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FOUNDATION DEffi HANDBOOK =-'

Reprinted from

HyDRocARBo[Fnocisslrvc

rye,?;.7ry.'

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.

Gutf pubtishing

D*.,-tl,

FOUNDA

Published b This reference manual has been reprinted from the regUlar issues of HYDROCARBON PROCESS+ruO. Otfrer- Handbooks Manuals in the series are:

LINES FOR BETTER MANAGEMENT I

N MANUAL

NSTRUM $tr' :l:i,

TME

D.ELARE STA

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FOUNDATION DESIGN HANDBOOK TABLE OF CONTENTS TOWER FOUNDATIONS .. Foundation Design For Stacks And Towers Simplified Design For Tower Foundations . Calculation Form For Foundation Design Use Graph To Size Tower Footings Simplified Design Method For lntricate Concrete Column Loading Unusual Foundation Design For Tall Towers Foundation Sizing Simplified . . .. Dowel Sizing For Tower Foundations . . Short Cuts To Tower Foundation Design VESSEL FOUNDATIONS . Foundation Design For 8-Legged Vessels Pressure Vessel Foundation Design . . .

.

,

.

.

COMPUTER FOUNDATION DESIGN How To Calculate Footing Soil Bearing By Computer

Concrete Support Analysis By Computer . .

FOUNDATIONS ON WEAK SOILS

....

.

.

Foundations On Weak Soils Graphs Speed Spread Footing Design Use Graph To Analyze Pile Supports .

.

Page No. 4 5 2',|

27 35

39 44 50 53 57 60

.61

.63 .70 . 7',1 .77 .84 .85 89 .93

\

3

TOWER FOUNDATIONS

#**.*"

-i

{}-@

Jng* dd-P

Foundation Design For Stacks and Towers The same principles ,pply in both stacks and towers. Use this method in making your calculations for either.

V. O. Morsholl, Tennessee Easlman Company, Kingspori,

Frorn the viewpoint of the f oundation designer,

stacks and tor.r.ers may be divided into tr,vo general classifications, deper-rding on the method utilized to maintain them in a vertical position; (a) Selfsupporting, which resist the overturning forces by the size, shape and weight of the foundation ; (b)

Guyed, in which the c,verturning forces are resisted by guy u,ires. It is obvious that the conditions alTectir-rg the design of founclations fc.,r these tr,r,o types u,ill not be the same, ar-rd that it is necessary to treat them separately. STA(--KS AND TOWERS are closely related as far as founrlation design is conccrnecl-in fact, the same principles apply. In the case of stacks, the

brick lining is a r,ariable load, corresponding to, and requiring the sanrc treatnrent as the liquid, insr-rlation, etc., in a tou.er. This cliscussior-r will be based on the design of tower forrndations, however, it should be kept in n-rind that it is also applicable to stacks.

3. Soil Loading (See Section 20 for complete definition of terms.)

The soil loading may be determined following formula:

S: S,* S, where

S: : Sr: 51

by

the

(l)

total unit soil loading (lbs.,/sq. ft.) unit soil loading due to dead load (lbs,,/sq. ft.) unit soil loading due to overturning mo,ment (lbs.,/sq. ft.)

4. Dead Load The dead load S, may be determined as follows:

^w a

Jrwhere

a: 'W:

-W: Wt :

Q)

area of base of foundation (sq. ft.)

total_

weight on soil (pounhj) calculated by

YY.,:l

W.

the following equation:

(3)

Minimum dead load (pounds), which is ihi

weight of the empty tower plus the weight of the foundation, including the earth fill on top

of the

2. Self-Supporting Tower There are two main considerations in designing the foundation for a self-supporting tower; (a) soii loading (b) stability. The foundation must be of such size and shape that the load on the soil below will not exceed the maximum load which it will safely support. The foundation must also maintain the tower in a vertical position, so that it will not be overturned by the maximum forces acting upon it. No direct method of calculating the size of the foundation has been developed, therefore, it must be determined by trial and error. A size is assumed, and the soil loading and stability calculated. If the results are not satisfactory, another assumption is made, and the calculations repeated.

Tenn.

base.

W.: Weight of auxiliary material and equipmenf supported by the tower and foundation

(pounds), which should include the liquid in the tower, insulation, platforms, pipin!, etc. (Does not include weight of tower)

Thls is o revised orticle whlch wos prevlously published

in the August, 1943 lssue of PETROLEUTI REFINER. All copies of thot lssue, oll reprints ond qll coples of thc l94O Process Hondbook, In which the origlnal qrticle wss reproduced, foiled to meet th. demqnd for thls englneering doto. When the outhor consldered tho revlslon he extended

the subiect to lnclude octuol deslgn of foundotlon typ6r commonly required In the erecllon of refinery vcsgctg.

fhe resuli ls o thorough study of o sublect whlch contlnues ro hold o forefront posiiion in refinery cnglneerlng. Reprinrs will be provided ln quantity sumcient to lnclude lhe demond thot hos extended Into constructlon fietds outslde of refining. Prlce $l.OO por copy.

5. Overturning Load The overturning load S, is the result of the overturning moment. Under ordinary conditions, the only force tending to overturn the tower is the r.vind pressure.

The magnitude of the wind pressure is obviously a fun-ction of the wind veloiity, which varies in different localities. In many instances laws have been enacted which state the wind velocity or wind pressure to be used for design purposes. The United States Wcather Bureau has proposed the following formula : p

:

0.004

B io v'

(4)

where -

wind pressure on a flat surface (pounds,/sq ft') : baromctric pressure (inches Hr) V : velocity of wind (miles per horrr) For a barometric pressure of 30 inches, the formula becomes:

;:

B

P: 0'004V'

(5)

It has been found that the wind pressure on a cylindrical tower is about 60 percent of that on a flat surface. For a cylindrical tower, theref ore, formula (5) becomes: (6) P": 0.0025v' where on the projected area of a cylinP": wind oressure(pounds per square foot). drical'tower

Wind Prcssurc

In most localities, zL wind velocity of 100 miles per hour is considered the ma'ximum. This giv.es 'u pr..rrr." of 25 pounds per square foot on the -the

tower, which is the figure generaliy used for design pu.rposes. I.t should be Emphasized, however, tiiat this figure is subject to variation in d.ifferent localities, and that local laws should not be overlooked in this connection' (Note : As a matter of interest,. the wind pres,ri. o., an octagon shaped stack is considered to be ?0 percent of that on a flat surface.). Figure 1 represents a .tower, mountg9 on a concrete-foundati,on. The 'arind pressure (P*) tends to rotate the tower and foundation about point A at the intersection of the vertical centerline and the base of the foundation. This rotating effect produces an overturning moment which can be calculated as follows: (7) Mr: P- L

oriiected area of

Foundotion ToP

where

M, : overturning moment about the base of the foundation (foot Pounds) p* : ioiat wina Laa (bounds) to be calculated

Gro

follows:

P- : o. D"H

Foundotion Bosc

as

(8)

L: ierrer irm of wind load (feet) to be calculated as follows:

H

(9) L-h,* , D.:diameter of tower measured over insulation ( feet) H -height of tower (feet) hr : height of foundation (feet)

It

FIGURE I Foundotion for self-supporting tover'

6

should be noted that alt dimensions are stated

i.,-ieet, giving the overturning moment (M3) in ioot pou"'at. ft',i. avoids the use of the excessively larse numbers encotlntered with the usual inch poind units. Care should be taken, however, to

use consistent units, that is foot pounds,

calculations.

in

all

load, o1-lhe soil resulting from ..Jl-: :l."rs, ormoment (y,) valies from p"oint

:1:_?":.t,i'ning potnt, and the maximum load (Sr) can lated as follows:

bi

to

calcu_

^Mr >2:z where Z

(

:

10)

section modulus the base of the found.ation. (Note: Z to be -of based a;-er.ion. i, i;;i.i

""

The value of (Z) can be expressed as follows:

,:* (11) where : I moment of inertia of ttre base of the foundation (based dimension, i,i-tJ"ti. c: distance fromonneutral a*ii oJ iJiria"tion l"s"

to point of maximum ,t..ss (fiefj. Having calculated- (Sr) and (Sr) as explained above, the total soil ioid ,na")-iilriimum dead

times,.(Sl.) must never be less than (Sr). In a perfectly bal_anced system, (S._) i; .q"ri to (Sr), in which case "*ritty Snro : SrSz:0 (l_d) a.bala".S..d sy:19* is ..Althou.gh,-such rarely pos_ sible, it is the ideal condition.' tt e- up*a.d'fo.ce at E due to the overturning -o*.r,i il-;_;;i; balanced by_the dead load, io tfrui-tfr. stress at r, ls zero. I he stress at F in such cases is the -maintain minimum which can exist ,iiif a

""a stable system. be emphasized.tha_t while (S,*) is fre_ ^__l!:,!.rld qu_enily greater than (Sr) it should nevei be less.

It should also be eri-rphasized tfrai stabilitv is based on the minimum dead ioaa ttre iwJ-;;tia the.soil -loading is based on the ma*)mum dead load (W).

.

load conditions can

b.e

determin.d

bt

This maximum soil load occurs 'r,t"q;"ii;"*(-ii if,. .agd ,;f designated as p, i. irequEntty ll:lgl,"1",ion, ,.toe reterred to as the ""Jft is pressure.,, obviout that the maximum toe pressure (S) should never exceed the safe bearing load of th; ;il in question.

I,

It

should

t. ,,o,"t,ilitigi is positive at point

,ng negative at E (Figu.l'f y.'tn oihe, *brds, rne wtnd load causes compressive stresses on the A, ihe *u*rn"* compres_ :gil t. the left of point and tensile stresses of equal 9:.Yt.ing.at f, :191 magnitude to_the right of A, the maximum tension

occurrrng at II.

Since the earth has no strength whatever in it is obvious that the suir of the stresses at any -point must be positive. fn other words, the base of the foundation. must exert a-comp.es.i.,re force on the soil over its entire ur"i, ott d.*ir" n tensile stress will be produced at E, which *"u.,i that the tower and fbundation wili be unstabte, and .likely to be overturned by the uiiion of the tension,

wind.

.Lt-*": shown by equation (1) that the maximum soll load. is equal to (S, + Sr). Since the value of 5, at .pornt.-b is negative, the minimum soil load (*jr:l obvlorsly occurs at point E) is (Sr_Sr). rs very lmportant to note that the condition ot-rtpoorest stability occurs immediately after the tower is mounted on the foundation, and before the insulation, platforms, piping, UquiJ, etc., are m ptace. In calculating the stability, 'therefore, (S,) must be replaced b"y (S,-) a, foiio*,,

Sr-: Wt a where Sr-: T_r.inimum-

(2-a)

-

S,

__ .421.2 cu. ft.

\

150 lbs.,/cu. ft.

Volume of earth'fill

if-='f

W:i{fu vv erg_nr or

:

tower

+

:

63,000

30,000 lbs.

+

3Z,7OO:125,700 tbs.

Insulation, platforms, piping,

(

l-a)

Therefore, in order that (S*i,) may always be positive, thereby assuring a'sti6fe .onaiiio., ut

"tt

16..

LIJ'rt'r:,1,.!'i;r'.',3,2,11?=39lr36io'l

_e-mpty

30,000

:63,666

W, will be as follows:

The minimum soil loading which can ever exist, therefore, is found to occuf at point E when the dead load is at its minimum value, and can be expressed as follows: Sr-

Since the ,r.., ,,,1""11tti'r,. o",o* grade; the foundation will be 6 ft. deep, with the- top 1 ft. above grade, making the bottom of the foundation 5 ft. below 'grade, br 1 ft. below the frost line. The foundation will be octagon shape, which is recommended for such cases, as it combines the features of stability, ease of construction and mini_ mum material better.than other shapes. The top course will have a short diameter of O ft. sin.l the tower is 4 ft. dia. and allowance must be made for founda.tion bolts, etc. The short diameter of the Dase wlll be assumed to be 13.5 ft. The thickness of 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 assumed to be 2 ft. Th_e weight of the foundation will be calculated as tollows (all slide-rule figures): octagon : 0,8f8 d,^1-0.828 X 6,:29.8 sq. ft. *l:l :f 6.ft. vofumeot top course : 4 ft. X 29.9- 119.2 cu. ft. r\rea ot base (octagor). (q): q.-g?g I3.5,: l5l sq. ft. Volume of base :2"1t. k i5i :30r;;lX i;. " Total. volume :119.2 *' jOZ: 4Zl i ;;.' i;. welght of concrete:

_wt

soil loading due to dead load (lbs.,/sq. ft.)

S*r,:

7. Example No. I Design the concrete founation for a tower 4 ft. dia. by 54 ft. high, including a 4 ft. skirt, and weighing.30,000 lbs. empty. The insulation, platforms and piping weigh 9b00 lbs., the maxirirum wind velociJy is 100 mlles per hour, and the frost line at the location of the proposed'installation is 4 ft. below grade. The maiimum safe soil loading is 2000 pounds per square foot.

etc. :

Water required to fili it t6i"".(4 ft. dia.) (50 ft. high) "

9,000 lbs.

:39,5001bs.

Total (W.) :4g3mlE; W:125,700 f 48,500: LZ4,2OO lbs. (from equation 3)

a: l5l sq. ft. 174,200|bs. ^ -1Sl:qlTI5': :

1155 lbs.,/sq.

ft.:

Maximum dead

load on soil (equation 2)

Allowing 2,, for the thickness of the insulation,

7

the effective diameter of the tower exposed to the action of the wind is 4.5 ft. A wind velocity of 100 miles per hour is equal to 25 pounds per sq. ft. of projected area. Therefore:

p.:25 lbs.,/sq. ft. 4.5 ft. H:- 54 ft. P*: 25 X 4.5 X 54 : 6080 lbs. (equation 8) hr: 6 ft. 54 ft. (equation 9) L:6* 7:33 : 6080 X 33:200,000 foot pounds (equation Mr Z: 0.1016 d' : 0.1016 X 13.53 :248.5

maxi-

mum soil load due to overturning moment'

The total maximum soil load (toe pressure) can

be calculated from equation (1) as follows:

:

1155

* 803:

1958 lbs.,/sq.

TABLE Elements Short Dlam. (Feet)

3.5 4

4.5 5

5.5 D

6.5 7

l.c

Area a

(sq. Ft.)

2. 691

46. 5

3. 105

40.5

52.8 59.5

9.5

74.5 82.8

I

10

10.5

66. 8

sl.2

12

100.0 109.8 119.5

12.5

140. 0

11

11.5 IJ

2.484

34. 8

8.5

8

1.863 2.O70

2.898

3.3r2 3.519

3.726 3.933

4.140 4.347 4.554 4.761 4. 968

r29.2

5.382 5.775

13.5

151.0

14 14. 5 15

162. 0

5. 796

174

6.003

186 199

0. 210

16 16. 5 17

2t2

226 240

77.5 18

18. 5 19 19. 5 21

21.5

23.5 24

24.5 25

25.5 26

26.5 27

27.5 28

28.5

3.78 4.06 4.59 4.87 5.13

74. tO 87. 30

3.5i

5.41,

101.60 117.60

5.94 6.22 6.48 7.03 6.76 7.30

154.10 774.50 227.00 198.00 248.50

5.b/

8.11 8.38

6.831

8..65 9.O2

342. 375. 416. 455.

6.624 7. 038

283 299

7. 866

8.073 8.280 8.487

8.694 8. 901

9.108 9.315 9.522 9.729 9. 936

497

10.143

518 539 560 582 603 526 650

10. 350

10.557 10. 764

10.971 11.178 11.385 17.592 11.799

9. 19

9.46 9.72

10. 55

731.00

to.25 10.81 11.08 11.35 11.62

11. 90

14. 01

14.88

15_ 15

15.42 15.68

17.3t

36 38 39 40

958 1015 1075 1134 1195 1260 7325

16.77 17.85

18, 38 18. 92

15.732

19.47 20.01 20.55

16. 146 16. 560

21 .65

14.904

15. 318

1085. 00

1400.00 1490.00 i585. 00

13.25 13.52 73.79 74.o7 14 .33

8.244

34

933. 00

1005.00

12. 98

72.43

30

13. 662 14. 076 14. 490

873.00

1145. 00 1240. 00 1320. 00

t2.420

902

811. 00

12. \7

15. 96 16. 23

12.434

497.(n

543.00

10.00

12. 006

1,2.2t3

00 00 00 00

590. 00 624. 00 652. 00

720 745 755 848

135. 00

277 .OO

695

31

52.00 62.70

6.417

.245 7.452 7.659

672

16.45

21.90 27.90 34.90 42.80

29

29.5

s.23 12. 68

309.20

7

348 365 383 401 420 438 458 177

.

7. 83

268

20

20.5

5.589

2.43 2.70 2.98 3.25

21. 09

Radius of o.772 0.900 'r

1.89 2.16

\3.2 16. 8

Bose

Sectlon of slde I to Extreme (Feet) lFlbffc(Feet) Modulus Z

7.45

l5r

1685. 00 1787, 00

1900.00

020

1.158 1. 286 1. 415

t.542 1.67

Usually it is found that the first assumption as to foundition size is not correct, in which case, another assumption is made, and the calculations repeated.

it is interesting to note that the soil loadinq

4000 lbs./sq. ft. Care should always- be taken, t.o ascertain the actual load carrying value of the soil at the site of construction.

8. EccentricitY

3021.00

2.44 2.57 2.70 2.83 2.96 3.09 3.47 3.60 3. 86

3.99

4.24 4.37 4.50 4.63 4.76 4. 89

5.02 5.14 5.27 5.40 5. 53 5. OO

5.78 5.92 6.04 6.17 6. 29 6. 43 6. 56 6. 68 6

.82

7.46 7.58

7.71,

3330. 00 3660. 00 3980. 00

a,23

4730. 00

9.26

4370.00

5130.00 5580.00

6020. 00

6500.00

vertical load. As explained previously, it is not necessary to calculate the eccentricity. in order to determine the stability- of the foundation. Several methods have been proposed, however, which make use of the eccentricity, and since there are terest.

2.t7

7.20

2740.OO

be noted that there are two forces acting on 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 eccentric

2.3r

.93

6. 94

2600.00

of

ft. allowed in this problem is rather low, as good clay soil will usually su-pport abou't 2000 lbs./sq.

2.06

1

2110.00 2220.00 2470.OO

tbs./sq. ft.

definite relationships between eccentricity and stability, they will be explained as a matter of in-

1. 80

2010. 00 2350. 00

:830

stable.

It will

LenAth lNeutral Arls

10.6

20.7 25.0 29.8

ft.

I

of Octogonol

125;700 lbs.

points, thus indicating that the foundation is

7)

From equation (10)

S

Sr-:

This is the dead load under the worst stability condition, and since it is greater than the overturning stress (Sr:803), the soil below the foundation-will always be under compression at all

D.

,,: l@,9%*j$@ :803 lbs./sq. rt. :

This loading is satisfactory, as the soil will

safely support 2000 lbs./sq. ft. From equation (2-a)

8. 48

a.75

9. 00 9. 51

L77

10. 04 10. 28

The eccentricity can be calculated as follows:

e: Mr where -W; e: eccentricity

(12) (feet)

Note : The value of (e) calculated by equation (12) is the maximum value, as the dead load-.(Wt) is minimum. The eccentricity for other conditions of dead loading may be obtiined by substituting the proper weight in place of (W,). It^ha; been ihown-by previous discussion that the following relationships exist:

^Wt -f5r-: ^Mr }z: _7_

z- cl-

(2-a) ( 10)

(11)

combining equations (10) and (11) cJ2-

M'c ( 13)

I

rearranging equation (12) Mr: Wte combining equations (12a) and 13) cJ'--

(12a)

Wtec I

(14)

It w.a_s shown by equation (1-a) that in to avoid tensile stress at E (which would

order

ocTAOOII

make

a = o.gzgdlc : o.s4td

the foundation unstable), the maximum value of (Sr) is as follows:

r = o.o55d+ z I o.Ioledr r: 0.257d

Sz: Sr* (ls) thus making the value of (S-sn) equal to zero, as shown by equation (1-d). .-Substituting the values obtained by equations (2-a) and (1a) in equation (15) Wtec Wt la ( 16) The value of I can be expressed as follows: rT

^-2 al

r:

radius of gyration of base (feet)

where -

uE)ucou

d=

(17)

substituting in equation (16) Wtec W, ar' a H_gnce, the maximum value conditions is

c = 0.577d

rI

(e)

for

o.o6d'

z = o,1o4d'

(18)

of

O,866d,2

r:

stable

o.e64d

r"

enax: I

(

le)

SAUAE

In the case of a circular foundation C:Td (20) Substituting in equation (19) 2f '-"" - d (21) the value.of (rr) in equation (21), ., Substituting the maxrmum value of (e) for a circular base is (*), ,n"r confirming the commen rule that in a stable foundation the resultant must fall within the middle-quarter of the diameter of a circular base. In the case of the octagon base usually used for

d= d2

c:

I TI' z : o.Ilgdt r: 0.289d OIRCIA

q: o.?g6{d2 c -.8-

tow.er foundations, the maximum allow;ble eccen_

tricity

r : o.o{gd+ z : o.og8dt 13-?r A

becomes

e^^,:0.122

d

(22)

The area surrounding the center of the base, within which the resullant causes a compressive

stress over the entire base, is known as the kernel or kern. . ,{t f9l.19ws, th.en, that the resultant must always

fall within the kern of the base in order to stability.

Eremenrs

llnll"loi,,*

(Axis A-A)

r**

assure

No. 1 (Section Z), it was shown that . In-exampl.e the foundation is stable, since the overturnins

stress (!r) i. less than the minimum dead loaE stress (S.1"). The_stability of this foundation will now be cal_ culated_ (as example No. 2) on the basis of the eccentricity for the purpose'of comparing the two

methods.

From equation (12) the eccentricity is

Mr e:=w;:

o.?o7d d4

200,000 foot pounds I25,700 Ibs.

-- 1.59 ft. Frorn equation (ZZ), the maximum permissible eccentrlclty 1s e^",: 0.722d: 0.122 X 13.5 : t.S{ 11. Ipasmuch as the actual eccentricity (1.S9) is less

than the maximum p_ermissible eccentricity (1.64) the foundation is stable, thus confirming ihe corr'_ clusion reached in Section ?. 9. Method of Calculating Soil Load From

Eccentricity calculate the soil loading (toe pressure) as a function of the eccentricity: This method will.b_e explained in order that it inay be compared with the method described in Seciions 3, 4 and. 5. Let (k) be a factor by which the dead load press.ure must be multiplied-to equal the soil loading due to overturning as follows I kSr:Sr (s7)

It is

possible

to

9

.)c ----J

FIGURE

FIGURE 3b

I

Substituting in equation (1) S: S,*kS,

or

S:S,(1+k)

(s8) (se)

From equation (2)

w c-.....:_ rr_ a

u,-

(2)

Mr-

Mr:

(10)

SrZ

(60)

Substituting the value of (Sr) from equation (57)

Mt-S,kZ

(61 )

therefore

^MrTZ Sr:

(62)

From equations (2) and (62) and

r'"-w-Mr a tvlr-a -

kZ

wkz

(63) (64)

Equation (64) can be written M, .W-_.kZ a

IO

(67)

ea k:- -Z

(68)

1-

-

In the case of an octagonal a:0.828 Z

:

(12)

(6s)

d'3 d8

(70)

(71)

Substituting these values in equation (68) 8.1 5e

koct.ro": --6-

U2)

Therefore, for an octagonal base, equation (69) may be written:

/ 8.15e \ s,(r+?/

(73)

For comparison, the maximum soil loading in example Nb. 1 will be calculate4 (as example No. 3) on^the basis of eccentricity' From previous calculations, it was found that: 200,000 foot pounds 174,200 pounds

Therefore by equation (65) 200,m0 e:-ffi:1.15

The 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

(66)

base,

0.01016

Mt: W:

The value of (e) calculated from equation (12) is maximum for'any particular foundation, which is the value governing s_tability. At the present moment we are concerned with the maximum soil loadins (toe pressure) which occurs---when the dead l"oad is maximum. It is therefore necessary to substitute (W) in place of (W,) as follows: Mr e: w-

e:-;KZ

soctsson:

From equation (12) M, e: wi

in both equations

Substituting this value of (k) in equation (59) / ee (69) S:S,\r+ zl)

Z

therefore

Since the ,.trn (#) occurs (65) and (66), it follows that and

From equation (10) c-

n

:

1155

X 1.693:

1955 pounds

per square foot.

This checks the value of 1958 pounds per square foot calculated (by slide rule) in Section 7, thus

/l

_._.___:t.-.

)c

---]n i

,-lt

FIGURE 3e

indicating that either method yields the

same

result.

10. Soil Loading at Any Point Having calculated the eccentricity, it is a simple matter to determine the unit stress on the soil at any point whose distance from the centroidal axis

is (c'). The unit stress on the soil, from equation (1), is as follows: (l) S: S, * S, Since the value of (Sr) for points to the right of the axis is negative, the value of (S) will be :

(1-b) S: S,-S, (see equation 1-a) Equations (1) and (1-b) can be combined as

obvious for this purpose that the same value of the dead load should be used in the calculation of the eccentricity (e), by means of equation (12). 11. Stresses

The steel shell is required to withstand the (a) the internal pressure; (b) the dead load; (c) the overturning moment due to the wind pressure. This discussion will be confined to the stress resulting from the wind stresses resulting from,

pressure.

It may be assumed, in determining the stress due to the wind pressure, that the tower is a vertical beam, and that the wind produces a bending moment. The ordinary formulas for determining bending moment and stress may therefore be applied, as follows :

*': n- (+)

follows:

S:SrtS,

(

1-c)

Mtc ^ __l_ S,: (2)

^

W a

(see equations 14 and, 17)

(23)

Wec' Zt'

Simplifying:

t:+(,*5)

Mt: bending moment about base of tower (foot

also

The value of (Sr) can be written: ^ -Wec' Sr: ^l Thereiore,

(26)

where

pounds).

From equation (2)

-_w ", -;

in Tower Shell

(24)

This value of (S) is the total unit stress at any point whose distance (in feet) from the centroidal axis is (c'). It is important to note that equations (1-a), (1a) and (1?) referred to above were used to determine the stability and the eccentricity under the poorest condition, which obviously occurs when the dead load is at its minimum. Equation (24) can be used to determine the stress under any dead load, therefore, equation (24) may be based on either (W) or (Ws) depending on the dead load for which the stress is to be determined- It is

(27)

where St

: unit stress in tower shell due to bending moment (Mt). (lbs./sq. ft.) Note: The unit stress in the tower shell (S,) is calculated in pounds per square foot in order to be consistent with the other calculations which are in foot-pound unts. Steel stresses, however, are ordinarily given in pounds per square inch. In order to convert the stress from (pounds/sq. ft.) as calculated, to (pounds/sq. in.) it is necessarv to divide (S,) by 1aa. The shell is a hollow cylinder, in which case:

c:-TD

Gg)

and

l:

64

where D

: Dr: when t:

and

(D'- D't)

outside diameter of tower (feet) inside diameter of tower (feet) thickness of shell (feet)

- Dr:11 Dr:D -2t D

Q9)

(30) (31 )

II

The maximum tensile stress per foot of circum-

Substituting in equation (29)

I: 64 [D'-

(D

(27)

St:

#rr'-

I

and

c in equation

D

32M.D

2

(D-2t)41 32

:

:@

(31)

-zt)'l

Substituting the values of Mt

ference to be resisted by anchor bolts is

32

- ,]D.-(D]ZDT

:

,rDt

-T-

MtD G2)

(38)

The load to be carried by each bolt can expressed

:r: q--

,lrDn

/ 4M,

-N \ 4Mt :ND"-

z.Do,

W. \ - rD, )

W"

(3e)

N

where

Sr: maximum load on each bolt (pounds).

gTpq-

(32-a)

The nut should always be tight, placing some initial tension on the bolt. Of course due allow-

4M.

z'Dl

(32-b)

The thickness of shel1 plate required to resist

the bending moment only, is therefore r

4Mt

(33) - z.D,St By multiplying the stress in pounds per square foot (S,) by the shell thickn.ess (t) the stress per foot of circumference is obtained as follows:

,S,:#

(34)

12. Foundation Bolts for Self-Supporting Tower The foundation or anchor bolts for a self-supporting tower are required to resist the overturning moment (Mr) resulting from the wind pressuie, after allowance has been made for the resistance offered by the weight of the tower. Obviously the t'esistance offered by the tower's weight is least effective when the minimum weight is acting. The anchor bolts should therefore be calculated for the condition existing when the tower is empty and without insulation, platforms, etc. This weight will be designated by (W"). It was shown that the maximum stress per foot of circumference due to the wind moment can be calculated by equation (3a). That equation gives the stress at the circumference of the shell, however, at the present moment it is desired to determine the bolt stress making it necessary to substitute (Ds) in place of (D). The sffess per ioot of bolt circle circumference can th€n be written: 4Mt

,rDF

(3s)

where

Dn: diameter of bolt circle (feet). The compressive stress per foot of circumference due to the weight of the tower is

.

t2

be

32MtD

This equation may be reduced to:

^5t:

rbsented as follows

MtD

The values of t2, t3 and ta are quite small and the three terms in the denominator containing them may be neglected without introducing appreciable error. For practical purposes, therefore, equation (3?) may be written as follows:

^s1:

4\{t W" (37) TDJ - qrDb Assuming that the number of bolts is represented by (N), each bolt will be required to carry the stress over a portion of the circumference rep-

____--1.

--------.,L-A-

l' \

\

Sl\..

--.LL/-ll./z'.r. /t'r'/ /

\ /',/ .,' ,'

Yi{iii i|i$i 4"M N.

E

t

.t

SLOPE

-r-t :ll lll

iH

'Tl

--'T',---___L----

(36)

O DRAIN

r| ccour

CHAMTER

l.lr

Llt

ltl

Llr

iil

x--6:ro'

'W"

;D;

ry,

FIGURE 4

ance for the initial tension should be made in determining the size of the bolt, and the strength of the bolt should be based on the area at the r6ot of the thread. An additional allowance, usually r/s", should be made for corrosion. The number of foundation bolts should never be less than 8, and should preferably be 1p or more, as the larger the number of boltj, the better the stresses are distributed, and the less danger resulting from a loose nut on one bolt. The bolt should be embedded in the concrete foundation in such a manner that the holding power of the concrete will be at least equal to thE full strength of the bolt. It is common practice to use a washer at the lower end, or to bend the end of the bolt to form an "L" for the purpose of anchoring the bolt in the concrete (iee'Figure ?).

fn

18. Guyed Towers cases where the tower is very high,

it

Pull on Guy Wires pull on the guy wire occurs when the wind blows along that wire, and each wire must be designed to take care of the entire wind reaction at the collar. - -Th" pull on the guy wire can be expressed as follows l 14.

_ The. maximum

The value of the reaction at the collar (R") may by calculating the moments about the base of the tower (the-top of the foundation). The wind moment was found by equation (2d) b.e determingd

to

be

(+)

"-

. The resisting moment arm at the collar is hr, therefore the reaction (R") may be calculated as follows:

-_n-(+) or

h,

(47)

P-H ,."__Zlr,_

(48)

where

h,:

height from top of foundation to collar (feet),

15. Foundations for Guyed Towers It.was shown by equation (1) that the total soil .toadrng. to be considered in the desisn of tower foundations, is the sum of (S,) whictiis the dead load, and (Sr) which is the toia due to the overturning, or wind moment. In the case of the guyed towers, there is no overturning.-an moment, hoiever, the wind pressure does have imporlant effeci ol the foundation, as the soil is required to resist th.e vertical component of the p,ril on the guy

w1res.

guyed t-owers, therefore, .beFor revised as follows:

, S:S+S"

(49)

wnere

Ss: unit soil loading

equation (1) must

d-ue

to the pull on the guy

(40)

(41) - Rr: Rc csc I ft": 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

where

(degrees )

The value of the angle a will usually lie between 30 and ?5 degrees. The vertical component of the pull on the guy wire can be expressed in any o? the follow'in! ways: R" )( cos d (42) R"Xcos0 Si,

R": (R" * Rt) cos o (4s) R",: vertical component of pull on guy wire (pounds) Rt: initial tension on wire (pounds).

wire. (pounds/sq ft.)

Rr:#-

the

where

is

sometimes found desirable to mainiain "stibititv b.y melgq of guy wires rather than a large foundition. Although it is not uncommon to find two or even three sets of guy wires on one stack, towers seldom have more than one set, and even these cases are rare. This discussion, therefore, will be confined to towers with one set of guy wires. Four .guy wires are usually useE ior each set, _ although in some instances three, and in others as many as six have been used. They are attached to a.rigid_ collar which is located it a point approximately 2/3 (sometimes %) of the tower height above the foundation.

-

the.sum of the pull due to wind pressure and initial tension as follows:

a-

RoXcotd

(43) (44)

It is important to note, however, that there is _ always some -initial tension on the guys which must be considered. This initial tensi-on mav be assumed to be 5000 lbs./sq. in., which amounis to 1900 lbs. lor /2" wires and ZSO tbs. for fu,, *ir.s. The weight of the wires may be neglected, when using these figures. The actual vertical component will be a function of the total pull on the guy wire, which is

The value of (S") can be determined as follows:

^R"

ra_

a

(s0)

From equation (2)

^w

a

(2)

Substituting in equation (49)

^

W+R. a

(sl)

16. Foundation Bolts for Guyed Towers The foundation bolts for guyed towers are required to resist the shearing iction of the wind pressure at the base of the tower. It is obvious that ample allowance should be made in the size of the bolts to provide for the initial tension due to tightening the nuts, and also for corrosion. The shear at the base of the tower, which must be resisted by the bolts, is equal to tlre horizontal reaction to the wind pressur-e at that point. This is equal to the difference between the^total wind

l3

at the collar and

can

wind reaction, or shear, at the of the tower. (pounds)

base

pressure and the reaction be expressed as follows:

Rr:P--Ro : horizontal

where Rr

$2)

17. Stress in Shell of Guyed Tower The wind pressure acting on a guyed tower pro-

duces

a negative bending moment at the collar,

has been widety published. There are really. two formulas; one ioi piles driven with a drop hammer, and another for piles driven with a steam hammer, as follows:

For drop hammer o_ 2Wrf . ^ - po+l.o

For steam hammer o_ 2Wrf

'-

where P

: Wh: f: ph:

+-=t=*

i--it+

(ss)

pof

(s6)

0.1

safe load which each pile

(pounds)

will support.

weight of hammer. (pounds) height of harrimer fall. (feet) penetration or sinking rrnder the last blow, on sound wood. (inches)

Care should be exercised

in driving piles,

to

to develop their full strength, but they should not be driven too much, as this practice results in splitting or breakassure that they are deep enough

FIGUR.E 5

and a positive bending moment between the base and the collar. The maximum values of these two moments can be calculated as follows : PM":-lE-(H-h,)'

(s3 )

r\.r,:-!*(r-+)

G4)

ing, and greatly reduces the load carrying capacity. Although piles have been driven with a center to center spacing as small as 2' 6", it- is strongly recommended that this distance be not less thbn 3' 0". Closer spacing disturbs the ground sufficiently to greatly reduce or destroy the frictional resistance.

The top of the piles should always be cut off below the water level, otherwise they will decay rapidly. The reinforced concrete cap is constructed on top of the piles in such a manner that the piles exlend about 6" into the concrete (see Figure 6).

where

Mc: negative bending moment at collar. (foot pounds)

Mn: maximum positive bending moment collar and base. (foot Pounds).

between

Having determined the bending moments, the in a given shell, or the shell thickness required 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). stress

19. Stresses

in Foundation

After having selected a foundation of such size and shape as- to fulfill the requirements of the problem-from the standpoint of stability and soil ioading, it becomes necessary to calculate the stresses in the foundation itself, to see that they do not exceed the allowab1e limits. The first step in this procedure is to determine

18. Piling

In cases where the safe soil loading is very low, it is sometimes found difficult to design an ordinary foundation which will not overload the soil. In such cases it is desirable to support the load

on piles rather than on the soil. Wooden piles are ordinarily used, and they vary greatly in length, depending on the nature of the soil. The diameter at the lower end is about 6"; and the diameter at the top is about 10" for piles not over 25 feet in length, and 72" for longer piles. Wooden piles generally depend on the frictional resistance of the ground for their load carrying capacity, as they have comparatively little strength as columns. The safe load which a pile will support varies greatly in different localities. Building laws sometimes g'overn the pile loading, and in such cases, the load is usualy about 20 tons per pile, although occasionally 25 tons is permissible. When conditions are not definitely known, however, the only safe procedure . is to drive a few piles for test purposes. The.dommon method of calculating the safe load is by means of what is known as the "Engineering News Formula," which

t4

FIGURE 6

the loading, which consists primarily of the upward reaction of the soil. Figure 3 represents the plan view of a typical (octagonal) foundation, and Figure 3a shows the loading diagram. In this diagram the dead load (Sr) is represented by the rectangle (jklm). The wind load (Sr), which is positive on one side of the centerline, is indicated by the triangle (-p*). On the opposite side of

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

The vertical shear, resulting from the upward reaction of the soil, producei diagonal t6nsion stresses in the foundation. The critical section lies at a distance from the face of the pedestal equal to the effective depth of the base, as indicatecl point (Zr) in Figure 3c. In ottrer words, the 9y foundation tends to break along line (ZZr).'The vertical shear to be resisted is equal to ihe net soil pressure on the part of the foundation outside the critical section. For design purposes, therefore, the load will be _ the area_ (oq.p), (Figure Bc) applied over the area.(a, b, (Figgre 3). Becaui6 of the irregular shape ofIg), the load diagram, its magnitude Ean be more conveniently calculated by breaking it up into its component parts, the totalioad (V")-being the sum of the individual loads, as follows:

port Prism -Rectangular Wedge Wedge pedge. Ivramid Pyramid shape or

o"'lt$r""lif*" a, b, u, i, ar tr g b, fu-, ar br ur tr ?r tr g b, fu].

outline In elevatlon

oqrv oqrv rvp

rvp rvp

as follows:

V" ,b'jdr 'd(80) where fd: 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 resist

dr

-

diagonal tension stresses (line a, br Figure 3). (inches) ratio of lever arm of resisting couple to depth (dr) (see Table 2). effective depth of base measured from top of base to centerline of reinforcing steel. (inches)

Erample No. 4. Check diagonal tension stresses in the foundation considered in example No. 1 : Figure unit soil loading due to weight of base and earth (see Section 7): Concrete base

Earth fiIl Total

Unit soil loading

Line (gf):67.1'(see Table Line (m, *1 :!:57'. Ltne

(ar b,)

Y 57 : g1

67.1

- --

1).

47.2"

: (b').

: -67.r--2 - 47.2-:9.95" 162 Lrne (a, t,): - 114 -24" -2- Table 2). Factor j: .87 (see Load (m, rl : 4lrf{:5651bs.,/sq. ft. Load (qr) : 565 + 522: 1,087 lbs.,/sq. ft. Line (gtr)

Calculate shear load (V")

47.2'xz"

xrffif_t

g.s5"

x24' x

47.2

xz4

#

xZ#44

9.95 X 238y.24:X2 3)/'144 Total (V")

: : : : :

8,550 lbs. 1,805 935 263

11,558 lbs.

Calculate unit stress in concrete (equation 80)

_ !o:

I 1.558 47

2

x

.87

x 2l :

l3'4 lbs''/sq' in'

This stress is satisfactory, as 40 lbs./sq. in. would be allowed (see Table 2).

oqrv

The unit stress (diagonal tension) resulting from this vertical shear load can be 'determined

j:

(mw): 162 Z :81" do:72". dt :21". dt:162" do:72+21+21:174" Line

of Slab Required for Punching Shear The thickness of the foundation slab (bottom course) must also be sufficient to withstand the tendency to shear along line (Z-Zr), (Figure 3c) 19b. Depth

at the edge of the pedestal. This shearing load may be determined as follows:

S,: S.* S.

(81)

The stresses in this case are not distributed over the foundation area, but are concentrated at the edge of the pedestal. Then Su: total maximum unit shearing load. (lbs. per lineal foot of pedestal perimEter). Sr : unit shearing l,oad due to dead load. (lbs. per lineal foot of pedestal perimeter). Sc: maximum unii shearinf load du6 to overturning_ moment, (lbs. per lineal foot of pedestal perimeter).

The value of (S*) can be determined by adding _ the weight supported by the pedesta[ to th;

weight of the pedestal itself, subtiacting the loadcarrying value of the soil directly under the p.edestal, and dividing the difference by the perimeter of the pedestal base as follows: -

63,000 lbs.

32,700|bs. D57oo

J'

tr..

95.700 lbs : ffi :

633 lbs

/sq' ft'

Total unit dead load (S,) (jm, figure 3c) : 1,155 lbs.,/sq. ft. Unit dead load due to weight of base : 633 and earth fill (jo) Net soil load (om) :-Wlbs./sq. ft. : 803 Maximum unit wind toad (S,), (mp) Maximum effective unit shear load (op) t,325 tbs.,zsq. ft.

where Wr: a, : ^ S.rr: Lr:

W"

+ W. + W,L,

(ap S"rr)

$2) of foundation pedestal (top course). (pounds) plan. area of foundation pedestal. (sq. ft.) weight_

maximum allowable unif soil loading. (pounds,/sq. ft.)

perimeter of foundation pedestal. (feet)

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

t5

or greater than (W" + W" * Wq),_the value of (Snl becomes zero, and (S.) will then be equal to (So). The'value of (Su) can be determined in a manner somewhat similar to that proposed in Section 12. In that section the overturning load was calculated as a function of the periphery of the foundation bolt

circle, by means of equations (27)

and (35). The bolt circle was assumed to be a hollow iylinder, the wall thickness being infinitely small, as compared with the diameter. In the determination of . the shear at the edge of the foundation pedestal, a similar procedure may be followed, substituting (M1) in place of (Ml), and appropriate values of (I) and (c) in iqui[ior, (ZZ)-, dtpending on the shape of the pedestal. -

Reduced

to their simplest forms, the equations

for the ordinary foundation shapes are as follows: Octagon

-16

Mt

-

Ilexagon

-ro

(83)

.814dp'?

Mt

-

(83a)

.832d,'

Souare

Mr

5r--ffip

(83b)

Circle

Jo-

(83c)

.7g5dn'

In these equations (dn) is the short diameter of

the pedestal (feet).

Once the shearing load (Sr) per foot of pedestal perimeter is knowri, it is a iimple matter to caliulate the unit stress in the concrete, by dividing (Sr) by the effective depth of the base, as follows:

-S, roTt

where

f"'

(84)

stress in concrete base - unit shear. (Pounds/sq. in.)

to

due

punching

Note: The factor 12 is introduced for the purfrom (pounds/lin' foot) to ?pounds/lin. inch) is- unit stress (fn) is in terms oi (pounds/sq. in.). pose of converting (S")

This stress is satisfactory, as 120 pounds/sq. in. is permissible. (See Table 2.) In the case of guyed towers, or stacks, the shear load due to overturning moment (Su) does not apply, but is replaced by

("*,) which is the load due to the pull on the guy wires, as follows:

^Rr S3lguyed)

- S.*

l9c. Reinforcement of Base for Upward Bending Reaction of Soil In designing the base of the foundation to resist the bending moment due to. the upward reaction of the soil, the critical section is located at line (ab), (Figure 3d) along one face of the pedestal (top course). The moments are therefore figured about line (ab), on the basis of the load on the trapezoid (abfg). The load which serves to produce the bending 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 do not contribute to the bending moment, they may be deducted from the total load when calculating the bending moment. The effective loading will therefore be represented by the area (o qr t, p) Figure 3e. The load is assumed to be applied at its center of gravity, and the moment figured about line (ab). Due to the irregular shape of the load diag'ram, it is difficult to locate the center of gravity, and it is therefore more convenient to break it up into its component parts (prisms, wedges, pyramids, etc.), and figure the moment of each part separately. Obviously, the total moment (Ms) will be the sum of the individual moments. In the case of the rectangular prism, the lever arm used in figuring its moment will be one half of the distance from point (a) to point (t), (Figure 3d). In the case of the wedges and pyramids, the lever arm will be two-thirds of the distance from point (a) to point (t). The individual components and their respective lever arms are as follows: h dor, rlc.3il

Outlltro

Erarnple No. 5. To illustrate the procedure, the punching shear will be calculated for the foundaiion considered in example No. 1. Calculate dead load shear (S.) by equation (82)

W.: W, a,:

30,000 lbs.

ft. X l50lbs. ft, 8:19.87 ft.

119.2 cu.

29.8 sq.

Ln:2.484 \

W. :48,500

: S.rr : -

S.-

:

lbs. 17,850 lbs.

2,000lbs./sq. ft.

(29.8

X

1,850

2,000)

lbs./lin. foot.

Calculate shearing stress due to overturning moment (S,) by equation (83) Mr: 200,(X)0 foot pounds (see section 7)' dn": C - 36, S': 200.m0 :6,820 Pounds/lin. foot

:ffifi0

S',: 1,850* 6,820:8,670 pounds/lin. foot. dr- 21" ttp- 8.670 pounds/sq. in. -::34.4 l?)(zl -

t6

(81)

(84)

(8la)

L

Rectangular

Prism

Outlino ln elevation, Fig.3c

abut

Lever Arm

Distance (at) QrfrVrO

2

Distance (at)

Wedge

atg

QrfrVrO

Wedge

bfu

QrfrVrO

Wedge

abut

frVrp

Pyramid

atg

frVrP

Pyramid

bfu

frVrP

3

Distance (at) 3

Distance (at) 3

Distance (at)

-

\2 XZ

X2 X2

Distance (at) X 2 3

In calculating the amount of reinforcement required. it is asJumed that the portion of the base 'designated (abut), (Figure 3d) acts as a cantilever beafi (of .eitattgula. cross-section) having a width equal io one fa.-ce of the pedestal (ab), a depth eciual to the effective depth of the base (d1) and alength equal to (at). Having ialculated the bending moment as, prostep is to check the depth posed ab-ove, the 19rt

of the base, and determine the amount of reinforcing steel required. These calculations are based on the commonly accepted formulas for reinforced concrete. (It should be noted that for this purpose it is more convenient to figure the momenti in terms of inch-pounds, as the stresses in concrete and steel are usually given as pounds per square inch, whereas in figuring soil lbading foot-pound units are used, as soil loading is uiually itated as pounds per square foot.) For balanced desi(tn, that is, conditions in which both concrete and steel are stressed to their full allowable capacity, the required depth (ds) of the base may be rletermined as follows:

a,:i#k_ where

(8s)

depth of base, measured from top of concrete to centerline of reinforcing steel. (inches) Mo: bending moment in base. linch-pounds) ' f": safe working stress, reinforcing steel in tension. (pounds per sq. in.) (1r

should therefore be placed within the limits of the beam width (ab). However, additional reinforcement should be installed to reinforce the base between the points (gt), and also at (uf),

us.rng_ the same type and spacing of bars as determined for the beam section (ab). This additional reinforcement insures that the entire area of the base is reinforced and weak spots eliminated. Obviously, the reinforcing- bars should extend entirely across the base. Also, there should be a set o{ reinforcing bars parallel to each of the axes, i.e., four sets of bars for an octagonal base, three sets Jor. a hexagon, etc., thus providing strength in all directions. There should be at least 3 inches of concrete below the reinforcing bars at the bottom of the base. Reinforcement in other parts of the foundation should be covered with not less than 2 inches

of concrete.

-

jb,: A":

: ("/",0, ) ratio of effective area of reinforcing 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 reinforcement in tension. (square inches)

If the design is

balanced, that is, the actual of the base (d1) is that calculated by equation (85), the value of (A") may be determined d-epth

as follows

:

A": b" dr p" (86) If the depth (dr) is greater than required by equation (85), in which case the steel is stressed to its full capacity but the concrete is understressed, the value of (A") becomes : .Mr ,'" (87) - f" jd, . If the depth (ds) is less than required by equation (85), it is recommended that ihe diminsions of the base be changed to give the required depth. In case circumstances make it impossible to increase (dt) to the required dimension, it will be

E.rample No. 6. Determine bottom reinforcement for the foundation referred to in example No. 1. Figure bending loads

Line(m,w):,72" -36" : 357 pounds/sq. ft. Load (m,.,, :!9i*4 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

-i?4

18.65

X 446X

45

144X3 Total (Mo)

'/)v

/\e/\

ft'

45"

X 2

45X2 3

:184,000in'-lbs

:

153,000

:

62,300

:

51,900.

:451,200 in.-lbs.

Check depth of base for balanced design (equa-

tion

(85)

: : j: b": f"

p.

necessary to increase the amount of reinforcement

used. The determination of the amount of reinforcement required for sdch special cases is beyond the scope of this article, and reference is made to the various publications dealing specifically with concrete design for further detaili. Having calculated the cross sectional area of steel required, a selection is made as to the diameter,.shape, number and spacing of bars which will give the required area. It ii recommended that the center-to-center distance be about 4 inches if possible, but not less than 2l times the bar diameter for round bars, or 3 times the side dimension for square bars. Generally speaking, a large number of small bars (1, s/s, o, % inch)-are preferable to a smaller number of larger bars. It should be borne in mind thaf the area of reinforcement determined above is the amount required for that portion of the foundation having a width equal to (ab), Figure 3d, which was assumed to be the cantilever beam carrying the entire bending load. This amount of reinforcement

45"

tql" X 879 Pounds'/sq' 18.65 X 45 45 X 2 - u4 x87ex--29.8 X 45 446 45 X 2 14442/\3

18,000 .0089 '87

29.8"

f"P"j:138.7 I - 4il:oo :10.5" 0t(hnrlnccd' : Y t.lg-z x, zg: Since the actual depth is 21 inches, whereas only 10.5 inches would be required, the concrete will be understressed, and the area of reinforcing steel should be calculated by equation (87). A"

:

451.200 .87

tsPoo

x

x 2l -

1 37

sq' in'

lJse )1 inch deformed square bars (.25 sq. in.

area).

Number required #+ :5.5. Use 6 bars within the width of beam (ab). 29.8" :4.96". l;se S-inch spacing,-enSpacing ? -tid" tirely acros. (gf), which will require {11 :13 bars per set. Four sets of bottom reinforcing bars will be required for the octagonal foundation.

t7

to Resist Stresses Due to UPlift As explained previously, the wind moment creates a positive sbil load on one side of the centerfi""-, utia i negative load on the opposite side' In oi-frlr-*otas, ihe action of the wind tends to lift tt fo""a"tibn on the negative side. This upward " oi "uplift" effect, G resisted by the -*gigh! i;;;; of ;i iht "o""i"t" base itself, and byItthe weight-betherefore ihe- earth fill on top of the base. cornes necessary to reinforce the top,of the base, resist the reiulting negative bending moment' to --tfr" pro"edure is qiit" ii-itur to thaf described soil'reaction (Section 1.9c)' The foi1t "'rp*ard (abfg), and the outline of the area the in ioua r.tt itt"o."tiiut beam carrying ihe load is (abut) as in S..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 unithe calculair#it iistrituted, thus simplifyingthe reinforcemoments, the ii;;:'Aiter figuring ment is determined 1n exactly the same manner as exptained in Section 19c, using the equation

Moment

19d. Reinforcem€nt

^-Mo f" jdt ^'-

2%!L;570X

E%q

:

4+Z

: :

Total (M")

119,000

in.-lbs.

99,500 218,500

in.lbs.

From equation (87a) 218,500 inch lbs.

in. within beam A":-l8p00X.87X22 -.636 sq.(29.8") width

lJse rft-inch deformed square bars,

at

10-inch

centers. 19e. Bond

In order for the reinforcement to be effective, the'-st-rength of the bond between concrete and be sufficient to permit the reinforce;;i;;t? stress *""1 t" develop its full strength. The bond following the of meais by ;;;- t" ialcul'ated formula:

V,

U: EJA; where ""-';:

(87a)

(88)

bond stress per unit of area of surface of bar' (Pounds)

>": 6erimeters of bars -within the limits or - ;;;;f it. beam width (ab)' (inches) Erample No. 8. Check bond stresses in example

In this case, (ds) is the depth of the base from the centerline'oi the upper layer of reinforcement ;;ih;-i;;i;* of the bise, and (M-) !s the bending (inch--pounds)' forces uplift the to a"" *o*.ttt Eramble No. 7. Determine top reinforcement to r.Sri "'pfitt in the foundation ieferred to in example No. 1. Weieht of concrete : 300lbs./sq. ft. 150 lbs./cu. ft.:r.Ztt. Weieht of earth :27Olbs./sq. ft. 60lbs.,/cu. ft. X 3 ft.

No. L. Bottom reinforcement V.

11.578

>.:6

lbs. (See examPle No' 4)

X.5 X 4:12" By equation (88)

u:-nffil:53 11.578

:T-701bs.,/sq. ft.

Totat

y57oX

+-

lbs.

Bond stress for bottom reinforcement is satisis permissible (see Table 2)'

fr.tlil "t ts por"as Top reinforcement IABLE 2

GonrlonE Apptylng

io

Figure shear

gz#y-

Foundotlon Dcslgn

Mlxture: Cement'

2 5

2 4

Sand. . .

CoarEeAEgreEat€''""' fb

Safe

baring load on concrete (lbs./sq' in') '

500

375

fi

Ultimate compressiv€ strength (lbs'/sq' in')

2.0(n

1.500

f.

Safe --ii. unit stress in

800

600

fa

unit stress in concr€te due to vertical Safe --"i".i-fai.co"al

ir).

40

30

fD

Safe

unit strss in concrete base due to nunchinc sher. (lbs./sq. in.)...'..'..

120

90

f.

Safe working stress, steel reinforcement in tedsion. (lbs./sq. in). . " '

18,000

18,000

i)

(inch-pounds).

15,600

16,000

ir,

(inch-pounds)

138.7

88.9

i

Ratio, lever arm

.87

.89

15

l5

(f.

(f.

qtreme fiber of concrete ompression)-(lbs./sq. in').." " "' tension) (lbs./sq'

F%J {"=oikJ *u

.

'

.

depth (dr).

of

resisting couple to

Ratio, modulus of elasticity of steel to that of concrete,

Ratio. efrective uea of tension reinforce ment to efiEctive ilea of concrete ' ' ' ' Safe bond stres --i6rje.&ti

(concrete

of surfae 45

m

* These figures may be slightly incresed by making "U"-bends on the

""a";l;|iif,l.i",i[f tipe-. riri tionJ

iiijre

of intqest.

t8

as most satisracrory ror rounda*?Ii;," is recommend_edniiture are presented as a matter ionstantJ ttiiiiii ilztr

:

947

:5,427 lbs.

Total (V.)

s427 $:- 6ffikT

(88)

-48lbs' 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 tower

rests on the cincrete pedestal) seldom cause any preaif1."ltu, but should lre checked as a -safetystress .rrlio".'tfte bearing stresses consist of due .the to the a""-1o *i"d pressuie, plus the stress dead load as follows: Bearing

60 75

:4,480 lbs.

57orbs.

2o:3X.5X4:6

.0056

to steel rein-

*i irnit of arm

of bar. (pounds)

.0089

X

2W-xsto

I

I

stress:

4M*/7tD"

+ (w'* W")'/z'D'

(37a)

(See Sections 11 and 12.)

ilquation (3?a) gives the bearing stress in po""tt p.t ii.,eil Toot of shell circumference'

These stresses are spread over the are-a of the base purposes the unit bear;"". tfr.tfore for prictical -determined as follows: i.,gjsttess can be

fr:

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

,r

(37b)

l2r*

in which

r,: width of the tower base ring. (inches) fb: unit compression stress on concrete. (pounds/sq. in.)

Equation (37b) may be modified slightly, depending on the exact shape and arrangement of ihe base ring (or base plate), but in the majority

this case are given for illustration only, the design has not been changed to take maximum advantage of the allowable stresses. The stresses in foundations of this type should not exceed those commonly accepted as good engineering practice in reinforced-concrete design, for the particular mixture of concrete used. As a matter of convenience Table 2 is presented to show allowable stresses and miscellaneous constants applying to two grades of concrete quite generally used for foundations. It is strongly recommended that the 1 :2 :4 mixture be used in practice, the figures for the 1 :2:5 mixture being shown primarily as a matter of interest. 19h. Suggestions and Recommendations

TOP

The calculations explained above provide for reinforcement to resist the stresses due to the

OF

PEDESTAL

various types of loading. It is good practice, however, to install additional steel as a means of tying the foundation together, to form an integral unit. The same size bars are used for this purpose as for the main slab reinforcement, and the designer must use his own judgment as to the number and location of the bars. Figure 4 represents what is considered good practice, and is offered

4.F4 ''a 'c .

.,SLEEVE NUT

'.4.'..y.4

4.4 " :.WELD

as a guide. In the case of very large foundations, consider-

FIGURE 7

cases it may be used reasonable accuracy.

of

in the

above form with

For guyed towers, equation (37b) becomes: 4M,

fr:

r

,r DF 1-

R"+W"+W" z'D,

(37c)

19g. Allowable Stresses in Foundation It is to be noted that in actual practice the depth of the base in the examples given above could be reduced, if desired. All of the stresses for diagonal

tension, punching shear, bending (upward ancl downward) and b-ond in the reinforcement are well below the-allowable values. As the examples in

able concrete and weight may be saved by constructing the pedestal with a hollow center, as illustrated in Figure 5. Of course, the inside form is left in place. It should be noted that the base slab extends all the way across, to provide protection and bond for the reinforcing bars. Foundations supported on piles should be so constructed as to allow the tops of the piles to extend about 6 inches into the base, with the bottom reinforcement about 2 inches above the piles. (See Figure 6.) Considerable inconvenience is sometimes encountered in setting the tower in place, due to the difficulty of lowering the heavy vessel over the foundation bolts without bending some of them or damaging the threads. Figure 7 illustrates a method of overcoming this difficulty. A sleeve nut is welded to the top of the bolt, and so placed that the top of the nut lies slightly below the surface of the concrete, with a sheet metal sleeve around it. The tower may then be placed in position without interference from the bolts. Stud bolts are next inserted through the lugs on the tower, and screwecl into sleeve nuts from the top.

Nomencloture Ar

: elfective cross sectional area of sion (square inches)

steel rcinforcement

ir

ten'

For balanced design (86) Ar: b" dr pr If depth (dr) is greater than required by equation (85) e"

:--14.0.

(87)

tr ldr For top reinlorcement of slab to resist uplift strcsses:

o.: ts-r1,, Jdt

c

c'

:

(87a)

face

assumed to dct as a cantilever beam resisting the'bendini stresses (line ab, Figure 3d) (inches)

section which serves to resist the diagona! tension stresses. (line ar br, Figure 3.) (inches) distance from neutral axis of foundation base to point of maximum stress. (feet) distauce from centroidal axis of foundation base to any point under consideratiofl. (feet) outside diameter of tower. (feet) inside diameter of tower. (feet) tower diameter measured over insulation. (feet) diameter oI Ioundation bolt circle. (ieet) short diarpeter of foundation base. (feet)

D: : Do: Du : dr : dc : short diameter of critical section for diagonal Dr

a : 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

of the oedestal)

b'- width of critical

dr

:

tension

stresses (see Figures 3 and 3c). (inches) effective depth of base of foundation, measured from top of base to centerline of reinforcing steel. (inches)

t9

dp : short diameter of foundation pedestal (feet) Ec : modulus oI elasticity of concrete.

E! : e

modulus of elasticity o{ reinforciflg steel. eccentricitY. (feet) This iactor ia the distance from the centroidal axis of the foundation to the point at which the resultant of the dead load and the rvind ioad intersects the base of the foundation. The eccentricity can be calculated as follows I

tr{r

(12)

Eouation (12) sives the eccentricitv at the condition of oobrest stabilirv, that i", with the miiimum dead load. This is the value which ordinarily is used for design purDoses, for maximum however. it is obvious that the eccentricity''ubstituting the a""a toia ".nditions ian be calculated by (Wt). The plaCe (W) (12) in of in equation value of marinum valtie rvhich it is possible for- (e) to have and still maintain the stability of the foundation is t2

em.x: emir :

(

Or

1e)

(19a)

-,

Values - of (e-"") for various foundation shapes are follorvs:

Octagon: ens:

Hexagon; em.':

Square: Circle:

0.122d 0.121d

: 0.1 18d em!:: .l-d emax

as

f"' : i" : fp

:

,,,

relation to the moment of inertia can be expressed as follows;

I: r

in. )

H: height of tower. (feet) hr: height of foundation. (feet) hr : height oI collar (to which the guy wires are attached) above

(68) 7.

k for various foundation o ? )-

";-'

: *'1'" Circle: ,.: *1'"

Square: L:

lever arm of

I:h,f

shapes are as follows

a 10jd,

d, is in inches. See Figure i.

Steel supplied -

:

--

7

No. -_uniformly) both ways across

Sum of bar perimeters.

: ___

in.2

-

_x_x_

:

:

lbs./in.:

Total Shear, lbs.

See Step 7.

d, is in inches, See Figure l. *Ref. ACI 318, Sec. 305

in.2 --

Step 11. Compute shear unit stress as a measure of diagonal tension along Section f-g, Figure 2. Area on

(4)

plan.

Stress indicated on stress diagram.

(5)

+

Volume of area-shess. Shear for

: s, a#

se

shape

Geometrical

(The numbers here are for defining the geometrical shapes o,n Figure 2. FilI in blanks at right for computation.)

:

(4)

-

S,,

2 s*' 2

t.l

(-+-

2

+ -

e u11

+

lbs

) ( _\

lbs

g

(-

'1

lbs

\

Ibs

\

lbs

-

s,o :t,.(?)(+)

(7)

4'-

I/_

_I_

s,, =f (1) r'r

(6)

4'-

--l

s.^

)

:

S,,

1"1,

1Cl^

(_

(-

12

shear for diagonal tension:Total of Compute diagonal tension shear unit stress, v:

m

\ /

_

12

V":Total

above------

- -)':--lbs --lbs

V",:Total shear across Section f-g (See above), m: Length of Section f-g (See FIG. 2) : ___ d" is in inches,

) (_)

Allowable v:

(__

See

FIG.

1.

)

-lbs/in.,z

Ref. ACI-318, Sec. 305.

32

:

Srr(e)2

:-l

\t

)

1"11-;

:/c-

(6)

(-

-)

:-(_2',(s)

diagonal tension

-lbs/inz

lbs inches

tn.

(150) + (100): Step 12. Compute flexure in top of pad due to uplift. W/:Figure 2, Section R-S. Wr:Uniform downward load on areas (1), (2) aod (3), See Figure 1, and using an average weight of reinforced -lbs/fu2 Figure 2. concrete : 150 lbs/cu. ft. and an average weight for : h1 (150) + hs (100) earth fill of 100 lbs/cu. ft.

AREA IN SQ. FT. (1) bXc

Xw,

XG/2*d)

x-x(2)

c2

(3)

: :

X LOAD, Ibs ft2 X ARM, ft.

MOMENT, ft.

r(-=-+-): XW/

:

X(c/3*d)

(----), x-

x(\5 ^ +-):/

DlXd

)(d/z

XW/

:

Mu, Total moment i:

Step 13. Compute steel required for flexure in top

of

pad.

:

ft.

d, is in inches. See Figure No.

IJse

Ibs.

1.

_.bars

at

--(Distribute uniformly across D. both ways)

12Mu 0-85 - - 12X fsjd, X X Mu: Moment -: due to uplift across Section R-S, Figure 1. Ars

lbs.

-0.85

--

Steel supplied

:

Number of bars

X

inches on center.

Area each bar. in.2

Step 14. Compute size of anchor bolts. Refer to Figure 1 and compute moment at base of tower as follows:

Mnw: Moment at base due to windr ft.

lbs.

ft.

lbs.

ft.

lbs.

ft.

Ibs.

M-*:

ft.

lbs.

-------:

ft.

lbs.

Mr:Tota1 moment at base:Mnw,*ME - - - - - - -: N : Number of bolts: _(,,N,, should never be less

ft.

lbs.

Mn:Moment due to reboiler from

Step 2

then B and preferably 12 or more)

Dn: Dihmeter of bolt

circle

:

ft.

Fl:Tensile force, due to Ma, per bolt:*(4 (wr/N)

:

(4X

M1,,/NDu)

-

-(-/-):_-._-lbs.

Ftr: initial tensile force due to tightening nut. - - -: (5,000 lbs. is suggested for "Fr") -/--X-.) F:Total maximum tensile force : Ft *F Fr - - - - - - -: Net Area (At root of thread) : F,/Allowable stress Size

of anchor bolt (Add /s' to

size determined above

for

lbs. lbs,

: --

corrosion.)

: Net Area of selected bolt. : See Tables

in.2

in. dia.

in.2

7 and 2 for detailing dimensions of anchor bolts, and for net areas to use in selecting bolt

xThis simplified formula is not exact, but is always on the safe

sizes.

side.

33

a o

o

i

o

()

n

long golvonized iron aleevss.

rJ'

UseJ'dao.

= No. Dio. use- J'dio. bolfs iE 3 t ies ol t2" o.c. ol *-borr -'o.c. y'-'borr olJ'o.c. Fin. Elov.

both woYs.

bolh

woYe.

EL EVAT IOT{

Step 15. Pedestal steel and reinforcing steel placement'

Thic cketch ia

for uss itl drofting finiehed drowing:'

For vertical steel in pedestal use greater of following two steel areas: (l) As: Net :

area of anchor bolts. in.z

(2) As: No. 5 bars at (max.) 6 inch

spacings.

tn.. No.

Use

formly around pedestal in octagon

As:

as

bars distributed unishown (Figure 3)

in.2

Anchor Bolts Detqiling dimensions. Tables I and 2 can be used for fabrication simply by marking or circling the desired bolt size and shaPe.

*Total

length:P*S+L+A

***Net AteA: ao, in,z ** Design basis for L: : Computations are made using allowable stress

PLAN

an

of 26,600 psi' Allow 10,000 psi for the hook

FIGURE 3-Foundation

details.

1 and develop the remainder of the bar strength in bond over length, L.

About the Author H.

ft ut fs'u' t' : rdp

is an

engineering group leader with Celanese Chemical Company, Pampa, Texas' He

Bernard

supervises

Shield

a dePartmental

grouP

hahdling mechanical design phases of plani alterations and expansions including project engineering, mechanical, electrical, and instrumentation. Mr. Shield holds a B.S. de-

gree in civil engineering from The IJrriversity of Texas. He worked with the Surface Water Branch of the USGS and instructed in the Civil Engineering Department of the Uni-

veriity of Texas before joining Celanese. He is a Registered Civil Engineer in the State of Texas, a member of Chi Epsilon, Tau Beta Pi, ASCE and TSPE.

Where:

psi.

fsrr: Allowable stress over gross area due to a":

hook

:

10,000 psi. Gross area of bolt, in.2 ACKNO\{LEDGMENT

There have been

rony fine articles published on t-his subject and I wish tc

have gained from them. In adglition, e@d @m;entr from Celanese engineers, particululy the late L'ldie {ay9{s and Willard North Carolina, 6f Chalotte. Johroon and womac soward Mth f"*.,'have contribuied to the developmqnt-o{ -thjs eim*-"i'fii[op. paper. I also wish io reognize the uork o[ our draftsroan, Don Staflord, in the drafting of this {orm,

*tr-"*I"a".- tfr. lnfo.uation which I

LITERATURE CITED

l Wilbur, W. E., "Foundations for Vertical Vessels," Pnrnoreuu

34,

34

fs: Allowable stress at root of thread : 26,600 psi. ao: Net area at root of thread, in.2 p:Allowable bond stress for 3,000 psi concrete:135

No. 6, 127 (1955).

RETTNEB'

Use Graph to size Tower Foofings Dimensionless numbers, computer

If M is increased further the structure must topple. So long as the allowable pressure is not exceeded, all of- these possible arrangementi are inherently stable. How_

calculated and plotted on graphs, simplify sizing of octagonal, square and rectangular spread footings.

J. Buchonon,

ever, at some stage in the sequence, the maximum soil pressure, th.at is the pressure af the extreme point on the leeward side, becomes equal to the allowable pressure. As the moment is increased further, the pressure at this point exceeds the allowable and the structure is in danger of toppling caused by differential settlement.

Newcastle University College,

Newcastle, N.S.W., Australia

Gnaprrs oF srMpLE dimensionless numbers may be used to size spread footings. These numbers describe the action ot the footing under a known Ioad system and allow the user to select a footing size that wiil maintain stability without exceeding a specified maximum allowable soil

For any particular footing shape (square, octagonal, etc.) and. orie.ntation, a pressure pattern as shown in urry o1e .of the diagrams_ of Figure 2 prescribes a unique relation bgtwegn W, M, the maximum pressure p, ,La the plan size of the footing-described by some character_ istic dimension L. any such case the dimensionless -beF-or formed from these

Deanng pressure.

. A typical footing arrangement is shown in Figure As in the usual

wLWsW

treatmeni the soil under tfre fo&;ng is When the moment

(M)

-M-, rrp, ]f,i_, etc._

1.

be perfectly elastic, and no credit is allowed for .1".

llk"" tne sorl tateraL support.

groups, which may

variables-

is negli_

have fixed values.

The graphs-of Figures Z and. 4 show the relationship ,between two of these groups,

w

/w

_w

r/ " and --_Pt.2 for footings having square and M

',

octagonal plan shapes and tne onentatrons shown. The terms of the first g.roup and the footing shape are the design data. Calculution tf this group and reflrence to the graph for the footing ,t apJ specined gives the value of the second group fiom it i"t tfr" size of the -

footing may be calculated. _ Figure 3 includes all the cases where the whole of the base of footing is loaded as in Figure 2(b) ; that is, -the where there is some pressure over the i"hol" of 'tLe lower face of the footing. The upper limit is at the point where the minimum pressure is 05 percent of the maximum.

B?9"9 this,point the effect of th" moment load may safely be neglected. Figure 4 describes

FIGURE l:Typical footing anangement for tall

towers.

gible compared with t_he dead weight (W), the soil bear_ ing pressure is uniform over the *f,ol" u.# of the base of the footing as shown in Figure 2(a). As the moment

increases, pressure diitribution'"h*.rg"r, as shown .r.iJ in Figure l!: 2(b), (c) and (d), until it reachei the extreme impossible)'case shown in Figure 2(e) 1111-r:,rr:ctice where the structure is just balanced on one corner of the footing and the bearing pressure is infinite at that point.

the cases where only part of the base is Ioaded fas in Figure 2(d) ]. The lower limits cor_ respond with cases where only about one tenth of the

base area is under load. Actual designs will rarely ap_ proach this condition or go beyond it. fne ,pp", iiroii, of Iigure 4 correspond, o1 cor.se, with the lower limits

of Figure

3.

given loading system on a footing, there is, except ? . Fo. Ior a ctrcular tooting some critical orientation of the axis of rotation which produces the highest maximum soil pressure. For both the square and the octason this orien_ tation is the axis passing through two verlices. The curves for the octagonal footing have been calcu-

35

(d)

(c)

(b)

(a)

FIGURE 2-Changes in soil pressure for tall towers'

lated on this basis. Similarly, for

tions, values of the parameters zl and p have been added to the graphs, where in Figure 3,

a completely unre-

strained structure on a square footing, the curves for the afi.""f axis should be used for calculating the minimum size of footing. If the struiture is restrained so that rotation about only orr" t*i, is possible, as for instance in the case of a pipe rack standird, a more economical design results if a to ;q;;t; footing is arranged so that this axis is p-arallel calcufor used one side. ThJ approprilte c,,rve is then i;;i.;;f the fooii"g tir". ln this orientation the required

minimum

and in Figure 4,

!:

to the axis of rotation' its - fieater'side perpendiculai well for design of recequally used b" fir" curves *uy q the plan aspect factor the using by footings tangular ratio of the footing. oi rotation ,* - dimension perpendiculal t9 the.axis diinension Parallel to the axls

In this

these cases,

L: dimension perpendicular to the axis of rotation Usually a will be greater than 1, but if for some other reason a'rectangular-footing must be laid out so that a is less than 1, tfie graphs may be used in the same way' The procedure then is to find the size of a square footing *hi"h, with the same loading, would produce a maximum bearing Pressure

"f I

matically by using the grouPs,

frtit is done auto-

M lwt

w{

-F-

and

Wa

5p

As is usual with dimensionless correlations, the units of the terms must be consistent. The length unit will usually be feet; the force unit may be pounds, kips, tons, or any other convenient' Typical sets are:

lb. kips tons

ft. ft. kiPs ft. ton

lb'/tt''

Ib.

kiPs/It'z

*o/t"

When the size of the footing base has been calculated

it is necessary to calculate the thickness and reinforcement necessary to resist the shears and bending momelts

in the footing itself. For

w /E *r, -'fr'YE

^ *,i**

is les thm

convenience

0.73, the situation

in

these calcula-

is

revened,

but in

36

connection

it

should be particularly noted that

*h"; " portion of the footing is unsupported by soil-re;;i;" tiere are shears and, more important, bending moments in the unsupported section of the slab in the oplotit" sense to those-usually considered-in the design' t'Iiere st."rses must be evaluated and the slab design may reouire modification to resist them (e'g' by the addition of iop bars to resist the reverse moment) ' In common with all other methods proposed for esti*uii"g footing size, the calculation must be a trial and !ro."tt.

The known data are usually:

"rro, . Structure deadweight ("mPty, working and under hydrostatic test conditions,

if required)

;

o Pedestal size and weight; o

this a

Depth of footing base below C."Pq (from knowledge of trost Iine leve-I or situation of desirable load bearing strata) ;

o Allowable maximum soil bearing pressure

(P)'

For full details of estimation of these see Brownell and Young2 or Marshall.s The moment load (M) 9a1 t{en be caTculated from the wind load and depth of footing base below ground. The total deadweight (W), however, comprises, i'esides the weight of the structure and the pedestal: o The weight of the footing itself, and

The weight of overburden above the footing' These can only be calculated when the footing plan size and thickness have been fixed. Thus it is necessary o

rectangulrfmting with uis of rotatiou parallcl to *"i side shoulil still be red.

ProPortion of the width of the

o Wind and other eccentric loads;

as

marked on the graPhs'

LWMP ft. ft. fr

ttne

footing under load'

Knowing the value of the appropriate p.arameter and of th" *rii*r* soil bearing pi"tt*", P, the load distribution over the lower face o1 the footing and the rethen be calcuili;;A thickness and reinforcement ormay Brown'l Marshall3 of methods Utea Uy the

size of the footing is somewhat less'* --irr r,rch a situition, however, an even more economical d*ig" *ry ,"rrlt from using a rectangular footing with

In

Pressure

'-;;i;;;;;;G

to guess initially a footing size so that (W) can be estimated, and then to refine this estimate by trial and error. Often, and particularly for deeply based footings, the siab thickness is of minor importance at this stage, since extra thickness of concrete onlv displaces overburden of not greatly different density. If the initial estimate of the thickness is reasonably good. final adjustment will have no great effect on the deadweight (W). It is usually desirable to compute separ.ately the footing _ size required for several critical load conditions. These are:

. Minimum weight and maximruu nind effect, e.g., in course of construction;

=l

;

o Working weight and maximum u'ind; o Test conditions-filied with tater ancl 50 percent of maximum wind moment.

The evaluation of the first and last of these depends on the design and method of construction, and no useful general rules can be given. A reduction in wind load for test conditions is allowed since it is most r.rnlikely that the test period and the maximum rvind would coincide. Since tho construction period is normall1, much longer than the 456

8t0

20

30

40

lwlE

60 80 t00

200 300 400

ur'p

FIGURE S-Relationship between two dim,ensionless numbers for all cases where some- pressure acts on lower base face. 0.5

0.5 o.4 0.3

l'

I

0.2

*l-'

o.rs

P

W: Weight of ,structure, footing and overburden M: Moment of wind load and any other eccentric loads about the center line ol the base of the

0. I0

0,09

'0.08 0.07 0.06 0.05

0.04

0.5

0.6

0.8 I

P: Maximum allorvable soil bearing pressure L: Characteristic length: square-length of side octagon-width

across flats

rectangle-length perpendicular to the axis of rotation (See Figure 5) PIan aspect tatio of rectangle

_ Length perpendicular to

axis of rotation

Length parallel with axis of rotation (See Figure 5 )

For other cases, a: 1 Minimum Dressure ,z (.tlgure Jl :: - _:+- pressure lvlaxlmum p (Figure 4) : Plgp-crrJion of the rvidth of the footing which is under load.

1.5 2 w

footing

3

4 5 6 78

/TE-

M/P

FIGURE 4-Relationship between two dimensionless numbers for cgses pressureicts on only part of Iower base face. ^w-here (i.e. Fie. 2d).

test, no such allowance is possible for min. wt. and max.

rvind effect.

The procedure to be followed is iliustrated in the fol-

lowing examples.

Exomple-Ocfogonol Tower Footing. A footing is to be designed to carry a. tower 54 feet high and 4 feet in diameter to be placed on soil for which the maximum allowable bearing pressure is 2,000 lb/ft., The frost line is 4 feet below grade and the pedestal

top is to be 1 foot above grade. The footing

base 1s

made

37

SIZING TOWER FOOTINGS. .

5 feet beiow grade, i.e., 1 foot beloru the frost line' The design" maximum wind velocity is 100 mph' The ma*"imum wind moment ahout the base of the footing is calculated to be 200,000 ft' lb' (M)'3

Restrained

=

0.318

L : 8.86 ft. so that rotation

78.6 It,2

abor'rt onlv one axis

\V PL:

30'000 lb' Empty Tower 9,000 lb' contents working and Appurtenances 40,000 lb' Water fitt for hydrostatic test

For a pad estimated to be 13'5 feet across flats and 1 foot thici< rvith an octagonal pedestal 6 feet across flats + i.", deep and ctay nit of density 90 1b'/ft'3, esti-

)\ -"

Lr -

0.38t)

T, :

8.1

. 65.8

ft.:

ft.

u,eights are: 63,000 lb. 33,000 lb.

Concrete Fi11

Then calculations for the three critical conditions are:

EmPtY Working Test 126,000 13s,000 i66,000 w (1b.) 200,000 200,000 100,000 M (lb. ft.) 2,000 2,000 2,000 P (lb./1t.2) \v i* 0.63 v-63 u.G.s v-G,s 1.66 r/ &3 M\ P : 5.54 : 15.11 :5'00 W ,f..r, nraoh) 0'405 0.585 0.423 PL2 ' 83 67.5 63 L2 U.5u5 0.423 0.105 11.9 12.61 t2.47 L (ft.) Thus the assumed size is too large and could be reduced. The next trial rvould assume a l2-foot octagon' and the minimum size would probably be somewhere

t

FIGURE 5-For rectangular footings, "a" is usually

greater

than one.

as in Exomple-Rectongutor Fooling. For the loads : (short 3 a assllme above, e*ample footing ,h" ,qrur" side parallel rvith axis):

]t - /'E M\ P

,o.i rl x

wu

:

PLT

:

L.

Unrestrained. Diagonal axis

25

:4.34

0.505 (from graphs)

7.5.

=:

1'18.5 ft2

0.505

L :

12.2

1:

+.1ft'

ft.

The rectangle required is l2'2 ft' x 4'1 ft' having pian +g.S ft"a, ,gri.rst 65.8 ft2 for the square footing

(from graPh)

"r"" under the same conditions.

In each of the above examples the maximum pressure *itt U" equal to the allowable, and the pressure distribution may te immediatelv sketched after finding the value of the parameter p o, ,i fronr the appropriate- graph'

Aboul the Aurhor lecturer in

Considering the essentially rare -and transitory occurrence of the riaximum moment load, the- basic assumption t"ginnitg are sufficient for most applica;;;-;i;he iio".. ffr" ,rsrr..r!tio, of perfectly elastic soil, however' is not entirelv sonlnd and in tritical cases the advice of a soil mechanics'expert should be sought'

chemicai engineering design at Newcastle l]niversitv College of the Universitv of New South Wales, Tighe's

Hil1, N.S.W', Australia' tle holds B.E. (Chem.) and M.E. (Chem') degrees from the University of Sydney'

Buchanan held Positions as a design engineer with I\'[onsanto Chemicals Ltd. and Union Carbide Ltd. in SydneY Prior to accePting

Mr.

his present Position'

I

o

* -/* :0.5t/-Ii:2.5 M!F

a

cl

t

bearing pressure of 2,000 Lb.lft2.

John Buchanan is

L

o o

Exomple-Squore Tower Fooling. A square footing is to be designed to carry a total estimated deadweight of 50,000 po.,nds and a maximum overturning moment of tOb,OOO Ib. ft. on soil having a maximum allorvable

#:0.318

-T

C

.9

near this figure.

38

is

possible.

Tower weights are as follorvs:

""J mated

?5

,-*

.

LITERATURE CITED 1

Brou'n, A.

141 (1963).

Buchanan

A'

HvorocangoN Psocrssrro &

Plrnorruu RsrrNet 42' No' 3'

",'d,.*"'"ir. L. E. and Young, E' H' in ^-"Process Equipment Design" )' ch#ri" ti, ^u"i'v*1. l"t " Wile"v' and Sons37;( 1959 No' 5 Desisn Suppl' (1958)' sMarshall. V. o. l'nrnorru*huu'"""

Simplified Design Method for lnfricate Concrete Column Loading Combined biaxial bending and axial Ioad on reinforced concrete columns present difficult design solutions. This method bypasses the usual tedious computations E. Czerniqk, The Fluor Corp., Ltd., Los Angeles

IIr,RB's A NElv AND srlvrplrrrED METr{oD of solving concrete column problems consisting of an axial load combined with diagonal bending. The method can be used to determine the combined stresses and the eccentric-load capacities of reinforced concrete columns from known or assumed positions of the neutral axis. The approach is unique because it provides greater accuracy with less computation than methods used up to now. It considerably simplifies the stress analysis of many structural components used in Hydrocarbon Processing Plants, e.g., pipe supports and rigid frame structures for supporting exchangers, compressors, etc. The method bypasses the usual, time consuming, tedious computations of principal axes as well as the need to rotate all computed properties about the principal axes. Significantly, the method is valid for both elastic and plastic stress distributions. ft thus unifies in one, simple approach the straightline and the ultimate-strength methods now used in reinforced concrete design. The methods of analytic geometry and the basic equilibrium equations from statics may be applied to a.variety of problems involving stress analysis. The term 'analytic, preceding'geometry' implies an analytical method, wherein all results are obtained algebraically, with any diagrams and figures serving merely as an aid in visualiz-

ing the problem. All given data must, therefore, be exin coordinates with respect to a suitable set of axes (preferably selected so as to make the coordinates as simple as possible). The procedure will be illustrated by the rather intricate problem of axial load combined with diagonal bending. In general, when bending in a concrete column occurs pressed

about both coordinate axes, and there is tension on part of the section, the effective portion of the reinforced concrete section (transformed area) resisting the applied load is not symmetrical about any axis. Though the unit stresses may still be expressed by the well known formula: P _.M*"*oMr",

Ar*I,

such a process is rather laborious because all the values

must be related to the principal axes through the centroid of the acting section. Thus, for each assumed neutral axis, one must repeat the numerous and tedious computations of : the centroid of the acting section; the orientation of the principal axes; moments of inertia about the principal axes; and, not the least, the calculation of load

eccentricities with respect to same principal axes. No wonder, then, that 'exact' solutions have been consistently avoided by practicing engineers. The technical literature, though abundant in advice on the 'how to' side of problem solving, is extremely meager when it comes to specific examples, except maybe for the most simple cases.

Should You Trusf Compulers? The increasing use of digital computers has somewhat improved the situation. Computer programs are now available that can accomplish the tiresome solution through successive approximations, at extremely rapid rates. Ilowever, when ihe engineer views the computer output sheet, he may sometimes bewilderingly wonder just how accurate these results really are and whether he could and should put his trust in the modern marvel of technical automalion 'design via computerization.' Needless to say, the engineer has no right, nor authority, to abdicate his responsibility for professional judgment. The responsibility for structural adequacy must always be his, irrespective of the methods or tools used to come up with the answer, be it a slide rule, desk calculator or a giant electronic computer. Ifence, if he is to make the most out of the new to-ol, he must possess some simple means for spot checking the machine. In the case of biaxial bending on concrete columns, the method outlined below could probably serve such a pu{pose.

Two Design Methods. The Building Code requirements

for reinforced concrete (ACI

31S.5G)

permiti columns

subjected to combined bending and axial load to be in-

vestigated by two methods:

o The so-called elastic method in which the straight line theory of flexure is used, except in regard to compressive reinforcement. o The ultimate strength method on the basis action.

of inelastic

Ultimate strength design is relatively new in American Codes, and hence some of the old timers may feel ill

at

ease 'r,r,ith new concepts and new

criteria.

It will be

shown, through illustrative examples, that the same approach applies throughout the full range, from elastic

to elastic-plastic and ultimate strength considerations. In the straight line stress distributionmethod. the code requires that colurfins in which the load p has analysis

an eccentricity greater than

/3 the column

depth

t

in

39

]NTRICATE CONCRETE COLUMN TOADING

...

,

either direction, the analysis should be based on the use of the theory for cracked sections, e'g', that the concrete does not resist tension. This e/t allowance does not apply in ultimate strength design. At ultimate loads, flexural tension

that

it

concret-e is insignificant, and the Code requires be comPletelY neglected.

in

filethod of Anolysis for Rectongulqr Sections' In the one case of rectangular sections, it is convenient to choose two with coincide axes corner as thelrigin and Iet the sides of the rectaigle' In Figure 1: O, B, C, and D are of the liven conirete section' Line QR desig,h" "o*"r, the neutral a*is 1line of zero strain), and intersects nates the x and y axes at a and b respectively' Let the toordinates of the eccentrically applied load, P, be i and !; and the coordinates of any given reinforcing bar, of area Ai be x1 and Yi' The" inte.cept form of the equation of the neutral axis QR is:

i**:' fi

Now, assuming that the stress at any point (xi,yi); is proportional ti its distance from the neutral axis, then Uy'*,it,iptying fo, the stress at origin (o, 9), by the ratio oi tf," dittr"Jes of point (x1y1) to that of (o,o) we obtain the general stress formula:

r,:t"\ (t -+a - ItJ b/ The engineer need not keep track of. the sign, as the stress foniula will automatically result in positive stress, or compression, for all points lying to the left of the neutral axis (see Figure 1) and negative stress, or tensron beyond the neutral axis' The coordinates of the centroid of the triangular area under compression, OQR, ate a/3 and b-/3' Hence, the average compression stress within the effective concrete section

will

simPly be:

x: a/3

l-In

origin.

In the general .case, when line concrete section (see Figure the outside partially QR'ir must be subtracted from triangles smaller more o.r"'o, fr, the over-all larger one. This is illustrated in the following

compression, is a triangle.

examples.

Reinforcing Steel Slress. The stress in the reinforcing bars is obtiined by multiplying the value (f1) in the seneral stress formula, by the modular ratio, n, the ratio It tt modulus of elasticity of steel to that of concrete' " 601 of the ACI Code gives the ratio, n, as equal Section

to 30,000/f'". H".rc",'thl

:

(Average Stress)

X

(Area)

:'/s oX /z ab: t" + f

rrci

axes are: fo

ab

(in x direction, about v

-e- X t:r.* f

(in Y direction, about x

axis)

axis)

(The reader may note that a2bf24 and abz/24 are simply ihe ,ralues of the section moduli of the effective concrete area, in the x and y directions respectively') InFigure 1, the intercepts of the neu-tral axis, line QR, are shoi., smaller than the corresponding dimensions of the section. Hence, the effective concrete area under

40

----t rrri

: '^'.\' nt^( t- f.:_ - +) b/ "

-

Ai

is:

F"r:f.rXA,

,irr"'A'

N : total number of bars.

Similarly the steel load moments about the coordinate axes

will

be:

M'"":

N

i ,",*, and M'.r: i=1i t-f

u",r,

The above formulas are completely general and the

he so wishes, may use diflerent diameters for thJ individual bars, and he may or may not arrange the bars with symmetry about either axis. However, since the same modular ratio n was apptied to both tension aird compression bars, we did piesuppose -that the bond between the steel and concrete remains intact, and they deform together under stress. That is, the steel in the comp.essio-n zone can withstand a stress only n times that in the concrete. In reality, this is not exactly so' Because engineer,

M'"r: ,+

any bar A1 designated by coordi-

and the load in said bar, having an area

where

can also be shown that, for equilibrium, the load F" is located at coordinates, af 4,bf 4;hence, the moments oi the compression load in the concrete, about the x and M'"*:

in

F":

It

y

stress is:

nates xi and yi

and the total compression load in the concrete: F"

rectangular sections, choose one corner as the

The total load carried by the reinforcing bars (tension and compression) is the summation of the loads in the individual bars:

f,,:fo O-%-/t):/s1"

v:b/3

FIGURE

if

I-

FIGURE 2-Example

about centerline.

l.

#8 Lorr

Find eccentric load p and moments

of plastic flow in the concrete, the compression bars are stressed more than indicated by elastic analysis. Codcs have recognized it, by assigning higher values to the reinforcing bars in the compression zone. Section 706(b) of the ACI Code requires that: "To approximate the effect of creep, the stress in compression reinforcement resisting bending may be taken at twice the value indicated by using the straight-line relation between stress and strain." Ilowever, in permitting this use of 2n the Code limits the stres.s in the compressive reinforcing to be equal to or less than the allowable stress in tension.

In the examples that follow, the stress in pressive reinforcement shall be made equal to: f",

:2nf, < f,

(Compressive reinforcement oniy)

F"r:Ai (fri-fr):A,

X , Y

a (c)

concrete and steel reach simultaneously maximum allow_ able values (balanced design). To determine the value of fo, compare:

a

l..-\

1+-'

__

hence, use: fo

b

:0.45 f'"

fi .45f'nn

(-f, ). Ffence, use:

ft

.

\

(+*+-,)

Exomple I. Find the maximum value of an eccentrically applied load P, and the moments about the centerlines of the column shown in Figure 2, when the neutral axis is the position indicated. f'" : 3,000 psi f, : 20,000 .in

ps1.

n

:

30.000

fo