC Frame System  Define Load case to include the Response Spectrum Analysis A. Lang © NORSAR – Kjeller (Norway) 2014 SAP2000 . Meslem & D. . Response spectrum will be applied as an acceleration in U1 (UX) direction using the previously defined curve EC-8-B A. Meslem & D..(2) Modal Response spectrum analysis SAP2000 – 3-Story RC Frame System SAP2000  Define Load case to include the Response Spectrum Analysis A number of ways to combine modes given direction including CQC.. SRSS.. Lang © NORSAR – Kjeller (Norway) 2014 .and others. Lang © NORSAR – Kjeller (Norway) 2014 SAP2000 . Meslem & D.(2) Modal Response spectrum analysis SAP2000 – 3-Story RC Frame System  Run Analysis A. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 SAP2000 .(2) Modal Response spectrum analysis SAP2000 – 3-Story RC Frame System  Display Base Reactions for the Response Spectrum (RS) case A. for the Response Spectrum (RS) case A. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 .(2) Modal Response spectrum analysis SAP2000 – 3-Story RC Frame System SAP2000  Moments. Shear Forces. Axial Forces... Meslem & D..for the Response Spectrum (RS) case A. Shear Forces. Axial Forces. Lang © NORSAR – Kjeller (Norway) 2014 ..(2) Modal Response spectrum analysis SAP2000 – 3-Story RC Frame System SAP2000  Moments. shall be based on an appropriate ground motion and shall be performed using accepted principles of dynamics. Lang © NORSAR – Kjeller (Norway) 2014 .  The result of a response spectrum analysis using the response spectrum from a ground motion is typically different from that which would be calculated directly from a linear dynamic analysis using that ground motion directly. since phase information is lost in the process of generating the response spectrum.(3) Linear time history analysis  Linear time history method of analysis. A. Meslem & D. when used. story forces.4 . A. the resulted design base shear (VB) shall be compared with a base shear (𝑉𝐵) calculated using a fundamental period TB.g. Lang © NORSAR – Kjeller (Norway) 2014  IS 1893-1:2002. • Irregular buildings – all framed buildings heigher than 12 m in Zones IV and V. displacements. Meslem & D.8. all the response quantities (e. and those greater than 40 m in height in Zones II and III.(3) Linear time history analysis IS 1893-1:2002   shall be performed for the following buildings: • Regular buildings – those greater than 40 m in height in Zone IV and V. 7. story shears and base reactions) shall be multiplied by 𝑉𝐵 /VB. and those greater than 90 m in height in Zones II and III. • Where VB is less than 𝑉𝐵 . Member forces. Lang © NORSAR – Kjeller (Norway) 2014 . Meslem & D.(3) Linear time history analysis SAP2000 – 3-Story RC Frame System SAP2000  Define ground motion to be used Linear Time History analysis in UX direction (LTH_UX) A. (3) Linear time history analysis SAP2000 – 3-Story RC Frame System SAP2000  Define ground motion to be used Linear Time History analysis in UX direction (LTH_UX) A. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 . Lang © NORSAR – Kjeller (Norway) 2014 .(3) Linear time history analysis SAP2000 – 3-Story RC Frame System SAP2000  Define ground motion to be used Linear Time History analysis in UX direction (LTH_UX) A. Meslem & D. Combination of effects of seismic action When the analysis is conducted the actions of the orthogonal components of ground motions shall be combined using approximate equations.Analysis in Y direction Compute Ey A. a) Horizontal components: Step 1: Compute the structural response of the structure under each component separately . Lang © NORSAR – Kjeller (Norway) 2014 – Analysis in X direction – Compute Ex . Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 . and EL stand for the response quantities due to Dead Load.7  DL  EL  3) 1.7  DL  IL 2) 1. A.9  DL  1. IL.5  DL  EL  4) 0.3 1) 1.Combination of effects of seismic action: ISN 1893-1:2002 Load factors for design of steel structures: The following combination shall be accounted for:  IS 1893-1:2002. Imposed Load and Earthquake Load. Meslem & D.5  EL The terms DL.2  DL  IL  EL  3) 1. 6.3  DL  IL  EL  Load factors for design of reinforced concrete and prestressed concrete structures: The following combination shall be accounted for: 1) 1.5  DL  IL 2) 1. Lang © NORSAR – Kjeller (Norway) 2014 . The building should be deisgned for: 1) 1. 6.7  DL  EL x  3) 1.3 Design horizontal earthquake load :  When the lateral load resisting elements are oriented along orthogonal horizontal direction.3  DL  IL  EL x  A. Example: Case of steel building.Combination of effects of seismic action: ISN 1893-1:2002  IS 1893-1:2002.7  DL  IL 2) 1. Meslem & D. and where lateral load resisting elements are oriented along UX direction. the structure shall be designed for the effects due to full design earthquake load in one horizontal direction at time. Meslem & D. and where lateral load resisting elements are not oriented along UX direction. Lang © NORSAR – Kjeller (Norway) 2014 . 6.3 Design horizontal earthquake load :  When the lateral load resisting elements are not oriented along orthogonal horizontal direction.7  DL  IL   3) 1.7  DL  EL x  0.3  EL y  2) 1. the structure shall be designed for the effects due to full design earthquake load in one horizontal direction PLUS 30% of the design earthquake load in the other direction Example: Case of steel building.3  DL  IL  EL x  0.3  EL y A. The building should be deisgned for: 1) 1.Combination of effects of seismic action: ISN 1893-1:2002  IS 1893-1:2002. 3ELx  0.Combination of effects of seismic action: ISN 1893-1:2002 Combination for two or three component motion:  When responses from the three earthquake components are to be considered. the responses due to each component may be combined using the assumption that when response from one component are 30% of their maximum. Meslem & D. Lang ELx 2  ELy 2  ELz 2 © NORSAR – Kjeller (Norway) 2014  IS 1893-1:2002.3ELx  EL y  0.3 .3ELz 3)  0.3EL y  ELz  Alternatively. 6.3EL y  0.3ELz 2)  0.  The response due earthquake force (EL) is the maximum of the following three cases: 1)  ELx  0. the response (EL) due to the combined effect of the three components can be obtained on the basis of Square Root of the sum of the Square (SRSS): EL  A. 3Ex Exception: For buildings satisfying the regularity criteria in plan and in which walls or independent bracing systems in the two main horizontal directions are the only primary seismic elements. the seismic action may be assumed to act separately and without combinations.3. 4. A. the response quantities can be combined as: E  Ex  0.3. Lang © NORSAR – Kjeller (Norway) 2014 .Combination of effects of seismic action: EN 1998-1:2004  EN 1998-1:2004.3E y E  E y  0. Meslem & D.5 a) Horizontal components: Step 2: Combine the response quantities using SRSS method: E  E2x  E2y Alternatively. 3Ex  0.3.3Ez E  0. 4. the vertical of the seismic action should be taken into account for the following cases: • for horizontal or nearly horizontal structural members spanning 20 m or more.Combination of effects of seismic action: EN 1998-1:2004  EN 1998-1:2004. • for horizontal or nearly horizontal pre-stressed components • for beams supporting columns • in base-isolated structures The analysis for determining the effects of the vertical component of the seismic action may be based on a partial model of the structure. • for horizontal or nearly horizontal cantilever components longer than 5 m.25g.3Ez A.3Ex  E y  0.3.5 b) Vertical components: If avg is greater than 0. Lang © NORSAR – Kjeller (Norway) 2014 E  0.3E y  0. The effects of 2 horizontal and vertical components will be combined by: E  Ex  0. which includes the aforementioned elements.3E y  Ez . Meslem & D. A. Lang © NORSAR – Kjeller (Norway) 2014 .Accidental/Torsional Effects Torsional effects created in a simple building configuration: Torsion is occuring because a uniformly distributed force is not resisted by a uniformly distributed lateral resistant. Meslem & D. 15627 RZ = 0. Lang Mode 1 UX = 0.144 UY = 0 RZ = 0.73509 UY = 0 RZ = 0 © NORSAR – Kjeller (Norway) 2014 .61726 Mode 1 UX = 0 UY = 0 RZ = 0. Meslem & D.08906 UY = 0.54874 Mode 1 UX = 0.7735 Mode 1 UX = 0.Accidental/Torsional Effects SAP2000 – 3-Story RC Dual System Modal Participating Mass Ratios A. lateral force acting on story i in direction j Mdi . edi to be used at floor i shall be taken as: 1. Meslem & D.9 bi .05  bi edi   or esi  0. Lang © NORSAR – Kjeller (Norway) 2014 bi x .Accidental/Torsional Effects IS 1893-1:2002  the design eccentricity.5  esi  0. j Qi.torsional moment applied at story i about its vertical axis z edi direction of seismic action x M z A. esi – Static eccentricity at floor i defined as the distance between centre of mass and centre of y rigidity. M di  edi  Qi . perpendicular to seismic action.Floor plan dimension of floor i. 7.05  bi  IS 1893-1:2002.j . Meslem & D. A. additive shears will be superimposed for a statically applied eccentricity of ±0.05bi.9 m4 In case of highly irregular buildings modeled as a system of lumped masses at the floor levels (with each mass having one degree of freedom). 7. with respect to the centre of rigidity. Lang © NORSAR – Kjeller (Norway) 2014 m3 m2 m1 .Accidental/Torsional Effects IS 1893-1:2002 Irregular Buildings  IS 1893-1:2002. j z Fi. y direction of seismic action x Ly M Mai = eai. width) perpendicular to seismic action eai.torsional moment applied at story i about its vertical axis z A. Lang © NORSAR – Kjeller (Norway) 2014 x .2 (1) Spatial models (3D): displace the theoretical center of mass M at story i by an accidental eccentricity eai for both directions of seismic motion/general building axes j of the structural model y eai.lateral force acting on story i in direction j Mai . 4.05 ∙ Li Li .floor dimension (length.j ∙ Fi.Accidental/Torsional Effects EN 1998-1:2004  to account for torsional effects predominantly depends on the model type (planar or spatial)  EN 1998-1:2004.j = 0. Meslem & D.j .3. 4.6  x )  Fi Le direction of seismic action Ly M → if two planar models are used: (1) increase accidental eccentricity eai by a factor of 2 or (2) double the factor d. Lang © NORSAR – Kjeller (Norway) 2014 Fi  d  Fi  (1  1.3.2.Accidental/Torsional Effects EN 1998-1:2004  to account for torsional effects predominantly depends on the model type (planar or spatial) (2) Planar models (2D):  EN 1998-1:2004. Meslem & D. so that: A. torsion cannot be considered in planar models → to overcome this.3.2  x )  Fi Le .4 → theoretically. action effects for each individual lateral force-resisting element are increased by a factor d : x Fi  d  Fi  (1  0. Lang © NORSAR – Kjeller (Norway) 2014 .Second-order Effects (P-Δ effects)  Structures in real life are flexible and can exhibit large lateral displacements in unusual circumstances. Meslem & D.  Gravity loading will influence structural response under significant lateral displacement.  P-Δ may contribute to loss of lateral resistance. A. and dynamic instability. The lateral displacements can be caused by wind or seismically induced inertial forces. ratcheting of residual deformations. Lang © NORSAR – Kjeller (Norway) 2014 .3.2.4. A.4. dr = is the design interstorey drift.Second-order Effects (P-Δ effects) EN 1998-1:2004  Second-order effects (P-∆ effects) need not be taken into account if the following condition is fulfilled in all storeys:  Ptot  d r  0. and h = is the interstorey height. evaluated as the difference of the average lateral displacements ds at the top and bottom of the storey under consideration and calculated in accordance with Chapter 4. Vtot = is the total seismic storey shear.10 Vtot  h  EN 1998-1:2004.2  = is the interstorey drift sensitivity coefficient. 4. Ptot = is the total gravity load at and above the storey considered in the seismic design situation. Meslem & D.  value of the coefficient θ shall not exceed 0. Meslem & D.3 A.2.1 < θ≤0.2. Lang © NORSAR – Kjeller (Norway) 2014 .4.Second-order Effects (P-Δ effects) EN 1998-1:2004  Second-order effects (P-∆ effects) need not be taken into account if the following condition is fulfilled in all storeys:  Ptot  d r  0. 4.θ). the second-order effects may approximately be taken into account by multiplying the relevant seismic action effects by a factor equal to 1/(1 .2  If 0.10 Vtot  h  EN 1998-1:2004. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 .Second-order Effects (P-Δ effects) Use P-Delta in SAP2000 A. Lang © NORSAR – Kjeller (Norway) 2014 .Second-order Effects (P-Δ effects) To mitigate second-order effects: two-story X-bracing or zipper columns are recommended A. Meslem & D. Meslem & D. • The duration of the accelerograms shall be consistent with the magnitude and the other relevant features of the seismic event underlying the establishment of ag.2T1 and 2T1.S for the zone under consideration. • in the range of periods between 0.Select and Scale Earthquake Records EN 1998-1:2004 The suite of recorded or simulated/artificial accelerograms should observe the following rules: • A minimum of 3 accelerograms should be used. • no value of the mean 5% damping elastic spectrum. where T1 is the fundamental period of the structure in the direction where the accelerogram will be applied. A. • The values are scaled to the value of ag. calculated from all time histories. Lang © NORSAR – Kjeller (Norway) 2014 . should be less than 90% of the corresponding value of the 5% damping elastic response spectrum. Select and Scale Earthquake Records EN 1998-1:2004 The parameters (that have the most influence on ground motion spectral shape) that need to be considered in selecting records : • Magnitude range of anticipated significant event; • Distance range of the site from the causative fault; • Site Condition (i.e. looking at the average shear velocity); • Basin effect (if basin exists) A. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 Contribution of Joint Regions A. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 Contribution of Joint Regions SAP2000 – 3-Story RC Frame System not considered considered A. Meslem & D. Lang © NORSAR – Kjeller (Norway) 2014 no Phone: (+47) 974 10 740 Web: www.no .Abdelghani Meslem. Kjeller abdelghani. PhD [email protected].