Staad Pro Advanced Training

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D D O O N N O O T T D D I I S S T T R R I I B B U U T T E E - - P P r r i i n n t t i i n n g g f f o o r r S S t t u u d d e e n n t t U U s s e e i i s s P P e e r r m m i i t t t t e e d d S S t t u u d d e e n n t t : : W W i i r r a a H H e e r r u u c c a a k k r r a a C C o o m m p p a a n n y y : : P P T T D D i i n n a a m m i i k k a a T T e e k k n n i i k k P P e e r r s s a a d d a a C C l l a a s s s s D D a a t t e e : : 0 0 9 9 - - O O c c t t - - 2 2 0 0 1 1 2 2

STAAD.Pro Advanced

Training

STAAD.Pro V8

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D D O O N N O O T T D D I I S S T T R R I I B B U U T T E E - - P P r r i i n n t t i i n n g g f f o o r r S S t t u u d d e e n n t t U U s s e e i i s s P P e e r r m m i i t t t t e e d d S S t t u u d d e e n n t t : : W W i i r r a a H H e e r r u u c c a a k k r r a a C C o o m m p p a a n n y y : : P P T T D D i i n n a a m m i i k k a a T T e e k k n n i i k k P P e e r r s s a a d d a a C C l l a a s s s s D D a a t t e e : : 0 0 9 9 - - O O c c t t - - 2 2 0 0 1 1 2 2 Copyright Information Copyright Information

Trademarks

AccuDraw, Bentley, the “B” Bentley logo, MDL, MicroStation and SmartLine are registered AccuDraw, Bentley, the “B” Bentley logo, MDL, MicroStation and SmartLine are registered trademarks; PopSet and Raster Manager are trademarks; Bentley SELECT is a

trademarks; PopSet and Raster Manager are trademarks; Bentley SELECT is a service mark ofservice mark of Bentley Systems, Incorporated or Bentley

Bentley Systems, Incorporated or Bentley Software, Inc.Software, Inc.

Java and all Java-based trademarks and logos are trademarks or registered trademarks of Sun Java and all Java-based trademarks and logos are trademarks or registered trademarks of Sun Microsystems, Inc. in the U.S. and other

Microsystems, Inc. in the U.S. and other countries.countries. Adobe, the Adobe logo,

Adobe, the Adobe logo, Acrobat, the Acrobat logo, Distiller, Acrobat, the Acrobat logo, Distiller, Exchange, and PostScript areExchange, and PostScript are trademarks of Adobe Systems Incorporated.

trademarks of Adobe Systems Incorporated. Windows, Microsoft and Visual Basic are

Windows, Microsoft and Visual Basic are registered trademarks of Microsoft Corporation.registered trademarks of Microsoft Corporation. AutoCAD is a registered trademark of

AutoCAD is a registered trademark of Autodesk, Inc.Autodesk, Inc.

Other brands and product names are the trademarks of their respective owners. Other brands and product names are the trademarks of their respective owners.

Patents

United States Patent Nos. 5,8.15,415 and 5,784,068 and 6,199,125. United States Patent Nos. 5,8.15,415 and 5,784,068 and 6,199,125.

Copyrights

IGDS file

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1. Introduction

This course is structured to look at the scop

This course is structured to look at the scope of the advanced features of e of the advanced features of STAAD.Pro. STAAD.Pro. ThereThere are many supplied examples & few hands on exercises to complete, each one containing are many supplied examples & few hands on exercises to complete, each one containing more STAAD commands and dealing with more complex problems.

more STAAD commands and dealing with more complex problems. This docum

This document has ent has been produbeen produced for ced for your byour benefit and enefit and assistance. assistance. This documThis document isent is copyright and no part of it is to be copied, reproduced electronically or otherwise without the copyright and no part of it is to be copied, reproduced electronically or otherwise without the prior written consent of

prior written consent of Bentley Systems, inc.Bentley Systems, inc.

2. Contents

Title page ………0 Title page ………0 1. 1. Introduction ...Introduction ... ... 00 2. 2. Contents ...Contents ... ... 00 3.

3. Zero Stiffness...Zero Stiffness... 2... 2 4.

4. Understanding Understanding Instabilities ...Instabilities ... 4... 4 5.

5. Seismic Analysis Seismic Analysis Using Using UBC UBC And And IBC IBC Codes Codes ... ... 1313 YRANGE 20 21

YRANGE 20 21 FLOAD 0.3 FLOAD 0.3 ... . 1515 FINISH ...

FINISH ... ... 1717 6.

6. Calculating Mode Calculating Mode Shapes, Shapes, FrequencieFrequencies And s And ParticipatParticipation ion factors factors ... ... 2020 7.

7. Response Spectrum Response Spectrum Analysis Analysis ... ... 2727 8.

8. Time History Time History analysis of analysis of a a structure for structure for seismic seismic accelerataccelerations ions ... ... 3939 9.

9. Time History Analysis Time History Analysis for for a a Structure subjected Structure subjected to to a a Harmonic Loading Harmonic Loading ... ... 4242 JOINT LOAD

JOINT LOAD ... ... ... 4343 PERFORM

PERFORM ANALYSIS ...ANALYSIS ... 44.... 44 10.

10. Time History Time History Analysis Analysis for for a a Structure subjected Structure subjected to to a a random random excitation excitation ... ... 4747 11.

11. Hands on Exercise 1 Hands on Exercise 1 – – Dynamic AnalysDynamic Analysis ...is ... ... 4848 1)

1) Structure Structure Wizard ..Wizard ... 48.... 48 2)

2) Add Add Properties and Properties and Supports ...Supports ... 49... 49 3)

3) Create Create Time Time History History Graphs Graphs ... ... 5050 4)

4) Create a Create a Time Time History History Loadcase Loadcase ... ... 5050 Masses

Masses ... 50... 50 Time

Time History History ... 50... 50 5)

5) Viewing Viewing Mode Mode Shapes Shapes ... 51.. 51 12.

12. P-Delta Analysis ...P-Delta Analysis ... 52... 52 13.

13. P-Delta analysis including stress P-Delta analysis including stress stiffening effect of stiffening effect of the KG the KG matrix matrix ... ... 5454 Purpose

Purpose ... ... 5454 Description

Description... ... 5454 14.

14. P-Delta analysis including Small P-Delta analysis including Small Delta ...Delta ... 56... 56 Purpose

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Description –

Description – Advanced Solver ...Advanced Solver ... 6... 611 17.

17. Modal Analysis including Modal Analysis including stress stiffening effect stress stiffening effect of of KG KG Matrix Matrix ... ... 6464 Purpose

Description... ... 6464 18.

18. Non Non Linear Linear Cable/Truss Analysis ...Cable/Truss Analysis ... 66... 66 19.

19. Hands on Exercise 3 Hands on Exercise 3 - - Non-Linear Truss analysis Non-Linear Truss analysis ... 6... 699 20.

20. Hands on Exercise 4 Hands on Exercise 4 - Non-Linear Cable - Non-Linear Cable analysis - I analysis - I ... ... 7070 21.

21. Hands on Exercise 5 - Hands on Exercise 5 - Non-Linear Cable analysis –II Non-Linear Cable analysis –II ... ... 7171 22.

22. Other STAAD Other STAAD features ...features ... 7... 744 OpenSTAA

OpenSTAAD D ... ... 7474 23.

23. Other STAAD.Pro Other STAAD.Pro Optional Optional modules. modules. ... 75... 75 STAAD.beava ...

STAAD.beava ... ... 7575 STAAD.found

STAAD.foundation ation ... ... 7575 Offshore

Offshore Loading Loading Program ...Program ... 75... 75 Section

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STAAD.Pro Advanced Training

3. 3. Zero

Zero Stiffness

Stiffness

Question :

Question : What does a zero stiffness warning message in the STAAD output fileWhat does a zero stiffness warning message in the STAAD output file mean?

mean?

Answer

Answer :: The procedure uThe procedure used by STAsed by STAAD in calculating AD in calculating displacements and displacements and forcesforces in a structure is the stiffness method. One of the steps involved in this method is the in a structure is the stiffness method. One of the steps involved in this method is the assembly of the global stiffness matrix. During this process, STAAD verifies that assembly of the global stiffness matrix. During this process, STAAD verifies that nono active degree of freedom (d.o.f) has a zero value, because a zero value could be a active degree of freedom (d.o.f) has a zero value, because a zero value could be a po

potetentntiaial l cacaususe e of of ininststababililitity y in in ththe e momodedel l alalonong g ththat at d.d.o.o.f. f. It It memeanans s ththat at ththee structural conditions which exist at that node and degree of freedom result in the structural conditions which exist at that node and degree of freedom result in the structure having no ability to resist

structure having no ability to resist a load acting along that d.o.f.a load acting along that d.o.f.

A warning message is printed in the STAAD output file highlighting the node A warning message is printed in the STAAD output file highlighting the node number and the d.o.f at which the zero stiffness

number and the d.o.f at which the zero stiffness condition exists.condition exists.

Question :

Question : What are examples of cases which give rise to these conditions? What are examples of cases which give rise to these conditions?

Answer :

Answer : Consider Consider a a frame frame structure structure where where some some of of the the members members are dare defined efined toto be

be trtrususseses. s. On On ththis is momodedel, l, if if a a jojoinint t exexisists ts whwherere e ththe e ononly ly ststruructcturural al cocompmpononenentsts connected at that node are truss members, there is no rotational stiffness at that connected at that node are truss members, there is no rotational stiffness at that nodenode along any of the global d.o.f. If the structure is defined as STAAD PLANE, it will along any of the global d.o.f. If the structure is defined as STAAD PLANE, it will result in a warning along the MZ d.o.f at that node. If it were declared as STAAD result in a warning along the MZ d.o.f at that node. If it were declared as STAAD SPACE, there will be at least 3 warnings, one for each of MX, MY and MZ, and SPACE, there will be at least 3 warnings, one for each of MX, MY and MZ, and pe

perhrhapaps s adaddidititiononal al wawarnrninings gs fofor r ththe e trtrananslslatatioionanal l d.d.o.o.f.f.

These warnings can also appear when other structural conditions such as member These warnings can also appear when other structural conditions such as member releases and element releases deprive the structure of stiffness at the associated releases and element releases deprive the structure of stiffness at the associated nodes along the global translational or rotational directions. A tower held down by nodes along the global translational or rotational directions. A tower held down by cables, defined as a PLANE or SPACE frame, where cable members are pinned cables, defined as a PLANE or SPACE frame, where cable members are pinned supported at their base will also generate these warnings for the rotational d.o.f. at supported at their base will also generate these warnings for the rotational d.o.f. at the supported nodes of the cables. In a SPACE frame structure, connections may be the supported nodes of the cables. In a SPACE frame structure, connections may be modeled in such a manner that all members meeting at any given node have a modeled in such a manner that all members meeting at any given node have a moment release along all 3 axes. The joint is thus deprived of any rotational moment release along all 3 axes. The joint is thus deprived of any rotational stiffness.

stiffness.

Solid elements have no rotational stiffness at

Solid elements have no rotational stiffness at their nodes. So, at all their nodes. So, at all nodes where younodes where you have only solids, these zero stiffness warning messages may appear.

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Question :

Question : Why Why are thare these ese warnings warnings and and not not errors?errors?

Answer :

Answer : The The reason reason why why these these conditions conditions are are reported reported as as warnings warnings and and notnot errors is due to the fact that they may not necessarily be detrimental to the proper errors is due to the fact that they may not necessarily be detrimental to the proper transfer of loads from the structure to the supports. If no load acts at and along the transfer of loads from the structure to the supports. If no load acts at and along the d.o.f where the stiffness is zero,

d.o.f where the stiffness is zero, that point may not be a that point may not be a trouble-spot.trouble-spot.

Question :

Question : What is the usefulness of these messages :What is the usefulness of these messages :

Answer :

Answer : A A zero zero stiffness stiffness message message can can be be a a tool tool for for investigating investigating the the cause cause ofof

instabilities in the model. An instability is a condition where a load applied on the structure is instabilities in the model. An instability is a condition where a load applied on the structure is not able to make its way into the supports because no paths exist for the load to flow through, not able to make its way into the supports because no paths exist for the load to flow through, and may result in a lack of equilibrium between the applied load and the support reaction. A and may result in a lack of equilibrium between the applied load and the support reaction. A zero stiffness message can tell us whether any of those d.o.f are obstacles to the flow of the zero stiffness message can tell us whether any of those d.o.f are obstacles to the flow of the load.

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4. 4. Understanding

Understanding Instabilities

Instabilities

Question :

Question : I have instability warning messages in my output file like that shownI have instability warning messages in my output file like that shown be

belolow. w. WhWhat at arare e ththesese?e?

***WARNING - INSTABILITY AT JOINT 26 DIRECTION = ***WARNING - INSTABILITY AT JOINT 26 DIRECTION = FXFX PROBABLE CAUSE SINGULAR-ADDING WEAK SPRING PROBABLE CAUSE SINGULAR-ADDING WEAK SPRING K-MATRIX DIAG= 5.3274384E+03 L-MATRIX DIAG=

K-MATRIX DIAG= 5.3274384E+03 L-MATRIX DIAG= 0.0000000E0.0000000E+00 EQN NO +00 EQN NO 127127 ***NOTE - VERY WEAK SPRING ADDED FOR STABILITY

***NOTE - VERY WEAK SPRING ADDED FOR STABILITY

**NOTE** STAAD DETECTS INSTABILITIES AS EXCESSIVE LOSS OF **NOTE** STAAD DETECTS INSTABILITIES AS EXCESSIVE LOSS OF SIGNIFICANT DIGITS

SIGNIFICANT DIGITS

DURING DECOMPOSITION. WHEN A DECOMPOSED DIAGONAL IS

DURING DECOMPOSITION. WHEN A DECOMPOSED DIAGONAL IS LESS THANLESS THAN THE

THE

BUILT-IN REDUCTION FACTOR TIMES THE

BUILT-IN REDUCTION FACTOR TIMES THE ORIGINAL STIFFNESS MATRIXORIGINAL STIFFNESS MATRIX DIAGONAL,

DIAGONAL,

STAAD PRINTS A SINGULARITY NOTICE.

STAAD PRINTS A SINGULARITY NOTICE. THE BUILT-IN REDUCTION FACTORTHE BUILT-IN REDUCTION FACTOR IS 1.000E-09

IS 1.000E-09

THE ABOVE CONDITIONS COULD ALSO BE CAUSED BY VERY STIFF OR

THE ABOVE CONDITIONS COULD ALSO BE CAUSED BY VERY STIFF OR VERYVERY WEAK

WEAK

ELEMENTS AS WELL AS

ELEMENTS AS WELL AS TRUE SINGULARITIES.TRUE SINGULARITIES.

Answer

Answer :: An instability is a An instability is a condition where a load applied on the structure is condition where a load applied on the structure is notnot able to make its way into the supports because no paths exist for the load to flow able to make its way into the supports because no paths exist for the load to flow through, and may result in a lack of equilibrium between the applied load and the through, and may result in a lack of equilibrium between the applied load and the support reaction.

support reaction.

Examples and causes of Instability Examples and causes of Instability : : Defining a member as a

Defining a member as a TRUSS when it needs shear and TRUSS when it needs shear and bending capacity. A framedbending capacity. A framed structure with columns and beams where the columns are defined as "TRUSS" structure with columns and beams where the columns are defined as "TRUSS" members is definitely a cause of instability. Such a column has no capacity to members is definitely a cause of instability. Such a column has no capacity to transfer shears or moments from the regions above it

transfer shears or moments from the regions above it to the supports.to the supports. When you declare all members connecting at

When you declare all members connecting at specific nodes to be truss specific nodes to be truss members, themembers, the alignment of the members must be such that the axial force from each member must alignment of the members must be such that the axial force from each member must be

be abable le to to mamake ke itits s waway y ththrorougugh h ththe e cocommmmon on nonode de to to ththe e ototheher r memembmberers. s. FoForr example, if you have 3 members meeting at a point, one of them is purely vertical example, if you have 3 members meeting at a point, one of them is purely vertical and the other 2 are purely horizontal, and they are all truss members, the axial force and the other 2 are purely horizontal, and they are all truss members, the axial force from the vertical member cannot be transmitted into the horizontal members. On the from the vertical member cannot be transmitted into the horizontal members. On the other hand, if they are frame members, the load will be transmitted into the other hand, if they are frame members, the load will be transmitted into the horizontals in the form of shear. This is an inherent weak point of trusses, and a horizontals in the form of shear. This is an inherent weak point of trusses, and a po

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A better option to calling a member a TRUSS is to define it as a frame member and A better option to calling a member a TRUSS is to define it as a frame member and use partial moment releases at its ends.

use partial moment releases at its ends.

Improper support conditions

Improper support conditions. When the supports of the structure are such that they. When the supports of the structure are such that they cannot offer any resistance to sliding or overturning of the structure in one or more cannot offer any resistance to sliding or overturning of the structure in one or more directions. For example, a 2D structure (frame in the XY plane) that is defined as a directions. For example, a 2D structure (frame in the XY plane) that is defined as a SPACE FRAME with pinned supp

SPACE FRAME with pinned supports and subjected to a orts and subjected to a force in the Z force in the Z direction willdirection will topple over about the X-axis. Another example is that of a space frame with all the topple over about the X-axis. Another example is that of a space frame with all the supports released for FX, FY or FZ.

supports released for FX, FY or FZ.

Connecting a very stiff member to a very flexible member

Connecting a very stiff member to a very flexible member. A math precision error. A math precision error is caused when numerical instabilities occur in the matrix decomposition (inversion) is caused when numerical instabilities occur in the matrix decomposition (inversion) pr

prococesess. s. OnOne e of of ththe e tetermrms s of of ththe e eqequiuililibrbriuium m eqequauatition on tatakekes s ththe e foform rm 1/1/(1(1-A-A), ), whwhereree A=k1/(k1+k2); k1 and k2 being the stiffness coefficients of two adjacent members. A=k1/(k1+k2); k1 and k2 being the stiffness coefficients of two adjacent members. When a very "stiff" member is adjacent to a very "flexible" member, viz., when When a very "stiff" member is adjacent to a very "flexible" member, viz., when k1>>k2, or k1+k2 .k1, A=1 and hence, 1/(1-A) =1/0. Thus, huge variations in k1>>k2, or k1+k2 .k1, A=1 and hence, 1/(1-A) =1/0. Thus, huge variations in stiffnesses of adjacent members are not permitted. Artificially high E or I values stiffnesses of adjacent members are not permitted. Artificially high E or I values should be reduced when this occurs. Math precision errors are also caused when the should be reduced when this occurs. Math precision errors are also caused when the units of length and force are not defined correctly for member lengths, member units of length and force are not defined correctly for member lengths, member pr

propoperertitieses, , coconsnstatantnts s etetc.c.

Excessive number of releases

Excessive number of releases. Releases completely deprive a . Releases completely deprive a member of any abilitymember of any ability to transmit a particular type of force or moment to the next member. Imagine for to transmit a particular type of force or moment to the next member. Imagine for example, a portal frame that looks like a table, with columns pinned at their base, example, a portal frame that looks like a table, with columns pinned at their base, and each column attached to 2 orthogonal beams at the

and each column attached to 2 orthogonal beams at the top.top.

If the beams are pinned connected to top of the column, it is customary to specify If the beams are pinned connected to top of the column, it is customary to specify releases on the beams along the lines

releases on the beams along the lines

2 3 START MX MY MZ 2 3 START MX MY MZ

The above release signifies that 100% of the resistance to

The above release signifies that 100% of the resistance to MX, MY and MZ has beenMX, MY and MZ has been switched off at the beam-ends. The beam is hence behaving as a simply supported switched off at the beam-ends. The beam is hence behaving as a simply supported be

beam am at at ththat at lolocacatitionon..

This condition, along with the pinned column base, deprives the column of any This condition, along with the pinned column base, deprives the column of any ability to transmit torsion to the base, leading to instability about the global MY ability to transmit torsion to the base, leading to instability about the global MY degree of freedom at the pinned support.

degree of freedom at the pinned support.

Improper connection between members.

Improper connection between members. When members cross each in space, if a When members cross each in space, if a connection exists between 2 members, that

connection exists between 2 members, that point of contact should be rpoint of contact should be r epresented byepresented by a common node between the members. Simply because lines appear to cross each a common node between the members. Simply because lines appear to cross each other in space, it doesn’t guarantee that STAAD will assume a connection between other in space, it doesn’t guarantee that STAAD will assume a connection between those members. The user has to ensure that. One tool for creating such common those members. The user has to ensure that. One tool for creating such common nodes is available under the Geometry menu. It is

nodes is available under the Geometry menu. It is called Intersect Selected Members.called Intersect Selected Members.

Duplicate nodes.

Duplicate nodes. They are 2 or more nodes, having distinct node numbers, but the They are 2 or more nodes, having distinct node numbers, but the same X, Y, Z coordinates. For example, if node number 5 has coordinates of (7, 10, same X, Y, Z coordinates. For example, if node number 5 has coordinates of (7, 10,

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duplicate. If you have 2 members, one attached to node 5, and the other to node 83, duplicate. If you have 2 members, one attached to node 5, and the other to node 83, then, those 2 members are not connected to each other at that point in space. Go to then, those 2 members are not connected to each other at that point in space. Go to Tools – Check Duplicate Nodes to detect and merge such sets of nodes into a single Tools – Check Duplicate Nodes to detect and merge such sets of nodes into a single node.

node.

Improper connection between members and plate elements.

Improper connection between members and plate elements. In the figure shown In the figure shown be

belolow, w, ththe e bebeam am gogoes es frfrom om nonode de 5 5 to to nonode de 6. 6. ThThe e elelememenent t is is coconnnnececteted d bebetwtweeeen n 2, 2, 3,3, 4 and 1. Thus, the beam has no common nodes with the element. No transfer of loads 4 and 1. Thus, the beam has no common nodes with the element. No transfer of loads is possible between these entities.

is possible between these entities.

In order for the above set In order for the above set of entities to be properly of entities to be properly connected, the element connected, the element would have to be broken would have to be broken into 2, and the beam too into 2, and the beam too needs to be split at node needs to be split at node 2, as shown below.

2, as shown below.

While there are no simple While there are no simple tools for splitting tools for splitting elements, using finer meshes

elements, using finer meshes of elements always helps. of elements always helps. See the Generate Plate Mesh See the Generate Plate Mesh and Generate Surface and Generate Surface Meshing options of the Meshing options of the Geometry menu. A beam in Geometry menu. A beam in the situation above may be the situation above may be br

brokoken en up up ininto to pipiececes es byby using means like Insert Node, using means like Insert Node, or Break Beams at Selected or Break Beams at Selected No

Nodedes, s, boboth th of of whwhicich h arare e inin the Geometry menu.

the Geometry menu.

Overlapping members

Overlapping members. When 2 members are collinear, and further, at least one of. When 2 members are collinear, and further, at least one of the nodes of one of those members happens to lie within the span of the other, but the nodes of one of those members happens to lie within the span of the other, but the 2 members are not connected at that node, those 2 members are considered as the 2 members are not connected at that node, those 2 members are considered as overlapping collinear members. In STAAD, the tool for detecting such members is overlapping collinear members. In STAAD, the tool for detecting such members is Tools – Check Overlapping collinear members.

Tools – Check Overlapping collinear members.

An example of 2 members which would qualify as overlapping collinear is: An example of 2 members which would qualify as overlapping collinear is:

STAAD SPACE STAAD SPACE UNIT FEET KIP UNIT FEET KIP JOINT

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MEMBER INCIDENCES MEMBER INCIDENCES 1 1 2; 2 2 3; 3 3 4; 1 1 2; 2 2 3; 3 3 4; 101 5 6 101 5 6 FINISH FINISH

Here, members 2 and 101 are overlapping collinear. Member 2 is entirely confined Here, members 2 and 101 are overlapping collinear. Member 2 is entirely confined within the span of member 101, and collinear, but they are not attached to each within the span of member 101, and collinear, but they are not attached to each other.

other.

Another example is: Another example is:

STAAD SPACE STAAD SPACE UNIT FEET KIP UNIT FEET KIP JOINT

JOINT COORDINATECOORDINATESS

1 0 0 0; 2 0 10 0; 3 10 10 0; 4 10 0 1 0 0 0; 2 0 10 0; 3 10 10 0; 4 10 0 0; 5 13 10 0; 6 -4 10 0;0; 5 13 10 0; 6 -4 10 0; MEMBER INCIDENCES MEMBER INCIDENCES 1 1 2; 2 2 3; 3 3 4; 1 1 2; 2 2 3; 3 3 4; 101 2 5 101 2 5 FINISH FINISH

Here, again, members 2 and 101 are overlapping collinear. But even though they are Here, again, members 2 and 101 are overlapping collinear. But even though they are connected to each other at node 2,

connected to each other at node 2, again member 2 is entirely confined within theagain member 2 is entirely confined within the span of member 101, and collinear.

span of member 101, and collinear.

Overlapping plates.

Overlapping plates. These are elements whose nodes intersect other elements at These are elements whose nodes intersect other elements at po

poinints ts ototheher r ththan an ththe e dedefifinened d nonodedes. s. ThThis is enentatailils s plplatates es whwhosose e bobounundadariries es wiwithth adjacent plates are not attached at the nodes or plates

adjacent plates are not attached at the nodes or plates within other plates (in the samewithin other plates (in the same pl

planane)e)..

The figure above represents such a The figure above represents such a condition. Elements 1 and 2 share only one condition. Elements 1 and 2 share only one common node which is node 4. Though the common node which is node 4. Though the drawing appears to indicate a common drawing appears to indicate a common boundary along nodes 4, 5 and 3, there is no boundary along nodes 4, 5 and 3, there is no connection along that boundary. From the connection along that boundary. From the Tools menu, choose Check Overlapping Tools menu, choose Check Overlapping Plates to detect such conditions in the model. Plates to detect such conditions in the model. The next figure shows what needs to be done The next figure shows what needs to be done

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to ensure proper connection. Our to ensure proper connection. Our original element 1 is converted to 3 original element 1 is converted to 3 triangular elements to accomplish it. triangular elements to accomplish it.

Question :

Question : If there are instability mIf there are instability messages, does it messages, does it mean my analyean my analysis results maysis results may be

be ununsasatitisfsfacactotoryry??

Answer :

Answer : There There are are many many situations situations where where instabilities instabilities are are unimportanunimportant t and and thethe STAAD approach of adding a weak spring is an ideal solution to the problem. For STAAD approach of adding a weak spring is an ideal solution to the problem. For example, sometimes an engineer will release the MX torsion in a single beam or at example, sometimes an engineer will release the MX torsion in a single beam or at the ends of a series of members such that technically the members are unstable in the ends of a series of members such that technically the members are unstable in torsion. If there is no torque applied, this singularity can safely be "fixed" by torsion. If there is no torque applied, this singularity can safely be "fixed" by STAAD with a weak torsional

STAAD with a weak torsional spring.spring.

Similarly a column that is at a pinned support will sometimes be connected to Similarly a column that is at a pinned support will sometimes be connected to members that all have releases such that they cannot transmit moments that cause members that all have releases such that they cannot transmit moments that cause torsion in the column. This column will be unstable in torsion but can be safely torsion in the column. This column will be unstable in torsion but can be safely "fixed" by STAAD with a

"fixed" by STAAD with a weak torsional spring.weak torsional spring.

Sometimes however, a section of a structure has members that are overly released to Sometimes however, a section of a structure has members that are overly released to the point where that section can rotate

the point where that section can rotate with respect to the rest of with respect to the rest of the structure. In thisthe structure. In this case, if STAAD adds a weak spring, there may be large displacements because there case, if STAAD adds a weak spring, there may be large displacements because there are loads in the section that are in the direction of the extremely weak spring. are loads in the section that are in the direction of the extremely weak spring. Another way of saying it is, an applied load acts along an unstable degree of Another way of saying it is, an applied load acts along an unstable degree of freedom, and causes excessive displacements at that degree

freedom, and causes excessive displacements at that degree of freedom.of freedom.

Question :

Question : If there are instability If there are instability messages, are there messages, are there any simple any simple checks to checks to verifyverify whether my analysis results are sa

whether my analysis results are sa tisfactory?tisfactory?

Answer

Answer :: There are 2 important checks that should be carried out if instabilityThere are 2 important checks that should be carried out if instability messages are present.

messages are present. a.

a. A static equilibrium check. This check will tell us whether all the appliedA static equilibrium check. This check will tell us whether all the applied loading flowed through the model into the supports. A satisfactory result would loading flowed through the model into the supports. A satisfactory result would require that the applied loading be in

require that the applied loading be in equilibrium with the support reactions.equilibrium with the support reactions. b.

b. The joint displacement check. This check will tell us whether the displacementsThe joint displacement check. This check will tell us whether the displacements in the model are within reasonable limits. If a load passes through a corresponding in the model are within reasonable limits. If a load passes through a corresponding

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unstable degree of freedom, the structure will undergo excessive deflections at that unstable degree of freedom, the structure will undergo excessive deflections at that degree of freedom.

degree of freedom.

One may use the PRINT STATICS CHECK option in conjunction with the One may use the PRINT STATICS CHECK option in conjunction with the PERFORM ANALYSIS command to obtain a report of both the results mentioned in PERFORM ANALYSIS command to obtain a report of both the results mentioned in the above checks. The STAAD output file will contain a report similar to the the above checks. The STAAD output file will contain a report similar to the following, for every primary load case that has been

following, for every primary load case that has been solved for :solved for :

***TOTAL APPLIED LOAD ( KG METE )

***TOTAL APPLIED LOAD ( KG METE ) SUMMARY (LOADING 1 )SUMMARY (LOADING 1 ) SUMMATION FORCE-X = 0.00 SUMMATION FORCE-X = 0.00 SUMMATION FORCE-Y = -817.84 SUMMATION FORCE-Y = -817.84 SUMMATION FORCE-Z = 0.00 SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE SUMMATION OF MOMENTS AROUND THE ORIGIN-MX= 291.23 MY= 0.00 MZ= -3598.50

MX= 291.23 MY= 0.00 MZ= -3598.50

***TOTAL REACTION LOAD( KG METE ) SUMMARY (LOADING 1 ***TOTAL REACTION LOAD( KG METE ) SUMMARY (LOADING 1 )) SUMMATION FORCE-X = 0.00 SUMMATION FORCE-X = 0.00 SUMMATION FORCE-Y = 817.84 SUMMATION FORCE-Y = 817.84 SUMMATION FORCE-Z = 0.00 SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE SUMMATION OF MOMENTS AROUND THE ORIGIN-MX= -291.23 MY= 0.00 MZ= 3598.50

MX= -291.23 MY= 0.00 MZ= 3598.50

MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 1) MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 1) MAXIMUMS AT NODE MAXIMUMS AT NODE X = 1.00499E-04 25 X = 1.00499E-04 25 Y = -3.18980E-01 12 Y = -3.18980E-01 12 Z = 1.18670E-02 23 Z = 1.18670E-02 23 RX= 1.52966E-04 5 RX= 1.52966E-04 5

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RY= 1.22373E-04 23 RY= 1.22373E-04 23 RZ= 1.07535E-03 8 RZ= 1.07535E-03 8

Go through these numbers to ensure that Go through these numbers to ensure that i.

i. The "TOTAL APPLIED LOAD" values and "TOTAL REACTION LOAD"The "TOTAL APPLIED LOAD" values and "TOTAL REACTION LOAD" values are equal and opposite.

values are equal and opposite. ii

ii.. The "MAXIMUM DISPLACEMENThe "MAXIMUM DISPLACEMENTS" are TS" are within reasonable limits.within reasonable limits. No

Note te ththat at frfrom om STSTAAAAD.D.PrPro o 202007 07 ononwawardrds s ththe e ststataticics s chchececk k reresusultlts s tatablblee

automatically appears in the post-processing mode – node – support reactions page. automatically appears in the post-processing mode – node – support reactions page.

Question :

Question : What is the mWhat is the meaning of this eaning of this message, "Probmessage, "Probable cause warningable cause warning-near-near singular"

singular"

Answer :

Answer : While While performing performing the the triangular triangular factorization factorization of of the the global global stiffnessstiffness matrix, a diagon

matrix, a diagonal matrix is computedal matrix is computed. . These compuThese computed diagonals are the samted diagonals are the same as ore as or smaller than the global stiffness m

smaller than the global stiffness matrix diagonals. atrix diagonals. If the computed dIf the computed diagonals becomeiagonals become zero then the matrix is sing

zero then the matrix is singular and the structuular and the structure is unstable. re is unstable. In STAAD In STAAD we say thatwe say that the structure is unstable/singular if any computed diagonal is less that (1.E-9) * (the the structure is unstable/singular if any computed diagonal is less that (1.E-9) * (the corresponding

corresponding stiffness matrix stiffness matrix diagonal). diagonal). Likewise in Likewise in STAAD STAAD we say we say that thethat the structure is nearly unstable/singular if any computed diagonal is less that (1.E-7) * structure is nearly unstable/singular if any computed diagonal is less that (1.E-7) * (the corresponding stiffness matrix diagonal).

(the corresponding stiffness matrix diagonal). If the overall results

If the overall results look OK, then ignore nearly singular messages.look OK, then ignore nearly singular messages.

Question :

Question : How to avoid instabHow to avoid instabilities if TRUSSES oilities if TRUSSES or RELEASEr RELEASES are the cause?S are the cause?

Answer

Answer :: There is a rather simple way to eliminate instabilities, especially if trussThere is a rather simple way to eliminate instabilities, especially if truss members are present or when MEMBER RELEASE commands are used and certain members are present or when MEMBER RELEASE commands are used and certain degrees of freedom are subjected to a

degrees of freedom are subjected to a 100% release.100% release.

In reality, connections always have some amount of force

In reality, connections always have some amount of force and moment capacity. Useand moment capacity. Use PARTIAL RELEASES to enable the connection to retain at least a very small PARTIAL RELEASES to enable the connection to retain at least a very small amount of capacity. This is a mechanism by which you can declare that, at the start amount of capacity. This is a mechanism by which you can declare that, at the start node or end node of a

node or end node of a member, rather than fully eliminating the stiffness for a member, rather than fully eliminating the stiffness for a certaincertain moment degree of freedom (d.o.f), you are willing to allow the member to have a moment degree of freedom (d.o.f), you are willing to allow the member to have a small amount of stiffness for that d.o.f. The advantage of this command is that the small amount of stiffness for that d.o.f. The advantage of this command is that the extent of the release

extent of the release is controlled by you.is controlled by you.

For example, if member 5, has a pinned connection at its

For example, if member 5, has a pinned connection at its start node, if you specifystart node, if you specify

5 START MY MZ 5 START MY MZ

it means MY and MZ are 100% released at the start node. But if you say, it means MY and MZ are 100% released at the start node. But if you say,

5 START MP 0.99 5 START MP 0.99

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you are saying that the

you are saying that the bending and torsional stiffnesses are 99% less bending and torsional stiffnesses are 99% less than what theythan what they would be fo

would be for a fully momenr a fully moment t resistant connection. Thresistant connection. Thus, the 1% avus, the 1% available stiffnessailable stiffness might be adequate to allow the

might be adequate to allow the load to pass load to pass through the node from one member to thethrough the node from one member to the other.

other.

So, this is what may be

So, this is what may be done :done : a.

a. Change the declaration of the truss Change the declaration of the truss members in your model frommembers in your model from

MEMBER TRUSS MEMBER TRUSS to to MEMBER RELEASE MEMBER RELEASE memb-list START MP 0.99 memb-list START MP 0.99 memb-list END MP 0.99 memb-list END MP 0.99 or or MEMBER RELEASE MEMBER RELEASE memb-list Both MP 0.99 memb-list Both MP 0.99 b.

b. Run the analysis. Check to make sure the instability warnings no longer appear.Run the analysis. Check to make sure the instability warnings no longer appear. Then check your nodal displacements.

Then check your nodal displacements. c.

c. If the displacements are large, reduce the extent of the release from 0.99 to sayIf the displacements are large, reduce the extent of the release from 0.99 to say 0.98.

0.98.

Repeat steps (b) and (c) by progressively reducing the extent of the release until the Repeat steps (b) and (c) by progressively reducing the extent of the release until the displacements are satisfactory. When they look reasonable, check the magnitude of displacements are satisfactory. When they look reasonable, check the magnitude of the moments and shear at the nodes of those members and make sure that the the moments and shear at the nodes of those members and make sure that the connection will be able to handle those forces

connection will be able to handle those forces and moments.and moments. STAAD.P

STAAD.Pro 2002 onwards, you can apply these ro 2002 onwards, you can apply these partial releases partial releases to individual momentto individual moment degrees of freedom. For example, you could say

degrees of freedom. For example, you could say

MEMBER RELEASE MEMBER RELEASE

memb-list Both MPX 0.99 MPY 0.97 MPZ 0.95 memb-list Both MPX 0.99 MPY 0.97 MPZ 0.95

This flexibility permits you to adjust just the specific degree of freedom that is the This flexibility permits you to adjust just the specific degree of freedom that is the pr

proboblelem m arareaea..

You can refer to Section 5.22.1 of the Technical Reference

You can refer to Section 5.22.1 of the Technical Reference Manual for details.Manual for details.

Question :

Question : Is there any graphIs there any graphical facility in STAAical facility in STAAD by which D by which I can examine I can examine thethe po

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