Oreta, A.W.C. (2004)."Understanding 2D Structural Analysis

Understanding 2D Structural Analysis:

Learning Modules in the Modeling and Analysis of Framed Structures using GRASP

Andres Winston C. Oreta Department of Civil Engineering De La Salle University Manila, Philippines

This project was funded by the University Research Coordination Office (URCO)

De La Salle University, Manila, Philippines 2004

Understanding 2D Structural Analysis: Learning Modules in the Modeling and Analysis of Framed Structures Using GRASP CONTENTS Preface

1 Understanding Structural Analysis

2 A Tour of GRASP

3 Loading Continuous Beams

4 Pattern Loading in Multistory Frames

5 Lateral Forces in Buildings

6 Pinned and Fixed Support Conditions

7 Soil Effects on Foundations

8 Support Settlements

9 Truss Analysis

10 Special Modeling Issues

About the Author

PREFACE An exploratory-type of instructional and learning material consisting of ten

modules about modeling and analysis of framed structures in 2D is presented.

Each module focuses on a specific issue on structural modeling and analysis

which is discussed with the aid of graphical and tabular results obtained from the

2D structural analysis software, GRASP. The set of learning modules is not a

substitute to a textbook on structural analysis. The theory is not presented. No

derivations or equations can be found. The student or reader must refer to the

textbooks for definitions, equations and techniques. Each chapter begins with

background information and a “case study”. The reader explores the issues

raised in the case study through the “Things to Do” activities or by simply

observing and analyzing the “Observation” and graphical and tabular results

presented in the module. Included in the modules are “Things to Try” exercises

and “Things to Ponder” comments on the analysis and design of structures.

Using the set of learning modules, the reader or student with the aid of a

structural analysis software like GRASP discovers important insights on the

response and behavior of structures due to variations in the parameters of the

model and configurations of the structure, changes in member and material

properties, and also changes in the restraint and loading conditions. Through

the graphical results, the student can visualize the phenomena and this would

accelerate his understanding of concepts through the experience of seeing and

interpreting solutions to various structural modeling and analysis problems. The

implication and relevance of the case study to the safe and reliable design of

structures are also discussed. Each chapter ends with a set of references and

reading materials related to the issue presented in the module. The student is

encouraged to perform the “Things to Try” exercises which are related to the

case study. Since GRASP provides direct feedback graphically and numerically,

the student can explore and have fun by simple modification of the configuration

of the structural model or loading condition and will expand his knowledge and

understanding about modeling, analysis and design of framed structures.

Understanding 2D Structural Analysis by A.W.C.Oreta 1 - 1

Architectural Functional Plans

Trial Sections

Revise Sections

Structural System

Modeling

Analysis

Final Design

Acceptable

Member Design

Detailing

Connection Design

CONCEPTUAL DESIGN

MODELING & ANALYSIS DESIGN & DETAILING

No

Yes

Figure 1.1 The Structural Design Process (ACECOMS ISCAAD Workshop Notes)

CHAPTER 1 UNDERSTANDING STRUCTURAL ANALYSIS ROLE OF STRUCTURAL ANALYSIS

Structural analysis is an integral part of any structural engineering project. The

structural design process of a typical structural engineering project may be

divided in to three phases as shown in Figure 1.1. The conceptual design phase


usually involves the formulation of the functional requirements of the proposed

structure, the preparation of the general layout and dimensions of the structures,

and the consideration of the possible types of structural system to be used. In the

modeling and analysis phase, a preliminary design of the structure is proposed

using trial sections. A model of the structure is developed and the loads that may

act on the structure are estimated. Structural analysis of the model is now

performed to determine the stresses or stress resultants in the members and the

deflections at various points of the structure. The results of the structural

analysis are used in design and detailing phase where the structural members

are designed to satisfy safety and serviceability requirements of the design codes.

If the code requirements are not satisfied, then the member sizes are revised and

a re-analysis of the model of the structure is carried out until all safety and

serviceability requirements are satisfied.

ANALYTICAL MODEL OF THE STRUCTURE

To determine the behavior and performance of a real structure, the structure

must be load tested. Testing and measurement of the real structure can only be

done after the structure has been built. However, load testing is not possible at

the planning and design stage of a new structure. Real structures can not be

analyzed. We need to model the structure and analyze the model of the structure

to determine approximately the response and behavior of the real structure due

to external loads and excitation (Figure 1.2). The response quantities of the

model which consists of internal forces and displacements are used in designing

the members of the real structure.

An analytical model is a simplified representation of a real structure. In

developing a model of a structure, certain idealizations about the real structure

must be made. How the members are supported and connected have to be

represented by simple models. Loads expected to occur during the lifespan of

the structure have to be estimated and applied to the model. The main objective


RESPONSE Stresses Strains Displacements Stress Resultants Support Reactions

STRUCTURE

EXCITATION Loads Vibration Settlements Thermal Changes

Structural Model

Figure 1.2 Structural Analysis

in modeling a structure is that the characteristics of the real structure must be

represented as accurately as practically possible by a mathematical model so

that the structural response predicted from the analysis of the model using

computer tools may be relevant to the real structure. It is therefore imperative

that the model represents the real structure with an appropriate likeness to

capture the desired response.

The process of modeling is more of an art than a science. The engineer, through

his practical experience and insight, must convert the real structure to an

appropriate model (Figure 1.3) by making simplifying assumptions with regards

to the type of structural model (3D or 2D), level of modeling (global or local),

choice of model type (frame, grid, membrane, plate or solid), choice of elements

(line, plate or solid), size and number of elements, type of restrains, properties of

members and type of loads and excitations.

Modeling of Structures in 2D If all the members of a structure and the loads acting on the structure lie on a

single plane, the structure is modeled as a plane or two-dimensional (2D)


(b) Solid Model (c) 3D Plate-Frame (d) 3D Frame

(a) Real Structure

(e) 2D Frame

Figure 1.3 Various Ways of Modeling a Structure (Anwar 2000)

structure. Figure 1.4 shows two-dimensional models of plane structures. In these

models, the members are represented as line elements with the line element

corresponding to the centroid of the member. Beams are horizontal members

used primarily to carry vertical loads – distributed and concentrated loads.

Beams resist external forces through bending moment and shear forces. Trusses

are structural members made by assembling short, straight members connected

by smooth pins at the joints and primarily designed to carry tensile and

compressive axial forces. The loads in a truss are assumed to act at the joints

which connect the members. Frames are composed of beams and columns that

are either pinned or fixed connected at the joints. When the joints connecting the

horizontal and vertical elements are fixed or rigid, the structure is said to be a

rigid frame. The forces developed internally in a frame member consist of axial

force, shear force and bending moment.


Figure 1.5 Continuous Bridge (http://nisee.berkeley.edu.ph/godden)

(a) Beam (b) Plane Truss

(c) Plane Frame

Figure 1.4 Two-Dimensional Models of Structures

Structures, in general, are three-dimensional. However, there are many actual

three-dimensional (3D) structures which can be divided into planar or two-

dimensional (2D) structures to simplify the analysis.

A continuous bridge (Figure 1.5) is

one example of a structure which

can be modeled as a plane

structure. The bridge deck

supported by the piers and

foundations and the traffic loads

carried by the deck may be

considered to lie on one plane and

the bridge deck can be modeled as

a beam.

The truss of a bridge can be analyzed as a 2D structure (Figure 1.6). The bridge

deck rests on beams called as stringers, which are then supported by floor


Figure 1.7 Multistory Building (http://nisee.berkeley.edu.ph/godden)

beams. The floor beams are

connected at their ends to the

joints on the bottom panels of

the two longitudinal trusses.

Thus, the weights of the

vehicles, bridge deck, stringers

and beams are transmitted to

the supporting trusses at their

joints; the trusses, in turn,

transfer the load to the

foundation. Since the truss and the applied loads at the joints of the truss lie on

one plane, the longitudinal truss can be treated as a plane truss.

A multistory building (Figure 1.7) which consists of interconnected beams,

columns, walls and footings may be modeled as a system consisting of several

rigid plane frames (Figure 1.8). At each story of a building, the floor slab rests on

floor beams, which transfer the floor loads including the weight of the slab and

beams to the girders of the rigid frames. The loads are then transferred from the

girders to the columns and then finally to the foundation. Since the applied loads

and the rigid frame consisting of

the girders, columns and

foundations all lie on one plane,

each frame can be analyzed

separately as a plane structure.

Buildings which are highly

symmetrical in plan and framing

system can ideally be

represented as a system of 2D

frames.

Figure 1.6 Truss Bridge

(http://nisee.berkeley.edu.ph/godden)


Figure 1.9 Space Structures (a) Tower (b) Dome (http://nisee.berkeley.edu.ph/godden)

Frame F1 Frame F2

Building Plan 3D Building made from F1 and F2

F1

F1

F2 F2 F1

Figure 1.8 A 3D Building as a System of Two Typical 2D Frames

Although many three-dimensional structures can be subdivided into plane

structures for the purpose of 2D structural analysis, some structures which are


Figure 1.10 Various Models for Supports

referred to as space structures (Figure 1.9) such as domes, transmission towers,

and highly unsymmetrical and irregular buildings are difficult to simplify into plane

structures because of the complexity of the arrangement of the structural

elements. A three-dimensional modeling and analysis has to be carried out for

these types of structures to accurately predict their behavior.

Modeling the Supports and Joints

One of the most critical

aspects in the modeling of a

structure is the representation

of the restraint conditions at

the supports or at the joints.

Depending on the type of

restraint expected in the

actual structure, the engineer

has to decide on what

appropriate model to use

(Figure 1.10). The restraint

conditions can be any of the

following:

o Roller - relative rotations at the joint and only translation parallel to the

plane of the roller are allowed

o Pinned - relative rotations at the joint are allowed but no translations

o Fixed or Rigid - rotation and translation are not allowed

o Flexible - spring models used to represent the relative stiffness of the

joint


Modeling the External Loads

In analyzing or designing a structure, it is necessary to determine the external

forces that are expected to occur during its design life. Depending on the type of

structure being analyzed, the following loads may have to be considered:

(a) Dead loads are forces acting vertically downward that represent the

weight of the structure and other permanent or fixed objects.

(b) Live loads are vertical forces that may or may not be present on the

structure at any given time. These loads are movable and can be applied

anywhere on the structure. Occupancy loads in buildings and truck loading

are examples of live loads.

(c) Wind loads are cause by the pressure or suction due to wind at a point on

a structure. These loads depend on various factors such as wind velocity,

dimensions and orientation of the structure and geographical location of

the structure.

(d) Earthquake loads are developed when the structure vibrates due to

ground excitation. Their magnitude depends on the type of ground

accelerations, mass and stiffness of the structure, soil properties and

location of structure with respect to seismic faults.

(e) Other Environmental loads such as temperature changes, differential

settlement of the foundation, vehicle loads, hydrostatic forces, soil

pressure, etc. have to be considered depending on the type of the

structure.

The various loads mentioned are determined approximately. In most cases,

these forces are modeled into two types of loads:

(a) Nodal loads – these are loads applied at the ends of the members or

nodes. The loads can be vertical or horizontal forces or moments.

(b) Member loads – these loads are applied directly on the members or

between the ends of the members. Various models of member loads are

shown in (Figure 1.11) :


Figure 1.11 Types of Member Loads

(1) Point or concentrated load

(2) Distributed load (rectangular, triangular, trapezoidal)

(3) Concentrated moment or couple

(4) Temperature

Load Combinations

A great number of different types of loadings act on a structure. These loads do

not act simultaneously on the structure. When these forces occur at the same

time, the design loads are usually determined using load combinations. The

combination which results to the worst condition is used in design. Load factors

are multiplied on the basic loads and these factors depend on the design method

being used. The basic load combinations can be found in the code. Examples of

combination of factored loads from the NSCP 2001 Section 203.3, when Load

and Resistance Factor Design (LRFD) is used, are:

o 1.4 DL

o 1.2 DL + 1.6 LL + 0.5 Lr

o 0.9 DL ± (1.0 EQ or 1.3 W)

On the other hand, when the Strength Design for concrete is used, NSCP 2001

Section 409.3 provides these load combinations:

o 1.4 DL + 1.7 LL

o 0.75 (1.4 DL + 1.7 LL + 1.7 W)


o 1.3 DL + 1.1 LL + 1.1 EQ

In these equations, the following notations were used: DL = dead load, LL = live

load, Lr = roof live load, W = wind load and EQ = earthquake load.

Results of Structural Analysis

The main objective of structural analysis is to determine the behavior and

response of the model of a structure. Various analytical methods are available -

from approximate methods such as slope-deflection or moment distribution

methods to the more refined finite element methods. The important structural

response quantities that any structural analysis procedure must produce are:

o Displacements at the nodes

o Shear forces at various sections (Figure 1.12)

o Bending moments at various sections (Figure 1.13)

o Axial forces at the ends of members (Figure 1.14)

o Reactions at the supports

The results of the structural analysis are usually presented in tabular or graphical

form (Figure 1.15). The engineer must be familiar with the sign convention used

so that he can properly interpret the results. These results are used in the

structural design of members assuring that safety and serviceability

requirements of the design codes are satisfied.

Figure 1.12 Shear Force

Figure 1.13 Bending Moment


Figure 1.14 Axial Force

Using the Computer as a Learning Tool

Software for structural analysis are now available commercially – from simple to

more sophisticated software and affordable to expensive ones (e.g. MicroFEAP,

GRASP, BATS, STAAD, ETABS, SAP2000). Some textbooks in structural

analysis (e.g., Kassimali 1999, Hibbeler 2000) also contain CD-ROM with

software. In this notes, GRASP, a user-friendly software is introduced for two-

dimensional analysis of framed structures to enhance the learning and

understanding of structural analysis. An advantage of using structural analysis

software is that more complex and larger structures may be analyzed and

designed by the students, which is not possible in the regular class in structural

analysis where the calculator or general math solvers are used by students in

their calculations. Another advantage of using software, especially those with

graphics, is that students can visualize the behavior of complex systems. The

software can be used to simulate a variety of structural and loading

configurations and to determine cause and effect relationships between loading

and various structural parameters, thereby increasing the students’

understanding on the behavior of structures. This develops the student’s “feel” to

real life problems.

Figure 1.15 Moment Diagram


An exploratory-type of instructional and learning material consisting of a set of

modules are presented in the succeeding chapters. Each module focuses on a

structural analysis issue which is presented through a case study. Included in the

modules are hands-on exercises and problems on two-dimensional analysis of

framed structures (beams, trusses and rigid frames). Using the set of learning

modules, the student with the aid of GRASP discovers the behavior of structures

due to variations in the parameters of the model and configurations of the

structure, changes in member and material properties, and also changes in the

restraint and loading conditions. Through the graphical results, students can

visualize the phenomena and this would accelerate their understanding of

concepts through the experience of seeing and interpreting solutions to many

different problems.

The set of learning modules is not a substitute to a textbook on structural

analysis. The theory will not be presented. No derivations or equations can be

found. The student must refer to the textbooks for definitions, equations and

techniques. Each module will focus on a specific issue. A case study on the

issue will be presented and walk through. The student by observing the graphical

results and by interpreting the numerical output discovers important insight and

can make conclusions. The implication and relevance of the structural analysis

issue to the safe and reliable design of structures are also discussed. The

module ends with a similar or related problem which the student has to solve

using GRASP. Since GRASP provides direct feedback graphically and

numerically, the student can explore and have fun by simple modification of the

configuration of the structural model or loading condition and will discover new

knowledge related to the structural analysis issue of the module.

CASE STUDY: How do I represent a real structure as a line model?

Structures can be modeled using line elements to represent the members. In

modeling the structure, locate the centroids of the members and draw the lines


CASE STUDY 1 : Modeling a Portal Frame

Reference: GRASP Help (Step by Step Examples) or User’s Manual (1997)

with respect to these centroids. For the portal frame shown, draw the model of

the structure by representing the members as line elements.


References

ACECOMS (AIT, Thailand) and Superior Software Solutions (Pakistan), GRASP

Version 1.0 User’s Manual, 1997

Anwar, N. (2000). “Structural Modeling,” ACECOMS News & Views, Jan-Jun

2000, pp. 8-10, AIT, Bangkok, Thailand

http://nisee.berkeley.edu.ph/godden, Godden Structural Engineering Slide

Library

Hibbeler, R.C. (2000). Structural Analysis, 4th Edition, Chapter 1, Prentice Hall,

New Jersey, USA

Kassimali, A. (1999). Structural Analysis, 2nd Edition, Chapters 1-2, Brooks-Cole

Publishing Co., USA

National Structural Code of the Philippines (NSCP 2001), Volume 1 : Buildings,

Towers, and Other Vertical Structures, Chapter 2 and Sections 409, Association

of Structural Engineers of the Philippines, Inc. (ASEP), Quezon City, Philippines

Schodek, D.L. (1998). Structures. Chapters 1-3, Prentice-Hall, Inc. New Jersey,

USA

Understanding 2D Structural Analysis by A.W.C. Oreta : A Tour of GRASP 2 - 1

CHAPTER 2

A TOUR OF GRASP

INTRODUCING THE SOFTWARE

GRASP stands for Graphical Rapid Analysis of Structures Program. It is a user-friendly

software for two dimensional analysis of framed structures which includes beams, trusses

and rigid frames. Especially developed for Windows, GRASP uses a Graphical User

Interface(GUI) which provides an interactive, easy to use, graphical environment for

modeling and analysis. GRASP is primarily based on a graphical means of interaction with

the user and can provide direct feedback and effect of modifications. The major features

of GRASP include:

• Modeling and analysis of multiple models in one file

• Presetting of default load cases and load factors

• Internal and automatic tracking of node numbers and member incidences

• Display the structural model at all times on the screen during analysis and

superimposition of the analysis results on the model after analysis

• A Structure Wizard provides a step-by-step guideline for the generation of multistory

structural models

• Supports SI, US and metric units and use of mixed units

• Apply loads on nodes and on members in multiple load cases

• Eight pre-defined types of cross-sections

• Set values of material properties

• Various restraint conditions including spring supports

• Apply member releases at the ends of members

• Diagram of results with values and tables

• View and print the analysis results for the full structure up to 20 sections for a

member


CASE STUDY: How do you model and analyze a rigid frame using GRASP?

In this example, you will model a two-story rigid frame shown in the figure. The frame will

carry uniform dead and live loads which will be applied fully on the beams. Wind loads will

also be applied at specified nodes. The basic loads will then be combined using the

following load combination cases:

o Dead Load (incl. Self Load) : 1.4 DL

o Combined Dead and Live Loads : 1.2 DL + 1.6 LL

o Combined Dead, Live and Wind Loads : 1.2 DL + 1.0 LL + 1.3 WL

Things to Do Modeling and analysis using GRASP can be divided into five general steps. Follow the

step-by-step procedure described by the figures for the following general steps.

1. Start analysis software and set basic parameters

2. Create geometry (in the figure assume dimensions are referred with respect to the

centroids)

3. Apply basic loads

4. Define load combinations

5. Perfom analysis and view the results

wDL = 15 kN/m , wLL = 7 kN/m

10 kN

Vertical Loading Wind Loading

6 kN


Frame geometry 3.0 m 3.0 m4.0 m

3.0 m

4.0 m

300 mm

300 mm

Column cross-section Modulus of Elasticity = 21 kN/mm2

Unit Weight = 24 kN/m3 Coefficient of Thermal Expansion = 11 x 10 -6/C

CASE STUDY 2

1200 mm

300 mm

Beam cross-section

100 mm

250 mm

Understanding 2D Structural Analysis by A.W.C. Oreta : A Tour of GRASP 2 - 4 1. Start analysis software and set basic parameters

Step 1-2 : Select the main unit system (SI, Metric or US) and other related measurement parameters (m, cm, mm; kg, k Not ton) from the options as shown in the dialog box above.

Step 1-1 : Select ‘units’ option from ‘Options’ menu to specify the working unit. To fix the working unit for future use ‘Customize’ option and select the system that you prefer.

Step 1-3: Select ‘Structure’ => ‘Materials’ and input material parameters. Change material properties of the default material if necessary. You may also add a new material by pressing ‘Add Material’.

Understanding 2D Structural Analysis by A.W.C. Oreta : A Tour of GRASP 2 - 5 2. Create geometry

Step 2-1: There are two ways of creating the geometry in GRASP. You may draw the model using the mouse and GRASP graphical tools or use the ‘Structure Wizard’. Let us use the second method for step by step and quick creation of typical building models. Select ‘Structure’ => ‘Frame’ => ‘Structure Wizard’.

Step 2-2: Select an appropriate typical frame based on bay width, story height and configuration. For this example, select type 2. If the frame that you want to model do belong to the four types, you can change the configuration of selected model later.

Step 2-3: Specify the number of bays (3) , number of stories (2) and typical values of bay width (3 m) and story height (4 m). You may also input the bay width (e.g., width = 4 m for bay 2) that is different from the default width.

Step 2-4: Select the type and specify the dimensions for a typical column.


Step 2-5: Select the type and specify the dimensions for a typical beam.

Step 2-6: Displaying the frame generated by Structure Wizard.

Step 2-8: To display the dimension line between selected nodes, select ‘Structure’ => ‘Add Dimensions’ and click any two nodes and double click at a position where the dimension line will be displayed . To delete, select ‘Del Dimensions’ and click each line.

Step 2-7: To define the type of supports, select ‘Structure’ => ‘Nodal Restraints’ and click at the node where restraints will be defined. Select the type of restraint.


Step 2-11: To change the coordinates of the top most nodes, go to ‘Structure’ => ‘Change Nodal Coordinates’ and then click the node and enter the new Y-coordinate (7 m)

Step 2-9: To view the Node and Member numbers or labels, select ‘View’ => ‘Node Numbers’ and ‘Member Numbers’.

Step 2-10: Here is the display of the frame with dimensions, member and node numbers. Observe that the height of the second story is 4.0 m. This must be changed to 3.0 m resulting to7.0 m as the total height of the frame.

Step 2-12: Display of the corrected model of the 2D frame.

Step 2-13: You may also view the outline of the members by selecting ‘View’ => ‘Member Outline’.To remove the node numbers, member numbers and member outline, select ‘View’ => ‘Node Numbers’, ‘Member Numbers’ and ‘Member Outline’.

Understanding 2D Structural Analysis by A.W.C. Oreta : A Tour of GRASP 2 - 8 3. Apply basic loads

Step 3-1: Let us first apply the uniform loads in the horizontal members. You may apply the loads one member at a time by simply clicking the specific member or to all horizontal members. Let us select all horizontal members. Select “Edit’ => ‘Select Member’ and click on all horizontal members while pressing the shift key. Note the change of color of the selected members.

Step 3-2: If you want to display only the horizontal members, select “View’ => ‘Members to Show’ => ‘Horizontal’. Let us now apply the Dead Load. Select ‘Dead Load’ option from the load cases, combination and envelope list (rightmost-top).

Step 3-3: Select ‘Loading’ => ‘Member Loads’ and click on the selected members.

Step 3-4: Press ‘Add’, select the appropriate member load and enter the magnitude of the load.


Step 3-5: Displaying the Dead Load on horizontal members.

Step 3-6: Select ‘Live Load’ option from the load cases, combination and envelope list (rightmost-top).

Step 3-7: Select ‘Loading’ => ‘Member Loads’ and click on the selected members. Press ‘Add’, select the appropriate member load and enter the magnitude of the load.

Step 3-8: Displaying the Live Load on horizontal members.


Step 3-10: Define the new load case – Wind Load and select the option ‘Basic Load Case’.

Step 3-11: : Select ‘Wind Load’ option from the load cases, combination and envelope list. Select ‘Loading’ => ‘Nodal Loads’ and click on the node where the loads will be applied.

Step 3-12: Input the magnitude and sign of the load. Follow the sign convention shown in the figures.To apply in opposite direction, use a negative value.

Step 3-9: To apply the Wind Load, let us add a load case. Select ‘Loading’ => ‘Add Load Case’.

Step 3-13: Displaying the Wind Loads

Understanding 2D Structural Analysis by A.W.C. Oreta : A Tour of GRASP 2 - 11 4. Define load combinations

Step 4-1: Define load combinations by pressing ‘Loading’ => ‘Add Load Case’.

Step 4-2: Specify the name of the load and select ‘Combination Load Case’.

Step 4-3: Input appropriate load factors for the defined load combination case.

Step 4-4 : Repeat the same steps (4-1 to 4-3) for the load combination case combining dead load and live load.


5. Analysis and Results

Step 4-5 : Repeat the same steps (4-1 to 4-3) for the load combination case combining dead load,

Step 5-1: Carry-out the analysis using ‘Perform’ => ‘Self Load Calculation’ and ‘Analysis’ in the menu option.


Step 5-2: To display graphical results, select the load case or load combination first.

Step 5-3: Select the type of result from the menu option ‘View’ => ‘Bending Moment’. Select ‘View’ => ‘Result Values’ if you want numerical values displayed in diagram.


Step 5-4 : Select the type of result from the menu option ‘View’ => ‘Shear Force’.

Step 5-5 : Select the type of result from the menu option ‘View’ => ‘Axial Force’.


Step 5-6 : Select the type of result from the menu option ‘View’ => ‘Reactions’.

Step 5-7: You can view the nodal displacements by simply pointing the mouse at a node or the member results by pointing the mouse at a member.


Step 5-9: Double click on any member to display the detailed results and

diagrams for that member.

Step 5-8: To view the displacements, select the type of result from the menu option ‘View’ => ‘Deflected Shape’


Step 5-10: Display the results in tabular form using ‘Tables’.

Step 5-11: Prepare the report using ‘File’ => ‘Report Set-up’.


Step 5-12: Select / Deselect the items to be included in the analysis report.

Step 5-13: Select the ‘File’ => ‘Print Preview Report’ to preview the

analysis report.


Step 5-14: Select the ‘File’ => ‘Print Preview Report’ to preview member results

Step 5 -15: Select the ‘File’ => ‘Print Preview Report’ to view graphical results

You may print a hard copy of the report by selecting ‘File’ => ‘Print Report’.

Understanding 2D Structural Analysis by A.W.C. Oreta : A Tour of GRASP 2 - 20 GRASP Toolbar

GRASP has a toolbar which provides shortcuts in using the software. The toolbar buttons

may be used instead of the commands in the menu.




GRASP Help and Manual If you want a step-by-step guide on

the use of GRASP, you may refer to

the document published by

ACECOMS (AIT, Thailand) and

Superior Software Solutions

(Pakistan), GRASP Version 1.0

User’s Manual, 1997 or click on the

“Help” button and detailed

information of various topics can be

found.

Understanding 2D Structural Analysis by A.W.C. Oreta : A Tour of GRASP 2 - 24 Things to Try : Test Your GRASP Skills. Analyze the following planar structures using

GRASP and fill in the blanks.

A. BEAM: The beam has a rectangular cross-section of 250 mm x 400 mm. Assume the

following material properties: Modulus of Elasticity, E = 21 kN/mm2 , Unit weight, γ = 24

kN/m3 and coefficient of thermal expansion, α = 12E-6.

1. The reactions at A are ____________ kN (vertical) and ___________ kN-m (moment).

2. The shear and moment at the left end in BC are ____________ kN and ____________

kN-m, respectively.

3. The maximum bending moment in member BC is about _____________ kN-m and is

located at ______________ m from point B.

B. PLANE TRUSS: All members are double angles 4 x 3 x 3/8, short legs back to back

(A = 4.97 in2; I = 3.84 in4 and ytop= 0.782 in, modulus of elasticity, E = 29,000 ksi,

unit weight, γ = 491 lb/ft3 and coefficient of thermal expansion, α = 6.5 x 10 -6/F)

4. The reactions at D are ___________ kips (horizontal) and ____________ kips (vertical).

5. The axial force in bar AE is ___________ kips and bar BE is ___________ kips.

6. The nodal displacements at C are _____________ in (horizontal) and _____________

in (vertical).

B C 12 m

20 kN/m

250 kN

12 m 4m 4m

A

4 @ 15 ft = 60 ft

1.5 kips1.2 kips

2.0 kips

A C

D

E F

G

B

20 ft

Understanding 2D Structural Analysis by A.W.C. Oreta : A Tour of GRASP 2 - 25 C. RIGID FRAME : Assume the following material properties: Modulus of Elasticity, E =

21 kN/mm2 , Unit weight, γ = 24 kN/m3 and coefficient of thermal expansion, α = 12E-6.

° LOADS : DL = 31 kN/m and LL = 10 kN/m applied in all beams

WL = 40 kN at the top level and 20 kN at the lower level

° LOAD COMBINATIONS: CASE NO. 1 = 1.4 D + 1.7 L

CASE NO. 2 = 1.05 D + 1.275 L + 1.275 W

° SECTION PROPERTIES

COLUMNS ( 400 mm x 400 mm) BEAMS (250 mm x 350 mm)

7. The reactions at the support F due to dead load (DL) are :

° Vertical Force = ____________ kN

° Horizontal Force = ____________ kN

° Moment = ____________ kN-m

4.0 m

5.0 m

6.0 m 4.0 m

A B

C D E

F G H

40 kN

20 kN


8. The end moments of the beam CD due to Combination Load Case 1 are :

° Moment at C = ____________ kN-m

° Moment at D = ____________ kN-m

9. The end moments of the beam CD due to Combination Load Case 2 are :

° Moment at C = ______________ kN-m

° Moment at D = ______________ kN-m

10. The axial force in the column CF for the different load cases are:

° P (WL only) = __________ kN

° P (DL only) = ___________ kN

° P (Combination Load Case 2) = __________ kN

11. The maximum left end moment for beam CD is _____________ kN-m and occurs at

loading case: ___________________ .

12. The maximum right end shear for beam AB is ____________ kN and occurs at loading

case : ___________________ .

13. The displacements at joint B due to wind load (WL) are :

o Horizontal : _______________ mm

o Vertical : _______________ mm

o Rotation: _______________ rad

14. The maximum span moment for beam CD due to Load Case 2 is ____________ kN-m.

Understanding 2D Structural Analysis by A.W.C. Oreta : Loading Continuous Beams 3 - 1

CHAPTER 3

LOADING CONTINUOUS BEAMS

BACKGROUND

Beams and girders are straight horizontal members in structures which resist forces

applied transversely to their lengths. This types of structural elements can be found

in buildings supporting the floor slabs or resting on columns. Bridge decks which are

frequently supported by piers

and abutments are usually

modeled as continuous beams.

Beams are primarily designed

to resist bending moments.

Shear forces in beams must

also be checked especially when the beams carry loads of large magnitude. The

important basic variables which affect the behavior of beams include the magnitudes

and arrangement of the loads, the nature of the support conditions and the section

properties. This chapter explores the effect of loading conditions on the internal

forces and moments in continuous beams.

CASE STUDY : How should live loads be placed to produce the maximum and minimum bending moments and shear forces in continuous beams?

The individual members of a structure must be designed for the worst

combination of loads that can reasonably be expected to occur during its useful

life. The internal forces developed in beams such as moments and shears are

caused by the combined effect of two types of loads: dead loads and live loads.

Dead loads, which include the weight of the beam, are constant and are placed

fully on the beams. On the other hand, live loads such as floor loads from human

occupancy or moving loads due to traffic can be placed on the beam in various

ways. Is the positioning of the live loads critical in the design of beams and


girders? What positions of live loads would produce maximum effects on a

continuous beam?

Things to Do 1. Model the three-span continuous beam shown in the figure using the given

material and section properties. Using the GRASP toolbar, click the button for

adding a member and draw graphically the geometry of the beam. Draw the

continuous beam four times as shown in Figure 3.1.

2. Apply the dead load (WDL = 20 kN/m) on all spans.

3. Apply live load (WLL = 12 kN/m) for four basic load cases shown in Figure 3.1.

4. Combine the dead load and the corresponding basic live load:

Service Load : DL + LL

Ultimate Load : 1.4 DL + 1.7 LL

5. Perform analysis and view graphical and tabular results.

3 @ 4.0 m = 12.0 m

250 mm

Beam cross-section

400 mm

Three-span Continuous Beam

Material Properties Modulus of elasticity = 20,500 N/mm2

Unit weight = 24 kN/m3

Coefficient of thermal expansion =

0.00099 / oC

CASE STUDY 3


Figure 3.1 Live Loading Cases and Deflection Curves

(a) Case 1: Full Live Load

(d) Case 4: Alternate Spans B (c) Case 3: Alternate Spans A

(b) Case 2: Adjacent Spans

Figure 3.2 Moment Diagrams due to Dead Load





Observation

One of the more interesting aspects of the behavior of statically indeterminate

structures such as continuous beams is the structure’s response under load. The

response quantities which are affected by the loading conditions are the deflections,

moments and shear forces. A three-span continuous beam when subjected to

uniform load applied similarly at all spans will bend with a deflection curve similar to

Figure 3.1 (a). The corresponding shape of the moment diagram for full loading

conditions will have a shape similar to the diagrams in Figure 3.2. This figure shows

the resulting moment diagram due to the dead load. Observe the location of the

maximum (negative and positive) moments. The maximum negative moments occur

at the internal pin supports while the maximum positive moments occur near the

midspan.

What is the effect of partially loading the spans of the continuous beam? Two types

of partial loading conditions are shown in Figure 3.1. In Figure 3.1 (b), two adjacent

spans are loaded with live load and the other span is not loaded. In Figures 3.1 (c)

and (d), on the other hand, the live load is placed at alternate spans with the

adjacent spans unloaded. These conditions reflect different loading patterns and

each loading pattern will affect the internal forces at various sections (e.g. near the

support or at midspan) of the beam. Figure 3.3 shows the resulting moment

diagrams for the four cases of live loading.

Maximum Negative Moment at a Support : Consider first the maximum negative

moment at the second support. An inspection of the moments associated with the

four cases reveals some curious results. The maximum negative moment at the

second pin support does not occur when the structure is fully loaded (Case 1) but

rather under a partial loading condition at adjacent spans (Case 2). The maximum

negative moment under full live loading condition is 19.2 kN-m while the maximum

negative moment when the adjacent spans are loaded is 22.4 kN-m. When the

dead and live loads are now combined either under service load (Figure 3.4) or

ultimate load (Figure 3.5), the same loading case produces the maximum negative


moment at that support. This means that the maximum negative moment at a

support will occur when the loads are placed on the two spans adjacent to that

particular support and the next span unloaded Hence, if the maximum negative

moment at the third pin support is desired, we must apply the live loads at the

second and third spans which are adjacent to the support while the first span is

unloaded. In case of more than three spans, the alternate spans must be loaded.

Maximum Span Moments : Observe the other two cases (Case 3 and 4) for

alternate span loading. Figure 3.1 (c) and (d) shows the deflection curves due to

alternate live loading. It can be seen that at the loaded spans, the curvatures are

positive or concave upwards and since the bending moments are proportional to

curvatures, the resulting span moment for the loaded spans are also positive.

Figures 3.3 (c) and (d) show that the maximum span moments in the loaded spans

do not also occur under full live loading condition (Case 1) but under partial loading

Figure 3.3 Moment Diagram due to Live Load





Figure 3.4 Moment Diagram for Service Load

Figure 3.5 Moment Diagram for Ultimate Load








(a) Full Live Loading Condition (Case 1)

(b) Alternate Live Loading Condition (Case 3)

Figure 3.6 Moment Diagram of First Span due to Live Load

when alternate live loading is applied. A comparison of the magnitudes of the

maximum positive moments for the first span is given in Figure 3.6 for Case 1 and

Case 3. Under full live load, the maximum span moment of the beam is about

15.360 kN-m. However, under alternate live load, the maximum span moment of the


Figure 3.7 Member Results for Service Load Condition

same beam is about 19.440 kN-m. This means that the maximum positive span

moment in a continuous beam may occur at the loaded span for the alternate

loading condition. Combining now the moments due to dead and live loads will

produce the maximum positive span moments in the loaded spans as shown in

Figures 3.4 and 3.5. Figure 3.7 shows member results for the Service Load

Condition. Compare the results for B-1, B-4, B-7 and B-10 which correspond to the

first span. The Maximum positive span moment is about 46 kN-m for B-7, a loaded

span in the alternate loading condition (Case 3).

Minimum Span Moments: Observe again the two cases (Case 3 and 4) for

alternate span loading. This time observe the unloaded spans of the continuous

beam. Notice that the curvatures in Figures 3.1 (c) and (d) are now negative or


Figure 3.8 Shear Diagram for Service Load

concave downwards meaning that the resulting span moments will be negative.

Figures 3.3 (c) and (d) clearly show that the moments in the unloaded span are

negative. If the moments due to dead and live loads are now combined for either

service load (Figure 3.4) or ultimate load (Figure 3.5) conditions, the resulting span

moments for the unloaded beams using the alternate span loading condition will be

minimum (which may be negative) since the signs of the moments due to dead and

live loads are not the same. This means that the minimum span moment in a

continuous beam may occur at the unloaded span for the alternate loading condition.

An inspection of the results for B-1, B-4, B-7 and B-10 in Figure 3.7 shows that the

minimum span moment is about 22 kN-m for B-10, an unloaded span in the alternate

loading condition (Case 4).

Maximum Shear at a Support: Compare the shear diagrams under service load condition for the four cases as

shown in Figure 3.8. Specifically, observe the shear forces at the second support.





Which case produces the maximum shear? The maximum shear at the second

support occurs under Case 2. It can be seen that the maximum shear at the support

occurs when the support is between two spans which are loaded using the adjacent

loading condition. Hence, placing the live load in the second and third spans will

result to a maximum shear in the third pin support. It is also interesting to note here

that the maximum shear at the end supports does not occur under full live load

condition but under partial loading condition (Case 3).

Things to Ponder

Loading conditions that produce the maximum effects on a structure are called

critical loading conditions. These conditions do not always occur when the live load

is placed fully on the structure. Partial loading conditions may produce the critical

moments and shear forces at a beam section. The occurrence of the critical values

do not simultaneously occur under one loading condition. Hence, various loading

arrangements have to be checked to determine what loading conditions are critical

on a structure.

In the design of beams , the code (e.g., NSCP 2001 sections 205) permits that the

arrangement of live load may be limited and states that “where uniform floor loads

are involved, consideration maybe limited to full dead load on all spans in

combination with full live load on adjacent spans and alternate spans.”

Things to Try

1. Analyze a continuous beam to obtain the maximum possible span and end

moments. Model a continuous RC beam of 250 mm x 400 mm rectangular section

consisting of four spans with pin supports. Each span has a distance of 5.0 m. The

beams will carry uniformly distributed vertical loads consisting of the dead load (WDL

= 20 kN/m) and live load (WLL = 15 kN/m). Consider various combination load cases

for dead load and live load. Use the basic load combination factors for dead and live


4 @ 5.0 m = 20 m

load specified in the NSCP 2001 section 203.3. Obtain the possible maximum and

minimum span and end moments. Observe also the effect of loading arrangement

on the shear forces.

2. Repeat the same steps in Exercise No. 1 for the beam shown below which has

fixed supports at both ends.

3. For the same continuous beams above, try changing the magnitude of the live

load and apply alternative live loadings. Observe if the span moments due to

dead and live loads becomes positive or negative. What is the ratio of live load to

dead load such that the combined effects produce a negative span moment?

4. For the same continuous beams above, change the distance between spans by

moving the second and fourth pin supports one meter towards the end supports

resulting to the span distances of 4.0 m – 6.0 m – 6.0 m – 4.0 m. Apply the same

4 @ 5.0 m = 20


loading conditions and observe the effect of spacing on the moments and shear

forces.

References and related readings


Towers, and Other Vertical Structures, Sections 203 and 205, Association of

Structural Engineers of the Philippines, Inc. (ASEP), Quezon City, Philippines

Nilson, A.H., Darwin, D. and Dolan, C.W. (2004). Design of Concrete Structures, 13th

Edition, Chapter 12, McGraw-Hill, Inc. NY, USA

Schodek, D.L. (1998). Structures. Chapter 8, Prentice-Hall, Inc. New Jersey, USA

Understanding 2D Structural Analysis by A.W.C. Oreta : Pattern Loading 4 - 1

CHAPTER 4

PATTERN LOADING IN MULTISTORY FRAMES

BACKGROUND

Design codes specify that every building and every portion of the structure must be

designed and constructed to sustain appropriate combinations of vertical loads and

lateral forces. The individual members of a building frame which consists of beams

and columns must be designed against loads which are reasonably expected to

occur during the structure’s useful life. The internal forces induced in the frame such

as moments, shears and axial forces are caused by the combined effect of both

vertical and lateral loads. Let us first consider the effect of vertical loads in a building

frame. The vertical loads which consist of dead and live loads are carried by the

horizontal members of the building. These loads are usually placed on the girders or

beams when a model of the structure is analyzed. The dead loads are constant and

are placed fully on the beams. On

the other hand, live loads such as

floor loads from human occupancy

can be placed in various ways, some

of which may result in larger effects

than others. Chapter 3 demonstrated

the effect of live load arrangement to

the moments and shear forces in

beams. These loading schemes also

apply to beams of rigid frames. However, the live loading schemes must be

extended to consider also the effect to the vertical elements or columns. This

chapter explores the various schemes that the live load can be placed on the

horizontal elements of a rigid frame and the corresponding effects on the internal

forces in the beams and columns.


CASE STUDY : How should live loads be placed to obtain the possible maximum internal effects in beams and columns of rigid frames?

A four-story rigid frame building with three bays will be analyzed. The girders or

beams will carry uniformly distributed vertical loads consisting of the dead load

(WDL = 15 kN/m) and live load (WLL = 7 kN/m).

Things to Do

1. Model the four-story rigid frame shown in the figure assuming fixed supports

and using the following material properties:

Modulus of elasticity = 20,500 N/mm2


Coefficient of thermal expansion = 0.00099 / oC

2. Place the dead load and live load on all spans.

3. Place the live load on the specified spans only as shown in Figures 4.1 to 4.6.

4. Apply seven combination load cases using the load factors of 1.4 for dead

load and 1.7 for live load:

DL & Full LL

DL & Alternate LL 1

DL & Alternate LL 2

DL & Adjacent LL 1

DL & Adjacent LL 2

DL & Column LL 1

DL & Column LL 2

5. Perform analysis and view graphical and tabular results.


350 mm

350 mm

Column cross-section

250 mm

400 mm

Beam cross-section

3 @ 5.0 m = 15.0 m

4 @3.0 m = 12.0 m

Frame geometry

CASE STUDY 4


Figure 4.2 Alternate Live Load No. 2

Figure 4.1 Alternate Live Load No. 1


Figure 4.3 Adjacent Live Load No. 1

Figure 4.4 Adjacent Live Load No. 2


Figure 4.5 Column Live Load No. 1

Figure 4.6 Column Live Load No. 2


Figure 4.7 Moment Diagram for Combined Dead Load and Full Live Load

Observation Maximum and Minimum Moments in Beams: From the results of Chapter 3 for

continuous beams, the same principles can be applied to rigid frames in determining

the maximum negative support moments, maximum span moments and minimum

span moments in the beams. Hence two alternate live loading cases (Figures 4.1

and 4.2) were applied to determine the possible maximum and minimum span

moments of the beams. On the other hand, two adjacent live loading cases (Figures

4.3 and 4.4) were applied to determine the maximum possible negative moments at

the supports of the beams. The span and end moments of the beams for each

combination load case can be compared with the combined dead and full live load

case in Figure 4.7.


Figure 4.8 Member Results for Beam B-17

To view the detailed results for a selected beam for the various load cases, prepare

a report and display the “Comparison of Results” using the “Print Preview Report”

as in Figure 4.8 (B-17) and Figure 4.9 (B-26) For these two beam examples, identify

the combination load cases where the maximum and minimum span moments occur.

For beam B-17, the minimum span moment (26.0404 kN-m) occurs under “DL &

Alternate LL 1” case while the maximum positive span moment (44.5745 kN-m )

occurs under “DL & Alternate LL 2” Case. Beam B-17 is an unloaded span for “DL &

Alternate LL 1” case and a loaded span for “DL & Alternate LL 2” case. The same

observation can be found for beam B-26 where the maximum positive span moment

(44.6703 kN-m) and the minimum span moment (24.5816 kN-m) occur when the

span is loaded (DL & Alternate LL 1) and unloaded (DL & Alternate LL 2),

respectively.


Figure 4.9 Member Results for Beam B-26

The maximum negative support moments, on the other hand, can be observed in

the other load cases particularly the adjacent live loading cases, especially if the

support is between loaded spans (e.g., the right end moment of B-17 of - 81.5235

kN-m is maximum at “DL & Adjacent LL 1” case).

Maximum End Moments of Columns: For columns, the largest moment occurs at

the top or bottom. Two loading cases (Figures 4.5 and 4.6) illustrate a live loading

arrangement which may produce the maximum possible column end moments. View

the moment diagrams for the combination load cases -“DL & Column LL 1 “ and “DL

& Column LL 2” (e.g., Figure 4.10) and compare the resulting column end moments

with those due to the combined dead and full live load in Figure 4.7. Observe

specifically the columns, C-8, C-9, C-11 or C-12 and the live load positions on the

beams connected to the columns at the top and bottom. For these columns, the


Figure 4.10 Moment Diagram for DL and Column LL No. 2

maximum end moments do not occur under full live loading condition, but under

partial loading condition given in Figure 4.5 or 4.6. A sample of the member results

is shown in Figure 4.11 for column C-8. It can be seen that the top and bottom

moments (-2.1564 kN-m and 1.5908 kN-m) due to the combined dead load and full

live load are relatively small compared to top and bottom moments (-13.6512 kN-m

and 14.1499 kN-m) for the “DL & Column LL No. 2” case. The same observation can

be made for the other columns (C-9, C-11 or C-12) if you view the detailed member

results. This means that If the spans exactly above and below a column are loaded,

the top and bottom moments of that column may produce the maximum end

moments. This is an alternative live loading arrangement that can be applied to

building frames to obtain the possible maximum end moments of a column.


Figure 4.11 Member Results for Column C-8

Things to Ponder Placing the live load fully on a structure does not always produce the critical

moments in beams and columns. Partial loading of the frame sometimes produces

the worst condition. Live loads can be placed in various ways on the structure ,

some of which will result in larger effects than others. By proper positioning of live

load, the maximum combined effect due to dead and live loads on the member can

be obtained - the effect can be both negative or positive. The load patterns in a

continuous frame that produce the maximum positive and negative moments in the

beams are different from those of the critical moments in the columns. Hence,

alternative live loadings must be considered separately for beams and columns to

obtain the worst combination of loads. The problem now results to predicting which

type of loading pattern produces the maximum internal forces. There are many


different live loading schemes that can be tried to obtain the maximum effects on the

structure. Trying all possible loading arrangements is of course not practical. By

considering the relative magnitude of the effects of the loading schemes, an

experienced designer can limit the analysis to a small number of significant cases.

In the design of beams , the code (e.g., NSCP 2001 sections 205 and 408.10)

permits that the arrangement of live load maybe limited and states that “where

uniform floor loads are involved, consideration maybe limited to full dead load on all

spans in combination with full live load on adjacent spans and alternate spans.” In

the analysis and design of concrete columns, on the other hand, the code (NSCP

2001 section 408.9 or ACI code 8.8) states that “columns must be designed to resist

the axial load from factored dead and live loads on all floors or roof and the

maximum moment from factored loads on a single adjacent span of the floor or roof

under consideration.” In the design of columns, the end moment and the axial force

are used simultaneously in proportioning the member and determining the amount

of steel (in the case of RC columns). However, the maximum values of the moment

and the axial force at the critical sections do not always occur at the same

combination load case. In one case, the moment may be large but the axial force is

small or vice versa. Which combination of moment and axial force results to the

most probable critical condition? This problem complicates the analysis process

enormously. Hence, the code (NSCP 2001 section 408.9.1), recognizing the

characteristic shape of the column strength interaction , specifies that in the design

of RC columns, “the loading condition giving the maximum ratio of moment to axial

load shall be considered.”

Things to Try

1. (a) Model the building shown using the following material properties:



Coefficient of thermal expansion = 0.00099 / oC


(b) Apply the basic load case for dead load (WDL = 15 kN/m) by placing the dead

load on all beams) and live load (WLL = 7 kN/m) by using full loading and several

pattern loading conditions.

(c) Combine the dead load and the corresponding basic live load using the load

factors of 1.4 for dead load and 1.7 for live load.

250 mm

400 mm

Beam cross-section

350 mm

350 mm


3 @ 5.0 m = 15.0 m

4 @3.0 m = 12.0 m

Frame geometry


(d) Perform analysis to obtain the maximum and minimum span moments, and

maximum negative support moments of the beams and the maximum end

moments of columns. Choose a beam and a column for a detailed discussion

about your observations.

References and related readings:


Towers, and Other Vertical Structures, Sections 203, 408 and 409, Association of


Hibbeler, R.C. (2000). Structural Analysis. 4th Edition, Section 9.11, Pearson

Education, Asia Pte. Ltd, New Jersey, USA



Schodek, D.L. (1998). Structures. Chapter 9, Prentice-Hall, Inc. New Jersey, USA

Understanding 2D Structural Analysis by A.W.C. Oreta : Lateral Forces in Buildings 5 - 1

CHAPTER 5

LATERAL FORCES IN BUILDINGS

BACKGROUND Lateral forces due to wind and earthquakes may act on a building during the lifespan

of the structure. Buildings when subjected to lateral forces would undergo horizontal

displacement or drift and must be checked against story drift limitations to prevent

the structure from collapsing laterally. The way a building resists lateral forces not

only influences the design of vertical members or columns but the horizontal

members or beams as well. How do lateral forces affect the behavior and response

of the structural members of building frames? This chapter aims to explore the effect

of lateral forces in the internal resultant forces of the structural members of rigid

frames.

CASE STUDY : How do lateral forces affect the bending moment of beams in rigid frames?

A two-story rigid frame with three bays will be analyzed. The frame will carry

uniformly distributed vertical loads consisting of the dead load and live load fully

applied on the beams. The lateral loads which will be assumed as earthquake

loads shall be applied at the 1st floor and 2nd floor levels of the frame in three

stages.

Things to Do

1. Model the two-story rigid frame shown in the figure using the following

concrete properties:



Coefficient of thermal expansion = 11 x 10-6 / oC


300 mm

300 mm

1200 mm

250 mm

300 mm

100 mm

Column cross-section Beam cross-section

Frame geometry 3.0 m 3.0 m4.0 m

3.0 m

4.0 m

CASE STUDY 5


2. Apply the basic load cases for dead load (DL), live load (LL) and the three

stages of lateral loads (EQ1, EQ2 and EQ3) applied from left to right.

3. Apply the following combination load cases :

Ultimate Load : 1.4 DL + 1.7 LL

Comb EQ1 : 1.3 DL + 1.1 LL + 1.1 EQ1

Comb EQ2 : 1.3 DL + 1.1 LL + 1.1 EQ2

Comb EQ3 : 1.3 DL + 1.1 LL + 1.1 EQ3

4. Perform the analysis and display graphical and tabular results.

5. Display the bending moment diagrams for the four combination load cases.

6. Display the bending moment diagrams for two beams (e.g., B-5 and B-6).

7. Display the member results comparison for a selected beam (e.g., B-5) using

the “Print Preview Report.”

Three Stages of Lateral LoadingLateral Load EQ1 EQ2 EQ3

2nd Level : F2 40 kN 80 kN 120 kN1st Level : F1 20 kN 40 kN 60 kN

wDL = 15 kN/m , wLL = 7 kN/m

F2

F1

Vertical Loading Lateral Loading


EQ2

Figure 5.2 Bending moment for combination load case : 1.3 DL + 1.1 LL + 1.1 EQ1

Figure 5.1 Bending moment for combination load case : 1.4 DL + 1.7 LL





Observation

Figure 5.1 presents the bending moment diagrams of the combination load case for

vertical loads only. This represents the effect of combining the dead load and live

load using appropriate load factors. Observe the bending moment diagrams of the

beams. The moments at the sections near the columns are negative, while the

moments near the midspan are positive.

Now observe the bending moment diagrams when the lateral forces are applied from

left to right. In Figure 5.2, the moments at the left ends of the leftmost beams (B-5

and B-12) in the first and second floors have changed from negative to positive. In

Figure 5.3, the moment at the left end of beam B-6 in the first floor also changed to

positive. Finally, in Figure 5.4, the moment at the left end of beam B-7, also in the

first floor, also changed to positive.

Figure 5.6 Bending moment for beam (B - 6)

Figure 5.5 Bending moment for beam (B -5)


This change in shape and sign of the moment diagram in beams in the first floor can

also be observed by displaying the respective bending moment diagrams of the

beams as shown in Figure 5.5 and 5.6 for two combination load cases: one case

when the frame was subjected to vertical loads only and the other case when the

lateral forces, EQ3, were applied.

Figure 5.7 shows the comparison of results for a selected beam (B-5) for the basic

load cases and combination load cases. Observe the moments at the ends for

various cases. Mzi, is the moment at the left end, while Mzj is the moment at the right

end. The moment at the left end is negative when the applied loads are vertical (.e.g.,

basic load case for dead load and live load or the combination load cases such as

service load, ultimate load). On the other hand, the moment at the left end is positive

when only lateral forces are applied (e.g., basic load cases EQ1, EQ2 and EQ3).

When the vertical loads and lateral loads are combined (e.g. Comb EQ1, Comb Eq2

and Comb EQ3) , the moment at the left end may be negative or positive, depending

Figure 5.7 Member results for beam (B-5)


on the magnitude of the lateral forces. For beam, B-5, for example, the left end

moment due to dead and live loads at ultimate condition is -14.7130 kN-m. This

moment changed to positive when combined with the moment due to lateral forces

and the final left end moment is 113.3387 kN-m. This means that a section of a

beam may resist both negative and positive moments when lateral forces act on the

building.

If you now observe the moment at the right end of the beams, what conclusions can

you make? If the lateral forces are applied from the right to left, what would you

expect about the moments of the left end of the beams? What about the span

moments of the beams?

Things to Ponder

Lateral forces must be applied on the plane structure in two directions to the left and

to the right since earthquakes or wind can come from any direction. The case study

illustrates the effect of lateral forces on the beam’s response, particularly on

bending moment at the supports of beams for rigid frames. Depending on the

direction of the lateral forces, the moments may become negative or positive when

combined with the vertical loads. The phenomenon where the type of the moment

changes from negative to positive and vice versa is referred to as “moment reversal”.

What is the implication of this phenomenon with respect to design of reinforced

concrete (RC) beams? In the RC design, longitudinal reinforcements are provided to

resist tension due to bending moment. When the moment is negative, tension occurs

at the top portion of the beam section and longitudinal reinforcements should be

provided near the top. On the other hand, when the moment is negative, tension

occurs at the bottom portion of the beam section and longitudinal reinforcements

should be provided near the bottom. In designing RC beams, we should consider

both negative and positive moments, meaning longitudinal reinforcements must be

provided at the top and bottom portion of the beam section.


The “Special Provisions for Seismic Design” for structural

concrete in the code (e.g. NSCP 2001 section 421.4 or ACI

code Chapter 21) addresses the phenomenon of moment

reversal in the design of flexural members of frames. Can

you locate the specific sections of the code where these

special provisions are ?

“At any section of a flexural member for top as well as bottom reinforcement, the

amount of reinforcement shall not be less than the minimum reinforcement and

the reinforcement ratio shall not exceed 0.025. At least two bars shall be

provided continuously both top and bottom.”

“Positive-moment strength at joint face shall not be less than one half of the

negative-moment strength provided at that face of the joint. Neither the negative

nor the positive-moment strength at any section along member length shall be

less than one fourth the maximum moment strength provided at face of either

joint.”

Bending moments are not the only internal forces that are affected by lateral forces.

Shear and axial forces are also significantly influenced by the lateral forces – in most

cases the combination loading with wind or seismic loads produces the worst

condition in a building. One other response and probably the most important

structural response that is affected by lateral forces is the horizontal displacement or

side sway. Buildings when subjected to lateral forces would undergo horizontal

displacement or drift and must be checked against story drift limitations to prevent

the structure from collapsing laterally. All these structural response quantities must

be checked against the effects of lateral forces.


Things to Try 1. Explore the same frame in the case study and observe the effect of lateral forces

to the other response quantities of the structure, such as :

Span moments in beams

Bending moment in the columns

Axial force in the columns

Shear in the beams

Shear in the columns

Write a report listing down your observations about the effect of lateral forces for

each structural response. Support your observations with figures and/or tables

for specific members. State the implications of your observations to design.

2. How much should the lateral force F2 be increased so that the moment at the left

end of all beams in the second level will also change to positive when dead, live

and lateral loads are combined?


Hibbeler, R.C. (2000). Structural Analysis. Chapter 7, Prentice-Hall, Inc. New Jersey,

USA


Towers, and Other Vertical Structures, Sections 409 and 421, Association of


Nawy, E. G. (1996). Reinforced Concrete: A Fundamental Approach, Chapter 15,

Prentice-Hall, Inc.



Schodek, D.L. (1998). Structures. Chapters 14, Prentice-Hall, Inc. New Jersey, USA

Understanding 2D Structural Analysis by A.W.C. Oreta : Pinned &Fixed Support Conditions 6 - 1

Figure 6.1 Models for Supports

CHAPTER 6

PINNED AND FIXED SUPPORT CONDITIONS

BACKGROUND Supports are used to attach structures to the ground to restrict their movement

due to external loads. The loads tend to move the structure; but the supports

prevent the movements by exerting reactions to neutralize the effects of the

forces; thereby keeping the structure under equilibrium. The type of reaction a

support exerts on a structure depends on the type of supporting device used and

the type of movement it

prevents. Figure 6.1

represents models of

supports for plane structures.

Consider first the idealized

models at the left portion of

the figure. A roller support

prevents translation normal

to the plane of the roller and produces a corresponding normal reactive force,

while a pinned or hinged support prevents translation in any direction but allows

rotation and thus produces reaction forces. A fixed support prevents rotation and

translation and thus produces reaction forces and a moment.

The pinned (or roller) and fixed support conditions are idealized models of

support conditions. What type of model for the support should you use when

you want to represent the actual support conditions? The answer to this question

depends on degree of constraints provided by the foundation. One factor which

affects the constraints at the support is the type and detail of the connection

between the column and the footing. Figure 6.2 shows two examples of

connections at the footing and the corresponding idealized models. The steel

column is welded to a base plate and the base plate is connected to a concrete


Pinned Fixed

Figure 6.2 Types of Connections

Figure 6.4 Fixed-Pinned Columns


footing by bolts. The degree of fixity

between the column and the footing will

depend on the bolted connections. Figure

6.3 is a detail of a rolling expansion

bearing which consists of a hinge on top

of a pedestal whose base rests on a

series of rollers. In the building in Figure

6.4, the frame consists of tapered

columns pinned at the base. As the top of

the columns is rigidly built into a stiff

beam, the columns are effectively fixed at the top and pinned at the base.

Schodek (1998) provides more examples of different types of connections and

idealized models. Another factor which affects the constraints or fixity at the

supports is the soil condition. A column supported on a relatively small footing

and resting on compressible soil may be assumed to be hinged at the end, since

such soils offer but little resistance to rotation of the footing. On the other hand, a

footing resting on solid rock, or a column supported by a pile foundation may be

assumed to have sufficient fixity to prevent rotation and a fixed support may be

assumed. Columns supported by a continuous foundation mat should likewise be

assumed fixed (Nilson et al 2004).

Figure 6.3 End Bearing Detail (http://nisee.berkeley.edu.ph/godden)


How important is the assumed model of the support in the behavior and

response of the structure? This chapter aims to explore the effects of the support

conditions on the response of a structure.

CASE STUDY : What are the implications of pinned and fixed support conditions to structural design? Two identical steel gabled frames with different support conditions similar to

Figure 6.2 will be analyzed subjected to two basic load cases – dead load and

wind load. Compare the behavior and response of the two structures.

Things to Do 1. Draw two identical frames with different support conditions - one frame

with pin supports and the other frame with fixed supports.

2. Apply dead load (WL = 0.5 k/ft) and display the diagrams for the bending

moment, shear and axial forces. 3. Apply the wind loads as shown acting on the windward and leeward walls

and the roofs. Display the diagrams for the bending moment, shear and

axial forces. 4. Apply combination load case : 0.9 DL + 1.3 WL

Observation

Displacements: Figures 6.5 and 6.6 show the deformed shapes of the two

frames due to dead and wind loads. Which frame has relatively larger

displacements? If you view the nodal displacements at the nodes, you will find

that the nodal displacements for the pinned supported frame are almost twice

that of the fixed supported frame. As an example, for the top node, the vertical

displacements due to dead load is about 2.4 in for the pinned case, while 1.28 for

the fixed case. On the other hand, for the same node the vertical displacements

due to wind load is about 12.0 in for the pinned case, while 6.40 for the fixed

case.


windwardwall

0.35 k/ft

Leeward wall

0.25 k/ft

Uplift = 3.0 k/ft Uplift = 3.0 k/ft

Wind Load

DL = 0.5 k/ft

Dead Load

90 ft

20 ft

15 ft

Column W 27 x 84

Rafter W 21 x 68

Section Properties (Ref. AISC manual) W 21 x 68 A = 20.0 in2 I = 1480 in4 d = 21.13 in W 27 x 84 A = 24.8 in2 I = 2830 in4 d = 26.69 in Material Properties (A36 steel) E = 29 x 103 ksi Specific weight = 0.284 lb/in3

Coefficient of Thermal Expansion = 6.5 x 10-6 /F

CASE STUDY 6


Figure 6.6 Wind Load and Deformation Diagram

Figure 6.5 Dead Load and Deformation Diagram

(a) Pinned (b) Fixed



Figure 6.7 Bending Moments due to Dead Load

Figure 6.8 Bending Moments due to Wind Load




Figure 6.9 Bending Moments for Combination Load Case : 0.9 DL + 1.3 WL

Bending Moments : Figures 6.7 and 6.8 shows the bending moment diagrams

for each frame for the two basic load cases. The bending moments for the

pinned-base frame are relatively larger than the fixed-base frame for both loading

conditions. The maximum end moment of the column for the pinned case due to

dead load is 259.6 kip-ft at the top end compared to 240.5 kip-ft at the bottom

end for the fixed case. The maximum end moment at the rafter due to dead load

is 259.6 kip-ft for the pinned case and only 196.7 kip-ft for the fixed case. Similar

observations can be found for the bending moments due to wind load. If the

loads are now combined using appropriate load factors as shown in Figure 6.9,

the end moments in the pinned case are about 12% more in the columns and

about 25% more in the rafters compared with the fixed case.



Figure 6.5 Shear Forces due to Dead Load


Figure 6.11 Shear Forces due to Wind Load




Figure 6.12 Axial Forces due to Dead Load

Figure 6.13 Axial Forces due to Wind Load

(a) Pin Supports (b) Fixed Supports



Shear Forces: Compare now the shear forces in Figures 6.10 and 6.11. The

magnitudes of the shear forces in the columns in the fixed-base condition are

greater than the pinned-base condition, but the shear forces in the rafters in the

pinned case are greater than the fixed case. This is true for both dead and wind

loading conditions.

Axial Forces: In Figures 6.12 and 6.13 are shown the comparison of the axial

forces. There is not much of a difference between the magnitudes of the axial

forces in the columns between the two frames, although the axial forces in the

fixed supported frame are slightly larger for the rafters.

What are the implications of the observations about the two gabled frames with

different support conditions? In the design of these structures, the size of the

members is determined based on the internal moments and forces. The size of

the rafters is usually determined based on the critical moments, while the size of

the columns is obtained for the combined effects of the moments and axial forces.

Based on the member size obtained, the shear requirements are checked. As

observed earlier, the maximum moments developed in the frame which has

fixed-base connections are relatively less than those developed in the pinned

supported frame. This means that the members of the fixed supported frame

may be designed with smaller sections. Moreover, there is a reduction in

deflections in the fixed case. However, to achieve these advantages of

minimizing moments and reducing deflections in the gabled frame using fixed

supports, special attention should be given in the design of the foundation so that

full fixity of the column will be achieved. Does this mean a fixed supported frame

is more superior than a pin supported frame? Not really! There are cases where

the design of the foundation is a problem and full fixity at the base is difficult to

achieve. In this case, a pinned-base connection may be the best overall solution.

Besides, there also advantages in a pinned supported frame. The foundation for

a pinned-base frame need not be designed to provide moment resistance.

Horizontal thrusts associated with vertical loads are usually smaller in a pinned


condition. Each specific design must be evaluated in its own context to see which

approach proves most desirable.

Things to Ponder The modeling of the supports of any structure should be based on actual

conditions at the site. The engineer should make sure what is assumed in the

modeling, analysis and design of the supports should be realized when the

structure is constructed. When full fixity is assumed in the modeling and analysis

of the structure, this condition should be assured when the design and

construction of the foundation is done. Similarly, the condition of no moment

resistance at the base should likewise be assured in the foundation design and

construction to simulate a pinned-base assumption. If the assumptions and

actual conditions are entirely different, the outcome may be catastrophic.

Things to Try

1. Analyze the frame shown. Compare the results between pinned-supported

frame and fixed-supported frame. Consider the effects of vertical loads (WDL =

15 kN/m and WLL = 7 kN/m) applied fully and earthquake loads applied at

each floor (F4 = 80 kN, F3 = 60 kN, F2 = 40 kN and F1 = 20 kN), separately.

And then combine the loads (1.4 DL + 1.7 LL and 1.3 DL + 1.1 LL + 1.1 EQ).

Assume the following material properties:



Coefficient of thermal expansion = 0.00099 / oC 2. Make sketches of possible details of column and footing connections for

reinforced concrete structures. When can the footing be assumed pinned or

fixed?


250 mm

400 mm

Beam cross-section

350 mm

350 mm


3 @ 5.0 m = 15.0 m

4 @3.0 m = 12.0 m

Frame geometry


References


Library

Kassimali, A. (1999). Structural Analysis, 2nd Edition, Section 3.3, Brooks-Cole

Publishing Co., USA

Nilson, A.H., Darwin, D. and Dolan, C.W. (2004). Design of Concrete Structures,

13th Edition, Section 12.5, McGraw-Hill, Inc. NY, USA

Schodek, D.L. (1998). Structures. Section 3-3-2, Prentice-Hall, Inc. New Jersey,

USA

Understanding 2D Structural Analysis by A.W.C. Oreta : Soil Effects on Foundations 7 - 1

Figure 7.2 Spring Models in GRASP

CHAPTER 7

SOIL EFFECTS ON FOUNDATIONS

BACKGROUND Foundations of structures are supported by the soil. The effect of the soil on the

behavior of the structure is significant especially for soft soils since the required

fixity between the column and footing may be difficult to realize. A simple isolated

footing may rotate, settle or

shift sideways by some

amount depending on the

load and soil conditions.

Modeling the foundation

considering the soil stiffness

falls between the pinned or

fixed conditions. When the

effect of the soil in the

structural model is

considered, this becomes a

“soil–structure interaction”

problem. One popular and simple approach of modeling the soil is by the used of

“springs”. An isolated footing or a pile foundation may be represented by three

springs – one for vertical settlement, one for rotation and one for lateral

movement (Figure 7.1).

GRASP provides an option

of representing the

constraint at a support by

springs as shown in Figure

7.2. You first choose a basic

support condition from the

six idealized models shown

Isolated Footing

Pile Foundation Figure 7.1 Modeling of Foundations (Anwar 1998)


at the left and then modify the restraint at one or more degrees of freedom by

spring models by inputting the appropriate spring stiffness. The stiffness of the

spring can be derived by the modulus of sub-grade reaction of the soil or by the

method suggested by Gazetas (1991) which is adapted by ATC-40 (1996), where

the footing dimensions, depth of embedment and soil properties (modulus of

elasticity, shear modulus, poisson’s ratio) are parameters. This chapter explores

the option of modeling foundations using springs and compares the results to the

idealized pinned or fixed conditions.

CASE STUDY : How can footings resting on soil be modeled to incorporate soil-structure interaction? Two identical steel gabled frames with different support conditions will be

analyzed. The supports will be represented by three springs – one frame

resting on dense soil and the other frame resting on soft soil. Two basic load

cases – dead load and wind load – will be applied. Observe the behavior and

response of the two structures supported by springs resting on two types of

soils and then compare the results with the frames supported by idealized

pinned and fixed supports in Chapter 6.

Things to Do 1. Draw two identical frames supported by three springs.

2. Input the stiffness of the springs for two types of soils : (a) dense soil and

(b) soft soil

3. Apply dead load on the rafters. 4. Apply the wind loads as shown acting on the windward and leeward walls

and the roofs. 5. Apply combination load case : 0.9 DL + 1.3 WL 6. Perform analysis and display the diagrams for the bending moment, shear

and axial forces. Compare the results with the case study in Chapter 6.


windwardwall

0.35 k/ft

Leeward wall

0.25 k/ft

Uplift = 3.0 k/ft Uplift = 3.0 k/ft

Wind Load

DL = 0.5 k/ft

Dead Load

90 ft

20 ft

15 ft

Column W 27 x 84

Rafter W 21 x 68

Section Properties (Ref. AISC manual) W 21 x 68 A = 20.0 in2 I = 1480 in4 d = 21.13 in W 27 x 84 A = 24.8 in2 I = 2830 in4 d = 26.69 in Material Properties (A36 steel) E = 29 x 103 ksi Specific weight = 0.284 lb/in3

Coefficient of Thermal Expansion = 6.5 x 10-6 /F

Dense Soil Kx = 4500 kip/in Ky = 1500 kip/in Kz = 200,000 kip-ft/rad Soft Soil Kx = 240 kip/in Ky = 100 kip/in Kz = 12,800 kip-ft/rad

CASE STUDY 7


Figure 7.4 Wind Load and Deformation Diagram


(a) Dense Soil

(b) Soft Soil

(a) Dense Soil

(b) Soft Soil


Figure 7.5 Bending Moments due to Dead Load

Observation Displacements: Figures 7.3 and 7.4 show the deformed shapes of the two

frames due to dead and wind loads. Which frame has relatively larger

displacements? If you view the nodal displacements at the nodes, you will find

that the frame resting on soft soil is more flexible and had displacements about

40% more than that of the frame resting on dense soil. Observe for example the

top node. The vertical displacements due to dead load is about 2.5 in for the soft

soil case, while 1.5 in for the dense soil case. On the other hand, the vertical

displacement due to wind load is about 12.6 in for the soft soil condition, while

7.6 in for the dense soil condition.

Bending Moments : Figures 7.5 and 7.6 shows the bending moment diagrams

for each frame for the two basic load cases. The bending moments for the frame

resting on soft soil are relatively larger than the dense soil condition for both

loading conditions. The magnitude of the maximum end moments of the rafters

(a) Dense Soil

(b) Soft Soil


Figure 7.6 Bending Moments due to Wind Load

Figure 7.7 Bending Moments for Combination Load Case : 0.9 DL + 1.3 WL

(a) Dense Soil

(b) Soft Soil

(a) Dense Soil

(b) Soft Soil



Figure 7.9 Shear Forces due to Wind Load

(a) Dense Soil

(b) Soft Soil

(a) Dense Soil

(b) Soft Soil



Figure 7.11 Axial Forces due to Wind Load

(a) Dense Soil

(b) Soft Soil

(a) Dense Soil

(b) Soft Soil


and columns for the soft soil case due to dead load is 246.6 kip-ft compared to

209.5 kip-ft for the dense soil case. On the other hand, The magnitude of the

maximum end moments of the rafters and columns for the soft soil case due to

wind load is 1,400 kip-ft compared to 1,100 kip-ft for the dense soil case. The

maximum end moments for the combined dead and wind loads in Figure 7.7, for

the soft soil case are about 13% more than the moments in the dense soil case.

In all loading cases, smaller moments at the bottom end of the columns occur in

the frame resting on soft soil.

Shear Forces: Compare now the shear forces in Figures 7.8 and 7.9. The

magnitudes of the shear forces in the columns in the dense soil condition are

greater than the soft soil condition, but the shear forces in the rafters in the soft

soil case are greater than the dense soil case. This is true for both dead and

wind loading conditions.

Axial Forces: In Figures 7.10 and 7.11 are shown the comparison of the axial

forces. There is not much of a difference between the magnitudes of the axial

forces in the columns between the two frames, although the axial forces in the

rafters for dense soil case are slightly larger than for the soft soil case. The axial

forces for both soil conditions are almost the same.

Comparing with pinned and fixed conditions: How do the results of the

analysis of the frames supported by spring models compare with the idealized

pinned-base and fixed-base conditions in Chapter 6? By simply comparing the

diagrams, we can see that the response of the frame resting under soft soil

conditions is similar to the pinned-base frame. The only difference between the

two models is that moments are developed at the bottom ends of the columns

for the spring model compared to zero moments for the pinned case. As a result,

the maximum end moments under the soft soil condition are slightly smaller than

the pinned-base condition. The response of the frame under the dense soil

condition is very similar to the fixed-base frame. However, the maximum end


moments of the columns for the frame under dense soil condition are slightly

smaller than the fixed-base condition. On the other hand, the rafter moments are

slightly larger than the dense soil case than the fixed-base case.

Things to Ponder A pinned-base support assumes zero moment resistance at the base, while a

fixed-base support assumes a rigid base connection. Foundations rest on soil

whose properties are variable from soft clay to hard rock. Depending on the soil

and load conditions, the actual restraint developed at the base may fall between

the pinned-base and fixed-base conditions. The modeling of the supports of any

structure should be based on actual conditions at the site. Spring models to

represent the soil may be a simple approach when soil-structure interaction is

considered. However, one of the problems that the designer should confront

when considering the soil effects is the soil property, particularly what appropriate

value of soil stiffness to use in the model. An unreasonable assumption of the

soil stiffness may lead to an unconservative design. In the absence of information

about the soil, the pinned-base or fixed-base models may be used appropriately

with the designer introducing additional safety factors in the design (e.g.,

introducing some moment at the base even if a pinned-base support is assumed).

Things to Try

1. Analyze the same gabled frame of the case study. Instead of using three

springs to model the soil, represent the support by a pin with a rotational

spring. Use the soil stiffness, kz values given for the dense and soft soil.

Compare the results of the “pin-rotational spring” supported frames with the

“three-spring” supported frames for both types of soil.



Anwar, N. (1998). “Modeling of Foundations,” ACECOMS News & Views, April –

June, page 17, AIT, Thailand

Gazetas, G. (1991). Foundation Vibrations, Principles and Practices. Prentice-

Hall, New Jersey, USA

Applied Technology Council (1996), “Foundation Effects”, Seismic Evaluation

and Retrofit of Concrete Buildings (ATC-40), Vol. 1, Chapter 10

De La Cruz, L., Florendo, C. and Santiago, H. (2003). The Effect of Soil-Spring

Modeling in 2D Frame Analysis. Undergraduate Thesis, De La Salle University,

Manila

Obrien E. and Keogh, D. (1999). Bridge Deck Analysis., Chapter 4, E & FN Spon,

London

Understanding 2D Structural Analysis by A.W.C. Oreta : Support Settlements 8 - 1

Figure 8.1 Defining Support Settlements

in GRASP

CHAPTER 8

SUPPORT SETTLEMENTS

BACKGROUND

The foundations supporting the structures can settle for a variety of reasons with

the most common being consolidation of the soil beneath a support. The larger

the load on the soil, the more likely is consolidation to occur. Rarely is the

amount of settlement exactly the same beneath for all supports. Differential

settlement is a common occurrence in structures which must be checked since

additional internal forces and moments are induced. This phenomenon of “soil-

structure interaction” is

usually incorporated in the

modeling of the structure by

representing the soil by

“springs”. Modeling of

foundations using springs was

presented in the previous

chapter. Another approach,

specifically to predict the

effect of differential settlement

is to introduce a prescribed amount of settlement in a foundation support.

GRASP has an option of introducing displacements (horizontal, vertical or

rotation) at a support (Figure 8.1). You first choose a basic support condition

from the six idealized models shown at the left and then modify the restraint at

one or more degrees of freedom by introducing a specified amount of

displacement. This chapter explores the effects of differential settlements of

support on the internal forces and moments of structures.


CASE STUDY: What are the effects of differential settlements of supports to the internal forces and moments of a rigid frame? A two-story rigid frame with fixed-base supports will be analyzed. Support

settlement will be introduced in two locations: (a) exterior support; and (b)

interior support. No loads will be applied. Observe the effects of the

settlement on the response of the members of structure when the settlement

occurs at the exterior and interior support.

Things to Do

1. Draw two identical frames with fixed-base supports as shown in Figure 8.2.

2. Input a 20 mm vertical displacement downwards at a support – at the

exterior support for one frame and at the interior support for the other

frame.

3. Perform analysis and display the diagrams for the deformation, bending

moment, shear and axial forces.

Observation

The deformation of the frames when a settlement of 20 mm is introduced at the

exterior support and interior support are shown in Figure 8.2(a) and Figure 8.2(b),

respectively. Observe the effect of the settlement in the members especially in

the beams supported by the columns where the support settlement was

introduced. Beams connected to columns along the line where the support

settled have relatively significant vertical and horizontal displacements. For

example, a 20 mm settlement at the interior support produces about 18 mm

vertical displacement and 6.8 mm horizontal displacement at the uppermost node

as shown in Figure 8.2(b). There is also a significant amount of sidesway

produced in the columns. Because of these relative displacements, curvatures

are developed in the members. Associated with these curvatures are internal


bending moments. Figure 8.3 shows the bending moment induced in the

members. Members with larger curvatures have larger induced moments. For

example, the beams in the first floor have the largest end moments about 213

300 mm

300 mm


1200 mm

250 mm

300 mm

100 mm

Beam cross-section

3.0 m 3.0 m4.0 m

3.0 m

4.0 m

Frame geometry

Modulus of Elasticity = 20, 500 N/mm2 Unit Weight = 24 kN/m3 Coefficient of Thermal Expansion = 0.00099/C

CASE STUDY 8


Figure 8.2 Deformation due to Support Settlement

kN-m and 345.6 kN-m for the frames with settlements at the exterior and interior

supports, respectively. The further the beams from the location of the settlement,

the smaller the induced moments. On the other hand, it can be observed that

curvatures in the columns in Figure 8.2 are larger for the upper columns resulting

to larger induced end moments than the lower columns in Figure 8.3. Figures

8.4 and 8.5 present the effect of the support settlement on the shear and axial

forces. The magnitude of shear forces are largest in the beams connected to the

columns supported by the foundation where the settlement occurred. On the

other hand, axial forces are significant in the columns near the support where

settlement was applied. Clearly, the greater differential settlement, the greater

the induce internal forces and moments in the beams and columns of the

structure. These internal effects on the frame may lead to failures in the design if

not anticipated. For this reason, special attention must be taken with the design

of foundations for rigid structures to minimize the risk.

(a) Exterior Support (b) Interior Support


Figure 8.4 Shear Forces due to Support Settlement

Figure 8.3 Bending Moments due to Support Settlement




Figure 8.5 Axial Forces due to Support Settlement

Things to Ponder Differential settlement of supports has

the effect of generating “sagging

moment” at the support which settles.

This is significant in rigid continuous

structures since the fixed-ended nature

of the connections restrains the

members from rotating and translating.

The increase in internal forces and

moments in the members may be

significant especially in soft soils.

Settlement of supports frequently occurs


Figure 8.6 Tilting Building


when the ground shakes during an earthquake (Figure 8.6). If the member is not

sufficiently sized to carry this increased internal moments and forces, the

member could potentially fail or become seriously overstressed.

Things to Try

1. Analyze the same rigid frame in the case study. Replace the fixed-base

supports by pinned-base supports. How does support settlement affect the

response of the members of a pinned-base supported frame? Is the effect of

differential settlement more pronounced in a pinned-based frame or a fixed-

base frame?

2. Compare the induced moments for two cases of the continuous beam shown

below:

a. Introduce vertical settlements of 20 mm at the roller and 40 mm at the

adjacent pin support.

b. Introduce a vertical settlement of 40 mm at the pin support adjacent to

the roller.

3 @ 4.0 m = 12.0 m

250 mmBeam cross-section

400 mm

Three-span Continuous Beam

Material Properties Modulus of elasticity = 20,500 N/mm2


Coefficient of thermal expansion =

0.00099 / oC



Bowles, J. (1982). Foundation Analysis and Design, Chapter 5, McGraw-Hill, Inc.,

New York, USA

Hibbeler, R. (2000). Structural Analysis, Section 10.5 and 11.5, Pearson

Education Asia Pte Ltd., Singapore

Obrien E. and Keogh, D. (1999). Bridge Deck Analysis. Chapter 3, E & FN Spon,

London

Orense, R. (2003). Geotechnical Hazards: Nature, Assessment and Mitigation,

Part 4, UP Press, Quezon City, Philippines

Schodek, D.L. (1998). Structures. Section 8-3-4 and 9-3-4, Prentice-Hall, Inc.

New Jersey, USA

Understanding 2D Structural Analysis by A.W.C. Oreta : Truss Analysis 9 - 1

CHAPTER 9

TRUSS ANALYSIS

BACKGROUND A truss is an assemblage of slender straight members arranged in triangles to

form rigid framework. The individual elements are typically assumed to be

connected at the joints with smooth pinned connections. The joint connections

are usually formed by bolting or

welding the ends of the members to

a gusset plate (Figure 9.1). Loads

and reactions are assumed to act at

the joints. As a result of the joint

loading and smooth hinge conditions,

the truss members are usually

designed to resist axial forces

(tension and compression) only. The

truss, through its stable configuration,

resists external loads by deflection of the structure which occurs only when one

or more of its members are deformed. Planar trusses lie in a single plane and are

often used to support roofs of buildings and bridges. This chapter presents an

analysis of a typical roof truss subjected to dead, live and wind loads.

CASE STUDY : Analysis of a Roof Truss The following case study illustrates the procedure for the analysis of a light steel

truss for a gable-form roof. The roof construction, truss configuration, and design

loads are shown. The truss will consist of double angles of A36 steel as

members. Joints will use gusset plates and bolts. Trusses are to be spaced 8 ft

on centers.

Figure 9.1 Detail of a Truss Joint



Dead Load and Live Load Computation: The weight of the truss is assumed to

be 20 lb/ft. The dead load on the truss is computed as follows: 20 lb/ft + (5 psf)(8

ft) = 60 lb/ft. Similarly, the live load on the truss is computed as : (15 psf)(8 ft) =

120 lb/ft. The joints loads corresponding to the dead and live loads are then

computed using tributary horizontal projection of the lengths of the truss

members. For example at the interior joint, the nodal force due to dead load is

(60 lb/ft)(10 ft) = 600 lb as shown in Figure 9.2.

Wind Load Computation: Wind pressure on roof surfaces depend on wind

velocity, wind direction, and the building’s exterior form and dimensions. The

computation of the wind pressure can be found in design codes (e.g. see NSCP

2001). The wind ordinarily induces a direct, inward pressure on surfaces of the

building facing the wind, and an outward suction pressure on surfaces on sides

No ceiling

4 @ 10 ft = 40 ft

10 ft

Steel deck + insulation at 5 psf

Glass sky light at 5 psf

Live Load at 15 psf

Trusses at 8 ft c/c

Wind Load 15 psf

Truss Geometry and Loads

Truss weight at 20 lb/ftCASE STUDY 9

No ceiling


opposite the wind. Pressures on roof surfaces depend partly on the slope. Flat

and near-flat surfaces tend to have uplift pressure. As the slope increases, the

roof surfaces facing the wind develop inward pressures. Since the wind direction

changes, the building has to be investigated for two wind directions – wind from

the left and wind from the right of the structure.

300 lb 300 lb

600 lb

600 lb600 lb

Dead Load applied at joints

600 lb 600 lb

1200 lb

1200 lb

1200 lb

Live Load applied at joints

Figure 9.2 Dead and Live Loading on Truss


Consider a wind force pressure of 15 psf acting normal to the roof surface. The

uniform wind force on the truss is computed as : (15 psf)(8 ft) = 120 lb/ft. By

multiplying the wind force (120 lb/ft) by the tributary length of the inclined

members, the joint loads can be computed. For example the nodal force at the

interior joint is (120 lb/ft)(10.54 ft) = 1,265 lb as shown in Figures 9.3 and 9.4 for

the wind from the left and from the right, respectively. The wind loads act

perpendicular to the surface of the truss. However, for the purpose of using the

forces in GRASP, the wind loads are resolved to vertical and horizontal

components.

Wind Loads (Left) Resolved to X & Y Components

600 lb

400 lb

400 lb

400 lb

200 lb

600 lb 600 lb

1200 lb

1200 lb

1200 lb

849 lb

849 lb

1265/2 lb

1265 lb

1265/2 lb

1265 lb Wind Left

Wind Loads Left on Truss

Figure 9.3 Wind Loads (Left) for Truss Analysis


Things to Do

1. Model the structure as a “truss.” Use the GRASP toolbars.

2. Assume the following material properties for steel:

modulus of elasticity = 29,000 ksi

unit weight = 491 lb/ft3

coefficient of thermal expansion = 6.5 x 10 -6/F

3. Assign the following sections for the truss members:

Top and bottom chord members : double angles 4 x 3 x 3/8, short legs

back to back (A = 4.97 in2; I = 3.84 in4 and ytop= 0.782 in)

Web members : double angles 3 x 2 x 3/16, short legs back to back

(A = 1.80 in2; I = 0.613 in4 and ytop= 0.470 in)

849 lb

849 lb

1265/2 lb

1265 lb

1265/2 lb

1265 lb

600 lb

800 lb

400 lb

400 lb

200 lb

600 lb 600 lb

1200 lb1200 lb

Wind Loads Right on Truss

Wind Right

Wind Loads (Right) Resolved to X & Y Components

Figure 9.4 Wind Loads (Right) for Truss Analysis


4. Apply the basic load cases for the vertical loads for dead load (DL) and

roof live load (Lr) as shown in Figure 9.2.

5. Apply the basic load cases for wind load (WL) : wind left and wind right as

shown in Figure 9.3 and 9.4, respectively.

6. Apply the combination load cases for allowable stress design (NSCP 2001

section 203):

DL + Lr

DL + WL left and DL + WL right

DL + 0.75(Lr + WL left) and DL + 0.75(Lr + WL right)

7. Choose structure to analyze as “truss” and perform analysis. Display

results for deformation and member forces.

Observation

Axial deformation of the truss members due to external loads result to joint

displacements. This results to the deformed shape of the truss. Figure 9.5 and

9.6 present the resulting deformed configuration of the truss due to dead loads

and wind loads from the left, respectively. Because of gravity loads, the

deflection of the truss is downwards (Figure 9.5). On the other hand, the joints of

the truss tend to move upwards due to uplift force produced by the wind loads

(Figure 9.6). The axial forces due to dead load and wind load from the left are

shown in Figures 9.7 and 9.8. Observe the type of axial force (positive for tension

and negative for compression) developed in the members. For example, the

bottom chord members resist tensile forces under gravity loads and compressive

forces under wind loads. On the other hand, diagonal web members carry

compressive forces under gravity loads and tensile forces due to wind loads. The

top chord members, however, carry only tensile forces. This means that a

member may resist a tensile force under one loading case and a compressive

force in another loading case. Truss members must be designed for both tensile

and compressive axial forces. Figure 9.9 shows the combined effect of the axial

forces for the two basic load cases. Using the preview report of GRASP, the


comparison of member results for the different loading cases can be viewed as

shown in Figure 9.10. Observe the axial forces including the sign for various load

cases. For what loading case does a specific member has the largest positive

and negative axial forces? Are there changes in the type of forces (stress

reversal)? From the results, you can obtain the design envelope which

represents the maximum and minimum values of the design forces. Figure 9.11

presents the maximum and minimum axial forces for some members. By

selecting the maximum and minimum forces, a member can now be designed for

tensile and/or compressive stresses. There are two methods for designing steel

members – the allowable stress design (ASD) and the load resistance factor

design (LRFD).



Figure 9.6 Deformations due to Wind Load (Left)



Figure 9.8 Axial Forces due to Wind Load (Left)

Figure 9.9 Axial Forces for Combined Load Case DL + WL (Left)


Figure 9.10 Member Results of Selected Members

Figure 9.11 Member Forces Envelope (Maximum and Minimum Forces)


Things to Try

1. Analyze the truss shown

subject to the same loading

conditions and using the

same member properties and

dimensions as given in the

case study. What is the effect of changing the arrangement of the web

members? Compare the results with those in the case study. Which truss

configuration resists larger axial forces? Which truss has smaller defections?

2. Compare the response of two types of roof trusses – a Pratt truss and a Fink

truss. What are the advantages and disadvantages of the each truss?

3. Analyze two types of bridge trusses – a Pratt truss and a Howe truss.

Compare the resulting axial forces due to various loads.


Ambrose, J. (1994). Design of Building Trusses. John Wiley & Sons, Inc.,

Canada


Library

Hibbeler, R. (2000). Structural Analysis, Chapters 3 and 13, , Pearson Education

Asia Pte Ltd., Singapore


Towers, and Other Vertical Structures, Sections 203, 204, 205 and 207,

Association of Structural Engineers of the Philippines, Inc. (ASEP), Quezon City,

Philippines

Understanding 2D Structural Analysis by A.W.C. Oreta : Special Modeling Issues 10 -1

CHAPTER 10

SPECIAL MODELING ISSUES

Modeling of structures is an art which is developed by a structural engineer

through experience and research. Every structure is unique. The same structure

may be modeled and analyzed by two engineers differently. The assumptions

made by each engineer may be different in one way or another. What is

important in structural analysis is that the model represents the real structure with

an appropriate likeness to capture the desired response. Presented in this

chapter are special topics in the modeling and analysis of structures that are

interesting to explore and investigate.

1. A Frame or a Truss? The typical simple analysis for the internal forces in a truss assumes smooth

“pin” joints meaning there is no moment resistance at these joints. The loads are

also assumed to act at these joints. As a result of these idealized loading and

joint conditions, the truss members become two-force members and carry only

axial forces. In reality, these ideal conditions may not be satisfied completely.

The truss member may not only carry axial forces but also bending moment and

shear forces. A pure truss action may not occur and the truss may function as a

rigid frame in resisting deformations. For what actual conditions do these occur?

(See Ambrose 1994).

(a) Semi-rigid and rigid joints: The connections at the joints of trusses rarely

are ideal pinned connections. Rigid or semi-rigid connections using welds or

bolts are common. Depending on the detail of the joint, a considerable

magnitude of moment resistance may be developed at the joint and

transferred to the members.

(b) Continuous chords: The members used in an actual truss are sometimes

continuous and come from one piece. Using a continuous top and bottom

chords is common in the design of trusses. If the chords are continuous


through any of the joints, the members will not act as two-force members but

will be subjected to both axial forces and bending moments.

(c) Member forces: Loads may not always be applied directly at a joint. Cases

where the roofing material is resting on the top chord or the ceiling is attached

at the bottom chord results to loads applied to the members. The truss

members will be subjected to combined bending and axial forces.

Things to Try

Select a typical roof truss configuration and analyze the truss for three

different cases:

(a) A pure truss with smooth pins

(b) A frame with rigid joints

(c) A frame with rigid joints but with moment releases at the ends of the web

members (GRASP has the option to introduce moment releases at the

end of members as shown in Figure 10.2)

Compare the resulting axial forces in the members among the three cases

and observe the moment developed in the members for the rigid frames.

Top & bottom chords maybe continuous members

Ceiling may be attached directly to bottom chord

Semi-rigid or rigid joints

Figure 10.1 Actual Conditions in a Truss


2. Shear Walls A shear wall-frame or dual system is commonly used as a structural framing

system in reinforced concrete buildings. A shear wall in a plane frame analysis

can be modeled by various techniques (Figure 10.3) in GRASP(Anwar and

Sharma 1997).

(a) Modeling the shear wall as a column: In this technique, the shear wall is

represented as a column with the column line coinciding with the centerline of

wall. The cross-section of the column is the same as the shear wall

dimensions. Connecting the beams to the column can be done two ways. The

length of the connecting beams may be taken from the center line of the

shear wall to the other end of the beam. Another approach is to divide the

beam into two segments. The first segment consists of the end portion of the

beam within the shear wall width and the other part is the main portion of the

beam outside the shear wall. The first part of the beam within the shear wall is

given an extra stiffness by modifying the cross-section dimensions of that

portion to be equal to the thickness of the wall and the height equal to the full

Figure 10.2 Introducing moment release results to zero moments in the web members of the rigid frame


story height. This stiffened end of the

beam may be considered as a rigid

end zone (Anwar and Sharma 1997).

(b) Modeling the shear wall as a truss: The shear wall is

represented as a truss or braced

frame consisting of columns and

diagonal members.

3. Lateral Stability Structures must be designed to resist

lateral forces due to wind and

earthquakes. Lateral displacements

must be limited to prevent large

deformations occurring in the structure.

The basic mechanisms for assuring

lateral stability depend on the type of

structural system. Buildings with

moment resisting space frames resist

lateral forces through frame action of the

columns and beams which are

Shear Wall – Frame System

Shear wall modeled as a column

Shear wall modeled as a truss

Figure 10.3 Modeling Shear Walls

Figure 10.5 X-Bracing of a Rigid Frame


Figure 10.4 Shear Wall – Frame

Interactive System (http://nisee.berkeley.edu.ph/godden)


connected by rigid joints. Another method used to assure stability is through

shear walls (Figure 10.4). These are rigid planar surface elements made of

reinforced concrete or masonry wall that inherently resist shape change because

of its high rigidity. Braced frames where diagonal members are used to form a

rigid framework have started to become popular especially in strengthening of

existing buildings against earthquakes (Figure 10.5). A dual system combines

moment-resisting frames and shear walls or braced frames to resist lateral forces.

Other types of lateral force resisting systems are listed in the codes (NSCP 2001

section 208).

Things to Try

Model a multistory building using three types of lateral force resisting systems

similar to Figure 10.6:

(a) rigid frame

(b) rigid frame with a shear wall

(c) rigid frame with one or two bays with bracing

Figure 10.6 Deformed Shape of a 2D Frame

Rigid Frame

Braced Frame Shear Wall - Frame


Analyze the frame when subjected to lateral forces and compare the response

quantities (lateral displacements, moments, shear and axial forces) of the

structures.

4. Construction Joints Because of construction difficulties, a long continuous beam out of one piece is

difficult to install. The continuous beam may consist of several pieces of beams

connected by construction joints. The joints need carry no moment and are

sometimes designed as simple pinned connections. Using construction joints

usually results to an assembly of statically determinate structures which function

together in a way that reflects the behavior of the continuous member (Figure

10.7). Where should the construction joints be located? The most ideal location

are points of inflection or points of zero moment when the beam is assumed to

be continuous. However placing the joints at the point of inflection is not always

possible. Design moments can be controlled by proper location of these joints.

Hence the effect of these joints on the bending of the structure must be

investigated (Figure 10.8)

(a) A Continuous Beam

(b) Construction Joints at the End Spans

(c) Construction Joints at the Middle Span

Figure 10.7 Use of Construction Joints in Continuous Members



Ambrose, J. (1994). Design of Building Trusses. Sections 7.7, 8.5, 11.1, John

Wiley & Sons, Inc., Canada

Anwar, N. and Sharma, B. (1997). “Modeling 2D Shear Walls”, ACECOMS News

& Views, July – Sept, pp. 4-5 & 17, AIT, Bangkok, Thailand


Library


Towers, and Other Vertical Structures, Sections 208, Association of Structural

Engineers of the Philippines, Inc. (ASEP), Quezon City, Philippines

Schodek, D.L. (1998). Structures. Sections 1-3-2, 8-4-4 and Chapter 14,

Prentice-Hall, Inc. New Jersey, USA

Figure 10.8 Controlling Moments using Construction Joints. (Zero moments occur at the construction joints)

Continuous Beam

Construction Joints at end spans

ABOUT THE AUTHOR

Andres Winston C. Oreta graduated D. Eng. and M. Eng. (Structural

Engineering) at the Nagoya University (Japan) and BSCE at UP Diliman. He is

presently a professor in civil engineering at the De La Salle University (DLSU) in

Manila, Philippines. He was the chair of the Civil Engineering Department in

1994-1997 and Director of the Engineering Graduate Studies in 1997 – 2000. He

has published papers in the ASCE Journal of Structural Engineering, Engineering

Structures, JSCE journals, JCI Transaction, Philippine Engineering Journal,

DLSU Engineering Journal, PATE Philippine Journal of Engineering Education

and Wiley’s Computer Applications in Engineering Education. His research

focuses on the areas of applications of artificial neural networks in civil

engineering, structural dynamics, earthquake engineering, reinforced concrete

structures and computer applications in engineering education. He can be

contacted at [email protected].

Documents

Oreta, A.W.C. (2004)."Understanding 2D Structural Analysis