Explanatory Examples on Indian Seismic Code IS 1893 (Part I)


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1 Doument No. :: IITKGSDMAEQ1V.0 inal Report :: A  Earthquake Codes IITKGSDMA Projet on Building Codes Explanatory Examples on Indian Seismi Code IS 1893 (Part I) by Dr. Sudhir K Jain Department of Civil Engineering Indian Institute of Tehnology Kanpur Kanpur
2 The solved examples inluded in this doument are based on a draft ode being developed under IITKGSDMA Projet on Building Codes. The draft ode is available at GSDMA.htm (doument number IITKGSDMAEQ05V3.0). This doument has been developed through the IITKGSDMA Projet on Building Codes. The views and opinions expressed are those of the authors and not neessarily of the GSDMA, the World Bank, IIT Kanpur, or the Bureau of Indian Standards. Comments and feedbaks may please be forwarded to: Prof. Sudhir K Jain, Dept. of Civil Engineering, IIT Kanpur, Kanpur 08016,
3 CONTENTS Sl. No Title Page No. 1. Calulation of Design Seismi ore by Stati Analysis Method 4. Calulation of Design Seismi ore by Dynami Analysis Method 7 3. Loation of Centre of Mass Loation of Centre of Stiffness Lateral ore Distribution as per Torsion Provisions of IS (Part I) 1 6. Lateral ore Distribution as per New Torsion Provisions Design for Anhorage of an Equipment Anhorage Design for an Equipment Supported on Vibration Isolator Design of a Large Sign Board on a Building Liquefation Analysis Using SPT Data Liquefation Analysis Using CPT Data 3 IITKGSDMAEQ1V.0
4 Example 1 Calulation of Design Seismi ore by Stati Analysis Method Problem Statement: Consider a fourstorey reinfored onrete offie building shown in ig The building is loated in Shillong (seismi zone V). The soil onditions are medium stiff and the entire building is supported on a raft foundation. The R. C. frames are infilled with brikmasonry. The lumped weight due to dead loads is 1 kn/m on floors and 10 kn/m on the roof. The floors are to ater for a live load of 4 kn/m on floors and 1.5 kn/m on the roof. Determine design seismi load on the struture as per new ode. [Problem adopted from Jain S.K, A Proposed Draft for IS:1893 Provisions on Seismi Design of Buildings; Part II: Commentary and Examples, Journal of Strutural Engineering, Vol., No., July 1995, pp ] y (1) () (3) (4) (5) (A) (B) 5000 (C) 5000 PLAN (D) x ELEVATION igure 1.1 Building onfiguration IITKGSDMAEQ1V.0 Example 1/Page 4
5 Solution: Design Parameters: or seismi zone V, the zone fator Z is 0.36 (Table of IS: 1893). Being an offie building, the importane fator, I, is 1.0 (Table 6 of IS: 1893). Building is required to be provided with moment resisting frames detailed as per IS: Hene, the response redution fator, R, is 5. (Table 7 of IS: 1893 Part 1) Seismi Weights: The floor area is sq. m. Sine the live load lass is 4kN/sq.m, only 50% of the live load is lumped at the floors. At roof, no live load is to be lumped. Hene, the total seismi weight on the floors and the roof is: loors: W 1 W W ( ) 4,00 kn Roof: W ,000 kn (lause7.3.1, Table 8 of IS: 1893 Part 1) Total Seismi weight of the struture, W ΣW i 3 4,00 + 3,000 15,600 kn undamental Period: Lateral load resistane is provided by moment resisting frames infilled with brik masonry panels. Hene, approximate fundamental natural period: (Clause of IS: 1893 Part 1) 0.09(13.8) / se The building is loated on Type II (medium soil). S rom ig. of IS: 1893, for T0.8 se, a g.5 ZI S A h a R g (Clause 6.4. of IS: 1893 Part 1) Design base shear VB A h W ,600 1,440 kn (Clause of IS: 1893 Part 1) ore Distribution with Building Height: The design base shear is to be distributed with height as per lause Table 1.1 gives the alulations. ig. 1.(a) shows the design seismi fore in Xdiretion for the entire building. EL in YDiretion: T Sa 0.09h.5; g A h 0.09 d 0.09(13.8) / 0.3 se 15 Therefore, for this building the design seismi fore in Ydiretion is same as that in the X diretion. ig. 1.(b) shows the design seismi fore on the building in the Ydiretion. EL in XDiretion: T 0.09h / d IITKGSDMAEQ1V.0 Example 1/Page 5
6 Storey Level Table 1.1 Lateral Load Distribution with Height by the Stati Method W i ( kn ) h i (m) W i h i (1000) Wihi W h i i Lateral ore at i th Level for EL in diretion (kn) 4 3, , , , Σ 1, ,000 1,440 1,440 X Y igure Design seismi fore on the building for (a) Xdiretion, and (b) Ydiretion. IITKGSDMAEQ1V.0 Example 1/Page 6
7 Example Calulation of Design Seismi ore by Dynami Analysis Method Problem Statement: or the building of Example 1, the dynami properties (natural periods, and mode shapes) for vibration in the Xdiretion have been obtained by arrying out a free vibration analysis (Table.1). Obtain the design seismi fore in the Xdiretion by the dynami analysis method outlined in l and distribute it with building height. Table.1 ree Vibration Properties of the building for vibration in the XDiretion Mode 1 Mode Mode 3 Natural Period (se) Mode Shape Roof rd loor nd loor st loor [Problem adopted from, Jain S.K, A Proposed Draft for IS: 1893 Provisions on Seismi Design of Buildings; Part II: Commentary and Examples, Journal of Strutural Engineering, Vol., No., July 1995, pp.7390] Solution: Storey Level i Table.  Calulation of modal mass and modal partiipation fator (lause ) Weight W i kn Mode 1 Mode Mode 3 ( ) 4 3, ,000 3, ,000 3, ,000 3, , ,797 3, ,490,900 4, ,007, ,944, ,411 1, , , ,868 3, ,67 4,335 Σ 15,600 11,656 9,40 ,905 8,8 1,366 11,60 [ wi φik ] g wi φik 11,656 14,450kN, kN 1, kN M k 9,40g g 8,8g g 11,60g g 14,45,000 kg 95,700 kg 16,100 kg % of Total weight 9.6% 6.1% 1.0% P k w φ i i ik ik w φ 11, ,40, ,8 39 1, ,60 It is seen that the first mode exites 9.6% of the total mass. Hene, in this ase, odal requirements on number of modes to be onsidered suh that at least 90% of the total mass is exited, will be satisfied by onsidering the first mode of vibration only. However, for illustration, solution to this example onsiders the first three modes of vibration. The lateral load Q ik ating at i th floor in the k th mode is Q ik A φ hk ik P k W i IITKGSDMAEQ1V.0 Example /Page 7
8 (lause of IS: 1893 Part 1) The value of A hk for different modes is obtained from lause Mode 1: T se; 1.0 ( S a / g) 1.16 ; 0.86 ZI A h1 ( S a / g) R (1.16) 5 Q i1 Mode : T 0.65 se; ( S a / g).5 ; φ W i1 i ZI A h ( S a / g) R (.5) ( 0.39) φ Q i1 Mode 3: T se; ( S a / g).5 ; i ZI A h3 ( S a / g) R (.5) (0.118) φ Q i3 i3 W W Table.3 summarizes the alulation of lateral load at different floors in eah mode. i i Table.3 Lateral load alulation by modal analysis method (earthquake in Xdiretion) loor Level i Weight W i ( ) kn i1 Mode 1 Mode Mode 3 φ Q i1 V i1 φ i Q i V i φ i3 Q i3 i3 4 3, , , , V Sine all of the modes are well separated (lause 3.), the ontribution of different modes is ombined by the SRSS (square root of the sum of the square) method V 4 [(155.5) + (88.8) + (31.9) ] 1/ 18 kn V 3 [(35.3) + (115.6) + (5.) ] 1/ 371 kn V [(508.) + (8.4) + (30.8) ] 1/ 510 kn V 1 [(604.) + (86.) + (14.6) ] 1/ 610 kn (Clause a of IS: 1893 Part 1) The externally applied design loads are then obtained as: Q 4 V 4 18 kn Q 3 V 3 V kn Q V V kn Q 1 V 1 V kn (Clause f of IS: 1893 Part 1) Clause 7.8. requires that the base shear obtained by dynami analysis (V B 610 kn) be ompared with that obtained from empirial fundamental period as per Clause 7.6. If V B is less than that from empirial value, the response quantities are to be saled up. We may interpret base shear alulated using a fundamental period as per 7.6 in two ways: 1. We alulate base shear as per Cl This was done in the previous example for the same building and we found the base shear as 1,404 kn. Now, dynami analysis gives us base shear of 610 kn whih is lower. Hene, all the response quantities are to be saled up in the ratio (1,404/610.30). Thus, the seismi fores obtained above by dynami analysis should be saled up as follows: Q kn Q kn Q kn IITKGSDMAEQ1V.0 Example /Page 8
9 Q kn. We may also interpret this lause to mean that we redo the dynami analysis but replae the fundamental time period value by T a ( 0.8 se). In that ase, for mode 1: T se; ( S a / g).5 ; A h1 ZI ( S a / g) R 0.09 Modal mass times A h1 14, ,300 kn Base shear in modes and 3 is as alulated earlier: Now, base shear in first mode of vibration 1300 kn, 86. kn and 14.6 kn, respetively. Total base shear by SRSS loor Level i Q (stati) 14.6 Table.4 Base shear at different storeys Q (dynami, saled) Storey Shear V (stati) 1,303 kn Notie that most of the base shear is ontributed by first mode only. In this interpretation of Cl 7.8., we need to sale up the values of response quantities in the ratio (1,303/610.14). or instane, the external seismi fores at floor levels will now be: Q kn Q kn Q kn Q kn Clearly, the seond interpretation gives about 10% lower fores. We ould make either interpretation. Herein we will proeed with the values from the seond interpretation and ompare the design values with those obtained in Example 1 as per stati analysis: Storey ShearV (dynami, saled) Storey Moment, M (Stati) Storey Moment, M (Dynami) kn 389 kn 611 kn 389 kn 1,907 knm 1,45 knm kn 404 kn 1,115kN 793 kn 5,386 knm 3,78 knm 97 kn 97 kn 1,41kN 1,090 kn 9.63 knm 7,70 knm 1 79 kn 14 kn 1,491 kn 1,304 kn 15,530 knm 1,750 knm Notie that even though the base shear by the stati and the dynami analyses are omparable, there is onsiderable differene in the lateral load distribution with building height, and therein lies the advantage of dynami analysis. or instane, the storey moments are signifiantly affeted by hange in load distribution. IITKGSDMAEQ1V.0 Example /Page 9
10 Example 3 Loation of Centre of Mass Problem Statement: Loate entre of mass of a building having nonuniform distribution of mass as shown in the figure m 4 m 100 kg/m 1000 kg/m 8 m A 0 m igure 3.1 Plan Solution: Let us divide the roof slab into three retangular parts as shown in figure.1 4 m 10 m 100 kg/m I 1000 kg/m II III 8 m ( ) 6 + ( ) 6 + ( ) Y ( ) + ( ) + ( ) 4.1 m Hene, oordinates of entre of mass are (9.76, 4.1) 0 m igure 3. Mass of part I is 100 kg/m, while that of the other two parts is 1000 kg/m.. Let origin be at point A, and the oordinates of the entre of mass be at (X, Y) ( ) 5 + ( ) 15 + ( ) 10 X ( ) + ( ) + ( ) 9.76 m IITKGSDMAEQ1 V.0 Example 3 /Page10
11 Example 4 Loation of Centre of Stiffness Problem Statement: The plan of a simple one storey building is shown in figure 3.1. All olumns and beams are same. Obtain its entre of stiffness. 5 m 5 m 5 m 5 m 10 m igure 4.1 Plan Solution: In the Xdiretion there are three idential frames loated at uniform spaing. Hene, the y oordinate of entre of stiffness is loated symmetrially, i.e., at 5.0 m from the left bottom orner. In the Ydiretion, there are four idential frames having equal lateral stiffness. However, the spaing is not uniform. Let the lateral stiffness of eah transverse frame be k, and oordinating of enter of stiffness be (X, Y). X k 0 + k 5 + k 10 + k m k + k + k + k Hene, oordinates of entre of stiffness are (8.75, 5.0). IITKGSDMAEQ1 V.0 Example 4 /Page11
12 Example 5 Lateral ore Distribution as per Torsion Provisions of IS (Part 1) Problem Statement: Consider a simple onestorey building having two shear walls in eah diretion. It has some gravity olumns that are not shown. All four walls are in M5 grade onrete, 00 thik and 4 m long. Storey height is 4.5 m. loor onsists of astinsitu reinfored onrete. Design shear fore on the building is 100 kn in either diretion. Compute design lateral fores on different shear walls using the torsion provisions of 00 edition of IS 1893 (Part 1). Y m 4m 4m C 4m A B 8m D 16m X igure 5.1 Plan Solution: Grade of onrete: M5 E N/mm Storey height h 4500 m Thikness of wall t 00 mm Length of walls L 4000 mm All walls are same, and hene, spaes have same lateral stiffness, k. Centre of mass (CM) will be the geometri entre of the floor slab, i.e., (8.0, 4.0). Centre of rigidity (CR) will be at (6.0, 4.0). EQ ore in Xdiretion: Beause of symmetry in this diretion, alulated eentriity 0.0 m Design eentriity: e d , and e d (Clause 7.9. of IS 1893:00) Lateral fores in the walls due to translation: KC CT 50.0 K + K kn C D K D DT 50.0 K + K kn C D Lateral fores in the walls due to torsional moment: K iri ir ( ed ) K r i i i A, B, C, D where r i is the distane of the shear wall from CR. All the walls have same stiffness, K A K B K C K D k, and r A 6.0 m r B 6.0 m IITKGSDMAEQ1 V.0 Example 5 /Page 1
13 r C 4.0 m r D 4.0 m, and e ±0. 4 m d Therefore, r ( e ) A AR d ( ra + rb + rc + rd ) k k ±. 31kN Similarly, BR ±. 31kN CR ± kn DR ± kn Total lateral fores in the walls due to seismi load in X diretion: A.31 kn B.31 kn C Max (50 ± ) kn D Max (50 ± ) kn EQ ore in Ydiretion: Calulated eentriity.0 m Design eentriity: ed m or m Lateral fores in the walls due to translation: K A 50.0 K + K AT kn A B K B BT 50.0 K + K kn A B Lateral fore in the walls due to torsional moment: when e d 3.8 m r k kn ( e ) A AR d ( ra + rb + rc + rd ) k Similarly, BR 1.9 kn CR kn DR 14.6 kn Total lateral fores in the walls: A kn B kn C kn D 14.6 kn Similarly, when e d 1. m, then the total lateral fores in the walls will be, A kn B kn C kn D 4.6 kn Maximum fores in walls due to seismi load in Y diretion: A Max (8.08, 43.07) kn; B Max (71.9, 56.93) 71.9 kn; C Max (14.6, 4.6) 14.6 kn; D Max (14.6, 4.6) 14.6 kn; Combining the fores obtained from seismi loading in X and Y diretions: A kn B 71.9 kn C kn D kn. However, note that lause also states that However, negative torsional shear shall be negleted. Hene, wall A should be designed for not less than 50 kn. IITKGSDMAEQ1V.0 Example 5/Page 13
14 Example 6 Lateral ore Distribution as per New Torsion Provisions Problem Statement: or the building of example 5, ompute design lateral fores on different shear walls using the torsion provisions of revised draft ode IS 1893 (part 1), i.e., IITKGSDMAEQ05V.0. Y m 4m 6m 4m C 4m A B 8m D 16m X igure 6.1 Plan Solution: Grade of onrete: M5 E N/mm Storey height h 4500 m Thikness of wall t 00 mm Length of walls L 4000 mm All walls are same, and hene, same lateral stiffness, k. Centre of mass (CM) will be the geometri entre of the floor slab, i.e., (8.0, 4.0). Centre of rigidity (CR) will be at (6.0, 4.0). EQ ore in Xdiretion: Beause of symmetry in this diretion, alulated eentriity 0.0 m Design eentriity, e d 0.0 ± ± 0. 8 (lause 7.9. of Draft IS 1893: (Part1)) Lateral fores in the walls due to translation: K C 50.0 K + K CT kn C D K D 50.0 K + K DT kn C D Lateral fores in the walls due to torsional moment: ir K r i i K iri i A, B, C, D ( e ) d where r i is the distane of the shear wall from CR All the walls have same stiffness, K A K B K C K D k r A 6.0 m r B 6.0 m r C 4.0 m r D 4.0 m r A ( ) ( e ) r + r + r + r k AR d A B C D kn Similarly, BR 4.6 kn CR 3.08 kn DR kn Total lateral fores in the walls: A 4.6 kn B kn C kn D kn k IITKGSDMAEQ1 V.0 Example 6 /Page 14
15 Similarly, when e d m, then the lateral fores in the walls will be, A kn B 4.6 kn C kn D kN Design lateral fores in walls C and D are: C D kn EQ ore in Ydiretion: Calulated eentriity.0 m Design eentriity, e d m or e d m Lateral fores in the walls due to translation: K A 50.0 K + K AT kn A B K B 50.0 K + K BT kn A B Lateral fore in the walls due to torsional moment: when e d 3.6 m r k kn ( e ) A AR d ( ra + rb + rc + rd ) k Similarly, BR 0.77 kn CR kn DR kn Total lateral fores in the walls: A kn B kn C kn D kn Similarly, when e d 0.4 m, then the total lateral fores in the walls will be, A kn B kn C 1.54 kn D kn Maximum fores in walls A and B A kn, B kn Design lateral fores in all the walls are as follows: A kn B kn C kn D kn. IITKGSDMAEQ1V.0 Example 6/Page 15
16 Example 7 Design for Anhorage of an Equipment Problem Statement: A 100 kn equipment (igure 7.1) is to be installed on the roof of a five storey building in Simla (seismi zone IV). It is attahed by four anhored bolts, one at eah orner of the equipment, embedded in a onrete slab. loor to floor height of the building is 3.0 m. exept the ground storey whih is 4. m. Determine the shear and tension demands on the anhored bolts during earthquake shaking. W p p CG 1.5 m Anhor bolt 1.0 m Anhor bolt igure 7.1 Equipment installed at roof Solution: Zone fator, Z 0.4 (for zone IV, Table of IS 1893), Height of point of attahment of the equipment above the foundation of the building, x ( ) m 16. m, Height of the building, h 16. m, Amplifiation fator of the equipment, a p 1 (rigid omponent, Table 11), Response modifiation fator R p.5 (Table 11), Importane fator I p 1 (not life safety omponent, Table 1), Weight of the equipment, W p 100 kn The design seismi fore Z x ap p 1+ IpWp h R kn < 0.1W 10.0kN p p ()( ) kn Hene, design seismi fore, for the equipment p 10.0 kn. IITKGSDMAEQ1V.0 Example 7/Page 16
17 The anhorage of equipment with the building must be designed for twie of this fore (Clause of draft IS 1893) Shear per anhor bolt, V p /4 10.0/4 kn 5.0 kn The overturning moment is M.0 (10.0 kn) (1. 5 m) ot 30.0 knm The overturning moment is resisted by two anhor bolts on either side. Hene, tension per anhor bolt from overturning is t (30.0) kn (1.0)() 15.0kN IITKGSDMAEQ1V.0 Example 7/Page 17
18 Example 8 Anhorage Design for an Equipment Supported on Vibration Isolator Problem Statement: A 100 kn eletrial generator of a emergeny power supply system is to be installed on the fourth floor of a 6storey hospital building in Guwahati (zone V). It is to be mounted on four flexible vibration isolators, one at eah orner of the unit, to damp the vibrations generated during the operation. loor to floor height of the building is 3.0 m. exept the ground storey whih is 4. m. Determine the shear and tension demands on the isolators during earthquake shaking. W p Vibration Isolator p CG 0.8 m 1. m igure 8.1 Eletrial generator installed on the floor Solution: Zone fator, Z 0.36 (for zone V, Table of IS 1893), Height of point of attahment of the generator above the foundation of the building, x ( ) m 13. m, Height of the building, h ( ) m 19. m, Amplifiation fator of the generator, a p.5 (flexible omponent, Table 11), Response modifiation fator R p.5 (vibration isolator, Table 11), Importane fator I p 1.5 (life safety omponent, Table 1), Weight of the generator, W p 100 kn The design lateral fore on the generator, Z x ap p 1+ IpWp h R p IITKGSDMAEQ1V.0 Example 8/Page 18
19 ( 1.5)( 100) kn kn 0.1Wp 10.0kN Sine the generator is mounted on flexible vibration isolator, the design fore is doubled i.e., 45.6 kn p 91. kn Shear fore resisted by eah isolator, V p /4.8 kn The overturning moment, M 91. kn 0.8 m ot ( ) ( ) 73.0 knm The overturning moment (M ot ) is resisted by two vibration isolators on either side. Therefore, tension or ompression on eah isolator, t ( 73.0) ( 1.)( ) 30.4 kn kn IITKGSDMAEQ1V.0 Example 8/Page 19
20 Example 9 Design of a Large Sign Board on a Building Problem Statement: A neon sign board is attahed to a 5storey building in Ahmedabad (seismi zone III). It is attahed by two anhors at a height 1.0 m and 8.0 m. rom the elasti analysis under design seismi load, it is found that the defletions of upper and lower attahments of the sign board are 35.0 mm and 5.0 mm, respetively. ind the design relative displaement. Solution: Sine sign board is a displaement sensitive nonstrutural element, it should be designed for seismi relative displaement. Height of level x to whih upper onnetion point is attahed, h x 1.0 m Height of level y to whih lower onnetion point is attahed, h y 8.0 m Defletion at building level x of struture A due to design seismi load determined by elasti analysis 35.0 mm Defletion at building level y of struture A due to design seismi load determined by elasti analysis 5.0 mm Response redution fator of the building R 5 (speial RC moment resisting frame, Table 7) δ xa 5 x mm δ ya 5 x mm (i) Dp δ xa δ ya ( ) mm 50.0 mm Design the onnetions of neon board to aommodate a relative motion of 50 mm. (ii) Alternatively, assuming that the analysis of building is not possible to assess defletions under seismi loads, one may use the drift limits (this presumes that the building omplies with seismi ode). Maximum interstorey drift allowane as per lause is IS : 1893 is times the storey height, i.e., Δ aa h sx D p Δ R( hx hy ) h aa sx 5 ( )(0.004) mm 80.0 mm The neon board will be designed to aommodate a relative motion of 80 mm. IITKGSDMAEQ1V.0 Example 9/Page 0
21 Example: 10 Liquefation Analysis using SPT data Problem Statement: Examples on IS 1893(Part 1) The measured SPT resistane and results of sieve analysis for a site in Zone IV are indiated in Table The water table is at 6m below ground level. Determine the extent to whih liquefation is expeted for 7.5 magnitude earthquake. Estimate the liquefation potential and resulting settlement expeted at this loation. Table 10.1: Result of the Standard penetration Test and Sieve Analysis Depth N 60 Soil Classifiation Perentage (m) fine 0.75 Poorly Graded Sand and Silty Sand 11 9 (SPSM) Poorly Graded Sand and Silty Sand (SPSM) Poorly Graded Sand and Silty Sand (SPSM) Poorly Graded Sand and Silty Sand (SPSM) Poorly Graded Sand and Silty Sand (SPSM) Poorly Graded Sand and Silty Sand (SPSM) Poorly Graded Sand and Silty Sand (SPSM) 6 Solution: Site Charaterization: This site onsists of loose to dense poorly graded sand to silty sand (SPSM). The SPT values ranges from 9 to 6. The site is loated in zone IV. The peak horizontal ground aeleration value for the site will be taken as 0.4g orresponding to zone fator Z 0.4 Liquefation Potential of Underlying Soil Step by step alulation for the depth of 1.75m is given below. Detailed alulations for all the depths are given in Table 10.. This table provides the fator of safety against liquefation (S liq ), maximum depth of liquefation below the ground surfae, and the vertial settlement of the soil due to liquefation (Δ v ). amax 0.4 g, M 7. 5, 3 γ sat 18.5 kn / m, w γ w 9.8 kn / m Depth of water level below G.L. 6.00m Depth at whih liquefation potential is to be 3 evaluated 1.75m Initial stresses: σ v kpa u ( ) kpa σ 0 ( ) u ' v σ v kpa Stress redution fator: r d z Critial stress ratio indued by earthquake: amax 0. 4g, M 7. 5 CSR eq w ' ( a / g) r ( σ / σ ) 0.65 maz d v v ( 0.4) 0.81 ( 35.9 /169.7) CSR eq Corretion for SPT (N) value for overburden pressure: ( N ) C N 60 C N 60 N ' ( 1/ σ ) 1/ 9.79 v IITKGSDMAEQ1V.0 Example 10/Page 1
22 igure  (for SPT data) igure  (for CPT data: in fator of safety alulation in olumn of page 4 this figure is wrongly ited as 6) igure 4 provides a plot for k m. Algebraially, the relationship is simply k m 10 k m M w subjeted to igure 6 igure 5 igure 8 IITKGSDMAEQ1V.0 Example 10/Page 1 A
23 C N 9.79 ( 1/169.7) 1/ ( N ) Critial stress ratio resisting liquefation: N, fines ontent of 8 % or ( ) CSR 0.14 (igure ) 7.5 Correted Critial Stress Ratio Resisting Liquefation: CSR CSR L 7. 5 k m k α k σ k m Corretion fator for earthquake magnitude other than 7.5 (igure 4) 1.00 for M 7. 5 w k α Corretion fator for initial driving stati shear (igure 6) 1.00, sine no initial stati shear k σ Corretion fator for stress level larger than 96 kpa (igure 5) 0.88 CSR L ator of safety against liquefation: S L CSRL eq Examples on IS 1893(Part 1) / CSR 0.1 / Perentage volumetri strain (%ε) or CSR CSR ( k k α k ) eql ( N ) eq / m σ 0.18 / (1x1x0.88) 0.1 % ε.10 (from igure 8) Liquefation indued vertial settlement (ΔV): (ΔV) volumetri strain x thikness of liquefiable level / m 63mm Summary: Analysis shows that the strata between depths 6m and 19.5m are liable to liquefy. The maximum settlement of the soil due to liquefation is estimated as 315mm (Table 10.) Table 10.: Liquefation Analysis: Water Level 6.00 m below GL (Units: Tons and Meters) Depth %ine σ v (kpa) ' σ v (kpa) N 60 N C ( ) 60 N r d CSR eq CSR eql CSR 7.5 CSR L S L % ε ΔV Total Δ IITKGSDMAEQ1V.0 Example 10/Page
24 Example: 11 Liquefation Analysis using CPT data Problem Statement: Prepare a plot of fators of safety against liquefation versus depth. The results of the one penetration test (CPT) of 0m thik layer in Zone V are indiated in Table Assume the water table to be at a depth of.35 m, the unit weight of the soil to be 18 kn/m 3 and the magnitude of 7.5. Table 11.1: Result of the Cone penetration Test Depth (m) q f s Depth (m) q f s Depth (m) q f s Solution: Liquefation Potential of Underlying Soil Step by step alulation for the depth of 4.5m is given below. Detailed alulations are given in Table 11.. This table provides the fator of safety against liquefation (S liq ). The site is loated in zone V. The peak horizontal ground aeleration value for the site will be taken as 0.36g orresponding to zone fator Z 0.36 a max /g 0.36, M w 7.5, 3 γ sat 18 kn / m, γ w 9.8 kn / m Depth of water level below G.L..35m Depth at whih liquefation potential is to be evaluated 4.5m Initial stresses: σ v kpa u (4.5.35) kpa σ 0 ( u ) kpa ' v σ v 0 93 Stress redution fator: 3 IITKGSDMAEQ1 V.0 Example 11 /Page 3
25 r d z Critial stress ratio indued by earthquake: CSR eq ' ( a / g) r ( σ / σ ) 0.65 maz d v v CSR eq ( 0.36) ( 81/ 59.93) Correted Critial Stress Ratio Resisting Liquefation: CSR L CSR eq k m k α k σ k m Corretion fator for earthquake magnitude other than 7.5 (igure 4) 1.00 for M 7. 5 w k α Corretion fator for initial driving stati shear (igure 6) 1.00, sine no initial stati shear k σ Corretion fator for stress level larger than 96 kpa (igure 5) CSR L Corretion fator for grain harateristis: K K M I I 4 for I I and I for I > 1.64 Q Q K M [( q σ ) P ]( P σ ) n v a a Examples on IS 1893(Part 1) [( ) ] ( ) v 4 3 (.19) (.19) 1.63(.19) (.19) Normalized Cone Tip Resistane: n ( q ) K ( P σ ) ( q P ) 1 N s a v 0.5 ( ) 1.64( ) ( ) q 1N s ator of safety against liquefation: q 1 70., or ( ) 77 N s CRR 0.11 (igure 6) S CRR / S liq liq CSR L Summary: 0.11/ Analysis shows that the strata between depths 01m are liable to liquefy under earthquake shaking orresponding to peak ground aeleration of 0.36g. The plot for depth verses fator of safety is shown in igure 11.1 a 0.5 The soil behavior type index, I, is given by ( 3.47 log Q) + ( 1. log ) I + I.19 ( 3.47 log 4.19) + ( 1. + log 0.903) Where, ( ) 100 σ f q v [ 9.7 /( ) ] and IITKGSDMAEQ1V.0 Example 11/Page 4
26 Table 11.: Liquefation Analysis: Water Level.35 m below GL (Units: kn and Meters) Depth σ v σ v ' r d q (kpa) fs (kpa) CSR eq CSR L Q I K (q1n)s CRR S liq IITKGSDMAEQ1V.0 Example 11/Page 5
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