Lecture-09 Introduction to Earthquake Resistant Analysis & Design of RC Structures (Part I) By: Prof Dr. Qaisar Ali Civil Engineering Department UET Peshawar www.drqaisarali.com 1 Topics Introduction Earthquake Design Philosophy Analysis for Seismic Loads References 2 1
Earth s Interior Introduction 3 Introduction Earthquake results from the sudden movement of the tectonic plates in the earth s crust. 4 2
Effect of Earthquake Introduction The movement, taking place at the fault lines, causes energy release which is transmitted through the earth in the form of waves. These waves reach the structure causing shaking. 5 Introduction Seismic Events around the globe Mostly takes place at boundaries of Tectonic plates Dots represents an earthquake 6 3
Introduction Types of Waves Generated Due to Earthquake Body Waves Surface Waves 7 Introduction Displacement due to Earthquake 8 4
Introduction Horizontal and Vertical Shaking Earthquake causes shaking of the ground in all three directions. The structures designed for gravity loading (DL+LL) will be normally safe against vertical component of ground shaking. The vertical acceleration during ground shaking either adds to or subtracts from the acceleration due to gravity. 9 Introduction Horizontal and Vertical Shaking The structures are normally designed for horizontal shaking to minimize the effect of damages due to earthquakes. 10 5
Introduction Earthquake Types with respect to Depth of Focus Shallow Depth of focus varies between 0 and 70 km. Deep Depth of focus varies between 70 and 700 km. 11 Introduction Earthquake characteristics with respect to distance from epicenter 1.0 T 10 sec 1 Hz f 0.1 Hz Near Field: 0 to 25 km Intermediate Field: 25 to 50 km Far Field: Beyond 50 km 0.3 T 1.0 sec 3.33 Hz f 1 Hz 0.05 T 0.3 20 Hz f 3.33 Hz 25 km Low period & high frequency field Epicenter Moderate period & low frequency field Large period & low frequency field 12 6
Introduction Resonance risk for structures w.r.t near, intermediate and far field earthquakes The natural time period of a structure is its important characteristic to predict behavior during an earthquake of certain time period (Resonance phenomenon). For a particular structure, the natural time period is a function of mass and stiffness {T = 2p (m/k)} T can be roughly estimated from: T = 0.1 number of stories 13 Introduction Resonance risk for structures w.r.t near, intermediate and far field earthquakes High rise Structure (Above 5 stories) Medium rise Structure (upto 5 stories) Epicenter Low rise Structure (upto 3 stories) 14 7
Introduction Earthquake Recording Seismograph Using multiple seismographs around the world, accurate location of the epicenter of the earthquake, as well as its magnitude or size can be determined. Working of seismograph shown in figure. 15 Introduction Earthquake Recording Richter Scale In 1935, Charles Richter (US) developed this scale. The Richter scale is logarithmic, So, a magnitude 5 Richter measurement is ten times greater than a magnitude 4; while it is 10 x 10, or 100 times greater than a magnitude 3 measurement. 16 8
Introduction Earthquake Recording Some of the famous earthquake records 17 Introduction Earthquake Occurrence 18 9
Introduction Seismic Zones 19 Introduction Importance of Architectural Features The behavior of a building during earthquakes depend critically on its overall shape, size and geometry, in addition to how the earthquake forces are carried to the ground. 20 10
Introduction Importance of Architectural Features At the planning stage, architects and structural engineers must work together to ensure that the unfavorable features are avoided and a good building configuration is chosen. 21 Introduction Other Undesirable Scenarios 22 11
Introduction Soft Storey 23 Earthquake Design Philosophy Performance level 24 12
Building Code of Pakistan In Pakistan, the design criteria for earthquake loading are based on design procedures presented in chapter 5, division II of Building Code of Pakistan, seismic provision 2007 (BCP, SP 2007), which have been adopted from chapter 16, division II of UBC-97 (Uniform Building Code), volume 2. 25 Lateral Force Determination Procedures The total design seismic force imposed by an earthquake on the structure at its base is referred to as base shear V in the UBC-97. The design seismic force can be determined based on: Dynamic lateral force procedure [sec. 1631, UBC-97 or sec. 5.31, BCP-2007]. Static lateral force procedure [sec. 1630.2, UBC-97 or Sec. 5.30.2, BCP 2007], 26 13
Dynamic Lateral Force Procedure UBC-97 section 1631 include information on dynamic lateral force procedures that involve the use of: Time history analysis. Response spectrum analysis. The details of these methods are presented in sections 1631.5 and 1631.6 of the UBC-97. 27 Dynamic Lateral Force Procedure Time History Analysis (THA) Lateral Displacement Ground acceleration T T 28 14
Dynamic Lateral Force Procedure a (ft/sec 2 ) Response Spectrum Analysis (RSA) Response D 1 a (ft/sec 2 ) T T s1 = 0.3 sec Response D 2 T Peak Response (T s1,d 1 ) (T s2,d 2 ) a (ft/sec 2 ) T T s2 = 1.0 sec Response T (T s3,d 3 ) T T T s3 = 2.0 sec D 3 Structural Time period 29 Dynamic Lateral Force Procedure Response Spectrum Analysis (RSA) UBC-97 Response Spectrum Curve (Acceleration vs. Time period) 30 15
Static Lateral Force Procedure = 31 Static Lateral Force Procedure The total design base shear (V) in a given direction can be determined from the following formula: Where, V = (C ν I/RT) W C ν = Seismic coefficient (Table 16-R of UBC-97). I = Seismic importance factor (Table 16-K of UBC-97 ) R = numerical coefficient representative of inherent over strength and global ductility capacity of lateral force-resisting systems (Table 16-N or 16-P). W = the total seismic dead load defined in Section 1630.1.1. 32 16
Static Lateral Force Procedure The total design base need not exceed [ V = (2.5C a I/R) W ] Where, C a = Seismic coefficient (Table 16-Q of UBC-97) The total design base shear shall not be less than [ V = 0.11C a IW ] In addition for seismic zone 4, the total base shear shall also not be less [ V = (0.8ZN ν I/R) W ] Where, N ν = near source factor (Table 16-T of UBC-97); Z = Seismic zone factor (Table 16-I of UBC-97) 33 Static Lateral Force Procedure Steps for Calculation of V : Step 1: Find Site Specific details. Step 2: Determine Seismic Coefficients Step 3: Determine Seismic Importance factor I Step 4: Determine R factor Step 5: Determine structure s time period Step 6:Determine base shear V and apply code maximum and minimum. Step 7: Determine vertical distribution of V. 34 17
Static Lateral Force Procedure Steps for Calculation of V : Step 1: Find Site Specific details. Following list of data needs to be obtained: Seismic Zone Soil type Past earthquake magnitude (required only for highest seismic zone). Closest distance to known seismic source (required only for highest seismic zone). 35 Static Lateral Force Procedure Steps for Calculation of V : Step 1: Site Specific details. i. Seismic Zone Source: BCP SP-2007 36 18
Static Lateral Force Procedure Steps for Calculation of V : Step 1: Find Site Specific details. ii. Soil Type As per UBC code, if soil type is not known, type S D shall be taken. 37 Static Lateral Force Procedure Steps for Calculation of V : Step 1: Find Site Specific details. iii. Past Earthquake magnitude: This is required only for seismic zone 4 to decide about seismic source type so that certain additional coefficients can be determined. iv. Distance to known seismic zone is also required to determine additional coefficients for zone 4. 38 19
Static Lateral Force Procedure Steps for Calculation of V : Step 2: Determination of Seismic Coefficients. C v : N v (required only for zone 4): 39 Static Lateral Force Procedure Steps for Calculation of V : Step 2: Determination of Seismic Coefficients. C a : N a (required only for zone 4): 40 20
Static Lateral Force Procedure Steps for Calculation of V : Step 3: Determination of Seismic Importance Factor. 41 Static Lateral Force Procedure Steps for Calculation of V : Step 4: Determination of R Factor. R factor basically reduces base shear V to make the system economical. However the structure will suffer some damage as explained in the earthquake design philosophy. R factor depends on overall structural response of the structure under lateral loading. For structures exhibiting good performance, R factor will be high. 42 21
Static Lateral Force Procedure Steps for Calculation of V : Step 4: Determination of R Factor. 43 Static Lateral Force Procedure Steps for Calculation of V : Step 5: Determination of structure s time period. Structural Period (By Method A, UBC 97): For all buildings, the value T may be approximated from the following formula: T = C t (h n ) 3/4 Where, C t = 0.035 (0.0853) for steel moment-resisting frames. C t = 0.030 (0.0731) for reinforced concrete moment-resisting frames and eccentrically braced frames. C t = 0.020 (0.0488) for all other buildings. h n = Actual height (feet or meters) of the building above the base to the nth level. 44 22
Static Lateral Force Procedure Steps for Calculation of V : Step 6: Determination of Base Shear (V). Calculate base shear meeting the following criteria: 0.11C a IW V = (C ν I/RT) W (2.5C a I/R) W For zone 4, the total base shear shall also not be less than: V = (0.8ZN ν I/R) W 45 Static Lateral Force Procedure Steps for Calculation of V : Step 7: Vertical Distribution of V to storeys. The joint force at a particular level x of the structure is given as: F x = (V F t )ω x h x / ω i h i (UBC sec. 1630.5) { i ranges from 1 to n, where n = number of stories } F t = Additional force that is applied to the top level (i.e., the roof) in addition to the F x force at that level. F t = 0.07TV {for T > 0.7 sec} F t = 0 {for T 0.7 sec} 46 23
Static Lateral Force Procedure Example: Calculation of V for E-W interior frame of the given structure. Structure is located in Peshawar. Soil type is stiff. SDL = Nil LL = 144 psf SDL = Nil LL = 144 psf f c = 4 ksi f y = 60 ksi 10 ft (floor to floor) SDL = Nil LL = 144 psf 25 ft 25 ft 10 ft 10 ft 25 ft 25 ft 20 ft 20 ft 20 ft Slab-Beam Frame Structure 47 Static Lateral Force Procedure Example: E-W interior frame 4 spans @ 25-0 l 2 = 20 3 spans @ 20-0 48 24
Static Lateral Force Procedure Example: Step 1: Site specific details. i. Seismic Zone: From seismic zoning map of Pakistan, Peshawar lies in seismic zone 2B (Z = 0.20) 49 Static Lateral Force Procedure Example: Step 1: Site specific details. ii. Soil Type: Stiff soil is classified as S D (stiff soil). 50 25
Static Lateral Force Procedure Example: Step 1: Site specific details. iii. Past earthquake magnitude: Not determined as it required for zone 4 only. iv. Distance to known seismic zone: Not determined as it is required for zone 4 only. 51 Static Lateral Force Procedure Example: Step 2: Determination of Seismic Coefficients. For seismic zone 2B, only C a and C v determination is required. 52 26
Static Lateral Force Procedure Example: Step 3: Determination of Seismic Importance Factor. I = 1.00 (Standard Occupancy Structures) 53 Static Lateral Force Procedure Example: Step 4: Determination of R Factor. R = 8.5 (Concrete SMRF) 54 27
Static Lateral Force Procedure Example: Step 5: Determination of Structure s time period. By method A: T = C t (h n ) 3/4 C t = 0.003; h n = 30 ft T = 0.003 (30) 3/4 = 0.384 sec 55 Static Lateral Force Procedure Example: Step 6: Determination of base shear (V). Base Shear (V) = {C v I/RT}W W (self weight of E-W interior frame + super imposed dead load) = 613 kips 25 % floor live load will also be added up (for warehouses, see UBC sec.1630.1.1.) W = 613 + 0.25 0.144 20 (4 25) = 685 kips 56 28
Static Lateral Force Procedure Example: Step 6: Determination of base shear (V). V = {C v I/RT}W = {0.40 1.00/ (8.5 0.384)} 685 = 83.94 kips The total design base need not exceed the following: V = (2.5C a I/R) W = {(2.5 0.28 1.00)/ (8.5)} 685 = 56.41 kips, The total design base shear shall not be less than the following: V = 0.11C a IW = 0.11 0.28 1.00 685= 21.098 kips, O.K. Therefore, V = 56.41 kip (8 % of seismic weight W) 57 Static Lateral Force Procedure Example: Step 7: Vertical distribution of V to storeys. F x = (V F t )ω x h x / ω i h i ω i h i = 228 10 + 228 20 + 228 30 = 13680 kip F 1 = (56.41 0) 228 10/ {(13680)} = 9.402 kip Storey forces for other stories are given in table below: Table Storey shears. Level x h x (ft) w x (kip) w x h x (ft-kip) w x h x /(Sw i h i ) F x (kip) 3 30 228 6840 0.5 28.21 2 20 228 4560 0.33 18.61 1 10 228 2280 0.166 9.36 Sw i h i = 13680 Check SF x =V = 56.18 kip OK 58 29
Static Lateral Force Procedure Example (Storey Forces): Same forces will be obtained for other E-W interior frame because it has same dimensions and loading conditions as of E-W interior frame considered. Half values shall be applied to E-W exterior frames. 59 Static Lateral Force Procedure Example (Storey Forces): 14.1 kips 28.21 kips Note: Base shear can also be computed for complete structure and then can be divided to different frames. 9.3 kips 18.61 kips 4.68 kips 9.36 kips 28.21 kips 14.1 kips 18.61 kips 9.3 kips 20 ft 20 ft 9.36 kips 4.68 kips 20 ft 25 ft 25 ft 25 ft 25 ft 60 30
Base Shear using UBC Response Spectra Spectral Acceleration (g s) Example: 2.5C a = 0.7 Line at T = 0.384 sec C v /T C a = 0.28 C v = 0.4 T s = C v /2.5C a = 0.57 sec T o = 0.2T s = 0.114 sec R = 8.5; W = 685 kips Now, T of given structure = 0.384 sec At T = 0.384 sec, Spectral acceleration = 0.7g V = W (a/g)/r = 56 kips C a = 0.28 0.114 0.57 Period (sec) Base shear computed here is same as computed using the static lateral force procedure. 61 Automated lateral force procedure of SAP2000 Steps for the given 3D structure are shown next. SAP2000 3D Model (20ft 15 ft) panels Seismic Zone: 2B Soil Type: SD Method A used for time period calculation Mass source: SDL only 62 31
Static Lateral Force Procedure 1. Automated Lateral Force Procedure of SAP2000 It is important to add SDL as Load for mass source with 3 rd option selected to avoid load to be taken two times. 63 Static Lateral Force Procedure 1. Automated Lateral Force Procedure of SAP2000 Case Study 2: Base shear calculation for E-W direction using SAP2000 automated lateral load feature and comparison with results of manually applied lateral loads. 64 32
Analysis for Seismic Loads Methods of Seismic (lateral load) Analysis Exact: FEM using SAP 2000, etc. This method was demonstrated in previous example Approximate lateral load analysis: This will be discussed next 65 ACI Requirements on Lateral Load Analysis Unlike ACI 6.4 which allows separate floor analysis for gravity loads, ACI R6.4 states that for lateral load analysis, a full frame from top to bottom must be considered. For Gravity Load For Lateral Load 66 33
Portal Method This is a method used to estimate the effects of side sway due to lateral forces acting on multistory building frame. This method is specialized form of point of inflection method. F 3 Side sway (Δ) F 2 F 1 67 Portal Method Prepositions: 1. The total horizontal shear in all columns of a given storey is equal and opposite to the sum of all horizontal loads acting above that storey. This preposition follows from the requirement that horizontal forces be in equilibrium at any level. F 3 H 31 H 32 H 31 + H 32 = F 3 F 2 H 21 H 22 H 21 + H 22 = F 3 + F 2 F 1 H 11 H 12 H 11 + H 12 = F 3 + F 2 + F 1 68 34
Portal Method Prepositions: 2. The horizontal shear is the same in both exterior columns. The horizontal shear in each interior column is twice that in exterior column. This preposition is due to the fact that interior columns are generally more rigid than exterior columns (interior column with larger axial load will require larger cross section). 6 H 3 = F 3 or H 3 = F 3 /6 H 3 = F 3 / 2n Where n= no. of bays And 2H 3 = F 3 / n F 3 F 2 F 1 H 3 2H 3 2H 3 H 2 2H 2 2H 2 H 3 H 2 H 1 2H 1 2H 1 H1 69 Portal Method Prepositions: 3. The inflection points of all members (columns and beams) are located midway between the joints except for bottom storey. F 3 Point of Inflection F 2 2h/3 F 1 h/3 Location of P.O.I depends on end restraints: 2h/3 (restraints with more resistance to rotation) h/3 (restraints with less resistance to rotation) At base (ideal hinge) 70 35
Portal Method Analysis Steps Step 1: Location of points of inflection on frame using preposition 3. F 3 F 2 F 1 l 1 l 2 l 3 71 Portal Method Analysis Steps Step 2: Determine column shears using proposition 1 and 2. F 3 H 3ext =F 3 /2n H 3int =F 3 /n H 3int =F 3 /n H 3ext =F 3 /2n F 2 H 2ext =(F 3 + F 2 )/2n H 2int =(F 3 + F 2 )/n H 2int =(F 3 + F 2 )/n H 2ext =(F 3 + F 2 )/2n F 1 H 1ext =(F 3 + F 2 + F 1 )/2n H 1int =(F 3 + F 2 + F 1 )/n H 1int =(F 3 + F 2 + F 1 )/n H 1ext =(F 3 + F 2 + F 1 )/2n n = number of bays l 1 l 2 l 3 72 36
Portal Method Analysis Steps Step 3a: Determine column moments from statics. F 3 M 3ext = H 3ext h/2 M 3int = H 3int h/2 M 3int = H 3int h/2 M 3ext = H 3ext h/2 F 2 F 1 H 3ext H 3int H 3int H 3ext M 3ext = H 3ext h/2 M 2ext = H 2ext h/2 H 2ext H 2int H 2int H 2ext M 2ext = H 2ext h/2 M 1ext = H 1ext h/3 H 1ext H 1int H 1int H 1ext M 1ext = H 1ext 2h/3 M 3int = H 3int h/2 M 2int = H 2int h/2 M 2int = H 2int h/2 M 1int = H 1int h/3 M 1int = H 1int 2h/3 M 3int = H 3int h/2 M 2int = H 2int h/2 M 2int = H 2int h/2 M 1int = H 1int h/3 M 1int = H 1int 2h/3 M 3ext = H 3ext h/2 M 2ext = H 2ext h/2 M 2ext = H 2ext h/2 M 1ext = H 1ext h/3 M 1ext = H 1ext 2h/3 h h h l 1 l 2 l 3 73 Portal Method Analysis Steps Step 3b: Determine beam moments from statics. Beam moments at a joint can be determined from equilibrium. The beam moments to the left (M BL ) and right (M BR ) of a joint can be determined from the following formulae: M BL = M col /m M BR = M col /m Where, m = # of connecting beams at a joint. M col = summation of column moments at a joint. 74 37
Portal Method Analysis Steps F 3 F 2 Step 3b: Determine beam moments from statics. M BL = M 3ext /1 M BL = M 3int /2 M 3ext M BR = M 3int /2 M 3int M3int M BR = (M 3int +M 2int )/2 M BL = (M 3int +M 2int )/2 M 2int Note: The direction of beam moment shall be opposite to the direction of column moment. F 1 l 1 l 2 l 3 75 Portal Method Analysis Steps Step 3c: Determine beam shear from statics. As the point of inflection is assumed to lie at mid span, the beam shear equals beam end moment divided by ½ beam span. 76 38
Portal Method Analysis Steps Step 3c: Determine beam shear from statics. F 3 M BL M BR M BL M BR P L =M BL /0.5l 1 P R =M BR /0.5l 1 P L =M BL /0.5l 2 P R =M BR /0.5l 2 P L P R F 2 P L P R P L P R P L P R F 1 P L P R P L P R P L P R l 1 l 2 l 3 77 Portal Method Analysis Steps Step 3d: Determine column axial force from statics. F 3 For a segment (abc for example), the axial force shall be arithmetic sum of beam shears within that segment, but in opposite direction. Axial force in lower storey shall be the sum of axial force in storey under question plus the axial forces in all above stories. a c P L P R P L P R P R b F 2 P L P R P L P R P L P R F 1 P L P R P L P R P L P R l 1 l 2 l 3 78 39
Portal Method Analysis Steps Step 3d: Determine column axial force from statics. F 3 P L3 P R3 +P L3 P L3 P R3 P L3 P R3 PL3 P R3 F 2 F 1 P L3 +P L2 P L2 P R2 P L2 P R2 P L2 P R2 Similarly all other column axial forces can be determined P L1 P R1 P L1 P R1 P L1 P R1 P L3 +P L2 +P L1 l 1 l 2 l 3 79 Portal Method (Case Study 1) Lateral load analysis for E-W Interior Frame of given 3D structure by portal method and its comparison with SAP2000. The objective of this study is to check the level of accuracy of portal method. 80 40
Portal Method (Case Study 1) SDL = Nil LL = 144 psf Given 3D structure. f c = 4 ksi f y = 60 ksi Note: Zone 2B SDL = Nil LL = 144 psf Slab = 7 SDL = Nil LL = 144 psf SDL = Nil LL = 144 psf 10 ft (floor to floor) 10 ft 10 ft 25 ft 25 ft 25 ft 25 ft 20 ft 20 ft 20 ft Slab-Beam Frame Structure 81 Portal Method (Case Study 1) F 3 =28.21 kip E-W Interior Frame F 2 =18.61 kip h=10 ft F 1 =9.36 kip h=10 ft h=10 ft l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 82 41
Portal Method (Case Study 1) F 3 =28.21 kip Step 1: Locate points of inflection. F 2 =18.61 kip F 1 =9.36 kip For Hinge l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 83 Portal Method (Case Study 1) F 3 =28.21 kip Step 2: Determine column shear. n = 4 F 2 =18.61 kip H 3ext =F 3 /2n H 3int =F 3 /n 3.5 7.05 7.05 7.05 3.5 H 2ext =(F 3 + F 2 )/2n 5.85 F 1 =9.36 kip H 1ext =(F 3 + F 2 + F 1 )/2n H 2int =(F 3 + F 2 )/n 11.7 11.7 11.7 H 1int =(F 3 + F 2 + F 1 )/n 5.85 7.00 14.0 14.0 14.0 7.00 l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 84 42
Portal Method (Case Study 1) Portal Method SAP 3D Step 2: Determine column shear (comparison with SAP). 3.5 (4) 7.05 (7) 7.05 (7) 5.85 (7) 11.7 (13) 11.7 (12) 7.00 (11) 14.0 (13) 14.0 (13) l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 85 F 3 = 28.21 kip F 2 = 18.61 kip F 1 = 9.36 kip Portal Method (Case Study 1) 17.5 17.5 29.3 29.3 Step 3a: Determine column moments. 3.5 7.05 5.85 35.25 35.25 58.5 58.5 70 140 35.25 35.25 58.5 M = H h/2 (for all stories except bottom) M = H h (for bottom storey) 7.05 7.05 3.5 11.7 11.7 11.7 58.5 140 35.25 35.25 58.5 58.5 140 17.5 17.5 29.3 5.85 29.3 70 7.00 14.0 14.0 14.0 l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 7.00 86 43
Portal Method (Case Study 1) Portal Method SAP 3D 17.5 (28) Step 3a: Determine column moments (comparison with SAP). 35.25 (39) 35.25 (37) 17.5 (20) 29.3 (45) 35.25 (33) 58.5 (69) 35.25 (32) 58.5 (65) 29.3 (28) 70 (111) 58.5 (62) 140 (133) 58.5 (56) 140 (129) l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 87 F 3 = 28.21 kip F 2 = 18.61 kip F 1 = 9.36 kip Portal Method (Case Study 1) 29.3 29.3 Step 3b: Determine beam moments. 58.5 58.5 M BL = M col /m M BR = M col /m 17.5 17.5 17.5 17.5 17.5 17.5 17.5 17.5 17.5 35.25 35.25 35.25 17.5 35.25 35.25 35.25 46.8 46.8 46.8 46.8 46.8 46.8 46.8 46.8 58.5 58.5 58.5 58.5 99 99 99 99 99 99 99 99 17.5 17.5 29.3 29.3 70 140 140 140 70 l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 88 44
Portal Method (Case Study 1) Portal Method SAP 3D 17.5 (19) Step 3b: Determine beam moments (comparison with SAP). 17.5 (14) 17.5 (16) 17.5 (14) 46.8 (46) 99 (99) 46.8 (37) 99 (68) 46.8 (41) 99 (81) 46.8 (37) 99 (70) l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 89 F 3 = 28.21 kip Portal Method (Case Study 1) Step 3c: Determine beam shear. P L = M BL /0.5l P R = M BR /0.5l 17.5 17.5 17.5 17.5 17.5 17.5 17.5 17.5 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4 F 2 = 18.61 kip F 1 = 9.36 kip 46.8 46.8 46.8 46.8 46.8 46.8 46.8 46.8 3.74 3.74 3.74 3.74 3.74 3.74 3.74 3.74 99 99 99 99 99 99 99 99 7.92 7.92 7.92 7.92 7.92 7.92 7.92 7.92 l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 90 45
Portal Method (Case Study 1) Portal Method SAP 3D Step 3c: Determine beam shear (comparison with SAP). 1.4 (1.8) 1.4 (1.5) 3.74 (4.4) 3.74 (4) 7.92 (9) 7.92 (7) l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft Portal Method (Case Study 1) F 3 = 28.21 kip F 2 = 18.61 kip F 1 = 9.36 kip Step 3d: Determine column axial loads. 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4 0 0 0 1.4 3.74 3.74 3.74 3.74 3.74 3.74 3.74 3.74 0 0 0 5.14 5.14 7.92 7.92 7.92 7.92 7.92 7.92 7.92 7.92 13.06 0 0 0 13.06 l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 92 46
Portal Method (Case Study 1) Portal Method SAP 3D Step 3d: Determine column axial loads (comparison with SAP). 1.4 (2) 0 (-0.5) 0 (0) 5.14 (7) 0 (-1.4) 0 (0) 13.06 (17) 0 (-4.4) 0 (0) l 1 =25 ft l 2 =25 ft l 3 =25 ft l 4 =25 ft 93 Portal Method (Case Study 1) Portal Method SAP 3D Similar comparison for 20 15 ft structure is shown below: 16 (16) 31 (22) 16 (13) 26 (26) 31 (20) 52 (38) 26 (19) 61 (68) 52 (36) 123 (77) l 1 =20 ft l 2 =20 ft l 3 =20 ft l 4 =20 ft 94 47
Portal Method (Case Study 1) Portal Method SAP 3D 16 (12) Similar comparison for 20 15 ft structure is shown below: 16 (8) 16 (9) 16 (8.5) 41 (29) 41 (22) 41 (25) 41 (22) 87 (64) 87 (40) 87 (50) 87 (43) l 1 =20 ft l 2 =20 ft l 3 =20 ft l 4 =20 ft 95 Portal Method (Case Study 2) Lateral Load Analysis of a frame corresponding to seismic demand in seismic zones 1 to 4 using Portal Method. 96 48
Portal Method (Case Study 2) SDL = 40 psf LL = 60 psf SDL = 40 psf LL = 60 psf SDL = 40 psf LL = 60 psf 10.5 ft 10.5 ft f c = 3 ksi f y = 40 ksi 10.5 ft (floor to floor) 20 ft 20 ft 20 ft 20 ft 15 ft 15 ft 15 ft Slab-Beam Frame Structure 97 Portal Method (Case Study 2) Zone 1 (Bending moments) C a = 0.12 C v = 0.18 R = 8.5 V = 20.04 kips 98 49
Portal Method (Case Study 2) Zone 2A (Bending moments) C a = 0.22 C v = 0.32 R = 8.5 V = 36.74 kips 99 Portal Method (Case Study 2) Zone 2B (Bending moments) C a = 0.28 C v = 0.40 R = 8.5 V = 46.76 kips 100 50
Portal Method (Case Study 2) Zone 3 (Bending moments) C a = 0.36 C v = 0.54 R = 8.5 V = 60.11 kips 101 Portal Method (Case Study 2) Zone 4 (Bending moments) C a = 0.44 C v = 0.64 R = 8.5 V = 73.47 kips 102 51
Portal Method (Case Study 2) Comparison (Interior Negative Beam Moment) Top Intermediate Bottom 103 Portal Method (Case Study 2) Comparison (Column Moment) Top Intermediate Bottom 104 52
References ACI 318-14 UBC-97 BCP SP-2007 Earthquake tips from IITK. Intermediate Bottom 105 The End 106 53