Calculation of earthquake base shear and the distribution of lateral forces at each story level using the Equivalent Lateral Force Procedure

 

 

Earthquake base shear and the lateral forces at each story level using the Equivalent Lateral Force Procedure

A six-story reinforced concrete (RC) residential building is located in Sylhet, Bangladesh, which lies in Seismic Zone 4 according to BNBC 2020. Each story of the building has a height of 3 m, and the depth of the foundation below ground level is 2 m. The plan dimensions of the building are 20 m × 20 m. The damping correction factor (η) is 1.2. The dead load, including the self-weight of the structural components, is 10 kN/m², while the live load is 2 kN/m² for the typical floors and 1 kN/m² for the roof. The site has an average SPT (N₃₀) value of 13.

In accordance with the provisions of BNBC 2020, determine the design earthquake base shear and the lateral forces at each story level using the Equivalent Lateral Force Procedure. The necessary reference tables and information’s are provided.

Solution:

Here,

Site class = SD (As SPT value = 13 <15) [From table 6.2.13]

Z= 0.36 (Sylhet-Seismic Zone IV)

I = 1 (Residential Building) [From table 6.2.17]

R = 8 (Special Moment Resisting Frame: As Located in Zone IV and Site class SD and Design Category D)

T = C_t (H_n)^m , C_t = 0.0466, m = 0.9 [From table 6.2.20]

 H_n = 6 x 3 + 2 = 20 m

Hence,

 T = 0.691

 

From table 6.2.16:

S = 1.35

T_B = 0.20

T_C = 0.80

T_D = 2.0

 

T_B  ≤ T ≤ T_C

So, C_S = 2.5 S η = 2.5 x 1.35 x 1.2 = 4.05 [ η = 1.2, Given]

 

Now,

Sa = 2/3 * ZI/R * Cs = 0.1215

Sa = 2/3 * ZIβS = 0.036 [ β= 0.11]

Hence, Sa = 0.1215 [As Maximum]

 

Calculating Seismic Weight, [Section 2.5.7.3]

 

Total Dead Load, DL = 10 x 20 x20 x 7 = 28,000 kN [ Ground Floor to Roof]

Live Load for Floor, LLF = 2 x 20 x 20 x 6 x 0.25 = 1,200 kN [ Ground Floor to fifth Floor]

Live Load for Roof, LLR = 1 x 20 x 20 x 1 x 0.25 = 100 kN

 

Total weight, W = 28000 + 1200 + 100 = 29300 kN

 

Base Shear, V = Sa x W = 0.1215 x 29300 = 3559.95 kN

 

Vertical distribution of lateral forces:

  

K = 1.0955 [Interpolating values between 1 and 2.5 for 0.691]

 

Per floor dead load = 10 x 20 x20 = 4,000 kN

Per floor Live Load = 2 x 20 x 20 x 0.25 = 200 kN

Roof Live Load = 1 x 20 x 20 x 0.25 = 100 kN

 

Typical Floors total load (Per Floor) = 4000 + 200 = 4200 kN

Roof total load = 4000 + 100 = 4100 kN

 

[4200 x 20^^1.0955 + 4200 x 17^^1.0955 + 4200 x 14^^1.0955 + 4200 x 11^^1.0955 + 4200 x 8^^1.0955 + 4200 x 5^^1.0955 + 4200 x 2^^1.0955 = 413595.35]

 

F_Roof = 3559.95 x =  939.57 kN

 

F_5th Floor = 3559.95 x =  805.52 kN

 

F_4th Floor = 3559.95 x =  651.18 kN

 

F_3rd Floor = 3559.95 x =  500 kN

 

F_2nd Floor = 3559.95 x =  352.74 kN

 

F_1st Floor = 3559.95 x =  210.62 kN

 

F_Ground Floor = 3559.95 x =  77.25 kN

 

Fig: Design lateral seismic forces