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Numerical problem of is code 1893: 2002

Numerical problem of is code 1893: 2002

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- 1. Earthquakes LoadsEarthquakes Loads As per IS 1893 : 2002As per IS 1893 : 2002
- 2. BuildingBuilding Step: 1 Design Horizontal Seismic Coefficient, Ah = ZISa / 2Rg where, Z = Zone Factor (Table 2) I = Importance Factor (Table 6) R = Response Reduction Factor (Table 7) Sa/g = Average Response Acceleration Coefficient (Fig. 2) Step: 2 Calculation of Seismic Weight , Step:3 Design Seismic Lateral Force : Vb = Ah . W W = Seismic weight of building (clause 7.4.2)
- 3. Design Horizontal Seismic CoefficientDesign Horizontal Seismic Coefficient It is the factor which is multiplied with Seismic wt. to determine design lateral force. Ah = ZISa / 2Rg 1) Z = Zone Factor Seismic Zone Z II 0.10 III 0.16 IV 0.24 V 0.36
- 4. MUSE 11BMUSE 11B
- 5. 2) I = Importance factor 3) R = Response Reduction Factor Sr. No. Structure I I Important services and community buildings 1.5 II All other building 1.0 Sr. No. Structure R I Ordinary RC moment resisting frame 3 II Special RC moment- resisting frame 5
- 6. 4) Sa / g = Spectral Acceleration Coefficient Step : 2 Seismic Weight Sr. No. Floor Load % I Upto 3 25 II Above 3 50
- 7. ExampleExample 10m 5m 10m Consider a single storey reinforced concrete office building which is located in Ludhiana. The soil conditions are medium stiff . The lumped weight due to dead loads is 12kN/m2 on roofs and a live load of 4 kN/m2 . Determine design seismic load on the structure as per new code.
- 8. Data GivenData Given • Zone factor (Z) = 0.24 ( Ldh in Zone IV) • Importance factor (I) = 1.0 (Office Building) • Response reduction factor = 5 (R.C.C. Frame) • Live Load = 4 KN/m2 • Dead load = 12KN/m2 +
- 9. Design Horizontal Seismic CoefficientDesign Horizontal Seismic Coefficient • Fundamental period = 2 sec • Soil Condition = medium stiff • Hence Sa / g = 2.5 Ah = ZISa / 2Rg = 0.24 x 1.0 x 2.5 / 5x2 = 0.06
- 10. G. N. D..E.C.G. N. D..E.C. Seismic WeightSeismic Weight Floor area = 10 X 10 m2 Live Load = 4 KN/m2 Only 50% of live load is to be lumped at roofs. Dead load = 12KN/m2 Seismic Weight = 100 X (4*0.5+ 12) = 1400KN
- 11. Seismic Shear forceSeismic Shear force Vb = Ah . W = 0.06 x 1400 = 84 KN
- 12. • Problem: A single storey R.C.C. building is situated at Jalandhar. The soil below foundation is hard strata. Determining seismic shear force acting on the building. Compare the seismic shear force with an identical building in Bangalore. take fundamental time period as unity and seismic weight of building = 1500 KN.
- 13. Case I: Building in JalandharCase I: Building in Jalandhar • Zone factor (Z) = 0.24 As Jalandhar in Zone IV • Importance factor (I) = 1.5 • Response reduction factor = 5 (R.C.C. frame building) • Spectral acceleration Sa/g = 1 (Hard Strata) • Seismic weight (W) = 1500 KN
- 14. Seismic Base ShearSeismic Base Shear Vb = Ah . W = ZISa / 2Rg x W = 0.24x 1.5x1x15000/2/5 = 540 KN
- 15. Case II: Building in BangaloreCase II: Building in Bangalore • Zone factor (Z) = 0.10 As Bangalore in Zone IV • Importance factor (I) = 1.5 • Response reduction factor = 5 (R.C.C. frame building) • Spectral acceleration Sa/g = 1 (Hard Strata) • Seismic weight (W) = 1500 KN
- 16. Seismic Base ShearSeismic Base Shear Vb = Ah . W = ZISa / 2Rg x W = 0.10x 1.5x1x15000/2/5 = 225 KN Result: Comparison of base shear values in the above two cases show that seismic design force in Jalandhar is 2.4 times that of Bangalore for identical type of building $ soil condition.
- 17. ExampleExample • Problem: Considering a four storey building of R.C.C. is located in Bhuj. The soil conditions are medium stiff and entire building is located on raft footing. The lumped weight due to dead load is 12 KN/m2 on floors and 10KN/m2 on roof. The floor are cater to carry a live load of 4KN/ m2 on floor and 1.5 KN/m2 on roof. Determine design seismic load on structure.
- 18. Data GivenData Given • Zone factor (Z) = 0.36 As Bhuj in Zone IV • Importance factor (I) = 1.0 • Response reduction factor = 5 (R.C.C. frame building) • Spectral acceleration Sa/g = 1 (Hard Strata)
- 19. Seismic weightSeismic weight The floor area is 15×20=300 sq. m. Since the live load class is 4kN/sq.m, only 50% of the live load is lumped at the floors. At roof, no live load is to be lumped. Hence, the total seismic weight on the floors and the roof is: Floors: W1=W2 =W3 =300×(12+0.5×4) = 4,200 KN Roof: W4 = 300×10 = 3,000 KN (clause7.3.1, Table 8 of IS: 1893 Part 1) Total Seismic weight of the structure, W = ΣWi = 3×4,200 + 3,000 = 15,600 KN
- 20. Design Horizontal Seismic CoefficientDesign Horizontal Seismic Coefficient • Fundamental period = 2 sec • Soil Condition = medium stiff • Hence Sa / g = 2.5 Ah = ZISa / 2Rg = 0.36 x 1.0 x 2.5 / 5x2 = 0.09
- 21. Seismic Shear forceSeismic Shear force Vb = Ah . W = 0.09 x 15600 = 1440 KN
- 22. Distribution of design force to each floor, Qi = Vb.wi.hi 2 / ∑ wj.hj 2 Qi = Design lateral force at floor i wi = Seismic weight of floor i hi = height of floor i measured from base n = number of floors
- 23. Lateral load distributionLateral load distribution Storey Level Wi hi wi.hi 2 wi.hi 2 / ∑ wj.hj 2 Lateral Force 4 3000 13.8 571.3 .424 611 3 4200 10.6 471.9 .350 501 2 4200 7.4 230 .171 246 1 4200 4.2 74.1 .055 79 1347.3 1000 1440

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