STAAD+PRO.nrc Verfication Manual 2005

STAAD.Pro 2005

QUALITY ASSURANCE PROGRAM

TEST VERIFICATION MANUAL

Document Revision No. 2005 March, 2006

Approved by: _______________________ ______________ Quality Assurance Manager Date

A Bentley Solutions Center www.reiworld.comwww.bentley.com/staad

http://www.reiworld.com/

STAAD.Pro 2005 is a suite of proprietary computer programs of Research Engineers, a Bentley Solutions Center. Although every effort has been made to ensure the correctness of these programs, REI will not accept responsibility for any mistake, error or misrepresentation in or as a result of the usage of these programs.

RELEASE 2005

© 2006 Bentley Systems, Incorporated. All Rights Reserved.

Published March, 2006

About STAAD.Pro

STAAD.Pro is a general purpose structural analysis and design program with applications primarily in the building industry - commercial buildings, bridges and highway structures, industrial structures, chemical plant structures, dams, retaining walls, turbine foundations, culverts and other embedded structures, etc. The program hence consists of the following facilities to enable this task. 1. Graphical model generation utilities as well as text editor based commands for

creating the mathematical model. Beam and column members are represented using lines. Walls, slabs and panel type entities are represented using triangular and quadrilateral finite elements. Solid blocks are represented using brick elements. These utilities allow the user to create the geometry, assign properties, orient cross sections as desired, assign materials like steel, concrete, timber, aluminum, specify supports, apply loads explicitly as well as have the program generate loads, design parameters etc.

2. Analysis engines for performing linear elastic and pdelta analysis, finite element analysis, frequency extraction, and dynamic response (spectrum, time history, steady state, etc.).

3. Design engines for code checking and optimization of steel, aluminum and timber members. Reinforcement calculations for concrete beams, columns, slabs and shear walls. Design of shear and moment connections for steel members.

4. Result viewing, result verification and report generation tools for examining displacement diagrams, bending moment and shear force diagrams, beam, plate and solid stress contours, etc.

5. Peripheral tools for activities like import and export of data from and to other widely accepted formats, links with other popular softwares for niche areas like reinforced and prestressed concrete slab design, footing design, steel connection design, etc.

6. A library of exposed functions called OpenSTAAD which allows users to access STAAD.Pro’s internal functions and routines as well as its graphical commands to tap into STAAD’s database and link input and output data to third-party software written using languages like C, C++, VB, VBA, FORTRAN, Java, Delphi, etc. Thus, OpenSTAAD allows users to link in-house or third-party applications with STAAD.Pro.

About the STAAD.Pro Documentation The documentation for STAAD.Pro consists of a set of manuals as described below. These manuals are normally provided only in the electronic format, with perhaps some exceptions such as the Getting Started Manual which may be supplied as a printed book to first time and new-version buyers. All the manuals can be accessed from the Help facilities of STAAD.Pro. Users who wish to obtain a printed copy of the books may contact Research Engineers. REI also supplies the manuals in the PDF format at no cost for those who wish to print them on their own. See the back cover of this book for addresses and phone numbers. 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 Manual 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 document 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. Technical Reference Manual 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. The documentation for the STAAD.Pro Extension component(s) is available separately.

Table of Contents

Part I - STAAD VERIFICATION PROBLEMS

1 A plane truss: support reactions due to a joint load 1

2 A beam supported on two springs with a point mass: period of free vibration

5

3 A cantilever beam: use of plate/shell elements. Deflection and support reactions due to a pressure load

9

4 A cantilever plane bent with an intermediate support: support reactions due to a load at the free end

13

5 A locomotive axle: deflection and stress at center 17

6 A 1x1 bay plane frame: maximum moment due to a uniform load on the horizontal member

21

7 A plane truss: joint deflection due to joint loads 25

8 A 1x1 bay plane frame: maximum moment due to a concentrated load on the horizontal member

29

9 A 1x2 bay plane frame: maximum moment due to lateral joint loads 33

10 Space frame: Forces and moments in members due to applied forces and a moment

37

11 A rigid bar suspended by two copper and one steel wires: stresses in wires due to a rise in temperature

41

12 A plane truss: joint deflection and member stress due to a joint load 45

13 Steel Design. Determine the allowable stresses (per 1989 AISC code) for the members. Also, perform a code check for these members based on the results of the analysis.

49

14 Concrete design as per ACI code. A plane frame is created with such loading as to create 138 Kip-Ft moment on beam and 574 Kip of axial load coupled with above moment on column.

61

Part II - STARDYNE VERIFICATION PROBLEMS

1 Uniform Beam With One Redundant Support. A uniform beam is clamped at one end and is simply supported at the other end. This arrangement of supports offers the minimum analytic difficulty for an indeterminate structure. The single load case is a uniformly distributed vertical force of 100 pounds per inch. Find the moment at the clamped end, the reaction at the simple support, and the mid-span deflection.

1

2 Indeterminate Frame. A statically indeterminate frame, consisting of three members, has a concentrated force applied at mid-span. The two vertical members are hinged at the ground support points. The reaction forces and, the bending moments at the ends of the horizontal beam will be determined.

5

3 Uniform Pressure On A Circular Plate. Deflections and stresses for a circular plate structure are determined, for a uniform pressure. The plate is simply supported around its entire periphery.

11

4 Uniform Pressure On A Rectangular Plate. A rectangular plate, 10in X 16in, , of constant thickness of 0.2in., is simply supported along all four edges. The applied loading is uniform pressure over the entire surface. The analysis results are lateral deflections and bending stresses.

17

5 Static Analysis Of Thermal Loading. The internal loads caused by a temperature change are analyzed in this example. The subject is a pipe line with two right angles. Both ends are completely restrained. Internal forces and moments will be computed for a temperature increase of 430 degrees F.

31

6 Imposed Deflections On Beam. A beam which is supported at three points (nodes 1, 2 and 3) has imposed translations specified at each of these points. Additionally, an imposed rotation is specified at node 1. There are no external loads applied to the structure. The static response of the structure due to these imposed deflections and rotation is calculated.

37

7 Natural Modes of a Simple Beam. The first five natural frequencies and the associated mode shapes are computed for the flexural motion of a simply supported beam. The HQR and LANCZOS solution methods are both tested.

41

8 Natural Frequencies Of A Circular Plate. A flat circular plate is simply

supported around the entire perimeter. The first six modes and their associated natural frequencies are to be computed using the HQR method offered by STARDYNE. This problem demonstrates that the natural frequencies of an axi-symmetric structure can be accurately computed utilizing a 180 degree model with the appropriate boundary conditions.

45

9 Natural Frequencies Of A Rectangular Plate. A flat rectangular plate is simply supported on all four sides. The first six modes and their associated natural frequencies are to be computed for this structure using the LANCZOS method offered by STARDYNE. This problem also demonstrates that the mesh refinement can be chosen to accurately calculate modes of interest based on the expected mode shapes.

51

10 Natural Modes Of A Framework. A three dimensional frame is analyzed for its natural frequencies and the associated mode shapes using the HQR method offered by STARDYNE.

57

11 Response Of A Simply Supported Beam To A Shock Spectrum. The supports of a simply supported beam are subjected to an acceleration time history. The maximum bending moment in the beam is computed for the first mode of the structure. This problem demonstrates the capabilities of STARDYNE to calculate the correct modal response of a structure utilizing response spectrum data.

63

12 Thermal loading on a Simply supported Rectangular Plate. A rectangular plate is simply supported on all four sides. The transverse and longitudinal bending moments as well as the deflections at several points on the plate are computed.

69

PART I

STAAD VERIFICATION PROBLEMS

STAAD Verification Problems 1

STAAD Verification 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


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42: 2 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD TRUSS :A PLANE TRUSS 2. * FILE: VER01.STD 3. * 4. * REFERENCE `STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO 5. * PAGE 346 PROBLEM NO. 3. THE ANSWER IS REACTION = 0.877P. 6. * THEREFORE IF P=10, REACTION = 8.77 7. * 8. UNITS INCH KIP 9. JOINT COORD 10. 1 0. 0. ; 2 150. 100. ; 3 150. 50. ; 4 300. 0. 11. MEMBER INCI 12. 1 1 2 ; 2 1 3 ; 3 2 3 ; 4 2 4 ; 5 3 4 13. MEMB PROP 14. 1 4 PRIS AX 5.0 ; 2 5 PRIS AX 3.0 ; 3 PRIS AX 2 15. CONSTANT ; E 30000. ALL 16. SUPPORT ; 1 4 PINNED 17. LOADING 1 18. JOINT LOAD ; 2 FY -10. 19. PERFORM ANALYSIS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 4/ 5/ 2 ORIGINAL/FINAL BAND-WIDTH= 2/ 2/ 4 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 4 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3653.1 MB 20. PRINT REACTION SUPPORT REACTIONS -UNIT KIP INCH STRUCTURE TYPE = TRUSS ----------------- JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z 1 1 8.77 5.00 0.00 0.00 0.00 0.00 4 1 -8.77 5.00 0.00 0.00 0.00 0.00 ************** END OF LATEST ANALYSIS RESULT **************

STAAD Verification Problems 3 21. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:42: 5 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ------------------------------------------------------------------------------------ 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb ------------------------------------------------------------------------------------


NOTES



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


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42: 8 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD PLANE :NATURAL FREQUENCY OF A BEAM SUPPORTED ON TWO SPRINGS 2. * FILE: VER02.STD 3. * 4. * REFERENCE: TIMOSHENKO,YOUNG,WEAVER, "VIBRATION PROBLEMS IN 5. * ENGINEERING", 4TH EDITION, JOHN WILEY & SONS, 6. * NEW YORK, 1974, PAGE 11, PROB 1.1-3 7. * THE ANSWER IN THE BOOK IS T = 0.533 SEC., VIZ., F = 1.876 CPS 8. * 9. UNIT POUND FEET 10. JOINT COORD ; 1 0. 0. ; 2 7. 0. ; 3 10. 0. 11. MEMB INCI ; 1 1 2 2 12. UNIT INCH 13. SUPPORT 14. 1 3 FIXED BUT MZ KFY 300. 15. MEMB PROP ; 1 2 PRIS AX 1. IZ 1. 16. CONSTANT ; E 30E6 ALL 17. CUT OFF MODE SHAPE 1 18. LOADING 1 1000 LB LOAD AT JOINT 2 19. JOINT LOAD ; 2 FY -1000. 20. MODAL CALCULATION 21. PERFORM ANALYS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 3/ 2/ 2 ORIGINAL/FINAL BAND-WIDTH= 1/ 1/ 5 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 7 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3653.1 MB NUMBER OF MODES REQUESTED = 1 NUMBER OF EXISTING MASSES IN THE MODEL = 1 NUMBER OF MODES THAT WILL BE USED = 1 *** EIGENSOLUTION: SUBSPACE METHOD *** CALCULATED FREQUENCIES FOR LOAD CASE 1 MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC) ACCURACY 1 1.876 0.53317 2.047E-16

STAAD Verification Problems 7 MASS PARTICIPATION FACTORS IN PERCENT -------------------------------------- MODE X Y Z SUMM-X SUMM-Y SUMM-Z 1 0.00100.00 0.00 0.000 100.000 0.000 22. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:42:11 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ------------------------------------------------------------------------------------ 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb ------------------------------------------------------------------------------------


NOTES



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


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


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42:14 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD SPACE FINITE ELEMENT VERIFICATION 2. * 3. * FILE : VER03.STD 4. * 5. * DEFLECTION OF A CANTILEVER PLATE UNDER UNIFORM PRESSURE. 6. * COMPARISON WITH ESTABLISHED FORMULA (WL^3/8EI) 7. * 8. UNIT KNS MMS 9. JOINT COORDINATES 10. 1 0 0 0 7 300 0 0 11. REPEAT 2 0 50 0 12. * 13. ELEMENT INCIDENCE 14. 1 1 2 9 8 TO 6 15. REPEAT 1 6 7 16. * 17. ELEMENT PROP 18. 1 TO 12 THICK 25.0 19. * 20. CONSTANT 21. E 210.0 ALL 22. POISSON STEEL ALL 23. * 24. SUPPORT 25. 1 8 15 FIXED 26. * 27. UNIT NEWTON 28. LOAD 1 5N/SQ.MM. UNIFORM LOAD 29. ELEMENT LOAD 30. 1 TO 12 PRESSURE 5.0 31. * 32. 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 = 21/ 12/ 3 ORIGINAL/FINAL BAND-WIDTH= 8/ 5/ 36 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 126 SIZE OF STIFFNESS MATRIX = 5 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.1/ 3653.1 MB 33. * 34. PRINT DISPLACEMENT LIST 14 JOINT DISPLACEMENT (CM RADIANS) STRUCTURE TYPE = SPACE ------------------ JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN 14 1 0.0000 0.0000 1.8159 0.0000 -0.0813 0.0000 ************** END OF LATEST ANALYSIS RESULT **************

STAAD Verification Problems 12 35. * 36. UNIT KN METER 37. PRINT REACTION SUPPORT REACTIONS -UNIT KN METE STRUCTURE TYPE = SPACE ----------------- JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z 1 1 0.00 0.00 -18.91 -1.54 5.47 0.00 8 1 0.00 0.00 -112.19 0.00 11.56 0.00 15 1 0.00 0.00 -18.91 1.54 5.47 0.00 ************** END OF LATEST ANALYSIS RESULT ************** 38. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:42:14 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ----------------------------------------------------------------------------------------- 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb -----------------------------------------------------------------------------------------



OBJECTIVE: To find 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


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42:17 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD PLANE :A CANTILEVER PLANE FRAME WITH AN INTERMEDIATE SUPPORT 2. * FILE VER04.STD 3. * 4. * REFERENCE: 'STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO 5. * PAGE 346 PROBLEM NO. 2. THE ANSWER IN THE BOOK AFTER 6. * RECALCULATION = 1.5 7. * 8. UNIT INCH KIP 9. JOINT COORD 10. 1 0. 0. ; 2 0. 10. ; 3 0. 20. ; 4 10. 20. 11. SUPPORT 12. 1 FIXED ; 2 FIXED BUT FY MZ 13. MEMB INCI 14. 1 1 2 3 15. MEMB PROP ; 1 2 3 PRIS AX 10. IZ 100. 16. CONSTANT ; E 3000. ALL 17. LOADING 1 18. JOINT LOAD ; 4 FY -1. 19. PERFORM ANALYSIS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 4/ 3/ 2 ORIGINAL/FINAL BAND-WIDTH= 1/ 1/ 6 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 8 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3653.0 MB 20. PRINT REACTION SUPPORT REACTIONS -UNIT KIP INCH STRUCTURE TYPE = PLANE ----------------- JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z 1 1 1.50 1.00 0.00 0.00 0.00 -5.00 2 1 -1.50 0.00 0.00 0.00 0.00 0.00 ************** END OF LATEST ANALYSIS RESULT **************

STAAD Verification Problems 15 21. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:42:20 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ------------------------------------------------------------------------------------ 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb ------------------------------------------------------------------------------------


NOTES



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.


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42:24 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD PLANE :A LOCOMOTIVE AXLE 2. * FILE: VER05.STD 3. * 4. * REFERENCE 'STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO 5. * PAGE 97 PROBLEM NO. 1 AND 2. 6. * ANSWERS ARE 3580 FOR MAX. STRESS AND 0.104 INCH FOR MAX. DEFLN. 7. * 8. UNIT INCH POUND 9. JOINT COORD 10. 1 0. 0. ; 2 13.5 0. ; 3 43. 0. ; 4 72.5 0. ; 5 86. 0. 11. SUPPORT ; 2 4 PINNED 12. MEMB INCI ; 1 1 2 4 13. MEMB PROP ; 1 TO 4 TABLE ST PIPE OD 10. ID 0. 14. CONSTANT ; E 30E6 ALL 15. LOADING 1 16. JOINT LOAD ; 1 5 FY -26000. 17. PERFORM ANALYSIS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 5/ 4/ 2 ORIGINAL/FINAL BAND-WIDTH= 1/ 1/ 4 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 11 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3653.0 MB 18. 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 0.0 1.00 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 0.0 1.00 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 0.0 1.00 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 0.0 1.00 0.0 0.0 0.0 0.0 441.4 0.0 ************** END OF LATEST ANALYSIS RESULT **************

STAAD Verification Problems 19 19. PRINT DISPLACEMENTS JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = PLANE ------------------ JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN 1 1 0.00000 -0.01138 0.00000 0.00000 0.00000 0.00086 2 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.00070 3 1 0.00000 0.01037 0.00000 0.00000 0.00000 0.00000 4 1 0.00000 0.00000 0.00000 0.00000 0.00000 -0.00070 5 1 0.00000 -0.01138 0.00000 0.00000 0.00000 -0.00086 ************** END OF LATEST ANALYSIS RESULT ************** 20. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:42:27 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ------------------------------------------------------------------------------------ 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb ------------------------------------------------------------------------------------


NOTES



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


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42:30 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD PLANE :A 1X1 BAY PLANE FRAME 2. * FILE VER06.STD 3. * 4. * REFERENCE: 'STRUCTURAL ANALYSIS' BY JACK C. MCCORMACK, 5. * PAGE 383 EXAMPLE 22-5, PLANE FRAME WITH NO SIDESWAY 6. * ANSWER - MAX BENDING = 44.4 FT-KIP 7. * 8. UNIT FT KIP 9. JOINT COORD 10. 1 0. 0. ; 2 0. 20. ; 3 20. 20. ; 4 20. 0. 11. MEMB INCI ; 1 1 2 3 12. MEMB PROP ; 1 2 3 PRIS AX 1. IZ 0.05 13. CONSTANT ; E 4132E3 ALL 14. SUPPORT ; 1 4 FIXED 15. LOADING 1 ; MEMB LOAD ; 2 UNI Y -2.0 16. PERFORM ANALYSIS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 4/ 3/ 2 ORIGINAL/FINAL BAND-WIDTH= 1/ 1/ 6 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 6 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3653.0 MB 17. PRINT FORCES MEMBER END FORCES STRUCTURE TYPE = PLANE ----------------- ALL UNITS ARE -- KIP FEET (LOCAL ) MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z 1 1 1 20.00 -3.33 0.00 0.00 0.00 -22.21 2 -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 44.44 3 -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 44.44 4 -20.00 -3.33 0.00 0.00 0.00 22.21 ************** END OF LATEST ANALYSIS RESULT **************


NOTES



OBJECTIVE: To find the joint deflection due to joint loads in a plane truss.



PROBLEM: Determine the vertical deflection at point 5 of 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


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42:36 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD TRUSS :A PLANE TRUSS 2. * FILE VER07.STD 3. * 4. * REFERENCE 'STRUCTURAL ANALYSIS' BY JACK MCCORMACK, PAGE 271, 5. * EXAMPLE 18-2. ANSWER - Y-DISP AT JOINT 5 = 2.63 INCH 6. * 8. UNIT FT KIP 9. JOINT COORD 10. 1 0 0 0 5 60 0 0 11. 6 15. 7.5 ; 7 30. 15. ; 8 45. 7.5 12. SUPPORT 13. 1 PINNED ; 3 FIXED BUT FX MZ 14. MEMB INCI 15. 1 2 6 ; 2 3 4 ; 3 4 8 ; 4 4 5 ; 5 1 2 16. 6 2 3 ; 7 3 6 ; 8 3 8 ; 9 3 7 17. 10 1 6 ; 11 5 8 ; 12 6 7 13 18. UNIT INCH 19. MEMB PROP 20. 1 TO 4 PRI AX 1.0 21. 5 6 PRIS AX 2. 22. 7 8 PRI AX 1.5 23. 9 10 11 PRI AX 3. 24. 12 13 PRI AX 4. 25. CONSTANT ; E 30E3 ALL 26. LOAD 1 VERTICAL LOAD 27. JOINT LOAD 28. 2 4 5 FY -20.0 29. PERFORM ANALYSIS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 8/ 13/ 2 ORIGINAL/FINAL BAND-WIDTH= 5/ 5/ 11 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 13 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3652.9 MB 30. PRINT DISPLACEMENTS LIST 5 JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = TRUSS ------------------ JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN 5 1 -0.72000 -2.63033 0.00000 0.00000 0.00000 0.00000 ************** END OF LATEST ANALYSIS RESULT **************


NOTES



OBJECTIVE: To find the maximum moment due to a concentrated load on the horizontal member in a 1x1 bay plane frame.


Educational Publishers, 3rd eEdition, 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


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42:42 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD PLANE :A 1X1 BAY PLANE FRAME 2. * FILE VER08.STD 3. * 4. * REFERENCE: 'STRUCTURAL ANALYSIS' BY JACK MCCORMACK, 5. * PAGE 385 PROB 22-6. 6. * ANSWER - MAX BENDING IN MEMB 1 = 69.4 KIP-FT 7. * 8. UNIT FT KIP 9. JOINT COORD 10. 1 0. 10. ; 2 0 30 ; 3 30 30 ; 4 30 0 11. SUPPORT ; 1 4 FIXED 12. MEMB INCI 13. 1 1 2 3 14. MEMB PROP ; 1 2 3 TAB ST W12X26 15. CONSTANT ; E 4176E3 16. LOAD 1 VERTICAL LOAD 17. MEMBER LOAD 18. 2 CON Y -30. 10. 19. PERFORM ANALYSIS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 4/ 3/ 2 ORIGINAL/FINAL BAND-WIDTH= 1/ 1/ 6 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 6 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3652.9 MB 20. PRINT FORCES MEMBER END FORCES STRUCTURE TYPE = PLANE ----------------- ALL UNITS ARE -- KIP FEET (LOCAL ) MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z 1 1 1 20.09 -3.74 0.00 0.00 0.00 -5.34 2 -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 69.44 3 -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 66.66 4 -9.91 -3.74 0.00 0.00 0.00 45.51 ************** END OF LATEST ANALYSIS RESULT **************


NOTES



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%


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42:49 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD PLANE :A 1X2 BAY PLANE FRAME 2. * FILE: VER09.STD 3. * 4. * REFERENCE: 'STRUCTURAL ANALYSIS' BY JACK MCCORMACK, 5. * PAGE 388, PROB 22-7. 6. * ANSWER - MAX MOM IN MEMB 1 = 176.4 K-F 7. * 8. UNIT FT KIP 9. JOINT COORD 10. 1 0 0 0 5 0 40 0 2 ; 2 20 0 0 6 20 40 0 2 11. SUPPORT ; 1 2 FIXED 12. MEMB INCI 13. 1 1 3 2 ; 3 3 5 4 ; 5 3 4 ; 6 5 6 14. MEMB PROP 15. 1 TO 6 PRI AX .2 IZ .1 16. CONSTANT ; E 4176E3 17. LOAD 1 HORIZONTAL LOAD 18. JOINT LOAD 19. 3 FX 20 ; 5 FX 10 20. PERFORM ANALYS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 6/ 6/ 2 ORIGINAL/FINAL BAND-WIDTH= 3/ 3/ 9 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 12 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3652.9 MB 21. PRINT FORCES MEMBER END FORCES STRUCTURE TYPE = PLANE ----------------- ALL UNITS ARE -- KIP FEET (LOCAL ) MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z 1 1 1 -22.26 15.06 0.00 0.00 0.00 178.01 3 22.26 -15.06 0.00 0.00 0.00 123.16 2 1 2 22.26 14.94 0.00 0.00 0.00 176.73 4 -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 34.49 5 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 35.34 6 -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 -157.65 4 -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 -64.93 6 -5.03 6.51 0.00 0.00 0.00 -65.24 ************** END OF LATEST ANALYSIS RESULT **************


NOTES



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

STAAD Verification Problems 38 **************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:42:55 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD SPACE :A SPACE FRAME 2. * FILE VER10.STD 3. * 4. * REFERENCE 'COMPUTER PROGRAMS FOR STRUCTURAL ANALYSIS' 5. * BY WILLIAM WEAVER JR. PAGE 146 STRUCTURE NO. 8. 6. * ANSWER - MAX AXIAL FORCE= 1.47 (MEMB 3) 7. * MAX BEND-Y= 84.04 (MEMB 3), BEND-Z= 95.32 (MEMB 3) 8. * 9. * STAAD VERIFICATION PROB NO. 10 10. * 11. UNIT INCH KIP 12. JOINT COORD 13. 1 0 120 0 ; 2 240 120 0 14. 3 0 0 0 ; 4 360 0 120 15. SUPPORT 16. 3 4 FIXED 17. MEMB INCI 18. 1 1 2 ; 2 3 1 ; 3 2 4 19. MEMB PROP 20. 1 2 3 PRIS AX 11. IX 83. IY 56. IZ 56 21. CONSTANT ; E 30000. ALL 22. POISS .25 ALL 23. LOAD 1 JOINT LOAD 24. JOINT LOAD 25. 1 FX 2. ; 2 FY -1. ; 2 MZ -120. 26. 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 = 4/ 3/ 2 ORIGINAL/FINAL BAND-WIDTH= 2/ 2/ 12 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 12 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3652.8 MB 27. PRINT ANALYSIS RESULT JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = SPACE ------------------ JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN 1 1 0.22267 0.00016 -0.17182 -0.00255 0.00217 -0.00213 2 1 0.22202 -0.48119 -0.70161 -0.00802 0.00101 -0.00435 3 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 4 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 SUPPORT REACTIONS -UNIT KIP INCH STRUCTURE TYPE = SPACE ----------------- JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z 3 1 -1.10 -0.43 0.22 48.78 -17.97 96.12 4 1 -0.90 1.43 -0.22 123.08 47.25 -11.72

STAAD Verification Problems 39 MEMBER END FORCES STRUCTURE TYPE = SPACE ----------------- ALL UNITS ARE -- KIP INCH (LOCAL ) MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z 1 1 1 0.90 -0.43 0.22 22.71 -17.97 -36.37 2 -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 96.12 1 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 -53.28 4 -1.47 0.71 0.48 37.02 84.04 -95.32 ************** END OF LATEST ANALYSIS RESULT ************** 28. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:42:58 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ------------------------------------------------------------------------------------ 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb ------------------------------------------------------------------------------------


NOTES



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


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:43: 1 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD PLANE :A RIGID BAR SUSPENDED BY TWO COPPER WIRES AND A STEEL WIRE 2. * FILE: VER11.STD 3. * 4. * REFERENCE: 'STRENGTH OF MATERIALS', TIMOSHENKO (PART 1), 5. * PAGE 30, PROB 9. 6. * THE ANSWERS ARE 19700 PSI AND 10200 PSI. 7. * 8. UNIT INCH POUND 9. JOINT COORD 10. 1 0. 20. ; 2 5. 20. ; 3 10. 20. 11. 4 0. 0. ; 5 5. 0. ; 6 10. 0. 12. MEMB INCI 13. 1 1 4 3 ; 4 4 5 5 14. MEMB PROP 15. 1 2 3 PRI AX 0.1 ; 4 5 PRI AX 1. IZ 100. 16. CONSTANT ; E 30E6 MEMB 2 4 5 17. E 16E6 MEMB 1 3 18. ALPHA 92E-7 MEMB 1 3 ; ALPHA 70E-7 MEMB 2 19. MEMB TRUSS ; 1 2 3 20. SUPPORT ; 1 2 3 PINNED 21. LOADING 1 VERT LOAD + TEMP LOAD 22. MEMB LOAD ;4 5 UNI Y -400. 23. TEMP LOAD ; 1 2 3 TEMP 10. 24. PERFORM ANALYSIS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 6/ 5/ 3 ORIGINAL/FINAL BAND-WIDTH= 3/ 3/ 10 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 12 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3652.8 MB ZERO STIFFNESS IN DIRECTION 6 AT JOINT 1 EQN.NO. 1 LOADS APPLIED OR DISTRIBUTED HERE FROM ELEMENTS WILL BE IGNORED. THIS MAY BE DUE TO ALL MEMBERS AT THIS JOINT BEING RELEASED OR EFFECTIVELY RELEASED IN THIS DIRECTION. ZERO STIFFNESS IN DIRECTION 6 AT JOINT 2 EQN.NO. 2 ZERO STIFFNESS IN DIRECTION 6 AT JOINT 3 EQN.NO. 3 ***WARNING - INSTABILITY AT JOINT 6 DIRECTION = FX PROBABLE CAUSE SINGULAR-ADDING WEAK SPRING K-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 DIGITS DURING DECOMPOSITION. WHEN A DECOMPOSED DIAGONAL IS LESS THAN THE BUILT-IN REDUCTION FACTOR TIMES THE ORIGINAL STIFFNESS MATRIX DIAGONAL, STAAD PRINTS A SINGULARITY NOTICE. THE BUILT-IN REDUCTION FACTOR IS 1.000E-09 THE ABOVE CONDITIONS COULD ALSO BE CAUSED BY VERY STIFF OR VERY WEAK ELEMENTS AS WELL AS TRUE SINGULARITIES.

STAAD Verification Problems 43 25. PRINT STRESSES LIST 1 2 3 MEMBER STRESSES --------------- ALL UNITS ARE POUN/SQ INCH MEMB LD SECT AXIAL BEND-Y BEND-Z COMBINED SHEAR-Y SHEAR-Z 1 1 .0 10150.8 T 0.0 0.0 10150.8 0.0 0.0 1.00 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 0.0 1.00 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 0.0 1.00 10150.8 T 0.0 0.0 10150.8 0.0 0.0 ************** END OF LATEST ANALYSIS RESULT ************** 26. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:43: 4 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ------------------------------------------------------------------------------------ 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb ------------------------------------------------------------------------------------


NOTES



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 in 2

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


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:43: 7 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD TRUSS :A PLANE TRUSS 2. * FILE VER12.STD 3. * 4. * REFERENCE: 'STRENGTH OF MATERIALS', (PART 1) BY TIMOSHENKO, 5. * PAGE 10, PROB 2. 6. * 7. * THE ANSWER IN THE BOOK , DEFLECTION = 0.12 INCH 8. * AND STRESS =10000 PSI 9. * 10. UNIT INCH POUND 11. JOINT COORD 12. 1 0. 0. ; 2 155.88457 -90. ; 3 311.76914 0. 13. SUPPORT ; 1 3 PINNED 14. MEMB INCI ; 1 1 2 2 15. MEMB PROP 16. 1 2 PRIS AX 0.5 17. CONSTANT ; E 30E6 18. LOAD 1 VERT LOAD 19. JOINT LOAD ; 2 FY -5000. 20. PERFORM ANALYSIS **WARNING** THE POISSON'S RATIO HAS NOT BEEN SPECIFIED FOR ONE OR MORE MEMBERS/ELEMENTS/SOLIDS. THE DEFAULT VALUE HAS BEEN SET FOR THE SAME. P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS/MEMBER+ELEMENTS/SUPPORTS = 3/ 2/ 2 ORIGINAL/FINAL BAND-WIDTH= 1/ 1/ 2 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 2 SIZE OF STIFFNESS MATRIX = 0 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3652.7 MB 21. PRINT DISPLACEMENTS LIST 2 JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = TRUSS ------------------ JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN 2 1 0.00000 -0.12000 0.00000 0.00000 0.00000 0.00000 ************** END OF LATEST ANALYSIS RESULT **************

STAAD Verification Problems 47 22. PRINT 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 10000.0 T 0.0 0.0 10000.0 0.0 0.0 1.00 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 0.0 1.00 10000.0 T 0.0 0.0 10000.0 0.0 0.0 ************** END OF LATEST ANALYSIS RESULT ************** 23. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:43:10 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ------------------------------------------------------------------------------------ 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb ------------------------------------------------------------------------------------


NOTES



TYPE: Steel Design.

REFERENCE: Attached step by step hand calculation as per 1989 AISC ASD code. Ninth Edition.

PROBLEM: Determine the allowable stresses (per 1989 AISC code) for the members of the structure as shown in figure. Also, perform a code check for these members based on the results of the analysis.

155 6

7 8

1 2

3 4

k

6 7

8 9

3

4

5

1

24'5'

11'10'

2k/ft (DL + LL)

Y

X

(WL)

16'

Members 1, 2 = W12X26, Members 3, 4 = W14X43 Members 5, 6, 7 = W16X36, Memb 8= L40404, Memb 9 = L50506

SOLUTION COMPARISON:

Governing ratios for the members Member 1 2 3 4 5 6 7 8 9 No.

Hand 1.1576 .914 1.117 0.936 .579 1.119 .668 1.024 .823 Calculation

STAAD. 1.157 .916 1.117 0.936 .582 1.119 .668 1.025 .823 PRO


VERIFICATION PROBLEM 13 HAND CALCULATION

Manual / Code refers to AISC Manual of Steel Construction, Allowable Stress Design, ninth edition. Steel Design - Member 1, Size W 12X26, L = 10 ft., a = 7.65 in2, Sz = 33.39 in3 From clause F1-2, page 5-45 of Manual, Lc = 6.85 ft. From observation Load case 1 will govern, Fx = 25.0 kip (compression), Mz = 56.5 k-ft Area of compressive flange = 6.49*0.38 = 2.466 sq.in. Allowable bending stress = Fb = 12. x 1000 x 1.0 = 20.1817 ksi [Clause F1-8 10 x 12 x (12.22/2.466) page 5-47 of Manual] (kl/r)y = 120/1.5038 = 79.8, so Fa = 15.38 ksi (Table C-36 page 3-16 of manual) fa = 25./7.65 = 3.268, fb = 56.5 x 12/33.39 = 20.31 ksi (kl/r)z = 120/5.1639 = 23.238, so F’ez = 276.54 ksi Try formula H1-1, page 5-54 of Manual 3.268 + 0.85 x 20.31 = 1.078 15.38 (1-3.268/276.54) x 20.1817 Try formula H1-2, page 5-54 of Manual. 3.268 + 20.31 = 1.1576 0.6 x 36 20.1817 Therefore formula H1-2 governs and ratio = 1.1576

STAAD Verification Problems 51 Member 2, Size W12X26, L = 5 ft., a = 7.65 in2, Sz = 33.39 in3 From observation load case 1 will govern,

Fx = 8.71 kip (compression), Mz = 56.50 k-ft Since L is less than Lc (6.85ft) [Clause F1-2 page 5-45 of Manual] Fb = 0.66x36 = 23.76 ksi (Clause F1-1 page 5-45 of Manual) (kl/r)y = 60/1.5038 = 39.90, so Fa = 19.19 ksi Table C-36 page 3-16 of Manual fa = 8.71/7.65 = 1.1385, fb = 56.5 x 12/33.39 = 20.31 ksi Since fa/Fa less than 0.15, apply formula H1-3, page 5-54 of Manual 1.1385 + 20.31 = .0593 + .8548 = 0.9141 19.19 23.76 Member 3, Size W14X43, L = 11ft., a = 12.6 in2, Sz = 62.7 in3 From observation load case 3 will govern, Fx = 25.5 kip (compression), Mz = 112.173 k-ft Refering to clause F1-2, page 5-45 of Manual. Lc = 8.4 ft. Therefore, Fb = 0.6 x 36 = 21.6 ksi (kl/r)y = 132/1.8941 = 69.69, so Fa = 16.46 ksi [Table C-36 page 3-16 of Manual] fa = 25.5/12.6 = 2.024, fb = 112.173 x 12/62.66 = 21.48 ksi since fa/Fa less than 0.15, use formula H1-3, page 5-54 of Manual


2.024 + 21.48 = 0.123 + 0.994 = 1.117 16.46 21.6 Member 4, Size W14X43, L = 4ft, a = 12.6 in2, Sz = 62.7 in3 From observation, load case 3 will govern, Fx = 8.75 kip (tension), Mz = 112.173 k-ft Lc = 8.4 ft Since L is less than Lc Fb = 0.66 x 36 = 23.76 ksi [Clause F1-1 page 5-45 of Manual] fa = 8.75/12.6 = 0.694, fb = 112.73x12/62.66 = 21.48 ksi Combined tension and bending, use formula H 2-1, page 5-55 of Manual. 0.694 + 21.48 = 0.032 + 0.904 = 0.936 0.6 x 36 23.76 Member 5, Size W16X36, L = 5ft, a = 10.6 in2, Sz = 56.49 in3 Lc=7.37 ft, [Clause F1-2 page 5-45 of Manual. From observation, load case 3 will govern. Fx = 14.02 kip (compression), Mz =57.04 k-ft Since L is less than Lc, Fb = 0.66 x 36 = 23.76 ksi (kl/r)y = 60./1.52 = 39.47, so Fa = 19.23 ksi [Table C-36 page 3-16 of Manual] fa = 14.02/10.6 = 1.32, fb = 57.04 x 12/56.5 = 12.12 ksi

STAAD Verification Problems 53 Since fa/Fa less than 0.15, use formula H1-3, page 5-54 of Manual 1.32 + 12.12 = 0.069 + 0.510 = 0.579 19.23 23.76 Member 6, Size W16X36, L = 16ft, a = 10.6 in2, Sz = 56.49 in3 From observation, load case 1 will govern. Forces at midspan are Fx = 5.65 kip (compression), Mz = 71.25 k-ft From Chapter F of the AISC ASD 9th ed. specs., with Cb = 1.0, sqrt(102,000Cb/Fy) =53.229 sqrt(510,000Cb/Fy) =119.02 L/rT = 192/1.79 = 107.26 53.229 < 107.26 < 119.02 Therefore Fb (as per F1-6, page 5-47 of Manual) [(2/3) – 36*107.26*107.26/(1530,000)]*36 = 14.25 ksi (Kl/r)y = 192/1.5203 = 126.29, so Fa = 9.36 [Table C-36 page 3-16 of Manual] fa = 5.65/10.6 = 0.533, fb = 71.25x12/56.49 = 15.14 ksi Since fa/Fa less than 0.15 use formula [H1-3, page 5-54 of Manual] 0.533/9.36 + 15.14/14.25 = 0.057 + 1.062 = 1.119


Member 7, Size W16X36, L =4ft, a = 10.6in2, Sz =56.49in3 Lc = 7.37ft (Clause F1-2 page 5-45 of Manual) From observation load case 3 will govern, Fx = 24.06 kip (tension), Mz = 62.96 k-ft From Clause F1-1, Fb = 0.66 Fy = 23.76 ksi = allowable compressive stress. Since section is in tension, Fb = 0.60X36 = 21.60 Ksi [Clause F1-5, page5-45 of Manual].

Choosing the larger of above 2 values, Fb = 23.76 Ksi fa = 24.06/10.6 = 2.2698, fb = 62.96X12/56.49 = 13.37 Since combined tension and bending, use formula H 2-1, page 5-55 of the AISC ASD 9th ed. specs. 2.2698 + 13.37 = 0.105 + 0.5627 = 0.6677 0.6 x 36 23.76 Member 8, Size L4x4x1/4, L = 7.071 ft, a = 1.94 in2 From observation load case 1 will govern, Fx = 23.04 kip (Comp.) Fa is computed as per page 5-310 of the AISC ASD 9th ed.specs. Qs = 1.34 – 0.00447*(4/0.25)*sqrt(36) = 0.9108 Qa = 1.0, Q = Qs * Qa = 0.9108 Cc = sqrt(2.0*pi*pi*E/(Q*Fy)) = sqrt(2.0*pi*pi*29000/(0.9108*36)) = 132.1241 Kl/r = 7.071 x 12 = 106.73 is less than Cc.

0.795 Hence, Fa = 11.6027 ksi (computed per equation 4-1)

STAAD Verification Problems 55 Actual compressive stress fa = 23.04/1.94 = 11.876 ksi Therefore, Ratio = fa/Fa = 11.876/11.602 = 1.024 Member 9, Size L5x5x3/8, L = 5.657 ft, a = 3.61 in2 From observation, load case 1 governs, Fx = 48.44 kip (Comp.) Fa is computed as per page 5-310 of the AISC ASD 9th ed.specs. Qs = 1.34 – 0.00447*(5/0.375)*sqrt(36) = 0.9824 Qa = 1.0, Q = Qs * Qa = 0.9824 Cc = sqrt(2.0*pi*pi*E/(Q*Fy)) = sqrt(2.0*pi*pi*29000/(0.9824*36)) = 127.2238 (Kl/r)min = 5.657 x 12 = 68.57 is less than Cc.

0.99 Hence, Fa = 16.301 ksi (computed per equation 4-1) Actual compressive stress fa = 48.44/3.61 = 13.418 ksi Therefore Ratio = fa/Fa = 13.418/16.301 = 0.823


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:43:14 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD PLANE VERIFICATION PROBLEM NO 13 2. * 3. * THIS DESIGN EXAMPLE IS VERIFIED BY HAND CALCULATION 4. * FOLLOWING AISC-89 CODE. 5. * 6. UNIT FEET KIP 7. JOINT COORD 8. 1 0 0 ; 2 25 0 ; 3 0 10 ; 4 25 11 9. 5 0 15 ; 6 25 15 ; 7 5 15 ; 8 21 15 10. MEMB INCI 11. 1 1 3 ; 2 3 5 ; 3 2 4 ; 4 4 6 12. 5 5 7 ; 6 7 8 ; 7 8 6 ; 8 3 7 ; 9 4 8 13. MEMB PROP AMERICAN 14. 1 2 TA ST W12X26 ; 3 4 TA ST W14X43 15. 5 6 7 TA ST W16X36 ; 8 TA ST L40404 ; 9 TA ST L50506 16. MEMB TRUSS ; 8 9 17. CONSTANT 18. E 4176E3 ALL 19. POISSON STEEL ALL 20. SUPPORT ; 1 2 PINNED 21. LOADING 1 DL + LL 22. MEMB LOAD ; 5 6 7 UNI Y -2.0 23. LOADING 2 WIND FROM LEFT 24. JOINT LOAD ; 5 FX 15. 25. LOAD COMB 3 ; 1 0.75 2 0.75 26. 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 = 8/ 9/ 2 ORIGINAL/FINAL BAND-WIDTH= 4/ 4/ 15 DOF TOTAL PRIMARY LOAD CASES = 2, TOTAL DEGREES OF FREEDOM = 20 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3652.7 MB 27. LOAD LIST 1 3 28. PRINT FORCES MEMBER END FORCES STRUCTURE TYPE = PLANE ----------------- ALL UNITS ARE -- KIP FEET (LOCAL ) MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z 1 1 1 25.00 -5.65 0.00 0.00 0.00 0.00 3 -25.00 5.65 0.00 0.00 0.00 -56.50 3 1 12.00 1.05 0.00 0.00 0.00 0.00 3 -12.00 -1.05 0.00 0.00 0.00 10.52 2 1 3 8.71 10.64 0.00 0.00 0.00 56.50 5 -8.71 -10.64 0.00 0.00 0.00 -3.29 3 3 15.83 -2.77 0.00 0.00 0.00 -10.52 5 -15.83 2.77 0.00 0.00 0.00 -3.34 3 1 2 25.00 5.65 0.00 0.00 0.00 0.00 4 -25.00 -5.65 0.00 0.00 0.00 62.15 3 2 25.50 10.20 0.00 0.00 0.00 0.00 4 -25.50 -10.20 0.00 0.00 0.00 112.17

STAAD Verification Problems 57 MEMBER END FORCES STRUCTURE TYPE = PLANE ----------------- ALL UNITS ARE -- KIP FEET (LOCAL ) MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z 4 1 4 6.50 -12.85 0.00 0.00 0.00 -62.15 6 -6.50 12.85 0.00 0.00 0.00 10.76 3 4 -8.75 -24.06 0.00 0.00 0.00 -112.17 6 8.75 24.06 0.00 0.00 0.00 15.95 5 1 5 -10.64 8.71 0.00 0.00 0.00 3.29 7 10.64 1.29 0.00 0.00 0.00 15.25 3 5 14.02 15.83 0.00 0.00 0.00 3.34 7 -14.02 -8.33 0.00 0.00 0.00 57.04 6 1 7 5.65 15.00 0.00 0.00 0.00 -15.25 8 -5.65 17.00 0.00 0.00 0.00 -0.75 3 7 10.20 4.50 0.00 0.00 0.00 -57.04 8 -10.20 19.50 0.00 0.00 0.00 -62.96 7 1 8 -12.85 1.50 0.00 0.00 0.00 0.75 6 12.85 6.50 0.00 0.00 0.00 -10.76 3 8 -24.06 14.75 0.00 0.00 0.00 62.96 6 24.06 -8.75 0.00 0.00 0.00 -15.95 8 1 3 23.04 0.00 0.00 0.00 0.00 0.00 7 -23.04 0.00 0.00 0.00 0.00 0.00 3 3 -5.41 0.00 0.00 0.00 0.00 0.00 7 5.41 0.00 0.00 0.00 0.00 0.00 9 1 4 26.16 0.00 0.00 0.00 0.00 0.00 8 -26.16 0.00 0.00 0.00 0.00 0.00 3 4 48.44 0.00 0.00 0.00 0.00 0.00 8 -48.44 0.00 0.00 0.00 0.00 0.00 ************** END OF LATEST ANALYSIS RESULT ************** 29. PARAMETER 30. CODE AISC 31. TRACK 1.0 ALL 32. CHECK CODE ALL STAAD.PRO CODE CHECKING - (AISC 9TH EDITION) ******************************************** ALL UNITS ARE - KIP FEET (UNLESS OTHERWISE NOTED) MEMBER TABLE RESULT/ CRITICAL COND/ RATIO/ LOADING/ FX MY MZ LOCATION ======================================================================= * 1 ST W12X26 (AISC SECTIONS) FAIL AISC- H1-2 1.157 1 25.00 C 0.00 56.50 10.00 ----------------------------------------------------------------------- | MEM= 1, UNIT KIP-INCH, L= 120.0 AX= 7.65 SZ= 33.4 SY= 5.3| | KL/R-Y= 79.8 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 20.18 | | FTZ= 21.60 FCY= 27.00 FTY= 27.00 FC= 15.26 FT= 21.60 FV= 14.40 | ----------------------------------------------------------------------- 2 ST W12X26 (AISC SECTIONS) PASS AISC- H1-3 0.916 1 8.71 C 0.00 56.50 0.00 ----------------------------------------------------------------------- | MEM= 2, UNIT KIP-INCH, L= 60.0 AX= 7.65 SZ= 33.4 SY= 5.3| | KL/R-Y= 39.9 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 23.76 | | FTZ= 23.76 FCY= 27.00 FTY= 27.00 FC= 18.61 FT= 21.60 FV= 14.40 | -----------------------------------------------------------------------

STAAD Verification Problems 58 ALL UNITS ARE - KIP FEET (UNLESS OTHERWISE NOTED) MEMBER TABLE RESULT/ CRITICAL COND/ RATIO/ LOADING/ FX MY MZ LOCATION ======================================================================= 3 ST W14X43 (AISC SECTIONS) FAIL AISC- H1-3 1.117 3 25.50 C 0.00 -112.17 11.00 ----------------------------------------------------------------------- | MEM= 3, UNIT KIP-INCH, L= 132.0 AX= 12.60 SZ= 62.7 SY= 11.3| | KL/R-Y= 69.7 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 21.60 | | FTZ= 21.60 FCY= 27.00 FTY= 27.00 FC= 16.46 FT= 21.60 FV= 14.40 | ----------------------------------------------------------------------- 4 ST W14X43 (AISC SECTIONS) PASS AISC- H2-1 0.936 3 8.75 T 0.00 -112.17 0.00 ----------------------------------------------------------------------- | MEM= 4, UNIT KIP-INCH, L= 48.0 AX= 12.60 SZ= 62.7 SY= 11.3| | KL/R-Y= 25.3 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 23.76 | | FTZ= 23.76 FCY= 27.00 FTY= 27.00 FC= 20.26 FT= 21.60 FV= 14.40 | ----------------------------------------------------------------------- 5 ST W16X36 (AISC SECTIONS) PASS AISC- H1-3 0.582 3 14.02 C 0.00 -57.04 5.00 ----------------------------------------------------------------------- | MEM= 5, UNIT KIP-INCH, L= 60.0 AX= 10.60 SZ= 56.5 SY= 7.0| | KL/R-Y= 39.5 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 23.76 | | FTZ= 23.76 FCY= 27.00 FTY= 27.00 FC= 18.39 FT= 21.60 FV= 14.40 | ----------------------------------------------------------------------- 6 ST W16X36 (AISC SECTIONS) FAIL AISC- H1-3 1.120 1 5.65 C 0.00 -71.25 8.00 ----------------------------------------------------------------------- | MEM= 6, UNIT KIP-INCH, L= 192.0 AX= 10.60 SZ= 56.5 SY= 7.0| | KL/R-Y= 126.3 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 14.24 | | FTZ= 21.60 FCY= 27.00 FTY= 27.00 FC= 9.36 FT= 21.60 FV= 14.40 | ----------------------------------------------------------------------- 7 ST W16X36 (AISC SECTIONS) PASS AISC- H2-1 0.668 3 24.06 T 0.00 62.96 0.00 ----------------------------------------------------------------------- | MEM= 7, UNIT KIP-INCH, L= 48.0 AX= 10.60 SZ= 56.5 SY= 7.0| | KL/R-Y= 31.6 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 23.76 | | FTZ= 23.76 FCY= 27.00 FTY= 27.00 FC= 18.87 FT= 21.60 FV= 14.40 | ----------------------------------------------------------------------- 8 ST L40404 (AISC SECTIONS) FAIL AISC- H1-1 1.024 1 23.04 C 0.00 0.00 0.00 ----------------------------------------------------------------------- | MEM= 8, UNIT KIP-INCH, L= 84.9 AX= 1.94 SZ= 0.8 SY= 1.7| | KL/R- = 106.7 CB= 0.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 0.00 | | FTZ= 0.00 FCY= 0.00 FTY= 0.00 FC= 11.60 FT= 21.60 FV= 0.00 | ----------------------------------------------------------------------- 9 ST L50506 (AISC SECTIONS) PASS AISC- H1-1 0.823 3 48.44 C 0.00 0.00 0.00 ----------------------------------------------------------------------- | MEM= 9, UNIT KIP-INCH, L= 67.9 AX= 3.61 SZ= 1.8 SY= 3.9| | KL/R- = 69.0 CB= 0.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 0.00 | | FTZ= 0.00 FCY= 0.00 FTY= 0.00 FC= 16.30 FT= 21.60 FV= 0.00 | -----------------------------------------------------------------------

STAAD Verification Problems 59 33. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:43:17 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ------------------------------------------------------------------------------------ 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb

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


NOTES



TYPE: Concrete design as per ACI code.

REFERENCE: CRSI Handbook and Notes on ACI-318 from ACI.

PROBLEM: A plane frame is created with such loading as to create 138 Kip-Ft moment on beam and 574 Kip of axial load coupled with above moment on column.

1

2 3

4

1

2

3

521.32k 521.32k

5.268 k / ft

20'

15'

GIVEN: Size of beam is 10 x 16 inch, column 14 x 16 inch. SOLUTION COMPARISON:

Area of Steel Area of Steel in beam in column

ACI notes 2.78 sq.in. X CRSI Handbook X 4.01% STAAD

2.792 sq.in 4.09% required 4.23% provided


**************************************************** * * * STAAD.Pro * * Version 2005 Bld 1003.US.NRC * * Proprietary Program of * * Research Engineers, Intl. * * Date= MAR 23, 2006 * * Time= 7:43:20 * * * * USER ID: Research Eng Ltd * **************************************************** 1. STAAD PLANE VERIFICATION FOR CONCRETE DESIGN 2. UNIT KIP FEET 3. JOINT COORDINATES 4. 1 0. 0. ; 2 0. 15. ; 3 20. 15. ; 4 20. 0. 5. MEMBER INCIDENCE 6. 1 1 2 ; 2 2 3 ; 3 3 4 7. UNIT INCH 8. MEMBER PROPERTY 9. 1 3 PRISMATIC YD 16. ZD 14. 10. 2 PRISM YD 16. ZD 10. 11. CONSTANTS 12. E CONCRETE ALL 13. POISSON CONCRETE ALL 14. SUPPORT 15. 1 4 FIXED 16. UNIT FT 17. LOADING 1 DEAD + LIVE 18. JOINT LOAD 19. 2 3 FY -521.32 20. MEMBER LOAD 21. 2 UNI GY -5.268 22. 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 = 4/ 3/ 2 ORIGINAL/FINAL BAND-WIDTH= 1/ 1/ 6 DOF TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 6 SIZE OF STIFFNESS MATRIX = 1 DOUBLE KILO-WORDS REQRD/AVAIL. DISK SPACE = 12.0/ 3652.6 MB 23. PRINT MEMBER FORCES MEMBER END FORCES STRUCTURE TYPE = PLANE ----------------- ALL UNITS ARE -- KIP FEET (LOCAL ) MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z 1 1 1 574.00 -13.69 0.00 0.00 0.00 -67.44 2 -574.00 13.69 0.00 0.00 0.00 -137.87 2 1 2 13.69 52.68 0.00 0.00 0.00 137.87 3 -13.69 52.68 0.00 0.00 0.00 -137.87 3 1 3 574.00 13.69 0.00 0.00 0.00 137.87 4 -574.00 -13.69 0.00 0.00 0.00 67.44 ************** END OF LATEST ANALYSIS RESULT ************** 24. UNIT INCH 25. START CONC DESIGN 26. CODE ACI 1999 27. TRACK 1.0 MEMB 2 28. FYMAIN 60.0 ALL

STAAD Verification Problems 63 29. FC 4.0 ALL 30. CLB 1.4375 ALL 31. DESIGN BEAM 2 ===================================================================== BEAM NO. 2 DESIGN RESULTS - FLEXURE PER CODE ACI 318-99 LEN - 20.00FT. FY - 60000. FC - 4000. SIZE - 10.00 X 16.00 INCHES LEVEL HEIGHT BAR INFO FROM TO ANCHOR FT. IN. FT. IN. FT. IN. STA END _____________________________________________________________________ 1 0 + 2-5/8 2-NUM.10 0 + 0-0/0 20 + 0-0/0 YES YES |----------------------------------------------------------------| | CRITICAL POS MOMENT= 125.53 KIP-FT AT 10.00 FT, LOAD 1| | REQD STEEL= 2.48 IN2, ROW=0.0185, ROWMX=0.0214 ROWMN=0.0033 | | MAX/MIN/ACTUAL BAR SPACING= 10.16/ 2.54/ 4.73 INCH | | REQD. DEVELOPMENT LENGTH = 48.52 INCH | |----------------------------------------------------------------| Cracked Moment of Inertia Iz at above location = 1837.17 inch^4 *** A SUITABLE BAR ARRANGEMENT COULD NOT BE DETERMINED. REQD. STEEL = 2.792 IN2, MAX. STEEL PERMISSIBLE = 2.873 IN2 MAX NEG MOMENT = 137.87 KIP-FT, LOADING 1 *** A SUITABLE BAR ARRANGEMENT COULD NOT BE DETERMINED. REQD. STEEL = 2.792 IN2, MAX. STEEL PERMISSIBLE = 2.873 IN2 MAX NEG MOMENT = 137.87 KIP-FT, LOADING 1 ___ 2J____________________ 240.X 10.X 16_____________________ 3J____ | | | | | | | | | 2#10H 3. 0.TO 240. | ||=========================================================================|| | | |___________________________________________________________________________| _________ _________ _________ _________ _________ _________ | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 2#10 | | 2#10 | | 2#10 | | 2#10 | | 2#10 | | 2#10 | | OO | | OO | | OO | | OO | | OO | | OO | | | | | | | | | | | | | |_________| |_________| |_________| |_________| |_________| |_________| ********************END OF BEAM DESIGN**************************

STAAD Verification Problems 64 32. DESIGN COLUMN 1 ==================================================================== COLUMN NO. 1 DESIGN PER ACI 318-99 - AXIAL + BENDING FY - 60000 FC - 4000 PSI, RECT SIZE - 14.00 X 16.00 INCHES, TIED AREA OF STEEL REQUIRED = 9.164 SQ. IN. BAR CONFIGURATION REINF PCT. LOAD LOCATION PHI ---------------------------------------------------------- 12 - NUMBER 8 4.232 1 END 0.700 (PROVIDE EQUAL NUMBER OF BARS ON EACH FACE) TIE BAR NUMBER 3 SPACING 14.00 IN ********************END OF COLUMN DESIGN RESULTS******************** 33. END CONC DESIGN 34. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= MAR 23,2006 TIME= 7:43:23 **** Information about the key files in the current distribution Modfication Date CRC Size (Bytes) File Name ------------ ----------- ---------- -- ------------------------------------------------------ ----- ----- ------ --------------------- 03/21/2006 0x28c1 12713984 SProStaad.exe 12/16/2004 0xca81 04558848 SProStaadStl.exe 09/19/2003 0x2fc0 00081970 CMesh.dll 03/01/2006 0xeb80 02486272 dbSectionInterface.dll 01/23/2001 0x9b40 00073728 LoadGen.dll 09/25/2003 0x6340 00704512 MeshEngine.dll 09/22/2003 0xce00 00069632 QuadPlateEngine.dll 12/22/2005 0x4181 00094208 SurfMesh.dll 09/30/2005 0x2dc1 00475136 aiscsections.mdb 01/05/2005 0x79c1 00319488 aiscsections_all_editions.mdb 01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb 01/05/2005 0xcac1 03651584 aitctimbersections.mdb 01/27/2005 0xeb01 00552960 aluminumsections.mdb 01/05/2005 0xcd01 00163840 australiansections.mdb 01/05/2005 0x6a41 00229376 britishsections.mdb 07/08/2005 0x9d41 00434176 bscoldformedsections.mdb 06/28/2005 0x8201 00327680 butlercoldformedsections.mdb 01/05/2005 0xabc0 00262144 canadiansections.mdb 05/31/2005 0x9e81 00450560 canadiantimbersections.mdb 05/05/2005 0x7f80 00409600 chinesesections.mdb 01/05/2005 0xd6c0 00600064 dutchsections.mdb 01/05/2005 0x1a00 00354304 europeansections.mdb 01/05/2005 0xd301 00202752 frenchsections.mdb 01/05/2005 0x11c1 00233472 germansections.mdb 01/05/2005 0x3c40 00264192 indiansections.mdb 01/05/2005 0xd540 00180224 iscoldformedsections.mdb 01/24/2006 0x2501 00221184 japanesesections.mdb 11/08/2005 0x9081 00376832 Kingspancoldformedsections.mdb 01/05/2005 0xb740 00174080 koreansections.mdb 02/03/2005 0xda00 00096256 lysaghtcoldformedsections.mdb 02/07/2005 0x9a00 00243712 mexicansteeltables.mdb 01/04/2006 0x8980 00413696 RCecoColdFormedSections.mdb 02/03/2005 0x9b40 00307200 russiansections.mdb 01/05/2005 0x9081 00206848 southafricansections.mdb 01/06/2005 0x9341 00194560 spanishsections.mdb 01/04/2006 0x8680 00223232 uscoldformedsections.mdb 01/05/2005 0xbac0 00149504 usersectionstemplate.mdb 01/05/2005 0x9d41 00141312 venezuelansections.mdb ------------------------------------------------------------------------------------------------------------------------------


NOTES STAAD reports that it is unable to find a suitable bar arrangement to satisfy the reinforcement requirement for the negative moment at the two ends of beam 2. However, this does not mean that it is impossible to come up with a bar arrangement. When STAAD looks for a bar arrangement, it uses only bars of the same size. It begins with the bar size corresponding to the parameter MINMAIN. If an arrangement is not possible with that bar, it tries with the next larger bar size. If all the permissible bar sizes are exhausted, the program reports that it could not come up with a bar arrangement. However, the user may be able to satisfy the requirement by mixing bars of various diameters. For example, 3 # 11 bars and 2 # 10 bars may satisfy the requirement. The program is not equipped with facilities to come up with such combinations of bar sizes.


NOTES

PART II

STARDYNE VERIFICATION PROBLEMS

STARDYNE Verification Problems 1

STARDYNE Verification Problem No. 1

Type: Uniform Beam With One Redundant Support. Purpose: Compare theoretical answers to the STARDYNE

solution. Reference: Timoshenko, S., “Strength of Materials”, Part I,

Second Edition, D. Van Nostrand Company, 1940, pages 181 and 182.

Problem: A uniform beam is clamped at one end and is

simply supported at the other end. This arrangement of supports offers the minimum analytic difficulty for an indeterminate structure. The single load case is a uniformly distributed vertical force of 100 pounds per inch. Find the moment at the clamped end, the reaction at the simple support, and the mid-span deflection.

STARDYNE Verification Problems 2 Model Two beam elements are sufficient for the analysis. The center node is included so that the mid-span deflection will be printed. Shear deformation is not included in the analysis. The beam has a solid circular cross-section, with a diameter of 1.5 inches. The material has an elastic modulus of 30x106 psi.

Theoretical Answers Moment at the clamped end:

( )( ) 50008

201008

qlM22

=== inch- pounds

Reaction at the simple support:

( )( ) 7502010083ql

83R === pounds


Mid-span deflection:

EIql

1921d

4

−=

where ( ) 24850.075.4

r4

I 44 =π

=π

= inches4

( )( )( )( ) ( )

011178.024850.01030

20100192

1d 6

4

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−= inches

STARDYNE Answers From the Equilibrium check :

1. The x3 Moment at node 1 is the clamped end moment, and is equal to 5000-inch pounds.

2. The x2 Force at node 3 is the simple support reaction of 750 pounds.

From the Deflections: The deflection at Node 2 is -.011178 inches.

Comparison of Theoretical and STARDYNE Answers

Item Theory STARDYNE

Clamped End Moment (inch-pounds)

5000. 5000.

Simple Support Reaction (pounds)

750. 750.

Mid-Span Deflection (inches) -.011178 -.011178

STARDYNE Verification Problems 4 Input file STAAD PLANE STARDYNE VER. PROB. 1 UNIT POUND INCH JOINT COORD 1 0 0 0 3 20 0 0 MEMB INCI 1 1 2 2 MEMB PROP 1 2 PRIS AX 1.767 IZ 0.2485 CONST E 30E6 ALL SUPP 1 FIXED 3 FIXED BUT FX MZ LOAD 1 MEMBER LOAD 1 2 UNI GY -100.0 PERF ANALY PRINT JOINT DISP LIST 2 PRINT SUPP REAC FINISH



Type: Indeterminate Frame. Purpose: Compare theoretical answers to the STARDYNE

solution. Reference: Timoshenko, S., “Strength of Materials”, Part I,

Second Edition, D. Van Nostrand Company, 1940, pages 188 thru 191.

Problem: A statically indeterminate frame, consisting of

three members, has a concentrated force applied at mid-span. The two vertical members are hinged at the ground support points. The reaction forces and, the bending moments at the ends of the horizontal beam will be determined.

STARDYNE Verification Problems 6 Model Five nodes and four beam elements form the computer model. Node 3, at the center of the horizontal member is required for the input of the applied load. For the restraints at Node 1 and 5, the X2 axis degress-of-freedom are input as being free to rotate. Shear deflection is not used for the beams. The analysis considers both axial and bending deformation of the beam elements, while the theory uses only bending deformation. The differences of the reactions and internal moments is infinitesimal.

Material Properties Elastic Modulus = 30 x 106 PSI Vertical Beams = 2 x 2 inches (Square) Horizontal Beam = 4 x 2 inches (Rectangular)


Cross-section Properties

Vertical Beams Area = 2 x 2 = 4 sq. inches

250.22649b

649J 44 === inches4

3333.1122

12bI

44

2 === inches4

I3 = 1.3333 inches4

Horizontal Beams Area = 4 x 2 = 8 sq. inches

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−−=

43

hb

1211

hb36.3

316

16hbJ

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−−=

43

42

1211

4236.3

316

162x4J 7.324 inches4

6667.212

2x4I3

2 == inches4

6667.1012

4x2I3

3 == inches4

Theoretical Answers Vertical reaction is:

5002

10002P

== pounds

STARDYNE Verification Problems 8 Horizontal reaction is:

=+

=+

3333.16667.10

120100

321

11000x8

120x1000

II

lh

321

1h8

Pl

1

227.5504 pounds

The corner bending moment is the horizontal reaction multiplied by the height, h: 27.5504 x 100 = 2755.04 inch pounds STARDYNE Answers From the Equilibrium Check, with the signs reversed, the reactions at Nodes 1 are: Horizontal reaction = 27.5497 pounds Vertical reaction = 500.0 pounds From the Beam End Loads in the Global System, the end moment for Beam 3, at Node 4, with the sign reversed, is: Moment MX3 = 2755.00 inch pounds Comparison of Theoretical and STARDYNE Answers

Item Theory STARDYNE Vertical Reaction (pounds) 500. 500. Horizontal Reaction (pounds) 27.5504 27.5497 Beam Moment (inch-pounds) 2755.04 2755.00


Input File STAAD SPACE STARDYNE VER. PROB. 2 UNIT POUND INCH JOINT COORD 1 0 0 0 ; 2 0 100 0 ; 3 60 100 0 ; 4 120 100 0 ; 5 120 0 0 MEMB INCI 1 1 2 ; 2 2 3 ; 3 3 4 ; 4 4 5 MEMB PROP 1 4 PRIS AX 4.0 IX 2.25 IY 1.3333 IZ 1.3333 2 3 PRIS AX 8 IX 7.324 IY 2.6667 IZ 10.6667 CONST E 30E6 ALL SUPP 1 5 FIXED BUT MZ LOAD 1 JOINT LOAD 3 FY -1000.0 PERF ANALY PRINT SUPP REAC PRINT MEMB FORCES LIST 1 FINISH


NOTES



Type: Uniform Pressure On A Circular Plate. Purpose: Compare theoretical answers from small deflection

plate theory to the STARDYNE solution. Reference: Timoshenko, S., and Woinowsky-Krieger, S.,

“Theory of Plates and Shells”, Second Edition, McGraw-Hill, 1959, pages 42 and 57.

Problem: Deflections and stresses for a circular plate

structure are determined, for a uniform pressure. The plate is simply supported around its entire periphery.

Note : While analyzing this model using the

“STARDYNE Advanced Analysis” option, the following settings have to be chosen for the “STARDYNE Element Type” : Quadrilateral - Thin Triangular - Thin

STARDYNE Verification Problems 12 Model The model consists of a triangular and quadrilateral elements. The quad plates are individually sized so that each plate has the same aspect ratio. This node spacing was selected because it gives the best agreement with the theoretical deflections and stresses.

Theoretical Answers The deflection at radius r is:

( )

⎟⎠⎞

⎜⎝⎛ −

ν+ν+−

= 2222

ra15

D64raqW

where a = circle radius q = uniform pressure

( )2

3

112EhD

ν−=

E = Elastic modulus h = plate thickness v = Poisson’s ratio


The radial bending stress at radius r is:

( )( )222r ra3

hq

83S −ν+=

and the tangential bending stress is:

( ) ( )[ ]ν+−ν+= 31r3ahq

83S 22

2t

The numerical values used for the above expressions are :

q = 0.0014 psi, a = 10.0 inches, E = 10x106 psi h = 0.02 inches v = 0.3


The deflections, perpendicular to the plate, at each node are compared in the table below. The deflections are in inches.

Theoretical STARDYNE Node Radius Deflection Deflection

1000 0 .121 734 .121 862 1 1.0 .120 221 .120 356

25 1.2916 .119 214 .119 351 49 1.6681 .117 539 .117 680 73 2.1544 .114 763 .114 906 97 2.7826 .110 176 .110 322

121 3.5938 .102 654 .102 802 145 4.6416 .090 460 .090 608 169 5.9948 .071 112 .071 251 193 7.7426 .041 588 .041 694 217 10.0 0 0

The stress comparison is based on evaluation of the theoretical stress at the average radius of each plate element. The only exception is the triangle plate at the center, where the 2/3 radius is used.


Element Radius Radial Stress, psi Tangential Stress, psi Theoretical STARDYNE Theoretical STARDYNE

1 .667 431.2 429.9 432.0 429.9 25 1.1458 427.4 428.0 428.9 429.2 49 1.47985 423.6 424.1 427.7 427.3 73 1.9113 417.3 417.8 424.0 423.8 97 2.4685 406.7 407.3 417.9 417.9

121 3.1882 389.1 390.0 407.8 407.9 145 4.11770 359.7 361.3 390.8 391.2 169 5.3182 310.6 313.3 362.7 363.4 193 6.86870 228.8 233.2 315.5 316.8 217 8.87130 92.26 99.68 236.9 239.2

Discussion: The difference in results between the theoretical solution and the STARDYNE solution can be accounted for by the following reasons. The theoretical solution assumes that the plate is clamped against rotation about the radial lines at its outer periphery. The only permissible rotation along the periphery is about the tangential lines. However the support condition in the STARDYNE model assumes that the plate is free to rotate about the global X1 and X2 axes. Also, while the example in the Reference has a true circular circumference, the STARDYNE model assumes the outer edge of the plate to be a series of secant lines instead of a true arc.


Input File STAAD SPACE STARDYNE VER. PROB. 3 UNIT POUND INCH JOINT COORD CYLINDRICAL 1 1 0 0 24 1 345 0 25 1.2916 0 0 48 1.2916 345 0 49 1.6681 0 0 72 1.6681 345 0 73 2.1544 0 0 96 2.1544 345 0 97 2.7826 0 0 120 2.7826 345 0 121 3.5938 0 0 144 3.5938 345 0 145 4.6416 0 0 168 4.6416 345 0 169 5.9948 0 0 192 5.9948 345 0 193 7.7426 0 0 216 7.7426 345 0 217 10.000 0 0 240 10.000 345 0 JOINT COORD 1000 0 0 0 ELEM INCI 1 1 2 1000 ; 2 2 3 1000 ; 3 3 4 1000 ; 4 4 5 1000 ; 5 5 6 1000 6 6 7 1000 ; 7 7 8 1000 ; 8 8 9 1000 ; 9 9 10 1000 ; 10 10 11 1000 11 11 12 1000 ; 12 12 13 1000 ; 13 13 14 1000 ; 14 14 15 1000 15 15 16 1000 ; 16 16 17 1000 ; 17 17 18 1000 ; 18 18 19 1000 19 19 20 1000 ; 20 20 21 1000 ; 21 21 22 1000 22 22 23 1000 ; 23 23 24 1000 ; 24 24 1 1000 25 1 2 26 25 TO 47 REP 8 24 24 48 24 1 25 48 REP 8 24 24 ELEM PROP 1 TO 240 TH 0.02 CONST E 10.0E6 ALL POISS 0.3 ALL SUPP * NODES ALONG PERIMETER 217 TO 240 FIXED BUT MX MY * INTERNAL NODES 1 TO 216 1000 FIXED BUT FZ MX MY

STARDYNE Verification Problems 16 LOAD 1 ELEM LOAD 1 TO 24 PR -0.0014 25 TO 240 PR 0.0014 PERF ANALY PRINT JOINT DISP LIST 1000 1 TO 217 BY 24 PRINT ELEM STRESS LIST 1 25 49 73 97 121 145 169 193 217 FINI



Type: Uniform Pressure On A Rectangular Plate. Purpose: Compare theoretical answers from small deflection

plate theory to the STARDYNE solution. Reference: Timoshenko, S., and Woinowsky-Krieger, S.,

“Theory of Plates and Shells”, Second Edition, McGraw-Hill, 1959, pages 113 thru 117.

Problem: A rectangular plate, 10in X 16in, , of constant

thickness of 0.2in., is simply supported along all four edges. The applied loading is uniform pressure over the entire surface. The analysis results are lateral deflections and bending stresses.


Model One inch node spacing is used in both directions to form the grid of the model. The origin is located at the center of one edge, to conform with the location used in the theoretical development. The only rotation degree-of-freedom that is active for the edge nodes is the one whose axis is parallel to the edge. At the four corner nodes, all 6 degrees-of-freedom are restrained. Theoretical Answers Deflection at any point on the plate is:

( ) ∑∞

=⎥⎦

⎤⎢⎣

⎡ π⎟⎠⎞

⎜⎝⎛ α

αα

+α

ααα+

−π

=,...3,1m

m

m

mm

m

mm

55

4

axmsin

by2sinh

cosh2by2

by2cosh

cosh2tanh21

m1

Dqa4y,xw

where q = uniform pressure, pounds per square inch a = dimension in x direction, inches b = dimension in y direction, inches

( )2

3

112EhD

ν−=

E = elastic modulus, pounds per square inch h= plate thickness in inches v = Poisson’s ratio

a2bm

mπ

=α

m = summation index, using only odd values The bending moment due to Sx stresses:

( ) ( ) ( ) ∑∞

=

π⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

ν−ν

−+πν−+−=,..3,1m

mmmmmm222

xa

xmsinKcosh12KsinhKBKcoshAmqa1xaqx

21y,xM

STARDYNE Verification Problems 20 and the bending moment due to Sy stresses:

( ) ( ) ( ) ∑∞

=

π⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

ν−ν

++πν−−−ν=,..3,1m

mmmmmm222

ya

xmsinKcosh12KsinhKBKcoshAmqa1xaqx

21y,xM

where

m

mm55m

coshtanh2

m2A

ααα+

π−

=

m55m

cosh1

m2B

απ=

a2bm

mπ

=α

a

ymKmπ

=

m = summation index, using only odd values The above moments are in units of bending moment per length. Bending stresses are computed from:

2x

xhM6S =

2y

yhM6S =

The numerical values used in the above deflection and moment expressions are: q = 1.0 psi a = 10. 0 inches b = 16.0 inches E = 1x106 psi h = 0.2 inches v = 0.3


Within the theoretical expressions, for deflections and moments, are infinite series with index m. The number of terms evaluated, for a particular set of coordinates x, y, was established by: 1. The maximum number of terms accumulated was 100. 2. Fewer than 100 terms were used if the ratio of the current term

to the first term (m = 1) was less than 1x10-16. Note : While analyzing this model using the

“STARDYNE Advanced Analysis” option, the following settings have to be chosen for the “STARDYNE Element Type” :

Quadrilateral - Thin

STARDYNE Verification Problems 22 Comparison of Theoretical and STARDYNE Answers Deflections for nodes along the X1 axis


Coordinates Node

Number X1 X2 Theoretical

Deflection (inches) STARDYNE

Deflection (inches)

161 0 0 0 0 162 1 0 0.035 912 0.035819 163 2 0 0.067 743 0.067568 164 3 0 0.092 472 0.092232 165 4 0 0.108 078 0.107796 166 5 0 0.113 406 0.113109 167 6 0 0.108 078 0.107796 168 7 0 0.092 472 0.092232 169 8 0 0.067 743 0.067568 170 9 0 0.035 912 0.035819 171 10 0 9.4 x 10-14 0

Deflections for nodes along the line X2 = 4


Coordinates Node

Number X1 X2 Theoretical Deflection

(inches)

STARDYNE Deflection (inches)

241 0 4 0 0 242 1 4 0.027003 0.026924 243 2 4 0.050824 0.050677 244 3 4 0.069231 0.069032 245 4 4 0.080800 0.080568 246 5 4 0.084740 0.084498 247 6 4 0.080800 0.080568 248 7 4 0.069231 0.069032 249 8 4 0.050824 0.050677 250 9 4 0.027003 0.026924 251 10 4 7.1x10-14 0


Deflections for nodes along the line X1 = 5


Coordinates Node Number X1 X2

Theoretical Deflection

(inches)

STARDYNE Deflection (inches)

186 5 1 0.111647 0.111353 206 5 2 0.106339 0.106054 226 5 3 0.097401 0.097132 246 5 4 0.084740 0.084498 266 5 5 0.068320 0.068117 286 5 6 0.048283 0.048136 306 5 7 0.025133 0.025055 326 5 8 2.5x10-15 0

Bending Stress in the X1 direction


Coordinates Quad

Plate Number

X1 X2 Theoretical

Bending Stress (psi)

STARDYNE Sx Stress (psi)

9 0.5 0.5 264.4 257.7 29 1.5 0.5 689.8 682.3 49 2.5 0.5 991.4 983.4 69 3.5 0.5 1184. 1175. 89 4.5 0.5 1277. 1269.

109 5.5 0.5 1277. 1269. 129 6.5 0.5 1184. 1175. 149 7.5 0.5 991.4 983.4 169 8.5 0.5 689.8 682.3 189 9.5 0.5 264.4 257.7 90 4.5 1.5 1242. 1233. 91 4.5 2.5 1169. 1161. 92 4.5 3.5 1058. 1050. 93 4.5 4.5 904.4 897.4 94 4.5 5.5 705.4 699.5 95 4.5 6.5 459.0 454.6 96 4.5 7.5 164.7 161.9

STARDYNE Verification Problems 28 Bending stress in the X2 direction


Coordinates Quad

Plate Number

X1 X2 Theoretical

Bending Stress (psi)

STARDYNE Sy Stress (psi)

9 0.5 0.5 134.6 132.3 29 1.5 0.5 367.2 364.2 49 2.5 0.5 546.8 543.2 69 3.5 0.5 669.1 665.0 89 4.5 0.5 731.0 726.7

109 5.5 0.5 731.0 726.7 129 6.5 0.5 669.1 665.0 149 7.5 0.5 546.8 543.2 169 8.5 0.5 367.2 364.2 189 9.5 0.5 134.6 132.3 90 4.5 1.5 726.3 721.8 91 4.5 2.5 714.4 709.5 92 4.5 3.5 689.3 683.9 93 4.5 4.5 641.1 635.1 94 4.5 5.5 554.3 547.8 95 4.5 6.5 406.2 399.6 96 4.5 7.5 166.2 159.7

Input File STAAD SPACE STARDYNE VER. PROB. 4. UNIT POUND INCH JOINT COORD 1 0 -8 0 11 10 -8 0 21 0 -7 0 31 10 -7 0 41 0 -6 0 51 10 -6 0 61 0 -5 0 71 10 -5 0 81 0 -4 0 91 10 -4 0 101 0 -3 0 111 10 -3 0 121 0 -2 0 131 10 -2 0 141 0 -1 0 151 10 -1 0 161 0 0 0 171 10 0 0 181 0 1 0 191 10 1 0 201 0 2 0 211 10 2 0 221 0 3 0 231 10 3 0 241 0 4 0 251 10 4 0

STARDYNE Verification Problems 30 261 0 5 0 271 10 5 0 281 0 6 0 291 10 6 0 301 0 7 0 311 10 7 0 321 0 8 0 331 10 8 0 ELEM INCI 1 1 2 22 21 TO 16 1 20 21 2 3 23 22 TO 36 1 20 41 3 4 24 23 TO 56 1 20 61 4 5 25 24 TO 76 1 20 81 5 6 26 25 TO 96 1 20 101 6 7 27 26 TO 116 1 20 121 7 8 28 27 TO 136 1 20 141 8 9 29 28 TO 156 1 20 161 9 10 30 29 TO 176 1 20 181 10 11 31 30 TO 196 1 20 ELEM PROP 1 TO 16 21 TO 36 41 TO 56 61 TO 76 81 TO 96 101 TO 116 - 121 TO 136 141 TO 156 161 TO 176 181 TO 196 TH 0.2 CONST E 1.0E6 ALL POISS 0.3 ALL SUPP 1 11 321 331 FIXED 2 TO 10 FIXED BUT MX 21 31 41 51 61 71 81 91 101 111 121 131 141 - 151 161 171 181 191 201 211 221 231 241 251 261 271 281 291 301 - 311 FIXED BUT MY 322 TO 330 FIXED BUT MX 22 TO 30 42 TO 50 62 TO 70 82 TO 90 102 TO 110 122 TO 130 - 142 TO 150 162 TO 170 182 TO 190 202 TO 210 222 TO 230 242 TO 250 - 262 TO 270 282 TO 290 302 TO 310 FIXED BUT FX FY FZ MX MY LOAD 1 ELEMENT LOAD 1 TO 16 21 TO 36 41 TO 56 61 TO 76 81 TO 96 101 TO 116 - 121 TO 136 141 TO 156 161 TO 176 181 TO 196 PR 1.0 PERF ANALY PRINT JOINT DISP LIST 162 TO 170 PRINT JOINT DISP LIST 242 TO 250 PRINT ELEM STRESS LIST 9 TO 189 BY 20 FINISH



Type: Static Analysis Of Thermal Loading. Purpose: Compare theoretical answers to the STARDYNE

solution. Reference: Seely, F.B., and Smith, J.O., “Advanced

Mechanics of Materials”, Second Edition, John Wiley and Sons, 1955, pages 494-497.

Problem: The internal loads caused by a temperature change

are analyzed in this example. The subject is a pipe line with two right angles. Both ends are completely restrained. Internal forces and moments will be computed for a temperature increase of 430 degrees F.

STARDYNE Verification Problems 32 Model Three beams and four nodes comprise the model. The built-in boundary condition, at both ends of the pipe, is realized by restraining all six degrees of freedom for nodes 1 and 4.

Section Properties PIPE :Outside diameter = 12 inches; Wall thickness = 0.8725 inches

( ) 50.30255.10124

Area 22 =−π

= sq.in

( ) 0.475255.101264

Inertia 44 =−π

= in4

J = 2 I = 950.0 in4


Theoretical Results The solution presented in the Reference consists of a set of simultaneous equations for the thermal expansion deflections. These equations are developed under the pretense that one end is fixed and the other end is free. See equations 582 and 583 of the Reference.

XEIPl

21

EIPl

127 y3x3 ∆=−

YEIPl

32

EIPl

21 y3x3 ∆=+−

where l = 240 inches E = 26.4 x 106 per sq. inch I = 475 inches4

*T*X ∆α=∆ length in X- direction = (7.26744x10-6) (430) (480) = 1.50 inches

*T*Y ∆α=∆ length in Y- direction = (7.26744x10-6) (430) (240) = 0.75 inches

STARDYNE Verification Problems 34 Upon substituting for l, E, and I, the two equations become: 8.064 Px - 6.912 Py = 18810.0 -6.912Px +9.216Py = 9405.0 and the solution is Px = 8980.47 pounds Py = 7755.86 pounds

Note that the theoretical solution depends only on the bending energy, and ignores the effect of axial energy. STARDYNE Results Forces and moments, at the two built-in ends, are taken from the Equilibrium Check, which is the final table in the output. Reverse the sign, from the Equilibrium Check, to find the support point reactions. The moment, at node 2, comes from the Beam Element Loads Table.


Comparison of Theoretical and STARDYNE Results

Item Theoretical STARDYNE Axial reaction 8 980.47 8 949.71 Shear reaction 7 755.86 7 729.66

Moment reaction 7 837 50. 781 152. Moment at Node 2 1 077 656. 1 073 970.

STARDYNE Verification Problems 36 Input File : STAAD SPACE STARDYNE VER. PROB. 5 UNIT POUND INCH JOINT COORD 1 0 0 0 ; 2 240 0 0 ; 3 240 240 0 ; 4 480 240 0 MEMB INCI 1 1 2 ; 2 2 3 ; 3 3 4 MEMB PROP 1 TO 3 PRIS AX 30.5 IX 950.0 IY 475.0 IZ 475.0 CONST E 26.4E6 ALL ALPHA 7.26744E-6 ALL SUPPORT 1 4 FIXED LOAD 1 TEMP LOAD 1 2 3 TEMP 430.0 PERF ANALY PRINT SUPP REAC PRINT MEMB FORCE FINI



Type: Imposed Deflections On Beam. Purpose: Compare theoretical answers to the STARDYNE

solution. Reference: Norris, C.H., and Wilbur, J.B., “Elementary

Structural Analysis”, Second Edition, McGraw-Hill, 1960, Example 13.20, page 442.

Problem: A beam which is supported at three points (nodes

1, 2 and 3) has imposed translations specified at each of these points. Additionally, an imposed rotation is specified at node 1. There are no external loads applied to the structure. The static response of the structure due to these imposed deflections and rotation is calculated.

STARDYNE Verification Problems 38 Model Four nodes and three beam elements are used for this model. The imposed deflections and rotations are: Node 1: X2 = -.12 inches, X6 = -.001 radians Node 2: X2 = -.48 inches Node 3: X2 = -.21 inches

Theoretical Results Numerical answers are presented for four items, in the above Reference. They are:

(a) The slope at the center support of -.0007 radians. (b) The slope at the right-hand support of .0026 radians (c) A bending moment, at the left-hand built-in support, of

3.15 x 106 inch pounds. (d) A bending moment, at the center support, of 3.3 x 106

inch pounds. Both the rotation angle and the moment sign conventions are reversed between the Reference and STARDYNE. Therefore, the algebraic sign on the above four results have been reversed. Also,


the units on the bending moments have been converted from kip feet to inch pounds. STARDYNE Results The elastic rotation angles, at the center support and the right-hand support, are obtained from the Deflection Table appearing in the output:

Location Node Rotation in Radians Center Support 2 -.0007 Right-Hand Support

3 .0026

The internal bending moments, for the beams, are listed in the Beam Element Loads Table:

Beam Number Node Moment (inch pounds) 1 1 3.15 x 106

1 2 3.30 x 106


Item Theoretical STARDYNE Rotation at Center Support -.0007 -.0007 Rotation at Right-Hand Support

.0026 .0026

Moment at Left-Hand Support

3.15x106 3.15x106

Moment at Center Support 3.30x106 3.30x106

STARDYNE Verification Problems 40 Input file STAAD PLANE STARDYNE VER. PROB. 6 UNIT POUND INCH JOINT COORD 1 0 0 0 ; 2 120 0 0; 3 300 0 0 ; 4 360 0 0 MEMB INCI 1 1 2 ; 2 2 3 ; 3 3 4 MEMB PROP 1 PRIS AX 1 IZ 1000.0 2 3 PRIS AX 1 IZ 3000.0 CONST E 30.0E6 ALL SUPP 1 FIXED 2 3 PINNED LOAD 1 SUPP DISP LOAD 1 FY -0.12 * ROTATION HAS TO BE SPECIFIED IN DEGREES. (0.001 RADIANS = 0.057296 DEGREES) 1 MZ -0.057296 2 FY -0.48 3 FY -0.21 PERF ANALY PRINT SUPP REAC PRINT MEMB FORCE PRINT JOINT DISP FINI



Type: Natural Modes of a Simple Beam. Purpose: Compare theoretical answers to the STARDYNE

solution. Reference: Timoshenko, S., “Vibration Problems in

Engineering”, Third Edition, D. Van Nostrand Company, Inc., 1955, page 322

Problem: The first five natural frequencies and the

associated mode shapes are computed for the flexural motion of a simply supported beam. The HQR and LANCZOS solution methods are both tested.


Model The simply supported beam is divided into twenty spanwise beam elements. At nodes 1 and 21, all degrees of freedom except the rotation about the X3 axis are restrained. For the remaining nodes, only the translation along X2 and the rotation about X3 are permitted. Both shear deformation and rotary inertia have been excluded from the model. The mass matrix is a diagonal matrix. Cross-section Properties Rectangular Section: 1inch Width x 2 inch Depth Area = 2 square inches

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−−=

43

ab

1211

ab36.3

316

16abJ

where a = 2 and b = 1 J = 0.4578 inch4


6667.012

2x1I3

2 == inch4

1667.012

1x2I3

3 == inch4

Theoretical Results The natural bending frequencies, for a uniform beam with hinged ends, are given by:

γπ

=AEIg

l2nf 2

2

n

where = natural frequency for mode n, in cycles per second nf l = span of the beam E = elastic modulus I = cross-section moment of inertia g = gravitational constant A = cross-section area = weight density γ The parameters used in the frequency equation are: l = 20 inches E = 10x106psi I = 0.6667 inches4

g = 386.4 inches per second2

A = 2.0 sq. inches = 0.1 pounds per cubic inch γ from which:

xnf 2n = 445.685674

STARDYNE Verification Problems 44 The table below shows the natural frequencies computed from the theoretical equation and the HQR and LANCZOS methods available within STARDYNE. Frequencies are in cycles per second.


STARDYNE Frequencies Mode

NumberTheoretical Frequency

HQR Lanczos

1 445.686 445.685 445.685 2 1782.74 1782.73 1782.73 3 4011.17 4011.03 4 011.03 4 7130.97 7130.12 7130.12 5 11142.1 11138.7 11138.7

Input file: STAAD PLANE STARDYNE VER. PROB. 7 UNIT POUND INCH JOINT COORD 1 0 0 0 21 20 0 0 MEMB INCI 1 1 2 20 MEMB PROP 1 TO 20 PRIS AX 2 IZ 0.6667 CONST E 10E6 ALL POISSON 0.3 ALL DENS 0.1 ALL CUT OFF MODE SHAPE 5 SUPP 1 21 FIXED BUT MZ 2 TO 20 FIXED BUT FY MZ LOAD 1 SELF X 1.0 SELF Y 1.0 MODAL CALC REQ PERF ANALY FINISH



Type: Natural Frequencies Of A Circular Plate Purpose: Compare theoretical answers to the STARDYNE

solution. Reference: Blevins, Robert D., “Formulas for Natural

Frequency and Mode Shape,” Van Nostrand Reinhold Company, 1979, Page 240.

Problem: A flat circular plate is simply supported around the

entire perimeter. The first six modes and their associated natural frequencies are to be computed using the HQR method offered by STARDYNE. This problem demonstrates that the natural frequencies of an axi-symmetric structure can be accurately computed utilizing a 180 degree model with the appropriate boundary conditions.

STARDYNE Verification Problems 46 Note : While analyzing this model using the “STARDYNE

Advanced Analysis” option, the following settings have to be chosen :

STARDYNE Element Type :

Quadrilateral - Thin Triangular - Thin

STARDYNE Analysis Type :

Analysis Type : Linear Static/Dynamic Analysis Modal Extraction : HQR

Model The 180 degree sector was modeled using radial lines at intervals of 15 degrees. Tangential lines were then located utilizing a relationship such that the aspect ratio of the quad-plate elements was approximately 1.0. All rotations normal to the plane of the plate were restrained. In-plane translations for all nodes were restrained because the theoretical solution does not consider in-plane effects. Rotations about the global X2 axis for the nodes at X1=0.0 were restrained because this is a symmetry boundary. X3 translation of all nodes on the outside radius were restrained to provide for the simply-supported condition. Quad-plate elements with only bending and transverse stiffness (STARDYNE Element Type = Thin) were used because of the conditions on the theoretical solution noted above. Likewise, tri-plate elements with only bending and transverse stiffness were used to model the inner elements.


It should be noted that the outside edge of the plate is a series of secant lines instead of a true arc. This will result in a loss of about 1% of the plate’s true mass or about .5% of the mass that is effective for this problem. Therefore, it is not unlikely that a few natural frequencies will be lower than the theoretical values instead of higher which is typical for a finite element analysis using plate elements. In addition, a true simple-support condition for this problem would require restraining the component of rotation that is radial to the outside edge.

Theoretical Answers From the reference case 2 in Table 11-1, the first six natural frequencies of the plate are described by the following equation:

( )2/1

2

3

2

2ij

ij112Eh

a2f ⎟⎟

⎠

⎞⎜⎜⎝

⎛ν−γπ

λ=

2ijλ = dimensionless parameter associated with the mode

indices i,j i = number of nodal diameters in this mode shape


j = number of nodal circles in this mode shape not counting the boundary

ν = Poisson’s ratio E = Elastic Modulus h = plate thickness γ = mass of plate per unit area a = radius of plate

The numerical values used for this example are:

ν = 0.30 E = 10.0x106 psi h = 0.10 inches γ =

( )( )( )

32f

52

fm

3m in/slb10x588.2

slb/inlb4.386in10.0in/lb10.0

−=−−

−

a = 10.0 inches with the numerical values used above

( ) ( )( )( )

( ) ( )( ) 467.93.0110x588.212

10.010x0.100.102

1112

Eha2

12/1

25

36

2

2/1

2

3

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛

−π=⎟⎟

⎠

⎞⎜⎜⎝

⎛ν−γπ −

cycles/sec

2ijλ is tabulated from the reference as follows:

Mode

Number 2ijλ Number of

Nodal Diameters (i)

Number of Nodal Circles

(j) 1 4.977 0 0 2 13.94 1 0 3 25.65 2 0 4 29.76 0 1

5(l) 3 0 6 48.51 1 1

(l) not tabulated in the reference



Mode Number

Theoretical Frequency (Hz)

STARDYNE Frequency (Hz)

1 47.12 46.31 2 132.0 130.5 3 242.8 241.5 4 281.7 282.1

5 * 379.0 6 459.3 463.5

*The reference did not tabulate a value of for the fifth mode of

the structure, hence a comparison with the theoretical value of this mode cannot be made.

2ijλ

Discussion All anti-symmetric mode shapes for the 360 degree circular plate were captured by the 180 degree model with a phase angle included in the calculation. Some of the difference between the theoretical and STARDYNE frequencies is attributed to the loss of mass due to the piecewise secant representation of the outer radius and, since this mass is about 1 percent lower than for a true circular plate, it is not surprising that the first few modes are lower than the theoretical solution.

Input file STAAD SPACE STARDYNE VER. PROB. 8 * NATURAL FREQUENCIES OF A CIRCULAR PLATE UNIT POUND INCH JOINT COORD CYLINDRICAL 1 1 -90 0 13 1 90 0 REP 2 1.244 0 0 REP 1 1.051 0 0 REP 1 1.367 0 0 REP 1 1.779 0 0 REP 1 2.315 0 0 JOINT COORD

STARDYNE Verification Problems 50 1000 0 0 0 ELEM INCI 1 1000 1 2 ; 2 1000 2 3 ; 3 1000 3 4 ; 4 1000 4 5 ; 5 1000 5 6 6 1000 6 7 ; 7 1000 7 8 ; 8 1000 8 9 ; 9 1000 9 10 ; 10 1000 10 11 11 1000 11 12 ; 12 1000 12 13 13 1 14 15 2 TO 24 REP 5 12 13 ELEM PROP 1 TO 84 TH 0.1 CONST E 10.0E6 ALL POISS 0.3 ALL DENS 0.1 ALL SUPP * CENTRE OF CIRCLE 1000 FIXED BUT FZ MX * INTERIOR NODES 2 TO 12 15 TO 25 28 TO 38 41 TO 51 54 TO 64 - 67 TO 77 FIXED BUT FZ MX MY * NODES ALONG CIRCUMFERENCE 79 TO 91 FIXED BUT MX MY * NODES ALONG DIAMETER EXCEPT THE TWO AT THE ENDS OF THE DIAMETER. 1 TO 66 BY 13 13 TO 78 BY 13 FIXED BUT FZ MX LOAD 1 SELF X 1.0 SELF Y 1.0 SELF Z 1.0 MODAL CALC REQ PERF ANALY FINI



Type: Natural Frequencies Of A Rectangular Plate. Purpose: Compare theoretical answers to the STARDYNE

solution. Reference: Blevins, Robert D., ”Formulas for Natural

Frequency and Mode Shape,” Van Nostrand Reinhold Company, 1979, page 258.

Problem: A flat rectangular plate is simply supported on all

four sides. The first six modes and their associated natural frequencies are to be computed for this structure using the LANCZOS method offered by STARDYNE. This problem also demonstrates that the mesh refinement can be chosen to accurately calculate modes of interest based on the expected mode shapes.

All dimensions are in inches.

STARDYNE Verification Problems 52 Note : While analyzing this model using the “STARDYNE

Advanced Analysis” option, the following settings have to be chosen :

STARDYNE Element Type :

Quadrilateral - Thin STARDYNE Analysis Type :

Analysis Type : Linear Static/Dynamic Analysis Modal Extraction : LANCZOS

Model A plate with an aspect ratio of 1.5 was used so that comparison could be made with theoretical results tabulated for plates in the reference. An equally spaced mesh was utilized in both the x and the y dimensions of the plate. The number of elements in each dimension was determined on the basis of the highest mode of interest. Since the number of half-waves in the sixth mode is 3 in the length dimension and 2 in the width dimension, a node spacing of 3.75 inches results in each half- wave being represented by four elements which means that no element will be expected to deform in double curvature. The simply supported edge condition requires that translation normal to the plane of the plate be restrained for these edge nodes. Rotations normal to the plate were restrained for all nodes.


Theoretical Answers From the reference case 16 in Table 11-4, the first six natural frequencies of the plate are described by the following equations:

( )2/1

2

3

2

2ij

ij112

Eha2

f ⎟⎟⎠

⎞⎜⎜⎝

⎛ν−γπ

λ=

2ijλ = dimensionless parameter associated with the mode indices

i, j i = number of half-waves in this mode shape along the horizontal

axis j = number of half-waves in this mode shape along the vertical

axis υ = Poisson’s ratio E = elastic modulus h = plate thickness γ = mass of material per unit area

STARDYNE Verification Problems 54 a = length of plate b = width of plate The numerical values used for this example are: υ = 0.30 E =30.0x106 psi h=0.2 inches

( )( )( )

32f

42

fm

3m in/slb10x460.1

slb/inlb4.386in20.0in/lb282.0

−=−−

=γ −

a = 45.0 inches b = 30.0 inches with the numerical values used above

( ) ( )( )( )

( ) ( )( ) sec/cycles9644.03.0110x460.112

20.010x0.300.452

1112Eh

a21

2/1

24

36

2

2/1

2

3

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛

−π=⎟⎟

⎠

⎞⎜⎜⎝

⎛ν−γπ −

2ijλ is tabulated from the reference as follows:

Mode Number

2ijλ Number of Half-

Waves in Length (i)

Number of Half-Waves in Width

(j) 1 32.08 1 1 2 61.69 2 1 3 98.70 1 2 4 111.0 3 1 5 128.3 2 2 6 177.7 3 2



Mode Number

Theoretical Frequency (Hz)

STARDYNE Frequency (Hz)

1 30.94 31.01 2 59.49 59.78 3 95.18 95.45 4 107.1 107.7 5 123.7 124.9 6 171.3 173.9

Discussion As was noted earlier, the node spacing was based on the highest mode of interest. It follows that the difference between the theoretical and STARDYNE frequencies generally increases with increasing mode sequence. Input file STAAD SPACE STARDYNE VER. PROB. 9. * NATURAL FREQUENCIES OF A RECTANGULAR PLATE UNIT POUND INCH JOINT COORD 1 0 0 0 13 45 0 0 REP ALL 8 0 3.75 0 ELEM INCI 1 1 2 15 14 TO 12 REP 7 12 13 ELEM PROP 1 TO 96 TH 0.2 CONST E 30.0E6 ALL DENS 0.282 ALL POISSON 0.3 ALL CUT OFF MODE SHAPE 6 SUPPORT * CORNER NODES 1 13 105 117 FIXED BUT MX MY

STARDYNE Verification Problems 56 * NODES ALONG Y=0 AND Y=30 2 TO 12 106 TO 116 FIXED BUT MX MY * NODES ALONG X=0 14 TO 92 BY 13 FIXED BUT MX MY * NODES ALONG X=45 26 TO 104 BY 13 FIXED BUT MX MY * INTERIOR NODES 15 TO 25 28 TO 38 41 TO 51 54 TO 64 67 TO 77 80 TO 90 - 93 TO 103 FIXED BUT FZ MX MY LOAD 1 SELF X 1 SELF Y 1 SELF Z 1 MODAL CALC REQ PERF ANALY FINI



Type: Natural Modes Of A Framework. Purpose: Compare the results from STARDYNE's

Householder-QR method with the results from EASE2 and ANSYS.

Reference: 1. Problem 1, from the ASME 1972 Program

Verification and Qualification Library 2. DeSalvo, G.J., and Swanson, J.A., “ANSYS Engineering Analysis System Examples Manual”, Swanson Analysis Systems, Inc., 1979, Example Problem No. 2. 3. Peterson, F.E., “EASE2 Elastic Analysis for Structural Engineering Example Problem Manual”, Engineering Analysis Corporation, 1981, Example 2.03.

Problem: A three dimensional frame is analyzed for its

natural frequencies and the associated mode shapes using the HQR method offered by STARDYNE.

Note: While analyzing this model using the “STARDYNE Advanced Analysis” option, the following settings have to be chosen:

STARDYNE Analysis Type: Analysis Type: Linear Static/Dynamic Analysis Modal Extraction: HQR


All dimensions are in inches.

Model The only element present in the model is the general purpose three dimensional beam. Cross-section Properties All the beam elements have the same cross section : Outside radius, Ro = 1.1875 inches Inside radius, Ri = 1.0335 inches

( )2i

2o RRArea −π= = 1.074532 square inches

Moment of Inertia, ( )4i

4o RR

4I −

π=


I = 0.665747 in4

Torsion constant = twice the inertia, for closed circular sections = 1.331494 in4

The expression for the shear flexibility factor is derived from the ratio of maximum shear stress to the average shear stress :

IbAQ

=α

Where the four items – A, Q, I, and b are properties of the half cross-section about the centerline. For the circular section they are :

Area = ( )2i

2o RR

21A −π=

( )3i

3o RR

32Q −=

( )4i

4o RR

8I −

π=

( )io RR2b −=

The final expression for the shear flexibility factor, for a circular tube section is :

( )( )io2

i2

o

3i

3o

RRRRRR

34

IbAQ

−+−

==α

α = 1.993620 Hence the shear area is entered as

STARDYNE Verification Problems 60 AY = Cross Section Area / α = 1.074532 / 1.99362 = 0.538985 sq.in

Comparison of Twenty-Four Natural Frequencies

Natural Frequency, Cycles per second Mode Number EASE2 ANSYS STARDYNE

1 111.53 111.52 111.28 2 115.96 115.95 115.85 3 137.61 137.60 137.22 4 218.03 218.02 215.89 5 404.23 404.23 404.43 6 422.72 422.70 422.81 7 451.75 451.72 451.78 8 554.07 553.99 549.23 9 735.81 735.70 733.88

10 762.44 762.32 758.87 11 852.72 852.57 851.70 12 894.26 894.08 892.64 13 910.40 910.21 893.42 14 917.18 916.98 911.20 15 940.02 940.02 932.66 16 960.27 959.98 956.78 17 971.44 971.15 964.40 18 977.22 976.92 967.66 19 1012.5 1012.2 981.93 20 1028.8 1028.4 1009.66 21 1123.9 1123.6 1070.99 22 1134.9 1134.5 1123.61 23 1164.4 1164.1 1149.83 24 1217.2 1216.7 1200.37


Discussion In both the References, the number of dynamic degrees of freedom has been reduced from 42 to 24, by means of the Guyan method. No such reduction is performed in the STARDYNE model. Input File STAAD SPACE STARDYNE VER. PROB. 10 * NATURAL FREQUENCIES OF A SPACE FRAME UNIT POUND INCH JOINT COORD 1 0 0 0 ; 2 27.25 0 0 3 0 10 0 ; 4 27.25 10 0 5 0 18.625 0 ; 6 8.625 18.625 0 ; 7 18.625 18.625 0 ; 8 27.25 18.625 0 9 0 18.625 8.625 ; 10 27.25 18.625 8.625 11 0 0 17.25 ; 12 27.25 0 17.25 13 0 10 17.25 ; 14 27.25 10 17.25 15 0 18.625 17.25 ; 16 8.625 18.625 17.25 17 18.625 18.625 17.25 ; 18 27.25 18.625 17.25 MEMB INCI 1 1 3 ; 2 2 4 ; 3 3 5 ; 4 4 8 5 5 6 7 8 5 9 ; 9 8 10 ; 10 9 15 ; 11 10 18 12 11 13 ; 13 12 14 ; 14 13 15 ; 15 14 18 16 15 16 ; 17 16 17 ; 18 17 18 MEMB PROP 1 TO 18 PRIS AX 1.074532 IZ 0.665747 IY 0.665747 IX 1.331494 - AY 0.538985 AZ 0.538985 CONST E 27.9E6 ALL POISS 0.3 ALL CUT OFF MODE SHAPE 6

STARDYNE Verification Problems 62 SUPPORT 1 2 11 12 FIXED LOAD 1 JOINT LOAD 3 4 13 14 FX 3.4517 FY 3.4517 FZ 3.4517 6 7 16 17 FX 3.4517 FY 3.4517 FZ 3.4517 9 10 FX 3.4517 FY 3.4517 FZ 3.4517 5 8 15 18 FX 9.7973 FY 9.7973 FZ 9.7973 MODAL CALC REQ PERF ANALY FINI



Type: Response Of A Simply Supported Beam To A Shock Spectrum.

Purpose: Compare theoretical answers to the STARDYNE

solution. Reference: 1. Biggs, John M., “Introduction to Structural

Dynamics,” McGraw Hill, 1964, pp. 256-263 2. Blevins, Robert D.,“ Formulas for Natural Frequency and Mode Shape,” Van Nostrand-Reinhold, 1979.

Problem: The supports of a simply supported beam are

subjected to an acceleration time history. The maximum bending moment in the beam is computed for the first mode of the structure. This problem demonstrates the capabilities of STARDYNE to calculate the correct modal response of a structure utilizing response spectrum data.

STARDYNE Verification Problems 64 Model The STARDYNE model consists of 11 nodes and 10 elastic beam elements. Node 1 is completely restrained with the exception of having rotational freedom in the X3 direction, the remaining nodes are restrained except for X1 and X2 displacements and X3 rotations. Node 11 is additionally restrained against displacements in the X2 direction to provide for the simple support condition . Only the contribution of the first mode of the structure is considered. x2

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10 11 x1

10 @ 24.0 inches

Theoretical Solution Material Properties: l = 240.0 inches E = 30 x 106 lb/in2

EI = 1.0 x 1010 lb-in2

m = 0.2 lb-sec2 / in2

h = 14.0 inches


From Reference 2, Table 8-1, page 108, the fundamental frequency of the beam is:

( )hz098.6

2.010x0.1

2402869.9

mEI

l2f

2/110

2

2/1

2

21

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛π

=⎟⎠⎞

⎜⎝⎛

πλ

=

The modal participation factor for the fundamental mode is:

( )

( )∫

∫

φ

φ=Γ l

0

2

l

0

dxxm

dxxm

Where φ(x) is the first mode shape = sinlxπ

∫

∫π

π

=Γ l

0

2

l

0

dxlxsinm

dxlxsinm

π=Γ

4

The maximum relative modal displacement is given by:

maxo

max uA Γ=

where:

( )max2so

..

maxo DLF

yu

ω= where

Hz098.6*2π=ω


g0.1yso

..= and

( )maxDLF = 1.648 at 6.098 Hz

therefore:

( )( )( ) ( )

5523.0098.62

4.386648.14A 22max =ππ

= inches

The bending moment 2

2

xuEIM

δδ

−=

Where u for the first mode lxsinA π

=

lxsinA

lxu

2

2

2

2 ππ−=

δδ

lxsin

lAEIM 2

2 ππ=

2

2

maxmax lEIAM π

= at x=l/2

( ) ( ) inlb10x351.946240

5523.010X1M 32

210

max −=π

=

Comparison of STARDYNE and Theoretical Answers

Bending Moment (kip-inch) Theoretical STARDYNE (Beam 5)

946.351 946.325


Input File : STAAD SPACE STARDYNE VER. PROB. 11 * RESPONSE OF A SIMPLY SUPPORTED BEAM TO A SHOCK SPECTRUM UNIT POUND INCH JOINT COORD 1 0 0 0 11 240 0 0 MEMB INCI 1 1 2 10 MEMB PROP 1 TO 10 PRIS YD 14.0 ZD 1.45777 AX 20.4082 IX 40.0 IY 3.6139 IZ 333.333 CONST E 30.0E6 ALL POISS 0.3 ALL DENS 3.78672 ALL CUT OFF MODE SHAPE 1 SUPPORT 1 FIXED BUT MZ 2 TO 10 FIXED BUT FX FY MZ 11 FIXED BUT FX MZ LOAD 1 SELF X 1.0 SELF Y 1.0 SPECTRUM SRSS Y 1.0 ACC DAMP 0.001 SCALE 386.4 0.15 1.648 ; 0.17 1.648 PERF ANALY PRINT MEMB FORCE LIST 5 FINI


NOTES



Type: Thermal loading on a Simply supported Rectangular Plate.

Purpose: Compare theoretical answers to the STARDYNE

solution. Reference: Timoshenko, S., and Woinowsky-Krieger, S.,

“Theory of Plates and Shells”, Second Edition, McGraw-Hill, 1959, pages 162 - 165.

Problem: A rectangular plate is simply supported on all four

sides. The transverse and longitudinal bending moments as well as the deflections at several points on the plate are computed.


All dimensions are in inches

Model The plate is modeled using 1 in. X 1 in. size elements. At the corner nodes, all the degrees of freedom are considered restrained. For the nodes along the four edges, rotation is permitted about that edge. Note: While analyzing this model using the ‘STARDYNE Advanced Analysis’ option, the following setting has to be chosen: STARDYNE Element Type:

Quadrilateral - Thin Theoretical Solution From the Reference, equation (j), the expression for deflection normal to the plate surface is:


( )

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

α

π

−

π

πν+α

−= ∑∞

= m,...5,3,1m33

2

cosha

ymcosh1

ma

xmsin

ha41tw

where a2bm

mπ

=α

From the Reference, equation (k), the expressions for bending moment per unit width are

( ) ∑∞

= α

ππ

πν−α

=,...5,3,1m m

2

x coshma

ymcosha

xmsin

h1tD4M

( ) ( ) ∑∞

= α

ππ

πν−α

−ν−α

=,...5,3,1m m

22

y coshma

ymcosha

xmsin

h1tD4

hD1tM

where α = Coefficient of Thermal Expansion t = Difference between the temperatures of the upper and lower

surfaces of the plate ν = Poisson’s ratio h = Plate thickness a = Dimension of the plate along the x1 axis b = Dimension of the plate along the x2 axis E = Elastic Modulus

( )2

3

112EhD

ν−=

STARDYNE Verification Problems 72 The numerical values used for this example are: α = 12.0E-06 / oF t = 450o F ν = 0.3 h = 0.3in. a = 12in. b = 16 in. E=10.0E6 psi

Comparison of STARDYNE and Theoretical Answers

Deflections along X = 6

Node Number

X Y Theoretical Deflection

(in)

STARDYNE Deflection

(in) 7 6 -8 0.00 0.00

20 6 -7 0.0897 0.0895 33 6 -6 0.1597 0.1593 46 6 -5 0.2132 0.2126 59 6 -4 0.2531 0.2525 72 6 -3 0.2818 0.2811 85 6 -2 0.3011 0.3004 98 6 -1 0.3122 0.3115

111 6 0 0.3158 0.3151 124 6 1 0.3122 0.3115


Deflections along Y = 1

Node Number

X Y Theoretical Deflection

(in)

STARDYNE Deflection

(in) 118 0 1 0 0 119 1 1 0.1004 0.1002 120 2 1 0.1794 0.1790 121 3 1 0.2387 0.2382 122 4 1 0.2799 0.2793 123 5 1 0.3042 0.3035 124 6 1 0.3122 0.3115 125 7 1 0.3042 0.3035 126 8 1 0.2799 0.2793 127 9 1 0.2387 0.2382 128 10 1 0.1794 0.1790 129 11 1 0.1004 0.1002 130 12 1 0 0

Bending Moments along Y = 0.5

Element Number

X Theoretical Moment (Pound-in/in)

STARDYNE Moment (Pound-in/in)

Y

Mx My Mx My 97 0.5 0.5 16.74 388.26 16.96 388.30 98 1.5 0.5 48.93 356.07 49.54 356.10 99 2.5 0.5 77.45 327.55 78.33 327.70

100 3.5 0.5 100.36 304.64 101.40 304.90 101 4.5 0.5 116.31 288.69 117.40 289.00 102 5.5 0.5 124.47 280.54 125.60 280.90 103 6.5 0.5 124.47 280.54 125.60 280.90 104 7.5 0.5 116.31 288.69 117.40 289.00 105 8.5 0.5 100.36 304.64 101.40 304.90 106 9.5 0.5 77.45 327.55 78.33 327.70 107 10.5 0.5 48.93 356.07 49.54 356.10 108 11.5 0.5 16.74 388.26 16.96 388.30


Bending Moments along X = 5.5

Theoretical Moment (Pound-in/in)

STARDYNE Moment (Pound-in/in)

Element Number

X Y

Mx My Mx My 6 5.5 -7.5 373.88 31.12 373.10 32.38

18 5.5 -6.5 311.64 93.36 311.70 94.49 30 5.5 -5.5 256.83 148.17 257.10 149.50 42 5.5 -4.5 211.14 193.87 211.70 195.10 54 5.5 -3.5 175.41 229.60 176.20 230.60 66 5.5 -2.5 149.47 255.53 150.50 256.20 78 5.5 -1.5 132.68 272.32 133.70 272.80 90 5.5 -0.5 124.47 280.54 125.60 280.90

102 5.5 0.5 124.47 280.54 125.60 280.90 114 5.5 1.5 132.68 272.32 133.70 272.80 126 5.5 2.5 149.47 255.53 150.50 256.20 138 5.5 3.5 175.41 229.59 176.20 230.60 150 5.5 4.5 211.14 193.87 211.70 195.10 162 5.5 5.5 256.83 148.17 257.10 149.50

Input File : STAAD SPACE STARDYNE VER. PROB. 12 * THERMAL LOADING ON A RECTANGULAR PLATE. UNIT POUND INCH JOINT COORD 1 0 -8 0 13 12 -8 0 REP 16 0 1 0 ELEM INCI 1 1 2 15 14 TO 12 REP 15 12 13 ELEM PROP 1 TO 192 TH 0.3 CONST E 10.0E6 ALL POISS 0.3 ALL ALPHA 12.0E-6 ALL SUPP 1 13 209 221 FIXED 2 TO 12 210 TO 220 FIXED BUT MX


14 TO 196 BY 13 26 TO 208 BY 13 FIXED BUT MY LOAD 1 TEMP LOAD 1 TO 192 TEMP 0 450 PERF ANALY PRINT JOINT DISP LIST 119 TO 129 7 TO 215 BY 13 PRINT ELEM FORCE LIST 97 TO 108 6 TO 186 BY 12 FINI


NOTES

STAAD+PRO.nrc Verfication Manual 2005

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Transcript of STAAD+PRO.nrc Verfication Manual 2005