Steel Design Guide
14
Staggered Truss Framing Systems
Neil Wexler, PE
Wexler Associates Consulting Engineers
New York, NY
Feng-Bao Lin, PhD, PE
Polytechnic University
Brooklyn, NY
AMERI CAN I NSTI TUTE OF STEEL CONSTRUCTI ON
Copyright � 2001
by
American Institute of Steel Construction, Inc.
All rights reserved. This book or any part thereof
must not be reproduced in any form without the
written permission of the publisher.
The information presented in this publication has been prepared in accordance with rec-
ognized engineering principles and is for general information only. While it is believed to
be accurate, this information should not be used or relied upon for any specific appli-
cation without competent professional examination and verification of its accuracy,
suitablility, and applicability by a licensed professional engineer, designer, or architect.
The publication of the material contained herein is not intended as a representation
or warranty on the part of the American Institute of Steel Construction or of any other
person named herein, that this information is suitable for any general or particular use
or of freedom from infringement of any patent or patents. Anyone making use of this
information assumes all liability arising from such use.
Caution must be exercised when relying upon other specifications and codes developed
by other bodies and incorporated by reference herein since such material may be mod-
ified or amended from time to time subsequent to the printing of this edition. The
Institute bears no responsibility for such material other than to refer to it and incorporate
it by reference at the time of the initial publication of this edition.
Printed in the United States of America
First Printing: December 2001
Second Printing: December 2002
AUTHORS
Neil Wexler, PE is the president of Wexler Associates, 225 Feng-Bao Lin, PhD, PE is a professor of Civil Engineer-
East 47th Street, New York, NY 10017-2129, Tel: ing of Polytechnic University and a consultant with Wexler
212.486.7355. He has a Bachelor s degree in Civil Engi- Associates. He has a Bachelor s degree in Civil Engineer-
neering from McGill University (1979), a Master s degree ing from National Taiwan University (1976), Master s
in Engineering from City University of New York (1984); degree in Structural Engineering (1982), and PhD in Struc-
and he is a PhD candidate with Polytechnic University, tural Mechanics from Northwestern University (1987).
New York, NY. He has designed more then 1,000 building
structures.
PREFACE
In recent years staggered truss steel framing has seen a Staggered trusses provide excellent lateral bracing. For
nationwide renaissance. The system, which was developed mid-rise buildings, there is little material increase in stag-
at MIT in the 1960s under the sponsorship of the U.S. Steel gered trusses for resisting lateral loads because the trusses
Corporation, has many advantages over conventional fram- are very efficient as part of lateral load resisting systems.
ing, and when designed in combination with precast con- Thus, staggered trusses represent an exciting and new steel
crete plank or similar floors, it results in a floor-to-floor application for residential facilities.
height approximately equal to flat plate construction. This design guide is written for structural engineers who
Between 1997 and 2000, the authors had the privilege to have building design experience. It is recommended that the
design six separate staggered truss building projects. While readers become familiar with the material content of the ref-
researching the topic, the authors realized that there was lit- erences listed in this design guide prior to attempting a first
tle or no written material available on the subject. Simulta- structural design. The design guide is written to help the
neously, the AISC Task Force on Shallow Floor Systems designer calculate the initial member loads and to perform
recognized the benefits of staggered trusses over other sys- approximate hand calculations, which is a requisite for the
tems and generously sponsored the development of this selection of first member sizes and the final computer
design guide. This design guide, thus, summarizes the analyses and verification.
research work and the practical experience gathered. Chapter 7 on Fire Resistance was written by Esther Slub-
Generally, in staggered-truss buildings, trusses are nor- ski and Jonathan Stark from the firm of Perkins Eastman
mally one-story deep and located in the demising walls Architects. Section 5.1 on Seismic Strength and Ductility
between rooms, with a Vierendeel panel at the corridors. Requirements was written by Robert McNamara from the
The trusses are prefabricated in the shop and then bolted in firm of McNamara Salvia, Inc. Consulting Structural
the field to the columns. Spandrel girders are bolted to the Engineers.
columns and field welded to the concrete plank. The exte-
rior walls are supported on the spandrel girders as in con-
ventional framing.
v
ACKNOWLEDGEMENTS
The authors would like to thank the members of the AISC McNamara Salvia, Inc. Consulting Engineers, who wrote
Staggered Truss Design Guide Review Group for their Section 5.1 Strength and Ductility Design Requirements.
review, commentary and assistance in the development of Bob s extensive experience and knowledge of structural
this design guide: design and analysis techniques was invaluable. Also thanks
to Esther Slubski who wrote Chapter 7 on Fireproofing.
J. Steven Angell Special thanks also go to Marc Gross from the firm of
Michael L. Baltay Brennan Beer Gorman Architects, Oliver Wilhelm from
Aine M. Brazil Cybul & Cybul Architects, Jonathan Stark from Perkins
Charles J. Carter Eastman Architects, Ken Hiller from Bovis, Inc., Allan
Thomas A. Faraone Paull of Tishman Construction Corporation of New York,
Richard A. Henige, Jr. Larry Danza and John Kozzi of John Maltese Iron Works,
Socrates A. Ioannides Inc., who participated in a symposium held in New York on
Stanley D. Lindsey special topics for staggered-truss building structures.
Robert J. McNamara Last but not least, the authors thank Charlie Carter, Steve
Robert W. Pyle Angell, Thomas Faraone, and Robert Pyle of the American
Kurt D. Swensson Institute of Steel Construction Inc., who have coordinated,
scheduled and facilitated the development of this design
Their comments and suggestions have enriched this guide.
design guide. Special thanks go to Robert McNamara from
vi
Table of Contents
Authors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v Chapter 5
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v Seismic Design. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi
5.1 Strength and Ductility Design
Requirements . . . . . . . . . . . . . . . . . . . . . . . . 29
Chapter 1
5.2 New Seismic Design Considerations
Staggered Truss Framing Systems . . . . . . . . . . . . . . . . 1
for Precast Concrete Diaphragms. . . . . . . . . 29
5.3 Ductility of Truss Members . . . . . . . . . . . . . . . 29
1.1 Advantages of Staggered Trusses. . . . . . . . . . . . 1
5.4 Seismic Design of Gusset Plates . . . . . . . . . . . 30
1.2 Material Description. . . . . . . . . . . . . . . . . . . . . . 1
5.5 New Developments in Gusset Plate
1.3 Framing Layout . . . . . . . . . . . . . . . . . . . . . . . . . 2
to HSS Connections . . . . . . . . . . . . . . . . . . . 31
1.4 Responsibilities . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.5 Design Methodology . . . . . . . . . . . . . . . . . . . . . 4
1.6 Design Presentation . . . . . . . . . . . . . . . . . . . . . . 4 Chapter 6
Special Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Chapter 2
6.1 Openings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Diaphragm Action with Hollow Core Slabs. . . . . . . . . 7
6.2 Mechanical Design Considerations . . . . . . . . . 33
6.3 Plank Leveling . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.1 General Information . . . . . . . . . . . . . . . . . . . . . . 7
6.4 Erection Considerations . . . . . . . . . . . . . . . . . . 33
2.2 Distribution of Lateral Forces . . . . . . . . . . . . . . 7
6.5 Coordination of Subcontractors . . . . . . . . . . . . 34
2.3 Transverse Shear in Diaphragm . . . . . . . . . . . . . 9
6.6 Foundation Overturning and Sliding . . . . . . . . 34
2.4 Diaphragm Chords . . . . . . . . . . . . . . . . . . . . . . 10
6.7 Special Conditions of Symmetry . . . . . . . . . . . 35
6.8 Balconies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Chapter 3
6.9 Spandrel Beams . . . . . . . . . . . . . . . . . . . . . . . . 35
Design of Truss Members. . . . . . . . . . . . . . . . . . . . . . . 15
3.1 Hand and Computer Calculations . . . . . . . . . . 15
Chapter 7
3.2 Live Load Reduction . . . . . . . . . . . . . . . . . . . . 15
Fire Protection of Staggered Trusses . . . . . . . . . . . . . 37
3.3 Gravity Loads. . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.4 Lateral Loads . . . . . . . . . . . . . . . . . . . . . . . . . . 17
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.5 Load Coefficients . . . . . . . . . . . . . . . . . . . . . . . 17
3.6 Vertical and Diagonal Members. . . . . . . . . . . . 19
3.7 Truss Chords. . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.8 Computer Modeling . . . . . . . . . . . . . . . . . . . . . 19
3.9 Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
Chapter 4
Connections in Staggered Trusses . . . . . . . . . . . . . . . . 25
4.1 General Information . . . . . . . . . . . . . . . . . . . . . 25
4.2 Connection Between Web Member
and Gusset Plate . . . . . . . . . . . . . . . . . . . . . . 25
4.3 Connection Between Gusset Plate
and Chord . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.4 Design Example . . . . . . . . . . . . . . . . . . . . . . . . 27
4.5 Miscellaneous Considerations . . . . . . . . . . . . . 27
vii
Chapter 1
INTRODUCTION
1.1 Advantages of Staggered Truss Framing Systems rate of one floor every five days. Once two floors are
erected, window installation can start and stay right behind
The staggered-truss framing system, originally developed at
the steel and floor erection. No time is lost in waiting for
MIT in the 1960s, has been used as the major structural sys-
other trades such as bricklayers to start work. Except for
tem for certain buildings for some time. This system is effi-
foundations and grouting, all  wet trades are normally
cient for mid-rise apartments, hotels, motels, dormitories,
eliminated.
hospitals, and other structures for which a low floor-to-floor
Savings also occur at the foundations. The vertical loads
height is desirable. The arrangement of story-high trusses in
concentrated at a few columns normally exceed the uplift
a vertically staggered pattern at alternate column lines can
forces generated by the lateral loads and, as a result, uplift
be used to provide large column-free areas for room layouts
anchors are often not required. The reduced number of
as illustrated in Fig. 1.1. The staggered-truss framing sys-
columns also results in less foundation formwork, less con-
tem is one of the only framing system that can be used to
crete, and reduced construction time. When used, precast
allow column-free areas on the order of 60 ft by 70 ft. Fur-
plank is lighter then cast-in-place concrete, the building is
thermore, this system is normally economical, simple to
lighter, the seismic forces are smaller, and the foundations
fabricate and erect, and as a result, often cheaper than other
are reduced.
framing systems.
The fire resistance of the system is also good for two rea-
One added benefit of the staggered-truss framing system
sons. First, the steel is localized to the trusses, which only
is that it is highly efficient for resistance to the lateral load-
occur at every 58 to 70 ft on a floor, so the fireproofing
ing caused by wind and earthquake. The stiffness of the sys-
operation can be completed efficiently. Furthermore, the
tem provides the desired drift control for wind and
trusses are typically placed within demising walls and it is
earthquake loadings. Moreover, the system can provide a
possible that the necessary fire rating can be achieved
significant amount of energy absorption capacity and duc-
through proper construction of the wall. Also, the elements
tile deformation capability for high-seismic applications.
of the trusses are by design compact sections and thus will
When conditions are proper, it can yield great economy and
require a minimum of spray-on fireproofing thickness.
maximum architectural and planning flexibility.
It also commonly offers the most cost-efficient possibili-
1.2 Material Description
ties, given the project s scheduling considerations. The
staggered-truss framing system is one of the quickest avail- A staggered-truss frame is designed with steel framing
able methods to use during winter construction. Erection members and concrete floors. Most often, the floor system
and enclosure of the buildings are not affected by prolonged is precast concrete hollow-core plank. Other options,
sub-freezing weather. Steel framing, including spandrel including concrete supported on metal deck with steel
beams and precast floors, are projected to be erected at the beams or joists, can be used.
With precast plank floors, economy is achieved by
 stretching the plank to the greatest possible span. 8-in.-
thick plank generally can be used to span up to 30 ft, while
10-in.-thick plank generally can be used to span up to 36 ft.
Specific span capabilities should be verified with the spe-
cific plank manufacturer. Therefore, the spacing of the
trusses has a close relationship to the thickness of plank and
its ability to span. 6-in.-thick precast plank is normally only
used with concrete topping.
Hollow core plank is manufactured by the process of
extrusion or slip forming. In both cases the plank is pre-
stressed and cambered. The number of tendons and their
diameter is selected for strength requirements by the plank
manufacturer s engineer based upon the design instructions
provided by the engineer of record.
The trusses are manufactured from various steels. Early
buildings were designed with chords made of wide-flange
Fig. 1.1 Staggered-truss system-vertical stacking arrangement.
sections and diagonal and vertical members made of chan-
1
Table 1.1 Material Guide
Section ASTM Fy (ksi)
Columns and Truss A992 or
Wide Flange 50
Chords A572
Web Members Hollow Structural A500 grade 46 or 50
(Vertical and Diagonal) Section B or C (rectangular)
Gusset Plates Plates A36 or A572 36 or 50
nels. The channels were placed toe-to-toe, welded with sep- welded studs as illustrated in Fig. 1.2. Alternatively, guying
arator plates to form a tubular shape. Later projects used or braces may also be used for temporary stability during
hollow structural sections (HSS) for vertical and diagonal construction.
members. The precast plank is commonly manufactured with 4,000
Today, the most common trusses are designed with W10 psi concrete. The grout commonly has 1,800 psi compres-
chords and HSS web members (verticals and diagonals) sive strength and normally is a 3:1 mixture of sand and
connected with gusset plates. The chords have a minimum Portland cement. The amount of water used is a function of
width of 6 in., required to ensure adequate plank bearing the method used to place the grout, but will generally result
during construction. The smallest chords are generally in a wet mix so joints can be easily filled. Rarely is grout
W10x33 and the smallest web members are generally strength required in excess of 2,000 psi. The grout material
HSS4�4�ź. The gusset plates are usually �-in. thick. is normally supplied and placed by the precast erector.
The trusses are manufactured with camber to compensate
for dead load. They are transported to the site, stored, and 1.3 Framing Layout
then erected, generally in one piece. Table 1.1 is a material
Fig. 1.3 shows the photo of a 12-story staggered-truss apart-
guide for steel member selection. Other materials, such as
ment building located in the Northeast United States. Its
A913, may be available (see AISC Manual, Part 2).
typical floor plan is shown in Fig. 1.4. This apartment build-
The plank is connected to the chords with weld plates to
ing will be used as an example to explain the design and
ensure temporary stability during erection. Then, shear stud
construction of staggered-truss-framed structures through-
connections are welded to the chords, reinforcing bars are
out this design guide. The floor system of this 12-story proj-
placed in the joints, and grout is placed. When the grout
cures, a permanent connection is achieved through the
Fig. 1.3 Staggered truss apartment building.
Fig. 1.2 Concrete plank floor system.
2
ect utilizes 10-in.-thick precast concrete plank. The stairs Dead Loads
and elevator openings are framed with steel beams. The
10 precast hollow core plank 75 psf
columns are oriented with the strong axis parallel to the
Leveling compound 5
short building direction. There are no interior columns on
Structural steel 5
truss bents; only spandrel columns exist. There are interior
Partitions 12
columns on conventionally framed bents.
Dead Loads 97 psf
Moment frames are used along the long direction of the
building, while staggered trusses and moment frames are
Plate Loads
used in the short direction.
Two different truss types are shown on the plan, namely 10 precast hollow core plank 75 psf
trusses T1 and T2. Fig. 1.5 shows truss T1B and Fig. 1.6 Structural steel 5
shows truss T2C. Truss T1B is Truss Type 1 located on grid Plate Loads 80 psf
line B, and T2C is Truss Type 2 located on grid line C. The
truss layout is always Truss Type 1 next to Type 2 to mini- Live Loads 40 psf
mize the potential for staggered truss layout errors. Each
truss is shown in elevation in order to identify member sizes Wall Loads
and special conditions, such as Vierendeel panels. Any spe-
Brick 40 psf
cial forces or reactions can be shown on the elevations
Studs 3
where they occur. The structural steel fabricator/detailer is
Sheet rock 3
provided with an explicit drawing for piece-mark identifi-
Insulation 2
cation. Camber requirements should also be shown on the
Wall Loads 48 psf
elevations.
Table 1.2 shows the lateral forces calculated for the
The loads listed above are used in the calculations that
building. For this building, which is located in a low-seis-
follow.
mic zone, wind loads on the wide direction are larger than
seismic forces, and seismic forces are larger in the narrow
1.4 Responsibilities
direction. So that no special detailing for seismic forces
would be required, a seismic response modification factor R The responsibilities of the various parties to the contract are
of 3 was used in the seismic force calculations. The distrib- normally as given on the AISC Code of Standard Practice
uted gravity loads of the building are listed below, where for Steel Buildings and Bridges. All special conditions
plate loads are used for camber calculations. should be explicitly shown on the structural drawings.
Fig. 1.4 Typical floor framing plan. Note: * indicates moment connections.
3
Table 1.2 Wind and Seismic Forces
(All Loads are Service Loads)
WIND (ON WIDE DIRECTION) SEISMIC (BOTH DIRECTIONS)
Lateral Story Lateral Load Story
Ś Ś
Ś
Ś
h h
Load Shear Service Shear
Floor V (kips) V (kips) (%) Vj (kips) Vw (kips) (%)
j w
Roof 107 107 9% 83
83 13%
12 105 212 18% 90
173 26%
11 103 315 27% 82
255 39%
10 103 418 36% 78
333 51%
9 103 521 45% 65
398 61%
8 98 619 54% 58
456 70%
7 96 715 62% 52
508
78%
6 93 808 70% 44
552 85%
5 91 899 78% 39
591 91%
4 86 985 86% 29
620 95%
3 84 1069 93% 21
641 98%
2 79 1148 100% 11
652 100%
Ground
1.5 Design Methodology are continuous members that do transmit moment, and
some moment is always transmitted through the connec-
The design of a staggered-truss frame is done in stages.
tions of the web members.
After a general framing layout is completed, gravity, wind,
The typical staggered-truss geometry is that of a  Pratt
and seismic loads are established. Manual calculations and
truss with diagonal members intentionally arranged to be
member sizing normally precede the final computer analy-
in tension when gravity loads are applied. Other geome-
sis and review. For manual calculations, gravity and lateral
tries, however, may be possible.
loads are needed and the member sizes are then obtained
through vertical tabulation.
1.6 Design Presentation
The design methodology presented in this design guide is
intended to save time by solving a typical truss only once The structural drawings normally include floor framing
for gravity loads and lateral loads, then using coefficients to plans, structural sections, and details. Also, structural notes
obtain forces for all other trusses. The method of coeffi- and specifications are part of the contract documents. Floor
cients is suitable for staggered trusses because of the repe- plans include truss and column layout, stairs and elevators,
tition of the truss geometry and because of the  racking or dimensions, beams, girders and columns, floor openings,
shearing behavior of trusses under lateral loads. This is sim- section and detail marks. A column schedule indicates col-
ilar to normalizing the results to the  design truss . umn loads, column sizes, location of column splices, and
Approximate analysis of structures is needed even in sizes of column base plates.
today s high-tech computer world. At least three significant The diaphragm plan and its chord forces and shear con-
reasons are noted for the need for preliminary analysis as nectors with the corresponding forces must be shown. It is
following: also important that the plan clearly indicate what items are
1. It provides the basis for selecting preliminary member the responsibilities of the steel fabricator or the plank man-
sizes, which are needed for final computer input and ufacturer. Coordination between the two contractors is crit-
verification. ical, particularly for such details as weld plate location over
2. It provides a first method for computing different stiffeners, plank camber, plank bearing supports, and clear-
designs and selecting the preferred one. ances for stud welding. Coordination meetings can be par-
3. It provides an independent method for checking the ticularly helpful at the shop drawing phase to properly
reports from a computer output. locate plank embedded items.
Theoretically, staggered-truss frames are treated as struc- In seismic areas, the drawings must also indicate the
turally determinate, pin-jointed frames. As such, it is Building Category, Seismic Zone, Soil Seismic Factor,
assumed that no moment is transmitted between members Importance Factor, required value of R, and Lateral Load
across the joints. However, the chords of staggered trusses Resisting System.
4
Fig. 1.5 Staggered truss type T1B. Note: [ ] indicates number of composite studs (� dia., 6 long, equally spaced).
5
Fig. 1.6 Staggered truss type T2C. Note: [ ] indicates number of composite studs (� dia., 6 long, equally spaced).
6
Chapter 2
DIAPHRAGM ACTION WITH HOLLOW-CORE SLABS
2.1 General Information diaphragm aspect ratio and by detailing it such that it
remains elastic under applied loads. From Smith and Coull
It is advisable to start the hand calculations for a staggered-
(1991), the lateral loads are distributed by the diaphragm to
truss building with the design of the diaphragms. In a stag-
trusses as follows:
gered-truss building, the diaphragms function significantly
different from diaphragms in other buildings because they
Vi = Vs + VTORS (2-1)
receive the lateral loads from the staggered trusses and
transmit them from truss to truss. The design issues in a
where
hollow-core diaphragm are stiffness, strength, and ductility,
as well as the design of the connections required to unload
Vi = truss shear due to lateral loads
the lateral forces from the diaphragm to the lateral-resisting
Vs = the translation component of shear
elements. The PCI Manual for the Design of Hollow Core
= Vw � GAi / ŁGAi (2-2)
Slabs (PCI, 1998) provides basic design criteria for plank
VTORS = the torsion component of shear
floors and diaphragms.
= Vw � eŻ#xiGAi / GJ (2-3)
Some elements of the diaphragm design may be dele-
where
gated to the hollow core slab supplier. However, only the
engineer of record is in the position to know all the param-
GAi = Shear rigidity of truss
eters involved in generating the lateral loads. If any design
ŁGAi = Building translation shear rigidity
responsibility is delegated to the plank supplier, the location
GJ = Building torsion shear rigidity
and magnitude of the lateral loads applied to the diaphragm
e = Load eccentricity
and the location and magnitude of forces to be transmitted
xi = Truss coordinate (referenced to the
Ż#
to lateral-resisting elements must be specified.
center of rigidity (CR))
An additional consideration in detailing diaphragms is
Vw = Story shear due to lateral loads
the need for structural integrity. ACI 318 Section 16.5 pro-
(see Table 1.2)
vides the minimum requirements to satisfy structural
integrity. The fundamental requirement is to provide a com-
Smith and Coull (1991) provide expressions for story
plete load path from any point in a structure to the founda-
shear deformations for a single brace as (Fig. 2.1):
tion. In staggered-truss buildings all the lateral loads are
transferred from truss to truss at each floor. The integrity of
�#ś#
V d3 L
(2-4)
"= +
each floor diaphragm is therefore significant in the lateral
ś#
EAg ź#
L2 Ad #
�#
load resistance of the staggered-truss building.
2.2 Distribution of Lateral Forces
The distribution of lateral forces to the trusses is a struc-
turally indeterminate problem, which means that deforma-
tion compatibility must be considered. Concrete
diaphragms are generally considered to be rigid. Analysis
of flexible diaphragms is more complex than that of rigid
diaphragms. However, for most common buildings subject
to wind forces and low-seismic risk areas, the assumption
of rigid diaphragms is reasonable. If flexible diaphragms
are to be analyzed, the use of computer programs with
plate-element options is recommended.
For the example shown in this design guide, a rigid
diaphragm is assumed for the purpose of hand calculations
and for simplicity. This assumption remains acceptable as
long as the diaphragm lateral deformations are appropri-
ately limited. One way to ensure this is to limit the
Fig. 2.1 Story shear deformation for single brace.
7
The hand calculations are started by finding the center of
where
rigidity, which is defined as the point in the diaphragm
about which the diaphragm rotates when subject to lateral
V = shear force applied to the brace
loads. The formula for finding the center of rigidity is
E = modulus of elasticity
(Smith and Coull, 1991; Taranath, 1997):
d = length of the diagonal
L = length between vertical members
(2-6)
x = Łxi GAi / ŁGAi
Ad = sectional area of the diagonal
Ag = sectional area of the upper girder
For staggered-truss buildings, the center of rigidity is cal-
culated separately at even floors and odd floors. Assuming
The shear rigidity GA is then computed as:
that the trusses of the staggered-truss building shown in
Figs. 1.5 and 1.6 have approximately equal shear rigidity,
Vh E h
GA = = (2-5)
GAi, per truss, the center of rigidity of each floor is calcu-
"
d3 /(L2 Ad ) + L / Ag
lated as follows (see Fig. 2.2):
Even Floors
where h is the story height. The overall truss shear rigidity
Truss xi (ft)
is the sum of the shear rigidities of all the brace panels in
T1B 36
that truss. The reader may use similar expressions to deter-
T1D 108
mine approximate values for GA in buildings where varia-
T1F 192
tions in stiffness occur.
Łxi = 336 xe = 336/3 = 112'
(a) Even Floor
(b) Odd Floor
Fig. 2.2 Center of rigidity for lateral loads.
8
where xe is the center of rigidity for even floors. 2.3 Transverse Shear in Diaphragm
Odd Floors
Planks are supported on trusses with longitudinal joints
Truss xi (ft)
perpendicular to the direction of the applied lateral load. To
T2C 72
satisfy structural integrity, the diaphragm acts as a deep
T2E 156
beam or a tied arch. Tension and compression chords create
T2G 228
the flanges, and boundary elements are placed around the
Łxi = 456 xo = 456/3 = 152'
openings. The trusses above are considered to act as  drag
struts , engaging the entire length of the diaphragm for
where xo is the center of rigidity for odd floors. The load
transferring shear to the adjacent trusses below (Fig. 2.3).
eccentricity is calculated as the distance between the center
Truss shear forces calculated in Table 2.3 are used to find
of rigidity and the location of the applied load.
the shear and moment diagrams along the diaphragm of the
bottom floor as shown in Fig. 2.4. Two torsion cases (+5%
ee = (264/2) - 112 = 20' even floors
and -5% additional eccentricities) are considered. The
eo = (264/2) - 152 = -20' odd floors
required shear strength of the diaphragm is calculated as
follows:
Adding 5% eccentricity for accidental torsion, the final
Vu = 1.7 � Ćh � V � 0.75
load eccentricity is calculated as follows:
= 1.7 �1.0 � 335 � 0.75
ee = 20 ą (5% � 264)
= 427k
= 33.2; 6.8 ft
eo = -20 ą (5% � 264)
where Ćh is the story shear adjustment coefficient (see Table
= -33.2; -6.8 ft
1.2 and Section 3.5 of this design guide), 0.75 is applied for
wind or seismic loads, and V = 335 k is the maximum shear
From this it is clear that for this example even and odd
force in the diaphragm as indicated in Fig. 2.4. The pro-
floors are oppositely symmetrical. The base torsion is cal-
vided design shear strength is calculated per ACI 318 Sec-
culated as the base shear times the eccentricity:
tion 11.3.
T = 1,148 � 33.2 = 38,114 ft-k
T = 1,148 � 6.8 = 7,807 ft-k
ĆVn = Ć Vc + Vs
( )
where the base shear of 1,148 k is from Table 1.2. The
ĆVc = Ć � 2 fc2 bd
above torsions have plus and minus signs. Again assuming
= 0.85 � 2 � 4000 � 6 � 0.8 �64 �12
()
that all trusses have the same shear rigidity GAi at each
= 396k
floor, the base translation shear component is the same for
all trusses:
where an effective thickness of 6 in. is used for the 10-in.-
Vs = 1,148/3 = 383 k
thick hollow core planks, and the effective depth of the
beam is assumed to be 80% of the total depth.
Next, the torsional rigidity GJ is calculated as shown in
Tables 2.1 and 2.2 for even floors and odd floors. The tor-
ĆVs = ĆAVF fy �
sional shear component varies and is added or subtracted to
the translational shear component. The results are summa-
where AVF is the shear friction reinforcement and � = 1.4 is
rized in Table 2.3, which is obtained by using Equations 2-1,
the coefficient of friction. Assuming one #4 steel bar is used
2-2, and 2-3. The second-to-last column in Table 2.3 shows
along each joint between any two planks,
the design forces governing the truss design. Note that the
design shear for the trusses is based on +5% or -5% eccen-
No. of planks = 64'/8' = 8 planks
tricity, where * indicates the eccentricity case that governs.
No. of joints = 8 - 1 = 7 joints
Table 2.3 also shows that the design base shear for trusses
AVF = 0.2 � 7 = 1.4 in2
T1B and T2G is 335 k, for trusses T1D and T2E is 380 k,
ĆVs = 0.85 � 1.4 � 60 � 1.4 = 100 k
and for trusses T1F and T2C is 634 k. We can now proceed
ĆVn = 396 + 100 = 496 k > 427 k
with the truss design for lateral loads, but we will first con-
(O.K.)
tinue to analyze and design the diaphragm.
9
Table 2.2 Torsional Rigidity, Odd Floors
Table 2.1 Torsional Rigidity, Even Floors
Truss
xi2
xi
Truss xi
xi2
T2C -80 6,400
T1B -76 5,776
T2E 4 16
T1D -4 16
T2G 76 5,776
T1F 80 6,400
Ł=12,192 ft2
Ł=12,192 ft2
Table 2.3 Shear Force in Each Truss due to Lateral Loads (Bottom Floor)
Design
V T=38,114 (ft-k) T=7,807 (ft-k) Shear
xi
s
Ś
Ś
V (kips) ecc
Truss VTORS Vi VTORS Vi Vi
T1B -76 383 -238 145 -48 335* 335 1.00
T1D -4 383 -13 370 -3 380* 380 1.13
T1F 80 383 251 634* 51 434 634 1.89
T2C -80 383 251 634* 51 434 634 1.89
T2E 4 383 -13 370 -3 380* 380 1.13
T2G 76 383 -238 145 -48 335* 335 1.00
2.4 Diaphragm Chords where constant 0.75 is applied for wind or seismic loads.
The calculated shear flows, fH, are shown in Fig. 2.4(a). For
The perimeter steel beams are used as diaphragm chords.
-5% additional eccentricity, similar calculations are con-
The chord forces are calculated approximately as follows:
ducted and the results are shown in Fig. 2.4(b). The shear
flows of the two cases are combined in Fig. 2.4(c), where a
H = M/D (2-7)
where
H = chord tension or compression force
M= moment applied to the diaphragm
D = depth of the diaphragm
The plank to spandrel beam connection must be adequate
to transfer this force from the location of zero moment to
the location of maximum moment. Thus observing the
moment diagrams in Fig. 2.4, the following chord forces
and shear flows needed for the plank-to-spandrel connec-
tion design are calculated:
With +5% additional eccentricity:
H = 5,223 / 64 � 0.75 = 61 k
fH = 61 / 72 = 0.85 k/ft
H = 5,223 / 64 � 0.75 = 61 k
fH = 61 /17 = 3.59 k/ft
H = 12,024 / 64 � 0.75 = 141 k
fH = 141 / 103 = 1.37 k/ft
Fig. 2.3 Diaphragm acting as a deep beam.
fH = 141 / 72 = 1.96 k/ft
10
value with * indicates the larger shear flow that governs. The connections of the beams to the columns must develop
These shear forces and shear flows due to service loads on these forces (H). The plank connections to the spandrel
the bottom floor are then multiplied by the height adjust- beams must be adequate to transfer the shear flow, fH. The
ment factors for story shear to obtain the final design of the plank connections to the spandrel are usually made by shear
diaphragms up to the height of the building as shown in the plates embedded in the plank and welded to the beams (Fig.
table in Fig. 2.5. The table is drawn on the structural draw- 1.2 and Fig. 2.6). Where required, the strength of plank
ings and is included as part of the construction contract doc- embedded connections is proven by tests, usually available
uments. Forces given on structural drawings are generally from the plank manufacturers. All forces must be shown on
computed from service loads. In case factored forces are to the design drawings. The final design of the diaphragm is
be given on structural drawings, they must be clearly spec- shown in Fig. 2.5.
ified.
The perimeter steel beams must be designed to support
the gravity loads in addition to the chord axial forces, H.
11
Fig. 2.4 Diaphragm shear force, moment, and shear flow (2nd floor).
12
Fig. 2.5 Diaphragm design.
Fig. 2.6 Detail for load transfer from diaphragm to spandrel beams.
13
14
Chapter 3
DESIGN OF TRUSS MEMBERS
3.1 Hand and Computer Calculations These tributary areas can also be verified from the mem-
ber loads as follows. Thus, considering the entire truss T1B,
The structural design of truss members normally begins
the tributary area is:
with hand calculations, which are considered to be approx-
imate and prerequisite to more detailed computer calcula-
TA = 64 � 36 � 2 = 4,608 ft2
tions. Computer analyses can be either two or three
dimensional using stiffness matrix methods with or without
The total dead load supported by the truss is:
member sizing. Some programs assume a rigid diaphragm
and the lateral loads are distributed based on the relative
WDL = 4,608 � 97 psf = 446.7 k
stiffness of the trusses. In other programs, the stiffness of
the diaphragm can be modeled with plate elements.
For member d1:
For truss design, hand and computer calculations have
both advantages and disadvantages. For symmetrical build-
Axial force T = 380 k � 97/(97 + 40)
ings, 2-D analysis and design is sufficient and adequate. For
non-symmetrical structures, 3-D analyses in combination = 269 k (see Fig. 3.3)
with 2-D reviews are preferred. The major advantage of a
Vertical component of T = 269/ 2
2-D analysis and design is saving in time. It is fast to model
= 190 k
and to evaluate the design results.
TA = 190 / 446.7 � 4,608 = 1,960 ft2
Hand calculations typically ignore secondary effects
such as moment transmission through joints, which may
This tributary area is the same as the one calculated pre-
appear to produce unconservative results. However, it is
viously. Similar calculations yield the tributary areas for
worthwhile to remember that some ductile but self-limiting
members d2 and d3.
deformations are allowed and should be accepted.
3.3 Gravity Loads
3.2 Live Load Reduction
Fig. 3.1 shows a one-story truss with applied gravity loads.
Most building codes relate the live load reduction to the
The members are assumed to intersect at one point. The ver-
tributary area each member supports. For staggered trusses
tical and diagonal members are assumed to be hinged at
this requirement creates a certain difficulty since the tribu-
each end. The top and bottom chords are continuous beams
tary areas supported by its vertical and diagonal members
and only hinged at the ends connected to the columns.
vary. Some engineers consider the entire truss to be a single
Because a diagonal member is not allowed to be placed in
member and thus use the same maximum live load reduc-
the Vierendeel panel where a corridor is located, the chords
tion allowed by code for all the truss members. Others cal-
cannot be modeled as axial-force members. Otherwise, the
culate the live load reduction on the basis of the equivalent
truss would be unstable. For hand calculation purposes, it
tributary area each member of the truss supports. Clearly,
member d1 in Fig. 1.5, which carries a heavy load, supports
an equivalent tributary area larger than that of member d3,
which carries a light load. Thus, assuming that web mem-
bers support equivalent floor areas, the following tributary
area calculations apply:
d1: TA = (7/2 + 9.5 � 2 + 9.5/2) 36 � 2
= 1,960 ft2
d2: TA = (7/2 + 9.5 + 9.5/2) 36 � 2
= 1,278 ft2
d3: TA = (7/2 + 9.5/2) 36 � 2
= 594 ft2
Fig. 3.1 Analysis of truss T1B gravity loads.
15
a. The method of joints.
is customary to convert the uniform loads to concentrated
b. The method of sections.
loads applied at each joint. It will be shown later that shear
forces in the chords have to be included in the hand calcu-
The reader is referred to Hibbeler (1998) or Hsieh (1998)
lations when lateral loads are applied. The chords are sub-
or any other statics textbook for in-depth discussion of each
ject to bending and shear, but the vertical and diagonal
method. Each method can resolve the truss quickly and pro-
members are not because they are two-force members.
vide the correct solution. Fig. 3.2 shows the truss solution
The truss model shown is  statically indeterminate . The
using the method of joints. It is best to progress the solution
truss can certainly be analyzed using a computer. However,
in the following joint order: L1, U1, L2, U2, etc. The fol-
reasonably accurate results can also be obtained through
lowing calculations are made for typical truss T1B subject
hand calculations. For gravity loads, the shear force in the
to full service gravity loads:
top or bottom chord in the Vierendeel panel vanishes
because of symmetry. The shear forces in the chords of
w = (97 psf + 40 psf) � 36' = 4.93 k/ft
other panels are very small and can be neglected. Based on
P1 = 4.93 � 9.5 / 2 = 23.41 k
this assumption, the truss becomes statically determinate
P2 = 4.93 � 9.5' = 46.83 k
and the member forces can be calculated directly by hand
P3 = 4.93 � (9.5 + 7)/2 = 40.67 k
calculations from statics. The best way to start the calcula-
tions is by finding the reactions at the supports. After the
The above concentrated loads are applied at the top and
reactions are determined, there are two different options for
bottom joints as shown in Fig. 3.1. The reactions at sup-
the further procedure.
ports are:
GRAVITY LOADS (KIPS)
LATERAL LOADS (KIPS)
Fig. 3.2 Truss solution method of joints.
16
R = (23.41 + 46.83 � 2 + 40.67) � 2 are all zero and thus the chord shear forces are also zero in
= 315.48 k these panels. Now we can proceed to find all the member
forces using the method of joints in the following order: U4,
The calculations then proceed for each joint as shown in L4, U3, L3, etc. The calculations are shown in Fig. 3.2.
Fig. 3.2. Here shear forces in the chord members are The above assumptions of zero moments in the chord mem-
excluded from the calculations because they are assumed bers are justified by comparing the results with those from
zero. The result of all the member forces of the typical truss the computer analysis. Fig. 3.4 shows the truss solution of
due to service gravity loads is summarized in Fig. 3.3. the bottom floor due to service lateral loads. Note that
while diagonals d1 and d2 have the same member force, the
3.4 Lateral Loads member force in diagonal d3 is larger because of the shear
force in that panel.
The allocation of lateral loads to each individual truss is
To verify these hand calculation results, the computer
done by the diaphragm based on the truss relative stiffness
analysis results due to gravity and lateral loads are included
and its location on the plan. Once the member forces due
in Fig. 3.5 and Fig. 3.6, respectively. The results are very
to lateral loads are calculated, they are combined with the
close to those from hand calculations.
gravity loads to obtain the design-loading envelope. The
member sizes are then selected to ensure adequate strength.
3.5 Load Coefficients
Fig. 3.4 shows the member forces due to design shear of
335 kips, which was computed in Table 2.3 for truss T1B of Once the member forces have been calculated for a typical
the bottom floor. Because the truss is anti-symmetrical truss, the design forces are computed for other trusses using
about its centerline for this load case, the horizontal reac- load coefficients. Load factors are then applied per LRFD
tion H at each support is 167.5 kips. Alternatively, the floor requirements.
diaphragm may distribute the horizontal shear force uni-
formly along the length of the top and bottom chords of the
truss, reducing the axial forces in these chords. The vertical
reaction at each support is:
R = (167.5 � 2 � 9.5) / 64.125 = 49.63 k
The moment and the axial force at midspan of each chord
in the Vierendeel panel are both zero because of geometri-
cal anti-symmetry. Considering half of the truss as a free
body and assuming the same shear force in the top and bot-
tom chords of the Vierendeel panel, the shear force can be
calculated as:
Fig. 3.3 Member forces of truss T1B due to gravity loads (kips).
V = 1 / 2 � (167.5 � 9.5) / 32.06
= 24.82 k
The chord end moment at joint U4 is equal to the shear
times half the panel length:
M = 24.82 � 7 / 2 = 86.87 ft-k
This end moment is also applied to the chord adjacent to
the Vierendeel panel. Assume the moment at the other end
of this chord is zero, the shear force in the member can then
be calculated as:
Fig. 3.4 Member forces of truss T1B (bottom floor) due to lateral loads (kips).
Notes: 1. Chord axial forces shown are actually in the concrete floor
V = (86.87 + 0) / 9.5 = 9.14 k
diaphragm.
2. Lateral forces are conservatively applied as concentrated loads at
This shear force is indicated in Fig. 3.4. It can further be
each end. Optionally loads may also be applied as distributed
assumed that the chord moments in the remaining panels
forces along the chord length.
17
Load coefficients are calculated as follows: ity dead and live loads that are used in the truss member
force calculations. The value of ĆL varies with load com-
Di = DT �ĆW �ĆL (3-1) bination cases. Load coefficient Ćecc is calculated in Table
Li = LT �ĆW �ĆL (3-2) 2.3, which is used to adjust wind and seismic forces for dif-
Wi = WT �Ćecc �Ćh (3-3) ferent design shear forces in different staggered trusses.
Ei = ET �Ćecc � Ćh (3-4) Load coefficient Ćh is computed in Table 1.2 that adjusts
story shears at different stories.
Subscript i indicates the member being designed and Showing below is an example of load coefficient calcu-
subscript T indicates the corresponding member of the orig- lations:
inally calculated typical truss, i.e., truss T1B. D, L, W, E are
the dead, live, wind, and earthquake forces, and the load DL = 97 psf, LL = 40 psf, and RLL
coefficients are defined as follows: = 20 psf (see Section 1.3)
Ćw = 1.0 for typical truss T1B
Ćw = Width or tributary area adjustment coefficient = (36 + 12) / 2 � (1 / 36)
ĆL = Load adjustment coefficient for load factor com- = 0.67 for truss T1D (see Fig. 1.4)
binations ĆL for load combination of 1.2DL + 1.6RLL
Ćecc = Truss eccentricity coefficient = (1.2DL + 1.6RLL)/(full service gravity loads)
Ćh = Story shear adjustment coefficient = (1.2 � 97 + 1.6 � 20) / (97 + 40)
= 1.083
The first two of the above coefficients are applied to Ćecc = 1.0 for typical truss T1B
gravity loads, and the later two to lateral loads. Load coef- = 380 / 335 = 1.13 for T1D
ficient Ćw is applied to a truss whose bay length is different = 634 / 335 = 1.89 for T1F (see Table 2.4)
from that of the typical truss. Load coefficient ĆL is the Ćh = (see Table 1.2 for Ćh value of each story)
ratio of a factored load combination to the full service grav-
Fig. 3.6 Computer analysis results of truss T1B
Fig. 3.5 Computer analysis results of truss T1B due to gravity loads.
of bottom floor due to lateral loads.
18
3.6 Vertical and Diagonal Members b. Wind
The detailed calculations for the design of diagonal member The maximum wind moment in the chords occurs in the
d1 in truss T1F of each floor using load coefficients are Vierendeel panel.
shown in Table 3.1, where load coefficients ĆL1, ĆL2, and
ĆL3 are applied to different load combinations. Truss T1F M = 86.87 ft-k (from Section 3.4 for typical truss
rather than typical truss T1B is intentionally selected as an T1B)
example here for explanation of how the load coefficients Ćecc � M = 1.89 � 86.87 = 164 ft-k
are applied. Five load combinations as specified in ASCE Mu = 164 � 1.3 = 213 ft-k
7 are considered in this table. A 50% live load reduction is
used in the design of the diagonal members. Numbers in The axial force applied to the chord due to the wind load
boldface in the table indicate the load case that governs. can be neglected as will be explained in Section 3.8. The
The governing tensile axial forces of the diagonal members above moment is also applied to the adjacent span, which
range from 412 k to 523 k for different floors. HSS 10�6� has a span length of 9.5 ft same as the span length used for
� is selected per AISC requirements for all the diagonal the gravity load moment calculation. The member forces of
members. the chords on the second story due to gravity and wind
loads are then combined as follows:
3.7 Truss Chords
Pu = 484 k
The designer must investigate carefully all load cases so as
Mu = 41 + 213 = 254 ft-k
to determine which load case governs. For this design
example for truss chords, it is found that the load combina-
It is observed that while wind loads vary with building
tion of 1.2D + 1.6W + 0.5L governs. The steel design must
heights, gravity loads do not. Thus, Table 3.2 is created and
comply with AISC Equation H1-1a.
the chord moments are calculated using coefficient Ćh of
each story as shown. The designed wide-flange sections per
Pu /(Ć Pn) + (8 / 9) � [Mux / (ĆbMnx)] [ 1.0 ]
AISC Equation H1-1a are also shown in the table. To facil-
where
itate the design calculations, the axial force and bending
Ć = 0.90 Tension
moment strengths of possible W10 members are calculated
= 0.85 Compression
first and listed in Table 3.3.
Ćb = 0.90 Bending
3.8 Computer Modeling
Calculations for gravity and wind loads are made sepa-
rately and then combined. When designing staggered truss buildings using computer
models (stiffness matrix solutions), the results vary with the
a. Gravity assumptions made regarding the degree of composite action
between the trusses and the concrete floor. The design
It is assumed that the chords are loaded with a uniformly
results are particularly sensitive to modeling because a bare
distributed load. Using a 50% live load reduction, the fol-
truss is more flexible than a truss modeled with a concrete
lowing are calculated for the chords of truss T1F on the sec-
floor. Upon grouting, the truss chords become composite
ond story:
with the concrete floor and thus the floor shares with the
truss chords in load bearing. Yet, a concrete floor, particu-
Ćw = 1.0 for truss T1F
larly a concrete plank floor, may not effectively transmit
M = 4.93 � 9.52 / 10
tensile stresses. Also, there is limited information on plank
= 44 ft-k (member end moments at joints)
and steel composite behavior. In addition, lateral loads are
P = 525 k (from Fig. 3.3)
assumed to be distributed to the trusses by the concrete
floor diaphragm and the participation of the truss chords in
Mu = ĆL M
distributing these forces may be difficult to quantify.
= [(1.2 � 97 + 0.5 � 20) / (97 + 40)] � 44
A reasonable approach to this problem is the assumption
= 41 ft-k
that the diaphragm is present when solving for lateral loads,
Pu = ĆL P
but is ignored when solving for gravity loads. This requires
= (1.2 � 97 + 0.5 � 20) / (97 + 40) � 525
working with two computer models one for gravity loads
= 484 k
19
Table 3.1 Design of Diagonal Member d1 of Truss T1F
DIAGONAL MEMBER d1, TRUSS T1F
WIND, kips SEISMIC, kips LOAD COMBINATIONS, kips
Floor Member Sizes
e
Roof 9% 12 13% 10 377 412 366 361 HSS 10�6�1/2
12 18 24 26 20 377 412 382 370 HSS 10�6�1/2
11 27 36 39 29 377 412 397 380 HSS 10�6�1/2
10 36 48 51 38 377 412 413 389 HSS 10�6�1/2
9 45 60 61 46 377 412 428 397
HSS 10�6�1/2
8 54 72 70 53 377 412 444 404
HSS 10�6�1/2
7 62 82 78 59 377 412 458 410
HSS 10�6�1/2
6 70 93 85 64 377 412 471 415 HSS 10�6�1/2
5 78 103 91 69 377 412 485 419 HSS 10�6�1/2
4 86 114 95 72 377 412 499 422 HSS 10�6�1/2
3 93 123 98 74 377 412 511 425
HSS 10�6�1/2
Rev.
1
2 100% 133 100 75 377 412 352 426
HSS 10�6�1/2
12/1/02
Ground 133 75 377 412 523 426
HSS 10�6�1/2
a
F in d1 of Typical Truss b c
70.2 39.9 380
T1B
Revs.
x 20)
Rev. 5/1/03 - corrected parenthesis
3
5/1/03 x 20)
12/1/02 - deleted stray text
20
 Table 3.2 Design of Staggered Truss Chords
TRUSS T1F
AISC
Ś
Floor Śh M M P Section
u,w u u
Eq. H1-1a
Roof 9% 19 484 W10�54
60
12 18 38 484 W10�54 1.0
79
11 27 58 484 W10�60
96
10 36 77 484 W10�60 1.0
118
9 45 96 484 W10�68
137
8 54 115 484 W10�68 0.99
151
7 62 132 484 W10�77
173
6 70 149 484 W10�77
190
5 78 166 484 W10�77 1.0
207
4 86 183 484 W10�88
224
3 93 198 484 W10�88
239
2 100 213 484 W10�88 0.97
254
Ground
M = M + M
u u,G u,W
M = Gravity load moment = 41 ft-k every story
u,G
M = Wind load moment.
u,W
Table 3.3 Section Strengths for Chord Design, F = 50 ksi
y
Section ĆcP (k) ĆbM (ft-k)
Ć
Ć
n nx
W10�112 1400 551
W10�100 1250 488
W10�88 1100 424
W10�77 961 366
W10�68 850 320
W10�60 748 280
W10�54 672 250
W10�49 612 226
W10�45 565 206
W10�39 489 176
W10�33 413 146
and the other for lateral loads, and then the results are com- from two floors. For a truss with hangers or posts (Truss
bined using load factors per code requirements. In combin- type T2), the first diagonal-to-column connection will carry
ing the results, it is assumed that any axial load actions from the accumulated load from three floors.
lateral loads are carried only in the concrete floor, but out- A difficulty exists in evaluating the actions imposed on
of-diaphragm-plane shear and moment actions from lateral the columns by the truss flexibility. The column design is
loads are resisted by the steel chords. best done using the shear and moment applied to the
columns obtained from construction loads (plate loads) on
3.9 Columns a bare truss. Column forces due to superimposed dead and
live loads and lateral loads are computed from a composite
The floor loads are delivered to the columns through the
truss.
truss-to-column connections. For trusses, the first diagonal
Since columns support large tributary areas, the maxi-
is responsible for carrying most of this load into the con-
mum live load reduction is permitted. For the purpose of
nection. Thus, for a typical one-story truss, the first diago-
this example, 50% reduction is assumed. The load combi-
nal-to-column connection will carry the accumulated load
21
nation of gravity loads, either 1.4D or 1.2D + 1.6L, governs Example:
the column designs. The following shows the design of col-
Only the dead loads of planks and structural steel are used
umn 1F. Refer to Fig. 1.4 for the column location and Sec-
to calculate column moments. Superimposed dead and live
tion 1.3 for the dead and live loads used in the calculations.
loads are applied after the erected planks act integrally with
the steel trusses. Additional column moments due to super-
Column Axial Force
imposed dead and live loads can be neglected because the
Tributary Area = 72 / 2 � 64 / 2 = 1,152 truss deformation caused by the superimposed loads is very
DL1 (plate loads only) = 80 psf � 1,152 = 92.2 k small as a result of the composite action of the truss and the
DL2 (all dead loads except exterior walls) planks. However, these superimposed loads will increase
= 97 psf � 1,152 = 111.7 k the column axial force.
RLL = 20 psf � 1,152 = 23 k
DL2 + RLL = 111.7 + 23 = 134.7 k "TS = � in. (assumed truss midspan deflection due to
weights of planks and structural steel)
Two Floors: L = 64'
c = 9'
DL1 = 92.2 � 2 = 184 k For the top and bottom chords of W10�54:
DL2 + RLL = 134.7 � 2 = 269 k
Exterior wall: 48 psf � 36' � 9' = 16 k per story "t = Ł PiLi / (EAi)
= [(9.5 � 12) / (29,000 � 15.8)]
Column Bending � (268.6 + 443.6 + 525 + 525/2)
� [80/(97 + 40)]
The truss axial deformation and downward deflection due
= 0.218 in.
to gravity loads force the column-to-truss joints to translate
and rotate. It is assumed that the truss moment of inertia is
"b = [(9.5 � 12) / (29,000 � 15.8)]
much larger than the columns. The assumed deformed
� (0 + 268.6 + 443.6 + 525/2)
shape of the columns due to joint rotation is shown in Fig.
� [80/(97 + 40)]
3.7(b). The member end moment caused by a unit rotation
= 0.142 in.
is calculated as 3EI/ c for this deformed shape. The mem-
ber end moment caused by a unit translation is 6EI/ c2 as
The chord axial forces used in the above calculations are
indicated in Fig. 3.7(a). The moment of the column due to
from Fig. 3.3. Try W12�65 for the column section.
gravity load is thus calculated as follows:
M = [6 � 29000 � 174/( 9 � 12)]
MCOL = MTRANS + MROT
� [(0.218 + 0.142)/(9 � 12)
MTRANS = 6EI ("t + "b) / c2
- 0.75/(64 � 12)]
MROT = -3EI� / c
= 661 in-k
= 55 ft-k
where
where moment of inertia Iy (rather than Ix)is used because
the columns bend about the weak axis. The column
� = 2"TS / L
moment calculated above is for the top story. For other sto-
6EI "t + "b "TS
ries, the moments can be calculated similarly and the results
( - )
4" MCOL =
are shown in Table 3.4. It is noted that axial deformations
c c L
Dt and Db are less in the bottom stories because of bigger
where
chord member sections. However, the column moments are
larger in the bottom stories because the column moment of
"t = Top chord axial deformation
inertias, Iy, are bigger in the bottom stories. These column
= Ł PiLi / EAi
moments are then combined with the axial forces using load
"b = Bottom chord axial deformation
factors for different load combinations. The results of the
= Ł PiLi / (EAi)
load combinations and the column sections selected based
"TS = Truss midspan deflection
on the most severe load case are shown in Table 3.4. Col-
L = Truss span
umn axial loads due to lateral loads, in this example, are
c = Column length
small and therefore, left out.
22
Fig. 3.7 Column deformed shaped due to chord displacement.
23
Table 3.4 Design of Column 1F
COLUMN 1F
Axial Force Moment Load Combinations
Floor Total 1.4 D 1.2 D + 1.6 L
Exterior
Section
Wall
DL1 DL2 + RLL DL2 + RLL Exterior M
u u u u
DL1 P M P
DL1 (4)
(3)
(1) (2) (5) Wall (6) (8) (9) (10) (11)
(7)
Roof 184 269 16 184 269 16 55 258 77 360 66 W12�65
12 16 184 269 32 258 380 W12�65
11 184 269 16 368 538 48 65 515 91 740 78 W12�87
10 16 368 538 64 515 759 W12�87
9 184 269 16 552 807 80 77 773 108 1120 92 W12�120
8 16 552 807 96 773 1139 W12�120
7 184 269 16 736 1076 112 82 1030 115 1499 98 W12�152
6 16 736 1076 128 1030 1518 W12�152
5 184 269 16 920 1345 144 97 1288 136 1879 116 W12�190
4 16 920 1345 160 1288 1898 W12�190
3 184 269 16 1104 1614 176 105 1546 147 2258 126 W12�230
2 16 1104 1614 192 1546 2278 W12�230
Ground 184 k 269 k 16 k 1288 k 1883 k 208 k 55 ft-k
Note: 1. DL1 includes the weights of precast planks and structural steel only.
2. DL2 includes all the dead loads except the weight of exterior walls.
3. All the columns bend about the weak axis (see Fig. 1.4).
4. The moments shown in the table are caused by the weights of precast plank and structural steel only.
5. Column (8) = 1.4 � Column (4); Column (9) = 1.4 � Column (7).
6. Column (10) = (1.2 � 97 + 1.6 � 20) / (97 + 20) � Column (5) + 1.2 � Column (6); Column (11) = 1.2 � Column (7).
24
Chapter 4
CONNECTIONS IN STAGGERED TRUSSES
4.1 General Information 2. Shear Strength of the HSS at Welds
The typical connection of web members to truss chords
ĆRn = ĆVn = Ć (0.6 Fy)(4Lwt)
consists of welded gusset plates. Since the truss is shop fab-
Ć = 0.9
ricated and transported in one piece, all connections are
shop welded (see Fig. 4.1). Only truss-to-column connec-
3. Strength of the Weld Connecting the Gusset Plate to
tions are bolted in the field (Fig. 4.2) when the truss is
the HSS
erected.
The HSS web member connection to the gusset plate is
ĆRn = ĆFwAw
often made by cutting a slot in the middle of the HSS sec-
Ć = 0.75
tion. The design methodology that follows is based upon
the recommendations listed in the AISC Hollow Structural
4. Shear Strength of the Gusset Plate
Sections Connections Manual (AISC, 1997). Shown in Fig.
4.1 is a typical slotted HSS to gusset plate connection. Seis-
ĆRn = ĆVn = Ć (0.6Fy1)(2Lwt1)
mic behavior and design of gusset plates was studied by
Ć = 0.9
Astaneh-Asl (1998), and will be discussed in Chapter 5.
The notations used in the above four limit state strength
4.2 Connection Between Web Member and Gusset Plate
expressions are as follows:
First, consider an HSS web member in tension. The design
strength of the connection between the HSS and the gusset Fu = specified minimum tensile strength of the
plate is the smallest value among the following four limit HSS, ksi
state considerations. Fy = specified minimum yield stress of the HSS, ksi
Ae = effective net area of the HSS, in2
1. Shear Lag Fracture Strength in the HSS = UAn
An = Ag - 2 t t1
ĆRn = ĆFuAe Ag = gross area of the HSS, in2
Ć = 0.75
Fig. 4.2 Truss to column connection.
Fig. 4.1 Slotted HSS and gusset plate connection.
25
The provided compression strength is calculated based
U = 1 - ( x / Lw) d" 0.9
Ż#
on simple column buckling procedures. The procedure
B2 + 2BH
x = assumes that both ends of the gusset plate are fixed and can
x = for rectangular HSS
Ż#
4(B + H)
sway laterally (See Fig. 4.3 and 4.4).
Lw = length of the weld to HSS, in
ĆcPn = ĆcAgFcr
e" 1.0 H (As a rule of thumb, Lw should not be
less than the HSS depth)
where
4Lw = total weld length
B = width of the HSS section, in
Ćc = 0.85
H = depth of HSS section, in
Pn = nominal compressive strength, kips
t = HSS wall thickness, in
Ag = gross area of gusset plate, in2. Whitmore s 30-
t1 = gusset plate thickness, in
degree effective width area (Whitmore, 1952;
Fw = nominal weld strength, ksi
Astaneh-Asl, Goel, and Hanson, 1981) should
= 0.60 FEXX (1.0 + 0.5 sin1.5 �)
be used for a large gusset plate.
FEXX = electrode classification number, i.e., minimum
Fcr = critical compressive stress, ksi
specified strength, ksi
2
� = angle of loading measured from the weld lon-
c
= for c d" 1.5
0.658 Fy1
gitudinal axis
2
(0.877 / c )Fy1
for c > 1.5
=
Aw = effective area of weld throat, in2
c = slenderness parameter
= 0.707 We (4 Lw)
We = effective weld size, in
Fy1
k
=
= Ww - 1/16
rĄ E
Ww = weld size, in
Fy1 = specified minimum yield stress of the gusset
plate, ksi
In case the HSS web member is in compression, in addi-
tion to the limit states (2), (3), and (4) stated above, the fol-
lowing limit state has to be considered as well.
5. Strength Based on Buckling of the Gusset Plate
Fig. 4.3 Forces to be considered at the
Fig. 4.4 Gusset plate in compression.
weld connecting gusset plate and chord.
26
Fy1 = specified minimum yield stress of the gusset Connection between the HSS and the gusset plate:
plate, ksi 1. Shear Lag Fracture Strength in the HSS
E = modulus of elasticity, ksi
k = effective length factor = 1.2 ĆRn = ĆFu Ae
= laterally unbraced length of plates, in
r = governing radius of gyration, in An = Ag - 2 t t1 = 13.5 - 2 � 0.465 � �
= 13.04 in2
= t1 / 12
Lw = 20 in
t1 = gusset plate thickness, in
Ż#x = (6 + 2 � 6 � 10) / 4 (6 +10)
= 2.44 in
4.3 Connection Between Gusset Plate and Chord
U = 1 - 2.44 / 20 = 0.88 < 0.9
The stress distribution in the weld connecting the gusset
ĆRn = 0.75 � 62 � 0.88 � 13.04
plate and the chord is much more complex. As shown in
= 534 k
Fig. 4.3, the weld is subject to shear force V = T cos �, ten-
sile force P = T sin � - C, and moment
2. Shear Strength of the HSS at Welds
M = T cos � � ev - P � eh
ĆRn = Ć (0.6Fy)(4 Lwt)
= T cos � � d / 2 - (T sin � - C) � eh
= 0.9 � 0.6 � 46 � 4 � 20 � 0.465
= 924 k
If no vertical external load is applied at the joint, tensile
force P = T sin � - C is zero. While shear force V causes
3. Strength of the Weld Connecting the Gusset Plate to
shear stress in the weld, tensile force P and moment M
the HSS
induce tensile or compressive normal stress. These stresses
must be combined vectorially. In design, a unit throat thick-
ĆRn = ĆFw Aw
ness of the weld is usually assumed in the stress calcula-
� = 0� (the load direction is parallel to the weld
tions. The maximum stress caused by the combination of
direction)
factored V, P, and M must be equal to or less than the
FEXX = 70 ksi
strength of the weld. The provided design strength of a unit
Fw = 0.6 � 70 � (1.0 + 0) = 42 ksi
length of weld is
Aw = 0.707 � (3/8 - 1/16) 4 � 20
= 17.68 in2
ĆRn = ĆFwAw = Ć(0.6 FEXX)(0.707 Ww)
ĆRn = 0.75 � 42 � 17.68
= 557 k
where Ć = 0.75. Meanwhile, the maximum shear stress
caused by the direct shear must be less than the shear
4. Shear Strength of the Gusset Plate
strength of the gusset plate. The provided design shear
strength of the gusset plate per unit length along the weld
ĆRn = Ć(0.6 Fy1)(2 Lwt1)
connection is given by
= 0.9 � 0.6 � 50 � 2 � 20 � �
= 540 k
ĆRn = ĆVn = Ć(0.6Fy1) t1
The smallest value among the four cases above governs,
where Ć = 0.9
i.e., ĆRn = 534 k, which is larger than Pu = 523 k.
Calculations also must be made for the connection
4.4 Design Example
between the gusset plate and the chord to ensure its strength
is adequate.
The connection design of diagonal member d1 in Truss T1F
to the chord of the second story is calculated in this exam-
4.5 Miscellaneous Considerations
ple.
Reinforcement of trusses can be accomplished using field
Diagonal member d1: HSS 10 � 6 � �
welded plates and channels (see Figs. 4.5 and 4.6). It is
advised to leave the chord web free of stiffeners, plates,
Pu = 523 k (in tension, see Table 3.1)
etc., so as to allow future sistering of channels to be fitted
in the web.
27
Fig. 4.5 Truss HSS reinforcement detail.
Fig. 4.6 Truss chord reinforcement detail.
28
Chapter 5
SEISMIC DESIGN
5.1 Strength and Ductility Design Requirements mic detailing is not required. In selecting an appropriate
natural period of the building to be used in calculating the
The staggered truss system provides excellent lateral resist-
base shear, it is recommended that the classification of the
ance in the transverse direction of the building (the direc-
structure type be assigned on the basis of the way the seis-
tion of the trusses). A separate lateral-force-resisting system
mic energy is dissipated. The presence of the braces in the
must be provided in the longitudinal direction of the build-
truss system does not influence the ductility of the system
ing. This longitudinal lateral-force-resisting system usually
since these elements and their connections are designed to
consists of perimeter moment frames on the exterior of the
avoid yielding under a seismic event.
building or bracing systems organized around the building
elevator cores and stair towers.
5.2 New Seismic Design Provisions for Precast
In designing a staggered-truss system for seismic forces,
Concrete Diaphragms
several questions related to the system behavior must be
addressed. Of primary importance is the selection of an According to Ghosh (1999) and Hawkins and Ghosh
appropriate seismic response modification factor R to be (2000), the Uniform Building Code has required that in
used in developing the base shear of the building in the regions of high seismicity (zones 3 and 4 i.e., when R is
staggered truss direction. Prior research both in the United taken greater than 3) cast-in-place topping slabs over pre-
States and Japan has indicated that the staggered-truss sys- cast planks must be designed to act as the diaphragm, with-
tem behaves as a combination of a braced system and a duc- out relying on the precast elements. The design philosophy
tile moment resisting system under the action of seismic is that a topping slab acts in the same manner as a cast-in-
excitation. Hanson and Berg (1973) have shown that with place reinforced concrete slab under in-plane lateral loads.
proper detailing, the Vierendeel panel in the center of the The Northridge earthquake showed that this is not the case.
truss can provide significant ductility in the structural In some cases the topping cracked along the edges of the
response. The presence of braces in the other panels con- precast elements and the welded wire fabric fractured.
centrates the energy dissipation to the Vierendeel panel. In Accordingly, the diaphragms may have become the equiva-
order to ensure that this is the primary source of energy dis- lent of untopped diaphragms with the connections between
sipation, the bracing diagonals and their connections should the planks, the chords, and the collectors not detailed for
be designed to remain nominally elastic under the assumed that condition.
seismic forces. Equal care must be taken in the design of the Unlike topped diaphragms, untopped hollow-core plank
diaphragm system, which is an integral part of the resisting diaphragms with grouted joints and chords have performed
system to transfer lateral forces. It is essential to design the satisfactorily in earthquakes and in laboratory tests. Thus,
diaphragm to keep the in-plane stresses below yield limits. according to Ghosh (2000), when diaphragms are designed
In regions of high-seismic activity (that is, when it is using appropriate forces to ensure continuity of load path,
desirable or required to provide a system with an R factor force development across joints, deformation compatibility,
greater than 3), researchers suggest that the behavior of the and connections of adequate strength, they may perform
staggered-truss system be evaluated utilizing time history well even untopped.
analysis enveloped with a spectrum for the site under con-
sideration. The ductility demands on the chords can then be 5.3 Ductility of Truss Members
evaluated directly from the analysis. The response charac-
Staggered trusses normally use rectangular HSS, which act
teristics of a staggered-truss structure that dissipates energy
mostly as concentrically braced frames (CBF). CBF s are
mainly through Vierendeel panels are similar to a ductile
characterized by a high elastic stiffness, which is accom-
moment frame or an eccentrically braced frame. This would
plished by the diagonal braces that resist lateral forces by
imply that an R factor of 7 or 8 could be used for the design
developing internal axial actions. Only the chords, which
in the transverse direction of the building.
span across Vierendeel panels at corridors and openings
In regions of moderate seismic activity, using a response
provide some flexibility and energy dissipation capacity by
modification factor equivalent to that used for ordinary
developing out-of-plane flexural and shear actions.
moment frames (R = 4.5) would be appropriate (Hanson
For braced frames, tension-only systems are not consid-
and Berg, 1973). However, given the limited data available,
ered to provide a sufficient level of energy dissipation capa-
the designer may select a more conservative value of R = 3
bility. They are assigned a small response modification
for the overall behavior. When R is taken as 3, special seis-
29
factor R, and are designed for larger forces than the nomi- above four zones has its own failure modes, and the gov-
erning failure mode within each of the zones should be duc-
nal design force to account for impact. Compression braces,
tile. In order to increase the global ductility of the braced
however, are susceptible to fracture failure induced by local
frame, the occurrence of the yielding in the four zones
buckling and subsequent material failure, unless stringent
should be in the following order: yielding of bracing mem-
seismic slenderness ratios are provided. Local buckling in
HSS reduces the plastic moment resistance and conse- ber, yielding of gusset plate, and yielding of connection ele-
ments such as welds. Yielding of bracing member results in
quently the axial compressive strength. Furthermore, the
large axial plastic deformations, which in turn results in
degree and extent of the local buckling at the plastic hinge
has a major influence on the fracture life of the brace. Pre- large ductility of the braced frame. On the other hand,
venting severe local bucking is the key to precluding pre- yielding of relatively short welds cannot provide large
global ductility.
mature material fracture. The onset of local bucking can be
The emphasis of this section is on the seismic design of
delayed until significantly into the inelastic range by reduc-
ing the width-to-thickness ratio b/t of the brace. In high- gusset plates. Failure modes of a typical gusset plate in the
order of ductility desirability are as follows (Astaneh-Asl,
seismic applications, from the AISC Seismic Provisions,
Fy
1998):
the b/t ratio for HSS should be limited to 110/ .
The exact contribution of Vierendeel panels to energy
1. Yielding of Whitmore s area of gusset plate
dissipation in staggered trusses has not been documented.
Subsequently, new resources for research on this topic are
This is most desirable failure mode of a gusset plate. Yield-
necessary.
ing caused by direct tension or compression can occur in the
Black (1980) suggested that the most efficient braces are
Whitmore effective width area (Whitmore, 1952; Astaneh-
tubular cross-sections with small k / r. He also suggested
Asl, Goel, and Hanson, 1982). The yield strength of the
that improved performance can be achieved by reducing the
failure mode is
b/t ratios of the rectangular brace HSS. Black ranked the
tested cross-sections in the following descending order of
Py = AgwFy
effectiveness for a given slenderness ratio:
where Agw = gross area of gusset plate as per Whitmore s
1. Round HSS brace
30-degree lines and Fy = specified minimum yield stress of
2. Rectangular HSS brace
the gusset plate.
3. I-shaped brace
4. T-shaped brace
5. Double-angle brace
2. Yielding of critical sections of gusset plate under
combined stresses
Black recommended that built-up members not be used
Critical sections of gusset plates can yield under a combi-
as braces for applications in which severe cyclic loading is
anticipated unless the members making up the built-up sec- nation of axial load, bending, and shear. To determine
strength of gusset plate subject to combined loads, the fol-
tion were adequately stitched together.
lowing interaction equation is suggested:
5.4 Seismic Design of Gusset Plates
2
�# ś# �# ś#
N M V
+ + d" 1.0
During an earthquake excitation, the gusset plates connect-
ś# ź# ś#
ĆNy ĆM ĆVy ź#
�# # �# #
p
ing bracing members should have sufficient ductility to
deform and provide the end rotation demands of the mem-
where N, M, and V are the axial force, bending moment, and
bers. To avoid brittle behavior of the structure, the gusset
shear force on the critical section; ĆNy, ĆMp, and ĆVy are the
plate connections should be governed by a yielding failure
axial load strength in yielding, plastic moment strength, and
mode rather than a fracture mode. This can be achieved by
strength in shear yielding, respectively.
designing the failure modes in a hierarchical order in which
the ductile failure modes such as yielding occur prior to the
3. Buckling of gusset plate
brittle failure modes such as fracture.
Whitmore's effective width area can be used to establish
Bracing members and sometimes gusset plates are the
buckling strength of a gusset plate subject to direct com-
most active elements during an earthquake. Four zones can
be identified in a bracing system: bracing member, connec- pression:
tion of the bracing member to the gusset plate, gusset plate,
Py = AgwFy
and connection of the gusset plate to the chord. Each of the
30
where Fcr is the cirtical stress acting on a 1-in.-wide strip
where Agv and Anv are the gross and net areas subject to
within the Whitmore effective width. The effective length
shear, and Agt and Ant are the gross and net areas subject to
factor K is suggested to be taken as 1.2 because of a possi- tension, respectively. Fu is the specified minimum tensile
bility of end of bracing member moving out of plane.
strength.
4. Buckling of edges of gusset plate
6. Fracture of net area of gusset plate
A gusset plate may buckle along its free edge as shown in
To ensure that this relatively brittle failure mode does not
Fig. 5.1. The edge buckling limits the cyclic ductility of the
occur prior to yielding of gusset plate, the following crite-
gusset plate. To prevent edge buckling under severe cyclic
rion is suggested:
loading, the following equation is proposed by Astaneh-Asl
(1998):
ĆnPn e"Ć(1.1RyPy)
Lfg
E
d" 0.75
where Pn = AnwFu. Anw is the net area of gusset plate along
t1 Fy
the Whitmore section.
where Lfg, t1, and E are free edge length, thickness, and
5.5 New Developments in Gusset Plate to HSS Connections
modulus of elasticity of the gusset plate, respectively.
Cheng and Kulak (2000) have determined on the basis of
tests that the slotted end of the HSS is stiffened significantly
5. Block shear failure
as the result of the constraint provided by the gusset plate.
Block shear failure is a relatively less ductile failure mode
In most of the physical tests, the geometries provided
and undesirable. To ensure that the strength of gusset plate
allowed yielding to occur in the gross section of the HSS
in block shear failure is greater than its strength in yielding,
without fracture in the net section of the connection region.
the following criterion is suggested:
However, use of a short weld length or the absence of trans-
verse welds across the thickness of the gusset plate weld
ĆnPbs e"Ć(1.1RyPy)
may increase the stress concentration sufficiently that frac-
ture will take place where the HSS enters the gusset plate.
where Ry is the ratio of expected yield strength to specified
In such a case, ductility will be reduced. Nevertheless, in all
yield strength. The values of Ry are given in AISC Seismic
the configurations investigated by them, the slotted HSS
Provisions for Structural Steel Buildings (1997). Pbs is the
exhibited considerable ductility, regardless of the location
nominal strength of gusset plate in block shear failure,
of fracture.
which can be calculated using the equations:
Based on tests and numerical analysis performed by
them, it was concluded that shear lag does not significantly
Pbs = 0.6RyFyAgv + FuAnt for FuAnt e" 0.6FuAnv
affect the ultimate strength of slotted tubular sections that
are welded to gusset plates. The shear lag expression given
Pbs = 0.6FuAnv + RyFyAgt for FuAnt < 0.6FuAnv
in Section 2.1(b) of the AISC Specification for the Design
of Steel Hollow Structural Sections underestimates the
strength of a slotted tube-to-gusset plate connection. Never-
theless, a transverse weld can be used across the thickness
of the gusset plate since it increases the ductility of the slot-
ted member significantly without incurring much extra cost.
See Cheng and Kulak (2000) for further information.
Fig. 5.1 Edge buckling of gusset plate.
31
32
Chapter 6
SPECIAL TOPICS
6.1 Openings Another way to create level floors is by using a leveling
compound such as gypcrete.
Where openings in the truss are needed for room access,
Vierendeel panels can be designed in the truss geometry
6.4 Erection Considerations
(Fig. 6.1). The effect of such a panel is one of shear trans-
fer across the opening through bending in the chords, which Staggered trusses are fabricated in the shop and delivered to
increases the flexibility of the truss and the bending the site, generally in one piece. Erecting staggered trusses in
moments in the chords to both vertical and lateral loads. busy downtown areas (such as New York City) does not
Larger members and rigid connections, at a premium, present a special challenge. The usual steel tolerances,
reduce this flexibility, but do not eliminate it. The problem which are listed in the AISC Code of Standard Practice, are
can be solved by approximate methods or by using a com- normally adequate for erection purposes, even for busy
puter program. The latter method is recommended for all downtown locations.
special framing conditions. Connections for members fram- For practical reasons, staggered truss buildings are at
ing a Vierendeel panel are shown in Figs. 6.1 and 6.2. least six stories in height and generally at most 25 stories in
height. Higher buildings are possible when the staggered
6.2 Mechanical Design Considerations trusses are supplemented with special wind and/or seismic
frames and reinforced diaphragm floors. The reason for the
Vertical chases are needed for distribution of plumbing
height limit is the staked geometry at the roof and second
pipes and ducts. When needed, they should be sized and
floor, which is by necessity, discontinuous.
located to eliminate the need for stagger. Staggering results
Economy is further achieved by lumping member sizes
in a cost increase. The common solution is to increase the
into a few groups. This will usually result in an increase in
chase size and to locate it midway between the truss mod-
steel weight, but a reduced overall cost due to the associated
ules.
savings in labor. Also, HSS sizes are used for vertical and
diagonal members with those member properties listed in
6.3 Plank Leveling
AISC Hollow Structural Sections Connections Manual
Precast plank is delivered to the site with camber. Fabrica- (AISC, 1997).
tion variances may result in adjacent planks having differ- Economy may also be achieved by using longer columns
ential camber. with fewer splices. Up to four stories tall columns have
Differential plank camber is best removed mechanically been used with no difficulty. The faster erection and
prior to grouting. This is accomplished by inserting reduced field labor more than compensates for the cost of
threaded rods with lock plates and nuts in the joints and the added material.
then forcing the planks together mechanically by turning Structural stability is mandatory during erection. Tempo-
the nuts. Grouting takes place after the planks are aligned. rary steel braces or tension cables are recommended. Plank
Fig. 6.1 Opening in truss using Vierendeel panel.
33
able to agree on dimensions. Such a meeting is best coordi-
weld plates can also be used. The plank weight may provide
nated by the general contractor prior to the start of fabrica-
bracing through friction at the interface with the truss
tion.
chords. However, with friction alone, erection tolerances
such as column plumbing may be jeopardized.
6.6 Foundation Overturning and Sliding
6.5 Coordination of Subcontractors
At foundation level, the codes require adequate safety fac-
tors against sliding and overturning. The safety factors vary
During construction, the steel and plank shop drawings may
with the building codes, but are usually 1.5. Sliding resist-
require special coordination. The plank manufacturer
ance is provided by friction of the footings and the base-
locates plank embedded items such as weld plates. The steel
ment slab against the soil, active pressure against the
fabricator locates steel stiffeners, or wedges, shims, etc. A
meeting between the steel and plank contractors is advis- foundation walls, grade beams and footings, battered piles,
Fig. 6.2 Details at Vierendeel openings.
Fig. 6.3 Precast cantilever balcony.
34
etc. The engineer of record prepares calculations and pro- plished with steel plates or with channels fitted within the
vides the details required to ensure sliding and overturning
web.
safety.
Overturning with staggered trusses is usually not a con- 6.8 Balconies
cern for mid-rise buildings.
Precast plank balconies are best manufactured using solid
slabs. Since hollow cores entrap moisture, solid slabs are
6.7 Special Conditions of Symmetry
best used for durability. Fig. 6.3 shows a method of attach-
The typical staggered truss is symmetrical about its center- ment of balconies.
line. Symmetry of geometry and symmetry of loads result
in reduced member sizes. Non-symmetry results in
6.9 Spandrel Beams
increased sizes most affected are the chords, which sup-
Spandrel beams support the exterior walls. Where precast
port out-of-plane actions. Such non-symmetry occurs at ele-
concrete planks are perpendicular to spandrel beams, the
vator machine rooms, at roof appurtances, at public spaces
spandrels support floor loads as well. The spandrel beams
on private floors and at large guest suites with access doors.
in the other direction support no floor loads. In addition, on
Pattern live loads often create non-symmetry. Pattern
the wide face of the building, the spandrels are an integral
loadings are created by skipping the loads on alternate bays.
part of the moment frame for resisting wind and seismic
With staggered trusses, if load patterns are created by skip-
loads. The design considerations for such frames are not
ping alternate bays, symmetry still remains. However, skip-
within the scope of this design guide. The exterior wall is
ping loads in alternate rooms on the same side of the truss
often eccentric with respect to spandrel beams and columns.
creates non-symmetry.
A field weld between the plank and the beam flange
Future changes in truss geometry or loading is possible.
strengthens the beam torsionally and enhances its ability to
Often, such modifications entail removal of diagonals and
span between columns (see Fig. 2.6).
reinforcing of the chords. Chord reinforcement is accom-
35
36
Chapter 7
FIRE PROTECTION OF STAGGERED TRUSSES
Fire safety is a fundamental requirement of building design ferent testing laboratories as accepted standards for a par-
and construction and fire resistance is one of the most vital ticular fire rating.
elements of all components of a structure. For economy in materials and construction time, gypsum
Qualifying criteria to meet these requirements are board and metal stud walls are preferred. Gypsum board
included in various building codes of national stature. type  X and light-gage metal studs in any of the approved
These are used as standards in different areas of the country configurations for a particular rating is acceptable. How-
and which may or may not be further regulated by the local ever, removals of portions of the wall, renovations or addi-
authorities having jurisdiction. The codes (and publishing tions with non-rated assemblies are issues that need to be
organization) are: considered to avoid possible future violations of fire rating
integrity when choosing this method.
- Standard Building Code (SBCCI) The other option is to protect each truss member with one
- Uniform Building Code (ICBO) of the following methods:
- National Building Code (BOCA)
" If the truss is to be enclosed and/or protected against
These code regulations are based on performance damage and without regard to aesthetics, gypsum-
achieved through the standard ASTM E119 test (Alternative based, cementitious spray-applied fireproofing is
Test of Protection for Structural Steel Columns). Due to the often the most economical option.
dimensional constraints imposed by the fire testing cham- " Intumescent paint films can be used where aesthetics
bers, specific fire tests for steel trusses that simulate actual are of prime concern, and visual exposure of the steel
conditions have not been performed. Therefore, individual truss design is desired. In addition this product is suit-
truss members are regarded as columns for the purpose of able for interior and exterior applications. Neverthe-
rating their fire resistance and the applicable code require- less, this method is often one of the most expensive at
ment will be applied for each member. the present time.
By definition, a staggered truss spans from floor slab to " For exterior applications and for areas exposed to traf-
floor slab. Slabs are typically pre-cast concrete and have a fic, abrasion and impact, a medium- or high-density
fire resistance rating. The truss and columns are other ele- cement-based formulation is suitable and can be
ments of this assembly requiring fire protection. There are trowel-finished for improved aesthetics.
basically two methods of providing fire protection for steel
trusses in this type of assembly: Whatever method is chosen, the designer must work in
close consultation with the product manufacturer by sharing
" Encapsulating it, in its entirety, with a fire-rated enclo- the specifics of the project and relating the incoming tech-
sure. nical information to the final design. Final approval must be
" Providing fire protection to each truss member. obtained from the local authorities having jurisdiction over
these regulations.
In the former, enclosure can be any type of fire-rated
assembly. Local regulation, however, might reference dif-
37
38
REFERENCES
Astaneh-Asl, Abolhassan, Seismic Behavior and Design of Hsieh, Yuan-Yu, Elementary Theory of Structures, Prentice
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Kirkham, William and Thomas H. Mille,  Examination of
ley, 1998.
AISC LRFD Shear Lag Design Provisions , Engineering
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Berg, G. and Hansen, R.P., A Seismic Design of Staggered
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gered Truss System Proves Economical For Hotels ,
2000.
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Nolson, A. H.,  Shear Diaphragms of Light Gage Steel ,
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Bruneau, Michael, Chia-Ming Uang and Whittaker,
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39


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