Foundation Design Handbook- Hydrocarbon Processing- 1974

5/22/2018 Foundation Design Handbook- Hydrocarbon Processing- 1974

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FOUNDATION DEffiHANDBOOK=-' Reprinted from HyDRocARBo[Fnocisslrvc . Gutf pubtishing


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FOUNDAPublished b

Manuals in the series are:This reference manual has been reprinted from theissues of HYDROCARBON PROCESS+ruO. Otfrer- regUlarHandbooks

LINES FOR BETTER MANAGEMENTI NSTRUM N MANUAL

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FOUNDATION DESIGN HANDBOOKTABLE OF CONTENTSTOWER FOUNDATIONS ..Foundation Design For Stacks And TowersSimplified Design For Tower Foundations .Calculation Form For Foundation Design .

Page N

Use Graph To Size Tower FootingsSimplified Design Method For lntricate ConcreteColumn LoadingUnusual Foundation Design For TallFoundation Sizing Simplified . . .. . TowersDowel Sizing For Tower Foundations . . ,Short Cuts To Tower Foundation DesignVESSEL FOUNDATIONS . .

223345556Foundation Design For 8-Legged VesselsPressure Vessel Foundation Design . . .

COMPUTER FOUNDATION DESIGNHow To Calculate Footing Soil Bearing By ComputerConcrete Support Analysis By Computer . . .FOUNDATIONS ON WEAK SOILS .... .Foundations On Weak Soils .Graphs Speed Spread Footing DesignUse Graph To Analyze Pile Supports .

.61

.63.70. 7.7.8.88.93

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TOWER FOUNDATIONS

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Foundation DesignFor Stacks and TowersThe same principles ,pply in both stacks and towers. Use this method i

making your calculations for either.

V. O. Morsholl, Tennessee Easlman Company, Kingspori, Tenn.Frorn the viewpoint of the f oundation designer,stacks and tor.r.ers may be divided into tr,vo generalclassifications, deper-rding on the method utilizedto maintain them in a vertical position; (a) Self-supporting, which resist the overturning forces bythe size, shape and weight of the foundation ; (b)Guyed, in which the c,verturning forces are re-sisted by guy u,ires. It is obvious that the con-ditions alTectir-rg the design of founclations fc.,rthese tr,r,o types u,ill not be the same, ar-rd that itis necessary to treat them separately.STA(--KS AND TOWERS are closely related asfar as founrlation design is conccrnecl-in fact, thesame principles apply. In the case of stacks, thebrick lining is a r,ariable load, corresponding to, andrequiring the sanrc treatnrent as the liquid, insr-rlation,etc., in a tou.er. This cliscussior-r will be based on thedesign of tower forrndations, however, it should bekept in n-rind that it is also applicable to stacks.

2. Self-Supporting TowerThere are two main considerations in designingthe foundation for a self-supporting tower; (a) soiiloading (b) stability. The foundation must be ofsuch size and shape that the load on the soil belowwill not exceed the maximum load which it willsafely support. The foundation must also maintainthe tower in a vertical position, so that it willnot be overturned by the maximum forces actingupon it.No direct method of calculating the size of thefoundation has been developed, therefore, it mustbe determined by trial and error. A size is as-sumed, and the soil loading and stability calcu-lated. If the results are not satisfactory, anotherassumption is made, and the calculations repeated.

3. Soil Loading(See Section 20 for complete definition of termsThe soil loading may be determined by thfollowing formula:S: S,* S, (whereS: total unit soil loading (lbs.,/sq. ft.)51 : unit soil loading due to dead load (lbs,,/sq. ftSr: unit soil loading due to overturning mo,ment(lbs.,/sq. ft.)

4. Dead LoadThe dead load S, may be determined as follows^wr- a Qwherea: area of base of foundation (sq. ft.)'W: total_ weight on soil (pounhj) calculated bthe following equation:-W: YY.,:l W. (3Wt : Minimum dead load (pounds), which is ihweight of the empty tower plus the weight othe foundation, including the earth fill on toof the base.W.: Weight of auxiliary material and equipmensupported by the tower and foundatio(pounds), which should include the liquid ithe tower, insulation, platforms, pipin , etc(Does not include weight of tower)Thls is o revised orticle whlch wos prevlously publishedin the August, 1943 lssue of PETROLEUTI REFINER. Allcopies of thot lssue, oll reprints ond qll coples of thcl94O Process Hondbook, In which the origlnal qrticle wssreproduced, foiled to meet th. demqnd for thls englneer-ing doto.When the outhor consldered tho revlslon he extendedthe subiect to lnclude octuol deslgn of foundotlon typ6rcommonly required In the erecllon of refinery vcsgctg.fhe resuli ls o thorough study of o sublect whlch contlnuesro hold o forefront posiiion in refinery cnglneerlng.Reprinrs will be provided ln quantity sumcient to lncludelhe demond thot hos extended Into constructlon fietdsoutslde of refining. Prlce $l.OO por copy.


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5. Overturning LoadThe overturning load S, is the result of the over-turning moment. Under ordinary conditions, theonly force tending to overturn the tower is ther.vind pressure.The magnitude of the wind pressure is obvi-ously a fun-ction of the wind veloiity, which variesin different localities. In many instances laws havebeen enacted which state the wind velocity orwind pressure to be used for design purposes.The United States Wcather Bureau has pro-posed the following formula :Bp : 0.004 io v' (4)where ;: wind pressure on a flat surface (pounds,/sq ft')B : baromctric pressure (inches Hr)V : velocity of wind (miles per horrr)For a barometric pressure of 30 inches, theformula becomes:

P: 0'004V' (5)It has been found that the wind pressure on acylindrical tower is about 60 percent of that on aflat surface. For a cylindrical tower, theref ore,formula (5) becomes:P":whereP": 0.0025v'

(6)wind oressure on the projected area of a cylin-drical'tower (pounds per square foot).

Wind Prcssurc In most localities, zL wind velocity of 100 milesper hour is considered the ma'ximum. This giv.es'u pr..rrr." of 25 pounds per square foot on theoriiected area of -the tower, which is the figuregeneraliy used for design pu.rposes. I.t should beEmphasized, however, tiiat this figure is subject tovariation in d.ifferent localities, and that local lawsshould not be overlooked in this connection'(Note : As a matter of interest,. the wind pres-,ri. o., an octagon shaped stack is considered tobe ?0 percent of that on a flat surface.).Figure 1 represents a .tower, mountg9 on a con-crete-foundati,on. The 'arind pressure (P*) tends torotate the tower and foundation about point A atthe intersection of the vertical centerline and thebase of the foundation. This rotating effect pro-duces an overturning moment which can be cal-culated as follows:whereM, : overturning moment about the base of thefoundation (foot Pounds)p* : ioiat wina Laa (bounds) to be calculated asfollows:P- : o. D"H (8)L: ierrer irm of wind load (feet) to be calculatedas follows:osc

FIGURE IFoundotion for self-supporting tover'

Mr: P- L

HL-h,* ,D.:diameter of tower measured( feet)H -height of tower (feet)hr : height of foundation (feet)

(7)FoundotionToPGroFoundotion

(9)over insulation

It should be noted that alt dimensions are statedi.,-ieet, giving the overturning moment (M3) inioot pou"'at. ft',i. avoids the use of the excessivelylarse numbers encotlntered with the usual inchpoind units. Care should be taken, however, to6


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^Mr2:zwhereZ : section modulus -of the base of the found.ation.(Note: Z to be based "" a;-er.ion. i, i;;i.iThe value of (Z) can be expressed as follows:,:*where (11)I : moment of inertia of ttre base of the founda-tion (based on dimension, i,i-tJ"ti.c: distance from neutral a*ii oJ iJiria"tion l"s"to point of maximum ,t..ss (fiefj.. Having calculated- (Sr) and (Sr) as explainedabove, the total soil ioid ,na")-iilriimum deadload conditions can b.e determin.d bt "q;"ii;"*(-iihis maximum soil load occurs 'r,t if,. .agd ,;fll:lgl,"1",ion, designated as p, ""J i. irequEnttyreterred to as the ,.toe pressure.,, ft is obvioutthat the maximum toe pressure (S) should neverexceed the safe bearing load of th; ;il in question.

use consistent units, that is foot pounds, in allcalculations...Jl-: :l."rs, or load, o1-lhe soil resulting from:1:_?":.t,i'ning moment (y,) valies from p"oint topotnt, and the maximum load (Sr) can bi calcu_lated as follows:times,.(Sl.) must never be less than (Sr). Iperfectly bal_anced system, (S._) i; "*ritty .qto (Sr), in which case

Snro : Sr- - Sz:0 (l_..Althou.gh,-such a.bala".S..d sy:19* is rarely psible, it is the ideal condition.' tt e- up*a.d'fo.at E due to the overturning -o*.r,i il-;_;;ibalanced by_the dead load, io tfrui-tfr. stressr, ls zero. I he stress at F in such cases is tminimum which can exist ""a ,iiif -maintainstable system.^__l :, .rld be emphasized.tha_t while (S,*) is frqu_enily greater than (Sr) it should nevei be lesIt should also be eri-rphasized tfrai ttre stabiliis based on the minimum dead ioaa iwJ-;;tiathe.soil -loading is based on the ma*)mum deaload (W).

7. Example No. IDesign the concrete founation for a towerft. dia. by 54 ft. high, including a 4 ft. skirt, anweighing.30,000 lbs. empty. The insulation, platforms and piping weigh 9b00 lbs., the maxirirumwind velociJy is 100 mlles per hour, and the frosline at the location of the proposed'installation i4 ft. below grade. The maiimum safe soil loadingis 2000 pounds per square foot.

Since the ,r.., ,,,1""11tti'r,. o",o* grade; thfoundation will be 6 ft. deep, with the- top 1 ftabove grade, making the bottom of the foundation5 ft. below 'grade, br 1 ft. below the frost lineThe foundation will be octagon shape, which irecommended for such cases, as it combines thefeatures of stability, ease of construction and mini_mum material better.than other shapes. The topcourse will have a short diameter of O ft. sin.the tower is 4 ft. dia. and allowance must be madefor founda.tion bolts, etc. The short diameter of theDase wlll be assumed to be 13.5 ft. The thicknessof the base.will depend on the bending and shear_l"C l.or.9: (see Sections .1-9 to 19h incl.), however,for the time being the thickness will be assumedto be 2 ft.Th_e weight of the foundation will be calculatedas tollows (all slide-rule figures):*l:l :f 6.ft. octagon : 0,8f8 d,^1-0.828 X 6,:29.8 sq. ft.vofume- ot top course : 4 ft. X 29.9- 119.2 cu. ft.r\rea ot base (octagor). (q): q.-g?g X I3.5,: l5l sq. ft.Volume of base :2"1t. k i5i :30r;;l i;. "Total. volume :119.2 *' jOZ: 4Zl i ;;.' i;.welght of concrete:__ .421.2 cu. ft. \ 150 lbs.,/cu. ft. :63,666 16..Volume of earth'fillW:i{fu if-='f LIJ'rt'r:,1,. 'i;r'.',3,2,11?=39lr36io'lvv erg_nr or _e-mpty tower : 30,000 lbs._wt : 30,000 + 63,000 + 3Z,7OO:125,700 tbs.W, will be as follows:Insulation, platforms, piping, etc. : 9,000 lbs.Water required to fili it " t6i"".-(4 ft. dia.) (50 ft. high) :39,5001bs.Total (W.) :4g3mlE;W:125,700 f 48,500: LZ4,2OO lbs. (from equation 3)a: l5l sq. ft.^ 174,200|bs.5': -1Sl:qlTI- : 1155 lbs.,/sq. ft.: Maximum deadload on soil (equation 2)

Allowing 2,, for the thickness of the insulation,

( 10)

- It should t. ,,o,"t,ilitigi is positive at pointI, ,ng negative at E (Figu.l'f y.'tn oihe, *brds,rne wtnd load causes compressive stresses on the:gil t. the left of point A, ihe *u*rn"* compres_:191 9:.Yt.ing.at f, and tensile stresses of equalmagnitude to_the right of A, the maximum tensionoccurrrng at II.Since the earth has no strength whatever intension, it is obvious that the suir of the stressesat any-point must be positive. fn other words, thebase of the foundation. must exert a-comp.es.i.,reforce on the soil over its entire ur"i, ott d.*ir" ntensile stress will be produced at E, which *"u.,ithat the tower and fbundation wili be unstabte,and .likely to be overturned by the uiiion of thewind..Lt-*": shown by equation (1) that the maximumsoll load. is equal to (S, + Sr). Since the value of5, at .pornt.-b is negative, the minimum soil load(*jr:l obvlorsly occurs at point E) is (Sr_Sr).-rt rs very lmportant to note that the conditionot poorest stability occurs immediately after thetower is mounted on the foundation, and beforethe insulation, platforms, piping, UquiJ, etc., are

m ptace. In calculating the stability, 'therefore,(S,) must be replaced b"y (S,-) a, foiio*,,Sr-:

whereSr-:Wta (2-a)

T_r.inimum- soil loading due to dead load(lbs.,/sq. ft.)The minimum soil loading which can ever exist,therefore, is found to occuf at point E when thedead load is at its minimum value, and can beexpressed as follows:S*r,: Sr- - S, ( l-a)Therefore, in order that (S*i,) may always bepositive, thereby assuring a'sti6fe .onaiiio., ut "tt7


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the effective diameter of the tower exposed to theaction of the wind is 4.5 ft. A wind velocity of 100miles per hour is equal to 25 pounds per sq. ft. ofprojected area.Therefore:p.:25 lbs.,/sq. ft.D. - 4.5 ft.H: 54 ft.P*: 25 X 4.5 X 54 : 6080 lbs. (equation 8)hr: 6 ft. 54L:6* 7:33 ft. (equation 9)Mr : 6080 X 33:200,000 foot pounds (equation 7)Z: 0.1016 d' : 0.1016 X 13.53 :248.5

From equation (10),,: l@,9%*j$@ :803 lbs./sq. rt. : maxi-mum soil load due to overturning moment'

The total maximum soil load (toe pressure) canbe calculated from equation (1) as follows:S : 1155 * 803: 1958 lbs.,/sq. ft.

This loading is satisfactory, as the soil willsafely support 2000 lbs./sq. ft.From equation (2-a)

TABLE IElements of Octogonol BoseRadius ofArea a(sq. Ft.)

o.7720.900'r . 020

This is the dead load under the worst stabilitycondition, and since it is greater than the over-turning stress (Sr:803), the soil below the foun-dation-will always be under compression at allpoints, thus indicating that the foundation isstable.Usually it is found that the first assumption asto foundition size is not correct, in which case, an-other assumption is made, and the calculationsrepeated.it is interesting to note that the soil loadinq of2000 lbs./sq. ft. allowed in this problem is ratherlow, as good clay soil will usually su-pport abou't4000 lbs./sq. ft. Care should always- be taken, t.oascertain the actual load carrying value of the soilat the site of construction.

8. EccentricitYIt will be noted that there are two forces actingon foundations of the type under consideration;(a) The dead load, acting in a vertical- direction;(b) the wind load, actingln a hotizontal direction.it e combined ac[io., oithese two forces, that is,their resultant, has thq same effect as an eccentricvertical load. As explained previously, it is notnecessary to calculate the eccentricity. in order todetermine the stability of the foundation. Severalmethods have been - proposed, however, whichmake use of the eccentricity, and since there aredefinite relationships between eccentricity and sta-bility, they will be explained as a matter of in-terest.The eccentricity can be calculated as follows:

Sr-: 125;700 lbs.l5r :830 tbs./sq. ft.

e: Mrwhere -W;e: eccentricity (feet)

ShortDlam.(Feet)LenAth lNeutral Arlsof slde I to Extreme(Feet) lFlbffc(Feet) SectlonModulus Z

1.892.16.547.4510.6\3.2

(12)

1.1581. 2861. 415t.5421.671. 801 .932.062.t72.3r2.442.572.702.832.963.093.473.603. 863.994.244.374.504.634.764. 895.025.145.275.405. 535. OO5.785.926.046.176. 296. 436. 566. 686 .826. 947.207.467.587.71,a,238. 48a.759. 009.269. 51L7710. 04

10. 28

2.432.702.983.253.5i3.784.064.594.875.135.41,5.b/5.946.226.487.036.767.307. 838.118.388..659.O29. 199.469.7210.00to.2510. 5510.8111.0811.3511.6211. 9012. \772.43

12. 9813.2513.5273.7974.o714 .3314. 0114.8815_ 1515.4215.6815. 9616. 2316.7717.3t17.8518, 3818. 9219.4720.0120.5521. 0921 .65

1.8632.O702.4842. 6912.8983. 1053.3r23.5193.7263.9334.1404.3474.5544.7614. 9685.3825.7755.5895. 7966.0030. 2106.4176.6246.8317. 0387 .2457.4527.6597. 8668.0738.2808.4878.6948. 9019.1089.3159.5229.7299. 93610.14310. 35010.55710. 76410.97111.17811.38517.59211.79912. 0061,2.2t3t2.42012.4348.24413. 66214. 07614. 49014.90415. 31815.73216. 14616. 560

16. 820.725.029.834. 840.546. 552.859.566. 874.582.8sl.2100.0109.8119.5140. 0r29.2151.0162. 01741861992t2226240268283299348365383401420438458177497518539560582603526650672695720745755848902958101510751134119512607325

4.555.5D6.57l.c88.5I9.51010.51111.51212.5IJ13.51414. 5151616. 51777.51818. 51919. 52020.52121.5

23.52424.52525.52626.52727.52828.52929.530313436383940

Note : The value of (e) calculated by equation(12) is the maximum value, as the dead load-.(Wt)is minimum. The eccentricity for other conditionsof dead loading may be obtiined by substitutingthe proper weight in place of (W,).It^ha; been ihown-by previous discussion thatthe following relationships exist:^Wt5r-: -f-^Mrz: _7_z- l-ccombining equations (10) and (11)M'cc- -J2- I

rearranging equation (12)Mr: Wte

combining equations (12a) and 13)Wtecc --'- I

(2-a)( 10)(11)

( 13)

(12a)

(14)

s.2312. 6816.4521.9027.9034.9042.8052.0062.7074. tO87. 30101.60117.60135. 00154.10774.50227.00198.00248.50277 .OO309.20342. 00375. 00416. 00455. 00497.(n543.00590. 00624. 00652. 00731.00811. 00873.00933. 001005.001085. 001145. 001240. 001320. 001400.001490.00i585. 001685. 001787, 001900.002010. 002110.002220.002350. 002470.OO2600.002740.OO3021.003330. 003660. 003980. 004370.004730. 005130.005580.006020. 006500.00


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It w.a_s shown by equation (1-a) that in orderto avoid tensile stress at E (which would makethe foundation unstable), the maximum value of(Sr) is as follows:ocTAOOIIa = o.gzgdl-c : o.s4tdr = o.o55d+z I o.Ioledr: 0.257d

Sz: Sr*thus making the value of (S-sn) equalshown by equation (1-d)..-Substituting the values obtained by(2-a) and (1a) in equation (15)

(ls)to zero, asequations

(18)for stable( le)

(20)

(21)

(22)

Wtec WtlaThe value of I can be expressed as follows:

T - ^-2 - alwherer: radius of gyration of base (feet)substituting in equation (16)Wtec W,ar' aH_gnce, the maximum value of (e)conditions is

r"enax: IIn the case of a circular foundationdC:T

Substituting in equation (19)2f'-"" - d

e^^,:0.122 d

Mr 200,000 foot poundse:=w;: I25,700 Ibs. -- 1.59 ft.

., Substituting the value.of (rr) in equation (21),the maxrmum value of (e) for a circular base is(*), ,n"r confirming the commen rule that in astable foundation the resultant must fall withinthe middle-quarter of the diameter of a circularbase.In the case of the octagon base usually used fortow.er foundations, the maximum allow;ble eccen_tricity becomes

( 16)(17)

uE)ucoud = O,866d,2c = 0.577dr I o.o6d'z = o,1o4d'r : o.e64dSAUAEd= d2c : o.?o7dd4I TI'z : o.Ilgdtr: 0.289dOIRCIAq: o.?g6{d2c -.8-r : o.o{gd+z : o.og8dt13 A-?rhe area surrounding the center of the base,within which the resullant causes a compressivestress over the entire base, is known as the kernelor kern.. ,{t f9l.19ws, th.en, that the resultant must alwaysfall within the kern of the base in order to assurestability.. In-exampl.e No. 1 (Section Z), it was shown thatthe foundation is stable, since the overturninsstress ( r) i. less than the minimum dead loaEstress (S.1").The_stability of this foundation will now be cal_culated_ (as example No. 2) on the basis of theeccentricity for the purpose'of comparing the twomethods.From equation (12) the eccentricity is

Eremenrs llnll"loi,,* r**(Axis A-A)than the maximum p_ermissible eccentricity (1.64the foundation is stable, thus confirming ihe corrclusion reached in Section ?.

9. Method of Calculating Soil Load FromEccentricityIt is possible to calculate the soil loading (toepressure) as a function of the eccentricity: Thismethod will.b_e explained in order that it inay becompared with the method described in Seciions3, 4 and. 5.Let (k) be a factor by which the dead load press.ure must be multiplied-to equal the soil loadingdue to overturning as follows IFrorn equation (ZZ), the maximum permissibleeccentrlclty 1se^",: 0.722d: 0.122 X 13.5 : t.S{ 11.Ipasmuch as the actual eccentricity (1.S9) is less (s7

9kSr:Sr


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.)c----J n

FIGURE 3bFIGURE I

Substituting in equation (1)S: S,*kS,or S:S,(1+k)From equation (2)wc-.....:_rr_ aFrom equation (10)

Mrc- -u,- ZthereforeMr: SrZSubstituting the value of (Sr) from(57)

Mt-S,kZtherefore^Mrr: TZFrom equations (2) and (62)

"-w-Mr'- a - kZand wkztvlr -aFrom equation (12)

M,e: wi

Mre: w-Equation (64) can be written

M, _.kZW- a

The value of (e) calculated from equation (12)is maximum for'any particular foundation, whichis the value governing s_tability. At the presentmoment we are concerned with the maximum soilloadins (toe pressure) which occurs---when thedead l"oad is maximum. It is therefore necessaryto substitute (W) in place of (W,) as follows:

Since the ,.trn (#) occurs in both equations(65) and (66), it follows thatKZe:-; (67)

and ea1- - -k: -Z (68)Substituting this value of (k) in equation (59)/ ee ) (69)S:S,\r+ zlIn the case of an octagonal base,

a:0.828 d'3 (70)Z : 0.01016 d8 (71)Substituting these values in equation (68)

8.1 5ekoct.ro": --6- U2)Therefore, for an octagonal base, equation (69)may be written:/ 8.15e \soctsson: s,(r+?/ (73)For comparison, the maximum soil loading inexample Nb. 1 will be calculate4 (as example No.3) on^the basis of eccentricity' From previous cal-culations, it was found that:Mt: 200,000 foot poundsW: 174,200 poundsTherefore by equation (65)

200,m0e:-ffi:1.15The maximum soil loading due to the dead load(Sr) was found to be 1155 pounds per square foot,and (dr) is equal to 13.5 feet.Substituting in equation (?3)/ 8.r5 x 1.15 \S:1155(r+ff/

S : 1155 X 1.693: 1955 pounds per square foot.This checks the value of 1958 pounds per squarefoot calculated (by slide rule) in Section 7, thus

(s8)(se)

(10)(60)

equation(61 )(62)

(63)(64)

(12)

(6s)

(2)

IO(66)


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FIGURE 3e

)c/l---]ni,-lt

_._.___:t.-.

indicating that either method yields the sameresult.10. Soil Loading at Any PointHaving calculated the eccentricity, it is a simplematter to determine the unit stress on the soil atany point whose distance from the centroidal axisis (c').The unit stress on the soil, from equation (1),is as follows:S: S, * S, (l)

Since the value of (Sr) for points to the right ofthe axis is negative, the value of (S) will be :S: S,-S, (see equation 1-a) (1-b)Equations (1) and (1-b) can be combined asfollows:S:SrtS,From equation (2)-_w", -;The value of (Sr) can be

^ Wec'Sr: - ^l (see equationsThereiore,^ W Wec'a Zt'Simplifying:t:+(,*5)

obvious for this purpose that the same value of thdead load should be used in the calculation of theccentricity (e), by means of equation (12).11. Stresses in Tower ShellThe steel shell is required to withstand thstresses resulting from, (a) the internal pressure(b) the dead load; (c) the overturning momendue to the wind pressure. This discussion will bconfined to the stress resulting from the windpressure.It may be assumed, in determining the stresdue to the wind pressure, that the tower isvertical beam, and that the wind produces a bending moment. The ordinary formulas for determining bending moment and stress may therefore beapplied, as follows :

*': n- (+)whereMt: bendingpounds).also ^ MtcS,: __l_moment about base of tower1-c)

(2)

(23)

(26(foot

whereSt : unit stress in tower shellmoment (Mt). (lbs./sq. ft.)Note: The unit stress in the tower shell (S,) iscalculated in pounds per square foot in order to beconsistent with the other calculations which arein foot-pound unts. Steel stresses, however, areordinarily given in pounds per square inch. Inorder to convert the stress from (pounds/sq. ft.as calculated, to (pounds/sq. in.) it is necessarto divide (S,) by 1aa.The shell is a hollow cylinder, in which case

Dc:-T Ggand l: 64 (D'- D't) Q9whereD : outside diameter of tower (feet)Dr: inside diameter of tower (feet)when t: thickness of shell (feet)and

(27due to bendingritten:

14 and, 17)

(24)This value of (S) is the total unit stress at anypoint whose distance (in feet) from the centroidalaxis is (c').It is important to note that equations (1-a), (1a)and (1?) referred to above were used to deter-mine the stability and the eccentricity under thepoorest condition, which obviously occurs whenthe dead load is at its minimum. Equation (24)can be used to determine the stress under anydead load, therefore, equation (24) may be based

on either (W) or (Ws) depending on the deadload for which the stress is to be determined- It is (30(31ID - Dr:11Dr:D -2t


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Substituting in equation (29)I: 64 [D'- (D -zt)'lSubstituting the values of I and(27)

Mt 32M.D- ,]D.-(D]ZDTt: #rr'- (D-2t)4132 MtD:32 MtD:@ G2)

The values of t2, t3 and ta are quite small andthe three terms in the denominator containingthem may be neglected without introducing ap-preciable error. For practical purposes, therefore,equation (3?) may be written as follows:^ 32MtDs1: gTpq-

This equation may be reduced to:^ 4M.5t: z'Dl

Assuming that the number of bolts is repre-sented by (N), each bolt will be required to carrythe stress over a portion of the circumference rep-rbsented as follows :

4\{t W"TDJ - qrDb (37)

,rDt-T-The load to be carried byexpressed ,lrDn / 4M, W. \--:r: -N \ z.Do, - rD, )4Mt W":ND"- NwhereSr: maximum load on each bolt (pounds).The nut should always be tight, placing someinitial tension on the bolt. Of course due allow-

stress per foot of circum-anchor bolts is

(38)each bolt can be

The maximum tensileference to be resisted by(31)c in equation

D2

(3e)

The thickness of shel1 plate required to resistthe bending moment only, is therefore4Mtr - z.D,St (33)

By multiplying the stress in pounds per squarefoot (S,) by the shell thickn.ess (t) the stress perfoot of circumference is obtained as follows:,S,:#

(32-a)

(32-b)

(3s)

circumfer-

(36)

(34)

12. Foundation Bolts for Self-Supporting TowerThe foundation or anchor bolts for a self-sup-porting tower are required to resist the overturn-ing moment (Mr) resulting from the wind pres-suie, after allowance has been made for the resist-ance offered by the weight of the tower. Obviouslythe t'esistance offered by the tower's weight isleast effective when the minimum weight is act-ing. The anchor bolts should therefore be calcu-lated for the condition existing when the toweris empty and without insulation, platforms, etc.This weight will be designated by (W").It was shown that the maximum stress per footof circumference due to the wind moment can becalculated by equation (3a). That equation givesthe stress at the circumference of the shell, how-ever, at the present moment it is desired to deter-mine the bolt stress making it necessary to sub-stitute (Ds) in place of (D). The sffess per iootof bolt circle circumference can thn be written:4Mt,rDFwhereDn: diameter of bolt circle (feet).

The compressive stress per foot ofence due to the weight of the tower is. 'W";D;t2

____--1.------ l'---.,L-A-

\--.LL/--ll./z'.r. \ Sl\../t'r'/ /\ /',/ .,' ,'Yi{iii i|i$i

Et

4"M

SLOPE

N. ry,

O DRAIN.t -r-tiHTl :llllll.lrLltltlLlriil r| ccourCHAMTER--'T',----___L---- x---

6:ro'FIGURE 4


13/96

ance for the initial tension should be made indetermining the size of the bolt, and the strengthof the bolt should be based on the area at the r6otof the thread. An additional allowance, usuallyr/s", should be made for corrosion.The number of foundation bolts should neverbe less than 8, and should preferably be 1p ormore, as the larger the number of boltj, the betterthe stresses are distributed, and the less dangerresulting from a loose nut on one bolt.The bolt should be embedded in the concretefoundation in such a manner that the holdingpower of the concrete will be at least equal to thEfull strength of the bolt. It is common practice touse a washer at the lower end, or to bend the endof the bolt to form an "L" for the purpose of an-choring the bolt in the concrete (iee'Figure ?).18. Guyed Towersfn cases where the tower is very high, it issometimes found desirable to mainiain "stibititvb.y melgq of guy wires rather than a large foundi-tion. Although it is not uncommon to find twoor even three sets of guy wires on one stack, tow-ers seldom have more than one set, and even thesecases are rare. This discussion, therefore, will beconfined to towers with one set of guy wires.

_ Four .guy wires are usually useE ior each set,although in some instances three, and in othersas many as six have been used. They are attachedto a.rigid_ collar which is located it a point ap-proximately 2/3 (sometimes %) of the towerheight above the foundation.14. Pull on Guy Wires

_ The. maximum pull on the guy wire occurs whenthe wind blows along that wire, and each wiremust be designed to take care of the entire windreaction at the collar.- -Th" pull on the guy wire can be expressed asfollows l

the.sum of the pull due to wind pressure andinitial tension as follows:R": (R" * Rt) cos owhereR",: vertical component of pull on guy(pounds)Rt: initial tension on wire (pounds).

The value of the reaction at the collar (R") mb.e determingd by calculating the moments abthe base of the tower (the-top of the foundatioThe wind moment was found by equation (2to be"- (+)

. The resisting moment arm at the collar istherefore the reaction (R") may be calculatedfollows:-_n-(+)h,or - P-H,."__Zlr,_whereh,: height from

(wire

((

top of foundation to collar (fee15. Foundations for Guyed Towers. It.was shown by equation (1) that the total stoadrng. to be considered in the desisn of towfoundations, is the sum of (S,) whictiis the deload, and (Sr) which is the toia due to the oveturning, or wind moment. In the case of the guytowers, there is no overturning moment, hoievethe wind pressure does have .-an imporlant effeol the foundation, as the soil is required to resth.e vertical component of the p,ril on the guw1res.. For guyed t-owers, therefore, equation (1) mube revised as follows:

, S:S+S" (4wnereSs: unit soil loading d-ue to the pull on the gwire. (pounds/sq ft.)Rr:#- (40)- Rr: Rc csc I (41)whereft": pull on .guy wire due to wind pressure(pounds)Ro: horizontal wind reaction at collar (pounds)d: angle that the guy makes with tlie vertical(degrees )

The value of the angle a will usually lie between30 and ?5 degrees.The vertical component of the pull on the guywire can be expressed in any o? the follow'inways:

The value of (S") can be determined as follow^R"a_ aFrom equation (2)

Substituting in equation (49)^ W+R.a

(s

(^w aR" )( cos dR"Xcos0- Si, a-RoXcotd

(s(42)(43)(44)

_ It is important to note, however, that there isalways some -initial tension on the guys whichmust be considered. This initial tensi-on mav beassumed to be 5000 lbs./sq. in., which amounis to1900 lbs. lor /2" wires and ZSO tbs. for fu,, *ir.s.The weight of the wires may be neglected, whenusing these figures.The actual vertical component will be a func-tion of the total pull on the guy wire, which is

16. Foundation Bolts for Guyed TowersThe foundation bolts for guyed towers are rquired to resist the shearing iction of the winpressure at the base of the tower. It is obviouthat ample allowance should be made in the sizof the bolts to provide for the initial tension duto tightening the nuts, and also for corrosion.The shear at the base of the tower, which musbe resisted by the bolts, is equal to tlre horizontareaction to the wind pressur-e at that point. Thiis equal to the difference between the^total winl3


14/96

pressure and the reaction at the collar and canbe expressed as follows:Rr:P--Ro $2)whereRr : horizontal wind reaction, or shear, at the baseof the tower. (pounds)17. Stress in Shell of Guyed TowerThe wind pressure acting on a guyed tower pro-duces a negative bending moment at the collar,

FIGUR.E 5

and a positive bending moment between the baseand the collar. The maximum values of these twomoments can be calculated as follows :

has been widety published. There are really. twoformulas; one ioi piles driven with a drop ham-mer, and another for piles driven with a steamhammer, as follows:For drop hammer

o_ 2Wrf .^ - po+l.oFor steam hammero_ 2Wrf'- pof 0.1whereP : safe load which each pile will(pounds)Wh: weight of hammer. (pounds)f : height of harrimer fall. (feet)ph: penetration or sinking rrndersound wood. (inches)

(ss)

(s6)support.

the last blow, onCare should be exercised in driving piles, toassure that they are deep enough to develop theirfull strength, but they should not be driven toomuch, as this practice results in splitting or break-ing, and greatly reduces the load carrying capacity.Although piles have been driven with a centerto center spacing as small as 2' 6", itis stronglyrecommended that this distance be not less thbn3' 0". Closer spacing disturbs the ground suffi-ciently to greatly reduce or destroy the frictionalresistance.The top of the piles should always be cut offbelow the water level, otherwise they will decayrapidly.The reinforced concrete cap is constructed ontop of the piles in such a manner that the pilesexlend about 6" into the concrete (see Figure 6).

19. Stresses in FoundationAfter having selected a foundation of such sizeand shape as- to fulfill the requirements of theproblem-from the standpoint of stability and soilioading, it becomes necessary to calculate thestresses in the foundation itself, to see that theydo not exceed the allowab1e limits.The first step in this procedure is to determine

P-M":-lE-(H-h,)' (s3 )r\.r,:- *(r-+) G4)whereMc: negative bending moment at collar. (footpounds)Mn: maximum positive bending moment betweencollar and base. (foot Pounds).Having determined the bending moments, the

stress in a given shell, or the shell thickness re-quired to reiist the bending moment -may be.-cal.-iulated by substituting the value of (M") or.(Mr)in place ot 1U,; in equations (32-b) and (33).18. PilingIn cases where the safe soil loading is very low,it is sometimes found difficult to design an ordi-nary foundation which will not overload the soil.In such cases it is desirable to support the loadon piles rather than on the soil.Wooden piles are ordinarily used, and they varygreatly in length, depending on the nature of thesoil. The diameter at the lower end is about 6";

and the diameter at the top is about 10" for pilesnot over 25 feet in length, and 72" for longer piles.Wooden piles generally depend on the frictionalresistance of the ground for their load carryingcapacity, as they have comparatively little strengthas columns. The safe load which a pile will supportvaries greatly in different localities. Building lawssometimes g'overn the pile loading, and in suchcases, the load is usualy about 20 tons per pile,although occasionally 25 tons is permissible.When conditions are not definitely known, how-ever, the only safe procedure . is to drive a fewpiles for test purposes. The.dommon method ofcalculating the safe load is by means of what isknown as the "Engineering News Formula," whicht4

FIGURE 6the loading, which consists primarily of the up-ward reaction of the soil. Figure 3 represents theplan view of a typical (octagonal) foundation, andFigure 3a shows the loading diagram. In this dia-gram the dead load (Sr) is represented by therectangle (jklm). The wind load (Sr), which ispositive on one side of the centerline, is indicatedby the triangle (-p*). On the opposite side of

+-=t=*i--it+


15/96

the centerline the wind load is negative, therebycounteracting a portion of the dead load (wlc).The actrral soil loading will therefore be iep.e-sented by the area (jkcp). However, the weighi ofthe bas-e, and of the earth fill above-the base"(areajkno -Figure 3b) do not exert any upward forie onthe foundation, and may therefo.e be ded.uctedfrom.the total load, for the present purpose. Theeffecttzte upward reaction will then be the area(oc, p) in Figure 3b.19a. Diagonal Tension

The vertical shear, resulting from the upwardreaction of the soil, producei diagonal t6nsionstresses in the foundation. The critical section liesat a distance from the face of the pedestal equalto the effective depth of the base, as indicatecl9y point (Zr) in Figure 3c. In ottrer words, thefoundation tends to break along line (ZZr).'Thevertical shear to be resisted is equal to ihe net soilpressure on the part of the foundation outside thecritical section._ For design purposes, therefore, the load will bethe area_ (oq.p), (Figure Bc) applied over thearea.(a, b, Ig), (Figgre 3). Becaui6 of the irregu-lar shape of the load diagram, its magnitude Eanbe more conveniently calculated by breaking it upinto its component parts, the totalioad (V")-beingthe sum of the individual loads, as follows:- shape or port o"'lt$r""lif*" outline In elevatlon-Rectangular Prism a, b, u, i, oqrvWedge ar tr g oqrvWedge b, fu-, oqrvpedge. ar br ur tr rvpIvramid ?r tr g rvpPyramid b, fu]. rvpThe unit stress (diagonal tension) resultingfrom this vertical shear load can be 'determinedas follows:

,- V"- 'd- b'jdr (80)wherefd: unit stress in_ concrete (in diagonal tension)due to vertical shear load. (pounds/sq. in.)V": vertical shear load, outside the criticil section(see Figures 3 and 3c). (pounds)b' : width of critical section which serves to resistdiagonal tension stresses (line a, br Figure 3).(inches)j: ratio of lever arm of resisting couple to depth(dr) (see Table 2).dr - effective depth of base measured from top ofbase to centerline of reinforcing steel. (inches)Erample No. 4. Check diagonal tension stressesin the foundation considered in example No. 1 :Figure unit soil loading due to weight of baseand earth (see Section 7):

162Line (mw): Z :81"do:72". dt :21". dt:162"do:72+21+21:174"Line (gf):67.1'(see Table 1).Line (m, *1 : :57'.67.1 Y 57Ltne (ar b,) - -- g1 : 47.2" : (b').67.r - 47.2Line (gtr) : - --2 -:9.95"162 - 114Lrne (a, t,):

-2- -24"Factor j: .87 (see Table 2).Load (m, rl : 4lrf{:5651bs.,/sq. ft.Load (qr) : 565 + 522: 1,087 lbs.,/sq. ft.

Calculate shear load (V")47.2'xz" xrffif_tg.s5" x24' x #47.2 xz4 xZ#449.95 X 238y.24:X23)/'144Total (V")

Calculate unit stress in concrete

: 8,550 lb: 1,805: 935: 263: 11,558 lb(equation 80

Maximum unit wind toad (S,), (mp)Maximum effective unit shear load (op)

63,000 lbs.32,700|bs.D57oo tr..95.700 lbs: ffi : 633 lbs /sq' ft'(jm, figure 3c) : 1,155 lbs.,/sq. ft.weight of base : 633:-Wlbs./sq. ft.: 803t,325 tbs.,zsq. ft.

_ I 1.558o: 47 2 x .87 x 2l : l3'4 lbs''/sq' in'This stress is satisfactory, as 40 lbs./sq. inwould be allowed (see Table 2).19b. Depth of Slab Required for Punching SheaThe thickness of the foundation slab (bottomcourse) must also be sufficient to withstand thetendency to shear along line (Z-Zr), (Figure 3cat the edge of the pedestal. This shearing load maybe determined as follows:

S,: S.* S. (81The stresses in this case are not distributed ovethe foundation area, but are concentrated at theedge of the pedestal.ThenSu: total maximum unit shearing load. (lbs. pelineal foot of pedestal perimEter).Sr : unit shearing l,oad due to dead load. (lbs. pelineal foot of pedestal perimeter).Sc: maximum unii shearinf load du6 to overturning_ moment, (lbs. per lineal foot of pedestaperimeter).

_ The value of (S*) can be determined by addingthe weight supported by the pedesta[ to th;weight of the pedestal itself, subtiacting the load-carrying value of the soil directly under thep.edestal, and dividing the difference by the pe-rimeter of the pedestal base as follows: -

- W" + W. + W,- (ap S"rr)J' - L, $2)whereWr: weight_ of foundation pedestal (top course).(pounds)^ a, : plan. area of foundation pedestal. (sq. ft.)S.rr: maximum allowable unif soil loading.(pounds,/sq. ft.)Lr: perimeter of foundation pedestal. (feet)

Obviously, if the value of (an S"rr) is equal tot5

Concrete baseEarth fiIlTotalUnit soil loadingTotal unit dead load (S,)Unit dead load due toand earth fill (jo)Net soil load (om)


16/96

or greater than (W" + W" * Wq),_the value of(Snl becomes zero, and (S.) will then be equalto (So).The'value of (Su) can be determined in a man-ner somewhat similar to that proposed in Sec-tion 12. In that section the overturning load wascalculated as a function of the periphery of thefoundation bolt circle, by means of equations (27)and (35). The bolt circle was assumed to be ahollow iylinder, the wall thickness being infinitelysmall, as compared with the diameter.In the determination of . the shear at the edgeof the foundation pedestal, a similar proceduremay be followed, substituting (M1) in place of(Ml), and appropriate values of (I) and (c) iniqui[ior, (ZZ)-, dtpending on the shape of thepedestal.- Reduced to their simplest forms, the equationsfor the ordinary foundation shapes are as follows:

Octagon-- Mt16 - .814dp'?Ilexagon-- Mtro - .832d,'SouareMr5r--ffipCircleJo- .7g5dn'

In these equations (dn) is the short diameter ofthe pedestal (feet).Once the shearing load (Sr) per foot of pedestalperimeter is knowri, it is a iimple matter to cal-iulate the unit stress in the concrete, by dividing(Sr) by the effective depth of the base, as follows:-S,ro- Ttwheref" - unit stress in concrete base' shear. (Pounds/sq. in.)Note: The factor 12 is introduced for the pur-pose of converting (S") from (pounds/lin' foot) to?pounds/lin. inch) is- unit stress (fn) is in termsoi (pounds/sq. in.).Erarnple No. 5. To illustrate the procedure, thepunching shear will be calculated for the founda-iion considered in example No. 1.Calculate dead load shear (S.) by equation (82)

a,: 29.8 sq. ft,Ln:2.484 \ 8:19.87 ft.W. :48,500 lbs.: 17,850 lbs.S.rr : 2,000lbs./sq. ft.

(84)due to punching

- (29.8 X 2,000)

RectangularPrismWedgeWedgeWedge

Outlltro h dor,rlc.3ilabut

atgbfu

abut

(8la)

2Distance (at) \23Distance (at) XZ3

Distance (at) X23Distance (at) X2

(83)

(83a)

(83b)(83c)

This stress is satisfactory, as 120 pounds/sq. in.is permissible. (See Table 2.)In the case of guyed towers, or stacks, the shearload due to overturning moment (Su) does notapply, but is replaced by("*,)which is the load due to the pull on the guy wires,as follows:

^Rr3lguyed) - S.* Ll9c. Reinforcement of Base for Upward BendingReaction of SoilIn designing the base of the foundation to resistthe bending moment due to. the upward reaction ofthe soil, the critical section is located at line (ab),(Figure 3d) along one face of the pedestal (topcourse). The moments are therefore figured aboutline (ab), on the basis of the load on the trapezoid(abfg). The load which serves to produce thebending moment in the base is the "unbalanced"upward reaction. Since the weight of the base,and the weight of the earth filI above the base donot contribute to the bending moment, they maybe deducted from the total load when calculatingthe bending moment. The effective loading willtherefore be represented by the area (o qr t, p)Figure 3e.The load is assumed to be applied at its centerof gravity, and the moment figured about line(ab). Due to the irregular shape of the load dia-g'ram, it is difficult to locate the center of gravity,and it is therefore more convenient to break it upinto its component parts (prisms, wedges, pyra-mids, etc.), and figure the moment of each partseparately. Obviously, the total moment (Ms) willbe the sum of the individual moments.In the case of the rectangular prism, the leverarm used in figuring its moment will be one halfof the distance from point (a) to point (t), (Figure3d). In the case of the wedges and pyramids, thelever arm will be two-thirds of the distance frompoint (a) to point (t).The individual components and their respectivelever arms are as follows:

Outlino ln eleva-tion, Fig.3c Lever ArmDistance (at)

W.: 30,000 lbs.W, - 119.2 cu. ft. X l50lbs.

QrfrVrOQrfrVrOQrfrVrO

frVrpfrVrPS.- Pyramid atgPyramid bfu Distance (at) X 2frVrP - 3: 1,850 lbs./lin. foot.

Calculate shearing stress due to overturningmoment (S,) by equation (83)Mr: 200,(X)0 foot pounds (see section 7)'dn": C - 36,200.m0S': :ffifi0 :6,820 Pounds/lin. footS',: 1,850* 6,820:8,670 pounds/lin. foot. (81)dr- 21"8.670t -::34.4 pounds/sq. in. (84)p- l?)(zl -t6

In calculating the amount of reinforcement re-quired. it is asJumed that the portion of the base'designated (abut), (Figure 3d) acts as a cantileverbeafi (of .eitattgula. cross-section) having a widthequal io one fa.-ce of the pedestal (ab), a deptheciual to the effective depth of the base (d1) andalength equal to (at).Having ialculated the bending moment as, pro-posed ab-ove, the 19rt step is to check the depth


17/96

where(1r -

of the base, and determine the amount of reinforc-ing steel required. These calculations are basedon the commonly accepted formulas for reinforcedconcrete. (It should be noted that for this purposeit is more convenient to figure the momenti interms of inch-pounds, as the stresses in concreteand steel are usually given as pounds per squareinch, whereas in figuring soil lbading foot-poundunits are used, as soil loading is uiually itatedas pounds per square foot.)For balanced desi(tn, that is, conditions in whichboth concrete and steel are stressed to their fullallowable capacity, the required depth (ds) of thebase may be rletermined as follows:

a,:i#k_

should therefore be placed within the limitsthe beam width (ab). However, additional rinforcement should be installed to reinforce thbase between the points (gt), and also at (ufus.rng_ the same type and spacing of bars as detemined for the beam section (ab). This additionreinforcement insures that the entire area of thbase is reinforced and weak spots eliminated.Obviously, the reinforcingbars should extenentirely across the base. Also, there should beset o{ reinforcing bars parallel to each of the axei.e., four sets of bars for an octagonal base, thresets Jor. a hexagon, etc., thus providing strengtin all directions.There should be at least 3 inches of concrebelow the reinforcing bars at the bottom of thbase. Reinforcement in other parts of the foundtion should be covered with not less than 2 incheof concrete.

E.rample No. 6. Determine bottom reinforcemefor the foundation referred to in example No. 1.Figure bending loads72"Line(m,w):, -36"

Load (m,.,, : 9i*4 : 357 pounds/sq. ft.Load (q, r,) : 357 + 522 - 879 pounds,/sq. ft.Load (v, p) : 803 - 357 :446 pounds,/sq. ft.Figure moment (M5)Line (ab) :29.8"Line (ta) :45" ,Line (gt) : 18.65"29.8',Y 45"-i?4 tql" X 879 Pounds'/sq' ft'18.65 X 45 45 X 2- u4 x87ex---29.8 X 45 446 45 X 214442/\3

45"X 2 :184,000in'-lbs: 153,000: 62,300

(8s)

(86)

(87)

Mo:f":depth of base, measured from top of concreteto centerline of reinforcing steel. (inches)bending moment in base. linch-pounds) 'safe working stress, reinforcing steel in ten-sion. (pounds per sq. in.)("/",0, ) : ratio of effective area of rein-forcing steel to effective area of concrete.ratio of lever arm of resisting couple to depth(d').rvidth of beam (line ab, Figure 3d). (inches)effective cross sectional area of steel reinforce-ment in tension. (square inches)

j-b,:A":If the design is balanced, that is, the actuald-epth of the base (d1) is that calculated by equa-tion (85), the value of (A") may be determinedas follows :A": b" dr p"If the depth (dr) is greater than required byequation (85), in which case the steel is stressedto its full capacity but the concrete is under-stressed, the value of (A") becomes :

.Mr,'" - f" jd, 18.65 X 446X 45144X3Total (Mo) '/)v 45X2/\e/\ 3 : 51,900.:451,200 in.-lbs. If the depth (ds) is less than required by equa-tion (85), it is recommended that ihe diminsionsof the base be changed to give the required depth.In case circumstances make it impossible to in-crease (dt) to the required dimension, it will benecessary to increase the amount of reinforcementused. The determination of the amount of rein-forcement required for sdch special cases is be-yond the scope of this article, and reference ismade to the various publications dealing speci-fically with concrete design for further detaili.Having calculated the cross sectional area ofsteel required, a selection is made as to the dia-meter,.shape, number and spacing of bars whichwill give the required area. It ii recommendedthat the center-to-center distance be about 4 inchesif possible, but not less than 2l times the bardiameter for round bars, or 3 times the side di-mension for square bars. Generally speaking, alarge number of small bars (1, s/s, o, % inch)-arepreferable to a smaller number of larger bars.It should be borne in mind thaf the area ofreinforcement determined above is the amount re-quired for that portion of the foundation havinga width equal to (ab), Figure 3d, which was as-sumed to be the cantilever beam carrying the en-tire bending load. This amount of reinforcement

Check depth of base for balanced design (equation (85)f" : 18,000p. : .0089j: '87b": 29.8"f"P"j:138.7 I - 4il:oo0t(hnrlnccd' : Y t.lg-z x, zg: :10.5"Since the actual depth is 21 inches, whereaonly 10.5 inches would be required, the concretwill be understressed, and the area of reinforcinsteel should be calculated by equation (87).

451.200A" : tsPoo x .87 x 2l - 1 37 sq' in'lJse )1 inch deformed square bars (.25 sq. inarea).Number required #+ :5.5. Use 6 bars withinthe width of beam (ab).29.8"Spacing ? :4.96". l;se S-inch spacing,-entirely acros. -tid" (gf), which will require {11

:13 bars per set. Four sets of bottom reinforcingbars will be required for the octagonal foundation


18/96

t7

19d. Reinforcemnt to Resist StressesDue to UPliftAs explained previously, the wind moment cre-ates a positive sbil load on one side of the center-fi""-, utia i negative load on the opposite side' Inoi-frlr-*otas, ihe action of the wind tends to lifttt " fo""a"tibn on the negative side. This upwardi;;;; oi "uplift" effect, G resisted by the -*gigh;i iht "o""i"t" base itself, and by the weight- ofihe- earth fill on top of the base. It therefore be-cornes necessary to reinforce the top,of the base,to resist the reiulting negative bending moment'--tfr" pro"edure is qiit" ii-itur to thaf describedfoi1t "'rp*ard soil'reaction (Section 1.9c)' Theioua r.tt in the area (abfg), and the outline of theitt"o."tiiut beam carrying ihe load is (abut) as inS..lio" 19c. Howevei, i., this case the load is the*Li"t,t Der square foot of concrete base, plus the*"iilt i i"r tor"t" foot of the earth fill, and is uni-ir#it iistrituted, thus simplifying the calcula-ii;;:'Aiter figuring the moments, the reinforce-ment is determined 1n exactly the same manner asexptained in Section 19c, using the equation

Moment2% L;570X +-E%q y57oX 4+ZTotal (M")

From equation (87a)218,500 inch lbs.A":-l8p00X.87X22

: 119,000 in.-lbs.: 99,500: 218,500 in.lbs.

-.636 sq. in. within beamwidth (29.8")

^-Mo^'- f" jdt

IABLE 2GonrlonE Apptylng io Foundotlon Dcslgn

lJse rft-inch deformed square bars, at 10-inchcenters.

19e. BondIn order for the reinforcement to be effective,the'-st-rength of the bond between concrete and;;i;;t? be sufficient to permit the reinforce-*""1 t" develop its full strength. The bond stress;;;- t" ialcul'ated by meais of the followingformula: V,U: EJA; (88)where""-';: bond stress per unit of area of surface of bar'(Pounds)>": ;;;;f 6erimeters of bars -within the limits or-it. beam width (ab)' (inches)Erample No. 8. Check bond stresses in exampleNo. L.Bottom reinforcement

V. - 11.578 lbs. (See examPle No' 4)>.:6 X.5 X 4:12"By equation (88)

11.578u:-nffil:53 lbs.Bond stress for bottom reinforcement is satis-fr.tlil "t ts por"as is permissible (see Table 2)'Top reinforcementFigure sheargz#y- X 57orbs.2W-xsto

Total (V.)2o:3X.5X4:6s427$:- 6ffikT -48lbs'

(87a)

: 300lbs./sq. ft.:27Olbs./sq. ft.:T-701bs.,/sq. ft.

In this case, (ds) is the depth of the base fromthe centerline'oi the upper layer of reinforcement;;ih;-i;;i;* of the bise, and (M-) s the bending*o*.ttt a"" to the uplift forces (inch--pounds)'Eramble No. 7. Determine top reinforcement tor.Sri "'pfitt in the foundation ieferred to in ex-ample No. 1.Weieht of concrete150 lbs./cu. ft.:r.Ztt.Weieht of earth60lbs.,/cu. ft. X 3 ft.Totat

:4,480 lbs.: 947:5,427 lbs.

(88)

* These figures may be slightly incresed by making "U"-bends on the""a";l;|iif,l.i",i[f *?Ii;," is recommend_ed as most satisracrory ror rounda-tionJ iiijre tipe-. - riri ionstantJ ttiiiiii ilztr niiture are presented as a matterof intqest.

The bond stress in top reinforcement is satis-factory, as ?5 pounds would be allowed.19f. Bearing Stresses

The bearing stresses (where th.e steel towerrests on the cincrete pedestal) seldom cause anyaif1."ltu, but should lre checked as a -safety pre-.rrlio".'tfte bearing stresses consist of .the stressa""-1o *i"d pressuie, plus the stress due to thedead load as follows:Bearing stress: 4M*/7tD" + (w'* W")'/z'D' (37a)

(See Sections 11 and 12.)ilquation (3?a) gives the bearing stress inpo""tt p.t ii.,eil Toot of shell circumference'These stresses are spread over the are-a of the base;"". tfr.tfore for prictical purposes the unit bear-i.,gjsttess can be -determined as follows:

Mlxture: Cement'Sand. . .CoarEeAEgreEat''""'I24 I25

fb Safe baring load on concrete (lbs./sq' in') ' 500 375fi Ultimate compressiv strength (lbs'/sq' in') 2.0(n 1.500f. Safe unit stress in qtreme fiber of concrete--ii. ompression)-(lbs./sq. in').." " "' 800 600fa Safe unit stress in concrte due to vertical--"i".i-fai.co"al tension) (lbs./sq' ir). ' 40 30fDf.

(f. i)(f. ir,iF%J

{"=oikJ

Safe unit strss in concrete base due tonunchinc sher. (lbs./sq. in.)...'..'.. 120 90Safe working stress, steel reinforcement intedsion. (lbs./sq. in). . " ' 18,000 18,000(inch-pounds). 15,600 16,000(inch-pounds) . . 138.7 88.9Ratio, lever arm of resisting couple todepth (dr). .87 .89Ratio, modulus of elasticity of steel to thatof concrete, 15 l5Ratio. efrective uea of tension reinforcement to efiEctive ilea of concrete ' ' ' ' .0089 .0056

*u Safe bond stres (concrete to steel rein---i6rje.&ti *i irnit of arm of surfaeof bar. (pounds) 6075 45m

t8


19/96

in which l2r*r,: width of the tower base ring. (inches)fb: unit compression stress on concrete.(pounds/sq. in.)Equation (37b) may be modified slightly, de-pending on the exact shape and arrangement ofihe base ring (or base plate), but in the majority

TOP OFPEDESTAL

fr:

FIGURE 7

of cases it may be used in the above form withreasonable accuracy.For guyed towers, equation (37b) becomes:

this case are given for illustration only, the desighas not been changed to take maximum advantagof the allowable stresses.The stresses in foundations of this type shounot exceed those commonly accepted as good engneering practice in reinforced-concrete design, fthe particular mixture of concrete used. As a mater of convenience Table 2 is presented to shoallowable stresses and miscellaneous constants aplying to two grades of concrete quite generaused for foundations. It is strongly recommendethat the 1 :2 :4 mixture be used in practice, thfigures for the 1 :2:5 mixture being shown primaily as a matter of interest.

19h. Suggestions and RecommendationsThe calculations explained above provide freinforcement to resist the stresses due to thvarious types of loading. It is good practice, however, to install additional steel as a meanstying the foundation together, to form an integrunit. The same size bars are used for this purposas for the main slab reinforcement, and the dsigner must use his own judgment as to the number and location of the bars. Figure 4 represenwhat is considered good practice, and is offereas a guide.In the case of very large foundations, consideable concrete and weight may be saved by costructing the pedestal with a hollow center,illustrated in Figure 5. Of course, the inside foris left in place. It should be noted that the basslab extends all the way across, to provide protetion and bond for the reinforcing bars.Foundations supported on piles should beconstructed as to allow the tops of the pilesextend about 6 inches into the base, with thbottom reinforcement about 2 inches above thpiles. (See Figure 6.)

Considerable inconvenience is sometimes ecountered in setting the tower in place, due to thdifficulty of lowering the heavy vessel over thfoundation bolts without bending some of theor damaging the threads. Figure 7 illustratesmethod of overcoming this difficulty. A sleevnut is welded to the top of the bolt, and so placethat the top of the nut lies slightly below thsurface of the concrete, with a sheet metal sleevaround it. The tower may then be placed in postion without interference from the bolts. Stubolts are next inserted through the lugs on thtower, and screwecl into sleeve nuts from the top

4M. W, + w",r 17+ ?r D,

4M, r,r DF 1- R"+W"+W"z'D,

(37b)

(37c)

Nomencloture

fr:19g. Allowable Stresses in FoundationIt is to be noted that in actual practice the depthof the base in the examples given above could bereduced, if desired. All of the stresses for diagonaltension, punching shear, bending (upward ancldownward) and b-ond in the reinforcement are wellbelow the-allowable values. As the examples in

Ar : elfective cross sectional area of steel rcinforcement ir ten'sion (square inches)For balanced designAr: b" dr pr (86)If depth (dr) is greater than required by equation (85)e" :--14.0. (87)tr ldrFor top reinlorcement of slab to resist uplift strcsses:o.: -r1,, (87a)ts Jdta : area of base of foundation (sq, It.)ar: plan area of foundation pedestal (sq. ft.)B : barometric pressure (inches Hg)ba: width of the critical section (equal to the width of thc face

of the oedestal) assumed to dct as a cantilever beam resist-ing the'bendini stresses (line ab, Figure 3d) (inches)

b'- width of critical section which serves to resist the diagonatension stresses. (line ar br, Figure 3.) (inches)c - distance from neutral axis of foundation base to pointmaximum stress. (feet)c' : distauce from centroidal axis of foundation base to any poinunder consideratiofl. (feet)D: outside diameter of tower. (feet)

Dr : inside diameter of tower. (feet)Do: tower diameter measured over insulation. (feet)Du : diameter oI Ioundation bolt circle. (ieet)dr : short diarpeter of foundation base. (feet)dc : short diameter of critical section for diagonal tensiostresses (see Figures 3 and 3c). (inches)dr : effective depth of base of foundation, measured from topbase to centerline of reinforcing steel. (inches)

.,SLEEVENUT4.F4 .''a 'c'.4.'..y.44.4 ":.WELD

t


20/96

dp : short diameter of foundation pedestal (feet)Ec : modulus oI elasticity of concrete.E - modulus of elasticity o{ reinforciflg steel.e : eccentricitY. (feet)This iactor ia the distance from the centroidal axis of thefoundation to the point at which the resultant of the deadload and the rvind ioad intersects the base of the foundation.The eccentricity can be calculated as follows I(12)

Pw: total wind load (pounds) to be calculated as followslP*. : pc Do H (8)p: wind pressure on a llat surface. (pounds per sq' ft.)Dc : wind pressure on projected area of a cylindrical torver.(pounds per sq. ft.)pb: penetration or sinking oI pile under the last hammer blow,on sound rvood. (inches)pr : ratio, effective area of reinforcing steel to effective area ofconcrete,

'":(A"/u"a')Rr: horizontal wind reaction or shear at base of tower, (poutrds)Rc : horizontal rvind reaction at collar. (pounds)Rs: pull on guy rvire due to rvintl pressure. (pounds)_R"I(s:S.n t or,Rg:cscdRv : vertical component of pull on guy wire. (pounds)R" : (Rs f Rt) cos dRt : initial tension on guy wire, (pounds)

Eouation (12) sives the eccentricitv at the condition ofoobrest stabilirv, that i", with the miiimum dead load. Thisis the value which ordinarily is used for design purDoses,however. it is obvious that the eccentricity for maximuma""a toia ".nditions ian be calculated by ''ubstituting thevalue of (W) in equation (12) in plaCe of (Wt). Themarinum valtie rvhich it is possible for- (e) to have and stillmaintain the stability of the foundation is

tr{r

t2em.x:-,

Oremir :

-

Circle:

Values of (e-"") for various foundation shapes are as r: radius of gyration of the base of the Ioundation (leet). Itsrelation to the moment of inertia can be expressed as follows;

( 1e)(19a)

(22)(22a)(22b)(22c)

(40)(41)(45 )

follorvs:Octagon: ens: 0.122dHexagon; em.': 0.121dSquare: emax : 0.1 18ddem :: .l-The value of (e) as calculated by equation (12), and basedon the minimum dead load (Wt) should ,r.i'.,'exceed thevalue calculated by equations (19) or (l9a).

F: barometric pressure. (inches Hg)f : height of hammer fall. (feet)fu: unit bearing stress on concrete. (pounds,/sq in ) (See equa'tions 37b and 37c)fc : safe unit stress in extreme {iber of coucrete (in compres'sion). (pounds/sq' in.)f"' : ultimate compressive strength oI c-oncrete .(nounds,/sq in )i" : iiiii it*.i in' concrete (in"diagonal tension) due to verticalshear load. (Pounds/sq. in.)fp : unit stress in concrete base due to punching shcar' (pounds,/sq. in. )

I: ar'r earranling:llt:1;

r* : rvidth o{ tower base ring. (inches)S: total unit soil loading. (pounds/sq. ft.)S: Sr * Su also,S:S'(1+k)

(17)(2s)

(l)(59)Sr: unit soil loading due to dead load. (pounds/sq. ft,)Srh: ulit soil loading due to minimum dead load (pounds/sq. ft.)to include the rveight of the empty tou'er, the loundation andthe earth 6ll only. It does not include insulation, platforms,piping, liquid, etc.Sz: unit soil loading due to oyerturning moment. (pounds/sq. ft.)Sa: total maximum unit shearing load, (pounds per lineal footof pedestal perimeter)Sr: Sa * Sr (81)Sr:unit shearing load due to dead load. (pounds per lineal footoi pedestal perimeter)- \Vs+.W"+Wp-(apSr:r')-{ * Lp,,, - t2d, (81 )fr : safe working stress, steel reinforcement in tension' (pounds,/sq. in. )H: height of tower. (feet)hr: height of foundation. (feet)hr : height oI collar (to which the guy wires are attached) abovefoundation. (feet)f : moment of inertia of the base of the loundation. (based ondimensions in feet)j : ratio of lever arm of resisting couple to dePth dr. (See

table 2)k : Iactor bv which the soil loading due to dead load must be-uiiiptiia to equal the soil loa-ding due to overtutning, as{ollows:kSr : Sz, also.ea

7.Values of k for various foundation shapes are as followsa 1


21/96

Simplified Design forTower FoundationsCurves reduce design time for octagonalreinforced concrete tower foundations byquick selection of base size, thickness, re-inforcement area and unit bond stressesAndrew A. Brown, Union Carbide Chemicals Co.South Charleston, W. Va.

DBsrcNBns oF FouNDATroNs have used many differentlocations for sections and beam widths to compute bondshear, bending moments, and diagonal tension shear.Since agreement on these important phases is not com-plete, this presentation uses The American ConcreteInstitute Building Code Requirements as a guide forreinforced concrete design and the allowable unit stressestherein. The usual assumptions are followed as to thebehavior of reinforced concrete and soils.For simplicitn the derivations of fornulas are based.on the inscribed circles of the octagonal base. This doesnot influence the accuracy of the final results. The foun-dation engineer is ever mindful of the fact that a sub-structure design based on inexact soil bearing determina-tions, concrete with variable strength, and loads whichcan be off 10 percent or more, is not very definite. Theapplication of good judgment coupled with experience ismore important than carrying out computations to moresignificant places than the informatior and assumptionswaftant.Foundqtion Size. As the size of the foundation is thefirst_design requirement after the permissible soil bearinghas been established, the formulas used for this determi-nation will be derived in that order.When the resultant of all forces acting on the founda-tion strike the base within the kern, the forces acting onthe. soil can be represented graphically by a right .eg.rlu.cylinder resting on an ungula of a right regulir cylinder.If it is on the edge of the kern the soil reaction forcesform an ungula whose base is a circle; when it is outsidethe kern, the ungula has a base in the form of a circularsegment (Figure 1). The volumes of these solids areequal to the total weight supported by the soil, and theirmoments about the center of the base are equal to themoments of the external forces acting on the foundationabout the same place. Then the eccentricity ,,e,, meas-ured from the center of the foundation equals

eao5a'ieoCsOE=o,-OLL

(M)which is equal to the moment of the forces acting on thebottom of the foundation divided by the total forcesacting on the base.

Resultant Within the Kern. For the condition wherethe resultant is within the kern (the area inscribed by a

lnscri bedCircle Equ i vo lenlSquo reACI 1208

Section for Bondond MornentACI 1204 (o),1205(c)FIGURE l-The resultant of all forces is within the kern

radius equal to /s of. the diameter of the circular foundation) the maximum soil pressure P is equal to the totheight of the right circuiar cylinder and ungula drawto graphically represent forces acting on the base.For this condition the maximum soil bearin

D_, - * (1+g\7R,\ D /and the minimum soil bearing equathe height of the soil pressure cylinder or

t: w (1- B" \z.Rr\ D /Let V equal volume of cylinder and ungala whicequals W, the total vertical load.To get the general formula, let the maximum soil beaing equal unity and h equal minimum soil bearing, thethe total load

w : 7R2h * rz'Rz (1 - h)For a value of |lers than r/s, the maximum soil bearD

ing (unity) can be computed in terms W and D. As aexample, forelD: .tl,pr: * (r + .B) and p*,o. :w .2WA (1-.8): i

external moment of ail forcepTotal vertical Ioad (W)

2


22/96

SIMPLIFIED DESIGN FOR TOWER FOUNDATIONS . . 'If 1.8 is reduced to unity or P, then P,,ir.: f , tn"

height of the right cylinder and the height of the ungalabecomes 8/e.Then .irpz/ 1 \ 5zgz*:-i(\r+L.o)-9-orcuR:Where C, is a coefficient which when multiplied bythe product of the maximum soil pressure and radiussquaied will give W, or the total volume of the cylinder

and ungula for this condition. For the values of f, , thecoeflicients C* were computed, Column 8 Table 1, andC values, Column 9, were obtained by dividing/e\C,R' by * (;/'

Resultant Outside the Kern. For the development ofthe equations for moments and total forces acting onthe base when the resultant force is outside the kern,refer to Figures I and 2, which show this condition'To get the volume of the ungula of height P, whosebase is bounded by the angle -+ d as measured from theX axis, we have dV : d A P'. dV is a volume whosearea of base is dA and height P' and is located a distanceR"o" { from the Y-Y axis. Then by similar triangles

P- (Cos d Cos a)pr - r r \v"." --- / : dA:2R Sin { dx and dx:R Sinddd.- (1-Cosa)

FIGURE 2-The resultant of all forces is outside the kern'c

e/DM/2v e/D.457.436.415.395.374.354.Bt4.294.275.256.238.220.202.185.168.153.138.125

.079.t67.2E4-403.583.8111.099t.4631.92t2.5053.2354.1705.3706.978.S411.6514.9919.60ao.tn

.0603.1045.1516.1988.2448.2879.3275.3626.3927.4172.4354.4493..4567.4557.4537.4408.4297.4t07.3927

36'52',45" 34',53'08',60'66" 25',78'28',84' 16',90'96" 44',r01'32',107'28',113" 35',120"126" 62',134" 26',143" 08',154" 09'

1 80"

.10.15.20.25.30.35.40.45.50.55.60.oD.70.75.80.85.90.951.00

vCvB2 Y /4(e/d\2

Reultant inside kern

Resultant outside or on edge of kern

t000.r0

2?5.250.225.200

.20 .30 .40 .50 .60 70.80901.0FIGURE 3-Curve used to determine soil bearing or diameter of foundation base'

r3.520.r4.75

22

fABLE l-Coeftlcients for Yorlous e/d Volues

.12 | 1.602 *l 27.82.11 | 1.6711 34.52.lo I r.zqsl le.ns.os I 1.8251 56.34.08 | t.grol zl.s,l.oz I 2.013 1102.7.06 I z.tzt 1117.3.os I 2.244 1224.4.04 i 2.880 1371.8.035 | 2.454 l5oo.8

.0660.1198.1823.2518.3269.4068.4S04.5773.6666.7679.8504.94401.0381.128t.2261.3121.4041.488t.57t


23/96

| .s0Il.aol.ru.70.65.so.55.50.45.40

35

.85.80.75-70,65.60.55.50.45.40-

.oo L.to .20 .25 30

Deduclion forFill ond Slob

ootU)oo

450ACt

R(.707-cos4

of Ung

for Sht205 (

.1y substitution,2R9P.dv : -aa- co;) (cos d - cos a )sin:t'dP

and V : a*rro1co, c - cos a) sin2 ddd- 2R2P1 [l'*, -cosa (+-+si^cc.,d )]"1 --Cosa)f lOr V (W) - CRzpl where C:2 f Sin3a , Sina665za-2e65alft-cos"ll 3-]r-- 2.--_J. ,,,The.moment o{ any ungula which represents the forcesapptred on the base of the foundation about the y_yaxis is the summation of the product of th" differentialvolumes, dV and R Cos {.

So dM - 2R3P1(1 _-Cos;; [(Cos d-Cosa) Cos CSin'z ddd]2R3P- la"o M :11-- c* O) ","*' d sin2 d - cos a cos c sin2 d) dc: ^ 3R'3 . [.---f /Lr,"or-r\-cosasin,cla1-Cosa) | B \4""'aY-Y)--3 l"2R3P.

( 1 - C"-",

FIG-U_RE 5-The soil reaction is the sum of the forces inshaded area.or M : CRBP, where C r: --1--l - Cos a)

_ By use of equations I and. Z, Columns Z, 4, 5 andof Table 1 were computed for values of K or angle alphColumn5 :#: +andColumn6:,V, .-

+ (z\''\D/_ Th: curves orr Figure 3 were plotted by using Column5, 6, 7 and g. From these .rrru", the size of tie foundation can be obtained for a permissible soil b"a.irg oconversely, the soil bearing can be computed for a kn"owfoundation. The formula for soil bearing iswP, : C",where W : yeight of foundation and equipment an" " : .eccentricity caused by wind ,.ro*"rrt, seismitorces, etc. andC : a numerical coefficient for the respective e/D

value.With the maximum soil bearing given^_ w" - Pr"'which locater* : C andD : +.The relation between K and f is sho*r, by the curveon Figure 4 which was formed by using Columns i and 5of Table 1.

Foundqtion Thickness. After getting the size of thebase, the next step is to determine its thickness. Sincethe missible maximum unit shear is 75 psi this is usu_

(

Cos a $i1s al I3Ja f Cos a Sin a- 2Cos3 a Sin a


24/96

SIMPLIFIED DESIGN FOR TOWER FOUNDATIONS .O *,orn of Footing ResistingSheor : C r R (See Toble 2 Col.l3). @ uo,rr. ol slress Prismlyhose Bose is o Tropezoid ond

Segmenl oI Circle . CP5R2(See lobte2 Col. 3)@ ,0,r." of stress Prism

.25 .30 .35 30 .45.5055.@ .70 .80 90 1.0 3.0 3.5 40 45 5.0 60 70 80 90

FIGURE 6-Curves for computing shear for diagonal tension.ally the controlling factor. In many designs this limitsthe strength of the concrete to 2,500 psi for the mosteconomical foundation. The soil reaction considered incomputing diagonal tension is the sum of the forces actingbetween 90 degree radial lines drawn from the center ofthe base through the two corners of the equivalent squareand bounded by section B-8. This section is parallel tothe side of the square at a distance "d," (depth of con-crete) from it. One can see by Figure 5 that these forcescan be represented by a section of an ungula whoseheight is Pa, a wedge whose base is a trapezoid abcd and(.707- Cos a) P.height of -]1

- cos a-and a force solid whose area

of base is the area of the trapezoid abcd and the segrnentof the circle whose chord is cd and of height Ps. Thisf,t Ilatter area tr L * -(1-2K')')R2. The r'orce is R2Pof,7r IL * -f , -2Kr)z_1 and is Column 3 of Table 2. (Seecurve 2, Figure 6.)When K is .1465 or less, this area is a segemnt of acircle. The volume of the section of the ungula can besolved by application of limits of 45 degrees for { incomputing the volume.

By integrating and substituting the values of the trigonometric functions for the 45 degree angle, the force V :, , '*'="t . [.1 I 7g5 - .1427 cos a](l-Cosa)'This formula provided the value in Column 4 Tablefor the various values of K (a).

The volume of the wedge is :R2P4(.707 - Cos 4)2(2 Cos a + 2.828)6(1-Cosa)

and for the respective K and 4 values the volumes arrecorded under Column 5, Table 2.These two columns are added (See Col. 6) and thresults are plotted producing Curve 3 of Figure 6.The width of the footing "b" for diagonal tension2R (1 - 2k). When a is 45 degrees or less, it is 2R Sin aThese values form Curve I of Figure 6 and are tabulatein Table 2, Column 13.

Bond-Bending Momeni. The slab is now investigateto determine the area of reinforcement and unit bonstresses. The moment of all forces to the right of SectioA-A, Figure 1, determines the area of steel, and the sumof these forces is the shear used in computing the bondSection A-A is located by passing a vertical plane througthe foundation along the side of the equivalent squareThe external forces acting on the base can be conceive2R2 P- (4v : (l -6;;d) )"ttt"' c cos d-Qe5 a sin'z P)dd

TABTE -Vqlus5 to Colculote Diogonol Tension, Bond, Moment ond Beqm widthI 2 3 4 5 6 7 E I IO 1t t2 13K d vCPsRz Sec, Ung.C'PnR2 WedgeC/PrR2 Col.4+5CPrR2 UngulaCPrR2 Seg, Cyl.CPsR2 Mom. Iuna. I tr,t.CPzRs I Seg. CyL 2RslncCR Octag,on I "b"CRICR

,10.15.20.25.30.35.40.45.50

36" 52',45" 34'530 08'60'66" ?,'',780 2884' 16',90"

.1635.2954.4254.5354.6254.6954.74ru.7754.7854

.0660.1198.1611.1860.2029.2144

.2302

:::::.0193.0547.0951.7726.2067.2357

.0660.1198.1804.2407.2980.3497.3958.4369.47&

.0668.1198.7823.2518.3269.4068.490l{.5773.6666

.1035.2955.4473.6142.7527.97807.7731.3891.577

.0076 I .0132-0203 I -0961.0422 I .oz2g.oz2s I .12s9.1140 I .rsor.1658 I .2829.2294 I Bsz4.8049 Ls195.3s22 I .6666

1.201.401.601.731.831.911.961.992.OO

1.28 I 1.201-43 I 1.40r.68 I 1.201.83 I 1.oo2.00 I .802.oo I .602.oo I .402.00 I .202.00 I .00Diagona Tension Bond Bending Mom. Width Beam

24


25/96

@ro,ur. of Stress prism Whose Bose is Segment ofCircle r used in Computing Sheor = Cp3R2@ Votr.. of Stress prism Which is on Ungut0 , usedin computing sr,.o, = c16R'@ *,0,n of Fooling ot Section A-A used lor Sheor ondEendirq Momenl = CR

@ ,or.nt :t"_tlT.r Prism (UngutolAbout SectionA-A,Fis.,=jl;]:-O Moment of Stress prism, Section ot CylindarAbout secrion o-o = '18t"

as being in the shape of an ungula of height pz and. asegmeqt of a right circular cylinder of height p3. Thesum of the two volumes is the shear force, and t-he sumof their moments about A-A is the bending moment thatdetermines the reinforcement. The weighis of the con_crete slab and earthen fill are deducted irom the verticalforces. This is easily accomplished by reducing the inten_sity of the uniform bearing load acting on the bottom ofthe base.The volumes for the ungula have been computedearlier for obtfiing soil bearing and those values for Kequal to or less than .5 are shown in Column 7 of Table2 and Curve B of Figure 7. (See Equation i for V.)The volume of the segment of the cylinder is equal tothe product of the area of the segmenl and p.. The areais easily computed by making use of the fact that themiddle ordinate is KD. Values for the respective K" areshown in Column 8, Table Z and. plottei as Curve A,Figure 7.. T " bending moment equation is derived by substitut_ing R (Cos f - Cos a) for R Cos { in the developmentof the formula for moment about the center of the foun-dation.

This gives Iut: fj* |""r"*d_-cosoj,si.,,p6p2Rsp1 [ r /r \n-c;;iL-- +(it'" 46-o)- 2cosaSinsd-cos2"(+--.Is;"r-)]:

By- substituting in this equation the trigonometric va1gs_ for the respective angles cor.esponding to the KColumn 9, was obtained and Curve D plottej on FigureThe moment on the forces whose configuration issegment of a cylinder (see Figures t and 2) is deriveas follows:

/adM : 2RrPs | {Cor O - Cos a) Sin2 d d C)":2RBPg:2RsPa

Cosa(p-+ sin 2 c)l- sirr 6L--s--_-l- sin, oL--.- +

30 3.5 404550 6.0 z0 8090

SinaCosza_aCosa ]"l

*: ,, 2R'.P, , [: r, * 4 cosz a;(1-Cos") LB' - asuae653a-4

The values obtained for the angles a (K) are noted inColumn 10 and form Curve E, Figure 7.The widths of the foundation at the sections are equato 2R Sin a and. are shown in Column 11, Table 2, andCurve C, Figure 7. Column 72 and. the dotted curve(Figure 8) indicate the width of beams for any octagonUse of Curves. As an illustration of application of the"l*9r: th-" following information is given: height of ves-sel, i12 feet; diameter, I feet; the anchor Lolt circlerequires a l0-foot octagonal pier; top of pier is one footabove grade and 6 feet, 6 inches above the bottom of thefoundation; permissible soil bearing 5,000 psf (pr) at 5feet, 6 inches below grade; wind pr"rrr.", Sb pri oi hori-zontal projection of the vessel.- . Operating weight of vessel, 200 kips; vessel empty, 100kips; and test weight, 300 kips.The diameter of the .base required under operatingconditions will be determined first.The moment of wind force about the bottom of thefoundation is 112 x.03 x B x 62.5 : 1680 krp_feet.

30 .3s .40.45.50.55.60 :o.ao .so toFIGURE 7-Curves for computing shear for bond a,,d bending moment for reinforcement.

$"",aSinsa-r*#-] 2s


26/96

SIMPLIFIED DESIGN FOR TOWER FOUNDATIONS . . .Estimate the weight of the foundation using a 22 foot,6 inch octagon, two feet thick.Pier: (82.8) (6.5) (.15) : Bl kipsSlab: (419 - 82.8) (.3) : 101FilI: 336.2 (.35) : 118

Total :300Weight of vessel (operating) : tg6W:SO6

1 680EccentricitY " :,|| : 3.36 feet, e2 : 11.3s00.000- (11.3) 3,000is obtained from the curve on Figure 3 as '153. Then

T) - 3'36 : 2l-9 ft..153Next try a 22 foot,0 inch octagon with a thickness of1 foot, 6 inches.The weight of concrete and fill becomes 280 kips andW : 480 kips. To compute the maximum unit bearing":#: g.s f""t# : +: '159.From the curve used above, C : 13.5 and

480.000P. - '""i, - , :2,900psf (3,000' This is considered to^ 1 r3.5(3.5),be near enough to the allowable soil bearing' To strivefor closer agreement is believed to be inconsistent withthe accurac of the established bearing value of the soilsand therefore would be a waste of time.The unit bearing under the foundation for test condi-tions and one-half of maximum wind load is found tobe 2,370 psf.To investigate the 1 foot, 6 inch slab for diagonaltension, the

"areaof the lO-foot octagonal pier-is usedto compute the side of the equivalent square of 9'1 feet'With t'he e/D of .159, K is found to be .88, by use ofFigure 4 and. KD : 19.35 {eet. The distance from the

Abouf the AufhorCharleston, W. Va. and an engineerin Design and Construction, IJnionCarbide Chemicals Company, SouthCharleston. Mr. Brofn's professionalexperience includes Several years inthe Bridge Department, State RoadCommission of West Virginia. Hehas been a consultant on bridges forseveral cities. During his 12 years ofactive duty in the United StatesNavv some of his billets were: De-sign'and Construction Ofiicer, FifthNaval District, RO in CC, Naval DruwMissile Test Center, Point Mugu, Califomia, and PublicWorks Ofiicer, Naval Station, San Juan, Puerto Rico,and Naval Air Station, Kaneohe Bay, Hawaii. He is amember of International Association for Bridge andStructural Engineers and has BSCE and CE degreesfrom West Virginia UniversitY.

Brown

center of equipment to the point where the diagonaltension is computed *#* l.l7 or 5.72' Then K'D :11.0 - 5.?2 :5.28 and P* :fji -^ ] 2,900 : 790 psf,- \ 1e.35 /and Ps : 2,900 - (790 + 625) : 1,475 psf. K' :5.28_- ,4"22.0By referring to the curvqs for computing shear fordiagonal tension (Figure 6) and using K : .24, thewidth of the footing resisting shear is 1.04 (11.0) :11.45 feet. The shearing force it : (.515) (1,475) (ll'?)+ (.23) (790) (11') : 92,000 + 22,000 : 114,000pounds. t : *=frffi a88) : 68 psi ( 75 psimaximum allowed.The section for computing bond and reinforcement istaken along the side of the equivalent square, A-A Figure1. Then K,D : r 1.o - (+) :6.45and K' : .2e3

", : (-tt-,f 2,eoo: e65 psrPs : 2,900 - (965 + 625) : 1'310 PsfBy use of the curves on Figure 7, the shear for bond,the'bending moment and width of beam are computedfor K' : .293.Width of beam: (1.82) (11) :20.0 ft. (circle): (1'96)(11) 0: 21.55 ft. (octagon)

: 120,600 J- 36,800: 157,400 lbs.Bending moment : (1.08) (e65) (113)

+ (1.83) (1,310) (113)l0: 139,000 + 319,000:458,000 ft.Jbs.458Area of steel required per foot : (20) (14) (1.44): 1.74 sq' in. Per ft. of width.Since wind forces contribute more than 25 percent othe moment, stresses can be increased one-third so thearea becomes (.75) (1.14) : .85 sq- in. A six-inch spac-ing each way of No. 6 bars - .88 sq. in. I0 : 4.7 inches1 5 7.400and bond stress is : lzoy 1rg 1.asn47;: 137 psi.

Some foundation engineers prefer to base the steel andbond requirements on the middle one foot wide stripUnder this condition the force for bond ;t : (1,310)(6.45) + (%) (6.45) (965) :8,450 + 3,110:11,560pounds.The bending moment ;t : (6.45'z) (%) (1,310) +(%) (6.+5'2) (965; : 27,300 + 13,400 : +0,7A0 ft'-lbs

(.75) &0.7)et : ffi : 1'51 sq' in', afive-inch spacingeachway of number 7 bars : 1.44 sq. in., l0 : 6.6 inches.11,560 : 142 psi.: (6.6) (.BB) (14)

r3.15) (96Shearforbond: (.76) (1,310) (11') +-

The design for the top of the slab reinforcement, "topbars," which are required by certain combinations oloading, is left for the reader. ##

(965) (11r)

10

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Calculation Form for Foundation DesignFor complete design of octagonalfoundations for stacks and towers orfor estimates only, this form willsolve the problem easily and quickly

Bernqrd il. Shield, Celanese Chemical Co.,Pampa, TexasIu rrre DESTGN oF FouNDATroNs and. structures forchemical plants, the structural engineer normally is nottoo concerned with a highly theoretical, or complicatedmathematical approach. From a practical standpoint, thedesign ,assumptions quite often are not accurate enoughto justify such an approach.Since the chemical industry is such a fast moving, oftenghySinS,_ arrd complex field, the design engineelr often

lacks sufficient time to make an accurate theoreticalanalysis or sometimes even a very thorough practicalanalysis. Quite often he must wade throug-h a lengthyarticle or text concerning an unfamiliar pioblem, Jr uproblem which he has not worked recently. Whiie thetime schedule suflers and other details of the job areneglected, he must set up the problem for practicalanalysis. For many problems of a rLpetitive nature, muchtime is consumed in setting up the sketches and frame_work for an analysis rather than in simply solving theproblem..Ifow many times have you heard the question, ,,Whenwill the foundation drawings be out?,, i hr.." heard it

many times, quite often as soon as a request for appro_priation for a new installation is approved. Faced withthis situation, the engineer must corxtantlv seek solutionsto his problems that will give safe and economical designsand use a minimum of his own time.. The following calculation form for octagonal founda-tions for towers and stacks was devised wiih this idea inmind.. .Y"^h"": used the prototype of this form quite success_fully for about seven years ind believe it is worthwhileto pass on to others. The form is largely self_explanatorywith the.nomenclature and design ,r,"ihod being explainedas th6 solution progresses.Design Bosis. The following general comments shouldbe of help in using the form 1t " firrt time. Moments arecomputed about the centroid of the base of the pad,ignoring any shifting of the neutral axis as loads- areapplied. Soil stresses are computed using the sectionT:9".1"r."j the base pad around its axis of slmmetry. Theslightly- higher soil stress which would be obtained byusing the section modulus around a diagonal is ignorej.Stresses causedty a moment in the base pld ur" coirputedaccording.to the ACI code by compuiing the momentalong a line which would coincide with'the side of asquare. of equivalent area to the pier. Two_way reinforce_ment is then provided similar to the normal method ofreinforcement for two-way reinforced footings.

fn,computations of forces, the area and stress diagraare divided into simpie geometrical shapes for ealecomputation. The design of tensile reinforcement inpier. is a practical rather than a theoretical approaAnchor. bo_lt lengths _and hooks are designed acco.dito the ACI code for _hooked piain bars. the length wdepend upon the design stress used for the boltsjso idesigner wishes to use a stress which differs from thshown on the anchor bolt table, he may easily chanthe length.If he desires additional safety, he may choose to uslower design stress for sizing the bolts #d ..r* the lenggiven in the table. I prefer using higher anchor bstresses than some designers, taking adriantage of theincrease in allowable stresses for*combinea- toaairrgwhich wind is a factor. This will of course give anihbolts which are smaller in diameter and longe-r in leng.f have a great deal of confidence in the reliability of"dsign stresses in steel but very little confidence in thelowable bond stress for a smooth bar. Many times anchbolts are installed without proper cleaning and wthread cutting oil all over them. So, who knows whbond stress will be developed?I believe much work remains to be done to devise. anprove by tests, a really good method of design for imanchor bolts. In the meantime, I prefer to rlse a desir.vhich I believe to be safe and economical, and ,".ogrthe right of other engineers to use their own criteriIt should be noted that the use of this form is niimited to the complete design of a foundation. Shouit be desired to obtain only the size of the foundatio

pad, for such things as estimating or layout, one need onproceed through.Step 5. Step 15 with Figure 3 are quiuseful to transmit information to a draflsman, and thanchor bolt tables are useful in fabrication of anchbolts.. Thg .nef time you have this type of design problemgive this form a try. It is easily ,e.rised fo, .p""Ll .*"rYou may lot appreciate its merits so much if yo, orrhave one foundation to design. If you have two or moreI think -you might begin to-like it. If you harre 50, yowill probably become downright fond oi itlProcedure. Considerable time and effort are usually required to make a detailed and accurate design fo. octagonal foundations for towers, tall reactors, preisu.e vessel-or stacks, particularly if the designer is'unfamiliar withthe problem. Consequently, a complete design is oftenot made, and this may lead to either an uniafe or ungcgnomigal design or both. This me

Documents

Foundation Design Handbook- Hydrocarbon Processing- 1974