Staad Basics

© r p clarke 2003 1

STAAD BASICS

- NOTES ON THE EFFECTIVE USE OF STAAD-PRO REL 3.1 -

- FOR STRUCTURAL ANALYSIS -

By R. P. Clarke


TABLE OF CONTENTS

1.0 SKELETAL STRUCTURES – STATIC LOADS1.1 SIGN CONVENTION 31.2 FUNDAMENTAL COMMANDS 51.3 EXAMPLE 6

2.0 CONTINUUM STRUCTURES – STATIC LOADS 9

2.1 SIGN CONVENTION 92.2 FUNDAMENTAL COMMANDS 112.3 EXAMPLE 11

3.0 SKELETAL STRUCTURES - DYNAMIC LOADS 14

3.1 FUNDAMENTAL COMMANDS 143.2 EXAMPLE 15


STAAD BASICS

The following is a description of fundamental considerations for the effective use ofSTAAD-PRO Release 3.1 for the analysis of structures. It must be mentioned howeverthat since STAAD is a computer program, blind faith should not be placed in STAAD orany other engineering program. This is due to the following factors. (1): The results areonly as good as the modeling of the structure in terms of load effects representation,effective structural systems, the connection behavior, and the material idealizations. (2):The procedures used in programs are not transparent to the user. (3): Computer programsusually have bugs. (4): Dependence on software can reduce the engineer's intuition of theactual behavior of the structure. It is therefore strongly recommended that until at leastone year's experience of continually using STAAD is obtained, and for importantstructures, parallel hand calculations for the analysis and design of the structure be doneas well.

For section 1.0 it is presumed that the reader understands Structural Theory and theStiffness Matrix Method of Structural Analysis. For section 2.0 it is presumed that thereader understands the fundamentals of Plate Theory, and the Finite Element Methodbased on the stiffness formulation. For section 3.0 it is presumed that the readerunderstands Structural Dynamic Analysis by the Lumped Mass Time History Method.

1.0 SKELETAL STRUCTURES – STATIC LOADS

1.1 SIGN CONVENTION

It is vital to understand the STAAD coordinate system in order to properly use STAAD.This is needed to ensure that the input data is as intended, and for the interpretation of theanalysis results.

Coordinate Systems:

Since STAAD uses the Matrix Displacement Method of structural analysis, there are 2Cartesian coordinate systems - the local and the global.

The geometry of the structure as a whole is defined by the nodes at the ends of thevarious structural members, and each node has a unique number. Each member also hasa unique number and the topology of the member is defined relative to the node numbersat its ends. This establishes the "MEMBER INCIDENCES" table.

The location of each node is defined relative to a global coordinate system. By default,the origin of the global coordinate system is at node number 1.


The location of points or sections within each structural member is defined relative to thelocal coordinate system with the origin at the left end node of the member viewedhorizontally. Each member has its own local coordinate system.

A. FORCES AT A SECTION OF A MEMBER

This applies to the sign of the quantity in the STAAD member stress diagram such as thebending moment diagram.

POSITIVE FORCE AT THE SECTION NEGATIVE FORCE AT THE SECTION

Bending:Mz

Axial:Fx

Shear:Fy

B. FORCES ACTING ON A MEMBER'S END

In STAAD this is called the "MEMBER END FORCES"

POSITIVE FORCE ON THE MEMBER NEGATIVE FORCE ON THE MEMBER

Bending:Mz

Axial:Fx

Shear:Fy

Torsion: Mx


1.2 FUNDAMENTAL ANALYSIS COMMANDS

Regardless of the structure being analysed, the following are fundamental steps andSTAAD command keywords shown in the brackets:

1. Define whether the problem is 2D or 3D (STAAD PLANE or SPACE)2. Define the length and force units (UNITS)3. Define the nodes and their locations (JOINT COORDINATES)4. Define the member and their nodes (MEMBER INCIDENCES)5. Define the section properties of the members, Ix, etc (MEMBER PROPERTY)6. Define the mechanical properties of the members such as the Young's modulus,

density, etc (CONSTANTS)7. Define the support conditions (SUPPORTS)8. Define the load cases (LOAD)9. Define the loads of each load case as member loads, joint loads, (or code loads)

(MEMBER LOAD or JOINT LOAD)10. Define the load combinations (LOAD COMB)11. Analyse the structure (PERFORM ANALYSIS)12. Define the output format (PRINT)13. Finish the run (FINISH)

These commands are stored automatically in STAAD in a file with the extension .std.This file is formatted as an ASCII text file which means that it can be edited outside ofSTAAD with a word processor or any other text editor. Therefore, you can also write theinput file independantly of STAAD and just refer to it when you enter STAAD to run theanalysis.

By using the "File" menu STAAD reads the .std file as its input and automatically createsan output file with the extension .anl. This file is also a text file and is useful forincluding in calculation reports.

STAAD also creates certain other output files for its internal use. STAAD creates adatabase for your analysis, .dbs, and files for the bending moments .bmd, displacements,.dsp, reactions, .rea, amomg others.

The aforementioned STAAD commands are incomplete by themselves - they are thekeywords of the commands. The complete commands follow a particular syntax tocompletely describe the problem. The example problem at the end of this section showsthe complete syntax for common commands.

Though you can use a totally character-based approach with STAAD, the most effectiveuse of STAAD is when you use the PRE-PROCESSOR of STAAD to write the .std foryou. The PRE-PROCESSOR is a set of functions within STAAD that you select fromthe menus of STAAD's GUI. Each of the command keywords presented in brackets


earlier has an icon or menu that you click on to create the commands in the .std file.From within STAAD you can see the contents of the .std file at any time, and the .anl fileas well after analysis. To do this you just click on the icon for each. This puts the file onthe screen and you can edit the file from there if you wish.

1.3 EXAMPLE 1:

The following is the .std file - ex1.std, for the analysis of a 3D 1-bay portal framecarrying a slab under an unfactored floor load of 6 kN/m2 and unfactored joint loads of30 kN at 2 of the upper joints. The command keywords are in BOLD.


STAAD SPACE EXAMPLE 1START JOB INFORMATIONJOB NAME EXAMPLE 1JOB CLIENT UWI STUDENTSJOB NO 024ENGINEER NAME rpcENGINEER DATE 09-Aug-03END JOB INFORMATIONINPUT WIDTH 79UNIT METER KNJOINT COORDINATES1 0 0 0; 2 0 4.572 0; 3 0 0 4.572; 4 4.572 0 4.572; 5 4.572 0 0;6 0 4.572 4.572; 7 4.572 4.572 0; 8 4.572 4.572 4.572;MEMBER INCIDENCES1 1 2; 2 3 6; 3 4 8; 4 5 7; 5 6 8; 6 2 7; 7 2 6; 8 7 8;MEMBER PROPERTY AMERICAN1 TO 4 PRIS YD 0.35 ZD 0.355 TO 8 PRIS YD 0.5 ZD 0.35CONSTANTSE CONCRETE MEMB 1 TO 8E 2.5e+007 MEMB 1 TO 8POISSON 0.17 MEMB 1 TO 8DENSITY 24 MEMB 1 TO 8ALPHA 1.2e-011 MEMB 1 TO 8DENSITY CONCRETE MEMB 1 TO 8SUPPORTS1 3 TO 5 FIXEDLOAD 1 FLOOR LOADSELFWEIGHT Y -1FLOOR LOADYRANGE 0 6 FLOAD -6LOAD 2 JOINT LOADJOINT LOAD2 6 FX 30LOAD COMB 3 FLOOR PLUS JOINT1 1.5 2 1.2PERFORM ANALYSISFINISH

Note that the above figure shows the loading for the factored loads. Also, by using theFLOOR LOAD command, STAAD automatically calculates the load on the beamssupporting the 2-way spanning slab. The self-weight of the members is automaticallycalculated by STAAD using the SELFWEIGHT Y -1 command.

If we wanted to know the internal forces at the ends of say the members 1 and 5 includedin the output data, we would put the following commands after the "PERFORMANALYSIS" command:

PRINT MEMBER FORCES LIST 1PRINT MEMBER FORCES LIST 5


The output data from the file ex1.anl that corresponds to these commands are as follows:

MEMBER END FORCES STRUCTURE TYPE = SPACE ----------------- ALL UNITS ARE -- KN METE

MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z

1 1 1 63.40 -2.92 2.92 0.00 -4.42 -4.42 2 -50.21 2.92 -2.92 0.00 -8.95 -8.95 2 1 -14.14 15.02 0.00 0.00 0.00 36.31 2 14.14 -15.02 0.00 0.00 0.00 32.37 3 1 78.13 13.64 4.38 0.00 -6.63 36.94 2 -58.34 -13.64 -4.38 0.00 -13.42 25.42

************** END OF LATEST ANALYSIS RESULT **************



5 1 6 2.92 25.10 0.00 0.00 0.00 8.95 8 -2.92 25.10 0.00 0.00 0.00 -8.95 2 6 14.98 -14.14 0.00 0.00 0.00 -32.37 8 -14.98 14.14 0.00 0.00 0.00 -32.29 3 6 22.36 20.69 0.00 0.00 0.00 -25.42 8 -22.36 54.62 0.00 0.00 0.00 -52.16



2.0 CONTINUUM STRUCTURES – STATIC LOADS

Continuum structures (plates, slabs, walls, shells, tanks, etc) are modelled in STAAD byusing finite elements. The following is with respect to the element of the STAAD librarywhich can be quadrilateral or triangular. Common rules for the use of finite elementmodelling apply and will not be repeated here and it is presumed that section 1.0 has beencovered.

2.1 SIGN CONVENTION

The sign convention is as follows:

Fx

Fy

Fxy

MEMBRANE FORCES

BENDING MOMENTS & TRANSVERSE SHEAR

MX

MY

MYX

MYX

MXY

MXY QX

QX

QY

QY


The diagrams above show the positive direction of the forces relative to the followinglocal coordinate system.

Hence for axial direct forces: tension is positive, for bending moments: hogging ispositive, and for transverse shear: down-to-the-left and up-to-the-right is positive.

Note that for non-rectangular and triangular elements, the x-y-z axes are not orthogonalto the edges or surfaces of the element. The x-axis is aligned with a line connecting themid-points of IL and JK, the z-axis is orthogonal to lines connecting the mid-points of IL-JK to those connecting the mid-points of IJ-KL, and the y-axis is orthogonal to the x andz axes so defined.

ELEMENT FORCE outputs are available at the centre node of the element, all cornernodes of the element, and at any user-specified point within the element.

The items included in the ELEMENT FORCE output are:

QX, QY Transverse shear forces stated as force per unit length per unitelement thickness.

FX, FY, FXY Membrane forces stated as force per unit length per unit element thickness.

MX, MY, MXY Bending moments stated as moment per unit length.

SMAX, SMIN Principal stresses stated as force per unit area.

TMAX Maximum in-plane shear stress stated as force per unit area.

X

Y

Z

Bottom surface

Top surface

I

J

K

L


ANGLE The orientation of the principal plane stated in degrees measuredanti-clockwise from the local x-axis.

The top and bottom surfaces are identified on the basis of the direction of the local z-axis.

2.2 FUNDAMENTAL COMMANDS

The fundamental commands for finite element analysis using STAAD closely followthose for the skeletal or frame member analysis. The following are the essentialdifferences:

• ELEMENT INCIDENCES command.• ELEMENT PROPERTY command.

Both frame members and finite elements can be used together in STAAD but theELEMENT INCIDENCES command must immediately follow the MEMBERINCIDENCES command.

The selfweight of the finite elements is converted to joint loads at the connected nodesand is not used as an element pressure load.

2.3 EXAMPLE 2

Analyse an uncovered reinforced concrete tank of dimensions 6.0m x 6.0m x 6.0m withwalls and base 200mm thick. The tank is filled with water and rests on rigid ground.

The following is the STAAD .std file for a model of the tank.

STAAD SPACE FINITE ELEMENT MODEL OF TANK STRUCTURESTART JOB INFORMATIONENGINEER DATE 10-Feb-04END JOB INFORMATIONUNIT MET KNSJOINT COORDINATES1 0 0 0; 2 0 1.5 0; 3 0 3.0 0; 4 0 4.5 0; 5 0 6.0 0; 6 1.5 0 0;7 1.5 1.5 0; 8 1.5 3.0 0; 9 1.5 4.5 0; 10 1.5 6.0 0; 11 3.0 0 0;12 3.0 1.5 0; 13 3.0 3.0 0; 14 3.0 4.5 0; 15 3.0 6.0 0;16 4.5 0 0; 17 4.5 1.5 0; 18 4.5 3.0 0; 19 4.5 4.5 0;20 4.5 6.0 0; 21 6.0 0 0; 22 6.0 1.5 0; 23 6.0 3.0 0;24 6.0 4.5 0; 25 6.0 6.0 0; 26 6.0 0 1.5; 27 6.0 1.5 1.5;28 6.0 3.0 1.5; 29 6.0 4.5 1.5; 30 6.0 6.0 1.5; 31 6.0 0 3.0;32 6.0 1.5 3.0; 33 6.0 3.0 3.0; 34 6.0 4.5 3.0; 35 6.0 6.0 3.0;36 6.0 0 4.5; 37 6.0 1.5 4.5; 38 6.0 3.0 4.5; 39 6.0 4.5 4.5;40 6.0 6.0 4.5; 41 6.0 0 6.0; 42 6.0 1.5 6.0; 43 6.0 3.0 6.0;44 6.0 4.5 6.0; 45 6.0 6.0 6.0; 46 4.5 0 6.0; 47 4.5 1.5 6.0;48 4.5 3.0 6.0; 49 4.5 4.5 6.0; 50 4.5 6.0 6.0; 51 3.0 0 6.0;


52 3.0 1.5 6.0; 53 3.0 3.0 6.0; 54 3.0 4.5 6.0; 55 3.0 6.0 6.0;56 1.5 0 6.0; 57 1.5 1.5 6.0; 58 1.5 3.0 6.0; 59 1.5 4.5 6.0;60 1.5 6.0 6.0; 61 0 0 6.0; 62 0 1.5 6.0; 63 0 3.0 6.0;64 0 4.5 6.0; 65 0 6.0 6.0; 66 0 0 4.5; 67 0 1.5 4.5;68 0 3.0 4.5; 69 0 4.5 4.5; 70 0 6.0 4.5; 71 0 0 3.0;72 0 1.5 3.0; 73 0 3.0 3.0; 74 0 4.5 3.0; 75 0 6.0 3.0;76 0 0 1.5; 77 0 1.5 1.5; 78 0 3.0 1.5; 79 0 4.5 1.5;80 0 6.0 1.5; 81 1.5 0 1.5; 82 1.5 0 3.0; 83 1.5 0 4.5;84 3.0 0 1.5; 85 3.0 0 3.0; 86 3.0 0 4.5; 87 4.5 0 1.5;88 4.5 0 3.0; 89 4.5 0 4.5;ELEMENT INCIDENCES SHELL1 1 2 7 6; 2 2 3 8 7; 3 3 4 9 8; 4 4 5 10 9; 5 6 7 12 11; 6 7 8 13 12;7 8 9 14 13; 8 9 10 15 14; 9 11 12 17 16; 10 12 13 18 17;11 13 14 19 18; 12 14 15 20 19; 13 16 17 22 21; 14 17 18 23 22;15 18 19 24 23; 16 19 20 25 24; 17 21 22 27 26; 18 22 23 28 27;19 23 24 29 28; 20 24 25 30 29; 21 26 27 32 31; 22 27 28 33 32;23 28 29 34 33; 24 29 30 35 34; 25 31 32 37 36; 26 32 33 38 37;27 33 34 39 38; 28 34 35 40 39; 29 36 37 42 41; 30 37 38 43 42;31 38 39 44 43; 32 39 40 45 44; 33 41 42 47 46; 34 42 43 48 47;35 43 44 49 48; 36 44 45 50 49; 37 46 47 52 51; 38 47 48 53 52;39 48 49 54 53; 40 49 50 55 54; 41 51 52 57 56; 42 52 53 58 57;43 53 54 59 58; 44 54 55 60 59; 45 56 57 62 61; 46 57 58 63 62;47 58 59 64 63; 48 59 60 65 64; 49 61 62 67 66; 50 62 63 68 67;51 63 64 69 68; 52 64 65 70 69; 53 66 67 72 71; 54 67 68 73 72;55 68 69 74 73; 56 69 70 75 74; 57 71 72 77 76; 58 72 73 78 77;59 73 74 79 78; 60 74 75 80 79; 61 76 77 2 1; 62 77 78 3 2;63 78 79 4 3; 64 79 80 5 4; 65 1 6 81 76; 66 76 81 82 71;67 71 82 83 66; 68 66 83 56 61; 69 6 11 84 81; 70 81 84 85 82;71 82 85 86 83; 72 83 86 51 56; 73 11 16 87 84; 74 84 87 88 85;75 85 88 89 86; 76 86 89 46 51; 77 16 21 26 87; 78 87 26 31 88;79 88 31 36 89; 80 89 36 41 46;ELEMENT PROPERTY1 TO 80 THICKNESS 0.20CONSTANTSE 20000000.0 ALLSUPPORTS1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81 TO 89 PINNEDLOAD 1ELEMENT LOAD4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 PR 15.3 7 11 15 19 23 27 31 35 39 43 47 51 55 59 63 PR 30.2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 62 PR 45.1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 PR 60.PERFORM ANALYSISPRINT JOINT DISPLACMENTS LIST 5 25 45 65PRINT ELEMENT FORCE LIST 9 TO 16DRAW ROTA X -20 Y 30 Z 20 STR 1FINISH

The following is part of the STAAD output for the displacements, element forces,stresses, etc, at the locations indicated in the input file.


63. PRINT JOINT DISPLACMENTS LIST 5 25 45 65

FINITE ELEMENT MODEL OF TANK STRUCTURE

JOINT DISPLACEMENT (CM RADIANS) STRUCTURE TYPE = SPACE ------------------

JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN

5 1 -0.0029 -0.0003 -0.0029 0.0001 0.0000 -0.0001 25 1 0.0029 -0.0003 -0.0029 0.0001 0.0000 0.0001 45 1 0.0029 -0.0003 0.0029 -0.0001 0.0000 0.0001 65 1 -0.0029 -0.0003 0.0029 -0.0001 0.0000 -0.0001

64. PRINT ELEMENT FORCE LIST 9 TO 16

FINITE ELEMENT MODEL OF TANK STRUCTURE

ELEMENT FORCES FORCE,LENGTH UNITS= KNS MET -------------- FORCE OR STRESS = FORCE/UNIT WIDTH/THICK, MOMENT = FORCE-LENGTH/UNIT WIDTH

ELEMENT LOAD QX QY MX MY MXY VONT VONB FX FY FXY

9 1 322.03 24.89 -6.52 6.89 6.55 2594.11 2279.42 39.19 177.12 58.53 TOP : SMAX= 1631.24 SMIN= -1360.09 TMAX= 1495.67 ANGLE= -22.0 BOTT: SMAX= 1395.94 SMIN= -1234.46 TMAX= 1315.20 ANGLE= -22.3

10 1 32.21 -104.02 22.55 27.88 5.06 4315.25 3841.25 -51.93 388.84 28.70 TOP : SMAX= 4952.95 SMIN= 2948.25 TMAX= 1002.35 ANGLE= -25.9 BOTT: SMAX= -2861.72 SMIN= -4365.67 TMAX= 751.97 ANGLE= -38.1

11 1 -61.01 -101.23 13.25 30.74 -0.05 4418.73 3607.97 -47.38 445.41 -21.84 TOP : SMAX= 5056.75 SMIN= 1939.18 TMAX= 1558.78 ANGLE= 0.5 BOTT: SMAX= -2034.12 SMIN= -4165.74 TMAX= 1065.81 ANGLE= -0.4

12 1 -39.29 -88.67 1.04 26.72 0.28 4287.26 3580.09 0.30 354.46 -14.06 TOP : SMAX= 4362.88 SMIN= 155.47 TMAX= 2103.71 ANGLE= -0.4 BOTT: SMAX= -154.15 SMIN= -3654.67 TMAX= 1750.26 ANGLE= -0.9

13 1 -2.80 -94.45 -5.15 -9.80 7.19 2416.03 2137.97 -39.19 144.56 145.84 TOP : SMAX= 181.92 SMIN= -2319.92 TMAX= 1250.92 ANGLE= 39.1 BOTT: SMAX= 2205.83 SMIN= 142.90 TMAX= 1031.47 ANGLE= 32.4

14 1 -76.30 -369.04 0.23 -22.18 5.07


3161.16 4043.74 51.93 520.22 27.13 TOP : SMAX= 286.87 SMIN= -3007.95 TMAX= 1647.41 ANGLE= 14.3 BOTT: SMAX= 3983.43 SMIN= -118.04 TMAX= 2050.74 ANGLE= 10.5

15 1 -17.52 -326.06 -5.54 -28.81 -0.23 3559.42 4387.81 47.38 438.41 -37.17 TOP : SMAX= -781.48 SMIN= -3885.23 TMAX= 1551.88 ANGLE= -1.3 BOTT: SMAX= 4760.39 SMIN= 877.91 TMAX= 1941.24 ANGLE= 0.0

16 1 4.67 -251.97 -8.13 -28.50 -0.26 3491.73 4139.45 -0.30 337.72 -17.83 TOP : SMAX= -1219.29 SMIN= -3937.89 TMAX= 1359.30 ANGLE= -1.2 BOTT: SMAX= 4612.27 SMIN= 1219.73 TMAX= 1696.27 ANGLE= -0.3

********************END OF ELEMENT FORCES********************

3.0 SKELETAL STRUCTURES – DYNAMIC LOAD

3.1 FUNDAMENTAL COMMANDS

The reader must first complete section 1.0. Only Time History dynamic analysis by theapplication of forcing functions to nodes or members is covered in this section (i.e. notground motion time history dynamics).

There are 2 issues to consider in the use of STAAD for dynamic analysis - how STAADidealises the distribution of mass and how to apply the forcing function.

STAAD distributes the mass via the SELFWEIGHT command. When this is done, themass is lumped at the nodes. If this is an inadequate model of the mass idealisation, themember concentrated load command CON can be used to tell STAAD that significantmasses are located there and their values. Also, the user can split the member into shorterlengths by inserting nodes along the member. In this case, the user can use the JOINTLOAD command to tell STAAD that a significant mass is located there and its mass.

The SELFWEIGHT command is placed as the first line (or lines) of the commands underthe relevant LOAD command. If the CON or JOINT LOAD commands are used, it isplaced after the SELFWEIGHT command.

To apply a forcing function in STAAD at the nodes, or at a particular location, you mustfirst define the type of forcing function using the DEFINE TIME HISTORY command.Note that a forcing function can only be applied at a node so if the user wishes to applythe function to a point along a member, a node must be placed there first.

You then use the TYPE “i” FORCE command along with its particular syntaxrequirements. Finally, under the relevant LOAD command and after the massidealisation commands (i.e. SELFWEIGHT, CON, JOINT LOAD), you use the TIMELOAD command. You can only use the TIME LOAD command in one load case.


3.2 EXAMPLE 3

A one-storey reinforced concrete structure of plan dimensions 10.0m x 4.572m supportsseveral loads: a mass of 120 kN at the mid-span of one of the long beams, point loads of30kN at the floor level in the long direction, and a floor load of 6.0 kN/m2 . Thecolumns are 0.3m x 0.3m and the beams are 0.45m deep x 0.3m wide. If the massvibrates at 2.5 Hz for 10 cycles, estimate the amplification factor for the bending momentin the beam under the mass?

The following is a STAAD model of the structure, loads, and assumed load factors.

STAAD SPACE EXAMPLE 3INPUT WIDTH 79UNIT METER KNJOINT COORDINATES1 0 0 0; 2 0 4.572 0; 3 0 0 4.572; 4 10 0 4.572; 5 10 0 0; 6 0 4.5724.572;7 10 4.572 0; 8 10 4.572 4.572; 10 5 4.572 4.572;MEMBER INCIDENCES1 1 2; 2 3 6; 3 4 8; 4 5 7; 6 2 7; 7 2 6; 8 7 8; 10 10 8; 11 6 10;MEMBER PROPERTY AMERICAN1 TO 4 PRIS YD 0.3 ZD 0.36 TO 8 10 11 PRIS YD 0.45 ZD 0.3CONSTANTSE 2.5e+007 MEMB 1 TO 4 6 TO 8 10 11POISSON 0.17 MEMB 1 TO 4 6 TO 8 10 11ALPHA 1.2e-011 MEMB 1 TO 4 6 TO 8 10 11DENSITY CONCRETE MEMB 1 TO 4 6 TO 8 10 11SUPPORTS1 3 TO 5 FIXED*INPUT THE TYPE OF FORCING FUNCTION HEREDEFINE TIME HISTORYTYPE 1 FORCEFUNCTION SINE*THE NEXT LINE IS A FORCE OF AMPLITUDE 120 KN VIBRATING AT 2.5 HZ FOR10 SECAMPLITUDE 120 FREQUENCY 2.5 CYCLES 10ARRIVAL TIME0DAMPING 0.05LOAD 1 FLOOR LOADSELFWEIGHT Y -1FLOOR LOADYRANGE 0 6 FLOAD -6LOAD 2 JOINT LOADJOINT LOAD2 6 FX 3010 FY -120*IN THE NEXT LINE YOU MUST INPUT THE LOAD DUE TO THE MASS THOUGH YOU*USE THE JOINT LOAD CMD IN THE TIME HISTORY CMDs TO TELL STAAD THAT AMASS


*IS THERE AND IN WHAT DIRECTIONS IT IS CAPABLE OF MOVINGLOAD 3 TIME HISTORY LOADSELFWEIGHT X 1SELFWEIGHT Y 1SELFWEIGHT Z 1*THE JOINT LOAD COMMAND IS USED TO TELL STAAD THE ACTIVE MASS AT AJOINTJOINT LOAD10 FY 120TIME LOAD*THE FORCING FUNCTION MUST BE APPLIED TO A JOINT HENCE THE FY10 FY 1 1LOAD COMB 4 FLOOR PLUS JOINT1 1.5 2 1.2LOAD COMB 5 FLOOR PLUS JOINT PLUS VIBRATION1 1.5 2 1.2 3 1.0PERFORM ANALYSISPRINT MEMBER FORCES LIST 11FINISH

The relevant STAAD output is:

CALCULATED FREQUENCIES FOR LOAD CASE 3

MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC)

1 3.050 0.32782 2 3.796 0.26346 3 3.899 0.25646

MASS PARTICIPATION FACTORS IN PERCENT --------------------------------------

MODE X Y Z SUMM-X SUMM-Y SUMM-Z

1 0.00 58.69 0.00 0.000 58.691 0.003 2 99.93 0.00 0.00 99.925 58.691 0.003 3 0.00 0.00 97.57 99.925 58.693 97.576

61. PRINT MEMBER FORCES LIST 11




11 1 6 23.95 68.81 0.00 0.00 0.00 73.21 10 -23.95 0.00 0.00 0.00 0.00 126.05 2 6 38.45 53.82 0.00 0.00 0.03 63.06 10 -38.45 -53.82 0.00 0.00 -0.03 206.01 3 6 -90.99 -232.35 0.13 0.00 0.34 -364.06 10 90.99 232.35 -0.13 0.00 -0.30 -797.72 4 6 82.06 167.79 0.00 0.00 0.03 185.50 10 -82.06 -64.58 0.00 0.00 -0.03 436.29 5 6 -8.93 -64.57 0.13 0.00 0.37 -178.56 10 -173.04 -296.93 0.13 0.00 0.27 1234.01


Hence from the output for load 5 and load 4, and at joint 10, the amplification factor is1234.01/436.29 = 2.82

This relatively high amplification occurs because the forcing function frequency of 2.5Hz is not far from the natural frequencies of the structure of 3.050 to 3.899 Hz.

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