british examples 2004

STAAD.Pro 2004

BRITISH EXAMPLES MANUAL

a division of netGuru, Inc. www.reiworld.com www.reel.co.uk

http://www.reiworld.com/

http://www.netguru.com/

STAAD.Pro 2004 is a proprietary computer program of Research Engineers, International (REI), a division of netGuru, Inc. The program and this document have been prepared in accord with established industry engineering principles and guidelines. While believed to be accurate, the information contained herein should never be utilized for any specific engineering application without professional observance and authentication for accuracy, suitability and applicability by a competent and licensed engineer, architect or other professional. REI disclaims any liability arising from the unauthorized and/or improper use of any information contained in this document, or as a result of the usage of the program.

RELEASE 2004

Copyright Research Engineers, International

Division of netGuru, Inc. Published August, 2004

About STAAD.Pro 2004 STAAD.Pro is a widely used software for structural analysis and design from Research Engineers International. The STAAD.Pro software consists of the following: The STAAD.Pro Graphical User Interface (GUI): It is used to generate the model, which can then be analyzed using the STAAD engine. After analysis and design is completed, the GUI can also be used to view the results graphically. The STAAD analysis and design engine: It is a general-purpose calculation engine for structural analysis and integrated Steel, Concrete, Timber and Aluminum design.

About the STAAD.Pro 2004 Documentation The documentation for STAAD.Pro consists of a set of manuals as described below. Getting Started and Tutorials : This manual contains information on the contents of the STAAD.Pro package, computer system requirements, installation process, copy protection issues and a description on how to run the programs in the package. Tutorials that provide detailed and step-by-step explanation on using the programs are also provided. Examples : This book offers examples of various problems that can be solved using the STAAD engine. The examples represent various structural analyses and design problems commonly encountered by structural engineers. Graphical Environment : This manual contains a detailed description of the Graphical User Interface (GUI) of STAAD.Pro. The topics covered include model generation, structural analysis and design, result verification, and report generation. This manual is generally provided only in the electronic form and can be accessed from the Help facilities of STAAD.Pro. Users who wish to obtain a printed copy of this book may contact Research Engineers. See the back cover of this book for addresses and phone numbers. Technical Reference : This manual deals with the theory behind the engineering calculations made by the STAAD engine. It also includes an explanation of the commands available in the STAAD command file. International Design Codes : This document contains information on the various Concrete, Steel, and Aluminum design codes, of several countries, that are implemented in STAAD. Generally, this book is supplied only to those users who purchase the international codes utilities with STAAD.Pro. OpenSTAAD : This document contains information on the library of functions which enable users to access STAAD.Pro’s input and results data for importing into other applications.

i

Introduction The tutorials in the Getting Started Manual mention 2 methods of creating the STAAD input data.

a. Using the facilities of the Graphical User Interface (GUI) modelling mode

b. Using the editor which comes built into the STAAD program

Method (a) is explained in great detail in the various tutorials of that manual. The emphasis in this Examples manual is on creating the data using method (b). A number of examples, representing a wide variety of structural engineering problems, are presented. All the input needed is explained line by line to facilitate the understanding of the STAAD command language. These examples also illustrate how the various commands in the program are to be used together. Although a user can prepare the input through the STAAD GUI, it is quite useful to understand the language of the input for the following reasons: 1) STAAD is a large and comprehensive structural engineering

software. Knowledge of the STAAD language can be very useful in utilizing the large number of facilities available in the program. The Graphical User Interface can be used to generate the input file for even the most complex of structures. However, the user can easily make changes to the input data if he/she has a good understanding of the command language and syntax of the input.

ii

2) The input file represents the user's thought about what he/she wants to analyze or design. With the knowledge of the STAAD command language, the user or any other person can verify the accuracy of the work.

The commands used in the input file are explained in Section 5 of the STAAD Technical Reference Manual. Users are urged to refer to that manual for a better understanding of the language. The procedure for creating the file using the built-in editor is explained further below in this section. Alternatively, any standard text editor such as Notepad or WordPad may also be used to create the command file. However, the STAAD.Pro command file editor offers the advantage of syntax checking as we type the commands. The STAAD.Pro keywords, numeric data, comments, etc. are displayed in distinct colors in the STAAD.Pro editor. A typical editor screen is shown below to illustrate its general appearance.

iii

To access the built-in editor, first start the program and follow the steps explained in Sections 1.3 and 1.4 of the Getting Started manual.

You will then encounter the dialog box shown in the following figure. In this dialog box, choose Open STAAD Editor.

iv

At this point, the editor screen will open as shown below.

Delete all the command lines displayed in the editor window and type the lines shown in bold in the various examples in this book (You don’t have to delete the lines if you know which to keep and where to fill in the rest of the commands). The commands may be typed in upper or lower case letters. For your convenience, the data for all the examples presented in this manual are supplied to you along with the program CD. You will find them in the folder location X:\spro2004\staad\examp\uk where “X:” is the drive, and “spro2004” is the name of the installation folder if you happened to go with the default during installation. The example files are named in accordance with the order they appear in this manual, namely, examp01.std for example 1, examp08.std for example 8, and so on.

v

The second part of this book contains a set of verification problems which compares the analytical results from the program with standard publications on the subject. They too are installed along with the examples. To view their contents in the editor, open the file you are interested in. Then, click on the STAAD editor icon, or, go to the Edit menu, and choose Edit Input Command File, as shown below.

vi

A new window will open up with the data listed as shown here:

To exit the Editor, select the File | Exit menu option of the editor window (not the File | Exit menu of the main window behind the editor window).

Table of Contents

Example ProblemsExample ProblemsExample ProblemsExample Problems

Example Problem No. 1 1111





























Verification ProblemsVerification ProblemsVerification ProblemsVerification Problems

Verification Problem No. 1 1111












PART – I

APPLICATION EXAMPLES

Description of Example Problems 1) Example problem No. 1 - Plane frame with steel design. After

one analysis, member selection is requested. Since member sizes change during the member selection, another analysis is done followed by final code checking to verify that the final sizes meet the requirements of the code based on the latest analysis results.

2) Example problem No. 2 - A floor structure (bound by global

X-Z axis) made up of steel beams is subjected to area load (i.e. load/area of floor). Load generation based on one-way distribution is illustrated in this example.

3) Example problem No. 3 - A portal frame type steel structure is

sitting on a concrete footing. The soil is to be considered as an elastic foundation.

4) Example problem No. 4 - This example is a typical case of a

load-dependent structure where the structural condition changes for different load cases. In this example, different bracing members are made inactive for different load cases. This is done to prevent these members from carrying any compressive forces.

5) Example problem No. 5 - This example demonstrates the

application of support displacement load (commonly known as sinking support) on a space frame structure.

6) Example problem No. 6 - This is an example of prestress

loading in a plane frame structure. It covers two situations: 1) The prestressing effect is transmitted from the member on which it is applied to the rest of the structure through the connecting members (known in the program as PRESTRESS load). 2) The prestressing effect is experienced by the member(s) alone and not transmitted to the rest of the structure (known in the program as POSTSTRESS load).

7) Example problem No. 7 - This example illustrates modelling of structures with OFFSET connections. Offset connections arise when the center lines of the connected members do not intersect at the connection point. The connection eccentricity is modeled through specification of MEMBER OFFSETS.

8) Example problem No. 8 - In this example, concrete design is

performed on some members of a space frame structure. Design calculations consist of computation of reinforcement for beams and columns. Secondary moments on the columns are obtained through the means of a P-Delta analysis.

9) Example problem No. 9 - A space frame structure in this

example consists of frame members and finite elements. The finite element part is used to model floor flat plates and a shear wall. Design of an element is performed.

10) Example problem No. 10 - A tank structure is modeled with

four-noded plate elements. Water pressure from inside is used as loading for the tank. Reinforcement calculations have been done for some elements.

11) Example problem No. 11 - Dynamic analysis (Response

Spectrum) is performed for a steel structure. Results of a static and dynamic analysis are combined. The combined results are then used for steel design.

12) Example problem No. 12 - This example demonstrates

generation of load cases for the type of loading known as a moving load. This type of loading occurs classically when the load-causing units move on the structure, as in the case of trucks on a bridge deck. The mobile loads are discretized into several individual immobile load cases at discrete positions. During this process, enormous number of load cases may be created resulting in plenty of output to be sorted. To avoid looking into a lot of output, the maximum force envelope is requested for a few specific members.

13) Example problem No. 13 - Calculation of displacements at

intermediate points of members of a plane frame is demonstrated in this example.

14) Example problem No. 14 - A space frame is analyzed for seismic loads. The seismic loads are generated using the procedures of the 1994 UBC Code. A P-Delta analysis is peformed to obtain the secondary effects of the lateral and vertical loads acting simultaneously.

15) Example problem No. 15 - A space frame is analyzed for

loads generated using the built-in wind and floor load generation facilities.

16) Example problem No. 16 - Dynamic Analysis (Time History)

is performed for a 3 span beam with concentrated and distributed masses. The structure is subjected to "forcing function" and "ground motion" loading. The maxima of the joint displacements, member end forces and support reactions are determined.

17) Example problem No. 17 - The usage of User Provided Steel Tables is illustrated in this example for the analysis and design of a plane frame.

18) Example problem No. 18 - This is an example which

demonstrates the calculation of principal stresses on a finite element.


usage of inclined supports. The word INCLINED refers to the fact that the restraints at a joint where such a support is specified are along a user-specified axis system instead of along the default directions of the global axis system. STAAD offers a few different methods for assigning inclined supports, and we examine those in this example.

20) Example problem No. 20 - This example generates the

geometry of a cylindrical tank structure using the cylindrical coordinate system.

21) Example problem No. 21 - This example illustrates the

modeling of tension-only members using the MEMBER TENSION command.

22) Example problem No. 22 - A space frame structure is subjected to a sinusoidal loading. The commands necessary to describe the sine function are demonstrated in this example. Time History analysis is performed on this model.

23) Example problem No. 23 - This example illustrates the usage

of commands necessary to automatically generate spring supports for a slab on grade. The slab is subjected to various types of loading and analysis of the structure is performed.

24) Example problem No. 24 - This is an example of the analysis

of a structure modelled using “SOLID” finite elements. This example also illustrates the method for applying an “enforced” displacement on the structure.


usage of compression-only members. Since the structural condition is load dependent, the PERFORM ANALYSIS command is specified once for each primary load case.

26) Example problem No. 26 - The structure in this example is a

building consisting of member columns as well as floors made up of beam members and plate elements. Using the master-slave command, the floors are specified to be rigid diaphragms for inplane actions but flexible for bending actions.

27) Example problem No. 27 - This example illustrates the usage

of commands necessary to apply the compression only attribute to automatically generated spring supports for a slab on grade. The slab is subjected to pressure and overturning loading. A tension/compression only analysis of the structure is performed.


input required for obtaining the modes and frequencies of the skewed bridge. The structure consists of piers, pier-cap girders and a deck slab.

29) Example problem No. 29 - Analysis and design of a structure

for seismic loads is demonstrated in this example. The elaborate dynamic analysis

Example Problem 1

1

Example Problem No. 1

Plane frame with steel design. After one analysis, member selection is requested. Since member sizes change during the member selection, another analysis is done followed by final code checking to verify that the final sizes meet the requirements of the code based on the latest analysis results.

Part I - Application Examples

Example Problem 1

2

Actual input is shown in bold lettering followed by explanation. STAAD PLANE EXAMPLE PROBLEM NO. 1

Every input has to start with the word STAAD. The word PLANE signifies that the structure is a plane frame structure and the geometry is defined through X and Y axes.

UNIT METER KN

Specifies the unit to be used. JOINT COORDINATES 1 0 0 ; 2 9 0 ; 3 0 6 ; 4 3 6 5 6 6 ; 6 9 6 ; 7 0 10.5 8 9 10.5 ; 9 2.25 10.5 ; 10 6.75 10.5 11 4.5 10.5 ; 12 1.5 11.4 ; 13 7.5 11.4 14 3 12.3 ; 15 6 12.3 ; 16 4.5 13.2

Joint number followed by X and Y coordinates are provided above. Since this is a plane structure, the Z coordinates need not be provided. Semicolon signs (;) are used as line separators to allow for input of multiple sets of data on one line.

MEMBER INCIDENCE 1 1 3 ; 2 3 7 ; 3 2 6 ; 4 6 8 ; 5 3 4 6 4 5 ; 7 5 6 ; 8 7 12 ; 9 12 14 10 14 16 ; 11 15 16 ; 12 13 15 ; 13 8 13 14 9 12 ; 15 9 14 ; 16 11 14 ; 17 11 15 18 10 15 ; 19 10 13 ; 20 7 9 21 9 11 ; 22 10 11 ; 23 8 10

Defines the members by the joints they are connected to. MEMBER PROPERTY BRITISH 1 3 4 TA ST UC356X368X129 ; 2 TA ST UC254X254X73 5 6 7 TA ST UB533X210X82 ; 8 TO 13 TA ST UB457X152X52 14 TO 23 TA ST UA100X100X8

Member properties are from the British steel table. The word ST stands for standard single section.

Example Problem 1

3 MEMB TRUSS 14 TO 23

The above command defines that members 14 through 23 are of type truss. This means that these members can carry only axial tension/compression and no moments.

MEMB RELEASE 5 START MZ

Member 5 has local moment-z (MZ) released at the start joint. This means that the member cannot carry any moment-z (i.e. strong axis moment) at node 3.

UNIT KN MMS CONSTANTS E 210. ALL DEN 76.977E-09 ALL POISSON STEEL ALL BETA 90.0 MEMB 3 4 UNIT METER

The CONSTANT command initiates input for material constants like E (modulus of elasticity), POISSON, etc. Length unit is changed from METER to MM to facilitate the input. The BETA command specifies that members 3 and 4 are rotated by 90 degrees around their own longitudinal axis. See section 1 of the Technical Reference Manual for the definition of the BETA angle.

SUPPORT 1 FIXED ; 2 PINNED

A fixed support is located at joint 1 and a pinned support at joint 2.


Example Problem 1

4

PRINT MEMBER INFORMATION LIST 1 5 14 PRINT MEMBER PROPERTY LIST 1 2 5 8 14

The above PRINT commands are self-explanatory. The LIST option restricts the print output to the members listed.

LOADING 1 DEAD AND LIVE LOAD

Load case 1 is initiated followed by a title. SELFWEIGHT Y -1.0

One of the components of load case 1 is the selfweight of the structure acting in the global Y direction with a factor of -1.0. Since global Y is vertically upward, the factor of -1.0 indicates that this load will act downwards.

JOINT LOAD 4 5 FY -65. ; 11 FY -155.

Load 1 contains joint loads also. FY indicates that the load is a force in the global Y direction.

MEMB LOAD 8 TO 13 UNI Y -13.5 ; 6 UNI GY -17.5

Load 1 contains member loads also. GY indicates that the load is in the global Y direction while Y indicates local Y direction. The word UNI stands for uniformly distributed load. Loads are applied on members 6, and 8 to 13.

CALCULATE RAYLEIGH FREQUENCY

The above command at the end of load case 1, is an instruction to perform a natural frequency calculation based on the Rayleigh method using the data in the above load case.

LOADING 2 WIND FROM LEFT MEMBER LOAD 1 2 UNI GX 9.0 ; 8 TO 10 UNI Y -15.0

Example Problem 1

5

Load case 2 is initiated and contains several member loads. * 1/3 RD INCREASE IS ACCOMPLISHED BY 75% LOAD LOAD COMB 3 75 PERCENT DL LL WL 1 0.75 2 0.75

The above command identifies a combination load (case no. 3) with a title. The second line provides the load cases and their respective factors used for the load combination. Any line beginning with the * mark is treated as a comment line.

PERFORM ANALYSIS

This command instructs the program to proceed with the analysis. LOAD LIST 1 3

The above command activates load cases 1 and 3 only for the commands to follow. This also means that load case 2 will be made inactive.

PRINT MEMBER FORCES PRINT SUPPORT REACTION

The above PRINT commands are self-explanatory. Also note that all the forces and reactions will be printed for load cases 1 and 3 only.

PARAMETER CODE BRITISH NSF 0.85 ALL BEAM 1 ALL KY 1.2 MEMB 3 4 RATIO 0.9 ALL

The PARAMETER command is used to specify steel design parameters such as NSF, KY, etc. Information on these parameters can be obtained from the manual where the implementation of the code is explained. The BEAM parameter is specified to perform design at every 1/12th point along the member length. The RATIO


Example Problem 1

6

parameter specifies that the ratio of actual loading over section capacity should not exceed 0.9.

SELECT ALL

The above command instructs the program to select the most economic section for ALL the members based on the results of the analysis.

GROUP MEMB 1 3 4 GROUP MEMB 5 6 7 GROUP MEMB 8 TO 13 GROUP MEMB 14 TO 23

Although the program selects the most economical section for all members, it is not always practical to use many different sizes in one structure. GROUPing is a procedure by which the cross section which has the largest value for the specified attribute, which in this case is the default and hence the AREA, from among the associated member list, is assigned to all members in the list. Hence, the cross sections for members 1, 3 and 4 are replaced with the one with the largest area from among the three.

PERFORM ANALYSIS

As a result of the selection and grouping, the member sizes are no longer the same as the ones used in the original analysis. Hence, it is necessary to reanalyze the structure using the new properties to get new values of forces in the members.

PARAMETER BEAM 1.0 ALL RATIO 1.0 ALL TRACK 1.0 ALL

A new set of values are now provided for the above parameters. The actual load to member capacity RATIO has been redefined as 1.0. The TRACK parameter tells the program to print out the design results to the intermediate level of descriptivity.

Example Problem 1

7 CHECK CODE ALL

With the above command, the latest member sizes with the latest analysis results are checked to verify that they satisfy the CODE specifications.

STEEL TAKE OFF

This command instructs the program to list the length and weight of all the different member sizes.

FINISH

This command terminates the STAAD run.


Example Problem 1

8 ***************************************************** ** STAAD.Pro ** Version Bld ** Proprietary Program of ** Research Engineers, Intl. ** Date= ** Time= ** ** USER ID: *****************************************************

1. STAAD PLANE EXAMPLE PROBLEM NO. 12. UNIT METER KN3. JOINT COORDINATES4. 1 0. 0. ; 2 9. 0. ; 3 0. 6. ; 4 3. 6.5. 5 6. 6. ; 6 9. 6. ; 7 0 10.56. 8 9. 10.5 ; 9 2.25 10.5 ; 10 6.75 10.57. 11 4.5 10.5 ; 12 1.5 11.4 ; 13 7.5 11.48. 14 3. 12.3 ; 15 6. 12.3 ; 16 4.5 13.29. MEMBER INCIDENCE

10. 1 1 3 ; 2 3 7 ; 3 2 6 ; 4 6 8 ; 5 3 411. 6 4 5 ; 7 5 6 ; 8 7 12 ; 9 12 1412. 10 14 16 ; 11 15 16 ; 12 13 15 ; 13 8 1313. 14 9 12 ; 15 9 14 ; 16 11 14 ; 17 11 1514. 18 10 15 ; 19 10 13 ; 20 7 915. 21 9 11 ; 22 10 11 ; 23 8 1016. MEMBER PROPERTY BRITISH17. 1 3 4 TA ST UC356X368X129 ; 2 TA ST UC254X254X7318. 5 6 7 TA ST UB533X210X82 ; 8 TO 13 TA ST UB457X152X5219. 14 TO 23 TA RA UA100X100X820. MEMB TRUSS21. 14 TO 2322. MEMB RELEASE23. 5 START MZ24. UNIT KNS MMS25. CONSTANTS26. E 210. ALL27. DEN 76.977E-09 ALL28. POISSON STEEL ALL29. BETA 90.0 MEMB 3 430. UNIT METER31. SUPPORT32. 1 FIXED ; 2 PINNED33. PRINT MEMBER INFORMATION LIST 1 5 14

MEMBER INFORMATION------------------

MEMBER START END LENGTH BETAJOINT JOINT (METE) (DEG) RELEASES

1 1 3 6.000 0.005 3 4 3.000 0.00 000001000000

14 9 12 1.172 TRUSS

************ END OF DATA FROM INTERNAL STORAGE ************

34. PRINT MEMBER PROPERTY LIST 1 2 5 8 14

MEMBER PROPERTIES. UNIT - CM-----------------MEMB PROFILE AX/ IZ/ IY/ IX/

AY AZ SZ SY1 ST UC356X368X129

164.00 40250.00 14610.00 152.6136.98 116.11 2263.78 792.73

2 ST UC254X254X73 93.10 11410.00 3908.00 57.6221.85 65.07 898.07 306.99

5 ST UB533X210X82 105.00 47540.00 2007.00 51.5250.72 49.61 1799.74 192.24

8 ST UB457X152X52 66.60 21370.00 645.00 21.3734.18 29.90 950.20 84.65

14 RA UA100X100X8 15.60 235.86 60.54 3.335.33 5.33 33.36 15.55


Example Problem 1

9 35. LOADING 1 DEAD AND LIVE LOAD36. SELFWEIGHT Y -1.037. JOINT LOAD38. 4 5 FY -65. ; 11 FY -155.39. MEMB LOAD40. 8 TO 13 UNI Y -13.5 ; 6 UNI GY -17.541. CALCULATE RAYLEIGH FREQUENCY42. LOADING 2 WIND FROM LEFT43. MEMBER LOAD44. 1 2 UNI GX 9.0 ; 8 TO 10 UNI Y -15.045. * 1/3 RD INCREASE IS ACCOMPLISHED BY 75% LOAD46. LOAD COMB 3 75 PERCENT DL LL WL47. 1 0.75 2 0.7548. PERFORM ANALYSIS

P R O B L E M S T A T I S T I C S-----------------------------------

NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 16/ 23/ 2ORIGINAL/FINAL BAND-WIDTH= 5/ 4/ 15 DOFTOTAL PRIMARY LOAD CASES = 2, TOTAL DEGREES OF FREEDOM = 43SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDSREQRD/AVAIL. DISK SPACE = 12.0/ 3570.0 MB

ZERO STIFFNESS IN DIRECTION 6 AT JOINT 9 EQN.NO. 21

LOADS APPLIED OR DISTRIBUTED HERE FROM ELEMENTS WILL BE IGNORED.THIS MAY BE DUE TO ALL MEMBERS AT THIS JOINT BEING RELEASED OREFFECTIVELY RELEASED IN THIS DIRECTION.

ZERO STIFFNESS IN DIRECTION 6 AT JOINT 11 EQN.NO. 31ZERO STIFFNESS IN DIRECTION 6 AT JOINT 10 EQN.NO. 37

*********************************************************** ** RAYLEIGH FREQUENCY FOR LOADING 1 = 4.25328 CPS ** MAX DEFLECTION = 2.54543 CM GLO X, AT JOINT 7 ** ***********************************************************

49. LOAD LIST 1 350. PRINT MEMBER FORCES

MEMBER END FORCES STRUCTURE TYPE = PLANE-----------------ALL UNITS ARE -- KNS METE

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

1 1 1 239.74 -7.78 0.00 0.00 0.00 -72.753 -232.17 7.78 0.00 0.00 0.00 26.06

3 1 179.67 85.56 0.00 0.00 0.00 334.513 -173.99 -45.06 0.00 0.00 0.00 57.37

2 1 3 150.36 -22.47 0.00 0.00 0.00 -26.067 -147.13 22.47 0.00 0.00 0.00 -75.07

3 3 128.85 -0.17 0.00 0.00 0.00 -57.377 -126.43 30.54 0.00 0.00 0.00 -11.73

3 1 2 258.36 0.00 -7.78 0.00 0.00 0.006 -250.79 0.00 7.78 0.00 46.69 0.00

3 2 244.54 0.00 -15.69 0.00 0.00 0.006 -238.86 0.00 15.69 0.00 94.13 0.00

4 1 6 142.82 0.00 -22.47 0.00 71.00 0.008 -137.14 0.00 22.47 0.00 30.13 0.00

3 6 141.66 0.00 -60.92 0.00 140.15 0.008 -137.40 0.00 60.92 0.00 133.97 0.00

5 1 3 -14.69 81.81 0.00 0.00 0.00 0.004 14.69 -79.39 0.00 0.00 0.00 241.80

3 3 -45.23 45.13 0.00 0.00 0.00 0.004 45.23 -43.32 0.00 0.00 0.00 132.67

6 1 4 -14.69 14.39 0.00 0.00 0.00 -241.805 14.69 40.54 0.00 0.00 0.00 202.57

3 4 -45.23 -5.43 0.00 0.00 0.00 -132.675 45.23 46.63 0.00 0.00 0.00 54.58


Example Problem 1

10 MEMBER END FORCES STRUCTURE TYPE = PLANE-----------------ALL UNITS ARE -- KNS METE


7 1 5 -14.69 -105.54 0.00 0.00 0.00 -202.576 14.69 107.96 0.00 0.00 0.00 -117.69

3 5 -45.23 -95.38 0.00 0.00 0.00 -54.586 45.23 97.20 0.00 0.00 0.00 -234.28

8 1 7 167.97 70.64 0.00 0.00 0.00 75.0712 -167.51 -46.26 0.00 0.00 0.00 27.17

3 7 173.02 43.51 0.00 0.00 0.00 11.7312 -172.68 -5.54 0.00 0.00 0.00 31.18

9 1 12 168.43 40.09 0.00 0.00 0.00 -27.1714 -167.97 -15.70 0.00 0.00 0.00 75.96

3 12 168.07 34.97 0.00 0.00 0.00 -31.1814 -167.72 2.99 0.00 0.00 0.00 59.15

10 1 14 188.13 -88.19 0.00 0.00 0.00 -75.9616 -187.67 112.57 0.00 0.00 0.00 -99.63

3 14 154.12 -61.66 0.00 0.00 0.00 -59.1516 -153.77 99.62 0.00 0.00 0.00 -81.91

11 1 15 188.10 -88.23 0.00 0.00 0.00 -76.0416 -187.64 112.61 0.00 0.00 0.00 -99.63

3 15 160.61 -70.51 0.00 0.00 0.00 -57.4216 -160.27 88.80 0.00 0.00 0.00 -81.91

12 1 13 182.74 36.38 0.00 0.00 0.00 -33.7315 -182.28 -11.99 0.00 0.00 0.00 76.04

3 13 123.15 38.72 0.00 0.00 0.00 -5.6815 -122.81 -20.43 0.00 0.00 0.00 57.42

13 1 8 185.13 48.70 0.00 0.00 0.00 30.1313 -184.66 -24.32 0.00 0.00 0.00 33.73

3 8 118.56 88.98 0.00 0.00 0.00 133.9713 -118.21 -70.69 0.00 0.00 0.00 5.68

14 1 9 -6.13 0.05 0.00 0.00 0.00 0.0012 6.24 0.05 0.00 0.00 0.00 0.00

3 9 29.87 0.03 0.00 0.00 0.00 0.0012 -29.79 0.03 0.00 0.00 0.00 0.00

15 1 9 4.76 0.05 0.00 0.00 0.00 0.0014 -4.54 0.05 0.00 0.00 0.00 0.00

3 9 -25.12 0.03 0.00 0.00 0.00 0.0014 25.28 0.03 0.00 0.00 0.00 0.00

16 1 11 -107.60 0.09 0.00 0.00 0.00 0.0014 107.82 0.09 0.00 0.00 0.00 0.00

3 11 -43.98 0.07 0.00 0.00 0.00 0.0014 44.14 0.07 0.00 0.00 0.00 0.00

17 1 11 -94.67 0.09 0.00 0.00 0.00 0.0015 94.88 0.09 0.00 0.00 0.00 0.00

3 11 -107.72 0.07 0.00 0.00 0.00 0.0015 107.88 0.07 0.00 0.00 0.00 0.00

18 1 10 -10.59 0.05 0.00 0.00 0.00 0.0015 10.81 0.05 0.00 0.00 0.00 0.00

3 10 26.60 0.03 0.00 0.00 0.00 0.0015 -26.43 0.03 0.00 0.00 0.00 0.00

19 1 10 12.32 0.05 0.00 0.00 0.00 0.0013 -12.21 0.05 0.00 0.00 0.00 0.00

3 10 -32.27 0.03 0.00 0.00 0.00 0.0013 32.35 0.03 0.00 0.00 0.00 0.00

20 1 7 -85.22 0.14 0.00 0.00 0.00 0.009 85.22 0.14 0.00 0.00 0.00 0.00

3 7 -95.44 0.10 0.00 0.00 0.00 0.009 95.44 0.10 0.00 0.00 0.00 0.00

Example Problem 1



21 1 9 -90.97 0.14 0.00 0.00 0.00 0.0011 90.97 0.14 0.00 0.00 0.00 0.00

3 9 -66.65 0.10 0.00 0.00 0.00 0.0011 66.65 0.10 0.00 0.00 0.00 0.00

22 1 10 -99.25 0.14 0.00 0.00 0.00 0.0011 99.25 0.14 0.00 0.00 0.00 0.00

3 10 -25.85 0.10 0.00 0.00 0.00 0.0011 25.85 0.10 0.00 0.00 0.00 0.00

23 1 8 -111.21 0.14 0.00 0.00 0.00 0.0010 111.21 0.14 0.00 0.00 0.00 0.00

3 8 5.03 0.10 0.00 0.00 0.00 0.0010 -5.03 0.10 0.00 0.00 0.00 0.00

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

51. PRINT SUPPORT REACTION

SUPPORT REACTIONS -UNIT KNS METE STRUCTURE TYPE = PLANE-----------------

JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z

1 1 7.78 239.74 0.00 0.00 0.00 -72.753 -85.56 179.67 0.00 0.00 0.00 334.51

2 1 -7.78 258.36 0.00 0.00 0.00 0.003 -15.69 244.54 0.00 0.00 0.00 0.00


52. PARAMETER53. CODE BRITISH54. NSF 0.85 ALL55. BEAM 1 ALL56. KY 1.2 MEMB 3 457. RATIO 0.9 ALL58. SELECT ALL

STAAD.Pro MEMBER SELECTION - (BSI )**************************

PROGRAM CODE REVISION V2.9_5950-1_2000

ALL UNITS ARE - KNS METE (UNLESS OTHERWISE NOTED)

MEMBER TABLE RESULT/ CRITICAL COND/ RATIO/ LOADING/FX MY MZ LOCATION

=======================================================================

1 ST UC305X305X118 PASS BS-4.3.6 0.770 3179.67 C 0.00 334.51 0.00

2 ST UC203X203X46 PASS ANNEX I.1 0.721 1150.36 C 0.00 75.07 -



5 ST UB406X178X67 PASS BS-4.3.6 0.855 114.69 T 0.00 241.80 3.00

6 ST UB406X178X67 PASS BS-4.3.6 0.875 114.69 T 0.00 247.44 0.75

7 ST UB406X178X67 PASS BS-4.3.6 0.829 345.23 T 0.00 234.28 3.00

8 ST UB356X127X33 PASS ANNEX I.1 0.625 1167.97 C 0.00 75.07 -


10 ST UB406X140X39 PASS BS-4.3.6 0.583 1188.13 C 0.00 99.63 1.75


Example Problem 1

12 ALL UNITS ARE - KNS METE (UNLESS OTHERWISE NOTED)


=======================================================================

11 ST UB406X140X39 PASS BS-4.3.6 0.583 1188.10 C 0.00 99.63 1.75


13 ST UB406X140X39 PASS BS-4.3.6 0.784 3118.56 C 0.00 133.97 0.00

14 RA UA50X50X4 PASS BS-4.7 (C) 0.799 329.87 C 0.00 0.00 0.00

15 RA UA40X40X3 PASS BS-4.7 (C) 0.754 14.76 C 0.00 0.00 0.00

16 RA UA45X45X6 PASS BS-4.6 (T) 0.887 1107.82 T 0.00 0.00 2.34

17 RA UA45X45X6 PASS BS-4.6 (T) 0.888 3107.88 T 0.00 0.00 2.34

18 RA UA60X60X5 PASS BS-4.7 (C) 0.774 326.60 C 0.00 0.00 0.00

19 RA UA40X40X3 PASS BS-4.7 (C) 0.816 112.32 C 0.00 0.00 0.00

20 RA UA40X40X6 PASS BS-4.6 (T) 0.888 395.44 T 0.00 0.00 0.00

21 RA UA45X45X5 PASS BS-4.6 (T) 0.882 190.97 T 0.00 0.00 0.00

22 RA UA50X50X5 PASS BS-4.6 (T) 0.865 199.25 T 0.00 0.00 0.00

23 RA UA65X50X5 PASS BS-4.6 (T) 0.857 1111.21 T 0.00 0.00 0.00

************** END OF TABULATED RESULT OF DESIGN **************

59. GROUP MEMB 1 3 4

GROUPING BASED ON MEMBER 1 (ST UC305X305X118 ) LIST= 1....60. GROUP MEMB 5 6 7

GROUPING BASED ON MEMBER 7 (ST UB406X178X67 ) LIST= 5....61. GROUP MEMB 8 TO 13

GROUPING BASED ON MEMBER 13 (ST UB406X140X39 ) LIST= 8....62. GROUP MEMB 14 TO 23

GROUPING BASED ON MEMBER 18 (RA UA60X60X5 ) LIST= 14....63. PERFORM ANALYSIS

** ALL CASES BEING MADE ACTIVE BEFORE RE-ANALYSIS. **ZERO STIFFNESS IN DIRECTION 6 AT JOINT 9 EQN.NO. 21



*********************************************************** ** RAYLEIGH FREQUENCY FOR LOADING 1 = 3.13919 CPS ** MAX DEFLECTION = 4.93130 CM GLO X, AT JOINT 7 ** ***********************************************************

64. PARAMETER65. BEAM 1 ALL66. RATIO 1.0 ALL67. TRACK 1.0 ALL68. CHECK CODE ALL

Example Problem 1

13 STAAD.Pro CODE CHECKING - (BSI )***********************




=======================================================================

1 ST UC305X305X118 PASS BS-4.3.6 0.809 3178.64 C 0.00 351.85 0.00

|---------------------------------------------------------------------|| CALCULATED CAPACITIES FOR MEMB 1 UNIT - kN,m SECTION CLASS 1 ||MCZ= 518.9 MCY= 234.3 PC= 2455.7 PT= 0.0 MB= 434.7 PV= 600.1|| BUCKLING CO-EFFICIENTS mLT = 1.00, mx = 1.00, my = 1.00, myx = 1.00 || PZ= 3975.00 FX/PZ = 0.04 MRZ= 516.4 MRY= 234.3 ||---------------------------------------------------------------------|







5 ST UB406X178X67 PASS BS-4.3.6 0.839 114.21 T 0.00 237.10 3.00


6 ST UB406X178X67 PASS BS-4.3.6 0.855 114.21 T 0.00 241.77 0.75


7 ST UB406X178X67 PASS BS-4.3.6 0.812 344.36 T 0.00 229.38 3.00

|---------------------------------------------------------------------|| CALCULATED CAPACITIES FOR MEMB 7 UNIT - kN,m SECTION CLASS 1 ||MCZ= 370.2 MCY= 63.0 PC=235263.6 PT= 1998.6 MB= 282.7 PV= 594.4|| BUCKLING CO-EFFICIENTS mLT = 1.00, mx = 1.00, my = 1.00, myx = 1.00 || PZ=349861.72 FX/PZ = 0.03 MRZ= 369.9 MRY= 63.0 ||---------------------------------------------------------------------|

8 ST UB406X140X39 PASS BS-4.3.6 0.415 1163.71 C 0.00 70.97 0.00



Example Problem 1



=======================================================================

9 ST UB406X140X39 PASS BS-4.3.6 0.452 1164.73 C 0.00 77.32 1.75


10 ST UB406X140X39 PASS BS-4.3.6 0.565 1186.58 C 0.00 96.64 1.75


11 ST UB406X140X39 PASS BS-4.3.6 0.565 1186.41 C 0.00 96.64 1.75


12 ST UB406X140X39 PASS BS-4.3.6 0.455 1175.44 C 0.00 77.83 1.75


13 ST UB406X140X39 PASS BS-4.3.6 0.764 3120.10 C 0.00 130.65 0.00


14 RA UA60X60X5 PASS BS-4.7 (C) 0.254 318.96 C 0.00 0.00 0.00


15 RA UA60X60X5 PASS BS-4.7 (C) 0.213 17.32 C 0.00 0.00 0.00


16 RA UA60X60X5 PASS BS-4.6 (T) 0.772 1106.66 T 0.00 0.00 2.34


Example Problem 1



=======================================================================

17 RA UA60X60X5 PASS BS-4.6 (T) 0.755 3104.25 T 0.00 0.00 2.34


18 RA UA60X60X5 PASS BS-4.7 (C) 0.677 323.28 C 0.00 0.00 0.00


19 RA UA60X60X5 PASS BS-4.6 (T) 0.204 328.12 T 0.00 0.00 1.17


20 RA UA60X60X5 PASS BS-4.6 (T) 0.587 381.10 T 0.00 0.00 0.00


21 RA UA60X60X5 PASS BS-4.6 (T) 0.645 189.17 T 0.00 0.00 0.00


22 RA UA60X60X5 PASS BS-4.6 (T) 0.697 196.33 T 0.00 0.00 0.00


23 RA UA60X60X5 PASS BS-4.6 (T) 0.731 1100.93 T 0.00 0.00 0.00




Example Problem 1

16 69. STEEL TAKE OFF

STEEL TAKE-OFF--------------

PROFILE LENGTH(METE) WEIGHT(KNS )

ST UC305X305X118 16.50 19.052ST UC203X203X46 4.50 2.033ST UB406X178X67 9.00 5.923ST UB406X140X39 10.50 4.015RA UA60X60X5 19.93 0.907PRISMATIC STEEL 0.00 0.000

----------------TOTAL = 31.931


70. FINISH

*********** END OF THE STAAD.Pro RUN ***********

**** DATE= TIME= ****

************************************************************* For questions on STAAD.Pro, please contact ** Research Engineers Offices at the following locations ** ** Telephone Email ** USA: +1 (714)974-2500 [email protected] ** CANADA +1 (905)632-4771 [email protected] ** UK +44(1454)207-000 [email protected] ** FRANCE +33(0)1 64551084 [email protected] ** GERMANY +49/931/40468-71 [email protected] ** NORWAY +47 67 57 21 30 [email protected] ** SINGAPORE +65 6225-6015/16 [email protected] ** INDIA +91(033)2357-3575 [email protected] ** JAPAN +81(03)5952-6500 [email protected] ** CHINA +86(411)363-1983 [email protected] ** ** North America [email protected] ** Europe [email protected] ** Asia [email protected] *************************************************************

Example Problem 1

17

NOTES


Example Problem 1

18

NOTES

Example Problem 2

19


A floor structure (bound by global X-Z axis) made up of steel beams is subjected to area load (i.e. load/area of floor). Load generation based on one-way distribution is illustrated in this example. In the case of loads such as joint loads and member loads, the magnitude and direction of the load at the applicable joints and members is directly known from the input. However, the area load is a different sort of load where a load intensity on the given area has to be converted to joint and member loads. The calculations required to perform this conversion are done only during the analysis. Consequently, the loads generated from the AREA LOAD command can be viewed only after the analysis is completed.


Example Problem 2

20

Actual input is shown in bold lettering followed by explanation. STAAD FLOOR A FLOOR FRAME DESIGN WITH AREA LOAD

Every input has to start with the word STAAD. The word FLOOR signifies that the structure is a floor structure and the structure is in the x – z plane.

UNIT METER KNS

Defines the UNITs.

JOINT COORDINATES 1 0 0 0 5 6 0 0 ; 7 1.5 0 3 8 3 0 3 ; 9 4 0 3 ; 10 4.5 0 3 ; 11 5 0 3 12 6 0 3 ; 13 0 0 7.5 ; 14 1.5 0. 7.5 15 3.5 0 7.5 16 5 0 7.5 ; 17 6 0 7.5 ; 18 0 0 8.5 19 6 0 8.5 ; 20 0 0 10.5 ; 21 6 0 10.5

Joint number followed by X, Y and Z coordinates are provided above. Since this is a floor structure, the Y coordinates are all the same, in this case zero. Semicolon signs (;) are used as line separators to allow for input of multiple sets of data on one line. Joints between 1 and 5 (i.e. 2, 3, 4) are generated in the first line of input taking advantage of the equal spacing between the joints (see section 5 of the Technical Reference Manual for more information).

MEMBER INCIDENCES 1 1 2 4 ; 5 7 8 9 ; 10 13 14 13 ; 14 18 19 15 20 21 ; 16 18 20 ; 17 13 18 ; 18 1 13 19 7 14 ; 20 2 7 ; 21 9 15 22 3 8 ; 23 11 16 ; 24 4 10 ; 25 19 21 26 17 19 ; 27 12 17 ; 28 5 12

Defines the members by the joints they are connected to.

Example Problem 2

21 MEMB PROP BRITISH 1 TO 28 TABLE ST UB305X165X40

Member properties are specified from the British steel table. The word ST stands for standard single section.

* MEMBERS WITH PINNED ENDS ARE RELEASED FOR MZ

MEMB RELEASE 1 5 10 14 15 18 17 28 26 20 TO 24 START MZ 4 9 13 14 15 18 16 27 25 19 21 TO 24 END MZ

The first set of members (1 5 10 etc) have local moment-z (MZ) released at the start joint. This means that these members cannot carry any moment-z (i.e. strong axis moment) at the start joint. The second set of members have MZ released at the end joints. Any line beginning with * mark is treated as a comment line.

UNIT MMS CONSTANT E 210. ALL POISSON STEEL ALL

The CONSTANT command initiates input for material constants like E (modulus of elasticity), POISSON, etc. E has been assigned as 210.0 KN/sq.mm. The built-in default for Poisson’s value for steel is used during the analysis.

UNIT METER SUPPORT 1 5 13 17 20 21 FIXED

A fixed support has been specified at the above joints. LOADING 1 14.5 KN/sq.m. DL+LL

Load case 1 is initiated followed by a title. AREA LOAD 1 TO 28 ALOAD -14.5

All the 28 members are subjected to an Area load of 14.5 KN/sq.m. The program converts area loads into individual member loads.


Example Problem 2

22 PERFORM ANALYSIS PRINT LOAD DATA

This command instructs the program to proceed with the analysis. The PRINT LOAD DATA command is specified to obtain a listing of the member loads which were generated from the AREA LOAD.

PARAMETERS CODE BRITISH BEAM 1 ALL DMAX 0.6 ALL DMIN 0.3 ALL UNL 0.3 ALL

The PARAMETER command is used to specify steel design parameters (see the manual where code specification information is provided). The BEAM parameter is specified to perform design at every 1/12th point along the member length. DMAX and DMIN specify maximum and minimum depth limitations to be used during member selection. UNL stands for unsupported length of the compression flange to be used for calculation of bending capacity.

SELECT MEMB 2 6 11 14 15 16 18 19 21 23 24 27

The above command instructs the program to select the most economical section from the British steel table for the members listed.

FINISH

The FINISH command terminates the STAAD run.

Example Problem 2

23

***************************************************** ** STAAD.Pro ** Version Bld ** Proprietary Program of ** Research Engineers, Intl. ** Date= ** Time= ** ** USER ID: *****************************************************

1. STAAD FLOOR A FLOOR FRAME DESIGN WITH AREA LOAD2. UNIT METER KNS3. JOINT COORDINATES4. 1 0. 0. 0. 5 6. 0. 0. ; 7 1.5 0. 3.5. 8 3. 0. 3. ; 9 4. 0. 3. ; 10 4.5 0. 3.6. 11 5. 0 3. ; 12 6. 0. 3. ; 13 0. 0. 7.57. 14 1.5 0. 7.5 ; 15 3.5 0. 7.58. 16 5. 0. 7.5 ; 17 6. 0. 7.5 18 0. 0. 8.59. 19 6. 0. 8.5 ; 20 0. 0. 10.5 ; 21 6. 0. 10.510. MEMBER INCIDENCES11. 1 1 2 4 ; 5 7 8 9 ; 10 13 14 13 ; 14 18 1912. 15 20 21 ; 16 18 20 ; 17 13 18 ; 18 1 1313. 19 7 14 ; 20 2 7 ; 21 9 1514. 22 3 8 ; 23 11 16 ; 24 4 10 ; 25 19 2115. 26 17 19 ; 27 12 17 ; 28 5 1216. MEMB PROP BRITISH17. 1 TO 28 TABLE ST UB305X165X4018. * MEMBERS WITH PINNED ENDS ARE RELEASED FOR MZ19. MEMB RELEASE20. 1 5 10 14 15 18 17 28 26 20 TO 24 START MZ21. 4 9 13 14 15 18 16 27 25 19 21 TO 24 END MZ22. UNIT MMS23. CONSTANT24. E 210. ALL25. POISSON STEEL ALL26. UNIT METER27. SUPPORT28. 1 5 13 17 20 21 FIXED29. LOADING 1 14.5 KN/SQ.M. DL+LL30. AREA LOAD31. 1 TO 28 ALOAD -14.532. PERFORM ANALYSIS PRINT LOAD DATA



LOADING 1 14.5 KN/SQ.M. DL+LL-----------MEMBER LOAD - UNIT KNS METE

MEMBER UDL L1 L2 CON L LIN1 LIN2

10 -7.250 GY 0.00 1.5011 -7.250 GY 0.00 2.0012 -7.250 GY 0.00 1.5013 -3.625 GY 0.00 1.0014 -21.750 GY 0.00 6.0015 -14.500 GY 0.00 6.0018 -10.875 GY 0.00 7.5019 -29.000 -25.375 GY20 -21.750 GY 0.00 3.0021 -25.375 GY 0.00 4.5322 -21.750 GY 0.00 3.0023 -14.500 -18.125 GY24 -21.750 GY 0.00 3.0027 -7.250 GY 0.00 4.5028 -10.875 GY 0.00 3.00



Example Problem 2

24 33. PARAMETERS34. CODE BRITISH35. BEAM 1 ALL36. DMAX 0.6 ALL37. DMIN 0.3 ALL38. UNL 0.3 ALL39. SELECT MEMB 2 6 11 14 15 16 18 19 21 23 24 27

STAAD.Pro MEMBER SELECTION - (BSI )**************************




=======================================================================

2 ST UB406X140X39 PASS BS-4.3.6 0.915 10.00 0.00 182.20 0.00

6 ST UB356X127X33 PASS BS-4.3.6 0.887 10.00 0.00 132.39 1.00

11 ST UB406X140X46 PASS BS-4.3.6 0.884 10.00 0.00 215.85 1.33

14 ST UB305X102X28 PASS BS-4.3.6 0.883 10.00 0.00 97.87 3.00

15 ST UB305X102X25 PASS BS-4.3.6 0.694 10.00 0.00 65.25 3.00

16 ST UB305X102X25 PASS BS-4.3.6 0.463 10.00 0.00 43.50 0.00

18 ST UB305X102X25 PASS BS-4.3.6 0.813 10.00 0.00 76.46 3.75

19 ST UB457X152X52 PASS BS-4.3.6 0.962 10.00 0.00 290.07 0.00

21 ST UB305X102X25 PASS BS-4.3.6 0.691 10.00 0.00 65.02 2.26

23 ST UB305X102X25 PASS BS-4.3.6 0.439 10.00 0.00 41.29 2.25

24 ST UB305X102X25 PASS BS-4.3.6 0.260 10.00 0.00 24.47 1.50

27 ST UB406X140X46 PASS BS-4.3.6 0.919 10.00 0.00 224.41 0.00


Example Problem 2

25 40. FINISH

******************************************************************************WARNING** SOME MEMBER SIZES HAVE CHANGED SINCE LAST ANALYSIS.

IN THE POST PROCESSOR, MEMBER QUERIES WILL USE THE LASTANALYSIS FORCES WITH THE UPDATED MEMBER SIZES.TO CORRECT THIS INCONSISTENCY, PLEASE DO ONE MORE ANALYSIS.FROM THE UPPER MENU, PRESS RESULTS, UPDATE PROPERTIES, THENFILE SAVE; THEN ANALYZE AGAIN WITHOUT THE GROUP OR SELECTCOMMANDS.

****************************************************************************


**** DATE= TIME= ****



Example Problem 2

26

NOTES

Example Problem 3

27


A portal frame type steel structure is sitting on concrete footings. The soil is to be considered as an elastic foundation. Value of soil subgrade reaction is known from which spring constants are calculated by multiplying the subgrade reaction by the tributary area of each modeled spring.

NOTE:

1) All dimensions are in meters. 2) Soil Subgrade Reaction – 41666.67 KN/m3

Spring constant calculation Spring of joints 1, 5, 10 & 14 = 2.4 x 0.3 x 41666.67

= 30000 KN/m Spring of joints 2, 3, 4, 11, 12 & 13 = 2.4 x 0.6 x 41666.67

= 60000 KN/m

Part I - Application Examples Example Problem 3

28

Actual input is shown in bold lettering followed by explanation. STAAD PLANE PORTAL ON FOOTING FOUNDATION


UNIT METER KNS

Specifies the unit to be used. JOINT COORDINATES 1 0.0 0.0 0.0 5 2.4 0.0 0.0 6 1.2 3.0 0.0 ; 7 1.2 6.0 0.0 8 7.2 6.0 0.0 ; 9 7.2 3.0 0.0 10 6.0 0.0 0.0 14 8.4 0.0 0.0

Joint number followed by X, Y and Z coordinates are provided above. Since this is a plane structure, the Z coordinates are given as all zeros. Semicolon signs (;) are used as line separators to facilitate specification of multiple sets of data on one line.

MEMBER INCIDENCES 1 1 2 4 5 3 6 ; 6 6 7 7 7 8 ; 8 6 9 9 8 9 ; 10 9 12 11 10 11 14

Defines the members by the joints they are connected to. MEMBER PROPERTIES BRITISH 1 4 11 14 PRIS YD 0.30 ZD 2.40 2 3 12 13 PRIS YD 0.60 ZD 2.40 5 6 9 10 TABLE ST JO254X203 7 8 TA ST UB305X165X40

First two lines define member properties as PRIS (prismatic) for which YD (depth) and ZD (width) values are provided. The program will calculate the properties necessary to do the analysis.

Example Problem 3

29

Member properties for the remaining members are chosen from the British steel table. The word ST stands for standard single section.

UNIT MMS CONSTANTS * E FOR STEEL IS 210 (KN/sq.mm.) AND FOR * CONCRETE 21 (KN/sq.mm.) E 210 MEMB 5 TO 10 E 21 MEMB 1 TO 4 11 TO 14 DEN 76.977E-09 MEMB 5 TO 10 DEN 23.534E-09 MEMB 1 TO 4 11 TO 14 POISSON CONCRETE MEMB 1 TO 4 11 TO 14 POISSON STEEL MEMB 5 TO 10

The CONSTANT command initiates input for material constants like E (modulus of elasticity), Density and Poisson’s ratio. Length unit is changed from METER to MMS to facilitate the input. Any line beginning with an * mark is treated as a comment line.

SUPPORTS 2 TO 4 11 TO 13 FIXED BUT MZ KFY 60 1 5 10 14 FIXED BUT MZ KFY 30

The supports for the structure are specified above. The first set of joints are supports restrained in all directions except MZ (which is global moment-z). Also, a spring having a spring constant of 60 KN/mm is provided in the global Y direction at these nodes. The second set is similar to the former except for a different value of the spring constant.

UNIT METER LOADING 1 DEAD AND WIND LOAD COMBINED

Load case 1 is initiated followed by a title. SELF Y -1.0

The selfweight of the structure is specified as acting in the global Y direction with a -1.0 factor. Since global Y is vertically upwards, the -1.0 factor indicates that this load will act downwards.


30 JOINT LOAD 6 7 FX 20.0

Load 1 contains joint loads also. FX indicates that the load is a force in the global X direction.

MEMBER LOAD 7 8 UNI GY -45.0

Load 1 contains member loads also. GY indicates that the load acts in the global Y direction. The word UNI stands for uniformly distributed load, and is applied on members 7 and 8, acting downwards.

PERFORM ANALYSIS

This command instructs the program to proceed with the analysis. PRINT ANALYSIS RESULTS

The above PRINT command instructs the program to print analysis results which include joint displacements, member forces and support reactions.

FINISH


Example Problem 3

31


1. STAAD PLANE PORTAL ON FOOTING FOUNDATION2. UNIT METER KNS3. JOINT COORDINATES4. 1 0.0 0.0 0.0 5 2.4 0.0 0.05. 6 1.2 3.0 0.0 ; 7 1.2 6.0 0.06. 8 7.2 6.0 0.0 ; 9 7.2 3.0 0.07. 10 6.0 0.0 0.0 14 8.4 0.0 0.08. MEMBER INCIDENCES9. 1 1 2 4

10. 5 3 6 ; 6 6 711. 7 7 8 ; 8 6 912. 9 8 9 ; 10 9 1213. 11 10 11 1414. MEMBER PROPERTIES BRITISH15. 1 4 11 14 PRIS YD 0.30 ZD 2.4016. 2 3 12 13 PRIS YD 0.60 ZD 2.4017. 5 6 9 10 TABLE ST JO254X20318. 7 8 TA ST UB305X165X4019. UNIT MMS20. CONSTANTS21. * E FOR STEEL IS 210 (KN/SQ.MM.) AND FOR CONCRETE 21 (KN/SQ.MM.)22. E 210 MEMB 5 TO 1023. E 21 MEMB 1 TO 4 11 TO 1424. DEN 76.977E-09 MEMB 5 TO 1025. DEN 23.534E-09 MEMB 1 TO 4 11 TO 1426. POISSON CONCRETE MEMB 1 TO 4 11 TO 1427. POISSON STEEL MEMB 5 TO 1028. SUPPORTS29. 2 TO 4 11 TO 13 FIXED BUT MZ KFY 6030. 1 5 10 14 FIXED BUT MZ KFY 3031. UNIT METER32. LOADING 1 DEAD AND WIND LOAD COMBINED33. SELF Y -1.034. JOINT LOAD35. 6 7 FX 20.036. MEMBER LOAD37. 7 8 UNI GY -45.038. PERFORM ANALYSIS




32 39. PRINT ANALYSIS RESULTS

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

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

1 1 0.0000 -0.1075 0.0000 0.0000 0.0000 -0.00032 1 0.0000 -0.1232 0.0000 0.0000 0.0000 -0.00023 1 0.0000 -0.1367 0.0000 0.0000 0.0000 -0.00024 1 0.0000 -0.1466 0.0000 0.0000 0.0000 -0.00025 1 0.0000 -0.1531 0.0000 0.0000 0.0000 -0.00016 1 0.5036 -0.1720 0.0000 0.0000 0.0000 -0.00327 1 1.0712 -0.1898 0.0000 0.0000 0.0000 -0.00458 1 1.0285 -0.2097 0.0000 0.0000 0.0000 0.00229 1 0.5240 -0.1902 0.0000 0.0000 0.0000 -0.0003

10 1 0.0000 -0.0897 0.0000 0.0000 0.0000 -0.000511 1 0.0000 -0.1210 0.0000 0.0000 0.0000 -0.000512 1 0.0000 -0.1504 0.0000 0.0000 0.0000 -0.000513 1 0.0000 -0.1759 0.0000 0.0000 0.0000 -0.000414 1 0.0000 -0.1969 0.0000 0.0000 0.0000 -0.0003



2 1 0.00 73.95 0.00 0.00 0.00 0.003 1 0.09 82.01 0.00 0.00 0.00 0.004 1 0.00 87.96 0.00 0.00 0.00 0.00

11 1 0.00 72.58 0.00 0.00 0.00 0.0012 1 -40.09 90.25 0.00 0.00 0.00 0.0013 1 0.00 105.55 0.00 0.00 0.00 0.001 1 0.00 32.25 0.00 0.00 0.00 0.005 1 0.00 45.92 0.00 0.00 0.00 0.00

10 1 0.00 26.90 0.00 0.00 0.00 0.0014 1 0.00 59.06 0.00 0.00 0.00 0.00



1 1 1 0.00 32.25 0.00 0.00 0.00 0.002 0.00 -22.08 0.00 0.00 0.00 16.30

2 1 2 0.00 96.03 0.00 0.00 0.00 -16.303 0.00 -75.69 0.00 0.00 0.00 67.82

3 1 3 0.00 -103.38 0.00 0.00 0.00 -92.634 0.00 123.71 0.00 0.00 0.00 24.50

4 1 4 0.00 -35.75 0.00 0.00 0.00 -24.505 0.00 45.92 0.00 0.00 0.00 0.00

5 1 3 261.08 -0.09 0.00 0.00 0.00 24.816 -258.66 0.09 0.00 0.00 0.00 -25.09

6 1 6 132.10 -56.74 0.00 0.00 0.00 -73.597 -129.67 56.74 0.00 0.00 0.00 -96.63

7 1 7 76.74 129.67 0.00 0.00 0.00 96.638 -76.74 142.70 0.00 0.00 0.00 -135.72

8 1 6 -36.65 126.56 0.00 0.00 0.00 98.689 36.65 145.80 0.00 0.00 0.00 -156.40

9 1 8 142.70 76.74 0.00 0.00 0.00 135.729 -145.12 -76.74 0.00 0.00 0.00 94.50

10 1 9 290.93 40.09 0.00 0.00 0.00 61.9012 -293.35 -40.09 0.00 0.00 0.00 58.38

Example Problem 3



11 1 10 0.00 26.90 0.00 0.00 0.00 0.0011 0.00 -16.73 0.00 0.00 0.00 13.09

12 1 11 0.00 89.32 0.00 0.00 0.00 -13.0912 0.00 -68.99 0.00 0.00 0.00 60.58

13 1 12 0.00 -134.12 0.00 0.00 0.00 -118.9613 0.00 154.45 0.00 0.00 0.00 32.39

14 1 13 0.00 -48.90 0.00 0.00 0.00 -32.3914 0.00 59.06 0.00 0.00 0.00 0.00


40. FINISH


**** DATE= TIME= ****



34 NOTES

Example Problem 4

35


This example is a typical case of a load-dependent structure where the structural condition changes for different load cases. In this example, different bracing members are made inactive for different load cases. This is done to prevent these members from carrying any compressive forces.


36

Actual input is shown in bold lettering followed by explanation. STAAD PLANE * A PLANE FRAME STRUCTURE WITH TENSION BRACING


UNIT METER KNS

Specifies the unit to be used. SET NL 3

This structure has to be analysed for 3 primary load cases. Consequently, the modeling of our problem requires us to define 3 sets of data, with each set containing a load case and an associated analysis command. Also, the members which get switched off in the analysis for any load case have to be restored for the analysis for the subsequent load case. To accommodate these requirements, it is necessary to have 2 commands, one called “SET NL” and the other called “CHANGE”. The SET NL command is used above to indicate the total number of primary load cases that the file contains. The CHANGE command will come in later (after the PERFORM ANALYSIS command).

JOINT COORDINATES 1 0 0 0 3 12. 0. 0. 4 0 4.5 0 6 12. 4.5 0. 7 6. 9. 0. ; 8 12. 9. 0.

Joint number followed by X, Y and Z coordinates are provided above. Since this is a plane structure, the Z coordinates are given as all zeros. Semicolon signs (;) are used as line separators, to facilitate specification of multiple sets of data on one line.

MEMBER INCIDENCE 1 1 4 2 ; 3 5 7 ; 4 3 6 ; 5 6 8 ; 6 4 5 7 8 7 8 ; 9 1 5 ; 10 2 4 ; 11 3 5 ; 12 2 6 13 6 7 ;14 5 8

Example Problem 4

37

Defines the members by the joints they are connected to. MEMBER TRUSS 9 TO 14

The above command defines that members 9 through 14 are of type truss. This means these members can only carry axial tension/compression and no moments.

MEMBER PROP BRITISH 1 TO 5 TABLE ST UB305X165X40 6 7 8 TA ST UB457X152X52 9 TO 14 TA LD UA150X150X10

Properties for all members are assigned from the British steel table. The word ST stands for standard single section. The word LD stands for long leg back-to-back double angle. Since the spacing between the two angles of the double angle is not provided, it is assumed to be 0.0.

UNIT MMS CONSTANTS E 210. ALL POISSON STEEL ALL

The CONSTANT command initiates input for material constants like E (modulus of elasticity), Poisson’s ratio, etc. Length unit is changed from METER to MMS.

SUPPORT 1 2 3 PINNED

PINNED supports are specified at Joints 1, 2 and 3. The word PINNED signifies that no moments will be carried by these supports.

INACTIVE MEMBERS 9 TO 14

The above command makes the listed members inactive. The stiffness contribution of these members will not be considered in the analysis till they are made active again.


38 UNIT METER LOADING 1 DEAD AND LIVE LOAD

Load case 1 is initiated followed by a title. The length UNIT is changed from MMS to METER for input values which follow.

MEMBER LOAD 6 8 UNI GY -4.5 7 UNI GY -6.75

Load 1 contains member loads. GY indicates that the load acts in the global Y direction. The word UNI stands for uniformly distributed load. The load is applied on members 6, 7 and 8.

PERFORM ANALYSIS

This command instructs the program to proceed with the analysis. It is worth noting that members 9 TO 14 will not be used in this analysis since they were declared inactive earlier. In other words, for dead and live load, the bracings are not used to carry any load.

CHANGES

The members inactivated earlier are restored using the CHANGE command.

INACTIVE MEMBERS 10 11 13

A new set of members are made inactive. The stiffness contribution from these members will not be used in the analysis till they are made active again. They have been inactivated to prevent them from being subject to compressive forces for the next load case.

LOADING 2 WIND FROM LEFT

Load case 2 is initiated followed by a title. JOINT LOAD 4 FX 135 ; 7 FX 65

Example Problem 4

39

Load 2 contains joint loads. FX indicates that the load is a force in the global X direction. Nodes 4 and 7 are subjected to the loads.

PERFORM ANALYSIS

This command instructs the program to proceed with the analysis. The analysis will be performed for load case 2 only.

CHANGE

The above CHANGE command is an instruction to re-activate all inactive members.

INACTIVE MEMBERS 9 12 14

Members 9, 12 and 14 are made inactive. The stiffness contribution of these members will not be used in the analysis till they are made active again. They have been inactivated to prevent them from being subject to compressive forces for the next load case.

LOADING 3 WIND FROM RIGHT

Load case 3 is initiated followed by a title. JOINT LOAD 6 FX -135 ; 8 FX -65

Load 3 contains joint loads at nodes 6 and 8. FX indicates that the load is a force in the global X direction. The negative numbers (-135 and -65) indicate that the load is acting along the negative global X direction.

LOAD COMBINATION 4 1 0.75 2 0.75 LOAD COMBINATION 5 1 0.75 3 0.75

Load combination case 4 involves the algebraic summation of the results of load cases 1 and 2 after multiplying each by a factor of 0.75. For load combinations, the program simply gathers the results of the component primary cases, factors them appropriately, and


40

combines them algebraically. Thus, an analysis in the real sense of the term (multiplying the inverted stiffness matrix by the load vector) is not carried out for load combination cases. Load combination case 5 combines the results of load cases 1 and 3.

PERFORM ANALYSIS

This command instructs the program to proceed with the analysis. Only primary load case 3 will be considered for this analysis. (As explained earlier, a combination case is not truly analysed for, but handled using other means.)

CHANGE

The above CHANGE command will re-activate all inactive members.

LOAD LIST ALL

At the end of any analysis, only those load cases for which the analysis was done most recently, are recognized as the "active" load cases. The LOAD LIST ALL command enables all the load cases in the structure to be made active for further processing.

PRINT MEMBER FORCES

The above PRINT command is an instruction to produce a report, in the output file, of the member end forces.

LOAD LIST 1 4 5

A LOAD LIST command is a means of instructing the program to use only the listed load cases for further processing.

PARAMETER CODE BRITISH BEAM 1 ALL UNL 1.8 ALL KY 0.5 ALL

Example Problem 4

41

The PARAMETER command is used to specify the steel design parameters (information on these parameters can be obtained from the manual where the implementation of the code is explained). The BEAM parameter is specified to perform design at every 1/12th point along the member length. UNL represents the unsupported length to be used for calculation of allowable bending stress. KY 0.5 ALL sets the effective length factor for column buckling about the local Y-axis to be 0.5 for ALL members.

CHECK CODE ALL

The above command instructs the program to perform a check to determine how the user defined member sizes along with the latest analysis results meet the code requirements.

FINISH



42


1. STAAD PLANE A PLANE FRAME STRUCTURE WITH2. * TENSION BRACING3. UNIT METER KNS4. SET NL 35. JOINT COORDINATES6. 1 0 0 0 3 12. 0. 0.7. 4 0 4.5 0 6 12. 4.5 0.8. 7 6. 9. 0. ; 8 12. 9. 0.9. MEMBER INCIDENCE10. 1 1 4 2 ; 3 5 7 ; 4 3 6 ; 5 6 8 ; 6 4 5 711. 8 7 8 ; 9 1 5 ; 10 2 4 ; 11 3 5 ; 12 2 612. 13 6 7 ; 14 5 813. MEMBER TRUSS14. 9 TO 1415. MEMBER PROP BRITISH16. 1 TO 5 TABLE ST UB305X165X4017. 6 7 8 TA ST UB457X152X5218. 9 TO 14 TA LD UA150X150X1019. UNIT MMS20. CONSTANTS21. E 210. ALL22. POISSON STEEL ALL23. SUPPORT24. 1 2 3 PINNED25. INACTIVE MEMBERS 9 TO 1426. UNIT METER27. LOADING 1 DEAD AND LIVE LOAD28. MEMBER LOAD29. 6 8 UNI GY -4.530. 7 UNI GY -6.7531. PERFORM ANALYSIS



32. CHANGES33. INACTIVE MEMBERS 10 11 1334. LOADING 2 WIND FROM LEFT35. JOINT LOAD36. 4 FX 135 ; 7 FX 6537. PERFORM ANALYSIS

38. CHANGE39. INACTIVE MEMBERS 9 12 1440. LOADING 3 WIND FROM RIGHT41. JOINT LOAD42. 6 FX -135 ; 8 FX -6543. LOAD COMBINATION 444. 1 0.75 2 0.7545. LOAD COMBINATION 546. 1 0.75 3 0.7547. PERFORM ANALYSIS

48. CHANGE49. LOAD LIST ALL

Example Problem 4

43 50. PRINT MEMBER FORCES



1 1 1 11.15 -0.90 0.00 0.00 0.00 0.004 -11.15 0.90 0.00 0.00 0.00 -4.06

2 1 -1.07 0.77 0.00 0.00 0.00 0.004 1.07 -0.77 0.00 0.00 0.00 3.47

3 1 68.13 -0.68 0.00 0.00 0.00 0.004 -68.13 0.68 0.00 0.00 0.00 -3.04

4 1 7.56 -0.10 0.00 0.00 0.00 0.004 -7.56 0.10 0.00 0.00 0.00 -0.44

5 1 59.46 -1.18 0.00 0.00 0.00 0.004 -59.46 1.18 0.00 0.00 0.00 -5.32

2 1 2 51.94 -0.07 0.00 0.00 0.00 0.005 -51.94 0.07 0.00 0.00 0.00 -0.32

2 2 46.14 0.53 0.00 0.00 0.00 0.005 -46.14 -0.53 0.00 0.00 0.00 2.40

3 2 129.19 -0.46 0.00 0.00 0.00 0.005 -129.19 0.46 0.00 0.00 0.00 -2.08

4 2 73.56 0.35 0.00 0.00 0.00 0.005 -73.56 -0.35 0.00 0.00 0.00 1.56

5 2 135.85 -0.40 0.00 0.00 0.00 0.005 -135.85 0.40 0.00 0.00 0.00 -1.80

3 1 5 13.68 -2.96 0.00 0.00 0.00 -5.297 -13.68 2.96 0.00 0.00 0.00 -8.01

2 5 -0.99 1.44 0.00 0.00 0.00 3.477 0.99 -1.44 0.00 0.00 0.00 3.02

3 5 47.37 -2.44 0.00 0.00 0.00 -6.167 -47.37 2.44 0.00 0.00 0.00 -4.81

4 5 9.52 -1.14 0.00 0.00 0.00 -1.377 -9.52 1.14 0.00 0.00 0.00 -3.75

5 5 45.79 -4.04 0.00 0.00 0.00 -8.597 -45.79 4.04 0.00 0.00 0.00 -9.62

4 1 3 31.40 0.97 0.00 0.00 0.00 0.006 -31.40 -0.97 0.00 0.00 0.00 4.38

2 3 103.80 0.21 0.00 0.00 0.00 0.006 -103.80 -0.21 0.00 0.00 0.00 0.93

3 3 -48.61 -0.59 0.00 0.00 0.00 0.006 48.61 0.59 0.00 0.00 0.00 -2.64

4 3 101.40 0.88 0.00 0.00 0.00 0.006 -101.40 -0.88 0.00 0.00 0.00 3.98

5 3 -12.91 0.29 0.00 0.00 0.00 0.006 12.91 -0.29 0.00 0.00 0.00 1.30

5 1 6 13.32 2.96 0.00 0.00 0.00 6.398 -13.32 -2.96 0.00 0.00 0.00 6.91

2 6 47.70 1.28 0.00 0.00 0.00 2.858 -47.70 -1.28 0.00 0.00 0.00 2.91

3 6 -1.29 -1.12 0.00 0.00 0.00 -2.148 1.29 1.12 0.00 0.00 0.00 -2.91

4 6 45.76 3.18 0.00 0.00 0.00 6.938 -45.76 -3.18 0.00 0.00 0.00 7.36

5 6 9.02 1.38 0.00 0.00 0.00 3.198 -9.02 -1.38 0.00 0.00 0.00 3.00

6 1 4 0.90 11.15 0.00 0.00 0.00 4.065 -0.90 15.85 0.00 0.00 0.00 -18.14

2 4 134.23 -1.07 0.00 0.00 0.00 -3.475 -134.23 1.07 0.00 0.00 0.00 -2.94

3 4 89.60 1.43 0.00 0.00 0.00 3.045 -89.60 -1.43 0.00 0.00 0.00 5.54

4 4 101.35 7.56 0.00 0.00 0.00 0.445 -101.35 12.69 0.00 0.00 0.00 -15.81

5 4 67.88 9.44 0.00 0.00 0.00 5.325 -67.88 10.81 0.00 0.00 0.00 -9.45




7 1 5 -1.98 22.41 0.00 0.00 0.00 23.756 1.98 18.09 0.00 0.00 0.00 -10.77

2 5 72.24 -1.12 0.00 0.00 0.00 -2.926 -72.24 1.12 0.00 0.00 0.00 -3.77

3 5 196.98 1.25 0.00 0.00 0.00 2.706 -196.98 -1.25 0.00 0.00 0.00 4.79

4 5 52.69 15.97 0.00 0.00 0.00 15.626 -52.69 14.40 0.00 0.00 0.00 -10.91

5 5 146.24 17.74 0.00 0.00 0.00 19.836 -146.24 12.63 0.00 0.00 0.00 -4.49

8 1 7 2.96 13.68 0.00 0.00 0.00 8.018 -2.96 13.32 0.00 0.00 0.00 -6.91

2 7 63.56 -0.99 0.00 0.00 0.00 -3.028 -63.56 0.99 0.00 0.00 0.00 -2.91

3 7 63.88 1.29 0.00 0.00 0.00 4.818 -63.88 -1.29 0.00 0.00 0.00 2.91

4 7 49.89 9.52 0.00 0.00 0.00 3.758 -49.89 10.73 0.00 0.00 0.00 -7.36

5 7 50.13 11.23 0.00 0.00 0.00 9.628 -50.13 9.02 0.00 0.00 0.00 -3.00

9 1 1 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

2 1 -156.47 0.00 0.00 0.00 0.00 0.005 156.47 0.00 0.00 0.00 0.00 0.00

3 1 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

4 1 -117.35 0.00 0.00 0.00 0.00 0.005 117.35 0.00 0.00 0.00 0.00 0.00

5 1 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

10 1 2 0.00 0.00 0.00 0.00 0.00 0.004 0.00 0.00 0.00 0.00 0.00 0.00

2 2 0.00 0.00 0.00 0.00 0.00 0.004 0.00 0.00 0.00 0.00 0.00 0.00

3 2 -111.16 0.00 0.00 0.00 0.00 0.004 111.16 0.00 0.00 0.00 0.00 0.00

4 2 0.00 0.00 0.00 0.00 0.00 0.004 0.00 0.00 0.00 0.00 0.00 0.00

5 2 -83.37 0.00 0.00 0.00 0.00 0.004 83.37 0.00 0.00 0.00 0.00 0.00

11 1 3 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

2 3 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

3 3 -136.68 0.00 0.00 0.00 0.00 0.005 136.68 0.00 0.00 0.00 0.00 0.00

4 3 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

5 3 -102.51 0.00 0.00 0.00 0.00 0.005 102.51 0.00 0.00 0.00 0.00 0.00

12 1 2 0.00 0.00 0.00 0.00 0.00 0.006 0.00 0.00 0.00 0.00 0.00 0.00

2 2 -91.64 0.00 0.00 0.00 0.00 0.006 91.64 0.00 0.00 0.00 0.00 0.00

3 2 0.00 0.00 0.00 0.00 0.00 0.006 0.00 0.00 0.00 0.00 0.00 0.00

4 2 -68.73 0.00 0.00 0.00 0.00 0.006 68.73 0.00 0.00 0.00 0.00 0.00

5 2 0.00 0.00 0.00 0.00 0.00 0.006 0.00 0.00 0.00 0.00 0.00 0.00

Example Problem 4



13 1 6 0.00 0.00 0.00 0.00 0.00 0.007 0.00 0.00 0.00 0.00 0.00 0.00

2 6 0.00 0.00 0.00 0.00 0.00 0.007 0.00 0.00 0.00 0.00 0.00 0.00

3 6 -76.80 0.00 0.00 0.00 0.00 0.007 76.80 0.00 0.00 0.00 0.00 0.00

4 6 0.00 0.00 0.00 0.00 0.00 0.007 0.00 0.00 0.00 0.00 0.00 0.00

5 6 -57.60 0.00 0.00 0.00 0.00 0.007 57.60 0.00 0.00 0.00 0.00 0.00

14 1 5 0.00 0.00 0.00 0.00 0.00 0.008 0.00 0.00 0.00 0.00 0.00 0.00

2 5 -77.85 0.00 0.00 0.00 0.00 0.008 77.85 0.00 0.00 0.00 0.00 0.00

3 5 0.00 0.00 0.00 0.00 0.00 0.008 0.00 0.00 0.00 0.00 0.00 0.00

4 5 -58.39 0.00 0.00 0.00 0.00 0.008 58.39 0.00 0.00 0.00 0.00 0.00

5 5 0.00 0.00 0.00 0.00 0.00 0.008 0.00 0.00 0.00 0.00 0.00 0.00


51. LOAD LIST 1 4 552. PARAMETER53. CODE BRITISH54. BEAM 1 ALL55. UNL 1.8 ALL56. KY 0.5 ALL57. CHECK CODE ALL

STAAD.Pro CODE CHECKING - (BSI )***********************




=======================================================================1 ST UB305X165X40 PASS BS-4.7 (C) 0.052 5

59.46 C 0.00 -5.32 4.502 ST UB305X165X40 PASS BS-4.7 (C) 0.118 5

135.85 C 0.00 -1.80 4.503 ST UB305X165X40 PASS BS-4.3.6 0.059 5

45.79 C 0.00 9.62 4.504 ST UB305X165X40 PASS BS-4.7 (C) 0.088 4

101.40 C 0.00 3.98 4.505 ST UB305X165X40 PASS BS-4.3.6 0.045 4

45.76 C 0.00 7.36 4.506 ST UB457X152X52 PASS BS-4.7 (C) 0.102 4

101.35 C 0.00 -15.81 6.007 ST UB457X152X52 PASS BS-4.7 (C) 0.147 5

146.24 C 0.00 19.83 0.008 ST UB457X152X52 PASS BS-4.7 (C) 0.050 5

50.13 C 0.00 9.62 0.009 LD UA150X150X10 PASS BS-4.6 (T) 0.072 4

117.35 T 0.00 0.00 0.0010 LD UA150X150X10 PASS BS-4.6 (T) 0.051 5

83.37 T 0.00 0.00 0.0011 LD UA150X150X10 PASS BS-4.6 (T) 0.063 5

102.51 T 0.00 0.00 0.0012 LD UA150X150X10 PASS BS-4.6 (T) 0.042 4

68.73 T 0.00 0.00 0.0013 LD UA150X150X10 PASS BS-4.6 (T) 0.036 5

57.60 T 0.00 0.00 0.0014 LD UA150X150X10 PASS BS-4.6 (T) 0.036 4

58.39 T 0.00 0.00 0.00



46 58. FINISH


**** DATE= TIME= ****


Example Problem 4

47

NOTES


48

NOTES

Example Problem 5

49


This example demonstrates the application of support displacement load (commonly known as sinking support) on a space frame structure.


50

Actual input is shown in bold lettering followed by explanation. STAAD SPACE TEST FOR SUPPORT DISPLACEMENT

Every input has to start with the word STAAD. The word SPACE signifies that the structure is a space frame structure (3-D) and the geometry is defined through X, Y and Z coordinates.

UNITS METER KNS

Specifies the unit to be used. JOINT COORDINATES 1 0.0 0.0 0.0 ; 2 0.0 3.0 0.0 3 6.0 3.0 0.0 ; 4 6.0 0.0 0.0 5 6.0 3.0 6.0 ; 6 6.0 0.0. 6.0

Joint number followed by X, Y and Z coordinates are provided above. Semicolon signs (;) are used as line separators. That enables us to provide multiple sets of data on one line.

MEMBER INCIDENCE 1 1 2 3 4 3 5 ; 5 5 6

Defines the members by the joints they are connected to. UNIT MMS MEMB PROP 1 TO 5 PRIS AX 6450 IZ 1.249E+08 IY 1.249E+08 IX 4.162E+06

Member properties have been defined above using the PRISMATIC attribute. Values of AX (area), IZ (moment of inertia about major axis), IY (moment of inertia about minor axis) and IX (torsional constant) are provided in MMS unit.

CONSTANT E 210. ALL POISSON STEEL ALL

Material constants like E (modulus of elasticity) and Poisson’s ratio are specified following the command CONSTANT.

Example Problem 5

51 SUPPORT 1 4 6 FIXED

Joints 1, 4 and 6 are fixed supports. LOADING 1 SINKING SUPPORT

Load case 1 is initiated followed by a title. SUPPORT DISPLACEMENT LOAD 4 FY –15

Load 1 is a support displacement load which is also commonly known as a sinking support. FY signifies that the support settlement is in the global Y direction and the value of this settlement is 15mm downward.

PERFORM ANALYSIS

This command instructs the program to proceed with the analysis. PRINT ANALYSIS RESULTS

The above PRINT command instructs the program to print joint displacements, support reactions and member forces.

FINISH



52


1. STAAD SPACE TEST FOR SUPPORT DISPLACEMENT2. UNITS METER KNS3. JOINT COORDINATES4. 1 0.0 0.0 0.0 ; 2 0.0 3.0 0.05. 3 6.0 3.0 0.0 ; 4 6.0 0.0 0.06. 5 6.0 3.0 6.0 ; 6 6.0 0.0 6.07. MEMBER INCIDENCE8. 1 1 2 39. 4 3 5 ; 5 5 6

10. UNIT MMS11. MEMB PROP12. 1 TO 5 PRIS AX 6450 IZ 1.249E+08 IY 1.249E+08 IX 4.162E+0613. CONSTANT14. E 210. ALL15. POISS STEEL ALL16. SUPPORT17. 1 4 6 FIXED18. LOADING 1 SINKING SUPPORT19. SUPPORT DISPLACEMENT LOAD20. 4 FY -15.21. PERFORM ANALYSIS



22. PRINT ANALYSIS RESULTS

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


1 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.00002 1 0.2737 -0.0012 -0.0323 -0.0002 0.0006 -0.00193 1 0.2735 -1.4975 -0.2735 -0.0018 0.0000 -0.00184 1 0.0000 -1.5000 0.0000 0.0000 0.0000 0.00005 1 0.0323 -0.0012 -0.2737 -0.0019 -0.0006 -0.00026 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

SUPPORT REACTIONS -UNIT KNS MMS STRUCTURE TYPE = SPACE-----------------


1 1 0.46 5.63 0.90 2780.50 -67.13 15492.894 1 0.43 -11.26 -0.43 15509.16 0.00 15509.166 1 -0.90 5.63 -0.46 15492.89 67.13 2780.50

Example Problem 5

53 MEMBER END FORCES STRUCTURE TYPE = SPACE-----------------ALL UNITS ARE -- KNS MMS


1 1 1 5.63 -0.46 0.90 -67.13 -2780.50 15492.892 -5.63 0.46 -0.90 67.13 94.34 -16878.26

2 1 2 0.46 5.63 0.90 94.34 -67.13 16878.263 -0.46 -5.63 -0.90 -94.34 -5305.18 16904.28

3 1 3 -11.26 -0.43 0.43 0.00 -16809.94 -16809.944 11.26 0.43 -0.43 0.00 15509.16 15509.16

4 1 3 0.46 -5.63 -0.90 -94.34 5305.18 -16904.285 -0.46 5.63 0.90 94.34 67.13 -16878.26

5 1 5 5.63 0.90 0.46 67.13 -16878.26 -94.346 -5.63 -0.90 -0.46 -67.13 15492.89 2780.50


23. FINISH


**** DATE= TIME= ****



54

NOTES

Example Problem 6

55


This is an example of prestress loading in a plane frame structure. It covers two situations: 1) From the member on which it is applied, the prestressing effect is transmitted to the rest of the structure through the connecting members (known in the program as PRESTRESS load). 2) The prestressing effect is experienced by the member(s) alone and not transmitted to the rest of the structure (known in the program as POSTSTRESS load).


56

Actual input is shown in bold lettering followed by explanation. STAAD PLANE FRAME WITH PRESTRESSING LOAD


UNIT METER KNS

Specifies the unit to be used. JOINT COORD 1 0. 0. ; 2 12. 0. ; 3 0. 6. ; 4 12. 6. 5 0. 10.5 ; 6 12. 10.5 ; 7 0. 15. ; 8 12. 15.

Joint number followed by X and Y coordinates are provided above. Since this is a plane structure, the Z coordinates need not be provided. Semicolon signs (;) are used as line separators, and that allows us to provide multiple sets of data on one line.

MEMBER INCIDENCE 1 1 3 ; 2 3 5 ; 3 5 7 ; 4 2 4 ; 5 4 6 6 6 8 ; 7 3 4 ; 8 5 6 ; 9 7 8

Defines the members by the joints they are connected to. SUPPORT 1 2 FIXED

The supports at joints 1 and 2 are defined to be fixed supports. MEMB PROP 1 TO 9 PRI AX 0.2044 IZ 8.631E-03

Member properties are provided using the PRI (prismatic) attribute. Values of area (AX) and moment of inertia about the major axis (IZ) are provided.

UNIT MMS CONSTANT E 21. ALL ; POISS CONC ALL

Example Problem 6

57

CONSTANT command initiates input for material constants like E (modulus of elasticity), Poisson’s ratio, etc. Length unit is changed from METER to MMS to facilitate the input.

LOADING 1 PRESTRESSING LOAD MEMBER PRESTRESS 7 8 FORCE 1350. ES 75. EM -300. EE 75.

Load case 1 is initiated followed by a title. Load 1 contains PRESTRESS load. Members 7 and 8 have a cable force of 1350 KNs. The location of the cable at the start (ES) and end (EE) is 75 MMs above the centre of gravity while at the middle (EM) it is 300 MMs below the c.g. The assumptions and facts associated with this type of loading are explained in section 1 of the Technical Reference Manual.

LOADING 2 POSTSTRESSING LOAD MEMBER POSTSTRESS 7 8 FORCE 1350. ES 75. EM -300. EE 75.

Load case 2 is initiated followed by a title. Load 2 is a POSTSTRESS load. Members 7 and 8 have a cable force of 1350 KNs. The location of the cable is the same as in load case 1. For a difference between PRESTRESS loading and POSTSTRESS loading, as well as additional information about both types of loads, please refer to section 1 of the Technical Reference Manual.

PERFORM ANALYSIS

This command instructs the program to perform the analysis. UNIT METER PRINT ANALYSIS RESULTS

The above command is an instruction to write joint displacements, support reactions and member forces in the output file.

FINISH




1. STAAD PLANE FRAME WITH PRESTRESSING LOAD2. UNIT METER KNS3. JOINT COORD4. 1 0. 0. ; 2 12. 0. ; 3 0. 6. ; 4 12. 6.5. 5 0. 10.5 ; 6 12. 10.5 ; 7 0. 15. ; 8 12. 15.6. MEMBER INCIDENCE7. 1 1 3 ; 2 3 5 ; 3 5 7 ; 4 2 4 ; 5 4 68. 6 6 8 ; 7 3 4 ; 8 5 6 ; 9 7 89. SUPPORT

10. 1 2 FIXED11. MEMB PROP12. 1 TO 9 PRI AX 0.2044 IZ 8.631E-0313. UNIT MMS14. CONSTANT15. E 21. ALL16. POISS CONC ALL17. LOADING 1 PRESTRESSING LOAD18. MEMBER PRESTRESS19. 7 8 FORCE 1350. ES 75. EM -300. EE 75.20. LOADING 2 POSTSTRESSING LOAD21. MEMBER POSTSTRESS22. 7 8 FORCE 1350. ES 75. EM -300. EE 75.23. PERFORM ANALYSIS



24. UNIT METER




1 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.00002 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

2 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.00002 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

3 1 0.1917 0.0000 0.0000 0.0000 0.0000 0.00042 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

4 1 -0.1917 0.0000 0.0000 0.0000 0.0000 -0.00042 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

5 1 0.1799 0.0000 0.0000 0.0000 0.0000 0.00082 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

6 1 -0.1799 0.0000 0.0000 0.0000 0.0000 -0.00082 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

7 1 -0.0014 0.0000 0.0000 0.0000 0.0000 0.00022 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

8 1 0.0014 0.0000 0.0000 0.0000 0.0000 -0.00022 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

Example Problem 6

59 SUPPORT REACTIONS -UNIT KNS METE STRUCTURE TYPE = PLANE-----------------


1 1 -30.59 0.00 0.00 0.00 0.00 80.492 0.00 0.00 0.00 0.00 0.00 0.00

2 1 30.59 0.00 0.00 0.00 0.00 -80.492 0.00 0.00 0.00 0.00 0.00 0.00



1 1 1 0.00 30.59 0.00 0.00 0.00 80.493 0.00 -30.59 0.00 0.00 0.00 103.07

2 1 0.00 0.00 0.00 0.00 0.00 0.003 0.00 0.00 0.00 0.00 0.00 0.00

2 1 3 0.00 62.59 0.00 0.00 0.00 121.895 0.00 -62.59 0.00 0.00 0.00 159.77

2 3 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

3 1 5 0.00 10.30 0.00 0.00 0.00 50.987 0.00 -10.30 0.00 0.00 0.00 -4.64

2 5 0.00 0.00 0.00 0.00 0.00 0.007 0.00 0.00 0.00 0.00 0.00 0.00

4 1 2 0.00 -30.59 0.00 0.00 0.00 -80.494 0.00 30.59 0.00 0.00 0.00 -103.07

2 2 0.00 0.00 0.00 0.00 0.00 0.004 0.00 0.00 0.00 0.00 0.00 0.00

5 1 4 0.00 -62.59 0.00 0.00 0.00 -121.896 0.00 62.59 0.00 0.00 0.00 -159.77

2 4 0.00 0.00 0.00 0.00 0.00 0.006 0.00 0.00 0.00 0.00 0.00 0.00

6 1 6 0.00 -10.30 0.00 0.00 0.00 -50.988 0.00 10.30 0.00 0.00 0.00 4.64

2 6 0.00 0.00 0.00 0.00 0.00 0.008 0.00 0.00 0.00 0.00 0.00 0.00

7 1 3 1371.41 -168.75 0.00 0.00 0.00 -326.214 -1371.41 -168.75 0.00 0.00 0.00 326.21

2 3 1339.41 -168.75 0.00 0.00 0.00 -101.254 -1339.41 -168.75 0.00 0.00 0.00 101.25

8 1 5 1287.12 -168.75 0.00 0.00 0.00 -312.006 -1287.12 -168.75 0.00 0.00 0.00 312.00

2 5 1339.41 -168.75 0.00 0.00 0.00 -101.256 -1339.41 -168.75 0.00 0.00 0.00 101.25

9 1 7 -10.30 0.00 0.00 0.00 0.00 4.648 10.30 0.00 0.00 0.00 0.00 -4.64

2 7 0.00 0.00 0.00 0.00 0.00 0.008 0.00 0.00 0.00 0.00 0.00 0.00



60 26. FINISH


**** DATE= TIME= ****


Example Problem 7

61


This example illustrates modelling of structures with OFFSET connections. OFFSET connections arise when the centre lines of the connected members do not intersect at the connection point. The connection eccentricity behaves as a rigid link and is modeled through specification of MEMBER OFFSETS.


62

Actual input is shown in bold lettering followed by explanation. STAAD PLANE TEST FOR MEMBER OFFSETS


UNIT METER KNS

Specifies the unit to be used. JOINT COORD 1 0. 0. ; 2 6. 0. ; 3 0. 4.5 4 6. 4.5 ; 5 0. 9. ; 6 6. 9.

Joint number followed by X and Y coordinates are provided above. Since this is a plane structure, the Z coordinates need not be provided. Semicolon signs (;) are used as line separators. This allows us to provide multiple sets of data in one line.

SUPPORT 1 2 PINNED

Pinned supports are specified at joints 1 and 2. The word PINNED signifies that no moments will be carried by these supports.

MEMB INCI 1 1 3 2 ; 3 3 5 4 5 3 4 ; 6 5 6 ; 7 1 4

Defines the members by the joints they are connected to. MEMB PROP BRITISH 1 TO 4 TABLE ST UC356X368X129 5 6 TA ST UB305X165X40 7 TA LD UA200X150X12

All member properties are from British steel table. The word ST stands for standard single section. LD stands for long leg back-to-back double angle.

Example Problem 7

63 UNIT MMS MEMB OFFSET 5 6 START 178. 0.0 0.0 5 6 END -178. 0.0 0.0 7 END -178.0 –152.0 0.0

The above specification states that an OFFSET is located at the START/END joint of the members. The X, Y and Z global coordinates of the offset distance from the corresponding incident joint are also provided. These attributes are applied to members 5, 6 and 7.

CONSTANT E 210. ALL POISSON STEEL ALL

Material constants like E (modulus of elasticity) and Poisson’s ratio are provided following the keyword CONSTANT.

LOADING 1 WIND LOAD

Load case 1 is initiated followed by a title. JOINT LOAD 3 FX 225.0 ; 5 FX 112.5

Load 1 contains joint loads at nodes 3 and 5. FX indicates that the load is a force in the global X direction.

PERFORM ANALYSIS

The above command is an instruction to perform the analysis. UNIT METER PRINT FORCES PRINT REACTIONS

The above PRINT commands are instructions for writing the member forces and support reactions to the output file.


64 FINISH

This command terminates a STAAD run.

Example Problem 7

65


1. STAAD PLANE TEST FOR MEMBER OFFSETS2. UNIT METER KNS3. JOINT COORD4. 1 0. 0. ; 2 6. 0. ; 3 0. 4.55. 4 6. 4.5 ; 5 0. 9. ; 6 6. 9.6. SUPPORT7. 1 2 PINNED8. MEMB INCI9. 1 1 3 2 ; 3 3 5 4

10. 5 3 4 ; 6 5 6 ; 7 1 411. MEMB PROP BRITISH12. 1 TO 4 TABLE ST UC356X368X12913. 5 6 TA ST UB305X165X4014. 7 TA LD UA200X150X1215. UNIT MMS16. MEMB OFFSET17. 5 6 START 178. 0.0 0.018. 5 6 END -178. 0.0 0.019. 7 END -178.0 -152.0 0.020. CONSTANT21. E 210. ALL22. POISS STEEL ALL23. LOADING 1 WIND LOAD24. JOINT LOAD25. 3 FX 225.0 ; 5 FX 112.526. PERFORM ANALYSIS



27. UNIT METER28. PRINT FORCES



1 1 1 -49.12 -20.25 0.00 0.00 0.00 4.593 49.12 20.25 0.00 0.00 0.00 -95.70

2 1 2 337.50 -25.13 0.00 0.00 0.00 0.004 -337.50 25.13 0.00 0.00 0.00 -113.07

3 1 3 -30.76 53.78 0.00 0.00 0.00 151.005 30.76 -53.78 0.00 0.00 0.00 91.03

4 1 4 30.76 58.72 0.00 0.00 0.00 170.716 -30.76 -58.72 0.00 0.00 0.00 93.52

5 1 3 299.03 -18.37 0.00 0.00 0.00 -52.034 -299.03 18.37 0.00 0.00 0.00 -51.62




6 1 5 58.72 -30.76 0.00 0.00 0.00 -85.556 -58.72 30.76 0.00 0.00 0.00 -88.04

7 1 1 -479.32 -1.95 0.00 0.00 0.00 -4.594 479.32 1.95 0.00 0.00 0.00 -9.61


29. PRINT REACTIONS



1 1 -362.63 -337.50 0.00 0.00 0.00 0.002 1 25.13 337.50 0.00 0.00 0.00 0.00


30. FINISH


**** DATE= TIME= ****


Example Problem 8

67


In this example, concrete design is performed on some members of a space frame structure. Design calculations consist of computation of reinforcement for beams and columns. Secondary moments on the columns are obtained through the means of a P-Delta analysis.

The above example represents a space frame, and the members are made of concrete. The input in the next page will show the dimensions of the members. Two load cases, namely one for dead plus live load and another with dead, live and wind load, are considered in the design.


68

Actual input is shown in bold lettering followed by explanation. STAAD SPACE FRAME WITH CONCRETE DESIGN


UNIT METER KNS

Specifies the unit to be used. JOINT COORDINATE 1 0 0 0 ; 2 5.4 0 0 ; 3 11.4 0. 0 4 0 0 7.2 ; 5 5.4 0 7.2 ; 6 11.4 0 7.2 7 0 3.6 0 ; 8 5.4 3.6 0 ; 9 11.4 3.6 0 10 0 3.6 7.2 ; 11 5.4 3.6 7.2 ; 12 11.4 3.6 7.2 13 5.4 7.2 0 ; 14 11.4 7.2 0 ; 15 5.4 7.2 7.2 16 11.4 7.2 7.2

Joint number followed by X, Y and Z coordinates are provided above. Semicolon signs (;) are used as line separators to facilitate input of multiple sets of data on one line.

MEMBER INCIDENCE 1 1 7 ; 2 4 10 ; 3 2 8 ; 4 8 13 5 5 11 ; 6 11 15 ; 7 3 9 ; 8 9 14 9 6 12 ; 10 12 16 ; 11 7 8 12 13 10 11 14 ; 15 13 14 ; 16 15 16 17 7 10 ; 18 8 11 ; 19 9 12 20 13 15 ; 21 14 16

Defines the members by the joints they are connected to.

UNIT MMS MEMB PROP 1 2 PRISMATIC YD 300.0 IZ 2.119E08 IY 2.119E08 - IX 4.237E08 3 TO 10 PR YD 300.0 ZD 300.0 IZ 3.596E08 IY 3.596E08 – IX 5.324E08 11 TO 21 PR YD 535.0 ZD 380 IZ 2.409E09 IY 1.229E09 – IX 2.704E09

Example Problem 8

69

All member properties are provided using the PRISMATIC option. YD and ZD stand for depth and width. If ZD is not provided, a circular shape with diameter = YD is assumed for that cross section. All properties required for the analysis, such as, Area, Moments of Inertia, etc. are calculated automatically from these dimensions unless these are explicitly defined. For this particular example, moments of inertia (IZ, IY) and torsional constant (IX) are provided, so these will not be re-calculated. The IX, IY, and IZ values provided in this example are only half the values of a full section to account for the fact that the full moments of inertia will not be effective due to cracking of concrete.

CONSTANT E 21.0 ALL POISSON CONC ALL UNIT METER CONSTANT DEN 23.56 ALL

The CONSTANT command initiates input for material constants like E (modulus of elasticity), Poisson’s ratio, Density, etc. Length unit is changed from MM to METER to facilitate input for DENsity. The built-in value for Poisson’s ratio for concrete will be used in the analysis.

SUPPORT 1 TO 6 FIXED

Joints 1 to 6 are fixed supports. LOAD 1 (1.4DL + 1.7LL)

Load case 1 is initiated followed by a title. SELF Y -1.4

The selfweight of the structure is applied in the global Y direction with a -1.4 factor. Since global Y is vertically upward, the negative factor indicates that this load will act downwards.


70 MEMB LOAD 11 TO 16 UNI Y -42.0 11 TO 16 UNI Y -76.5

Load 1 contains member loads also. Y indicates that the load is in the local Y direction. The word UNI stands for uniformly distributed load.

LOAD 2 .75 (1.4DL + 1.7LL + 1.7WL)

Load case 2 is initiated followed by a title.

REPEAT LOAD 1 0.75

The above command will gather the load data values from load case 1, multiply them with a factor of 0.75 and utilize the resulting values in load 2.

JOINT LOAD 15 16 FZ 40.0 11 FZ 90.0 12 FZ 70.0 10 FZ 40.0

Load 2 contains some additional joint loads also. FZ indicates that the load is a force in the global Z direction.

PDELTA ANALYSIS

This command instructs the program to proceed with the analysis. The analysis type is P-DELTA indicating that second-order effects are to be calculated.

PRINT FORCES LIST 2 5 9 14 16

Member end forces are printed using the above PRINT commands. The LIST option restricts the print output to the members listed.

Example Problem 8

71 START CONCRETE DESIGN

The above command initiates a concrete design. CODE BRITISH TRACK 1.0 MEMB 14 TRACK 2.0 MEMB 16 MAXMAIN 40 ALL

The values for the concrete design parameters are defined in the above commands. Design is performed per the BS 8110 Code. The TRACK value dictates the extent of design related information provided in the output. MAXMAIN indicates that the maximum size of main reinforcement is the 40 mm bar. These parameters are described in the manual where British concrete design related information is available.

DESIGN BEAM 14 16

The above command instructs the program to design beams 14 and 16 for flexure, shear and torsion.

DESIGN COLUMN 2 5

The above command instructs the program to design columns 2 and 5 for axial load and biaxial bending.

END CONCRETE DESIGN

This will end the concrete design. FINISH



72


1. STAAD SPACE FRAME WITH CONCRETE DESIGN2. UNIT METER KNS3. JOINT COORDINATE4. 1 0.0 0.0 0.0 ; 2 5.4 0.0 0.05. 3 11.4 0.0 0.0 ; 4 0.0 0.0 7.26. 5 5.4 0.0 7.2 ; 6 11.4 0.0 7.27. 7 0.0 3.6 0.0 ; 8 5.4 3.6 0.08. 9 11.4 3.6 0.0 ; 10 0.0 3.6 7.29. 11 5.4 3.6 7.2 ; 12 11.4 3.6 7.2

10. 13 5.4 7.2 0.0 ; 14 11.4 7.2 0.011. 15 5.4 7.2 7.2 ; 16 11.4 7.2 7.212. MEMBER INCIDENCE13. 1 1 7 ; 2 4 10 ; 3 2 8 ; 4 8 1314. 5 5 11 ; 6 11 15 ; 7 3 9 ; 8 9 1415. 9 6 12 ; 10 12 16 ; 11 7 8 1216. 13 10 11 14 ; 15 13 14 ; 16 15 1617. 17 7 10 ; 18 8 11 ; 19 9 1218. 20 13 15 ; 21 14 1619. UNIT MMS20. MEMB PROP21. 1 2 PRISMATIC YD 300.0 IZ 2.119E08 IY 2.119E08 -22. IX 4.237E0823. 3 TO 10 PR YD 300.0 ZD 300.0 IZ 3.596E08 IY 3.596E08 -24. IX 5.324E0825. 11 TO 21 PR YD 535.0 ZD 380 IZ 2.409E09 IY 1.229E09 -26. IX 2.704E0927. CONSTANT28. E 21.0 ALL29. POISSON CONC ALL30. UNIT METER31. CONSTANT32. DEN 23.56 ALL33. SUPPORT34. 1 TO 6 FIXED35. LOAD 1 (1.4DL + 1.7LL)36. SELF Y -1.437. MEMB LOAD38. 11 TO 16 UNI Y -42.039. 11 TO 16 UNI Y -76.540. LOAD 2 .75(1.4DL + 1.7LL + 1.7WL)41. REPEAT LOAD42. 1 0.7543. JOINT LOAD44. 15 16 FZ 40.045. 11 FZ 90.046. 12 FZ 70.047. 10 FZ 40.048. PDELTA ANALYSIS



Example Problem 8

73 49. PRINT FORCES LIST 2 5 9 14 16

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


2 1 4 302.06 -18.38 -2.86 0.00 3.42 -23.5110 -293.67 18.38 2.86 0.00 6.88 -42.65

2 4 243.86 -15.23 -31.79 1.32 57.70 -20.2810 -237.56 15.23 31.79 -1.32 56.75 -34.54

5 1 5 1289.48 -2.85 -2.99 0.00 3.57 -6.0411 -1278.80 2.85 2.99 0.00 7.18 -4.23

2 5 1015.60 -4.64 -65.16 1.13 122.65 -9.0711 -1007.58 4.64 65.16 -1.13 111.93 -7.63

9 1 6 758.87 19.98 -2.77 0.00 3.31 21.2812 -748.19 -19.98 2.77 0.00 6.66 50.63

2 6 620.14 12.66 -65.97 0.16 124.64 11.6312 -612.13 -12.66 65.97 -0.16 112.84 33.93

14 1 11 -40.63 434.78 0.00 -0.26 -0.01 497.7712 40.63 316.45 0.00 0.26 0.00 -142.77

2 11 -34.51 326.81 2.92 -1.03 -11.89 375.1012 34.51 236.61 -2.92 1.03 -5.62 -104.51

16 1 15 61.00 378.46 0.00 0.03 0.00 141.6216 -61.00 372.77 0.00 -0.03 0.00 -124.55

2 15 45.73 283.67 0.36 -0.13 -1.21 105.7216 -45.73 279.76 -0.36 0.13 -0.95 -93.99


50. START CONCRETE DESIGN51. CODE BRITISH

PROGRAM CODE REVISION V1.5_8110_97/152. TRACK 1.0 MEMB 1453. TRACK 2.0 MEMB 1654. MAXMAIN 40 ALL55. DESIGN BEAM 14 16

====================================================================

B E A M N O. 14 D E S I G N R E S U L T S - FLEXURE

LEN - 6000. mm FY - 460. FC - 30. SIZE - 380. X 535. mm

LEVEL HEIGHT BAR INFO FROM TO ANCHORmm mm mm STA END

-------------------------------------------------------------------

1 35. 5- 20 MM 740. 6000. NO YES|----------------------------------------------------------------|| CRITICAL POS MOMENT= 257.07 KN-M AT 3500. mm, LOAD 1|| REQD STEEL= 1353. mm2, ROW=0.0067, ROWMX=0.0400,ROWMN=0.0013 || MAX/MIN/ACTUAL BAR SPACING= 190./ 45./ 58. mm ||----------------------------------------------------------------|

2 487. 4- 32 MM 0. 1760. YES NOCOMP. 3- 12 MM (REQD. STEEL= 338. SQ. MM)

|----------------------------------------------------------------|| CRITICAL NEG MOMENT= 497.75 KN-M AT 0. mm, LOAD 1|| REQD STEEL= 3155. mm2, ROW=0.0155, ROWMX=0.0400,ROWMN=0.0013 || MAX/MIN/ACTUAL BAR SPACING= 164./ 37./ 147. mm ||----------------------------------------------------------------|

3 497. 7- 12 MM 4740. 6000. NO YES|----------------------------------------------------------------|| CRITICAL NEG MOMENT= 142.76 KN-M AT 6000. mm, LOAD 1|| REQD STEEL= 707. mm2, ROW=0.0035, ROWMX=0.0400,ROWMN=0.0013 || MAX/MIN/ACTUAL BAR SPACING= 183./ 37./ 41. mm ||----------------------------------------------------------------|


74 B E A M N O. 14 D E S I G N R E S U L T S - SHEAR

PROVIDE SHEAR LINKS AS FOLLOWS|----------------------------------------------------------------|| FROM - TO | MAX. SHEAR | LOAD | LINKS | NO. | SPACING C/C ||----------------|------------|------|-------|-----|-------------|| END 1 2000 mm | 434.8 kN | 1 | 8 mm | 26 | 79 mm || 2000 3500 mm | 184.4 kN | 1 | 8 mm | 6 | 249 mm || 3500 5000 mm | 191.2 kN | 1 | 8 mm | 6 | 249 mm || 5000 END 2 | 316.4 kN | 1 | 8 mm | 9 | 124 mm ||----------------------------------------------------------------|

___ 11J____________________ 6000.X 380.X 535_____________________ 12J____| ===============||||===================== 7No12 H 497.4740.TO|6000 || 4No32 H 487.| |0.TO 1760 | | | | | | | | || 26*8 c/c 79 | | | | | | 9*8 c/c124| | || | | | | | | | | | | | | | | | || 5No20 H |35.|740.TO 6000 | | | | | | | | || ==================================================================|||___________________________________________________________________________|___________ ___________ ___________ ___________ ___________

| | | | | | | | | ooooooo || OOOO | | OOOO | | | | | | 7T12 || 4T32 | | 4T32 | | | | | | || | | | | | | | | || | | | | | | | | || | | 5T20 | | 5T20 | | 5T20 | | 5T20 || | | ooooo | | ooooo | | ooooo | | ooooo ||___________| |___________| |___________| |___________| |___________|

====================================================================

B E A M N O. 16 D E S I G N R E S U L T S - FLEXURE

LEN - 6000. mm FY - 460. FC - 30. SIZE - 380. X 535. mm

LEVEL HEIGHT BAR INFO FROM TO ANCHORmm mm mm STA END

-------------------------------------------------------------------

1 41. 4- 32 MM 0. 6000. YES YES2 497. 7- 12 MM 0. 760. YES NO3 495. 4- 16 MM 5240. 6000. NO YES

REQUIRED REINF. STEEL SUMMARY :-------------------------------SECTION REINF STEEL(+VE/-VE) MOMENTS(+VE/-VE) LOAD(+VE/-VE)( MM ) (SQ. MM ) (KN-METER)

0. 0.0/ 700.6 0.00/ 141.61 0/ 1500. 264.3/ 0.0 31.96/ 0.00 1/ 0

1000. 876.1/ 0.0 174.23/ 0.00 1/ 01500. 1527.2/ 0.0 285.21/ 0.00 1/ 02000. 2064.2/ 0.0 364.88/ 0.00 1/ 02500. 2432.0/ 0.0 413.25/ 0.00 1/ 03000. 2571.9/ 0.0 430.32/ 0.00 1/ 03500. 2454.9/ 0.0 416.09/ 0.00 1/ 04000. 2105.6/ 0.0 370.57/ 0.00 1/ 04500. 1581.4/ 0.0 293.74/ 0.00 1/ 05000. 938.8/ 0.0 185.61/ 0.00 1/ 05500. 264.3/ 0.0 46.18/ 0.00 1/ 06000. 0.0/ 611.1 0.00/ 124.54 0/ 1

Example Problem 8

75 B E A M N O. 16 D E S I G N R E S U L T S - SHEAR

PROVIDE SHEAR LINKS AS FOLLOWS|----------------------------------------------------------------|| FROM - TO | MAX. SHEAR | LOAD | LINKS | NO. | SPACING C/C ||----------------|------------|------|-------|-----|-------------|| END 1 2000 mm | 378.5 kN | 1 | 8 mm | 26 | 79 mm || 2000 3000 mm | 128.1 kN | 1 | 8 mm | 4 | 249 mm || 3000 4000 mm | 122.4 kN | 1 | 8 mm | 4 | 249 mm || 4000 END 2 | 372.8 kN | 1 | 8 mm | 26 | 79 mm ||----------------------------------------------------------------|

___ 15J____________________ 6000.X 380.X 535_____________________ 16J____||======== =========||| 7No12 H 497. 0.TO 760 4No16 H 495.5240.TO|6000 || | | | | | | | | | || 26*8 c/c 79 26*8 c/c 79| | || | | | | | | | | | || 4No32 H 41. 0.TO 6000 | | | | | |||=========================================================================|||___________________________________________________________________________|___________ ___________ ___________ ___________ ___________

| ooooooo | | | | | | | | oooo || 7T12 | | | | | | | | 4T16 || | | | | | | | | || | | | | | | | | || | | | | | | | | || 4T32 | | 4T32 | | 4T32 | | 4T32 | | 4T32 || OOOO | | OOOO | | OOOO | | OOOO | | OOOO ||___________| |___________| |___________| |___________| |___________|

********************END OF BEAM DESIGN**************************

56. DESIGN COLUMN 2 5

====================================================================

C O L U M N N O. 2 D E S I G N R E S U L T S

FY - 460. FC - 30. N/MM2 CIRC SIZE 300. MM DIAMETER

AREA OF STEEL REQUIRED = 1586. SQ. MM.

BAR CONFIGURATION REINF PCT. LOAD LOCATION----------------------------------------------------

8 16 MM 2.244 2 EACH END(ARRANGE COLUMN REINFORCEMENTS SYMMETRICALLY)

====================================================================

C O L U M N N O. 5 D E S I G N R E S U L T S

FY - 460. FC -30. N/MM2 SQRE SIZE - 300. X 300. MM,

AREA OF STEEL REQUIRED = 2499. SQ. MM.

BAR CONFIGURATION REINF PCT. LOAD LOCATION----------------------------------------------------

4 32 MM 2.776 2 EACH END

********************END OF COLUMN DESIGN RESULTS********************

57. END CONCRETE DESIGN


76 58. FINISH


**** DATE= TIME= ****


Example Problem 9

77


The space frame structure in this example consists of frame members and finite elements (plates). The finite element part is used to model floor slabs and a shear wall. Concrete design of an element is performed.


78

Actual input is shown in bold lettering followed by explanation.

STAAD SPACE * EXAMPLE PROBLEM WITH FRAME MEMBERS AND FINITE ELEMENTS

Every STAAD input file has to begin with the word STAAD. The word SPACE signifies that the structure is a space frame and the geometry is defined through X, Y and Z axes. The second line forms the title to identify this project.

UNIT METER NEWTON

The units for the data that follows are specified above.

JOINT COORD 1 0 0 0 ; 2 0 0 6 REP ALL 2 6 0 0 7 0 4.5 0 11 0 4.5 6 12 1.5 4.5 0 14 4.5 4.5 0 15 1.5 4.5 6 17 4.5 4.5 6 18 6 4.5 0 22 6 4.5 6 23 7.5 4.5 0 25 10.5 4.5 0 26 7.5 4.5 6 28 10.5 4.5 6 29 12 4.5 0 33 12 4.5 6 34 6 1.125 0 36 6 3.375 0 37 6 1.125 6 39 6 3.375 6

The joint numbers and their coordinates are defined through the above set of commands. The automatic generation facility has been used several times in the above lines. Users may refer to section 5 of the Technical Reference Manual where the joint coordinate generation facilities are described.

MEMBER INCI *COLUMNS 1 1 7 ; 2 2 11 3 3 34 ; 4 34 35 ; 5 35 36 ; 6 36 18 7 4 37 ; 8 37 38 ; 9 38 39 ; 10 39 22 11 5 29 ; 12 6 33 *BEAMS IN Z DIRECTION AT X=0

Example Problem 9

79

13 7 8 16 *BEAMS IN Z DIRECTION AT X=6.0 17 18 19 20 *BEAMS IN Z DIRECTION AT X=12.0 21 29 30 24 *BEAMS IN X DIRECTION AT Z = 0 25 7 12 ; 26 12 13 ; 27 13 14 ; 28 14 18 29 18 23 ; 30 23 24 ; 31 24 25 ; 32 25 29 *BEAMS IN X DIRECTION AT Z = 12.0 33 11 15 ; 34 15 16 ; 35 16 17 ; 36 17 22 37 22 26 ; 38 26 27 ; 39 27 28 ; 40 28 33

The member incidences are defined through the above set of commands. For some members, the member number followed by the start and end joint numbers are defined. In other cases, STAAD's automatic generation facilities are utilized. Section 5 of the Technical Reference Manual describes these facilities in detail.

DEFINE MESH A JOINT 7 B JOINT 11 C JOINT 22 D JOINT 18 E JOINT 33 F JOINT 29 G JOINT 3 H JOINT 4

The above lines define the nodes of super-elements. Super-elements are plate/shell surfaces from which a number of individual plate/shell elements can be generated. In this case, the points describe the outer corners of a slab and that of a shear wall. Our goal is to define the slab and the wall as several plate/shell elements.

GENERATE ELEMENT MESH ABCD 4 4 MESH DCEF 4 4 MESH DCHG 4 4


80

The above lines form the instructions to generate individual 4-noded elements from the super-element profiles. For example, the command MESH ABCD 4 4 means that STAAD has to generate 16 elements from the surface formed by the points A, B, C and D with 4 elements along the edges AB & CD and 4 elements along the edges BC & DA.

UNIT MMS MEMB PROP 1 TO 40 PRIS YD 300 ZD 300

Members 1 to 40 are defined as a rectangular prismatic section with 300 mm depth and 300 mm width.

ELEM PROP 41 TO 88 TH 150

Elements 41 to 88 are defined to be 150 mm thick.

CONSTANT E 21000 ALL POISSON CONCRETE ALL

The modulus of elasticity and Poisson’s ratio are defined above for all the members and elements following the keyword CONSTANT.


Joints 1 to 6 are defined as fixed supported.

UNIT KNS METER LOAD 1 DEAD LOAD FROM FLOOR ELEMENT LOAD 41 TO 72 PRESSURE -10.0

Load 1 consists of a pressure load of 10 KNS/sq.m. intensity on elements 41 to 72. The negative sign (and the default value for the axis) indicates that the load acts opposite to the positive direction of the element local z-axis.

Example Problem 9

81 LOAD 2 WIND LOAD JOINT LOAD 11 33 FZ -90. 22 FZ -450.

Load 2 consists of joint loads in the Z direction at joints 11, 22 and 33.

LOAD COMB 3 1 0.9 2 1.3

Load 3 is a combination of 0.9 times load case 1 and 1.3 times load case 2.

PERFORM ANALYSIS

The command to perform an elastic analysis is specified above.

LOAD LIST 1 3 PRINT SUPP REAC PRINT MEMBER FORCES LIST 27 PRINT ELEMENT STRESSES LIST 47

Support reactions, members forces and element stresses are printed for load cases 1 and 3.

START CONCRETE DESIGN CODE BRITISH DESIGN ELEMENT 47 END CONCRETE DESIGN

The above set of command form the instructions to STAAD to perform a concrete design on element 47. Design is done according to the British code. Note that design will consist only of flexural reinforcement calculations in the longitudinal and transverse directions of the elements for the moments MX and MY.

FINI

The STAAD run is terminated.


82


1. STAAD SPACE2. * EXAMPLE PROBLEM WITH FRAME MEMBERS AND3. * FINITE ELEMENTS4. UNIT METER NEWTON5. JOINT COORD6. 1 0 0 0 ; 2 0 0 6.07. REP ALL 2 6.0 0 08. 7 0 4.5 0 11 0 4.5 6.09. 12 1.5 4.5 0 14 4.5 4.5 0

10. 15 1.5 4.5 6.0 17 4.5 4.5 6.011. 18 6.0 4.5 0 22 6.0 4.5 6.012. 23 7.5 4.5 0 25 10.5 4.5 013. 26 7.5 4.5 6.0 28 10.5 4.5 6.014. 29 12. 4.5 0 33 12. 4.5 6.015. 34 6.0 1.125 0 36 6.0 3.375 016. 37 6.0 1.125 6.0 39 6.0 3.375 6.017. MEMBER INCI18. *COLUMNS19. 1 1 7 ; 2 2 1120. 3 3 34 ; 4 34 35 ; 5 35 36 ; 6 36 1821. 7 4 37 ; 8 37 38 ; 9 38 39 ; 10 39 2222. 11 5 29 ; 12 6 3323. *BEAMS IN Z DIRECTION AT X=024. 13 7 8 1625. *BEAMS IN Z DIRECTION AT X=6.026. 17 18 19 2027. *BEAMS IN Z DIRECTION AT X=12.028. 21 29 30 2429. *BEAMS IN X DIRECTION AT Z = 030. 25 7 12 ; 26 12 13 ; 27 13 14 ; 28 14 1831. 29 18 23 ; 30 23 24 ; 31 24 25 ; 32 25 2932. *BEAMS IN X DIRECTION AT Z = 12.033. 33 11 15 ; 34 15 16 ; 35 16 17 ; 36 17 2234. 37 22 26 ; 38 26 27 ; 39 27 28 ; 40 28 3335. DEFINE MESH36. A JOINT 737. B JOINT 1138. C JOINT 2239. D JOINT 1840. E JOINT 3341. F JOINT 2942. G JOINT 343. H JOINT 444. GENERATE ELEMENT45. MESH ABCD 4 446. MESH DCEF 4 447. MESH DCHG 4 448. UNIT MMS49. MEMB PROP50. 1 TO 40 PRIS YD 300 ZD 30051. ELEM PROP52. 41 TO 88 TH 15053. CONSTANT54. E 21000 ALL55. POISSON CONCRETE ALL56. SUPPORT57. 1 TO 6 FIXED58. UNIT KNS METER59. LOAD 1 DEAD LOAD FROM FLOOR60. ELEMENT LOAD61. 41 TO 72 PRESSURE -10.062. LOAD 2 WIND LOAD63. JOINT LOAD

Example Problem 9

83 64. 11 33 FZ -90.65. 22 FZ -450.66. LOAD COMB 367. 1 0.9 2 1.368. PERFORM ANALYSIS



69. LOAD LIST 1 370. PRINT SUPP REAC

SUPPORT REACTIONS -UNIT KNS METE STRUCTURE TYPE = SPACE-----------------


1 1 8.18 74.29 10.38 15.47 0.00 -12.193 7.47 68.32 10.88 17.77 0.05 -11.04

2 1 8.18 74.29 -10.38 -15.47 0.00 -12.193 7.29 65.42 -7.75 -9.88 0.23 -10.99

3 1 0.00 211.43 67.76 -10.53 0.00 0.003 0.00 791.68 476.07 12.76 0.00 0.00

4 1 0.00 211.43 -67.76 10.53 0.00 0.003 0.00 -411.15 336.67 31.06 0.00 0.00

5 1 -8.18 74.29 10.38 15.47 0.00 12.193 -7.47 68.32 10.88 17.77 -0.05 11.04

6 1 -8.18 74.29 -10.38 -15.47 0.00 12.193 -7.29 65.42 -7.75 -9.88 -0.23 10.99


71. PRINT MEMBER FORCES LIST 27



27 1 13 0.67 -11.94 -0.06 6.77 0.05 -21.6214 -0.67 11.94 0.06 -6.77 0.05 3.71

3 13 22.11 -10.57 -0.72 6.29 0.53 -19.5114 -22.11 10.57 0.72 -6.29 0.55 3.65


72. PRINT ELEMENT STRESSES LIST 47

ELEMENT STRESSES FORCE,LENGTH UNITS= KNS METE----------------

FORCE OR STRESS = FORCE/UNIT WIDTH/THICK, MOMENT = FORCE-LENGTH/UNIT WIDTH

ELEMENT LOAD SQX SQY MX MY MXYVONT VONB SX SY SXYTRESCAT TRESCAB

47 1 17.15 4.52 -10.44 -13.35 1.273310.85 3278.36 -11.85 -16.82 5.073705.87 3666.78

TOP : SMAX= -2666.32 SMIN= -3705.87 TMAX= 519.77 ANGLE= 20.7BOTT: SMAX= 3666.78 SMIN= 2648.07 TMAX= 509.36 ANGLE= 20.4

3 14.96 4.17 -9.54 -11.99 0.993078.51 2876.09 -46.18 -59.92 180.343479.60 3149.22

TOP : SMAX= -2369.52 SMIN= -3479.60 TMAX= 555.04 ANGLE= 26.5BOTT: SMAX= 3149.22 SMIN= 2487.70 TMAX= 330.76 ANGLE= 7.2


84 **** MAXIMUM STRESSES AMONG SELECTED PLATES AND CASES ****

MAXIMUM MINIMUM MAXIMUM MAXIMUM MAXIMUMPRINCIPAL PRINCIPAL SHEAR VONMISES TRESCASTRESS STRESS STRESS STRESS STRESS

3.666784E+03 -3.705870E+03 5.550363E+02 3.310846E+03 3.705870E+03PLATE NO. 47 47 47 47 47CASE NO. 1 1 3 1 1

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

73. START CONCRETE DESIGN74. CODE BRITISH

PROGRAM CODE REVISION V1.5_8110_97/175. DESIGN ELEMENT 47

ELEMENT DESIGN SUMMARY-BASED ON 16mm BARS-----------------------------------------MINIMUM AREAS ARE ACTUAL CODE MIN % REQUIREMENTS.PRACTICAL LAYOUTS ARE AS FOLLOWS:FY=460, 6No.16mm BARS AT 150mm C/C = 1206mm2/metreFY=250, 4No.16mm BARS AT 250mm C/C = 804mm2/metre

ELEMENT LONG. REINF MOM-X /LOAD TRANS. REINF MOM-Y /LOAD(mm2/m) (kN-m/m) (mm2/m) (kN-m/m)

47 TOP : 195. 0.00 / 0 195. 0.00 / 0BOTT: 205. -10.44 / 1 301. -13.35 / 1

***************************END OF ELEMENT DESIGN**************************

76. END CONCRETE DESIGN77. FINI


**** DATE= TIME= ****


Example Problem 9

85

NOTES


86

NOTES

Example Problem 10

87


A tank structure is modeled with four-noded plate elements. Water pressure from inside is used as loading for the tank. Reinforcement calculations have been done for some elements.

Tank Model

Deflected Shape


88

Actual input is shown in bold lettering followed by explanation. STAAD SPACE FINITE ELEMENT MODEL OF TANK * STRUCTURE

Every input has to start with the word STAAD. The word SPACE signifies that the structure is a space frame (3-D) structure.

UNITS METER KNS

Specifies the unit to be used. JOINT COORDINATES 1 0. 0. 0. 5 0. 6. 0. REPEAT 4 1.5. 0. 0. REPEAT 4 0. 0. 1.5. REPEAT 4 -1.5. 0. 0. REPEAT 3 0. 0. –1.5. 81 1.5. 0. 1.5. 83 1.5. 0. 4.5. REPEAT 2 1.5. 0. 0.

Joint number followed by X, Y and Z coordinates are provided above. The REPEAT command generates joint coordinates by repeating the pattern of the previous line of joint coordinates. The number following the REPEAT command is the number of repetitions to be carried out. This is followed by X, Y and Z coordinate increments. This is explained in section 5 of the Technical Reference Manual.

ELEMENT INCIDENCES 1 1 2 7 6 TO 4 1 1 REPEAT 14 4 5 61 76 77 2 1 TO 64 1 1 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

Example Problem 10

89 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 connectivities are input as above by providing the element number followed by joint numbers defining the element. The REPEAT command generates element incidences by repeating the pattern of the previous line of element nodes. The number following the REPEAT command is the number of repetitions to be carried out and that is followed by element and joint number increments. This is explained in detail in section 5 of the Technical Reference Manual.

UNIT MMS ELEMENT PROPERTIES 1 TO 80 TH 200.0

Element properties are provided by specifying that the elements are 200.0 mm THick.

CONSTANTS E 21.0 ALL POISSON CONC ALL

Material constants like E (modulus of elasticity) and Poisson’s ratio are provided following the keyword CONSTANTS.

SUPPORT 1 TO 76 BY 5 81 TO 89 PINNED

Pinned supports are specified at the joints listed above. No moments will be carried by these supports. The expression “1 TO 76 BY 5” means 1, 6, 11, etc. up to 76.

UNIT METER LOAD 1 ELEMENT LOAD 4 TO 64 BY 4 PR 50.0 3 TO 63 BY 4 PR 100.0


90 2 TO 62 BY 4 PR 150.0 1 TO 61 BY 4 PR 200.0

Load case 1 is initiated. It consists of element loads in the form of uniform PRessure acting along the local z-axis.

PERFORM ANALYSIS

This command instructs the program to proceed with the analysis. UNIT MMS PRINT JOINT DISPLACEMENTS LIST 5 25 45 65 PRINT ELEM FORCE LIST 9 TO 16

Joint displacements for a selected set of nodes and element corner forces for some elements are written in the output file as a result of the above commands. The forces printed are in the global directions at the nodes of the elements. The LIST option restricts the print output to that for the joints/elements listed.

START CONCRETE DESIGN

The above command initiates concrete design. CODE BRITISH DESIGN SLAB 9 12

Slabs (i.e. elements) 9 and 12 will be designed and the reinforcement requirements obtained. In STAAD, elements are typically designed for the moments MX and MY at the centroid of the element.

END CONCRETE DESIGN

Terminates the concrete design operation. FINISH


Example Problem 10

91


1. STAAD SPACE FINITE ELEMENT MODEL OF TANK STRUCTURE2. UNIT METER KNS3. JOINT COORDINATES4. 1 0.0 0.0 0.0 5 0.0 6.0 0.05. REPEAT 4 1.5 0.0 0.06. REPEAT 4 0.0 0.0 1.57. REPEAT 4 -1.5 0.0 0.08. REPEAT 3 0.0 0.0 -1.59. 81 1.5 0.0 1.5 83 1.5 0.0 4.5

10. REPEAT 2 1.5 0.0 0.011. ELEMENT INCIDENCES12. 1 1 2 7 6 TO 4 1 113. REPEAT 14 4 514. 61 76 77 2 1 TO 64 1 115. 65 1 6 81 7616. 66 76 81 82 7117. 67 71 82 83 6618. 68 66 83 56 6119. 69 6 11 84 8120. 70 81 84 85 8221. 71 82 85 86 8322. 72 83 86 51 5623. 73 11 16 87 8424. 74 84 87 88 8525. 75 85 88 89 8626. 76 86 89 46 5127. 77 16 21 26 8728. 78 87 26 31 8829. 79 88 31 36 8930. 80 89 36 41 4631. UNIT MMS32. ELEMENT PROPERTIES33. 1 TO 80 TH 200.034. CONSTANTS35. E 21.0 ALL36. POISSON CONC ALL37. SUPPORT38. 1 TO 76 BY 5 81 TO 89 PINNED39. UNIT METER40. LOAD 141. ELEMENT LOAD42. 4 TO 64 BY 4 PR 50.043. 3 TO 63 BY 4 PR 100.044. 2 TO 62 BY 4 PR 150.045. 1 TO 61 BY 4 PR 200.046. PERFORM ANALYSIS



47. UNIT MMS


92 48. PRINT JOINT DISPLACEMENTS LIST 5 25 45 65



5 1 -0.0103 0.0006 -0.0103 0.0002 0.0000 -0.000225 1 0.0103 0.0006 -0.0103 0.0002 0.0000 0.000245 1 0.0103 0.0006 0.0103 -0.0002 0.0000 0.000265 1 -0.0103 0.0006 0.0103 -0.0002 0.0000 -0.0002


49. PRINT ELEM FORCE LIST 9 TO 16

ELEMENT FORCES FORCE,LENGTH UNITS= KNS MMS--------------

GLOBAL CORNER FORCESJOINT FX FY FZ MX MY MZ

ELE.NO. 9 FOR LOAD CASE 111 -8.8359E+01 -5.3753E+01 1.8154E+02 1.3170E+05 -4.0959E+04 1.0868E+0412 -8.4106E+01 -6.1780E+00 -1.5632E+02 7.9661E+04 5.9853E+04 -1.7299E+0417 1.4404E+02 4.6568E+01 -1.6787E+02 1.2673E+05 -1.0931E+05 2.2151E+0416 2.8428E+01 1.3363E+01 1.4265E+02 1.4820E+05 5.2581E+04 -1.5720E+04

ELE.NO. 10 FOR LOAD CASE 112 -1.8792E+02 6.1780E+00 -4.0552E+01 -7.9660E+04 1.0833E+05 4.7652E+0413 -2.0044E+02 -3.6136E+01 -6.3740E+01 6.9905E+04 1.0284E+05 -5.8259E+0418 2.3039E+02 -1.4028E+01 3.0854E+01 5.8249E+04 -5.3646E+04 6.1978E+0417 1.5797E+02 4.3985E+01 7.3437E+01 8.3445E+02 -1.0826E+03 -5.1372E+04

ELE.NO. 11 FOR LOAD CASE 113 -2.1198E+02 3.6136E+01 -7.6885E+01 -6.9905E+04 1.2229E+05 5.7388E+0414 -2.3157E+02 -1.4563E+01 -2.4018E+01 2.6342E+04 1.0499E+05 -6.4697E+0419 2.0999E+02 -3.2387E+01 8.4179E+01 -1.5610E+04 -2.4207E+04 6.5236E+0418 2.3355E+02 1.0814E+01 1.6725E+01 -3.1068E+04 -5.1714E+04 -5.7927E+04

ELE.NO. 12 FOR LOAD CASE 114 -1.7545E+02 1.4563E+01 -6.0357E+01 -2.6342E+04 1.0566E+05 4.3868E+0415 -1.7875E+02 8.2356E-07 -2.8125E+01 -4.0310E-02 9.6044E+04 -4.9015E+0420 1.6418E+02 -2.4213E+00 6.5563E+01 -2.8717E+04 -2.4535E+04 4.5288E+0419 1.9002E+02 -1.2142E+01 2.2919E+01 -1.0980E+03 -4.4444E+04 -4.0140E+04

ELE.NO. 13 FOR LOAD CASE 116 -1.0527E+02 -8.1796E+01 -1.6646E+01 -1.3732E+04 2.6331E+04 1.5180E+0417 -3.5139E+01 -6.7286E+01 -7.8110E+01 -7.8741E+03 -3.1163E+03 -2.4809E+0422 1.8422E+02 2.6896E+01 8.1528E+01 -1.7008E+04 6.2736E+04 3.0020E+0421 -4.3808E+01 1.2219E+02 1.3228E+01 3.3487E+04 5.6183E+04 -2.0392E+04

ELE.NO. 14 FOR LOAD CASE 117 -2.6686E+02 -2.3267E+01 -2.2121E+02 -1.1969E+05 1.1350E+05 5.4029E+0418 -2.5280E+02 -5.3068E+00 -1.4951E+02 3.3215E+04 5.0983E+04 -5.3096E+0423 2.8137E+02 5.5470E+01 2.2661E+02 -8.0444E+04 2.0370E+05 6.3346E+0422 2.3829E+02 -2.6896E+01 1.4411E+02 5.1267E+04 1.8790E+05 -6.4279E+04

ELE.NO. 15 FOR LOAD CASE 118 -2.1115E+02 8.5202E+00 -1.7932E+02 -6.0396E+04 5.4377E+04 4.9044E+0419 -2.3106E+02 2.8931E+01 -1.4843E+02 4.3665E+04 4.6709E+04 -5.2522E+0424 1.9361E+02 1.8019E+01 1.6502E+02 -7.1533E+04 1.9615E+05 4.9770E+0423 2.4860E+02 -5.5470E+01 1.6273E+02 6.3390E+04 1.9439E+05 -4.6292E+04

ELE.NO. 16 FOR LOAD CASE 119 -1.6895E+02 1.5597E+01 -1.2741E+02 -2.6957E+04 2.1943E+04 2.7426E+0420 -1.6418E+02 2.4213E+00 -1.2181E+02 2.8717E+04 2.4535E+04 -4.5288E+0425 1.4616E+02 1.3697E-06 1.1804E+02 -5.5252E+04 1.6211E+05 5.5252E+0424 1.8697E+02 -1.8019E+01 1.3119E+02 5.9154E+04 1.6525E+05 -3.7391E+04

Example Problem 10

93 50. START CONCRETE DESIGN51. CODE BRITISH

PROGRAM CODE REVISION V1.5_8110_97/152. DESIGN SLAB 9 12

ELEMENT DESIGN SUMMARY-BASED ON 16mm BARS-----------------------------------------MINIMUM AREAS ARE ACTUAL CODE MIN % REQUIREMENTS.PRACTICAL LAYOUTS ARE AS FOLLOWS:FY=460, 6No.16mm BARS AT 150mm C/C = 1206mm2/metreFY=250, 4No.16mm BARS AT 250mm C/C = 804mm2/metre

ELEMENT LONG. REINF MOM-X /LOAD TRANS. REINF MOM-Y /LOAD(mm2/m) (kN-m/m) (mm2/m) (kN-m/m)

9 TOP : 260. 0.00 / 0 387. 25.20 / 1BOTT: 341. -24.50 / 1 260. 0.00 / 0

12 TOP : 260. 0.00 / 0 1570. 90.22 / 1BOTT: 260. -0.43 / 1 260. 0.00 / 0

***************************END OF ELEMENT DESIGN**************************

53. END CONCRETE DESIGN54. FINISH


**** DATE= TIME= ****



94 NOTES

Example Problem 11

95


Dynamic analysis (Response Spectrum) is performed for a steel structure. Results of a static and dynamic analysis are combined. The combined results are then used for steel design.


Example Problem 11

96

Actual input is shown in bold lettering followed by explanation. STAAD PLANE RESPONSE SPECTRUM ANALYSIS


UNIT METER KNS

Specifies the unit to be used. JOINT COORDINATES 1 0.0 0.0 0.0 ; 2 6.0 0.0 0.0 3 0.0 3.0 0.0 ; 4 6.0 3.0 0.0 5 0.0 6.0 0.0 ; 6 6.0 6.0 0.0

Joint number followed by X, Y and Z coordinates are provided above. Since this is a plane structure, the Z coordinates are all the same, in this case, zeros. Semicolon signs (;) are used as line separators to allow for input of multiple sets of data on one line.

MEMBER INCIDENCES 1 1 3 ; 2 2 4 ; 3 3 5 ; 4 4 6 5 3 4 ; 6 5 6

Defines the members by the joints they are connected to. MEMBER PROPERTIES BRITISH 1 TO 4 TA ST UC254X254X73 5 TA ST UB305X165X54 6 TA ST UB203X133X30

Properties for all members are assigned from the British steel table. The word ST stands for standard single section.

SUPPORTS 1 2 FIXED

Fixed supports are specified at joints 1 and 2.

Example Problem 11

97 UNIT MMS CONSTANTS E 210.0 ALL POISSON STEEL ALL DEN 76.977E-09 ALL

Material constants such as E (modulus of elasticity), Poisson’s ratio and DENsity are specified above. Length unit is changed from METER to MMS to facilitate the input.

CUT OFF MODE SHAPE 2

The number of mode shapes to be considered in dynamic analysis is set to 2. Without the above command, this will be set to the default which can be found in section 5 of the Technical Reference Manual.

* LOAD 1 WILL BE STATIC LOAD UNIT METER LOAD 1 DEAD AND LIVE LOADS

Load case 1 is initiated followed by a title. Prior to this, the length unit is changed to METER for specifying distributed member loads. A line starting with an asterisk (*) mark indicates a comment line.

SELFWEIGHT Y -1.0

The above command indicates that the selfweight of the structure acting in the global Y direction is part of this load case. The factor of -1.0 is meant to indicate that the load acts opposite to the positive direction of global Y, hence downwards.

MEMBER LOADS 5 CON GY -25.0 1.8 5 CON GY -37.5 3.0 5 CON GY -25.0 4.2 5 6 UNI Y -22.5

Load 1 contains member loads also. GY indicates that the load is in the global Y direction while Y indicates local Y direction. The word UNI stands for uniformly distributed load while CON stands


Example Problem 11

98

for concentrated load. GY is followed by the value of the load and the distance at which it is applied.

* NEXT LOAD WILL BE RESPONSE SPECTRUM LOAD * WITH MASSES PROVIDED IN TERMS OF LOAD. LOAD 2 SEISMIC LOADING

The two lines which begin with the asterisk are comment lines which tell us the purpose of the next load case. Load case 2 is then initiated along with an optional title. This will be a dynamic load case. Permanent masses will be provided in the form of loads. These masses (in terms of loads) will be considered for the eigensolution. Internally, the program converts these loads to masses, hence it is best to specify them as absolute values (without a negative sign). Also, the direction (X, Y, Z etc.) of the loads will correspond to the dynamic degrees of freedom in which the masses are capable of vibrating. In a PLANE frame, only X and Y directions need to be considered. In a SPACE frame, masses (loads) should be provided in all three (X, Y and Z) directions if they are active along all three. The user has the freedom to restrict one or more directions.

SELFWEIGHT X 1.0 SELFWEIGHT Y 1.0

The above commands indicate that the selfweight of the structure acting in the global X and Y directions with a factor of 1.0 is taken into consideration for the mass matrix.

MEMBER LOADS 5 CON GX 25.0 1.8 5 CON GY 25.0 1.8 5 CON GX 37.5 3.0 5 CON GY 37.5 3.0 5 CON GX 25.0 4.2 5 CON GY 25.0 4.2

The mass matrix will also consist of terms derived from the above member loads. GX and GY indicate that the load, and hence the resulting mass, is capable of vibration along the global X and Y directions. The word CON stands for concentrated load.

Example Problem 11

99

Concentrated forces of 25, 37.5, and 25 kNs are located at 1.8m, 3.0m and 4.2m from the start of member 5.

SPECTRUM CQC X 1.0 ACC DAMP 0.05 SCALE 9.806 0.03 1.00 ; 0.05 1.35 0.1 1.95 ; 0.2 2.80 0.5 2.80 ; 1.0 1.60

The above SPECTRUM command specifies that the modal responses be combined using the CQC method (alternatives being the SRSS method, ABS method, etc.). The spectrum effect is in the global X direction with a factor of 1.0. Since this spectrum is in terms of ACCeleration (the other possibility being displacement), the spectrum data is given as period vs. acceleration. Damping ratio of 0.05 (5%) and a scale factor of 9.806 are used. The scale factor is the quantity by which spectral accelerations (and spectral displacements) must be multiplied by before they are used in the calculations. The values of periods and the corresponding accelerations are given in the last 3 lines.

LOAD COMBINATION 3 1 0.75 2 0.75 LOAD COMBINATION 4 1 0.75 2 -0.75

In a response spectrum analysis, the sign of the forces cannot be determined, and hence are absolute numbers. Consequently, to account for the fact that the force could be positive or negative, it is necessary to create 2 load combination cases. That is what is being done above. Load combination case no. 3 consists of the sum of the static load case (1) with the positive direction of the dynamic load case (2). Load combination case no. 4 consists of the sum of the static load case (1) with the negative direction of the dynamic load case (2). In both cases, the result is factored by 0.75.

PERFORM ANALYSIS PRINT MODE SHAPES

This command instructs the program to proceed with the analysis. The PRINT command instructs the program to print mode shape values.


Example Problem 11

100 PRINT ANALYSIS RESULTS

Displacements, reactions and member forces are recorded in the output file using the above command.

LOAD LIST 1 3 4 PARAMETER CODE BRITISH SELECT ALL

A steel design in the form of a member selection is performed based on the rules of the British code. Only the member forces resulting from load cases 1, 3 and 4 will be considered for these calculations.

FINISH


Example Problem 11


1. STAAD PLANE RESPONSE SPECTRUM ANALYSIS2. UNIT METER KNS3. JOINT COORDINATES4. 1 0.0 0.0 0.0 ; 2 6.0 0.0 0.05. 3 0.0 3.0 0.0 ; 4 6.0 3.0 0.06. 5 0.0 6.0 0.0 ; 6 6.0 6.0 0.07. MEMBER INCIDENCES8. 1 1 3 ; 2 2 4 ; 3 3 5 ; 4 4 69. 5 3 4 ; 6 5 6

10. MEMBER PROPERTIES BRITISH11. 1 TO 4 TA ST UC254X254X7312. 5 TA ST UB305X165X5413. 6 TA ST UB203X133X3014. SUPPORTS15. 1 2 FIXED16. UNIT MMS17. CONSTANTS18. E 210.0 ALL19. POISSON STEEL ALL20. DEN 76.977E-09 ALL21. CUT OFF MODE SHAPE 222. *LOAD 1 WILL BE STATIC LOAD23. UNIT METER24. LOAD 1 DEAD AND LIVE LOADS25. SELFWEIGHT Y -1.026. MEMBER LOADS27. 5 CON GY -25.0 1.828. 5 CON GY -37.5 3.029. 5 CON GY -25.0 4.230. 5 6 UNI Y -22.531. * NEXT LOAD WILL BE RESPONSE SPECTRUM LOAD32. * WITH MASSES PROVIDED IN TERMS OF LOAD.33. LOAD 2 SEISMIC LOADING34. SELFWEIGHT X 1.035. SELFWEIGHT Y 1.036. MEMBER LOADS37. 5 CON GX 25.0 1.838. 5 CON GY 25.0 1.839. 5 CON GX 37.5 3.040. 5 CON GY 37.5 3.041. 5 CON GX 25.0 4.242. 5 CON GY 25.0 4.243. SPECTRUM CQC X 1.0 ACC DAMP 0.05 SCALE 9.80644. 0.03 1.00 ; 0.05 1.3545. 0.1 1.95 ; 0.2 2.8046. 0.5 2.80 ; 1.0 1.6047. LOAD COMBINATION 348. 1 0.75 2 0.7549. LOAD COMBINATION 450. 1 0.75 2 -0.7551. PERFORM ANALYSIS PRINT MODE SHAPES




Example Problem 11

102 NUMBER OF MODES REQUESTED = 2NUMBER OF EXISTING MASSES IN THE MODEL = 8NUMBER OF MODES THAT WILL BE USED = 2

*** �EIGENSOLUTION�: SUBSPACE METHOD ***

CALCULATED FREQUENCIES FOR LOAD CASE 2

MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC) ACCURACY

1 5.179 0.19308 2.209E-102 19.437 0.05145 6.482E-08

The following Frequencies are estimates that were calculated. These are forinformation only and will not be used. Remaining values are either abovethe cut off mode/freq values or are of low accuracy. To use thesefrequencies, rerun with a higher cutoff mode (or mode + freq) value.



3 50.784 0.01969 2.898E-104 57.172 0.01749 4.004E-10

MODE SHAPES-----------

JOINT MODE X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN

1 1 0.00000 0.00000 0.00000 0.000E+00 0.000E+00 0.000E+002 1 0.00000 0.00000 0.00000 0.000E+00 0.000E+00 0.000E+003 1 0.58328 0.00218 0.00000 0.000E+00 0.000E+00 -3.863E-034 1 0.58328 -0.00218 0.00000 0.000E+00 0.000E+00 -3.863E-035 1 1.00000 0.00251 0.00000 0.000E+00 0.000E+00 -2.797E-036 1 1.00000 -0.00251 0.00000 0.000E+00 0.000E+00 -2.797E-03

MODE SHAPES-----------

JOINT MODE X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN

1 2 0.00000 0.00000 0.00000 0.000E+00 0.000E+00 0.000E+002 2 0.00000 0.00000 0.00000 0.000E+00 0.000E+00 0.000E+003 2 -0.07067 0.00282 0.00000 0.000E+00 0.000E+00 -2.798E-034 2 -0.07067 -0.00282 0.00000 0.000E+00 0.000E+00 -2.798E-035 2 1.00000 0.00400 0.00000 0.000E+00 0.000E+00 -9.742E-036 2 1.00000 -0.00400 0.00000 0.000E+00 0.000E+00 -9.742E-03

RESPONSE LOAD CASE 2

CQC MODAL COMBINATION METHOD USED.DYNAMIC WEIGHT X Y Z 9.889183E+01 9.889183E+01 0.000000E+00 KNSMISSING WEIGHT X Y Z -3.077453E-04 -9.889183E+01 0.000000E+00 KNSMODAL WEIGHT X Y Z 9.889152E+01 7.885937E-34 0.000000E+00 KNS

MODE ACCELERATION-G DAMPING---- -------------- -------

1 2.74098 0.050002 1.36730 0.05000

Example Problem 11

103 MASS PARTICIPATION FACTORS IN PERCENT BASE SHEAR IN KNS-------------------------------------- ------------------

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

1 98.20 0.00 0.00 98.197 0.000 0.000 266.17 0.00 0.002 1.80 0.00 0.00 100.000 0.000 0.000 2.44 0.00 0.00

---------------------------TOTAL SRSS SHEAR 266.18 0.00 0.00TOTAL 10PCT SHEAR 266.18 0.00 0.00TOTAL ABS SHEAR 268.61 0.00 0.00TOTAL CQC SHEAR 266.18 0.00 0.00




1 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.00002 0.0000 0.0000 0.0000 0.0000 0.0000 0.00003 0.0000 0.0000 0.0000 0.0000 0.0000 0.00004 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

2 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.00002 0.0000 0.0000 0.0000 0.0000 0.0000 0.00003 0.0000 0.0000 0.0000 0.0000 0.0000 0.00004 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

3 1 -0.0042 -0.0283 0.0000 0.0000 0.0000 -0.00172 2.4239 0.0091 0.0000 0.0000 0.0000 0.00633 1.8148 -0.0144 0.0000 0.0000 0.0000 0.00354 -1.8211 -0.0280 0.0000 0.0000 0.0000 -0.0060

4 1 0.0042 -0.0283 0.0000 0.0000 0.0000 0.00172 2.4239 0.0091 0.0000 0.0000 0.0000 0.00633 1.8211 -0.0144 0.0000 0.0000 0.0000 0.00604 -1.8148 -0.0280 0.0000 0.0000 0.0000 -0.0035

5 1 0.0163 -0.0390 0.0000 0.0000 0.0000 -0.00172 4.1560 0.0104 0.0000 0.0000 0.0000 0.00463 3.1293 -0.0214 0.0000 0.0000 0.0000 0.00224 -3.1048 -0.0371 0.0000 0.0000 0.0000 -0.0047

6 1 -0.0163 -0.0390 0.0000 0.0000 0.0000 0.00172 4.1560 0.0104 0.0000 0.0000 0.0000 0.00463 3.1048 -0.0214 0.0000 0.0000 0.0000 0.00474 -3.1293 -0.0371 0.0000 0.0000 0.0000 -0.0022



1 1 23.48 185.52 0.00 0.00 0.00 -21.572 133.09 59.00 0.00 0.00 0.00 250.113 117.43 183.39 0.00 0.00 0.00 171.404 -82.21 94.89 0.00 0.00 0.00 -203.76

2 1 -23.48 185.52 0.00 0.00 0.00 21.572 133.09 59.00 0.00 0.00 0.00 250.113 82.21 183.39 0.00 0.00 0.00 203.764 -117.43 94.89 0.00 0.00 0.00 -171.40



1 1 1 185.52 -23.48 0.00 0.00 0.00 -21.573 -183.37 23.48 0.00 0.00 0.00 -48.85

2 1 59.00 133.09 0.00 0.00 0.00 250.113 59.00 133.09 0.00 0.00 0.00 149.17

3 1 183.39 82.21 0.00 0.00 0.00 171.403 -93.28 117.43 0.00 0.00 0.00 75.24

4 1 94.89 -117.43 0.00 0.00 0.00 -203.763 -181.78 -82.21 0.00 0.00 0.00 -148.52


Example Problem 11



2 1 2 185.52 23.48 0.00 0.00 0.00 21.574 -183.37 -23.48 0.00 0.00 0.00 48.85

2 2 59.00 133.09 0.00 0.00 0.00 250.114 59.00 133.09 0.00 0.00 0.00 149.17

3 2 183.39 117.43 0.00 0.00 0.00 203.764 -93.28 82.21 0.00 0.00 0.00 148.52

4 2 94.89 -82.21 0.00 0.00 0.00 -171.404 -181.78 -117.43 0.00 0.00 0.00 -75.24

3 1 3 70.53 -43.63 0.00 0.00 0.00 -65.875 -68.38 43.63 0.00 0.00 0.00 -65.03

2 3 9.07 8.95 0.00 0.00 0.00 3.885 9.07 8.95 0.00 0.00 0.00 27.13

3 3 59.70 -26.01 0.00 0.00 0.00 -46.495 -44.49 39.44 0.00 0.00 0.00 -28.42

4 3 46.10 -39.44 0.00 0.00 0.00 -52.325 -58.09 26.01 0.00 0.00 0.00 -69.12

4 1 4 70.53 43.63 0.00 0.00 0.00 65.876 -68.38 -43.63 0.00 0.00 0.00 65.03

2 4 9.07 8.95 0.00 0.00 0.00 3.886 9.07 8.95 0.00 0.00 0.00 27.13

3 4 59.70 39.44 0.00 0.00 0.00 52.326 -44.49 -26.01 0.00 0.00 0.00 69.12

4 4 46.10 26.01 0.00 0.00 0.00 46.496 -58.09 -39.44 0.00 0.00 0.00 28.42

5 1 3 -20.16 112.84 0.00 0.00 0.00 114.734 20.16 112.84 0.00 0.00 0.00 -114.73

2 3 0.00 49.47 0.00 0.00 0.00 148.414 0.00 49.47 0.00 0.00 0.00 148.41

3 3 -15.12 121.73 0.00 0.00 0.00 197.354 15.12 121.73 0.00 0.00 0.00 25.26

4 3 -15.12 47.53 0.00 0.00 0.00 -25.264 15.12 47.53 0.00 0.00 0.00 -197.35

6 1 5 43.63 68.38 0.00 0.00 0.00 65.036 -43.63 68.38 0.00 0.00 0.00 -65.03

2 5 0.00 9.04 0.00 0.00 0.00 27.136 0.00 9.04 0.00 0.00 0.00 27.13

3 5 32.73 58.07 0.00 0.00 0.00 69.126 -32.73 58.07 0.00 0.00 0.00 -28.42

4 5 32.73 44.50 0.00 0.00 0.00 28.426 -32.73 44.50 0.00 0.00 0.00 -69.12


53. LOAD LIST 1 3 454. PARAMETER55. CODE BRITISH56. SELECT ALL

Example Problem 11

105 STAAD.Pro MEMBER SELECTION - (BSI )**************************




=======================================================================



3 ST UC203X203X46 PASS BS-4.3.6 0.551 446.10 C 0.00 69.12 0.00

4 ST UC203X203X46 PASS BS-4.3.6 0.551 359.70 C 0.00 69.12 0.00

5 ST UB457X191X82 PASS BS-4.3.6 0.898 315.12 T 0.00 197.35 0.00



57. FINISH



****************************************************************************


**** DATE= TIME= ****



Example Problem 11

106

NOTES

Example Problem 12

107


This example demonstrates generation of load cases for the type of loading known as a moving load. This type of loading occurs classically when the load-causing units move on the structure, as in the case of trucks on a bridge deck. The mobile loads are discretized into several individual immobile load cases at discrete positions. During this process, enormous number of load cases may be created resulting in plenty of output to be sorted. To avoid looking into a lot of output, the maximum force envelope is requested for a few specific members.


108

Actual input is shown in bold lettering followed by explanation. STAAD FLOOR A SIMPLE BRIDGE DECK

Every input has to start with the word STAAD. The word FLOOR signifies that the structure is a floor structure and the geometry is defined through X and Z axis.

UNITS METER KNS

Specifies the unit to be used. JOINT COORDINATES 1 0 0 0 6 7.5 0 0 R 5 0 0 9.0

Joint number followed by X, Y and Z coordinates are provided above. Since this is a floor structure, the Y coordinates are all the same, in this case, zeros. The first line generates joints 1 through 6. With the repeat (R) command, the coordinates of the next 30 joints are generated by repeating the pattern of the coordinates of the first 6 joints 5 times with X, Y and Z increments of 0,0 & 9 respectively.

MEMBER INCIDENCES 1 1 7 6 7 1 2 11 R A 4 11 6 56 31 32 60

Defines the members by the joints they are connected to. The fourth number indicates the final member number upto which they will be generated. Repeat all (abbreviated as R A) will create members by repeating the member incidence pattern of the previous 11 members. The number of repetitions to be carried out is provided after the R A command and the member increment and joint increment are defined as 11 and 6 respectively. The fifth line of input defines the member incidences for members 56 to 60.

MEMBER PROPERTIES BRITISH 1 TO 60 TA ST UB305X165X40

Example Problem 12

109

Properties for all members are assigned from the British steel table. The word ST stands for standard single section.

SUPPORTS 1 TO 6 31 TO 36 PINNED

Pinned supports are specified at the above joints. A pinned support is one which can resist only translational forces.

UNITS MMS CONSTANTS E 210.0 ALL POISSON STEEL ALL DEN 76.977E-09 ALL UNIT METER KNS

Material constants like E (modulus of elasticity), Poisson’s ratio and DENsity are specified above following a change in the units of length from METER to MMS.

DEFINE MOVING LOAD TYPE 1 LOAD 90.0 90.0 45.0 DISTANCE 3.0 1.5 WIDTH 3.0

The characteristics of the vehicle are defined above in METER and KNS units. The above lines represent the first out of two sets of data required in moving load generation. The type number (1) is a label for identification of the load-causing unit, such as a truck. 3 axles (90 90 45) are specified with the LOAD command. The spacing between the axles in the direction of movement (longitudinal direction) is specified after the DISTANCE command. WIDTH is the spacing in the transverse direction, that is, it is the distance between the 2 prongs of an axle of the truck.

LOAD 1

Load case 1 is initiated. SELF Y -1.0


110

Selfweight of the structure acting in the negative (due to the factor -1.0) global Y direction is the only component of load case 1.

LOAD GENERATION 10 TYPE 1 2.25 0. 0. ZI 3.0

This constitutes the second of the two sets of data required for moving load generation. 10 load cases are generated using the Type 1 vehicle whose characteristics were described earlier. For the first of these load cases, the X, Y and Z location of the reference load (see section 5.31.1 of the Technical Reference Manual) have been specified after the command TYPE 1. The Z Increment of 3.0m denotes that the vehicle moves along the Z direction and the individual positions which are 3.0m apart will be used to generate the remaining 9 load cases.

PERFORM ANALYSIS PRINT LOAD

The above command instructs the program to proceed with the analysis and print the values and positions of all the generated load cases.

PRINT MAXFORCE ENVELOP LIST 3 41 42

A maximum force envelope consisting of the highest forces for each degree of freedom on the listed members will be written into the output file.

FINISH


Example Problem 12


1. STAAD FLOOR A SIMPLE BRIDGE DECK2. UNITS METER KNS3. JOINT COORDINATES4. 1 0.0 0.0 0.0 6 7.5 0.0 0.05. R 5 0.0 0.0 9.06. MEMBER INCIDENCES7. 1 1 7 68. 7 1 2 119. R A 4 11 6

10. 56 31 32 6011. MEMBER PROPERTIES BRITISH12. 1 TO 60 TA ST UB305X165X4013. SUPPORTS14. 1 TO 6 31 TO 36 PINNED15. UNITS MMS16. CONSTANTS17. E 210. ALL18. POISSON STEEL ALL19. DEN 76.977E-09 ALL20. UNIT METER KNS21. DEFINE MOVING LOAD22. TYPE 1 LOAD 90.0 90.0 45.0 DISTANCE 3.0 1.5 WIDTH 3.023. LOAD 124. SELF Y -1.025. LOAD GENERATION 1026. TYPE 1 2.25 0. 0. ZI 3.027. PERFORM ANALYSIS PRINT LOAD



LOADING 1-----------

SELFWEIGHT Y -1.000

ACTUAL WEIGHT OF THE STRUCTURE = 124.391 KNS

LOADING 2-----------

MEMBER LOAD - UNIT KNS METE


8 -90.000 GY 0.7510 -90.000 GY 0.753 -45.000 GY 3.002 -45.000 GY 3.005 -45.000 GY 3.004 -45.000 GY 3.003 -22.500 GY 4.502 -22.500 GY 4.505 -22.500 GY 4.504 -22.500 GY 4.50


112 LOADING 3-----------



3 -45.000 GY 3.002 -45.000 GY 3.005 -45.000 GY 3.004 -45.000 GY 3.003 -45.000 GY 6.002 -45.000 GY 6.005 -45.000 GY 6.004 -45.000 GY 6.003 -22.500 GY 7.502 -22.500 GY 7.505 -22.500 GY 7.504 -22.500 GY 7.50

LOADING 4-----------



3 -45.000 GY 6.002 -45.000 GY 6.005 -45.000 GY 6.004 -45.000 GY 6.00

19 -90.000 GY 0.7521 -90.000 GY 0.7514 -22.500 GY 1.5013 -22.500 GY 1.5016 -22.500 GY 1.5015 -22.500 GY 1.50

LOADING 5-----------




LOADING 6-----------




Example Problem 12

113 LOADING 7-----------




LOADING 8-----------




LOADING 9-----------




LOADING 10-----------





114 LOADING 11-----------





28. PRINT MAXFORCE ENVELOP LIST 3 41 42

MEMBER FORCE ENVELOPE---------------------

ALL UNITS ARE KNS METE

MAX AND MIN FORCE VALUES AMONGST ALL SECTION LOCATIONS

MEMB FY/ DIST LD MZ/ DIST LDFZ DIST LD MY DIST LD FX DIST LD

3 MAX 81.16 0.00 3 0.03 0.00 40.00 0.00 1 0.00 0.00 1 0.00 0.00 1

MIN -31.34 9.00 3 -504.76 9.00 50.00 9.00 11 0.00 9.00 11 0.00 9.00 11

41 MAX 73.48 0.00 10 9.18 1.50 50.00 0.00 1 0.00 0.00 1 0.00 0.00 1

MIN -18.36 1.50 11 -147.24 0.75 100.00 1.50 11 0.00 1.50 11 0.00 1.50 11

42 MAX 0.30 0.00 1 9.17 0.00 50.00 0.00 1 0.00 0.00 1 0.00 0.00 1

MIN -0.30 1.50 1 -134.84 1.50 100.00 1.50 11 0.00 1.50 11 0.00 1.50 11

********** END OF FORCE ENVELOPE FROM INTERNAL STORAGE **********

29. FINISH*********** END OF THE STAAD.Pro RUN ***********

**** DATE= TIME= ****


Example Problem 12

115

NOTES


116

NOTES

Example Problem 13

117


Calculation of displacements at intermediate points of members of a plane frame is demonstrated in this example.

The dashed line represents the deflected shape of the structure. The shape is generated on the basis of displacements at the ends plus several intermediate points of the members.


118

Actual input is shown in bold lettering followed by explanation. STAAD PLANE TEST FOR SECTION DISPLACEMENT


UNIT METER KNS

Specifies the unit to be used. JOINT COORDINATES 1 0 0 ; 2 0. 4.5 ; 3 6 4.5 ; 4 6 0.

Joint number followed by X and Y coordinates are provided above. Since this is a plane structure, the Z coordinates need not be provided. Semicolon signs (;) are used as line separators which allows us to provide multiple sets of data on one line.

MEMBER INCIDENCE 1 1 2 ; 2 2 3 ; 3 3 4

Defines the members by the joints they are connected to. MEMBER PROPERTY BRITISH 1 3 TABLE ST UC203X203X46 2 TABLE ST UB305X165X40

Member properties are specified from the British steel table. The word ST stands for standard single section.

UNIT MMS CONSTANTS E 210.0 ALL POISSON STEEL ALL

Material constants like E (modulus of elasticity) and Poisson’s ratio are provided after the length unit is changed from METER to MMS.

Example Problem 13

119 SUPPORT 1 FIXED ; 4 PINNED

A fixed support is specified at Joint 1 and a pinned support at Joint 4.

UNIT METER LOADING 1 DEAD + LIVE + WIND

Load case 1 is initiated followed by an optional title. JOINT LOAD 2 FX 25.0

Load 1 contains a joint load of 25KN at node 2. FX indicates that the load is a force in the global X direction.

MEMBER LOAD 2 UNI GY -45.0

Load 1 contains a member load also. GY indicates that the load is in the global Y direction. The word UNI stands for uniformly distributed load.

PERFORM ANALYSIS

This command instructs the program to proceed with the analysis. PRINT MEMBER FORCES

The above PRINT command is self-explanatory. * * FOLLOWING PRINT COMMAND WILL PRINT * DISPLACEMENTS OF THE MEMBERS * CONSIDERING EVERY TWELFTH INTERMEDIATE * POINT (THAT IS TOTAL OF 13 POINTS). THESE * DISPLACEMENTS ARE MEASURED IN GLOBAL X * Y Z COORDINATE SYSTEM AND THE VALUES * ARE FROM ORIGINAL COORDINATES (UNDEFLECTED * POSITION) OF CORRESPONDING TWELFTH * POINTS.


120 * * MAX LOCAL DISPLACEMENT IS ALSO PRINTED. * THE LOCATION OF MAXIMUM INTERMEDIATE * DISPLACEMENT IS DETERMINED. THIS VALUE IS * MEASURED FROM ABOVE LOCATION TO THE * STRAIGHT LINE JOINING START AND END * JOINTS OF THE DEFLECTED MEMBER. * PRINT SECTION DISPLACEMENT

Above PRINT command is explained in the comment lines above. FINISH


Example Problem 13


1. STAAD PLANE TEST FOR SECTION DISPLACEMENT2. UNIT METER KNS3. JOINT COORDINATES4. 1 0.0 0.0 ; 2 0.0 4.5 ; 3 6.0 4.5 ; 4 6.0 0.05. MEMBER INCIDENCE6. 1 1 2 ; 2 2 3 ; 3 3 47. MEMBER PROPERTY BRITISH8. 1 3 TABLE ST UC203X203X469. 2 TABLE ST UB305X165X40

10. UNIT MMS11. CONSTANTS12. E 210.0 ALL13. POISSON STEEL ALL14. SUPPORT15. 1 FIXED ; 4 PINNED16. UNIT METER17. LOADING 1 DEAD + LIVE + WIND18. JOINT LOAD19. 2 FX 25.020. MEMBER LOAD21. 2 UNI GY -45.022. PERFORM ANALYSIS



23. PRINT MEMBER FORCES



1 1 1 121.84 1.66 0.00 0.00 0.00 33.552 -121.84 -1.66 0.00 0.00 0.00 -26.08

2 1 2 23.34 121.84 0.00 0.00 0.00 26.083 -23.34 148.16 0.00 0.00 0.00 -105.03

3 1 3 148.16 23.34 0.00 0.00 0.00 105.034 -148.16 -23.34 0.00 0.00 0.00 0.00


24. *25. * FOLLOWING PRINT COMMAND WILL PRINT26. * DISPLACEMENTS OF THE MEMBERS CONSIDERING EVERY27. * TWELFTH INTERMEDIATE POINT (THAT IS TOTAL OF 13 POINTS).28. * THESE DISPLACEMENTS ARE MEASURED IN GLOBAL X Y Z29. * COORDINATE SYSTEM AND THE VALUES ARE FROM30. * ORIGINAL COORDINATES (UNDEFLECTED POSITION)31. * OF THE CORRESPONDING TWELFTH POINTS.32. *


122 33. * MAX LOCAL DISPLACEMENT IS ALSO PRINTED. THE34. * LOCATION OF MAXIMUM INTERMEDIATE DISPLACEMENT35. * IS DETERMINED. THIS VALUE IS MEASURED FROM ABOVE36. * LOCATION TO THE STRAIGHT LINE JOINING START AND37. * END JOINTS OF THE DEFLECTED MEMBER.38. *39. PRINT SECTION DISPLACEMENT

MEMBER SECTION DISPLACEMENTS----------------------------UNIT =INCHES FOR FPS AND CM FOR METRICS/SI SYSTEM

MEMB LOAD GLOBAL X,Y,Z DISPL FROM START TO END JOINTS AT 1/12TH PTS

1 1 0.0000 0.0000 0.0000 0.0250 -0.0037 0.00000.0982 -0.0074 0.0000 0.2188 -0.0111 0.00000.3858 -0.0148 0.0000 0.5983 -0.0185 0.00000.8555 -0.0222 0.0000 1.1563 -0.0260 0.00001.5000 -0.0297 0.0000 1.8855 -0.0334 0.00002.3120 -0.0371 0.0000 2.7785 -0.0408 0.00003.2842 -0.0445 0.0000

MAX LOCAL DISP = 0.78661 AT 225.00 LOAD 1 L/DISP= 572

2 1 3.2842 -0.0445 0.0000 3.2831 -0.7367 0.00003.2820 -1.3906 0.0000 3.2809 -1.9446 0.00003.2798 -2.3527 0.0000 3.2787 -2.5847 0.00003.2777 -2.6264 0.0000 3.2766 -2.4789 0.00003.2755 -2.1593 0.0000 3.2744 -1.7006 0.00003.2733 -1.1513 0.0000 3.2722 -0.5758 0.00003.2712 -0.0541 0.0000


3 1 3.2712 -0.0541 0.0000 3.5396 -0.0496 0.00003.6669 -0.0451 0.0000 3.6659 -0.0405 0.00003.5494 -0.0360 0.0000 3.3302 -0.0315 0.00003.0213 -0.0270 0.0000 2.6353 -0.0225 0.00002.1853 -0.0180 0.0000 1.6839 -0.0135 0.00001.1440 -0.0090 0.0000 0.5784 -0.0045 0.00000.0000 0.0000 0.0000


************ END OF SECT DISPL RESULTS ***********

40. FINISH


**** DATE= TIME= ****


Example Problem 13

123

NOTES


124

NOTES

Example Problem 14

125


A space frame is analyzed for seismic loads. The seismic loads are generated using the procedures of the 1994 UBC Code. A P-Delta analysis is peformed to obtain the secondary effects of the lateral and vertical loads acting simultaneously.

13

9

5 6 7 8

10 11 12

14 15 16

17

4321

18 19 20

21 22 23 24

25 26 27 28

29 30 31 32

33 34 35 36

37 38 39 40

41 42 43 44

45 46 47 48

49 50 51 52

54 5553 56

57 58 59 60

61 62 63 64

X

Y

Z


126 STAAD SPACE EXAMPLE PROBLEM FOR UBC LOAD

Every input has to start with the word STAAD. The word SPACE signifies that the structure is a space frame.

UNIT METER KNS

Specifies the unit to be used. JOINT COORDINATES 1 0 0 0 4 10.5 0 0 REPEAT 3 0 0 3.5 REPEAT ALL 3 0 3.5 0

The X, Y and Z coordinates of the joints are specified here. First, coordinates of joints 1 through 4 are generated by taking advantage of the fact that they are equally spaced. Then, this pattern is REPEATed 3 times with a Z increment of 3.5 m for each repetition to generate joints 5 to 16. The REPEAT ALL command will then repeat 3 times, the pattern of joints 1 to 16 to generate joints 17 to 64.

MEMBER INCIDENCES * beams in x direction 101 17 18 103 104 21 22 106 107 25 26 109 110 29 30 112 REPEAT ALL 2 12 16 * beams in z direction 201 17 21 204 205 21 25 208 209 25 29 212 REPEAT ALL 2 12 16 * columns 301 1 17 348

Defines the members by the joints they are connected to. Following the specification of incidences for members 101 to 112, the REPEAT ALL command command is used to repeat the pattern and generate incidences for members 113 through 136. A similar logic is used in specification of incidences of members 201 through 212

Example Problem 14

127

and generation of incidences for members 213 to 236. Finally, members incidences of columns 301 to 348 are specified.

MEMBER PROPERTIES BRITISH 101 TO 136 201 TO 236 PRIS YD 0.40 ZD 0.30 301 TO 348 TA ST UB457X152X52

The beam members have prismatic member property specification (YD & ZD) while the columns (members 301 to 348) have their properties called from the built-in BRITISH steel table.

CONSTANT E STEEL MEMB 301 TO 348 E CONCRETE MEMB 101 TO 136 201 TO 236 DENSITY STEEL MEMB 301 TO 348 DENSITY CONCRETE MEMB 101 TO 136 201 TO 236 POISSON STEEL MEMB 301 TO 348 POISSON CONCRETE MEMB 101 TO 136 201 TO 236

In the specification of material constants, the default built-in values are used. The user may see these values with the help of the command PRINT MATERIAL PROPERTIES following the above commands.


Indicates the joints where the supports are located as well as the type of support restraints.

DEFINE UBC LOAD ZONE 0.2 I 1.0 RWX 9 RWZ 9 S 1.5 CT 0.032 SELFWEIGHT JOINT WEIGHT 17 TO 48 WEIGHT 7.0 49 TO 64 WEIGHT 3.5

There are two stages in the command specification of the UBC loads. The first stage is initiated with the command DEFINE UBC LOAD. Here we specify parameters such as Zone factor, Importance factor, site coefficient for soil characteristics etc. and,


128

the vertical loads (weights) from which the base shear will be calculated. The vertical loads may be specified in the form of selfweight, joint weights and/or member weights. Member weights are not shown in this example. It is important to note that these vertical loads are used purely in the determination of the horizontal base shear only. In other words, the structure is not analysed for these vertical loads.

LOAD 1 UBC LOAD X 0.75 SELFWEIGHT Y -1.0 JOINT LOADS 17 TO 48 FY -7.0 49 TO 64 FY -3.5

This is the second stage in which the UBC load is applied with the help of a load case number, corresponding direction (X in the above case) and a factor by which the generated horizontal loads should be multiplied. Along with the UBC load, deadweight is also added to the same load case. Since we will be doing second-order (PDELTA) analysis, it is important that we include horizontal and vertical loads in the same load case.

LOAD 2 UBC LOAD Z 0.75 SELFWEIGHT Y -1.0 JOINT LOADS 17 TO 48 FY –7.0 49 TO 64 FY –3.5

In load case 2, the UBC load is being applied in the Z direction. Vertical loads too are part of this case.

PDELTA ANALYSIS PRINT LOAD DATA We are requesting a second-order analysis by specifying the command PDELTA ANALYSIS. PRINT LOAD DATA is used to obtain a report of all the applied and generated loadings.

Example Problem 14

129 PRINT SUPPORT REACTIONS FINISH

The above commands are self-explanatory.


130


1. STAAD SPACE EXAMPLE PROBLEM FOR UBC LOAD2. UNIT METER KNS3. JOINT COORDINATES4. 1 0 0 0 4 10.5 0 05. REPEAT 3 0 0 3.56. REPEAT ALL 3 0 3.5 07. MEMBER INCIDENCES8. * BEAMS IN X DIRECTION9. 101 17 18 103

10. 104 21 22 10611. 107 25 26 10912. 110 29 30 11213. REPEAT ALL 2 12 1614. * BEAMS IN Z DIRECTION15. 201 17 21 20416. 205 21 25 20817. 209 25 29 21218. REPEAT ALL 2 12 1619. * COLUMNS20. 301 1 17 34821. MEMBER PROPERTIES BRITISH22. 101 TO 136 201 TO 236 PRIS YD 0.40 ZD 0.3023. 301 TO 348 TA ST UB457X152X5224. CONSTANT25. E STEEL MEMB 301 TO 34826. E CONCRETE MEMB 101 TO 136 201 TO 23627. DENSITY STEEL MEMB 301 TO 34828. DENSITY CONCRETE MEMB 101 TO 136 201 TO 23629. POISSON STEEL MEMB 301 TO 34830. POISSON CONCRETE MEMB 101 TO 136 201 TO 23631. SUPPORT32. 1 TO 16 FIXED33. DEFINE UBC LOAD34. ZONE 0.2 I 1.0 RWX 9 RWZ 9 S 1.5 CT 0.03235. SELFWEIGHT36. JOINT WEIGHT37. 17 TO 48 WEIGHT 7.038. 49 TO 64 WEIGHT 3.539. LOAD 140. UBC LOAD X 0.7541. SELFWEIGHT Y -1.042. JOINT LOAD43. 17 TO 48 FY -7.044. 49 TO 64 FY -3.545. LOAD 246. UBC LOAD Z 0.7547. SELFWEIGHT Y -1.048. JOINT LOAD49. 17 TO 48 FY -7.050. 49 TO 64 FY -3.551. PDELTA ANALYSIS PRINT LOAD DATA



Example Problem 14

131 LOADING 1-----------

SELFWEIGHT Y -1.000


JOINT LOAD - UNIT KNS METE

JOINT FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM-Z

17 0.00 -7.00 0.00 0.00 0.00 0.0018 0.00 -7.00 0.00 0.00 0.00 0.0019 0.00 -7.00 0.00 0.00 0.00 0.0020 0.00 -7.00 0.00 0.00 0.00 0.0021 0.00 -7.00 0.00 0.00 0.00 0.0022 0.00 -7.00 0.00 0.00 0.00 0.0023 0.00 -7.00 0.00 0.00 0.00 0.0024 0.00 -7.00 0.00 0.00 0.00 0.0025 0.00 -7.00 0.00 0.00 0.00 0.0026 0.00 -7.00 0.00 0.00 0.00 0.0027 0.00 -7.00 0.00 0.00 0.00 0.0028 0.00 -7.00 0.00 0.00 0.00 0.0029 0.00 -7.00 0.00 0.00 0.00 0.0030 0.00 -7.00 0.00 0.00 0.00 0.0031 0.00 -7.00 0.00 0.00 0.00 0.0032 0.00 -7.00 0.00 0.00 0.00 0.0033 0.00 -7.00 0.00 0.00 0.00 0.0034 0.00 -7.00 0.00 0.00 0.00 0.0035 0.00 -7.00 0.00 0.00 0.00 0.0036 0.00 -7.00 0.00 0.00 0.00 0.0037 0.00 -7.00 0.00 0.00 0.00 0.0038 0.00 -7.00 0.00 0.00 0.00 0.0039 0.00 -7.00 0.00 0.00 0.00 0.0040 0.00 -7.00 0.00 0.00 0.00 0.0041 0.00 -7.00 0.00 0.00 0.00 0.0042 0.00 -7.00 0.00 0.00 0.00 0.0043 0.00 -7.00 0.00 0.00 0.00 0.0044 0.00 -7.00 0.00 0.00 0.00 0.0045 0.00 -7.00 0.00 0.00 0.00 0.0046 0.00 -7.00 0.00 0.00 0.00 0.0047 0.00 -7.00 0.00 0.00 0.00 0.0048 0.00 -7.00 0.00 0.00 0.00 0.0049 0.00 -3.50 0.00 0.00 0.00 0.0050 0.00 -3.50 0.00 0.00 0.00 0.0051 0.00 -3.50 0.00 0.00 0.00 0.0052 0.00 -3.50 0.00 0.00 0.00 0.0053 0.00 -3.50 0.00 0.00 0.00 0.0054 0.00 -3.50 0.00 0.00 0.00 0.0055 0.00 -3.50 0.00 0.00 0.00 0.0056 0.00 -3.50 0.00 0.00 0.00 0.0057 0.00 -3.50 0.00 0.00 0.00 0.0058 0.00 -3.50 0.00 0.00 0.00 0.0059 0.00 -3.50 0.00 0.00 0.00 0.0060 0.00 -3.50 0.00 0.00 0.00 0.0061 0.00 -3.50 0.00 0.00 0.00 0.0062 0.00 -3.50 0.00 0.00 0.00 0.0063 0.00 -3.50 0.00 0.00 0.00 0.0064 0.00 -3.50 0.00 0.00 0.00 0.00


132 LOADING 2-----------

SELFWEIGHT Y -1.000




17 0.00 -7.00 0.00 0.00 0.00 0.0018 0.00 -7.00 0.00 0.00 0.00 0.0019 0.00 -7.00 0.00 0.00 0.00 0.0020 0.00 -7.00 0.00 0.00 0.00 0.0021 0.00 -7.00 0.00 0.00 0.00 0.0022 0.00 -7.00 0.00 0.00 0.00 0.0023 0.00 -7.00 0.00 0.00 0.00 0.0024 0.00 -7.00 0.00 0.00 0.00 0.0025 0.00 -7.00 0.00 0.00 0.00 0.0026 0.00 -7.00 0.00 0.00 0.00 0.0027 0.00 -7.00 0.00 0.00 0.00 0.0028 0.00 -7.00 0.00 0.00 0.00 0.0029 0.00 -7.00 0.00 0.00 0.00 0.0030 0.00 -7.00 0.00 0.00 0.00 0.0031 0.00 -7.00 0.00 0.00 0.00 0.0032 0.00 -7.00 0.00 0.00 0.00 0.0033 0.00 -7.00 0.00 0.00 0.00 0.0034 0.00 -7.00 0.00 0.00 0.00 0.0035 0.00 -7.00 0.00 0.00 0.00 0.0036 0.00 -7.00 0.00 0.00 0.00 0.0037 0.00 -7.00 0.00 0.00 0.00 0.0038 0.00 -7.00 0.00 0.00 0.00 0.0039 0.00 -7.00 0.00 0.00 0.00 0.0040 0.00 -7.00 0.00 0.00 0.00 0.0041 0.00 -7.00 0.00 0.00 0.00 0.0042 0.00 -7.00 0.00 0.00 0.00 0.0043 0.00 -7.00 0.00 0.00 0.00 0.0044 0.00 -7.00 0.00 0.00 0.00 0.0045 0.00 -7.00 0.00 0.00 0.00 0.0046 0.00 -7.00 0.00 0.00 0.00 0.0047 0.00 -7.00 0.00 0.00 0.00 0.0048 0.00 -7.00 0.00 0.00 0.00 0.0049 0.00 -3.50 0.00 0.00 0.00 0.0050 0.00 -3.50 0.00 0.00 0.00 0.0051 0.00 -3.50 0.00 0.00 0.00 0.0052 0.00 -3.50 0.00 0.00 0.00 0.0053 0.00 -3.50 0.00 0.00 0.00 0.0054 0.00 -3.50 0.00 0.00 0.00 0.0055 0.00 -3.50 0.00 0.00 0.00 0.0056 0.00 -3.50 0.00 0.00 0.00 0.0057 0.00 -3.50 0.00 0.00 0.00 0.0058 0.00 -3.50 0.00 0.00 0.00 0.0059 0.00 -3.50 0.00 0.00 0.00 0.0060 0.00 -3.50 0.00 0.00 0.00 0.0061 0.00 -3.50 0.00 0.00 0.00 0.0062 0.00 -3.50 0.00 0.00 0.00 0.0063 0.00 -3.50 0.00 0.00 0.00 0.0064 0.00 -3.50 0.00 0.00 0.00 0.00

************************************************************ ** X DIRECTION : Ta = 0.455 Tb = 0.285 Tuser = 0.000 ** C = 2.7500, LOAD FACTOR = 0.750 ** UBC TYPE = 94 ** UBC FACTOR V = 0.0611 X 1078.45 = 65.91 KNS ** ************************************************************

************************************************************ ** Z DIRECTION : Ta = 0.455 Tb = 1.103 Tuser = 0.000 ** C = 2.7500, LOAD FACTOR = 0.750 ** UBC TYPE = 94 ** UBC FACTOR V = 0.0611 X 1078.45 = 65.91 KNS ** ************************************************************

Example Problem 14

133 JOINT LATERAL TORSIONAL LOAD - 1

LOAD (KNS ) MOMENT (KNS -METE) FACTOR - 0.750----- ------- ---------

17 FX 0.449 MY 0.00018 FX 0.568 MY 0.00019 FX 0.568 MY 0.00020 FX 0.449 MY 0.00021 FX 0.568 MY 0.00022 FX 0.687 MY 0.00023 FX 0.687 MY 0.00024 FX 0.568 MY 0.00025 FX 0.568 MY 0.00026 FX 0.687 MY 0.00027 FX 0.687 MY 0.00028 FX 0.568 MY 0.00029 FX 0.449 MY 0.00030 FX 0.568 MY 0.00031 FX 0.568 MY 0.00032 FX 0.449 MY 0.000

----------- -----------TOTAL = 9.083 0.000 AT LEVEL 3.500 METE






134 JOINT LATERAL TORSIONAL LOAD - 2

LOAD (KNS ) MOMENT (KNS -METE) FACTOR - 0.750----- ------- ---------

17 FZ 0.449 MY 0.00018 FZ 0.568 MY 0.00019 FZ 0.568 MY 0.00020 FZ 0.449 MY 0.00021 FZ 0.568 MY 0.00022 FZ 0.687 MY 0.00023 FZ 0.687 MY 0.00024 FZ 0.568 MY 0.00025 FZ 0.568 MY 0.00026 FZ 0.687 MY 0.00027 FZ 0.687 MY 0.00028 FZ 0.568 MY 0.00029 FZ 0.449 MY 0.00030 FZ 0.568 MY 0.00031 FZ 0.568 MY 0.00032 FZ 0.449 MY 0.000






++ Adjusting Displacements 8:39:50


Example Problem 14

135 52. PRINT SUPPORT REACTIONS



1 1 -2.22 43.36 0.06 0.07 0.00 5.752 0.45 41.43 -3.03 -5.61 0.00 -0.49

2 1 -3.40 65.55 0.06 0.07 0.00 7.022 0.01 56.12 -3.00 -5.62 0.00 -0.02

3 1 -3.42 64.35 0.06 0.07 0.00 7.062 -0.01 56.12 -3.00 -5.62 0.00 0.02

4 1 -3.12 57.12 0.06 0.07 0.00 6.732 -0.45 41.43 -3.03 -5.61 0.00 0.49

5 1 -2.31 62.76 -0.01 -0.01 0.00 5.962 0.45 73.81 -3.14 -5.82 0.00 -0.49

6 1 -3.51 85.18 -0.01 -0.01 0.00 7.262 0.01 88.52 -3.11 -5.83 0.00 -0.02

7 1 -3.53 83.95 -0.01 -0.01 0.00 7.292 -0.01 88.52 -3.11 -5.83 0.00 0.02

8 1 -3.21 76.95 -0.01 -0.01 0.00 6.942 -0.45 73.81 -3.14 -5.82 0.00 0.49

9 1 -2.31 62.76 0.01 0.01 0.00 5.962 0.45 65.90 -3.14 -5.80 0.00 -0.49

10 1 -3.51 85.18 0.01 0.01 0.00 7.262 0.01 80.60 -3.11 -5.81 0.00 -0.02

11 1 -3.53 83.95 0.01 0.01 0.00 7.292 -0.01 80.60 -3.11 -5.81 0.00 0.02

12 1 -3.21 76.95 0.01 0.01 0.00 6.942 -0.45 65.90 -3.14 -5.80 0.00 0.49

13 1 -2.22 43.36 -0.06 -0.07 0.00 5.752 0.45 59.06 -3.11 -5.75 0.00 -0.49

14 1 -3.40 65.55 -0.06 -0.07 0.00 7.022 0.01 73.79 -3.08 -5.76 0.00 -0.02

15 1 -3.42 64.35 -0.06 -0.07 0.00 7.062 -0.01 73.79 -3.08 -5.76 0.00 0.02

16 1 -3.12 57.12 -0.06 -0.07 0.00 6.732 -0.45 59.06 -3.11 -5.75 0.00 0.49


53. FINISH


**** DATE= TIME= ****



136

NOTES

Example Problem 15

137


A space frame is analyzed for loads generated using the built-in wind and floor load generation facilities.

1 2 3

4 5

6 7 8

9 10 11

12 13

14 15 16

17 18 19

20 21

22 23 24

1 23

4 5

6 7 8

9 10 11

12 13

14 15

16

17 18

19

20 21

22 23

24

25 26

27

28

29

3031

32

33 35

34

X

Y

Z


138 STAAD SPACE - WIND AND FLOOR LOAD GENERATION

This is a SPACE frame analysis problem. Every STAAD input has to start with the command STAAD. The SPACE specification is used to denote a SPACE frame.

UNIT METER KNS

The UNIT specification is used to specify the length and/or force units to be used.

JOINT COORDINATES 1 0 0 0 2 4 0 0 3 9 0 0 4 0 0 4 5 4 0 4 6 0 0 8 7 4 0 8 8 9 0 8 REPEAT ALL 2 0 3.5 0

The JOINT COORDINATE specification is used to specify the X, Y and Z coordinates of the JOINTs. Note that the REPEAT ALL command has been used to generate JOINTs for the two upper storeys each with a Y increment of 3.5 m.

MEMBER INCIDENCES * Columns 1 1 9 16 * Beams in the X direction 17 9 10 18 19 12 13 20 14 15 21 22 17 18 23 24 20 21 25 22 23 26 * Beams in the Z direction 27 9 12 ; 28 12 14 ; 29 10 13 ; 30 13 15 ; 31 11 16 32 17 20 ; 33 20 22 ; 34 18 21 ; 35 21 23 ; 36 19 24

Example Problem 15

139

The MEMBER INCIDENCE specification is used for specifying MEMBER connectivities.

MEMBER PROPERTIES BRITISH 1 TO 16 TA ST UB457X191X74 17 TO 26 TA ST UB457X152X52 27 TO 36 TA ST UB457X152X52

Properties for all members are specified from the built-in BRITISH steel table. Three different sections have been used.

CONSTANT E STEEL ALL DENSITY STEEL ALL POISSON STEEL ALL

The CONSTANT specification is used to specify material properties. In this case, the built-in default values have been used.

SUPPORT 1 TO 8 FIXED BUT MX MZ

The SUPPORTs of the structure are defined through the SUPPORT specification. Here all the supports are FIXED with RELEASES specified in the MX (rotation about global X-axis) and MZ (rotation about global Z-axis) directions.

DEFINE WIND LOAD TYPE 1 INTENSITY 1.0 1.5 HEIGHT 3.5 7.0 EXPOSURE 0.90 YRANGE 6.0 8.0 EXPOSURE 0.85 JOINT 9 12 14

When a structure has to be analysed for wind loading, the engineer is confronted with the task of first converting an abstract quantity like wind velocity or wind pressure into concentrated loads at joints, distributed loads on members, or pressure loads on plates. The large number of calculations involved in this conversion can be avoided by making use of STAAD’s wind load generation utility. This utility takes wind pressure at various heights as the input, and converts them to values that can then be used as


140

concentrated forces known as joint loads in specific load cases. The input specification is done in two stages. The first stage is initiated above through the DEFINE WIND LOAD command. The basic parameters of the WIND loading are specified here. All values need to be provided in the current UNIT system. Each wind category is identified with a TYPE number (an identification mark) which is used later to specify load cases.

In this example, two different wind intensities (1.0 KN/sq. m and 1.5 KN/sq. m) are specified for two different height zones (0 to 3.5m and 3.5 to 7.0m). The EXPOSURE specification is used to mitigate or magnify the effect at specific nodes due to special considerations like openings in the structure. In this case, two different exposure factors are specified. The first EXPOSURE specification specifies the exposure factor as 0.9 for all joints within the height range (defined as global Y-range) of 6.0m – 8.0m. The second EXPOSURE specification specifies the exposure factor as 0.85 for joints 9, 12 and 14. In the EXPOSURE factor specification, the joints may be specified directly or through a vertical range specification.

LOAD 1 WIND LOAD IN X-DIRECTION WIND LOAD X 1.2 TYPE 1

This is the second stage of input specification for the wind load generation. The term WIND LOAD and the direction term that follows are used to specify the WIND LOADING in a particular lateral direction. In this case, WIND loading TYPE 1, defined previously, is being applied in the global X-direction with a positive multiplication factor of 1.2 .

LOAD 2 FLOOR LOAD @ Y = 3.5M AND 7M FLOOR LOAD YRANGE 3.4 3.6 FLOAD –5.0 XRANGE 0.0 4.0 ZRANGE 0.0 8.0 YRANGE 3.4 3.6 FLOAD –2.5 XRANGE 4.0 9.0 ZRANGE 0.0 8.0 YRANGE 6.9 7.1 FLOAD –2.5

In load case 2 in this problem, a floor load generation is performed. In a floor load generation, a pressure load (force per unit area) is converted by the program into specific points forces

Example Problem 15

141

and distributed forces on the members located in that region. The YRANGE, XRANGE and ZRANGE specifications are used to define the area of the structure on which the pressure is acting. The FLOAD specification is used to specify the value of that pressure. All values need to be provided in the current UNIT system. For example, in the first line in the above FLOOR LOAD specification, the region is defined as being located within the bounds YRANGE of 3.4 – 3.6 m, XRANGE of 0.0 - 4.0 m and ZRANGE of 0.0 - 8.0 m. The –5.0 signifies that the pressure is 5.0 KN/sq.m. in the negative global Y direction. The program will identify the members lying within the specified region and derive MEMBER LOADS on these members based on two-way load distribution.

PERFORM ANALYSIS PRINT LOAD DATA

We can view the values and position of the generated loads with the help of the PRINT LOAD DATA command used above along with the PERFORM ANALYSIS command.

PRINT SUPPORT REACTION FINISH

Above commands are self-explanatory.



1. STAAD SPACE - WIND AND FLOOR LOAD GENERATION2. UNIT METER KNS3. JOINT COORDINATES4. 1 0 0 05. 2 4 0 06. 3 9 0 07. 4 0 0 48. 5 4 0 49. 6 0 0 8

10. 7 4 0 811. 8 9 0 812. REPEAT ALL 2 0 3.5 013. MEMBER INCIDENCES14. * COLUMNS15. 1 1 9 1616. * BEAMS IN THE X DIRECTION17. 17 9 10 1818. 19 12 1319. 20 14 15 2120. 22 17 18 2321. 24 20 2122. 25 22 23 2623. * BEAMS IN THE Z DIRECTION24. 27 9 12 ; 28 12 14 ; 29 10 13 ; 30 13 15 ; 31 11 1625. 32 17 20 ; 33 20 22 ; 34 18 21 ; 35 21 23 ; 36 19 2426. MEMBER PROPERTIES BRITISH27. 1 TO 16 TA ST UB457X191X7428. 17 TO 26 TA ST UB457X152X5229. 27 TO 36 TA ST UB457X152X5230. CONSTANT31. E STEEL ALL32. DENSITY STEEL ALL33. POISSON STEEL ALL34. SUPPORT35. 1 TO 8 FIXED BUT MX MZ36. DEFINE WIND LOAD37. TYPE 138. INTENSITY 1.0 1.5 HEIGHT 3.5 7.039. EXPOSURE 0.90 YRANGE 6.0 8.040. EXPOSURE 0.85 JOINT 9 12 1441. LOAD 1 WIND LOAD IN X-DIRECTION42. WIND LOAD X 1.2 TYPE 143. LOAD 2 FLOOR LOAD AT Y = 3.5M AND 7M44. FLOOR LOAD45. YRANGE 3.4 3.6 FLOAD -5.0 XRANGE 0.0 4.0 ZRANGE 0.0 8.046. YRANGE 3.4 3.6 FLOAD -2.5 XRANGE 4.0 9.0 ZRANGE 0.0 8.047. YRANGE 6.9 7.1 FLOAD -2.548. PERFORM ANALYSIS PRINT LOAD DATA



Example Problem 15

143 LOADING 1 WIND LOAD IN X-DIRECTION-----------



1 4.20 0.00 0.00 0.00 0.00 0.004 8.40 0.00 0.00 0.00 0.00 0.006 4.20 0.00 0.00 0.00 0.00 0.009 8.93 0.00 0.00 0.00 0.00 0.00

12 17.85 0.00 0.00 0.00 0.00 0.0014 8.93 0.00 0.00 0.00 0.00 0.0017 5.67 0.00 0.00 0.00 0.00 0.0020 11.34 0.00 0.00 0.00 0.00 0.0022 5.67 0.00 0.00 0.00 0.00 0.00

LOADING 2 FLOOR LOAD AT Y = 3.5M AND 7M-----------



17 -0.156 GY 0.1717 -0.469 GY 0.3917 -0.781 GY 0.6317 -1.094 GY 0.8817 -1.406 GY 1.1317 -1.719 GY 1.3817 -2.031 GY 1.6317 -2.344 GY 1.8817 -2.344 GY 2.1217 -2.031 GY 2.3717 -1.719 GY 2.6217 -1.406 GY 2.8717 -1.094 GY 3.1217 -0.781 GY 3.3717 -0.469 GY 3.6117 -0.156 GY 3.8329 -0.156 GY 0.1729 -0.469 GY 0.3929 -0.781 GY 0.6329 -1.094 GY 0.8829 -1.406 GY 1.1329 -1.719 GY 1.3829 -2.031 GY 1.6329 -2.344 GY 1.8829 -2.344 GY 2.1229 -2.031 GY 2.3729 -1.719 GY 2.6229 -1.406 GY 2.8729 -1.094 GY 3.1229 -0.781 GY 3.3729 -0.469 GY 3.6129 -0.156 GY 3.8319 -0.156 GY 0.1719 -0.469 GY 0.3919 -0.781 GY 0.6319 -1.094 GY 0.8819 -1.406 GY 1.1319 -1.719 GY 1.3819 -2.031 GY 1.6319 -2.344 GY 1.8819 -2.344 GY 2.1219 -2.031 GY 2.3719 -1.719 GY 2.6219 -1.406 GY 2.8719 -1.094 GY 3.1219 -0.781 GY 3.3719 -0.469 GY 3.6119 -0.156 GY 3.8327 -0.156 GY 0.1727 -0.469 GY 0.3927 -0.781 GY 0.6327 -1.094 GY 0.88


144 MEMBER LOAD - UNIT KNS METE


27 -1.406 GY 1.1327 -1.719 GY 1.3827 -2.031 GY 1.6327 -2.344 GY 1.8827 -2.344 GY 2.1227 -2.031 GY 2.3727 -1.719 GY 2.6227 -1.406 GY 2.8727 -1.094 GY 3.1227 -0.781 GY 3.3727 -0.469 GY 3.6127 -0.156 GY 3.8319 -0.156 GY 0.1719 -0.469 GY 0.3919 -0.781 GY 0.6319 -1.094 GY 0.8819 -1.406 GY 1.1319 -1.719 GY 1.3819 -2.031 GY 1.6319 -2.344 GY 1.8819 -2.344 GY 2.1219 -2.031 GY 2.3719 -1.719 GY 2.6219 -1.406 GY 2.8719 -1.094 GY 3.1219 -0.781 GY 3.3719 -0.469 GY 3.6119 -0.156 GY 3.8330 -0.156 GY 0.1730 -0.469 GY 0.3930 -0.781 GY 0.6330 -1.094 GY 0.8830 -1.406 GY 1.1330 -1.719 GY 1.3830 -2.031 GY 1.6330 -2.344 GY 1.8830 -2.344 GY 2.1230 -2.031 GY 2.3730 -1.719 GY 2.6230 -1.406 GY 2.8730 -1.094 GY 3.1230 -0.781 GY 3.3730 -0.469 GY 3.6130 -0.156 GY 3.8320 -0.156 GY 0.1720 -0.469 GY 0.3920 -0.781 GY 0.6320 -1.094 GY 0.8820 -1.406 GY 1.1320 -1.719 GY 1.3820 -2.031 GY 1.6320 -2.344 GY 1.8820 -2.344 GY 2.1220 -2.031 GY 2.3720 -1.719 GY 2.6220 -1.406 GY 2.8720 -1.094 GY 3.1220 -0.781 GY 3.3720 -0.469 GY 3.6120 -0.156 GY 3.8328 -0.156 GY 0.1728 -0.469 GY 0.3928 -0.781 GY 0.6328 -1.094 GY 0.8828 -1.406 GY 1.1328 -1.719 GY 1.3828 -2.031 GY 1.6328 -2.344 GY 1.8828 -2.344 GY 2.1228 -2.031 GY 2.3728 -1.719 GY 2.6228 -1.406 GY 2.8728 -1.094 GY 3.12

Example Problem 15



28 -0.781 GY 3.3728 -0.469 GY 3.6128 -0.156 GY 3.8318 -0.122 GY 0.2118 -0.366 GY 0.4918 -0.610 GY 0.7918 -0.854 GY 1.1018 -1.099 GY 1.4118 -1.343 GY 1.7218 -1.587 GY 2.0418 -1.831 GY 2.3518 -1.831 GY 2.6518 -1.587 GY 2.9618 -1.343 GY 3.2818 -1.099 GY 3.5918 -0.854 GY 3.9018 -0.610 GY 4.2118 -0.366 GY 4.5118 -0.122 GY 4.7931 -0.122 GY 0.2131 -0.366 GY 0.4931 -0.610 GY 0.7931 -0.854 GY 1.1031 -1.099 GY 1.4131 -1.343 GY 1.7231 -1.587 GY 2.0431 -1.831 GY 2.3531 -6.250 GY 2.50 5.5031 -1.831 GY 5.6531 -1.587 GY 5.9631 -1.343 GY 6.2831 -1.099 GY 6.5931 -0.854 GY 6.9031 -0.610 GY 7.2131 -0.366 GY 7.5131 -0.122 GY 7.7921 -0.122 GY 0.2121 -0.366 GY 0.4921 -0.610 GY 0.7921 -0.854 GY 1.1021 -1.099 GY 1.4121 -1.343 GY 1.7221 -1.587 GY 2.0421 -1.831 GY 2.3521 -1.831 GY 2.6521 -1.587 GY 2.9621 -1.343 GY 3.2821 -1.099 GY 3.5921 -0.854 GY 3.9021 -0.610 GY 4.2121 -0.366 GY 4.5121 -0.122 GY 4.7930 -1.831 GY 1.6530 -1.587 GY 1.9630 -1.343 GY 2.2830 -1.099 GY 2.5930 -0.854 GY 2.9030 -0.610 GY 3.2130 -0.366 GY 3.5130 -0.122 GY 3.7930 -6.250 GY 0.00 1.5029 -0.122 GY 0.2129 -0.366 GY 0.4929 -0.610 GY 0.7929 -0.854 GY 1.1029 -1.099 GY 1.4129 -1.343 GY 1.7229 -1.587 GY 2.0429 -1.831 GY 2.3529 -6.250 GY 2.50 4.00




22 -0.078 GY 0.1722 -0.234 GY 0.3922 -0.391 GY 0.6322 -0.547 GY 0.8822 -0.703 GY 1.1322 -0.859 GY 1.3822 -1.016 GY 1.6322 -1.172 GY 1.8822 -1.172 GY 2.1222 -1.016 GY 2.3722 -0.859 GY 2.6222 -0.703 GY 2.8722 -0.547 GY 3.1222 -0.391 GY 3.3722 -0.234 GY 3.6122 -0.078 GY 3.8334 -0.078 GY 0.1734 -0.234 GY 0.3934 -0.391 GY 0.6334 -0.547 GY 0.8834 -0.703 GY 1.1334 -0.859 GY 1.3834 -1.016 GY 1.6334 -1.172 GY 1.8834 -1.172 GY 2.1234 -1.016 GY 2.3734 -0.859 GY 2.6234 -0.703 GY 2.8734 -0.547 GY 3.1234 -0.391 GY 3.3734 -0.234 GY 3.6134 -0.078 GY 3.8324 -0.078 GY 0.1724 -0.234 GY 0.3924 -0.391 GY 0.6324 -0.547 GY 0.8824 -0.703 GY 1.1324 -0.859 GY 1.3824 -1.016 GY 1.6324 -1.172 GY 1.8824 -1.172 GY 2.1224 -1.016 GY 2.3724 -0.859 GY 2.6224 -0.703 GY 2.8724 -0.547 GY 3.1224 -0.391 GY 3.3724 -0.234 GY 3.6124 -0.078 GY 3.8332 -0.078 GY 0.1732 -0.234 GY 0.3932 -0.391 GY 0.6332 -0.547 GY 0.8832 -0.703 GY 1.1332 -0.859 GY 1.3832 -1.016 GY 1.6332 -1.172 GY 1.8832 -1.172 GY 2.1232 -1.016 GY 2.3732 -0.859 GY 2.6232 -0.703 GY 2.8732 -0.547 GY 3.1232 -0.391 GY 3.3732 -0.234 GY 3.6132 -0.078 GY 3.8323 -0.122 GY 0.2123 -0.366 GY 0.4923 -0.610 GY 0.7923 -0.854 GY 1.1023 -1.099 GY 1.4123 -1.343 GY 1.7223 -1.587 GY 2.0423 -1.831 GY 2.3523 -1.831 GY 2.65

Example Problem 15



23 -1.587 GY 2.9623 -1.343 GY 3.2823 -1.099 GY 3.5923 -0.854 GY 3.9023 -0.610 GY 4.2123 -0.366 GY 4.5123 -0.122 GY 4.7936 -0.122 GY 0.2136 -0.366 GY 0.4936 -0.610 GY 0.7936 -0.854 GY 1.1036 -1.099 GY 1.4136 -1.343 GY 1.7236 -1.587 GY 2.0436 -1.831 GY 2.3536 -6.250 GY 2.50 5.5036 -1.831 GY 5.6536 -1.587 GY 5.9636 -1.343 GY 6.2836 -1.099 GY 6.5936 -0.854 GY 6.9036 -0.610 GY 7.2136 -0.366 GY 7.5136 -0.122 GY 7.7926 -0.122 GY 0.2126 -0.366 GY 0.4926 -0.610 GY 0.7926 -0.854 GY 1.1026 -1.099 GY 1.4126 -1.343 GY 1.7226 -1.587 GY 2.0426 -1.831 GY 2.3526 -1.831 GY 2.6526 -1.587 GY 2.9626 -1.343 GY 3.2826 -1.099 GY 3.5926 -0.854 GY 3.9026 -0.610 GY 4.2126 -0.366 GY 4.5126 -0.122 GY 4.7935 -1.831 GY 1.6535 -1.587 GY 1.9635 -1.343 GY 2.2835 -1.099 GY 2.5935 -0.854 GY 2.9035 -0.610 GY 3.2135 -0.366 GY 3.5135 -0.122 GY 3.7935 -6.250 GY 0.00 1.5034 -0.122 GY 0.2134 -0.366 GY 0.4934 -0.610 GY 0.7934 -0.854 GY 1.1034 -1.099 GY 1.4134 -1.343 GY 1.7234 -1.587 GY 2.0434 -1.831 GY 2.3534 -6.250 GY 2.50 4.0024 -0.078 GY 0.1724 -0.234 GY 0.3924 -0.391 GY 0.6324 -0.547 GY 0.8824 -0.703 GY 1.1324 -0.859 GY 1.3824 -1.016 GY 1.6324 -1.172 GY 1.8824 -1.172 GY 2.1224 -1.016 GY 2.3724 -0.859 GY 2.6224 -0.703 GY 2.8724 -0.547 GY 3.1224 -0.391 GY 3.3724 -0.234 GY 3.61




24 -0.078 GY 3.8335 -0.078 GY 0.1735 -0.234 GY 0.3935 -0.391 GY 0.6335 -0.547 GY 0.8835 -0.703 GY 1.1335 -0.859 GY 1.3835 -1.016 GY 1.6335 -1.172 GY 1.8835 -1.172 GY 2.1235 -1.016 GY 2.3735 -0.859 GY 2.6235 -0.703 GY 2.8735 -0.547 GY 3.1235 -0.391 GY 3.3735 -0.234 GY 3.6135 -0.078 GY 3.8325 -0.078 GY 0.1725 -0.234 GY 0.3925 -0.391 GY 0.6325 -0.547 GY 0.8825 -0.703 GY 1.1325 -0.859 GY 1.3825 -1.016 GY 1.6325 -1.172 GY 1.8825 -1.172 GY 2.1225 -1.016 GY 2.3725 -0.859 GY 2.6225 -0.703 GY 2.8725 -0.547 GY 3.1225 -0.391 GY 3.3725 -0.234 GY 3.6125 -0.078 GY 3.8333 -0.078 GY 0.1733 -0.234 GY 0.3933 -0.391 GY 0.6333 -0.547 GY 0.8833 -0.703 GY 1.1333 -0.859 GY 1.3833 -1.016 GY 1.6333 -1.172 GY 1.8833 -1.172 GY 2.1233 -1.016 GY 2.3733 -0.859 GY 2.6233 -0.703 GY 2.8733 -0.547 GY 3.1233 -0.391 GY 3.3733 -0.234 GY 3.6133 -0.078 GY 3.83


Example Problem 15

149 49. PRINT SUPPORT REACTION



1 1 -9.58 -11.46 -0.01 0.00 -0.01 0.002 0.80 25.70 0.16 0.00 0.00 0.00

2 1 -6.70 3.60 0.01 0.00 0.00 0.002 -0.15 52.75 0.26 0.00 0.00 0.00

3 1 -5.01 7.95 0.00 0.00 0.00 0.002 -0.71 49.30 1.16 0.00 0.00 0.00

4 1 -20.54 -28.14 0.00 0.00 0.00 0.002 1.72 66.85 0.00 0.00 0.00 0.00

5 1 -12.05 27.97 0.00 0.00 0.00 0.002 -1.61 117.68 0.00 0.00 0.00 0.00

6 1 -9.58 -11.46 0.01 0.00 0.01 0.002 0.80 25.70 -0.16 0.00 0.00 0.00

7 1 -6.70 3.60 -0.01 0.00 0.00 0.002 -0.15 52.75 -0.26 0.00 0.00 0.00

8 1 -5.01 7.95 0.00 0.00 0.00 0.002 -0.71 49.30 -1.16 0.00 0.00 0.00


50. FINISH


**** DATE= TIME= ****



Example Problem 15

150

NOTES

Example Problem 16

151


Dynamic Analysis (Time History) is performed for a 3 span beam with concentrated and distributed masses. The structure is subjected to “forcing function” and “ground motion” loading. The maxima of joint displacements, member end forces and support reactions are determined.


152 STAAD PLANE EXAMPLE FOR TIME HISTORY ANALYSIS

Every input file has to start with the word STAAD. The word PLANE signifies that the structure is a plane frame.

UNITS CMS KNS

Specifies the units to be used. JOINT COORDINATES 1 0.0 0.0 0.0 2 0.0 120.0 0.0 3 0.0 240.0 0.0 4 0.0 360.0 0.0

Joint number followed by the X, Y and Z coordinates are specified above.

MEMBER INCIDENCES 1 1 2 3

Incidences of members 1 to 3 are specified above. MEMBER PROPERTIES 1 2 3 PRIS AX 100.0 IZ 833.33

All the members have "PRISMATIC" property specification. Since this is a PLANE frame, Area of cross section "AX", and Moment of Inertia "IZ" about the Z axis are adequate for the analysis.

SUPPORTS 1 4 PINNED

Pinned supports are located at nodes 1 and 4. CONSTANTS E 2850 ALL DENSITY 25E-6 ALL POISSON CONCRETE ALL

The material constants defined include Young's Modulus "E", density and Poisson’s ratio.

Example Problem 16

153 UNIT NEWTON METER DEFINE TIME HISTORY TYPE 1 FORCE 0.0 –20.0 0.5 100.0 1.0 200.0 1.5 500.0 2.0 800.0 2.5 500.0 3.0 70.0 TYPE 2 ACCELERATION 0.0 0.1 0.5 –0.25 1.0 –0.5 1.5 –0.9 2.0 –1.3 2.5 –1.0 3.0 –0.7 ARRIVAL TIMES 0.0 DAMPING 0.075 There are 2 stages in the command specification required for a time history analysis. The first stage is defined above. First the characteristics of the time varying load are provided. The loading type may be a forcing function (vibrating machinery) or ground motion (earthquake). The former is input in the form of time-force pairs while the latter is in the form of time-acceleration pairs. Following this data, all possible arrival times for these loads on the structure as well as the modal damping ratio are specified. In this example, the damping ratio is the same (7.5%) for all modes.

LOAD 1 STATIC LOAD MEMBER LOAD 1 2 3 UNI GX 500.0

Load case 1 above is a static load. A uniformly distributed force of 500 Newton/m acts along the global X direction on all 3 members.

LOAD 2 TIME HISTORY LOAD SELFWEIGHT X 1.0 SELFWEIGHT Y 1.0 JOINT LOAD 2 3 FX 4000.0 TIME LOAD 2 3 FX 1 1 GROUND MOTION X 2 1

This is the second stage in the command specification for time history analysis. This involves the application of the time varying load on the structure. The masses that constitute the mass matrix of the structure are specified through the selfweight and joint load commands. The program will extract the lumped masses from these


154

weights. Following that, both the "TIME LOAD" and "GROUND MOTION" are applied simultaneously. The user must note that this example is only for illustration purposes and that it may be unlikely that a "TIME FUNCTION" and a "GROUND MOTION" both act on the structure at the same time. The Time load command is used to apply the Type 1 force, acting in the global X direction, at arrival time number 1, at nodes 2 and 3. The Ground motion, namely, the Type 2 time history loading, is also in the global X direction at arrival time 1.

PERFORM ANALYSIS

The above command initiates the analysis process. PRINT JOINT DISPLACEMENTS

During the analysis, the program calculates joint displacements for every time step. The absolute maximum value of the displacement for every joint is then extracted from this joint displacement history. So, the value printed using the above command is the absolute maximum value for each of the six degrees of freedom at each node.

UNIT KNS METER PRINT MEMBER FORCES PRINT SUPPORT REACTION

The member forces and support reactions too are calculated for every time step. For each degree of freedom, the maximum value of the member force and support reaction is extracted from these histories and reported in the output file using the above command.

FINISH

Example Problem 16


1. STAAD PLANE EXAMPLE FOR TIME HISTORY ANALYSIS2. UNITS CMS KNS3. JOINT COORDINATES4. 1 0.0 0.0 0.05. 2 0.0 120.0 0.06. 3 0.0 240.0 0.07. 4 0.0 360.0 0.08. MEMBER INCIDENCES9. 1 1 2 3

10. MEMBER PROPERTIES11. 1 2 3 PRIS AX 100.0 IZ 833.3312. SUPPORTS13. 1 4 PINNED14. CONSTANTS15. E 2850 ALL16. DENSITY 25E-6 ALL17. POISSON CONCRETE ALL18. UNIT NEWTON METER19. DEFINE TIME HISTORY20. TYPE 1 FORCE21. 0.0 -20.0 0.5 100.0 1.0 200.0 1.5 500.0 2.0 800.0 2.5 500.0 3.0 70.022. TYPE 2 ACCELERATION23. 0.0 0.1 0.5 -0.25 1.0 -0.5 1.5 -0.9 2.0 -1.3 2.5 -1.0 3.0 -0.724. ARRIVAL TIMES25. 0.026. DAMPING 0.07527. LOAD 1 STATIC LOAD28. MEMBER LOAD29. 1 2 3 UNI GX 500.030. LOAD 2 TIME HISTORY LOAD31. SELFWEIGHT X 1.032. SELFWEIGHT Y 1.033. JOINT LOAD34. 2 3 FX 4000.035. TIME LOAD36. 2 3 FX 1 137. GROUND MOTION X 2 138. PERFORM ANALYSIS



MORE MODES WERE REQUESTED THAN THERE ARE FREE MASSES.NUMBER OF MODES REQUESTED = 6NUMBER OF EXISTING MASSES IN THE MODEL = 4NUMBER OF MODES THAT WILL BE USED = 4



156 CALCULATED FREQUENCIES FOR LOAD CASE 2


1 3.087 0.32397 3.022E-162 11.955 0.08365 3.224E-163 443.457 0.00226 3.599E-164 768.090 0.00130 1.599E-16

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

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

1 100.00 0.00 0.00 100.000 0.000 0.0002 0.00 0.00 0.00 100.000 0.000 0.0003 0.00100.00 0.00 100.000 100.000 0.0004 0.00 0.00 0.00 100.000 100.000 0.000

TIME STEP USED IN TIME HISTORY ANALYSIS = 0.00139 SECONDSNUMBER OF MODES WHOSE CONTRIBUTION IS CONSIDERED = 2WARNING-NUMBER OF MODES LIMITED TO A FREQUENCY OF 240.0 DUE TO THE DT VALUE

ENTERED.TIME DURATION OF TIME HISTORY ANALYSIS = 2.999 SECONDSNUMBER OF TIME STEPS IN THE SOLUTION PROCESS = 2159

BASE SHEAR UNITS ARE -- NEWT METE

MAXIMUM BASE SHEAR X= -2.777268E+03 Y= 0.000000E+00 Z= 0.000000E+00AT TIMES 2.054167 0.000000 0.000000

39. PRINT JOINT DISPLACEMENTS



1 1 0.0000 0.0000 0.0000 0.0000 0.0000 -0.00412 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0084

2 1 0.4002 0.0000 0.0000 0.0000 0.0000 -0.00202 0.8420 0.0000 0.0000 0.0000 0.0000 -0.0042

3 1 0.4002 0.0000 0.0000 0.0000 0.0000 0.00202 0.8420 0.0000 0.0000 0.0000 0.0000 0.0042

4 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.00412 0.0000 0.0000 0.0000 0.0000 0.0000 0.0084


40. UNIT KNS METER41. PRINT MEMBER FORCES



1 1 1 0.00 0.90 0.00 0.00 0.00 0.002 0.00 -0.30 0.00 0.00 0.00 0.72

2 1 0.00 1.39 0.00 0.00 0.00 0.002 0.00 -1.39 0.00 0.00 0.00 1.67

2 1 2 0.00 0.30 0.00 0.00 0.00 -0.723 0.00 0.30 0.00 0.00 0.00 0.72

2 2 0.00 0.00 0.00 0.00 0.00 -1.673 0.00 0.00 0.00 0.00 0.00 1.67

3 1 3 0.00 -0.30 0.00 0.00 0.00 -0.724 0.00 0.90 0.00 0.00 0.00 0.00

2 3 0.00 -1.39 0.00 0.00 0.00 -1.674 0.00 1.39 0.00 0.00 0.00 0.00


Example Problem 16

157 42. PRINT SUPPORT REACTION



1 1 -0.90 0.00 0.00 0.00 0.00 0.002 -1.39 0.00 0.00 0.00 0.00 0.00

4 1 -0.90 0.00 0.00 0.00 0.00 0.002 -1.39 0.00 0.00 0.00 0.00 0.00


43. FINISH


**** DATE= TIME= ****



158

NOTES

Example Problem 17

159


The usage of User Provided Steel Tables is illustrated in this example for the analysis and design of a plane frame. User provided tables allow one to specify property data for sections not found in the built-in steel section tables.


160

Actual input is shown in bold lettering followed by explanation. STAAD PLANE EXAMPLE FOR USER TABLE

Every input file has to start with the command STAAD. The PLANE command is used to designate the structure as a plane frame.

UNIT METER KNS

The UNIT command sets the length and force units to be used. JOINT COORDINATES 1 0. 0. ; 2 9 0 ; 3 0 6 0 6 9 6 0 7 0 10.5 ; 8 9 10.5 ; 9 2.25 10.5 ; 10 6.75 10.5 11 4.5 10.5 ; 12 1.5 11.4 ; 13 7.5 11.4 14 3.0 12.3 ; 15 6.0 12.3; 16 4.5 13.2

The above set of data is used to provide joint coordinates for the various joints of the structure. The cartesian system is being used here. The data consists of the joint number followed by global X and Y coordinates. Note that for a space frame, the Z coordinate(s) need to be provided also. In the above input, semicolon (;) signs are used as line separators. This allows the user to provide multiple sets of data on one line.

MEMBER INCIDENCES 1 1 3 ; 2 3 7 ; 3 2 6 ; 4 6 8 ; 5 3 4 6 4 5 ; 7 5 6 ; 8 7 12 ; 9 12 14 10 14 16 ; 11 15 16 ; 12 13 15 ; 13 8 13 14 9 12 ; 15 9 14 ; 16 11 14 ; 17 11 15 18 10 15 ; 19 10 13 ; 20 7 9 21 9 11 ; 22 10 11 ; 23 8 10

The above data set contains the member incidence information or the joint connectivity data for each member. This completes the geometry of the structure.

Example Problem 17

161 START USER TABLE

This command is utilized to set up a User Provided steel table. All user provided steel tables must start with this command.

TABLE 1

Each table needs an unique numerical identification. The above command starts setting up Table no. 1. Upto twenty tables may be specified per run.

UNIT CM WIDE FLANGE

This command is used to specify the section-type as WIDE FLANGE in this table. Note that several section-types such as WIDE FLANGE, CHANNEL, ANGLE, TEE etc. are available for specification (See section 5 of the Technical Reference Manual).

BEAM250 32.2 25.5 0.6 10.2 0.85 3400 150 6.2 15.7 15.4 BEAM300 47.5 30.4 0.72 12.4 1.07 7160 335 14.3 21.9 23.8 BEAM350 64.6 35.6 0.73 17.2 1.15 14150 970 22.8 26.0 35.5

The above data set is used to specify the properties of three wide flange sections. The data for each section consists of two parts. In the first line, the section-name is provided. The user is allowed to provide any section name within twelve characters. The second line contains the section properties required for the particular section-type. Each section-type requires a certain number of data (area of cross-section, depth, moment of inertias etc.) provided in a certain order. For example, in this case, for wide flanges, ten different properties are required. For detailed information on the various properties required for the different section-types and their order of specification, refer to section 5.19 in the STAAD Technical Reference Manual. Without exception, all required properties for the particular section-type must be provided.


162 TABLE 2 UNIT CM ANGLES L30305 3.0 3.0 0.5 0.58 1.0 1.0 L40405 4.0 4.0 0.5 0.78 1.33 1.33 L50505 5.0 5.0 0.5 0.98 1.67 1.67

The above command and data lines set up another user provided table consisting of angle sections.

END

This command signifies the end of the user provided table data set. All user provided table related input must be terminated with this command.

MEMBER PROPERTIES 1 3 4 UPT 1 BEAM350 2 UPT 1 BEAM300 ; 5 6 7 UPT 1 BEAM250 8 TO 13 UPT 1 BEAM250 14 TO 19 UPT 2 L30305 20 TO 23 UPT 2 L40405

In the above command lines, the member properties are being assigned from the user provided tables created earlier. The word UPT signifies that the properties are from the user provided table. This is followed by the table number and then the section name as specified in the user provided table. The numbers 1 or 2 following the word UPT indicate the table from which section names are fetched.

MEMBER TRUSS 14 TO 23

The above command is used to designate members 14 to 23 as truss members.

Example Problem 17

163 MEMBER RELEASE 5 START MZ

The MEMBER RELEASE command is used to release the MZ moment at the start joint of member no. 5.

CONSTANTS E STEEL DENSITY STEEL ALL POISSON STEEL ALL BETA 90.0 MEMB 3 4

The above command set is used to specify modulus of elasticity, density, Poisson’s ratio and beta angle values. Built-in default value of steel is used for the material constants.

UNIT KNS METER

The force unit is reset to KNS, and length unit to METER using this command.

SUPPORT 1 FIXED ; 2 PINNED

The above command set is used to designate supports. Here, joint 1 is designated as a fixed support and joint 2 is designated as a pinned support.

LOADING 1 DEAD AND LIVE LOAD SELFWEIGHT Y -1.0 JOINT LOAD 4 5 FY -65. ; 11 FY -155. MEMB LOAD 8 TO 13 UNI Y –13.5 ; 6 UNI GY –17.5

The above command set is used to specify the loadings on the structure. In this case, dead and live loads are provided through load case 1. It consists of selfweight, concentrated loads at joints 4, 5 and 11, and distributed loads on some members.


164 PERFORM ANALYSIS

This command instructs the program to execute the analysis at this point.

PARAMETER CODE BRITISH BEAM 1.0 ALL NSF 0.85 ALL KY 1.2 MEMB 3 4

The above commands are used to specify parameters for steel design.

CHECK CODE MEMBER 3 19 SELECT MEMBER 20

This command will perform a code check on members 3 and 19 per the BRITISH steel design code. A member selection too is performed for member 20. For each member, the member selection will be performed from the table that was originally used for the specification of the member property. In this case, the selection will be from the respective user tables from which the properties were initially assigned. It may be noted that properties may be provided (and selection may be performed) from built-in steel tables and user provided tables in the same data file.

FINISH

This command terminates a STAAD run.

Example Problem 17

165


1. STAAD PLANE EXAMPLE FOR USER TABLE2. UNIT METER KNS3. JOINT COORD4. 1 0. 0. ; 2 9 0 ; 3 0 6 0 6 9 6 05. 7 0 10.5 ; 8 9 10.5 ; 9 2.25 10.5 ; 10 6.75 10.56. 11 4.5 10.5 ; 12 1.5 11.4 ; 13 7.5 11.47. 14 3.0 12.3 ; 15 6.0 12.3 ; 16 4.5 13.28. MEMBER INCIDENCES9. 1 1 3 ; 2 3 7 ; 3 2 6 ; 4 6 8 ; 5 3 4

10. 6 4 5 ; 7 5 6 ; 8 7 12 ; 9 12 1411. 10 14 16 ; 11 15 16 ; 12 13 15 ; 13 8 1312. 14 9 12 ; 15 9 14 ; 16 11 14 ; 17 11 1513. 18 10 15 ; 19 10 13 ; 20 7 914. 21 9 11 ; 22 10 11 ; 23 8 1015. START USER TABLE16. TABLE 117. UNIT CM18. WIDE FLANGE19. BEAM25020. 32.2 25.5 0.6 10.2 0.85 3400 150 6.2 15.7 15.421. BEAM30022. 47.5 30.4 0.72 12.4 1.07 7160 335 14.3 21.9 23.823. BEAM35024. 64.6 35.6 0.73 17.2 1.15 14150 970 22.8 26.0 35.525. TABLE 226. UNIT CM27. ANGLES28. L3030529. 3.0 3.0 0.5 0.58 1.0 1.030. L4040531. 4.0 4.0 0.5 0.78 1.33 1.3332. L5050533. 5.0 5.0 0.5 0.98 1.67 1.6734. END35. MEMBER PROPERTIES36. 1 3 4 UPT 1 BEAM35037. 2 UPT 1 BEAM300 ; 5 6 7 UPT 1 BEAM25038. 8 TO 13 UPT 1 BEAM25039. 14 TO 19 UPT 2 L3030540. 20 TO 23 UPT 2 L4040541. MEMBER TRUSS42. 14 TO 2343. MEMBER RELEASE44. 5 START MZ45. CONSTANTS46. E STEEL ALL47. DEN STEEL ALL48. POISS STEEL ALL49. BETA 90.0 MEMB 3 450. UNIT KNS METER51. SUPPORT52. 1 FIXED ; 2 PINNED53. LOADING 1 DEAD AND LIVE LOAD54. SELFWEIGHT Y -1.055. JOINT LOAD56. 4 5 FY -65. ; 11 FY -15557. MEMB LOAD58. 8 TO 13 UNI Y -13.5 ; 6 UNI GY -17.559. PERFORM ANALYSIS


166 P R O B L E M S T A T I S T I C S-----------------------------------


ZERO STIFFNESS IN DIRECTION 6 AT JOINT 9 EQN.NO. 21LOADS APPLIED OR DISTRIBUTED HERE FROM ELEMENTS WILL BE IGNORED.THIS MAY BE DUE TO ALL MEMBERS AT THIS JOINT BEING RELEASED OREFFECTIVELY RELEASED IN THIS DIRECTION.


60. PARAMETER61. CODE BRITISH62. BEAM 1 ALL63. NSF 0.85 ALL64. KY 1.2 MEMB 3 465. CHECK CODE MEMBER 3 19

STAAD.Pro CODE CHECKING - (BSI )***********************




=======================================================================

* 3 ST BEAM350 FAIL BS-4.8.3.3.1 2.472 1246.69 C 52.86 0.00 -

* 19 ST L30305 FAIL BS-4.7 (C) 1.654 118.19 C 0.00 0.00 0.00


66. SELECT MEMB 20

Example Problem 17

167 STAAD.Pro MEMBER SELECTION - (BSI )**************************




=======================================================================

20 ST L30305 PASS BS-4.6 (T) 0.970 162.38 T 0.00 0.00 0.00


67. FINISH



****************************************************************************


**** DATE= TIME= ****



168

NOTES

Example Problem 18

169


This is an example which demonstrates the calculation of principal stresses on a finite element.

Fixed Supports at Joints 1, 2, 3, 4, 5, 9, 13 Load intensity = 2000 Kn/sq.m in - Y direction


170

Actual input is shown in bold lettering followed by explanation. STAAD SPACE SAMPLE CALCULATION FOR * ELEMENT STRESSES

Every input has to start with the word STAAD. The word SPACE signifies that the structure is a space frame (3-D structure).

UNIT METER KNS

Specifies the unit to be used. JOINT COORDINATES 1 0 0 0 4 9 0 0 REPEAT 3 0 0 3

Joint number followed by X, Y and Z coordinates are provided above. The REPEAT command is used to generate coordinates of joints 5 to 16 based on the pattern of joints 1 to 4.

ELEMENT INCIDENCE 1 1 5 6 2 TO 3 REPEAT 2 3 4

Element connectivities of elements 1 to 3 are defined first, based on which, the connectivities of elements 4 to 9 are generated.

ELEMENT PROPERTIES 1 TO 9 THICK 0.25

Elements 1 to 9 have a thickness of 0.25 m. CONSTANTS E CONCRETE ALL POISSON CONC ALL

Modulus of Elasticity and Poisson’s ratio of all the elements is that of the built-in default value for concrete.

Example Problem 18

171 SUPPORT 1 TO 4 5 9 13 FIXED

"Fixed support" conditions exist at the above mentioned joints.

LOAD 1 ELEMENT LOAD 1 TO 9 PRESSURE -2000.0

A uniform pressure of 2000 Kn/sq.m is applied on all the elements. In the absence of an explicit direction specification, the load is assumed to act along the local Z axis. The negative value indicates that the load acts opposite to the positive direction of the local Z.

PERFORM ANALYSIS

The above command instructs the program to proceed with the analysis.

PRINT SUPPORT REACTION

The above command is self-explanatory. UNIT MMS PRINT ELEMENT STRESSES LIST 4

Element stresses at the centroid of the element are printed using the above command. The output includes membrane stresses, shear stresses, bending moments per unit width and principal stresses. The change of length unit from metre to mms indicates that the values will be printed in KN and MMs units.

FINISH



172


1. STAAD SPACE SAMPLE CALCULATION FOR2. * ELEMENT STRESSES3. UNIT METER KNS4. JOINT COORDINATES5. 1 0 0 0 4 9 0 06. REPEAT 3 0 0 37. ELEMENT INCIDENCE8. 1 1 5 6 2 TO 39. REPEAT 2 3 410. ELEMENT PROPERTIES11. 1 TO 9 THICK 0.2512. CONSTANT13. E CONCRETE ALL14. POISSON CONC ALL15. SUPPORT16. 1 TO 4 5 9 13 FIXED17. LOAD 118. ELEMENT LOAD19. 1 TO 9 PRESSURE -2000.020. PERFORM ANALYSIS






1 1 0.00 -1219.61 0.00 -391.50 0.00 391.502 1 0.00 8767.59 0.00 -26682.03 0.00 -524.213 1 0.00 37675.70 0.00 -88168.02 0.00 2985.304 1 0.00 35166.52 0.00 -66475.37 0.00 -24033.655 1 0.00 8767.59 0.00 524.21 0.00 26682.039 1 0.00 37675.70 0.00 -2985.30 0.00 88168.02

13 1 0.00 35166.52 0.00 24033.65 0.00 66475.37


22. UNIT MMS

Example Problem 18

173 23. PRINT ELEMENT STRESSES LIST 4

ELEMENT STRESSES FORCE,LENGTH UNITS= KNS MMS----------------



4 1 0.02 -0.02 2111.84 10726.22 4553.971.21 1.21 0.00 0.00 0.001.22 1.22

TOP : SMAX= 1.22 SMIN= 0.01 TMAX= 0.60 ANGLE= -23.3BOTT: SMAX= -0.01 SMIN= -1.22 TMAX= 0.60 ANGLE= -23.3

**** MAXIMUM STRESSES AMONG SELECTED PLATES AND CASES ****MAXIMUM MINIMUM MAXIMUM MAXIMUM MAXIMUM

PRINCIPAL PRINCIPAL SHEAR VONMISES TRESCASTRESS STRESS STRESS STRESS STRESS

1.217976E+00 -1.217976E+00 6.017490E-01 1.210802E+00 1.217976E+00PLATE NO. 4 4 4 4 4CASE NO. 1 1 1 1 1


24. FINISH


**** DATE= TIME= ****



174

Calculation of principal stresses for element 4 Calculations are presented for the top surface only.

SX = 0.0 kN/mm2 SY = 0.0 kN/mm2 SXY = 0.0 kN/mm2 MX = 2111.84 kN-mm/mm MY = 10726.22 kN-mm/mm MXY = 4553.97 kN-mm/mm S = 1/6t2 = 1/6*2502 = 10416.67mm2 (Section Modulus)

σx = 67.1041684.21110.0S

MXSX +=+

= 0.2027 kN/mm2

σy = 67.1041622.107260.0S

MYSY +=+

= 1.0297 kN/mm2

τxy = 67.1041697.45530.0S

MXYSXY +=+

= 0.4372 kN/mm2

TMAX x yxy=

−+

22

4( )σ σ

τ

24372.04

)0297.12027.0(2

TMAX +−

=

= 0.6018 kN/mm2

Example Problem 18

175

SMAX = ( )x y TMAXσ σ+

+2

= 6018.02)0297.12027.0( ++

= 1.218 kN/mm2 say 1.22 kN/mm2

SMIN = ( )x y TMAXσ σ+

−2

= 6018.02)0297.12027.0( −+

= 0.0144 kN/mm2 say 0.01 kN/ mm2

Angle xy

x y

=−

−12

21tanτ

σ σ

−−= 0297.12027.0

4372.0*21tan21

= -23.3o


176

NOTES

Example Problem 19

177


This example demonstrates the usage of inclined supports. The word INCLINED refers to the fact that the restraints at a joint where such a support is specified are along a user-specified axis system instead of along the default directions of the global axis system. STAAD offers a few different methods for assigning inclined supports, and we examine those in this example.


178 Actual input is shown in bold lettering followed by explanation.

STAAD SPACE INPUT WIDTH 79


UNIT METER KN

Specifies the unit to be used for data to follow. JOINT COORDINATES 1 0 5 0; 2 10 5 10; 3 20 5 20; 4 30 5 30; 5 5 0 5; 6 25 0 25;

Joint number followed by X, Y and Z coordinates are provided above. Semicolon signs (;) are used as line separators. That enables us to provide multiple sets of data on one line.

MEMBER INCIDENCES 1 1 2; 2 2 3; 3 3 4; 4 5 2; 5 6 3;

Defines the members by the joints they are connected to. UNIT MMS KN MEMBER PROPERTY AMERICAN 4 5 PRIS YD 800 1 TO 3 PRIS YD 750 ZD 500

Properties for all members of the model are provided using the PRISMATIC option. YD and ZD stand for depth and width. If ZD is not provided, a circular shape with diameter = YD is assumed for that cross section. All properties required for the analysis, such as, Area, Moments of Inertia, etc. are calculated automatically from these dimensions unless these are explicitly defined. The values are provided in MMS unit.

Example Problem 19

179 CONSTANTS E CONCRETE ALL POISSON CONCRETE ALL DENSITY CONCRETE ALL

Material constants like E (modulus of elasticity) and Poisson’s ratio are specified following the command CONSTANTS.

UNIT METER KN SUPPORTS 5 INCLINED REF 10 5 10 FIXED BUT MX MY MZ KFX 30000 6 INCLINED REFJT 3 FIXED BUT MX MY MZ KFX 30000 1 PINNED 4 INCLINED 1 0 1 FIXED BUT FX MX MY MZ

We assign supports (restraints) at 4 nodes - 5, 6, 1 and 4. For 3 of those, namely, 5, 6 and 4, the node number is followed by the keyword INCLINED, signifying that an INCLINED support is defined there. For the remaining one - node 1 - that keyword is missing. Hence, the support at node 1 is a global direction support. The most important aspect of inclined supports is their axis system. Each node where an inclined support is defined has its own distinct local X, local Y and local Z axes. In order to define the axis system, we first have to define a datum point. The support node and the datum point together help define the axis system. 3 different methods are shown in the above 3 instances for defining the datum point. At node 5, notice the keyword REF followed by the numbers (10,5,10). This means that the datum point associated with node 5 is one which has the global coordinates of (10m, 5m, 10m). Coincidentally, this happens to be node 2. At node 6, the keyword REFJT is used followed by the number 3. This means that the datum point for support node 6 is the joint number 3 of the model. The coordinates of the datum point are hence those of node 3, namely, (20m, 5m and 20m).


180

At node 4, the word INCLINED is merely followed by 3 numbers (1,0,1). In the absence of the words REF and REFJT, the program sets the datum point to be the following. It takes the coordinates of node 4, which are (30m,5m,30m) and adds to them, the 3 numbers which comes after the word INCLINED. Thus, the datum point becomes (31m, 5m and 31m). Once the datum point is established, the local axis system is defined as follows. Local X is a straight line (vector) pointing from the support node towards the datum point. Local Z is the vector obtained by the cross product of local X and the global Y axis (unless the SET Z UP command is used in which case one would use global Z instead of global Y and that would yield local Y). Local Y is the vector resulting from the cross product of local Z and local X. The right hand rule must be used when performing these cross products. Notice the unique nature of these datum points. The one for node 5 tells us that a line connecting nodes 5 to 2 is the local X axis, and is hence along the axis of member 4. By defining a KFX spring at that one, we are saying that the lower end of member 4 can move along its axis like the piston of a car engine. Think of a pile bored into rock with a certain amount of freedom to expand and contract axially. The same is true for the support at the bottom of member 5. The local X axis of that support is along the axis of member 5. That also happens to be the case for the supported end of member 3. The line going from node 4 to the datum point (31,5,31) happens to be coincident with the axis of the member, or the traffic direction. The expression FIXED BUT FX MX MY MZ for that support indicates that it is free to translate along local X, suggesting that it is an expansion joint - free to expand or contract along the axis of member 3. Since MX, MY and MZ are all released at these supports, no moment will be resisted by these supports.

Example Problem 19

181 LOAD 1 DEAD LOAD SELFWEIGHT Y -1.2 LOAD 2 LIVE LOAD MEMBER LOAD 1 TO 3 UNI GY -6 LOAD COMB 3 1 1.0 2 1.0 PERFORM ANALYSIS PRINT STATICS CHECK

3 load cases followed by the instruction for the type of analysis are specified. The PRINT STATICS CHECK option will instruct the program to produce a report consisting of total applied load versus total reactions from the supports for each primary load case.

PRINT SUPPORT REACTION

By default, support reactions are printed in the global axis directions. The above command is an instruction for such a report.

SET INCLINED REACTION PRINT SUPPORT REACTION

Just earlier, we saw how to obtain support reactions in the global axis system. What if we need them in the inclined axis system? The “SET INCLINED REACTION” is a switch for that purpose. It tells the program that reactions should be reported in the inclined axis system instead of the global axis system. This has to be followed by the PRINT SUPPORT REACTIONS command.

PRINT MEMBER FORCES PRINT JOINT DISP FINISH

Member forces are reported in the local axis system of the members. Joint displacements at all joints are reported in the global axis system. Following this, the STAAD run is terminated.



1. STAAD SPACE2. INPUT WIDTH 794. UNIT METER KN5. JOINT COORDINATES6. 1 0 5 0; 2 10 5 10; 3 20 5 20; 4 30 5 30; 5 5 0 5; 6 25 0 257. MEMBER INCIDENCES8. 1 1 2; 2 2 3; 3 3 4; 4 5 2; 5 6 3

10. UNIT MMS KN11. MEMBER PROPERTY AMERICAN12. 4 5 PRIS YD 80013. 1 TO 3 PRIS YD 750 ZD 50015. CONSTANTS16. E CONCRETE ALL17. POISSON CONCRETE ALL18. DENSITY CONCRETE ALL20. UNIT METER KN21. SUPPORTS22. 5 INC REF 10 5 10 FIXED BUT MX MY MZ KFX 3000023. 6 INC REFJT 3 FIXED BUT MX MY MZ KFX 3000024. 1 PINNED25. 4 INC 1 0 1 FIXED BUT FX MX MY MZ27. LOAD 1 DEAD LOAD28. SELFWEIGHT Y -1.230. LOAD 2 LIVE LOAD31. MEMBER LOAD32. 1 TO 3 UNI GY -634. LOAD COMB 335. 1 1.0 2 1.036. PERFORM ANALYSIS PRINT STATICS CHECK



STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 1DEAD LOAD

***TOTAL APPLIED LOAD ( KN METE ) SUMMARY (LOADING 1 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = -696.00SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE ORIGIN-MX= 10439.93 MY= 0.00 MZ= -10439.93

***TOTAL REACTION LOAD( KN METE ) SUMMARY (LOADING 1 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = 696.00SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE ORIGIN-MX= -10439.93 MY= 0.00 MZ= 10439.93

MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 1)MAXIMUMS AT NODE

X = -7.99237E-01 5Y = -2.49498E+00 3Z = -7.99237E-01 5RX= -2.66161E-03 4RY= -9.86155E-16 4RZ= 2.66161E-03 4

Example Problem 19

183 STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 2LIVE LOAD

***TOTAL APPLIED LOAD ( KN METE ) SUMMARY (LOADING 2 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = -254.56SUMMATION FORCE-Z = 0.00


***TOTAL REACTION LOAD( KN METE ) SUMMARY (LOADING 2 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = 254.56SUMMATION FORCE-Z = 0.00



X = -2.97411E-01 5Y = -9.31566E-01 3Z = -2.97411E-01 5RX= -1.18888E-03 4RY= -3.70449E-16 4RZ= 1.18888E-03 4



SUPPORT REACTIONS -UNIT KN METE STRUCTURE TYPE = SPACE-----------------


5 1 215.36 288.60 215.36 0.00 0.00 0.002 86.45 94.77 86.45 0.00 0.00 0.003 301.81 383.37 301.81 0.00 0.00 0.00

6 1 -212.20 286.84 -212.20 0.00 0.00 0.002 -85.19 94.06 -85.19 0.00 0.00 0.003 -297.39 380.91 -297.39 0.00 0.00 0.00

1 1 -3.15 60.21 -3.15 0.00 0.00 0.002 -1.27 32.84 -1.27 0.00 0.00 0.003 -4.42 93.05 -4.42 0.00 0.00 0.00

4 1 0.00 60.33 0.00 0.00 0.00 0.002 0.00 32.89 0.00 0.00 0.00 0.003 0.00 93.22 0.00 0.00 0.00 0.00


39. SET INCLINED REACTION40. PRINT SUPPORT REACTION



5 1 415.30 59.81 0.00 0.00 0.00 0.002 154.54 6.79 0.00 0.00 0.00 0.003 569.83 66.60 0.00 0.00 0.00 0.00

6 1 410.64 60.94 0.00 0.00 0.00 0.002 152.67 7.25 0.00 0.00 0.00 0.003 563.31 68.19 0.00 0.00 0.00 0.00

1 1 -3.15 60.21 -3.15 0.00 0.00 0.002 -1.27 32.84 -1.27 0.00 0.00 0.003 -4.42 93.05 -4.42 0.00 0.00 0.00

4 1 0.00 60.33 0.00 0.00 0.00 0.002 0.00 32.89 0.00 0.00 0.00 0.003 0.00 93.22 0.00 0.00 0.00 0.00



184 41. PRINT MEMBER FORCES

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


1 1 1 -4.46 60.21 0.00 0.00 0.00 0.002 4.46 89.73 0.00 0.00 0.00 -208.74

2 1 -1.79 32.84 0.00 0.00 0.00 0.002 1.79 52.01 0.00 0.00 0.00 -135.58

3 1 -6.25 93.05 0.00 0.00 0.00 0.002 6.25 141.75 0.00 0.00 0.00 -344.32

2 1 2 300.10 75.79 0.00 0.00 0.00 125.953 -300.10 74.15 0.00 0.00 0.00 -114.37

2 2 120.47 42.75 0.00 0.00 0.00 76.773 -120.47 42.10 0.00 0.00 0.00 -72.12

3 2 420.57 118.55 0.00 0.00 0.00 202.713 -420.57 116.25 0.00 0.00 0.00 -186.49

3 1 3 0.00 89.61 0.00 0.00 0.00 207.024 0.00 60.33 0.00 0.00 0.00 0.00

2 3 0.00 51.96 0.00 0.00 0.00 134.894 0.00 32.89 0.00 0.00 0.00 0.00

3 3 0.00 141.58 0.00 0.00 0.00 341.904 0.00 93.22 0.00 0.00 0.00 0.00

4 1 5 415.30 59.81 0.00 0.00 0.00 0.002 -344.24 40.69 0.00 0.00 0.00 82.79

2 5 154.54 6.79 0.00 0.00 0.00 0.002 -154.54 -6.79 0.00 0.00 0.00 58.81

3 5 569.83 66.60 0.00 0.00 0.00 0.002 -498.77 33.90 0.00 0.00 0.00 141.61

5 1 6 410.64 60.94 0.00 0.00 0.00 0.003 -339.58 39.55 0.00 0.00 0.00 92.64

2 6 152.67 7.25 0.00 0.00 0.00 0.003 -152.67 -7.25 0.00 0.00 0.00 62.77

3 6 563.31 68.19 0.00 0.00 0.00 0.003 -492.25 32.30 0.00 0.00 0.00 155.41


42. PRINT JOINT DISP



1 1 0.0000 0.0000 0.0000 0.0026 0.0000 -0.00262 0.0000 0.0000 0.0000 0.0012 0.0000 -0.00123 0.0000 0.0000 0.0000 0.0038 0.0000 -0.0038

2 1 0.0005 -2.4510 0.0005 0.0007 0.0000 -0.00072 0.0002 -0.9139 0.0002 0.0003 0.0000 -0.00033 0.0008 -3.3649 0.0008 0.0011 0.0000 -0.0011

3 1 -0.0363 -2.4950 -0.0363 -0.0007 0.0000 0.00072 -0.0146 -0.9316 -0.0146 -0.0003 0.0000 0.00033 -0.0509 -3.4265 -0.0509 -0.0011 0.0000 0.0011

4 1 -0.0363 0.0000 -0.0363 -0.0027 0.0000 0.00272 -0.0146 0.0000 -0.0146 -0.0012 0.0000 0.00123 -0.0509 0.0000 -0.0509 -0.0039 0.0000 0.0039

5 1 -0.7992 -0.7992 -0.7992 0.0023 0.0000 -0.00232 -0.2974 -0.2974 -0.2974 0.0007 0.0000 -0.00073 -1.0966 -1.0966 -1.0966 0.0031 0.0000 -0.0031

6 1 0.7903 -0.7903 0.7903 -0.0024 0.0000 0.00242 0.2938 -0.2938 0.2938 -0.0008 0.0000 0.00083 1.0841 -1.0841 1.0841 -0.0032 0.0000 0.0032


Example Problem 19

185 43. FINISH


**** DATE= TIME= ****



186

NOTES

Example Problem 20

187


This example generates the geometry of a cylindrical tank structure using the cylindrical coordinate system. The tank lies on its side in this example.


188

In this example, a cylindrical tank is modeled using finite elements. The radial direction is in the XY plane and longitudinal direction is along the Z-axis. Hence, the coordinates in the XY plane are generated using the cylindrical coordinate system.

STAAD SPACE UNIT KIP FEET

The type of structure (SPACE frame) and length and force units for data to follow are specified.

JOINT COORD CYLINDRICAL

The above command instructs the program that the coordinate data that follows is in the cylindrical coordinate system (r,theta,z).

1 10 0 0 8 10 315 0

Joint 1 has an 'r' of 10 feet, theta of 0 degrees and Z of 0 ft. Joint 8 has an 'r' of 10 feet, theta of 315 degrees and Z of 0 ft. The 315 degrees angle is measured counter-clockwise from the +ve direction of the X-axis. Joints 2 to 7 are generated by equal incrementation the coordinate values between joints 1 and 8.

REPEAT 2 0 0 8.5

The REPEAT command is used to generate joints 9 through 24 by repeating twice, the pattern of joints 1 to 8 at Z-increments of 8.5 feet for each REPEAT.

PRINT JOINT COORD

The above command is used to produce a report consisting of the coordinates of all the joints in the cartesian coordinate system. Note that even though the input data was in the cylindrical coordinate system, the output is in the cartesian coordinate system.

ELEMENT INCIDENCES 1 1 2 10 9 TO 7 1 1 8 8 1 9 16 REPEAT ALL 1 8 8

Example Problem 20

189

The above 4 lines identify the element incidences of all 16 elements. Incidences of element 1 is defined as 1 2 10 9. Incidences of element 2 is generated by incrementing the joint numbers of element 1 by 1, incidences of element 3 is generated by incrementing the incidences of element 2 by 1 and so on upto element 7. Incidences of element 8 has been defined above as 8 1 9 16. The REPEAT ALL command states that the pattern of ALL the elements defined by the previous 2 lines, namely elements 1 to 8, must be REPEATED once with an element number increment of 8 and a joint number increment of 8 to generate elements 9 through 16.

PRINT ELEMENT INFO

The above command is self-explanatory. FINISH


190


1. STAAD SPACE2. UNIT KIP FEET3. JOINT COORD CYLINDRICAL4. 1 10 0 0 8 10 315 05. REPEAT 2 0 0 8.56. PRINT JOINT COORD

JOINT COORDINATES-----------------COORDINATES ARE FEET UNIT

JOINT X Y Z

1 10.000 0.000 0.0002 7.071 7.071 0.0003 0.000 10.000 0.0004 -7.071 7.071 0.0005 -10.000 0.000 0.0006 -7.071 -7.071 0.0007 0.000 -10.000 0.0008 7.071 -7.071 0.0009 10.000 0.000 8.500

10 7.071 7.071 8.50011 0.000 10.000 8.50012 -7.071 7.071 8.50013 -10.000 0.000 8.50014 -7.071 -7.071 8.50015 0.000 -10.000 8.50016 7.071 -7.071 8.50017 10.000 0.000 17.00018 7.071 7.071 17.00019 0.000 10.000 17.00020 -7.071 7.071 17.00021 -10.000 0.000 17.00022 -7.071 -7.071 17.00023 0.000 -10.000 17.00024 7.071 -7.071 17.000


7. ELEMENT INCIDENCES8. 1 1 2 10 9 TO 7 1 19. 8 8 1 9 16

10. REPEAT ALL 1 8 8

Example Problem 20

191 11. PRINT ELEMENT INFO

ELEMENT INFORMATION-------------------

ELEMENT INCIDENCES THICK POISS E G AREANO. (FEET)

1 1 2 10 9 0.000 0.000 0.00 0.00 65.05622 2 3 11 10 0.000 0.000 0.00 0.00 65.05623 3 4 12 11 0.000 0.000 0.00 0.00 65.05624 4 5 13 12 0.000 0.000 0.00 0.00 65.05625 5 6 14 13 0.000 0.000 0.00 0.00 65.05626 6 7 15 14 0.000 0.000 0.00 0.00 65.05627 7 8 16 15 0.000 0.000 0.00 0.00 65.05628 8 1 9 16 0.000 0.000 0.00 0.00 65.05629 9 10 18 17 0.000 0.000 0.00 0.00 65.0562

10 10 11 19 18 0.000 0.000 0.00 0.00 65.056211 11 12 20 19 0.000 0.000 0.00 0.00 65.056212 12 13 21 20 0.000 0.000 0.00 0.00 65.056213 13 14 22 21 0.000 0.000 0.00 0.00 65.056214 14 15 23 22 0.000 0.000 0.00 0.00 65.056215 15 16 24 23 0.000 0.000 0.00 0.00 65.056216 16 9 17 24 0.000 0.000 0.00 0.00 65.0562

******************END OF ELEMENT INFO******************

12. FINISH


**** DATE= TIME= ****



192 NOTES

Example Problem 21

193


This example illustrates the modeling of tension-only members using the MEMBER TENSION command.


194

This example has been created to illustrate the command specification for a structure with certain members capable of carrying tensile force only. It is important to note that the analysis can be done for only 1 load case at a time. This is because, the set of “active” members (and hence the stiffness matrix) is load case dependent.

STAAD PLANE EXAMPLE FOR TENSION-ONLY MEMBERS

The input data is initiated with the word STAAD. This structure is a PLANE frame.

UNIT METER KNS

Units for the commands to follow are defined above. SET NL 3


JOINT COORDINATES 1 0 0 ; 2 0 3.5 ; 3 0 7.0 ; 4 5.25 7.0 ; 5 5.25 3.5 ; 6 5.25 0

Joint coordintes of joints 1 to 6 are defined above. MEMBER INCIDENCES 1 1 2 5 6 1 5 ; 7 2 6 ; 8 2 4 ; 9 3 5 ; 10 2 5

Incidences of members 1 to 10 are defined.

Example Problem 21

195 MEMBER TENSION 6 TO 9

Members 6 to 9 are defined as TENSION-only members. Hence for each load case, if during the analysis, any of the members 6 to 9 is found to be carrying a compressive force, it is disabled from the structure and the analysis is carried out again with the modified structure.

MEMBER PROPERTY BRITISH 1 TO 10 TA ST UC152X152X30

All members have been assigned a UC section from the British table.

CONSTANTS E STEEL ALL POISSON STEEL ALL

Following the command CONSTANTS, material constants such as E (Modulus of Elasticity), and Poisson’s ratio are specified. In this case, the built-in default value of steel is assigned.

SUPPORT 1 6 PINNED

The supports are defined above. LOAD 1 JOINT LOAD 2 FX 70 3 FX 45

Load 1 is defined above and consists of joint loads at joints 2 and 3.

PERFORM ANALYSIS

An analysis is carried out for load case 1.


196 CHANGE MEMBER TENSION 6 TO 9

One or more among the members 6 to 9 may have been inactivated in the previous analysis. The CHANGE command restores the original structure to prepare it for the analysis for the next primary load case. The members with the tension-only attribute are specified again.

LOAD 2 JOINT LOAD 4 FX -45 5 FX -70

Load case 2 is described above. PERFORM ANALYSIS CHANGE

The instruction to analyze the structure is specified again. Next, any tension-only members that become inactivated during the second analysis (due to the fact that they were subjected to compressive axial forces) are re-activated with the CHANGE command. Without re-activation, these members cannot be accessed for any further operations.

LOAD 3 REPEAT LOAD 1 1.0 2 1.0

Load case 3 illustrates the technique employed to instruct STAAD to create a load case which consists of data to be assembled from other load cases already specified earlier. We would like the program to analyze the structure for loads from cases 1 and 2 acting simultaneously. In other words, the above instruction is the same as the following:

Example Problem 21

197 LOAD 3 JOINT LOAD 2 FX 70 3 FX 45 4 FX -45 5 FX -70

PERFORM ANALYSIS

The analysis is carried out for load case 3.

CHANGE LOAD LIST 1 2 3

The members inactivated during the analysis of load 3 are re-activated for further processing. At the end of any analysis, only those load cases for which the analysis was done most recently, are recognized as the "active" load cases. The LOAD LIST command enables the above listed load cases to be made active for further processing.

PRINT ANALYSIS RESULTS FINI

The analysis results are printed and the run terminated.



1. STAAD PLANE EXAMPLE FOR TENSION-ONLY MEMBERS2. UNIT METER KNS3. SET NL 34. JOINT COORDINATES5. 1 0 0 ; 2 0 3.5 ; 3 0 7.0 ; 4 5.25 7.0 ; 5 5.25 3.5 ; 6 5.25 06. MEMBER INCIDENCES7. 1 1 2 58. 6 1 5 ; 7 2 6 ; 8 2 4 ; 9 3 5 ; 10 2 59. MEMBER TENSION

10. 6 TO 911. MEMBER PROPERTY BRITISH12. 1 TO 10 TABLE ST UC152X152X3013. CONSTANTS14. E STEEL ALL15. DEN STEEL ALL16. POISSON STEEL ALL17. SUPPORT18. 1 6 PINNED19. LOAD 120. JOINT LOAD21. 2 FX 7022. 3 FX 4523. PERFORM ANALYSIS



**START ITERATION NO. 2

**NOTE-Tension/Compression converged after 2 iterations, Case= 1

24. CHANGE25. MEMBER TENSION26. 6 TO 927. LOAD 228. JOINT LOAD29. 4 FX -4530. 5 FX -7031. PERFORM ANALYSIS



32. CHANGE33. LOAD 334. REPEAT LOAD35. 1 1.0 2 1.036. PERFORM ANALYSIS



37. CHANGE38. LOAD LIST 1 2 3

Example Problem 21




1 1 0.0000 0.0000 0.0000 0.0000 0.0000 -0.00082 0.0000 0.0000 0.0000 0.0000 0.0000 0.00053 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0001

2 1 0.2412 0.0135 0.0000 0.0000 0.0000 -0.00042 -0.1648 -0.0475 0.0000 0.0000 0.0000 0.00043 0.0234 0.0000 0.0000 0.0000 0.0000 0.0000

3 1 0.3722 0.0136 0.0000 0.0000 0.0000 -0.00022 -0.3423 -0.0608 0.0000 0.0000 0.0000 0.00043 0.0151 0.0000 0.0000 0.0000 0.0000 0.0000

4 1 0.3423 -0.0608 0.0000 0.0000 0.0000 -0.00042 -0.3722 0.0136 0.0000 0.0000 0.0000 0.00023 -0.0151 0.0000 0.0000 0.0000 0.0000 0.0000

5 1 0.1648 -0.0475 0.0000 0.0000 0.0000 -0.00042 -0.2412 0.0135 0.0000 0.0000 0.0000 0.00043 -0.0234 0.0000 0.0000 0.0000 0.0000 0.0000

6 1 0.0000 0.0000 0.0000 0.0000 0.0000 -0.00052 0.0000 0.0000 0.0000 0.0000 0.0000 0.00083 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001



1 1 -114.92 -106.67 0.00 0.00 0.00 0.002 0.08 106.67 0.00 0.00 0.00 0.003 -0.05 0.00 0.00 0.00 0.00 0.00

6 1 -0.08 106.67 0.00 0.00 0.00 0.002 114.92 -106.67 0.00 0.00 0.00 0.003 0.05 0.00 0.00 0.00 0.00 0.00



1 1 1 -30.23 0.26 0.00 0.00 0.00 0.002 30.23 -0.26 0.00 0.00 0.00 0.89

2 1 106.67 -0.08 0.00 0.00 0.00 0.002 -106.67 0.08 0.00 0.00 0.00 -0.29

3 1 0.00 0.05 0.00 0.00 0.00 0.002 0.00 -0.05 0.00 0.00 0.00 0.16

2 1 2 -0.25 0.21 0.00 0.00 0.00 0.203 0.25 -0.21 0.00 0.00 0.00 0.55

2 2 29.82 -0.44 0.00 0.00 0.00 -0.773 -29.82 0.44 0.00 0.00 0.00 -0.76

3 2 0.00 -0.05 0.00 0.00 0.00 -0.143 0.00 0.05 0.00 0.00 0.00 -0.04

3 1 3 44.79 -0.25 0.00 0.00 0.00 -0.554 -44.79 0.25 0.00 0.00 0.00 -0.76

2 3 44.79 0.25 0.00 0.00 0.00 0.764 -44.79 -0.25 0.00 0.00 0.00 0.55

3 3 45.05 0.00 0.00 0.00 0.00 0.044 -45.05 0.00 0.00 0.00 0.00 -0.04

4 1 4 29.82 0.44 0.00 0.00 0.00 0.765 -29.82 -0.44 0.00 0.00 0.00 0.77

2 4 -0.25 -0.21 0.00 0.00 0.00 -0.555 0.25 0.21 0.00 0.00 0.00 -0.20

3 4 0.00 0.05 0.00 0.00 0.00 0.045 0.00 -0.05 0.00 0.00 0.00 0.14




5 1 5 106.67 0.08 0.00 0.00 0.00 0.296 -106.67 -0.08 0.00 0.00 0.00 0.00

2 5 -30.23 -0.26 0.00 0.00 0.00 -0.896 30.23 0.26 0.00 0.00 0.00 0.00

3 5 0.00 -0.05 0.00 0.00 0.00 -0.166 0.00 0.05 0.00 0.00 0.00 0.00

6 1 1 -137.81 0.00 0.00 0.00 0.00 0.005 137.81 0.00 0.00 0.00 0.00 0.00

2 1 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

3 1 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

7 1 2 0.00 0.00 0.00 0.00 0.00 0.006 0.00 0.00 0.00 0.00 0.00 0.00

2 2 -137.81 0.00 0.00 0.00 0.00 0.006 137.81 0.00 0.00 0.00 0.00 0.00

3 2 0.00 0.00 0.00 0.00 0.00 0.006 0.00 0.00 0.00 0.00 0.00 0.00

8 1 2 -53.30 0.00 0.00 0.00 0.00 0.004 53.30 0.00 0.00 0.00 0.00 0.00

2 2 0.00 0.00 0.00 0.00 0.00 0.004 0.00 0.00 0.00 0.00 0.00 0.00

3 2 0.00 0.00 0.00 0.00 0.00 0.004 0.00 0.00 0.00 0.00 0.00 0.00

9 1 3 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

2 3 -53.30 0.00 0.00 0.00 0.00 0.005 53.30 0.00 0.00 0.00 0.00 0.00

3 3 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

10 1 2 114.31 -0.41 0.00 0.00 0.00 -1.095 -114.31 0.41 0.00 0.00 0.00 -1.06

2 2 114.31 0.41 0.00 0.00 0.00 1.065 -114.31 -0.41 0.00 0.00 0.00 1.09

3 2 69.90 0.00 0.00 0.00 0.00 -0.025 -69.90 0.00 0.00 0.00 0.00 0.02


40. FINI*********** END OF THE STAAD.Pro RUN ***********

**** DATE= TIME= ****


Example Problem 21

201 NOTES


Example Problem 21

202 NOTES

Example Problem 22

203


A space frame structure is subjected to a sinusoidal (dynamic) loading. The commands necessary to describe the sine function are demonstrated in this example. Time History analysis is performed on this model.


Example Problem 22

204 STAAD SPACE *EXAMPLE FOR HARMONIC LOADING GENERATOR

Every STAAD input file has to begin with the word STAAD. The word SPACE signifies that the structure is a space frame and the geometry is defined through X, Y and Z axes. The comment line which begins with an asterisk is an optional title to identify this project.

UNIT KNS METER


JOINT COORDINATES 1 0 0 0 ; 2 5 0 0 ; 3 5 0 5 ; 4 0 0 5 5 0 7 0 ; 6 2.5 7 0 ; 7 5 7 0 ; 8 5 7 2.5 9 5 7 5 ; 10 2.5 7 5 ; 11 0 7 5 12 0 7 2.5 ; 13 2.5 7 2.5

The joint number followed by the X, Y and Z coordinates are specified above. Semicolon characters (;) are used as line separators to facilitate input of multiple sets of data on one line.

MEMBER INCIDENCES 1 1 5 ; 2 2 7 ; 3 3 9 ; 4 4 11 ; 5 5 6 ; 6 6 7 7 7 8 ; 8 8 9 ; 9 9 10 ; 10 10 11 ; 11 11 12 ; 12 12 5 13 6 13 ; 14 13 10 ; 15 8 13 ; 16 13 12

The members are defined by the joints they are connected to.

UNIT MMS MEMBER PROPERTIES 1 TO 4 PRIS YD 600 ZD 600 5 TO 16 PRIS YD 450 ZD 450

Members 1 to 16 are defined as PRISmatic sections with width and depth values provided using the YD and ZD options. The UNIT command is specified to change the units for length from METER to MMS.

Example Problem 22

205 SUPPORTS 1 TO 4 PINNED

Joints 1 to 4 are declared to be pinned-supported.

CONSTANTS E CONCRETE ALL DENSITY CONCRETE ALL POISSON CONCRETE ALL

The modulus of elasticity (E), density and Poisson’s ratio are specified following the command CONSTANTS. Built-in default values for concrete are used.

DEFINE TIME HISTORY TYPE 1 FORCE * FOLLOWING LINES FOR HARMONIC LOADING GENERATOR FUNCTION SINE AMPLITUDE 30 FREQUENCY 60 CYCLES 100 * ARRIVAL TIMES 0.0 DAMPING 0.075

There are two stages in the command specification required for a time-history analysis. The first stage is defined above. Here, the parameters of the sinusoidal loading are provided.

Each data set is individually identified by the number that follows the TYPE command. In this file, only one data set is defined, which is apparent from the fact that only one TYPE is defined.

The word FORCE that follows the TYPE 1 command signifies that this data set is for a forcing function. (If one wishes to specify an earthquake motion, an ACCELERATION may be specified.)

The command FUNCTION SINE indicates that instead of providing the data set as discrete TIME-FORCE pairs, a sinusoidal function, which describes the variation of force with time, is provided.


Example Problem 22

206

The parameters of the sine function, such as FREQUENCY, AMPLITUDE, and number of CYCLES of application are then defined. STAAD internally generates discrete TIME-FORCE pairs of data from the sine function in steps of time defined by the default value (see section 5.31.6 of the Technical Reference Manual for more information). The arrival time value indicates the relative value of time at which the force begins to act upon the structure. The modal damping ratio for all the modes is set to 0.075.

UNIT METER LOAD 1 MEMBER LOAD 5 6 7 8 9 10 11 12 UNI GY -10.0

The above data describe a static load case. A uniformly distributed load of 10 kN/m acting in the negative global Y direction is applied on some members.

LOAD 2 SELFWEIGHT X 1.0 SELFWEIGHT Y 1.0 SELFWEIGHT Z 1.0 JOINT LOAD 8 12 FX 15.0 8 12 FY 15.0 8 12 FZ 15.0 TIME LOAD 8 12 FX 1 1

This is the second stage of command specification for time history analysis. The 2 sets of data specified here are a) the weights for generation of the mass matrix and b) the application of the time varying loads on the structure.

The weights (from which the masses for the mass matrix are obtained) are specified in the form of selfweight and joint loads.

Following that, the sinusoidal force is applied using the "TIME LOAD" command. The forcing function described by the TYPE 1

Example Problem 22

207

load is applied on joints 8 and 12 and it starts to act starting at a time defined by the 1st arrival time number.

PERFORM ANALYSIS PRINT ANALYSIS RESULTS FINI

The above commands are self explanatory. The FINISH command terminates the STAAD run.


Example Problem 22


1. STAAD SPACE EXAMPLE FOR HARMONIC LOADING GENERATOR2. UNIT KNS METER3. JOINT COORDINATES4. 1 0 0 0 ; 2 5 0 0 ; 3 5 0 5 ; 4 0 0 55. 5 0 7 0 ; 6 2.5 7 0 ; 7 5 7 0 ; 8 5 7 2.56. 9 5 7 5 ; 10 2.5 7 5 ; 11 0 7 57. 12 0 7 2.5 ; 13 2.5 7 2.58. MEMBER INCIDENCES9. 1 1 5 ; 2 2 7 ; 3 3 9 ; 4 4 11 ; 5 5 6 ; 6 6 7

10. 7 7 8 ; 8 8 9 ; 9 9 10 ; 10 10 11 ; 11 11 12 ; 12 12 511. 13 6 13 ; 14 13 10 ; 15 8 13 ; 16 13 1212. UNIT MMS13. MEMBER PROPERTIES14. 1 TO 4 PRIS YD 600 ZD 60015. 5 TO 16 PRIS YD 450 ZD 45016. SUPPORTS17. 1 TO 4 PINNED18. CONSTANTS19. E CONCRETE ALL20. DENSITY CONCRETE ALL21. POISSON CONCRETE ALL22. DEFINE TIME HISTORY23. TYPE 1 FORCE24. * FOLLOWING LINES FOR HARMONIC LOADING GENERATOR25. FUNCTION SINE26. AMPLITUDE 30 FREQUENCY 60 CYCLES 10027. *28. ARRIVAL TIMES29. 0.030. DAMPING 0.07531. UNIT METER32. LOAD 133. MEMBER LOAD34. 5 6 7 8 9 10 11 12 UNI GY -10.035. LOAD 236. SELFWEIGHT X 1.037. SELFWEIGHT Y 1.038. SELFWEIGHT Z 1.039. JOINT LOAD40. 8 12 FX 15.041. 8 12 FY 15.042. 8 12 FZ 15.043. TIME LOAD44. 8 12 FX 1 145. PERFORM ANALYSIS



NUMBER OF MODES REQUESTED = 6NUMBER OF EXISTING MASSES IN THE MODEL = 27NUMBER OF MODES THAT WILL BE USED = 6

*** EIGENSOLUTION: SUBSPACE METHOD ***

Example Problem 22

209 CALCULATED FREQUENCIES FOR LOAD CASE 2


1 1.863 0.53664 2.094E-142 1.864 0.53639 2.672E-143 2.214 0.45158 2.114E-144 18.318 0.05459 1.696E-085 19.303 0.05181 8.423E-106 23.509 0.04254 1.940E-08




7 28.606 0.03496 1.253E-088 30.175 0.03314 1.232E-069 31.831 0.03142 6.180E-09

10 34.539 0.02895 2.737E-0611 41.121 0.02432 3.384E-0712 82.426 0.01213 2.484E-0513 83.366 0.01200 4.212E-0714 83.479 0.01198 2.918E-0515 84.414 0.01185 2.249E-0616 102.892 0.00972 3.888E-0517 103.028 0.00971 6.272E-0618 103.040 0.00970 2.230E-0519 103.764 0.00964 1.376E-0620 117.529 0.00851 6.936E-0521 159.295 0.00628 3.567E-0522 192.872 0.00518 6.055E-05



1 100.00 0.00 0.00 99.996 0.000 0.0002 0.00 0.00100.00 99.996 0.000 99.9983 0.00 0.00 0.00 99.996 0.000 99.9984 0.00 0.00 0.00 99.996 0.000 99.9985 0.00 43.46 0.00 99.996 43.456 99.9986 0.00 0.00 0.00 99.999 43.456 99.998

TIME STEP USED IN TIME HISTORY ANALYSIS = 0.00139 SECONDSNUMBER OF MODES WHOSE CONTRIBUTION IS CONSIDERED = 6TIME DURATION OF TIME HISTORY ANALYSIS = 1.665 SECONDSNUMBER OF TIME STEPS IN THE SOLUTION PROCESS = 1199

BASE SHEAR UNITS ARE -- KNS METE

MAXIMUM BASE SHEAR X= -1.669197E+00 Y= 2.384186E-07 Z= 1.862645E-09AT TIMES 0.127778 0.062500 0.327778


Example Problem 22




1 1 0.0000 0.0000 0.0000 -0.0001 0.0000 0.00012 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0001

2 1 0.0000 0.0000 0.0000 -0.0001 0.0000 -0.00012 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0001

3 1 0.0000 0.0000 0.0000 0.0001 0.0000 -0.00012 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0001

4 1 0.0000 0.0000 0.0000 0.0001 0.0000 0.00012 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0001

5 1 0.0001 -0.0045 0.0001 0.0002 0.0000 -0.00022 0.0408 0.0001 0.0000 0.0000 0.0000 0.0000

6 1 0.0000 -0.0472 0.0000 0.0001 0.0000 0.00002 0.0408 0.0000 0.0000 0.0000 0.0000 0.0000

7 1 -0.0001 -0.0045 0.0001 0.0002 0.0000 0.00022 0.0408 -0.0001 0.0000 0.0000 0.0000 0.0000

8 1 0.0000 -0.0472 0.0000 0.0000 0.0000 0.00012 0.0410 -0.0005 0.0000 0.0000 0.0000 0.0000

9 1 -0.0001 -0.0045 -0.0001 -0.0002 0.0000 0.00022 0.0408 -0.0001 0.0000 0.0000 0.0000 0.0000

10 1 0.0000 -0.0472 0.0000 -0.0001 0.0000 0.00002 0.0408 0.0000 0.0000 0.0000 0.0000 0.0000

11 1 0.0001 -0.0045 -0.0001 -0.0002 0.0000 -0.00022 0.0408 0.0001 0.0000 0.0000 0.0000 0.0000

12 1 0.0000 -0.0472 0.0000 0.0000 0.0000 -0.00012 0.0410 0.0005 0.0000 0.0000 0.0000 0.0000

13 1 0.0000 -0.0583 0.0000 0.0000 0.0000 0.00002 0.0410 0.0000 0.0000 0.0000 0.0000 0.0000



1 1 2.15 50.00 2.15 0.00 0.00 0.002 -0.42 -1.15 -0.02 0.00 0.00 0.00

2 1 -2.15 50.00 2.15 0.00 0.00 0.002 -0.42 1.15 0.02 0.00 0.00 0.00

3 1 -2.15 50.00 -2.15 0.00 0.00 0.002 -0.42 1.15 -0.02 0.00 0.00 0.00

4 1 2.15 50.00 -2.15 0.00 0.00 0.002 -0.42 -1.15 0.02 0.00 0.00 0.00



1 1 1 50.00 -2.15 2.15 0.00 0.00 0.005 -50.00 2.15 -2.15 0.00 -15.04 -15.04

2 1 -1.15 0.42 -0.02 0.00 0.00 0.005 1.15 -0.42 0.02 0.00 0.12 2.92

2 1 2 50.00 2.15 2.15 0.00 0.00 0.007 -50.00 -2.15 -2.15 0.00 -15.04 15.04

2 2 1.15 0.42 0.02 0.00 0.00 0.007 -1.15 -0.42 -0.02 0.00 -0.12 2.92

3 1 3 50.00 2.15 -2.15 0.00 0.00 0.009 -50.00 -2.15 2.15 0.00 15.04 15.04

2 3 1.15 0.42 -0.02 0.00 0.00 0.009 -1.15 -0.42 0.02 0.00 0.12 2.92

4 1 4 50.00 -2.15 -2.15 0.00 0.00 0.0011 -50.00 2.15 2.15 0.00 15.04 -15.04

2 4 -1.15 0.42 0.02 0.00 0.00 0.0011 1.15 -0.42 -0.02 0.00 -0.12 2.92

Example Problem 22

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


5 1 5 2.09 25.00 0.06 1.32 -0.07 16.366 -2.09 0.00 -0.06 -1.32 -0.07 14.89

2 5 -0.10 -1.05 -0.06 -0.02 0.08 -2.546 0.10 1.05 0.06 0.02 0.06 -0.08

6 1 6 2.09 0.00 -0.06 -1.32 0.07 -14.897 -2.09 25.00 0.06 1.32 0.07 -16.36

2 6 0.10 -1.05 -0.06 -0.02 0.06 -0.087 -0.10 1.05 0.06 0.02 0.08 -2.54

7 1 7 2.09 25.00 0.06 1.32 -0.07 16.368 -2.09 0.00 -0.06 -1.32 -0.07 14.89

2 7 -0.04 0.14 0.08 -0.38 -0.08 0.148 0.04 -0.14 -0.08 0.38 -0.12 0.21

8 1 8 2.09 0.00 -0.06 -1.32 0.07 -14.899 -2.09 25.00 0.06 1.32 0.07 -16.36

2 8 -0.04 -0.14 -0.08 0.38 0.12 -0.219 0.04 0.14 0.08 -0.38 0.08 -0.14

9 1 9 2.09 25.00 0.06 1.32 -0.07 16.3610 -2.09 0.00 -0.06 -1.32 -0.07 14.89

2 9 0.10 1.05 0.06 0.02 -0.08 2.5410 -0.10 -1.05 -0.06 -0.02 -0.06 0.08

10 1 10 2.09 0.00 -0.06 -1.32 0.07 -14.8911 -2.09 25.00 0.06 1.32 0.07 -16.36

2 10 -0.10 1.05 0.06 0.02 -0.06 0.0811 0.10 -1.05 -0.06 -0.02 -0.08 2.54

11 1 11 2.09 25.00 0.06 1.32 -0.07 16.3612 -2.09 0.00 -0.06 -1.32 -0.07 14.89

2 11 0.04 -0.14 -0.08 0.38 0.08 -0.1412 -0.04 0.14 0.08 -0.38 0.12 -0.21

12 1 12 2.09 0.00 -0.06 -1.32 0.07 -14.895 -2.09 25.00 0.06 1.32 0.07 -16.36

2 12 0.04 0.14 0.08 -0.38 -0.12 0.215 -0.04 -0.14 -0.08 0.38 -0.08 0.14

13 1 6 0.12 0.00 0.00 0.00 0.00 -2.6413 -0.12 0.00 0.00 0.00 0.00 2.64

2 6 0.00 0.00 0.10 0.17 -0.12 0.0013 0.00 0.00 -0.10 -0.17 -0.13 0.00

14 1 13 0.12 0.00 0.00 0.00 0.00 -2.6410 -0.12 0.00 0.00 0.00 0.00 2.64

2 13 0.00 0.00 -0.10 -0.17 0.13 0.0010 0.00 0.00 0.10 0.17 0.12 0.00

15 1 8 0.12 0.00 0.00 0.00 0.00 -2.6413 -0.12 0.00 0.00 0.00 0.00 2.64

2 8 -0.03 0.24 0.00 0.00 0.00 0.7613 0.03 -0.24 0.00 0.00 0.00 -0.17

16 1 13 0.12 0.00 0.00 0.00 0.00 -2.6412 -0.12 0.00 0.00 0.00 0.00 2.64

2 13 0.03 0.24 0.00 0.00 0.00 -0.1712 -0.03 -0.24 0.00 0.00 0.00 0.76



Example Problem 22

212 47. FINI *********** END OF THE STAAD.Pro RUN ***********

**** DATE= TIME= ****


Example Problem 22

213

NOTES


Example Problem 22

214

NOTES

Example Problem 23

215


This example illustrates the usage of commands necessary to automatically generate spring supports for a slab on grade. The slab is subjected to pressure loading and analysis of the structure is performed. The numbers shown in the diagram below are the element numbers.


Example Problem 23

216

STAAD SPACE SLAB ON GRADE

Every STAAD input file has to begin with the word STAAD. The word SPACE signifies that the structure is a space frame and the geometry is defined through X, Y and Z axes. The remainder of the words form a title to identify this project.

UNIT METER KNS


JOINT COORDINATES 1 0.0 0.0 13.33 2 0.0 0.0 12.0 3 0.0 0.0 9.39 4 0.0 0.0 6.78 5 0.0 0.0 4.17 6 0.0 0.0 2.17 7 0.0 0.0 0.0 REPEAT ALL 3 2.83 0.0 0.0 REPEAT 3 2.67 0.0 0.0 REPEAT 5 2 0.0 0.0 REPEAT 3 2.67 0.0 0.0 REPEAT 3 2.83 0.0 0.0

For joints 1 through 7, the joint number followed by the X, Y and Z coordinates are specified above. The coordinates of these joints is used as a basis for generating 21 more joints by incrementing the X coordinate of each of these 7 joints by 2.83 metres, 3 times. REPEAT commands are used to generate the remaining joints of the structure. The results of the generation may be visually verified using the STAAD graphical viewing facilities.

ELEMENT INCIDENCES 1 1 8 9 2 TO 6 REPEAT 16 6 7

The incidences of element number 1 is defined and that data is used as a basis for generating the 2nd through the 6th element. The incidence pattern of the first 6 elements is then used to generate the

Example Problem 23

217

incidences of 96 (= 16 x 6) more elements using the REPEAT command.

UNIT CM ELEMENT PROPERTIES 1 TO 102 TH 14.0

The thickness of elements 1 to 102 is specified as 14 cms following the command ELEMENT PROPERTIES.

UNIT METER CONSTANTS E CONCRETE ALL POISSON CONCRETE ALL

The modulus of elasticity (E) and Poisson’s Ratio are specified following the command CONSTANTS. The built-in default value for concrete is used.

SUPPORTS 1 TO 126 ELASTIC MAT DIRECTION Y SUB 1570.

The above command is used to instruct STAAD to generate supports with springs which are effective in the global Y direction. These springs are located at nodes 1 to 126. The subgrade modulus of the soil is specified as 1570 KN/cu.m. The program will determine the area under the influence of each joint and multiply the influence area by the subgrade reaction to arrive at the spring stiffness for the "FY" degree of freedom at the joint. Additional information on this feature may be found in the STAAD Technical Reference Manual.

PRINT SUPP INFO

This command will enable us to obtain the details of the support conditions which were generated using the earlier commands.


Example Problem 23

218

LOAD 1 WEIGHT OF MAT & EARTH ELEMENT LOAD 1 TO 102 PR GY –74.2

The above data describe a static load case. A pressure load of 74.2 kN/sq.m. acting in the negative global Y direction is applied on all the 102 elements.

LOAD 2 'COLUMN LOAD-DL+LL' JOINT LOADS 1 2 FY -965. 8 9 FY -485. 5 FY -1373. 6 FY -2746. 22 23 FY -1824. 29 30 FY -912. 26 FY -2414. 27 FY -4828. 43 44 50 51 71 72 78 79 FY -1368. 47 54 82 FY -1175. 48 55 76 83 FY -2350. 92 93 FY -912. 99 100 FY -1824. 103 FY -2166. 104 FY -4333. 113 114 FY -485. 120 121 FY -965. 124 FY -1216. 125 FY -2431.

Load case 2 consists of several joint loads acting in the negative global Y direction.

LOADING COMBINATION 101 TOTAL LOAD 1 1. 2 1.

A load combination case, identified with load case number 101, is specified above. It instructs STAAD to factor loads 1 and 2 by a value of 1.0 and then algebraically add the results.

Example Problem 23

219

PERFORM ANALYSIS

The analysis is initiated using the above command.

UNIT CM LOAD LIST 101 PRINT JOINT DISPLACEMENTS LIST 33 56 PRINT ELEMENT STRESSES LIST 34 67

Joint displacements for joints 33 and 56, and element stresses for elements 34 and 67, for load case 101, is obtained with the help of the above commands.

FINISH



Example Problem 23


1. STAAD SPACE SLAB ON GRADE2. UNIT METER KNS3. JOINT COORDINATES4. 1 0.0 0.0 13.335. 2 0.0 0.0 12.06. 3 0.0 0.0 9.397. 4 0.0 0.0 6.788. 5 0.0 0.0 4.179. 6 0.0 0.0 2.17

10. 7 0.0 0.0 0.011. REPEAT ALL 3 2.83 0.0 0.012. REPEAT 3 2.67 0.0 0.013. REPEAT 5 2 0.0 0.014. REPEAT 3 2.67 0.0 0.015. REPEAT 3 2.83 0.0 0.016. ELEMENT INCIDENCES17. 1 1 8 9 2 TO 618. REPEAT 16 6 719. UNIT CM20. ELEMENT PROPERTIES21. 1 TO 102 TH 14.022. UNIT METER23. CONSTANTS24. E CONCRETE ALL25. POISSON CONCRETE ALL26. SUPPORTS27. 1 TO 126 ELASTIC MAT DIRECTION Y SUBGRADE 1570.28. PRINT SUPP INFO

SUPPORT INFORMATION (1=FIXED, 0=RELEASED)-------------------UNITS FOR SPRING CONSTANTS ARE KNS METE DEGREES

JOINT FORCE-X/ FORCE-Y/ FORCE-Z/ MOM-X/ MOM-Y/ MOM-Z/KFX KFY KFZ KMX KMY KMZ

1 1 0 1 0 1 00.0 1477.3 0.0 0.0 0.0 0.0

2 1 0 1 0 1 00.0 4376.5 0.0 0.0 0.0 0.0

3 1 0 1 0 1 00.0 5798.2 0.0 0.0 0.0 0.0

4 1 0 1 0 1 00.0 5798.2 0.0 0.0 0.0 0.0

5 1 0 1 0 1 00.0 5120.7 0.0 0.0 0.0 0.0

6 1 0 1 0 1 00.0 4631.9 0.0 0.0 0.0 0.0

7 1 0 1 0 1 00.0 2410.4 0.0 0.0 0.0 0.0

8 1 0 1 0 1 00.0 2954.7 0.0 0.0 0.0 0.0

9 1 0 1 0 1 00.0 8752.9 0.0 0.0 0.0 0.0

10 1 0 1 0 1 00.0 11596.5 0.0 0.0 0.0 0.0

11 1 0 1 0 1 00.0 11596.5 0.0 0.0 0.0 0.0

12 1 0 1 0 1 00.0 10241.3 0.0 0.0 0.0 0.0

13 1 0 1 0 1 00.0 9263.9 0.0 0.0 0.0 0.0

14 1 0 1 0 1 00.0 4820.8 0.0 0.0 0.0 0.0

Example Problem 23

221 SUPPORT INFORMATION (1=FIXED, 0=RELEASED)-------------------UNITS FOR SPRING CONSTANTS ARE KNS METE DEGREES


15 1 0 1 0 1 00.0 2954.7 0.0 0.0 0.0 0.0

16 1 0 1 0 1 00.0 8752.9 0.0 0.0 0.0 0.0

17 1 0 1 0 1 00.0 11596.5 0.0 0.0 0.0 0.0

18 1 0 1 0 1 00.0 11596.5 0.0 0.0 0.0 0.0

19 1 0 1 0 1 00.0 10241.3 0.0 0.0 0.0 0.0

20 1 0 1 0 1 00.0 9263.9 0.0 0.0 0.0 0.0

21 1 0 1 0 1 00.0 4820.8 0.0 0.0 0.0 0.0

22 1 0 1 0 1 00.0 2871.1 0.0 0.0 0.0 0.0

23 1 0 1 0 1 00.0 8505.5 0.0 0.0 0.0 0.0

24 1 0 1 0 1 00.0 11268.7 0.0 0.0 0.0 0.0

25 1 0 1 0 1 00.0 11268.7 0.0 0.0 0.0 0.0

26 1 0 1 0 1 00.0 9951.8 0.0 0.0 0.0 0.0

27 1 0 1 0 1 00.0 9002.0 0.0 0.0 0.0 0.0

28 1 0 1 0 1 00.0 4684.5 0.0 0.0 0.0 0.0

29 1 0 1 0 1 00.0 2787.6 0.0 0.0 0.0 0.0

30 1 0 1 0 1 00.0 8258.0 0.0 0.0 0.0 0.0

31 1 0 1 0 1 00.0 10940.9 0.0 0.0 0.0 0.0

32 1 0 1 0 1 00.0 10940.9 0.0 0.0 0.0 0.0

33 1 0 1 0 1 00.0 9662.3 0.0 0.0 0.0 0.0

34 1 0 1 0 1 00.0 8740.1 0.0 0.0 0.0 0.0

35 1 0 1 0 1 00.0 4548.2 0.0 0.0 0.0 0.0

36 1 0 1 0 1 00.0 2787.6 0.0 0.0 0.0 0.0

37 1 0 1 0 1 00.0 8258.0 0.0 0.0 0.0 0.0

38 1 0 1 0 1 00.0 10940.9 0.0 0.0 0.0 0.0

39 1 0 1 0 1 00.0 10940.9 0.0 0.0 0.0 0.0

40 1 0 1 0 1 00.0 9662.3 0.0 0.0 0.0 0.0

41 1 0 1 0 1 00.0 8740.1 0.0 0.0 0.0 0.0

42 1 0 1 0 1 00.0 4548.2 0.0 0.0 0.0 0.0

43 1 0 1 0 1 00.0 2437.9 0.0 0.0 0.0 0.0

44 1 0 1 0 1 00.0 7221.9 0.0 0.0 0.0 0.0

45 1 0 1 0 1 00.0 9568.1 0.0 0.0 0.0 0.0

46 1 0 1 0 1 00.0 9568.1 0.0 0.0 0.0 0.0

47 1 0 1 0 1 00.0 8450.0 0.0 0.0 0.0 0.0

48 1 0 1 0 1 00.0 7643.5 0.0 0.0 0.0 0.0

49 1 0 1 0 1 00.0 3977.6 0.0 0.0 0.0 0.0


Example Problem 23



50 1 0 1 0 1 00.0 2088.1 0.0 0.0 0.0 0.0

51 1 0 1 0 1 00.0 6185.8 0.0 0.0 0.0 0.0

52 1 0 1 0 1 00.0 8195.4 0.0 0.0 0.0 0.0

53 1 0 1 0 1 00.0 8195.4 0.0 0.0 0.0 0.0

54 1 0 1 0 1 00.0 7237.7 0.0 0.0 0.0 0.0

55 1 0 1 0 1 00.0 6546.9 0.0 0.0 0.0 0.0

56 1 0 1 0 1 00.0 3406.9 0.0 0.0 0.0 0.0

57 1 0 1 0 1 00.0 2088.1 0.0 0.0 0.0 0.0

58 1 0 1 0 1 00.0 6185.8 0.0 0.0 0.0 0.0

59 1 0 1 0 1 00.0 8195.4 0.0 0.0 0.0 0.0

60 1 0 1 0 1 00.0 8195.4 0.0 0.0 0.0 0.0

61 1 0 1 0 1 00.0 7237.7 0.0 0.0 0.0 0.0

62 1 0 1 0 1 00.0 6546.9 0.0 0.0 0.0 0.0

63 1 0 1 0 1 00.0 3406.9 0.0 0.0 0.0 0.0

64 1 0 1 0 1 00.0 2088.1 0.0 0.0 0.0 0.0

65 1 0 1 0 1 00.0 6185.8 0.0 0.0 0.0 0.0

66 1 0 1 0 1 00.0 8195.4 0.0 0.0 0.0 0.0

67 1 0 1 0 1 00.0 8195.4 0.0 0.0 0.0 0.0

68 1 0 1 0 1 00.0 7237.7 0.0 0.0 0.0 0.0

69 1 0 1 0 1 00.0 6546.9 0.0 0.0 0.0 0.0

70 1 0 1 0 1 00.0 3406.9 0.0 0.0 0.0 0.0

71 1 0 1 0 1 00.0 2088.1 0.0 0.0 0.0 0.0

72 1 0 1 0 1 00.0 6185.8 0.0 0.0 0.0 0.0

73 1 0 1 0 1 00.0 8195.4 0.0 0.0 0.0 0.0

74 1 0 1 0 1 00.0 8195.4 0.0 0.0 0.0 0.0

75 1 0 1 0 1 00.0 7237.7 0.0 0.0 0.0 0.0

76 1 0 1 0 1 00.0 6546.9 0.0 0.0 0.0 0.0

77 1 0 1 0 1 00.0 3406.9 0.0 0.0 0.0 0.0

78 1 0 1 0 1 00.0 2437.9 0.0 0.0 0.0 0.0

79 1 0 1 0 1 00.0 7221.9 0.0 0.0 0.0 0.0

80 1 0 1 0 1 00.0 9568.1 0.0 0.0 0.0 0.0

81 1 0 1 0 1 00.0 9568.1 0.0 0.0 0.0 0.0

82 1 0 1 0 1 00.0 8450.0 0.0 0.0 0.0 0.0

83 1 0 1 0 1 00.0 7643.5 0.0 0.0 0.0 0.0

84 1 0 1 0 1 00.0 3977.6 0.0 0.0 0.0 0.0

Example Problem 23



85 1 0 1 0 1 00.0 2787.6 0.0 0.0 0.0 0.0

86 1 0 1 0 1 00.0 8258.0 0.0 0.0 0.0 0.0

87 1 0 1 0 1 00.0 10940.9 0.0 0.0 0.0 0.0

88 1 0 1 0 1 00.0 10940.9 0.0 0.0 0.0 0.0

89 1 0 1 0 1 00.0 9662.3 0.0 0.0 0.0 0.0

90 1 0 1 0 1 00.0 8740.1 0.0 0.0 0.0 0.0

91 1 0 1 0 1 00.0 4548.2 0.0 0.0 0.0 0.0

92 1 0 1 0 1 00.0 2787.6 0.0 0.0 0.0 0.0

93 1 0 1 0 1 00.0 8258.0 0.0 0.0 0.0 0.0

94 1 0 1 0 1 00.0 10940.9 0.0 0.0 0.0 0.0

95 1 0 1 0 1 00.0 10940.9 0.0 0.0 0.0 0.0

96 1 0 1 0 1 00.0 9662.3 0.0 0.0 0.0 0.0

97 1 0 1 0 1 00.0 8740.1 0.0 0.0 0.0 0.0

98 1 0 1 0 1 00.0 4548.2 0.0 0.0 0.0 0.0

99 1 0 1 0 1 00.0 2871.1 0.0 0.0 0.0 0.0

100 1 0 1 0 1 00.0 8505.5 0.0 0.0 0.0 0.0

101 1 0 1 0 1 00.0 11268.7 0.0 0.0 0.0 0.0

102 1 0 1 0 1 00.0 11268.7 0.0 0.0 0.0 0.0

103 1 0 1 0 1 00.0 9951.8 0.0 0.0 0.0 0.0

104 1 0 1 0 1 00.0 9002.0 0.0 0.0 0.0 0.0

105 1 0 1 0 1 00.0 4684.5 0.0 0.0 0.0 0.0

106 1 0 1 0 1 00.0 2954.7 0.0 0.0 0.0 0.0

107 1 0 1 0 1 00.0 8752.9 0.0 0.0 0.0 0.0

108 1 0 1 0 1 00.0 11596.5 0.0 0.0 0.0 0.0

109 1 0 1 0 1 00.0 11596.5 0.0 0.0 0.0 0.0

110 1 0 1 0 1 00.0 10241.4 0.0 0.0 0.0 0.0

111 1 0 1 0 1 00.0 9263.9 0.0 0.0 0.0 0.0

112 1 0 1 0 1 00.0 4820.8 0.0 0.0 0.0 0.0

113 1 0 1 0 1 00.0 2954.7 0.0 0.0 0.0 0.0

114 1 0 1 0 1 00.0 8752.9 0.0 0.0 0.0 0.0

115 1 0 1 0 1 00.0 11596.5 0.0 0.0 0.0 0.0

116 1 0 1 0 1 00.0 11596.5 0.0 0.0 0.0 0.0

117 1 0 1 0 1 00.0 10241.4 0.0 0.0 0.0 0.0

118 1 0 1 0 1 00.0 9263.9 0.0 0.0 0.0 0.0

119 1 0 1 0 1 00.0 4820.8 0.0 0.0 0.0 0.0


Example Problem 23



120 1 0 1 0 1 00.0 1477.3 0.0 0.0 0.0 0.0

121 1 0 1 0 1 00.0 4376.5 0.0 0.0 0.0 0.0

122 1 0 1 0 1 00.0 5798.2 0.0 0.0 0.0 0.0

123 1 0 1 0 1 00.0 5798.2 0.0 0.0 0.0 0.0

124 1 0 1 0 1 00.0 5120.7 0.0 0.0 0.0 0.0

125 1 0 1 0 1 00.0 4631.9 0.0 0.0 0.0 0.0

126 1 0 1 0 1 00.0 2410.4 0.0 0.0 0.0 0.0


29. LOAD 1 'WEIGHT OF MAT & EARTH'30. ELEMENT LOAD31. 1 TO 102 PR GY -74.232. LOAD 2 'COLUMN LOAD-DL+LL'33. JOINT LOADS34. 1 2 FY -965.35. 8 9 FY -485.36. 5 FY -1373.37. 6 FY -2746.38. 22 23 FY -1824.39. 29 30 FY -912.40. 26 FY -2414.41. 27 FY -4828.42. 43 44 50 51 71 72 78 79 FY -1368.43. 47 54 82 FY -1175.44. 48 55 76 83 FY -2350.45. 92 93 FY -912.46. 99 100 FY -1824.47. 103 FY -2166.48. 104 FY -4333.49. 113 114 FY -485.50. 120 121 FY -965.51. 124 FY -1216.52. 125 FY -2431.53. LOADING COMBINATION 101 TOTAL LOAD54. 1 1. 2 1.55. PERFORM ANALYSIS



56. UNIT CM57. LOAD LIST 10158. PRINT JOINT DISPLACEMENTS LIST 33 56



33 101 0.0000 -10.5859 0.0000 -0.0256 0.0000 0.054356 101 0.0000 -12.2476 0.0000 0.0647 0.0000 0.0282


Example Problem 23

225 59. PRINT ELEMENT STRESSES LIST 34 67

ELEMENT STRESSES FORCE,LENGTH UNITS= KNS CM----------------



34 101 -0.02 -0.03 2.43 18.01 40.712.22 2.22 0.00 0.00 0.002.54 2.54

TOP : SMAX= 1.58 SMIN= -0.96 TMAX= 1.27 ANGLE= -39.6BOTT: SMAX= 0.96 SMIN= -1.58 TMAX= 1.27 ANGLE= -39.6

67 101 0.28 0.03 63.64 40.28 88.134.97 4.97 0.00 0.00 0.005.44 5.44

TOP : SMAX= 4.31 SMIN= -1.13 TMAX= 2.72 ANGLE= 41.2BOTT: SMAX= 1.13 SMIN= -4.31 TMAX= 2.72 ANGLE= 41.2





60. FINISH


**** DATE= TIME= ****



Example Problem 23

226

NOTES

Example Problem 24

227


This is an example of the analysis of a structure modelled using “SOLID” finite elements. This example also illustrates the method for applying an “enforced” displacement on the structure.


Example Problem 24

228

STAAD SPACE *EXAMPLE PROBLEM USING SOLID ELEMENTS

Every STAAD input file has to begin with the word STAAD. The word SPACE signifies that the structure is a space frame and the geometry is defined through X, Y and Z axes. The comment line which begins with an asterisk is an optional title to identify this project.

UNIT KNS MET


JOINT COORDINATES 1 0.0 0.0 2.0 4 0.0 3.0 2.0 5 1.0 0.0 2.0 8 1.0 3.0 2.0 9 2.0 0.0 2.0 12 2.0 3.0 2.0 21 0.0 0.0 1.0 24 0.0 3.0 1.0 25 1.0 0.0 1.0 28 1.0 3.0 1.0 29 2.0 0.0 1.0 32 2.0 3.0 1.0 41 0.0 0.0 0.0 44 0.0 3.0 0.0 45 1.0 0.0 0.0 48 1.0 3.0 0.0 49 2.0 0.0 0.0 52 2.0 3.0 0.0

The joint number followed by the X, Y and Z coordinates are specified above. The coordinates of some of those nodes are generated utilizing the fact that they are equally spaced between the extremities.

ELEMENT INCIDENCES SOLID 1 1 5 6 2 21 25 26 22 TO 3 4 21 25 26 22 41 45 46 42 TO 6 1 1 7 5 9 10 6 25 29 30 26 TO 9 1 1 10 25 29 30 26 45 49 50 46 TO 12 1 1

The incidences of solid elements are defined above. The word SOLID is used to signify that these are 8-noded solid elements as opposed to 3-noded or 4-noded plate elements. Each line contains the data for generating 3 elements. For example, element number 1 is first defined by all of its 8 nodes. Then, increments of 1 to the

Example Problem 24

229

joint number and 1 to the element number (the defaults) are used for generating incidences for elements 2 and 3. Similarly, incidences of elements 4, 7 and 10 are defined while those of 5, 6, 8, 9, 11 and 12 are generated.

CONSTANTS E 2.1E7 ALL POIS 0.25 ALL DENSITY 7.5 ALL

Following the command CONSTANTS above, the material constants such as E (Modulus of Elasticity), Poisson's Ratio, and Density are specified.

PRINT ELEMENT INFO SOLID LIST 1 TO 5

This command will enable us to obtain, in a tabular form, the details of the incidences and material property values of elements 1 to 5.

SUPPORTS 1 5 21 25 29 41 45 49 PINNED 9 ENFORCED

The above lines contain the data for supports for the model. The ENFORCED support condition is used to declare a point at which an enforced displacement load is applied later (see load case 3).

LOAD 1 SELF Y -1.0 JOINT LOAD 28 FY -1000.0

The above data describe a static load case. It consists of selfweight loading and a joint load, both in the negative global Y direction.

LOAD 2 JOINT LOADS 2 TO 4 22 TO 24 42 TO 44 FX 100.0


Example Problem 24

230

Load case 2 consists of several joint loads acting in the positive global X direction.

LOAD 3 SUPPORT DISPLACEMENT 9 FX 0.0011

Load case 3 consists of an enforced displacement along the global X direction at node 9. The displacement in the other enforced support degrees of freedom will default to zero.

UNIT POUND FEET LOAD 4 ELEMENT LOAD SOLIDS 3 6 9 12 FACE 4 PRE GY -500.0

In Load case 4, a pressure load of 500 pounds/sq.ft is applied on Face # 4 of solid elements 3, 6, 9 and 12. Face 4 is defined as shown in the following table :

SURFACE JOINTS FACE NUMBER f1 f2 f3 f4

1 front Jt 1 Jt 4 Jt 3 Jt 2 2 bottom Jt 1 Jt 2 Jt 6 Jt 5 3 left Jt 1 Jt 5 Jt 8 Jt 4 4 top Jt 4 Jt 8 Jt 7 Jt 3 5 right Jt 2 Jt 3 Jt 7 Jt 6 6 back Jt 5 Jt 6 Jt 7 Jt 8 The above table, and other details of this type of loading can be found in section 5.32.3.2 of the STAAD.Pro 2003 Technical Reference manual.

UNIT KNS MMS LOAD 5 REPEAT LOAD 1 1.0 2 1.0 3 1.0 4 1.0

Example Problem 24

231

Load case 5 illustrates the technique employed to instruct STAAD to create a load case which consists of data to be assembled from other load cases already specified earlier. We would like the program to analyze the structure for loads from cases 1 through 4 acting simultaneously. In other words, the above instruction is the same as the following:

LOAD 5 SELF Y -1.0 JOINT LOAD 28 FY -1000.0 2 TO 4 22 TO 24 42 TO 44 FX 100.0 SUPPORT DISPLACEMENT 9 FX .0011 ELEMENT LOAD SOLIDS 3 6 9 12 FACE 4 PRE GY -500.0

LOAD COMB 10 1 1.0 2 1.0

Load case 10 is a combination load case, which combines the effects of cases 1 & 2. While the syntax of this might look very similar to that of the REPEAT LOAD case shown in case 5, there is a fundamental difference. In a REPEAT LOAD case, the program computes the displacements by multiplying the inverted stiffness matrix by the load vector built for the REPEAT LOAD case. But in solving load combination cases, the program merely calculates the end results (displacements, forces, reactions) by gathering up the corresponding values from the individual components of the combination case, factoring them, and then algebraically summing them up. This difference in approach is quite important in that non-linear problems such as PDELTA ANALYSIS, MEMBER TENSION and MEMBER COMPRESSION situations, changes in support conditions etc. should be handled using REPEAT LOAD cases, not load combination cases.

PERFORM ANALYSIS PRINT STATICS CHECK

A static equilibrium report, consisting of total applied loading and total support reactions from each primary load case is requested along with the instructions to carry out a linear static analysis.


Example Problem 24

232 PRINT JOINT DISPLACEMENTS LIST 8 9

Global displacements at nodes 8 and 9 are obtained using the above command.

UNIT KNS METER PRINT SUPPORT REACTIONS

Reactions at the supports are obtained using the above command.

UNIT NEWTON MMS PRINT ELEMENT JOINT STRESS SOLID LIST 4 6

This command requests the program to provide the element stress results at the nodes of elements 4 and 6. The results will be printed for all the load cases. The word SOLID is used to signify that these are solid elements as opposed to plate or shell elements.

FINISH


Example Problem 24

233


1. STAAD SPACE EXAMPLE PROBLEM USING SOLID ELEMENTS2. UNIT KNS MET3. JOINT COORDINATES4. 1 0.0 0.0 2.0 4 0.0 3.0 2.05. 5 1.0 0.0 2.0 8 1.0 3.0 2.06. 9 2.0 0.0 2.0 12 2.0 3.0 2.07. 21 0.0 0.0 1.0 24 0.0 3.0 1.08. 25 1.0 0.0 1.0 28 1.0 3.0 1.09. 29 2.0 0.0 1.0 32 2.0 3.0 1.0

10. 41 0.0 0.0 0.0 44 0.0 3.0 0.011. 45 1.0 0.0 0.0 48 1.0 3.0 0.012. 49 2.0 0.0 0.0 52 2.0 3.0 0.014. ELEMENT INCIDENCES SOLID15. 1 1 5 6 2 21 25 26 22 TO 316. 4 21 25 26 22 41 45 46 42 TO 6 1 117. 7 5 9 10 6 25 29 30 26 TO 9 1 118. 10 25 29 30 26 45 49 50 46 TO 12 1 120. CONSTANTS21. E 2.1E7 ALL22. POIS 0.25 ALL23. DENSITY 7.5 ALL25. PRINT ELEMENT INFO SOLID LIST 1 TO 5

ELEMENT NODE-1 NODE-2 NODE-3 NODE-4 NODE-5 NODE-6 NODE-7 NODE-8

1 1 5 6 2 21 25 26 222 2 6 7 3 22 26 27 233 3 7 8 4 23 27 28 244 21 25 26 22 41 45 46 425 22 26 27 23 42 46 47 43

MATERIAL PROPERTIES.--------------------ALL UNITS ARE - KNS MET

ELEMENT YOUNG'S MODULUS MODULUS OF RIGIDITY DENSITY ALPHA

1 2.1000002E+07 0.0000000E+00 7.5000E+00 0.0000E+002 2.1000002E+07 0.0000000E+00 7.5000E+00 0.0000E+003 2.1000002E+07 0.0000000E+00 7.5000E+00 0.0000E+004 2.1000002E+07 0.0000000E+00 7.5000E+00 0.0000E+005 2.1000002E+07 0.0000000E+00 7.5000E+00 0.0000E+00

27. SUPPORTS28. 1 5 21 25 29 41 45 49 PINNED29. 9 ENFORCED31. LOAD 132. SELF Y -1.033. JOINT LOAD34. 28 FY -1000.036. LOAD 237. JOINT LOADS38. 2 TO 4 22 TO 24 42 TO 44 FX 100.040. LOAD 341. SUPPORT DISPLACEMENT42. 9 FX .001144. UNIT POUND FEET45. LOAD 446. ELEMENT LOAD SOLIDS47. 3 6 9 12 FACE 4 PRE GY -500.049. UNIT KNS MMS51. LOAD 5


Example Problem 24

234 52. REPEAT LOAD53. 1 1.0 2 1.0 3 1.0 4 1.055. LOAD COMB 1056. 1 1.0 2 1.058. PERFORM ANALYSIS PRINT STAT CHECK



ZERO STIFFNESS IN DIRECTION 4 AT JOINT 9 EQN.NO. 22LOADS APPLIED OR DISTRIBUTED HERE FROM ELEMENTS WILL BE IGNORED.THIS MAY BE DUE TO ALL MEMBERS AT THIS JOINT BEING RELEASED OREFFECTIVELY RELEASED IN THIS DIRECTION.

Note - Some or all of the rotational zero stiffness warnings may be due to solidelements in the model. Solids do not have rotational stiffnesses at

nodes.ZERO STIFFNESS IN DIRECTION 5 AT JOINT 9 EQN.NO. 23ZERO STIFFNESS IN DIRECTION 6 AT JOINT 9 EQN.NO. 24

STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 1

***TOTAL APPLIED LOAD ( KNS MMS ) SUMMARY (LOADING 1 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = -1090.00SUMMATION FORCE-Z = 0.00


***TOTAL REACTION LOAD( KNS MMS ) SUMMARY (LOADING 1 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = 1090.00SUMMATION FORCE-Z = 0.00



X = -1.12983E-03 23Y = -1.01204E-02 28Z = 1.12983E-03 7RX= 0.00000E+00 0RY= 0.00000E+00 0RZ= 0.00000E+00 0


***TOTAL APPLIED LOAD ( KNS MMS ) SUMMARY (LOADING 2 )SUMMATION FORCE-X = 900.00SUMMATION FORCE-Y = 0.00SUMMATION FORCE-Z = 0.00


***TOTAL REACTION LOAD( KNS MMS ) SUMMARY (LOADING 2 )SUMMATION FORCE-X = -900.00SUMMATION FORCE-Y = 0.00SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE ORIGIN-MX= -0.01 MY= -899999.96 MZ= 1799999.92

Example Problem 24

235 MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 2)

MAXIMUMS AT NODEX = 2.22892E-02 4Y = 7.83934E-03 4Z = 9.49033E-04 10RX= 0.00000E+00 0RY= 0.00000E+00 0RZ= 0.00000E+00 0


***TOTAL APPLIED LOAD ( KNS MMS ) SUMMARY (LOADING 3 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = 0.00SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE ORIGIN-MX= 0.00 MY= 0.00 MZ= 0.00


SUMMATION OF MOMENTS AROUND THE ORIGIN-MX= 0.03 MY= -0.20 MZ= -0.04


X = 1.10000E-01 9Y = -1.21497E-02 6Z = 1.61372E-02 24RX= 0.00000E+00 0RY= 0.00000E+00 0RZ= 0.00000E+00 0







X = 3.17652E-05 50Y = -3.35288E-04 28Z = -3.17652E-05 50RX= 0.00000E+00 0RY= 0.00000E+00 0RZ= 0.00000E+00 0


Example Problem 24

236 STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 5



***TOTAL REACTION LOAD( KNS MMS ) SUMMARY (LOADING 5 )SUMMATION FORCE-X = -900.00SUMMATION FORCE-Y = 1185.76SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE ORIGIN-MX= -1185760.43 MY= -899999.85 MZ= 2985760.30


X = 1.10000E-01 9Y = -1.66887E-02 12Z = 1.62734E-02 4RX= 0.00000E+00 0RY= 0.00000E+00 0RZ= 0.00000E+00 0


60. PRINT JOINT DISPLACEMENTS LIST 8 9



8 1 0.0000 -0.0010 -0.0008 0.0000 0.0000 0.00002 0.0200 0.0001 0.0000 0.0000 0.0000 0.00003 0.0193 -0.0049 0.0089 0.0000 0.0000 0.00004 0.0000 -0.0003 0.0000 0.0000 0.0000 0.00005 0.0393 -0.0062 0.0081 0.0000 0.0000 0.0000

10 0.0200 -0.0009 -0.0009 0.0000 0.0000 0.00009 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

2 0.0000 0.0000 0.0000 0.0000 0.0000 0.00003 0.1100 0.0000 0.0000 0.0000 0.0000 0.00004 0.0000 0.0000 0.0000 0.0000 0.0000 0.00005 0.1100 0.0000 0.0000 0.0000 0.0000 0.0000

10 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000


62. UNIT KNS METER63. PRINT SUPPORT REACTIONS



1 1 16.05 74.37 -16.05 0.00 0.00 0.002 -72.24 -232.67 42.18 0.00 0.00 0.003 -202.27 -30.20 -119.24 0.00 0.00 0.004 1.52 6.63 -1.52 0.00 0.00 0.005 -256.94 -181.87 -94.63 0.00 0.00 0.00

10 -56.19 -158.30 26.13 0.00 0.00 0.005 1 0.00 135.25 -31.85 0.00 0.00 0.00

2 -62.32 11.42 -0.05 0.00 0.00 0.003 -1641.00 743.48 -228.79 0.00 0.00 0.004 0.00 11.97 -2.98 0.00 0.00 0.005 -1703.32 902.13 -263.67 0.00 0.00 0.00

10 -62.32 146.68 -31.90 0.00 0.00 0.00

Example Problem 24

237 SUPPORT REACTIONS -UNIT KNS METE STRUCTURE TYPE = SPACE-----------------


21 1 31.85 135.25 0.00 0.00 0.00 0.002 -159.92 -450.84 0.00 0.00 0.00 0.003 -334.17 -292.36 -187.70 0.00 0.00 0.004 2.98 11.97 0.00 0.00 0.00 0.005 -459.26 -595.97 -187.70 0.00 0.00 0.00

10 -128.07 -315.58 0.00 0.00 0.00 0.0025 1 0.00 251.52 0.00 0.00 0.00 0.00

2 -138.00 9.51 0.00 0.00 0.00 0.003 -1919.80 524.87 -1097.52 0.00 0.00 0.004 0.00 21.34 0.00 0.00 0.00 0.005 -2057.79 807.24 -1097.52 0.00 0.00 0.00

10 -138.00 261.03 0.00 0.00 0.00 0.0029 1 -31.85 135.25 0.00 0.00 0.00 0.00

2 -170.27 431.34 0.00 0.00 0.00 0.003 390.27 51.20 384.26 0.00 0.00 0.004 -2.98 11.97 0.00 0.00 0.00 0.005 185.18 629.76 384.26 0.00 0.00 0.00

10 -202.12 566.59 0.00 0.00 0.00 0.0041 1 16.05 74.37 16.05 0.00 0.00 0.00

2 -72.24 -232.67 -42.18 0.00 0.00 0.003 -89.12 -273.99 -159.85 0.00 0.00 0.004 1.52 6.63 1.52 0.00 0.00 0.005 -143.78 -425.66 -184.46 0.00 0.00 0.00

10 -56.19 -158.30 -26.13 0.00 0.00 0.0045 1 0.00 135.25 31.85 0.00 0.00 0.00

2 -62.32 11.42 0.05 0.00 0.00 0.003 -43.04 -75.25 -23.76 0.00 0.00 0.004 0.00 11.97 2.98 0.00 0.00 0.005 -105.36 83.40 11.12 0.00 0.00 0.00

10 -62.32 146.68 31.90 0.00 0.00 0.0049 1 -16.05 74.37 16.05 0.00 0.00 0.00

2 -81.35 226.24 45.03 0.00 0.00 0.003 -77.83 207.38 119.24 0.00 0.00 0.004 -1.52 6.63 1.52 0.00 0.00 0.005 -176.75 514.62 181.84 0.00 0.00 0.00

10 -97.40 300.61 61.08 0.00 0.00 0.009 1 -16.05 74.37 -16.05 0.00 0.00 0.00

2 -81.35 226.24 -45.03 0.00 0.00 0.003 3916.95 -855.13 1313.37 0.00 0.00 0.004 -1.52 6.63 -1.52 0.00 0.00 0.005 3818.02 -547.89 1250.77 0.00 0.00 0.00

10 -97.40 300.61 -61.08 0.00 0.00 0.00


65. UNIT NEWTON MMS66. PRINT ELEMENT JOINT STRESS SOLID LIST 4 6

ELEMENT STRESSES UNITS= NEWTMMS-------------------------------------------------------------------------------

NODE/ NORMAL STRESSES SHEAR STRESSESELEMENT LOAD CENTER SXX SYY SZZ SXY SYZ SZX-------------------------------------------------------------------------------

4 1 21 -0.088 -0.280 -0.098 0.001 -0.003 0.0004 1 25 -0.076 -0.204 -0.076 -0.003 -0.003 0.0054 1 26 -0.008 -0.214 -0.008 0.004 0.004 0.0054 1 22 -0.011 -0.280 -0.002 0.009 -0.011 0.0094 1 41 -0.095 -0.311 -0.095 -0.008 -0.008 -0.0054 1 45 -0.098 -0.280 -0.088 -0.003 0.001 0.0004 1 46 -0.002 -0.280 -0.011 -0.011 0.009 0.0094 1 42 0.011 -0.301 0.011 -0.016 -0.016 0.014

4 1 CENTER -0.046 -0.269 -0.046 -0.003 -0.003 0.005S1= -0.041 S2= -0.051 S3= -0.269 SE= 0.223DC= 0.707 -0.021 0.707 -0.707 0.000 0.707


Example Problem 24

238 ELEMENT STRESSES UNITS= NEWTMMS-------------------------------------------------------------------------------


4 2 21 0.176 1.021 0.284 0.217 0.014 0.0054 2 25 0.154 -0.006 0.022 0.251 0.014 -0.0294 2 26 -0.028 0.053 -0.015 0.253 0.016 -0.0024 2 22 -0.054 1.031 0.103 0.219 0.012 -0.0364 2 41 0.189 1.034 0.321 0.258 0.038 0.0294 2 45 0.162 -0.006 0.054 0.223 -0.010 -0.0054 2 46 -0.225 -0.016 -0.051 0.221 -0.008 -0.0264 2 42 -0.247 0.976 0.071 0.255 0.036 -0.060

4 2 CENTER 0.016 0.511 0.099 0.237 0.014 -0.015S1= 0.606 S2= 0.101 S3= -0.082 SE= 0.617DC= 0.372 0.928 0.014 -0.106 0.027 0.994

4 3 21 0.090 0.518 0.188 0.499 0.506 0.0404 3 25 -0.089 -0.574 -0.129 0.661 0.143 -0.1234 3 26 0.525 -0.328 0.365 0.565 0.047 0.3274 3 22 0.538 0.597 0.183 0.403 0.602 0.1654 3 41 0.215 0.951 0.255 0.011 0.589 0.1234 3 45 0.228 0.435 0.129 -0.152 0.060 -0.0404 3 46 -0.133 0.355 0.298 -0.056 -0.036 0.2444 3 42 -0.313 0.705 -0.076 0.107 0.685 0.082

4 3 CENTER 0.132 0.332 0.152 0.255 0.324 0.102S1= 0.703 S2= 0.041 S3= -0.128 SE= 0.760DC= 0.425 0.744 0.516 0.809 -0.055 -0.586

4 4 21 -0.008 -0.024 -0.008 -0.001 -0.001 0.0004 4 25 -0.008 -0.022 -0.008 -0.001 -0.001 0.0004 4 26 0.001 -0.022 0.001 -0.001 -0.001 0.0004 4 22 0.001 -0.024 0.001 -0.001 -0.001 0.0004 4 41 -0.008 -0.026 -0.008 -0.001 -0.001 0.0004 4 45 -0.008 -0.024 -0.008 -0.001 -0.001 0.0004 4 46 0.001 -0.024 0.001 -0.001 -0.001 0.0004 4 42 0.001 -0.026 0.001 -0.001 -0.001 0.000

4 4 CENTER -0.004 -0.024 -0.004 -0.001 -0.001 0.000S1= -0.003 S2= -0.004 S3= -0.024 SE= 0.021DC= 0.705 -0.070 0.705 -0.707 0.000 0.707

4 5 21 0.170 1.235 0.366 0.716 0.515 0.0454 5 25 -0.019 -0.806 -0.191 0.908 0.153 -0.1474 5 26 0.490 -0.512 0.343 0.822 0.066 0.3304 5 22 0.474 1.324 0.285 0.630 0.602 0.1384 5 41 0.300 1.649 0.472 0.259 0.618 0.1474 5 45 0.283 0.125 0.087 0.067 0.050 -0.0454 5 46 -0.360 0.035 0.238 0.153 -0.036 0.2284 5 42 -0.548 1.354 0.007 0.345 0.704 0.036

4 5 CENTER 0.099 0.551 0.201 0.487 0.334 0.092S1= 1.000 S2= 0.093 S3= -0.243 SE= 1.114DC= 0.469 0.795 0.386 -0.452 -0.160 0.878

4 10 21 0.088 0.741 0.186 0.218 0.011 0.0054 10 25 0.078 -0.210 -0.054 0.247 0.011 -0.0244 10 26 -0.036 -0.161 -0.022 0.257 0.020 0.0034 10 22 -0.065 0.751 0.102 0.228 0.001 -0.0274 10 41 0.093 0.724 0.225 0.249 0.030 0.0244 10 45 0.064 -0.286 -0.034 0.220 -0.009 -0.0054 10 46 -0.227 -0.296 -0.062 0.210 0.001 -0.0164 10 42 -0.236 0.675 0.082 0.240 0.020 -0.046

4 10 CENTER -0.030 0.242 0.053 0.234 0.011 -0.011S1= 0.376 S2= 0.054 S3= -0.165 SE= 0.472DC= 0.498 0.867 0.012 -0.064 0.023 0.998

Example Problem 24



6 1 23 0.317 0.428 0.402 -0.043 -0.126 -0.0606 1 27 -0.082 -1.708 -0.082 -0.098 -0.098 -0.0056 1 28 -0.670 -1.819 -0.670 -0.552 -0.552 -0.0056 1 24 -0.160 0.428 0.146 -0.497 0.329 0.0516 1 43 -0.108 -0.163 -0.108 -0.181 -0.181 -0.1156 1 47 0.402 0.428 0.317 -0.126 -0.043 -0.0606 1 48 0.146 0.428 -0.160 0.329 -0.497 0.0516 1 44 -0.253 -0.052 -0.253 0.273 0.273 0.106

6 1 CENTER -0.051 -0.254 -0.051 -0.112 -0.112 -0.005S1= 0.032 S2= -0.046 S3= -0.341 SE= 0.341DC= 0.619 -0.484 0.619 -0.707 0.000 0.707

6 2 23 -0.032 0.112 -0.001 0.030 -0.002 0.0166 2 27 -0.001 -0.025 -0.046 0.073 -0.013 -0.0276 2 28 -0.096 -0.003 -0.065 0.083 -0.003 -0.0356 2 24 -0.085 0.177 0.109 0.040 -0.012 -0.0786 2 43 -0.152 0.158 0.052 0.136 -0.023 -0.0056 2 47 -0.140 -0.041 -0.013 0.092 0.008 -0.0496 2 48 -0.496 -0.105 -0.119 0.082 0.019 -0.0146 2 44 -0.464 0.136 0.076 0.125 -0.033 -0.057

6 2 CENTER -0.183 0.051 -0.001 0.083 -0.007 -0.031S1= 0.081 S2= -0.001 S3= -0.213 SE= 0.263DC= 0.314 0.928 -0.202 -0.060 0.232 0.971

6 3 23 -0.274 -0.053 -0.004 -0.033 -0.047 0.0416 3 27 -0.314 -0.056 -0.102 0.064 0.030 -0.0566 3 28 0.182 0.057 0.061 0.040 0.006 0.0216 3 24 0.190 0.028 0.065 -0.057 -0.023 -0.0766 3 43 0.064 -0.048 0.069 -0.003 -0.031 0.0566 3 47 0.072 0.094 0.019 -0.100 0.014 -0.0416 3 48 -0.014 0.013 -0.012 -0.076 -0.010 0.0066 3 44 -0.053 -0.160 -0.056 0.021 -0.007 -0.091

6 3 CENTER -0.018 -0.016 0.005 -0.018 -0.009 -0.017S1= 0.014 S2= -0.001 S3= -0.042 SE= 0.051DC= -0.484 0.038 0.874 -0.507 0.802 -0.316

6 4 23 0.000 -0.024 0.000 0.000 0.000 0.0006 4 27 0.000 -0.024 0.000 0.000 0.000 0.0006 4 28 0.000 -0.024 0.000 0.000 0.000 0.0006 4 24 0.000 -0.024 0.000 0.000 0.000 0.0006 4 43 0.000 -0.024 0.000 0.000 0.000 0.0006 4 47 0.000 -0.024 0.000 0.000 0.000 0.0006 4 48 0.000 -0.024 0.000 0.000 0.000 0.0006 4 44 0.000 -0.024 0.000 0.000 0.000 0.000

6 4 CENTER 0.000 -0.024 0.000 0.000 0.000 0.000S1= 0.000 S2= 0.000 S3= -0.024 SE= 0.024DC= -0.707 0.000 0.707 0.707 -0.002 0.707

6 5 23 0.010 0.463 0.397 -0.046 -0.174 -0.0036 5 27 -0.397 -1.813 -0.230 0.039 -0.081 -0.0886 5 28 -0.585 -1.789 -0.674 -0.429 -0.550 -0.0186 5 24 -0.055 0.609 0.320 -0.514 0.294 -0.1036 5 43 -0.196 -0.077 0.014 -0.049 -0.236 -0.0646 5 47 0.334 0.458 0.323 -0.134 -0.020 -0.1496 5 48 -0.363 0.312 -0.291 0.335 -0.489 0.0436 5 44 -0.770 -0.101 -0.233 0.420 0.233 -0.042

6 5 CENTER -0.253 -0.242 -0.047 -0.047 -0.128 -0.053S1= 0.019 S2= -0.210 S3= -0.351 SE= 0.323DC= -0.102 -0.422 0.901 0.819 -0.550 -0.165


Example Problem 24



6 10 23 0.285 0.540 0.401 -0.013 -0.128 -0.0446 10 27 -0.083 -1.733 -0.129 -0.025 -0.111 -0.0326 10 28 -0.766 -1.822 -0.735 -0.469 -0.555 -0.0396 10 24 -0.245 0.605 0.255 -0.457 0.317 -0.0286 10 43 -0.259 -0.005 -0.055 -0.046 -0.204 -0.1206 10 47 0.262 0.388 0.304 -0.034 -0.034 -0.1086 10 48 -0.350 0.323 -0.279 0.411 -0.479 0.0376 10 44 -0.717 0.084 -0.177 0.399 0.240 0.049

6 10 CENTER -0.234 -0.203 -0.052 -0.029 -0.119 -0.036S1= 0.015 S2= -0.206 S3= -0.298 SE= 0.278DC= -0.070 -0.472 0.879 0.822 -0.526 -0.217

67. FINISH


**** DATE= TIME= ****


Example Problem 25

241


This example demonstrates the usage of compression-only members. Since the structural condition is load dependent, the PERFORM ANALYSIS command is specified once for each primary load case.


Example Problem 25

242

This example has been created to illustrate the command specification for a structure with certain members capable of carrying compressive force only. It is important to note that the analysis can be done for only 1 load case at a time. This is because, the set of ‘active’ members (and hence the stiffness matrix) is load case dependent.

STAAD PLANE * EXAMPLE FOR COMPRESSION-ONLY MEMBERS

The input data is initiated with the word STAAD. This structure is a PLANE frame. The second line is an optional comment line.

UNIT METER KNS

Units for the commands to follow are specified above.

SET NL 3


JOINT COORDINATES 1 0 0 ; 2 0 3.5 ; 3 0 7.0 ; 4 5.25 7.0 ; 5 5.25 3.5 ; 6 5.25 0

Joint coordinates of joints 1 to 6 are defined above.

Example Problem 25

243 MEMBER INCIDENCES 1 1 2 5 ; 6 1 5 ; 7 2 6 ; 8 2 4 ; 9 3 5 ; 10 2 5

The members 1 to 10 are defined along with the joints they are connected to.

MEMBER COMPRESSION 6 TO 9

Members 6 to 9 are defined as COMPRESSION-only members. Hence for each load case, if during the analysis, any of the members 6 to 9 is found to be carrying a tensile force, it is disabled from the structure and the analysis is carried out again with the modified structure.

MEMBER PROPERTY BRITISH 1 TO 10 TA ST UC152X152X30

Members 1 to 10 are assigned the UC152X152X30 section from the British steel table.

CONSTANTS E STEEL ALL DEN STEEL ALL POISSON STEEL ALL

Following the command CONSTANTS, material constants such as E (Modulus of Elasticity), density and Poisson’s ratio are specified. In this case, the built-in default value of steel is assigned.

SUPPORT 1 6 PINNED

Joints 1 and 6 are declared as pinned-supported.


Example Problem 25

244 LOAD 1 JOINT LOAD 2 FX 70 3 FX 45

Load 1 is defined above and consists of joint loads in the global X direction at joints 2 and 3.

PERFORM ANALYSIS

The above structure is analyzed for load case 1.

CHANGE MEMBER COMPRESSION 6 TO 9

One or more among the members 6 to 9 may have been in-activated in the previous analysis. The CHANGE command restores the original structure to prepare it for the analysis for the next primary load case. The members with the compression-only attribute are specified again.

LOAD 2 JOINT LOAD 4 FX -45 5 FX -70

In load case 2, joint loads are applied in the negative global X direction at joints 4 and 5.

PERFORM ANALYSIS CHANGE

The instruction to analyze the structure is specified again. Next, any compression-only members that were inactivated during the second analysis (due to the fact that they were subjected to tensile axial forces) are re-activated with the CHANGE command. Without the re-activation, these members cannot be accessed for further processing.

Example Problem 25

245 LOAD 3 REPEAT LOAD 1 1.0 2 1.0

Load case 3 illustrates the technique employed to instruct STAAD to create a load case which consists of data to be assembled from other load cases already specified earlier. We would like the program to analyze the structure for loads from cases 1 and 2 acting simultaneously. In other words, the above instruction is the same as the following:

LOAD 3 JOINT LOAD 2 FX 70 3 FX 45 4 FX -45 5 FX -70

PERFORM ANALYSIS

The analysis is carried out for load case 3.

CHANGE

The members inactivated during the analysis of load case 3 are re-activated for further processing.

LOAD LIST ALL

At the end of any analysis, only those load cases for which the analysis was done most recently, are recognized as the "active" load cases. The LOAD LIST ALL command enables all the load cases in the structure to be made active for further processing.

PRINT ANALYSIS RESULTS

The program is instructed to write the joint displacements, support reactions and member forces to the output file.

FINISH



Example Problem 25


1. STAAD PLANE2. * EXAMPLE FOR COMPRESSION-ONLY MEMBERS3. UNIT METER KNS4. SET NL 35. JOINT COORDINATES6. 1 0 0 ; 2 0 3.5 ; 3 0 7.0 ; 4 5.25 7.0 ; 5 5.25 3.5 ; 6 5.25 07. MEMBER INCIDENCES8. 1 1 2 59. 6 1 5 ; 7 2 6 ; 8 2 4 ; 9 3 5 ; 10 2 5

10. MEMBER COMPRESSION11. 6 TO 912. MEMBER PROPERTY BRITISH13. 1 TO 10 TA ST UC152X152X3014. CONSTANTS15. E STEEL ALL16. DEN STEEL ALL17. POISSON STEEL ALL18. SUPPORT19. 1 6 PINNED20. LOAD 121. JOINT LOAD22. 2 FX 7023. 3 FX 4524. PERFORM ANALYSIS





25. CHANGE26. MEMBER COMPRESSION27. 6 TO 928. LOAD 229. JOINT LOAD30. 4 FX -4531. 5 FX -7032. PERFORM ANALYSIS



33. CHANGE34. LOAD 335. REPEAT LOAD36. 1 1.0 2 1.037. PERFORM ANALYSIS


38. CHANGE39. LOAD LIST ALL

Example Problem 25




1 1 0.0000 0.0000 0.0000 0.0000 0.0000 -0.00052 0.0000 0.0000 0.0000 0.0000 0.0000 0.00073 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0001

2 1 0.1648 0.0475 0.0000 0.0000 0.0000 -0.00032 -0.1946 -0.0135 0.0000 0.0000 0.0000 0.00033 0.0141 0.0031 0.0000 0.0000 0.0000 0.0000

3 1 0.2960 0.0609 0.0000 0.0000 0.0000 -0.00032 -0.2959 -0.0135 0.0000 0.0000 0.0000 0.00023 0.0093 0.0082 0.0000 0.0000 0.0000 0.0000

4 1 0.2959 -0.0135 0.0000 0.0000 0.0000 -0.00022 -0.2960 0.0609 0.0000 0.0000 0.0000 0.00033 -0.0093 0.0082 0.0000 0.0000 0.0000 0.0000

5 1 0.1946 -0.0135 0.0000 0.0000 0.0000 -0.00032 -0.1648 0.0475 0.0000 0.0000 0.0000 0.00033 -0.0141 0.0031 0.0000 0.0000 0.0000 0.0000

6 1 0.0000 0.0000 0.0000 0.0000 0.0000 -0.00072 0.0000 0.0000 0.0000 0.0000 0.0000 0.00053 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001



1 1 -0.13 -106.67 0.00 0.00 0.00 0.002 114.87 106.67 0.00 0.00 0.00 0.003 10.38 0.00 0.00 0.00 0.00 0.00

6 1 -114.87 106.67 0.00 0.00 0.00 0.002 0.13 -106.67 0.00 0.00 0.00 0.003 -10.38 0.00 0.00 0.00 0.00 0.00



1 1 1 -106.67 0.13 0.00 0.00 0.00 0.002 106.67 -0.13 0.00 0.00 0.00 0.45

2 1 30.22 -0.21 0.00 0.00 0.00 0.002 -30.22 0.21 0.00 0.00 0.00 -0.73

3 1 -6.94 0.03 0.00 0.00 0.00 0.002 6.94 -0.03 0.00 0.00 0.00 0.09

2 1 2 -29.91 0.24 0.00 0.00 0.00 0.383 29.91 -0.24 0.00 0.00 0.00 0.45

2 2 0.15 -0.12 0.00 0.00 0.00 -0.083 -0.15 0.12 0.00 0.00 0.00 -0.33

3 2 -11.49 -0.03 0.00 0.00 0.00 -0.083 11.49 0.03 0.00 0.00 0.00 -0.03

3 1 3 0.12 -0.15 0.00 0.00 0.00 -0.454 -0.12 0.15 0.00 0.00 0.00 -0.33

2 3 0.12 0.15 0.00 0.00 0.00 0.334 -0.12 -0.15 0.00 0.00 0.00 0.45

3 3 27.79 0.00 0.00 0.00 0.00 0.034 -27.79 0.00 0.00 0.00 0.00 -0.03

4 1 4 0.15 0.12 0.00 0.00 0.00 0.335 -0.15 -0.12 0.00 0.00 0.00 0.08

2 4 -29.91 -0.24 0.00 0.00 0.00 -0.455 29.91 0.24 0.00 0.00 0.00 -0.38

3 4 -11.49 0.03 0.00 0.00 0.00 0.035 11.49 -0.03 0.00 0.00 0.00 0.08


Example Problem 25



5 1 5 30.22 0.21 0.00 0.00 0.00 0.736 -30.22 -0.21 0.00 0.00 0.00 0.00

2 5 -106.67 -0.13 0.00 0.00 0.00 -0.456 106.67 0.13 0.00 0.00 0.00 0.00

3 5 -6.94 -0.03 0.00 0.00 0.00 -0.096 6.94 0.03 0.00 0.00 0.00 0.00

6 1 1 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

2 1 137.81 0.00 0.00 0.00 0.00 0.005 -137.81 0.00 0.00 0.00 0.00 0.00

3 1 12.51 0.00 0.00 0.00 0.00 0.005 -12.51 0.00 0.00 0.00 0.00 0.00

7 1 2 137.81 0.00 0.00 0.00 0.00 0.006 -137.81 0.00 0.00 0.00 0.00 0.00

2 2 0.00 0.00 0.00 0.00 0.00 0.006 0.00 0.00 0.00 0.00 0.00 0.00

3 2 12.51 0.00 0.00 0.00 0.00 0.006 -12.51 0.00 0.00 0.00 0.00 0.00

8 1 2 0.00 0.00 0.00 0.00 0.00 0.004 0.00 0.00 0.00 0.00 0.00 0.00

2 2 53.66 0.00 0.00 0.00 0.00 0.004 -53.66 0.00 0.00 0.00 0.00 0.00

3 2 20.72 0.00 0.00 0.00 0.00 0.004 -20.72 0.00 0.00 0.00 0.00 0.00

9 1 3 53.66 0.00 0.00 0.00 0.00 0.005 -53.66 0.00 0.00 0.00 0.00 0.00

2 3 0.00 0.00 0.00 0.00 0.00 0.005 0.00 0.00 0.00 0.00 0.00 0.00

3 3 20.72 0.00 0.00 0.00 0.00 0.005 -20.72 0.00 0.00 0.00 0.00 0.00

10 1 2 -44.55 -0.31 0.00 0.00 0.00 -0.825 44.55 0.31 0.00 0.00 0.00 -0.81

2 2 -44.55 0.31 0.00 0.00 0.00 0.815 44.55 -0.31 0.00 0.00 0.00 0.82

3 2 42.29 0.00 0.00 0.00 0.00 -0.015 -42.29 0.00 0.00 0.00 0.00 0.01


41. FINISH


**** DATE= TIME= ****


Example Problem 25

249

NOTES


Example Problem 25

250

NOTES

Example Problem 26

251


The structure in this example is a building consisting of member columns as well as floors made up of beam members and plate elements. Using the master-slave command, the floors are specified to be rigid diaphragms for inplane actions but flexible for bending actions.


Example Problem 26

252

STAAD SPACE *MODELING RIGID DIAPHRAGMS USING MASTER SLAVE

Every STAAD input file has to begin with the word STAAD. The word SPACE signifies that the structure is a space frame and the geometry is defined through X, Y and Z axes. The second line is an optional title to identify this project.

UNITS KIP FT

Specify units for the following data.

JOINT COORD 1 0 0 0 4 0 48 0 REPEAT 3 24 0 0 REPEAT ALL 3 0 0 24 DELETE JOINT 21 25 37 41

The joint numbers and coordinates are specified above. The unwanted joints, created during the generation process used above, are then deleted.

MEMBER INCI 1 1 2 3 ; 4 5 6 6 ; 7 9 10 9 ; 10 13 14 12 13 17 18 15 ; 22 29 30 24 ; 25 33 34 27 34 45 46 36 ; 37 49 50 39 ; 40 53 54 42 43 57 58 45 ; 46 61 62 48 ; 49 2 6 51 52 6 10 54 ; 55 10 14 57 ; 58 18 22 60 61 22 26 63 ; 64 26 30 66 ; 67 34 38 69 70 38 42 72 ; 73 42 46 75 ; 76 50 54 78 79 54 58 81 ; 82 58 62 84 ; 85 18 2 87 88 22 6 90 ; 91 26 10 93 ; 94 30 14 96 97 34 18 99 ; 100 38 22 102 ; 103 42 26 105 106 46 30 108 ; 109 50 34 111 ; 112 54 38 114 115 58 42 117 ; 118 62 46 120

The MEMBER INCIDENCE specification is used for specifying MEMBER connectivities.

Example Problem 26

253

ELEMENT INCI 152 50 34 38 54 TO 154 155 54 38 42 58 TO 157 158 58 42 46 62 TO 160 161 34 18 22 38 TO 163 164 38 22 26 42 TO 166 167 42 26 30 46 TO 169 170 18 2 6 22 TO 172 173 22 6 10 26 TO 175 176 26 10 14 30 TO 178

The ELEMENT INCIDENCE specification is used for specifying plate element connectivities.

MEMBER PROPERTIES AMERICAN 1 TO 15 22 TO 27 34 TO 48 TA ST W14X90 49 TO 120 TABLE ST W27X84

All members are WIDE FLANGE sections whose properties are obtained from the built in American steel table.

ELEMENT PROP 152 TO 178 THICK 0.75

The thickness of the plate elements is specified above.

CONSTANTS E STEEL MEMB 1 TO 15 22 TO 27 34 TO 120 DENSITY STEEL MEMB 1 TO 15 22 TO 27 34 TO 120 POISSON STEEL MEMB 1 TO 15 22 TO 27 34 TO 120 BETA 90.0 MEMB 13 14 15 22 TO 27 34 TO 39 E CONCRETE MEMB 152 TO 178 DENSITY CONCRETE MEMB 152 TO 178 POISSON CONCRETE MEMB 152 TO 178

Following the command CONSTANTS above, the material constants such as E (Modulus of Elasticity), Poisson's Ratio, and Density are specified. Built-in default values for steel and concrete


Example Problem 26

254

for these quantities are assigned. The orientation of some of the members is set using the BETA angle command.

SUPPORTS 1 TO 17 BY 4 29 33 45 TO 61 BY 4 FIXED

The supports at the above mentioned joints are declared as fixed.

SLAVE DIA ZX MASTER 22 JOINTS YR 15.0 17.0 SLAVE DIA ZX MASTER 23 JOINTS YR 31.0 33.0 SLAVE DIA ZX MASTER 24 JOINTS YR 47.0 49.0

The 3 floors of the structure are specified to act as rigid diaphragms in the ZX plane with the corresponding master joint specified. The associated slave joints in a floor are specified by the YRANGE parameter. The floors may still resist out-of-plane bending actions flexibly.

LOADING 1 LATERAL LOADS JOINT LOADS 2 3 4 14 15 16 50 51 52 62 63 64 FZ 10.0 6 7 8 10 11 12 18 19 20 30 31 32 FZ 20.0 34 35 36 46 47 48 54 55 56 58 59 60 FZ 20.0 22 23 24 26 27 28 38 39 40 42 43 44 FZ 40.0

The above data describe a static load case. It consists of joint loads in the global Z direction.

LOADING 2 TORSIONAL LOADS JOINT LOADS 2 3 4 50 51 52 FZ 5.0 14 15 16 62 63 64 FZ 15.0 6 7 8 18 19 20 FZ 10.0 10 11 12 30 31 32 FZ 30.0 34 35 36 54 55 56 FZ 10.0 46 47 48 58 59 60 FZ 30.0 22 23 24 38 39 40 FZ 20.0 26 27 28 42 43 44 FZ 60.0

Example Problem 26

255

The above data describe a static load case. It consists of joint loads that create a torsional loading on the structure.

LOADING 3 DEAD LOAD ELEMENT LOAD 152 TO 178 PRESS GY -1.0

The above data describe a static load case. It consists of plate element pressure on a floor in the negative global Y direction.

PERFORM ANALYSIS

The above command instructs the program to proceed with the analysis.

PRINT JOINT DISP LIST 4 TO 60 BY 8 PRINT MEMBER FORCES LIST 116 115 PRINT SUPPORT REACTIONS LIST 9 57

Print displacements at selected joints, then print member forces for two members, then print support reactions at selected joints.

FINISH



Example Problem 26


1. STAAD SPACE2. *MODELING RIGID DIAPHRAGMS USING MASTER SLAVE3. UNITS KIP FT5. JOINT COORD6. 1 0 0 0 4 0 48 07. REPEAT 3 24 0 08. REPEAT ALL 3 0 0 249. DELETE JOINT 21 25 37 41

11. MEMBER INCI12. 1 1 2 3 ; 4 5 6 6 ; 7 9 10 9 ; 10 13 14 1213. 13 17 18 15 ; 22 29 30 24 ; 25 33 34 2714. 34 45 46 36 ; 37 49 50 39 ; 40 53 54 4215. 43 57 58 45 ; 46 61 62 48 ; 49 2 6 5116. 52 6 10 54 ; 55 10 14 57 ; 58 18 22 6017. 61 22 26 63 ; 64 26 30 66 ; 67 34 38 6918. 70 38 42 72 ; 73 42 46 75 ; 76 50 54 7819. 79 54 58 81 ; 82 58 62 84 ; 85 18 2 8720. 88 22 6 90 ; 91 26 10 93 ; 94 30 14 9621. 97 34 18 99 ; 100 38 22 102 ; 103 42 26 10522. 106 46 30 108 ; 109 50 34 111 ; 112 54 38 11423. 115 58 42 117 ; 118 62 46 12025. ELEMENT INCI26. 152 50 34 38 54 TO 15427. 155 54 38 42 58 TO 15728. 158 58 42 46 62 TO 16029. 161 34 18 22 38 TO 16330. 164 38 22 26 42 TO 16631. 167 42 26 30 46 TO 16932. 170 18 2 6 22 TO 17233. 173 22 6 10 26 TO 17534. 176 26 10 14 30 TO 17836. MEMBER PROPERTIES AMERICAN37. 1 TO 15 22 TO 27 34 TO 48 TA ST W14X9038. 49 TO 120 TABLE ST W27X8439. ELEMENT PROP40. 152 TO 178 THICK 0.7542. CONSTANTS43. E STEEL MEMB 1 TO 15 22 TO 27 34 TO 12044. DENSITY STEEL MEMB 1 TO 15 22 TO 27 34 TO 12045. POISSON STEEL MEMB 1 TO 15 22 TO 27 34 TO 12046. BETA 90.0 MEMB 13 14 15 22 TO 27 34 TO 3947. E CONCRETE MEMB 152 TO 17848. DENSITY CONCRETE MEMB 152 TO 17849. POISSON CONCRETE MEMB 152 TO 17851. SUPPORTS52. 1 TO 17 BY 4 29 33 45 TO 61 BY 4 FIXED54. SLAVE DIA ZX MASTER 22 JOINTS YR 15.0 17.055. SLAVE DIA ZX MASTER 23 JOINTS YR 31.0 33.056. SLAVE DIA ZX MASTER 24 JOINTS YR 47.0 49.058. LOADING 1 LATERAL LOADS59. JOINT LOADS60. 2 3 4 14 15 16 50 51 52 62 63 64 FZ 10.061. 6 7 8 10 11 12 18 19 20 30 31 32 FZ 20.062. 34 35 36 46 47 48 54 55 56 58 59 60 FZ 20.063. 22 23 24 26 27 28 38 39 40 42 43 44 FZ 40.065. LOADING 2 TORSIONAL LOADS66. JOINT LOADS67. 2 3 4 50 51 52 FZ 5.068. 14 15 16 62 63 64 FZ 15.069. 6 7 8 18 19 20 FZ 10.070. 10 11 12 30 31 32 FZ 30.071. 34 35 36 54 55 56 FZ 10.072. 46 47 48 58 59 60 FZ 30.073. 22 23 24 38 39 40 FZ 20.0

Example Problem 26

257 74. 26 27 28 42 43 44 FZ 60.076. LOADING 3 DEAD LOAD77. ELEMENT LOAD78. 152 TO 178 PRESS GY -1.080. PERFORM ANALYSIS



82. PRINT JOINT DISP LIST 4 TO 60 BY 8

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


4 1 0.23216 0.04609 8.13263 0.00108 -0.00056 -0.000082 1.49676 0.04919 6.87442 0.00090 -0.00363 -0.000463 0.02679 -0.19716 -0.32921 0.00792 -0.00041 -0.00625

12 1 0.23216 0.02166 8.45739 0.00159 -0.00056 0.000142 1.49676 0.02716 8.96702 0.00166 -0.00363 0.000003 0.02679 -0.86713 -0.09027 0.07454 -0.00041 0.00495

20 1 0.06978 -0.00054 8.13263 0.00120 -0.00056 -0.000252 0.45046 0.00140 6.87442 0.00103 -0.00363 -0.000313 -0.09268 -0.88242 -0.32921 0.00452 -0.00041 -0.07454

28 1 0.06978 -0.07792 8.45739 -0.00058 -0.00056 0.000242 0.45046 -0.07823 8.96702 -0.00059 -0.00363 0.000283 -0.09268 -21.50252 -0.09027 0.04716 -0.00041 0.04703

36 1 -0.09261 0.02065 8.13263 0.00102 -0.00056 0.000302 -0.59584 0.01536 6.87442 0.00088 -0.00363 0.000363 -0.21215 -0.86781 -0.32921 -0.00503 -0.00041 -0.07452

44 1 -0.09261 0.08468 8.45739 -0.00057 -0.00056 -0.000282 -0.59584 0.08128 8.96702 -0.00059 -0.00363 -0.000313 -0.21215 -21.51350 -0.09027 -0.04712 -0.00041 0.04704

52 1 -0.25499 -0.06556 8.13263 0.00245 -0.00056 -0.000022 -1.64214 -0.06312 6.87442 0.00207 -0.00363 0.000173 -0.33161 -0.19363 -0.32921 -0.00649 -0.00041 -0.00791

60 1 -0.25499 -0.02115 8.45739 0.00162 -0.00056 -0.000142 -1.64214 -0.02678 8.96702 0.00167 -0.00363 0.000013 -0.33161 -0.86677 -0.09027 -0.07468 -0.00041 0.00504


83. PRINT MEMBER FORCES LIST 116 115

MEMBER END FORCES STRUCTURE TYPE = SPACE-----------------ALL UNITS ARE -- KIP FEET


116 1 59 0.00 4.73 0.00 0.00 0.00 198.8443 0.00 -4.73 0.00 0.00 0.00 -85.38

2 59 0.00 5.10 0.00 -0.01 0.00 208.2343 0.00 -5.10 0.00 0.01 0.00 -85.83

3 59 0.00 129.34 0.00 0.32 0.00 1407.2743 0.00 -129.34 0.00 -0.32 0.00 1696.94

115 1 58 0.00 7.70 0.00 -0.01 0.00 322.1442 0.00 -7.70 0.00 0.01 0.00 -137.41

2 58 0.00 8.32 0.00 -0.01 0.00 336.8842 0.00 -8.32 0.00 0.01 0.00 -137.13

3 58 0.00 125.39 0.00 0.34 0.00 1173.8242 0.00 -125.39 0.00 -0.34 0.00 1835.58



Example Problem 26

258 84. PRINT SUPPORT REACTIONS LIST 9 57

SUPPORT REACTIONS -UNIT KIP FEET STRUCTURE TYPE = SPACE-----------------


9 1 -6.98 -54.60 -54.87 -470.13 0.01 50.692 -28.54 -69.17 -58.55 -500.63 0.03 231.313 -14.10 1732.37 92.25 487.26 0.00 70.68

57 1 7.65 53.36 -54.76 -469.56 0.01 -55.902 31.74 68.14 -58.52 -500.47 0.03 -257.513 -11.82 1731.53 -91.91 -483.96 0.00 51.09


86. FINISH


**** DATE= TIME= ****


Example Problem 26

259

NOTES


Example Problem 26

260

NOTES

Example Problem 27

261


This example illustrates the usage of commands necessary to apply the compression only attribute to spring supports for a slab on grade. The spring supports themselves are generated utilizing the built-in support generation facility. The slab is subjected to pressure and overturning loading. A tension/compression only analysis of the structure is performed. The numbers shown in the diagram below are the element numbers.


Example Problem 27

262

STAAD SPACE SLAB ON GRADE * SPRING COMPRESSION EXAMPLE

Every STAAD input file has to begin with the word STAAD. The word SPACE signifies that the structure is a space frame and the geometry is defined through X, Y and Z axes. An optional title to identify this project is provided in the second line.

SET NL 3

This structure has to be analysed for 3 primary load cases. Consequently, the modeling of our problem requires us to define 3 sets of data, with each set containing a load case and an associated analysis command. Also, the supports which get switched off in the analysis for any load case have to be restored for the analysis for the subsequent load case. To accommodate these requirements, it is necessary to have 2 commands, one called “SET NL” and the other called “CHANGE”. The SET NL command is used above to indicate the total number of primary load cases that the file contains. The CHANGE command will come in later (after the PERFORM ANALYSIS command).

UNIT FEET KIP JOINT COORDINATES 1 0.0 0.0 40.0 2 0.0 0.0 36.0 3 0.0 0.0 28.167 4 0.0 0.0 20.333 5 0.0 0.0 12.5 6 0.0 0.0 6.5 7 0.0 0.0 0.0 REPEAT ALL 3 8.5 0.0 0.0 REPEAT 3 8.0 0.0 0.0 REPEAT 5 6.0 0.0 0.0 REPEAT 3 8.0 0.0 0.0 REPEAT 3 8.5 0.0 0.0

For joints 1 through 7, the joint number followed by the X, Y and Z coordinates are specified above. The coordinates of these joints is used as a basis for generating 21 more joints by incrementing the X coordinate of each of these 7 joints by 8.5 feet, 3 times. REPEAT

Example Problem 27

263

commands are used to generate the remaining joints of the structure. The results of the generation may be visually verified using the STAAD graphical viewing facilities.

ELEMENT INCIDENCES 1 1 8 9 2 TO 6 REPEAT 16 6 7

The incidences of element number 1 is defined and that data is used as a basis for generating the 2nd through the 6th element. The incidence pattern of the first 6 elements is then used to generate the incidences of 96 more elements using the REPEAT command.

UNIT INCH ELEMENT PROPERTIES 1 TO 102 TH 8.0

The thickness of elements 1 to 102 is specified as 8.0 inches following the command ELEMENT PROPERTIES.

CONSTANTS E 4000.0 ALL POISSON 0.12 ALL

The modulus of elasticity (E) and density are specified following the command CONSTANTS.

SPRING COMPRESSION 1 TO 126 KFY

The above two lines declare the spring supports at nodes 1 to 126 as having the compression-only attribute. The supports themselves are being generated later (see the ELASTIC MAT command which appears later).

UNIT FEET SUPPORTS 1 TO 126 ELASTIC MAT DIRECTION Y SUBGRADE 12.0


Example Problem 27

264

The above command is used to instruct STAAD to generate supports with compression-only springs which are effective in the global Y direction. These springs are located at nodes 1 to 126. The subgrade reaction of the soil is specified as 12 kip/cu.ft. The program will determine the area under the influence of each joint and multiply the influence area by the subgrade reaction to arrive at the spring stiffness for the "FY" degree of freedom at the joint. Units for length are changed to FEET to facilitate the input of subgrade reaction of soil. Additional information on this feature may be found in the STAAD Technical Reference Manual.

LOAD 1 'WEIGHT OF MAT & EARTH' ELEMENT LOAD 1 TO 102 PR GY -1.50

The above data describe a static load case. A pressure load of 1.50 kip/ft acting in the negative global Y direction is applied on all the elements.

PERFORM ANALYSIS PRINT STATICS CHECK CHANGE

Tension/compression cases must each be followed by PERFORM ANALYSIS and CHANGE commands. The CHANGE command restores the original structure to prepare it for the analysis for the next primary load case.

LOAD 2 'COLUMN LOAD-DL+LL' JOINT LOADS 1 2 FY -217. 8 9 FY -109. 5 FY -308.7 6 FY -617.4 22 23 FY -410. 29 30 FY -205. 26 FY -542.7 27 FY -1085.4 43 44 50 51 71 72 78 79 FY -307.5 47 54 82 FY -264.2

Example Problem 27

265

48 55 76 83 FY -528.3 92 93 FY -205.0 99 100 FY -410.0 103 FY -487.0 104 FY -974.0 113 114 FY -109.0 120 121 FY -217.0 124 FY -273.3 125 FY -546.6 PERFORM ANALYSIS PRINT STATICS CHECK CHANGE

Load case 2 consists of several joint loads acting in the negative global Y direction. This is followed by another ANALYSIS command. The CHANGE command restores the original structure once again for the forthcoming load case.

LOAD 3 'COLUMN OVERTURNING LOAD' ELEMENT LOAD 1 TO 102 PR GY -1.50 JOINT LOADS 1 2 FY -100. 8 9 FY -50. 5 FY -150.7 6 FY -310.4 22 23 FY -205. 29 30 FY -102. 26 FY -271.7 27 FY -542.4 43 44 50 51 71 72 78 79 FY -153.5 47 54 82 FY -132.2 48 55 76 83 FY -264.3 92 93 FY 102.0 99 100 FY 205.0 103 FY 243.0 104 FY 487.0 113 114 FY 54.0 120 121 FY 108.0


Example Problem 27

266

124 FY 136.3 125 FY 273.6 PERFORM ANALYSIS PRINT STATICS CHECK

Load case 3 consists of several joint loads acting in the upward direction at one end and downward on the other end to apply an overturning moment that will lift off one end. The CHANGE command is not needed after the last analysis.

LOAD LIST 3 PRINT JOINT DISPLACEMENTS LIST 113 114 120 121 PRINT ELEMENT STRESSES LIST 34 67 PRINT SUPPORT REACTIONS LIST 5 6 12 13

A list of joint displacements, element stresses for elements 34 and 67, and support reactions at a list of joints for load case 3, are obtained with the help of the above commands.

FINISH


Example Problem 27


1. STAAD SPACE SLAB ON GRADE2. * SPRING COMPRESSION EXAMPLE3. SET NL 34. UNIT FEET KIP6. JOINT COORDINATES7. 1 0.0 0.0 40.08. 2 0.0 0.0 36.09. 3 0.0 0.0 28.167

10. 4 0.0 0.0 20.33311. 5 0.0 0.0 12.512. 6 0.0 0.0 6.513. 7 0.0 0.0 0.014. REPEAT ALL 3 8.5 0.0 0.015. REPEAT 3 8.0 0.0 0.016. REPEAT 5 6.0 0.0 0.017. REPEAT 3 8.0 0.0 0.018. REPEAT 3 8.5 0.0 0.020. ELEMENT INCIDENCES21. 1 1 8 9 2 TO 622. REPEAT 16 6 724. UNIT INCH25. ELEMENT PROPERTIES26. 1 TO 102 TH 8.028. CONSTANTS29. E 4000.0 ALL30. POISSON 0.12 ALL32. SPRING COMPRESSION33. 1 TO 126 KFY35. UNIT FEET36. SUPPORTS37. 1 TO 126 ELASTIC MAT DIRECTION Y SUBGRADE 12.039. LOAD 1 'WEIGHT OF MAT & EARTH'40. ELEMENT LOAD41. 1 TO 102 PR GY -1.5043. PERFORM ANALYSIS PRINT STATICS CHECK




STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 1'WEIGHT OF MAT & EARTH'

***TOTAL APPLIED LOAD ( KIP FEET ) SUMMARY (LOADING 1 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = -7740.00SUMMATION FORCE-Z = 0.00



Example Problem 27

268 ***TOTAL REACTION LOAD( KIP FEET ) SUMMARY (LOADING 1 )

SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = 7740.00SUMMATION FORCE-Z = 0.00


MAXIMUM DISPLACEMENTS ( INCH /RADIANS) (LOADING 1)MAXIMUMS AT NODE

X = 0.00000E+00 0Y = -1.50000E+00 1Z = 0.00000E+00 0RX= 9.51342E-10 121RY= 0.00000E+00 0RZ= -4.19726E-10 80


44. CHANGE46. LOAD 2 'COLUMN LOAD-DL+LL'47. JOINT LOADS48. 1 2 FY -217.49. 8 9 FY -109.50. 5 FY -308.751. 6 FY -617.452. 22 23 FY -410.53. 29 30 FY -205.54. 26 FY -542.755. 27 FY -1085.456. 43 44 50 51 71 72 78 79 FY -307.557. 47 54 82 FY -264.258. 48 55 76 83 FY -528.359. 92 93 FY -205.060. 99 100 FY -410.061. 103 FY -487.062. 104 FY -974.063. 113 114 FY -109.064. 120 121 FY -217.065. 124 FY -273.366. 125 FY -546.668. PERFORM ANALYSIS PRINT STATICS CHECK


STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 2'COLUMN LOAD-DL+LL'



***TOTAL REACTION LOAD( KIP FEET ) SUMMARY (LOADING 2 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = 13964.90SUMMATION FORCE-Z = 0.00



X = 0.00000E+00 0Y = -1.09725E+01 120Z = 0.00000E+00 0RX= 7.89606E-02 99RY= 0.00000E+00 0RZ= 9.69957E-02 6


Example Problem 27

269 69. CHANGE71. LOAD 3 'COLUMN OVERTURNING LOAD'72. ELEMENT LOAD73. 1 TO 102 PR GY -1.5074. JOINT LOADS75. 1 2 FY -100.76. 8 9 FY -50.77. 5 FY -150.778. 6 FY -310.479. 22 23 FY -205.80. 29 30 FY -102.81. 26 FY -271.782. 27 FY -542.483. 43 44 50 51 71 72 78 79 FY -153.584. 47 54 82 FY -132.285. 48 55 76 83 FY -264.386. 92 93 FY 102.087. 99 100 FY 205.088. 103 FY 243.089. 104 FY 487.090. 113 114 FY 54.091. 120 121 FY 108.092. 124 FY 136.393. 125 FY 273.695. PERFORM ANALYSIS PRINT STATICS CHECK

**START ITERATION NO. 2**START ITERATION NO. 3**START ITERATION NO. 4**START ITERATION NO. 5


STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 3'COLUMN OVERTURNING LOAD'



***TOTAL REACTION LOAD( KIP FEET ) SUMMARY (LOADING 3 )SUMMATION FORCE-X = 0.00SUMMATION FORCE-Y = 10533.10SUMMATION FORCE-Z = 0.00



X = 0.00000E+00 0Y = 2.83669E+01 120Z = 0.00000E+00 0RX= -1.22268E-01 120RY= 0.00000E+00 0RZ= 1.09786E-01 125


97. LOAD LIST 3


Example Problem 27

270 98. PRINT JOINT DISPLACEMENTS LIST 113 114 120 121



113 3 0.00000 19.17264 0.00000 -0.09579 0.00000 0.06945114 3 0.00000 14.53915 0.00000 -0.09437 0.00000 0.06506120 3 0.00000 28.36691 0.00000 -0.12227 0.00000 0.10056121 3 0.00000 22.49737 0.00000 -0.11615 0.00000 0.08912


99. PRINT ELEMENT STRESSES LIST 34 67

ELEMENT STRESSES FORCE,LENGTH UNITS= KIP FEET----------------



34 3 -4.50 -6.74 2.45 7.99 6.96188.81 188.81 0.00 0.00 0.00202.25 202.25


67 3 37.83 6.21 -57.38 5.58 43.511303.44 1303.44 0.00 0.00 0.001449.91 1449.91






100. PRINT SUPPORT REACTIONS LIST 5 6 12 13

SUPPORT REACTIONS -UNIT KIP FEET STRUCTURE TYPE = SPACE-----------------


5 3 0.00 148.06 0.00 0.00 0.00 0.006 3 0.00 168.10 0.00 0.00 0.00 0.00

12 3 0.00 149.08 0.00 0.00 0.00 0.0013 3 0.00 153.60 0.00 0.00 0.00 0.00


Example Problem 27

271 101. FINISH


**** DATE= TIME= ****



Example Problem 27

272

NOTES

Example Problem 28

273


This example demonstrates the input required for obtaining the modes and frequencies of the skewed bridge shown in the figure below. The structure consists of piers, pier-cap girders and a deck slab.


Example Problem 28

274 STAAD SPACE FREQUENCIES OF VIBRATION OF A SKEWED BRIDGE

Every STAAD input file has to begin with the word STAAD. The word SPACE signifies that the structure is a space frame and the geometry is defined through X, Y and Z axes. The remainder of the words forms a title to identify this project.

IGNORE LIST Further below in this file, we will call element lists in which some element numbers may not actually be present in the structure. We do so because it minimizes the effort involved in fetching the desired elements and reduces the size of the respective commands. To prevent the program from treating that condition (referring to elements which do not exist) as an error, the above command is required.

UNIT METER KN The units for the data that follows are specified above.

JOINT COORDINATES 1 0 0 0; 2 4 0 0; 3 6.5 0 0; 4 9 0 0; 5 11.5 0 0; 6 15.5 0 0; 11 -1 10 0 25 16.5 10 0 REPEAT ALL 3 4 0 14

For joints 1 through 6, the joint number followed by the X, Y and Z coordinates are specified first. Next, using the coordinates of joints 11 and 25 as the basis, joints 12 through 24 are generated using linear interpolation. Following this, using the data of these 21 joints (1 through 6 and 11 through 25), 63 new joints are generated. To achieve this, the X coordinate of these 21 joints is incremented by 4 meters and the Z coordinate is incremented by 14 meters, in 3 successive operations.

Example Problem 28

275

The REPEAT ALL command is used for the generation. Details of this command is available in Section 5.11 of the Technical Reference manual. The results of the generation may be visually verified using STAAD.Pro's graphical viewing facilities.

MEMBER INCI 1 1 13 ; 2 2 15 ; 3 3 17 ; 4 4 19 ; 5 5 21 ; 6 6 23 26 26 34 ; 27 27 36 ; 28 28 38 ; 29 29 40 ; 30 30 42 ; 31 31 44 47 47 55 ; 48 48 57 ; 49 49 59 ; 50 50 61 ; 51 51 63 ; 52 52 65 68 68 76 ; 69 69 78 ; 70 70 80 ; 71 71 82 ; 72 72 84 ; 73 73 86

The member connectivity data (joint numbers between which members are connected) is specified for the 24 columns for the structure. The above method, where the member number is followed by the 2 node numbers, is the explicit definition method. No generation is involved here.

101 11 12 114 202 32 33 215 303 53 54 316 404 74 75 417

The member connectivity data is specified for the pier cap beams for the structure. The above method is a combination of explicit definition and generation. For example, member 101 is defined as connected between 11 & 12. Then, by incrementing those nodes by 1 unit at a time (which is the default increment), the incidences of members 102 to 114 are generated. Similarly, we create members 202 to 215, 303 to 316, and, 404 to 417.

DEFINE MESH A JOINT 11 B JOINT 25 C JOINT 46 D JOINT 32 E JOINT 67 F JOINT 53 G JOINT 88 H JOINT 74


Example Problem 28

276

The next step is to generate the deck slab which will be modeled using plate elements. For this, we use a technique called mesh generation. Mesh generation is a process of generating several "child" elements from a "parent" or "super" element. The above set of commands defines the corner nodes of the super-element. Details of the above can be found in Section 5.14 of the Technical Reference manual. Note that instead of elaborately defining the coordinates of the corner nodes of the super-elements, we have taken advantage of the fact that the coordinates of these joints (A through H) have already been defined or generated earlier. Thus, A is the same as joint 11 while D is the same as joint 32. Alternatively, we could have defined the super-element nodes as A -1 10 0 ; B 16.5 10 0 ; C 20.5 10 14 ; D 3 10 14 ; etc.

GENERATE ELEMENT MESH ABCD 14 12 MESH DCEF 14 12 MESH FEGH 14 12

The above lines are the instructions for generating the “child” elements from the super-elements. For example, from the super-element bound by the corners A, B, C and D (which in turn are nodes 11, 25, 46 and 32), we generate a total of 14X12=168 elements, with 14 divisions along the edges AB and CD, and 12 along the edges BC and DA. These are the elements which make up the first span. Similarly, 168 elements are created for the 2nd span, and another 168 for the 3rd span. It may be noted here that we have taken great care to ensure that the resulting elements and the piercap beams form a perfect fit. In other words, there is no overlap between the two in a manner that nodes of the beams are at a different point in space than nodes of elements. At every node along their common boundary, plates and beams are properly connected. This is absolutely essential to ensure proper transfer of load and stiffness from beams to plates

Example Problem 28

277

and vice versa. The tools of the graphical user interface may be used to confirm that beam-plate connectivity is proper for this model.

START GROUP DEFINITION MEMBER _GIRDERS 101 TO 114 202 TO 215 303 TO 316 404 TO 417 _PIERS 1 TO 6 26 TO 31 47 TO 52 68 TO 73 ELEMENT _P1 447 TO 450 454 TO 457 461 TO 464 468 TO 471 _P2 531 TO 534 538 TO 541 545 TO 548 552 TO 555 _P3 615 TO 618 622 TO 625 629 TO 632 636 TO 639 _P4 713 TO 716 720 TO 723 727 TO 730 734 TO 737 _P5 783 TO 786 790 TO 793 797 TO 800 804 TO 807 _P6 881 TO 884 888 TO 891 895 TO 898 902 TO 905

END GROUP DEFINITION

The above block of data is referred to as formation of groups. Group names are a mechanism by which a single moniker can be used to refer to a cluster of entities, such as members. For our structure, the piercap beams are being grouped to a name called GIRDERS, the pier columns are assigned the name PIERS, and so on. For the deck, a few selected elements are chosen into a few selective groups. The reason is that these elements happen to be right beneath wheels of vehicles whose weight will be used in the frequency calculation.

MEMBER PROPERTY _GIRDERS PRIS YD 0.6 ZD 0.6 _PIERS PRIS YD 1.0

Member properties are assigned as prismatic rectangular sections for the girders, and prismatic circular sections for the columns.

ELEMENT PROPERTY YRA 9 11 TH 0.375


Example Problem 28

278

The plate elements of the deck slab, which happen to be at a Y elevation of 10 metres (between a YRANGE of 9 metres and 11 metres) are assigned a thickness of 375 mms.

UNIT NEWTON MMS CONSTANTS E 21000 ALL POISSON CONCRETE ALL

The Modulus of elasticity (E) is set to 21000 N/sq.mm for all members. The keyword CONSTANTS has to precede this data. Built-in default value for Poisson's ratio for concrete is also assigned to ALL members and elements.

UNIT KNS METER CONSTANTS DENSITY 24 ALL

Following a change of units, density of concrete is specified.

SUPPORTS 1 TO 6 26 TO 31 47 TO 52 68 TO 73 FIXED

The base nodes of the piers are fully restrained (FIXED supports).

CUT OFF MODE SHAPE 65 Theoretically, a structure has as many modes of vibration as the number of degrees of freedom in the model. However, the limitations of the mathematical process used in extracting modes may limit the number of modes that can actually be extracted. In a large structure, the extraction process can also be very time consuming. Further, not all modes are of equal importance. (One measure of the importance of modes is the participation factor of that mode.) In many cases, the first few modes may be sufficient to obtain a significant portion of the total dynamic response. Due to these reasons, in the absence of any explicit instruction, STAAD calculates only the first 6 modes. This is like saying that the command CUT OFF MODE SHAPE 6 has been specified.

Example Problem 28

279

(Versions of STAAD prior to STAAD.Pro 2000 calculated only 3 modes by default). If the inspection of the first 6 modes reveals that the overall vibration pattern of the structure has not been obtained, one may ask STAAD to compute a larger (or smaller) number of modes with the help of this command. The number that follows this command is the number of modes being requested. In our example, we are asking for 65 modes by specifying CUT OFF MODE SHAPE 65.

UNIT KGS METER LOAD 1 FREQUENCY CALCULATION SELFWEIGHT X 1.0 SELFWEIGHT Y 1.0 SELFWEIGHT Z 1.0 * PERMANENT WEIGHTS ON DECK ELEMENT LOAD YRA 9 11 PR GX 200 YRA 9 11 PR GY 200 YRA 9 11 PR GZ 200 * VEHICLES ON SPANS - ONLY Y & Z EFFECT CONSIDERED ELEMENT LOAD _P1 PR GY 700 _P2 PR GY 700 _P3 PR GY 700 _P4 PR GY 700 _P5 PR GY 700 _P6 PR GY 700

_P1 PR GZ 700 _P2 PR GZ 700 _P3 PR GZ 700 _P4 PR GZ 700 _P5 PR GZ 700 _P6 PR GZ 700

The mathematical method that STAAD uses is called the eigen extraction method. Some information on this is available in Section 1.18.3 of the STAAD.Pro Technical Reference Manual. The


Example Problem 28

280

method involves 2 matrices - the stiffness matrix, and the mass matrix. The stiffness matrix, usually called the [K] matrix, is assembled using data such as member and element lengths, member and element properties, modulus of elasticty, Poisson's ratio, member and element releases, member offsets, support information, etc. For assembling the mass matrix, called the [M] matrix, STAAD uses the load data specified in the load case in which the MODAL CAL REQ command is specified. So, some of the important aspects to bear in mind are : 1. The input you specify is weights, not masses. Internally,

STAAD will convert weights to masses by dividing the input by "g", the acceleration due to gravity.

2. If the structure is declared as a PLANE frame, there are 2

possible directions of vibration - global X, and global Y. If the structure is declared as a SPACE frame, there are 3 possible directions - global X, global Y and global Z. However, this does not guarantee that STAAD will automatically consider the masses for vibration in all the available directions.

You have control over and are responsible for specifying the directions in which the masses ought to vibrate. In other words, if a weight is not specified along a certain direction, the corresponding degrees of freedom (such as for example, global X at node 34 hypothetically) will not receive a contribution in the mass matrix. The mass matrix is assembled using only the masses from the weights and directions specified by the user.

In our example, notice that we are specifying the selfweight along global X, Y and Z directions. Similarly, a 200 kg/sq.m pressure load is also specified along all 3 directions on the deck.

Example Problem 28

281

But for the truck loads, we choose to apply it on just a few elements in the global Y and Z directions only. The reasoning is something like - for the X direction, the mass is not capable of vibrating because the tires allow the truck to roll along X. Remember, this is just a demonstration example, not necessarily what you may wish to do.

The point we wish to illustrate is that if a user wishes to restrict a certain weight to certain directions only, all he/she has to do is not provide the directions in which those weights cannot vibrate in.

3. As much as possible, provide absolute values for the weights.

STAAD is programmed to algebraically add the weights at nodes. So, if some weights are specified as positive numbers and others as negative, the total weight at a given node is the algebraic summation of all the weights in the global directions at that node and the mass is then derived from this algebraic resultant.

MODAL CALCULATION REQUESTED

This is the command which tells the program that frequencies and modes should be calculated. It is specified inside a load case. In other words, this command accompanies the loads that are to be used in generating the mass matrix. Frequencies and modes have to be calculated also when dynamic analysis such as response spectrum or time history analysis is carried out. But in such analyses, the MODAL CALCULATION REQUESTED command is not explicitly required. When STAAD encounters the commands for response spectrum (see example 11) and time history (see examples 16 and 22), it automatically will carry out a frequency extraction without the help of the MODAL .. command.


Example Problem 28

282 PERFORM ANALYSIS

This initiates the processes which are required to obtain the frequencies. Frequencies, periods and participation factors are automatically reported in the output file when the operation is completed.

FINISH This terminates the STAAD run.

Example Problem 28

283


1. STAAD SPACE FREQUENCIES OF VIBRATION OF A SKEWED BRIDGE2. IGNORE LIST4. UNIT METER KN5. JOINT COORDINATES6. 1 0 0 0; 2 4 0 0; 3 6.5 0 0; 4 9 0 0; 5 11.5 0 0; 6 15.5 0 07. 11 -1 10 0 25 16.5 10 08. REPEAT ALL 3 4 0 14

10. MEMBER INCI11. 1 1 13 ; 2 2 15 ; 3 3 17 ; 4 4 19 ; 5 5 21 ; 6 6 2312. 26 26 34 ; 27 27 36 ; 28 28 38 ; 29 29 40 ; 30 30 42 ; 31 31 4413. 47 47 55 ; 48 48 57 ; 49 49 59 ; 50 50 61 ; 51 51 63 ; 52 52 6514. 68 68 76 ; 69 69 78 ; 70 70 80 ; 71 71 82 ; 72 72 84 ; 73 73 8616. 101 11 12 11417. 202 32 33 21518. 303 53 54 31619. 404 74 75 41721. DEFINE MESH22. A JOINT 1123. B JOINT 2524. C JOINT 4625. D JOINT 3226. E JOINT 6727. F JOINT 5328. G JOINT 8829. H JOINT 7430. GENERATE ELEMENT31. MESH ABCD 14 1232. MESH DCEF 14 1233. MESH FEGH 14 1235. START GROUP DEFINITION36. MEMBER37. _GIRDERS 101 TO 114 202 TO 215 303 TO 316 404 TO 41738. _PIERS 1 TO 6 26 TO 31 47 TO 52 68 TO 7340. ELEMENT41. _P1 447 TO 450 454 TO 457 461 TO 464 468 TO 47142. _P2 531 TO 534 538 TO 541 545 TO 548 552 TO 55543. _P3 615 TO 618 622 TO 625 629 TO 632 636 TO 63944. _P4 713 TO 716 720 TO 723 727 TO 730 734 TO 73745. _P5 783 TO 786 790 TO 793 797 TO 800 804 TO 80746. _P6 881 TO 884 888 TO 891 895 TO 898 902 TO 90548. END GROUP DEFINITION50. MEMBER PROPERTY51. _GIRDERS PRIS YD 0.6 ZD 0.652. _PIERS PRIS YD 1.054. ELEMENT PROPERTY55. YRA 9 11 TH 0.37557. UNIT NEWTON MMS58. CONSTANTS59. E 21000 ALL60. POISSON CONCRETE ALL62. UNIT KNS METER63. CONSTANTS64. DENSITY 24 ALL66. SUPPORTS67. 1 TO 6 26 TO 31 47 TO 52 68 TO 73 FIXED69. CUT OFF MODE SHAPE 6571. UNIT KGS METER72. LOAD 1 FREQUENCY CALCULATION73. SELFWEIGHT X 1.074. SELFWEIGHT Y 1.075. SELFWEIGHT Z 1.077. * PERMANENT WEIGHTS ON DECK78. ELEMENT LOAD79. YRA 9 11 PR GX 200


Example Problem 28

284 80. YRA 9 11 PR GY 20081. YRA 9 11 PR GZ 20083. * VEHICLES ON SPANS - ONLY Y & Z EFFECT CONSIDERED84. ELEMENT LOAD85. _P1 PR GY 70086. _P2 PR GY 70087. _P3 PR GY 70088. _P4 PR GY 70089. _P5 PR GY 70090. _P6 PR GY 70092. _P1 PR GZ 70093. _P2 PR GZ 70094. _P3 PR GZ 70095. _P4 PR GZ 70096. _P5 PR GZ 70097. _P6 PR GZ 70099. MODAL CALCULATION REQUESTED

100. PERFORM ANALYSIS


NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 579/ 584/ 24ORIGINAL/FINAL BAND-WIDTH= 496/ 24/ 150 DOFTOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 3330SIZE OF STIFFNESS MATRIX = 500 DOUBLE KILO-WORDSREQRD/AVAIL. DISK SPACE = 23.4/ 3141.8 MB, EXMEM = 568.1 MB

** WARNING: PRESSURE LOADS ON ELEMENTS OTHER THAN PLATE ELEMENTSARE IGNORED. ELEM.NO. 101







1 1.636 0.61111 1.344E-162 2.602 0.38433 4.254E-163 2.882 0.34695 1.213E-154 3.754 0.26636 0.000E+005 4.076 0.24532 1.733E-166 4.373 0.22870 1.205E-157 4.519 0.22130 2.821E-168 4.683 0.21355 5.253E-169 5.028 0.19889 6.835E-16

10 7.189 0.13911 2.229E-1611 7.238 0.13815 4.397E-1612 7.363 0.13582 8.499E-1613 10.341 0.09671 2.155E-1614 10.734 0.09316 1.999E-1615 11.160 0.08961 1.480E-1516 11.275 0.08869 5.436E-1617 11.577 0.08638 3.438E-1618 11.829 0.08454 4.939E-1619 11.921 0.08388 4.863E-1620 12.085 0.08275 1.578E-1621 12.488 0.08007 2.954E-1622 13.677 0.07311 1.232E-1623 14.654 0.06824 6.437E-1624 14.762 0.06774 2.114E-1625 15.125 0.06612 6.042E-1626 17.308 0.05778 3.076E-1627 17.478 0.05721 3.017E-1628 17.747 0.05635 8.777E-1629 19.725 0.05070 4.737E-1630 19.921 0.05020 4.644E-16

Example Problem 28

285 MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC) ACCURACY

31 20.536 0.04869 0.000E+0032 20.618 0.04850 2.168E-1633 20.845 0.04797 0.000E+0034 21.146 0.04729 0.000E+0035 21.426 0.04667 4.015E-1636 21.801 0.04587 1.939E-1637 22.070 0.04531 3.784E-1638 23.153 0.04319 1.719E-1639 23.518 0.04252 3.332E-1640 23.985 0.04169 3.204E-1641 24.655 0.04056 3.032E-1642 25.469 0.03926 9.944E-1643 26.002 0.03846 1.499E-1544 26.422 0.03785 1.069E-1445 26.808 0.03730 2.180E-1546 27.305 0.03662 7.416E-1547 27.776 0.03600 2.950E-1448 28.972 0.03452 1.320E-1349 29.550 0.03384 6.248E-1450 29.804 0.03355 2.278E-1351 30.992 0.03227 3.947E-1252 31.501 0.03174 7.134E-1253 31.690 0.03156 1.366E-1154 32.009 0.03124 3.471E-1255 32.574 0.03070 1.186E-1056 32.863 0.03043 6.803E-1157 34.101 0.02932 1.277E-1058 34.923 0.02863 3.012E-0959 35.162 0.02844 2.272E-0960 35.411 0.02824 9.878E-0961 35.927 0.02783 1.082E-0862 36.529 0.02738 4.701E-0963 38.585 0.02592 2.902E-0764 38.826 0.02576 6.399E-0765 39.494 0.02532 2.540E-07




66 40.013 0.02499 4.013E-0667 40.518 0.02468 3.403E-0668 40.924 0.02444 6.088E-0669 41.523 0.02408 6.986E-0670 42.110 0.02375 9.985E-0671 42.318 0.02363 2.559E-0772 42.701 0.02342 1.154E-0573 42.891 0.02331 1.442E-0574 44.395 0.02253 4.132E-0575 45.025 0.02221 7.723E-0576 45.134 0.02216 6.369E-05

PARTICIPATION FACTORS



1 0.01 0.00 99.04 0.012 0.000 99.0422 99.14 0.00 0.02 99.151 0.000 99.0613 0.00 0.23 0.00 99.151 0.229 99.0624 0.00 3.27 0.00 99.151 3.496 99.0625 0.00 0.04 0.05 99.151 3.536 99.1126 0.05 0.04 0.02 99.202 3.575 99.1357 0.00 26.42 0.00 99.204 30.000 99.1358 0.00 25.59 0.00 99.204 55.587 99.1369 0.53 0.15 0.19 99.735 55.740 99.326

10 0.00 0.13 0.00 99.736 55.871 99.32611 0.00 0.06 0.00 99.736 55.927 99.326


Example Problem 28

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

12 0.00 0.04 0.00 99.736 55.969 99.32613 0.00 0.00 0.56 99.740 55.969 99.88914 0.00 0.01 0.01 99.740 55.979 99.89815 0.00 0.37 0.01 99.740 56.349 99.90916 0.00 0.00 0.01 99.740 56.354 99.92317 0.00 0.00 0.00 99.741 56.354 99.92318 0.01 0.00 0.01 99.753 56.355 99.93719 0.00 0.00 0.00 99.753 56.358 99.93720 0.00 0.01 0.00 99.754 56.364 99.93921 0.00 0.00 0.00 99.754 56.364 99.94022 0.01 0.00 0.00 99.765 56.364 99.94023 0.00 0.00 0.01 99.766 56.368 99.94724 0.00 0.01 0.00 99.766 56.383 99.94825 0.00 0.00 0.02 99.767 56.384 99.96526 0.00 0.18 0.00 99.767 56.562 99.96527 0.00 0.03 0.00 99.768 56.591 99.96528 0.00 1.09 0.00 99.768 57.676 99.96529 0.06 0.53 0.00 99.829 58.205 99.97030 0.00 3.53 0.00 99.832 61.732 99.97131 0.02 0.01 0.00 99.852 61.737 99.97132 0.00 3.48 0.00 99.854 65.214 99.97133 0.00 0.09 0.00 99.854 65.308 99.97134 0.00 0.01 0.00 99.855 65.319 99.97235 0.00 0.42 0.00 99.856 65.736 99.97236 0.00 0.00 0.00 99.856 65.736 99.97237 0.00 2.19 0.00 99.856 67.926 99.97238 0.00 0.01 0.00 99.856 67.932 99.97239 0.00 0.03 0.00 99.856 67.967 99.97240 0.00 0.00 0.00 99.856 67.969 99.97241 0.00 0.15 0.00 99.856 68.116 99.97342 0.00 4.79 0.00 99.856 72.905 99.97343 0.00 0.07 0.00 99.857 72.980 99.97344 0.00 0.05 0.00 99.857 73.033 99.97345 0.00 0.02 0.00 99.857 73.057 99.97346 0.00 0.17 0.00 99.857 73.226 99.97347 0.00 0.00 0.00 99.858 73.226 99.97348 0.00 0.38 0.00 99.858 73.610 99.97349 0.00 0.00 0.00 99.858 73.611 99.97350 0.00 0.00 0.00 99.858 73.611 99.97351 0.00 0.01 0.00 99.858 73.622 99.97352 0.00 0.00 0.00 99.858 73.623 99.97453 0.00 0.11 0.00 99.858 73.730 99.97454 0.00 0.01 0.00 99.859 73.736 99.97455 0.00 0.02 0.00 99.859 73.758 99.97456 0.00 0.06 0.00 99.859 73.823 99.97457 0.00 0.01 0.00 99.859 73.831 99.97458 0.00 0.14 0.00 99.860 73.968 99.97559 0.00 0.00 0.00 99.860 73.970 99.97660 0.00 0.04 0.00 99.861 74.006 99.98161 0.00 0.14 0.00 99.865 74.144 99.98462 0.00 0.26 0.00 99.868 74.407 99.98563 0.00 0.06 0.00 99.868 74.467 99.98564 0.00 0.30 0.00 99.868 74.764 99.98565 0.00 0.15 0.00 99.869 74.911 99.986

Example Problem 28

287

102. FINISH


**** DATE= TIME= ****



Example Problem 28

288

Understanding the output: After the analysis is complete, look at the output file. (This file can be viewed from File - View - Output File - STAAD output). (i) Mode number and corresponding frequencies and periods

Since we asked for 65 modes, we obtain a report, a portion of which is as shown:

CALCULATED FREQUENCIES FOR LOAD CASE 1 MODE FREQUENCY PERIOD ACCURACY (CYCLES/SEC) (SEC) 1 1.636 0.61111 1.344E-16 2 2.602 0.38433 0.000E+00 3 2.882 0.34695 8.666E-16 4 3.754 0.26636 0.000E+00 5 4.076 0.24532 3.466E-16 6 4.373 0.22870 6.025E-16 7 4.519 0.22130 5.641E-16 8 4.683 0.21355 5.253E-16 9 5.028 0.19889 0.000E+00 10 7.189 0.13911 8.916E-16 11 7.238 0.13815 0.000E+00 12 7.363 0.13582 0.000E+00

Example Problem 28

289

(ii) Participation factors in Percentage

MASS PARTICIPATION FACTORS IN PERCENT MODE X Y Z SUMM-X SUMM-Y SUMM-Z 1 0.01 0.00 99.04 0.012 0.000 99.042 2 99.14 0.00 0.02 99.151 0.000 99.061 3 0.00 0.23 0.00 99.151 0.229 99.062 4 0.00 3.27 0.00 99.151 3.496 99.062 5 0.00 0.04 0.05 99.151 3.536 99.112 6 0.05 0.04 0.02 99.202 3.575 99.135 7 0.00 26.42 0.00 99.204 30.000 99.135 8 0.00 25.59 0.00 99.204 55.587 99.136 9 0.53 0.15 0.19 99.735 55.740 99.326 10 0.00 0.13 0.00 99.736 55.871 99.326 11 0.00 0.06 0.00 99.736 55.927 99.326 12 0.00 0.04 0.00 99.736 55.969 99.326

In the explanation earlier for the CUT OFF MODE command, we said that one measure of the importance of a mode is the participation factor of that mode. We can see from the above report that for vibration along Z direction, the first mode has a 99.04 percent participation. It is also apparent that the 7th mode is primarily a Y direction mode with a 26.42 % participation along Y and 0 in X and Z. The SUMM-X, SUMM-Y and SUMM-Z columns show the cumulative value of the participation of all the modes upto and including a given mode. One can infer from those terms that if one is interested in 95% participation along X, the first 2 modes are sufficient. But for the Y direction, even with 10 modes, we barely obtained 60%. The reason for this can be understood by an examination of the nature of the structure. The deck slab is capable of vibrating in several low energy and primarily vertical direction modes. The out-of-plane flexible nature of


Example Problem 28

290

the slab enables it to vibrate in a manner resembling a series of wave like curves. Masses on either side of the equilibrium point have opposing eigenvector values leading to a lot of cancellation of the contribution from the respective masses. Localized modes, where small pockets in the structure undergo flutter due to their relative weak stiffness compared to the rest of the model, also result in small participation factors.

(iii) Viewing the mode shapes

After the analysis is completed, select Post-processing from the mode menu. This screen contains facilities for graphically examining the shape of the mode in static and animated views. The Dynamics page on the left side of the screen is available for viewing the shape of the mode statically. The Animation option of the Results menu can be used for animating the mode. The mode number can be selected from the “Loads and Results” tab of the “Diagrams” dialog box which comes up when the Animation option is chosen. The size to which the mode is drawn is controlled using the “Scales” tab of the “Diagrams” dialog box.

Example Problem 28

291

NOTES


Example Problem 28

292

NOTES

Example Problem 29

293


Analysis and design of a structure for seismic loads is demonstrated in this example. The elaborate dynamic analysis procedure called time history analysis is used. In this model, static load cases are solved along with the seismic load case. For the seismic case, the maximum values of displacements, forces and reactions are obtained. The results of the dynamic case are combined with those of the static cases and steel design is performed on the combined cases.

3.50m

1.80m

1.70m

1.90m

3.00m

2.80m

1.80m

3.00m


Example Problem 29

294

Actual input is shown in bold lettering followed by explanation. STAAD SPACE DYNAMIC ANALYSIS FOR SEISMIC LOADS

Every STAAD input file has to begin with the word STAAD. The word SPACE signifies that the structure is a space frame and the geometry is defined through X, Y and Z axes. The remainder of the words form a title to identify this project.

UNIT METER KNS


JOINT COORDINATES 1 0 0 0 ; 2 0 3.5 0 ; 3 0 5.3 0 ; 4 0 7 0 REPEAT ALL 1 9.5 0 0 REPEAT ALL 1 0 0 3 17 1.8 7 0 ; 18 4.6 7 0 ; 19 7.6 7 0 REPEAT ALL 1 0 0 3

For joints 1 through 4, the joint number is followed by the X, Y and Z coordinates as specified above. The coordinates of these joints are used as a basis for generating 12 more joints by incrementing the X & Z coordinates by specific amounts. REPEAT ALL commands are used for the generation. Details of these commands are available in Section 5.11 of the Technical Reference manual. Following this, another round of explicit definition (joints 17, 18 & 19) and generation (20, 21 & 22) is carried out. The results of the generation may be visually verified using STAAD.Pro's graphical viewing facilities.

MEMBER INCIDENCES 1 1 2 3 REPEAT 1 3 4 7 9 10 9 10 13 14 12 13 4 17; 14 17 18; 15 18 19; 16 19 8 17 12 20; 18 20 21; 19 21 22; 20 22 16 21 2 10; 22 4 12; 23 6 14 24 8 16; 25 3 17; 26 7 19; 27 11 20; 28 15 22; 29 18 21

Example Problem 29

295

A mixture of explicit definition and generation of member connectivity data (joint numbers between which members are connected) is used to generate 29 members for the structure.

START GROUP DEFINITION MEMBER _VERTICAL 1 TO 12 _XBEAM 13 TO 20 _ZBEAM 21 TO 24 29 _BRACE 25 TO 28 END GROUP DEFINITION

The above block of data is referred to as formation of groups. Group names are a mechanism by which a single moniker can be used to refer to a cluster of entities, such as members. For our structure, the columns are being grouped to a name called VERTICAL, the beams running along the X direction are assigned the name XBEAM, and so on.

MEMBER PROPERTIES CANADIAN _VERTICAL TA ST W310X97 _XBEAM TA ST W250X39 _ZBEAM TA ST C200X17 _BRACE TA ST L150X150X13

Member properties are assigned from the Canadian steel table. The members which receive these properties are those embedded within the respective group names. The benefit of using the group name is apparent here. Just from the looks of the command, we can understand that the diagonal braces are being assigned a single angle. The alternative, which would be

25 TO 28 TA ST L150X150X13 would have required us to go to the graphical tools to get a sense of what members 25 to 28 are.


Example Problem 29

296 UNIT KNS MMS CONSTANT E 200 ALL

The Modulus of elasticity (E) is set to 200 kN/sq.mm for all members. The keyword CONSTANTS has to precede this data.

UNIT KGS METER CONSTANT DENSITY 7800 ALL POISSON STEEL ALL BETA 180 MEMB 21 22

Density and Poisson for all members is set using the above commands. The BETA angle for the channels along the left edge is set to 180 so their legs point toward the interior of the structure.

SUPPORTS 1 5 9 13 PINNED

The bottom ends of the columns of the platform are pinned supported.

CUT OFF MODE SHAPE 30 The above command is a critical command if one wishes to override the default number of modes computed and used in a dynamic analysis. The default, which is 6, may not always be sufficient to capture a significant portion of the structural response in a response spectrum or time history analysis, and hence the need to override the default. This command is explained in Section 5.30 of the Technical Reference manual.

UNIT METER DEFINE TIME HISTORY TYPE 1 ACCELERATION READ EQDATA.TXT ARRIVAL TIME 0.0 DAMPING 0.05

Example Problem 29

297

There are two stages in the command specification required for a time-history analysis. The first stage is defined above. Here, the parameters of the earthquake (ground acceleration) are provided. Each data set is individually identified by the number that follows the TYPE command. In this file, only one data set is defined, which is apparent from the fact that only one TYPE is defined. The word FORCE that follows the TYPE 1 command signifies that this data set is for a ground acceleration. (If one wishes to specify a forcing function, the keyword FORCE must be used instead.) Notice the expression "READ EQDATA.TXT". It means that we have chosen to specify the time vs. ground acceleration data in the file called EQDATA.TXT. That file must reside in the same folder as the one in which the data file for this structure resides. As explained in the small examples shown in Section 5.31.4 of the Technical Reference manual, the EQDATA.TXT file is a simple text file containing several pairs of time-acceleration data. A sample portion of that file is as shown below.

0.0000 0.006300 0.0200 0.003640 0.0400 0.000990 0.0600 0.004280 0.0800 0.007580 0.1000 0.010870

While it may not be apparent from the above numbers, it may also be noted that the geological data for the site the building sits on indicate that the above acceleration values are a fraction of "g", the acceleration due to gravity. Thus, for example, at 0.02 seconds, the acceleration is 0.00364 multiplied by 9.806 m/sec^2 (or 0.00364 multiplied by 32.2 ft/sec^2). Consequently, the burden of informing the program that the values need to be multiplied by "g" is upon us, and we shall be doing so at a later step.


Example Problem 29

298

The arrival time value indicates the relative value of time at which the earthquake begins to act upon the structure. We have chosen 0.0, as there is no other dynamic load on the structure from the relative time standpoint. The modal damping ratio for all the modes is set to 0.05.

LOAD 1 WEIGHT OF STRUCTURE ACTING STATICALLY SELFWEIGHT Y -1.0

The above data describe a static load case. The selfweight of the structure is acting in the negative global Y direction.

LOAD 2 PLATFORM LEVEL LOAD ACTING STATICALLY FLOOR LOAD YRA 6.9 7.1 FLOAD -500

Load case 2 is also a static load case. At the Y=7.0m elevation, our structure has a floor slab. But, it is a non-structural entity which, though capable of carrying the loads acting on itself, is not meant to be an integral part of the framing system. It merely transmits the load to the beam-column grid. There are uniform area loads on the floor (think of the load as wooden pallets supporting boxes of paper). Since the slab is not part of the structural model, how do we tell the program to transmit the imposed load from the slab to the beams without manually converting them to distributed beam loads ourselves? That is where the floor load utility comes in handy. It is a facility where we specify the load as a pressure, and the program converts the pressure to individual beam loads. Thus, the input required from the user is very simple - load intensity in the form of pressure, and the region of the structure in terms of X, Y and Z coordinates in space, of the area over which the pressure acts. In the process of converting the pressure to beam loads, STAAD will consider the empty space between criss-crossing beams (in plan view) to be panels, similar to the squares of a chess board. The load on each panel is then tranferred to beams surrounding the panel, using a triangular or trapezoidal load distribution method.

Example Problem 29

299 LOAD 3 DYNAMIC LOAD * MASSES SELFWEIGHT X 1.0 SELFWEIGHT Y 1.0 SELFWEIGHT Z 1.0 FLOOR LOAD YRANGE 6.9 7.1 FLOAD 500 GX YRANGE 6.9 7.1 FLOAD 500 GY YRANGE 6.9 7.1 FLOAD 500 GZ

Load case 3 is the dynamic load case, the one which contains the second part of the instruction set for a dynamic analysis to be performed. The data here are

a. loads which will yield the mass values which will populate the mass matrix

b. the directions of the loads, which will yield the degree of freedom numbers of the mass matrix for being populated.

Thus, the selfweight, as well as the imposed loads on the non-structural slab are to be considered as participating in the vibration along all the global directions.

GROUND MOTION X 1 1 9.806 The above command too is part of load case 3. Here we say that the seismic force, whose characteristics are defined by the TYPE 1 time history input data, acting at arrival time 1, is to be applied along the X direction. We mentioned earlier that the acceleration input data was specified as a fraction of “g”. The number 9.806 indicates the value which the accleration data, as read from EQDATA.TXT are to be factored by before they are used.


Example Problem 29

300 LOAD COMBINATION 11 (STATIC + POSITIVE OF DYNAMIC) 1 1.0 2 1.0 3 1.0 LOAD COMBINATION 12 (STATIC + NEGATIVE OF DYNAMIC) 1 1.0 2 1.0 3 -1.0

In a time history analysis, the member forces FX thru MZ each have a value for every time step. If there are a 1000 time steps, there will be 1000 values of FX, 1000 for FY etc. for that load case. Not all of them can be used in a further calculation like a steel or concrete design. However, the maximum from among those time steps is available. If we wish to do a design, one way to make sure that the structure is not under-designed is to create 2 load combination cases involving the dynamic case, a positive combination, and a negative combination. That is what is being done above. Load combination case no. 11 consists of the sum of the static load cases (1 & 2) with the positive direction of the dynamic load case (3). Load combination case no. 12 consists of the sum of the static load cases (1 & 2) with the negative direction of the dynamic load case (3). The user has discretion on what load factors to use with these combinations. We have chosen the factors to be 1.0.

PERFORM ANALYSIS The above is the instruction to perform the analysis related calculations. That means, computing nodal displacements, support reactions, etc.

PRINT ANALYSIS RESULTS The above command is an instruction to the program to produce a report of the joint displacements, support reactions and member end forces in the output file. As mentioned earlier, for the dynamic case, these will be just the maximum values, not the ones generated for every time step. If the user wishes to see the results for each time step, he/she may do so by using STAAD's Post-processing facilities.

Example Problem 29

301 LOAD LIST 11 12 PARAMETER CODE CANADA CHECK CODE ALL

A steel design - code check - is done according to the Canadian code for load cases 11 and 12.

FINISH


Example Problem 29

302


1. STAAD SPACE DYNAMIC ANALYSIS FOR SEISMIC LOADS3. UNIT METER KNS4. JOINT COORDINATES5. 1 0 0 0 ; 2 0 3.5 0 ; 3 0 5.3 0 ; 4 0 7 06. REPEAT ALL 1 9.5 0 07. REPEAT ALL 1 0 0 38. 17 1.8 7 0 ; 18 4.6 7 0 ; 19 7.6 7 09. REPEAT ALL 1 0 0 3

11. MEMBER INCIDENCES12. 1 1 2 313. REPEAT 1 3 414. 7 9 10 915. 10 13 14 1216. 13 4 17; 14 17 18; 15 18 19; 16 19 817. 17 12 20; 18 20 21; 19 21 22; 20 22 1618. 21 2 10; 22 4 12; 23 6 1419. 24 8 16; 25 3 17; 26 7 19; 27 11 20; 28 15 22; 29 18 2121. START GROUP DEFINITION22. MEMBER23. _VERTICAL 1 TO 1224. _XBEAM 13 TO 2025. _ZBEAM 21 TO 24 2926. _BRACE 25 TO 2827. END GROUP DEFINITION29. MEMBER PROPERTIES CANADIAN30. _VERTICAL TA ST W310X9731. _XBEAM TA ST W250X3932. _ZBEAM TA ST C200X1733. _BRACE TA ST L150X150X1335. UNIT KNS MMS36. CONSTANT37. E 200 ALL39. UNIT KGS METER40. CONSTANT41. DENSITY 7800 ALL42. POISSON STEEL ALL43. BETA 180 MEMB 21 2245. SUPPORTS46. 1 5 9 13 PINNED48. CUT OFF MODE SHAPE 3051. UNIT METER52. DEFINE TIME HISTORY53. TYPE 1 ACCELERATION54. READ EQDATA.TXT55. ARRIVAL TIME56. 0.057. DAMPING 0.0560. LOAD 1 WEIGHT OF STRUCTURE ACTING STATICALLY61. SELFWEIGHT Y -1.063. LOAD 2 PLATFORM LEVEL LOAD ACTING STATICALLY64. FLOOR LOAD65. YRA 6.9 7.1 FLOAD -50067. LOAD 3 DYNAMIC LOAD68. * MASSES69. SELFWEIGHT X 1.070. SELFWEIGHT Y 1.071. SELFWEIGHT Z 1.073. FLOOR LOAD74. YRANGE 6.9 7.1 FLOAD 500 GX75. YRANGE 6.9 7.1 FLOAD 500 GY76. YRANGE 6.9 7.1 FLOAD 500 GZ78. GROUND MOTION X 1 1 9.80680. LOAD COMBINATION 11 (STATIC + POSITIVE OF DYNAMIC)81. 1 1.0 2 1.0 3 1.0

Example Problem 29

303 83. LOAD COMBINATION 12 (STATIC + NEGATIVE OF DYNAMIC)84. 1 1.0 2 1.0 3 -1.086. PERFORM ANALYSIS







1 0.694 1.44194 3.742E-162 1.216 0.82260 1.218E-153 1.368 0.73103 3.847E-164 1.566 0.63849 1.321E-155 2.080 0.48072 2.496E-156 3.032 0.32978 1.253E-157 4.198 0.23820 1.634E-168 4.270 0.23420 1.580E-169 5.540 0.18051 9.384E-16

10 5.545 0.18033 3.746E-1611 5.711 0.17510 2.649E-1512 12.753 0.07841 2.833E-1613 12.761 0.07836 2.971E-1514 15.185 0.06585 0.000E+0015 15.242 0.06561 1.983E-1516 16.375 0.06107 0.000E+0017 16.383 0.06104 7.897E-1518 45.949 0.02176 1.164E-1319 45.973 0.02175 1.692E-1020 48.961 0.02042 5.151E-1421 48.989 0.02041 4.764E-1022 52.125 0.01918 1.764E-1523 52.272 0.01913 1.106E-1424 55.107 0.01815 1.082E-1225 57.060 0.01753 1.160E-1326 57.075 0.01752 2.809E-1027 65.692 0.01522 1.990E-1328 65.856 0.01518 2.962E-1129 86.634 0.01154 1.969E-1230 86.707 0.01153 4.119E-08

The following Frequencies are estimates that were calculated. These arefor

information only and will not be used. Remaining values are either abovethe cut off mode/freq values or are of low accuracy. To use thesefrequencies, rerun with a higher cutoff mode (or mode + freq) value.



31 88.023 0.01136 1.720E-1132 88.471 0.01130 2.539E-1233 88.729 0.01127 5.326E-0834 89.364 0.01119 2.525E-1235 89.756 0.01114 1.000E-1136 90.001 0.01111 5.809E-0737 154.025 0.00649 1.039E-1038 154.031 0.00649 2.453E-0939 160.638 0.00623 9.101E-1140 160.646 0.00622 2.348E-0741 165.740 0.00603 5.445E-1342 165.742 0.00603 1.303E-1043 188.250 0.00531 2.892E-10


Example Problem 29

304 MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC) ACCURACY

44 188.259 0.00531 2.628E-0945 192.465 0.00520 8.740E-1046 192.474 0.00520 5.365E-0847 350.643 0.00285 2.069E-0848 350.843 0.00285 7.427E-0949 351.878 0.00284 4.450E-07




1 0.00 0.00 85.32 0.000 0.000 85.3192 98.02 0.00 0.00 98.023 0.000 85.3193 0.00 0.00 0.00 98.023 0.000 85.3244 0.00 0.00 0.03 98.023 0.000 85.3535 0.00 0.00 13.46 98.023 0.000 98.8156 0.00 0.00 0.00 98.023 0.000 98.8157 0.00 0.00 0.00 98.023 0.000 98.8158 0.00 0.00 0.01 98.023 0.000 98.8269 0.00 51.82 0.00 98.023 51.816 98.826

10 0.00 0.00 0.00 98.023 51.816 98.82811 0.00 0.00 0.23 98.023 51.816 99.06112 1.71 0.25 0.00 99.728 52.068 99.06113 0.00 0.00 0.00 99.728 52.068 99.06114 0.00 0.00 0.39 99.728 52.068 99.45515 0.00 0.00 0.52 99.728 52.068 99.97716 0.03 7.05 0.00 99.758 59.122 99.97717 0.00 0.00 0.00 99.758 59.122 99.97718 0.00 11.15 0.00 99.760 70.271 99.97719 0.00 0.00 0.00 99.760 70.271 99.97720 0.23 0.10 0.00 99.991 70.374 99.97721 0.00 0.00 0.00 99.991 70.374 99.97722 0.00 0.00 0.00 99.991 70.374 99.97723 0.00 0.00 0.00 99.991 70.374 99.97724 0.00 0.00 0.00 99.991 70.374 99.97725 0.00 1.67 0.00 99.991 72.039 99.97726 0.00 0.00 0.00 99.991 72.039 99.97727 0.00 0.00 0.01 99.991 72.039 99.98928 0.00 0.00 0.01 99.991 72.039 100.00029 0.00 0.26 0.00 99.995 72.295 100.00030 0.00 0.00 0.00 99.995 72.295 100.000

TIME STEP USED IN TIME HISTORY ANALYSIS = 0.00139 SECONDSNUMBER OF MODES WHOSE CONTRIBUTION IS CONSIDERED = 30TIME DURATION OF TIME HISTORY ANALYSIS = 31.160 SECONDSNUMBER OF TIME STEPS IN THE SOLUTION PROCESS = 22435

BASE SHEAR UNITS ARE -- KGS METE

MAXIMUM BASE SHEAR X= -9.454792E+03 Y= 4.703766E+01 Z= -3.820288E-06AT TIMES 5.809722 2.484722 2.762500

Example Problem 29




1 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.00012 0.0000 0.0000 0.0000 0.0000 0.0000 0.00153 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0173

11 0.0000 0.0000 0.0000 0.0000 0.0000 -0.015612 0.0000 0.0000 0.0000 0.0000 0.0000 0.0189

2 1 -0.0284 -0.0011 0.0000 0.0000 0.0000 0.00002 -0.4150 -0.0050 -0.0004 -0.0001 0.0000 0.00043 5.7155 0.0046 0.0000 0.0000 0.0000 -0.0141

11 5.2720 -0.0015 -0.0005 -0.0001 0.0000 -0.013712 -6.1590 -0.0107 -0.0005 -0.0001 0.0000 0.0145

3 1 -0.0245 -0.0015 -0.0002 0.0000 0.0000 -0.00012 -0.3585 -0.0075 -0.0123 0.0000 -0.0001 -0.00123 7.9309 0.0070 0.0000 0.0000 0.0000 -0.0100

11 7.5480 -0.0021 -0.0124 0.0000 -0.0001 -0.011312 -8.3139 -0.0160 -0.0124 0.0000 -0.0001 0.0088

4 1 -0.0030 -0.0016 0.0000 0.0000 0.0000 -0.00012 -0.0453 -0.0075 0.0004 0.0002 0.0000 -0.00203 9.3438 0.0023 0.0000 0.0000 0.0000 -0.0083

11 9.2955 -0.0068 0.0004 0.0002 0.0000 -0.010412 -9.3920 -0.0115 0.0004 0.0002 0.0000 0.0061

5 1 0.0000 0.0000 0.0000 0.0000 0.0000 -0.00012 0.0000 0.0000 0.0000 0.0000 0.0000 -0.00143 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0174

11 0.0000 0.0000 0.0000 0.0000 0.0000 -0.018912 0.0000 0.0000 0.0000 0.0000 0.0000 0.0158

6 1 0.0256 -0.0011 0.0000 0.0000 0.0000 0.00002 0.3726 -0.0050 -0.0004 -0.0001 0.0000 -0.00023 5.7319 -0.0046 0.0000 0.0000 0.0000 -0.0141

11 6.1301 -0.0107 -0.0005 -0.0001 0.0000 -0.014412 -5.3337 -0.0015 -0.0005 -0.0001 0.0000 0.0139

7 1 0.0202 -0.0015 -0.0002 0.0000 0.0000 0.00012 0.2943 -0.0075 -0.0123 0.0000 0.0001 0.00133 7.9447 -0.0069 0.0000 0.0000 0.0000 -0.0100

11 8.2592 -0.0160 -0.0124 0.0000 0.0001 -0.008612 -7.6303 -0.0021 -0.0124 0.0000 0.0001 0.0114

8 1 -0.0027 -0.0016 0.0000 0.0000 0.0000 0.00012 -0.0399 -0.0077 0.0004 0.0002 0.0000 0.00213 9.3451 -0.0025 0.0000 0.0000 0.0000 -0.0082

11 9.3025 -0.0118 0.0004 0.0002 0.0000 -0.005912 -9.3876 -0.0068 0.0004 0.0002 0.0000 0.0105

9 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.00012 0.0000 0.0000 0.0000 0.0000 0.0000 0.00153 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0173

11 0.0000 0.0000 0.0000 0.0000 0.0000 -0.015612 0.0000 0.0000 0.0000 0.0000 0.0000 0.0189

10 1 -0.0284 -0.0011 0.0000 0.0000 0.0000 0.00002 -0.4150 -0.0050 0.0004 0.0001 0.0000 0.00043 5.7155 0.0046 0.0000 0.0000 0.0000 -0.0141

11 5.2720 -0.0015 0.0005 0.0001 0.0000 -0.013712 -6.1590 -0.0107 0.0005 0.0001 0.0000 0.0145

11 1 -0.0245 -0.0015 0.0002 0.0000 0.0000 -0.00012 -0.3585 -0.0075 0.0123 0.0000 0.0001 -0.00123 7.9309 0.0070 0.0000 0.0000 0.0000 -0.0100

11 7.5480 -0.0021 0.0124 0.0000 0.0001 -0.011312 -8.3139 -0.0160 0.0124 0.0000 0.0001 0.0088

12 1 -0.0030 -0.0016 0.0000 0.0000 0.0000 -0.00012 -0.0453 -0.0075 -0.0004 -0.0002 0.0000 -0.00203 9.3438 0.0023 0.0000 0.0000 0.0000 -0.0083

11 9.2955 -0.0068 -0.0004 -0.0002 0.0000 -0.010412 -9.3920 -0.0115 -0.0004 -0.0002 0.0000 0.0061

13 1 0.0000 0.0000 0.0000 0.0000 0.0000 -0.00012 0.0000 0.0000 0.0000 0.0000 0.0000 -0.00143 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0174

11 0.0000 0.0000 0.0000 0.0000 0.0000 -0.018912 0.0000 0.0000 0.0000 0.0000 0.0000 0.0158

14 1 0.0256 -0.0011 0.0000 0.0000 0.0000 0.00002 0.3726 -0.0050 0.0004 0.0001 0.0000 -0.00023 5.7319 -0.0046 0.0000 0.0000 0.0000 -0.0141

11 6.1301 -0.0107 0.0005 0.0001 0.0000 -0.014412 -5.3337 -0.0015 0.0005 0.0001 0.0000 0.0139


Example Problem 29

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


15 1 0.0202 -0.0015 0.0002 0.0000 0.0000 0.00012 0.2943 -0.0075 0.0123 0.0000 -0.0001 0.00133 7.9447 -0.0069 0.0000 0.0000 0.0000 -0.0100

11 8.2592 -0.0160 0.0124 0.0000 -0.0001 -0.008612 -7.6303 -0.0021 0.0124 0.0000 -0.0001 0.0114

16 1 -0.0027 -0.0016 0.0000 0.0000 0.0000 0.00012 -0.0399 -0.0077 -0.0004 -0.0002 0.0000 0.00213 9.3450 -0.0025 0.0000 0.0000 0.0000 -0.0082

11 9.3025 -0.0118 -0.0004 -0.0002 0.0000 -0.005912 -9.3876 -0.0068 -0.0004 -0.0002 0.0000 0.0105

17 1 -0.0027 -0.0262 0.0000 0.0000 0.0000 -0.00022 -0.0402 -0.3688 0.0008 0.0001 0.0000 -0.00253 9.3268 -1.3999 0.0000 0.0000 0.0000 -0.0033

11 9.2839 -1.7950 0.0008 0.0001 0.0000 -0.005912 -9.3696 1.0049 0.0008 0.0001 0.0000 0.0006

18 1 -0.0028 -0.0632 0.0000 0.0001 0.0000 0.00002 -0.0426 -0.9657 0.0000 0.0038 0.0000 -0.00033 9.3292 -0.0733 0.0000 0.0000 0.0000 0.0086

11 9.2838 -1.1022 0.0000 0.0039 0.0000 0.008312 -9.3746 -0.9555 0.0000 0.0039 0.0000 -0.0089

19 1 -0.0030 -0.0291 0.0000 0.0000 0.0000 0.00022 -0.0452 -0.4118 0.0009 0.0001 0.0000 0.00253 9.3267 1.4627 0.0000 0.0000 0.0000 -0.0031

11 9.2785 1.0219 0.0009 0.0001 0.0000 -0.000412 -9.3749 -1.9036 0.0009 0.0001 0.0000 0.0057

20 1 -0.0027 -0.0262 0.0000 0.0000 0.0000 -0.00022 -0.0402 -0.3688 -0.0008 -0.0001 0.0000 -0.00253 9.3268 -1.3999 0.0000 0.0000 0.0000 -0.0033

11 9.2839 -1.7950 -0.0008 -0.0001 0.0000 -0.005912 -9.3696 1.0049 -0.0008 -0.0001 0.0000 0.0006

21 1 -0.0028 -0.0632 0.0000 -0.0001 0.0000 0.00002 -0.0426 -0.9657 0.0000 -0.0038 0.0000 -0.00033 9.3292 -0.0733 0.0000 0.0000 0.0000 0.0086

11 9.2838 -1.1022 0.0000 -0.0039 0.0000 0.008312 -9.3746 -0.9555 0.0000 -0.0039 0.0000 -0.0089

22 1 -0.0030 -0.0291 0.0000 0.0000 0.0000 0.00022 -0.0452 -0.4118 -0.0009 -0.0001 0.0000 0.00253 9.3267 1.4627 0.0000 0.0000 0.0000 -0.0031

11 9.2785 1.0219 -0.0009 -0.0001 0.0000 -0.000412 -9.3749 -1.9036 -0.0009 -0.0001 0.0000 0.0057

SUPPORT REACTIONS -UNIT KGS METE STRUCTURE TYPE = SPACE-----------------


1 1 60.21 991.51 0.96 0.00 0.00 0.002 873.02 3562.50 -19.70 0.00 0.00 0.003 -2345.85 -3297.44 0.00 0.00 0.00 0.00

11 -1412.61 1256.57 -18.74 0.00 0.00 0.0012 3279.08 7851.45 -18.74 0.00 0.00 0.00

5 1 -60.21 991.66 0.96 0.00 0.00 0.002 -873.02 3562.50 -19.70 0.00 0.00 0.003 -2381.55 3294.01 0.00 0.00 0.00 0.00

11 -3314.79 7848.17 -18.74 0.00 0.00 0.0012 1448.31 1260.15 -18.74 0.00 0.00 0.00

9 1 60.21 991.51 -0.96 0.00 0.00 0.002 873.02 3562.50 19.70 0.00 0.00 0.003 -2345.84 -3297.44 0.00 0.00 0.00 0.00

11 -1412.61 1256.57 18.74 0.00 0.00 0.0012 3279.08 7851.45 18.74 0.00 0.00 0.00

13 1 -60.21 991.66 -0.96 0.00 0.00 0.002 -873.02 3562.50 19.70 0.00 0.00 0.003 -2381.55 3294.01 0.00 0.00 0.00 0.00

11 -3314.79 7848.17 18.74 0.00 0.00 0.0012 1448.31 1260.15 18.74 0.00 0.00 0.00

Example Problem 29

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


1 1 1 991.51 -60.21 0.96 0.00 0.00 0.002 -655.10 60.21 -0.96 0.00 -3.37 -210.74

2 1 3562.50 -873.02 -19.70 0.00 0.00 0.002 -3562.50 873.02 19.70 0.00 68.97 -3055.59

3 1 -3297.44 2345.85 0.00 0.00 0.00 0.002 3297.44 -2345.85 0.00 0.00 0.00 8210.46

11 1 1256.57 1412.61 -18.74 0.00 0.00 0.002 -920.17 -1412.61 18.74 0.00 65.59 4944.13

12 1 7851.45 -3279.08 -18.74 0.00 0.00 0.002 -7515.04 3279.08 18.74 0.00 65.59 -11476.79

2 1 2 629.59 -60.21 5.88 0.01 -8.90 210.743 -456.58 60.21 -5.88 -0.01 -1.69 -319.13

2 2 3562.50 -873.02 112.18 0.24 -79.32 3055.593 -3562.50 873.02 -112.18 -0.24 -122.60 -4627.03

3 2 -3297.41 2244.72 0.00 0.00 0.00 -8210.463 3297.41 -2244.72 0.00 0.00 0.00 12250.95

11 2 894.68 1311.48 118.06 0.24 -88.22 -4944.133 -721.67 -1311.48 -118.06 -0.24 -124.30 7304.79

12 2 7489.50 -3177.95 118.06 0.24 -88.22 11476.793 -7316.49 3177.95 -118.06 -0.24 -124.30 -17197.10

3 1 3 188.28 183.19 5.87 -0.01 1.68 304.114 -24.88 -183.19 -5.87 0.01 -11.66 7.31

2 3 38.88 2836.07 111.53 -0.35 121.90 4573.314 -38.88 -2836.07 -111.53 0.35 -311.50 248.00

3 3 6870.06 -8868.55 0.00 0.00 0.00 -12282.614 -6870.06 8868.55 0.00 0.00 0.00 -2794.08

11 3 7097.21 -5849.30 117.40 -0.36 123.58 -7405.184 -6933.81 5849.30 -117.40 0.36 -323.16 -2538.77

12 3 -6642.90 11887.81 117.40 -0.36 123.58 17160.034 6806.30 -11887.81 -117.40 0.36 -323.16 3049.39

4 1 5 991.66 60.21 0.96 0.00 0.00 0.006 -655.25 -60.21 -0.96 0.00 -3.38 210.74

2 5 3562.50 873.02 -19.70 0.00 0.00 0.006 -3562.50 -873.02 19.70 0.00 68.95 3055.59

3 5 3294.01 2381.55 0.00 0.00 0.00 0.006 -3294.01 -2381.55 0.00 0.00 0.00 8335.43

11 5 7848.17 3314.79 -18.74 0.00 0.00 0.006 -7511.76 -3314.79 18.74 0.00 65.58 11601.76

12 5 1260.15 -1448.31 -18.74 0.00 0.00 0.006 -923.74 1448.31 18.74 0.00 65.58 -5069.10

5 1 6 629.74 60.21 5.88 -0.01 -8.89 -210.747 -456.73 -60.21 -5.88 0.01 -1.69 319.13

2 6 3562.50 873.02 112.15 -0.23 -79.31 -3055.597 -3562.50 -873.02 -112.15 0.23 -122.55 4627.03

3 6 3293.96 2279.65 0.00 0.00 0.00 -8335.437 -3293.96 -2279.65 0.00 0.00 0.00 12438.80

11 6 7486.19 3212.89 118.03 -0.23 -88.21 -11601.767 -7313.19 -3212.89 -118.03 0.23 -124.24 17384.95

12 6 898.28 -1346.41 118.03 -0.23 -88.21 5069.107 -725.28 1346.41 -118.03 0.23 -124.24 -7492.64

6 1 7 202.37 -179.47 5.87 0.01 1.68 -302.908 -38.97 179.47 -5.87 -0.01 -11.66 -2.19

2 7 254.77 -2793.04 111.52 0.33 121.92 -4569.208 -254.77 2793.04 -111.52 -0.33 -311.50 -178.97

3 7 -6518.57 -9035.23 0.00 0.00 0.00 -12471.408 6518.57 9035.23 0.00 0.00 0.00 -2888.75

11 7 -6061.43 -12007.74 117.39 0.34 123.59 -17343.508 6224.83 12007.74 -117.39 -0.34 -323.16 -3069.90

12 7 6975.71 6062.73 117.39 0.34 123.59 7599.308 -6812.31 -6062.73 -117.39 -0.34 -323.16 2707.59


Example Problem 29



7 1 9 991.51 -60.21 -0.96 0.00 0.00 0.0010 -655.10 60.21 0.96 0.00 3.37 -210.74

2 9 3562.50 -873.02 19.70 0.00 0.00 0.0010 -3562.50 873.02 -19.70 0.00 -68.97 -3055.59

3 9 -3297.44 2345.84 0.00 0.00 0.00 0.0010 3297.44 -2345.84 0.00 0.00 0.00 8210.46

11 9 1256.57 1412.61 18.74 0.00 0.00 0.0010 -920.17 -1412.61 -18.74 0.00 -65.59 4944.13

12 9 7851.45 -3279.08 18.74 0.00 0.00 0.0010 -7515.04 3279.08 -18.74 0.00 -65.59 -11476.79

8 1 10 629.59 -60.21 -5.88 -0.01 8.90 210.7411 -456.58 60.21 5.88 0.01 1.69 -319.13

2 10 3562.50 -873.02 -112.18 -0.24 79.32 3055.5911 -3562.50 873.02 112.18 0.24 122.60 -4627.03

3 10 -3297.41 2244.72 0.00 0.00 0.00 -8210.4611 3297.41 -2244.72 0.00 0.00 0.00 12250.95

11 10 894.68 1311.48 -118.06 -0.24 88.22 -4944.1311 -721.67 -1311.48 118.06 0.24 124.30 7304.79

12 10 7489.50 -3177.95 -118.06 -0.24 88.22 11476.7911 -7316.49 3177.95 118.06 0.24 124.30 -17197.10

9 1 11 188.28 183.19 -5.87 0.01 -1.68 304.1112 -24.88 -183.19 5.87 -0.01 11.66 7.31

2 11 38.88 2836.07 -111.53 0.35 -121.90 4573.3112 -38.88 -2836.07 111.53 -0.35 311.50 248.00

3 11 6870.06 -8868.55 0.00 0.00 0.00 -12282.6012 -6870.06 8868.55 0.00 0.00 0.00 -2794.08

11 11 7097.21 -5849.30 -117.40 0.36 -123.58 -7405.1812 -6933.81 5849.30 117.40 -0.36 323.16 -2538.77

12 11 -6642.90 11887.81 -117.40 0.36 -123.58 17160.0312 6806.30 -11887.81 117.40 -0.36 323.16 3049.39

10 1 13 991.66 60.21 -0.96 0.00 0.00 0.0014 -655.25 -60.21 0.96 0.00 3.38 210.74

2 13 3562.50 873.02 19.70 0.00 0.00 0.0014 -3562.50 -873.02 -19.70 0.00 -68.95 3055.59

3 13 3294.01 2381.55 0.00 0.00 0.00 0.0014 -3294.01 -2381.55 0.00 0.00 0.00 8335.43

11 13 7848.17 3314.79 18.74 0.00 0.00 0.0014 -7511.76 -3314.79 -18.74 0.00 -65.58 11601.76

12 13 1260.15 -1448.31 18.74 0.00 0.00 0.0014 -923.74 1448.31 -18.74 0.00 -65.58 -5069.10

11 1 14 629.74 60.21 -5.88 0.01 8.89 -210.7415 -456.73 -60.21 5.88 -0.01 1.69 319.13

2 14 3562.50 873.02 -112.15 0.23 79.31 -3055.5915 -3562.50 -873.02 112.15 -0.23 122.55 4627.03

3 14 3293.96 2279.65 0.00 0.00 0.00 -8335.4315 -3293.96 -2279.65 0.00 0.00 0.00 12438.80

11 14 7486.19 3212.89 -118.03 0.23 88.21 -11601.7615 -7313.19 -3212.89 118.03 -0.23 124.24 17384.95

12 14 898.28 -1346.41 -118.03 0.23 88.21 5069.1015 -725.28 1346.41 118.03 -0.23 124.24 -7492.64

12 1 15 202.37 -179.47 -5.87 -0.01 -1.68 -302.9016 -38.97 179.47 5.87 0.01 11.66 -2.19

2 15 254.77 -2793.04 -111.52 -0.33 -121.92 -4569.2016 -254.77 2793.04 111.52 0.33 311.50 -178.97

3 15 -6518.57 -9035.23 0.00 0.00 0.00 -12471.4016 6518.57 9035.23 0.00 0.00 0.00 -2888.75

11 15 -6061.43 -12007.74 -117.39 -0.34 -123.59 -17343.5016 6224.83 12007.74 117.39 0.34 323.16 -3069.90

12 15 6975.71 6062.73 -117.39 -0.34 -123.59 7599.3016 -6812.31 -6062.73 117.39 0.34 323.16 2707.59

Example Problem 29



13 1 4 -183.19 -0.63 -0.01 0.00 -0.01 -7.3117 183.19 69.57 0.01 0.00 0.03 -55.87

2 4 -2836.07 -523.62 -0.67 0.09 -0.29 -248.0017 2836.07 1311.12 0.67 -0.09 1.50 -1178.27

3 4 9460.47 6869.92 0.00 0.00 0.00 2794.0817 -9460.47 -6869.92 0.00 0.00 0.00 9571.78

11 4 6441.21 6345.66 -0.68 0.09 -0.30 2538.7717 -6441.21 -5489.23 0.68 -0.09 1.53 8337.63

12 4 -12479.72 -7394.18 -0.68 0.09 -0.30 -3049.3917 12479.72 8250.61 0.68 -0.09 1.53 -10805.92

14 1 17 60.21 126.72 0.00 -0.03 0.01 66.5318 -60.21 -19.50 0.00 0.03 -0.01 138.18

2 17 873.02 2212.50 -0.01 -1.73 0.27 1194.9218 -873.02 -675.00 0.01 1.73 -0.23 2341.33

3 17 -868.53 -3418.30 0.00 0.00 0.00 -9964.6618 868.53 3418.30 0.00 0.00 0.00 394.20

11 17 64.71 -1079.08 -0.02 -1.76 0.27 -8703.2018 -64.71 2723.80 0.02 1.76 -0.23 2873.71

12 17 1801.77 5757.52 -0.02 -1.76 0.27 11226.1118 -1801.77 -4112.80 0.02 1.76 -0.23 2085.30

15 1 18 60.21 -6.02 0.00 0.03 0.01 -138.1819 -60.21 120.90 0.00 -0.03 -0.01 -52.20

2 18 873.02 -450.00 0.01 1.61 0.23 -2341.3319 -873.02 2137.50 -0.01 -1.61 -0.26 -977.42

3 18 821.78 -3427.33 0.00 0.00 0.00 -394.2019 -821.78 3427.33 0.00 0.00 0.00 -9888.38

11 18 1755.02 -3883.35 0.01 1.64 0.23 -2873.7119 -1755.02 5685.74 -0.01 -1.64 -0.27 -10918.00

12 18 111.46 2971.32 0.01 1.64 0.23 -2085.3019 -111.46 -1168.93 -0.01 -1.64 -0.27 8858.76

16 1 19 -179.47 59.30 0.01 0.00 -0.03 41.368 179.47 13.46 -0.01 0.00 0.01 2.19

2 19 -2793.04 1170.23 0.64 -0.08 -1.48 971.978 2793.04 -307.73 -0.64 0.08 0.27 178.97

3 19 -9658.26 6518.57 0.00 0.00 0.00 9496.548 9658.26 -6518.57 0.00 0.00 0.00 2888.75

11 19 -12630.76 7748.10 0.64 -0.09 -1.51 10509.878 12630.76 -6812.84 -0.64 0.09 0.28 3069.90

12 19 6685.75 -5289.04 0.64 -0.09 -1.51 -8483.218 -6685.75 6224.30 -0.64 0.09 0.28 -2707.59

17 1 12 -183.19 -0.63 0.01 0.00 0.01 -7.3120 183.19 69.57 -0.01 0.00 -0.03 -55.87

2 12 -2836.07 -523.62 0.67 -0.09 0.29 -248.0020 2836.07 1311.12 -0.67 0.09 -1.50 -1178.27

3 12 9460.47 6869.92 0.00 0.00 0.00 2794.0820 -9460.47 -6869.92 0.00 0.00 0.00 9571.77

11 12 6441.21 6345.66 0.68 -0.09 0.30 2538.7720 -6441.21 -5489.23 -0.68 0.09 -1.53 8337.63

12 12 -12479.72 -7394.18 0.68 -0.09 0.30 -3049.3920 12479.72 8250.61 -0.68 0.09 -1.53 -10805.92

18 1 20 60.21 126.72 0.00 0.03 -0.01 66.5321 -60.21 -19.50 0.00 -0.03 0.01 138.18

2 20 873.02 2212.50 0.01 1.73 -0.27 1194.9221 -873.02 -675.00 -0.01 -1.73 0.23 2341.33

3 20 -868.53 -3418.30 0.00 0.00 0.00 -9964.6521 868.53 3418.30 0.00 0.00 0.00 394.20

11 20 64.71 -1079.08 0.02 1.76 -0.27 -8703.2021 -64.71 2723.80 -0.02 -1.76 0.23 2873.71

12 20 1801.77 5757.52 0.02 1.76 -0.27 11226.1121 -1801.77 -4112.80 -0.02 -1.76 0.23 2085.30


Example Problem 29



19 1 21 60.21 -6.02 0.00 -0.03 -0.01 -138.1822 -60.21 120.90 0.00 0.03 0.01 -52.20

2 21 873.02 -450.00 -0.01 -1.61 -0.23 -2341.3322 -873.02 2137.50 0.01 1.61 0.26 -977.42

3 21 821.78 -3427.33 0.00 0.00 0.00 -394.2022 -821.78 3427.33 0.00 0.00 0.00 -9888.38

11 21 1755.02 -3883.35 -0.01 -1.64 -0.23 -2873.7122 -1755.02 5685.74 0.01 1.64 0.27 -10918.00

12 21 111.46 2971.32 -0.01 -1.64 -0.23 -2085.3022 -111.46 -1168.93 0.01 1.64 0.27 8858.76

20 1 22 -179.47 59.30 -0.01 0.00 0.03 41.3616 179.47 13.46 0.01 0.00 -0.01 2.19

2 22 -2793.04 1170.23 -0.64 0.08 1.48 971.9716 2793.04 -307.73 0.64 -0.08 -0.27 178.97

3 22 -9658.26 6518.57 0.00 0.00 0.00 9496.5416 9658.26 -6518.57 0.00 0.00 0.00 2888.75

11 22 -12630.76 7748.10 -0.64 0.09 1.51 10509.8716 12630.76 -6812.84 0.64 -0.09 -0.28 3069.90

12 22 6685.75 -5289.04 -0.64 0.09 1.51 -8483.2116 -6685.75 6224.30 0.64 -0.09 -0.28 -2707.59

21 1 2 -4.92 -25.51 0.00 0.00 0.01 -12.2710 4.92 -25.51 0.00 0.00 -0.01 12.27

2 2 -131.89 0.00 0.00 0.00 0.24 -10.3610 131.89 0.00 0.00 0.00 -0.24 10.36

3 2 0.00 0.00 0.00 0.00 0.00 0.0010 0.00 0.00 0.00 0.00 0.00 0.00

11 2 -136.80 -25.51 0.00 0.00 0.24 -22.6310 136.80 -25.51 0.00 0.00 -0.24 22.63

12 2 -136.80 -25.51 0.00 0.00 0.24 -22.6310 136.80 -25.51 0.00 0.00 -0.24 22.63

22 1 4 5.88 -25.51 0.00 0.00 0.00 -11.6612 -5.88 -25.51 0.00 0.00 0.00 11.66

2 4 112.20 -562.50 0.00 0.00 0.06 -311.5912 -112.20 -562.50 0.00 0.00 -0.06 311.59

3 4 0.00 0.00 0.00 0.00 0.00 0.0012 0.00 0.00 0.00 0.00 0.00 0.00

11 4 118.08 -588.01 0.00 0.00 0.06 -323.2512 -118.08 -588.01 0.00 0.00 -0.06 323.25

12 4 118.08 -588.01 0.00 0.00 0.06 -323.2512 -118.08 -588.01 0.00 0.00 -0.06 323.25

23 1 6 -4.92 25.51 0.00 0.00 0.01 12.2714 4.92 25.51 0.00 0.00 -0.01 -12.27

2 6 -131.85 0.00 0.00 0.00 0.23 10.3614 131.85 0.00 0.00 0.00 -0.23 -10.36

3 6 0.00 0.00 0.00 0.00 0.00 0.0014 0.00 0.00 0.00 0.00 0.00 0.00

11 6 -136.76 25.51 0.00 0.00 0.23 22.6314 136.76 25.51 0.00 0.00 -0.23 -22.63

12 6 -136.76 25.51 0.00 0.00 0.23 22.6314 136.76 25.51 0.00 0.00 -0.23 -22.63

24 1 8 5.88 25.51 0.00 0.00 0.00 11.6616 -5.88 25.51 0.00 0.00 0.00 -11.66

2 8 112.16 562.50 0.00 0.00 0.06 311.5916 -112.16 562.50 0.00 0.00 -0.06 -311.59

3 8 0.00 0.00 0.00 0.00 0.00 0.0016 0.00 0.00 0.00 0.00 0.00 0.00

11 8 118.04 588.01 0.00 0.00 0.06 323.2516 -118.04 588.01 0.00 0.00 -0.06 -323.25

12 8 118.04 588.01 0.00 0.00 0.06 323.2516 -118.04 588.01 0.00 0.00 -0.06 -323.25

Example Problem 29



25 1 3 361.18 27.93 0.01 0.00 0.02 15.0117 -311.73 24.42 -0.01 0.00 -0.05 -10.66

2 3 5115.96 14.97 0.66 -0.11 0.91 53.7217 -5115.96 -14.97 -0.66 0.11 -2.53 -16.65

3 3 -14988.80 171.48 0.00 0.00 0.00 31.7517 14988.80 -171.48 0.00 0.00 0.00 392.88

11 3 -9511.66 214.39 0.66 -0.11 0.93 100.4817 9561.11 -162.03 -0.66 0.11 -2.58 365.57

12 3 20465.94 -128.57 0.66 -0.11 0.93 36.9817 -20416.50 180.93 -0.66 0.11 -2.58 -420.19

26 1 7 348.22 29.75 -0.01 0.00 -0.02 16.2219 -298.78 25.52 0.01 0.00 0.05 -10.84

2 7 4937.68 20.54 -0.62 0.10 -0.84 57.8319 -4937.68 -20.54 0.62 -0.10 2.43 -5.46

3 7 14900.46 -166.51 0.00 0.00 0.00 -32.8619 -14900.46 166.51 0.00 0.00 0.00 -391.85

11 7 20186.37 -116.23 -0.63 0.10 -0.86 41.2019 -20136.92 171.49 0.63 -0.10 2.47 -408.15

12 7 -9614.56 216.80 -0.63 0.10 -0.86 106.9119 9664.01 -161.54 0.63 -0.10 2.47 375.56

27 1 11 361.18 27.93 -0.01 0.00 -0.02 15.0120 -311.73 24.42 0.01 0.00 0.05 -10.66

2 11 5115.96 14.97 -0.66 0.11 -0.91 53.7220 -5115.96 -14.97 0.66 -0.11 2.53 -16.65

3 11 -14988.80 171.48 0.00 0.00 0.00 31.7520 14988.80 -171.48 0.00 0.00 0.00 392.88

11 11 -9511.66 214.39 -0.66 0.11 -0.93 100.4820 9561.11 -162.03 0.66 -0.11 2.58 365.57

12 11 20465.94 -128.57 -0.66 0.11 -0.93 36.9820 -20416.50 180.93 0.66 -0.11 2.58 -420.19

28 1 15 348.22 29.75 0.01 0.00 0.02 16.2222 -298.78 25.52 -0.01 0.00 -0.05 -10.84

2 15 4937.68 20.54 0.62 -0.10 0.84 57.8322 -4937.68 -20.54 -0.62 0.10 -2.43 -5.46

3 15 14900.46 -166.51 0.00 0.00 0.00 -32.8622 -14900.46 166.51 0.00 0.00 0.00 -391.85

11 15 20186.37 -116.23 0.63 -0.10 0.86 41.2022 -20136.92 171.49 -0.63 0.10 -2.47 -408.15

12 15 -9614.56 216.80 0.63 -0.10 0.86 106.9122 9664.01 -161.54 -0.63 0.10 -2.47 375.56

29 1 18 0.00 25.51 0.00 0.00 0.00 0.0621 0.00 25.51 0.00 0.00 0.00 -0.06

2 18 -0.03 1125.00 0.00 0.00 0.00 3.3421 0.03 1125.00 0.00 0.00 0.00 -3.34

3 18 0.00 0.00 0.00 0.00 0.00 0.0021 0.00 0.00 0.00 0.00 0.00 0.00

11 18 -0.03 1150.51 0.00 0.00 0.00 3.4021 0.03 1150.51 0.00 0.00 0.00 -3.40

12 18 -0.03 1150.51 0.00 0.00 0.00 3.4021 0.03 1150.51 0.00 0.00 0.00 -3.40


89. LOAD LIST 11 1290. PARAMETER91. CODE CANADA92. CHECK CODE ALL


Example Problem 29

312 STAAD.PRO CODE CHECKING - (CAN/CSA-S16.1-94)********************************************

ALL UNITS ARE - KNS MET (UNLESS OTHERWISE NOTED)


=======================================================================

1 ST W310X97 PASS CSA-13.8.1B 0.320 1273.70 C -0.64 112.55 3.50

2 ST W310X97 PASS CSA-13.8.1B 0.465 1271.75 C 1.22 168.65 1.80

3 ST W310X97 PASS CSA-13.9.A 0.462 1265.14 T 1.21 168.28 0.00

4 ST W310X97 PASS CSA-13.8.1B 0.324 1173.67 C -0.64 -113.77 3.50

5 ST W310X97 PASS CSA-13.8.1B 0.470 1171.72 C 1.22 -170.49 1.80

6 ST W310X97 PASS CSA-13.9.A 0.465 1159.44 T 1.21 -170.08 0.00

7 ST W310X97 PASS CSA-13.8.1B 0.320 1273.70 C 0.64 112.55 3.50

8 ST W310X97 PASS CSA-13.8.1B 0.465 1271.75 C -1.22 168.65 1.80

9 ST W310X97 PASS CSA-13.9.A 0.462 1265.14 T -1.21 168.28 0.00

10 ST W310X97 PASS CSA-13.8.1B 0.324 1173.67 C 0.64 -113.77 3.50

11 ST W310X97 PASS CSA-13.8.1B 0.470 1171.72 C -1.22 -170.49 1.80

12 ST W310X97 PASS CSA-13.9.A 0.465 1159.44 T -1.21 -170.08 0.00

13 ST W250X39 PASS CSA-13.9.A 0.860 12122.38 T -0.01 105.97 1.80

14 ST W250X39 PASS CSA-13.8.2+ 0.850 1217.67 C 0.00 110.09 0.00

15 ST W250X39 PASS CSA-13.8.2+ 0.850 1117.21 C 0.00 107.07 3.00

16 ST W250X39 PASS CSA-13.9.A 0.840 11123.87 T -0.01 103.07 0.00

17 ST W250X39 PASS CSA-13.9.A 0.860 12122.38 T 0.01 105.97 1.80

18 ST W250X39 PASS CSA-13.8.2+ 0.850 1217.67 C 0.00 110.09 0.00

19 ST W250X39 PASS CSA-13.8.2+ 0.850 1117.21 C 0.00 107.07 3.00

20 ST W250X39 PASS CSA-13.9.A 0.840 11123.87 T 0.01 103.07 0.00

21 ST C200X17 PASS CSA-13.9.A 0.009 111.34 T 0.00 -0.22 0.00

22 ST C200X17 PASS CSA-13.8.1C 0.143 111.16 C 0.00 -3.17 0.00

23 ST C200X17 PASS CSA-13.9.A 0.009 111.34 T 0.00 0.22 0.00

24 ST C200X17 PASS CSA-13.8.1C 0.143 111.16 C 0.00 3.17 0.00

25 ST L150X150X13 PASS CSA-13.8.1B 0.539 12200.22 C 0.03 4.12 2.48

26 ST L150X150X13 PASS CSA-13.8.1B 0.538 11197.48 C -0.02 4.00 2.55

27 ST L150X150X13 PASS CSA-13.8.1B 0.539 12200.22 C -0.03 4.12 2.48

28 ST L150X150X13 PASS CSA-13.8.1B 0.538 11197.48 C 0.02 4.00 2.55

29 ST C200X17 PASS CSA-13.8.1C 0.492 110.00 C 0.00 -11.19 1.50

Example Problem 29

313 93. FINISH


**** DATE= TIME= ****



Example Problem 29

314

NOTES

PART – II

VERIFICATION PROBLEMS

aaaaaaaaaaaaaaaaa

1

Verification Problem No. 1Verification Problem No. 1Verification Problem No. 1Verification Problem No. 1

OBJECTIVE: To find the support reactions due to a joint load in a plane truss.

REFERENCE: Timoshenko, S., “Strength of Materials,” Part 1, D. Van

Nostrand Co., Inc., 3rd edition, 1956, page 346, problem 3.

PROBLEM: Determine the horizontal reaction at support 4 of the

system.

COMPARISON:

Support Reaction, Kips

Solution R4 Theory 8.77 STAAD 8.77 Difference None

Part II – Verification Problems

2


1. STAAD TRUSS VERIFICATION PROBLEM NO. 12. *3. * REFERENCE `STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO4. * PAGE 346 PROBLEM NO. 3. THE ANSWER IS REACTION = 0.877P.5. * THEREFORE IF P=10, REACTION = 8.776. *7. UNITS INCH KIP8. JOINT COORD9. 1 0. 0. ; 2 150. 100. ; 3 150. 50. ; 4 300. 0.

10. MEMBER INCI11. 1 1 2 ; 2 1 3 ; 3 2 3 ; 4 2 4 ; 5 3 412. MEMB PROP13. 1 4 PRIS AX 5.0 ; 2 5 PRIS AX 3.0 ; 3 PRIS AX 214. CONSTANT15. E 30000. ALL16. POISSON STEEL ALL17. SUPPORT ; 1 4 PINNED18. LOADING 119. JOINT LOAD ; 2 FY -10.20. PERFORM ANALYSIS



21. PRINT REACTION

SUPPORT REACTIONS -UNIT KIP INCH STRUCTURE TYPE = TRUSS-----------------


�1 1 8.77 5.00 0.00 0.00 0.00 0.004 1 -8.77 5.00 0.00 0.00 0.00 0.00


22. FINISH

3


OBJECTIVE: To find the period of free vibration for a beam supported on two springs with a point mass.

REFERENCE: Timoshenko, S., Young, D., and Weaver, W., “Vibration

Problems in Engineering,” John Wiley & Sons, 4th edition, 1974. page 11, problem 1.1-3.

PROBLEM: A simple beam is supported by two spring as shown in

the figure. Neglecting the distributed mass of the beam, calculate the period of free vibration of the beam subjected to a load of W.

Y

L

W = 1000 lbs

1 1 2 K

A B 2

3

K X

GIVEN: EI = 30000.0 ksi A = 7.0 ft B = 3.0 ft.

K = 300.0 lb/in. COMPARISON:

Solution Period, sec Theory 0.533 STAAD 0.533 Difference None


4


1. STAAD PLANE VERIFICATION PROBLEM NO 22. *3. * REFERENCE 'VIBRATION PROBLEMS IN ENGINEERING' BY4. * TIMOSHENKO,YOUNG,WEAVER. (4TH EDITION, PAGE 11, PROB 1.1-3)5. * THE ANSWER IN THE BOOK IS T = 0.533 SEC., VIZ., F = 1.876 CPS6. *7. UNIT POUND FEET8. JOINT COORD ; 1 0. 0. ; 2 7. 0. ; 3 10. 0.9. MEMB INCI ; 1 1 2 2

10. UNIT INCH11. SUPPORT12. 1 3 FIXED BUT MZ KFY 300.13. MEMB PROP ; 1 2 PRIS AX 1. IZ 1.14. CONSTANT15. E 30E6 ALL16. POISSON STEEL ALL17. CUT OFF MODE SHAPE 118. LOADING 1 1000 LB LOAD AT JOINT 219. JOINT LOAD ; 2 FY -1000.20. MODAL CALCULATION21. PERFORM ANALYS




*** �EIGENSOLUTION�: SUBSPACE METHOD ***CALCULATED FREQUENCIES FOR LOAD CASE 1


1 1.876 0.53317 0.000E+00




1 0.00100.00 0.00 0.000 100.000 0.000

22. FINISH

5


TYPE: Deflection and moments for plate-bending finite element.

REFERENCE: Simple hand calculation by considering the entire structure as a cantilever beam.

PROBLEM: A simple cantilever plate is divided into 12 4-noded finite elements. A uniform pressure load is applied and the maximum deflection at the tip of the cantilever and the maximum bending at the support are calculated.

GIVEN: Plate thickness = 25mm,

Uniform pressure= 5N/sq.mm


6

HAND CALCULATION:

Max. deflection = WL3/8EI, where WL3=(5x300x100) x (300)3 = 405x1010 8EI=8x(210x103N/sq.mm)x(100x253/12) = 21875x107 Deflection = 18.51mm Max. moment = WL/2 = (5x300x100)x300/2 = 22.5x106N.mm = 22.5KN.m

SOLUTION COMPARISON:

Max. Defl. Max Moment Hand calculation 18.51 mm 22.50 kNm STAAD 18.20 mm 22.50 kNm

7


1. STAAD SPACE FINITE ELEMENT VERIFICATION2. UNIT MM KN3. JOINT COORDINATES4. 1 0 0 0 7 300 0 05. REPEAT 2 0 50 06. ELEMENT INCIDENCE7. 1 1 2 9 8 TO 68. REPEAT 1 6 79. ELEMENT PROP10. 1 TO 12 THICK 25.011. CONSTANT12. E 210.0 ALL13. POISSON STEEL ALL14. SUPPORT15. 1 8 15 FIXED16. UNIT NEWTON17. LOAD 1 5N/SQ.MM. UNIFORM LOAD18. ELEMENT LOAD19. 1 TO 12 PRESSURE 5.020. PERFORM ANALYSIS



21. PRINT DISPLACEMENT LIST 14



14 1 0.0000 0.0000 1.8159 0.0000 -0.0813 0.0000


22. UNIT KN METER23. PRINT REACTION



1 1 0.00 0.00 -18.91 -1.54 5.47 0.008 1 0.00 0.00 -112.19 0.00 11.56 0.0015 1 0.00 0.00 -18.91 1.54 5.47 0.00


24. FINISH


8

NOTES

9


OBJECTIVE: To find the support reactions due to a load at the free end of a cantilever plane bent with an intermediate support.


Nostrand Co., Inc., 3rd edition, 1956, page 346, problem 2.

PROBLEM: Determine the reaction of the system as shown in the

figure.

COMPARISON:

Reaction, Kip

Solution RX Theory 1.5 STAAD 1.5 Difference None


10


1. STAAD PLANE VERIFICATION PROBLEM NO. 42. *3. * REFERENCE 'STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO4. * PAGE 346 PROBLEM NO. 2. THE ANSWER IN THE BOOK AFTER5. * RECALCULATION = 1.56. *7. UNIT INCH KIP8. JOINT COORD9. 1 0. 0. ; 2 0. 10. ; 3 0. 20. ; 4 10. 20.

10. MEMB INCI11. 1 1 2 312. MEMB PROP ; 1 2 3 PRIS AX 10. IZ 100.13. CONSTANT14. E 3000. ALL15. POISSON CONCRETE ALL16. SUPPORT17. 1 FIXED ; 2 FIXED BUT FY MZ18. LOADING 119. JOINT LOAD ; 4 FY -1.20. PERFORM ANALYS



21. PRINT REACTION

SUPPORT REACTIONS -UNIT KIP INCH STRUCTURE TYPE = PLANE-----------------


�1 1 1.50 1.00 0.00 0.00 0.00 -5.002 1 -1.50 0.00 0.00 0.00 0.00 0.00


22. FINI

11


OBJECTIVE: To find deflections and stress at the center of a locomotive axle.

REFERENCE: Timoshenko, S.,“Strength of Materials,” Part- 1, D. Van

Nostrand Co., 3rd edition, 1956. page 97, problems 1, 2. PROBLEM: Determine the maximum stress in a locomotive axle (as

shown in the figure) as well as the deflection at the middle of the axle.

59 in13.5 in 13.5 in

X

PP

Y

GIVEN: Diameter = 10 in., P = 26000 lb, E = 30E6 psi COMPARISON:

Stress (σ), psi, and Deflection (δ), in

Solution σ δ Theory 3575.* 0.01040 STAAD 3575. 0.01037 Difference None None

* The value is recalculated.


12


1. STAAD PLANE VERIFICATION PROBLEM NO. 52. *3. * REFERENCE 'STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO4. * PAGE 97 PROBLEM NO. 1 AND 2. ANSWERS ARE 3580 FOR MAX. STRESS5. * AND 0.104 INCH FOR MAX. DEFLECTION.6. *7. UNIT INCH POUND8. JOINT COORD9. 1 0. 0. ; 2 13.5 0. ; 3 43. 0. ; 4 72.5 0. ; 5 86. 0.

10. MEMB INCI ; 1 1 2 411. MEMB PROP AMERICAN ; 1 TO 4 TABLE ST PIPE OD 10. ID 0.12. CONSTANT13. E 30E6 ALL14. POISSON STEEL ALL15. SUPPORT ; 2 4 PINNED16. LOADING 117. JOINT LOAD ; 1 5 FY -26000.18. PERFORM ANALYSIS



19. PRINT MEMBER STRESSES

MEMBER STRESSES---------------ALL UNITS ARE POUN/SQ INCH

MEMB LD SECT AXIAL BEND-Y BEND-Z COMBINED SHEAR-Y SHEAR-Z

1 1 .0 0.0 0.0 0.0 0.0 441.4 0.01.00 0.0 0.0 3575.3 3575.3 441.4 0.0

2 1 .0 0.0 0.0 3575.3 3575.3 0.0 0.01.00 0.0 0.0 3575.3 3575.3 0.0 0.0

3 1 .0 0.0 0.0 3575.3 3575.3 0.0 0.01.00 0.0 0.0 3575.3 3575.3 0.0 0.0

4 1 .0 0.0 0.0 3575.3 3575.3 441.4 0.01.00 0.0 0.0 0.0 0.0 441.4 0.0


20. PRINT DISPLACEMENTS

JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = PLANE------------------


1 1 0.00000 -0.01138 0.00000 0.00000 0.00000 0.000862 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.000703 1 0.00000 0.01037 0.00000 0.00000 0.00000 0.000004 1 0.00000 0.00000 0.00000 0.00000 0.00000 -0.000705 1 0.00000 -0.01138 0.00000 0.00000 0.00000 -0.00086

************** END OF LATEST ANALYSIS RESULT **************21. FINISH

13


OBJECTIVE: To find the maximum moment due to a uniform load on the horizontal member in a 1x1 bay plane frame.

REFERENCE: McCormack, J. C., “Structural Analysis,” Intext

Educational Publishers, 3rd edition, 1975, page 383, example 22 - 5.

PROBLEM: Determine the maximum moment in the frame.

GIVEN: E and I same for all members. COMPARISON:

Moment, Kip-ft

Solution MMax Theory 44.40 STAAD 44.44 Difference Small


14


1. STAAD PLANE VERIFICATION PROBLEM NO. 62. *3. * REFERENCE 'STRUCTURAL ANALYSIS' BY JACK C. MCCORMACK,4. * PAGE 383 EXAMPLE 22-5, PLANE FRAME WITH NO SIDESWAY5. * ANSWER - MAX BENDING = 44.4 FT-KIP6. *7. UNIT FT KIP8. JOINT COORD9. 1 0. 0. ; 2 0. 20. ; 3 20. 20. ; 4 20. 0.

10. MEMB INCI ; 1 1 2 311. MEMB PROP ; 1 2 3 PRIS AX 1. IZ 0.0512. CONSTANT13. E 4132E3 ALL14. POISSON STEEL ALL15. SUPPORT ; 1 4 FIXED16. LOADING 1 ; MEMB LOAD ; 2 UNI Y -2.017. PERFORM ANAL



18. PRINT FORCES

MEMBER END FORCES STRUCTURE TYPE = PLANE-----------------ALL UNITS ARE -- KIP FEET


1 1 1 20.00 -3.33 0.00 0.00 0.00 -22.212 -20.00 3.33 0.00 0.00 0.00 -44.44

2 1 2 3.33 20.00 0.00 0.00 0.00 44.443 -3.33 20.00 0.00 0.00 0.00 -44.44

3 1 3 20.00 3.33 0.00 0.00 0.00 44.444 -20.00 -3.33 0.00 0.00 0.00 22.21


19. FINISH

15


OBJECTIVE: To find the joint deflection due to joint loads in a plane truss.



PROBLEM: Determine the vertical deflection at point 5 of the plane

truss structure shown in the figure.

GIVEN: AX 1-4 = 1 in2, AX 5-6 = 2 in2, AX 7-8 =1.5 in2, AX 9-11 = 3 in2, AX 12-13 = 4 in2, E = 30E3 ksi COMPARISON:

Deflection, in.

Solution δ5 Theory 2.63 STAAD 2.63 Difference None


16


1. STAAD TRUSS VERIFICATION PROBLEM NO. 72. *3. * REFERENCE 'STRUCTURAL ANALYSIS' BY JACK MCCORMACK, PAGE4. * 271 EXAMPLE 18-2. ANSWER - Y-DISP AT JOINT 5 = 2.63 INCH5. *6. UNIT FT KIP7. JOINT COORD8. 1 0 0 0 5 60 0 09. 6 15. 7.5 ; 7 30. 15. ; 8 45. 7.5

10. MEMB INCI11. 1 2 6 ; 2 3 4 ; 3 4 8 ; 4 4 5 ; 5 1 212. 6 2 3 ; 7 3 6 ; 8 3 8 ; 9 3 713. 10 1 6 ; 11 5 8 ; 12 6 7 1314. UNIT INCH15. MEMB PROP16. 1 TO 4 PRI AX 1.017. 5 6 PRIS AX 2.18. 7 8 PRI AX 1.519. 9 10 11 PRI AX 3.20. 12 13 PRI AX 4.21. CONSTANT22. E 30E3 ALL23. POISSON STEEL ALL24. SUPPORT25. 1 PINNED ; 3 FIXED BUT FX MZ26. LOAD 1 VERTICAL LOAD27. JOINT LOAD28. 2 4 5 FY -20.029. PERFORM ANALYSIS




JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = TRUSS------------------


1 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.000002 1 -0.12000 0.18000 0.00000 0.00000 0.00000 0.000003 1 -0.24000 0.00000 0.00000 0.00000 0.00000 0.000004 1 -0.48000 -0.89516 0.00000 0.00000 0.00000 0.000005 1 -0.72000 -2.63033 0.00000 0.00000 0.00000 0.000006 1 -0.00820 0.24000 0.00000 0.00000 0.00000 0.000007 1 0.29758 -0.12000 0.00000 0.00000 0.00000 0.000008 1 0.06578 -0.83516 0.00000 0.00000 0.00000 0.00000


31. FINISH

17


OBJECTIVE: To find the maximum moment due to a concentrated load on the horizontal member in a 1x1 bay plane frame.


Educational Publishers, 3rd edition, 1975, page 385, problem 22 - 6.

PROBLEM: Determine the maximum moment in the structure.

GIVEN: E and I same for all members COMPARISON:

Moment, Kip-ft

Solution MMax Theory 69.40 STAAD 69.44 Difference Small


18


1. STAAD PLANE VERIFICATION PROBLEM NO. 82. *3. * PLANE FRAME WITH SIDESWAY. REFERENCE 'STRUCTURAL ANALYSIS'4. * BY JACK MCCORMACK. PAGE 385 PROB 22-6.5. * ANSWER - MAX BENDING IN MEMB 1 = 69.4 KIP-FT6. *7. UNIT FT KIP8. JOINT COORD9. 1 0. 10. ; 2 0 30 ; 3 30 30 ; 4 30 0

10. MEMB INCI11. 1 1 2 312. MEMB PROP AMERICAN13. 1 2 3 TAB ST W12X2614. CONSTANT15. E 4176E316. POISSON STEEL ALL17. SUPPORT ; 1 4 FIXED18. LOAD 1 VERTICAL LOAD19. MEMBER LOAD20. 2 CON Y -30. 10.21. PERFORM ANALYSIS



22. PRINT FORCES



1 1 1 20.09 -3.74 0.00 0.00 0.00 -5.342 -20.09 3.74 0.00 0.00 0.00 -69.44

2 1 2 3.74 20.09 0.00 0.00 0.00 69.443 -3.74 9.91 0.00 0.00 0.00 -66.66

3 1 3 9.91 3.74 0.00 0.00 0.00 66.664 -9.91 -3.74 0.00 0.00 0.00 45.51


23. FINISH

19

Verification Problem Verification Problem Verification Problem Verification Problem No. 9 No. 9 No. 9 No. 9

OBJECTIVE: To find the maximum moment due to lateral joint loads in a 1x2 bay plane frame.



PROBLEM: Determine the maximum moment in the frame.

GIVEN: E and I same for all members. COMPARISON:

Moment, Kip-ft

Solution MMax Theory 176.40 STAAD 178.01 Difference 0.91%


20


1. STAAD PLANE VERIFICATION PROB NO. 92. *3. * MULTIPLE LEVEL PLANE FRAME WITH HORIZONTAL LOAD.4. * REFERENCE 'STRUCTURAL ANALYSIS' BY JACK MCCORMACK,5. * PAGE 388, PROB 22-7. ANSWER - MAX MOM IN MEMB 1 = 176.4 K-F6. *7. UNIT FT KIP8. JOINT COORD9. 1 0 0 0 5 0 40 0 2 ; 2 20 0 0 6 20 40 0 2

10. MEMB INCI11. 1 1 3 2 ; 3 3 5 4 ; 5 3 4 ; 6 5 612. MEMB PROP13. 1 TO 6 PRI AX .2 IZ .114. CONSTANT15. E 4176E316. POISSON STEEL ALL17. SUPPORT ; 1 2 FIXED18. LOAD 1 HORIZONTAL LOAD19. JOINT LOAD20. 3 FX 20 ; 5 FX 1021. PERFORM ANALYS



22. PRINT FORCES



1 1 1 -22.26 15.06 0.00 0.00 0.00 178.013 22.26 -15.06 0.00 0.00 0.00 123.16

2 1 2 22.26 14.94 0.00 0.00 0.00 176.734 -22.26 -14.94 0.00 0.00 0.00 122.10

3 1 3 -6.51 4.97 0.00 0.00 0.00 34.495 6.51 -4.97 0.00 0.00 0.00 64.93

4 1 4 6.51 5.03 0.00 0.00 0.00 35.346 -6.51 -5.03 0.00 0.00 0.00 65.24

5 1 3 9.91 -15.75 0.00 0.00 0.00 -157.654 -9.91 15.75 0.00 0.00 0.00 -157.44

6 1 5 5.03 -6.51 0.00 0.00 0.00 -64.936 -5.03 6.51 0.00 0.00 0.00 -65.24


23. FINISH

21

Verification ProbleVerification ProbleVerification ProbleVerification Problem No. 10m No. 10m No. 10m No. 10

OBJECTIVE: To find the maximum axial force and moment due to load and moment applied at a joint in a space frame.

REFERENCE: Weaver Jr., W., “Computer Programs for Structural

Analysis,” page 146, problem 8.

PROBLEM: Determine the maximum axial force and moment in the space structure.

GIVEN: E = 30E3 ksi, AX = 11 in2 IX = 83 in4 IY = 56 in4 IZ = 56 in4 COMPARISON:

Solution FMax (kips) MY,Max (kip-in) MZ,Max (kip-in) Reference 1.47 84.04 95.319 STAAD 1.47 84.04 96.120 Difference None None Small


22


1. STAAD SPACE VERIFICATION PROB NO. 102. *3. * REFERENCE 'COMPUTER PROGRAMS FOR STRUCTURAL ANALYSIS'4. * BY WILLIAM WEAVER JR. PAGE 146 STRUCTURE NO. 8.5. * ANSWER - MAX AXIAL FORCE= 1.47 (MEMB 3)6. * MAX BEND-Y= 84.04, BEND-Z= 95.319 (BOTH MEMB 3)7. *8. UNIT INCH KIP9. JOINT COORD

10. 1 0 120 0 ; 2 240 120 011. 3 0 0 0 ; 4 360 0 12012. MEMB INCI13. 1 1 2 ; 2 3 1 ; 3 2 414. MEMB PROP15. 1 2 3 PRIS AX 11. IX 83. IY 56. IZ 5616. CONSTANT ; E 30000. ALL17. POISS .25 ALL18. SUPPORT19. 3 4 FIXED20. LOAD 1 JOINT LOAD21. JOINT LOAD22. 1 FX 2. ; 2 FY -1. ; 2 MZ -120.23. PERFORM ANAL



24. PRINT ANALYSIS RESULT



1 1 0.22267 0.00016 -0.17182 -0.00255 0.00217 -0.002132 1 0.22202 -0.48119 -0.70161 -0.00802 0.00101 -0.004353 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.000004 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000

SUPPORT REACTIONS -UNIT KIP INCH STRUCTURE TYPE = SPACE-----------------


3 1 -1.10 -0.43 0.22 48.78 -17.97 96.124 1 -0.90 1.43 -0.22 123.08 47.25 -11.72

23

MEMBER END FORCES STRUCTURE TYPE = SPACE-----------------ALL UNITS ARE -- KIP INCH


1 1 1 0.90 -0.43 0.22 22.71 -17.97 -36.372 -0.90 0.43 -0.22 -22.71 -34.18 -67.36

2 1 3 -0.43 1.10 0.22 -17.97 -48.78 96.121 0.43 -1.10 -0.22 17.97 22.71 36.37

3 1 2 1.47 -0.71 -0.48 -37.02 15.69 -53.284 -1.47 0.71 0.48 37.02 84.04 -95.32


25. FINISH


24

NOTES

25


OBJECTIVE: A rigid bar is suspended by two copper wires and one steel wire. Find the stresses in the wires due to a rise in temperature.


Nostrand Co., 3rd edition, 1956, page 30, problem 9. PROBLEM: Assuming the horizontal member to be very rigid,

determine the stresses in the copper and steel wires if the temperature rise is 10º F.

GIVEN: Esteel = 30E6 psi, Ecopper = 16E6 psi αsteel = 70E-7 in/in/°F, αcopper = 92E-7 in/in/°F AX = 0.1 in2 MODELLING HINT: Assume a large moment of inertia for the

horizontal rigid member and distribute of the concentrated load as uniform.

COMPARISON:

Stress (σ), psi

Solution σSteel σCopper Theory 19695 10152 STAAD 19698 10151 Difference Small Small


26


1. STAAD PLANE VERIFICATION PROB NO 112. *3. * THIS EXAMPLE IS TAKEN FROM 'STRENGTH OF MATERIALS' BY4. * TIMOSHENKO (PART 1), PAGE 30, PROB 9.5. * THE ANSWERS ARE 19695 PSI AND 10152 PSI.6. *7. UNIT INCH POUND8. JOINT COORD9. 1 0. 20. ; 2 5. 20. ; 3 10. 20.

10. 4 0. 0. ; 5 5. 0. ; 6 10. 0.11. MEMB INCI12. 1 1 4 3 ; 4 4 5 513. MEMB PROP14. 1 2 3 PRI AX 0.1 ; 4 5 PRI AX 1. IZ 100.15. CONSTANT ; E 30E6 MEMB 2 4 516. E 16E6 MEMB 1 317. POISSON 0.15 ALL18. ALPHA 92E-7 MEMB 1 3 ; ALPHA 70E-7 MEMB 219. MEMB TRUSS ; 1 2 320. SUPPORT ; 1 2 3 PINNED21. LOADING 1 VERT LOAD + TEMP LOAD22. MEMB LOAD ;4 5 UNI Y -400.23. TEMP LOAD ; 1 2 3 TEMP 10.24. PERFORM ANALYSIS



ZERO STIFFNESS IN DIRECTION 6 AT JOINT 1 EQN.NO. 1



***WARNING - INSTABILITY AT JOINT 6 DIRECTION = FXPROBABLE CAUSE SINGULAR-ADDING WEAK SPRINGK-MATRIX DIAG= 6.0000004E+03 L-MATRIX DIAG= 0.0000000E+00 EQN NO 10***NOTE - VERY WEAK SPRING ADDED FOR STABILITY

**NOTE** STAAD DETECTS INSTABILITIES AS EXCESSIVE LOSS OF SIGNIFICANT DIGITSDURING DECOMPOSITION. WHEN A DECOMPOSED DIAGONAL IS LESS THAN THEBUILT-IN REDUCTION FACTOR TIMES THE ORIGINAL STIFFNESS MATRIX DIAGONAL,STAAD PRINTS A SINGULARITY NOTICE. THE BUILT-IN REDUCTION FACTORIS 1.000E-09

THE ABOVE CONDITIONS COULD ALSO BE CAUSED BY VERY STIFF OR VERY WEAKELEMENTS AS WELL AS TRUE SINGULARITIES.

27

25. PRINT STRESSES



1 1 .0 10150.8 T 0.0 0.0 10150.8 0.0 0.01.00 10150.8 T 0.0 0.0 10150.8 0.0 0.0

2 1 .0 19698.3 T 0.0 0.0 19698.3 0.0 0.01.00 19698.3 T 0.0 0.0 19698.3 0.0 0.0

3 1 .0 10150.8 T 0.0 0.0 10150.8 0.0 0.01.00 10150.8 T 0.0 0.0 10150.8 0.0 0.0

4 1 .0 0.0 0.0 0.0 0.0 1522.6 0.01.00 0.0 0.0 3.8 3.8 1477.4 0.0

5 1 .0 0.0 0.0 3.8 3.8 1477.4 0.01.00 0.0 0.0 0.0 0.0 1522.6 0.0


26. FINISH


28

NOTES

29


OBJECTIVE: To find the joint deflection and member stress due to a joint load in a plane truss.


Nostrand Co., Inc., 3rd edition, 1956, page 10, problem 2. PROBLEM: Determine the vertical deflection at point A and the

member stresses.

L=180”A=0.5 in 2

L=180”A=0.5 in2

P=5000 lb

30° 30 °

GIVEN: AX = 0.5 in2, E = 30E6 psi COMPARISON:

Stress (σ), psi and Deflection (δ), in.

Solution σA δA Theory 10000. 0.12 STAAD 10000. 0.12 Difference None None


30


1. STAAD TRUSS VERIFICATION PROBLEM NO 122. *3. * THIS EXAMPLE IS TAKEN FROM 'STRENGTH OF MATERIALS'4. * (PART 1) BY TIMOSHENKO, PAGE 10 PROB 2.5. * THE ANSWER IN THE BOOK , DEFLECTION = 0.12 INCH6. * AND STRESS =10000 PSI7. *8. UNIT INCH POUND9. JOINT COORD

10. 1 0. 0. ; 2 155.88457 -90. ; 3 311.76914 0.11. MEMB INCI ; 1 1 2 212. MEMB PROP13. 1 2 PRIS AX 0.514. CONSTANT15. E 30E616. POISSON 0.15 ALL17. SUPPORT ; 1 3 PINNED18. LOAD 1 VERT LOAD19. JOINT LOAD ; 2 FY -5000.20. PERFORM ANALYSIS




JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = TRUSS------------------


1 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.000002 1 0.00000 -0.12000 0.00000 0.00000 0.00000 0.000003 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000


22. PRINT STRESSES



1 1 .0 10000.0 T 0.0 0.0 10000.0 0.0 0.01.00 10000.0 T 0.0 0.0 10000.0 0.0 0.0

2 1 .0 10000.0 T 0.0 0.0 10000.0 0.0 0.01.00 10000.0 T 0.0 0.0 10000.0 0.0 0.0


23. FINISH


32

british examples 2004

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