Verification of cantilever beam analysis in STAAD

Never having used STAAD for modal analysis before, I referred to Example 15 of the STAAD verification manual which demonstrates how the natural frequency of a simply supported beam . 

For my requirement , I first I attempted to model a simple cantilever beam in STAAD so I could compare the results with manually calculated values. 

For information my input file is shown along the  message.

 I modelled the beam with the section properties and material properties as below

Ax & Az = 0.56m2 

Ixx=Izz = 1.30415 m4

E=205KN/mm2

l = 22m

POISSON 0.3
DENSITY 325.3 kN/m3

I divided the beam up with a series of node points to ensure the mass of the beam was uniformly distributed .

I fully restrained the beam at one end and all other intermediate nodes were free.

 I get the following results: -

Mode

Natural frequency, Hz

(Manual analysis)

Natural frequency, Hz

(STAAD analysis, basic)

1

4.43

4.393

2

27.79

27.479

3

77.79

28.313

4

152.44

76.806

5

84.819

I am looking for an explanation on following issues

1. In analysis by STAAD i am getting only 5 modes of frequency .How to increase the number modes? 

( i tried increasing cutoff modes also tried increasing number of nodes but no change in result)

2. Mode shape 3 (28.313 Hz) & 5 (84.819) given by STAAD analysis is the axial mode. How to avoid the axial modes in analysis.

Input file :

STAAD PLANE
START JOB INFORMATION
ENGINEER DATE 09-Sep-14
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 22 0; 3 0 11 0; 4 0 5.5 0; 5 0 16.5 0; 6 0 2.75 0; 7 0 8.25 0;
8 0 13.75 0; 9 0 19.25 0; 10 0 3.66667 0; 11 0 4.58333 0; 12 0 0.916667 0;
13 0 1.83333 0; 14 0 6.41667 0; 15 0 7.33333 0; 16 0 9.16667 0; 17 0 10.0833 0;
18 0 11.9167 0; 19 0 12.8333 0; 20 0 14.6667 0; 21 0 15.5833 0; 22 0 17.4167 0;
23 0 18.3333 0; 24 0 20.1667 0; 25 0 21.0833 0;
MEMBER INCIDENCES
1 1 12; 2 3 18; 3 4 14; 4 5 22; 5 6 10; 6 7 16; 7 8 20; 8 9 24; 9 10 11;
10 11 4; 11 12 13; 12 13 6; 13 14 15; 14 15 7; 15 16 17; 16 17 3; 17 18 19;
18 19 8; 19 20 21; 20 21 5; 21 22 23; 22 23 9; 23 24 25; 24 25 2;
DEFINE MATERIAL START
ISOTROPIC STEEL1
E 2.06e+008
POISSON 0.3
DENSITY 325.3
ALPHA 1.2e-005
DAMP 0.03
END DEFINE MATERIAL
MEMBER PROPERTY AMERICAN
1 TO 24 PRIS AX 0.56 AZ 0.56 IX 1.30415 IZ 1.30415
CONSTANTS
MATERIAL STEEL1 ALL
CUT OFF MODE SHAPE 20
SUPPORTS
1 FIXED
LOAD 1
SELFWEIGHT X -1
SELFWEIGHT Y -1
MODAL CALCULATION REQUESTED
PERFORM ANALYSIS
PERFORM ANALYSIS PRINT MODE SHAPES
FINISH

I would appreciate any assistance anyone could give me with this. 

Parents Reply
  • In the beam model, the area and moment of inertia have to be equivalent to those that one would compute for the cross section of the solid cylinder. I used the Section Wizard program to calculate them and obtained

    Area = 138019 sq.cm

    Largest moment of Inertia = 1517414822 cm^4

    I assigned these values to the beam model using the command

    UNIT CMS

    MEMBER PROPERTY AMERICAN

    1 TO 24 PRIS AX 138019 IZ 1517414822

    With this modification, the frequency of the lowest mode of the beam model is 3.01 cycles/sec. The lowest frequency of the solid model is 2.96 cycles/sec - a 2% difference.

    The shape of the innermost elements that look like wedges is not correct. Along the two triangular faces, you have a node along the mid-point of one of the sides of the triangles. A screenshot of one of those elements is shown.

    The mid-side node needs to be eliminated and the element incidence needs to be specified using the sequence

    A-B-C-A-D-E-F-D

    where A, B and C are the vertices of the triangle at one end of the wedge, and D, E and F are the vertices at the other end.

    The modified beam model is attached.



    cantilever Beam model - modified.std
Children