Design of Welded Structures - Omer W. Blodgett

FIGURE 23

The resultant vertical monrement (h,) of the load, at the left end of tlre beam, is -

i.:;1c11 segment of the beam bends under its individoal bcnding momcnt and its angle change causes the cnd of the hcam to drflect. S w l'igwe 72. Tho total deflection at the end of the beam eqnals the sum of the deliections at the end of the beam caused hy the angle change of each segment of the 1,cam. Set, Figurta 23.

The following tapcved beam is 30' long. It has 1" X 10" fiarrge plates and a 'h" thick \voh. It is 11" deep at the ends and 33" deep at centerline. It supports two 58kip loads at the 'h points. Find the maximum deflection of tlic Iicam. See l.'igiire 24. Divide the length of thc btmn into 12 equal segn~ents.'1%~greater the nnmher of segments or divisions, the more :iccrirnte will he the answer. Normally 10 divisions \vor~ldgive a fairly acmratc result (Fig. 25)

eflecticn by Bending

/

2.5-13

Moment diagram FIGURE 25

Here

5

-

8. DESIGNING FOR MULTIPLE LOADS

10''

and A,,>+,, = E Z

h'ft>x,> I" ~

The moment of inertia of each scgment (I,) is taken at tlle sectional centroid of the segment. ~h~ formllla L.ompont~l,tsM,, x,, and 1,: are easier to hwdle in t&le form:

Normally, thr calculation of the maximum deflection of members suhjected to bending loads is very comples. Tilt: poiirt of maximum deflection must first be found; t h w , from this; the mnxim~~m deflcctiorr is found. I h I ~ ~ stlx,r.re s arc no more than two loads of cqnal vahir. and rqnal distance from thc ends of the hczm (Fig. 26), existing l ~ e a mtat~lesin handbooks do not cover this pn)hlem.

FIGURE 26

Total vertical d14iection -

For <~x,tniplc, most Ixarnc have mole than two loads (Fig. 2 7 ) . 7'11~maxi~nunidcflcction risrrally docs not nwnr at the rniddlr~or ccntcrlinr of tlic bcarn (Fig. 28). T\\w things can 1w donc to siniplify this problcrn. First, consider only thc deficction at the middle or centerline of the mtmher, rather than the maximum ~lrfii:(~tio~i at sornc point which is dilficnlt to determine. This is justified, sirlce the dcflcction at midpoint or centerline is almost as great as tllc masimum deflection,

2.5-14

/

Load & Stress Analysis

FIGURE 27

I

I

M c ; x i m m deflection

deflection ( A ) at tht* centcrlinc, each individual load, taken one at a time, will rcqnire the member to have a certain section ( I I , I:, etc.). The moment of inertia ( I ) of the beam section required to support all of the vertical loads within this allowable vertical ddlection ( A ) will equal the sum of the individual moments of inertia (I,,) required for the several loads. 4ny torqnc or cor~plc,appiicd horizontal to the beam will cause it to d&ct vcitically. This can be lrandlcd in the same manner. The required moment of inertia of the member (I,,) lor vnch torqric acting selxmtdy is found and ;tddcd into !hi! total ri.qr~irement for the pl-operty of the section ( I ) . 'The following two formulas may be w e d to find the individual properties of the section ( I , , ) :

for each force

D e f l e c t ~ o no! middle

FIGURE 28

the grcatest dcviatiori coming within I. or 2% of this value. For esnrnple, a simply supported h a m with a single concentrated load at the one-quarter point has a deflection at centerline r 98.5'1: of the maximum deflection. Secondly, a simple method of adding the rtxpirtd moments of inertia required for each individual load a n b e used. For a given size member, Figure 29, it is found that each load, taken one at a time, will cause a certain amount of deflection at the middle or centerline. The total deflection at the cerrterlinc will equal the sum of these individr~aldeflwtioris anisod by each load. This principle of adding dcflcctions may be used in a reverse nralmcr to find the required section of the meniher ((I,Fignre 30. For a given allowable

The two formrilas have been simplified into the fonnulas given below in which the expression K, now produces n constant ( 4 or B ) which is found in Table 3.

FIGURE 29

FIGURE 30

FIGURE 31-Required

Moment of Inertia to Resist Bending

2.5-16

/

Load & Stress Analysis TABLE 3-Values

af Constants ( A and B) for Simplified Formulas (16 and

for each force

for euch couple

Consider the continuous beam represented by the diagram at Figure 32a. The problem here is to find the reactions of the supports for various positions of the load (P,). According to hlaxwell's theorem, the ddectian at point 1 (A?,) due to the load ( P b ) at point x, Figure 32b. eqnids the dc4ection at point x (A,) due to the same amount of load ( P C ) applied to point 1, Figure 32c. There is a similar relationship between an applied load or moment and the resulting rotation of a real beam. Figures 32b and 32c constitute a simple reversal

The value of K,, is equal to the ratio a,/L, where a,, is thc distance from the point at which the specific force or couple is applied to the nearest point of support. I, is the span or length of beam between supports. From the value of K for arry givcn load ( P ) , the substitute constant A or B is obtained from Table 3. \tihen a force is applied to the member, use the constant A aud substitute into the first formula. When a m i ~ p l eis applied to the member, use the constant B and substitute into the secoud formula. A shorter method would be to make use of the nomograph in Figure 31. 9. INFLUENCE LINE FOR REACTIONS

.\lax\veli's Theorem of Reciprocal Deflections may be usrd to 6nd the reactious of a continuous beam or frame, and is especially adaptable to model analysis.

17)

FIGURE 32

eflection by Bending

of points at which the pressure is applied. This concept supplies a very useful tool for finding influence lines lor reactions, deflections, moments, or shear. In this case, the interest is in reactions. To find the value of the reaction ( R , ) at the lefthand support in Figure 32a, the support is rcnloved; this causes the left end to deflect ( A b ) ,as at Figure 32b. 111 order to restore the left end to its initial position, an upward reaction ( P C ) must be applied, as in Figme 32c. In extending h4awwell's theorem of reciprocal defiections to Figure 32b and Figure 32c, it is noticed: if P, = P .

23-17

as the reaction in question, the resulting ddledion curve becomes the plot of the reaction as the load is moved across the Icngtli of the beam. This is called an "iniiuaice cnrve". Considering the conditions of tlie rwl beam representlted by Figwe 32a, the reaction ( R , ) at point 1 due to a load (P,) at point x will be proportional to the ratio of the two ordinates at points x and 1 of the deflection ciirve. In other words:

then A,, = A,

For continuous beams of constant cross-section,

However, in order to return the beam to the initial condition of Fignre 32a, Ad must be reduced until it i:quals A,. To do this the upward reaction ( P C )must be rednccd by the factor: Ah/Ad And since A, = A,, this reduction factor becomes Ae/Ad.

.'. RI =

/

A or, usmg Fignre 32a Ad

Yb-2

This means that if the model bcam (as in Fig. 32c) is displaced in the same direction and at the same point

a ~viremodel may be set u p on a drawing board, with

the wire beam supported by thumb tacks spaced so as to represent the supports on the real beam. See Figure 33. A load diagram of the real beam is shown at the bottom. Notice that the thumb racks used for supports of the wire must be located vertically so as to function in the opposite direction to reactions on the real beam. The point of the model beam at the reaction in question ( R , ) is raised upwlu.d some convenient distanct,, for example 'h" or l", and the deflection curve of the wire beam is traced in pencil. This is shown immediately l ~ l o wthe model. The final value for tlie reaction ( R I ) is equal to

Thumb tacks

FIGURE 33

2.5-18

/

Load & Stress Analysis Deflection curve of the wire model is shown Erst and then the load diagram of the real beam.

I

FIGURE 34

the sum of the actual applied forces mnltiplied by the ratio of their ordinates of this curve to the original displacement at RI. The influence curve for the central reaction (Rn) may also be fouud in the same manner. See Figure 34.

Problem 3

1

A continuous beam has 5 concentrated loads and 4 supports. The problem is to find the reactions at the supporzs. The reactions are found by comparing the ordinates of the deflection curve of a wire representing the beam. See F i y r e 35, where the critical dimensions appear on the (upper) load diagram. For the ends, reactions R, and R4, displace the end of the wire a given amount as shown. The portion of each applied load ( P ) to be transferred to the reaction RI is proportional to the ordinate of the deflection curve under the load ( P ) and the given displacement at R,. For the interior reactions Rz and R3, displace the wire a given amount at Rn. From the ordinates of this

FIGURE 35

Deflection by Bending

/

2.5-19

FIGURE 36

deflected wire, determine the ratios of each applied load ( Y ) for the reaction at Rlr. The cornputation of forces for the rcaetions R, and R, is as follows:

R2 =

+ ,695 PI + 1.11 P, + 5 6 Pa - ,352 P4 ,296 Po

Reactions, either horizontal (11) or vertical ( V ) at the supports, may he found by displacing the frame at the support a given amount in the direction of the desired reaction. Sce Figure 36. The outline of the displaced model frame is traced in pencil, and this becomes the curve showing the infinenee of any load (at any point) upon this reartion. The displacement of each point of the model frame ( A ) u~herca load is applied is measured in the same direction as the application of the load, and the resulting reaction may- be computed from the following: horizontal reuction

+

=z

.695(2000#) f 1.11(2000#) .56(1000#) - .352(15OO#) - .296(1500#)

=

-C 3198 lbs

vertical reaction

Rcactious R:, and R, can be found in like manner. Application to Frames

This same method may be extended to the analysis of frames. If the frmr has a ninstant r ~ o m e n t of inertia, a stiff wire may lie bent into the shapc of the frame. If the frame has a variable mornwt of inertia, the model may be made of a sheet of plastic or cardboard proportioned to the actual moments of inertia.

Moments at the ends of the frame (or at any point in the frame) may he found by rotating the point in question a given angle (+,) and again drawing the resulting displaced model frame. See Figure 37. The displacement of each point of the model f r a n c (A) where a load is applied is measured in the same direction as the application of the load, and the resulting moment may be c o m p u t ~ dfrom the following: moinent at left-hand support

FIGURE 37

l t is necessary to displace the model a considerable distance in order that some accuracy may be obtained in the readings. Therefore, some error may be introduced because the final shape of the frame may alter the real load conditions. This error can be reduced greatly by me~suringthe displacements between one

2.5-20

/

Load

& Stress Analysis

(a) Measuring dirplacerrrrni of model frame from initial condition i o disploced condition

( b ) Measuring displacement of model frame from one displaced condition to an equal and opposite displaced

condition

FIGURE 38

condition and the opposite condition. See Fignre 38. This method of equal to opposite displacement may also be applied to monrents in which the frame is rotated an equal ill both directions, and mcasnrcments taken from one extrclne to the other.

E FOR

DEFLECTION

I u like manner, the use of a wire model based on Maxwell's Theorem of Reciprocal Dcflec~ionis useful in finding the dcflectitnis of a bean1 under various loads or under a moving load. If a 1-lb load is placed at a particular point on a beam, the resrilting ddection curve becomes the plot of the deflection ( A ) at this point as the 1-lb load is moved across ihc length of the beam. This is called the influence line for deflection at this particular point. T A B L E 4-lncremenral Paint

-----Total

Deflections of R e d B e a m Ordinate

Deflecsion At Free End (In.)

To dettel-mine the ddlection of the overhung portion s Asof this trailer, Fignre 39, under the v a r i o ~ ~loads. sume a cross-section moment of inertia ( I ) of 2 X 11.82 in.' Using the standard beam formula for this type of beam, the deflection of the free (right) end is deteimined for a 1-lb load placed at that point:

A wire model of this beam is held at the two supports (trailer hitch and the wheel assembly) with tbnnrh tacks on a drawing board. The outer end is displaced an amount equal to 3.25 on a snitable scale. The dt4ection c t m e is traced in pencil from this disp l a c ~ dwire beam. The ordinates of this resulting deflection ciirvc become the actual deflections at the free md as the I-lb load is moved across the length of the beam. Multiplying each of the loads on t!ie real beam by the ordinate at that point gives the deflection at the free end cansed by enc?~load on the real beam. See Table 4. Summing these incremental deflections gives tluc total deflection:

~

3300 lbr

-2.360"

A = 2.36" upward

DeSlection by B e n d i n g

+ 3.2:

Drawing boord

FIGURE 39

Erection of the 32-story Commerce Towers in Kansas City, Missouri war speeded with the aid of modern semi-automatic orc welding. Field use of self-shielding cored electrode quodrupled the rote of weld metal deposition. The weldor shown here is moking o field splice of two sections of the heovy building column.

/

2.5-21

2.5-22

/

L o a d a n d Stress A n a l y s i s

Complex antenna systems needed in age of space communications are sensitive to bending deflections caused by high wind loads. Good engineering, including the specification of high strength steels and rigid welded connections, is essential to the satisfactory performance of such structures. In the parabolic antenna dish shown, 6400 sq fi of expanded metal mesh are welded to a space frame of tubular welded trusses.

S E C T I O N 2.6

1. NATURE OF SWEAR DEFLECTION F,

Shear stresses in :a buam section cause a displacement or sliding action on a plane normal to the axis of the beam, as shown in the right hand view of F i y e I. This is unlike the dofledion resulting from bending in a beam, which is shown in the left hand view of Figure 1. Normally deflection due to shear in the usual beam is ignored hecansc it r~presents a very small percentage of the entire dt4ection. Figure 2 shows that the deflection due to shear increases linearly as the length of the beam increases, whereas the deflection

.--.-. -----.

L b J

------

b--i+-i

FIG. 1

Deflection in beam caused by bending moment, left, and by shear, right.

Length of cantilever beom (1)

FIG. 2 Deflection caused by shear increases linearly as length of beam, but that caused by bending increases as the third power of beam length. 2.61

/

2.6-2

Load

Stress Analysis the member and also tihe value of the shear stress (7). Figure 3 shows the shcar stress-strain diagram which is similar to the usual stress-strain diagram, altE~ough the shear yield strength is much lower than the tensile yield strength of the same material. After the shear +d strength is reached, the shear strain (t,) ir~creases rapidly and the shear strength iricreases because of strain hardening.

i

r, = 0.3 [Poisson's ratio]

INlNG SHEAR DEFLECTION 0

Y0

I 0 10 Sheot stroin

I

I

0 20

0 30

,) in in

[i

FIG. 3 Shear stress-stroin diagram.

due to bending irxreases vcry rapidly as a third power of the length of the beam. For this reason the de8ection due to shear is not an import:int factor except for extremely short spans where drAcctiorr due to bending drops off to a vcry sm;iIl valnc. The deflection due to shear is dependent entirely on the shear distribution across the cross-section of

The theory of deflection caused by shear stress is rather simple. However, the actual determination of th,e shear stresses and their distribution across the heam section (which two factors cause the deflection) i~ more difficrllt. In all cases, some kind of a form factor ( a )must be drtemlined, and this is simply a matter of expr~,ssingthe distribution of shear stress throughout the web of the scction. Since there is pmctically no shear stvcss in the flange area, this particular area has negligible effcct on the deflection due to sheas ( A , ) . The following formulas arc vdid for several types of hcams and loading:

Shear deflection of cantilever beom with concentrated load

Sheor stres

(7)

o r aiea beyond neutral onti

dirtorice between center of c~rovityof this aiea and neutral o x i s of entrre croii~iection

y

I :

A

= total ore" of section

I =- moment of ineitio of section t

FIG. 4

= tofol thickncis

Form f a c t ~ rfor shear deflection in built-up beams.

of web

Shear D e f l e c t i o n i n Beams

/

2.6.3

simply suppurled bcnn~;uniform load ( w )

simply nrpported bcmn, conr entrutecl load ( P )

FIG. 5 Beam sections for which Eq. 5 applies.

confilecw bcant; uniform loud (u:) The slope of the deflection curve ( 0 ) is equal at each cross-seetioil to the shearing strain ( E , ) at the centroid of this cross-section. cr is a factor with which the avcrage shearing strcss ( ) must be multiplied in order to obtain thc shearing stress ( T ) at the centl.oic1 of the cross-sections. On thi.s hasis, the form factor ( a ) for an I heam or hox beam would be: where:

P = total load. lbr A = area. of entirc sectkn~ E, = modulus of elasticity in shear (steel = 1 2 , O W ) O O psi) w = distribntcd load, lbs/linc:ar in.

Welding was used extensively in the fabrication and erection of this steelframed, 8-story, bolconized apartment building which features cantilevered cross beams in the upper stories. The building wor designed basically as a rigid structure with moin beoms designed plastically and light X-braces used to accommodate wind moments. The welded steel design cost 16@/sq ft less thon a reinforced concrete building would hove.

where Figure 5 :ipplies. Don't compnto area ( A ) in this forruula b r c a ~ ~ site will canct.1 out when used in the formul:is for shear ddlection.

2.6-4

/

Load and Stress Analysis

Both shop and field welding were used extensively in building the Anaheim Stadium, home of the Lor Angeles Baseball club-the Angels. The steelwork was designed as an earthquake-resistant frame, with high moment carrying capacity i n both directions. Having very good torsional resistance in addition to bending strength in both directions, the tapered box section frames can be located more widely (45' centers along straight sides) and eliminate the need for conventional cross-bracing between bents.

OMENT METHOD FOR CURVED CANTILEVER BEAM

In Sect. 2.5, Fignres 20 to 2.3, the arca moment method was used to find the dtflrction of a straight cantilever beam of variable section. This same mcthod may be estcndt~l to a cwvcd cnntilr,vi-r heam of variable scdion. As beforc, tho Bram is divided into 10 scgtnents of oclual length ( s ) and the nmnent of inertia ( I , > ) is determined for mch sepinmt. See Fi:nre 1. The moinent applied to :my segment of the hcam is equal to the applied force ( P ) mnltiplied by the distance (X,) to the segment, inc;~surcd~ I - U I X and at right angles to the line passing tlirongh and in the same direction as the load (1'). This moincnt (M,,) a p p l i d to the sc,‘-merit causes it to rotate !O , , ) . and-

The resulting deflection (A,) at the point of the

FIG. 1 To find deflection of curved cantilever beam of variable section, first divide it into segments of equal length.

beam where the deflection is to be determined is e q n d to the angle of mtation of this segment(@,,) mnltiplicd by the distance (Y,) to the segment, measured from and at right angles to the line passing through and in the s a n e direction as the dt~sired (leSlection(A)

F,

I,,

E I,.

The dist:n~crs X I Y ) and the moment of inertia ( 1 , ) arc dr.titrrnmin,rd for each of the 10 segments and placed in table form. In most cases, the dt,flectior~to hc dctmnined is in line with the applied form so that thcsc. two di.stnnws :Ire equal and the formula 11ew)mt:s-

The valucs of X,,"/l,, w e ionnd and totaled. From this the total defiection (A) is fuortd:

2.7-2

/

Load & Stress Analysis

........................ ( 4 )

Segment

A symmetrical beam forming a single continuous arc, for example, is comparable to two equal cantilever beams connected end to end. Thus, the prediction of dcflection in a curved beam can be approached in a manner similar to finding the deflection in a straight cantilever beam.

4 5 6 7 8 9 10

i

23 29 32 32 29 23 15 5

216 358 550 800 800 550 358 216 119

1

1.04 1.48 1.53 1.28 1.28 1.53 1.48 1.04

i

21

The total vertical deflection ( A ) is needed on a curved beam that will carry a maximum load ( P ) of 100,000 lbs. See Figure 2. Given the segment length ( s ) = 10" and the various values of X, and I., complete the computation.

Deflection of Curved Beams Solving for defleclion by using formula

PS

x:

A =-E-C

7

first colculote value of X;/I,

by using stiffness nomograph grophicolly find value of P X ~ / E I , for use in fzimuio

a= s z - El

FIG. 2 For deflection of simple curved beam, use Eq. 4 or nomograph, Fig. 3.

1

FIGURE 3-Deflection

of Curved Beam (Stiffness Nomograph)

Total load (P) on Curved Beam I bs

Moment arm (X,)

3-

Feet

1,000,000 A

Inches Moment of inertio of section (I,)

i n.4 -I

Deflection o f curved beom

where

I

X,=

50 in.

Ii

M u l t ~ p l ythe sum of these values by "st' to get total deflec!inn of t h e curved beam

'i

.00000,

2.7-4

/

Load & Stress Analysis

By using the stiH~wssnomograph, Figure 3, the computation can he collsiderahly shortened with no significant loss of accuracy The nomograph is based on the modified formula:

1

P X,,' E,,,

1

. . . . . . . . . . . . . . . . . . . . . . .( 5 )

Ileadirig are obtained from the nomograph for P X / I for each segment and cntered in the last column of the tahle. These are then addcd and their sum mdtiplied by s to give the total vertical deflection.

I

Problem 2

I

Use the same heam examplc as in Problem 1, the same valzrrs for l', s. X,, and I,,; and the s u m form of table. Complrte the compiitation.

Engineers of the Whiskey Creek Bridge i n No. California specified that the 300' welded steel girders across eoch span utilize three types of steel in order to meet stress requirements economically while maintaining uniform web depth and thickness and uniform flange section. High strength quenched and tempered sieel was prescribed for points of high bending moment, A-373 where moments were low, and A-242 elsewhere.

SECTION 2.8

1. NATURE OF IMPACT LOADING

Impact loading resnlts not only from :ictual impact (or blow) of a moving body against the member, but by any sudden application of the load (Fig. 1 ) . It may occur in any of the following methods: 1. A direct impact; risnally by another member or an external body moving with considerable velocity, for example: ( a ) A pile clrivcr hammer striking the top of a pile. ( b ) The die striking the workpiecr in x drop forge press or punch press. ( c ) A large rock dropped from a height onto a tn1ck. 2. A d d e n npjilicution of force, witliont a blow being involved. ( a ) The sudden crcation of a force on a inember as during the explosive stroke in an engine, the ignition or misfirr of a niissile motor \&en moui~tedon a test stand. ( b ) The suddm~moving of a force onto a member, as wlicn a lit~uvyloadrd train or trilck moves rapidly o w r a bridge deck, or a heavy rock rolls from the b11cl;et of a shovel onto a truck without any appreciable drop in height.

Heovy rock :oiled from shovel onto frome without ony initial drop in height:

h=O F=2W

3. The inertia of the mcml~crrmisting high acceleration or deceleration. ( a ) Rapidly ret:iprocating levers. ( b ) A machinr sohject to earthquake shocks or explosives in wa1-fare. ( c ) The bniking of :I heavy trailer. 2. APPROACH TO DESIGN PROBLEM

In many cases it is ditficult to evaluate impact forces 'lwantitatively. The analysis is grnerally more qualitative and requires recognition of all of the factors involved and tlwir inter-relationship. The &.signer can follow one of two metilods: I . IWimate the m ~ x i ~ n u m force exerted on the resisting mrmher hy ;ipplying an impact factor. Colisider this fol.ce to bo a static loitd and use in standard design formulas. 2. Estimate tlic cncrgy to hc : ~ b s o r h ~by d the resisting memhrr, and design it as an o~crgy-absorbing member. The propwtics of the nraterial and the dimer~sions of the resisting memhcr that give it maximum resistance to an energy load, we quite differcnt fi-om those that give the member maximum resistance to a static load.

Sudden ignition of missile;

Fort moving, boded wogon p a r i n g over supporting

or niisrile miifires and

beom:

then re-ignites

F = between W and 2 W

FIG. 1 Types of impact loading.

F = 2 T (thrust)

2.8-2

/

Load & Stress Analysis

KiNETiC ENERGY (E,,) is the omount of work a body can do by virtue of its motion.

POTENTIAL ENERGY (ED) is the omount of work o body can do by virtue of its position.

if the supporti~igmember is flexible ond deflects, this addi'ional movement must be considered as port of the total height the body con foil.

E = -F d t is also the amount of work a body con do by virtue of its ;tote of strain or deflection.

Spring

d-

-

2

FIG. 2 Formulas for kinetic energy and potential energy.

3. INERTIA FORCES

g = ac~deration of gravity (386.4 in./sec2 or 32.2 ft/se6

Inertia is the propcrty of a member which causes it to remain at rest or in in~iiormmotion miless acted on by some external force. Inertia force is the resisting force which inust be overcome in order to cause the member to accrlcrste or decelerate, equal hut opposite to--

where:

W, = weight ~f member. Ibs a = acceleration or decelerntion of member, in./sec2 or ft/scc2

In tool rondtot1

4. IMPACT FORCES

A moving body st]-iking a member produces a force on the member due to its deceleration to a lower velocity or perhaps to zero velocity:

Wb = weight of body, lbs a = dreelrrntion of body, i n . / s e ~or~ ft/sec2 g = acceleration of gravity (386.4 in./se+ or 32.2 ft/sec2)

At ,niiont of irnpoct

Maximum deflection of member o n d body

FIG. 3 Efiect of member's inertia.

t

esigning for impact Loads

Fortunatrly the mernber will dtbflcct slightly and allow a certain time for thc moving body (W,) to come to rest, therehy reducing this impact force ( F ) . Since the time interval is usually ~nrknown, the above formul;~ cannot h r wed directly to find the force ( F ) . I h v e v e r , it is us~rally possible to solve for this force by finding thc nlnount of kinetic energy ( E L )or potential encrgy (E,,) that must be absorbed by th,e memlwr (Fig. 2 ) . This applied cncrgy ( E k ) or (E,,) rnay then be set equal to tbc energy ( U ) dxorhed by the member within a given stress (a), see Table 2.

5. POTENTIAL ENERGY OF V O N MEMBE (See Figure

LE I-Basic

/

2.8-3

Laws Used in Analvris of l n r ~ a c t Angular

d

perpcndiculnr dis:onre f r o m center of rolotion to line of force

i

rodivi oi point for which be i o u n d

w in to

3)

Potential encrgy of falling bodv ( W , ) :

Potential energy received by deHt:cted member:

Then:

but K

=r

F

-

A

being the spring constant of the beam

to a load and & s i p as tl~onghit were a stcady load. As the weiglrt of the snpporiing nriw&r ( V ) increases, this inlp;~ctfactor of ( 2 ) becomrs less. In a similar nmuler, it is possible to exprrss the resultant impact dcflsdion ill tc:rms of s k d y load deflection. or since V

.-:

\/ 2 g b

If the body ( W,,) is suddenly applied to the member witliont any appreciable drop in height ( h = 0 ) , the lnaximi~mforce dne to inqx~ctis twice that of the applied load ( W , , ) :

T ~ I I Sit, is cominor~practice to apply an impact factor

6. EFFECT OF MEMBER'S INERTIA

If the weight ( ) of the stipporting mcmbcr is relatively high, some of the applied encrgy will be ;~bsorlx.d became of the imrtin of the rnemher to mov~m:snt.A good txa~npleis the cfkct of the mass of

2.8-4

/

Load & Stress Analysis TABLE 2-Impact

Formulos for Common Member-Load Conditions

Energy stored in member, may be set equol to kinetic energy Bending

D u= simp y suppor e concentroted lood uniform section

simp y suppor e uniform lood uniform section

(Coefficient = ,1667)

Bending

concentroted lood uniform section

(Coefficient

=

=

,26671

(Coefficient

=

,1000)

W

uniform lood uniform section

,1667)

(Coefficient

U=-

uy1I L 10 E c 2 10 E

concentroted load uniform section

(Coefficient

=

uniform lood uniform section

,1667)

simply supported concentrated load variable section so o uniform section

(Coefficient

=

.500)

(Coefficient

=

,1000)

=

constant volue (Coefficient = ,3333)

Torsion

I9

0,' R L U = ------

2 E,,,,t

where R

=

E, = shear round shoft

torsion01 resistance

modulus of elasticity open section (Coefficient

=

,250)

(Coefficient

=

,500)

Designing for impact Loads

a concrete bridge deck in reducing the impact forces transferred into the member supporting it. If the applied energy is expressed in terms of the velocity of the body ( V ) , the reduced velocity (V,) at instant of irnnact is-

Wb

-- weight of the body

W,

2

If the applied cnergy is expressed in terms of the height of fall of the body ( h ) , the reduced velocity (V,,) may be expressed in terms of a reduced effective height ( h , ) :

7. ENERGY-ABSORBING CAPACITY OF MEMBER

equivalent weight of the member

If the member were compact and anc cent rated at a point, the entire weight of the member would be effective in rtducing the velocity of thc body. However: the supporting mernber is spread ont in the form of a beam or frame and therefore only a portion of its weight is effective in moving along with the body and slowing it down. Tinmhenko shows the portion of the weight of the member to be used is: Simply supported beam with concentrated load at midpoint

Cantilever beam with ~mmzntratedload at end

W,

2.

This represents the effective height the body would have to fall in order to have the reduced velocity (V,) at the instant of inpact with the member.

where:

*

/

.;-

,236 W,,,

The reduccd k i ~ ~ e t ienergy c (E,) applied to the n~embmcansing stress and deflection wonld be

Ek = (WI,

+ W,)--Ve2 - -

Wb

The allou~ableenergy load, or load that can be absorbed elastically (without plastic deformation) by the mernber in bending, is basically-

where (k) is a constant for a specific type of beam with a specific t y p of loading. Table 2 shows the application of this formula to various member and load conditions, with numerical values substituted for the ( k ) factor. Obse~vation shows that the critical property of ,, 2 I the section is --,,while that of the material is -.~L c2 I;' 8. lMPACT PROPERTIES OF MATERIAL The two most important properties of a material that indicate its ability to absorb energy arc obtained from the stress-strain diagram (Fig. 4).

V2

2 g

Unit Stress

0

5 Unit strain [ r /

..

FIG. 4 Stress-strain diagram: basis far material's impact properties.

D

2.8-6

/

Load & Stress Analysis

The modulus of resilience ( u ) of a material is its capacity to absorb energy within its elastic range, i.e. without permanent deformation. This is represented on the tensile stress-strain diagram by the area under the crIn7e defined hy the triangle 0 A B, having its apex A at the elastic limit.

Section Properry

I

Steady load rtiength

I S= :. -Impact load strength

I

Since the absorption of energy is actually a volumetric property, the u in (in.-lhs/in.") = u in psi. When impart loadmg exceeds the elastic limit (or y e l d stren@h) of the material, it calls for toughness in the material rather than resilience. The ultimate energy resistance ( n u ) of a material indicates its toughness or ability to resist fracture nnder impact loading. This is a measure of how well the material absorbs rnergy without fracture. A material's ultimate energy resistance is represented on the stressstrain diagram by the total area OACD under the curve. Here point 4 is at the material's yield strength (cry) and point C ;it its ultimate strength ( r , , ) . For ductile steel, the uliimate energy resistance is approxiniately-

where: 6.

=.dtiinate miit elongation, in./in

Since the absorption of energy is actually a volumetric property, u,, in (in.-lb~/in.~)= u,, in psi. Impact properties of common &sign materials are charted in Tablc 3. 9. IMPACT PROPERTIES OF SECTION The section property which is needed to withstand impact loads or to absorb energy in bending is I/?. This is very important because as moment of inertia ( I ) increases with deeper sections, the distance from the neutral axis to the outer fiber ( c ) increases ~ L Sits square. So, increasing only the depth of a section will increase the section's moment of inertia but with little or no increase in impact property. For example, suppose there is a choice between these two beams:

I

Beom A 12" WF 65# Beam

Beam B

I

533.4 i n ?

533.4

533.4

= 88.2 --

24" WF76#

Beam

2096.4 in?

in?

2096.4 1 1.96

----- - 175 i n ? 2096.4 111.96!'

-

14.6 i n 2

The new be:m ( B ) with twice the depth, has about 4 times the bending stiffness ( I ) , and 2 times the steady load strength ( I / c ) , but for all practical puryoses there is no increase in the impact load strength (I/cY). In this example, there would b e no advantage in changing from ( A ) to ( B ) for impact. 10. IMPROVING ENERGY ABSORPTION CAPACITY The basic rule in designing members for maximum energy absorption is to have the maximum volume of the member subjected to the maximum allowable stress. If possible, this maximum stress should be uniform on every cubic inch of the member. I . For any given cross-section, have the maximum amount of the area stressed to the maximum allowable. In the case of beams, place the greatest area of the section in the higher stressed portion at the outer fibers. 2. Choose sections so the member will be stressed to the maximum allowable stress along the entire length of the member. For a member snbjected to iinpact in axial tension, specifying a constant cross-section from end to end will uniformly stress the entire cross-section to the maximum value along the full length.

Designing for Impact Loads

/

2.8-7

TABLE L l m p a c t Properties o f Common Design Materials

Material

1-1 Steel

Alloy Steel

Gray Coif lion Malleable Cost l i o n

200,000

..

230,000

/

30x10'

0.12

6.000

20,000

15 X 10'

0.05

20.000

50,000

23 X 10'

0.10

-

* Bored

667.0 ..

1.2.. 17.4

22,000

70 3,800

on integmtor-rneoruicd area under rtierr-rtroin curve.

A beam can be designed fol- constant bending stress along its entire longth; by making it of variable depth. Although the cross-section at any point is not uniformly stressed to the maximum value, the outer fiber is stressed to the maximum value for the entire length of the member.

FIGURE 5

In Table 3 the member in tension (No. 4 ) has t h e e times the energy-absorption capacity of the simple beam with a concentrated load (No. 1). This is because the tensile member (No. 4 ) has its entire cross-section nnifor~nly stressed to maximum for its full length. In contrast, the maxinn~~n bending stress in beam No. 1 is at thc outer fibers only; and this bending stress decreases away from the central portion of the beam, being zero at the two ends. Notice that decreasing the depth of the beam at its supports, so the n~aximnrnbending stress is uniform along the entire lcngth of the hram, doubles the energy absorbing ciipacity of the beam. See (1) and (9). For a steady load, doubling the length of a beam will double the resnlting bending stress. However, for an impact load, doitbling the length of the beam will reduce the resulting impact stress to 70.7% of the original. Two identical rectangular beams can theoretically absorb the same amonnt of energy and are just as strong under impact loading. The section property

which detemiines this is I/?, and this is constant for a given rectanqular area repaniless of its position.

2.

/

Load & Stress Analysis

stress due to impact. 4. In a simple tensile bar of a given uniform cross-section, increasing the length (1) will not alter tho static stress yet it will decrease the stress due to impact.

11. NOTCH EFFECT ON ENERGY ABSOR CAPAC lTY

FIGURE 7

The two tensile bars shown in Figure 5 have equal strength under steady loads; yet, the bar on the right, having uniform cross-section, is able to absorb much more energy and can withstand a greater impact Isad. Summary

1. The property of the section which will reduce the impact stress in tension is increased volume ( A L ) . 2. The property of the section which will rcduce the impact stress in a simple beam is:

3. In a simple beam, a decrease in length ( L ) will decrease the static stress, but will increase the

In Figure 8, diagrams e and f represent the energy absorbed along the length of a member. The total energy absorbed corresponds to the area under this diagram. Assume the notch produces a stress concentration of twice the average stress ( d ) . Then for the same maximnm stress, the average stress will be reduced to % and the energy absorbed ( f ) will be of the energy absorbed if no notch were present ( e ) . For a stress concentration of three times the average stress, the enorgy absorbed will be t k Notched bar impact test results are of limited value to the design engineer, and can be misleading: ( a ) The test is highly artificial in respect to severe notch condition and manner of load condition. ( b ) The results can be altered over a wide range by changing size, shape of notch, striking velocity, and temperature. ( c ) The test does not simulate a load condition likely to be found in service. ( d ) The test docs not give quantitative values of the resistance of the material to energy loads.

I 101

Tensile member with notch

Tensile member. unifbrm section

/Stress

7

FIGURE 8

1

a t notch

,,Sfrerr in member

k!

Designing for Impact Loads

/

2.8-9

12. GUIDES TO DESIGNING FOR IMPACT LOADS 1. Design the mr.m5er as an energy-absorbing system, that is have the maximum volume of material stressed to the highest working stress; this increases the energy absorbed. 2. For any given cross-section of the member, have the maximum area subjected to the maximum allowable stress; also stress the entire length to this value. 3. The property of thc section which will reduce the impact stress in tcnsion is increased volume ( A L ) . 4. The property of the section which will reduce the impact stress in bending is increased I/+. 5. Increasing the length ( L ) of a beam will increase the static stress, but will decrease stress due to impact. 6. Increasing the length ( L ) of a tensile member of uniform cross-section will not change the static stress, but will decrease stress due to impact. 7. Use the basic formula, or those shown in Table 3, as a guide to select the required property of section and property of material. 8. Select material that has a high modulus of resilience n

9.

10. 11.

12.

13.

14.

15.

- --

us2 -. Materials having lower modulus 9. F

of elasticity ( E ) generally have lower values of yield strength (us),and this latter value is more important becanse it is squared. Therefore steels with higher yield strengths have higher values of modulus of resilience and are better for impact loads. The material should be ductile enough to plastically relieve the stress in any area of high stress corrccntration; and have good notch toughness. Thc: material shonld have high fatigue strength if the impact load is vepeatedly applied. The material should have good notch toughness, and for low temperatnre service, a low transition temperatnre. Reduce stress concentrations to a minimum and avoid a b n ~ p tchanges in section. If possible, place material so that the direction of hot rolling (of plate or bar in steel mill) is in line with impart force. For inertia forces, decrease the weight of the member, while maintaining proper rigidity of the member for its particular use. This means lightweight, well-stiffened members having sufficient moment of inertia ( I ) should be used. One aid against possible inertia forces caused by the rapid movcment of thc member due to explosive energy, earthquakes, etc., is the use of

FIGURE 9

flexible supports, to decrease the ac~eleration and/or deceleration of the member.

I

Problem 1

Accelerating a load

k

Beam

FIGURE 10

Find the load placed on the supporting beam for a hoisting unit in the shaft of a mine if the 5000-lb load ( W 2 ) is accelerated upward to a velocity ( V ) of 1800 feet per minute in 5 seconds ( t ) . The dead weight of the hoisting unit is 1000 lbs ( W , ) .

2.8-10

/

Load & Stress An~lyris

acceleration a = V2

on trailer have failed, and stops from a speed of 60 miles pcr hour within 15 seconds.

- V1

= 6 ft/sec'

deceleration

force of accelcratio?~

= 931 lbs

force of deceleration

total load on beam

F = -W a g

+ ( 5 0 0 0 ) i(931)

WI i- wa i- Fa = (1000) = 6931 lbs

- (40'000) (5.86) - (32.2) r

Asmme the truck brak~sthe trailer, because brakes

V = 60 MPH 6-------

W

: 40.000

7275 lbs

The king pin on the fifth wheel, connecthe the trailer to the tractor must be designed to transfer this force.

lbi

FIGURE 1 1

F = 7275

FIGURE 12

Designing for Fatigue Loads 1. ENDURANCE LIMIT When the load on a member is constantly varying in value, or is repeated at relatively high frequency, or constitutes a complete reversal of stresses with each operating cycle, the material's endurance limit must be suhstitnted for the ultirnate strength where called for by design formulas. Under high load valnes, the variable or fatigue mode of loading reduces the material's effective ultimate strength as the nnmbcr of cycles increases. At a given high stress value, the material has a definite service or fatigue life, expressed as N cycles of operations. Conversely, at a given nnmber of service cycles the material has a definite allowable fatigue strength. The end:~raiicc limit is the maximum stress to which the material can be subjected for a given service life.

2. NATURE OF FATIGUE LOADING Fatigue failure is a progressive failure over a period of time which is started hy a plastic movement within a localized region. Although the average unit stresses across the entire cross-section may be below the yield point, a non-uniform distribution of these stresses may cause them to exceed the yield point within a small area and catlse plastic movement. This eventually produces a minnte crack. The localized plastic movement f u r t h ~ raggravates tlie non-iuiiform stress ditribution, and frrrther plastic movcment causes the crack to prog e . s . The stress is important only in that it causes the plastic nrov~ment. Any fatigne test usnally shows considerable scatter in the resnlts obtained. This resnlts from the wide range of time required hcfore the initial crack develops

in the specimen. Once this has occurred, the subsequent time to nltimate failnre is fairly well confined and proceeds in a rather uniform manner. The designrr when first encountering a fatigue loading probleni will often use the material's endurance limit or fatignc strength value given in his engineering handbook, ~vithnut f d l y considcring what this value represents and how it was obtained. This procadure wnld lead to scrioiis trouble. There are many types of fatigue tests, types of loading, and types of specimens. Theoretically the fatigue value used by the designer should be determined in a test that exactly duplicates the actual service conditions. The sample used should preferably be identical to the member, the tcsting machine should reproduce the actnal scrvice load, and tlie fatigue cycle and frequency should be the same as wonld be enconntcri~din actlid scrvice. For example, if thc problem is a butt xvdd in tension, the allowable fatigue strength used in thc design must come from data obtained from loading a hntt weld in axial tension on a pulsating type of fatigne testing machine, with the same range of stress. 3. ANALYZING THE FATIGUE LOAD Fignre 1 illustrates a typical fatigue load pattern, the cnrve represeuting tlie applied stress at any given moment of time. There are two ways to represent this fatigue load: 1. As a niwn or average stress (v,,,) with a superimposed variable stress (r,, ). 2. As a stress varying from maximum value (IT,,,,,) to a minimum (IT,),,,,!.Here, the cycle can be represented by the ratio--

FIGURE 1

I

Time ----+

2.9-2

/

Load & Stress Analysis

One approach to this problem is to let the variable stress (u,.) be the ordinate and the steady or mean stress (urn) be the abscissa. When the mean stress (urn) is zero, see Figure 2, the varihle stress (u,) becomes the value for a complete reversal of stress ( u r ) . This value would have to be determined by ,experimental testing, and becomes point b in the diagram. When there is no variation in stress, i.e. a steady application of stress, u, becomes zero, and the maximum resulting mean stress (u,,) is equal to the ultimate stress for a steady load (u,,); this becomes point a.

FIGURE 2 FIGURE 3

where:

= fatigue strength for a complete reversal of stress u v = variable stress which is superimposed upon steady stress USI= ultimate strength under stead load (Some set u, equal to the yiel strength, u 7 ) u r n = mean stress (average stress) "2

conservative values; almost all of the test data will lie just outside of this line. From similar triangles it is found that-

g

A line connecting points b and a will indicate the relationship between the variable stress ( u , ) and the mean stress (u,,) for any type of fatigue cycle, for a given f a t i y e life ( N ) . This straight line d l yield

A Goodman diagram, Figure 3, is constructed from Figure 2 by moving point a vertically to a height q u a 1 to u,,; in other words, line a-c now lies at a 45" angle. It can be shown by similar triangles that the same relationship holds:

esigning for Fatigue Loads

The Goodman diagram of Figure 3 may bc modified so that the ordinate becomes the maximum stress (urn,,) and the abscissa becomcs the minimum stress (urnin);see Figure 4. It can be proved that all three diagrams yi'eld the same results. The American Welding Society (Bridge Specification) uses this last type of diagram to illustrate their fatigue data test results. If the maximum stress (urn,,) lies on line a-b, this value is found to be-

/

2.9-3

These "dependable values" have been reduced to some extent below the minimum values obtained in the test. A factor of safety is applied to obtain allowable values; these are shown hy dotted lines. This is expressed as a formula along with a value which should not b e exceeded. In this case, the maximum allowable is 18,000 psi. This formula represents thc slanting line, but a maximum value must be indicated so that it ir not carried too far. Figure 6 illustrates several types of fatigue cycles, with conesponding K values to be used in the fatigue strength formulas. ABLE MAXIMUM STRESS

urnin where K = -

urnax

The next diagram, Figure 5, is constructed with the values for complete reversal ( a , ) and the ultimate strength ( u , ) for butt welds in tension. The fatigue data from test results are also plotted. Notice the values lie on or slightly above these straight lines for service life ( N ) of 100,000 cycles and that of 2 million cycles.

Fatigue strength formulas, for determining the allowable maximum stress for a given service life of N cycles, are presented in Table 1for A7 mild steel, A373 and A36 steels, in Table 2 for A441 steel, and in Table 3 for T-1, quenched and teropered high yield strength steel. Reqnircd fatigue life or number of cycles will vary but usually starts at several hundred thousand cycles. It is assumed that by the time the value of several million cycles is reached, the fatigue strength has

@

Allowable values-

+10

+20

Minimum stress, ksi

FIGURE 5

+30

+40

100,000 cycler 2,000,000 cycles

----

+50

+60

+

/

2.9-4

Load & Stress Analysis

The constant (k) will vary slightly with the specimen; however, 0.1:: has been widely used for butt welds and 0.18 for plate in axial loading (tension and/or compression ) . The curve in Fignre 7 illustrates the general increase in fatigue life when the applied fatigue stress is reduced. As an cxarnple, in this case, reducing the fatigue stress to 75% of its normal value will in general increase the fatigue life about nine times.

leveled off and further stress cycles wor~ldnot produce failure. For any particular specimen and stress cycle there is a relationship between the fatigue strength (o-) and fatigue life ( N ) in number of cycles before failure. The followmg empirical formula may be used to convert fatigue ytrengths from one fatigue life to another:

where:

= fatigue strength for fatigue life N.

Test data indicates a fatigue life of N, = 1,550,000 cycles when the member is stressed to oa = 30,000 psi. What would be the fatigue strength at a life of 2,000,000 cycles?

ub --. fatigue strength for fatigue life Nb N, = fatigue life for fatigue strength o-, Nb = fatigue life for fatigue strength ub

LE 1-Allowable Fatigue Stress For A7, A373 and A36 Steels and Their But Not to Exceed

2,000,000

cycler

0= ---------

Bane Metol in Tension Connected By Fillet Weldr But not to exceed

c

10.500 I-213K

psi

2 P, I( psi

Bare Metal

Compression Connected By Fillet Weidi

0=

Butt Weid in Tension

T '

18.000 K psi 1- -

Pt

psi

2

But1 Weid Comprersion

d-

-Butt Weid in Shear

--

.

@ r

9.000 = ---K psi 1- 2

= 18,000 psi I .8K

-

0= -7

10.000

K psi i --

fillet Weids 0,

=

---

0 --

Leg Sire

K = minlrnox Adopted from AWS Bridge Specificotionr. P. = Allv.voble unit compressive stress far member. Pt = Allowable unit tensile stress for member.

2

PI psi

esigning for Fatigue Loads

"II/ -+

"I

mln

1

+ -

Time

-

Time

-+ FIGURE 6

=

Time ---+-

4 : "I-

+ max

/

K = +

1

(steady)

-

min

=

-

min

=0

K=fl

mln

=

K = - 1

+ 1/2 m m

- max

K =

f 1/2

[complete

reversal)

-I

FIGURE 7

For butt welds, k

Increase in fatigue life

2.

= .I3

2.9-6

/

Load 8, Stress Analysis LE 2-Allowable For A441 Steel

I

2.W0.000 cycler

Fatigue Stress md i t s Welds

600,000 cycler

100,WO cycler

But Not to

Exceed

23 pRI

Base Metal in Tension Connected By Fiilet Weidr

.- psi

PCpsi

Bore Metal Compression Connected By Fillet Weldr

PCpsi 1

- i/z R

Bun Weld in Tension

Pt psi

Butt Weld In Shear

*

@

Fiilet Weidr w = leg liz

f = 8800 i-fiR

iblin.

= 10,400 w iblin.

Adapted from AWS Bridge Specificotionr. if SAW-I, use 8800 R = .in/mox load p, = Allowable unit tenriie rtiesr for member P, = Allowable unit cornpierrive rtierr for member

*

TABLE 3-Allowable Fatigue Stress uenched ond Tempered Sfeels of High Yield Strength

. Fiilet Weld w = leg size

n

-

f I

-

lap.TT

6 360 W i .A0 K

9,900 w f = ------Ibrlin. I - .75 K

= 2-Ibrlin.

-

I

f = I

f = 26.1600 Ibrlin.

14,500 W Ibrlin. .6O K

1

-

I

Above valves adopted from "The Fabrication and Design of Structurer of 7-1 Steei" by Gilligon and Englond, United Stater Steel Cotporotion.

Designing for Fatigue Loads

/

FIGURE 8 FATIGUE NOMOGRAPH

Given: Test data indicates a butt-weld fatigue life of N, = 1,550,000 cycles when the member is stressed to a, = 30,000 psi Find: The weld's fatigue strength (ab]at 2,000,000 cycles (N,)

and since the butt weld's k factor is .13, the nomograph indicates

-ah -- 96.8% 0-

or a,

= 30,000 X

96.8% = 29,000 psi

The anti-log of this is 0.96740; hence:

Since:

55 =

( )

(For butt welds, k = 0.13) or:

(2-)

and:

Wb

3= 0,

ii

Using logarithms* for the right hand side:

= 0.13(log 0.775) = 0.13(9.88930 - 10) (add 8.7 to left side and = 1.285fX9 - 1.3 8.7 - 8.7 subtract 8.7 from right side) 9.985609 -10.0

+

= 29,020 - psi at Nb = 2,000,000 cycles) The nomograph, Figure 8, further facilitates such conversion and permits quickly finding the relative allowable stress for any required fatigue life provided the fatigue strength at some one fatigue life is h o w n and that the constant k value has been established. Conversely, the relative fatigue life can be readily found for any given stress and any constant ( k ) . * A log-log slide rule could be used to find the value of 0.775 raised to the 0.13 power.

/

Load

&

Stress Analysis

5. RELATIVE S E V E R I T Y OF FATlG In Figure 9, the allowable fatigue stress is the vertical axis (ordinate) and the type of fatigue stress cycle ( K = min/max) is the horizontal axis (abscissa). The extreme right-hand vertical line ( K = 1) represents a stcady stress. As we proceed to the left, the severity of the fatigue cycle increases; finally at the extreme left-hand axis ( K = - I ) there is a complete reversal of stress. This is just one method of illustrating fatigve stress conditions. The important thing to be noticed here is that actual f a t i y e strength or allowable fatigue values are not reduced below the steady stress condition until the type of cycle (K = min/max) has progressed well into the fatigue type of loading. In the case of 2 ndlion cycles, the minimum stress must drop down to '/z of the maximum stress before there is any reduction of allowable strength. In the case of 100,000 cycles, the minimum stress can drop to zero before any reduction of allowable strength takes place. Even at these levels, the member and welds would be designcd as though they were subjected to a steady load. The stress cycle must extend into a wider range of fluctuation before it becomes necessary to use lower fatigue allowables.

+

In other words, a fatigue problem occurs only if1. Stress is very high, 2. Anticipatrd service extends for a great number of cycles, 3. Stress fiuctnates over a wide range. And it generally rcquires a11 three of these situations occurring simultaneously to produce a critical fatigue condition worthy of consideration. The allowable fatigue strength values obtained from the formulas in Table 1 take all three of these into consideration, and it is believed they will result in a conservative design.

Several formulas are available for this consideration but very little actual testing has been done on this. In many cases there is not very good agreement between the actual test and the formulas. 1. Principal-stress theory -

2. Maximum shear-stress theory-

we = $\r(w,

-- v , ) ~+- 4 7.2

FIG. 9 Severity of fatigue depends on stress value and range of fluctuation, as well as service life.

esigning for Fatigue Loads

/

TABLE &-Fatigue Strength o# Butt Summary of Results, Using %-In. Carbon-Steel Plates FATIGUE STRENGTH I N 1000's OF PSI

-

TENSION TO A N EQUAL COMPRESSION

Derctiptian

of

-N=

Specimen

1

Reiniorcement On Sties, Relieved

22.3

O

-

-

N=

N=

100,OW Ar Welded

--

2,000,000

1

14.4

N= 2,000,000

100.000

1

33.1

TENSION TO TENSION if> AS GREAT

I

22.5

--

100.000

1

53.3

-

N= 2,000,000

N=

/

36.9

1

43.7

31.9

/ 1

Reinforcement Mochined Off Not Strera Relieved Reinforcement Mochined Off Stress R e l w e d

28.9

I

24.5 26.8

Reinforcement Ground Off Not Stress Relieved

1

1

16.6

1

/ /

1

48.8 49.4

44.5

1 1

1

28.4 27.8 26.3

/ 1

1

1

1

I

42.6

Plain Piole Mill Scoie On Ploin Plote Mili Scole Machined O i i ond Surioce Poished

Bun Weld. Reinforcement and Miii Scoie Mochined Off ond Suiioce Polished

3. Shear-stress-invariant theoryme

=

vux2-

+

UxU7

where: uo, = fatigure strength in (x) direction a,, = fatigue strength in ( y ) direction

-

+ 3 rxy2

fCTy2

4. Combined bending and torsion. Findley corrected shear-stress theory for anistropy-

a, and uy = applied stresses 7. INFLUENCE OF JOINT Any abrupt change of section along the path of stress Bow will reduce the fatigue strength. It is not welding that effects a reducing of the fatigue strength but the resultant shape or geometry of the section. It is for this reason that fillet welds have lower fatigue strength. simply because they are used in lap joints and all lap joints including riveted joints have lower fatigue strength.

is the ratio of fatigue strength in pure where ub/?bending to that in pure tension.

5. Combined tensile stresses. Gough suggests-

TABLE 5-Effect

of Transverse Attachments On Fatigue Strength

-

100,000 cycler 2,000,000 cycler

25.800 psi

1

1

I

I

22,800 psi

I

25,400 pri

I

18.9OOpri

I

76

U

22,900 p d 13,100 pi

Stress

Analysis

By means of Table 4, we can see that removing the reinforcement of a butt weld increases its fatigue strength to that of unwdded plate, also that stress relieving the weld has no appreciable effect on its fatigue strength. Table 5 illustrates the effect of transverse fillet welds upon the fatigue strength of plate, this is %" plate. The attachment causes an abrupt change in section, and this reduces the fatigue strength of the plate. It is believed these results could be duplicated by machining these joints out of solid plate, without any welding.

ING FOR FATIGUE LOADING I. Usually a member is stressed to the full maximum value for only a portion of its fatigue life or cycles. For most of its fatigue life, the member is stressed to a much lower value, and not to its full rated capacity; hence, most fatigue loading is not as severe as it may first appear. Consider actual stress rather than average stress. Reduce if possible the range of stress without increasing the maximum or average stress. 2. Fatigue loading requires careful fabrication, smooth transition of sections. Avoid attachments and openings at locations of high stress. Avoid sharp comers. Use simple butt weld instead of lap or T fillet weld. Grinding the reinforcr:ment off of butt welds will increase the fatigue strength. This weld will have about the same fatigue strength as unweldt-d plate. Grinding, however, should not be specified unless essential, sincc it does add to the final unit cost. Avoid excessive reinforcement, undercut, overlap, lack of penetration, roughness of weld. Avoid placing weld in an area which flexes. Stress relieving the weld has no appreciable &ect upon fatigue strength. DBiculties are sometimes caused by the welds being too small, or the members too thin. 3. Under critical loading, place material so that the direction of rolling (of plate in stml mill) is in line with force, because the fatigue strength may be higher in this direction than if placed at right angles with the direction of rolling. See Figure 10. 4. Where possible, form member into shape that it tends to assume under load, and hence prevent the resulting Aexial movement. 5. Avoid operating in the critical or resonant fre-

quency of individual member or whole structure to avoid excessive amplitude. 6. Perhaps consider prestressing a beam in axial compression. This will reduce the tensile bending stress and lessen chance for fatigue failure even though the compressive bending stress is increased to some extent. 7. Avoid eccentric application of loads which may cause additional flexing with each application of load. 8. Stiffeners decrease flexibility of panel and result in better fatigue strength, unless they cause a more abrupt change of section. 9. A rigid frame type of structwe or statically indeterminate type of structure may be better than a simple structure since the load is shared by other members; hence, the structure is less likely to collapse immediately if a fatigue failure starts in one member. 10. Avoid biawial and triaxial stresses, avoid restrained internal sections.

~ e c o m e ~ d emethod d if fatique

Direit8on

or impact ioodinq

of hot rollinq

ofsheets insteel mills e e c o w e n d of Least on boitom h d f or thlrd,or w h o i e t a n k , s h e e t s be run lenqthwise with tonk FIG. 10 Grain direction of sheet or plate should be in line with force, for greater fatigue strength.

S E C T I O N 2.1

Torsional loading is the application of a force that tends to cause the member to twist about its simxtural axis. Torsion is usually referred to in terms of torsional moment or torque ( T ) , which is basically the product of the externally applied force and the moment alm or force arm. The moment arm is the distance of the centerline of rotation from the line of force and perpendicular to it. This distancc often equals the distance from the member's center of gravity to its outer fiber (radius of a round shaft, for example), but not always. The principal deflection caused by torsion is measured by the angle of twist, or by the vertical movement of one comer of the frame. Steel, in rolled structural shapes or built-up sections, is very efficient in resisting torsion. With steel, torsionally rigid sections are easily developed by the use of stiffeners. Here are the three basic rules for designing sbuctural members to make the best use of steel where torsional loads are a problem: 1. Use closed sections where possible. 2. Usc diagonal bracing. 3. Make rigid end connections.

When a round shaft is subjected to a twisting or torsional moment (torque), the resulting shear stress in the shaft is-

where:

= shear stress, psi c = distauce from centcr of section to outer fiber T = torque, in.-lhs. J = polar moment of inertia of sedion, = IX I7 = 21 T

+

The angular twist of a round shaft is-

where: B = over-all angular twist of shaft, in radians (1 radian = 57.3" approx.)

I, = length of shaft, in iuches E, = modulus of elasticity in shear (steel E, = 12,000,000 psi) In most cases, the desiper is interested in holding the torsional moment within the material's elastic limit. Where the torsional strength of a round shaft is required (i.e. the stress it can take without failure), the polar section modulus is J/c, and the allowable torque is thns-

T =

T,,--J

c

where, lacking test data, the ultimate shear strength of steel ( 7 , ) is assumed to be in the order of 75% of the material's ultimate tensile strength. The above three formulas are true for sdid round or tubular round shafts. For non-circular sections the shear stresses are not uniform, 'and therefore the standard torsional formulas no longer hold. 3. TORSIONAL RESISTANCE Valucs of torsional resistance (K)-stiffness factorhave h e m estahlish~d for various standard sections and provide more reliable solutions to torsional rigidity problems. Values of R are exprssed in inches to the fourth power. Table 1 shows the formulas for shear stress and torsional resistance of various sections. The formulas for solid rectaiigitlar sections caU for valurs, of a and ,8, which are derived froin the ratio of section width ( h ) to depth ( d ) , as shown in the table. Actual tests show that the torsional resistance ( R ) of an open section made up of rectangular areas, nearly equals the sun1 of the torsional resistances of all thc individual rectangular areas. For example, the torsional resistance of an I benm is approximately

2.10-2

/

Lood & Stress Analysis

/

Angle of twist

Conventional

FIGURE 1 poior moment

equal to the sum of the torsional resistances of the two flanges and weh (Fig. 1). Figure 2 shows the results of twisting an I beam made of three equal plates. Calculated values of twist by using the conventional polar moment of inertia ( J ) and the torsional resistance ( R ) are compared with the actual results. This shows greater accuracy by using torsional resistance ( R ) . This means that the torsional resistance of a flat

Resistonce

FIGURE 2

TABLE I-Torsional

Properties of Various Sections -..

. -

Section

ifor mlid

/

b = 1.00 1.50 1.75 d rectangular---

jsections

a

2 0 8 1.31 ,239

Shear stress

2.00

2.50

.246

.258

(for steel) R4orsionai Resistance

3.00

4.00

6

Use t4.s

[&yo& bracing

Rz3.542

l of

ning for Torsional Loading

!

Angle of twist

!

/

2*T&

sisting torsion is ;i closed square or rectangular tubular section. Tablr 2 provides formulas for dstexmining the torsional rt~sistarrce( R ) of various closcd tubular sections. It also provides tire basic fomiulas for detemining the shear stress ( T ) at any given point along the sidewall of any closed section regardless of configuration or variation of thiclaicss, and for determining the section's torsional resistance ( R ) . T l k poorest sertions for torsional loading are open sections, flat plates, angle sections, channel sections, Z-bar sectioris, T-har sections, I-beam sections, and tubular sections which have a slot.

FIGURE 3

plate is approximately thc same whether it is used as such or is formed into an angle, channel, open tube section, ctc. This is illustrated in Figure 3. Samples of different sections made of 16-gage steel are subjected to torsion. The flat section twists 9". The same piece of steel formed into a channel ( b ) twists 9%". When rolled into a tube with an open beam ( e ) , it twists 11". When the same section is made into a closed section ( d ) by placing a single tack weld in thc middle of the open seam, the torsional resistance increases several hundred times. When the tube becomes a closed section, the torsional stresses are distributed more evenly over the total area, thus permitting a greater load. Notice the emor in using polar moment of inertia ( J ) for the angle of twist of open sections, and the good agreement by using torsional resistance (R).

FIGURE 5

After the R values of all areas in a built-up section have becn added together, their sum is inserted into the following formula or n modification of it:

Torque ( T ) in in.-lbs may be obtained from one of the formulas in Table 3, such asT

63,000 X IIP RPM

=. --

where: The solid or tubular round closed scction is best for torsional loading since the shear strmses are uniform around the circumference of the member. Next to a tubular section, the best section for re-

FIGURE 4

HP = horsepower RPM = speed of revolution P I- applied force, lbs e = moment arm of force (the perpendicular distance from the center of rotation to the line of force)

As an example, consider the torsional resistance of a closed round tube and one that is slotted. The tube has an O.D. of 4", and I.D. of 3", a length of 100f', and is subjected to a torque of 1000 in.-lbs.

/

2.10-4

Load

Stress Analysis

Case 1

CA

From Table 1,the torsional resistance of the closed round tube is found to be-

R = 0.0982 (dyi - dl4) = 0.0982 (4' -31)

FIGURE 6

and the angular twist isCase 2

= 0.000485 radians, or 0.0278" LE 2-Torsional

From Table 1, the tors~ollalresistance of the slotted round tube is found to bcResistance ( ) of Closed Tubular Sections -=

== = = == =

0 = L

=;

f =

enclosed within mean dimensions. length of p ~ r t i c u i a rsegment o i section overage thickness of segment a t point Is! sheoi rtresi a t point (i! torsion01 resistance, in4 modulus of eloiticity in rheor (steel = i2.000.000i onguloi twist lrodionr) length of member (inches! unit shear force

-

R

esigning for Torsional Loading

1.0472 t:! d

/

LE 3-Formulas ~ O PDetermining Safe Torque Under Various

Based on tangential load: and the angular twist isBased on horsepower transmitted:

r

-- 0.018 radians,

NP = 63,030 X --

RPM

or 1.04" Based on strength of shaft:

Thus, the tube witlmut the slot is many times more rigid than the slotted tube.

1

Problem 2

1

where S, = 25,000

Two 6" X 2" X 10%-lb chaniids are to be used in making a 100"-long frame, which will be subjected to a torque of 1000 in.-lbs. In what relationship to each other will these channels offer the greatest resistaxe to twist?

2945 dz4 - dl4 T = -dz Based on safe twist of shaft (.08"/ft):

Case 1

These two channels when separated but fastened together by end plates do not have much torsional resistance.

Based on fillet weld leg size around ihaft or hub:

Based on butt weld size around hub:

T = 20,420 d2 t

Case 2

FIGURE 7

When these two channels are securely fastened back to back, there is suitable n:sistance to any slip or movemcnt due to horizontal shear. Here the two webs are considercd as one solid web, and the top and bottom flanges are considered solid.

From Tdhk 1 . the \.due of R for each of the flanges is found to b e Rl = 0.0306 in4 and that of each web isRP = 0.0586 in.' and thus the total angular twist is-

= 0.0348 radians, or -2.0"

FIGURE 8

/

2.10-6

Load & Stress Analysis

From Table 1, thc value of R for each of the two conlposite flanges is found to heR1 = 0.066 and that of the composite web is-

R? = 0.459 in4 m d thus the total angular twist is-

= 0.0141 radians, or 0.81" ~ohichis much less than in Case 1 Case 3

If these two channels were welded toe to toe to form a box section, the, torsional resistance would be greatly increased.

From Table 2, the value of Fi for a box section is found to be-

and the angular twist is-

= 0.00027 radians, or 0.015" which is far less than in Case 2, which in turn was much better than Case 1.

Torsional Resistance Nomograph A panel or other member may be sufficiently resistant to deflection by bending, and yet have very low torsional resistance.

The nomogl-aph, Figure 10, permits the designer to quickly find the torsional resistance of a proposed design. The total torsional resistance of a built-up design equals the sum of the resistances offered separately by the memhers. On this nomograph: Line 1 = Type of section, or element of a built-up scdion. Obscrve caution as to meaning of letter symbols. For a solid rectangular section use the ratio of wiclth ( a ) divided by thickness ( b ) ; for a hollow rectangular section use width ( b ) divided by depth ( c ) . Line 2 = Dimmsion ( a ) , in. Line 3 = Pivot line Line 4 = Dimension ( b ) , in. Line 5 = Torsional resistance of the section ( R ) , i n 4 Thcse values for cacli crlement are added together to give tho total torsional resistance of the section, and the resistances of the sections are added to give the total torsional resistance of the frame or base. This is used in the design formula for angular twist, or in the next nomograph, Figure 14. In the ease of a member having a built-up crosssection, such as a T or I beam, read the Figure 10 nomograph for the R value of each element or area making up the section. Start at vertical Line 1 in the nomograph, using the scale to the right of it that expresses the rectangular element's a/b ratio. i n the case of solid squares or rounds, and closed or open round tubes, go dil-cetly to the point on the scale indicated by the visnal represontation of the crosssection. Notice that the meaning of a and b varies. In the case of a rectangnlar element, a is the longer dimensidn; hut in the case of a hollow rectangle, (I is the wall or plate thickness. The valuc of a or b on Lines 1, 2 and 4 must correspond, according to the type of section or element for which torsional rcsistance ( R ) is sought. For hollow rwtangnlar sections (of uniform wall or plate thickotxs j, use the scale along the left of vertical Line 1 that expresses the ratio b/c. Here b the section's width and c = its depth.

-

U M SHEAR STRESS IN BUILT-UP SECTIONS The maximum shear stress of a rectangnlar section in torsion lies on thc surface at the center of the long side. For the maximum shear stress on a narrow rectangular section or section element-

eaigning for Torsional Loading

/

2.10-8

/

Load & Stress Analysis

I

I

Problem 3

A 6" X 2" X 10%-lb channel is subjected to a torque of T = 1000 in.-lbs. Find the shear stress along the web. See Figure 13. Applying the fotmula for rectangular sections from Table 1, find the torsional resistance of each of the two identical 2" X %" flanges ( R , ) and of the gr X 5/16" web ( R 2 ) :

FIGURE 11

where:

Q, = unit angular twist of whole section (each element twists this amount), in radians/linear inch of member t = thickness of rectangular section

R = torsional resistance of entire member, not necessarily just this one flat element

Then:

This formula can be used for a flat plate, or the flat plate of a built-up section not forming a closed section (i.e. channel, angle, T- or 1-beam section). 111 such a built-up open section, the unit angular twist (4) of the whole member is first found:

= 2,580 psi

I and then the maximum shear stress in the specific rectangular element.

'1

FIGURE 12

7-

Problem 4

1

Two 6" X 2" X 10%-lb cha~melsare welded toe to toe, to form a short box section. This is subjected to a torque of T = 100,000 in.-lbs. Find the horizontal shear stress at the toes and the amount of groove welding required to hold these channels together for this torsional load. See Figure 14. From Table 2, the shear stress at mid-length of the short side is found to be-

T

where:

b =6 d = 4

Shear stresses tend to concentratc at re-entrant corners. In this case, the maximum stress valne should be used and is-

- % = 5.625" - XB = 3.6875"

[A] = bd 100,000 2(5.625 X 3.6817%

- ..

where a = inside corner radius.

= 6420 psi

, FIGURE 13

Designing for Torsional Loading

/

2.1

I

Two

6" x 2" x 10B# channels

FIGURE 14

-t 9" brick = 140 lhs/sq ft Since the wall is 12' high, this is a load of 1680 lhs/lincar ft or 140 lbs/linear in. Or, use w = 1.56 Ihs/lin in. to include beam weight. 4" limestone

The horizontal shear force is then-

f =7.t 1

6420 X ,375

= 2410 lbs/linear inch

bending resistance (monwnt of inertia)

Since weld metal is good for 13,000 psi in shear, the throat or depth of the continuous hutt weld must hetorsional resistance

The groove weld connecting the channels must have a throat depth of at least 3/16". - Of course, if the torsional load is applied suddenly as an impact load, it would be good practice to add a safety factor to the computed load. This would then necessitate a deeper throat for the hutt weld.

= 442 in." The eccentricity of the dead load applies torque to the beam. From torsional member diagrams in Reference Section 8.2:

Check the following built-up spandrel beam supporting a wall 12' high, made of 4" of limestone and 9" of

brick. The heam's span is 20', and the dead load of the wall is applied 6" off the beam's centerline.

uniform torque

t = 150 lhs/in. X 6" = 900 in.-lbs/in.

FIGURE 15

angular tzoist at center of beam

k b '

= 8.5"

-"f

= ,00122 radians (or .07")

/

2.10-10

Load

Stress Analysis total shear stress

torque at end

= 4100 psi

Then to determine the required size of Ulet weld between flange and web:

torsional shcar stress T

=

T 2 [A1 ts

OK -

where:

t, = thickness of single web

= 1410 psi unit shear force from torque ft=.it

= (1410) ('12) = 700 lbs/in. unit shear force along N.A. from bending

v

FIGURE 17

= w L/2 = (150)(120) = 18,000 Ibs

unit shear force at weld from bending

unit shear force at weld fvom torque ft

= 700 lbs/in.

total unit shear force at weld

FIGURE 16

f, = f t 4- f,, = (700) + (900) = 1600 ibs/in. required k g size of fillet weld (E70)

:

18,000)(10 -. -. -

x 4.5 + 1 x 2.0)

w = -

(449.3) ( 2 webs)

actual force allowable force

total unit shear force on beam web (each) f, =

ft

i-

ft,

= (700) $- (860) = 1560 ibs/in.

However, because of the 1"flange, AWS Bldg. 212, AWS Bridge 217 and AISC 1.17.4 would require a

x,,, h .

for

Tht, r i n i l e s of torsicii~ wliicli dctermin:: iht: bcst sectioi~sfur resisting twist apply to built-II~fx~~lir~es. Just 1 1 t h torsio1t:il rcsist:rnw of the section i s i:qil:~l to the total of tlii. r~~sisi:~rices of its itidi~i1111;ilarms, so is the torsional n'sist:u~ceof a fr:tme approxix~i:i!cIy equ;tl to thr totd I-r-ds!;rnce of it: jnrlivid~l:il p r i s . Tlte tcasional rcsidance of the fvnmc nhos,: litrigit u d i d rn~.nibersart: two chan11i:ls wo~ildbe :ippn)simatply eq11,llto twin, the torsi~ioalrvsistancc of wch channel section, Figure 18. T ~ I(lista~iw , betwwr thost. mentbers fur purpose of this ~,x;~mplt: is considered to have, no effect. Sincc t h ~ c. l o s ~ dsectioir is best for rcsistirig twist, the torsional resistmce of this frame coulil be greatly increased by making t h o channels into rectangular box sections through the addition of plate.

A frame is made of two 6" standard pipes, spaced 24" between centers, and having a length of 60". Tl~isframe supports a 10-hp motor. running at 1800 lljm and driving a pump. Find the approximate twist of the frame undcr tho load.

Porrioslcll Loa

FIGURE 18

Then, adding together the X of each tube, the angular twist is:

= 0.0000156 radians, or 0.00089" Maxinmm deflection in the frame is the vertical displacement ( A ) , which is the product of ailgular twist ( 8 ) arid frame width ( W ) between centers:

FIGURE 19

The, 6" standard p i p h:is O.D.

=

ROHi",

In finding

tiit.

I fi.C;i?!ir' and I.D. lorsioi~alr!%st:~nce of each

tube:

The torque is easily found:

2.10-12

/

Load

Stress Analysis

FIGURE 20

4

t

members

The longitudinal members are now considered to make up a frame of their own. 'When the vertical force (PL) applied at the corncr rcaches the proper value, the frame will deflect vertically the given distance ( A ) and each longitudinal member will twist (81,). The same separate analysis is also made of the transverse members.

TL .'. PL = and \V

TT PT = --- L

and substituting for PL and PT A E, n.r Rr A E,n~ RL PI, = and PT = W I," W2 L

Since the external force ( P ) applied at the comer is the sum of these two forces:

By observation we findA=&W--@TL Then:

A &=-and&=W

A L

Using the common formula for angular twist& =

TL 1.~-. ----E, n~ RI.

and 81. =

and s~~bstituting for 0,. and BT

A W -

.TI. L E, IIL RI.

TT W E, n~ KT

-

A = 'pr l\rW aItd ~... I, C, rrr R r

Then:

T,,

.

-

- . . . . . (4)

where:

L = length of whole framc, in. W = width of \vhole frame, in.

RI. = torsional resisttince of longitudinal member, in."

--

A I<*IIL fir. W L

TT

Since the applied torque is-

TL = PL W and TT = PT L

A E IIT RT W 1,

-5

RT = torsional resistance of transverse member, in,4 nr* = numbor of longitudinal mcmbers n r = number of transverse mcmbcrs

P = load applied at comer, 1bs

esigning Cor Torsional Loading

/

2.1

FIGURE 21

E, = modulus of elasticity in shear (steel: 12 X lo6), psi A = vertical deflection, in.

some deflection due to bending of all the members, and this would slightly i~lcrcasethe over-all deflection of the frame. For simplicity this has been neglected in this analysis.

It can be seen that the torque on a given member is actually produced by the transverse forces supplied by the cross members attached to them. These Fame forces subject the cross members to bending. In other words, the torque applied to a member equals the end moment of the crosq member attached to it. There is

TABLE

4--Torsional

-Ln: PLW

-

To illustrate the use of the preceding deflection formula, consider a small elcvator frame 15" wide and 30" long, made of standard 3" channel, Figure 21. Find the

Resistance of Frame and Various Sections

Deflection of Frame Under Torsional Lood

A =

pzzq

Torsional Resistance of Common Sections

1 +

nT]:

d

L

R=

2 t t, ( b - t)2(d - t p

bt

+ d t,-

t2 - t12

1

2.10-14

/

Load

Stress Analysis Wall load

3%" X 13" box sectlo

8" X 8" box sectcon

FIGURE 22

vertical deflection of the unsupported comer when under a load of 5 lbs. Using the appropriate formula from Table 4, torsional resistance of the U channel cross-section 1s -

R = --

2 ht,"

dtS3 3

2 -

2 (1.875) (.3125)3 3

htr3

+

+ dtw3 3

3( 1875)3 3

centerlines of the longitudinal members is 34.75", and the latter are 82" long. Determine: a ) The approximate vertical deflection of the unsupported comer, b ) the shear stress in lougitudinal and transverse members, and c ) the size of the connecting weld between the longitudinal and transverse members. torsional resistance of longitudinal membe~s

Substituting actual values into formula #4:

FIGURE 23

The actual deflection when tested wasA = .030"

The struchlral frame of Figure 22, simply supported at three comers, is designed to support a 17-kip load at its unsupported comer. Here the width between

2 b2 d? -b + -d ti, td ) ~ - ~ ( 3 (113/4)= - (3) (11~47 ( 1 (%) = 137.5 ia4

RL =

--

+

erigning for Torsional Loading

long side of its cross-section is

tolswnal resistance of transoerse member (only one in this example)

/

2.1

-

= 3820 psi shear stress in transverse member In a similar manner it is found that the applied torque on the transverse member is -

FIGURE 24

See formula development, p. 2.10-12

oertical deflection of frame L. .- - 1 A = P -W - "s

[+di

+

( 34%)(82) - -(-17,000) --

(12 X 10")

= .35"

Since the cross-section of the transverse member is a hollow rectangle of uniform thiclaess, the shear stress at mid-length along either side of the section 1s -

yT]

i

1

( 2 ) (137.5) ( 343h)

+

11) (298.3) (82)

I

(438,500) Z(7.5 X 9.5)(%) = 6160 psi --

size of connecting fillet weld

shear stress in longitudinal member The applied torque on only one longitudinal member is -

Treating the weld as a line -

TL = A Ea nL R1' See formula development, p.2.10-12 W L (35) (12 X 10" . (l)(l37.5) ..- . (31%)(82) = 202,500 in.-lbs, each member The shear stress at midpoint of the longitudinal member, on the short side of its cross-section is FIGURE 25

= 2300 psi and the shear stress at midpoint of the member, on the

ming for Torsional Loading

Resolving combined forces on weld at point of greatest effect -

,.

, ,

/

-17

L~ngthwisememoers and cross members are subject \, to twisting action of the \ s h e a r h y stresses

Transverse member

/-\

FIGURE 28

diagonal

\ , brace)

There i5 no twisting action on 45'diagonal member since s h e a r components cancel out Only dm gonal tensibn comprass/on a r e formed, which place member in bending> I / member is very r i g i d . V

Since 11,200 lbs is the accepted allowable load per linear inch of fillet weld having a 1" leg size, the minimum leg size for this application is 0

3560 =11,200

-

(E70-weld allowable)

The two main stresses on a member under torsional loading are ( 1 ) transverse shear stresses and ( 2 ) longitudinal shear stresses. These two stresses combine to produce diagonal tensile arid compressive strcsses which are maximum at 45" At 45', the transverse and longitudinal shear stresses cancel each other. Therefore, there is no twisting stress or action on a diagonal member placed at 45" to the frame. In a frame made up of flat members, the transverse shear stresses cause the longitudinal members to twist. The iongitudinal shear stresses cause the cross braces and end members to twist. On a diagonal mcmber at 45" to axis of twist, the transverse and Iougitudinal shear stress components are opposite in direction to each other and cancel out, but in line with this member they combine to produce diagonal tensile and compressive stresses wlueh tend

FIGURE 29

to cause bending rather than twisting. See Figure 29. Since these two shear stresses cancel out, there is no tendency for a diagonal member placed in this direction to twist. The diagonal tcnsih: and compressive stresses try to cause this diagonal member to bend; but being very resistant to bending, the diagonal member greatly stiffens the entire frame against twisting.

Stiffening the Braces

Previous experience in designing longitudirral side mcinhcrs for bending is now used to design these diagonal n~embers. It is important that the diagonal members have a high moment of inertia to provide suAicient stiffness so there will bo no f:~ilurrfrom local buckling, under srvcre torsional loads. Since the diagonal brace is not subjected to any twisting action, it is not necessary to use a closed box section. For short diagonal braces, use a simple flat bar. The top and/or hottom panel of the frame will stiffen this to some extcnt (Fig. 30). As the nilsupported length of the diagoilal brace becomes longer, it may becomc necessary to add a flange (Fig. 31). This is

/

Load & Stress Analysis

done by flanging one edge of the brace or using an angle kar or T section. The flange of the brace may also be stiffcncd to keep it from buckling. For opcn frames with no Aat panel, it is better to use a channel or I beam section having two flanges (Fig. 32).

FIGURE 31

elative f ffectiveness of Tests were made on scale models of typical machine frames to illustrate increase in resistance to twist as a result of the diagonal bracing.

FIGURE 32

FIGURE 33

esigning for Torisonal Lcadin

FIGURE 34

The top frame in Figure 33 has conventional cross bracing at 90' to side members. It twisted 9". The above frame is little better in resistance to twist than a flat sheet of the same thiclmess, as shown in the middle. The plain sheet twisted 10". The bottom frame has diagonal braces at 45" with side members. It twisted only 'A0. I t is 36 times as resistant to twisting as the first frame, yet uses 6% less bracing material.

A = . (' F, Y3 48 E I

% = =A - - ' La - - 12FEY3I L, --

T Since T = F L, then F = L

.'.B i

j

1

(See Figure 34) An approximate indication of the angular twist of a frame using double diagonal bracing (in the form of ;m X ) may be made by the following procedure. Here each brace is treated as a beam.

T Y:j

= 6 -E I- L 2

Since Y =

\

fl L

B = T(\/?;)'L' 6EIL2 -

f

i

3 E I

~

~

T L ~ T - L T L Hence Ea R 3 E I -E,R 3 E I = 5.3 1 'and R r= fi E8 For fixed ends, R = 21.2 I For the usual frame, the following is suggested: also 8. =

\

(simply supported)

--

which appeared in Table 1. Therefore: For a double diagonal brace use R = 10.6 1 and substitute this value into the standard

-

esigning for Torsional ~ o a d i n g

/

FIGURE 38

NECTIONS

Case 2 (Diagonal bracing) since this is "doublez3bracing, the ~ for this type of frame is used -

~ 1 formula b l

R = 10.6 I First find the moment of inertia for the cross-section of a brace, which is a simple rectangle, assuming the brace also is %" X 10":

~ When a member having an open section is twisted, the cross-section warps (see b, in Fig. 37) if ends of the mcmber are free. The flanges of these members not only twist, hut they also swing outward (see c), allowing the member to twist more. If the ends of the flanges can be locked in place in relation to each other, this swinging will be prevented.

I = -b d3

CONNECTIONS

12

where b = the section width (plate thickness), and d = the section depth

then substituting into the formula for R -

The angular twist on the frame is then-

= .0000152 radians or .00087"

OF TORSI

There are several methods of locking the flanges together. The simplest is to weld the end of the member to the supporting member as in ( d ) . If the supporting member is then neither thick enough nor rigid enough, a thin, squiue plate may he welded to the two Banges at the end of the member ( e ) . Another method is to use diagonal braces between the two flanges at the two ends of thc member ( f ) . Either of these methods reduces the angular twist by about %. Members having a box section, when butt welded directly to n primary member, have the fully rigid end connections required for high torsional resistance.

I

Problem 10

/

A 12" WF 27-lb beam, 25' long, with a uniFormly distributed load of 8 kips, is supported at each end by a box girder. See Figure 38. If the beam is continuously welded to these girders, estimate a ) the resulting end

2.10-22

/

Load & Stress Analysis

FIGURE 39

moments in the beam, b ) the torsional stresses in the girder, and c ) the weld size required to hold the box girder together.

8. = 0

torsional resistance of box girder

R =

2 b2 d2 b d

(See Figure 39)

,+t,

Me - L, - -W 12

- (8") (25'12X

2(13.33)2(10%)2 - (13.33) (10%) -+pBq (%) = 910 in.4 -

12")2

-. 200 in.-kips torque on box girder See Sect. 8.2 Torsional Member Formulas.

Torque in the central section of the box girder support is equal to the end moment of the supporting beam. end moment of beam See Sect. 8.1 Beam Formulas.

Determine what torque must be applied to the central section of the supporting box girder to cause it to rotate the same amount as the end rotation of the supported beam, if simply supported (0, = ,0049 radians) : If the beam is simply supported without any end restraint, the end moment (Me) is zero, and the slope of the beam at the end is -

= ,0049 radians Now, if the ends of the beam are so restrained that it cannot rotate, the end moment becomes

-

8% =

T L, r~

A moment-rotation chart shows the relationship; see Figure 40. A straight line represents the end moment ( M e ) and end rotation (8,) of the supported beam

esigning for Torsional Loading

/

2.18-23

FIGURE 40

under all conditions of end restraint. A similar straight line, but in the opposite direction, represents the applied torque ( T ) and angular rotation ( 8 ) at the central section of the supporting box girder. These two lines arc plotted, and where they intersect is the resulting end moment (Me) or torque ( T ) and the angular rotation ( 8 ) :

Me = T = 190 in,-kips 0,

torsional shear force on fillet weld

f i = rb tb = (1830)(%) = 690 lbs/lin in. which must be transferred by the ellet weld joining the top and bottom plates to the side channels, lo make up the box girder.

= ,0002 radians l~orizontulshear force on fillet weld due to bending

torsional shear stresses in box girder

FIGURE 42 d = I 03/a"

FIGURE 41

= 1830 psi

1

Half of the $-kip load goes to each end of the beam, or a Ckip load is applied to the central section of each box girder. And V = 2 kips.

/

2.10-24

Load & Sfress Analysis

(4.875) (594,) - (F) (468) ( 2 welds)

= 54 ibs/lin in. FIGURE 43

total shear force on weld

f = f,

+ f,

+

= (690) (54) = 744 lbs/lin in.

torsional resistance of suppoiting beam

required leg size of fillet weld (E70 weldsj 0

= --actual force

allowable force

= ,066" (continuous)

torque on suppoi-ting beam

However, AWS and ASSC would require a minimum fillet weld leg size of 3/1," (See Section 7.4). If intetmittent fillet welds are to be used, the length and spacing of the welds would be%

Detelmine what torque must be applied to the central section of this supporting beam for it to rotate the same amount as the end rotation of the supported beam, if simply supported (0, = ,0049 radians):

= calculated leg size of continuous weld

actual leg size of intermittent weld used

= 35% or use

3"

- 8"

Alternate Design

As a matter of interest, consider the support to be provided by a 1 0 WF 39-lb beam. (See Figure

43)

The moment-rotation diagram, Figure 44, shows the resulting end moment on the supported beam to be 4.67 in.-kips. Thus, this beam could be connected as a

FIGURE 44

Rototion (81, radians

esigning for Torsional Loading

simply snpported beam with just vertical welds on the web si~fficicntto carry the 4-kip shcar reaction. Thc end restraint is ahout 2.3%.

Mcmhrane analogy is a very :isefnl method to mderstand the behavior of open st,ctions mhrn subjected to torsion. To make nsc of this method; holes are cut into a thin plate making the outline of varions shaped sections. it m e m b r n e material si~chas soap film is spread owe tbc open surface and air prcssure is applied to the film. The mathematical expressions for the slope and volrnnc of this membranr or film cowring the openings rt:presenting diffr:rimt cross-sections are tho samt: as the expressions for the shcar stressas and torsional resistance of the actual member being studied. Tt is from this t p e of analysis that formulas for various types of open sections subjected to torsion have been developed and confirnred. If several outlin<,s are cut into the thir plate and the same pressure applied to each membrane, the following will b e tnie:

/

2.10-25

1. The volumes under tht: membranes will be proportional to the torsional resistances of the corresponding srctions. 2. The slope of the membrane's surface at any imint is -propor-tional to the shear stress of the section ;it this point. 3. A narrow section (thin plate) has practically the same torsional resistance rcgardltss of the shape of tht: scction it is formed into. Notice a, h, and c in Figure 45. For a given area of section, the volume under the membrane rcmains the same regardless of the sIi;ipr of the section. It is possihlt? to dctcrminc the torsional resistance oE these opcrr st:ctions by comparing them witli a standard circle on this same icst plate whose torsio~ialresistance can readily he calculated. fly comparing thc memhrarrc of the slottcd open tube, ( c ) in Figure 15, to that of the mt,mhrane of the ( c ) , it is I-cadily seen why the closed tnhe closed t~~brx is several hundred times morr. resistant to tu-ist, when it is renrembcred that the v o l ~ ~ munder e the membrane is proportional to the torsiol~alresistance.

FIGURE 45

2.10-26

/

Load and Stress Analysis

Modern structural steel shops ore equipped with highly efficient equipment for the welding of fabricated plate girders. Here an automatic submerged-arc welder runs o transverse splice in 7/8" web plote to full width, with the oid of a small runout tab previously tacked in place.

This automatic submerged-arc welder mounted on o track-mounted, gantry type monipulotor runs o web-to-flange fillet weld the full 84' girder length. Welding generators travel with the monipulotor.

Structural members are often subject to combined loading, such as axial tension and transverse bending. These external forces induce internal stresses as forces of resistance. Even without combined loading, there may be combined stress at points within the member. The analysis of combined stresses is based on the concept of a cubic unit taken at any point of intersection of three planes perpendicular to each other. The total forces in play against these planes result in proportionate forces of the same nature acting against faces of the cube, tending to hold it in equilibrium. Since any member is made up of a multitude of such cubes, the analysis of stresses at a critical point is the key to analysis of the member's resistance to combined extemal forces.

.6:

ESS

Biaxial and triaxial stresses are tensile and compressive stresses combined together. Combined stresses are tensile and compressive stresses combined together. Principal pZanes are planes of no shear stress.

Prirzipal stresses are normal strcsses (tensile or compressive) acting on these principal planes. These are the greatest and smallest of all the normal stresses in the clement. Normal stresses, either tensile or compressive, act normal or at right angles to their reference planes. Shear stresses act parallel to their reference planes.

I Normal stress

FIGURE 1

These stresses may be represented graphically on Mohr's circle of stress. By locating the points (cr,, 7.1) and (u,, 7.1) on a graph, Figure 2, and drawing a circle through these two points, the other stresses at various planes may be determined. By observation of Mohr's circle of stress, it is found that-

Stress in Member

FIGURE 2

Sheor stress

Mohr's Circle of Stress

.I14

/

Load

Stress Analysis

FIGURE 4

FIGURE 5

In this case, US and anare principal stress% ad, and u a , since they act on planes of zero shear stress. For any angle of rotation on Mohr's circle of stress, the corresponding planes on which these stresses a d in the member rotate through just half this angle and in the same direction. 180" from usin Notice in Figure 3, U, lies at Mohr's circle of stress, and the plane ( b ) on which 0-2 acts in the member lies at 90" from the plane ( a ) on which u, acts. 90" from u, and Notice in Figure 4, T,,, lies at the plane ( b ) on which T,,,,, acts in the member lies at 45" from the plane ( a ) on which usacts. In this case US and u3are principal stresses because there is no applied shear on these planes. This is a simple method to graphically show how stresses within a member combine; see Figure 5. On the 7,) graph, right, locate the two stress points (+ US,

and (+ rrz, - T I ) and draw a circle through these points. Now determine maximum normal and shear stresses. By observation of Mohr's circle of stress, it is found that-

+

+

+

+

+

The above formula for the maximum shear stress ( T ~ ~isXtrue ) for the flat plane considered; however, there are really two other planes not yet considered and their maximum shear stross could possibly be greater than this value. This is a very common mistake among engineers. To be absolutely sure, when dealing with biaxial

Analysis of Combined Stresses

/

FIGURE 6

stresses, always let the third normal stress he zero instead of ignorulg it, and treat the problem as a triaxial stress problem. The example in Figurc 2 will now be reworked, Figure 6, and the third normal stress ( u l ) will he set equal to zero. Here, u3 = -k 12,000 psi u2 =

+ 8,000 psi

T~

r2

-

Circle 1 Tmax

=

2

8,000-0 - 2

= 4,000 psi 0 It is seen that, in this example, the maximum shear stress is 6,000 psi, and not the 2,000 psi vali~c that would usually be found from the conventional formulas for biaxial stress. 3. TRIAXIAL STRESS COM (See Figure 7) STRESS The three principal stresses (ex,,, u?,, r a p ) are given by the t h e e roots (u,,) of this cubic equation: Vy3

u 3

-

UZ

= 2,000 psi Circle 2

- (Vi$. VZ t G - S ) ~ :

+(u~+ V ~ G-aVa -1-

- (ulG-9?+ 2

VlUi

i,iii:i

- TS' - ~3~
4 ; '

- m,.;,"

~ 7=2 0)

.( 4 )

For maximum shear stress, w e the two principal stresses (cr,,)whose algebraic diffrrmce is the grcatest. The maximum shear stress (r,,,,,) is equal to half of this diflerence. *Since a, b, and c are coefficients of this equation: a =-(u

x + u 2 + ~ )

b = Flu*

+

c = ulr,* f

= 6,000 psi

t"

- ci Tmax -= 0-2

=0

On graph, right: Locate stress points (mi) ( u a ) , and draw three circles through these points, Now determine the three maximum shear stresses. There are three values for the maximum shear stress, each equal to half of the difference betweell two principal (normal) stresses. The plane of maximum shear stress (shaded in the following sketches) is always at 45' to the planes of principal stress. (US)

Circle 3

cgr3 ~

2

+ - 7? - 7 2 - 72 + u8r3*- uiu2u3- 2 UlUj ~

2

71727:~

'Solution of Cubic Equation from "Practical Solution of Cubic Equations': G. L. Sullivan, MACHINE DESIGN, Feb. 21, 1957.

2.11-4

/

Load & Stress Analysis

T

+ 0.2

3

The ambiguous sign is opposite to the sign of Q (approximate, but very accurate). For either Case 1 or Case 2 The additional two roots (u2,, u3,,)of the general cubic equation are calculated by solving for u, using the exact quadratic: 0-2-t(a+ul,)up--=

C

0

TIP

FIGURE 7

Determine the maximum normal and shear stress in this web section, Figure 8:

Then calculateN3 K = - as a test ratio.

0 3

Q"

Case 1

+

When ( 1 K) is positive (one real root) or when ( 1 f K) is zero (three real roots, two of which are equal) calculatb

FIGURE 8

where: 0-1

=0

TI

US

= - 13,650 psi

T~

- 14,500 psi

TZ

u8 =

and compute the root-

= 11,000 psi =0 =0

Substituting these values into the general cubic equation:

+

-

uD3 ( - 13,650 - 14,500)uD2

Case 2

+

When ( 1 K) is negative (three real and uneaual roots) calcnlate-

[( u,"

- 13,650) ( - 14,500) - (11,000)2]o;, = 0

28,150 a, f 76,925,000 = 0

A

T=q=x and compute the root-

the tbree principal normal stresses are-

=0 uz, = - 25,075 psi Ul,

u3, =

- 3,075 psi

Analysis o# Combined Stresses

0,=

-

/

= - 13,650 psi and 71 = 11,000 psi

14,500 psi ond ri = 11,000 psi

a,

FIGURE 9

b 4 z p =and taking one-half of the greatest difference of two principal stresses: rmax

25,075 - 0 2

=

= 12,535 psi

I

1-

25,075 psi (rnox)

Problem 2

1

For the beam-to-girder network represented by Figure 10, assume the combination of stresses represented by Figure 11.

These various values are shown diagramed on Mohr's Circle of Stress, Figure 9. Checking Effect of Applied Stresses

The Huber-Mises formula is convenient for checking the effect of applied stresses on the yielding of the plate. If a certain combination of normal stresses (UX and u,) and shear stress (r,,) results in a critical stress (uc,)equal to the yield strength ( u ) of the steel when tested in uniaxial tension, this combination of stresses is assumed to just produce yielding in the steel.

FIGURE 11

Here:

4'

-

FIGURE 10

/

2.11-6

Load & Stress Analysis

The apparent factor of yielding is

This seems reasonable and under these conditions, the beam flange could be groove welded directly to the edge of the girder flange without trying to isolate the two intersecting flanges.

actual testing of members under various combinedload conditions, and from this a simple formula is derived to express this relationship. If points a and b are the ratios produced by the actual loads, point c represents the combination of these conditions, and the margin of safety is indicated by how close point c lies to the interaction curve. A suitable factor of safety is then applied to these values. Combined Bending ond Torsion

, Pure bending

ENGTH UNDER CQM !NED LOADING A very convenient method of treating combined loadings is the interaction method. Here each type of load is expressed as a ratio of the actual load (P,M,T) to the ultimate load (P,,M,,T,) which would cause failure if acting alone. axial load

bending load

torsional load Pure torsion

In the general example shown in Figure 12, the effect of two types of loads (x) and ( y ) upon each other is illustrated.

FIGURE 13

Combined Axial Looding and Torsion interaction curve

R, = constant R, = variable

0

.2

.4

.6

.8

1.0

R.

FIGURE 12

The value of R, = 1 at the upper end of the vertical axis is the ultimate value for this type of load on the member. The value R, = 1 at thc extreme right end of the horizontal axis is the ultimate value for this type of load on the member. These values are determined by experiment; or when this data is not available, suitable calculations may be made to estimate them. The interaction curve is usually determined by

FIGURE 14

Analysis of Combined Stresses Combined Axial Compression rrnd Bending

in this case, the axial compression will cause additional deflection, which in turn increases the moment of the bending load. This increase can easily be taken care of by an amplification factor (k). See Figures 15 and

/

The bending moment applied to the member (chosen at the cross-section where it is maximum) is then multiplied by this amplification factor (k), and this value is then nsed as the applied moment ( M ) in the ratio:

For sinusoidal initial bending moment curve

FIGURE 15

FIGURE 17

For constont bending moment

P

P

FIGURE 16

bending

Here: The chart in Figure 18 is used to determine the amplification factor ( k ) for the bending moment

FIG. 18

Amplification

factor (k) for bending moment on beam also subject to axial compression.

k

1

2

.3

.4

?/PC,

5

7

2.11-8

/

Load

Stress Analysis

-

Top panel

Transverse load w = 185 lbs/in

width b = 56" thickness t = $6"

FIGURE 19

applied to a beam when it is also subject to axial compression. The resulting combined stress is found from the following formula:

Obtaining the amplification factor ( k ) for the sinusoidal bending moment from the curve, Figure 18A loading platform is made of a %" top plate and a 10-gage bottom shect. The whole structure is in the form of a truss, Figure 19.

The actual applied moment due to extra deflection is found to be--

Determinotion ot comb and bending) in top co

With L = 16%" A = 21 in."

The resulting combined stress formula being-

I = ,247 in.4 First the critical load-

of which there are two components: ( a ) the compressive stress above the neutral axis of the top panel being-

= 272,000 lbs

Oc

126,000

=21 + =:

Then the ratio-

x6)

11,600( .247

14,800 psi

( b ) and the tensile stress below the neutral axis of the top panel being-

The bending moment-

= 2,800 psi

Determindion 08 Focfor of

The ultimate load values for this member in compression alone and in bending alone are unknotm, so the following are used. For compression alone

-

Elastic

L

*Since - = 150 (where r = radius of gyration) r assume P, = PC, = 272,000 lbs For bending alone-The plastic or ultimate bending moment is--

Plastic

FIGURE 20

These ultimate values are represented on the following interaction curve, Figure 21. Plotting the present load values at a against the curve, indicates there is about a 2:1 factor of safety before the top compression panel will buckle.

T h i s Ljr ratio of 150 is high enough so we can assume the ultimate load carrying capacity of the column (Pa) is about equal to the critical value (P..). If this had been an extremely short column ( w r y low Ljr ratio), the critical value (Pa.) could be quite a bit higher than the actual ultimate value (Pa).

Mu = 64,900 in-lbs

"M"-Applied bending moment, x 1000 in-lbs

FIG. 21 interaction Curve for Problem 3

.11-10

/

Load & Stress Analysis

The Air Force Academy Dining Hall (seating the entire student body) at Colorado Springs was b u i l t on the ground and jacked into position atop columns. The complexity of joints, the heavy cantilevered construction and large lateral forces offered unique problems in combined stresses. W e l d i n g was the only practical approach to the complex connections required to join members of this three-dimensional truss sysiem.

1. CAUSES 01: BUCKLING

Buckling of flat plates may be experienced whon the plate is excessively stressed in compression along opposite edges, or in shear unifo~mly distributed around all edges of the platc, or a combination of both. This uecessitates cstablishirrent of values for tile critical buckling stress in co~nprcssior~ (u,,) and in shear (r,,.).

The critical comprcssive stress of a plate when subject to compression ( r e ,can ) be found from the following:

represented by the portion of the curve C to D in Figure 2. If the rrsr~ltirig value ( u )is above the proportional limit (u,,).indicated by the portion of the curve A to C:, hr~cklingis s:dd to he ine1;rstic. Here, the tangent modulirs (I?,) n ~ u s tbe used in some form to replacc Young's or secant modulus ( E ) in the fomxola for detcrminiug u,,,.. This problem can he simplified by limiting the maximum value of the critical buckling stress ( u c r ) to the yield strength ( u ? ) . However, the value of the critical bncklirrg stress (u,,) may 1)c calculated if required. Above the proportional limit (o,), the ratio E = ~ J isEno longer constaut, hut varies, depending upon

LE 1-Compression

Load on Ware

/ d u e s for Plate Factor

S u p p w i (long ploter)

(k

to be Ured in Farmulo

Critical Strsrr on Plate to cause Bucklins (o',,)

r'm =

$07

FIGURE 1

where:

E = modulus of elasticity in compression (Steel = 30,000,000 psi) t = thickuess of plate, inches b = width of plntv, inches a = length of platc, inches v = Poisson's ratio (for steel, usually -- 0.3) k = constant; depends upon plate shape b/a and support of sides. See Tables 1 and 3. If the resulting critical stress ( u ) from this formula is below the proportional limit (u,,), buckling is said to he clnstic and is confined to a portion of the plate away from the supported side; this does not mean complete collapse of the plate at this stress. This is

Bleich, "Buckling Strength of Metal Structures," p. 330

2.12-2

/

Load

Stress Analysis

the type of steel (represented by its stress-strain diagram) and the actual stress under consideration (position on the stress-strain diagram). See Figure 3. Above the proportional limit (u,,), the modulus of elasticity ( E ) must he multiplied by a factor (A) to give the tangent modulus (E,). The tangent modulus (Ei) is still the slope of the stress-strain diagram and Et = U / E , but it varies. If it is assumed that the plate is "isotropic" (i.e., having the samc properties in both directions x and y ) , the critical buckling lorrnula hecomes-

I

where:

A=+

If it is assumed that the plate has "anisotropic" behavior (i.e. not having the same properties in both directions x and y), the tangent modulus ( E t ) would he used for strases in the x direction when the critical stress (u,,) is above the proportional limit ( u n ) .However, the modulus of elasticity ( E ) would he used in the y direction because any stress in this direction would be bek~wthe proportional limit (up). In this case, the above formula #2 would he conservative and

the following would give better results:

For steel, this becomes-

If the critical buckling stress (u,,) is less than the proportional limit (up)then A = Ei/E = 1 and formola #4 could he used directly in solving for critical stress (u~,). However, if the critical huckling stress ( u ) is greater than the proportional limit (u,), then A < 1 and formula $4 cannot be used directly. It would be better to divide through by and express the formula as-

?'x

From the value of re,/\/T;;formula #6 will give the value of re,.Obtain proper value for the plate factor ( k ) from Table 1 or 3.

Curve for A7 Steel n, = 33,000 psi

oe = 25,000 psi inelastic

= 2.70

10

20

30

40

50

(

$1

60

FIG. 2 Buckling $tress curve for plater in compression.

Buckling of Plates

/

2.1

FIG. 3 Stress-strain diagram showing where tangent modulus need be applied to determine critical stress.

Determining Tangent Modulus Factor

fX)

Then, multiply through by

Bleich in "Buckling Strength of Metal Structures", p. 54, gives the following expression for this factor ( X = E,/E):

where:

rr, = yield point u, =3 propostional limit u,, = critical buckling stress

TABLE 2-Shear

Load on Plate

If we use a ratio of-

the expression hecornes-

I--- "

-4

I -Sleich,

I ',Buckling Strength of Metal Stiucturer." p. 395

2.12-4

/

Load & Stress Analysis

where:

See Figure 2 for curves representing these formulas applied to the critical buckling compressive stress of plates of A7 steel (u, = 33,000 psi).

E = modulus of elasticity in compression (Steel

= 30,000,000 psi) t = thickness of plate, inches

UCKLlNG OF PLATES UNDER SWEAR

b = width of plate, inches

The critical buckling shearing stress (T,,) of a plate when subject to shear forces ( T t ) may be expressed by the formula in Figure 4 (similar to that used for the critical buckling stress for plates in edge compression).

a = length of plate, inches ( a is always the larger of the plate's dimensions)

v = Poisson's ratio (for steel, usually = 0.3) k = mnst,int, depends upon plate shape b/a an? edge restraint, and also accounts for the moduluz. of elasticity in shear (E.). See Tables 2 and 3. It is usual practice to assume the edges simply supported. Shear yield strength of steel ( T ) is usually consid1 ered as - of the tensile yield strength (o,), or .58 uy Since

"

ular Plates Supported On 4 Sides Between Stiffeners mnd e Knee Between Stiffeners) Volver for Plate Factor (k) to be Used in Formulw 3, 4, 5, ond 6

Load CImnrrrslan

when 1

a,

k = 4

when a

;z 1

.

k

= 7.7

k

=

when

2

1nir

i

7.7

+ 33 (1 - 0i9

Crilicol Stress 7'(.

and

S',,

Buckling of Plates

2

v2)(i)

As before in the buckling of plates by compres-

TABLE 3-Critical

Stress #or Rectangular Continued

-

lpltes Supported On

-

Volusr for Plate Factor (k) to be Used in Formulas 3, 4, 5, and 6 when $21

where

l +'a

4

il = 3

when

%5.5

l

'I3

k =( where

n =

[- + q7]

+ )i 4 a' + 5.34 !az + 1)'

I

a

when

where

'i

5.34

=

+ 41a' 77

when

% s = s l

n2~fl+3

k = 3.85

where

q

=

+' -6 I'a- -

2 -

9

when

fiCa(1 k

where

=

1

n =6

2

4

2 + -~r 9 ' a

2.12-5

siort, in the irrelastic rangc the critical stress ( u , , ) exceeds the proportional limit (u,,), and the tangent modulus (E,) is introduced by the factor ( h = Et/E). Therefore, folmulas #5 and #6 would be used also in the buckling of plates by shear. Proper values for the plate factor ( k ) are obtained from Table 2, for purc shear load, and Table 3, for shear load comhined with compression.

Since the plate constant ( k ) can be adjusted to ' 3 factor, this becomescontain the 1 k + E ccr= 12(L -

/

n

m

J

I

2.12-6

/

Load & Stress Analysis TABLE &-Buckling Stress Formulas 1Compressionl

4. SUMMARY FOR DETERMINING CRITICAL BUCKLING STRESS OF PLATE

1. The value of the plate factor ( k ) to be used in formula #5 comes from Tables 1 , 2 or 3, adapted from "Buckling Strength of Metal Strl~ctures", Bleich, pp 330, 395, 410. 2. Solve for u,,/ L I Tfrom formula #5.

Portion 01 Curve

,

' a. If u,,/\ I-h = uD,this is the value of ucx,SO go to step 4. b. If u , , . / \ T > u,,, go to step 3. 3. Insert this value (u,,/ \ / x ) into formula #6, and solve for the critical buckling stress (uc,). 4. After the critical stress ( ) has been determined, the critical buckling stress of the given plate (u',, or r',,) is determined from the relationship shown in the right-hand column of Tables 1, 2, or 3.

5. BUCKLING STRESS CURVES (Compression) In regard to plates subjected only to compression or only to shear, H. M. Priest and J. Gilligan in their "Design A4anual for High Strength Steels" show the cwrve 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 criticd stress values in the elastic buckling region by 25% to more nearly conform to actual test r&ults. 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-Cilligan curve for comparison.

/

\&

.I - .".I 3820 \,% to

BtoC

to

Crilicol Buckling Compressive Stress (nF,) Determined by

o,, = 1.8

5720

-x

-

c

/

Foclar --''I

where:

"

=

0 " .

-

\/-."

n

bit --

v'k

-7-

4770

-

-

\/G

0,

6. =

The horizontal line ( A to B ) is the limit of the yield strength ( u , ) . Here uc, is assumed equal to u,. The curve from B to C is expressed by-

ucr= 1.8 uy - n

(b/t)

' I

1

where: n = -. 4770

The curve from C to D is 75% of the critical bockling stress formula, Figure I, or: k r 2 E U", = .75 12 ( 1 -.2,(t)'

[%I

4434 - -

All of this is expressed in terms of the factor

FIG. 5

Buckling stress curves for plates in edge compression.

Ratio C r i t i c o l b u c k l i n g compressive

b/t s-r

stress [n,,) for A-7 steel having 0,

= 33,000 psi

.I24 /

Lood & Stress Analysis

TABLE 5-Factors

-

of b i t are recognized, Tablc 7 , extended to higher yield strengths, lists these limiting values of b/t.

for Eucklina Formulas

of Steel

4770 3820 -.

5720 -

\ 6

--- \/<

7. EFFECTIVE lDTH OF PLATES I COMPRESSION

The 20" X %" plate shown in Figure 7, simply supported along both side<, is ~ubjectcdto a compressive load

Simply supported sides

/ A-7 steel o,

b

= 33,000 psi

= 20"

= %" k = 4.0 t

--Limiiins Yield Strength

Side Conditions One simply supported; the other free

Both simply supported

Values of

33,000

13 & 16

12

12

1 1 & 13 .-. 44

-

-

40

40

36

34

32

.-

.--

50,000 33,000

. 50,000

--

FIGURE 7

AiSC-American institute of Steel Construction AASHO-American Ariociotion of State Highway Ofiicials AREA-Amercon Railway Engineers Arrociotion

Factors needed for the formulas of curves in Figure 5, for steels of vario~isyield strengths, are given in Table 5. Figu-c 6 is just an enlargement of Figure 5, with additional steels having yield strengths from 33,000 psi to 100,000 psi. For any given ratio of plate width to thickness ( b / t ) , the critical buckling stress ( u ) can be read directly from the curves of this figure.

A suitable lactor of safety must be used with these values of b/t since they reprcscnt ultimate stress valnes for buckling. Some structural specifications limit the ratio b i t to a maximum value (point B ) at which the critical buclding stress ( u ) is equal to the yield strength (u,). 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

Under these conditions, the critical buckling compressive stress (u,,) as found from the curve ( a , = 33.000 psi) in Figure 6 isu,, = 12,280 psi

LE 7-Usual Yield Strength or psi

I

Limiting Valuer of b/t

One Edge Simply Supported; the Other Edae Free

I

Both Edger Simply Supported

This value may also he found fro~nthe fonnulas in Tahle 4. is 40.0 and thus exceeds

Since the ratio

the value of 31.5 for point C, the ioilnwing formnla must be used-

= 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 be-

Since k = 4.0 (both sides simply supported), the ratio-

Since the plate thickness t = %" width, b = 42.0 t or h = 10.5". This is the rffcctivc width of the plate which may be stressed to the yield point (o;)before ultimate collapse of the tmtirc plate. The total comprcssive load at this state of loading would be as shown in Figure 9. The total comprcssive load here would be-

Another method makes no aIlowai~cefor 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 be-

FIGURE 8

The ovcr-all plate shonld not ~vllapsesince the portion of the plate along tbe supported sides could still be loaded np to the yield point (cr,) before ultimate collapse. This portion of the plate, called the "effectivti width" can be dete~minedby finding the ratio h/t when (u,,)is set equal to yield strength (u,)or point B. From Figure 6 we find-

or from Table 4 we find-

FIGURE 9

2.12-10

/

Load

Stress Analysis

"

Critical buckling rheor stress

= 33,000 psi

for A.7 i t e e l hovng a,

FIG. 10 Buckling stress curves for flate plates in shear.

. BUCKLING STRESS CURVES (Shear)

uckline 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.

Portion Of Curve

I

Fccaor

Critical Bvckling Sheor Strerg (T,,) Determined by

--

Vi

(3).

% , w e is expressed in t e r n

%e

Table 8. Comparison of Figure 10 and Table 8 with Figure 5 and Table 4 reveals the parallelism of critical buckling stress for compression u ) and for shear (~cz).

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 fmmulas of curves in Figure 11 are given in Table 9. For any value of

(yi) .-

the critical buckling shear

~,

stress ( r e r ) 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

(3) .

to the value at

~~,

point B, .r,, = T? and it will not be necessary to compute the critical shear stress (r,,). Assuming the edges are simply supported, the value of k = 5.34 4(b/a)" Then using just the three values of b/a as 1 ( a square panel), "I (the length twice the w-idth of panel) and zero (or infinite length), the required b/t 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 &that the critical buckling stress in shear is given directly as (T,,). In T a b l a 2 and 3 it is given &st as ( u ) and then changed to (T,,).

+

TABLE 9-Factors Yiald Strength of Steel ar7 pri

for Buckling Formulas

blt Corresponding g, ,,,,in+ 6 - far point \i k Shearing Yield Vk 3820 5720 Strength - -- -r, = 3 8 o, psi

%

-

V 77

V ~7

c

(Shear) =

3 ' % 4770

Buckling of Plates

/

2.12-11

TABLE 10-Maximum Values 06 b/t To Avoid Formulas Maximum Value3 of b / t to Hold r,, to (Panels with limply rvpported edges) Tensile Yield Strength PI psi

7,

b/o = 1 ($quore panel

Foul edges - rimply supported

k = 5 34 Four edges

10

20

30 Rotto

40

-

+

4[b/oj2

fixed

k = 8 98

+ 5.60(b/oj2

50

60

&i

FIG. 11 Buckling stress curves (plates in shear) for various steels.

10

.12-12

/

Load and Stress Anolysis

United Airlines hangar a t 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) i s completely shop welded. Large plate girders like this one are stiffened to prevent web buckling due to edge compression. Contilevered welded plate girders weigh 125 tons.

1. COMPRESSIVE STRESS

Comprcssivo loarli~lg of a mcrnb(ar when a p p l i e d (axially) touctintric with thc ccnter of gravity of the member's cross-s(,ction, results in compressive stresses distribi~ttduniformly across tlir srctior~.This comprt+ sive unit stress is -

A short column (slendc,rnisss ratio L/r g u a l to aborit unity or less) tlmt is over1o;idrd in comprt.ssion may fail hy crusliir~g.From a desigri standpoint; short omp press ion nirrnhcrs pxsont little problt:rn. It is important to hold tlw compressive unit strcss within the material's colnpressirc strength. For stccl, the \-ield and nltimate strengths are considered to bc tlrc same in compression as in tension. Any liolcs or opcni~igsin the section in the path of force tmnsl;ition will u.cakm t l ~ crnemlxr, rriiless sucli openings arc cuinp1atci)- filled iiy wiothcr member that \vilI wrry its sllarc- of the load. Excessive comprc.ssirm of long columns may cause failure by buckling. As cornpressiw lo:iding of a long colmnn is increased, it r i w ~ t u d l ycalms some ecc1.ntricity. This in turn sets np ;I bending monwnt, causing the column to deficct or bucklc sliglltiy. Tliis deflection incre;isrs thc ecc~mtricityand this thc h i d i n g moment. Tliis may pnigrcss to whwe t11c bending moment is incre:ising at a rate greatel- than the ina-case in load, and ilie ct~luiirnsoou fails by buckling.

2. SLENDERNESS RATIO As t l ~ cme~nberbecomcs longer or nmre slender, there is rnorc of a trndenty for dtirnntc failure to be caused by brickling. The most nintmon usay to indicate this t t d e n c y is the slenden~essratio which is equal to1, r where L = i ~ n s u ~ p o r t elength d of mcniher r = tile least radius of gyration of the section

and-

I f tlw rrrmnbcr is made longcr, wing the same cross-s<.ctirin ;iud tlw sanrc conrprtxive I d , the res u l t i ~ ~cori~pn:ssivr g strr.ss u'iil rt?maili the same, dtho~iglitlic tmdrmcy for buckling will increase. The ilcwd~~nwss ratio im.re;lsc\s as the radius of hyration US thi. section is n d u c i d or :is the length of the memhcr is incrrwwd. 'nie allowable compressive load which may h~ applied to tbr member deue:~ses as the slendimiess ratio inweasi:~. The various columr~ formulas (Tablcs 3 and 4 ) givr tlic allowable :werage cornprmsive stress ( 5 )for the culomn. Tlwy do not give the actual unit%ess devr~lopdin t l v column 11y tlir luad. Tlie unit stress resulting trorn tltiw forniu1;is may he multiplied by the cross-sectioniil arc:\ of tlir column to give the alliiwddr load \ ~ l ~ i ma); c l ~ be supported. 3. RADIUS OF GYRATION

Tlie riidius ol gyration ( r ) is thr' distance from the rreutral axis of a section to an imaginaiy point at which the w l d c awn of the section wrild be concentrated arrd still llavi, the same amonnt of inertia. I t is found hp the erpressimi: r = f l l / ~ : in tli? dosigir of ulrsy~n~nctrical sections to be used as mlumns, tht. le:rst r;tdius of gyration (r,,%,,) of the section must h t kriowrr in ordcr to make nse of the siendt~rnrssratio (l./r) in tlrc coliimn fo~mulas. If the sc.cti~ir~ in question is not a standard rolled srr.tion tlrr priipc~tiesof which are listed in steel handkioks, it will hi: uectwtry to m n p u t e this least radius of gyration. Sincr the bast radius of gyration is -

the minimrim li~orneiitof inertia of the section must fir dvtermilned. M i n i m u m Moment of Inertia

T h ( ~m;isiinllin moment of iiicrtin (I,,;;,,) ;ind t l ~ emininxim monierit of inertia (I,,,,) of a cross-scction are

3.1-2

/

Column-Reloted Design

and, applying fnrinula # I from Section 2.3, the distance nf neutral axis x-x from its parallel axis XI-XIis -

V

XM =ZA =

NA,., FIGURE 1

to locate neutral axis y-y:

Y

A

found on principal axes, 90" to each other.

W x I " ..r NA,.,

I

Problem I

-. -

..

Total

Knowing I,, I,, and I,, it will he possible to find I,;,.

21.0 12.0 = - 1.75"

-

M ; s

~

6.0 12.0 -

~

XM = -7,A

0

1

,-

.

+12.09.0

- +

+ 9.0 -

75"

product of inertin It will he nwessary to find the product of inertia ( I ) of the scction. This is the area ( A ) times the product of distances d, and d, as shown in Figure 3.

I

Locate the (neutral) x-x and y-y axes of the offset T section shown in Figure 2:

(Set: Figure 3 on facing page). In finding the moment of inertia of an area about a given axis (I, or I,), 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, and d, hecnnse [lie product of these two could be either positive or negative and this will determine the sign of the resiilting product of inertia. The total product of inertia of the d i o l e scction, which is the sum of the values of the individual areas, will depend upon these signs. Areas in diagonally opposite quadrants will have prodncts of ineltia 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 entirc larger section is -

FIGURE 2 FIGURE 4

to locate neutral axis x-x:

where: a and b = dimensions of rectangle ( = A) d and c = distance of area's center of gravity to the x-x and y-y axes (= d, and d,)

where d = distance from center of gravity of element area to parallel axis (here: XI-XI)

The product of ir~crtiaof a T or angle section is (See Figure 5 ) .

-

Analysis of Compression

l x = A d:

I

Moment of inertlo obout x - x axis

l i t Quadrant

A d : -

3rd Quadrant 1

XY

3.1-3

iXy : A dx dy Product of i n e ~ t i u about x-x and y ~ yu x c i

Moment of inertia about y y 0x1s

2nd Quodront i" Y = -A dx dy

lxy = + A dx dy

/

= iAdxdy

dih Quadrant iX"

= -A dx d

FIGURE 3 Y

Xow use formula given previously for product of inertia of such 21 section:

x

Ixy = --

r

+

a d t (d-&)(a -t) 4 (a d)

+

( 4-).( 5 ) ( % ) ( 5- 2.5)(4 ..~ 4 (4 5) 3.125 in.'.. -~~

+

+

+ %-~)

Y

FIGURE 5

Here, determine sign by mspection.

1

I 25" Y

Determine the product of inertia of this offset 'r section about the x-x and y-y axes:

I,, = ZA ( & ) ( d l ) = 2.5 ( - + 1) ( - 1 - ,555) =

=

t

1.388

+ 1.737

+ 3.125 in."

+ 2 (-

1.25) (-- ,695)

FIGURE 6

I

/

3.1-4

Column-Related Design

L)ett.rn~iiiethe minimnm radius of gyration of the ~ previously (Fig. 2 ) and reoffset T s e c t i o ~shown peated licrc:

minimum radius of gyration

FIGURE 7

As a matter of interest, this r,,,!,,is about axis x'-x', the angle ( 0 ) of which is-

tan20 =

-.

moment of inertia about axis x-x

-

,, x '6 Total

;')

NA,.,

=

'--.:-~--+ -

! 6 - 3 . 5 - I . I + 73.5t18.00 pzlimp I= I , $ ~ 2 . 0

ZM - -- .21.0 . - - 1.75'' and A - 12.0

.

-

(See sketch below).

Iy

20 = -- 46.4" or 0 ~and 0 = 66.8"

+ 133.6"

-+-

Any ultimate buckling could be expected to occur ahout this axis (x'-x').

1 minimum moment of inertio

2 I,,

-sx -

Problem

4

1

Thc clian~icl section, Figtire 8, is to be used as a collinrn. Determine its radins of gyration about its X-x axx. ['sing the conventional formulas for the properties of the section -

Analysis of Compression

FIGURE 8

/

3.1-5

FIGURE 9

Mean dimensions b and d are used, Figure 9.

area of the section

A = bd - bldl = ( 6 ) ( 4 )

-

(5.5)(3.75) rx =

= 3.375 i n 2

-2-

b

+ 2d

distance of neutral axis

The exact value obtained from this formula for r is 1.279". The value obtained by using the conventional formula is 1.281". Assuming a possible error of 1 : one part in 1000 for every operation of the slide rule, it would be possible to get an answer as high as 1.283" and as low as 1.275". This represents an error of about Y4 of the error using the conventior~alfonnulas with slide rule. The time for this last calculation was 2 minutes. oment of Inertia About Any Axis Y

radius of gyration

FIGURE 10

-

If a slide rule had been used, assuming a possible error of one part in 1(K)O for every operation, this ms\var co111dbe as high as 1.336" and as low as 1.197". This represents an error of 4.3% and - 6.6%. For this reason it is necessary, when using these conventional formulas, to make use of logarithms or else do the n:ork longhand. To do this rcquires about 30 min~ites. The radios of gyration \ d l now he found directly, using thc properties of thin sections, treating them as a line. Sce Table 2. Section 2.2.

t

X

Sometimes (as in Problcm 3 ) the moment of inertia of a sedion is nedehl about an axis lying at an angle ( 0 ) with the cor~ventionalx-x axis. This may be found by using the prodt~etof inertia ( 1 of the section about the conventional axes (x-x and p-y) \?,ith the moments of i~wrtizi ( I , ) and ( I , ) about these same axes in the following formnla:

3.1-6

/

Column-Related Design

/ pinned

FIGURE 11

4. CRITICAL COMPRESSIVE STRESS

stress by dividing by the cross-sectional area of the column. Since A = I/r2, this hecomes -

The critical load on a column as given by the Eulerformula is -

where L, = eiivctive length of column. This can be changed into terms of average critical

Bccause this formula gives excessively high values for short columns, Engesscr modified it by substituting the tangent modulus (13,) in place of the usual Young's modulus of elasticity ( E ) . The modified formula then becomes -

where:

Et

-

tangent modulus of elasticity, corresponding to the modulus of elasticity when stressed to ffw

r =: least radius of gyration of the cross-section L, == effective length of the column, cwrresponding to the length of a pinned column that would have the same critical load. See Figure 11. The Ihgesser fonnula is also called the Tangent hlodnlus formula and chccks well with expcrimeutal values. 5. TANGENT MODULUS

s,roin

, r I. :n,lnxid3

FIGURE 12

Use of the Tangent Modulns formula necessitates a stress-strain curve (preferably in compression) of the materid. See Figure 12, stress-strain cnrve for a quenched and tcmpered steel in compression. IVhereas the usual Young's modulus of elasticity represents a fixed value for stccl (30 X 10') according to the ratio

Analysis o(. Compression

/

Slenderness Ratios: Quenched & Tempered Steel

TABLE 1

TABLE 2 ingcsrer

of curve

lioeloriir bending:

of stress lo t r a i n lwlow the propoi-tiol~allimit, the tangent moditlus of t ~ l x t i d ytiikrs into corisidrrntion the cliirnciirrq eifwt of p1;tstic strain h ~ y o n dthis point correspo~~tlinq to the actual s t r ~ ~ ilrr.ol\-cd. s Notice; in Figure 1% tlw hrokrw lirws rt:pres~:nting the slopc For various v a l i r ~of ~ tangmi modulus of elasticity ( & ) , iu this case from 1 X 10" psi up to 30 X 10". Tiit: c:omp~-iwiv~: strrss le\wl ( r r , . ) at which a given E, mluc applies is di:tcrrnine
i t

"<$ ( 0

,*

FIGURE 13

iCOOW."<.

."'O

I.,;,

FIGURE 14

3.1-7

/

3.1-8

Column-Related Design

and the critical slerrdcrncss ratio (I.,,/r) is determined for ~ w i m t s~ d t r c sof strcss ( c r , ) , restilting in Tables I and 2 for qucnchcd and teinpered steel only. Table 1 givcs rorrwporlding \;altles of slendemcss ratio (I&) for given v;ilucs of strcss (u,,) above the proportio~i:il limit of ;I quimched and tempcred steel. firlow tlrc m;itrrinl's propor-tiod limit, the use of Yot~ug's modt~lris (I.:) or tangrnt modiilos ( E , ) provide the sanrr vdite. Tablc 2 for qtienclied and tempcred stecl givcs ihc slerr~lernt*ssratio (L,/r) for stress levtals (cr,) \viihin the prip~-tied portion of the stress-strain cur\,tr. Si~iccthe o t i i i Eitler fornrula for cr,, iipplics here, this portioir of the crirvc is often called tho Eolcr curve

stress (cr) ~

d t 1 ;ipplyiug ~

.~

a factor of safety of 1.8.

7. BASIC FORMULAS FOR COMPRESSION MEMBERS In "lirirklirrg Strmgtli oi h?ctal Stnicttlrcs," page 53, 131r~icIrintnxliir~sa pit-:rbolic formtila to csprcss this tangent rnodiiltrs i:i~rvt. for comprcssiorr, i3y applying a factor of s;tfrty (F.S.). this Iitwmcs thc allowable cot~ipri~ssiv~~ strws, l ' l r t ~ hasir paralwlk: formula thus rnuilificd is -

6. PLOTTING ALLBWA LE STRESS CURVE These val~tesfrom Tabltx I and 2 arc now plotted to i'orm the cilrvc in Figurr 11.The Eulcr portion of the curve is cxtmded upward hy a hrokcn line to indicate the variance that would 11c o b t x i n d by continuing to use the Euler formula beyond tlie proportional limit. This must be kept in mirid in designing compn%ion members having a low slenderness ratio (L/r). A few test results are also sliown to indicate the close relationship hetwem thr Modulus formula and actual valiies. Note that a correporiding wrvf has been plotted below the main citrve, r r p x w i ~ t i i ~ the g allowable

TABLE 3-Allowable

Compressive Stress (AISC)

Ronga of

Average Allowable compra*,ive

LI -Values

unit

o-,

z

proportiond limit

u

-:::

yirld point

:

factor of s;~f(t!-

F.S.

Any rcsidiml coniprcssive strcss (ir,,) in the member tends to lo\ver tlie 171-oportional limit ( c r ) os straight-liw prirtio~~ of thc stri,ss-stmirr ciirvc, in rompuessioir: \vitlioirt 2iff1,rting tl~t. yicld p i n t For the purposr of tlic ;thov<-fonnula, it is assumed that

Also assriming this value of residual co~npressive strcss is ahorit half of the yield point, or cr,, = '12 cr,, Formula 1113 becomes:

st,e,s (C) -

-

4

.ii"

F.S.

,

.. ., .

This fonniil;~Im)viilcs a paralxrlic curve, starting at a slrndrr~iessr;rtio of' ( r = = 0 ) with V R ~ I I C S at yield stress ( r r , ) , ;urd mtc~iiclingdown to one-half of this strrss wli<~i-i. it hwonics taiig(wt \villi the I1uler curw ;it the i q > l j ( litnit ~ of (~lastic11n1diny. The slcrtdtwlcss ratic ;it tllis point is:

where:

I-,. ..

2:3.925 for stecl

. . . ( 15)

l1)ovc~tliis slc~tdcrrlt~s ratio, the 1:ulcr is I I ~ :

fonnula

r

to that of t h foctor of iilfe?y !F.S., is ieniion ( i s . =. 1.67). For ior8gei ~ o l u m n r , the iofrty of foctoi ormenier giadunlly to o m o x m u v of F.S. = 1.92.

For very rtiort :aluinns. mmbcrs n

K

F.S.

--

effective length factor

-.

\,'cry

Analysis of Compression

/

3.1-

MAXIMUM WIDTH-TO-THICKNESS RATIOS

For Elements of Members Under Axsol Compieriicn or Compreiiior Due to Bendzng Adopted from 1961 AISC, Sec 1 9 . 1 o r d I 9.2

with ~ e p o ~ o t o r

FIGURE 15

The above rotioi of b ' t may be exceeded i f , b y u i n g n the coliuiationr a w d t h equol to the maxmum of these limits, the cornpierrive itreis value obtamcd ti wtthln the oliowobe sties

8. AlSC FORMULAS FOR COMPRESSION EMBERS The AISC I r a i~rcorpoi-ati~d (1'363) tliesc h s i c column Corincil for~nitlas rlrdorsd hy the C:olumn R ~ w : i r c l ~ Report in its spc.eifientimis for structrrr-al buildings. The slcnderrlcss ratio w111:r~the liulcr and parabolic portions of the citrvc intrl-scct, Formula 15, lias been dcsigiiated in tlre AISC Specification as ( C c ) . This is also i~iwrl>oriitrdinto F o r r n ~ ~ l13. a AISC itses n \ d ~ i cof ti =: 298,0O0,000psi (instcad of the r~sual30,000,000 psi) for tlre itrodrilus of elasr t i of the curve, ticity of t e c l . For the I Fornn~la16, AISC uses a factor of safety of 1.92. Tlre rcs~iltitig ncw 41S(: wlitnin forrr~ulas arc sho\v~rin Tol~lc.3. l';il~lcs6 tlrro~~gliL1 give the AISC comprcwion ailo~vablesfor several strengths of structural stccl.

For v;ir-iow conditioils of colri~nn cross-section, Figure 15: there is a limiting ratio of element width to thickricxs ( b / t ) . This ratio is rqressed as being C Y ~ I I ~toI or 11,s~than ( - ) a rwt~iinw l r l i . divicl~dby t1r.e sqlt:ir<, root of the, ii?;itcrinl's yicld strmgtl~.The r r l a t d 'l'ablc 4 pcrlirifs (lirrbct reading of ;I cornprcssion ~~lmnmt's b/t ratio for v:iriorrs yield strcngtl~so i strel. At times it may he desiral,lo to exceed the limiting 11-t I-atio of ;in clrwirwt. 'This air1 11e done if, in the calc~ll;rtiotis,substitriting t l ~ esirort:,r ~nasimurnwidth allowed (by t l r Fig. 15 limits) wo~ildgive n coi~iprcssivc ~ m i tstrtw valiic within the ;illo\r.iblr stress. To 111'111 i n visii:ilizilrg rt,l:rtivc s:rvings in iirctal liy the ttsv of lrighrr-strrngtli steels, I'igui-c 16 indicates tlrc :illowable comprcssivc s t r c ~ i g t l( ~I T ) o1)tained from ttic Tahlc 3 formttl:rs for 8 difFertm< yield strengths. Notirc tlmt tlw adv:rrrtagc of the higlrer strengths drops oil 3s the coltmn becomes marc slender.

I

3.1-10

/

Column-Related Design

TABLE 4-Limiting

b / t Ratios of Section Elements Under Compression

Limits of Ratio of Width to Thickness of Compression Elements for Different Yield Strenrrths of Steel

8,000

44.0

va, 10,000 --

55.0

b'a,

4 2 .

1

i

I I)I 1 1

39.0

37.3 . ~.~ .~

- -

52.6

48.7

~

47.1

46.6

~

34.1

32.6

31.4

p~

~

44.7

42.6

40.8

~I 1

39.2

26.6

33.4

/

25.9

32.4

25.3

/

31.6

Round off to the neoreit whale nurnbcr. * Quenched and tempered iteelr: yield strength at 0.2% aifret.

Allowable Compressive Stress ( g ) based on I963 AISC Sec 1.5.13

whfch

20

40

60

80

100

120

Slenderness iotio [ L l r ]

FIGURE 16

is

for steel of

36,000pis

160

180

200

Analysis o f Compression

If the allowable stress curve of quenched and tcrnpwed steel (Fig. 14) werc now sr~pe~imposed on this graph, thc e w n greater, strength advantnge of quenched and tenrpcred stcel at lower slcndcrness ratios would be rradily apparent. The allowable compressive unit stress ( u ) for a given sicndcrncss ratio (KL/r), from unityihrough 200; is q n i ~ k l yread from Tahlcs 6 thnjngli 11 for stwls of various yield strengths. .%hove KL/r of 130, the higlier-strength stcels offer no advantage as to allowable con~prcssivestress (-u ) . Above this point, nse Table 7 for the. rnorc economical steel of 36,000 psi yield strength. TABLE 5-AASHO

A-7 and A373

I

MEMBERS

'Tal)l(. 5 givt,s t h A A S t f O fonnoias, which are applic:iblc to bridge design. As a matter of gcnrr;~linterest, the colnmn formula r:sta'tjlishrd for use of qur:rrchrd and temperod steel on the Carqninez Strait Sridge (California) is -

Rigid Ends ond C o n c e n t r i c h o d s

j/4" m d under ii,

= 50,000 psi

Steel skeleton for 10-story Buffalo, New York apartment building features unique shop-welded construction. Principal erection element is a "bent" consisting of a 50' floor girder or "needle beam" threaded through the web of column section near each end and welded. Girder is supported mainly by on angle bracket or "saddle" previously welded to the column web. Girders cantilever out as much as 13' from column.

3.1-1 1

9. OTHER F O R M U L A S FOR COMPRESSlON

A l f o w a b l e Stress (.or Compression

Having

/

0,

= 46,000 mi

I

over 1%" to 4" a, = 42,000 ori

3.1-12

/

Column-Related Design

TABLE 8--82,000

psi yield steel

LE 9-45,800

psi yield steel

TABLE 10-46,000

psi yield steel

3.1-14

/

Column-Related Design

1. INTRODUCTION

The preceding Section 3.1 covers the general Analysis of Compression, along with an evaluation of the methods for determining stress aliowables. This present section deals more specifically with the aciual design of colmnns and other omp press ion members. For purposes of illustration, t h e term "column" is uscd quite liberally. This is due partly to much of the material having been originally developed expressly for columns. However, the information is generally applicable to all compression members.

These values are determined for the column or coli~mnsin qr~cstion(IJL,.), as well as for any beam or other restraining member lying in the piane in which buckling of the column is being considered (IJL,). The moments of inertia ( I , and I,) are taken about an axis perpendicular to the plane of buckling being considered. The values of G for each end ( A and B) of the column are determined:

2. RESTRAINT A N MEMBER

Section 3.1 explained how a compression member's slendei~icssratio (L/r) relates to its buckling strength. The degree of end restraint on a member results in its having an effective length wvl~ichmay vary considerably from its actual unbraced longth. This ratio ( K ) of effective length to actual unbraced length is wed as a multiplier in determining the dfedive length (L,) of a compression member.

TABLE 'f-Effective

Length (L.

of

Compression Members

Buckled shape o f member i shown by doshed line

where:

L = actual length of the column L, = effective length of the column to b e used in column formulas K = effective length factor Table 1 lists theoretical values of K and the Column Research Council's corresporrding recommended values of K for the effective length (L,) of columns under ideal conditions. Where End Conditions Can't Be Classified In actual practice it will be more difficult to classify the end conditions. If classification is doubtful, the Column Research Council recommends the following method based on the relative stiffness of connecting beams and columns. The stiffness factor of any member is given as I/L, its moment of inertia divided by its length.

Theoreticol K value

--

Recommended design volue when ideal cond:tionr o m ooaroximoird

,totion fixed

translation fixed

,ration free

translotian fixed

,tation fixed

translotion free

3totion free

tronrlation free

End condition

K' moy be greater than 2.0 **Top end ossvmed truly rotation iree From "Guide to Design Criteiio for Metol Cornpierrion Members" 1960, p. 28, Column Rereorch Council

3.2-2 /

Column-Related Design

094 0 boo d

999999 9

m (3

m (3

1

1

1

1

1

I

1

Ill I I I i I

oqo q q d o & n cu 8 a'&, l l l t l l ,

1

I

I

I

I

0mcof:wIq

d-00000 0 I

I l l I I

I l l

9 -

N

M

-

$Q'D" c u

9

9

0GXOICW 0 d-

I

-

N

d

0 I

I l l 1 1 1

I

?i I

0

I

I

0

0

I

I

Design of Compression Members

where:

I, = the total for the columns meeting at -

'' L"

Ig

the joint considered.

-

-

the total for the beams or restraining members meeting at the joint considered.

For a column end that is supported, but not fixed, the moment of inertia of the support is zero, and the resulting value of G for this end of the column would be z.However in practice, unless thc footing were designed as a frictionless pin, this value of G would be taken as 10. If the column end is fixed, the moment of inertia of the support is c c , and the resulting value of G for this end of the column would be zero. However in practice, there is some movement and G may he taken as 1.0. If the beam or restraining member is either pinned ( G = o: ) or fixed against rotation ( G = 0) at its far end, further refinements may be made by multiplying the stiIfness ( I / L ) of the beam by the following factors:

I

Problem I

sidesway permitted far end of beam pinned = 0.5 For any given column, knowing the values (GA and G,) for each end, the nomograph, Figure 1, may be used to determine the value of K so that the effective length ( L , ) of the column may be found: L,=KL This nomograph is taken from the Column Research Council's "Guide to Design Criteria for Metal Compression Members", 1960, p. 31. The nomograph was developed by Jackson & Moreland Division of United Engineers and Constructors, Inc.

3. STRENGTH OF A very convenient method of treating combined loadings is the interaction method. (Also see Sect. 2.11, Analysis of Combined Stresses.) Here each type of

Find the effective ltmgth factor ( K ) for column A-B under the following conditions: Sidesway

FIGURE 3

FIGURE 2

Here:

= ,260 GB = o ~ use ; 10 From the nomograph read K = .76

3.2-3

sidesway prevented far end of beam pinned = 1.5 far end of beam fixed = 2.0

I

Sldesway prevented

/

Here:

= ,620 Gg = zero; use 1.0 From the nomograph, read K = 1.26

/

3.2-4

Column-Related Design

1 .o

Margin of hafety R, = constant

R, = vanoble

.2

0

6

.4

8

1 .O

R, FIGURE 4

FIGURE 5

load is expressed as a ratio of the actual load to the ultimatc load which would cause failurt. if acting alone.

vertical axis is the ultimate value for this type of load on the meniber when acting alone. The value of R, = 1 at the extnxme right end of the horizontal axis is the ultimate value for this type of load on the member when acting alone. These ultimate values are determined by experiment; or w h m this data is not available, suitable calculations may be made to estimate these values. The interaction curve is usually determined by actual testing of members undcr various combinedload conditions. From this, a simple formula is derived to fit the cnrve and express this rclationship. If points a and b are tlrc ratios produccd by the actual loads, point c represents the combination of these conditions. Thc margin of safety is indicated by how close point c lies to the irrteraction curve. A suitable factor of safety is then applicd to these values. Figure 5 illustrates this for axial compression and bending. IIowever, the applied bending moment ( M I ) c a w s the column to bend, and the resulting displacement or eccentricity induces a secondary moment from the applied axial force. See Figure 6. Assume that the moment ( M i ) applied to the column is s i ~ ~ a o i d in n l nati~re;Figure 7. A siniisoidal moment applied to a pinned end member rcsults in a sinllsnidal deflection curve, whose maximum deflection is equal to -

axial load

P R. = -Pa, bending load

M Rb = Mu

torsionul load T Rt = T" In the general example shown in Figure 4, the effect of two types of loads ( X and Y ) upon each other is illustrated. The value of R, = 1 at the upper end of the

-

M "?ox

Since the critical Euler load is Applied moment

Induced secondary moment

FIGURE 6

Resultant moximum moment

esign of Compression Members

/

3.2-5

this becomes

When the axial load ( P ) is also applied to this deflected column, a secondary moment is induced and this is also sinusoidal in nature, its maximum value being -

Applied sinvsoidol moment

FIGURE

FIGURE 8

Resulting deflection Curve

7

The interaction Formula #4 then becomes -

+

This slightly higher moment (M2 MI) will in the same manner produce a slightly greater deflection (A2 A l ) , etc. Each successive increment in deflection becomes smaller and smaller. The final values would be -

+

(ultimate load condition) Each ultimate load condition factor in the above formula is equal to the corresponding factor for working conditions multiplied by the factor of safety ( n ) ; or

<

= 1 and since M,.

= MI

+ P Ail,,,

then

where: subscript, is for working loads subscript A is for allowable loads Accommodating Increased Moment Due to Deflection

Notice:

This increase in the moment of the bending load caused by deflection is easily taken care of in the basic interaction formula by an amplification factor ( k ) : so:

?r2

E

ue = -

/&\"

3.2-6

/

Column-Related Design

Or, on a stress basis -

the Euler stress (u,) divided by the factor of safety ( n ) . The term (v',) is used here in place of AISC's (Ffe).

where: o; = computed axial stress

ub = computed compressive bending stress at point considered a. axial stress permitted if there is no - =: allowable bending moment; use largest (L/r) ratio, regardless of plane of bending u, = allowable compressive bending stress permitted if there is no axial force. (AISC Sec. 1.5.1.4)

AISC uses E = 29,000,000 psi and n = 1.92 in the above. Here: r, = radius of gyration about an axis normal to the plane of bending

The AISC Specification Sec. 1.6.1 uses the same amplification factor. They use the term (F',) which is TABLE 2-Euler

L, = actual unbraced length of column in the plane of bending

Stress Divided By Factor of Safely

..

7,410

7,300

7,200

7.100

7,010

6.910

6,820

6,730

6,460

6,380

6,300

6.220

6,140

6,060

5,980

5.910

5.620

5,550

5,360

5,290

5,230

4.930

5,490 ... - . 4,880

5.420

4,990

4,820

4,770

4,710

4,660

4,460

4,410

4,360

4,320

4,270

4,230

4.180

4,010

3,970

3.930

3,890

3,850

3.810

140

7,620

7,510

i 50

6,640

6.550

i 60

5,830

5,760

5.690

170

5,170

5.110

5,050

180

4,610

190

4,140

4,560 4.510 .- ....... 4,090 4,050

200

3,730

....

-.

1

is = octuol unbioced length of column in the plane of bending iB

= radius of gyration about the oxir of bending

......

-. -

-

3,770

---

Design of Compression Members

/

of $ for Several Load Conditions

TABLE 3-Value

According to AISC Sec. 1.5.6, this value (o',) may be increased 'h for wind loads. Table 2 lists the values of 5', (Eulcr stress divided KLb ratios from 20 to 200. by factor of safety) for y-

Core

'b

These values apply for all grades of steel, hut are based on the conservative factor of safety = 1.92. The derivation of the amplification factor has been based on a member with pinned ends and a sinusoidal moment applied to it. In actual practice these conditions will vary; however this factor will be reasonably good for most conditions. AISC Sec. 1.6.1 applies a second factor ( C , ) to adjust for more favorable conditions of applied end moments or transverse loads. applied end momeflts

applied tranmerse load7

where: MI and M2are end moments applied to the column.

MI 5 M2, and the ratio (MI/M2) is positive when the column is bent in a single curve and negative when bent in reverse curve. AISC 1963 Cornmentori

.. -

(see Table 3 for values J, and C,, for several load conditions)

(AISC Formula 6 )

Here: A = maximum ddection due to transverse load

When

L = actual length of member also used in deflection ( A ) calculation M = maximum moment between supports due to transverse load

the amplification factor must he used

AlSC Formulas For Checking

Formula #8 now becomes-

When

the influence of the amplification factor is generally small and may be neglected. Hence the following formula will control:

(AISC Formula 7a) This formula provides a check for column stability.

3.2-8

/

Columm-Related Design -

-7

In this exompie: A36 steel L i r = 80

AISC formula 7b

3 = 15,360

-

0, 0: 0,

1

= 22,000 = 23,300 = 36,000

FIGURE 9

, old AISC formulo Bending compressive stress (ab)

I t is an attempt to estimate thc total bending stress in the central portion of the column and to hold the axial compressiw stress down to a safe level. As L/r increases, this formnla will reduce the axial load carrying capacity of the column. This is because the Euler stress (o;,)decreases as L/r increases. As C , increases, caused by a less favorable condition of applird and moments or transverse forces, Formnlx #I1 will reduce the axial load carrying capacity of the column. The end of the member also must satisfy tho straight-line interaction formula:

1

( AISC Formula

7h)

I

In this formula, the allowable for compression (u,) - is for a column having a slendemcss ratio of L/r = 0, hence r, -- = .60 IT,. This formula provides a check for the limiting stress at the ends of the column, and as such applies

only at braced points. k Figure 9 is an example of the relationship of AISC Formulas 7a and 7h in the design of a specific member, nnder varions loading conditions. For bending moments applied about both axes of the column, these formulas become:

(AISC Formula 6 )

I I

(AISC Formula 7a)

(AISC Formula 7b)

esign of Compression Members

4. DESIGN OUTLINES

The design procedure is simplified by iollowing the appropriate outline in Tables 4, 5, or 6. Table 4 applies to compression mcinbers under combined loading (interaction problems). Table 5 applies to open-sectioned LE 4---Design

:ategory@ olumnr with computed mu:iioals ianimum a t the ends with no .onweire loading, and iderway i s pievented

.tegopy@ ornprersian members with iditionol tronrveire lood;; example a c o m p i ~ i s i v e iord of o truss with onsverse loading between ippoitf (panel pointr!.

Tionrvene l o o d i

v

No tionriotion of iointr

Sideway orevented

Siderwov permitted

).

= max deflection due to tianiveiie looding

= mox aoment between rupporir due to trans. loading ie K L i n computing or

-

ie

Check # i l and # 1 2 using o,

Check # I i

Ma

= -

using a,

S

Li, i n computing moments (M)

#12

Check #Ii

M? , -s

M,

a, =

"ring

S

Check # i 2

ou

I

(AISC Forrnulo 7oi

--

a,. oh ond .60 or moy be increoied

S e i iSd! (AISC Forrnulo 7b)

3.2-9

n~cinbersnnder coinpression in bending. Table 6 applies to hox members tinder compression in bending. Earh of these tl~ldrscategorize the mt:mber-load conditions \\.hi& innst be satisfied, ; u ~ dthen presents the nqoircd for~nulaswith which to determine the ailowablc cornpressive stress.

Outline for Compression Members Under Combined Loading flnteroction Problems)

category@ Coiumni i n framer with computed momcnli moximum ot the ends with no transverse loading, and d e r w o y is permitted. Were the latcrol stability 01 the iinme depends upan the bending r t i f f n e i i of its members.

/

$6 i o r wind

(AISC

-

M

2 S

3.2-10

/

Column-Related Design

TABLE 5-Design

Outline for Compression Members Under Compression In Bending

Members Which Are Symmetrical About An Axis i n Plane of Bending And Having Some Lateral Support of Compression Flange

Comprer%ionelement8 which are not "campoct" but meet the lollowing AISC Sec 1.9 ieqvirementi

l i i n addition. lateral sirppait of iamprerriui flange does not exceed: A7. A373. A36 steels 13 bc Other stronger steels

Having on axis of symmetry in the plane of its web: AISC 1.5.1.4.5

when

L

T5

-and compression elements meet the following AISC Sec i 5 . 1 4 . 1 "rampoct section" requirementi:

40 don't need AiSC Foirnuio 4

* This.mtio may be exceeded if the Lending stress, using

a

u d t h not cxcecding this limit, is within the allowable stress. t For "oompact" columns (AISC Scc. 1.5.1.4.1) which are s y n rnetrical about an axis in the plane of bending, with the above lateral support of its con~piessionRange and 0. = .15 a, use 90% of the moments applied to the ends of the column if caused by the gravity loads of the connecting beams. f For rolled sections, an upward variation of 3% may be tolerated.

In

Tables 5 and 6: L = unbraced length of the compression flange br = width of ~vmpressionflange d = depth of member treated as a beam r = radius of gyration of a Tee section comprising the compression flange plus % of the web area; about an axis in the plane of the web. For shapes symmetrical about their x axis of bending, substitution of r, of the entire section is conservative At = area of the compressiort flange MI is the smaller and Mp the larger bending mo-

ment at the ends of the nilbraced length, taken abont the strnng axis of the member, and where MI/& is the ratio of end moments. This ratio is positive when Mi and M2 have the same sign, and negative when they have different signs. When the bending moment within an unbraced length is larger than that at both ends of this length, the ratio shall be taken as unity.

(but not more than 2.3 can conswvatively be taken as 1.O)

Design of Compression Members

/

3.2-1 1

TABLE 6-Design Outline for Box Members Under Compression I n Bending embers Which Are Symmetrical About An Axis I n Plcrne o l Bendinq

No AlSC limit on laterol rupport of compresdon flange beiowe box section is torsionally rigid

And if lotero! support doer not exceed: A7, A373. A36 steels 13 bz Other rtronger steels 2400 h i

Compiession elements which ore not "campad" but meet the following AlSC Sec 1.9 requirements (1.5.1.4.31

b/t = 3000

*

%

B/t =

8000 * %

"'

20,000,000 A, d <,

And compoiiion eiernrntr meet the following AISC Sec 1.5.1.4.1 "compoct section" iequliernentr; b/t

I600 t 5 ---G

B/1

5

6000 .-

V T

d" 5 *( t~ - fl

1

- 1.43

5

-2)

but need not be lerr than Note: All notes from Table 5

V T --

-.

apply equolly to this gs

Toble 6.

= .60

-=

a,

a,

.66 a,

8000

-

t

TABLE 6 A

*

36,000

42,000

45.000

46,000

50,000

55.000

60,000

65,000

90,000

95,000

100.000

20.000

22,000

25,000

27,000

27,500

30,000

33,000

36,000

39,000

54,000

57,000

60,000

22,000

24,000

28,000

29,5001 30,500

33,000

36,500

39.500

62,700

66,000

yield strength of steel

Allowable

n

= ,605,

bending

e = .66 s,

ttres

fl 3000 -

9.5

6000 -

t o exceed:

~

13,300

\K 2400

Lotero! support

6

of compression flonge of "compact" -

20,000,000 Ar

sectionr not t o exceed:

a~ 4

---- -. C. =

1.18.2.3: mox. longitudino! rpocing between intermittent fillet welds ottoching compression flonge t o girderr

S

~

4Wo - t

E

~

43,O0Oi 59,400

. -

thickncrr rotio not

1

~-

1

1600 Width-to-

*

*

33.000

1

2

*Quenched & Tempered Steels: yield strength at 0.2% offset Round off to nearest whole number

3.2-12

/

Column-Related Design

5. BUILT-UP COMPRESSION MEMBERS

The basic requirements of welds on built-up compression members, as specified by AISC, are summarizkd by Figures 10, 11, 12, and 13.

Welding a t t h e ends of built-up compression members bearing on base plates or milled surfaces ( A I S C 1.18.2.2):

Weld odequote to transfer any calculated force

Continuous fillet weld at end of all elements in contact with each other (AISC 1.18.2.2)

FIGURE 10 Bearing or base plate or milled surfaces

Plate in contact with a shape (AISC 1.18.2.3):

Two rolled shapes in contact with each other (AISC 1.18.2.3):

n of

Compression

lates and Lacing

,,.'8.2.5) 0"d (1

182.61

Main comprrsion member built-up from plates or shapes and czrrying a calculated force:

P

FIGURE 14

The spacing of lacing must be such (AISC 1.18.2.6) that -

S of elemc-nt =

(if whoL member

rl

Single B!oiiriy

Wlwn the, su:i~~irrzbctuwri intcrn~i.ttr:nt melds

For sil~globracing:

Double C1:ocin~

FIGURE 16

For doubk bracing:

/

3.2-14

Column-Related Design

Design laciug bar for axial compressive force ( F ) :

Typical Built-Up Compression Members

Figure 18 slrows a number of examples of compression members built up from common shapes by means of welded construction. As indicated in lower views, perforated plates are often substituted for lacing bars for aesthetic effect. (AISC 1.18.2.6)

\vhere: n = number of bars carrying shear ( V ) Determine nllowablr compressive stress ( u a from - ) one of the following two formulas:

( A I S C Formula 1 )

I

Problem 2

/

To cheek the design of the following built-up section for the hoist of a boom. The 15' column is fabricated from A36 steel by welding four 4" x 3%" x 'h" angles together with lacing bars.

I

(Use Tables 6 through 14, Section 3.1)

u. =

ua from Form. #15 -

. . . . . . . . . . . . .(17)

On continuous cover plates with access holes ( A I S C 1.18.2.7):

" F o r double brace, use .70 L,

Use net section for cornpierrion

I

Design of Compression Members

FIGURE 18-Typical Built-up

Compression Members

moment of inertia of built-up sodion about axis 2-2

properties of each corner angle A = 3.5 in."

r, I, I, x

= .72" = 5.3 in.* = 3.8 in.* = 1.0" y = 1.25"

least radius of gymtion

moment of inertia of built-up section about axis 1-1 1, = 4(3.5)(5.75j2 4(5.3)

+

= 484

/

3.2-15

v

slendcrncss ratio

= 2% P = (.02) (278.Ck) = 5.57"2 bars)

The axial force on each bar isThen from Tahle 7 in Sect. 3.1, the allowable comprmsive stress is uc Z= 19,900 psi and the allowable compressive load is-P=u,A

-

= (19,900) (14) = 278.6 kips Check slenderness ratio of single 4" x 3%" x %" angle between bracing:

= 22.4

<

30.6

OK -

(AISC Sec. 1.18.2.6)

The unsupported length of the lacing bar between connecting welds is -

The least radius of gyration of the %" x l/z" bar is obtained tliusly A = 'I* in."

And the slenderness ratio of the lacing bars is -

= 56.3

< 140

OK single lacing -

(AISC Sec. 1.18.2.6) From Tahle 7 in Sect. 3.1, the allowable compressive stress on thc bas is uc = -

17,780 psi

The allowable compressive force on the bar is FIGURE 20

F-U
= (17,780) (.25) = 4.49 > 3.22"

OK

Design of Lacing

AISC specifies that lacing bars b e proportioned to resist a shearing forw normal to the axis of the member and equal to 2% of the total compressive force on the member (Sec. 1.18.2.6):

If each end of each bar is connected to thc angles by two 1%''long %<:/,,j" (ETO) Met welds, this will provide an allowable forco of -

F = 2 X I.% X 2100 lhs/in = 6.3k > 4.4Sk OK

---.

Design o$ Compression Members

I

Problem 3

/

3.

1

A multi-story building, having no interior columns, has a typical welded built-up cdumn with the section shown in Figure 21. A36 steel and E70 welds arc employed. The following three load conditions are recognized: Case C deod ood iive loodr with wind in x-x direction P = 2800 kips Mr = 250 ft-kips M, = i 200 ft-kips

Care A dead and iive loodr no wind

with in v-v ~ . . . wind . direction P = 2700 kips M, = 2200 ft-kips M, = 0 ~~

P = 2500 k i p

Mr = 250 ft-kips M, = 0

properties of the 14" W F 426# section A = 125.25 in.'

moment of inertia about x-x Let reference axis be a-a here L

Outride face of column

FIGURE 21 L

I

25625

Total

=:

74,507 in.'

=

+ 2.84"

+

(from a-a)

moment of inertia about y-y

I

727

---A Allowcrble

+ 76.570

Stresses

The various axial compressive stresses 'and bending stresses on the built-up cohlmn are checked according to Formulas #I1 and 12 (AISC Sec. 1.6.1, Formulas 6, 7a, and 7b). When wind loads arc included, the basic allowable stresses are increased by %. provided the resulting section is not less than that required for dead load, live loads, and any impact (AISC Sec. 1.5.6). Compression members are considered "compact" when syn~metricalabout an axis in the plane of bending, with lateral support of the column's compression flange not exceeding a distancc equal to 13 times its width (A36 steel) (AISC Sec. 1.5.1.4.1). For "compact" columns, the engineer can use just 90% of moments applied to ends of the column if caused by gravity loads on connecting beams (no wind loads) and ua 5 .15 u,, (AISC See 1.5.1.4.1). If the section is not "compact", AISC Formulas 4 and 5 must be used to determine the allowable com. pressive bending stress (ubr and -ub,). check for lateral support LC = maximum unbraced length of compression flange for "compact" section

/

3.2-18

Column-Related Design

1

I = B m 11"= 5720 L 13'

About strong axis [x-x)

13'

@ 4;

36" W 3 0 0 4 I -

L

- U 9 O . d = 405 8

/'

50'

1 - 74,507 1n4 = 5720LL 13' End V ~ e wof Bldg

-

'Not

@ fixed

4

FIGURE 22 (a)

Therefore it is a "compact" section and following can be used: rbr = -uby= .66 uy or 24,000 psi Euler stress (I+',,) and (u',,) About strong axis (x-x):

check for "compact" section Bange half, width to thickness ( a ) outer flange plate

From Table 2, read u',, = 133,750 psi. About weak axis (y-y) :

( b ) inner WF section

From Table 2, read d,, = 50,400 psi. ullowahle axial compressive stress

check web depth to web tlzicknes

dm 34" = 22.6 -Actual ,t 1%

8000 but need not be less than V T

but need not be less than 42.1 42.1

>

22.6

OK

Sidesway being permitted, from the nomograph (Fig. I):

I< = 3.65 and L,=KL = (3.65)(13' x 12") = 569"

Design ot Compression Members

/

FIGURE 22 (b)

CASE A Dead and Live Loads; KO Wind

Sidesway being permitted, from the nomograph (Fig. 1):

moment at

support

K = 2.1 and L.=KL

= 2.1 (13' x 12") = 328" FIGURE 23

This value of r, = 54.4 governs, and from Table 7 in Sect. 3.1 (A36 steel) cr* = 17,970 psi

applicd loark P 2500 kips M, = 250 ft-kips

--

M, = 0 Column Analysis

The following three analyses of the column (Cases A, B, and C ) are for columns with computed moments maximum at the ends with no transverse loading and with sideswny being permitted. This would be catezory A on Table 4. In this case (6, = .85) for both axes (x-x) and (y-y).

applied stresses

= 9760 psi

3.2-19

3.2-20

/

Coiumn-Related Design

- (250

x 1000 X

12)(23.50) ( 74,507 )

= 947 psi (max at 4" x 20" flange fk )

u,, = 0

= .15, .9M, can b e x e d (Sec 1.5.1.4.1): but u = ---sib0 in this case, u* 17,970 - '54 = 5 4 > .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

X 1.33

ubx= 24,000

X 1.33

allowable stresses

- = 17,970 psi

--

US

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 )

vex=

133,750

X 1.33

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)

= 24,000 psi

u',, = 133,750 psi

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

M, = 2200 ft-kips

applied stresses P - .-2700 lo0O 256.25 =K

My = 0

= 10,520 psi

FIGURE 24

checking against Fornizrlti $15 (;1lSC 7b) -. m,,

0.6

w,

h

+ F,,

--(10,520) -- .--

(22,000

x

--.I I

1.33)

.

.

Obi

i\4g o ~

li

I .()

V,,? ..

(8330) ~2-t.(nx)x~~ i . 3 3 )

,621

<

1.0 OK

14.500 1'"

Y

of W F section)

or 1.5.',00 pi (irr;ix :it outrr c r f ~ c of . 4" X

CASE C Dead and I i v c I,o;ds; Wind in S Direction /

; m i x ;it fi;nigc

We cailrlot I I W .Y hl, bec:iuw: wind loading is involved; h m < vf~ill\ ; h e of (M,) ;md (M,) must be used. allowable s t r t w r s o;, = 17,070 ;,< 1.33

= 24,000 uby= 24,000

>:

1.33

utCx z z 13X750 5

Wind in addition (SK 15.6) No wild in this direction

cr,,

= 50,400 >< 1 3

Wind in this direction (Sec 1.5.6) No wind in this direction Wind in this direction

checkins apiirlst I'oi.rnr11a #11 JAISC 7nj

FIGURE 25

applied loads P 2Y00 kips M, 2,70 ft-kips M, 1200 ft-kips

:

10,920 psi

2wR)

olumn-Related Design

4" X

20" R

J Torque box -V

FIGURE 26

Torsion on Built-Up Column

One item left to investigate in the built-np column is the twisting action applied to it. In Case C, the wind in the x-x direction causes a moment of M, = 1200 ft-kips because of the restraint of the spandrel beams. ( 1 ) One way to analyze this problem is to assume that this moment (M,) is resisted by the elements (the 14" W F section and the 4" X 20'' flange plate) of the built-up column in proportion to their moments of inertia about axis y-y. See Figure 26. Since:

The moment resisted by the 4" X 2 0 flange plate is-

= 346 ft-kips = 4,153,000 in.-lbs This moment is to be transferred as torque from the 13" W F section to the 4" X 20" plate through a

torque box, made by adding %"-thick plates to the built-up column in line with the beam connections. This torque box is checked for shear stress; Figure 27.

= 6600 psi OK ( 2 ) Another method of checking this twisting action is to consider the moment (M,) as applying torque to the built-up column. See Figure 28. This applied moment may be considered as two flange forces: in this case, 411 kips iu the upper and the lower flanges of the spandrel beam, but in opposite directions. Since these forces are not applied at the "shear center" of the column, a twisting action will be applied to the column abont its longitudinal axis within the region of the beam connection where these forces are applied; there is no twisting action along the length of the column in between these regions. Since an "open" section such as this built-up

Design of Compression Members

FIGURE 27

Shear oxis

I I

Torsue box

FIGURE 28

No twisting action

Twisting oction o section

/

3.2-23

3.2-24

/

Column-Related

P = 1000 lbs A

column offers very little torsional resistance, two plates will be added within this region to form a closed section about the shear axis to transfer this torque. See Figure 29. If this torque had to be transferred from one floor to the next, these plates would havc to be added the full length of the column. How-ever, this torque is only within the region of the connecting beams which apply these forces, hence plates are only added within this short distance.

this 1-kip force will be applied in the opposite direction. Treating this short section of the built-up column as a bcam, the shear forces due to this I-kip force will he analyzed on the basis of shear flow. In an open section it is not difficult to do this because there is always one or more starting points, the unit shear force a t the outer edges alwtys being zero. But in a closed section such as this, it is necessary to assume a certain value (usually zero) at some convenient point, in this case at the midpoint of the web of the W F section. The unit shear forces are then found, starting from this point and working all the way around the section using the general formula-

q2 = q, In our analysis of the column under Case C loading conditions, a transverse force of 1 kip was assumed to be applied in line with the web of the W F section of the built-up columu (this is the position of the spandrel beams). This cross-section is in the plane of the top flange of the spandrel beam. Just below this, in the plane of the lower flange of the spandrel beam,

+

V a y I

where: V = transverse force applied to srction (Ibs)

I = moment of inertia of built-up section about the axis normal to the applied force (h4) a = area of portion of sectiou considered ( i n 2 ) y = distance between center of gravity of this

erign 06 Compression Members

area and the neutral axis of the boilt-up section (in.) % = unit shcar force at the start of this area

This work is shown as Computation A. Relow, in Figure 30: the total shear force ( Q ) in the various areas of this section are found; thcse are indicated by arrows. This work is shown as Computation R. By Computation C, thesr shear forces are seen to producc an unbalanced moment oi 70.519 in-lhs, which if nnresisted will cause this section of the colurnn to twist.

3.2-25

In order to couiiterbalancc thiq moment, a negative moment of the same value is set up by a constant shear force flow of-

(lbs/in. ) q a = unit shear force at the end of this area (Ibs/m. )

/

q = -51.1

1bs per linear inch

When this is sr~perimposedupon the original shear flow, Figure 30, we obtain the final %ow shown in Figure 31. The resulting shear stress ( r ) is obtained by dividing the unit shear force ( q ) by the thickness of the section. Also the valucs must bc increased because the actoal forcc is 111 kips instcad of 1 kip, the work and resulting shcar stresses are shown as Computation D. Sce Figuw 37 also. These shear stresses seem reasonable.

FIGURE 30

FIGURE 31

olumn-Related Design Computation A = 0

1.q.

0

v L 1 ~ - o +

- I+0=.q.2

(1000)(7.83 X 1.8751(3.921 = 11,491.

+

5,01 =

5.01

3. q a = 0

0 Y + v=0 + I qr + q,' = 5.01 + 0

4. qe' = q s

=

5. q."

6 . q a = q,"

7. qir = qa

=

8. qp

" " -+I

39.49

17,24 =

17.24

17.24 =

+

22.25

110003(8.35 X 3.03W.83) 11.491.

+

= 22.25

+ Ta

+

i10001l28.64 X '/2)!9.095) 11,491.

= 22,25 = 39,49

+

+

17,24 =

39,49

= 0

0

9. qc' = q. 10. qr" = qt'

+ 7= 0 + V o Y

+ qt

l1000)1.905 X 4)(9.548) = 11,491.

+ 50.82 = 11000)(9.095 X 4114.5481= = 53.81 + 11,491.

+

2.99 =

2.99

= 2.99

+

= ".q

II. q.

IlOOO)i8.35 X 3.03%!7.83) = 11,491.

53.81 53,81

+

X 8.351 = 329.7

#

14,40 =

68,2,

Computation B

= (3 X

12. Om. 13. Qsa

= (% X

0

+ ]/5

X 5.01) 15.66 = 26.1

+

17.24 X 8.35)

14. Qas' = 39.49 X 1.265 = 50.0 15. Q a t

=

I: V

(36 ==

X 68.21

#

+ 39.49 2

#

39'49 + 50'82 X 28.64 = 1293.2 2

16. Q u = Check

22.25

+% X

#

53.81) 18.19 = 1153.4

#

0

Computation C Now, take moments about

@

The unboionred moment ir 70,519 in-ibs

Make

2 M, moment of

= 0 o constant shear force flaw, which must be added

- 70,519.

The resulting aheor force is

-

Where [A] = aieo enclored by renterline of web, flonges, and [A] = (15.66)(8.35) (18.19)(28.64) = 651.7 in'

+

This giver the true rhea? flow (Fig. 31).

to iarm o negative

Design of Compression Members

(@

/

3.2-29

1450 psi

FIGURE 34

FIGURE 32

Sharp reentrant corner

FIGURE 35 FIGURE 33 Reentrant Corners

(Figures

33 and 34)

The only other concern on this built-up construction is the sharp reentrant corner at points ( d ) and (f). Timoshenko in "Theory of Elasticity", p. 259, indicates thc following shear stress increase for a reentrant comer:

In structural steel. any stress concentration in this area probably would be relieved through plastic flow and could he ncglectcd nnlcss fatigne loading were a factor or there were sonic amount of triaxial stress along with impact loading. Of course if a fillet weld could be made on this inside corner, it would eliminate this problem. See Fignre 35. This is possible in this case, because these plates for the torque box ;ire not vcry long and the welding operator could reach in from each end to make this weld.

3.2-28 /

Column-Related Design ELDS FOR FABRICATED COLUMN

The melds that join the web of a built-up column to its inside WF seetion and its ontside flange plate, me subject to longitudinal shear forces resulting from the changing moment along the length of the column. As an example, continue with the conditions stated for the preceding Problem 3. The bending force in the flanges of the girder applied to the colunm is found by dividing this moment (M,) by the depth of the girder:

- .2200

ft-kip X 12" 35"

= 754 kips Thc point of contraflexure, or zero moment, is assumed at about midheight of the column. The horizontal force at this point, or bansverse shear in the column, may be found by dividing half of the moment applied to the column at the connection by about onehalf of the column height. This assumes half of applied

moment cnters upper colt~mnand half enters 1owcr column.

F,, = --M

'6 h

-

1100 ft-kip 6.5'

= 170 kips The moment and shear diagrams for the column when loaded with dead and live loads and wind in the y-y direction (Case H ) are given in Figure 38. This shear diagram indicates the transverse shear within the region of the beam connection is Vz I= 584 kips, and that in the remaining length of the column is V, = 170 kips. The size of the connecting weld shall be determined for the larger shear within the region of the beam connection, and for the lower shear value for the remaining length of the column. The minimum fillet weld size is aiso dependent on the maxi~num thickness of plate joined (AWS Building Article 212 a 1, and AISC Sec. 1.17.4).

Wind

i

T

J -*

Midheight

h

Shear diagram Moment diagram

FIGURE 36

This is also o picture of the amount and location of the connecting welds to hold column together

Design of Compression

embers

/

3.2-29

maximum thiclmess of plate here is 17/e", and the minimum size of fillet weld for this thickness is W' (AWS Bldg Art 212 and AISC: Sec. 1.17.4). IIcncc use

-

Weld

33"

@

in line with the beom connection

- ( , 5s l k )(SO) - (21.84)

-

(74,507) ( 2 wc4ds)

6860 lbs/in.

6860 leg size w = -----11.200

weld

@

for the remaining length of the column

V1 = 170L or 29% of Vz

where: A = 256.25 in.'

I, = 74,507 in.* The following allowable shear force for the fillet weld will be used: f = 11,200w (A36 steel and E70 weld metal)

We will not reduce the shear carrying capacity of the 61let weld due to the axial compressive sbess on it. weld

@ in the way of the beum connection

hence use 29% of thc above leg size, or leg s i ~ ew = ,178" or 3/16"; however, the maximum thickness of plate here is 4" and the minimum size of fillet weld for this thickness is 'h" (.4WS Bldg Art 212 and AISC Sec. 1.17.4). Hence use M". When the column is subjected to the dcad and live loads and wind in the x-x direction, bending is about the y-y axis. Here the inside and outside portions of the colurrni arc continuous throughout the crosssection of the colimm, and the connecting welds do not transfer,any force; hence, the weld size as dctermined above for Case R would control. should be further increased Perhaps w'ld within the region of t e beam connection, to transfer the horizontal forces of the hcam end moment back into the column web. The horixontal stiffeners in the colurnn at this point, however, would undoubtedly take care of this.

9

7. SQUARE AND RECTANGULA SECTIONS FOR COLUMNS

-

(584k) (125.25) (1515) ( 74,507) ( 2 welds)

7450 leg size w = -----11,200

= weld

.665" or use W

@ for the remaining length of the column

Vs = 170" or 29% of V2 hence use 29% of the leg size or ,192". However, the

Square and rectangnlar tubnlar shapes are now being hot rolled from A7 (33,000 psi yield) and A36 (36,000 psi yield) steel at about the same price as other hotrolled sections. These sections have exceptionally good compressive and torsional resistance. See Tables 7 and 8 for dimensions and properties of stock sizes. Many cngineers feel that the round tnhular section is the best for a column since it has a rather high radius of gyration in all directions. This is much better than the standard W F or I sections, which have a much lower radius of gyration about the weaker y-y axis.

3.2-30

/

Column-Related Design

Unfortunately the usually higher cost of round tubular sections prohibits their universal use for columns. However, a sqnare tube is slightly better than the round section; for the same outside dimensions and cross-sectional area the square tube has a larger radius of gyration. This of course would allow higher corn. pressive strcsses. Consider thc following two sections, 12' long, made of A36 steel:

For another rxamplc, consider the following A36 Techon:

FIGURE 40

FIGURE 38

FIGURE 41

FIGURE 39

x 4" square tubing

3%" extra-heazjy pipe

4"

A = 3.678 in.'

A = 3.535 in.'

Wi = 12.51 lbs/ft

W, = 12.02 lbs/ft

r,,, = 1.31"

r,,bi,,= 1,503"

ue = 11,670 psi -

u, = -

13,500 psi

In this example, the square ttlbe has 3.9% less wcifiht and yet has an allou&le load 11% greater. Its radius of gyration is 14.7% greater.

- = 15,990 psi

uc

-

uc -

19,460 psi

P = (15,990) (9.71) P : (19,169) (9.18) = 155,0k = 184.3" The 32-lb/ft 10" square tubular section has a radius of gyration which is more than twice that about the weak y-y axis of the 33-lb/ft 1 0 W F section. This results in an allowahlr compri:ssive load 19% grcater. The second advantage to the square and rcctanguIar sections is thc flat surface they offer for connections. This results in the simplest and most direct type of joint with minimum preparation and wclding. Also by closing the ends, there would be no maintenance problem. It is common practice in many tubular structures not to paint the inside.

3.2-32

/

Column-Related Design

Four all-welded multilayer Vierendeel trusser make up the exposed frame of the beautiful Rare Book Library of Yale University. Weldfabricated tapered box sections are used in the trusses. Good planning held field welding to o minimum, the trusses being shop built in sections. Here, a cruciform vertical member of the grilled truss is field spliced.

S E C T I O N 3.3

1. BASIC REQUIREMENTS

Rase plates are reqnirtd on the ends of columns to distribute the concentrated compressive load ( P ) of the column over a much larger area of the material which supports the column. The base plate is dimensioned on the assumption that the overhanging portion of the base plate acts as a cantilever beam with its iixed end just inside of the column edges. The upwnrd bending load on this cantilever beam is considered to be uniform and cqual to the bearing pressure of the supporting material.

area (A). Tablc 2 lists standard sizes of rolled plate used for bearing plates. 3. Determine overhanging dimensions m and n, the projection of the plate beyond the assumed (shaded) rectangle against which the load ( P ) is applied.

4. Use the larger value of m or n to solve for required plate thickness ( t ) by one of the following formulas:

Derivafion of Formula # I critic$ Section

in Bending

The primary fnnction of the plate thickness is to provide sufficient resistance to the bending moment ( M ) on the overhang of the plat(, just beyond the rectangular area contacted by the column. Treating this over-

LE

1-Masonry

Bearing Allowabler

(AlSC Sec 1.5.5)

FIGURE 1

AISC suggests the following method to determine tho reqnired thiclmess of bearing plate, using a maximum bending stress of .75 cry psi (AISC Scc 1.51.4.8): 1. Determine the required minimum base plate area, A = P/p. The column load ( P ) is applied uniformly to the base plate within a rectangular area (shaded). The dimensions of this area relative to the column section's dimensions are .95 d and .SO h. The masonry foundation is assumed to have a unifonn bearing pressure ( p ) against the full area ( A = B x D ) of tho base plate. See Table 1 for allowable vah~esof p. 2. Detmmine plate dimensions f3 and D so that dimensions m and n are approximately equal. As a guide, start with the square root of required plate

=

On sandstone and limestone

p

On brick i n cement m o r t o i

p = 250 pi1

400 psi

On full oiea of concrete support

p

On ?$ orec of concrete support

p = 0.375 f',

= 0.25

f'.

where f', ir the specified iarnpicirion strength of the concrete a t 2 8 doys !In this text, a', ir used as equivalent t o AISC'i Pi.)

LE 2-Standard Sizes of Rolled Plate For Bearing Plates I

/

2 8 x 3

1 4 X Ill2 1 6 X 1'12 16 % 2 20 i2

2 8 X 3% 32X3'/2 32 Xi4 36 X 4

20 20 24 24 24

36 40 40 44 44

X 2112 X 3 X 2 X 21/2 X 3

X 4% X 4112 X 5 X 5 X 51/2

4 4 x 48 X 4 8 x 48 X 52 X 52 X 52 X 56 X 56 X 56 X

6 5% 6 6Il2 6 6% 7 61/2 7 8

6 0 x 7 60 X 7 % 6 0 x 8 66 X 7% 66 X 8 66 X 8% 66 X 9 72 X 8 72 X S1/2 72 X 9

7 2 X 9!j 7 2 X 10 7 8 X 9 78 X I0 8 4 X 9l/2 84 X 10

3.3-1

3.3-2 /

Column-Related Design

FIGURE 2

hang (m or n ) as a cantilever beam with M being maximum at the fixed or column end: bending moment p m'

M = ---- parallel to thc column's x-x axis and 2

M=-

2

parallel to the column's y-y axis

bending stress in plate where, assuming a 1" strip:

I

t' -S = (I") 6

and by substitution:

6 p m V p m m ' and 2 u

t = m

plates over 2" hut not over 4" in thickness may be straightened by pressing; or, if presses are not available, by planing for all bearing surfaces (except as noted under requirement 3 ) to obtain a satisfactory contact bearing; rolled steel bearing plates over 4" in thickness shall be planed for all hearing surfaces (except as noted under requirement 3 ) . "2. Column bases other than rolled steel bearing plates shall he planed for all bearing surfaces (except as noted undcr reqnirement 3 ) . "3. The bottom snrfaces of bearing plates and column hnses which are grouted to insure full bearing contact on fonndations need not be planed." The above reqnirements assume that the thinner base plates are sufficiently smooth and flat as rolled, to provide full contact with milled or planed ends of column bases. Thicker plates (exceeding 2") are likely to be slightly bowed or cambered and thus need to be straightened and/or made smooth m d flat. 2. STANDARD DETAILING PRACTICE

u

)r Formnla #1

(similarly for dirncnsion n ) Finishing of Bearing Surlaces

AISC Sce 1.21.3 prescribes that colunin base plates he finished as follows: "1, liolled steel bearing plates, 2" or less in thickness, map be used withont planing, provided a satisfactory contact bearing is obtained; rolled steel bearing

Fignre 2 shows typical column bases. Note the simplicity of these designs for arc-welded fabrication. Designs a and h are intendcd for where column and base plate are erected separately. The angles are shop welded to the column, and the column field welded to the base plate aftcr erection. Design c is a standard of fabrication for light colnmns. Hwe the base plate is first punched for anchor bolts, then shop welded to the colnmn. If the end of the colnmn is milled, there must be just sufficient welding to thr. base plate to hold all parts

Column Bases

/

3.3-3

securely in place (ATSC Sec 1:15.8). If the end of the colu~nnis not milled, the connecting weld must be large enough to carry the co~npressiveload. Welding Practices

In most cases, during fabrication, the columns are placed horizontally on a rack or table with their ends overhanging. The base plate is tack welded in place (Fig. 3 ) , using a square to insure proper alignment, a d is then finish welded. As much as possible of the welding is done in the downhand position because of the increased welding speed through higher welding currents and larger electrodes. After completing the downhand welding, along the outside of the top flange, the column is rolled over and the downhand welding is applied to the other flange.

FIGURE 4

It is possible to weld thc base plate to the column without turning. Sce Figure 4. With the web in the vertical position and the flangm in the horizontal position, the top flange is weldcd on the outside and the lower flange is welded 011 the inside. This will provide sufficient welding at the flanges without further positioning of the column.

(a) Base plate shop welded to column.

FIGURE 3

3. ANCHOR ATTACHMENTS TO COLUMN BASES Anchor bolt details can be separated into two general classes. First, those in which the attachnrents serve only for erection purposes and carry no important stresses in the finished structure. These include all columns that have no uplift. The design of these columns is governed by direct grnvity loads and slenderness ratios set up by specifications for a givcn column formula. IIere the columns can be shop welded ctirectly to the base plate, unless the detail is too cumbersome for shipment. The anchor bolts preset in the masonry are made to engage the base plate only. See Figure 5a. I.arge base plates are usually set and levelled separatcly bclore hcginning column erection. In this case d i p angles may hz shop welded to the column web or Nanges, and in field creetion the anchor bolts engage both base plate and clip angle. See Figure 5b. Secondly, those in which the attachments are designed to resist a direct tcnsion or bending moment, or some combination in which the stability of the

(b) Bose plate shipped separate-attaching angles shop welded to column.

3.3-4

/

Column-Related Design

finished structure is dependent on the anchor attachments. These include all columns having direct loads combined with bending stresses, caused by the eccentric applications of gravity loads or horizontal forces; for example, wind, cable reactions, sway or temperature, etc. These are found in everyday practice in such structures as mill buildings, hangers, rigid frames, portals and towers, crane columns, etc. In large structures that extend several hundred feet between expansion joints in each direction, the columns at ends and corners of thc structure may be plumb only at uormal temperature. As temperatures rise and fall, milled-end bearing conditions at edges or corners of the column base may prove very unsatisfactory, even though shop work were pcrfect. Such columns should have anchor bolt details designed to hold the column firmly fixcd, in square contact with the base plate. The combined efiects of the direct load and overturning moments (due to wind, cranc runway, etc.) can always be considered by properly applying the direct load at a givcn c c ~ e n t r i ~ i t yeven , though the bending stresses sometimes occur in two directions simultaneously. Design of the anchor bolts resolves itself into a problem of bending and direct stress.

If there is any appreciable uplift on the column, angles may be welded to the base of the column and anchored by means of hold-down bolts. Under load., the angle is subject to a bending action, and its thickness may be determined from this bending moment. Trcating the cross-section of the angle as a frame, the problem is to know the end conditions. Some engineers treat the horizontal leg as a cantilever beam, fixed at one end by the clamping action of the hold-down bolts. See Figure 6. This is not quite a true picture because there is some restraint offered by the other leg of the angle.

Otlrer engineers have assumed the horizontal leg of the angle acts as a beam with both ends fixed. In this case the resnlting moment at either end of the portion being considered, the heel of the angle or the cnd at the bolt, is only half that indicated by the previous approach. St:e Figure 7.

However, it might be argued that the vertical leg is not completely fixed and that this will increase the moment in thc horizontal leg near the bolt. The fo1low~ing analysis, made on this basis, is probably more nearly correct. See Figure 8.

FIGURE 8

FIGURE 6

1. Considering first just one angle and temporarily ignoring the eRect of the other, the upper end of the vertical leg if not restrained would tend to move in horizontally (A,,) when an uplift force (P,) is applied to the column.

Column Bases

/

3.3-5

3. Combining the initial moment resulting from the uplift force (1) and the secondary moment resulting from the restraint offered by the opposite angle (2):

The resulting moment is

M = P, b and area of moment diagram X moment arm AhY = E I

2. Since the opposite angle does provide restraint, a horizontal force (PI,) is applied to pull the vertical leg back to its support position. The resulting moment is Substituting into the previous equations:

M = P,, d and

at the heel of the angle, and

which is the critical moment and is located at the holddown bolts. Required Thickness of Angle

area 1 X moment arm I A,, = E I drea 2 X moment arm 2 tE I

The leg of the angle has a section modulus of-

or required thickness of where: M S =u Since the horizontal movement is the same in each direction:

or, see Figure 9, where the vertical leg of the angle is welded its full lcllgth to the column ~rovidinga fixedend condition (Case A ) ; here formula #3 applies-

or where, the vertical leg of the angle is welded only

3.3-6

/

Column-Related

FIGURE 9

at its toe to the column (Case B); here formula #5 applies1, 13b

+ d) u

Allowable Stresses Table 3 presents the allowable stresses for holddown bolts used in building (AISC) and in bridge (AASHO) TABLE 3-Allowable

Stresses for Hold-Down Bolts

Aliowoble unit tension and $heor itrerier on baltr and threaded ports (psi of unthieoded body oieo): Tension Sheor AlSC 1.5.2.1 (Building) psi psi A307 boltr ond threaded parts of A7 ond A373 rteei 14,000 10.000 A325 boltr when threading ir excluded from shear planer 40,000 15.000 A325 bolts when threading excluded fiom rheor ~ l a n e r 40,000 22,000 A354, Grode BC, boltr when thieoding ir not excludcd from rhear ploner 50.000 20,000 A354. Grode K , Y h e n threading excluded from rheor planer 50,000 24,000 - -. AASHO 1.4.2 (Bridge) psi tension - boitr ot root of threod 13,500 11.000 shear - turned bolts 20,000 beoring - turned bolts tffeitive beorjog o m o of o pin or bolt iholl be its diometer multipiicd by the thickness of the metal on which it beoir.

construction. Aim included are dimensions of standad bols. (Tablc 3.4). 5. BASE PLATE F

R C O L U M N LOADE

MOMENT When a moment ( k t ) is applied to a column already srihjectcd to an axial compressiveforcc (P,), it is more couwbnicnt to exprcss this combined load as the same axial forcc ( P C )applicd at some eccentricity ( e ) from the neutral axis of the column. t-e+

, (4

lbl FIGURE 10

In either representation, there is a combination of axial compressive strcss arid bending stress acting on a cross-section of the column See Figure 11. Multiplying this stress by the width of the Range (or the thickness of thc web) over which the stresses are applied, gives the following force distribution

Column Bases

/

TABLE 3A--Standard Bolt Dimenrionr

Compressive stress

=

$

Bending stress

" = -P, e

S

Total stress

P

a='+-

A

FIGURE 12 PC e S

FIGURE 11

across the depth of the column. This force is transferred to the base plate. See Figure 12. This assumes that the column flanges are welded directly to the base plate.

FIGURE 13

If anchor hold-down bolts transfer the tensile forces, thenThe column is usually set with the eccentricity ( c ) lying within the plane of the column web ( a d s y-y), as in Figure 11. Thus the column Aangcs will carry most of the resulting forces because of their having relatively greater cross-sectional arca, and being located in areas of higher stress. See Figure 14.

3.3-8

/

Column-Related Design

FIGURE 14 FIGURE 15

If the eccentricity ( e ) is less than % D, there is no uplift of the base plate at the surFace of the masonry support (Figure 15): section modulus of base plate

stress in base plate UT

=

TI

compression t T? bending

When the eccentricity ( e ) exceeds % D, there is uplift on the base plate which is resisted by the anchor hold-down bolts. The beariug stress on the masonry support is maximum at thc extrcme edge of the bearing plate. It is assumed this stress decreases linearly back along the plate for a distance (Y); however, there is some qucstion as to how far this extends. One problem analysis approach treats this section as a reinforced concrete beam.

There are three equations, and three unknowns ( P t ) , (V, and (5,): l.;r;V=O MYu,B-Pt-P,=O

and

where: cr, = pressure supplied by masonry supporting material 2. 2 M = 0 (About N.A. of column)

aud

......... (Qb)

FIGURE 16

3. Representing the elastic behavior of the concrete support and the steel hold-down bolt (see Figure 17) :

Column Bases

/

Also

where: A, = total area of steel hold-down bolts under tension us = stress in steel bolt Es = ah in s t d bolt E. = modulus of elasticity of steel bolt then and: 7, o: = stress in concrete rt support a A, Pt eC = strain in concrete 6 = - Twc-n support E, = modulus of elasticity of concrete and from similar triangles SUPP0l.t D n = modular ratio of - - Y + f elasticity, steel to = 2 Y concrete

FIGURE 17

Solve for Y:

;

* * *

This reduces t ( t -

Substituting formula #10 into formula #8a:

or to express it in a manner to facilitate repetitive use, let-

Substituting formula #9a into formula #11:

3.3-10

/

Column-Related Design

From this assumption, the overhang of the hearing plate, i.e. the distance from the column flange to the plate's outer edge, is seen to equal the effective bearing length.

then-

There are several ways to solve this cubic equation. Perhaps the easiest is to plot a few points, letting Y = simple whole numbers, for example, 9, 10, etc., and reading the value of Y on the graph where the curve crosses zero. Having found the effective hearing length (Y) in this manner, formula #9b can be used to solve for the tensile force (P,) in the hold-down bolts. Formula #10 then gives the amount of bearing stress in the masonry support.

FIGURE 19

FIGURE 18

Another approach to determining the effective bearing length, involving less work, assumes the same triangular distribution of bearing forces from the supporting masonry against the bearing plate. However, the center of gravity of the triangle, or the concentrated force representing this triangle, is assumed to be fixed at a point coinciding with the concentrated compressive force of the wlnnln flange. See Figure 18.

Figure 19 shows a column base detail. The columns have a maximum load of 186 kips, and receive no uplift under normal wind. See Figure 19. Under heavier wind load and in combination with temperature, they may receive up to 20 kips dircct uplift. See Fibwe 20. Four bolts are provided, attached by means of 6" X 6" X %" clip sngles, 11" long on a 4" gauge. To be effective, the angles must carry this load on the anchor bolts into the column web. This causes a bending moment on the outstanding legs of the angles. Analysis follows that for formula if3. The bolt tension fixes the toe of the angle against the base plate and causes LUI inflection point between the bolts and the vertical leg of the angle, so that the bolt load is cantilevered only about halfway.

To compute the bending stress in the angles:

FIGURE 20

diagram

where: ub = s t r e s in outer fibers

bending fl,

M = hcnding moment c -- distance to neutral axis I = moment of inertia Since:

M =S,

-

(l0.000* x -4") (78 in.')

shear f"

P

= 19,400 psi Hence, thc. dfstail with %" angles is OK for this load.

Check Welds to Column Web The angles are welded to the column web with 'h" fillet wclds; this will now be checked. The heel of the angle is in coinpression against the wt:b of the column and is equivalent to an additional weld across the bottom for rcsisturg moment. On this basis, the section rnodulus of the weld is calculated. For simplicity, the weld is treated as a line without any cross-sectional area. From Table 5 ; Sect. 7.4, the section modulus of a rectanzular connection is:

leg size of (170) fillet weld

= .actual -

force -allowable force

= .06" but 3k"thick angle requires a minimum of Ydl (Table 3, Section 7 3 ) . If it is dcsircd to incrrasc the anchor bolt capacity of the d i p angle &tail, tllicker arrgles should be used with large plate w~ishcrs on top of the angle. The ;ittaclrmc~lts s h o ~ ~ lbe d maclc to the column flanges, sincc the welds arc more accessible there and the bolts Iiave better leverage.

and liere:

Normally, section modulus is expressed as inches to the third power; however, here where the weld has no area, thc rcsultirrg swtion modulus is expressed as iiiches squared. When a stmdard bending formula is used, the answer ( ) is strcss in lhsjin.\ however, when this new section modulns is used in the bcnding formula, the answer ( f ) is forcc on the weld in lbs/linear in.

To ilhistrnte how the colnmn Aange can lx: checked to clctcmiine whcther or not it is too tliin, considcr a clip angle mchored with two 1%" bolts centered 2?W out l'rorn the face of the cohimii flange; see Figure 21. The angle is att;iclied to the column flange by fillet ~velclsacross the top a i d down each side. The capacity of thc two lx~ltsat 14,000 psi allowable stress on nntlircaded area (AISC Sec 1.5.2) is2 (1.2") (14,otl()) =: 31,400 lbs

> 28,500 lbs OK --

Tlie hending nioment on tire ~ c l dis) , )71,250 in.-lbs (28,500 lbs) ( ~ ~ h =

n-Related Design

-

zootal tor, weld. At the ends of the angle, - the force (915)(3) couple is - --- -- - 1370 lbs centered 1" below the 2 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 resistauce in the column web. The bending moment in the column flange is computod as follows: Force along top of angle = 915 X 5.5 = 5040 lbs

M, = 5040 X 2.75 = 13,860 in.-lbs M, = 1370 X 5.5 = 7,535 -i d b s Total M 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 ihe welded connection is found to be--

= 21,395 i d b s

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 thick. flange 1% This is calculated as follows:

,"

= 78 in.= (See Problem 1 ) The bending force is-

= 45,300 psi 51,250 in.-lbs 58 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 Aange heyond the yield point, the cvlnmn Aange will deflect outward sufficiently to relieve these stresses and cause a redistribution. Thr resultant stresses in the weld metal on the toe of the clip angle will be concentrated opposite the column web.

FIGURE 22

Column Bares

Thus, the capacity of this anchor bolt detail is limited by thc bending strength of the column flange even alter the clip angle has bccn satisfactorily stiffened. The force back through the column web is:

F = (915 lbs/in.) (11") = 12,800 lbs

+2

(1370 lbs)

A 'h" fiUet weld 3 inches long on the top of the angle opposite the column web will satisfactorily resist the force couple:

F = (3") (5600 lbs/in.) = 16,800 ibs. OK --

E70 welds

For greater anchor bolt capacities than shown in Figure 22, either horizontal stiffeners or diaphragms shonld be provided to prevent bending of the column flanges.

e =

/

( 175,000)(12) -

( 130,000)

= 16.15" The load on the bolts is(9.49) F = .(130,000) (15.66) = 78,800 lbs The area of the thrce lWrdia. bolts in the unthreaded body area isA = (3)(2.074)

= 6.22 in.2 The tensile stress in the bolts is:

u = -(78;800) (6.22) = 12,700 psi

<

14,000 psi

OK -

(AISC Sec 1.5.2) A rather simple detail, whereby a wide-flanged channel scrves as a stiEener, is shown in Figure 23. This detail was used with three lSk"dia anchor bolts on a 14" X 87-lb mill building column designed to resist a wind bending moment of 175,000 ft-lbs combined with a direct load downward of 130,000 lbs. The tension on the bolts is determined by taking moments about the right-hand wmpression flange of the colrrvnn after first determining the eczatricity at which the direct lond will cause a moment of 175,000 ft-lbs about the centerline of the column. The eccentricity is-

FIGURE 23

The compression Aange reaction ( R ) is the sum 01 the 130,000-lb c:Arrmn load plus the 78,800-lb pull of the anchor bolts, or 208,800 lbs. The 13" ship channels are st:t up just clear of the bearing on the base plnte so that the end of the column will take the compressive load of 208,800 lbs without overloading channels. Bearing stress on masonry

The hearing stress on the masonzy support is maximum at the extreme edge of the bearing plate, and is assumed to decrease linearly back along the plate. This bearing stress would resemble a triangle in which

3.3-14

/

Column-Related Design

Hence, the distance from the compressive force of the Range out to the edge of the bearing plate (in oth,er words, the overhang of the bearing plate) equals 'h the effective distance of the bearing support. See Figure 24.

@ 8

@

= 24" Anchor hold-down bolts ore inactive on compression side

@

a r m of triangle

A=%u,Y

=

PC

+ Pt

effectice beuring length of base plate (from formula # 8 )

1

= 23.2"

I

Y

and - - 7.73" overhang 3 .'. D = 7.73" 13.31" f 7.73" = 28.77" or use 28%''

= .25 (3000 psi) = 750 psi

+

Bolt load

78.Ek FIGURE 24

the altitude is the maximum hearing stress at the edge of the plate, and the base of the triangle is the effective bearing length ( Y ) against the plate. (See short method described on page 10.) Since the area of this triangle has a center of gravity % Y h e k from the altitude, the bearing pressnre may be resolved into a concentrated force at this point. This point will be assumcd to lie wh'ere the column flange's concentrated compressive load of 208,800 1hs is applied.

FIGURE 25

The load on the bolts is supported by the top flange of the 13" channel, reinforced by four 3%" X 'ii' s t B cner plates welded between the channel flanges. See Figurc 23. The two interior plates each support a full bolt load of '/, (78,800 Ibs) or 26,300 lbs. Thesc stiffeners are attached to the channel web with four I" X intermittent fillet welds on each side of the plate, and to both flanges by continl~ous3$,j" fillet welds on each side of the plate. See Figmo 25. The welds at the chnnncl flanges transmit the moment to the channel flangcs, and the welds at the channel web support most nf the shearing load. Thc 2" eccentricity of the bolt load to column Range is trar~sposedto a force couple acting on the channel flanges. This couple is obtained by dividing

Column Bases the momeut by the depth of the stiffeners:

This is a hori~ontalload acting at right angles to the column flange. I t is delivered as four concentrated loads at the tops of stiffeners and then carried horizontally by the channel flange to a point opposite the column web where it is attached to the column with a 2%'' x M" fillet weld. 2%" X 5600 lhs/in. = 14,000 lbs. The concentrated load valucs are 2015 lbs at each end stiffener for one-half a bolt load, and 4030 lbs at each interior stiffener.

3.3-15

For simplicity, this analysis has assumed that the effective bearing length ( 1 ' ) was such that the center of gravity of the triangular bearing stress distribution, C.G. a t % Y, lies along the centerline of the column Bange where the comprcssive force of the colunm is applied.

\With the same column base detail as in Problem 3, we will now m e the original derivation for this effective bearing length ( Y ), treating the analysis as a reinforced concrete beam and solving the resulting cubic equation. The work may takc longrr, hut rcsults are more accurate. See Figun: 26, temporarily ignoring the anehorbolt channel attachments.

The total moment on the flanges is: (2,015) (7.5) = 15,200 in.-lbs (4,030) (2.5) = 10,100 -in.-lbs

M = 25,300 in.-lbs I t causes a bending stress in the channels 4" X %" top flange section of approximately-

= 15,800 psi To keep the channel section from sliding parallel to the column flange, the direct vertical pull of the continuous fillet bolts is supported by two 13" X welds between the edge of the cnlumn flanges and the web of the 13" channel section. The shear on these welds is-

FIGURE 26

The problem in Figure 23 has been analyzed on the basis of simple levers with the compression load concentrated on the colnmn flange. It ignores the compression are:> 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"

D =z 283/4" B = W

/

3.3-16

Column-Related

four %" X

Tensile stress in bolts

3%" R 's

FIGURE 27

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

n =

Plotting these three points, the curve is observed to pass through zero at-

Y = 13.9" -

from formula #13 (cubic ~ q u a t i o n j Y3+K1YZ+K2Y+K3=0

which is the effective bearing length.

where:

from fornula #9b

.1=3(~-$)

=3

28%

2 (16.15 - -

= 5.33 6 n A, K' = -B (f -6

+ e)

(lOj(6.22) (9 24

+

16.15)

= 392

which is the tensile load on the hold-down bolts. from formula #8b

= lOiiO psi Therefore.~ substitutinr" into formula &13:

E3

+ 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

-

Y:

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

Column Bases

/

3.3-17

+

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:

-

(199.98 (27.36

= 6.93" distance of K.A. to rcf. axis y-y

.'. c =

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

+ 2 1 n")

= 8220 + 4300

+ .42 n )

I

I

Dirtonce: C.G. to ref. mi. y y

4.688

12,150 psi

Moment

= .21n2 -- 4.615

of web

-

=

+n

2

Poition

6.93" distance of N.A. to outer fiber

.--

-

Column flonge

4.344

Channel web

3.812

Chonnel stiffenen

2.00

-

--

Fict Totol

-t

Second Totol

+

By substituting value of n = 6.93":

27.36

42.83

1 86.05 ~-

6.00

22.87

7.25

14.50

. 87.19 29.00

+ .42 n 30.27

199.98

+ .21

'n

210.07

.... 7.92

2789.93

/

3.3-18

Column-Related Design

tensile stress in hold-down bolts

M c PC ut = - I A

= 15,500 psi

where c is distance of N.A. from extreme fiber of tensile area

- 4,300 =

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-

11,200 pd

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-

5.15''

a

"

-6.W

(8220 psi)

= 6110 psi

856 lbs/linear inch

r

+ 4300 psi

+ 4300psi

=

OJ

-

856 11,200 for E7O welds ('firhlr 5, Sect. 7.4) = ,076" or use (Table 2, Sect. 7.4)

7 -

$iG"h

With this 1r:g size, intermittent welds can be used instead of contiriuous wdding-

10,410psi

elding Channel Assembly to Column Ftonge

Sa =

i a =

d212b 3ib

+ dl

S, = bd

+ dl

1131212 X 14.5 f 13) 3114.5 13) 2 86.1 in.

+

M

S,

-

f.

L

i. =

+

ib2

+ isn

= f(2020,Zf;3050) - 3670 i b d i n . 0

242.2 in.'

actvol farce cliawabie force

=

(36701 111.200) t E70%

-

= ,328'' or 5/16" A

-

v

= L

f

f123.4001 2(13 1-14.5) 2240 i b d i n .

fx-. +(211;;;;2 fez

-

2350 Ibsiin.

=

uctual force . ollawoble force

-

-

3 56.3 in." is

Sr -( 1 74.2001 156.31

3100 l b d i n .

v

fe

L I123.4001

.+

.-

"= -

- ,210.' or *,,A

-

-

114.51li3)

--

185.9 in."

M S, -.-I 174,2001 (185.91 - 935 ibslin. v = L

=

-

-

2 (13) 4750 ibslin.

+

+ - \/1937t;4260iz

f , = V fb2 -.

5680 Ibrlin. aituol force aliowabie farce 156801 111.2001

-

.506" or X"

1123,400) 2 114.5)

- 4260 I b d i n .

i? = V' f,,2 i*z = '"i3100li i475012

.- -(23501

i11.2001

3 (131" -

i" = --

720 Ibdin.

d "8(

d2 -.

M

Sx, -I 174,2001 (242.2)

==

"

-

(13)' 3

M

f. =

7

=

+

s ," -

--

. . -

(123.4001 2(i3) 04.5i 3050 i b d i n .

v'

.-

-

v

-

ll4.5)(13)

-

186.11 2020 lbslin.

i. = -

-

-

f

l174.2001 - -

d2 + 3

A

(11

=

/.?

4360 ibnlin.

octucl force oilowobie force

Column Bases

/

3.3-19

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.

I

F

1

Column Flange

The average compressive stress on the channel web is-

= 3700

+ 4300

=

8000 psi

total compressive force on channel assembly

F = 48,000 =r

+ 4(18,850)

123,400 Ibs

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 28

FIGURE 29

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

+ (48,000 1bs) (3116")

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.

-

8.25 in."

bending moment on bat-

rcsulting bending stress

= 18,000 psi

3.3-20

/

Column-Related Design

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

= 112 in.' S,

( 1288)

= --

(9.5) = 135.5 in."

maximum bending force on ueld

shear force on weld

resultant force on weld

FIGURE 30

Take area moments about the base line ( y-y) :

--

2weldrX15"

30

Total

36

15.3

405.0

-. 5467.5

414.0

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"

I 562.5 6048

required flkt weld size 3000 a =113J0

WI

E70 allowable

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

olumn Bares

y

/

stiffeners moy be

FIGURE 31

\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-

3.3-22 /

Column-Related Design

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"

t =

t b,-Li

or:

4-

= 50' FIGURE 35

I

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:

P = A w

= (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

Since:

M = a S FIGURE 36

3.3-24

/

Column-Related

Bendtng stress [a)

Shear force (f)

PI

(4

FIGURE 37

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

=M I

Cb

= 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

= 4370 psi

l/g (24,630) =----

(11,200)

-

E70 allowable

= ,545"or use %/,Br'[l --is

(The minimum fillet wcld leg size for 6" plate .)

WB

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

Column Bases

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)

/

3.3-25

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

FIGURE 38

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

/

Column-Related

FIGURE 39

'

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:

7

same as plate

= 15,800 psi

shear jorce on one weld

f. = 4310 lbs/in. required effective throat

i j using bevel ioint

Y6 ,ik

+

t = t, '/a'' = ,273'' 4- W"

= ,398" raot face (land) = ll/z"

- 2(.39Wr)

= ,704'' or use --W'

if using 1 joint

k-%"4

However, in this example, the column flange thickness of 3" would require a %" fillet weld to be used. Brackets t o column flange edges

1 %"

t = t,

"

= ,273"

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.)

6" = 73.9"

to be deducted: 516.5&- 73.9' = 4426&

FIGURE 40

FIGURE 41

COLUMN BASE PLATE DIMENSIONS (AISC, 1963)

-

II c

L--?Q

COLUMN BASE PLATES Dimensions for maximum column loads

-

/

For

/

'Or

-

-

Base nlaien, ASTM 1116. h 27 iri Cuacilic, <', 30UO or,

Bare OaiPs, hSTM A16. F,, = 27 kr, coacrsts. l , 3OM nri

Wt. P"

Fi.

tn.

ib.

61:16

1w

W

161 133 120 106 99 92 85 79

72 65

58 53 50

45 40

I12 100 89

4 X 14%

W

77

72 66 60 54 49

45 1 X 12 W

39

1 X 10 W:

67 58 48 40 35

1 x 8 W

33

31

28 24 20 17 ~. Note' i SIlO Wht

-This

and following toblei prenenisd here by cauttery of American Institute of Steel Conltruction.

COLUMN BASE PLATES Dimensions for maximum column loads

~-

mT . 1.-

.1-

. J-

Column Bases

/

3.3-2

2.3-30 /

Column-Related

E .

2

xk

-

"

5! a

X;

-

a

x x- r x sa x z

" a

S r ,

I F

Y . L D

xm

x

a

L

D

F

XS -

F

x x

-

-

a

F

Xm

XF

"

s

8

X-

X-

?

ases

/

3.3-31

Column base plates for the 32-story Commerce Towers, Kansas City, Mo., were shop-fabricated and shipped separately. At the site they were positioned and bolted to the concrete. The heovy columns were then erected ond field welded to base plates. This was facilitated by use of semi-automatic arc welding with self-shielding cored electrode wire. Process quadrupled the speed of manual welding and produced sounder welds.

Ten-ton weldments were required for tower bases on lift bridges along the St. Lawrence Seaway. Edges of attaching members were doublebeveled to permit fuil penetration. Iron powder electrodes were specified for higher welding speeds and lower costs. Because of high restraint, LH-70 (low hydrogen) E7018 electrodes were used on root posses to avoid cracking, while E6027 was used on subsequent passes to fill the ioint.

3.3-32

/

C o l u m n - R e l a t e d Design

In designing a scenic highway bridge with 700' arch span, near Santa Barbora, Col., engineers called for tower columns to be anchored to the concrete skewbacks b y means of 1%" prestressing rods. The bottom of the column is slotted to accommodate the base, an "eggbox" grill made up of vertical plates welded together and to the box column. The towers suppoFt heavy vertical girder loads but also safely transmit horizontal wind and seismic loads from the deck system to the foundation.

1. INTRODUCTION

2. TYPES OF SPLICES

.41SC specifies that, where full-milled tier-huitding coliirnns are spliced, there shall be suflkicnt welding to hold them securdy in place. These connections shall be proportioned to rcsist any horizontal shear forces, and any tension that would be developed by specified wind forces acting in conjnnction with 75% of the ea1c:ulated dead load strws and no live load, if this condition will prodrice more tension than full dead load and live load applied. (AISC: Sec 1.15.8). Figures 1 and 2 show various designs of column splices mhicli diminate punching of the columns. Note that these details require only llandling and punching of small pieces of angles or plates v&ch are easily carried to, and welded to, the columns in the shop. The details provide for temporary bolted connections in the field prior to making the permanent welded connections. Sometimes the column connections are placed about mid\~ayin height, in order to get the connection away from the regiorr of heavy bending moment caused by windloads, etc. The resnlt is a. ~wnncctionsufficient to hold tlre columns in place and designed for horizontal shear m d axial compresion only

In Fignre I ( a j, a plate m d two :mgles are punched or, if nccessary, drilled. The plate is shop welded to the top of the lower colrrmn. The two angles are shop welded to the web at the lower end of the npper rolnmn. The r r p p ~column ~ is erected on top of the l o u w column and eroction bolts are inserted. The tipp'r c~lnrnnis then &.Id \velded to the mnnecting plat?. Where additional clc.ar;mce is needed for erection of beams framing into the web o i the lower c,olumn, it might be nccessary to shop weld the plate to the upper col~m~rr and tlicn field weld in the overlicnd position to the lo\vcx colun~n. If tlre nppm and lowcr colnrnns differ in size, the conn
FIG. 1-Typical

Column Splices

.Q-2

/

Column-

Cc) FIG. 2 - Typical tions and welded thereto. In the field the colun~n sections arc bolted temporarily prior to welding, as indicated at ( d ) . In Figure 2 ( a ) the ends of both column sections are first milled for a square bearing surface. Then the two lower ewction splice angles are shop welded on opposite sides of the web of the heavier w l u m ~ i section, so as to project past the end of the column. The outstanding legs of these angles are provided with holes for erection bolts to engage the outstanding legs of the other two angles that are shop welded to tlie 11pper column section. In this type of detail where lighter con~iectingmaterial projects from heavy main sections, care should he taken in handling to prevent damage to the lighter material. The flangcs on the lower end of tlie upper column section are partially beveled or "J" grooved, and this partial pelletration groove joint is then welded in thc field. The ~ u r p o s eof the angles is to splice and hold the two adjacmt columns together kemporarily while they are being fir:ld welded. These erecting angles may be placed horizontally LE 1-Allowables

for Weld M e t a l in Partiol-Penetration Groove Welds For Field Splices of Columns E70 Welds SAW-2

E60 Welds SAW-I

some or piate some os plate .-

campreision

1:

lendon ironsverse to crosssection oi throat oiea

--

..... .

~

~

1

2:

.~~

AWS Building Por 20510: a n d AiSC Scc

1.5.3

15.800 psi

15,800p

Column Splices

on ttw ~ r v hof thc colririms. Figwr '(b). Trie advantage of this position is that tlwy do not i ~ t i : n dhcyond the eiids of the coliimri for possihlc drm~iigedr~ringtransit or ertrtion. F m r plntcs xi-e piiiichctl, timi shop welded het\veen the A:inges of tlrc two colurrin sections as sliown i r ~ Figure 2 ( c ) , lraving enorlgh space hetween the back of the, platc.s and tlic colt~mn web to insert a \vrtmcl~.Two splicc plates art: also punchcd and shop wrldcd to the l o u - < ~ coli~mn sc~ctiorihcfore siripping to the crcction sit<,. .After bolting in the field as indicated, the permanent coii~iwtitmis in:& by welding. Tlic splico in Figr~rc2(1l) is similar to that at ( a ) but is lor coiinccting two coliimiis of differrnt sizes. The flanges 01' tht. r~ppcrcolumlr lic inside of thc flanges of the 1ov;ol- colririm Rcfort, shop \vtIding tile erecting arlglcs, spiicc phies art, first shop fillet welded to the ir1sid.r. face of ihc ilarigc 01' the lowcr colrunn. Tliey are milled with tlw lowcr col~imn section. As an altrrnate to this, spliw plates with their lower edges prepared for wcldiiig are slrop fillet wi-oided to the outside face of tlrt. fiarigcs on the uppcr column. In case only m ~ cside of the i d u m n is acccssil)le, far example wlim new stecl is crccted adjacent to an old stmcture, a cornhination of this procedure may bc usrd. Placc the lowcr splice plates on the inside face of the lo\vcr coliinln a i d the ~ ~ p p esplice r plate oii tlw oi~tsideface of thc r~ppcrcolr~mn;See Figure 2 ( d ) . In this m a ~ i ~ i eall r fiold w\.olds on both a)!umn flmgcs can he maclt from the one side. \t'Iisrc splice platc,s arc used and filler plates are needed l~ecanseof the diflerencc in sizes of the upper and lower columns, these plates are welded to the iippcr coliiinii. Sre Figiirt, 2(tS).This allows the greater amount of \velding to ljc d o i ~ cin the shop where larger electrodes arrd higher \velding currents used in

Column Splicer

FIG. 3-Pariiul-Penelrotion

the fiat position result in higher welding speeds and lower cost. After erection the splice plate is Iicld welded to the lower column. Two attaching plates are shop welded to the upper end of the lower column. The column may be hoisted by attachiug the cable to the erection holes of these plates. After erecting the upper columns, these plates are field welded to the upper column.

/

Welds

unwe!ded portion, these field splices should never be subject to radiographic inspection.

. EXAMPLES Figure 4 illusbates a typical field splice used on columns of the Detroit Bank & Trust Buiiding in Detroit, htichigan. These fabricated columns were spliced by partial-penetration bevel joints in the column

3. WELD ALLOWABLES Both the AWS Building Code and the AISC Specifications allow partial-penebation groove welds, either a bevel or a J preparation, to be used on column field splices. For a J joint, the effective throat (t,) is equal to the actual throat (t). For a beveled joint, the eflective throat (t,.) equals the actual throat ( t ) less '/a". This reduction in throat is made because the weld may not extend all the way down into the very root of the joint. The Ye" reduction is very conservative. No reduction is made in the throat of the J preparation because there is no problem in reaching the root of the joint. .4 beveled joint is usually flame cut along the end of the column flange. A J groove must be machined or else gouged out by tbc air carbon-arc procvss. Although it may seem that the beveled groove might require more weld metal because it must be Ya" deeper than required, the J groove on the other hand must start with a %" radius and an included angle of 45". There may be no reduction in the amount of weld metal by using the J groove; see Figure 3. A decision on joint design should be made only after all factors are carefully evaluated. Since it is impossible to properly read radiographs of this partial penetration groove joint, because of the

BUILT-UP COLUMN

BUILT- U P COLUMN

FIELD SPLICE

FIG. 4-Typical column splice on Detroit Bank & Trust Building.

3.

/

Coiumn-Related

flanges. These A36 steel columns were welded with E70 low-hydrogen electrodes. Notice the schedule of u-eld sizes. The angle were sliop welded to column ends and field bolted during erection, using high-tensile bolts. These bolts were left in place and carried any horizontal shear in the direction of the column web, hence no field u&ling was required on the web of the columns. Figure 5 illustrates the field splice of columns in the Michigan Consolidated Gas Co. Building in Detroit, Michigan. These fabricated A36 steel box-shaped columns were field welded with E70 low-hydrogen electrodes. Partial-penetration J-groove welds were used on all four flanges around the periphery of the column. Notice the schedule of weld sizes.

BUILT- UP COLUMN

BUILT- UP COLUMN FIELD SPLICE

FIG. 5-Typic01 column splice on Michigan Con. solidoted Gos Co. Building.

FIG. &-Typical column splice in sections of some depth. Plate on the web i s for bolting to facilitate erection.

FIG. 7-Field

splicing of column flanges, using vapor-shielded arc welding process.

Figure 1 illustrates a suggested detail for a pin connection at the cnd of a built-up compression member of an arch bridge, subject to a reaction of 90 kips. The next step is to compute the thickness of the connecting plate. This is based oil the minimum required bearing area of the plate because of the pin reaction against the plate, Figurc 3. The 90,000-lb load is divided by the allowable bearing pressure, which in this case is 24;000 psi assuming no rotation, (AASHO 3.4.2) and the minimum bearing area comes out to be

FIGURE 1

There are many approaches to this type of problcm and, of courso, many solutions. This is simply one analysis and one solution. One of the design requirements in this i~artierllarexample is to have a smoothappearing surface on the outside or faeia side of the arch compression member, Notice in tlie sketch of thc cross-section of the built-up compression member, Figure 1, that the center of gravity is ,935'' in from the outer face. By selecting an attaching plate of sufficient tliickuess for its ccnter of gravity to line up with the compressio~~ ~nernber'scenter of gravity, the compression load will be transferred in a direct line without any eccentricity. Thc bearing pin is subjected to a double-shear load: 90,OOO lhs on two areas, or 45,000 lbs cach. See Figure 2. .4c(:ording to AASI-IO (Scc 3.4.2), the allowable stress on this pin is 13,500 psi.

= 3.33 in.+equired

pin area

or -use a 2%"-&a pin having A = 3.98 ia2

--

Smce thc pm's diameter has been computed to be 2%", the rcquired plate thickness to make u p tbis bearing area would bt-

= 1.67" hut .~ nse -~ 2"-thick plate ..~..~~

since this will also line up with the center of gravity of the compression member (CG = ,933'). The next step is a simple determination of the required depth ( d ) of this courrecting plate. See Figure 4. In this analysis, some structural designers consider this connecting plate as a beam supported at the center, or pin. and withstanding the tvmpression loads transmitted from the compression tncmber. In most cases, the compression load (here 90 kips) is assumed to bc ~ q u a l l ydistributed tllroughout the

3.5-2

/

Column-

various parts of the compression memher by the ratio of the individual areas to the total area. Accordingly, the compression load carried by each angle wodd h e -

= 16.9 kips

Since the required section modulus is in tenns of ( d ) : M:=,,S

"

-- (288,000 in.-lhs).. . (20,000 psi)

and the compression load carried by the 5h" X 20" web plate would beSince

= 56 kips throughout its entire width. Dividing this load by 2G" results in a uniform load of-

= 2.8 kips/linear in. Treat this connecting plate as a cantilever beam from the centcrline with these two loads: ( 1 ) the concentrated load of 16.9 lups at 8.75" from center, and ( 2 ) the unifonn load of 2.8 kips/in. for a distance of lo". The resulting bending moment i.; then computed:

= 288 in.-kips

FIGURE 4

and the minimum depth of upper plate is found to b e d = 6.58" or 7" deep beyond the pinhole would he sufficient.

. FINALIZING + The final detail has k e n sketched in Figure 5. The outer leg of each angle might be triinmed back slightly so as to fit to the 2" connecting plate. Whether this is cut back or not, there will be a loss of 25h1t of the angle leg. This area ( A = 2 X 'h" X 2.625" = 2.625 is made up by additional attaching stiEening plates. These have been chosen to be two %" X 3" plates ( A = 4.5 in.2) and two 'h" X 13h" bars ( A = 1.375 i n 2 ) . The total added area is thus 5.875 square inches. The entire built-up compression member has an area of 20 square inches. These additional attaching plates simply mean that the cross-sectional area in contact with the 2" cunnecting plate is in excess of the required 20 square inches. After the compression member has been welded, its end might he nulled to provide a Bat, smooth surface for bearing against the 2" plate. If this is done, the entire section would not have to be welded 100% all the way through. Under these conditions, it is suggested that a bevel he made part way through these plates of the compression member and that a groove weld be made on the outside. Reinforcing Ellet welds should then be made on the inner side of this compression member where it co~mectswith the 2" plate.

FIGURE 5 lost at connection; replaced by adding

/

Column-Relate

Bearing-pin connections like those shown on this bridge over Michigan's John C. Lodge Expressway must be designed to transfer the compression load without eccentricity. Note simplicity and beauty of the welded rigid frame employed i n this bridge design.

1.

In the past, when engineers required steel columns of heavier section than those commercially available, they designed the columns to be made by riveting eover plates to the flanges of 14" WF rolled sections. See Figure l ( a ) . The cover plates w e e si7ed to produce the required additional section area. In recent years, fabricating shops have simply substituted fillet welds for rivets and produced the same column section; Figure l ( b ) . This practice has presented a design problem in getting an efficient transfer of tensile force from the beam flange through the cover plate into the column without pulling the cover plate away from the column flange. The cover plate, being attached only along its two outer edges, tends to bow outward; Figure 2. This results in uneven distribution

FIGURE 3

FIGURE 2

of forces on the beam-to-column weld.

The best design is a completely welded built-up column, Figure l ( c ) . This gives the exact section required without any increase in welding, and there is no problem in transferring tensile forces from the beam flange through the column.

fillet welds are usually used. When their size becomes too large, they are replaced with some type of groove weld because iess weld metal is required.

FIGURE 5

FIGURE 4

For very large column sections, 4 plates can be welded together to form a box section; Figure 3 ( a ) . Sometimes a web plate is added to this box for additional area in the lower part of a building; Figure 3(b). Moving up the bnilding, the point is reached where this web plate can be omitted without changing the outer section dimeiisions.

There are two general requirements for the welds holding the plates of the columns together; Figure 4. a. The cntire length of the column must have sufficient welds to witlistand any longitudinal shear resulting from moments applied to the column from wind or beam loads; Figure 4 ( a ) . Notice at the left the rathcr h ~ wchange in moment along most of the colu~nnlength. b. Within the region where the beams connect to th,e col~imn, this longitudinal shear is much higher because of the abrupt cl~angein moment within this region; Figme 4 ( b ) . .41so the tensile force from the beam flange will be t r a n s f c ~ ~ ethrough d a portion of this weld. Thcse two conditions require heavier welds in the connection region. Varioris t n x s of welds are employed in fabricating: a. Fillet u;el& (Fig. 5) require no plate preparation. They can be madc to any size simply by making more passes. However, since the amount of weld metal varies as the square of the leg size, these welds can require a large amount of weld metal for the larger sizes. For nominal size welds (approx. 'h" to %"),

b. Bevel and Vcc groove welds (Fig. 6 ) require joint edges of the plate to be beveled, usually by the oxygen cutting process. On larger size welds, this additional preparation cost is offset by the reduction in weld metal required. AWS and AISC deduct the first %" of weld to compensate for any slight lack of penetration into the very bottom of the bevel joint, if welded manually.

FIGURE 6

c. J und l T groove u d d s ((Fig. 7 ) require the plates to be googed or machined. Machining is seldom used in thc structural field, although air carbon-arc gouging is becoming more popular. The J and U welds may not require as much weld metal as the bevel or Vee weld. AWS and MSG allow the full throat or depth of groove to be used.

FIGURE 7

groove. Here the cfiective throat ( t , ) will equal the throat of the groove ( t ) minus 'In", see Fig. 8 ( a ) .

hE 2-Partial-Penetration

/

I

Groove

~ " ' j b....a.~.>J

depth of leg sire of fillet weld

a

FIGURE 8

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

1

of force'

7 = 13,600 psi I i = 9.- 6 0 0 ~ ~

= 15,800 psi !=11,300" 7

PARTIAL PENETRATION GROOVE WELDS

sheor

1r,

13,600 pi

##tension transverse t o axis o i wcld

o = 13,600

tension poiallel to axis of weld

1 some or piote

! 1

=

15,800 psi

1=

15.800 psi

i

I o

scme or p o r e

COMPLETE PENETRATION GROOVE WELDS tellion compression bending

I

some o i piate

1

same oi piote

'

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

force

-

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

Groove

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

( coiurnn carinecfion

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 ) .

no

Additlono! beveling in region of beom to (column COnneCTiOn

Region of beom to connection

,,,column

Weld o

FIGURE 12

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.

FIGURE 13

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

FIGURE 14

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,)

FIGURE 15

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.

FIGURE 16

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&

+I

along

angle

+,

along

angIe

4,

along

GI-@

@-@ @-@

&

@-@

@-@ = J

@-m -e az + 36 t2 a

or distance

5-t A 3 t

and sirice aZ

+ 36 tz

@--a- \/ -.

a'

A

+I=

+2=2+1=-

Distance

\/

a 6 t

a2

4- 36

t2

Now find the angle changes (+) along the hinges at ultimate load:

0-0, 0-@, @-a &

With reference to Figure 17:

= -6 t a

or distance

+ 36 t2 Sectiori x-x

FIGURE 17

p Columns

, ;

,$,=

+3

A + +,, = 6 at

I'

allowable force

&Ctt3;-;;~-

e x t e n d work = internal work

internal ~ o r k ..-

= M. [ m , 2 ( 2 a + b ) + ~ b + m , * d a ~ + 3 8 t ' ] A Applying a load factor of 2, and using the yield strength (u,),the allowable force ( P ) which may be applied to the plate would be-

Example where the plastic moment (M,), in in.-lbs/liuear inch is-

Here:

t = 31/2'' a = 5 "

b = 14" u = 22,000 psi

T t

calculated tensile force on beam flange = 386 kips The allowable force:

FIGURE 18

- (3%)(38 kips/sq in.) 12

external work

= P" A

FIGURE 19

= 1178 kips

>

386 kips

OK -

Equitable Life Asmrcmce Building Colurnns for tlm Equitab1.e Life Assmxnce building in Sun Francisco, an earthquake area, were built and erected in 3-story lcngtl~s. The columns were uniformly tapered :$$, in./ft from the base to the 14th story.

Exterior columns started with a 42" web at the bottom, tapering to a 1.2" web at the 14th story level; Figure 20. Flanges were 18" X 3" at the h e . The tapered columt~swere fabricated by welding two flange plates and a web together. L-shaped columns were used at the corners of the building. C.IL. House The 32-story C.I.L. House in Montreal, Caiiada has the heaviest TI"section columns ever constructed. The fabricated columns weigh as much as 2,000 lbs/ft. A t y p i ~ dcolumn, Figure 21, consists of two 7%" X 28" fiange plates welded to a 5" X 16%" web plate.

FIGURE 20

FIGURE 21

FIGURE 22

Automatic submerged-arc welding was used in fabricating these columns; Figure 22. Simple continuous fillet welds of about 3/4" leg size join the column flanges to the web. Because of the greater forces $thin the beam-to-column connection region, these welds were increased in size by beveling the web. The depth of the bevel for this double beveled T-joint varied with the forces to be transferred, hut ranged from a minimum of Yz" on each side of the web up to 100%. Less than 10% of these groove welds required 100% beveling. The grooved joints extended in length slightly above and below the depth of the connecZing beam and ranged in length from 2' to 5'. Joint preparation involved beveling with oxygen cutting equipment at a 22" to 30" angle to the correct depth. After tacking the Range to the web, the weldor lightly air carbon-arc gouged the bottom of the joint prior to welding to open it up for the root pass; the result was a modified J-groove. The columns, 2 stories high, range from 22' to 34' in length. Flange and web plates were clamped in heavy fixtures to maintain proper alignment during welding; Figure 22. After tack welding, trunnions were

Designing Built-Up Columns

/

3.611

FIGURE 23

attached to the column ends so that all welds could be deposited in the Bat position. The columns with trunnions attached were then transferred to the automatic welding unit. After preheating to the correct temperature, using natural gas torches, the shorterlength groove welds were made first. The remaining length of unwclded ailurnn was then fillet welded. After welding, trumions were removed and the column ends machine facod to proper length. Connection plates were attached after machining, with most weids positioned downhand to achieve maximum wclding spced. Preheating preceded the manual welding of these plates in position, using low-hydrogen electrodes; Fi@re 23. Inland Steel Building & N o ~ t l r Carolina h7alional Bunk Building

Elimination of interior colnmns in a building designed for wclded coiistruction is not unique, but

FIGURE 24

nsually reqnires the design and fabri~lltionof special colnrn~~s; Figure 24. The ~ o l u m ndesign on the right was used in the Inland Stet:] Bidding in Chicago. The inner portion of the built-up colt~rnnis a standard WF section; the outer portion is a flat plate from 1" to 3" thick. A web plate, From %" to 1'W thick, joins thcse tw-o segments. Notice that a section of the main girder was shop welded to the fabricated column. Dotted lines show the spandrel bean~sand remainder of the girder that were fidd welded to prorluce a rigid vonnection. The main ginlrrs span 60'. On t t left ~ is n typic;il column from the North Carolina National Rank B~~ilding in Ch;dotte. A spccially rolled V\'F sertion is tlic innin s q p e n t of this column. Wing plates have bix+*rriaddrd to one flange and a mvcr plate to the othel- to d w d o p the necdvd ~.olrimn proportics. The m;iin girrl<~-sand spandrels (dotted sections) were later att:~chedby field welding.

3.612

/

Column-Related Design

Fabrication of special colnmn seetiom demand low cost, high production assembly and welding techniques. Submerged-arc automatic welding is uscd extensivcly in fabricating these columns. The welding head, Figure 25, is mounted on a universal, track traveling type welding ma~ripuletor.The manipulator,

ifux recovery nnit, and welding generators are mounted on a self-propelled carriage having a G5 ft track travel distance. Two identical welding fixtures are positioned parallel to and on either side of the carriage track. This has rcduced handling time for setup and repositioning of the columns. During fabrication of columns for the North Carolina National Rank Rnilding, they were placed in a specially designed trunnion fixture; Figure 26. This stood the columl~son end. Shop welding of connection dotails could then be performed in the fiat and horizontal position. This, facilitated use of semi-automatic, submerged-arc welding and minimized weld costs.

Commerce Towers Building

FIGURE 26

Columns of similnr section configuration were used in the 32-floor Commerce Towers Building in Kansas City. Here, heavy floor loading due to the modern electronic business machines to be installed necessitated very heavy sections. Column sections were built up by first welding plates into an I section and a T section, and then joining th'e end of the T section web to the middle of the I section web. The typical column length is 34' and the lower columns use 5" Aange plates and 5" web plates. Tanden-arc automatic sobmerged-are welding was used in joining the Aange plates to web; Figure 27. The basic weld was a Yz" fillet deposited at 32-36 ipm. Preheat torches ran ahead of the arc. In joining together the I and T sections, they are assembled in an air-clamping fixtnre and tack welded; Figure 28. Automatic submerged-arc welding is then used, with the fixtwe on a rail-mounted carriage.

esigning Built-up Columns

41/811

On this project in Detroit, Michigan, the engineer originally detailed the fabricated columns to the 17th Soor as built-up box sedions, flush around the outside periphcry. U-groove welds were to be used; Figure 29(a). This would have meant grooving the platcs for the entire length of the column. Tile falxicator, chose to set one set of plates slightly in or out; Figurc 2 9 ( h ) . This w-odd allow use of continuous fillet welds for the basic welding. The fabricator obtained pcrmission to exceed the original outside column dimension in one direction by '1'4''. Any further adjustment was prccluded because of the already deiailed curtain walls, etc. The original outside dirncnsions of the columns were 18" X 22" to the 5th floor, 18" X 20" to the 11th floor, 18" X 19'' to the 13th floor, and 18" X 18" to the 17th floor. Above the 17th floor, W F sections were used. The modified box section on the lower floors were then built u p from two 18%" X 4S/s'' flange plates, with two 12%'' X 4%" web plates recessed slightly to permit tlie fillet welding. Above the 5th floor, the

3.6-13

FIGURE 28

FIGURE 27

First Federal Savings b Loan Co. Building

/

444

% L

smaller plates were set out slightly. In general, these full-length welds were 'h" fillets; with %'' fillets for plates 2%" or less in thickness. This eliminated plate preparation except for short distances in the region of the beam-to-column connections. Here the plates were previously beveled, to the required depth, varying from 3/8" to 5/1,j'' depending upon load requirements. The typical joint consisting of the beveled groove weld topped by the continuous fillet weld extended 9" above and below the beam-to-column connection.

. FIELD

SPLICES

Partial-penetration groove welds; either single bevel or single J, may he used for the field splicing of columns. The information presented previously under ''PartialPenetration Groove Welds" will apply here. Attaching angles shop-welded to the coh~mns serve to temporarily hold the column sections in alignment. For the II colum~i in Figure 30, using high tensile bolts, this connection was considered sufficient to transfer any horizontal shear force across the

3.6-14

/

Column-Reloted

BUILT- UP

BUILT-UP COLUMN

COLUMN

I I

I I

2. & under

BUILT- UP

BUILT- UP COLUMN FIELD SPLICE

COLUMN FIELD 5PLICE

LA FIGURE 30

FIGURE 31

web in this dir<,ction. Tlx. colnmn field splice, consisting of two sirrglr bevel. partial-penetration groove welds, wonld transfer any horizontd shear in the other direction.

For the box colnmn in Figure 31, the column ficld splice consisted of a partial-penetration J groove weld on all four sides of the column. These four welds would transfer any horizontal shear in the column splice. The attaching angles here were used simply to facilitate erection. Partial-penetration welds on colornn splices pemut fast semi-automatic welding techniques to be used in the field. In the Commerce Towers project, semiautomatic arc welding with self-shielding, cored electrode permitted dqxxition of 100 lbs/mam/8-hour day; Figure 3%.

. COPICLUSION The full econoinic impact of welded steel built-up columns in constn~ctionof t d l multi-story btiildings, can be realized by carefully considering the major cost factors. These are colnmn design, placement of welds, joint design, weld size, and procedure. The dominating objective is the fullest use of automatic arc welding lncthods in the shop, with an extension of these henefits into the field l ~ yusc of semi-automatic arc welding for beam-to-column connections and for field splices.

FIGURE 32

Designing B u i l t - U p C a l u m ~ s

Built-up columns are a key design feature of the 28-story Michigan Consolidated Gas Co. Building in Detroit. Welding was considered to be the only procticol method for fabricating these columns which carry a maximum load of approximately 6800 kips. Photo shows a field splice of the column, revealing the shop beveling that facilitated welding. Clip angler shown are for temporary use during erection.

Typical splice

for builbup column

Alternate splice

Typicof splice

for built-up ralumn

for WF colvmn

Splice details from the Michigan Consolidated project show how maximum use was made of material at minimum weight.

/

-1 5

/

Column-Refated Design

Automatic submerged-arc welding was used extensively i n shop fabricating the unique and complex built-up columns for the 500' space tower which overlooked the Seattle World's Fair. Approximately 50% of oll shop welding was with the submerged-arc process: 25% with selfshielding cored wire, semi-automatically; and the remainder manual stick electrode. At the top of the tower is a fivestory observatory and restaurant, The structure required 3400 tons of structural steel.

Therefore, the required flange area isPlate girders arc fabricated for requirements which exceed those of a rolled beam, or a rolled beam with added cover plate. The usual welded plate girder is made of two flange plates fillet welded to a single web plate. Where needed, web stiffeners are attached to one or both sides of the web. Box girders are made of two Range plates fillet welded to two web plates. Internal stiffening of these is accomplished with diaphragm plates. The flange-area method is used to get an approximate dimension of the girder. This assumes that the flanges will carry all the bending moment and the web will carry all the shear forces. The required web area is-

where:

V

-- vertical shear

applied to cross-section to be

considered T

= allowable shear stress on web section

The formula for required flange area is derived froin properties of the girder:

where.

\

M = bending moment applied to section u = allowable bending stress d = distance between centers of gravity of flange plates This method will require some approximate knowledge of what the girder depth should be an+ some adjustment of the resulting figures before the design is finalized.

The previous AISC specification held the depth of girders to a minimum value of 1/24 of the span. The Commcutary on the new AISC specifications suggests, as a guide, that the girder depth should not exceed the following: Floors: Roof purlins:

u,/ 800,000 times the span

/ 1,000,000 times the span

IT,

This translates into the Table 1 limiting values of depth-to-length lor girders used in floors. These values are for general guidance only.

For simplicity, this assumes web depth is equal to ( d ) , the distance between the centers of gravity of the two flange plates.

Also, * Q u e n c h e d & iempcred steels: Yield strength ot 0.2% ofiret.

Camprerrion elements which ore not " i o n ~ p a c t ' but meet the ioilowing AlSC Sec 1.9 requirements-

box girder tension

(1.5.1.4.31

a = .60 o, ~~~

compression

.-

.

(1.5.1.4.3)

. ..

(AiSC Formula 5)

(AiSC Formuio 4) Use the larger of if

L -<

@ or @ but not @

to exceed .60 a,

40, don? need to w e

reduction i n o l o w a b l e compressive bending stress due to possible ioteroi displacement of web. (1.10.6) d < when 2~ tw

24,000 cry

s,,= oliowoble cornprerston stress from obave

(AiSC Farmulo l li this limit, is *?his ratio may be cxweded i f the camproiaive bending stress, using a width not within the oi!owable stress, i n e above toble does not include tho higher bending rtreir to = .66 <,I for ":ompart" sections because most fabricated piate and box girders will exceed the widtli-thickness ratio of ' ' c o ~ O C ~ C ~ 5ections. ''

ending Stresses

Table 2 suinm;irizes tho AISC allowable bending stresses for plate a n d box girders. I

In Table 2: L = s p a n or iinlxaccd length of compression flange r = radius of gyriitiml of ;I Tw section comprisin7 tlic cornprcssioir flarigc plus 1/6 of the ~ ~ ;rrea, 1 3about the y-y axis (in the plane of thc we11). For girders symmetrical about tlieir I-x ;%\isof hending, substitution of r, of ilrc: enlirc section is conse~ative At = area of the compression flange

o-,

= aiiowable compressive bending stress from nhn.,a ---""

MI is the smaller, und Mz is the larger bending moment at tl,e ends of the onbraced length ( L ) , taken about the strong axis of the member. M1/M2 is the

ratio of these end moments. When MI and Mz have

Bme Girders for

the same signs, this ratio is positive; when they have different signs, it is negative. When the bending moment within an irnhraced Imgth is larger than that at both nids of this Icngth, the ratio is taken as imity. Figrire 1 is a graph showing t11c valrie of C,, for any given ratio of MI/M2. When the bending moment within an rillbraced length is larger than that at both ends of this length, the ratio shall be takcn as unity, and C,, becomes 1.0.

Loads applied to beams and girders cause bending moments dong the Icngth of the member. When these moments are non-uniform along the length of the member, both horizontal and vertical sliear stresses are set u p because shear is equal to the rate of change of moment. The horizontal shear forces worild cause the f l a n ~ e of a platc girder to slide past the web if it were not

for t h e fillet welds joining them. Thrse liorixontal and vc:rtical sllear st~rssescombine and prodwc both dingo~ral tcnsiori and compression, c;wh at 45" to the shmr strcsscs. i n steel structr~res, trmsion is not thc problrrri; howev-er, the diagonal eomprcssion could be high enough to cause the wtkb to bwk1,c. StiRmrrs arc used to prevent the web from buckling in ribgions o i high shear s t r ~ s . The ratio of wch thickness to clear depth of web in the oldrr spccificatioiis \rxs bascd on predications of file plat^, buckling tlrcory: tire wch being subjected to shear throughoi~tits daptl~,and to rompressi\,e beridi i ~ g,iiesscs o w r a portion of its depth, See Figure 2. The plate buckling tlrcory assumes the portion of the web 11etwt~~ii stiil'c~l~rs to be an isolated plate; l~owovcr, in thr plate girdcr, the web is part of a built-rip n~emher.When tllc critical buckling strcss in the wcb is rcacbcd, the gilder does not collapse. This is because the flangcs carry all of the bendirrg moment,

1 -----w-

Diogonol compresson f i om sheor forces

FIGURE 2

,C . ,... ,-.,--., . -,...-. Compresswe

bending sireis

FIGURE 3 Transverse stiffeners act or compression struts

the buckled web then serves as a tension diagonal, and the transverse stiffeners hecome the vertical compression struts. This in e%ect makes the plate girder act as a truss. See Figure 3. The carrying capacity of the plate girdcr is greater under this analysis, being cqual to that supported by the beam action shear (Fig. 2) and that supported by the diagonal tension field in the web (Fig. 3 ) . AISC Formulas 8 and 9 will meet this requirement. These formulas appear further along on this page. ABSC Specifications

Intermediate stiffeners are not required when the ratio (d,/t.) is less than 260 and the maximum web shear stress is less than that pcimitted by AISC Formula 9 (AISC 1.10.5.3). Figure 4 partially s~.immarizesthe AISC specifications for intermdiate stiffeners. These requirements apply: 1. If single stfillers are used, they must be welded to cumprcssion flange (AISC 1.10.5.4). 2. Intermediate stiffeners may be cut short of tension flange for a distance less than 4 t, when not

Cut short of tension flange 4 t,

<

needed for bearing (AISC 1.10.5.4). 3. For intcmmittent falet welds, clear spacing ( s ) between lengths of weld must L 16 t , and L 10'' (AISC 1.10.54). 4. Welds joining stiffeners to web must be SUEci.ent to transfer a total unit shear force of-

f, = d,

(AISC 1.10.5.4)

This shear force to be transferred may be reduced in same proportion that the largest computed shear stress ( T ) in the adjacent panel is less than that allowed by AISC Formula S (AISC 1.10.5.4). 5. If lateral bracing is attached to stiffener, u d d s connecting stiflrner to colnpression flange must be SUEcirnt to transfer a horizontal force ( F ) = 1%of flange force (AISC 1.10.5.4). Wlreri intermediate stiffeners are required, their maximunr spacing ( a ) depends on three items: a/&, d,/t,., m d shear stress ( 7 ) . The largest average web shear stress (T,, = V/A,) in any panel between transverse intcrmdiate s t i f h e r s shall not exceed the following (AISC 1.10.5.2):

f, = d,

FIGURE 4

late Girders for

<

when C,

1.0

which you will notice is the same as

5 0 a,C,

r

or (.347 a?C,). The expression (.60 u,) is recognized

. . (3a) This provides an allowable shear stress ( T ) up to about .35 u, and takes advantage of tension field action. when 6,

> 1.0 or when no stifenem

are used

This provides an allowable shear stress ( 7 )within the range of ,347 o; to .40 o; and does not take advantage of tension field action.

as the basic allowable tensile stress and

," -,

as ( 7 ~ ) .

For greater depth to thickness of web (d,/kV) and greater stiffener spacing (a/d,), the values of (C,) will become lower. Thir will result in lower values for the allowable shear stress in the web. For these conditions, AISC Formnla 8 has an additional factor which takes advantage of the increased carrying capacity provided by the diagonal tension field and results in a higher shear allowable. When C, = 1, this factor becomes zero and AISC Formula 8 becomes Formula 9. The ratio a/d, shall not exceed (AISC 1.10.5.3):

nor

where: a = clear distance between transverse stiffeners, in. d, = clear distance between flanges, in. t, = thickness of web, in. a, = yield strength of girder steel, psi when C,

<

when C,

> .8

.8

when a/d,

< 1.0

when a/d,

>

1.0

Above, the one C, formula picks up exactly where the other leaves off. The value of C, may be read directly from the nomograph, Figure 5, without separately c~mputingthe value of k. Both ASIC Formulas 8 and 9 contain a basic factor

These arbitrary values provide a girder which will facilitate handling during fabrication and erection. When a/d, exceeds 3.0, its value is taken as infinity. Then AISC Formula 8 reduces to AXSC Formula 9 and k = 5.34 (AISC 1.10.5.2). This work can be greatly simplified by using the appropriate AISC Table 3 for the speci6c yield point of steel. See AISC's "Specification for the Design, Fabrication and Erection of Structural Steel for Buildings" and Bethlehem Steel Corp's Steel Design File on "V Steels-Recommended Allowable Stresses for Building Design." In end panels and panels containing large holes, the smaller dimension ( a or d,) shall not exceed (AISC 1.1053)-

where web:

T

is the computed average shear stress in the

i t is necessary that the stiffeners have sufficient cross-sectional area for them to act as compressive struts to resist the vertical component of the tension field in the web. This cross-sectional area, in square inches, of intermediate stifFeners when spaced in accordance with

0

----------arge o e

one w ,

FIGURE 6

= allowable web shear stress from AISC

AISC Formula 8 (total area when in pairs) must not be less than (AISC 1.10.5.4)-

. T .

Fonnnlas 8 or 9 u,,== -

allowahle bending tensile stress

It can he shown that this formula will result ina ) full bending tensile stress allowable, if the concurrent shear stress is not (Treater than 60% of the full allowahle value, or b ) full shear stress allowable, if the concnrrent bending tensile stress is not greater than 75% of the full allowable value. See Tahle 6B €or abbreviated Fonnula 12 l o w e for a specific yield strength of steel.

(7)

-

(See the appropriate AISC Table 3 ) where: yield of web steel y = _ --..point ----

y~eldpoint of stiffener steel

D = 1.0 for a pair of stiffeners 1.8 for a single angle stiRen'er 2.4 for a single plate stiffener When the greatest shear stress ( T ) in a panel is less than that permitted by AISC Formula 8, this area (A,) requirement may be reduced in like proportion (AISC 1.10.5.4). The moment of inertia of a pair of stiffeners or a single stiffener, with reh:rence to an axis in the plane of the web, shall not he less than (AISC 1.10.5.4)-

Concentrated loads cause high compressive stress at the web toe of the fillet along a distance of N K for end reactions, and N 2K for interior loads. If there a r e a 0 boaring stiffeners, this compressive stress shall not exceed (AISC 1.10.10.1)-

+

+

for cnd reactions

+

t,(N K) = ' (AISC Formula 14)

-

See Tabies 3, 1,and 5. Plate girder webs, subjected to a combination of hending tensile stress and shear stress shall be checked according to the following interaction formula:

for interior loads

n < 75u, t,JN 2K) = ' ( AlSC Formula 13)

+

where: T

v

= computed average wcb shear stress = -A,

. . . . . . . . . . . (10a)

. . . . . . . . . . .( l o b )

Also, the sum of the compressive stresses from concentratsd and distributed loads on the compression edge of the web plate not supported directly by bearing stSeners shall not exceed (AISC 1.10.10.2)-

ABLE 3-Minimum

Moment 04 Inertia of intermediate Stiffener

f

i d,

I.

d,

,.

= > p14 50

I.

d,

I.

d,

I.

d,

1,

oment of inertia of

-11t-

n-i-

ese Volues tor St Sides 04 Girder

Width

of bor (d)

t B( Inertia of Sing

se Vatu'ues tor St Sides BP Girder

1

Angle size

,,, 1

1/*11

1

Thickness of ongle rtiffener0)

j

1

1

y2

1 x 1

3j~-

(

XP

1

1/4"

FIGURE 7

if flange restrained against rotation

if JEange not restrained against rotation

Concentrated loads and loads distributed over a partial length of panel shall be divided by either the product of the web thickness and the girder depth or the length of panel in which the load is placed, whiehever is the smaller panel dimension. Any other distributed loading, in lbs/iinear in. of length, shall be divided by the web thickness. If the above stress limits are exceeded, bearing stiffeners shall be placed in pairs at unframed ends and at points of concenixated loads, Figure 8.

12 t* [or less]

Bearing stiffeners with the above sections of web are designed as columns (AISC 1.10.5.1). These requirements apply: 1. Bearing stiffeners shall extend almost to edge of flange (AISC 1.10.5.1). 2. Bearing stiiFcners shall have close bearing against flange or flanges to which load is applied (AISC 1.10.5.1). 3. Clear spacing of internittent fillet w e < 16 t, < 10'' (AISC 1.10.5.4.). 4. Deduct leg of fillet weld or corner snipe for width of stilfenev (b,) effective in bearing at 90% o; (AISC 1.5.1.5.1). If parts have different yield strengths, use the lower value. 5. The limiting ratio of stiffener width to thickness shall be-

b" ts

--

-

3300

s(AISC 1.9.1)

6. Use I,, 2 3/4 dw for slenderness ratio (L,/r) of coltrmn section to determine allowable compressive stress (AISC 1.10.5.1); r is figured about an axis in the plane of the web.

4 25

tw [or less)

[or less) [a) Single pair of

st~ffenersat end

[b] Single pair of stiffeners - interioi

(c) Double pair of stiffeners - interior

FIGURE 8

(or less)

(d) Double pair of stiffeners at end

FIGURE 9

a = area of flange held by welds

If intermittent fillet welds are used in plate or box girders, their longitudinal clear spacing shall not exceed-

y = distance between center of gravity of fiange area held by welds, and neutr a1 axis of entire section 1 = moment of inertia of entire section

tension flange (AISC 1.18.3.1)

n = number of fillet welds holding flange area, usually 2 welds

(12) compression flung8 (AISC 1.18.2.3)

Table 6 summarizes the principal AISC specifications in (,asy to use form, permitting direct readout of the limiting value for the specific yield strength steel being used.

The longitudinal shear force on fillet weld hetween flange and web is-

f = V a y Ibs/linear in. I n where: V = external shear on section

Design a welded plate girder to support a 120-kip uniformly distribnted load, and a 125-kip concentrated load at midspan; Figure 11. Girder is to be simply supporkl, have a span of 50', and have sufficic.nt lateral support for its compressive flange. Usc A36 steel and E70 or SA-2 weld metal.

Plate Girders Cor

125 ki r s uniformly distributed

FIGURE 11

L = 50' = 600"

bending moment for the unifoim load,

shear V = 122.5 kips Design Procedure

for the concentrated load,

1. Design the girder web for the shear requirements, assuming it held to a depth of 66".

= 97,750 in.-kips

Total M

c

* Quenched

/

ACSC Formula

& tempered ileels; yield brength ot 0.2% offset.

4

1

(1.5.1.4.5)

A15C Formula

12

(1.10.7)

i r = averoge shear sties in web

v

=-

A,

irder-Rekted Design

/

Consider the following average shear stress ( T ~ , ) and maximum panel length ( a ) for various web thicknesses (t,):

remaining moment of inertia required of flanges

If = I, - I= = (44,880) - (7487) = 37,393 in.4 and since

If = 2 Af cr2

c, = 33''

+ W'

area of flange required Although the Y4" thick web would result in a reasonable shear stress of 7430 psi, the greatest stiffener spacing ( a ) allowed would be 97% of the web depth (d,); this would require more intermediate stiffeners. It would be more practical, in this example, to increase the web thickness to x6", thus allowing a greater distance between stiffeners.

= 16.67 in.2 or use two 17" X I" flange plates. final properties of girder

I = 2 (17 in.=)(33.5")2 = 46,766 in.'

2. Design the flange to make up the remainder of the moment requirements. Assume a bending stress of about cr = 21,000 psi.

>

= 1375 in".

+ (%6")(66")3 12

44,880 i n 4 OK -

1320 im3 OK

-

actual bending stress in girder

section modulus required of girder

-

(27,750 in.-kips) (21,000 psi)

= 13W in.3 distance from neutral axis of girder to outer fiber assuming a flange thickness of about 1" 6

= 'h d, = (33") = 34"

= 20,200 psi reduced allotuahle compressiue bcnding stress in jiange due to possible lateral displacement of the web in the compression region (AISC 1.10.6) U L ,

+ ti + (1")

total moment of inertia required of girder

It=Sc

= (1320)(34) = 44,380 in.4

5 -

21,347 psi

>

20,200 psi actual OK -

where:

u,,= allowable bending stress - .60 o;

- 22,000 psi

=

V,

FIGURE 12

122.5 kips

kd Shear diagram

3. Design the Wansverse intermediate stiffeners. Figure 12 is a shear diagram of the girder.

Since the girder web's ratio is-

end panel distance between intermediate stiffeners (AISC 1.10.5.3)

and the ratio of panel width to web thickness is-

= 211

d,/t,

the maximum allowable shear stress ( T ) to be carried by the girder, web and the total area of stiffener (A.) to resist this shear are found from Table 3-38 in the following manner:

45.6" or use 45"

nzazimum shear just inside of this stiffener V = (12.2.5 kips

- 62.5 kips)

)

+ 62.5 kips

= 155.6 kips maximunl spacing between remaining intermediate stiffeners (AISC 1.10.5.3) Within the above limited area of the 1;uger AISC table, the values in the four corner cells are read directly from the AlSC tal~lc.Then the rerpirt,d values obtained by interpolation ax: filled into (he center cell. Within each cell, the upper value is the allowable shear stress ( 7 ) and the lower value is the required area of stiffener (A,). Thus, for our problem: T

required number of panels WO" - 2(45") = 510"

so use 6 panels of a = 85" each.

check the allowable shear stress in the web and determine required area of stiffener

= 8.0 kips or S O N psi > 5950 psi OK

width of stificner (if using t, = 3h")

-. -- (2.16)

2(?18) . .

= 2.88" or use 3?i"

I

Since: A,

-

2bS t,

/

4.1-16

Girder-Related

wherever the ealciilated shear stress exceeds 60% of that allowed according to AISC Formulas 8 and 9. The allou-able shcar stress was found to be T = 8000 psi and 60% of this would be 4800 psi. ~This would correspond to a shear form of

also check AISC Sec 1.9.1:

b. t,

3% -%

V =7 A ,

required moment of inertia

= (4800 psi) (%, X 66) = 99.0 kips and would occur at x = 125". The bending moment at this point is-

actual moment of inertia

I, =

(2

x

31%''

+ %,")""

-

12

and the bending stress is4. Determine the size of fillet weld joining intermediate stigeners to thr girder web.

unit shear force per h e a r inch of stiffener

13,750 -- --

in.-kips

1375 in."

= 10,000 psi

or f, = 1140 ibs/in. for a single fillet weld (one on each side). leg size of fillet weld

= ,102" or use "./I6 or, for a % = -

"/10"

I t is only when the shear stress exceeds 60% of the allowablc that the allowable bending stress must be reduced according to AISC Formula 12. Since the calculated bending stress at this point (x = 125") is only 10,000 psi or 45% of the allowable, and it rapidly decreases as we approach the ends, there will be no problem of the combined bending tensile stress and shear stress exceeding the allowable values of AISC Formula 12. 6. Determine the size of fillet weld joining flanges to the girder web, Figure 13.

,r

cont~nuousfillet intermittent fillet weld

.102"

9{#''

[\

= 58.8% or use v 3 . 5 or or, for a 3'4' intermittent fillet weld

--TJ&-

5. Check the combined bending tensile stress and shear stress in the girder web according to FIGURE 13

(AISC Fonnula

12)

FIGURE 1 4

force on add

portion of i ~ e bacting with stiffeners to form column ,

-

.

( 122.5 kips ) (17 i n 2 )(33.5") (46,776 h 4()2 welds)

= 746 lbs/in.

17" FIGURE 15

leg size of fillet weld W

I-

= 746

11,200

= ,066" but because of 1" thick flange plates, use

-

Xo"

-

12 t,

12 (%,") = 3%''

Bearing Stiffeners

6. Check to see if bearing stifleners are needed at the girder ends (AISC 1.10.10.1); Figure 14. compressitje stress at web toe of girder fillet

= -

-

-

:= (3%") ( = 1.17 in.' required a r m of bearing stiffeners

R

t d y -t K) (122.5 kips) --~

~:h6(lo" + ~

= 34,700 psi

>

27,000 psi, or .75 uy

This stress is too high; bearing stiffeners are needed. Try a singlc pair and treat the stiffeners along with a portion of the web ns a (.ohinn. Assume an acccptable cotnpressive stress of about 20,000 psi. 7. Determine size of bearing stiffeners. sectional ureu required to cawy this stress

-

awn of this web portion

(122.*kips) (20,000 psi)

6.10

-

1.17 = 4.93 in."

If stiffeners extend almost the full w-idth of the flange, a wkltli of 7" will be needed on each side. A, = 2 (7") t,

= 4.93 in."

= ,352 or use %" thiclrness 8. Check stiffener proiile for resistance to compression (AISC 1.9.1).

?'Iris ratio is too hizh, so m e hearing stiffeners.

11

pair of 7" x 7/10"

9. Check this bearing s t i f h e r area as a coliimn; Figure 16.

force

f

on tceld (treufilzc weld us a linej

= -R-

L ( 122.5 ....-kips -- ) . .. (264")

leg size of fillet weld

11. Check bearing stress in these stiffeners.

beoring area of stiffenel. (less comer snipes) (7" -- 1") 7/;8" = 2.62 h2each

FIGURE 16

bearing stlass in stiffener

= 106.8 in.'

(122.5 kips) -. 2("62) = 23,400 psi < 27,000 psi or .75 u? ...-

A = (7$6)(14%0) -I- ( 3 7 4 6 ) ( % e ) = 7.3 in."

OI(

12. In a similar manner, cheek the bcaring stifEencr at cciiterliire for resistance to 125-kip load. I f irsing the same s t i f h e r size as at ends, Figure 17:

slorderncss ratio & __ 3h(66") r (4.6") '

= 10.6

allowable comprcssivc stress u =: 21,100 psi, from Table 6 in Section 3.1 and R = u A = (21,100) (7.3) = 154.0 kips > 122.5 kips actual OK

-

1.0. Determine tlie size of fillet weld joining bearing stiflencvs to tl-ic girder web.

length of weld

L := 4 d, = 4 (66")

= 264"

FIGURE 17

= 106.8 in.' A = (7/,,")(14%0") = 8.56 in.'

+ (7.8"

-

7/16")(%0")

/

elded Plate Girders (or Buildings

4.1-1

Uniformly distilbutrd loud of 120 kius

FIGURE 18

rillolonblc r:o?npressice stress ugninst web edge assiiming flange is not restrained :ig:rinst rotation

nllotcablr compressitje stress

v = 21,000 psi, from Table 6 in Section 3.1

-C .- 990 psi

and

rictual presszrrc of uniform loiid against web edge

F = u A

= (21,000) (8.56) = 179.5 kips > 125.0 kips actual

=

OK

:= 640 psi

so use the same ainouiit of fillet welding as before. heriring stress in center

stiffener

F u = 4

.-

(125 kips)

2(7"

(5,") < 27,000 psi

- 1")

= 23,800 psi

or .75 u,

(120 kips) . (600") ( 24,")

.

OK -

13. C11ei:k the compressive stresses from the uniforirily dis~rihi~tid 1o;id of 120 kips on thc comprwsion edgr of tlw a.eb pint? (AlSC 1.10.10.2). See Figure 18.

lntermediote stiffener

OK -

990 psi allowable

14. Consolidate these findings into the final girder design, Figure 19. As a matter of interest, rcducing the web thickness to Yn" would have saved about 143 lbs in stml. I-Iowwer, this would have required 13 pairs of stiifeiiers instead of 9 pairs, Figure 20. The additional cost in fitting and welding the extra 4 pairs of stiffeners probably would exceed any savings in steel. Increasing tbc web thiclmcss to %" would only rednce tlre iiuinber of stilfeners by 2 pair, Figure 21. However, this would iiicre:ise the weight by 287 lbs.

Bearing stiffeners 2 - R 7 : ' ~ V,&,' Bearng sttffeners

<

125 kips

FIGURE 19

/

Bearing stiffeners

2 - R 7" x Vt;/ia"

intermediate stiBenei

1

66" X 36'' web 16" X I " flange

FIGURE 20

~nni~u -.

66" X %" web

I

Many tinrt.s access Irolcs must be cut into the wrbs of beams a d girders for dnct u.ork, etc. If snirrciently large, they must be reinforced in some manncr. Sinrr: the flanges carry most of tlrr bending forces, the loss of web arca docs not p w n t much of a problem. Howrver, sincc thc shear ( V ) is carried for the most part by the web, any reduction of web area must Be checked. See Fignre 22. If the hole is located at m i d s p o ( b ) , the shear is minimum and may have little cffcct on the strcngth of the girder. If the liolc is located near the support in a region of liigh shear, tfic additional bending stresses produced hy this s11c;tr milst bc added to the conventional bending stresses froin tile applied beam load. See Figure 23. An irrsidt: Iiorizontal f l a u g ~may be added to the Tee scction in (11-dm to give it sufficient bending btrength, or sufficient comprc~ssive buckling strciigtli.

i

I

FIGURE 21

When this is done, it must he rernem1)cred that this Range bemmes a part of the Tee arm 311~1is subjrctrd to tlie same axial tension ( F , , ) and con,prt:ssio~ (F,) force causcd hy tho bending mon-rvnt ( M , ) from the external loiiding. Tlievefore, tliis flange must extend iar monglr beyond tire web opcwing to effectively transfer this portion of tlic axial force hack into the main web of the girder; see Figure 24. Of course in the region of low inomflit ( \ 4 x ) , this iixial f o r c ~may he low a ~ r dnot req~iiretliis extra length of Range.

Applied load FIGURE 23

FIGURE 22

If t11cw ;ii.wss Iiolcs in tlic \rcb are close cnoi~gh togctlicr. the portion of the wrh between the holes beha\-cs iir tlw s;rii-re rriatiner as tlrc vcrtical mcmbrrs of a V i c ~ r r n d dtruss. Scr Figure 25. I'nlcss the bmding stress at the corner of the accrss hole is r;ltl~crlow, rt:inforcernctit of this corner sho111db(. consic!r:cd: 1. liccalisc 01' t l r ? a h n ~ p tchange in section, there scwral times the average stress is a stress co~iceritriiti~~~r valrie. Sec Figlire 76. 2. Tlw Tce scrtion at this inside corner behaves similar to a ciirved lxwm i n t1i;rt thc neiitrxl axis shifts in ton-srrd this i n r greatly increasing the bendiiig stressrs on this i~iward face. This increase is gre:itcr with a smnllcr r;?diiis of corner. 111 tlie us~ial aiialysis of a Vicre~rdeel truss, the horizontal slicar ( ) along the neutral axis of the

6.1-24

/

Miscellaneous Structure Design

A,

_

b ~~~~

(L-

~

log, -~11. 11,

11%)

STEP 7: Determine Properties of tlrf Elastic .4rea area of elastic urea

- i L

4-

1 = 200"

Topered beam

Moment of inertia

1

1 1 = c,

134.30"

Elastic orea

0 FIGURE 30

cornpl-ession must be checked against biickling according to AlSC l.
is 11ividt:d b c t w ~ ~these n two sections in proportion to tlioir depths. For Tccs of equal deptli, Vt = V,, - . 2 VI. The top ~ n bottom d Tee sections must be capable of withstanding this conihiiiecl bending stress, and the vcrtical slirar. A flange may be added arornld the edge of the web openi~rgto gi\.c LIie Tee section snfficient strength for the bending inomcxt. An aciditioiial plate may bo addcd to the \vch of the Tee to give it sufficient strengtli for the wrtical shear ( V ) .

"

7. COVER PLATES

FIGURE 29

If tlitx resnlting bcriding st]-ess in the stem is excessive, it must he rcinfort:ed by an insi(1e flangr

or stiffener. Cornim of the liolr slio~ildalways llc round m d snrootli. A ~~iininir~in cornel- r;idii~s of Y is recommeridrd \rIwn i l ~ ehole is not stifl'iwd lisirally it is assu~ncd Lhi: point of conti-:18cxnre of t l ~ arnmncnt in thr top and l~ottennpovtioirs prodiiced by tlir shear ( I ' , ) and (I?,,) is allout ~nitlst~tion of the Irole ( g ) . It is also assnmed t l ~ ctotal vcrtical shcar

It mi) bc ;idv;rntageous in some cases to use parti;il-lengtli cover plates in the beari~igregions of a beam or girder, to reduce tlie required thickness of the iiaiige plate extending from end-to-end of the mrcnher. Related disciissioli will hc ioiind forther along in this tmt iinder Section 4.3 on 1Vi:ided Plate Girders for 131-idges(sac Topic 12) ;ind iinder St.ction 6.1 on Design of Rigid Franics (see Topic 3 ) . 'The te,niiination of partial-lcngth cover platcs for Ix~ildings is govcme:d by I S ( : SIC 1.10.4. The foll o w i ~ ~l~u,-agr;il~lis g sunnn:~rin~ tlmc reqniremeiits. COVIY pli~tcssliiill crtcnd beyond the Pit1-tii~l-1~11gtli tlicorctic:11 nit-oil' point for ;i distmci~ a ) , di4inetl Iwlo~r.Tliis e~str~~itl(d pnrtinii ( a ' ) s11;iil he attaclicd to tlie h<.;ini or girtlev \vitlr siiffii.iw~t fillet wcids to d e i h p the uwri- pIati.'s pc~rtioiioS tho bending force

FIGURE 30

tote Girders for Buil

in the heam or girdel. at ihe theoretical cut-off point d r i c h is equal to-

s t r t i o ~(~a ' ) milst ),<' i~icnxsed,or the aciual end of t l ~ ccavcr plat<, rnlist ht, ~ ~ s t c ~ to ~ ~adpoint r d of lower momclit. The lcwgtl~ ( r r ' ) l l l ~ : l ~ r l ~ < Yfrom l the actual end of t l i ~cowr platc shall l1c: 1. A distnnco eipinl to the* width of the cover plntc when t l i t ~ tis~ a colrtiiir~orlsfillri weld i:qual to or larger tlmr 34 oi' tlw pl;itc, tliickitcss across the end I thr plate :ind (xmtin~tr,cl\\.<,Ids along hoth edges (11 the cover plate iu tlw Itygth ( i t ' ) , 2. A distancr r q u d to 1 % timrs the width of the w v c ~plate h i tiicw is ;I coiiiinnoi~s fillet weld smaller tlian 3h of tlw plirtr. thickness across the end of the plate and continnd \v<,ldsalong hoth edges of the (,over plate ill the lmgth ( a ' ) . 3. A disimce eyud to 2 tiines the width of the cover platc wl~enthere is no weld across the end of the plate but continuous wclds along both edges of the cover plate in the lengtll (a' j .

Q = statical moment of cover plate area ahout neutral axis oi covrl--plated beam section

I = nromel~t of inertia of cover-plated beam section The moment, coinputed by equating

gis

to the

L

c:ipaeity of the connecting fillet welds in this distance ( a ' ) fxom the actoal c.nd of the cover plate, must equal or exceed the moment at the theoretical cut-off point. Otbel-wise, the size of the fillet welds in this teiminal

?."+I

I

B e n d ~ r ,stress ~ from oppiied beam load

Top secton

Resuking bending stress

FIGURE 31

4.1-24

/

Girder-Related Design

M, I

=

moment ot n n r i end of teiminol develooment :f beyond theoietz

/ M,

=

moment at theoretical

I Moment dtagram development

cut-off point

6Theorelicol cut~offpoint t

J

I

I I

I

I I

I

8

I

I

J

f

If

If

terminal development starts at theoretical cut-08 point

inner end of teiminal development lies beyond theoreticol cut-off point

, End weld

,~Y -M----

-+F = \

f

=

x;i-,

..

I

Inner end of terminal development End weld - r F

I

=

Inner end of terminal development

I %Theoretical

M,a y

cut-off point

1

f

=

A+,-:

vw. 7

~Endweld+F=

Theoreticnl cut-off p o t i i t d - 1 !

Inne terminal developmen

FIGURE 32

h~4

Mi 0 y

7-

elded Blare Girder for Buildings

/

4.1-

elded Plate Girders $or

rrildimgs

/

4.1-27

/

Girder-Related Design

elded Plate Girders for Buiidings

/

4.8-2

4.1-30

/

Girder-Related Design

Access holes cut in girder web must be reinforced. In regions of high bending moment, flonges must extend far enough beyond web o p e n i ~ gto effectively transfer forces into moin web of girder. Semi-automatic welding, with self-shielding cored electrode wire, is used here in ottaching reinforcements at double the speed of manual welding.

Every plate girder must havc several properties: 1. Sufficient strength, as measured by its section moduins ( S) . 2. Sufficient stiffness, as measurcd by its moment of incrtia ( I ) . 3. Ability to carry the shcar forces applied to it, as measured by its web area (.4,). 4. Ability to withstand web buckling, as indicated by the empirical relationship of the web depth to web thickness-

,-

In some cases, the depth ( d ) must be held within a certain maximum value. Also, the choice of Aange and web plates should not result in any n11usua1 fabricating diEcuIties. An "efficimt" girder will satisfv all of these requircments with the minimum weight. An "econon~ical" girder will satisfy these same requirements and in addition will be fabricated for the least cost for the whole structure. This may not necessarily be the iowest weight design. Most structural texts sr~ggesta method of girder design in which some assumption is made as to the depth, usually from % , to I/,, of the girder length ( a rninimum of ? h 5 ) .Knowing the web depth, the wcb thickness is the11 found. This is kept above the value required for web area (A,-) to satisfy the shear forces and also to insure that the ratio K = d,/t, will be below the proper value. Table 1 lists the AASHO (Bridge) limiting values of K == d,/t, for common materials, with or without transverse stiffeners.

-

with any advantages of tlrc altered design, such as increased head room, less fill a t bridge approaches, ete. In order to simplify the derivation of the efiicient girder, it u-ill he necessa~yto assume the depth of the web plate (d,) is also the distance between the centers of gravity of the two Range plates as well as the overall depth of thc girdcr. Sec Fignre I. In the case of welded plate girders where the thiclmcss of flange plates is vory small compared to the girder's depth, this assumption doesn't introduce very much of an error while greatly simplifying the procedure and resulting fom~ulas. The moment of inertia of the girder section is-

S

I = --

+

A, d

dw3 6 K

01

S dW2 Af = - also d, 6 K

A, = t, d, =

A't.

2. DESIGN APPROACH It might he well to investigate thc efficient girder design on the basis of minimum weight. If done simply, it would offer a good guide or starting point in any design of a girder. An estimate of weight that is obtained quickly would allow the designer to deviate from the efficient depth to a more shallow girder when necess a y . He would then balance off the additional weight

-

d/2 -

dW2 K

I

I

Assume: dw

= d, = db

FIG. 1 Girder description

TABLE 1-Limiting

Ratios cf Web Depth to Thickness

=

d,

-

web depth . -

- web

II I

Mild Steci

I

A373, A36

i No tronrverre

KC60

stiffenerr

AASHO (Bridgar)

thickne3r

I 11 1

Low Ailoy Steel A441 or Weldable A242

/

46 000 pri held

50,000 psi yield

K

K - 5 2

5

53

/i

(1.6.80)

Longitudinal stiffener with ironweire rfiffeneri

Therefore, the total girder area isd,' 2 S At=2A,+A,"=---m+X d,"

Also, the total area of the girder isd,"

Now differentiate with respect to the depth (d,) and set equal to zero:

K

... I A , ]

. . . . . . . . . . . . . . . . . . . . . . . . .(4)

This indicates that the efficient girder has half its weight in the w& and half in the flanges. Based on steel weighing 3.4 lhs/linear ft/sq in. of section area, the efficient girder's weight is-

.(2) also

Figure 2 contains two curves showing the weights and depths of girders for a given set of requirements; in this case a section modulr~sof S = 5,000 in." Curve '4 gives the weight (Wc, ibs/lin ft) and depth (&, inches) of the girder for any given value of K. These two values come from Formulas 2 and 5 :

Since

S Af =--d,"

- 32Kd," d,

6.8 dWY and W, = ... K

dTY " G K

These combi~leto formd*-' GK

. . . . . . . . . . . . . . . . . . . . . . . . . .( 3 )

1-

. . . . . . . . . . . . . . . . . . . . . . . .(6)

which is the weight of girder not including weight of

Weight of e f i c i e n t girder for

different values of

K =

d

Z tW

m i n w e g h t for mclxmurn value of I<:

W,= 9.80

FIG. 2 Relationship of efficient girder weight and depth for given requirements (here, 5=5,000 in.").

70

80

90

100

110

120

130

Deplh of web ( d W jin.

stiffeners. It is seen that larger values of K result in lower weight (Wt) and increased depth (d,?) of girder. Conversely, lower val~iesof K will produce heavier and more sl-tallow girders. This represents the lowest weight design for any given value of K. Assuming the weight of stiffeners will be 20% of the web waight, and since in the efficient girder, the web represents half of the girder weight, the stiffeners would increase the girder weight by lo%, or-

\-

,.*.,...............,,,, (7)

which is the weight of girder including weight of s t 8 eners.

Effect of Changing Dinlensions In an efficient girder the depth of which is determined by Formula 2-

the weight decreases as the ratio (I<) increases; hence use as large a K ratio as is possibic (see Table 1). Once the flange area ( A f ) is determined, the actual profile

of the flangc (thicknms to width) has almost no .effect on the resulting girder weight (Wt). Occnsionaily the girder depth may be restricted because of head room or some othcr reason. The shallow-depth wcb thrii innst he thickcr in order to make UP the wch area required for the shear forces; in this case, it may he possible to further increase the web thicl;~less, \,cry slightly, to arrive at 1/60 of its clear depth and thus eliminate thc transverse stiffeners. If this is thc case, the decision not to use stitfeners should be made at the start of the design rather than later. For example, See Figure 3. Hcre on the left side, the efficient girder using stiffeners ( K = 170) \veighs 188 ibs/linear ft. Taking this same dcsign and incrrasing the web thickness to 1/60 of its dcpth to cliininate the stiffeners, would increase its weight to 328 Iibs/lincnr ft, or 1.74 times. On the other hand if the emcient depth is first determined using no stifimrrs ( K = HI), the weight is increased to only 243 ibs/linenr ft, or 1.29 times. In this particular case, the design which eliminated the stiffeners at the start (right-hand girder) weighs only 74% as much as the dcsign which eliminated the stiffeners after the dcpth was determined (center girder). The graph in Figure 4 show-s the direct effcct of changing web depth. Changing the combination of flange diniensions, but using same depth of web (d,)

Percent of e f i c i e n t depth used

(doid,)

FIG. 4 Effect of changing web depth on girder weight.

must he used. 2. For web thickncss, use

c

d,, 2

1 -

+ tf

3. Check the resulting values for

. . to use valrws of t, and d , that will provide the highcsf allowable valrre of I<. If resulting A , erjnals or excccds the given r q u i r c d value, procecd to Stcp 4 of Method A; if not; jump to Step 3A of Method H. T .I?

4. Kow compute the web's moment 01 inertia:

5. Select a flango tliicklress and wmpirte the distance from the cntire section's neutral axis to the outor fiber ( c ) , and tlrcn coinputc c,:

FIG. 5. Girder description.

3

=a,

SUE

Efficient PBaiie Girders

6. With this, compute the section's total reqnired moment of inrrtia:

7. Now select a flange width from the following: J Since:

and use the next larger corrvenicnt plate width for flange width ( b r ) . 8. Then c h e ~ k ip

I,

-- 2 1 1 ~tt cr2 and = I, + 1, and

Tliis final value of section modulus ( S ) must equal or exceed the value initially stated as a requiroment to resist the bending moment. overpls Design 08

Girder

If the xveb a r m (A,) cornpted hack in Stop 3 does not equal or exceed the givcn required arno~urt; take these addition;rl steps before proceeding with Step 4 of Mrthod A. 3A. Calculate the web thickness (t,) and web depth (d,>,) from the required web area (A,T) and rrrpired depth-to-tliickness ratio ( K ) , wing the iollowing formulas:

and

3R. Usi~ig this its a glide, adjnst the thickness ( L ) and drpth (d,, i of thr web plate to satisfy the ahove coiiditio~~s a r ~ dalso the following:

/

4.

a.hich rnttst q n a l or bc l c ~ sthan tlic maximum allowable \ d r w of K. Ha\irrg s<,lccttd d, i t i d t,v, n,tum to Step 4 of Metlrod 4 and follow t h r o ~ ~ gtoh completion (Step 8 ) . Short-Cut iVomographs

The first nomograph, Figurt: 6, will quickly give tlir girdcr's cflicimt u r b deptl~its wrll as its estimated wriglrt (lhs/lin f t ) On this r~ornograph: Line 1 =. r q u i r d section n~odrilus( S ) I.,inr 2 reqnired ratio of web (ltytln to web thicknrss ( K ) Linc 3 = (read:) rlficient web depth (d,) I h e 4 = required vatio of web & y t h to web thickrrrss ( K ) Line 5 = (read:) estimated weight of girder (W,) Line 6 = (rrad: ) d1~1wableshear carried by web ( V ) on the, basis of r = 11,000 psi (bridges)

-

l f the right-hmd line G shonld indicate an allowable shear value ( V ) for the efficient web which is lcss than the a(,tuaS value, thc girder design must he hascd on the shcar-carrying capacity of the web. This is done by going to the second nomograph, Figurt* 7: Here: Line 1 = actrial shear value which ~nirst be carried hy the usel.) ( V ) Line 2 = requii-cd vatio of web depth to web thickness ( K ) Line 3 = (red) we17 thickness to be used (t,,.) Line 4 = reqnirtd ratio of wab depth to web thickness (K) Lint: 5 = ( m t d : ) wvb daptlr to 11c used (d,) The weight of this slicar design may he estimated by the third noniograplr, Figure 8. Two valnes of weight are obtained; tlrc:c rnrlst be added together. Ilrre, for first weight: Lint l a = rrqnircd section rnodr~lus ( S ) Lint %I web drptlr ( d ) Line 3 == (,-cad:) cstir~iatedweight (W,) For t l ~ cst~mnd\wight: 1,inc. l b := shrnr to be c;~rriwlby w-el> (V) Line 2)) =: allo\vd~leshwr stress ( T ) 1,irrc 3 := (rcad:) esti11n;rtcd weight (W,) The slim of thcsc two weights still does not inclnde the weights of stifi'mers if required.

>

t, d, = A," which must equal or exceed the rcqnirrd value of A,v ( = V/r); and

Problem 1 Design a hridge girder for the follo\ving loads:

Ivl V

7~500ft-kips

--

600 kips

For A36 steel, AASIlO Sec 1.6.75 (see Table 1) requires the K ratio of web depth to thickness (d,/t,) to be not more than K = 170 using transverse stiffeners. Then:

(7500) (12) --

(18 ksi)

-

(600) (11 ksi)

= 54.5 in.2 Following the suggested outline for designing an efficient girder:

= 16.65" or use 17.0'' wide

x

2" thick Range plates

8. Then, to find properties of the actual proposed section:

or use an

1l/lau

thick web, 1 1 V deep

3. Check these proposcd dimensions:

= 160

<

Then, to find the weight of this designed girder:

170 O.K.

2 A, = 2(2")(17") =

A, = t, d,

= (11/16) (110) = 75.6 in.2

>

54.5 in.'

O.K.

= 76,255 in.4 5. Let flange thickness be t, = 2":

68.0

A, = (11,/16")(110") = 75.6 143.6 in.'

.'. Wt = 488 Zbsllin f t of girder, on the basis of steel's weighing 3.4 lbs/lin ft/inl.' of cross section, To show that this does result in the n~inimunl girder weight, nine other combinations have been figured, from a web depth of 70" up to 120", as shown by Cwe B in Figure 2. In the example just worked, the various dimensions were rounded off to the next

FIG. 8

Weight of Plate Girder When Design Is Governed by Sheor

size fraction based on available plate. The actual plate girder t:xample using a web depth of 110" weighed 488 lhs/ft, yet the efkient girder for this same depth should weigh 473 lbs/ft. Four other combinations of flange Jimensions were figured, using the same web depth (d, = 108.45"), but there was little difference in girder weight. The thinner and wider flanges result in a very slight rednction in weight.

is increased to V = 1000 kips. This will illustrate the work to he done where shear ( V ) would govern the design. Here:

V A" = -i -

( 1000)

- (11ksi) = 90.9 in.'

Consider the same girder in which the shear load

Following the suggested outline:

Efficient Plate Girders

/

5. Let flange thiekness be tt = 2":

In the previous problem, this led to a web 11/16" X 110"; however-

In this case the '%6" X 110" web plate has insufficient area to carry the shear load. So, switching to Method B:

= 12.65"

or use a W-thick web plate.

or use 13" wide x Y thick flange plates 8. Then, to find propertics of the actual proposed section:

or use a 124" deep web plde 3B. Check:

A,

= ,t d, = (3/q)(1%) = 93.0 in.'

> 90.9 i a 2

OK

Now returning to the basic Method A outline:

Then, to find the weight of this designed plate girder:

4.2- 12

/

Girder-Related 2nd Nomograph

WL = 462.8 lbs/lin

ft

142.0 in." of girder

If the shear value is increased to V = 1000 kips as in Problem 2, this exceeds the allowable value of 750 kips mad from the &st nomograph. Therefore, shear governs the design and the second nomograph must be used. Given:

Find the approximate web dimensions and weight for the same girder, using the nomographs, Figures 6, 7 and 8.

V = 1000 kips

read: ,t

= ,725" or use Vi"

Given:

Given:

S = 5000 in."

K =-d,

= 170

tw

read:

read: d, = 126" or use 124"

d = 108" Given:

Given: S = 5000 in.Y d = 124"

read: Wt = 470 lbs/ft

read:

and:

Given:

V = 750 kips allowable

Wt =

V = 1000 kips T

Using an actual depth of 110" as in Figure 1 would increase this estimated weight to 483 lhs/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 llext fraction, 11/16".

P 275 lhs/ft

= 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.

If the valuc of u,, resulting from the above formula is eqnal to the yield point of the steel in nni-axial tension (what is commonly called the yield strength, u r ) , it is assumed this conhination of stresses will just produce yielding in the miiterial. Hence, the nse of this formula will give some indication of the factor of safety against yielding.

Transverse intermediate stiffcners shall preferably be in pairs. They may be either single or double, and be plates or invertrd tees. When stiffcners are used on only one side of the web, they shall be welded to the compression fiange to give it proper support. The nioment of inertia of the transverse stiffener shall not be less than-

where: d ' J = 2 j - . ~ - 2 0 = 5 a.

(a) Cross-sections of test specimens

. . . . . . . . . . . . . . .(3)

I -= minimum required moment of inertia of stilfcner, in." a, = required clear distance befween hansversc stiffeners, in. a, = ach~aldear distance between transverse stiffeners, in. d, = uninpporied d q t h of web plate between flanges, ID. t, = web thickness, in. When transverse stiffeners are in pairs, the moment of inertia shall he taken about the centerline of the weh plate. When single stiffeners are nsed, the moment of inertia shall be taken about the face in contact with the wcb plate. The width of n plate stiffener shall not he less than 16 times its thickness, and not less than 2" plus 1/30 of the girder depth. The distanct~ bctwcen transverse stiffeners shall not exceed1. 12 feet 2 the clear nnsupportcd depth of the web (d,)

where: T

(b) Comporison: ultimote ond critical loads of bending rests

FIG. 1 Eiiect of web thickness on ultimate carrying copocity of the girder.

= average unit shear stress in the web's crosssection at the point considered, psi

4. LONGITUDINAL STIFFENERS MASH 1.6.81 ) The longitudinal stiffener shall lie along a line 1/5 d,

l a t e Girders tor

(a) Longitudinol stiffeners on inside of girder

FIG. 2

Placing longitudinal stiffeners on outride of girder and transverse stiffeners inside saves fobricafing time.

/ Longitudinal stiffener

Longitudinal and tionsverse stiffeners do not inteisec,

(b) Longitudinal

from thc compression flange. Its moment of inelZia shall not be less than-

These stiffeners do not nwessarily have to bc continuous, but may be cut where they intersect transversc intermediate stiffeners if they lie on the same sidc of the web.

5. BEARING STIFFENERS Transverse stiffeners shall I>F w e d over the end bcarings or along the Icngth of tire girder wherc concentrated loads must he carried, and shall hc designed to transmit thc n:actions io the web. They shall extend as nearly as ixaeticahlc to ihr oi~teredge of the flange, hut not to excwd 12tiilncs their thiclcness. (AASIIO 1.6.17) Some ixidges have longitudinal stiffeners on the inside of the girders, otircrs orl the outside. If the longituclinal stiiIrr~ers arc on the inside, along with the transverse stiffeners, it loaves the ontside of the girder smooth; Figure 2 ( a ) . This, of course, means the iongi-

stiffeners on outside of girder

tudinal stiffener mnst he cut into short lengths and then inserted betwccsr the transverse stiffeners. This results in inrreascd welding tirnc and production costs. Some states havc used longitudi~riilstiffenci-s on the outside and transvrrsib on the insidc; Figure 2(b). Tf~ismethod saves on fabricating time and aL~oallows the use of automatic welding trchniques to join the Iongihldinal stiffeners to thc girder web, thereby substantially incrrasing welding speed. C OF STIFFENERS AASIIO (2.10.32) will allow the welding of stiffeners or attachments transversc to a tension flange if the bending stress is 75% or less than the al1owal)le. .4WS Bridge (225 c j will allow the welding of stiffeners or attachments transverse to a telrsion flanga if thc bending strrLssin the f h g e is held to within those of the fatigot. formulas ( I j, ( R ) , or ( 5 ) for the welding of atiachnrents hy fillet vmlds; s w Section 2.9, Fable 1. Figure 3 illustrates the eflcct of transverse attachments wclded to a plate when tested from tcnsion to an cqual compression (I< = -I ) .*

.~ -

"Fatigue 'Tests of Weliid Joints in Structural Steel Plates", Bull. 327, University of Illinois, 1941.

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 machined out of solid plate without any welding. This same problem cxists in the machining of stepped shafts used in large high-speed t u r l k e s and similar equipment. Figure 4 illustrates the effcct of welding transverse stiffeners to tension flanges.* Tests, again a t 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 incrsased 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 Illinois** took into consideration not only the bending stress in the flange, but also the resulting principal tensile stress in the wcb 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 ol maximum bending stress of the beans. Failure stailed at the lower termination of the fillet weld conrlecting ths stiffener to the web. When the bottom of thr stiflerrer was also welded to the tension flange, failure started at the toe of the fillet weld connecting the stiffesner to the beam flange. After the flange had failad, the crack wonld progrrss upward into the web. Ilerz, the failures usnally occurred in the maximum moment section of thq heilm. This test indicated fairly good correlation when the results were considerod in terms of the principal tensile strcssis (including the effect of shear) rather than simply the bending stress. The 'angle of the fatigue failme in the web generally was found to be about

" "Flexural Strength of Steel Uenms", Bull. 377, University of Illinois, 1948. ** "Fatigue in Weldcd Beams and Girders" W. H. Munse & J. E. S t a h e y e r , Highway Research Board, Bull. 315, 1982, p 45.

_

min. K-*.0 2000,000CYCLE5 FIG. 4 Effect of welded intermediate stiffener on tension flange.

INTERMEDIATE

WELDED

m

COMPREJSIOU R A N E E AND TO UPPER

I&400psi

26,600 psi.

JZ.700 psi.

safe

WE

'A"

TYPE '8'

TYDE

P.

Girders for

TYPE I T

T/Pt 'E'

TYPE 'F.

(a) Details of various stiffener types

FIG. 5 Effect of stifiener type on fatigue strength of member.

(b) Sigma-n diagram for maximum principol tensile stress at failure section.

20% less than the computed angle of the principal strcss. AASHO Specifications (2.10.32) state that transverse intermediate stiffeners shall fit sufficiently tight to exclude water after painting. Some insprctors interpret a tight fit to he onc in which the s t i f h c r s must be forced into position. Many fabricators frel this is an unnecessary dcterrent since it takes extre time to force the edges of tlie flanges apart to allow the stiffeners to be inserted. There err two gencra! methods of fitting these stiffeners to the plate girder (Fig. 6 ) : 1. Use a stiRener that does not fit too tight. Push it tightly against thc tension flange. \Vt,ld it to the girder web and to the compression flange. With this method, tlie fitting of the stiffener will comply with the above AASHO spec.;; yet it is not welded to the tension flange, nor is it a problem to insert. An alternate mcthod is to2. Use a stiffener which is cut short about 1".Fit it against the compression flange and weld it to the web. If it is a single s t i f h e r , also weld it to the campression flange:. It is not v d d e d to the tcnsinn flange. Experience indicates thc 1" gap at the lower tcnsion

FIG. 6 Fit of stiffeners to girder.

4.3-6

/

Girder-Related

flange will present no maintenance problem. Although this does not cornply with the above AASHO requirement, many girders for higl~waybridges are fitted with stiffeners in this manner. Plate girder research at Lchigh University* has indicated the stiffener does not have to contact the tension flange to develop the ultimate capacity of the girder. They recommended the stiffeners be cut short a,described in the alternate method above (2). The distance between the lower and tension flange and the stiffener is set at 4 times the wcb thickness; see their recommcndations in Figure 7. There is no clear-cut answer as to whether continuous or intermittent fillet welds should be used to attach the stilfencr to the web. The latest research at Illinois on stifIeners indicated that fatigue failurcs occurred at the terminations of fillet welds, regardless of whethrr they were continuous or intermittent. Naturally, a continuous weld \ d l have fewer tcnninations, hence fewer aaras for potential fatigue cracks. Where lwge, intormittent fillet welds are specified, %" for example, roplacement with %" continuous fillet welds made by automatic welding equipnrent achieves a considerable saving in cost. Where small intermittent *"Strength of Plntc Cirdcrs", Hrmio Thurlimm, AISC Proceedings 1958; "Plate Giriicr Rcsr:rrch", Konrad Resler & Bruno Thurlirnan, AISC Proceedings,, 1059.

fillet welds are specified, 'h" possibly, savings from the introduction of continuous welds and automatic equipment become qumtionable. With thin, deep web plates, a smaller size weld may tcnd to reduce distortion. In this case, automatic welding would be of benefit, provided this substitution of continuous welds for intermittent welds does not increase weld length to any major extent. 7. FLANGE-TO-WEB WELDS

These welds hold the flanges to the web of the plate girder. They are located in areas of bending stresses and must transfer longitudinal shear forces between Ranges and web. Some restraining action may develop with thick flange plates, but any resulting transverse residual stress should not reduce the weld's load-cawing capacity. This bcing parallel loading, the actual contour or shapf: of the fillet weld is not as critical as long as the minimum throat dimension is maintained. Shop practice today usually calls for submerrgedarc automatic welding equipment to make these welds. For the usual thickness of web plate, the two fillet welds per~etrate deeply within the web and intersect as in Figure 8(1>), giving complete fusion even though simple fillet \welds are called for, as in ( a ) . A few

one a two sided

FIG. 7 Summary of design recommendations relative to girder stiffeners

e l d e d H a r e Girders @or

FIG. 8 Flange-to-web welds.

states recognize this perlctration and are now detailing this weld with cornplctr fusion. 'Tlris proves no problem on the rrormal web thicknas. In thc futurr, however, if the same detail is showrr on much thickcr web plates, the fabricator will have to use a double-bevel edge preparation to obtai~rthe intersretion ( c ) , w e n thongh detail ( d ) is sufkient. It sho~ddnot he necessary to detail groove welds for this ioiot from a dcsign standpoint. Selection of a groove T-joint design should be Ilased on a cost comparison with filkt wrlds. The groovid l'-joilit requires abont ?b the arnonn? of weld metal compared with fillet welds (assuming full-strength welds). However, the grooved joint has the extra cost of PI-eparing thi. double hevcl. In respect to the physical perfonnaiice of cither tiit>fillet or the groovd T-joint design, tests liave been made, hy .4. Ncum:mrr, of these \velds nnder fatigue hending from 0 to tcnsion, K 0, at 2 111illioncycles.*

No ciifkrenec was iirdicat~cdfor thr: fatigue strength of the beam using cither joint dcsign, with both types dernopstrntiilg a f:ttiguc strcugth iri the beam of 22,000 to 24,000 psi (hvirding strcss); Figure 9.

From a dcsign sta~xlpoin?,thm: welds may be quite small. Their achrd size is usually established by the minimum allowable leg size for the thickness of For Vorious Plate Thicknesses (AWS) THICKNESS O F THICKER PLATE T O BE J O I N E D

THKU

over

-

%

inch in.

y4 i n

*IW"

$5

Need not

M I N I M U M LEG SIZE O F FILLET WELD* 3 / 1 6 in. in. 5/16 in.

lh

in,

*hi" 1% in. Ovoi 1% in. thru 21,: i n . Over 21/4 in. t h r u 6 in. Over 6 in. over

/ I

1

),'a in. '/>in. % in.

the t h i r k n c s i of the i h i n n e i plotb

-

the fiangc plat(%.T;rblc 7 lists tile minimum size of fillets for various platc tliickuwses as established by rlM'S Sprdficntions. 1,cg sizc ilicri'ases to take care of thc fastt,r cooling rate and grisatc.r rcstrairlt that exists in thicker platcs. On tliickcr plates. with rrniltiple pass wclcls, it is desirable to gel as nindr h w t input into the first pass as possible. This means 1iight:r ucldiiig currents and siower urlding spwds. L.ou--11ydrogcn olt:ctrodes are bettor'for manual wcldirrg in this work. 'The lmv-hydrogm characteristics of a submerged-arc wclding deposit gives this welding mrthod ;I si~nilaradvaiitagt:. "Discussion at the Syinposium on Fatigue of Wuided Structiircs" The British WtMing Joonial, August, 1900. ~~

FIG. 9 Both weld types showed same fatigue strength.

/

Girder-Related

TABLE 3-Allowable Shear Forces O n Fillet Welds For Various Fatigue Loodings

1

100,000 CYCLES

8800 o f = -i - -

K

600,000 CYCLES

lb/in.

1 I

2,000,000 CYCLES

K

2

2

but rhoil not exceed f

axial normal stress from the bending, applied to the fillet weld, would increase the maxin~umshear stress applied to the tlrroat. For a given applied normal stress (u), the resulting ~naximwnvaluc for the allowable force ( f ) which may be applied to the fillet weld of a given leg size (a)under parallel loading is expressed by the formula:-

= 8.800

f =

o (€60 or SAW i welds) 10.400 o (E70 or SAW 2 welds1 --

--

Where.

( E N or S.4W-1 welds)

MiNlMUM (sheor (V) opplied to girder1 MAXIMUM w = leg size of fiile, K =

Determination of Combined Stress

mbined stresses in a fillet weld between the and flanges is seldom considered for the following reasons: 1. The maximum bending strcss for a simply supported girder docs not occur at the same region as the maximum shear force. For a continuous girder, however, the ncgative moment and shear force are high in the same region near the support, and perhaps the combined forces in this fillet weld should be checked. 2. The maximum bending stress in the outer surface of flange is always designed for something less than the allowable (Bridge code = 18,000 psi). The weld lies inside of the flange and is stressed at a lower value. Ex: If the weld is in an area of 15,000 psi bending stress, this additional normal stress would reduce, theoretically, the allowable shear force for the weld from f = 8800 w to f = 7070 w, or about 80% of what it would be if just horizontal shear were considered (E60 or SAW-1 welds). 3. Usually these welds must be larger than design requirements because of the minimum weld size specifications listed above. Nevertheless, if desirable to determine the combinell stresses, it can be theoretically shown that the

(I370 or SAW2 welds) This formulatio~~ still pennits the maximum shear stress ren~ltingfrom the combined shear stresses to be held within thc allo\vable of T = 12,400 psi ( I 3 0 or SAW-1 welds) or 14,700 psi (E70 or SAW-2 welds). Allowable Fatigue Strength

Table 3 contains tho formulas for establishing the albwahle shear foucc that may hc applied to fillet welds under various conditions of fatiguc loading.

8.

FLANGE B U T T J O I N T S

In nearly all welded plate girdms, the flange is a single plate. These plates are stcpped down as less area is required. A smooth transition is made between the two, by reducing either the thickness or width of the larger flange to comqxmd to that of the smaller. When this tra~xitionis rnade in thickness, the end of the larger flange is hevelcd by a flame-cutting torch. There is a practical limit to the angle of bevel, but this slope, according to AWS Bridge Specifications, should not be greater than 1" in 2l%"(an angle of 23"). On the Calcasieu River bridge, this slope was decreased to about 1" in 6" (an angle of about 9%"). Transitions also e m be made by varying the surface contour of

FIG. 10 Plate bevels made by flame cutting.

(a) Beveling end of flange plate for groove butt held

(b) Beveling end of flange plate for tronsition in thickness.

Fatigue Strengths

1

in 100,000 CYCLES

1

Suit

loinfs

600,000 CYCLES

BUTT WELD I N TENSION (not to exceed 18.000 psi1

1

I

2,000,WO CYCLES

-- - .7 K

i - 8 K

BUTT WELD IN COMPRESSION

inor to exceed p1

(a) Straight-line transition in width

Where: (p) is t h e allawobie ian;piciiivi member involved.

BUTT (WELD

(b) Curved transition in width

FIG. 11 Method of transition in width affects weld's allowable fatigue values.

the groove welds. The usrial method of flame rutting a bevel in the preparation of a wcldcd joint is to cnt down through the surfaw of the plate at the proper angle. lkcause of the wide angle needed for this transition in thickness, it is often better to flame-cut back from the edge of the plate after the flange platc has been cut to length. Scc Figure 10. When the transition is made in width, the end of the wider flange is cnt back at an angle, again with the flame-cutting torch. There is no prohlcm in cntting in this matn~er,and any slope rnay be used; many tinrcs 1 in 12, hot usually a maxiinom slope of 1 in 4. Often this tapar m;ry extend back for several feet. Gent~ally,it is fctt that the straight-line transition in width is sufieient, ;md in the case crf fatigue loading the allowable fatigue va1ut.s for butt groove welds in tension or compressior~ are used. See Figure 11. If a curve tangent to thc edgr of the rtarrow flange at the point of twinination is used, it may be assumed the flanges h a w eqnal widths. Thus, for equal plate thicknesses and with the \veld reinforeern::nt removed, the butt groove meld may he assigned the same allo\vable strcss as the tiangc plat
K =

MINIMUM (bending -

itrerr far t h e

stress or

MAXIMUM

bending moment1

rather than in thickrsess. This advantage undoubtedly would bt: greater if the transition in width wert: made more gradual; however, both methods are sound and acceptzible. Fatigut, values for these transitions are found in Figure 12. A l l a w a b i e F a t i g u e Strengths

Croove wt~ldsin hntt joints of equal platc thiekness. if the rcinforcmnent is finished smooth with the surface, rnay hc ;rllowcd the same fatigue strength under any type of fatigne loading as the base metal. For plates of nnrrpal thickness where the transition slope is not grcata than 1 in W 2 , the formulas found in Table 4 may bc used.

transition in thickness I

I

FIG. 12 Making a transition in flange width rother than thickness has a slight advantage in fatigue strength.

4.3-10 /

Girder-Reloted Design FIG. 13-Summary

of Bridge Plate-Girder Specifications AWS & AASHO

Neutroi axis of girder

ARV OF BRIDGE SPECIFlCATlO

9. Use transverse intermediate s t i f h e r preferably in pairs on opposite sides of web. If only one side of web, wcld ends to compression flange and intermittent weld to weh (1.6.80, 22%). 10. The minimum moment of inertia of transverse intermediate stinener shall be (1.6.80)-

In order to aid thc bridge rrrgineer in designing a welded plate girder, the pertinent .4WS and AASHO Specifications liavt; been brought together into a single drawing, Figiire 13, and related text, below. The corresponding numbers are inclrided so the engineer may refer back to the original speciiicntions. This summary can also serve as a checkoff list, so that nothing will he inadvertently omitted. where: The following requirements apply: 1. Extend bearing stinener as near as practical to outer edge of flange. Proportion for hearing. Welds to web must transmit end reaction. (1.6.79) d .-actual , distance between stiffeners, in. 2. Width of bearing stiffener mist not exceed 12 d, = required distance bctween stiifeners, in times stiffener thicliness ( 1.6.17). 3. Space (horizontal) longit~idinalstiffener Si, ~ 1 , ~ d , = w-eb depth, in. from compression Range (1.6.81). t, = web thickness, in. 4. Dimension longitudinal stiiicncr for required moment of inertia, usingT = average shear stress in web

about edge of stiffencr (1.6.81). 5 Mill or grind bcnring stiffener ends For even bearing to iiange. StifFcner may be welded without rnilling to comprrssion flange, or to tcnsion flange if less than 75% terrsile strength (2.10.32). 6. Do not wcld transverse intermediate stiffener to tension flange if stressed over 75% (2.10.32) or unless stress is within that of fatiguc formulas 1, 3 or 5 of Art. 228 ( 2 2 5 ~ ) . 7. Fit intermediate stitYcner tight to flnnges to excludc water aftm painting (2.10.32). 8. Consider placing intermtdiatt: stiffeners at points of conccntrated load to transmit reactions to the web (1.6.80).

11. Girder ffange shall not extend beyond 12 times its thickness (1.6.17). 12. Ilistance betwem stiffeners must not exceed

12', d,, or l ~ o o O

\F T '

(1.6.80)

13. All shop groove butt welds in flange and web plates shall be made before final litting and welding into girder (404f). V a y 14. Web-to-flange lillet weld leg size = 17,600 1 15. Width of tr'msverse intermediate stiffeners must not exceed 16 times stiffmcr thickness, or 2" plus K O of girder depth. Also, deflection due to live load plus impact shall not exceed 1/800 of the span; for cantilever arms, 1/300 of the span (1.6.10).

lute Girders for

-8 8

MINIMUM WEB THICKNESS (twl

/

i f long. ond tiani. stiffeners

t , ~=

i 340

- - dil

Also. ratio oi depth to length of span shall prefcrably not be less than :;is; for lowor depth the saction shall be incrcxrscd so that the maximum dt:flection will not 1)e grcatcr than if this ratio llad not b w n cxceeded ( 1.6.11). Also, wrh thiekr~cssshall meet requirements given FIG. 14-Maximum

dapthr up to %'inel.

depths

wrr

3be+O 72"

dmths over 72"

tr

=

dSv

290

I1

" '

tr

= - I- d x 280

in the above t a l h for the more coiilrnolr steels.

ENSlONAh TOh Tho dimensional tolcrmces ill Figurc 14 have been set lip for welded plate girdcrs by the AWS Bridge Specifications.

Dimensional Tolerances AWS

=i + C- &*

iml.

/

Oerio,ion

Frm

407

Flotncrr of Gird" Web in a b q i h

Between Stiffeners a a h g t h Eouol to Depth oC Girder

ALLOWABLE

THICKNESS

Fignre 15 illustratt:~several types of diaphragms used, and rcliresent the extremes in designs and fabrication. Diaphragm ( a ) , although so simple in design that no shop welding is rqnired, must be fitted and welded in the field. Diaphragm ( h ) , although mnch more complicated, may he mass-produced in the shop: The anglcs are shcared to length; and the plates are shcared and pnnched. Thcse are placed into a simple fixture and welded together at low cost. Thc field crection is simpler, since the ciiaplu~~gms are put into position, held by an ervction bolt, and then weldcd into place.

. COVER

PLATES

Using A-441 sted (previonsly A - a ? ) , it may he adYantngeous in some cascs to use two plates, a flange plate and a covcr plate, to make np the flange. This will pcrmit use of thinner plates and take advantage stresses. This stcd has the of the higher allo\n~~ble following allowable tension in mcmbers subject to bending:

FIG. 15 Diaphragms

%"

and under

over ?/," to

over 1%"

Ilh" to

i j

4"

27.000 psi 24.000 psi 22.000 psi

Many methods have bcrn suggested for twinination of cover plates. Thc existence of at lcast four conditions which affect this makes it irnpossiblc to recommend one specific covrr plate m d which will hcst meet all conditions. First, the tensile forces, assnmed to be uniformly distributed across the width of the cover plate, sllould be transferred simply and directly into the corresponding flange of the rolled beam withoi~tcansiug any stress concentmiion in the beam flange. In general, a large tmnsversc fillet wrltl across the end of the cover plate dors this in tlv, simplest manner. Second, there must bc a very gradual change in the beam sertion at the mid of the cover plate, in order to develop a similar gradual change in bending stress of the beam. Any abrupt change in beam section

used in modern bridges: (a) angles cut to length and dropped into place; (b) Shop welded diaphragm, field welded to girder stiffener; (c) angler ottoched to siiffeners; and (d) channel welded to web ond stiffeners.

elded Plate Girders for Bridges

/

4.3-1

FIG. 16 Cover plates extending beyond width of beam flange.

- 8ulietin No. 377 will rcducr the bcam's fatigue strength. This would tend to favor a gradual tapered w-idth ithat the end of the cover plate. Third, some caution slhould he txerciscd relative to terminating the cowr plate in the narrow zone of the flange that is in direct line of the beam web. This is a rigid portion with little chance for localizrd yielding to pnwmt the build-up of possible high stress concentration. [:ntrrilz, the selectt.d joint should be rconomically practical to make and answer functional rtrquircments. For cxample: 1. Continuons welds may be needed to provide a positivc seal and prevent moisturc from entering underneath the plate and causing connection deterioration. 2. Ilinimum appcai-ai-icc stanrlnrds may eliminate solno joint designs. Early fatigur tosting at the University of Illinois* on rolled lwams \\-it11covrr platcs indicatcd that:

1 . In geiwral, continnous fillet welds were better than intermittent fillet welds for joining cover plates to the beam Aange. 2. On covar plates extending beyond the width of the heam flangr and conncctcd with longitudinal continuous fillet welds, adding a "/,, fillet weld across the end of the cover plate produced a slight increase in fatigue strength (from 8900 psi to 9300 psi at 2 million cycles). Omitting thc welds for a distance at each corwr of the cover plate increased this valnc up to 11,000 psi; see Figure 16 Thc intersection of the longiturlil~aland transverse fillet welds conld present a point of wrakness if not properly made. This "cross-over" usually results in a very shallow concave weld. By eliminating this weld for 1" back from cach comer, the fatigue strength is incrcased. This does not apply if the cover plate lies within the brain flange, since the weld does not have to " C ~ ~ I S S O V ~ . "

,xij"

.

* Bid1 No. 377, J a n 1'348.

FIG. 17 Cover plates lying within width of beam flange.

I / , loo

no

t e h m a d e w i t h the : r a n s v e r ~ efillet w t l d l e f t off - Bulletin No. 377

U n i v ~ r s i t yof n l i n o i s

FIG. 18 Effect of cover plate terrninaiion on fatigue strength. Calculations cover plate and based on 4" x filiei weld.

%"

3. For cover plates lying witl-rin tho width of thc bmm flangv, incrraseil fillet i \ d d sizt across tiic end of the covrr p l n i ~pr(x111cd o gradual increase in fatigut. strength. h ";,;" fillc~tweld iiad n strength of "3100 psi at 2 millioii cycles. a :ib" fillet weld 11,000 psi, and a 3/h" X 1'' fillet weld tip to 12.600 psi. This piilrticular size of (wvcr plate \ms not testid with the transverse fillet \I-ild omittrd; scc Figriri. 17. Tiir latmt work reportcd at thc University of Floi-ids on stcady 10:rding of 18'' WF XI# 1)eoms with 5" "s" covcr p1:ites showcd that th(, beam flange within the. wvrr-plated I-egion was stressed Iouw when a ad' fillet weld W;IS pIiiv(id acwss the end of the covcr plat? as coinp;u-cd to that wit11 no tmnsvarse \veld. 'i'hc trarrsvorsc wt~ldnlso prod~rccda more uniform distribution of s t r ~ s sacrnss tllc covcr pliitc as \ v ~ l las the ],cam Aairgc, and dlowed tlic platc to pick up its share of tlic, 11mm lorcv in a shorter distance However, all of these factors occlir within the cover-plated ragion of grcatrr stvtion modulus and lower hcnding stress, so this is not vcry scrions.

1/4"

What is inore important is thc effect the transverse weld and shape of tlrc cover plate's end has on ~ adjacent to where the thr s t x s i i l ~ J I C I I C ~ I Iflange covczr plate is nttaehed. This is the region of lower section modol~isand higher bending strcss and is much more critic;rl than any regirnl within the cover plate. The drawing, Figure, 18, illristratcs variations of cover plate tcrnrin:~tiorts.*7 ' 1 1 ~data stiinlnarizes recent tests on t h fatigiic strciigth of l~rainswith partial cover plates. m n d i ~ c t i dnt tlrr i!nivcmity of illiirois. Although llle comnioll inr~tllod of tcrrniliatirig the cover plate dircctly across thr Hmgc wit11 a transverse fillet weld is satisfiicton, and ;rcceptable hy the AWS Bridge Specifications, this data worild sccm to indicate that tapering thc end of tire cover platc and eliminating transverse welds across the end slightly increases the fatigue strength.

~

"

"Fatigue in Welded Beams and Girtleis", W. Ii. Mume and

1. C. S t a l h e y e r , lfighway Rescarch Board, Bull. 315, 1962, p. 45.

lare Girders for

ridges

/

43-15

higher s t r e s s conccn~ration ,n beam flange w i t h smaller transverse f i l l e t weld

FIG. 19 Effect of transverse fillet weld size on fatigue strength. cflangc. of beam

It should be noted that a small 'A' fillet weld was used across the end of the 'h" thick cover plate. The results might have been different if a larger transverse weld had heen used. Most states require continuous welds on cover plates and across their ends, thereby limiting the selection to termination types u or b. Since the data indicates that tapering has little effect, final selection between o or b would have to h e made on the basis of some other factor such as appearance, or lower dead weight. In summary, it would appear that the short section of the transverse weld across the end of the cover plate directly over tha web of the beam ( I ) is restrained and ( 2 ) wlien tested under severe fatigue loading may reduce the fatigue strength of the connection unless it is made large. A large transverse fillet weld, especially in this central section, would more uniformly transfer this force through the surface of the beam Aange into the end of the cover plate. See Figure 19.

Summary 06 Cover Plate Speciticationr (AWS Art. 225) I l i e 4WS Bridge Specifications limit the thickness of cover plates to I'h times the thickness of the Aange to which it is attached (225 e 1). For partial-length cover plates, their end shall extend beyond the "theoretical e n d (theoretical cutoff point) which is determined by the allowable stresses from fatigue formulas ( I ) , ( 3 ) , or ( 5 ) of Section 2.9, Table 1. The ends of thc cover plate shall extend beyond this "theoretical end" a sufficient distance to allow "terminal development" (ti-ansfer of cover plate bending force into the beam aange) by either of the following two methods: A. With square ends and a continuous transverse

fillet weld across the and and along both edges of the cover plate, the minimum tenninal devrlopment length measnred from the actual end of the cover plate to the tlicoretical m d or cut-off point shall be 1%times the width of the cover plate. B. With f a p w c d cuds having no transverse wcld across the end but welds along both tapered edges, tapered heyorid the terminal rnd to a width not greater than ?6 the width, but not ICSS than 3", the tennilla1 development length sllall be 2 times the width of the cover platit. Nonnally the inner end of the tcrminal development lerigth will lir :it the theori:ticrtl cut-off point; see Figun: 9.0, ( A ) and ( R ) . However, the cover plate may be extended farther so that tlie distance between the actual knd the theon:tical cut-off point exceeds the requircd t t ~ m i n a ldeveloprrlent length. In tlus case only the r~rjniredtci-minnl development length shown in ( A ) and ( 8 ) shall be used for the length of connecting weld when determining weld size, rather than the actual length hctween the actual and theoretical cut-off point; see (A') and ( R ' ) . Fillet welds bctween terminal de\&p~nents along the cover plated length, shdl be continuous and be designed to transfer the horizontal shear forces:

-

(for mch weld, there are 2 welds along the edge of the cover plate) Fillet welds within the terminal development zone (between the inner crid of the terminal development and the actual end of tbe covcr plate) shall be continnous and be dcsigncd to trnnsfer the cover plzte portion of the bending force in the beam at the inner

-16

/

Girder-Related Deri

teirn,nol development if beyond cut-off point

I

Momenf d i o g ~ m

I

Theoreticol cut-off point

%

,

I

Cover plated beam

1

FIG. 20 Relationship of terminal development to weld size. Required terminal development length (A and 0 ) is used rather thon actual length (A' a n d B') beiween actual and theoretical cut-off poinis.

k Y

f =

1

21

M, Y = -

End weld: F I 3

I

!

& c,,,~ -- - p-

C " .

I

J

j.

Clih

,, Cut-off

II

I

I

IAi , ,

W-4

end of term~noldevelopment A, rv,, a y , -End weld: F = --7----

%,

I

end of the terminal development length (usually the theoretical cut-off point):

1-

,

(0)

. . . . . . . . . . . . . . . . . . . . . . . .(8)

Cut-0ff "=ner

end of terminal development

-M Z ~ Y I

M y u = --I

(A')

. . . . . . . . . . . . . . . . . . . . . . . . . . (0 ) Inner end of termtnol develo~ment

where:

V = vertical shear at section of beam under consideration a

W

Cover R

i

I

(0'1

-- area of cover plate connected by the 2 fillet welds

Cut-off point Inner end of terminol development

y = distance between C. 6. of cover plate and the N.A. of the total section

I = moment of incrtia of the total section MI = moment applied to beam at the section of the theoretical cut-off point Ma = moment applied to bean at the section of the inner end of the tem~inaldevelopment The allowable to be used for these fillet welds would come from formulas ( l o ) , (14), or (18) of Table 1, Section 2.9, and shall conform to the minimum

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

whrre d = depth of beam (feet)

FORCE VALUE It has been pointed out* that the sloping bottom flange of the parabolic haunch has a vertical componcnt of its compressive force and this will reduce the shear stress (r',.) in the girder web in this region. In addiiion, the concave compression flange produces a radial compressive stress ( u ? ) 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 wcb near the support. This is because the slope of the bottom fiange is small in that area. Also, the convex compressive flange produces a radial tensile stress (u,)in the web, w-hich is greater than the radial compressive stress in the parabolic haunch. This is because of the sharper curvature of the fish belly haunch. I? is seen by observation of the Huber-Mises formula that both of these factors will result in the yield criterion (we,) having a lower value in the ease of the parabolic haunch. This result con~paredwith the yield strength of the steel (in uniaxial tension) would indicate a higher factor safety.

uer =

(Huber-Mises Formula) ur2 - u, u). u?

d

+

+ 37,y2

Haunched girders do not present much increase in cost for welded construction for longer spans. The web plates are normally trimmed by Aame 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 Aange plates are simply pulled into place against the curved web. Although the bansverse stiffeners u~ouldvary 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 (F,,) 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 stictiol~modulus of the girder. This method will produce a more accurate value. From this value, the actual force in the Aange (F,) may be found, as well as the vertical componcnt (F,) of this force: Fh and F2 - - - -d cos B cos B M F, = Fh tan 6 = d tan 0 This vertical componcnt (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 thc flange force (F,) 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

Parabolic Haunch

*

"Design of the Bridge Over the Quinnipiac River" by Roman Wolchuk.

FIGURE 1

Resistance of web

FIGURE 2 eslstance o f bottom flange due to its vertical component of tensile force

Fv = Fh tan B

Simply Supported Girder Staoighf os Curred See Figure 2. Here the external shear is-

V = A. rw

M +tan B d

and the modified shear is-

M

=V--tan0 d In this case the vertical component is subtracted from the web shear.

ontinuous Parabolic Hounched Girder

Sce Figure 3. Here the external shear isV

=

A, rv

M +tan 0 d

and the niodified shear is-

M d

= V - - tan 0 In this case the vertical component is subtracted from the web shear.

FIGURE 3 Resistance of bottom

f cornprerrive force

FIGURE 4

I

------+ 4Resistonce of web due to its shear

FIGURE 5

FIGURE 4

Fish Belly liaunch

between the fish belly haunch and the parabolic haunch in the area of the compression 5ange near the support.

See Figure 4. Here the e x t e n d shear is-

V = Aw

T~

Parabolic Haunch

M tan B -d

and the modiIied shear is-

See Figure 6. Conditions include the following: Use of A431 steel

M = 55,000 ft-kips V = 1200 kips I, = 3,979,000 in.'

In this case the vertical component is add& to the web shear. osatinuous

Fish

See Figure 5. Mere the cxternal shear is-

In this case the flange force has no vertical component; hence, there is no reduction of shear in the web.

Check the haunched girder section ( a t poini of support) shown in Figure 7, to detennine the difference

FIGURE 7

Analysis of Porobolic Haunch

stress in U:I&ut lower jfli~nge (it support

aocrage bending stress i n louer flange

FIGURE 9

= 21,150 psi compressini~ .-

Range forces

F, = c*Af = (21,150)(25/8 = 2,000 kips

x

p~ ~~~~

= 20,900 psi, eompressio~~ 36) These stresses in Figure 10. Irft-]land side, must now be rotated 10" to line 1113 with the sloping ilange in order that the radial cornpressive stress may be added. This is shown on the right-hand side of Figure 10. '%is may 11e analyzed by one of two methods:

F, = F,, Van B = (2000) (.l763) = 353 kips F =

(55,000 X 12)(126) (0,979;000)

- - -

I. Graphically, using Molrr's circle of stress: (Fig. 11) a ) Dmw thc gi\,en st]-cssrs (w,', u,', and 7') at the two points (a') nrid (b') h ) Constrni:t a circlc thro~ighthese two points c ) Rotate clockwise ilirongli a n angle o f 20 or 10"

F,, cos B

d ) Read the ncw stresses (c,, u,, and 1

2030 kips

2. Analytically; woi-k is 1)rrforrncd as follows:

slzar stress in web Siiice the external shear isV = A,,.

Tw z

7,"

+ F,

V - F, A"?

7 )

or

0

=-

FIGURE 11

v7=k-n = (10,450)

-

sin

p

cos

p =

7

-

(11,500)

= 1050 psi, tension radial force of l o o m compression @nge againat w e b

,0886 ,9961

= m sin /3 = (11,540) (.0886) = 1020 psi

n == m cos

FIGURE 12

/3

= ( 11,540) (.9961) = 11,500 psi o ; = k + n = (10,450)

+ (11,500)

= 21,950 psi, compression

= 846 ibs/linear in.

resultant radial cornpresshjc stress in uocb

FIGURE 14

This produccs the final sircss condition o f :

At this point: crx = cr,, & F, = Fh stress in weh or lower flange from bending moment

d-Y U,

=

- 21,950 psi

= 20,900 psi, compression

FIGURE 13

average stress in lotoer ftange from bending moment critical stress Using the IIuber-Miscs formula: ,--~

ucr,, = V urZ- 0; O;

-twTZ+ 3 ~~

T~~~

-.. - . = \ (-21,Q50)2-(-21,050)(-180)+(-180)' ..

+3(1020)Z

=29,000 - psi

This results in an indicated factor of safety against y~eldingof-

= 21,150 psi force in lower flange from bending moment

F, = Uf Af = (21,150)(2% '/a 36) = 2000 kips radial tensile force of lower compression flange against web

Analysis of Fish

NOW wing the same load conditions on t l ~ cfish belly hannch with the same web and flange dimensions:

ridge Plate Girders

restiltur~tradio1 trnsile stress in web

r;

2420 psi f

_

. .f,, cos 8

~~~~

but the distance d o n g this s l q w for inch is-

I

W

l~orizoni~il ~

i" --

cos H

6930 psi combining strcssrs to f t l d fhi, critical dress

Using tlic IIuhi-r-Lliscs formula:

This rcs~iltsin an iidicatcd factor of safety against yiolding of-

F.S. =- u, IT, r

It is apparrnt fmm this that tha paxii1,olic haunch lins a sligl~tly lowcl. criticirl stress and, ihiwforr, a slightly 11ighi.r fiwtor of sixfcty. 3. WELDS CONNECTING SLOPING FLANGE TO WEB

s o that t11c s11mr f ~ ) r won t h i ~wt,ld iilwrg this sloping Hang(, is obt;iiried froin i h ~ :ii)ovr: , fonnrilii for the lrorizontnl flange, using the rnoiliiird v;rlue of \":

Erection view of New York State Thruway bridge shows haunched girders. Siraightness and true camber of the lower fianges are apparent. Note veriicoi stiffeners and suspended (235') span bearing suriaces at girder junctions.

Portion of 295' span of bridge on Connecticut Turnpike being settled onto supporting piers. Note continuous parabolic haunched girder construction.

1. RECENT PROJECTS

Today, it is accepted practice to design and fabricate plate girders with horizontal curves when necessary. Several such bridges or freeway overpasses have been built within the past several years. A series of 4 lines of curved welded plate girders with 90' spans are a part of the Pasadena-Golden State Freeway's interchange in the Los Angeles area, Figure 1. These have a curve radius of 400'. They were fabricated in Kaiser Steel's plant at Montrhello. e One of Milwaukee's new expressways has a section of 4 continuous spans with n total lengtli of 345' in which tlie two orrtcr girders have a 9' horizontal curve and the 2 inner girders are straight. Bristol Steel & Iron Works, Bristol, Tennessee, rt:cently fabricated several curved girders for the Southwest Freeway-Inner Loop in Washington, D. C.

2. DESIGN AND FABRICATION Although there are torsional stresses within the curved girder, usually the degree of curvature is not overly high and these additional stresses arc- offset by the diaphragms connecting the girders. The number of diaphragms has occasionally been increased for this reason, and sometimes the allowable stresses have been reduced sligl~tly.

FIG. 1 Welded plate girders, having a 400' radius of curvature, dominate the interest in Los Angeles interchonge of Pasadena-Golden Stale Freeway. Curving girders permit economies in deck system b y keeping overhangs uniform from end to end of curve.

Curved flange plates are laid out by offsets and flame cut from plate. By cutting both edges at thc samc time, there is no bowing from any unbalanced shrinkage tBect of the flame cutting. The web plates do not have to he prcfonned, usually being rasily pulled into alignment along the ct.nterline of the flanges. Caution must he ustad in placing attaching plates for thr diaphragms to the webs and flanges. The proper angle for these plates may vary along the length of the girder. Shear attachments are added mainly to accomplish composite action between the concrete dcck and steel girder, and thereby increase torsional rigidity. During erection, a pair of curved girders is usually attached togethcr by moans of the dinpluagms and then hoisted into position as a unit.

4.5-2

/

Girder-Related

FIG. 2 Bridge plate girders being weld fabricated. With flanges flame-cuf on a curve, weight of the rolled web is utilized in making i t conform to desired radius.

FIG. 3 A two-span continuous box girder and curved ramp construction provided the answer to space iimitotions in reaching elevated parking area at busy New York terminal complex. Smooth, clean lines, without outside stiffeners, demonstrate oesthetic possibilities inherent in welded design.

The use of tapercd girders has hecome widespread, especially in the frarning of roofs ovrr large ;ireas where it is desirable to minimize ihe number of interior colnmns or to clirniriatc them ;iltogelhcr. They permit placing maximum girdrr depth whm: it is needed, while rpducing tho dcpih consiclrrably ;it points whcrr it is not necdcd. T a p r e d girders are fahricatcd either 1) by welding two flange piates to a t;ipcmd \vch plate, or 2 ) by cutting a rolled WE' b u m kmgthwisc along its wcb at an angle, tnrning onn half r d for m d 2 arid then wt:lding the two h;ilm>s back togct1tt.r again along tila web. Sea Figure 1..

Gtlniber can he built into the tapend girder when required. Wlien thc girder is made from WF beams, each half is clamped into the propcr canher during asscmhly. Then the h i t joint dong the web is groove welded while the girder is held in this shape. Sincr the weld along the 1)earn web lies along the nentral axis, no bcnding or distortion will result from welding, and the girder nil1 retain the shape in wltich it is held (luring wtlding. When the girder is made of two h n g e plates and a tapered web, the proper caniber can ?it: ohpained by simply ciitting thr wel, to the p q w r wmbrr outline. The flange plat<,s during nssombly are then pulled tightly against the web, into the proper camlier. The four flllot welds joining the flanges to the web are l>alanced about the ncuiral axis of the girder and as a result there shodd he no distortion p r o ? h n .

slxm dc*sigil,the central span can use the tapered flange lip, forming thcb slop!: of the roof; the two ndjaccnt spans usc the taperrd liangc rlo\vn to provide a flat roof, hnt tiltrd to in. t l ~ swn<. slope as the cmrtml swtior~. Th? pro1)lrnl of 1ntt.r;il srrpport for the top ?omprisioii Wnng~s of tapcrrd girders is 110 different than with other lirnms and gird:,rs. C~~nrrally the roof deck is s:rfficit?iitly rigid to function as a di:ipl~ragm, ;md it's only neetxary to attach the deck to the top flange. Tl~ci-e'sappnrcntly no advantage in clrsigning with a rodticcd stross :illow:~blc, in aieord:mce with AISG Foi-inol:is 4 or 5, in order to pwinit a greater distance between bracing points ;kt thc top ilange. Whmc iapcred girdrrs are critical, Section 5.11 on Rigid Fr;nn<>Kriccs g o c ~into more detail rclative to stresstss (elastic design). Bcca~lscof the rrduccd dcptli at the ends of thi.

Application of Tapered

When the tapered girders are used with the sloping flange at the top, their t a p r in both dircetions from the ridge will provide t l ~ cslo:,e needed for drainage. By varying the depth ai the ends of successive girders, the deck can bc canted to drain tow:lrd roof boxes in thc valleys betwecn adjacent galikd spans and at flanking parapet walls. For flat roofs, the girders are inverted, with their tapercd flange down. Thcre art: inany combinations of roof framing systcins possible. For example, on a three-

FIGURE I

eided Structures Required depth,

Required depth,

(b) Tapered girder

(a) Conventional beam

FIGURE 2 Curve of required section modulus [S)has same shape as moment diagram for uniform load on simply supported beam Moment d i a g ~

(b) Tapered girder

(a) Conventional beam

FIGURE 3

tap(:red girders, their connection to supporting colum~ls may offer little resistance to horizontal forces. For this rcason, sonre knce braces may l x required ur~lessthe roof deck or a positive system of bracing in the plane of the roof is stiff enorigh to transmit these forces to adeqr~atclybraced walls. At first glance, there appears to bc quite a weight saving in tapered girdcr; how?ver, this is not always as great as it might seem:

First, the flange arca remains the same; the only weight saving is in the web. See Figure 2. Second, the depth of the tapered girder at midspan r n ~ ~he s t iricrcased over that of the conventional straight beam to he sofficicnt at thc critical section (about "4 span). This is necessary to dcvelop the required section rnodulus along the full length of the tapered girder. This will slightly offset the initial weight saving in the? wcb. See Figure 3.

FIG. 4 For flat roofs, tapered girders are used inverted, with tapered flange downward. Frequently the girder i s tilted to provide a slope to the roof or roof section.

itders

/

4.63

2. DETERMINING CRlTlCAL AND SLOPE The critical depth scction of a tapered girder is that section in which the actual depth of the girder just equals the minimum d ~ p t hrequired for the moment. It would be the highest skessed section of the girder in bending. In the case of a uniformly loaded, simply supported girder, its sloping flange must be tangent to the repired-depth cilrve at this point in order for the beam to havc sufficient depth along its length. Setting thc slope of the tapered girder flange so that the critical section is located at the V4 span will result in about the minimum wcb weight. See Figure 5. The properties of this critical section are-

This formula for section n~oduluscan be simpliIied with little loss in accaracy, by letting-

I-

dw = dl = dl,

[depth between

C G , of flanges)

df

I

I

If the section modulus required to resist the bending moment is known, the required beam depth ( d ) is solved for: FIGURE 6

FIG. 7 Tapered g i r d e r s used with the tapered flange a t 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.

Also, at x = L/4:

For a simply mpported, unitormly loaded, tapered girder-

w

4

= 50 lbsiin uniform lood

L

and:

I

b

= d,

4.-4 4

d, = d, -

tan B

tan 0

1

. . . . . . . . . . . . . . . . . . .( 5 )

. . . . . . . . . . . . . . . . . . .(6)

Since loading on the girdcr is not always ouiform, the ~ h o v cformulas do not always apply. Table 1 summarizcs the working formulas to use for various conditions of loading, as w d l as locating the critical depth.

FIGURE 8

or to find the depth in one step-

To find the slope of the critical-depth curve forined by points d, along the girdcr length, this expression for depth (d,) is digcrentiated with respect to the distanccs ( x ) :

3. --w (I, d d 2 t, u 8 = ---x = - dx A,

- %I -. ( L - x)

w

It is simpler to find the slope at Y4 span, letting x

= L/4:

Fignre 9 shows the effects of placing lnultiple loads npon a simply-s~rpportrdtapered girder. These effccts on the hending rnonxnt and the critical depth of the girder can be explained as follows: In the case of the single contcntrated load ot midspan, the critical dcpth section is :it midspan, and the maximum slope is 8. * In the casi: of 2 cqual conccntratcd k~adsapplied at 'h points, the critical depth section is at the p i n t s of lotid application and the m;~xinnlrnslope is 0. .lssi~rning the slope n w e to pivot itbout this criticnl depth section, any slope lcss than this value \ r m ~ l dcause ihc dcpih at the end to incrmisi. at twice the rate at which the depth at centerline is dctrcasing. Sincc such a shift would incrt:;isr the web weight, this maximum slope vahle of 0 should be nsed initially. If morc dt:pth is ncedcd at tllc end because of higher vcrtical shear, do this by pivoting about this critical depth section. This will rcwlt in thc least increasv in \vet) wcight. It can IIC shown that, nnder this condition, the rcst~liingdepth at centerlinc will be-

rn d, =

3 d,

-

-- d,,. . . . . . . . . . . . . . . . . . . . . . . . ( 7 )

* In tho case of 3 equal concentratsd loads applied at 'A points, the critical depth section will he chosen at U4 spa^ Thr slope of the girder mnst lie sonrcwhcrc bctwcen 0 and 4. For any ar~glel~etwecnthese two values, the wcight of the web will rcmain tlw same

8.1-12

/

Reference Design Formulas

Influence Lines Effect of position of force (F) upon moments Ma, MI, M2 and upon kmax

0

.1

.2

.3

.4

.5

6

.7

Position jo)of applied force F

.8

.9

1.0

ratio of tccb's depth to thickness

S = -M u

(1687.5) --

( 22,ooq

= 76.7 in." To use an "eficient" section (Sect. 4.2, Topic 2), the efficimt depth would be-

It would be prefcra1,le not to have to use transverse intermiitcwt stiflei~crs.1,ooking in Section 4.1 on i'latc Girders for Rnildings, Topic 2, it is secn that these stiffeners are not n:rjuircd if:

a,

The ratio K = ,- is less than 260 rw b ) The shear stress (7) does not exceed that of AlSC Formula 9. This means the values of K and shear sircss (7) shall fall within the values of the right-hand column of AISC Tahlc 3-36, in Section 4.1, page 25. Assume a value of K = 70 at the end of the girder; herc the shear ( V ) is highest. Assume a value of K = 170 at midspnn; here the shear ( V ) is very low. This means at 34 span (the critical section under consideration) K would fall halfway between these two values, or K = 120. :I)

And from Table 3-36 in S<.ct. 4.1; since wi& no stiireners a/& = n (over 3 ) , allowable shcar is 7 = 5000 psi. nctual shear stress 7 = -

v A, (7.5 kips)

(3/16)(24) = 1670 psi < 5000 psi

=

OK

required slope of tapered girder

= ,0852 radians, or 4.88' required depth of web

therefore, the eficicnt depth

= d,

+ -L4 tan

= (24.0) = 24.0 = 36.8"

0

+ -(- m ) (.08538) 4

+ 12.8

required flung6 area (&cicnt section)

=. 2.4 in.' or usc Yz" of which is Af = 2.5 i a 2

x

S' flange, the area

u e b thickness

eck Shear Stress at End

A, = 3/16 (11.2)

= 2.1 in.'

= .W' or use a 3/,6" thick plate. Then-

(15 kips) (2.1)

= 7140 psi

flcre:

Since the reqoircd section modulus of the critical section at '/a span is-

K = - d, t,

S = 76.7 in."

- (11.2) -(

3/16)

an 18" M;F 50.11, bcam could be used

= 60, and from Table AISC 3-36 in Section 4.1, page 25 it is dctelmined tha't no stiffeners are required.

A, = (.57)(7.5)

Check Section of Midspan

= 4.27 in.'

IC = d, t, -

properties of this rolled beurn

d, = 18.00 - 2(.57)

= 16.86" 8 = 89.0 in."

(36.8)

rn)

= 196

<

260 OK

shear stress at h ' span \I

Also, practically no shear here.

M9 = 8-W L2

-

(50) (600)' 8

= 1240 psi OK slope of tapered girder

= 2250 in.-kips S,

= At d .

tu dW2 + --6

= (2.5) (36.8)

cr*

+

(3/.L6) (36.8)2 6

= ,05415 radians or 3.10'

= 134.4 in." Ma ss

'e-

( 2250 .- --

in.-kips) (134.4 i n 3 )

= 16,750 psi

<

22,000 psi OK dg

= d,

+ L tan 0

Alternate Design

d, = d,

- L4

tan 0

FIGURE 1 1

Before going further, check the sheav stvess at i h c end of beamTo make this tapered girder by splitting a W F rolled beam, and weiding back together aftcr reversing one-half end for end.

A, = t, d,

= ( 3 8 )(8.8-i) = 3.17 in.'

ing point for flame cutting the WF beam to prepare a tapered girder. -

(15R)

- (3.17)

Check Girder Section at

= 4730 psi OK depth of beam

Also, practically no shear here.

sturling poini of cut

= 2.0014 a and 26.12 a=2.0014 = 13.06" or use the dimension ( a = 130") to determine the start-

= 13,500 psi OK EFLEGTlON OF TAPE

The area-moment method may be used with good results to find the deflection of tapered girders, where no pori~onof the rnember has a constant mo~nentof

FIG. 12 Turn o n e - h o f f end for end, and submerged-arc weld this web ioint without rpecid edge p r e p a r a t i o n . Trim ends.

/

4.6-10

Girder-Relaled Design

Depth of web ot end

d

=

11.2"

For each division, the moment of inertia (In), moment (M,),and distance to the end ( x ) are determined and listed in table form.

of centerline

dw

=

36.8"

FIGURE 14

FIGURE 13

inertia. This method is described under Topics 5 and 7 of Section 2.5 on Deflection by Bending.

Here, for each segment:

To compute the deflection of the tapered girder shown in Figure 13. This girder has a uniform load of 50 lbs/in., and a length of 50' or 600". Usin the area-moment method, the distance of point from the tangent to point equals the moment of the area under the moment diagram taken about point @ , divided by the EI of the section. Divide the girder into 10 equal lengths ( s = 60" long). The greater the number of divisions, the more accurate the anywer will be.

Since:

.3,76"

14.26"

346.in."

427.5in.-k

37.2

90"

18.88"

19.3W

669.in.'

-

i147.5in.-k

154.6

150"

24.00"

24.50"

l I17.in.'

1687.5 in.-k

226.7

210"

29.12"

29.62"

1702.in.'

2047.5in.-k

253.2

270"

34.24..

34.74..

2439.h."

2227.5 in.-k

246.7

--

-

-

Total

+

918.4

*

-

, 3.0 in.=

t* = Tbe above formula, in this problem, reduces to:

Since:

and:

Dramatic savings can be obtained from an often forgotten design conccpt. The opcn-w-ob expanded bcam has already paid substmtial dividends for various engineering firms. It shonld hc considertd on many more projects. The opening up of ii rolled beam i ~ i c r e ~ s ciis s section moclnlus and rnorrient of incrtia, results iir greater strmgth a i d rigidity. Thc reduction in bcam wright has a chain cfFiict on savings throughout ihc structure. The open-wcb expanded beam is made economically by flime cutting a ri)l:ed henrn's weh in a zig-zag patiwn along its ccnt~~rlinc. S
Rolled beam cut along web

FIG. 2 Use semi-automatic arc welding to rejoin the two halves. A 100% fully penetrated butt weld can often be mode with a single pass on each side of web withoiit beveling.

relatively easy on ;I template-rrjuipped machine. The !is<: of stm-automatic arc wclding to rcjoin the two hnlvcs onablcs good, soirnd welds to be made faster, more economically. M7i>ldingis confined to a portion af tho web's total length. A 100% fully penetratcd butt wcld c m usually be made with a single pass on each side of thc wrh, without prior beveling of the cdges. See Figure 2.

Welded back together to produce open-web expanded beam FIG. 1 Result: a deeper beam, stronger and stiffer than the original. Design starts with a lighter beam for immediate savings in material and handling costs. It often eliminates need for heavy built-up beam.

Starting the design with a lighter rolled heam realizes immediate savings in rnatcrial and handling costs. There is no waste material with this mcthod. It often clirniitatcs the ~ i c c dfor a hmvy built-up beam. In the design of hnildings, the web opcning is frequcrrtly oscd for duct work, piping, etc. which conventio~iallyare suspmtled below the bearrr. See Figure 2. On this Basis for cqnivalcnt strength, open-web expandcd 1x:nms usnally permit a reduction in the distance between wiling hclow and floor dmve and thus providcs savir~gsin Iniilding hcight. Oxygen Barnc cutting of the light heam wcb is

18" W 5 0 2 ~opened up to 27" Duct work inside Wetghs 65%, saves 3" tn h e q h f

FIG. 3 Opening in web used for duct work, piping, etc., normally suspended below beam. For equivalent strength, open-web expanded beam usually reduces distance between ceiling below and floor above.

Cutting the zig-zag paitmn along a slight angle to the beam axis results in a tapercd open-wih cxpanded heam. See Figlire 4. This has many applica?ions in roof framing, etc.

/

4.7-2

Girder-Related

FIG. 4 Cutting the zig-zag patlern along an axis at slight angle to the beam results in tapered open-web expanded beam. This has many applications in roof framing, etc. Tied together with plates

Two opm-wd> i,xp:ud(:d bcams can sornctirncs be nested togatl~erto form a coltrrnn l~avirig;I liiglr rnoment of inertia alxnrt lrotll its x-x and y-y ; i x t ~Sce . Figilre 5.

EOMETRY OF C U T I N G PATTE The zig-zag cutting pattern and the rrsiiltirig geometly of the web cut-or~thelp determine prtipr,rties of the section.

FIG. 5 Two open-web expanded beams can sometimes be nested logether to form a column having a high moment of inertia about both its x-x and y-y axes.

si~fficientto kecp thc horizontal shcar stress along the web's nentral axis Eroiri txcrciling the allowable; see Figurc 7.

f

Cut W benm olong rig-zag line

FIGURE 6

rn tan

& =-

In gencrd, the angle ( 4 ) will be within about 45" rninirnrrm and about 70" n~axirnu~n, with 45" arid 60" beir~gmost commonly used. This angk: must be

FIGURE 7

The distancc ( c ) may 11r varied to provide the prop"- \wh opmirig for duct work, ctc., and/or the pro pi:^ dist:iircc for ~ ~ l d i l lbetween g openings. Set: Figure 8. I-Iov~~\.cr, as this distarrce ( e ) increasrs, the b(wding strt:ss witl~inthe Tce st,ctioir dtrc to the applied Thus. t h c is a limit to bow shear forw ( V ) iricr~~ases. largr ( e ) rnay be.

irders

/

4.7-3

Auulied load

Looded open-web expcnded beam

dShear dicgiom

Moment diogrorn .-

FIGURE 8

. RESISTANCE

FIGURE 9

10 A

Since the bmrn flanges carry most of tbr 1)rncliiig load, the loss of well area is not much of n prn!~lc.m as far as m o r t i t ~ is ~ tc o n c ~ w ~ cHowever, ~l. sl~ear(V) is carried by the web, and must he considered. .4t cuch we!> op2ning, two Tee scciions act as members of a framr in resisting vertical shcar forces. At midspnn b , Figure 9, the shcar ( V ) is minimum and may have little rftnct on the beam's strength. Approaching the srrppori in the rcgion of high shear a , the hcnding strcss produced by this shear on tlic shallow Tee st~dionmust be added to the conventional bending stress f n m the applied beain load. The bending moment due to shear is diagrammed in Figure 10. Usually, thr point of inficction in top and

bottom Teo scctions doc, to thc rnommt prodi~ccdhy shear. is ;iss~irnr~l to ba at inid-scction of the opcning (c.'2). It is furtl~cr :mnmed that thc total vertical shcnr ( V ) at this point is divided ccpally bctmcen tbcst: two Tre scctions. sincc they arc of rqunl depth. Actually, thc dcsign and st]-cssbchzrvior of an opcnweh expandid heam or girder is wry similar to that of a Vicrci~dm>ltruss. Thc primary d('sig11 consider:itions ;in. as follows: I . The top and bottom portions of the girder are suhjectcd to coinpression arid tixnsion bcnding stresses from ihr m:iin bcnding moment. u,,= hf/S,,. Thcrr must be ;r continuity of thew sections tl~rougboutthe girder lcrngtk to transfer tlrcsr stresses. In addition, the comprssiio~portion most hc cl~cckedfor lateral sup-

FIGURE 10

4.7-4

/

GirderPoint of inflection

4 0, k

t' s

/t~ompression

-

Bending stress of beom section due to load on beam

Resultant (total)

Bending stress of Tee section

bending stress ( 0 )

due to application of vertical sheor a t point of inflection

FIGURE 11

port, niinimurn width-to-thickness ratio, and ;rllowahle compressive stwss; scc the left end of Figure 11. 2. The vertical shcar ( V ) in tho girder i q carried by tile web, and producrs vertical shear stresscs in the wch section, both in the solid portion of the web, and in tho stein of thc Tee scctiou of tlic open portion. 3. In the open portion of the web, the vertical shear ( V ) is divided equally between the top and bottom Tee sections (assuming same depth of Tee sections). Assuming the shear is applied at the midopening, it will produce a bending moinent 011 the cantilevrrtd Tee section; see the right-hand end of Figure 11. The resulting secondary bending stresses

must be added to those of the main bending moment, Item 1. If needed, a flange may be added around the inside of the web opening to give the Tee sections added strength. 4. The horirontal sliear force (V,,) applied at tho solid portion of tlic web along the girder's neutral axis may subjcct this portion to buckling. SIX: Figure 20. The resulting co~nprcssivebending stress on this unreinforced web scction is important because of the possibility of this w:b scction buckling under this stress. 5. The solid portion of the web may trnnsfcr a vertical axial force (compressive or tensile) tqiial to one-half of thc cliffermce between the applied vertical shears (V,) aild (Va) at thc cnd of any given unit panel of the girder. See Figure 27. 6. There should be 100% web depth at the points

stiffeners m;iy h r needed at the of s ~ ~ p p o rBcauing l. t:n& of th? ~ i r d c rw1icr.e rmctions an: applied. 4. TOTAL BENDING STRESS IN

t

? ' h ~ main bending stress ( r , )Itern 1, acting on ;I section where tile open Tre swtion stabs. is assunred to increasc linearly to a rnaxiinu~uat tlw outer fihcr. To this stress must be :~ddwlor snbtractc& depcndiilg up011 signs, the secondary boridii~gstress (u.~,),Itein 3. See cerrtr;il portion of f:iguw 11. ~t point

@

Second:rry hendir~gstrcss at stem of Tee due to vertical shear ( V j at Section , atidcd to main bending stress at stcm of l'ce d11c to inair1 moment ( M ) at Section @ :

@

A* point

@

Secondary brnding stnlss at fiange of Tee due to vertical shcar (V) at Section @ , added to the main bending stress at flange of Tee* du? Lo main moment (M)at Section @ :

Open-Web Expended Beams ond Girders Research at the University of Texas- indicated these main bending strcsses in the Tee scacfion do not increase linearly to a maximinn at thc ontar film of the flange, hut in some casrs the revt:rse is true; the stress along the stem of the Tce scction is high(,r than that at the outer f i l m of thc flange. For this rcason, in their analysis, they calci~latcdthe bending forcc 17 = M d , using the moment ( M ) on thc girder at Section ~

"

d = Distancf: h~:twe~:nneutral axes of Tcc srction db -: Dcpth of original beam d, = Depth of cxpandt:d girder e = Lnrgth of Tee swtion, also lcngth of solid web srction along nrntra! axis of girder. h -= Height of crlt, or distance of expansion AT = CIOSS-s~ctionid area of TI:? section I, = Moment of inertia of open section of cxp:indrd girdw s - Section . modolns of flange of T m section S , = Section modulns ol strm of Tcc stxction

"Experimmtul Investigntions of Espencicd Stsd Bwrms", by M. 11. .4ltflliscl1; Tlirsis; Aug. 195%. "Stress Distribution in Enpandml Strel B r a n d ' , by R. W. Lidwig; Tlrcsis: Jan. 1957. "An Invrsiipntion of Wclilrrl Open Web lixpaiald Beams", by Altfilliscli, Cooke, and Toprac: AWS Jouiml, Feb. 1957, p 77-s.

X

X

A36 steel C, = .40

Secondary bending stress In,) from applied shear, ksi Near

t

4.7-5

DEFINITIONS OF SYMBOLS

d

.- ..~~-~

/

(AISC a,)

Neor support

[high sheor)

(high moment)

FIGURE 12

4.7-6

/

Girder-Related Design

the point of inflection of the Tee scction. This is convenient because it is the same section at which we assume the vertical shcar ( V ) is applicd for the secondary txnding stress. They also assume this force ( F ) is ~miformlydistributed across the Tee scction. This simplifies the calculations, since for a given unit panel only onc section must he considered for both the applied moment ( M ) and the applied shear ( V ) . This is Section @ at the point of inflection of the Tee section. Also, only one total bending stress is required for this sectinn-the maximum secondary bending stress at the stem added to the average main bending stress. It does not require calculating at two different points-the stem at Section @ and the flange at Section @

M since F = A d

Buckling Due to Axiof Compression

The Tee section, because it is subjected to axial compresiooli, also nnut bc checked against hnckling according to AISG Sec 1.9.1. See Figure 13, and see Table 1 of limiting ratios for steels of various yield strengths.

i

FIGURE 13

Tee Section Stiffened by Flange Welded Around Web Opening < 3000

Tee Section Unstiffened < 3000hi - -. ..~

and

b, --

T-&

The main bending stress (v,,) and secondary h m d ing stress ( m y ) may be considered according to AISG Interaction Formulas 6, 7a, and 7b. These are shown graphically in Figure 12. (Note that .41SC refers to main bending stress as u;,and to secondary bending stress as u,,.)

tf

-

Number of Poinfs t o Check Along Girder's Length

It w11l hr dmrablc to chcck the proposed dcsign at only a limited number of points to determine initially whether it will work. Total bending stress

t

Support

r

\ vr

'/i Span Point Along Length of Beom

FIGURE 14

Midspon

eams and Girderr

/

4.7-7

Total bending stress

'/a Span Pomt Along Lengih of Beam

Support

Midipan

FIGURE 15

Referring to Fignre 11; notice the bending stress (u,,) from the applied moment is assumed to be maximuin at the outer fibers of the flange. The bending stress (VT) from the applied shear is greatest at thc stem of the Teo because its section modnlns (S,) is less than the section inod~~lus at the outer flange ( S f ) For this reason, combinations of bending stresses must be ~.onsidr.reda t the outer fibers of the flange as \wil as the stem of the Tee. In Figure 14, thc total trending strcsses at the outer fiber of the flarige as well us at the stem of the Tee section arc plotted along the length of thc beam. This data is from a typical &sign l>rohlcm. ln this case, the vertical shear :it the support is = 25 kips. 111Fignre 15, tlic example has hren rcworkcd with cliffererit span lengths. and with diiferent applied nniform loading so that thr bcnding moment (and thc bending stress dnc to this moment) rtmains the same. Tho sliorter sixins reqnirt: an incrt:ased load, lierice increased shotir ( V ) . The longer spans require ;ilowcr load, hcner dncreawd sllear ( V ) . Notice in Figure 15, tllat for short beams with higlrcr shcar form relativc to bending moment, this curve for tlir total hending stress (moment and s11t:ar) will rise on the left-hand sidc, a i d tlic point of maximrnn strcss will movr to the left. or ncar the, support. Of conrse there is a limit to how short and how high the vertical shrar ( V ) ma)- he, bocausc this type of open web construction docs weaken thc web for shear. For

TABLE I-limiting

Ratios of Section Elements Under Compression

\'

very high slit:ar loads, tllc opcwing in the exp;\nd(d web would dofcat its pnrpose, and a stantlard solid wch l m u n 01- girder slio~~ld he u s d For longer spans, with rd;itivdy lower sbrar force to bending moment, this c ~ ~ r v\vill c lower. shifting thr point of inaximum stress to tho right, or near the niidspm. An altcmatc mctliod to finding the bending stress dircctly from the a p p l i d momcnt ( M ) is to convert the moment ( M ) into a concentrated fnrct: ( F ) applied at the centcr of grxvity of the Tec scction and assume it to bc uniformly distribotcd across the section. See Figure 18.

FIGURE 16

ation of bending stresses due: to applied shear and ;litplied momcnt lirs sonitxhcrt. hchvccn I ) the support (region of high vcrtical shcar) and 2 ) the midsparr (rcgion of high hcnding moment). This point of maxinmrn stress is indicated in Figure 17 by an arrow. Unless the hcarn is cxaniincd as in Figure 17 for t.he maximum stress all the way between the support arid micispan, it would he well to check a third point in addition to the support and midspan. A conveniont point wonld 11e at 'A span.

Then:

This bending stress is the s a ~ n eat the outor flange of the Tce section 21s well as the inner stern. It is now only necessary to add the g r u t e r bmding stress from the applied shcar ( V ) of the Tee section. Therefore, the smaller section modulus at the stem of the Tee section will he nscd, and only one st:t of total stress values will be considered. In Fignre 17, the applicd inomcnt ( M ) has been converted into a concentrated force ( F ) applicd a t the center of gravity of thc Tee section and assumed to he uniforndy distributed across the section. This ilh~stratesthat the point of maximum combin-

Thcre are threc mrthods of checking the horizontal shcar stress along the beam's neutral axis (N.A.): 1. Use: the n~nvcntionalformula for shear stress, Totol bending stress

FIGURE 17

I I i

0 Support

%a

&

%6

X

%a

% Span Point Along Length of Beom

%

%a

% Midrpan

pea-Web Expanded Beams and Girders

(

/

= VI- t9 ) . Then . increase this stress by the ratio of overall web ;egment to net web scgment (s/e) to account for only a portion (e/s) of the web along the nentral 'axis being solid. assuming the web to be solid

7

1

Assuming that

V,

+ V2 = V,, the average vertical

2 shear at this point, this becomes-

and

t. e

. . . . . . . . . . . . . . . . . . . . . . . . (.6 )

FIGURE I8

EB BUCKLING DUE TO HORIZONTAL SHEAR FORCE

2. Treat a top segment of the beam as a free body acted upon by the bending moment forcc. The difference in this force from one end of the segment to the other is transferred out as horizontal shear along the neutral axis into the similar section below. This horizontal shear force is then divided by the net area of the solid portion of the web section along the neutral axis. See Figure 19. By substitution: V!, =

Mz - MI which acts along distance ( e ) . d

This horizoiltal shear force is then divided by the net area of the solid web section ( e t w ) to give the shear stress:

3. Using the same free body, Figure 19, take momcnis about point ( y):

The web of a conoentional plate girder may have to have transverse intermediate stiffeners to keep it from bnckling due to the diagonal compressive stresses resulting from the applied shear stresses. If stiffeners are used, the girder will have a higher carrying capacity. This is because the web, evrn though at the point of buckling, is still able to carry the diagonal tensile stresses, while the stiffener will transfer the compressive forces. The web of the girder then functions as the, web of a truss. However, in the open-web expanded girder, treated as a Vierendeel trrrss, the opcn portion prevents any tension acting in the web. Therefore, a transverse stiffener on tho solid web section will not function as the vertical compression member for truss-like action. Since this solid portion of the web is isolated to some extent, the horizontal shear force (V,) applied along the neutral axis of the honm will stress this web portion in bending. The simplest method of analysis would be to consider a straight section ( I ? ) , Figure 20. However, the resulting bending stress acting vertically would somehow have to he rcsolved about an axis parallel to the

FIGURE 20

sloping ~ l g of c tliis tapervd web srction. Onc method by which tapered l~camsarid knees are analyzed is the 'it'cdge hlethod, or-iginally pn~posed by W. R. Osgood arid iatcr modified 7' y H. C. Olander (ASCE Transaction paper 2698, 1954). With this method, Figure 21, tlic non-parallel sides are extcnded out to where they intwscct; this becomes point 0.From this point as a rrnter, an arc is dr;iwn tlirough the wedge section reprmenting tlir scvtion ( a ) to be considered. Tlie section modnlus of this curved section is determined. Thc actual forces and rnoments npplied to the member are then transferred old to point 0. The? horizontal force (V,,) \vill cansc a moment at point 0. It can l ~ oshown that these forces and moments acting at point 0 i~allsethe bvnding stresses on t l ~ e curved section ( a ) of the wedge; sce Figurc 22. Moment acting on curved section ( a ) :

Radial bending stress on this ct~rvtdsection ( a ) :

FIGURE 21

where

Since: ni + f or cos 0

p =---

Since:

= 2 p 0 and t,v a2 S = - - . .. 6

FIGURE 22

eb Expanded Beams and Girders

/

4.7-1 1

buckling

+I FIGURE 23 Moment applied to member

Thereforc, the radial bending strcss along cu~.ved section ( a ) :

It can b r shown that the curvt:d section ( a ) haviug the greatest bending strvss ( u ) occurs at a distance of:

This value of ( m ) will be Icss than ( h ) and may be used in the following Formula 12 if ( e ) docs not exceed these values-

1 for

0 = 30', e 5 1.58 111

For most drsigns, this wo~ildhe true and Formula 12 could be rlscd directly witlmut first solving for ( m ) in For~nula11. This vahw of ( m ) for thc position of the greatest bending strcss may 11c inserted 1,ack into Fonnula 10, and tlre following will give the grratrst hending stress along ( a ) :

section, resulting iii ii~crcwcof the seco~iclarybending stress in the Tee st,cticin juT). -4s an alternative to incrmsing distanw ( e ) , it u-odd he possiblc to stiifen the outcr edge, of this wedge portion of the web by adding a flangt: around the edge of the hole in the well in the particular panel which is overstressed. Allowable Compressive Bending

Tlicre ;m, two srrggt-stions for determining tht: allowable coinpressivt~bending stress along the sloping edge of the wrrlgc scction of thr web: I. Trmt this srv.?ion as a prismatic member and apply ALSG Scc. 1.5.1.4.5 Formula 4; sce Figure 23. ATSC Fonnula @ for allowable compressive stress:

when:

and

Scc additional 11otrs; Section 3.1. I I : - hi? i l l thc ahow formula, C ,

-

2.83;

I I I I ~siiicc it carmot rxcwd 2.3 therefore C,, = 2.3 and AISC Formilla @ becomesTlir next step is to drterrnine the allowable cornpressive I~endingstress (d. If thc above bcnding strrss in the solid portiori of the web ( u ) is excessive, it might be possil~lcto incrrasc the distance ( e ) . IIowever, this will also increasc the length of the Tev

Scc: Tal11c 2 for viiliirs of Form111:i 13 for various stecls. 2. As an alterl~ntcmethod, treat this as a canti-

4.7-12

/

Girdea-Related

where:

TABLE 2-Allowable Compressive Stress On Wedge Section of Qpen-Web Girder For Various Steels

Co~lsiderthe oriicr fibcr of this cmtilcver as an element in coinprt~ssion.Using the resrrlting (I&) ratio, determine the allon-ablc c<~nlpressivestress from the AISC tables. l e Shear Stress

From eithrr 1'ormol;i 13 or the ahovr. Mcthod 2; we obtain the allov-ablr compressive bmding stress ( u ) . Since V,,=: 7 t, c and holding the inaxiinum bending stress (u,) of Formula 1.2 to the allowable (u ) , we obtain the followitrg-

= -3. v,, t m- -.0 4 t, e 0"

or: lever beam, and measure its unsupported length ( L ) from the point of inflection ( e ) to thc support; sec Figure 24.

-

6 hupport

"h

FIGURE 24 Reverse top holf end for

Formrda 14 for nlloud>le shear stress ( 7 ) has b e ~ m simulifird for various anrrles of cut (~ 0 ,, ) : see Table 3. If the allowablr: shaar stress ( I )in this web scction is hcld within the value shown in Formula 14, no f111.ther chock of web lxickling dut. to the comprtissive bending., stress will have to 1)e made, nor 1\41 this edrre ., have to be reinforced with a flancrc. ., To kcrp the resulting shcar stress within this allow;lble, either ( t , ) or (t.) may have to be increased; see Figurc 25. % ,

end

Reverse top holf end far end

1

I

t

Support

+ie,it-

YA point

-+lk

4 2 e , i C

ez

FIGURE 25

I % point

1

Support

/

Open-Web Expanded Beoms and Girders

Adjusting t h e Distance of Cut (el

The clistance ( e ) may be varied to providc the proper strength of the web, or the proper opening for duct work; sec Figure 8. Howevcr, as this distance ( e ) increases, the secondary brnding stress within the Tee section due to the applied shear force ( V ) also increases. In otht,r words, ( c ) must he snfficiently large to provide proper strength in thc web section, pet must be small enough to provide proper Ixnding strength in the Tec scsction. In both cascs, these s t n s c s are eauscd directly by the applied vc:rtic:11 shrnr (\!) on the mon~her.This lxcom<~s nioro critical m a r the, snpports whwe the shcar is thr: highest. Largcr trial W F beam sections arcp choscn rlntil the v:he of ( c ) will satisfy both conditions. It would be possible to gradually w r y the s k e of to thc ceuterlinc; howthe openinzs from the support .. ever, this \vould be diiiicult to fabricate. If this is desired. it might he better to use t\vo dimcrisions of horizontal cut ( e l ) and (e,), altcrmating them and reversing their order at the s/' point. See Figure 25. This would allow a larger value of ( e l ) for the strength of the web and a smaller value of (e2) for the strength of the Tcc section, near the support in the region of high shear ( V ) . In the central region of the girder between the 'I4 points where the shear ( V ) is onc-half of this valrre or less, these values \viil reverse, resulting in the smaller value of ( e 2 ) for the web and the larger value of ( e l ) for the Tee. The top portion of the cut W F bexm would be cut in half and each half turned end for end. This will require a butt groove weld. However, this top section is in comp~-essionand the requirrment for the wald will not be as severe as though it were in the bottom tensile chord. It might be possible to make this compression butt joint by fillot welding splice bars on cach side of the Tee section. This lap joint would transfer the compressivc force; the splice bars u~ouldapply additional stiffness and therefore a higlrcr allowable compressi\v strrss for this Tee section at midspan.

FIGURE 26

TABLE 3-Allowable S h e w Stress For Various Angles of Cut

n

6 = 4 s

_

r s

,

,8225 o.

This cutting pattcn~ results iii the hole at the wntcrlinc having twice the: lcngth as the othrrs. Howeve,-. this is the. ~rcgionof (mly high momont ( M ) ; there is almost no shear ( \ ' ) . This section should be snfficient if it car devclop the requirrd compn.ssion from the main btmding load.

The edge of the wedge section of the weh may be strengthmed against buckling due to the horizontal shear force, by adding a flange aroiind the web opening. Set: Figure 26. Here: S

.= At a

+

t , a' 6

Inscrling this into Formula 7, we gct-

It can he shown that the value of (in) for the position of thi~gn3at<.stbending strrss is-

FIGURE 27

This value of ( m ) could then he 1rst:d in Formula 12 for the bending stress. This \vould give the following formula for the greatest bending stress:

where:

EB BUCKLING DUE

TO CO

Any dircct triursvt~rsi~ loiid :ippiied to the upper flangc of the open-weh girdcr is carried as vertical shear on t11c web. Sce I'igmo 27. Since this resisting shear is eqt~allydividod b<,twet:nthe top and hottonr Tee sectioli chords, half of this transverse load applied to a unit 17mel scginent of thc girdcr (distance s) must he transfi:rrtd as compr~ssiorr dou~nthrough thc solid portion ( < * )ot' the web into the bottom chord. If it is felt that this solid wet) section, acting as a column, cimnot handle this forw; it could he reinforced with a transvcrsc (vcrtical) stifl'cner. Usually this force, one-half of tho applied trnnsvcrse load with tbc segmcnt ( s ) , is small. Thus, the resulting cornpressivc stress within this web section ( e ) is low, and stiffwing is not usually required. Compressivc stress in web section ( c ) :

The allow;~blrcompressivc stress would he found in the AlSC tnblt~s;wing V

Moment d i o g s

= 25k

Shear diogrom --

FIGURE 28

eb Expatided Beams and Girders

8. GENERAL OUTL1NE FOR DESIGN OF OPEN-WEB EXPANDED GIRDER

/

4.7-15

in order to keep tlw vertical shear stress in the stem of thc Tec swtion within the allowable:

l)esib% of an open-web cxpandcd girder will be faciliiated by following the design outline bclow. Its application is dernonstratd by working a. typical design l r o h l ~ ~ ml h: i g n an opcn-web expanded girdcr with a span of 38 ft to support a nilifor~nly-distril~~~te(l load of 50 kip. Ilrsign on tlic basis or wing ,436 steel and Eli0 welds, and angle of cut d, = 15". Sre Figure 28.

h

5

d,, - 2 d,"

STEP I . Determine the expanded girder's required

5ection snodulns (S,) at midspan for the main bending moment:

STEP 2. For the relationship of the cspanded girder'?

depth to that of the original beam, let-

STEP 5. Then

Assume it = about 1.5 STEP 3. Select a trial WF beam having a section modulus of-

1.30 Sh = ~= 86.4 i n ? (use this as a guide) 1.5 Try an 18" W F 5 0 # / f t h a m , liaving S , = 89.0 in."

Now, refigure K, using the S,, of the actual selcctect beam:

STEP 6. Dctcr-mine the allowable compressive bending strws on wedge srction of web, using modified AISC

SEC1.5.1.4.5 liormula

0:

STEP 4. Determine the height of ct it ( h ) and rol off to the nearest rnch or fraction of a n inch:

"

However, ( h ) cannot exceed the following value

Could nsstimv shear ( V ) is :itmiit !IS% 01 nmximiinr shear (at the snpporl) liei.;iosc first panel will be away from tha point of SiippOrt. Howcwr, bcciiise wt. arc not ;at the support, thwc will l w some main hmdirrg s t i i s s e s lo hr nd&d lo thiw: sctondaiy hi,ndiiig s t n s c s in thc 'Tec s c c t i m from applied slienr ( V ) , tlmce, it would be hrttrr lo use i d 1 vaiuo of shear

(V).

4.7-24

/

Girder-Related Design

Open-web expanded beam serves os longitudinal roof girder in the Tulsa Exposition Center. It provides the needed high moment of inertia, at minimum weight, and eliminates lateral wind bracing. Below, weldor is shown making connections of beam to the tapered box columns.

tlis shrar ( V ) at this ''i point is rcduccd to about lralf of that ;it thr slqiport, thc distancc ( e , ) may hc dor~ble thiit of (c,:) and still not inuxisc tlit rcsnlting secondary heiiding sticss (rr.,).Th<,ri,forrs,K:%=- c 2 / i s i sl~onld not t)l> Iws thas1 ?i. Using the t\vo dirntnsions ( e , ) and ( ~ 2 ) the ; ;ihovc formulas btworne:

P 9. Now determirir. the properties of expanded girder:

. At the suppoft, cheek the secondary bending stress:

FIGURE 32

A , ~ =A,

IMl

=

+

A,

-

*, ( 4. +

b t,

+ d,t,]

FIGURE 33

= 5.861 in.'

T

LtB i n z i

-= 105.53 in.l

UT

( W )( 8 ) ~ -- .- ~= 20,300 psi .1(2:16)

=-

Tht. :illowahlo cornpi-t~ssivohi~iiciingstress nxiy be found in ;i similar inaniicr to tliat of Step 6, exccpt the unsupportrd Icngth l i ~ r is o (o). At the support, thlw is no main bending mainexit,

TABLE 4 - 4 0 s Various Steels a = 22.000 - 14.44

hence no axial coinpressive force acting on this Tee section. The allo\vablc stress here is-

or, fsosn Table 4 of vdue5 for dllrelent steels-

cr = 22,000 -

-

3.81

= 20,200 psi

STEP I I . At midspan of girdel; check the main bending seess:

(as a compressiv(: or tensile stress)

a,

"50"" " . . - U)200 , psi (24.08) (5.861 ) --

1

(~ h ) 'c t,~

= 22.000 - 3.61

(.:;.I'

v d u r of ( h ) ; howcvcr, this will greatly increase the sea,i~d:uy bending stress ( Q ) of Step 1.0, since it reduces the depth ( d + ) of the Tce section. In this case ~indoubtcdly,the WF hcam selected cannot be used and a larger WF Iimm must h r tried. If tile miin Bmding stress ( a , , ) is within thi?r. allo\\~ablc,hut thr sectmd:rry lwnding stress ( m ) in Step 10 excocds the allo\viihle, ( u , ~ may ) be greatly rednced by decri-asing ( h ) with jnst a slight increase in ( u , , ) . Strtwes (u,,)and (u.,.) may bc considered according to AlSC interaction formulas @ , @ and @ , shown grap11ic:tlIy in Figurc 12. As a matter oi interest: Table 5 shows that decrmisiilg ( h ) rrsnits in a largc decrcase in the secondary bending strcss (u.,.)and n slight increase in the main bending stress ( u , , ) If (11) cannot bc rcduccd bccaosc (u,,) is close to the nllo\val~le, m e two different size holes, (el) and ( e 2 ) .Pn~\.ide;1 larger vnlne of distance ( e , ) for the compressive bending strcss in the \vcdge section of the woh, but a lower valw of ( e l ) for the cantilevered Tee section. TABLE 5

or (as a bending stress)

STEP 12. If the main hellding stress ( m ) in Step 11

is excessive, it niay be redlicecl slightly with a higher

earns and Girders

/

STEP 13. h/iake any adjustments necessary to facilitate fabrication. See the text immediately foilowing this design outline.

STEP 14. After the girder is detailed, the stresses may be rechecked in view of marc exact valrm of (V,) and (M,) since the cxact positions of the pancls are not known. .Also, i t may be well to check additioiial points between the point of support and midspan. SPC Figwe 34 and Tablc 6.

FIGURE 35

s

.

FIGURE 36

The practical aspects of structural fabrication may mean some adjustincut of original girder design is required. ame

number. The distance left over ( z ) on each side is-

Size Holes Are to be Used

If openings in the web are to be of uniform size for the full lcngth of the girder, that is c, = e2, and the open-web expanded girdrr is to he synzmetrical about its centerline, let n I : skumber d unit panek and use as a starting point in measuring a unit panel either: ( a ) Cmterline of wedge web section. Figure 35, or ( b ) Ceuterlinr of open Tee section, Figure 36 Divide the length of thc reqnired girder (L,) by the length of one unit panel ( s ) to get the number of units ( n ) . Then reduce ( n ) to the nearest whole

FIGURE 34

Since the length of the open-web expanded girder 1sL , = n s + 2 z the length of thr W F beam to he cut is-

L,?

--

-

(u

+ % ) s -+ 2 z

The extra length of WF beam required isL,

- L, =

% s

Girder-Related

FIGURE 37

I n either case ( a ) or ( h ) , there probably will be a small hole left in the girder at the ends which must be filled. The simplest method is to add one or a pair of web doubling bars or plates at each end to cover and lap over the holes. See Figurc 38. Web doubler plate

It m ~ g h tbe po?cihle to adjust the value of ( e ) so that the panels w l l fit exactly into the length of the girder (L,), See Figure 40. Here: L , = n s + e

Web doubler bar

= e ( 2 n i - I ) f 2nhtan4 First, determme the number of holes ( n ) from the following formula and round off to the nearest whole number-

FIGURE 38

-

If the same size holes are to be used, that is c, the c i d e r is not to be symmetrical about its centerline, then start a unit panel right at one end of the girder. The othcr end may have a partial hole in the web which will have to be ~vvered.The only advantage to this method is that just one end will have a bolo in the web to be covered. See Figure 39.

- e?; and -

, - f

L, = (n

+

Y2)s

+e

Second, find ihc required vairle of ( e ) from the iollowinlg formula- 2 n h tan e = .I>, .. 2 1 1 1 1

Hole in web on this end must be covered

/

+

,

., , , , , ,,..,, ,,,(20)

FIGURE 39

eb Expanded Beams and Girders

/

4.9-21

nsc

FIGURE 40

L,=ns+e=e(2n

This arljiistcd valiic of ( c ) xin not be less t h m that of S t q 8 in the dmign nirtiii~c,nor exceed the ~ ; i l u cwhich wonld rtwlt in ;iu extcssive secondary heridiug stress (w.,)i n Stcp 10. I f Ditferent Size Holes Are to be Used If distances ( e , ) and ( e 2 ) are riot to be the same, and the girder is symmetri<:nl ahout its cmteriine, then the following method may be employed. I13 ordcr to easily fabricate this type of opeu-web zirder. it is necwsarv to be nhie to rotate each tow half about the % point. This prcsents two possibiliticscase ( a ) rotation at the ' 4point a b m t the larger dirncnsion ( e , ) , and case (I)) rotation at the ?k point aboiit the sn~allerdimension [. c.. .). See Figure 41. Let ( n ) = number of holcs in thc web, counting the cvntc4ine hole as two holes. Iktermine the approxilnate number of holes from'2

. . . . . . . . . . (21) Case (a). There are an odd number of holes in each half, therefore: Adjust ( n ) so it is a rnuitiple of 2 only, and solve for (e,) from the foliowing-

+ 1)+2nhton+ I,

-

(11

-1)

h tan

+

. . . . . . . (2%)

Cn,w (1)). There artL an evcn number of holes in each half, therefore: Adjnst ( n ) so i t is a multiple of 4, and rolvc for (,,) fmin the fol1<,wing-

1 el

I, - (11 . - 1) . h tan . K:t) n (I

-+-

~

. . . . . . . . . . ( 22b)

In both case ( a j ;md cast, (I,) this resulting value of ( e l ) shonld not 131. less th;m tliat obtained in Step 8 and that jl~stused in Poi-mula 11 to find (11).

. TAPERED

BPEN-

c~~~~~~~~

pattern axis at a slight angle to the axis of tho heain results in a tapered girder. Sec Figiire 12. In ordcr to have ilie dcrper scction at the midspan, it is nwx!ssary to crit the top portion in half and revcrse these two top halvcs. The cut could be made in ihc lower portion; howc\:cr this is in tension, and a simpl<,r \w4d a)iild be ni;idc in the compression or top portion,

/

-

Girder-Related

Reverie top half end for end

Reverse top half end for end Q

FIGURE 42

In iapercd open-web expanded girders, the axial forcc in the chord which slopes has a vcrtical component (F, =. F , tan a ) ; here ( F , = M/d). Whenever this chord changes direction, for example at the midspan of the girdcr, this vertical component must be considered. It will be carried as shear in the web members back to the suppol-t, and in this case has a sign opposite to that of the main shear ( V ) . Hence, its effect is to reduce the shear over most of the girder's length, but to increase it in the midspun region. The modified shear bccomes-

This means there is a vcrtical shift of the initial shear diagram on each half of the gir-der, so that the central portion to be checked which initially had zero F,) as shear ( V = 0) now has a shear valuc (V'

-

. . . .

I

wcll as the maximnin hrnding moment. See Figrirt: 43. A transvcrsr stiffener at the point where the sloping flange changes direction would transfer the vertical component of the flange efficim~tlyinto the wob. The greater the change in slope, the more important this would bccome. If there is a panel opcning at this point, the Tee section must resist this vertical component in bending (in this cxample, the top Tec section). This is similar to the arlalysis of the secondaiy 1)cnding stress (ul.) due to the shear applied to the Tee section at midopening whrrre each half behaved as a cantilever beam. See Figure 44. However, in this ease, the cantilever beams have fixed ends ( a t the centerline of the girdcr); rrsdting in one-half the bending monient and stress. (This half length Tee section is treated ns a beam fixed at one end arid guidrd at the other end, with a concentrated load.)

.

Girder with Iood - vertical componenl (F,] causes shear in web

11

Shear diagram from applied load

Diagram of modified shear

V'

FIGURE 43

= v - F,

earns a n d Girders

Q FIGURE 44

The open-web expanded rolled beam is sometimes a n economical substitute for o heavy built-up plate girder.

In the 21-story Washington Bldg., open-web expanded beams led to significant savings in construction costs.

/

4.7-23

4.7-24

/

Girder-Related Design

Open-web expanded beam serves os longitudinal roof girder in the Tulsa Exposition Center. It provides the needed high moment of inertia, at minimum weight, and eliminates lateral wind bracing. Below, weldor is shown making connections of beam to the tapered box columns.

The concrete floor may be attached to the top flanges of the steel girders or beams by the use of suitable shear connectors. These allow the slab to act with the steel and form a composite heam having greater strength and rigidity. The concrete slab lxcomes part of the compression flange of this composite element. As a result, the neut+al axis of the section will shift upward, making the bottom flange of the bcam more effective in tension. By such an arrangement, beam cross-sections and weight can he reduced. Since the concrete already serves as part of the floor, the the only additional cost will be the shear connectors. The types of shear eounectors in use today take various shapes and sizes. Some typical ones are shown in Figure 1. In addition to transmitting the horizoutal shear forces from the slab into the steel beam making both beam and slab act its a unit, the shear connector provides anchorage for the slab. This prevents any tendency for it to separate from the beam. While providing for these functions, conllector placement must not present difficulty in the subsequent placing of reinforcing rods for the concrete slab. Because of lower shop costs and better conditions,

FIG. 1 Representation of five common types of shear connectors welded to top flange of steel girder to anchor an overlayer of concrete, Only short portions of connectors are sketched.

It

v

it is more economical to install t h e e connectors in the shop. However, this may be offset by thr possibility of damage to them during shipping, and by the difficulty presented to walkiug along the top flanges during er~vtion before the slab is poured. For the latter reasons, there is a growing bend toward geld installation of connectors. The previous APSC Specifications had no infonnation on the use of shcar attachments for use in a m posite construction. If shear attachments were to be used, AASIlO allowables were followed. These require the use of rather long fonnnla~to detcnnine the in&vidual factor of safety to b e used on the connector. It also made a difference whether the beam was to be shored or not shored during the placing of the concrete floor.

Facsor of Safely The ncw AISC Specifications recognize the use of shear attachments and, as a result of recent research on this subjcct, has taken a more liberal stand on this. The design work has been greatly reduced, and no longer is it necessary to compute the factor of safety. A more liberal factor of safety is now included in the shear co~rnectionfoimulas. The use of shoring is no longer a factor in the design calculations of the connector, since it has heen found that the ultimate load carrying

Steel beam

(a) Slab

(b) Slab o n one side

on both sides of beam

capacity of the composite beam is umffrcted whether shores have or have not been used.

of beam

dicate possible combinations of rollcd beam and concrete slab.

hear Connector Spacing

AASHO requires the determination of shear connector spacing, which may vary along the length of the beam. Now AISC requires just one determination of spacing, and this value is used throughout the length of the beam, greatly simplifying the worli. This is because the allowables are such that at ultimate loading of the composite beam, some of the comlectors will yield before the others. This moverncnt provides a redisbibution of shear transfer so that all connections are ultimately loaded uniformly, hence uniform spacing is allowed. Composile Section Properties

Effective slab width (B)

shb on both sides of beam, Figure 2(aj B 'h beam span a 5 M distance to adjacent beam a

A further help is a series of tables listing properties of possible combinations of rolled beams with typical concrete slab sections, similar to tables in wide use for available rolled beam sections. These new tables have been published in the AISC "Manual of Steel Constnzction,'' Sixth Edition, 1963, and in Bethlehem Steel Co.'s "Properties of Composite Sections for Bridges and Buildings." The new tables eliminate the various calculations for composite sections. A simple calculation will indicate the required section modulus of the composite section, and a quick reference to the lablcs will in-

k

In order to get the transfonned area of the concrete floor, it is necessary to decide how large a width of the concrete acts along with the steel beam to form the composite section. This is known as the effective width (B) of the slab. AISC (1.11.1) requires the foUowing:

--i

5 8 times least thickness of slab ( k )

slab on one side of beam, Figure 2jb)

5 B 5

B

B

4/12

beam span

'/z distance to adjacent beam

5 6 times ieast thickness of

slab (t,)

This effective width of concrete is now transformed into an equivalent steel section, having the same thickness as the concrete (t,), but having a width equal to I/n that of the concrete. See Figure 3. Here n, the modular ratio, is the ratio of the moduh~s of elasticity of the stcel to that of the concrete. From this transformed section, the various properties of the section may be determined.

I = moment of inertia of transfonned section, in.' S = section modulus for thc extreme tension fibers of the steel beam (bottom flange), in.3

i

~ i s t d n c eto outer fiber of tension flange

_k

FIGURE 3

Beams may be totally encased within the floor slab as a Tee section in which the top of the beam is at least 1%" below the top and 2" above the bottom of the slab, and encased with at least 2" of concrete around the sides of the beam. With thcse conditions,

ear Attachments

shear attachments are not iised (AISC 1.11.1). if no temporary shores are used, the total bending strass in thc telrsioil flange of tlre ciicnsed stcel hearn is figorod under two conditions: 1. The steel hcani acting alone for any dead loads applied prior to hardening of the concrete. 2. The steel beam acting with the concrete for any live loads and additional dead loads applied after hardening of the concrete. The henin shall be so proportioned that. the above stress nnder either condition does not cxcecd .66 5," ( M S G 1.11.2.1j. If temporary shores are used, the tension steel flange of thc enc;ised beam acting with thc concrete slab to folm the composite section shall be designed at u = .66 a;,* to carry all dead and livc loads applied

-

after hardening of tlrc concrete. If shcar attacl~mc~its are used, encasemcxt is not needed and it ilocs not inntier in the design whcthi:r temporary shores are used or not used. in either casc, the steel tension Range acting with the concrnte s1;rb to f o m thc coml~osites'~tion shall hc dcsigned at cr = .66 uJ"to carry all of tile lands (AISC 1.11.2.2). If no temporary shoring is used, the section modnlus of the cwnpositc section (S,) in rcgard to thc tension Range of the bcani shall not cxcced thr following:

I

1

( M S C Forniula 17)

where: S, = section modulus of composite section (rela-

';If steel section is not compact: a = .60 c,.

tive to its tcilsion steel flange) esign of Section

or Composite &onstructiora With Shew Attachments (I.I1.2.2)

Encored Beams (1.11.2.1) (no iheor ottach,nanti)

Section Modulus

Section Modulus Used

- -

With

Sharing

-

Withoul

and

Shoring

a,.

Me

= --

s.,

+ Ms.

5 66 -

0,

( .60 a,

'

I * o = .66

a, ior "compact" beams; otherwise a

= 6 0 or

(AISC formulo 17)

I

4.8-4

/

1

Girder-Related Design

-4

n

L

k "r 4 Uitimate iood condition

Within elostic limit

FIGURE 4

S, = section tnodnlus of steel beam (relative to its tension flange)

M, = dead-load moment prior to hardening of concrete

beam, is equal to the total horizontal forces ( F , ) from bending acting on either the slab or the beam. See Figure 5.

MI, = moment .due to live and additional dead Ioad alter hardening of concrete Table 1 summarizes these requirements for encased beams without shear attachments and for composite beams with shear attachments.

where:

B = eifective width of slab t, = thickness of slab ft - compressive strengtl~of concrete A, == cross-soctional area of steel beam A, = cross-st:ctional area of effective concrete slab cr, = yield strength of steel

.

Farces Carried by Connectors

For elastic design, the horizontal unit shear force is obtained from the well-known fonnula:

V a y f = ---I However in the new AISC Specification for building applications, the cl(2sign is based on thc shear connectors allowing the composite beam to reach ultimate Ioad. In the usual con~positebeam, the ultimate load is reachcd aftcr the full dapth of the steel beam reaches yicld stress in tcnsion. This forcc is resisted by the ~mnpressivearea of the concrete slab. See Figure 4. The total horizontal shear (V,,) at ultimate load to be transferred from concrete slab to steel beam between section of maximum moment and ends of the

Figure 6 diagrams the bending inoment that results in horizontal forces; compression in the concrete slab and tension in tlie steel beam. Thcse horizontal ultimate forces are then reduced hy a factor of safcty of 3, and concrete is taken at 85% of its strength. These formulas become:

4 t=-

rn v ,, =

3 5 f', .A,

-

. . . . . . . . . . . . . . * . . . . . .( 2 )

(AISC For~nnla18)

f:

k-%-I

F

(a) Neutral axis lies within steel beam

VI,

=

FJ,

=b

"

y

i

(b) Neutral axis lies within concrete slab

Vh

t, f',

FIGURE 5

= Fh =

A. my

Shear Ateochments

I

-

Moment dmgm

I

-i =I I I

L L L L

+ } F, = f: b t

7,-

-37

4-

(compresrioo)

F, = A, o, (tension)

FIGURE 6

I (AISC Formula

19) 1

The smaller of the two values above (V,i) is taken as the total horizontal shear force to be carried by all of the connectors betw-cen the point of maximum moment and the ends of the beam, or between the point of maximum moment and a point of contraflexure in continuous beams. The number of shear connectors needed within this region is found by dividing the above force ( V h ) by the allowable ( q ) for the type of connector used.

formulas, but applied a factor of safety of 2 and these became allowable loads for the conncctors. In thc meantime additicmal testing has indicated the cvnnectors to have greater strength than previously thought. Although AISC did not pnblish these final formulas with their constants (10, they did produce Table 1.11.4 of values for allowable loads on some of the typical standard shear conncctors. See Table 2. Working back from this table, the basic formulas for allowable loads on shear connectors would be the following:

Allowable loads

Formulas have been established to give the useful capacity of three types of shear connections. These are used hy AASHO in the bridge field with the proper values of ( K ) :

TABLE 2-Allowable orizontol Shear Load (q), Kips (Applicable Only to Sfone Concrete)

If'.= 3,000 if',= 3,500 if',

.= 4,000

Conneder

%"

channel q

=:

K2.( h

+ % t)w

(Ibs/channel)

where: w = channel length in inches spiral q = KB da

GC

( ~ b s / t u mof spiral)

Later the Joint ASCE-ACT Committee on Composite Construction recommended these same basic

diom. X 2'. hooked heoded stud i/B" diom. X 2%" hooked heoded stud %" diam. X 3" hooked headed stud '/s" diom. X 31/2" hooked headed stud 3" channel, 4.1 lb. 4" choonel, 5.4 ib. S' channel, 6.7 lb. I/2" diom. spiral bar 1/8" diom, spiral boi

or

or or or

o = length of chonnei in inches.

FIGURE 7

live load momcnt (6)

ML = -WI. - L 8

-

These will enable the engineer to compute the value for a shear cvnnector not covered in the AISC table. The connectors may be spaced evenly along this region and shall have at least 1" of concrete cover in all directions.

pziL7-l Check the composite beam of Figure 7, and its shear connectors. The following art? given conditions: 36" W F 150-lb beams on 7' centers, with a 6" thick concrete slab A36 steel, E70 welds, and 3000 psi cvncrete A nnifonnly distribnted live load of 240 kips Span of 40' between supports E n = = 10 (modular ratio) E,

"

dead load moment Steel beam = 6,000 lbs Concrete slab = 20,160 Ibs Total WD = 26,160 lbs

(240,000) (480) 8

proiection of conmete slab a S S t ,

5 8(6") 5 46"

a

5 'h distance to adjacent beam 5 %(84 - 12) 5 36" < 48'' OK

efjifectitic width of concrete flange acting with beam

B 5 -

beam span 5 %(40) Y4

5 - 10' or 120"

B = 2a

+ hi

= 2(36) = 84"

<

+ (12) 120"

OK

and width of transformed concrete area is 84" B/n =- = 8.4" 10

M ( dirtmce from reference axis N.A. = A to iicutral axis)

properties of steel beam section 36" WF 150-lb beam

I = 9012.1 i n 4 S = 502.9 in.%

A. = 44.15 in."

db = 35.84" bi = 11.972" tg = ,940" t, = ,625" properties of compo.rite section k

I

8

4"4,/

= 670 in."relative to bottom tension Range in steel beam)

Tro~isformed concrete a i e o

1

-

check bending stress in hcam Check the tensile be~ldingst~essin bottom flange of steel beam. From Table 1VB

= M,

+ 5,

S" (1570 -

+

(14,400) ( 670 = 23,800 psi < .GG uy

check secMon modulus Since no shores are to b e used, a further requirement is that the section modulus of the composite section shall not exceed-

I

+ 0.35 s, 4 400)] 5 [1,135 + 0.35 ( (11570 ) 1.35

FIGURE 8

,,., ,,

(5019)

5 2220 i n 3

Taking reference section (y-y) through the beam's center of gravity:

SCc

= 670 i n J

< 2290 in.3

OK

horizontal sheuli~ The horizontal shear to he transferred by connectors will be the smaller of the following two values: T", II

.-.

'

85 f', A, 2

.85 . -- -

(3000)(6 x54) 2

= 642.6 kips

= 794.9 kips

So, w e Vh = 642.6 kips

length of FUet weld

L = 2 x 1W' Use %" x 4" studs. From Table 2, q = 11.5 kips per stud.

= 20" force on weld

number of studs

leg size of weld (E70) or 60 studs from centerline to each end of beam. If using 2 rows of studs, use 28 lines on each end of girder. approximate spacing S

240" (half length) = 28 (studs)

= ,205" or use %"

h

Check: Welding lo .94" thick flange calls for minimum weld size of %6" , but the weld need not exceed thickness oi the thinner part joined, which is the channel. Hence, use % 6"

Place first line of studs at h ' of this space (or 4%") from end of beam; from there on give all studs full spacing (89/16"). hannel Connectars Use 4" 5.4-lb channel of 10" length. From Table 2, q = 4.6 w = 4.6 (10)

= 4.6 kips per channel number of channels n = - Vh q - (642.6) - (4.6)

= 14 channels from centerline to each end of beam, or 28 channels per beam.

Use %" diameter bar. From Table 2, q = 17.8 kips per t u n . number of turns

. = ! !v cl

approximate spacing S

=

24G" (half length) 14 (channels)

and use M of this or 8'k" for spacing first channel from end of beam. To compute the required size of connecting weld:

F = 46 kips, each cl~annel

= 36.1 from end to cnd or 37 turns from centerline to e a d ~end of beam. approximate pitch S

240" = -- (half length) 37 (turirs) = 6.49" or use 67/,,"

To compute the required connecting welds (E70), assume weld size is equivalent to a %" fillet weld (has same throat). Force on the weld is-

h

f = 11,200 o = 11,200 (%) = 4200 lbs/in.

we'd

lcngth of uvld at each turn of spiral

L = -9 f (17.8 kips) -

( 4 2 0 lhs/iin.)

= 3.18" or 1%'' on each sue

Application of one type of proprietary shear connector for composite construction, providing equivalent strength with less steel tonnage. Connectors welded to beams makes concrete slab integral with supporting member.

-OK

Lightweight stud welders permit shear connectors to be attached to girder flanges at high speed. Studs are the most papular form of attachment #or anchoring concrete floor slab to the steel girders, permitfing steel and concrete to act together for greater strength and rigidity.

Concrete roadway dccks may be attacl~edto the top flanges of s t 4 girders or benms by the use of suitable shear connectors. T h t w coi~riectorsallow the slabs to act with tlw steel and form a coinpositr heam having greater sbcngth and rigidity. 7% cont:rete slah becomes part of thc compression flangc of this composite dcmerrt. As a rcsnlt; the neutral axis of the section will shift upward, making the bottom flange of the beam more cfft4vi. in tcnsion. By such an anangcment, beam cross-scction and \veight can be reductd Since the concrete already serves as part of the floor, the only additional cost will he tho shear connectors. The types of shcar connectors in use today take various shapcs and sizos. Sonrc typical ones arc shown in Figure 1. In nddiijon to iransmitting t h ~horizontal shear forces from the slab into thc steel heam making both beam arid slab act as a unit; the shear connrctor provides anchorage for the slab. This prevents any tendency for it to separate from the beam. While providing for these functions, connector placement must not present difiiculty in the subsequent placing of reinforcing rods for the conmete slab. Because of lower shop costs and better conditions,

FIG. 1 Representotion of five common types of shear connectors welded to top flange of steel girder to anchor an overlayer of concrete. Only short portions of conneciors are sketched.

it is more cronomical to install thew connectors in the shop. Ilorvcver, this may be oiTsct by the possibility of damage to them t l u r i ~ ~shipping, g arrd by the tiifficulty presented to walking along the top flanges durixg i.rection before the slab is poure(1. For the latter reasons, there is a growing trmd toward ficld ir~stallation of cori~rectors. Erection procedures influenci. the d e i g n of the composite hwm. If thr girder or beam has proper tenrporasy support during construction, its d c s i p can be bastd on the dead loads plus live loads being carried by the composite section after the concrete has attained 75% of its %-day strength. If the girdcr is not shored, then the steel alone must he designed to support the entire dead load during the curing period, and the composite section designed For ;my live, impact, and additional dead loads. This usually requires greater steel cross-section than is required for thr eoniposite design using temporary shoring. Howcvor, in bridge construction this savings in stoel usually cannot offst:t the high shoring costs for the long spans iiivolved. As a result, most bridges are designed withont shoring. 111the negative moment regk~iisat the supports of continuous boams. the concrctc slah would hc stressed in tension a i d cannot be considered offodive in the design. Some bridge designers assume the reinforcing

1

Concrete slob/ Steel beam

(b) Slab on one side of beam

(a) Slab on both sides of beam

FIGURE 2

steel in this area to be effective in tension when proper shear attaclments are continl~edthroughout the area. This approach slightly reduces the beam's crosssectional area.

modular ratio, is the ratio of the modulus of elasticity of the steel to concrete. From this transformed section, the various section properties may be determined: m = statical moment = A, d, of concrete about neutral axis of composite section

Shear connectors should have at least 1" of concrete cover in all directions. They should b e designed for only the portion of the load carried by the composite section. horizontal shear

where: Vb = horizontal shear of steel flange, at junction of slab and beam, lbs/linear in. V, = total external shear a&ing on composite section after concrete has attained 75% of its 28-day strength, ibs m = statical moment of transformed concrete area about neutral axis of composite section, or the statical moment of the area of reinforcement embedded in slab for negative moment, in." I, -. moment of inertia of transformed composite section transformed area In order to get the transformed area of the concrete deck, it is necessary to decide how large a width of the concrete acts along with the steel beam to form the composite section. This i~ known as the effective width ( B ) of the slab (AASHO 1.9.3). Thus effective width of concrete is now transformed into an equivalent steel section, having the same thickness as the concrete (t,), but having a width equal to l / n that of the concrete. See Figure 3. Here n, the

I, = moment of inertia of transformed composite section, in.* S = section modulus for the extreme tension fibers of the steel beam (bottom flange), in.3 The moment of inertia of the transformed concrete section (I,) may be read directly from Table 1, the section modulus ( S ) from Table 2, and the coefficient value of m/I, for horizontal shear (V,)) from Table 3. Tables 1, 2 and 3 are from "Composite Construction in Steel and Concrete" by Viest, Fountain and Smgleton, McGraw-H111. where: n =: E,/E, = 10, the modular ratio B = effective slab width t = slab thickness design load (umking value) for one shear connector

Distance to outer fiber

1

of tenr~onflange

FIGURE 3

where:

Q = useful capacity of one shear connector, beyond which the connector permits an appreciable slip between concrete slab and steel beam, lbs F.S. = factor of safety Note: f', = 28-day compressive strength of concrete For most conditions, the uscful capacity ( Q ) of the she.= connector may be read dii-ectlp from Table 4, 5, or 6 which makc it unnecessary to work the above formulas.

useful capacity of one shear connector

factor of safety The factor of safety to be used in coinputlng the allowable design load for one shear corincctor, is obtained from the following formula*:

. . . . . . . . . . . . . . . . . . . . . . . .(7) "

AASIIO (1.95) now allows as an alternnte, a factor safety of 4 in lieu of calculating it with the above formula.

omen(. 04 Inertia, Transformed Gomposih Seckion Modular inlio n =

lo, h =

dfective slab width,

... 1

a

,

t = slab thickness

........

TABLE 2-Section

b = effective

Modular ratio n = 1 1 st"?:I,r,<,"

?.,.*,~,"

-.

I

i-n,<

Modufus. I Beam slob width,

,,%,,,,,>I,,. .Y,,. "I

t = slob thiikners ,,,.

=,,,,,,,,,~ , $ =

,>*A ,I,,

8

b - l r t

. :, W,.$,ii i :,B ?Bi

wr

W P ?d

it,

80 W Y i l i

WFZJ:

I 6

:so W F

iOI

O F 18: WF ,L W,?,
91,

I" :3<,

l 0 W I 110

:
U P 101 2: W P Y< 2:

2' Wi- ,Mi

WF

?i

$4

9,

W I U

9°F

1II

2%wi 7s Z i W P 68 2, W P ' 8 i8WF C e 18

W*

56

i B W F 60 16

Wf ,m

I8 W F 48 iii WF I 0 16 W P 36 34

wi

1.i W f

From "comporite C ~ r f r v n i o n in Steel m d Concrete" by Fountoin & Sineleton. Copyright Q 1958. McGraw-Hill B w k pany. Ured by permirrion.

vier*, com-

:
-

i

I i Fiom "Comporiie Conrtruction in Steei end Concrete" by Viest Fountain & Copyright Q 1958. MiGraw-Hill Book corn: Ured by iiermiision.

pmy.

Singleton.

TABLE 4.-Useful Capacity, Q, of One Stud Conneckw. Ibs. (h/d > 4.2)

where:

CONCRETE S l k t h G T ~ .

1 1

Scm d o .

I

pr

6,500 9,300 12,600-

7,100 7,600 r , 2 0 0 10,200 I 1,000 1 1,700 1 6.000 % 13.800 ~.... 15.000 --Note: A faitor of ioiety must be opplisd to t h e obove useful copacity, Q, to orrive at t h e working volue, q.

TABLE 5-Useful Capacity, Q, er Turn o( Spiral Connector

where:

MD, = max. moment caused by dead loads acting on composite section MD, = max. moment caused by dead loads acting on steel beam alone Mr. = max. moment caused by live load S, = section moddus of composite beam for extreme tension fibers S. = section modulus of steel beam for extreme tension fibers 3-Coefficient Modulor ratio n = 10,

b =

m/i,

for Horizontal Shear

ciiective slob width,

t = dab

Spiral wire dia. in.

Note:

..

C O N C R E T E STRENGTH, f',, - ...-.. -.

r

i--c0

p~--

0

0

psi.

... 7-..4000

A factor of safety must be applied to the above useful coaacitv, Q. to oirivs at t h e war4ina volue. s.

VD --- vertical shcar caused by dead load acting on composite section

VL

I=

vertical shear causcd by live load

spacing of slzcer connectom

thichnesr

where: s = spacing or pitch of shear connectors in the direction of beam axis, in. n = number of shear connectors at one transverse beam cross-section

q = capacity of one connector, Ibs Vh = horizontal shcar to be transferred, lbs The spacing of shear connectors sllould not exceed 24". ESlGN

OF CONNECTING WELDS

Welds joining shear connectors to beams should be designed to the allowable fatigue force (f,?.), for the range ( K ) of shcar stress and the working load ( q ) of the connector. See Table 7. where: K = min. shear ( V ) max. shear ( V ) h a m "Comporiie Cons
= leg s i ~ eof fillet weld, in. f, = allowable force on fillet weld, lbs/lin. in w

4.9-6

1

/

Girder-Related Design

Problem 1

I

Stud Connectors

To determine the working load ( q ) , spacing (s), and weld length ( L . ) for each of several typcs of shear connectors, for a typical composite section. In the building field, the total horizontal shear force to be carried by the shear connectors is based on the total bending force in either the concrete or the steel section resulting from the maximum positive moment on the beam. It is assumed this force will be transferred from the concrete slab into the steel beam by the connectors along a distance from the point of maximum positive moment ont to the end of the beam, for simply supported beams; or from the point of maximum positive moment out to the point of contrafiexure, for continuous beams. In the bridge field, this shear transfer is based on the vertical shear applied to the beam. In most cases this value will vary along the beam's length. For this reason, more than one section may have to be checked when the size and number of shear connectors are determined. This example considers just one point of application, the section near the pier supports, and assumes certain conditions:

Use W dia, x 4" studs. From Table 4, Q = 10.2 kips/stud. working load q =- Q F.S.

spacing of connectors (use 4 studs per transverse section)

weld length Complete contact surface of stud is joined to beam. No calculation of weld length is necessary. hannel Connectors

Use a 4" 5.4-lb channel of 10" length. From Table 6, L = 49.6 kips/channel.

Q

working load q

FIGURE 4

L

.=

F.S. -

Q

spacing of connectors

f,' = 3000 psi (concrete) m =4 i n

1, F.S. = 3.81 V,,, = 49.6 kips

(See Table 3 )

V,,, = 5.06 kips calculating for horizontal shear

= 10.75"

or use 10%''

allowable force on weld Assume fillet leg size of w = 3/1$" 600,000 cycles: V", I< = Vm,,

(+EL06 .-

kips) (+46.6 kips)

and N =

force on weld Pissumc fillet leg size of w = %" and cycles:

= 1.4 kips/in. of weld

- ( i 5 . 0 6 kips) ($49.6 kips)

required ueld length f

= 9.3"

<

20" actually used

N = 600,000

W -

-

7070 o K

(From Table 7 )

OK -

This indicates most channels are overwelded.

= 2.8 kips/in. of weld

Use %'' dia rod. From Table 5, Q = 21.31 kips/tum.

= 2.0" or I" each side.-in- .contact area .. -. .. -working load

q =-uQS

4.61" or use 4K"/turn-

Studs are widely used in both building and bridge work as shear connectors for composite construction. Quickly attached by efficient orcwelding equipment, studs serve to anchor the concrete slab to the steel beams. The composite beam provides high strength at lower cost.

Typical scenes of modern bridge work featuring composite construction. Prior to pouring the concrete deck, studs are ottached to girder flanges by specialized arc-welding equipment. Connectors allow the concrete slob to act with the steel.

1. REINFORCED CONCRETE

Many hridge designs use reinforced concrete slabs for floors. These may be suppoited by stringers and floor beams of the bridge. When no iloor beams are present, the concrctc floor is supported directly on top of the primary longitudinal members. On deck-type bridges, with the concrete floor resting on the top flange or top chord of the longitudinal member, the concrete slab may be anchored to the steel by means of shew attachmcnts. In this manner, the concretc floor becomes an integral part of the steel member in compression. This composite construction is rccugnized by most structural authorities as an effective means of insuring economy (particularly in steel tonnage); of promoting shallow depth and more graceful shuchual lines, and of improving the rigidity of bridges. Typical savings produced with composite construction alone are in the range of 8 to 30% by weight of steel. To be effective, of course, the concrete must always be in compression to prevent cracks in the pavement. Some types of shear attachments are shown in Figure 1. See Section 4.9 on Shear Attachments for

arid results in a savings in the amount of steel and cost of the bridge. 2. Snow does not remain on the grid floor; hence, grids greatly lower snow rcmoval cost during the wintcr. 3. Since s11ow and rain do not remain on the grid floor, therc is no reason for a crown for drainage purpose"^ This si~nplificsconstruction costs. 4. For the same reason, scrlppers and drains are riot required. 5. Tlrr grid flooring a n be installed easily and quickly. Sometimes a light concrete layer is applied to the steei grid.

FIGURE 2

3. STEEL PLATE Steel plate welded to the hridge structure and properly stiffoned has been used for flooring. By welding a cornparativcly thin steel plate to the top flange of longi-

FIGURE 1

Steel grids may be used for floors for the following reasons: 1. Reduced dead weight of flooring. This reduces the required size of stringers, floor beams, and girders

FIGURE 3

tudinal members, a built-up section is produced which greatly increases the strength and stiffness of the member. This has sometimes been called "battledcclc flooring".

. TYPICAL

FLOOR SYSTEMS

The design in Figure 4 ( a ) utilizes a steel grid floor in order to reduce the dead weight of the structure. The steel grid rests on the main girders and the longitudinal stringers. The floor beams are set lower so that the stringers, when placed on top, will be flush with the top of the girder. Brackets ,ue shop welded to the girders to receive the floor beams. The top bracket plate is slightly narrower than the flange of the floor beam, and the bottom bracket plate is slightly wider than the flange of the Boor beam. This is so that downhand fillet welds may be used in the field connection of the floor beams to the girders. With a little extra care in shipping and creeting, it would be possible to shop weld the railing and like attachments to the girders and further reduce the field welding.

The floor system in Figure i ( b ) is made up of two longitudinal steel girdcrs with a concrete floor attached to the girders by means of shc'u connections. Althongh spiral shear cmncctiuos are shown here, this composite beam could b e made by using any type of shear attachments. Shrar attachments can also be used on the floor beams. in the design in Figure 5 ( a ) , the top portion of the girders hclps to form the curb. For this reason, the floor bcams mnst be lowered, so as to get the bridge floor helow the top flange of the girders. To keep this floor level down, the stringers nnl Between the floor beams and their top flanges are flush with the top flanges of the floor beams. Although this produces a very compact and A c i e n t design, it does involve a little more fitting and welding than the previous floor designs. A vcry popular design today is the continuous girder deck bridge, Figure 5 ( h ) . Several plate girders are placed side-by-side with sufficient cross bracing. A composite concrete Boor is attached to the top of the girders by means of shear connectors. For short spans, rolled beams are used with cover plates added

FIGURE 4

FIGURE 5

F l o o ~Systems for Bridges

at points of high moment. For longer spans, deeper plate girders are fabricated. For a more efficient design, these girders are deeper at points of high moment. The outside girders usually have their intermediate stiffeners placed on one side only, the inboard side, so that they have a more pleasing appearance. Box girders have been used for bridges; usually two or more are used. They may be joined by several metl~ods.The example in Figure 6 ( a ) uses floor heams flush with the top of the box girder, on which is placed a concrete floor attachcd with shear connectors.

/

4.10-3

with floor beams extending outward to support the bridge Boor. In Figure 6 ( b ) , longitudinal stringers are sapported on the Woor beams, and the floor rests on tticse. It has evcn been suggested that a similar design could he made from a large diameter fabricated pipe section.

5. TORSIONAL RESISTANCE Designers arr coming to realize the importance of designing bridge floors, etc.; with more inherent lateral stability and torsional resistance.

When a simple inernher is subjected to a torsional Box girder constrnction has sevcral :tdvantagcs. It moment, shmr stresses occi.ir; one set being at right presents a flat surface for otlicr r~tt;tch~ncnts; hmce, anglrs to the axis of the member and the other set the floor b c a m do not havc to be copad whcn they are welded to the girder. Then, is irss of a c ~ l ~ o ~ i o nltmgthwise. In I'igr~rr. 7, shear forces ( b ) act at right angles to the lengthwise member and causc it to twist. problem because of thl. flat srirfaces. Also, since tlie A fiat scction 01- any opcn section offers vcry little box girder ends may hc s c a l d off, the illside is protected. Perhaps the grmtcst advnntaga is the tremanrosistanec to twist. Thv cross membcrs are subjected dous incrcase in torsioi~iil rcsistanee offered by tlw to thr slicx forccs ( ; I ) and. likewise, twist. If a diagonal closed box section. It ;tiso lias good lateral stability. mrmber is placcd in the: strncture, both shear forces These torsional and l a t ~ r a lst;ihiliiy proper tit.^ nrc br( a ) mid ( b ) act O I I it. 'fowcver, the coinponents of coining recognized advantagis. and morc bridge engithesr forws. acting at right angles to the diagonal neers are making use of tlrim Some designs havr made use of a single box girder,

FIGURE 8

member, cancel each other out, so there is no twisting action applied to the member. These forccs do combine to place tension and compression in line with the member, thus placing the diagonal member in bending for which it is very rigid. Welding can be used to very good advantage in diagonal bracing. Figure 8 is From a bridge designed by Camilo Piccone and ermteted over the Rio Blanco River in Mexico. It is based on an earlier design of Thomas C . Kavanagh. The floor makes use of diagonal members which procluce a grid type structure, extremely resistant to twisting and lateral movement.

Thermal chianges in temperature cause certain physical changes in the size and shape of all construction materials and in their completed strudures. The changes are in proportion to the dimensions of the structure, the coefficients of expansion for the materials, and the number of degrees of temperature change. The structure contracts with the cold and expands with the heat, so a typical bridge might be approximately 1" longer per 100 linear feet in the summer than in the winter. It will also have daily and shorttime changes of a lesser degree in proportion to every change in temperature and it will have additional movements from the elastic deflections of the structure.

These changes in lcngth can be compensated for hy corresponding drformations within the structure itself. This is because changing the stress in the stnicture will also cause it to change in length in proportion to its modulus of dasticity. However, it is usually more economical to u s e expansion joints since the forces that are required to deform a structure are very large. Masonry materials such as stone and concrete compress elastically but will not stretch. Therefore, they are iilcely to crack when subjected to the stresses of temperature contraction. For these reasons and others, most structures are dosigned with provision for expansion joints at intcrvals to take care of the uormal movements of expansion and contraction and to relieve the thermal forces. Many types of joints in common use have been designed to do this, varying from open joints, simple planes of wealmess, 'md elastite joints such as are commonly used in pavements, to the long interlocking fingered castings and sliding bar joints used in bridge work. One Example

The all-welded expansion joint shown in Figure 9 is similar to those in the deck of a large bridge built in recent years. This joint is made entirely kom rolled structural plates m d angles at a great saving in cost by welding. It is typical of many cases wherein welding has

FIGURE 9

Door Systems for

not only simplified and improved bridge deck designs b ~ i thas also reduced the cost of the installation to corlsiderahly less than half the estimated cost of convrntional type of segmental cast steel fingered joints. The joint as shown provides for 16" of movement computed at the rate of 1%'' per 100' for the 1200' length of stnicturc. The joint (Fig. 9 ) is made in two halves, each half being symmetrical by rotating 1180" with respect to the other half. Thc joint integral with the curbs, extends the full width of the 24' roadway in one piece. This

-5

teeth. The slight side taper of %" in the length of the tooth adds to the clearance as the teeth are pulled apart. The 18" length of i w i h is dctwmined by adding 1" ciearancr at extremc expansion movements, plus a minimum lap of I" whcn the bridge is fully contracted to the 16" of required movement. The treth are spaced on 4" centers. This spacing is as small as practical in order to distribnte the loads from the roadway sm-fitce over as many treth as possible. It is also desirable in ordcr to avoid having large l~olesbetween the teeth when the joint is open. The

LAYOUT OF FLAME CUT TEETH

FIGURE 1 1

is fabricatcd to fit t l curvature ~ of the roadway crown. The intwlocking teeth which form the top surfaces on both sides ol thc joint are flame-cut in a single operation from a common 28" x I" x 24' plate as shown in the layout of Figure 10. The cut is made just wide enough to insure finish on both edges of the cut and to give proper clearance fur the final meshing of the

upper surfaces of the ends of the teeth are ground down and rounded slightly to insure a smooth transition of the loads from one side of the joint to (he other. The joint shown in Figure 9 is designed to support 16,000-lb Ii-20 tmck wheel loads with 100% impact. This load is distributed equally to each of five adjacent

4.10-6

/

Girder-Related

teeth and is assumed to be applied on a contact area 3" long, centered I.'?" from the end of the teeth. While in this extreme position, the teeth on only one side of the joint support the entire load. On this basis the depth of the web, the thickness of the plates, and other proportions are determined to support these load requirements. The unusually long cantilevered projection of the teeth is reduced by snppolting the teeth directly on an auxiliary end cross beam. The cross beams in turn are supported from the end flwr beams at 10'-3" intervals by means of cantilevered stringer brackets. The floor beams span 35' center-to-center of busses, and the trusses are supported on expansion rocker or roller bridge shoes. The strength of the tecth in this case is obtained by continuor~slygroove or fillet welding 5" x 'h" x 1'8%" vertical web plate ribs to the underside of each tooth, as shown in Figure 11. Thc rear ends of these ribs are anchored for uplift by groove welding to the back of the 7" x 4" x %" slab closure angle. This angle is continuously welded to the I" surface plate, and serves also as a latesai distribution beam between the plate anchors. Plate anchors composed of 5' x %" x 1'3" web plates are welded to the rear of the joint opposite the web of every fourth taoth. These plates are spaced at 16" centers, and each plate engages two Y4" jacking bolts to the flange of the floor beam. These bolts serve both as erection bolts for setting the joint to elevation and grade, and as anchor bolts to hold down the rear of the joint against uplift caused by traffic. The plate anchors lap with the main longitudinal reinforcement

bars in the slab for continuity, and the end of the concrete casts into the pocket formed by the surface plate and the 7" x 4" x %" angle. The vertical leg of the 7" x 4" angle is flame cut to fit the curve of the roadway crown before welding to the 1" plate. This helps to hold the joint in proper shape. The ribs are all held together at the bottom by welding to the 5" x %" continuous plate bolted to the anxiliary cross beam. The entire joint should be assembled in the shop with the cross brams :uid the field holes drilled to insure a proper fit in the field. Field erection consists simply of setting the bridge shoes the proper distances apart, shimming the end cross beams to proper grade, and a final adjustment of the jacking bolts and the bolts to the cross beams. The concrete slab is then cast up to the joint around the anchors and cured, and the joint is ready for traffic. One complete 24' joint as shown in Figure 9 weighs 6250 lhs. This compares to an estimated weight of 8500 lbs for a conventional cast stecl fingered joint. This comparison indicates that the welded detail accomplishes a saving in metal weight of 26%, in addition to rpplacing expensive cast steel metal with rolled structural material. The relative cost of rolled metal is much less per pound.

A very important type of floor construction is the orthotropic deck, in which all elements of the structure work together. Having principal application in the bridge field, orthotropic constn~ction will be covered separately in the following Section 4.11.

1. THE ORTHOTROPlC DESIGN CONCEPT

There is a growing interest in this country in the use of orthotropic bridge design and construction, a system now commonly used in Europe. With conventional bridge structures, the three main elc?ments-longitiidinal main girders, transverse floor beam, and lighter longitudinal stringers or stiffenersall act indeperldently of each other. Usually an 8" thick concrete floor distributes the applied loads; see Figure 1(A). In contrast, a11 elements of the orthotropic structure work together; see Figure 1(B). This new system uses a thin steel deck plate across the entire width and length of the bridge, and this serves as the top flange plate of the (1)longitudinal main girders, (2) transverse floor beams, and (3) lighter longitudinal stiffcners. The deck plate also contributes to the torsional resistance of the stiffeners when it forms a closed section. I-Iaving a common top fiange member, all three elements act and load up together in the most efficient manner. The steel deck plate is topped with a light 1%"thick asphalt wearing surface for complete elimination of the heavy concrete floor. The combined orthotropic deck st~uctureacts as a single plate or membrane with three separate sectional

properties: hending resistance about the x-x axis (transverse to the length of the bridge), bending resistance about the y-y axis (parallcl to the bridge), and torsional rcsisiance about the y-y axis. A corrcentrated load placed upon the deck plate is distributed over a wide area to several adjacent floor beams. The longitudir~al stiffeners below this load act as beams on elastic supports. With increasing load, the rather fiexible deck and stiffeners spread the load over a greater area. This action has been confirmed by many tests on modcls as well as actual bridges. In the tests of the model of one bridge, the computed test load corresponding to maximum allowable design stress was 2.06 tons. The computed ultimate load was 5.6 tons. During testing, measurements indicated there was perfect dastic behavior up to an actual load of 4.1 tons. When loaded above the dastic limit, there was no rapid and unrestrained increase in deflection as is customary in the usual bending of beams; rather the deflections increased linearly just a little faster than the applied load. At a load of 48 tons, a crack started to appear in the stiffener region, and at 56 tons this had spread over the entire depth of the stiffener. This test hldicated an apparent factor of safety of 27 to 1. With optimum use of welding, orthotropic construc-

/ $ " a ~ p h a i tsurface

Conventiono I1 Brldyr

,-. 'Noor

FIGURE 1

beam

tion rt-sults in the bridge superstruclur? ns~inillyweighing only half as rnrrch as woi~ldrmrlt froin any other design system. This weight :a\,ing is such a tremenclorrs advwtagr on lorig span bridges, that ortliotropic desig:r is rapidly replacing truss dcssign on a11 European bridgt.s having spans of 100' or more, and shoiild do thc same in this coiintry. AISC hns piiidishcd ail cxccllcnt design i~xirrunloil "Orthotropic S t r d Plate Deck Bridges" by Roman Wolcliirk (1963). It contains thcory, methods of design, and sr~ggcstcddetails of orthotropic bridges. This typo of hridge design ivor~ldbe impractic;il without the i.xtensive usc of welding. The miles of welded joints afford a good opportimity to sub-

~ ~ s s o r hthe l e sections (or anion~aticdobvnhand wilding and rnodrrn fabricating methods. Sincc riumi~rorssidmtical dwk sections are rrquinrd, they may ix. set up in ;i jig and autorn:~tirnlly suhmrrged-:trc w e l d d with ~nir~imr~in time and cost.

. JOINING

LONGITU lNAL STIFFENERS TO

ECK PLATE In Er~mpean orthotropic hridgt- design, longitudinal stiflcners :ire cominonl!. of trixprzoidal cmss-scction for torsional rigidity. .4mwican &%signinterest appears to favor this approach; sre Figure 2. -4ltho11gh riot too clear on the slwtrh of the Port hlann bridge, the edge

Hollow Trapezoidal Ribs and Connecting Welds

FIG. 2-Typical

interrupted

Mannheim-Ludwigshafen Continuous

1

Wesei Porta

floor beam

rib

4"

I

AiSC Standard (initial] (Feb 1960)

Web of floor beom. 11

-

floor beam - 1 I Port Monn

L6%'4 Poplar Street St. Louis (proposed)

They considered both interrupted and continuous trapezoidal ribs

FIGURE 3

ca

of the stiffener was cut square without any bevel. It was shown in tests by the f~lbricntorthat a single pass madi. with the aut.om;itic srihrrrerged-arc wclder would prodilce a sound weld with tbroat grmter than stsener thickness; see Figure 3. The torsional resistance of any closed tubular section, as indicated by Figure 4, is:

where:

[A] = tg = t, bR= b, =

-

area encloscd hy the: trapczoid thickness of deck plate thickness of stiffrwcr width of deck p!:ite within region of stiffcner umlevoloped width of stiffener

In designing the Port Monn Bridge in British Columbia, Canada, engineers specified orthotropic deck construction for maximum weight reduction ond dollar economy. Deck plate is stiffened by longitudinal troughshoped stringers formed by press-brake. Welding of stringers to tronsverse beams is done by a progressive ossembly technique for near continuous-flow production.

...

FIGURE 4

l'hv &sign Manniil fur Orthotropic Sttel Plate I h c k liridgcs innlti~licsthis torsional rc~sistairce ( R ) by a rcdnction factor ( p ) which lias Ivwr dciermincd hy trsting of varims shapes of stiffcm:rs. %is factor is afFcctrd by the shape. of thc stiffcncr. Stiffcncrs can readily fomreri to the trqwmidal shape oil a prcss hr;rke. Recmse of the torrnagc re(psired, it might hc more eco~iomical to pnrchase a spt.ciai irrill-rolled srvtion for the stiii'cnms; see Figure 5. T h ~ nthc outer portions of tlw platr w-kith which become webs of the lmilt-np trapmoid scction are rollcd thiimcr, m d tlic ccntral portion is left thicker for tho lowrr fiangc,. This places the inatcrial where required: f o r t h r(.ducing the bridge uvight and tonnage of stccl required. The plate conld bc rollcd to the final trapeztrid section, thiis ciiminating the braking operntion Imigths of this scction would nest and preseiit no problem in shipping. Another rt+hment \vould bc to pnwicie slightly greater tliickness at web cxtreniities so as to give more hearing against the deck plate and greater throat to the connecting weld.

Thitker section

FIGURE 5 Back~ngbar,

Y Two splices every /5'

Z 9r40ve welds

FIGURE 6

3. FIELD SPLICE OF LONGITUDINAL STIFFENERS There are two basic methods for detailing the intersection of longit~~dinal stiffeners and transverse floor beams; see Figure 6. ( A ) Following the common European practice, the floor beam webs run continuous and stiffeners are but to fit between the beams. The stiffeners are thus limited to about 15' ui length, and the main bending stresses of the structure in the stiffeners must be transferred trausversely through the w ~ hof each floor beam by means of groove welds ( T joint). There niight be a question of the possibility of a lamination in the web opening up because of the transverse force applied through it. This method requires a large nnmber of field groove welds to be made in the vertical and overhead position. There are 2 welds at each heam per stiffener. ( B ) An alternate method would be to have the trapezoid stiffeners run continuous throughout the length of the structure, with webs of the floor beams cut ont to fit around the stiffeners. This would clirninate any questions as to the safe transfer of main bending stresses.

This method wonld grcatly rcdnce the required field welding. For exaniple, the stiffcnrrs could be shop fabricated into 60' lengths; this would require just a single groove meld in the field every 60'. This would be a single groove butt joint in contrast to the 2 groove welds at each floor heam required by Alethod A. The critical field welding thus u~ouldbe only % of that required by Method A. In a translation of a German paper, "l'atigue Tests on Ilollow Rib C:onncctions" by FI. Hansch and C:. Mullcr, rcsnlts of fatigne testing three different dctails of longitndinal stiifeners were snmmarizcd: 1. The longitndinal stiffeners were internrpttd at the transverse floor beam wcbs and joined by fillet welds to the webs of t l ~ efloor boam. 2. The longitlldinnl stifl'mers cverc interrupted at the floor heam wehs, but \ \ w e \vrldcd with single bevel groove wclds to thc webs of tlif. floor beams. 3. The longitudinal stilfencrs ran continuously tln-ough the floor beam webs. The results sholved the continuous stiffener (1) to have the highest fatigue strength, cr = 28,000 psi, when tested with a stress range of

-

'Thc shape of tllc closed t~ibiiliir k ~ n g i t \ ~ d i n d stiffrnm- tcsted hail no appri.ciahlc rficct lipon the tcst I-csulls. Cold formirig of tlrta stiflcnrrs had no c i I t ~ t . T h q rcconimend thxt thr dcsigncr place the firld splice of tlw stiilcncrs iri low-stressed regions. 4. SHOP FABRICATE

SUBASSEMBLIES

l t is possihle to lal~ricatenrzirly the rmtir~.drck of the bridge, in sections. r~ndcroptimum shop nmditions and thcrchy miriimizp the amo~mtof fir.id w~icling.This includes dcck swtims lying 1xhw11 tlit. mnin box girders. and ;my swtiorls to h r c;intilrv
f i n d means of transpoi-t. in some cases :I barge. 1':acii longittrdinal joint ol thc top d ~ pink k can he made ~ using x \i,it!i a two-p:ws ivi~ld;o ~ i epass on r d sidc sr~hmergerl-arc:~utorii;iticwclder. This joint is a simple sq~~arc*-bistt joint witliout i~nyb;icLing bar, a ~ rcquircs ~ d ~ the first pass, rio l>r:vi:ling of plat[. edqm. : \ l t ~making tllr fonr floor bcams 21re r i i i ~ n ~ ~ welded all~ in place. Each bran1 consists of ;I hottom flmge p!atc and a a e h having t r ; i p r ~ o i ~ l<,litorlts d ;ilong the top edge to fit :~rorn~d e w h stifk:rir:r. With the, tr;~nsvrrsc.foor hc~imwelded in pl:icc;

FIGURE 7

v2 l S e p e ~on s deck

FIGURE 8

E

4.1 1-6

/

Girder-Related

the entire nnit can be turned over without undue strain on the incompleta butt weld. A second pass is taken to complete the automatic welding of the longitndinal joints, all in the dovvlihand position; see Figure S ( B ) . The result is a complete dock unit, 27' X 60', weighing abont 29 tons, to be hoisted from the barge into position between the two main box girders. The Port Mann bridge d t ~ panels k were fabricated and wc4dcd in the shop as units 65' wide, the width of the drck lying in between the main longitudinal girders, and 25' long, the distance 1)etwt:en tlie imnsvmse floor beams. Thesc panels weighed bctwecn 32 and 36 tons, drperiding upon the deck plate thickness. In Europe, panels up to 58' X 18' have bccn Fabricated and transported by barges to the site. The Save River bridge had prefabricated panels weighing 27.5 tons. The Mannheim-Lndwigshafen tiridge was erected in panels 18.5' wide and 60' long. The Severin bridge in Cologne was erected in panels 62.8' wide and 47 to 54' long.

. FIELD

FIGURE 9

hand position; see Figure 9. Longitudinal stiffn,crs would be field spliced by n~anr~ally groove welding tlic hntt joint using a light hacking bar placed on the inside of the trapezoid, very similar to pipe welding. The upper edge of the stiffener could be notched at this joint so a backing bar can run contin~iously across the deck to facilitate automatic welding of the deck piate transverse joint. Under these conditions, tlic joints of deck plate and stiffeners shciild be offset at least 2", as shown in Figlire 10, so each deck unit can he lowered down u-ithout interference of the backing bars.

ERECT10

The ontire superstructure probably wo~ildhe erected in units, starting from a pier support and cantilevering out. A travelirig crane coirld place tlic individual units. For any givm scgment of the span, the main longitudinal box girders would be put into position first. The field splice of tho top flange deck plate should be weidrd bec;rnse the l'h" thick asphalt floor to be applicd leaves little room [or splice plates and bolts. The erection bolts probably shoold bc on the girder webs. The girder's bottom flange may vary from %" to 3 or 4" thick platc, and could be spliced by field welding because field bolting of this thick plate would be costly. Transverse shrinkage of the weld on the $5'' dack platc witliin this 1x1s girder is estimate1 at ahout .03", and shrinkage of the groovc wcld of 3 3" bottom flange plate at about .10". Under this condition, a suggested I ~ ~ - ~ ~ c e disu ito - c weld the bottom flange to about ?* completiorr, thrw weld the top deck simultaneous with welding the remaining % of the bottom flange. In this maiinrr. botl~ilangcs shonld pull in togctller evenly. T l ~ rnest stcp would be rrrction of the sohassembled d w k unit hctucen these two main box girders.

Wit11 a dcck unit raised into place, tlie ends of each floor beam would hc field w e l d d to the main box girdcrs. The two lor~gitudinaljoints and one t r a n s v a x joint of thc l / ~ ' r deck platc siiould he weldrd in a single p s s with :I submcrgcd-arc tractor. Plates should be imrtially hrveled at the top and a backing bar i i s d so that iull-penetration u d d s can be made in the down-

ckiny b a r f o r ~ t I ' f f e n e r

e

Mew akcksection about t o be lowered inp/a.ce FIGURE 10

If there is :my doubt ahont thr fit-up of multiple stilfcnm for field splicing, (wls of the stiffrncrs can br left un\veldc(I to the deck plate for about a foot. This will permit thrm to ho i~idividuallyaligned horizontally for welding. If slxcific dimensions OF the stiffener indicates a possiblc prohlcm in accssihility for the wcldor in niaking the ficld splices, the deck plate can be left sliort by about 10" from cxch m d of tlir section; see Fignre 11. This worild also allow tlir back of the joints on the illside of tlic trapezoid stitrcner to bc root gonged and R mot or back p x s inadc. .4 20" wide deck platc section wolild thcn hc inserted, and two transvcrsc groove xelds made. This would doi~blcthe icngth of translwsc welds for splicing the deck plates; ho\wver, all of this wclcli~~g would hc automatic, singlc pass work. Ends of the stiifenrrs \vonld then 11c overhcad welded to this deck insert; as shown i n Fignrc 11. An alterriate way to field splice the trapezoidal stiifenrrs is to place the 1x:vrl on the inside and a backing bar on the outsidc; tire weliior then makcs all the splices while working from the top of tlie deck.

ecks

transverse eutonktic weld

o f deck

\ F;eldsplicc of stiffener

Deck R + stiffemer serving es the top

fluye,

1 4.11-7

js in

v

Deck E

stiffener servioj as the t o p +

f l m y e , is in tension

comprcssioo

FIGURE 12

This type of inspvction should he limited to critical joints \\hicli the Engineer should seli~ct.Fatigue conditions that reduce the allo\v:ible stress in design may indicate such a nwd; for rxample, groove wcldcd h t t joints snhject to tension, 3 \vide mngr of stress, ;I high stress, and a large ntirnher of cycles. As the factors that produce fatigue loadi~ig;il-e reduced, the necessity for mdiographic inspection is likewise rcdoced. If all of 1111. groove \vcicls in the deck plate are madt~by the suhmergcd-arc antomatic proccss, proprr procedures car1 1)c cstahlishad to insure good mvlding. d costly radiographic This should eliniinatc the ~ i t lor inspection ol tllesr \velds, altho~ighlinritrd spot chocks conld IF mad<>. Any ficld spli~i.in the lower flanga of the main box girticrs in ;i rcgion of pwitivc moment, rnight be inspected by radiograph)-. Fi,.ld splir:t!s in the longitudinal stiifcners must be considcrr.d from the type of loading: 1. The stilfecer si>rves along with the deck plate as the top f l a n g ~of thc main structure, and as such is subjected to tension in the negative rnornent region

near tlic pier supports. liowcver, this comes front the dred load of thr strvcturc and any live load sprt:;rd over ;t rather Izirge arm, thus the range of strtw varintion and the n ~ m ~ l w ofr strcss c.yr1t.s would hc ri:lativcly small; S
.I14

/

Girder-Related Design

eld splice in siiffencr

I L -----------------A

Deck @ in thsion ;bottom

Moment diaqram

Concentrated wheel load

Max. moment (due t o

?-

concentrated l o u d ) on deck section

Infhence /;nes showing shift of maximum moment as the concentrated toed moves along span. --

FIGURE 13

of the stiffrner. This hole can he filled later by- welding, or by tapping it and screwing a pipe plug into it.

. WELDOR

QUALIFICATION

In addition to the standard .4\1'S u d d o r qualification test, it would he well for those men assigned to field weld the stiffeners to &st weld a test joint of this splice in position. This can be givcn a visnal inspection, including sawing of the joint at one or more points and etching to determine if proper fusion was obtained. It might bo well to consider weldors who have had some experience in pipe welding.

I

Problem 1

to the shrin1r;lgc of the welds; see Figlire 14. To find the prnpcrties of this section, seiect reference axis ( x - x ) along ~ ~ n t l t r ~ ~sill-face c a t h of deck plate. This is almost through tlrc ciwtcr of gravity of the 2 welds, and tlw resr~ltingdistance to the ncritrxl nsis ( n ) will also hc the disiancc I~etwern the neutral axis aim1 the ccnter of gravity of wt.lds ( d ) .

= (279.87)

--

(-3.5.412)2 (Proin Table A ) ( a 7 9j

I

An orthotropic deck is to be fabricated in units 104" wide containing 4 trnPoidal stifleners cach 13" wide and on 11" centers. The stiffeners are weldcd to the 3 n /8 deck plate along their edges. If these nnits are 30' long, cstimate the amount of bending or camber due

-

-2.19"

also = d

/

Orthotropic Bridge Decks

4.1 1-

-L-N e u t r a l oxis

% (%") ('L") = 1 / 2 4

FIGURE 14

TABLE A

bending or camber

' "I

A = 005 A

-1-

I , = 30' = 360"

d L'

I n orckr to find t l i ~propi.rty of tlnis 111iilt-tip section, it is newss:rry to lmrw tlit: properties of the i ~ r cof a cil& whicli fornns tlie roui~dhuttom portion.

,585" ( r ~ i d swonld go iip this ;Imolmt)

Tl~is1~1e;inswhen tlie 30' long unit is upside clown for \ w l d i ~ r ~the . fixturt, should be curvcd suEcicntly to pull tlir. central section of thl. unit down by this arnouut ( ,585").

/

Problem 2

1

t r t i i i I I I n I t .\lami bridge in British Cohirnhia coiisists of tr;n,c;.oiclal stillpni:rs with r o 1 1 1 1 bottoins spac(d OII 2-1" cpntcrs and dcck plat<,. Tl~cs(.dtrck se<,tir~ns n-eldtd to u 'r" to i t \\-id,. th
FIGURE 16

k

11 m ~ Iw i d ~ n v t~l ~i a tIlic f(1l111wi11g is trii~.: h

2 t r H

dl

lg

~ t r t [ O ; I z s i ~ t ? ~

k

t n t r 0 1 riiiity

)

2

\il,~'

8

11

irder-Related Design

TABLE B

In this example:

= (323.35)

- (-38'76)2 -

(19.27)

(From Table B )

0 = 72.45" or 1.263 radians

.

bentling or camber A = .: 0 0 5 A d U ~

L = 25' = 30G"

-1

These values will now be used in finding the properties of the built-up section. To find these properties, select reference axis (x-x) along the w~derneathsurface of the deck platc. This is almost through the center of gravity of the 2 welds, and the resulting distance to the nentral axis ( n ) mill also be the distancc between the neutral axis and the center of grnvity of welds ( d ) .

= .48" (ends would go lip this amount) This means when the 25' long unit is upside down for wclding, tiit. fixture shoi~ldbe cnrved sufficiently to pnll the central section of the :mit down by this amount or about irY,

FIG. 1 Multiple burning torches cut heavy steel plaie to be used in fabricated bridge girders.

1. PLATE PREPARATION

Flange plates may be ordered as bars rolled to the proper width and thickness. No further prepamtion is rcquired excppt cntting to proper length and beveling the ends for thc butt joint. Some fabricators will flame cut the flange plates from wide plates; Fignre 1. Since there is some shrink;tqe due to the &%mecntting opwation, the flalrge will have a swoep or bend if it i? cut along just one side. For this reason the flange is rnadc by cutting alorrg both sides, usually with a cutting unit having mnultiple torches which are cut at thc same time. For girders with a horizontal curve, the flange plates arc flame cut to the proper cnrve.

2. FIT-UP A N D ASSE Fabricators having fnll-automatic, submerged-arc weld-

ing hc;rils usrx~llyfit thc flanges to the web ; ~ n dthen cornplcte thc fillct wrlc1ii1g. Platc gii-dcrs may be fitted a ~ assembled d by one of the follo~vingpl-occdures: First, one flange is laid fiat on the floor. A chalk Un,: is markcd along tlrc wrrtrrlinc of the flangc and srndl right-angle clips tack weldt~iat intervals along the Inngth of the flangc w a r this ceutcrline. See Fignrc 2. Next, thc web is plaetd vertically on the flange and temporarily siipportd with :~ngl
.?2-2

/

Cirder-

FIGURE 3

FIGURE 4

FIGURE 6

clamping method (such as wedges, screws, jacks, or in some cases compressed air) is used to force the flangc tight against the edge of the web. Thcse fixtures automatically hold thc flange in propcr vertical alignlnent. If thc wch is thin and very deep, caution must he used so that exccssive prcssllre is not used against the flanges because this may bow the web upward. See Fignre 4. Since the Ranges arc: vcrtical in the fixture, when the pressure is rcloascd and the web straightens out, the flange3 may rotate ;md not be parallel. I-Iaunched or fishbelly girders are usually asscmhlcd with the web horizontal in this manner. However, some fishhclly girders that ,are not too deep have hem assembled upside down with the web vertical. Sec Figure 5. What would be the stmight top flange is placed on the bottom of the fixture, and the web is positioned vertically. What would he tho bottom flange is asscmhlcd on top, and its own weight is usually sufficient to pull it down against the cnrved edge of thr web with little additional force or heating.

3. CONTINUOUS

If r o l l d hams with cover plates, plate girders, and/or hox girdcrs arc symmetrical, the firnr fillet welds will be well balanccd about the neutral axis of the section. Rtwuse of this, there should ho very little distortion or bowing of the gil-der. Sre Figure 6. The seqilcnce for antomatic wcldi~rgto produc? the four fillet welds can he varicd without major dfcct on distortion. In most cases the welding seqnence is hasetl on the type of fixturt used and the method of rnoving the girder from one welding position to another in the shop. In Figurc 7, the fabricator has two fixhlres to hold the girder assembly at an inclined angle. Thcse fixtures lie on each side of the automatic weldrr which nxns lengthwise on a track. Since, it is more difficnlt to cornpletely tnrn tho girdcr ovcr, the scqucnce must be designcd to do this as low times as possible. In Fignre 7 ; the girder assembly is first placed

ricatiom o f Plate

ideas

/

4.12-3

FIGURE 8

a

in the left fixturc and \veld is made. The ncxt casiest stcp is to pick up the g i r h with the crane hook(:d to the upper and swing it over to the is made on thc samt: flailgr right fixture. Heris but opposite side of the veb. Now the girder rr~risthe picked up, laid down on the flor~r,turned over, and placid hack into one of the fixtures where weld @ is madc in thc flat position. Findly the girder is picked and suvng over to the other fixture where weld 4 is made. In Figure 8, the fabricator uscs a set of trunnions on the cnd of the girder asstmbly, or places the girder within a serirs of eircdar hoops, so that the girdor may he revolved. After weld @ is com lctrd, the girder is turned complctely over and wcld is made. Now the welding head must be moved over to the back

b

&

FIGURE 9

side of the girdi.r and wcld @ is m;&. Finally the girder is hmwd coinpk~tr~lyover 2nd wi:ld @ is made. The dilfcrcncc in the above sripience of wrldhg pnsses dcpends twtirtily on thc fixtoring zind methods ustd rzither [ h m any &ect on distortion. 4. ANGULAR DISTORT10 STIFFENERS

Usually thr flangr-to-w-eb fillet welds have been tomplmd; the trmsvcrse stiEoncrs ;,re fitted and wcldcd into the girder; Figure 9. If the flanges arc? thin and wide, the girders may exhibit some angular distortion of thc flange platis. Ifthis has occiirrcd, thl. Aangcs may have to be forced

4.12-4

/

Girder-Related

apart before the stiffeners can be inserted between them. The following formula will holp in estimating the amount of angular distortion of the flanges:

/=oa?:Yo_/

FIGURE 10

girder before welding the flanges to the web. This is easily done since the unwelded flanges are flat (not distorted). With the girder weh in the horizontal position, the semi-automatic welders are used to make the fillet welds between the flange and web as well as the stiifenen in the same set-up. The corners of t l ~ cstiffeners are snipped so that the flange-to-web fillet weld may be continued in back of the stiffeners. Quite often all of this welding is completed in a single pnnel area before moving to the next. The girder is then turned over and the welding completed on the other side.

5. POSITION OF WELDING TABLE A

See Table A for value of 1) corresponding to actual leg of weld (a).

-,406

- -,543 .... .~ --

,688

. 1.000

...

AASHO bridge specifications (2.10.32) state that these stiffeners shall fit sufficiently tight after painting that they will exclude water. In addition, no attachments should be welded to the tension flange if it is stressed ahove 73% of the allowable. Some interpret the AASHO specikation to mean a force fit; this is costly and not necessary. The following procedure will comply with this: 1. Use a loose stifEener so it may b e fitted easily. 2. Push this tight against the tension flange. 3. Weld this to the web of the girder. 4. Weld this to the compression flangc. Some states have not been concerned with this tight fit and have cut the stiffeners short by about 1";these have been pushed tight against the compression flange and welded to the web, If just a single stiffener is used, it is also welded to the compression flange. The recent plate girder research at Lehigh University found that the stiifenrrs do not have to be against the tension flange in order to develop the full capacity of the girder. The new AlSC specifications follow this in allowing transverse inte~mcdiate stiffeners to be cut sl~ortat the tension flange by a distance equal to 4 times the web thickness. Fabricators having scmi-automatic welding equipment sometimes insert the transverse stiffeners into the

The girder may he positioned with the web at an angle betwoen 30" and 45" with the horizon, pcrnlitting the welds to be deposited in the flat position. This position is desirable, since it makes welding easier and slightly faster. It also pelmits hctter control of bead shape and the production of larger welds in a single pass when necessary. For example, the Iargcst single-pass fillet weld made in the horizontal position is about .X6''with a single wire, and %" with tandem arc; whereas in the flat position this single-pass weld may be about 3/4" u-ith either process. For a 1/4" or Gr' fillet weld, the position in which the weld is made, whether horizontal or flat, would not make mnch difference. If a %'' or ''%1 fillet weld is required, the fabricator has several choices. If the girder may be positioned with the web vertical, this will allow both welds on the same flange to be completed without moving the girder. See Figure l l ( a ) . If the fabricator has two welding heads, these two welds may be made simultaneously, thus reducing the overall welding time. However, this horizontal position does limit the maximum size of the weld which may be made in a single pass. If the fabricator has a single-wire automatic head, he must make this fillet weld in two passes. If he has a tandem setup, this weld can be made in a single pass with less welding timr. By tilting the girder at an angle, either a single wire or tandem heads can make this weld in a single pass; however, only one of the welds can be made a t one time. See Figure i l ( b ) . I t would b e necessary to rotate the girder for each weld with increased handling time. A fabricating shop with two automatic welding heads can make two fillct welds on the girder simultaneously. To do this, the shop must decide between two method^ of positioning the girder; Figure 12. It might be argued that method ( a ) should he used

(a)Two welds-multiple

(b) O n e weld-single pars

posr

FIGURE 12

-Y

Y-

lbl

becausr tlw girder is in~ichmorr rigid about this axis (x-x) m d thrrcforr: would d d r c i less as a result of the first two welds on tlir hottom Aarigc. However in method ( h ) tile weld is next to the neutral axis (y-y) of the girder. Its distance to this axis is rnr~ch less than that in ( a ) , and therefore it would have very little hending efi'ect on the girder. Since this is a thick ffange, therc may be concern about gcttiilg a large cnongh fillct weld to provide enough welding licat for thc mass of flange plate. Tlier:rcfore, it might also he argued that method ( a ) would provide douhle the amount of heat input on the flange. .4ctmlly then: should he little diffcrence between these rncthods in the efFect of wcld shrinkage after all of the welds have heen made

plate on cadi flmgr, this shrinkage on top and bottom flimges of the beam will halnncc and ihe beam will not distort. liowcvm-, if there is a cover plate on just the bottom flange, the unbalanced shrinkage will cause the centcr of tlw beam to how upward; in other words, it will increase thc camber of the beam. The cauihr~ingthat resoits from this unbalanced welding can be estimated by the following formula:

where: A

-:

total cross-sectional area of welds, sq. in.

6. COVER PLATES FOR BEA Many times, rolled bnams mnst have cover plates added to their flanges for increased sircngtl~.Usually two cover plates are added, keeping the section symmetrical a l ~ o uthe t horizontal axis. For composite b e a m having shear attachments on the top flange so that the concrete floor x t s compositely with the bean, a cover plate may he added to the bottom ffange for increased strength. All of tiiesc hcams mnst have a certain amount of camber. The u-clds conuecting thc cover plates to the beam Aange tend to shrink upon cooling. With a cover

~ & e r piote

Neutral axis of

of weid oteo

4.12-6

/

Girder-Reloted Design

If more comber is needed

Position of beam in service

Welded in this position

If less comber is needed

(a) When cover plate is less than flange width

FIGURE 14

position

in service

(b) When

If less camber i s needed

cover plare is greater than flange width

d = distrrnce from the center of gravity of welds to the neutral axis of the section, inches

L = length of the beam, inches I = moment of inertia of the section, This may be more or less than the final desired camber, Figure 14. If this camber due to welding is excessive, the beam must be snpported in such a manner that it tends to sag in the opposite direction before welding. If the camber due to welding is not enough, then the beam must sag in the same direction before welding. A good experienced shop man will support the beam either near its ends or near its midpoint so as to control the direction and extent to which the beam bends before it is welded. If the cover plate docs not extend to the full width of bottom fiange, it must be welded with the beam upside down, Figure 14(a). Supporting this beam near its ends will increase the final camber, and supporting the beam near its midpoint will decrease the final camber. If the cover plate extends beyond the bottom flange, it must be welded in this position and just the opposite technique must b e used in supporting it; Figure 14(b). The fillet welds holding this cover plate to the

beam should b e intem~ptedat the comer, if it is wider than the beam flange, as shown in Figure 15.

9. SHOP WELDING VS H E L D WELDING It is practical to do as milch welding in the shop as possible and to makc only those weids in the field that can't be made in the shop. The following two sections on the Field Welding of Buildings (Sect. 4.13) and of Bridges (Sect. 4.14) include some recomrnendaiions on shop welding specific connection joints. Cover plate

1Don't hook weld round corner; will not hove full throat

'~olled

beam

FIGURE 15

Hardwood bloiks

FIGURE 2

. ERECTION

Several methods of ieinporarily fastening these connrctions have heen used. Tack welding alone may br u~lsatisfactoryl~ecauseit does not malie :~llomvancefor plnn~hingthe hnilding before final welding. Clamping the beams to the colnmn scat is not ahvays safe, althmgh this hiis h w n itscd for "sito erection" of lighter strncttirrs; see Figure 2. The steel is ordered cut to length and delivered to the site of erection. Trmporary se;lt angles are clamped onto the colutnrr at the proper position, and a temporary lug clampc~lot~totlrr: top flange of thc btwn. The 11eam is hoisted into position and set npon

the. temporary seat angle of the coiornn. A tie bolt is thert s c r m d on to hold the beam in proper alignment with the colrnnn. Next, the hcam is weldcd directly to the colrmm, and any tcrnporary lugs then disconnccied and used over again. Saxe rrwtion clips, which arc w ~ l d ~todthe beam mds and the colrrmr~,have h c m ilsed with success; s w Figi1ri.s 3 and 4. Thcse rrnits mnsist of a forged steel clip and scat. The clip is shop wrldcd to the end of the bcnm, and tlit: scat is shop wrided at the p r o p a position on the column During erwtion; the beam is placcd in position so tllat the clips drop down into the sent. An adjnstnble clip has h e m devclopcd to take care of possible poor fit-up between the beam

FIGURE 3

FIGURE 4

HEL

and t h cohimn. It is rt~comniriidedthat th:. wor1;ing lo;id on any onc s u t sho~rld not c m w l 10.000 11,s. i f n gcatcr erection load is to h~ ciirriid. sucli iis a hoavy plat? girder or truss, it is r ~ ~ c ~ ~ m r n ~that ~ n i ltw-o e d or mow swts be used, side by sidc. The use of a feu. wcction l~oltslras 11cm found to br a satisfactory incoris of trrirlior;irily fastening b ~ f o r e\wldirig. fiolting n u y br: donr dire<:tly to main inc~~il~ei-s. It is I t s rostly to plmclr sinall attaclnnents for erection holts than to niovc hcavy main mr'ml~crs into the putich sliop for plriiching. Many tinios; holcs ;irv llatnc ciit iri thc ends of lisams for r:rt:ction bolts. In Figure 5 ( a ) , a sm:rll ronnt.ction plate is shop wel
FIGURE 6

hrmn cnd into pmprr ulignrnmt with the connection. I-iowevt!r, \vith tlit. :iccurxy of placing the welding stilds arid laying oril t l ~ ccorrrsponding slottcd Iiolcs so ;is to allow for sonrc horiz~~ntal ;tdjirstmcmt. tl~crt,should lic. little diificolty.

i'lrlrnl~ingof a 11dili1rgrisrially stiirts amrrrd un rlevator shdt or srrvicr core. This is rrsoally centrally Iocatcd ;1nd has grt,atrr lirii~ii~g. The butt wtxlds of the hram ; i d girdcr fl:iiigc,s to t h ~ supportirrg column \\,ill h a w sonr~.Iralisvrrsr siiriokirgr. It is ~ ~ ~ ~ c c s sthat ; i r y this shrii1k;ige be mti~n;it(dand t h ~ ,joint o p m r d r ~ pb y this amount bnfor~,\w,l(iing. Otherwis(>,this shrinkage will accunir~liitc~:dong the lcrlgth or width of the buildii~ga n d hiiild up to a sizal~k;mount. Sce Figuw "1. .A good r,stim:~te01' this transwrsc shrinkage is-

whcrc:

A,

-c

'I'he a-oss-s~~cti~~i?:iI arm of tbc wcld may 11e cornp i t d by hreaki~rg it (lo\r-rr into st;~r~dard arms; that is, rcctnnglrs for root opening. triariglt:~ for ilicludr~d imglo of !~rsvt+l,arid par:il,olas for wold I-einforeenre~it. This c:ilcdutio~i can he grmtly shortened hy making i ~ s cof starrtlard tal,lc giviiig thc wright oi weld mst:iI for v:irious joints; risc T:~blc6 in Section 7.5. It is only necessary to divid~:t h ~ wvalues by 3.4 to arrive ;it the arca of the weld. This \dire is then placed into one of the above For~n~ilas for shrinkage.

I

Problem 1

To dctrrmin~~ thr shrinkage dFccts in making the welds indicat~din Figilre 0. The ginlrr with a 1%" flange is to be \veld(d to :I colrirnn. The joint has a 'h" root olwning, an included angle of 45", and uses a backing bar. From Table 6 in Scction 7.5, the weight of weld metal is 5.93 lbs/ft. m d has an are;? of-

cross-scction;il area of weld

Before welding, open up joints to increase distonce between faces of columns to allow for weld rhrinkoge Beam or girder -~fter

I

welding, welds j-w lli shrink and pull columns back to proper distonce

FIGURE 9

.I34

/

Girder-Related

FIGURE 10

The transverse shrinkage is-

Using 'A'' fillet wclds on the w& will result in vcry littlc transverse shrinkage. The average width of a '/4" fillet weld is ?'V, and 10% of this is .012" or about 10% of the shrinkage of the flange h t t welds. In this example, thr joint of the girder Wangcs

FIGURE 11

would be opened up an cxtra '/a" on rach rnd of the girder so that the distance Letwecn the faces of the two n)lnmns is ?%" greater thiin the detail calls for. After w-c,lding. tllc two joints shonld shrink snificient to tiring the two columns back to the, dcsircd spacing. This shrinki~gccoiild he checked after w-elding and this v a h c adjnstcd.

Thr box coh~mnsin thc building shown in Figure

FIGURE 12

FIGURE 13

10, wcr? fai~ricatrdb y \wlding togt.ihcr four ailglcs. After they werc cn.ctcd; 21 short :niglc, scction was rrmovrd and a iong srciioii oE tlw girtltr- was slippic! into position within the colnmr~.Later the anglc swtion was put back. Thr ends of the hcams were coped back so they coirid b(5 slippcd into plaw with their top Aangc rwiing on thc top flangr of the girdcrs; Fignre 11. 4 short seat angle shop wclded io the girder web supporicd t h r lower hcarn fi;~nge.This r(~srr1tcd in a very fast crectiot~ proccdurr without the rise of crection bolls. Latcr the hottom beam Bang<. was field w c l d d to the girder web, wing the seat angle as $1backing strap.

FIGURE 15

FIGURE 14

-4 plate was placcd between the top bsam flangcs and tlir giudnr. Thc top Hangcs of tlir 1)cams w t w hntt groovc wclclc~Iiogr~ther,nsing the plate as a hacking strap. The plirtc was then fillct welded to thc heum Bangcs. A long cover plate \rm them vdrird h, ilic 1)c~im 8angi:s l o tnkc care of the incrcnsed negatiue inonimlt of the. b w m at this support point. 1V1)ticcthat this t y y . of w i ~ l d ~connection l rn;ilir,s the 11t.am contin~ious, t h m r l ~ yrrclncii~gits rcqnircd size. At the same time, it cloos no: tie the top ilangcs of the ),earn to the girder, which rniglit pridncc some l~iasialstressrs. All of the ficld w ~ l d i n gsho\vn lrerc was done in thr flat position, groatfy specding lip the crtbction \velding.

4.13-8

/

Girder-Related Design

FIGURE 16

Welding is iised quite extensively on rigid frames. Figure 12 shows the sliop fabrication a r ~ dwclding of sectior~sof a large rigid fi-amc. For small structures, the entire frame is fabricated and erected in one piwe. For larger strtictures, the frame may be divided into two or more sections and assemhlcd at the job site and eroctcd. Figun:s 13 and 14 show the construction of a rigid-frame freight ttmninal area, and the upright portions of the framc hcing ~infoadedfrom the railcar and hoisted into position by thc rail crane. Later tlie central portions of the arch were put into position. Welding macliincs, also on flat cars, were brought in and the field joints welded. Frames for tile Long Beach Ihrhor Sired were

;rssemhlrd on the groin~d,E'igort 15. The scctions wcre Inid out on wood blocks and jacked u p to proper position arid cliccked with n transit. The field joints were tlicn mcl~icd.T h e crawler crmcs picked tlie elitire frame np and pl;ictd it in j~osition. Some of the Elcld welding which was in:rcccssihl(~wherr on the ground, such as the back side of tlie web lxitt joint, was completed in the air. 4. WELDING OF JOISTS AND FLOORlNG

Welding is used univrrsally in tlrc attschmcnt of openvr& joist to heams. This becomes a simple matter of laying the joist on the heam at the proper place and l;rtcr wclding in thc flat position. A considerable amount oi light-gai~gt,stecl roof dt&ing is used on top of joists or beams. This is easily and qnickly attached by means of wddirig in thr flat position. The use of both openweb joist a i d sterl decking is shown in Figure 16. Flotx dccking of bravier gauge has been used as :I support for any of several iloor materials. Welding is used in the flat positio~l to fasten this steel deck to Imams of the steel strrlctorr. Many timcs this deck is designed to take the horizontal forces on the structure caused by wind or t:arthquakr.

5 . WELDOR PLATFORMS

FIGURE 17

It does not take much in tha scaffolding to support a weldor and his equipment. Many of t h r joints can bo reached without any platform; the weldor simply works off of the beam or works from a ladder. For welds below the beam, it may be necessary to put up a platform. Figure 17 shows a rectangular wooden platform with four ropes att;lcbed to it. The platform is fastened to the steel structure at the proper

Icvx~lby tbc ropes. Altliorigli tliis type of platform is sclf-contained, it is reiher hmi,y; cspcciallp for onc mall. Figrrrta IS slio\vs ;i sirnplr.r scafiold for a sin~ilar position in thr i d strii(.trirr. It is lighter 2nd easier Cor one mini lo set lip. Two wood pl:iiks have ropcs f;ist~wcdat tlicir ~ d s tlic ; miws art, tird to steel grab hcmks. Tho hooks, siipporting thc wood planks, are droppc~lowi- tlw ti111flange, of' tlic h r m , and the other two plmks arc, put into pl;icc. This platform can hc irscd oil all h i r n s lr;lvi~~g approsim:ltcly the same dcptli \rithoi!t ;in?. fririhcr :!djiisln?~~ntin the rope length. It c;in hr. r i s d in nlmost any coiidition. L~snallya weldor's lieilwr or one fmm thr crcctirrg crew will set np thc necessary sc;~Roidingal~codof time so there will he no delay in nelcli~ig. On large structnrfr u&h liavc ronnections reyniring quite a bit of w r l d i ~ ~;..tg the connections, it may help to rist: a woldor's cage \hphichhooks over the top flange of the bcanis and is pnt in place by the dcnick. This is SIIOWII in F i g u r ~19. I11ose cages can be c o v ~ w 011 ~ l tlircc sid1.s to f m n a windbrcok when used or, the ontsidr of t1i1. stwi strr~ctni-r.The weldor is not awnn, he is working :it 8 great height whon he is inside this shieltled cage.

FIGURE 19

FIGURE 18

Semi-automatic welding, using selfshieiding cored electrode, being employed i n making beam-tocolumn connections on WilshireArdmore Building in 10s Angeles.

Semi-ai~tomotic welding speeding erectio n of 32-story Commerce Towers in Kansas City, Missouri. Making weided girder connections in the open was facilitated by use o f lightweight compact gun and cointinuously-fed, self-shielding cored edectrode.

1 . BUTT JQlNTS 111 butt groove weldir~gthe cnds of Bang<,plates, some thoxght s l h ~ l dtx sivcn to thr kproper iype of joint. J and U joints require lh(: l r u t amolmt of weld metal; however, these joint typrs gmrrally require the plates to he preparcd by planing or milling which is impractical in most structtird fabricating shops. This limits the preparation to flame beveling, giving n V joint. In the V joint, less wcid ~ r ~ e t is a l necessary as thc inclndcd angle is dwrc:~std. Howevcr, as this angle decrcascs, thc. root opciiing mnst he increased in order to get the clrctrode down into thta joint and producc a sou~idweld at the root of the joint. Obviously, the on(: tends to o i h t the other slightly in rtspect to the amount of weld m&l necded. On thicker plates, the joint with the smaller inc,lr~ddangle arid larger root opcning, rtquires the least weld metal. If a h c k i n g strap is usrd. any arnourit of root openiiig within rcason can he tolerated, and ail of the welding most he done on thc same side; in other words, a single-V joint. If a backing strap is not rmployed, this root oprning must bc held to nhout '/ar'. This enables the root pass to bridge tlir gap and not f d l t l ~ r o ~ ~ g l i . The welding may be done on one side only, single-V; or it may be (lorre on both sides, double V. In cithor case, the joint is Imck-goug~xi from the opposite side to thc root bcfore depositing additional wcld metal on thc other side. This xi11 insure sound nictal throoghout tho rutire joint. Single-V joints may be acceptable if the plates are riot too thick; for thicker plates. double-V joints ;ire prderred sincr they reqriire less wcld metal. Kcmernher thtrt ;i singlc-V joint will pl-oduce more ang~rlardistortion 'This incrcnses rapidly ;is the Range ttlickrress iricreases.

Fieid Splicing

t'i~sld splicrs u s ~ ~ : i larr l ~ lomtnl on ;I siuglc>.plane. Slaggering the h t t iwlds 01 fiariges ;md wrbs will not irnl~roveperforru;irice of the giu~kr.It is much casier lo ~".c~pax:lhe joints ;uid maintain proper fit-up by flarnc-cntting :ind lxvrling whni a11 ure iocatrd in the snme plnnc. Sce Figure 2. Tlrcrr is :in advantage to haviiig estci~dedthr: fillet welds of l1:ii~gcs to the web d l the way to tlic wry crid of the girdcr. This provides h c t t ~ rsupport when thc flanges arc clamped togcther for temporary sl~pportdi~riiigerection. Most welding sqnonces for ficld splices of beams a r d girders arc*hasvd on tbc iollowing general outline

I

Manual-Flat

Shop Splicing

Shop splices in flange and web plates shoi~ldbe rnade before tht: girder is fitted together and wolded, providing the resnlting scctiotrs are riot too long or hcavy to h:uidle. These shop splices do not have lo lie in a single plane, hut are pl;~cedwhere they arc most convenient, or where a transition in section is clcsired. in the shop, flange plates can he turned over e;isily as woldir~gprogresses, so that on thicker plate? double-V joints would be osed. They require the least

L

w hwr

I

r powdcr E-6024 T .?S emps d 40% OF #/.6.?/lb. rydmgrn iron p*dw E~6018 /80 ompr k 30% O f %55/1b Sam; ;-Automatic -Rat 500 amps d 60% OP I. 05/16.

FIG. 1 Relative

cost of flange butt welds.

either the single-V or double-V type, depending on the flange thickness and the method of welding used. For higher welding speeds, such as when using iron powdered manual electrodes, or scmi-automatic, or fullyautomatic bubrnerged-arc welding, more of the welding would be done in the flat position, with less in the overhead position. It must be remembered that a single-V jcint will result in more angular distortion, and this increases

FIG.2 Three methods of preporing edges of girders for field welding. Placing the three welds in three different planes makes it difficult to get close fit. It is easier to lay out a i l three butt welds in same pione. Placing two flange welds i n the some plane and slighdy offseeing the weld i n the web offers o method of supporting one girder on the other during erection. in which both Aanges and web are alternately welded to a portion of their depth, after secnring with sufficient tack welds; see Figure 3. 1. Weld a portion of the thickness of both fianges (about 'h to %), full width. 2. Weld a portion of the thickness of the n e b (about M ) , full width. 3. Complete the welding of the Aanges. 4. Complete thc welding of the web. For deep webs, the vertical welding is sometimes divided into two or more sections, and a baekstep method is used; Figure 4. This will result in a more uniform trausverse shrinkage of this joint. Most butt joints used in field splicing the webs are of the single-V type. For thicker webs, perhaps above M", a double-V joint is used in order to reduce the amount of welding required and to balance the welding about both sides to ciirninate any angular distortion. Most flange butt joints to be field welded are

FIG. 3 &oth flanges and web are alternorely welded.

Direction o i welding: vertical up

FIG. 4 For deep webs, use back-step sequence.

rapidly as flange thickness increases. A double-V joint with half of the welding on both the top and bottom of the joint is best as far as distortion is concerned, but it may require a considerable amount of overhead welding. For this reason the AWS Prequalified Joints allow the double-V joint to be prepared so that a maximum weld of 3/a of the flange thickness is on top, and the remaining 'A on the bottom; Figure 5. This will give some reduction in the overall amount of weld metal, and yet reduce the amount of overhead welding. Table 6 in Section 7.5 givcs the amount of weld metal required (lbs/ft of joint) for the various AWS Prequalified Joints. This wiil aid in making a better choice of the actual details for the best overall joint. For the double-V butt joint for the flange, the State of Texas allows the field weldor to place the overhead pass in i l x bottom side of the joint first, and then after cleaning the top side to place the next pass in the flat position. Their thinking is that while some overhead weldillg is needed regardless of the sequence used, this procrdure eliminates a11 of the back chipping or back gouging in the overhead position. If the welding is done properly, there should be less clean-up required.

(0) Single-V groove joint. Simplest preporation. Tendency for ongulor distortion.

(b) Double-V groove joint. For thicker plate, reduces amount of weld metal. I$ welds alternote between top and bowom, there's no ongulor distortion. Unless plate is turned over, will require overhead welding on the bottom.

EB AT SPLICE Considerable questioning has been directed toward whether the web should have coped holes to aid in field welding butt joints in the flange. The disadvantage of the cwped holes must be carefully weighed against the advantages of making a sounder weld in the flange. Tcsts on 12" deep girders at the Unkwsity of lllinoisr have shown that the field splice having welds

* "Fatigue in Welded Beams and C,irdors", W. H. Miinre & J. E. Stallmeyer; Highway Research Board, Bulletin 315, 1962, p 45.

(c) When plates cannot be turned over, the amount of overhead welding con be reduced by extending the top portion of the double V to a moximum of 3/4 plate thickness. FIGURE 5

FIG. 6 Results of ioiigue tests on welded beoms with splices.

in a single plane and wing coped holes has a fatigue strength of about 83% of the corresponding splice with no coped holes s t 100,000 cycles, and about 90% at 2,O(K),000 cycles. See Figure 6. Knowing these figures represent the maximum reduction in fatigue strength because of the coped holes, it is felt these holes will do more good than harm since they insnrt. the best possible weld in the butt joint of the flanges, The reduetion in fatigue strength dne to coped holes on much deeper plate girders woirld seem to he less, since the reduction in section modulus ascribable to the coped hole would he mr~chless. Of course, any notch effect of the coped hole wo111d still be present. If necessary, tbis bole can he filled by u&hg after the hutt joint of the flanges is comp1t:ted.

Good fit-up is essential to the development of efficient welding procedures. This means proper alih~mentand correct root opening. Placement of flange and web butt spliccs in tire same plane greatly increases the ability to achieve correct root opening when the girder is pulled into alignment. Figure 7 ilh~strates a misaligned double-V butt joint in a girder flange at the point of transition. Note the offset of the joint preparation makes it difficult to reach the root of the joint and deposit a sound weld

FIGURE 7

throrlghont the entire joint. The flange joints should be checked for alignment throughout their entire length before weiding. This illustrated condition can exist at the ffange exiremitics even though perfect alignment exists in the web area. Accidental tilt of the Aanges during fabrication, mishandling during movement to the job site, or even a difference in warpage of the two flanges can cause this condition. The warpage problem increases with the size of web-to-flange fillet weld and decreases as the flange thickness increases. Various methods exist for correcting this condition. Figure 8 illustrates one such method. When the p l a t ~ sare not too thick, small clips can be welded to the edgc of one plate. Driving a steel wedge hetwcen each clip and the other plntc will bring both edges into alignment. Welding the clips on just one side greatly simplifies their removal. Figure 9 illnstrates still another method wlucb is used comn~onlywhen problems develop in respect to misalibaed thicker flanges. Here (top sketch) a heavy

FIG.8 Weld clip along one edge only, so it may be removed eosily with o hammer. Drive steel wedge below clip until piare edges are in alignment.

(a) Plates forced into alignment and held there by means of strongbocks. Pressure is opplied by means oC wedge driven between yoke and strongback.

(b) For heavier plates, pressure may be applied by means of bolts temporarily welded to the plate. Strongback is then pulled tightly against the plote.

bar or strongback is pulled up against the misaligned plates by driving steel. wedges between the bar and attached yokes. An alternate method (lower sketch) involves the welding of bolts to the misaligned plate ;ind then drawing the plate up against the strongback by tightening u p on the bolts. 4. RUN-OFF +A S OR EXTENSION

Rutt joints of stress carrying members should, where possible, be welded with some type of nm-off bar attadled to the ends of the joint to make it oasicr to obtain good quality weld metal at the ends. In general the bar should have a similar joint prcpnration to that being welded: gonging or chipping may be osed to provide the depth of groove. For automatic: eldi ding, the bars should have s~lfieientwidth to support the flux osed during welding. These bars are {isu~allyremoved after welding. A flat run-off bar may not give proper support for weld metal to keep the top comers of the plate from melting b:ick at the mds; Figure lO(a), i f the bars were placed high m o i ~ g hfor this, they would be above the groovt: of the joint and \vould interfere with proper welding at the ends; the welding wire (if automatic welding) \v[.ould have to drop down into the groove at the start and climb out at the other end very quickly, undoul~tedlysticking; F i y r e 1 0 ( b ) . The flat run-off bar in Figure 1 0 ( c ) for manual welding does not give proper support or maintain the

FIGURE 10

sides of the welded joint at the ends as welding progresses and requires special effort on the part of the welding operator to build these ends ilp. The types of run-off bars illustrated in Fignre 11 w o d d give the proper equivalent joint detail at the ends.

FIGURE l l

Steel sulky seat aids weldors on bridge construction. Float a t left lacks stability in windy weather, while sulky a t right enables operator to sit comfartably and safely.

Shop weld-fabricated girders of variable depth provided important economies and facilitated erection of Thompson'r Bridge near Gainesville, Georgia.

Determining Weld Size

/

7.4-5

s 2 4 d d 2 t*

+

<

2% tw

Spacing and Sire of SIof

L

s

w

2 t*

10 t,

+ X8" 5 2% t,

s,24w ST, 2 2 L r 2 t* 4. PARTIAL-PENETRATION GROOVE Partial-penetmtion groove welds are allowed in the building field. They have many applications; for example, field splices of cohimns, br~ilt-upbox sections for trnss chords, etc. For the V, J or U grooves made by manual welding, and all joints made by snhmcrged-arc welding, it is assirn~ctlthe hottom of the joint can he rcached rasily. So. thc effective throat of the weld ( t , ) is equal to the ;ictlinI throat of the prepared groove ( t ) . See Figure 13. If a hevcl groove is tvclded manually, it is assumed that the wcldor may not ( p i t r reach the bottom of the groove. Thcrefore, AWS and AISC deduct 36" from the p r c p r c d groove. IIere the effective throat ( t , ) will q ~ a the l throat of the groove ( t ) minus %". See Figure 1 3 ( a ) .

(a) Single bevel joint

(b) Single J joint

FIGURE 13

Tension applied parallcl to the weld's nsis, or compression in any direction, has the same allowable stress as the plate.

Tension applied transverse to the weld's axis, or shear in any direct~on,has a reduced allowable stress, e q d to that for the throat of a corresponding fillet weld. Jnst as fillet wolds have a minimnm size for thick plates because of fast cooling and greater restraint, so partial-penetration groove welds have a mininium cffective throat ( t , ) which should be used t, > =

where: t, = thickness of thinner plate

a. Primary welds transmit the entire load at the particular point where they are located. If the weld fails, the member fails. The weld must have the same property as the member at this point. In brief, the weld becomes the member at this point. b. Secondary welds simply hold the parts together, thus forming the member. In most cases, the forces on these welds are low. c. Parallel welds have forces applied parallel to their axis. In the ,case of fillet welds, the throat is stressed only in shear. For an cqnal-legged fillet, the maximum shear stress occurs on the 45" throat. d. Transverse welds ]lave forces applied transversely or at right angles to their axis. In the casc of fillet welds, the throat is strcssed both in shear and in tcl~sionor comprrwion. For an wpal-lcggcd fillet weld, the m;iximum shear stress occurs on the 67'h" throat, and the masin~umnormal stress ocmrs on the 22%" throat.

Flexible connedlon No i e i t i o l n t , R =

0

Moment diogrnrn

-

Full reitroint, R = 10096

Moment dioqiom

Fully Rigid

(3 Poitiol reitroint

Moment diogiam .-

Moment diogiam

FIGURE i

I,<, <~q~l:ll, 01- hl

~ :: I < , : I'Y I>, This \ ~ l I ~ i l~)ro(lli?
earn-fa-Column Connections

/

5.1-3

4. RIGID CONNECTIONS (Elortic Design)

5. PLASTIC-DESIGN CONNECTIONS .l'hc nsc of \veld~!dcrninc<.tionsbased on plastic design 11;is scvci-;:I ;~dvantages: I. i\ more a c w r a t ~ in
OF WELDED CONNECTIONS

Onc Ivay to lvttcr undcrstaiid tbc behavior of a Iirain-to-wliirrrlr cc~iir~cctio~i ~ n ~ d load, t ~ r and its loadc;trrying c;~p;~city, is to plot it on n rnor~lcnt-rotntior~ chart; sce Figure 2. The vertical ;tsis is tlw c t ~ dmomci~tof thr b w m ,

Beam Ice at working load

End rotation (0,). rodions

FIGURE 2

\vhii.Ii is ; i p p l i d to tlii c.mlr~i.ction.T h e liori2o11t:il axis is tile rcsiiltilli: rot:ition iii rndi;lns. kisically tilis is

coinpicti,ly rcstl-ai11r4 (0,. -- 0 ) , in othcr words f ~ x l ~ & chKii11, ~ ~ d ;llld is t Y ~ l l dto-

cqimtion cxpi-essiiig ihc rcsnltir~g eiid moii~cnt ( > I v ) alrd e n d r.ot;itio~i 1 0 , ) . inr a iiiiifor~iiiy l o a d d bciun and ;my cild r.cstr;~i~,l froiri ( ~ ~ m i p l c rigid tr to siniply s u p p t 1 ~ 1is: ,

I'oint h is thc tvrd rotalion whcll thc collrlcction has no ri,struint (.\I,. :- 0 ) . in other i r i ~ r d sa si~iiplebeam. and is cqnal to-

( b ) 0,. = :

This is a straiglil line, 1i:i~ilrgpuints o and b on

thc cliart. Point a i s the e ~ l dmoinci~twhen t l ~ cwilncclio~iis

;i

\\' L'> ~~- . 24 il I

-~ ,~

For. inwe:iscd loads on the l m m ~ t, h e beam linr iiioYrs o ~ i parallrl t lo tlic first line. wit11 corri?qxmdii~gl) incrc:ised valncs of end i n o ~ i m r t ( ) and thc end rotation ( 0 , ) . This (d;ishrd) sccond 11cam linc or, t h e

Beam-to-Column Connections c,)l:~rt~ q > r < ~ s m tllc ~ t xldition s of ii saft>ty f w t o - , ~!I~IcI is ,,sll~lIly 1.67 t c 1 2 ti17it.s t1iat of ti,,, &st d t i d i is l>:ls.g i w s [ I t ( , riw~liii~ig L I I I C ~ tS h givcw 1~1;id1:rom this it is ~ W I 111n\: I r~jt;itio~i tlie he;~in'sl x ~ l i i i i i ~~ ~l qr ~ c r ~on i l sits c o n ~ i r c t i o r ~ . It is ;issnni:~il. i r ~illis w x , . ilic IXYII~I is sylnin<.lri~ ~ ~tllc I IS;IIII<,. S i , d l y l i ~ l c ~m c ld the, t \ w end ~ T I I I ~ ~iw<, 113 this \ m y hilt11 < w l s will react si~nil;irly. O I I>I. \?t,ry CIII-vc1 r ( , l ~ ~ - t w , nat sk x i l ~ l c~~~ I I I I I K : ~ At Imv I ~ I O I ~ I ( it~ I safvly I~ >,ic,lcls ( M I ) mid allo\vs t!~c>W ~ I ~ ti^ to rot;itiz ( 0 , 1 . This is typical of top :i~?glt, slid ti)]) pI:1tt, <.i>lllll~c.. <:~l1lll~~ctiolls. \v1+1fr~llllillg;lllgl~~s, c Notiw, N ~ > I I with tIr<~sc~ t i srll r g h I i s ~ I - ~ Ifkxihli~ I I ~ ~ eor~~~c~i:tions, SOIIIC. < ! I K ~ I I I < I I I I ~ , J I&~w s

-

Simply supported beam designed for R = 0

Fixed end heom designed for R = 100%

FIGURE 3

5.1-5

w d tlw h ; ; m I i r ~ t >ill 1%; lo;>d l-<,htivt~ to 1hc.i~crossilig q (11ill<.IXYUII lin,~at w o r k i ~ ~10;id. nit, Z ~ ~ . I L I Cr tI>Is l l ~ t 5 ,I{ t t > ~ t i r l~gI ~ I - W~ O I I~ I X I PC O ~ I iic.ctions OI: :ill IS" \ii;F S'i* I K W I arc ~ I s1iou.n in Figtirr 4, 'i\,r(! w ~ d i t i m si ~ r ;irnl i r ~ c , 11 is 11:isid oil a ~t c t w t < ~ of t h e ~ W I II I ~ Vh ; ~ sthis I I I O I I X W ~of 1,; \Y I , m i l ;i rc4r:iint ~ v i Ir I t i t i 1 s ~ s I s central portion ! ~ f the l i c a n ~ . t l ~ i r kplatt:, 3" widi. :it the T o p plat? ;II is ;I 5,;'' r l c c s~~c:timi,: I I I ~has ;I i : r w s - s w t i w d nri.;~ of ii,, :~ .>J.$ ill.: It is \vid(,r~eclto 6" :st tht, h1tt-\vc~1dcd ~ O I I I I W ~ ~ O T11is II. i w ~ r ~ d i os1101!1d n ~ w d yield i a! :11w11t f -= A,, I I , - ( 4 ) (:30.000) (18) r- ,553 i ~ ~ . - k i p MI iii..lsti.;rint of Ii 100% thc ~ ~ I I I I I W ~ ~ O IcIn' Ss w \?oiild !MW inters c c t ~ >the d 11w1mI i w 1) illst shml of t11c R = 505 v a l ~ ~ ~ . i ~the s bema at 'l'lit~c\votild tlri.11 1w ;I slight m ~ ~ s t r -of ct~i~tc~-Iin(,. Top platr, %:3 is "d' tliid< ~ I I 7W' I ~ i\,ide, h:ivirig a cross-swtioi~dXI-ixnof 1,, =: (5.56irr.' This grmteiasva p~-odiii.<*s ;i nloi-c rigid c m ~ i ~ r c t i owith n greater ~ c t Ii I ~ I I (solid) s h o ~ s ri.str;iint. I slightly more f i i i t tliali tlrir c a l c ~ ~ l a t c dc o r w (clotted). The m t r a llr\\:ibility probat~ly conies fro111 i t Itj\vcr portion of thc cinii~cction sonic r n ~ ~ w i i i ct~~i ~tii:, which has ~ ~ 1 short st p ; ~ a l l i , l fillct \velds joining tht: lower i i m g e of tile lmirn to t l ~ cscxt. A butt wrld pl;icrd d i r i d y :I~I-ossthc ciid of this lowtrr fialigc to the colnirur. i~ndoobtcdlywoiilci lwing the rigidity of t!ic coii~ic~ciion c u r w u p alii~ost t ( ~t h t of the caloilated curve. ~

'

/

'

elded-Connection Design

.-

A,:

.34IN a

FIGURE 4

I

Figure 5 illwtr;it~.stlie additional rrstraining action provided by column flange stiffcncrs. Both connections x 6 top plates. I r e 'jl<,'' Corinection # I has column stiffeners. In the case of the beam designed for a moment of '/;? \V L ( R =I 100%: down to R = 3%),it would sllpply a restraint of ahout R = 70.2%. Connection li2 lias no column s t i f f e ~ ~ m r s d loses sufficicnt rigidity so that tire hram dcsigned for a moment of ' j W L, (1% = 100% do\\-n to 11 r= 50% ) will be overstressed. This is bcc;i~lse tli(, connection restraint wotdd hc only ;iborit 1% = 45%. This sho\vs the i~nportmceof proper stiflrning.

7. FACTORS IN CONNECTION DESlG Tbr following iterns grcittly :dFrct the cost of wrldcd st!-octt~rnlstccl and ~ a l i ~ l ohet overlooke(1. In order to takc f111l ad\,aiitngc of n-cldcd wnstri~rtion,they mnst he consid~wd. oment Transfer

The hc~rdil~g forws from thc r n d momcnt lie dmost I ' Sthe 1)eiim. Tlien~osteffecentir-ely \vitI~ixt h ~~I ~ I I I ~ of tivc and diri,ct mrtliod to tralisfi~ihcsc forces is solne t y p of flangc weld. The rrlativc n ~ ~ v i of t s thret: types are discussed llert~.

Beam-to-Column Connections

/

5.1-7

y#:* BEnM FOR

8

.W1

.#04

.OOb

901

.DM

.NI

.OW

.Or6

.Old

.#LO

ROT4TION (6$,RADIANS

In iFiglire 6. tllc flai~gi.s;:re dii-i.i.tly i ~ o n i i w t ~tol 1,) I I N ~ I Sof gi-o(,v(~xi-(;Ids. This is ilic inosl th<. ro11111ii~ 01 iriiiisiwrii~g forws :uid rtqiiir(,s tlie d i r w t 11i~tliii~1 lmst : I I I K I WofI ~~ I ~ \ ( I I I I ~

FIGURE 6

Tlw h ~ ~ k i i istrip g illst ~ I P ~ O I VC Y I ( of ~ i11? f h ~ g v s d l o w s t l i ~ ,\s<~l(l to lw m x l ~~vilhin ~ r ~ x s o i ~ ~fit-~ql. ~lil~~ 21s long :IS t l i c w is :I p r q w r root op(wirig. 'l'hr.ri. is littli, prnvisioi~iiii- r ) v r r - r r ~of ~ i ilir i.oIiiiiiir For diir~~~iisirms is-11iih ~ri:~?:11~. ;is iii~ich :is <,xi.~.ssi\-i, ii\~c2r-n~ri. tlw i\:~irgi,sof ilw 11e:rrri in;iy h v t . 11;idi. i l l t h ii<,ki, ill orc!<,r to proto \I(. fImii(,-c~~t ~ n i o p i ~ ~ i ~ oi g r. I I I I ~ ~ tht2 I ~ I , v i d ~ ,the, r n i ~ ~ i r ~ i iroot c!sct,ssiw I I I I I ~ I Iivill ~ i~icrcose Ilic :unoiiilt of w(>Iilir~g r e q i ~ i r ~ 1n1t d , llw joiiit is still possil~l~:, I t is iisii:rlly niorr costly lo i.111 tlw lic:un to m z c t le11gI11: i l l :uI,liti~xi thvw is t 1 1 c ~ cost of 11mviiiig t h e I h Iw:uri to ~1~11gIh is ~ ~ o s l land y not f h ~ ~ g t hiillilig s. ~ C ( , ( I I I I I ~ ~ I , I > I ~
FIGURE 7

FIGURE 8

elded-Connection Design

FIGURE 19

The stiffcning of the latter connc:ction is mainly dependent on thc thickness of the stem of the Tee stiffener, tlie Ranges of the colnmn being too Ear away to offer much resistance. The column wcb is ably assisted in preventing rotation at the connection by the flanges of the splitbeam Tee stiffeners.

~ n a l y s i sof this plate by incans 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 colomn flange can be considered as acting as two plates, both of type ARCD; sec Figure 19. The beam flange is assnmrd to place a line load on each of these plates. The effective length of the plates ( p ) is assumed to b e 12 t,. and the plates are assumed to be fixed at the ends of this length. The plate is also assumed to he fixed adjacent to the column web. where: m = w,

+ 2 ( K - t,)

For the wide-fiangr colrimns and beams used in pactical connections, it has h e n found that ci varies within the range of 3.5 to 5. A conservative figure would be-

P, = 3.5 u, t,' The force carried by the central rigid portion of thc column in linc with the web is-

ekded-Connection Design In Fig111-e12, a shopwcldetl seat provides support fol- tllc dcad load oT the b ~ ; n n .The 1re:rm is lit~ldi l l place hy inwns of erection holts tlrrm~ght l ~ rlmttoirr flangc. In Figure 1:3, a slrop-\icldrd plate on the columii provides temporary support Tor thc be;irn. Erwtion holts

Tl~is:illows t11e beam to slip easily into place during wcction. O11e typo of S:ise (,lip is adjnstaldc aild allows ;r movement of :i
FIGURE 13

tl~rouglithe beam wcb hold the heam in position. An anglo could be used i n s t t d of tlw platc. Altho~iglrtliis ~ o t ~ increase ld the matari:11 cost slightly, it would be easier to install and hold i n proper alignment dnring welding. Sometimi:s a small seat is shop welded to the column, as sho~vn,to give support wliilc the ercction bolts are being installed. If the beam is supported on a seat, the elevation at thc top of the beam may vary hccause of possible ovi:r-run or nnder-run of the beam. If thc beam is supported by a web connection, this may be laid out from t1r1. top of thr beam so as to eliminate this problem. Saxe erection clips, Figarc 14, are made of forged stet31 and are readily \vel&aIrle. The clip is shop welded

FIGURE 15

FIGURE 14

to the nrder side of the beam Hitnge and the seat is sbop welded in tile proper position on the column.

FIGURE 16

earn-to-Column Connections

/

5.1-1 1

FIGURE 17

FIGURE 18

Usc the neu-c.r 1 0 6 strcl for a 1 0 5 liighcr strtsss allowt addi:IIIIP and ahout 5 to ?-; s;ivirrgs in stw1 ; ~ little, tioir;tl rmit pricr3 iu s t t d EiO \i-clds 1 l : i i - ( 2 16%' highhtar allowal~lefor fillot welds. Ie stress for Use ;I 10% l~iglicr : i l l o ~ ~ a l ~herrdi~lg "compact ben~iis";u == .66 ui irrstcad ol .CiO u,, and for ~ity,ativt:moment rrgior~ ;it srrpports use only 90% of tlic tm~rrriwt (-4ISC Scc 1.5.1.1.1). Mnriy cmnwtiorrs prrrvid~, a dircct m d etFrctive transfr,r of iorct,s and yet arc too costly irr preparation, fitting ; ~ n dwt>ldiug. '\I:trim~~nrr c o ~ ~ o misy obtained wlreii a joint is

rlf.signcd for w<.lrling. It is not siiflicicnt to apply ~ ~ ' l d i t to i g a riv~'tedor b~rltcddesign. Us<, rigid, r,ontinrimis connectiotrs for a more ef& ('icnt structlrl-r,. This will rcdrrct. the beam weight nnd 1 s 1 1 y r t d ~ i c e s tlw overall weight of the completc strurtrtrc. Use plastic design to r d u c e steel weigl~t hclow that ol simple f r a n h g . :irrri r e d w e tlrc design t i m : below that of conve~itionalelastic rigid design. Thc grratcst portior~ of wclding on a co~ineciion should 11c d w e in the shop and in tlie flat position. As much ;is possillle. rnisc~.llaneo~~s plates u s t ~ Iin conri(:ctions, soch as scat angles, stiffelrers on coiritnris, etc.. s l ~ o r ~ lk xl asscmhlcd. f i t t t d ntrd weldcd in the sliop in the flat position. proper n~~cessibility for Tlir ronncctiot~t l ~ i ~ soff~m t welding; whetlrcr clo~rcin h o p or field. This is c s l w ci;rlly true of bc;rms fr:iming into the wcRs of coloinns. I'roper fit-tip must hc obtained for l m t wel~lirip. Care must be i ~ s e din layout of tlie conncrtion, fl:rmi. vutting thc h c ~ i mto the pnlpcr irngtll. preparation of thr joint, aiid crc,ctiilg t h rrlcnlber to tile propcr position a i d alig~~rncnt. Coo11 wwkmanslrip, resrilting in good fit-up pays on.

General

Weldor makes continuous beam-to-column connection on Inland Steel Co.'s office building in Chicago. At this level, the column cross-section is reduced, the upper column being stepped back. Spandrel beam is here joined to column by groove welds. The weldor, using low-hydrogen electrodes, welds into a backing bar. Run-off tabs are used to assure full throat size from side to side of flange.

For New York's 21-story 1180 Avenue of the Americas Building, welded construction offered imp o r t a n t weight reductions a n d economy, quiet and fast ereciion. Maximum use of shop welding on connections minimized erection time.

tlic same s l o p as tlie loaded beam, the point of contact moves back !2.,c) When designing a flexible seat angle, it is important to understand how it is loaded, and how it reacts to its load. See Figure 1

FIGURE 2

FIGURE 1

The outstanding ( t o p ) k g of the seat angle is snbject to bending stresses, and will deflect downward (1,a). Tlre vertical reaction ( R ) on tha connecting weld of the angle results in direct shcar (1,b) arid in heirding forces ( 1,c). If the seat angle is too thin, the top of the connecting weld tends to tear, because only this portion of the weld resists the hcnding action. Wit11 thicker angles, the whole lcngth of the conricciilig weld would carry this bending lo:~d (Fig. 1,d). The top leg of the seat angle is stressed in bending by tile rmction ( R ) on the end of the henm which it supports I t is necessary to determine the point at which this force is applied on the leg in order to get the moment arm of the force. See F ~ g u r e2. A simply snpported beam is pIacct1 on the seat angle (2,s). Because of the loading on tlie beam, thc bean] deflects and its ends rotate (2.b). Consequently the point of contact of the rcnction ( R ) tends to move outward. This increase in moment arm incrcases thr bending moment on the seat, causing the leg of the angle to dc5ect downward. As the deflected leg takes

if tht: Icg of the angle were macle thicker, it woiild deflect less. Conseqircntly, the! point of cont:ict u~on1d extend farther o ~ i talong tire leg, i-lrns irlcrcasing the bending Inoment. If the angle were made too thick, this hearing reaction would h e concentrated :ind might overstress the heam web in bearing. If the angle \vex: nradc too thin, it would deilect too easily and the point of wntact would shift to the end of the beam, therehy not pmdncing snfficimt iengtltil of contact for proper srippt~stof the beam web.

Definitions of Symbols w

= k g sim of fillet wdd, inchcs

= yield strrnpth of material u d , psi a = clearanve betwccn column and tnd of heam, usually 55" b =- width of sent angle, inthcs v -- rrrommt arm of reaction ( K ) to witical section of iiorizo~italk g of scat angb,, inrhc,s ee = distance of n:actim (N) to liack of fIexible sciit noglc, inclrei t = thickiicss of mil mgli,, inches t, = thicknrss of h.nm wrb, inches I( = vertical distancc from liottoin of b e a n flar~gcto lop of fillet of beam web, ohtainei! f n m steel liandbook, irichcs Lr = liorizuotal lrl: of sest an&., inches L, = vertical 1i.g of scat angle, also Icrigtl~of vcitical i.onnectirig wckl, i n c h N = miriinnim bcaring lcngth Ii s: vcrtical henring reaction :at mil r i i hcarn, kips

a,

re;wtion is applied to the arrglr, so that the eccentricity or moment arm ( e j of the 1o;id may he known. AISC (Sec. 1.10.10) specifics that the compressive stress at the web toe of the fillet ol a beam withoot hearing stiffcners shall riot exceed u = .75 uspsi. This stress is located at distance K up from bottom face of flange. See Figure 3.

Nomograph No. 1 (Fig. 1 ) for A 36 steel will give the \,due of cr for Nesihlr seats or e, for stiffmid scats. (Stiffened scat brackets are discussed fnrther in the following section.) Known \ d u e s needed for use of this nomograph are the cnd reaction ( R ) of tlre beam in kips, the thicliriess of thc beam web ( k ) ,and the distance frmn the hottom of the hearn flange to the top of the fillet ( K ) , obtained from any steel handbook.

FIGURE 3

For mcl reactiom, the following formula is given:

R ........... itv (N

not over .75 rr, psi + K T == (AlSC See 1.10.10)

S t e p 2: Dtstcrmirle thc required thickness of the angle ( t ) to provide sofficiait bending resistance for the giver1 heam reaction ( R ) .

......(lj

+

K ) may This means that the web scctioi~ ( N bc stressed to u = .75 cr, psi. This planc lies at the top of the toe of the fillet of the beam web, or at height K. This can he projected down at 15" to the h u e of the hram flange to get the minimum bearing length ( N ) . lt is assnmed the hearing renction ( R ) may bt: centered midway ;dong this length ( N ) . 3. SEAT ANGLE AlSC (Steel Constn~ctionManual), recommends the following method for finding the required size of the scat angle. Thc point of critical bending moment in the auglc k g is assr~medto he at the tangent of thc. fillet of the ontstandirlg leg of the angle. This is approxiI t c1y- :!6"in from the inside face of the vertical leg, for most angles rrscd as seat anglcs. S t e p I : Determine the point where the beam

Fmm this we gct-

R e

0-

.I:

1 s t = rr S ~=: --

b t' -

6

.

Since the ontstarrding leg of the angle acts as a be:rm with partially restrainrrd ends, tlre AISC ;\I:inuai (1956, 11 263) allows a hending stress of u = 24,000 psi for A7 or A373 stacl. For A36 stecl, a value of u = 26,000 psi will br 11set1. Tliis thcn hctomes: A7 or A373 Steel

A 3 6 Steel

Flexible Seat An

FIGURE %Thickness

of Flexible Seat For A36 Steel

NOMOGRAPH NO. 2

Flexible Seat Angles

LE

To solve directly for ( t ) , the forlnula + 9 may bc, prlt into the following form:

]

A7 or A373 Steel

I

A36 Steel

1-Values of For A36 Steel

/

R/b

THICKNESS OF SEAT ANGLE i t )

I

Knowing tlic values of A and e,, the tliickncss of the scat aiigle ( t ) may he found from the above formnla. Ko~nographNo. 2 (Fig. 5 ) for A36 steel makes w e of Formiila $9m d will give values of seat angle tlrickiicss ( t ) . Tlic width of thc svat :nigIc (11) is knowil sincr it is nsn;rlly made to estcnd at lcast %" on ench side of the beam Rango. -4 linc is dr:t\in from this valne ol ( h ) throrrgh the d r r e of ( R ) to the vt.rtical axis A-A. The rrqtiired thickness of the angle, ( t ) is foiind at thc intcwc.ction of a Irorizontal line through !-A and a vcrtifiil line through tlic givrn vaiw of In case these 1inr.s intcrswt hetwcen t\vo values of angle thickncss; ilic lnrgcr value is wed as the answer. 'Tal~lc1 will give \.;iht.s of R j b in tcrms of smt angle tliickncss ( t ) and eccentricity ( r , ) . Table 1 is for 436 steel. Step 3: 1)ctc~rrninr~ the horizontal length of the seat angle leg (I,,,). This mrrst bc srtiiicicnt to permit vasy ercrtion and pro\5da aniplc distance for the coiluccting \velds and rrcction bolts on the hottoin flange of tlic heam. This lniniinr~m lcngth is: ( ( 2 , ) .

I'

:

1

-

K

1

. . . . . . . . . . . . . . . . . . . . . . . . (12)

Step 4: I>rtcrminv the vmtical length (I,,) of tlir eoniiecting fillc~t\vdd7 for a givm leg size of weld ( o ) This will deii,rrninc the r q i r i r ( d lengtli of the seat

angle's vertirnl Icg, k i n g assumed equal. horiron/ul forcc on weld Moment (each weld) also: FIGURE 6

P =

'"L

(fb)

( % Li)

-

-

K

-9

~

( q )== I' (31LT)

omneetion Design

TABLE 2-Values or R / o Par A36 %eel & 270

From this: 2.25 R el f, = L,'

R -. Reoction, w

ljertical force on weld

n~.. I

kips

Leg r i m iillet weld

L',

22.4 L', 20.25

+

exl

VERTICAL LEG LENGTH OF SEAT ANGLE (Lr)

rrsultant force on weld

leg sizc of fillct weld actual force -o =-allowable force A7, A373 Sreel; E60 Weld*

.R - -~ -~ 0

-

19.2 L,'

1

A36 Steel;

R0

E7O Weldr

22.1 L,"

m-;~ 2 o . z es

I ~-

. . . .(14)

Since there are a Limited number of rolled angles available (for example, L = 9",S",7", 6",5", 4", etc.) it might be well to select a vcrtical leg length (L,) = vertical weld Iengt11, and solve for the required leg size of fillct weld (w). Nomograph No. 3 is based on formula #14 and will give the required length of the vertical connecting wekl (L,) and its leg size ( o ) if the other vah~es( R and e l ) are known. (The weld length is assrimed equal to the seat's leg length.) Nomograph No. 3 is for A36 steel and E'iO welds. Table 2 will give values of R/o in terms of vertical leg length of the seat angle (L,) and ecccotricity (e,) Table 2 is for A36 steel, and E70 welds. 4. APPLYING CONNECTING

The two vertical fillet welds should be "hooked around the top portion of the seat anglc for a distance of about twice the leg sizc of the fillet weld, or about K", provided the width of column flange exceeds the width of seat angle. A horizontal 6llct meld across the top of the seat angle would greatly increase its strength; however, it might interfere with thc end of the beam during erwtion if the hcam were too long or the column too deep in section. When width of the seat angle exceeds the width of the colunrn flange, coimecting fillet welds arc placed along the toes of tbc flange on the back side of thc

angle. These seats may line up on opposite sides of a supporting web, either web of coliunn or w-t,h of girder, if the leg size of the fillet wcld is hcld to 3/4 of the web thickness when determining the lcngth (L,) of the weld. This will prevent the web within this length of coniwction from being stressed in s1it:ar in excess of a value equivalent to 3/4 of the allowable tension.

Hook weld around corner of seat angle

Don't hook w e d oround corner; will not have full throat

Seat Angle Width

Seat Angle Width Gieotei t h a n Column Flange

Less tho" Column Flange

FIGURE 8

A fiexihie top angle is usually used to give sufiicient horizontal stabi1it)- to the bcim. It is not assumed to carry uny of the l ~ c a mrc:ietio~-i.The most common is a 4" s 1" x %" arsglo, which will not restrain the beam end from rotating under load. Aftor the h t ~ i mis twcted, this top angle is field melded orsly alorsg its two tocs. For beam flanges 5" and less in \vidtli, the top angle is usually cot 4" long; for beam Hanges over 4" in width, the angle is usually cut 6" long. In straight tmsion tests of top connecting angies at Leisigh University, the 4" s 4" x 'A" all& p~illedout as much 2s: 1.98" before failul-c, which is ahout 20 times

greater than us11al11.r e p i r c d under noimal load conditions. Notice in the following figure, that the greatest mo\~e~ncnt or rotation occurs in the fillet \veld cxmnecting the upper icg of the mgle to the column. It is important that this weld be made full size. This trst ulsu inriicatnd that a return of the fillet weld around the: ends of the an& :it the column cqml to about '14 of thc log length rcsulted in the greatest strength askc1 mo\irJinelit hrfon: failure.

1

Hook oround

W C o l u m n flange Greatest rotofion occurs

FIGURE 10

I FIGURE

9

Problem 1

/

Design a ficrible seat angle to support a 12" WF 27# heam, having an end rcwtiou of R = 30 kips. Use A36 steel, ETO welds.

le Seat Angles

/

5.2-

horizontal leg of scat nngle (1

-

j I ,= I~:

+-

N

(?4)

+

1

(3.82)

4.32'' or 4%" min. . ~~

A 5" anglc. 1" thick, is not rollrd. T h e only 7" and CJ" iing1t.s n d l d h a w a 1" liorizor~trrlleg which is not sufficient. 'This leaves just the 6" and 8" angles. a ) Using a 6" r 6"x 1" srat m g l c I,, = 6" FIGURE 1 1

ihickncss of s e a t unglri .-

,461 or rise -~

V2"

-.

b ) I'sirig a 8" i6" x 1" scat mgle

L, = 8"

Thc structural dcsigner might bc incliired to selrct the, 6" s li" r 1" angle himrisc of th(, obvioris saving in

\veigl~t.T h e shop man knowing that tbc ?i,;" fillet weld in ( h ) is a single-pass wcld and can be made very fast, wlirrens the %" fillet weld in ( a ) is a three-pass

Don't hook weld more than fir' i '

Ploce top angle on

-4Angle leilgth /-.FIGURE 12

weld, would select thc 8" x 6" x 1" angle ( b ). He knows that the cross-sectional area of a fillet weld, and therefore its wcighi, varies as the square of the leg size. He figures the ratio of the leg sizes for ( a ) and for ( h ) to he 8 to 5. This ratio squared produces 64 to 25, or as far as he is concerned 2k times the amount of weld metal. From Table 1, K/b := 30/S = 3.75. Using ef = 2.4" would give this value if t = I". (Here R/b = 4.22)

(From Anicncan In~tituteof Sled Construction)

SEATED BEAM COMNECTIOMS Welded-EGOXX & E70XX electrodes TABLE Vlll

.

.

aitm:li bulm m ma* (optiundi. AUownbir londa in Table VIII-A ue Nnnunol b e m snthach is &XI on 9,. x t b c k , r h i r l i pnvbdes for pssible miU u n d s m n b k m (o

'LA'.

From Table 2, using et = 2.4" a ) If L, = ti", R / o = 65.2 or leg size of fillet weld, 30 - 4 65.2

~ or ' use W'

b ) If L, = 8", R / o = 107.0 or leg size of fillet weld,

30 101.0

o .= -~,-= ,280 or use

x6"

. ..

S E C T I O N 5.3

ene

e

e

Sinci the inarirniiir~ strcss, Ilihcn the r'action load ( K ) rcqnircs a tliiclacss of angle greater than tlie alailable sections, a stifferred seat bracket may be usvd. T h t w are t\vo alml>ws: ( A ) in uhich the scat stiffener is at riglit angles to the web of the heam, arid ( B ) in u-hie11 the seat stilfener is in line with the web of the beam. For analysis, the stiEener of Type ( A ) is considered an eccentrically loadcd colnmn with the rmction load applicd at a f i x e d point. 'rli~.mtirin~imstrcss is the sum of the direct load and brnding alfccts. The line of action of thc- comprt%ivc lo;rd is approxim;~trlyparallel to thc outer edge of thc stiiFeni,r. Tlic criticd crossl the area and scction of the stiffcirw ( t o hc u s ~ lor section modillus) is at r-ight ;tnglrs to thr linc of action of the load. The arca a i d sccti:io moduli~si m A = t X = t L,, sin

X = Ll, sin tp

FIGURE 1

4

thc rcqiiireil thickncss of thc bracket web is-

c1cti:rThe thickness of tlic imcket wrh car miircd qnicldy f n m Noniograpli No. .4 (Fig. 2 ) for 1\36 steel; tliis is h:wd on formnla +I. The v i ~ t i c d line at thr left is for v;iln~~s of load eccnrtricity (c,) and length of ontsianding braclict li'g (L,]).Tlie ~ l m t line is for thr angle 1)ctvwn the sidc of ilic brad& 1x7

PROBLFW FIND THICKNESJ Of ITIffENED SEA7 FOR THE FOLLObW6 CONDITNNS. Lh'8. e * 4.5B . 90' P ; 58 KIPS (EN0 REACWON) REM t

. v6

/NCH (STIFFENER ISNCI(NE;S)

S OF STIFFENE EB

2.

If tlie beam rests in line with the bracket stiffener, Type B, Figure 3, the bearing length ( Y ) of tlie be;~m (AISC See 1.10.10) is-

and this would he tlie miniriium valuc allowed.

If the bracket is made up of plates, AISC rccnrnmcnds that thc wc4ds conncctiiig the top plate to the wcb of the stiffcnrr should lhave s t r e ~ ~ g tequivalent h to tile horizol~tal n&ls between thc bracket and the column support. The depth of the stiffener is determined by thc vertical lcngth of w.&l (L,) retpired to connect the bracket. Thc lcilgth of the 1)rackct top plate (I,,,)s l i o ~ k hr i sufficient for it to rxtcnd at loast beyond t l ~ chearing Icmgth of the beam ( N ) . The stiffened scat bracket is shop welilcd to the siipporting m(.mbcr in the flat or downhand position. IJsually the top portion of the bracket is welded on the underside only, and tllc useb of tlic stiiiemr is rvt:lded both sides, full Icngth. By placlng the weld on the underside of the bracket, it docs not interfere in any way with the beam which it supports. Sorne rngineers do 11ot like the notch effect of this fillet weld's root to be at the outer fiber of the connection, and would prcfer to place this fillet wclcl on top of tlie bracket; this can be done.

FIGURE 3

The eccentricity jc,) of the reaction load is-

e, = L,,-

-

This value of load eccentricity (e,) can be quickly found by using Nomograph No. 1 (Fig. 4 in previous Sect. 5.2). Sonictimrs it is figured as 80% of tho bracket's outstandirlg leg length (I,,,). The eccentrically loaded column forniula ( + I ) is seldom used in this case because it will result in an excessively thick bracket web or stiffener. This is becatise the formula is based upon stress only and does not take into consideration some yielding of the bracket wliich will causc t11c point of application of the load to shift in toward the support, this n:dncing the moment arm arid t~endingstress. AISC Maniml, page 4-39 recomnic~ids for A36 brnckct material that the bracket wcb's thichiess be at least equal to 1.33 tinics the requii-ed fillet weld size (E70 welds). Also it should not be lcss than the supported beam web thickness for 47, A373 and A36 beams, m d not less than 1.4 times the beam web thickness for A242 and A441 beams. For stiffcncd seats in line on oppositc sides of the colnmn web, the fillet weld size should not esceed % the column web thickness when determining its length

(L).

The folIo\ving method is uscd to detennine the leg size of the connecting fillet weld ( w ) . For simplicity the length of the llorizo~italtop weld is assumed to be a certain prrccntage of the vertical weld lcngth (I,,). The top weld length is usoally less than the bracket width, and the vrrtical weld Icngth is assuinttd equal to the vertical length of the bracket. This analysis uses the value of 0.4 I., for the top weld as it is a more i m n n o ~ ~uscd i y value, although any reasonable value rniyiit be used, Figure 4.

'hus it can b o shown &at: rwutrul oris of connecting t ~ e l d

section 111odt~2t1.s of connerlzng weld

S, = 0.6 LT2 (top)

titfened Seat Brackets

-- 2.4 L,

R.--

B =

A,

13 =

23.04 w

1

= 1% =

A36 Steel; ETO Welds

AT, A373 Steel; EEO Welds

bending force on wcld

f

5.3-5

where:

length of connccling zticld A,

/

I

K - - 26.88 w

-' By knowing the value of R a d e,, the (mwwer may solve directly for I,,. The lcngth of connecting \vrticai weld (I,,) miy he dctcrmined quickly from Nomograph No. 5 (Fig. 5 ) for A36 steel arid E7O welds; this is based on Eonnula gi.The wclded consiectioii is assmntxl to ewtrmd horizontally 0.2 1. on ctich side of the bracket web. The ~nasirnnmk g size of fillet weld ( w ) is held to % of the stiifener ttsickncss. Ilra\v n h e from \wid size ( w ) through thr re:iction ( R ) to the vertical line ( U ) . The rtqi~irecl lcngth of weld (I>,), vcrtical length of stiffener (I,), is found at the intcrsectiols of a horizontal line through ( D ) and tt vertical lisle throligh the: given d r r e of (e,). For stiffcner brackets which have a top width ( b ) other than 30% of the depth (L,), the Table I ionnulas may be isscd.

K 2.4 I,,

.. .

resultant force on u.eld

-

leg size of fillct weld actual force = .. ..-

I

or

allo\z.able force

Problem 1

I

Design a bracket to support a beam with an end reaction of 58 kips. Tho beam lies at right angles to tile bracket. Use A36 strt%land E70 welds. See Figure 6. Using Nomograph No. 4: vertical weld length (L,.)

f,-

90"

R = 58 kips TABLE I-Fillet

&,

ii = 0.4 L, ~~

~

A36 Sfeel K E70 Welds

~ 7 A373 , Steel K E60 Weldr

B m c b t Width

~

~~~

~-

23.04 l',

fi.1-z

W

=

- ~ \ 26.88

-~

/ ~

~~

~ 1 6 . 0 e', ~ + . ~

r=Lm ~~

~

J ~ 2 ~ + ~ 4 . 0 6 ~ ~28.00

i? = 0.6 Lr b = 0.7 1,

I

- JL',

= .

,.~-~ -~ w

=

24.96 L'r R

+

12.57

c2,

JL'-+

1 1 01 $ 2 ~ .

w

=

29.1

JLzv +

11.37~2~ W = . - R J ~ 2 v + 30.24

12.57

+ ex6

i1.37+eP,

FIGURE 7

Using Nomograph No 1 (Fig. 4, Sect. 5.2): FIGURE 6

read the required stiffener thickness ast = y1*'' Using Nomograph No. 5: 0

=

X6"

read the hearing Icngth and 1o:rd ecceiitriciky as-

N -. 1.54''

(t =

R = 58 kips e, = 4.5" read the required vertical length of the stiffener as-

Lv

- -

(if L,, = 4")

e, = 3.23" Since t = using

13"

R

-

1

XG", use t = Y4.' -

plate.

Komogral;hxo,5 : 58 kips

~-xiCq

= 3.37" for o = %"; read L, = 10"

Design a bracket to support a 2N', 65# I-beam with an end reaction of 55 kips. The beam lies in line with the brackct. Use A36 steel and E70 welds.

fi,r

e,

--

3;o",

Use tile qi/;O"

read L, = 11" fillrt \w111 wit11 a length of 11".

FIGURE 8

Stiffened Seat

(From Amcrican lrrstiti~teof Stcel Constniction)

STIFFENED SEATED BEAM CONNECTIONS Welded&E60XX or E70XX electrodes TABU X

Aiiow;%ld<~ ioed* in 'i'&ir X iirr b i d iiu tile urr o f I!M)XX eintmdes. For 1170X'i ciiwLiinics. multi&y Labulri hndn i>y 1.16, o r enter the U ~ t i l ev i t i i M'?o of iliv aivae rwrfion. Note Advrntagc, may & t r k e n a f t h e higher nUownhle umlt dias of F:70XX rl.airodw onlv i f hih bracket md r u p ~ u r t i n grnemkwz am &STM 3 6 . A242 or A441 m a k i r i .

1

'02.

~11.

1 235

i 1s

~~~~~~

1

! 159. 191. 1223 1 6 0 i o i 1 2% .

Xi 168

"insusb y=pal *jldl m*nr *-lh i i o r x r i r c ! 16. a rnlrihr ,mnio *,,a 86% os //lb/l

0",1

wnmn rmrx u=no
rrc vsra

I f the reaction values ola

beam are not shorn on mnbrad draainm. the mn-

onnestion Design

Beam-to-coiumn connection being mode on the Colorado State Services Building in Denver. Operator i s anchoring the beam to o stiffened seat bracket by downhond welding, using iron powder electrode.

Extensive use of modern structural techniques and welding processes speeded erection of Detroit Bonk & Trust Co. Building. Stiffened seat bracket can be seen a t upper left. Angle clip to facilitate field splicing of column lengths shows immediately above.

2. ANALYSIS OF FIELD

1. GENERAL REQUIREMENTS

\Vrh framing angles are usually shop welded to tht: web of the beam. cstending abont S'' beyond the end uf the beam, m d field nv1dr.d to the supporting member. Erection bolts are risually plactd near the bottom of the angle, so they do not restrain the beam end from rotating under load. For deeper girders, the erection bolts may he placed near the top of tlie angle for better stability during erection. If theri IS '. concern about any restraining action, the bolts may he removed after field welding. The thickness of the framing angles must be limited to that which will allow snfficient flexibility, otherwise the connection wonld rcstrain the end of tho simply supported beam from rotating and thns would load up in end moment. AISC has a table of typical framing mgle corini~ctions.It lists 3" and 1" angles of ' 4 6 ' ' to - , , ,,;"thickness. Whcn thicker angles are used the leg against the supporting iix~mbrrmust be iricreascd in ;ihout the same proportion as the thickness in order to maintain the same order of fit.xihility. The analysis of this type of connection is divided into two parts: a ) the field weld of the angle to the supporting member and b ) the shop weld of the angle to the web of the beam.

Hook weld around top; not to exceed % leg of angle, usually M"

When the reaction ( R ) is applicd, the franring ang1t.s tend to twist or rotate, pressing against each other at the top, and swinging away from rach other at the hottom. It is assumed the two angles bear against each of their length. other for a vertical distance eqrlal to The remaining % of the lengtli is resisted by thr connecting welds. It is assunid d s o that these forces on the x ~ l d sincrease linearly, rcaching a masimiun (f,,) at tlie bottom of tlie conncctior~.Figure 1. horizontal forcc on weld Applid monient flom load =: Resisting moment of weld

R

-"

2

L,) = - P I,, 3

where L,, = leg length ot angle orp

.75 R L,,

I>7

From force triangle, fin&

I

FIGURE 1

P = 95 ( f , , ) ( 3 & L,)

FIGURE 2-Framing

Angles and Size of Field Welds For A36 Steel & E70 Weldr NOMOGRAPH NO. 6

FIELD

PRO&EM.

R

FIND TN£ 1 f h G 7 f l ( I ~ /Of ThE P%AM/#6 AN6lt W :%,(.VIE OF f/€UWfZD,

R =IS KIPS (END .mcr/uN, 3 - (LEG Jilt O i A N G l t ) RE40 1, ; /2"/i€rV67// OFRNGLE, in

2 i

I

2;

3'

7' (LEG SIZE OF IINGLE) 5

3;' 4

L,

6

5'

/

Web Framing Angles

TABLE t-Valuer of R/w For Field Weld of Framing Angle to Support For A36 Steel & P70 Welds

From thme two equations, detcrminef

1,

--

9 R I,,, 5 L,'

-wR

~

-.

Reaction, kips ~

~~

~~

~~

~~

~

2" 4'.

~~... ~~

flf,,'

~~~~~

.... .

5 112 "".)

=

4 f,'

~

i+

~~~~

Leg sire of fillet weld

.. ,. .~

f,

5.4-3

4 3 1

Leg o f Angle ( L i d ,~

~~~

~~

3" 30

4"

/

22

5"

1

I9

~. . ~~

1 1 6

~~~

7"

6"

/

I4

8"

1

12

..

(2L,I<

)2

or:

I

-

-

f

_

I -

~.Y 2LY2

I

d , .

2

+

I A7, A373 %eel; E60 welds] R -I"

~

19.2 L;' -

~p~

f~y-' +

A36 Steel; E70 Welds

R

.~.-

12.96 L;,"IA72

--

12.96 I,,,?

(,)

..

.

20.1 LV2 ~~

.

I (I?)

,? 12.96 L,,'

Be sure thc supporting plate is thick enough for this resrilting weld size ( a ) Thc~two vrrticnl \aelds comx'cting framing anglcs to supporting incnibrr should be "hookcd" around the top of the nnglcs for a dist;aice of about twice the leg size of the \r-eld, or about 'i".(Origi~mltests indicated that a distance not to cxwrd 'A of thc ;iriglr's leg lcngth 11ciped thc carrying capxcity of thy connection.) Nomograph S o . 6 (Fig. 2 ) may be used for the f i l l l i n g This nomograph is for A36 steel and EiO \velds. In the chart on thc: right-hand siclc, from the point of intersection of the angle's leg size (LI,) and the length of the angle (L,),draw a horizontal lint! to the \ ~ r t i c a laxis 1.7.15. From this point, draw a line throng11 the rc;lction ( H ) to the left-hand axis. Read tht. leg sizc ( w ) of thc field weld on this axis. Table 1, for A36 steel and EiO welds, gives valucs of R / o in terms of leg size of angle (L,7) and length of angle (L,). AISC, Sect 1.17.5 specifies that the leg size of a fillet weld used in calculating its lcngth (L,) should not came the web of the snpporting member to be overstressed in shear. For n single pair of framing angles on just one side of the supporting web, assume thc leg size of the

fillct weld not to exceed 1.3 t , . For t ~ v opairs of framing angles, o m on endl side of the supporting wish. nssilme ihr leg size of the fillet weld not to escw%lI!? t,v. and !1.3 = 2 x % ) mny be Ti~escfaciors of )i'( djusted lor the oxact type of steel used l ~ yreferring to Table 2.

R

Assume !h" set bock

FIGURE 3

elded-Connection Design

In Figure 3, analysis of the shop weld sho\vs-

rcsultunt force on outer end of connecting weld

k fk

FIGURE 4

*

twisting (horizontal)

leg size of fll& weld

ttristing (oertical)

1-

f

forcc on melds ~-allowable force

actual

=

. . . . . . . . . . . . . (4)

2 -

~

A7, A373 Steel; E60 W e l d r A36 %eel; E70 Weldr

) ..

shcur (oertical)

"

=

0

. . . . . . . . . . . . . . . . . . . . . . . . .( 5 )

bL,

0

9600

Unfortunately there

TABLE 2-Maximum

A7 A373

Steel

A36

WELD

METALS A N D STEEL

A242, A441 O w , 1%" To 4"

thickness

OY

33.000

T

13,000

..............

To

-

f

E60 o r SAW-I

-

9,600 i o .

o/t

5

,667

42,000

14,500

--

weld

36,000

17.000 .............. E70 or SAW9 E70 or SAW-2 .. 11.200 w 1 1.200 w .648

, j/,"-

Over

I

-

,

46.000 . . . 18.500

E70 or SAW-2

*

759

Web thickness i t u ) over

e r less

50,000 -

.

20,000

L70 or SAW-2 -11,200 u 1 1.200 w -. ,826 ,893

Then: Moximvrn leg size of fillet weld t o use in rolculoting veiticol length

k g size

%"

-

17)

is no way to simplify these

Leg Size to Use in Calculating Vertical Lengrh of Weld

FOR VARIOUS C O M B I N A T I O N S OF Given these conditions:

= -11,200

Web Framing Angles

/

5.4-5

/

5.4-6

Welded-Connection Design

Leis than % thick"

/

If edge is butit up to ensure full thioot of weld

%'*.' thick or more

I

-

FIGURE 6

formulas into one workahlr formula. It is necessary t11 work out eai.11 step l~trtiltlw final restilt is ohtair~od. The leg size of this shop w&i nray h r dctcrmined quickly by rncans of Kornograph No. 7 (Fig. 5 ) , for 436 stcrl mil Ii70 wclds. In thc c11;rrt on t l ~ right-hand r side, from thc point of iuterstctim of the anglc's h i zontal lrg length ( I ) and its vrrticnl length (L,) draw a l~orizontaliinc to thl, vrrtical x i s F-F. Fmm ttlis point, draw ;I liiic through the reaction ( R ) to the left-11:ind axis, Read ttlc leg size ( w ) of the shop wcld along thp left-11;nnd svalc of this axis. I f the nomogr;iph is u s d f m n l~xft to right to i,stahlish ;in arrglr six,. be sill-<.that the leg size of tbt. fillct wcld docs not cxcc~xla v;rloc which vould overstress the web of t / ~ (hiwm ' in s11~:rr (AISC SCC1.17.5) by producing ~ I I Oshort a lorgtl~of connecting weld (I>,). The follou.irrg limits apply to the fillet weld leg size ( w ) rclativc to thr thiclmess of the heam web (:IS usr:d in c;ilctll:tti~rg tlw wrticnl length of connecting weld ) : A7, A373 Steel and E60 Weld /T

--

10.000 , x i ) ( f , = 9600 w Ihs/in.)

e q t d to or csi,ct~istl~is\ d r ~ cfor~ndjust opposite the resulting lrr: sizc of the wi:ld. l limiting shmr v;ilire (.406 Somt: rnginwrs f i ~ this stt.c.1, r L. 14,500 psi) is to ins~~i-I, that thc wcb of thr hearrr d w s not bllckli., and that a higlrcr allowable vdnr iniglrt 11e IISCYI hcrt., pcr11;ips 3/r of the allowable ttwsilc strength. In this rase thr ~n:rxim~imlcg sirc of thr weld would he Ireld to ?/r of thc web thick~rcss.

1w =:

TABLE 3-Values of R/w For Shop Weld of framing Angle T o Beom W e b For A 3 6 Steel & E70 R

I

l-lo\\:rve~-,tlir acti~alleg six: of the fillrt wcld used may exceed this value. Tahlr 2 reflccts thr limiting \.aloe of w = Zi t,. AISC holds to this limit for shop weld of the ;rnglr to the beam (.4ISC M;rmlaf. pages 4-25). Notice tlrc left-Iran11 :isis of Nomograph KO. 7 also gives the millill-urn \veb thickncss of tlrc h(mn in order to hold its sbcnr sti-css ( 7 ) within 14.500 psi. Illst 11c sure the ttctr~ai\?-~II tl~icknessof t l ~ esupported hpain is

1 .......................... .(9)

I S C (Scc 1.17.5) sp.rifics the m a s i m ~ ~ kmg size of fillct \wid rr1:itive to :rrrglc plate thickncss to be as shown in Figtirc 6, l';rhlc 3 \\-ill give ~,;ill~cs of R/w in trrms of leg size of angle (L,,,) ; x r d lmgtli of :rngle (I,,). Table 3 is for i t I S i t 3 6 stwl. and I T 0 w~.lds.

go

A36 S t e e l ond E70 Weld

% t,

.-.

. .

Reaction, kips .................

-

Leg sire o i fillet weld

W e b Framing Angles

.4s i n d i a l t d hv, Firnrr~ 3 and the rolatcrl weld ,~> ;m:rlysis, thc fillct welds con~i(vtirlganglc to heam w d ~ should hi, Irookd aroriud tlir ends of tllr anglc. top mcl bottom, for the distance ( h ) to t l ~ cend of tlic bcam wcti. They sh011l11not ire continned aronnd the c i ~ dof the wrb, Fignrr 7.

&,--

-

/

5.4-7

-

Shoo Weld of Fromins Anale t o Beam W e b

Nomograph No, 7 shows t l l n t for a rwction ( K ) of 58 kips. ;rn alrgle leg (I,,,) of 3" and l<>rlgth(I,,.) of ly, a fill(,t wrl(l ( w ) ~,,llllldhi, rt!(illireil, Ilellce llsc G". 3,r %, fralning ar,gles, 94t;,, weld to crilunm and Sk" shop weld to boain web. 4. STANDARD

EB FRAMING ANGLE CONNECTIONS

Don t hook weld I "round tthis oiound h i i edge

Hook weld orovnd

FIGURE 9 FIGURE 7

To design ;i wch framing ;aiglc cnnrlrctioll to sl~pporta 70" 85,1 1 bcan~,Ira\-ing an rnd rt,action of R = 58 kips. Use A36 steel uiid J+:X)u & k . Sct: Figlire 8. Field Weld of Framing Angle t o Column

Nornograpli No. fi s l ~ o \ xthat ~ for a %" fillct wcld reaction j R ) of 58 kips and a11 mgle \uth a 1c.g (I,,,) of 3". its lciigtli ( I J , ) si-lonld br 101,L". However, for a %i,i'' fillct weld ( w ) the angk l t ~ ~ g t(I,.) h violrid only li;tvc to be ir~creasedto 12". ( w ); a

Tnbie 3 giws tlw I S C allowil~lelo:~ds (kips) on n ~ f i hframing mglc conncctioris. rlsirig A%, r12.42 and A141 s t t ~ l s; ~ n di-70 \n.t,lcls. T111, talilc givcs the capacity ;und sizc of (Shop) \'(,Id -4 coi~ncciingthe framing angle to the hcnm web. and of (Ficld) Weld I3 co11neoting tlic framing arigle to the h a m slipport.

pzL-q

To s?lcct a \wIi frmning ariglc roi?ncctioii for n 16'' H 263 Ir;rm (0.75" \vrh tlrickiirss and T =: 11") of

h3.41 stwl, \\-it11rmd rwciion of R :1 05 kips. Usc I 3 0 wcl~ls.Allowal~lcs h r x is 20 ksi. This h
FIGURE 8

\

'

.

51.4-8

/

Welded-Connection Design

TABLE L S t a n d a r d Web Framing Angle Connections From American lnstitutt of S t e e l Cotistruction

Weid&;..

T1

FRAMED BEAM CONNECTIONS

,,,I

WeIded~&E6OXX electrodes

( ,

'@I.&

1

S

t

,. -

,

B J07 &i

I,,. #,,A

% , ",

7>,,

FRAMED BEAM CONNECTIONS Welded E 7 O X X electrodes

~ c ~ , . z 2-7

.

TABLE V!

t

1 hriir

Llr"

,MI,

i

%,,

16:' St'

i s 71,

!X L?: 19:

LIB i!8

1% 116 >,L

) ! I

'4,

iar

:*?*-51:

ih7 li4

O,?X%,,

:00

4

cP'

ixii:,

-r

I*?>

,,

I?? I41

~

i

r

;

?"

is 39 ~ x 4h 39

4XIY'. LXi*l,

)i

ill

i!" 81 4

)., i.,

:,,

n

0 i;i

!lil

R i

i,

i:o

l,6

12 I

+IS

3 $5 ;

>,,

,:

3

'

1" i

i.il%# 3, i x i ,

i*iXb~

I

lXlX*46

29 dB

!d i

irlr%+ 1YiX'I

t,'

3

"

48

39 ii

1 x 3 ~ : ~ ~68

bl b

'

.

ir

'9

.

iR

I*?*%

irix;i

19 li

5,: !

i > , ! ik t

di

ii 6

1Xir;

1)

0 il ;

,$

>>,

%<

!*iY

I& i

:r

3

,

," >,... ? a:

A

, :

4'

,

39 ?Y

40 I

i*!Y+ i*iX'i,

:i 5 34 6 ?d i

3 , I<,<%

! 0

"'

"7

a

7

.: 5

16 4 :a 6 1"

li 9 ,&,,*,

'%*"-

A

)1

,

!%

!V

1"

Web Framing Angles nf 38.4 kips for a weld size of o = ?ip," and anglc length of L, : : 10" sliglitly excteds the rmction. The corrcsponding (Field) W d d R, nsing w :'h", also is satisfactory. Sinct. the beanis r c q u i r d wweh thickncss is 0.31'' while tlic actual ivcb thickncss is 0 . W , the indicatcd 3" x 3" x 5/,(1'' is d l right. If the beam is rnade of A36 steel, this conncction's capacity will bc rcdueed in the ratio of 0.25/0.29 of actual to rcyrriml web thickness. The r t d t i n g capacity of 33.1 kips is less than the reaction. The nest larger connection with apparently sui6cient capacity sllows that (Shop) Weld A's capacity is -17 kips, using same angle section hut an angle lcngtl~of L, = 12". Applying the multiplier of O.2.5/0.!?9 redr~ccstho capcity of the connection to 40.5 kips, which excw:ls the end reaction.

FIG. 10-Dauble-web

/

5.4-9

framing angle

5. SINGLE-PLATE OR TEE CONNECTION O N BEAM WEB In the previous dcsign of the field weld, connecting a pair of web framing angles to the supporting column or girdcr, it was assumrd that the reaction ( R ) applied eccentric to ench angle, rtsdted in a iendeocy for the angles to twist or rotate. In doing su, thcy would press togcthcr at thc top and swing :way from each other at the bottom, this bring r m i s t ~ dby the welds. These forces arc in ;rddition to thc vertical fol-ces c a ~ ~ s eby d thc reaction ( R ) ; see Figure 10. IIowrver, in both the single-pl;~tewcb connection and the Tce-st3ction tyl~i.,this portion of thc conrrection welded to the col~nnn is solid. Thns, there is no tendmcy for this sprcding action which must be rcsisted by the welds. These vcrtical field welds to the

FIG. 12-Flat

FIG. 11-Single

plate or Tee

co111rnn \voold be designed then for just the vcrticd rcaction (11); see Figure 11. In the shop ~ d ofd the singlc plate to the web of the honm, Figrirc 13, this donlde vertical weld wonld be designcd for just tlic vrrtical reaction ( R ) . There is not enough rc~witrieity to considcr any bending action.

plate used for flexible connection on web of beam.

5.4-10

/

Welded-Connection Design

Tee section used far flexible connecttaii on web of beam

FIG. 13-Tee

section used for flexible connection on

In the shop n r l d of thc T w connection to thc web of tlrc bram. Figure 13, the size and limgth of thr fillct .ivclti w o ~ k lbe dt,tcrniinid inst as in the cast, of the doublr-ncl) fran~inz;nrgirs. (:xct.pt thew is jnst a single fii1t.t weld in this casc rather than two; so, for n given cos~ncrtion,this wonk1 can-y just half of thc rcwtion of the corresponding donblt:-anglc connection. 6. DIRECTLY-WELDED WEB CONNECTION To sec how this typc of connection hch;ivcs, consider the follo\ving 18" WF 85# beam, simply supported, 15' soan. with a unifonnlv distributcd load of 139 kios.' the same hcnm and load u s d in the grricral discwsion on behavior of connt.ctions in Srct. 5.1, Topic 6. If only thc wvl) is to h e nvklrd to the cohn~rn,tho n ~ n s th m c stifficit>ntlength ( L , ) so that the a & cent \vi,b of the hwim will not hc overstressed in s11o;tr. For A373 stecl

web

of beom.

FIGURE 14

fillct tcelrl in slzetir; portillel load 2(96(10w)l, := t, 13,000 I,

1-

A

== 10.2", or

U.SC

11"

The leg size, of this fillrt weld rnrist hc t ~ p to d the thickness, ~ , uporl ~ stanci:lrd ~ ~ ;lllcl,,.~t~,les, d if it is to matrh the :~liowahirstrength of this web sectioii in shear as wcll as tcrrsion.

l.~ic~o;illy, tmnsvcrw fillct welds arc: ;ibout ?$ stron-ei than p;tr;iild fillet wclds; this can i>e pnn.ril by thmry as d l as twtiiig. 'This m a n s h,r trmsverst, h;rtis, tlri; 1i.g s i i c wodd bc 3 of tlw platr tliirkn<,ss, iiist ;is in l);ti;i!Ii,l luaiis. Iiowevcr, ~ , r I d i n gcodrs do not ;,s yct i-wognim tliir; :ind for code work, f i 1 1 rvl,~ds ~ ~ for tr.ir,svcrsc io;,~is.jiri,,,id bc. ii,itrIi~ ccliiiil to ilic pl;itc tiiiclaiess.

Web Framing Angles

I FIGURE 16

-

. the k g size of this fillet weld is increased by this nmount. The moment-rotation chart, Figwe 17, shows the beam line for this pnrtictilar bcmn lcngtli and load; and the actual connection curve taken from test data at 1,ehigh University. In testing this co~meetion,thc heam \veh showed initial signs of yielding adjaccnt to the lo\ver m d s of the weld at a monirnt of 3fi0 in.-kips. At a moment of 660 in.-kips, point ( a ) , thew wcrc indications that the beam ~ v e along l ~ the full length of the weld had yielded. At a moment of 870 in.-kips: 110th \velds cracked slightly

/

5.4-11

at the top; this point is ~narkeclwith an "X" on the curvc. With furlhcr cracking of the weld and yielding in tho beam wch, thc lo\vm finngc of the beam rolltadcd thc colurnn, point ( h ) , arid this resultrd in irlrreascd stiffness. Thr- inoment built tip to a ~ i ~ a x i m ~ n n of 1918 in.-kips, and t b e ~gr:idually ~ fell off as tire \ d d continued to tear. Notice in this partic~ularcminplr. the web w o d d h a w yicldrd the in11 Icngtli of the \wid at design lo;~d. The \veld s t : ~ r t d to crwk whcn the corrnt.ction h:d rotatrri ;ihout ,011 r;idi:ins; this woold corrcspond to a horizo~~tal inovenncnl of .OV at t l ~ ctop portion oi the wold. Cornpaw this s ~ n ; ~:inro~~nt ll of mov~:mcntwith that ol~tainrcli r ~t l ~ ctop conrn,ctiiiq plait: c~x:u~~ple of Figure 4 n-hicb 1i;irl thc zihility to pnll out 1.6" Iwforc failirrg. This diicctly \velricd wrh conrieetion (Fig. 18)

FIGURE 17

eided-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, hut many times the rotational ability. The use of side platrs, Fignre 21, would allow a wide variation in fit-up, b.ut in general they are no better than the directly welded web connection. Unless the plates are as thick as tile beam web, the resulting connecting fillet welds will he smaller and will rednce the strength of the conncction

FIGURE 19

FIGURE 20

is not as dependable as a top connet:ting plate designed to picld at working load (Fig. 19) or aither flexible web framing angles (Fig. 20) or flexible top angle. Also rcmember this highly y i e l d d web section, in the case of the directly welded wcb connection, must still snpport or carry the vertical reaction ( R ) of the beam, whereas in the top plate connection, the support of tire beam at the bottom seat is still sound no matter what happens to the top plate. Figwe 17 ~vouldindicate the directly welded web connection rosoits in an end moment of M, = 720 in: kips, or an end restraint of-

FIGURE 21

Fteld weld

shop weld

/ 'Field

weld

1 FIGURE 22

1

Field weld only on toe of ongle

Web Framing Angles

/

5.4-13

In the tests at Leliigl~University, the corresponding connection on the 18" WF 85# beam (S26"-thick web) nsed :$6" thick side plates with fillet welds. They failed at a lower load. If 'htr thick side plates with %" fillet welds had been used, they undoubtedly- wonld have becn as strong as the directly welded wch connection.

7. ONE-SIDED WEB CONNECTlONS A single web framing angle nsed by itself is not recom-

mended; 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 resdts in a maximum force on the weld of about 4 times that of the double-angle connection; see Fignn:s 23 and 24. It might be argued that in the conventional doubleangle connection, the fieId 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 mould greatly decrease its strength. Any additional welding on the single anglc, 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 ccnnection. In the original resenrcb at 1,chigh University on welded connections, this single-angle connection wit11 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 cnd moment built up above the strength of t l ~ cconnection.

5.4-14

/

Welded-Connection Design

Web framing angles ore commonly shop welded to the supported beam. To facilitate erection, bolts are used in joining the other member until the web framing angle con 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.

1. DESIGN PLATE TO BE STRESSED AT WELD

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 See. 1.15.4); see Figure 1. For a simply supported beam, uniformly loaded, this maximum movement ( e ) ~vo~ild be:

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.

where:

b

e

p Beam

I

~

=:

movement, in inches

L = 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.

(length of beam) FIGURE 1

( 2 loads

@ % points 4 loods @ 1/, points

6

Uniformly distributed load

5

5 loods @ points 3 loads @ % points

.4

1 load at Z

v6

.3 .2

.I

10

20

30

40

50

60

70

80

Length of rimply supported beom (L), feet (orruming beam to be stressed to u = 20,000 at FIGURE

2

90

C )

100

elded-Connection Design E 6024 weld metal 6010 weld metal 80

I

I

I

I

I

I

5

10

15

20

25

I

30

35

I

40

Elongation, % in 2"

FIG. 3 Stress-strain diagram for weld metal and beam plate.

ASTM specifies the following minimum percent of elongation as measured in an 8" gage length for structural steels:

This minimum value of 2m 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 of the capacity of this top plate to yield. Figure 3, a stress-strain diagram, shows that a miid 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 specin~en necking down and yklding. This is similar to the behavior of a top connecting plate which yields plastically under load.

x5

2. TOP PLATE FOR S I

There is some question as to what value should be uscd 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:

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 plate being designed to develop about 23% of the restraining moment of the

----FIGURE 4

ind

Top PIaies gar Simple Beams

beam using the standard allowable bending stress. The standard bending stress allowed here would be limited to u = .60 u,. (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 ( u 7 ) .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 retuming 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 -- ----

~-

A36, A441 Sleek; E70 Weld -....-

. . .( 2 )

IND BRACING Wind moments applied to simple beam c~nnections present an additional problem. Some means to transfer these wind moments must be provided in a connection which is designed to be Rexible. 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 stondard ollowobler

/

5.5-3

ethod I. The wind connections, designed to resist the assumed moments, are adequate to resist the moments induced by the gravity loading and the wind at the increased unit stresses allowable, or fhod 2. The wind connections, if welded and if design& 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 % 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 % stress increase, nor less than that required by Sec. 1.7, (repeated Ioading) if it is applicable. Since we are discussing Type 2 construction (simple framing) the initial basic allowable stress is 60 u,, not .66 u?

pz-tizq 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 V3 increase in the standard stress allowable (or u = .80 u,). This 4; increase may also be applied to. the connecting welds (AISC See. 1.5.3, & 1.5.6). The fillet welds connecting the lower Range 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

1" X W' backing bar

FIGURE 5

F = MI rtandord allowabl

(gravity)

d,

elded-Connection Design

Fillet weld at 1'/3 stondord allowabl when loaded with F

In the alternate design of the top plate shown at upper right in Figure 6, the reduced section ( W ) is designed for the force resulting from the end moment caused by the combination of the gravity and wind moments at a 'h increase in the standard allowables. It will reach yield at a 25% increase in load ( F ) . The wider section at the groove weld (1% W ) will reach 1%5- or .SO u, when the reduced section has reached this yield value.

I

Method 2

1

The top plate (Fig. 7) is designed to carry the force resulting from the wind moment (M,) using a % increase in the standard allowables: u = (1%)$0 up : :.80 up

The top plate must be capable of yielding plastiAt standard allowables when reduced section is at yield in7)

M, (grovity)+M,(windl FIGURE 6

cally to relieve larger moments induced by gravity loading, figuring the connecting welds at standard allowable~.*This is the same method for figuring the connecting welds of top connecting plates for simply supported beams without wind loads. The reduced section will reach yield stress (u,) at a 25% increase in load ( F ) . The wider section at the groove weld (1% W ) will reach standard allowables ( 8 0 u,) at this time. In case there should be a reversal in wind moment, the top plate must be thick enough to safely withstand any compressive load without buckling. It is recommended that the top plate's thickness of its length ( L ) between welds. be held to at least This will provide a slenderness ratio (L/r) of 83; and corresponds to about 80% of the allowable compressive strength for a short column (L/r ratio of 1 ) .

x4

'This weld altowable by AISC i s not clcar; AISC srmply says welds shall not be overstressed when plate is at yield. M~n~murn lensth of reduced section between welds

1" X W' backing At 1 % a when loaded with [F) ivind moment M, Fillet weld at standard olloviobles when reduce

(wind) db

FIGURE 7

Top Plates tor Simple Beams & Wind

/

5.5-5

(gravity) moment as a simply supported beam:

= 300 in.-kips on connection at each end

FIGURE 8

= 21.3 kips

Where: 1x --

The rcduccd section of the top plate is designed to carry this force at yield stress (u,):

Wt" --

12

and

raditw of gyration

- (21.3

kips) (36,000 psi)

-

.59 in."

or use a 1%'' x W' plate

Connecting Welds at Standard Allowables

slenc1erncss ratio

For the groove weld to the cwlnmn flange, this plate is widened to 1%W, orwidth = 1% (1%)

= 2.V or use 3.0" For the fillet welds to the beam flange, use 5/,," fillets at an allowable force of4. EXAMPLE OF TOP PLATE DESIGN-

wltn WIND

MOMENT

A 14" M7F 38# beam is simply supported and loaded urtiformly with 296 Ibs/in. on a 15-ft span. Based on these beam-load conditions, the masimum bending = 1200 in.-kips. Use A36 moment at center is M steel and E70 welds. Wind moment on each end is M,T = 600 in.-kips.

Beam conditions here:

(See Figure 9.)

14" W F 38# beam

b = 6.776'' db = 14.12" ti = ,513"

S = 54.6 in."

If there were no wind load, the above connection might hc designed for about 25% of the present

FIGURE 9

f, = 11,200 0 = 11,200 ( X e ) = 3500 lhs per linear inch

Force on top plate isF = -

M db

The length of this weld is-

= 63.8 lcips

-

( 6 % in.2)(36,000 psi) (3500 lhs/in. )

The top plate is designed for this force at fS higher allowahles:

-

This would he 13h" across the end, and 2%" along the sides. efhod 1 for Additional

This connection will now he designed for the additional wind moment of M, = 600 in.-kips, using Method 1.

(63.8 kips) 1%(22,000 ps?K

= 2.18 in.2 or use -- a 3%"x %' plate A, = 2.19 in."

22.8 i n . 0-. K

The connecting welds are figured at % higher allowable~: For the fillet welds at the beam flange, use M" fillets. The standard allowahle force is f, = 11,200 cd = 11,200 ( M ) = 5600 lhs per linear inch. The length of this weld is-

-

FIGURE 10

Beam conditions here: 14" W F 35# beam

tc = ,513"

S = 54.6 in.3 Total moment on the connection isM = M, M, = 300 in.-kips 600 in.-kips

= 900 in.-kips

This weld length would be distributed 3%'' across the end, and 2%" along the side edges of the top plate. The above connection may be cut from bar stock without the necessity of flame 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 vield stress, the connecting - welds would be stressed only up to the wind allowable or % higher, hence u = 3 0 u,. Accordingly, the plate is widened here to 1W =

-

b = 6.776' db = 14.12"

+

(63.8 kips)

in (5600)

+

(See Figure 11.) The length of the fillet weld, using M" fillet welds and allowable of f, = 5600 lhs/in., would he-

Top Plates for Simple Beams & Compression top

R

Tensi

on top

Wind moment

FIGURE 13

FIGURE 11

reduced section at yield ( ) and fillet weld at 'h higher allowable

L,=- F 1%f, (2.19 im2)(36,000 - -

psi)

1%(5600)

= 42.5 kips The reduced section of the plate is designed to canr)- this at 44 higher allowable:

=10.55" This would he 3%" across the end, and 3%"along the side edges of the plate. Applying Method 2 for Additional

--(42.5 kips) - 1%(22,000j

= 1.45 in.2 or use 3" by 36'

plate

The plate must now be modified so that larger moments induced by the gravity loading can be relieved by plastic yielding of tlre top plate, designing the connecting welds at standard allowablcs. The plate is widened at the groove weld to 1%W = 1%( 3 ) = 5.c". For the connecting fillet weids to the beam flange, use %" fillets:

f, = 11,200 0

= 11,200 (%)

w

= 4200 lbs per linear inch FIGURE 12

Temporarily ignoring the gravity load, the top plate is designed to carry the wind load, M, = 600 h k i p on each end.

The length of this weld is-

L, = -F f,

1.5 in.') (36,000 psi) =L (4200)

elded-Connection Design

Beam

10,550 psi] Connection (28.330)

& =

+

600

it?-k

M,

= - 600 in-k

FIGURE 14

FIGURE 15

This would be 3" of weld across the end, and 5" along each side.

5. EXAMINING THIS EXA To better understand how this wind connection operates, this example will be examined, using Method 2. 1. The cmm~ctioq~ is Erst designed for the wind moment of M, = 600 in.-kip at % increase in the standard allow-ables applied to each end of the beam. The wind moment will cause a bending stress in the beam of-

2. Now the gravity load can be gradrially 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 corrcspond 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 froin gravity load is (hcated as a fixed end beam)-

= 10,990 psi (See Figure 14.) The corresponding stress in the top connecting plate is-

-

w* --L' also =cr, A, d Me, = --12 the portion of thc gravity load to be added here is-

(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 stress in this beam end due to gravity load is then added to the initial wind moment diagram: (See Figure 16.)

Top Plates for Simple Beams & Wind

/

5.5-9

FIGURE 16

- Mc2

,490 psi

3

FIGURE 17

8010 pri

',= - 517.6 in-

M

,

Ma

= - 762.8 in-k

9480 psi \

Connection (24,430 psi)

13,970 psi

FIGURE 18

At this point, the right-hand connection reaches yield stress (u,. = 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 296 - 60.2 = 235.8 lhs/in.) can now be applied, w, 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-

-

- (235.8)(180)2 Me* = 8 8 955 in.-kip

wz L"

or a bending stress of

Mez = (955 in.-kips)

ub2= -Sb

(54.6 in.3)

= 17,490 psi Also since M

w*

L*

=16

ub at 9 = $5 (17,490) = 8750 psi

'll~ese stresses are then added to the previous moment diagram; Figure 18.

elded-Connection Design

FIGURE 19

2670 ps,

Beam Connection 3660 PSI 4650 psi (36,000 psi) 990 psi

FIGURE 20

Connection Beom (36,000 psi) 4650 p

4650 psi

FIGURE 21

(36,000 psi)

The corresponding stress in the top plate isA lower design wind moment will not require as large a top connecting plate. The smaller plate will yield sooner and it is possible that the h a 1 gravity load would cause both end connections to yield. Consider the same pl-obkm as previously but with the wind moment reduced to M, = 200 in.-kip, applied to each end of the beam. The required top plate is designed for this wind moment:

= .48 in.' or . use a1" xM"-.plate ---(This very small top plate is used here only for illustrative purposes.) A, = 5 0 in.'

>

M,-- - (200 in.-kips) ---d A,, - (14.12) ( 5 0 )

~~

~~

~~~

= 28,330 psi

A portion of the gravity load is added, treating the bcarn as having fixed ends, until the riglit hand connection reaches yield stress. This would be an additional stress in the connection plate of: 36,000 - 28,330 = 7670 psi. This would correspond to :I stress in the beam of: (.l29) (7670 psi) = 990 psi. See Fignre 20. Since the allowable moment on this end conncction resulting from gravity load is-

.48 in.'

This moment will cause a bending stress in the beam of-

M,

the portion of the gravity load to be addert here is-

ub =-

Sb

--(200 .

in.-kips) (54.6 in.:')

= 3660 psi

W, = See Figure 19.

12 u,,A, d, .- - 12 (7670)(.50)(14.12) (180)' L'"

Top Plates for Simple At this point, the right-hand connection reaches yield stress (u, = 36,000 psi) even though the end of the beam is stressed to only 5 = 4650 psi. In this example, if the remainder of the gravity load were applied, the left-hand connection would go over the yield point. For this reason only enough of the gravity load will be added to bring the lcft-hand connection just to yield, treating the beam as having one fixed end on the left and simply supported on the right. See Figure 21. To reach yield stress in the left connection, the stress in the beam must increase from 2670 psi compression in upper flange to 4650 psi tension, or 7320 psi. This would correspond to an applied gravity load of:

so

cr,=

8(7320 psi) (54.6 i n 3 ) (180)"

M(7320)

= 3660 psi This now leaves a gravity load of ws to be applicd, treating the beam as having simply snpportcd ends since their connections have both reached yield stress. The remaining gravity load:

/

5.5-11

Since:

= 13,150 psi Ibis stress in the bcam is added to the preceding moment diagram; see Figure 22: The total ue = 17,310 psi 7.

-

ind

<

22,000 psi

OK

AD APPLIED FIRST, THEN

In the preceding examination of the wind connection, the wind was applied &st and thcn the gravity load. This is the seqnence of design followed in Method 2. The cross-sectional area of the top plate is determined by wind only, and then the connecting welds are designed so that larger monlents induced by the gravity loading under actual conditions of restraint may cause the plate to yield plastically. Of course in actual practice, the gravity load is applied first and thcn the wind may be encountered secondly. The same problem will now be examined in this order of loading. The b e a n with the gravity load is considered as simply supported; however, the top plate which must resist the wind moment does restrain the end of the heam to some extent. ?he larger the plate, the greater the restraint, this will also increase the end moment rcstilting from the gravity load. It is necessary to get some indication of the restraining action of the connection so that the cnd moment from the gravity load may be known.

FIGURE 22

elded-Connection Design To do this, a simple moment-rotation diagram is constructed for both the loaded beam and the connection. The resulting conditions are represented by the point of intersection of these two lines or curves. In the Lel~ighresearch of connections, the actual test results of moment-rotation of the connections were plotted on this type of diagram; in this example the properties of this top plate connection are computed, and will be fairly accurate since practically all of the movement will occur in the reduced portion of the top plate. Connection l i n e

FIGURE 23

I

v-v--w

I

also

M, = upA, db

If the bottom of the beam is securely anchored and the top plate is relatively small, Figure 23, rotation map be assumed to occur about a point near the bottom of the beam. As the top plate becomes larger, offering more restraint, this point of rotation moves up. If the top plate has the same size as the beam flange, Figure 24, rotation may be assumed to be at mid-height of the beam. Since movement ( e ) depends upon the over-all elastic elongation of the top plate, and for simplicity length (L,) is shown only as the length of the reduced portion, there is some elongation in the widened section as well as in the reduced section within the fillet welded zone. For this reason the value of the calculated rotation ( 8 ) in this example will be doubled. Two points will determine the connection line. Since this line passes through the origin or zero load, it is only necessary to have a second point; for simplicity this second point will be a yield conditions. At yield:

- --(36,000

bottom of beam

-

psi) (4.5") (30 x 10') ((14.12")

= ,382 x

radians

This value will be doubled because of elastic elongation of other portions of the plate:

0,

=:

,764 x

radians

and: FIGURE 24

M, = a, A, db =r

(36,000 psi) (1.5 in.2)( 14.12")

= 762 in.-kips Beam L i n ~ G r a v i t yload, uniformly loaded

I

I

I

mld-he~ghtof beam

G-Q-VS where L, = length of plate section between welds, inches e Since 0 =-and db

e =:

E

Lp

It is necessary to have two points to detelmine this beam line on the moment-rotation chart: ( a ) the end moment ( M e ) if fully restrained

Top Plates for Simple Beams & Wind

/

5.5-13

With the gravity load only on the beam, this would indicate that the end moment7 would be Me = 720 in.-kip. This would leave:

= 800 in.-kips ( b ) the end rotation ( 8 , ) if simply supported

This would correspond to a bending stress at the end of the beam of-

See Figure 26.

= 13,200 psi

The stress at centerline of the beam would bewhere L = length of beam in inches

= 8800 psi U b

As before K =-

U"

Connection a t yield in,)

in the connecting plate would be-up =

moment, Me = 720 in-kips

13,200 psi .388

= 34,020 psi Now the wind load is gradually applied equally to both ends until the right-hand connection reaches yield. This would occur when the stress in the connecting plate is increased from 31,020 psi to 36,000 psi, or an increase of 1980 psi. This would correspond to a wind moment of-

Connection line

End rotation (OJ, X 10-3 radians

= ,388 so that the stress

6.24

FIGURE 25

13,200 psi i 3,200 psi

Connection (34,020 psi)

FIGURE 26

elded-Connection Design M,

= 42 in-kips

J

Connection (1980 psij

"

FIGURE 27

.-

8800 psi

Beam

7

12,430 psi

L

FIGURE 28

FIGURE 29

-

19,020 psi

pea..

13.970 psi Connection (36,000 psij

FIGURE 30

Top Plates far Simple Beams & Wind

8. ALTERNATE GRAPHICAL SOLUTION

See Figure 27

This same example can he illustrated in a slightly different manner. The right-hand connection and beam end is on the right of Fignre 31; the left-hand connection and its l~earnond is on the left. As b i h r e , the beam line with gravity load only is constructed for both ends. This hcnm line represents the moment at the end caused by the gravity load, the actual value of the lnolnerrt depends on the effect of the connection. A wind mornent would be represented by a horizontal line throngh the actual value of the moment. It would not he influenced by the connection iinless it exceeds the yield of the connection; then the portion of the wind moment carried would be limited by the yield of the coimection. 4 n y wind moment superimposed on the gravity load will shift the beam line vertically up or down depending on the sign of the wind moment. By observation, the right-hand connection can be

Adding this wind moment diagram to the initial gravity moment diagram gives I'ignrt: 28. There now is left a wind mornent of 600 42 = 5% in.-kip to be applied to each end, but since the right-hand connection has reached yield stress, the remaining moment of 2 x 558 = 1116 in.-kip must he added to the left end of the beam.

= 20,440 psi

.= 52,680 psi (compressio~~) to be added to the 32,040 psi in tension already in the left-hand connecting plate

"3

a

5.5-15

Adding this last wind moment diagram to the diagram in Figure 28 gives thc final diagram, Figure 30.

And stress in the beam is-

= 770 psi

/

J' b

= - 13,970 p s ~

oe

= - 36,000 psi

G M- =

- IOOOl -- 900.-

i

'6.24

x

- 762 in-kips 1

lo3

End rototion [OJ, X 10-3 radians

I

j.Add wind moment of negotive to right end of

%Add

M, = 4 2 in-kips beom; connection reaches yield

wind moment of positive

M, = 1 I56 in-kips to left end of beam [compression] Right End

Left End

FIGURE 31

I

elded-Connection Design increased another 42.0 in.-kip from wind, then it will reach yield and no further moment can be applied. Since the applied wind moment was 600 in.-kip on cach end, this will leave a balance of 2 x 800 in.-kip 42 in.-kip = 1156 in.-kip to be carried entirely by the left-hand connection. To do this, the beam line on the left of Figure 31 will be lowered vertically 1158 in.-kip; see the dotted line. This will inkrsect the connection curve (extcnded into the positive moment region) at an end moment of Me = 320 in.-kip. This will correspond to a bending stress in the beam end of 6050 psi, and in the connection plate of 15,600 psi. In this case, the connection curve h:~dto be extended downward into the positive moment region in order to intersect the new beam line. This indicates a $ moment and reverses the stress in the plate, now compression, arid the bottom of the beam connection is now in tension. The previous examination of this problem indicated a bcnding stress in the left end of the beam of cri,= 8010 psi; this examination indicates a stress of ul, := 6050 psi. Why should there be a difference? The previous examination stopped after the first end momcnt due to gravity load was determined and then for simplicity from then on considered the connection as perfcctly rigid, whereas this examination considered the elastic properties of the connecting plate all the way through the problem. This last approach would be a

+

? .-

-

little more accurate. This same prohlem was pi-eviously worked with a reduced w i d moment of M , = 200 in.-kip applied to each end. Figure 32 shows how this can be worked graphically. This is an intrresting prohlcm since the lower wind lnorncnt reqnires a smaller top plate, with ?/3 the cross-sectional area, hence 'h the strength, and the gravity load ca~isedthe plate to yield plastically at both rnds even before any wirid load is applied. This is represented by the black dot where the beam line (without wind) intersects with the connection curve. When the wind moment is added, the right conncction is alrt:ady at yield and can carry no additio~ral moment, therefore the mtirc v.ind moment of 2 x 200 in.-kip = 400 in-kip must be carried by the left-hand connection. Accordingly the beam line is lowered vertically a distance of 400 in.-kip: see thc dashed line. As this is lowered. tbt. resulting moment (M,,) and rotation (0,) of the connection (black dot) slide down parallel to the clnstic portion of tlrc connection line until it intersocts with this new beam line (white dot). In Figure 33 these final conditions representing the heam with gravity load and wind load are represented with black dots. If the wind were now removed, the left beam line moves npw-ard 200 in.-kip 2 n d the right beam line movt.s dou.11 200 in.-kip, tho new conditions being represented by the white dots. For a complete reversal of wind, this operation is again repeated and is represented by the broken lines.

+

9004

A-

Gravity iood; no w i n d

Me = - 330 in-ktpr 8. = - 3.8 x l o 3 iudionr

400 in-kips-..

Gravity load with wind

M.

= + 30

in~kiar

- ruu - 800

\@'

<624

2

-

X 10"

Right connectton 1s a t yieid and con toke no odditionol moment; hence, odd wind moment of 21 2001 iwkiar = 400 in-kips to ieft end

+

+

Right End

Left. End . . . .

~~

FIGURE 32

Y

T o p Plates (or S i m p l e Beams

4

Left End

Right End

FIGURE 33

Typicol scene in structurol shop with weldors

attaching stiffeners in place on curved knees. Proper use of welding results in significant savings in structural steel weight and in fabricating costs.

&

elded-Connection D e s i g n

Welded continuous connections were used extensively i n the Hartford Building in Son Francisco. Photo shows the use of short Tee sections welded in place under ends of girders to provide deeper section o t the point of moximum negative moment. Note thot columns ore weld fobricoted. The small ongle supports steel roof decking.

--f

1. ANALYSIS OF CONNECTION

k

H

-4 / i , k

Me!--

-4 top connccting $ate designed to hc stressed only below its yicld point may he used as a semi-rigid connection. The reduced poltioil of the plate is detailed to have sufficient length ( L ) for elastic elongation of this section to provide the proper amount of joint rotation. See Figure 1. Analysis of this type of connection reqriires locating the center of rotation. 'This depmds on the relative stiffness of the top hottom portions of the conneetion. For the more flcxible type of semi-rigid connection, rotation will occur closcr to thc bottom of the beam; see Figire 2. For thc more rigid ccorir~cction,rotation will occur closer to the rnidhcight of the beam; see Figure 3.

Column flonge stdfenerr may be

1

4

e

+

I

Rotation about bottom of beom FIGURE

2

Rotation about mid-height of beam FIGURE 3

The rmisting mornc,nt of the connection is-

. . . . . . . . . . . . . . . . . . . . . . . . (1)

Alternote detail

iequiied for joints of high reitioint

FIGURE 1

elded-Connection Design

My = A,

or d; (top plate at yield)

I

t 8, [octuol)

0, (flexible beam)

and the required cross-sectional area of the top plate is-

This connection line breaks at the yield point, or becomes horizontal at:

I M ~= A,, u).d b I . . . . . . . . . . . . . . . . . . . . . . . . ( 5 ) The rotation of the connection, assuming rotation about midheight of the beam is-

mi

8, = -- and

The slope of this connection line is-

The actual conditions of moment (M,) and rotation ( & ) are found at the intersection of the beam line and the connection line; see Figure 4. Table 1 shows the moments ( M ) and end rotation ( 8 ) for various load and beam conditions. The total centerline moment (ZMr ) and total end moment (ZM,.) of a beam with any combination of the Table 1 loads equals the sum of the individual values resulting from each type of load. When designing a beam for a given end restraint ( R ) , the resulting maximnm moment at centerline for which the beam is designed (MI,) equals the difference between the maximum centerline moment ( M y ) when R = 0 and the actual end moment ( R M,) for the given value of R. See Figure 5.

This can also be found by totaling the individual

7-

Simply Supported

R = O

ZM*

iR = 100%

Beom with desired end restraint (R)

FIG. 5 Moment diagrams for different restraints (R).

Top Plates for Semi-Rigid Connections

W

@

+1.M*(

# -

W

c

1)

Simply supported, w t h load

Apply negative moment at ends to

@

bring up to horizonto1 position

4

Final end moment for louded beom

@

$5

equal to oppiied moment in jb)

/I = M, L '' 2 E l ~~

W Me(#

~~~~

4) Me

@

Fixed end, ended beom

Me 0 = .. L... " 2 E l

Simply supported, lauded beam

W

FIGURE 6 TABLE l-Moments

and End Rotation for Various Load/Beam Conditions

M.

~~. --

End Mornen!

i x e o €id

W L

10

9.2-2

/

Joint

In order to evaluate the weldability of steels, a limited kno\vledge of the basic arc welding process is advisable. Welding consists of joining two pieces of metal by establishing a metnllurgical bond between them. Many different welding processes may be used to produce bonding through the application of pressure and/or through fnsion. Arc welding is a fusion process. The bond between the mptals is produced by reducing to a molten state the surfaces to be joined and then allowing the metal to solidify. When the molten metal solidifies, union is completed. In the arc welding process, the intense heat required to reduce thr inetal to a liquid state is produced by an electric arc. The arc is formed between the work to be wt~ldedand a metal wire or rod called the elcctrode. The arc, which produces a Welding Machme AC or DC Power Source and Controls Electrode Holder 7

\Ground

Cable

I

temperature of about 6500°F at the tip of the electrode, is formed by bringing the electrode close to the metal to he joined. The tremendous heat at the tip of the electrode melts filler metal and base metal, thus liquifying them in a common pool called ;I crater.* As the arens solidify, the metals are joined into one solid homogeneous piece. By moving the electrode along the scam or joint to be welded, the surfaces to be joined are welded together along their entire length. The electric arc is the most widely used source of energy for the intense heat required for fusion * F o r soinc applications, filler metal is deposited b y a consumnblc w e l d i n g electrode; for others, a "nonmnsumable" elcctrode supplies the heat a n d s separate welding rod the

filler metal.

wclding. The arc is an electrical discharge or spark sustziined in a gap in the electrical circuit. The resistance of the air or gas in the gap to the passage of thc current, transforms the electrical energy into heat at extremely high temprmtures. Electrical power consists of amperes and voltage. The amount of energy available is the product of the amperes and the voltage flowing through the circuit and is meastired in watts and kilowatts. The energy used is affected h y such variables as the constituents in &ctrode coatings, the typc of current (-46 or DC), the direction of cul-rent flow, and many others. In all modern arc welding processes, the arc is shielded to control the complex arc phenomenon mid to improve the physical properties of the weld deposit. This shielding is accomplished through varions techniques: a chemical coating on the electrode wire, inert gases, granular flux compoi~nds, and metallic salts placed in thc core of the electrode. Arc shielding varies with the type of arc welding process used. In all cases, however, the shielding is intended: 1) to protect the molten metal from the air, oither with gas, vapor or slag; 2) to add alloying and fluxing ingredients; ,and 3 ) to control the melting of the rod for more effective use of the arc energy.

Gaseous Shield

The arc welding process requires ;I continuous supply of electric cnrrent suflicient in amperage :md voltage to maintain an wrc. 'l'his currcnt may be either altcmating (AC) or dircct ( D C ) , but it must be provirlecl through a source which can be controlled to satisfy the variables of the welding 11roces" :mmnerage and voltage.

Top Piales for Semi- igid Connections

/

5.6-5

The length of the reduced portion of the top plate will be made I. = 7". Thc slope of thc connection line:

Me L 6' --2 E I

(762) (180) 2(30 x lO"(289.6) = 7.9 x 10-Qadians -

-

M c - A,, --

8,

d,,? E 2 I,

Design top plate for an end moment of 75% M . = .75 (762 in.-kips) = 571 in.-kips. Cross-sectional area of top plate:

M A "-- -- u d,,

-

This connfction line can also be constructed by solving for end momcot (M,.) and end rotation (8,) when stressed to yield, u. = 33,000 psi:

(571,000) (eo,ooo) ( m 8 6 j

M, - A,, u> db = (2.06) (33,000) ( 13.86)

= 2.06 in." or use a ?k" x 5%'' plate, having A, = 2.06

FIGURE 9

FIG. 10. Moment Capacity of Top Plate Connection.

KIP - INCHES

WIDTH OF TOP CONNECTIN6 &?

0 16 IS I4

N 12

N

f

PIVOT POINT IN THIS

EXAMPLE

10

9 8 7 6 5

AISC S K 1.5. I. 4.1 IF COMPACT (SEC 2-61 AND AT NEGATIVE MO&E'IVl CAN USE 90% APPUED EffD MOMENT

6 00

TDO COO

500

so0 400

400

Top Plates for Semi-Rigid Connections

WL M, = -= 12

/

5.6-7

750

M

=

410 in-kips

FIGURE 1 1

0'

,y M =---

A, db

= 21,400 psi This calculated connection line is shown as a dotted line in Figure 9. It rises to a moment of M = 943 in.. kips at which time the top plate should reach yield stress. From tlicre on, this plate will yield plastically and build up a higher resistance as it work hardens. It would finally reach the ultimate tensile strength of the plate unless some other portion of the connection would fail first. Superimposed upon this graph in soiid lines are the actual test results of this particular connection, from the paper "Weided Top Plate Ream-Column Connections" by Pray and Jcnsen, AWS Welding Journal, July 19.55, p 338-s. The beam lines of the particular example are shown as broken lines in tbc figure. Notice that the beam line at working load intersects the connection curve (point a ) well within the capacity of the connection. The second beam line at 1% working load also is well within the ultimate capacity of the connection (point b ) . Holding the length of the reduced portion of the top plate to L = 7" has resulted in an end moment of M = 680 in.-kips instead of the 75% value or M = 571 in.-kips as originally planned. This is a restraint of R = 89.3% instead of R = 75%. A lower restraint coulcl he obtained by increasing the length of the reduced portion ( L ) of the top plate. However with the present conneetion the top plate has sufficient strength:

90% M used at negative moment; (AISC Sec 1.5.1.4.1)

<

22,000 psi

OK -

(AISC Sec 1.5.1.4.1) Notice also that the connection curve lies quite a distance abovc the R = 50% point of the beam line. Since the beam is desigued on the basis of R = SO%, the connection could drop down to this valuc before the beam \i:onld be o\;erstressed. The moment capacity of a proposed top plate connection can be readily obtained from the nomograph, Figure 10.

2. CONNECTION BEHAVIOR UNDER ASYMMETRICAL CONDITIONS In the usual analysis of a connection made by superimposing a beam line on a connection curve, it is assumed that the beam is symmetrically loaded and has identical connectioris on both ends. This is illustrated in Figure 11, where the member is a 14" W F 43# beam, and:

W = 50 kips L = 15 ft I = 429 in.* When these conditions of symmetrical loading and identical connections do not exist, the following niethod may be used to better understand the behavior of the connection under a given load. The above beam and load value will be used.

5.6-8

/

Welded-Connection Design

Step I. Start at the left end ( a ) of the beam with the right end ( b ) held fixed. The left end ( a ) is first held fixed (0, = 0) and the end moment (M,) determined; the left end is then released and simply snpported (M, = 0) and the end rotation (0,) determined. See Figure 12. FIGURE 14

@ fixed

released; simply supported

@

0" --4 8 E l

- 2.62

-

x 10.l

Step 2. Thus with the right end held fixed (u,= O), the rcsulting moment at the right end ( h ) consisting of the initial momcnt and the additional moment d71e to moven~entof the left end ( a ) , is-

FIGURE 12

From these two points (M, = 750 in.-kips and 0. the bcam line for the left end ( a ) is drawn, Figure 13. Upon this is superimposed the connection line, and the point at which it intersccts the beam line represents the actual cnd moment and end rotation after the connection has allowed the bcam end to move.

= 2.62 x 10-"radians),

750 in-k

Now the left end ( a ) of the beam is held fixed at 8,

= -1.6 x 1 0 " while the right end ( b ) is released

and simply supported (M, = 0 ) and the end rotation (Bb) determined. See Figure 15.

750 in-k

Beom line Connection curve

l.- 2.62 x - 1.6 X 70-3

1 Left end

@

simply supported

~ i ~end@ h t

held fixed

[Leftm

held fixed at

@

= - 1.6 x 1 0 7

FIGURE 1 5

FIGURE 13

This relaxing or movement of the left end ( a ) , from 0, = 0 to 0, = 1.6 x radians, causes the fixed opposite end ( b ) to increase in end moment (M,,). This increase may be found by the following: If a uniformly loaded beam is supported by fixed ends which have previously rotated (0, and O,), the two end moments (M, and Mb) are-

R

From: Mb =

.+

-2

E I O,, E -I 0 + 4." L L

when:

M b = 0 and

@, = -1.6

x 10-8

L - -W - -12

T o p Plates for Semi-Rigid Connections

/

5.6-9

tlic rotatioti of the beam at the right end ( b ) , if simply supported a i d no restraint from the connection, would be:

These two points ( M h = -979

and 0b = +3.42

x 10- " ) detcrmine tile beam line for the right end ( b ) ;

Figure 16. Its intersection with the connection curve represents the actual end moment and end rotation after the comxction has allowed the end to move. FIGURE 17

Left m d held fired of

due to niovemen!

Step 4. W l l r ~the ~ left end ( a ) is simply supported (M, = 0 ) , tlic end rotation wonld be 0, = -3.67 s lo--:'. Releasing the left end ( a ) allows it to rotate to 8, = -2.25 x 10 " .

to x -2.25 x 10-Qn the left end catrses the right moment to increase to Mi, = -472 in.-kips. When the right end ( b ) is simply supported (M. = O), the end rotation Releasing the right end would be Bb = $3.74 x ( b ) allows it to rotate to Ob = +2.3 x lo-'. Step 5. This movement 8, from -1.6

Left end

@ FIGURE 16

Step 3. As before, this movement of the right end causes an in( b ) from B,, = O to Oh = +2.1 x crease in the moment on the left end ( a ) ; Figure 16, left.

From :

when: 8, = -1.6

x 10

"

and 0, = $2.1 s LOW3

the moment on the left ~ n (da ) is fomid to be FIGURE 18

This ontirc procedure is repeated until the corrections bccomc very small, Figures 17 and 15.

Step 6. This movement of BI, from 1-2.1 x 10Qo

5.6-10

/

Welded-Connection Design

+2.3 x 10-%n the right eud causes the left moment to increase to M a = -43.5 in.-Mps. When the left end ( a ) is simply s~~pportctl (M, = 0), the cnd rotation would be H, = -3.76 x lo-". Releasing the left end ( a ) allows it to rotate to 0, = -2.40 x 10P3. Step 7. This movement of 8, from --2.25 x 1O"o -2.40 x 10-%on the left end causes the right moment to incrcase to M,,= --428 in.-kips. When the right end ( b ) is simply supported (MI, = 0 ) , the end rotation would be HI, = +3.80 x 10 -:'. Releasing the right end ( b ) allows it to rotate to: B,, = . +2.40 x lP3. Conclusion: The final end conditions resulting from this sequential handling of the givrn connection and beam loading a r e

Reference to Fignre 11 shows that thrse are the same values as obtained when thc beam was considered to be symtnetrically loaded with identical conditions on both ends. 3. BEHAVIOR OF CONNECTIONS STRESSED ABOVE YIELD The same method wed prcvionsly may also be applied to connections that arc stressed above their yield points and thus yield plastically. See Figure 19, using same beam as before.

moment (XI,.) is applied at the snpported end and the resulting end rotation (H,.) is fonnd at this same end, Figure 20. Here:

In this particular example:

With the particular scale used in the original construction of Figure 19,

1" = 4 x 10 :' radians or 1 radian = % x lo1 inch and 1" = 400 in-kips = 400,000 in.-lbs or 1 in.-lb = % x

inch

The slope of this beam line is-

or an angle of 70.7", Figure 21

'4

Connection curve

750 in-k

tj,

FIGURE 19

To simplify this ar~alysis, two changes will be made. First. In computing the two points of the beam line ( M , ) for fixed ends and (8,) for this end simply supportcd, it is noticed that these same values can be obtained by considering the beam as fixed at one end and sqqmrted at the other; with no gravity load. A

FIGURE 20

Beom lhne determiried by Meand o

Me

y supported

He FIGURE 21

Another method of constructing this slope is to use a convcnient valne of H, for example, 0, = 5 x 10 '. The corresponding end moment would be-

T o p Prates for Semi-Rigid Connections

/

These two values are plotted on the figure and the slope determined by protractor, Figure 22. Since the slope of the beam line remains constant, it won't he necessary to compute the value of 6, for the simply supported end for each step. Second. Instead of computing the end moment after it has been increased by the angle movement on the other end of the beam, it is seen that the actual increase in moment is-

This may be drawn on the figure from any convenient value of 6, and Me. Any given increase in 6. is laid off horizontally on this line, and the increase in moment (M,) is measured off as the vertical distance and added to the moment on the opposite end of the beam. See Figure 23.

8, FIGURE

Application of Method

33,000 psi) at a moment of 423 in.-kips. With additional movement, the plate will strain harden and its resisting moment will very gradually increase. This accounts for the slight rise in the connection line above the point of initial yield.

This method is now used on the same 14" WF 43# beam, uniformly loaded with 50 kips on a 15-ft span; Figure 24. The connection is made with a top connecting plate, X6'' x 3", which is stressed to yield (cr =

I FIGURE 23

22

Fixed end /

I

/

/

!

1

/

Increase in moment on opposite end rimply supported

V

_I

Movement of left end (Ox) Left end

Right end @

@

-k FIGURE

24

1

2

3

x

4

5

1p3 go

Left end @

6

7

Change in 0,

7

6

5

x

4

3

2

1

10-3 8, Right end@

elded-Connection Design

O n the Ainsley Building in Miami, weldor is completing fillet weld on top connecting plate, leaving an unwelded length 1.2 times the plate width. PIote i s beveled and groove welded to the column.

Wcldirig is most effici~:utin structures 11t.sipt.dtor full contiriiiity. This typr of dcrign builds ir~tothe structure the inh(rmt strength w1iir.h comes from (ontinuous action of ;ill members. Lo;~(lsare easily rvdistrihuted when ovcrluading occurs oil wrtain mi-mbcrs. Tliic type of desigri rr;ilizrs a weight swing in the beams sirrw a negativc mo~ncntacts ;it the supports, thus redui:ilij: the positive moment at tlit: center of the span hy i11e same amount. C;oiitiuuc~us couuc~ctioiw ;11so t ; i h ac1v:mtage of what ired fct h ~ a: 20% iucr-r,:!s(: in t l ;illow.il?le ~ i~ending str(,ss irk the negativ~,rnonwrit rr:giorr new the support. This is ;iccompli:,htd tlinn~gh;I 10% iucreue irr bending ;111wwables for "crrrnp:~~'t" swtims. :md using a 10% reduction in the: ncg;~:ivc rnornt.nt. 'i'liis ridurtion iu negrtive moment is :illuwcd iur 'coi~rpact" st,(:,tions, provi
FIGURE 1

FIGURE 2

5.7-1

Alternate method of b u t t welding top flange connecting plate t o column flange using

FIGURE 3

placed between the conriocting plate and the beam flange to r ~ r s ~ ~ar ccomplete-pcrit:tr:~tioll : groove weld to the column. This eliminates b;rck gouging and welding an overlicad pass on the other side. Reducing Welding Requirements It is possiblc to design the seat stiffener to carry all of

the end reaction, eliminating any vcrlical u ~ l d i n gin the field. This reduccs the ficld \I-ekling to just dowithand groove \vekiing of the heam flanges to tho column. Where good fit-up can be assurtd, the beam fianges are beveled from the top side and groove welded in the field directly to the colurnn Aange. The beam web

is cut hack about 1" a r ~ dfillet welded to the wcb connecting platc. Some fabricating shops have jigs so that colr~mns can be elcvnted into a vertical position. This allows muclr of t l ~ cshop welding on tho connecting plates to he made in the downhand position. Cover Plates

When addcd at crids of beams to carry the extra negative momcnt; covcr plates must be welded to the column for continuity; Figure 4. Shop wclding tllc cover platas to the beam, with the lower beam flange and the upper cover piatc left

FIGURE 4

Beam-$0-Column Continuous Connections

/

5.7-3

nnbeveled, prodriccs a type of "J" groove for the weld corinecting them to the column flange. If column-flangc stiKensr plates are needed in this case, they should be of about the same thickness as the beam Bange and cover plate combined. The ~ ~ s rsingle ~al thick stiffener in line with tach heam flange can be replaced with two platr:s, each having half tbe required thickness. This means working with lighter connecting matt:ri;~S and using two groove welds, each being half the size of the original singlc groove weld, which reduces the amonnt of welding on the stiffeners by half. 2. A N A L Y Z I N G NEED FOR C O L U M N STIFFENERS

If the flange of the supporting column is too flexible, the forcrs transmitted by the connr,cting flanges will load the outstanding portion of the column Range as a cantilever beam and cause it to deflect slightly; Figure 5. As this ~Icflectioutakes place it reduces the stress in the outer ends of the hum-to-column connecting weld, thereby loading up the center portion of the weld in line with the column web. It was previously thought that unless the column Aange is extremely rigid (thick), flange stiffeners must be added to the colr~mnin line with the beam's top

FIGURE 6

and bottom flanges (or their connecting plates). Snch stiffeners k w p thc column flange from deflecting and load the u d d uniformly. However, recent resmrcl~ at Lehigh University indicates that iri most cases thc deciding factor is a crippling of the column web; Figure 6. If the column web is thick enough, stiffeners are not required. Buckling of Column W e b Due t o Compressive force of Lower Beam FIange

FIGURE 5

A test was set up, Figure 7, to evaluate effects of the lower flange of the beam in compression against the column. Two bars, one on each side of the column, relx':sentt,d the cross-section of the beam flange. The test member was placed in a testing machine and loaded under compression. In all cases, yielding began in the fillet of the

FIGURE 7

5.7-4

/

Welded-Connection Design

column just inside the column flange, and directly beneath the bars. Yielding progressed into the column web by means of lines radiating from this point to the column " K line, at a maximum slope of 1 to 2%. This progressed for some distance. I\ slight bending of the column Ranges was noticed at about 80% of the failure load. Figure 8 shows an analysis of this.

FIGURE 10

A test was set up, Figure 9, to evaluate effects of the upper flange of the beam in tension against the column. Two plates, one on each side of the column and welded to it, represented the cross-section of the beam flange. The member was pulled in a tensile testing machine,

Dimensions of both the column flange and the connecting plates were varied in order to study the effect of different combinations of colulnns and beams. First yielding was noticed in the fillet of the column just inside the column flange, and directly beneath the attaching plates, at about 40% of the ultimate load. With fnrther loading, yielding proceeded into the column web, underneath the colnmn flange parallel to the attaching plate, and into the cohnnn flange from the center of the conrrecting welds, and parallel to the colnmn web. Aftcr ultimate loading, some members failed by cracking of the central portion of the connecting weld directly over the column web, some by cracking in the inside fi1lt.t of the column, and some by cracking in the inside fillet of the column, and some by a tearing out of material in the column flange.

FIGURE 9

FIGURE 11

Column web

+-/~/c FIGURE 8 Overlooding of Column Flange Due t o Tension Force of Upper Beom Flange

earn-to-Column Continuous Connections

/

5.7-5

FIGURE 12

Stondord Stitfeners When some type of wrb stiffening is required, the standard horizontal flange stiffcners are an eiticicrit way to stiffen the column web. Figure 10 shows this type under test. A Tee section flamo cot from a standard wideflangr section may be lisnd for stiffening, Figure 11. The stem of the Tee section is welclcd to tire colrrmn web for a short distanw in from the m t l s . This could be entircly shop welded, all of it being clo~icin thr flat position, pmsibly using a sc,mi-automatic wdder. This type stiffnrer would h a w nt~mcrousadvantages i n fomway beam connections. The bmms rrormally framing into the columi~web wonld now butt against this ilat surface with good :~cc<:ssil~ility, Tllc flnngcs of the beam coiild be beveled 45" and then easily groove welded in the field to tllis sltrface, using hacking straps. Thcre wonld be no 0 t h conrir:cting or attaching p1att.s to be used. In effect this part of tlie coriucctiu~~ would be identical to the connection used for beams framing to colnmn flanges. See Figures 28, 29 and 30 and related text for speciiications of stiffeners applicable to clastic design. Effect of E c c e n i ~ i cStiffeners

In a four-way beam-to-column colmection, the column

flanges may be stiffened by the connecting plates of the beam framing into the column web. It may be that the bcam framing to the column flange is of a different drpth. This in effect will provide eccentric stiffeners, Figure 12. The lower part of Figure 12 shows how this was testcd. It was found that an eccentricity of 2" provided only al)out 65%:of the stiffening provided by concentric stiffont.rs, and an eccentricity of 4" provided less than 20%. Three metliods of framing beains of different depths on opposilc flanges of columns are shown in Figure 13. 3. TEST COMPARISON OF STIFFENER TYP The following is adapttxl from "We1dt:d Interior BeamTo-Columrr (:orinections", AISC 1959, wllicli summarized lcsls o11 various connections. Figure 14 represents a dirwt beam-to-column connection. Iiero tllc column has no stiffening and is not as stiff against rotation as tllc 16'' W F 36# beams which frame to the colu~nn. This arrattgeinent showed high stress conccntratio~isat thc ccnt~vof the bc:im tension flanges, and therefore at the celitcr of the connecting groovc weld.

elded-Connection Design

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 thc 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 stifIencrs in the elastic range transmitted only ahout y/,, of the forces coming from the beam fangcs and the column web transmitted % 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

I

+

d

4

I

20,000 psi w-

Stress distribution in tension flonge

FIGURE 14

Beam-to-Column Continuous Connections

/

5.7-7

FIGURE 15

FIGURE 16

tensile stresses occumed at midflange. The con~lectionin Figure 17 was stronger than its two-way counterpart. This evidently shows that the stiffening action provided hy two beams framing into the column web strengthens the connection more than

it is weakened by the triasial stresses. The connections of Figure 18 involving (EastWest) beams weldcd dircctly to the column Ranges proved stiffer than (he com~ection of (Nori-11-South) beams to the Tee stiffeners.

FIGURE 17

FIGURE 18

elded-Connection Design

FIGURE 19

The stiffcning of the latter connc:ction is mainly dependent on thc thickness of the stem of the Tee stiffener, tlie Ranges of the colnmn being too Ear away to offer much resistance. The column wcb is ably assisted in preventing rotation at the connection by the flanges of the splitbeam Tee stiffeners.

~ n a l y s i sof this plate by incans 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 colomn flange can be considered as acting as two plates, both of type ARCD; sec Figure 19. The beam flange is assnmrd to place a line load on each of these plates. The effective length of the plates ( p ) is assumed to b e 12 t,. and the plates are assumed to be fixed at the ends of this length. The plate is also assumed to he fixed adjacent to the column web. where: m = w,

+ 2 ( K - t,)

For the wide-fiangr colrimns and beams used in pactical connections, it has h e n found that ci varies within the range of 3.5 to 5. A conservative figure would be-

P, = 3.5 u, t,' The force carried by the central rigid portion of thc column in linc with the web is-

earn-to-Column Continuous Connections Setting this total force equal to that of the beam's tension Hange:

/

5.7-

If the tliickness of the colnmn web (w,,) meets the ;hove rrqnircmrnt. column stiiicners ;we not neoded in linc with the coniprcssion fla~~gcs of the 1)ram. If the ;~ctuelti~ickn(,ssof the column xvob (w,) is less than this value, tlie \veb must be stiffened in some manner.

. HORIZONTAL

STIFFENERS

FIGURE 20

Reducing the strength of this column region by 20% and making the conservative assumption that m/b, = .15, this reduces to the following:

If the thickness of the column flange (t,) meets the above requirement, colnmn s t i f h e r s are not needed in line with the tension Rangcs of the beam. If the actud thickness of the column iiange (t,) is less than this valne, stiffeners are needed.

5. ANALYSIS OF STIFFENER RE I N COMPRESSION REGION OF CONNECTION (Elastic Design) It is assrimed i11e coucentrated compression force from the beam flange spreads out into the column web at a slope of 1 in 2% m~tilit reaches the K line or web toe of the fillet; see Figure 8. Equating the resisting force of the column web to the applicd force of the beam flange, assuming yield stress-

FIGURE 21

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-

A,

2; A,

- w,.

(tb

4-5

K,.)

I

where: A, - total cross-scctioual area of pair of stiffeners

To prevent buckling of the stiffcner-

where: b, = total width of pair of stifleners If the stiflcner 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

elded-Connection Design

eBective as when in direct line. The stiffener thickness (t,) fourd from the above formula s l i o ~ ~ lthan d he mi~ltipliedby 1.70 to giw thc actual required value.

7. VERTICAL STIFFENERS

Becaust~tlw vertical stiffelicrs (~lsuallyTees) are placed at the outer d g c s ol the column ilnngc. they are assulncd to Be half as d f r ~ t i v eas tl~ouglip1:iccd noar the colurnli wch. It is :rssumcd the corlcc~~tr:it
+2

x

% tr (ti,

+ 5 Kc)

u,

-

A* u,

or

To prevent buckling of the stiffrner-

1

PraHem 1

1

As an example of applying the preceding analysis of the tension region of a connection, we will analyze a connection which, wliel~tested to failure, performed well; see Figure 23.

FIGURE 22

w W A O i t column

FIGURE 23

Beam-to-Column Continuous Connections

where:

+2

j(1 % e )

5.7-1 1

region of the colnmn stiffcncr's flange must eqnal or exceed the force of the beam's tension iiange, or:

m = w, f 2 (K - t,)

= (.390)

/

- (.606)] Provided both column stiffener and beam have same yield strength:

4.28 2 3.00 If w e nsed the conservativc formula:

O.K.

but the initial design called for t, = ,606'' and the connection tested O.K. 8. CONNECTIONS THROUGH VERTICAL TEE STIFFENERS

Tests have shown that when thc beam flange extends the full width of the connecting plate, Figure 24, about 3~ of the flange force is carried by tho central portion of the plate. Each of the two outer edges carry about of this force. Figure 25 comes from test data of Lehigh University. Notice in the East-West beams, thc flange of which extends almost the full width of the colun~n

Since:

A = -

h (I

-

(2.72)

--

.58

x,

- (4.69)

p = -P

9 - (7.27) - (4.69)

= 1.55 and:

-

The total force which car] hc carried by the tcnsion

FIGURE 24

elded-Connection Design

FIGURE 25

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 3$, of the load. The report "Welded Interior Beam-To-Column Connections", AISC 1959, mentiolis that "from strain gagc readings it was calculated that the vertical plate stiffeners in the elastic range each transmitted only about 3/1,ths of the forces coming from the beam

flanges and the web transmitted %ths." Of course, the same would not bc true in the KorthSouth beams becaose they do not extend the full width of the flange of the Tee stiffener. As a resitit, most of this force rniist be transfrrucd 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 hcnding out along the flange of the Tee section. Weld Size: Stiffener Stem t o Column W e b

On the basis of these tests at Lehigh University, on coniiectioris where the beam flange extends the full

FIGURE 26

~

earn-to-Column Continuous Connections

width of the stifTener flange, we will assume that % of the beam flange force is carried by the stem portion of the connection. See Figorc 26. Because of the stiffening effect of the beam web and the stem of the a~nncctingplate, tliis ccnlral (stem) portion of the connection will load u p in bending. This assumes it rotates as a unit aboi~ta 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 3 r d lower edges of the connection. Treating the weld group as a line, the section modulus is cqual t o -

M S,

% M D 3

(D" - g")

f b = - =

The leg size of this weld would be found by dividing this value by the allowable for the particular weld metal.

1 h 7 , A373 Steel; E60 Weldr

f = 9600 o

5.7-13

Weld Size: Stiffener Flange to Column Flange

The Tec stiffcwxs may be joined to the column flanges by a ) fillet welds, b ) groove welds, or c ) corncr welds. The groove welds ( b ) were used in the Lehigh Research of this connection.

(4 The resulting maximum unit bending force at the top portion of the weld on the stem is-

/

(bl

(4

FIGURE 27

Since tests on full-width flanges showed that the of two outer edges ol the connection carry about the flange force, we will assume that each outer weld must carry 'h of the flange force. See Figure 28. These welds will be pulled with an axial force of K F. We may assume the same distribution of force through the coniiecting plate at a slope of 1 to 2% into the connecting welds. This will provide an effective 5 t, to carry this force. length of weld of tb The unit force on this weld is-

+

A36, A441 Steel; E70 Welds

The leg size of the fillet weld, or throat of groove weld, is detelmined by dividing this unit force by the suitable allowable. The effect of the vertical shear load ( V ) on these

Here:

FIGURE 28

5.7-14

/

Welded-Connection Design

welds could Le checked by using the elktire length of the welds. Ilowever, this would represent little additional force on t i m e wolds.

5. As a guide, the stiffener should satisfy this condition:

Proportioning t h e Tee Stiffener

Tho following will be helpful in selecting a Tee stiffener section for this type of connection, where the bcam flange equals the full width of the stiirener flange:

or an approximation on the conservative side:

idth < Stiffener Flange idth Where the beam flange docs not extend the full width of the co~~necting plat(:, the stem portion oi the connection is assumed to carry the entire moment. Therefore thc maximum bending force on the top portion of this weld will be-here Beam Flange

FIGURE 29

1. The thickness of the stiffener flange (t,) must be suificieut 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 stiffencr flange (b,) must be sufficient for it to reach to the column flanges.

3. The thickness of the stifiener stem (w,) should be about the same as the beam flange thickness (t,). FIGURE 30

4. Tho depth of the stiffener (d,), as measured through the stem portion, must be sufkient for it to extend from the face of the column web to the outer edge of the column flange.

The same items as before are used to proportion thr Tee stiffener, except in items 1 and 5 where the full vzilue of thc: heam flange's section area is used instead of 3/4 of tbis value. These formulas bccome-

earn-to-Column Continuous Connectians

I

Problem 2

/

52-25

The wcld on the 1je;rm's wcb niiist he able to stress the well in benLing to yicld (u,) tlirongho~~t its 'i~tirc dcptlr; see the bcndiiig strrss riistribntion in Fignre 5 The weld mnst also lic able to tr:insfer the vertical slrear.

1

To dcsigrr a fiilly wcldtd bcnm-to-volnmn conncction for a 11" WF bram to all 8" W F coliimr~to transfer 1lCU in.-kips anti a vertic:il an end moment of M shear of V == 20 kips. The solution of this problem will be considered with sevcn variations. Use A36 steel and E70 welds.

--

leg size of fillct u e l d @

=

actual force ~-~~~ allowable force ~

~

~p

I-lowever, since the beam web is welded to a ,433" thick flange of the column, the minimum size for this fillet wt.ld would be % G ' r ; see Section 7.4, Table 3. FIGURE 31

WELD SIZE TO DEVELOP ULTIMATE LOAD Hero:

M = 1100 in.-kips V = 20 kips Thc welding of both thc flailgrs and thc we11 along its full dvpth enahlcs thc lieam to d t v l o p its iull plastic moment, thus allowing the "compact" beam to be strcsscd 10% higher in brnding, or c . = .66 c,.This also allows the encl of the bcam, atid its welded connectioli, to be designrd for '30% of tlie elid snonwnt due to gravity 1o;iding. (AISC Sec l.5.1.4.1 and Sec 2.6)

= 23,700 psi

<

.66 u,

<

24.000 psi

The next qi~estion is what size fillet wold would be required to develop the bcam web to yield stress. The forcc 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 allo\vable values for transverse loading. (l~amllt~l load) 2(11,200 w )

2

(tr:insvi:rsc load-tension) t, ( G O u s ) = t, 22,000

OK -

(transverse load)

(transverse load-tcnsion)

5.7-16

/

Welded-Connection Design

For plastic design concepts, basrd on ultimate loading, the allowable for the fillet weld would be increased by the factor 1.67 (AISC Sec 2.7). This is the same increase used for the member ( 3 0 u, up to u y ) , hence the same relationship betwcen weld size and plate thickness will still hold. Based on AWS Code allowables (for parallel loading), this fillet weld on the web of the beam would have to be equal to the web thickness.

t, = .27W or use o

=:

Ya"

However since it is known a fillet weld ( o = j/4 t,") will outpull the web, a fillet weld will be used here.

FIGURE 33

Here:

M =. 1050 in.-kips V = 20 kips If this cant~lpverbeam had an end moment of M

-- 1050 in.-kips instead ol the previous 1100 in.-kips:

= 21,600 psi FIGURE 32

M = 1100 in.-kips V = 20 kips Thc welding of the Ranges and full depth of the web enables the bcam to develop its full plastic moment, allowing the "compact" beam to be stressed 10% higher in bending, or u = .6G ui' In this casc the beam cantilevehs out from the support so that 110 10% redoction in the negative moment can be made.

-

= 22,700 psi

<

.G6 u,

<

24,000 psi

.GO ur < 22,000 psi

OK -

In this case the bending stress is within .GO u,, and the beam and connection must be able to develop a bending resistance q u a 1 to the product of the beam's section modulus and yield point stress (scc Fig. 27) rather than the full plastir moment. As a result it is not necessary to weld the web for its full depth. For detormining the minimum length of the fillet weld on the web, assume the leg size to not exceed % tTV= ZiJ (287") = ,192". This will provide sufficient length of wcld so the bcam web at the connection will not he overstrossed in shear. (AISC Sec 1.17.5) The minimum ler~gthof fillel weld on each side of the web is-

-

(1100 in-kips) (4S.5 in.")

<

(20 kips) - -~ 20 kips 2(11,200 w ) - - 2(11,200) (.192)

OK

The fillet weld on the web of the beam is figured as in met~Ioct@

,"

If 3.: fillet welds are used (next size smaller than . l W r ) , their length w o d d he-

Beam-to-Column Continuous Connections

L, =

v

/

5.7-17

bending stress in beam

-

2 f,"

(20 kips) 2(11,200)(~,)

.--

.9 (1100 in.-kips) - -.. .-. - - -- . (41.8 in.a)

= 4.75" Hence use

Xe"

5" long on both sidcs

< 4.65".

= 23,700 psi

Since the size of this weld used in detcrmining its length was held to 24 of the wcb thickness, it is unnecessary to check the resulting shear stress in the web at this connection. Ho.rvever, to illustrate this, it will be checked here: 7web

=

<

OK .

v

.66 a, < 24,000 psi OK -

bending force on top connecting plate

.9 (1100 --in-kips) . . - -13.86'

= 71.5 kips

-

AT" (20 kips)

section area of top connecting plate

= 5)(.2S7) = 14,000 psi

<

.40 u7 < 14,500 psi

OK -

(71.5 kips) (24,000 psi)

or use a 5%" x %" plate, the section area of which is-

If %" fillet welds are used to connect top plate to upper flange of beam: f, = 11,200 ( % )

= 4200 lbs/linear inch length of fillet weld

FIGURE 34

(71.5 - -- .

kips) (4200 ibs/in.)

Here:

The wdding of the flanges and fnll depth of the web enables the l ~ e a mto dcvclop its fn11 plastic moment, allowing the "conrp;lct" beam to be stressed 10% higher in bcnding, or .66 u,. This also allows the end of thc beam, and its wcldcd connection, to be designed for '30% nf the end moment dne to gravity loading. (AISC Scc 1.5.1.4.1 and Scc 2.6)

-

or use 5?'zr' of weld across the end, and return 6" along each side, fnr a total weld length of 17M". Tho lower flange of the beam is groove butt welded dircctly to the colnrnn flange; and, since the wcb framing anglc carries thc shear reaction, n o fnrther work is reqnired on this lower portion of the connection. The seat angle simply serves to provide temporary snpport for the beam during erection and a hacking for the flange groove weld. The fillet eld on the web of the beam is figured as in method 1 .

6

or use a 5" x''4'3

plate, the section area of which is-

A, = 3.75 in.'

>

3.54 in.'

OK

-

If M" fillet welds are used to connect the top plate to the upper flange of the beam: f, = 11,200 (3h)

= 4200 lbs/linear inch length of fillet weld Top plate: 8%" x 3" x Stiffener:

5"

x

E"

3" x %"

-

(78.0 kips) (4200 lbs/in.)

FIGURE 35

or use 5" of weld across the plate end and return 7" along each side, to give a total weld length of 19'' > 18.6" OK

-

V = 20 kips

DESIGN O F BOTTOM SEAT

In this particular connection, the shear reaction is taken as bearing through the lower ilange 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 hcld to u = .& u, or Iu = 22.000 psi for A36 steel. (ATSC See 1.5.1.4.1) bending stress in beam

= 20,200 psi

<

.60 a, < 22,000 pso

bending force in top connecting p h t e

= 78.0 kips section urca of top connecting plate

-

(78.0 kips) (22,000 psi)

= 3.54 i n 2

FIGURE 36

The shcar reaction ( V ) by itself, applied to the bracket, produces a bending moment in the seat. This causes a tensile force in thc seat bracket's top plate and connecting welds. In the usual simple bcam type construction, this moment must bc considered in addition to the shcar reaction when determining the required size of connecting weld on the seat. In a continuous beam, the negative moment produces a compressive force in the lower flange which, in most cases, will offset the tensile force mentioned above.

Beam-to-Column Con@inuous Connections

/

5.7-1

As a result, the welds connecting the seat bracket will be designed only to resist the vertical shear force (V).

web crippling from end reactions R =: 75 u, (AISC Sec 1.10.10) t g S 7 ~ )

or:

-

-

(20 kips) .75[36,000 psi) (.313") ~

~~

-

1.0

IIcnce the top plate of the seat must extend to at least M" gap 4-1.37" = 1.87" and have a width at least 1" greater than the beam's flange width ( b ) = 1" -t 6.776 = 7.776"; or use an 8%" x 3" x 'h" plate. The 3" dimension would allow room for erection bolt. seat stiffener The thickness of the seat stiffener (t,) should be slightly grcater than that of the bmm web (t, = .313"), or use a ?8" plate. For determining the minimum length of the Blct weld on the stifrener, :lssunie the leg size to not exceed YJ t --. .. ( ) 1 lh". This krcps the stiffener at the connection from being ovrrstr(:ssed in shear. (AISC Sec 1.17.5) Thus, the niinimum lengtl~oi fillet weld on cach side of the stiffener is-

FIGURE 37

be groove welded to the column flange. Instead, the top plate of the seat bracket will be extended to provide sufficient length of fillet weld. If Ys" fillet welds are used along the edge of the ,513" thick beam flange:

-

(78.0 kips) -2(11,200) (3/8)

- -

= 9.3" or use 9%'' Therefore, allowing for 'h" fit-up gap, use a 10" x 8l/2" x 'W top plate for the seat.

(20 kips) =2 7 T m o w

Because the column flange to which this weld is placed is ,433" thick, the miliimum fillet weld size would be $/16". Hence, use:

-

20 kips 2(11,200) ( :$;6)

or use welds of %," leg ~ i 7 cand 5" long, m d of course the stiffcner must be 5" deep. In this case, the lower flange of the beam will not

FIGURE 38

/

5.7-20

Welded-Connection Design

In this case the connection is made through the Tee stiffeners of the column. Since the beam flange is nearly as wide as thc stifIen~rflange, the crntrd stem portion of the stiffener is designed for % of the moment and each outer edge of the stifiielrer flange for 'h of the moment. The welding of the upper and lower portions of the stem to the column web is sufficient to stress the beam web up to yield (in bending) through its full depth. Thus, the beam may develop its full plastic moment. This allow,^ the "compact" beam to be stressed at u = .66 o;,and also to he designed for only 90% of the end moment. (AISC Sec 1.5.1.4.1 and Sec 2.6)

011this basis use Tee section cut from an 8" WF 48j: beam; see Figure 39.

t,

= ,683''

-7r - -Ii

DETAIL THE TEE STIFFENER

FIGURE 39

CHECK SIZE O F WELDS ON STIFFENER STEM

2x

5.*w,K,

* w,

b,, tb 5

2 -

x(6.733) (.387) 5

2

.39 -

(t,

-+ 5 K,)

= 314 beam flange area = x bb t,>

+lib

-'-TI

maximum bending force At top of weld on stem. Use % of the moment (M ).

For simplicity, use a conservative value:

= 6500 lbs/linear inch

earn-to-Column Continuous Connections

=

5.7-21

CHECK EFFECT O F SHEAR

leg size of fillet ucld @

/

~ctniil force . ..allowable force

The vertical shear oi 20 kips was not considered on the welds bccausr of the great length of welding. This conld be cl~eckedout.

-

ossumcd total length of welding

L CHECK WELDS AT OUTER EDGES OF STIFFENER Use Ih of the moment ( M )

2 D i ;4 (t,,

= : 2 (9.18) = 61.2"

+ 5K,)

+4

(387

-+ 5 x I%,)

unit shear force an zocld

= 327 lbs/linear inch For fillet welds. this would represent an additional leg size of-

For partial-penetration groove welds, this would represent an additional throat of-

force on weld

6270 lbs/Iinear in. if fillet welds, leg size

=

actnal Eorce allowable force

These additional weld sizes are neglected in this exainple. If they had bccn appreciably larger, they would have been added to the weld sizes already obtained for bending. 9. LARGE HEAVILY LOADED BEAM-TOCOLUMN CONNECTlON

if partial-perwtratian single-bevel groove welds, throat

size t = - actnal forccallo\vable force

actual throat ist = t, +

Y4"

= ,397" $ Yd' = ,647" or use

l%B"

It might be wcll to consider the hasic transfer of forces through a beam-to-cohrmn connection. A forcc applied transverse or at right angles to a member is transft:rrr:d almost wholly into the portions of that mi~mherwlricl~ lie piirallel to this f o r m See Figure 40. In the design of some connections, the portion of this force ( F ) transfcrrcd into any given element of the built-np member has been assumed to be proportionate to the stiffness or moment of inertia of this element compared to thc total. Soe Figure 41. An axial force in n member can transfer out at one end either as an axial forrc (norinal stress, either tensile or compressive) or orit sidcwnys into an adjacent member as shear.

5.7-22

/

Welded-Connection Design

FIGURE 40

column web, left-hand s t i f h e r , and into flange of opposing beam. ~ e / d lo s column wch and flange must he designed for this force. Although the total length of welding on the stiffener would be figlwed for this force, actually most of the force would bc carried by the transverse weld hetwern the stiffener and the column web. Under ultimate loading, we can assume the transverse portion will have yielded and the force will he uniformly distributed. Shear Transfer

FIGURE 41

Tensile Transfer

FIGURE 43

FIGURE 42

Tensile force from right-hand beam flange transfers directly as tension through the right-hand stiffener,

Tensile force from beam flange transfers directly as tension into stiffener and then out as shear into the column flanges. Parallel welds to column flanges must be designed for this force, unless another stiffener is placrd on the opposite sidc of the coluinn wob to back up this stiffener.

Beam-to-Column Continuous Connections

/

5.7-23

ing the colinnn nlust be transierrcd into the column flanges as a shear transfer. Assinne 211 > M2.

Tensile Transfer

FIGURE 46 FIGURE 44

Tensile force from hcain flallge ti-ansfm clirectly as tension throngh both stiffeners and web of colu~nn into otllcr 1)carn Aanga. Transverse welds bet\rwn column flanges wid stiffcnisrs intist be designed for tl~isform ( F ) less that which passm directly into the web f n ~ mthe flange. 1'ar;illcl welds hetwccri stiffeilcrs and cohrmn web transfer no force. Comprrssion portion of beam connection wo1.11d keep stiifcner from buckling. Shear Transfer

The tensile force F? of tlw flange of the left-hand beam will t r a ~ l s f ~asr tension into the stiffener, then throngh the transverse welds along the column web into the other stiffenwj and into the flange of the other beam. The unba1;ulced tensile force (F, - Fa) of the flange of the right-hand beam will tr8nsfi.r as tension into the right-lmid stiffener, and half of this through the transverse wi.lds of thc coluinn web into thc lefthand stilkner. This unbalanced tensile force in these stiffeners now transfers through the parallel welds as shown into the flanges of !lie u~lnmns. Welds to column wcb must bc designed for the FI i- -Fz ~ - .~ balanced force, or 1% 17 + F2 =

+

Welds to column flange must be designed for the unbalanced force or F, - Fa. Distribution of Tensile Force

F=2Fs+Fw

There is some problem in estiiiiating the portion of the tensile force in the beam flange transferring directly into the web of the column and into thc colnmn stiffeners.

FIGURE 45

Tensile force from beam flange transfers directly as tansion into s t i f h e r s nnd colnmn web. The tensile force in thc stiffeners thni transfers ont as shear thmugh the parallel welds intr; coluinii web. Trmsversr welds l~etweenu1111mnflanges on the beam side tmd stiffenrrs mint bc drsigned for this force (I?) less that wl~ichpasses directly illto !he web from the flange. Pamilel welds to coluinn web must be designed for this same force. Any unbalanced lnomciit ( M = MI - Ma) enter-

FIGURE 47

At first glance it \vonld swm reasonable to assnme this force wonlrl be divided according to tlle width of the stiffeners ( b , j and thickness of column web (t,").

5.7-24

/

Welded-Connection Design

fabricated column d = tb+5K,

'column web

FIGURE 48

However, this column web scction is not limited to the thickness of the bcam 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 colun~nweb through which force is distributed, is obtained as follows:

FIGURE 50

rolled column d =ti, + 5 K ,

-4 K. /t

Since: A,

area i~f colnrnn web over which force is distributed = d t,

A, = area of one stiffener (there is a pair)

(web) F, = F (Aq. 2 2 As) (stiffener) F, = F

(A,%. :2

A)

Combined Stress in Stiffener (See Figure 51.) On the left-hand figure, tho shear stress (T,,) results from the unbalanced East-West mumcrrts. This causes the diffcrence in tensile beam flange force (FI-F2)to be transferred as shear in the stiffeners into the colnmn flanges. Although coi~servativein this particular analysis, it is assumed the small section in the stiffener to b e checked lies outside of the path which the East-West tensile flange force will travel; hence us = 0.Actually some of this tensile force will spread out into this region, and this would result in lower principal stress. In eithcr case, it would be checked by the following formilla:

On thc right-hand figure, it is assumed the small section to be chrcked is not snbjected to any shear stress, just biaxial tensile stress. In this case, the use of the formula results in the principal stresses being e q d to the applied tensile stresses. This does not result in any higher stress.

earn-to-Column Continuous Connections

Mohr'r Circle of Stress FIGURE 51

To check beam-to-column connection shown in Figure 52 (next page) for weld sizes.

flange fo~.ce: 24" WF 160# beam ,

1.135''

M = u S = (22,000 psi) (413.5 in.3) = 9097 in.-kips d -- 24.72" - 1.135" = 23.59"

(9097 -~ in-kips ) (23.59") = 3% kips -

flange force: 21" WF 73# beam

M = c S (22,000 psi) ( 150.7 i x 3 )

= : 3315 in.-kips d = 21.01" - .74"

= 20.50"

/

5.7-25

elded-Connection Design

FIGURE 52

F = 386 kips

FIGURE 53

earn-to-Column Continuous Connections

FIGURE 54

= 162, kips (See Figure 53.) distvihution of ber~mforce Depth of coli~rnllweb t h o u g h which beam force is transferred is-

If I" ho~.izonlalplate stificners are uscdA,

.-

(10%)(1)

= 10.5 in.'

/

5.7-27

5.7-28

/

Welded-Connection Design

Figure 55 sl~owsthc forces on the various welds for \vhic11 size must be determined. u:ckE size: stifienw to colunm flange; case@ and

= ,344" or

%"

B

if shop weld,

but 3%" plate would need "z"

@

B-

In the shop, fillet welds would he used, because they can be made on both sides of the stilTener. For field welding, use 45" single bevel groove weld because it wonk1 be difficult to weld underside ovcrhead.

weld size: stifcncr to column web; case c and d

I

I FIGURE 55

= 246 kips

= ,605" or

5/arr$

if shop weld

(2" plate needs min. of %''

\)

For field weld, use 45" single hevel groove weld.

= 70 kips Figure 54 diagrams this distribution of beam force for four situ:%tiutls.Only onc need he considered for any one problem. Ilowever, in this example we will detail the welds so they can carry any combirration of forces from any of these four situations.

weld size: heam flange to stiffener; case @and

@

eom-to-Column Continuous Connections

/

5.7-2

= .6Yr or %'' check combinmi stress in stiffener; cuse @

1

Problem 4

)

To cllrck tlrt. wi4d size joining the flange and web of the bnilt-up w e l d d column i i r Figures 57 arid 58.

@ weld

on column bettoem floors

= 1310 lbs/in. lorrgitudinal shear on weld

FIGURE 56

= .1c" use -& V ' but bccausc of 3%" plate, .

@ ueld

on column within, beam connection

= G6GO psi

= 3860 psi

FIGURE 57

L

Moment diagram

elded-Connection Design k-93/,..9Y

I,

=

16.815

ind

The resultant forcr on the weld is-

( a ) If fillet welds are used, tlre rerjoired leg size

( b ) If partial penetlxtiou J-groove welds are used, the requircd throat ist

-- 10.460 13,800 -A-

= ,622" and the root face is-

( c ) If partial penetration bevel groove welds are used. the reauired t h r o ~ tis -

= ,662" t = t,.

FIGURE 58

The transverse force must he :~dded to this. A portion ol tlic beam Range forct, must he transferred through this ilal~ge-to-webweld witliiri thc distance d -- ti, 5 K, = 18.64"; the rem;~i~idur of this force is transferred clirectly throilgh the horizontal stiffeners:

+

= 2-17 kips This is a unit force on the weld of-

-C %"

and the root face is-

10. ADDITIONAL STIFFENING OF WEB W I T H I N BEAM-TO-COLUMN CONNECTION

In wscs of mnisually h i g unb:~liince of applied inomelits l o a c o l u m ~ ~it, iniglrt br well to check the rrsrilting sli
Beom-to-Column Continuous Connections

/

5.7-31

FIGURE 59

are tr;ilisIiwrd into tlw colu~nnwel) within the connection rcgioo as shmr. It c:m be assurni,d that xilost of tbib vertical shear force ( V ., , \ of thr beain weh is tra~~sferred diucctlv into the flange of the supportiilg cohim~iarid does not enter the web of tile corin(,ctioi~. The Iiorizontal shear force (V,) of the upper columr~ will he translrrred through the web of the connection illto tlie luw.er column if caused by wind; or out across the beam to the adjacent column if ca~rsed by gravity load. ~

\

Analysis of Required Web Thickness

The unit shear force applied to thc web of the connection is-

= .V- = - F, - - - Vp d

dc

-

Mi dud,

Vq d,

The resulliilg unit shear stress in the web of the comcction isT = - -v

wi

-

1 ME w ( d

d.

Using plastic design concepts, the applied moment (MI) will become tlic plastic moment. For this valuc, thc allowable shear stress ( 7 ) will be based on the yield streiigtli of the steel. The value for the shear

FIGURE 60

Thcsr: rcsuiting vcrtical ;imI liorizontnl shear forces cause a diagonal coin]?uessive force to act on the web oi tlic co~inection;xnd, if the \vcb is too thin cornpared to its width or depth, it may suEer some buckling action. SFC Figlire 61. Thc following a~lalysis,based on plastic design concepts, rmay be used to chwk iliis condition.

FIGURE 61

/

5.7-32

Welded-Connection Design

Resisting moment at ollowoble [a] FIGURE

stress at yield ( T , . ) may be found 11). usi~igthe Mises yield criterion: 0;.

=

J uX2- ur uy + q24

Q

62

Or assuming that a conservative shape factor, f

rxy

In this application of pure shmir, u, and o; = 0, and setting the critical value (rr,,.) q u a 1 to yield (cry), we obtain:

hence:

Reststing plastic moment

h'f 1\1,

Z = 1.12 =-

; -2

M , = 1.12 M,,

s

and My = o; S

Formula 2 may bc reduced to-

If tbc actual thidmess of the web in the connection 7

=

T i

6

-

\

I

,,I

d,,&

-

( v ) is equal to or greater than this required valiie

")d,:

( I Y I ) , 110 additional stiffening of the web would be necessary. If t l ~ cweb tl~iik~iess is less tiinn this value, it must be stiffened by some metllod.

Methods of Stiffening Web in Connection

The horizontal s h r forw ( V4) of the upper column acts in the oppuite dirwtion to ( F 1 ) and thus r e d r ~ u st h : slirai- valili. in tilt: wr.h of t f ~ cconnection; so this portiou < ~ o i ~be i d neglected for siinplieity. This formrtla t h m bccomrs:

A wch doubler p h t e could be added to makc up this difkrcnce bt.twer~i actual aud rcqiiired v;ilucs of web thickness.

Web doubler plate

Tlw plastic mornmt ( h i l ) is obtainnd hy multiplying tiir plastic swtion rno~lriliis( % ) of thc bmm by the yield slrt?ngtli (v,.) of the stt~>l. l'ha plastic si:ctio~i mo11olri.s for all rolled sections is availal~li~ in s e w d strr.1 malii~;~ls. The plastic sectiori rn(~dii111sof a n~eided plate girder (Fig. 62) is obtained Slam the following formula: Z = b t (d

- t)

-1-

W

4

(d

- 2ty

. . . . .(3)

FIGURE 63

'I% most co~ninmsoliitio~r is to usr: n pair of diagonal stiffe~wrs. Thcir cross-swtional area would

earn-to-Column Continuous Connections

/

5.7-33

FIGURE 64

depend on the comprr~ssiveforce they must carry, over and above that carried by thc web. See Figure 64. The horizontal force applied to the connectiol~is-

w, = miiiimoin reqiiircd web thickness, from Formula 2 or 4 w2

d,

The horizontal shrar force resisted by the web is-

-

nctiii~l\v1,1) thickness of connection

:

length of diagonal of conncdiosi area

11. COPE HOLES

The rcsulting lioriLontal component applied to the diagonal stiffener is-

The force on the di;igonal stiffener is-

and the required total area of hoth stiifmers is-

also

also

A

.

\/ .

d,"-

d,'

(

,-

a'---

\

)

d, . . (-\ v , . 3

w?)

in other words going from a givm stress down to zero, etc. For a mow n;irron rmge of stress, for example K :z~ I/*, going from a givm strcss down to just oneI d , etc., tlierc was almost no diiference with or without copc holes.

elded-Connection Design Provides accesiibility

Provides orcersibrlity

( for root govg~ng \

L '

for welding

Bending stress ot

No backing bar used; joint must be back gouged

plostic moment (Mpj

for welding Bocking bar used; no back gouging needed

FIGURE 65

Plastic dwig~n is not risd rirrder fatigw loading conditions, so therc shonld he less concern here about thc need for cope holcs and tl~eirrcsnlting cffcct on tlie connectiori's stnmgth. Cope holes \voulcl prol)olily not result in any npj,rr,ii;ible loss in plastic strcngth. The additiond inomcmt brouglrt abont liy t r h v i n g tlic \veh to be stressd to yicld strcngth uftcr the outer filxrs once reach yic*ldis ahont 105, and tlie cope liole represents 2% \ - e n srnall portion of tliis wch scction. Ifcnce, the rcd~rction in strmgth ca~r.scd by the cope liole should lic only n small fraction of the 10%. Along the sariie liric of thought, any minor lack of weld pc~rictration dne to this lack of accessibility with no copt, hole \\-onld not be as critical. 111 going throrigli tlw original test rcports of wcldcd coirncctioiis for plastic ilcsign. thwc ;ire rinany boam-tocolumn connections or knccs in wlrich no cope holes were used. In the AISC report, "\17eldcd Interior ReannTo-Colnnin Connections" cope liolcs were nscd and a detail of these sliowi; s w Figure 66. Notice that back-

%," Flame-cut cope hole

ing bars were rrscd a i d the holes were not later filled with n.cld metal. 111 plastic design_ cx>pe holrs w e not rerlniri~d to prn\,idc the weld quality rr:qniueti, althongh t h y would make it easirr for tlic wc4ding opcrntor. And, if they arc osccl, they \von't Slaw a dctrimental effect on the strength of tlrr. connection if lelt ul~filid. Thc cope hole hclps more for ;iccc.ssibility of the groove \veld on the lowcr flange if weldrd in position. In most cases tliis would be an amr of negatiw inomcnt and this \r-rsld would he un&,r compression, so this should not be as critical as the timion weld on the upper flangc. IF the rnmihrr c:~ir~ld he tm-lied o w ~ for r shop welding, both fl:nigt~coi~ldbe h r w l t d from tlie outside end copc holes nwild not he nredcd: sre Fignrr 67.

EAMS CONTINUOUS THROUGH COLUMN (COLUMN CUT OFF)

Bocking bar

FIGURE 66

On orrc-story mnstrnction, it is qnitc common to nlitnin continnity of the hcam by allowing it to run continrioi~slyover tlic toil of thc colninn for two or more spans. Freqnently the splice in tlie hcnm is carried out to the point of coiitrailexure.

eom-to-Column Continuous Connections

/

5.7-35

FIGURE 68

FIGURE 69

Figure 68 ( a ) shows the Iwam resting on a plate shop wckled t o t h e top of t h e column. In most cases fillet welds made in tho iIo\vnliarld or Hat position will be sufficient, since there is usrially very little moment which must be tmmisfcrrrd from tlic heam into t h e column. Figriro 68 ( b ) sliows a similar connection m a d e in t h e hcarn ;ind t11c girilcr which sripports it. Figiircs 69 ( a ) and (1-1) s l ~ o wthis mctliod c x t r n d d to multi-story constriiction. In hoth cascs, stiffming plates art, sliop m c l ~ l dill betwwrl tiit. Rarlgcs of the beam, in liilo with the voliirnn fiangcs. so that the compressivt~ lo:id m;iy be t r a n s f c r r d diret.tly from one colrlrnn flange to the other.

PLATES FOR CONTINUOUS FRA Cover pliites nrl. sonwtimcs i ~ s e din vonnrr.tio~r with rolltd helims in or&r to illwc;ise t h c strcrigtlr ( S ) or stif%~iess( I ) p n ~ p c r t i c sof thc he;rm. Uiiless niiiiin~r~m wcigl~tis ;I rcnl factor; t l ~ cusr of covcr p1:itc.s on simply siipporttd 11c~nnismight not 11c justified in l i ~ d d i n gvo~istri~ction since the savings in strcl inight riot o f h t the a d ~ l i t i o i ~ acost l of fabricntirig a n d wclding the vovrr pl;it<. to tho hcam. 'I'his is becaust tilt, c o w r plate initst c x k n d quite a dist;incc to hot11 si11i.s of t h e beam centerline. Notice in the r r a m ple showr for uniform loacii~ig,Figure 70 ( a ) . that t h e covcr plat? must extend 70.7% of the beam's length ( c ) .

Becausi: of this grtmt Ic~igtli.the wcight reduction is only 8.79. On contiii~~oiis g i r d w ;in11 tieams, however, there is a r t d ndvantagr in using covcr platcs since t h e iiicr
Chi11-is h:iv<~IXYW t l t ~ v ~ ~ 1 i q 11y ~ c ~ whicli tl tht, dc,sigiit>~c:m I-i.ndily fiiid crnistiiirts to I I W ill
etded-Connection Design Sjmply supported beam

Continuous beom-fixed ends

uniform load

uniform load

(aj

idi Moment diagram

Moment diagram

weinht

--

100.096

weight

=

100.0%

(fl

(c) iengtli of cover ps= 7 0 7% weight cover

L

length of c o v e r k s

= 91.0%

ies increase

weight

S by 1

cover

= 70.246

ps

increase S by 1

FIGURE 70

( 2 ) "Moment I>istribulion", J . h l . Gmr:, 1963; D. Van Nostrand Co. 29 cl~;rrts for hcnrns with cover p1att:s at cnds; 42 i.1l;n.t~for tapered hca~ns. For methods of c;rlctrlnting t1it:se design factors, see Scctio~~ 6.1, on Dcsigrr of Rigid Frames.

I Example

A frame is to he rlesigncd to support a nnifonn load ol 2.4 kip"/E Tlirtv spans of 20' c:lclr ;,st: s r ~ p p o r t dby fnur col~irnns,12' Iiiglr. The beams arc 12" \VF 2 7 g beams, r r i n f o ~ c dn,itli .?6" s 5" coves plates for a distance of 2' on wclr si& of the intcrior srrpports. The colrimns arc S" \Z'F 31fi scctims. S w Figure 71. * I urt! The section proprrtics of tht: rolled lbeani, I.'& 72, without and wit11 cowr plates are as follows:

= 56

= 18 3% L

in."

earn-to-Column Continuous Connections

, %" X 5" cover

/

5.7-37

Fks

12" WF 27# beams

8" WF 31# columns

I

I

FIGURE 71 weight of this continuous b e a m with rover pleter= 1750 Ibl.

weight o f equivdent simple beom construction

= 3480 lbr.

FIGURE 73

NTINUOUS CON

FIGURE 74 (a)

FIGURE 74 (b)

FIGURE 74 (c)

5.7-38

/

Welded-Connection Design

Multi-Story- Dormitory Buildinq-

Shop rabvicated and w e l d e d con+inuous beam t w o interior columns. Assembly e r e c t e d as single unit.

FIGURE 75

FIGURE 76

B e a m - T o - C o l u m n C o n t i n u o u s Connections

/

Girder terminating at a column and not continuing through loads the column web i n shear in the region of the beam connection. This couses high diagonal compressive stresses, and diagonal stiffeners ore used to resist the tendency of the web to buckle.

Typical column joint to develop continuity in both directions. The column is cut off at this point. The main girder (left to right) has 100% continuity, no joint; column stiffeners on girder webs are shop welded. The cross beams are provided continuity by the use of o welded top plote extending right across the upper girder flange. The column for the floor above is positioned on top of this connecting plate, temporarily held by angles shop-welded to the column web, and then permanently field welded along the flanges to the connecting plote.

5.7-39

elded-Conneciiion Design

Actual service conditions on beam-to-column continuous connections were simulated in this experimental setup at Lehigh University's Fritz Engineering Laboratories. Here, the column is subjected to compresrive axial lood by the main press ram while the beam stubs are loaded individually by means of hydraulic cylinders.

1. INTRODUCTION Bcams may be made continuous through their girder supports by any of the methods illustrated in Figure 1. In Figure 1 ( a ) , the beam flange and part of the web below are cut back so that this flange can he bntt welded directly to the edge of the girder flange, with top surfaces of both members on the same level. In ( h ) , ( c ) and ( d ) , the beam web is cut hack just below the top flange so that this top flange rests on the top flange of the girder. This allows a very easy method of erection. Additional plates are used in ( c ) along the top after the top beam flanges have been welded to the girder. This gives the necessary increased area for the

negative moment over the support, and reduces the beam size for the reinainder of the span. Sometimes a small seat is placed below the beam; as in ( e ) and ( f ) . This facilitates erection and also serves as a backing strip for tile groove weld on the lower beam flange. Top connecting plates are used in ( e ) and ( f ) . These also serve 3s covcr plates to increasc the stiffness ( I ) or strength ( S ) properties at ends of the beam. If beams are offset, Fignre 2, the top connecting plate can be adjusted to tie both together with the girder. At exterior columns, Figure 3, the top connecting plate is cut in the shape of a Tee so as to tie in spandrel beams, girder and column.

FIGURE 1

5.8-2

/

Welded-Connection

FIGURE 2

FIGURE 3

"4' FIGURE 5

TERSECTING FLANGES

Should the intersecting flanges of beams and girders be isolated or may they be welded directly togeththa?

TYPICAL BAY

(1) For example, assume the girder to be simply supported, and the beams welded for continuity to the girders.

FIGURE 6

Design the girder as simply supported. Use 14" WF 68# beam having S = 103.0 in.s FIGURE 4 2oL

-

Consider the bay, Figure 5, with a dead live load of 200 lbs/ft2. On this basis each beam would have a 20-kip load uniformly distributed; each main girder would have three concentrated forces of 20 b p s applied at quarter points.

1L Ma = w --ij---

(WL) (24C") 6

= 2400 in.-kips

20X

20"

Beam-to-Girder Comtimuour Connections

/

5.8-3

Here: -

(2400 in.-kips) (103.0 in.")

= 23,300 psi compression Since the girder in itself provides very little end restraint for the intersecting beams which it supports, the beams will be designed as simply supportcd even though their flanges are welded to the girder. Use a 10" WF 25jf hcam having S = 26.4 in." However, if two beams framing on opposite sides of a girder are loaded, their ends will bc restrained and their end moments must be considered.

= 41.5 kips

= 15,900 psi These two biaxial stresses, a, = - 23,300 psi and u, = $- 15,900 psi, will &ect the yield properties of the girder's top flange within the region where the beam flange is attached. A plate subjected to uniaxial tensile stress, or stress in one direction only, will have a certain critical stress (uc,)above which the plate will yield plastically. In this case, this stress point is referred to as the yield strength.

uniaxial stress

However, if in addition, there is a compressive stress applied at right angles, this will allow the plate to yield easier and at a lower load.

The resulting flange farces and stresses can be diagrammed as in Figure 7.

bioxiol stress

A convenient method to check the effect of the applied stresses upon the yielding of the plate is the Huber-Mises formula. If for a certain combination of normal stress ( u ) and ( ) and shear stress (T~,.), the resulting value of critical stress (u,,) is equal to the yield strength of the steel when tested in uniaxial tension, this combination of stresses is assumed to just produce yielding in the steel.

= 36,600 psi

FIGURE 7

This would indicate the top flange of the girder is on the verge of yielding, and the tensile flange of the beam should be isolated from the biaxial compressive

elded-Connection Design

stress. This may he done by one of several methods, Figure 8. ( 2 ) Now assume the girder to be fixed at the ends and the beams welded for continuity to the girders.

0-1

90 M =S - .W -

(1500 in.-kips) (62.7 in.3)

~~~p

= 21,500 psi (Only need S = 56.2 in.3, but this is the lightest 14" WF section.) M2 = + M, L 48 (60k)(24W')

FIGURE 9

=+ 48 = + 300 in.-kips M =2 S - (300 in.-kips) (62.7 in."= 4780 psi Ma=+- W L 16 = + (SOk) (240") 16 = + 900 in.-kips 0-2

Design the girder as having fixed ends. Use 14" WF 43# beam having S = 62.7 in."

6

Moment diagrom

us =

M3

-

S

- (900 -in.-kips) . .

(62.7 i n 3 )

= 14,350 psi

Beam-(a-Girder Continuaus Connections

/

5.

ELDING OF TAPERED FLANGES

a,

=

-

a,

=

+ 15,900 psi

a,,=

14,350 psi

Jux2-a,~,+a~2+3~xy

- / (-14,350)'-( = 21,600 psi

-14,350) (15,900) +15,9002

The apparent factor of yielding is-

This seems reasonable, and under these conditions the beam flange could be butt welded directly to the edge of the girder flange without trying to isolate the two intersecting flanges.

Figure 10 shows the method for butt welding wideflange rolled beams which have a slightly tapered flange to the edge of a girder flange. By using a light %" x 1" backing bar, it may be hammered as it is tack welded so that it will be tight against the joint. Figure I1 shows the method for butt welding wide-flange rolled beams with a slightly tapered Bange to a flat plate. By using a light YB" x 11' backing bar, it may be hammered as it is tack welded so that it will be tight against the joint. If there is any criticism in doing this, the followi~tg should be remembered. This type of butt welded joint on the wide-flange beans with a slightly tapered flange presents a smoother transition in section and transfer of beam flange force, than the widely used type of (beamto-columnj top connecting plate shown in Figure 12 which is accepted. In this case (Fig. 12) the flange force must work

groove bun weld, ond oiro server or run-off tob ot outer edge

FIGURE 10

5.8-6

/

Welded-Connection Design

groove butt weid, ood olio server or run-off tab at outer edge

FIGURE I 1

!

/

~ o connecting p plate

FIGURE 12

itself up through the connecting fillet welds into the top plate, and then out throngh the groove butt weld into the supporting member. Although there is a transverse Gllet weld across the end of the top plate, much of the flange force must spread out along the edge in order to enter the fillet welds along the side of the plate. These connections stood up very well under testing and showed they could develop the full plastic moment of the beam.

eom-to-Girder Continuous Connections

LES OF CONTINUOUS CONNECTIONS

FIG. 13

Beomr framing to girder web.

/

elded-Connection D e s i g n

Welded connections ore used throughout the Ainsley Building in Miomi. Here, the beams ore given continuity b y connecting top flonges, using strop plotes reaching ocross the girder. Lower flanges ore butt welded to the web on both sides.

Continuous welded connections were used extensively in building the 7-story Horvey's Deportment Store in Nashville, Tenn. Here cross beoms ore given continuity through the moin floor girders by meons of a 1" thick cover plote ond a bottom support plote, wider thon the beom flange. This type of connection eliminotes any need for beveling plates and loying groove welds.

SECTION 5.9

1. INTRODUCTION

.'

In trnsses of proper arc welded design, gusset plates are generally eliminated. Tensiorl members in the welded design are lighter bcwuse the entire crosssection is effective, and the amount of extraneous detail metal is reduced to a minimum. Welded trusses may be designed in various ways, using T shapes, 13 and W F sections, etc. for chords. The diagonal members are t~snallyangles. Various tll,es of welded truss designs are illustrated in the following: 1. Perhaps the simplest lype of truss construction is made of angle shapes and Tee's. In this example, the bottom and top chords are made of T sections, with angle sections fur the diagonals. This is easy to fabricate and weld because the s t ~ t i o n slap each other and fillet welds are used, Fi y r e I .

FIGURE 1

2. For a heavier trnss, the vertical member can be an I 01 WF section. The web of this member, in the examplt~illustrated, is slotted to fit over the stem of the T section. The T section is used for both the top and bottom chord members. The diagonal members are made of a double set of angles, Figure 2. 3. Some trusses make use of T sections for their diagonal members. The flanges of the diagonal members must be slotted to fit over the stem of the T section used for the top and bottom chords. The stem of the diagonal is also cut back and butt welded to the stem of the top and bottom chords, Figure 3. 4. Quite a few tn~ssesarc made of WF sections completely: both top and bottom chords as well as

5.9-2

/

Welded-Connection Design

diagonal a i d vertical members. This allows loads to be placed anywhere along the cop and bottom chords because of their high bending strength. (With the conventional truss design, loads must be placed only at points where diagonal or vertical members connect to the chord mcmbers.) Almost all of the \velds are on the flanges of the top and bottom chords, and since these are fiat surfaces, there is no difkvlt fitting of the members to make these connections, Figure 4. 5. Where longer lengths of connecting fillet welds are required, a simple flat plate may be butt welded directly to the stem of the horizontal T chord, without any joint preparation. This weld is then chipped or ground flush in the area where web members will connect, Figwe 5.

-

FIGURE 5

6. Sometimes heavier trusses are made of WF sections with the web of the top and bottom chords in the horizontal position. The welding of these members would consist mainly of butt welding the, flanges together. Under severe loading, gusset plates may be added to strengthm~the joint aud reduce the possibility of concentrated stresses, Figure 6.

7. It is now possible to obtain hot-rolled square and rectangular tubular sections in A36 steel at about the

same price as other hot-rc>lled sections. This type of section has many advantages. It has good resistance to bending, and has high moment of inertia and section modulus in both directions. It offers good streugth in compression because of high radius of gyration in both directions. It is very easy to join by welding to other similar swtions because of its flat sides. For lighter loads, fillet welds are sufficient. These sections offer good torsional resistance; this in tun1 provides greater lateral stability under compression, Figure 7.

8. Rormd tubular sections or pipe have certain advantages in truss construction: good bending resistance, good compressive strength, and good torsional resistance. There is no rusting problem on the inside if they are scaled at the ends by welding, hence only the outside must be painted. Although it is more difficult to cut, fit, and weld the pipe sections t o g ~ a e r this , is not a problem for fitters and weldors experienced in pipe fabrication and welding. Pipe is used extensively in Europe for trusses. In this country it has been used for some mill buildings, special trusses for material handling bridges, extremely large dragline booms, off-shore drilling rigs, etc., Figure 8.

Design of Trusses TABLE 1-Effect

f = 9,600 w f = 11,200 w

5.

of Eccentric Loading

If consider momeni MI

Welded connection

/

= - Pe

If neglect moment

A7, A373 iteel & E60 welds A36 steel R E70 welds

There are many methods by which to join the various pipe sections together in a truss. In this case, the pipe is cut back and a gusset plate is used to tie them together. A gusset plate also provides additional stiffness to the pipe within the connection arra. However, they tend to cause an unwen stress distribution within the pipe, with rather high strmses in line with the gusset plate. See Fignre 9. These closed sections, with less surface area exposed to the elements, are less subject to corrosion than are open sections; in practically all cases they are left unpainted on the inside. It is only necessary to see that the ends are scaled hy welding.

2. EFFECT OF ECCENTRlC LOADING It can be shown that, with mcrnhers hack to hack, or separated with a gusset plate, the connections will supply a restraining cnd moment:

1

,

Since this moment is equal and opposite to the moment due to thc eccentric loading [ M = P e ) , they will cancel. As a result there will he no moment through-

.

FIGURE 10

elded-Connection Design

FIGURE 1 1

out the length of the member and it will remain straight. However, this moment ( M , ) is carried by the connecting welds in addition to thcir axial load (P). This moment is usually ncglected in the design of the welded connection, because of the difficulty in determining the length of weld ( L ) when it is considered. Further, there usually is not much clifferenee in the actual length of the required weld whether it is considered or not.

Here: e = y = .94" d = 4" w

= g6"

P = 53.4 kips since: F

(a) if the moment ( M e ) is neglected:

(See Figure lo.) Assuming A373 steel and E60 welds, AT = 2.67 in.= P=uAT

= (20,000) (2.67) = 53.4 kips

h

leg size of fillet u:eld @=%ti = ?4 (.425)

= ,3185" or

%,Ir

h

total length of weld LT =

Xo

and from this we find L = 8". (This value was found by plotting several valucs,of L on graph paper and selecting that L value which gave the closest value of P = 53.4 kips.) This would give a total length of 20" of % 6r' weld. In this case, the extra work involved in considering the moment did not pay for the very slight overstress in the weld when the moment was neglected. If only one member is used, and the plate to which it is attached is not very rigid, this restraining end moment will not be sct up. The member will then have a moment due to the eccentric load ( M = P e ) , in addition to its axial load ( P ) . See Figure 12.

P -kips . -( 9 6 ) kips/in.

uxiul tensile stress in member

P u =A This would be distributed 4" across the end, returning 6.9' on the sides, or use 7" long on each side. This would give a total length of 18" of %B'' weld.

bending stress

( b ) If the moment ( M e ) is considered:

Since the distance to the outer tensile fiber ( c ) and the distance of the st-ction's center of gravity from the base line ( y ) are equal, and since the eccentricity of

(See Figure 11.)

Design of Trusser

/

5.

Moment d i o g ~ m

of section (obtained

m steel handbook)

FIGURE 12

loading ( e ) is nearly tqual to these, it is assumed for simplicity that c = e r y. Therefore, the total (maximum) stress is-

In this particular case, the additional moment due to the eccentrically applied axial load reduces the mt:mber's allowable load carrying capacity by 40%. This far exceeds any reduction in the strength of the welded connection due to this moment. Thus, the connection will be on the conservative side. Conclusions:

or the maximum axial load ( P ) for a given allowable stress (G-)is-

For the ST 4" 19.2# member used in the previous example, Figure 10, this additional moment due to eccentricity of loading would reduce the member's allowable axial tensile force to:

( a ) If the attaching plate is very flexible and offcrs no restraining action at the end of the member, the full moment ( M = P e ) must b e added to the member and no moment added to the connection. In other words, the connection is designed for the transfer of thc axial force only. ( b ) If the a t t a c h g plate is rigid cnougl~so there is no end rotation of the member, this moment is not added to the member, but must be added to the connection. Evcn in this example, if the moment were also figured to he added to the connection, at thc reduced load of P = 32 kips, it would not require as much weid as in the previous case:

e = .94"

= 32 kips

FIGURE 13

P = 32 kips

( b ) calctdatcd allowable load:

since:

p =

f

i($)'+

.

1

1

(ld

L)l

+

(diJ

= 32 kips From this we find L = 4.4" or = 4%". (This value was found by plotting several values of 1, on graph paper and selecting that which gave the closest value of P = 32 kips.) This would give a total length of 13" of %/,," weld. This is another case where theory would indicate a much higher reduction in the carrying capacity of a connection than actual testing shours. The following lap joints wcre welded and pulled to failure.

h

( a ) calculatcd allouable load:

= 7500 lbs

Theory would indicate that, in the above samples, increasing the eccentricity ( e ) from '/a" up to 1" would dccrcase the strength of the welds by 60%. Yet, the actual test results showed: ( a ) f = 11,260 Ibs/in. ( b ) f = 10,380 ibs/in. or that this large increase in ecxentricity ( e ) , from V4" to l", only decreased the strength by 8.7%. The reasons for neglecting this eccentricity in the detailing of most connections may be summarized as folIows: 1. In the usual welded connection, the eccentricity is not vely large, and in these cases the thcoretical reduction in strength due to the additional moment induccd by the eccentricity is not very much. 2. Actual test results indicate a much smaller decrease in strength due to this eccentricity than theory would indicate. Also these test pieces were very short; the nsnal member would be much longer and, if any-

FIGURE 14

Design of Trusses

/

5.9-9

FIGURE 15

FIGURE 16

FIGURE 17

thing, would minimize this problem. 3. The eccentric loading would effect a reduction in strength of the member several times greater than any reduction in the strength of the welded connection. 4. It is very time-consuming to include this moment in consideration of the connection. AISC Sec 1.15.3 requires that welds at the ends of any member transmitting axial force into that member shall have their center of gravity line u p with the gravity axis of the member unless provision is made for the effect of the resulting eccentricity. However, except for fatigue loading conditions, fillet welds connecting the ends of single angles, double angles, and similar types of members (i.e. having low center of gravity or neutral ixis, relative to attaching surface) need not be balanced about the neutral axis of the member. 3. DISTRIBUTION A N D TRANSFER OF FORCES It is assumed that the axial forces in a member are uniformly distributed throughout the various elements of the cross-section. See Figure 15, where: A* = area of flange

A, = area of web AT = total arca of section If the force in some element of a member cannot be transferred directly through the connection, this portion of the force must work its way around into another element of the member which can provide this transfer. See Figure 16. This decrease in axial force ( F ) of one element of a member is accomplished through a transfer in shear ( V ) into another element. See Figure 17. The length of this shear transfer (I,) must be sufficient so that the resulting shear stress ( 7 ) within thk area does not exceed the allou,able. This area may also have to be reinforced with doubler plates so it call safely carry this increased axial force. If we assume uniform distribution of axial stress through the cross-section of the following member, then the web arca has a force of P,. (See Figure 18.) Shear transfer from web:

V, = P, = u A,

and

elded-Connection Design

= ,270''

,t

- 2.67 in2 ( A, = 0.99 in2 T

J

Web

3/8/1V FIGURE 18

P, =

5

A,

= (20,000) (.99) = 19.8 kips

-

This force in the web area (P, = 19.8 kips) must be transferred down into the flange by shear (V,), and out into the conncction. Theoretically, if the section is not to be stressed above its allowable, this shear transfer (V,) must take place within a length bounded by the connecting welds. If this is true, then this 19.8-kip force in the web, transferred as shear through a length of 5%" where the flange joins the web, causes a shear stress in the section (a-a) of:

(19.8 kips) (% . 0') (5%)

= 13,330 psi

+

13,000 psi (A373 steel)

This is close cnough. However, if it were higher, it would indicate that one of the following conditions exists: a. The shear transfer takes place over a greater distance and, beyond the welds, must travel this short distance in the flange as additional tension until the weld is reached. It thus slightly overstresses the section (b-b) in tension.

8"W 31 .#

F = 125 kcpr

>

t, = W' doubler plater

FIGURE 19

esign of Trusses

/

5.9-9

FIGURE 20

b. The shear transfer does take place within this 5%" length, and slightly ovcrstresses this section (a-a) in shear. In most cases the welded c o n n t d o n will provide sufficient length (a-a) for the proper transfer of thme forces from one portion of the member to another.

I

Problem 1

The k g size of these parallel welds would be based upon the force on the weld:

]

To detail an attachment to the tension member shown in Figure 19. If wc assume the total axial tensile force ( F = 125 kips) is divided among the two flanges and web of the beam by the ratio of their areas to the total area, then the force in the flange which must be transferred out is-

=

actual force allowable force

-

= ,194'' or use

Y4"

(A373 steel; E60 weld)

( b ) If the doubler plates are 7" wide and are welded directly to the inside of the flanges of the WF section, the flange force (F* = 47.5 kips) will transfer directly through the parallel welds. See Figure 21. If the leg size of these parallel fillet welds is o = %", the length of these welds would be-

= 47.5 kips ( a ) If the doubler plates are 6" wide, this flange force (F, = 47.5 kips) must first transfer into the beam web along the length ( L ) as shear, V = 47.5 kips. This length ( L ) must be-

L =

v t,"

-

(17.5 kips) 2(Y600) ('h)

. .

= 4.95" or use 5" Transverse Forces Any transmme component of a force applied to a memher is carried by those dements of the member which lie parallel to this force. In other words, a vertical force applied to an I beam with the web vertical is camed as

(See Figure 20.)

T

(47.5 kips) ( 2 8 8 ) (13,000) = 12.7" or 12%" -

h- 5,. -4

FIGURE 21

5.9-10

/

Welded-Connection Design

shear almost entirely by the web. If the web is horizontal, this force is carried as shear almost cntirely by the two flanges. See Figure 22. In a truss connection subject to a moment (for example, a Vierendeel Trnss), the applied moments, if unbalanced, cause shear forces ( V ) around the periphery of the connection web. The resulting diagonal compression from these shear forces can buckle the web if it is not thick enough. See Figure B. The Law of Force and Reaction states that in a member constrained by its supports, an applied force at any point sets up at this point an equal, collinear. opposite reaction. This of course assumes the memher to be a rigid body, that is one which does not change its shape or dimensions. In the following member which is supported, the applied force ( F ) has two components: horizontal (F,,) and vertical (F,). The result is two reactions in the member: vertical (R,) in the web stiffener, and horizontal ( R , , ) for the most part in the lower flange. See Figure 24. In order for one of thcse components of the applied force to bc transferred into another member, it is nwessary for the othcl- cwmponcnt to be transfmed also. Figure 25 illustrates this. If either one of the form components cannot be carried (F, in this example,

FIGURE 22

because therc is no stiffener), there will be little or no transfer of the other component (hcrt: F,,) even though there is a member or slemmt present to do this. In other words the amount of a force component (here F,,) which may be transferred into the member depends on the ability of the connection to transfer the

FIGURE 23

Diagonal compression on web o f connection due to shear forces from unbolonced moment

FIGURE 24

Design of Trusses

/

5.9-11

FIGURE 25

- Stiffeners FIGURE 26

other component (here F ) . Of cor~rsethe applied force (17) will bc reduced also_ and under thcse conditions some other portion of this member must transfer it. In this case the web of member A will transfer thi? halancc of the force ( F ) .

ud~ereK = the distancr from the outer face of the flange to thc web toe of the fillet. This value for all rolled scctions may bo found in any steel handbook. tt = thickness of the flange of the cor~necting member which supplies the compressive fo1-cc.

Determining Need for Stiffeners

No~mallystiffeners woold b e 21dded to a mcmber in which largc concentrated transvrrse forces are applied. IIowrver, for smaller mcmbers with lower forces, thesc stiffeners are sometimes left off in truss ronncctions. It is difficult to know under what conditions this might have to bc stifiened. In n:cent research at 1,rhigh liniversity or1 "Welded lntcrior Ream-to-Column Connections", short scctiolis were tested imder trarrsversc comprrssion as uell as tension, with 2nd without stiffoners. See Figure 37. It was foond that the compressive force applied over a narrow section ( t r ) of inemher's flange spread out over a wide section of the wc11 by the time the net web thickness was reached. A conservative valw for this distar~ceis given as: (te

+ 5K)

Although thcre usas no axial compression applied to the member in this test, on subsequent work involving actual beam-to-colr~mnconnections, axial compression was sin~ultanronslyapplied. See Figure 28. It was found that an axial compressive stress of ahout l.fi5 times the working stress (14,500 psi), or u 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 strcss was increased to twice the working stress or u = 29000 psi with no indicalion of trouble in the conncction. From this, they concluded that the minimum web thickness of thc c o h ~ ~ nfor n which stiffeners are not required is found from the following:

--

elded-Connection Design

$9

Bar represents connecting flange

I Toe of of web

(0)

FIGURE 27

Test to determine Compression region criterion

(b) Test to determine Tension region criterion

w

t*bb 2t, + 5K

This research, concrmed with the application of .. concentrntcd flange forces applied to flanges of W F members, was directed toward beam-to-column conncctions. However, it does seem reasonable to use this as a guide for the distribution of Range forccs in tnrss connertions. 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 thicknrss of the column flange ( t , ) determined whether stiffeners were required. On the basis of their tests, they made the following analysis. Analysis of Tension Region of Connection

The following is adaptrd from "Welded Interior Beamto-Column Connections", AlSC 1959.

FIGURE 28

Design of Trusses

/

5.9-13

FIGURE 29

The column flange can be considered as acting as two plates, both of type ABCD; see Figure 19. The beam flange is assumed to place a line load on each of these platcs. The effective lmgth of the plates ( p ) is assumed to be 12 t, and the plates are assumed to be fixed a t the ends of this leugth. The plate is also assumcd to be fixed adjacent to the column web. See Figure 29. where:

+ 2(K

m = wTC

- tc)

For the wide-flange 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 bc-

The force carricd by the central rigid portion of the column in line with the web isAnalysis of this plate by means of yield line theory leads to the uitima
5.9-14

/

Welded-Connection Design Application t o Truss Connections

- bb tb - .12 b,, tb t, - 5.6

If the column flange has this thickness, stiffeners are not required as far as the tension area is concerned. We might cany this thought one step further and apply it to a tension flange which connmts to the member at an angle other than 9O0, such as in a truss connection. See Figure 30. resistance of supporting flange (t,)

P = (.SO) us tb (.15 bb)

+

.'.

(.go)

U,

ITy

t,, (.I5 bb)

bb tb (sin

-+ (.SO)

a -

.12)

Here: t =

tb sin d,

-;--.

The vertical component of the web force of member directly into the web of member @ within the distance of d sin d,

(.180) 7 uY h2

7 IT^ tC2 = bb tb us sin

+

@ transfers

pull of tension flange (tb)

PI = b, t ,

This Lchigh 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 flange force of @occurs in the web of membc@within distance of 5K). See Figure 31. (t

Within the region b-c, these compressive stresses in the web of member @ overlap and would be added. a

. . . . . . . . . . . . .(4)

+ 5.)-

F , sin d, + (&)w

FIGURE 31

/

Design of Trusses

5.9-15

FIGURE 32

or n =

sin' (h 1

t,,

+

F ! sm (h 5K

+Fw]! d

'..'''

(5)

Another method would b e to assume ultimate load conditions, with all pzrts involved, stressed to yield. Using the previous formula ( 5 ) :

The vertical component of the web force of member @ t~ansfersdirectly into the web of member @ within the distance - d. sin 4 The compressive stress within this section would b~

I where:

(T:,=--

force - F, sin d, d area sin 4

6

Within the re ion (b-c), these compressive stresses in the member overlap and would be added: or w

2 sin2 d, rt,,

b,, tb 5K sin 4

+

+]

(r

(

6

)

If the thickness of the web ( w ) of member @ satisfies this fonnula, stiffeners are not r e q ~ ~ i r e dNor. mally, member @ will not be stressed up to its allowable in compn:ssion, so that this shorter method of checking stiffener requirements is on the conservative side.

=

..

Ff sin d,

F, sin2 d,

~~

(--,ti-+ 5K ) w + h, sm r6

+

force

Ul

= 'uea =

Ff sin

4)

(7)

+a ;.--

Now if ultirnatc load conditions are assumed, that is all parts involved are stressed to yield:

I

4. VERTICAL STIFFENERS

If Formula 8 should indicate that stiffeners are required, the same method of analysis may he extended to get an expression for the cross-sectional area of the vertical stiffeners. S c e Figme 32. It is assumed the transfer of the flange force of member @ occurs in the web of member @ within the distance ( t 5K) as well as in the flange stseners. The compressive stress within this section would be--

t,

-

bt tr,

UY

sin

where:

F, = h, t, o; F, = d WD u,

4

'-(&+sK

+

d w, ur sinP 4 d w

and the required cross.sectionaj area of a pairof stiffenbecomes:

,,,

h, ts ' ,_Wbb tli sin 6 w - wb sin2 4

-

sin d,

(

5

elded-Connection Design

FIGURE 33

5. LONGfTUDlNAL STIFFENERS

The type of connection shown here may b e 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 of was found that thcse plates each carried about the applied compression, while the central web section loaded up and carried the remaining %. 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 sopported a t three points, the two ends and the center, of the the hvo outer supports each will carry only load and at center 56 of the load. If the outer supports are pushed in for 3;of the beam length toward the center, all three reactions will be equal. By setting the stiffening plates about 5 bb 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 lies only to the distribuand has nothing to do tion in the web of member with these side plates, the Lehigh researchers for sim. plicity 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: force 0-1= area

x0

The compressive stress in the weh of member @ due to the vertical component of the web force of member @ is: 0-2

F , sin d, force .-. =-

area

.-d .

sin d,

w

T h e s stresses are added together.

xo

Now if ultimate load conditions are assumed, that is all parts involved are stressed to yield: where:

F, = bb tb as F, = d wb u,

@

and the required thickness of the two vertical plate stiffeners becomes:

/

Design of Trusses

5.9-17

@

FIGURE 35

These plates must have sufficient welds connecting them to the lower Bange because the compressive force of member A enters here. Since fillet welds cannot be placed on t e inside, this would incan 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 silnple fillet weld along the top edge of thc plate to the upper flange would bc sufficient. This discussion and resulting formulas will allow the connection to be d~atailedwithout 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 ~n~npression, a more efiicient 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 bc determined from thesc computed forces. These ideas will now be applied to various parts of a truss connection.

Q

ponent to entpr the lowcr flange of @ . This forcc: ( F ) , now in the stiffener, gradually transfers into the web of @ as shear, from section a-a to section b-h.

-

This unit shear force is equivalent to v = p7 -The weld bctween stiffeners and web of memb$@ would bc designed to transfer this shwr force ( V ) , F i y r e 35.

6. STIFFENING ACTUAL TRUSS CONNECTIONS The vertical cnmponent (F,) from the flange enters the stiffener and passes into the web of shear, V = F,, along section a-a. The horizontal comfrom the flange of @ enters the lower . The weld bet\veu stiffener and web would be designed to transfer this shear form ((V, Figure 34. The force ( F ) from the flange of @ transfers directly into the stiffener, leaving no horizontal com-

The force ( F ) from the flange of @ enters thc stiffcner, and is transferred through to the opposite end. The vertical component (F,) miters the flange of , and the horizontal component (F,,) enters the

elded-Connection Design upper flange of @ . No shear force is transferred throu h the weld between stiffener and web of member . Only enough weld is required near midsection of stiffener to keep it from buckling, Figure 36.

6

concentrated force into the web is to he 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 forees can be effectively distributed into the web of @ without overstressing it.

( 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 d the left flange and the vertical force in the right flange of be transferred around through the web of of two vertical stiffeners. See Figure 40. ( a ) Cheek the size of the connecting welds on the flanges of @ . FIGURE 37

The force ( F ) from the flange of @ enters the stiffener, and is transferred through to the opposite end. The vertical component (F,) is taken by the second stiffener as (F,), and the horizontal component (F,,) is taken by the upper flange of @ , Figure 37. In these last two eases, it is assumed that no portion of the force ( F ) in the stiffener is transferred into the ) . The welding of the stiffener would be web of @ similar to the previons case, that is Figure 37.

unit force on f m g e fillet welds

-

(138 kips) 2(10)

= 6.9 kips/linear inch leg size of flange fillet welds

= .72" or use''/3

(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

F f" = t; -

(74 kips) Z(17.5)

= 2.11 kips/linear inch FIGURE 38

leg size of web fillet welds OR

If there are no flange stiffeners on member A and no advantage of the precceding distribution of the

2.11 9.6 = .22"

=

-

Design cf Trusses

/

5.9-19

FIGURE 39

FIGURE 40

However, the minimum fillet weld to be attached w, = %u". (AISC to the 1.063"-thick flange would be Set 1.17.4) ( c ) Determine required sectional area of vertical stiffeners.

-. -

=:

(97 kips) (29.7 ksi)

(AISC Sec 1.5.1.5.2)

3.27 i a 2 , or use two %" x-5" stiffeners

Their A, = 3.75 in.2

> 3.27

OK -

( d ) Check the size of connecting welds to transfer this force (F,) as shear into the web of B .

unit force on stiffener-to-web fillet we& kips f = 97 -4(12.6) = 1.92 kips/linear inch leg size of fiUet welds

( e ) Check the vertical shear stress along a-a. T =

v -

A,

See Figure 41

(97 kips) (.660) (12.62) = 11,650 psi < 13,000 psi < .40 ur OK (AISC See 1.5.1.2)-

5.9-20

/

Welded-Connection

edges of the upper and lower flanges of @ . ( g There is one more item to check; consider point x in the figure below. It is necessary that the vertical component of the right flange of @ be transferred into the left flange of @ , and yet its horizontal componmt be transferred into the lower flange of @

b

FIGURE 41

( f ) Check the horizontal shear stress along b-b in the web of @ arallel to the welded connection . Thk length is about 20". betwen @ and

&)

The total horizontal mmponent from transferred into @ is 248 kips. The @ ha5 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 ?OM.

FIGURE 43

can only transmit Theoretically, the flange of and @ . There an axial force ( F ) bcttween point would be no problem if these 3 flanges met at a common point.

(151 kips)

=y

m2q

= 11,430 psi

<

OK 13,000 psi < .40 u, (AISC Sec 1.5.1.2)-

As a result no stiffening of the web of @! is required as far as shear is cunce~ned.If these shcar stresses exceed the allowable, the web of the connection could be reinforced with a doubler plate, eithcr on the web itself, or separated slightly and welded to the

FIGURE 44

In order for the flangc of @ to take the vertical component (F,) from the flange of @ at @ , it is necessary that the horizontal component (F,,) also

FIGURE 42

Design of Trusses

/

5.

FIGURE 45

be taken a t this point and somehow carricd up into the lower flange of @ . Likewise, in order for the f a n e of @ to take , it is neccsrrry the horizontal component (F.) at that the vertical component also be takcn at this point and carried into the flange of @. There are several methods by which this may be done.

(FY)

($

If the shcar transfer ( V ) hetwcen thcse two stiffenws exceeds the allowable of the web of @ , a doublcr plate may h c added to the web; or a plate 1~myhe set out on each side to box in this area.

.--. . --:.,. %

(5,)

of the flange of @ into the web of @ so that the horizontal cam onent (F,,) could be transferred . into the Range of

6

FIGURE 46

In this substructure for an offshore drilling rig, the truss connections carry iorge concentrated transverse forces. Vertical flange stiffeners are required to prevent web buckling. The triangular "gusset" is welded in to enclose the ores for greater protection against corrosion in addition to stiffening.

5.9-22

/

Welded-Connection Design

Ka"

FIGURE 47

Another solution of the same problem would be to check the stiffener requirements using the Lehigh research for beam-to-column connections as a guide for the distribution of the forces through the connection, ( a ) See if the web thickness ( w ) of @ is sufficient for stiffeners not to be required; Figure 47.

( b ) Check the tension flange of @ where it joins the flange of @ , as to the necessity of stiifcners to transfer the flange force; Figure 48.

t , = . 4 0 a x = .40 J (10.075) (.as) = 1.05" < 1.063" OK - .On this basis. stiffeners would not b e needed onposite this flange of @ where it joins the bottom flange of (@ ( c ) Check the tension flange of @ where it joins the flange of member @, Figme 49.

w 2 1.18" required

>

,660" actual

On this basis some stiffeners would be required.

t"

=

4

G

h, t, (sin

a

:

- .12)

Design of Trusses

/

5.9-23

FIGURE 50

FIGURE 51

On this basis, stiffeners would not h e required on of member @ . Either tical flange stiffeners or longihidinal flange stiffeners can be used to provide added stilhl-ness for the compressive force of @ .

longiturlinul flange stifcners

@ opposite this flangc

oertical flange stifiencrs w h,,t,, sin -~ w - w,, sin"

+

,663 -2

(.660)(10.345) (1.118)(.707) . 'r (.go)-(.6% j .(.707 )" 1.118

2 53'' .-

~~~~

- 7.03 in.' 2

so use two pairs of

3/4"

x 5" stiffeners.

~..-~~-

-

+ 5 x 1%~

or use a pair of 35" x 12%" x 36" stiffeners.

5.9-24

/

Welded-Connection Design

7. TYPICAL TRUSS PROBLE

FIGURE 52

Properties of Members Used in Problem 3

- 168'

shear

Check the details of this connection, using A373 steel and E60 welds.

resulting maximum normal stress

(See Figure 53.)

( a ) Consider the moment and vertical shear on section a-a.

M = F d = (16SV - 14L)(1(Y') = 1540 in-kips V = 154 kips

= 10,980 psi The resulting bending stress of u = 8,000 psi at the outer fiber is for a horizontal edge. If this edge slopes ($), the resulting fiber stress along this edge may be found from the following:

bending (See Figure 54.)

Design of Trusses

/

5.9-25

= 5730 psi n = 8000 psi

FIGURE 53

FIGURE 54

at top edge of gusset plate Q, = 12. cos 12" = 977 u =

8 000 = 8,390 psi (compression) ,977"

b

( c Consider the vertical weld between connection plate C and member @ . The forces applied on the left side of this weld are-

-2--

at bottom edge of gusset plate @ = 30"

u =

cos 30" = ,865

8 000

2;--= 10,700

,865'

]

FIGURE 55

psi (tension)

( b ) Consider the transfix of the vortical component (I.',) of the truss members @ and @ duough gusset plate @ and into the web of column @ within the connection length of 43" a s shear. From this vertical component (F,), deduct the portion to be carried by the right flange of A . (This does not have to enter the web of column .) This portion carried by the right flange can be determined by the ratio of the flange area to the total section area. The forcc taken by this flange is-

8

'\\

f, = 1.76k/in

\\ fr

M = (168" - 14")(7.03") = 1082-in.-kips V = 154 kips section morlwlz~sof ucld connection

bending forcc on weld

shcar force on zccld

= 55.5 kips This leaves 154 - 55.5 = 98.5 kips to pass into the web (some of which will enter into the left Range). The resulting shear stress within this 43" length of web is:

-

d

v

f, = = 2.51 kips/in.

=

(1.76)'

-+ (1.79)2

lcg size of fillet weld 5,490 psi

<

( 168 kips) (20.00 in.')

= (2.51) . (9.6)

15,000 psi < .40 u, OK (AISC See 1.5.1.zj--

This transfcr can ho made while still keeping the flangc compressive stress within the unifonn stress of-

=

rcwrltant forcc on weld

= 8400 psi

."1"

or use

X6''

( d ) Flange plates, %" by 4%", are welded onto to extend the flange of @ back a sufficient distance. The cornpressivc force in the flange of @ is-

@

elded-Connection Design

On this basis, the stress in each of these flange plates is: LT

kips) ..-. ( 2 " )(%") (4%') = 13,100 psi OK -

=

--(78

must be taken by @ alone. The cross-sectional area of @ is A = 1 5 . B in2. For the same stresy in , this would require the same cross-sectional area, or 15.88 in." and a net width of

0 -

The force from an adjacent pair of these plates is transferred into @ as double shear. There is sufficient width; see Figure 52. ( f ) At swtion c-c halfway along the flange lates, it is assumed that half of the flange force of has been transferred out into @ :

6

For the two Range plates, this reduction would Ieave-

(200")

0. @

(39.1")

- 122.0 kips

to be taken by

For the same stress, this would require an area

FIGURE 56

This shear stress in

-2

of-

is-

and a net width of-

=

(78.0 kips)

qiT7)v)

= 2600 psi

<

13,000 psi

OK -

size of connecting zcelds

&,

=

( 1 5 6 k/in. ) - .IW or use (9.6 k/in.)

x6,,h

However, tho .4WS as well as the AtSC would require a %" fillet .. wald bccause of the %" plate. ( e ) At section 1,-b at the termination of the flange plates, wc will assume the 200-kip compressive forcc

There is sufficient width; see Figure 52. ( g ) Another section which might be checked is along d-d. The ioads on this section are the direct comprcssiivr load of the colnmn @ , a shearin? force from the tension in the lowcr chord mcmber , and a bmding moment from the eccentricity of both the colun~rr A and the hottomchord h o d @ . This critical section ( - d ) is placed as high as possible above the lomcr chord (@ without intrwxpting the stiffening elements of the conneciion. In this case it is placed Y' ahove the ce~nterlineof mcmber @ . The propcrtics of this built-up cross-section are

9

FIGURE 57

Design of Trusses

computed and the eccentricities determined. For simplicity in this compntation, the reference axis (x-x) is placed along the conterline of the column A

G-

A

14" WF 68#

1

20.00

d

1

0

M=:Adl=:Md

1

0

1

0

I,

/

724.1

=(233'0.3.0) -- = ( 127.5)

a% psi

(compression)

+

=%I This is a total compressive stress of 3050 % 7470 psi, and a shear stress of 7 6 4 psi at the outer edge of the connection plate @ . The resultant maximum normal (compressive) stress at the edge of the plate is-

= 12:800 psi

From this: c = (14.06

5.9-27

bending

0.

Member

/

+ 17) - (7.03 + 5.36)

= 18.67"

Check the outer edge of this plate colnmn.

@

as a

radius of gyration Applbd Loads

r = ,289 t

t=

(.289)(%) = ,181"

The unbraccd length of this edge is L = IS", and

and the corresponding allowable compressive stress isu = 14,130 psi

>

12,285 psi OK ( A I S C ~ ~1.5.1.3.1) C

If the calculated compressive stress had exceeded this allo\vahle, a flange could have been added along this one outer adge to give it sufficient stiffness against lx~ckling. Plate @ will have i/lBt' b 4'' flange plates to extend the flanges of member along a distance of 12". W' fillct wclds will ire sufficient to attach these plates, tl~issize being n q o i l d because of the %" plate. No further checking is necessary because, by observation, the lB-kip force is much less than the 200-kip force of mrlnber @ and the same amount of plate @ is available.

6

FIGURE 58

compression

F, = 16Sk - 14k = 154 kips cr

-

F -T = 5050 psi (compression)

A

shear

F,,= 126&

Determine the leg sizs of the four fillet welds connecting the two %' gusset plates to the vertical leg of a tower. A373 steel, $360 \velds. See Figme 59. The horizontal component of the 350-kip force of the diagonal mcmhrr (10" W F 100#) is transferred back to the horizontal member (248 kips) through the

5.9-28

/

Welded-Connection Design

3.87"

I

-4 kWeld group

horizontal force

gusset plate. The only force transferred through this connecting weid to the vertical memhcr (14" WF 136#) connecting weld to the vertical member (14" TVF 136+) is the 248-kip vertical force acting 3l/2" away from thc crnter of gravity of the welded connection. Trcot the weld group as a line:

tuisting nction vcrticnl force:

vertical sllear

resultant

actud force (, -- allowable force

l:iowcvcr, .\\.I5 and becanse of 1:; ,".ilqge.

FIGURE 59

A373 steel

A, = 1.83 in2

X 6%" X 12"

SAa"

R

FIGURE 60

I

Problem 5

1

Determine the weld sizes on this connection. A373 steel, I160 welds. ( a ) Find the reqnired size of fillet weld hchvren member @ and connecting platcs @ The total length of connrcting weld is-

L = 1(W)

+ 2(6.08")

= 36.0"

force of weld f

F L

= (95 kips) = 2,641' upsjiri (36")

= (2.64 k/in.) (9.6 kjin.)

- 07-" J

or use

x6"

Check the length of web @ within the conncction along section x-x, requirrd to transfer the force of as shear. the web @ ont into the flanges of force i n wck

F,

'351,

(7.37)

= 23.6 kips

( b ) Find tha reqnired s k e of fillet weld between flanges of @ and platcs @. The total length of connecting wrld is-

L = 1(3'/2") jorcc on u&l

+ 2(12")

== 38.0"

elded-Connection Design

Design of Trusses

Trusses were essential to the all welded froming of the steel and gloss Phillis Wheotley Elementary School in New Orleans. The school was erected off the ground on two rows of concrete piers, plus exposed steel supporting columns under end trusses of the contilevered classroom wings. This provides both open and sheltered play area beneath the structure.

The roof supporting space frame that tops the Upjohn Co.'s Kolomozoo office building is of welded angle construction. A system of subassembly jigs focilitated the holding of alignment during fabricotion of the giant frome sections. Nearly all joints are welded downhand.

/

5.9-31

5.9-32

/

Welded-Connection Design

Main load-carrying element in the world's largest ore reclaimer, at Kaiser's Eagle Mountain mine in California, is a 170' long welded truss of triangular cross-section. Tubular construction is used where practical for extra strength and torsional resistance, and in order to keep weight to a minimum. Closeup below shows welded cluster where vertical and diagonal members meet the top chord.

1. INTRODUCTION

Tubular construction is bcginning to be used to a greater extent in this country, although for many years it has been an accepted method in Europe where it is used extensively. Although the advantages of thc tube have been known for a long time, it was the introduction of welding to the connections which made its extensive use possible. The tube represents an efficient section, having good properties in all directions. There is no problem in maintaining the inside of the tube against corrosion and in most cases this is loft unpainted. The welded connections seal the tube against any moisture entering and prcvents the circulation of air, hence any rusting very soon stops and equilibrium is reached. The joints represent the intersections of curved surfaces, and therefore extra care and time is involved in cntting the pipe to prepare the joints. Usually these are flame-cut, although there are abrasive cut-off saws which make a series of straight cuts and provide good fit-up and there arc shears with special tools which allow the end of the tube to be sheared. Fully automatic flame-cutting machines have been built which may b e preset for the inner diameter of the tube to be cut, the outside diameter of the tube which it intersects, and the angle of intersection. This will very quickly provide the proper cut, at the proper bevel, and results in close fit-up of the joint. Recently steel mills have introduccd square and rectangular tubing; these of course, are much easier to connect because of their flat surfaces.

Weld @ does not hove i o be made os carefully becoure fillet weld @ provides addition01 strength

FIGURE 1

( 2 ) Allows the inttmecting pipe members to be cut short and the gusset platc caries the cntire load back to the main member. In some cases, the web membcrs are shop fabricatcd and welded into assrmhlies. This facilitates field erection and wrlding hecause only vertical wclds between the main pipe member and gusset plate are still reqnired.

weld

Man pipe

2. GUSSET PLATES

Gusset plates have been used in pipe connections for at least 3 reasons: (1) Provides additional length of fillet weldii~gto the pipe Most pipe is not very thick. For example, 4" standard pipe Bas a V4" thick wall. Unless extra care is used in cutting, beveling, and fitting, it is easier to use fillet wclds rather than try to make 100%penetration groove welds on thin-wall pipe.

FIGURE 2

( 3 ) Providcs a dirtzt transf(:r of force through a main p i p member when othtr members connect on nppositr. sides of thc ni~mber.This may hc done if it is felt that the maill menrhcr has too low a thicktiess ( t ) to diameter ( d ) ratio and would need additional stiffness.

elded-Connection Design

FIGURE 3

Anothm solution to this problem would he to add a "slwve" or "collar" around the main rncinher u'itlrin this cr)nrwction mnc so that it ~ v o i ~ have l d the required thichrws. Tt a;onld ljc possible to insert hy welding, a short lcngtli of thicker tiibing within this zone. Ustially the inain pipe members must be butt welded together somewl~creto provide the required irngth, and this weld could be located at tliis position. See Figure 4.

If the wail thicknrss, bevel, and fit-up of thc pipr: arc sufficient for 100% pcnctxation groovr \velds to be made, there should hc no rtaason lor gusset platss. In most cases, with proper care, groovc welds could be made easily. Although gi~ssctplates arc used in pipe connections, they tend to stiffcn the pipe and, as a resiilt, concentratr the stress in thc pipo at the end of the plate. See Figure 5. Ii has been si~ggestedthat, if gusset plates are to be nscd, tlwy he t:rperrd at their ends so as to have less stincning effect on the pipe and thus provide a more even distribution of stTess within the pipe at this connection. Under static loads, any reasonable stress concentration in the pipe near thc termination of the gusset plate woi~ldprobably bo reduced by some localized plastic yielding; so, this \vould not be a prohlem. However, grisset plates should be avoidcd for connrctions subject to fatigue loading. 3. ORDER OF ASSEMBLY When web mtambers intersect at a connt7ction,normally the tensile member is first welded completely all the wny aroimd to the maill niemhcr. Then the compression member is cut back to overlap the tensile member, and

FIGURE 4

FIGURE 5

G i e o i e i stress concentration

More uniform stress

Connections for Tubular Condruction

1

/

5.10-3

Tensile member

FIGURE 6

this is wi.ld~xlto hoth of thesc mt~mbel-s.Evmy effort is i r d e to obtain tlic hest tcnsile connection; Figure 6. This is not quite as important as it first sounds sirm most of the vertical co~nponentin the tension member is tr:msfrrred directly into tho compression memho. through thp \velds of this overlapping portion ( b ) witlro~rtw e r passing throngh the wcld connecting t h tension ~ ~ irmnhcr to ihr main horizontal member ( a ) . Thc portion of the x ~ l d( a ) in the overlapped area connecting the tcnsion m(,mhcr to the main member is snhjcctd to two Sorces: tension from the tensile, rnernbw, and r:ompression fmrn the cornpression memher sincc it pushes agitilist this overlapped portion of thc tensile member. One forcc offsets the othrr, so that vt:ry little of any vcrtical force mrlst he casried by this portion of the weld at (a). jwt the horizontal force into thc top rnoml~er. Fignrvs 7 :tnd 8 descrihc a trst condncted at the Vniwrsity of Chlifon~ia,"~ic~scarchon Tubular Conrirctions in St]-uctnrd LVork" ]. C;. Uouwkamp, WRC ji.71, .hog. 1961. This test shows the effect that overlapping the intwsocting web members has on the strength of the joint. It is seen that a more negative rccentrieity of the connection ( c ) resnlts in more overlapping of the web ~ncmbers and greater stifl'ness of the main member. With this grcat ovcriapping of thc u e b members, the tr:tnsfr.r of the vertird component 01 the diagonal web mmnhrr into thc vertical iwh momber will occur before it miters thc main horizontal chord inember. The above

test shows this connection to have the highest strength, actually slightly higher than the tube itself, which in a separate test pulled a t an average of 260 kips. Eotice all three of the above tests failed in the tube wall adjacent to the connecting weld. 4. APPLICABLE BRITISH SPECIFICATIONS

The following is taken from Addition No. 1 (Nov 1953) to H.S. 449 (l.948), British Standards Institution: Sealed tubes or sealed box sections, for exposed structures shall not be thilmer than ,160"; for nonexposed structures this limit is .128", and not less thant

= . l o VE-

D = outside diameter of pipe t = thickrms of pipe

The angle betwwn intersecting pipe shall not be less tharr 30"; otherwise the strength of the connection shall be demonstrated. A cmnplete ptw:tration goove weld may be used regardless of the ratio of the diameters of the intersecting pipes. If the ratio of the diameter of the pipes is less than 'h, fillet welds may ho used. If this ratio is '/A or greater, a combination of fillct welds for a portion of the joint and groove welds for the remainder may br. nssd. Pipes eonrrected end to and shall be groove u-alded. In a fillct u ~ l or d a combination of fillet and groove

eided-Connection Design

FIG. 80 This pipe connection (Fig. 7a) hod o positive eccentricity of the diameter of the lorger pipe. Its ultimote lood was 137 kips.

v4

FIG. 8b This pipe connection (Fig. 7b) had no eccentricity. There's o slight overlopping of the connection. Its ultimote lood was 209 kips.

FIG. 8c This pipe connection (Fig. 7c) hod o negotive eccentricity of $f4 the diorneter of the lorger pipe. Because of larger ornount of overlapping, its ultimate load was 277 kips.

Connections for Tubular Construction

/

5.10-5

weld, the allowable stress on the t11ro:lt shall not exceed the allowable shear stress of the pipe. In a groove weld, the allownblc tensile, compressive, or shear strcss on the throat shall not exceed that of the pipe.

5. DESIGN OF TUBULAR TRUSS CONNECTIONS The application of tuitular construction to a truss arrangement is typified by the following problem. Here the loading is similar to that on the connection which was the snbject of Problem 3, in the preceding Section 5.9.

To design an eEcirnt connection on this tubular truss, Figure 9. ( a ) First c h c k the allowable loads on the various selectcd pipe sections against thc actual loading. Member

L -. r

@

-12"

(432) (4.38)

Std pipe t = l/s" A = 14.58 i n 2 r = 4.38" (rodius of gyrotion)

= 98.7 and the allowable is rr = 12,520 psi P = u A

= (12,520 ) (14.58) = 182 kips > 168 kips Member

OK

FIGURE 9

@

and the allowable is

IT

( b ) Use a W gusset plate on this connection, resulting in Figure 10.

= 16,660 psi

P Z U A

= (16,660) (14.58) = 243 kips > 200 kips

moment applied to pipe

OK

M,, = ( 1 B k ) ( 7 . 8 W ) = 990 in.-kip also M, = (154")(6%")

= (20,000) (7.165) = 145.3 kips > 126 kips

= 982 in.-kips OK --

assumed oalue of e e=12t

= 12 ( % ) = 4%"

5.10-6

/

Welded-Connection Design

FIGURE 10

maximum unit force jrudial) applied to 1" ring section of pipe @

This represents a worse condition than actually exists.

t? 6

s =-

(36)' --

6

= 1.98 kips

FIGURE 11

Althonglt there is just a single radial force ( f ) acting on tlw p i p shell, assume there is an equal force on the oppositr side of the shell, resisting this force.

= .0Z3 in." M,,, ( a t force f ) = k f r = (.318)(1.98(6) = 3.78 in.-kips M ( r = S ( 3.78) -(.023) = 164,000 psi Excessive. Heca~iseof t h < w excessive bending stresses within the pipe shell resdting from the moment applied by

Connections for Tubular Construction

thc connecting pl:rtc, somc mcans of stiffcning the pipe \tithin this arra must lie ~ ~ s c T11
I-

-

60" oroove weld on %" liner also j p n s pipe member. Weld lies d o n g neutral o x i i of plpe, so this becomes built-up section to resist bending

%"

3/l"-thi~kstiffening lhner around p,pe

FIGURE 12

5.10-7

Since 1.17" - %" (present thickness of @ ) = ,745" requirrd additkina1 thicla~css.or add :I %'-thick wrap-arorind shcet around this pipc @ in the arcit of tht. connwtion. See Figllrc 12. ( 2 ) :\notht~rpossiblr solntion \tould hr to add to the wall thickness at top and bottom of the eonnt:etion.

= ,210 in."

Do not nerd cirrurnferent~ol fillet welds oround either end of %" liner unless to ieol the ends

/

..

,

,, X

10" wrap-oround R FIGURE 13

elded-Connection Design

FIGURE 14

= 2.77 in.:'

s

=:

m

t'

where:

~

6

w = width of stiffening ring

:== l.29" rqnircd, and since i.29" - 3h'' = .915", sldd a 1" x 10" plat? &-rapped aromd tlle pipe

@

An alt(m~atrmt,thod woi~ld11e to rise %'' fillet weld all the way aroimd the m d of the pipe @ :

at t h top ~ and bottom of the coirncction.

( c ) I)etcrmii~cthe amonnt of required a m ~ c c t i n g wtrld between pipc @ and gnsset plate @ For dotermining the minimum length of connecti011 ( 1 .) to hold slrtw strcss ( r ) within the sillowablt~, use the following lr~avimnrnleg size of weld:

plate 4 I, 9600 w = 4 t,

1. r

FIGURE 15

/

Connections $or Tubular Construction

. :

,i.5.6", or 17.8'' o ~ rcadi sidc of the $6'

giisset

5.10-9

ilo tliis witliout any difhdty. If fillct \wlds are to be irsed instead of groove iwlds, t b i secmid cut or i~cvclis only needcd at rctwtrant corrlcrs of th(, joint or whew the nnglc 11rtwec11 the siirf:iccs of tha iirtcrstbctirrg pipes is less than 90".

plate.

If thc transvrrsc \veld is 12" long, this leavcs 27.8 or 8" on cach sidn.

- 12 = 15.8".

6. TEMPLATES FOR PlPE CONNECTIONS Althougli pipc f;ibricatitrg shops have shop nwn who an. cxpt~rimctdin laying out m d prrprirrg t h i w joints hy making thvir O\VII tcmplntw this is somt.thing new for most strl~ot:ir;ilshops. It may hc, m w x a r y to supply templates for th? morc critical pipe joillts w11cre a jiussct plat? is riot spccifietl. Thwt: arc t;~I,los of ordirratrs amilahlc for most standard pipe sizt,s ;ind given angk,s of intersection (ED, 30' , 45", 6O", and $10") . tIou;cvt.r, t h ( w may hc of little, v d u e btcause otlicr rormd tubi~lars(:ctions ma); be ustd which are ilot standard pipe sizes, : i r d in \<%I1not iitw'sstructur;d work thc arlgl? of itltcrs~~ctiorl sarily be one of the ahovc. For good fit-rip. it is nt,cessary that the inirnr radius ( 1 , ) of the snrallvr pi ,tz @ and the outer radius ( r z ) of the larger pipe irrtersc.ct along a curve which forms the root of the joint. Followitig is a suggcstrd metlmd for making templates which will cover a11 possible connections at any angfr of intwscction, any :irnount of offset, and any possiblt combin;rtion of pipe sizes. This template will allow the c t d of the srndlcr pipe to he cnt for proper fit-lip against thr surface of the larger pipe. 111 struct u r d work, it is not ncccssary lo cut a hole into the side of thc 1;irgcr pipe at the conn?ctionl as is done in p r c s s ~ r epiping so a srcor~dtr~nplatcis not needed for this cut. The inner radius ( r , ) of thc~sinallcr p i x .\ and the outer radius ( r , ) of thc largcu pipe H .me med to makr the template. This is done gr;ipliicnlly or ;~rr;~lyti~~;rlly, as explained a in\:lx~ragrapfrsfurther. Tlrc tcniplatc is mad? of soiue type of iiravy p;ipcr. It is nirapp(d arorlnd i h t pipe to he cut, at the propcr location. The c?ntc,r of tlris tcnrl)l:rtc rdg? is transfcrrcd onto the pipe with chdk. Thr: rhalkcd curvi. on the pipeis tlien marked with a st:rics of c:~ntcrpmch marks. Tht: pipe is thrn flainr-cut along this cl~rve,krrping the torch tip trorni;rl or at right mglcs to the surface of the pipe. This \vill prtid~ice the p r o p ~ rcurve for the joi~rtas far as tho inside of tlrc pip(. is <,or~cerncd. It is then necessary to brvt,l the edgo of this pipe back from the outsidr, jnst torrclring tliis inside cut to pro\.ide the raquirrd inclrldd angle, for the groove weld. A good expcrie~rcrdflamc-cutting operator will

60

TABLE 1-Properties

of Polar Angles

1 2 POSITIONS mrifion

1

a

I

rina

TABLE 2-Properties

I

sin2 a

I

or

(8)

I-cor a

of Polar Angles

5.10-10

/

0

Wetded-Connection Design

@

0

@

0

@

0

FIGURE

@

c

4

3

TEMPLA76

~ DEVELOPMENT

~

~

@

@

OF PIPE A

FIGURE 16

Graphical Method of Making Template Refercnct>sat-? t o vir\\.s ( a ) , ( h ) , arrd ( c ) of Figrrrr 16. 1. Ilraw a sidc vie\v of the cormcction. figure ( a ) . Drnw an end v i t v of the i,onnection. fig~rrc(11). 2. l a y off pipe @ into a g i \ w ti~~mhc,r of q r a l swtions, for (~x;inrplc16, ;ind nitmlwr these 1 ; 3, ctc. tlrrr~rlghto 1G. llraw iitlrm throiigh thcsc points p;rrallcl to thc axis of pipc @ in Imth f i g ~ i r ~ s . 3. \Vhere tllrse prirrilld liiics of pipc @ intcmcct pip? @ , in figrlrc ( b ) , rnnkc points ( D ) .

"

5. II'hcr~, th~sc*p;lrnll<,l liu<,s of pipe @ intersect c i ~ r r i q m ~ d i l rp;~r,illi.! g lini3sof pis" @ , in figiirc ( i i ) , m i l - k points ( I : ) . Krirnlwr thrsr points in accordalrce \rith the origirinl divisiiw of t111. pipe @ . 6. 111 ~ ; ~ J I I T ( c ) , lay ofr l i 1 1 1 ~x-x, m1t1;d to t11c ortt<,~ circrimfrrrtiw of pip<, @ , and d i ~ i d ointo 113 q u a 1 scgm(x~~ts.

@

Connections tor Tubular Construction

/

5.10-1 1

FIGURE 17

7 . In fignrc [ a ) , dr:iw refermcts line %-Z at right aoglos to tht: itxis of pipc @ and thnil~glitlw vertex of the coni~c~ction angle. From this line Z-Z mtSasure thc 01-dinatr c1istam.c (11) to thc various intrrsccting points ( E ) . I.ay t h t w distattcr:~( h ) off v(zrtically {I-om linc Z-Z in figrise ( c ) . Do this for all the points and draw a curw through thc upper cxtrcmitii~sof ilicss vt,rtical h e s . This Becornrs the tcmplatc for cutting pipe @ , figure ( c ) .

or:

h =

r., [A] i- -- .r ~ ' stn 4 tan 4 [R]

[R] = 1

. .. . . .. .. . . (3)

- cos rn

r, = inner radios of s m d r r intrrsccfing pipe

Analytical Method Thc follon-ing f ~ x t n i ~ lwill n give thr value of the ordi~ ~ a(11 t rj tor ;my polar position ( m ) d o n g tlw s~riallrr pip? This mcthrid of finding d t fly formda c~Iirnin;rti~s thr mapping of f i g ~ ~ (r a~)s and ( b ) in the 'grapliic:il 1nclI1~1 of 1:igum lfi.

6 =- irnglc of interswtion h<>twcc~i ases of pipes 11

z

ordiiinte of thc trmplate for the smallt:r pipe for any 1)osition ( m )

Tahlrss 1 and 2 will give the rlcwssary valrrcs for sin U - . sill' o , a 1 ~ 1 - cos m for the viirio~~s pt~lar mgles ( a ) for r i t h 12 positions or 16 positions of the pipr. I f Formula 3 is h) be ~ c d tile , followiirg nornograph, Figr~reIS, will give vollies of [ A ] . Valurs of / 13 1 may I J ~fo1111din Tables 1 and 2. ~

I'ri~cticdly all s t r ~ ~ c t p~i ~ p ,r cdo ~ ~ ~ ~ e c tu-ill i o n 11avc s no offs~t,:I = 0; :ind this ljrcomcs-


I

Problem 2

~~

I

I 8 r2ll = -

r7 .-.-

.

~

~~

~

2'

m

Foa thts tuhnlar coiinwtion w p n w n t e d in Figure 16,

~-

-;-I

tan

4( 1

- cos a )

iiitersrets the sn~;rllt.rpipc A , inside ri?clius r, =-: Y, 3", at an thc 1;trgw pipe R . rmlsirk r;idi~is r2 a~rgtcof is'',and ~ . i t l:in l oifset of a = 2".

5.10-12

/

Welded-Connection Design

Connections for Tubular Construction

Following are the ordinates ( h ) for the various positions figured both graphically (see Figure 16) and analytically (with Formula 1 ) . This tahle shows close agreement between the two sets of values. porltion

1

graphical

/

onolyticd

I

/

5.10-13

1

Problem 3

in the connection represented in Figure 19, the axes of these three intersecting pipes lie on a common plane; there is no offset ( a = 0). A template is re uired to cut pipe @ which intersects both pipes and @ . The inner radius of pipe @ is 2", the outer radius of pipe @ is 3", and the outer radius of pipe @ is 21%".The graphical work is shown in Figure 19. Notice that the finished template is made of two portions, that due to the intersection with pipe @ , and that due to intersecting p i p @ .

&)

I

1

Problem 4

In this example, the nomogaph (Fig. 18) will be used to find the ordinates ( h ) for a template to he used in cutting the smaller pipe of a two-pipe ~xmnection. The smaller pipe A has an inside radius of rn = 2", the larger pipe B has an outside radius of r2 = 3", and the angle of their intcrscction is $ = 60'.

8

A sheet of paper is laid out. A straight line X-X is drawn across the paper, parallel to the long edge and %" or 3" from this edge. Starting from the left edge of the paper, measure off a distance on this line equal to the outer circumference of the smaller pipe A and mark this on the line. This can be done in two ways; the circumference of the pipe may be figured by knowing the outside diameter of the pipe, or this paper may be wrapped around the outside of the pipe and marked where this edge of the paper overlaps. The easiest way to divide this line (which represents the circumference) into equal segments is to fold the left edge of the paper back toward the right until it lies directly on top of this mark, then fold this flat upon itself. This divides the circumference into two equal parts. Now fold this edge hack toward the left until it lies directly over this fold, and fold down. Do the same for the similar portion on the bottom. This now divides the circumference into four equal parts. Open the paper and divide each of these quarter sections into three equal parts and number each of these vertical lines from 1 to 12. If 16 positions are to be used, divide each of these quarter sections into four equal parts and number from 1 to 16. Lay off the comesponding ordinates ( h ) on these lines. Draw a curve through these points and cut along this curve; the lower portion of the paper is the template.

sin 60' = .8660 tan 60' = 1.7321 Formula ( 3 )

The results are shown below in table form. As o matter of interest, the values computed by Formula (2) are listed on the extreme right and indicate the reasonable accuracy of the nomograph.

5.10-14

/

Welded-Connection Design

FIGURE 19

FIGURE c

TEMPI ATE : D€YElOPM€NI OF PIPE A

ConnecPions for

7. BOX SECTlONS The squara and rwtangiilar hox sections, in which tubing has more recently hecome available at competitive prices, eliminate the, prohlcm of fit-up that is associated with the i r ~ i ~ nst:ctions. d With box sections, the tnd of the sm:~Ilertuhc can be simply sawed with a single cut at the reqnired angle. Field erection of box sections is easily siiiqMied by the use of Saxe clips, Figure 20. The clip and its seat are shop u-elded to the two intersecting members. Usually t l ~ cclip is welded to tlie inside of the box Ixam whm: it is loss \wlnorahle l o damage during shipment to the projrct site. The clip also furrctions as 2% seat to help in support of the beam. Tliis allows the joint to be made without any attachments on the ontside, and produces a pleasing appearance.

Tubular

ansfruction

Soxe seat ihop

Sone clip shop

welded to face of columri

welded to inside of box beam

I

5.1

of tubular box beom, oliowing use of simple fillet weld "round \ , outside, Ideal for exposed steel

,

L FIGURE 20

Square a n d rectangulor structurol tubing, now ovoiloble in many standard sizes, tends to simplify desigv and focilitote erection. Both shop a n d field connections a r e g e n e r a l l y more easily mode thon when using round tubing.

/

5.10-16

/

Welded-Connection

Space frame roof on the combined worehovse ond mochine shop in Bethlehem Steel Co.'s reseorch complex offers on interesting silhouette (ot top). Roof frome is formed by eleven 96'-span welded pipe trusses braced aport b y inclined pipe struts ond orched structural members. The result is a very rigid structure, olthough temporary stiffening with steel chonnels wor required during erection.

Conneceionr for Tubular Construction

Typical connections to facilitate erection of structure using square tubing for columns. Columns hove equally high strength in both x and y directions, plus excellent torsional resistance. Connections combine welding ond erection bolting.

/

5.1

elded-Connection Design

Unique roof suspension system combiner with "tubular" design of members and weld fabrication to provide vast unobstructed area and light oiry atmosphere to the Tulsa (Oklahoma) Exposition Center. In photo above, slag is being chipped from root pass on splice of built up box-section roof girder, preporatory to making main fill passes.

1. GENERAL REQUIREMENTS

The knee is an irnport;~ntpart of a rigid frame and some thought should be given to its design. The knee of any rigid frame must be capabl~aof1. Transfining the end nmnent from the beam into the colomn. 2. Transftming thc vertical shear at the end of the beam into the colurnn. 3. Transferring the horizontal shcar of the column into the beam.

(a] Squoie corner

A kner differs from the usual straight beam in these rqxx!ts: 1. The lic~~tral axis shifts toward the inrrt.r flange, causing nn incrcnse in the i l s d l x d i n g Forces at this point. 2. Axial Range forws must change dircction, causing radial forces to he set 11p. 2. EVALUATlON OF KNEE TYPES IQuro 1 illustratt~sthc fin. principlil types of knees for rigid frames.

[b) Square c o i n e r w ~ t hbracket

(c) Topered haunch

je) Curved h o m c h

( d ) Tapered hclutich FIGURE 1

elded-Connection Design

FIGURE 2

,0002

,0004

,0006

.0008

0010

,0012

,001 4

.a016

.[

Unit ongulor rotation (+). rodiondin.

I t might he thought that the simple square type of knee connection would naturally he as rigid as the connc.:cting members, since it is a continuation of the same section. In many cases, this is true. However, stress causes strain, and the accumulation of strain over a distance results in a movement of some kind: deuection, angular movement, etc. This means that the sharp comer of this joint increases the stress in this region by several times. This stress concentration resdts in a higher strain and, therefore, grcater movement in this local region. With the square type of knee in which just Uange stiffeners are addcd, it is difficult to cxcccd the stiffness of the member. In most cases it will just equal the

member, and in some cases it will he less. Figure 2 shours moment-rotation curves of various knee connections.* The vcrtical axis is the applied moment; the horizontal axis is the rcstsolting rotation of the connection. The vertical height of the curve represents the maximwn or ultimate strength of the connection. The slope of the straight portion of the curve represents the stiffness of the connection, with the more nearly vertical con7es being the stiffer. The right-hand extremity of the curve represents the rota*Figure 2 adapted from "Connections for Wolded Continuous Portal Frames", Sccdle, Tripractsoglon, and Johnston; AWS Journal; Part I July 1951, Part 11 August 1951, and Part I11 November 1952.

/

igid-Frame Knees (Elastic)

5.1 1-3

llililillliilliiiillllilllilii

Frame under load

v

v

Point of inflection; zero moment

Moment diagE

FIGURE 3

,

Portion of knee in testlng machine, subject to compressive force (F) to duplicate actual load conditions in frame

points of reflection; no moment opplied

v

v

tional capacity of the connection. Notice that the square-comer knee is the most flexible. It falls slightly short of the beam itself, but it does have the greatest rotational capacity. Tapered haunch knees (not shown here) and those with the additional bracket have greater stiffness and higher mornent capacity, but less rotational capacity. The curved knees are the most rigid, have the highest moment capacity, and have a rotational capacity somewhere between the simple square corner and the haunched knee. As the radius of curvature of this inner flange is increased, the stiffness and moment capacity

F

increase slightly, with slightly lower rotational capacity. Another purpose of the hannched and curved knees is to rnovr the connection to the beam back into a region of lower mon~entso that the beam will not be overstressed in bending. The dimensions of the test knm are so chosen that they ~ x t e n dout to the point of inflection (zero moment) of an a c t ~ ~ framc; al Figurc 3. In this manner, the ttxting machine applies a compressive force ( F ) which becomes the component of the two forces V (vrrtical) and H (horizontal) which would actually he applied to the knee at the frame's point of infl~ction.

/

5.1 1-4

Welded-Connection Design

3. SHEAR IN CONNECTION An axial force (tensile or compressive) can transfer sideways out of one elemcnt of a ~nemberas shear. For example, the tensile force from the beam flange will transfer down through the connection web as shear into the supporting column; Figure 4.

I

1-

. . . . , . , , . . . . . . , . . . . (1)

If this shear stress exceeds the allowable for the web, it must be rcdnced by increasing the web thickness within the connection area. Or, a pair of diagonal stiffeners must be added to transfer some of this flange form as a diagonal component. One method of detailing this connection is to calculatc the imrtion of the flangr force which may be transferred as shear within the web by stressing it to the allowable. Then, diagonal stiffeners are detailed to transfer whatever flange force remains. Anotlicr method is to assume that the shortening of the diagonal stiffener under the compression component is equal to the diagonal shortening of the web due to tire shear strrss. From this, the resulting shear stress ( r w )in the web and thc compressive stress ( U S ) in the diagonal stiffener may be found for any given set of conditions.

e F b

A

I

shear into the connection web within the distance equal to the depth of the connecting member, the resulting shear stress within this counection web is-

...

FIGURE 4

Derivation of Stress Values

where the flange force in the beam is-

The final diagonal dimension ( d l ) of the web, due to shear action on the web, will bed 2 = d" dC2- 2 db d, cos (90' - y )

+

and the flange force in the column is-

but cos (90"

- y) = -

Assuming this flange force ( F ) is transferred as

.

cos 90" cos ( y ) sin 90" sin ( y )

+

sin y

.

--Initial conditions of stiffener and web

y

= T/G= E

Final conditions of web

FIGURE 5

Finol conditions of stiffener

/

Rigid-Frame Knees (Elastic)

For small strains

(t,)

and the compressive stress in the diagonal 3tiEener is-

and angles ( y ) -

sin ( 7 ) = tan ( y )

u. = 2.5

- 6

Hence: d12 = c$h2

I i

4

d2

dc L 2
7

and

E, 7

d,%- 2 dh d" E, &--sin 0 db == dC tan 0 cos 0

d, =

5.1 1-5

hut

T

. . . . . . . . . . . . . . . . . .( 2 )

sin 0 cos 0

Now we go back to the flange force ( F ) since it causes this load on the connection region. The flange force of the beam is equal to the shear force carried by the web plus the horizontal component of the compressive force carried by the diagonal stiffener.

--

dl =

d"-

-

d, cos 0

e

sin2 0 cos? 0 1

4. &'-

-2

T

ER

sin 0 E, cos 0

T --

"2 C

sin 0 cos 0

Subytih~ting( 2 ) into ( 3 ) givesF =

Thc final dimension of the diagonal stiffener (dz), due to compression, will be-

=

7

t, d,

+ (2.5 T sin 0 cos 0 )

T [tW dc

+ 2.5 A,

A, cos 0

sin 0 cos2 01

or, the shear stress in the connection web is-

I

Since the movemeot-

A : .=

E

d,

7

= t, d,.

F

+ 2.5 A,

sin 0 cosY 0

Also, from (2)T

u x

= 2.5 sill 0 cos 0

Substituting this into ( 3 ) -

-

2.5 sin 0 cos 0

Since diagonal stiffener and web are attached, the final diroension of diagonals in each case must be equal:

=

itw6

( 2.5 sintw 0d,nu 0 .

+ rsAs cos 0

-t

4, cos 0

or; the compressivr. stress in the diagonal stiffener is0,=

-

. . . , , . . . .(5) t,- d" -.i A, cos 0 2.5 sin 0 cos 0

Squaring both sides: T

1 - 2 =-

Some knees are more complex than those described here and analysis most consider factors that are covered more adequately in Section 2.12, Buckling of Plates.

sin 0 cos 0 =

or

I 7

--

E,

sin 0 cos 0 =

Since for steel: E = 30,000,000 psi E, 12,000,000 psi .'. E = 2.5 E.

-

To check stiffener requiremcnts on the square knee connection shown in Fignrc 6, for the loads indicated. A36 steel and E i 0 welds are used.

elded-Connection Design of stiffeners required ~ e c t i o marea l

- -(26.4)

( 22.0)

= 1.2 in.' (pair) Also required:

b,/t, = 17 Hence, nse a pair of ?*it'x 3" diagonal stiffcners. -

--

.

Checking this size against the requirements: -L

14" UZ 84+ tw

I

column

= 451"

A, = 2 x %" x 3"

= 3.0 in.' > 1.2 in.'

OK -

FIGURE 6

Here: ethod 2 Plastic Design (See Sect. 5.12)

required thicknrss of connection web 14.18 = ,561 cos 0 = :25.33 20.99 tan 0 = = 1.480 14.18 flange force on the beam

-

This exceeds the actual web thickness of t, ,451". so stiffening is required. 117.6 kips

ethod 1

h o r i z ~ n t dcomponent carried by u e b in shear

F, = T t, d, = (14,500) (.45l)(14.18) = 92.8 kips

= 5.64 in.' (pair)

Use a pair of -~ ----

This leaves (117.6 - 92.8 =) 1.4.8 kips to he carried hy the horizontal component of the comprcssivc force on the diagonal stiffener. compresrice force on stiffcrm

= 26.4 kips

~~~

~.

V4" ~

x 4" diagonal stiffeners.

.. .

Checking this size against the requirements: As - .-

1" = 6.0 in.? > 5.64 i n . 0 K 9

3/21

=

Rigid-Frame Knees (Elastic) ethod 3 Start with a pair of 'A'' x 3" diagonal stiffeners and, assuming both diagonals contract the same amount urrdrr load, check stresses in web and stiffener.

shear stress in web 7

=

F

t, d,

+ 2.5 A,

--

sin 0 cos2 0

F f, = 2 lbs/linear in. of web . . . . . . . . . . . . (6) ri

F

+ A,

t, d, 2.5 sin 0 cos 0

cos 0

= 14,200 psi As a matter of interest, increasing the size of the diagonal stiffener to 3/" x 4" would decrease these stresses toT 06

5.1 1-7

in direction of the compressive Range force is accomplished by means of a diagonal stifleuer; Figure i ( b ) In the curved haunch, this change in direction of the axial force is uniform along the curved edge of the flange and resnlts from radial con~pressiveforces in thc web; Fignre 7 ( a ) . The force in the inner flangc of the knee is greater than the force in the outer fiangc because: it has a smaller radius of curvature. Iisually this inner flange is the compression flange; therefore, this is the region to be checked for stiffening requiremonts using the following formula for radial compressive forces in the web.

compressiue stress in diagonal u. =

/

= 11,400 psi = 13,250 psi

4. COMPRESSIVE FORCES IN CONNECT1 WEB

In this case, the unit radial force (f,) is a function of the compressive force (F,) in the flange and the radius of curvature ( r , ) of the flange. This action is similar to the radial pressure applied to the rim of a pulley by the tensile forces in the belt. As the radius of curvature decreases, these forces increase. As this change in direction of the flange becomes more abrupt, as in a square or tapered haunch, these radial forces are concentrated into a single force. And, they must be resisted by a diagonal stiffener; Figure 5@). The axial force in the flange is assumed to be uniformly distributed across the width, therefore the radial pressure or stress is-

An axial force is able to change its direction if suitable resisting components of force are available. In the square or tapered haunch, this abrupt change

Diagonal resisting

A F, Rodiol cornpresslve

F< = a A,

FIGURE 7

/

5.1 1-8

Welded-Connection Design

When applied to the flange, this radial stress will load any cross-section as a cantilever beam, since it i s supported only along its centerline by the web; Figure

Also:

M=utS - ut t12

-6

8.

h

&

ut t f 2 6

inner flange

0- ti

-

b12 8 r, 3 u bf2

Where:

S=--

or

1" tf2

and

F T X b - b + 4

From this relationship, it is seen that in order to hold the transverse tensile stress (u,)to a value not exceeding the axial compressive stress of the flange (u), the following must be held:

FIG. 8 Cross-section of lower flange and web.

. . . . . . . . . . . . . . . . . . . . . . (81

The bending moment along the centerline of 1the beam flallge due tn th;.*" ~ ~ a i . 1lnla Y ~ ~ Ihe. I v

-

...

."--

a-.*.-.

YX bi2 - --

-

\ - I

If this value is exceeded, stiffeners would be used between the inner compressive curved flange and web.

8

G--tf - .

hr2 ri 8

bl = width of flange tf = thickness

of flange

ri = radios of curvature of inner flange

ut = transverse tensile stress in flange a = axial compressive stress in flange

Radiai compressive force exerted by web ,

FIGURE 9

II

/

J

\ Tionweire tensile stress due to bending of .,..lf

Rigid-Frame Knees (Elastic)

/

5.1 1-9

FIGURE 10

This analysis assumes a uniform distribution of stress across the cross-section of the flange. If this is based on plastic design, the plastic section modulus ( 2 ) is used instead of section modulus (S), where

The transverse tensile bending stress (u,) in the curved flange is found in the following formula; the value of p comes from the graph, Figure 10.

Then ( 7 ) becomes the following:

Bleich has carried this analysis a little further; see Figure 9.* Because of the slight yielding of the flange's outer edge, there is a non-uniform distribution of flange stress ( u ) . This compressive stress is maximum in line with the web. In the following formula, the value of a comes from the graph, Figure 10.

If this value is too high, stiffeners should be welded between this flange and the web. These keep the flange from bending. These stiffeners usually need not extend all the way between flanges, but may be a serics of short triangular plates connecting with the curved flange. The unit radial compressive force (f,) which acts transverse to the connecting fillet welds between the curved flange and the web is found irom-

* Fmm "Design of

Rigid Frame Knees" F. Bleich, AISC

2

Inner face of Ronge

curvature

'"""'

on inner flange

Sirerr

(b/

1 FIGURE 11

elded-Connection Design

I

F f, = - Ibs/linear in. ( 2 welds)

6. LOCATING SECTION OF HAU CHECK Most theories concerning the strength of knees differ only in the placing of the neutral axis, and in locating the resul~ing section for determining the section modulus.

F RADlUS OF CURVATURE ON STRESS IN INNER CORNER A straight beam has an infinite radius of curvature ( r = m ). As the b e d becomes curved, this radius decreases, and thc 11eutral.axis-na longer coincides with the center of gravity, but shifts toward the inner face. See Figure 11 ( a ) . Because of the shift of the neutral axis, the bending stress in the inner flange increases greatly while the bending stress in the outer flange decreases. This increase at the inner flange becomes more severe as the radius of curvature decreases. In a squarsknee connection, this radius of curvature is provided by only the reinforcement of the bevel groove weld or fillet weld on this inside comer; Figure 11 ( b ) . For this reason, the square knee may not quite develop the full plaqtic moment of the connecting member unless it is somehow reinforced. If for some reason a reversal in moment should be applied to the knee and the inner face of the lnee is subjected to tension instead of the usual compression, it is important that this be a good sound weld. This is especially true at the surface of the weld. If the knee is loaded up to its plastic moment, the metal within the section below the weld is stressed up to its yield strength. During this time, the weld undergoes a considerable amount of plastic yielding and some strain hardening. The weld metal does have the ability to elongate about 28% as measured in 2" before failure. However, this zone in which the yielding is confined is very narrow, being the width of the weld. Conscquently, the overall movement of the connection due to plastic yielding of the weld is very low, although s&~cient. In this case almost all of the weld's ability to elongate may be used in developing the plastic moment of the connection. Any defect in the weld which would lower its ductility would probably prevent the connection from reaching its plastic moment. The knee could have greater strength and rotational capacity if this inner face were changed to a haunched or curved knee section. In testing these square knees in tension, plastic moment was reached when this weld was of good quality. Fortunately most knees are stressed in compression at this inner comer, without any tendency for this weld to fail.

FIGURE 12

One method, Figure 12, uses straight sections normal to the axis of either the beam or column. The section modulus is dctcrmined about an axis through the center of gravity of the section. The resulting stress in the inner flange is increased by the factor

where 4 is the slope of the flange. Although this method is easy, it might indicate excessively high stresses when the flange has a rather steep slope.

FIGURE 13

Another method, Figure 13, is to extend the centerlines of the beam and column to intersed in the knee. Straight sections are used, and the section modulus is determined about an axis lying on this centerline. This will give conservative values for the stress in the sloping flange. Because of this, no factor is used for the stress on the sloping flange. A more accurate but longer method, Figure 14, is based on a curved section forming a wedge beam by

Rigid-Frame Knees (Elastic)

5.1 1-1 1

Here:

v = r sin ( 2

a;)

. . . . . . . . . . . . . . . . . . . . .(13)

Bending Stress in Curved Flange (See Figure 16.) Here:

b = a cos d, FIGURE 14

fa = W. R. Osgood* and aodified by H. C. Olander.**

fa

-------

cos d,

-

* "Theory of Flexure for Beams with Nonparallel Extreme Fibers''

by W. R. Osgood, ASME Vol. 61, 1939. **"Stresses in the Comers of Rigid Frames" by H. 0. Olander, ASCE Transactions Paper 2698, 1953. Method of Using a Straight Cross-Section

Dimension of Straight Section The dimensions of a straight sec%ion (A-B) of the haunch may be found from the following:

fb a,,= b x I"

- - fa

cos d,

x

1 a cos d,

--

FIGURE 15

FIGURE 16

5.1 1-12

/

Welded-Connection Design P'

H

(b) Tapered knee

(a) Curved knee

FIGURE 17

. . . . . . . . . . . . . . . . . . . . . . . . . .(15)

Here:

\,

d- + P=sinT2 a )

Wedge Method of Determining Section

The wedge method may be used on any beam section whose flanges are not parallel. A curved section (A-R) is constructed where the stresses are to be checked. This is normal to both gauges and has a radius ( p ) the center of which lies on the straight flange. See Figure 17. The transverse force ( P i ) , axial force (P,'), and moment (M') acting at the apex ( C ) of the wedge are found. See Figure 18.

=

d tan ( 2

-1

(d,l

FIGURE 18

a

)

r[l-cos ( 2 a ) ] sm ( 2 a )

- r[l-cos

( 2 cc ) ] sin ( 2 a )

. . . ..(16) . . . . .(17)

. . . . . . . . . . . . . . . . . . . . . . . (18) . . . . . . . . . . . . . . . . . . . . . . . . .(19)

Rigid-Frame Knees (Elastic)

transuerse force applied to wedge at point C

(P{ = pi cos

a

- Pa sin

cc

1 .. . ... . .

.( 2 0 )

axial force applied to wedge a t point C

. (21) moment about point C

I",-

M' = 3- P t m

+ Pa-

. .. . . . .

. (22)

These applied forces result in various stresses on

(a) Resisting horizontal bending stress

5.1 1-13

the curved haunch section, as described in following paragraphs. Mameot

(M') Applied t o Wedge M e m b e r

The horizontal bending stresses (u,,) resulting from the applied rnommt ( M ' ) , Figure 19(a), may be replaced with its two components: radial bmtiing stress (a,) and tangential shcar stress ( T ) , Figure 1Y(b). In Figure lY(c) arc shown the resulting stresses. It is seen in taking moments about the apex ( C ) of the wedge that all of the radial bending stresses pass throngh this point and cannot contribute to any moment. The tangential shear stresses along the curved section ( A - B ) acting normal to, and at a distance ( p )

(b) Components

of bending stress

(c) Resisting radial bending stress (a,) normal to curved section (A-8); also tangenfiol shear stress (7)

FIGURE 19

(a) Resisting horizontal bending stress and vertical shear stress

/

(b) Components of these two stresses FIGURE 20

(c) Resisting rodial bending stress (u,)normal to curved section (A-B)

5.11-14

/

Welded-Connection Design

u ~ i produce l an eqnal and oppo~itemoment. The value of this tangentla1 shear force ( V ) acting on this cr~rved section (A-B) may he found from the following:

moment applied to section A-B

normal stress on inner flange Transverse Force IP,') Applied t o Wedge Member

The applied transverse force (P,') results in horizontal bending stresses (cr,,) as well as vertical shear stresses; Figure 20(a). These two stresses may he completely replaced with a single component, radial bending stress (err); Figure 2 0 ( b ) . The results are shown in Figure 20(c). Notice that no tangential shear stresses are present. Axial Force

(Pi)Applied

normal stress on outer flange

I

Problem 2

1

t o Wedge Member

The axial force (P,') applied a t 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 I : Check Lower Curved Flange (Figure 2 3 )

Summary

The effects of all these forces applied to the wedge member may he summarized as follows: shear stress on section A-B

(a) Resulting axial stress

properties of haunch section (1-1) Use reference axis (y-y) through centcrline of web plate.

(b) Components of axial stress FIGURE 21

(c)

Resisting radial stresses (crJ

Rigid-Frame Knees (Elastic)

FIGURE 22

FIGURE 23

/

5.1 1-15

Ided-Connection

average stress in lower curved flange at (1-1) Uf

P M ct =A 1-'(150 kips) (100" x 100 kips) (!Z3.1%") -- --- (41.6%) + (15,153 in.") = 18,870 psi (compression)

force i n flange

F, = ur At = (18,870) (10) = 188.7 kip? radial pressure of flange againsi web

transverse bending stress in flange 0-t

= p urnax = (.70) (19,600) = 13,760 psi

These stresses are a little high, so radial stiffeners will be added between the lower curved flange and the web. STEP 2: Check Nounch Section tor Bending Stress Using O l a n d e r i wedge method and curved section (A-B) (See Figure 24.)

Here: sin 1.8" = ,30902 cos 18" = .95106 tan 18" = ,32492 sin 9" = .I5643

radial compressive stress in web

cos 9" = ,98769 18" = ,31417 radians

dimensions of wedge section (ABC)

-

(1887 Ibs/in. ) (W')

= 3774 psi The outer edges of the lower curved flange will tend to bend away from the center of curvature under this radial pressure, and will cause an uneven &hihution of flange stress. The maximum flange stress will be-

I

Haunch

I I

section

and the transverse bending stress in the flange will h e -t

=p

Urn,,

The values of 10. In this case,

a

-

and we find= .96

and /3 are obtained from Figure

p = .70

Hence:

maximum flange stress

= 19,660 psi

Wedge section (ABC)

Rigid-Frame Knees (Elastic)

, 8 e ~ 8 c ' ~

,

M'

=

Point of inflection in beom (M

/

= 0)

Point of inflection in column [M = 0)

+ 14,456 in-kips

Hounch section (A-8)

FIGURE 24

= .:.--d +-;- r sm 2 a sm 2-

(1 - cos 2 = )

= 161.79 + 15.84 = 177.63" d, = p 2 a = (177.63) (.31417 radians) = 55.81''

n

d tan 2 a

r sin 2-

(I-cos2a)

5.1 1-17

elded-Connection Design

m = u - n

= 25"

-

-

-

138.04"

113.04"

properties of haunch section (A-BJ Use reference axis (y-y) through centerline of web plate.

FIGURE 26

Total

shear stresses in section (A-B)

v = -M ' P

- ( 14,456) -

(177.63)

= 81.35 kips 7

- V[Atyr = --V Q -

+

Awywl 1 tw I tw (81.35) (7.5 x 28.975 14.3 x 14.3) - .(19,686) (35)

+

-

= 1800 psi moment applied to section (A-B) Find forces applied at apex ( C ) of wedge section (ABC) :

M = M' - Ptr p = (114,456) - (132.5) (177.6) -

transverse force at C

Ptf = Pt cos a - P, sin a = (150) (.98769) - (100) (.15643) = 132.5 kips

in.-kips

nonnal stress on inner flange

- - (123.5) -+ (-9082) (26.46)

axial force at C

+ Pt sin a = (100)(.98769) + (150)(.15643)

Pa' = P, cos

-9082

a

= 123.5 kips

(44.53)

- -

( 19,686)

15,000 psi

normal stress on outer flung?

moment about C .-

- (123.5) - (-9082) (29.35)

(44.53) = - 10,300 psi

(19,686) --

-+

These forces result in the following stresses on the haunch section (A-B) o£ the wedge (see Figure 26):

As an alternate method Clzeck Haunch Section for Bending Stress Using Conventional Struigllt Section (A-B) (See Figure 27.)

Rigid-Frame Knees (Elastic)

/

5.1 1-19

v = 3(

U

=

-

FIGURE 27

c , = 28.872"

Here:

v = r sin ( 2 a ) = ( 100) ( ,30402)

= d, = =: =

c, = 26.008"

moment applied to section

M = (150) (55.902) = 8385.3 in.-kips

30.902"

d +r ( 1

-

cos 2

a

)

(50) --I- (100) (.O488) 54.88"

properties of hnrrnch section (A-B)

o = - 13,800

o' = 15,280

Use reference asis ( y - y ) through cmtcrlirre of web. Plate

A

y

M

= A-y

I, z Mey

FiGURE 28

I

tensile bending nnrl axial stress in outw flange

-+ 10,550 psi, tension compressii;e bending and uxiol stress normal to section in inner flange

-

-

13,800 psi, corriprcssion

5.11-20

/

Welded-Connection Design

(b) Bending stresses in haunch

(a) Bending stresses in haunch using curved wedge sections, bored on Olander method

using conventional straight sections

FIGURE 29

STEP 4: Summary

stress n o m ~ n lto axis of czmed flange u' = -.

LT

cos' 2

a

13,800 (.95106)?

= 15,260 psi, compression ~

Fignre 19 summarizes the stresscs at several sections of the haunch for botlr the \vetlge mt:tliod and the conventional method using straight sections. Tho \vedga inetliod gives results that check close with experirriental results, although it does require more time. The conventional rnctliod nsing straight sections in which the stress oil the inward c~invedflange is increased to acconnt for the sloping flange is easier. No\vever, rrotc that it does give lriglier values for the steeper S I O ~ R .

1. INTRODUCTION TO PLASTIC DESIGN

Thc allowabli. stress used on steel structnres in hendiug is .GO rr,. a percantage of the steel's yield strength (AISC Scc 1.5.1.4). A steel structure desigued on this basis may carry an overload as great as 1.67 times the designed load before the most stressrd fiber reaches the yield point. Katurally, this does not represent the maximum lond-carrying capacity of the structure, nor does it indicate. the reserve strength still in the structure. Plastic design does not make use of the conventional allowable stresses, but rather thc calculated ultimate load-carrying capacity of the structure. With this method, the given load is increased by 1.70 times the given live and dead load for simple and continuous beams, 1.83 times the given live and dead load for continuous frames, and 1.40 times these loads when acting in conjunction with 1.40 times any specified wind or earthquake forces. Then the members are designed to ccary this load at their ultimate or plastic strength. Some yielding must take place before this ultimate load L reached; however, under normal working loads, yielding will seldom occur. For the past 25 years, a considerable amount of research, both in Europe aud the United States, has been devoted to the ultimate load-carrying capacity of steel structures. For about 15 years, extensive work on fiill-scale structures has beerr going on at Lehigh University under the joint sponsorship of the Structural Committee of the Welding Research Council and the American Institute of Steel Construction. Much has been learned as a result of this work.

thew is a strzrigllt-line relationship. It is assu~ncdthat the I~endingstresses arc zero along the ntwtral axis of the bram and incrcase linearly until thry are maximum at the outrr fihcrs. This is illustrated at the top of the figure. At poiirt ( A ) , the maximum outer fiber bending stress has reached 23,000 psi. .4t point ( B ) , this stress has reached the yield point, or 36,000 psi, and yielding at the outer fiber starts to take place. In couventional design, this point is assuined to be the ultimatc load on the member; however, this mrve shows there is still son~emore vcserve strength left in the beam. As the beam is still further loaded, as at ( C ) , the outer fibers are not stressed higher, but thc fibers down inside the beam start to load to the yield point, as in ( D ) . At this point, tlie beam becomes a plastic hinge; in other words, it will undergo a considerable amount of angle change with very little further increase in load. M, is the moment yield point ( B ) , and M, is the

4 36 Sfeei ojor Conclusions

The ultimate load-carrying capacity of a beam section is much greater than the load at yield point. For many years, it has been known that a beam stressed at its outer fibers to thc yield point still had a considerable amount of reserve strength before final rupture or collapse. Consider Figure 1. In this graph for A36 steel, the vertical axis is the applied moment ( M ) , the horizontal axis is the resulting angle of rotation (+). Within the elastic limit ( B ) ,

8

?L

---

$6 iangla FIGURE 1

of rotation)

5.12-2

/

Welded-Connection Design no

load

Mr

*

!st plastic hrnga formed

a t center

0

r-+-l

becomes arch

hinge

plaastic

hinge

hinge

FIGURE 2

plastic moment which causes the beam at point ( D ) to act as a plastic hinge. For a rectangular cross-section, the plastic moment (M,,) is 1.5 times the moment at yield point (M,). For the standard rolled W F sections, this plastic moment (M,,) is usually taken as 1.12 times the moment at yicld point (M,). The multiplier varies for other sectional configurations. Redistribution of moments causes other plastic hinges to form. In Figure 2, a rigid frame with pinned ends is loaded with 21 concentrated load at midspan. The frame w-ith no load is shown in ( a ) . The frame is loaded in ( b ) so that its maximum bending stress is 22,000 psi, the albwnble. Notict: from the bending diagram that the moment at n~idspanis grratcr than thc momcnts at the ends or knees of the frame. The three marks at midspan show the moment M where u = 22,000 psi, or allou-able; My where u = 36,000 psi, or yield point; and M,, at plastic hinge. Notice at the left knee how much more the moment can be increased before a plastic hinge is formed. In ( c ) the load has h e m increased until a plastic hinge has becn formed at midspan. The knees of the frame in this example have only reached about half of this value. Even though, with conventional thinking,

this heam has scrvcd its us(~fnlness,it still will not fail hecnusc the txvo krwcs an- still intact and the frame now 1 ) ~ ~ ) mae thrt-r-hinged s arch, the other two hinges bcing the original pinrrcd rnds. Further loadirig of the frame may he continued, as in ( d ) , with the kners loading u p ~rntilthey become plastic hingcs, as in ( e ) . Orlly when this point is rc;ichrd would the whole frarnc fail. This condition is rt.fcrrd to as mcchaiiism; that is, the structure would dcforni a p p ~ ~ ~ $ a l :with l y only the slightest increase in load. This entire hin~c,adion takrs place in u small portion of the uoai1ol)L: clnngution of the membcr. In the lo\\-er portion of Figure 3 is a stress-strain curve showing the amouut of movement which may be used in the plastic range. This may seem large, but it is a very small portion of the u h d e cnrvt., as shown in the upper portion of tha figure, which is carried out to 25% d o n g t lon. ' The working load is nzultiplicd by a factor of safety (1.85) to give thr? ultimate load. The dcsign of the structurc is bused on this ultimate load. In order to establish a proper factor of safety to use in connection with thr ultimate, load, as found in the plastic method of design, it worrld bc w d l to consider the loading of a simply supported beam with a concentrated load applied at its midpoint. This is shown in Figure 4. The moment diagrams for this beam are shown for the three loads: the inomcnt M causing a bending stress of 22,000 psi; thc moment My causing 36,000 psi or yield point; and the moment M, causing a plastic hinge. Here, for A36 stcel: Allowable hending stress = 22,000 psi Yield stress = 36,000 psi = 67% above @ Plastic hinge occurs 12% above @

2 ,k---Plastic Ranye-2 t *)

1.30

I

t

2;

I

ZC

,

-

-

3 i

I

1

:I

o.,.,o-a

.G

I

I

I

, - zo./oQ P i b rio-2 0 Strain E /"/in

FIGURE 3

elded Connections for Plastic Design

Momant

/

5.12-3

= GOO in." So, use 36" W F 18% beam with S = 621 in."

Diagram

FIGURE 4

Hence:

@ ( . 6 7 ) ( 1 . 2 ) = 1.88 of @ Thus. the true load factor of safct\, of thn simide beam is -.1.88. In convei~tionaldesign, it is assumed that thc ultimate load is the value which causes the lxam to he stressed to its yield point at the point of maximum stress. This would bc represented in the figure by the moment at In conventional design, if the allowable bending stress is 22;000 psi and thr yicld point of the (A3G) steel is assumed to bc 36,000 psi, the designer is actually using a factor of safety of 1.67. l3y means of plastic design, the ultimate load is approximately 12% higher (in the case of a WF beam) than the load which causes the yield point to be reached. Therefore, the factor of safrty for plastic design on the same basis would be (1.67) (1.12) = 1.88.

( b ) The elastic design, rigid frame is shown in Figrire 6. Its span is 80' end its height is 20'. Tllcre are several ways to s o h for the bending moments on this frame.

@.

m Example

To illustrate plastic design, a hcam will be designed using thrcc difftarent mcthods: ( a ) simple beam, ( b ) elastic dcsign, rigid frame, and ( c ) plastic design, rigid frame Thc bcarn will have a span of 80' and carry a conctmtratd load of 55 kips at midspan. For simplicity the d w d load \vill hr rit$ectcd. ( a ) The siml~lysripportcd beam is shown in Figure 5 with its monicnt diagmm. The mnxirnum momcnt for~nulais found in any beam table. From this, the reqnired srction modulns ( S ) is found to bc 600.0 in.3, using an a1lownl)le load of 22.000 psi in b e n d i ~ ~ This g. beam may be made of a 36" WF b r a n which woighs 182 Ibs/ft. Simpla Beam

l

moment

FIGURE 6

In this exn~nplcthc momrnt at midspan would be-

(,iS.O00) (SO x 12) 7 (22,OW) = 343 in.:' --

So, use a 30" WF 124# beam with S = 354.6 in.3

diagram FIGURE 5

The redundant or unknown horizontal force at the pinned end of the frame is first found. Then, froin this, the moment diagram is drawn and the maximum moment found. The required section ~nodulus( S ) of the frame is determined from this maximum moment.

elded-Connection Design

This is foiind to be 343 in." wliich is 55% of that required for the single beam. This hfam could be made of a 30" W F beam having a weight of 124 lhs/ft. ( c ) The plastic design, rigid framc is shown in Figure 7. With this method, the possible plastic hinges are found which could caust: a mcchanis~nor the condition whcrrby the strocti~rebeyond a certain stress point wonld deform appreciably with only the slightest increase in load. These points of plastic hinge, in this example, are at the midpoint and the two ttiids, and are assigned the \.due of M,,. An expression is needed from which this value hl, can be found.

Plastic dasign

PL /-M= 7

1. More accurately indicates the true carrying capacity of the structure. 2. Reqnires less steel than conventional simple beam constri~rtion and, in most cases, results in a saving over tlie use of conventional elastic design of rigid frames. 3. Requires lvss design time tlian does elastic design of rigid framing. 4. Result of years of research and testing of fullscale structures. 5. Has the backing of the American Institute of Steel Construction. 2. D E S I G N R E Q U I R E M E N T S OF T H E M E M B E R Loads (AISC Src. 2.1)

The applied loads shall be increased by the following factor: 1.70 livc and dcad loads on simple and continuous beams 1.85 live and dead loads on continuous frames 1.40 loads acting in conjunction with 1.40 times any wind and earthqualie forces

FIGURE 7

Here:

Columns (AISC Sec. 2.3)

Columns in continnous frames where side-sway is not prevented shall he proportioned so that: I

I

(AISC formula 20)

= 1017.5 ft-kips So, use a 27" W F 114# beam, with plastic moment (M,) of 1029 ft-kips. (See AISC Manual of Steel Construction, Plastic Section Modulus Table.) In this case, it is noticrd that the altitude of the overall triangle in the moment diagram, which is M,, plus M,,, is also thc same as that of the moment diagram of a simply supported beam with a concentrated load at its midspan, Figure 5. This can be fount1 in any P L beam table. Hence, M, plus M , is set equal to 4 .

using for P the ultimate load which is the working load times 1.85. This works out to M,, = 1017.5 ft-kips as the ultiinnte load plastic momcut, at centrriine and at the two beam cnds.

*

*

*

Summary of Advantages

As a summary, here are some of the advantages of plastic desibm:

where:

= unbr;~ced lmgtli of column in the plane nornral to tliat of tlie ~nntin11011s frame r = radins of gyration of coluinn about an axis n o r ~ n dto the p1;ine of the continuous frame Stte the nomograph, Fignro 8, lor convenience in reading thc limiting value of L/r directly from the vnl~scsof P and P,. The AISC fominlas ( 2 1 ) , (221, and (23) give tlie effective moment (M,,), which a giviw sllape is capable of resisting in terms of its full plastic moment (M,,) when it snpports an axial force (1') in addition to its moment. See Table 1. The maximum axial load (1') shall not csveed .60 P, or .60 u, A,, where A, = cross-sectional area of the column. I,

FIGURE 8-Limiting

Slenderness Ratio of Columns in Continuous Fromes (Plastic Design), Sideswoy Permitted

LIMIT OF(%) FOR COLUMNS IN CONTINUOUS FRAMES WHERE JIDE SWff Y 15 NOT PREVENTED

EXAMPLE : P = 1000 lC Py = 4000'

READ jk

=

35

elded-Connection Design TABLE 1-Allowable End Moments Relative To Full Plastic Moment of Axially-Loaded Members

case?

when

P/Py

5 0.15

pz-=-q when

P/P,

> 0.15

AISC iormula

when

-

L -

0

< 60 and

< .I5

then

Notes: See Tcbler 2-33. 3-33. 2-36 ond 1-36 for volvei o i B, G, K ond J

TABLE 2-33 (AISC Table 4-33)

TABLE 3-33 (AISC Table 5-33) FOX

sa

XSI

S r E c i r l r " rirLo

miai srrxr

elded Connections for Plastic Design

/

5.12-7

TABLE 3-36 (AISC Table 5-36)

TABLE 2-36 (AISC Table 4-36)

run 36 rsi mi.<,P,lo riais ixrxxr n F i , .

Dl.

A.,

dl

. .M.

M, .M

. .$I.

ill.

M.

l , X .'11,

i.0:iB 1 019

(3

1 070

44 6.3

1 ""I: 1 106

91

16

1 122

92

4:

l i4li 1 138 i 176 1 is:>

Y:,

in

94 95

43 50

96 Y i 9" '9 100

61

in*

!>C

102 103 104 10s

17

1 310 1 310

10

1 :XI l 371 1 892

62 Sii 54

66

ix

611

106

6,

107 108

Ci

62

i 1 1 1

('13

132 211

271 1 290

1 1 I l

418 416 (56

ti0

Ui

'7" i 601

ill

F*,

1 623

,I2

1 b46

Lii

67 ti* 69 70

118

71

i 64,

72

1 6,;s 1 591 i.716

ice

iii ill

$17 ii6 ii9

izii

-

If L/r > 120, the ratio of axial load ( P ) to plastic load (P,) shall be-

I (AISC formula 24) 1 Shear (AISC Sec. 2.4)

61

73 71 75

is70 1 143 1 (117

1.742

76

1.mw

77 78 79 80

1 "94 1 S21 1 848 1 R76

- -

Assuming depth of web = 9 5 d (depth of memh r r ) , the shear on web section at ultimate load isV" = t,(.95 d ) m,

Webs of columns, beams, and girders not reinforced by a web doubler plate or diagonal stiffeners shall be so proportioned that:

FIGURE 9

inimum Width-to-Thickness Rufios (AISC Scc. 2.6) When subjected to cornpression involving plastic hinge rotation under ultimate loading, section clcments shall be so proportioned that:

5.12-8

/

Welded-Connection Design

(AISC formula 26) but nerd not be

1t.s

than

-1

where: r, = radius of gyration of meniher about its weak axis

M = the lesser of the moments at the ends of the rrnbraced segment M -- the end momcnt ratio, positive when the segment is brnt in single curvature and negative when bcnt in double curvature

-

FIGURE 10

In the usual square frame. plastic hinges would ultimately form at maximum negative moments at the coiners, and at thc maximum positive moment near the center of the span. However, a tapered haunch may develop a plastic liingc at the comer and also at the point wliere tlie Iraunch connects to the straight portions of the rafter or colunm because of the reduced depth of the momber. These also become points where lateral bracing must he provided. 3. BASIC REQUIREMENTS OF WELDED CONNECTIONS

and when beam or g ~ r d r ris s~~hjected to axial force ( P ) and plastic bending moment (P, ) at ultimate load,

See nomograph, Figure 11, for convenient direct reading of d,,/t,\, ratio from values of P and P,. Lateral Bracing (AISC Sec. 2.8)

Plastic hinge locations associated with all but the last failure mrchanism shall be adequately braced to resist lateral and torsiold displacement. Laterally unsupported distance ( L ) from such braced hinged locations to the nmrest adjacent point on the frame similarly braced shall b ~ -

Coilr~ectionsare an important part of any steel structure desihpcd according to plastic &:sign concepts. The connection must allow the members to reach their full plastic moments with sufficient strength, adeqrratc rotational ability, and proper stiffness. Thcy must be capable of resisting momcnts, shear forces, and axial loads to which tb(~ywould he sul~jwtedby the ultimate loading. Stiffcncrs may be rcquiretl to preserve the flange continuity of intcrrupted mmnhcrs at their junction with othcr mm1hrrs in :I continuous frame. A basic reqiiiremtwt is that the web of the resulting cor~nectioo mirst provide adtynate resistance against blickling from ( a ) Shear-the diagonal compressive force resulting fl-om shear forces applied to the u ~ from b thc colrr~cctingflangt,~,which in turn are stressed by the end moment of the member, and ( b ) Thrust-any conceiitratrd compressive force applied at the rdge of the web from a11 intersecting flange of a member, this force rcsl~ltingfrom the end moment of that member. See Figure 12. In addition to mveting the above requirements, the connection should be so designed that it may be economically fabricated and w ~ h l c d . Groove welds and fillet welds shall be proportioned

Welded Connections (or Plastic Design

/

5.12-9

5.12-10

/

Welded-Connection Design

(0)

Web resisting shear FIGURE 12

FIGURE 13

elded Connections for Plastic Design

/

5.12-1 1

to rrsist thc forces produced at ultimate load, using an increase of 1.67 ovcr tiic standard a1low:lblcs (AISC Sec. 2.7). Followiiig p;~g~:scover first thr (itssign of simple two-way r(x:tangillar corr1t.r conn(~ctions,tapered lrarrr~clti:s,and cnrvtd I~aurrcht~s. Kext, tlrr design of beam-to-colrirnn <,mtncctions,wI~i,thertlirce-way or four-way, is d c d t with. r\nalysis mcl dcsign oi a particirl;~rconntsction may not always br as simple as those ill~islrat(d011 tlicsc pages. Figure 13 slrows some other typical welded connections.

4. STRAIGHT CORNER C FIGURE

The forces in the flanges of' both rnornbcrs at the connection resulting from the moment (hl,) are transferred into the contirction .iwh as shc:tr ( V ) . Some of the vertical shear in thc hoam (V,) :md the horizontal shear in the column (V,) will also be transferred into the cotincction web. IIowr\er, in most cases these values are small compared to those resulting from the applied moment. Also, in a simple comer connection, these are of opposite sign and tend to reduce the actual shear valiio in the connection. 111this analysis, only the shear resulting from the applied rnomcnt is considered in the web of the connection.

15

resulting shear stress in connection web

The values for the shear stress at yield ( 7 , ) may be found by using the Mises criterion for yielding-

uCr=

J u? - u,

my

+ uY2+ 3 rri2

In this application of pure shear, u, and u, = 0 and setting the critic;rl value (u,,) equal to yield ( u , ) , we obtain-

Diagonal compression

Fc

Hence,

=v

connecilon

The nornograph, Figure 16, will facilitate finding this reqnired wcb thickness. FIGURE 14

In ihc above: The miitimiirn wi,h thickness rtqriir-1.d to assuro clots not huckle from that the web of thc mniitr~i~liorr the shear forces set "1' 1)) llw rnrimcltt applied to thc corinection (M,,), inay hi. fo~irrdfroin the following: unit shear force applied to mn~rer:fionn:eh

M,, =: phstii. rnontc.nt at connection, in-lbs d ,, . . ,.It ptli of lwam, in. d,. w

..:

d
w, = rcijtiircd web thicktress in connection area, in.

uF = yic,lcl str<.nglhor s t d , psi

elded-Connection Design

Welded Connections for Plastic Design

AISC uses an effective depth of the heam and column as 9.5% of their actual drpths to allow for the presence of plastic strain in thr flanges. due to concurrent bending. .Applying this reduction to both the depth of the beam ( I ) and the col~imn (d,), and also expressing the applic:d plastic moment (M,) in ftIbs rather than in.-lbs, this formula hccomes:

Here M,, = plastic moment, f t : h For most wide flange ( W F ) stxctions, the web thickness ( w ) will he less than the required value (w,) above, and some form of stiffening will be required.

A sy~lin~ctl-icd pair of diagonal stiffeners may be addcd to this comcction to pirwmt the. wr,h imm hi~ckling. These stiffeners rr.sist cnoogh of the flangc forw ( F ) that the l-esdtirrg shcar ( ) xl~p1irdto this wrh is rechlccd sufficiently to prevcnt hnckling. Stiffen~rshaving a thickr~esscqod to that of the rolled section flange of the heam or column nonually will be adequate, although this thickness will he greater than required. Thc minimum thickness of ilks stifrcner may be f o ~ m dfrorn the following: of the hearn is Thn horizontal flange iorce ( resisted by the combined effect of tlla web shear ( V ) and thc horizontal component of the mrnpressive force ( P ) in the stiffener.

where

A web doubler plate, or a pair, may be used to bring the total web thickness up to the minimum (w,) obtained above. Welds should be arranged at the edges of doubler plates so as to transfer the shear forces directly to the boundary stiffeners and flanges.

and since

&=

11_Ji

5.12-13

Diogonol Stiffeners

Web Doubler Plate

plate

/

-1 e[ cos

M

--L dl, u,. -

-&II.. . w d

.(12)

where

e

= angle of diagonal stiffener with horizon,

FIGURE 17

FIGURE 18

elded-Connection Design

A.

7 . :

area of a pair of diagonal stifhers, A, = b, t,

In t h ~ :usnal detailing of the corlrrtxtion, the reqtiirtvl wcb thickness (u,,) is first fonntl. The actual wab tlrirknrss ( w ) of course is known, therefore it would he simpler to change this fotmula into the following so that the reqriircd area of thc tiiagonal stiffener may be found from these two values (w,)and ( w ) : Girder

Column

21" WF62#

14" WF841f

14" W 84#

From Formula 10,

dr

column

--

14.18''

1

di1 = 20.99..

and substituting this into Formula 12,

A, =

1 M, . cos B [[dl o, --

FIGURE 19

fl

and since

The required web reinforcement is determined as follows:

-

cos B = d, d, c ( 4 3 2 ft-kips x 12) > - ( 20.99 i.) (14.18") (36 ksi) .-

2 0.837"

web furnished by the 14" W F 84# col~l~nm= 0.451" -effective web to be furnished by stiffeners 2 0.386" This reinforcement may be provided by one of two possible types of stiffcners as noted below. or could use t, = t*

also in all cases

(a)

Web Doubler

Plate

The additional web plate must be suflicient to develop the required web thickness. The welds should be arranged at the edges so as to transmit the shear forces directly to the boundary stiffeners and Ranges. Plate must be ,386" thick, or use a 5
For full strength, stiffeners should be welded across their ends wit11 either fillet welds or groove welds, and to the connection web with continuous fillet welds.

To design a 90" connection for a 21" M7F 62# roof girder to a 14" W F 8411; column. Use A36 steel and E70 welds. Load from girder: M, ultimatc plastic moment = 432 ft-kips.

'- Web Plate Doubler FIGURE 20

elded Connections for Plastic Design

/

---l-T 21" W 6 2 2

F = 432 X 12 14.18 =441k

1 I

stiffeners

4" X '!4/ia"

L -- -

1 I

20.99"

\I

- - --Y

FIGURE 21

fbj Diagonol Stiffener

The diagonal stiffener will resist the diagonal component of the flange load as a compression strut. The flange force to be carrind by the stiffener is the portion that exceeds the amount carried by thc web. Assuming the bending moment to be carried entirely by the flanges, the compressive force in the diagonal stiffener is compnted as in Fignre 21. Multiply this diagonal compressive force of 441 kips by the ratio of the additional thickness needed to that already in the web: 441

(g)

8 = tan-'

db dc

-

= tan-' 1.48

and cos 55.03" = ,560

= 204 kips h r a on diagonal stiffener

= 5.65 in.%eedcd in the stiffener If b. = 8",thcn -

204 kips 36 ksi

= 5.65 in.%eeded in the stiffener or use a pair of %" x 4" stiffeners, As = 6.0 > 5.65 OK Now solve this portion of the problem by using Formula 3:

= ,707" or use

3/4"

Or use two plates, 3/4" x 4", for the diagonal stiffeners. Check their width-to-thickness ratio:

5.12-16

/

Welded-Connection Design

Welds for Stiffener

E70 We1d.s & An6 Plate

Only nominal fillct welding is required between stitfener and connection web to rcsist buckling. These welds are rised simply to hold thc stiffeners in position. Welding at terminations of the stiffener should be sufficient to transfer forces. To dt:velop the full capacity of the stiffener, it may be butt welded to the comers, or full-strength fillet welds niay be used. Thp required leg size of fillet weld to match the ultimate capacity of the stiffener would be-

Hence, use %" leg fillet welds acmss the cnds of the stiffener. It may bc simplcr to make the cross-sectional area of these diagonal stifIenrrs equal to that of the flange of the member whose web they reinforce.

5. HAUNCHED CONNECTIONS

EGO Welds & A7, A373 Plate 2(9600 w j 1.67 = t, 33,000

o = 1.03 t,

Haunched connections, Figure 23, are sometimes used in order to nlore nearly match the moment requirements of a frame. This produccs a deecpr section in the region of maximum momcnt, extending back until the moment is rcduced to a value which the rolled section is capable of carrying. In this manlier a smaller rolled section may be used for the remainder of the frame. This has been a rather standard practice in the conventional elastic rigid frame. Haunched knecs may exhibit poor rotational ability if the knee buckles laterally before the desired design conditions have b'en reached. The haunch connection should be proportioned with si~fficientstrength and buckling resistance so that a plastic hinge may be formed at the end of the haunch where it joins the rolled member.

I FIGURE 23

W e l d e d Connections for

Plastic Design

Lehigh University's extensive research in plastic design included the testlng to destruction of full-scale structures such as this 40' gabled frome

Plastic design of this 8-acre rubber plant simplified mathematical analysis of the structure ond moment distribution. Two results: a uniform factor of safety and a saving of 140 tons of structural steel.

/

5.12-17

5.12-18

/

Welded-Connection Design

FIGURE 24

Lower flange of beam

FIGURE 25

Welded Connections for Plastic Design

/

5.12-19

(See Figures 24 ;md 25, facing page) Thickness of Top Flange and W e b of Haunch

The thickness of the top flange nrrd the web of the haunch slmild be> at least equal to the tl~icknessof the rolled beam to which it connects.

flangc'sthiicknms. Silrcl, this is the tc~rrsionflange, it will be same or thiinrer than the lower (co~npression)flange. It can he siionii that the pl;~sticswtion mo~iolos [ Z ) of an 1. section is: tension

Thickness of Lower Flange of Haunch

The lower flangc of the haunch must be increased in thickness so that when it is stressed to the yield point (u,), its horizontal component will be equal to the force in the lowcr beam flange stressed to yield. The force in the sloping lowcr flange of the haunch a t the plastic moment (M,,) is-

Tc = u, hh ti, The component of this force (T,) in line with and against the force in the beam Aange is-

T = T, cos p

Stress distribution at plastic moment (M,]

= u, b,, t,, cos p

FIGURE 26

and this must match the force ( T ) in the lower Aange of the rolled beam, or:

resisting plastic moment of section

T = u, bl, ti, cos p must equal T = u, bn tb 4ssuming the same flange width for the haunch as the beam, i.e. h , = bbr gives-

d,, - 2 t + 2 ( dl, - 2 t )(--i-) I

since Transverse Stiffcners

T, = T, sin p or

us b, t,, = cr, bl, ti, sin

p

Assuming the same flange width for the stiffener as the beam, i.e, h,, = b,,, gives-

AlSC suggests making the total area of these stiffeners not less than % of the haunch flange area (AISC Commenta~yp 37, item 4 ) R e g u i r d h u n c h Section

Section (1-1j, in the region of high moment, should be checked. The two flanges may vary in thickness, so for simplicity and a conservative value use the upper

This increased plastic section modulus may be obtained by: 1. Increasing the dtyth (d,,) and holding the Aange area constant, or 2, Increasing the iia~igethickness ( t ) and holding the depth ( d l , ) constant. By assuming that (d,, - t ) is equal to ( d , - 2 t ) , and solving for the expression (d,, - 2 t ) , it is found from the above formula that:

/

5.12-20

Welded-Connection Design

Fmm this, the requind depth (d,,) of the 1l:imich may brt io~inrlfor any vehie of pl:tstic s < , c h nniodulus (Z). Thc 1i;iiinch s~,ctiorimiist h? :rhlc to dcvt~lopthe plastic inor~iciltat any lmint ;rloiig its imgth:

,

z

u

J . . . . . . . . . . . . . . . . . . . . . . . . . . (22)

The cornniwtary of the AISC specifications scts the following lirnils for latvral I~racing. The taper of' tli<, hau~jcli may br snch that the resrilhig hei~dingstrcss 21t plnstic Ionding, wheu cornp ~ i t ( db y using thc plastic inodiil~is ( % ) , is approxiat Ilotll cnds @ & @. If this is the mately a t yield ( q ) (Li,): caw, then Ii~nitihr ur~l~r;rccd l~~rigth

or at any scctiou (x-x)-

Usuall\- just tlw two ends of thr: haunch must he checked. This would l x scction (1-1) at ille haunch point ( H ) , and scctioli (2-2) ;it the connection to the rolled beam. The latter finding will also dictnte the required section moriirlus of the straiglht beam, since its highest moment will occur at section (9-2). Heedlr~' points orit that. if the moment is assumed to incrrnsr linerally Srorn the point of inAi:ction ( 0 )to the haunch point ( H ) , and the distance ( 0 - R ) from the point of inflection to the end of the rolled beam is 3 d, then the critical section will always be along (2-2) if the angle P of the txpcr is greater than 12 ,; if this angle is less than 12", then section (1-1)must also be checked. Laterol

(LJ: . . . . . . . . . . . . . * (26) but

Stability

Bracing should be placed at the extremities and the common intersecting points of the compression flange.

-

If thc bcntiing strcss :it orrc rnd is approsiinately at yield (cry), using the pl;isiic modulus ( Z ) _ and at the other end is I t s than yicld (my c: u s ) when using the secton ~nodulus S , limit the ~inhraced length

*"Plastic Design of Steel Frames" Lynn S. Ueedle; John S. Wiley & Sons, publishers.

If the bending stress colnputed on the basis of section modulirs ( S ) is less than yield (us< IT,)at all then transverse sections of the haunch from @ to 0, check to sce that greatest co~nputedstrrss:

Resisting shear forces in web of section ABCD

CD = -

tan

FIGURE 27

dh

(a + yj

Welded Connections Cor Plostic Design

/

5.12-21

I

F& = A', q FIGURE 28

\vhero:

At

area of top (temion) flange of haunch

:

A, =- total area of a pail- of diagonal stiffeners Diagonal Stiffeners

(2) Based on compressive forces a t

Tlra tapered liamch has an extra-large web in the bend of the knee. This is subject to buckling, and should he strengthmi,d by di:~gonal stiff(m.rs. The required stiffener scction arca sho~ildbe figtrrcd from tbr compressive force on the web diagonal r ( d t i n g fronr tllc larger of two forces: ( a ) the itmsilc forces on the outer ilange of the I ~ a u m hat point @, and ( b ) the compressive forces on the inner flange of the haunch at point &f

-

( 1 ) Based on tensile forces at

The compressive force in the diagonal stiffener is found in a similar mnnncr as before; the horizontal components of the forces in tho inner flanges are set in cqrlilihriurn. See Figure 28.

4-A,

r, cos

u

4-A,?

ussin

P2

- A,, uscos (PI

@

The comprcssive force in the diagonal stiffcrier is fouird by taking thc sum of the horizontal components of the Forces in tllc outer Aanges nlid setting them equal to zero. Sce Figure 27. n.,, dt, At u, cos y -

A 0

4 --

A,, cos

. E -

+y )

=0

(6, 4- --y )- A,,? sin PL . . . . (29)

If .4, = !I,, = A,.,, tlris hecomes-

+

cOS

y

- A, a; cos

a

=0

or

A - 4 -

cos y t ( G

;) - (n-tiirli

w,,

ill, cos y
)z;;-

(31 W h e n outer (tensile) flanges form right ongfe

If the beam and colrirnn are at right mgles to each other, y = 0. See Figme 29. and O , = Pi = P2 a z

15"

A,

A,, = .4e2

r:

elded-Connection Design

Thc modifird formulas above may also be used for cor~venience in finding the stiffener requirement of gablo frames, b u t will provide a more conservative value. Summary of Tapered Haunch Wr

2

th 2 -

W

L

t cos

,

~

~

p

Based on load from ttwsion flange-

fi~,

A, - 0.82 \v,,d, Rased on load from vomprcssion flange-

FIGURE 29

A, 2 also b, -

JT

,17

I A, 2 O A , - 0 . 8 2

w,, d , , ]

2 0 . 4 , (cos p

also

--

sin /3)]

2 ti, sin p

b', 2

3/s

b, = 17

tbhb

Zl, = b t [dl, - t )

+4 WI,

(d,, - 2 t)"

M

2

rr,

Check laferal stability of compression flange

based on comprcssivc forces in inner flange

LA,

t, tt,

. . . . . . . . . . .(31)

p - sin p )

t. =

Then the preceding two formulas reduce to the following: based on tensile forces in outer flanges and shear reststance of web

A, (cos

(32)

( a ) ii both ends of haunch @ or @ are stressed to yield (c,)using Z

Welded Connections for Plastic Design

/

5.12-23

NNECTIONS

Thickness of Lower Flange ot Haunch

Tlic lower flange of tlie ht~unclimust b r increased in thickness so that when it is stirssed to yield (u,),its component along the bcem axis is equal to the force in the lower beam flange when stressed to yield. Here: ,8 = angle between tangents of given section and

beam flange r = radius of curvature of inner flange

d, = depth of curved haunch at any section (x-x)

= d2 -1 '(1 x = r sin p,

-

cos

PA)

It is seen in F i p r e 331 that thc moment resdting from ultimatr: loading gradn:dly incrrvxrs ont to the corner of thc 1i:iurrch. IIowever, tlle dcpth of the ha~inch and therefore its hi~ndingstress also increases toward tlie corner, so that the critical scction (x-x) witliin tho h;nmih will occ~irat some distancc ( x ) or some angle (P.) El-om section 2-2. For most curvod h;runclics, this angle (p,) will he about 12". Thickness of Top Flange and Web of Haunch

The thickness of the top fiatige arid of thc web of the hatinch should be at lenst cijn;iI to t i m e fmtiires of the rolled beam to which it ronnrds. If bcniline.,stress \I; at @, u? =- < c,,then the onter flange thickness S of tlie hnnncl~ ( t ) does not have to ( w e e d the bram flange thickness (ti,) (AISC Commentary).

FIGURE 32

5.12-24

/

W e l d e d - C o n n e c t i o n Design

As in tlie tapcred haunch, the phstic st,ction motl~ilus ( 2 ) at m y given point ( X ) is:

Z, = b,, ti, (d,

- ti,)

+4 W

"id,

- 2

.(35)

For any givcn depth (d,), the pkistic section moclulus (Z,) may be increased by increasing the flange thickness ( t,,). Assuming the web thickucss aud ilange width of tlie curved llaunch is at least equal to that of the beam, the required thiclcness of the lower flaugc would be:

Z, = bl, ti, (d,

-- t,)

+ -' (d, - 2 t,,)2 4 W

FIGURE 34

The l I S C Cominentary (Sec. 2.7) recommends that thc thickuess of this inner flange of the curved halnich should be-

This is based on a '30" knee (outer flanges fonn a right angle), which is the most conservative. Thc radiiis of curvature may be increased above this limit if additional poiuts of snpport are added to decrease the critical arc length ( C ) . The unbraccd length between uoints of lateral support must be held tou

where values Tor ( m ) come from the graph, Figure 33.

where C = r $

4 =

radian measure

.l

3

4 n

5 = a/d

6

7

this lirrlit, the .. . tliidness of thc m w d iuucr ilauge n u s t he iucrc:xsed hy-

FIGURE 33

Here:

or the final tlriikrcss will bc-

a = distance from point of inflection ( M = 0 ) of the column to the point of plastic moment (M,,) in the haunch d = depth of coh~mlisection In order to prevent local buckling of the curved inner flange, limit the radius of curvature t o -

An ;iltern;it<. metliod wol~ld he to increase the width of the iiilwr fl;tilgc ( h i , ) to a minimurn of C/6 --

-

* ASCli

Commmt:iry on Plmtic Design in Steel, p. 116

Welded Connections for Plastic

esign

/

5.12-25

FIGURE 35

without decreasing the original Hange thickness (t,,):

A, u, = 2 A, u, sin (20.5" - y / 4 )

u Diagonal Stiffeners

@

(1) Based on compressive forces a t An approximate value of the comprcsive force .. appliod to the diagonal stiEener as a rcsult of the compressive forces in the ciirvcd inner Ransc may be made by treating the curved hauncli as a tapercd haunch. Sec Figure 35.

( 2 ) Based on tensile forces a t

@

The compressive force in tile diagonal stiflener taking tlIc llouizontal cornpo~ientsof these is forrild telisilc aarlSe forces, and settirig them equal to zero. Sen Figure 36. cos

ws ds

-

tan(

-uy

co$ y + y-) A,fl u,cos

a:

Resisting sheor forces in web of section ABCD

FIGURE 36

=0

5.12-26

/

W e l d e d - C o n n e c t i o n Design

FIGURE 37

Radial compressive force exerted

Transverse tensile stress due to bending

of flange

A

>

cos y

A,

-

W, di,

COS a

where:

II

. . . (43)

= area of top (tension) flange of haunch A. = total area of a pair of diagonal stiffeners At

Radiol Support of Lower Flange

The radial components of force in the curved inner flange tcnd to lxsh tlre flange in toward the web, and to bend the flange as shown in Fignre 37(h). Because of the slight yielding of the outer edge of the flange, there is a non-uniform distribntion of ihc flange strcss ( r ) Figure , 37(a). This stress is maximum in line wit11 the web. There is also a tl.a~~sversc tensile stress across the onter face of this flange, Figure 3 7 ( b ) . The unit radial form ( f , ) acting on the curved inner flange from the axial coinprcssive force (F,) within the flange, Figure 38, is-

FIGURE 38

F f, = 2 (Ibs/cir inch) r Trcating 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 (f,), Figure 39:

or unit load ( p ) on section:

p =

r , th

7--

Welded Connections for Plastic Design

/

5.12-27

ness (h,/t,) of thp mrved inner flange to the following, wlrichcver is thc sm;iller:

P~~ovicic s:ifimers : ~ t:ind midway bctween the two points of talqpicy. Make the total cross-sectional area of the pair of diagoiial stiffcncrs at their midpoint not less than % of the inner curved flange area. FIGURE 39

Summary of Curved Haunch Requirements

thicklness of outer flange ( t ) ) ti, web of lraunch (w,,)

2

wb tb cos p = (1 -t m ) t

thickness of curved inner flange ( i s ) 2; (based on tcnsile fiange)

.

A, g cos y cos "

I

A, - -- WI,di,

(based on compressive flange)

2

2~~ sin

(&-I

90 - y

and

.. Therefor(. limit the ratio of flange width to thick-

M >, <

Ii bending stress at @ u2 =- S

cr,, then

elded-Connection Design

.

outer flange tliickni+ss ( t ) does not have to exceed beam flang~:(ti>).

BEAM-TO-COLUMN CONNECTIONS (Multiple Span)

W e b Resisting Shear

Othrnvist~,usc additional lateral support to decrease arc length ( C ) . Asswr~eci-itical section (x-S) at-

p, =

12O

then

When the moments in two beams franring into an intcrior coliimri iliiier by a larger amormt, this difference in mommt d l c ; i ~ ~ largc s c shear forces to act on the conrrc,ction \veh. Tile \vvb must be cliccked to see if it has sutficiiwt thickness; if not, it must be reinforcrd with either a wi,h doublcr plate or diagonal stiffeners. (See Figure 41.) hori~omtuld ~ e u rapplied on connection web aloi~gtop portion

and

z

x

= F,

M 2 - =.-

C 5 6 b ,

-

PI

-

V,

wliere: shear resisted by connection web

C = r +

+ = radian measure

along top portion

= w d, r y

Otherwisc, increase the thickness of the curved flange to-

or increase the width of the curved inner flange to-

without decreasing the flange thickness.

where: V, = horizontal h e a r force in the column above

the connection, lbs

FIGURE 41

elded Connections for Plastic Design

/

5.12-29

FIGURE 42

M, and M2 = moinrmts in bt::~ms (1) and ( 2 ) , in.-lbs. d,

-

depth of coliiuin, in.

dl and d? = dcpth of beams ( I . ) and ( 2 )

w = tliickness CIE connection w b , i n

Stiffencis are qriitc ofiw required on members in line with the coniixessiorr ilii~qeswhich act against them, to ixeveni crippling of the web where the concentrated coinpressivc force is al~plied. Wliera a heam sripportsa column, or a column slipports a beam, on just one fiange, the stiffeners on its web net:d only estmd inst bcyoncl its neutral axis.

If it is assumcd that: 1. tlrc column li(:igiit (11) has a point of inflectioi~ at mid-hlbight, 2. the d q ~ t hof the larger beam ( d 2 ) is %r, of the column bight ( I I ) , or less, 3, tlic yield strength of the stcel is u, = 33,000 psi, and 4, the unbalancrd morncnt ( h l ) is expressed in fooi-kips, tiris for~nulawill rcdwe to t11:: folloiving:

The method of detcrininirig the value of M is illustxatcd in Figrire 42, eb Resisting Thrust

'nie following formii1;is will indicate w1x:n stiileners are required, and also the oecesswy sim of t i m e stiffei~ors: 1. \Z't,l~still(mws arc required adjacent to the beam

2. Wr.1, stificlri.rs ;ir(, rcqnirrd adjaccnt to the beam co~npmssio~~ ILiiig~if-----

. . . . . . . . . . . . . . . . . . . . . . . . . . . . (49) where: FIGURE 43

*

=

A,

..

t1,+5Kc

5.12-30

/

Welded-Connection Design frncrs arc only I r d i as effrctive, since they lie at the (Inter edge of tlic flange.

FIGURE 45

If horizontal flalige platc stiffeners are uscd, F i g ~ ~ r e 45, their dimensions are found I'rorn the following: ts > .AI

- I\,

(ti, b.

-+-

5 Kc) . . . . . . . . . . . . (50)

also

where : A, =

llb

X ti,

w,. = reqiiircd tltickness of connection wch

( S r r Swtion 5.7 on ( h t i i l u o r ~ sCoi~ncciions for fut-tlrrr cxpl;~tintion) If vertical plate stiiieners are used, Figure -46: they s l i d r l he proportioned to carry tlic excess of beam flaiigc force ovcr that wllich the column web is able to can-y. It is assti~liedthe beam iiange estends alniost thc fnli width of the co111mn Ranges, and that the stif-

FIGURE 46

(See S<,ction 5.7 on Continrious Coiinections for hrtlicr exp1:ui:rtion.) The niimngi-apli, Figrirt: 47, inn) bc riscd to find the dist:riice ( t , , -4- 5 K,) ovcr whicli the corrw~itr:~ted foi-c:e f n m tlic beait~fiairg~sp~-c.:tdsnut into tllc c o l ~ ~ m n web, In tllc case of ;I hilt-ui, colntnn, use the flangc 5 t,) from iliickncss (t,.) and find the distance (t, thc ~iomogrnpli. 'I'his value of (t,, - 1 5 K,) or (t,, - 5 t,.) can then be used in fitiding [he required wc11 tliickncss (w,) from the nomogr;ipli, Figure 46.

+

Welded Connections for Plastic

es%n

/

5.12-31

FIGURE 4GThickness of Connection Web To Resist Thrust of Compression Flange.

z.7

L?6 2,5

If COLUMN FLANGE THICKNEJS EXCEEDS THIS VALUE OF (tc )STIFFENERS ARE NOT REQUIRED OPPOSITE TENSION BEAM FL4NGE

7

IF WEB THICKNESS OF COLUMN ( w c ) EXCEEDS THIS REQUIRED YALUE( +)STIFFENERS ARE NOT REQUIRED OPPOSITE COMPRESSION BEAM FLANGE

<

.

, ,

.

I

..

.'

14"W34 *BEAM TO 8'' W35* COLUMN Af = 6.75 x.453 = 3.058 INz HENCE NEED STIFFENERRS TENSON FLANGE bb45%) = 4.8 (FROM PREVIOUS NOMOGRAPH) READ wr = .64 ACTUAL /5 ,315 "

HENCE NEED STIFFENERS COMPRESSION FLANGE

I

wr

ed Connections for Plastic Design

M

/

460 ft-kips - 250 ft-kips

r-:-

:= 210 ft-kips Is r<.inforum~r,i?i nwcssnrv at this interior connection? h/fornet~tsat iiltiliiat<>load arc, sllovr-n t,clo\v. A36 steel

and

M

I

-

0

lips

+-

50 ft-kips

= 210 ft-kips required thickness of conncction u:eb

-

f i ( 2 1~. 0 it-kips x 12) (21.13)(i3.81j(36 ksi). ~

~

= ,416" onclusions (Fig.

FIGURE 49

50)

( a ) This req~lil-cdn e b thickness would be satisfied if the beam were ;illowed to lun through the column. This would give :i web thickness of ,430". OK ( b ) If tlw column were to run continuons through the beam, as illustrated above, then a l/4" doubler plate would be required in this conncction area to make up the difference in thickness. ( c ) Another choice v.ould be to use a pair of diagonal stiffeners having thc following cross-sectional arw

beam dimcrwioru d, = 21.13"

bb

-

8.27"

= ,430'' tb = ,685"

Wb

Or use a pair of 3" by 36'' stifhers, which checks out as-

the area of

+

A, = %" ( 2 x 3" ,339") = 2.38 in." 1.03 iu."

column clirnensioru d, = 13.81" w, = ,339"

OK -

.41so, thc required thickness is-

b, = 8.031" Kc =: 1X6" diagonal of cairncr.tioil web dB ==

m --

= a-13'

T-

13.812

= 23.18"

The necessary web thickness will be determined by the AISC requirements for webs irr the connection region, The algebraic sums of thc clockwise and counter-clockwise moments on opposite sides of the coinlcction are:

In adrtitiou to this, the web of the column must be checked against buckling from the conceutrated compressive forces applied by the beam flanges. If the web thickness exceeds the following value, stiffeners are not needed opposite beam compression flange :

5.12-34

/

Welded-Connection Design

(a) Run beam through column Add plate stiffeners across beam, in line with column flanges to transfer column load

(b) A

(c) A pair of 3" x 3/8" diagonal stiffeners (d) A pair of 4" x horizontal flange plate stiffeners

Y2"

Y4"

doubler plate

(dl A pair of 4" x 1Y2" horizontal flange plate stiffeners

(e) A pair of Tee vertical stiffeners cut from 21" WF 1 1 2 g or 5/8" plote Tee section also provides the necessary additional web material fol this connection.

FIGURE 50

elded Connections for Plastic Design

/

5.12-35

which is found from the following formula:

Since w, = .339", some additional stiffening is required. There are two solutions.

and this checks against the following requirement-

( d ) Horizontal flange plufr stiffeners, the required thickness of which is found from the following formula:

but the following is called for-

This T section ro111d be name cut from a 12" WF 112# section, which has a flange thickness of ,865" ( w t need ,517") and a flange width of 13.00" (we need a t least 13.635"). Othemdse, it could be fabricated from %" thick plate welded together. Summary

IIencc, use a pair of 4" x K" horizontal plate stiffeners. ( e ) Vertical stiffeners, the required thickness of

There are four possible methods of making this connection, Figure 50. E;rch uses a combination of the preceding solutions to stiifm the connection weh so it may safely transmit thr s l ~ r forces ~ ~ r rcsnlting from the unbalarrced monicnt as well as to prevent buckling from the concentrated comprt:ssive forces applied by the beam.

/

Welded-Connection Design

Shop-fabricated Vierendeel trusses lowered steel requirements and reduced time for erection of Hamburgers clothing store in Baltimore. Here a weldor is connecting a corner bracket between web member and bottom chord of the truss, using low-hydrogen electrode for root passes.

1. ADVANTAGES OF VlERENDEEL TRUSSES

A Vicrendeel truss is in effect a rigid frame. It differs from the simple truss (Sect. 5.9), but it also differs in some respects from the usual rigid frame (Sect. 5.11). Although the Vierendeel truss has been used widely in European bridge design, the relatively high cost of riveted construction precluded its early popularity in this country. Modem welding processes have changed the economics and several structures using the welded Vierendeel truss have been built here in recent years. Currently the major field for welded Vierendeel trusses is in building design; Figure 1. For example, they have been used as roof supports to carry the extra load of a superstructure, as exterior floor-high members for rigid support of heavy masonry walls, and in exterior wall grid systems for aesthetic value as well as construction advantages. In exterior use, the large panel areas provide adequate window area to be 6lled in by glass or translucent materials; chord and web members arc sometimes faced with masonry. When used as interior members,

the web openings permit savings in space since piping, conduits, and duds may be fed through them. Some Vierendeel trusses are fabricated from widefiange beams, as shown at the top in Figure 2. Here the top and bottom chord members, as well as the verticals, are standard rolled beams. Additional plates are used to join these members. At the center in Figure 2, the vertical rolled sections are extended ail the way to the top and bottom members. A triangular gusset section or bracket is insected on each side of the connection. These gussets are flame cut from standard rolled sections, usually having the same flange width as the other members. This is a simpler method and therefore is widely used. However, it does not result in as smooth stress distribution at points of high bending moment as does a design with crwed comers. Another method of achieving these curved comers is illustrated at the bottom in Figure 2. Here the truss is Aame cut from flat plate with flanges welded to it around the web openings and across top and bottom edges. Also see Open-Web Expanded Beams, Section 4.7.

5.13-2

/

Welded-Connection Design

FIGURE 2

2. BASIC C O N N E C T I O N REQUIREMENTS In the usual rigid-frame design certain assumptions are made: the beams and columns deflect, and the connections rotate; but within the connection itself, there is no appreciable movement. Of course the connection does undergo some movement (not to be confused with rotation). However, the distances over which this movement takes place are small mmpared with the lengths of the beams and ~wlurnns.Consequently the movement

within the joint has little effect on the h a 1 moment distribution in the frame. The Vierendeel truss on the other hand is more compact; for example, the lengths of the vertical members often are relatively shorter. See Figure 3. The more massive ~ ~ n n e c t i o nthus s occupy a larger portion of this frame than most others. Any angular movement of vertical members due to yielding within the connection itself will greatly increase the moments in horizontal members. There is no method of computing or predicting how much the connection will yield; therefore, every effort must be made to provide a connection at least as rigid as the adjoining members. It might be thought that the simple square type of connection would naturally be as rigid as the members, since it is a continuation of the same section. In many cases this is true. However, it might be well to remember that stress causes strain, and the accumulation of strain over a distance resulb in appreciable movement of some kind: deflection, angular movement, etc. The sharp comer of this connection increases the stress in this area by several times. This stress concentration results in a higher strain 2nd therefore greater movement in this small area. Since only flange stiffeners are added to this square-comered connection, it is difficult to exceed the stiffness of the member. In most cases, it will just equal the member, and in some cases it will be less. 3. PLASTIC D A T A H A S A P P L I C A T I O N There is little test data on the connections used in the Vierelldeel truss. However, data available on the plastic design of comer connections or knees will be helpful.

FIG. 3-In this building addition, use of Vierendeel trusses will provide a columnfree orea of about 30' x 60' for large trucks and trailers to load and unload communicotions equipment.

elded Connections for Vierendeel Trusses

/

5.13-3

FIGURE 4

1

,0002

,0004

.0006

,0008

,0010

Unit angular rotation

Figure 4 shows moment-rotation curves of various comer connections.* The vertical axis is the applied moment; the horizontal axis is the resulting rotation of the connection. The vertical height of the curve represents the maximum or ultimate strength moment of the connection. The slope of the straight portion of the cuwe represents the stiffness of the connection, with the more nearly vertical curves representing the stiffer connections. The right-hand extremity of the curve represents the rotational capacity of the connection. In plastic design, it is necessary that the connection

-

*Figure 1 adapted from "Connections for Welded Continuous Portal Frames", Beedle, Topractsoglou and Johnston; AWS Journal; Part I July 1951, Part I1 August 1951, and Part 111 November 1952.

,0012

.0014

,0016

.C

(4); rodions/in.

havc high rotational capacity in addition to exceeding the moment capacity of the member. In Vierendeel trusses, it is more important that the connection have a stiffness equal to or exceeding that of the member, and a high moment capacity in order to safely carry accidental overloads. Here the extra rotational capacity would not be as important because it is an elastic design rather than a plastic design. In Figure 4 notice that the square-comer conoection is the most flexible. It falls slightly short of the beam itself, but does have the greatest rotational capacity. The comer with the bracket has greater stiffness and higher moment capacity, hut less rotational capacity. Tapered haunch knees, not shown here, were found

5.13-4

/

Welded-Connection Design

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 corner 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. UARE 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 membcr. 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 Bange forces.

The unbalanced moment about a connection will cause shear forces around the periphery of the conneo tion web. Fignre 6. The vertical shear force and the horizontal shear force will result in a diagonal compressive force applied to the mmection web. Unless the web has sufficient thickness or iri re~nforced,it may buckle. According to plastic design (and this may be used in elastic d e s i p ) , the required thickness of the joint web must b e -

and:

FIGURE 6

Connections for V i e r e n d e e l 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 equals the yield shear stress (Tp),orand ?;r = 7,.

FIGURE 7

t, = thickness of connection web, inches

f, = unit shear force, lbs/Iinear inch = T t, dh = depth of horizontal member, inches d, = depth of vertical nmnber, inches

Web doubler plate

FIG. 8 Methods of obtaining web thickness to meet requirement of Formula #2. (a)

W e b of connection reinforced with web doubling plate

Diagonal stiffener

(b)

W e b of conneaion reinforced with diagonal stiffeners

(c)

W e b of connection reinforced with longitudinal stiffeners

5.1 3-6

/

Welded-Connection Design

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 flat plates standing vertically h&ween 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 ( u ) . This stress is maximum in line with the web. In addition there is a transverse tensile bending stress ( u , ) 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 o: and p come from the graph, Figure lo.*

longitudinal tensile stress in flange

5. CURVED-KNEE CONNECTIONS Tensile stress (urn,.,) 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)

transuerse tensile bending stress in fhnge

......................... M

radial force

.............................( 5 ) The radial force (f,) acts transverse to the m e t welds connecting the flange and the web. *From "Design of Rigid Frame Knees", by F. Bleicb, AISC.

Rad~oit e n s k force (f,j

FIGURE 9

elded Connections for Vierendeel Trusses

/

5.1

Values

FIGURE 10

ARY OF REQUIREMENTS Here is a summary of the general requirements for these Vierendeel truss connections: 1. The bottom chord is in tension and the connections here must provide continuity of the member for this tensile force; the top chord is in compression and the connections here must provide continuity of the member for this compressive force. For these reasons, the inside flanges of the horizontal chords should be made continuous throughout the connection. 2. There may be some axial tension or compression in the vertical member, but this is usually of a smaller magnitnde. 3. Large moments are applied by the horizontal and vertical legs to each connection. 4. A pair of connections, one above the other, tend to he restrained from rotation by the vertical member which connects them. The rotation of these connections

Vierendeel trusses in this addition to the New England Life Insurance CO. home office building permined orchiteci to match window openings in original buildings, yet accomplish significant savings in steel ond in floor spoce. Design also provided stiffer construction, reducing d a n g e r of cracked masonry.

Valuer

due to deflection of horizontal and vertical members is taken into consideration when the truss is designed. However, yielding within the connection itself is not considered in the design and this could alter the moment distribution of the truss, therefore it is important tlrat the connection have equal or greater s@ness than the members connecting to it. 5. The web of the connection must be stiffened against buckling due to the high shear stress resulting from the unbalanced moment of the two horizontal members connecting at the joint. This difference in moment is equal to the moment applied by the vertical member also connected there. This web must either have sufficient thickness or be reinforced with a doubler plate or some type of stiffeners. 6. Flange stiffeners should be used whenever there is an abrupt change in direction or curvature of the flange.

5.13-8

/

Welded-Connection Design

Use of Vierendeel trusses here provided a column-free area of about 30' x MY for lorge trucks ond trailers to load and unload communications equipment.

1. METHODS OF A

There is no single best method to analyze statically indi~tcrminatcstni~.tiircs.l'lirre arc- many mcthods, and muny comhirintions and adaptations of these methods. One mc~tliodmap h r siinpli~and (pick, h i t can only be used to a lirnitcd cxtt:rrt. :'mother mcthod may have wide applicatioir, h t I>(: so lai~orious that it is not used inuch. Most tcxts on stritically indctcnniriate structures start ont witli the various m(*tliotlsof dctcrmi~iingdeflections of thr s t r i ~ c t ~ ~ rThcy c. them corrsider the analysis of thesr strnctnrixs. Thc nicthoc1s of finding deffoctions arc siniple tools which may he usad in tho analysis of ihc strr~ctni-c. 1Prvrc arc nctunlly ahout five basic, mrll uscd methods for the :tn:ilysis of sieticnlly inclrtrrnrinate structures tmcorrnt~ndin rigid frame designing: I. Least Work Mt:thod 2. Central Mctliod 3. SIopv I>~.flrction hlctliod 4. hlo~nentand Shcar Ilistribution Mrthod 5. Column Analogy hlcthod All of these mrthods. when applied to continuous beams and frames, give the resulting bending mements at various points along tlw structure. In order to proct:cd this far to get the n,sulting niornents 011 thc stnictorc, it is first ncwssary to assum? thc moments of inertia of the rn~mhcrs.This is 11suallya good guess or appn~xiinatioir,Tlren, from t h < w resr~ltingbcnding mornmts, the rncmt>er is built up. If the find reqiiirctl m o ~ n c i ~oft ir~crti;~ is more than that which u-as started with, thc work mlrst be repcatcd; or adjusted, using this niwcr v:llne. I11 sonrc tnetlmds orily the ratios of the mrious rnorncrits of incrtia rrctd he wed. ethod of Least W o r k

The nictlrocl of least work dqmids on tlrc follow-ing. It is coi~sidcrcd that a strncti~rr.will deform under the nppliattion of n lo:ld, i r i such a inanncr that the intrrri:d \\-ork of drforin;itii~nwill hi. held to a minimum. This inethod may be oiitlini~las follows: 1. Cut tile strutiuro so that it hcconrc~sstatically det~mniiiatr. 2. 1 ' 1 1 ~ uuknown monic~lts or forces become the reduridar~tsor unknown qnantities. 3. Set up an equation for the internal work of the

stri~ct~rre in tt,rms of these r e d i d a n t s . 4. A derivative of this is then set tyn:rl to zcro, and this will give thr minimum vdue of this redundant forct,. General Method

The gmrral mcthud consists of the following: 1. Cut the stniehire at the reduildnnt or urrlmown forcr. 2. Di:ter~ninethe oprning of this gap caused by tho givm h a d (while cut). Several methods may be usad to find this dr&ction. 3. Apply a rcdondat~tforw to close this gap. 4 Fmm t h ~ .given 1 1 d s m d this redundant force, makc np a momc.nt diagr;m and design thc sirncture from this. For more thm one rt&uidant force, cut a11 members at thew rednndar~t forccs arid close the gaps simulii~nrously. To usr the general method, the designer must be able to find d14Irctions in Step 2. Sorne of the methods for finding deflections are as follows: ( a ) R c d Work ( b ) Castigliano's Theorem ( c ) Virtual Work ( d ) Arca Moment ( c ) Conjugat~tDearn ( f ) ilngle Wc,ights ( g ) Willot-Mohr Diagram Scvcral of these int~thodsan. described in Section 2.5 or) Dt.i3rction by liendir~gand will not he discussed h~w. Slope Deflecfion Method

In the general mcthod jnst outlined; the redundant or i~nknoa-nforccs : i d moments are fonnd. In a similar rnimnor, it is possiblr to solve for thc onknown joint rot;ltions and dcflrctions. As soon as these are found, tho end moments may he dctrnnintd and these cornh i n d mith the original niomcnts from the applied load. oment and Shear Distribution Method

The moment distrihntiorr or 1I:irdy Cross method consists of lrolding thc joints in a frame fxed so that they cannot rot:ite. Tlrr: m d moments of cach loaded membcr are forind f ~ m nstandard hram diagrams in handbooks. Thcn, onc ;it a timc, a joint is rcltvscd, allowed

6.1-2

/

Miscellaneous Structure Design

(b) Elastic area of a n a l o ~ a u rcolumn

(a) Actual Frame FIGURE 1

to rotate, and the11 fixed again. This relcase causes a new distribution of the moment about this point, and somc of this change is carried over into the next joint. This proccdmc is followed for each joint in the entire frame, and then the whole process is repeated over all the joints as often as required rmtil these corrections become very small. This method is outlined as follows: I. Fix the joints from rotation and find the moments, trcating the member as a simple structure. 2. Remove thr joint restraints one at a time, and balance moments about the joint. This unbalanced moment is then distributed about the joint. 3. Some of this distributed moment is then canied over into the other end of the member. 4. This is repeated until the unbalanced moments become very small. The final moments are then used to design the structure.

corrective moments which must b e added to the statically determinate moments of the "cut" frame in order to bring the frame back to its original shape and condition before it was "cut". This is outlined as follows: 1. Determine properties of the elastic area: area, center of gravity or elastic center, and moments of inertia about the two axcs (x-x and y-y). 2. Cut the frame to make it statically determinatc. Use moment diagram from applied loads as a load ( M / E I ) on thc clastic arca of the analogous column. 3. Determine axial "stress" and the two bending "stresses" of the analogous column. These become corrective moments which must bc added to the statically deternminatc moment of Step 2 to give the final moments of the statically indeterminate frame. 4. From these moments, find the redundant forces at the cut portion of the frame.

Column Analogy M e t h o d

2. COLUMN ANALOGY METHOD

The outline or over-all shape of the given frame is considered as a column cross-section, called an elastic area. Thc length of each portion of this elastic area is equal to the actual length of the corresponding member of the frame. The width of each portion of this elastic area is equal to the 1/EI of the corresponding member of the frame. The properties of this elastic area are determined: area, center of gravity or elastic center, and moments of inertia about the two axcs (x-x and y-y). The statically indeterminate frame must be cut, usually at one of the supports, so that it becomes statically determinate. Under this condition, the moment diagram caused by the applied loads is constructed and then treated as a load (M/EI) applied to the elastic area of the analogous column. Just as an eccentrically loaded column has an axial compressive strcss and bending stresses about the two axes (x-x and y-y), so the analogous column has "stresses" at any point equal to the axial compressive "stress" and the two bending "stresses". These resulting "stfesscs" of the analogous column are the

The outline of the givol frame is considered to be a column cross-section, called an elastic area; Figure 1. The length of each member in the elastic area is considered cqual to the actual length of the corresponding member of the actual frame. The width of each member in the elastic area is equal to 1/EI of the corresponding member of the frame. It is seen by Figure 1 that for a pinned-end frame the moment of inertia of the flexible pin is zero. Hence the width of the elastic area at this point is

and the elastic area at this pinned end would equal m . For a fixed end, the moment of inertia at this rigid support is assumed to be m. The resulting width oE the elastic area at this point is-

Design of Rigid Frames

and thc v1:rstic :Ira at this fixed end would he zero. The, ('1astic area, with its dinwnsions riow known

L 111,ight =: 11

I,mgth

is now t r c ~ ~ t c Iikc d any other cross-stxction, and its propertics d~~termirird. i i i ~ ~ ~encis: iid I

Area

( 2 coluinns)

ill<,t\vo pitintd ends isrid thcst. lir :ri the, extreme ends of 1111, svction ;~lroutaxis x-x.

(beam)

T l ~ cstatimlly intlct~~miirrate frarrrr, I'igl~re 2 ( a ) , lnlrst h a w sonic portiorr cut, r~srially;it orrc of the stipports, so t11;1ti t tiwornr:s st;itifi~lIydt,tr:rrnin;rtc, Figrrl-c 2(1)). ITiliicr this eoirditiori, the. f,r,~iding momcnt diagram npi11ii.d 1o:irls is coristruct~~d,i;igur<, c ; i r ~ s ~by~ t I 2 j c ) . 'I'his is tlwn tr(.nttd ;rs ;I 1o:rd (hl,jCT) applied to tht ixl;~stiearm of ilrc :rnaIogous column, Figur(. 3(a). J I I S;is ~ an ccc~mlric;rllyloaded iollnnn has an axial load mri tilting mom~iils(kt, = I-' y, and M, = P x),

(pinned ends)

Elastic Center

The elastic center is fo~indas though it were the center of gravity of the elastic area. axis x-x

Taking moments aliout the base line, it is seen that the. i,l;~sticaxis s-x of the c1:rstic area must pass through the frame basc since, in the nrialogoos column, thc. p i n o d ends h a w irifi~ritc( x ) mea. This may be proved by mathcmatically determining tiit, elastic cmtcr of gravity:

[ o ) Statically indeterminate frame

(b) One support rut t o make

axis y-y

frame itotically determinate

By ohscr~viiorr,it is sccn that thc y-y axis wo~rld pass thmugh the wnter of this elastic art:;^ h~.causcof section symmetry. oment of inertia

I,, = 2

6.1-3

Apply Load to Elastic dreo

:

h this cx;r~nrileof

/

(

( 2 coluinns)

-b

(-I<: I[ ) 1. 19 + 2 (i) 0 (beams)

(pinned ends)

li) Moment

Since the infinite ekistic arcn a t thr pin lics along the clnstic axis x-x, it will hnvc no effwt upon I,.,. I p ~= > V-, since there is ;in i n h i t c elirstic sea at

clioijrom for the

stoticnlly delertninote frome

FIGURE

2

6.1-4

/

Miscellaneous Structure Design

(a) Analogous column loaded with

M E l

(b) Actual column with eccentric lood (P)

FIGURE 3

so the analogons column has an axial load and tilting moments. Consider the moment diagram dividcd by EI as the load about the two axes (x-x and y-y) through thc elastic center:

Just as the eccentrically loaded column has stresses at any point equal to the axial compressive stress plus the two bending stressesu = ua (axial) t IT, (bending,.,) i o;. (bending,.,)

axial loud on anologous column

P - -.

a -. b 2 E IL

moment about axis x-x on analogous column

-

so thc analogons column has "stresses" at any point equal to thr axial "stress" (ma) plus the two bcnding "strcsscs" (us & u,). Thrse are the correspouding corrective momonts (M,, M,, & M,.) which must be app l i d to the statically determinate moments of the "cut" franic in Figurc 2 ( b ) lo bring thc frame back to its original shape and condition, Figure 2 ( a ) .

P a b h -2 E I,,

moment about axis y-y on unalogous column

=0 -

P a b ( b - a ) (4b 12 E I, L

+ 4a - 3L)

(See Figure 4.)

Design of Rigid Frames

/

Y

FIG. 6 No corrective moment to be added here.

FIG. 4 No corrective moment to be added here.

when c, = 0

M, = 0

(See Figure 5.)

. .. . . .. . . . . . . . . ( 3 ) Since I,., = oo

M, = 0

(See Figure 6.)

Thc final moincnt on thc frame will be as given in Figure 7.

FIG. 5 Corrective moment to be added here.

FIGURE 7

6.1-6

/

Miscellaneous Structure Design

-

________PI

48'

FIGURE 8

I

-

I

48'

Y

-------q

FIGURE 9

Member

Calumnr

Find the moments ( M ) and the other rt!dunttant forces (I3 and V ) of the following frame, having fixed ends, by mcans of the Column Analogy hlethod; Figure 8. This frame must he transformed into the analogous column, and the properties of this eq~iivalcntelastic area determined; Figure 9.

Rafters

+5.0

+ 147.5

Total

433

+260

2856

= $ 2.2' measured from reference axis (x'-x' )

Use a reference axis (x'-x') through the top of the column.

= 2856 - 325 = 2531 i n 4

Design of Rigid Frames

--

- lo'

/

Moment of inertis of rafter about its own center of gravity

FIGURE 10

distance from elastic ccntrr (y-y) to outer fiber

distance from elastic center (x-x) to outer fiber (bottom) c, = -15 (top)

c, = +I0

axis y-y of 'Iristic centcr

(right side) c, = +24

- 2.2 = -17.2' - 2.2 =

(left side)

+ 7.8'

Cutting Frame So I t Becomes Statically Determinate

(Sce Figure 11.)

H" observation it is seen that this passes through thc cmtwlinc of the frame: I,~,

-

2(7.5)(24)'

+ 2(26)(1Z2) + 2(1248)

( 2 cohlmns)

= 18.624 i n 4

( 2 rafters)

c, = -24

Thc frame is now crlt so that it becomes st:itically determinate. The resulti~rgmoment diagram, divided hy the real momt,nt of irwrtia ( I ) , is trratcd 3s a load upon thc nnalogous coh~rrrrior elastic nrca. (We don't divide by E here bcc;ruse E is constant; for stecl, E = 30 x 10") This may be done in several ways, principally:

FIGURE 11

6.1-8

/

Miscellaneous Structure Design

Method A: Cut the frame at @. With the load applied at 0, the raftcr cantilevers out from @. The end moment at @, M = -60,000 ft-lbs, is also applied to the left column 0-0.(Sec Figure 15.)

=

+ 4,571,700

My., = ( -- 450,000) ( -- 24)

- (-390,000)

(-20)

=i 18,600,000 correction moment at @

--.I." --

the three loads on elastic area

-

+ (--390,000) (--.53)

My.,= (-450,000)(-9.7)

A. Cut thc right fixed end support at @. The portion of the rafter to the left of the applied load becomes a statically determinate cantilever beam. B. Release the ends of the rafters at @ and 0. This becomes a statically determinate simply supported haunched beam.

-840,000

-

P

M 7 .-x c ! F

LX

.4

c -

Mr., c, 1.w

-

+-

- 67,570 ft-lbs

W TO DETERMINE C RRECTIVE MOMENTS (Diagrams Apply to Option A) The moment diagram divided by the moment of inertia of the statically doterminant frame is considered to he the load on the elastic area of the analogous column. ( E is constant.)

These loads, in turn, result in 3 types of resisting stresses":

"

P

O"

=

cx =

a

M..

c7

LX

or

=

----M,., c, 1Y.Y

The resultant "stress" at any point of the elastic area may he found from the conventional stress in an eccentrically-loaded column:

P ; =a

M F . 7 CY

1,-

Mx-x cx

$ 7

These "stresses" are the correcting moments, which must be applied to the original moments of the statically determinate frame to produce the final moments of the statically indeterminate frame. FIGURE 12

This total load on the clastic area may be broken down into 3 loads:

X

a. Axial load, P b. Moment, M,

.,about axis x-x

c. Moment, M,.,,

about axis y-y

FIG. 14 Correcting moments

/

Design of Rigid Frames

6.1-9

FIGURE 15

final moment = original moment

-

correction moment

MI = - 60,000 $- 67,620 = --1- 7570 ft-lbs

corrcciion moment at @ .

-

-840,000 --

~

2531 (+18,600,000) (0) + .. .. 18,624

I c = - 2.2'

- P - A .-

-

+ ?.E~~(2 +. My.,r, L 1,~~

---840,000

67

- 24,

cx

(-2.2) + ( +4,571,700) 2,531P-.-

(. 4-18,800,000) -24) + .. (-18,624

-

-

final moment

correction moment at @

P

original moment

-

corrtvtion moment

correction moment at @ I' A

= + 1550 ft-lbs final moment Mh = - 1550 ft-lbs

40,480 ft-lbs

-

Ix~,

- -

- A

= +

+

hl, 1,

,. c,

+ 7.8'

.+ (+4,571,700). (4-7.8) . . . -.

67

correction moment at @

=:

P + . M'"L c + -M,, c,~. A Lx L,

-

c

r- 3.8' c, 7-- 1.7' -

-

4..

Ms.,c, I,~,

:

1~

h , I , , c, I,%y

c, = - 2.2'

.

+ 7460 ft-lbs

-1

final moment

ME =

-- 7460

ft-lbs

(

=

- 19,640

ft-lbs

+18,600 000) (,!?A)

+ -78,624

6.1-10

/

Miscellaneous Structure Design

+ 19,460' '

FIGURE 16

final momcnt

M6 =

H. = .

+ 19,640 ft-lbs

The- final moments of the statically indeterminate frame are di;rgramn~ed in Fignre 16. ~~

.MI h

7570 ft-lhs 4.191 - = 1806 lbs

-.

Horizontal Redundant Force To find the horizontal rednndant force ( H ) at the base of the column, first find the point of inflection (zero moment) in thc column. Then find the horizontal force required at this point to equal the end moment at the base of the column.

FIGURE 17

Vertical Rmciion To find the vertical reaction ( V ) at the base of the column, take the inomcnts about the base of the opposite column and set them equal to zero. (See Figure 18.)

Method 8 : Relcase ends of the rafters at @ and @, so that the rafter hecomes simply snpportcd and statically indeterminate. (See Figure 19.)

the three loads on elastic area

Design of Rigid Frclmes

/

6.1-1 1

FIGURE 18

M,.

+ (97,500) (-k4.47) + (390,000) (+5.3) + (300,000) (+4.47) = + 4,576,650

= ( j-292,500) (+1.13)

M,., = (+292,%0) (-16)

final moment MI = 0

(97,500) (-8) (390,000)(+6)

+ (390,000)(-6) +

c, = - 2.2'

-*+- -.P

correction moment at @ M,~, --2 L

-- +1,170,000

m

-

c

X

+ -M,~,

67

c,

( --17.2) + (+4,576,650) 2531

- 7600 ft-lbs

Ix.x

17-s

=

~~~~-~~~

(-4 (-24) + --' ---!... 680 000) ;, 18,621

M ,., c,

M,.,c,

+

7,;

+1,170,000 (+4.576,650) - L ' .-

c, = - 17.2'

+

original moment - correction moment

correction moment at @

- - 4,680,000

- -P - A

r

+ 7600 = + 7600 ft-lbs

+ 19,520 ft-lbs

final moment

MI. =

FIGURE 19

- 19,520

ft-lbs

(-2.2) 2531 (-4,680,000) ( -24) ~+ - .. 18,624

/

6.1-12

Miscellaneous Structure Design

final moment

correction moment at @

c, =

+ 2.8'

= -PA+ - M,.,Is.xc,

+ +

Me = 0 19,670 = 19,670 ft-lhs Alternate Method

( ---4,680,000)(-12) - 18,624

+=

+ 25,540 ft-lbs

final moment

Ma = =

+ 45,000 - 25,540 + 19,4% ft-lbs

corrcction moment at @ c, =

---+: P - A

+ 7.8'

M,~, c, + . M,., c, LX 17.y

It is possible to work this problem in a slightly different manner. As heforc1. Determine the properties of the elastic area. 2. Cut the frame to make it statically determinate, as before. 3. Dividing the moment diagram of this cut frame by the moment of inertia of the corresponcling members of the frame, treat it as the load on the elastic area. ( E is constant.) 4. Find the resulting three parts of this load on the elastic area; that is, a. Load, P b. Moment, M,., c. Moment, My., Then find the three corrective actioi~s-fixed end moment (MI,), liorizontal force ( A ) , and vertical force (V)-which must bc applied at the base of the frame to bring it back to the original shape and condition of the statically indeterminate frame. Find these from the following formulas:

final moment

Mg =

=

+ 30,000 - 31,560 -

1560 ft-lbs

correction moment at @ -

P + M;" CY A x-I

+

My-,,

~2

c, = - 2.2' c, = 24'

17.1

= i7450 ft-lbs final moment

Mg =

-

7450 ft-lbs

correction moment at @

c, = - 17.2' P M c M,., c, - - + 2>xA + -:.~ - A Ix.x Ir.y +1,170,000 (+4,576,630) (-17.2) 2531 67 (+N) + (-4,680.000) SS,6% - - 19,670 ft-lbs -

+

Figure 20 shows their application to solution of the immediate problems. The resulting moments ahout the frame for each of these mrrective actioxs are dctermincd and placed for cnnvenicnce in tahlr form. This facilitates totaling then1 to produce the final moments at any point of the statically indeterminate framc. See Figure 21. 3. FIXED END MOMENTS, STIFFNESS FACTORS, A N D CARRY-OVER FACT0 When some type of moment distribution is used for the analysis of continuous frames, it is necessary to know the following: I. Fixed end moments (Mi,) of the beam. 2. Stiffness factor ( K ) for each end of the beam so the distribution factors may be determined. 3. Carry-over factor ( C ) of a moment from one end of the beam to the other end. These items may he found from already-developed charts, or by use of the column analogy method which

Design of Rigid Frames

/

"+---X

-

compression

H= 1806 M,=ticl=[-1806)[-17.2)=+31,062'ti

M = -

i ..A~ M, = Vc,: FIGURE 20

/

6.1-13

6.1-14

/

Miscellaneous Structure Design

0

0 -60.000

M,

-60.000

..~

0 0 ~

~~~

@

0

0

0

0 0

----

~

P Mrr = - +12,537 A -

M3 =

+ 19,465'*

H V

. a

Total

y x

+12,537 +12.537 1-12.537 +12,537 ~. -.. 3,974 - 5.057 -14,008 +31.062 3,974 -~ . .. -~ -23.969 +23,969-11.~4-~--~ 123,969 .+---. .. 7,SM -19,520 +19,465 - 1,551

.+

/

1

+

M,

= - 7458'

+

i12.537

--

i +31.062

~ ,' g f i

1

23,969

8

FIGURE 21

is applicable to any type of beam, Figure 22. The cover-plated beam is representative of any beam in which there is an abrupt change of scction . . . and of mome,nt of inertia. The other two common conditions in which there is an abrupt change of scction are 1) where plate of heavier thickness is used for the flanges for a short distance nt the ends of the beam, and 2 ) where short lengths of smaller beams are used below the regular beams to reinforce them a t and near the points of support.

Charts have been developed by which the designer can readily find constants to use in determining stiffness factors, carry-over factors, and fixed-end moments for beams. Sources include: I. Bull. 176, R. .4.Caughy and R. S. Cebula; Iowa Engineering Experiment Station, Iowa State College, .4mes, Iowa. 36 charts for beams with cover platcs at ends.

Prismatic

-

-

Constants to Help Cwlculwte Finwf Moments

-

Cover plated beam

Topered beam

/

\

Hounched beom

@=+ FIGURE 22

Design of Rigid Frames

2. "hloment l)istrih~rtion," J. M. r I ; D. Van Nostrnnd Co.. 378 pages; 29 clixts for braox \vith covcr plates at ends; -12 chal-ts lor tapwcd hmrns. 4. F I N D I N G FIXED END M O M E N T S BY C O L U M N ANALOGY

Roftving hack to Topic 2, Thc Coliiml~ Analogy h4ethod. the outlinc of tlie hem1 is ronsidercd to be the cross-section of ;I colmnn ( o r elastic zrrcn). See I'igurc 23.

/

6.1-15

3. The rtwiltirig "strcssts" at thc ends @ :rnd

@

I~i,comcthc correction momrnts which n111st he added ti? ilia mommt of the "simply-s~rppol-[rd" hemr to transform it back to the original fixed-wd, statically indeterminate bcarn. Sinw i n this case \VI. started o r ~ with t zcro cnd m n m ~ n t sfor thc "si1nply-srrplx1rt~~d" hcam, these corri.ction ~iioinrwtstbcr~1)rcornc thc fixed m d moments of t l ~ efinal I-igid hram: M,,, at c.nd @

k-

hir,, at cnd

@

Stiffness Factor by Column Analogy --

"

Real

beam

The stiffness factor ( K ) is a measurt? of tlw resistance of the mrmbrr against t ~ l drotation. i t map be defined as the moment necrassary to produce a irnit end rotation at the same end, wl~ilethe opposite end is held fixed: K* = MA Carry-Over Factor by Column Analogy

Y

Elostic area

FIGURE 23

The length of the clastic area is equal to the length of the veal beam, and the ~ i d t hat any point of the elastic awa is equal to tht: l / E I of the r r d beam at the corresponding point. Since \vex arc: draling with steel, the modulus of elasticity ( I ? ) is constant and will drop out of tlrc calculations. As the depth and moment of inertia of the real beam increases, the i.lastic area decreases.

8

For any applied moment ( M A ) at A , the resulting moment ( M A ) at the other rnd is determined. The carry-over factor is the ratio of thcse two moments: MAIJ C* = - -MA In both of these two c;lscs, Stifl'ness Factor and Carry-Over Factor, the fixed-end beam is rr.leased at one end @ :md rotatrd througli a unit angle chango ( 4 ) . The restilting end moments ( M A ) at @ and (M,,,*) at @ are found.

Thc following dcs~gn procedure may then he followc d. 1. Detcmmi~rethe propcrtics of thc clastic area: a. Area of the cl;~sticarm ( A ) h. Iacation of axis y-y through the elastic center of thc elastic ;area. c. Distanct. iron1 the elastic center ( y - 1 - ) to the outer fiber-s of thc cltistic nrca ( c . , ) tmd ( c , , ) . d. Momcnt of inertia of the elastic area (I,.,).

2. Kelexse both ends @ and @ of the fixtdcnd beam and draw the moment diagram of this "simplysupportcd" hcam. Use this moment diagram, divided by El, as thc load upon tht: elastic area (analogous column).

FIGURE 24

This unit ;iilqIc rotation is a p p l i ~ das a single load at thc nutcr edgy of the clastic :rrm (analogous column), just as ;in
6.1-16

/

Miscellaneous Structure Design

5. COLUMN ANALOGY METHOD APPLIED TO

BEAMS HAVING ABRUPT CHANGE OF SECTiON The Columtr Analogy Method will now bc uscd to find the Iixcd end moments, stiffness factor, and thc carryover factors for a fixed-end beam with cover plates at one end, supporting a uniform load (us).The technique would bc applicd similarly to any beam having abrupt change of section. Figure 25 diagrams the rcal loaded beam, at top, and the elastic area of an analogous column, below. On this dastic area,

TABLE l-Column Load of the "nil mgle ihonge

=

Ana

y:

Unit Angle Rotation

1

-

x

@

elortic aieo jonalogous

Unit m g I e chonge i!oodl ploced at

M ,., at

@

0

Load F placed ot M ,., = F a*

A

= I

c*

ot

A

@

"*

=

--F A

+

M,-,

ca

F - --

3-7

tood F placed at M, = F ca

,

Ot

0

@

A

@

+ *F c Z 17.7

Design of Rigid Frames

/

length = actual length of beam

STEP 1: Dcterrnnre Properties of thi5 E l a \ t r ~Area

moment of inertia jl,~,j

area

elastic center (y-yj Take momwts about

@ STEP 2: Iktermine the Fixed End Moments the bcain are releascd so snpported. This moment diagram now becon~es the load on the elastic area, Figure 26.

TABLE 2-Loads

and Their Eccentricity

Load (Pal or (Pi1 o f portion of marner! diagram

Distance t o c q of this load

L = a + b

w =

u n i t uniform laud (Ibr/in.!

Distance to CG of this o o d

8

w = unit uniform load lIbi/in.!

6.1-18

/

Miscellaneous Structure Design

Moment d i a g m

M,

=

= (L - x ) 2

Load d i a g m m jM,/l,) on elortic a i m

Now the load of a unit angle change ( 4 ) ) is a plied to the elastic area at the other outer edge and thc resulting end moment (M,*) at n is found. Notice that the end mommt (Mu,) at A is equal to (M~L,,) at @ which is already found.

8

MY.? = 1 ci3 FIGURE 26

at

@ Mn

"uxiuP' load (P)

--

1 A

--

1 cn2 *. L - Y

From thcse three values ( M A ) ,(MAB)and (M"), the iollowing may he found: stiffness juctor at

P = P,

+ PI,

K, = M, stiffness factor at

8 @

K, = M, fixed end moments

curry-over fucto~, @ to

@

This load ( P ) and (M,..,,) on the elastic arca causes "stresses" similar to those on an eccentrically loaded column. These "stresses" become the correction moments, or in this case the end moments of the fixedend beam.

STEP 3: Determine Stiffness and Cnrry-Over Factors

A load of a unit angle change (+) is applicd to the , and the resulting elastic area at the outer edge end moments (MA) at @ (Mar,) at @ are found.

M,.,

= 1 c*

FIGURE 27

Design of Rigid Frames TABLE 3-Design

/

8.1-19

Summary: Beam Cover Plated A t One End

C,, Cu,e ,

+

eb

conridered t o be

are

I+)

Fixed End Mamentr

End Momentz Resulting from Treating Angulor Ratotion or

o

Load

Stiffness Foctoir

Carryover Foctois

6. COLUMN ANALOGY METHOD APPLIED TO BEAMS H A V I N G GRADUALLY VARYING SECTION Summary

cxnll.,ple of tilt: uIlifor,ll~y.~o~I~jc~C~,fixt.&e,l(i ljeam slsnril,ar~edas Lvith l,latc.s at one t,l,d , l l n y in T a l k 3. Modified E x a r n p h

Althonglr the woi-k is not shown, the s:nnc3 fixcxl-cnd tmnn n.it11 coucr pl;ttcs at both r d s , uniformly loatled, may bc sommarizcd as in Table 1. (Sce lrcrt page)

The lollowing method may hc rised to find the fixed end monicnts. stiffncss riictors, and carry-over factors of 1jc~a1n~\r.hicIi 11avt~constantly varying moments of incrtin, s~iclr21s 1 ~ 1 1 n c h r d;ind taprred bmnts, Figurt: "" 21.

A hcanr tvbich tapws along a straight line (in otlwr woi-(Is. its clvpth i n w x s < , s 1inr;rrly don:: t l ~ : letigth of tlw brarn; se? Fig. 25, t o p ) will liavc a ~nomcntof iiwrtia ( I ) ulrich docs not increase linearly

6.1-20

/

Miscellaneous Structure Design

TABLE 4--Design

Summary: Beam Cover Ploted

At Both Ends

6I L

3a1

iiiiiiiiii

M, at A

2

M,,o~B

End Moment. Resulting from Treoting Angulor Rotation an o Load

Stiffness Factors

Carry-Over Factors

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) linc. However, this slight error may be reduced by breaking the beam into two parts (see Fig. 28, bottom) and assumii~ga stsaight line variation of the moment of inertia between the thrce points @, @, and This is represented by the dashed line in Figure 28, center.

0.

STEP I : Determine Properties of the Elastic Area

area of elastic area Ax Az -

a log 1, I" - IA I* b 10 log Ic - In In

moment of elnstic area

M*.//&A

=

(

a D

-

*, about axis A-A

) (

-

-

-

moment of ehstic area A, about axis B-B

I, - I~ /B-B

-

I,%log,

-

Design of Rigid Fromes

/

6.1-21

FIGURE 28

distance from C.G. of elastic area A, to axis A-A

moment of inertia of elastic urea A, about axis A-A

MaJ

distance from C.G, of elastic area A, to axis B-B

Ma./ moment of incrtda of elastic area A, ubout axis B-B monzent of elastic urea A, about axis A-A

total moment of ekstic area about axis A-A

MA.*

MAz/

/*-A

elastic center ( y - y )

-4- MA^/ /A-A

Since thrsc moments of inertia can't be added, not being taken about tho same axis, it will be necessary to shift axis 13-Ii and axis A-A to the elastic center y-y. If axis A-A is always taken at thc shallow end of the tapered beam, negative sigrrs will be avoided in the calculations.

/

6.1-22

Mircelloneous Structure Design

momcnt of inertia of clustic area A, ubout axis y-y Using the parallel axis theorem:

=I

I

-I-

/A.A

A, c,"

/x.x

- A,

Tax/ = IAJ

..

women (A{) applied to elastic area about its elastic crnlcr

Q~

/A-A

/x-x

Now we wish moments of inertia of A, about the elastic axis y-y, and again using parallel axis iheorem-

I*J

= TAX/

or I&/

= IAx/

/m

andLJ

-fAX(CA-

Ix.=

/Y-Y

- A,

+ AI(cb - c,)"

cX2

/A-A

=I*=/

/F.F

+A,c~(cA-~~,)

/&A

moment of inertia of elastic area A, about axis y-y in same manner-

where: total moment of inertia of elastic area

IF.7=

1AX/

/w

4- IAS/ /w

STEP 2: Determine the Fixcd 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 ( M / I ) on the elastic area which is treated as a column. The axial load ( P ) applied to the elastic area is equal to the total M/I. This axial load applied at some distance from the elastic center of the elastic area causes a moment ( M ) on tht! elastic area. 'Both of these loads cause "stresses" on the elastic area. Tlrc following applies if the designer can assume a-unifonn load ( w ) :

and the total moment-

axial loud (P) applied to elustic area

STEP 3: Determine Stiffness and Carry-Over Factors

MA= and,

P = P,

+ P.

-A1- + -I,,c Z

Design of Rigid Frames

/

6.1-23

Elastic oiea

FIGURE 29

stifness factor at

@

Ka = MA stiffness factor at

@

Then procecd first to find formula elements made up of these properties:

Kc = M, carry-ooer factor,

Cao =

@

to

@

I, = log, (2540) log, - - - log, 3.9276 IA (fi467)

Mac -MA

carry-ouer factor,

@

log, n = 2.3026 loglo n

to

@

Cca = - -MAC Mc

For the uniformly-loaded lmrm shown at top in Figure 30, having fixed crids, find the fixod end moments, stiffness factor, and carry-over f:ictors. At center in Figtire 30, the solid cmve is the actual moment of inertia ( I ) as it varics along the length of the henm. The dashed Iine is the assumed straight-line variation in moment of inertia along the two halves of the tapered beam. The following properties arc established:

Ic = log, (5930) log, -= log,, 2.3346 11, (2530 )

6.1-24

/

Miscellaneous Structure Design

A,

_

b ~~~~

(L-

~

log, -~11. 11,

11%)

STEP 7: Determine Properties of tlrf Elastic .4rea area of elastic urea

- i L

4-

1 = 200"

Topered beam

Moment of inertia

1

1 1 = c,

134.30"

Elastic orea

0 FIGURE 30

Design of Rigid Frames

I../

--

11,

,>

(( I < " I n ) )<( ( I . -

- (

1

2

/

6.1-25

-

3 113)

6.1-26

/

Miscelloneous Structure Design

(IH - I A ) ( 2 12

- 7 In IA +

"I

11 I*") - la" loge1,

STEP 2: Determine tlic Fixed End Moments at

@

M,, =

P A

-

4-

Mr., % 17.F

STEP 3: Determine Stilfness and Carry-Over Factors

+ [a(2 L - 3 a )

-

c A ( L -- 2 a ) ]

stifncss factor at

@

KA = MA = 25.67

Design o# Rigid Frames

/

FIGURE 31

a

@

stiffness factor at

Problem 3

Kc = Mo = 74.59 --

@

carry-ouer factor,

to

@

Cat = - MAC MA

-

- -.(-

21.18) (25.67) ~~~

= -,825 carry-over jactor,

@

to

@

Cca =

-

MAC hi,.

-

-

(-21.18).. ~. (74.59)

=,284

For the hannchod hcam at top in Figure 31, having 6xcd ends, find thv fixed end inorncnts (rmiformly loaded). stifrntxs factors. and carry-over factors. Dn:ak h r n into sections :md me ~irinrri?alintcgralion. Thc elastic ; i r a could bc d i v i d d into rectangular aroas, as ;it ccnicr in Figr~re 31, iinil the resirlting propcrtirs of ~ l l crlastic area found in this mmncr. Of courst~s o m ~tmor will lx, introdirccd 1xr:c;mse these rectang~~lnr arras do not qriitf cqr~iilthe actual curve of the clastic a]-ea. I-Iowt:vr~r,as thc ~lnrnhrrof divisions is incrcastd, this error will dccreasc. \Vithoi~tany ;rdditional \wrk, the following mcthod ~villmorc r~rarlyfit the outline of ihc elastic area and will r ( ~ u l tin lcss error. See iowor dingr:~m, Figure ,31. The ( r i r v d portion n.ithin thc clastic a r w is dividcd into irimgrrlar areas. It is noticed that a pair of tri-

6.1-28

/

Miscellaneous Structure Design

angular areas share the same altitude and since the division in lmgth ( s ) is the same, they will have t l ~ e samo area. Therefon:, the cmtcr 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 ineltia along its length).

moinmt of inertia

STEP 1: Determine the Properties of the Elastic Area

elastic center urc,a ( A ) of section @of MJ1, diagram A =

w a'

-

12

(a+3b)

ccnfcr of gravity of section @

= 4644

Momenf[Mjhgram of uniform load

-x = 73h'"

1 = 775.46

8 I

I '

!

I

Y I

Elastic center

I = 1824.71 1 = 142062 1 = 1071.54 1 = 882.33 Moment of inertia

FIGURE 32

Design of Rigid Frames

other propertics of M , / I , rliagrun~ These ;we s h o r n in the table above.

STEP 2: Determine the Fixed End Moments

/

6.1-2

KA = MA = 17.12 .~~~~ ~. .. stiffness fucior nt

@

KO = M O =~26.35~.

7. READY-TO-USE DESIGN CONSTANTS 'he follo\ving 36 charts-appearing on the following pages-giw the fi.~ctf cnd rnomcnls, sliifiirw facirirs, and c a r r y ~ v c rfactors for h a m s wit11 almipt &tnges in nromt.nt of inrrtia ;it111 may hc med for bmnrs with covrr plates. T h ~ ywwc drut4opi.d hy R. A. Canghy, Profrssor of Civil Enginei.ring, Iowa State Collcgc, mcl Ricl1:rrd S. (:cl~l;r.flc:rcI, Engi~~ccring I>t*p:irtmimt,St. M;lrtiii's (X,llrgc~, 7Iicst. charts appmrtil in Dull. 176 of tlse Iowa Enginwring Kxprrirnent Station.

6.1-30

/

Miscellaneous Structure Design

11.000

io.000

9.000

KAE

or KBA

in terms

terms

EI, L

7.000

6.000

5.000

4.000

ID

1.4

1.8

2.2

Chad 1 . Stiffnerr factors ot either end o f ryrnmetr~colbeom

Chart 2. Stiffnerr foctorr ot small end o f unrymmehicol beom.

Ken

in terms

Chorf 3. Stifinell foitot3

large end of vnrymmclricol bcom.

Chart 4. Coiry-over failors for rymmetiicol beam from either end to the other.

Design of Rigid Frames

Chart 5. Carry-over faclorr for unrymmctiical beam from m a l l end to large end.

/

Chort 6 . Carry-over foctarr for u n y m r t r i c o l b r o m from i h g e end to rmall end.

Mm in terms of PL

Chort 7. Fired-end moments of left end of rymmebical beam far concentrated lood o t . l point.

Chart 8. Fixed-end rnomenfr l o a d o f .2 point.

left endof rymmetricoi beom for concentrated

6.1-32

/

Miscellaneous Structure Design

Mns

in terms of PL

Choit 9. Fired-end moment$atleftendof rymmetiicol beom forconcentrated load ot . 3 point.

Chart 10. Fixed-end moment, a t left end of rymmetiicol beom for conceotroted b o d a t 4 point.

MAB in terms of PL

Choit 11. Fixed-end moments ot lefl end o f ,yrnmetrical beom for concent r d e d b o d at .5 point.

Chart 12. Fixed end moments a1 left end of symmetricol beam for roncentrmted loed ot .6 point.

Design of Rigid Frames

/

6.1-33

MA, in terms of PL

Chort 1 3 . Fixed-end moment* at left end of rymmetricol b e a m for ioncen

trated lond

at

.7 point.

Churl 14. Fixed-end moment? 0 1 lei1 end of rymmrtri