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
𝑇𝑎=0.0488 (143/ 4
) =0.353 Sec
8
Step 6b: Calculate time-period by FE model
𝑇𝑥=0.287 𝑆𝑒𝑐
Since Tx<Ta So select time period Tx=0.287 Sec
9
Step 7: Determine the Seismic Response Coefficient
= 0.1845
𝐶 𝑠=
𝐶 𝑣 𝐼
𝑅 𝑇
𝑉 = 𝐶 𝑠 𝑊
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)
4th
2100 14 29400 0.4 326.12
3rd
2100 10.5 22050 0.3 244.59
2nd
2100 7 14700 0.2 163.06
1st
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
Thank you

Lec 26 - Eqauvalent Static lateral force procedure.pptx

  • 1.
    Reinforced Concrete Design-II Lecture26: Equivalent lateral static procedure Semester – Fall 2020 Dr.Tahir Mehmood
  • 2.
    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.
    3 Step 1: DetermineSeismic Zone Factor Z Step 2 : Determine Soil Profile Type
  • 4.
    4 Step 3: Determinethe Occupancy Categories and Importance Factor I
  • 5.
    5 Step 4: Classifythe Structural System and determine the Response Modification Factor R
  • 6.
    6 Step 5: DetermineGround Response Coefficients and
  • 7.
    7 Step 6a: DetermineFundamental period T 𝑇𝑎=0.0488 (143/ 4 ) =0.353 Sec
  • 8.
    8 Step 6b: Calculatetime-period by FE model 𝑇𝑥=0.287 𝑆𝑒𝑐 Since Tx<Ta So select time period Tx=0.287 Sec
  • 9.
    9 Step 7: Determinethe Seismic Response Coefficient = 0.1845 𝐶 𝑠= 𝐶 𝑣 𝐼 𝑅 𝑇 𝑉 = 𝐶 𝑠 𝑊 Step 8 Determine the Base Shear x8400 =1549.8 KN One story weight = 2100KN
  • 10.
    10 But V cannot exceed by following Equation =815.3 KN =304.9 KN Remember that:
  • 11.
    11 Step 8: VerticalDistribution 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.
    12 Floor Weight ),KN,m Force (KN) 4th 2100 14 29400 0.4 326.12 3rd 2100 10.5 22050 0.3 244.59 2nd 2100 7 14700 0.2 163.06 1st 2100 3.5 7350 0.1 81.53 8400 73500 1 815.3
  • 13.
  • 14.