Special Report

Lateral Stability of LongPrestressed Concrete Beams

Part 1

by

Robert F. MastChairman of the BoardABAM EngineersA Member of the Berger GroupFederal Way. Washington

As the span of prestressed concretebeams has been growing over the

years, lateral stability during handlingand shipping has become increasinglyimportant. Many designers only con-sider lateral stability of the finishedstructure, which is seldom a problemonce the beam is integrated with a flooror deck. The problems of stability dur-ing construction are left to the fabrica-tors and contractors.

The lateral buckling formulas in mosttextbooks are not adequate to deal withthe special cases of a beam suspendedon cables or a beam on "springy" sup-ports. This paper deals with these spe-cial cases and provides the background

and derivations of the proposed formulas.The material is prepared in two parts.

Part 1 deals with the lateral bendingstability of beams when suspended fromlifting loops. This method was first de-veloped by the author in 1963 and wasthe basis of the PCI Design Handbook'provisions for lateral stability duringIifting, although it appears in modifiedform in the Handbook.

Part 2 extends the analysis of lateralbending stability to the more generalcase of beams whose supports provideelastic restraint to rolling. This includesbeams supported on elastomeric padsand on trucks and trailers, and includesthe effects of superelevation.

34

SynopsisA theory for the lateral bending sta-

bility of prestressed concrete girdersfree to roll at the supports is pre-sented. The factor of safety is depen-dent on the height of the roll axis, theinitial lateral eccentricity, the lateralstiffness, and the maximum permissi-ble tilt angle of the beam. The theoryis compared to the PC! Design Hand-

book and to field experience. Methodsfor improving the lateral stability oflong beams are discussed. A numeri-cal example is included to demon-strate the proposed method. A simplecomputer program is furnished tosolve more general cases. Derivationsof some of the major equations aregiven in an Appendix.

CONCLUSIONS1. When a beam hangs from lifting

points, it may roll about an axisthrough the lifting points.

2. The stability and safety of a hang-ing beam are dependent on fourquantities:e l = initial lateral eccentricity

of the center of gravity ofthe beam with respect tothe roll axis

y, = height of the roll axis abovethe center of gravity of thebeam

zo = theoretical lateral deflec-tion of the center of gravityof the beam, computedwith the full dead weightapplied Iaterally

9m S = maximum permissible tiltangle of the bean

These quantities may be reduced totwo dimensionless ratios,and 9mu. f9 i (where 9, = e i lyr , the

initial roll angle ofa rigid beam).3. The net factor of safety of a hanging

beam, after accounting for initialimperfections, is the lesser factor ofsafety calculated from the follow-ing two equations:

FS=-.u i B' (1)

FS = Bm s (I — x°) (2)As ` rlr

4. Eq. (5.2.3) of the PCI DesignHandbook is a reasonably con-servative approximation of Eq. (1).

5. Several methods are available forimproving the lateral stability ofhanging beams. The most commonand effective method is to move thelifting point in from the end by asmall amount.

PCI JOURNAL/January-February 1989 35

BACKGROUND

Classic studies of lateral buckling ofbeams are reported in Timoshenko 2 andRoark. s These analyses are based on theassumption that the beams are rigidlyrestrained from rotation at the supports.Buckling is caused by the middle part ofthe span twisting relative to the support,creating a sideways deflection. Thistype of buckling is important in steel

I-beams, which have low torsional stiffness.

The torsional stiffness of an I-beamvaries as the cube of the thickness of theweb and flanges. Concrete I-beams,with relatively thick webs and flanges,are 100 to 1000 times stiffer in torsionthan steel I-beams. As a result, lateralbuckling of the type described by Timo-shenko is seldom critical in a concretebeam. But, when the supports have roll

CENTER OF GRAVITY OF THE CURVEDBEAM ARC LIES DIRECTLYBENEATH THE ROLL AXIS AXIS

Fig. 1 a. PERSPECTIVE OF A BEAM FREE TO ROLL AND DEFLECT LATERALLY

LIFTING LOOPSDEFLECTION OF BEAMDUE TO BENDINGABOUT WEAK AXIS

IliCOMPONENT OF S,^$WEIGHT ABOUTWEAK AXIS---

CENTER OF MASS OFDEFLECTED SHAPEOF THE BEAM — --

w^ROLL AXIS

CENTER OF GRAVITY IjOF CROSS SECTION SI sn

1 AT LIFTING POINT

w

Fig. lb. END VIEW

Fig. 1c. EQULERRJM DIAGRAM

Fig. 1. Equilibrium of beam in tilted position.

36

flexibility, the beam may roll sideways,producing lateral bending of the beam.

This is the cause of most lateral sta-bility problems of long concrete I-beams. Muller' gave solutions for thecritical buckling load of beams on sup-ports that have roll flexibility. A similarapproach is given by Libby.' Swann andGoddeng showed how numerical inte-gration may be applied to find the buck-ling load of beams on elastic supports.

The approaches given in Refs. 4through 6 may he greatly simplified byassuming the beam to be rigid in torsion.For concrete I-beams with webs andflanges 6 in. (150 mm) or more in thick-ness, the torsional stiffness of the beamswill normally be much greater than theroll stiffness of the supports when thebeam is hanging (see Appendix E). Theassumptions of torsional rigidity for thebeam transforms the problem from abuckling problem to a simple bendingand equilibrium problem.

BASIC THEORY OFROLL EQUILIBRIUM

When a beam hangs from flexiblesupports such as lifting loops, it is free toroll, The center of rotation is the point atwhich the flexible support joins therigid body. This is normally at the topsurface of a concrete beam. A line pass-ing through the center of rotation (rollcenter) at each support forms a roll axis.

If the beam were perfect, it wouldhang in a plumb position, with the cen-ter of gravity of the beam directly be-neath the roll axis. But, sweep toler-ances and lifting loop placement toler-ances always cause the center of gravityof the beam to be slightly to one side orthe other of the roll axis. This causes thebeam to tip about the roll axis by a smallangle, 8, where:

9 3 = initial roll angle, radians, of arigid beam

= e i /y,[more precisely, tan (e;/t,)]e ! = initial eccentricity of the center

of gravity from the roll axis

i,.= distance from the center of grav-ity to the roll axis, measuredalong the (original) vertical axisof the beam

The slight tipping of the beam causesa component of the beam weightW to beapplied about the weak axis of the beam.This component is W sin 9, and it causesa lateral deflection of a flexible beam,which further shifts the center of gravityof the mass of the beam. This causes anincrease in the roll angle 0, whichcauses further lateral load componentand further deflection, etc. Dependingon the lateral stiffness of the beam, itmay reach equilibrium at a roll angle B

slightly larger than 8, or 0 may increaseto the point where the lateral bending issufficient to destroy the beam. The lat-eral stiffness necessary to prevent fail-ure may be computed as follows.

The final equilibrium position of thehanging beam is shown in Fig. 1. Thebeam is assumed to be uniformly tippedby an angle 0. The component of thedead weight acting about the weak axis,W sin 0, has caused an additional lateraldeflection z of the center of gravity ofthe mass of the now curved beam. Tofind the equilibrium angle 6, one mustfind z, but z is determined by the weightcomponent W sin 0, which is itself de-pendent on 0.

TUe problem may be solved by firstcomputing a theoretical deflection z o ofthe center of gravity of the mass of thebeam with the full weight W appliedabout the weak axis. Then, because theweak axis component of the weight isWsin 0, z may be found from z = 7 o sin 0.

The midspan deflection of a uniformlyloaded simple span beam may be com-puted by the well-known formula:'-'

5 wd'(5.2.2)

^" – 384 E1„

where p,, is the weak axis deflection and1,, is the weak axis moment of inertia.But, z, is the distance to the center ofgravity of the deflected arc of the beam,not the maximum deflection of the arc;'

PCI JOURNAL/January-February 1989 37

1.0

0.8

0.6x

ECD

0.4

0.2

0 -I,0 0.2 0.4 0.6 0.8 1.0

LEo/Yr

Fig. 2. Factors of safety.

zb is approximately 2 of $ . More pre- sufficiently small (say 0.2 radian or less)eisely: so that the approximation 0 = sin 0 -- tan

0 may be used. The equilibrium equa-_ 1 u^1 4 (3)

„° 120 EI –0.64/3,,

The derivation of Eq. (3) is given inAppendix F. Note that the quantity z, isa fictitious quantity because most beamswould fail if the frill weight were ap-plied laterally. But, do is used to computethe smaller quantity, z = za sin 0.

The equilibrium equation (see Fig.le) may now he written:

tan8= (zo sinB+e i)/yr(4)

For a given beam and span, Y r isknown, zo may he computed, and e f maybe assumed. The only unknown is 0,which may he found by successive ap-proximations. For most applications, 0 is

tion then simplifies to:

9 = e f (5)Yr — zo

This may also be written (recallingthat 8 i = e;lyr):

D = Bf (5a)

The quantity 11(1 – ze /y r) may bethought of as a multiplier that increasesthe tilt of the beam and is dependent onthe lateral elastic properties of thebeam. Note that as zo approaches y,., thedenominator approaches zero and themultiplier becomes very large. When zais equal to y,., the beam is totally on-

38

stable even if the initial imperfectionsare virtually -zero, and this representsthe limiting (critical) case for lateralbuckling stahility.

FACTORS OF SAFETYFor stability, the height of the roll

center yr must he greater than za , and theratio yr z. may be thought of as the fac-tor of safety against lateral buckling in-stability:

FS = Jr (6)zo

Eq. (6) gives the gross factor of safetyagainst total instability for a near perfectbeam. Beams with initial imperfectionsmay fail before total instability isreached, as there is a limit on the angle 0that the lateral bending strength of thebeam can tolerate. This maximum angleis defined as B„, and (yrl zo)cuui isdefined as the ratio of y,.!, whichmakes 0 = 0ma.r . From Eq. (5a):

Yt 1 (7)zo rrin i I —

Eq. (7) gives the critical value of theratio ;J ,./ , which would cause the tiltangle 0 of the beam to he equal to theangle O,,, which would cause failure inlateral bending. The actual ratio of yT / 7o

must exceed that given by Eq. (7) by afactor of safety:

1/d ^oFS =

(Yr'—'u)cre^ica^

Substituting Eq. (7) into the above:

FS = !!r (1 – i` I

I (1)za 8maa

Note that when e, = 0, i.e., no imper-fections, 0, = 0 and Eq. (1) reduces toEq. (6).

The Eqs. (1) and (6) defining the fac-tor of safety were derived assuming theimportant parameter to be the lateral

elastic properties of the beam repre-sented by o . The effect of a and O,,,,,,,was taken to be a modifying effect on thebasic stability represented by y,l To in

Eq. (1). The basic stability is representedby fir' za , but it is reduced by the quantity(I – o t /B ), accounting for the effects ofinitial imperfections.

If the beam is stiff laterally (and thusz„ is small), the factor of safety may nothe as large as indicated by Eq. (1). Evenvery stiff beams have a maximum toler-able roll angle 9,rtrtr beyond which thebeam would fail in lateral bending. Inthis case, the effect of the initial eccen-tricity would be the dominant effect,and it would be more logical to definethe factor of safety as the ratio of B,„a,.1 B.

Assume:

FS –

Substituting Eq. (5a) for 0:

FS = $ rrtaa(2)^1 — x0

J

Eq. (2) is very similar to Eq. (1), but inEq. (2), 8mar / 0i is the main parameterand the quantity (1 — z o ly,) is the modi-fier. Swann" gives an equation identicalto Eq. (2) The true factor of safety is thelower of that given by Eqs. (1) and (2).These equations give equal factors ofsafety when yr 1 zo = 9,,,a,. /0^. Fig. 2 showsa plot ofEqs. (1) and (2).

COMPARISON TOPCI DESIGN HANDBOOK

The material presented on p. 5-14 ofthe PCI Design Handbook, Third Edi-tion, was based on the above consiclera-tions, but is presented in a simplifiedform:

FS = Yy (5.2.3)

Eq. (5.2.3) is similar to Eq. (6), butwith q, replacing y, and 8„ replacing zo,

PCI JOURNAL/January-February 1989 39

Because y, is approximately equal to y,,and ° is approximately 2A of j3,„ Eq.(5.2.3) produces a factor of safety about¼ that of Eq. (6). However, Eq, (5.2.3)produces results approximately equiv-alent to Eq. (1) in a worst case situation,as may be seen by considering a numer-ical example.

Assume a 120 ft (36.6 m) beam with y,= 36 in. (914 mm), 3 in. (76 mm) ofcamber, and 93„ – 18 in. (457 mm). ByEq. (5.2.3):

FS= f3 18 2

Assume a lifting loop placement toler-ance of 1 in. (25.4 mm) and a sweep of %in./10 ft, or 1.5 in. (38 min). Themaximum value of e„ measured to thecenter of gravity of a parabolic arc, is 2in. (51 mm). Assume the camber isparabolic, so that yT, the distance fromthe center of gravity to the roll center, isy, – 2/3 x camber, or 34 in. (864 mm).The quantity ° – 0.64 /3 y or 11.5 in. (292mm). Assume the roll angle is limited to0.2 radian at failure, i.e., Bmar ° 0.2•From Eq. (1):

FS= -1– I– 0, 34 1– 204z°H,n„x 11.5 ' 0.02

= 2.1

The above example with e, = 2 in. (51mm) represents a worst case situation.For this case, Eq. (5) gives 6 = 0.089radian or 5.1 degrees when the beam islifted, This- excessive tilt would givewarningthate, is excessive.

One normally computes the strengthof a member assuming the member to hestraight and true_ The effect of toler-ances is covered by the safety factor.The gross safety factor on a straight andtrue member, computed from Eq. (6), is:

FS =yr/k°

This produces a total safety factorabout 1.5 times that given by the PCIDesign Handbook; the PCI Design

Handbook method has a "hidden" factorof safety of 1.5. Tolerances normally offeet the strength of a member by a fewpercent. The effect of tolerances can bemuch more drastic in the analysis of lat-eral bending stability. The hidden factorof safety accounts for this. The effect oftolerances may be evaluated explicitlyby the use of Eq. (1).

The values of 6m°r and 0, vary fromcase to case. The above example gives aworst case value for O. The quantity H,,,°,r

is determined by the lateral bendingstrength of the beam, which is depen-dent on the amount of precompressionin the top flange. Imper and Laszlo"have suggested using temporary post-tensioning in the top flange; this im-proves O,,,, and the factor of safety.

EFFECT OF LIFTINGPOINT LOCATION

Locating the lifting point even a smalldistance in from the end can dramat-ically improve the lateral bending sta-bil ity. Not only is the deflection reducedby approximately the fourth power ofthe net span, but z° is improved evenfurther, as the weight in the overhang-ing ends is on the opposite side of theroll axis.

Anderson ? and Imper and Laszlo9show how the midspan deflection is im-proved as the lifting points are moved infrom the end. Fig. 3 shows the effect onz° of moving the lifting points in from theends. The equation for z° was obtainedby integrating the shape of the deflec-tion curve to find its centroid.

z°1211(-L1,1–,:11,-+3a-1,+±a-)

O 5

(8)

When a = 0:

= w14

zO 120E1(3)

40

1.0

0.8D

Nzw= 0.6

IMJO10

SW 0.4LL

0O

I0.2

0

7* +_ 1

U 002 q .04 0.06 0.08 0.10 0.12 0 14 0 16

all = RATIO OF OVERHANG TO OVERALL LENGTH

Fig. 3. Reduction of with overhangs.

The slope 8o at the support is:

^°Z4EI (l,' —6a 2l) (9)

Note that moving the lifting points 6percent of the span virtually halves zaand doubles the factor of safety. Ofcourse, one must not overdo it as the topfiber stresses should remain compres-sive. Fortunately, very long concretegirders normally have top compressionand can tolerate being lifted a short dis-tance from the end.

APPROXIMATECOMPUTATION OF 10

Eq. (8) may readily be used to find zpfor a uniform beam with equal over-hangs. For cases that do not quite meetthese conditions, an approximate pro-cedure may be used to find z^.

1. Compute the midspan deflection f3„with the full dead weight appliedlaterally.

2. Assume the slope at the supports, Bo= 3.2 1,Il, .

3. Compute the distance to the cen-troid of the uniform load betweensupports as 2/3 of,.

4. Compute the distance to the cen-troid of the overhangs, assumingthe overhangs are straight, at aslope equal to O.

5. Estimate the centroid of any othermasses (i.e., end blocks, stub dia-phragms) on the beam.

6. Combine the above components tofind the zo of the total mass of thebeam.

MISCELLANEOUS EFFECTSEnd blocks normally have a stabil-

izing effect, as their weight is close tothe supports and decreases the tendencyto roll. In most cases, the weight of endblocks may be safely (and conserva-tively by 5 to 10 percent) disregarded tothe computation of `o.

Camber raises the centroid of the

PC i JOURNAUJanuary-February 1989 41

mass of the beam, and thus decreasesthe distance Yr between the roll axis andthe centroid of the mass. It is sufficientlyaccurate to assume the centroid of themass is shifted upward by 2/3 of themid span camber.

Beams are sometimes lifted using in-clined cables. The critical buckling loadPr'r is:

EI,,Per – I z

The quantity z. will be magnified by(approximately) the quantity (1 -P/Pcr ), where P is the horizontal com-ponent of the tension in the inclinedcable, multiplied by a factor of safety.

HOW TO INCREASEBEAM STABILITY

1. Move the lifting points inward.This is by far the most effectivemeans of improving lateral stabil-ity while hanging. As shown inFig. 3, moving the lifting points afew percent of the length maymore than double the factor ofsafety. Top stresses must bechecked. Imper and Laszlo" havesuggested using temporary post-tensioning in the top flange wherenecessary to control top stresses.

2. Raise the roll axis. This may bedone by providing a yoke attachedto the beam at the lifting point, orby using a pair of inclined liftingloops. Either method can effec-tively raise the roll axis (and in-crease yr ) by a foot or two.

3. Increase the modulus of elasticity,E. Because E varies as the squareroot of the concrete strength, in-creasing f, and therefore E resultsin only a slight improvement inlateral stability.

4. Add bracing. This commonly usedmethod is one of the less effectivemethods. The quantity zp is deter-mined by the lateral stiffness of the

member, and bracing using V in.(12.7 mm) strands adds very littleto the lateral stiffness. Bracingusing angles or other rolled sec-tions with substantial steel area canbe much more effective. Bracingcan add significantly to the lateralbending strength and thus to 0....This improves safety when high B

angles are encountered, such as onsuperelevated curves. This will bediscussed in Part 2.

5. Modify the beam cross section.This usually cannot he done on aparticular project. When the beamshape can be changed, it is impor-tant to realize that the bottomflange contributes just as much tolateral stability as the top flange. Infact, adding material to the bottomflange is more beneficial because itlowers the center of gravity and in-creases yr as well as I,,. Further-more, the bottom flange is undercompression and not as subject toloss of stiffness through cracking asis the top flange.

EXAMPLEThe example in Appendix B repre-

sents an extreme case. It is taken fromreal life. In 1963, the author designed abarrel shell roof of approximately 150 ft(45.7 m) span. The valley beam (whichwas actually the tension flange of thebarrel shell) was a prestressed I-girdernormally used on bridge spans of about100 ft (30 m). The beam was stretched to145 ft (44.2 m), and was heavily pre-stressed throughout its depth using highstrength concrete, Camber wasapproximately zero. The beam wascarefully checked by classical lateralbuckling formulas and found to hesatisfactory.

Lifting loops were located 135 ft (41.2m) center to center. The first beam to belifted was 42 hours old and was handledwithout incident. The second beam, 18

42

Flours old, tilted immediately on beinglifted from the form pan, and bent side-ways approximately 1 ft (300 mm). Thebeam was immediately set down on theplant floor; fortunately, it righted itselfinstead of rolling over. It was straightand stable once it was resting on bunkson the floor.

The theory presented here was de-veloped to explain what went wrong.The factor of safety against lateralbending, computed after the fact, wasalmost exactly 1.Q. (The initial eccen-tricityea is not known for this case, but itis believed to have been quite small.)Apparently, the slight difference in E ofthe two beams caused one to be stableand the other to buckle.

After the lateral bending phenomenonwas discovered and analyzed, the liftingloops were moved to 120 ft (36.6 m)center to center with a 12.5 ft (3.8 m)overhang at each end. The remainder ofthe beams were handled without inci-dent, although the computed gross fac-tor of safety for this condition, 1.9, isprobably slightly less than desirable.(These beams were never shipped overthe road; they were delivered by bargeand laterally braced against the barge.)

The recomputation was done by theapproximate method to demonstrate itsuse. In this case, it would be about aseasy to use the more exact method,which gives H^, = 14.95 in. (381 mm) andFS=2.

WHAT FACTOR OFSAFETY IS NECESSARY?The necessary factor of safety cannot

be determined from mathematicalderivations; it must be determined fromexperience. The PCI Design Handbooksuggests a factor of safety of 2, but thisproduces an actual gross factor of safetyof about 3. The 1963 experience withthe 145 ft (44.2 m) beams appeared toindicate that a gross factor of safety of 2was adequate when the initial eccen-tricity e l due to tolerances was very

small. Imper and Laszlo9 suggest using afactor of 1.5 for yard handling and 1.75for field handling, with the factor beingbased on B3. This produces gross factorsof 2.3 and 2.7, respectively, based on zp.

The computation of a net factor ofsafety requires a knowledge of e t andBm„r . The initial eccentricity e i may beassumed, based on the worst case com-bination of permissible tolerances, aswas done in the earlier example. How-ever, that maximum eccentricity ejwould have caused the beam to hang at a5.1 degree angle when first lifted, whichshould serve as a warning that qualitycontrol needs to be tightened in order toreduce et.

The determination of H.,,,. also in-volves some difficulties. Using a goodcomputer program for ultimate strengthin biaxial bending, the maximum lateralbending moment in combination withthe vertical bending moment may befound, and thus O,,,^ at ultimate load.Unfortunately, once the lateral momentexceeds the cracking strength, the stiff-ness decreases and zo increases, calcu-lated on a cracked section.

A conservative approach is to compute8,,,s based on the lateral moment,which, when combined with the verticalmoment, produces a tension in the topcorner equal to the modulus of rupture.The "right” value of 9,,, . probably liesbetween that computed by this ap-proach Lind by the ultimate strength ap-proach. This will be discussed in moredetai I in Part 2.

SUMMARYA simple method for the analysis of

the lateral stability of hanging beamshas been presented. The method per-mits the evaluation of the effects of ini-tial imperfections. Several methods forimproving the lateral stability aresuggested. In Part 2, the analyticalmethod will be extended to the case ofbeams on flexible supports such astrucks and neoprene pads.

PCI JOURNAL,^January-February 1989 43

REFERENCES1, PCI Design Handbook, Third Edition,

Prestressed Concrete Institute, Chicago,Illinois, 1985.

2. Timoshenko, S., Advanced Strength ofMaterials, Part II, Third Edition, D. VanNostrand, Princeton, New Jersey, 1956.

3. Roark, R. J., Formulas for Stress andStrain, Fourth Edition, McCraw-Hill,New York, N.Y., 1965.

4. Muller, J., "Lateral Stability of PrecastMembers During Handling and Placing,"PCI JOURNAL, V. 7, January-February1962, pp. 21-31.

5. Libby, J. R., Modern Prestressed Con-crete, Second Edition, Van NostrandReinhold, New York, N.Y., 1977.

6. Swann, R. A., and Codden, W. G., "The

Lateral Buckling of Concrete Beam'Lifted by Cables," The Structural En-gineer (London), V. 44, No. 1, January1966, pp. 21-33.

7. Anderson, A. R., "Lateral Stability of LongPrestressed Concrete Beams," PCIJOURNAL, V. 16, No. 3, May-June 1971,pp. 7-9.

8. Swann, R. A., Reader's Comment to articleon "Lateral Stability of Long PrestressedConcrete Beams," PCI JOURNAL, V. 16.No. 6, November-December 1971, pp.85-87.

9. Imper, R. R., and Laszlo, C., "Handlingand Shipping of Long Span BridgeBeams," PCI JOURNAL, V. 32, No. 6.November-December 1987, pp. 86-101.

NOTE: Discussion of this paper is invited. Please submityour comments to PCI Headquarters by October 1, 1989.

44

APPENDIX A - NOTATION

a = length of overhangb = widthe { = initial eccentricity of the center

of gravity of beamE = modulus of elasticityFS – factor of safety against lateral in-

stability or failureG = shear modulusI y = lateral moment of inertiaKe = rotational spring constant of sup-

portl – overall length1 1 = length between supportsM = momentM = torsional momentP = axial loadPer = critical compressive buckling

loadr = Ke1Wt – thickness

w = weight ofbeam per unit lengthW = total weight of beamx = lengthy = deflectionyT = height of roll axis above center of

gravity of beamyt = distance from center of gravity of

cross section to top of beam= lateral deflection of center of

gravity of beamzo = lateral deflection of center of

gravity of beam with the fulldead weight applied laterally

a = superelevation angle,3„ = midspan lateral deflectionA9 = twist angle9 = roll angleBf = initial roll angle = e fly,Ba = slope at support, when full dead

load is applied laterally

PCI JOURNAL/January-February 1989 45

APPENDIX B - EXAMPLE

Given

l = 145 ft; w = 0.61 kip/ftEnd blocks are 5 ft long.Added w = 0.53 kip/ft1, = 135 ft center lifting pointsa=5ftE 4 55UQ ksiI„= 15,000 in. 4; Jr = 30.1 in.

Required

Compute zo and the factor of safety whilehanging from loops. Include effect ofend blocks (see Fig. B1).

Solution

For unilorni load, use Eq. (8):

zO

12Fl1[101`^a21,+3 9 1 ' + 5 51

0.61 kip/ft (4,422, 781,000 ft 3)(1728 in.3lft3)12 (5,500 kip/in. 2)(15,000 in. 4)(145 ft)

W = 145 (0.61) = 88.45 kipsMoment about roll axis:M = `z°W = 2872 kip-in.

Effect of end blocks:From Eq. (9):

9° = t j (1,' – a6 ^l, )2

0.61 kip/ft (135'ft3 -6.5 2 x 135 ft')_

24 (5,500 kip/in.") (15,000 in.4)

x (144 in.2/ft2)B° = 0.1083 radian

z° to cg of end blocks = – 0, (1 to cg of endblock)

= 32.48 in.

END BLOCKS^Hr

eO i

a=5'

135' ai51

Fig. 81. Computation of z,.

ASSUME 5' END BLOCKSTRAKIHT

IhJ

9 0 +

a = L^ 120' g z12.5'

Fig. B2. Revised computation of zp.

46

0.10R3 (2.5 ft) (12 in./ft)

° =- 3.25 inn.

W =2x5x0,53=5.3kipsM about roll axis = -°W = - 17 kip-in.

For total 1)camn:

- _ EM's _ 2872 - 17 kip-in.SW's 88.45 + 5.3 kips

= 30.46 in.

For nocamber,y,.= y r = 30.1 in.

FS = i!30.1

1.0 (!)

E. °30.46For further details, see the text.

Revise 1 1 to 120 ft (a = 12.5 ft) to improvestability. Recompute zo and FS by ap-proximate method (see Fig. B2).

For center 120 ft:

z= 384E1 (51; - 24a2)

0.61 kip/1t (120 2ft2 ) (5 x 1201 - 24 x 12.52)ft21, = x (1728 in, a/$')$ 384 (5500 kip/in. 2 ) (15,01)0 in.')

= 32.70 in.

= 3/3°= 21.80in.

W= 120(0.61)=73.2 kipsM about roll axis = 73.2 (21.80)

= 1596 kip-in.0. = 3.2 f3^I1, = 3.2 x 32.701120.x 12Bo = 0.0727 radianFor end 12.5 ft:zo = - 00 (a12)= -0.0727 (75 in.)

= -5.45 in.W 2 x 12.5 x 0.61 15.25 kips

M about roll axis = 15.25 (-5.45)= -83 kip-in.

For end blocks:_ -0 (Ito cg of end block)

No = -0.0727 (120 in.) _ -8.72 in.W = 5.3 kipsM about roll axis = -46 kip-in.

For total beam:_ E. M's _ 1596 - 83 - 46

° L. W's 93.75

z° = 15.46 in.

FS = T̂ = 30.1 _ 1.9415.46

For further details, see the text.

PCI JOURNAL(January-February 1989 47

APPENDIX C- A PREVIEW OF PART 2

When a beam is supported on flexiblesupports such as bearing pads or truckand trailer, a similar situation occurs inwhich there is a tendency for the beamto roll about a roll axis. In this case, theroll center is below the beam, and te r isnegative (see Fig. Cl).

When the roll axis is beneath thecenter of gravity, the support must becapable of providing resistance to rota-tion. This resistance is expressed as anelastic rotational spring constant Ka.Taking moments about the roll axis (seeFig. Cl):

W [ (z. sin O cos B + e) cos 0 — yr sin 01

=K 9 (8— a)

(10)

where a is the superelevation angle orslope angle of the support.

Using the approximations 0 — sin Btan 9 and cos 0 ==1:

W (Zo9+et — yre) = Ke(B-- a) (11)

Let r = K a /W. The quantity r has aphysical interpretation. It is the heightat which the weight W could he placedto cause neutral equilibrium with the

o

LATERALDEFLECTIONOF BEAM ^ I 1

CG OFDEFLECTED . a - o sink «eBEAM --1

W^ 4 t

ROLL AXIS —

M = I( B (B- 0C

MOMENT INSPRINGS

EVATION

O-= ANGLEAT SPRINGS

(2a sin a ei) case- Yr sineAI SI'HINGS

Fig. CI. Equilibrium of slender beam on flexible support.

48

w!

FOR NEUTRALEQUILIBRIUM,W (r9)= M -6(Ka)

r = Ke/W

M = 8(KO)

Fig. C2. Definition of r.

spring (see Fig. C2). For neutralequilibrium, the overturning momentwill just equal the resisting momentwhen the rod supporting W is displacedby a small angle. The quantity r mighthe called the radius of stability.

Solving Eq. (11) For 0:

ar+e;(12)r+1r —za

When r is very large, i.e., the supportis very stiff, 0 approaches a, the tiltangleof the support. Whenr = 0:

eI

Yr -a

This formula is identical to Eq. (5).Eq. (12) is an expanded version of Eq,(5).

The definition of B„ the initial rollangle of a rigid beam, may be expandedby setting zo equal to zero in Eq. (12):

ar+ei(13)

r +y,.

Expanded versions of Eqs. (1) and (2)may he derived from Eq. (11) as follows,with B,as defined in Eq. (13):

FS= tir (1 0i \ (14)Z o 8maa

FS = 0Bi' (1– (15)r +°^ f^rl

Part 2 will give the derivation of Eqs.(14) and (15) and describe their use.This requires the determination of Kaand r, the location of the roll axis, andthe determination of O,, .

PCI JOURNAUJanuary-February 1989 49

APPENDIX 0-SIMPLE BASIC PROGRAM

A simple BASIC program is shown(Fig. D1) which allows one to quicklyevaluate the effect on the factor of safetycreated by varying the "A" distance, Theprogram computes the factor of safetyagainst cracking of the top flange. Thefactor of safety against failure may beconsiderably higher, but the computa-tion of 9 and Ho at failure is a morecomplex problem.

The Iogic of the program is verystraightforward. The user may readilyalter the program to suit his/her needs. Afew of the lines in the program are ex-plained as follows:Line 100 "A" is the distance from the

end of the beam to the liftingpoint.

Line 110 The initial eccentricity e F isestimated at one half the PCIsweep tolerance plus '/4 in. forlifting loop placement. Theuser may wish to modify thisstatement, to reflect his ex-perience with actual toler-ances. The actual initial ec-centricities being experi-enced may he evaluated byfinding the values of i which

produce the tilt angles 0 actu-ally being observed whilebeams are hanging.

Line 120 Eq. (8).Line 140 Assumes maximum lateral

moment is that which pro-duces tension equal to themodulus of rupture in the topcorner of the top flange.

Line 160 Estimate of camber.Line 170 B, = e Il sr.Line 175 Eq. (5).Line 190 Eq. (1).Line 195 Eq. (2).Line 300 Allows one to revise "A" dis-

tance without re-enteringother data.

Line 320 Allows one to change e1.

This program often produces quitelow factors of safety because the com-puted factor of safety is that againstcracking. In most cases, the factor ofsafety against failure will be higher. Ad-ditional data from the field are neededto find 0,,, qr at failure or, conversely,what factor of safety against cracking isadequate. Part 2 will discuss B,,,,, ,r at fail-ure in more detail.

50

1 PROGRAM ROLLINPUT "AREA, SQ IN";ACINPUT "DEPTH, IN";UINPUT "YT, IN";YTINPUT "VERTICAL MOMENT OF INERTIA, IN^4";IINPUT "TOP FLANGE WIDTH, IN";BINPUT "LATERAL MOMENT OF INERTIA, IN^4";IYINPUT "OVERALL LENGTH, FT";LINPUT "CONCRETE STRENGTH, PSI"; FCPINPUT "CONCRETE DENSITY, PCF";GAMMAINPUT "PRESTRESS FORCE, KIPS";PINPUT "HEIGHT OF PRESTRESS FORCE ABOVE SOFFIT AT MIDSPAN., IN" ;YS

80 E=33AGAMMA^1.5*SQR(FCP1/100085 W=AC*GAMMA/144/100090 ST=I/YT95 ECC=D-YS-YT100 INPUT "A DISTANCE, FT";A105 L1=L -2'4A110 EIN=L1/10'.125/2+1/4120 ZZEROBAR=W/(12*E+IY*L)*(L1^5/10-A^2AL1^3+3*A^4*L1+6kA^5/5)*1728125 MG=Wk(L1^2/8-A^2/2)*12130 FTOP=P/AC+(-P*ECC+MG)/ST140 MLAT=(7.5'SQR(FCP)/1000+FTOP)*IY/(B(2)150 THETAMAX=MLAT/MG160 CAMBER= (0.11*P*ECC-5/48*MG)AL1^2/(EI)k144*1.85165 YR=YT-2/3*CAMBER170 THETAIN=£IN/YR175 THETA=EIN /(YR-ZZEROBAR)190 FS1=YR/ZZEROBAR*(1-THETAIN /THEPAMAX)195 FS2=THETAMAX/THETAIN*(1-ZZEROBARIYR)200 PRINT

PRINT "CAMBER = ";CAMBERPRINT "INITIAL ECCENTRICITY = ";EINPRINT "ZZEROBAR = ";ZZEROBARPRINT "MG = ";MGPRINT "MAX M LAT = ";MLATPRINT "THETA INITIAL - ";THFTAINA180/PIPRINT "THETA ";THETA*180/PIPRINT "THETA MAXIMUM = ";THETAMAX*180/PIPRINT "F. S. EQ #1 = ";FSIPRINT "F. S. EQ 42 = ";FS2PRINT "THE ABOVE FACTORS OF SAFETY ARE AGAINST CRACKING OF

TOP FLANGE"PRINT "THE FACTOR OF SAFETY AGAINST FAILURE MAY BE HIGHER"PRINT

300 INPUT "NEW 'A DISTANCE? Y OR N" ;AIF A$="Y" THEN

GOTO 100END IF

320 INPUT "REVISE INITIAL ECCENTRICITY? Y OR N";ESIF E$="Y" THEN

INPUT "INITIAL ECCENTRICITY, IN' ;EINGOTO 120

END IF9999 STOP32767 END

Fig. D1. Simple BASIC computer program.

PCI JOURNAUJanuary-February 1989 51

APPENDIX E -TORSIONAL STIFFNESS

Examine the validity of the assump-tion of torsional rigidity by analyzing thebeam described in Appendix B of Ref. 9.The beam is a PCI BT-72 on a span of136 Et. Assume it to be tilted on an 8 per-cent slope. The end reaction is 56.1 kips,which produces a torsional moment Mtof 159 kip-in. when tilted on an 8 per-cent slope. The torsional constant,Ibt 313, is estimated by idealizing thebeam as four rectangles (see Fig. E1).

26(8.25) 313 = 4,866 in.363.75(6) 313 = 4,5902(18)(4.5)13 = _ 1,(14Ibt ,3 = 10,550 in.3

Assume the shear modulus, G, to be2000 ksi. The torsion is maximum at theend, varying to zero at midspan, over alength of 68 ft. The twist, OB, betweenend and midspan is:

M1 x /2c

08=GIbt313

_ 159(68)(12)/2

20(10,550)AO = 0.0037 radians = 0.18 degrees

R = 56•1 k Ti-- .08(35.4) = 2.83"EA END ------:i—

MIl

t

.08 RADIANS

Fig. El. PCI BT-72 beam on 8 percent slope.

This is small in comparison to otheruncertainties such as fabrication toler-ances, and the beam may be assumed tobe torsionally rigid.

52

APPENDIX F -- DERIVATION OF EQUATION FOR z

The quantity z ° is the deflection of thecenter of mass of the deflected shape ofthe beam, with the self weight appliedlaterally. The quantity z° may be found byintegrating the deflected shape of the1)eam.

– Y odx

l

The equation for the deflected shape of abeam without overhangs may be derivedfrom the deflection equations given on p.5-15 of the PCI Design Handbook.' Alter-nately, the formula can be found from astandard structural design manual.

= 24 1 (x — 21x 3 + lax)u

w x' _ lx` lar 2 'J y dx

= 24E1 v 5 2 2°

w i',

120E;1„

Therefore, zo = w1

1 (3)120EI,

Similarly, the equations for the de-flection curve of overhanging beamsgiven on p. 5-15 of the PCI DesignHandbook may be integrated to produceEqs. (8) and (9).

METRIC (SI) CONVERSIONS1 ft = 0.3048 m 1 kip 4.45 kN1 in. = 25.4 mm 1 kip-in. = 0.113 kN -m

PCI JOURNAL/January-February 1989 53