V. NSR-10 Static Seismic Short Structure

Calculate the base shear and horizontal equivalent loads of a two-story steel structure per the NSR-10 code. Additional, verify the fundamental period using the Rayleigh method.

Reference

Colombian Association of Earthquake Engineering. NSR-10 Colombian Regulation of Earthquake Resistant Construction. 2010. Bogata B.C., Colombia.

Details

A two-story structure in Florencia, Colombia is subjected to seismic loads. The lateral load resisting system is a structural steel moment-resisting frame. The total building height is 6 m. The lump mass of each floor above ground is 150 kN.

The soil profile at the site is moderately stiff rock (Profile B). The structure is use group II.

Validation

Seismic parameters:

Acceleration coefficient, A_a = 0.2 (Region 4 from Figure A.2.3-2; by city in Appendix A-4)
Velocity coefficient, A_v = 0.15 (Region 3 from Figure A.2.3-3; by city in Appendix A-4)
Importance factor, I = 1.1 (Use group II from Table A.2.5-1)
Short period amplification factor, F_a = 1.0 (Table A.2.4-3)
Intermediate period amplification factor, F_v = 1.0 (Table A.2.4-4)

From Table A.4.2-1:

C_t = 0.072

α = 0.8

The approximate fundamental period of the structure:

T_a = C_th^α = 0.072 (6)^0.8 = 0.302 sec.

T_{0} = 0.1 \frac{A_{v} F_{v}}{A_{a} F_{a}} = 0.1 \frac{0.15 \times 1.0}{0.2 \times 1.0} = 0.075 sec.

T_{C} = 0.48 \frac{A_{v} F_{v}}{A_{a} F_{a}} = 0.48 \frac{0.15 \times 1.0}{0.2 \times 1.0} = 0.36 sec.

T₀ < T_a < T_C, therefore:

S_{a} = 2.5 A_{a} F_{a} I = 2.5 \times 0.2 \times 1.0 \times 1.1 = 0.55

The horizontal base shear:

V_{s} = S_{a} \times W = 0.55 \times 300 = 165 kN

The exponent in the formula for vertical distribution, k = 1.0, as T_a < 0.5 sec

The vertical distribution of the seismic force (as per A.4.3-3):

Floor	Weight, w_i (kN)	Height, h_i (m)	$w_{i} {h_{i}}^{k}$	$\frac{w_{i} {h_{i}}^{k}}{\sum w_{i} {h_{i}}^{k}}$	C_vx (kN)
1	150	3	450	0.333	55
2	150	6	900	0.667	110
∑	300	-	1,350	1.0	165

Period Using Rayleigh Method

The Rayleigh method period of the structure is dependant on the horizontal deflection of the structure.

Moment of inertia used: I = 178×(10)^-6 m⁴
Modulus of elasticity: E = 200 GPa
Stiffness term, EI = 178 × 200 = 36,500 kN·m²

Deflection at the top (floor 3):

δ_{3} = \frac{P_{2} {(\frac{h}{2})}^{2}}{6 E I} (3 h - \frac{h}{2}) + \frac{P_{3} h^{3}}{3 E I} = \frac{5 P_{2} h^{3}}{48 E I} + \frac{P_{3} h^{3}}{3 E I}

= \frac{5 (55) {(6)}^{3}}{48 (35,600)} + \frac{110 {(6)}^{3}}{3 (35,600)} = 0.035 + 0.222 = 0.257 m

Deflection at the mid-height (floor 2):

δ_{2} = \frac{P_{2} {(\frac{h}{2})}^{3}}{3 E I} + \frac{P_{3}}{6 E I} [2 h^{3} - 3 h^{2} (\frac{h}{2}) + {(\frac{h}{2})}^{3}] = \frac{P_{2} h^{3}}{24 E I} + \frac{5 P_{3} h^{3}}{48 E I}

= \frac{P_{2} h^{3}}{24 E I} + \frac{5 P_{3} h^{3}}{48 E I} = \frac{55 {(6)}^{3}}{24 (35,600)} + \frac{5 (110) {(6)}^{3}}{48 (35,600)} = 0.014 + 0.070 = 0.083 m

From the STAAD.Pro output, the horizontal deflections at floors 1 and 2 are 1.2836 cm and 3.9332 cm, respectively. The fundamental period of the structure calculated using the Rayleigh method is then:

T = 2 π \sqrt{(\frac{1}{g}) \frac{\sum w_{i} {δ_{i}}^{2}}{\sum C_{cx,i} δ_{i}}} = 2 π \sqrt{(\frac{1}{9.81}) \frac{150 {(0.083)}^{2} + 150 {(0.257)}^{2}}{55 (0.083) + 110 (0.257)}} = 1.158 sec.

Results

Result Type	Reference	STAAD.Pro	Difference
Fundamental period, T_a (sec.)	0.302	0.302	none
Seismic base shear, V_s (kN)	165	165	none
Horizontal force at floor 1, C_vx,1 (kN)	55	55	none
Horizontal force at floor 2, C_vx,2 (kN)	110	110	none
Fundamental period by Rayleigh method (sec.)	1.158	1.164*	negligible

* A second load case is used in the STAAD.Pro model with lateral loads equal to the seismic weight at each node. As the Rayleigh method for fundamental frequency (and, inversely, period) is thus load dependant, STAAD.Pro uses the weight as a force instead of the equivalent static forces to evaluate the Rayleigh period. As seen in the second load case, when lateral joint loads of 150 kN, STAAD.Pro calculates the fundamental period using the Rayleigh method as 1 / 0.85884 Hz = 1.164 sec.

STAAD.Pro Input

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\06 Loading\NSR\NSR-10 Static Seismic Short Structure.STD is typically installed with the program.

Note: The structure is modeled as a simple two story column for the purposes of this example.

The default values of the CT and ALPHA parameters are used (which are for steel moment-resisting frames).

STAAD SPACE NSR-10 STATIC SEISMIC LOADS
START JOB INFORMATION
ENGINEER DATE 13-May-22
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 3 0; 3 0 6 0;
MEMBER INCIDENCES
1 1 2; 2 2 3;
DEFINE MATERIAL START
ISOTROPIC STEEL
E 1.99947e+08
POISSON 0.3
DENSITY 76.8191
ALPHA 6.5e-06
DAMP 0.03
G 7.7221e+07
TYPE STEEL
STRENGTH RY 1.5 RT 1.2
END DEFINE MATERIAL
MEMBER PROPERTY AMERICAN
1 2 TABLE ST W14X43
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 FIXED
DEFINE REFERENCE LOADS
LOAD R1 LOADTYPE Mass  TITLE REF MASS MODEL
JOINT LOAD
2 3 FX 150
2 3 FY 150
2 3 FZ 150
END DEFINE REFERENCE LOADS
DEFINE COLOMBIAN 2010  LOAD
AA 0.2 AV 0.15 FA 1 FV 1 I 1.1 CT 0.072 PX 0.5 PZ 0.5
LOAD 1 LOADTYPE Seismic-H  TITLE SEISMIC
COLOMBIAN LOAD X 1
LOAD 2 LOADTYPE Dead  TITLE MASS
JOINT LOAD
2 FX 150
3 FX 150
CALCULATE RAYLEIGH FREQUENCY
PERFORM ANALYSIS PRINT LOAD DATA
PRINT ANALYSIS RESULTS
FINISH

STAAD.Pro Output

**************************************************************
   *                                                            *
   *  COLOMBIAN 2010 SEISMIC LOAD :                             *
   *                                                            *
   *  TIME PERIODS FOR X DIRECTION:                             *
   *   Ta =  0.302 Tb =  1.164 Tuser = 0.500                    *
   *   TIME PERIOD USED (T)       =  0.302                      *
   *  LOAD FACTOR                 =  1.000                      *
   *  DESIGN BASE SHEAR =  0.550 X     300.00 =     165.00  KN  *
   *                                                            *
   **************************************************************

    JOINT            LATERAL          TORSIONAL            LOAD -    1
                     LOAD (KN  )      MOMENT (KN  -METE) FACTOR -   1.000
    -----            -------          ---------

       2     FX       55.000    MY        0.000
                 -----------        -----------
          TOTAL =     55.000              0.000 AT LEVEL     3.000 METE

       3     FX      110.000    MY        0.000
                 -----------        -----------
          TOTAL =    110.000              0.000 AT LEVEL     6.000 METE

   **********************************************************
   *                                                        *
   * RAYLEIGH FREQUENCY FOR LOADING     2 =    0.85884 CPS  *
   * MAX DEFLECTION = 40.44359 CM   GLO X,  AT JOINT     3  *
   *                                                        *
   **********************************************************