STAAD PRO TRAINING MANUAL

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STAAD.Pro

TRAINING MANUAL ADVANCED TOPICS

A Bentley Solutions Center

www.reiworld.com

www.bentley.com/staad

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STAAD. Pro is a suite of propr ietary computer programs of

Research Engineers, a Bentley Solutions Center. Although

every effort has been made to ensure the correctness of theseprograms, REI will not accept responsibility for any mistake,

error or misrepresentation in or as a result of the usage of

these programs.

© 2006 Bentley Systems, Incorpor ated. All Rights Reserved.

Published October, 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 connectiondesign, 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 ManualThis 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

Modeling Problems

• Zero Stiffness Conditions

• Understanding Instabilities

Dynamic Analysis

• Seismic Analysis using UBC and IBC codes

• Calculating mode shapes, frequencies, participation factors

• Response Spectrum Analysis

• Time History Analysis for seismic accelerations

• Time History Analysis subjected to a harmonic loading

• Time History Analysis subjected to a random excitation

Mat Foundations

• Automatic Spring Support Generation

• Modeling soil supports as compression only

• Viewing soil pressure diagrams and intensities

Load Generation• Moving Loads

• Floor Loads

• Wind Loads





Zero Stiffness





1

Question : What does a zero stiffness warning message in the STAAD output

file mean?

Answer : The procedure used by STAAD in calculating displacements and

forces 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 no active degree

of freedom (d.o.f) has a zero value, because a zero value could be

a potential cause of instability in the model along that d.o.f. It

means that the structural conditions which exist at that node and

degree of freedom result in the structure having no ability to resist

a load acting along that d.o.f.

A warning message is printed in the STAAD output file

highlighting the node number and the d.o.f at which the zero

stiffness condition exists.

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

Answer : Consider a frame structure where some of the members are defined

to be trusses. On this model, if a joint exists where the only

structural components connected at that node are truss members,there is no rotational stiffness at that node 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 SPACE, there will be at least 3 warnings, one for each of

MX, MY and MZ, and perhaps additional warnings for the

translational d.o.f.

These warnings can also appear when other structural conditions

such as member 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 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 the supported nodes of the cables.



STAAD.Pro Training Manual – Advanced Topics

2

In a SPACE frame structure, connections may be 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 anyrotational stiffness.

Solid elements have no rotational stiffness at their nodes. So, at all

nodes where you have only solids, these zero stiffness warning

messages may appear.

Question : Why are these warnings and not errors?

Answer : The reason why these conditions are reported as warnings and not

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 d.o.f where the stiffness

is zero, that point may not be a trouble-spot.

Question : What is the usefulness of these messages :

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

of instabilities in the model. An instability is a condition where a

load applied on the structure is not able to make its way into thesupports 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 zero stiffness message can tell us whether

any of those d.o.f are obstacles to the flow of the load.



Understanding Instabilities





1

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

shown below. What are these?

***WARNING - INSTABILITY AT JOINT 26 DIRECTION = FX

PROBABLE CAUSE SINGULAR-ADDING WEAK SPRING

K-MATRIX DIAG= 5.3274384E+03 L-MATRIX DIAG= 0.0000000E+00 EQN

NO 127

***NOTE - VERY WEAK SPRING ADDED FOR STABILITY

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

SIGNIFICANT DIGITSDURING DECOMPOSITION. WHEN A DECOMPOSED DIAGONAL IS

LESS THAN 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 STIFFOR VERY WEAK

ELEMENTS AS WELL AS TRUE SINGULARITIES.

Answer : 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, and may result in a lack of

equilibrium between the applied load and the support reaction.

Examples and causes of Instability :

Defining a member as a TRUSS when it needs shear and bending

capacity. A framed 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 transfer shears or

moments from the regions above it to the supports.

When you declare all members connecting at specific nodes to be

truss members, the alignment of the members must be such that the

axial force from each member must be able to make its way

through the common node to the other members. For 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




2

members, the axial force 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 thehorizontals in the form of shear. This is an inherent weak point of

trusses, and a potential cause of instability.

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.

Improper support conditions. When the supports of the structureare such that they 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 SPACE FRAME with pinned supports and subjected to a force in

the Z direction will topple over about the X-axis. Another example

is that of a space frame with all the supports released for FX, FY

or FZ.

Connecting a very stiff member to a very flexible member. A math

precision error is caused when numerical instabilities occur in the

matrix decomposition (inversion) process. One of the terms of the

equilibrium equation takes the form 1/(1-A), where A=k1/(k1+k2);

k1 and k2 being the stiffness coefficients of two adjacentmembers. 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 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 units of length and force are not defined correctly

for member lengths, member properties, constants etc.

Excessive number of releases. Releases completely deprive a

member of any ability 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, and each

column attached to 2 orthogonal beams at the top.




3

If the beams are pinned connected to top of the column, it is

customary to specify releases on the beams along the lines

2 3 START MX MY MZ

The above release signifies that 100% of the resistance to MX, MY

and MZ has been switched off at the beam-ends. The beam is

hence behaving as a simply supported beam at that location.

This condition, along with the pinned column base, deprives thecolumn of any ability to transmit torsion to the base, leading to

instability about the global MY degree of freedom at the pinned

support.

Improper connection between members. When members cross each

in space, if a connection exists between 2 members, that point of

contact should be represented by 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 those members. The user has to ensure that. One tool for

creating such common nodes is available under the Geometry

menu. It is called Intersect Selected Members.

Duplicate nodes. 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, 0), and node 83 also has

coordinates of (7, 10, 0), node 5 and 83 are considered 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 Tools – Check Duplicate Nodes to detectand merge such sets of nodes into a single node.

Improper connection between members and plate elements. In the

figure shown below, the beam goes from node 5 to node 6. The

element is connected between 2, 3, 4 and 1. Thus, the beam has no

common nodes with the element. No transfer of loads is possible

between these entities.




4

In order for the above set of entities to be properly connected, the

element would have to be broken into 2, and the beam too needs to

be split at node 2, as shown below.

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

meshes of elements always helps. See the Generate Plate Mesh and

Generate Surface Meshing options of the Geometry menu. A beam

in the situation above may be broken up into pieces by using

means like Insert Node, or Break Beams at Selected Nodes, both of

which are in the Geometry menu.




5

Overlapping members. 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 2 members are not connectedat that node, those 2 members are considered as overlapping

collinear members. In STAAD, the tool for detecting such

members is Tools – Check Overlapping collinear members.

An example of 2 members which would qualify as overlapping

collinear is:

STAAD SPACE

UNIT FEET KIP

JOINT COORDINATES

1 0 0 0; 2 0 10 0; 3 10 10 0; 4 10 0 0; 5 13 10 0; 6 -4 10 0;

MEMBER INCIDENCES

1 1 2; 2 2 3; 3 3 4;101 5 6

FINISH

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 other.

Another example is:

STAAD SPACE

UNIT FEET KIP

JOINT COORDINATES1 0 0 0; 2 0 10 0; 3 10 10 0; 4 10 0 0; 5 13 10 0; 6 -4 10 0;

MEMBER INCIDENCES

1 1 2; 2 2 3; 3 3 4;

101 2 5

FINISH

Here, again, members 2 and 101 are overlapping collinear. But

even though they are connected to each other at node 2, again




6

member 2 is entirely confined within the span of member 101, and

collinear.

Overlapping plates. These are elements whose nodes intersect

other elements at points other than the defined nodes. This entails

plates whose boundaries with adjacent plates are not attached at

the nodes or plates within other plates (in the same plane).

The figure above represents such a condition. Elements 1 and 2

share only one common node which is node 4. Though the drawing

appears to indicate a common boundary along nodes 4, 5 and 3,

there is no connection along that boundary. From the Tools menu,

choose Check Overlapping Plates to detect such conditions in the

model. The next figure shows what needs to be done to ensure

proper connection. Our original element 1 is converted to 3

triangular elements to accomplish it.




7

Question : If there are instability messages, does it mean my analysis results

may be unsatisfactory?

Answer : There are many situations where instabilities are unimportant and

the 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 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 STAAD with a weak torsional spring.

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 torsion in the column. This column

will be unstable in torsion but can be safely "fixed" by STAAD

with a weak torsional spring.

i i l d d i




8

Sometimes however, a section of a structure has members that are

overly released to the point where that section can rotate with

respect to the rest of the structure. In this case, if STAAD adds aweak spring, there may be large displacements because there 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 freedom, and causes excessive displacements at

that degree of freedom.

Question : If there are instability messages, are there any simple checks to

verify whether my analysis results are satisfactory?

Answer : There are 2 important checks that should be carried out if

instability messages are present.

a.

A static equilibrium check. This check will tell us whether allthe applied loading flowed through the model into the

supports. A satisfactory result would require that the applied

loading be in equilibrium with the support reactions.

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

that degree of freedom.

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 the above checks. The STAAD

output file will contain a report similar to the following, for every

primary load case that has been solved for :

STAAD P T i i M l Ad d T i




9

***TOTAL APPLIED LOAD ( KG METE ) SUMMARY (LOADING

1 )SUMMATION FORCE-X = 0.00

SUMMATION FORCE-Y = -817.84

SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE ORIGIN-

MX= 291.23 MY= 0.00 MZ= -3598.50

***TOTAL REACTION LOAD( KG METE ) SUMMARY

(LOADING 1 )

SUMMATION FORCE-X = 0.00

SUMMATION FORCE-Y = 817.84

SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE ORIGIN-

MX= -291.23 MY= 0.00 MZ= 3598.50

MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 1)

MAXIMUMS AT NODEX = 1.00499E-04 25

Y = -3.18980E-01 12

Z = 1.18670E-02 23

RX= 1.52966E-04 5

RY= 1.22373E-04 23

RZ= 1.07535E-03 8

Go through these numbers to ensure that

i. The "TOTAL APPLIED LOAD" values and "TOTAL

REACTION LOAD" values are equal and opposite.

ii . The "MAXIMUM DISPLACEMENTS" are within reasonable

limits.

STAAD Pro Training Manual Advanced Topics




10

Question : What is the meaning of this message, "Probable cause warning-

near singular"

Answer : While performing the triangular factorization of the global stiffness

matrix, a diagonal matrix is computed. These computed diagonals

are the same as or smaller than the global stiffness matrix

diagonals. If the computed diagonals become zero then the matrix

is singular and the structure is unstable. In STAAD we say that

the structure is unstable/singular if any computed diagonal is less

that (1.E-9) * (the corresponding stiffness matrix diagonal).

Likewise in STAAD we say that the structure is nearly

unstable/singular if any computed diagonal is less that (1.E-7) *

(the corresponding stiffness matrix diagonal).

If the overall results look OK, then ignore nearly singular

messages.

Question : How to avoid instabilities if TRUSSES or RELEASES are the

cause?

Answer : There is a rather simple way to eliminate instabilities, especially if

truss members are present or when MEMBER RELEASE

commands are used and certain degrees of freedom are subjected

to a 100% release.

In reality, connections always have some amount of force and

moment capacity. Use 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 node orend node of a member, rather than fully eliminating the stiffness

for a certain 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 extent of the release is

controlled by you.

STAAD Pro Training Manual – Advanced Topics



STAAD.Pro Training Manual Advanced Topics

11

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

node, if you specify

5 START MY MZ

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

you say,

5 START MP 0.99

you are saying that the bending and torsional stiffnesses are 99%

less than what they would be for a fully moment resistant

connection. Thus, the 1% available stiffness might be adequate to

allow the load to pass through the node from one member to the

other.

So, this is what may be done :

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

from

MEMBER TRUSS

to

MEMBER RELEASE

memb-list START MP 0.99

memb-list END MP 0.99

or

MEMBER RELEASE

memb-list Both MP 0.99

b. Run the analysis. Check to make sure the instability warnings

no longer appear. Then check your nodal displacements.

c. If the displacements are large, reduce the extent of the release

from 0.99 to say 0.98.




g p

12

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 the moments and shear at the

nodes of those members and make sure that the connection will be

able to handle those forces and moments.

STAAD.Pro 2002 onwards, you can apply these partial releases to

individual moment degrees of freedom. For example, you could

say

MEMBER RELEASE

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 problem area.

You can refer to Section 5.22.1 of the Technical Reference Manual

for details.

Question : Is there any graphical facility in STAAD by which I can examine

the points of instability?

Answer : Yes, there is. Go to the Post processing mode. If instabilities arepresent, the Nodes page along the left side should contain a sub-

page by the name Instability. If you click on this, two tables will

appear along the right hand side.

The upper table lists the node number, and the global degrees of

freedom at that node which are unstable. A zero for a d.o.f

indicates that all is well, and, 1 indicates it is unstable. Click on

the row and the node and all members connected to it will be

highlighted in the drawing.

The lower table has all of the joints in the order that gives the

stiffness matrix the minimum bandwidth which minimizes the

running time. When a joint is unstable, it means that the joint and

some or all of the joints before it in the list form an unstable

structure. That is, even fixing every subsequent joint in the list

would not make it stable.




13

If the instability is at the last joint [or sometimes the last joint and

one other joint], then the whole structure is free in that direction.

Note that the instability is reported at the last joint in the list that

is on the unstable component. If a column is pinned at the base

and floor connections are released in global My, the column will

be torsionally unstable, but only one joint on the column will be

reported as unstable and it could be any joint on the column.




14



Seismic Analysis UsingUBC And IBC Codes





1

Basic principle

When a building is subjected to an earthquake, it undergoes

vibrations. The weights of the structure, when accelerated along

the direction of the earthquake, induce forces in the building.

Normally, an elaborate dynamic analysis called time history

analysis is required to solve for displacements, forces and

reactions resulting from the seismic activity. However, codes like

UBC and IBC provide a static method of solving for those values.

The generalized procedure used in those methods consists of 3

steps

Step 1 : Calculate

Base Shear = Factor f * Weight W

where "f" is calculated from terms which take into consideration

the Importance factor of the building, Site Class and soil

characteristics, etc. W is the total vertical weight derived from

dead weight of the building and other imposed weights.

Step 2 : The base shear is then distributed over the height of the

building as a series of point loads.

Step 3 : The model is then analyzed for the horizontal loads

generated in step 2.


2



2

The input required in STAAD consists of 2 parts.

Part 1, which appears under a heading called

DEFINE UBC LOAD

or

DEFINE IBC LOAD

contains the terms used to compute "f" and "W" described in step

1.

Part 2, which appears within a load case, contains the actual

instruction to generate the forces described in step 2 and analyze

the structure for those forces.

Let us examine this procedure using the example problem shown

below.

STAAD SPACE

SET NL 5

The structure is defined as a space frame type. The maximum

number of primary load cases in the model is set to 5.

UNIT KIP FEET

JOINT COORD

1 0 0 0 ; 2 0 10 0 ; 3 13 10 0 ; 4 27 10 0 ; 5 40 10 0 ; 6 40 0 0

7 0 20.5 0 ; 8 20 20.5 0 ; 9 40 20.5 0

REPEAT ALL 1 0 0 11

Joint coordinates are specified using a mixture of explicit

definition and generation using REPEAT command.


3



3

MEMBER INCI

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

21 10 11 25 ; 26 10 12 ; 27 13 15 ; 28 11 16 ; 29 16 17 30 ; 31 18

14

32 11 17 ; 33 14 17

41 2 11 44

45 7 16 47

51 1 11

52 10 2

53 2 16

54 11 7

55 6 14

56 15 5

57 5 18

58 14 9

Member incidences are specified using a mixture of explicit

definition and generation.

MEMBER PROPERTIES

1 5 8 11 21 25 28 31 TA ST W14X90

2 3 4 22 23 24 TA ST W18X35

9 10 29 30 TA ST W21X5041 TO 44 TA D C12X30

45 TO 47 TA D C15X40

6 7 26 27 TA ST HSST20X12X0.5

51 TO 58 TA LD L50308

12 13 32 33 TA ST TUB2001205


4



4

Various section types are used in this model. Among them are

double channels, hollow structural sections and double angles.

CONSTANTS

E STEEL ALL

POISSON STEEL ALL

DENSITY STEEL ALL

Structural steel is the material used in this model.

SUPPORT

1 6 10 15 FIXED

Fixed supports are defined at 4 nodes.

MEMBER TENSION

51 TO 58

Members 51 to 58 are defined as capable of carrying tensile forces

only.

UNIT POUND

DEFINE UBC ACCIDENTAL LOAD

ZONE 0.3 I 1 RWX 2.9 RWZ 2.9 STYP 4 NA 1 NV 1

SELFWEIGHT

FLOOR WEIGHT

YRANGE 9 11 FLOAD 0.4

YRANGE 20 21 FLOAD 0.3

There are two stages in the command specification of the UBCloads. The first stage is initiated with the command DEFINE UBC

LOAD. Here we specify parameters such as Zone factor,

Importance factor, site coefficient for soil characteristics etc. and,

the vertical loads (weights) from which the base shear will be

calculated. The vertical loads may be specified in the form of

selfweight, joint weights, member weights, element weights or

floor weights. Floor weight is used when a pressure acting over apanel has to be applied when the structural entity which makes up

the panel (like a aluminum roof for example) itself isn’t defined as


5



5

part of the model. The selfweight and floor weights are shown in

this example. It is important to note that these vertical loads are

used purely in the determination of the horizontal base shear only.

In other words, the structure is not analyzed for these vertical

loads. LOAD 1

UBC LOAD X

This is the second stage in which the UBC load is applied with thehelp of load case number, corresponding direction (X in the above

case) and a factor by which the generated horizontal loads should

be multiplied. Along with the UBC load, deadweight and other

vertical loads may be added to the same load case (they are not in

this example). PERFORM ANALYSIS PRINT LOAD DATACHANGE

A linear elastic type analysis is requested for load case 1. We can

view the values and position of the generated loads with the help

of the PRINT LOAD DATA command used above along with the

PERFORM ANALYSIS command. A CHANGE command should

follow the analysis command for models like this where the

MEMBER TENSION command is used in conjunction with UBC

load cases.

LOAD 2

UBC LOAD Z

We define load case 2 as consisting of the UBC loads to be

generated along the Z direction. The structure will be analyzed for

those generated loads.

PERFORM ANALYSIS PRINT LOAD DATA

CHANGE

The analysis instruction is specified again.


6



LOAD 3

SELF Y -1.0

FLOOR LOAD

YRANGE 9 11 FLOAD -0.4

YRANGE 20 21 FLOAD -0.3

In load case 3 in this problem, we apply 2 types of loads. The

selfweight is applied in the global Y direction acting downwards.

Then, a floor load generation is performed. In a floor load

generation, a pressure load (force per unit area) is converted by theprogram into specific points forces and distributed forces on the

members located in that region. The YRANGE (and if specified,

the XRANGE and ZRANGE) values are used to define the region

of the structure on which the pressure is acting. The FLOAD

specification is used to specify the value of that pressure. All

values need to be provided in the current UNIT system. For

example, in the first line in the above FLOOR LOADspecification, the region is defined as being located within the

bounds YRANGE of 9-11 ft. Since XRANGE and ZRANGE are

not mentioned, the entire floor within the YRANGE will become a

candidate for the load. The -0.4 signifies that the pressure is 0.4

Kip/sq. ft in the negative global Y direction.

The program will identify the members lying within the specified

region and derive MEMBER LOADS on these members based on

two-way load distribution.

PERFORM ANALYSIS

CHANGE


LOAD 4

REPEAT LOAD

1 1.0 3 1.0

Load case 4 illustrates the technique employed to instruct STAADto create a load case which consists of data to be assembled from

other load cases already specified earlier. We would like the


7



program to analyze the structure for loads from cases 1 and 3

acting simultaneously.

PERFORM ANALYSIS PRINT STATICS CHECK

CHANGE


LOAD 5

REPEAT LOAD2 1.0 3 1.0

In load case 5, we instruct STAAD to create a load case consisting

of data to be assembled from cases 2 and 3 acting simultaneously.

PERFORM ANALYSIS PRINT STATICS CHECK

CHANGE


LOAD LIST 4 5

PRINT JOINT DISPLACEMENTS

PRINT SUPPORT REACTIONS

PRINT MEMBER FORCES LIST 51 TO 58

Various results are requested for just load cases 4 and 5.

FINISH

The STAAD run is terminated.


8



Question : When I specify vertical weights under the DEFINE UBC LOAD

command, why do I have to specify them again under the actual

load case? Won't STAAD be double-counting those weights?

Answer : Generally, all code related seismic methods follow a procedure

called static equivalent method. That is to say, even if seismic

forces are dynamic in nature, they can be solved using a static

approach.

That means, one has to first come up with static loads. These are

calculated usually using an equation called

H = constant x V

where H is the horizontal load which is calculated.

V is the applied vertical load.

In STAAD, the V has to be defined under commands like

DEFINE IBC LOAD

or

DEFINE IBC LOAD

There, they are defined in the form of selfweight, joint weight,

member weight, etc. The data specified over there is used just to

compute the V. Hence, once the H is derived from the V, the V is

discarded. If a user wants the structure to be analysed for the

vertical loads, they have to be explicity specified with Load cases.That is what you'll find in example 14. Load cases 1 & 2 contain a

horizontal load and a vertical load. The horizontal load comes

from the UBC LOAD X and UBC LOAD Z commands. The

vertical load comes from selfweight, joint load commands.

So, there is no double counting.


9



Question : We would like to know what Ta and Tb in the static seismic base

shear output stand for. We know that both are computed time

periods, but we would like to know why there are two values for it.

Answer : The UBC and IBC codes involve determination of the period based

on 2 methods - Method A and Method B. The value based on

Method A is called Ta. The value based on Method B is called Tb.

Question : What is the difference between a JOINT WEIGHT and a JOINT

LOAD?

Answer : The JOINT WEIGHT option is specified under the DEFINE UBC

LOAD command and is used merely to assemble the weight values

which make up the value of "W" in the UBC equations. In other

words, it is the amount of lumped weight at the joint and a fraction

of this weight eventually makes up the total base shear for thestructure.

A JOINT LOAD on the other hand is an actual force which is

acting at the joint, and is defined through the means of an actual

load case.

Question : When using the "ACCIDENTAL" option in the "DEFINE UBC

LOAD" command, it appears that for the mass displacement along

a given axis STAAD.Pro only considers the displacement in one

direction rather than a plus or minus displacement. Is this true?

You can verify this by adding the "ACCIDENTAL" option to

Example Problem 14 and comparing the reactions.

Answer : Use the "ACC f2" option as explained in the command syntax in

section 5.32.12 of the Technical Reference manual. You can

specify a negative value for f2 if you want the minus sign for the

torsional moments. You will need STAAD.Pro 2003 to use this.


10

Q ti



Question : How do I display the Load values of an IBC2000 load case?

Answer : First run the analysis. Then go to the View menu, choose StructureDiagrams. Click on the Loads and Results tab. Select the load case

corresponding to the IBC load command. Switch on the checkbox

for Loads, click on OK.



Calculating Mode Shapes, Frequencies

AndParticipation factors



1

In STAAD there are 2 methods for obtaining the frequencies of a



In STAAD, there are 2 methods for obtaining the frequencies of a

structure.

1. The Rayleigh method using the CALCULATE RAYLEIGH

FREQUENCY command

2. The elaborate method which involves extracting eigenvalues

from a matrix based on the structure stiffness and lumped

masses in the model.

The Rayleigh method in STAAD is a one-iteration approximate

method from which a single frequency is obtained. It uses the

displaced shape of the model to obtain the frequency. Needless to

say, it is extremely important that the displaced shape that the

calculation is based on, resemble one of the vibration modes. If

one is interested in the fundamental mode, the loading on the

model should cause it to displace in a manner which resembles thefundamental mode. For example, the fundamental mode of

vibration of a tall building would be a cantilever style mode, where

the building sways from side to side with the base remaining

stationary. The type of loading which creates a displaced shape

which resembles this mode is a lateral force such as a wind force.

Hence, if one were to use the Rayleigh method, the loads which

should be applied are lateral loads, not vertical loads.

For the eigensolution method, the user is required to specify all the

masses in the model along with the directions they are capable of

vibrating in. If this data is correctly provided, the program extracts

as many modes as the user requests (default value is 6) in

ascending order of strain energy. The mode shapes can be viewed

graphically to verify that they make sense.


2

Eigenvalue extraction method



Eigenvalue extraction method

The input which is important and relevant to the analysis of a

structure for frequencies and modes – using the eigenvalue

extraction method is explained below. These are explained in

association with an example problem provided at the end of this

section.

1. The DENSITY command

One of the critical components of a frequency analysis is the

amount of "mass" undergoing vibration. For a structure, this mass

comes from the selfweight, and from permanent/imposed loads on

the building. To calculate selfweight, density is required, and is

hence specified under the command CONSTANTS.

2. The CUT OFF MODE SHAPE command

Theoretically, a structure has as many modes of vibration as the

number of degrees of freedom in the model. However, the

limitations of the mathematical process used in extracting modes

may limit the number of modes that can actually be extracted. In a

large structure, the extraction process can also be a very time

consuming process. Further, not all modes are of equal importance.

(One measure of the importance of modes is the participation

factor of that mode.) In many cases, the first few modes may be

sufficient to obtain a significant portion of the total dynamic

response.

Due to these reasons, in the absence of any explicit instruction,

STAAD calculates only the first 6 modes. (Versions of STAAD

prior to STAAD/Pro 2000 calculated only 3 modes by default).

This is like saying that the command CUT OFF MODE SHAPE 6

has been specified.

If the inspection of the first 6 modes reveals that the overall

vibration pattern of the structure has not been obtained, one mayask STAAD to compute a larger (or smaller) number of modes with

the help of this command. The number that follows this command


3

is the number of modes being requested. In our example, we are



g q p

asking for 10 modes by specifying CUT OFF MODE SHAPE 10.

3. The MODAL CALCULATION REQUESTED command.

This is the command which triggers the calculation of frequencies

and modes. It is specified inside a load case. In other words, this

command accompanies the loads which are to be used in

generating the mass matrix.

Frequencies and modes have to be calculated when dynamic

analysis such as response spectrum or time history analysis are

carried out. But in such analyses, the MODAL CALCULATION

REQUESTED command is not explicitly required. When STAAD

encounters the commands for response spectrum (see example 11)

and time history (see examples 16 and 22), it automatically will

carry out a frequency extraction without the help of the MODAL ..command.

4. The MASSES which are to be used in assembling the MASS

MATRIX

The mathematical method that STAAD uses is called the subspace

iteration eigen extraction method. Some information on this isavailable in Section 1.18.3 of the STAAD.Pro Technical Reference

Manual. The method involves 2 matrices - the stiffness matrix, and

the mass matrix.

The stiffness matrix, usually called the [K] matrix, is assembled

using data such as member and element lengths, member and

element properties, modulus of elasticity, poisson's ratio, member

and element releases, member offsets, support information, etc.


4

For assembling the mass matrix, called the [M] matrix, STAAD



uses the load data specified in the load case in which the MODAL

CAL REQ command is specified. So, some of the important

aspects to bear in mind are:

i. The input you specify is weights, not masses. Internally,

STAAD will convert weights to masses by dividing the

input by "g", the acceleration due to gravity.

ii . If the structure is declared as a PLANE frame, there are 2

possible directions of vibration - global X, and global Y. If

the structure is declared as a SPACE frame, there are 3

possible directions - global X, global Y and global Z.

However, this does not guarantee that STAAD will

automatically consider the masses for vibration in all the

available directions.

You have control over and are responsible for specifying

the directions in which the masses ought to vibrate. In

other words, if a weight if not specified along a certain

direction, the corresponding degrees of freedom (such as

for example, global Z at node 34) will not receive a

contribution in the mass matrix. The mass matrix is

assembled using only the masses from the weights anddirections specified by the user.

In our example, notice that we are specifying the

selfweight along global X, Y and Z directions. Similarly,

the element pressure load is also specified along all 3

directions. We have chosen not to restrict any direction for

this problem. If a user wishes to restrict a certain weight to

certain directions only, all he/she has to do is not provide

the directions in which those weights cannot vibrate in.

iii. As much as possible, provide absolute values for the

weights. STAAD is programmed to algebraically add the

weights at nodes. So, if some weights are specified aspositive numbers, and others as negative, the total weight


5

at a given node is the algebraic summation of all the



weights in the global directions at that node.

STAAD SPACE

* EXAMPLE PROBLEM FOR CALCULATION OF MODES AND

FREQUENCIES

UNIT FEET KIP

JOINT COORDINATES

1 0 0 0; 2 0 0 20; 3 20 0 0; 4 20 0 20; 5 40 0 0; 6 40 0 20; 7 0 15 0;

8 0 15 5; 9 0 15 10; 10 0 15 15; 11 0 15 20; 12 5 15 0; 13 10 15 0;

14 15 15 0; 15 5 15 20; 16 10 15 20; 17 15 15 20; 18 20 15 0;

19 20 15 5; 20 20 15 10; 21 20 15 15; 22 20 15 20; 23 25 15 0;

24 30 15 0; 25 35 15 0; 26 25 15 20; 27 30 15 20; 28 35 15 20;

29 40 15 0; 30 40 15 5; 31 40 15 10; 32 40 15 15; 33 40 15 20;34 20 3.75 0; 35 20 7.5 0; 36 20 11.25 0; 37 20 3.75 20; 38 20 7.5 20;

39 20 11.25 20; 40 5 15 5; 41 5 15 10; 42 5 15 15; 43 10 15 5;

44 10 15 10; 45 10 15 15; 46 15 15 5; 47 15 15 10; 48 15 15 15;

49 25 15 5; 50 25 15 10; 51 25 15 15; 52 30 15 5; 53 30 15 10;

54 30 15 15; 55 35 15 5; 56 35 15 10; 57 35 15 15; 58 20 11.25 5;

59 20 11.25 10; 60 20 11.25 15; 61 20 7.5 5; 62 20 7.5 10; 63 20 7.5 15;

64 20 3.75 5; 65 20 3.75 10; 66 20 3.75 15; 67 20 0 5; 68 20 0 10;69 20 0 15;

MEMBER INCIDENCES

1 1 7; 2 2 11; 3 3 34; 4 34 35; 5 35 36; 6 36 18; 7 4 37; 8 37 38;

9 38 39; 10 39 22; 11 5 29; 12 6 33; 13 7 8; 14 8 9; 15 9 10; 16 10 11;

17 18 19; 18 19 20; 19 20 21; 20 21 22; 21 29 30; 22 30 31; 23 31 32;

24 32 33; 25 7 12; 26 12 13; 27 13 14; 28 14 18; 29 18 23; 30 23 24;

31 24 25; 32 25 29; 33 11 15; 34 15 16; 35 16 17; 36 17 22; 37 22 26;

38 26 27; 39 27 28; 40 28 33;

ELEMENT INCIDENCES SHELL

41 7 8 40 12; 42 8 9 41 40; 43 9 10 42 41; 44 10 11 15 42;

45 12 40 43 13; 46 40 41 44 43; 47 41 42 45 44; 48 42 15 16 45;49 13 43 46 14; 50 43 44 47 46; 51 44 45 48 47; 52 45 16 17 48;

53 14 46 19 18; 54 46 47 20 19; 55 47 48 21 20; 56 48 17 22 21;


6

57 18 19 49 23; 58 19 20 50 49; 59 20 21 51 50; 60 21 22 26 51;

6 23 9 2 2 62 9 0 3 2 63 0 3 6 26 2



61 23 49 52 24; 62 49 50 53 52; 63 50 51 54 53; 64 51 26 27 54;

65 24 52 55 25; 66 52 53 56 55; 67 53 54 57 56; 68 54 27 28 57;

69 25 55 30 29; 70 55 56 31 30; 71 56 57 32 31; 72 57 28 33 32;

73 18 19 58 36; 74 19 20 59 58; 75 20 21 60 59; 76 21 22 39 60;

77 36 58 61 35; 78 58 59 62 61; 79 59 60 63 62; 80 60 39 38 63;

81 35 61 64 34; 82 61 62 65 64; 83 62 63 66 65; 84 63 38 37 66;

85 34 64 67 3; 86 64 65 68 67; 87 65 66 69 68; 88 66 37 4 69;

MEMBER PROPERTY

1 TO 40 PRIS YD 1 ZD 1

ELEMENT PROPERTY

41 TO 88 THICKNESS 0.5

CONSTANTS

E CONCRETE ALLDENSITY CONCRETE ALL

POISSON CONCRETE ALL

CUT OFF MODE SHAPE 10

SUPPORTS

1 TO 6 FIXED

UNIT POUND FEET

*MASS DATA AND INSTRUCTION FOR COMPUTING FREQUENCIES

AND MODES

LOAD 1

SELFWEIGHT X 1.0

SELFWEIGHT Y 1.0

SELFWEIGHT Z 1.0

ELEMENT LOAD

41 TO 88 PR GX 300.0

41 TO 88 PR GY 300.041 TO 88 PR GZ 300.0


7

MODAL CALCULATION REQUESTED



MODAL CALCULATION REQUESTED

PERFORM ANALYSIS

FINISH

Understanding the output :

After the analysis is completed, look at the output file. This file

can be viewed from File - View - Output File - STAAD output.

i. Mode number and corresponding frequencies and periods

Since we asked for 10 modes, we obtain a report which is as

follows:


8

ii . Participation factors in Percentage



In the explanation above for the CUT OFF MODE command,

we said that one measure of the importance of a mode is the

participation factor of that mode. We can see from the abovereport that for vibration along X direction, the first mode has

a 90.89 percent participation. It is also apparent that the 4th

mode is primarily a Y direction mode due to its 50.5 %

participation along Y and 0 in X and Z.

The SUMM-X, SUMM-Y and SUMM-Z columns show the

cumulative value of the participation of all the modes up to

and including a given mode. One can infer from those terms

that if one is interested in 95% participation along X, the first

5 modes are sufficient.

But for the Z direction, even with 10 modes, we barely

obtained 0.6%. The reason for this can be understood by a

close examination of the nature of the structure. Our model

has a shear wall which spans in the YZ plane. This makes the

structure extremely stiff in that plane. It would take a lot of

energy to make the structure vibrate along the Z direction.

Modes are extracted in the ascending order of energy. The

higher modes are high energy modes, compared to the lower

modes. It is likely that unless we raise the number of modesextracted from 10 to a much larger number - 30 or more -


9

using the CUT OFF MODE SHAPE command, we may not

be able to obtain substantial participation along the Z



be able to obtain substantial participation along the Z

direction.

Another unique aspect of the above result are the modes

where all 3 directions have 0 or near 0 participation. This is

caused by the fact that the vibration pattern of the model for

that mode results in symmetrically located masses vibrating

in opposing directions, thus canceling each other's effect.

Torsional modes too exhibit this behavior. See the next item

for the method for viewing the shape of vibration. Localized

modes, where small pockets in the structure undergo flutter

due to their relative weak stiffness compared to the rest of the

model, also result in small participation factors.

iii. Viewing the mode shapes

After the analysis is completed, select Post-processing from

the mode menu. This screen contains facilities for graphically

examining the shape of the mode in static and animated

views. The Dynamics page on the left side of the screen is

available for viewing the shape of the mode statically. The

Animation option of the Results menu can be used for

animating the mode. The mode number can be selected fromthe "Loads and Results" tab of the "Diagrams" dialog box

which comes up when the Animation option is chosen. The

size to which the mode is drawn is controlled using the

"Scales" tab of the "Diagrams" dialog box.


10

How are modes, frequencies and the other terms are calculated



The process of calculating the MODES and FREQUENCIES is

known as Modal Extraction and is performed by solving theequation:

ω2 [ m ] { q } - [ K ] { q } = o

Where

[ m ] = the mass matrix (assumed to be diagonal, i.e., no

mass coupling)

ω = the natural frequencies (eigenvalues)

{ q } = the normalized mode shapes (eigenvectors)

Frequency (HZ or CPS) = ω /2π

The solution method used in STAAD is the Subspace iterationmethod.

Please note that various nomenclature is used to refer to the

normal modes of vibration. (Eigenvalue, Natural Frequency,

Modal Frequency and Eigenvector, Mode Shape, Modal Vector,

Normal Modes, Normalized Mode Shape.

Generalized Weight and Generalized Mass

Each eigenvector {q} has an associated generalized mass defined

by

Generalized Mass (GM) = { q }T

[ M ] { q }

Generalized Weight (GW) = GM * g


11

Participation Factors - A participation factor (Qi) is computed

for each eigenvector for each of the three global (Xi) translational



g g ( )

directions. N is the number of modes.

Q

q w

GWi

j,i j,i j

N

=

∑=

( )( )1

Modal Weights - The modal weight for each mode is (GW)(Q i²) .

The summation of modal weights for all modes in a given direction

is equal to the Base Shear which would result from a one g baseacceleration. The sum of the modal weights for the computed

modes may be compared to the total weight of the structure (only

the weight that has not been lumped at supports). The difference

is the amount of weight missing from a dynamic, base excitation,

modal response analysis. If too much is missing, then rerun the

eigensolution asking for a greater number of modes.

STAAD prints the "MASS PARTICIPATION FACTOR IN

PERCENT" for each mode. This is the modal weight of a mode as

a percentage of the total weight of the structure. Also a running

sum for all modes is given so that the last line indicates the percent

of the total weight that all of the modes extracted would represent

in a 1g base excitation.


12





Response Spectrum Analysis



1

Description

R l f i f i l d



Response spectra are plots of maximum response of single degree

of freedom (SDOF) systems subjected to a specific excitation. Forvarious values of frequency of the SDOF system and various

damping ratios, the peak response is calculated.

Structures normally have multiple degrees of freedom (MDOF).

The dynamic analysis of a MDOF system having "n" DOF involves

reducing it to "n" independent SDOF systems. The modal

superposition method is used and the maximum modal responses

are combined using SRSS, CQC and other methods available in

STAAD.

The command syntax for defining response spectrum data is

explained in Section 5.32.10.1 of the Technical Reference manual.

It is important to understand that once the combination methods

like SRSS or CQC are applied, the sign of the results is lost.

Consequently, results of a spectrum analysis, like displacements,

forces and reactions do not have any sign.

Because spectrum analysis requires modes and frequencies, the

mass data and other details explained in the chapter on calculatingmodes and frequencies are all applicable in the case of spectrum

analysis also. In other words, the mode and frequency calculation

is a pre-requisite to performing response spectrum analysis.


2

Calculation of Base Shear in a Response Spectrum Analysis

Th b h f i d f i di ti t d i



The base shear, for a given mode for a given direction, reported in

the response spectrum analysis is obtained as

A * B * C * D

where

A = Mass participation factor for that mode for that direction

B = Total mass specified for that direction

C = Spectral acceleration for that mode

D = direction factor specified in that load case

A is calculated by the program from the mass matrix and mode

shapes

B is obtained from the masses specified in the response spectrum

load case

C is obtained by interpolating between the user provided values of period vs. acceleration and multiplying the resulting value by the

SCALE FACTOR.

D is specified by the user

Bending Moment Diagram for a load case that involves the

Response Spectrum Analysis

In a response spectrum analysis in STAAD, the member forces are

computed accurately only at the 2 ends of the member. The sign of

these forces cannot be determined due to the fact that the method

used to combine the contribution of modes does not allow for the

determination of the sign of the forces. Further, these force values

do not necessarily indicate whether these forces occur at the same

instant of time.


3

In order to draw the bending moment diagram, one needs to know

the moments at the intermediate section points on the member. In

order to calculate these section force values the forces at the



order to calculate these section force values, the forces at the

member ends have to be used. However, due to the special natureof these end force values as described in the paragraph above, it

makes no sense to calculate the intermediate section forces based

on the end force values.

Due to this reasoning, the bending moment diagram simply cannot

be drawn accurately for the response spectrum loading. STAAD

merely plots a straight line that joins the bending moment values at

the start and end joints of the member which are as mentioned

earlier, absolute (positive) values. Current versions of STAAD do

not let the user draw the diagram at all from certain places such as

the Member Query.

Comparison of results of a spectrum analysis (which uses the

UBC spectrum data) with the results of an equivalent UBC

static analysis

For the following reasons, this comparison isn't meaningful :

1. In a spectrum analysis, the number of modes to be combined

is a decision made by the engineer. If 100% participationfrom the modes isn't utilized in the displacement calculation,

it is obvious that the results will be only approximate.

2. In a spectrum analysis, the contribution from the various

modes is combined using an SRSS method or a CQC method,

both of which are only approximate methods. One very

important drawback of both these methods is that the sign of

the displacements and forces cannot be determined. Also, the

results can vary significantly depending on the type of

method used in the combination.

3. In the UBC method, only a single period is used. Normally,

the assumption is that this period is associated with a mode

that encompasses a significant portion of the overall response

of the structure. This may not necessarily be true in reality. If


4

more than one mode is required to capture the overall

response of the structure, that fact is not brought to light in

the UBC static equivalent approach



the UBC static equivalent approach.

4. The UBC static equivalent method involves several

parameters such as Importance factor, soil structure

coefficient, etc. which are incorporated through an emperical

formula. In a response spectrum analysis, there is no facility

available to incorporate these factors in a direct manner.

Due to these reasons, a direct comparison of the results of a

spectrum analysis and a static equivalent approach is not

recommended.

Question : What is the Scale Factor (f4) that needs to be provided when

specifying the Response Spectra?

Answer : The spectrum data consists of pairs of values which are Period vs.

Accn. or Period vs. Displacement. The acceleration or

displacement values that you obtain from the geological data for

that site may have been provided to you as normalized values or

un-normalized values. Normalization means that the values of

acceleration or displacement have been divided by a number

(called normalization factor) which represents some reference

value. One of the commonly used normalization factors is 'g', the

acceleration due to gravity.

If the spectrum data you specify in STAAD is a normalized

spectrum data, you should provide the NORMALIZATION

FACTOR as the SCALE FACTOR. If your spectrum data is un-normalized, there is no need to provide a scale factor(Another way

of putting it is that if you provide un-normalized spectrum values,

the scale factor is 1, which happens to be the default value also.)

Make sure that the value you provide for the SCALE FACTOR is

in accordance with the length units you have specified. (A common

error is that if the scale factor is 'g', users erroneously provide 32.2

when the length unit is in INCHES.)


5

STAAD will multiply the spectral acceleration or spectral

displacement values by the scale factor. Hence, if you provide a

normalized acceleration value of 0.5 and a scale factor of 386.4



o a ed acce e at o va ue o 0.5 a d a sca e acto o 386.

inch/sq.sec., it has the same effect as providing an un-normalizedacceleration value of 193.2 inch/sq.sec. and a scale factor of 1.0.

Question : What is the Direction Factor that needs to be provided when

specifying the Response Spectra?

Answer : The Direction factor is a quantity by which the spectraldisplacement for the associated direction is multiplied.

For example, if the command reads as

SPECTRUM SRSS X 0.7 Y 0.5 Z 0.65 DISP DAMP 0.05SCALE 32.2

the following is done:

1. For each mode, the period is determined.

2. Corresponding to the period, the spectral displacement for

that mode is calculated by interpolation from the input pairs

of period vs. spectral displacement. Call this "sd"

3. Calculate the spectral displacement for each direction by

multiplying "sd" by the associated Direction factor.

The X direction spectral displacement = sd * 0.7

The Y direction spectral displacement = sd * 0.5The Z direction spectral displacement = sd * 0.65

These factored values are then multiplied by

a. the mode shape value corresponding to that degree of

freedom,

b. participation factor.

Call the result T(m) where "m" stands for the mode number.


6

Once the T(m) is determined for all modes, subject them to the

SRSS calculation. That will provide the node displacement

corresponding to that degree of freedom.



p g g

Question : The results of the response spectrum load case are always positive

numbers. Why? How do I know that the positive value is always

critical, especially from the design standpoint?

Answer : In a spectrum analysis, the contribution of the individual modes is

combined using methods such as SRSS or CQC to arrive at theoverall response. The limitation of these methods is that the sign

of the response cannot be determined after the method is applied.

This is the reason why the output you get from STAAD for a

response spectrum analysis are absolute values.

One way to deal with the problem is to create 2 load combination

cases for each set of load cases you wish to combine. For example,if the dead load case is 1, and the spectrum load case is 5, you

could create

LOAD COMB 101 1.1 5 1.3

LOAD COMB 111 1.1 5 -1.3

and use the critical value from amongst these 2 load combination

cases for design purposes. What you accomplish from this process

is that you are considering a positive effect as well as the negative

effect of the spectrum load case.


7

Question : In the Technical Reference manual section 5.32.10.1, you state: "

Note, if data is in g acceleration units, then set SCALE to a

conversion factor to the current length unit (9.81, 386.4, etc.)"



co ve s o acto to t e cu e t e gt u t (9.8 , 386. , etc.)

What does "g acceleration units" mean?

Related question : What is the Scale Factor (f4) that needs to be provided

when specifying the Response Spectra?

Answer : The spectrum data consists of pairs of values which are Period vs.

Accn. or Period vs. Displacement. The acceleration ordisplacement values that you obtain from the geological data for

that site may have been provided to you as normalized values or

un-normalized values. Normalization means that the values of

acceleration or displacement have been divided by a number

(called normalization factor) which represents some reference

value. One of the commonly used normalization factors is 'g', the

acceleration due to gravity.

If the spectrum data you specify in STAAD is a normalized

spectrum data, you should provide the NORMALIZATION

FACTOR as the SCALE FACTOR. If your spectrum data is un-

normalized, there is no need to provide a scale factor(Another way

of putting it is that if you provide un-normalized spectrum values,

the scale factor is 1, which happens to be the default value also.)

Make sure that the value you provide for the SCALE FACTOR is

in accordance with the length units you have specified. (A common

error is that if the scale factor is 'g', users erroneously provide 32.2

when the length unit is in INCHES.)

STAAD will multiply the spectral acceleration or spectraldisplacement values by the scale factor. Hence, if you provide a

normalized acceleration value of 0.5 and a scale factor of 386.4

inch/sq.sec., it has the same effect as providing an un-normalized

acceleration value of 193.2 inch/sq.sec. and a scale factor of 1.0.


8

Question : STAAD allows me to use SRSS, ABS, CQC, ASCE4-98 & TEN

Percent for combining the responses from each mode into a total

response. The CQC & ASCE4 methods require damping. But,



p Q q p g ,

ABS, SRSS, and TEN do not use damping unless Spectra-Period

curves are made a function of damping. Why?

Answer : The spectral acceleration versus period curve is for a particular

value of damping. So the user has selected a damping when he

selects the acceleration curve. The damping on the SPECTRUM

command only affects the calculation of the closely spaced modalinteraction matrix which SRSS, ABS, and TEN do not use.

Question : I have some doubts in how to use the Spectrum command.

First of all, dead loads are always applied in the Y axis direction

(downwards). When I’m going to run a spectrum analysis and I use

the same dead loads, do I have to modify the direction of the

loads?

Answer : The load data you provide in the load case in which the

SPECTRUM command is specified goes into the making of the

mass matrix. The mass matrix is supposed to be populated with

terms for all the global directions in which the structure is capableof vibrating. To enable this, the loads must be specified in all the

possible directions of vibration.

Consequently, the load case for response spectrum might look

something like this :

LOAD 20 SPECTRUM IN X DIRECTION*

SELFWEIGHT X 1

SELFWEIGHT Y 1

SELFWEIGHT Z 1


9

MEMBER LOAD



274 TO 277 UNI GX 1.36272 466 998 UNI GX 4.13

313 314 474 477 UNI GX 6.29

274 TO 277 UNI GY 1.36

272 466 998 UNI GY 4.13

313 314 474 477 UNI GY 6.29

274 TO 277 UNI GZ 1.36

272 466 998 UNI GZ 4.13

313 314 474 477 UNI GZ 6.29

JOINT LOAD

420 424 FX 47.32

389 TO 391 FX 560

420 424 FY 47.32

389 TO 391 FY 560

420 424 FZ 47.32

389 TO 391 FZ 560

SPECTRUM CQC X 1 ACC SCALE 9.81 DAMP 0.07

0.025 0.14; 0.0303 0.1636; 0.05 0.2455; 0.0625 0.2941;

0.0769 0.3479; 0.0833 0.3713;

0.1 0.3713; 0.125 0.3713; 0.1667 0.3713; 0.1895

0.3713; 0.25 0.2815; 0.2857 0.2463;

0.3333 0.2111; 0.4 0.1759; 0.5 0.1407; 0.6667 0.1056;1 0.0704; 2 0.0344; 10 0.001372;

LOAD 21 SPECTRUM IN Z DIRECTION

SPECTRUM CQC Z 1 ACC SCALE 9.81 DAMP 0.07

0.025 0.14; 0.0303 0.1636; 0.05 0.2455; 0.0625 0.2941;

0.0769 0.3479; 0.0833 0.3713;


10

0.1 0.3713; 0.125 0.3713; 0.1667 0.3713; 0.1895

0.3713; 0.25 0.2815; 0.2857 0.2463;

0.3333 0.2111; 0.4 0.1759; 0.5 0.1407; 0.6667 0.1056;



1 0.0704; 2 0.0344; 10 0.001372;

Question : Can I specify a different spectrum for each of the 3 directions

(x , y or z)?

Answer : Yes.

Question : Can I decide how many modes I want to include in the spectrum

analysis?

Answer : Use the command CUT OFF MODE SHAPE. Refer to example

problems 11, 28, 29, etc.

Question : In the results, what are the dynamic, missing, and modal weights?

Answer : The dynamic weight line contains the total potential weight for base

shear calculations. Missing Weight is the amount of weight

missing in the modes; Modal weight is the total weight actually

used in the modes. If you algebraically add up Dynamic &

Missing, you should get Modal.

SRSS MODAL COMBINATION METHOD USED.

DYNAMIC WEIGHT X Y Z 8.165253E+02 8.165294E+02 8.165276E+02 POUN

MISSING WEIGHT X Y Z -4.118054E+01 -3.292104E+02 -4.840284E+02 POUN

MODAL WEIGHT X Y Z 7.753447E+02 4.873190E+02 3.324991E+02 POUN


11

Question : What is meant by MASS PARTICIPATION FACTORS IN

PERCENT?



Answer : When the weight of the building is accelerated in a certain

direction, it produces a force in that direction. That force can be

broken down into small parts, with each part coming from a

specific mode. The sum of the values of these parts is called the

base shear.

The percentage of the weight of the building, participating in thevibration in a mode in a specific direction is called the

PARTICIPATION factor. It is a reflection of the "part" of the base

shear, generated by that mode in that direction.

Question : I am a little bit confused with the response spectrum analysis

results. Refer your Example 11 results. The support reactions that

we are getting are the same for both the supports for load cases 3

& 4. In combining lateral loads (response spectrum loading in this

case) with vertical loads, one support should have less force than

the other. At one support, the vertical reaction from the lateral load

case will add to that from the vertical load case, and, at the other,

it will get subtracted. Why do I not see that in the results?

Answer : The support reaction values from a response spectrum analysis

(like any other results from a response spectrum analysis) are

absolute quantities. Consequently, the reactions from case 2, which

is the spectrum case, are both equal and have the same sign. The

primary reasons for this are

a. when the numbers are subjected to the SRSS, CQC or othermethods, their sign is lost

b. the values do not necessarily reflect the result at the same

instant of time.

When you combine these results with those from the dead load

case, it leads to the same value at both supports.


12

If you want the results to truly reflect the sign, use a static

equivalent method like that stipulated by the UBC code.

Alternatively, perform a time history analysis where the sign of the



values is obtained for each time step.

Question : Is it possible to get the vertical distribution of the total base shear

in a response spectrum analysis, like one can for a UBC analysis?

Answer : Unfortunately No. Since the values from a response spectrum

analysis are absolute quantities (numbers without sign), there is noreasonable way to obtain it. You may add up the shears in the

columns above that level for an approximate estimate.

Question : Can you please let me know if we can print nodal acceleration

from response spectra runs? If so, how do I print the data in the

report format or display it in the Post-Processing mode?

Answer : Add the word SAVE at the end of the SPECTRUM command. A

.ACC file will be created.

There is unfortunately no facility available for displaying it in the

post-processing mode. However, since the ACC file is simply a

text file, you can open it using any text editor, and in Excel too. InExcel, you can use the graph generation facilities for plotting it.

Question : In a response spectrum analysis using the STAAD.Pro, the base

shear is not matching with the summation of the support reaction

values in that direction. Why? Also, which values should be

taken for designing the foundation? the base shear value or the

support reaction value? If it is the base shear value then what is

the method generally used to distribute this base shear to all the

supports?

Answer : The results are statistical, SRSS, CQC, etc. The numbers are all

peak positive values. Since each of the reactions at the time of

peak base shear could be less than that reaction's peak and couldbe positive or negative, it is likely that the peak base shear will be

much less than the sum of the peak reactions.


13

There is no way to distribute the base shear to the supports. Even

if you could, that would not be the peak reaction at the support, the

reaction printed by STAAD is the peak value. If there are several

f i j i h k l h



components of reaction at a joint, these are peak values that mayhave occurred at different times.

Question : The base shear reported by STAAD does not match with the

Summation of Support Reactions in the relevant direction. I want

to know the reason for the same.

Answer : When the SRSS method is used, all results from a Response

Spectrum analysis are a result of a square root of a sum of the

squares (SRSS) of the desired output quantity from each mode.

The reactions within a single mode may have equal and opposite

reactions of the various supports such that the base shear for that

mode is near zero. Therefore the contribution of that mode to aSRSS of all the modal base shears will be nearly zero.

However, in that same mode, a particular support may have a large

reaction value. So when that value is SRSSed with that supports

reaction value from all the other modes, that same mode may be a

major contributor to the final result for the support reaction while

that mode contributes little to the base shear.

Of course if all the support reactions in all of the modes have the

same sign, then the answers will be close.

Question : I am getting a large Difference in the results ( Base shear ) of

between Seismic Coefficient Method (UBC) Response SpectrumMethod. Can you explain why? Also, the CQC method produces a

higher base shear than the SRSS method.

Answer : If the base shear is spread over many frequencies, the Response

Spectrum method will result in a base shear that is much lower

than an absolute sum of the base shears of all the modes. The

theory of SRSS combination is that the peak value from each mode

will occur at a different time and is statistically independent. In


14

STAAD 200x the base shear is also printed using Absolute Sum

combination which assumes that the modes are all in phase and

peaks occur at the same time. You will note that in many problems

th b l t lt i h hi h th th SRSS lt I



the absolute sum result is much higher than the SRSS result. Ibelieve that the UBC approach is closer to the absolute response

since a static case is entirely in phase.

For close spaced eigenvalues the CQC method will amplify the

response of those modes as compared to the SRSS method.

Question : I am trying to correlate the relationship between the base shears

and the Global Support Reactions. For example, on the attached

model, the total base shear in the x-direction does not add up to the

total reaction in the x-direction for the dynamic load case. I'm

thinking that STAAD solves a reaction for each mode and

subsequently sums them in either SRSS or CQC, but I am trying to

justify in my mind why the total base shear in the X direction isnot also the total Global Reaction in the X direction. Could you

try to explain?

Answer : Every individual output result value in a response spectrum

analysis is independent and all results are absolute (positive).


15

Lets say you have two modes and 4 supports in the x direction.

Then for the SRSS combination method the results are computed

as follows:



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

***

Support# Mode 1 Mode 2 Sum of Squares Square root

Reaction Reaction SRSS

1 10. -15. 325 18.0

2 -5. 19. 386 19.63 17. 43. 2138 46.2

4 -3. -12. 153 12.4

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

SRSS Base Shear 19 35 1586 96.2

(Sum of

Reactions)

1586 = 39.8

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

***

Note that SRSS base shear (39.8) does not equal the sum of theSRSS reactions (18.0+19.6+46.2+12.4=96.2). In effect the

procedure says that the maximum likely reaction value at each

support is as shown. However the maximum likely sum is the Base

shear as shown. This is due to the fact that the individual

maximums would not occur at the same time and not necessarily

with the same sign. So the base shear magnitude is usually much

less than the sum of the reactions.

Question : For Load case 1, I have

SPECTRUM SRSS X 1 ACC SCALE 0.9806 DAMP 0.050.03 0.8702; 0.05 1.0752; 0.1 1.5876; 0.15 2.1; 0.3 2.1; 0.5

2.1; 0.7 1.5;0.9 1.1667; 1.1 0.9545; 1.3 0.8077; 1.5 0.7; 1.7 0.6176; 1.90.5526;


16

2.1 0.4762; 2.3 0.397; 2.5 0.336; 2.7 0.2881; 2.9 0.2497; 3.10.2185;3.3 0.1928; 3.5 0.1714; 3.7 0.1534; 3.9 0.1381; 4.1 0.1249;

4 3 0 1136;



4.3 0.1136;4.8 0.0911; 6 0.0583; 7 0.0429; 8 0.0328; 10 0.021; 200.0053; 30 0.0023;

For Load case 2, I have

SPECTRUM SRSS Z 1 ACC SCALE 0.9806 DAMP 0.050.03 0.8702; 0.05 1.0752; 0.1 1.5876; 0.15 2.1; 0.3 2.1; 0.52.1; 0.7 1.5;0.9 1.1667; 1.1 0.9545; 1.3 0.8077; 1.5 0.7; 1.7 0.6176; 1.90.5526;2.1 0.4762; 2.3 0.397; 2.5 0.336; 2.7 0.2881; 2.9 0.2497; 3.10.2185;3.3 0.1928; 3.5 0.1714; 3.7 0.1534; 3.9 0.1381; 4.1 0.1249;4.3 0.1136;4.8 0.0911; 6 0.0583; 7 0.0429; 8 0.0328; 10 0.021; 200.0053; 30 0.0023;

For Load case 3, I have

SPECTRUM SRSS X 1 Z 1 ACC SCALE 0.9806 DAMP 0.050.03 0.8702; 0.05 1.0752; 0.1 1.5876; 0.2 2.1; 0.3 2.1; 0.52.1; 0.7 1.5;0.9 1.1667; 1.1 0.9545; 1.3 0.8077; 1.5 0.7; 1.7 0.6176; 1.90.5526;2.1 0.4762; 2.3 0.397; 2.5 0.336; 2.7 0.2881; 2.9 0.2497; 3.1

0.2185;3.3 0.1928; 3.5 0.1714; 3.7 0.1534; 3.9 0.1381; 4.1 0.1249;4.3 0.1136;4.8 0.0911; 6 0.0583; 7 0.0429; 8 0.0328; 10 0.021; 200.0053; 30 0.0023;

Load combination 5 is an SRSS of 1 & 2.

LOAD COMBINATION SRSS 5 Überlagerung


17

1 1.0 2 1.0

Should load case 5 produce the same answers as load case 3?



Answer : Load case 1 means the earthquake is acting in the X direction at an

intensity of say 100%.

Load case 2 means the earthquake is acting in the Z direction at an

intensity of say 100%.

Then, load case 3 means the earthquake is acting at a 45 degree

angle to the X and Z directions at an intensity of 141.414%.

Load combination 5 will not produce the same result as load case

3. An earthquake with a 100% intensity in X and another with a

100% intensity in Z is not the same as one with a 141.4% intensity

at a 45 degree angle to X and Z. The combination methods such asSRSS or CQC are not linear.

Another reason for the difference has to do with the Direction

factor.


18





Time History analysis of a structure

for

seismic accelerations



1

Time history analysis is an extension to the process of calculating

modes and frequencies in the sense that it occurs after those are

calculated.

The input which is relevant to the time history analysis of a



The input which is relevant to the time history analysis of a

structure for seismic accelerations is explained below.

There are two stages in the command specification required for a

time-history analysis.

Stage 1 : The first stage is defined as shown in the followingexample. Here, the characteristics of the earthquake, the arrival

time, and damping are defined.

Example :

UNIT METER

DEFINE TIME HISTORY

TYPE 1 ACCELERATION SCALE 9.806

READ EQDATA.TXT

ARRIVAL TIME

0.0

DAMPING 0.05

Each data set is individually identified by the number that follows

the TYPE command. In this file, only one data set is defined,

which is apparent from the fact that only one TYPE is defined.

The word ACCELERATION that follows the TYPE 1 command

signifies that this data set is for a ground acceleration. (If one

wishes to specify a forcing function, the keyword FORCE orMOMENT must be used instead.)

Notice the expression "READ EQDATA.TXT". It means that we

have chosen to specify the time vs. ground acceleration data in the

file called EQDATA.TXT. That file must reside in the same folder

as the one in which the data file for this structure resides. As

explained in the small examples shown in Section 5.31.4 of the

Technical Reference manual, the EQDATA.TXT file is a simple


2

text file containing several pairs of time-acceleration data. A

sample portion of that file is as shown below.

0.0000 0.0063000.0200 0.003640



0.0200 0.003640

0.0400 0.000990

0.0600 0.004280

0.0800 0.007580

0.1000 0.010870

While it may not be apparent from the above numbers, it may alsobe noted that the geological data for the site the building sits on

indicate that the above acceleration values are a fraction of "g",

the acceleration due to gravity. Thus, for example, at 0.02 seconds,

the acceleration is 0.00364 multiplied by 9.806 m/sec^2 (or

0.00364 multiplied by 32.2 ft/sec^2). Consequently, the burden of

informing the program that the values need to be multiplied by "g"

is upon us. We do that by specifying the term “SCALE 9.806”

alongside “TYPE 1 ACCELERATION”.

The arrival time value indicates the relative value of time at which

the earthquake begins to act upon the structure. We have chosen

0.0, as there is no other dynamic load on the structure from the

relative time standpoint. The modal damping ratio for all the

modes is set to 0.05.

Stage 2 :

UNIT POUND FEET

LOAD 3 DYNAMIC LOAD CASE

SELFWEIGHT X 1.0

SELFWEIGHT Y 1.0

SELFWEIGHT Z 1.0

ELEMENT LOAD

41 TO 88 PR GX 300.0

41 TO 88 PR GY 300.041 TO 88 PR GZ 300.0


3

Load case 3 is the dynamic load case, the one which contains the

second part of the instruction set for a dynamic analysis to be

performed. The data here are

a. loads which will yield the mass values which will populate



y p p

the mass matrix

b. the directions of the loads, which will yield the degree of

freedom numbers of the mass matrix for being populated.

Thus, the selfweight, as well as the imposed loads on the structuralslab are to be considered as participating in the vibration along all

the global directions. This information is identical to what is

specified in the situation where all that we are interested is

frequencies and modes.

GROUND MOTION X 1 1

The above command too is part of load case 3. Here we say that

the seismic force, whose characteristics are defined by the TYPE 1

time history input data, acting at arrival time 1, is to be applied

along the X direction.

Example:

LOAD 1

Mass data in weight units

GROUND MOTION direction Type# Arrival Time#

PERF ANALFINISH


4





Time History Analysis for a Structure

subjected to a

Harmonic Loading



1

A sinusoidal loading is one which has the characteristic of

repetitiveness, as in the case of a tower at the top of which are two

radar antennas which cause a rotational type of dynamic loading

with a specified rotation rate and a nominal turning circle.

A i id l l di ll b d ib d i th ti



A sinusoidal loading usually can be described using the equation.

φ)t(ωsin0F(t)F +=

In the above equation,

F(t) = Value of the force at any instant of time "t"

F = Peak value of the force

ω = Frequency of the forcing function

φ = Phase angle

A plot of the above equation is shown in the figure below.

Definition of input in STAAD for the above forcing function

As can be seen from its definition, a forcing function is a

continuous function. However, in STAAD, a set of discrete time-

force pairs is generated from the forcing function and an analysis

is performed using these discrete time-force pairs. What thatmeans is that based on the number of cycles that the user specifies

for the loading, STAAD will generate a table consisting of the

magnitude of the force at various points of time. The time values

are chosen from time '0' to n*tc in steps of "STEP" where n is the

number of cycles and tc is the duration of one cycle. STEP is a

value that the user may provide or may choose the default value

that is built into the program. Users may refer to section 5.31.4 of the Technical Reference Manual for a list of input parameters that


2

need to be specified for a Time History Analysis on a structure

subjected to a Sinusoidal loading.

A typical example of input specification for the above is shown below. Some typical input that normally appears pr ior to these

commands is also included



commands is also included.

UNIT KIP INCH

DEFINE TIME HISTORY

TYPE 1 FORCE

FUNCTION SINEAMPLITUDE 10.8 FREQUENCY 47 PHASE 30 CYCLES 150

TYPE 2 FORCE

FUNCTION COSINE

AMPLITUDE 12.3 FREQUENCY 28 PHASE 40 CYCLES 200

ARRIVAL TIME

0.0 3.0

DAMPING 0.06

There are two stages in the command specification required for a

time-history analysis. The first stage is defined above. Here, the

parameters of the sinusoidal loading are provided.

Each data set is individually identified by the number that follows

the TYPE command. In this file, two data sets are defined, which

is apparent from the fact that two TYPEs are defined.

The word FORCE that follows the TYPE n command signifies that

this data set is for a forcing function. (If one wishes to specify an

earthquake motion, an ACCELERATION may be specified.)

The command FUNCTION COSINE indicates that instead of

providing the data set as discrete TIME-FORCE pairs, a sinusoidal

function, which describes the variation of force with time, is

provided.

The parameters of the cosine function, such as FREQUENCY,

AMPLITUDE, and number of CYCLES of application are thendefined. STAAD internally generates discrete TIME-FORCE pairs


3

of data from the sine function in steps of time defined by the

default value (See section 5.31.6 of the Technical Reference

Manual for more information). The arrival time value indicates the

relative value of time at which the force begins to act upon the

structure. The modal damping ratio for all the modes is set to

0 075



0.075.

LOAD 1 DEAD LOAD

SELF Y -1.0

The above is a static load case.

LOAD 2 LOADING FOR TIME HISTORY ANALYSIS

SELF X 1.0

SELF Y 1.0

SELF Z 1.0

JOINT LOAD

10 FX 7.5

10 FY 7.5

10 FZ 7.5

TIME LOAD

7 FX 1 1

14 FZ 2 1

17 FZ 2 2

The above is the second stage of command specification for time

history analysis. The 2 sets of data specified here are a) the

weights for generation of the mass matrix and b) the application of

the time varying loads on the structure.

The weights (from which the masses for the mass matrix areobtained) are specified in the form of selfweight and joint loads.

Following that, the sinusoidal force is applied using the "TIME

LOAD" command. The forcing function described by the TYPE 1

load is applied on joints 7 it starts to act starting at a time defined

by the 1st arrival time number. At joint 14, the TYPE 2 force is

applied along FZ, also starting at arrival time number 1. Finally, at


4

joint 17, the TYPE 2 force is applied along FZ, star ting at arrival

time number 2.

LOAD COMB 3

1 1.2 2 1.4



The static and dynamic load cases are combined through the above

case.

PERFORM ANALYSIS

PRINT SUPPORT REACTIONS

PRINT MEMBER FORCES

PRINT JOINT DISPLACEMENTS

The member forces, support reactions and joint displacements are

calculated for every time step. For each degree of freedom, the

maximum value of these values is extracted from these historiesand reported in the output file using the above commands.

How modes, frequencies and the other terms calculated

The process of calculating the MODES and FREQUENCIES is

known as Modal Extraction and is performed by solving the

equation:

ω2[ m ] { q } - [ K ] { q } = o

Where

[ m ] = the mass matrix (assumed to be diagonal, i.e., nomass coupling)

ω = the natural frequencies (eigenvalues)

{ q } = the normalized mode shapes (eigenvectors)

Frequency (HZ or CPS) = ω/2π

The solution method used in STAAD is the Subspace iterationmethod.


5

Please note that various nomenclature is used to refer to the

normal modes of vibration. (Eigenvalue, Natural Frequency,

Modal Frequency and Eigenvector, Mode Shape, Modal Vector,

Normal Modes, Normalized Mode Shape.

Generalized weight and generalized mass



g g

Each eigenvector {q} has an associated generalized mass defined

by

Generalized Mass (GM) = { q }T

[ M ] { q }

Generalized Weight (GW) = GM * g

Participation Factors - A participation factor (Qi) is computed

for each eigenvector for each of the three global (Xi) translational

directions. N is the number of modes.

Q

q w

GWi

j,i j,i j

N

=∑=

( )( )1

Modal Weights - The modal weight for each mode is (GW)(Q i²) .

The summation of modal weights for all modes in a given directionis equal to the Base Shear which would result from a one g base

acceleration. The sum of the modal weights for the computed

modes may be compared to the total weight of the structure (only

the weight that has not been lumped at supports). The difference

is the amount of weight missing from a dynamic, base excitation,

modal response analysis. If too much is missing, then rerun the

eigensolution asking for a greater number of modes.

STAAD prints the "MASS PARTICIPATION FACTOR IN

PERCENT" for each mode. This is the modal weight of a mode as

a percentage of the total weight of the structure. Also a running

sum for all modes is given so that the last line indicates the percent

of the total weight that all of the modes extracted would representin a 1g base excitation.


6





Time History Analysis for a Structure

subjected to a

random excitation



1

A random excitation is a force which varies with time, and not

necessarily in an orderly fashion. An example of the same is a

blast loading.

The only difference between this type of loading and the

sinusoidal loading is that the force versus time data has to be

d fi d li i l d h DEFINE TIME HISTORY d



defined explicitly under the DEFINE TIME HISTORY command.

An example of it is shown below.

UNIT METER KNSDEFINE TIME HISTORY

TYPE 1 FORCE

0.00001 -0.000001 0.005 -650 0.01 -800 0.015 -800 0.02 -800

0.025 -800

0.03 -700 0.035 -350 0.04 -250 0.045 -500 0.05 -730 0.055 -600

0.06 -350 0.065 -280 0.07 -450 0.075 -600 0.08 -550 0.085 -440

0.09 -415 0.095 -410 0.1 -420

ARRIVAL TIME

0.0

DAMPING 0.07

For a blast type of loading, there will be a sudden spike in the

value of the force over a very short period of time.

DEFINE TIME HISTORY

TYPE 1 FORCE

0.0 0.0 0.1 80.0 0.2 0.1 0.35 0.0 0.4 0.0 1.0 0.0

ARRIVAL TIMES

0.0

DAMPING 0.05


2





Mat foundations



1

Description

STAAD has the ability to generate supports for structures like

slabs on grade, which also go by the name mat foundations. A mat

foundation is a large concrete slab sitting on soil. The support for

the structure is the soil itself. The resistance of the soil is

represented through a term called Modulus of Subgrade Reaction,



ep ese ted t oug a te ca ed odu us o Subg ade eact o ,

the definition of which may be found in many textbooks on

foundation analysis.

The general approach to solving such problems is to sub-divide theslab into several plate elements. Each node of the meshed slab will

then have an influence area or a contributory area, which is to say

that soil within the area surrounding that node acts like a spring.

The influence area is then multiplied by the subgrade modulus to

arrive at the spring constant. Subgrade modulus has units of force

per length^3. So, the spring will have units of force/length.

The problem with using this method is that, for irregularly-shaped

or large slabs with many nodes, computing the influence area for

each node can become quite tedious and time-consuming. The

model below exemplifies the problem.


2



This is where the Foundation type of support can be useful.

STAAD will calculate the influence areas of all the nodes by itself

and derive the spring constants for you. In STAAD, we refer to

facility as SPRING SUPPORT GENERATION .

STAAD has two options for such supports:

a) The ELASTIC MAT option

b) The PLATE MAT option

The ELASTIC MAT option :

When the spring support generation facility was first introduced in

STAAD, it was based on this method. In fact, this was the only

method available until and including STAAD.Pro 2002 Build 1004.

This method calculates the influence area of the various nodes

using the Delaunay triangle method.


3

The distinguishing aspect of this method is that it uses the joint-

list that accompanies the ELASTIC MAT command to form a

closed surface. The area within this closed surface is then

determined and the share of this area for each node in the list is

then calculated.

Hence, while specifying the joint-list, one should make sure that



these joints make up a closed surface. Without a proper closed

surface, the area calculated for the region may be indeterminate

and the spring constant values may be erroneous. Consequently,

the list should have at a minimum, 3 nodes.

While forming the closed surface, namely, a polygon, the sides of

the polygon have to be assembled by lining up points along the

edges. The edge detection aspects of this method are very sensitive

to out-of-straightness, which may occur if the coordinates of the

nodes aren't precise to a significant number of digits.

Also, the internal angle formed by 2 adjacent lines connecting 3

consecutive nodes in the list should be less than 180 degrees,

which is to say that, the region should have the shape of a convex

polygon.

Failure to form straight edges and convex polygons can lead to

erroneous influence area values and consequently, erroneous

spring constants. This is the limitation of this feature.

The example below explains the method that may be used to get

around a situation where a convex polygon is not available.

For the model comprised of plate elements 100 to 102 in the figure

below, one wishes to generate the spring supports at nodes 1 to 8.

However, a single ELASTIC MAT command will not suffice

because the internal angle between the edges 1-8 and 8-7 at node 8

is 270 degrees, which violates the requirements of a convex

polygon.


4

So, one should break it up into 2 commands:

1 2 3 8 ELASTIC MAT DIREC Y SUBG 200.

3 4 5 6 7 8 ELASTIC MAT DIREC Y SUBG 200.



Joints 3 and 8 will hence get the contribution from both of the

above commands.

Because this method uses nodes to generate contours, it may be

used whether the mat is defined using plates, or solids. This is the

advantage of this method.


5

The PLATE MAT option :

If the foundation slab is modeled using plate elements, the

influence area can be calculated using the principles used in

determining the tributary area of the nodes from the finite element

modeling standpoint. In other words, the rules used by the program

in converting a uniform pressure load on an element into fixed end



actions at the nodes are used in calculating the influence area of

the node, which is then multiplied by the subgrade modulus to

obtain the spring constant. This feature has been available since

STAAD.Pro 2002 Build 1005.

The advantage of this method is that it overcomes one of the major

limitations of the Delaunay triangle method, which is that the

contour formed by the nodes of the mat must form a convex hull.

Example

SUPPORTS

17054 TO 17081 PLATE MAT DIR YONLY SUBGRADE 5000.0

PRINT

YR -.01 0.01 PLATE MAT DIR YONLY SUBGRADE 5000.0

The first of the above 2 commands instructs STAAD to internally

generate supports for the nodes at the corners of plate elements

17054 TO 17081.

The second example instructs STAAD to internally generate

supports for the nodes at the corners of plate elements which lie in

the global XZ plane bound by the YRANGE value of -0.01 and

+0.01 length units.

Another advantage of the PLATE MAT method is that it enables us

to view soil pressure contours beneath the base of the slab. After

the analysis, go to the post-processing mode, and click on the

Plates page. In the selection box for choosing the type of result to

plot, choose base pressures. This is not currently available with the

ELASTIC MAT method.


6

Question : How do I tell STAAD that my soil spring is effective only in

COMPRESSION, and should not be considered when it goes into

tension?

Answer : This may be done by using the ELASTIC MAT or PLATE MAT

command in conjunction with the SPRING COMPRESSION

command. The program iteratively solves the problem so that the



p g y p

final answer reflects the condition corresponding to actual contact

between slab & soil. Example problem 27 illustrates this.

Question : Is it possible to get a report which shows the influence area

generated by STAAD for each support node?

Answer : Yes. Use the PRINT option available with the ELASTIC MAT or

PLATE MAT commands. This will produce a report of the

influence areas. An example of such a report is shown below.

To get a report of the spring constants themselves, use the

command

PRINT SUPPORT INFORMATION


7

Question : Is it possible to find out the base pressure at each node for each

load case?

Answer : Yes. In the post-processing mode, go to the Node – Base pressurepage. A table will appear along the right side of the screen

showing these values. The Summary tab will show the maximum

and minimum pressure along with the associated node for each of



the 3 global directions.

Question : How does subgrade modulus differ from soil bearing capacity?

Answer : A soil must be capable of carrying the loads it is subjected to,

without undergoing a shear failure, or excessive settlements. This

capacity is referred to as the soil bearing capacity.

The modulus of subgrade reaction is a measure of the stiffness of

soil if it were to behave like a spring. It is the relationship betweenbearing pressure and soil deflection.

The modulus of subgrade reaction is the quantity by which the

influence area of a support node is multiplied by to get the

equivalent spring constant which can be used at the analysis stage.

One would provide this as an input item when one wishes STAAD

to generate spring supports using the ELASTIC MAT command, asexplained in section 5.27.3 of the STAAD.Pro Technical Reference

manual.

At the end of the mat foundation analysis, the maximum soil

pressure you get from STAAD’s soil pressure diagram should be

within the limits of the soil’s bearing capacity. If the actual

pressure exceeds the capacity, it is an indication of failure.


8

Question : If you have the value for soil bearing pressure, how do you use

that to come up with the subgrade modulus that STAAD uses for

elastic mat definitions?

Answer : One doesn't use the bearing capacity of soil to determine the

subgrade modulus. Instead, it is a separate attribute of soil. If you

have a look at the text book "Foundation Analysis and Design" by



Joseph Bowles, you will find a few sections devoted to that topic,

with specific values listed for specific types of soil.

The basic difference between these 2 attributes is that, bearingcapacity (or bearing pressure) is the pressure at which the soil

fails, either in shear or compression. It hence has units of force per

unit area. Subgrade Modulus on the other hand is a measure of the

"spring constant" of soil. It is the distance that a unit area of soil

would deflect under a unit load.



Generating loads from

moving load-causing units



1

This type of loading occurs classically when the load-causing

units move on the structure, as in the case of trucks on a bridge

deck. The mobile loads are discretized into several individual

immobile load cases at discrete positions.

Defining the input data

There are 2 stages for specifying these types of loads.



g y g y

Stage 1 is as shown in the example below.

DEFINE MOVING LOAD

TYPE 1 LOAD 119.6 108.3 94.5 DISTANCE 1.778 1.5 WIDTH 1.8

TYPE 2 LOAD 34.9 34.9 34.9 34.9 DISTANCE 1.3 1.3 1.3 WIDTH 1.7

The above lines represent the first out of two sets of data required

in moving load generation. The type number (1) is a label for

identification of the load-causing unit, such as a truck. 3 axles (119.6 108.3 94.5) are specified with the LOAD command. The

spacing between the axles in the direction of movement

(longitudinal direction) is specified after the DISTANCE

command. Since there are 3 axles, there are 2 spacings between

them. WIDTH is the spacing in the transverse direction, that is, it

is the distance between the 2 prongs of an axle of the truck. For

the TYPE 2 truck, there are 4 axles and 3 spacings.

LOAD 1

SELF Y -1.0

Load case 1 is a static load case.

LOAD GENERATION 75

TYPE 1 -3.278 0. 4. XINC 1.5

TYPE 2 -3.9 0. 6. XINC 1.5

This constitutes the second of the two sets of data required for

moving load generation. 75 load cases are generated using the

Type 1 and Type 2 vehicles whose characteristics were describedearlier. For the first of these load cases, the X, Y and Z location of


2

the reference load (see section 5.31.1 of the Technical Reference

Manual) have been specified after the command TYPE 1 and TYPE

2 respectively. The X Increment of 1.5ft denotes that the vehicle

moves along the X direction and the individual positions which are

1.5ft apart will be used to generate the remaining 74 load cases.

The basis for determining the number of load cases to generate, 75

in the example above, is as follows :



As seen in Section 5.31.1 of the Technical Reference manual, the

reference wheel is on the last axle. The first load case which is

generated will be the one for which the first axle is just about to

enter the bridge. The last load case should be the one for which the

last axle is just about to exit the bridge. Thus, the total distance

travelled by the reference load will be the length of the vehicle

(distance from first axle to last axle) plus the span of the bridge.

Let us call this term "D".

If we want the vehicle to move forward in 1.5 feet increments

(each 1.5 foot increment will create a discrete position of the truck

on the bridge), it would required (D/1.5+1) cases to be generated.


The load generation commands are followed by the PERFORMANALYSIS command. The PRINT LOAD DATA option is used to

obtain a report in the output file of the values and positions of the

generated loads.


3

Question : I want to move a crane along a beam. How do I use the moving

load generation for this case?

Answer : Use the same procedure as in the case of a bridge. Set the WIDTHvalue to zero.

Question : Could you tell me how I can display the generated moving loads

graphically? I want to see whether I enter and generate the moving



graphically? I want to see whether I enter and generate the moving

loads correctly.

Answer : If you wish to obtain the position of the concentrated loads

generated from a moving vehicle, this is what you can do.

First, make sure the input file does have the commands required to

generate loads from a vehicle. Example 12 is a good reference.

Then, run the analysis. After the analysis is successfullycompleted, the "Select Load" drop down list box will contain

individual load case numbers for each generated load case. For

example, if your sequence of load data is

LOAD 1

LOAD 2

LOAD 3LOAD GENERATION 30

then, after the analysis, the load selection box will list them as

LOAD 1

LOAD 2

LOAD 3

LOAD GENERATION, LOAD # 4



etc.


4

Select those cases, and switch on the load display icon. Or, right

click the mouse on the drawing area. Select Structure Diagrams. In

the Loads and Results tab, switch on the check box for Loads,

select the load case from the list, and click on Apply. Keep

changing the load case, and keep clicking on Apply.

Question: Is there any way to generate a moving load on an inclined member

?



Answer : Yes you can. Have a look at Section 5.32.12 of the Technical Ref

Manual. You will find an option called YRANGE. So, have theload located at an elevation below the lower node of the member,

and provide a YRANGE which will enable the program to apply

the load on members whose longitudinal axis lie in the range

between the lower and upper ends of the inclined member.

However, there is no guarantee that it will work every time.

Question : How do I define the moving load data through an external file?

Answer : See example below :

Example : When data is provided through the external file

"MOVLOAD"

Data in the input file

UNIT KIP FEET

DEFINE MOVING LOAD FILE MOVLOAD

TYPE 1 AXLTYP1

TYPE 2 AXLTYP2 1.25


5

Data in the external file "MOVLOAD"

AXLTYP1

10 20 15

5.0 7.5

6.0

AXLTYP2

20 20

10



10

7.5


6





Pressure loads on panels – Floor loads



1

Question : I am modeling a steel building consisting of columns and beams.

The floor slab is a non-structural entity, which, though capable of

carrying the loads acting on it, is not meant to be an integral part

of the framing system. It merely transmits the load to the beam-

column grid.

There are uniform area loads on the floor (think of the load as

wooden pallets supporting boxes of paper). Since the slab is not

part of the structural model is there a way to tell the program to



part of the structural model, is there a way to tell the program to

transmit the load to the beams without manually figuring out the

beam loads on my own?

Answer: STAAD's FLOOR LOAD option is ideally suited for such cases.

This is a facility where you specify the load as a pressure, and the

program converts the pressure to individual beam loads. Thus, the

input required from the user is very simple - load intensity in the

form of pressure, and the region of the structure in terms of X, Y

and Z coordinates in space, of the area over which the pressure

acts.

In the process of converting the pressure to beam loads, STAAD

will consider the empty space between crossing beams (in plan

view) to be panels, similar to the squares of a chessboard. The load

on each panel is then transferred to beams surrounding the panel,

using a triangular or trapezoidal load distribution method.

Users can verify the accuracy of the values of the joint and

member loads generated by the FLOOR LOAD and AREA LOAD

option by using the command


The output file will contain the values of the generated loads. If

the values are not what you expect, you may directly specify the

JOINT LOADs and MEMBER LOADs on those members instead

of using the FLOOR LOAD option to generate loads for those

members.


2

STAAD also provides an option called ONEWAY load if the

distribution is desired along the shorter direction of a panel instead

of a 2-way action. This and additional information on the FLOOR

LOAD facility is available in example problem 15 in the examples

manual, and section 5.32.4 in the STAAD.Pro Technical Referencemanual.

Question : Are there any graphical tools to examine the individual panels the

program considers in processing the floor load command?



program considers in processing the floor load command?

Answer : Yes. Click the right mouse button, and select Labels. UnderLoading Display Options, Display Floor Load Distribution will

show the division of panels into influence areas based on a color-

coded scheme (see figure below). Display Floor Loads will show

the triangular and trapezoidal loading on the individual members

around each panel.


3

Question : When does one use FLOOR LOAD and when does one use

ELEMENT LOAD?

Answer : When modeling a grid system made up of horizontal beams and the

slabs which span between the beams, there are 2 approaches one

may take :

1. Model the beams only, and do not include the slabs in the

model However the large in-plane stiffness of the slab may



model. However, the large in plane stiffness of the slab may

be taken into account by using the master-slave relationship

to tie together the nodes of the deck so that a rigid diaphragmeffect is simulated for the horizontal plane at the slab level.

2. Include the slabs by modeling them using plate elements.

The question that arises is, how does one account for the

distributed loading (load per area of floor) which is present on top

of the slab?

If you model the structure using method (1), the load can be

assumed to be transferred directly on to the beams. The slab-beam

grillage is assumed to be made up of a number of panels, similar to

the squares of a chessboard. The load on each panel is then

transferred to beams surrounding the panel, using a triangular or

trapezoidal load distribution method. You can do this in STAAD

by defining the load intensity in the FLOOR LOAD command. In

other words, the pressure load on the slabs (which are not included

in the model) are converted to individual beam loads by utilizing

the FLOOR LOAD facility.

In method (2), the fact that the slab is part of the model makes itvery easy to handle the load. The load can be applied on individual

elements using the ELEMENT LOAD facility. The connectivity

between the beams and elements ensures that the load will flow

from the plates to the beams through the columns to the supports.


4

Question : I have a floor made up of several panels. The floor consists of

straight-line edges but with a concave face and a convex face, like

a boomerang.

The total floor area is 381 sq.m. and I am applying a floor load of

1 t/sq.m. on the entire area. Thus, expecting a total load of 381 t.

From analysis I get total load as 810.2 t which is not correct.

When I try to apply floor load to individual panel I get nearly the



expected load. But when the floor load is applied on group of

panels or on entire area, graphically it shows wrong distribution of load and total load is also not correct.

Answer : The problem you mention is one of the limitations of the floor load

routine. If you have a floor whose shape contains a mixture of

concave and convex edges, break up the floor load command into

several parts, as you have done. This will force the program to

localize its search for panels and the solution will be much better.If you don't do this, the entire floor will end up being treated as

one giant panel with unsatisfactory results.

The example below illustrates a case where the floor has to be sub-

divided into smaller regions for the floor load generation to yield

proper results. The internal angle at node 6 between the sides 108

and 111 exceeds 180 degrees. A similar situation exists at node 7

also. As a result, the following command

LOAD 1

FLOOR LOAD

YRANGE 11.9 12.1 FLOAD -0.35

will not yield acceptable results.


5

Instead, the region should be subdivided as shown in the

following example

LOAD 1

FLOOR LOADYRANGE 11.9 12.1 FLOAD -0.35 XRANGE -0.1 15.1

ZRANGE -0.1 8.1

YRANGE 11.9 12.1 FLOAD –0.35 XRANGE 4.9 10.1

ZRANGE -7.9 16.1




6





Wind load generation



1

Description

Wind load generation in STAAD.Pro is a facility, which takes as

input wind pressure, and height ranges over which those pressures

act and generates nodal point loads on windward and leeward sidesof buildings. This may be found in sections 5.31.3 and 5.32.12 of

the Technical Reference manual, and in example problem 15 of the

Examples manual.

Until and including STAAD.Pro 2003, this feature is capable of



g , p

generating loads on panel type of exposed faces. So, the basis of

this generation is that the program first identifies panels – regions

circumscribed by members and the ground – and assumes that the

wind pressure acts on the panel area. So, the force on that panel is

calculated by multiplying the pressure by the panel area.

Consequently, this type of load generation is applicable to

“closed” structures such as office buildings where the component

constituting the panels could be a glass façade, or walls made of

wood or other material that was not considered to be part of the

structural model.

This facility has been enhanced in STAAD.Pro 2004 by

considering lattice type open structures also.

Defining the input data

There are 2 stages for specifying these types of loads.

Stage 1 is as shown in the example below.

UNIT KIP FEET

DEFINE WIND LOAD

TYPE 1

INT .015 .022 .026 .028 HEI 10. 30. 60. 100.

EXPOSURE 1.2 YR 0. 75.

The numbers which follow the word INTENSITY are the wind

pressures. The first intensity acts from the ground (the datum) to


2

10 ft, the second from 10 ft to 30 ft, and so on. EXPOSURE

factors, which are magnification or reduction factors for the

resulting generated loads should be specified if their value is

different from 1.0. Here, all nodes between 0 and 75 feet are

assigned a value 1.2.

LOAD 1 DW

SELFWEIGHT Y -1.

LOAD 2 WIND



WIND LOAD X -1. TYPE 1

LOAD 3 WINDZ

WIND LOAD Z 1. TYPE 1

The second part of the command consists of the actual load

application and is done through the WIND LOAD command as

shown above.

Question : What is the significance of the TYPE command and the number

that follows?

Answer : STAAD permits the definition of several different wind loads, each

with certain characteristics. In order to distinguish the wind load

having one set of characteristics from another wind load with a

different set of characteristics, each wind load is identified using a

TYPE command followed by an identification number. In other

words, the TYPE command and the number are entirely a creation

of the user. They are not terminologies that the user will find from

any code or handbook that provides guidelines on loading for

structures. The advantage of this feature is that the user is nowable to communicate to the program information such as that the

wind pressure is different at different heights, the structure has

openings at certain heights and so on.


3

Question : What is Wind Intensity?

Answer : Wind intensity as required for input in STAAD is merely the wind

pressure in units of Force per unit area. The user is required to

compute the pressure from any coefficients that codes require.

Question : Does the wind load command in STAAD take into account any

wind codes like ASCE 7? Does it take into account the drag factor,

or shape factor for different shapes like angle, channel etc.



Answer : WIND LOAD generation in STAAD is not based on any code. It

also does not take into account any of the other factors you

mentioned. It is based purely on the concept of influence areas of

nodes and multiplying them by the user defined wind pressures for

the respective heights. Any reduction or magnification of the

resulting force is achieved by multiplying the generated values by

exposure factors for nodes.

Influence area of a node is defined as the region surrounding a

node over which any wind pressure acting over that area is

transmitted entirely into that node as a concentrated force.

Influence area is equal to influence length multiplied by influence

height where:

influence length is half the distance from the joint to the

joints to the left and to the right of the joint and

influence height is the distance from the joint to the joints at

the top and to the bottom.

Multiply the influence area of each joint by the wind intensity and

the exposure factor for the joint. This will yield the concentrated

horizontal force for the joint. The exposure factor becomes useful

for situations where the entire panel area is not effective due to the

presence of openings or needs to be magnified due to a curvilinear

shape of the load bearing panel.


4

Question : If I have wind speeds from different directions acting on a tower

having (round shaped) discs fitted to it, how can I make the

software take the discs into account which are also exposed to

wind?

Answer : The influence area calculation will work correctly if and only if the

exposed area is parallel to one of the global planes. A region

which is curvilinear in shape cannot be handled by the program.

Question : Can the wind force be generated in the Y direction?



Question : Can the wind force be generated in the Y-direction?

Answer : No. Wind forces can be computed for horizontal (X and Z)

directions only.

Question : What if the windward face of the structure is inclined to the X and

Z axes, viz., not perpendicular to X or Z axes?

Answer : The feature works best when the panels are parallel to one of the

global planes. The program does have some capability for

generating loads on inclined planes too. However, if the user finds

the results unsatisfactory, other load generation methods like the

"ELEMENT LOAD JOINTS" option may be used.

Question : What influence do finite elements have on wind load generation?

Answer : The presence or absence of elements, along or perpendicular to the

direction of wind has no effect on wind load generation. Wind load

generation is possible only with panels surrounded by members as

described above. If the panels are already defined using plate

elements, apply the load using the ELEMENT PRESSURE optioninstead of using wind load generation for those panels.


5

Question : I have three questions. 1) How can I tell STAAD that the load is

acting on the LEEWARD side and not on the WINDWARD side of

the building? 2) How do I specify that a load acts from east to west

instead of west to east? 3) How do I specify a suction load instead

of one which bears against the structure?

Answer : The command syntax accommodates all of the above. For example,

along the X direction, the following four types are allowed.

WIND LOAD +X +f



WIND LOAD -X +f WIND LOAD +X -f

WIND LOAD -X +f

See the figure below for the meaning of the four commands.

X or Z - f -X or -Z - fX or Z X or Z

X or Z

X or Z + f -X or -Z + fX or Z

Y

Y

Y

Y


6

Question : Can STAAD perform wind load generation on open-lattice

structures?

Answer : In STAAD.Pro 2004, the wind load generation facility has been

enhanced for generating loads on open structures too. These are

structures like electrical transmission towers, in which the region

between members is “open” allowing the wind to blow through.

For those, the program first calculates the exposed surface area of

individual members of the model. Then, that exposed area is

multiplied by the wind pressure and divided by the member length



p y p y g

to arrive at a uniform distributed member load. It is assumed thatall members of the structure within the specified ranges are

subjected to the pressure and hence, they will all received the load.

The concept of members on the windward side shielding the

members in the inside regions of the structure does not exist for

open structures.

Documents

STAAD PRO TRAINING MANUAL