Lec 26 - Eqauvalent Static lateral force procedure.pptx

lecture

1. Reinforced Concrete Design-II Lecture 26: Equivalent lateral static procedure Semester – Fall 2020 Dr.Tahir Mehmood

2. Case study Building Case study building compose of 4 stories Story height is 3.5 m Total height is 14 m All stories are to be designed as beam-slab system

3. Step 1: Determine Seismic Zone Factor Z Step 2 : Determine Soil Profile Type

4. Step 3: Determine the Occupancy Categories and Importance Factor I

5. Step 4 : Classify the Structural System and determine the Response Modification Factor R

6. Step 5 : Determine Ground Response Coefficients and  

7. Step 6a : Determine Fundamental period T Type equation here.     =0.353 Sec

8. Step 6b : Calculate time-period by FE model   Since Tx<Ta So select time period Tx=0.287 Sec

9. Step 7: Determine the Seismic Response Coefficient   = 0.1845       Step 8 Determine the Base Shear x8400 =1549.8 KN   One story weight = 2100KN  

10. But V can not exceed by following Equation = 815 .3 KN   = 304.9 KN   Remember that:

11. Step 8 : Vertical Distribution of Base Shear into Lateral Forces Ft = 0 Since Tx =0.287 ( Ft =0 for T<0.7 sec) = (2100x14)+(2100x10.5)+(2100x7)+(2100x3.5) =73500 KNm   V= 815 .3

12. Floor Weight ),KN ,m Force (KN) 4 th 2100 14 29400 0.4 326.12 3 rd 2100 10.5 22050 0.3 244.59 2 nd 2100 7 14700 0.2 163.06 1 st 2100 3.5 7350 0.1 81.53 8400 73500 1 815.3 Floor Force (KN) 4 th 2100 14 29400 0.4 326.12 3 rd 2100 10.5 22050 0.3 244.59 2 nd 2100 7 14700 0.2 163.06 1 st 2100 3.5 7350 0.1 81.53 8400 73500 1 815.3

13. 326.12 KN 244.59 KN 163.06 KN 81.53 KN

14. Thank you