V. Column Pushover Displacement

Compare theoretical solution to a pushover analysis to the STAAD.Pro solution.

Reference

Hand calculation.

Problem

Cantilever member model

A transverse load P is applied to a cantilever member and increased until the member fails.

Member length = 24 inch

Section = Wide flange W16X77

Material = Steel

Expected yield strength = 36 ksi

Plastic Rotation (in radians) vs. Moment (in in·kips) plot for moment hinge

Results from STAAD.Pro

Note: Pushover analysis requires the STAAD.Pro Advanced Analysis license.

The analysis results are saved at an interval of 0.1 inch deflection of the cantilever tip. The following results are displayed in the Postprocessing workflow by selecting the Layouts > Pushover-Graphs tool in the Dynamics group on the Results ribbon tab.

Capacity curve (Displacement at Control Joint, inches vs. Base Shear, kips) as calculated by STAAD.Pro

Table 1. Tabular form of the different points in Capacity curve as reported by STAAD.Pro
Load Step	Displacement in	Base Shear kip
1	0	0.153
2	0.001	1.153
3	0.078	57.837
4	0.136	61.953
5	0.243	63.979
6	0.349	65.979
7	0.455	67.979
8	0.562	69.979
9	0.668	71.979
10	0.774	73.979
11	0.880	75.979
12	0.986	77.979
13	1.065	79.479
14	1.065	21.510

Results from hand calculation

At Load Step 3

Equation of elastic deflection at cantilever tip:

δ_{z} = \frac{P L^{3}}{3 E I} + \frac{P L}{G A_{v}}

where

P	=	57.837 kip
L	=	24 in.
E	=	29,000 ksi
I	=	138.0 in.⁴
G	=	11,154 ksi
A_v	=	10.4323 in.²

Thus elastic deformation:

δ_{z} = \frac{57.837 \times 24^{3}}{3 \times 29, 000 \times 138} + \frac{57.837 \times 24}{11, 154 \times 10.4323} = 0.066 + 0.012 = 0.078 in.

At Load Step 13

Equation of elastic deflection at cantilever tip

δ_{z} = \frac{P L^{3}}{3 E I} + \frac{P L}{G A_{v}}

where

79.479 kip

(The other values are as listed for Load Step 3)

Thus elastic deformation at cantilever tip:

δ_{z} = \frac{79.479 \times 24^{3}}{3 \times 29, 000 \times 138} + \frac{79.479 \times 24}{11, 154 \times 10.4323} = 0.092 + 0.016 = 0.108 in.

Plastic rotation = 0.04 radian.

Since STAAD.Pro assumes small displacements (sinθ = θ), plastic deformation at cantilever tip

δ_z _plastic = L × θ = 24 × 0.04 = 0.96 inch

Total deflection = δ_z _elastic+ δ_z _plastic = 0.108 + 0.96 = 1.068 inch

Comparison

Table 2. Comparison of results
Load Step	Force P (free end) kips	Deflection (free end) in		Percent Difference
Load Step	Force P (free end) kips	STAAD.Pro Advanced Analysis	Hand Calculation	Percent Difference
3	57.837	0.078	0.078	none
13	79.479	1.065	1.068	negligible

Note: The deflection results from STAAD.Pro shown in the above table were obtained in the Postprocessing workflow.

STAAD Input

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\07 Nonlinear Analysis\Column Pushover Displacement.STD is typically installed with the program.

STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 20-Jul-07
END JOB INFORMATION
INPUT WIDTH 79
UNIT INCHES KIP
JOINT COORDINATES
1 0 0 0; 2 0 24 0;
MEMBER INCIDENCES
1 1 2;
DEFINE MATERIAL START
ISOTROPIC STEEL
E 29000
POISSON 0.3
DENSITY 0.000283
ALPHA 6.5e-06
DAMP 0.03
END DEFINE MATERIAL
MEMBER PROPERTY AMERICAN
1 TABLE ST W16X77
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 FIXED
DEFINE PUSHOVER DATA
FRAME 2
FYE 36.000000 ALL
HINGE PROPERTY MOMENT
TYPE 1
A 0 0 B 1 1 C 11 1.3 D 11 0.32 E 15 0.32 YM 1480 YR 0.004
IO 2 LS 7 CP 10.5
HINGE TYPE 1 ALL
GNONL 1
DISP Z 2 JOINT 2
LOADING PATTERN 1
LDSTEP 1
SPECTRUM PARAMETERS
DAMPING 2.0000
SC 4
SS 1.1
S1 1.6
SAVE LOADSTEP RESULT DISP 0.100000
END PUSHOVER DATA
LOAD 1 LOADTYPE Gravity   
SELFWEIGHT Z 1 
LOAD 2 LOADTYPE Push   
JOINT LOAD
2 FZ 1
PERFORM PUSHOVER ANALYSIS
FINISH

STAAD Output

            P R O B L E M   S T A T I S T I C S
            -----------------------------------
     NUMBER OF JOINTS          2  NUMBER OF MEMBERS       1
     NUMBER OF PLATES          0  NUMBER OF SOLIDS        0
     NUMBER OF SURFACES        0  NUMBER OF SUPPORTS      1
           Using 64-bit analysis engine.
           SOLVER USED IS THE IN-CORE ADVANCED MATH SOLVER
   TOTAL      PRIMARY LOAD CASES =     2, TOTAL DEGREES OF FREEDOM =       6
   TOTAL LOAD COMBINATION  CASES =     0  SO FAR.
 MORE MODES WERE REQUESTED THAN THERE ARE FREE MASSES.
 NUMBER OF MODES REQUESTED              =     6
 NUMBER OF EXISTING MASSES IN THE MODEL =     3
 NUMBER OF MODES THAT WILL BE USED      =     3
   ***  EIGENSOLUTION : ADVANCED METHOD ***
      STAAD SPACE                                              -- PAGE NO.    3
               CALCULATED FREQUENCIES FOR LOAD CASE       3
       MODE            FREQUENCY(CYCLES/SEC)         PERIOD(SEC)
         1                     306.367                  0.00326
         2                     544.518                  0.00184
         3                    1865.405                  0.00054
            MODAL WEIGHT (MODAL MASS TIMES g) IN KIP          GENERALIZED
      MODE           X             Y             Z              WEIGHT
         1       0.000000E+00  0.000000E+00  7.674960E-02    7.674960E-02
         2       7.674960E-02  0.000000E+00  0.000000E+00    7.674960E-02
         3       0.000000E+00  7.674960E-02  0.000000E+00    7.674960E-02
 MASS PARTICIPATION FACTORS 
                     MASS  PARTICIPATION FACTORS IN PERCENT
                     --------------------------------------
           MODE    X     Y     Z     SUMM-X   SUMM-Y   SUMM-Z
             1     0.00   0.00 100.00    0.000    0.000  100.000
             2   100.00   0.00   0.00  100.000    0.000  100.000
             3     0.00 100.00   0.00  100.000  100.000  100.000
   ***WARNING: MEMBER #       1 HAS FAILED IN " DEFORMATION-CONTROLLED ACTION. 
   *** WARNING : STRUCTURE HAS REACHED AN UNSTABLE STATE WHERE
                 A VERY SMALL INCREASE IN PUSH LOAD IS CAUSING LARGE DISPLACEMENT.