Design of Welded Structures
DESIGN OF WELDED STRUCTURES BY o m er W. Blodgett THE -JAMES F. LINCOLN ARC WELDING FOUNDATION CLEVELAND OHIO P
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DESIGN OF
WELDED
STRUCTURES BY
o m er W. Blodgett
THE -JAMES F. LINCOLN
ARC WELDING
FOUNDATION CLEVELAND
OHIO
Published as a Service to Education
by THE JAMES F. LINCOLN ARC WELDING FOUNDA liON First Printing June 1966 Fourteenth Printing May 1991
Special acknowledgment is herewith made to Watson N. Nordquist who has contributed much to the editing and organization of the material from which this manual has been prepared
Trustees of the Foundation:
Dr. Donald N. Zwiep, Chairman, Worcester Polytechnic Institute, Worcester, Massachusetts John T. Frieg, Trustee, Cleveland, Ohio Leslie L. Knowlton, Trustee, Arter & Hadden, Cleveland, Ohio Officers:
Secretary-Richard'S.
Sabo, Cleveland, Ohio
Library of Congress Catalog Card Number: 66-23123 Printed in U.S.A.
Permission to reproduce any material contained herein will be granted upon request, provided proper credit is given to The James F. Lincoln Arc Welding Foundation, P. O. Box 17035, Cleveland, Ohio, 44117. Copyright 1966 by The James F. Lincoln Arc Welding Foundation The serviceability of a product or structure utilizing this type of information is and must be the sole responsibility of the builder/user. Many variables beyond the control of The James F. Lincoln Arc Welding Foundation affect the results obtained in applying this type of information. These variables include, but are not limited to welding procedure, plate chemistry and temperature, weldment design, fabrication methods and service requirements.
PREFACE WELDED STRUCTURAL CONNECTIONS have long been used in the construction of buildings, bridges, and other structures. The first welded buildings were erected in the '20s-the greatest application being in low-level buildings of many types. The American Welding Society first published specifications for welded bridges in 1936. But early progress came slowly. During that year, 1936, The James F. Lincoln Arc Welding Foundation was created by The Lincoln Electric Company to help advance the progress in welded design and construction. Through its award programs and educational activities, the Foundation provided an exchange of experience and gave impetus to the growing application of welding. Thus, within the last decade and particularly the past few years, unitized welded design has become widely accepted for high-rise buildings and bridges of nobler proportions in addition to the broad base of more modest structures. Now, the Foundation publishes this manual for further guidance and challenge to architects, structural engineers, fabricators and contractors who will build the structures of tomorrow ... and to the educators who will prepare young people for these professions. This material represents an interpretation of the best in accumulated experience of all who have participated in prior Foundation activities. The author has coordinated this with a continuing study of current welding research conducted both in the United States and Europe, and against a background of participation on various code-writing committees. Much of the direct instructional information that resulted has been pretested in over 70 structural seminars attended by over 4000 engineers. The production of this manual has spanned several years during which constant effort was made to eliminate errors. The author will appreciate having called to his attention any errors that have escaped his attention and invites correspondence on subjects about which the reader may have questions. Neither the author nor the publisher, however, can assume responsibility for the results of designers using values and formulas contained in the manual since so many variables affect every design.
Secretary The June 1966
James
F. Lincoln
Arc Welding
Foundation
CREDITS The author and the publisher gratefully acknowledge the organizations and individuals who have contributed photographs or other illustrative material: Allied Steel Corporation Allison Steel Mfg. Co. Allison Structural Steel Co. American Bridge Division, U.S. Steel Corporation American Institute of Steel Construction American Iron & Steel Institute American Welding Society Barber-Magee & Hoffman John F. Beasley Construction Co. Bethlehem Fabricating Co. Bethlehem Steel Corporation J. G. Bouwkamp Burkhardt Steel Company The California Co. California State Division of Highways Canadian Welding Magazine J. A. Cappuccilli, Architect Column Research Council Connecticut State Highway Dept. Dinwiddie Construction Company Dominion Bridge Company, Ltd. Dominion Structural Steel Co., Ltd. B. M. Dornblatt & Associates, Inc. Dreier Structural Steel Co. Edmundson, Kochendoerfer & Kennedy Engineering News-Record Englert Engineering Company Flint Steel Corporation Frankel Steel Company General Electric Company, Industrial Heating Dept. David R. Graham & Associates Granco Steel Products Co. Harley, Ellington, Cowin & Stirton, Inc. Haven-Busch Co. Herzberg & Associates Hewitt-Robins, Inc.
Nathan N. Hoffman Hoyle, Doran & Berry Inland Steel Company Jackson & Moreland Division, United Engineers and Constructors, Inc. Kaiser Steel Corp. Kansas City Structural Steel Co. Felix M. Kraus, Consulting Engineer Lehigh Construction Company Lehigh University, Fritz Engineering Laboratory Robert Charles Lesser, Architect R. C. Mahon Company P. H. Mallog Co. McGraw-Hill Book Co. Midwest Steel & Iron Works Nelson Stud Welding Division, Gregory Industries, Inc. New England Construction Magazine Pacific Car & Foundry Co. Pacific Iron and Steel Corporation Phillips-Carter-Osborn, Inc. Pittsburgh-Des Moines Steel Co. H. Platt Company Port of New York Authority Product Engineering Magazine Republic Steel Corporation Joseph T. Ryerson & Sons, Inc. Van Rensselaer P. Saxe, Engineer Schact Steel Construction, Inc. Steel Joist Institute Tennessee Gas Pipeline Co. United States Steel Corporation Vermont Structural Steel Co. Paul Weidlinger, Consulting Engineers Welding Engineer Magazine Welding Research Council West Coast Steel Works Minoru Yamasaki-Smith, Hinchman & Grylls
In certain subject areas, the author has made adaptations of work done by earlier investigators, to wit: Friedrich Bleich "Buckling Strength of Metal Structures" McGraw-Hill Book Co., New York, N. Y.
S. Timoshenko "Theory of Elasticity" McGraw-Hill Book Co., New York, N. Y.
Raymond Roark "Formulas for Stress and Strain" McGraw-Hill Book Co., New York, N. Y.
S. Timoshenko and S. Woinowsky Krieger "Theory of Plates and Shells" McGraw-Hill Book Co., New York, N. Y.
F. R. Shanley "Strength of Materials" McGraw-Hill Book Co., New York, N. Y.
S. Timoshenko and James Gere "Theory of Elastic Stability" McGraw-Hill Book Co., New York, N. Y.
The publisher regrets any omissions from this list, and would appreciate being advised about them so that the records can be corrected.
TABLE OF Part One Introduction to Welded Construction
1.1
Properlies of MCiteriols
2.1
Properli
BUIlt· Up Tension Members
2.2 2.3
Ana lysis of Bending
24
Deflection by Bendin Deflection of Curved Beams
2.5 2.6 2.7
D (gning fOr Impact loads
2.8
Designing fot Fatigue loads
2.9
Designmg for Torsional loadin
2.10
of Section
Shear Deflection in Beams
INTRODUCTION
Pa rt Two
LOA D & STRESS ANALYSIS
Analysi
of Combined Stresses
2.11
Bucklin
of Plate.
212
Anolysi
of Compre sion
3.1
Design ot Compr sslon Members
Part Three
Column Bas s
3.2 3.3
Column Splice
3.4
Baorlng-Pin Connections
3.5 3.
COLUMN-RELATED DESIGN
Designing Built Up Column
Welded Plate Girders for Building Efflclenl Plate Girders
4.1 4.2
Welded Plate Girders for Bridges
4
Bridge Pia e Girder witI Variable Depth
4.4
Girders on
0
Ho izontol Curve
Open Web Expcmde Shear
4.5
4.6
Topered Gird rs Beams and Girders
.7
Part Four
ttcchment fIJr Composite
Construe ion -Build n9
4.8
Shear Attachments for Composite Constru in-Bridges
4.9
Floo r Systems for Bridg s
4.10 411
Otthotropic Bridge Deck Fabrica tion of Plate Girders and Cover Plated Beams Field Welding of Buildings Field We ld ing of Bridg
4.12
413 4.14
GIRDER-RELATED DESIGN
CONTENTS lumn Conneetlens
5.1
~
a ke
5.2 5.3
W
F mrng Angle~
5.4
Top
on
Beam- •
Angl
t
ti fene
eat
ding Plates or
imple Beams and Win
Part Five
WELDED-CON NECTION DESIGN
55
Bracing
Pletes for
Top Connectin
mr-Rieid onn clio s Beam to-Column ontinuau Connedio
B am- -Girder Continuo
ign of
5.
Conn die
5
u e
Conn dlo s f r Tubul r Connection nees (Elas ic
Rigid-Frame
Part Six
MISCELLANEOUS STRUCTURE DESIGN
5.10 5.11
e 'gn)
W Idad Conn ction .. for Plastic De
Welded C
necn n for Vi r n
el
I
n
5.12
russe
5.13
D sign of Rigid Frames (Elastic De ign)
6.1
Op n We r Joi Reinf rdng 80
6.2
Ho
to Stiffen
Tanks. Bins an
o
63
Panel
0
H pp rs
6.5
6.6
igo of Han ers and Suppa
f
S Iectio
ruc1ural St el or
Welded Constructi
Part Seven
JOINT DESIGN AND PRODUCTION
W Idebility and Welding Procedure
72 7.3
Oet rmi
geld
I
rze
7.4 7.5
Estimating Walding Cost WeI
In
0
Painting &
Weld
REFERENCE DESIGN FORMULAS
7.1
Joint Design
Contr I
Part Eight
5.7
E)l"s ine Str duro
7.6
hrmkag
7.7
and Oi torhan
orroslon of
u Iity
Ided SI cures
nd Inspe h n
Bea . Diagr ms and F rmula Trio
Memb
rOd rams
nd
7.8 7.9
8.1 rrnulcs
8.2
LIST OF SYMBOLS AND DEFINITIONS a -
tJ. = E
v
I = =
CT
CTb CTy 7'
o a
= = =
angular acceleration (radians/sec/sec); included angle of beam curvature (degrees); form factor perpendicular deflection (in.), bending (tJ.b ) or shear (tJ..) unit strain, elongation or contraction (in./in.) unit shear strain (in./in.) Poisson's ratio (steel = 0.3 usually); unit shear force leg size of fillet weld (in.); rate of angular motion about an axis (radians/sec) unit angular twist (radians/linear inch); included angle; angle of rotation sum normal stress, tensile or compressive (psi); strength (psi) bending stress (psi) yield strength (psi) shear stress (psi); shear strength (psi) angle of twist (radians; 1 radian = 57.3 degrees); angle of rotation (radians); slope of tapered girder; any specified angle
area of section beyond plane where stress is desired or applied (in. 2 ) ; length of plate (in.); acceleration or deceleration (ft/min, ft /sec}, clear distance between transverse stiffeners of girder (in.) b width of section (in.); distance of area's center of gravity to reference axis (in.) c distance from neutral axis to extreme fiber (in.); distance of elastic center from reference axis d depth of section (in.); moment arm of force ( in. ); distance (In.), distance between centers of gravity of girder flanges (in.) clear distance between girder flanges (in.) eccentricity of applied load (in.); total axial strain (in.); moment arm of force (m.): effective width (m.), length of Tee section in open-web girder (in.) f force per linear inch of weld (Ibs/In.): horizontal shear force (Ibsym.), (vectorial) resultant force (Ibsytn.), allowable strength of weld (Ibsyin.) fc' = compressive strength of concrete (psi) g acceleration of gravity (386.4"/sec2) h = height; height of fall; distance of expansion on open-web girder (in.) k any specified constant or amplification factor m mass; statical moment of transformed concrete (composite construction) n distance of section's neutral axis from reference axis (in.); number of units in series p = internal pressure (psi) q allowable force on shear connector r radius (In.), radius of gyration s length of curved beam segment (in.); clear distance between ends of increments of weld (in.)
u
w x
y
thickness of section (In.}, time (min.); time interval (sec) material's tensile modulus of resilience (In-Ib/In." ) material's ultimate energy resistance (in.-lb/in. 3 ) uniformly distributed load (Ibs/Iinear inch) length of moment arm (curved beam) distance of area's center of gravity to neutral axis of entire section (in.)
area (in. 2 ) ; total area of cross-section stiffness factor used in moment distribution; any specified constant E modulus of elasticity, tension (psi); arc voltage (volts) E. modulus of elasticity in shear (psi) E, tangential modulus of elasticity (psi) E, kinetic energy Ep potential energy F total force (Ibs ): radial force (lbs) I moment of inertia (in.'); welding current (amps) J polar moment of inertia (in.'); heat input (joules/in. or watt-sec/in.) K ratio of minimum to maximum load (fatigue I; ratio of web depth to web thickness; distance from outer face of beam flange to web toe of fillet (in.}; thermal conductivity; any specified constant L length of member (in. or ft.); span between supports (In.) L, effective length of column M bending moment (in.-Ibs) M, applied bending moment (in.-lbs) M, = plastic moment at connection (in.-lbs) N number of service cycles; minimum bearing length of beam on seat (m.) P concentrated load (lbs) Q shear center; statical moment of cover plate area about neutral axis of cover-plated beam section R reaction (lbs); torsional resistance of member (in}); weld cooling rate (0 F/ sec) S section modulus (in. 3 ) = I/c T torque or twisting moment (in.-lbs); temperature (0 F) U = stored energy V = vertical shear load (Ibs ), shear reaction; velocity; volume; arc speed (in./min) W = total load (Ibs ), weight (Ibs ), total width (in.) Y effective bearing length on base plate (in.) Z plastic section modulus (in. 3 ) A C
e.G.
center of gravity HP horsepower N.A. neutral axis RPM = revolutions per minute
SECTION 1 .1
Introduction to Welded Construction 1. WELDING'S IMPORTANCE TO STRUCTURAL FIELD Welding has been an imp ort ant fac tor in our economy. The progress made in we lding equipment an d electrodes, the advancing art and science of designing for welding, and the growth in trust and acc eptance of wel ding have combined to make welding a powerful implement for an expanding construction industry. More an d more buildings and bridges are being b uilt according to th e precep ts of goo d weld ed d esign . The economies inhere nt in we ld ing are helping to offset evolu tionary increases in the p rices of materials and cost of labor. In ad d ition, th e shortened production cycles, mad e poss ib le by welding, have he lped effect a quickening in th e pace of new construction. W eld ed construction has p aid off ha ndsomely for ma ny archit ect s, struct ural engineers, con tractors, and th eir client-cu sto mers. It will become incr easingly important as more peop le ac q uire a greater d epth of knowl ed ge an d expe rien ce with it.
2. RECOGNITION OF WELDING The widespread recognitio n of welding as a safe means of making structu ral connections h as come about only after yea rs of dili gent effort, p ioneering action b y the more p rogressive eng ineers an d builders, and heavy documentati on of research findings and su ccesses attaine d.
Today, there just ar en 't many men in industry who speak disparagingly of welding. Most regulatory agencies of local and fed er al government now accept welded joints which meet th e requirements imposed by cod ewriting bodies such as the American Institute of Steel Construction and the American W elding Society. With this acceptance, there remains however a considerable task of ed ucation and simple dissemination of information to achieve maximum effi ciency in the application of welded des ign . And , th ere is even a continuing need for more th orough understanding of weld ing by codewriting bodies who fail to use the full strength of welded joints.
3. WHY WELDED CONSTRUCTION? T here are many reasons for using weld ed design an d construction, but probab ly the two basic ones are 1) we lded design offers th e opportunity to achieve more efficient use of materials, and 2) th e speed of fabrication and erection can hel p compress production schedules, enabling the en tir e industry to b e more sensitive and react fast er to rapidly shifting market needs.
Freedom 01 Design Welding permits the architect and structural engineer com p lete freedom of design-freedom to develop and us e modern economical d esign pri nciples, freedom to
FIG. 1 Indicative of the design freedom offered by unitized welding design, the Yale Rare Book library's four outside waifs are each a 5-story high Vierendeel truss. Each is a network of Greek-type crosses. The structure is alf welded-shop and field.
1.1-1
~ l~~t 1.1-2
/
Introduction
employ the most elementary or most daring concepts of form, proportion and balance to satisfy the need for greater aesthetic value. Just about anything the designer may envision can now be given reality . because of welding. Welded construction imposes no restrictions on the thinking of the designer. Already, this has resulted in wide usage of such outstanding design advancements as open-web expanded beams and girders, tapered beams and girders, Vierendeel trusses, cellular floor construction, orthotropic bridge decks, composite floor construction, and tubular columns and trusses.
con nections, resultin g in reduced beam depth and we igh t. This reduced b eam depth can noticeably low er th e ove rall height of a building. The weight of th e structure and th er efore static loadin g is gr eatly reduced . This saves column steel, walls and partitions, fa cia , and re d uced found ation requirem ents. 'Welded conne ctions ar e w ell suited to the new field of p lastic design, res ulting in furth er appreciable weight savings over con ventiona l rigid fram e d esign. Sav ings in transportation , h andling time, and erection are proportional to the weight savings.
Weld Metal Superior to Base Metal
Available Standards
A welded joint basically is one-piece construction. All of the other methods of connecting members are mechanical lap joints. A properly welded joint is stronger than the material joined. The fused joints create a rigid structure in contrast to the nonrigid structure made with mechanical joints. The compactness and calculable degree of greater rigidity permits design assumptions to be realized more accurately. Welded joints are better for fatigue loads, impact loads, and severe vibration.
Arc w eldin g, either in the shop or in the field, has been used lon g enoug h to h ave b een proved thoroughly dep endabl e. The AWS an d AISC ha ve set up dep endab le sta nda rds for all phases of structur al activity. These sta nda rds ar e b ack ed up b y yea rs of res earch and ac tua l testing. They simplify th e d esign of welded con nections an d facilitate ac cep tance by purchasers and inspe ctors.
Weldi ng Saves Weight, Cuts Costs Connecting steel plates are reduced or eliminated since they often are not required. Welded connections save steel because no deductions need be made for ho les in the plate: the gross section is effective in carrying loads. They oHer the best method of making r igid
Other Advantages Less time is required on det ailin g, layout and fabrica tion since fewer pi eces are used . Punching or drilling, an d reaming or coun tersinking are eliminated-a substantial saving on large projects. T he typical weld ed joint produces a smooth, uncluttered conn ecti on th at ca n b e left exp osed, without det ract ing from th e appear ance of the structure. W elded
FIG. 2 The a thlet ic unit of Lad ue Jr. High School (Missouri) features an all-welded steel lame lla roof fram e spa nning 252 ', expressing the strength of one-piece welded construction.
~W'1"
Introduction to Welded Construction joints exhibit less corrosion and require little or no maintenance. The smooth welded joints also make it easier to install masonry, facia and other close fitting members, often reducing th e thickness of walls or Boors in buildings. Structures can be erected in relative silence, a definite asset in building in downtown areas , near office buildings or hospitals.
4. HOW GOOD IS A WELD? Many engineers are unaware of the great reserve of strength that wel ds have, and in many cases this is not recognized by code bodies. Notice in Table 1 that the minimum yield strengths of the ordinary E60xx electrodes are about 50% higher than the corresponding values of th e A7, A373 an d A36 structural steels with which they would be used.
1
,
1.1-3
Inspection and Quality Mu ch mon ey is spent annua lly by industry an d govern ment in ob ta ining an d insp ecting for a specified weld qua lity . Usua lly th e we ld qua lity specified is obtai ne d , but too often th e qu alit y specified h as little or no relation to service requ irements. W eld s th at meet the actual service requirem ents , at th e least possible cost, are the result of1 ) proper d esign of connec tions an d joints, 2 ) good weld ing procedure, 3 ) good w eld or technique an d workma nsh ip, and 4) intelligent, responsible inspecti on. In th e following examples (Figures 3, 4, 5 and 6) test sp ecim en s exhib it undercut, undersiz e, lack of fusion, and porosity. In spite of th ese adverse conditions,
TABLE 1-Comparison of Typical Weld Me ta ls and Steels Yield Strength
Minimum Tensile Strength 62,000 psi 67,000 62,000 62,000 72,000
Minimum
Material AWS A5.1 & ASTM A233 Weld Metal (a s welded)
E6010 E6012 E6024 E6027 E70xx
50,000 psi 55,000 50,000 50,000 60,000
ASTM Steel s
A7 A373 A36 A441
33,000 32,000 36,000 42,000 46,000 50,000
60,000 to 75,000 58,000 to 75,000 58,000 to 80,000 63,000 67,000 70,000
Many of the commercial E60xx electrodes also meet E7 0xx specifications. Used on the same A7, A373 an d A36 steels, th ey have ab out 75% h igh er yield strength th an th e steel. There are nume rous reasons why we ld metal has higher strength than the corresponding plate. T he two most important are: 1. T he core wire used in the electrode is of premium steel, held to closer specifications than the plate. 2. There is complete shielding of the molten metal during welding. This, plus the scavenging and deoxidizing agents and other ingredients in the electrode coating, produces a uniformity of crystal structure and physical properties on a par with electric furnace steel. Because of these, properly deposited welds have a tremendous reserve of strength or factor of safety, far beyond what industry specifications usually recognize. But even without a reduced safety fac tor, there is a considerab le cost advantage.
1/2
H
PtATE
-+ I _ II..
II . 1J ..-
~OO~Z~Z~S REDUCTION IN PLATE SE (Tp, go to step 3. 3. Insert this value ((Terl VX) into formula #6, and solve for the critical buckling stress ((T er ) • 4. After the critical stress ((Ter) has been determined, the critical buckling stress of the given plate ( (T're or r' er) is determined from the relationship shown in the right-hand column of Tables 1, 2, or 3.
B to C
/v-x.
=
cry
i.a e,
-
(fer
a c r- =
3820
5720 - - to-~
y;;
bit
V
where: cT,3 n =--
4770
C to D
5720
y;;;cT,
and over
n cT
[ 4434 =c
n
Vk"
r
(~~)
The horizontal line (A to B) is the limit of the yield strength ((Tl')' Here (Tel' is assumed equal to cry. The curve from B to C is expressed by-
5. BUCKLING STRESS CURVES (Compression) I
In regard to plates subjected only to compression or only to shear, H. M. Priest and J. Gilligan in their "Design Manual for High Strength Steels" show the curve patterns, Figure 5 (compression) and Figure 10 (shear). They have divided the buckling curve into three distinct portions (A -B, B-C, and C- D), and have lowered the critical stress values in the elastic buckling region by 25% to more nearly conform to actual test results. Values indicated on this typical curve are for ASTM A-7 (mild) steel, having a yield strength of 33,000 psi. The buckling curve (dashed line) of Figure 2 has been superimposed on the Priest-Gilligan curve for comparison. Ocr
==
(Tel'
~
-
~
On
I ~
20,000
V> V>
Q)
(bit) V k-
See Table 4.
VI bit = 1.8 o; - n {k
"~Buckling curve of Fig. 2 © \. o = ~434 ]
0.-
cr
E
au
2
~~)
FIG. 5 Buckling stress curves for plates in edge compression.
?' \)C)\) ,,:>,,:>.
20,000
10,000 10
20
30
40
50
60
70
80
I
90
(k = 4)
@
Both Supported 10
5 I I
I
I
1 supported and 1 free
I
I
I
20
15 I
I
I
I
I
I
I I I
I
35
30
25 I
I
I
I
i
I
I
45
40 I
i
Ratio bit
FIG. 6 Buckling stress curves (plates in edge compression) for verlous- steels.
I
i
I
I
I {k=.425)
2.12-8
I
Load & Stress Analysis
TABLE 5-Factors for Buckling Formulas Yield Strength of Steel fr, psi
(bit ) yk"
for Point B
(bit Yk )
for Point C
y--;;-3 n=--'4770
3820 =-..;u:,
5720 =yU;
33,000
21.0
31.5
1260
35,000
20.4
30.6
1370
40,000
19.1
28.6
1680
45,000
18.0
27.0
2000
50,000
17.1
25.6
2340
55,000
16.3
24.4
2700
60,000
15.6
23.4
3080
70,000
14.4
21.6
3470
80,000
13.5
20.2
4740
90,000
12.7
19.1
5660
100,000
12.1
18.1
6630
of bit are recognized. Table 7, extended to higher yield strengths, lists these limiting values of bit.
7. EFFECTIVE WIDTH OF PLATES IN COMPRESSION The 20" X W' plate shown in Figure 7, simply supported along both sides, is subjected to a compressive load.
Simply supported sides
A-7 steel
o; b t k
TABLE 6-Limiting Values of bit (Code) Side Conditions One simply supported; the other free Both simply supported
Yield Strength fr, psi
AISC
AASHO
AREA
33,000
13 & 16
12
12
50,000
11 & 13
-
-
33,000
44
40
40
50,000
36
34
32
Factors needed for the formulas of curves in Figure
additional steels having yield strengths from 33,000 psi to 100,000 psi. For any given ratio of plate width to thickness (bit), the critical buckling stress ( (Tor) can be read directly from the curves of this figure.
6. FACTOR OF SAFETY A suitable factor of safety must be used with these values of bit since they represent ultimate stress values for buckling. Some structural specifications limit the ratio bit to a maximum value (point B) at which the critical buckling stress ((Tor) is equal to the yield strength ( (Ty). By so doing, it is not necessary to calculate the buckling stress. These limiting values of bit, as specified by several codes, are given in Table 6. In general practice, somewhat more liberal values
= v."
= 4.0
bit _
80 - 40
--W--{4-
FIGURE 7
AISC-Americon Institute of Steel Construction AASHQ--Americon Associotion of State Highway Officials AREA-Amercan Railwoy Engineers Association
5, for steels of various yield strengths, are given in Table 5. Figure 6 is just an enlargement of Figure 5, with
= 33,000 psi = 20"
Under these conditions, the critical buckling compressive stress ((Tor) as found from the curve (rTy = 33,000 psi) in Figure 6 is(Tor = 12,280 psi
TABLE 7-Usual Limiting Values of bit Yield Strength fr, psi
One Edge Simply Supported; the Other Edge Free
Both Edges Simply Supported
33,000
13.7
42.0
35,000
13.3
40.8
40,000
12.5
38.2
45,000
11.7
36.0
50,000
11.1
34.2
55,000
10.6
32.6
60,000
10.1
31.2
70,000
9.4
28.8
80,000
8.8
27.0
90,000
8.3
25.4
100,000
7.9
24.2
Buckling of Plates This value may also be found from the formulas in Table 4. Since the ratio
~~
is 40.0 and thus exceeds
the value of 31.5 for point C, the following formula must be used-
~
Since k = 4.0 (both sides simply supported), the ratio--
b
= 12,280 psi At this stress, the middle portion of the plate would be expected to buckle, Figure 8. The compressive load at this stage of loading would beU
21.0v'k
= 42.0
Yk
P = A
2.12-9
= 21.0
t = _ [4434]2_ U cr bit - [4434]2 40
/
= (20" X 114") 12,280
= 61,400 lbs
Since the plate thickness t = V4" width, b = 42.0 t or b = 10.5". This is the effective width of the plate which may be stressed to the yield point ((7'y) before ultimate collapse of the entire plate. The total compressive load at this state of loading would be as shown in Figure 9. The total compressive load here would beP = Ai
(7'1
+
A2
(7'2
(lOlh X V4)( 33,(00)
+ (9lh
X 114) ( 12,280)
= U5,BOO lbs Another method makes no allowance for the central buckled portion as a load carrying member, it being assumed that the load is carried only by the supported portion of the plate. Hence the total compressive load would beP = Ai
(7'1
(10lh X V4) (33,000) B6,6oo lbs Locol buckling
FIGURE 8
The over-all plate should not collapse since the portion of the plate along the supported sides could still be loaded up to the yield point (u y) before ultimate collapse. This portion of the plate, called the "effective width" can be determined by finding the ratio bit when ((Tcr) is set equal to yield strength (u y) or point B. From Figure 6 we find-
-bt = 42.0 or from Table 4 we find-
FIGURE 9
2.12-10
/
Load & Stress Ana lysis
'" '" '" (; 10,000
~
Cll
.s:
'"
10
20
30
.
R0110
50
40
70
60
bit
IT
Critical buckling shear stress (T«), for A-7 steel having
0y
= 33,000 psi
FIG. 10 Buckling stress curves for flate plates in shear.
8. BUCKLING STRESS CURVES (Shear)
TABLE 8-Buckling Stress Formulas (Shear)
The Priest & Gilligan curve, corresponding to Figure 5, when applied to the buckling of plates in shear is shown in Figure 10. The curve is expressed in terms of (
bit VIZ ) .
See
Table 8. Comparison of Figure 10 and Table 8 with Figure 5 and Table 4 reveals the parallelism of critical buckling stress for compression (CTer ) and for shear
Portion of Curve
Factar
bit
v'k
A to B
o t o3820 --
B to C
3820 5720 - - to--
Critical Buckling Shear Stress (T,,) Determined by
v;:;
'T e r
v;:;
VT;
T"
where: n
= 1.8
For any value of (
~~)
C to D
R
[4434
and over
T"
=
bIt n Y1{
T, -
4770
5720
-v;: T,
7"7
= --'
( 'Ter ) .
Figure 11 is just an enlargement of Figure 10, with additional steels having yield strengths from 33,000 psi to 100,000 psi. Factors needed for the formulas of curves in Figure 11 are given in Table 9.
=
r
~.
the critical buckling shear
stress ('T er) can be read directly from the curves of this figure. A suitable factor of safety must be used with these values since they represent ultimate stress values for buckling. By holding the ratio of
(~)
to the value at
point B, 'Ter = 'Ty and it will not be necessary to compute the critical shear stress ('Ter ) . Assuming the edges are simply supported, the value of k = 5.34 4(b/a)2 Then using just the three values of h/a as 1 (a square panel), 1fl (the length twice the width of panel) and zero (or infinite length), the required bit value is obtained from Table 10 for steels of various yield strengths. The plate thickness is then adjusted as necessary to meet the requirement. Notice in Figure 10 and Table s.. that the critical buckling stress in shear is given directly as ('T er ). In Tables 2 and 3 it is given first as (CTer ) and then changed to ('Ter) •
+
TABLE 9-Factors for Buckling Formulas (Shear) Yield Strength of Steel 0", .. psi
bit
bit
'V;; Corresponding --= for point B Yk for point C n=-' Shearing Yield Yk 4770 Strength 3820 5720 T, = .58 0", psi Y-;:; YT,
33,000
19,100
27.6
35,000
20,300
40,000
23,200
45,000 50,000
41.4
550
27.6
40.2
610
25.1
37.6
740
26.100
23.6
35.4
880
29,000
22.4
33.6
1030
55,000
31.900
21.4
32.1
1200
60.000
34,800
20.5
30.7
1360
70,000
40,600
19.0
28.4
1680
80,000
46,400
17.7
26.6
2100
90,000
52,200
16.7
25.1
2500
100,000
58,000
15.9
23.8
2920
Buckling of Plates
I
2.12-11
TABLE lO-Maximum Values of bIt To Avoid Formulas Maximum Values of bit 10 Hold 7'0. 10 (Panels with simply supported edges) Tensile Yield Strength (1', psi
60,000
oy
= 100,000
Oy
= 90,000
=
7',
=
=
b/a 1 b/a 0 b/a '12 (square panel) (panel with length (ponel with twice the width) infinite length)
33,000
84.5
69.6
63.9
35,000
82.0
67.6
62.0
40,000
76.7
63.2
58.0
45,000
72.3
59.6
54.7
50,000
68.6
56.5
51.9
55,000
65.4
53.9
49.5
60,000
62.6
51.6
47.4
70,000
58.0
47.8
43.9
80,000
54.2
44.7
41.0
90,000
51.1
42.1
38.7
100,000
48.5
40.0
36.7
- !i t tt
I
•I
a
"
50,000
a. '" '"
~
I = 80,000
Oy
= 70,000
Oy
I = 60,000
40,000
..."
~
Oy
30,000
0y
0y
= 33,000 pSI
_Oy
0 Q)
o;
~ OJ
o,
c -'"
::> 0
20,000
v 's:
b
--- -
!J -
Four edges - simply supported
q\J\J'§) ,
k
\J\J\J
= 5.34 +
4(b/a)'
Four edges - fixed
CO'\),
= 8.98 + 5.60(b/a)'
ClClCl k 1Cl, ClClCl bCl, R)Cl ~~~ ClClCl ~Cl' ClClCl ~ry, ClClCl ~Cl' s:lCl ~ s:l
r
v
.LJ
I
= 55,000 I = 50,000 = 45,000 = 40,000 = 35,000
Oy
t t
I
V
":lry, Cl~ f":J":l'
10,000
10
20
30 Ratio
40
50
bit
IT
FIG. 11 Buckling stress curves (plates in shear) for various steels.
60
70
2.12-12
/
Load and Stress Analysis
Ul)ited Airlines hangar at San Francisco features double-cantilevered roof over areas into which large jet aircraft are wheeled, nosing up to the 3-story inner "core" for servicing. Center girder section half (at left) is completely shop welded. large plate girders like this one are stiffened to prevent web buckling due to edge compression. Cantilevered welded plate girders weigh 125 tons .
SECTION 3.1
Analysis of Compression 1. COMPRESSIVE STRESS Compressive loading of a member when applied (axially) concentric with the center of gravity of the member's cross-section, results in compressive stresses distributed uniformly across the section. This compressive unit stress is-
~ ~
'"
(1)
A short column (slenderness ratio L/r equal to about unity or less) that is overloaded in compression may fail by crushing. From a design standpoint, short compression members present little problem. It is important to hold the compressive unit stress within the material's compressive strength. For steel, the yield and ultimate strengths are considered to be the same in compression as in tension. Any holes or openings in the section in the path of force translation will weaken the member, unless such openings are completely filled by another member that will carry its share of the load. Excessive compression of long columns may cause failure by buckling. As compressive loading of a long column is increased, it eventually causes some eccentricity. This in turn sets up a bending moment, causing the column to deflect or buckle slightly. This deflection increases the eccentricity and thus the bending moment. This may progress to where the bending moment is increasing at a rate greater than the increase in load, and the column soon fails by buckling.
2. SLENDERNESS RATIO As the member becomes longer or more slender, there is more of a tendency for ultimate failure to be caused by buckling. The most common way to indicate this tendency is the slenderness ratio which is equal toL r
where L r
and-
I,
~l
I
:..
(2)
If the member is made longer, using the same cross-section and the same compressive load, the resulting compressive stress will remain the same, although the tendency for buckling will increase. The slenderness ratio increases as the radius of gyration of the section is reduced or as the length of the member is increased. The allowable compressive load which may be applied to the member decreases as the slenderness ratio increases. The various column formulas (Tables 3 and 4) give the allowable average compressive stress (a-) for the column. Thcy do not give the actual unit-Stress developed in the column by the load. The unit stress resulting from these formulas may be multiplied by the cross-sectional area of the column to give the allowable load which may be supported.
3. RADIUS OF GYRATION The radius of gyration (r) is the distance from the neutral axis of a section to an imaginary point at which the whole area of the section could be concentrated and still have the same amount of inertia. It is found by the expression: r = VTlA. In the design of unsymmetrical sections to be used as columns, the least radius of gyration (rmin) of the section must be known in order to make use of the slenderness ratio (L/r) in the column formulas. If the section in question is not a standard rolled section the properties of which are listed in steel handbooks, it will be necessary to compute this least radius of gyration. Since the least radius of gyration isr min --
~Imin A
(3)
the rrummum moment of inertia of the section must be determined,
unsupported length of member
Minimum Moment of Inertla
the least radius of gyration of the section
The maximum moment of inertia (1 1ll " , ) and the minimum moment of inertia (I min) of a cross-section are
3.1-1
/
Column-Related Design
and, applying formula # 1 from Section 2.3, the distance of neutral axis x-x from its parallel axis Xl-Xl is-
y
NAx-x
x
=
IM IA -
I mox _ min
Ix +
-
(Ix -
2
+
I
2
.... (
4)
xy
Knowing Ix, Iy, and I xy it will be possible to find I m ln •
NAy-y
=
= - 1.75"
+
,
~
_
M
d 1.5 0
I .
_
-
--
_
found on principal axes, 90° to each other.
.
A I" X 6" 6.0 6" X I" 6.0 Total -..+ 12.0 L
y
I y) 2
21.0 12.0
to locate neutral axis y-y:
FIGURE 1
I y -+- ~ I 2 -"
-
_
----+~-+----
x
-
3.1-2
+
IM 9.0 IA = ----r2.()
+
+ 9.0 0 + 9.0
-
.75"
product of inertia It will be necessary to find the product of inertia ( I x J· ) of the section. This is the area (A) times the product of distances d, and d, as shown in Figure 3.
Problem I
Locate the (neutral) x-x and y-y axes of the offset T section shown in Figure 2:
(See Figure 3 on facing page). In finding the moment of inertia of an area about a given axis (Ix or Iy), it is not necessary to consider the signs of d, or d,.. However, in finding the product of inertia, it is necessary to know the signs of d x and d, because the product of these two could be either positive or negative and this will determine the sign of the resulting product of inertia. The total product of inertia of the whole section, which is the sum of the values of the individual areas, will depend upon these signs. Areas in diagonally opposite quadrants will have products of inertia having the same sign. The product of inertia of an individual rectangular area, the sides of which are parallel to the x-x and y-y axes of the entire larger section is-
y
f--a---t..L
c~+ d
x
x
Ixy
a
b:YJ
y
FIGURE 2 FIGURE 4
to locate neutral axis x-x: 6" X I" I" X 6" Total -..+ where d
A 6.0 6.0 12.0
-
where: d 0 3.5
M -
0 21.0 21.0
distance from center of gravity of element area to parallel axis (here: xl-xd
a and b d and c
=
dimensions of rectangle (
=
A)
distance of area's center of gravity to the x-x and y-y axes (= d x and dy )
The product of inertia of a T or angle section is(See Figure 5).
Analysis of Compression
y
x---+------'- x
x----t----- x y
y
Y
Ix = A d~
Iy = A d'y
Moment of inertia about x-x axis
Moment of inertia about y-y axis
Ix y = A d x dy
Product af inertia about x-x and y-y axes
y
y
3.1-3
y
y
x ----t----'----- x
/
y
y
y'
~-dy
+d +d x x
x
+d x x
x
x
x
x
x
-d x
y
ctt--d y
y
1st Quadrant Ixy
+A d x dv
-d x
2nd Quadrant
+ dy-t!J
y
Y
3rd Quadrant
Ixy = -A d x d y
Ixy
4th Quadrant Ixy = -A dx d y
+A d x d y
FIGURE 3 y
Now use formula given previously for product of inertia of such a section:
~---II-- d ------J .L t
T
adt(d-2c)(a+t) 4 (a d)
+
(4) (5) (liz) (5 Ixy
_ a d t (d - 2c) (a 4 (a + d)
+
4 (4
t)
+ 3.125
+
2.5) (4
+
1/2 )
5)
in."
y
_-----5"------+-1
FIGURE 5
Here, determine sign by inspection.
I
Problem 2
T
+ .555"
W'
1rr-+7f-4--+---'----x 1" r/!-of!--+- - 1.25"
I
f-
Determine the product of inertia of this offset T section about the x-x and y-y axes:
r., =
y
4" -.695"
!A (d,,)( d, )
= 2.5 (+ 1)( + .555) + 2 (- 1.25) (- .695)
+ 1.388 + = + 3.125 in.
1.737 4
FIGURE 6
-jvd!--1.945" y
3.1-4
/
Column-Related Design
55.25 + 25.25 2
I
Problem 3
~(55.25
Determine the minimum radius of gyration of the offset T section shown previously (Fig. 2) and repeated here:
40.25 -
2
_~5.25r
+ (15.75)2
21.75
18.50 in." minimum radius of gyration
rml
n
_lImAill
= ""
~ 18.50 12.0
6"
I
1.24" FIGURE 7
As a matter of interest, this rlllill is about axis x'-x', the angle (()) of which is-
tan 20 moment of inertia about axis x-x
=
(See sketch below).
2 (15.75) 55.25 - 25.25 46.4 0 or + 133.6 0
20 and ()
-
I,
21.0 12.0
- - - = - 1.75" and
NAx. x
=I
-
M2
A
= 92.00
-
Any ultimate buckling could be expected to occur about this axis (x' -x').
moment of inertia about axis y-y
1" X 6" 6" X 1" Total -
----~
6.0 6'
IT
+0
ry2~~ - -
=
I -
M
9.0 . 0 + 9.0
+
M2
A =
+
32.00 -
y
x'
I I Ig ----l3'j18.00, o .50 + 32.00 .-
9.0 'l2.()
NAy •y I,
-t:::
d_~ ~~
A
+ 66.8°
55.25 in."
36.75
1.05
- - ----.
,,
,,
+
,,
,,
X - - -_ _-"\-
1
"co ~ (Le/,12
80
o
~
-+---"I--.~
I TT2 E,
-
rr2 E
--
cr" (L e/r)2
.....................1'• 1
\
110
I
25 X106
d-
=-.1
120
1\
I
Euler
V \ I ~I ~\
140
.L,
r-,V V ;;;:T .... N
Allowable stress (foetor of safely l.B)
--
13'
O"bI
OK
>
13'
KxLx _ rx
t-.:
0
0
FIGURE 2
[o]
hang (m or n) as a cantilever beam with M being maximum at the fixed or column end:
bending moment p m2
M = -2- parallel to the column's x-x axis and n2 M = P 2 parallel to the column's y-y axis bending stress in plate
M
where, assuming a I" strip:
28,500 lbs
The bending moment on the weld is-
(28,500 lbs ) (21h") = 71,250 in.-lbs
OK
3.3-12
I
Column-Related Design
zontal top weld. At the ends of the angle, the force . (915)(3) couple IS 2 = 1370 lbs centered 1" below the
57,000# Uplift On Columns
top toe of the angle. See Figure 22. This is the force on each of the vertical welds at ends of the angle. Since these forces are not resisted by anything but the flange, they have to be carried transversely by bending stresses in the flange until they reach the resistance in the column web. The bending moment in the column flange is computed as follows: Force along top of angle Mh
= 5040
My
=
14,250# FIGURE 21
As in the previous example, the heel of the angle is in compression against the web of the column and is replaced with an equivalent weld. The welds are treated as a line, and the section modulus of the welded connection is found to be--
= (11) 6) = 78 in.2
=
Total M
= 21,395 in.-Ibs
(6") (1'li6")S 12
3
.1625 in,"
7,535 in.-Ibs
Mc
O"b
(See Problem 1)
= -1_
The bending force isfb
= 5040 lbs
= 13,860 in-lbs
1370 X 5.5
d2
+ (~2
5.5
If we assume a 6" wide strip of the column flange to resist this load, this moment will cause a bending stress of 45,300 psi in the 14" WF 87-lb column with a flange 1 YI 6" thick. This is calculated as follows: 1-
Sw=bd+
X 2.75
= 915 X
(21,395) (171g2 ) (.1625)
= 45,300 psi
M
= Sw
71,250 in.-Ibs 78 in.2
= 915 lbs/in, all along the top edge of the angle, pulling outward on the column flange. This is the force on the hori-
Obviously, since this stress distribution along the welds is capable of bending the column flange beyond the yield point, the column flange will deflect outward sufficiently to relieve these stresses and cause a redistribution. The resultant stresses in the weld metal on the toe of the clip angle will be concentrated opposite the column web.
12,800#
11"
'-------i~ II
II
T1370#
311
I-L- 6x4q-
28,500#
I 22I".....- - - 6 "---1--2 I"-1I 1 2
FIGURE 22
Column Bases Thus, the capacity of this anchor bolt detail is limited by the bending strength of the column flange even after the clip angle has been satisfactorily stiffened.
+2
16.15" The load on the bolts is-
(1370 lbs)
78,800 lbs
A lh" fillet weld 3 inches long on the top of the angle opposite the column web will satisfactorily resist the force couple: (3") (5600 Ibs jin. ) 16,800 lbs.
The area of the three 1%" dia. bolts in the unthreaded body area isA
E70 welds
Problem 3
(3)(2.074)
= 6.22 in. 2
OK
The tensile stress in the bolts is:
For greater anchor bolt capacities than shown in Figure 22, either horizontal stiffeners or diaphragms should be provided to prevent bending of the column flanges.
I
( 130,000) (9.49) ( 15.66)
F
= 12,800 lbs
F
(T
=
(78,800) (6.22) 12,700 psi
I
Q.
..
~
Q
11>
.. .. o..
~
Q.
11>
ta
C
Q. ~
IQ III
:
Cl
110 20.0
~
5
120 19.7
s:
~
""s: "~
o7 oR
0.9
1 0
1 2
1.8
1.411.6
2.0
I 2.5
3.0
over 3
130 18.2 140 17.3 0.5
150 17.1 -e 1.4 160 16.9 22 16 7 ~ 2.9 E 180 16.3 "C 0 4.1 '1. 200 15.8 59 220 15.4 7.3 I 240 1-51 ~
" r-no '"
'"
'"
8.3
20.0 19.1 17.9 17 1 16.7 0.6 1.2 20 0 18 3 17.3 16.9 16 3 15 R 0.4 1.:1 2 1 25 200 18 1 17.2 16 7 16 3 15 4 14.6 09 1 9 2.6 :1.1> 4.4 18.5 172 16.7 16 2 15 4 14.4 13.6 09 2.2 32 4.5 5 5 59 17.4 16.8 16.2 15.4 14.7 13.7 128 7 0 0.4 2.0 3 4 5.0 6 1 6.9 17.0 16.4 156 14 R 14.1 13.1 12.2 1 .5 3.1 5.1 6 5 74 79 7.9 16.7 15.9 151 14.3 13 6 12 6 11.8 2 . .5 4.7 6.5 7 7 1>.4 H.7 86 16.4 15 5 14 6 13.9 13.3 12.2 11.4 3.6 6.0 r.« H.fi 92 94 92 16.0 15.1 14.3 13.6 13 0 11.9 11.1 49 7.1 1>5 94 9 I> 9.9 9.7 15.6 14.8 14.0 13.3 12.7 11.7 10.8 6.0 1>.0 9 2 100 10.4 104 10.0 15 4 I 14 5 13 8 13.1 12.5 106 1 1 . 5 10.4 7.0 1>.8 10.0 105 10 9 110.1> 14.9 14.1 13.4 12.8 122.!11.1 103 8.4 9.9 10.8 11.4 11.fil11.4 :10.9 '-1';-14.6 138 13.2 12.5 11.9! 10.9,10.0 9.5 10 8 11.6 12 0 12.1'11.8f· 3 j14.3,113 6\13.0 12.3 11.7 10.3 11 .5 12.1 12.4 12 ..5
16.4 1.4 15 3 2.8 13.9 46 12.9 5.9 12.1 6.9 11.5 7 7 110 8.3 10.7 88 10.3 .9.2 10.1 9.6 9.9 9.8 9.5
0.2
0.4
161 1.6 14.8 3.1 13.4 4.6 12.3 5.8 11.5 6 7 10.9 7.4 10.4 7.9 10.0 8.4 9.7 8.7 9.5
15.8 1.6 14.3 3.1 12.9 4.5 11.8 5.6 11.0 6.4 10.4
9.0
7.0
9.9 7.5 9.5 7.9 9.2 8.2 89 85 8.7 8.7
OJ
0.8
0.9
10
12
1.3
1.4
1.5
1.6
1.8
20
2.5
18.00
18.00
18.00
18.00
1800
18.00
18.00
18.00
18.00
18.00
18.00
1800
18.00
17.95
18.00
18.00
]8.00
18.00
18.00
18.00
17.93
17.47
17.09
16.79
16.53
16.13
15.84
15.45
g:~
18.00
0.6
i
3.0
2.8
18.00
18.00
17.19
1800
15.54 0.1 0.1
17 00
100 1800 v c
£
,
17.38
110
, -
1
I
14.85
1422
13 52
2.4 30
15.08
14.50
18 22
36 45
4.0 5.1 13J2 5.6 6.9
13.04 6.8 8.6
15.40
14.85
14.06
13.31
12.66, 12.06
OJ
26 3.3
10 .84
150
0.9
s:
ro
15.19
68 8.5 12.98 7.8 9.7
14.19
13.40 72
12.71
13.15
12.48
~:~ I i:~
160 170
1 5.0 63 14.52 13JO 6.2 7.8
I ]~ ~1
en 180
28
5.1 63
14.82
13.91
9.0'
10.8
3.4 4.3
13.30
1291
12.57
t~
4.9 6.1
5.1 6.4
11 I
i
8.82
5.77
5.2 65 876 5.8 7.2
4.6 5.8 8.15 5.1 6.3
4.92
~.i ! ~:6
i
9.63
~:~
6.6 8.2 9.13 7.1 8.9
82
8.1
1~~4 11~~3
.
10.2
10.2
10.0
9.8
I 9.4
]0.76
10.40
1~:~6
~:~5
!
89
88
11.1
11.0
108
i
10.5
,I
13.9
12.8
9.3 11.6
9.2~
9.66 10.9 13.6
10.6
9.42 11.3 14.2
111
9.75 i 9.44
9.1
1 8.9
~~9
~:60
110.0
I 84 8.89
11.4 111.1 ~5 950 1 9.18 8.62
i.·~
11
i.t Ilg.·g
UI 12.34 2.8
7.83
6.6 8.2
8.41 7.9 9.9
7.49 6.9 8.6
8.14 I 8.2 i 10.3
6.87 I-
4.24
7.61 5.4
6.2 7.8
76 9.5
8.31
9.68
8.25
8.74
10.27
2.2 '
68 7 18 5.7 7.1
3.70
1 325 i I
I 2.88
I
10.6
i
i 208
110.1
13 3
1
I
i 1.72
9.04 1 8.70m T 10.7 13.4, ' 13.8
i
\\('\1
1 144
I
"V
Q
(l)
.. ..
(l) 1.23
10.40
F,.
In third h'jri/.ol1t"llitlt, uulirutr- re-quired g-ro.":-'
5000 in." OK
Then, to find the weight of this designed plate girder:
4.2-12 2 At Aw
.'. w,
/
Girder-Related Design
= 2(2")(13") = = (314" ) ( 120") = = 482.8 Ibs/lin
~52.0
90.0 142.0 in. 2
It of
girder
2nd Nomograph
=
If the shear value is increased to V 1000 kips as in Problem 2, this exceeds the allowable value of 750 kips read from the first nomograph. Therefore, shear governs the design and the second nomograph must be used. Given:
Problem 3
V
Find the approximate web dimensions and weight for the same girder, using the nomographs, Figures 6, 7 and 8.
1000 kips
= 170
K read: tw
1sf Nomograph
Given:
Given:
S
5000 in."
K
dw = 170 tw
read:
Given:
Given: dw tw
170
read:
= 4701bs/ft
and:
V
170
3rd Nomograph
d = 108"
w,
K
d w = 126" or use 124"
read:
K
.725" or use 3;4"
S
5000 in."
d
124"
read:
Wt
)
275 lbs/ft
Given:
750 kips allowable
Using an actual depth of 110" as in Figure 1 would increase this estimated weight to 483 lbs /Ft as read on the nomograph. In Problem 1, the weight was computed to be 488 lbs/ft, this slight increase is due to the increase in web thickness from the required .638" to the next fraction, 11/16".
V
1000 kips
T
11,000 psi
read:
Wt
+
210 lbs/ft Total = 485 lbs/ft )
In Problem 2, the weight was computed to be 482.8 lbs/ft,
SECTION 4.3
Welded Plate Girders for Bridges
1. INTRODUCTION Plate girders are fabricated for requirements which exceed those of a rolled beam, or a rolled beam with added cover plates. The usual welded plate girder is made of two flange plates fillet welded to a single web plate. Box girders are made of two flange plates and two web plates. They have extremely high torsional strength and rigidity. Plate girders are proportioned by their moments of inertia. See preceding Section 4.2 on Efficient Plate Girders. AASHO Specifications govern in the Bridge field, with AWS Specifications generally governing welded joint details. This particular section brings together these two Specifications, with interpretation and supplementary recommendations being added for the designer's guidance. AASHO (1.6.11) limits the minimum ratio of the depth of beams and plate girders to 1/25 of their length. For continuous spans, the span length shall be considered as the distance between dead-load points of contraflexure.
2. PLATE GIRDER WEBS AASHO Specifications (1.6.75 & 1.6.80) require that the thickness-to-depth ratio of girder webs be not less than the values indicated in Table 1. The above ratio of web thickness to clear depth is based on predications of the plate buckling theory, the web plate being subjected to shear throughout its depth and to compressive bending stresses over a portion of its depth. The plate buckling theory assumes the panel portion of the web to be an isolated plate; however, in the plate girder, the web is part of a built-up member. When the critical buckling stress in the web is reached, the girder does not collapse. The flange plates carry all of the bending moment, the buckled web serves as a tension diagonal, and the transverse stiffeners become the vertical compression members. This has the effect of making the girder act as a truss. Research at Lehigh University tested, among other things, the effect of the web thickness on the ultimate carrying capacity of the girder; see Figure 1. It was found that the ultimate load carrying capacity of the
girder, expressed as the ratio of the ultimate load (PH) to load causing yield stress (P y) was directly proportional to the restraint offered by the compression flange. The more torsionally flexible flange (wide and thin) resulted in the lower strengths, and the more torsionally rigid flange (tubular) resulted in higher strengths. Differences in web slenderness ratio produced little effect on the ultimate load carrying capacity of the girder for the same compression flange. Although the tubular type of compression flange was used to obtain a torsionally rigid flange, it is not recommended for actual bridge practice. However, the concrete floor slab directly on top of the usual compression flange offers a similarly high torsional restraint, as well as good lateral bracing. Designers in Europe, as well as Canada, are not held to this (Table 1) fixed ratio of web thickness to web depth. One exception made in the United States is the Quinnipiac River Bridge in New Haven, Conn., which used thin longitudinally stiffened webs. Instead of using an arbitrary ratio of web thickness to depth, the design was based on the elastic stability of the web from information by Mosseiff and Lienhard*. The design also considered safety against yielding based on a yielding criterion obtained from the following Huber-Mises formula:
* "Theory of Elastic Stability Applied to Structural Design" ASCE paper 2120.
TABLE 1-Minimum Girder. Web Thickness- To-Depth Ratios (AASHO 1.6.80, 1.6.75) A-441 Steel A·7, A-373 A-441 Steel & A·36 Steels 46,000 psi yield 50,000 psi yield No stiffeners
1/60
1/52
1/50
- ~
I ntermediate transverse
stiffeners
1/170 ~
Longitudinal stiffeners
1/340
1/145 - - - t--~
-
1/290
1/140 --_.
t------
1/280
4.3-1
4.3-2
/
Girder-Related Design
If the value of (Ter resulting from the above formula is equal to the yield point of the steel in uni-axial tension (what is commonly called the yield strength, (T y ), it is assumed this combination of stresses will just produce yielding in the material. Hence, the use of this formula will give some indication of the factor of safety against yielding.
~Y2"1
r---=--i ~- 3" ~4
I/4 "
"
I~- f-
I .. ~8
~4
1 .. ~8
50"
I.. ·
·
(2)
50"
IJ=
2.5
¥-
20 =
51
(3)
III
IT
I
aal~~;2J
where:
1 "
3~6- f50"
Transverse intermediate stiffeners shall preferably be in pairs. They may be either single or double, and be plates or inverted tees. When stiffeners are used on only one side of the web, they shall be welded to the compression flange to give it proper support. The moment of inertia of the transverse stiffener shall not be less than-
II =
12"
7/ 16"
3. TRANSVERSE IN"rERMEDIA"rE S"r1FFENERS (AASHO 1.6.80)
I = minimum required moment of inertia of stiHener, in.4
-
(a) Cross-sections of test specimens ?/Py
a, = required clear distance between transverse stiffeners, in. a,
actual clear distance between transverse stiHeners, in.
dw
unsupported depth of web plate between flanges, in.
t w = web thickness, in. When transverse stiffeners are in pairs, the moment of inertia shall be taken about the centerline of the web plate. When single stiffeners are used, the moment of inertia shall be taken about the face in contact with the web plate. The width of a plate stiffener shall not be less than 16 times its thickness, and not less than 2" plus 1/30 of the girder depth. The distance between transverse stiffeners shall not exceed1. 12 feet 2. the clear unsupported depth of the web (d w )
1.0
a
3'1 a Girder
GI
G2
200
ar =
~
tw
I
(4)
where:
G3
'f00
(b) Comparison: ultimate and critical loads of bending tests
T
= average unit shear stress in the web's crosssection at the point considered, psi
4. LONGITUDINAL STIFFENERS (AASHO 1.6.81)
FIG. 1 Effect of web thickness on ultimate carry-
ing capacity of the girder.
The longitudinal stiffener shall lie along a line 1/5 d:!,
Welded Plate Girders for Bridges
/
4.3-3
-----17 .......
Longitudinal stiffener
.......
/" .......
.......
.......
.......
"~
" "....... ....... .......
Facio side of girder is clean ,/
"
"
,/
"
,/
....... ....... ....... .......
,/
.......
,
-----~
(0) Longitudinal stiffeners on inside of girder
FIG. 2 Placing longitudinal stiffeners on outside of girder and transverse stiffeners inside saves fabricating time.
. . ,---------------7 . . . . ....... " tzzm ....... ....... "" ,,~
-
Longitudinal stiffener
.......
Longitudinal and transverse stiffeners do not intersect
""
,/
,,
"" " "
,/
,/ ,/
,,/
'"
.......
.......
.......
....... .......
,
,, " ~--------------~
(b) Longitudinal stiffeners on outside of girder
from the compression Bange. Its moment of inertia shall not be less than-
1 I
= dw twS (2.4
~
-
0.13)
I
(5)
These stiffeners do not necessarily have to be continuous, but may be cut where they intersect transverse intermediate stiffeners if they lie on the same side of the web.
5. BEARING S1'IFFENERS Transverse stiffeners shall be used over the end bearings or along the length of the girder where concentrated loads must be carried, and shall be designed to transmit the reactions to the web. They shall extend as nearly as practicable to the outer edge of the Bange, but not to exceed 12 times their thickness. (AASHO 1.6.17) Some bridges have longitudinal stiffeners on the inside of the girders, others on the outside. If the longitudinal stiffeners are on the inside, along with the transverse stiffeners, it leaves the outside of the girder smooth; Figure 2(a). This, of course, means the longi-
tudinal stiffener must be cut into short lengths and then inserted between the transverse stiffeners. This results in increased welding time and production costs. Some states have used longitudinal stiffeners on the outside and transverse on the inside; Figure 2(b). This method saves on fabricating time and also allows the use of automatic welding techniques to join the longitudinal stiffeners to the girder web, thereby substantially increasing welding speed.
6. WELDING OF S"rlFFENERS AASHO (2.10.32) will allow the welding of stiffeners or attachments transverse to a tension Bange if the bending stress is 75% or less than the allowable. AWS Bridge (225 c) will allow the welding of stiffeners or attachments transverse to a tension Bange if the bending stress in the Bange is held to within those of the fatigue formulas (1), (3), or (5) for the welding of attachments by fillet welds; see Section 2.9, Table 1. Figure 3 illustrates the effect of transverse attachments welded to a plate when tested from tension to an equal compression (K = -1).· • "Fatigue Tests of Welded Joints in Structural Steel Plates", Bull: 327, University of Illinois, 1941.
4.3-4
/
Girder-Related Design
5~'t
K' min __ ,
max.:
1lt1t
~
a
100,000 CYC'LES
2~800 PSI
25. 400
2,000,000 cytLE5
Z~ 8001'4;
18,900
psi
. . 5/J?1
p6;
~ ,.
22,900 13.'00
psi psi
FIG. 3 Effect of transverse attachments on fatigue strength of member.
Some engineers have felt this reduction in fatigue strength is due to the transverse fillet welds; however, it is caused by the abrupt change in section due to the attachment. It is believed these plates would have failed at about the same value and location if they had been machined out of solid plate without any welding. This same problem exists in the machining of stepped shafts used in large high-speed turbines and similar equipment. Figure 4 illustrates the effect of welding transverse stiffeners to tension flanges." Tests, again at the University of Illinois, were made from tension to zero tension in bending (K 0) and at 2 million cycles. Eliminating the weld between the stiffener and the tension flange increased the fatigue strength of the beam. In addition, leaving the weld off the lower quarter portion of the web in the tension region gave a further increase in fatigue strength. Later tests at the University of Illinoisv" took into consideration not only the bending stress in the flange, but also the resulting principal tensile stress in the web at critical locations, such as the termination of the
connecting fillet weld of the stiffener. See Figure 5. It was discovered that the fatigue failure in the stiffener area did not necessarily occur at the point of maximum bending stress of the beam. Failure started at the lower termination of the fillet weld connecting the stiffener to the web. When the bottom of the stiffener was also welded to the tension flange, failure started at the toe of the fillet weld connecting the stiffener to the beam flange. After the flange had failed, the crack would progress upward into the web. Here, the failures usually occurred in the maximum moment section of the. beam. This test indicated fairly good correlation when the results were considered in terms of the principal tensile stresses (including the effect of shear) rather than simply the bending stress. The angle of the fatigue failure in the web generally was found to be about
=
* "Flexural Strength of Steel Beams", Bull. 377, University of Illinois, 1948. ** "Fatigue in Welded Beams and Girders" W. H. Munse & J. E. Stallmeyer, Highway Research Board, Bull. 315, 1962, P 45.
~/'"
VJ·"~· STIFFENERS
I~"W 36 LB.
x·!!!!!1· max.=0
8'-6"
2.000,000 CYCLES
1 T'r ~J If"
INTEAMEOIA1S STIFFENER'
INTEANlI!DIATe ST/~~ENEIU WftDED 7D we~
AND
"ll /Jl1TH ,.LANtJ e
IB.400P5;
.
WELDED 1t) WEB AND Tr>P FLANflE ONL Yo
Z',600 psi.
IZ"
INTERMEDIATE >STIFFENERS
lNaoeD7D COMPt:tESS/OAJ RAN6E AND Tr> UPPER ~... OFWE/J.
.32,700 p5i.
FIG. 4 Effect of welded in· termediate stiffener, on tension flange.
Welded Plate Girders for Bridges
.t,,;
7
'1,0
4,3-5
n-
d
Il ~ V
'",
/
rT 3/16
~
'",
I-.-J '",
NOWEIJ}
'Nl \ ff.JI
h
ff TYPE~"
\
TYPE Zl'
TYPE 'c'
TyPE '8'
,
NOWELO
NOWEI.D
j
'",
/-"'"
TYPE'F
(a) Details of various stiffener types
ISTIFFENER "TYPE IAIs IC1DIEI~1 I .sYMBOL. I_ ¢ "1+1 01+1 FIG. 5 Effect of stiffener type on fatigue strength of member.
0
' '1•
-1-
. ·IF=~I··
1
Cut-off point
f-
Theoretical cut-off point
j
I
.;-
:J
(B')
~2W----l
+
Inner end of terminal development
fillet weld size of Table 2. AASHO (1.6.74) specifies that the length of any cover plate added to a rolled beam shall not be less than(2d
+ 3)
feet
where d = depth of beam (feet)
SECTION 4.4
Bridge Plate Girders With Variable Depth 1. TYPES OF HAUNCHED GIRDERS
2. NEED FOR MODIFIED SHEAR FORCE VALUE
It has been pointed out" that the sloping bottom flange
of the parabolic haunch has a vertical component of its compressive force and this will reduce the shear stress ( T Xy) in the girder web in this region. In addition, the concave compression flange produces a radial compressive stress ((Ty) in the web depending on the radius of curvature of the flange. In contrast, the fish belly haunch provides no appreciable reduction in shear in the critical portion of the web near the support. This is because the slope of the bottom flange is small in that area. Also, the convex compressive flange produces a radial tensile stress ((Ty) in the web, which is greater than the radial compressive stress in the parabolic haunch. This is because of the sharper curvature of the fish belly haunch. It is seen by observation of the Huber-Mises formula that both of these factors will result in the yield criterion ((Tcr) having a lower value in the case of the parabolic haunch. This result compared with the yield strength of the steel (in uniaxial tension) would indicate a higher factor safety. (Huber-Mises Formula )
Haunched girders do not present much increase in cost for welded construction for longer spans. The web plates are normally trimmed by flame cutting, so that a gradual curve would add little to the cost. In most cases the curved flange plates can be added without prior forming, the flat flange plates are simply pulled into place against the curved web. Although the transverse stiffeners would vary in length, this should be no problem. The flange can still be automatically fillet welded to the web by placing the web in the horizontal position. The portable automatic welder would then ride against the curved flange.
The horizontal force (Fh) in the sloping flange is equal to the bending moment at that section divided by the vertical distance between the two flanges:
Or, this force may be found by multiplying the flange area by the bending stress in the flange using the section modulus of the girder. This method will produce a more accurate value. From this value, the actual force in the flange (F x) may be found, as well as the vertical component (Fv) of this force: M d cos 8 and
Fv
=
F h tan 8
= dM
tan 8
This vertical component (F-) acting along with the shear force in the web resists the external shear (V) at this section. Modified shear is the resulting shear force in the web after the vertical component of the flange force (Fv) is substracted or added, depending upon whether it acts in the same direction or opposite direction as the shear in the web.
Fish Belly Haunch • "Design of the Bridge Over the Quinnipiac River" by Roman Wolchuk.
Parabolic Haunch
FIGURE 1
4.4-2
/
Girder-Related Delign Compression
I
11::"1 ~~ .,,;,10","
of web due to its shear
---..:
4-----""""",",,::::"
d
:~Fh=M h. ~
I
F.
) Fh tan 0
=
Simply Supported Girder Straight or CUlTed Bottom Flange See Figure 2. Here the external shear isV = A.,
'Tw
+
M
d
A.,
cosO Resistance of bottom flange due to its vertical component of tensile force -.-
FIGURE 2
Continuous Parabolic: Haunc:hed Girder See Figure 3. Here the external shear isV
tan 9
=
A.,
+ dM
'Tw
tan 9
and the modified shear is-
and the modified shear is-
V'
d F F _ _h_
o
V' -
'Tw
A.,
'Tw
V In this case the vertical component is subtracted from the web shear. Tension
n
tan 9
In this case the vertical component is subtracted from the web shear.
I
I
-
M
d
•
~~ I 1 1~"';'Io","Ofweb ~ due to its shear
r
FIGURE 3
~
Resistanceof bottom ~ flange due to its I ~ -. vertical component ......",,,.....Iooo:::;;,,. . .,I+~ compressive force d
:
(F
I
I
Compression
Fh F.
•
I III
1D ~
~ Resistance of
_
R
d..... I ~::::Io-''''''':=..I
web
due to its shear
Effect of bottom
~ flange due to its / :If ~ vertical component
i~ yr • fJ
Fh
of tensile force F
FIGURE 4
Bridge Plate Girden With Variable Depth T
J
"
/
4.4-3
I
~_fie~n~S'O~nj'I ..
~
1D ~
- - - A-t... Resistance of web
d
due to its shear
___
I I --jt:::::=~.1 +-- Fx
FIGURE 5
= Fh= dM
Compression I
FIGURE 6
Fish Belly Haunch
Simply Supported Haund.ed Girder See Figure 4.
V
=
Aw
Tw -
d tan
between the fish belly haunch and the parabolic haunch in the area of the compression flange near the support. See Figure 6.
Here the external shear isM
Parabolic Haunch
Conditions include the following: 8
Use of A441 steel M = 55,000 It-kips
and the modified shear isV' = Aw
Tw
V
=
1200 kips
Ix = 3,979,000 in."
"rr=-r
.l1-36"-.I In this case the vertical component is added to the web shear.
Continuous Fish Belly Haunched Girder See Figure 5. Here the external shear isV = Aw
2W'T
128%"
127+'16"
_ _ _ 21'
J_
T
In this case the flange force has no vertical component; hence, there is no reduction of shear in the web.
I
Problem 1
I
Check the haunched girder section (at point of support) shown in Figure 7, to determine the difference
FIGURE 7
4.4-4
/
Girder-Related Design
Analysis of Parabolic Haunch
stress in web at lower flange at support
I I I I I
..£-
-r'
= 4890 psi
11D~
1-.. I
~ I
I F.
FIGURE 8
or:
F,
= 2030
Fh
= 2000k
Me crt = -1-
2,000 kips
(2000) (.1763) 353 kips
20,900 psi, compression These stresses in Figure 10, left-hand side, must now be rotated 10° to line up with the sloping flange in order that the radial compressive stress may be added. This is shown on the right-hand side of Figure 10. This may be analyzed by one of two methods: 1. Graphically, using Mohr's circle of stress: (Fig. 11) a) Draw the given stresses (o;', cry', and 'T') at the two points (a') and (b") b) Construct a circle through these two points c) Rotate clockwise through an angle of 28 or 20° d) Read the new stresses (crxs cr~., and 'T)
2000 .9848
= 2030 kips
2. Analytically; work is performed as follows:
shear stress in web Since the external shear is-
AI\'
'Tw
k
(55,000 X 12) ( 126 ) ( 3,979,(00)
crt At (21,150) (2% X 36)
AI\'
= 353
Me -I
flange forces
v=
IF
k
FIGURE 9
(55,000'" X 12) (127%6") 3,979,000
= 21,150 psi compression
Fh
= 353
8~+·
average bending stress in lower flange
_
k
Fh F' _______
'Tw
+
= V -
k
+ cr' y
2 20,900
F; or
2 10,450
F v and tan a
'Tw
crx'
'T'
%(crx'
+
1200 - 353 (252 X lYt6)
4890 %(20,900)
4890 psi
.4680
cry')
Bridge Plate Girders With Variable Depth
7"'
:
/
4.4-5
= 4890 psi 0'
~
----+ - - -+- - - . - ,
FIGURE 10
I
I
I b'
a' = 20,900 psi
ax = 21,950 psi
x
p=
a~ = 0
5.08°
7"'
2() = 20°
ex: =
25~08° ,---*;--
4-\~
_
0
= 4890 psi
ax
= 1050 psi
7"
= 1020 pSI
.
ax = - 21,950 psi 7"
= 1020 psi
a:
FIGURE 11
7"
a
=
25.08°
b' -" = 20,900 psi
= 4890 psi
fry
cos
f3 = .9961 m
(11,500)
= 1050 psi, tension
= 5.08°
f3 = .0886
n
( 10,450) -
f3 = 25.08° - 20° sin
= k -
radial force of lower compression flange against web
= V k + (r') 2 = V (10,450)2 + = 11,540 2
48902
r = m sin {3
= (11,540) ( .0886) = 1020 psi FIGURE 12
n=mcos{3
= (11,540) (.9961) = 11,500 psi (Tx
= k
=
+n
(10,450)
f
+
(11,500)
= 21,950 psi, compression
(2030) (200 X 12) = 846 lbs /Iinear in.
4.4-6
/
Girder-Related Design
resultant radial compressive stress in web
/
(846) (1 X 17ia)
= 100'
r
1230 psi, compression
8=?--t FIGURE 14
This produces the final stress condition of: At this point:
O"x
=
O"h
& F;
=F
h
stress in web or lower flange from bending moment «,
Mc = -1-
(55,OOO'k X 12) (126) 3,979,000 20,900 psi, compression FIGURE 13
average stress in lower flange from bending moment critical stress Using the Huber-Mises formula:
Mc
O"t
= -1(55,000'k X 12) (127%a) 3,979,000
=
= '" (-21,950)2-( -21,950) (-180) +( -180)2 +3( 1020)2 = 22,000 psi
21,150 psi
force in lower flange from bending moment This results in an indicated factor of safety against yielding of-
FlO
=
O"t At
(21,150) (20/8 X 36) FS =
= 2000 kips
O"y O"cr
( 42,(00) (22,000)
1.90
Analysis of Fish Belly Haunch Now using the same load conditions on the fish belly haunch with the same web and flange dimensions:
radial tensile force of lower compression flange against web f
=
Fx r
(2000) (100 X 12) 1670 Ibs/Iinear in.
Bridge Plate Girders With Variable Depth
resultant radial tensile stress in web
f l1 = d~ Ibs/In,
=tw
( 1670) (1 X lYIO)
Where the Range slopes, the modified vertical shear (V') must be used. The shear component along the slope will be-
= 2420 psi
f
shear stress in web
x
V
T
__ f l1_ cos (J
-
but the distance along this slope for every horizontal inch is-
= Aw -
4.4-7
Approximate value:
f
U'7
/
( 1200) (252 X 1 YI 0
)
Sx
= 6930 psi combining stresses to find the critical stress ay =
I" = cos (J
so that the shear force on the weld along this sloping Range is obtained from the above formula for the horizontal Range, using the modified value of V':
+ 2420 -... T
= 6930
.--L..-...,~
ax
= - 20,900
FIGURE 15 FIGURE 16
Using the Huber-Mises formula:
= '\I (20,900)"-( -20,900) ( +2460) + = 25,100 psi
r,
(+2460)2+ 3( 6,930)2
f
fx Sx
cos
(J
SI1
cos
r, Sll
(J
This results in an indicated factor of safety against yielding of-
F.S. -
sr, U'er
-
or the approximate:
( 42,000) (25,100)
= 1.67 It is apparent from this that the parabolic haunch has a slightly lower critical stress and, therefore, a slightly higher factor of safety.
3. WELDS CONNECTING SLOPING FLANGE TO WEB In any girder, the horizontal shear force in the connecting weld between the web and the horizontal Range is found from the following formula:
f =
Vay rn-
lbs/In.
I
f =
V' ern
Ibs/in.
where:
f = shear force on weld, lbs/linear in. external shear on the section, lbs. V modified shear on section if sloping Range, lbs V' area of Range held by weld, in. 2 a y= vertical distance between center of gravity of Range held by weld, and neutral axis of section, in. I moment of inertia of section, in." n number of welds connecting web to Range d distance between C.J. of Ranges, in.
4.4-8
/
Girder-Related Design
Erection view of New York State Thruway bridge shows haunched girders. Straightness and true camber of the lower flanges are apparent. Note vertical stiffeners and suspended (235') span bearing surfaces at girder junctions.
Portion of 295' span of bridge on Connecticut Turnpike being settled onto supporting piers. Note continuous parabolic haunched girder construction.
SECTION 4.5
Girders on a Horizontal Curve
1. RECENT PROJECTS Today, it is accep te d practice to design an d fabricat e plate girders with horizontal curves when n ecessary . Several such bridges or freew ay ove rpasses have b een built within th e pa st several years. • A ser ies of 4 lines of curve d we lde d pl at e girde rs with gO' spans are a p art of th e Pa sad en a-Golden State Freeway's interchange in th e Lo s Angeles area, Figure 1. These h ave a curve radius of 400' . They w er e Iabricated in Kaiser Steel's pl ant at Monteb ello. • On e of Milwaukee's n ew expressways has a section of 4 con tinuous spa ns with a tot al lengt h of 345' in which th e two outer girders hav e a go h orizontal curve and th e 2 inn er girders ar e straight. • Bristol Steel & Iron Works, Bristol , T ennessee, recently fabricated several curved girde rs for th e Southwest Freeway-Inner Loop in W ashington, D. C.
Curved flan ge pl at es are laid out by offsets and flam e cut from plate. By cutt ing both edge s at th e sam e tim e, th er e is no bo win g from any unbalanced shrinkag e effect of th e flam e cutting. The web plates do not have to b e prefo rm ed , usually b eing easily pulled into alig nme nt along th e centerline of th e flang es. Caution mu st b e used in placing attaching plates for th e diaphragms to th e webs and flang es. The proper an gle for th ese plates ma y vary along the length of th e girder. Shear attachme nts ar e added mainly to accomp lish com posite action b etween th e concrete deck and steel girde r, and th er eby increase torsional rigidity. Durin g erection, a pair of curved girders is usuall y attache d together by means of th e diaphragm s and th en hoisted into po sition as a unit.
2. DESIGN AND FABRICATION
Although th er e ar e torsional stre sses w ithin th e curved girder, usually the degree of curva ture is not overl y high and th ese additional stresses are offset by the diaphragms connecting th e girders. The number of diaphragms has occasionally b een in cr eased for thi s reason, and sometimes the allo wable stresses have b een reduced slightly.
FIG. 1 Welded plate girders, having a 400' radius of curvature, dominate the interest in Los Angeles interchange of Pasadena-Golden State Freeway. Curving girders permit economies in deck system by keeping overhangs uniform from end to end of curve. \ ',
',
\
\\ \
•
4.5-1
4.5 -2
/
Girder-Related Design
FIG. 2 Bridge plate girders being weld fabricated. With flanges flame-cut on a curve, weight of the rolled web is utilized in making it conform to desired radius.
FIG. 3 A two-span continuous box girder and curved ramp construction provided the answer to space limitations in reaching elevated parking area at busy New York terminal complex. Smooth, clean lines, without outside stiffeners, demonstrate aesthetic possibilities inherent in welded design.
SECTION 4.6
Tapered Girders 1. FABRICATION AND USE The use of tapered girders has become widespread, especially in the framing of roofs over large areas where it is desirable to minimize the number of interior columns or to eliminate them altogether. They permit placing maximum girder depth where it is needed, while reducing the depth considerably at points where it is not needed. Tapered girders are fabricated either 1) by welding two flange plates to a tapered web plate, or 2) by cutting a rolled WF beam lengthwise along its web at an angle, turning one half end for end, and then welding the two halves back together again along the web. See Figure l.
Camber When Required Camber can be built into the tapered girder when required. When the girder is made from WF beams, each half is clamped into the proper camber during assembly. Then the butt joint along the web is groove welded while the girder is held in this shape. Since the weld along the beam web lies along the neutral axis, no bending or distortion will result from welding, and the girder will retain the shape in which it is held during welding. When the girder is made of two flange plates and a tapered web, the proper camber can be obtained by simply cutting the web to the proper camber outline. The flange plates during assembly are then pulled tightly against the web, into the proper camber. The four fillet welds joining the flanges to the web are balanced about the neutral axis of the girder and as a result there should be no distortion problem.
span design, the central span can use the tapered flange up, forming the slope of the roof; the two adjacent spans use the tapered flange down to provide a Hat roof, but tilted to continue the same slope as the central section. The problem of lateral support for the top compression flange of tapered girders is no different than with other beams and girders. Generally the roof deck is sufficiently rigid to function as a diaphragm, and it's only necessary to attach the deck to the top flange. There's apparently no advantage in designing with a reduced stress allowable, in accordance with AISC Formulas 4 or 5, in order to permit a greater distance between bracing points at the top flange. Where tapered girders are critical, Section 5.11 on Rigid Frame Knees goes into more detail relative to stresses (elastic design). Because of the reduced depth at the ends of the
En-
-~cutI
f----r ! d
I E=--= = ---~cut = L=-------}m/d~
Application of Tapered Girders When the tapered girders are used with the sloping flange at the top, their taper in both directions from the ridge will provide the slope needed for drainage. By varying the depth at the ends of successive girders, the deck can be canted to drain toward roof boxes in the valleys between adjacent gabled spans and at flanking parapet walls. For Hat roofs, the girders are inverted, with their tapered flange down. There are many combinations of roof framing systems possible. For example, on a three-
FIGURE 1
4.6-1
4.6-2
/
Design of Welded Structures
(a) Conventional beam
(b) Tapered girder FIGURE 2
........L..J..J..J...L.J...l..I..J...L.J...l..I.....L...L.L..J...J....L...L.L..LJ....L...L.L..J...J....L...L.L..LJU.J>.
Curve of required section modulu s (5) ha s sa me shape a s moment diag ram for uniform load on simply supported beam
Moment diagram
(a) Conventional beam
(b) Tapered g ird e r FIGURE 3
tapered girders , th eir connec tion to supporting columns may offer little resistance to horizontal forces. For thi s reason, some knee braces may be required unless the roof deck or a positive syst em of bracing in th e pl an e of the roof is stiff enough to transmit th ese forc es to ad equately braced w alls. At first glance, th ere appears to b e quite a w eight saving in tapered girder ; however , thi s is not always as great as it might seem:
First, th e flang e area remains the same; th e only weight saving is in the w eb. See Figure 2. Second, th e depth of th e tapered girder at midspan mu st b e increased over that of th e conventional straight beam to be sufficient at the critical section (about ¥4 span ). This is necessary to develop the required section modulus along th e full length of the tapered girder. This wiII slightly offset th e initial weight saving in th e web . See Figure 3.
FIG. 4 For flat roofs, tapered g ird ers are used inverted, with ta pered fla ng e downward . Freq ue ntly the g irder is filte d to p rovid e a slope to the roof or roof se ction.
Tapered Girders
/
4.6-3
2. DETERMINING CRITICAL DEPTH AND SLOPE
Th e critical depth section of a tapered girder is that section in which th e actua l depth of th e girder just equals th e minimum depth required for th e mom ent. It would be th e highest stressed section of th e girde r in bending. In th e case of a uniformly loaded , simply supported girder, its sloping flange must be t angent to th e r equired-depth curve at this point in order for th e b eam to have sufficient depth along its length. Setting th e slope of th e tap ered girde r flange so that the critical section is located at th e % span will result in about th e minimum web weight. See Fi gure 5. Th e properties of this critical section are-
I -S
=
t 2 A((d 2)
+
At d t2
l d b2
---a;;-
~ 79.8%
~ .30 L
78.6% I
I
.25 L
I
w 12d
I
- --- - - - - - - -
.20 L
t w d.,
6 db
FIGURE 5
This formula for section modulus can be simplified with little loss in acc uracy, b y letting-
b.lw"oT
e.G. of flanges)
df
dw
1
____ ----OJ 85.5%
~80 ~
tw
Af [depth
c-------_
db
(de h lt (overall of web) de pth)
177';'77~~7'7;~
FIGURE 6
FIG. 7 Tapered girders used with the tapered flange at the top provide for roof drainage in both directions from the ridge . Multi-span designs often call for combinations of girders having tapered flange up and others having tapered flange down.
=
dw
d,
I
At d w
S
db
+~
I·
(1)
If th e sec tion modulus required to resist the bending mom ent is known, th e required b eam depth (d ) is solved for:
6 S tw
o
oIlo
t TABLE 1 2
3
.
5
cone. loads
cone. loads
cone. loads
cone. loads
~
~
~
t ***** t
cone. load
f critical depth
depth dw
t
01
1/2 L
01
crilical
1
~c:tr +
slope
3 Pl 21w . For any angle between these two values, the weight of the web will remain the same
4.6--6
/
Girder-Related Design p
"~
For fully rigid connections the actual moments must be found by one of several methods, and the beams and their connections designed for the proper moments and shear forces. The connections must have sufficient rigidity to hold virtually unchanged the original angles between connecting members. The rigidity of a connection is also influenced by the rigidity of its support. For beams framing into column flanges, a decrease in rigidity will occur if the column flanges are too thin, or if stiffeners are not used between the column flanges in line with the beam flanges. For a single beam framing into a column web, a decrease in rigidity may occur unless the beam flange is also welded directly to the column flanges or attached with suitable connecting plates.
5. PLASTIC-DESIGN CONNECTIONS The use of welded connections based on plastic design has several advantages: 1. A more accurate indication of the true carrying capacity of the structure. 2. Requires less steel than conventional simple beam construction. In many cases, there is a slight saving over conventional elastic design of rigid frames. 3. Requires less design time than does elastic design of rigid frames. 4. Tested by several years of research on full-scale structures. 5. Backed by the AISC. So far, plastic design connections have been largely restricted to one-story structures, and to applications where fatigue or repeat loading is not a problem. See separate Sect. 5.12 in this manual for a full discussion of Welded Connections for Plastic Design.
6. BEHAVIOR OF WELDED CONNECTIONS One way to better understand the behavior of a beam-to-column connection under load, and its loadcarrying capacity, is to plot it on a moment-rotation chart; see Figure 2. The vertical axis is the end moment of the beam,
5.1-4
/
Welded-Connection Design
\ \
\ \ \ \ \ \ \ \ CQ> E
o
E -0
c:
w
\ ~ Beam line at 1~ working load
'\
\
~ \
\
Beam line at working load
_-----0
\
End rotation (8e), radians
FIGURE 2
which is applied to the connection. The horizontal axis is the resulting rotation in radians. Basically this is another type of stress-strain diagram. Superimposed upon this is the beam diagram. The equation expressing the resulting end moment (Me) and end rotation ((Je), for a uniformly loaded beam and any end restraint from complete rigid to simply supported, is: 1
M e -_ _ 2EL ~!J__ WLI 12
This is a straight line, having points a and b on the chart. Point a is the end moment when the connection is
completely restrained ((Jo = 0), in other words a fixed-end beam, and is equal to( a) Me =
WL ------rr-
Point b is the end rotation when the connection has no restraint (M,. = 0), in other words a simple beam, and is equal to(b)
(Je
=-
WU
24 E 1
For increased loads on the beam, the beam line moves out parallel to the first line, with correspondingly increased values of end moment (Me) and the end rotation ((}e). This (dashed) second beam line on the
Beam-to-Column Connections
chart represents the addition of a safety factor, and is usually 1.67 to 2 times that of the first which is based on the working load. The point at which the connection's curve intersects the beam line, gives the resulting end moment and rotation under the given load. From this it is seen how the beam's behavior depends on its connection. It is assumed, in this case, the beam is symmetrically loaded and the two end connections are the same. In this way both ends will react similarly. Curve 1 represents a flexible connection. At a very low moment it safely yields (M 1 ) and allows the connection to rotate (0 1 ) , This is typical of top angle connections, web framing angles, and top plate connections small enough to yield. Notice, even with these so-called flexible connections, some end moment does set up. Curve 2 represents a semi-rigid connection. One type is the top connecting plate so detailed that under working load it elastically yields sufficiently to provide the necessary rotation of the connection, and yet has sufficient resistance to develop the proper end moment. Although thick top angles have been suggested for service as semi-rigid connections, they are impractical to design and fabricate with the desired built-in restraint. Curve 3 represents a rigid connection, using a top connecting plate detailed to develop the full end moment. Since no elastic yielding is needed or desired, the plate is made as short as practical. All three of these connections have ample reserve carrying capacity, as shown by where their curves inter-
Case a
M(. _
-
W L
8
18" 'IF 85#
W== 139 ~ fl-uuqUU ~-----15' k
Simply supported beam designed for R == 0
Case b
M. == -
WL
12
18" 'IF 85#
~ Fixed end beam designed for R == 100%
FIGURE 3
/
5.1-5
sect the beam line at 1% load relative to their crossing of the beam line at working load. The actual results of testing three top plate connections on an 18" WF 85# beam are shown in Figure 4. Two conditions are considered, as shown by the load diagrams, Figure 3. Beam line a (in Figure 4) is based on a design moment of 1Js W L at centerline, i.e. simply supported. Beam line al is for a load Ph times that of the working load. Beam line b is based on a design moment of Yt2 W L at the ends, i.e. fixed ends, and will support a 50% greater load. Beam line b, is for a load 1% times that of the working load. Both of these two beam lines stop at R = 50%, because at this restraint the center of the beam now has this moment of Yt 2 W L and a restraint lower than this would overstress the central portion of the beam. Top plate # 1 is a 'X 6" thick plate, 3" wide at the reduced section, and has a cross-sectional area of All = .94 in." It is widened to 6" at the butt-welded connection. This connection should reach yield at about M = All a, db = (.94) (33,000) (18) = 558 in.-kip. The actual value from the test is about M = 600 in.-kip. Above this moment, the plate yields and due to strain hardening will have increased resistance. The ultimate moment should be about twice this yield value, or about M = 1200 in.-kip. The resulting restraint is about R = 34.5%, a little too high for the beam to be classed as simply supported. Top plate #2 has the same %6" thickness, but has a 6" width throughout its length. It has double the cross-sectional area, All = 1.88 in. 2 As expected, it is twice as rigid. It should reach yield at about M = 1110 in.-kip. The actual is about M = 1000 in.-kip. The restraint is about R 58%. Notice if the beam had been designed for a moment of "Yt2 W L, i.e. a restraint of R = 100%, the connection's curve would have intersected the beam line b just short of the R = 50% value. There would then be a slight overstress of the beam at centerline. Top plate #3 is 'l's" thick and 7W' wide, having a cross-sectional area of All = 6.56 in. 2 This greater area produces a more rigid connection with greater restraint. The actual connection curve (solid) shows slightly more flexibility than the calculated curve (dotted). The extra flexibility probably comes from some movement of the lower portion of the connection which has just short parallel fillet welds joining the lower flange of the beam to the seat. A butt weld placed directly across the end of this lower flange to the column, undoubtedly would bring the rigidity of the connection curve up almost to that of the calculated curve.
=
5.1-6
/
Welded-Connection Design
WEB DF CflLUM'"
BUT7 WILD
F#.~~~~~~~~~'-"'-9 "LL67
WELD
6DOO
r
j"lf7
70P
I-
AI' • 1#.5" IN a
\ \
\~ \
.011'
.tHD
"II.
..,.
.0/.
.o/a
.0.0
ROTATION (8e ), RADIANS
FIGURE 4 Figure 4 from: "Tests of Miscellaneous Welded Building Connections"; Johnston & Diets; AWS Welding Journal Jan. 1942 and "Report of Tests of Welded Top Plate and Seat Building Connections"; Brandes & Mains; AWS Welding Journal Mar. 1944
Figure 5 illustrates the additional restraining action provided by column Bange stiffeners. Both connections use 'YIn" x 6" top plates. Connection # 1 has column stiffeners. In the case of the beam designed for a moment of 7{2 W L (R = 100% down to R = 50%), it would supply a restraint of about R 70.2%. Connection #2 has no column stiffeners and loses sufficient rigidity so that the beam designed for a moment of %:l W L (R = 100% down to R = 50%) will be overstressed. This is because the connection restraint would be only about R = 45%. This shows the importance of proper stiffening.
=
7. FACTORS IN CONNECTION DESIGN The following items greatly affect the cost of welded structural steel and cannot be overlooked. In order to take full advantage of welded construction, they must be considered.
Moment Transier The bending forces from the end moment lie almost entirely within the Banges of the beam. The most effective and direct method to transfer these forces is some type of Bange weld. The relative merits of three types are discussed here.
Beam-to-Column Connections
/
5.1-7
'"WMN FLANGE $TIFFE/'IU.s
®
ROTIITION
(Be).
IB"W"
Itrw
RADIANS
FIGURE 5 Figure 5: From "Tests of Miscellaneous Welded Building Connections"; Johnston & Deits; AWS Welding Journal, Jan. 1942
In Figure 6, the flanges are directly connected to the column by means of groove welds. This is the most direct method of transferring forces and requires the least amount of welding.
.
.
FIGURE 6
The backing strip just below each of the flanges allows the weld to be made within reasonable fit-up, as long as there is a proper root opening. There is little provision for over-run of the column dimensions which may be as much as .± l/s". For excessive over-run, the flanges of the beam may have to be flame-cut back, in the field, in order to provide the minimum root opening. For under-run, the excessive root opening will increase the amount of welding required, but the joint is still possible. It is usually more costly to cut the beam to exact length; in addition there is the cost of beveling the flanges. :Milling the beam to length is costly and not recommended because the over-run or under-run of the column having a tolerance of .± 118 " would reduce this accuracy in fit-up.
5.1-8
/
Welded-Connection Design
In Figure 7, a top connecting plate is shipped loose and, for proper fit-up, is put in place by the weldor after the beam is erected. A greater tolerance can be allowed in cutting the beam to length, and any method can be used (circular cutoff saw, Harne-cutting,
.
.
FIGURE 7
etc.) without subsequent beveling of the Ranges. The beams frequently are ordered from the steel supplier cut shorter than required: %" ± If4". Sometimes beams are ordered still shorter, allowing a cutting tolerance of ± %". This greatly reduces the cost of cutting and preparation. This type of connection requires the extra connecting plate, which must be cut to size and beveled. It doubles the amount of field welding on the top Range. It also can interfere with metal decks placed on top of the beam. Occasionally, the top plate is shop welded to the top Range on one or both ends of the beam. This decreases the amount of field welding but eliminates the fit-up advantage. The beam's bottom Range may be groove welded directly to the column if sufficient root spacing is obtained, even though the edge of the Range is not beveled. Although this is not an AWS Prequalified Joint, it is widely used, perhaps because the bottom Range weld is in compression. One disadvantage is that the beam length must be held accurately. As the beam Range increases in thickness, the required root spacing must increase. The bottom seat also serves as a backing strip. Sometimes for additional strength, the Range is fillet welded to this plate for a short distance along its
FIGURE 8
length. All of this field welding is done in the Rat position. In Figure 8, the lower Range is not groove welded directly to the column; instead, the bottom seat plate is extended farther along the beam, and is fillet welded to the beam Range. These welds are designed to transfer the compressive force of the Range back into the column. All of the field welding is done in the Rat position. This connection requires a little more care in handling and shipping so that these longer plates are not damaged. This also requires a little more weight of connecting material. A beam that is "compact" (AISC Sec. 2.6) permits a 10% higher bending stress,
L,
I'"
>
FIGURE 3
5.4-4
/
Welded-Connection Design resultant force on outer end of connecting weld
In Figure 3, analysis of the shop weld shows-
1Y
n
2b
+ L,
I
/ /
If"f("'09'h of weld
/
~@ (2b
r
/
/ /
•
+ Lv)3
/f
/
/ /
12
FIGURE 4
twisting (horizontal) - T c,. _ R (Lb - n) Cv fb - T - 2 Jw
. (3 ) leg size of fillet weld
twisting (vertical) f v!
=r=
~
(L'J:
n)~
actual force on welds allowable force
en =
(4)
A7, A373 Steel; E60 Welds A36 Steel; E70 Welds
shear (vertical) f V2
=
en
R/2 2b +L
(5) v
=
fr
en =
9600
f,
(7)
11,200
Unfortunately there is no way to simplify these
TABLE 2-Maximum Leg Size to Use in Calculating Vertical Length of Weld FOR VARIOUS COMBINATIONS OF WELD METALS AND STEEL Given these conditions: Steel
A7 A373
A242, A441
A36
thickness
Over 1112" To 4 11
Over %" To l'h"
%"
or less
dy
33,000
36,000
42,000
46,000
50,000
T
13,000
14,500
17,000
18,500
20,000
weld f wit ~
E60 or SAW-l 9,600 w
.667
E70 or SAW·2 11,200 w
.648
E70 or SAW·2
E70 or SAW·2
E70 or SAW·2
11,200 w
11,200 w
11,200 w
.759
.826
.893
Then: Maximum leg size of fillet weld to use in calculating vertical length leg size w
Web thickness (t w ) over -
114"
.375
.386
.329
.325
.280
~6"
.468
.482
.412
.378
.350
¥a"
.562
.579
.494
.454
.420
K6"
.656
.676
.576
.529
.490
112"
.750
.772
.659
.605
.560
U6"
.844
.868
.741
.681
.630
¥a"
.937
.964
.824
.756
.700
FIGURE 5---Size of Shop Weld of Framing Angles For A36 Steel & E70 Welds
NOMOGRAPH NO. 7
F L-30"
MIN, WEB THICKNES.s TO HOLD .sHEil/? STRESS
LEG 5/ZE OF fILLET WELD CV
-,
BELOW 14J500
r 7 •
'&
L'Z""
f'.51
/
L'Z4'
SHOP WELD
/~ --/~
L 'ZO'
I"
R
--/~
L 'If,"
Ito:
.9
7"
.8
..
-'%
s.: ~
S·
L'14"
tee
--~
14C L'lr
IN
.7
Jri
1.'18"
KIPS
:lee
~.
~.
L'Zl"
3 "
~.-[--/~" r
~.
L 'ur f----
ICC
-.6
30
5 ..
1"8
80
"
,-10"
L-9"
-fO
.5-:Z.
~
-" .>
""
7:J/" "
--~i'
4
""
""
""
""
"
""
I.-~"
50
IT
..
"
~. ~/>"3-&" -~~6'
."
30
a
L'~"
3 :::I Ia
1..'5"
20
PIlO8LEM: FINO JIZE Of SHOP WELD fOR THE FOLLOWING FRAME ANGLE. L.. • 3" (LEG SIZE OF ANGLE) L • ie: (LENGTH OF ANGLE) R • 58 KIPS (END REACTION) READ"'" Y.$. (SIZE OF SHOP WELO)
~
t1)
c-r:
40
~
-
L'~"
c
:::I Ia
if ........
I
2 1" 2
J. Ln • LEG
Ji" SIZE OF ANGLE
4"
VI
t
5.4-6
I
Welded-Connection Design
I
Wmcll.
= t
I
I
Wm o x
1/
16" l
I
Wmcll.
I
== t
I
----.i / 16 "
~t~ ~t~I~t~6±
~tESI Less than
== t -
V. thick"
If edge is built up to ensure full throat of weld
W' thick or more
FIGURE 6
formulas into one workable formula. It is necessary to work out each step until the final result is obtained. The leg size of this shop weld may be determined quickly by means of Nomograph No. 7 (Fig. 5), for A36 steel and E70 welds. In the chart on the right-hand side, from the point of intersection of the angle's horizontal leg length (L h ) and its vertical length (Lv) draw a horizontal line to the vertical axis F-F. From this point, draw a line through the reaction (R) to the left-hand axis. Read the leg size (w) of the shop weld along the left-hand scale of this axis. If the nomograph is used from left to right to establish an angle size, be sure that the leg size of the fillet weld does not exceed a value which would overstress the web of the beam in shear (AISC Sec 1.17.5) by producing too short a length of connecting weld (Lv). The following limits apply to the fillet weld leg size (w) relative to the thickness of the beam web (as used in calculating the vertical length of connecting weld) : A7, A373 Steel and E60 Weld
equal to or exceeds this value found just opposite the resulting leg size of the weld. Some engineers feel this limiting shear value (A36 steel, T = 14,500 psi) is to insure that the web of the beam does not buckle, and that a higher allowable value might be used here, perhaps 3/4 of the allowable tensile strength. In this case the maximum leg size of the weld would be held to 3/4 of the web thickness.
Iw =
w < tw 2
13,~0 X
9600
;16
'A 3/16
'A
'A :'/16
'A /16
14 14 14
3x3x%
3x3Xo/Hi
.41 .33 .25
.32 .26 .19
.3Q .24 .18
92.9 74.3 55.7
12 12 12
3X3x 7/ 16 3X3X% 3X3Xo/Hi
.41 .33 .25
.32 .26 .19
.30 .24 .18
78.3 62.6 47.0
Vlo
10 10 10
3X3x'iib 3X3X% 3X3xo/16
.41 .33 .25
.32 .26 .19
.30 .24 .18
64.0 51.2 38.4
0/16
9 9 9
3X3X'l'16 3X3X% 3X3X'Y16
.41 .33 .25
.32 .26 .19
.30 .24 .18
57.1 45.6 34.2
8 8 8
3x3X'l'Hi 3x3x¥a 3X3xo/I6
.41 .33 .25
.32 .26 .19
.30 .24 .18
50.2 40.1 30.1
%, 'A
7 7 7
3X3X~'i6
.41 .33 .25
.32 .26 .19
.30 .24 .18
43.5 34.8 26.1
5/16
3X3X%
6 6 6
.41 .33 .25
.32 .26 .19
.30 .24 .18
37.0 29.6 22.2
5/16
3X3X%
5 5 5
3X3X'l'16 3x3X¥a 3x3x 5/ 16
.41
.33 .25
.32 .26 .19
.30 .24 .18
30.7 24.5 18.4
4 4 4
3X3X 1/ 16 3X3X% 3X3xo/I6
.41 .33 .25
.32 .26 .19
.30 .24 .18
24.6 19.7 14.8
3X3Xo/I6 3x3X'l'16 3x3xo/16
~/1I5
'A
:>;16
'Capacity Kips
'0.
"Minimum Web l tuckness for Welds A
A36
A242 and A441
F.=14.5
F,~18.5
F,=20.0
32 32 32
4X3X'l'16 4X3X% 4X3Xo/16
.48 .39 .29
.38 .30 .23
.35 .28 .21
30 30 30
4x3X'l'16 4X3X¥a 4X3X 5/ 16
.48 .39 .29
.38 .30 .23
.35 .28 .21
% 0/16
28 28 28
4x3x 11 16 4x3x¥a 4x3X 5/ 16
.48 .39 .29
.38 .30 .23
.35 .28 .21
%
26 26 26
4X3x'l'16 4x3x% 4x3X 5/ 16
.48 .39 .29
.38 .30 .23
.35 .28 .21
24 24 24
4x3x 1/ 16 4x3x¥a 4x3x 5/ 16
.48 .39 .29
.38 .30 .23
.35 .28 .21
22 22 22
4X3X 1/]6 4x3x% 4X3x 5/Hi
.48 .39 .29
.38 .30 .23
.35 .28 .21
20 20 20
4X3X 1/ 16 4x3x% 4X3x 5/ 16
.48 .39 .29
.38 .30 .23
.35 .28 .21
18 18 18
4X3X 1116 4x3x% 4X3x"'llti
.48 .39 .29
.38 .30 .23
.35 .28 .21
16 16 16
3x3x 7/ 16 3X3X% 3x3x'V16
.48 .39 .29
.38 .30
.23
.35 .28 .21
14 14 14
3x3x 7/ 16 3X3X% 3X3X 5/ 16
.48 .39 .29
.38 .30 .23
.35 .28 .21
12 12 12
3X3X 11 16 3X3x% 3X3XVI6
.48 .39 .29
.38 .30 .23
.35 .28 .21
10 10 10
3x3X'l'16 3x3x% 3X3X'V16
.48 .39 .29
.38 .30 .23
.35 .28 .21
9 9 9
3x3x 7/ 16 3X3X% 3X3xV16
.48 .39 .29
.38 .30 .23
.35 .28 .21
8 8 8
3x3x 7/ 16 3x3x% 3x3x 5/ 16
.48 .39 .29
.38 .30 .23
.35 .28 .21
7 7 7
3x3x'l'16 3x3x J/a 3x3x 5/ 16
.48 .39 .29
.38 .30 .23
.35 .28 .21
6 6 6
3x3x 7/16 3X3x% 3x3x 5/ 16
.48 .39 .29
.38 .30 .23
.35 .28 .21
5 5 5
3X3X~/115
3x3xYa 3x3X 5/16
.48 .39 .29
.38 .30 .23
.35 .28 .21
4 4 4
3X3x 7/ 16 3X3X% 3x3X116
.48 .39 .29
.38 .30 .23
.35 .28 .21
245 204 163
%
227 189 151
%
209 174 139 191 159 127
0/10
173 144 115
V16
155 129 103
5/10
136 114 90.9
5/10
5/16
'A 5/16
%
'A
'A 'Is 'A %
'A %
'A
118 98.4
Sjlo
78.7
'A
III
'0.
"Sue
Angle Size (ASTM A36)
%
%
92.9 74.3
5/16
93.1 77.6 62.1
%
'A 5/16
'A I
UWhen a beam web IS less than the rn.ntmom. mulfiply the connection capacity furnished by welds A by the rauc of the actual ttucknes s to the tabulated minimum ttnckne s s. Thus, It ~/Ifi~ weld A, wtth a connection capacity of 54.9 kips and a 10" long angle, is being considered for a beam 01 web thickness .270", ASTM Al6, the connection capacity must be multiplied by .270/.41, giving 36.l kips. ~When beam material is ASTM A7 or AlB. wllh F r .. 13.0 ksi, minimum web Ihicknessesto develop ~/lfi", '/4· and 1/,6· welds A are 46", .37" and .28" respectively. (Should the thickness 01 material to whrch connection angles are welded exceed the limits set bV AISC Specrbcetron. Sect 1.17.4. for weld sizes sp ecrfie d. Increase the weld size as required. but not to exceed the angle thickness. '/For welds on outstanding legs, connection capacity may be limited by the shear capacity of the See exarnntes (d) supporting member as stipulated by AISC Specttrcauon, Sect. 1.17.5. and (e), pages 111·26,4·27.
'A 0/10
'A '/Io 5/16
'A 116
3/16
'A 3116
'A 3116 5/16
%
3/]6 5/16
'A
YI6
74.9 62.5 50.0
5/16
57.1 47.6 38.0
0/16
48.4 40.4 32.3
5/16
%
'A ji,
'A
%
'A
40.0 33.4 26.7
5/16
32.0 26.7 21.3
5/16
24.5 20.4 16.3
V16
17.6 14.7 11.8 11.6 9.7 7.8
%
'A %
'A ¥, 'A % 5/16
'A %
0/16
'A
I ,
"When a beam web is less than the rnrrurnu ni, multiply the connection capacity furnished by welds A by t he ratio of the actual ttuc koess (0 the tebutatec minimum thrc s ne ss Thus. if ~(16" weld A, with a connection capacity0150.2 kips and an 8"long angle, is being con stder e d for a beam of web thickness lOS". ASrM A36, the conneclton capacity ruu st be multiplied by .305/.48, grvmg 31.9 kips. hShould the thickness of material to which connection angles are welded exceed the limits set by AISC Specilication, Sect 1.17.4. lor weld Slzes apecrt-ed. increase the weld size as required. but not to exceed the angle thickness. (For welds on outstanding legs, connection cepecrtv may be urn.te d by the shear capacity of the See exe mptes (d) supporting members as stipulated by AISC Specrttcanon. Sect. 1 17.5 and {e). pages 4-26. 4·27. Note 1: Capacities shown In ttus table apply only when matena! welded IS ASTM A36, A242 or A441. Use appropriate capactte s from Table V when beam or supporting rneterterrs ASTM A7 or A373.
Web Framing Angles
of 38.4 kips for a weld size of CI) = %6" and angle length of L, = 10" slightly exceeds the reaction. The corresponding (Field ) Weld B, using CI) = 1/4", also is satisfactory. Since the beam's required web thickness is 0.21" while the actual web thickness is 0.25", the indicated 3" x 3" X %6" is all right. If the beam is made of A36 steel, this connection's capacity will be reduced in the ratio of 0.25/0.29 of actual to required web thickness. The resulting capacity of 33.1 kips is less than the reaction. The next larger connection with apparently sufficient capacity shows that (Shop) W eld A's capacity is 47 kips, using same angle section but an angle length of L, = 12". Applying the multiplier of 0.25/0.29 reduces the capacity of the connection to 40.5 kips, which exceeds the end reaction.
/
5.4-9
R
FIG. lO-Double-web framing angle.
q I
I
I I
I I I I
I I I I
5. SINGLE-PLATE OR TEE CONNECTION ON BEAM WEB
I I
:' I I I I
In the previous design of the field weld, connecting a pair of web framing angles to the supporting column or girder, it was assumed that the reaction (R) applied eccentric to each angle, resulted in a tendency for the angles to twist or rotate. In doing so, they would press together at the top and swing away from each other at the bottom, this being resisted by the welds. These forces are in addition to the vertical forces caused by the reaction (R); see Figure 10. However, in both the single-plate web connection and the Tee-section type, this portion of the connection welded to the column is solid. Thus, there is no tendency for this spreading action which must be resisted by the welds. These vertical field welds to the
I I
I I
, R
FIG. ll-Single plate or Tee.
column would be designed then for just the vertical reaction (R); see Figure 11. In the shop weld of the single plate to the web of the beam, Figure 12, this double vertical weld would be designed for just the vertical reaction (R). There is not enough eccentricity to consider any bending action.
Field weld to supporting column or web of supporting girder
~t:~~~:~i
~~'?"'7"~
\
, I I
L.
:
.J
Q)
Flot plate used for flexible connection on web of beam
FIG. 12-Flat plate used for flexible connection on web of beam.
5.4-10
/
Welded-Connection Design
r - - - - - --, I I
, ,
: Beam web
:
I
,
I I
I
IL
0
...JI
Tee section used for flexible connection on web of beam
FIG. 13-Tee section used for flexible connection on web of beam.
In the shop weld of the Tee connection to the web of the beam, Figure 13, the size and length of the fillet weld would be determined just as in the case of the double-web framing angles, except there is just a single fillet weld in this case rather than two; so, for a given connection, this would carry just half of the reaction of the corresponding double-angle connection.
FIGURE 14
fillet weld in shear; parallel load 6. DIRECTLY-WELDED WEB CONNECTION
2 ( 9600w ) L = t w 13,000 L
To see how this type of connection behaves, consider the following 18" WF 85# beam, simply supported, 15' span, with a uniformly distributed load of 139 kips, the same beam and load used in the general discussion on behavior of connections in Sect. 5.1, Topic 6. If only the web is to be welded to the column, the weld must have sufficient length (Lv) so that the adjacent web of the beam will not be overstressed in shear. For A373 steel T
= .40
O'y
13,000 psi
R
FIGURE 15
fillet weld in tension; transverse load .526"
( 69.5 k ) ( .526") (13 ksi) 10.2", or use 11" The leg size of this fillet weld must be equal to the web thickness, based upon standard allowables, if it is to match the allowable strength of this web section in shear as well as tension.
2(9600w)L = t w 20,000 L
*1w =
tw
I
* Actually, transverse fillet welds are about lh stronger than parallel fillet welds; this can be proved by theory as well as testing. This means for transverse loads, the leg size would be % of the plate thickness, just as in parallel loads. However, welding codes do not as yet recognize this; and for code work, fillet welds for transverse loads would be made equal to the plate thickness.
Web Framing Angles A
•
T
r-, "- W,V
~ A
;;-
"--FIGURE 16
If there is a gap between the beam and the column, the leg size of this fillet weld is increased by this amount. The moment-rotation chart, Figure 17, shows the beam line for this particular beam length and load, and the actual connection curve taken from test data at Lehigh University. In testing this connection, the beam web showed initial signs of yielding adjacent to the lower ends of the weld at a moment of 360 in.-kips. At a moment of 660 in.-kips, point (a), there were indications that the beam web along the full length of the weld had yielded. At a moment of 870 in.-kips, both welds cracked slightly 2500
--+ 13
I-
?l6"
I?"
c..
2000
-¥
"f"~ '0
FIGURE 18
~
I~ IcP
X 3" reduced It.
1-------
1918 In-kips mox
J
18" we 85# beam
\0-
",; 1500
c Q)
E E 1000
o
--(-t'/I--~ ~ I
@~
f\
"I
\
I
I
J~Firstin crack weld
1F====:::::j 18" 'v'F
~" 1"
5oo-rl \\
85# beam
W'V
I
.002
.004
.006
.008
.010
5.4-11
at the top; this point is marked with an "X" on the curve. With further cracking of the weld and yielding in the beam web, the lower flange of the beam contacted the column, point (b), and this resulted in increased stiffness. The moment built up to a maximum of 1918 in.-kips, and then gradually fell off as the weld continued to tear. Notice in this particular example, the web would have yielded the full length of the weld at design load. The weld started to crack when the connection had rotated about .Oll radians; this would correspond to a horizontal movement of .06" at the top portion of the weld. Compare this small amount of movement with that obtained in the top connecting plate example of Figure 4 which had the ability to pull out 1.6" before failing. This directly welded web connection (Fig. 18)
18" 'v'F 85# beam
11 "
/
.012
.014 .016
End rotation (Oe), radians FIGURE 17
.018 .020
5.4-12
/
Welded-Construction Design This restraint is a little high to be classed as simply supported. The same top plate connection is shown in dotted lines on Figure 17; it has about the same stiffness, but many times the rotational ability. The use of side plates, Figure 21, would allow a wide variation in fit-up, but in general they are no better than the directly welded web connection. Unless the plates are as thick as the beam web, the resulting connecting fillet welds will be smaller and will reduce the strength of the connection.
FIGURE 19
l
A
!..Am
FIGURE 20
is not as dependable as a top connecting plate designed to yield at working load (Fig. 19) or either flexible web framing angles (Fig. 20) or flexible top angle. Also remember this highly yielded web section, in the case of the directly welded web connection, must still support or carry the vertical reaction (R) of the beam, whereas in the top plate connection, the support of the beam at the bottom seat is still sound no matter what happens to the top plate. Figure 17 would indicate the directly welded web connection results in an end moment of Me = 720 in.kips, or an end restraint ofR
r=
A
A
720 in.-kips W16 in.-kips FIGURE 21
35.8%
Field weld Shop weld
Shop weld Field weld only on toe of angle
Field weld
FIGURE 22
Web Framing Angles In the tests at Lehigh University, the corresponding connection on the 18" WF 85# beam (.526"-thick web) used 0/16" thick side plates with %6" fillet welds. They failed at a lower load. If Ih" thick side plates with ¥2" fillet welds had been used, they undoubtedly would have been as strong as the directly welded web connection.
/
5.4-13
Weld w
7. ONE-SIDED WEB CONNECTIONS A single web framing angle used by itself is not recommended; see Figure 22. Use of only a single vertical fillet weld to join the angle to the supporting member imposes a greater eccentricity upon the connection. This results in a maximum force on the weld of about 4 times that of the double-angle connection; see Figures 23 and 24. It might be argued that in the conventional doubleangle connection, the field weld is subject only to
FIGURE 23
FIGURE 24
vertical shear because the stiffness of the angles largely prevents any twisting action on the connection even though the analysis is based upon this twist as shown in Figure 23. However, there is no doubt that the single-angle connection has this twisting action which would greatly decrease its strength. Any additional welding on the single angle, such as vertically along its heel or horizontally across the top and bottom edges, would make it rigid and prevent it from moving under load. This would cause the end moment to build up and greatly overstress the connection. In the original research at Lehigh University on welded connections, this single-angle connection with a single vertical weld was never tested. Single angle connections welded both along the sides and along the ends were tested, but as already mentioned, they did not have enough flexibility, and the end moment built up above the strength of the connection.
5.4-14
/
Welded-Connection Design
Web framing angles are commonly shop welded to the supported beam. To facilitate erection, bolts are used in joining the other member until the web framing angle can be permanently welded to it. The erection bolts can be left in, or removed if there is any concern that they will offer restraint. Note the use of box section column, in this case it being hot rolled square structural tubing.
SECTION
~._5
Top Connecting Plates For Simple Beams and Wind Bracing 1. DESIGN PLATE TO BE STRESSED AT YIELD A top connecting plate if designed to be stressed at its yield will provide a flexible connection, suitable for a simple beam and easily adapted to carry the additional moment due to wind. Since this flexibility is due to plastic yielding of the plate, the portion of its length which is to yield should be at least 1.2 times its width.
The plate should be capable of plastically yielding a distance equivalent to the movement of the end of the top beam flange as it rotates under load if the connection were to offer no restraining action (AISC Sec. 1.15.4); see Figure 1. For a simply supported beam, uniformly loaded, this maximum movement (e) would be:
I......··.. '-------------2
U"
(12 L) 3E
3,~~
(1)
where: e
Beam
L
1+------- L - - - - - - - - (length of beam) FIGURE 1
= =
movement, in inches length of beam, feet
The graph in Figure 2 illustrates what this movement would be as a function of beam length, under various load conditions. There is no problem in detailing a top plate to safely yield this much, providing there are no notches which might act as stress risers and decrease the plate's strength. Any widening of the plate for the connecting welds must be done with a smooth transition in width.
.6 -+---,---r----,-----,.----,-----,,--,--,--..,..----,
aE
.5--r---+---1----I---+---Jf---+---+--+----t::;,.~7t""" l
J:J
'0
c"'.4~-_+-__+-__+-_+-----1f_-t_-h~~-+____::;~ Gl
2 loads @ V3 points 4 loads @ '.4 points { Uniformly distributed load 5 loads @ 1/6 points 3 loads @ 1,4 points 1 load at t
Gl
EJ:
g~ .~; E goc -_
N .~
I
.3 .2 -r ---+---+---1-----;;JiiIf'''--:::;ool'''''---II----+---+--+------I
0
-g Gl
.1 -fc'--_+___::;Of/I'F:7"'~-_+-----1f_-t_-+--+--+_-__j
10
20 30 40 50 60 70 80 90 Length of simply supported beam (L), feet (assuming beam to be stressed to C1 = 20,000 at t )
100
FIGURE 2
5.5-1
5.5-2
/
Welded-Connection Design E 6024 weld metal E 6012 weld metal . . E6010weldmetal / /Mild steel
/
r 80
70
-"" 0;;; e' -; ~
en
,
.;:::- '":::--
60 50 40 30 20 10
/
II
-,0:
I 7"
t...---
5
10
15
20 Elongation, % in 2"
35
30
25
40
FIG. 3 Stress-strain diagram for weld metal and beam plate.
2. TOP PLATE FOR SIMPLE BEAMS
ASTM specifies the following minimum percent of elongation as measured in an 8" gage length for structural steels: A7
There is some question as to what value should be used for the end moment in the design of the top plate for simple beams. Any top plate will offer some restraint, and this will produce some end moment. Lehigh researchers originally suggested assuming simple beam construction (AISC Type 2) to have an end restraint of about 20%. On this basis, the end moment for a uniformly loaded beam would be:
21%
A373
21%
A36
20%
A242
18%
A44J
18%
This minimum value of 20% for A36 steel would represent a total elongation of 20% X 8" = 1.6" within the 8" length. Notice in Figure 2 that a simply supported beam, uniformly loaded, with a span of 20 feet would rotate inward about .106", so that this particular beam would utilize only 7{5 of the capacity of this top plate to yield. Figure 3, a stress-strain diagram, shows that a mild steel base plate will yield and reach maximum elongation before its welds reach this yield point. The test specimen in Figure 4 shows that ample plastic elongation results from the steel tensile specimen necking down and yielding. This is similar to the behavior of a top connecting plate which yields plastically under load.
Me
=
WL (.20) 12
W L
= ----eo
and this is 13.3% of the beam's resisting moment. Heath Lawson ("Standard Details for Welded Building Construction", AWS Journal, Oct. 1944, p. 916) suggests designing the top plate (simple beam construction) for an end moment of about 25% of the beam's resisting moment. This would correspond to an end restraint of about 37.5%, which approaches the range of "semi-rigid" connections. In Figure 5 the end of the top connecting plate is beveled and groove welded directly to the column, the groove weld and adjacent plgte being designed to develop about 25% of the restraining moment of the
~After pUlling~ 1Y2 W (before " " "
I
l
i
- --...,
>
.-!-
I ?
~
w
>
\
°
--\--- Reduced cross-sectio n after
orlglna~ nec kiIn d own cross-section ~ 9 °
>
1
FIGURE 4
Top Plates for Simple Beams & Wind beam using the standard allowable bending stress. The standard bending stress allowed here would be limited to CT = .60 CTy • (Type 2, simple framing). Just beyond the groove weld section, the plate is reduced in width so that the same load will produce a localized yield stress (CT r ). The length of this reduced section should be at least 1.2 times its width to assure ductile yielding. This plate is attached to the beam flange by means of a continuous fillet weld across the end and returning a sufficient distance on both sides of the plate to develop the strength of the groove weld at standard allowables: A7, A373 Steels; E60 Welds
f = 9600
Cl)
lbs/Iinear in.
... (2)
A36, A441 Steels; E70 Weld
f = 11,200
Cl)
lbs/Iinear in.
3. TOP PLATE FOR WIND BRACING Wind moments applied to simple beam connections present an additional problem. Some means to transfer these wind moments must be provided in a connection which is designed to be flexible. Any additional restraint in the connection will increase the end moment resulting from the gravity load. AISC Sec 1.2 provides for two approximate solutions, referred to hereafter as Method 1 and Method 2. In tier buildings, designed in general as Type 2 construction, that is with beam-to-column connections (other than wind connections) flexible, the distribution of the wind moments between the several joints of the frame may be made by a recognized empirical method provided that either:
At standard allowables (0 .60
=
or)
/
5.5-3
Method J. The wind connections, designed to resist the assumed moments, are adequate to resist the moments induced by the gravity loading and the wind loading at the increased unit stresses allowable, or Method 2. The wind connections, if welded and if designed to resist the assumed wind moments, are so designed that larger moments induced by the gravity loading under the actual condition of restraint will be relieved by deformation of the connection material without over-stress in the welds. AISC Sec. 1.5.6 permits allowable stresses to be increased If.J above the values provided in Sec 1.5.1 ( steel), and 1.5.3 (welds), when produced by wind or seismic loading acting alone or in combination with the design dead and live loads, on condition that the required section computed on this basis is not less than that required for the design dead and live load and impact, if any, computed without the If.J stress increase, nor less than that required by Sec. 1.7, (repeated loading) if it is applicable. Since we are discussing Type 2 construction (simple framing) the initial basic allowable stress is .60 CTy , not .66 CTy •
I Method 1 The top plate (Fig. 6) is designed to carry the force resulting from the end moment caused by the combination of the gravity and wind moments, and at a If.J increase in the standard stress allowable (or CT = .80 CTy). This If.J increase may also be applied to. the connecting welds (AISC Sec. 1.5.3, & 1.5.6). The fillet welds connecting the lower flange of the beam to the seat angle must be sufficient to transfer this same load. The top plate must have the ability to yield plastically if overloaded (last paragraph of AISC Sec. 1.2).
(" Minimum length of reduced
i
section between welds
1.2 W 1" X Va" backing bar
FIGURE 5 F
_ M. (gravity)
-
db
5.5-4
/
Welded-Connection Design
Length of unwelded section
For the groove weld to the column flange, this plate is widened to I%W, or-
(24 t) (.289 t)
width = Ph (1%)
= 83
= 2.9" or use 3.0"
.~
.. ~
For the fillet welds to the beam flange, use fillets at an allowable force of-
4. EXAMPLE OF TOP PLATE DESIGNWITH WIND MOMEN1" A 14" WF 38# beam is simply supported and loaded uniformly with 296 lbsjin. on a 15-ft span. Based on these beam-load conditions, the maximum bending moment at center is M = 1200 in.-kips. Use A36 steel and E70 welds. Wind moment on each end is M w = 600 in-kips.
Beam conditions here:
.59 in."
Connecting Welds at Standard Allowables
slenderness ratio
r -
21.3 kips
P':
.
' 12W'1•
1 2WJ
•
"...,........-.1.
rr,
/: 3"
~ I
vlV
"~/%"
A.
n1%"
,T
.
(See Figure 9.)
14" WF 38# beam b
= 6.776"
S
= .513" = 54.6 in."
If there were no wind load, the above connection might be designed for about 25% of the present
511/'
~
V
IF J-F
db = 14.12"
db = 14.12" tr
2.18 in.2
OK
The connecting welds are figured at ~ higher allowabIes: For the fillet welds at the beam flange, use :!h" fillets. The standard allowable force is fw = 11,200 Cd = 11,200 (lh) = 5600 lbs per linear inch. The length of this weld is--
12W'1•
jK?"'" \ ~"V
•
A
•
U'
(63.8 kips) 1~ (22,000 psi >--
This would be 1%" across the end, and 2W' along the sides.
5"
higher
F
A p = 1~
= 6.74"
I•
~
Lw =
F 11,3
fw
_
(63.8 kips)
-
1~
(5600)
= 8.54" This weld length would be distributed 3W' across the end, and 2W' along the side edges of the top plate. 'The above connection may be cut from bar stock without the necessity of Harne cutting any reduced section in it. This is a good connection and is in widespread use. The connecting groove weld and fillet welds are strong enough to develop the plate to yield plastically if necessary due to any accidental overload of the connection. Some engineers prefer to widen this plate at the groove weld so that if the plate should have to reach yield stress, the connecting welds would be stressed only up to the wind allowable or 1,3 higher, hence U' = .80 U'1' Accordingly, the plate is widened here to 1¥4W = 1¥4 (3lh) = 4%". (See Figure 11.)
+ 600 in.-kips
The length of the fillet weld, using W' fillet welds and allowable of fw = 5600 Ibs/in., would be-
Top Plates for Simple Beams & Wind 5"
A.
3Y2"
~\
>%"
A.
y
1
T
.L
5.5-7
3Y2"
I• •I• -I 4 =HI"
/
J
R.\
I
Y2"V
-r db
F
= 14.12"
+
J..F
I~
Mw
T
1 ~mrrrrTTT1mrn-.L
LU..U...........................
-~
T Wind moment
, , ,
FIGURE 13
Mw F = db
FIGURE 11
F
L" = 1% f w
= 42.5 kips The reduced section of the plate is designed to carry this at % higher allowable: -
(2.19 in.2)(36,OOO psi) 1% (5600) -
(600 in.-kips) (14.12" )
reduced section at yield (U'7) and fillet weld at % higher allowable
10.55"
F
This would be 3;2" across the end, and 3W' along the side edges of the plate.
Ap
= 1% U' (42.5 kips) 1% (22,000)
Applying Method 2 for Additional Wind Moment
.l
I-
4Y2"
I
--
5"
I
5" I
•
"
or use 3" by ;2" plate
-I
I
Ap = 1.50 in. 2
l-L
l-r !1-
jF db
1.45 in. 2
OK
= 14.12"
~F
The plate must now be modified so that larger moments induced by the gravity loading can be relieved by plastic yielding of the top plate, designing the connecting welds at standard allowables. The plate is widened at the groove weld to 1% W = 1% (3) = 5.0". For the connecting fillet welds to the beam flange, use %" fillets:
f w = 11,200 w
=
" -, .,
>
3"
ri~"yw,V ~
1.45 in. 2
11,200 (%)
= 4200 lbs per linear inch FIGURE 12
Temporarily ignoring the gravity load, the top plate is designed to carry the wind load, Mw = 600 in.-kip on each end.
The length of this weld is-'T
_
'-'Vi
-
F _ fw -
= 12.9"
(1.5 in.2 ) (36,000 psi) (4200)
5.5-8
/
Welded-Connection Design
Beam
10,990 psi (28,330)
M."
+
600 in-k
Mw
=-
600 in-k
Beam
Jt _r
10,990 psi
(0,""';00
FIGURE 14
(28,330)
FIGURE 15
This would be 3" of weld across the end, and 5" along each side.
Ub L e t K --~_ Sb - up
5. EXAMINING THIS EXAMPLE
10,990
To better understand how this wind connection operates, this example will be examined, using Method 2. 1. The connection is first designed for the wind moment of M; = 600 in.-kip at If.1 increase in the standard allowables applied to each end of the beam. The wind moment will cause a bending stress in the beam of-
.388
(600 in.-kips) (54.6 in.")
28,330
2. Now the gravity load can be gradually added, treating the beam as having fixed ends, until the righthand connection reaches yield stress. This would be an additional stress in the connecting plate of: 36,000 28,330 = 7670 psi. This would correspond to a stress in the beam end of: (.388) (7670 psi) = 2980 psi. (See Figure 15.) Since the allowable moment on this end connection resulting from gravity load is (treated as a fixed end beam)-
10,990 psi (See Figure 14.) The corresponding stress in the top connecting plate is-
Up
M = db A p
_ (600 in.-kips) - (14.12)(1.5) = 28,330 psi
Note that the connection will not yield until a stress of 36,000 psi is reached.
the portion of the gravity load to be added here isw. =
12 up Ap d L2
_ 12( 7670) (1.5) (14.12) (180 )2
= 60.2 lbs/in. The stress in this beam end due to gravity load is then added to the initial wind moment diagram: (See Figure 16.)
Top Plates for SimDle Beams & Wind
I
5.5-9
} M{Beam 13,970 psi e 1
Connection (36,000)
FIGURE 16
8750 psi
-M e2
17,490 psi
FIGURE 17
10,240 psi 8010 psi
Me
L
..... .....
- 517.6 in-ki
-,
......
Beam) 9480 psi
MeR
......
""
Connection (24,430 psi)
""
FIGURE 18
At this point, the right-hand connection reaches yield stress (uy = 36,000 psi) even though the beam end is stressed to only U = 13,970 psi. 3. The remainder of the gravity load (W2 = W WI = 296 - 60.2 = 235.8 lbsjin.) can now be applied, treating the beam as having one fixed end on the left and simply supported on the right. See Figure 17. The resulting end moment here is_
W2 L2
8
= 955 in.-kip
(235.8)( 180)2 8
=-
762.8 in-k
Beam 13,970 psi
{ Connection (36,000 psi)
or a bending stress of Ub2
M e2 --
_
Sb
(955 in.-kips) ( 54.6 in.a)
= 17,490 psi Also since M 4E. = Ub
at -'m....,
200 +E'-+---+-design Beam line at7 load-f~=-"'+-"...... 100
t, , .001
.002
.003
.004 .005
I",J,. . 1-
t+ I-',-+---i
.006 .007 .008 .009 .010 End rototion (8.), radians FIGURE 9
,
.011
.012
.013
.014
FIG. 10. Moment Capacity of Top Plate Connection.
K,P - INCHEJ 10,*0
MOMENT
@$)
(cr.2Z,fXJOpal)
la, 000
20,000
16,000
18,000MIIMEIIT
1+,"'0
16,000 ~
",00
''',000(cr:If,DDO,.I)
117,A~73
IMt::: Wta'h 0- I
WIDTH OF TOP CONNECTIN6 ~
STEEL
I.
J;~
DEPTH Of BEAM
16
"'"I I /I
,
......
10
~
B
...~ ............ II{,
7
6/I'.-.6JI"
\:'
........
..... .....
.... 20.......
I
5
8'1
6,1H
6
.
',44.
> Section of column tested
>
FIGURE 7
5.7-4
/
Welded-Connection Design
column just inside the column flange, and directly beneath th e bars. Yielding progressed into the column web by means of lines radiating from this point to the colum n "K" line, at a maximum slope of 1 to 2%. This pro gressed for some distance. A slight bending of the column flang es was noticed at about 80% of the failure load. Figure 8 shows an analysis of this.
Beam compression flange
~
t
FIGURE 10
Column web A
Dimensions of both the column flang e and the connecting plates wer e vari ed in order to study the effect of different comb ina tions of columns and beams. First yielding was noticed in th e fillet of the column just inside the column flang e, and directly beneath the attaching plates, at about 40% of ultim ate load. With furth er loading, yielding proceed ed into the column web, underneath th e column flange parallel to th e att aching plate, and into th e column flange from the ce nte r of th e conne cting welds, and p arallel to the column web. Aft er ultimate loading, some members fail ed by cracking of th e central portion of the connecting weld dir ectl y over th e column web, some by cr acking in th e insid e fillet of th e column, and some by cra cking in th e insid e fillet of th e column, and some by a tearing out of material in th e column flange.
tpe
FIGURE 8
Overloading of Column Flan ge Due to Tension Force of Upper Beam Flange A test was set up, Fi gure 9, to evaluate effects of the upper flange of th e beam in tension against the column. Two plates, on e on each sid e of the column and welded to it, represented th e cross-sect ion of th e beam flange . The member was pulled in a tensile testing machine.
t > Section of column tested
>
1FIGURE 9
FIGURE 11
Beam-to-Column Continuous Connections
A
/
5.7-5
"
>
~
"
L
1. FIGURE 12
~
>-
>
:-
'I Standard Stiffeners
When some type of web stiffening is required, the standard horizontal Bange stiffeners are an efficient way to stiffen the column web. Figure 10 shows this type under test. A Tee section flame cut from a standard wideflange section may be used for stiffening, Figure 11. The stem of the Tee section is welded to the column web for a short distance in from the ends. This could be entirely shop welded, all of it being done in the Bat position, possibly using a semi-automatic welder. This type stiffener would have numerous advantages in fourway beam connections. The beams normally framing into the column web would now butt against this Bat surface with good accessibility. The flanges of the beam could be beveled 45° and then easily groove welded in the field to this surface, using backing straps. There would be no other connecting or attaching plates to be used. In effect this part of the connection would be identical to the connection used for beams framing to column flanges. See Figures 28, 29 and 30 and related text for specifications of stiffeners applicable to elastic design. Effed of Eccentric Stiffeners In a four-way beam-to-column connection, the column
"
/"
flanges may be stiffened by the connecting plates of the beam framing into the column web. It may be that the beam framing to the column Bange is of a different depth. This in effect will provide eccentric stiffeners, Figure 12. The lower part of Figure 12 shows how this was tested. It was found that an eccentricity of 2" provided only about 65% of the stiffening provided by concentric stiffeners, and an eccentricity of 4" provided less than 20%. Three methods of framing beams of different depths on opposite flanges of columns are shown in Figure 13.
3. TEST COMPARISON OF STIFFENER TYPES The following is adapted from "Welded Interior BeamTo-Column Connections", AISC 1959, which summarized tests on various connections. Figure 14 represents a direct beam-to-column connection. Here the column has no stiffening and is not as stiff against rotation as the 16" WF 36# beams which frame to the column. This arrangement showed high stress concentrations at the center of the beam tension flanges, and therefore at the center of the connecting groove weld.
5.7-6
/
Welded-Connection Design
A
=-'
\:::
>
A
A
~
:.
:. ~
~
>
:"\
'J
\::
r: t"\
-
I-'
A
A
A
(b)
(a)
A
>
~
FIGURE 13
However, it was noted that no weld failures occurred until after excessive rotation had taken place. The stiffeners here in Figure 15 provide the equivalent of beam flanges to the columns, and the columns become as stiff against rotation as the beams framing to the column. The stress distribution on the compression flanges were uniform on the whole, while in the tension areas the stresses were somewhat higher in the center. In Figure 16 the column is shown stiffened by a pair of wide-flange Tee sections. As a result the columns are as stiff against rotation as the beams framing into the columns.
From strain gage readings it was calculated that each of the vertical plate stiffeners in the elastic range transmitted only about %6 of the forces corning from the beam flanges and the column web transmitted o/s of the forces. Placing these stiffener plates closer to the column web might have improved the distribution. However, since the prime purpose of this type of connection is to afford a convenient four-way connection, the plate usually needs to be positioned flush with the edge of the column flange. ' The stress distribution was uniform in both flanges at the working load. At 1.5 of the working load, high
Zero
~
~
+a
.1
:>
II
>
~
-- 20,000 psi ..A
Stress distribution in tension flange
V
.
~ 16" VF 36# beam
r
>
A
~
A
FIGURE 14
Beam-to-Column Continuous Connections
/
5.7-7
:.
:.
>
A
:1 II
:.
II II II II
>
II II II
II II A
II
A
A
FIGURE 15
FIGURE 16
tensile stresses occurred at midflange, The connection in Figure 17 was stronger than its two-way counterpart. This evidently shows that the stiffening action provided by two beams framing into the column web strengthens the connection more than
it is weakened by the triaxial stresses. The connections of Figure 18 involving (EastWest) beams welded directly to the column flanges proved stiffer than the connection of (North-South) beams to the Tee stiffeners.
I"®" I >
>
>
>
>
>
@
.
.,
., v
®
I ® I
-
v
"
I
II II I
:.
>
>
>
>
>
> II
II
"
·v
FIGURE 17
v
v
FIGURE 18
5.7-8
/
Welded-Connection Design
t K
t
FIGURE 19
The stiffening of the latter connection is mainly dependent on the thickness of the stem of the Tee stiffener, the Ranges of the column being too far away to offer much resistance. The column web is ably assisted in preventing rotation at the connection by the Ranges of the splitbeam Tee stiffeners.
Analysis of this plate by means of yield line theory leads to the ultimate capacity of this plate being- .
where:
4. ANALYSIS OF STIFFENER REQUIREMENTS IN TENSION REGION OF CONNECTION (Elastic Design)
Let: The following is adapted from "Welded Interior Beamto-Column Connections", AISC 1959. The column Range can be considered as acting as two plates, both of type ABCD; see Figure 19. The beam Range is assumed to place a line load on each of these plates. The effective length of the plates (p) is assumed to be 12 t e and the plates are assumed to be fixed at the ends of this length. The plate is also assumed to be fixed adjacent to the column web. where: ill
q
h p
We
+2
be -
(K -
te )
T1 -
fi
~ [~
132
+ SA
-
fi]
~ q
A=!.q For the wide-Range columns and beams used in practical connections, it has been found that ci varies within the range of 3.5 to 5. A conservative figure would be-
ill
2 The force carried by the central rigid portion of the column in line with the web is-
Beam-to-Column Continuous Connections Setting this total force equal to that of the beam's tension flange: tr,
~
>
wC
tc
t
-.I r+-
Af
T
~ t ., Ldc~
.(> bb
f
::-
6. HORIZONTAL STIFFENERS
~
17
>
T
~
~
t
~
r' I~
?
A
t
db
>1
:-
5.1-9
If the thickness of the column web (We) meets the above requirement, column stiffeners are not needed in line with the compression flanges of the beam. If the actual thickness of the column web (We) is less than this value, the web must be stiffened in some manner.
tb
t
/
t'
>
+
1
:>
A
1
> >
FIGURE 20
----.fi
Reducing the strength of this column region by 20% and making the conservative assumption that m/b,
= .15, this reduces to the following:
FIGURE 21
If the thickness of the column flange (t e ) meets the above requirement, column stiffeners are not needed in line with the tension flanges of the beam. If the actual thickness of the column flange (t e ) is less than this value, stiffeners are needed.
5. ANALYSIS OF STIFFENER REQUIREMENTS IN COMPRESSION REGION OF CONNECTION (Elastic Design) It is assumed the concentrated compression force from
the beam flange spreads out into the column web at a slope of 1 in 2% until it reaches the K line or web toe of the fillet; see Figure 8. Equating the resisting force of the column web to the applied force of the beam flange, assuming yield stress-
we
(t b
+5K
e)
Iwe~dh-I
U"y
> At
U"y
or
Equating the resisting force of the column web and a pair of horizontal plate stiffeners to the applied force of the beam flange at yield stress-
where: A.
= total cross-sectional area of pair of stiffeners
To prevent buckling of the stiffener-
where: b. = total width of pair of stiffeners
If the stiffener is displaced not more than 2" from alignment with the adjacent beam flange (as in Fig. 12), it may still be used if considered about 60% as
5.7-10
/
Welded-Connection Design
effective as when in direct line. The stiffener thickness (t.) found from the above formula should then be multiplied by 1.70 to give the actual required value.
Because the vertical stiffeners (usually Tees) are placed at the outer edges of the column flange, they are assumed to be half as effective as though placed near the column web. It is assumed the concentrated beam flange force spreads out into the vertical stiffener in the same manner as the column web. Equating the resisting force of the column web and a pair of vertical Tee stiffeners to the applied force of the beam flange at yield stress-
7. VER·nCAl STIFFENERS
T~
~---I:~!....J L-L.!4'---' -.i
fry
= At
fry
or
A
I
~ -
I~
1
t. > - 30
>
I
1- Problem 1 As an example of applying the preceding analysis of the tension region of a connection, we will analyze a connection which, when tested to failure, performed well; see Figure 23.
---A.---. FIGURE 22
12"
w:-
40# column
I I I I
JI
Il 1(-----
•
390" .
I 1
II
I 1
:Y16"
I I I
1
1 I
.606 " I I
,
I
I---~r_-----__r~..
FIGURE 23
6.992"
16"
w:-
36# beam
Beam-to-Column Continuous Connections where: m =
+ 2 (K - te) (.390) + 2 [(1 %6)
q = _
be -
2
-
(.606)]
Provided both column stiffener and beam have same yield strength;
m
(10.92) 2
(.428)(1.553)
+ 2 (4.94)(.606)2 >
(6.992)(.428) 4.28 > 3.00 O.K. If we used the conservative formula;
(1.55)
.= 4.69" b, - m h = 2
t.>O.4~ > .40 "';'"(-:-6-.9-92--:'")"""':""(.-428--:-)
(6.99)
(1.55)
> .692"
2
but the initial design called for t e = .606" and the connection tested O.K.
= 2.72" p=12te
8. CONNECTIONS THROUGH VERTICAL TEE STIFFENERS
= 12 (.606)
= 7.27"
Tests have shown that when the beam flange extends the full width of the connecting plate, Figure 24, about % of the flange force is carried by the central portion of the plate. Each of the two outer edges carry about %6 of this force. Figure 25 comes from test data of Lehigh University. Notice in the East-West beams, the flange of which extends almost the full width of the column
Since: h q (2.72)
= (4.69) = .58
f3='£.q (7.27)
= (4.69) = 1.55 and:
~ [ Vf3~ + 8 A -
f3]
= (1:5) [ '1'(1.55)2
+8
TJ =
(.58) -
(1.55)
J
= .387 i+~ f3 TJ 2 - 21 A
4
(f.55) 2 _
5.7-11
region of the column stiffener's flange must equal or exceed the force of the beam's tension flange, or:
We
= = 1.553"
/
(1.55)
+ f387) (.387) (.58)
= 4.94 The total force which can be carried by the tension
FIGURE 24
5.7-12
/
Welded-Connection Design
16" 'IF 36 # beam
.606"
.516"
100% F
®
...----. 44%
-
100% F
34% FIGURE 25
® 100% F
flange, 44% of the force is transferred through the web of the connection even though it is only about half as thick as the stiffener plates. This corresponds well with the idea that the flange of the column in this region is similar to a two-span beam on three supports with a uniform load; in this case the center reaction is % of the total load, and the two outer supports each carry :% (\ of the load. The report "Welded Interior Beam-To-Column Connections", AISC 1959, mentions that "from strain gage readings it was calculated that the vertical plate stiffeners in the elastic range each transmitted only about o/t6ths of the forces coming from the beam
flanges and the web transmitted %ths." Of course, the same would not be true in the NorthSouth beams because they do not extend the full width of the flange of the Tee stiffener. As a result, most of this force must be transferred into the web or stem of the Tee stiffener since any portion of this force reaching the outer edges of the column flange must be transferred as bending out along the flange of the Tee section.
Weld Size: Stiffener Stem to Column Web On the basis of these tests at Lehigh University, on connections where the beam flange extends the full
FIGURE 26
~
I
I I I
I I
l J.o'''
,"
Beam-to-Column Continuous Connections width of the stiffener flange, we will assume that % of the beam flange force is carried by the stem portion of the connection. See Figure 26. Because of the stiffening effect of the beam web and the stem of the connecting plate, this central (stem) portion of the connection will load up in bending. This assumes it rotates as a unit about a point at midheight. The bending force on the weld is zero at this neutral axis and increases linearly to a maximum value at the upper and lower edges of the connection. Treating the weld group as a line, the section modulus is equal to-
5.7-13
/
Weld Size: Stiffener Flange to Column Flange The Tee stiffeners may be joined to the column flanges by a) fillet welds, b) groove welds, or c) comer welds. The groove welds (b) were used in the Lehigh Research of this connection. '/
'Il
1 T//
'/
(a)
777
'/
'////
(e)
(b)
The resulting maximum unit bending force at the top portion of the weld on the stem is-
FIGURE 27
The leg size of this weld would be found by dividing this value by the allowable for the particular weld metal.
Since tests on full-width flanges showed that the two outer edges of the connection carry about %6 of the flange force, we will assume that each outer weld must carry % of the flange force. See Figure 28. These welds will be pulled with an axial force of If.J F. We may assume the same distribution of force through the connecting plate at a slope of 1 to 2lh into the connecting welds. This will provide an effective length of weld of t b 5 t, to carry this force. The unit force on this weld is-
A7, A373 Steel; E60 Welds
f = 9600
Ct.l
+
A36, A441 Steel; E70 Welds
f = 11,200
Ct.l
The leg size of the fillet weld, or throat of groove weld, is determined by dividing this unit force by the suitable allowable. The effect of the vertical shear load (V) on these
Here:
t,
f
FIGURE 28
H +--
----
T\
tb + 5 t,
~,
tb !f.lF
T db
---. T\ ---. ---. t + 5 t, ---. ~' ---. b
1 YJ F
5.7-14
/
Welded-Connection Design
welds could be checked by using the entire length of the welds. However, this would represent little additional force on these welds.
Proportioning th.· Tee Stiffener The following will be helpful in selecting a Tee stiffener section for this type of connection, where the beam flange equals the full width of the stiffener flange:
FIGURE 29
5. As a guide, the stiffener should satisfy this condition:
or an approximation on the conservative side:
Where Beam Flange Width < Stiffener Flange Width Where the beam flange does not extend the full width of the connecting plate, the stem portion of the connection is assumed to carry the entire moment. Therefore the maximum bending force on the top portion of this weld will be-
1. The thickness of the stiffener flange (t.) must be sufficient to transfer the tensile force of the beam flange. In this case 3/4 of the beam flange will be used.
2. The width of the stiffener flange (b,) must be sufficient for it to reach to the column flanges.
3. The thickness of the stiffener stem (w,) should be about the same as the beam flange thickness (t b ) . FIGURE 30
4. The depth of the stiffener (dB)' as measured through the stem portion, must be sufficient for it to extend from the face of the column web to the outer edge of the column flange. be We ----
2
The same items as before are used to proportion the Tee stiffener, except in items 1 and 5 where the full value of the beam flange's section area is used instead of % of this value. These formulas become-
Beam-to-Column Continuous Connections
5.
I
I
5.1-15
The weld on the beam's web must be able to stress the web in bending to yield (U"~.) throughout its entire depth; see the bending stress distribution in Figure 25. The weld must also be able to transfer the vertical shear.
w.
Problem 2
/
I
To design a fully welded beam-to-column connection for a 14" WF beam to an 8" WF column to transfer an end moment of M = llOO in.-kips and a vertical shear of V = 20 kips. The solution of this problem will be considered with seven variations. Use A36 steel and E70 welds.
unit force on this weld from the vertical shear
(20 kips) 2( 13.86 - 2 x .387) 770 lbs/in.
leg size of fillet weld W
actual force = allowable force (770) (1l,200 ) .069"
However, since the beam web is welded to a .433" thick flange of the column, the minimum size for this fillet weld would be %6"; see Section 7.4, Table 3. FIGURE 31
WELD SIZE TO DEVELOP ULTIMATE LOAD Here: M
1100 in.-kips
V
20 kips
The welding of both the flanges and the web along its full depth enables the beam to develop its full plastic moment, thus allowing the "compact" beam to be stressed 10% higher in bending, or U" = .66 U"y. This also allows the end of the beam., and its welded connection, to be designed for 90% of, the end moment due to gravity loading. (AISC Sec 1.5.1.4.1 and Sec 2.6)
The next question is what size fillet weld would be required to develop the beam web to yield stress. The force in question results from bending, so it is transverse to the weld. The AWS allowables for fillet welds are based on parallel loading, AWS has not set up any allowable values for transverse loading.
weld
uieb plate
vs
(parallel load)
(transverse load-tension)
2( 1l,200 w) > t w (.60 U").) = t w 22,000 w 2': .982 t w
.9 M
U"
= -S-
or
.9( llOO in.-kips) (41.8 in.")
23,700 psi
18.6" OK
Here:
M
1100 in.-kips
V
20 kips
DESIGN OF BOTTOM SEAT
In this particular connection, the shear reaction is taken as bearing through the lower flange of the beam. There is no welding directly on the web. For this reason it cannot be assumed that the web can be stressed (in bending) to yield through its full depth. Since full plastic moment cannot be assumed, the bending stress allowable is held to U" .60 U"y or U" 22,000 psi for A36 steel. (AISC Sec 1.5.1.4.1)
=
=
N+ K---1 V
= 20 kips I /
,,
/
(compression)
bending stress in beam M
u"=s (1100 in.-kips) (54.6 in.") 20,200 psi
M2 • F
F,
F,
stiffener
FIGURE 46 FIGURE 44
Tensile force from beam flange transfers directly as tension through both stiffeners and web of column into other beam flange. Transverse welds between column flanges and stiffeners must be designed for this force (F) less that which passes directly into the web from the flange. Parallel welds between stiffeners and column web transfer no force. Compression portion of beam connection would keep stiffener from buckling. Shear Transfer
~
t
.,.
F ..1.
A
The tensile force F 2 of the flange of the left-hand beam will transfer as tension into the stiffener, then through the transverse welds along the column web into the other stiffener, and into the flange of the other beam. The un balanced tensile force (F1 - F 2) of the flange of the right-hand beam will transfer as tension into the right-hand stiffener, and half of this through the transverse welds of the column web into the lefthand stiffener. This unbalanced tensile force in these stiffeners now transfers through the parallel welds as shown into the flanges of the columns. Welds to column web must be designed for the F1 F2 balanced force, or lh F F2 = 2 .
+
+
Welds to column flange must be designed for the unbalanced force or F 1 - F 2 •
Distribution of Tensile Force There is some problem in estimating the portion of the tensile force in the beam flange transferring directly into the web of the column and into the column stiffeners.
w
t 1 F ~~
FIGURE 45
1tF..
F
Tensile force from beam flange transfers directly as tension into stiffeners and column web. The tensile force in the stiffeners then transfers out as shear through the parallel welds into column web. Transverse welds between column flanges on the beam side and stiffeners must be designed for this force (F) less that which passes directly into the web from the flange. Parallel welds to column web must be designed for this same force. Any unbalanced moment (M = M 1 - M2 ) enterI
FIGURE 47
At first glance it would seem reasonable to assume this force would be divided according to the width of the stiffeners (b.) and thickness of column web (t w ) .
5.7-24
/
Welded-Connection Design
Column flange
fabricated column
d = tb
+5K
Since: area of column web over which force is distributed = d t w
Aw
A. = area of one stiffener (there is a pair)
FIGURE 48
However, this column web section is not limited to the thickness of the beam flange since there is some spreading out of this force in the web. This might be assumed to occur at a slope of 1 to 2%.
FIGURE 49
The effective depth of the column web through which force is distributed, is obtained as follows:
(web) F w
w
F (
Aw
+A 2 A.
)
A. ) ( stiffener) F. _ F ( Aw 2 A.
+
Combined Stress in Stiffener (See Figure 51.) On the left-hand figure, the shear stress (TX)") results from the unbalanced East-West moments. This causes the difference in tensile beam flange force (FI-F 2) to be transferred as shear in the stiffeners into the column flanges. Although conservative in this particular analysis, it is assumed the small section in the stiffener to be checked lies outside of the path which the East-West tensile flange force will travel; hence a"x = O. Actually some of this tensile force will spread out into this region, and this would result in lower principal. stress. In either case, it would be checked by the following formula: a"x +
a"max
a"y
+
2
or "
..
Beam flange
-,
Column flange Column web
FIGURE 50
rolled column
d = tb
+ 5 K,
On the right-hand figure, it is assumed the small section to be checked is not subjected to any shear stress, just biaxial tensile stress. In this case, the use of the formula results in the principal stresses being equal to the applied tensile stresses. This does not result in any higher stress. a"max
_
-
a"x +a"y
2
+ ~( a"x - 2
or
I a"max
a"x
or
a"y
I
a"y
)2
+
T xy
2
Beam-to-Column Continuous Connections
T
/
5.7-25
~
xy
+1
(J.
=0
+T
+ T. y -
(J
+
-H-----=i~--F----.
(J
-
T
Mohr's Circle of Stress
Mohr's Circle of Stress
FIGURE 51
I Problem
3
M
I
F=d
To check beam-to-column connection shown in Figure 52 (next page) for weld sizes.
(9097 in-kips) (23.59/1 ) 386 kips
flange force: 24" WF 160# beam
flange force: 21" WF 73# beam
1.135 "
74"
F = 386 kips
t
+ 24" VF
.
21" VF
160# beam
r-i? F
73# beam
d = 20.50"
-~
M=(J"S
M
= (22,000 psi) (413.5 In.") = 9097 in.-kips d
= 24.72" = 23.59"
1.135"
(J"S (22,000 psi) (150.7 in.") 3315 in.-kips
d
21.24" -
= 20.50"
.74"
= 162 kips
5.7-26
/
Welded-Connection Design
24" W 160# beam
I~~----24"----_
3%" I
I ...J
L..-_ _.......
FIGURE 52
24" W 160# beam
/ / /
-- --f
------\ \
/
d = tb
/
,,
;:.
\
5 tc
___1___
,,
I
::>
+
tb
\
/ /
/
- -
-
1-----------
-
\ \
/
,
~
/
\
/
\
\~-
-
- - - - - - -- I
1\
/
A
tc
A
~
,..--
I
I
I
I
4--
386 kips
-
\
/
>
... F -
\
/
4--
f t
4--
:L __
~ 4--
4--
4--
4--
4--
4--
?
I
H-
f----
4--
: 4--
4--
4--
--, :- .. r -- f----
I-
4-- 4--
4--
4-- 4 -
246 k
4--
4-- 4--
70
~
14--
I
I
I I
I I
'--
'--A
70 k
4--
__J
•
k
> >
FIGURE 53
Beam-to-Column Continuous Connections
+-
386k
1 "- lV
.r.: 1~
11111111111 IIII
I
-"f-
1I I. 1- ----i.J I
I
+- +-- ~
;>
I
81 k 162k_~
~ • +-- .f-81 k ~ L
...
...
~~
FIGURE 23
\J l
V
Diagonal compression on web of connection due to shear forces from unbalanced moment
j
V
® Stiffeners
Stiffeners ~~
FIGURE 24
II Li- __
r
.. I
II II II II II II II I I I I
_-L.J
Design of Trusses
/
5.9-11
® Rv
!
Rh II
I I
FIGURE 25
I I I I I I LJ
1
I I I I I I I
I I I I
I
I
uI
I
....
:-
~
-z,
Stiffeners
.
~ ~
FIGURE 26
v
"Y
1
other component (here F v). Of course the applied force (F) will be reduced also, and under these conditions some other portion of this member must transfer it. In this case the web of member A will transfer the balance of the force (F). Determining Need for Stiffeners Normally stiffeners would be added to a member in which large concentrated transverse forces are applied. However, for smaller members with lower forces, these stiffeners are sometimes left off in truss connec~ tions. It is difficult to know under what conditions this might have to be stiffened. In recent research at Lehigh University on "Welded Interior Beam-to-Column Connections", short sections were tested under transverse compression as well as tension, with and without stiffeners. See Figure 27. It was found that the compressive force applied over a narrow section (t-) of member's Bange spread out over a wide section of the web by the time the net web thickness was reached. A conservative value for this distance is given as: (tr
+ 5K)
1 where K
tr
=
the distance from the outer face of the flange to the web toe of the fillet. This value for all rolled sections may be found in any steel handbook. thickness of the flange of the connecting member which supplies the compressive force.
Although there was no axial compression applied to the member in this test, on subsequent work involving actual beam-to-column connections, axial compression was simultaneously applied. See Figure 28. It was found that an axial compressive stress of about 1.65 times the working stress (14,500 psi), or (F = 24,000 psi, had little effect on the strength of the connection. At the end of each test with the final loads left on the beams, this axial compressive stress was increased to twice the working stress or (F = 29,000 psi with no indication of trouble in the connection. From this, they concluded that the minimum web thickness of the column for which stiffeners are not required is found from the following:
5.9-12
/
Welded-Connection Design
Bar represen ts connecting flange
(0) Test to determine Compression region criterion
FIGURE 27
'"
Plate represents .............. connecting flange
(b) Test to determine Tension region criterion
This research, concerned with the application of concentrated flange forces applied to flanges of WF members, was directed toward beam-to-column connections. However, it does seem reasonable to use this as a guide for the distribution of flange forces in truss connections. This will then provide an indication of the stresses in the chord resulting from the flange force of the connecting member. In the test of the tension area, they found that the thickness of the column flange (te) determined whether stiffeners were required. On the basis of their tests, they made the following analysis.
I
T E, ~
t, +5K :::
-i.E I
Analysis 01 Tension Region 01 Connection The following is adapted from "Welded Interior Beamto-Column Connections", AISC 1959.
jp FIGURE 28
~ • t,
Design of Trusses
/
5.9-13
FIGURE 29
·v
The column Bange can be considered as acting as two plates, both of type ABeD; see Figure 19. The beam Bange is assumed to place a line load on each of these plates. The effective length of the plates (p) is assumed to be 12 t, and the plates are assumed to be fixed at the ends of this length. The plate is also assumed to be fixed adjacent to the column web.
See Figure 29, where: m
=
q
=
h =
We
+ 2(K
be -
te )
m
2
b, -
TJ =
~[ v' ~2 + 8A - ~ ]
f3
E. q
A=~ q
For the wide-Bange columns and beams used in practical connections, it has been found that Cl varies within the range of 3.5 to 5. A conservative figure would be---
m
2
P = 12 t e
The force carried by the central rigid portion of the column in line with the web is-
Analysis of this plate by means of yield line theory leads to the ultimate capacity of this plate beingSetting this total force equal to that of the beam's tension Bange: where:
±+~ f3 TJ 2-!l A
Reducing the strength of this column region by 20% and making the conservative assumption that m/b, .15, this reduces to the following:
=
5.9-14
/
Welded-Connection Design
(.80) a; t b (.15 bb) t
2
_
bb t b -
• -
+ (.80)
7 a; t.2 = a; b, t b
Application to Truss Connections This Lehigh work for beam-to-column connections will now be applied as a guide for determining the distribution of compressive forces in a truss connection. It is assumed that this transfer of the Hange force of (6) occurs in the web of membeZ®within distance of (t 5K). See Figure 31.
.12 bb t b 5.6
I
or t. ~ .40 ~I
(3)
+
If the column flange has this thickness, stiffeners are not required as far as the tension area is concerned. We might carry this thought one step further and apply it to a tension Hange which connects to the member at an angle other than 90°, such as in a truss connection. See Figure 30. resistance of supporting flange (t.)
P = (.80)
(Ty
+
t b (.15 bb)
(.180) 7
(Ty
.'. (.80)
or
I
t. > =
(Ty
®
t.2
(Ty
t b (.15 bb)
~bb
t The vertical component of the web force of member @ transfers directly into the web of member within the distance of d sin 1>
pull of tension flange (t b )
P 1 = bb t b
Here:
t b (sin5.60::
+ -
t.2 = bb t b
(.80) 7
.12)
Within the region b-c, these compressive stresses in the web of member overlap and would be added.
®
(Ty
(Tr
I
sin
0::
F r sin (4) (
_
1>
Si~b 1> + 5K )w
+
F w sin
(s: 1» w
P - - - - - - ---"1- __
_
1
FIGURE 30
P
® I
r-
= PI sin a
0 I
.,
®
FIGURE 31 w
II
'I
II
'I
,I
,I
II
1>
Design of Trusses
/
5.9-15
€--~-_ Flange
stiffeners
L
K
T FIGURE 32
or I
CT
2
= sin 4> [ t b
-----w-
+
F r sin 4> 5K
+
dF w ]
I
(5 )
Another method would be to assume ultimate load conditions, with all parts involved, stressed to yield. Using the previous formula (5): where:
The vertical component of the web force of member @ transfers directly into the web of member within the distance d sin 4>
®
The compressive stress within this section would be-CT2
Fr Fw
=
force area
F w sin 4> d sin 4> w
Within the r~on (b-c ), these compressive stresses in the member ® overlap and would be added: or w
~
2
sin 4> [tb
+b5~\in
4>
+ wbJ ..... (6)
®
If the thickness of the web (w) of member satisfies this formula, stiffeners are not required. Normally, member @ will not be stressed up to its allowable in compression, so that this shorter method of checking stiffener requirements is on the conservative side.
+F F sin 4> ( t ) sin 4> + 5K w + b. t, r
CT
=
r
where: F r = bb t b F w = d Wb
If Formula 6 should indicate that stiffeners are required, the same method of analysis may be extended to get an expression for the cross-sectional area of the vertical stiffeners. See Figure 32. It IS assumed the transfer of the flange force of member (6) occurs in the web of member within the distance (t 5K) as well as in the flange stiffeners. The compressive stress within this section would b~ CTl
=
force area
(
Si~r 4> + 5K )
w
+ b. +
F r sin 4>
b. t.
CTy CTy
t,
d Wb
CTy
sin 2 4>
dw
®
(Si~r 4> + 5K )w +
sin 2 4> d w (7)
Now if ultimate load conditions are assumed, that is all parts involved are stressed to yield:
4. VERTICAL STIFFENERS
+
w
and the required cross-sectional area of a pair of stiffeners becomes:
(~+
5K
)w
(8)
5.9-16
~
/
Welded-Connection Design
(8)
... -
...
I---,-!Lep +5K-I .
W
•
-
Sin
~
IT
-
t,
..1.. K
~JY
/
iIII
// // /
tb
//" I // II // II /// II
-,
About
II
//(
..., ,-w b
II
C
T
'I II II
FIGURE 33
II II II
II II II
II
I
II
1
I-~-I 5. LONGITUDINAL STIFFENERS IThe type of connection shown here may be reinforced with two stiffeners placed parallel to the web, and welded to the flanges of member @. See Figure 33. In the Lehigh test of this type of stiffening for beam-to-column connections, these plates were added along the outer edges of the flange so that beams framing in the other direction could be attached directly to them without extending within the column section. It was found that these plates each carried about %6 of the applied compression, while the central web section loaded up and carried the remaining 0/8. For this reason the recommendation was made to assume these plates to be about half as effective. It is interesting to remember that when a beam is supported at three points, the two ends and the center, the two outer supports each will carry only %6 of the load and at center 0/8 of the load. If the outer supports are pushed in for 7t of the beam length toward the center, all three reactions will be equal. By setting the stiffening plates about 7t b, in from the edge of the flange of member @, as shown above, it seems reasonable to assume they will carry a greater load and can be considered as effective as the web. Although the K value a~Fes only to the distribution in the web of member ® and has nothing to do with these side plates, the Lehigh researchers for simplicity assumed the same distribution in the plates. The compressive stress in the web and the two side stiffeners due to the vertical component of the flange force of member @ is:
The compressive stress in the web of member @ due to the vertical component of the web force of member @ is: force area
F w sin cf> d ---w sin cf>
F w sin 2 cf> dw These stresses are added together.
F, sin cf>
CT
(
Si~b cf> + 5K
) ( w
+F
CTI
(
Si~b cf> + 5K )
w
F f sin cf>
+ 2 ( Si~b cf> + 5K
sin 2 cf> dw
where: bb t b Ff F w = d Wb
+ 5K ) (
w
+ 2 t. +
CTy CTy
)
d Wb
2 CTy sin cf> dw
and the required thickness of the two vertical plate stiffeners becomes:
>
F f sin cf> -
w
) ... ( 9)
Now if ultimate load conditions are assumed, that is all parts involved are stressed to yield:
®
force = area
+ 2 t.
) t,
b_ 2(_t sin cf>
+ 5K)(w -
w, sin2 cf»
.(10)
I
Design of Trusses
T 1
5.9-17
FIGURE 34
~
II
®
II I
II II
LLy-1" __ylJ
I
T
®
FIGURE 35
These plates must have sufficient welds connecting them to the lower Range because the compressive force of member (A) enters here. Since fillet welds cannot be placed on"t6e inside, this would mean a rather large fillet weld on the outside. It may be more economical to bevel the plate and use a groove weld. In this example, the vertical compressive force is transferred from the plate down into the vertical member @; thus a simple fillet weld along the top edge of the plate to the upper Range would be sufficient. This discussion and resulting formulas will allow the connection to be detailed without computing the actual stresses. It is based on providing a connection as strong as the members. Since member will normally not be stressed to its full allowable compression, a more efficient connection would probably result if the actual stresses were computed, using these guides on distribution. Instead of providing full-strength welds, their size would then be determined from these computed forces. These ideas will now be applied to various parts of a truss connection.
®.
ponent to enter the lower Range of This force ( F ), now in the stiffener, gradually transfers into the web of as shear, from section a-a to section b-b.
®
This unit shear force is equivalent to v =
~:
The weld
®
between stiffeners and web of member would be designed to transfer this shear force (V), Figure 35.
F -------, F y
I I
I
0
~
I
_ _ II
F h
6. STIFFENING ACTUAL TRUSS CONNECTIONS The vertical component (F v) from the Range of.-iA) enters the stiffener and passes into the web of as shear, V F v, along section a-a. The horizontal component (F0.. from the Range of @ enters the lower flange of ® . The weld between stiffener and web of member would be designed to transfer this shear force (V), Figure 34. The force (F) from the Range of transfers directly into the stiffener, leaving no horizontal com-
00-
=
®
0
FIGURE 36
The force (F) from the Range of @ enters the stiffener, and is transferred through to the opposite end. The vertical component (F v) enters the Range of and the horizontal component (Fh) enters the
©'
5.9-18
/
Welded-Connection Design
®.
upper flange of No shear force is transferred thro~ the weld between stiffener and web of member ® . Only enough weld is required near midsection of stiffener to keep it from buckling, Figure 36.
concentrated force into the web is to be taken, then the conservative method may be used. Thus, it is assumed that the flange force must first be transferred as shear into the web of the same member before it is transferred through the connecting weld into member This weld may have to be made larger because of this additional force, Figure 38. If this flange force (F) is high, a web doubler plate might have to be used so that these forces can be effectively distributed into the web of @ without overstressing it.
®.
F, ..;x.:----_+. F,
®
Stiffener
1 Problem 2A
I
Consider the connection of Figure 39, using A373 steel and E60 welds. In this case a portion of the vertical component of @ is transferred directly into It will be assumed that the vertical component of the left flange A and the vertical force in the right flange of C will be transferred around through the web of B by means of two vertical stiffeners. See Figure 40. (a) Check the size of the connecting welds on the flanges of @ .
©.
l FIGURE 37
unit force on flange fillet welds
The force (F) from the flange of @ enters the stiffener, and is transferred through to the opposite end. The vertical component ( F v) is taken by the second stiffener as (F.), and the horizontal component Figure 37. (F,,) is taken by the upper flange of In these last two cases, it is assumed that no portion of the force (F) in the stiffener is transferred into the web of The welding of the stiffener would be similar to the previous case, that is Figure 37.
ft =
(138 kips) 2( 10)
®'
®.
F
[
6.9 kips/linear inch
leg size of flange fillet welds 6.9 9.6 .72" or use 3J4" (or use a groove weld)
® +--
(b) Check the size of the connecting welds on the web of @ , which has a force of 74 kips.
unit force on web fillet welds II
fw =
F
[
II II II
I I I
II II
L+--r+U FIGURE 38
If there are no flange stiffeners on member B and no advantage of the preceeding distribution of the
(74 kips) 2(17.5)
= 2.11 kips/linear inch leg size of web fillet welds 2.11 9.6 .22"
Design of Trusses
/
5.9-19
14" \¥' 136#
® 300k
•
- - - - - - - - - - - - - - - - 7 1 - - - - - - - - - - - - - . . - 548k " "
"
I
I I
10"\¥' 100# 10" \¥' 60#
FIGURE 39
118k ---l~====:::;::;t======:::;;:;;;;:::::====~I-- 2l5
k
®
FIGURE 40
However, the minimum fillet weld to be attached to the 1.063"-thick flange would be Ww = %6". (AISC Sec 1.17.4)
(c) Determine liequired sectional area of vertical stiffeners.
A. =
OOFv . U'y (97 kips) (29.7 ksi)
= 3.27 in. 2 , or use two Their A.
(AISC Sec 1.5.1.5.2) %I"
= 3.75 in. > 3.27 in. 2
2
x 5" stiffeners OK
( d) Check the size of connecting welds to transfer this force (Fv) as shear into the web of B.
unit force on stiffener-to-web fillet welds 97 kips f - 4(12.6)
=
1.92 kips/linear inch
leg size of fiUet welds 1.92 9.6 .20" or use
l(4"
~
( e) Check the vertical shear stress along a-a. 7'
V Aw
See Figure 41.
(97 kips) (.660) (12.62) OK 11,650 psi < 13,000 psi < .40 U'r (AISC Sec 1.5.1.2)-
5.9-20
/
Desig~
Welded-Connection
®.
---------~----------------
edges of the upper and lower flanges of (gl. There is one more item to check; consider point Q0 in the figure below. It is necessary that the vertical component of the right flange of be transand yet its horiferred into the left flange of zontal component be transferred into the lower flange of
©'
®
®.
I I
I I
®
I
FIGURE 41
(f) Check the horizontal shear stress along b-b ~arallel to the welded connection in the web of between and l!V . This length is about 20".
®
®
The total horizontal component from (A) to be is 248 kips. The lower'tl'ange of transferred into has a compressive force of 215 kips on the right end and 118 kips on the left end. This means it will pick up 215 - 118 = 97 kips from @ Hence, a force of 248 - 97 = 151 kips is to be transferred into the web of over a distance of 20".
®
®
®
T
FIGURE 43
Theoretically, the flange of @ can only transmit There an axial force (F) between points.B and would be no problem if these 3 flanges met at a common point.
0 .
V
Aw (151 kips) (.660) (20) 11,430 psi
(.707)
A..[ 'I'
2 [
tb
+
bb
5K sin cP
(
1.118
w > 1.16/1 required
tb
+
Wb
]
10.345)( 1.118) + 5(1l7ie)(.707)
>
+
.685
J
©
(b) Check the tension flange of where it joins the flange of as to the necessity of stiffeners to transfer the flange force; Figure 48.
®'
t,
®
>
I
V
.40 bb t b .40 y~-:-(1--=-0.--=-07=5-:-)(:-.683=-=-::-:)
1.05/1
= 30
..., f \ ,\"~'Cf> ", 1 "
FIGURE 54
at top edge of gusset plate
=
\~ ..~
'T',
7"
FIGURE 53
If>
f __ fh _ - cos -J..
" " , .
-J..
1" cos ~'
I \ \
0=0
/
_ 8400 . pSI
w
=
(2.51) (9.6)
=. 261"
or use
(d) Flange plates, O/S" by' 4%", are welded onto back a sufficient disistance. The compressive force in the flange of
© to extend the flange of ® F = 200k (10.028) (.618) ( 15.88)
78
=.
°
®
ki
ps
5.9-26
I
Welded-Connection Design
On this basis, the stress in each of these flange plates is: (78 kips) (2") ( o/s") ( 43/4" ) 13,100 psi
OK
The force from an adjacent pair of these plates is transferred into as double shear.
©
©
must be taken by alone. The cross-sectional area is A = 15.88 in 2 • of For the same stress in this would require the same cross-sectional area, or 15.88 in.", and a net width of
®
©'
W --
15.88 o/s
= 25.4"
There is sufficient width; see Figure 52. (f) At section c-c halfway along the Hange,.Rlates, it is assumed that half of the flange force of @ has been transferred out into
©:
rz, r"
(200k)
(10.028) (.618 ~ ( 15.88)
- 390 ki -. ps
For the two flange plates, this reduction would leave( 2OO k) - 2 (39.1 k) - 122.0 kips to be taken by
©.
For the same stress, this would require an area
FIGURE 56
©
This shear stress in T-
_
of( 122.0k) A = (12.52 ksi)
is-
F 4 L t
and a net width of-
(78.0 kips) 4(12") (o/s")
= 2600 psi
182 kips Member
L
r -
168 kips
OK FIGURE 9
® (238) (4.38)
= 54.3
(b) Use a %" gusset plate on this connection, resulting in Figure 10.
and the allowable is o: = 16,660 psi
P=erA
moment applied to pipe
(16,660) (14.58) 243 kips Member
@
P
erA
>
200 kips
OK
(126 k ) (7.86") 990 in.-kips
( 154 k ) ( 6%" )
(20,000) (7.265) 145.3 kips
Mh
>
126 kips
982 in.-kips OK
assumed value of e e = 12 t 12 (%) 41f:t'
5.10-6
/
Welded-Connection Design
~ e
©
48"
d -
--J
6%"
FIGURE 10
e
...L
maximum unit force (radial) applied to 1" ring section of pipe @
fb
=
(d
+
6 Mh e) (d + 2e)
6(990) (48 + 4Jh) (48
This represents a worse condition than actually exists.
5_
+
9)
1.98 kips
t2
6 (%)2 6
= .023 in." M m ax (at force f) = k f r
= (.318) (1.98( 6) = 3.78 in.-kips M
(T-S FIGURE 11
( 3.78) (.023 )
= 164,000 psi Although there is just a single radial force ( f) acting on the pipe shell, assume there is an equal force on the opposite side of the shell, resisting this force.
Excessive
Because of these excessive bending stresses within the pipe shell resulting from the moment applied by
Connections for Tubular Construction the connecting plate, some means of stiffening the pipe within this area must be used. There are several possibilities. (1) One possible solution is to put a casing around the pipe so as to increase its wall thickness. This will provide sufficient section modulus so that the resulting bending stress is reduced to an allowable value. (Assume a = 18,000 psi.)
S
M tr
(3.78 ) ( 18,(00)
S t
t2 6
Y6S
V6
(.210)
1.12" required wall thickness Since 1.12" - 3fs" (present thickness of @ ) .745" required additional thickness, or add a %"-thick wrap-around sheet around this pipe @ in the area of the connection. See Figure 12. ( 2) Another possible solution would be to add to the wall thickness at top and bottom of the connection.
--------~-----..~
© 48" 38"
Do not need circumferential fillet welds around either ~ end of 3/4" liner unless to
1" X 10" wrap-around It
seal the ends
FIGURE 13
60" groove weld on 3/4" liner also joins pipe member. Weld lies along neutral axis of
F
section to resist bending
(%" gusset
Mil
d (990 in.-kips) (38")
It
26.1 kips W'-thick stiffening liner around pipe
FIGURE 12
5.10-1
or
.210 in."
pipe, so this becomes built-up
/
M
k F r, ( .318) (26.1 k) ( 6") 49.8 in.-kips
5.10-8
/
Welded-Connection Design
®
5" Pipe
%" gusset It
FIGURE 14
t~
M
S
w
/@
(a) Resisting horizontal bending stress and vertical shear stress
5.11-13
the curved haunch section, as described in following paragraphs.
FIGURE 19
(J+~
/
(b) Components of these two stresses FIGURE 20
(c) Resisting radial bending stress (0"1') normal to curved section (A-B)
5.11-14
/
Welded-Connection Design
will produce an equal and opposite moment. The value of this tangential shear force (V) acting on this curved section (A-B) may be found from the following:
I
V
~'I·
(23)
Transverse Force (P/) Applied to Wedge Member The applied transverse force (P,") results in horizontal bending stresses (CTIl) as well as vertical shear stresses; Figure 20( a). These two stresses may be completely replaced with a single component, radial bending stress (CT r); Figure 20 (b ). The results are shown in Figure 20 ( c ). Notice that no tangential shear stresses are present. Axial Force
(PD')
moment applied to section A-B
IM =
+ Pt pl·
M'
(25)
normal stress on inner flange [CTr =
+If-+ ¥I ......·......·....·
(26)
normal stress on outer flange [CTr = -
!f +~I ......·........·..
(27)
Problem 2
Applied to Wedge Member
The axial force (PD ' ) applied at the apex of the wedge member, causes radial stresses to occur along the curved section (A-B); Figure 21. There are no tangential shear stresses from this force, because they cancel out.
To check stresses and stiffener requirements on the knee connection shown in Figure 22, for the loads indicated. A36 steel and E70 welds are used. STEP J: Check Lower Curved Flange (Figure 23)
Summary The effects of all these forces applied to the wedge member may be summarized as follows:
shear stress on section A-B
~ V=
'
properties of haunch section (1-1) Use reference axis ( y-y) through centerline of web plate. Plate
%" x
P
10"
112" x 48.25" 1" x 10"
Toto I
~ ----+
(a) Resulting axial stress
A
y
7.50
+24.500
M
=
A"y
0
0
10.000
-24.625
-246.25 -
62.50
Tangential shear stresses cancel out
(b) Components of axial stress
FIGURf-21
= M"y
I.
-
+ 183.75 +4502.
24.125
41.625
Iy
(c) Resisting radial stresses (CTr)
0
4681
-
+6064 15.247
Rigid-Frame Knees (Elastic)
/
5.11-15
-I
__- - - - - - - - 2 2 5 " - - - - - - - -.... 10" X 3;''' flange
, - 10" X 3;4" flange
W' web
r-----------------
I
I I I I
---------------_ ..--db = 50" \H = 150"
~===::::;::::====±::=:::::J
I
Point of inflection
1" flange 150" r = 100"
~centerof curvature
FIGURE 22
£-----100"
'>
1 -. '1 ..
r;============::::jCD-+ F,
___ --J~
---:--f-Neutral
-
cxrs
FIGURE 23
I
"~I5"
-
= 10& k
H= 150
11 t
i --
y
50.0"
-r r
1501 "
~5'
10" X 1" Haunch section (1 -
1)-
+ I,
-
(15,247) 15,153 in."
M2
NA
A
M
-T ( -62.50) ( 41.625)
( -62.50):! ( 41.625)
-- 1.501 Cr
23.125
1)
5.11-16
/
Welded-Connection Design
average stress in lower curved flange at (1-1) (Tt
_~+ M
-
A
I
43
55
zoo
30
p
0.386"
or
I~
d. (w r
-
V3
w)
................. (14)
or could use
{a}
t. = t r also in all cases
I
~
t cos hf3 x
, . (34)
5.12-24
/
Welded-Connection Design
~-== CD ---
As in the tapered haunch, the plastic section modulus (Z) at any given point (X) is:
t
--- ------
0- - -
,
\
',;., th
For any given depth (d x ) , the plastic section modulus (Zx) may be increased by increasing the flange thickness (tit). Assuming the web thickness and flange width of the curved haunch is at least equal to that of the beam, the required thickness of the lower flange would be:
I'
'.f, cf>
'
\ I
I
,,\
r-r- 0
T -0
Z;
=
bit tit (d, -
tit)
+
~II (d, -
t:
2 t lt )2
II
1 FIGURE 34
d x 2 bit _ Z 4 x bit - WI'
(36) r
~
6 bh
I
*
(38 ) 0
The AISC Commentary (Sec. 2.7) recommends that the thickness of this inner flange of the curved haunch should be-
I tit ~
(l
+
m) t
I
(37)
where values for (m) come from the graph, Figure 33.
This is based on a 90 knee (outer flanges form a right angle), which is the most conservative. The radius of curvature may be increased above this limit if additional points of support are added to decrease the critical arc length (C). The unbraced length between points of lateral support must be held to--
Ic ~
6 bit
I
(39)
.5
~~III
E
.4 .3
where
C
rc/>
.2
c/>
radian measure
.1
3
4
5
n
= aid
6
7
FIGURE 33
Here: a
If the unbraced length (C) exceeds this limit, the thickness of the curved inner flange must be increased by0.1
(~I
-
6 ) tit
or the final thickness will bedistance from point of inflection (M = 0) of the column to the point of plastic moment (M p ) in the haunch
d = depth of column section In order to prevent local buckling of the curved inner flange, limit the radius of curvature to--
An alternate method would be to increase the width of the inner flange (bit) to a minimum of C/6
* ASCE
Commentary on Plastic Design in Steel, p. 116.
Welded Connections for Plastic Design
o22.5° -
/
/
5.12-25
y/4
Fc = A,c ay
F = A a
~~~' &"
'y
&1'
o /
/ /
CD
/
/
45° - y/2
CD
90° -
y
FIGURE 35
without decreasing the original flange thickness (th )
Ib
h
>
~ I'
:
(41 )
2 Ac
ITy
2 A"
S10
or
I As
~
'
Diagonal Stiffeners (1) Based on compressive forces at
@
An approximate value of the compressive force applied to the diagonal stiffener as a result of the compressive forces in the curved inner flange may be made by treating the curved haunch as a tapered haunch. See Figure 35.
'Y /4)
sin (22.so -
(90° 4-
'Y )\
(2) Based on tensile forces at
( 42)
©
The compressive force in the diagonal stiffener is found by taking the horizontal components of these tensile flange forces, and setting them equal to zero, See Figure 36. At
ITy
cos 'Y -
w,
tan(
a:
a,
ITy
~-
+ 'Y) V3 -
Vw = wh(CD)T y ~
©-;b~,~ -,Y ...... a
+ Y'
. . ... J \,d ...
®
,h
... \ ------'~®
Resisting shear forces in web of section ABCD dh CD = -,-ta-n-,('a-+-,----y. ). FIGURE 36
As
IT y
cos 'Y cos
a:
0
5.12-26
/
Welded-Connection Design
FIGURE 37
Radial compressive force exerted by web.;J
or
Transverse tensile stress due to bending of flange
, I
A • --
A.
>
A t cos Y cos a:
I
V3tan
rAt -.rs3 tan( d ~ cos y
Wh
(a:
h
a:
+ y)
+ y) J
cos
a:
... (
43 )
I
(b)
rnj
tnrm
where:
At
area of top (tension) flange of haunch
A.
total area of a pair of diagonal stiffeners
Radial Support of Lower Flange The radial components of force in the curved inner flange tend to push the flange in toward the web, and to bend the Hange as shown in Figure 37 (b). Because of the slight yielding of the outer edge of the flange, there is a non-uniform distribution of the flange stress ( U"), Figure 37 ( a). This stress is maximum in line with the web. There is also a transverse tensile stress across the outer face of this flange, Figure 37 ( b ). The unit radial force (f r ) acting on the curved inner flange from the axial compressive force ( Fe) within the flange, Figure 38, isf r = Fe (lbsjcir inch) r
Treating a 1" slice of this flange supported by the web of the haunch as a cantilever beam and uniformly loaded with this unit radial force (fr ) , Figure 39:
FIGURE 38
Z -
th2 4
r,
Fe
U"y
bh t h r
r
or unit load (p) on section: U"y
P =
r
th
I
Welded Connections for Plastic Design
5.12-27
ness (br/t x ) of the curved inner flange to the following, whichever is the smaller: .................... (45)
Provide stiffeners at and midway between the two points of tangency. Make the total cross-sectional area of the pair of diagonal stiffeners at their midpoint not less than 34 of the inner curved flange area. FIGURE 39
Summary of Curyed Haunch Requirements thickness of outer flange (t ) > t b web of haunch
_
th(~)2 M -_ cry 2 r 2 -
(Wh)
>
Wb
thickness of curved inner flange (t ) > _tb_ h = cosf3 (1 m) t
+
=
also
(based on tensile flange)
A. > bh < ~ 8 r = 4 2
cos 'Y cos ex:
[At _ V3 tan d(ex: Wh
h
]
+ 'Y)j
(based on compressive flange)
or
A. >
Therefore limit the ratio of flange width to thick-
2Ac sin ( 90 4- 'Y )
If bending stress at
®
Haunch
FIGURE 40
and
5.12-28
/
Welded-Connection Design
outer flange thickness (t) does not have to exceed beam flange (til)'
6. BEAM-TO-COLUMN CONNECTIONS (Multiple Span)
Web Resisting Shear When the moments in two beams framing into an interior column differ by a larger amount, this difference in moment will cause large shear forces to act on the connection web. The web must be checked to see if it has sufficient thickness; if not, it must be reinforced with either a web doubler plate or diagonal stiffeners. (See Figure 41.)
Otherwise, use additional lateral support to decrease arc length (C). Assume critical section (x-x) at-
fix
= 12°
then
horizontal shear applied on connection iceb along top portion and
F2
Zx >
Mx
M2 d2
x
where: r
F1
-
M1
_
V4
_
a,
4
shear resisted by connection web
1>
along top portion
radian measure
w de
Ty
Otherwise, increase the thickness of the curved flange totIl
[1
+
0.1
(~
-
6)
or
J
or increase the width of the curved inner flange to-
bll >
or ......... (46)
~ where:
without decreasing the flange thickness.
bh tx
24-
n'
ZZ
(DASHED LINE)
EXAMPLE:
tb
BUILT UP COLUMN = 1.135 (24" W" /60'8EAM)
Kc
..
Z5 l'
3"1" (3'" FlANGE.""~ ~ Ws.LD)
o, +5 tc.)
.. /8. ~5"
..... ."
..
Q
C
(I) II>
IA' :::I
Z7
H
READ
..
(I)
II>
I
EXAMPLE: ROLLED COLUMN tb -.4S3 (/4 w:- 34- BEAM) Kc .. (8· W" 35- COLtJMN) READ (t b -I- SKc) = 4.8"
:::I :::I
'Z/
I
...
0
'ZO
I
(DOTTED LINE)
n n
19
!--I I,'
tb+5 1; M= SO/k
"\
21" W
\
68# beam
\ \
(~o,.
\
M=:J
\
d, \ \ \
•
\ \
M = 160'k \
\J
1
w- =
14" W 48# column
I~ FIGURE 49
O'y
y'""3(210 ft-kips x 12) (21.13) ( 13.81 ) (36 ksi)
:>
J
V3M
db dc
.416"
Conclusions (Fig. 50) (a) This required web thickness would be satisfied if the beam were allowed to run through the column. This would give a web thickness of .430". OK (b) If the column were to run continuous through the beam, as illustrated above, then a lj.{' doubler plate would be required in this connection area to make up the difference in thickness. ( c) Another choice would be to use a pair of diagonal stiffeners having the following cross-sectional area:
d. (w r -
beam dimensions db
21.13"
bb
8.27"
w,
.430"
(23.18) (.416 -
y3
Or use a pair of 3" by 3!s" stiffeners, the area of which checks out as-
column dimensions
3!s" (2 x 3"
2.38
13.81" .339"
t, >
1%6"
diagonal of connection web
-I d + d b
in. 2
>
+
.339")
1.03 in. 2
OK
Also, the required thickness is-
8.031"
d.
.339)
1.03 in. 2
.685"
db
we)
y3
2
..; 21.13 2
e
+
b.
17
2 x 3" 17
2
13.812
23.18"
Web Resisting Shear The necessary web thickness will be determined by the AISC requirements for webs in the connection region. The algebraic sums ofthe clockwise and counter-clockwise moments on opposite sides of the connection are:
.35"
>
-J
.--..---.. .-.--....--F
;.
>
;.
>
>
/
=~
.:>
/
/
>
/
Stiffeners added
FIGURE 4
to behave similarly. The curved knees are the most rigid, have the highest moment capacity, and have a rotational capacity somewhere in between the simple square comer and the haunched knee. As the radius of curvature of this inner flange is increased, the stiffness and moment capacity increase slightly, with slightly lower rotational capacity.
4. SQUARE CONNECTIONS When the flanges of one member intersect the flange of another, stiffeners should be added in line with the intersecting flanges. The stiffeners transfer the forces of the flange back into the web of the other member. See Figure 5. These flange forces are distributed as shear into the web along the full web depth. This will prevent the web from buckling due to the concentrated flange forces.
The unbalanced moment about a connection will cause shear forces around the periphery of the connection web, Figure 6. The vertical shear force and the horizontal shear force will result in a diagonal compressive force applied to the connection web. Unless the web has sufficient thickness or is reinforced, it may buckle. According to plastic design (and this may be used in elastic design), the required thickness of the joint web must be--T
tw
=
f,
=
Fh
d, -
~ db d,
and:
It
w
d
~ I
(1 )
T
h
, ...
FIGURE 6
Welded Connections for Vierendeel Trusses
/
5.13-5
M = algebraic sum of clockwise and counterclockwise moments applied by members framing to opposite sides of the joint web boundary at ultimate load, inch-pounds
- -- -
For a panel subjected to shear forces and having a ratio of width to thickness up to about 70 (the connection webs will almost always be within this value), the critical shear stress (Tcr) equals the yield shear stress (Ty),or-
FIGURE 7
and
T cr
where: t.. = thickness of connection web, inches f" = unit shear force, lbsjlinear inch = T tw
or:
Ty
db = depth of horizontal member, inches d, = depth of vertical member, inches
.. '................•..... (2)
/
>
> Web doubler plate
>
Web doubler plate
>
/
1/
-h
V
-- ~
V ///. "/://///
FIG. 8 Methods of obtaining web thickness to meet requirement of Formula #2.
A
A
(a)
A
Web of connection reinforced with web doubling plate
Diagonal stiffener
(b) Web of connection reinforced with diagonal stiffeners
>
>
~ I->
1\
(c)
longitudinal stiffener fillet or groove welded
A
Web of connection reinforced with longitudinal stiffeners
5.13-6
/
Welded-Connection Desigfll
If the thickness of the connection web should be less than this required value, AISC in their work on Plastic Design (which may also be used in Elastic Design) recommends adding either (a) a doubler plate to the web to get this required thickness, see Figure 8, or (b) a pair of diagonal stiffeners to carry this diagonal compression, the area of these stiffeners to be sufficient for just the additional requirements. It seems reasonable that (c) a pair of longitudinal stiffeners extending through the connection area would be sufficient to resist this web shear. These stiffeners would be Hat plates standing vertically between flanges of the chord member and welded to the flanges near their outer edges.
of Figure 9. Because of the slight yielding of the flange's outer edge, there is a non-uniform distribution of flange stress (0"). This stress is maximum in line with the web. In addition there is a transverse tensile bending stress (O"t) in the curved flange. If this value is too high, stiffeners should be welded between this flange and the web. These keep the flange from bending and pulling away from the web. These stiffeners usually need not extend all the way between flanges, but may be a series of short triangular plates connecting with the curved flange, In the following formulas, the values of factors ex and f3 come from the graph, Figure 10,·
longitudinal tensile stress in flange
Iu_. = ~ I'·
5. CURVED-KNEE CONNECTIONS Tensile stress (0"mean) in the inner flange of a curved knee tends to pull the flange away from the web, and to bend the curved flange as shown at the lower right
(3)
transverse tensile bending stress in flange
I
= f3
----._----.-
A
at
B
STEP 3: Determine Stiffness and Carry-Over Factors
A load of a unit angle change (ep) is applied to the elastic area at the outer edge (A), and the resulting end moments (M A) at @ a;d (MAIl) at are found.
®
FIGURE 27
b
Design of Rigid Frames
6.1-19
/
TABLE 3-Design Summary: Beam Cover Plated At One End w
A=
0(0
+
e.
b
,Y
CB
I
2 h)
b'
+
I,
I,
2 A
=
,,
I
en -
1_e.
--,,-( a 2 a
eh
-a2 =
)'
+ -I,b
(
b )' e. - 2
-~(~-~) ~2 ~:
(30
+
1
h)
1
+
I
C:t, ChI ea eb are considered to be (+)
------J
Fixed End Moments at
At0
++ ~
G) M,.
Mre =
=
P.
+
Ph
--~-
A
+
M y _y
CB
1,-,
End Moments Resulting from Treating Angular Rotation as a load 1
M.
=
~I
1n-----I-M-AA 1,-,
1
~
Mn
1
;;:+
en'
1,-,
Stiffness Factors at
CD I
KB =
MB
Carry-Over Factors from
carry-over factor, ellA
0
®
to
to
G)
1
CnA
@
MAli = - M --g
Summary
This example of the uniformly-loaded, fixed-end beam with cover plates at one end may he summarized as in Table 3. Modified Example
Although the work is not shown, the same fixed-end beam with cover plates at botli ends, uniformly loaded, may be summarized as in Table 4. (See next page)
from
CD to
01
ellA
~:n
1
6. COLUMN ANALOGY METHOD APPLIED TO BEAMS HAVING GRADUALLY VARYING SECTION The following method may be used to find the fixed end moments, stiffness factors, and carry-over factors of beams which have constantly varying moments of inertia, such as haunched and tapered beams, Figure
27. A beam which tapers along a straight line (in other words, its depth increases linearly along the length of the beam. see Fig. 28, top) will have a moment of inertia (1) which does not increase linearly
6.1-20
/
Miscellaneous Structure Design TABLE
~Design
Summary: Beam Cover Plated At Both Ends
w
It
La _ 12
wo"
w b
121;
here
I
M._. =
b3
12
It
3 b)
+
(6 oa
b"
+
It
0+
~ (4
P.
++1
2 a
A =
6 a b
+
bal
0
Fixed End Moments
End Momenta Resulting from Treating Angular Rotation as a Load
MA=MR=
A
La I +-4 1._.
and
@
- 1
J A
~:-. I
4
Stiffness Factors
at
0
I
KA= KR =
MA
I
Carry-Over Factor. CAB =
but will have a slight curve (see Fig. 28, center, solid line). This curve approaches a straight line as the beam becomes less tapered. Although a slight error will be introduced, it will greatly simplify the analysis if we assume this moment of inertia distribution to be a straight (dotted) line. However, this slight error may be reduced by breaking the beam into two parts (see Fig. 28, bottom) and assuming a straight line variation of the moment of and inertia between the three points @, This is represented by the dashed line in Figure 28, center.
®'
©.
STEP 1: Determine Properties of the Elastic Area
CRA =
_
I
MAR MA
area of elastic area Ax
I
AZ =
I
a B
_
b e -
I
I A
IB
oge~
Ie I loge-I B
B
moment of elastic area Ax about axis A-A
x/ MA /A-A
_ (
-
h -a I A
)2 (
IB
-
IA
-
I
I B)
I B loge
~:)
I A og, I A
moment of elastic area A z about axis B-B
MAi~_B
= ( Ie
~
I
B
)
2 (
Ie -
IB
-
Design of Rigid Frames
I
I [===========~=J ,./ ..
I.,......-----l---------:j __ a
I -I
6.1-2.1
©
®
-
/
= + tI=I+IC-!BZ~I f
I
IA
B
A
X
"fgp..ered beam
I.
a
•
B
b
L
I
IA
(,
.
d:."~.;"l;~.:..~.;.-.-... 1
rt:=X~a
_.
i
B
'c
.r-Z~ Moment of inertia
(])
®
1
T1
Elastic center
®/t
I
IA
L
ep
I
0r- c , j 0 ~
cA
®j
I
------*----
c.
~
Cc - - - - ' - - - - - -
(])
Elastic area
FIGURE 28
distance from
e.G. of elastic area A" to axis A-A
moment of inertia of elastic area Ax about axis A-A
MAxi
IA-A Ax
distance from
e.G. of elastic area A. to axis B-B MAsI IB-B
A.
moment of inertia of elastic area A. about axis B-B
moment of elastic area A. about axis A-A
MuJ = I A·A
a A.
+
MAsI I B-B
total moment of elastic area about axis A-A M A •A =
MAxi IA.A
elastic center (y-y) M A •A
-AL -
CA
+
MA.I IA-A
Since these moments of inertia can't be added, not being taken about the same axis, it will be necessary to shift axis B-B and axis A-A to the elastic center y.y. If axis A-A is always taken at the shallow end of the tapered beam, negative signs will be avoided in the calculations.
6.1-22
/
Miscellaneous Structure Design
moment of inertia of elastic area Ax about axis y-y Using the parallel axis theorem:
+
IAxl = lAx! I A-A /..
momen (M) applied to elastic area about its elastic center
Mxl
Ax c/
_ ~[-
IY,y -
2
o,
+
(L + CA) IAxl I A-A CA L
-
IAxl = lAx! I A-A
r.:
Now we wish moments of inertia of Ax about the elastic axis y-y, and again using parallel axis theorem-
+ Ax(CA -
IAxl
IAxl
I x -x
In or IAxl
TAxi
Iy-y
-
Ax c/
where:
o.
cx ) 2
+
Ax(CA -
= (Ill a I
Cx )2
-
A
r[
In -; I
A )
(2 In2
7 In I A + 11 IA2) -
IA3 lOge::]
/A-A
and IAxl
IAxl lA-A
Iy-y
+ Ax CA( CA -
MzI _
2 cx)
IN -
~[ -
Qz +
(L -
3 a + CA) IAzl
/B-B
2
+[a(2 L -
3a) -
cA(L -
2a)]
moment of inertia of clastic area A, about axis y-y
IB-B
+a
in same mannerIAzI
IAzl
Iy-y
IB-B
+ Az [(cz + b -
CA)2 -
MAzl
(L -
a) (a -
J
CA) Ay
ez2] where:
total moment of inertia of elastic area
r.;
= lAx!
+ IAz/
In
I y-y
STEP 2: Determine the Fixed End Moments The moment diagram from the applied load on the real beam is divided by the moment of inertia (I) of the real beam, and becomes the load (Mil) on the elastic area which is treated as a column. The axial load (P) applied to the elastic area is equal to the total rvI/I. This axial load applied at some distance from the elastic center of the elastic area causes a moment (M) Oil the elastic area. Both of these loads cause "stresses" on the elastic area. The following applies if the designer can assume a .uniform load (w):
__v\' ( 2
L MAxi lA-A
P = ~vrJ- IAzl z
2
L
IB-n
+
(L -
M y _y = Mx!
I y-y
IAxl ) lA-A
+
MzI
I y-y
fixed end moments at
@ M re =
at
axial load (P) applied to elastic area Px -
and the total moment-
P
-A -
M y • y CA
-'o-I~ y-y
© M re =
P
A
+
2 a) MAzl
IB-B
+
a (L -
STEP 3: Determine Stiffness and Carry-Over Factors a) Ay]
MA =
Al +
CA2 I y-y
and, p
Design of Rigid Frames Mz
2
L
M=~+~
=
[-
W
2
Z2
+ z(L
- 2 0)
+ a(L -
/
6.1-23
all
'2~
t:= (6)
x
4
a
0
..
r
z
~
Moment
diagram
CD
~ ~0
©
~:~ ~"~ ------j0• ~c.~~© cA
Cc
.1
a
b
Elastic area FIGURE 29
Mo =
1 ec2 -A +-1s-s
stiffness factor at
lAB
@
©
carry-over factor,
@
to
©
loge n = 2.3026 loglo n IB (2540) loge ~ = loge (6467)
~A:
carry-over factor,
COA = _
©
to
Iog,
~:e
(5930) loge (2540)
h - I A = (2540)
=
loge 2.3346
(646.7)
= 1893.3
100"
h 2 = 6.4516 X 100
646.7
IB3
4.1822
t, r;=
= .84780
For the uniformly-loaded beam shown at top in Figure 30, having fixed ends, find the fixed end moments, stiffness factor, and carry-over factors. At center in Figure 30, the solid curve is the actual moment of inertia (I) as it varies along the length of the beam. The dashed line is the assumed straight-line variation in moment of inertia along the two halves of the tapered beam. The following properties are established:
IA2
loge 3.9276
= 1.3680
@
Problem 2
IA
le2 = 3.5163 x 107
Then proceed first to find formula elements made up of these properties:
Ko = Me
= = =
Ie = 5930
108
h = 2540
stiffness factor at
a
X
b = 100"
KA = MA
CAO = _
= 3.678
X
105
L
= =
1.6386 X 1010 100"
Ie -
In = (5930) -
(2540)
= 3390 (100) (1893.3 )
a
.052813 ( (h -a I A ) ( (h -a I A ) ( (h -a
IA
)2 )3 = )4 =
2.7892 X 10-3 1.4731 X 10-4 .77800 X 10-5
6.1-24
/
Miscellaneous Structure Design
b (Ie -
b ( (Ie - In)
=
((Ie~In)r
loge I
)
A
.072252
.8722 X 10- 3
)'J
lA
( .052813) ( 1.3680)
.029499
( (Ie -b)" In)
h
a (h -
( 100 )
(3390)
h)
A, =
2.5728 X 10-;'
b Ie (Ie _ h) loge 1;
(.029499) (.84780) .025008
= .75895 X 10-0
A = Ax
STEP 1: Determine Properties of the Elastic Area
+
Az
( .072252)
area of elastic area
+
(.025008)
= .097260
©
®
(6)
1"X10"
~:I= L_-=::====~=======-J I ~ ~,,:t~ 1-L soI
: dw
-
20"
I
dw
10"
1 - - 0 - - - - - L = 200" - - - - - - - - - ~l
I9~ered beam
18
= 2540 in 4
/'·1 Ie = 5930 in 4
A
I
=
-.t
6
4
.7_in_ L
-+
-r- I·
1 . I Moment of inertia
r Elastic area
FIGURE 30
Design of Rigid Frames
=
moments of elastic area
M/~_A =
~
((I B
= (2.7892
=
I A) ) 2
(h -
IA
10- 3 )( 1893.3 -
X
I A loge
-
646.7
X
~:)
IA.I = In-B
=
(.8722
2
(I 0
-
10- 3 )( 3390 -
X
IIl
(
b (Io -
IB )
)3 ((Io -
In) (Ie 2
1.3680)
I
10
In agel;
-
+I + 6.4516
2540 X .84780)
=
X 106 X .84790 )
67.02
Ax ( 2.8132) (.072252 )
(.072252)(65.70)(65.70 - 2 X 38.937)
= 110.16
MA.I IB-B
c. =
+
= (167.93)
= 38.937"
+
IA.I = IAzl IY·Y IB-n
A. (1.0762 ) ( .025008)
Az
[(Cz
+b
-
=
A.a
+
+ .025008) 843.04 +-\OQ -
(67.02
-
MA.I IB-B
= 170.28
= (.025008)( 100) +
(1.0762)
IAxl
I s-s
Y-Y
M A-A = MAxi I A-A
+
MAzl
+
= (110.16)
I A-A
+
= (2.8132) = 6.3902
=
(3.5570) px .
MA -A
CA =-~
=
W
2 W
= 65.70"
= r, =
= 134.30" (
197.36
CA
(65.70)
a (La -
) 3 ((In IA
)
I A )( I B 2
+
-
3 IA )
I A 2 loge
~: )
= ( 1.4731 X 10-4) (1893.3 ~ 599.9
+ 4.1822
X 10
5
X 1.3680)
I s-s
+
W
2
IAxl ) IA-A
2.8132 -
167.93)
(L24 A. - IA.IIn-n )
= 91.53 p = P,
X
-
W
= ; (2~-=-
.025008 _
67.02 )
W
+ r,
= (197.36 w)
=
(170.28)
(L MAxiIA-A
= 2" (200
(200) -
IAzl
280.44
_ (6.3903) - (.097260)
= I A-A -
=
I
= 3..5570
= L -
C.2]
CA)2 -
= 43.038"
IAxl
loge ~:)
B2
MAxi IA-A
c, =
Cc
3 Ill)
(2.5728 X 10- 5 )C3390)~-1690)
)
= 1.0762
MA.I I A-A
6.1-25
167.93
2.8132
MA.I IB-B = ( (1 0 - b I B ) )
/
288.89 w
+
(91.53 w)
65.70)2 43. 04
2J
6.1-26
/
Miscellaneous Structure Design
+ 3994
=
w
STEP 2: Determine the Fixed End Moments at
F8~3.3
(.7780 x 10- 5 ) -
7(2450) (646.7) -
=
[2 x 6.4516 x 10
+
11(4.1822
3.678
X
X
®
6
P My.y CA Mfe=-AI y-y
105 ]
108 ( 1.3680) ]
(288.89 w) ( .09726)
(3994 w) (65.70) (280.44)
11,878
= 2034.5 w Q z -_ ( (Ie _b I ) ) B
rl11 (Ie -6 +
7 Ie h
(.75895 -
4
X
10- 6 )
h) ) (2 I
11 I B2 )
7(5930)(2540) -
~: ]
at
©
x 3.5163 x 107
11(6.4516
1.6386
2
I B 3 loge
-
~90 )[2
[(
e
X
X
106 )
+
(288.89 w) (.09726)
]
1010 ( .84780) ]
(3994 w) ( 134.30) (280.44)
= 4882.8 w
= 4827
M,l
-
w 2
/y-y -
STEP 3: Determine Stillness and Carry-Over Factors
l
o, +
; [-
+
11,878
(265.70)167.93 -
=Mz/
_ /y_y -
1 ( .09726)
W [ - Qz
2
= 25.67
65.70(200) (2.8132)J
2113 w
MAe
+
(L -
3 a
+ CA)
=
1
A -
[a(2 L -
MAi~'B
+a
3 a) -
(L -
CA(L a)(a -
2 a)]
CA)
Az
= -
J
= ;[- 4827 + (-34.30)67.02 + 10,000( 1.0762) + 343,000(0.25008~ My.y = Mx /
/y-y
= (-
+
M./
+
1 ec -+ -1A s-r
(6107 w)
+
74.59
stiffness factor at
/y-y
2113 w)
21.18
1 ( .09726)
=
W
( 65.70) ( 134.30) (280.44)
2
Me =
= 6107
CA Co
----y;:;-
1 ( .09726)
I Az/ /B-B
+
+
(65.70 2 ) (280.44)
@
(134.302 ) (280.44 )
Design of Rigid Frames
@
---±-I T
dw
= 11"
©
®
I
I:
a
L
b
= 200"
6.1-27
1" X 10"
"'f---'-
]
~14
= 120"
/
~ 19"
2
= 80"==j
Haunched beam
•
.0010905
--:.----:.----..;,....------r'
~~~:jt~::f..0008068
.0006189
14-------- 130" Elastic area
I L
.L = .0012896 A
I
+ --------r, .0009332 •
_ 20"
20"
20"
----~ 180"
20"
CD[
, .0007039 .0001,481 t
l
.0004375
Elastic area
--------.-1 193Y3 FIGURE 31
stiffness factor at
©
Problem 3
ICc = Mc = 74.59 carry-over factor, CAC = -
@
to
©
MAC
MA
(- 21.18) (25.67)
= .825 carry-over factor, Co...
©
=_
M~~
-
(-21.18) (74.59)
-
= .284
to
@
For the haunched beam at top in Figure 31, having fixed ends, find the fixed end moments ( uniformly loaded), stiffness factors, and carry-over factors. Break beam into sections and use numerical integration. The elastic area could be divided into rectangular areas, as at center in Figure 31, and the resulting properties of the elastic area found in this manner. Of course some error will be introduced because these rectangular areas do not quite equal the actual curve of the elastic area. However, as the number of divisions is increased, this error will decrease. Without any additional work, the following method will more nearly fit the outline of the elastic area and will result in less error. See lower diagram, Figure 31. The curved portion within the elastic area is divided into triangular areas. It is noticed that a pair of tri-
6.1-28
/
Miscellaneous Structure Design
v'
A (area)
Section
0 ®
(.0012896) 120 112(.0012896)
20
(c)
112 (.0009332)
40
0
112(.0007039)
40
(e)
112 (.0005481)
40
CD
112 (.0004375)
20
Total ~
= = = = = =
60
.154752 .012896
iy'
M(moments)
126%
=
I.
M'Y'
9.2851
557.11
185.70
1.6335
206.91
.29
.018664
140
2.6130
368.38
1.24
.014078
160
2.2525
360.40
.94
.010962
180
1.9732
355.17
.73
.004375
193%
.8458
163.52
.09
2199.69
18.6036
1.21573
moment of inertia
angular areas share the same altitude and since the division in length (s) is the same, they will have the same area. Therefore, the center of gravity of the two triangles lies along their common altitude. (This graphical method is applicable to any beam with a non-uniform change in moment of inertia along its length).
( 18.6036)2 (.21573 )
(2199.7) 2199.7 -
STEP 1: Determine the Properties of the Elastic Area
1604.2
=
595.5
595.5
elastic center area (A) of section 00f Mx/lx diagram
M
if A
( 18.6036) (.21573)
=
w a2
12 (a
w ( 120) 2 (120 12
86.23"
L -
CA
(200) -
+ 3 b) +3
557.10 w (86.23)
center of gravity of section
113.77"
®r
M = 4644.5 w
a = 120"
I
;;
-.5800
V
v
./
./
I/,
~
.>
II
VI II V
I
/
V-
6.00
1/
I.
-.6000
Y
7.00
.15
1....-
N/
•30
t>
.35
/,
1-- I-
KBA
'j,.
,
/,
~--
8.00
t2
I-pLV
Y
in terms Ell of L
kf30
' / .29--
-.6400
.35
2~
I I
I
I
Ie
f- t-
3.0
Chart 2. Stillness f cctors at small end 01 unsymmetrical beam.
~-L~~P:j
~-
2.6
I.
AI 9.00
2.2
I 2 r= __
Chart 1. Stillness foctors at either end 01 symmetrical beam.
I
f./1/
!jill ~
I
.10
I-
,'lV
p·.o5
1-1-
/"
/v
VI/ V/V V I...'lV
4.100
I-
~ ~t/-
fl
V-
.IV
4..300
I
1--"-
l.d~v
~v ~/
4.200
-~o ;=
./
V
VII
VII
f.-
5.000
V
4.400
f.-
II:
.20
i-r
V
.15
'j
6.000
V-
KA B
jV -.5000
II' IIJ
1.4
I.e
2.2
2.6
3.0
I
2 r=--
I.
Chart 4. Carry-over fcctors lor symmetrical beam Irom either end to the other.
Design of Rigid Frames
-(ttttj ~
1--1.I-
-.7BOO
__
- I--
A
l~ Y t-- .35
~B
~
'
~ ~
-L--
-.7400
6.1-31
---I-
1-L1-
j j j
I
/
.25
V
'/
1/ VI/
-.7000
V
1/ I- t-I- -- t-- t-t-
V
I/
-.6600 I - t- - l -
1/
I
l-
t-+
1/
v
rill
~
lI;
I-- I-t--.6200 - l -
V
y-
t>
I.
- ."J V" t--l/
t-'-
I--
20 I
y-
j
-
1/
.10
II
_.5BOO
IL- t--
f-Y
0;
-
t->
I----"
/
if J'I
- ---
.-
-,--
/"
/'
'II
-.5400
-.4600 H-+--I--,
--
-.4400
t-r- I--
- -+---- - --- ,
4
p » 0.5
_C----
IXI W/I/
1-+-+-
-.4200
_f-
I----"
F/ :...--V
-.5000 ~ 1.0
--
1.4
1.8
2.2
2.6
1.4
3.0
1.8
I 2 r= __ I, Chart 5. Carry-over factors for unsymmetrical beam from srno!l end to large end.
Chart 6. Carry-over factors for unsvrnme tricc l beam from large end to srnoll end.
r-r-
I--
1--1-1-1--1--1.-
'8
1--1-1-1--1--1-1--1--1--
PL
-.
.093 01-- +-1--
~.
-I
L
r t - t1--1-- I-
+=P=-
IJ
f--J
~p
t-jJ.-c
17-'
1560 I1-1--
1--1-- j -
t-t- t-
1--1--
1--1-- I--
I-
-
~-
\7r--lr -t- ·-tt-
./
MA B
--
10
V
.0890
in terms of PL
1--- -
~
--
/
V
-0 V
V
v: I-"
/
IV
V
If/ Vv
P'
I-
14
-l-
1360
r
26
--
I.
10
14
.~
;k- S;:i