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6. 4 7 10 11 12 14 16 18 20 21 23 IITK-GSDMA-EQ21-V2.Examples on IS 1893(Part 1) CONTENTS Sl. 9. 3. 11. 10. 5. 2.0 . Title Calculation of Design Seismic Force by Static Analysis Method Calculation of Design Seismic Force by Dynamic Analysis Method Location of Centre of Mass Location of Centre of Stiffness Lateral Force Distribution as per Torsion Provisions of IS 1893-2002 (Part I) Lateral Force Distribution as per New Torsion Provisions Design for Anchorage of an Equipment Anchorage Design for an Equipment Supported on Vibration Isolator Design of a Large Sign Board on a Building Liquefaction Analysis Using SPT Data Liquefaction Analysis Using CPT Data Page No. 8. 4. 7. No 1.
Vol. The R.Examples on IS 1893(Part 1) Example 1 – Calculation of Design Seismic Force by Static Analysis Method Problem Statement: Consider a four-storey reinforced concrete office building shown in Fig. July 1995. The soil conditions are medium stiff and the entire building is supported on a raft foundation.0 Example 1/Page 4 . [Problem adopted from Jain S. The building is located in Shillong (seismic zone V). pp. Determine design seismic load on the structure as per new code. frames are infilled with brick-masonry. No. 1.K.1.73-90 ] y (1) (2) (3) (4) (4) (5) (A) (B) 3 @ 5000 (C) (D) 4 @ 5000 x PLAN 3200 3200 3200 4200 ELEVATION Figure 1.2.22.1 – Building configuration IITK-GSDMA-EQ21-V2. The lumped weight due to dead loads is 12 kN/m2 on floors and 10 kN/m2 on the roof.5 kN/m2 on the roof. C. The floors are to cater for a live load of 4 kN/m2 on floors and 1. Journal of Structural Engineering. Part II: Commentary and Examples”. “A Proposed Draft for IS:1893 Provisions on Seismic Design of Buildings.
the total seismic weight on the floors and the roof is: Floors: W1=W2 =W3 Roof: W4 = 300×10 = 3.000 = 15.1. Since the live load class is 4kN/sq.7. Fig.09(13. Hence.09 h d ΣW i = 3×4.3 of IS: 1893 Part 1) Force Distribution with Building Height: The design base shear is to be distributed with height as per clause 7.600 kN = 0.0 (Table 6 of IS: 1893).0 × 2. I. R.Examples on IS 1893(Part 1) Solution: Design Parameters: For seismic zone V. S a g = 2.1 gives the calculations.0 Example 1/Page 5 . Fig.09 Therefore.5 = 2×5 = 0. W = =300×(12+0.09(13.32 sec Fundamental Period: Lateral load resistance is provided by moment resisting frames infilled with brick masonry panels.36 × 1.5×4) = 4. At roof. 2 of IS: 1893.2(b) shows the design seismic force on the building in the Y-direction. is 1.28 sec. Table 1. m.600 = 1. Table 8 of IS: 1893 Part 1) Total Seismic weight of the structure.m. = 0. of IS: 1893 Part 1) EL in X-Direction: Sa g Ah = 2. no live load is to be lumped. for T=0.5 ZI S a = Ah 2R g 0. only 50% of the live load is lumped at the floors.8) / 15 = 0.6.5.2(a) shows the design seismic force in X-direction for the entire building. Hence.4. is 5.440 kN (Clause 7. for this building the design seismic force in Y-direction is same as that in the Xdirection. (Table 7 of IS: 1893 Part 1) Seismic Weights: The floor area is 15×20=300 sq. Hence. EL in Y-Direction: T = 0. 1.200 + 3. 1.28 sec The building is located on Type II (medium soil).09 (Clause 6.09 × 15. T = 0.8) / 20 = 0.36 (Table 2 of IS: 1893).5. the zone factor Z is 0.200 kN = 0. From Fig.000 kN (clause7. Building is required to be provided with moment resisting frames detailed as per IS: 13920-1993. approximate fundamental natural period: (Clause 7.2. the importance factor. the response reduction factor.1.3.2 of IS: 1893 Part 1) Design base shear VB = AhW = 0. Being an office building.09h / d IITK-GSDMA-EQ21-V2.
and (b) Y-direction.3 471.2 571.8 10.3 0.Design seismic force on the building for (a) X-direction.2 -.347.200 4.000 4.1 – Lateral Load Distribution with Height by the Static Method Storey Level Wi (kN ) hi (m) Wi hi2 × (1000) ∑W h Wi hi2 2 i i Lateral Force at Level for EL direction (kN) X Y 611 504 246 79 1.440 ith in 4 3 2 1 Σ 3.350 0.6 7.0 74.1 1.200 4.9 230.055 1.4 4.000 Figure 1.200 13.0 Example 1/Page 6 .440 611 504 246 79 1.171 0. IITK-GSDMA-EQ21-V2.Examples on IS 1893(Part 1) Table 1.424 0.
904 0.000 0. the dynamic properties (natural periods.905 3.000 -3. Obtain the design seismic force in the X-direction by the dynamic analysis method outlined in cl.716 0.411 4.860 Mode Shape 1. in this case.700 kg 6.822 g g 1.822 1.366 3.000 3.0 .831 -0.000 3.000 0.000 kg 92.064 3.5) Storey Level i 4 3 2 1 Σ Mk i Weight Wi (kN ) Mode 1 1.8.656 = 1.366 = 0.868 -2. The lateral load Qik acting at ith floor in the kth mode is Qik = Ahk φ ik Pk Wi Example 2/Page 7 IITK-GSDMA-EQ21-V2.620 % of Total weight Pk = ∑w φ ∑w φ i i ik 2 ik It is seen that the first mode excites 92.016 Mode 3 3.145 1.620 g g = 14.797 3.000 -0.K.216 -0.000 196 2.200 4.6562 14. Table 2.22.000 907 -2.921 Mode 2 3.402 1.905 = −0.441 Mode 2 0.4.016 [Problem adopted from. pp.402 =95. Vol. will be satisfied by considering the first mode of vibration only. Journal of Structural Engineering.701 -0.944 -3.Examples on IS 1893(Part 1) Example 2 – Calculation of Design Seismic Force by Dynamic Analysis Method Problem Statement: For the building of Example 1.384 4.8.329 8.600 2 ik [∑ w φ ] = g ∑w φ i 2 ik 11.900 1.6% 11. codal requirements on number of modes to be considered such that at least 90% of the total mass is excited.118 11.441 3.2 -. However. for illustration.3662 161kN = 11.432 2.000 4.000 0.2.656 3.701 -0.490 -2.0% 1. solution to this example considers the first three modes of vibration. Hence.9052 957kN = 8.153 817 9.73-90] Solution: Table 2.200 15.5 and distribute it with building height.574 1. “A Proposed Draft for IS: 1893 Provisions on Seismic Design of Buildings. and mode shapes) for vibration in the X-direction have been obtained by carrying out a free vibration analysis (Table 2.831 -0.335 11.1 – Free Vibration Properties of the building for vibration in the X-Direction Natural Period (sec) Roof 3rd Floor 2nd Floor 1st Floor Mode 1 0.000 2.Calculation of modal mass and modal participation factor (clause 7.267 1.000 0. July 1995.620 3.904 0. 7.4.921 Mode 3 0.716 0.852 11. Part II: Commentary and Examples”.100 kg 1.450kN = 9.6% of the total mass.402 g g 2.1). No.563 8.45.1% − 2.822 = 16.200 4.000 -0.007 1.216 -0.240 9.574 1.265 1. Jain S.
1 .000 -88.9 -5.860 sec.2)2+ (28.0 (S a / g ) = = 1.2.09 × (−0.2 604.09 Qi 3 = 0. 7.574 1.2).329) × φi 2 × Wi = T2 = 0.2 1.30).8.000 -0.240 × φ i1 × Wi Mode 2: Qi1 Mode 3: ZI (S a / g ) 2R 0. We may interpret “base shear calculated using a fundamental period as per 7.3)2+ (115.09 × (0.3 summarizes the calculation of lateral load at different floors in each mode.30 = 320 kN = 610 kN (Clause 7.2)2]1/2 = 371 kN V2 = [(508.441 155.0418 Qi1 = 0.6 -28.4a of IS: 1893 Part 1) The externally applied design loads are then obtained as: Q4 = V4 = 182 kN Q3 = V3 – V4 = 371 – 182 = 189 kN Q2 = V2 – V3 = 510 – 371 = 139 kN Q1 = V1 – V2 = 610 – 510 = 100 kN (Clause 7.2 -30.5 .2) + (86.5) 2×5 = 0.5) 2×5 = 0.8)2]1/2 = 510 kN V1 = [(604. 0.36 × 1 = × (1.200 1.2 1.9 -37.8 14.200 4.404/610 = 2. all the response quantities are to be scaled up in the ratio (1. Mode 1: Ah 2 T1 = 0.8 -115.4.6 Since all of the modes are well separated (clause 3.118) × φ i 3 × Wi Table 2.3.2 -0.904 0.0 155.9)2]1/2 = 182 kN V3 = [(352.6) ] 2 2 2 1/2 Clause 7.3 508.09 = 0.200 4.6.4. dynamic analysis gives us base shear of 610 kN which is lower.8.4 31.3 – Lateral load calculation by modal analysis method (earthquake in X-direction) Floor Level i 4 3 2 1 Weight Wi (kN ) Mode 1 φ i1 Q i1 V i1 Mode 2 φ i2 Q i2 V i2 Mode 3 φ i3 Q i3 V i3 3. This was done in the previous example for the same building and we found the base shear as 1.000 4. Now.5 .8 -0.716 0.5. Thus.4)2+ (30.4. ( S a / g ) = 2.4 86.2 requires that the base shear obtained by dynamic analysis (VB = 610 kN) be compared with that obtained from empirical fundamental period as per Clause 7.016 31.831 -0.16 .8 155.6)2+ (5.Examples on IS 1893(Part 1) (clause 7.000 0.4.921 114.265 sec. the contribution of different modes is combined by the SRSS (square root of the sum of the square) method V4 = [(155. the response quantities are to be scaled up.216 -26.5 196.5 352.6 -88. Table 2.701 87.8 0.8.5 c of IS: 1893 Part 1) The value of Ahk for different modes is obtained from clause 6.8.6” in two ways: 1.1 -25. the seismic forces obtained above by dynamic analysis should be scaled up as follows: Q4 = 182 × 2.86 ZI Ah1 = (S a / g ) 2R 0.30 = 435 kN Q2 = 139 × 2.6 45. ZI Ah 3 = (S a / g ) 2R 0.9 96. If VB is less than that from empirical value.36 × 1 = × (2.0 Example 2/Page 8 . T3 = 0. ( S a / g ) = 2.5f of IS: 1893 Part 1) IITK-GSDMA-EQ21-V2.145 sec.2) + (14. We calculate base shear as per Cl. Hence.8)2+ (31.404 kN.5)2+ (88.36 × 1 = × (2.0418 × 1.30 = 419 kN Q3 = 189 × 2.16) 2×5 = 0.
For instance. M (Static) 1.450 × 0.907 kNm 5.14 = 297 kN 2.303 kN Notice that most of the base shear is contributed by first mode only.270 kNm 12.28 sec).632 kNm 15.6 kN. for mode 1: T1 = 0.Examples on IS 1893(Part 1) Q1 = 100 × 2.14 = 404 kN 2.782 kNm 7.2 + 14.14).09 = 1.750 kNm Notice that even though the base shear by the static and the dynamic analyses are comparable. scaled) 389 kN Storey Moment. In that case.2 kN and 14.09 Modal mass times Ah1 = 14.303/610 = 2.090 kN 1. there is considerable difference in the lateral load distribution with building height. respectively. base shear in first mode of vibration =1300 kN. M (Dynamic) 1.386 kNm 9.300 kN 2.30 = 230 kN = 1. For instance. We could make either interpretation.491 kN Storey ShearV (dynamic. In this interpretation of Cl 7. the storey moments are significantly affected by change in load distribution. = ZI (S a / g ) 2R =0.0 Example 2/Page 9 . IITK-GSDMA-EQ21-V2.245 kNm 3.304 kN Storey Moment.2.14 = 214 kN Base shear in modes 2 and 3 is as calculated earlier: Now.5 . 86. and therein lies the advantage of dynamic analysis. We may also interpret this clause to mean that we redo the dynamic analysis but replace the fundamental time period value by Ta (= 0.4 – Base shear at different storeys Floor Level i 4 3 2 1 Q (static) Q (dynamic. we need to scale up the values of response quantities in the ratio (1.530 kNm 793 kN 1.6 Table 2. Herein we will proceed with the values from the second interpretation and compare the design values with those obtained in Example 1 as per static analysis: 2 2 2 = 1300 + 86. Ah1 ( S a / g ) = 2. the external seismic forces at floor levels will now be: Q4 = 182 Q3 = 189 Q2 = 139 Q1 = 100 × × × × 2.28 sec. Total base shear by SRSS Clearly.115kN 1. scaled) Storey Shear V (static) 611 kN 504 kN 297 kN 79 kN 389 kN 404 kN 297 kN 214 kN 611 kN 1.8.14 = 389 kN 2.412kN 1. the second interpretation gives about 10% lower forces.
Examples on IS 1893(Part 1) Example 3 – Location of Centre of Mass Problem Statement: Locate centre of mass of a building having non-uniform distribution of mass as shown in the figure 3. 4. coordinates of centre of mass are (9. while that of the other two parts is 1000 kg/m2. and the coordinates of the centre of mass be at (X.1 10 m I 4m 1200 kg/m2 1000 kg/m2 III 8m II Y= (10 × 4 × 1200) × 6 + (10 × 4 × 1000) × 6 + (20 × 4 × 1000) × 2 (10 × 4 × 1200) + (10 × 4 × 1000) + (20 × 4 × 1000) = 4. Let origin be at point A.76.2 Mass of part I is 1200 kg/m2.1 10 m 4m 1200 kg/m2 1000 kg/m2 8m A 20 m Figure 3.1) 20 m Figure 3.1 m Hence.1 –Plan Solution: Let us divide the roof slab into three rectangular parts as shown in figure 2. Y) (10 × 4 × 1200) × 5 + (10 × 4 × 1000) × 15 + (20 × 4 × 1000) × 10 (10 × 4 × 1200) + (10 × 4 × 1000) + (20 × 4 × 1000) = 9.76 m X = IITK-GSDMA-EQ21 –V2.0 Example 3 /Page10 . .
Y).0). IITK-GSDMA-EQ21 –V2. However. 5. In the Y-direction.. Let the lateral stiffness of each transverse frame be k.0 m from the left bottom corner. All columns and beams are same. there are four identical frames having equal lateral stiffness. 5m 5m 5m 5m Figure 4. X = k × 0 + k × 5 + k × 10 + k × 20 = 8.1. the spacing is not uniform. at 5.1 –Plan 10 m Solution: In the X-direction there are three identical frames located at uniform spacing. the ycoordinate of centre of stiffness is located symmetrically.Examples on IS 1893(Part 1) Example 4 – Location of Centre of Stiffness Problem Statement: The plan of a simple one storey building is shown in figure 3.0 Example 4 /Page11 . i.75 m k+k+k+k Hence. and coordinating of center of stiffness be (X. Obtain its centre of stiffness.e. Hence.75. coordinates of centre of stiffness are (8.
All four walls are in M25 grade concrete. k. Floor consists of cast-in-situ reinforced concrete. calculated eccentricity = 0.4 .0.0). (8.0 m rB = -6. 4. Centre of rigidity (CR) will be at (6. Compute design lateral forces on different shear walls using the torsion provisions of 2002 edition of IS 1893 (Part 1). spaces have same lateral stiffness. EQ Force in X-direction: Because of symmetry in this direction. All the walls have same stiffness.05 × 8 = −0.0 m Design eccentricity: ed = 1.0 m ed = 0. It has some gravity columns that are not shown. B .0 kN KC + K D KD F = 50.e.4 (Clause 7.5 m.0 − 0.0. KA = KB = KC = KD = k. Y 2m 4m C 4m 4m A B 8m D 16m Figure 5. Centre of mass (CM) will be the geometric centre of the floor slab. and rA = -6. and IITK-GSDMA-EQ21 –V2.0 Example 5 /Page 12 .0 kN FDT = KC + K D Lateral forces in the walls due to torsional moment: K i ri (Fed ) FiR = K i ri2 i = A . 200 thick and 4 m long. Design shear force on the building is 100 kN in either direction. and hence.9.05 × 8 = 0.0).5 × 0. i.2 of IS 1893:2002) Lateral forces in the walls due to translation: KC FCT = F = 50.Examples on IS 1893(Part 1) Example 5 –Lateral Force Distribution as per Torsion Provisions of IS 1893-2002 (Part 1) Problem Statement: Consider a simple one-storey building having two shear walls in each direction.0 + 0. D ∑ where ri is the distance of the shear wall from CR. 4. Storey height is 4..C .1 – Plan X Solution: Grade of concrete: M25 E = 5000 25 = 25000 N/mm2 Storey height h = 4500 m Thickness of wall t = 200 mm Length of walls L = 4000 mm All walls are same.
4.54 kN Total lateral forces in the walls due to seismic load in X direction: FA = 2.2 m.92 kN FC =51.0 m Design eccentricity: ed = 1.05 × 16 = 1.0 + 0.62) = 14.77= 71. FBR = ± 2.0 kN KA + KB KB FBT = F = 50.62 kN.62.62) = 14.93 = 43.9.62 kN Total lateral forces in the walls: FA = 50 .54 ) = 51. when ed = 1.0 m rD = -4. wall A should be designed for not less than 50 kN.2 m Lateral forces in the walls due to translation: KA F = 50.93 = 56.92 kN FC = -14.92= 28.62 kN FDR = 14.54 kN. and ed = ±0.62.07 kN FB = 50 + 6.8 m FAT = IITK-GSDMA-EQ21-V2. 4.92 kN ( rA k (Fed ) = 2 r + r + rC2 + rD k 2 A 2 B ) - FAR rA k (Fed ) = 2 2 rA + rB2 + rC2 + rD k ( ) Similarly.5 × 2. Combining the forces obtained from seismic loading in X and Y directions: FA = 43.0 Example 5/Page 13 .21.4.54 kN EQ Force in Y-direction: Calculated eccentricity= 2.54 ) = 51.62 kN FD = 4.93 kN FC = . FC = Max (14. However.0 m.54 kN FD =51. FA = 50 – 6.92 kN FCR = -14.62 kN.0 kN KA + KB Lateral force in the walls due to torsional moment: when ed = 3.05 × 16 = 3. negative torsional shear shall be neglected”.07) = 43.07 kN FB =71.62 kN FD = 14.31 kN FC = Max (50 ± 1.62 kN Maximum forces in walls due to seismic load in Y direction: FA = Max (28. then the total lateral forces in the walls will be.8 m or = 2. = ± 2. FB = Max (71.31 kN FB = 2.62 kN Similarly. note that clause 7.Examples on IS 1893(Part 1) rC = 4. Hence. 56.0 − 0. 43. FBR = 21.92. FAR = 21.93) = 71.54 kN FDR = ± 1.1 also states that “However.08.4 m Therefore.31 kN FCR = ± 1.08 kN FB = 50 +20.92 kN.31 kN Similarly. FD = Max (14.54 kN FD = Max (50 ± 1.07 kN.
0).92 kN Example 6 /Page 14 .0. k..62 kN FB = .Examples on IS 1893(Part 1) Example 6 – Lateral Force Distribution as per New Torsion Provisions Problem Statement: For the building of example 5.4.e.8 (clause 7. and hence.0). B .0 m FAR = ( rA k (Fed ) 2 r + r + rC2 + rD k 2 A 2 B ) Design eccentricity. i.0 m FiR = i = A . compute design lateral forces on different shear walls using the torsion provisions of revised draft code IS 1893 (part 1). same lateral stiffness.. D K i ri (Fed ) ∑ K i ri 2 where ri is the distance of the shear wall from CR All the walls have same stiffness. ed = 0. calculated eccentricity = 0.08 kN FD = 50-3.08 = 46.08 = 53.0.1 × 8 = ±0. 4. IITK-GSDMA-EQ05-V2. Centre of rigidity (CR) will be at (6.08 kN FDR = -3. Y 2m 4m 4m C 6m 4m A B 8m D 16m Figure 6. i. FBR = 4.0 ± 0.08 kN Total lateral forces in the walls: FA = 4.0 kN KC + K D Lateral forces in the walls due to torsional moment: IITK-GSDMA-EQ21 –V2.1 – Plan X Solution: Grade of concrete: M25 E = 5000 25 = 25000 N/mm2 Storey height h = 4500 m Thickness of wall t = 200 mm Length of walls L = 4000 mm All walls are same. KA = KB = KC = KD = k rA= -6.2 of Draft IS 1893: (Part1)) Lateral forces in the walls due to translation: KC F = 50.0 = .9.0 m rC= 4.62 kN FC = 50+3.62 kN Similarly.0 kN FCT = KC + K D KD FDT = F = 50.0 m rB= -6.62 kN FCR = 3.C . 4.0 m rD= -4. Centre of mass (CM) will be the geometric centre of the floor slab.4. EQ Force in X-direction: Because of symmetry in this direction.0. (8.e.
05 kN.77= 70.1.6 m FAT = FAR = 20.31 kN FC = 1.1 × 16 = 0.1 × 16 = 3.6 m or ed = 2.77 kN ( rA k (Fed ) = 2 r + r + rC2 + rD k 2 A 2 B ) - IITK-GSDMA-EQ21-V2.77 kN Design lateral forces in all the walls are as follows: FA =47. FBR = 20.0 − 0.85 kN FD = -13.08 = 53.69 kN FB = 50+2.77 kN FCR = 13.77 kN FC = 13. then the lateral forces in the walls will be.77= 29. FB =70.31= 53. Lateral forces in the walls due to translation: KA F = 50.0 kN KA + KB KB F = 50.85 kN FDR = -13.54 kN Maximum forces in walls A and B FA =47.Examples on IS 1893(Part 1) Similarly. then the total lateral forces in the walls will be. when ed= 0.0 kN FBT = KA + KB Lateral force in the walls due to torsional moment: when ed= 3.31= 47.08 = 46.54 kN FD = .05 kN FD =53.4.92 kN FD = 50+3. FA = 50-2.8 m.08kN Design lateral forces in walls C and D are: FC= FD= 53.85 kN Similarly.8 kN Total lateral forces in the walls: FA = 50-20. when ed= .62 kN FB = 4. FA = .0 + 0.05 kN EQ Force in Y-direction: Calculated eccentricity= 2.0.4 m Similarly.69 kN.77 kN FC =53.69 kN FB =70.0 m Design eccentricity.4 m. ed = 2.23 kN FB = 50+20.62 kN FC = 50-3.0 Example 6/Page 15 .
Height of the building.5 = 9. Height of point of attachment of the equipment above the foundation of the building.2 m.0 × 4) m = 16.2 m. Wp = 100 kN IITK-GSDMA-EQ21-V2.2 +3. Determine the shear and tension demands on the anchored bolts during earthquake shaking. for the equipment Fp =10.1) is to be installed on the roof of a five storey building in Simla (seismic zone IV).24 ⎛ 16. Example 7/Page 16 .1– Equipment installed at roof Solution: Zone factor. x = (4.0 m Figure 7. one at each corner of the equipment. design seismic force.24 (for zone IV. Z = 0.0 (1)(100 ) kN ⎜1 + ⎟ 2 ⎝ 16. It is attached by four anchored bolts.5 (Table 11). a p = 1 (rigid component.2 ⎠ 2. embedded in a concrete slab.5 m Anchor bolt Anchor bolt 1.0 The design seismic force Z ⎛ x⎞ a Fp = ⎜1 + ⎟ p I pW p 2 ⎝ h ⎠ Rp = 0. Importance factor Ip = 1 (not life safety component. except the ground storey which is 4.6 kN < 0. Wp Fp CG 1. h = 16.0 m.0kN Hence.0 kN.2 ⎞ 1. Response modification factor Rp = 2.1W p = 10. Floor to floor height of the building is 3. Table 11). Table 2 of IS 1893). Table 12). Amplification factor of the equipment.Examples on IS 1893(Part 1) Example 7 – Design for Anchorage of an Equipment Problem Statement: A 100 kN equipment (Figure 7.2 m. Weight of the equipment.
3.0)(2) =15.4 of draft IS 1893) Shear per anchor bolt.0 kN The overturning moment is M ot = 2.0 kN) × (1.0 kN-m The overturning moment is resisted by two anchor bolts on either side.0/4 kN =5.13. Hence.0 Example 7/Page 17 .0 × (10. V = 2Fp/4 =2 × 10.5 m) = 30. tension per anchor bolt from overturning is Ft = (30.Examples on IS 1893(Part 1) The anchorage of equipment with the building must be designed for twice of this force (Clause 7.0) kN (1.0kN IITK-GSDMA-EQ21-V2.
except the ground storey which is 4.0 m. Determine the shear and tension demands on the isolators during earthquake shaking. It is to be mounted on four flexible vibration isolators.8 m 1. Z = 0.Examples on IS 1893(Part 1) Example 8 – Anchorage Design for an Equipment Supported on Vibration Isolator Problem Statement: A 100 kN electrical generator of a emergency power supply system is to be installed on the fourth floor of a 6-storey hospital building in Guwahati (zone V). Table 12).2 m. x = (4.1 – Electrical generator installed on the floor Solution: Zone factor.36 (for zone V. Table 11).2 m. to damp the vibrations generated during the operation. one at each corner of the unit.0 Example 8/Page 18 .5 (flexible component. Response modification factor Rp = 2. Weight of the generator. Fp = Z ⎛ x ⎞ ap I pW p ⎜1 + ⎟ 2 ⎝ h ⎠ Rp IITK-GSDMA-EQ21-V2.5 (life safety component. Height of the building. a p = 2. Table 11). Table 2 of IS 1893). Amplification factor of the generator. Wp = 100 kN The design lateral force on the generator.2 + 3.2 + 3. Wp Fp Vibration Isolator CG 0 .2 m. Height of point of attachment of the generator above the foundation of the building.0 × 5) m = 19.0 × 3) m = 13. Importance factor Ip = 1.2 m Figure 8. h = (4. Floor to floor height of the building is 3.5 (vibration isolator.
2 kN Shear force resisted by each isolator.8 kN The overturning moment. Fp = 2 × 45.0 Example 8/Page 19 .0kN Since the generator is mounted on flexible vibration isolator.. Therefore.8 m ) = 73.6 kN = 91.2 kN ) × ( 0.4 kN IITK-GSDMA-EQ21-V2.5 = 45. tension or compression on each isolator.1Wp = 10.2 )( 2 ) = 30. the design force is doubled i.2 ⎠ 2.2 ⎞ 2.5)(100 ) kN ⎜1 + ⎟ 2 ⎝ 19. M ot = ( 91.0 ) kN (1.0 kN-m The overturning moment (Mot) is resisted by two vibration isolators on either side. Ft = ( 73.5 (1.36 ⎛ 13.e.6 kN 0.Examples on IS 1893(Part 1) = 0. V = Fp/4 = 22.
004 times the storey height.004) mm = 80.1 is IS : 1893 is 0. (ii) Alternatively.0 m. respectively.0 m Deflection at building level x of structure A due to design seismic load determined by elastic analysis = 35.0 mm =5 (12000.0 mm Response reduction factor of the building R = 5 (special RC moment resisting frame. From the elastic analysis under design seismic load.. Height of level x to which upper connection point is attached. Maximum interstorey drift allowance as per clause 7.0) mm = 50.0)(0. i.0 m and 8. assuming that the analysis of building is not possible to assess deflections under seismic loads.0 m Height of level y to which lower connection point is attached.Examples on IS 1893(Part 1) Example 9 – Design of a Large Sign Board on a Building Problem Statement: A neon sign board is attached to a 5-storey building in Ahmedabad (seismic zone III). Table 7) (i) D p = δ xA − δ yA = (175. hx = 12.0 – 8000. Solution: Since sign board is a displacement sensitive nonstructural element. It is attached by two anchors at a height 12. it is found that the deflections of upper and lower attachments of the sign board are 35. Δ aA = 0.0 mm and 25. it should be designed for seismic relative displacement. hy = 8.0 mm Deflection at building level y of structure A due to design seismic load determined by elastic analysis = 25.11.e.0 mm IITK-GSDMA-EQ21-V2.0 mm.0 mm The neon board will be designed to accommodate a relative motion of 80 mm. Find the design relative displacement.0 – 125.0 Example 9/Page 20 . δ yA = 5 x 25 = 125.0 mm Design the connections of neon board to accommodate a relative motion of 50 mm. one may use the drift limits (this presumes that the building complies with seismic code).004 hsx D p = R (hx − h y ) Δ aA hsx δ xA = 5 x 35 = 175.
75 6.24 .81 × (235.81 Critical stress ratio induced by earthquake: a max = 0. Table 10.24g corresponding to zone factor Z = 0.79 1 / σ v' ( ) 1/ 2 Example 10/Page 21 .75 N 60 9 17 13 18 17 15 26 Soil Classification Poorly Graded Sand and Silty Sand (SP-SM) Percentage fine 11 16 12 8 8 7 6 Poorly Graded Sand and Silty Sand (SP-SM) Poorly Graded Sand and Silty Sand (SP-SM) Poorly Graded Sand and Silty Sand (SP-SM) Poorly Graded Sand and Silty Sand (SP-SM) Poorly Graded Sand and Silty Sand (SP-SM) Poorly Graded Sand and Silty Sand (SP-SM) evaluated = 12.5 .2 kPa σ v' = (σ v − u 0 ) = 235.65 × (0.7 kPa Stress reduction factor: Step by step calculation for the depth of 12.24 Liquefaction Potential of Underlying Soil σ v = 12.65 × (a maz / g ) × rd × σ v / σ v' ( ) CSReq = 0.5 CSReq = 0.2. Determine the extent to which liquefaction is expected for 7.75 × 18. and the vertical settlement of the soil due to liquefaction (Δv). The peak horizontal ground acceleration value for the site will be taken as 0.8 = 66.015 × 12.75m is given below. Detailed calculations for all the depths are given in Table 10. = 6.9 / 169.5 kN / m 3 .24 g .1.75 3. The water table is at 6m below ground level. M w = 7. g γ sat = 18.75 15. maximum depth of liquefaction below the ground surface. The SPT values ranges from 9 to 26.75 12.75 9.9 kPa u 0 = (12.75 − 6. M w = 7.0 (N )60 = C N × N 60 C N = 9.8 kN / m 3 Depth of water level below G.18 Correction for SPT overburden pressure: (N) value for a max = 0.5 magnitude earthquake.1: Result of the Standard penetration Test and Sieve Analysis Depth (m) 0. γ w = 9.7 ) = 0.24) × 0.L.015 z = 1 − 0.Examples on IS 1893(Part 1) Example: 10 Liquefaction Analysis using SPT data Problem Statement: The measured SPT resistance and results of sieve analysis for a site in Zone IV are indicated in Table 10.75m Initial stresses: Solution: Site Characterization: This site consists of loose to dense poorly graded sand to silty sand (SP-SM). This table provides the factor of safety against liquefaction (FSliq).00m Depth at which liquefaction potential is to be IITK-GSDMA-EQ21-V2. rd = 1 − 0.9 − 66.75 18.00) × 9. Estimate the liquefaction potential and resulting settlement expected at this location.5 = 235. The site is located in zone IV.75 = 0.2 = 169.
18 = 0.13 0.30 1.15 FS L 1.10 (from Figure F-8) Liquefaction induced vertical settlement (ΔV): = 1.9 180.67 Percentage volumetric strain (%ε) For ( N )60 = 13 .18 / (1x1x0.12 / 0.18 CSR L 0.00 26.5 k m kα kσ k m = Correction factor for earthquake magnitude other than 7.20 0.15 0.00 8.34 0.15 0.Examples on IS 1893(Part 1) C N = 9.14 0.00 13.8 221.09 0.18 0.82 0.9 69.15 0.88) = 0.00 for M w = 7.67 0.94 0.21 0.7 ) 1/ 2 = 0.00.11 0.14 0.90 0.66 ( N )60 18 20 12 15 13 10 17 rd 0.18 0.4 117.88 0.063 0.0 / 100 = 0.051 0.75 × 17 = 13 Critical stress ratio resisting liquefaction: FS L = CSR L / CSReq = 0.00 15.00 7.75 11.76 0.15 0.22 0.99 0.25 0.2: Liquefaction Analysis: Water Level 6.22 CSR7.90 0.00 18.14 (Figure F-2) Corrected Critical Stress Ratio Resisting Liquefaction: For CSReql = CSReq / (k m kα kσ ) = 0.14 0.12 0.069 0.18 0.79 (1 / 169.0 Example 10/Page 22 .70 0.7 195.1 × 3.5 k α = Correction factor for initial driving static shear (Figure F-6) (ΔV) = volumetric strain x thickness of liquefiable level = 2.13 0.12 Factor of safety against liquefaction: (N )60 = 0.9 69. fines content of 8% CSR7.2) = 0.00 CN 2.75 3.27 0.32 0.75 9.00 17.88 = 0.81 0.5 = 0.10 2.85 0.00 12.86 0.50 1.315 IITK-GSDMA-EQ21-V2.21 (N 1 )60 = 13 CSRL = CSR7.4 235.18 CSReql 0.9 291.14 × 1 × 1 × 0.00 17.18 0.00 8.4 124.75 6.075 0.6 169.75 18.00 16.72 Total Δ CSReq 0.17 0.00 1.5 (Figure F-4) %ε = 2.00 13.75 0.5 0.9 13.50 0. The maximum settlement of the soil due to liquefaction is estimated as 315mm (Table 10.75 15. since no initial static shear kσ = Correction factor for stress level larger than 96 kPa (Figure F-5) Analysis shows that the strata between depths 6m and 19.16 0.75 CSR L = 0.00 m below GL (Units: Tons and Meters) σv Depth %Fine (kPa) ' σv (kPa) N 60 9.00 6.90 2.9 0.83 %ε 2.063m = 63mm Summary: = 1.5m are liable to liquefy.15 0.67 2.057 0.5 143.75 12.88 Table 10.70 ΔV 0.4 346.
05 48.595 0.248 0.359 0.50 8.50 11.69 112.74 9.00 15.36.00 12.05 46.39 36.00 17.94 56.155 0. Detailed calculations are given in Table 11.275 0.50 14. The peak horizontal ground acceleration value for the site will be taken as 0.231 0.135 0.02 fs 0.93 53.60 62.2. The results of the cone penetration test (CPT) of 20m thick layer in Zone V are indicated in Table 11.00 10.75 91.50 18. The site is located in zone V.08 fs 0.282 0.50 6.235 0.00 19.30 51.99 48.43 64.375 Solution: Liquefaction Potential of Underlying Soil γ sat = 18 kN / m 3 .00 3.00 1.392 0.50 5.208 0.50 3.16 115.00 5.49 77. the unit weight of the soil to be 18 kN/m3 and the magnitude of 7.28 20.35m Depth at which liquefaction potential is to be evaluated = 4.35 m.00 8.357 0.027 0.193 0.297 0.00 9.161 0.90 104.5 − 2.00 2.00 6.50 4.159 0.50 9.60 46.69 49.69 70.8 kN / m 3 Depth of water level below G.50 1. Mw=7.00 qc 46.94 57. This table provides the factor of safety against liquefaction (FSliq).68 45.5 × 18 = 81.36g corresponding to zone factor Z = 0.93 kPa Stress reduction factor: amax/g = 0.130 0.00 4.219 1.39 54.652 0.099 0. IITK-GSDMA-EQ21 –V2.50 15.50 13.233 0.1.0 Example 11 /Page 23 .50 10.00 18.329 0.00 qc 144.70 51.Examples on IS 1893(Part 1) Example: 11 Liquefaction Analysis using CPT data Problem Statement: Prepare a plot of factors of safety against liquefaction versus depth. γ w = 9.50 17.8 = 21.24 fs 0.93 55.L.36 Initial stresses: σ v = 4.184 0.46 39.184 0.07 kPa σ v' = (σ v − u 0 ) = 81 − 21.291 0.00 11.50 2. Table 11.132 0.00 13.94 63.5m Step by step calculation for the depth of 4.218 0.50 10.144 0.00 qc 45.58 74.75 53.5.77 47.1: Result of the Cone penetration Test Depth (m) 0.50 12.07 = 59.346 0.5.185 0.62 150.217 0.58 41.49 39.00 16.181 Depth (m) 7.50 19.193 0.31 95.39 58.11 33.50 7.602 0.5m is given below.35) × 9.129 0.50 20.256 0.58 52.281 0. = 2.50 16.173 Depth (m) 14.00 kPa u 0 = (4. Assume the water table to be at a depth of 2.
= 1.19 Where.64(101.581(2.19 ) − 17.7 / (3369 − 81)] × 100 = 0.32 = 0.19 ) M + 33.00 .000765 z = 1 − 0.0 4 for I c ≤ 1.93) = 0.34 Summary: Analysis shows that the strata between depths 0-1m are liable to liquefy under earthquake shaking corresponding to peak ground acceleration of 0.903 and IITK-GSDMA-EQ21-V2.5 (3369 101.64 The soil behavior type index.35) = 70.36g.22 + log F )2 (3.903)2 = 2. I c .93) = 42.403I c + 5.64 and 3 2 K c = −0.5 k α = Correction factor for initial driving static shear (Figure F-6) CRR =0.11 / 0.63(2.000765 × 4.65 × (0.5 CSReq = 0.88 = 1.Examples on IS 1893(Part 1) rd = 1 − 0. F = f (q c − σ v ) × 100 F = [29.11 (Figure F-6) FS liq = CRR / CSR L FS liq = 0.65 × (a maz / g ) × rd × σ v / σ ( ) K c = −0. is given by Ic = Ic = (3.5 (Figure F-4) (q c1N )cs = 1.63I c M + 33.32 × 1 × 1 × 1 = 0. The plot for depth verses factor of safety is shown in Figure 11.77 .19 ) − 21.5 = 0.00 for M w = 7.22 + log 0.47 − log Q )2 + (1.93)0.1 = 1.36) × 0.35 59.32 Correction factor for grain characteristics: K c = 1 .403(2.47 − log 42.32 Corrected Critical Stress Ratio Resisting Liquefaction: Normalized Cone Tip Resistance: (qc1N )cs ′ ) (q c Pa ) = K c (Pa σ v n CSRL = CSReq k m kα kσ k m = Correction factor for earthquake magnitude other than 7.19 ) + 5.75 I c − 17.77 Factor of safety against liquefaction: For (q c1N )cs = 70.0 Example 11/Page 24 .75(2. since no initial static shear kσ = Correction factor for stress level larger than 96 kPa (Figure F-5) = 1.88 for I c > 1.64 2 CSReq = 0.35] × (101.19 4 3 0.35 59.19)2 + (1.997 Critical stress earthquake: ratio induced by ' v ′) Q = [(q c − σ v ) Pa ](Pa σ v n Q = [(3369 − 81) 101.997 × (81 / 59.581I c − 21.00 CSR L = 0.
55 0.59 0.90 21.23 0.83 182.35 0.01 1.50 9.71 85.26 0.00 333.00 σv ' 9.57 0.22 0.50 20.88 1.21 1.84 56.70 23.47 1.33 1.72 0.33 1.50 11.00 1.68 1.29 0.36 0.36 0.26 0.80 17.18 1.24 0.48 0.67 IITK-GSDMA-EQ21-V2.90 18.06 1.00 1.50 62.72 0.36 0.99 0.31 1.60 25.23 0.53 129.50 9.00 27.23 0.00 162.14 FSliq 434.95 1.48 51.21 1.34 0.00 180.40 0.52 0.50 6.25 1.79 1.10 0.19 0.39 44.46 0.00 243.73 0.13 113.96 11.36 0.25 0.00 14.00 126.38 0.93 187.19 86.47 qc (kPa) 14431 9549 3928 2062 15093 5550 1074 911 3369 7069 4970 5143 6494 5724 4546 3939 3668 4530 5105 4639 5805 4894 6375 5393 5360 6239 5458 5208 4660 4677 4758 4199 4894 5669 11290 10449 7775 9158 7416 11502 fs (kPa) 65.50 18.00 19.22 0.00 306.92 5.26 44.35 88.47 416.39 0.53 47.53 0.45 0.24 0.20 60.10 0.08 0.00 0.03 rd 1.03 109.26 0.23 0.44 0.38 0.00 225.34 0.16 58.34 0.83 1.62 59.80 15.46 55.26 0.23 76.53 88.10 0.00 261.56 0.00 1.00 13.48 0.50 16.30 24.36 0.31 1.56 46.00 324.08 2.10 0.46 0.29 0.24 0.00 1.50 2.00 270.28 0.00 27.00 207.06 (qc1N)cs 242.99 1.50 1.00 1.50 CSReq 0.60 28.60 72.63 92.10 13.13 0.23 0.35 0.00 1.21 F 0.21 53.00 135.28 0.29 0.00 11.24 0.77 96.00 72.98 1.0 Example 11/Page 25 .99 0.63 174.78 0.63 51.00 1.95 1.73 178.30 0.37 0.73 1.23 0.43 33.63 57.00 198.45 83.35 0.00 45.34 0.73 137.72 1.42 0.48 0.87 65.00 234.50 17.23 0.00 117.79 41.27 0.10 1.50 0.27 0.53 2.36 0.30 0.00 18.35 0.64 0.92 2.40 1.80 19.00 1.39 0.53 68.25 0.40 0.93 146.21 CSRL 0.67 57.31 0.81 1.13 1.79 62.90 18.94 68.83 1.57 64.31 31.40 39.24 1.36 0.04 227.28 0.28 0.26 1.73 0.28 0.26 0.10 0.88 1.14 0.29 41.10 0.53 52.00 2.65 0.34 0.00 2.00 9.00 1.53 170.28 36.00 1.11 1.60 0.00 32.11 2.50 3.10 15.09 0.99 0.06 160.63 0.50 23.48 68.90 102.00 100.00 90.18 50.75 0.35 0.68 36.83 59.00 99.78 434.67 0.02 2.78 43.24 0.33 1.80 84.11 0.00 2.09 60.46 75.46 0.11 0.34 Q 241.63 1.95 2.56 59.68 72.10 0.19 2.36 0.00 1.00 8.11 0.06 2.09 0.29 0.10 0.28 91.00 63.50 35.10 0.90 14.06 2.33 0.29 0.34 0.40 0.17 1.83 2.27 0.00 153.20 13.26 0.93 36.50 19.00 15.20 28.00 7.00 108.73 Kc 1.22 1.01 1.11 0.35 m below GL (Units: kN and Meters) Depth 0.34 0.25 0.25 0.50 15.50 20.16 0.33 0.50 7.00 47.25 0.23 0.50 14.67 0.17 83.12 0.41 0.37 0.32 0.33 0.61 0.32 0.67 45.64 70.93 105.35 0.39 1.23 117.23 0.33 80.35 0.00 279.00 54.38 0.57 0.10 13.00 12.00 36.34 0.00 18.23 0.00 0.23 36.02 87.28 0.30 16.03 150.23 158.45 0.12 0.34 0.33 121.30 23.99 0.01 2.43 0.00 6.43 125.00 351.48 76.00 0.15 42.29 0.23 0.43 1.83 100.44 63.23 105.62 58.2: Liquefaction Analysis: Water Level 2.37 0.02 1.00 17.04 36.00 18.63 58.30 0.17 1.33 0.25 0.90 12.00 144.10 0.26 0.92 2.00 360.96 2.81 60.92 65.24 1.93 1.51 0.10 0.61 0.37 0.32 0.36 0.32 0.13 0.00 43.38 0.00 36.38 1.38 0.43 166.27 0.00 288.75 1.08 3.90 0.40 43.24 55.03 68.50 8.35 0.63 0.29 0.54 226.22 53.38 0.35 68.19 1.50 5.28 0.09 0.50 13.68 0.45 56.24 0.33 0.22 0.00 252.85 72.37 0.36 0.45 0.10 0.30 0.70 35.22 0.20 34.13 154.37 0.00 1.00 1.47 CRR 100.00 216.00 10.43 84.28 0.39 0.28 0.00 315.73 55.65 1.97 2.11 0.10 0.72 0.09 48.93 Ic 1.00 3.51 0.14 0.13 72.63 133.37 0.00 81.00 171.23 1.11 0.27 1.10 27.27 0.30 1.29 1.50 10.55 1.89 1.50 18.27 1.00 1.01 68.00 189.00 4.09 0.61 71.51 61.00 1.70 37.40 29.00 5.55 79.91 159.02 49.13 1.00 σv 9.00 1.69 0.00 342.90 53.34 0.20 21.50 19.24 0.63 0.83 141.23 62.23 0.26 0.30 29.93 64.92 1.71 0.15 0.50 12.00 1.48 0.00 297.70 59.50 4.02 1.00 16.11 100.10 21.66 40.89 80.30 1.49 0.99 0.31 5.55 0.30 0.40 13.31 0.64 1.10 18.10 0.07 1.33 162.99 55.29 0.30 0.73 96.Examples on IS 1893(Part 1) Table 11.12 0.97 2.10 0.
5 1.0 0 0.Examples on IS 1893(Part 1) Factor of Safety 0.0 1.5 2.1: Factor of Safety against Liquefaction IITK-GSDMA-EQ21-V2.0 Example 11/Page 26 .0 3 5 8 Depth (m) 10 13 15 18 20 Figure 11.
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