V. IBC 2000 Static Seismic
Calculation of base shear and its distribution along the height for equivalent lateral force method in IBC 2000.
Problem
- Short period acceleration, SDS = 0.06 (SDS 0.06)
- 1-Second period acceleration, SD1 = 1.36 (SD1 1.36)
- Site class D (SCLASS 4)
- Building period coefficient = 0.073 (CT 0.073)
- Importance factor = 1.0 (I 1.0)
- Response modification factor X = 3 (RX 3)
The seismic weight for each floor is assumed as 200 kN applied as joint weight of 50kN to nodes 2 3 5 6 8 9 11 To 20. Seismic Load as per IBC 2000 specifications are generated along horizontal direction global X. The base shear and its distribution along height reported by STAAD.Pro is verified against hand calculation.
Calculations
Height = hn = 16m, Ta = CT × hnx = 0.073 × 160.75 = 0.584,
Since SD1 = 1.36 > 0.4, Cu = 1.2
Cu × Ta = 1.2 × 0.584 = 0.701
T (from output file) = 1.286, Cu × Ta = 0.701 < 1.286, Tused = 0.701
Csx = SDS/(Rx/I) = 0.06 / (3/1) = 0.02
Csx upper limit = SD1 / (Tused × (Rx/I)) = 1.36 / (0.701 × (3/1)) = 0.6469
Csx used = 0.2167
Seismic Base Shear V = Cs × W = 0.2167 × (200 × 4) = 173.3 kN
Lateral seismic force Fx = Cvx × V
where= |
Since Tused = 0.701 sec > 0.5 Sec distribution of base shear along story levels would have the value of exponent k linearly interpolated between 1 (less than equal to 0.5 sec) and 2 (greater than equal to 2.5 sec) i.e. k = 1.1004.
STAAD Input
The file C:\Users\Public\Public Documents\STAAD.Pro 2023\Samples \Verification Models\06 Loading\IBC\IBC 2000 Static Seismic.STD is typically installed with the program.
STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 18-May-15
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 4 0; 3 4 4 0; 4 4 0 0; 5 0 8 0; 6 4 8 0; 7 0 0 5; 8 0 4 5;
9 4 4 5; 10 4 0 5; 11 0 8 5; 12 4 8 5; 13 0 12 0; 14 4 12 0; 15 0 12 5;
16 4 12 5; 17 0 16 0; 18 4 16 0; 19 0 16 5; 20 4 16 5;
MEMBER INCIDENCES
1 1 2; 2 2 3; 3 3 4; 4 2 5; 5 5 6; 6 6 3; 7 2 8; 8 3 9; 9 5 11; 10 6 12;
11 7 8; 12 8 9; 13 9 10; 14 8 11; 15 11 12; 16 12 9; 17 5 13; 18 6 14;
19 11 15; 20 12 16; 21 13 14; 22 13 15; 23 14 16; 24 15 16; 25 13 17; 26 14 18;
27 15 19; 28 16 20; 29 17 18; 30 17 19; 31 18 20; 32 19 20;
DEFINE MATERIAL START
ISOTROPIC CONCRETE
E 2.17185e+07
POISSON 0.17
DENSITY 23.5616
ALPHA 1e-05
DAMP 0.05
TYPE CONCRETE
STRENGTH FCU 27579
END DEFINE MATERIAL
MEMBER PROPERTY AMERICAN
1 3 4 6 11 13 14 16 TO 20 25 TO 28 PRIS YD 0.3 ZD 0.3
2 5 7 TO 10 12 15 21 TO 24 29 TO 32 PRIS YD 0.3 ZD 0.25
CONSTANTS
MATERIAL CONCRETE ALL
SUPPORTS
1 4 7 10 FIXED
DEFINE IBC 2000
SDS 0.06 SD1 1.36 S1 1.3 IE 1 RX 3 RZ 4 SCLASS 4 CT 0.073
JOINT WEIGHT
2 3 5 6 8 9 11 TO 20 WEIGHT 50
*SELFWEIGHT 1
*MEMBER WEIGHT
*2 5 7 TO 10 12 15 UNI 10
LOAD 1 LOADTYPE Seismic TITLE EL-X DIR
IBC LOAD X 1
PERFORM ANALYSIS PRINT LOAD DATA
PRINT ANALYSIS RESULTS
FINISH
STAAD Output
************************************************************ * IBC 2000 SEISMIC LOAD ALONG X : * * CT = 0.073 Cu = 1.200 * * TIME PERIODS : * * Ta = 0.584 T = 1.286 Tuser = 0.000 * * TIME PERIOD USED (T) = 0.701 * * LOAD FACTOR = 1.000 * * DESIGN BASE SHEAR = 1.000 X 0.217 X 800.00 * * = 173.33 KN * ************************************************************ JOINT LATERAL TORSIONAL LOAD - 1 LOAD (KN ) MOMENT (KN -METE) FACTOR - 1.000 ----- ------- --------- 2 FX 3.907 MY 0.000 3 FX 3.907 MY 0.000 8 FX 3.907 MY 0.000 9 FX 3.907 MY 0.000 ----------- ----------- TOTAL = 15.628 0.000 AT LEVEL 4.000 METE 5 FX 8.377 MY 0.000 6 FX 8.377 MY 0.000 11 FX 8.377 MY 0.000 12 FX 8.377 MY 0.000 ----------- ----------- TOTAL = 33.508 0.000 AT LEVEL 8.000 METE 13 FX 13.088 MY 0.000 14 FX 13.088 MY 0.000 15 FX 13.088 MY 0.000 16 FX 13.088 MY 0.000 ----------- ----------- TOTAL = 52.351 0.000 AT LEVEL 12.000 METE 17 FX 17.962 MY 0.000 18 FX 17.962 MY 0.000 19 FX 17.962 MY 0.000 20 FX 17.962 MY 0.000 ----------- ----------- TOTAL = 71.846 0.000 AT LEVEL 16.000 METE