Published as a Seroice to Education by S F. LINCOLN ARC WELDING FOUN First Printing 5,000 June 1966 Second Printing 10,000 November 1966 Third Printing 15,000 August 1967 Fourth Printing 15,000 July 1968 Fifth Printing 10,000 May 1972 Sixth Printing 10,000 February 1974 Seventh Printing 10,000 October 1975 Eighth Printing 10,000 July 1976
Special acknowledgment is herewith made to Watson N. Nordquist who has contributed much to the editing a d organization of the material from which this manual has been prepared
rustees of the Foundation:
E. E. Dreese, Chairman; The Ohio State University, Columbus, Ohio T. V. Koykka, Partner, Arter and Haddcn, Cleveland, Ohio
R. C . Palmer, Vice President, Central National Bank, Cleveland, Ohio fficerr:
S. Sabo, Cleveland, Ohio
Price: in U.S.A. (Postage included)
Ocerseas and Quantity Prices Upon Request FB-37
Library of Congress Catalog Card Alumbe?: 66-23123 Printed in
Permission to reprodnce any material contained herein will be granted upon request, providcd proper credit is given to The James F. Lincoln Arc Welding Foundation, P. 0. Box 3035, Cleveland, Ohio, 44117. Copyright 1966 by The James F. Lincoln Arc Welding Foundation
WELDED STRUCTURAL CONNECTIONS have long been used in the coristrnction of buildings, bridges, and other strnctures. Tho first \\&led buildings were erectcd in the '20s-the greatest application being in low-level buildings of many types. The American Welding Society first puhlislxd specifications for welded bridges in 1936. Hut earl!. progress came slowly. During that ycar, 1936, The Jalncs F. Lincoln Arc Welding Foundation was created by The Linwln Electric Company to help advance the progress in welded dcsign and construction. Through its award programs and educational activities, the Foundation providcd an exchange of experience and gave impehls to the growing application of welding. Thus, within the last decadc and particularly the past few years, unitized welded design llas 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 manna1 for fi~rther guidance and cl~allengeto architects, strtrctural engineers, fabricators and contractors who will build the structures of tomorrow . . . and to the educators who will prepare young people for thest: professions. This material represents an interpretation of the best in accumulated esperiencc of all w11o have participated in prior Foundation activities. The autlior has coordinated this \vith a continuing study of current welding research conducted hoth in the United States and Eumpe, and against a background of participation on various code-writing cominittees. Much of the direct instructional information that resulted has been pretested in over 70 structural seminars attended by over 4000 engineers. Tho prodnction of this manual has spanned several years during \&ch constant effort \vas made to eliminate errors. The author will appreciate having callcd to his attentiorr any errors that have escaped his attention and inliitcs corrr~~pcmdei~ce on subjects about which the reader may have questions. Neither the author nor the pblisher, howover, can assume responsibility for the results of designers using values and forniulas contained in the manual since so many variables affect every design.
The Jomer F. Lincoln Arc Welding Foundation
ITS The author and the publisher firatefully aclpic bridgr, design, long accepted in Elirope:, is coming into pnminencc i l l Amrrica as a major approach to reduction of bridge costs. This coucmpt calls
Large bridge sections are shop-fobricoted, shipped to the site, and lifted into position. This lowers erection costs ond compresses the project tirnetoble.
for tlrt, romplctc deck to act as a nnit. Olihotropic dcsign could not he rxecr~tedwithout welding. 8. W E L D E D C O N S T R U C T 1 0 STRUCTURES
\V&iing h;is f;wilit;itcd thr. design and construction of :t grrat mriety of strnctirres with the mntrmporary look. liven \v;itcr to\vcrs hiivc takm oir o bcauty that complenients adjncrrrt ar(.hitccturt,. Stadiums for big-lc;igt~csports clubs and for Eigxunc collcgrs arta lmiring Ilcavily on wc~lding.Among tllcsc arr can make tremendous savings tl~roogh tlic design of special stnictirral members, This irrcludes members lraving complex cross-sectional coriGguration and hybrid mcmbers that are a mix of steels iraving different analyses. Modern strnctui-nl fabricating shops have fixtures for assembling plates into colum~rsanti girders, manipulators for welding aotom:itically, ;md positioners for supporting ~nernhersso that att:icIring plates may bc welded in thc flat position. Welding developments in the past few- ycars h a w greatly incrcased welding speeds, while assuring high quality welds. In sribmerged-arc welding the use of multiple arcs, will1 two and three, welding heads has
trmn~endor~slyincreased welding s p ~ e d s . Contintioris wire processes for semi-meclranized welding for both shop am1 field applications have substantially increasd prodoctivity. Mtich progress has been made in automatic manipr~lators, enabling the welding head to be p i t into proper aligninel~twith the joint of the membcr in a matter of scconds. This alignment is automatically maintainctl along the length of the joir~tduring wclding. Thtrse manipulators represent :I major cost rcductioii pssildity. As the size of the struclurc increases, tlrc total arc time on a n.eldcd job becomcs a decreasingly smaller percentagre given by the steel handbook as -
Load & Stress Analysis
The handbook value of I, = 206.8 in.' can he inserted dircctIy into the following table, for the I, of this WF section C . By adding areas aud their properties:
4. SECTION MODULUS 6)
?he section modulus ( S ) is found by dividing the moment of inertia ( I ) by the distance ( c ) from the neutral axis to the outermost fiber of the section:
0 . +16.25& 18 WF 96# . -.-
distance of neutral aris from reference axis
moment of inertia about neutral axis
Since this distance ( c ) can be measured in two directions, there are actually two values for this property, although only the rnalirr value is usually available in tables of rolled sections because it results in the greater stress. If the section is symmetrical, these two values are equal. Section modulus is a mcasurement of the strength of the beam in bending. In an unsymmetrical section, the outer face having the greater value of ( c ) will have the lower value of section modulus ( S ) and of course the greater stress. Since it has the greatcr stress, this is the value needed. With some typical sections it is not necessary to solve first for moment of inertia (I). The section modulus can be computed directly from the simplified formulas of Table 1. In many cases, however, the moment of inertia ( I ) must he found before solving for section modulus ( S ) . Any of the previously described methods may be applicable for determining the moment of inertia.
distance from N.A. to outer fiber Q,
= 18.00 - ,925 = 17.075"
section modulus (see Topic 4 which follows)
Using a welded "T" section as a problem in finding the section modulus, its neutral axis is first located, Figure 7. Using the standard formula ( j r l ) for determining the distance ( n ) of the neutral axis from any reference axis, in this case the top horizontal face of the iiange:
Properties of Sections
Sum of moments n = - - -MA - Total area of section
Next, the section's moment of inertia is determined, using the elements method (Figure 8 ) :
of strength nnder torsional loading of round solid bars and closed tubular shafts. This value is slightly higher than the required I = 700 in.' because depth of section was made d = 15" instead of 14.9". Finally, the section modulus ( S ) is determined:
7. TORSIONAL RESISTANCE (R) Torsiond resistance ( K ) has largely replaced the less accurate polar moment of inertia in standard design formula for angular twist of open sections. I t should be employed where formulas have been developed for the type of section. Thcsc are given in the later Section 2.10 on Torsion.
5. RADIUS OF GYRATION (r)
8. PROPERTIES OF THIN SECTIONS
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 he concentrated and still have the same moment of inertia. This property is used primarily in solving column problems. It is found by taking the square root of the moment of inertia divided by the area of the section and is expressed in inches.
Because of welding, increasingly greater use is being found for structural shapes having thin cross-sections. Thin sections may be custom roll-formed, rolled by small specialty steel producers, brakc-formed, or fabricated by welding. Propt:rties of these sections are needed by the designer, but they are not ordinarily listed among the standard rolled sections of a steel handbook. FJropcrties of thin sections customarily are found by the standard formulas for sections. With a thin section, the inside dimension is almost as large as the ontside dimension; and, in most c a m , the property of the section varies as the cubes of these two dimensions. This means dealing with the differcnce between two very large numbers. In order to get any accuracy, it would be necessary to calculate this out by longhand or by using logarithms rather than use the usual slide rulc. To simplify the problem, the section may be '.treated as a line", having no thickness. The property of the "line", is t l ~ e nmultiplitd by the thickness of the section to giva the approximate value of the section property within a very narrow tolerance. Table 2 gives simplified formnlas for nine properties of six different cross-sections. In this table: d = mean depth, b = mean widtli of the section, and t = thickness.
The polar monrcnt of inertia ( J ) equals the sum of any two moments of inertia about axes at right angles to each other. The polar moment of inertia is taken about an axis whiclr is perpendicular to the plane of the other two axes.
Polar moment of ine~tiais used in determining the polar section modulus (J/c) which is a measure
Load & Stress Analysis
TABLE 2-Properties of T h i n Secthns Where thickness (I) is small, 6 = mean width, and d = mean depth of section
2(b+d) down from top
dZ b + 2d down from top
= add t / 2 to c for S )
The error in calculating the moment of inertia by this Line Method varsus the conventional formula is represented by the curve in Figure 9, using a square tu1,ular section as an example. As indicated, the error increases with the ratio of section thickness ( t ) to depth ( d ) .
An excellent example of the savings in design time offcrcd by use of the Line Method exists as (column) Problem 4 in Section 3.1. Table 3 givrs the most important properties of additional thin sections of irregular but common configurations.
Properties o f Sections
FIG. 9 Possible error in using Line Method is minimal with low ratio of section thickness to
Ratio: thickness jtj t o depth
For additional formulas and reference tables, see "Light Cage Cold-Formed Steel Design Manual" 1962, American Iron & Stet4 Institute. 9. SHEAR AXlS AND SHEAR CENTER
Since the bending moment decreases as the distance of the load from thc support increases, bending force f, is slightly less than force f2, and this difference (fy - f l ) is transferred inward toward the web by the longitudinal shear force (f.). See Figure 10.
This force also has an equal component in the transverse dircction. A transverse force applied to a beam sets up transverse (and horizontd) shear forces within the secticn. See Figure 11. In the case of a symmetrical section, A, a force ( P ) applied in line with the principal axis (y-y) does not result in any twisting action on the mcmber. This
Load & Stress Analysis TABLE 3-Properties of Typical Irregular Thin Sections Where thickness f t ) is srnaN, b = mean width, and d = mean depth of secfian
shear force flow in the section
Properties of Sections
is because the torsioud moment of the internal transverse shear forces (4) is equal to zero. On the othcr hand, in the case of a11 unsymmetrical section, U, the internal tra~xverse shear forces (4) form a twisting moment. Thercfore, the force ( P ) must bo applicd eecccntrically at a proper distance ( e ) along the shcnr axis, so that it forms an exteinal toi-sional monierit which is equal and opposite to ti-.,' intrrtui torsional momimt of the transverse shear forces. If this pr~rc;iutionis not taken, tlrcre will be a I ivisting action ;ippli:d to the member \vhich will twist under load, in addition to bending. Sec Figure 12. Any axis of symmetry will also be a shear axis. 'There will be two shear axes and thcir intersection forms the shear ccnter (Q). A force, if applicd at the shear center, may be at any angle in the plane of the cross-section and there will be no twisting moment on the member, just transverse shear and bending. As stated pre\:iously, rniless forces which are applied transvetse to a int>rnbcr also pass through the shear axis, the mcmher \?;ill be subjected to a twisting moment as well as bending. .As a result, this beam should be considered as follows:
Reference oxis y-y
1. The applicd force I' should be resolved into a forcc P' of ttic same \dire passing through the shear
ccntor ( Q ) and parallel to the origin:~lapplied force P. P' is then resolved into the two components at light angles to each other and p;rrallel to the principal axes of thc section. 2. A twisting moinmt ( T ) is produced by the applied force ( P ) about the shear center ( Q ) . The stress from tlw twisting moment ( T ) is computed separately and t h m silparimposed upon thc stresses of the two rrct:ingular componrnts of force P'. This means that the shear center must be located. Any axis of symmetry will be onc of the shear axes. For open sections lying on one common neutral axis (y-y), the location of the other shear axis is -
Notice the similarity between this and the following:
Load & Stress Analysis
which is used to find the neutral axis of a built-up section. Just as the areas of individual parts are used to find the neutral axis, now the moments of inertia of individual areas are used to find the shear axis of a composite section, Figure 13. The procedure is the same; select a reference axis (y-y), determine I, for each member section (about its own neutral axis x-x) and the distance X this member section lies from the rcfcrence axis (y-y). The resultant ( e ) from the formula will then bc the distance from the chosen reference axis (y-y) to the parallel shear axis of the built-up section.
Here, at point M:
Locating Other Shear Centers
or, since areas have a common (x-x) neutral axis:
Normally Q might be assumed to be at the intersection of the centerlines of the web and the flange. The James F. Lincoln Arc Welding Foundation also publishes collections of awardwinning papers describing the best and most unique bridges, buildings and other structures in which modem arc welding is used effectively.
Properties of Sections
Figure I7 suggests an approach to locating shear axes of some other typical sections.
Structural steel for Gateway Towers, 26-story Pittsburgh apartment building was erected in tiers of three floors each b y two derricks. Shop and field welding combined to facilitate erection; nearly 15 tons of electrode were used.
Load and Stress Analysis
Eighty-foot hollow steel masts and suspension cables help support the continuous roof framing system of the 404' x 1200' Tulsa Exposition Center. Welds holding brockets (orrow) to which cables are anchored are designed to withstand the high tensile forces involved in such a structure.
I . TENSILE STRESS The simplest type of loadirrg on a member is tension. A tensile load applied (axially) in line with the center of gravity of the section will result in tensile stresses distributed uniformly across the plane of the crosssection lying at right angles to the line of loading. The formula for the stress is -
tricity, will introduce some bending stresses. These must he combined with the original tensile stresses.
2. TENSlLE STRAIN The nnit clongation or strain of the member under tension is found by the following relationship:
P = the tensile force applied to the member A = area of cross-section at right angles to line of force crt
= unit tensile stress
A tensile load that is not applied in line with the center of gravity of the section, but with some eccen-
= unit elongation (tensile strain) = unit tensile stress
E = modulus of elasticity (tension) The total elongation or displacement is cqual to this unit strain ( E ) multiplied by the length ( L ) of the member. Elongation =
A welded tensile coupon (test specimen) measures Yz" x 1'P at the reduced section, and has two punch marks 2" apart with which to later measure elongation. Just after the test is started, a load of 10,000 lbs is reached. Find (1) the unit tensile stress on the reduced section, and (2) the total elongation as measured within the two marks.
. I, = 0.000444 . 2" = 0.00089'' in 2" E
In any calcr~lation for strain or elongation it is understood that the stresses are held below the yield point. Beyond the yield point, the relationship of stress to strain is no longer proportional and the fomula does not apply.
ELDING OF BUILT-UP TENSION ME
FIGURE %W - edlnig of Built-up Tension Members
Any force applied transversely to the structural axis of a partially supported member sets up bending moments ( M ) along the length of the member. These in turn stress the cross-sections in bending. As shown in Figure 1, the bcnding stresses are zero at the neutral axis, and are assumed to increase linearly to a maximum at the outer fiber of the section. The fibers stressed in tension elongate; the fibers stressed in compression contract. This causes each soction so stressed to rotate. The cumulative effcct of this movement is an over-all deflection (or bending) of the member.
The cantilever beam shown in Figure 1 is in tension along the top and in compression along the bottom. In contrast, the relationship of the applied force and the points of support on the member shown in Figure 2 is such that the curve of deflection is inverted, and the member is in tension along the bottom and in compression along the top.
Within the elastic range (i.e. below the proportional elastic limit or the yield point), the bending stress (u,) at any point in the cross-section of a beam is -
A4 = bending nlornent at the section in question, in.-lbs
I = moment of inertia of the section, in.* c z:distarlce from neutral axis to the point at which stress is drsiretl, in. ub = bending stress, may he tension or compression, psi TABLE 1-Beam T y p e d Beam
1 a d d e d end
M = - -PL 3 Fined end
Loud & Stress Anulysis
n = 1.47" 1, = 62.6 in' P = 10,000 lbs
FIGURE 3 5,000 ibs
The bending moment ( M ) may be determined from standard beam diagrams. Table 1 lists several of these, along with the formulas for bending moment, shear, and deflection. A more complete presentation is included in the Hcfsrcnce Section on Beam Diagrams. Normally there is no interest in knowing what the bending stresses are somewhere inside a beam. Usually the bending strrss at the outer fiber is needed because it is of ~naximumvalue. In an unsymmetrical section, the distance c must hr taken in the correct direction across that portion of the section which is in tension or that portion which is in compression, as desired. Ordinarily only the maximum stress is needed and this is the stress at the outer fiber under tension, which rests at the greater distance c from the neutral axis.
The top portion of the benm being in compression,
= 5,870 psi (compression)
A standard rolled '"I? section (ST-6" wide flange, 80.5 lbs) is used as a bcam, 100" long, supported on each end and bearing a concentrated load of 10,000 Ibs at the middle. Find the maximum tensile and maximum compressive bending stresses. Figure 3 shows the cross-section of this beam, together with its load diagram. Referring to Tahlc I, the formula for the bending moment of this type of bcam is found to be-
PL and therefore 4
Find the maximum deflection of the previous beam under the sainr loading. From the beam diagrams, Table 1, the appropriate iormula is found to he -
L:' and therefore 48 E 1 ( 10,000) (100)" flr6-2q
Since thc bottom portion of the beam is stressed in tension, substituti~igappropriate known values into the formula:
21,845 . -. psi - (tension)
In addition to pure bending stresses, horizontal shear stress is often present in beams, Figure 5. I t depends
Anolysis of Bending
on vertical shear and only occurs if the bending moment varies ;dong the beam. (Any beam, or portion of the bcam's length, that has uniform bending moment has no \wtical shear and thrtreforc no horizontal shcar). Unlike: bending stress, thc horizontal shear stress is zero at thc onter fibers of the beam and is maximum at the neutral axis of the beam. It tends to cause one part of the heam to slide past the olhex. The horizontal shear stress at any point in the cross-section of a beain, Figure 6, is -
The following values also are known or determined to he where:
V == extem;il vrrtical shcar on bt:am, lhs
I = moinznt of incrtia of whole section, in.i t = tbickncss of scctioil at plane whtm stress is a
desird, in. arca of section hiyxid planc where stress is desired, in."
y = distance of wntcr of gravity of area to neutral axis of entire section, in.
( a ) Substituting the above values into the formula, the horizontal shear strcss ( 7 ) is found:
] =1196 psi ( b ) Since the shear force is borne entirely by the web of the " T , the horizontal shear force ( f ) depends on the thickness of the web in the plane of interest:
f = T t 'and thus = I196 X 0.905 FIGURE
Assume that the "T' beam in our previous example (Problem 1) is fabricated by wclding. Under the same load conditions, ( a ) Find thc horizontal shear stress in the plane wherc the weh joins the flange. ( b ) Then find thc size of co~itinuo~is fillet welds on both sides, joining the web to the flange. From the beain diagrams, Table 1, the appropriate formula for vrrtical shear ( V ) is found to ber V = - and thus 2
= 1080 Ihs/in. There are two M e t welds, one on each side of the "T" joining the flange to the web. Each will have to support half oi the shear force or 540 ibs/in. and its leg size would be:
This would be an extremely small continuous fillet meld. Bascd upon the AWS, the minimum size fillet weld for the thicker 1.47" plate would be 5/16". If manual mtermittent fillet welds are to be used, the percentage of the length of the joint to be welded would he:
Load & Stress Analysis
Amalysis of Bending
fillct weld would satisfy this requirement because it- resnlts in 25% of the length of the joint being \vt4ded. 3. QUICK METHOD FOR FINDING REQUIRED SECTION MODUL S (STRENGTH) OR OMEN+ OF INERTIA (STIFFNESS) To aid in designing members for lxnding loads, the following two nomog;aphs have been consirncted. The first nomograph drtermincs the reqnirtd strength of a straight beam. Tlir st:cond nomograph deteimines the required stiffness of thc beam. In both nornographs sewral types of beams are included for conccntratod loads as well as nniform
loack. The length of the hmm is sl~ownboth in inches and in feat, tllc loi~din p n n d s . 111 the first n o ~ n o g a p h ( i . 8) an allo\v:rl,li: I~endingstreLss ( u )is shown rind the strmgth pi-operty of the hcem is read as seelion modulus ( S ) . In the s w o l ~ dnomograph (Fig. 9 ) an allowable imit deflection (A/I,) is shown. This is the resulting dc4ecti11ii of the 11e:ini dividtd by the lt~ngtliof the 11e;rm. The stiffness PI-opertyof thc haam is read as monicrit of irrri-tia ( I ) 13y using thrse nomogr:~phs thc designer can quiicidy find tliv required swtion moduhrs ( s t r c ~ ~ g t h ) or rno~nmtof irwrtia (stilhcssj of the be;rm. We can thcrr refcr to a stecl handbook to choose a steel sectiori that will meet these rcqrrirc~ments. If he wisli .15To full value of h4, must be used.
= 8330 psi (max at 4"
We cannot use .9 M,, because wind loading is involved; hence full value of M, must be used.
allowable stresses ua .- = 17,970
Since it is a "compact" section laterally supported witlun 13 times its compression flange width (Sec 1.5.1.4.1): oi. = uby= .66 us
X 20'' flange ifi )
M7ind in addition (Set 1.5.6) Wind in this direction (Set 1.5.6) Wind in this direction (Scc 1.6.1 and 1.5.6)
checking against Formula #14 (AISC 7a)
0.60 u, = 22,000 psi checking against Fornzula # I 4 (AISC 7a)
Here C, = .85 because sidesway is permitted
checking against Formula #15 (AISC 7 b )
CASE B Dead and Live Loads; Wind in Y Direction applied loads p = 2700 kips
applied stresses P - .-2700 lo0O 256.25 =K
checking against Fornizrlti $15 (;1lSC 7b) -. m,,
(8330) ~2-t.(nx)x~~ i . 3 3 )
under the web of the column and illustrates the prohlcrn where the channel flange of the anchor bolt attachment does not bear against the base plate.
Here: e = 16.15" f = 9"
Tensile stress in bolts
1 E = 10 (E, = 3000 psi) Ec 15h" bolts A. = 3 (2.074) = 6.22 in.' (bolts under tension) Q, = 130 kips
Compression stress at outer
edge of channel st~ffcners
Plotting these three points, the curve is observed to pass through zero at-
from formula #13 (cubic ~ q u a t i o n j Y3+K1YZ+K2Y+K3=0
which is the effective bearing length.
from fornula #9b
= 5.33 6 n A, K' = -B (f -6
which is the tensile load on the hold-down bolts. from formula #8b
= lOiiO psi Therefore.~ substitutinr" into formula &13:
+ 5.33 Y2 + 392 Y - 9160 = 0
Letting Y = +lo, --1-12, and +15 provides the following solutions to the cubic equation as the function of
which is the bearing pressore of the masonry support against the bearing plate. If the anchor hold-down bolt detail is milled with the column base so that it ht:ars against the base &ite, it must be made strong enough to support the portion
of the reaction load (PC P,) which tends to bear upward against the portions of the bolt detail outside the colu~nnflange. This upward reaction on the compre.ssion side (PC P,) is much larger than the downward load of the bolts on the tension side (P,). The area of section effective in resisting this reaction includes all the area of the compression material-column Bange, portion of column web, the channel web, and stiffeners-plus the area of the anchor bolts on the tension side. See shaded area in Figure 27. The anchor bolts on the compression side do not act because they have no way of transmitting a compressive load to the rest of the cohunn. In like manner, the column flange and web on the tension side do not act because they have no way of transmitting a tensile stress across the milled joint to the base plate. The tension flange simply tends to lift off the base plate and no stress is transmitted in the tensile area except bv the hold-down bolts attached to tllc column.
Determining moment of inertia
To determine the moment of inertia of this effective area of section, the area's neutral axis must he located. Properties of the elements making up this effective area are entered in the table shown here. Moiamts are taken about a reference axis (y-y) at the outermost edge of the channel stiffeners on the compression side (Fig. 27). See Section 2.2 for method. Having obtained the 1st totals of area ( A ) and moment ( M ) , solve for the location ( n ) of the neutral axis relative to the reference axis:
= 6.93" distance of K.A. to rcf. axis y-y
Now, having the value of n, properties of the effectivr portion of the column woh can he fixed and the table completed. With the 2nd totals of area ( A ) , momcot ( R ) , and also ~noinentsof inertia. (I, I,), solve for the moment of inertia about the neutral axis (In):
Smce the concentrated compressive load (P,) is applied at an wxent~icity( e ) of 16.15" to provide for the wind moment of 175,000 kips, the moment arm of the 130-kip load is9.15" from face of column gauge
5.15" from outer edge of channel stiffeners 12.08" from neutral axis of effective area compressioc stvess a t outer edge of channel stiffeners
Dirtonce: C.G. to ref. mi. y y
6.93" distance of N.A. to outer fiber
By substituting value of n = 6.93":
tensile stress in hold-down bolts
M c PC ut = - I A
where c is distance of N.A. from extreme fiber of tensile area
This co~npressiveforce on cach channel stiffener is transferred to the c11aiinr:l wcb by two vertical fillet welds, each 11" long. The force on (:a& weld is tllus-
total force in hold-down bolts
P* = A, 0-t = (6.22) (1 1,200)
= 69.6 kips e!ds Attaching Stitfeners t o Channel
and the rtqnired Gl1t.t wcld Icg size is-
Compressive force is carried by each of the four channel stiffeners. The average compressive stress on these stiffeners is-
856 lbs/linear inch
856 11,200 for E7O welds ('firhlr 5, Sect. 7.4) = ,076" or use (Table 2, Sect. 7.4)
With this 1r:g size, intermittent welds can be used instead of contiriuous wdding-
elding Channel Assembly to Column Ftonge
1131212 X 14.5 f 13) 3114.5 13) 2 86.1 in.
= f(2020,Zf;3050) - 3670 i b d i n . 0
actvol farce cliawabie force
(36701 111.200) t E70%
f123.4001 2(13 1-14.5) 2240 i b d i n .
uctual force . ollawoble force
Sr -( 1 74.2001 156.31
M S, -.-I 174,2001 (185.91 - 935 ibslin. v = L
2 (13) 4750 ibslin.
5680 Ibrlin. aituol force aliowabie farce 156801 111.2001
i? = V' f,,2 i*z = '"i3100li i475012
Sx, -I 174,2001 (242.2)
(123.4001 2(i3) 04.5i 3050 i b d i n .
186.11 2020 lbslin.
octucl force oilowobie force
5. USE OF WtNG PLATES
or a total length of 4%" of 3/16" fillet welds on each side of each stiffener.
Id Connecting Channel Assembly t o
When large wing plates are uwd to increase the leverage of an anchor bolt, the detail sho~rldalways be checked for weakness in bearing against the side of the column flange.
The average compressive stress on the channel web is-
total compressive force on channel assembly
The fillet welds connecting the assembly to the column flange must transfer this total compressive force into the column flange. There are four ways to weld this, as shown in Table 4. Assume the welds cany all of tlie compressive force, and ignore any bearing of the channel against the column Aange.
Figure 29 illustrates a wing-plate type of column base dotail that is not limited with respect to size of bolts or strength of colnmn flange. A similar detail, with bolts as large as 4%'' diameter, has been used on a large terminal project. The detail shown is good for four 2Yd'-dia. anchor bolts. Two of these bolts have a gross area of 6.046 in.' and are good for 84,600 lbs tension at a stress of 14,000 psi. In this detail, the bolt load is first carried laterally to a point opposite the column web by the horizontal bar which is 5%'' wide by 3" thick. section modulus of section a-a
First find the moment applied to the weld, Figure 28, which applies in each case of Table 4:
M = 4(18,850 lbs) (2.187") = 174,200 1%-lbs
Then, making each weld pattern in turn, treat the weld as a line to find its section modulus (S,), the maximl~mbending force on the weld (f,), the vertical shear on the wcld (f,), thc resultant force on tlie weld (f,), and the required weld leg size (o). Perhaps the most efficient way to weld this is method ( d ) in which two transverse 'h" fillet welds are placed across the column Aange and channel flange, with no longitudinal welding along the channel web.
bending moment on bat-
rcsulting bending stress
At the center of the 3" bar, the bolt loads are snpported by tension and compression forces in the 1" thick web platcs above and below the bar. The web plates are attached to the column flange, opposite the column web, by welds that carry this moment and shear into the column. The shear pnd moment caused by the anchor bolt forces, which are not in the plane of the weld, determine the size of the vertical welds. The welds extend 15" above and 3" below the 3" transverse bar. The properties and stresses on the vertical welds are figured on the basis of treating the welds as a line, having no width. See Figure 30.
section moddus of weld
maximum bending force on ueld
shear force on weld
resultant force on weld
Take area moments about the base line ( y-y) :
moment of inertia about N.A M" I, = I, I, - A
= 11.5" (up from base line y-y) distance of N.A. from outer fiber cbotbm = 11.5"
required flkt weld size 3000 a =113J0
This requircs continuous fillet welds on both sides for the full length of the 1" vertical web plate. If greater weld strength had been required, the 1" web platc could be made thicker or taller. For bolts of ordinary size, the upper portion of the plates for this detail can be cut in one piece from colnmn sections of 14" flanges. This insures fnll continuity of the web-to-flange in tension for carrying the bolt loads. By welding across the top and bottom edges of the liorizontal plate to the column flange, the required thickness of flange plate in bending is reduced by having support in two dircctions. 6. TYPICAL COLUMN BASES
In ( a ) of Figure 31, small brackets are .groove butt
stiffeners moy be
\voided to the oirtcr edges of thc colnmr Annges to develop greatcr moment resistance for the attachment to the bas? plate. This will help for moments about either the x-x or the y-y :tsis. A single bovel or single V joint is preparcd by beveling just the edge of the brackets; no hcveling is done on the column flanges. For colnnrn flanges of nominal thickness, it might he easier to simply add two brackets, fillet welded to the base of the column; see ( h ) and ( c ) . No beveling is required, and handling and assembling time is reduced hecat~seonly two additional pieces are requirod. In ( b ) thc bracket plates are attached to the face of the coluin~rflange; in ( c ) the p1atr.s are>attached to the outer edge of the column Nange. In any rolled section used as a column, greater berrtling strength and stiifiress is obtained about the x-x nxis. If the moment is ahont the x-x axis, it would be better to attach the additional plates to the face of the column as in ( b ) . This will provide a good transverse fillet across the n)lumn flange and two longitudinal fillet welds along the outer edge of the column flange with good acct%ssihilityfor melding. Thc attaching plates and the welds connecting thein to the base plate are in tho most effcct~vcposition and location to transfer
this moment. The only slight drawback is that the attaclring plntcs will not stiffen the overhung portion of the base plate for the hending due to tension in the hold-down bolts, or due to the upward hearing pressure of the masonry support. Mowevrr if this is a problem, smxll hrackrxts shown in dottrd lines may be easily added. The plates can he fillet wrlded to the outer edges nf thc column flange as in ( c ) , although there is not good accessibility for the welds on the inside. Some of these inside fillet welds can be made before the unit is assembled to the base plate. For thick Ranges, clctail ( a ) might represent the lrast amount of \velding and additioml plate material. Short lengths of pipe have been welded to the outer edge of the cohnnn flange to develop the necessary moment for the hold-down bolts; see ( d ) . The length and leg size of the attaching fillet welds are sufficicnt for thc moment. In ( e ) two channels with additional stiffeners are w c l d d to the cohnnrr flanges for the required moment from the hold-down bolts. By setting this channel assenibly back slightly from the milled end of the column, it does not have to be designed for any bear-
A 14" WF 426# column of A36 steel is to carry a compressive load of 2,000 kips. Using a bearing load of 730 psi, this would require a 30" X 60" base plate. Use E70 welds.
ing, but just the tension from the hold-down bolts. If this assembly is set flush with the end of the column and milled to bear, then this additional bearing load must be considered in its design. Any vertical tensile load on the assembly from the holddown bolts, or vertical bearing load from the base plate (if iu contact), will produce a horizontal force at the top which will be applied transverse to thc column flange. If the column flange is too thin, then horizontal plate stiffeners must be added between the column flanges to eflectively transfer this force. These stiffeners are shown in ( e ) by dotted lines. In ( f ) built-up, hold-down bolt supports are welded to the column flanges. These may be designed to any size for any value of moment. In (g), the attaching plates have been extended out farther for very high moments. This particular detail uses a pair of channels with a top plate for the hold-down bolts to transfer this tensile force back to the main attaching plates, and in turn back to the column. One of the many possible details for the base of a built-up crane runway girder column in a steel mill is shown in Figure 32. Two large attaching plates are fillet welded to the flanges of the rolled sections of the column. This is welded to a thick basc plate. Two long narrow plates are next welded into the assembly, with spacers or small diaphragms separating them from the base plate. This provides additional strength and stiffness of the base plate through beam action for the forces from the hold-down bolts. Short sections of I beam can also be welded across the ends between the attaching plates. 7 . HIGH-RISE REQUIREMENTS
Columns for high-rise buildings may use brackets on their base plates to help distribute the column load out over the larger area of the base plate to the masonry wpport.
For simplicity, each set of lxackets together with a portion of the base plate formed by a diagonal line from the outer comer of tlir plate hack to the coh~snn flange, will be assrsmcd to resist the bearing pressure of tho masonry snpport; see Figure 34. This is a conservative analysis because the base plate is not cut along these lines and thcse portions do not act independently of each other.
This portion of the assembly occupies a trapezoidal area; Figure 35. / + h i = 167"
where: a = 7 5 a, (ALSC L.J.1.4.8)
= 5.51" or use 6"-thick plate Check bending stresses & shear stresses in base plate bracket section
Start with lYzf'-thick brackets ( 2 x 1M" = 3" flange thickness) at right angles to face of column flange. Find moment of inertia of the vertical section through brackets and base plate, Figure 37, using the method of adding areas:
= (690 in." (750 psi) = 516 kips
moment of inertia about N.A.
Determining thickness of base plate
To get an idea of the thickness of the base plate ( t ) , consider a 1" wide strip as a uniformly loaded, continuous beam supported at two points (the brackets) and overhanging at each end. See Fignre 36. From beam formula #6Bh in Section 8.1: -w a2 M,, (at support) = 2
M = a S FIGURE 36
Bendtng stress [a)
corresponding shear stress iu brackets
distance of N.A. to outer fiber cb = 9.27"
= 8400 psi OK shear force at face of 6" base plate (to be transferred through fillet welds)
bending stresses Vb
= 24,630 llx/in. ( t o be carried by four fillet welds at 1%" thick brackets) leg size of mch fillet w d d joining base plate to brackets W
= ,545"or use %/,Br'[l --is
(The minimum fillet wcld leg size for 6" plate .)
Determining vertical weld requirements
= 9770 psi OK n~aximumshear forcc at neutral axis
In determining fitlet weld sizes on the usual beam seat bracket, it is often assumed that the shear reaction is uniformly distributed along the vertical length of tho bracket. The hvo unit forces resulting from shear and bending are then resolved together (vectorially added), and the resultant force is then divided by the allowable force for the fillet weld to give the weld size. This is of course conservative, because the maximum unit bending force does not occur on the fillet weld at the
same region as does the maximum unit shear force. However the analysis docs not take long:
bending force on weld
f, = u t = (9770 psi) (1%") = 14,660 lbs/in. (one bracket and two fillet welds ) or
= 7330 lbs/in. (one fillet weld)
vertical shear force on weld (assuming unifolm distribution)
Alternate method. In cases where the forces are high, and the requirement for welding is greater, it would be wcll to look further into the analysis in order to reduce the amount of welding. In Figure 37, it is seen that the maximum unit force on the vertical wt:ld due to bending moment occurs at the top of the bracket mnnection ( b ) in a rcgion of very low shear t~msfcr.Likewise the maximum unit shear force occurs in a region of low bending moment ( c ) . In the following analysis, the weld size is determined both for bending and for shear, and the larger of these two values are used:
ccrtical shear requirement (maximum condition at N.A.) fl = 25,200 lbs/in. to be carried by four fillet welds
resultant force on weld 0
required leg size of certical fillet weld 0
actual force allowable force
actual force allowable force
= ,562" or %,/,," bending requirement (maximum condition at top of bracket)
= -. actual force allowable force
Hence use the larger of the two, or 3/4" fillet welds. .4lthough this altrrmate method required a slightly smaller fillet weld (.654") as against (.758"), they both endod up at %' wheu they were rounded off. So, in this particular example, there was no saving in rising this method. Column stiffeners
A rather high eompr~~ssive force in the top portion of these brackets is applied horizontally to the column Range. It would hs wcll to add stiifenors behveen the column flanges to transfer this force from one bracket through the column to thc opposite column flange; Figure 38. It might he argncd that, if the brackets are milled to brar against the column flanges, the bearing area may then be considered to carry the compressive horizontal force bctwecn the bracket and the column flange. Also, the connecting welds may then be considered to
Unit sheor force on weld
between bracket ond column flange
carry only the vertical shear forces. See Figure 39, left. If the designer questions whether the weld would load up in compression along with the bearing area of the bracket, it should be remembered that weld shrinkage will slightly prcstrrss the weld in tension and, the end of the bracket within the weld region in compression. See Figure 39, right. As the horizontal compression is applied, the weld must first unload in tension before it would be loaded in compression. In the meantime, the bracket bearing area continues to load up in compression. This is very similar to standard practice in welded plate girder design. Even though the web is not milled along its edge, it is fittpd tight to the flange and simple fillet welds join the hvo. In almost all cases, these welds are designed just for the shear transfer (parallel to the weld) between the web and the flange; any distributed floor load is assnmed to transfer down through the flange (transvrrse to the weld) into the cdge of the web which is in contact with the flange. Designers believe that even if this transverse force is transferred through the weld, it does not lower the capacity of the fillet weld to transfer the shear forces. Refer to Figure 37(b) and notice that the bending action provides a horizontal compressive force on the vertical connedng wclds along almost their entire length. Only a vcry small lcngth of the welds near the base plate is subjected to horizontal tension, and these forces are very small. The maximum tensile forces occur within the base plate, which has no connecting welds. shear force on certical weld (assuming uniform distribution) 516.5k fs - .------4 x 30"
= 4310 lbs/in. (one weld) t;crtical weld size (assuming it to transfer shear force only)
Slight tensile prestress in weld before load is applied
bnt 3" thick column flange would require a minimum lhr' h (Table 2, Sect. 7.4). If partial-penetration groove welds are used (assuming a tight fit) the following applies: allowables (E70 welds) compression: shear:
shear jorce on one weld
f. = 4310 lbs/in. required effective throat
i j using bevel ioint
t = t, '/a'' = ,273'' 4- W"
= ,398" raot face (land) = ll/z"
if using 1 joint
However, in this example, the column flange thickness of 3" would require a %" fillet weld to be used. Brackets t o column flange edges
root face (land) = 1Yz" - 2(.273") = ,954" or use '/8"
The base section consisting of the brackets attached to the edge of the column flanges, Fignre 40, is now considered in a similar manner. From Illis similar analysis, thc brackets will be made of 1W-thick plate. Figure 41 shows the resulting column base detail.
A portion of the shear transfer represented by the shear force di~tributionin Figure 37 ( c ) lies below a line through the top surface of the base plate. It might be reasoned that this portion u a ~ l dbe carried by the base plate and not the vertical connecting welds between tire bracket and the colnmn flange. If so, this triangular arcs would approximately represent a shear force of ?5. (24,63O#/in.)
to be deducted: 516.5&- 73.9' = 4426&
COLUMN BASE PLATE DIMENSIONS (AISC, 1963)
COLUMN BASE PLATES Dimensions for maximum column loads
Base nlaien, ASTM 1116. h 27 iri Cuacilic, J
depth of leg sire of fillet weld
P~uiial-penetrationgroove welds are :illowed in the Building field. They have many applications; for cxample, field splices of coliiinns, built up columns, built-up box sections for truss chords, etc. If a vee J or U groove is used, it is assumed the welder can easily reach the bottom of the joint. Thus, the effective tliroat of the weld (t,?) is equal to the actual throat of the prepared groove ( t ) , see Fig. 8 ( b ) If a bevel groove is used, it is assumed that the weldor may not quite reach the bottom of the groove, therefore AWS and AISC drdu'ct %" from the prepared
FILLET WELDS tor any direction
7 = 13,600 psi I i = 9.- 6 0 0 ~ ~
= 15,800 psi !=11,300" 7
PARTIAL PENETRATION GROOVE WELDS
##tension transverse t o axis o i wcld
tension poiallel to axis of weld
1 some or piote
scme or p o r e
COMPLETE PENETRATION GROOVE WELDS tellion compression bending
low hydrogen E60 8 S A W 1 rruy be ,,red for fillet weldr & p a i t i o l penctrofion groove weldr on A242 or A441 steel. (at the lower ollowobfe r = 13,600 psii or other members subject $ dyfor pieilr or connections a i p r i l l a r i l y 10 axial camprrs5ion stress
f b i per l i n i n r rmch
weight of weld m e f a
upper volue A?, A373 ifre! & €60 welds l o w : d u e H36. A441 steel & E70 weldr Ibs per foot. -
Tcnsion applied parnllel to the weld's axis, or compression in any direction, has the same allowable stress as the plate. Tension appliccl transverse to the weld's axis, or shear in any dinxtion. has reduced allowable stress, equal to that for tlir throat of a correspanding fillet wcld. Jnst as fillet wclds have a minimum size for thick platcs becaiise of fast cooliiig and greater restraint, so partial-perietmtion .:move welds have a minimum effective throat ( t , ) ofTABLE 3-Portiol-Perpetration
Reinforced by o
leg size of fillet weld
where: t, - thicknc~sof t h i ~ ~ n eplate r
LD METAL REQUIREMENTS
Table 1 lists the AWS and AISC allowable stresses in welds iiscd on Buildings. Vaiucs for both partialpcnetratio~iand full-pciletration groove welds and for fillet welds are included. Table 2 tr:inslates the Table 1 values into allowable forces (Ibs/linear in.) and required weld metal (Ibs/ft) h r fillet wclds and scveral types of partial-penetration groove welds. These values cover weld sizes from M" to 3". Table 3 provides allowable forces for partial-penetration groove welds reinforced by a fillet weld. Table 4 directly compares a number of joints to carry a giveil force, illustrating their relative requirements in weight of weld metal. LE 4-Joints
1st value force ibs per lineor Inch A7, A373 s i t e l & E6O welds 2nd value loice ibr per lineor inch A36, A441 steel & E70 weids 3rd valve weight of weld metol lbr per foot
to Carry Force o# 2
IlillMG WELD TYPES
There arc! several w:rys in which different types of welds can be combined in economically fabricating built-up colunins to meet the two basic rtqiiirements: a ) welds from end-to-end of column to withstand longitudinal shear resulting from (wind and beam load) applied momonts, and 11) hcavier wclds in connection rcgions to withstand higher longihdinal shear due to abrupt change in moment, and to carry tensile force from the beam flange. The following cases illustrate combinations that permit optimum use of automatic welding:
The wcb plate is txvdcd to the proper drpth on all 4 edges dong tllc ci~tirclength. Croovr weld ( a ) is iirst made :iIoiig the rntire Icngill. Second, fillet weld ( b ) is made over tilo groove \veld within the connection region to hiilg it rip to the propvr size.
Region of beam to
Region of beam to column connection
Beveled only within coni?ection region FIGURE 11
The web plate is beveled to tbc proper depth dong short lcngths within tho connection region. First, groove weld ( h ) is made flrish with ihe surface within the connwtior~ region. Second, fillot weld ( a ) is made along thc entire lmgih of the column. FIGURE 9
If the weld sizes are not too large, the column may be first fillet welded with -hw!ld ( a ) along its entire length. Second, additional passes are made in the mnnection region to bring the fillet weld a p to the proper size for weld ( b ) .
Additlono! beveling in region of beom to (column COnneCTiOn
Region of beom to connection
Double beveled entire length FIGURE 10
Thc web plate is beveled to tbc, jxopcr depth on all 4 edges along the entire lcygth. \Vithin the co~snection I-egioii, the \vcb is furti~crl w ~ i , l i ~toI a dcepcr depth. First, groove weld ( b ) is I I I X ~ C within the connection region until the plate edge is built rip to the heigllt of the first bevel. Second, groove weld ( a ) is made ;iIong the entire Icrigtli.
In colunin bos sections, J and U gl-oovo welds may be substituted fol- bevel and Vee grnovixwelds if the fabricator is eqnipped to gouge anti profrrs to do so rather than bevel. Since bevelirrg is a cutting method, the plates must bc beveled before :rsscmbling them together. Gouging, Irowevcr, may be done either before or after assembling. Further, heavy J or U groove welds normally n q ~ i i n :less wcld mctai than the bevel or V1.1: groove wvlds. Some fabricators, in making hrilt-isl> box sections; have ass(~mb1ed;illd liglitly t:r& weldrd the plates together witliont ;iny prqxration; Figurc 1 3 ( a ) . The joints are next air carbon-arc g o n g ~ lto the desired depth for very slsort distalices and fr~rthertack welded; Figure 1 3 ( b ) . Next, the longer distances in between.
tack welds are air carbon-arc gonged. Whcn this is completed, the entire length is ;~utomatically submerged-arc welded togeth'er; Figure 13(c).
At first glance it might he thought that the rcquiremciits for a bc;trn Range welded to the flange of a 1111iit-upbox colmsin, Figure 14(;r), would be similar to the beam h i g c nelded to the flange oi an I shaped colnrnn, Figun: 14(b). This is because the box colnmn flangc is treated as ;I beam simply supported at its two miter edges, Figmr 14(c); it has the same maximum bending nroment as the W F columz flange treated as a beam supported at its center, Figure 1 4 ( d ) . The follo\virig analysis of a beam flauge welded to
a box column, Figure 1 5 ( a ) , is based upon a simila analysis of a line forcc applied to a cover-plated WF column, i'igure 15(1)). The latter analysis was made by Dr. T. R. Higgins, llirrctor of J",ngint:ering and Research of the AISC. The following assrlmptions are made: 1. The length of the box column Aange resisting this line forcc is limittd to a distance equal to 6 times its thickness abovc arid below the application of the line force. See Figure 16. 2. T l ~ eedge welds oirer no restraining action to this Clmge plate. 1.11 oilier words, these two edges are just supported. The nppcr and lower bo~uidaryof this portion of the column flange are fixed. 3. The tensile line force applied to this Aange area is urriiormly distributed. At ultimate load (P,l), it is assunied that this roctirngdar plate has failed as a mechanism with plastic binges Forming along thc dotted lines. The internal work done by the resisting plate eq11aIs the summation of the plastic moments (M,)
ini~ltipliedhy tlie angle change (6)along these edges. 'I'he exten~alwork done equals the ultimate load (P,,) rnrrltiplied by the virtual displacement ( A ) . By setting those two exprrssions equal to each othcr; it is possible to solve for the ultimate load (P,,) which may be applied to this portion of the flange plate.
At ultimate loading (P,), plastic moments (M,) will build up along the dashed lines (Fig. 16) to form plastic hinges. The iutemal work done, when this plate is pulled out, will be the plastic moment (M,) multiplied by the corresponding angle changes (+) along these lengths: an&
